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  <row Id="1" PostTypeId="1" AcceptedAnswerId="3" CreationDate="2012-04-25T18:24:34.340" Score="36" ViewCount="34355" Body="&lt;p&gt;It seems to me that the addition of electrons and protons as you move across a period would cause an atom to become larger. However, I'm told it gets smaller. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2017-07-16T07:06:26.797" Title="Why do atoms generally become smaller as one moves left to right across a period?" Tags="&lt;electrons&gt;&lt;atoms&gt;&lt;periodic-trends&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2" PostTypeId="1" AcceptedAnswerId="56" CreationDate="2012-04-25T18:25:27.290" Score="22" ViewCount="3785" Body="&lt;p&gt;My understanding is that $\mathrm{NaCl}$ is an ionic compound, in which $\mathrm{Cl}$ becomes (effectively) $\mathrm{Cl^-}$ and $\mathrm{Na}$ becomes $\mathrm{Na^+}$. So I understand why I would get a &quot;sea&quot; of particles that would stick together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the above mean that it will have a face centered cubic structure with the ions held in place so rigidly?&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="66065" LastEditDate="2018-08-23T13:43:44.867" LastActivityDate="2018-08-23T13:43:44.867" Title="How does NaCl maintain its crystalline structure?" Tags="&lt;ions&gt;&lt;crystal-structure&gt;&lt;ionic-compounds&gt;&lt;solid-state-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3" PostTypeId="2" ParentId="1" CreationDate="2012-04-25T18:27:18.427" Score="32" Body="&lt;p&gt;As you move from left to right across a period, the number of protons in the nucleus increases.  The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons.  These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.&lt;/p&gt;&#xA;" OwnerUserId="21" LastActivityDate="2012-04-25T18:27:18.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4" PostTypeId="1" AcceptedAnswerId="13" CreationDate="2012-04-25T18:33:07.480" Score="36" ViewCount="25737" Body="&lt;p&gt;A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is &lt;em&gt;minimum&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we have $$\Delta G=\Delta G^\circ + RT\ln Q$$, where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there &lt;em&gt;was&lt;/em&gt; a change in $G$ and $\Delta G &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="194" LastEditDate="2017-02-09T15:02:37.887" LastActivityDate="2020-09-26T18:49:48.763" Title="Gibbs free energy-minimum or zero?" Tags="&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5" PostTypeId="2" ParentId="2" CreationDate="2012-04-25T18:37:11.060" Score="5" Body="&lt;p&gt;A crystal forms to balance the forces of repulsion and attraction amongst the ions. Some ionic compounds form different structures depending on how the are solidified (whether they are created through rapid or slow cooling, for example). Silica (SiO2) is an example of an ionic compound found with &lt;a href=&quot;http://en.wikipedia.org/wiki/Silica#Crystal_structure&quot; rel=&quot;nofollow&quot;&gt;many different structures.&lt;/a&gt; When there is enough energy available to the ions, say at higher temperatures, they can move from these ideal, symmetric and balanced geometries and temporarily maintain a less favorable geometry. An ionic liquid is constantly doing this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crystal's specific structure depends on the size of the ions, because their size affects how the ions pack together.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="123" LastEditDate="2012-06-05T07:27:40.817" LastActivityDate="2012-06-05T07:27:40.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6" PostTypeId="2" ParentId="4" CreationDate="2012-04-25T18:39:16.133" Score="13" Body="&lt;p&gt;Gibbs free energy is a measure of how much &quot;potential&quot; a reaction has left to do a net &quot;something.&quot; So if the free energy is zero, then the reaction is at equilibrium, an no more work can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be easier to see this using an alternative form of the the Gibbs free energy, such as $\Delta G = -T\Delta S$.&lt;/p&gt;&#xA;" OwnerUserId="10" LastEditorUserId="10" LastEditDate="2012-04-25T18:49:03.220" LastActivityDate="2012-04-25T18:49:03.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7" PostTypeId="1" AcceptedAnswerId="12" CreationDate="2012-04-25T18:44:29.807" Score="45" ViewCount="13246" Body="&lt;p&gt;The nitration of &lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-dimethylaniline with $\ce{H2SO4}$ and $\ce{HNO3}$ gives mainly the &lt;em&gt;meta&lt;/em&gt; product, even though $\ce{-NMe2}$ is an &lt;em&gt;ortho&lt;/em&gt;,&lt;em&gt;para&lt;/em&gt;-directing group. Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/59GEW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/59GEW.png&quot; alt=&quot;Reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="16683" LastEditDate="2016-09-09T06:51:46.393" LastActivityDate="2016-09-09T07:08:09.893" Title="Why does nitration of N,N-dimethylaniline occur at the meta position?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8" PostTypeId="2" ParentId="1" CreationDate="2012-04-25T18:45:30.110" Score="17" Body="&lt;p&gt;Remember, the 'size' of an atom has nothing to do with the size of the nucleus. It has to do with the size of the valence shell (which itself is not well-defined*).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we neglect change in electrical attraction, the size should stay the same—a shell is a shell and it need not 'expand' to accomodate electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as we add more protons and electrons, the attraction between the nucleus and shell increases and the shell contracts. Thus the atom gets smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Shells reach 'til infinity, so it's better to define the size on the basis of 'the electrons of the outermost shell will be within this region x% of the time.' This only changes this answer a tiny bit. Now, we say 'the probability of finding the electrons closer to the center increases due to increased nuclear charge.'&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="16" LastEditDate="2012-04-26T15:55:06.287" LastActivityDate="2012-04-26T15:55:06.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9" PostTypeId="5" CreationDate="2012-04-25T18:45:50.773" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-25T18:45:50.773" LastActivityDate="2012-04-25T18:45:50.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10" PostTypeId="4" CreationDate="2012-04-25T18:45:50.773" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-25T18:45:50.773" LastActivityDate="2012-04-25T18:45:50.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11" PostTypeId="1" AcceptedAnswerId="19" CreationDate="2012-04-25T18:48:54.483" Score="26" ViewCount="6502" Body="&lt;p&gt;Re-crystallization is a nice way of purifying a product, but choosing a suitable solvent, if you can't rely on the literature, seems like a lot of trial-and-error.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there any general rules on which kind of solvents could be used for re-crystallization? &lt;/li&gt;&#xA;&lt;li&gt;Which criteria should one use when trying to recrystallize a compound for which no literature on usable conditions are available? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7" LastEditorUserId="15429" LastEditDate="2017-05-30T02:43:38.433" LastActivityDate="2017-05-30T02:43:38.433" Title="Are there any general rules for choosing solvents for recrystallization?" Tags="&lt;purification&gt;&lt;solvents&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12" PostTypeId="2" ParentId="7" CreationDate="2012-04-25T18:51:23.513" Score="39" Body="&lt;p&gt;In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31644/how-is-the-nh3-substituent-meta-directing-in-electrophilic-substitution&quot;&gt;electron-withdrawing &lt;em&gt;via&lt;/em&gt; the inductive effect&lt;/a&gt;. This discourages attack at the electron-poor &lt;em&gt;ortho&lt;/em&gt; position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the conditions I know for that experiment, you get a mixture of &lt;em&gt;para&lt;/em&gt;- and &lt;em&gt;meta&lt;/em&gt;-product, but no ortho-product due to steric hindrance.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2016-09-09T07:08:09.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13" PostTypeId="2" ParentId="4" CreationDate="2012-04-25T18:56:19.917" Score="25" Body="&lt;p&gt;I think your question really arises from some confusion about what &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; represents. In general, &lt;span class=&quot;math-container&quot;&gt;$\Delta X$&lt;/span&gt; for a thermodynamic quantity &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is the change of &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; along some process. You could make it clear by actually writing &lt;span class=&quot;math-container&quot;&gt;$\Delta G(\text{A}\rightarrow\text{B})$&lt;/span&gt; where A and B are before and after states. (We'll note that, in the general case, &lt;span class=&quot;math-container&quot;&gt;$\Delta X$&lt;/span&gt; depends on the path take from A to B, making this notation improper. If &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is a function of state, though, you're good to go.)&lt;/p&gt;&#xA;&lt;p&gt;However, in the equation you quote:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G = \Delta G^0 + RT \ln Q$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;the &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; is a &lt;em&gt;free energy of reaction&lt;/em&gt; and should thus be denoted &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r G$&lt;/span&gt;, with the correct equation being:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The free energy of reaction is defined as &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Thus, this &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r G$&lt;/span&gt; is not the variation of &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; over the entire reaction, which would be the &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; of the system between the start of the reaction and the equilibrium.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;PS: I think &lt;a href=&quot;http://www.chem.arizona.edu/%7Esalzmanr/480a/480ants/g&amp;amp;react/g&amp;amp;react.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorDisplayName="user7951" LastEditDate="2020-09-26T18:49:48.763" LastActivityDate="2020-09-26T18:49:48.763" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="14" PostTypeId="5" CreationDate="2012-04-25T18:59:49.693" Score="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://goldbook.iupac.org/A00493.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC GoldBook&lt;/a&gt;, an atom is the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Smallest particle still characterizing a chemical element. It consists of a nucleus of a positive charge (Z is the atomic number and e the elementary charge) carrying almost all its mass (more than 99.9%) and Z electrons determining its size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This basic unit of &lt;a href=&quot;https://en.wikipedia.org/wiki/Baryon#Baryonic_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryonic matter&lt;/a&gt;, that consists of a dense &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;central nucleus&lt;/a&gt;, i.e. &lt;a href=&quot;/questions/tagged/protons&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;protons&amp;#39;&quot; rel=&quot;tag&quot;&gt;protons&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/neutrons&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;neutrons&amp;#39;&quot; rel=&quot;tag&quot;&gt;neutrons&lt;/a&gt;, surrounded by a cloud of negatively charged &lt;a href=&quot;/questions/tagged/electrons&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;electrons&amp;#39;&quot; rel=&quot;tag&quot;&gt;electrons&lt;/a&gt; is electrically neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that questions about nuclear physics are most suited on &lt;a href=&quot;https://physics.stackexchange.com/questions/tagged/nuclear-physics&quot;&gt;Physics.SE&lt;/a&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="9961" LastEditDate="2018-08-02T00:20:27.863" LastActivityDate="2018-08-02T00:20:27.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15" PostTypeId="4" CreationDate="2012-04-25T18:59:49.693" Score="0" Body="Smallest particle still characterizing a chemical element. It consists of a small nucleus charged positively, carrying almost all of the atom's mass, with electrons  surrounding it. This tag should be applied to questions that specifically concern atoms or their properties. For the charged particles, please use [ions] instead. If your question is specifically about [protons], [neutrons], or [electrons], use those tags instead." OwnerUserId="43" LastEditorUserId="9961" LastEditDate="2018-08-02T00:20:56.017" LastActivityDate="2018-08-02T00:20:56.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16" PostTypeId="1" AcceptedAnswerId="22" CreationDate="2012-04-25T19:00:39.463" Score="24" ViewCount="682" Body="&lt;p&gt;I often hear that water gets purified by being in a silver vessel, which sounds plausible because of bactericidal feature of silver. What doesn't sound plausible, though, is the way it's explained: that silver releases ions into the water. Since silver is a noble metal, why would any &quot;reaction&quot; at all occur with something as neutral as water?&#xA;Is the above explanation nonsense? Does the &quot;disinfection&quot; of water happen only on contact with the metal?&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="4945" LastEditDate="2017-09-26T09:21:28.540" LastActivityDate="2017-09-26T09:21:28.540" Title="What is the origin of the bactericidal properties of silver in water?" Tags="&lt;ions&gt;&lt;purification&gt;&lt;silver&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17" PostTypeId="1" CreationDate="2012-04-25T19:03:31.093" Score="17" ViewCount="1387" Body="&lt;p&gt;In environmental chemistry, most of the organophosphorous compounds we worry about are actually organophosphate compounds. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there something about the $\ce{C-P}$ bond that makes it less stable (and thus less persistent in the environment) than a $\ce{C-O-P}$ bond?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="35" LastEditorUserId="15429" LastEditDate="2017-05-31T08:32:34.287" LastActivityDate="2017-05-31T08:32:34.287" Title="Stability of organophosphorous vs organophosphates?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18" PostTypeId="1" AcceptedAnswerId="65" CreationDate="2012-04-25T19:05:18.847" Score="29" ViewCount="8802" Body="&lt;p&gt;Organic compounds are typically defined as “molecules containing carbon”. &lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_compound&quot;&gt;Wikipedia states&lt;/a&gt; that there for some historical (read: non-logical) reasons,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon, and cyanides, as well as the allotropes of carbon such as diamond and graphite, are considered inorganic.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thus wonder: &lt;strong&gt;is &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot;&gt;activated charcoal&lt;/a&gt; (also known as activated carbon) typically classified as an organic or inorganic material?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I follow the list of exceptions given by Wikipedia, it should be organic (it's not an allotrope of carbon, in particular), but I get the feeling that most people in the field of porous materials would classify it as inorganic. So, I'm looking for an authoritative reference on this question.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="11367" LastEditDate="2017-01-14T14:59:27.527" LastActivityDate="2017-11-01T09:47:50.833" Title="Is activated carbon classified as organic or inorganic?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19" PostTypeId="2" ParentId="11" CreationDate="2012-04-25T19:08:09.370" Score="17" Body="&lt;p&gt;Generally speaking, the best solvent will be dependent on the impurity that you are trying to remove.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solvent must dissolve both the desired compound and the impurity at a high temperature, but only the desired compound at lower temperatures.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_product&quot;&gt;solubility product&lt;/a&gt; of the impurity, as well as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Common-ion_effect&quot;&gt;common ion effect&lt;/a&gt; should both be taken into consideration.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-04-25T19:08:09.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20" PostTypeId="5" CreationDate="2012-04-25T19:08:09.623" Score="0" Body="&lt;p&gt;Ion is an atom or molecule in which the total number of electrons is not equal to the total number of protons, giving it a net positive or negative electrical charge. An ion consisting of a single atom is an atomic or monatomic ion; if it consists of two or more atoms, it is a molecular or polyatomic ion.&lt;/p&gt;&#xA;&lt;p&gt;The name was given by physicist &lt;a href=&quot;https://en.wikipedia.org/wiki/Michael_Faraday&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Faraday&lt;/a&gt; for the substances that allow a current to pass (&amp;quot;go&amp;quot;) between electrodes in a solution when an electric field is applied. It is the transliteration of the Greek participle ἰόν, ión, &amp;quot;going&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/I03158.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An atomic or molecular particle having a net electric charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="124" LastEditorUserId="9961" LastEditDate="2021-05-29T15:42:31.733" LastActivityDate="2021-05-29T15:42:31.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21" PostTypeId="4" CreationDate="2012-04-25T19:08:09.623" Score="0" Body="For questions about ions - atomic or molecular particles having a net electric charge. Do NOT use this tag just because your question involves ions but is not about the ions themselves (as in electrochemistry, etc.)." OwnerUserId="43" LastEditorUserId="9961" LastEditDate="2018-07-12T18:19:39.273" LastActivityDate="2018-07-12T18:19:39.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22" PostTypeId="2" ParentId="16" CreationDate="2012-04-25T19:19:23.213" Score="21" Body="&lt;p&gt;Silver is not as inert as gold. Tarnish is the name we give to the phenomenon when silver metal is oxidized and becomes a salt. Surfaces made of silver tend to disinfect themselves pretty quickly. As for disinfecting water poured into a silver cup, I imagine that would take a little longer since you have to wait for silver to diffuse away from the surface and into the solution. But even very trace levels of silver can have strong antimicrobial effects.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2012-04-25T19:19:23.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23" PostTypeId="2" ParentId="16" CreationDate="2012-04-25T19:22:03.257" Score="11" Body="&lt;p&gt;As far as I know, a chunk of solid silver will not spontaneously react with water.  But if you pass an electrical current through silver electrodes immersed in water,  the silver will be oxidized according to the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O(l) + 2Ag(s) -&amp;gt; 2Ag+(aq) + H2(g) + OH- (aq)}\qquad E^\circ=-1.63\ \mathrm V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will get you the ions you need.  Incidentally, Russia's MIR space station and the ISS use electrolytically dissolved silver ions to purify drinking water.&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorDisplayName="user7951" LastEditDate="2015-10-29T01:34:42.580" LastActivityDate="2015-10-29T01:34:42.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24" PostTypeId="1" AcceptedAnswerId="57" CreationDate="2012-04-25T19:24:31.880" Score="26" ViewCount="2799" Body="&lt;p&gt;What software is available out there to calculate the equilibrium in a set of reactions in aqueous solution? In particular, I'm interested in software general enough to simulate things like titration curves involving acido-basicity, complex formation, precipitation…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A web search turns up plenty of academic papers giving algorithms to find the equilibrium of such systems, but I'd like to find software that actually performs it. In particular, it would be really helpful if there existed such software that included a decent database of reaction constants for some well-known reactions (pKa for common acide-base couples, complexation and precipitation constants, etc.).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; Apparently I was unclear: I'm not looking for molecular simulation codes, but for software that can calculate chemical equilibrium in solution from initial concentrations and a database of equilibrium constants for each possible reaction. The kind of thing teachers can use to predict titration curve and demonstrate the effect of various initial concentrations to students.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="11" LastEditDate="2012-04-26T03:37:20.003" LastActivityDate="2016-07-11T21:43:51.030" Title="What software can calculate aqueous solution equilibria?" Tags="&lt;equilibrium&gt;&lt;software&gt;&lt;aqueous-solution&gt;&lt;acid-base&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="25" PostTypeId="1" AcceptedAnswerId="93" CreationDate="2012-04-25T19:33:07.527" Score="19" ViewCount="487" Body="&lt;p&gt;How can one choose which group has more shifting tendency in 1,2 carbocation rearrangement? The obvious order is via the stability of the carbocation of the group. But, phenylic groups have high shifting tendency--and a phenylic carbocation is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of predicting these? I was thinking that it will be related to the delocalization in the triangular intermediate formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I'm talking about comparing &quot;migratory aptitude&quot; of the $R$ groups in a system similar to $&amp;gt;C(+)-C(R_1R_2R_3)$&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2012-04-26T07:49:24.057" LastActivityDate="2015-03-31T10:29:37.923" Title="Precedence of 1,2 carbocation rearrangement" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26" PostTypeId="1" AcceptedAnswerId="233" CreationDate="2012-04-25T19:44:55.943" Score="15" ViewCount="7288" Body="&lt;p&gt;The reagent &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/440140?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow noreferrer&quot;&gt;APTES&lt;/a&gt; is a fairly common &quot;ink&quot; for microcontact printing, a technique that forms covalent bonds between the silanols found on the surface of the glass and the silane in the the APTES. It's also been demonstrated that the mono-layer then polymerizes somewhat, forming bonds between neighboring silanols.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the mechanism by which the printing and polymerization occur? &lt;/li&gt;&#xA;&lt;li&gt;What is the reason for using APTES instead of a more reactive trichlorosilane?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="45" LastEditorUserId="15429" LastEditDate="2017-06-02T00:56:36.793" LastActivityDate="2017-06-02T00:56:36.793" Title="What is the mechanism of APTES mono-layer formation on glass substrates?" Tags="&lt;surface-chemistry&gt;&lt;reaction-mechanism&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27" PostTypeId="1" AcceptedAnswerId="67" CreationDate="2012-04-25T19:49:30.720" Score="18" ViewCount="3481" Body="&lt;p&gt;I've heard quite a few times that the chromate and permanganate have a $d^3s$ configuration. Also, their colors arise due to a rapid switching of electrons between the oxygen and metal atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand the 'rapid switching' part--it's obvious why it can give color, but I fail to see why there is a need for such a switching-what's so special about $\mathrm{Cr}$ and $\mathrm{Mn}$? (I also do not know what the switching exactly is)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation of $d^3s$ would be appreciated, though not necessary.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="78" LastEditDate="2012-04-26T07:49:53.167" LastActivityDate="2014-02-27T13:13:53.763" Title="Color of chromate and permanganate" Tags="&lt;hybridization&gt;&lt;color&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29" PostTypeId="1" AcceptedAnswerId="777" CreationDate="2012-04-25T19:56:25.523" Score="15" ViewCount="456" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitamin_B12_total_synthesis&quot; rel=&quot;noreferrer&quot;&gt; total synthesis &lt;/a&gt; of vitamin B-12 by Robert Burns Woodward and Albert Eschenmoser, is over 30 years old. At its time, it was considered a landmark in the field.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With current developments (e.g. olefin metathesis, palladium catalysis, etc.) is it possible to optimize this synthesis? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;optimize&quot; I mean reduce the number of steps in the synthesis.&lt;/p&gt;&#xA;" OwnerUserId="52" LastEditorUserId="4231" LastEditDate="2015-02-22T23:21:49.157" LastActivityDate="2015-02-22T23:21:49.157" Title="Can recent developments improve the total synthesis of B-12?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30" PostTypeId="1" AcceptedAnswerId="34" CreationDate="2012-04-25T20:00:09.247" Score="40" ViewCount="13855" Body="&lt;p&gt;It is known that impurities in a desired isolated product lower the melting point of the mixture, even if the impurities' melting point is much higher than the desired product. Why is that so? &lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="7448" LastEditDate="2015-12-14T12:19:42.757" LastActivityDate="2015-12-14T12:19:42.757" Title="Why do impurities lower the melting point of an isolated substance?" Tags="&lt;thermodynamics&gt;&lt;mixtures&gt;&lt;melting-point&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31" PostTypeId="1" CreationDate="2012-04-25T20:06:18.680" Score="25" ViewCount="1126" Body="&lt;p&gt;In a crystal structure I've determined, a triazole ring on my ligand appears to be stacking with a tyrosine (top in picture):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Irazz.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Irazz.jpg&quot; height=&quot;300&quot; alt=&quot;stacking interactions in a protein&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is also an amide, courtesy a glutamine, near it (bottom).  Is it likely this terminal amide is engaging in &amp;pi;-&amp;pi; interactions with the triazole or just weaker dispersion forces?  How would I quantify how significant the interaction is?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="16683" LastEditDate="2018-12-25T13:34:52.310" LastActivityDate="2018-12-25T13:36:09.873" Title="How can I determine if there are π-π interactions between an amide and an aromatic ring in a protein?" Tags="&lt;biochemistry&gt;&lt;carbonyl-compounds&gt;&lt;heterocyclic-compounds&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33" PostTypeId="1" AcceptedAnswerId="51" CreationDate="2012-04-25T20:09:33.027" Score="24" ViewCount="5695" Body="&lt;p&gt;What conditions promote a nucleophile to undergo the Michael reaction over the nucleophilic attack at the alpha carbon of the carbonyl group in an alpha-beta-unsaturated ketone? I'm looking for an answer that considers organic and inorganic nucleophiles (like organocuprates/organolithium compounds).&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="78" LastEditDate="2012-04-26T01:51:36.790" LastActivityDate="2013-10-19T19:34:30.590" Title="Michael reaction or nucleophilic attack at the alpha carbon of an alpha-beta-unsaturated ketone?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;thermodynamics&gt;&lt;reaction-control&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34" PostTypeId="2" ParentId="30" CreationDate="2012-04-25T20:10:35.037" Score="32" Body="&lt;p&gt;It's a very general statement, but it's not always true. I'll explain why it's often true, and give a counter-example at the end.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your majority component B and the impurity (let's call it A) form a binary system. In most cases, such binary mixtures exhibit a solid–liquid phase diagram as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Og92.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image taken from &lt;a href=&quot;http://research.pbsci.ucsc.edu/chemistry/li/teaching/chem268/Interpretation%20of%20Phase%20Diagram.pdf&quot; rel=&quot;noreferrer&quot;&gt;these lecture notes&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This binary phase diagram has pure A on the left, pure B on the right. A and B form, somewhere, a eutectic. It is the point here at concentration &lt;em&gt;e&lt;/em&gt; and temperature &lt;em&gt;y&lt;/em&gt;. Because the existence of a eutectic point is guaranteed for any A/B binary system, and because the eutectic corresponds to a lower temperature, your liquidus curve decreases with increasing impurity concentration, and the impurity thus lowers the melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not all binary mixtures form a eutectic. In the words of &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Not all binary alloys have a eutectic point; for example, in the silver-gold system the melt temperature (liquidus) and freeze temperature (solidus) both increase monotonically as the mix changes from pure silver to pure gold.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The corresponding phase diagram is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.crct.polymtl.ca/fact/phase_diagram.php?file=Ag-Au.jpg&amp;amp;dir=SGTE&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fidq0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-04-26T03:12:52.453" LastActivityDate="2012-04-26T03:12:52.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="38" PostTypeId="2" ParentId="31" CreationDate="2012-04-25T20:48:39.247" Score="23" Body="&lt;p&gt;It seems that NickT was looking for an experimental solution. My post deals, however, with a computational solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would I quantify how significant the interaction is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Determining the interaction energy between two defined monomers such as your aromatic triazole and amide is a rather straightforward process.  This process is referred to as the supramolecular approach.  I'll point you to a paper that analyzes the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja025896h&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzene dimer&lt;/a&gt;.  This method is strictly a computational one so a bit of knowledge in computational chemistry is necessary.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;The Supramolecular Approach&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The What&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supramolecular approach boils down to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_{\mathrm{int}} =  E_{\mathrm{dimer}} - (E_{\mathrm{mon1}} + E_{\mathrm{mon2}})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have some interaction energy (&lt;span class=&quot;math-container&quot;&gt;$E_{\mathrm{int}}$&lt;/span&gt;) determined from the difference of a dimer energy (&lt;span class=&quot;math-container&quot;&gt;$E_{\mathrm{dimer}}$&lt;/span&gt;) and the sum of the two monomers (&lt;span class=&quot;math-container&quot;&gt;$E_{\mathrm{monomer}}$&lt;/span&gt;).  If both monomers were equivalent (say, you were interested in the interaction energy of the benzene dimer where each monomer was a benzene ring), you could simplify the summation to two times the energy of one monomer (&lt;span class=&quot;math-container&quot;&gt;$2\times E_{\mathrm{mon}}$&lt;/span&gt;). In your particular case, you have two different monomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole idea is if I have two interacting molecules, I can determine the energy of each molecule individually (as if they were separated at infinite distance) as well as their complex.  So as you bring these molecules closer and closer together, the energy starts to go down (the interaction energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The How&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Your geometry is from a crystal structure.  Do NOT modify this geometry.  You will want to keep everything exactly as is.  This means you don't want to optimize your system.  You do not want to eyeball monomer placement.  Take everything from your known structure and be careful not to change it otherwise it can ruin this process.  We are modifying the structure by truncating and capping but intermolecular parameters must stay the same for whatever it is you are trying to model.  You will want to optimize your cap meaning, run an optimization on your capped-monomer but freeze everything but the thing you are using to cap.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define your monomers.  You will need to determine what part of your 'dimer' system is important for describing this weak interaction.  I recommend keeping the aromatic ring and truncating the ring with something similar to what is being truncated.  You could cap your monomer with a hydrogen or a methyl group for example.  If the piece you've cut out is highly polarizable, cap with something with a similar property.  If your truncated piece is neutral in charge, cap with something that is neutral.  You get the idea.&lt;/li&gt;&#xA;&lt;li&gt;Define your dimer.  Your dimer is simply a combination of your two defined monomers.&lt;/li&gt;&#xA;&lt;li&gt;Determine the method you want to implement.  Wavefunction based or density functional theory (DFT) will work for this.  The former will be prohibitively expensive. &lt;/li&gt;&#xA;&lt;li&gt;Define your basis set.  For aromatic systems your best bet would be to use Dunning-Hunzaga's correlation consistent family of basis sets.  I recommend using aug-cc-pVTZ for good results.  Whatever you decide, be sure your basis set includes polarization and diffuse functions.  The suggested basis set does this (augmented means diffuse on all atoms whereas pVTZ means 'polarized-valence triple zeta'). &lt;/li&gt;&#xA;&lt;li&gt;Determine the energies of your monomers and dimer.  You will want to run a single-point energy calculation on your monomers and your dimer. Optimize your monomers first but freeze all atoms except those you added to cap your monomer.&lt;/li&gt;&#xA;&lt;li&gt;Determine the interaction energy.  Plug your energies into the equation given above and determine the interaction energy.  Convert to whatever units you wish to use (I prefer kJ/mol but most people use kcal/mol so you may want to use that).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.  Looks like an interesting project to say the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A quick note about the 'valence' basis set:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically core electrons do not participate in any sort of interesting chemistry (they are so much lower in energy than the valence electrons that they do not play a role in things like chemical bonding).  When you perform these energy computations, you will be invoking the frozen-core approximation (default on many quantum software packages but be sure to check the documentation first).  The pVTZ basis set expands the basis functions out to triple-zeta quality for valence electrons only.  The core electrons are given &lt;em&gt;s&lt;/em&gt;-type functions and that is it.  If you ever find yourself in a situation where the core electrons are important to your system of interest, you would want to use the pCVXZ basis sets (where X = D,T,Q, etc.) which stands for polarized core-valence X zeta.  Here the core electrons are given the full set of X-zeta quality basis functions.  However this means you've just made your computation much more expensive.  The frozen-core approximation is typically used.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="16683" LastEditDate="2018-12-25T13:36:09.873" LastActivityDate="2018-12-25T13:36:09.873" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="40" PostTypeId="2" ParentId="27" CreationDate="2012-04-25T21:35:26.457" Score="2" Body="&lt;p&gt;&quot;Color&quot; is what we see when electrons transition from a high-energy orbital to a less energetic one, if the energy associated with that transition happens to produce a photon within the visible spectrum. It might be helpful to stop thinking of electrons as being on either the metal or the oxygen, and to consider that in these cases electrons occupy molecular orbitals, which can encompass more than one atom. In this case, it seems likely that one of the orbitals in question might have higher electron density around the metal, while the other could have higher electron density around oxygen.&#xA;Why are Cr and Mn special? I would interpret it as coincidence that the oxides of these metals are prone to electronic transitions that generate photons which register as vibrant colors to our eyes.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2012-04-25T21:35:26.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="41" PostTypeId="2" ParentId="29" CreationDate="2012-04-25T21:47:34.650" Score="7" Body="&lt;p&gt;A bibliographic for total syntheses of Vitamin B&lt;sub&gt;12&lt;/sub&gt; or any other cobalamin turns out to be fruitless. In fact, all references turn out to be to the famous historical synthesis, but I could not find much work (any work, in fact) on total syntheses, or even proposals for new synthesis routes. Not much was achieved since then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is partly due to lack of practical need. There are many molecules out there that are challenging to synthesize, and Vitamin B&lt;sub&gt;12&lt;/sub&gt; is currently produced at the industrial level by bacterial fermentation. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitamin_B12#Synthesis_and_industrial_production&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The species &lt;em&gt;Pseudomonas denitrificans&lt;/em&gt; and Propionibacterium shermanii are more commonly used today. These are frequently grown under special conditions to enhance yield, and at least one company […] used genetically engineered versions of one or both of these species. Since a number of species of &lt;em&gt;Propionibacterium&lt;/em&gt; produce no exotoxins or endotoxins and are generally regarded as safe (have been granted GRAS status) by the Food and Drug Administration of the United States, they are presently the FDA-preferred bacterial fermentation organisms for vitamin B&lt;sub&gt;12&lt;/sub&gt; production.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A good recent reference on the subject of its synthesis would be: K. C. Nicolaou and E. J. Sorensen, “Vitamin B12” in &lt;em&gt;Classics in Total Synthesis&lt;/em&gt;, VCH: New York, 2003.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-04-25T21:47:34.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="42" PostTypeId="5" CreationDate="2012-04-26T00:01:26.030" Score="0" Body="&lt;p&gt;An organic molecule is a molecule containing carbon, typically with significant amounts of hydrogen, oxygen, and nitrogen but also containing other stable (usually main group, non-metallic) elements such as the halogens, sulfur, phosphorous and others. Organic molecules were historically the molecules isolated from nature, such as the amino acids and sugars, but the definition has since been expanded to include ‘man-made’ organic molecules.&lt;/p&gt;&#xA;&lt;p&gt;Organic chemistry is the branch of chemistry that deals with these organic molecules, and the reactions of these molecules to form new ones which may be useful as pharmaceuticals, materials, dyes etc.&lt;/p&gt;&#xA;&lt;h2&gt;Definition:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://www.acs.org/content/acs/en/careers/college-to-career/areas-of-chemistry/organic-chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The American Chemical Society&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Organic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which include not only hydrocarbons but also compounds with any number of other elements, including hydrogen (most compounds contain at least one carbon–hydrogen bond), nitrogen, oxygen, halogens, phosphorus, silicon, and sulfur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Organic chemistry is a chemistry subdiscipline involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms.&#xA;The range of chemicals studied in organic chemistry include hydrocarbons (compounds containing only carbon and hydrogen), as well as myriad compositions based always on carbon, but also containing other elements, especially oxygen, nitrogen, sulfur, phosphorus (these, included in many organic chemicals in biology) and the radiostable elements of the halogens.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Applicability of the &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; tag:&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; should be applied to any questions dealing with organic molecules, whether relative to structure, reactivity or any other property.&lt;/li&gt;&#xA;&lt;li&gt;The field of organic chemistry is vast, and as such the &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; tag is not very descriptive by itself (it is the most used tag on chemistry.se), therefore additional tags should be added to help narrow down the scope of the question. In general, at least two other tags will be appropriate: a first tag dealing with the topic of the question such as &lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt;, and a second tag dealing with the molecule, for which a series of functional group tags exist such as &lt;a href=&quot;/questions/tagged/carbonyl-compound&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;carbonyl-compound&amp;#39;&quot; rel=&quot;tag&quot;&gt;carbonyl-compound&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Related tags:&lt;/h2&gt;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following which may be useful as additional tags on questions identified as being related to organic chemistry:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/synthesis&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/stereochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/nomenclature&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nomenclature&amp;#39;&quot; rel=&quot;tag&quot;&gt;nomenclature&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/chirality&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chirality&amp;#39;&quot; rel=&quot;tag&quot;&gt;chirality&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Further reading:&lt;/h2&gt;&#xA;&lt;p&gt;The literature of Organic Chemistry is vast, however, there are a handful of highly respected books that are a good starting point for the subject:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Clayden, J.; &lt;em&gt;Organic Chemistry&lt;/em&gt;; Oxford University Press: Oxford, 2012.&lt;/li&gt;&#xA;&lt;li&gt;March, J.; &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt;; Wiley: New Jersey, 2013&lt;/li&gt;&#xA;&lt;li&gt;Carey, F.; &lt;em&gt;Advanced Organic Chemistry- Part A&lt;/em&gt;; Springer: New York, 2007&lt;/li&gt;&#xA;&lt;li&gt;Carey, F.; &lt;em&gt;Advanced Organic Chemistry- Part B&lt;/em&gt;; Springer: New York, 2007&lt;/li&gt;&#xA;&lt;li&gt;Robert T. Morrison, Robert N. Boyd; &lt;em&gt;Organic Chemistry - 6th edition&lt;/em&gt;; Pearson: 1992&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12" LastEditorUserId="95133" LastEditDate="2020-08-17T16:14:44.663" LastActivityDate="2020-08-17T16:14:44.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="43" PostTypeId="4" CreationDate="2012-04-26T00:01:26.030" Score="0" Body="Use this tag for questions relating to organic molecules and their properties (structure of organic molecules, spectroscopic properties, reaction mechanisms, stereochemistry etc). DO NOT use this tag as the only tag in your question, as this tag by itself cannot appropriately classify your question. Always use this tag in addition to other more specific tags." OwnerUserId="30" LastEditorUserId="42741" LastEditorDisplayName="user95" LastEditDate="2017-07-12T11:32:43.410" LastActivityDate="2017-07-12T11:32:43.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="44" PostTypeId="2" ParentId="2" CreationDate="2012-04-26T00:11:33.207" Score="11" Body="&lt;p&gt;If I'm reading your question rightly, I think the answer to your question is &quot;geometry.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, sodium and chlorine atoms form the &lt;em&gt;particular&lt;/em&gt; structure of face-centered cubic because the relative size of the ions -- small for sodium large for chlorine -- makes that arrangement less stressful (energetically favorable). However, if you change the relative sizes of the ions, face-centered cubic is no longer guaranteed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cesium chloride is an example. It's not really body-centered since it has two different kinds of ions, but the larger size of cesium ions gives it the same form as body centered, only with &lt;a href=&quot;http://www.ilpi.com/inorganic/structures/cscl/index.html&quot;&gt;cesium and chloride ions alternating&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2012-04-26T00:31:30.907" LastActivityDate="2012-04-26T00:31:30.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="45" PostTypeId="1" AcceptedAnswerId="362" CreationDate="2012-04-26T00:27:07.043" Score="15" ViewCount="4252" Body="&lt;p&gt;One synthesis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quinolone&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;quinolones&lt;/strong&gt;&lt;/a&gt; begins with the formation of an &lt;strong&gt;ethyl ethoxymethylenemalonate&lt;/strong&gt;, as seen in this &lt;em&gt;Organic Syntheses&lt;/em&gt; &lt;a href=&quot;http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv3p0395&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2taiJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been asked if the &lt;strong&gt;malonate&lt;/strong&gt; derivative would be formed if &lt;strong&gt;methylmalonate&lt;/strong&gt; was treated with &lt;strong&gt;trimethyl orthoformate&lt;/strong&gt; with a catalytic amount of $\mathrm{H^+}$, &lt;em&gt;without&lt;/em&gt; heating. The nature of the solvent isn't mentioned, but I guess the reaction is carried out in a non-aqueous medium with an &lt;a href=&quot;http://www.amberlyst.com/sac.htm&quot; rel=&quot;noreferrer&quot;&gt;ion-exchange resin&lt;/a&gt;, since I think &lt;strong&gt;trimethyl orthoformate&lt;/strong&gt; acts as a water trap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is that there shouldn't be a significant quantity of product formed. We're dealing with an &lt;strong&gt;activated methylene&lt;/strong&gt;, why not simply use a base and then carry on with our addition? I can't imagine how, under normal circumstances, the elimination product &lt;strong&gt;ethyl ethoxymethylenemalonate&lt;/strong&gt; will be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this reaction happen in a mechanistically reasonable way?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-04-26T01:13:27.047" LastActivityDate="2012-05-12T21:35:12.047" Title="Acid catalysed synthesis of ethyl ethoxymethylenemalonate" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="46" PostTypeId="1" AcceptedAnswerId="355" CreationDate="2012-04-26T01:04:34.733" Score="22" ViewCount="4586" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2/how-does-nacl-maintain-its-crystalline-structure&quot;&gt;This question&lt;/a&gt; on NaCl crystalization actually got me wondering: are there any ionic amorphous solids? Like ionic crystals are crystalline materials of electrostatically-attracted ions, can ions form an amorphous phase? I can see no reason why not, but I cannot think of any example either…&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2014-12-15T10:32:55.510" Title="Are there any ionic amorphous solids?" Tags="&lt;solid-state-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="47" PostTypeId="2" ParentId="18" CreationDate="2012-04-26T01:22:53.597" Score="5" Body="&lt;p&gt;Charcoal is the product of burning, dehydrating, and evaporating the volatile substances from formerly living matter.  Since activated carbon is made by processing charcoal, it would be considered an organic product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a bit of a grey area here, admittedly.  Iron is typically considered inorganic, but it's absolutely critical to life.  In the context of studying hemoglobin, some experts speak about this iron as organic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the distinction between &quot;organic&quot; vs &quot;inorganic&quot; is less meaningful than it used to be.  &lt;/p&gt;&#xA;" OwnerUserId="87" LastActivityDate="2012-04-26T01:22:53.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="48" PostTypeId="2" ParentId="46" CreationDate="2012-04-26T01:32:22.980" Score="4" Body="&lt;p&gt;The following is not a proof, but an idea that would imply that it is not possible to have such a material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any ionic &quot;sea&quot; of atoms, there are two types of situations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The atoms have enough energy to move freely, disregarding the attractive and repulsive forces of the other ions in the sea. This would cause a fluid, as the particles are not maintaining a position anywhere.&lt;/li&gt;&#xA;&lt;li&gt;They do not have that much energy, and therefore they will be impacted by the attractive and repulsive charges in the &quot;sea.&quot; It then follows that there is an optimal configuration that will stabilize all of these forces. While I suppose that it might be possible for the optimal configuration not to be periodic (and thus not crystalline) I would tend to believe that it is not possible (it would seem from tessellations that if one assumes some basic characteristics, they must be regular). In a similar vein, I believe that it is not possible to have degrees of freedom in such an optimized state (though there may of course be more than one locally optimal state possible) as each atom is at a local minimum of energy, and it lacks the ability to &quot;move up&quot; the gradient (if it had more energy, it would be a liquid).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I therefore believe that it is improbable that such things exist.&lt;/p&gt;&#xA;" OwnerUserId="10" LastActivityDate="2012-04-26T01:32:22.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="49" PostTypeId="2" ParentId="30" CreationDate="2012-04-26T02:02:10.093" Score="19" Body="&lt;p&gt;Thermodynamically, you're considering the chemical potentials ($\mu$) of the liquid and solid(s), specifically the temperature where they're equal.  In a mixture, the potential is lower as the disorder (entropy) has increased, so all things equal, it will favor the liquid over a purer solid where there can be more disorder.  Chemical systems seek to lower their potential through spontaneous chemical changes (e.g. of phase), minimizing the free energy ($G$).  Equilibrium occurs where $dG = 0$.  In a constant temperature, constant pressure system,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dG =\sum_i\mu_idn_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu$ is the chemical potential of a given species (some compound in some phase) and $n$ is the amount of that compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming our solid generated is pure (valid sometimes, not always), our reaction (freezing) is&#xA; $$\require{mhchem} \ce{A_{(l)} -&amp;gt; A_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the relationship between the two species is equal and opposite (generating one mole of solid consumes one mole of liquid).&#xA;$$-dn_{(l)} = dn_{(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;dG = 0 &amp;amp;= \mu{(l)}dn_{(l)} + \mu{(s)}dn_{(s)}\\&#xD;&#xA; &amp;amp;= \mu{(l)}dn_{(l)} - \mu{(s)}dn_{(l)}\\&#xD;&#xA;0 &amp;amp;= \mu{(l)} - \mu{(s)}\\&#xD;&#xA;\mu{(s)} &amp;amp;= \mu{(l)}\\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the solid is pure (prior assumption, note: $\star$ denotes pure compound), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu{(s)} = \mu^\star{(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ideal mixture (assumes interactions between all components are equal) with mole fraction $\chi_A$ and $T$ being the freezing point of the mixture,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If non-ideal, a general $a$ term is used $\mu{(l)} = \mu^\star{(l)} + RT \ln a_A$).  All together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\mu^\star{(s)} &amp;amp;= \mu{(s)} = \mu{(l)} = \mu^\star{(l)} + RT \ln \chi_A\\&#xD;&#xA;\mu^\star{(s)} - \mu^\star{(l)} &amp;amp;= RT \ln \chi_A\\&#xD;&#xA;-\Delta G^\star_{m, fus} &amp;amp;= RT \ln \chi_A\\&#xD;&#xA;-(\Delta H^\star_{m, fus} - T \Delta S^\star_{m, fus}) &amp;amp;= RT \ln \chi_A\\&#xD;&#xA;\frac{\Delta S^\star_{m, fus}}{R} - \frac{\Delta H^\star_{m, fus}}{RT} &amp;amp;= \ln \chi_A\\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mixture is pure (freezing point at $T^\star$), $\chi_A = 1$, so $\ln \chi_A = 0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;-\frac{\Delta H^\star_{m, fus}}{RT^\star} + \frac{\Delta S^\star_{m, fus}}{R} &amp;amp;= 0\\&#xD;&#xA;\frac{\Delta S^\star_{m, fus}}{R} &amp;amp;= \frac{\Delta H^\star_{m, fus}}{RT^\star} \\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combined...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\frac{\Delta H^\star_{m, fus}}{RT^\star} - \frac{\Delta H^\star_{m, fus}}{RT} &amp;amp;= \ln \chi_A\\&#xD;&#xA;\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{1}{T} - \frac{1}{T^\star}\right) &amp;amp;= \ln \chi_A\\&#xD;&#xA;\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{T^\star - T}{TT^\star}\right) &amp;amp;= \ln \chi_A\\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $T \approx T^\star$ and if we define $\Delta T$ as the change in equilibrium temperature (melting point) from the pure substance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\frac{\Delta H^\star_{m, fus}}{R}\left(\frac{\Delta T}{T^{\star2}}\right) &amp;amp;= \ln \chi_A\\&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be taken further to derive the coefficient for the freezing point depression, but from here we can see that because $\Delta H$, all $T$'s, and $R$ are positive and $\ln \chi_A$ is guaranteed to be negative ($\chi_A$ must be less than zero), $\Delta T$ must be negative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To recap the assumptions, this is an ideal mixture and the solid formed is pure.  Addressing the first, if non-ideal, we can use the general $a_A$ instead of $\chi_A$.  The mole fraction can never exceed 1, but I'm not sure about $a$; if it does exceed 1, then there would be a freezing point &lt;em&gt;elevation&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second, if the solid formed is a mixture (e.g. a metal alloy), the that throws another wrench in the works that I don't definitively know how to address.  I believe it would then be based on the difference of interaction between the components in liquid phase versus solid phase as well as the relative concentration in each.&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditDate="2012-04-26T21:03:16.393" LastActivityDate="2012-04-26T21:03:16.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="50" PostTypeId="2" ParentId="24" CreationDate="2012-04-26T02:06:16.773" Score="6" Body="&lt;p&gt;I've not tried it out, but implementations of &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp004368u&quot;&gt;ReaxFF&lt;/a&gt; may be suitable for your purposes. It performs atomistic finite-temperature simulation of reaction mixtures using a forcefield and as such is more general than your request but it seems like it could do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's available in &lt;a href=&quot;http://www.scm.com/Products/Overview/ReaxFF.html&quot;&gt;ADF as an addon&lt;/a&gt; (note the &lt;a href=&quot;http://www.scm.com/Products/images/ReaxFFmovie.png&quot;&gt;nice demonstration&lt;/a&gt; of the time evolution of a combustion process), in &lt;a href=&quot;http://lammps.sandia.gov/&quot;&gt;LAMMPS&lt;/a&gt; (again, a &lt;a href=&quot;http://lammps.sandia.gov/workshops/Feb10/Aidan_Thompson/ReaxFF_LAMMPS_2010.pdf&quot;&gt;cool demonstration&lt;/a&gt; of the concentration of species changing over time as a shockwave propagates through a crystal) and probably a lot of other packages.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-04-26T02:17:00.790" LastActivityDate="2012-04-26T02:17:00.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="51" PostTypeId="2" ParentId="33" CreationDate="2012-04-26T02:19:03.160" Score="22" Body="&lt;p&gt;An &lt;strong&gt;$\alpha,\beta$-unsaturated ketone&lt;/strong&gt; is electrodeficient at the $\beta$ position. This can be seen if you draw the resonance structures of such a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\beta$ carbon is thus a good site for nucleophilic attack. But, as you know, carbonyls are also prone to nucleophilic attack. To discriminate between the two, you need to look at how the reaction is &lt;em&gt;controlled&lt;/em&gt;, either &lt;em&gt;thermodynamically&lt;/em&gt; or &lt;em&gt;kinetically&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a kinetically controlled reaction, the product that is formed fastest predominates. In a thermodynamically controlled reaction, the predominant product is the energetically favored one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Michael addition is a 1-4 addition, where a nucleophile attacks the $\beta$ carbon, and produces the thermodynamically favored product. On the other hand, a 1-2 reaction (on the carbonyl) gives the kinetic product, and is obtained at low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the 1-4 product thermodynamically more stable? Because the resulting product benefits from keto-enol tautomerism, which results in lowering the energy of the system. Usually, the more resonance forms a compound has, the more its electrons are delocalized, the more stable it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the resonance forms of the 1-4 and 1-2 products, and see.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You asked for specific affinities of different organometallics in 1-4/1-2 additions. My knowledge is that organocuprates ($\mathrm{R-CuLi}$) will perform Michael additions, and that organolithians seem to prefer 1-2 addition. Also, according to &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-4-2.html&quot;&gt;this source&lt;/a&gt;, Grignard reagents do not seem to have a preference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My take on this is that the cuprate is less reactive, and therefore can form the thermodynamic product, whereas the lithium reagent is so destabilized that it reacts right away.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-04-26T02:38:00.733" LastActivityDate="2012-04-26T02:38:00.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="52" PostTypeId="2" ParentId="33" CreationDate="2012-04-26T02:30:27.713" Score="10" Body="&lt;p&gt;It must first be understood that Michael Addition is a special type of conjugate (1,4) addition where the nucleophilic species is an enolate. This will undergo 1,4 addition to the A-B unsaturated carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organocuprates add in a 1,4 fashion. The mechanism behind this is not fully understood, but it is known that radical intermediates are involved. Organolithium agents add in a 1,2 fashion. The reason for this is that the organolithium is very close to a carbanion and it will attack the most positive species in the molecule. In this case, the carbonyl carbon is the most electrophilic and so the organolithium will attack there. For reference, organomagnesium compounds are fickle and can attack either 1,2 or 1,4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alpha carbon is the carbon adjacent to the carbonyl carbon. The alpha carbon is actually quite nucleophilic (which is aldol condensation, alpha-alkylation, and alpha halogenation can occur). So, a nucleophile will attack either the partially positive berta carbon or the carbonyl carbon.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2012-04-26T02:30:27.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="53" PostTypeId="2" ParentId="7" CreationDate="2012-04-26T02:35:01.187" Score="21" Body="&lt;p&gt;In these acidic conditions, the lone pair on the nitrogen will abstract a proton of the acid in an acid-base fashion. This protonated amine will then be considered a meta-directing group. It is a deactivator that will draw electron density away from the ring (because nitrogen, a relatively electronegative atom, will have a positive charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to get around this is to N-acylate. This can be done by first reacting the mono-substituted amine with acetic chloride ($\ce{CH3COCl}$) with pyridine to get an amide. That way, the ring is still activated and the lone pair on the amide nitrogen will not interfere with acids. It will direct ortho-para (para favored due to sterics).&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorDisplayName="user7951" LastEditDate="2015-10-02T22:39:30.967" LastActivityDate="2015-10-02T22:39:30.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="54" PostTypeId="1" CreationDate="2012-04-26T02:54:13.763" Score="50" ViewCount="5183" Body="&lt;p&gt;Many of us have experienced the failure of nitrile gloves when exposed to chloroform. What's going on at a mechanistic level when this occurs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that the chloroform dissolves some of the polymer into its constituent monomers, but I've never heard anything more definite than that. Is it a similar mechanism to what happens when chloroform is left too long in a plastic bottle?&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="537" LastEditDate="2012-08-20T07:46:59.283" LastActivityDate="2014-06-23T18:29:42.917" Title="How does chloroform degrade nitrile gloves?" Tags="&lt;polymers&gt;&lt;safety&gt;&lt;solvents&gt;&lt;halides&gt;&lt;gloves&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="55" PostTypeId="2" ParentId="46" CreationDate="2012-04-26T03:08:40.070" Score="11" Body="&lt;p&gt;The difficulty with creating an ionic bond glass is that creating true glasses depends on a combination of directional bonding between atoms (or molecules) and a rate of cooling that doesn't give those bonds time to move into optimal positions. The result is a bit like piling together sticky spheres that become locked before they can find an optimal stacking arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure ionic bonds in contrast are more like slippery spheres that are attracted towards each others' centers, but don't have any particular preferences about which parts of their surfaces make contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, the most likely result by far from very rapid cooling of a molten ionic solid will be to produce micro or nanocrystalline ionic solids that may superficially resemble glasses, but which under X-ray or neutron diffraction crystallography would prove actually to have crystal structure at a very fine scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, you nonetheless can come up with a recipe for &lt;em&gt;trying&lt;/em&gt; to make a truly glassy ionic solid, especially since there's no such thing as a pure ionic bond. You would want to cool a very thin layer of molten salt at at an incredibly high rate, at least millions of degrees per second if I recall glassy metal cooling rates rightly. For ionic, you would also want the result to be pretty close to absolute zero, since whatever degree of &quot;directional stickiness&quot; you can get out of ionic solids is likely to be less even than metallic bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you &lt;em&gt;really&lt;/em&gt; want to try to make such a thing as a serious project, you of course must check the literature in detail. My rule of thumb is that if you dig deep enough, there are almost no decently plausible ideas that you can think of in the physical sciences that someone has not at least written a paper on in some journal, and maybe even done some experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, what you are asking is an intriguing idea for a real experiment, and it wasn't that many years ago that both metallic glasses and Penrose-tiled quasicrystals were new ideas.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2012-04-26T03:08:40.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="56" PostTypeId="2" ParentId="2" CreationDate="2012-04-26T05:23:46.687" Score="35" Body="&lt;p&gt;Crystals have inspired a great many chemists because they are fascinating &lt;em&gt;for a good reason&lt;/em&gt;. Not only are they aesthetically pleasing, but they serve as an excellent subject to tour a variety of theoretical subjects important for understanding high-level chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystalline materials are made up of periodic structures. We’re only going to primarily focus on binary compounds where there is &lt;strong&gt;not&lt;/strong&gt; a high degree of covalency. There are several ways to think about this problem, but let’s start with the melting of a crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that at some definite temperature a highly ordered crystal will melt into a liquid. Those of us familiar with the language of equilibrium thermodynamics might recognize that the change in free energy for this phase change can be written, at constant temperature, as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G_\text{liquid} - G_\text{crystal} = H_\text{liquid} - H_\text{crystal} - T ( S_\text{liquid} - S_\text{crystal} ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G = \Delta H - T \Delta S $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we suppose that this process is spontaneous then we would say that the change in Gibbs’ free energy is negative, i.e. $\Delta G &amp;lt; 0$. This is true if and only if,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H &amp;lt; T \Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally we interpret this as saying that there is a thermally-driven increase in entropy when we melt a highly ordered crystal into a liquid which more than offsets the energy cost associated with the enthalpies of the interactions holding that crystal together. A chemist tends to learn early on that the reverse is not necessarily true: at some definite temperature a perfect crystal &lt;strong&gt;rarely&lt;/strong&gt; forms from the liquid. This inability to just heat up any substance and always produce a perfect crystal by cooling illustrates how crystal formation is a case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_versus_kinetic_reaction_control&quot; rel=&quot;noreferrer&quot;&gt;kinetic- rather than thermodynamic- control&lt;/a&gt;. So the process by which you form your crystal could possibly result in a different crystal structure. Sometimes crystal structures change just by altering the temperature of the chamber you’re measuring the crystal structure in! Now neither of these cases apply to sodium chloride to the best of my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of an ionic crystal such as sodium chloride is a delicate balance between electrostatic attraction and Pauli repulsion. Electrostatic attraction says that between two different charges, $q_+$ and $q_-$, there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Coulomb force&lt;/a&gt; given by, $$F= \frac{k q_+ q_-}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the distance between the two charges. If one plays with the numbers then it’s easy to see that at short distances the force is strongest, but there is a limit to how close they may come together. Eventually a repulsive force due to a quantum mechanical principle called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;noreferrer&quot;&gt;Pauli Exclusion Principle&lt;/a&gt; overpowers the attraction. An equilibrium results in which the atoms sit a certain distance from one another so that, if you will humor me, the “forces” between them balance out. This is why we traditionally represent crystal packing using marbles with a unique radii. The radii of the hard marble represents where the Pauli repulsion overpowers the attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say, “Sure, we have these kinetic, electrostatic and quantum mechanical factors to consider, but how do these help with the final crystal structure?” Hold your horses, we’re getting there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A famous mathematician and scientist thought about the most efficient ways to pack spheres of the same size together. By most efficient I mean this in terms of what FedEx considers efficient, fitting things together into the smallest possible volume. This is also what electrostatics want. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_conjecture&quot; rel=&quot;noreferrer&quot;&gt;Kepler&lt;/a&gt; suggested that the best way to pack spheres with this in mind can be maximized in two ways: &lt;a href=&quot;http://en.wikipedia.org/wiki/Face-centered_cubic&quot; rel=&quot;noreferrer&quot;&gt;fcc&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_close_packing&quot; rel=&quot;noreferrer&quot;&gt;hcp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This conjecture, however, rests on all the spheres being uniform. We can’t assume this is the same for the atoms in table salt because the cation, sodium, is said to be smaller than the chloride anion. A tool that is useful as a &lt;em&gt;guide&lt;/em&gt; for helping predict the structure is based off of the relative size of the cation, $r_\text C$, to the anion, $r_\text A$. This radius ratio, $r_\text C / r_\text A$, is mostly only useful for simple binary species. These $r_\text C$ and $r_\text A$ are tabulated as ionic radii in many chemistry books; they are contrived by various rules proposed by researchers, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Linus_Pauling&quot; rel=&quot;noreferrer&quot;&gt;Linus Pauling&lt;/a&gt;, applied to the experimentally determined interionic distances. In this system we find it useful to discuss how many neighbors an anion or a cation has of the opposite charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordination number ($\mathrm{CN}$) tells us how many neighbors a type of atom has around it. A certain $r_\text C/r_\text A$ gives us a feel for the probable CN. From one source I was able to calculate the ratio for $\ce{NaCl}$ as $r_{\ce{Na+}}/r_{\ce{Cl-}} = 0.564$. Most textbooks will say that if your ratio is $0.414{–}0.732$, then you have $\mathrm{CN}=6$. I’ll show you how to calculate the minimum value for $\mathrm{CN}=6$. The easiest way to obtain the maximum is to obtain the minimum for $\mathrm{CN}=8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, if we put a small cation on a $XY$ plane and surround it in the manner shown below then we could also place two circles in the $Z$-axis above and below the central cation for $\mathrm{CN}=6$. (I won’t show those two in the $Z$-axis for clarity.) We would say that a given $\mathrm{CN}$ is stable only if the spheres are all touching each other. We would say that for the ratios drawn that a $\mathrm{CN}=8$ is not stable because our cation’s sphere would be too small and wouldn’t touch all eight of its neighbors. A higher density is always preferred, so $\mathrm{CN}=4$ is not preferred when $\mathrm{CN}=6$ is stable. &lt;img src=&quot;https://i.stack.imgur.com/H6TV6.png&quot; alt=&quot;for $\mathrm{CN}=6$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, $\angle \mathrm{DAC} = 45^{\circ}$. Moreover, we can say that $\overline{\mathrm{DC}} = r_\text A$ and $\overline{\mathrm{AC}} = r_\text A + r_\text C$. Trigonometry will tells us that $$\sin 45^{\circ} = \frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}$$ Substituting in those values just determined,&#xA;$$\frac{\sqrt{2}}{2} = \frac{r_\text A}{\overline{r_\text A + r_\text C}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $r_\text C/r_\text A$ gives,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_\text C/r_\text A=\frac{2}{\sqrt{2}} -1 = 0.414$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the lower bound desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can say that the $\mathrm{CN}=6$ for the cation. Similarly you can figure out that $\mathrm{CN}=6$ for the anion. This is why we say it has a 6:6 coordination pattern, which is indicative of a sodium chloride type structure. Cesium chloride, on the other hand, is 8:8 and indicative of its structure type. We tend to say that these patterns are consistent with a certain crystal structure because all the pieces fit together consistently. This geometric concept is based purely on sphere packing and is not without its limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more quantitative way to figure out which geometry is preferred is to calculate the electrostatic energies for the different geometries. A Russian by the name of Kapustinskii derived a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kapustinskii_equation&quot; rel=&quot;noreferrer&quot;&gt;formula&lt;/a&gt; to do this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{lattice} (\mathrm{kJ/mol}) = \frac{\alpha}{r_\text C + r_\text A} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you need to put it in terms of the radius ratio with a constant $r_\text A$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I should note that this is evil algebra: $ \frac{a/b}{d/b + 1}= \frac{a}{b+d} $ where $a=\alpha$, $b=r_\text A$, $d=r_\text C$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{lattice} = \frac{\alpha/r_\text A}{\frac{r_\text C}{r_\text A} + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically $\alpha$ is different for a change in the crystal structure type and from there you compare which has the preferred energy. You can reproduce the geometric trends using this more quantitative model. There are problems with this approach too, such as the neglected dipole and quadrupole and covalency. Yes, even in $\ce{NaCl}$ there is some covalent character. I think we say that $\ce{NaCl}$ $67\ \%$ ionic character. These become important when two different crystal structures are close in energy or when more exact calculations are required. &lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T12:18:07.707" LastActivityDate="2015-05-20T12:18:07.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="57" PostTypeId="2" ParentId="24" CreationDate="2012-04-26T05:41:25.713" Score="13" Body="&lt;p&gt;&lt;a href=&quot;http://www.eawag.ch/research_e/surf/Researchgroups/sensors_and_analytic/chemeql.html&quot;&gt;ChemEQL&lt;/a&gt; is pretty nice in some instances and free, but it looks like it has been mostly abandoned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of &lt;a href=&quot;http://code.google.com/p/titrator-delphi/&quot;&gt;Titrator&lt;/a&gt;, but it only runs on Windows and I've never had  a chance to run it. It's also free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard good things about &lt;a href=&quot;http://www2.lwr.kth.se/English/OurSoftware/vminteq/&quot;&gt;Visual MINTEQ&lt;/a&gt;, but it is Windows only and I've never had a chance to run it. Supposedly well maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also &lt;a href=&quot;http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/&quot;&gt;PHREEQC&lt;/a&gt;, which sort of works on stuff other than Windows. I hear it is harder to use, but probably worth it.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-04-26T05:41:25.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="58" PostTypeId="5" CreationDate="2012-04-26T06:19:06.757" Score="0" Body="&lt;p&gt;Thermodynamics is the study of various forms of energy in observable systems. As applied to chemical reactions, thermodynamics is used to study and explain the behaviour of chemical reactions in terms of the energy contained within chemical bonds of a substance, and which must be input into or is output from the reaction as a requisite or consequence of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different questions that come under this tag involve one or more of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;free energy, enthalpy, internal energy or entropy of a gas or their change&lt;/li&gt;&#xA;&lt;li&gt;state and path functions&lt;/li&gt;&#xA;&lt;li&gt;heat capacity&lt;/li&gt;&#xA;&lt;li&gt;expansion and compression processes or their graphical representation&lt;/li&gt;&#xA;&lt;li&gt;various standard enthalpies of reaction (sublimation/neutralization/etc.)&lt;/li&gt;&#xA;&lt;li&gt;spontaneity of a reaction&lt;/li&gt;&#xA;&lt;li&gt;Hess's law/Born-Haber cycle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and other related topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions about statistical mechanics or &quot;microscopic derivation of thermodynamical laws&quot; invoking thermodynamics are suited for &lt;a href=&quot;https://physics.stackexchange.com/tags/thermodynamics/info&quot;&gt;Physics.SE&lt;/a&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-03-14T13:39:45.183" LastActivityDate="2018-03-14T13:39:45.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="59" PostTypeId="4" CreationDate="2012-04-26T06:19:06.757" Score="0" Body="Applicable to questions about heat, energy, work, and their interconversion in chemistry. See the tag wiki for a detailed list of topics. Questions tagged may also be tagged with [enthalpy], [energy], [free-energy] where appropriate." OwnerUserId="94" LastEditorUserId="5026" LastEditDate="2018-03-14T13:41:58.060" LastActivityDate="2018-03-14T13:41:58.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="60" PostTypeId="5" CreationDate="2012-04-26T06:19:43.577" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-26T06:19:43.577" LastActivityDate="2012-04-26T06:19:43.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="61" PostTypeId="4" CreationDate="2012-04-26T06:19:43.577" Score="0" Body="Questions relating to the environment in terms of Chemistry." OwnerUserId="94" LastEditorUserId="94" LastEditDate="2012-04-26T15:57:46.640" LastActivityDate="2012-04-26T15:57:46.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="62" PostTypeId="1" CreationDate="2012-04-26T06:48:23.603" Score="14" ViewCount="397" Body="&lt;p&gt;In the hydrogenation of unsaturated compounds with hydrogen gas and a catalyst, the choice of palladium on carbon is able to hydrogenate alkenes and alkynes, but is unable to hydrogenate aromatic compounds. Using rhodium or Raney nickel, however, allows one to hydrogenate aromatics. How does the choice of catalyst affect what kind of unsaturated compounds can be hydrogenated?&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastEditorUserId="7448" LastEditDate="2015-12-14T12:17:55.773" LastActivityDate="2015-12-14T12:17:55.773" Title="How does the choice of metal (oxide) catalyst affect the range of unsaturated compounds that can be hydrogenated?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="63" PostTypeId="1" AcceptedAnswerId="75" CreationDate="2012-04-26T06:53:50.587" Score="84" ViewCount="3862" Body="&lt;p&gt;The most notable characteristic of polytetrafluoroethylene (PTFE, DuPont's Teflon) is that &lt;em&gt;nothing&lt;/em&gt; sticks to it. This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing indeed sticks to Teflon, how might one coat an object (say, a frying pan) with PTFE?&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastEditorUserId="33991" LastEditDate="2016-11-21T12:45:47.087" LastActivityDate="2020-07-06T19:28:10.677" Title="If nothing sticks to Teflon, how does Teflon stick to frying pans?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="64" PostTypeId="1" CreationDate="2012-04-26T07:00:40.530" Score="24" ViewCount="1790" Body="&lt;p&gt;Solvent extraction in a separation funnel is a very common method in preparative organic chemistry. But sometimes you don't get a nice phase separation between the organic and the water phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the possible causes that can prevent the formation of two distinct phases, and how can you force a phase separation when you encounter such situations?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="121" LastEditDate="2012-06-21T20:15:08.043" LastActivityDate="2012-06-21T20:15:08.043" Title="What can I do if I don't get a phase separation between my organic and water phases?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="65" PostTypeId="2" ParentId="18" CreationDate="2012-04-26T07:54:25.113" Score="25" Body="&lt;p&gt;While nomenclature is of particular interest to organic chemists to specify an exact compound, the classification of X into broad category Y or Z isn't a precise science, and not really of practical use.  The article cites a textbook by Seager to this effect, stating &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The distinction between &quot;organic&quot; and &quot;inorganic&quot; carbon compounds, while &quot;useful in organizing the vast subject of chemistry... is somewhat arbitrary&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Even if you find a source that says &quot;charcoal is (in)organic&quot;, you may just as well find one stating the opposite.  Just like the coal from which it may have been produced, it was once biomass and decidedly organic, but so was graphite and diamond, or CO&lt;sub&gt;2&lt;/sub&gt; and CO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;2&amp;minus;&lt;/sup&gt;. I think it's overly pedantic and unproductive to try to come up with definitive judgements for these decidedly edge case scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, it's just a chemical on the shelf, what one &lt;em&gt;does&lt;/em&gt; with it is far more relevant.  I don't use it on a daily basis, but it seems more like a tool than a reagent.  The fact it contains carbon seems beside the point; it's value isn't in the chemical composition but rather its extraordinary adsorptive properties.&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2012-04-26T07:54:25.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="66" PostTypeId="2" ParentId="64" CreationDate="2012-04-26T07:58:54.857" Score="6" Body="&lt;p&gt;Well, shaking a sep-funnel is a great way of creating a horrible intractable emulsion with no clear phase boundary, so don't do that (also, the stopper might pop off as soon as you set it down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for breaking an emulsion, if you have the time, just leave the sep-funnel somewhere and wait a few days. I seem to recall the first and last time I shook a sep-funnel the lab tech added some salt to the water/diethyl ether emulsion however I'm not clear on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consider sacrificing part of your product by drawing the funnel off to the very base of where the emulsion begins, draining the emulsion layer entirely and then taking the remainder. If you have blobs of one phase in the other, you may be able to remove them with a pipette.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-04-26T07:58:54.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="67" PostTypeId="2" ParentId="27" CreationDate="2012-04-26T08:21:50.023" Score="11" Body="&lt;p&gt;By &quot;rapid switching&quot; they technically mean &lt;a href=&quot;http://goldbook.iupac.org/L03520.html&quot;&gt;Ligand-to-Metal Charge Transfer&lt;/a&gt; (LMCT). A more modern framework is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ligand_field_theory&quot;&gt;Ligand Field Theory&lt;/a&gt;. I would have to teach a class to fully explain it in those terms, but I'll try to explain it in terms of this &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot;&gt;hybridization&lt;/a&gt; you brought up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chemical bond implies a higher probability of finding electrons between two bound nuclei. &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;Atomic orbitals&lt;/a&gt; describe the electron density for individual nuclei. A bond between two nuclei necessitates some physical overlap of the relevant atomic orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is in this way that we say that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_combination_of_atomic_orbitals&quot;&gt;linear combination of atomic orbitals&lt;/a&gt;, say two consisting of a $2s$ and $2p$ orbital, results in a molecular orbital that describes the molecule. We would write the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave_function&quot;&gt;wave function&lt;/a&gt; for the complex as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of &quot;mixing&quot; is just a matter of adjusting the coefficients, $C_n$. And so we might say that the bond in this made-up molecule is $sp$ hybridized. $sp^3$ hybridized would mean the following kind of wave function:&#xA;$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p) \pm C_3 \psi(2p) \pm C_4 \psi(2p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $sd^3$ could mean a wave function of the following nature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi = C_1 \psi( (n+1) s) \pm C_2 \psi(n d) \pm C_3 \psi(nd) \pm C_4 \psi(nd) $$ where $n=3$ in the case of Mn or Cr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one performs the proper &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital&quot;&gt;molecular orbital&lt;/a&gt; calculations on the valence $3d$, $4s$, and $4p$ for manganese or chromium and the $2s$ and $2p$ for oxygen in tetrahedral symmetry then you can draw the molecular orbital energy diagram for your complex once you match your eigenvalues to the correct trace of the appropriate transformation matrix... and you can hear a bunch more jargon on it, or I can explain the main point of our models: why we think MnO$_4$ $^-$ is purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds in, specifically permanganate  (I have never actually done the calculations on dichromate, although they should be similar in principle) are between Mn$^{+7}$ and four O$^{-2}$ in the geometry shown &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPermanganate-anion-2D.png&quot;&gt;here&lt;/a&gt;. The electronegativity of oxygen clearly dictates most of the electron density and, as a result, has them sitting on ligands (oxygen) instead of the metal. An electronically excited state can be reached with light &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_%28electromagnetic_radiation%29&quot;&gt;absorption&lt;/a&gt; in the range of 500-600 &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavelength&quot;&gt;nm&lt;/a&gt; light due to the relative weakness of the interactions between the ligand and metal. If they were stronger interactions then it would take more energy to promote electrons into a higher energy state and the color would be shifted into the UV. A colorwheel tells you that light absorption in the 500-600nm range should be approximately violet, which is what we see for the color of this particular complex. Anyway, this excited state means some electrons temporarily move from the ligand to the metal, resulting in a LMCT band.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-04-26T08:21:50.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="68" PostTypeId="2" ParentId="63" CreationDate="2012-04-26T08:37:06.947" Score="13" Body="&lt;p&gt;Before putting Teflon on a pan, the pan is scratched hardly leaving some tiny holes where the hole opening is smaller than the hole inner size (like a bottle). When Teflon is cast into that, it physically can't get out. That's also why when you scratch a Teflon pan, things start sticking to it.&lt;/p&gt;&#xA;" OwnerUserId="28" LastEditorUserId="10" LastEditDate="2012-04-26T15:56:21.230" LastActivityDate="2012-04-26T15:56:21.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="69" PostTypeId="1" AcceptedAnswerId="78" CreationDate="2012-04-26T09:19:35.163" Score="35" ViewCount="21723" Body="&lt;p&gt;I'm currently taking VCE (Victorian Certificate of Education) Chemistry classes, and we're currently studying the interpretation of spectra produced by Hydrogen NMR (Nuclear Magnetic Resonance) spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When studying the spectra of High Resolution &lt;sup&gt;1&lt;/sup&gt;H NMR, the peaks representing the different Hydrogen environments are split into multiplets based on the protons surrounding these environments. There has been a considerable amount of confusion in my classes over the actual principles / rules of thumb on how to calculate the multiplets for a particular environment of a known chemical (i.e.; known structue), based off the 'n + 1' rule.&lt;br&gt;&#xA;(i.e.; an environment with n neighbours will be split into n+1 multiplets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are absolute on the principles that;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Peaks of a particular Hydrogen environment are not split by neighbouring protons in equivalent environments. &lt;/li&gt;&#xA;&lt;li&gt;OH does not, and is not split by, it's neighbouring environments.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, immense confusion arised over whether the following principle was correct.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Peaks of a particular Hydrogen environment will only be split by the protons in neighbouring environments once for each type of neighbouring environment.&lt;br&gt;&#xA;e.g.  The middle &quot;CH&lt;sub&gt;2&lt;/sub&gt;&quot; environment in &quot;CH&lt;sub&gt;3&lt;/sub&gt;–CH&lt;sub&gt;2&lt;/sub&gt;–CH&lt;sub&gt;3&lt;/sub&gt;&quot; will only have 4 peaks;&lt;br&gt;&#xA;Although it has 6 neighbouring protons, they are two lots of the same environment (CH&lt;sub&gt;3&lt;/sub&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a class, we found numerous examples from different text books and sources that provide examples of ¹H NMR spectra which did not clarify the matter; Some considered all neighbouring protons as neighbours, others discriminated on the repeated neighbouring environments. For example, the CH&lt;sub&gt;2&lt;/sub&gt; in CH&lt;sub&gt;3&lt;/sub&gt;–CH&lt;sub&gt;2&lt;/sub&gt;–CH&lt;sub&gt;3&lt;/sub&gt; was sometimes split into 4 peaks or 7 peaks, depending on the source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Chemistry teachers contradicted each other on the matter.&lt;br&gt;&#xA;There was repeated self corrections made by the teachers, such that now nobody really knows whether this principle is correct or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there anybody that has the correct information on the matter?&lt;br&gt;&#xA;Is there a reasonable explanation behind this strange lack of correlation,&lt;br&gt;&#xA;or is there a common misconception about multiplet splitting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately; How many multiplets should the CH&lt;sub&gt;2&lt;/sub&gt; in CH&lt;sub&gt;3&lt;/sub&gt;–CH&lt;sub&gt;2&lt;/sub&gt;–CH&lt;sub&gt;3&lt;/sub&gt; have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that if there is a complicated explanation that I am currently only at Year 12 VCE level, so links to resources I can pursue would be extremely helpful! It's been established that VCAA (an authority for the education system in Australia) ensures that the chemicals featured in the exams for NMR analysis will not be of a structure so as to allow the ambiguity above.)&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="30" LastEditDate="2012-04-27T12:53:13.700" LastActivityDate="2015-11-06T04:20:59.657" Title="Splitting of multiplets in ¹H NMR spectra?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="70" PostTypeId="2" ParentId="64" CreationDate="2012-04-26T10:07:35.507" Score="8" Body="&lt;p&gt;One such cause, where separation does not take place is the nature of the organic solvent. Its better to check the miscibility of the solvent with water. There are several websites, where one can get this info. For example &lt;a href=&quot;http://www.erowid.org/archive/rhodium/pdf/solvent.miscibility.pdf&quot;&gt;1&lt;/a&gt; and &lt;a href=&quot;http://www.sigmaaldrich.com/chemistry/solvents/water-center/miscibility-immiscibility.html&quot;&gt;2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with hydrophobic solvent, if the problem persists, one can add NaCl and then check the separation. &lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2012-04-26T10:07:35.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="71" PostTypeId="1" CreationDate="2012-04-26T10:19:00.100" Score="6" ViewCount="362" Body="&lt;p&gt;I am curious about the timescales of protonation and deprotonation of solvent systems. As a followup, how is this affected when the proton source is separated by an organic phase? For instance, it is known that the pH inside a liposome can be maintained from the outside. What are the timescales of protonation when the protons need to be transported across various micelles thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://makemelonger.com&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="57" LastEditDate="2012-04-26T17:53:08.173" LastActivityDate="2014-10-16T21:50:16.173" Title="Timescale of Protonation and Deprotonation across micelle layers" Tags="&lt;kinetics&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="72" PostTypeId="1" AcceptedAnswerId="318" CreationDate="2012-04-26T10:23:25.990" Score="16" ViewCount="3024" Body="&lt;p&gt;Our work is faced with the issue of determining the charge of compounds that are dissolved in chloroform. One could realistically determine the charge of the relevant moieties knowing the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of those groups. However, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How would you determine or calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of an organic solvent? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An alternative way of asking this question is how would one calculate the concentration of hydronium ions or free protons in a nonaqueous solution.&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="41328" LastEditDate="2019-01-07T06:51:06.433" LastActivityDate="2019-01-07T06:51:06.433" Title="Determining the pH of an organic solvent" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="74" PostTypeId="2" ParentId="24" CreationDate="2012-04-26T10:30:53.247" Score="6" Body="&lt;p&gt;After some more search, I have found two pieces of software who do titration simulation, but are (as far as I can tell) only available in French:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.linux-france.org/prj/xem/&quot;&gt;Xem&lt;/a&gt; and (I can't believe what I'm about to do)… &lt;a href=&quot;http://translate.google.com/translate?sl=fr&amp;amp;tl=en&amp;amp;js=n&amp;amp;prev=_t&amp;amp;hl=en&amp;amp;ie=UTF-8&amp;amp;layout=2&amp;amp;eotf=1&amp;amp;u=http%3A%2F%2Fwww.linux-france.org%2Fprj%2Fxem%2F&amp;amp;act=url&quot;&gt;Google translation&lt;/a&gt; of the same page&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jeanmarie.biansan.free.fr/dozzzaqueux.html&quot;&gt;Dozzzaqueux&lt;/a&gt; and &lt;a href=&quot;http://translate.google.com/translate?sl=fr&amp;amp;tl=en&amp;amp;js=n&amp;amp;prev=_t&amp;amp;hl=en&amp;amp;ie=UTF-8&amp;amp;layout=2&amp;amp;eotf=1&amp;amp;u=http%3A%2F%2Fjeanmarie.biansan.free.fr%2Fdozzzaqueux.html&amp;amp;act=url&quot;&gt;translation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-04-26T10:30:53.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="75" PostTypeId="2" ParentId="63" CreationDate="2012-04-26T10:49:14.380" Score="66" Body="&lt;p&gt;It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www2.dupont.com/Teflon/en_US/products/qanda_cookware_care.html#careq1&quot;&gt;&lt;strong&gt;“If nothing sticks to Teflon®, then how does Teflon® stick to a pan?&quot;&lt;/strong&gt;&lt;/a&gt;&lt;br/&gt;&#xA;  Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that makes it adhere to the metal surface of a pan.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And from &lt;a href=&quot;http://www2.dupont.com/Teflon/en_US/products/cookware_how_it_works.html&quot;&gt;this other webpage&lt;/a&gt; of theirs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7aAt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primer (or primers, if you include the “mid coat” in the picture above) adheres to the roughened surface, often obtained by sandblasting, very strongly: it's chemisorption, and the primer chemical nature is chosen as to obtain strong bonding to both the metal surface. Then, the PTFE chain extremities create bonds with the primer. And thus, it stays put.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-04-26T10:49:14.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="76" PostTypeId="1" CreationDate="2012-04-26T11:42:15.200" Score="7" ViewCount="748" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hofmeister_series&quot; rel=&quot;noreferrer&quot;&gt;Hofmeister Series&lt;/a&gt; is a useful metric in understanding how to precipate proteins. However, wikipedia and many other source indicate that the order of the series is rather empirical. I'm curious about the variety of explanations of the mechanism of the series as well as the application of the series to other non-protein macromolecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/psJGL.png&quot; alt=&quot;Hofmeister Series&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2012-06-12T22:19:39.063" Title="Molecular explanation of the Hofmeister Series" Tags="&lt;ions&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="77" PostTypeId="2" ParentId="76" CreationDate="2012-04-26T11:52:18.933" Score="7" Body="&lt;p&gt;The molecular origin of the Hofmeister series is not well understood and still a very active area of research. For example, see the upcoming &lt;a href=&quot;http://www.rsc.org/ConferencesAndEvents/RSCConferences/FD160/index.asp&quot;&gt;Faraday discussion&lt;/a&gt; conference. The syllabus for the conference says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although it is now clear that the Hofmeister series is intimately connected with ion hydration in homogeneous and heterogeneous environments and with ion pairing, the molecular origin of these effects has been poorly understood. Biochemists and physical chemists have been typically using the term Hofmeister series to put a label on ion specific behaviour in various environments, rather than to reach a molecular level understanding and, consequently, an ability to predict a particular effect of a specific salt ion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(So, this is not an answer, but rather too long for a comment…)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-04-26T12:14:03.103" LastActivityDate="2012-04-26T12:14:03.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="78" PostTypeId="2" ParentId="69" CreationDate="2012-04-26T12:34:46.190" Score="24" Body="&lt;p&gt;Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain like CH&lt;sub&gt;3&lt;/sub&gt;CH&lt;sub&gt;2&lt;/sub&gt; the coupling between the CH2 and the CH3 hydrogens would be a $\mathrm{^3J}$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $\mathrm{^4J}$-couplings or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In propane the hydrogens of both CH&lt;sub&gt;3&lt;/sub&gt; groups are three bonds away from the hydrogens of the CH&lt;sub&gt;2&lt;/sub&gt; group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH&lt;sub&gt;3&lt;/sub&gt; group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH&lt;sub&gt;3&lt;/sub&gt; groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH&lt;sub&gt;2&lt;/sub&gt; hydrogens. The result of that is a splitting into a septet (7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="30" LastEditDate="2012-04-27T13:00:23.583" LastActivityDate="2012-04-27T13:00:23.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="79" PostTypeId="1" AcceptedAnswerId="86" CreationDate="2012-04-26T14:15:53.693" Score="44" ViewCount="27882" Body="&lt;p&gt;When treated with hot, concentrated acidic $\ce{KMnO4}$, arenes are oxidised to the corresponding carboxylic acids. For example, toluene is oxidised to benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to examine how this happens, using the mechanism of oxidation of double bonds via cyclic intermediate as a reference, but I can't manage to cook up a satisfactory one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an older book, I have read that there is no (known) mechanism for many organic oxidation reactions. I'm inclined to think that this may have changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there a mechanism for this? If so, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what are the hurdles in finding this mechanism? For example, what problems are there with other proposed mechanisms (if they exist)?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="16683" LastEditDate="2018-03-25T17:06:36.323" LastActivityDate="2018-03-25T17:06:36.323" Title="Mechanism of arene side chain oxidation by permanganate" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;organic-oxidation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="82" PostTypeId="2" ParentId="69" CreationDate="2012-04-26T16:03:44.180" Score="12" Body="&lt;p&gt;In addition to mad scientists answer I would like to explain what the quantum mechanical phenomenon is that leads to a J-Coupling. The interesting part about the &lt;em&gt;J&lt;/em&gt;-coupling for chemists is that this coupling is a proof for the presence of an electron bond between the two coupling nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coupling is based on two principles: a Fermi contact between the nucleus and an electron, the Pauli exclusion principle that two electrons must not have the same set of quantum numbers. In this case if one electron spin is up, then the other electron in this bonding orbital must be down. The other electron couples then to the second nucleus by another Fermi contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain of couplings is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleus 1 ---Fermi contact---&gt; electron 1 ----Pauli exclusion principle----&gt; electron 2 ---Fermi contact----&gt; Nucleus 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This total chain is the effect which is called &lt;em&gt;J&lt;/em&gt;-coupling. In a handwaving fashion one could say that through the first nucleus feels through the coupling if the second nucleus is up or down. This is shifting the energy levels a bit which can be then found in the spectrum as a shift in the resonance position. As the energy conservation principle holds one line is shifted to lower energies and the other by the same amount to higher energy leading to a doublet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OH proton is actually split as well by the &lt;em&gt;J&lt;/em&gt;-coupling, however the line width of the OH resonance is bigger than the coupling constant, so the doublet is smeared out.&lt;/p&gt;&#xA;" OwnerUserId="111" LastActivityDate="2012-04-26T16:03:44.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="86" PostTypeId="2" ParentId="79" CreationDate="2012-04-26T19:40:29.973" Score="38" Body="&lt;p&gt;Some general information on &lt;strong&gt;side-chain oxidation in alkylbenzenes&lt;/strong&gt; is available at &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/arenes/other.html&quot; rel=&quot;noreferrer&quot;&gt;Chemguide&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An alkylbenzene is simply a benzene ring with an alkyl group attached&#xA;  to it. Methylbenzene is the simplest alkylbenzene.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by&#xA;  an alkaline solution of potassium manganate(VII) (potassium&#xA;  permanganate).&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Methylbenzene is heated under reflux with a solution of potassium&#xA;  manganate(VII) made alkaline with sodium carbonate. The purple colour&#xA;  of the potassium manganate(VII) is eventually replaced by a dark brown&#xA;  precipitate of manganese(IV) oxide.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The mixture is finally acidified with dilute sulfuric acid.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Overall, the methylbenzene is oxidised to benzoic acid.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KGsmw.gif&quot; alt=&quot;AlkylBenzene&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Interestingly, any alkyl group is oxidised back to a -COOH group on&#xA;  the ring under these conditions. So, for example, propylbenzene is&#xA;  also oxidised to benzoic acid.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2OzZ5.gif&quot; alt=&quot;Benzoic acid&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the mechanism, a Ph.D. student at the University of British Columbia did his doctorate on the mechanisms of permanganate oxidation of various organic substrates.&lt;sup&gt;1&lt;/sup&gt; Quoting from the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It was found that the most vigorous oxidant was the permanganyl ion ($\ce{MnO3+}$), with some contributing oxidation by both permanganic acid ($\ce{HMnO4}$) and permanganate ion ($\ce{MnO4-}$) in the case of easily oxidized compounds such as alcohols, aldehydes, or enols.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The oxidation of toluene to benzoic acid was one of the reactions investigated, and a proposed reaction mechanism (on pp 137–8) was as follows. In the slow step, the active oxidant $\ce{MnO3+}$ abstracts a benzylic hydrogen from the organic substrate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\begin{align}&#xA;\ce{2H+ + MnO4- &amp;amp;&amp;lt;=&amp;gt; MnO3+ + H2O} &amp;amp; &amp;amp;\text{(fast)} \\&#xA;\ce{MnO3+ + PhCR2H &amp;amp;-&amp;gt; [PhCR2^. + HMnO3+] &amp;amp; &amp;amp;\text{(slow)}} \\&#xA;\ce{[PhCR2^. + HMnO3+] &amp;amp;-&amp;gt; PhCR2OH + Mn^V} &amp;amp; &amp;amp;\text{(fast)} \\&#xA;\ce{PhCR2OH + Mn^{VII} &amp;amp;-&amp;gt; aldehyde or ketone} &amp;amp; &amp;amp;\text{(fast)} \\&#xA;\ce{aldehyde + Mn^{VII} &amp;amp;-&amp;gt; benzoic acid} &amp;amp; &amp;amp;\text{(fast)} \\&#xA;\ce{ketone + Mn^{VII} &amp;amp;-&amp;gt; benzoic acid} &amp;amp; &amp;amp;\text{(slow)} \\&#xA;\ce{5 Mn^V &amp;amp;-&amp;gt; 2Mn^{II} + 3Mn^{VII}} &amp;amp; &amp;amp;\text{(fast)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The abstraction of a benzylic hydrogen atom is consistent with the fact that arenes with no benzylic hydrogens, such as &lt;em&gt;tert&lt;/em&gt;-butylbenzene, do not get oxidised.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reference&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Spitzer, U. A. The Mechanism of Permanganate Oxidation of Alkanes, Arenes and Related Compounds. Ph.D. Thesis, The University of British Columbia, November 1972. &lt;a href=&quot;https://doi.org/10.14288/1.0060242&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.14288/1.0060242&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="114" LastEditorUserId="16683" LastEditDate="2017-08-08T11:46:58.740" LastActivityDate="2017-08-08T11:46:58.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="87" PostTypeId="1" CreationDate="2012-04-26T22:10:45.390" Score="25" ViewCount="3117" Body="&lt;p&gt;I have a solution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;copper acetate&lt;/a&gt; and I would like to play around with the ligands to get different colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; The copper acetate was made through mixing vinegar (5% acetic acid), &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}_{\textrm{(s)}}$&lt;/span&gt;. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit] I neglected to mention that the experiment simultaneously produced a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 inch}$&lt;/span&gt; deposit of what appears to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper_carbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Copper Carbonate&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Verdigris&quot; rel=&quot;nofollow noreferrer&quot;&gt;Verdigris&lt;/a&gt; on the bottom of the 1-Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former &lt;span class=&quot;math-container&quot;&gt;$\pu{1L}$&lt;/span&gt; is now &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 L}$&lt;/span&gt; and has begun precipitating crystals (like the ones on the wiki page). Fun fact: During the Renaissance, glacial &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;acetic acid&lt;/a&gt; was made by &lt;a href=&quot;http://alchymie.ca/radicalvinegar.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;dry distilling metal acetates&lt;/a&gt; and primarily copper(II) acetate.[/edit]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question(s):&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color? &lt;/li&gt;&#xA;&lt;li&gt;Will heating or cooling the solution change the color and/or ligands?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="114" LastEditorUserId="61578" LastEditDate="2019-12-19T18:04:12.820" LastActivityDate="2019-12-19T18:04:12.820" Title="How may copper acetate ligands be manipulated to change colors?" Tags="&lt;coordination-compounds&gt;&lt;home-experiment&gt;&lt;color&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="88" PostTypeId="1" AcceptedAnswerId="173" CreationDate="2012-04-27T03:04:14.023" Score="20" ViewCount="1790" Body="&lt;p&gt;One frequent argument against the safety of the synthetic sweetener aspartame is that it can be hydrolyzed, with one of the hydrolysis products being methanol, which is known to be toxic. Of course, usually not much aspartame is added in food anyway to yield a sufficient amount of methanol or its oxidation product formaldehyde, but this does beg the question: why wasn't, say, the ethyl ester chosen to be developed as an artificial sweetener? Is there a structure-activity relationship between the alcohol used to esterify the dipeptide and the sweetness of the product?&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastEditorUserId="10185" LastEditDate="2015-09-12T19:36:16.023" LastActivityDate="2019-11-01T19:54:40.690" Title="During the development of aspartame, why was the methyl ester chosen?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;&lt;esters&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="89" PostTypeId="1" AcceptedAnswerId="95" CreationDate="2012-04-27T05:36:41.457" Score="18" ViewCount="1529" Body="&lt;p&gt;The nucleobases in DNA and RNA are generally present in the keto-form, and not in the enol-form. As an interesting historical note, James Watson and Francis Crick did initially believe them to adopt the enol-form, which isn't compatible with the correct model of the DNA they later proposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are some reports indicating that in certain RNA structures guanosine and uridine form essentially a Watson-Crick basepair where one of the two bases is present in the enol-form, instead of the usual G-U wobble base pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything known on how large the energy difference between the keto- and the enol-form of those bases actually is? &lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="10185" LastEditorDisplayName="user95" LastEditDate="2015-09-24T17:16:29.140" LastActivityDate="2015-09-24T17:16:29.140" Title="How large is the energy difference between keto- and enol-form of guanine and uracil?" Tags="&lt;organic-chemistry&gt;&lt;structural-biology&gt;&lt;tautomer&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="90" PostTypeId="2" ParentId="17" CreationDate="2012-04-27T09:17:04.210" Score="10" Body="&lt;p&gt;&lt;strong&gt;Setting the scene for your question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Environmental Chemistry there are several pathways a chemical might travel through including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;being subjected to the biochemistry of an animal&lt;/li&gt;&#xA;&lt;li&gt;photodegradation&lt;/li&gt;&#xA;&lt;li&gt;chemical degradation (perhaps it slowly reacts with water, etc.)&lt;/li&gt;&#xA;&lt;li&gt;uptake as a metabolite in plants&lt;/li&gt;&#xA;&lt;li&gt;degradation by soil microorganisms  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and this is ignoring aquatic and air pathways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, within a mouse or a fly, a possible pathway for degradation for some insecticide is cleavage of the P-O bond. How they have determined this in the past is noteworthy; a synthetic chemist would &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopic_labeling&quot;&gt;isotope label&lt;/a&gt; an atom in the molecule, say $^{33}$&lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_phosphorus&quot;&gt;P&lt;/a&gt;-labeled or $^{14}$&lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_carbon&quot;&gt;C&lt;/a&gt;-labeled, and he purified it before finally determining its &lt;a href=&quot;http://en.wikipedia.org/wiki/Specific_activity&quot;&gt;specific activity&lt;/a&gt;. Then a biochemist would dose a mouse with a solution of the compound dissolved in an appropriate solvent, such as olive oil. Urine and feces were collected after administration. An aliquot of the collected urine, for example, would have its radioactivity measured using &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_scintillation_counting&quot;&gt;scintillation counting&lt;/a&gt;. The organo-soluble species would be extracted from the rest of the urine and the distribution of radioactivity could then be determined between the aqueous and organic layers. Then one could investigate what were present by &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_layer_chromatography&quot;&gt;TLC&lt;/a&gt; if you're old school -- they would actually run, via cochromatography, all the possible metabolite products they could independently think of and synthesize to identify a TLC spot! (That is a lot of work.) These days we can cheat with &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_chromatography%E2%80%93mass_spectrometry&quot;&gt;HPLC-MS&lt;/a&gt;. So, you might ask, &quot;are results from this generalizable?&quot; I have to tell you that they generalize in a very rough way. Enzymes are responsible for all the nifty biochemistry and they tend to be rather selective towards a given substrate and, in many cases, &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29&quot;&gt;chirality&lt;/a&gt; even matters! What they are picky about is a tough problem that is tailored to each individual enzyme. One can make all kinds of generalizations about the chemical nature of the bonds and some possible predictions from it, but those won't necessarily hold up in an animal model with a different molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So nitty-gritty mechanisms might not be what we want to look at. Instead we'll step back and take a broader view of this landscape. We need to look at bigger items in terms of physical properties, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Persistence: what is its half-life in soil, etc. This is a matter of degradation, whereas the next is all about movement of an intact molecule. A simple chemical way to test this is how they resist hydrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Mobility: over time, does the compound tend to sit or spread itself around?&#xA;For example, factors that influence this are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sorption&quot;&gt;sorption&lt;/a&gt;: how much does our chemical bind to our soil?&lt;/li&gt;&#xA;&lt;li&gt;water solubility: how easy it is to move in water for our chemical?&lt;/li&gt;&#xA;&lt;li&gt;vaporizability: how likely is the material to evaporate away from its source?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A famous organophosphate is &lt;a href=&quot;http://en.wikipedia.org/wiki/Malathion&quot;&gt;malathion&lt;/a&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_coefficient&quot;&gt;partition coefficient&lt;/a&gt;, according to the wiki page, is 2.36. This implies its solubility is greater in organics than water, but some of it will most certainly dissolve in water. I think, based on its size, it is reasonable to assume it has a negligible vapor pressure too. These factors suggest it will have a low mobility: it will tend to stick to organics, won't appreciably evaporate away and will take quite a lot of water to wash it away. Contrast this with a famous organochlorine, &lt;a href=&quot;http://en.wikipedia.org/wiki/DDT&quot;&gt;DDT&lt;/a&gt;, which has an even higher partition coefficient (on the order of 10, so even more sticky and less soluble in water) and an even higher mass (so probably a lower vapor pressure) and thus an even lower mobility. Indeed, organophosphates largely replaced organochlorines as pesticides, in part, because they persist for shorter periods of times in the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these organophosphates were tested by us and their history is relevant. World Wars I and II ushered in a new era of chemical warfare agents and pesticides, notably of the classes of organochlorines, organophosphates and carbamates. Some organophosphates hardly persist at all, such as tetraethyl pyrophosphate, but they are pretty toxic to mammals. Many other organophosphates were screened for low toxicity towards mammals and increased persistence. We actually sought longer-lasting half-lives for these compounds! It turns out that we see alkyl phosphates with shorter lives than the longer-lived aryl species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We tend to see certain trends within the different organophosphates, such as the water solubility being greater for the less complex and more polar compounds. An ethyl- is less soluble than a methyl- derivative and an aryl-substituted species get partition coefficients approaching those of the organochlorines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;All this is fine and interesting, but it doesn't have much to do with my direct question,&quot; you might say. And most of the above just sets the scene for my answer to your question. Your specific query, I believe, requires us to go back to a nitty-gritty detail which is really only relevant in the case of your comparison between a phosphate (P-O-C) and an organophosphorus compound (P-C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer to your question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the biochemistry of most life, phosphates are pretty ubiquitous. As a result, many enzymes have evolved to manipulate them in many different forms; one class of these enzymes is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphatase&quot;&gt;phosphatases&lt;/a&gt; and they assist dephosphorylation reactions. They certainly cannot manipulate just any ol' phosphate (hence why more complex phosphates have longer residues). We could say organophosphates are more &quot;natural&quot; than your typical organophosphorus compound. There is, therefore, a greater chance of having a tool -- an enzyme -- that can pick your more &quot;natural&quot; looking compound apart than your harder to find P-C bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've indicated, these phosphates are actually pretty stable. It's why DNA is built using them! But when you compare the bond lengths between P-C and P-O and O-C, there's a decent chance of finding that a P-C bond length is longer than a P-O or O-C. This leads me to believe that there are many cases in which P-O and O-C are actually stronger than P-C! The same could easily be said in the other way too! This, I think, is why we have to step-back and look at the bigger picture: the exact physical properties vary considerably from one compound to another and paint a landscape that says the issue is complicated. Perhaps, instead of it being a mere thermodynamic issue dealing with a comparison of the general bonding energies, it has more to do with the kinetic advantage of having enzymes equipped to slice the P-O-C rather than P-C. Of course, your mileage may vary depending on your specific examples.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2012-04-27T09:37:08.273" LastActivityDate="2012-04-27T09:37:08.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="91" PostTypeId="1" AcceptedAnswerId="96" CreationDate="2012-04-27T09:18:09.013" Score="8" ViewCount="231" Body="&lt;p&gt;I am working on metal and its ion. The problem is, I would like to know whether the metal has been converted to its ionic form or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hypothesis for the analysis is that metals are good conductors and ions possess redox potential. Therefore, by measuring the conductance and red-ox potential is it possible to  differentiate whether I have metal or metal ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Based on the hypothesis, can I expect a positive result?&lt;/li&gt;&#xA;&lt;li&gt;Apart from this, is there any other technique where I can qualitatively / quantitatively measure  metal and its ion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="92" LastEditorUserId="7448" LastEditDate="2015-04-24T15:23:43.100" LastActivityDate="2015-04-24T15:23:43.100" Title="How to differentiate a metal from its ion?" Tags="&lt;ions&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ClosedDate="2017-09-06T23:04:25.657" ContentLicense="CC BY-SA 3.0" />
  <row Id="92" PostTypeId="1" AcceptedAnswerId="98" CreationDate="2012-04-27T09:46:33.863" Score="19" ViewCount="1380" Body="&lt;p&gt;Whilst we are taught to represent benzene as below (A) due to its delocalised electrons, however when two benzene rings share carbons (e.g. naphtalene) it seems to be more commonly represented in textbooks as two conjugated rings (B).  Whilst I realise that the result is the same, what is the advantage gained by drawing them like this rather than (C)?  Is it simply a matter of clarity?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c1fY8m.png&quot; alt=&quot;Benzene Ring&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;B:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ePDgvm.png&quot; alt=&quot;Naphtalene Kekule's Form&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;C:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AKN2wm.png&quot; alt=&quot;Naphtalene&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Whilst draing those in Marvin I wondered if it was as the delocalisation between the shared carbons isn't shown?  I.e. it looks like the electrons are delocalised across both rings separately rather than as one big ring?&lt;/p&gt;&#xA;" OwnerUserId="118" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-05-31T08:33:06.870" Title="Why are arenes with conjoined benzene rings drawn as they are?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;structural-formula&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="93" PostTypeId="2" ParentId="25" CreationDate="2012-04-27T10:23:45.690" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The obvious order is via the stability of the carbocation of the group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you meant the migrating groups stability?&lt;br&gt;&#xA;This is not what textbooks say.&lt;br&gt;&#xA;Reactions are ruled by the $ΔG$ of the activated complex. &#xA;This may be close  to educts, or to products, or some halfway &#xA;state or a real intermediate minimum is on the reaction path. &#xA;So, in some cases the model of a separate carbenium ion &#xA;(in a ionic pair maybe) is useful, sometimes you need to &#xA;think of a three-center bond state. &lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="-1" LastEditDate="2015-03-31T10:29:37.923" LastActivityDate="2015-03-31T10:29:37.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="94" PostTypeId="2" ParentId="92" CreationDate="2012-04-27T10:37:42.460" Score="6" Body="&lt;p&gt;(Too long for a comment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With drawing it as in diagram B, although we lose the information that the electrons are in reality delocalized, it makes for easier electron accounting when we start &quot;pushing electrons&quot; (that is to say, in illustrating the mechanism of a reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that representing electron delocalization in Lewis structures requires crutches like resonance structures or that circle/circle arc business. At least you can say that there's an electron pair in a single bond. How many electrons are there in the circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't deal much in the illustration of reaction mechanisms, I suppose the first and third drawings are okay. My point is that those circles don't really play nice with arrows.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastActivityDate="2012-04-27T10:37:42.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="95" PostTypeId="2" ParentId="89" CreationDate="2012-04-27T12:44:09.593" Score="14" Body="&lt;p&gt;I am not an expert in the field, but a quick literature check turns up a good amount of papers on the topic. In particular, I found &lt;a href=&quot;http://www.wag.caltech.edu/publications/sup/pdf/526.pdf&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; [1] which nicely answers the question for guanine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vm3Yr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wodwl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the gas phase, the most stable enol tautomer has a 0.9 kcal/mol free energy difference with the most stable keto tautomer. That free energy difference is very exacerbated &lt;strong&gt;in aqueous solution&lt;/strong&gt;, with now a &lt;strong&gt;∆G° = 8.7 kcal/mol&lt;/strong&gt;. We also notice that the most stable keto tautomer is not the same in the gas phase and in solution, and that both keto and enol have many tautomers close in free energy, showing the limits of the simple keto vs. enol line of thought.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding uracil, the first reference that comes up in a bibliographic search is &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.560560517/abstract&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; [2]. I only quote the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effects of the solvent on the tautomeric equilibria of cytosine and uracil are studied using Onsager's reaction field model in the framework of density functional theory. […] Our results are in good agreement with available experimental results and confirm that the polarization of the solute by the continuum has important effects on the absolute and relative solvation energies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;pKa Values of Guanine in Water: Density Functional Theory Calculations Combined with Poisson-Boltzmann Continuum-Solvation Model&lt;/em&gt;, Y. H. Jang et al., &lt;em&gt;J. Phys. Chem. B&lt;/em&gt; 2003, DOI: 10.1021/jp020774x&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Solvent effects in density functional calculations of uracil and cytosine tautomerism&lt;/em&gt;, L. Paglieri et al, &lt;em&gt;Quantum Chem.&lt;/em&gt;, 2004, DOI: 10.1002/qua.560560517&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-04-27T13:48:32.393" LastActivityDate="2012-04-27T13:48:32.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="96" PostTypeId="2" ParentId="91" CreationDate="2012-04-27T13:46:41.990" Score="11" Body="&lt;p&gt;If you're looking for an electrochemical method for identifying ionization states, I would suggest &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_voltammetry&quot;&gt;&lt;strong&gt;cyclic voltammetry&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like this quote from &lt;a href=&quot;http://www2.truman.edu/~jmccormi/Chem475/Laboratory/InorgCmpds.htm&quot;&gt;online lectures notes&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cyclic voltammetry has been called the &quot;electrochemical equivalent of spectroscopy&quot; because it can be used to determine oxidation state, E° of each redox process that the compound can undergo and can even be used to study the kinetics of the redox process.  Cyclic voltammetry is one of the standard characterization methods in most inorganic laboratories.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-04-27T13:46:41.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="97" PostTypeId="2" ParentId="91" CreationDate="2012-04-27T14:09:05.483" Score="12" Body="&lt;p&gt;Other than the excellent answer by F'x, there is also actual spectroscopy, specifically UV/Vis spectroscopy. This'll only work if the metal has transitions in the right range, though -- transition metals, lanthanides and actinides are usually fine, &lt;em&gt;s&lt;/em&gt; or &lt;em&gt;p&lt;/em&gt;-block metals and metalloids, probably not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you're looking for the absorption/transmission spectrum of the ion and the metal separately, and how much of each there is in a combined spectrum. It's not strictly quantitative, but you can get a good idea of the ratios.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-04-27T14:09:05.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="98" PostTypeId="2" ParentId="92" CreationDate="2012-04-27T14:37:23.277" Score="17" Body="&lt;p&gt;Questions about &lt;em&gt;why&lt;/em&gt; we represent things in one way or another are hard to answer in general, because they appeal to one’s sense of taste… &lt;em&gt;de gustibus non est disputandum&lt;/em&gt;. Here, I want to state the position and the reasoning of IUPAC on this matter. The relevant publication on this are the IUPAC 2008 Recommandations, titled &lt;em&gt;Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008)&lt;/em&gt; [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section GR-6 of the report deals with “Aromatic rings and other types of electron delocalization”. I quote part of it (emphasis is mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;GR-6.5 Curves should only be used when delocalization is being represented&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Curves represent delocalization, yet lots of structures have some delocalized elements. Although all such elements could be depicted with curves, there is little gained by doing so. On the contrary, the &lt;strong&gt;&lt;em&gt;arbitrary use of curves can draw the viewer’s attention to insignificant portions of a structural diagram&lt;/em&gt;&lt;/strong&gt;, and away from areas that are chemically more important such as an active site or a reactive group. Accordingly, &lt;strong&gt;&lt;em&gt;curves should only be used when the delocalization is specifically being highlighted as an important feature of the structure&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When curves are not used, any alternating configuration of double bonds is acceptable within the further constraints discussed in GR-3.5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7XW0b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text then goes further to note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is &lt;strong&gt;&lt;em&gt;generally not acceptable to use curves in two adjacent fused rings&lt;/em&gt;&lt;/strong&gt;, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008)&lt;/em&gt;, Pure Appl. Chem., 2008 DOI: 10.1351/pac200880020277&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS: hmm, the leader for the IUPAC report works at CambridgeSoft…&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-04-27T14:46:07.307" LastActivityDate="2012-04-27T14:46:07.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="100" PostTypeId="2" ParentId="92" CreationDate="2012-04-27T20:18:17.987" Score="16" Body="&lt;p&gt;Both ways of drawing benzene should mean the same thing to you. Use one or the other depending on what you want to express.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kekulé structure, &lt;strong&gt;B&lt;/strong&gt;, is more useful when you're interested in a reaction mechanism, or to visualize bonding once aromaticity is lifted. Here is an intermediate way to draw it, which is aesthetically more pleasing (to me):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yt8mo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt; structures are called Clar structures, but they first appeared in reference 1. Actually, &lt;strong&gt;C&lt;/strong&gt; is nonsensical. A circle represents a sextet of $\pi$ electrons, like those of benzene. Naphtalene has only 10 delocalized electrons. After all, it is aromatic, and thus possesses 4n+2 $\pi$ electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the rules for drawing Clar structures:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Never inscribe circles in adjacent benzenoid rings.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;On ignoring rings with circles, all remaining benzenoid rings must have a Kekulé structure, that is, must have either one or two double&#xA;  bonds, as in 1; a ring can be “empty” as in 2 but there must be no&#xA;  unpaired electrons.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Subject to the above constraints, a Clar structure must have the maximum number of benzenoid rings with inscribed circles. When a&#xA;  choice is available within a linear string of benzenoid rings, the&#xA;  circle can occupy any of the available rings, and this may be&#xA;  indicated by an arrow.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Read reference 2 for more information on Clar structures and their use in theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;J.W.Armit and R.Robinson, Polynuclear heterocyclic aromatic types. II . J. Chem. Soc. (1925):1604–1610.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alexandru T. Balaban (2004): &lt;a href=&quot;http://dx.doi.org/10.1080/10406630490424124&quot; rel=&quot;noreferrer&quot;&gt;Clar Formulas: How to Draw and How not to Draw Formulas of Polycyclic Aromatic Hydrocarbons&lt;/a&gt;, Polycyclic Aromatic Compounds, 24:2, 83-89&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-04-27T20:18:17.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="101" PostTypeId="2" ParentId="87" CreationDate="2012-04-27T20:52:39.583" Score="8" Body="&lt;p&gt;Copper(II) is almost always blue, so I'm not sure how much luck you'll have trying to change the color. &lt;span class=&quot;math-container&quot;&gt;$\ce{CuCl2}$&lt;/span&gt; is one of the few solutions I can think of off the top of my head that isn't blue - it's more greenish-blue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CuCl2}$&lt;/span&gt; can be made via electrolysis of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; and copper plates. You could also make &lt;span class=&quot;math-container&quot;&gt;$\ce{CuCl2}$&lt;/span&gt; if you add hydrochloric acid and some &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="61578" LastEditDate="2019-11-03T06:37:14.160" LastActivityDate="2019-11-03T06:37:14.160" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="102" PostTypeId="2" ParentId="69" CreationDate="2012-04-27T21:05:13.170" Score="7" Body="&lt;p&gt;While the above answers quite clearly explain the theory, I thought I would provide some clarification and citations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is indeed propane you were taking the spectrum of, then you would get a heptet (7 peaks, in the standard pascals triangle pattern). Lets make another example though&#xA;R-CH2-CH2-CH2-R'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If R = R' then you get a quintet (5 peaks) in a pascal's triangle. &#xA;However, if R is not the same as R' then you will get a triplet of triplets, that is, three distinct triplets with the same J-coupling. This may or may not be easy to discern, depending on what R and R' are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some useful sources:&#xA;These two papers can give you some great examples to examine, and are useful enough I always keep them on  my USB key.&#xA;&lt;a href=&quot;http://dx.doi.org/10.1021/jo971176v&quot; rel=&quot;nofollow&quot;&gt;NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities; Hugo E. Gottlieb, Vadim Kotlyar, and Abraham Nudelman; J. Org. Chem. 1997, 62, 7512-7515&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/om100106e&quot; rel=&quot;nofollow&quot;&gt;NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist; Gregory R. Fulmer,Alexander J. M. Miller, Nathaniel H. Sherden, Hugo E. Gottlieb, Abraham Nudelman, Brian M.  Stoltz, John E. Bercaw, and Karen I. Goldberg; Organometallics 2010, 29, 2176–2179&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditDate="2015-11-06T04:20:59.657" LastActivityDate="2015-11-06T04:20:59.657" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="103" PostTypeId="2" ParentId="88" CreationDate="2012-04-27T21:27:27.513" Score="11" Body="&lt;p&gt;The most likely reason to use methanol instead of ethanol is cost. &lt;sup&gt;[&lt;a href=&quot;http://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_3_ORLANDO_08-96_0880.pdf&quot;&gt;1&lt;/a&gt;]&lt;/sup&gt;  Ethanol is roughly twice as expensive as methanol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although both aspartame and the ethanol analog can be produced with reasonably high yield&lt;sup&gt;[&lt;a href=&quot;http://www.patentstorm.us/patents/4730076/description.html&quot;&gt;2&lt;/a&gt;]&lt;/sup&gt;, the ethanol analog has not yet been produced commercially, so it would be even more expensive to develop industrial scale production of the ethanol analog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to address the toxicity part of the question, aspartame is not at all toxic.  It is most stable at lower pH (~4.3) which is in the range of carbonated beverages.  So methanol is not generally produced while the food is in storage, and the body gets rid of methanol as it is produced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more note, aspartame &lt;em&gt;can&lt;/em&gt; break down via heating.  This is why food that is intended to be baked is not usually sweetened with aspartame (or it is mixed with saccharine).  Although the fear here is not toxicity so much as cake that isn't sweet!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_3_ORLANDO_08-96_0880.pdf&quot;&gt;http://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_3_ORLANDO_08-96_0880.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.patentstorm.us/patents/4730076/description.html&quot;&gt;http://www.patentstorm.us/patents/4730076/description.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-04-27T21:27:27.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="104" PostTypeId="2" ParentId="62" CreationDate="2012-04-27T21:59:54.787" Score="7" Body="&lt;p&gt;The currently accepted hydrogenation mechanism for metal like palladium is the Horiuti-Polanyi mechanism. (&lt;a href=&quot;http://www.springerimages.com/img/Images/Springer/PUB=Springer_US-Boston/JOU=11244/VOL=2008.48/ISU=1-4/ART=2008_9052/MediaObjects/MEDIUM_11244_2008_9052_Fig7_HTML.jpg&quot;&gt;image&lt;/a&gt;)  This mechanism works with good yield for alkenes and alkynes.  For an aromatic compound, the activation energy is much higher because the aromaticity of the target molecule is being broken by the first hydrogenation.  Indeed, 1,3-cyclohexadiene is very unstable and will revert to benzene and hydrogen instead of continuing to cyclohexane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition of a Lewis Acid makes palladium an acceptable catalyst for the hydrogenation of benzene, however.  An intermediate compound with the benzene and the LA is first formed, and hydrogen is added to this intermediate.  &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.200803850/abstract&quot;&gt;This paper&lt;/a&gt; discusses the mechanism of hydrogenation in the presence of a Lewis Acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rhodium catalysts work slightly differently.  They generally have ligands attached that help stabilize the intermediates.  You can see a simplified mechanism &lt;a href=&quot;http://203.199.213.48/343/1/29.pdf&quot;&gt;here&lt;/a&gt; and there are papers discussing specific catalysts &lt;a href=&quot;http://pubs.acs.org/doi/full/10.1021/om990713u&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://evans.harvard.edu/pdf/evans155.pdf&quot;&gt;here&lt;/a&gt;.  Basically, in the case of benzene, the relatively unstable 1,2-cyclohexadiene is held in place on the catalyst until it has been completely hydrogenated.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-04-27T21:59:54.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="105" PostTypeId="1" CreationDate="2012-04-28T02:38:07.870" Score="22" ViewCount="2483" Body="&lt;p&gt;It is known that although only the &lt;em&gt;(S)&lt;/em&gt;-enantiomer of the infamous sedative thalidomide possesses teratogenic properties, it is not very useful to administer the pure &lt;em&gt;(R)&lt;/em&gt;-enantiomer since it is racemized within the body. Are there other known examples of drugs that are stereoconverted &lt;em&gt;in vivo&lt;/em&gt;? How does the body perform the conversion?&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastEditorUserId="41890" LastEditDate="2017-11-27T17:59:45.360" LastActivityDate="2017-11-27T17:59:45.360" Title="What are known examples of drugs that racemize/stereoconvert in vivo, and how are they converted?" Tags="&lt;biochemistry&gt;&lt;stereochemistry&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="106" PostTypeId="2" ParentId="105" CreationDate="2012-04-28T05:27:54.430" Score="15" Body="&lt;p&gt;&lt;em&gt;Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral.  A reversal back to the ketone results in a mix of &lt;em&gt;(R)&lt;/em&gt; and &lt;em&gt;(S)&lt;/em&gt; enantiomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w1k3n.png&quot; alt=&quot;Chemical diagram for s-thalidomide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the body, this tautomerization is generally catalyzed by basic amino acids.  Specifically, albumin is the main catalyst in humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this is beyond the scope of your question, the reason that only &lt;em&gt;(S)&lt;/em&gt;-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=HibwXSgofi0C&amp;amp;pg=PA330&amp;amp;lpg=PA330&amp;amp;dq=albumin+thalidomide+racemization&amp;amp;source=bl&amp;amp;ots=8aLGIAbZa9&amp;amp;sig=uaQhWP7tAzUScQiQ0AuRom-v09k&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=mHubT43TOeGD6AGG1oXuDg&amp;amp;ved=0CB8Q6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Reddy - &lt;em&gt;Chirality in Drug Design and Development&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/9860497&quot; rel=&quot;noreferrer&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/9860497&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/9499573&quot; rel=&quot;noreferrer&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/9499573&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12" LastEditorUserId="12" LastEditDate="2012-04-28T18:04:51.787" LastActivityDate="2012-04-28T18:04:51.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="107" PostTypeId="5" CreationDate="2012-04-28T09:48:27.670" Score="0" Body="&lt;h1&gt;Definition (&lt;a href=&quot;http://goldbook.iupac.org/M03804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook: mechanism of a reaction&lt;/a&gt;)&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;mechanism&lt;/strong&gt; &lt;em&gt;of a reaction&lt;/em&gt;&lt;br&gt;&#xA;  Synonym: reaction mechanism&lt;br&gt;&#xA;  A detailed description of the process leading from the reactants to the products of a reaction, including a characterization as complete as possible of the composition, structure, energy and other properties of reaction intermediates, products, and transition states. An acceptable mechanism of a specified reaction (and there may be a number of such alternative mechanisms not excluded by the evidence) must be consistent with the reaction stoichiometry, the rate law and with all other available experimental data, such as the stereochemical course of the reaction. Inferences concerning the electronic motions which dynamically interconvert successive species along the reaction path (as represented by curved arrows, for example) are often included in the description of a mechanism. It should be noted that for many reactions all this information is not available and the suggested mechanism is based on incomplete experimental data. It is not appropriate to use the term mechanism to describe a statement of the probable sequence in a set of stepwise reactions. That should be referred to as a reaction sequence, and not a mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Related definitions:&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reaction &lt;a href=&quot;http://goldbook.iupac.org/I03096.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermediate&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/T06468.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;transition state&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/T06468.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rate law&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/R05174.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction path&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/G02630.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gibbs energy diagram&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/T06470.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;transition state theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;elementary reaction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Further reading&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reaction mechanism (Wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Rate_Laws/Reaction_Mechanisms/Reaction_Mechanisms&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reaction mechanisms (UC Davis ChemWiki)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Jie Jack Li: Name Reactions. A Collection of Detailed Reaction Mechanisms. Springer Berlin Heidelberg: 2003. &lt;a href=&quot;http://link.springer.com/book/10.1007%2F978-3-662-05336-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1007/978-3-662-05336-2&lt;/a&gt; (&lt;a href=&quot;http://books.google.co.jp/books?id=6mZJ3084ouAC&quot; rel=&quot;nofollow noreferrer&quot;&gt;google books&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Aplicability of the tag&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This tag should be applied to questions, which seek information on elementary steps of chemical reactions. It may be applied to hypothetical and/or proposed mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Related tags&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/transition-state-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;transition-state-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;transition-state-theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/reaction-coordinate&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-coordinate&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-coordinate&lt;/a&gt;&lt;a href=&quot;/questions/tagged/reaction-control&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-control&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-control&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt;&lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/kinetics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;kinetics&amp;#39;&quot; rel=&quot;tag&quot;&gt;kinetics&lt;/a&gt;&lt;a href=&quot;/questions/tagged/physical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;physical-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="402" LastEditorUserId="9961" LastEditDate="2017-06-21T13:35:24.877" LastActivityDate="2017-06-21T13:35:24.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="108" PostTypeId="4" CreationDate="2012-04-28T09:48:27.670" Score="0" Body="The detailed (and as complete as possible) description of plausible or observable pathways leading from reactants to products of a chemical process. This tag should be applied to questions about these processes. It may be applied to hypothetical and/or proposed mechanisms." OwnerUserId="43" LastEditorUserId="2599" LastEditDate="2017-10-19T07:18:58.367" LastActivityDate="2017-10-19T07:18:58.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="109" PostTypeId="5" CreationDate="2012-04-28T09:49:32.847" Score="0" Body="&lt;p&gt;Questions that are related to any living organism should have this tag. However, chemicals that are needed in life, e.g. proteins, enzymes, etc. can also have this tag, provided that the question being asked is concerned with how these proteins, enzymes, etc. affect living organisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Organism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on Organism&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Biochemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on Biochemistry&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://pubs.acs.org/journal/bichaw&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACS publications about Biochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="94" LastEditorUserId="15751" LastEditDate="2017-06-25T13:03:15.670" LastActivityDate="2017-06-25T13:03:15.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="110" PostTypeId="4" CreationDate="2012-04-28T09:49:32.847" Score="0" Body="This tag is for questions concerning biochemical methods (e.g. electrophoresis) or those concerning biochemical mechanisms or research. Do not use this tag if your question is merely about compounds often used in areas related to biochemistry or associated with these. These may fall under organic chemistry or the appropriate compound’s functional groups’ tags." OwnerUserId="43" LastEditorUserId="402" LastEditDate="2017-04-08T01:06:02.397" LastActivityDate="2017-04-08T01:06:02.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="111" PostTypeId="5" CreationDate="2012-04-28T09:52:37.853" Score="0" Body="&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/C01023.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy,  $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:&#xA;  $$\begin{align}&#xA;\Delta G_\mathrm r&amp;amp;=\Delta G_\mathrm r^\circ+R\cdot T\cdot\ln K=0\\&#xA;\Delta G_\mathrm r^\circ&amp;amp;=−R\cdot T\cdot\ln K&#xA;\end{align}$$&#xA;  The equilibrium constant, $K$, is given by the mass-law effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common reactions involving equilibria are &lt;a href=&quot;/questions/tagged/acid-base&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;acid-base&amp;#39;&quot; rel=&quot;tag&quot;&gt;acid-base&lt;/a&gt; reactions, like the incomplete dissociation of acetic acid,&#xA;$$\ce{H3C-COOH + H2O &amp;lt;=&amp;gt; H3COO- + H3+O},$$&#xA;&lt;a href=&quot;/questions/tagged/precipitation&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;precipitation&amp;#39;&quot; rel=&quot;tag&quot;&gt;precipitation&lt;/a&gt; reactions, like &#xA;$$\ce{AgNO3(aq) + HCl(aq) &amp;lt;=&amp;gt; AgCl v (s) + H+(aq) + NO3-(aq)},$$ &#xA;the formation of &lt;a href=&quot;/questions/tagged/coordination-complexes&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;coordination-complexes&amp;#39;&quot; rel=&quot;tag&quot;&gt;coordination-complexes&lt;/a&gt;, or ligand exchange reactions, like&#xA;$$\ce{[Cu(H2O)_{6}]^{2+}~(aq) + 4 NH3~(aq) &amp;lt;=&amp;gt; [Cu(NH3)_{4}(H2O)_{$n$}]^{2+}~(aq) + ($6-n$) H2O},$$&#xA;among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is closely related to &lt;a href=&quot;/questions/tagged/physical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;physical-chemistry&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/thermodynamics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;thermodynamics&amp;#39;&quot; rel=&quot;tag&quot;&gt;thermodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorDisplayName="user7951" LastEditDate="2018-02-26T17:37:13.723" LastActivityDate="2018-02-26T17:37:13.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="112" PostTypeId="4" CreationDate="2012-04-28T09:52:37.853" Score="0" Body="The state in which both reactants and products are present at concentrations which have no further tendency to change with time." OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-29T16:55:57.337" LastActivityDate="2012-04-29T16:55:57.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="113" PostTypeId="5" CreationDate="2012-04-28T09:53:19.643" Score="0" Body="&lt;p&gt;&lt;strong&gt;aromatic&lt;/strong&gt; from the &lt;a href=&quot;https://goldbook.iupac.org/A00441.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the traditional sense, 'having a chemistry typified by benzene'.&lt;/li&gt;&#xA;  &lt;li&gt;A cyclically conjugated molecular entity with a stability (due to delocalization) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.&lt;br&gt;&#xA;  See also: Hückel $(4n + 2)$ rule, Möbius aromaticity&lt;/li&gt;&#xA;  &lt;li&gt;The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel $(4n + 2)$ rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hückel $(4n + 2)$ rule&lt;/strong&gt; from the &lt;a href=&quot;https://goldbook.iupac.org/H02867.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain $(4n + 2)$ π-electrons (where $n$ is a non-negative integer) will exhibit aromatic character. The rule is generally limited to $n = 0–5$. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons $\ce{(CH)_{$m$}}$ where $m$ is an integer equal to or greater than 3 according to which $(4n + 2)$ π-electrons are contained in a closed-shell system. Examples of systems that obey the Hückel rule include:&lt;br&gt;&#xA;  &lt;img src=&quot;https://goldbook.iupac.org/SVG/H02867.png&quot; alt=&quot;H02867&quot;&gt;&lt;br&gt;&#xA;  Systems containing $4n$ π-electrons (such as cyclobutadiene and the cyclopentadienyl cation) are 'antiaromatic'.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Möbius aromaticity&lt;/strong&gt; from the &lt;a href=&quot;https://goldbook.iupac.org/M03957.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A monocyclic array of orbitals in which there is a single out-of-phase overlap (or, more generally, an odd number of out-of-phase overlaps) reveals the opposite pattern of aromatic character to Hückel systems; with  $4 n$ electrons it is stabilized (aromatic), whereas with  $4 n + 2$ it is destabilized (antiaromatic). In the excited state  $4 n + 2$ Möbius π-electron systems are stabilized, and  4 n systems are destabilized. No examples of ground-state Möbius π-electron systems are known, but the concept has been applied to transition states of pericyclic reactions [See: aromatic]. The name is derived from the topological analogy of such an arrangement of orbitals to a Möbius strip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;antiaromatic compounds&lt;/strong&gt; from the &lt;a href=&quot;https://goldbook.iupac.org/A00382.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compounds that contain  $4 n~ (n ≠ 0)$ π-electrons in a cyclic planar, or nearly planar, system of alternating single and double bonds, e.g. cyclobuta-1,3-diene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See also:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://goldbook.iupac.org/A00442.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;aromaticity (Gold Book)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aromaticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;aromaticity (Wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="5026" LastEditDate="2018-03-15T13:29:31.973" LastActivityDate="2018-03-15T13:29:31.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="114" PostTypeId="4" CreationDate="2012-04-28T09:53:19.643" Score="0" Body="For questions about the properties and reactions of aromatic compounds. This tag should also be applied to antiaromatic compounds. This tag should not be applied to general questions about resonance, use the [resonance] tag instead." OwnerUserId="43" LastEditorUserId="5026" LastEditDate="2018-03-15T13:30:48.420" LastActivityDate="2018-03-15T13:30:48.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="115" PostTypeId="5" CreationDate="2012-04-28T09:54:25.883" Score="0" Body="&lt;p&gt;An &quot;inorganic compound&quot; is technically defined as a chemical substance based on or primarily consisting of some other element than carbon. It is the antonym of an &quot;organic substance&quot;, which is a substance that &lt;em&gt;is&lt;/em&gt; primarily based on carbon, which along with hydrogen and oxygen, comprises 93% of the &quot;biomass&quot; of the Earth, providing a natural dividing line between chemicals generally of interest in &quot;life sciences&quot;, and those of interest elsewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inorganic chemistry, therefore, is &quot;the sub-discipline of the field of chemistry that studies interactions, reactions, and substances primarily involving elements other than carbon&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition would seem to create a disproportionate divide in favor of inorganic chemistry over organic, given that there are 117 other known or theorized elements besides carbon, 83 of which are &quot;primordial elements&quot; stable enough to have been around before the formation of the Earth. However, the versatility and flexibility of carbon in forming a wide array of diverse molecules, and the disproportionate number of sub-fields pursued in modern chemistry that are interested in organic substances tend to balance the scales or even tip the balance toward organic chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the inclusion in both disciplines of molecules &lt;em&gt;based&lt;/em&gt; on carbon, but &lt;em&gt;including&lt;/em&gt; other elements that react in unique ways, blurs the dividing line between the substances of primary interest to both fields. While 93% of the mass of a human is hydrogen, carbon, and oxygen, the remaining 7% consists of almost 60 other elements, of which about 20 are considered essential for human life. Among the most important are nitrogen, calcium, sodium, potassium, chlorine, sulfur, phosphorous, iron, copper, and zinc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of particular interest to both fields is organometallics, a class of chemicals in which a metal is bound to a carbon atom. The properties of the metal (very often the ability to exist in several oxidation states, as with iron or zinc) can be leveraged in combination with the properties of the organic functional group in biochemical reactions ranging from organic synthesis (the organometallic will often deposit its organic functional group onto a larger molecule, often in exchange for a simpler moiety such as an oxygen or hydroxyl group) to drug interaction mechanisms (the method by which the drug exhibits its desirable behavior, often by complexing with a cellular mechanism such as a neuroreceptor to activate or disable it).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-04-07T00:35:07.697" LastActivityDate="2017-04-07T00:35:07.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="116" PostTypeId="4" CreationDate="2012-04-28T09:54:25.883" Score="0" Body="Inorganic compounds generally do not have C-H bonds, while organic compounds do have such bonds. The distinction between inorganic and organic chemistry, however, is far from absolute. " OwnerUserId="43" LastEditorUserId="5084" LastEditDate="2014-08-11T05:12:52.883" LastActivityDate="2014-08-11T05:12:52.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="117" PostTypeId="5" CreationDate="2012-04-28T09:55:12.483" Score="0" Body="&lt;p&gt;Synthesis is a purposeful execution of chemical reactions to obtain the desired product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Syntheses come in all tastes and sizes, from a single reaction to tedious, multi-step processes, involving purification of numerous intermediates. Almost the entire division of natural product chemistry is devoted to the stereoselective synthesis of the vast natural product library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, long and complex syntheses are preceded by retrosynthetic analyses, deciding which steps to perform in which order, where to cut the molecule apart into fragments and which carbon-carbon bonds need to be formed during this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Applicability of the synthesis tag&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/synthesis&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis&lt;/a&gt; should be applied to all questions relating to a synthesis, from short 2 step sequences, to long 50 step total syntheses&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The tag may be applied to both published syntheses, or hypothetical ones&lt;/li&gt;&#xA;&lt;li&gt;The tag may be applied in both the forward sense (syntheses), or the reverse sense (retrosynthesis). &lt;/li&gt;&#xA;&lt;li&gt;The tag should not be applied to simple reactions such as the formation of hydrogen by acidic oxidation of metals.&lt;/li&gt;&#xA;&lt;li&gt;The tag should not be applied to &lt;a href=&quot;/questions/tagged/synthesis-golf&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis-golf&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis-golf&lt;/a&gt; questions. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Depending on the nature of the synthesis, &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt; should also be applied to the question to further define the area of synthesis being discussed. &lt;/li&gt;&#xA;&lt;li&gt;Additional tags such as &lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/stereochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereochemistry&lt;/a&gt; may also be applied, if the question is asking about a specific aspect of the synthesis, e.g. &lt;em&gt;'why does step 5 give the 1,2-anti stereochemistry from the aldol reaction'&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related tags:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/synthesis&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following which may be useful as additional tags on questions identified as being related to organic chemistry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&#xA;&lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt;&#xA;&lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt;&#xA;&lt;a href=&quot;/questions/tagged/stereochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereochemistry&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="21296" LastEditDate="2017-06-07T20:56:56.107" LastActivityDate="2017-06-07T20:56:56.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="118" PostTypeId="4" CreationDate="2012-04-28T09:55:12.483" Score="0" Body="Synthesis is a purposeful execution of chemical reactions to obtain the desired product.&#xD;&#xA;Applies to long and complex natural product syntheses as much as to short one or two-step syntheses. This also covers synthetic problem sets and retrosynthetic analyses. " OwnerUserId="43" LastEditorUserId="21296" LastEditDate="2017-06-07T20:57:35.063" LastActivityDate="2017-06-07T20:57:35.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="119" PostTypeId="5" CreationDate="2012-04-28T09:55:52.607" Score="0" Body="&lt;p&gt;Chemical kinetics includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In elementary physical chemistry, the study of chemical kinetics involves studying a few &quot;orders&quot; of kinetics (zero, first, second, pseudo-order); their half-lives; concentration or log[concentration] vs. time graphs; Radioactive decay and growth kinetics are also frequently covered. &lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="5026" LastEditDate="2018-03-13T15:42:17.433" LastActivityDate="2018-03-13T15:42:17.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="120" PostTypeId="4" CreationDate="2012-04-28T09:55:52.607" Score="0" Body="The study of rates and steps in mechanisms of chemical processes." OwnerUserId="43" LastEditorUserId="9961" LastEditDate="2016-07-07T03:51:06.173" LastActivityDate="2016-07-07T03:51:06.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="121" PostTypeId="5" CreationDate="2012-04-28T09:56:22.543" Score="0" Body="&lt;p&gt;A polymer is a large molecule, or macromolecule, composed of many repeated subunits. Because of their broad range of properties, both synthetic and natural polymers play an essential and ubiquitous role in everyday life. Polymers, both natural and synthetic, are created via polymerization of many small molecules, known as monomers. Their consequently large molecular mass relative to small molecule compounds produces unique physical properties.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-16T16:54:39.407" LastActivityDate="2018-03-16T16:54:39.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="122" PostTypeId="4" CreationDate="2012-04-28T09:56:22.543" Score="0" Body="For questions about synthetic or natural polymers, their chemical or physical properties, and their synthesis, etc." OwnerUserId="43" LastEditorUserId="5026" LastEditDate="2018-03-16T16:53:24.960" LastActivityDate="2018-03-16T16:53:24.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="123" PostTypeId="5" CreationDate="2012-04-28T09:57:46.723" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T09:57:46.723" LastActivityDate="2012-04-28T09:57:46.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="124" PostTypeId="4" CreationDate="2012-04-28T09:57:46.723" Score="0" Body="For questions about purification - the process of rendering something pure, clean of foreign or undesired elements. Not to be confused with [metallurgy] or [chromatography]." OwnerUserId="43" LastEditorUserId="5026" LastEditDate="2018-03-17T15:50:14.933" LastActivityDate="2018-03-17T15:50:14.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="125" PostTypeId="5" CreationDate="2012-04-28T09:58:31.447" Score="0" Body="&lt;p&gt;Redox is the portmanteau of reduction and oxidation and therefore refers to reactions in which electrons are transferred and in consequence the &lt;a href=&quot;/questions/tagged/oxidation-state&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;oxidation-state&amp;#39;&quot; rel=&quot;tag&quot;&gt;oxidation-state&lt;/a&gt; of the involved elements changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally it stems from two concepts, that involved electron transfer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;oxidation&lt;/strong&gt; (&lt;a href=&quot;https://goldbook.iupac.org/O04362.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;)  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').&lt;/li&gt;&#xA;  &lt;li&gt;An increase in the oxidation number of any atom within any substrate.&lt;/li&gt;&#xA;  &lt;li&gt;Gain of oxygen and/or loss of hydrogen of an organic substrate.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation of an organic substrate that can be rationally dissected into steps or primitive changes. The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons or hydroxide ions, or by nucleophilic substitution by water or its reverse and/or by an intramolecular molecular rearrangement. This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1. For example the oxidation of methane to chloromethane may be considered as follows:&lt;br&gt;&#xA;  &lt;img src=&quot;https://goldbook.iupac.org/img/svg/O04362.png&quot; alt=&quot;oxidation scheme&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;reduction&lt;/strong&gt; (&lt;a href=&quot;https://goldbook.iupac.org/R05222.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;)&lt;br&gt;&#xA;  The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general a reduction cannot take place without an oxidation, however it is often advisable to consider both half reaction separately. A common example for redox processes is the $\ce{Cu|Cu^{2+}||Zn^{2+}|Zn}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;galvanic cell&lt;/a&gt;:&#xA;\begin{align}&#xA;\ce{Cu^{2+} + 2e- &amp;amp;&amp;lt;=&amp;gt; Cu}\\&#xA;\ce{Zn            &amp;amp;&amp;lt;=&amp;gt; Zn^{2+} + 2e- }\\\hline&#xA;\ce{Cu^{2+} + Zn  &amp;amp;&amp;lt;=&amp;gt; Cu + Zn^{2+}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow noreferrer&quot;&gt;Redox (Wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wiley.com/college/boyer/0470003790/reviews/redox/redox.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Redox Reactions (Wiley.com)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2017-04-07T10:59:21.303" LastActivityDate="2017-04-07T10:59:21.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="126" PostTypeId="4" CreationDate="2012-04-28T09:58:31.447" Score="0" Body="Redox is the portmanteau of reduction and oxidation. This tag should be used in questions where both the reduction and the oxidation process are under focus. This tag is not limited to inorganic chemistry. If the question is specifically about either reduction or oxidation (not both) of an organic compound, use [organic-reduction] or [organic-oxidation] instead." OwnerUserId="43" LastEditorUserId="5026" LastEditDate="2018-03-18T10:35:56.960" LastActivityDate="2018-03-18T10:35:56.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="127" PostTypeId="5" CreationDate="2012-04-28T09:59:54.940" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T09:59:54.940" LastActivityDate="2012-04-28T09:59:54.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="128" PostTypeId="4" CreationDate="2012-04-28T09:59:54.940" Score="0" Body="A liquid, solid, or gas that dissolves another solid, liquid, or gaseous solute, resulting in a solution that is soluble in a certain volume of solvent at a specified temperature." OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-30T14:41:02.633" LastActivityDate="2012-04-30T14:41:02.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="129" PostTypeId="5" CreationDate="2012-04-28T10:00:34.317" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T10:00:34.317" LastActivityDate="2012-04-28T10:00:34.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="130" PostTypeId="4" CreationDate="2012-04-28T10:00:34.317" Score="0" Body="Isomeric molecules that have the same molecular formula and sequence of bonded atoms (constitution), but that differ only in the three-dimensional orientations of their atoms in space." OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-30T14:41:06.423" LastActivityDate="2012-04-30T14:41:06.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="131" PostTypeId="5" CreationDate="2012-04-28T10:01:59.103" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T10:01:59.103" LastActivityDate="2012-04-28T10:01:59.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="132" PostTypeId="4" CreationDate="2012-04-28T10:01:59.103" Score="0" Body="The application of nuclear magnetic resonance in NMR spectroscopy with respect to hydrogen-1 nuclei within the molecules of a substance, in order to determine the structure of its molecules." OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-30T14:27:41.277" LastActivityDate="2012-04-30T14:27:41.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="133" PostTypeId="5" CreationDate="2012-04-28T10:03:01.847" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T10:03:01.847" LastActivityDate="2012-04-28T10:03:01.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="134" PostTypeId="4" CreationDate="2012-04-28T10:03:01.847" Score="0" Body="A chemical reaction that results in the formal migration of a hydrogen atom or proton, accompanied by a switch of a single bond and adjacent double bond. " OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-30T14:41:19.450" LastActivityDate="2012-04-30T14:41:19.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="135" PostTypeId="5" CreationDate="2012-04-28T10:03:28.837" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T10:03:28.837" LastActivityDate="2012-04-28T10:03:28.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="136" PostTypeId="4" CreationDate="2012-04-28T10:03:28.837" Score="0" Body="Compounds in which at least some of bonds have ionic character stronger than covalent or metallic. Many compounds called salts are ionic compounds but not all of them." OwnerUserId="43" LastEditorUserId="9961" LastEditDate="2015-08-17T06:35:57.520" LastActivityDate="2015-08-17T06:35:57.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="137" PostTypeId="5" CreationDate="2012-04-28T10:04:03.983" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;pH&lt;/strong&gt; (source: &lt;a href=&quot;http://goldbook.iupac.org/P04524.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Goldbook&lt;/a&gt;)&lt;br&gt;&#xA;  The quantity $\mathrm{p}\ce{H}$ is defined in terms of the activity of hydrogen(1+) ions (hydrogen ions) in solution:&#xA;  $$\mathrm{p}\ce{H} = −\lg\bigr[a(\ce{H+})\bigr] = −\lg\bigr[m(\ce{H+})~\gamma_\mathrm{m}(\ce{H+}) / m^\circ\bigr],$$&#xA;  where  $a(\ce{H+})$ is the activity of hydrogen ion (hydrogen 1+) in aqueous solution,  $\ce{H+ (aq)}$,  $\gamma_\mathrm{m}(\ce{H+})$ is the activity coefficient of $\ce{H+ (aq)}$ (molality basis) at molality  $m(\ce{H+ (aq)})$, and  $m^\circ = 1~\mathrm{mol\cdot{}kg^{−1}}$ is the standard molality.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Notes:  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\mathrm{p}\ce{H}$ cannot be measured independently because calculation of the activity involves the activity coefficient of a single ion. Thus it can be regarded only as a notional definition.&lt;/li&gt;&#xA;  &lt;li&gt;The establishment of primary $\mathrm{p}\ce{H}$ standards requires the application of the concept of 'primary method of measurement', assuring full traceability of the results of all measurements and their uncertainties. Any limitation in the theory of determination of experimental variables must be included in the estimated uncertainty of the method.&lt;/li&gt;&#xA;  &lt;li&gt;The primary method for measurement of pH involves the use of a cell without transference, known as the Harned cell:&lt;br&gt;&#xA;  $\ce{Pt(s) | H2(g) | Buffer~S, Cl−(aq) | AgCl(s) | Ag(s)}$&lt;br&gt;&#xA;  The equation for this cell can be rearranged to give:&lt;br&gt;&#xA;  $−\lg[ a(\ce{H+})\gamma\ce{Cl^{−}})] = \frac{E − E^\circ}{\mathcal{R} T \ln 10 / F} + \lg[m(\ce{Cl^{−}}) / m^\circ],$&lt;br&gt;&#xA;  where $E$ is the potential difference of the cell and  $E^\circ$ is the known standard potential of the $\ce{AgCl | Ag}$ electrode. Measurements of $E^\circ$ as a function of  $m(\ce{Cl^{−}})$ are made and the quantity $a(\ce{H+})\gamma(\ce{Cl^{−}})$ (called the acidity function) is found by extrapolation to  $m(\ce{Cl^{−}}) / m^\circ = 0$. The value of $\gamma(\ce{Cl^{−}})$ is calculated using the Bates–Guggenheim convention based on Debye–Hückel theory. Then $\lg[a(\ce{H+})$ is calculated and identified as $\mathrm{p}\ce{H}(\ce{PS})$, where $\ce{PS}$ signifies primary standard. The uncertainties in the two estimates are typically  $0.001$ in $\lg[a(\ce{H+})\gamma(\ce{Cl^{−}})$ and  $0.003$ in $\mathrm{p}\ce{H}$. Materials for primary standard buffers must also meet the appropriate requirements for reference materials, including chemical purity and stability, and applicability of the Bates–Guggenheim convention for the estimation of  $−\lg[\gamma(Cl^{−})]$ This convention requires that the ionic strength be  $\leq 0.1~\mathrm{mol\cdot{}kg^{−1}}$. Primary standard buffers should also lead to small liquid junction potentials when used in cells with liquid junctions. Secondary standards,  $\mathrm{p}\ce{H}(\ce{SS})$, are also available, but carry a greater uncertainty in measured values.&lt;/li&gt;&#xA;  &lt;li&gt;Practical $\mathrm{p}\ce{H}$ measurements generally use cells with liquid junctions in which, consequently, liquid junction potentials,  $E_j$, are present. Measurements of $\mathrm{p}\ce{H}$ are not normally performed using the $\ce{Pt|H2}$ electrode, but rather the glass (or other $\ce{H+}$-selective) electrode, whose response factor ($\mathrm{d}E/\mathrm{d}\mathrm{p}\ce{H}$) usually deviates from the Nernst slope. The associated uncertainties are significantly larger than those associated with fundamental measurements using the Harned cell. Nonetheless, incorporation of the uncertainties for the primary method, and for all subsequent measurements, permits the uncertainties for all procedures to be linked to the primary standards by an unbroken chain of comparisons.&lt;/li&gt;&#xA;  &lt;li&gt;Reference values for standards in $\ce{D2O}$ and aqueous-organic solvent mixtures exist.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-04-08T01:49:58.043" LastActivityDate="2017-04-08T01:49:58.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="138" PostTypeId="4" CreationDate="2012-04-28T10:04:03.983" Score="0" Body="pH (or the potential of hydrogen) is a scale used to measure how acidic or basic an aqueous solution is." OwnerUserId="43" LastEditorUserId="402" LastEditDate="2017-04-02T07:05:16.053" LastActivityDate="2017-04-02T07:05:16.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="139" PostTypeId="5" CreationDate="2012-04-28T10:04:57.587" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-04-28T10:04:57.587" LastActivityDate="2012-04-28T10:04:57.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="140" PostTypeId="4" CreationDate="2012-04-28T10:04:57.587" Score="0" Body="A common organic aromatic 6-carbon chemical compound." OwnerUserId="43" LastEditorUserId="43" LastEditDate="2012-04-29T16:55:45.247" LastActivityDate="2012-04-29T16:55:45.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="141" PostTypeId="1" AcceptedAnswerId="338" CreationDate="2012-04-28T14:23:08.087" Score="52" ViewCount="54310" Body="&lt;p&gt;I'm looking at the melting temperature of metallic elements, and notice that the metals with high melting temperature are all grouped in some lower-left corner of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-block. If I take for example the periodic table with physical state indicated at &lt;span class=&quot;math-container&quot;&gt;$\pu{2165 K}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LTl0N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that (apart from boron and carbon) the only elements still solid at that temperature form a rather well-defined block around tungsten (which melts at &lt;span class=&quot;math-container&quot;&gt;$\pu{3695 K}$&lt;/span&gt;). So what makes this group of metals melt at such high temperature?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="61578" LastEditDate="2019-12-24T06:59:09.860" LastActivityDate="2019-12-24T07:10:21.383" Title="What makes some metals melt at higher temperature?" Tags="&lt;melting-point&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="142" PostTypeId="1" AcceptedAnswerId="186" CreationDate="2012-04-28T17:01:22.647" Score="29" ViewCount="7482" Body="&lt;p&gt;Many diagrams of orbitals I've seen involve electrons moving through a center point—where the nucleus is. How can this be? Clearly they don't actually pass through the nucleus, so what does happen?&lt;/p&gt;&#xA;" OwnerUserId="16" LastActivityDate="2019-12-24T10:06:25.763" Title="How do orbitals coexist with a nucleus?" Tags="&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="144" PostTypeId="2" ParentId="142" CreationDate="2012-04-28T17:43:01.303" Score="12" Body="&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals, for example, have a nodal plane where the nucleus is, which means the electron density is zero there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An orbital doesn't represent a path the electrons take when moving. An orbital is a region of probability. To make things clear and definite, when we draw an orbital we only draw the region where 95% (for example) of the probability lies. The fact that the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals have a nodal plane simply means that the probability of finding an electron on that plane vanishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Positivism&quot; rel=&quot;nofollow noreferrer&quot;&gt;positivist&lt;/a&gt; would thus consider an electron's trajectory around a nucleus nonsensical, since by the uncertainty principle, we can never measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of orbitals is for visualizing electron density - where &lt;em&gt;can&lt;/em&gt; the electron be, most of the time? It's a very useful way of interpreting phenomena such as chemical reactivity (think of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt;) or stability (think of benzene's &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals).&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="61578" LastEditDate="2019-12-24T06:54:10.353" LastActivityDate="2019-12-24T06:54:10.353" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="145" PostTypeId="2" ParentId="142" CreationDate="2012-04-28T17:44:50.767" Score="16" Body="&lt;p&gt;Electrons can be considered both particles and waves (&lt;a href=&quot;http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality&quot; rel=&quot;noreferrer&quot;&gt;wiki&lt;/a&gt;).  Basically, considering electrons to be particles is insufficient to explain many of the observed phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a particle wouldn't really pass through the nucleus, but a wave certainly could.  Take, for example, the 2p orbital, which is centered around the nucleus with two lobes.  The wave function can let us visualize where an electron is more likely to be found: (from &lt;a href=&quot;http://courses.chem.psu.edu/chem210/mol-gallery/hydrogen-atom/plots/2p-wave-full.jpg&quot; rel=&quot;noreferrer&quot;&gt;PSU.edu&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xstrd.jpg&quot; alt=&quot;2p Wave Function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability density function is found by squaring the wave function, and the PDF shows where the electron is likely to be observed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XX575.gif&quot; alt=&quot;2p Electron Density Function&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(sorry for the small size)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from this diagram, there is probability 0 of actually observing an electron at the nucleus although the electron must pass through the nucleus from side to side, which gets to the heart of your question.  The PDF deals mostly with the particle nature of the electron, because it shows where it is likely to observe one.  When the electron is exhibiting wave features, however, it can pass through the nucleus without actually being found there.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best analogy I can give is if you wave a jump rope up and down, you can't really isolate the wave in the jump rope, but it is still clearly there.  Long story short, a particle doesn't pass through the nucleus, but a wave can and does.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-04-28T17:44:50.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="146" PostTypeId="2" ParentId="142" CreationDate="2012-04-28T17:48:38.687" Score="5" Body="&lt;p&gt;Remember, orbitals don't show where electrons &lt;em&gt;are&lt;/em&gt;, they show the probability density of where a measurement would find them. Likely, at a scale close enough to the nucleus (where the probability of the electron being is exceedingly small to begin with, the electron's actual wavefunction is affected by the nucleus in such a way that either they can't overlap or the chance is even more vanishingly small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a bit rusty, but I believe every orbital but the 1s has a node at or on a plane through the nucleus anyway. &lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditDate="2012-04-28T18:03:17.310" LastActivityDate="2012-04-28T18:03:17.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="147" PostTypeId="2" ParentId="141" CreationDate="2012-04-28T19:06:21.397" Score="7" Body="&lt;p&gt;I think there are two properties at play here: atomic velocity and stability of the lattice structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that temperature is a measure of the average kinetic energy of the molecules, so $v \propto \sqrt{\frac{T}{m}}$, or to achieve the same velocity, $T \propto m$. So at the same temperature, two heavier atoms will move more slowly past each other than two lighter molecules, giving them more time to interact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now recall that in a solid, molecules are held in a lattice by intermolecular forces, and in a liquid, atoms have enough energy that the forces between them are no longer strong enough to hold the atoms in a lattice. So the more stability a substance's atoms gain from being in their solid, crystalline arrangement, the higher its melting point will be.  Now, why would the metals in that area gain more stability in their solid forms? I'm fairly sure the answer has to do with the completeness of orbitals and half-orbitals.  I'm not sure of the particulars of the delocalized electron cloud in metals, but I think it's likely that it allows these metals to in some manner fill or empty their incomplete orbitals. &lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="402" LastEditDate="2017-04-16T23:45:56.240" LastActivityDate="2017-04-16T23:45:56.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="148" PostTypeId="1" AcceptedAnswerId="150" CreationDate="2012-04-28T20:10:53.783" Score="40" ViewCount="74378" Body="&lt;p&gt;I know that the electron configuration of vanadium is $[\ce{Ar}]\mathrm{4s^2 3d^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the electrons in the 3d subshell are paired. Once it loses these three electrons, shouldn't the remainder of the electrons be paired? How can $\ce{V^{3+}}$ be paramagnetic if it loses all its unpaired electrons?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="16683" LastEditorDisplayName="user95" LastEditDate="2017-08-24T05:48:18.413" LastActivityDate="2019-12-24T08:08:16.993" Title="Why is the vanadium(3+) ion paramagnetic?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="150" PostTypeId="2" ParentId="148" CreationDate="2012-04-28T21:24:26.313" Score="39" Body="&lt;p&gt;In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-block (a.k.a. the &lt;em&gt;transition metals&lt;/em&gt;) obey one special rule:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;In general, electrons are removed from the valence-shell &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbitals before they are removed from valence &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals when transition metals are ionized.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt;(I took this formulation from &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php#ox&quot; rel=&quot;noreferrer&quot;&gt;these online lecture notes&lt;/a&gt;, but you will find equivalent statements in your textbooks.)&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, what that does mean is that if you remove electrons from vanadium(0), you will remove the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt; electrons before you remove the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt;-electrons. So, you have the following electronic configurations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{V}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]} \mathrm{4s^2 3d^3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{V^2+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]} \mathrm{4s^0 3d^3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{V^3+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]} \mathrm{4s^0 3d^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{V^4+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]} \mathrm{4s^0 3d^1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{V^5+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]} \mathrm{4s^0 3d^0}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And thus, &lt;span class=&quot;math-container&quot;&gt;$\ce{V^3+}$&lt;/span&gt; is paramagnetic, because it has two unpaired &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt;-electrons. In fact, all the ions above are paramagnetic, except &lt;span class=&quot;math-container&quot;&gt;$\ce{V^5+}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-12-24T08:08:16.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="151" PostTypeId="1" CreationDate="2012-04-28T22:03:40.027" Score="66" ViewCount="30152" Body="&lt;p&gt;When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm really looking for is &lt;strong&gt;why&lt;/strong&gt; the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does this not happen with the other columns as well?&lt;/strong&gt; Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="16683" LastEditDate="2016-12-04T17:28:02.937" LastActivityDate="2019-12-24T06:28:57.773" Title="Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?" Tags="&lt;orbitals&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="152" PostTypeId="1" CreationDate="2012-04-28T22:14:02.010" Score="86" ViewCount="123510" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tnn32.gif&quot; alt=&quot;Diagram of P orbital vs. S orbital distance from nucleus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry book explains that even though electrons in the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; orbital are closer to the nucleus on average, electrons from the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?&lt;/p&gt;&#xA;" OwnerUserId="53" LastEditorUserId="61578" LastEditDate="2019-12-24T05:53:12.150" LastActivityDate="2019-12-24T06:20:40.313" Title="Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus?" Tags="&lt;energy&gt;&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="7" CommentCount="2" FavoriteCount="43" ContentLicense="CC BY-SA 4.0" />
  <row Id="153" PostTypeId="1" CreationDate="2012-04-29T00:57:13.443" Score="28" ViewCount="63458" Body="&lt;p&gt;There are some stereochemical reactions that result in the presence of enantiomers. When moving forward with a practical organic synthesis, how does one usually separate them in order to continue with one of the enantiomers?&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2012-04-29T01:11:34.800" LastActivityDate="2019-07-18T05:59:47.317" Title="How do you separate enantiomers?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="154" PostTypeId="2" ParentId="153" CreationDate="2012-04-29T01:16:01.643" Score="14" Body="&lt;p&gt;The physicochemical properties of enantiomers being the same (except for optical activity), one needs to make use of the chirality of the molecules to separate them. This is done by &lt;em&gt;asymmetrizing&lt;/em&gt; the environment of the molecules. I'll explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say you have a racemic mixture of a chiral carboxylic acid. To effectively differentiate both enantiomers, you can use an enantiopure amine to form a salt with your carboxylic acid. You've then created two diastereomers from the combination of chiral molecules. Since the physical properties of diastereomers are different, you can then separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to do this would be to crystallize your salts, extract/wash them, then pass them through a flash chromatography column, and then get rid of the amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to use a chiral chromatography column, which would effectively have the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also analytical techniques that can differentiate between enantiomers, for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Micellar_electrokinetic_chromatography&quot;&gt;MEKC&lt;/a&gt; used with chiral cyclodextrins.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think you've gotten the point that, to separate enantiomers, you need to create diastereomers.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-04-29T01:16:01.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="155" PostTypeId="2" ParentId="153" CreationDate="2012-04-29T01:18:12.137" Score="28" Body="&lt;p&gt;There are several ways that enantiomers can be separated, but none of them are particularly simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first way to separate them is &lt;strong&gt;chiral chromatography&lt;/strong&gt;.  In chiral chromatography, silica gel is bonded to chiral molecules to form what is called a chiral stationary phase.  The enantiomers will then separate as they run down the column because one of the enantiomers will interact more strongly with the column and &quot;stick&quot; in place.  Chiral sugars (ex. cellulose) are frequently used in chiral chromatography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second common method is to react the enantiomers with another chemical to form &lt;strong&gt;diastereomers&lt;/strong&gt;.  While enantiomers are identical in terms of chemical properties, diastereomers are not.  Diastereomers can be created by reacting a mixture of both the enantiomers with another chiral molecule, such as &lt;em&gt;s&lt;/em&gt;-brucine, which is commonly used because it is cheap.  Diastereomers have different chemical properties (for example melting points), so it is much easier to separate them.  Then, after separation, the enantiomers can be recovered from the single diastereomer.  &lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="7220" LastEditDate="2016-03-09T00:12:44.770" LastActivityDate="2016-03-09T00:12:44.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="156" PostTypeId="2" ParentId="152" CreationDate="2012-04-29T01:29:06.790" Score="11" Body="&lt;p&gt;This is not a thorough answer. It holds when comparing orbital energies of the same shell number &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; (in your case 2) in atoms containing a low number of electrons. &lt;a href=&quot;https://chemistry.stackexchange.com/q/148/78&quot;&gt;This question&lt;/a&gt; contains more information on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The letter of the orbital (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;, etc.) is a nameplace for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Azimuthal_quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;azimuthal quantum number&lt;/a&gt; (orbital angular momentum, &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;). The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbital has &lt;span class=&quot;math-container&quot;&gt;$\ell=0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; has &lt;span class=&quot;math-container&quot;&gt;$\ell=1$&lt;/span&gt;. The angular momentum being higher, the energy also is.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="61578" LastEditDate="2019-12-24T06:20:40.313" LastActivityDate="2019-12-24T06:20:40.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="157" PostTypeId="2" ParentId="152" CreationDate="2012-04-29T06:28:38.573" Score="12" Body="&lt;p&gt;Because of the penetration, that is, because the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; electron can spend time near the nucleus, it is less effectively shielded by the core electrons than a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; electron. Because it is less effectively shielded, a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; electron experiences a higher effective nuclear charge and is held closer to the nucleus than a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; electron which gives the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; orbital a lower energy.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="61578" LastEditDate="2019-12-24T05:47:54.797" LastActivityDate="2019-12-24T05:47:54.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="158" PostTypeId="1" CreationDate="2012-04-29T10:17:08.590" Score="9" ViewCount="277" Body="&lt;p&gt;The most stable cyclohexane form is the chair conformation but on the other hand, the bigger the side-chain of the cyclohexane is, its hydrogen atoms become more equatorial rather than axial, which brings them closer to the cyclohexane's hydrogens.&#xA;Shouldn't they arrange in a less energetic, but a more stable conformation?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="16" LastEditDate="2012-05-09T20:43:57.423" LastActivityDate="2015-05-13T19:05:05.990" Title="Why is it that the bigger the side-chain is, its hydrogen atoms tend to be more equatorial?" Tags="&lt;organic-chemistry&gt;&lt;equilibrium&gt;&lt;structural-formula&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="160" PostTypeId="2" ParentId="151" CreationDate="2012-04-29T15:47:43.587" Score="25" Body="&lt;p&gt;As I understand this, there are basically two effects at work here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you populate an &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals and putting one in an &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital increases as you fill the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, pairing electrons in one orbital (so, adding the second &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the effect isn't strong enough to avert fluorine having a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s^2}$&lt;/span&gt; occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d}$&lt;/span&gt; instead of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4f}$&lt;/span&gt; orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6s^2}$&lt;/span&gt; occupation, but is for gold - more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="61578" LastEditDate="2019-12-24T06:28:57.773" LastActivityDate="2019-12-24T06:28:57.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="161" PostTypeId="2" ParentId="152" CreationDate="2012-04-29T15:53:51.847" Score="21" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems like it should be the average distance that matters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It is the average energy that matters. Note that this stuff about &quot;spends so much time here and so much there...&quot; is really just a (not particularly good) way of describing a quantum-mechanical wave function's absolute square. The electron is actually never at any particular place in the orbital, rather it's &lt;em&gt;everywhere at the same time&lt;/em&gt; – until you decide to measure its precise position (thereby destroying the orbital, that has to do with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heisenberg_uncertainty_principle&quot;&gt;Heisenberg uncertainty principle&lt;/a&gt;), in which case there would be a certain &lt;em&gt;probability&lt;/em&gt; of finding it here or there. That probability can be calculated from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavefunction&quot;&gt;wave function&lt;/a&gt; $\psi(\mathbf{x})$ by taking its absolute square, which gives the radial distribution function in your plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of any quantum state is determined by the expectation value of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29&quot;&gt;Hamiltonian operator&lt;/a&gt;, which very roughly means &quot;measure the energy at &lt;em&gt;all&lt;/em&gt; places in space, multiply with the probability the electron is found at that place, and add all those values up&quot;. That should sound quite intuitive. In actuality it's more complicated: the electron also has some kinetic energy at every place in space, which is itself &lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum_operator#Definition&quot;&gt;encoded in the wave function&lt;/a&gt; – but you need to consider its &lt;em&gt;complex phase&lt;/em&gt;, which is a whole extra dimension that you simply don't see in your plots and certainly not in the average distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the exact energies for atomic orbitals is a lot of maths that probably doesn't belong here, but it's possible to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen-like_atom#Non-relativistic_wavefunction_and_energy&quot;&gt;carry it out&lt;/a&gt;*, and it should be obvious now that the result doesn't necessarily have much to do with simple &quot;energy at the average distance&quot;. The quantities such as angular momentum mentioned in CHM's answer can be used to make it a bit simpler again (the Hamiltonian &lt;em&gt;separates&lt;/em&gt; into a radial and spherical part), but that alone doesn't really explain much I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;*As a matter of fact, it's &lt;em&gt;not&lt;/em&gt; possible to calculate it exactly for anything more complicated than a hydrogen atom! You have to do it with certain approximations.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2012-04-29T18:26:54.653" LastActivityDate="2012-04-29T18:26:54.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="162" PostTypeId="1" CreationDate="2012-04-29T17:23:10.420" Score="8" ViewCount="5349" Body="&lt;p&gt;Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect?&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-06-14T19:38:16.523" Title="Availabilty of magnesium ions in phosphate buffer" Tags="&lt;solubility&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="163" PostTypeId="1" AcceptedAnswerId="166" CreationDate="2012-04-29T18:16:56.693" Score="23" ViewCount="38097" Body="&lt;p&gt;I was tasked with figuring out whether carbon or nitrogen has a more negative electron affinity value. I initially picked nitrogen, just because nitrogen has a higher $Z_\mathrm{eff}$, creating a larger attraction between electrons and protons, decreasing the radius, causing a higher ionization energy, and therefore decreasing the electron affinity value, but I was actually wrong, and the solutions manual explains it as this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;As you go from C to N across the Periodic Table you would normally expect N to have the more negative electron affinity. However, N has a half-filled p subshell, which lends it extra stability; therefore, it is harder to add an electron.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there any major exceptions to the rules when comparing electron affinity? I'm hesitant to use nitrogen as an exception, because I don't know how far it extends. If nitrogen has a more positive EA than carbon, does that also extend to boron, aluminium, or phosphorus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I later found that this also applies when comparing silicon and phosphorus. The explanation given was the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exceptions should be noted when comparing electron affinities? Are there any at all? And how far does the exception with atoms with half-filled p subshells extend?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="16683" LastEditDate="2016-01-08T10:16:56.963" LastActivityDate="2017-02-08T05:17:34.607" Title="Are there any major exceptions when comparing electron affinity?" Tags="&lt;electronic-configuration&gt;&lt;periodic-trends&gt;&lt;electron-affinity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="164" PostTypeId="1" AcceptedAnswerId="165" CreationDate="2012-04-29T18:27:30.440" Score="32" ViewCount="10906" Body="&lt;p&gt;Most periodic tables only feature one Hydrogen atom, on the top of the first group. But some, like the one I was given, also show Hydrogen in the 7th group, to left of Helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are there two Hydrogen atoms? What's the difference between the two, and why do we work with the left one more often than the right?&lt;/p&gt;&#xA;" OwnerUserId="85" LastEditorUserId="12" LastEditDate="2012-04-29T18:32:30.557" LastActivityDate="2015-06-20T21:32:17.557" Title="Why are there two Hydrogen atoms on some periodic tables?" Tags="&lt;atoms&gt;&lt;periodic-trends&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="165" PostTypeId="2" ParentId="164" CreationDate="2012-04-29T18:30:29.547" Score="29" Body="&lt;p&gt;The two hydrogens are the same, but some periodic tables show hydrogen in both places to emphasize that hydrogen isn't really a member of the first group or the seventh group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is a diatomic gas in it's elemental state, which is different from the other group one metals (and similar to the group seven elements).  At the same time, hydrogen usually loses its electron, similar to sodium and the other group one metals.&lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-04-29T18:30:29.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="166" PostTypeId="2" ParentId="163" CreationDate="2012-04-29T19:48:29.740" Score="18" Body="&lt;p&gt;This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing takes place (difference between energy levels of 2p and 3s is greater than electron pairing energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron pairing effects have significant impact to physical properties of coordination complexes (like color and magnetic properties).&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2012-04-29T19:48:29.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="168" PostTypeId="1" CreationDate="2012-04-29T23:35:26.400" Score="23" ViewCount="5951" Body="&lt;p&gt;Overall ring strain seems to be a big issue when it comes to organic chemistry. That is why cyclopentane may be in an &quot;envelope form&quot; or why cyclobutane may be in a kinked, kite form. Both of these example molecules are not planar. So, why is epoxide (oxacyclopropane) a stable molecule, if at all?&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="10185" LastEditDate="2015-03-19T18:02:40.900" LastActivityDate="2015-03-19T18:02:40.900" Title="What makes an epoxide stable?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="169" PostTypeId="2" ParentId="88" CreationDate="2012-04-29T23:54:45.723" Score="14" Body="&lt;p&gt;Aspartame’s sweetening properties were discovered by accident. the G.D. Searle company was trying to develop drugs to treat ulcers in the 1960s when one of the scientists accidentally contaminated his finger and licked it. Knowing how many versions of molecules get tested in drug development, my guess is that an ethyl ester version was made, and probably failed due to improper taste or some other property. Aspartame is already “sparingly soluble” in water; I assume the ethyl ester would be even less soluble, making it less useful for sweetening drinks. However, the amount of aspartame that would have to be dissolved is so small I don’t know if it would make a difference. It's probably taste related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for toxicity concerns, it takes a lot of methanol to cause toxicity, Wikipedia claims 10 mL can cause permanent blindness, and 30 mL can be fatal, but the amount of aspartame a human would need to ingest to produce 10 mL worth of methanol would be ridiculous.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T18:30:25.837" LastActivityDate="2015-05-11T18:30:25.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="170" PostTypeId="2" ParentId="168" CreationDate="2012-04-30T00:01:17.050" Score="5" Body="&lt;p&gt;You can't quite compare rings consisting only of carbon atoms with rings with oxygens in them. In the latter, the $\ce{C-O}$ bonds are polar, the electrons are closer to the oxygen atom which means the acuteness at the $\ce{C}$ atoms is not as much of a problem as it would be in a hydrocarbon ring.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="418" LastEditDate="2015-03-19T07:08:03.253" LastActivityDate="2015-03-19T07:08:03.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="171" PostTypeId="2" ParentId="168" CreationDate="2012-04-30T02:31:25.053" Score="22" Body="&lt;p&gt;Well, you have to ask: &lt;em&gt;how&lt;/em&gt; stable? and &lt;em&gt;compared to what&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopropane&quot;&gt;cyclopropane&lt;/a&gt;, which also contains a three-membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from &lt;a href=&quot;http://chemistry.illinoisstate.edu/standard/che38037/handouts/380.37thermo.pdf&quot;&gt;these lecture notes&lt;/a&gt;, which indicate that &lt;strong&gt;&lt;em&gt;“ethylene oxide has less ring strain than cyclopropane”&lt;/em&gt;&lt;/strong&gt;. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-05-01T11:31:37.960" LastActivityDate="2012-05-01T11:31:37.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="172" PostTypeId="1" AcceptedAnswerId="183" CreationDate="2012-04-30T04:27:49.763" Score="13" ViewCount="299" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi-pi_interaction&quot;&gt;wikipedia and the references given therein&lt;/a&gt;, $\pi\cdots\pi$ stacking interactions are the result of interaction between the quadrupole moments of two aromatic rings, rationalising the stabilisation of perpendicular and offset-parallel association modes of the benzene dimer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventional wisdom is that the supramolecular complexes of buckminsterfullerene and functionalised corannulenes (the so-called buckycatchers) are stabilised by $\pi\cdots\pi$ stacking. How can the quadrupole interaction explanation be reconciled with the pseudo-spherical symmetry of buckminsterfullerene? Surely the interaction must be entirely Van der Waals in nature?&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="7448" LastEditDate="2015-04-24T15:26:03.893" LastActivityDate="2015-04-24T15:26:03.893" Title="Are buckminsterfullerene-corannulene complexes actually stabilised by π stacking?" Tags="&lt;theoretical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="173" PostTypeId="2" ParentId="88" CreationDate="2012-04-30T05:23:03.123" Score="20" Body="&lt;p&gt;This question raises more important issues than just the technical &quot;why methyl ester,&quot; so I'll address those too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja01038a046&quot; rel=&quot;nofollow noreferrer&quot;&gt;This report&lt;/a&gt; says it is approximately &lt;span class=&quot;math-container&quot;&gt;$10\times$&lt;/span&gt; less sweet (see Table VI on page 2689 and the entry '&lt;span class=&quot;math-container&quot;&gt;$\ce{Asp-Phe-OEt}$&lt;/span&gt;' with &quot;&lt;span class=&quot;math-container&quot;&gt;$++$&lt;/span&gt;&quot; being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A toxicology class I once took can mostly be summed up by a cliche phrase, &quot;the dose makes the poison.&quot; Even too much of a &lt;strong&gt;good&lt;/strong&gt; thing is by definition a &lt;strong&gt;bad&lt;/strong&gt; thing, for example too much &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_toxicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxygen&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_intoxication&quot; rel=&quot;nofollow noreferrer&quot;&gt;water&lt;/a&gt;, or even the &lt;a href=&quot;http://www.epa.gov/sunwise/uvandhealth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sun&lt;/a&gt; can lead to death. We also say that too little of any compound won't cause &lt;em&gt;measurable&lt;/em&gt; harm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a &lt;a href=&quot;http://en.wikipedia.org/wiki/CNS_depressant&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNS depressant&lt;/a&gt;. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for an enzyme, which metabolize ethanol. This enzyme is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcohol_dehydrogenase&quot; rel=&quot;nofollow noreferrer&quot;&gt;alcohol dehydrogenase&lt;/a&gt; (ADH), which leads to biochemical pathways that allow the safe removal of ethanol. However, methanol is a different case. ADH instead make formaldehyde, which is then converted to &lt;a href=&quot;http://en.wikipedia.org/wiki/Formic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;formic acid&lt;/a&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldehyde_dehydrogenase&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALDH&lt;/a&gt;. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fire_ant&quot; rel=&quot;nofollow noreferrer&quot;&gt;fire ant&lt;/a&gt;. In low quantities it is fairly benign. &lt;a href=&quot;http://en.wikipedia.org/wiki/Formaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formaldehyde&lt;/a&gt; is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly &lt;span class=&quot;math-container&quot;&gt;$\pu{1.8 g}$&lt;/span&gt; of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, &lt;span class=&quot;math-container&quot;&gt;$\pu{8 hours}$&lt;/span&gt; after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~&lt;span class=&quot;math-container&quot;&gt;$\pu{200 lbs}$&lt;/span&gt; person ingests &lt;span class=&quot;math-container&quot;&gt;$\pu{1.8 g}$&lt;/span&gt; of aspartame. Let's stop for a second and put this in terms of something we can imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any reliable numbers, but let's just say that for a &lt;span class=&quot;math-container&quot;&gt;$\pu{12 oz}$&lt;/span&gt; diet pop there are &lt;span class=&quot;math-container&quot;&gt;$\pu{200 mg}$&lt;/span&gt; of aspartame. That means for every liter of pop, there are approximately &lt;span class=&quot;math-container&quot;&gt;$\pu{564 mg}$&lt;/span&gt; of aspartame. That means &lt;span class=&quot;math-container&quot;&gt;$\pu{3 L}$&lt;/span&gt; of pop would get me to &lt;span class=&quot;math-container&quot;&gt;$\pu{1.8 g}$&lt;/span&gt; of aspartame. That means I would need to drink more than &lt;span class=&quot;math-container&quot;&gt;$\pu{3 L}$&lt;/span&gt; in one sitting to overwhelm the excretion systems. This suggests an &lt;a href=&quot;http://en.wikipedia.org/wiki/Acute_toxicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;acute poisoning&lt;/a&gt; isn't very plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. &quot;Sure, there is no acute toxicity, but what happens in the long-term with a low dose?&quot; For the answer to this question, we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty&quot; rel=&quot;nofollow noreferrer&quot;&gt;uncertainties&lt;/a&gt; in their measurements and this is just the general nature of science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledge base. Once in a while the general community is &lt;a href=&quot;http://en.wikipedia.org/wiki/Missoula_Floods#Flood_hypothesis_proposed&quot; rel=&quot;nofollow noreferrer&quot;&gt;proven wrong about something long thought to be true&lt;/a&gt; and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="61578" LastEditDate="2019-11-01T19:54:40.690" LastActivityDate="2019-11-01T19:54:40.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="174" PostTypeId="1" AcceptedAnswerId="451" CreationDate="2012-04-30T07:11:18.607" Score="23" ViewCount="14722" Body="&lt;p&gt;Since the steam pressure of ammonia is higher than that of water, I would expect distillation to be a reasonable way of separating a mixture of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in industrial applications known to me there's always a step of stripping the ammonia chemically from the vapor. Why is this necessary?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="70883" LastEditDate="2019-06-10T17:38:32.507" LastActivityDate="2020-12-14T21:58:21.803" Title="Why is distillation not a viable way to separate ammonia from water?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;water&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="175" PostTypeId="2" ParentId="16" CreationDate="2012-04-30T07:56:14.837" Score="8" Body="&lt;p&gt;As to the above answers I also want to include the mechanism of action of silver as an antimicrobial agent. The exact mechanism of action of silver as an antimicrobial agent is not known and the current hypothesis is silver will converted to silver ions and this positively charged ions will attack the cell membrane, DNA or proteins which are negatively charged thus inhibiting the growth of microorganisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www.worldwidewounds.com/2004/february/Cooper/Topical-Antimicrobial-Agents.html&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="30" LastEditDate="2012-05-06T21:49:05.340" LastActivityDate="2012-05-06T21:49:05.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="176" PostTypeId="2" ParentId="162" CreationDate="2012-04-30T08:58:03.420" Score="-1" Body="&lt;p&gt;I don't have a source (at the moment) but it is accepted that high phosphate buffers do chelate/sequester Mg&lt;sup&gt;2+&lt;/sup&gt; ions. Pyrophosphate is particularly infamous for this. During RNA transcription experiments, the solution will get cloudy which is due exactly to this phenomena. It is also seen in cell-free protein synthesis which requires a steady concentration of Mg&lt;sup&gt;2+&lt;/sup&gt;, that the increased concentration of the phosphate buffer requires an increased amount of Mg&lt;sup&gt;2+&lt;/sup&gt; to maintain the same effective free Mg&lt;sup&gt;2+&lt;/sup&gt; concentration [&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/18854819&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;].&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="30" LastEditDate="2012-04-30T11:19:16.593" LastActivityDate="2012-04-30T11:19:16.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="177" PostTypeId="5" CreationDate="2012-04-30T12:12:18.730" Score="0" Body="&lt;p&gt;From the &lt;a href=&quot;https://iupac.org/projects/project-details/?project_nr=1999-001-1-200&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Red Book &lt;strong&gt;2005&lt;/strong&gt;, IR-3.5&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, the elements of groups 3–12 are the d-block elements. These elements are also commonly referred to as the transition elements, though the&#xA;  elements of group 12 are not always included; [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These metals, like all metals, are ductile and have the capability to conduct electricity as well as heat.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-20T17:53:52.123" LastActivityDate="2018-03-20T17:53:52.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="178" PostTypeId="4" CreationDate="2012-04-30T12:12:18.730" Score="0" Body="For questions about the characteristic physical properties, chemical reactions, etc. of d-block elements, group 3-12, or their compounds. Do not use this tag if your questions is about general properties of metals, use [metal] instead. Also see [rare-earth-elements]." OwnerUserId="94" LastEditorUserId="5026" LastEditDate="2018-03-20T17:54:20.460" LastActivityDate="2018-03-20T17:54:20.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="180" PostTypeId="1" AcceptedAnswerId="181" CreationDate="2012-04-30T19:28:42.007" Score="23" ViewCount="78829" Body="&lt;p&gt;For example, why does for example oxygen turn into gas at a much lower temperature than water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have anything to do with the molecular structure? A water molecule does have a more complex structure than oxygen, though the R-410A (a mixture of two gases commonly used in heating pumps) is much more complex than water, and it boils at -48.5 degrees Celsius.&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="30" LastEditDate="2012-04-30T22:32:33.587" LastActivityDate="2015-04-24T11:50:05.667" Title="Why do different substances have different boiling points?" Tags="&lt;molecules&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="181" PostTypeId="2" ParentId="180" CreationDate="2012-04-30T20:01:59.850" Score="32" Body="&lt;p&gt;The boiling point of a liquid depends on the intermolecular forces present between the atoms or molecules in the liquid since you must disrupt those forces to change from a liquid to a gas. The stronger the intermolecular forces, the higher the boiling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two oxygen molecules are attracted to each other through London dispersion forces (induced temporary dipoles between the molecules) while water molecules are attracted to each other by hydrogen bonding (attraction of the + dipole on H in one molecule to the – dipole on an oxygen in an adjacent molecule) that is relatively strong. (Hydrogen bonding is an important intermolecular force for molecules where H is directly covalently bonded to F, O or N, which are quite electronegative and thus form bond with H with a relatively strong dipole.) London dispersion forces become more important for atoms and molecules with more electrons. Dipole–dipole attractions are also important in some molecules. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="30" LastEditDate="2012-05-04T12:05:37.963" LastActivityDate="2012-05-04T12:05:37.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="183" PostTypeId="2" ParentId="172" CreationDate="2012-04-30T22:56:59.627" Score="6" Body="&lt;p&gt;In short, I think the answer is that although the overall symmetry of the buckminsterfullerene is almost spherical, the quadrupole–quadrupole interaction with other molecules with aromatic cycles are local (in addition to the ever-present van der Waals interactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to think of a way to exemplify that, and drew the 2D picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6EWZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this 2D model, the “double bonds” of the circular structure represent the individual aromatic rings of your buckminsterfullerene (&lt;strong&gt;which isn't &lt;em&gt;superaromatic&lt;/em&gt;, i.e. electrons are not delocalized over its entire structure&lt;/strong&gt;; think may be a key point), bearing each a local quadrupole moment (symbolized in the red/blue thingy). The other particle with a quadrupole is close to one of the ball's quadrupoles, and this local interaction dominates. In fact, other interactions maybe be stabilizing or not, depending on the orientation of the individual quadrupoles, but they are weaker that the local quadrupole–quadrupole interaction.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-04-30T22:56:59.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="186" PostTypeId="2" ParentId="142" CreationDate="2012-05-01T01:42:40.653" Score="18" Body="&lt;p&gt;In the definition of an s shell, you will find that its $\ell$ number is zero. In classical terms, that corresponds to an orbit with zero orbital or angular momentum — which for a large object is a clear impossibility. For an electron, it gives the peculiar result that any electron in any s shell is, classically speaking, moving back and forth through the nucleus, rather than around it. So, in one curious sense, what you just asked is exactly what happens: the classical analogy is that the electrons do go through the nucleus, which is why they have such a lovely spherical symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the answer, however, is that electrons can't go through the nucleus unless they are hugely more energetic than the ones found in a typical small-nucleus atom. Clearly there's a bit of a paradox going on there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution of the paradox is that very-low-mass charged particles must be treated by quantum rules. So, for example, rather than the electron behaving like a well-defined particle, it behaves like a standing wave. That standing wave can in turn be thought of as two simultaneous versions of the electron, one going (for example) clockwise and the other counterclockwise around. (The real situation has an infinite number of such components; I'm just picking out a single pair of them that demonstrate the principle.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these components can furthermore be thought of as being refracted by the powerful spherical charge field of the nucleus, curving around it without every striking it. This refraction is not the same as an attraction. In fact, it is this refraction effect that prevents the density of the electron cloud from reaching infinity at the nucleus — that is, from striking the nucleus. If you think of how a tank of water can cause a beam of light to bounce off the surface instead of entering the tank — and that's a terrible analogy, I know, I know — you can at least get some idea of how an increasing &quot;optical density&quot; towards a central point might keep light away rather than bringing it closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for an electron that behaves &quot;like&quot; two waves going both clockwise and counterclockwise, the combined waves curve around the nucleus rather than striking it. This is a very quantum sort of event, since for a classical object such &quot;splitting&quot; of the object is simply not possible and the object simply dives straight into the source of attraction. But if objects are light enough, that kind of particle-like behavior simply ceases to be available to the object. Instead, you get waves that neatly and with perfect spherical symmetry curve around the nucleus, never obtaining enough energy (which makes it more particle-like) to connect directly with that nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, notice that electrons in s (and other) shells necessarily combine multiple paths all at the same time. For each &quot;image&quot; of the electron that is traveling clockwise, there must also be an exactly balancing &quot;image&quot; of the same electron traveling counterclockwise, so that the two images always balance out to zero orbital momentum. What an amazing thing that is! And an important thing, too, since it's what makes chemistry possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, good question, even if it really is more of a physics question per se than a chemistry question. But it's such an important chemistry question! It's like asking how the engine that powers the car works. You can accept it as a given that all cars and vehicles have engines, and that they all work in a certain way. However, sometimes it's nice to dive in a little deeper and try to understand why these peculiar things do the things that make chemistry possible — that is, how the engine really works.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="23561" LastEditDate="2018-09-17T19:05:36.380" LastActivityDate="2018-09-17T19:05:36.380" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="187" PostTypeId="1" AcceptedAnswerId="188" CreationDate="2012-05-01T07:29:48.457" Score="13" ViewCount="453" Body="&lt;p&gt;If dry ammonia gas is passed through anhydrous copper sulfate, will it turn blue (due to the formation of tetraamminecopper(II) complex)? Or will silver chloride form diamminesilver(I) complex in liquid ammonia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above complexes do not get formed in non-aqueous mediums, why is it so?&lt;/p&gt;&#xA;" OwnerUserId="140" LastActivityDate="2015-10-29T10:46:21.060" Title="Is aqueous medium necessary for complex formation? if so, why?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="188" PostTypeId="2" ParentId="187" CreationDate="2012-05-01T08:07:33.827" Score="10" Body="&lt;p&gt;It should be noted that ammonia is a mildly polar solvent and that (for instance) you can prepare $\ce{[Cu(NH3)6]^2+}$ by reaction with liquid ammonia&lt;sup&gt;1&lt;/sup&gt;. The description of the geometry (two pages earlier) of $\ce{CuSO4.5H2O}$ (namely containing a square planar $\ce{[Cu(H2O)4]^2+}$ unit with two additional $\ce{[SO4]^2-}$ above and below the plane) suggests that ammine ligands could fulfill a similar role to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that passing ammonia gas over a solid will probably mean that if the reaction happens it will only happen near grain surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your second suggestion, Housecroft and Sharpe&lt;sup&gt;1&lt;/sup&gt; identifies a reaction producing $\ce{[Ag(NH3)4]+}$ from $\ce{Ag2O}$ in liquid $\ce{NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Housecroft, C. E.; Sharpe, A. G. &lt;em&gt;Inorganic Chemistry,&lt;/em&gt; 2nd Edition; Pearson Prentice Hall, 2005; pp. 635–637, 693.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorDisplayName="user7951" LastEditDate="2015-10-29T00:46:06.933" LastActivityDate="2015-10-29T00:46:06.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="189" PostTypeId="1" AcceptedAnswerId="192" CreationDate="2012-05-01T08:39:30.903" Score="26" ViewCount="4795" Body="&lt;p&gt;Over the course of my studies, I have switched largely from using Z-matrix representations of molecular geometries in calculations to Cartesian representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The software that I use now makes it easy to add the sorts of constraints/restraints/transits that I would have previously used Z-matrices for, and I know that Z-matrix geometries can be problematic in large molecules&lt;sup&gt;*&lt;/sup&gt; where minute changes in a bond angle or dihedral (due, for instance, to rounding errors/low-quality gradients) can result in large movements in peripheral atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;What pros or cons exist for either geometry definition that I don't know about? What circumstances recommend one representation over another?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Or small molecules with silly Z-matrices.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-12T19:25:58.433" Title="Pros and cons of Cartesian vs. Z-matrix representations of molecules?" Tags="&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="191" PostTypeId="1" AcceptedAnswerId="193" CreationDate="2012-05-01T15:30:42.923" Score="12" ViewCount="746" Body="&lt;p&gt;Pennsylvania State University provides irradiated $\ce{NaCl}$ (180,000 rads of gamma radiation) to teachers for a very entertaining demonstration. The salt is orange after irradiation. When it is placed on a hot plate, you see a flash of visible light and the salt returns to it normal color. The explanation provided with the salt is that &quot;The heat allows the electrons that were trapped in the crystal structure (when the energy was deposited) to return to their original position.&quot; This language evokes a picture of a particle electron in a crystal jail for me. The explanation goes on to say that the energy is released as a photon of visible light. (I understand this part of the explanation.) So, my question is, how are electrons trapped in a crystal structure? What keeps them in an excited state?&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="4231" LastEditDate="2015-02-21T22:26:50.687" LastActivityDate="2015-02-21T22:26:50.687" Title="How is the electron &quot;trapped in the crystal&quot; in irradiated NaCl?" Tags="&lt;electrons&gt;&lt;nuclear&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="192" PostTypeId="2" ParentId="189" CreationDate="2012-05-01T15:59:11.963" Score="23" Body="&lt;p&gt;&lt;strong&gt;Cartesian Space&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function.  To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of.  The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Internal Coordinates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z-matrices use a different approach.  When dealing with Z-matrices, we keep track of the relative positions of points in space.  Cartesian space is 'absolute' so to speak.  A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity.  However, consider a two atom system.  The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule.  An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1).  However, say we increase the distance between the hydrogen atoms.  We now have altered the molecule in such a way that the properties of that molecule has changed.  What did we change?  We simply changed the bond length, one variable.  We increased the distance between the two atoms by some length R.  With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D).  Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules).  For linear molecules we keep tabs on 3N-5 coordinates.  When performing complex computations, the less you have to keep track of, the less expensive the computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Symmetry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following molecule, H2O.  We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent.  When using some sort of optimizing routine, you may want to specify symmetry in your system.  With a Z-matrix, the process is very straightforward.  You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond.  Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry.  With Cartesian space this is not guaranteed.  Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Picking the Right One&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix.  I warn you that specifying your program to optimize using Cartesian coordinates makes your calculation much more expensive.  I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to use Z-matrices on systems that are rather small.  If dealing with systems with high symmetry, Z-matrices are almost essential.  They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error.  If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2012-05-02T13:17:38.150" LastActivityDate="2012-05-02T13:17:38.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="193" PostTypeId="2" ParentId="191" CreationDate="2012-05-02T00:56:03.857" Score="8" Body="&lt;p&gt;What you're describing is thermoluminescence, which is used as a paleochronology method (aka Trapped Electron (TE) dating, afaik). The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermoluminescence_dating&quot;&gt;wikipedia article&lt;/a&gt; has a good synopsis. Simply, electrons get bound to potential wells generated in the radiation-defected structure. When you heat the crystal up you allow it to anneal and return to a more regular structure. The defected crystal is stuck in an energetic local minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do this trick with X-rays. Calcite (or was it aragonite?) exposed to X-rays will turn light blue for a while.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-02T00:56:03.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="194" PostTypeId="1" CreationDate="2012-05-02T04:32:11.023" Score="13" ViewCount="2678" Body="&lt;p&gt;To my understanding, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cationic_polymerization&quot;&gt;cationic polymerization&lt;/a&gt; is a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Living_polymerization&quot;&gt;living polymerization&lt;/a&gt; but my teacher's notes and Wikipedia seem to suggest they're &lt;a href=&quot;http://en.wikipedia.org/wiki/Living_cationic_polymerization&quot;&gt;different&lt;/a&gt;. I can't see the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for anionic polymerization. From what I've studied so far, the difference must be subtle, if there is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a difference?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="7" LastEditDate="2012-05-04T05:07:35.890" LastActivityDate="2016-02-29T05:46:39.007" Title="Living-cationic vs cationic polymerization" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="195" PostTypeId="5" CreationDate="2012-05-02T04:52:00.230" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-02T04:52:00.230" LastActivityDate="2012-05-02T04:52:00.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="196" PostTypeId="4" CreationDate="2012-05-02T04:52:00.230" Score="0" Body="Cyclohexane is a cycloalkane with the molecular formula $\ce{C6H12}$. This tag should be applied to questions about the properties and reactions of cyclohexane and its derivatives. " OwnerUserId="124" LastEditorUserId="402" LastEditDate="2017-05-02T22:59:14.337" LastActivityDate="2017-05-02T22:59:14.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="197" PostTypeId="5" CreationDate="2012-05-02T04:57:00.193" Score="0" Body="&lt;p&gt;A metal is an element, compound, or alloy that is a good conductor of both electricity and heat, characterized by an opaque lustrous appearance and the absence of a bandgap.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="23561" LastEditDate="2018-07-02T08:03:16.663" LastActivityDate="2018-07-02T08:03:16.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="198" PostTypeId="4" CreationDate="2012-05-02T04:57:00.193" Score="0" Body="For questions about metals in general and their physical or chemical properties. For characteristic properties and reactions of d and f block metals specifically, use the [transition-metals] and [rare-earth-elements] tags respectively instead." OwnerUserId="124" LastEditorUserId="5026" LastEditDate="2018-03-15T13:28:15.707" LastActivityDate="2018-03-15T13:28:15.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="199" PostTypeId="5" CreationDate="2012-05-02T04:58:48.707" Score="0" Body="&lt;p&gt;At the melting point the solid and liquid phase exist in equilibrium. The melting point of a substance depends (usually slightly) on pressure and is usually specified at standard pressure.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:11:16.307" LastActivityDate="2012-05-05T14:11:16.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="200" PostTypeId="4" CreationDate="2012-05-02T04:58:48.707" Score="0" Body="The melting point of a solid is the temperature at which it changes state from solid to liquid." OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:10:16.653" LastActivityDate="2012-05-05T14:10:16.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="201" PostTypeId="5" CreationDate="2012-05-02T05:02:01.760" Score="0" Body="&lt;p&gt;These are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see &lt;a href=&quot;/questions/tagged/lewis-structure&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lewis-structure&amp;#39;&quot; rel=&quot;tag&quot;&gt;lewis-structure&lt;/a&gt; if you have a question involving the use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Lewis_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;lewis structures&lt;/a&gt;. It's &lt;a href=&quot;http://meta.chemistry.stackexchange.com/questions/2972/structural-diagrams-tag-disorder&quot;&gt;preferable&lt;/a&gt; &lt;a href=&quot;https://chemistry.meta.stackexchange.com/a/2984&quot;&gt;not&lt;/a&gt; to include this tag where you've already applied &lt;a href=&quot;/questions/tagged/lewis-structures&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;lewis-structures&amp;#39;&quot; rel=&quot;tag&quot;&gt;lewis-structures&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="4945" LastEditDate="2018-03-19T04:52:31.487" LastActivityDate="2018-03-19T04:52:31.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="202" PostTypeId="4" CreationDate="2012-05-02T05:02:01.760" Score="0" Body="This tag should be applied to questions about structural diagrams - diagrams used to show the crude location of possible existent bonds and lone pairs of electrons in a chemical species. Do not use this tag if your question is about Lewis structures, use [lewis-structures] instead." OwnerUserId="124" LastEditorUserId="5026" LastEditDate="2018-03-19T04:52:48.777" LastActivityDate="2018-03-19T04:52:48.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="203" PostTypeId="5" CreationDate="2012-05-02T05:08:28.850" Score="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://goldbook.iupac.org/E01975.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Elementary particle not affected by the strong force having a spin quantum number $\frac12$, a negative elementary charge and a rest mass of 0.000 548 579 903 13 u.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="4945" LastEditDate="2015-01-20T09:50:46.393" LastActivityDate="2015-01-20T09:50:46.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="204" PostTypeId="4" CreationDate="2012-05-02T05:08:28.850" Score="0" Body="Electrons are subatomic particles with the symbol e−. They have a negative electric charge (-1 elementary charge. )" OwnerUserId="124" LastEditorUserId="402" LastEditDate="2017-04-02T07:04:43.560" LastActivityDate="2017-04-02T07:04:43.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="205" PostTypeId="5" CreationDate="2012-05-02T05:09:47.040" Score="0" Body="&lt;p&gt;Knowledge of the electron configuration of different atoms is useful in understanding the structure of the periodic table of elements. The concept is also useful for describing the chemical bonds that hold atoms together. In bulk materials, this same idea helps explain the peculiar properties of lasers and semiconductors.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="402" LastEditDate="2017-04-08T16:00:36.090" LastActivityDate="2017-04-08T16:00:36.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="206" PostTypeId="4" CreationDate="2012-05-02T05:09:47.040" Score="0" Body="In atomic physics and quantum chemistry, the electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals." OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:07:35.063" LastActivityDate="2012-05-05T14:07:35.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="207" PostTypeId="5" CreationDate="2012-05-02T05:10:40.737" Score="0" Body="&lt;p&gt;This tag is intended for questions relating to software used by chemists specifically to perform or expedite tasks related to chemistry (&lt;a href=&quot;http://meta.chemistry.stackexchange.com/a/75/22&quot;&gt;relevant meta discussion&lt;/a&gt;). Examples include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule_editor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quantum chemistry software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule_editor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecule editors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_design_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular design software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chromatography_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chromatography software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_spectrometry_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass spectrometry software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Category:Molecular_dynamics_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular dynamics software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Category:Molecular_modelling_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular modelling software&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Laboratory_information_management_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laboratory information management system (LIMS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions relating to generic types of software that merely happen to be used by chemists (such as word processors, mathematical/statistical software, etc.) may be more appropriate to ask in a different site.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="9961" LastEditDate="2017-06-06T20:08:06.513" LastActivityDate="2017-06-06T20:08:06.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="208" PostTypeId="4" CreationDate="2012-05-02T05:10:40.737" Score="0" Body="For questions pertaining to software used by chemists for professional or academic purposes." OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-05-31T19:16:54.977" LastActivityDate="2013-05-31T19:16:54.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="209" PostTypeId="5" CreationDate="2012-05-02T05:16:02.627" Score="0" Body="&lt;p&gt;Silver is a metallic chemical element with the chemical symbol Ag (Latin: argentum) and atomic number 47. It is the second element of the copper group and is, therefore, a d-block element or as commonly referred to a transition element or &lt;a href=&quot;/questions/tagged/transition-metal&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;transition-metal&amp;#39;&quot; rel=&quot;tag&quot;&gt;transition-metal&lt;/a&gt;.  (compare &lt;a href=&quot;http://www.iupac.org/nc/home/publications/iupac-books/books-db/book-details.html?tx_wfqbe_pi1[bookid]=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Red Book &lt;strong&gt;2005&lt;/strong&gt;, IR-3.5&lt;/a&gt;)&lt;br&gt;&#xA;Unrecognized by the IUPAC is the still sometimes used name &quot;coinage metals&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\scriptstyle\Tiny&#xA;\begin{array}{l|cccccccccccccccccc}&#xA;\text{Gr.} &amp;amp;1 &amp;amp;2 &amp;amp;3 &amp;amp;4 &amp;amp;5 &amp;amp;6 &amp;amp;7 &amp;amp;8 &amp;amp;9 &amp;amp;10 &amp;amp;11 &amp;amp;12 &amp;amp;13 &amp;amp;14 &amp;amp;15 &amp;amp;16 &amp;amp;17 &amp;amp;18\\&#xA;\text{Per.}\\\hline&#xA;1&amp;amp; \ce{^1H}	&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^2He}\\&#xA;2&amp;amp; \ce{^3Li}&amp;amp; \ce{^4Be}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^5B}&amp;amp; \ce{^6C}&amp;amp; \ce{^7N}&amp;amp; \ce{^8O}&amp;amp; \ce{^9F}&amp;amp; \ce{^{10}Ne}\\&#xA;3&amp;amp; \ce{^{11}Na}&amp;amp; \ce{^{12}Mg}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^{13}Al}&amp;amp; \ce{^{14}Si}&amp;amp; \ce{^{15}P}&amp;amp; \ce{^{16}S}&amp;amp; \ce{^{17}Cl}&amp;amp; \ce{^{18}Ar}\\&#xA;4&amp;amp; \ce{^{19}K}&amp;amp; \ce{^{20}Ca}&amp;amp; \ce{^{21}Sc}&amp;amp; \ce{^{22}Ti}&amp;amp; \ce{^{23}V}&amp;amp; \ce{^{24}Cr}&amp;amp; \ce{^{25}Mn}&amp;amp; \ce{^{26}Fe}&amp;amp; \ce{^{27}Co}&amp;amp; \ce{^{28}Ni}&amp;amp; \color{\orange}{\ce{^{29}Cu}}&amp;amp; \ce{^{30}Zn}&amp;amp; \ce{^{31}Ga}&amp;amp; \ce{^{32}Ge}&amp;amp; \ce{^{33}As}&amp;amp; \ce{^{34}Se}&amp;amp; \ce{^{35}Br}&amp;amp; \ce{^{36}Kr}\\&#xA;5&amp;amp; \ce{^{37}Rb}&amp;amp; \ce{^{38}Sr}&amp;amp; \ce{^{39}Y}&amp;amp; \ce{^{40}Zr}&amp;amp; \ce{^{41}Nb}&amp;amp; \ce{^{42}Mo}&amp;amp; \ce{^{43}Tc}&amp;amp; \ce{^{44}Ru}&amp;amp; \ce{^{45}Rh}&amp;amp; \ce{^{46}Pd}&amp;amp; \color{\red}{\ce{^{47}Ag}}&amp;amp; \ce{^{48}Cd}&amp;amp; \ce{^{49}In}&amp;amp; \ce{^{50}Sn}&amp;amp; \ce{^{51}Sb}&amp;amp; \ce{^{52}Te}&amp;amp; \ce{^{53}I}&amp;amp; \ce{^{54}Xe}\\&#xA;6&amp;amp;\ce{^{55}Cs}&amp;amp; \ce{^{56}Ba}&amp;amp; \text{*}&amp;amp; \ce{^{72}Hf}&amp;amp; \ce{^{73}Ta}&amp;amp; \ce{^{74}W}&amp;amp; \ce{^{75}Re}&amp;amp; \ce{^{76}Os}&amp;amp; \ce{^{77}Ir}&amp;amp; \ce{^{78}Pt}&amp;amp; \color{\orange}{\ce{^{79}Au}}&amp;amp; \ce{^{80}Hg}&amp;amp; \ce{^{81}Tl}&amp;amp; \ce{^{82}Pb}&amp;amp; \ce{^{83}Bi}&amp;amp; \ce{^{84}Po}&amp;amp; \ce{^{85}At}&amp;amp; \ce{^{86}Rn}\\&#xA;7&amp;amp; \ce{^{87}Fr}&amp;amp; \ce{^{88}Ra}&amp;amp; \text{**}&amp;amp; \ce{^{104}Rf}&amp;amp; \ce{^{105}Db}&amp;amp; \ce{^{106}Sg}&amp;amp; \ce{^{107}Bh}&amp;amp; \ce{^{108}Hs}&amp;amp; \ce{^{109}Mt}&amp;amp; \ce{^{110}Ds}&amp;amp; \color{\orange}{\ce{^{111}Rg}}&amp;amp; \ce{^{112}Cn}&amp;amp; \ce{^{113}Uut}&amp;amp; \ce{^{114}Fl}&amp;amp; \ce{^{115}Uup}&amp;amp; \ce{^{116}Lv}&amp;amp; \ce{^{117}Uus}&amp;amp; \ce{^{118}Uuo}\\\hline&#xA;\end{array}\\&#xA;\scriptstyle\Tiny&#xA;\begin{array}{lc} &#xA;\text{* Lanthanoide}&amp;amp; \ce{^{57}La}&amp;amp; \ce{^{58}Ce}&amp;amp; \ce{^{59}Pr}&amp;amp; \ce{^{60}Nd}&amp;amp; \ce{^{61}Pm}&amp;amp; \ce{^{62}Sm}&amp;amp; \ce{^{63}Eu}&amp;amp; \ce{^{64}Gd}&amp;amp; \ce{^{65}Tb}&amp;amp; \ce{^{66}Dy}&amp;amp; \ce{^{67}Ho}&amp;amp; \ce{^{68}Er}&amp;amp; \ce{^{69}Tm}&amp;amp; \ce{^{70}Yb}&amp;amp; \ce{^{71}Lu}\\&#xA;\text{** Actinoide}&amp;amp; \ce{^{89}Ac}&amp;amp; \ce{^{90}Th}&amp;amp; \ce{^{91}Pa}&amp;amp; \ce{^{92}U}&amp;amp; \ce{^{93}Np}&amp;amp; \ce{^{94}Pu}&amp;amp; \ce{^{95}Am}&amp;amp; \ce{^{96}Cm}&amp;amp; \ce{^{97}Bk}&amp;amp; \ce{^{98}Cf}&amp;amp; \ce{^{99}Es}&amp;amp; \ce{^{100}Fm}&amp;amp; \ce{^{101}Md}&amp;amp; \ce{^{102}No}&amp;amp; \ce{^{103}Lr}	\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Silver&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.webelements.com/silver/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Webelements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2017-05-03T00:11:38.370" LastActivityDate="2017-05-03T00:11:38.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="210" PostTypeId="4" CreationDate="2012-05-02T05:16:02.627" Score="0" Body="Silver is a metallic chemical element with the chemical symbol Ag (Latin: argentum) and atomic number 47." OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:09:57.293" LastActivityDate="2012-05-05T14:09:57.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="211" PostTypeId="5" CreationDate="2012-05-02T05:18:56.190" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-02T05:18:56.190" LastActivityDate="2012-05-02T05:18:56.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="212" PostTypeId="4" CreationDate="2012-05-02T05:18:56.190" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-02T05:18:56.190" LastActivityDate="2012-05-02T05:18:56.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="213" PostTypeId="5" CreationDate="2012-05-02T05:19:39.390" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Recrystallization_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Recrystallization&lt;/a&gt; is a procedure for purifying compounds. By dissolving both impurities and a compound in an appropriate solvent, either the desired compound or impurities can be removed from the solution, leaving the other behind&lt;/p&gt;&#xA;" OwnerUserId="23561" LastEditorUserId="23561" LastEditDate="2018-10-06T18:49:45.167" LastActivityDate="2018-10-06T18:49:45.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="214" PostTypeId="4" CreationDate="2012-05-02T05:19:39.390" Score="0" Body="For questions related to dissolving and subsequently precipitating as substance. The purpose may be for purifying, creating polymorphs, or other reasons." OwnerUserId="124" LastEditorUserId="23561" LastEditDate="2018-10-06T06:44:12.970" LastActivityDate="2018-10-06T06:44:12.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="215" PostTypeId="1" AcceptedAnswerId="226" CreationDate="2012-05-02T06:51:04.140" Score="13" ViewCount="2773" Body="&lt;p&gt;I've always seen benzyne (benzene with a triple bond) classified as an &quot;intermediate&quot;. I really don't see why it needs to be an intermediate, though. The possible reasons I can come up with are flimsy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It has an unstable &lt;em&gt;sp&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;–&lt;em&gt;sp&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; $\pi$ bent bond: we have bent bonds with more angle strain in cyclopropane, which is not an intermediate. Though those bonds are $\sigma$ bonds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It immediately reacts with itself to form a dimer, biphenylene, in the absence of other reagents: there are many other molecules which spontaneously dimerize (eg $\ce{NO2}$ at certain temperatures), yet they are not classified as &quot;intermediates&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, what makes benzyne an &quot;intermediate&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally (but possibly too broad), what makes a molecule an &quot;intermediate&quot;?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="73" LastEditDate="2012-06-11T16:22:20.893" LastActivityDate="2016-01-08T15:07:39.930" Title="Why is benzyne an intermediate?" Tags="&lt;organic-chemistry&gt;&lt;bent-bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="216" PostTypeId="1" AcceptedAnswerId="219" CreationDate="2012-05-02T07:34:06.360" Score="27" ViewCount="2862" Body="&lt;p&gt;You can buy NMR tubes in a huge variety of qualities, with an equally huge difference in price between the cheapest and the most expensive NMR tubes. They are usually rated for a specific spectrometer frequency, e.g. 300 MHz+ or 600 MHz+.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between these different grades of NMR tubes? And how large is the effect on the quality of the spectra if a lower-grade tube is used?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7475" LastEditDate="2016-05-06T20:02:56.707" LastActivityDate="2017-10-03T11:20:41.680" Title="What benefits do higher-rated NMR tubes provide?" Tags="&lt;nmr-spectroscopy&gt;&lt;equipment&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="217" PostTypeId="1" AcceptedAnswerId="227" CreationDate="2012-05-02T08:41:17.453" Score="4" ViewCount="13359" Body="&lt;p&gt;In solid materials, there are two kinds of bonds &lt;strong&gt;between&lt;/strong&gt; polymers: primary and secondary. Primary bonds can be also called cross-linked bonds. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the difference between the two (strength, type, response to stress/strain, etc)? &lt;/li&gt;&#xA;&lt;li&gt;How do they compare to bonds inside a polymer?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="144" LastEditorUserId="15429" LastEditDate="2017-05-31T16:10:19.817" LastActivityDate="2017-05-31T16:10:19.817" Title="What is the difference between cross-linked bond and secondary bond in polymers?" Tags="&lt;polymers&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="218" PostTypeId="2" ParentId="215" CreationDate="2012-05-02T09:17:59.607" Score="9" Body="&lt;p&gt;Intermediate is a middle step in the chemical process: reactants -&gt; intermediate(s) -&gt; products. Benzyne is formed from benzene and is highly reactive. Thus, it is never the first or last step of the process. To my knowledge benzyne is rarely observed as a product, because its reacting fast with other molecules because its triple bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes a molecule an intermediate? It is a highly reactive &quot;middle product&quot; of a reaction.&#xA;See &lt;a href=&quot;http://academic.scranton.edu/faculty/cannm1/advancedorganic/advancedorganicmodule.html&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Aryne&quot;&gt;there&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="30" LastEditDate="2012-05-02T11:15:39.970" LastActivityDate="2012-05-02T11:15:39.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="219" PostTypeId="2" ParentId="216" CreationDate="2012-05-02T09:57:56.090" Score="26" Body="&lt;p&gt;A general rule is: crap goes in, crap comes out. A large-sample low-field 1D NMR at room temperature is usually only minimally affected by using a cheap NMR tube. There are important differences though and I’ll highlight a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first distinction between prices is what the tube is constructed from: quartz obviously costs more than borosilicate. Why would a chemist ever use the more expensive quartz? You can &lt;a href=&quot;http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion&quot;&gt;heat/cool quartz faster&lt;/a&gt; (nice for thermal studies), the UV cutoff is lower (think 190 nm opposed to 320 nm) which is important for photolysis, you can work with quartz at higher temperatures (around 1300 °C instead of 250 °C), and the purity of quartz is better controlled than your typical Pyrex. There are different grades of quartz, fused and synthetic, and there are different grades of borosilicate, such as the high-quality Pyrex or the lower-quality Class B, each comes with its own limitations as far as purities are concerned and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three more important parameters have to do with the manufacturing of your tube are: concentricity, camber and wall thickness. Lower quality tubes will tend to have less precision and accuracy over each of these parameters and as a result your sample may wobble while spinning (introducing problems such as modulation sidebands). A particularly bad tube can hit your RF coils and cause damage to your probe over time slowly or quickly if it is ignoring any reasonable standard – even more apparent for a tube at this level of “quality” is that it may be easier to break while acquiring your sample and we all should be aware of how much fun that is for everybody involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shimming can deal with impurities present in the glass (such as ferric oxide) and increased impurities in the glass/inhomogeneities will result in taking longer to get a good shim. Time is money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of these things have lower tolerances in more complex experiments and at higher fields. It really does depend on your particular experiment and what you’re hoping to get out of it. &lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T13:10:54.543" LastActivityDate="2015-05-20T13:10:54.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="220" PostTypeId="1" AcceptedAnswerId="223" CreationDate="2012-05-02T15:56:30.590" Score="14" ViewCount="71540" Body="&lt;p&gt;Given an organic compound, is there any way to decide if it is volatile (or compare volatility--everything is volatile in the end)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volatility is due to the tendency to evaporate. The conclusion I can draw from this is that a more volatile compound will be lighter and/or will have less intermolecular forces. But this doesn't seem to work--I recall that methanol is &lt;em&gt;less&lt;/em&gt; volatile than ethanol ([citation needed]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are there any general guidelines for this or is it another experimental property?&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2020-07-22T10:59:16.383" Title="General rules for deciding volatility" Tags="&lt;physical-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="221" PostTypeId="2" ParentId="220" CreationDate="2012-05-02T19:03:43.887" Score="3" Body="&lt;p&gt;In chemistry and physics, volatility is the tendency of a substance to vaporize. Volatility is directly related to a substance's vapor pressure. At a given temperature, a substance with higher vapor pressure vaporizes more readily than a substance with a lower vapor pressure.(Taken from Wikipedia)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;But this doesn't seem to work--I recall that methanol is less volatile than ethanol&lt;/code&gt;&lt;br&gt;&#xA;I think you have got this wrong. According to the wikipedia page of &lt;a href=&quot;http://en.wikipedia.org/wiki/Methanol&quot; rel=&quot;nofollow&quot;&gt;Methanol&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethanol&quot; rel=&quot;nofollow&quot;&gt;Ethanol&lt;/a&gt; the vapour pressure of Methanol &gt; Ethanol thus Methanol is more volatile than Ethanol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volatility can also be predicted on the type of intermolecular and intra molecular bonding and of course on the BP of the compound.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-02T19:11:15.297" LastActivityDate="2012-05-02T19:11:15.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="222" PostTypeId="1" AcceptedAnswerId="224" CreationDate="2012-05-02T21:13:50.033" Score="14" ViewCount="3876" Body="&lt;p&gt;The main chemistry prep room at my institution (and also at the ambulance station) is labelled with the standard green compressed gas sign.  However, there is a small '2' included within the diamond.  None of the chemistry technicians, tutors or ambulance staff seem to know what it is referring to and I can't find any reference of it on the internet.  Would anyone here happen to know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9q3n8.jpg&quot; alt=&quot;Example of Compressed Gas (2) Hazard Sign&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The related products of this &lt;a href=&quot;http://www.directa.co.uk/site/scripts/product_browse.php?product_id=8174&amp;amp;category_id=481&quot; rel=&quot;noreferrer&quot;&gt;online provider&lt;/a&gt; show that other hazards have different numbers but some also have two (e.g. flammable).  Is there a standard that is being referred to here?&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2012-06-11T16:25:22.183" Title="What is the meaning of the '2' on some Compressed Gas Hazard signs in the lab?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="223" PostTypeId="2" ParentId="220" CreationDate="2012-05-02T21:45:52.577" Score="10" Body="&lt;p&gt;This answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/180/why-do-different-substances-have-different-boiling-points&quot;&gt;this question&lt;/a&gt; applies to your query. Many physical properties depend on the intermolecular forces experienced in the liquid or solid. These intermolecular forces are hydrogen bonding (strongest), dipole-dipole attractions (in polar molecules), and London dispersion forces (weakest of the forces when comparing molecules of the same size, but dependent on the number of electrons in the substance). Stronger intermolecular forces would make the substance less volatile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the first response, methanol is more volatile than ethanol. Methanol and ethanol would both have both hydrogen bonding (a relatively strong type of dipole-dipole attraction) and London dispersion forces. Because ethanol has more electrons (because it is a bigger molecule, but not necessarily because it is a heavier molecule) it would have more London dispersion forces in comparison to methanol, so, with stronger intermolecular forces, it would be less volatile. So, this example follows the general guidelines--figure out the intermolecular forces present in the molecules you want to compare and the molecule with the weakest cumulative intermolecular forces will be the most volatile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intramolecular bonding (covalent bonds, when you're talking about molecules) does not affect physical properties. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-05-02T21:45:52.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="224" PostTypeId="2" ParentId="222" CreationDate="2012-05-02T21:58:20.670" Score="14" Body="&lt;p&gt;The 2 refers to DOT (Department of Transportation) hazard class 2. &lt;a href=&quot;http://www.drs.illinois.edu/transportationandshipping/definitions.html&quot; rel=&quot;noreferrer&quot;&gt;This reference&lt;/a&gt; describes the different DOT hazard classes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As summarized in the comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Class 2 (compressed gas) is divided into three subclasses (2.1 - flammable, 2.2 - non-flammable, non-toxic and 2.3 - toxic). I imagine the diamonds for Class 2 don't contain the subclass number because there is a different diamond colour/symbology for each subclass (as opposed to for instance 5.1 (oxidising agent) and 5.2 (organic peroxide), which have the same symbol and colours).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="133" LastEditorUserId="73" LastEditDate="2012-06-11T16:25:22.183" LastActivityDate="2012-06-11T16:25:22.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="225" PostTypeId="2" ParentId="152" CreationDate="2012-05-03T01:40:53.457" Score="12" Body="&lt;p&gt;I think the simplest way to explain what is going on is by way of a classical analogy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture an elastically bound ball that whizzes back-and-forth along the $x$ axis. Next, whack the moving ball hard with a $y$-aligned impulse. The result is an elliptical or figure-8 path, one for which the maximum radius of motion along the $x$ axis will, in a perfect world, remain unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The single-axis ball is the classical analog to an $s$ orbital, and the elliptical or figure-8 ball corresponds to a $p$ orbital. Notice that for a given maximum radius, the $p$ orbital will include &lt;em&gt;two&lt;/em&gt; orthogonal oscillation components, $x$ and $y$, and so will always have more energy than an $s$ (or $x$-only) oscillation with the same maximum radius. That also means that you can construct an $s$ orbital that extends beyond the maximum range of the $p$, yet will still have less energy than the $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or even more succinctly, think of a rocket with a fixed energy budget. The $s$ orbital is using all of its energy to achieve the maximum possible &quot;launch altitude&quot; from the nucleus. It can get away with that because in quantum land a light object can &lt;a href=&quot;https://chemistry.stackexchange.com/questions/142/how-do-orbitals-coexist-with-a-nucleus/186#186&quot;&gt;head straight back towards the nucleus without ever hitting it&lt;/a&gt;. In contrast, the $p$ orbital diverts some of its available energy into achieving a more conventional concept of an orbit, which makes its maximum radius lower than that of the do-or-die $s$ electron. I suspect that the textbook was trying to convey this &quot;short cut&quot; aspect of the $s$ orbital, rather than a special attraction &lt;em&gt;per se&lt;/em&gt; near the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. - &quot;Conventional orbit&quot; is a loaded phrase in this context. The peculiarities of the quantum world require orbital solutions that look like figure eights and other loopy forms, rather than conventional elliptical orbits. These loopy forms are all spherically distorted standing-wave solutions, which can in turn be thought of as akin to distorted multiple loops on a skip rope. These skip ropes rotate in the complex plane, though, not in ordinary space.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2012-05-11T14:49:36.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="226" PostTypeId="2" ParentId="215" CreationDate="2012-05-03T15:26:34.893" Score="8" Body="&lt;p&gt;As Juha says, &lt;em&gt;o&lt;/em&gt;-benzyne (to distinguish from e.g. &lt;em&gt;p&lt;/em&gt;-benzyne which is generated by the Bergman reaction) is an intermediate, in the sense of &lt;em&gt;o&lt;/em&gt;-benzyne itself not being the product of interest; rather it is subsequently reacted with a trapping agent (e.g. a diene for Diels-Alder reactions) within the reaction mixture, and it is the adduct that is isolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there have been isolation experiments conducted, where &lt;em&gt;o&lt;/em&gt;-benzyne is trapped in such a way that it is barred from reacting; for instance, &lt;a href=&quot;http://dx.doi.org/10.1021/ja00799a060&quot;&gt;Chapman and coworkers&lt;/a&gt; isolated benzyne in an argon gas matrix at 8 K for subsequent spectroscopic study. More recently, &lt;a href=&quot;http://dx.doi.org/10.1002/anie.199713471&quot;&gt;Warmuth&lt;/a&gt; studied benzyne by trapping it within a so-called &lt;em&gt;hemicarcerand&lt;/em&gt;; the benzyne complex, kept at 173 K, is stable enough that NMR experiments can be performed on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://dx.doi.org/10.1071/CH10179&quot;&gt;this survey article by Wentrup&lt;/a&gt; for more references.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastActivityDate="2012-05-03T15:26:34.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="227" PostTypeId="2" ParentId="217" CreationDate="2012-05-03T22:48:57.730" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.uwosh.edu/faculty_staff/mihalick/materials/Chapter5.pdf&quot;&gt;This reference&lt;/a&gt; indicates that primary bonds in a polymer are the intramolecular bonds (the covalent bonds) that hold the polymer together. Cross-linking different chains of polymer together does occur through a covalent bond, such as a disulfide bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary bonds that help to give the polymer its physical properties are intermolecular forces, such as hydrogen bonding for polymers that contain hydroxyl or primary or secondary amine groups. These secondary bonds can be disrupted by changing solvent conditions or heating, for example, which wouldn't disrupt the primary bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I suppose that you could say, loosely, that there are &quot;two kinds of bonds &lt;strong&gt;between&lt;/strong&gt; polymers in solid material&quot; if you mean between the polymer chains before they have an opportunity to cross-link (by a change in pH, for example). After they cross-link, the physical properties of the bulk sample of the polymer will still be affected by secondary bonds, the intermolecular interactions, &lt;strong&gt;between&lt;/strong&gt; the cross-linked chains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary bonds will be stronger and can only be affected by chemical changes, the secondary bonds (intermolecular forces) can be affected by changing the physical conditions. &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-03T22:48:57.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="228" PostTypeId="1" AcceptedAnswerId="229" CreationDate="2012-05-04T00:13:18.073" Score="50" ViewCount="32565" Body="&lt;p&gt;Diborane has the interesting property of having two 3-centered bonds that are each held together by only 2 electrons (see the diagram below, from &lt;a href=&quot;http://en.wikipedia.org/wiki/Diborane&quot;&gt;Wikipedia&lt;/a&gt;).  These are known as &quot;banana bonds.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming there is some sort of bond hybridization transpiring, but the geometry doesn't seem like it is similar to anything I'm familiar with Carbon doing.  What sort of hybridization is it, and why don't we see many (any?) other molecules with this bond structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9L9j2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2012-05-04T07:06:35.503" LastActivityDate="2017-05-11T16:49:49.517" Title="What makes banana bonds possible in diborane?" Tags="&lt;hybridization&gt;&lt;bond&gt;&lt;bent-bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="229" PostTypeId="2" ParentId="228" CreationDate="2012-05-04T01:10:20.807" Score="36" Body="&lt;p&gt;Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Diborane-2D.png/320px-Diborane-2D.png&quot; alt=&quot;enter link description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing  $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane/229#comment250_229&quot;&gt;It seems that there are a lot more compounds with 3c2e geometry&lt;/a&gt;. I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there &lt;em&gt;are&lt;/em&gt; Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So the basic reason for this happening is due to an incomplete octet wanting to fill itself.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &quot;banana&quot; is not necessarily only for &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_two-electron_bond&quot; rel=&quot;noreferrer&quot;&gt;3c2e&lt;/a&gt; bonds. Any &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent_bond&quot; rel=&quot;noreferrer&quot;&gt;bent bond&lt;/a&gt; is called a &quot;banana&quot; bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sneaks off and checks Wikipedia.&lt;/em&gt; Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2012-05-04T11:30:52.697" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="230" PostTypeId="1" AcceptedAnswerId="234" CreationDate="2012-05-04T09:59:35.740" Score="19" ViewCount="27525" Body="&lt;p&gt;If you characterize the chemical bonds to two categories physical and chemical bonds, how do you do it? Aren't all bonds &lt;strong&gt;&lt;em&gt;chemical and physical&lt;/em&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the freedictionary.com, chemical bond:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any of several forces, especially the ionic bond, covalent bond, and metallic bond, by which atoms or ions are bound in a molecule or crystal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the physical bond then? How should the atom be bonded to molecule if these bonds are not the ones. What are these &quot;several forces&quot; excactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the answer I would like to see the lists of chemical and physical bonds.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="656" LastEditDate="2012-09-28T01:06:24.543" LastActivityDate="2019-01-30T18:43:13.900" Title="What is the difference between physical and chemical bonds?" Tags="&lt;bond&gt;&lt;atoms&gt;&lt;molecules&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="231" PostTypeId="2" ParentId="230" CreationDate="2012-05-04T10:32:31.033" Score="2" Body="&lt;p&gt;All bonds are physical, of course. But not all bonds are chemical – as a somewhat stupid example consider a magnet attached to your fridge, which is in a way a physical bond but doesn't have much to do with chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition you quoted is fine, and leaves quite clearly those kinds of bonds that are not included as &quot;&lt;em&gt;the&lt;/em&gt; physical bonds&quot;. For instance, Van der Waals forces that do not bind molecules into a rigid crystal structure, particularly London dispersion forces, are usually not considered chemical bonds, neither are the forces that hold together atomic nuclei or neutron stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, none of these bind molecules together; if you restrict it to that you're right that every bond is physical and chemical at the same time.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2012-05-04T10:32:31.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="233" PostTypeId="2" ParentId="26" CreationDate="2012-05-04T10:44:52.827" Score="6" Body="&lt;p&gt;Here's two links (&lt;a href=&quot;http://www.ipen.br/biblioteca/2004/10220.pdf&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/la061240g&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;) that show the simple scheme on forming monolayers. The essential figure is also shown here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dia31.png&quot; alt=&quot;Here is the essential figure from the latter paper if you don&amp;#39;t have access&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question... I can only point out &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/poc.372/pdf&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; that describes which properties change, if you make the molecules more bulky or add chloride, etc. I would guess that APTES has well known properties and they are &quot;neutral&quot; regarding common applications. &lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="30" LastEditDate="2012-05-07T06:56:41.420" LastActivityDate="2012-05-07T06:56:41.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="234" PostTypeId="2" ParentId="230" CreationDate="2012-05-04T11:23:11.930" Score="21" Body="&lt;p&gt;In short: the definition of a chemical bond is not unique and a clearly-drawn line. The simplest and most common definition is &lt;strong&gt;the sharing of electrons between two or more nuclei&lt;/strong&gt;. In contrast, other interactions are often said to be &lt;em&gt;intermolecular&lt;/em&gt; (which is somewhat more specific than the term “physical”.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a longer commentary, I see can have five different types of definition of the chemical bond (vs. intermolecular interactions).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's start from the beginning, in this case using the words of Linus Pauling, winner of the 1954 Nobel Prize for “determining the nature of the chemical bond linking atoms into molecules”. In &lt;a href=&quot;http://books.google.com/books?id=L-1K9HmKmUUC&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;The Nature of the Chemical Bond&lt;/em&gt;&lt;/a&gt; (1960), he gives the following definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The Chemical Bond Defined&lt;/strong&gt;.—We shall say that &lt;em&gt;there is a chemical bond between two atoms or groups of atoms in case that the forces acting between them are such as to lead to the formation of an aggregate with sufficient stability to make it convenient for the chemist to consider it as an independent molecular species.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A bond is what links atoms into molecules&lt;/strong&gt;, and molecules are defined at the discretion of the chemist. You can find the same definition still in use in some high-school textbooks, but it isn't very helpful…&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The complete opposite: &lt;strong&gt;consider all interactions as chemical bonds&lt;/strong&gt;, whose strenght can vary. I actually hadn't heard that one before I researched textbooks to write this answer, but you can find it in some textbooks, like &lt;a href=&quot;http://books.google.com/books?id=bLgkS8pE1N0C&amp;amp;pg=PT175&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chemical bond is the physical phenomenon of chemical species being held together by attraction of atoms to each other through sharing, as well as exchanging, of electrons and is a phenomenon that is fully described by the laws of quantum electrodynamics. In general, strong chemical bonds are found in molecules, crystals or in solid metal and they organize the atoms in ordered structures. Weak chemical bonds are classically explained to be effects of polarity between molecules which contain strong polar bonds. Some very weak bond-like interactions also result from induced polarity London forces between the electron clouds of atoms, or molecules. Such forces allow the liquification and solidification of inert gases. At the very lowest strengths of such interactions, there is no good operational definition of what constitutes a proper definitional &quot;bond&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This view has some grounding, because all interatomic interactions stem from the behaviour of the system's electrons (in addition to nuclei–nuclei Coulombic forces). However, it does not allow to make a strong distinction between interactions whose energies differ by orders of magnitude. Chemists like molecules, and they like categorizing things between intra- and inter-molecular, as it's a nice model (making it easier for our mind to handle).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can classify interactions by energy: decide that &lt;strong&gt;chemical bonds are those interactions that have an energy higher than a certain threshold&lt;/strong&gt;, let's say 50 kJ/mol. This makes things clean, and makes sure that you can easily classify interactions. However, the choice of a threshold is problematic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, what I believe is the most common description is to look at &lt;strong&gt;the nature of the interaction&lt;/strong&gt;, and classify it following a certain &lt;strong&gt;convention&lt;/strong&gt;. The two other answers so far have focused on this part and listed the various “usual” types of bonds and intermolecular interactions, so I won't say more on that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I said five types, right? Well, the fifth is mine, of course. Not only mine, but that of the &lt;em&gt;New Oxford American Dictionary&lt;/em&gt; as well, which I quite like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;chemical bond&lt;/strong&gt;&lt;br&gt; a strong force of attraction holding atoms together in a molecule or crystal, resulting from the sharing or transfer of electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Short and powerful. What I like in that is that it gives a general prescription, allowing one to argue individual cases and not based too much on convention. What are the features of a chemical bond? Well, it has to be an attractive force between atoms, sure… but I think the most relevant criterion of all is &lt;strong&gt;sharing (or transfer) of electrons&lt;/strong&gt;. That is, after all, what chemistry is about: description of electronic clouds around two or more atoms. And I think when this criterion is applied to the list of interaction types commonly classified, it works quite well (whithout being dogmatic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what I like in it is that a given interaction type can be considered one way or another depending on its strength. The best example for that may be the hydrogen bond, which is the archetype of the almost-chemical-bond…&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastEditorUserId="41328" LastEditDate="2019-01-30T18:43:13.900" LastActivityDate="2019-01-30T18:43:13.900" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="235" PostTypeId="1" AcceptedAnswerId="237" CreationDate="2012-05-04T12:08:59.357" Score="13" ViewCount="219" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neil_Bartlett_(chemist)&quot; rel=&quot;nofollow&quot;&gt;Neil Bartlett&lt;/a&gt; (1932–2008) first synthesized $\ce{XeF4}$ (and $\ce{XeF6}$) in 1962.  In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Xenon_tetrafluoride#Synthesis&quot; rel=&quot;nofollow&quot;&gt;synthesis&lt;/a&gt;, a nickel chamber is used, and heated to 400&amp;deg;C, causing the formation of $\ce{NiF4}$, which Wikipedia claims is not part of the reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the nature of the noble gas, xenon, one might imagine that the fluorines would need to  could be &quot;forced&quot; onto the $\ce{Xe}$ atom at the same time, otherwise with each addition, the molecule would be even less energetically likely.  I was trying to ascertain in what geometric form the four fluorine atoms would configure themselves to make this addition possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had first assumed that the nickel was being used as a catalyst, and the fluorines could be plucked from its surface accordingly, but this does not jive with the description of the synthesis above.  According to the response below, one synthesis of the molecule does rely on forming fluoride radicals on a catalyst, but it's mysterious that the synthesis above does not mention this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there is no catalyst, how are the fluorines at 400&amp;deg;C oriented to allow the addition of 2 molecules at presumably the same time?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F - F&#xA;.   .&#xA;.   .&#xA;F - F  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(and is that even possible?)&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="24052" LastEditDate="2016-03-11T23:51:43.723" LastActivityDate="2016-03-11T23:51:43.723" Title="What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride?" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="236" PostTypeId="1" AcceptedAnswerId="339" CreationDate="2012-05-04T12:23:10.227" Score="21" ViewCount="894" Body="&lt;p&gt;I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphine&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{PH3}$&lt;/a&gt; has &lt;em&gt;nearly&lt;/em&gt; pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a &lt;em&gt;nearly&lt;/em&gt; pure $s$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While answering &lt;a href=&quot;https://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane&quot;&gt;this question&lt;/a&gt;, I realised that it's not that easy predicting the numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the same image of $\ce{B2H6}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/63/Diborane-2D.png/320px-Diborane-2D.png&quot; alt=&quot;enter link description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, seeing the 97&amp;deg;, I thought &quot;well, the inner $\ce{B-H}$ bonds will be almost pure $p$&quot;. But, with that, I couldn't figure out where the 120&amp;deg; came from, because that is for perfect $sp^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then realised that I was being stupid and it wasn't that simple--just because you have a 120&amp;deg; doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one generally go about predicting such &quot;nonuniform hybridizations&quot; if one knows the bond angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2018-09-03T19:02:31.187" Title="Non-integer hybridization" Tags="&lt;hybridization&gt;&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="237" PostTypeId="2" ParentId="235" CreationDate="2012-05-04T12:37:03.660" Score="7" Body="&lt;p&gt;The answer appears to be right there in the linked synthesis details - the nickel is non-catalytic and is only there to generate a passivated surface that won't degrade in the high-temperature fluorine atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The plot thickens!&lt;/strong&gt; &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0021951770902010&quot;&gt;this paper&lt;/a&gt; suggests that xenon may in fact abstract fluorine off the passivated nickel fluoride surface due to observed zeroth order kinetics, which lends some serious credence to jonsca's original inference of Ni being catalytic. This will take some more investigation...&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-04T14:43:04.470" LastActivityDate="2012-05-04T14:43:04.470" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="238" PostTypeId="1" AcceptedAnswerId="239" CreationDate="2012-05-04T18:31:33.973" Score="23" ViewCount="1610" Body="&lt;p&gt;I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll explain via an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to these questions:&#xA;How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they &lt;em&gt;themselves&lt;/em&gt; wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd like a clear explanation of how/why the symmetry breaks.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="30" LastEditDate="2012-05-04T20:48:13.663" LastActivityDate="2012-08-28T02:15:57.997" Title="Symmetry lost in orbitals?" Tags="&lt;orbitals&gt;&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="239" PostTypeId="2" ParentId="238" CreationDate="2012-05-04T21:42:55.920" Score="7" Body="&lt;p&gt;Symmetry is not actually broken that badly. If you take any one symmetry axis (or plane or...), it exists in both cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets do a thought experiment: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the spherical case you have only one electron. Your only symmetry axis goes through the electron and the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;In the case of three electrons, the symmetry axis goes through one of the electrons and the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;Doing the above collapses the system in to one possible configuration from the many possible ones (you're measuring quantum system, Heisenberg, etc.). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, &lt;strong&gt;before&lt;/strong&gt; you measure the location of one of the electrons, you don't know the symmetry axis and thus the two cases are equally symmetric. The probability of finding the electron number one at the &quot;north pole&quot; of the atom is equally probable in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that defining the symmetry axis is purely theoretical (mathematical) construction. The symmetry axis exists even if you don't do the measurement. See AcidFlask's comment below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'd like to highlight, that measuring electron states is &lt;strong&gt;becoming&lt;/strong&gt; possible: &lt;a href=&quot;http://phys.org/news177582885.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news177582885.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditDate="2012-05-11T13:41:51.183" LastActivityDate="2012-05-11T13:41:51.183" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="240" PostTypeId="1" AcceptedAnswerId="269" CreationDate="2012-05-04T22:11:46.910" Score="18" ViewCount="9089" Body="&lt;p&gt;Let's say I synthesized or isolated a chiral molecule and want to know the absolute configuration (R or S) of that molecule. I could obviously solve the structure by X-Ray crystallography, but that's a lot of work and I might not want to do that or might not have access to the necessary equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other ways exist to determine the absolute configuration of a molecule experimentally?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="42741" LastEditDate="2017-05-31T16:26:12.040" LastActivityDate="2017-05-31T16:26:12.040" Title="How do I determine the absolute configuration experimentally?" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="241" PostTypeId="2" ParentId="240" CreationDate="2012-05-05T02:30:37.103" Score="13" Body="&lt;p&gt;If your compound has been synthesized before, you can measure its optical activity and compare it with the data in the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your compound hasn't been synthesized before, it &lt;em&gt;can&lt;/em&gt; be possible to differentiate between two enantiomers with NMR by measuring the spectra of your molecule in a chiral environment. Here's a &lt;a href=&quot;http://journal.library.iisc.ernet.in/vol201001/Burkhard.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;review article&lt;/a&gt; on how to resolve enantiomers with NMR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, diastereomers &lt;em&gt;can&lt;/em&gt; be differentiated easily. By looking at your molecule, you can (sort of) predict which spectrum it should give rise to by using the concepts of &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-08-symmetry.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;enantiotopic, homotopic, heterotopic and diastereotopic&lt;/a&gt; protons. We had to analyse and solve 2D-NMR spectra (&lt;sup&gt;1&lt;/sup&gt;H and &lt;sup&gt;13&lt;/sup&gt;C) of diastereomers last year, and to a beginner, I have to say it was a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, it boils down to forming diastereomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at these diasteromers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gV9VGm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gV9VGm.jpg&quot; alt=&quot;Without getting into much detail&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've got finals coming up), it's easy to see that the hydroxyl will shift the ethyl's CH&lt;sub&gt;2&lt;/sub&gt; peaks towards lower field (higher ppm) in the second isomer. The interpretation of chiral media NMR spectra would have to be similar. My impression is that the &lt;em&gt;acquisition&lt;/em&gt; of chiral media spectra is more involved, but the interpretation isn't. I might be totally off track, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone might want to share a more thorough answer.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="15429" LastEditDate="2017-05-31T16:20:40.213" LastActivityDate="2017-05-31T16:20:40.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="242" PostTypeId="1" AcceptedAnswerId="4555" CreationDate="2012-05-05T04:30:00.273" Score="7" ViewCount="1413" Body="&lt;p&gt;I remember seeing a periodic table that had the top-left corner of the d-block shaded and marked as &quot;brittle&quot;. If I recall correctly, the elements were $\ce{Sc,Ti,V,Cr,Mn,Y,Zr,Nb,La}$. I think (sorry, I can't find much on it on the net).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triangular shaded area is a clear indicator of a groupwise and period wise trend (similar to how the non-metals form a triangle due to the electron affinity/Ionization potential trends).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I ask, what are the trends leading to the brittleness of a metal, and, more importantly, &lt;em&gt;why&lt;/em&gt; are these trends there (usually trends can be explained in terms of simpler things like atomic size/shielding/etc)?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="10185" LastEditDate="2015-09-18T18:05:57.157" LastActivityDate="2015-09-18T18:05:57.157" Title="Brittle d-block metal trend" Tags="&lt;transition-metals&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="243" PostTypeId="1" AcceptedAnswerId="245" CreationDate="2012-05-05T05:22:12.237" Score="12" ViewCount="6934" Body="&lt;p&gt;During nitration of phenol, both &lt;em&gt;para&lt;/em&gt;- and &lt;em&gt;ortho&lt;/em&gt;-nitrophenols will be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R5OXE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way in which we can synthesize &lt;em&gt;para&lt;/em&gt;-nitrophenol  only?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="10185" LastEditDate="2016-05-25T09:33:09.070" LastActivityDate="2016-12-06T20:10:04.617" Title="How can we synthesize para-nitrophenol" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;&lt;phenols&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="244" PostTypeId="2" ParentId="243" CreationDate="2012-05-05T07:53:54.330" Score="10" Body="&lt;p&gt;This can be done by first reating the phenol with $\ce{HNO2}$ Which gives us this  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3YaTt.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then reacting with dil $\ce{HNO3}$ we get 90% yield of para nitro phenol  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because $\ce{N=O}$ has a very strong $- I$  effect thus it concentrates the electrons more towards the para position so on reacting with dil $\ce{HNO3}$ we get max yield of para  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S - Another way of getting &lt;strong&gt;only&lt;/strong&gt; para is using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Protecting_group#Alcohol_protecting_groups&quot; rel=&quot;nofollow noreferrer&quot;&gt;protecting agent&lt;/a&gt; but  i cant recall the reaction now. I will post it later &lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T09:04:05.943" LastActivityDate="2012-05-05T09:04:05.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="245" PostTypeId="2" ParentId="243" CreationDate="2012-05-05T08:57:45.400" Score="10" Body="&lt;p&gt;You can always fractionally distillate the two. In this case, since the boiling points have a large difference, simple distillation works as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not too good with &quot;collateral damage&quot; of reactions, so if a reaction affects an extra group that it shouldn't, please let me know)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Claisen_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claisen rearrangement&lt;/a&gt; to protect the ortho position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll see if I can get a chemdrawing software later)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take phenol, add $\ce{Na+ CH2=CH-CH2-}$ (you may use lithium instead of sodium--easily prepared from $\ce{CH2=CH-CH2-X}$). You will get $\ce{Ph-O-CH2-CH=CH2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat it, the group migrates to ortho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/1b/Aromatic_Claisen_Rearrangement_Scheme.png/400px-Aromatic_Claisen_Rearrangement_Scheme.png&quot; alt=&quot;enter link description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat, and now you have a phenol with two $\ce{CH2=CH-CH2\bond{-}}$ groups at each ortho position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now nitrate. The nitro group will go to para, since $\ce{R\bond{-}}$ groups are &lt;em&gt;weakly&lt;/em&gt; activating/orthopara directing, whereas $\ce{-OH}$ is strongly activating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, vigorously oxidize with acidic $\ce{KMnO4}$. The $\ce{CH2=CH-CH2\bond{-}}$ groups become $\ce{-COOH}$ groups, and the other stuff stays intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, decarboxylate it. I'm not exactly sure of it, but we can do this via:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Treating with $\ce{NaOH}$ or soda lime.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(or) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jf00029a001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Treating with Vit. C&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(few more from Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Barton_decarboxylation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barton decarboxylation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Decarboxylation&quot; rel=&quot;nofollow noreferrer&quot;&gt;(check the stuff here)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The acidic and basic media may have some adverse effects on the compound, but I doubt it.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.253" LastActivityDate="2012-05-05T10:23:05.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="246" PostTypeId="1" AcceptedAnswerId="247" CreationDate="2012-05-05T09:14:34.527" Score="2" ViewCount="431" Body="&lt;p&gt;I would like to know about a Chemdrawing sotware for 2-d molecular structures. it should be mainly linux based (but if it supports other OS's then that would be great)&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="22" LastEditDate="2012-05-23T05:08:44.603" LastActivityDate="2017-03-19T18:00:23.090" Title="A Chemdrawing software?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ClosedDate="2017-03-19T18:10:55.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="247" PostTypeId="2" ParentId="246" CreationDate="2012-05-05T10:24:11.147" Score="4" Body="&lt;p&gt;I'd really reccomend the free &lt;strong&gt;&lt;a href=&quot;http://www.chemaxon.com/products/marvin/marvinsketch/&quot; rel=&quot;nofollow&quot;&gt;Marvin suite&lt;/a&gt;&lt;/strong&gt; from ChemAxon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MarvinSketch (the actual drawing package) hasn't disappointed me in the past and has a wide feature set.  In addition to just being able to draw compounds in Lewis format, the software includes plugins to name what you have drawn (systematic/traditional), predict properties, change atom and bond properties, generate stereoisomers and much more that I haven't got the experience to do anything more than experiment with!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view what you have drawn in Lewis form and also as a 3D structure should you wish to.  It supports many chemical drawing filetypes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemaxon.com/products/marvin/marvinsketch/&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is the product page, including brochure and technical details.  It is Java based so works on Mac, Windows, Linux and is even available &lt;a href=&quot;http://www.webqc.org/moleculareditor2.php&quot; rel=&quot;nofollow&quot;&gt;on the web&lt;/a&gt; (but is quite slow to generate properties etc. - I'd reccomend doing this locally).&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2012-05-05T10:24:11.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="248" PostTypeId="2" ParentId="246" CreationDate="2012-05-05T10:29:41.260" Score="2" Body="&lt;p&gt;Go to this &lt;a href=&quot;http://en.bio-soft.net/chemdraw.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;, it has links to many chemical drawing softwares&#xA; but I will recommend &lt;a href=&quot;http://www.cambridgesoft.com/software/ChemDraw/&quot; rel=&quot;nofollow&quot;&gt;Chemdraw&lt;/a&gt; and &lt;a href=&quot;http://www.acdlabs.com/resources/freeware/&quot; rel=&quot;nofollow&quot;&gt;Chemsketch&lt;/a&gt;. Chemdraw is a commercial version but chemsketch is a freeware and it works on both windows and  linux operating system. I am currently using Chemsketch and it is very good.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="30" LastEditDate="2012-05-23T08:22:34.420" LastActivityDate="2012-05-23T08:22:34.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="249" PostTypeId="1" AcceptedAnswerId="371" CreationDate="2012-05-05T11:02:55.277" Score="18" ViewCount="2507" Body="&lt;p&gt;Besides elucidating or verifying a chemical structure, NMR can also be used e.g. for quantifying a mixture of different chemicals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the quality of the spectrum and the specific substance, integrating the NMR signals for the same molecule can result in significant variations from the actual ratio of nuclei, which is known from the chemical structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factors do affect the quantification of NMR signals in a proton 1D spectrum? What precautions should one take during acquisition and processing of NMR spectra to make sure that the spectra can be analyzed quantitatively?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-05-05T13:40:15.647" LastActivityDate="2015-07-07T09:46:55.017" Title="What factors are important for quantitative analysis of a proton 1D-NMR spectrum?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="250" PostTypeId="1" AcceptedAnswerId="251" CreationDate="2012-05-05T11:47:25.267" Score="26" ViewCount="3856" Body="&lt;p&gt;I once had an Orgo TA refer to a diazo compound as &quot;diazo-boom-boom&quot; (the technical term).  I have always been curious as to the reason behind the instability and reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Diazo&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some of the most stable diazo compounds are α-diazoketones and α-diazoesters since the negative charge is delocalized into the carbonyl.  In contrast, most alkyldiazo compounds are explosive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is it about the alkyldiazos that makes them so much more unstable?  There doesn't appear to be any bond strain or other factors.  Naively, I can say that having a resonance structure should make it marginally more stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Dc3w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="7475" LastEditDate="2015-10-02T21:12:20.013" LastActivityDate="2015-10-02T21:12:20.013" Title="What makes diazo compounds so unstable and explosive?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;explosives&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="251" PostTypeId="2" ParentId="250" CreationDate="2012-05-05T13:18:51.557" Score="17" Body="&lt;p&gt;Well, your question is equivalent to &lt;em&gt;“what is it about α-diazoketones that makes them so much more stable?”&lt;/em&gt;, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/op5nu.png&quot; alt=&quot;resonance structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N&lt;sub&gt;2&lt;/sub&gt;, which is a very stable compound… the reaction is thermodynamically very favourable.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-05-05T13:28:54.083" LastActivityDate="2012-05-05T13:28:54.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="252" PostTypeId="1" AcceptedAnswerId="253" CreationDate="2012-05-05T13:56:09.307" Score="9" ViewCount="2409" Body="&lt;p&gt;It seems that a rinsing agent helps the dishes in the dishwasher dry much better than they do without it (which I recently noted on my new dishwasher). What molecules are used in the composition of these rinsing agents? How do they help the plates dry?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="7475" LastEditDate="2015-10-30T17:33:46.583" LastActivityDate="2018-11-08T21:22:47.973" Title="What's the chemical nature of a dishwasher's rinsing agent?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="253" PostTypeId="2" ParentId="252" CreationDate="2012-05-05T14:09:53.073" Score="10" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.rbnainfo.com/MSDS/US/Finish-Dishwasher-Cleaner.pdf&quot; rel=&quot;nofollow&quot;&gt;MSDS&lt;/a&gt; for Jet Dry (a common brand name) it contains:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sodium 2-phenylpropane-2-sulfonate (Sodium cumene sulfonate) &lt;/li&gt;&#xA;&lt;li&gt;Propane-1,2-diol &lt;/li&gt;&#xA;&lt;li&gt;Citric acid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think it may be safe to assume that the citric acid is in there for a &quot;clean&quot; smell and also to soften the water (thanks to the commenter).  The sulfonate is presumably a surfactant to help break the surface tension of the water, and promote evaporation, and the propylene glycol may help serve as a coolant (and is favored due to low toxicity).&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2012-08-10T05:33:48.773" LastActivityDate="2012-08-10T05:33:48.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="254" PostTypeId="5" CreationDate="2012-05-05T14:18:42.790" Score="0" Body="&lt;p&gt;A chemical bond is an interaction between atoms, which results in the release of energy when created. The same can be told about intermolecular interactions, but they are usually weaker and, perhaps more importantly, their covalent component is low to none. Ionic component is complementing to covalent, and can be attributed even to symmetric diatomic molecules, but without at least minuscule covalent character, the bond would be rather intermolecular interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is one of the most fundamental concepts of chemistry, along with chemical reactions, which result in bonds being broken and/or formed.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2020-04-03T14:56:05.833" LastActivityDate="2020-04-03T14:56:05.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="255" PostTypeId="4" CreationDate="2012-05-05T14:18:42.790" Score="0" Body="For questions about chemical bonds (ionic/covalent/etc.), their properties (length, energy, angle, etc.) or their relation to other molecular properties (intermolecular forces/dipole moment/etc.). Do not confuse this tag with [lewis-structures], [dipole], [intermolecular-forces], etc." OwnerUserId="1322" LastEditorUserId="9961" LastEditDate="2020-04-03T01:35:59.633" LastActivityDate="2020-04-03T01:35:59.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="256" PostTypeId="5" CreationDate="2012-05-05T14:19:26.907" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-05T14:19:26.907" LastActivityDate="2012-05-05T14:19:26.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="257" PostTypeId="4" CreationDate="2012-05-05T14:19:26.907" Score="0" Body="Hybridisation is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the qualitative description of atomic bonding properties." OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:22:54.140" LastActivityDate="2012-05-05T14:22:54.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="258" PostTypeId="5" CreationDate="2012-05-05T14:22:12.643" Score="0" Body="&lt;p&gt;From the &lt;a href=&quot;http://goldbook.iupac.org/M04002.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An electrically neutral entity consisting of more than one atom ($n &amp;gt; 1$). Rigorously, a molecule, in which  $n &amp;gt; 1$ must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is, therefore, a necessary requirement to be a molecule, that the &lt;a href=&quot;/questions/tagged/atoms&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;atoms&amp;#39;&quot; rel=&quot;tag&quot;&gt;atoms&lt;/a&gt; are covalently bound to each other, i.e. &lt;a href=&quot;/questions/tagged/covalent-compounds&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;covalent-compounds&amp;#39;&quot; rel=&quot;tag&quot;&gt;covalent-compounds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag should be applied to questions about certain molecules or classes of them. Other tags should also be applied to further narrow the field, for example &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/physical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;physical-chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2017-04-09T16:22:14.077" LastActivityDate="2017-04-09T16:22:14.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="259" PostTypeId="4" CreationDate="2012-05-05T14:22:12.643" Score="0" Body="An electrically neutral entity consisting of more than one atom (n &gt; 1). Rigorously, a molecule, in which n &gt; 1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state. This tag should be applied to question about certain molecules or a class of them." OwnerUserId="124" LastEditorUserId="402" LastEditDate="2017-04-09T16:22:25.853" LastActivityDate="2017-04-09T16:22:25.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="260" PostTypeId="5" CreationDate="2012-05-05T14:22:42.270" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-05T14:22:42.270" LastActivityDate="2012-05-05T14:22:42.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="261" PostTypeId="4" CreationDate="2012-05-05T14:22:42.270" Score="0" Body="The boiling point of a substance is the temperature at which the vapor pressure of the liquid equals the environmental pressure surrounding the liquid" OwnerUserId="124" LastEditorUserId="124" LastEditDate="2012-05-05T14:23:18.087" LastActivityDate="2012-05-05T14:23:18.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="262" PostTypeId="2" ParentId="243" CreationDate="2012-05-05T16:05:17.647" Score="6" Body="&lt;p&gt;The current set of solutions are rather too complicated, I think. Here's what I'd do: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make phenyl acetate with acetic anhydride and phenol.&lt;/li&gt;&#xA;&lt;li&gt;Nitrate your ester with the usual conditions; the bulky acetyl group makes &lt;em&gt;o&lt;/em&gt;-nitration less likely.&lt;/li&gt;&#xA;&lt;li&gt;Separate out your &lt;em&gt;p&lt;/em&gt;-nitrophenyl acetate&lt;/li&gt;&#xA;&lt;li&gt;Hydrolyze your nice ester.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a number of bulky &lt;em&gt;protecting groups&lt;/em&gt; one could use for the phenolic -OH, but acetic anhydride is rather cheap in my neck of the woods, and it's what I'm accustomed to.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastActivityDate="2012-05-05T16:05:17.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="263" PostTypeId="1" AcceptedAnswerId="281" CreationDate="2012-05-05T17:30:35.327" Score="25" ViewCount="6132" Body="&lt;p&gt;I have come across many reactions that are zero order reactions but at one glance I'm unable to tell if they are zero order or not.  Is there any criteria that can be used to identify these reactions or are they purely experimentally determined?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="4945" LastEditDate="2017-06-01T10:32:58.750" LastActivityDate="2017-06-24T15:30:55.947" Title="How to identify zero order reactions?" Tags="&lt;kinetics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="264" PostTypeId="1" CreationDate="2012-05-05T17:32:54.903" Score="17" ViewCount="1159" Body="&lt;p&gt;One of the drugs I work with is a beta-lactam (4-membered ring with an amide bond) fused to a sulfone ring, tazobactam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/colHO.png&quot; alt=&quot;skeletal diagram of tazobactam&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's relatively stable in water; the lactam is not significantly hydrolyzed without being catalyzed.  When the lactam is hydrolyzed, however, typically by an enzyme, and the lactam ring opened, the sulfone ring also opens cleaving the bond between the sulfur and bridgehead carbon.  A &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/bi0022745&quot; rel=&quot;noreferrer&quot;&gt;paper by Kuzin&lt;/a&gt; gives the following reaction scheme (this is part of figure 2, click for the whole thing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ABosK.png&quot; rel=&quot;noreferrer&quot;&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/WLbOi.png&quot; alt=&quot;Kuzin&amp;#39;s reaction scheme&quot;&gt;&#xA;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and says of it (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Ser70-bound moiety (2) is thought to be the initial intermediate [...] &lt;em&gt;Ring opening after departure of the sulfinate from C5&lt;/em&gt; produces a reactive imine (3) and the more inert tautomeric enamine forms (4, 5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But why is the sulfuryl&lt;/strong&gt; (sulfone, sulfonyl(?)...) &lt;strong&gt;leaving only when the lactam is opened and not spontaneously beforehand?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other fused bicyclic beta-lactam drugs don't open the other ring, where in lieu of the sulfone there is commonly a thioether (as in penicillin).  The sulfone group seems more electron hungry, but I don't see how the broken lactam bond affects this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard some people suggest that the lone pair on the nitrogen is able to make this happen, but why doesn't it then happen in the intact drug?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="41890" LastEditorDisplayName="user95" LastEditDate="2017-11-27T18:10:41.053" LastActivityDate="2017-11-27T18:10:41.053" Title="Why does the sulfone ring in tazobactam open when the lactam is hydrolyzed?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;&lt;pharmacology&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="265" PostTypeId="2" ParentId="263" CreationDate="2012-05-05T17:42:32.817" Score="17" Body="&lt;p&gt;Determining the rate of the kinetics of a reaction isn't done with only one measurement, or looking at only one graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What zero order kinetics means, is that the rate of the reaction is &lt;em&gt;independent&lt;/em&gt; of the concentration of chemical species. To determine the kinetics of a process, you need to take a series of measurements while altering &lt;em&gt;one&lt;/em&gt; variable, namely the concentration of one of the chemical species under investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In zero order kinetics, you should observe that as you change the concentration of reactant, for example, the rate of the reaction remains constant.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-05-05T17:42:32.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="266" PostTypeId="2" ParentId="250" CreationDate="2012-05-05T17:58:32.203" Score="-2" Body="&lt;p&gt;Very short answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = \Delta H - T\Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is an org-chem question, no need to solve the equation numerically. As you might know, $\ce{N2_{(g)}}$'s standard formation enthalpy ($\Delta H_\text{f}^\circ$) is 0 – at standard $T$ and $p$, $\ce{N2_{(g)}}$ is gaseous. Therefore, its enthalpic contribution is 0. Its $S^\circ$ is $191.32\ \mathrm{\frac{J}{mol\ K}}$, which indicates that, at 273.15 K, $\Delta G^\circ_\text{f}$ is not only negative, but also large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically, $\ce{N2_{(g)}}$ is favored. Take a look at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=nitrogen+phase+diagram&quot; rel=&quot;nofollow&quot;&gt;nitrogen's phase diagram&lt;/a&gt;, generated by Wolfram Alpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've just noticed you're considering &lt;em&gt;diazonium ions&lt;/em&gt;, not &lt;em&gt;azo&lt;/em&gt; compounds. For the reasons F'x explained, the ions are more prompt to explosion than are the azo compounds. I think azo compounds are used in the fabrication of CDs, or so my teacher told us... They're supposedly light sensitive.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T19:22:13.307" LastActivityDate="2015-05-11T19:22:13.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="267" PostTypeId="1" AcceptedAnswerId="388" CreationDate="2012-05-05T19:49:18.397" Score="28" ViewCount="906" Body="&lt;p&gt;Oscillating reactions are a funny aspect of chemistry. I have tried to find various simplified kinetic models of oscillating reactions such as the Belouzov-Zhabotinsky, the Briggs–Rauscher or the Bray–Liebhafsky reactions in order to be able to study them. However, the models I have found so far are either too complicated (8 or more species considered), or loose chemical significance. For an example of the second case, Ball's 1994 model&lt;sup&gt;[1]&lt;/sup&gt; corresponds to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AYhm1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/byTzq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this is useful and introduces a nice circularity in the model, creating the possible feedback, it has lost all chemical sense — by which I mean that no correspondance can be established between the species of this model and a real oscillating system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: &lt;strong&gt;&lt;em&gt;what's the simplest known chemically-meaningful model of an oscillating reaction?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] Ball, P. 1994 &lt;em&gt;Designing the molecular world: Chemistry at the frontier&lt;/em&gt;. Princeton, NJ: Princeton University Press.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="10185" LastEditDate="2015-09-28T13:34:36.630" LastActivityDate="2018-08-17T20:49:50.670" Title="What's a minimal yet chemically-meaningful kinetic system for an oscillating reaction?" Tags="&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="268" PostTypeId="2" ParentId="264" CreationDate="2012-05-05T23:42:40.253" Score="5" Body="&lt;p&gt;Here's a draft:&lt;/p&gt;&#xA;&lt;p&gt;First, read &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ejtc.34/pdf&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; paper. Considering we're under alkaline conditions (blood pH between 7.35-7.45) we'll take their alkaline mechanism proposal.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Mechanism&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6xuZB.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is obviously incomplete, with good reason. Without knowing what the active site of the enzyme looks like (except for Ser70), it's difficult to devise of a mechanism. I could imagine another alkoxy attacking the sp&lt;sup&gt;3&lt;/sup&gt; carbon between S and N, and shifting two electrons to the sulfur thus generating your SO&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt; in one step, but that would require a proper active site.&lt;/p&gt;&#xA;&lt;p&gt;The geometry of the molecule inside the active site of the enzyme is obviously optimal for this reaction to proceed. I'd like to know more. For some reason my school doesn't give me access to this paper.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-05-05T23:42:40.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="269" PostTypeId="2" ParentId="240" CreationDate="2012-05-06T01:39:38.127" Score="11" Body="&lt;p&gt;As you alluded to in your post, &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;noreferrer&quot;&gt;XRC&lt;/a&gt; has a long history for absolute configuration assignment. The drawback is more than resources, collaborators exist for such reasons, but the hard part is actually growing a suitable crystal. You also need heavy enough atoms present, usually introduced via derivatization if not already present. It can takes years, sometimes, to obtain a suitable crystal.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_rotation&quot; rel=&quot;noreferrer&quot;&gt;OR&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_dichroism&quot; rel=&quot;noreferrer&quot;&gt;CD&lt;/a&gt; have been used in the past to predict configuration. For example, we don't teach it these days but there were a variety of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultraviolet-visible_spectroscopy_of_stereoisomers&quot; rel=&quot;noreferrer&quot;&gt;chirality rules&lt;/a&gt; developed to do this on CD. These were mostly empirical in nature and as a result had many exceptions. Today there are ab initio time-dependent density functional theory calculations which are trying to make accurate enough calculations to be useful in this arena.&lt;/p&gt;&#xA;&lt;p&gt;NMR obviously has some ancillary methods to address this, but I don't know anything about them.&lt;/p&gt;&#xA;&lt;p&gt;Stereocontrolled reactions are another way to help deduce the structure, but that is time and effort consuming as a synthetic endeavor and particularly useless for anybody in the pharmaceutical industry. Syntheses are also error prone.&lt;/p&gt;&#xA;&lt;p&gt;If you want to know what the pharmaceutical industry does when it cannot do XRC, then look no further than failed drugs. As an example, &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chir.10171/abstract&quot; rel=&quot;noreferrer&quot;&gt;gossypol&lt;/a&gt; was being evaluated as male birth control. It failed due to a certain lack of reversibility, among other things. Anyway, the technique used to determine its conformation and absolute configuration is called vibrational circular dichroism (&lt;a href=&quot;http://en.wikipedia.org/wiki/Vibrational_circular_dichroism&quot; rel=&quot;noreferrer&quot;&gt;VCD&lt;/a&gt;). I should also mention at this point that Raman optical activity (ROA) is another method that is useful in some instances.&lt;/p&gt;&#xA;&lt;p&gt;Modern DFT program suites allow pretty reasonable predictions for VCD spectra taking on the order of a day, these calculations are much simpler than those required for good OR or CD. These calculated spectra are compared with experimental and from here the absolute configuration can be assigned. The quality of the calculations can be easily be determined by comparing the calculated and measured IR spectra.&lt;/p&gt;&#xA;&lt;p&gt;The really basic idea is that there are different interactions of right- and left-circularly polarized light during vibrational excitation. The difference in these responses is plotted versus wavenumber. Without the right instrumentation, this technique is really hard, but one can purchase equipment to make these pretty straightforward to measure. There is a lot more you can do with it, but Big Pharma definitely uses VCD when XRC is inaccessible.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-05-06T01:39:38.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="270" PostTypeId="1" AcceptedAnswerId="516" CreationDate="2012-05-06T05:05:18.427" Score="11" ViewCount="2102" Body="&lt;p&gt;Periodic acid (per-iodic acid, $\ce{HIO4}$ ) cleaves vicinal diols(to carbonyls) via a cyclic intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the cyclic intermediate, it needs a syn configuration for the $\ce{-OH}$ groups for the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly which configurations are allowed on cyclohexane derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take 1,2-dihydroxycyclohexane(boat conformer). If the two hydroxy groups are both on the axial positions (&lt;em&gt;trans-&lt;/em&gt; ), then obviously the reaction will not take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about when the two are &lt;em&gt;cis-&lt;/em&gt; (one axial, one equatorial), and when they are both equatorial (&lt;em&gt;trans-&lt;/em&gt; )?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My original question had to do with diastereomers of cyclic glucose--aldohexoses--but I realised that the extra hydroxy groups make the reaction more complicated)&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2020-12-19T17:20:29.040" Title="Effect of periodic acid on cyclohexane derivatives" Tags="&lt;stereochemistry&gt;&lt;cyclohexane&gt;&lt;conformers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="271" PostTypeId="1" AcceptedAnswerId="272" CreationDate="2012-05-06T05:41:17.493" Score="12" ViewCount="17865" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_coefficient&quot;&gt;partition coefficient&lt;/a&gt; is a very useful metric in characterizing compounds particularly drugs. What chemical properties influence that partitioning? Are chemical moieties additive in their LogP? What if the compound has both hydrophobic and hydrophilic ends?&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2018-09-05T20:30:17.040" Title="What are the major factors that influence LogP" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="272" PostTypeId="2" ParentId="271" CreationDate="2012-05-06T07:55:39.200" Score="4" Body="&lt;p&gt;The Wikipedia page you linked to has some &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_coefficient#Prediction&quot; rel=&quot;nofollow&quot;&gt;interesting information&lt;/a&gt;, that could at least get you started:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has been shown that the &lt;strong&gt;log P of a compound can be determined by the sum of its non-overlapping molecular fragments&lt;/strong&gt; (defined as one or more atoms covalently bound to each other within the molecule). Fragmentary log P values have been determined in a statistical method analogous to the atomic methods (least squares fitting to a training set). In addition, &lt;strong&gt;Hammett type corrections are included to account of electronic and steric effects&lt;/strong&gt;. This method in general gives better results than atomic based methods, but cannot be used to predict partition coefficients for molecules containing unusual functional groups for which the method has not yet been parameterized (most likely because of the lack of experimental data for molecules containing such functional groups).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-06T07:55:39.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="273" PostTypeId="2" ParentId="271" CreationDate="2012-05-06T10:03:39.690" Score="5" Body="&lt;p&gt;Partition Coefficient, represented as P and defined as: &lt;strong&gt;P = [Organic phase] / [Aqueous phase]&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &quot;[]&quot; represents concentration in the  phase and &lt;strong&gt;$\mathrm{\log (P)= \log_{10}(Partition Coefficient)}$&lt;/strong&gt; if log P value is higher then its more lipophilic ;those drugs were more absorbed by cells and elimination of those drugs from body will be less .The partition coefficient is a ratio of concentrations of un-ionized compound between the two solutions so for a chemical which is having both hydrophobic and hydrophilic ends also have log P values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg: &lt;a href=&quot;http://wenku.baidu.com/view/fb1abe68561252d380eb6ef0.html#&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oleic acid&lt;/a&gt;: $\mathrm{\log (P)= 7.64}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chemicals which are hydrophopic(lipophilic) tends to have more Log P values and hydrophilic(lipophobic) chemicals have less log P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg: pentane($\mathrm{\log (P)= 3.45}$) [lipohilic], 1-pentanol ($\mathrm{\log (P)= 1.51}$) [lipophobic]&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="23561" LastEditDate="2018-09-05T20:30:17.040" LastActivityDate="2018-09-05T20:30:17.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="274" PostTypeId="2" ParentId="267" CreationDate="2012-05-06T14:17:46.637" Score="17" Body="&lt;p&gt;I'm not sure if this is what you're looking for (it may well still be too abstract), but oscillations can occur in the Gray-Scott system, which is&#xA;$$&#xA;A + 2B \to 3B\\&#xA;B \to P,&#xA;$$&#xA;where $P$ is an inert product, and the reaction is assumed to take place in a flow reactor that provides a supply of A, giving rise to the dynamics&#xA;$$\begin{align}&#xA;\frac{\mathrm da}{\mathrm dt} &amp;amp;= f(1-a) - ab^2;\\&#xA;\frac{\mathrm db}{\mathrm dt} &amp;amp;= ab^2 - (f+k)b,\\&#xA;\end{align}&#xA;$$&#xA;where $f$ is a rate determined by the flow reactor, $k$ is the rate of the $B\to P$ reaction, and the rate constant of the autocatalytic reaction has been set to 1 without loss of generality by scaling $f$ and $k$ relative to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the appropriate choice of the parameters $f$ and $k$, oscillation can happen because $B$'s concentration increases autocatalytically, but then it overshoots its food source (i.e. the concentration of A), which then builds up again, allowing the cycle to repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might not like the trimolecular step, but I've found you generally get similar behaviour if you split it up into something like&#xA;$$&#xA;A + B \to C + B\\&#xA;B + C \to 2B\\&#xA;B \to P.&#xA;$$&#xA;(Simply using $A+B\to 2B$ doesn't work, because its kinetics don't have the right sort of nonlinearity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that this has an advantage over the Ball model that you posted in that it obeys the laws of thermodynamics. (At least from what you showed in your question, it looks to me that Ball's model only oscillates because the reverse reactions have been neglected, and if they weren't then it would have to go to equilibrium, because it's a closed system.) It makes explicit that you need a power source (the supply of $A$) for oscillation, and illustrates the connection between oscillatory behaviour and autocatalytic kinetics.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorDisplayName="user7951" LastEditDate="2018-08-17T20:49:50.670" LastActivityDate="2018-08-17T20:49:50.670" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="275" PostTypeId="1" AcceptedAnswerId="277" CreationDate="2012-05-06T14:57:17.110" Score="23" ViewCount="4180" Body="&lt;p&gt;Historically, thiols $\ce{-SH}$ were known as &quot;mercapto-&quot; due to their strong propensity to bond with mercury.  Certain drugs still carry this designation, though &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercaptopurine&quot; rel=&quot;noreferrer&quot;&gt;mercaptopurine&lt;/a&gt; has an $\ce{=S}$ group instead of a thiol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the thiol group bond so strongly with mercury in particular? Why don't alcohols make such a strong bond with $\ce{Hg}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's known, was there any historical reason why it was important to be able to &quot;capture&quot; mercury?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="9961" LastEditorDisplayName="user95" LastEditDate="2017-04-22T13:15:08.660" LastActivityDate="2017-04-22T13:15:08.660" Title="Why do thiols have such a propensity for bonding with mercury?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;organosulfur-compounds&gt;&lt;hsab&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="276" PostTypeId="1" AcceptedAnswerId="337" CreationDate="2012-05-06T15:14:30.183" Score="11" ViewCount="1800" Body="&lt;p&gt;In a thesis I am reading, it is said that one of the reasons for using plane-wave basis sets for first-principles molecular dynamics (aka &lt;em&gt;ab initio&lt;/em&gt; MD) is that the Pulay forces&lt;sup&gt;[1,2]&lt;/sup&gt; that arise from an MD using atomic basis sets are computationally expensive to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I understand that having additional terms means more code to write, i.e. they make writing the software harder to write, is it true that they are CPU-intensive to compute? The criterion I would use to quantify this subjective statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that you have already calculated the energy and forces at that point, you have already computed a large number of  integrals required for this task and involving basis functions: overlap integrals, terms of the form $\left\langle\phi_\alpha\left|\hat{A}\right|\phi_\beta\right\rangle$ where operator $\hat{A}$ is either the hamltonian, its gradient, or any other operator necessary for the calculation of energy or forces. Would the calculation of Pulay forces require any more integrals to be evaluated, or can it be straightforwardly computed from those previously-calculated integrals alone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] P. Pulay, &lt;em&gt;Molec. Phys.&lt;/em&gt; 19, 197 (1969)&lt;br&gt;&#xA;[2] See also slide 6 of &lt;a href=&quot;http://tfy.tkk.fi/~asf/physics/lectures/PDF/lect11.pdf&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-05-09T19:28:23.260" LastActivityDate="2012-06-18T13:15:20.500" Title="Are Pulay forces expensive to compute?" Tags="&lt;computational-chemistry&gt;&lt;basis-set&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="277" PostTypeId="2" ParentId="275" CreationDate="2012-05-06T15:33:05.277" Score="18" Body="&lt;p&gt;Well, sulfur in general, and thiolates in particular, are known for bonding very strongly to gold, lead and mercury. In the particular case of mercury, which was known very early in the history of chemistry (indeed since the times of alchemy), this is the reason thiols are often called &quot;mercaptans&quot;, from the Latin &lt;em&gt;mercurium captans&lt;/em&gt; (capturing mercury).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Pearson's HSAB theory (hard and soft acids and bases), the reason the S-Hg bond is be stronger than the O-Hg can be explained because S&lt;sup&gt;2-&lt;/sup&gt; is a soft basis and Hg(II) is a soft acid, making a good fit… while O&lt;sup&gt;2-&lt;/sup&gt; is harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, regarding the historical use of thiols to bind mercury (and lead to), let's just say a chelating agent for mercury and other toxic compounds was quite useful as an antidote. You can read a very nice book on the historical use of mercury and lead in John Emsley's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0192805991&quot;&gt;&lt;em&gt;The Elements of Murder&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-06T15:33:05.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="278" PostTypeId="1" AcceptedAnswerId="291" CreationDate="2012-05-06T16:10:24.420" Score="16" ViewCount="410" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Related (very similar, but here I want a mechanism) &lt;a href=&quot;https://physics.stackexchange.com/q/21827/7433&quot;&gt;https://physics.stackexchange.com/q/21827/7433&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Molar_conductivity#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kohlrausch law&lt;/a&gt;, all ions contribute to the conductivity of an electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I understand it, the mechanism of conduction in an electrolyte is thus:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ions migrate in solution&lt;/li&gt;&#xA;&lt;li&gt;These ions get reduced or oxidized at the electrodes and converted to electrons&lt;/li&gt;&#xA;&lt;li&gt;These electrons continue down the wire, leading to an increased/maintained conductivity/current&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But, this mechanism doesn't work for ions which do not get redoxed--movement of ions in the solution cannot be translated to movement of electrons in the wire and thus it seems (to me) that conductivity should &lt;em&gt;not&lt;/em&gt; increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ions like $\ce{NO3-}$ have a comparable $\lambda$ (ionic molar conductivity)--so they experimentally do increase conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism for conduction via these ions?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2012-06-28T05:08:09.347" Title="How does conductivity work for non-redoxed ions?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;conductivity&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="279" PostTypeId="1" AcceptedAnswerId="280" CreationDate="2012-05-06T17:15:42.013" Score="10" ViewCount="2977" Body="&lt;p&gt;In order to calculate the $K_\mathrm{sp}$ of lead(II) iodide, I mixed $\pu{6.0 mL}$ of $\pu{0.01 M}$ $\ce{Pb(NO3)2}$, $\pu{6.0 mL}$ of $\pu{0.02 M}$ $\ce{KI}$, and $\pu{8.0 mL}$ of water, and a precipitate (barely) formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I calculated $[\ce{Pb^2+}]$ to be $\pu{0.0036 M}$ and $[\ce{I-}]$ to be $\pu{0.0060 M}$, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 1.1 \times 10^{-7}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the actual value (at $\pu{25 ^\circ C}$) is $7.9\times 10^{-9}$ (Skoog et al. &lt;em&gt;Fundamentals of Analytical Chemistry&lt;/em&gt;), so my result is off by a factor of $\sim 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure I had clean equipment and measured carefully and accurately. Why is there a discrepancy? Is my maths wrong, or are there some inaccuracies in my experiment?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="16683" LastEditDate="2017-09-05T06:51:56.487" LastActivityDate="2017-09-05T06:53:35.723" Title="Experimental calculation of solubility product constant" Tags="&lt;experimental-chemistry&gt;&lt;solubility&gt;&lt;precipitation&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="280" PostTypeId="2" ParentId="279" CreationDate="2012-05-06T19:13:26.017" Score="6" Body="&lt;p&gt;It's probably an experimental error. The 'barely' may be important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you redone the experiment? Doing an experiment more than once is integral to getting good values.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="16683" LastEditDate="2017-09-05T06:53:35.723" LastActivityDate="2017-09-05T06:53:35.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="281" PostTypeId="2" ParentId="263" CreationDate="2012-05-06T20:12:36.007" Score="16" Body="&lt;p&gt;To your question of whether zero-order reactions can be recognized “at one glance” and from an non-experimental criterion, &lt;strong&gt;the answer is clearly no&lt;/strong&gt;. The only theoretical way of firmly establishing the order of a reaction (if it has one) is to fully calculate its kinetic law (or &lt;em&gt;rate equation&lt;/em&gt;), which can only be done if you know the scheme of the reaction, i.e. the list of all &lt;em&gt;elementary reactions&lt;/em&gt; involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one case where you can take a shortcut, namely if you know (or strongly suspect) that the reaction in question is an elementary reaction, for example, if it involves no catalyst, proceeds in a single step with only one transition state and no intermediate. If that is the case, then the order of the reaction is determined by the stoichiometry ($\ce{A –&amp;gt; X}$ is first order, $\ce{2 A -&amp;gt; X}$ or $\ce{A + B -&amp;gt; X}$ are second order). However, this does not apply in your case, because &lt;strong&gt;elementary reactions cannot have zero order&lt;/strong&gt;!&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4945" LastEditDate="2017-06-01T10:26:33.767" LastActivityDate="2017-06-01T10:26:33.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="282" PostTypeId="5" CreationDate="2012-05-06T23:43:37.333" Score="0" Body="&lt;p&gt;Please see the full homework policy in this Meta discussion: &lt;a href=&quot;https://chemistry.meta.stackexchange.com/q/141&quot;&gt;How do I ask homework questions on Chemistry Stack Exchange?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="15751" LastEditDate="2017-06-25T13:03:10.567" LastActivityDate="2017-06-25T13:03:10.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="283" PostTypeId="4" CreationDate="2012-05-06T23:43:37.333" Score="0" Body="DEPRECATED. DON'T USE THIS TAG." OwnerUserId="10" LastEditorUserId="39539" LastEditDate="2017-05-16T08:36:25.703" LastActivityDate="2017-05-16T08:36:25.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="284" PostTypeId="2" ParentId="279" CreationDate="2012-05-07T00:50:33.303" Score="3" Body="&lt;p&gt;How long you observed your mixture? It is possible that you had not reached the equilibrium required for the solubility product constant to be valid. Allowing time for the system to reach equilibrium might have resulted in the appearance of more precipitate that would indicate that you had added too much material. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-05-07T01:53:52.843" LastActivityDate="2012-05-07T01:53:52.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="285" PostTypeId="1" AcceptedAnswerId="292" CreationDate="2012-05-07T03:30:35.577" Score="22" ViewCount="1723" Body="&lt;p&gt;My understanding is that the NBO methodology transforms the molecular orbital picture of a molecule into a collection of pairwise interactions (bond orbitals). This is obviously handy for didactic and comprehension purposes as it distils what might be a very complex and non-local MO picture into manageable lobes conforming to the introductory theories of chemical bonding (VB, orbital hybridisation) taught in high-school and early university courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does natural bond orbital theory have utility outside of this role?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2016-01-28T17:11:18.640" Title="What is natural bond orbital theory used for?" Tags="&lt;hybridization&gt;&lt;bond&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="286" PostTypeId="1" AcceptedAnswerId="287" CreationDate="2012-05-07T04:15:24.710" Score="6" ViewCount="150" Body="&lt;p&gt;When I was reading through the ADF-BAND tutorials, one of the toy systems presented was a 1-D periodic structure involving 3 collinear hydrogen atoms. The tutorial pointed out that, topologically speaking, this is cylindrically symmetrical (more specifically, it's ring symmetrical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the instance of a 2-D structure, can the calculation be considered a model for the surface of a torus (this seems logical), a sphere (I doubt this because if you fit a rectilinear grid to a sphere you end up with two poles and dissimilar meridians and parallels), or other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus questions: Has anyone used periodic calculations to model electronic structure/chemistry on the surface of a sphere or torus? Can you introduce a curvature term to account for these structures having a finite size?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-07-21T16:48:11.303" Title="Is a 2-D periodic structure isomorphic with the surface of a torus, a sphere, neither or both?" Tags="&lt;computational-chemistry&gt;&lt;symmetry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="287" PostTypeId="2" ParentId="286" CreationDate="2012-05-07T06:51:17.807" Score="5" Body="&lt;p&gt;Yes, a 2D-periodic space can be mapped to a torus, but &lt;a href=&quot;https://math.stackexchange.com/questions/66789/if-a-rectangular-grid-where-left-right-and-top-bottom-wrap-can-be-mapped-o&quot;&gt;that's more a question for the math.SE&lt;/a&gt;…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your bonus question, why would there be? What would you do with it? Molecular structures are intrinsically 3D, so I don't see what you would do in a 2D (periodic or not) space? Even when we talk about planar or pseudo-2D structures (buckyball, nanotube, etc.) they are 3D objects with 3D electronic densities and wavefunctions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; 3D structures that are periodic in two dimensions and finite in the other one can be studied by many computational chemistry codes. They are often referred to as &lt;em&gt;slab calculations&lt;/em&gt; or &lt;em&gt;surface calculations&lt;/em&gt;. The most common issue is that of Coulombic interaction (or Poisson equation solver), which typically requires special treatment in the 2D case.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:09.340" LastActivityDate="2012-05-07T09:43:58.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="288" PostTypeId="2" ParentId="279" CreationDate="2012-05-07T08:56:47.163" Score="6" Body="&lt;p&gt;I don't have time to do the calculations manually, but inputing your problem data into a &lt;a href=&quot;https://chemistry.stackexchange.com/q/24/30&quot;&gt;chemical equilibrium software&lt;/a&gt; (here Dozzaqueux) reveals that &lt;strong&gt;the precipitate you observe is Pb(OH)&lt;sub&gt;2&lt;/sub&gt;&lt;/strong&gt;, while no PbI&lt;sub&gt;2&lt;/sub&gt; should form. The former is white, while the later is yellow; can you tell us what color was your precipitate?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2012-05-07T08:56:47.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="289" PostTypeId="1" CreationDate="2012-05-07T09:20:39.427" Score="10" ViewCount="7853" Body="&lt;p&gt;Can derive the quantity of a substance from a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_chromatography%E2%80%93mass_spectrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;GC/MS&lt;/a&gt; report if I know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the ratio to another substance in the data, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the quantity of that second substance, and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the molecular mass of both substances. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The site was very dense and vague regarding how GC/MS test work, but fairly upfront regarding how they derive their ratios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the numbers (slightly tweaked, but more or less in proportion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substance A: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantity: $300 \:\pu{mg}$ (Know already, not part of GC/MS report)&lt;/li&gt;&#xA;&lt;li&gt;Mol. Mass: $500 \:\pu{g/mol}$ (found in external source)&lt;/li&gt;&#xA;&lt;li&gt;Peak Proportion: 3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Substance X: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantity: Unknown&lt;/li&gt;&#xA;&lt;li&gt;Mol. Mass: $425 \:\pu{g/mol}$ (again, external source)&lt;/li&gt;&#xA;&lt;li&gt;Peak Proportion: 1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To be clear, the ratio of 1:3 means that substance X has a horizontal peak $\frac13$ of substance A.  I'm not clear on what the x-axis is actually a measurement of, which is part of why I'm at a loss. The site does make it very explicit that the ratios are not directly proportional  to mass ratio (so if it found a ratio of 3:1 of glucose to arsenic, this doesn't mean the substance is 75% glucose and 25% arsenic, only that the glucose &quot;peaks&quot; 3 times higher, which I've taken to mean &quot;3 times the oomph&quot;, but that may be incorrect, as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I know that the molecular mass($\rm M$) of substance X is 0.85 of substance Y, can I derive the actual mass of substance X using the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\mathrm{qty}_X=\frac{\mathrm{qty}_A\times(M_X/M_A)}{\mathrm{peak}_A/\mathrm{peak}_X}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with data being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\rm qty_X=\frac{300\:\pu{mg}\times(425/500)}{3/1}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\rm qty_X=(300\:\pu{mg}\times 0.85)/3\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and finally the result of $85 \:\mathrm{mg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess in the end there are 3 questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this even how GC/MS results work?&lt;/li&gt;&#xA;&lt;li&gt;If so, is my assumption to derive the mystery quantity using the molecular mass correct?&lt;/li&gt;&#xA;&lt;li&gt;Is the math itself in order? (I'm specifically worried that I should invert either the mass ratio or the peak ratio or both).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, if the answer to the first questions is no, then my true question is: can I derive the quantity of substance X with the given data, and if so, what would be the right approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone is curious for some context, I need to know the actual quantity of substance X as I know that, by mass, it has a threshold between harmless and toxic, so just knowing substance X is $\frac13$ &quot;peak&quot; of substance A doesn't let me know if I should let my dog/child/self ingest it.&lt;/p&gt;&#xA;" OwnerUserId="152" LastEditorUserId="23561" LastEditDate="2018-09-05T20:24:25.227" LastActivityDate="2018-09-05T20:24:25.227" Title="Gas chromatography/mass spectrometry: Can I use peak ratios and molecular mass to derive quantity?" Tags="&lt;physical-chemistry&gt;&lt;chromatography&gt;&lt;mass-spectrometry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="290" PostTypeId="1" CreationDate="2012-05-07T15:15:11.740" Score="17" ViewCount="2894" Body="&lt;p&gt;Is there a formal definition or algorithm that can take a list of reactions and tell me whether it contains a set of species that can produce itself autocatalytically? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the following reaction is autocatalytic:&#xA;$$\ce{&#xA;A + B \to 2 B&#xA;}$$&#xA;B appears on both sides of the equation, so it's a catalysis reaction, and there is an extra B produced on the right, so it catalyses its own production from A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the following reaction scheme can also be considered to contain autocatalysis:&#xA;$$&#xA;\ce{B + C \to D + E\\&#xA;A + D \to 2C}&#xA;$$&#xA;If this system starts out with plenty of A and B, together with one molecule of C, then the C molecule can react with a B to produce a D, which can then react with an A to produce two C's — whereas, without the initial &quot;seed&quot; molecule, no reactions will take place. (I'm neglecting reverse reactions here of course, but hopefully it gets the point across.)  The autocataltic cycle can easily be seen in the following diagram, where circles represent species and boxes represent reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQJMA.png&quot; alt=&quot;autocatalysis diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another non-obvious example of autocatalysis is a reaction scheme of this form:&#xA;$$\ce{&#xA;A + C \to C + D\\&#xA;B + D \to D + C + E}&#xA;$$&#xA;Here, C and D each catalyse the production of one another from the precursors A and B. I'm aware of Kauffman's definition of reflexively autocatalytic sets, which applies to schemes of this form, but would miss an autocatalytic cycle of the kind shown above, since it doesn't consist of single-step catalysis reactions. So I'm looking for a definition of autocatalysis that would include both cases, as well as (hopefully) other, more complex schemes that I haven't yet thought of.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="30" LastEditDate="2012-05-27T16:34:27.170" LastActivityDate="2012-07-16T23:55:41.313" Title="Definitions of autocatalysis" Tags="&lt;theoretical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="291" PostTypeId="2" ParentId="278" CreationDate="2012-05-07T17:27:24.953" Score="14" Body="&lt;p&gt;This can arise due to ion-ion interactions brought on by Coulombic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proper explanation is rooted in &lt;a href=&quot;http://en.wikipedia.org/wiki/Transport_phenomena&quot;&gt;transport phenomena&lt;/a&gt;. Unfortunately chemists are largely not taught this, but chemical engineers do get the opportunity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under standard conditions, ions in a solution will take a &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_walk&quot;&gt;random walk&lt;/a&gt; during diffusion. This results in no net movement of our ions over some span of time. Such a system perturbed by an applied electric field will cause ions to preferentially move in one direction, we say the ions will drift. We say they have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Drift_velocity&quot;&gt;drift velocity&lt;/a&gt;. These velocities are affected by a number of parameters, as there is still continual collision, but we can say that the mobilities will be unequal for ions of unequal sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different mobilities mean concentration gradients form and charge separation too. Typically we use an electroneutrality field as a reasonable approximation to allow analytic solutions for various equations. This field depends on all the ionic fluxes present and as such will affect the total measurable current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is simple: the introduction of any ion will cause a perturbation in the field that all the ions encounter in solution.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-05-07T17:27:24.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="292" PostTypeId="2" ParentId="285" CreationDate="2012-05-08T01:33:50.227" Score="13" Body="&lt;p&gt;Yes, along with a family of related methods. I honestly don't fully understand the mechanics behind it, but the orbitals obtained with the NBO method can be used for a treatment called Natural Resonance Theory, which gives you an idea of what sort of Lewis-structure resonance components there are and in what weightings, as well as bond indices, and a measure of covalency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three papers from the developers of the technique, covering &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291096-987X%2819980430%2919:6%3C593%3a%3aAID-JCC3%3E3.0.CO;2-M/abstract&quot;&gt;theory&lt;/a&gt;, &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291096-987X%2819980430%2919:6%3C610%3a%3aAID-JCC4%3E3.0.CO;2-U/abstract&quot;&gt;usage&lt;/a&gt;, and &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291096-987X%2819980430%2919:6%3C628%3a%3aAID-JCC5%3E3.0.CO;2-T/abstract&quot;&gt;some more examples&lt;/a&gt;. (I'd probably start with the usage paper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, along with NBO, there are the Natural &lt;em&gt;Atomic&lt;/em&gt; Orbitals which are often lumped in because they're also computed by the NBO program, and Natural Population Analysis which you can use the Natural Atomic Orbitals to do. These give you an idea of what sort of atomic orbital character is contributing to your molecular orbitals, and also atomic charges and net spins.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2012-05-08T03:06:02.620" LastActivityDate="2012-05-08T03:06:02.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="293" PostTypeId="1" AcceptedAnswerId="352" CreationDate="2012-05-08T04:08:36.293" Score="14" ViewCount="132" Body="&lt;p&gt;Looking at something like $\ce{YBa2Cu3O7}$ which was one of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Superconductivity&quot; rel=&quot;nofollow&quot;&gt;first cuprate superconductors&lt;/a&gt; to be discovered, I'm always curious how the selection of these substances as likely superconductors comes about?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with optimal crystalline structure, unpaired electrons, or is it similar to the way semiconductors are designed with doping to create &quot;holes&quot;?  Or are these superconductors simply determined experimentally based on small changes of what has worked in the past? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question: I've always been curious as to how they stumbled upon Yttrium for this particular case.  It's not something I think of as being carried in ready supply.  Would this have been after YAG lasers were already around? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="24052" LastEditDate="2016-03-11T23:22:10.360" LastActivityDate="2016-03-11T23:22:10.360" Title="Is there a theory behind selecting elements that may be successful in potential superconductors?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="294" PostTypeId="1" AcceptedAnswerId="296" CreationDate="2012-05-08T08:41:16.933" Score="8" ViewCount="1144" Body="&lt;p&gt;How can I write a Z-matrix for the following complex?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  O==C==O  H--O&#xA;               \&#xA;                H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have managed to write:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O     &#xA;C   1  a  &#xA;O   2  a     1  180 &#xA;H   3  d     2  180    1  ?&#xA;O   4  b     3  180    2  ?&#xA;H   5  b     4    t    3  ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;but I don't know how to specify the dihedral angles indicated with &lt;code&gt;?&lt;/code&gt;. Are they all zero? Or completely unspecified? If so, will I be able to do a coordinate optimization with these variables or will it be a problem?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-06-08T02:36:47.860" Title="Z-matrix of a series of 5 colinear atoms" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="295" PostTypeId="1" AcceptedAnswerId="588" CreationDate="2012-05-08T08:48:42.150" Score="9" ViewCount="118" Body="&lt;p&gt;I'm looking into uses of zeolites as water absorbers for heat storage and absorbtion cooling. However, I don't find hard numbers on the likes of absorption enthalpies at different temperatures, saturation moistures, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an institution, chemical suppliers are traditionally reluctant to sell to private individuals. I found zeolitic absorbers in few catalogues, but only with a descriptor of the type, no further data. One company I asked (that sell zeolites in amounts of kg to private customers) was unwilling to share data on their product (They're an engineering company doing product developement with zeolitic absorbers for heating/cooling etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only found (sketchy) data on single zeolites in working papers, but no comprehensive source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: seeing that vendors don't share the info, where do I find datasheets of zeolites?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="30" LastEditDate="2012-05-08T09:22:04.510" LastActivityDate="2012-06-04T08:34:29.767" Title="Where do I find datasheets of zeolites?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="296" PostTypeId="2" ParentId="294" CreationDate="2012-05-08T08:52:09.770" Score="14" Body="&lt;p&gt;You cannot have bond angles of 0 or 180 degrees in a Z-matrix. This is because the dihedral becomes degenerate. To solve this, you can add '&lt;em&gt;dummy atoms&lt;/em&gt;' which provide an auxiliary point of reference to remove the 180 degree angle. These atoms are usually denoted $\ce{X}$ or $\ce{Xx}$ in computational chemistry suites. Usually, dummy atoms are placed at 90 degrees to a pair of atoms and the third collinear atom is then defined with a bond angle of 90 degrees and a dihedral of 0 or 180 degrees with respect to the previous 3 atoms.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To illustrate what I'm talking about, here's a schematic of your pair of molecules, with three dummy atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7D0Ix.png&quot; alt=&quot;An illustrated Z-matrix&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could have a Z-matrix like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O   1&#xA;C   1    a&#xA;Xx  2  1.0  1  90&#xA;O   2    b  3  90     1  180&#xA;Xx  4  1.0  2  90     3  0&#xA;H   4    c  5  90     2  180&#xA;Xx  6  1.0  4  90     5  0&#xA;O   6    d  7  90     4  180&#xA;H   8    e  6  104.5  7  180&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Notes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are no hard and fast rules about how many you should use (obviously we could simply define subsequent dihedrals in terms of the leftmost dummy atom) however it's best to strive for locality so that small changes in some variable don't correspond to large displacements elsewhere, in which case rounding errors or large gradients can ruin your day. You could also have defined all of the dihedrals in your molecule with respect to that convenient non-collinear hydrogen, however this is the Z-matrix equivalent of spaghetti code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will note that I use different letters for each variable. This is because forcing the $\ce{H-O}$ and $\ce{C=O}$ bonds to be of equal length is probably an unrealistic constraint for a geometry optimisation of this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that dummy atoms do not take part in the electronic structure of a molecule, so feel free to put them in anywhere and set the dummy atom 'bond' length to whatever you want. 1.0 Ångström is traditional as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;P.S.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cmbi.ru.nl/molden/&quot; rel=&quot;noreferrer&quot;&gt;Molden&lt;/a&gt; has a complete, if idiosyncratic graphical Z-matrix editor.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-08T09:42:31.440" LastActivityDate="2012-05-08T09:42:31.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="297" PostTypeId="1" AcceptedAnswerId="305" CreationDate="2012-05-08T09:15:16.000" Score="18" ViewCount="1684" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solvated_electron&quot; rel=&quot;noreferrer&quot;&gt;Solvated electrons&lt;/a&gt; have a long lifetime in ammonia solutions, but their counterparts in water (called hydrated electrons) have a much smaller lifetime, of the order of microseconds in very pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7veZ6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;(image from: Boero et al, &lt;em&gt;Phys Rev Lett&lt;/em&gt; &lt;strong&gt;2003&lt;/strong&gt;, 90, 226403)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What properties of the solvent account for these very different lifetimes? Both are polar protic solvents, so what other(s) factor(s) could be involved here?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2015-05-20T12:02:10.797" Title="What dictates the lifetime of a solvated electron in a given solvent?" Tags="&lt;solvents&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="298" PostTypeId="1" AcceptedAnswerId="304" CreationDate="2012-05-08T09:20:10.113" Score="33" ViewCount="2326" Body="&lt;p&gt;I've always been a bit uncomfortable with the concept of more than two electrons in a single orbital-like region(probability-wise) which occurs in resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to disobey Pauli's exclusion principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take benzene for example. It has a delocalized cloud of electrons. These all had the &quot;original&quot; (as in, if resonance didn't occur) quantum numbers $n=2,l=1,m=\text{(1,0,or -1; but it must be the same for all)},m_s=\pm\frac12$. But, due to delocalization, they're now all in one extra-large orbital-like region (by orbital-like, I mean that the pi cloud is similar to an orbital in the sense that it is also a probability cloud which defines the probabilities of finding one or more electrons at different coordinates). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's obvious that the original quantum numbers are insufficient here due to Pauli's exclusion principle. They may not even exist for these electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest thing to do would be to &quot;propose&quot; (not exactly propose, since I &lt;em&gt;guess&lt;/em&gt; the Dirac equation can &lt;em&gt;prove&lt;/em&gt; their existence) an extra quantum number, which can take twelve values. Or a combination of a few extra quantum numbers that have the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this raises the question: do the electrons belonging to the pi cloud &quot;retain&quot; their &quot;original&quot; quantum numbers as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I want to know how benzene and other resonating compounds are described quantum mechanically, with focus on quantum numbers. (links to papers are fine, probably necessary here--but please explain the gist of the link)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd prefer the answer in terms of simpler quantum mechanics, though I'm not averse to the idea of some extra reading.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2012-05-08T09:59:46.460" LastActivityDate="2019-06-17T23:12:10.877" Title="Pauli exclusion principle and resonance" Tags="&lt;quantum-chemistry&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="299" PostTypeId="1" AcceptedAnswerId="301" CreationDate="2012-05-08T12:17:57.360" Score="13" ViewCount="2027" Body="&lt;p&gt;In many places I've seen the &quot;extra&quot; bond in benzyne being labelled as $\ce{sp^2-sp^2}$ overlap or distorted (not parallel) $\ce{p\pi-p\pi}$ overlap. But I've failed to see &lt;em&gt;why&lt;/em&gt; we can't have a normal, parallel $\ce{p\pi-p\pi}$ overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In benzyne, however, the p orbitals are distorted to accommodate the triple bond within the ring system, reducing their effective overlap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That makes no sense to me — why can't a normal, parallel $\ce{p\pi-p\pi}$ overlap be &quot;accommodated&quot; within the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the reason for the distorted $\pi$ bond in benzyne?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fine with answers involving a bit of QM/MO theory.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="5026" LastEditDate="2018-04-04T16:24:57.420" LastActivityDate="2018-10-06T15:08:27.857" Title="Why is the benzyne triple bond distorted?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="300" PostTypeId="1" AcceptedAnswerId="311" CreationDate="2012-05-08T12:38:17.257" Score="19" ViewCount="664" Body="&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8f/EofRu%28bz%292.png/300px-EofRu%28bz%292.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hapticity#Changes_in_hapticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;From here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see why hapticity changes here, but not why it is necessary for the benzene ring to bend. The resonance energy of benzene is pretty large, and there's no balancing increase in stability elsewhere--unless the decrease of hapticity increases stability, which I doubt (more delocalization=more stability, not vice versa). Also, the concept of a &quot;bent benzene&quot; is a bit strange to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what exactly is going on in this reaction and why it happens.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;@F'x mentioned that the bent benzene may just be a way of drawing the molecule to clearly show the change of overlap. If this is the case, I'd still like to know how the hapticity changed (because my initial logic behind change of hapticity is invalid if the structure doesn't actually bend).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.270" LastActivityDate="2015-12-05T23:03:33.827" Title="Why does benzene bend in this reaction?" Tags="&lt;redox&gt;&lt;aromatic-compounds&gt;&lt;coordination-compounds&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="301" PostTypeId="2" ParentId="299" CreationDate="2012-05-08T13:40:28.917" Score="10" Body="&lt;p&gt;First, Wikipedia doesn't really say what is the extent of this change. Here are superimposed a benzene and a benzyne molecule (own HF calculation, don't ask for the details):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7OBgD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon atoms involved in the triple bond are displaced by less than $\pu{0.2 Å}$. In benzene, we have with this level of calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C–C}$ distance = $\pu{1.39 Å}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C–H}$ distance = $\pu{1.08 Å}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C–C–C}$ angle = $120^\circ$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;while in benzyne, we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C–C}$ triple bond = $\pu{1.22 Å}$&lt;/li&gt;&#xA;&lt;li&gt;other $\ce{C–C}$ bonds = $\pu{1.38 Å}$ (almost all equal)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, merely because the triple bond is stronger, the distance will be shorted and the benzyne structure is distorted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding orbitals, well, the p orbitals of the carbon atoms involved in the triple bond are “naturally” directed towards where the hydrogens would be. Slightly distortion of the system is possible, and indeed evidence above on the structure, but they will not be parallel, as that would disrupt the structure too much and diminish the $\sigma$ bond. So it's a competition between: making the new p orbitals “more parallel” increases overlap, but it weakens the $\sigma$ bonding pattern, which is much stronger. As such, while some distortion happens, you stay mostly close to the benzene-type orbitals.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5026" LastEditDate="2018-04-04T16:26:42.073" LastActivityDate="2018-04-04T16:26:42.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="302" PostTypeId="2" ParentId="298" CreationDate="2012-05-08T15:41:07.520" Score="34" Body="&lt;p&gt;To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) and a set of antibonding molecular orbitals (which are higher in energy). Bonding orbitals are generated from sets of $p_{z}$ orbitals which are mostly or totally in phase, whereas the highest energy antibonding orbital of the set is composed of $p_{z}$ orbitals that are totally out of phase with respect to their adjacent neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, for instance, is the MO that is generated from the totally in-phase set (I calculated this at a spin-restricted RI-BP86/6-311G* level of theory in ORCA&lt;sup&gt;(1)&lt;/sup&gt; and visualised the isosurfaces in VMD&lt;sup&gt;(2)&lt;/sup&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1XkF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this MO is shared equally across all of the carbons, i.e. it is not localised. However it only contains 2 electrons and thus satisfies the exclusion principle. (F'x mentions spin-orbitals - this is where the orbitals are split on the basis of their spin, which entails a single electron per orbital. Benzene is a closed-shell singlet so we would expect the $\alpha$ and $\beta$ spin orbitals to be spatially indistinguishable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Not to sound like a walking advertisment for Housecroft and Sharpe, but H&amp;amp;S has a great visual introduction to MO theory.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;P.S.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It was suggested in chat that I expand upon the generated orbitals - here's one of the two highest occupied molecular orbitals, from the same calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0rFd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there's a big node along one of the mirror planes. This orbital is higher in energy than the earlier example because it's composed of two antiphase sets of 3 $p_{z}$ orbitals, so this orbital has antibonding character for two of the carbon-carbon pairs. It should be easy to extrapolate to the highest lying MO of this type - antibonding all the way around.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) Neese, F. ORCA – an ab initio, Density Functional and Semiempirical program package, Version 2.6. University of Bonn, 2008&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Humphrey, W., Dalke, A. and Schulten, K., 'VMD - Visual Molecular Dynamics', J. Molec. Graphics 1996, 14.1, 33-38.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-08T16:41:33.950" LastActivityDate="2012-05-08T16:41:33.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="303" PostTypeId="2" ParentId="289" CreationDate="2012-05-08T23:48:03.063" Score="4" Body="&lt;p&gt;This sounds like A being an &lt;a href=&quot;http://en.wikipedia.org/wiki/Internal_standard&quot; rel=&quot;nofollow&quot;&gt;internal standard&lt;/a&gt; for the determination of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal standards are used to get rid of (small) multiplicative variation due to certain influences, e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in GC, the injected volume&lt;/li&gt;&#xA;&lt;li&gt;in optical spectroscopy, the optical path length or the illuminated volume.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sometimes the calculation part of the idea is called &lt;em&gt;normalization&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The English Wiki page for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Response_factor#Chromatography&quot; rel=&quot;nofollow&quot;&gt;Response factor&lt;/a&gt; explains the internal standard (and calculations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important part is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One of the main reasons to use response factors is to compensate for the irreproducibility of manual injections into a gas chromatograph (GC). Injection volumes for GC's can be 1 microliter (µL) or less and are difficult to reproduce. Differences in the volume of injected analyte leads to differences in the areas of the peaks in the chromatogram and any quantitative results are suspect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you switch your whole calibration to response normalized to the response of the internal standard (= signal analyte / signal internal standard) and the result is concentration analyte / concentration of internal standard = molar amount analyte / molar amount internal standard. With the known amount/concentration of internal standard you can calculate back to the analyte amount/concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can read German, here's  &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/3/anc/croma/datenauswertung.vlu/Page/vsc/de/ch/3/anc/croma/datenauswertung/quantitativ/innererstandard/innerstandardm80ht0801.vscml.html&quot; rel=&quot;nofollow&quot;&gt;another explanation and HPLC example calculation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skoog &amp;amp; Leary's Instrumental Analysis says that with a suitable internal standard, accuracy of &amp;lt; 1 % is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if your A is an internal standard, then, yes, that is a proper technique.&#xA;Note, however, that arsenic does not sound like an internal standard for glucose (nor vice versa): the internal standard should be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;as similar as possible to the analyte&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;but the signal should not overlap with other substances&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for GC/MS that could even be an isotope labeled version of the analyte for maximum chemical similarity. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2012-05-10T11:21:56.613" LastActivityDate="2012-05-10T11:21:56.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="304" PostTypeId="2" ParentId="298" CreationDate="2012-05-09T01:37:12.347" Score="15" Body="&lt;p&gt;I very much like the first answer, especially the superb visuals!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I will nonetheless add a simpler approach to answering your question, which is this: Pauli exclusion holds as long as there is &lt;em&gt;something&lt;/em&gt; in space or in spin (spin being the pairs of electrons) that firmly distinguishes two &quot;stable&quot; states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that means is that your initial intuition is essentially correct if &lt;em&gt;all&lt;/em&gt; the electron pairs shared exactly the same molecular orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question in general terms is that electron pairs settle into molecule-sized orbitals (molecular orbitals) that are physically spread out over the entire ring, but which nonetheless have unique geometric features that keep them distinguished in real space. I would add that they are even more unique when represented in an abstract space called momentum space, where the geometric differences translate into actual distances between the electron pairs. These abstract distances in momentum space satisfy the Pauli exclusion principle just as well as physical separation in real space, so that even very similar looking orbitals can remain distinct under Pauli exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details and some easy visual analogies, you can read more in the addendum below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The same general problem of how to distinguish electron pairs to meet Pauli exclusion exists in metals. In that case, the conduction electrons are free to drift over the entire macroscopic metal. The result is an even more extreme need than in benzene for the pairs to distinguish themselves firmly from each other, since for conduction electrons &lt;em&gt;all&lt;/em&gt; of these gazillions [technical term] of pairs must find &quot;unique places&quot; in which they can maintain their unique identity. That's tough!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it turns out that these two problems of benzene electrons and metallic conduction electron pairs are related in a rather deep fashion, since a benzene ring can be interpreted (quite correctly) as a special tiny case of a one-dimensional metal. That means that if you understand what happens conceptually in metals, you are well on the way to understanding all forms of hybridization in organic molecules. So, let's take a quick look at how that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation for electron pairs in metals is most easily understood by imagining the pairs moving back and forth along the $x$ length of a very narrow, highly conductive, and unusually regular metallic wire. Next, borrow the $y$ and $z$ plane to represent the complex plane (complex phase of a wave function). The amplitude for finding an electron from the pair becomes the radius of the complex plain at each point along $x$. For stable electron pair states, these radii and their phases will exhibit a lot of continuity as you move their as along the $x$ length of the wire. They will also rotate at the same speed in the complex plane of $y$ and $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact... if you calculate the stable or &quot;stationary&quot; states possible in such a configuration, you will find that they bear a striking resemblance both visually and mathematically to skip rope solutions, that is, to the ways that you can twirl a skip rope so that it maintains stable, unmoving (the &quot;stationary&quot; part) loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's look at that situation as a good analog solution to the quantum equations. Anyone can make one large loop, and that in fact represents the electron pair state with the lowest possible energy along the length of the wire, or the base state. Incidentally, the volume enclosed by the skip rope is proportional to the square of the radius, which as I mentioned earlier is the &quot;amplitude&quot; of the wave function. So... that means that if you simply look at the volume enclosed by the skip rope as it turns, you pretty much instantly know what the probability of finding an electron from the pair is. In the case of the base state, it means that the electron pair will tend to be found in the center of the wire, and almost never at the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry such varying probabilities of finding electrons are usually represented as densities in ordinary space, without the benefit of the dynamics of the complex plane. And for good reason, since for anything except wires, trying to show the complex plane would result in five-dimensional spaces. Those are tough to visualize! Instead, the idea of a &lt;em&gt;phase&lt;/em&gt; is attached to tell the relative positions of the rotating points in the unseen complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skip rope in the one-dimensional wire makes the idea of phase a lot easier to visualized. For example, good skip-ropers can easily create double loop skip ropes. These correspond to the first excited state of the electron pair, one with a bit more energy added (you have to twirl the rope faster) and two distinct loops. Phase just means this for such a double loop: When one loop is towards you, the other is opposite of you. You can label one such loop positive and one negative, but of course it's all arbitrary, and really just a technique for roughly describing those invisible but rope-like complex plane dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the volume-equals-probability rule still applies to the double loop. That means that in the first excited state of the electron pair, the odd are good and roughly equal for finding it in the left loop or right loop, but close to zero at the ends &lt;em&gt;and&lt;/em&gt; at the center. So, this really is a unique state for the excited electron pair, one that is quite distinct from the first single-loop or base state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's an interesting bit of mathematics, Fourier transforms, in which &lt;em&gt;any&lt;/em&gt; very long sinusoidal shape can always be distinguished very sharply from any other sinusoidal, even if they only differ slightly in wavelength. People use this idea all the time, since in the case of radio waves (which are sinusoidal waves, mostly) it's called tuning a radio!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uniqueness of long sinusoidal waves is relevant here because it turns out that Pauli exclusion is met equally well if two pairs of electrons are separated from each other in &lt;em&gt;frequency&lt;/em&gt; space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, another way of saying all of this is that when electrons are &quot;delocalized&quot; either in bonds or in metals -- it really is the same idea -- they can achieve uniqueness by bumping around &lt;em&gt;either&lt;/em&gt; in ordinary space &lt;em&gt;or&lt;/em&gt; in a more abstract frequency space called momentum space. Both work equally well. The &quot;extra quantum number&quot; you speculated about in your question could be equated to this idea of additional separation in momentum space for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the benzene orbitals maintain Pauli exclusion uniqueness by finding specific stable solutions that are similar but geometrically distinct, and which are even more unique when interpreted in terms of frequencies or loop counts. Those equate to a quite different-looking, distance-like separations in the abstract space called momentum space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to finish up, what about that extreme case I mentioned of metals that have an electron or two for &lt;em&gt;every&lt;/em&gt; atom in what may be a very large crystal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a truly fascinating case, a sort of &quot;benzene gone bad&quot; scenario in which the need for Pauli exclusion uniqueness pushes the electron pairs into absolutely bizarre behaviors, ones very important to modern (and ancient) technologies: Some of the electrons are forced to get very hot. Now that's weird!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is easy to see with the above background. The first pair of electrons in a silver wire is very, very cold, near absolute zero, moving so slowly that it forms only a single loop in the center of the wire. To maintain Pauli exclusion -- the momentum space version of it to be precise -- the next pair must look like a skip rope with two loops, and the third three, and so forth. You get the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But each such loop has more energy, even if only by a little bit! So what happens when you have a gazillion such electron pairs all vying for uniqueness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone gets pushed very, &lt;em&gt;very&lt;/em&gt; high up in frequency, and thus also in energy, while other lurk in the deeper and colder regions of this pile of moving electrons. The result is almost like a sea in momentum space, and that is in fact what it is called: A Fermi sea, with cold electron pairs at the bottom, and some truly &lt;em&gt;hot&lt;/em&gt; ones at the surface. Fermi seas are also referred to by the more common phrase of &quot;conduction bands,&quot; meaning the range of energies that the electrons occupy in order to maintain Pauli uniqueness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see Fermi surfaces all the time, since it is &lt;em&gt;only&lt;/em&gt; the hottest electrons in a metal that interact with photons and bounce them back at you. You are in effect looking directly at a hot Fermi surface every time you look into a silver or aluminum backed mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How hot? Well, believe it or not, X-ray hot. A typical piece of ordinary shiny metal, say a fork or spoon, would kill you from X-ray exposure if all of its hot Fermi surface and near-surface electrons could suddenly fall back into a &quot;ground state&quot; of just one loop. They are tamed and prevented from doing so by all those other cooler electrons that refuse to let them fall to the bottom of the Fermi sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with that I'll close. I realize the parts about metals are a bit remote from the original chemistry of benzene rings. But then again, as you get into longer and longer carbon chains such as the ones that carry off electrons from chlorophyll molecules, you may find that recognizing the &quot;almost metallic&quot; nature of multiple pairs and the way they interact in momentum space can be a good shorthand for understanding how some pretty complicated forms of large-molecule dynamics play out.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2012-05-09T22:34:41.060" LastActivityDate="2012-05-09T22:34:41.060" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="305" PostTypeId="2" ParentId="297" CreationDate="2012-05-09T07:09:37.460" Score="8" Body="&lt;p&gt;For those who have never had the pleasure of personally doing this, see &lt;a href=&quot;http://www.youtube.com/watch?v=o9KT7lM1UHU&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; video. It has been known since 1807 that dissolving sodium in liquid ammonia results in a beautiful color. It was originally thought the color was due to some familiar complex instead of a solvated electron. A similar phenomenon happens with other alkali metals in ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research in the field says that it takes, at least, forty some ammonia molecules to solvate a given electron. Metstable cavities form and their stability could depend highly on electrostatic interactions to solvate our electron. Ammonia will slowly react by evolving hydrogen gas,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NH3 + 2 e- -&amp;gt; H2 + 2 NH2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differences in solvent can play a huge role in the stability of our solvated electron. An analogous decomposition occurs in water,&#xA;$$\ce{2 H2O + 2 e- -&amp;gt; H2 + 2 HO-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Famously the latter occurs faster than the former, see &lt;a href=&quot;http://www.youtube.com/watch?v=HY7mTCMvpEM&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; in comparison to my earlier link. We could say that these differences in rate reflect different stabilities of our solvated electron. Although, the addition of an appropriate catalyst to our ammonia will result in rapid evolution of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more quantitative description for the difference in energies is obtained by measuring the UV-vis spectrum for a solvated electron in both water and ammonia. One will quickly notice that the band appears at higher energies in water than it does in ammonia and there are considerable differences in the band shapes (the band is wider in ammonia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So Why Do They Differ&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here I have to make the disclaimer that no current theories quantitatively reproduce the observed phenomena, such as the absorption spectrum for a solvated electron. This question has no established/accepted answer as it is an ongoing area of research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The self-ionization of water has an equilibrium constant on the order of $K = 10^{-14}$ and that of ammonia is on the order of $10^{-30}$. Hydronium formation occurs in the case of water and ammonium in the case of ammonia. Hydronium has a considerably lower $\mathrm{p}K_\text{a}$ than ammonium and so it’s reasonable to see why a reaction in water would be more likely with an electron. (Ammonia rarely produces a weakly acidic species, but water often produces a strongly acidic complex.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly deeper reason may be that water forms more ordered local domains/structures in solution than ammonia and this influences the rate by resulting in a larger entropy of activation when the hydrolyzed electron breaks up these larger structures, recall that $k \propto \exp (\Delta S ^\ddagger /R) $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just speculation though.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T12:02:10.797" LastActivityDate="2015-05-20T12:02:10.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="306" PostTypeId="1" AcceptedAnswerId="309" CreationDate="2012-05-09T09:38:00.300" Score="11" ViewCount="971" Body="&lt;p&gt;In “periodic table”, the adjective is related to the noun &lt;em&gt;period&lt;/em&gt;, and &lt;a href=&quot;http://www.etymonline.com/index.php?term=periodic&amp;amp;allowed_in_frame=0&quot; rel=&quot;noreferrer&quot;&gt;comes from&lt;/a&gt; Ancient Greek &lt;em&gt;περίοδος&lt;/em&gt; through French &lt;em&gt;périodique&lt;/em&gt;. In “periodic acid”, it is formed from the prefix &lt;em&gt;per-&lt;/em&gt; and &lt;em&gt;iodic&lt;/em&gt; (like &lt;em&gt;peroxide&lt;/em&gt; and &lt;em&gt;permanganate&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wiktionary.org/wiki/periodic#Etymology_1&quot; rel=&quot;noreferrer&quot;&gt;Wiktionary&lt;/a&gt; lists their respective UK pronunciations of as &lt;code&gt;/pɪə(ɹ).iˈɒdɪk/&lt;/code&gt; and &lt;code&gt;/ˌpɜːraɪˈɒdɪk/&lt;/code&gt;, markedly different: &lt;code&gt;pɪə&lt;/code&gt; (as in &lt;em&gt;piece&lt;/em&gt;) vs. &lt;code&gt;pɜː&lt;/code&gt; (as in &lt;em&gt;perfume&lt;/em&gt;); then &lt;code&gt;i&lt;/code&gt; (as in &lt;em&gt;it&lt;/em&gt;) vs. &lt;code&gt;aɪ&lt;/code&gt; (as in &lt;em&gt;eye&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-05-15T21:19:56.680" LastActivityDate="2017-11-08T13:15:53.947" Title="Should one pronounce “periodic” the same in “periodic acid” and “periodic table”?" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="307" PostTypeId="1" CreationDate="2012-05-09T09:39:06.290" Score="13" ViewCount="3181" Body="&lt;p&gt;Turbostratic graphite is graphite in which there is quenched rotational disalignment between adjacent graphene sheets, i.e. one sheet is rotated with respect to its neighbor. I suppose this could be considered a crystallographic defect of sorts. How do these turbostratic layers form, and can they possess long range order, such as glide or helical symmetry (e.g. similar to a twisted nematic phase)?&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="7448" LastEditDate="2015-01-01T18:33:05.330" LastActivityDate="2015-01-01T18:33:05.330" Title="How are turbostratic graphite phases formed?" Tags="&lt;solid-state-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="308" PostTypeId="1" AcceptedAnswerId="370" CreationDate="2012-05-09T09:48:40.473" Score="6" ViewCount="20957" Body="&lt;p&gt;What is the difference between melting and dissolving? I am looking some general features. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below (or at the) melting point of ice, or similar phenomena. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now assuming that the ice is dissolving and melting at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words which reaction energy is higher in the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xD;&#xA;\ce{H_2O(s) &amp;amp;\to H_2O(l)\\&#xD;&#xA;H_2O(s) + n EtOH(l) &amp;amp;\to H_2O \centerdot (EtOH)_{n}}&#xD;&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, are there substances that release more energy in dissolvation than consume in melting?&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="11" LastEditDate="2012-05-13T14:03:12.820" LastActivityDate="2015-10-26T18:53:50.203" Title="What is the difference between melting and dissolving?" Tags="&lt;solvents&gt;&lt;solubility&gt;&lt;melting-point&gt;&lt;phase&gt;" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="309" PostTypeId="2" ParentId="306" CreationDate="2012-05-09T09:54:00.267" Score="12" Body="&lt;p&gt;I am an Australian English speaker and yes, this distinction is made in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly the only reasonable opportunity to use the &lt;code&gt;pɪə&lt;/code&gt; pronunciation in the name of a compound is in the case of the entertaining molecule &lt;em&gt;periodane&lt;/em&gt;, which is actually named after the periodic table. This &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.20948/abstract&quot; rel=&quot;noreferrer&quot;&gt;molecule&lt;/a&gt; (and &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.21322/abstract&quot; rel=&quot;noreferrer&quot;&gt;later&lt;/a&gt; a number of different plausible isomers) was identified computationally by a methodology called 'mindless chemistry' which optimises randomly generated molecular graphs. &lt;em&gt;Periodane&lt;/em&gt; is a stable configuration of each atom on the second row of the periodic table, with the exception of neon (although some people are &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0166128008007677&quot; rel=&quot;noreferrer&quot;&gt;working&lt;/a&gt; on that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qpeqX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-09T09:57:59.730" LastActivityDate="2012-05-09T09:57:59.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="310" PostTypeId="2" ParentId="306" CreationDate="2012-05-09T09:57:37.353" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not a professional chemist, just a student, but here are some observations from which I can speculate:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note-- in this post,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;/ˌpɜːraɪˈɒdɪk/&lt;/code&gt;==&lt;code&gt;per-iodic&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;/pɪə(ɹ).iˈɒdɪk/&lt;/code&gt;==&lt;code&gt;peer-iodic&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(since IPA is annoying to read)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, it's written as &quot;per-iodic&quot; acid many times. Which means it must be pronounced as &quot;per-iodic&quot; by whoever who reads/writes those texts. So we have &lt;em&gt;some&lt;/em&gt; evidence showing the use of &quot;per-iodic&quot; as a pronunciation. I don't see any such evidence for &quot;peer-iodic&quot;, though one of my teachers used it. I myself initially used &quot;peer-iodic&quot;, but once I realized what the acid actually was, I switched to &quot;per-iodic&quot;.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-09T09:57:37.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="311" PostTypeId="2" ParentId="300" CreationDate="2012-05-09T10:31:02.763" Score="14" Body="&lt;p&gt;That diagram is misleading because it’s technically in reference to bis(hexamethylbenzene)ruthenium(0). This is a famous example by the late and great E.O. Fischer. This Ru(0) complex is 18e, three methyl resonances were observed for the bent ring at −10 °C. Crystallography later confirmed this. This structure is surprising to the new student, perhaps, because the classic 18e bis(benzene)chromium is sandwiched between two planar arenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an organic chemistry question. It is deep in the realm of organometallic chemistry, with a proper explanation requiring a molecular orbital energy diagram and then one has to invoke a Jahn Teller instability argument that requires lowering symmetry from $D_{6d}$ or $D_{6h}$. Suffice it to say, stability is increased by changing the hapticity.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T19:32:02.057" LastActivityDate="2015-04-24T19:32:02.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="312" PostTypeId="1" AcceptedAnswerId="490" CreationDate="2012-05-09T10:35:08.833" Score="16" ViewCount="2202" Body="&lt;p&gt;I have a degree in chemical engineering and remember having seen a inductively coupled plasma mass spectrometer (ICP-MS) during the laboratory part of a chemistry course I took while at university, but it is many years ago. Now I would like to know what a typical output from a ICP-MS looks like and how I interpret the results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no picture at the wikipedia page but some frequency domain like plots can be found with a Google search and I understand the Fourier transform and how it is used in number theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the ICP-MS results anything like the spectrum from the Fourier transform of some signal?&lt;/p&gt;&#xA;" OwnerDisplayName="anon" LastEditorDisplayName="anon" LastActivityDate="2012-05-22T18:33:27.227" Title="What do typical results from inductively coupled plasma mass spectrometry (ICP-MS) look like, and how do I interpret them?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="313" PostTypeId="1" CreationDate="2012-05-09T13:33:52.940" Score="5" ViewCount="488" Body="&lt;p&gt;Mercury is toxic. It forms amalgams and inorganic mercury compounds/salts are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is on toxicity. I am interested in its toxicity, not the mode of action (as it will be part of biology SE)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is amalgam toxic? (sort of yes / no)&lt;/li&gt;&#xA;&lt;li&gt;Mercury salts are toxic? (sort of Yes / no)&lt;/li&gt;&#xA;&lt;li&gt;If mercury salts / amalgams are non-toxic, what makes them non-toxic. Is it because of the ionic nature / other metal or any other reason?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="92" LastEditorUserId="11" LastEditDate="2012-05-10T06:36:13.793" LastActivityDate="2012-05-14T00:05:50.780" Title="Mercury, amalgams and mercury compounds" Tags="&lt;metal&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2012-05-23T05:01:08.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="314" PostTypeId="1" AcceptedAnswerId="315" CreationDate="2012-05-09T13:41:11.803" Score="16" ViewCount="20925" Body="&lt;p&gt;I have been reading about resolution in mass spectrometry and there are a few things I do not understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolving power is determined by $m/(m_2-m_1)$ at full width half height of a peak. The higher the value obtained the better. This is where I struggle to form a link between the resolution of the mass spectrometer and the resolving power. In image (a) below the resolution of the mass spectrometer is 1000 and would have a very low mass resolution value, but in image (b) the resolution is 5000 and the nominal peaks masses are separated, therefore providing a high mass resolution value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can a mass spectrometer with a resolution of 5000 resolve the nominal masses and the mass spectrometer with a resolution of 1000 cannot? And how does the mass resolution come into play? Also, what do the values of 1000 and 5000 mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xKyH9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="168" LastEditorDisplayName="user7951" LastEditDate="2015-03-30T16:43:34.380" LastActivityDate="2015-03-30T16:43:34.380" Title="Resolution in Mass Spectrometry" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="315" PostTypeId="2" ParentId="314" CreationDate="2012-05-09T18:16:57.430" Score="14" Body="&lt;p&gt;I'm not sure I completely understand what exactly your problem is, but:&lt;/p&gt;&#xA;&lt;p&gt;You want to resolve to signals at approx. m/z = 1060.2 and m/z = 1061.2.&#xA;So &lt;span class=&quot;math-container&quot;&gt;$\Delta m = 1$&lt;/span&gt; The required resolution is: &lt;span class=&quot;math-container&quot;&gt;$\frac{m}{\Delta m} = \frac{1061}{1} \approx 1060$&lt;/span&gt;: The spectrometer in (a) has only 1000 which is too low. The spectrometer in (b) has 5000 which is quite more than needed.&lt;/p&gt;&#xA;&lt;p&gt;Maybe it is easier to think of the reciprocal: &lt;span class=&quot;math-container&quot;&gt;$\frac{\Delta m}{m}$&lt;/span&gt; which tells you how far two signal need to be apart so your spectrometer can resolve them.&lt;/p&gt;&#xA;&lt;p&gt;In (a) this is &lt;span class=&quot;math-container&quot;&gt;$\frac{\Delta m}{m} = \frac{1}{1000} = \frac{1.06}{1060}$&lt;/span&gt; two signals can be resolved if they are at least 1.06 m/z apart at m = 1060. I.e. if they were slightly further apart than they are.&lt;/p&gt;&#xA;&lt;p&gt;The spectrometer in (b) can resolve signals that are at least  &lt;span class=&quot;math-container&quot;&gt;$\Delta m/z =  0.21$&lt;/span&gt; apart at  &lt;span class=&quot;math-container&quot;&gt;$m/z = 1060$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;FWHM and resolution&lt;/h3&gt;&#xA;&lt;p&gt;The FWHM is basically part of the resolution. The &lt;span class=&quot;math-container&quot;&gt;$\Delta m$&lt;/span&gt; is the m/z difference between two signals that are just resolved.&lt;/p&gt;&#xA;&lt;p&gt;Have a look at the &lt;a href=&quot;http://goldbook.iupac.org/R05318.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC's definitions of resolution in mass spec&lt;/a&gt;. There are several approaches to specify what &amp;quot;just resolved&amp;quot; means.&lt;/p&gt;&#xA;&lt;p&gt;The FWHM can be used directly as the &lt;span class=&quot;math-container&quot;&gt;$\Delta m$&lt;/span&gt; of the &amp;quot;peak width definition&amp;quot;.&#xA;If instrument A has &lt;span class=&quot;math-container&quot;&gt;$5\times$&lt;/span&gt; the FWHM than instrument B, then resolution of A  &lt;span class=&quot;math-container&quot;&gt;$\approx \frac{1}{5}$&lt;/span&gt; resolution of B.&lt;/p&gt;&#xA;&lt;p&gt;I'm more used to the &amp;quot;10 % valley&amp;quot; which roughly equals the full with at 5 % height: you specify how much signal is allowed at the minimum between the two peaks, e.g. 10 %.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;It is important to realize that these two definitions differ by almost a factor 2!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here are some results of &amp;quot;simulated&amp;quot; peaks at different resolutions and &lt;span class=&quot;math-container&quot;&gt;$\Delta m/z$&lt;/span&gt; (all gaussian, which of course doesn't need to be the case in reality). I use the full width at 5% = 10 % valley definitions for the resolution.&lt;/p&gt;&#xA;&lt;p&gt;The first row is resolution 5000 with the, second row is resolution 1000. You see that the two signals are much better resolved than in your example picture above. Which means that above another resolution definition was used, possibly &lt;span class=&quot;math-container&quot;&gt;$\Delta m$&lt;/span&gt; = FWHM. In any case, a resolution of 1000 should be &lt;em&gt;almost sufficient&lt;/em&gt; to resolve m/z 1060.2 from 1061.2.&lt;/p&gt;&#xA;&lt;p&gt;The third row shows two signals of equal height with &lt;span class=&quot;math-container&quot;&gt;$\Delta m/z $&lt;/span&gt; = 1 = FWHM. According to the resolution definitions above, this is a resolution of ca. 520. Note that while you can say that there are at least two peaks, you cannot take the two maxima as the m/z of the two underlying signals (grey).&lt;/p&gt;&#xA;&lt;p&gt;The fourth row is almost the same, but I multiplied the second signal by 56% to account for the frequency of having bradykinin with one &lt;span class=&quot;math-container&quot;&gt;$^{13}$&lt;/span&gt;C. Looks quite similar to your picture.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/luF3q.png&quot; alt=&quot;example calculation for mass spec resolution&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, have a look at &lt;a href=&quot;http://www.chromatographyonline.com/lcgc/Column%3A+MS+in+Practice/Debating-Resolution-and-Mass-Accuracy/ArticleStandard/Article/detail/88264&quot; rel=&quot;noreferrer&quot;&gt;M. P. Balogh: Debating Resolution and Mass Accuracy, LC•GC Europe, 17(3), 152–159 (2004)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-05-12T15:51:10.757" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="316" PostTypeId="1" AcceptedAnswerId="336" CreationDate="2012-03-16T10:41:09.000" Score="57" ViewCount="3230" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot;&gt;Density Functional Theory&lt;/a&gt; courses, one is often reminded that &lt;a href=&quot;http://en.wikipedia.org/wiki/Kohn%E2%80%93Sham_equations&quot;&gt;Kohn-Sham orbitals&lt;/a&gt; are often said to bear no any physical meaning. They only represent a noninteracting reference system which has the same electron density as the real interacting system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are plenty of studies in that field’s literature that given KS orbitals a physical interpretation, often after a disclaimer similar to what I said above. To give only two examples, KS orbitals of &lt;a href=&quot;http://www.roaldhoffmann.com/pn/modules/Downloads/docs/444s.pdf&quot;&gt;H&lt;sub&gt;2&lt;/sub&gt;O&lt;/a&gt;&lt;sup&gt;[1]&lt;/sup&gt; and &lt;a href=&quot;https://wiki.fysik.dtu.dk/dacapo/Solution_3#kohn-sham-wavefunctions-of-a-co-molecule&quot;&gt;CO&lt;sub&gt;2&lt;/sub&gt;&lt;/a&gt; closely resemble the well-known molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I wonder: &lt;strong&gt;What good (by virtue of being intuitive, striking or famous) examples can one give as a warning of interpreting the KS orbitals resulting from a DFT calculation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] “What Do the Kohn-Sham Orbitals and Eigenvalues Mean?”, R. Stowasser and R. Hoffmann, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;121&lt;/em&gt;, 3414–3420.&lt;/p&gt;&#xA;" OwnerUserId="30" OwnerDisplayName="F'x" LastEditorUserId="7448" LastEditDate="2015-12-14T12:21:31.010" LastActivityDate="2015-12-14T12:21:31.010" Title="Striking examples where Kohn-Sham orbitals clearly have no physical meaning" Tags="&lt;computational-chemistry&gt;&lt;orbitals&gt;&lt;density-functional-theory&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="317" PostTypeId="1" AcceptedAnswerId="319" CreationDate="2012-05-09T21:02:06.917" Score="3" ViewCount="8450" Body="&lt;p&gt;Is it possible for a dissolution reaction of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{X(s) -&amp;gt;[solvent] X (solv)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{M_mX_x (s) -&amp;gt;[solvent] m M^x+ (solv) + x X^m- (solv)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be endothermic, i.e. &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r H &amp;gt; 0$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="16683" LastEditDate="2019-03-10T23:01:59.353" LastActivityDate="2019-03-10T23:02:15.087" Title="Can dissolution reactions be endothermic?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="318" PostTypeId="2" ParentId="72" CreationDate="2012-05-09T21:08:02.863" Score="8" Body="&lt;p&gt;I'm not sure why you're wanting to exert so much effort to find the charge of compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to know the charge, determine it more directly. Electrophoresis can allow you to do so. A charged species is put in a capillary tube that is in between a cathode an an anode. When a charged species is subjected to an electric field, it will experience a force proportional to its charge and the field strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag forces present in the solvent are also a consideration in this setup and they depend on the &quot;size&quot; of your ion as well as viscosity of your solvent. One also has to worry about Joule heating, as an increase in temperature will cause poor reproducibility. Picking a suitable detector is fun, but hopefully you can get by with the common UV-vis detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you can find a suitable setup, know the radius of your ion and know the time between injection of your sample and its arrival at a detector, you can at least determine its effective charge.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2012-05-10T01:17:56.577" LastActivityDate="2012-05-10T01:17:56.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="319" PostTypeId="2" ParentId="317" CreationDate="2012-05-09T21:10:39.607" Score="8" Body="&lt;p&gt;You will find the answer in probably any textbook on chemical thermodynamics, or in a thermochemistry chapter in most physical chemistry textbooks, or in Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_change_of_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;enthalpy of dissolution&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dissolution can be viewed as occurring in three steps:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.&lt;/li&gt;&#xA;  &lt;li&gt;Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding&lt;/li&gt;&#xA;  &lt;li&gt;Forming solvent-solute attractions (exothermic), in solvation.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;The value of the enthalpy of solution is the sum of these individual steps. Dissolving ammonium nitrate in water is endothermic. The energy released by solvation of the ammonium ions and nitrate ions is less that the energy absorbed in breaking up the ammonium nitrate ionic lattice and the attractions between water molecules. Dissolving potassium hydroxide is exothermic, as more energy is released during solvation than is used in breaking up the solute and solvent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it can have any sign. The linked page also gives data for common compounds in water, some of them endothermic, some of them exothermic. Ammonium nitrate, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4NO3}$&lt;/span&gt;, is a good example of a salt which dissolves in water in an endothermic reaction: if you dissolve some &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4NO3}$&lt;/span&gt; in water, the drop in the temperature will be noticeable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding why the substance is soluble why its dissolution is endothermic, you have to remember that the reaction takes place if &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r G$&lt;/span&gt; is favourable (i.e. negative), and &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm r G = \Delta_\mathrm r H - T \Delta_\mathrm r S$&lt;/span&gt;. Overall the free energy must be negative for dissolution to occur (on a thermodynamic basis; kinetics are another issue), not the enthalpy.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2019-03-10T23:02:15.087" LastActivityDate="2019-03-10T23:02:15.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="320" PostTypeId="5" CreationDate="2012-05-09T21:33:09.197" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-05-09T21:33:09.197" LastActivityDate="2012-05-09T21:33:09.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="321" PostTypeId="4" CreationDate="2012-05-09T21:33:09.197" Score="0" Body="Density Functional Theory" OwnerUserId="177" LastEditorUserId="177" LastEditDate="2012-05-09T21:45:56.973" LastActivityDate="2012-05-09T21:45:56.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="322" PostTypeId="2" ParentId="158" CreationDate="2012-05-09T23:45:39.070" Score="6" Body="&lt;p&gt;It's not exactly clear from your question what distances you consider, but I have taken snapshots of a methylcyclohexane with the methyl group in the equatorial and axial positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ME9Ab.png&quot; alt=&quot;enter image description here&quot;&gt;      (equatorial)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnPW4.png&quot; alt=&quot;enter image description here&quot;&gt;    (axial)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance of the methyl $\ce{H}$ atoms to the nearest cyclohexane $\ce{H}$ atoms are displayed, and you'll see that indeed the axial position has a shorted $\ce{H–H}$ distance of ~ 2 Å, while equatorial position has longer $\ce{H–H}$ distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as others have remarked in the comments: only the groups (and $\ce{H}$ atoms) directly on the cyclohexane ring are said to be in axial or equatorial position. Atoms of side chains do not follow this terminology, in particular for side chains that can rotate freely as it means nothing there.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T19:05:05.990" LastActivityDate="2015-05-13T19:05:05.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="323" PostTypeId="2" ParentId="289" CreationDate="2012-05-10T03:15:51.057" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Is this even how GC/MS results work?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As cbeleites said the method you described is a proper technique but not likely to be appropriate given the information you cited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In GC/MS you should have two sets of information. The first is the GC Total Ion Chromatograph (TIC) which will have time as the x-axis and response (abundance) as the y-axis. For each retention time on the TIC there will be a corresponding Mass Spectra (MS). In the MS the x-axis is m/z (ion) and y-axis is also response (abundance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different compounds have different responses so if you inject the exact same amount of two different compounds you could get a much larger response from one compared to the other. For example tramadol gives much higher response compared to hydrocodone. This is why you want your internal standard to be structurally similar to your analyte. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantification of a compound is often done by running 3 - 5 calibrators at known concentrations in order to make a calibration curve. Once an acceptable calibration curve is made the sample along with controls can be fit onto the curve to calculate the amount of the compound of interest in the sample and controls. If the controls are correct then the sample value can be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a calibration curve you need an internal standard in each calibrator, control and sample. You can then make the curve either with ion ratios between paired internal standard and analyte ions OR if your internal standard response is consistent in all the calibrators, controls and sample you can use the GC peak response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion if you only have one GC/MS run with the two dissimilar compounds in it you will not be able to easily calculate the quantity of the second compound based on the GC response of the first unless you have additional information that was not listed in this question.&lt;/p&gt;&#xA;" OwnerUserId="180" LastActivityDate="2012-05-10T03:15:51.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="324" PostTypeId="1" AcceptedAnswerId="325" CreationDate="2012-05-10T04:16:03.393" Score="19" ViewCount="781" Body="&lt;p&gt;Many computational chemistry packages permit the calculation of vibrational and electronic spectra. These spectra are obtained as a set of discrete eigenvalues however they are often convolved with some distribution to give a continuous spectrum that is realistic for a finite temperature, rather than a sequence of impulse functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ADF package allows the user to convolve the spectrum with either a Gaussian or Cauchy (Lorentzian) function. My understanding is that the latter affords a more realistic line broadening, however the Gaussian must be there for a reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which spectral broadening scheme is preferred? Why the choice?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-10T04:38:45.743" Title="When simulating spectral line broadening, which convolution is preferred?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="325" PostTypeId="2" ParentId="324" CreationDate="2012-05-10T04:38:45.743" Score="15" Body="&lt;p&gt;I'm not familiar with the computational packages -- I'm an experimentalist, not a theoretician. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example for why both might be present, we can turn to gaseous infrared spectroscopy:&#xA;one will quickly find that the line-widths in vibration-rotation spectra depend on pressure. At lower pressures, say less than a torr, Doppler-broadening is the main mechanism. This is Gaussian in nature. While at higher pressures the broadening is instead due to collisions, which are Lorentzian. You'll find combinations of the two and so forth depending on the specifics of your case.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-05-10T04:38:45.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="326" PostTypeId="1" AcceptedAnswerId="327" CreationDate="2012-05-10T09:19:26.863" Score="11" ViewCount="6945" Body="&lt;p&gt;The identification of point groups of a molecule is usually done following a strict scheme, either manually or algorithmically. In all textbooks I could find, however, the first step of the scheme is actually not explicited: in this example scheme drawn from Housecroft and Sharpe,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W05SA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see that there is a very unhelpful step &lt;em&gt;“Does this molecule have $I_h$, $O_h$ or $T_d$ symmetry?”&lt;/em&gt;. It is implied that one would recognize e.g. an icosahedral molecular compound on sight. However, I wonder: how can one strictly identify these “special” point groups? What set of rules is to be followed (again, either manually or algorithmically)?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4945" LastEditDate="2014-06-12T08:45:10.093" LastActivityDate="2014-06-12T08:45:10.093" Title="How does one recognized Td/Oh symmetry in molecules?" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="327" PostTypeId="2" ParentId="326" CreationDate="2012-05-10T09:57:54.940" Score="9" Body="&lt;p&gt;You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this untitled &lt;a href=&quot;http://troglerlab.ucsd.edu/GroupTheory224/Chap1B.pdf&quot;&gt;document&lt;/a&gt;, which I presume&lt;sup&gt;1&lt;/sup&gt; is by W.C. Trogler, the elements are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$, $3\sigma{_h}$, $6\sigma{_d}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_{h}$: $E$, $6C_{5}$, $10C_{3}$, $15C_{2}$, $i$, $6S_{10}$, $10S_6$, $15\sigma$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you don't need to check for the identity symmetry $E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're a visual sort of person a symmetry spot check is to mentally superimpose a tetrahedron, cube or dodecahedron over the molecule and see whether the view down the surface normal of each face is identical. Cubes and octahedra are duals of each other, as are dodecahedra and icosahedra. The tetrahedron is self-dual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, H&amp;amp;S does not list chiral forms of these point groups, probably because they are so rarely encountered, however &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chir.20545/abstract&quot;&gt;researchers&lt;/a&gt; have come up with molecules that obligately satisfy $T$, $I$ and $O$ symmetry&lt;sup&gt;2&lt;/sup&gt; (I have not yet read the paper).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; I would be indebted to anyone who can furnish me with a full citation to this work, and confirm the authorship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; Narasimhan, S. K., Lu, X. and Luk, Y.-Y. (2008), Chiral molecules with polyhedral T, O, or I symmetry: Theoretical solution to a difficult problem in stereochemistry. Chirality, 20: 878–884.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-10T10:53:12.220" LastActivityDate="2012-05-10T10:53:12.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="328" PostTypeId="2" ParentId="308" CreationDate="2012-05-10T14:23:51.073" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is dissolving always exothermic? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it is not always exothermic. For example, dissolving $\ce{NaOH}$ is an exothermic reaction, while dissolving $\ce{NH4NO3}$ is an endothermic reaction (This is from my personal experience)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is melting always endothermic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it always is (except for the exception F'x gave). This is because during melting the body to be melted takes in &quot;latent heat of fusion&quot; from the surroundings, thus bringing down the temperature.  The latent heat of fusion is required to break the intermolecular attraction forces that define a solid. Breaking such bonds requires energy, hence the reaction is endothermic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can alcohol dissolve ice?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lower alcohols like methanol and ethanol may be able to dissolve ice. They can effectively form hydrogen-bonds with water, but higher alcohols may not be able to do that.&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="22" LastEditDate="2012-05-17T13:27:31.373" LastActivityDate="2012-05-17T13:27:31.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="330" PostTypeId="1" AcceptedAnswerId="602" CreationDate="2012-05-10T17:26:31.220" Score="26" ViewCount="23583" Body="&lt;p&gt;I've noticed that adding a chunk of ice cream to soda makes the soda fizz slightly near the soda-ice cream interface. I thought it was a physical effect due to the temperature, but adding ice has no effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It still can be a physical effect due to solubility, or it may be a chemical effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know which it is, and the details of the mechanism.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2015-05-15T00:35:47.793" Title="Why does ice cream make soda fizz?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="331" PostTypeId="2" ParentId="330" CreationDate="2012-05-10T18:45:27.933" Score="16" Body="&lt;p&gt;Soda contains $\ce{CO2}$. Over time the gaseous $\ce{CO2}$ comes to the surface. The ice cream increases the rate of this  $\ce{CO2}$ from the soda causing increased fizzing. This increased rate is a result of the solvation of ice cream particles in the liquid. The particles serve as nucleation sites for the growth of gas bubbles. The bubbles get large quiet fast thus you observe fizzing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of experimentation I found the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Effect of adding milk&lt;/strong&gt; &amp;mdash; There is a bit of extra bubbling and a curdy precipitate is formed. (Couldn't dare to drink it.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Effect of adding salt&lt;/strong&gt; (&lt;a href=&quot;http://chemistry.about.com/cs/howtos/a/aa020404a.htm&quot; rel=&quot;nofollow&quot;&gt;yes ice cream does contain salt&lt;/a&gt;) &amp;mdash; Lots of bubbling around the salt. After the salt settles to the bottom, bubbles collect around it become big and then come to the surface. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Only thing left to try is sugar and vanilla essence but I ran out of soda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we come to know that the fizzing effect may be due to milk or salt or a combo of both.  &lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorDisplayName="user10259" LastEditDate="2015-05-15T00:35:47.793" LastActivityDate="2015-05-15T00:35:47.793" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="333" PostTypeId="2" ParentId="308" CreationDate="2012-05-10T21:00:16.370" Score="5" Body="&lt;p&gt;Some additional information relevant to this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to use &quot;forces&quot; and &quot;particles&quot; in this explanation to generalize it. Particles can be ions (ionic compounds held together by strong electrostatic forces), metal atoms (held together by electrostatic forces from attraction of the metal atom nuclei for the &quot;sea of electrons&quot;) or molecules (for covalent compounds held together in the solid or liquid by intermolecular forces: London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Note: we are not talking about the intramolecular covalent bonds in this explanation.) So, solids and liquids are held together in that form by forces between the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that melting is always endothermic (except for the exceptions given by F'x...which I also did not know about) is that when you melt a substance, you have to break up some of the forces holding the particles together in the solid so that they can move past each other and form a liquid. This takes energy, the &quot;heat of fusion&quot; (usually given on a per mole basis) mentioned above. The stronger the forces, the more energy it would take to melt the substance. As long as you are melting a substance and holding the temperature at the melting point (i.e. not adding enough energy to raise the temperature) then you will have an equilibrium between the solid and liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is some confusion in the terms you are using. &lt;strong&gt;Desolvation&lt;/strong&gt; means removing the solvent from a solute-solvent mixture. (This can happen in crystallization, for example.) &lt;strong&gt;Solvation&lt;/strong&gt; (which I think you're asking about) is the interaction of solvent particles with solute particles. &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dissolution_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;Dissolving or dissolution&lt;/a&gt;&lt;/strong&gt; is the process in which a solid, liquid, or gas forms a solution with a solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as answered by Ashu, dissolution and be either exothermic or endothermic as he's noted above. This is true because in order to dissolve a substance in a solute, it takes energy to break up the solvent-solvent interactions, it takes energy to break up the solute-solute interactions, and you get energy back from new solvent-solute interactions. If you have stronger attractions between the solvent and solute than you had originally, then the process releases energy and is exothermic. The same concepts can be applied to mixing two liquids...which is usually referred to by saying they are &quot;miscible&quot; instead of speaking of one dissolving in the other. Ethanol and water are miscible in all proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can alcohol dissolve ice (&quot;at or just below the melting point of ice&quot; as phrased in your original question)? There are a couple of ways to answer this. If you put ice in ethanol at 0 degrees (its melting point) it will melt because the melting point of ice in a mixture of water and ethanol is lower than the melting point of pure water. The melting point depends on the proportions of water and ethanol.(That is to say, as some melted, it could not refreeze as you would expect to happen in an equilibrium mixture, so it would continue to melt.) See &lt;a href=&quot;http://en.wikipedia.org/wiki/Colligative_properties&quot; rel=&quot;noreferrer&quot;&gt;colligative properties&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've given two equations,        H2O(s)→H2O(l)&#xA;                                   H2O(s)+nEtOH(l)→H2O⋅EtOHn&#xA;In either case, you will need to provide the heat of fusion to melt the ice. So the question you asked is really about the energy of solvation of water in ethanol. When you mix water and ethanol, the flask feels warmer, so this is an exothermic process; you get more energy back than you put into it because of good interactions (hydrogen bonding) between water and ethanol particles. That solvation energy would provide some of the energy needed to melt the ice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the broader question &quot;can something happen at a given temperature&quot;, you're generally asking if it happens spontaneously. To answer this you'd have to calculate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Gibbs Free Energy&lt;/a&gt;. This would take into account not only the energy considerations spoken of above, but also entropy considerations (which would matter here). If the Gibbs Free Energy value is negative, then the process or reaction is spontaneous at the temperature given. So, the answer to which would be the preferred process would also depend on entropy considerations.  &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-10T21:00:16.370" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="334" PostTypeId="2" ParentId="238" CreationDate="2012-05-10T22:17:00.007" Score="15" Body="&lt;p&gt;What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and placing an electron in this orbital does not break rotational symmetry.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-10T22:17:00.007" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="335" PostTypeId="2" ParentId="285" CreationDate="2012-05-10T22:26:40.307" Score="9" Body="&lt;p&gt;The entire premise of NBO theory is to rotate canonical molecular orbitals into orbitals that have maximal chemical meaning, in the sense that is defined in the NBO papers. I am not aware of any purpose of NBOs and their corresponding natural charges beyond interpretation.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-10T22:26:40.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="336" PostTypeId="2" ParentId="316" CreationDate="2012-05-10T22:29:33.703" Score="34" Body="&lt;p&gt;When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has &lt;em&gt;proved mathematically&lt;/em&gt; that they mean anything. However, it has been &lt;em&gt;empirically observed&lt;/em&gt; that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory. In fact, the reference in the OP lends evidence to precisely this latter viewpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that orbitals are &quot;good&quot; or &quot;bad&quot; is not really that meaningful in the first place. A basic fact that can be found in any electronic structure textbook is that in theories that use determinantal wavefunctions such as Hartree-Fock theory or Kohn-Sham DFT, the occupied orbitals form an invariant subspace in that any (unitary) rotation can be applied to the collection of occupied orbitals while leaving the overall density matrix unchanged. Since any observable you would care to construct is a functional of the density matrix in SCF theories, this means that individuals orbitals themselves aren't physical observables, and therefore interpretations of &lt;em&gt;any&lt;/em&gt; orbitals should always be undertaken with caution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the premise of this question is not quite true. The energies of Kohn-Sham orbitals are known to correspond to ionization energies and electron affinities of the true electronic system due to Janak's theorem, which is the DFT analogue of Koopmans' theorem. It would be exceedingly strange if the eigenvalues were meaningful while their corresponding eigenvectors were &lt;em&gt;completely&lt;/em&gt; meaningless.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2012-05-10T22:39:09.353" LastActivityDate="2012-05-10T22:39:09.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="337" PostTypeId="2" ParentId="276" CreationDate="2012-05-10T23:02:59.013" Score="9" Body="&lt;p&gt;Yes, you need additional quantities beyond the minimum necessary for calculating energies and the Hellmann-Feynman piece of the force &lt;em&gt;when the wavefunction is not variational&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a very rough sketch of why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Puláy forces arise from applying the chain rule for calculating forces and was first discussed in the context of applying the Hellman-Feynman theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that the force is the change of energy with changes in (nuclear) coordinate and the energy is the expectation value of the (electronic) Hamiltonian, $E = \left&amp;lt;\psi\left|H\right|\psi\right&amp;gt;$. Applying the chain rule to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-F = \nabla E = \left&amp;lt;\psi \left\vert \nabla H \right\vert \psi\right&amp;gt; +  2\left&amp;lt; \nabla\psi \left\vert H \right\vert \psi\right&amp;gt;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is what you get from Hellman-Feynman and is the expectation of a one-electron operator of the first you have listed. The second term goes away in Hellmann-Feynman &lt;em&gt;only because it assumes the wavefunction is variational&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you expand out the orbitals in the Slater determinant in some AO basis $\chi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi(r_1, r_2, ..., r_N) = \phi(r_1) \wedge \phi(r_2) \wedge \dots \wedge \phi(r_N) $$&#xA;$$ \phi(r_1) = \sum_i c_i \chi_i(r_i) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then it is clear that $\nabla\psi$ generates terms which are MO coefficient derivatives (usually denoted $c_i^x$) and also AO derivatives ($\chi_i^x$). If the wavefunction were variational, then by definition all these derivatives are zero. If you work through the algebra, you will find that there are some new quantities that need to be calculated in order to work out the non-Hellmann-Feynman theorem, most notably the one-sided AO overlap derivative matrices $\left&amp;lt;\chi_i^x\vert\chi_j\right&amp;gt;$ which do not normally show up otherwise. These terms turn out to be critical for getting correct &lt;em&gt;ab initio&lt;/em&gt; molecular dynamics since the wavefunction is rarely (if ever) variational in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are tricks one can do to reduce the cost of calculating MO coefficient derivatives , but they cannot be obtained for free in general. And there's really no way around calculating the overlap derivative terms.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2012-05-11T04:58:03.057" LastActivityDate="2012-05-11T04:58:03.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="338" PostTypeId="2" ParentId="141" CreationDate="2012-05-10T23:21:35.063" Score="32" Body="&lt;p&gt;Some factors were hinted, but let me put them in an order of importance and mention some more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;metals generally have a high melting point, because &lt;strong&gt;metallic interatomic bonding by delocalized electrons&lt;/strong&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; having only a few electrons for this &quot;electron sea&quot;) between core atoms is pretty effective in those pure element solids compared to alternative bonding types (ionic &lt;span class=&quot;math-container&quot;&gt;$\pu{6-20 eV/atom}$&lt;/span&gt; bond energy, covalent 1-7, metallic 1-5, van-der-Waals much lower). Also, ionic lattices like &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; have a higher lattice and bonding energy, they have weak interatomic long-range bonding, unlike most metals. They break apart or are easily solvable, metals are malleable but don't break, the electron sea is the reason for their welding ability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the crystal structure and mass play an inferior role among your filtered elements (just look up the crystal structure of those elements), as metallic bonding is not directional unlike covalent bonding (orbital symmetry). Metals often have half filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; bands (stronger delocalized than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{f}$&lt;/span&gt;) at the Fermi-edge (meaning high conductivity) and therefore many delocalised electrons which can move into unoccupied energy states yielding the biggest electron sea with half or less fill bands.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;noble metals like &lt;span class=&quot;math-container&quot;&gt;$\ce{Au,Ag}$&lt;/span&gt; have a full &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; orbital, therefore low reactivity/electronegativity and are often used as contact materials (high conductivity because of &quot;very fluid&quot; electron sea consisting only of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital electrons. Unlike tungsten with half or less occupied &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals they show no interatomic &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d-d}$&lt;/span&gt; bonding by delocalized &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-electrons, and more importantly, &lt;strong&gt;a half filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbital contributes 5 electrons to the energy band, while a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; only 1, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; only 3, the electron sea is bigger among the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-group.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;&quot;packaging&quot; of core atoms in the lattice&lt;/strong&gt; (interatomic distance) among the high &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; atoms (compared to &lt;em&gt;e.g.&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt;) is denser (more protons, stronger attraction of shell electrons, smaller interatomic radius), means stronger interatomic bonding transmitted by the electron sea: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6a5pa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see here that in each series (&lt;span class=&quot;math-container&quot;&gt;$\ce{Li,\ Na,\ K}$&lt;/span&gt;) the melting points rise to a maximum and then decrease with increasing atomic number (lacking unoccupied energy states for delocalized &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-electrons), bigger electron sea being here a stronger factor than a bit more dense packaging. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boron as a semi-metal shows metallic and covalent bonding, Carbon strong directional covalent bonding and is able to build a network of bonds unlike other non-metal elements showing covalent intramolecular bonding, &lt;em&gt;e.g.&lt;/em&gt;, in diatomic molecules but not strong intermolecular bonding in macromolecules because of lacking unpaired electrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there are some bigger trends for melting points explaining the high melting points of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-metals, but also some minor exceptions to the rule like &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="165" LastEditorUserId="61578" LastEditDate="2019-12-24T07:10:21.383" LastActivityDate="2019-12-24T07:10:21.383" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="339" PostTypeId="2" ParentId="236" CreationDate="2012-05-10T23:24:30.047" Score="18" Body="&lt;p&gt;Classical hybridization theory does not allow for noninteger hybridizations. However, &lt;em&gt;ab initio&lt;/em&gt; calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the other coefficients are very small, the ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would yield a number that &lt;em&gt;could&lt;/em&gt; be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2012-05-11T05:01:54.907" LastActivityDate="2012-05-11T05:01:54.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="340" PostTypeId="1" CreationDate="2012-05-11T00:58:04.223" Score="17" ViewCount="4925" Body="&lt;p&gt;We have a well that delivers hard water and we do not have a water softener. (There was a “green sand filter” that utilizes permanganate to oxidize other metals on the system, but it has been disconnected.) Analysis of the well water shows the presence of iron in noticeable amounts. No other metal ions were specifically identified. The system does have an ozonator to oxidize Fe(II) to Fe(III). Over the last couple of years, the porcelain toilet bowls have developed a blue ring at the air/water interface. When you scrub on the blue ring, it doesn’t feel like there is a deposit present (like normal hard water deposits). The ring is unresponsive to ammonia (household), strong acid (6 M HCl) or strong base (Drano), as well as to hydrogen peroxide (30 %).  (We haven’t tried a good reducing agent yet.) So, my question is, does anyone have an idea of what is causing the blue color? Is it possible that porcelain reacted chemically with salts, urea, uric acid, oxygen, etc?  Not an earth-shattering theoretical question, but I’m chemically curious as to what this could be if anyone has any ideas. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="7475" LastEditDate="2016-09-30T22:08:04.037" LastActivityDate="2016-09-30T22:08:04.037" Title="Not your ordinary hard water deposit" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="341" PostTypeId="1" AcceptedAnswerId="353" CreationDate="2012-05-11T03:21:30.083" Score="4" ViewCount="11529" Body="&lt;p&gt;Brands like Brita and Pur (in the U.S.) have made a name for themselves for the ability of their product (essentially a large-pore filter with activated carbon/charcoal) to extract the added chlorine from tap water.  I visualize $\ce{C}^*$ as a vast network of channels in which ions are &quot;trapped&quot;.  I had assumed that most cations and anions were trapped effectively, but it seems to favor chlorine and iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Activated carbon does not bind well to certain chemicals, including alcohols, glycols, strong acids and bases, metals and most inorganics, such as lithium, sodium, iron, lead, arsenic, fluorine, and boric acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would it not bind effectively to fluorine?  Is it an issue of atomic radius?  Does $\ce{F-}$, with a radius of &lt;a href=&quot;http://boomeria.org/chemlectures/textass2/table10-9.jpg&quot; rel=&quot;nofollow&quot;&gt;0.136 nm&lt;/a&gt; sneak by, while $\ce{I-}$ and $\ce{Cl-}$ at 0.181 and 0.216 nm, repectively, get caught up in the matrix?  Why do the cations get passed through?&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="133" LastEditDate="2012-05-12T02:36:41.360" LastActivityDate="2016-05-18T15:12:41.803" Title="Why does activated carbon preferentially adsorb anions?" Tags="&lt;ions&gt;&lt;everyday-chemistry&gt;&lt;filtering&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="342" PostTypeId="1" AcceptedAnswerId="346" CreationDate="2012-05-11T05:24:51.460" Score="25" ViewCount="1079" Body="&lt;p&gt;The ideal gas equation (daresay &quot;law&quot;) is a fascinating combination of the work of dozens of scientists over a long period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I encountered Van der Waals interpretation for non-ideal gases early on, and it was always somewhat in a &quot;closed-form&quot;&#xA;$$\left( p + \frac{n^2a}{V^2} \right)(V - nb) = nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $a$ being a measure of the charge interactions between the particles and $b$ being a measure of the volume interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understandably, this equation is only still around for historical purposes, as it is largely inaccurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fast-forwarding to the 1990s, &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_state#Elliott.2C_Suresh.2C_Donohue_equation_of_state&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a listing of one of the more current manifestations (of Elliott, Suresh, and Donohue):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p V_\mathrm{m}}{RT} = Z = 1 + Z^{\mathrm{rep}} + Z^{\mathrm{att}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the repulsive and attractive forces between the molecules are proportional to a shape number ($c = 1$ for spherical molecules, a quadratic for others) and reduced number density, which is a function of Boltzmann's constant, etc (point being, a lot of &quot;fudge factors&quot; and approximations are getting thrown into the mix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than seeking an explanation of all of this, I am wondering whether a more &quot;closed form&quot; solution lies at the end of the tunnel, or whether the approximations brought forth in the more modern models will have to suffice?  &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="16683" LastEditDate="2016-10-20T22:26:07.830" LastActivityDate="2019-09-21T02:30:37.020" Title="Why are equations of state for a non-ideal gas so elusive?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;intermolecular-forces&gt;&lt;equation-of-state&gt;" AnswerCount="6" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="344" PostTypeId="1" AcceptedAnswerId="345" CreationDate="2012-05-11T09:29:08.547" Score="8" ViewCount="1171" Body="&lt;p&gt;When drawing a skeletal formula, what is the difference between an angular version and a linear version?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was asked to draw the Z isomer of Resveratrol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sKlxu.png&quot; alt=&quot;(E)-Resveratrol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which I drew:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5gRaA.jpg&quot; alt=&quot;My (Z)-Resveratrol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the markscheme states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;skeletal structure must be correct &lt;strong&gt;and angular not linear&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't come across the difference between the two before and can't find anything on google to suggest one.  Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. What am I missing here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their version:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2qYhs.jpg&quot; alt=&quot;Their (Z)-Resveratrol&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="118" LastEditorUserId="5026" LastEditDate="2018-02-25T08:35:12.077" LastActivityDate="2018-02-25T08:35:12.077" Title="What is the difference between an angular and linear skeletal formula?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="345" PostTypeId="2" ParentId="344" CreationDate="2012-05-11T09:40:40.060" Score="8" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mnxor.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are $sp^2$ carbons and should have a triangular planar 120&amp;deg; geometry. The angles you've drawn are clearly not 120&amp;deg; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've not exactly come across the terminology before, but I believe this is the distinctions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In a &quot;linear&quot; formula, you disregard the actual bond angles and just try to keep everything in a line.&lt;/li&gt;&#xA;&lt;li&gt;In an &quot;angular&quot; formula, you draw everything according to the bond angles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not sure of this, though.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-11T09:40:40.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="346" PostTypeId="2" ParentId="342" CreationDate="2012-05-11T14:15:08.570" Score="18" Body="&lt;p&gt;At the end of the tunnel, you're still trying to approximate the statistical average of interactions between individual molecules using macroscopic quantities. The refinements add more parameters because you're trying to parametrise the overall effect of those individual interactions for every property that is involved for each molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're never going to get a unified &quot;parameter-free&quot; solution for those without going down to the scale of the individual molecules (e.g. &lt;em&gt;ab initio&lt;/em&gt; molecular dynamics), as far as I can tell.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-05-11T14:15:08.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="347" PostTypeId="1" CreationDate="2012-05-11T16:54:45.533" Score="22" ViewCount="27972" Body="&lt;p&gt;When testing for pH for common household cleaning products, such as dishwasher fluid, I noticed that majority of them tend to be basic? Why is that? Do acid make good cleaning solutions as well?&lt;/p&gt;&#xA;" OwnerUserId="196" LastEditorUserId="196" LastEditDate="2012-05-11T17:03:31.013" LastActivityDate="2013-04-14T14:10:26.117" Title="Why are the majority of cleaning solutions basic?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="349" PostTypeId="2" ParentId="342" CreationDate="2012-05-11T22:37:00.503" Score="8" Body="&lt;p&gt;I'll add to Aesin’s answer that in this case, the burden of proof rests on the side of an analytical (or closed-form) equation of state. Statistical mechanics explicitly guarantees that there is a relationship between $p$, $V$ and $T$, i.e. that they are not independent state variables. However, no further generic statement can be made about it, and only by making appropriate approximations can one actually write an actual equation of state (EOS). One such EOS is the ideal gas law, others are as you have cited, but if one were to prove the existence of a closed-form generic EOS for fluids, that would be a major upheaval of the current understanding of thermodynamics of fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an overview of the thermodynamics of fluids, you can refer to your favorite statistical mechanics or statistic thermodynamics textbook (my own preference goes to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1891389157&quot;&gt;McQuarrie&lt;/a&gt;). For a introduction to issues specific to the liquid state (but also valid in general for fluids far from the ideal gas behaviour), I would recommend &lt;a href=&quot;http://books.google.fr/books/about/Theory_of_Simple_Liquids.html?id=Uhm87WZBnxEC&amp;amp;redir_esc=y&quot;&gt;Theory of Simple Liquids&lt;/a&gt;, by Hansen and McDonald.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-11T22:37:00.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="350" PostTypeId="2" ParentId="342" CreationDate="2012-05-12T04:36:39.340" Score="21" Body="&lt;p&gt;It's been known since 1941 that the answer to your question is in the negative, i.e. that there will never be a closed form equation of state for a nonideal gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1941 Mayer and Montroll developed what is now known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mayer_cluster_integral&quot;&gt;cluster expansion&lt;/a&gt; for the partition function of a nonideal gas whose particles have pairwise interactions. This cluster expansion provides a 1-1 correspondence between various integrals over the interaction potential and virial coefficients in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Virial_equation#Virial_equation_of_state&quot;&gt;Kamerlingh Ohnes&lt;/a&gt; equation of state which has in principle an infinite number of terms in it. Therefore, it would not be considered a closed form equation of state. In practice the virial coefficients are known to decay so it is usually safe to truncate the expansion for practical calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could extend the cluster expansion to three-body and higher-order interaction potentials but this will not change the fundamental argument above.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-12T04:36:39.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="351" PostTypeId="2" ParentId="152" CreationDate="2012-05-12T05:13:55.800" Score="40" Body="&lt;p&gt;General chemistry textbooks tend to explain atomic structure exceedingly poorly using a hodgepodge of obsolete concepts. Your chemistry book provides such a typical example - the notion of penetration only makes sense in the ancient &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model#Refinements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bohr-Sommerfeld&lt;/a&gt; model that has been obsolete since the discovery of quantum mechanics! The idea was that orbits of electrons in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; states were more elliptical than those in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; states because they had more angular momentum, and so on average there will be some fraction of the time where &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; electrons would be closer to the nucleus than those in the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbits, &lt;em&gt;i.e.&lt;/em&gt; they would &lt;em&gt;penetrate&lt;/em&gt; the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now know that electrons do not actually have classical orbits and that the Bohr-Sommerfeld model is wrong. However there remains one correct statement of fact, which is that electrons in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals by definition have more angular momentum than those in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbitals (The orbital angular momentum is &lt;span class=&quot;math-container&quot;&gt;$\ell=1$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ell=0$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;). Nonetheless it is completely misleading, since &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; orbitals are only well-defined in hydrogen and other ions with only one electron, and furthermore that &lt;em&gt;all orbitals of the same principal quantum number are degenerate in energy in hydrogenic systems&lt;/em&gt;. Thus angular momentum alone cannot be the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only due to electron-electron interactions that the orbitals that correspond to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; in multi-electron atoms split and become nondegenerate. The effect of having more angular momentum therefore turns out to be only indirectly responsible via the different many-electron interactions occurring in the analogs of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; orbitals in many-electron atoms. The simplest explanation that is even remotely correct is the screening effect, where the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; electrons by their very presence make the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; electrons feel a lower effective nuclear charge than if they were absent. A more precise quantum mechanical statement is that Slater's variational treatment of the nuclear charge in hydrogenic orbitals results in a lower effective nuclear charge for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; electrons than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; orbitals. There is unfortunately no good &lt;em&gt;a priori&lt;/em&gt; physical principle for this; it is merely a statement of a result from a quantum mechanical calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; electrons are higher in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s}$&lt;/span&gt; ones due to screening effects that result from electron-electron interactions. Explanations involving penetration of classical orbits or purely angular momentum considerations are wrong, even though they tend to show up in general chemistry textbooks.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="61578" LastEditDate="2019-12-24T06:16:19.883" LastActivityDate="2019-12-24T06:16:19.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="352" PostTypeId="2" ParentId="293" CreationDate="2012-05-12T05:30:17.090" Score="6" Body="&lt;p&gt;Suffice to say, nobody really knows, and a Nobel prize surely awaits the first people who can explain the relationship between chemical composition and high-temperature superconductivity.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-12T05:30:17.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="353" PostTypeId="2" ParentId="341" CreationDate="2012-05-12T05:32:12.670" Score="8" Body="&lt;p&gt;This answer applies to carbon filtration of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to most of the sources I found, activated carbon binds to most substances through London dispersion forces (from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot;&gt;Wikipedia article)&lt;/a&gt;. This should mean that it adsorbs larger molecules and non-polar molecules preferentially since they would have larger dispersion forces. As your quote indicates, it does not bind polar molecules like &quot;alcohols, glycols, strong acids and bases&quot; well. &lt;a href=&quot;http://www.calgoncarbon.com/documents/ACWhatitisHowitworks.pdf&quot;&gt;This source&lt;/a&gt; also lists neutral, non-polar compounds, along with organics and compounds with low water solubility as being effectively removed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you may have misread or misinterpreted the sentence about binding &quot;chlorine and iodine&quot; as saying that it traps chloride, iodide, and (by extension) other anions well. I can't find data to support this statement. It traps &lt;strong&gt;molecular iodine&lt;/strong&gt; well, and as a large (in terms of electrons), non-polar molecule, that makes sense. But it removes &lt;strong&gt;molecular chlorine&lt;/strong&gt; $\ce{Cl2}$ through a &lt;a href=&quot;http://www.resintech.com/pdf/ActivatedCarbonFiltration.pdf&quot;&gt;chemical reaction&lt;/a&gt; by reducing it to the chloride ion $\ce{Cl-}$ which is then soluble in water and flows through the filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find a couple of sources (&lt;a href=&quot;http://www.tigg.com/what-is-activated-carbon.html&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.ag.ndsu.edu/pubs/h2oqual/watsys/wq1029.pdf&quot;&gt;2&lt;/a&gt;) that said &quot;activated charcoal&quot; has a slight electropositive charge that helps it to attract negatively charged species, which would fit with your statement that it preferentially attracts anions. (However, they don't list what kind of negative species they're talking about and nitrate, sulfate, and fluoride ions were specifically excluded). The idea that it will remove ionic compounds seems to be contradicted by the statement in at least a couple of sources (&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_filtering&quot;&gt;one given here&lt;/a&gt;) that carbon filters don't remove minerals, salts, and dissolved inorganic compounds   &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-05-12T06:52:08.780" LastActivityDate="2012-05-12T06:52:08.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="354" PostTypeId="2" ParentId="189" CreationDate="2012-05-12T06:06:19.760" Score="15" Body="&lt;p&gt;Ring systems (like benzene) are the canonical example of when Z-matrices go awry. A Z-matrix cannot contain all the bond coordinates of the ring. One either has to suffer an intrinsically asymmetric description of a highly symmetric system, which is both intellectually unsatisfying and can lead to practical numerical convergence issues arising from broken symmetries, or otherwise define one or more dummy atoms in the Z-matrix, which is then no longer a minimally redundant description of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of coordinate system really depends on the intended calculation. There are more than two choices of coordinate system that can be used, by the way. While internal coordinates are often erroneously considered synonymous with Z-matrices, there are in fact many other internal coordinate systems that are not Z-matrices, such as pairwise distance coordinates or the various redundant internal coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some specific examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Redundant internal coordinates are the most efficient known coordinate systems for carrying out geometry optimizations. Roughly speaking, the redundancy is useful for avoiding singularities in nonredundant systems like Z-matrices, and minimizing the correlations (lack of independence) between coordinates that occur in coordinate systems like Cartesian coordinates that result in large off-diagonal cross terms in the Hessian matrix. You can find more details in the original literature which is cited in any quantum chemistry package's user manual.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are coding up analytic gradients, these tend to be simplest in Cartesian coordinates because you don't have to worry about curvilinear effects in the Hessian matrix. Non-Cartesian coordinates have extra terms in gradient expressions arising from Jacobians; these can be fairly expensive to calculate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Z-matrices themselves are often useful when creating interpolations along a specific internal coordinate like a specific torsional mode, because they are an internal coordinate system that is not redundant and hence allow various internal coordinates to be varied independently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2012-05-12T19:25:58.433" LastActivityDate="2012-05-12T19:25:58.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="355" PostTypeId="2" ParentId="46" CreationDate="2012-05-12T07:15:59.860" Score="11" Body="&lt;p&gt;I assume you meant amorphous solid phases, since you can always melt any ionic compound to form an essentially amorphous liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;amorphous salts&quot; brings up a fair number of results in the primary literature, such as &lt;a href=&quot;http://jcp.aip.org/resource/1/jcpsa6/v74/i9/p4774_s1&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.springerlink.com/content/ul260m3750607j23&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0013468699003357&quot;&gt;this&lt;/a&gt;. The defining characteristic of such amorphous solid phases is that they exhibit no measurable melting transition, i.e. no latent heat of fusion can be measured in a calorimetry experiment. There also examples of amorphous semiconductor oxides and amorphous metal oxides and amorphous metals - feel free to search for these and see for yourself. So yes, they do exist, and in fact are of research interest.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-12T07:15:59.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="356" PostTypeId="1" AcceptedAnswerId="368" CreationDate="2012-05-12T07:43:46.190" Score="8" ViewCount="701" Body="&lt;p&gt;In crystallography, we know that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Burgers_vector&quot; rel=&quot;noreferrer&quot;&gt;Burgers vector&lt;/a&gt; is parallel to the dislocation direction in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dislocation#Screw_dislocations&quot; rel=&quot;noreferrer&quot;&gt;screw dislocation&lt;/a&gt;, and perpendicular in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dislocation#Edge_dislocations&quot; rel=&quot;noreferrer&quot;&gt;edge dislocation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I have in my lecture notes &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Face-centered_cubic&quot; rel=&quot;noreferrer&quot;&gt;FCC&lt;/a&gt; cannot have edge dislocations&quot; but there is no explanation of why, aside from &quot;in FCC, the Burger's vector belongs to the family of the dislocation direction vectors&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what this means. &lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="73" LastEditDate="2012-11-14T19:55:05.330" LastActivityDate="2012-11-14T19:55:05.330" Title="Why are there no edge dislocations in a face-centered cubic (FCC) lattice?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="357" PostTypeId="2" ParentId="298" CreationDate="2012-05-12T08:05:36.850" Score="17" Body="&lt;p&gt;The short answer to your question is that no, they don't all occupy the same orbital but yes, they occupy very similar regions in space. What tends to be confusing in these pictures is it sometimes unclear that the pictures are sometimes of the molecular orbitals and sometimes of the &lt;em&gt;electron density&lt;/em&gt;, which is an observable quantity and quite another thing altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The six pi electrons occupy three different molecular orbitals of very different shapes, as can be seen &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch11/ch11-4-1.html&quot;&gt;here&lt;/a&gt;, for example, and in Richard Terrett's answer. However, their corresponding &lt;em&gt;probability densities&lt;/em&gt; can be superposed onto similar regions of &lt;em&gt;real&lt;/em&gt; space. There is no conflict with Pauli exclusion because you can see (and calculate!) that the orbitals are orthogonal by virtue of having an appropriate number of nodes. While &lt;em&gt;on average&lt;/em&gt; the electrons look like they are all in the same region of space, if you measure the positions at any time, no two electrons will ever occupy exactly the same point in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your discussion of orbital and spin angular momenta is completely irrelevant. Angular momentum operators $L_z$ and $L^2$ commute with atomic Hamiltonians because of rotational and Runge-Lenz invariance; however, they do not commute with molecular Hamiltonians because molecules have neither of these symmetries. Therefore, molecular orbitals are not indexable by the same set of quantum numbers as can be done for atomic orbitals. This is not because of Pauli exclusion, but because the Hamiltonian, $L_z$ and $L^2$ no longer form a complete set of commuting observables. (Note that Pauli exclusion must also hold in atoms where this latter statement is true!) Nor is there any need to introduce new fictitious quantum numbers.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2012-05-12T08:23:13.880" LastActivityDate="2012-05-12T08:23:13.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="358" PostTypeId="2" ParentId="347" CreationDate="2012-05-12T08:30:13.083" Score="10" Body="&lt;p&gt;Dishwashing agents are usually anionic detergents—i.e. they are alkylbenzenesulfonates.&lt;br&gt;&#xA;They consist of two parts:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)  &lt;/li&gt;&#xA;&lt;li&gt;Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, when in water, they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ .  $\ce{Na+}$ is from a strong base (making it a weak conjugate acid), while $\ce{RSO3-}$ is from a weak acid (making it a weak conjugate base), thus the resulting solution is &lt;strong&gt;basic&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sulphonate ion bonds to water, while alkylbenzene bonds to oil, protein, or fat . Thus fats, oils, proteins, and grease dissolve in water  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the mechnism of a dishwashing agent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j30zD.jpg&quot; alt=&quot;enter image description here&quot;&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the ingreadients mentioned on a dishwashing bottle in my house:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V3MWQm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V3MWQm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, sodium dodecylbenzenesulphonate (an alkylbenzenesulfonate) is one of the main ingredients.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do acid make good cleaning solutions as well?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes they can make a good cleaning solution, but not as a dishwashing agent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cleaning_agent#Acidic&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; ,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="124" LastEditorUserId="22" LastEditDate="2012-05-16T14:41:38.370" LastActivityDate="2012-05-16T14:41:38.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="359" PostTypeId="1" AcceptedAnswerId="360" CreationDate="2012-05-12T08:32:17.987" Score="14" ViewCount="6482" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_equation&quot;&gt;Hammett plot&lt;/a&gt; is commonly invoked in organic chemistry to reason about the plausibility (or implausibility) of various reaction mechanisms. The vertical axis is essentially the logarithm of an equilibrium constant (or rate constant) measured relative to a hydrogen functional group, which I understand. However, I am at a complete loss for understanding what is being plotted on the horizontal axis, which conventionally are denoted $\sigma$ and $\sigma^\pm$. What are these parameters and how are their values determined for various functional groups? In practice it seems that people look them up in tables, but where do the values in these tables come from?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2014-01-29T09:29:36.217" Title="How are $\sigma$ and $\sigma^\pm$ determined in Hammett plots?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="360" PostTypeId="2" ParentId="359" CreationDate="2012-05-12T11:31:30.760" Score="13" Body="&lt;p&gt;Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't used because steric effects would overlay with the electronic effects) as a guide and plotted them against the logarithm of the reaction rates.&#xA;So, he based his scale on the following reaction reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iIk9R.png&quot; height=&quot;75&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later Hammett decided not to use the $pK_{\mathrm{a}}$s themselves for his correlation but defined a new parameter, which he called $\sigma$.&#xA;This $\sigma$ shows how electron-donating or -withdrawing a group is relative to $\ce{H}$ as a ratio of&#xA;the $\log K_a$s or the difference of the $pK_{\mathrm{a}}$&#xA;between the substituted benzoate and benzoic acid itself.&#xA;If the acid required to determine $\sigma$ for a new substituent was not available, $\sigma$ could be determined by correlation with other reactions.&#xA;A different value of $\sigma$ for any given substituent was needed for the meta and para positions (sometimes called $\sigma_{\mathrm{m}}$ and $\sigma_{\mathrm{p}}$, respectively). &#xA;The equation for $\sigma$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sigma_{\ce{X}} = \log \left( \frac{ K_{\mathrm{a}}(\ce{X-C6H4COOH}) }{ K_{\mathrm{a}}(\ce{H-C6H4COOH}) } \right) = pK_{\mathrm{a}}(\ce{X-C6H4COOH}) - pK_{\mathrm{a}}(\ce{H-C6H4COOH})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in short notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sigma_{\ce{X}} = \log \left( \frac{ K_{a} (\ce{X}) }{ K_{a} (\ce{H}) } \right) = pK_{\mathrm{a}}(\ce{X}) - pK_{\mathrm{a}}(\ce{H})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{X}$ is the functional group, i.e. the EWG or EDG, whose effect on the reaction rate shall be evaluated.&#xA;It is also possible to determine the $\sigma$ values via the logarithm of the ratio of the rate constants for the aforementioned reaction with substituent $\ce{X}$ (in short notation: $k_{\ce{X}}$) and with $\ce{X}=\ce{H}$ (in short notation: $k_{\ce{H}}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sigma_{\ce{X}} = \log \left( \frac{ k_{\ce{X}} }{ k_{\ce{H}} } \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;AcidFlask&lt;/code&gt;'s comment reminded me that I forgot to say something about $\sigma^{+}$ and $\sigma^{-}$ values.&#xA;As mentioned above the &quot;normal&quot; $\sigma$ values are based upon the difference of the $pK_{\mathrm{a}}$ between a substituted benzoate and benzoic acid itself.&#xA;But for benzoates one cannot draw resonance structures that delocalize the negative charge onto the benzene ring via the $\pi$ electron system.&#xA;Yet, many reations of interest create negative or positive charges that can be stabilized by delocalization via resonance with the substituent.&#xA;For these reactions, one finds that Hammett plots using $\sigma$ values have considerable scatter.&#xA;Therefore, two new substituent effect scales were produced, one for groups that stabilize negative charges via resonance ($\sigma^{-}$), and one for groups that stabilized positive charges via resonance ($\sigma^{+}$).&#xA;The $\sigma^{-}$ scale is based upon the ionization of para-substituted phenols (I've found different accounts for this. According to the German Wikipedia the $\sigma^{-}$ scale is based upon the ionization of para-substituted anilins.),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFNlD.png&quot; height=&quot;75&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which groups like nitro can stabilize the negative charge via resonance.&#xA;The $\sigma^{+}$ scale is based upon the heterolysis ($\mathrm{S}_{\mathrm{N}}1$) reaction of para-substituted cumyl chlorides (phenyldimethyl chloromethanes),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RIWUR.png&quot; height=&quot;75&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which groups like amino can stabilize the positive charge via resonance.&#xA;The $\sigma^{\pm}$ values can then be evaluated by the aforementioned equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\sigma_{\ce{X}}^{\pm} = \log \left( \frac{ K_{a}^{\pm} (\ce{X}) }{ K_{a}^{\pm} (\ce{H}) } \right) = \log \left( \frac{ k_{\ce{X}}^{\pm} }{ k_{\ce{H}}^{\pm} } \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but with the rate constants $k_{\ce{X}}^{\pm}$ and $k_{\ce{H}}^{\pm}$ (or the equilibrium constants $K_{a}^{\pm} (\ce{X})$ and $K_{a}^{\pm} (\ce{H})$) measured for their respective defining reactions.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-01-29T09:29:36.217" LastActivityDate="2014-01-29T09:29:36.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="362" PostTypeId="2" ParentId="45" CreationDate="2012-05-12T14:17:31.253" Score="9" Body="&lt;p&gt;Ok, I will try my hand on the mechanism. In my opinion the reaction should work with a catalytic amount of $\ce{H+}$ since the key step for the reaction is the hydrolysis of the orthoester. This is an entropically very favourable reaction since from one reactant molecule you get two product molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wNjTZ.png&quot; height=&quot;80&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting molecule is extremely reactive. So it can easily react with the enolized malonate. The enolization might be facilitated by the $\ce{ZnCl2}$ and/or the cat. $\ce{H+}$.&#xA;(I'm not entirely sure whether the $\ce{ZnCl2}$ really plays the role I've given it below in this reaction. So, if anyone knows good arguments against it, please let me know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pzc8W.png&quot; height=&quot;115&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protonated product can then take part in an $\mathrm{E}1$ elimination reaction to yield ethyl ethoxymethylenemalonate. The carbocation which results when $\ce{EtOH}$ leaves the molecule is stabilized by the neighboring ethoxy group. The hydrogen on the central carbon atom is very acidic, since it is placed between three carboxylate groups, so it can even be removed by rather non-basic molecules, e.g. by an $\ce{EtOH}$, a malonate or an othoester molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E7SKs.png&quot; height=&quot;100&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for the acid catalysed mechanism. Now for the question, why one doesn't simply use a base to facilitate the addition of the activated methylene compound. Well, this is due to the fact, that orthoesters are quite stable against basic hydrolysis. So, under basic conditions you don't get the carbonyl component of the addition reaction.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2012-05-12T21:35:12.047" LastActivityDate="2012-05-12T21:35:12.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="363" PostTypeId="1" AcceptedAnswerId="365" CreationDate="2012-05-12T17:21:28.227" Score="10" ViewCount="117" Body="&lt;p&gt;QTPIE is a &lt;a href=&quot;http://www.slideshare.net/acidflask/constructing-a-rigorous-fluctuatingcharge-model-for-molecular-mechanics&quot;&gt;fluctuating charge model&lt;/a&gt; that substantially improves modeling of polarization and charge transfer. I was curious whether it has ever been applied to 1D metals such as polyacetylene to provide new perspectives on topics such as &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v29/i4/p2109_1&quot;&gt;Peierls' condensation and phase solitons&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2012-05-12T19:13:38.807" Title="Has QTPIE ever been applied to 1D metals such as polyacetylene?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="364" PostTypeId="2" ParentId="347" CreationDate="2012-05-12T18:08:37.357" Score="10" Body="&lt;h2&gt; Basic dishwashing agents? &lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Some dishwashing agents contain substantial amounts of NaOH, see below. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sometimes NaClO solutions are used as disinfectant for dishwashing. &#xA;&lt;br/&gt;$\require{mhchem}$&#xA;As $\ce{Cl- + ClO- + 2 H+ &amp;lt;=&amp;gt;&amp;gt; Cl2 ^ + H2O}$, they are kept basic (and $\ce{Cl-}$free, but $\ce{3 ClO- -&amp;gt; 2 Cl- + ClO3-}$ and $\ce{2 ClO^- -&amp;gt; 2 Cl^- + O2 ^}$) for stabilization.&#xA;&lt;br/&gt;&#xA;&lt;em&gt;@Ashu asked whether I'm really sure, so here's an example: with 15 - 30 % KOH + 15 - 30 % NaOCl, pH 14 &quot;Liquid intensive cleaner for dishwashing machines&quot; (&lt;a href=&quot;http://www.pack2go.de/media/uploads/G%20300%20Geschirrsp%C3%BClmittel%20fl%C3%BCssig%20Desinfektion.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;technical data&lt;/a&gt;, &lt;a href=&quot;http://www.pack2go.de/media/uploads/G%20300%20Sicherheitsdatenblatt.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;saftety sheet&lt;/a&gt; both German, though)&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Soaps (chemical meaning) are salts of fatty acids. Na- and K-soaps are good for cleaning and as salts of weak acids and strong bases they are basic.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt; Dishwashing liquid chemistry? &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we approximate the food rests you want to clean away during the dishwashing as mixture of carbohydrate, proteins and lipids, then your dishwashing liquid needs that take care of those 3 substance classes in water.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;carbohydrates are hydrophilic - no problem here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;amphiphils are needed to clean lipophilic substances like lipids, oils, greases as &lt;a href=&quot;https://chemistry.stackexchange.com/a/358/160&quot;&gt;Ashu explained already&lt;/a&gt;. They can take care of quite a bit of protein as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;basic solutions usually cause a faster hydrolysis of the amide bonds in proteins than acids. Usually quite a bit faster: NaOH-solutions immediately give a &quot;soapy&quot; feeling while e.g. HCl doesn't. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As far as I know here (Germany), basic liquids for &lt;em&gt;household&lt;/em&gt; dishwashing machines are uncommon/seldom/not used at all (I'm not so sure about the tablets). The actually used solutions are less corrosive (for both the machine and your fingers if you more or less accidentally decide to use it for dishwashing by hand).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.poisons.co.nz/fact.php?f=26&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page from a New Zealand poison centre&lt;/a&gt; talks about the &lt;em&gt;&quot;dishwasher [...] powder or tablets [...] are often highly alkaline&quot;&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For dishwashing liquids where you are thought/supposed to put your hands in, the pH is usually neutral to slightly acidic, so you need to scrub a bit more, but don't dissolve your skin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt; Acids for cleaning? &lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Acids (usually acetic acid or citric acid, maybe HCl) are used in &quot;household chemistry&quot; to clean limescale (CaCO$_3$). But that usually isn't the problem of your dirty dishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dishwashing machines use ion exchangers to soften the water (or phosphate in the dishwashing agent), so you don't get limescale or lime soaps inside the machine.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2012-05-14T21:34:05.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="365" PostTypeId="2" ParentId="363" CreationDate="2012-05-12T19:13:38.807" Score="8" Body="&lt;p&gt;How flattering that someone asked this question ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is yes, I've tried QTPIE on series of conjugated molecules like the oligoacetylenes and oligoacenes. The results, however, are fairly dismal, as can be seen in the preprint. The problem is that in these empirical charge models, the polarizability grows faster than the size of the molecule, which means that these models predict infinite polarizabilities and charge transport behavior for bulk materials, which is obviously nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is quite intimately related to that of self-interaction error in approximate density functional theories, which also exhibit similar problems.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-12T19:13:38.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="366" PostTypeId="2" ParentId="308" CreationDate="2012-05-12T19:32:36.653" Score="2" Body="&lt;p&gt;To summarize the differences and similarities, here is a list (I will update the list, if new comments arrive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;differences&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;melting is (almost) always exothermic, dissolving can be endo- or&#xA;exothermic&lt;/li&gt;&#xA;&lt;li&gt;melting is within one substance, dissolvation is between solute and&#xA;solvent that are different substances&lt;/li&gt;&#xA;&lt;li&gt;melting is a collective phenomenon that requires $N_a$ particles, dissolvation can happen one atom at the time &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;similarities&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In both cases, bonds are broken by crossing an energy barrier&lt;/li&gt;&#xA;&lt;li&gt;phase of matter changes&lt;/li&gt;&#xA;&lt;li&gt;You can create a situation where it is difficult to say whether the&#xA;substance is melting or dissolving.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditDate="2012-05-12T20:01:10.673" LastActivityDate="2012-05-12T20:01:10.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="367" PostTypeId="1" AcceptedAnswerId="14076" CreationDate="2012-05-12T21:15:40.380" Score="5" ViewCount="532" Body="&lt;p&gt;What kind of bonds silica bead forms with polystyrene? Silica is now as ~100 $\mu$m spherical particles in bulk polystyrene that is made by melting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I bond the silica particles&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;more tightly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;more loosely&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;to the polystyrene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be generalized to any nano/micro particle and plastics? I am thinking now metal oxides and other thermoplastics.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="4945" LastEditDate="2014-05-22T09:11:39.963" LastActivityDate="2015-08-15T10:49:03.857" Title="How is silica bead bonded into polystyrene?" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="368" PostTypeId="2" ParentId="356" CreationDate="2012-05-12T21:46:05.207" Score="3" Body="&lt;p&gt;I have to stretch my memory to remember how this goes (and read a bit from &lt;a href=&quot;http://www.tf.uni-kiel.de/matwis/amat/def_en/kap_5/backbone/r5_4_1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a &lt;strong&gt;partial&lt;/strong&gt; (edge)dislocation (partial dislocations have their own name). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to go this through by yourself and check that I didn't make any mistakes.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2012-05-12T21:46:05.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="369" PostTypeId="1" CreationDate="2012-05-12T21:46:57.587" Score="5" ViewCount="724" Body="&lt;p&gt;X-ray photoelectron spectroscopy (&lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;XPS&lt;/a&gt;) is a particularly useful technique for examining the surfaces of all sorts of materials. Organic monolayers are particularly difficult to characterize completely via this technique due to the fact that many carbon peaks do not differ much by chemical shift. Traditionally quantitative information is obtained via a lot of guess-work and forced peak fitting. It would be better to predict peaks shapes and peak positions beforehand to aid in the interpretation and prevent a terribly mis-guided interpretation. How well can theory help the non-specialist with this? I wouldn't have the time to develop a program to do this either. Preferably one could point me to the right direction in terms of a research paper/software.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2012-05-12T21:53:41.230" LastActivityDate="2013-07-10T17:57:05.960" Title="Predict XPS Spectrum of Organic Monolayers on a Semiconductor" Tags="&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="370" PostTypeId="2" ParentId="308" CreationDate="2012-05-13T05:03:50.637" Score="7" Body="&lt;p&gt;Juha invited me to write a summary (see comment on my previous answer ) of the differences between melting and dissolving. I’ll try to outline this roughly in the same order as his. I'm giving this as a new answer since my last answer was quite long as it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Differences&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Melting and dissolving are completely different processes on the molecular/atomic level that could not be mistaken for each other if you could observe what was happening at that scale.&lt;/li&gt;&#xA;&lt;li&gt;If you cool the liquid that arose from melting the solid, to a temperature below it’s melting point, you would see the entire sample solidify. If you cool the solution (dissolved solid + solute) to below the melting point of the solid (solute) in it, you would see no change. (Unless you had a saturated or nearly saturated solution.)&lt;/li&gt;&#xA;&lt;li&gt;Melting requires only a single substance and energy input while dissolving requires a solvent and a solute that are compatible (“like forces”). (This is actually a pretty huge difference and would potentially affect all of the physical and chemical properties.) Dissolving a solid can be either endo- or exo-thermic.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Phase change&quot; (wording from the previous summary)  In each case you end up with a liquid. Melting caused a phase change (composition of the substance didn’t change) while dissolving a solid in a liquid is not considered a phase change since a change in composition occurred.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Similarities&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In each case, forces between the particles that comprise the solid are disrupted and that takes energy. (Whether it’s chemical bonds or intermolecular forces depends on the process and on the solid and on your definitions. (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds&quot;&gt;See this question.&lt;/a&gt;) &lt;strong&gt;But&lt;/strong&gt; melting (rare exception noted in previous comments) is &lt;strong&gt;endothermic&lt;/strong&gt; and dissolving can be either endo- or exo-thermic.&lt;/li&gt;&#xA;&lt;li&gt;In each case you end up with a liquid. Macroscopically, if you walked into a room and saw the liquid on the table, it would be difficult to say whether this liquid came from a solid that had melted or a solid that had dissolved in a solute and made a solution.&#xA;&lt;em&gt;But it would be very easy to determine which you had experimentally in a “dozen” different ways.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Both melting and dissolving require interaction among groups of atoms, molecules, or ions. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are probably more differences that could be given (how to handle thermodynamic calculations, complexity of the system etc.), and possibly more similarities, but that's enough for me on this topic. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2012-05-13T05:03:50.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="371" PostTypeId="2" ParentId="249" CreationDate="2012-05-13T10:43:24.290" Score="10" Body="&lt;p&gt;The signals you get in a simple 1D proton spectrum are in most cases roughly quantitative, though there are some exceptions and some aspects you need to consider if you need high accuracy.&lt;/p&gt;&#xA;&lt;h3&gt;Acquisition&lt;/h3&gt;&#xA;&lt;p&gt;There are several things you need to keep in mind while setting up the experiment to ensure that the resulting NMR spectrum will be quantitative.&lt;/p&gt;&#xA;&lt;p&gt;The spectrum should be of high quality, any artifacts and noise will affect the quantification.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Relaxation delay&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The relaxation delay is the time between experiments, if it is too short some protons won't relax completely back to the equilibrium state. This affects the intensity of the NMR signal and will cause the integrals of those signals to be off.&lt;/p&gt;&#xA;&lt;p&gt;The relaxation delay (including the acquisition time) should be around &lt;span class=&quot;math-container&quot;&gt;$5 \cdot \mathrm{T_1}$&lt;/span&gt;. You can either estimate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{T_1}$&lt;/span&gt; or measure it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;sup&gt;13&lt;/sup&gt;C satellites&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;sup&gt;13&lt;/sup&gt;C satellites are around 1% of the total signal, if you need high accuracy you should use decoupling to get rid of those.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Other aspects&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;You also need to ensure that the pulses used in the experiment have a reasonably flat excitation profile for the area you're looking at. This is usually no problem for standard 1D proton experiments, but might be a problem for other kinds of experiments or if you have an unusually long pulse and/or and unusually wide distribution of your chemical shifts.&lt;/p&gt;&#xA;&lt;h3&gt;Processing&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;Baseline&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The baseline should be absolutely flat and at exactly zero, any error there will lead to large errors in the quantification. Using a digital filter during acquisition is a good idea for that (&lt;code&gt;baseopt&lt;/code&gt; option for Bruker spectrometers), this will ensure a very flat baseline at zero, as long as no other problems distort the baseline.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Phasing&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The phase-correction should be exact, it might be necessary to perform a manual phase-correction.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-05-13T10:43:24.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="372" PostTypeId="1" CreationDate="2012-05-13T14:42:28.543" Score="-5" ViewCount="1460" Body="&lt;p&gt;I have the following question to solve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tungsten, $\ce{W}$, and chlorine, $\ce{Cl}$, form a series of compounds with the following compositions:&#xA;  \begin{array}{rr}&#xA;\text{Mass % W}  &amp;amp; \text{Mass % Cl}\\&#xA;\hline&#xA;    72.17     &amp;amp;  27.83\\&#xA;    56.45     &amp;amp;  43.55\\&#xA;    50.91     &amp;amp;  49.09\\&#xA;    46.36     &amp;amp;  53.64\\&#xA;\end{array}&#xA;  If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one gram of tungsten, chlorine has mass&#xA;$$&#xA;\begin{align}&#xA;\frac{27.83}{72.17} &amp;amp;= 0.3302~\mathrm{g}\\[3pt]&#xA;\frac{43.55}{56.45} &amp;amp;= 0.7715~\mathrm{g}\\[3pt]&#xA;\frac{49.09}{50.91} &amp;amp;= 0.9643~\mathrm{g}\\[3pt]&#xA;\frac{53.64}{46.36} &amp;amp;= 1.157~\mathrm{g}.\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\frac{0.7715}{0.3302}\approx\frac{7}{3}$, $\frac{0.9643}{0.3302}\approx3$, and $\frac{1.157}{0.3302}\approx\frac{7}{2}$, the number of atoms of chlorine for a given mass of tungsten are respectively in the ratio $6:14:18:21$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a molecule of each compound contains only one tungsten atom, the formulas are $\ce{WCl_{6}}$, $\ce{WCl_{14}}$, $\ce{WCl_{18}}$, and $\ce{WCl_{21}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="211" LastEditorUserId="4945" LastEditDate="2018-08-10T13:05:19.577" LastActivityDate="2018-08-10T13:05:19.577" Title="How to compute molecular formula?" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="373" PostTypeId="2" ParentId="372" CreationDate="2012-05-13T17:31:13.887" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We don't have a homework policy yet, but I'm quite sure that it will be against providing full answers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You're supposed to take into account molar masses. The ratios in a formula are &lt;em&gt;molar&lt;/em&gt; ratios, so you need to use those, not mass ratios.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Read up a bit on emperical formulae, as Janice suggested.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-13T17:31:13.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="374" PostTypeId="1" AcceptedAnswerId="375" CreationDate="2012-05-13T18:08:52.127" Score="53" ViewCount="68457" Body="&lt;p&gt;In the standard brown ring test for the nitrate ion, the brown ring complex is:&#xA;$$\ce{[Fe(H2O)5(NO)]^{2+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, iron has stable oxidation states +2 and +3. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the $\ce{NO}$ to −1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorDisplayName="user7951" LastEditDate="2018-04-26T06:01:38.117" LastActivityDate="2018-04-26T06:01:38.117" Title="Is iron in the brown ring compound in a +1 oxidation state?" Tags="&lt;coordination-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="375" PostTypeId="2" ParentId="374" CreationDate="2012-05-13T20:09:43.000" Score="25" Body="&lt;p&gt;Your basic assumption is incorrect: the iron in [Fe(H&lt;sub&gt;2&lt;/sub&gt;O)&lt;sub&gt;5&lt;/sub&gt;(NO)]&lt;sup&gt;2+&lt;/sup&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrate_test#Brown_ring_test&quot;&gt;Fe(III)&lt;/a&gt;, and the ligand is NO&lt;sup&gt;–&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-13T20:09:43.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="376" PostTypeId="2" ParentId="152" CreationDate="2012-05-13T22:01:03.963" Score="64" Body="&lt;p&gt;I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different &lt;strong&gt;angular momentum quantum number&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; but with the same &lt;strong&gt;principal quantum number&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; (&lt;em&gt;e.g.&lt;/em&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3s}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3p}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt;) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?&#xA;Although &lt;code&gt;AcidFlask&lt;/code&gt; already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.&#xA;I will split my answer in three parts: The first will address the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3s}$&lt;/span&gt; states are lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3p}$&lt;/span&gt; states (which are in turn lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt; states).&lt;/p&gt;&#xA;&lt;h2&gt;&lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy of the hydrogen atoms energy eigenvalues&lt;/h2&gt;&#xA;&lt;p&gt;The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kepler_problem&quot; rel=&quot;noreferrer&quot;&gt;Kepler problem&lt;/a&gt; known from classical mechanics.&#xA;This potential (aka Kepler potential) has the form &lt;span class=&quot;math-container&quot;&gt;$\frac{\kappa}{r}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the distance between the nucleus and the electron, and &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; is a proportionality constant.&#xA;Now, it is known from physics that symmetries of a system lead to conserved quantities (&lt;a href=&quot;http://en.wikipedia.org/wiki/Noether%27s_theorem&quot; rel=&quot;noreferrer&quot;&gt;Noether Theorem&lt;/a&gt;).&#xA;For example from the rotational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;. But while the length of the angular momentum vector is fixed by &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; there are still different possibilities for the orientation of its &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-component, characterized by the &lt;strong&gt;magnetic quantum number &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;&lt;/strong&gt;, which are all energetically equivalent as long as the system maintains its rotational symmetry.&#xA;So, the rotational symmetry leads to the &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;-degeneracy of the energy eigenvalues for the hydrogen atom.&#xA;Analogously, the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the &lt;span class=&quot;math-container&quot;&gt;$SO(4)$&lt;/span&gt; symmetry.&#xA;The system's &lt;span class=&quot;math-container&quot;&gt;$SO(4)$&lt;/span&gt; symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.&#xA;(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)&#xA;This dynamical symmetry conserves the &lt;a href=&quot;http://en.wikipedia.org/wiki/Runge-Lenz&quot; rel=&quot;noreferrer&quot;&gt;Laplace-Runge-Lenz vector&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\hat{\vec{M}}$&lt;/span&gt; and it can be shown that this conserved quantity leads to the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-independent energy spectrum with &lt;span class=&quot;math-container&quot;&gt;$E \propto \frac{1}{n^2}$&lt;/span&gt;. (A detailed derivation, though in German, can be found &lt;a href=&quot;http://www.desy.de/%7Ejlouis/Vorlesungen/PS_06/vortrag_09.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;h2&gt;Why is the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy of the energy eigenvalues lifted in multi-electron atoms?&lt;/h2&gt;&#xA;&lt;p&gt;As the &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, &lt;em&gt;e.g.&lt;/em&gt;, by applying a magnetic field, the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure &lt;span class=&quot;math-container&quot;&gt;$\frac{\kappa}{r}$&lt;/span&gt; form.&#xA;This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.&#xA;(Other factors, like spin and relativistic effects, also lead to a lifting of the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy even in the hydrogen atom.)&lt;/p&gt;&#xA;&lt;h2&gt;Why do states with the same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; but lower &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; values have lower energy eigenvalues?&lt;/h2&gt;&#xA;&lt;p&gt;Two effects are important here:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;centrifugal force&lt;/strong&gt; puts an &amp;quot;energy penalty&amp;quot; onto states with higher angular momentum.&lt;span class=&quot;math-container&quot;&gt;${}^{1}$&lt;/span&gt; So, a higher &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; value implies a stronger centrifugal force, that pushes electrons away from the nucleus.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part &lt;span class=&quot;math-container&quot;&gt;$R(r)$&lt;/span&gt; of the wave function &lt;span class=&quot;math-container&quot;&gt;$\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0&#xA;\end{equation}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad&#xA;\end{equation}&lt;/span&gt;&#xA;The radial part experiences an additional &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-dependent potential &lt;span class=&quot;math-container&quot;&gt;$V^{\ell}_{\mathrm{cf}} (r)$&lt;/span&gt; that pushes the electrons away from the nucleus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Core repulsion&lt;/strong&gt; (Pauli repulsion), on the other hand, puts an &amp;quot;energy penalty&amp;quot; on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum&lt;span class=&quot;math-container&quot;&gt;${}^{1}$&lt;/span&gt;. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; value feel an additional effect from core orthogonalization.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The &amp;quot;accidental&amp;quot; &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt;-degeneracy of the hydrogen atom can be described as a balance&#xA;between centrifugal force and core repulsion, that both act against the nuclear Coulomb&#xA;attraction.&#xA;In the real atom the balance between centrifugal force and core repulsion is broken,&#xA;The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.&#xA;Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, &lt;em&gt;i.e.&lt;/em&gt; lower &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; values. So, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3s}$&lt;/span&gt; states are lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3p}$&lt;/span&gt; states which are in turn lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt; states.&lt;/p&gt;&#xA;&lt;p&gt;Of course, one has to be careful when using results of the hydrogen atom to describe effects in multielectron atoms as &lt;code&gt;AcidFlask&lt;/code&gt; mentioned. But since only a qualitative description is needed this might be justifiable.&lt;/p&gt;&#xA;&lt;p&gt;I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-12-24T06:09:50.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="377" PostTypeId="2" ParentId="313" CreationDate="2012-05-14T00:05:50.780" Score="4" Body="&lt;p&gt;Toxicities are often greatly influenced by how you are exposed and in what dose.&lt;/p&gt;&#xA;&lt;p&gt;Mercury is a fascinating metal. You can find it as the pure metal, a salt or in an organic form. In its standard state it is a liquid and is rather volatile. Large amounts are naturally released into the atmosphere, but humans contribute as well, for example: burning fossil fuels, such as coal, will release it into the atmosphere.&lt;/p&gt;&#xA;&lt;p&gt;The forms of mercury are not set in stone it is quite able to switch forms from the free metal, to the salt, to the organic by processes in nature.&lt;/p&gt;&#xA;&lt;p&gt;Orally consuming some liquid metal isn't the worst possible route of exposure; mercury was at one time even used as a laxative. The only absorption in the GI tact is thought to be due to mercury vapor. Skin exposure is undesirable, but we know that inhaling the vapor is more toxic because it is soluble in lipids. Upon entering the bloodstream, some makes it way past the blood/brain barrier (hence its neurotoxicity) and some is converted to Hg(II) by &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalase&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalase&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Mercury's inorganic salts have a long useful history where some were used to treat syphilis, some for lights and some for hat making and so on. &amp;quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mad_hatter_disease&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mad as a hatter&lt;/a&gt;&amp;quot; won its name due to chronic exposure to Hg(NO&lt;span class=&quot;math-container&quot;&gt;$_3$&lt;/span&gt;)&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;, which was used in hat making. This salt isn't very lipid soluble, but more absorption does occur when swallowed than the free metal. These salts tend to build up in the kidneys after absorption and high exposure will cause renal failure. Not much will pass through the blood/brain barrier, but chronic exposure will eventually allow some across and cause neurotoxicity. They also tend to bind with proteins containing sulfur and this can cause some cell death. As a general rule, Hg(I) salts are less toxic than Hg(II), presumably due to the decreased solubility of Hg(I). This does not mean they cannot cause harm: mercuric chloride was used in a factory and the wastes were discharged into a nearby sea, anaerobic bacteria methylated the mercury, this methylmercury was absorbed by fish that were later eaten and &lt;a href=&quot;http://en.wikipedia.org/wiki/Minamata_disease&quot; rel=&quot;nofollow noreferrer&quot;&gt;hurt&lt;/a&gt; many people.&lt;/p&gt;&#xA;&lt;p&gt;As an organic, mercury compounds tend to be considerably more lipid soluble and therefore can readily be absorbed or cross the blood/brain barrier, in comparison to the free metal. A terrible example is dimethylmercury, which claimed the life of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Karen_Wetterhahn&quot; rel=&quot;nofollow noreferrer&quot;&gt;inorganic chemist&lt;/a&gt; who was accidentally exposed to some she was using in an NMR experiment.&lt;/p&gt;&#xA;&lt;p&gt;I do not know much about its amalgams aside from gold/silver. I believe most of the toxicity was seen in gold refineries due to the fact that the amalgam is heated to remove the mercury, which leaves behind gold, and results in the release of mercury vapor.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-05-14T00:05:50.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="378" PostTypeId="1" AcceptedAnswerId="391" CreationDate="2012-05-14T03:44:19.323" Score="21" ViewCount="489" Body="&lt;p&gt;In &lt;a href=&quot;http://www.springerlink.com/content/g83871qw83n26152/abstract/&quot; rel=&quot;noreferrer&quot;&gt;a series of papers in the early 1980s&lt;/a&gt;, Michael Springborg explored an interpretation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution&quot; rel=&quot;noreferrer&quot;&gt;Wigner phase space function&lt;/a&gt; as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Springborg 6D phase model model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ADDENDUM 2012-05-14: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a world dominated by Hilbert spaces, the orbitals of chemistry are interesting because they are one of the few places in physics where a 3D spatial image of the underlying quantum situation is consistently retained. (My recollection is that Schrodinger liked such low-dimensionality approaches, since they emphasized waves that could be visualized.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, the problem of course is that you need both position $q$ and momentum $p$ to get an accurate picture of the overall electron state. Using a space-only 3D $q$ representation forces the analysis of molecules to flip back and forth at rather arbitrary thresholds between localized (position space $q$) and delocalized (momentum space $p$) views of the electrons in the molecules or metal crystals in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That bothers me, both from a physics perspective and from a computational integrity perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that there exists a very deep and profound physics symmetry between $q$ and $p$ in terms of how effects such as Pauli exclusion applies to fermions. I find that simple symmetry nothing short of amazing, because to me there's just nothing intuitive about the idea that half-unit spin would result in such extraordinarily similar behaviors in these quite different spaces. At some level it seems to be one of those cases of &quot;that's just the way it works.&quot; So, it seems pretty reasonable that accurate modeling of orbitals of many diverse sizes likely requires this symmetry to be captured accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true, this older work by Springborg (he still heads a research group BTW) strikes me as a possible opportunity to for taming some models, making them smoother and more sane over a much broader range of molecular sizes. That's because Springborg's 6D electron density functions -- if they work OK in broader contexts -- might help eliminate any need for abrupt switches between &quot;mostly position&quot; or $q$ 3D views (classic &quot;electron clouds,&quot; whatever those really mean) and &quot;mostly momentum&quot; or $p$ 3D views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place to test whether 6D $(q,p)$ electron distributions might provide higher computational stability would be in modeling a range of lengths for a long-chain carbon polymer such as polyacetylene. (I always pick polyacetylene for its simplicity, but there are other good choices.) Springborg appears not to have explored that domain, since as best I can tell his papers addressed mostly simpler bonds. (I think; I don't have them.) The &lt;a href=&quot;https://chemistry.stackexchange.com/a/365/83&quot;&gt;instabilities of approximate density functional theories&lt;/a&gt; &lt;em&gt;may&lt;/em&gt; (or may not) be related to $(q,p)$ transition instabilities. That's not a question I can or should try to address, as there are folks out there who know that domain orders of magnitude better than I ever will.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ADDENDUM 2012-05-16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned &quot;algorithmic opportunity&quot; and wondered what I meant myself by that, since as @AcidFlask reminds the $q$ and $p$ basis sets are fully interchangeable in terms of representing any stationary wavefunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is (again, I think) one of accumulating uncertainty. Picture doing some kind of calculation in the $q$ representation. For a highly localized electron, you should get a pretty solid answer, but for something like a delocalized electron in a conductive polymer, a higher level of error would be likely (not a necessity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, Fourier transform the result into $p$, and do the next iteration of whatever it is you are trying to calculate. There the nature of the errors should change, but in the case of a delocalized electron, it's likely to work out a bit more accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You reverse transform that and repeat, with the objective of converging to a solution that looks pretty stable in both spaces. That, I would argue, is the &quot;algorithmic opportunity&quot; part of the 6D argument: that is, that by incorporating both views, you can come to an &lt;em&gt;overall&lt;/em&gt; solution that diverges less from reality when applied to a broad range of molecular orbital sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there problems with that analysis? Oh my yes. You'd have to be nuts to do complete Fourier transforms twice per full iteration of your convergence, and even then, the devil will be in the details about whether it even &lt;em&gt;works&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the operative word was &quot;opportunity,&quot; that is, such an analysis hints that there &lt;em&gt;may&lt;/em&gt; be away to use the dual perspective to create more stable approximation methods. To really work the algorithm would necessarily &quot;live&quot; a lot closer to a continuous 6D representation with constraints than a per-iteration transform back and forth. Can that be done? I don't know; as I said, folks like AcidFlux know this area orders of magnitude better than I ever will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, for any non-mathematical readers (AcidFlux, shoo, go away now!) who may have wandered in, the section below provides a simple, easy-to-imagine visual analogy to explain why location and momentum both play a major role at the level of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a very flexible membrane with many embedded holes (but the holes don't stretch). Label one side of the membrane &quot;ordinary space&quot; (abbreviated $q$, don't ask why), and the other side &quot;momentum space&quot; (labeled $p$, and I told you not to ask).  Next, put water balloons with fixed quantities of water in them through each hole, glued to the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $q$ side of each balloon now represents the &quot;real space&quot; representation of the water balloon. If you squeeze that part very tightly, it will shrink down almost to a point, so it looks more like a particle than a balloon. You think you have &quot;captured&quot; that balloon n a small space, but beware! On the other side, the $p$ side of the membrane, the balloon has gotten comparatively huge! So your point-like &quot;capture&quot; of the balloon on the $q$ side was more illusion than reality, since in the broader view that includes both sides, all you really did is shift a lot of the balloon from the $q$ world into the hidden $p$ side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that you don't really &lt;em&gt;need&lt;/em&gt; to see both sides at once, because each side of the balloon fully determines the size of the other side. From the amount of water on one side you can &lt;em&gt;always&lt;/em&gt; calculate exactly how much water is in the balloon on the other, hidden side. And the opposite is also true! That is, if you know how big each balloon is on the $p$ side, you also automatically know how much of the balloon is on the $q$ side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That kind of equivalence between two views that may appear at first to be quite different is an example a &quot;basis set.&quot; A basis set has enough &quot;pieces&quot; in its tool kit to allow you to represent something &lt;em&gt;completely&lt;/em&gt;, in the sense that you have all of the information you will ever need to know about the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, complete is not always the same as convenient. For example, if you try to stack together the balloons as they appear on the $q$ side, you will quickly find that the hidden $p$ side of the balloons can make it very difficult. The balloons just won't pack as easily together as their forms on the $q$ side seem to suggest they would. The problem, of course, is that you have those huge $p$ sides of the balloons interacting in a very different way over in the $p$ or &quot;momentum space&quot; side of the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not that any information is missing! After all, since the $q$ side provides a complete basis set for describing the balloons, you &lt;em&gt;know&lt;/em&gt; that when a balloon is small there that it must necessarily be large in $p$. The trick, then, is make sure that you don't have to recalculate the size on the balloons on the other side every single time you try to do something with them on the $q$ side. That's inefficient at best, and could easily turn into a significant source of errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's a bit simpler and certainly a lot less calculation if you preserve a simultaneous image of &lt;em&gt;both&lt;/em&gt; sides of the membrane, and change them only when you have to, e.g. after squeezing some of the balloons on either (or both) the $q$ and $p$ sides. Doing that is inherently redundant, sure, since you only &lt;em&gt;need&lt;/em&gt; to have one side to recreate the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you've ever done some serious algorithm design, you know that &quot;caching&quot; mostly persistent results can be a great way to reduce calculation times, sometimes hugely. More subtly, it can also help stabilize the problem if it turns out that you have complementary error modes across your different &quot;views&quot; (basis sets) of the final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, put that all together, and that's what I'm trying to suggest &lt;em&gt;may&lt;/em&gt; also be true for the physics of electrons, which similarly like to slosh around (as in Planck's Theory of Slosh,&quot; and yes &lt;em&gt;of course&lt;/em&gt; I just now made that up!) and spend some of their time point-like in one space (called $q$ or position space; ordinary, real space), or alternatively in another more abstract space (called $p$ or momentum space). Interestingly, just as with the balloons, they bump into and push each each other in almost exactly the same way in &lt;em&gt;both&lt;/em&gt; of these two spaces. Particles that do that are called fermions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are fermions so turf-protective in &lt;em&gt;both&lt;/em&gt; of these quite different spaces? I really don't think anyone really knows, at least not with the kind of depth that is as satisfying as when chemistry started coming together and making deep sense. It's just one of those little mysteries of how the universe works, one (there are several actually) that's still sitting there thumbing its nose at us and saying &quot;ha ha, gotcha!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's cool! What's a universe without a few mysteries still hanging around, after all?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2012-05-17T11:56:12.983" Title="Is the Springborg 6D phase space model used in modern molecular orbital modeling?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;electronic-configuration&gt;&lt;software&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="379" PostTypeId="1" AcceptedAnswerId="381" CreationDate="2012-05-14T09:14:14.890" Score="8" ViewCount="238" Body="&lt;p&gt;I'm reading a book on carbon nanotubes (P.J.F. Harris, Carbon Nanotube Science, to be specific) at the moment and puzzling about Stone-Wales or 5775 defects. This is where two adjacent carbons undergo a 90-degree rotation about their mutual centroid and generate a defect consisting of two five membered rings and two seven membered rings, as I've illustrated below (This is only supposed to be qualitative, and is on a finite PAH fragment rather than a nanotube, but you should get the idea):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLFKf.png&quot; alt=&quot;Stone-Wales defect on PAH&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I note that tropylium ($\ce{[C7H7]^+}$) is aromatic as a cation, and the ever famous cyclopentadienide ($\ce{[C5H5]^-}$) is aromatic as an anion, so what I'm wondering is: do the 5 and 7-membered rings in a Stone-Wales defect pick up either formal or partial negative and positive charges to maintain aromaticity in a fashion consistent with the rest of the structure?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;P.S.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Big props to F'x, who attempted electronic structure calculations on the PAHs depicted, regrettably without SCF convergence.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-14T12:20:29.573" LastActivityDate="2012-05-14T14:43:10.057" Title="What does the charge distribution around a Stone-Wales defect look like?" Tags="&lt;theoretical-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="380" PostTypeId="1" CreationDate="2012-05-14T11:33:53.190" Score="16" ViewCount="4622" Body="&lt;p&gt;My Sister asked me this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Why is it that when we chew gum,  it is soft and mushy to begin with , but slowly gets firmer and firmer like after 20 mins of chewing?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think  it is because when we initially chew gum it has plenty  of sugar causing rapid salivation and hence mushiness and the rubber polymer is very flexible to begin with and slowly looses its elasticity hence becomes firmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the chemistry involved ?&lt;/p&gt;&#xA;" OwnerUserId="199" LastEditorUserId="402" LastEditDate="2016-10-09T18:50:59.287" LastActivityDate="2019-12-06T00:37:21.643" Title="Why does gum get harder to chew after a while?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="381" PostTypeId="2" ParentId="379" CreationDate="2012-05-14T14:18:18.603" Score="7" Body="&lt;p&gt;FWIW, here are some Hückel calculations on the 5775 PAH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm showing the first 7 energy levels. Each molecular graph presents the contribution of each atom to the electron density ($\mathrm{Abs}[c_i]^2$). This is not normalized. I do not use wavefunctions, only the coefficients multiplying them, assuming&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi_{\mathrm{mol}} = \sum_i^n c_i\psi_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this approximation (HMO), the first 7 energy levels will be occupied (14 pi electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will first give you pictures of each energy level, and then a linear combination of each. Take it for what it is, as there is much information lost in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, this simply &lt;em&gt;suggests&lt;/em&gt; that indeed, the system will behave as if composed of tropylium and cyclopentadienyl ions. But this is not publication quality calculation either =)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2bDc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are compared the linear combination of the first 7 energy levels of the 5775 and 6666 systems. It is clear that electron distribution is not the same in both PAHs, which hints that my calculations are not so far fetched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5775&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/itewem.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6666&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bt6rDm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-05-14T14:43:10.057" LastActivityDate="2012-05-14T14:43:10.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="382" PostTypeId="2" ParentId="380" CreationDate="2012-05-14T14:20:24.810" Score="4" Body="&lt;p&gt;I would be surprised if there was reticulation of the gum's polymers upon contact with your saliva. According to &lt;a href=&quot;http://www.gumbase.com/on-line/Home/Confectionery/Products/GUMBASE.html&quot; rel=&quot;nofollow&quot;&gt;this manufacturer&lt;/a&gt;, plasticizers are used, and the plasticizers &lt;a href=&quot;http://en.wikipedia.org/wiki/Gum_base&quot; rel=&quot;nofollow&quot;&gt;appear to be fats&lt;/a&gt;. This would mean that, upon chewing, the plasticizers would be removed from the gum, rendering it unpleasantly harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plasticizers are used to render a polymer more pliable, more elastic, than it naturally is. This could get its own question, imo.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-05-14T14:20:24.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="383" PostTypeId="1" AcceptedAnswerId="387" CreationDate="2012-05-14T15:32:43.253" Score="13" ViewCount="19570" Body="&lt;p&gt;I googled a bit about what $\ce{AgNO3 + AlCl3}$ will yield and found out the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3AgNO3 + AlCl3 -&amp;gt; Al(NO3)3 + 3AgCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? I know that $\ce{Ag}$ is higher up in the reactivity series than $\ce{Al}$ but that does not make sense to me in this problem. Is it because $\ce{Cl}$ is just one when $\ce{NO3}$ is multiple?&lt;/p&gt;&#xA;" OwnerUserId="221" LastEditorUserId="46306" LastEditDate="2018-02-20T08:10:19.190" LastActivityDate="2020-04-30T13:44:51.923" Title="Reaction between silver nitrate and aluminum chloride" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="384" PostTypeId="2" ParentId="383" CreationDate="2012-05-14T15:48:20.040" Score="5" Body="&lt;p&gt;To analyse such situations, you first must realize that nearly everything is present as ions. For example, if you took the reaction $\ce{NaCl +KBr-&amp;gt;NaBr +KCl}$, the &quot;direction&quot; it goes in has no real meaning since $\ce{Na+}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Br-}$ are present as ions. There is no actual &lt;em&gt;reaction&lt;/em&gt; going on here, just a mixing of ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we have something slightly different. $\ce{AgCl}$ precipitates here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3AgNO3_{(aq)} + AlCl3_{(aq)} -&amp;gt; Al(NO3)3_{(aq)} + 3AgCl v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the $\ce{NaCl}$ case above, $\ce{NO3-}$ and $\ce{Al^{3+}}$ are present as ions on both sides--so nothing to worry about there. We can safely remove them from the equation. This reduces the reaction to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag+ + 3Cl- -&amp;gt; 3AgCl v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have reduced the question to &quot;why does $\ce{AgCl}$ precipitate?&quot;. The easiest way to decide is by looking at thermodynamic stability--solid silver chloride is more stable. We can experimentally prove this, or use standard enthalpies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions proceed in the direction of more stability, so the above reaction goes forward.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course, I sort of &quot;assumed&quot; that $\ce{AgCl}$ precipitates in the beginning--this was so that I wouldn't have to complicate stuff. Nevertheless, we can, in a similar way, experimentally prove that the others are found in ionized form and generally do not precipitate. There are some nice arguments why $\ce{AgCl}$ is less soluble than $\ce{AgNO3}$  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/410/why-is-ceagcl-less-soluble-than-ceagno3&quot;&gt;in this post&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-05-17T13:18:27.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="385" PostTypeId="1" AcceptedAnswerId="386" CreationDate="2012-05-14T17:37:27.707" Score="7" ViewCount="150" Body="&lt;p&gt;For example — &quot;high purity chloride salt of Zinc $(\ce{ZnCl2.2H2O})$&quot; or &quot;various concentrations of $\ce{FeCl2.4H2O}$&quot;. What does the number before $\ce{H2O}$ mean?&lt;/p&gt;&#xA;" OwnerUserId="222" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T01:25:30.137" LastActivityDate="2017-02-16T18:04:18.233" Title="How to understand this form of writing the solution: (some salt • n H₂O)?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="386" PostTypeId="2" ParentId="385" CreationDate="2012-05-14T17:50:55.550" Score="7" Body="&lt;p&gt;The salt's crystal lattice's repeating unit is constituted of &lt;strong&gt;n&lt;/strong&gt; molecules of salt and &lt;strong&gt;m&lt;/strong&gt; molecules of water. Such salts are called hydrates. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrate&quot;&gt;Wikipedia&lt;/a&gt; even has a nice picture of a hydrated vs non hydrated salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have anhydrous ferrous chloride, as well as ferrous chloride &lt;em&gt;tetrahydrate&lt;/em&gt;. The number indicates how many water molecules are present in the crystal lattice's unit cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has just crossed my mind: be careful when preparing solutions of metal complexes to make sure you know what compound you're working with. One gram of anhydrous salt contains more equivalents than a gram of hydrated salt. This information should be obvious on the bottle. If not, ask a lab tech if they can identify the hydrated/anhydrous salt by memory.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-05-14T17:58:05.200" LastActivityDate="2012-05-14T17:58:05.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="387" PostTypeId="2" ParentId="383" CreationDate="2012-05-14T18:49:39.130" Score="9" Body="&lt;p&gt;Le Châtelier's principle only states that a system previously at equilibrium will want to &lt;em&gt;stay&lt;/em&gt; at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is &lt;em&gt;initially not at equilibrium&lt;/em&gt;, and therefore we don't need his principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the reaction proceeds because the formation of AgCl&lt;sub&gt;(s)&lt;/sub&gt; is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO&lt;sub&gt;3&lt;/sub&gt;, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO&lt;sub&gt;3&lt;/sub&gt; since it is insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is a type of metathesis reaction. It can be called &lt;em&gt;double displacement reaction&lt;/em&gt; or &lt;em&gt;salt metathesis reaction&lt;/em&gt;, depending on who you're talking with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers make an error, I think. Both suggest that the &lt;em&gt;solubility product&lt;/em&gt; of AgCl is of importance in determining &lt;em&gt;why&lt;/em&gt; the reaction takes place. My opinion is that the argument, &lt;strong&gt;as presented&lt;/strong&gt;, is circular, in the following manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;vs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;AgCl is insoluble in water, therefore its solubility product is low.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that the fact AgCl is insoluble in water implies that it has a low K&lt;sub&gt;sp&lt;/sub&gt; in water. If someone can &lt;em&gt;predict and quantify&lt;/em&gt; the solubility product of AgCl in water &lt;em&gt;from first principles&lt;/em&gt;, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the K&lt;sub&gt;sp&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag&lt;sup&gt;+&lt;/sup&gt; and Cl&lt;sup&gt;-&lt;/sup&gt; ions &lt;em&gt;from different sources&lt;/em&gt; (i.e. in different solutions) put into contact will spontaneously form a precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you maybe thinking this argument is also circular, but it's not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have access to experiment, which clearly informs us that AgCl is &lt;em&gt;insoluble&lt;/em&gt; in water. We can confidently assume (but we might be wrong!) that the AgCl bond is &lt;em&gt;less&lt;/em&gt; ionic than the AgNO&lt;sub&gt;3&lt;/sub&gt; bond, since it &lt;em&gt;practically does not dissociate/has a low K&lt;sub&gt;sp&lt;/sub&gt; in water&lt;/em&gt;. A bond with a greater covalent character is more stable, since the electrons are &lt;em&gt;shared&lt;/em&gt; and not &lt;em&gt;transfered&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H&lt;sub&gt;2&lt;/sub&gt;O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you can see, I cannot provide you with a &lt;em&gt;thorough&lt;/em&gt; answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the &lt;em&gt;why&lt;/em&gt;. Maybe you can do your own research from the answers provided here and come back with a more &quot;elaborate&quot; answer.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-05-15T04:18:40.037" LastActivityDate="2012-05-15T04:18:40.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="388" PostTypeId="2" ParentId="267" CreationDate="2012-05-14T20:05:21.170" Score="8" Body="&lt;p&gt;After some more research, the &lt;strong&gt;&lt;a href=&quot;http://www.scholarpedia.org/article/Oregonator&quot; rel=&quot;noreferrer&quot;&gt;Oregonator&lt;/a&gt;&lt;/strong&gt; appears to be &lt;strong&gt;“the simplest realistic model of the chemical dynamics of the oscillatory Belousov-Zhabotinsky (BZ) reaction”&lt;/strong&gt; or any other oscillatory reaction I could look up. It is described in Field and Noyes (1974) [1], and take its name from the University of Oregon where these researchers worked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is described by the following set of reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H15NN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each entity correspond to a chemical species in the real system: X = HBrO&lt;sub&gt;2&lt;/sub&gt;, Y = Br&lt;sup&gt;–&lt;/sup&gt;, Z = Ce(IV), A = BrO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;–&lt;/sup&gt;, B = CH&lt;sub&gt;2&lt;/sub&gt;(COOH)&lt;sub&gt;2&lt;/sub&gt;, and P = HOBr or BrCH(COOH)&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] R. J. Field, R. M. Noyes, “Oscillations in Chemical Systems IV. Limit cycle behavior in a model of a real chemical reaction”, &lt;em&gt;J. Chem. Phys.&lt;/em&gt; 60 (1974) 1877-84.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-14T20:05:21.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="389" PostTypeId="1" AcceptedAnswerId="628" CreationDate="2012-05-14T20:55:55.227" Score="7" ViewCount="319" Body="&lt;p&gt;In this molecule, which is 3,5-dimethyl-4-carboxypyrazolate (unless I got it wrong):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFC2U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think there is free rotation around the bond indicated in red, because one can write a resonance structure with significant reasonable weight where a double bond would be located between the two groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how can I go further than that and qualitatively describe the energy barrier for the rotation. I have found (&lt;a href=&quot;http://dx.doi.org/10.1021/jp0529467&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1021/jo991146n&quot; rel=&quot;nofollow noreferrer&quot;&gt;there&lt;/a&gt;) that in the case of benzoates, the barrier is in the range 3–6 kcal/mol. Should I expect it to be higher or lower in the above molecule?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2014-10-30T16:30:28.830" LastActivityDate="2014-10-30T16:31:29.590" Title="Rotation energy barrier for carboxypyrazolate" Tags="&lt;conformers&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="390" PostTypeId="2" ParentId="383" CreationDate="2012-05-14T22:12:33.190" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;indeed this is &lt;strong&gt;no&lt;/strong&gt; redox problem, oxidation states do not change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;instead it is a phase transition equilibrium &#xA;$$\ce{Ag+ + Cl- &amp;lt;=&amp;gt; AgCl v}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you marked this question as homework:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are two experiments that you can do to check that it is really a &lt;strong&gt;dynamic&lt;/strong&gt; equilibrium (though much on the side of the precipitate)&#xA;$$\ce{Ag+_(aq) + Cl(aq)- &amp;lt;=&amp;gt;&amp;gt; AgCl_(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.&#xA;&lt;br/&gt;&#xA;Here are some ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$&lt;br/&gt;have you ever done this? If not, you should really try. It is really fast. &lt;/li&gt;&#xA;&lt;li&gt;reduce the $\ce{Ag+}$&lt;/li&gt;&#xA;&lt;li&gt;oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.&#xA;&lt;br/&gt;&#xA;(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;F'x pointed us to a paper &lt;a href=&quot;http://www.springerlink.com/content/g455t63482504217/&quot; rel=&quot;nofollow&quot;&gt;Paradigms and Paradoxes: The Solubility of AgCl in Water&lt;/a&gt;. As it is behind a paywall, I'll try to summarize a few points out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it's actually kind of an overdue homework question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and&#xA;  KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are some interesting points for the question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is &lt;em&gt;better&lt;/em&gt; solvated than $\ce{Na+}$ or $\ce{K+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -&amp;gt; AgCl_{(s)}}$) is more negative than for NaCl and KCl.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -&amp;gt; AgCl_{(s)}}$ is comparable to that for the alkali chlorides. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My personal conclusions:&lt;/em&gt; It's a really though question. I'll stick to the experiment with measurable properies such as equilibrium constants. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, insolubility of AgCl is no surprise to me: after all, I'm used to the &lt;em&gt;fact&lt;/em&gt; for years. And given the insolubility of AgBr and AgI, I'd guesstimate unsoluble even though AgF is soluble (after all $\ce{F-}$ is usually different from the other 3). But I find the particular situation far away from any textbook kind of reasoning that should introduce general chemical concepts to students. Unless the message is supposed to be that reality is complicated... (which OTOH allows for the beautiy of reality and makes it interesting).&#xA;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="24052" LastEditDate="2016-03-13T19:56:55.937" LastActivityDate="2016-03-13T19:56:55.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="391" PostTypeId="2" ParentId="378" CreationDate="2012-05-15T05:00:42.017" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Alas, the problem of course is that you need both position q and momentum p to get an accurate picture of the overall electron state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not true in quantum mechanics; it is sufficient to characterize the wavefunction $\left\langle x |\psi\right\rangle$ in position space or $\left\langle k|\psi\right\rangle$ in momentum space. Which is chosen is entirely a matter of convenience: most molecules have the most compact description in position space, while most crystals have compact descriptions in momentum space. It can be proven that position alone, or momentum alone, form a complete basis separately. In fact, position-momentum uncertainly says that an electron state cannot be characterized completely in both position and momentum simultaneously. (This, by the way, has &lt;em&gt;nothing&lt;/em&gt; to do with fermionic symmetries and Pauli exclusion, which is another quantum-mechanical axiom unto itself. The same would have held for looking at bosonic particles also.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, however, true that a probability density in phase space $(q,p)$ is needed to completely characterize a classical system. &lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution&quot;&gt;Wigner's function&lt;/a&gt; was introduced to address this quantum-classical correspondence and is regarded as the quantum analogue of a classical probability distribution in phase space. However, it is also well known that it can take on &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_probability&quot;&gt;negative values&lt;/a&gt; and therefore cannot be interpreted in the usual way as a probability distribution. This affects all methods using the Wigner function: in the Springborg paper in the OP, there are regions of negative &quot;probability density&quot; in the lithium hydride molecule being analyzed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suffice to say, the question of how to deal with negative probabilities is an extremely controversial subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the Wigner function has found uses in &lt;em&gt;ab initio&lt;/em&gt; molecular dynamics when a classical phase space density is required to sample dynamical trajectories from the potential energy surface generated by quantum electronic states. The regions of negative probability are usually neglected in such procedures with seemingly no loss of accuracy (to my knowledge). I am not aware of other common usages in electronic structure theory. (The other problem brought up in the OP about the inaccuracies in DFT are conventionally attributed to incorrect modeling of electron correlation effects, which is a matter of incorrect treatments of fermion symmetries and its consequences when projected down into coordinate or momentum space.)&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2012-05-15T05:00:42.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="392" PostTypeId="1" CreationDate="2012-05-15T06:03:07.537" Score="-16" ViewCount="30288" Body="&lt;p&gt;What is the temperature of pure oxygen being combusted? I was wondering about it.&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="-1" LastEditDate="2016-03-26T19:25:06.813" LastActivityDate="2016-03-26T19:25:06.813" Title="Temperature of pure oxygen combustion" Tags="&lt;combustion&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2012-05-23T03:57:30.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="393" PostTypeId="2" ParentId="392" CreationDate="2012-05-15T06:10:48.493" Score="9" Body="&lt;p&gt;Oxygen is not a fuel by itself, it's only the oxidant. Combustion involves a fuel (acetylene, for example) and an oxidizer (here, O&lt;sub&gt;2&lt;/sub&gt;). So if you have &lt;em&gt;no fuel&lt;/em&gt;, but &lt;em&gt;all oxidizer&lt;/em&gt;, then what's left to burn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion is just a name we give to certain reactions. The point to get here is that the burning and its associated temperature is dependent on the reactants' chemical nature and stoichiometry. When a flame has a higher temperature than another, it means that both reactions give products that have different enthalpies of formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greater the difference between the enthalpy of the products and reactants, the greater the flame temperature.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-05-15T06:15:57.100" LastActivityDate="2012-05-15T06:15:57.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="394" PostTypeId="2" ParentId="392" CreationDate="2012-05-15T06:14:44.870" Score="8" Body="&lt;p&gt;Pure oxygen does not burn. Oxygen is a &lt;em&gt;supporter&lt;/em&gt; of combustion, not a combustible material(fuel) itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion reactions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{combustible stuff} \ce{+ O2 -&amp;gt; } \text{various oxides} + \text{heat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that nitrous oxide ($\ce{N2O}$) is also a supporter of combustion, which means that you can use it to carry out a combustion reaction without oxygen--but this is only because $\ce{N2O}$ generates oxygen itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since combustion is a reaction with oxygen, what's left for oxygen to oxidise? Itself? Not likely. You need a fuel to release heat for a combustion reaction.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="1431" LastEditDate="2014-04-10T17:34:42.720" LastActivityDate="2014-04-10T17:34:42.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="395" PostTypeId="1" CreationDate="2012-05-15T07:30:12.580" Score="8" ViewCount="24535" Body="&lt;p&gt;This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as $\ce{H2SO4}$,$\ce{ HNO3}$, $\ce{HCl}$, and $\ce{NaOH}$ not suitable primary standards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will their higher or lower pH affect the accuracy of the results of a titration?&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="22" LastEditDate="2012-05-15T08:44:58.100" LastActivityDate="2018-09-05T20:31:37.187" Title="Why are strong acids and bases not suitable as primary standards?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="396" PostTypeId="2" ParentId="392" CreationDate="2012-05-15T09:06:38.323" Score="12" Body="&lt;p&gt;Other answers have emphasized the fact that oxygen is not a fuel, but an oxidizing agent. I'll try to underline another point: &lt;strong&gt;there is no uniquely defined “combustion temperature” for a chemical compound&lt;/strong&gt;. This is unlike phase transitions like melting and ebullition, which happen at fixed temperature for a given pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of a combustion process depends on many parameters, including the heat of combustion, the ratio of air to fuel, the heat capacity of both and the initial temperature. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_flame_temperature&quot;&gt;adiabatic flame temperature&lt;/a&gt; is one of such combustion temperatures under certain conditions. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Autoignition_temperature&quot;&gt;autoignition temperature&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Flash_Point&quot;&gt;flash point&lt;/a&gt;, on the other hand, are characteristics of a given chemical compound; they are not, however, combustion temperatures.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-15T09:06:38.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="397" PostTypeId="1" AcceptedAnswerId="398" CreationDate="2012-05-15T12:32:57.183" Score="10" ViewCount="3569" Body="&lt;p&gt;In many tests for various radicals, a &quot;&lt;code&gt;&amp;lt;color&amp;gt;&lt;/code&gt; precipitate&quot; or &quot;&lt;code&gt;&amp;lt;color&amp;gt;&lt;/code&gt; turbidity&quot; is mentioned. For example, lead acetate with chromate ions gives a yellow precipitate, but hydrogen sulphide with potassium dichromate gives a yellow(or white) &quot;turbidity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the reaction itself, the turbidity is noted as a precipitate with a down arrow, but I'm wondering if there is any difference between the two. It looks like turbidity refers to a colloid/suspension being formed, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The physical differences/distinguishing features between turbidity and precipitate, if any.&lt;/li&gt;&#xA;&lt;li&gt;The observable differences/distinguishing features between the two, if any.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-15T13:09:13.450" Title="What's the difference between precipitate and turbidity?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="398" PostTypeId="2" ParentId="397" CreationDate="2012-05-15T12:52:58.370" Score="15" Body="&lt;p&gt;My understanding is that turbidity is an adjective describing the presence or degree of particulate matter suspended in a liquid (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Turbidimetry&quot;&gt;turbidimetry&lt;/a&gt;), whereas a precipitate is a particulate that can result in turbidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbidity generated by a precipitate is a function of its ability to form a stable suspension. If the precipitate simply drops to the bottom of a container or readily flocculates, the resultant turbidity will be low. The turbidity of a suspension is also going to be a function of the solvent and the nature and concentration of dissolved ions - a dramatic example of this is the rapid clarification of muddy freshwater at river deltas mixing with salt water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the word turbidity is being used as a noun, it is probably referring to a light-occluding suspension.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-15T13:09:13.450" LastActivityDate="2012-05-15T13:09:13.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="399" PostTypeId="2" ParentId="395" CreationDate="2012-05-15T14:32:36.623" Score="3" Body="&lt;p&gt;There's nothing particular about strong acids or strong bases that disqualifies them from being a primary standard, but most of them are too unstable or difficult to work with to be useful as a primary standard.  A primary standard should have these qualities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;High purity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stability in presence of air&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Absence of any water of hydration which might vary with changing humidity and temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cheap&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dissolves readily to produce stable solutions in solvent of choice&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A larger rather than smaller molar mass &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now for the examples you gave:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4}$ is hygroscopic, so it's concentration will change quite readily in air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HNO3}$ is very hygroscopic as well.  The $\ce{NO3-}$ anions can also start redoxing other things which isn't particularly useful for titrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$ is a gas, so it will be difficult to determine the concentration of the solutions (you'd have to compare it against another primary standard), and high-concentration solutions are unstable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaOH}$ is hygroscopic as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Georg pointed out below, one generally needs high-purity solid compounds so they can be weighed accurately (and in addition to the cons above, these substances are difficult to find/weigh in solid form).  I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="21" LastEditDate="2012-05-16T00:23:52.867" LastActivityDate="2012-05-16T00:23:52.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="400" PostTypeId="1" AcceptedAnswerId="428" CreationDate="2012-05-15T15:04:32.327" Score="15" ViewCount="1449" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Slush_hydrogen&quot;&gt;Slush hydrogen&lt;/a&gt; is a mixture of liquid and solid hydrogen &lt;strong&gt;at the triple point&lt;/strong&gt; considered as a possible vehicle fuel. What is the need of having it at the triple point? Couldn't any other set of thermodynamic conditions consistent with liquid–solid equilibrium work as well?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2019-12-19T00:28:06.993" Title="Why is slush hydrogen at the triple point?" Tags="&lt;thermodynamics&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="401" PostTypeId="1" AcceptedAnswerId="403" CreationDate="2012-05-15T17:09:01.730" Score="4" ViewCount="130" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Accessible_surface_area&quot; rel=&quot;noreferrer&quot;&gt;Solvent-Accessible Surface Area (SASA)&lt;/a&gt; is a valuable metric for looking at protein folding and protein-protein interactions. However, this measurement is typically done by calculating the SASA from a solved (and generally static) structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical probes like diazirine and hydroxyl radicals show some bias regarding where they tend to bind. I'm realizing while I'm writing this question that NMR is a perfectly valid method to determine a solution based structure and then calculate the SASA. Similar strategies has also been used to examine structured RNAs. I'm curious about the variety of these methods and how accurate they are.&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="30" LastEditDate="2012-05-15T18:49:33.780" LastActivityDate="2012-05-15T19:03:10.487" Title="Solution based measurement of Solvent-Accessible Surface Area of macromolecules" Tags="&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="402" PostTypeId="1" AcceptedAnswerId="414" CreationDate="2012-05-15T17:16:41.673" Score="7" ViewCount="5341" Body="&lt;p&gt;I am not planning to do these types of experiments but I would like to know the order of magnitude of the time it takes to run an LC-MS/MS experiment. Let's assume that I have the sample prep (protein) finished and I just need to load the sample into the LC. How long would I need to wait before I would get MS/MS data?&lt;/p&gt;&#xA;&lt;p&gt;I know that for a typical HPLC experiment, the sample takes around 20 minutes to go through the column.&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorDisplayName="user7951" LastEditDate="2020-10-07T18:49:29.667" LastActivityDate="2020-10-07T18:49:29.667" Title="How long does it take to run an MS/MS experiment" Tags="&lt;mass-spectrometry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="403" PostTypeId="2" ParentId="401" CreationDate="2012-05-15T19:03:10.487" Score="2" Body="&lt;p&gt;As you noted, the solvent-accessible surface area is a geometric concept, and as such is most easily calculated using computational geometry tools from a given molecular (or crystalline) structure. However you obtain this atomistic structure (X-ray, neutrons, molecular simulation, NMR, …) is of course of significant importance to the resulting area, but the calculation method will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, however, a few studies which have attempted to link the solvent-accessible surface area to directly measurable properties, and in particular to spectroscopic properties. A quick search of the literature reveals three examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://komiveslab.ucsd.edu/reprints/truhlar2.pdf&quot; rel=&quot;nofollow&quot;&gt;Amide H/&lt;sup&gt;2&lt;/sup&gt;H Exchange MALDI-TOF Mass Spectrometry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pitt.edu/~asher/homepage/spec_pdf/UV%20Raman%20Determination%20of%20the%20Environment%20and%20Solvent.pdf&quot; rel=&quot;nofollow&quot;&gt;UV Raman&lt;/a&gt; directly on the exposed residues&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022283609011978&quot; rel=&quot;nofollow&quot;&gt;Using diazirine as a probe&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-15T19:03:10.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="406" PostTypeId="2" ParentId="395" CreationDate="2012-05-16T00:25:16.580" Score="9" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;strong &lt;strong&gt;Bases&lt;/strong&gt; in general are not suitable as they will react with $\ce{CO2}$ from the air. This is actually more problematic than hygroscopicity, as it means that you also cannot store diluted $\ce{NaOH}$ solution (openly) as standard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instead, $\ce{Na2CO3}$ or $\ce{KHCO3}$ are much better alternatives (for &lt;em&gt;strong&lt;/em&gt; acids).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;acids&lt;/strong&gt; are either a gas ($\ce{HCl}$) or are produced from gas. I don't see how you can produce a conentration reliably enough for use as primary standard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;even if you would manage to produce saturated $\ce{HCl}$, it's much more difficult and less reliable than known alternatives (see below).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;but that wouldn't work e.g. for $\ce{H2SO4}$ as $\ce{SO3}$ is soluble in $\ce{H2SO4}$ (Oleum).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{HNO3}$: redox reactions are not only not particularly useful, but actually harmful as they use up acid equivalents:&#xA;$\ce{NO3- + 2e- + 2 H+ -&amp;gt; NO2- + H2O}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;also here, a solid alternative (KH-Phthalate) is available. If you don't have that, you may get away with oxalic acid which is still better than the acids suggested here. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorUserId="23561" LastEditDate="2018-09-05T20:31:37.187" LastActivityDate="2018-09-05T20:31:37.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="409" PostTypeId="1" AcceptedAnswerId="439" CreationDate="2012-05-16T09:30:40.663" Score="23" ViewCount="45600" Body="&lt;p&gt;Please take a look at the following &lt;a href=&quot;http://www.youtube.com/watch?feature=player_embedded&amp;amp;v=f95bNFVZkh0&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?feature=player_embedded&amp;amp;v=f95bNFVZkh0&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kV1dL.png&quot; alt=&quot;Ferrofluid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a new project, and I need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any special preparation for the glass needed?&lt;/p&gt;&#xA;" OwnerUserId="229" LastEditorUserId="4945" LastEditDate="2014-07-29T09:39:44.360" LastActivityDate="2014-11-18T13:17:34.263" Title="what kind of suspension liquid should be used with ferrofluid (so it does not stain the glass)" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="410" PostTypeId="1" AcceptedAnswerId="427" CreationDate="2012-05-16T11:27:58.700" Score="27" ViewCount="17905" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Related: &lt;a href=&quot;https://chemistry.stackexchange.com/q/383/22&quot;&gt;Reaction between silver nitrate and aluminum chloride&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Experimentally, $\ce{AgCl}$ is insoluble in water, but $\ce{AgNO3}$ is soluble. They're pretty common in a lab (well, $\ce{AgCl}$ is a common precipitate)--so I think most of us know this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Fajans%27_rules&quot; rel=&quot;noreferrer&quot;&gt;Fajan's rules&lt;/a&gt;, on the other hand, larger anion $\implies$ more polarization/covalent character $\implies$ less solubility. But, $\ce{NO3-}$ is the larger anion, yet $\ce{AgNO3}$ is more soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any theoretical reason for this?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="69490" LastEditDate="2019-08-12T22:22:42.390" LastActivityDate="2019-08-12T22:22:42.390" Title="Why is silver chloride less soluble than silver nitrate?" Tags="&lt;ions&gt;&lt;solubility&gt;&lt;aqueous-solution&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="411" PostTypeId="1" AcceptedAnswerId="413" CreationDate="2012-05-16T11:56:52.213" Score="6" ViewCount="1309" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hematite&quot; rel=&quot;nofollow&quot;&gt;Hematite&lt;/a&gt; is composed of $\ce{Fe2O3}$, and is paramagnetic, whereas &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetite&quot; rel=&quot;nofollow&quot;&gt;magnetite&lt;/a&gt; is $\ce{Fe3O4}$ and is diamagnetic.  Magnetite's nature is due to the presence of both $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ (Wikipedia even goes so far as to call it $\ce{FeO}\cdot \ce{Fe2O3}$, but explains that this is not a &quot;solid solution&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can surmise that in the crystalline solid, there are divalent and trivalent cations dispersed within.  I can also assume that there might be some sharing of electrons between the oxygens and each of these cations.  One would assume this also happens within the hematite crystal, so what is it about &quot;tossing&quot; the divalent cation into the mix of the crystalline structure that makes the magnetite diamagnetic?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the charge differential between the two cations cause a permanent dipole, if so, why don't all of the small dipoles simply cancel each other out? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="4231" LastEditDate="2015-03-04T21:07:09.103" LastActivityDate="2015-03-04T21:07:09.103" Title="What is it about the relationship between the Fe²⁺ and Fe³⁺ in magnetite that makes it diamagnetic?" Tags="&lt;crystal-structure&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="412" PostTypeId="2" ParentId="411" CreationDate="2012-05-16T12:04:50.853" Score="3" Body="&lt;p&gt;Magnetite is &lt;em&gt;not&lt;/em&gt; diamagnetic. &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Magnetite&quot; rel=&quot;nofollow&quot;&gt;It is ferrimagnetic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Ferrimagnetism&quot; rel=&quot;nofollow&quot;&gt;Ferrimagnetism&lt;/a&gt; requires layers of unequal but opposite magnetic moments in the crystal lattice. The two iron ions have different magnetic moments, and they probably are arranged nicely in the lattice, so the ferrimagnetic property comes from there.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-16T12:04:50.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="413" PostTypeId="2" ParentId="411" CreationDate="2012-05-16T12:21:22.660" Score="7" Body="&lt;p&gt;Magnetite has a spinel structure with two types of $\ce{Fe}$ sites: octahedral and tetrahedral, respectively in green and orange below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Kd5n.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in this structure, oxygen atoms are the small red balls).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This structure is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Spinel#Spinel_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;spinel&lt;/a&gt;, with $\ce{Fe^{2+}}$ ion in tetrahedral coordination and $\ce{Fe^{3+}}$&gt; ions in octahedral sites. The coupling between atoms from these two lattices are &lt;a href=&quot;http://en.wikipedia.org/wiki/Superexchange&quot; rel=&quot;nofollow noreferrer&quot;&gt;superexchange interactions&lt;/a&gt;, which result in an antiparallel alignment of spins. However, because the magnetic moments of the two lattices are not equal, their addition results in a net magnetic moment, even though they are of opposition alignment. This is called &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ferrimagnetism&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferrimagnetism&lt;/a&gt;&lt;/strong&gt;, and in practice most of its macroscopic properties are similar to &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ferromagnetism&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferromagnetism&lt;/a&gt;&lt;/em&gt;, though they stem from a completely different microscopic magnetic ordering.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-03-04T21:07:05.177" LastActivityDate="2015-03-04T21:07:05.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="414" PostTypeId="2" ParentId="402" CreationDate="2012-05-16T12:22:52.670" Score="10" Body="&lt;p&gt;The MSMS part is the detection of the LC in this case (nb. you don't need to do LC-MSMS, you can do direct infusion quite happily if you don't need to separate anything), so the actual experiment wouldn't need to take much longer than any other LC (as CHM says).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, calibration + set-up of these things, and extensive cleaning afterwards, can make it a day's job to run just a few samples (even direct infusion). If you don't have much information on the system already, you may also find that you need to do a 'sighting' run, to work out which ions you need to examine more closely (MSMS). And then there is data processing.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all: quick, if you know exactly what you want, and someone is running everything for you. If you want to argue that it is an easy job, you can, but be aware that this is not always gong to be the case.&lt;/p&gt;&#xA;" OwnerUserId="231" LastActivityDate="2012-05-16T12:22:52.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="415" PostTypeId="2" ParentId="54" CreationDate="2012-05-16T12:27:37.003" Score="6" Body="&lt;p&gt;Close but no cigar. It does dissolve the polymer, but certainly not into its constituent monomers. That would be one heck of a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you choose &lt;a href=&quot;http://www.rheothing.com/2012/03/rubber-gloves-theyre-not-all-same.html&quot; rel=&quot;noreferrer&quot;&gt;the proper gloves&lt;/a&gt; for whatever you are working with.&lt;/p&gt;&#xA;" OwnerUserId="232" LastEditorUserId="30" LastEditDate="2012-05-16T12:41:43.210" LastActivityDate="2012-05-16T12:41:43.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="416" PostTypeId="1" AcceptedAnswerId="419" CreationDate="2012-05-16T13:24:21.133" Score="2" ViewCount="3271" Body="&lt;p&gt;I was told that &quot;Sodium Tallowate, sodium cocoate and sodium palm kernelate are the 'natural' forms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_dodecyl_sulfate&quot; rel=&quot;nofollow&quot;&gt;SLS&lt;/a&gt;&quot;.  Is this true?  I thought &quot;natural&quot; SLS would be SLS derived from a natural product such as coconut oil.&lt;/p&gt;&#xA;" OwnerUserId="233" LastActivityDate="2018-04-16T04:17:08.503" Title="Is soap the natural form of Sodium Lauryl Sulfate?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="417" PostTypeId="2" ParentId="410" CreationDate="2012-05-16T13:27:09.697" Score="2" Body="&lt;p&gt;I'm not much sure about this. I too was much curious about precipitates and this is similar to what I had read somewhere on the net (can't find the site, got it from Wikipedia)  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solubility occurs under dynamic equilibrium, which means that solubility results from the simultaneous and opposing processes of dissolution and phase joining (e.g., precipitation of solids). The solubility equilibrium occurs when the two processes proceed at a constant rate.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So according to this the process of &lt;em&gt;phase joining&lt;/em&gt; is very very more than &lt;em&gt;dissolution&lt;/em&gt; in $\ce{AgCl}$ thus it is insoluble in water, but in $\ce{AgNO3}$ it is just the opposite.  &lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="30" LastEditDate="2012-05-16T14:20:08.413" LastActivityDate="2012-05-16T14:20:08.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="419" PostTypeId="2" ParentId="416" CreationDate="2012-05-16T14:18:13.520" Score="6" Body="&lt;p&gt;Conventional soaps are the sodium salts of long chain carboxylic acids (aka fatty acids). These are formed by the hydrolysis of triglycerides, a process known as saponification. This process breaks the triglyceride at its ester linkages and yields three fatty acid molecules and glycerol molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_dodecyl_sulfate&quot; rel=&quot;noreferrer&quot;&gt;aforementioned wikipedia article&lt;/a&gt; indicates, sodium lauryl sulfate is formed by refunctionalising the fatty acid &lt;em&gt;lauric acid&lt;/em&gt;. In this case, the $\ce{R-OOH}$ (carboxylic acid) head group has been replaced with an $\ce{R-OH}$ (alcohol) group, which is in turn replaced by $\ce{R-OSO3^{-}}$ (sulfate). Here I am using $\ce{R}$ to refer to the long alkyl chain of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether sodium lauryl sufate occurs naturally, however the other molecules you mention are different in the sense that they have carboxylate ($\ce{R-COO^{-}}$) head groups whereas sodium lauryl sulfate has the sulfate head group ($\ce{R-OSO3^{-}}$). Both classes of molecule will act as surfactants as they have polar head groups which are hydrophilic, and nonpolar alkyl tails which have an affinity for fats and oils.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Lauric acid derivatives spotter's guide&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(These were drawn using the nifty &lt;a href=&quot;http://www.ctan.org/pkg/chemfig&quot; rel=&quot;noreferrer&quot;&gt;ChemFig&lt;/a&gt; package for &lt;a href=&quot;http://en.wikipedia.org/wiki/TeX&quot; rel=&quot;noreferrer&quot;&gt;(La)TeX&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FZBlE.png&quot; alt=&quot;Some molecules&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last two (sodium laurate and sodium lauryl sulfate) are sodium salts. Both carboxylic acids ($\ce{R-COOH}$) and hydrogen alkyl sulfates ($\ce{R-OSO3H}$) are fairly willing to ionise to give the corresponding carboxylate ($\ce{R-COO^{-}}$) or sulfate ($\ce{R-OSO3^{-}}$). Sodium ions happily associate with these anions, balancing the negative charge.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-16T15:03:24.867" LastActivityDate="2012-05-16T15:03:24.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="420" PostTypeId="2" ParentId="410" CreationDate="2012-05-16T15:25:02.327" Score="14" Body="&lt;p&gt;I'm afraid this is rather a non-answer (or why is it so difficult to answer this)...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pretty much all nitrates are soluble. This is often explained by the exceptionally good delocalization of the negative charge. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, silver salts in general aren't well soluble (I recall only fluoride, nitrate and perchlorate as soluble. Sulfate, carbonate, oxide, sulfide (of course), even the stochiometric cyanide (if I remember correctly) isn't soluble.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So in the case of the nitrate, usually good solubility of nitrates seem to win over the usually not so good solubility of silver salts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the Chloride, it is maybe similar: the Chloride stands between the good solubility of AgF and the low solubility of AgBr. Just here, &quot;unsoluble&quot; wins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(There's also - related to Fajan I think - the &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;HSAB concept&lt;/a&gt; which tries to extend also to hard-hard unsoluble salts [e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{CaF2}$&lt;/span&gt;] -  However, for the case here it is  about equally unhelpful as Fajan. I think those two concepts work better in a row of similar ions, like Ag-halogenides)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think the difficulty in explaining/predicting from theory may come from the fact that there are contrary but strong effects to be considered, so it is hard to predict what exactly the result will be (loss of significance: subtracting two almost equal large numbers, you may not even be able to be sure about the sign of the result).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;pre&gt;&#xA;substance     solubility / (mol/l)&#xA;AgF               1.4 ⋅ 10¹   = 1.4e1&#xA;AgNO3         1 - 5   ⋅ 10⁰   =   1e0 - 5e0&#xA;AgCl              5   ⋅ 10⁻⁴  =   5e-4&#xA;AgBr              7   ⋅ 10⁻⁷  =   5e-7&#xA;AgI               9   ⋅ 10⁻⁹  =   9e-9&#xA;Ag2S              1.7 ⋅ 10⁻¹⁷ = 1.7e-17&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The solubility between different halogenides varies 10 orders of magnitude between fluoride and iodide. That corresponds to a 10fold change in &lt;span class=&quot;math-container&quot;&gt;$\Delta_RG^0$&lt;/span&gt;. A factor 10 in energy is like bond strength covalent vs. H-bond or like rotational vs. vibrational vs. electronic transition of a molecule. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have a look at the paper that I linked from the other question. It is actually focused on &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2019-08-12T07:43:33.450" LastActivityDate="2019-08-12T07:43:33.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="421" PostTypeId="2" ParentId="402" CreationDate="2012-05-16T16:37:07.253" Score="4" Body="&lt;p&gt;You can adjust your run times to your preference, based on your conditions.  For example, I am doing LC-MS/MS on steroids (low molecular weight compounds) extracted from human urine samples through a liquid-liquid extraction. It only takes me 10 min per sample from the time the run starts until the very end of it.&lt;/p&gt;&#xA;" OwnerUserId="235" LastActivityDate="2012-05-16T16:37:07.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="422" PostTypeId="2" ParentId="410" CreationDate="2012-05-16T17:11:03.543" Score="12" Body="&lt;p&gt;In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found data that the crystal lattice energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; is -916.3 kJ/mol &lt;a href=&quot;https://web.archive.org/web/20150416071628/http://www.drbateman.net/asa2sums/sum4.1/sum4.1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;(experimental)&lt;/a&gt;, while the lattice energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO_3}$&lt;/span&gt; is -820 kJ/mol. (Since lattice energy is defined as the energy released when the ions join to form a solid, it is always negative. Reverse the sign for the amount of energy you need to put in to break the crystal apart into separate ions in the gas phase.) From this data you can see that it takes quite a bit more energy to break apart the &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; ions than it does the &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt; ions, presumably because the chloride ion is smaller and more tightly held. If the enthalpy of hydration value is about the same for the two cases (I couldn't find actual values), then it may not be enough to overcome the extra energy needed to break up the ions.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, both enthalpy of hydration and crystal lattice energy are important considerations for solubility.There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in &lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lattice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added this paragraph to the answer after putting it in the comments below first. &#xA;Calculation from enthalpy of hydration data (-850.7 for &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; and -794.4 for &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; says that &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl−}$&lt;/span&gt; is the more soluble ion, by -61.3 kJ/mol (since the contribution from solvation of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; would be the same in each case). Nitrate would be the less soluble ion (from energy concerns) in agreement with Fajan's rules. The enthalpy of hydration data was obtained from problems in a chemistry book. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="69490" LastEditDate="2019-08-10T10:51:46.900" LastActivityDate="2019-08-10T10:51:46.900" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="423" PostTypeId="1" AcceptedAnswerId="495" CreationDate="2012-04-24T03:31:38.560" Score="21" ViewCount="10863" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/N-hydroxysuccinimide&quot; rel=&quot;nofollow noreferrer&quot;&gt;N-Hydroxysuccinimide&lt;/a&gt; (NHS) chemistry is commonly used in chemical biology to react with primary amines like lysine and the N-terminus of proteins. I was curious how labile NHS esters are to other nucleophiles like secondary amines and sulfhydryls. This knowledge will be helpful to know the extent of side products when I do a bioconjugation experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/48/Disuccinimidyl_suberate.png/200px-Disuccinimidyl_suberate.png&quot; alt=&quot;Disuccinimidyl suberate&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="57" OwnerDisplayName="bobthejoe" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.280" LastActivityDate="2016-01-21T17:30:26.563" Title="Side reactions of NHS Chemistry" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="424" PostTypeId="2" ParentId="410" CreationDate="2012-05-16T18:44:51.187" Score="3" Body="&lt;p&gt;Look at the crystalline structure of $\ce{AgCl}$ vs $\ce{AgNO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorides usually form face centred cubic crystals while nitrates form trigonal planar crystals. Nitrate is a triangular molecule with a positive charge on the nitrogen and a (2–) charge on the oxygens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In FCC each ion interacts with six ions of opposing charge in an octahedral arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with a nitrate the nitrate ions are only interacting with 2 ions of opposing charge above and below the 2/3 negative charge on the oxygens point to neighboring positively charged nitrogen. This arrangment is far more unstable and in silver nitrate ions are far more likely to be whisked away by ion-dipole interactions.&lt;/p&gt;&#xA;" OwnerUserId="237" LastEditorUserId="4231" LastEditDate="2015-03-04T20:33:37.840" LastActivityDate="2015-03-04T20:33:37.840" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="425" PostTypeId="1" CreationDate="2012-05-16T22:18:53.013" Score="16" ViewCount="1953" Body="&lt;p&gt;In its pure form (–)-sparteine is an only-slightly-yellow clear, viscous liquid. Yet, after only a week of being kept in dry, normal atmosphere (in a round bottom flask covered by a septa), some of it reacts with &lt;em&gt;something&lt;/em&gt; to form a sticky, orange-brown substance. I've had to distill the sparteine I have twice due to whatever reaction is occurring, and I would like to know exactly what the reaction is so I don't have to distill every time I need it. The literature I found mentioned that it readily absorbs carbon dioxide in a 1-1 molar ratio, and I know that tertiary amines do sometimes react with CO&lt;sub&gt;2&lt;/sub&gt;, but I haven't been able to piece it together as something that could happen in ambient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gjcNs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="240" LastEditorUserId="19616" LastEditDate="2015-10-02T16:09:21.640" LastActivityDate="2015-10-02T16:09:21.640" Title="By what reaction does (–)-sparteine decompose in ambient conditions?" Tags="&lt;organic-chemistry&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="427" PostTypeId="2" ParentId="410" CreationDate="2012-05-17T05:05:55.400" Score="22" Body="&lt;p&gt;In the comment to my previous answer, you asked for a theoretical reason for the solubilities, not considering energy data. Since I know from energy considerations that the issue is not the solvation of the anions, I can present a reason based on the strength of the ionic bond in the two compounds. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; (as well as others) states the bonding in $\ce{AgCl}$ has an unusually high covalent character which makes it a tighter bond. The $\ce{Ag+}$ ion and the $\ce{Cl-}$ ion are close to the same size (with the silver ion being smaller), so they can approach each other quite closely. In silver nitrate, the $\ce{NO3-}$ ion is larger and does not allow as close an approach as the chloride ion, so the bond is weaker, easier to break up, and the salt is more soluble. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="-1" LastEditDate="2018-05-23T02:08:41.617" LastActivityDate="2018-05-23T02:08:41.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="428" PostTypeId="2" ParentId="400" CreationDate="2012-05-17T05:34:04.240" Score="8" Body="&lt;p&gt;I'll have a stab at this one, despite the fact I've forgotten almost the totality of engineering thermodynamics, as I do recall seeing these papers before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the references&lt;sup&gt;1&lt;/sup&gt; in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Slush_hydrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; describes the production of slush hydrogen through the auger technique, contrasting it with the freeze-thaw method. Specifically, the auger method can be performed above the triple point pressure of hydrogen, whereas the freeze-thaw method uses the latent heat of vaporisation to freeze some of the hydrogen. In this technique the vaporisation of some of the hydrogen (prompted by the application of a vacuum) at the triple point causes some of the liquid hydrogen to freeze, generating a crust of solid hydrogen which is then broken up to make slush.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, the triple point conditions are only required to make the freeze-thaw process go, and are not required for storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The auger technique, also described in the report&lt;sup&gt;1&lt;/sup&gt;, seems to work on the basis of shaving solid hydrogen which is refrigerated with liquid helium. Voth compares the technique to a conventional ice shaver (albeit much, much colder!). In this instance there's no requirement for triple point conditions, which is given as one of the advantages of the technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Haberbusch and McNelis&lt;sup&gt;2&lt;/sup&gt; give us this sad quote, in reference to a 1991 paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;According to Hans, projected annual demand for slush hydrogen in the year 2010 is 584 million pounds to support a fleet of 20 space-planes&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day, maybe...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) Voth, R.O.; &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19780018251_1978018251.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Producing Liquid-Solid Mixtures of Hydrogen in an Auger&lt;/a&gt;; National Bureau of Standards Department of Commerce; 1978, pp. 1-16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Haberbusch, M.S. and McNelis, N.B.; &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19970026089_1997043543.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparison of the Continuous Freeze Slush Hydrogen Production Technique to the Freeze-Thaw Technique&lt;/a&gt;; NASA Technical Memorandum 107324, 1996, p. 1&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="61578" LastEditDate="2019-12-19T00:28:06.993" LastActivityDate="2019-12-19T00:28:06.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="429" PostTypeId="1" AcceptedAnswerId="433" CreationDate="2012-05-17T08:33:20.303" Score="16" ViewCount="22094" Body="&lt;p&gt;Lithium and magnesium are Group 1 and Group 2 elements respectively. Elements of these groups are highly ionic, and I've never heard of them forming significantly covalent &lt;em&gt;inorganic&lt;/em&gt; compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet these elements form a variety of organometallic compounds ($\ce{PhLi}$, the whole family of Grignard reagents, etc). Organometallic compounds have significant covalent character (i.e., the bond can be called covalent)  in the carbon–metal bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's so special about carbon that makes these elements form covalent bonds?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2012-05-18T03:51:01.170" LastActivityDate="2017-03-15T11:59:58.713" Title="Why do Magnesium and Lithium form *covalent* organometallic compounds?" Tags="&lt;organic-chemistry&gt;&lt;ionic-compounds&gt;&lt;organometallic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="430" PostTypeId="1" AcceptedAnswerId="449" CreationDate="2012-05-17T08:43:30.900" Score="16" ViewCount="953" Body="&lt;p&gt;I've been asked this question a few times, and while I &lt;em&gt;think&lt;/em&gt; I know the answer, I'd like  to know more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite, as we know, is a sheet polymer. Since polymers are bound to be finite by physical considerations, graphite must have &quot;edges&quot;. My question is, what happens at these edges? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can cook up these possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A graphene sheet &quot;folds back&quot; on itself, thus forming a tube and taking care of &lt;em&gt;most&lt;/em&gt; of the edge carbons. But this would make graphite less slippery, so I doubt this is the case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Random elements/compounds from the environment at time of synthesis latch on to the edges, taking care of the valency of Carbon&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The edge carbons form double/triple bonds amongst themselves&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd like to know more about this. I'm also curious what happens to the resonance of graphene at the edges.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="7448" LastEditDate="2014-12-31T19:46:31.767" LastActivityDate="2014-12-31T19:46:31.767" Title="Edges of graphite" Tags="&lt;inorganic-chemistry&gt;&lt;resonance&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="431" PostTypeId="1" AcceptedAnswerId="432" CreationDate="2012-05-17T09:00:22.050" Score="45" ViewCount="10295" Body="&lt;p&gt;Oxygen is a rather boring element. It has only two allotropes, dioxygen and ozone. Dioxygen has a double bond, and ozone has a delocalised cloud, giving rise to two &quot;1.5 bonds&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, sulfur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of sulfur to catenate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, sulfur does not have a stable diatomic allotrope at room temperature. I, personally would expect disulfur to be &lt;em&gt;more&lt;/em&gt; stable than dioxygen, due to the possibility of $\mathrm{p}\pi\text{-}\mathrm{d}\pi$ back-bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why do sulfur and oxygen have such opposite properties with respect to their ability to catenate? &lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="16683" LastEditDate="2016-10-22T14:36:41.467" LastActivityDate="2016-11-24T19:46:10.843" Title="Why does sulfur, but not oxygen, catenate?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="432" PostTypeId="2" ParentId="431" CreationDate="2012-05-17T12:54:07.337" Score="40" Body="&lt;p&gt;First, a note: while oxygen has fewer allotropes than sulfur, it sure has &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropes_of_oxygen&quot; rel=&quot;noreferrer&quot;&gt;more than two&lt;/a&gt;! These include $\ce{O}$, $\ce{O_2}$, $\ce{O_3}$, $\ce{O_4}$, $\ce{O_8}$, metallic $\ce{O}$ and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the values of the &lt;a href=&quot;http://faculty.ncc.edu/LinkClick.aspx?fileticket=L2sV5SDzIhE%3D&amp;amp;tabid=1871&quot; rel=&quot;noreferrer&quot;&gt;single and double bond enthalpies&lt;/a&gt;:&#xA;$$\begin{array}{ccc} \hline&#xA;\text{Bond} &amp;amp; \text{Dissociation energy / }\mathrm{kJ~mol^{-1}} \\ \hline&#xA;\ce {O-O} &amp;amp; 142 \\&#xA;\ce {S–S} &amp;amp; 268 \\&#xA;\ce {O=O} &amp;amp; 499 \\&#xA;\ce {S=S} &amp;amp; 352 \\ \hline&#xA;\end{array}$$&#xA;This means that $\ce{O=O}$ is stronger than $\ce{S=S}$, while $\ce{O–O}$ is weaker than $\ce{S–S}$. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the reason for the weaker $\ce{S=S}$ double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the $\mathrm{3p}$ orbitals overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: $\ce{Se=Se}$ has an even weaker &lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html&quot; rel=&quot;noreferrer&quot;&gt;bond enthalpy of $\ce{272 kJ/mol}$&lt;/a&gt;. There is more in-depth discussion of the relative bond strengths in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/47056/&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not particularly stable, it's actually also possible for oxygen to form discrete molecules with the general formula $\ce{H-O_n-H}$; water and hydrogen peroxide are the first two members of this class, but $n$ goes up to at least $5$. These &quot;hydrogen polyoxides&quot; are described further in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/35574/&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2016-11-24T19:46:10.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="433" PostTypeId="2" ParentId="429" CreationDate="2012-05-17T13:22:12.580" Score="17" Body="&lt;p&gt;The character of the bond is determined by the electronegativity of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of bonds as purely ionic or covalent is not always correct - usually it is more correct to say that a bond has ionic or covalent characteristics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So comparing the difference in electronegativities gives us the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{cc}\hline&#xA;\text{Difference in electronegativity} &amp;amp; \text{Type of bond} \\ \hline&#xA;&amp;lt; 0.4 &amp;amp; \text{Non-polar covalent} \\&#xA;0.4 \mathrm{-} 1.7 &amp;amp; \text{Polar covalent} \\&#xA;&amp;gt;1.7 &amp;amp; \text{Ionic} \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the upper end of the polar covalent spectrum, the bonds frequently have both covalent and ionic characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For organometallic compounds, the difference in electronegativity between Li and C is $1.57$.  So while this is still in the polar covalent range, it is also close to ionic.  Similarly, the difference between Mg and C is $1.24$ - again, a very polar covalent bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the difference between H and C ($0.35$) - a non-polar covalent bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, the thing that is &quot;special&quot; about carbon is that it has a fairly low electronegativity compared to the chalcogens and halogens.  Granted, bonds with carbon are also going to be weaker than in say LiCl, but that's what makes organometallic compounds actually work to form carbon-carbon bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The electronegativity values came from &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements&quot; rel=&quot;noreferrer&quot;&gt;wikipedia's great chart&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="12" LastEditorUserId="16683" LastEditDate="2017-03-15T11:59:58.713" LastActivityDate="2017-03-15T11:59:58.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="434" PostTypeId="2" ParentId="430" CreationDate="2012-05-17T14:06:01.500" Score="5" Body="&lt;p&gt;Defects in graphite result from oxidation. They are $\ce{ OH, COOH}$ and epoxy groups, as in  this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUNPU.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://cnx.org/content/m29187/latest/&quot; rel=&quot;noreferrer&quot;&gt;http://cnx.org/content/m29187/latest/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the edges of graphite have the same chemical nature, and thus have these defects.&lt;/p&gt;&#xA;" OwnerUserId="174" LastEditorUserId="22" LastEditDate="2012-05-17T14:15:13.373" LastActivityDate="2012-05-17T14:15:13.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="435" PostTypeId="2" ParentId="54" CreationDate="2012-05-17T14:54:00.697" Score="27" Body="&lt;p&gt;Nitrile gloves are made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrile_rubber&quot;&gt;&lt;strong&gt;nitrile rubber&lt;/strong&gt;&lt;/a&gt;, or &lt;strong&gt;poly(butadiene/acrylonitrile)&lt;/strong&gt;. This polymer is &lt;a href=&quot;http://www.polysciences.com/Core/Display.aspx?pageId=98&amp;amp;categoryId=114&amp;amp;productId=454&quot;&gt;highly soluble&lt;/a&gt; in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it &lt;a href=&quot;http://onlinelibrary.wiley.com/store/10.1002/ceat.200900268/asset/97_ftp.pdf;jsessionid=3685250CB0326943BAD0364311FC7D5D.d04t03?v=1&amp;amp;t=h2bwx35v&amp;amp;s=7690887daf7950a04640bf1898f100a03673ee4b&quot;&gt;permeates easily&lt;/a&gt; through NBR, meaning we can expect the dissolution to be fast in addition to thermodynamically &#xA;favourable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I would not expect the mechanism here to be any different from that of any polymer dissolution by a good solvent. The solvent will permeate through the polymer, intercalate between polymer chains, and solvate them (inducing swelling). Once solvated, the network of polymer chains looses its mechanical properties and they can fully separate. &lt;em&gt;(I wish I could find a good existing illustration for that part, but I can't right now… If anyone can, feel free to edit!)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-17T14:54:00.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="436" PostTypeId="1" AcceptedAnswerId="437" CreationDate="2012-05-17T15:28:06.873" Score="15" ViewCount="713" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen&quot;&gt;Ortho- and parahydrogen&lt;/a&gt; are two forms of the $\ce{H2}$ molecule that are distinguished by their pairing or antipairing of nuclear spins, giving rise to metastable singlet (ortho-) and triplet (para-) states. This constitutes an example of nuclear spin isomerism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the thermal properties of these two species are different, my understanding is that their chemical properties are essentially identical. Are there any examples of this kind of isomerism having nontrivial implications for chemical reactivity?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-17T19:43:21.013" Title="Are there any examples of nuclear spin isomers having consequences for chemical reactivity?" Tags="&lt;physical-chemistry&gt;&lt;hydrogen&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="437" PostTypeId="2" ParentId="436" CreationDate="2012-05-17T19:43:21.013" Score="10" Body="&lt;h1&gt;Yes!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Papers in the field of preparation or selection of nuclear spin isomers have introductions that &lt;a href=&quot;http://dx.doi.org/10.1016/S0009-2614(00)00445-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;justify the work by potential applications&lt;/a&gt; and some of them include chemical reactions, either under that name or under the more descriptive terminology of “reactive collisions”. While no specific cases are cited as examples, the potential exist but only in limited situations such as bimolecular reaction in &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_beam&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular beams&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have seen, the canonical reference is &lt;a href=&quot;http://dx.doi.org/10.1080/00268977700101861&quot; rel=&quot;nofollow noreferrer&quot;&gt;“Detailed symmetry selection rules for reactive collisions”&lt;/a&gt;, &lt;em&gt;Mol. Phys.&lt;/em&gt; &lt;strong&gt;1977&lt;/strong&gt;, 34, 477–504. This is the theoretical paper laying out the foundations of the effect of nuclear spin isomerism on reactions. I don't have access to the full text, but here is the abstract:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;                   &lt;img src=&quot;https://i.stack.imgur.com/CJgaV.png&quot; alt=&quot;Paper abstract&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It confirms what I suggested in my &lt;a href=&quot;https://chemistry.stackexchange.com/questions/436/are-there-any-examples-of-nuclear-spin-isomers-having-consequences-for-chemical#comment661_436&quot;&gt;earlier comment&lt;/a&gt;: the different nuclear spin isomers have different &lt;a href=&quot;http://en.wikipedia.org/wiki/Rovibrational_coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;rovibrational states&lt;/a&gt;, and accounting for effect changes the selection rules indicating which collisions are reactive or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One specific example which I found is &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevLett.78.3844&quot; rel=&quot;nofollow noreferrer&quot;&gt;“Observation of Ortho-Para H3+ Selection Rules in Plasma Chemistry”&lt;/a&gt;, &lt;em&gt;Phys. Rev. Lett.&lt;/em&gt; &lt;strong&gt;1997&lt;/strong&gt;, 78, 3844–3847:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Large dependences of the intensity ratios on the parent gases have been observed, indicating the spin memory of protons even after chemical reactions. The results clearly demonstrate the existence of selection rules on nuclear spin modifications in chemical reactions theoretically anticipated by M. Quack.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their results are obtained on the formation of H&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; ions from a H&lt;sub&gt;2&lt;/sub&gt; plasma.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2012-05-17T19:43:21.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="439" PostTypeId="2" ParentId="409" CreationDate="2012-05-17T21:40:34.330" Score="10" Body="&lt;p&gt;I don't have a full answer, but a hint at how to choose your liquid. You want the ferrofluid not to mix with it, so your fluid has to be immiscible with the ferrofluid solvent (or carrier fluid; ferrofluids are colloidal suspensions). In the case of your “FerroFluid EFH-1”, the solvent is a light mineral oil, so you should go with a polar solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you want the ferrofluid solvent not to touch the glass, to avoid staining. In order to do so, you need your liquid to wet the bottle more than the mineral oil, i.e. to have a smaller &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot;&gt;contact angle&lt;/a&gt;. You can play on both the bottle material and nature of the fluid to achieve this. I suggest protic polar solvents, which should wet regular glass well: water, ethanol, isopropanol, acetic acid, …&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-17T21:40:34.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="440" PostTypeId="1" CreationDate="2012-05-18T04:27:12.047" Score="28" ViewCount="758" Body="&lt;p&gt;The qualifier &lt;em&gt;'free'&lt;/em&gt; seems to be ubiquitously attached to discussion of radicals as highly reactive species with unpaired spins. What, precisely (or imprecisely, as the case may be) does &lt;em&gt;'free'&lt;/em&gt; really mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;sub&gt;Free as in free speech? Free beer? Freedom of movement? Kinetic accessibility?&lt;/sub&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="11367" LastEditDate="2016-12-13T01:01:24.630" LastActivityDate="2016-12-13T03:48:22.320" Title="What makes a radical 'free'?" Tags="&lt;terminology&gt;&lt;history-of-chemistry&gt;&lt;radicals&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="441" PostTypeId="1" CreationDate="2012-05-18T05:46:01.417" Score="10" ViewCount="259" Body="&lt;p&gt;I'm attempting to perform atomistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Structural_alignment&quot; rel=&quot;noreferrer&quot;&gt;optimal structure alignment&lt;/a&gt; on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="30" LastEditDate="2012-05-18T11:10:51.873" LastActivityDate="2012-06-12T20:48:52.390" Title="How can I perform structure alignment on an inorganic species?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="442" PostTypeId="1" AcceptedAnswerId="443" CreationDate="2012-05-18T10:55:36.910" Score="16" ViewCount="4801" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_(chemistry)#Naming_conventions&quot; rel=&quot;noreferrer&quot;&gt;R/S naming system&lt;/a&gt; for stereocenters relies on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules&quot; rel=&quot;noreferrer&quot;&gt;Cahn–Ingold–Prelog priority rules&lt;/a&gt; to rank the four substituents of a stereocenter; the R or S name is then attributed depending on the spatial orientation of the ranked substituents. While this is mostly used for asymmetric carbon atoms, it holds true of other stereocenters with four substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, nitrogen and phosphorus atoms can also act as stereocenters if the inversion of their configuration is blocked (e.g. because they're in a cyclic molecule) or slowed enough (e.g. because of bulky substituents). Apparently, these stereocenters are also labelled in R/S fashion. Examples I found include &lt;a href=&quot;http://en.wikipedia.org/wiki/Tr%C3%B6ger%27s_base&quot; rel=&quot;noreferrer&quot;&gt;Tröger's base&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KQNxz.png&quot; alt=&quot;Tröger&amp;#39;s base&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(above is the (5S,11S)-enantiomer, below is the (5R,11R)-enantiomer),&#xA;and &lt;a href=&quot;http://en.wikipedia.org/wiki/DIPAMP&quot; rel=&quot;noreferrer&quot;&gt;DIPAMP&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WaclT.png&quot; alt=&quot;DIPAMP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having said that, I can't find any reference as to how the R/S naming scheme is applied to these tetrahedral stereocenters which don't have four substituents, but a lone electron pair and three substituents instead. I assume that the same scheme can be followed if one considers the lone pair as the lowest-ranked substituent, but I can't find any canonical reference as to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how are such stereocenters named? What authoritative reference can I find on this topic?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-18T12:13:14.923" Title="How are non-carbon stereogenic centers named (S/R)?" Tags="&lt;nomenclature&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="443" PostTypeId="2" ParentId="442" CreationDate="2012-05-18T11:53:35.603" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;but I can't find any canonical reference as to that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.co.in/webhp?sourceid=chrome-instant&amp;amp;ie=UTF-8#hl=en&amp;amp;output=search&amp;amp;sclient=psy-ab&amp;amp;q=Cahn-Ingold-Prelog%20lone%20pair&amp;amp;oq=&amp;amp;aq=&amp;amp;aqi=&amp;amp;aql=&amp;amp;gs_l=&amp;amp;pbx=1&amp;amp;fp=c59d3fc9b9c4309b&amp;amp;bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&amp;amp;biw=1067&amp;amp;bih=548&quot; rel=&quot;nofollow noreferrer&quot;&gt;Look harder&lt;/a&gt; ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be that the lone pair has lowest priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;OK, I can't find any &lt;em&gt;authoritative&lt;/em&gt; reference to that (I don't have access to the &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.196603851/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;original paper&lt;/a&gt;), but I've got plenty of others:&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, IUPAC &lt;a href=&quot;http://www.iupac.org/publications/pac/1976/pdf/4501x0011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Except for hydrogen, ligancy, if not already four, is made up to&#xA;  four by adding &quot;phantom atoms&quot; which have atomic number zero&#xA;  (0) and are thus always last in order of preference.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This has&#xA;  various uses but perhaps the most interesting is where nitrogen&#xA;  occurs in a rigid skeleton, as for example in a-isosparteine (14);&#xA;  here the phantom atom can be placed where the nitrogen lone pair&#xA;  of electrons is; then N-i appears as shown alongside the formula,&#xA;  and the chiraiity (R) is the consequence; the same applies to N-i6.&#xA;  Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides (see example to Rule E—4.9).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When ligancy is less than four, we obviously have some lone pairs hanging around (except in Hydrogen--which has been excluded, and some other cases like carbocations--which are planar anyway)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the lone pair becomes a &quot;phantom atom&quot; and has 0 priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I wonder what happens to these rules if you have &lt;em&gt;two&lt;/em&gt; lone pairs, one of them &lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_atom#Muonic_atoms&quot; rel=&quot;nofollow noreferrer&quot;&gt;muonic&lt;/a&gt; ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Check the &lt;a href=&quot;https://chemistry.stackexchange.com/revisions/443/1&quot;&gt;first revision of this post&lt;/a&gt; for my original, non-authoritative references&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2012-05-18T12:13:14.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="444" PostTypeId="1" CreationDate="2012-05-18T17:47:14.853" Score="176" ViewCount="152613" Body="&lt;p&gt;According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can an atom have more than 8 valence electrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not possible, why can't an atom have more than 8 valence electrons?&lt;/p&gt;&#xA;" OwnerUserId="247" LastEditorUserId="18002" LastEditDate="2015-10-17T01:04:21.283" LastActivityDate="2017-12-15T20:44:09.893" Title="Can an atom have more than 8 valence electrons? If not, why is 8 the limit?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;valence-bond-theory&gt;" AnswerCount="8" CommentCount="0" FavoriteCount="78" ContentLicense="CC BY-SA 3.0" />
  <row Id="445" PostTypeId="2" ParentId="444" CreationDate="2012-05-18T18:20:47.433" Score="91" Body="&lt;p&gt;Yes, it can. We have molecules which contain &quot;superoctet atoms&quot;. Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PBr5, XeF6, SF6, HClO4, Cl2O7, I3- ,  K4[Fe(CN)6], O=PPh3 }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost &lt;em&gt;all&lt;/em&gt; coordination compounds have a superoctet central atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-metals from Period 3 onwards are prone to this as well. The halogens, sulfur, and phosphorus are repeat offenders, while &lt;em&gt;all&lt;/em&gt; noble gas compounds are superoctet. Thus sulfur can have a valency of +6, phosphorus +5, and the halogens +1, +3, +5, and +7. Note that these are still covalent compounds -valency applies to covalent bonds as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this isn't usually seen is as follows. We basically deduce it from the properties of &lt;a href=&quot;http://enwp.org/Atomic_orbital&quot; rel=&quot;noreferrer&quot;&gt;atomic orbitals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_Principle&quot; rel=&quot;noreferrer&quot;&gt;aufbau principle&lt;/a&gt;, electrons fill up in these orbitals for period $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n\mathrm{s}, (n-2)\mathrm{f},(n-1)\mathrm{d},n\mathrm{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(theoretically, you'd have $(n-3)\mathrm{g}$ before the $\mathrm{f}$, and so on. But we don't have atoms with those orbitals, yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the outermost shell is $n$. In each period, there are only eight slots to fill in this shell by the Aufbau principle - 2 in $n\mathrm{s}$, and 6 in $n\mathrm{p}$. Since our periodic table pretty much follows this principle, we don't see any superoctet atoms usually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the $\mathrm{d,f}$ orbitals for that shell &lt;em&gt;still exist&lt;/em&gt; (as empty orbitals) and can be filled if the need arises. By &quot;exist&quot;, I mean that they are low enough in energy to be easily filled. The examples above consist of a central atom, that has taken these empty orbitals into its hybridisation, giving rise to a superoctet species(since the covalent bonds add an electron each)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cooked up a periodic table with the shells marked. I've used the shell letters instead of numbers to avoid confusion. $K,L,M,N$ refer to shell 1,2,3,4 etc. When a slice of the table is marked &quot;M9-M18&quot;, this means that the first element of that block &quot;fills&quot; the ninth electron in the M(third) shell, and the last element fills the eighteenth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click to enlarge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kY6qF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kY6qFm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Derivative of &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:White_periodic_table.svg&quot; rel=&quot;noreferrer&quot;&gt;this image&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are a few irregularities, with $\ce{Cu}$, $\ce{Cr}$, $\ce{Ag}$, and a whole bunch of others which I've not specially marked in the table.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="402" LastEditorDisplayName="user15489" LastEditDate="2017-04-02T20:30:02.417" LastActivityDate="2017-04-02T20:30:02.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="446" PostTypeId="2" ParentId="440" CreationDate="2012-05-18T19:21:32.527" Score="4" Body="&lt;p&gt;A &lt;a href=&quot;http://www.questia.com/library/encyclopedia/radical-in-chemistry.jsp&quot; rel=&quot;nofollow&quot;&gt;radical&lt;/a&gt; is a group of atoms that tend to stay together and act as a persistent chemical unit in reactions. The methyl group $CH_3$ is a common example. My own favorite radical is ammonium $NH_4$, which behaves strikingly similarly to an alkali metal such as lithium or sodium, so much so that you can even create amalgams of ammonium with mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A free radical is what you get if the radical group is capable of existing on its own. The ammonium radical is an example, and one of the more stable ones. Methyl can be free also, but tends to be far more reactive and so shorter lived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free radicals always have &quot;dangling bonds,&quot; since they are by definition parts of larger molecules that have been broken off of those molecules. However, that doesn't mean they are any more reactive than are free elemental atoms, ammonium again being a good example.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2012-05-18T19:21:32.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="447" PostTypeId="2" ParentId="444" CreationDate="2012-05-18T20:38:59.640" Score="55" Body="&lt;p&gt;&lt;em&gt;In chemistry, and in science in general, there are many ways of explaining the same empirical rule. Here, I am giving an overview that is very light on quantum chemistry: it should be fairly readable at a novice level, but will not explain in its deepest way the reasons for the existence of electronic shells.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The “rule” you are citing is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;octet rule&lt;/strong&gt;&lt;/a&gt;, and one of its formulations is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;atoms of low (&lt;em&gt;Z&lt;/em&gt; &amp;lt; 20) atomic number tend to combine in such a way that they each have eight electrons in their valence shells&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll notice that it's not specifically about a &lt;em&gt;maximal&lt;/em&gt; valence (i.e. the number of electrons in the valence shell), but a &lt;em&gt;preferred&lt;/em&gt; valence in molecules. It is commonly used to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_structure&quot; rel=&quot;noreferrer&quot;&gt;Lewis structure&lt;/a&gt; of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the octet rule is not the end of the story. If you look at hydrogen (H) and helium (He), you will see that do not prefer an eight-electron valence, but a two-electron valence: H forms e.g. H&lt;sub&gt;2&lt;/sub&gt;, HF, H&lt;sub&gt;2&lt;/sub&gt;O, He (which already has two electrons and doesn't form molecules). This is called the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule#Exceptions&quot; rel=&quot;noreferrer&quot;&gt;duet rule&lt;/a&gt;&lt;/strong&gt;. Moreover, heavier elements including all &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_metal&quot; rel=&quot;noreferrer&quot;&gt;transition metals&lt;/a&gt; follow the aptly-named &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/18-Electron_rule&quot; rel=&quot;noreferrer&quot;&gt;18-electron rule&lt;/a&gt;&lt;/strong&gt; when they form metal complexes. This is because of the quantum nature of the atoms, where electrons are organized in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;noreferrer&quot;&gt;shells&lt;/a&gt;: the first (named the K shell) has 2 electrons, the second (L-shell) has 8, the third (M shell) has 18. Atoms combine into molecules by trying in most cases to have valence electrons entirely filling a shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are elements which, in some chemical compounds, break the duet/octet/18-electron rules. The main exception is the family of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;hypervalent molecules&lt;/strong&gt;&lt;/a&gt;, in which a main group element nominally has more than 8 electrons in its valence shell. Phosphorus and sulfur are most commonly prone to form hypervalent molecules, including $\ce{PCl5}$, $\ce{SF6}$, $\ce{PO4^3-}$, $\ce{SO4^2-}$, and so on. Some other elements that can also behave in this way include iodine (e.g. in $\ce{IF7}$), xenon (in $\ce{XeF4}$), and chlorine (in $\ce{ClF5}$). (This list isn't exhaustive.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/5242/16683&quot;&gt;Gavin Kramar's answer&lt;/a&gt; explains how such hypervalent molecules can come about despite apparently breaking the octet rule.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2017-08-14T15:49:56.633" LastActivityDate="2017-08-14T15:49:56.633" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="448" PostTypeId="2" ParentId="440" CreationDate="2012-05-18T20:52:10.107" Score="18" Body="&lt;p&gt;I liked this question because I had never thought much about it. However, it's not such a mystery because the answer is on &lt;a href=&quot;http://en.wikipedia.org/wiki/Radical_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's “radical” page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Historically, the term radical was also used for &lt;strong&gt;bound parts of the molecule&lt;/strong&gt;, especially when they remain unchanged in reactions. These are &lt;strong&gt;now called functional groups&lt;/strong&gt;. For example, methyl alcohol was described as consisting of a methyl &quot;radical&quot; and a hydroxyl &quot;radical&quot;. Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons. However, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the &lt;em&gt;free&lt;/em&gt; in &lt;em&gt;free radical&lt;/em&gt; was originally used to distinguish it from a radical bound to a carbon skeleton.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Digging into historical chemistry texts reveals that “free radical” becomes really widely used around 1930:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/ngrams/graph?content=free+radical&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=0&amp;amp;smoothing=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a8bMx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some quotes for earlier works use it, though it does not appear so frequently that it may be considered a common expression. In other words, at that time simply the juxtaposition of the adjective &lt;em&gt;free&lt;/em&gt; with &lt;em&gt;radical&lt;/em&gt;, became synonymous with &lt;em&gt;group&lt;/em&gt; (or today's noun &lt;em&gt;substituent&lt;/em&gt;). For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“This hydrocarbon was formerly supposed to be the free radical of the methyl group, but no methyl compounds have in any way been derived from it, whilst by the action, of chlorine upon it, ethyl chloride is obtained as the product” (1870, &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=8K8XAAAAIAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lessons in elementary chemistry&lt;/a&gt;&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Although, however, the free radical methylene is unknown, many of its compounds have been prepared, of which some, such as the haloid ethers, &amp;amp;c.” (1884, &lt;em&gt;A Treatise on Chemistry&lt;/em&gt;, Henry Enfield Roscoe and Carl Schorlemmer)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though more quotes could be dug and provided, the ones I've read support this idea that free radical initially meant “the chemical group as a molecule by itself”.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="402" LastEditDate="2016-12-13T03:48:22.320" LastActivityDate="2016-12-13T03:48:22.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="449" PostTypeId="2" ParentId="430" CreationDate="2012-05-18T20:59:31.580" Score="8" Body="&lt;p&gt;It does appear that the graphene sheets fold back on themselves. &lt;a href=&quot;http://www.lehigh.edu/%7Esvr204/newdata/mypapers/191-200.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; reports that the graphene sheets that make up both natural and synthetic graphite, double back over each other with &amp;quot;nano-arches&amp;quot;.&#xA;This is an &lt;img src=&quot;https://i.stack.imgur.com/1dMyp.png&quot; alt=&quot;illustration&quot; /&gt; from the paper.&lt;/p&gt;&#xA;&lt;p&gt;There are also SEM and TEM micrographs of a graphite crystal that are pretty interesting.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-05-18T20:59:31.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="450" PostTypeId="2" ParentId="444" CreationDate="2012-05-18T21:27:38.720" Score="43" Body="&lt;p&gt;This question may be difficult to answer because there are a couple of definitions of valence electrons. Some books and dictionaries define valence electrons as &quot;outer shell electrons that participate in chemical bonding&quot; and by this definition, elements can have more than 8 valence electrons as explained by F'x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some books and dictionaries define valence electrons as &quot;electrons in the highest principal energy level&quot;. By this definition an element would have only 8 valence electrons because the $n-1$ $d$ orbitals fill after the $n$ $s$ orbitals, and then the $n$ $p$ orbitals fill. So, the highest principal energy level, $n$, contains the valence electrons. By this definition, the transition metals all have either 1 or 2 valence electrons (depending on how many electrons are in the $s$ vs. $d$ orbitals).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ca with two $4s$ electrons would have two valence electrons (electrons in the 4th principal energy level).&lt;/li&gt;&#xA;&lt;li&gt;Sc with two $4s$ electrons and one $3d$ electron will have two valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;Cr with one $4s$ electron and five $3d$ electrons will have one valence electron.&lt;/li&gt;&#xA;&lt;li&gt;Ga with two $4s$ electrons, ten $3d$ electrons, and one $4p$ electron would have three valence electrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By the other definition, they could have more since they have more &quot;outer shell&quot; electrons (until the $d$ shell is filled). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &quot;highest principal energy level&quot; definition for valence electrons allows you to correctly predict the paramagnetic behavior of transition metals ions because valence electrons (the $d$ electrons) are lost first when a transition metal forms an ion. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="30" LastEditDate="2012-05-30T14:56:08.743" LastActivityDate="2012-05-30T14:56:08.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="451" PostTypeId="2" ParentId="174" CreationDate="2012-05-18T22:54:31.217" Score="13" Body="&lt;p&gt;Having embarrassed myself with my first answer, let me try this again. Ammonia and water hydrogen bond exceptionally well. In fact, the $\ce{O-H...N}$ value is 29 kJ/mol while the $\ce{O-H...O}$ value is 21 kJ/mol. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot;&gt;Ref&lt;/a&gt;) So, if there is any water around, ammonia would be happy to become solvated as they recondense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, ammonia is a weak base and reacts with water (which is not usually the case when you're doing a distillation). $$\ce{NH3 + H2O &amp;lt;=&amp;gt; NH4+ + OH-}$$ So, if there's any water around as the ammonia is distilling and they recombine, they would also react which would make it difficult to remove the last of the water. (Some reaction would occur because the lowest energy point of the equilibrium is not at reactants or products.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as noted in the comments, distillations to prepare dry ammonia use a scrubber to remove the water. &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-18T22:54:31.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="453" PostTypeId="1" AcceptedAnswerId="455" CreationDate="2012-05-19T00:13:41.737" Score="16" ViewCount="317" Body="&lt;p&gt;Chiral molecules tend to absorb one circular polarization of light more than the other. Is this ever used to isolate a particular enantiomer?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="194" LastEditDate="2017-06-01T21:11:07.557" LastActivityDate="2017-06-01T21:11:07.557" Title="Is circularly polarized light ever used to isolate particular chiralities?" Tags="&lt;stereochemistry&gt;&lt;spectroscopy&gt;&lt;chirality&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="454" PostTypeId="2" ParentId="453" CreationDate="2012-05-19T01:13:48.143" Score="5" Body="&lt;p&gt;The optical rotation is certainly used to follow the separation of optical isomers (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diastereomer&quot;&gt;enantiomers&lt;/a&gt; if they are mirror images, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Diastereomer&quot;&gt;diastereomers&lt;/a&gt; if you have more than one stero-center and the two isomers are not mirror images). You can see how the optical rotation changes as you use the normal separation techniques (chromatography, distillation, crystallization, etc.). If you know the true value for the substance, you can determine the purity. But I don't know how polarized light would be used to do the actual separation.  &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-19T01:13:48.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="455" PostTypeId="2" ParentId="453" CreationDate="2012-05-19T04:25:36.743" Score="12" Body="&lt;p&gt;Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases increase the rate of breakdown of the enantiomer that absorbs that particular circular polarization (left or right), resulting in an increase in the relative concentration of the other enantiomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the first page of a paper on asymmetric photolysis using circularly polarized ultraviolet light &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00453a018#&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="83" LastActivityDate="2012-05-19T04:25:36.743" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="456" PostTypeId="1" AcceptedAnswerId="514" CreationDate="2012-05-19T08:27:36.243" Score="22" ViewCount="7237" Body="&lt;p&gt;I'm familiar with the concept of a reaction coordinate from high school chemistry as some generic conformational parameter that all of the intermediate states of a chemical reaction lie upon. Recently, though, I've seen some presentations where chemists presented plots of the free energy with respect to &lt;strong&gt;two&lt;/strong&gt; reaction coordinates, and my understanding is that the plot was of a measurement (I could be wrong, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly does the reaction coordinate mean? What does it mean to have more than one reaction coordinate for the same reaction? Is the free energy as a function of the reaction coordinate a measurable quantity?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditDate="2015-05-17T00:29:24.103" LastActivityDate="2017-10-29T19:06:39.370" Title="What does &quot;reaction coordinate&quot; mean? What does it mean when a reaction has TWO reaction coordinates?" Tags="&lt;theoretical-chemistry&gt;&lt;free-energy&gt;&lt;reaction-coordinate&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="459" PostTypeId="1" CreationDate="2012-05-20T02:23:30.930" Score="10" ViewCount="201" Body="&lt;p&gt;While reading &lt;a href=&quot;http://www.lehigh.edu/~svr204/newdata/mypapers/191-200.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; (suggested by @Janice &lt;a href=&quot;https://chemistry.stackexchange.com/a/449/22&quot;&gt;here&lt;/a&gt;), I came across the following line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surface reconstruction is known to be extremely important for Si, the material of  20th century electronics. However, surface reconstruction has been much less studied for graphite. One of the reasons might be graphite's limited use in electronics because &lt;strong&gt;fabrication of ordered graphite substrates is difficult.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can sort of guess why &quot;fabrication of ordered graphite substrates is difficult&quot; — I'm not too sure though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, I'd like to know more about it. Exactly &lt;em&gt;why&lt;/em&gt; is such fabrication difficult?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-10-30T04:13:43.910" Title="Why is fabrication of ordered graphite substrates difficult?" Tags="&lt;inorganic-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="461" PostTypeId="2" ParentId="444" CreationDate="2012-05-20T12:23:02.573" Score="96" Body="&lt;p&gt;&lt;strong&gt;2017-10-27 Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[NOTE: My earlier notation-focused answer, unchanged, is below this update.]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the same bonding mechanisms that enable formation of greater-than-8 valence shells also enable alternative structural interpretions of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. &lt;a href=&quot;https://chemistry.stackexchange.com/a/445/83&quot;&gt;Manishearth's excellent answer&lt;/a&gt; explores this issue in much greater detail than I do here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic (&quot;astronomical” might be a more apt word).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggests that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why my overall answer to this question is simply &quot;yes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Question: Why are octets special?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, the reason is that for any energy level  above the special case of the $n=1$ shell (helium), the &quot;closed shell&quot; orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus to avoid the larger energy penalties imposed by Pauli exclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A final note&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A hint: If you have ever watched people create a two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original 2014-ish Answer (Unchanged)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is intended to supplement &lt;a href=&quot;https://chemistry.stackexchange.com/a/445/83&quot;&gt;Manishearth's earlier answer&lt;/a&gt;, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are &quot;borrowing&quot; each electron. A dot with an arrow means that that the electron &quot;belongs&quot; mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are examples using table salt (ionic) and oxygen (covalent):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tjb6K.gif&quot; alt=&quot;salt and oxygen in donation notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has &quot;donated&quot; its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IIDSF.gif&quot; alt=&quot;calcium sulfate and sulfite in donation notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and &quot;borrow&quot; entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a more ambiguous example is sulfur hexafluoride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aU9gk.gif&quot; alt=&quot;sulfur hexafluoride in donation notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule#History_and_controversy&quot; rel=&quot;noreferrer&quot;&gt;some controversy even today&lt;/a&gt; about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2014-04-04 Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ckKXP.png&quot; alt=&quot;covalent and u-bond versions of donation notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second &lt;em&gt;u-bond&lt;/em&gt; notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates &lt;em&gt;both&lt;/em&gt; electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beggar atom at the top of the U gets to use &lt;em&gt;both&lt;/em&gt; of the electrons for free, which in turn means that &lt;em&gt;two&lt;/em&gt; of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken as a whole, the atom at the bottom of a u-bond is saying &quot;I don't like it, but if you are &lt;em&gt;that&lt;/em&gt; desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon monoxide with its baffling &quot;why does carbon suddenly have a valence of two?&quot; structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j3Wbb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j3Wbb.png&quot; alt=&quot;Carbon monoxide in u-bond notation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon &quot;share&quot; one of the electron pairs from oxygen's already-full octet. Carbon ends up with four line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologues, one of the consequences of which is that they have nearly identical physical chemistry properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mMR6h.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mMR6h.png&quot; alt=&quot;Collection of hypervalent molecules&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2017-10-31T23:39:12.327" LastActivityDate="2017-10-31T23:39:12.327" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="462" PostTypeId="1" AcceptedAnswerId="465" CreationDate="2012-05-20T13:20:28.940" Score="8" ViewCount="92" Body="&lt;p&gt;Just some while ago I was burning a box of cardboard in my backyard and I saw a very curious phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw that as the cardboard box started burning it started to curl.  Up after a while as I came back to check if the fire was out I saw that though the cardboard box wasn't a box any longer but individual sides/faces of the box had retained their ridges and texture though they had tuned greyish white and crumbly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why did the cardboard start to curl up as it began to burn?&lt;/li&gt;&#xA;&lt;li&gt;Why did it retain its texture if not shape and size?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="199" LastEditorUserId="30" LastEditDate="2012-05-20T14:18:33.943" LastActivityDate="2012-05-20T16:04:17.413" Title="Retention of surface texture after combustion" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="464" PostTypeId="1" AcceptedAnswerId="542" CreationDate="2012-05-20T14:12:48.507" Score="22" ViewCount="1785" Body="&lt;p&gt;Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot;&gt;inversion&lt;/a&gt; at room temperature. On the other hand, the free energy barrier for &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphine&quot;&gt;phosphines&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfonium&quot;&gt;sulfoniums&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfoxide&quot;&gt;sulfoxides&lt;/a&gt; are high enough that they are optically stable: the rate of racemization is slow at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what effect is responsible for this difference in behaviour (the larger energy barrier). I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4325" LastEditDate="2016-02-21T00:41:39.903" LastActivityDate="2017-02-02T01:05:25.553" Title="Why don't trigonal S and P compounds undergo inversion at room temperature?" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="465" PostTypeId="2" ParentId="462" CreationDate="2012-05-20T15:45:47.597" Score="7" Body="&lt;p&gt;The curling up has a simple reason: the burning was not uniform. Paper contains water (really!), and heating it makes it lose the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, paper is a bad conductor of heat. When you burn it, inevitably the flame will be mainly on one side(the side that's pointing up). The flame will transfer heat to nearby regions of paper, and evaporate some (or possibly all) of the water. The loss of water leads to a contraction of the cellulose fibers in the paper (I'm not too sure of this, comments appreciated). But, this only happens on one side of the paper. Even though paper is razor-thin, the fact that it still has two &quot;faces&quot; is important here. Now, due to the contraction, the paper curls in that direction (similar to how a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bimetallic_strip&quot;&gt;bimetallic strip&lt;/a&gt; works). That's it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for retaining texture, is something I'm not too sure about. The burning reaction throws up a lot of soot, but a lot of the products stay right where they are. This results in the charred/burnt paper occupying the previous position of the cellulose. While charred paper isn't a polymer and thus isn't too strong, it can &quot;mimic&quot; the initial structure just by occupying the position of the cellulose. But, unlike cellulose, if we try bending it, it will crumble since there are no long polymer &quot;chains&quot;/intermolecular forces holding it together. It's like the difference between a house of cards and a similar &quot;house&quot;, made of a single piece of rubber. The structure is the same, but the former case is only staying together because it had the &quot;luck&quot; to exist in a stable configuration. Disturbing it will destroy it.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="30" LastEditDate="2012-05-20T16:04:17.413" LastActivityDate="2012-05-20T16:04:17.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="466" PostTypeId="1" AcceptedAnswerId="627" CreationDate="2012-05-20T16:15:12.207" Score="7" ViewCount="831" Body="&lt;p&gt;Given two crystal space groups, how can one determine if they are in a group-subgroup relationship? The specific case at hand is &lt;strong&gt;P 21/m 2/m 2/a&lt;/strong&gt; (aka &lt;strong&gt;P m m a&lt;/strong&gt;, &lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/large/051az1.htm&quot; rel=&quot;nofollow&quot;&gt;#51&lt;/a&gt;) and &lt;strong&gt;P m m 2&lt;/strong&gt; (&lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/large/025az1.htm&quot; rel=&quot;nofollow&quot;&gt;#25&lt;/a&gt;), but knowing how to solve this question in a general manner will sure prove helpful in the future as well…&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="73" LastEditDate="2013-09-17T01:01:43.507" LastActivityDate="2013-09-17T01:01:43.507" Title="Are these two space groups in a group–subgroup relationship?" Tags="&lt;crystal-structure&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="467" PostTypeId="2" ParentId="464" CreationDate="2012-05-20T20:22:39.310" Score="6" Body="&lt;p&gt;Neither of the two enantiomers is actually an energy eigenstate of the molecule. If the right-handed and left-handed enantiomers are states $|R\rangle$ and $|L\rangle$ respectively, then the ground and first-excited energy eigenstates are going to be something like $|g\rangle=\left( |R\rangle +|L\rangle \right)/\sqrt{2}$ and $|e\rangle=\left( |R\rangle -|L\rangle \right)/\sqrt{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because each enantiomer is in a superposition of energy eigenstates, a molecule initially in one chiral state will oscillate back and forth from one to the other, with a frequency determined by the difference in energy between $|g\rangle$ and $|e\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the difference in energy between  $|g\rangle$ and $|e\rangle$ is determined by the barrier between the $|R\rangle $ and $|L\rangle$ states: the higher and wider the energy barrier, the slower the oscillation. Since P and S are much larger than N, the energy barrier between the $|R\rangle $ and $|L\rangle$ states is larger, and the frequency of oscillation between $|R\rangle $ and $|L\rangle$ is much lower.&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2012-05-20T20:22:39.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="468" PostTypeId="1" AcceptedAnswerId="469" CreationDate="2012-05-21T02:28:40.930" Score="17" ViewCount="3288" Body="&lt;p&gt;I just bought a heat bag from store and it has special property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is made up of transparent plastic bag and transparent unknown liquid inside when sold. There is a tiny circular metal piece in there. If you bend the metal piece, crystalization will occur and release quite a lot of heat. The crystalization will spread out and change all the liquid into crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the hardened plastic bag cool down, you need to put it into boiling water and the crystals will dissolve and become liquid again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is more interesting, I tried to heat it up but not letting all the crystal dissolve, take it out of the boiling water and assume it is ready for the next use. However, the half-dissolved liquid/crystal turns into all crystals again (without me bending the metal piece). I am guessing the transformation was not complete and the process was reverted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did the Japanese invent this material?&lt;/p&gt;&#xA;" OwnerUserId="253" LastEditorUserId="7448" LastEditDate="2015-10-30T16:53:36.247" LastActivityDate="2021-01-13T06:54:06.697" Title="What material are heat bags made of?" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="469" PostTypeId="2" ParentId="468" CreationDate="2012-05-21T03:01:10.720" Score="13" Body="&lt;p&gt;The bag contains a supersaturated solution of some salt with a highly negative enthalpy of crystallisation, which could be something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_acetate&quot;&gt;sodium acetate&lt;/a&gt; ($\ce{Na^{+}CH3CO2^{-}}$). The dissolved salt would really like to come out of solution but lacks a nucleation site. Bending the metal disk produces a transient site where crystallisation can begin, and from there are chain reaction results, causing all of the solute to crystallise and release the heat associated with this crystallisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your guess regarding the half-dissolved crystals is correct - the crystals that have not dissolved act as &lt;a href=&quot;http://en.wikipedia.org/wiki/Seed_crystal&quot;&gt;seed crystals&lt;/a&gt; for further crystal growth. You need to dissolve all of the crystals to prevent the process from spontaneously reversing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the popularity of this kind of reaction for modern self-heating food/drink packaging in Japan, I suspect the discovery of this kind of reaction is lost to antiquity, as for instance sodium acetate can be made by simply mixing salt and vinegar. I recall reading that the highly exothermic reaction of calcium oxide and water was used in self-heating cans of food back in World War I.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-21T07:33:38.330" LastActivityDate="2012-05-21T07:33:38.330" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="471" PostTypeId="1" AcceptedAnswerId="473" CreationDate="2012-05-21T08:54:54.317" Score="8" ViewCount="8883" Body="&lt;p&gt;I am currently searching for a portable dehumidifier but notice that most of them come with a indicator that actually using the Blue Silica Gel, which I believe that it may contain Cobalt Chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would like to ask, &lt;br&gt;&#xA;(1) if that is the case, would the &lt;strong&gt;portable dehumidifier become a harmful product&lt;/strong&gt;?&lt;br&gt;&#xA;(2) would an &lt;strong&gt;Orange Silica Gel contain other harmful chemical&lt;/strong&gt; that is currently not found?&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="68033" LastEditDate="2018-09-18T06:52:19.117" LastActivityDate="2018-09-18T06:52:19.117" Title="Is Blue Silica Gel more harmful than the Orange ones?" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="472" PostTypeId="1" AcceptedAnswerId="10363" CreationDate="2012-05-21T09:35:14.873" Score="16" ViewCount="302" Body="&lt;p&gt;Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5HJGq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bidentate-binuclear complex&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PyORS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monodentate-binuclear complex (both from &lt;a href=&quot;http://arsenic.tamu.edu/about/course/mod2/notes/pg7.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? The first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps.  &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="21964" LastEditDate="2016-07-27T22:42:18.120" LastActivityDate="2016-07-27T22:42:18.120" Title="What are the mechanisms for capture of As(III) and As(V) by magnetite from water?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="473" PostTypeId="2" ParentId="471" CreationDate="2012-05-21T09:50:58.380" Score="7" Body="&lt;p&gt;I had a poke around in an effort to determine the dye used in orange silica gel and was able to find two material safety datasheets (MSDSes) for orange silica gels which both gave different empirical formulae for the dye (&lt;a href=&quot;http://www.hysilicagel.com/images/MSDS%20for%20Silica%20Gel%20Orange.pdf&quot;&gt;$\ce{C25H30ClN3}$&lt;/a&gt; and &lt;a href=&quot;http://www.interraglobal.com/Portals/0/pdf/msds/msds-sSORB-yellow-indicating.pdf&quot;&gt;$\ce{C20H14O4}$&lt;/a&gt;), which suggests that there are multiple formulations of orange/yellow silica gel. Both of these dyes are organic molecules (as opposed to transition metal salts), and now their possible identities are given at the bottom of this response. Suffice to say that orange silica gels were designed specifically to avoid the use of cobalt chloride, which is fairly toxic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a document&lt;sup&gt;1&lt;/sup&gt; from the US National Parks Service indicates, the chief danger of cobalt chloride is its environmental toxicity if it enters soil or water through improper disposal. Cobalt salts are often rather evil, and this is actually reflected in the name, which is ostensibly an alteration of 'kobold'. However as long as the beads are kept in a suitable enclosure such as a filtered dessicator plug, they should not present a hazard. Inhalation of dust from the silica gel beads, direct handling of beads or cobalt chloride-impregnated materials or complete inundation of the dessicant (which could dissolve some of the cobalt chloride) are three possible ways the cobalt chloride could present a direct hazard, but the disposal of blue silica gel-containing products in landfill and subsequent leaching of cobalt chloride is probably the chief concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the orange beads go, as I was not able to determine the specific indicator molecule(s) I can't ascertain their toxicity, suffice to say that they are by design less harmful than cobalt chloride and appear to be generally considered safe under &lt;em&gt;normal use&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;P.S. &lt;sub&gt;(finally, after chrome crashing multiple times)&lt;/sub&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;F'x points out that $\ce{[C25H30N3]^{+}Cl^{-}}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_violet&quot;&gt;crystal violet&lt;/a&gt;, which is not super toxic but for which there is evidence of carcinogenicity&lt;sup&gt;2&lt;/sup&gt; and which is considered an environmental pollutant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C20H14O4}$ may very well be &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenolphthalein&quot;&gt;phenolphthalein&lt;/a&gt;, which is considered to be a carcinogen by some authorities, most notably appearing on the California Prop. 65 listing&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the idea is that cobalt chloride is less desirable than either of these dyes. Don't eat the beads, and wash your hands!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;a href=&quot;http://www.nps.gov/museum/publications/conserveogram/02-15.pdf&quot;&gt;U.S. National Park Service; Cobalt Indicating Silica Gel Health and Safety Update; &lt;strong&gt;2005&lt;/strong&gt;, pp. 1-3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0272059085901721&quot;&gt;Littlefield, N.A.; Blackwell, B.-N.; Hewitt, C.C.; Gaylor, D.w.; Chronic toxicity and carcinogenicity studies of gentian violet in mice; Fundamental and Applied Toxicology, &lt;strong&gt;2004&lt;/strong&gt;, pp. 902-912&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) &lt;a href=&quot;http://oehha.ca.gov/prop65/prop65_list/files/P65single031612.pdf&quot;&gt;Chemicals Known to the state to cause cancer or reproductive toxicity; State of California EPA; March 16 &lt;strong&gt;2012&lt;/strong&gt;, p. 17&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-21T11:05:53.280" LastActivityDate="2012-05-21T11:05:53.280" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="474" PostTypeId="1" CreationDate="2012-05-21T15:17:55.847" Score="13" ViewCount="196" Body="&lt;p&gt;I've heard some people mention that this or that program isn't &quot;up to date&quot; with respect to the borders used in the Ramachandran plots to classify &amp;phi;/&amp;psi; as being in the most favorable/acceptable/disallowed regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How &lt;em&gt;are&lt;/em&gt; the accepted regions defined?  What is the current accepted paper that covers this?&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2012-06-11T10:21:41.840" Title="What are the border definitions in the Ramachandran plot?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="475" PostTypeId="1" AcceptedAnswerId="533" CreationDate="2012-05-21T20:01:05.743" Score="10" ViewCount="17401" Body="&lt;p&gt;In cell culture, dimethylsulfoxide (DMSO) is added to prevent the formation of ice crystals which may lyse the cells. Exactly how does dimethylsulfoxide act as a cryoprotectant?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="19616" LastEditDate="2015-10-23T18:08:12.570" LastActivityDate="2019-04-24T18:46:34.320" Title="How does dimethylsulfoxide serve as a cryoprotectant?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="476" PostTypeId="2" ParentId="216" CreationDate="2012-05-21T21:08:01.280" Score="10" Body="&lt;p&gt;There are additional aspects concerning NMR tubes. Besides their precision of wall thickness and concentricity the material itself can be of great importance.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;sup&gt;1&lt;/sup&gt;H and &lt;sup&gt;13&lt;/sup&gt;C-NMR&lt;/h2&gt;&#xA;&lt;p&gt;Borsilicate (pyrex) glass will be fine for standard applications. For high throughput measurements these NMR-Tubes rated for 400 MHz can be disposed after usage avoiding any cross contamination. These tubes have de facto no background signal for these nuclei. Good quality tubes rated at 400 MHz can be used at higher field machines, too, however one has to be aware that some degradation in signal quality can occur. This has to be cross checked with higher rated tubes.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;sup&gt;29&lt;/sup&gt;Si-NMR&lt;/h2&gt;&#xA;&lt;p&gt;In this case both borsilicate and quartz tubes will produce a glass hump, a broad signal resonating at the position where typically Q-groups are found. A possible solution is the usage of PTFE (teflon) liners, which are tubes made of plastic. However one has to be aware that the probe head itself contains quartz tubes which are built in to support the RF coils. Only changing the tube material could still lead to a significant background signal.&lt;/p&gt;&#xA;&lt;p&gt;In addition special care has to be taken concerning gradient-equipped probes as the gradient assembly can contain silicon as well.&lt;/p&gt;&#xA;&lt;p&gt;Changing the NMR-tube can help, however it is worthwhile to measure just the empty probe to get an impression how big the background of the tube is.&lt;/p&gt;&#xA;&lt;p&gt;There are spectroscopic means to at least partially suppress these background signals like relaxation filters or using a DEPTH sequence.&lt;/p&gt;&#xA;" OwnerUserId="111" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-05-21T21:08:01.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="477" PostTypeId="1" AcceptedAnswerId="478" CreationDate="2012-05-21T21:21:17.737" Score="17" ViewCount="2805" Body="&lt;p&gt;When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given composition after equilibrium is re-established? &lt;/p&gt;&#xA;" OwnerUserId="258" LastEditorUserId="30" LastEditDate="2012-05-21T21:38:20.413" LastActivityDate="2018-07-08T18:22:04.810" Title="Is enthalpy additive in mixing?" Tags="&lt;thermodynamics&gt;&lt;mixtures&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="478" PostTypeId="2" ParentId="477" CreationDate="2012-05-21T21:52:43.253" Score="12" Body="&lt;p&gt;The enthalpy is not additive under isobar, adiabatic conditions for non-ideal mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change in the enthalpy is called the &lt;strong&gt;excess enthalpy&lt;/strong&gt; (enthalpy of mixing). It can be computed with models such as the Margules, van Laar, Wilson, NRTL, and UNIQUAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These model give equations for the &lt;strong&gt;excess Gibbs energy&lt;/strong&gt;. The excess enthalpy can be derived from these as described in &lt;em&gt;6.2 Fundamental Relations of Excess Functions&lt;/em&gt; in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0139777458&quot;&gt;Molecular Thermodynamics of Fluid-Phase Equilibria (3rd Edition)&lt;/a&gt;. (F'x already referred to this excellent book.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to be careful with the conflicting sign convetions in the literature.&lt;/p&gt;&#xA;" OwnerUserId="259" LastEditorUserId="259" LastEditDate="2012-05-22T17:15:48.947" LastActivityDate="2012-05-22T17:15:48.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="479" PostTypeId="2" ParentId="477" CreationDate="2012-05-21T23:56:04.470" Score="6" Body="&lt;p&gt;As Ali said, unless the mixture is &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_solution&quot;&gt;ideal&lt;/a&gt; the enthalpy of mixing (or excess enthalpy) is not zero. This is the basis for a whole field of thermodynamics (and in particular, statistical thermodynamics): &lt;strong&gt;solution theories&lt;/strong&gt;. To quote only a few of the most famous ones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;VST (Vacancy Solution Theory) &lt;sub&gt;&lt;em&gt;[I don't have a good reference at hand right now, I'll try to come back and edit it in if noöne beats me to it]&lt;/em&gt;&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-random_two-liquid_model&quot;&gt;NRTL&lt;/a&gt; (Non Random Two Liquid)&lt;/li&gt;&#xA;&lt;li&gt;SAFT (Statistical Associating Fluid Theory) and &lt;a href=&quot;http://www.sklogwiki.org/SklogWiki/index.php/SAFT_%28statistical_associating_fluid_theory%29&quot;&gt;its variants&lt;/a&gt;: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie00104a021&quot;&gt;original paper&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie0102201&quot;&gt;a sort of review&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this literature, you'll frequently encounter the name of &lt;a href=&quot;http://www.cchem.berkeley.edu/jmpgrp/&quot;&gt;John M. Prausnitz&lt;/a&gt;, whose book are considered classics (though I have not read them myself), including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books/about/Molecular_Thermodynamics_of_Fluid_Phase.html?id=VSwc1XUmYpcC&quot;&gt;Molecular Thermodynamics of Fluid-Phase Equilibria&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books/about/The_Properties_of_Gases_and_Liquids.html?id=9tGclC3ZRX0C&quot;&gt;The Properties of Gases and Liquids&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The above are for mixture of molecular fluids… polymer solutions are a bit of a special case of this, and specific models have been developed for them, the best-known being &lt;a href=&quot;http://en.wikipedia.org/wiki/Flory%E2%80%93Huggins_solution_theory&quot;&gt;Flory–Huggins&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, you did not specify what kind of &lt;em&gt;computation&lt;/em&gt; of the mixing enthalpy you were considering, so I have to add that several molecular simulation techniques have been developped to addres this question. For a basic introduction on this topic, I would recommend Singh and Gubbins’ &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ba-1983-0204.ch004&quot;&gt;review in &lt;em&gt;Molecular-Based Study of Fluids&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-21T23:56:04.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="480" PostTypeId="1" AcceptedAnswerId="481" CreationDate="2012-05-22T00:52:30.203" Score="-1" ViewCount="4286" Body="&lt;p&gt;A portable dehumidifier requires the consumer to plug it into an outlet to regenerate its crystal. When the crystal is thus &quot;regenerated&quot;, the indicator will change from pink color to its original blue color. However, I noticed that there is usually a colorless gas (or smell - it smells differently from the surrounding air) accompanying this change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is the gas (or smell) from the portable dehumidifier (when plugged into a outlet for re-generation of its crystal) harmful?&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="30" LastEditDate="2012-05-22T07:26:39.637" LastActivityDate="2012-05-22T07:26:39.637" Title="Is the smell coming out from a portable dehumidifier safe?" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-03-10T02:23:45.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="481" PostTypeId="2" ParentId="480" CreationDate="2012-05-22T01:29:34.963" Score="4" Body="&lt;p&gt;Can you explain what the crystal is? Based on the common colors you describe and the &quot;Regeneration&quot; I would think that the &quot;crystal&quot; you are describing is just a basic Desiccant (like a silica gel, it's hygroscopic, it absorbs water). Most desiccants are pretty harmless (never heard of any issues with common ones used in consumer goods). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You really need to add more information to your description. What is the make (in case someone wants to look it up)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest danger I can think of with a portable dehumidifier is bacterial/fungal growth. Both of these can give off a smell, it could be that when &quot;regenerating the crystal&quot; (which if it's a desiccant like I think would mean heating it, is actually &quot;cooking&quot; the growth creating a smell. (But this is me trying to image what could potentially cause that smell, not what is happening)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, not enough information to reach any conclusion. (But I do think you are describing a desiccant). &lt;/p&gt;&#xA;" OwnerUserId="261" LastEditorUserId="30" LastEditDate="2012-05-22T07:25:56.143" LastActivityDate="2012-05-22T07:25:56.143" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="483" PostTypeId="2" ParentId="477" CreationDate="2012-05-22T04:48:23.960" Score="4" Body="&lt;p&gt;This is intended as an addendum to the other answers, not a full answer in itself.&lt;span class=&quot;math-container&quot;&gt;$\newcommand{\b}[2]{\ce{#1\bond{...}#2}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's say we have a solution of molecule A, and a solution of molecule B. I'm denoting the strength of intramolecular forces by &lt;span class=&quot;math-container&quot;&gt;$\b AA,\b BB$&lt;/span&gt; etc.&lt;/p&gt;&#xA;&lt;p&gt;Whenever we create a &lt;em&gt;real&lt;/em&gt; (i.e. non-ideal) solution, we &lt;a href=&quot;http://en.wikipedia.org/wiki/Raoult%27s_law#Real_solutions&quot; rel=&quot;nofollow noreferrer&quot;&gt;classify the solution on the basis of deviation from Raoult's law&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;h2&gt;Positive deviation&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta V_\text{mix}&amp;gt;0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{mix}&amp;gt;0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$VP&amp;gt;VP_\text{expected}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\b AB&amp;lt;\b AA, \b BB$&lt;/span&gt;.  (cohesive forces are stronger than adhesive forces)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Example: water and benzene.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here, the adhesive forces are weaker. This leads to a net &amp;quot;expansion&amp;quot; of the mixture, and, it is &lt;strong&gt;endothermic&lt;/strong&gt; due to the fact that we are partially replacing strong cohesive &amp;quot;bonds&amp;quot; with weaker, adhesive ones.&lt;/p&gt;&#xA;&lt;h2&gt;Negative deviation&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta V_\text{mix}&amp;lt;0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{mix}&amp;lt;0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$VP&amp;lt;VP_\text{expected}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\b AB&amp;gt;\b AA, \b BB$&lt;/span&gt;.  (cohesive forces are weaker than adhesive forces)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Example: ethanol and chloroform.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Here, the adhesive forces are stronger. This leads to a net &amp;quot;contraction&amp;quot; of the mixture, and, it is exothermic due to the fact that we are partially replacing weak cohesive &amp;quot;bonds&amp;quot; with stronger, adhesive ones.&lt;/p&gt;&#xA;&lt;p&gt;So yes, the enthalpy is not additive for non-ideal solutions.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-07-08T18:22:04.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="484" PostTypeId="1" AcceptedAnswerId="486" CreationDate="2012-05-22T07:04:00.050" Score="25" ViewCount="1950" Body="&lt;p&gt;What software does the PDB use to generate these &quot;cel-shaded&quot; protein models with outlines around foreground atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rcsb.org/pdb/education_discussion/molecule_of_the_month/images/149-Leptin_1ax8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hwdmMl.jpg&quot; alt=&quot;Leptin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Leptin: &lt;a href=&quot;http://www.rcsb.org/pdb/101/motm.do?momID=149&quot; rel=&quot;noreferrer&quot;&gt;May 2012 Molecule of the Month&lt;/a&gt; by David Goodsell)&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="3748" LastEditDate="2014-05-29T20:11:20.300" LastActivityDate="2014-05-29T20:11:20.300" Title="What software is used to generate the PDB molecule of the month images?" Tags="&lt;biochemistry&gt;&lt;software&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="485" PostTypeId="1" CreationDate="2012-05-22T07:14:06.810" Score="16" ViewCount="2422" Body="&lt;p&gt;&lt;a href=&quot;http://kinemage.biochem.duke.edu&quot; rel=&quot;noreferrer&quot;&gt;Molprobity&lt;/a&gt; and some other protein structure validation tools report a &lt;em&gt;C&amp;beta; deviation&lt;/em&gt; statistic and offer plots for it (example below).  Apparently if the C&amp;beta; is greater than 0.25 &amp;Aring;, some attention should be given to the residue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistic is given in units of angstroms, but I'm not sure what it's actually referring to.  With the Rama plot, it's the peptide dihedral angles, but what is this?  And why are the graphs 2-D despite the one dimension given?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hxzOd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxzOd.png&quot; alt=&quot;example C-beta graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="11" LastEditDate="2012-05-22T16:57:39.873" LastActivityDate="2014-11-22T01:42:52.683" Title="What is Cβ (C-beta) deviation?" Tags="&lt;biochemistry&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="486" PostTypeId="2" ParentId="484" CreationDate="2012-05-22T08:16:57.660" Score="28" Body="&lt;p&gt;These illustrations are by &lt;a href=&quot;http://mgl.scripps.edu/people/goodsell/&quot; rel=&quot;noreferrer&quot;&gt;David Goodsell&lt;/a&gt;, and carry his famous look. Whilst I was not able to determine what specific software he uses, &lt;a href=&quot;http://www.bmc.med.utoronto.ca/molecule/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;noreferrer&quot;&gt;VMD&lt;/a&gt; is a fairly powerful program (that F'x has contributed patches to!) capable of &lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/minitutorials/glslgoodsell/&quot; rel=&quot;noreferrer&quot;&gt;achieving this cel-shaded look&lt;/a&gt; with minimal postprocessing by using a GLSL Fresnel shader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How to do it with VMD and Blender&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the first instance, I tried using the GLSL shader Goodsell style in VMD on the &lt;a href=&quot;http://www.rcsb.org/pdb/explore/explore.do?structureId=1ron&quot; rel=&quot;noreferrer&quot;&gt;PDB file 1RON&lt;/a&gt;, a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlAKk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like &lt;a href=&quot;http://developer.download.nvidia.com/assets/gamedev/files/sdk/11/FXAA_WhitePaper.pdf&quot; rel=&quot;noreferrer&quot;&gt;FXAA&lt;/a&gt;, which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Next step - Blender&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into &lt;a href=&quot;http://www.blender.org&quot; rel=&quot;noreferrer&quot;&gt;Blender&lt;/a&gt;. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0N3Sy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. &lt;strong&gt;Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'&lt;/strong&gt;. I like the subtly shaded look better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/50zHb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5bTa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LoDSG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Invert and level, set blendmode to multiply and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xUzcn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not perfect, but not bad.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Bonus&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's the shadeless version, which is closer to Goodsell's ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrTrc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Journey Continues!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Poking around with the Toon shaders in Blender, I have found a super easy way to make Goodsell-esque figures!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you configure your material like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/heCym.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a shader that will rapidly shade to black at angles far from normal with a lightsource. To make this work, you should disable environment lighting and position a bright point light behind the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will note that I have adjusted the Toon size to 0.98, which will create a shadow just shy of perpendicular to the light source normal. You will also note that I have set the specularity to zero, which gives an approximately flat finish to the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Render it up and you will get something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BmKnL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-22T12:00:04.010" LastActivityDate="2012-05-22T12:00:04.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="487" PostTypeId="1" AcceptedAnswerId="489" CreationDate="2012-05-22T13:59:43.987" Score="20" ViewCount="927" Body="&lt;p&gt;I know that $\ce{H+}$ is not possible in water and it is present as $\ce{H3O+}$. But later on I come to know that even $\ce{H3O+}$ is not possible and that it is present as $\ce{H9O4+}$.&#xA;Why does this happen? What give that compound so much stability that is not present in $\ce{H3O+}$ or $\ce{H+}$?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T19:21:05.717" LastActivityDate="2015-05-13T19:22:48.600" Title="Stability of H⁺ ion" Tags="&lt;ions&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="488" PostTypeId="2" ParentId="487" CreationDate="2012-05-22T14:35:17.870" Score="11" Body="&lt;p&gt;I think that for this type of question a bit of context as to why you are asking the question, would be useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for instance you are in an Intro Chem class I would say the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All chemical systems tend towards their most stable state (in the absence of external energy being applied).&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+}$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $\ce{H^+}$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By forming $\ce{H_3O^+}$ the single proton $\ce{H^+}$ is able to spread its positive charge out over not just itself but the entirety of the bonded $\ce{H_2O}$ molecule. (Note that because Oxygen can be seen in resonance to distribute the negative charge associated with its loan pairs it is particularly good at stabilizing the extra positive charge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats better than  $\ce{H_3O^+}$? Getting even more water molecules to share the burden of the extra positive charge and producing $\ce{H_9O_4^+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the short answer I would give to your question would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main factors: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system. &lt;/li&gt;&#xA;&lt;li&gt;The Oxygen in particular (being able to share the negative charge of its lone pair) can &quot;handle&quot; the extra positive charge better than the lone proton ($\ce{H^+}$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope that helps. Because in my opinion, beyond this explanation the answer gets very complicated and as many have pointed out, broad.&lt;/p&gt;&#xA;" OwnerUserId="261" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T19:22:48.600" LastActivityDate="2015-05-13T19:22:48.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="489" PostTypeId="2" ParentId="487" CreationDate="2012-05-22T15:16:23.267" Score="19" Body="&lt;p&gt;The formation of bonds nearly always leads to a decrease in energy. This is desirable, since less energy $\implies$ stability, so wherever it is possible, bonds tend to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H+}$ has an empty $s$ orbital. $\ce{H2O{:}}$ has a lone pair. These easily form something akin to a dative bond ($\ce{H2O{:}\bond{-&amp;gt;}H+}$), giving $\ce{H3O+}$. Note that there is no single positively charged hydrogen--the charge is distributed over the entire molecule. Distribution of charge leads to more stability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\ce{H9O4+}$ (called the &quot;Eigen cation&quot;) is even stranger. You can imagine it like this: Three $\ce{H2O}$ molecules approach $\ce{H3O+}$ (remember, this cation is positively charged and water has a lone pair which is attracted to positive charge). Basically, charge-dipole attraction. Of course, the lone pair will &quot;point&quot; towards the hydronium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, these charge-dipole attraction &quot;bonds&quot; resonate. The hydrogens of the central hydronium &quot;detach&quot; from the main molecule and attach to the peripheral ones. In reality, the electron cloud is delocalised and we have partial bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lsbu.ac.uk/water/images/h904res.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fThdf.gif&quot; alt=&quot;resonance structures of the Eigen cation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the charge is distributed even more. We can have other species like $\ce{H7O3+}$, $\ce{H15O7+}$, etc. Note that there are crowding issues as well - this is why $\ce{H_{103}O_{51}+}$ &lt;em&gt;probably&lt;/em&gt; doesn't exist.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Basically, the &quot;original&quot; $\ce{H+}$ is being &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation&quot; rel=&quot;noreferrer&quot;&gt;solvated&lt;/a&gt;. The solvation delocalises the charge and forms extra bonds, reducing the net potential energy and increasing stability. That's basically it. It's similar to what happens when you dissolve &lt;em&gt;any&lt;/em&gt; ionic compound in water.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="3748" LastEditDate="2014-05-29T20:33:18.127" LastActivityDate="2014-05-29T20:33:18.127" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="490" PostTypeId="2" ParentId="312" CreationDate="2012-05-22T18:33:27.227" Score="11" Body="&lt;p&gt;I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, but other molecules and clusters are possible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to attempt an answer to your question, the standard output of an ICP-MS will be presented as a mass spectrum (intensity versus m/z or intensity versus amu). There are a couple of sample spectra in the brochures published by instrument makers, see for instance &lt;a href=&quot;http://www.bdal.com/uploads/media/ICP-MS_aurora_M90_brochure_ebook.pdf&quot;&gt;the Bruker brochure&lt;/a&gt; or &lt;a href=&quot;https://static.thermoscientific.com/images/D14386~.pdf&quot;&gt;a Thermo application note&lt;/a&gt;. How this mass spectrum is obtained from the detector will lead to the type of data that is acquired: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For a quadrupole mass filter, as well as for magnetic sector instruments, two modes of operation are possible. Either a single mass is selected, and therefore ont will only get a trace of the abundance of a given mass as a function of time (chromatography or on-line sample analysis). Otherwise the mass filter can be scanned, leading to a mass spectrum.&lt;/li&gt;&#xA;&lt;li&gt;For time of flight detectors, the mass spectrum is built by the measurement of a flight time for an ion to reach the detector. Therefore the mass spectrum is converted from a time dependent signal to a spectrum.&lt;/li&gt;&#xA;&lt;li&gt;Finally there have been some attempts to couple an ICP ion source with a Fourier transform mass spectrometer. As far as I know, these are not commercial instruments. In such instruments, the ions oscillate at m/z dependent frequencies, and the recorded image current is converted to a spectrum through the use of Fourier transform.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="263" LastActivityDate="2012-05-22T18:33:27.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="493" PostTypeId="1" CreationDate="2012-05-23T00:51:07.113" Score="17" ViewCount="529" Body="&lt;p&gt;Assume a cubic diamond crystal with hydrogen-terminated surface bonds. The terminating hydrogen atoms will form pairs that are geometrically close together in alternating diagonal pairs (think of the rectangles on corrugated walking steel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, picture a formaldehyde molecule being pushed oxygen-first into one of these terminating hydrogen pairs. If the oxygen atom can be made to react with the hydrogen pair, a water molecule would be expelled and the remaining $\ce{CH2}$ group would presumably bond to the two bonds of the diamond lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you repeat this process over the entire cubic face, the result will be a new layer of diamond that is terminated with the same paired hydrogen atoms as the original layer. The entire layer-growth process thus in principle could be repeated indefinitely, resulting in a diamond grown at or near room temperature from an aqueous or other low-temperature solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Can anyone think of a fundamental reason why the above process, or some variant of it, could &lt;em&gt;not&lt;/em&gt; be used to grow diamond crystals at room temperature? (A &quot;variant&quot; might start with carbohydrate units larger than $\ce{OCH2}$, for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One or more geometrically sophisticated catalyst molecules would be required. Each catalyst molecule would snag, carry, and orient a formaldehyde molecules onto some part of the textured surfaces provided by the cubic diamond face, such as the face itself or along a growth edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this a long time ago in high school, but my recollection is that the reaction should be mildly exothermic. But even if it is not, energy-source molecules (think of the role of ATP in living systems) could provide energy to enable an endothermic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... thoughts, anyone? Is there any deep reason why this approach to room-temperature or near-room-temperature diamond synthesis is clearly &lt;em&gt;not&lt;/em&gt; possible?&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="84" LastEditDate="2012-05-23T08:51:17.890" LastActivityDate="2013-01-16T05:00:12.833" Title="Is it possible that diamond can be grown from aqueous solutions?" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="494" PostTypeId="1" CreationDate="2012-05-23T11:46:46.517" Score="7" ViewCount="17463" Body="&lt;p&gt;Why is calcium fluoride insoluble in water? $\ce{NaCl}$ is soluble, but $\ce{CaF2}$ is not. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="270" LastEditorUserId="62703" LastEditDate="2018-04-12T02:31:03.840" LastActivityDate="2020-08-13T13:14:32.153" Title="Why is calcium fluoride insoluble in water?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="495" PostTypeId="2" ParentId="423" CreationDate="2012-05-23T14:35:00.050" Score="10" Body="&lt;p&gt;The reactivity of NHS activated esters has mostly been desribed quite anecdotically. They are susceptible to hydrolysis in water, the kinetics of which is dependent on the nature of the NHS activated group (ester vs carnonate vs carbamate) as well as on the pH of the reaction buffer&lt;sup&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0165-022X%2898%2900027-X&quot;&gt;1&lt;/a&gt;&lt;/sup&gt;, which is one way to show that primary amines are not the only nucleophile which can react with NHS-activated esters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the chemical groups that have been observed to react with NHS activated esters, I did find a couple of examples upon looking up this issue. For instance, the Zenobi group have attempted a quite systematic study on the groups that can react with doubly NHS activated cross-linkers. Citing the abstract&lt;sup&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/jms.1544&quot;&gt;2&lt;/a&gt;&lt;/sup&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As soon as additional cross-linkers were attached or loops were formed, other amino acids were also involved in the reaction. In addition to the primary amino groups, serine, threonine and tyrosine showed significant reactivity due to the effect of neighboring amino acids by intermediate or permanent Type-1 cross-link formation. The reactivity is highly dependent on the pH and on adjacent amino acids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an earlier work, the reactivity of a biotinylation NHS-ester based reagent was assessed on a single peptide,  [D-Lys(6)]gonadotropin releasing hormone. A similar reactivity of -OH groups was observed, with in addition that of the arginine guanidinium group&lt;sup&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0196-9781%2897%2900225-8&quot;&gt;3&lt;/a&gt;&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to the O-acylation of Ser(4) and Tyr(5) in this peptide, we have also identified a novel biotinylation of the Arg(8) side chain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other groups have also observed some reactivity with cysteins (sulfhydryl groups), but in a situation where the reagent is included within a ligand of the target protein. One can assume, as do the authors, that specific positionning of the ligand within the binding site can lead to binding with a cystein group&lt;sup&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/0014-5793%2895%2900637-O&quot;&gt;4&lt;/a&gt;,&lt;a href=&quot;http://dx.doi.org/10.1111/j.1432-1033.1995.tb20359.x&quot;&gt;5&lt;/a&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these earlier work, one cas conclude, in answer to your question that although primary amines are the most reactive group for NHS-ester derivatization, other groups present in peptide side-chains (-OH for tyrosine, serine, threonine, guanidinium for arginine and sulfhydryls for cysteins).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last issue raised in the question remains that of the secondary amine groups. It is widely accepted that pH influences strongly the kinetics of NHS derivatization, mostly through protonation of the amine groups which cannot act as nucleophiles in protonated state. This has been put to use in controlling the buffer pH in order to favor N-terminal reaction vs lysine side chain reaction. Although, in my experience, it is not at all a straightforward experiment and it is highly dependent on the nature of the derivatization group. One could hypothesize that since secondary amines are more basic then primary amines, the reactivity of secondary amines towards NHS should be reduced compared to primary amines, but I have no experimental or litterature proof for this part of the answer.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2012-05-23T14:35:00.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="496" PostTypeId="1" AcceptedAnswerId="505" CreationDate="2012-05-23T15:05:16.407" Score="11" ViewCount="832" Body="&lt;p&gt;It's possible to select a perfluorinated hydrocarbon that is immiscible in another organic solvent, forming a distinct phase boundary. Such systems may become miscible at elevated temperature which is exploited in the design of some catalytic processes&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that many fluorous solvents are immiscible with organic solvents? Are their polarities sufficiently different or is there another reason?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Housecroft, C.E.; Sharpe, A.G.; &lt;em&gt;Inorganic Chemistry 2e&lt;/em&gt;; Pearson Prentice Hall; &lt;strong&gt;2005&lt;/strong&gt;, pp. 798-799&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-24T06:26:34.577" Title="Why do biphasic systems of fluorous and organic solvents form?" Tags="&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="498" PostTypeId="1" AcceptedAnswerId="504" CreationDate="2012-05-23T16:05:31.497" Score="13" ViewCount="2988" Body="&lt;p&gt;It seems from a brief search of the literature that 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) is a prototypical room temperature ionic liquid (RTIL) that has been studied extensively.  However, numerous -- almost limitless -- combinations of cations and anions that give rise to ionic liquids are possible.  Why does it seem that the imidazolium cations are so extensively studied?  Is one possible reason the fact that $\text{EMI}^+$ is aromatic and thus affords the ionic liquid a high degree of stability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, why does it seem that $\text{BF}_4^-$ (along with, perhaps, $\text{PF}_6^-$) is &lt;strong&gt;the&lt;/strong&gt; prototypical ionic liquid anion?  I have seen studies of other anions (for example, $\text{CF}_3 \text{SO}_3^-$ and $(\text{CF}_3 \text{SO}_2)_2 \text{N}^-$), but $\text{BF}_4^-$ seems to be one of the most popular.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2018-10-16T13:20:01.230" Title="Why is 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) often considered a prototypical room temperature ionic liquid?" Tags="&lt;ions&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="499" PostTypeId="1" AcceptedAnswerId="502" CreationDate="2012-05-23T17:16:27.303" Score="14" ViewCount="6501" Body="&lt;p&gt;Does &lt;em&gt;liquid&lt;/em&gt; iron have a constant specific heat capacity?  Everywhere I've looked (save one sourceless online reference to $c_p = 611\ \mathrm{J\ kg^{–1}\ K^{–1}}$ &lt;a href=&quot;http://wiki.answers.com/Q/What_is_specific_heat_capacity_of_iron&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;), I find no value listed for liquid iron.  Is this because the heat capacity is temperature-dependent?  Or is this related to a measurement problem?  Or does an accepted constant value exist, but I'm just a poor researcher?&lt;/p&gt;&#xA;" OwnerUserId="273" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T00:47:22.937" LastActivityDate="2016-07-02T00:16:25.443" Title="Is the specific heat capacity of liquid iron constant?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="500" PostTypeId="1" AcceptedAnswerId="503" CreationDate="2012-05-23T18:49:47.653" Score="5" ViewCount="268" Body="&lt;p&gt;I am not a chemist. I hope I will be specific enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there are two chemical species $\ce{A}$, $\ce{B}$ with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at temperature $t &amp;lt; T_r$, no reaction occurs between $\ce{A}$ and $\ce{B}$ (in any combination).&lt;/li&gt;&#xA;&lt;li&gt;at $t\ge T_r$, $\ce{A}$ interacts with itself to create $\ce{A_2}$, $\ce{B}$ reacts with itself to create $\ce{B_2}$, and $\ce{A}$ and $\ce{B}$ are reacting to create $\ce{AB}$.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{A_2}$, $\ce{B_2}$ and $\ce{AB}$ are never reacting.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In experiment, we first mix $\ce{A}$ and $\ce{B}$ in temperature $t&amp;lt;T_r$. Amounts of species mixed are $a$ for $\ce{A}$, $b$ for $\ce{B}$. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What amounts of $\ce{A_2}$, $\ce{B_2}$, and $\ce{AB}$ can be expected to be produced?&lt;/li&gt;&#xA;&lt;li&gt;To obtain the amounts, should probability theory be used? E.g., amount of $\ce{AB}$ equals to probability that species $\ce{A}$, $\ce{B}$ will interact (&quot;collide&quot; or similar interpretation).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assume the rates of the reactions are equal.&lt;/p&gt;&#xA;" OwnerUserId="274" LastEditorUserId="402" LastEditDate="2017-10-13T19:33:21.917" LastActivityDate="2017-10-13T19:33:21.917" Title="What is the probability of 3 possible products, between two chemical species?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="501" PostTypeId="2" ParentId="499" CreationDate="2012-05-23T20:43:36.497" Score="7" Body="&lt;p&gt;Heat capacities are usually dependent on temperature except for ideal gases. You will generally find the values tabulated at specific temperature and pressure values. For instances for which one is interested in heat capacity as a function of temperature, this is usually approximated through a regression equation. The NIST JANAF Themochemical Tables use the Shomate equation with 5 coefficients, of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C&lt;sub&gt;p&lt;/sub&gt;° = A + B t + C t&lt;sup&gt;2&lt;/sup&gt; + D t&lt;sup&gt;3&lt;/sup&gt; + E / t&lt;sup&gt;2&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where t is T (in K) / 1000, and A, B, C, D and E are regression coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one looks at the &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot;&gt;Condensed phase thermochemistry data page for iron&lt;/a&gt;, one can find the relevant coefficients (A, B, C, D and E in the above equation) and thus calculate heat capacity of liquid iron in the 1809 - 3133 K range. Since all coefficients except A are close to zero, for liquid iron heat capacity is nearly constant in this range.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2012-05-23T21:16:46.923" LastActivityDate="2012-05-23T21:16:46.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="502" PostTypeId="2" ParentId="499" CreationDate="2012-05-23T20:55:17.683" Score="8" Body="&lt;p&gt;Okay, let's answer bit by bit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Measurement of heat capacity of liquids: this isn't particularly hard to do, and in fact it's much easier to obtain $C_p$ experimentally than $C_V$… however, once you know one, the other can be determined from it and some other thermodynamic properties. Different types of calorimeters are used for liquids, but the best-known is probably the &lt;a href=&quot;http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Heat%20energy/text/Specific_heat_capacity_measurement/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Callender and Barnes apparatus&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;“Does an accepted value exist?”&lt;/em&gt; and &lt;em&gt;“Is the specific heat of liquid iron temperature-dependent?”&lt;/em&gt; — Well, I found a good paper which includes measurements and comparison with previously published values: &lt;a href=&quot;http://www.springerlink.com/content/q636918771156554/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Thermophysical Properties of Liquid Iron&lt;/em&gt;&lt;/a&gt;, M. Beutl, G. Pottlacher and H. Jäger, &lt;em&gt;Intl. J. Thermophys.&lt;/em&gt;, &lt;strong&gt;1994&lt;/strong&gt;, 15, 1323.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;            &lt;img src=&quot;https://i.stack.imgur.com/z5Glp.png&quot; alt=&quot;Fig. 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which confirms that considering that &lt;strong&gt;heat capacity is roughly constant in the 1800–4000 K range&lt;/strong&gt;. The authors thus extract the constant pressure specific heat capacity as &lt;strong&gt;825 J/kg/K&lt;/strong&gt;, apparently in reasonable agreement with previous work that they cite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BAtoA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://insertingsomeverticalspacing.com&quot; rel=&quot;nofollow noreferrer&quot;&gt; &lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;“Or am I a poor researcher?”&lt;/em&gt; — Well, the above paper is the first result return by Google for a search for &lt;em&gt;&quot;specific heat&quot; and &quot;liquid iron&quot;&lt;/em&gt;, so I'll let you be the judge of that ;-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-23T20:55:17.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="503" PostTypeId="2" ParentId="500" CreationDate="2012-05-23T21:04:23.553" Score="8" Body="&lt;p&gt;Well, if you assume the rates are known and the reactions' order follows from stoechiometry (e.g. if they are elementary reactions), you can put the chemical kinetics into simple equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm da}{\mathrm dt} = -k_1 a(t)^2 - k_3 a(t)b(t) $$&#xA;$$\frac{\mathrm db}{\mathrm dt} = -k_2 b(t)^2 - k_3 a(t)b(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($t$ here being time, not temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing initial amounts or concentrations $a_0=a(t=0)$ and $b_0=b(t=0)$, you can pretty much integrate  the system to find out what happens.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit —&lt;/em&gt; solving this system for $k_1=k_2=k_3$ yields the quantities of AA, AB and BB at infinite time to be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$aa = \frac{a_0^2}{a_0+b_0}$$&#xA;$$ab = \frac{a_0 b_0}{a_0+b_0}$$&#xA;$$bb = \frac{b_0^2}{a_0+b_0}$$&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorDisplayName="user7951" LastEditDate="2016-10-31T18:55:22.343" LastActivityDate="2016-10-31T18:55:22.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="504" PostTypeId="2" ParentId="498" CreationDate="2012-05-24T01:58:26.963" Score="10" Body="&lt;p&gt;This is a very good question. These compounds have been studied for long enough that most people working with them &quot;know&quot; the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the &quot;right&quot; properties, and 2) they are easy to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning these compounds having the &quot;right&quot; properties:&#xA;Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt;) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt; is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt; is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning their ease of synthesis:&#xA;The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt; anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF&lt;sub&gt;4&lt;/sub&gt;) is added. Silver chloride, AgCl, is virtually insoluble in water (&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;sp&lt;/sub&gt; = 1.77 x 10&lt;sup&gt;-10&lt;/sup&gt;) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt;). AgBF&lt;sub&gt;4&lt;/sub&gt; and AgPF&lt;sub&gt;6&lt;/sub&gt; are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;-&lt;/sup&gt;. Who wants to do more synthesis just to make their solvents?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-24T01:58:26.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="505" PostTypeId="2" ParentId="496" CreationDate="2012-05-24T05:25:47.523" Score="12" Body="&lt;p&gt;Fluorous solvents have the odd property of being both hydrophobic and lipophobic (&lt;a href=&quot;http://en.wikipedia.org/wiki/Organofluorine_chemistry&quot;&gt;Ref&lt;/a&gt;) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. (&lt;a href=&quot;http://www2.chemistry.msu.edu/courses/cem958/FS04_SS05%5CStewartHart.pdf&quot;&gt;Ref&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie0308745&quot;&gt;Ref&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=6MzGgfWZAIMC&amp;amp;pg=PA240&amp;amp;lpg=PA240&amp;amp;dq=solubility%20of%20organic%20solvents%20in%20fluorous%20solvents&amp;amp;source=bl&amp;amp;ots=0eS9j7ysiB&amp;amp;sig=XIpj4mJh2tawr9cVPdtvl9Rj3ZA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=zdC9T6jME6qsiAKM2_X1DQ&amp;amp;ved=0CEwQ6AEwBA#v=onepage&amp;amp;q=solubility%20of%20organic%20solvents%20in%20fluorous%20solvents&amp;amp;f=false&quot;&gt;This reference&lt;/a&gt; suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-05-24T06:26:34.577" LastActivityDate="2012-05-24T06:26:34.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="506" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2012-05-24T16:34:50.367" Score="10" ViewCount="2308" Body="&lt;p&gt;Can someone explain the nomenclature of 7-amino-4-methylcoumarin, pictured below (&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sigma/a9891?lang=en&amp;amp;region=US&quot; rel=&quot;noreferrer&quot;&gt;from the Sigma-Aldrich website&lt;/a&gt;)?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fCOKR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alpha-carbons of the methyl- and amino-groups here are separated by a minimum of three C's, whereas the numbers 4 and 7 are separated by two digits. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2012-05-24T17:30:46.633" Title="Explain the nomenclature of 7-amino-4-methylcoumarin" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="507" PostTypeId="1" CreationDate="2012-05-24T17:04:16.290" Score="4" ViewCount="2846" Body="&lt;p&gt;While trying to understand the solution of a problem given in my textbook, I realized I'm having some difficulty with the solution. The problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ionization constant of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$3.2 \times 10^{-4}$&lt;/span&gt;. Calculate the degree of dissociation of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; in its &lt;span class=&quot;math-container&quot;&gt;$\pu{0.02 M}$&lt;/span&gt; solution. Calculate the concentration of all species present &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; in the solution and its PF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the solution of this problem, the equation is given as &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HF + H2O &amp;lt;=&amp;gt; H3O+ + F-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration at the time of equilibrium are given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{[HF]}   &amp;amp;= 0.02 - 0.02x, &amp;amp;&#xA;\ce{[H3O+]} &amp;amp;= 0.02x, &amp;amp;&#xA;\ce{[F- ]}  &amp;amp;= 0.02x&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why are we not adding the contribution of water to the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions?&lt;/li&gt;&#xA;&lt;li&gt;Why are we assuming that the value &lt;span class=&quot;math-container&quot;&gt;$0.02x$&lt;/span&gt; is dissociated from &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; and not just &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="277" LastEditorUserId="4945" LastEditDate="2020-02-17T09:58:10.560" LastActivityDate="2020-02-17T10:09:31.003" Title="Why is water ignored in the ionic equilibrium of HF?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="508" PostTypeId="2" ParentId="506" CreationDate="2012-05-24T17:20:14.660" Score="7" Body="&lt;p&gt;This is one of the tricky parts in the nomenclature of fused cyclic systems. Excerpt from page 180 of the 1998 &lt;a href=&quot;http://www.iupac.org/publications/pac/1998/pdf/7001x0143.pdf&quot; rel=&quot;noreferrer&quot;&gt;IUPAC Nomenclature of fused and bridged fused ring systems&lt;/a&gt;&#xA;(I have added bold to the piece of importance to answer your question): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The numbering of the peripheral atoms in a particular orientation starts from the uppermost ring. If there is more than one uppermost ring the ring furthest to the right is chosen. Numbering starts from the non-fusion atom most counterclockwise in the ring selected, and numbering proceeds in a clockwise direction around the system, including fusion heteroatoms. &lt;strong&gt;Fusion carbon atoms are not included&lt;/strong&gt;; each fusion carbon atom is given the same number as the immediately preceding position modified by a roman letter a, b, c, etc. If&#xA;  heteroatoms are identified by replacement nomenclature all atoms retain the numbers of the parent&#xA;  hydrocarbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If one considers coumarin, this is not the IUPAC nomenclature but the numbering rule remains the same. Below is an example of the numbering scheme for coumarin which summarizes why the substituents are numbered by this manner on coumarin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/15mtt.png&quot; alt=&quot;coumarin with numbering scheme&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2012-05-24T17:20:14.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="510" PostTypeId="2" ParentId="506" CreationDate="2012-05-24T17:30:46.633" Score="6" Body="&lt;p&gt;A quick perusal of the internet reveals that there are at least four numbering schemes used for coumarin derivatives. In each case, elision, duplication or out-of-sequence numbering creates a gap that accounts for the odd numbering of substituents.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Skip over the fused ring carbons&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Some numbering schemes do not concern themselves with the carbons shared between the rings, as seen &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1389556705000304&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (Fig. 1). This is common in PAH numbering schemes and is recommended by IUPAC&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3sBjk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4a and 8a&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As seen &lt;a href=&quot;http://www.springerimages.com/Images/Chemistry/1-10.1007_s11224-006-9097-y-1&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, 4 is followed by 4a and 8 is followed by 8a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BvOxQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4' and 8'&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the &lt;em&gt;a&lt;/em&gt; modifier can be replaced with a prime, as seen &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0143720899000121&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (Fig. 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1IYrq.png&quot; alt=&quot;prime notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;9 and 10&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0020169310005487&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; coumarin-like structure denotes the fused ring carbons 9 and 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tNZde.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) International Union of Pure and Applied Chemistry; &lt;em&gt;NOMENCLATURE OF FUSED AND BRIDGED FUSED RING SYSTEMS&lt;/em&gt;; Pure and Applied Chemistry; &lt;strong&gt;1998&lt;/strong&gt;, 70, pp. 143-216&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-24T17:30:46.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="511" PostTypeId="2" ParentId="507" CreationDate="2012-05-24T17:56:16.657" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Why are we not adding the contribution of water to the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write down the dissociation constant equation with and without &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;.&lt;br&gt;&#xA;What changes? &lt;/li&gt;&#xA;&lt;li&gt;Can you find a formulation of the dissociation constant where it seems natural not to include &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;? &lt;/li&gt;&#xA;&lt;li&gt;Read up on &quot;activity&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Why are we assuming that the value &lt;span class=&quot;math-container&quot;&gt;$0.02x$&lt;/span&gt; is dissociated from &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; and not just &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Write down and solve the equation with &lt;span class=&quot;math-container&quot;&gt;$0.02 x$&lt;/span&gt; and with &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; only. What is the difference?&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="4945" LastEditDate="2020-02-17T10:09:31.003" LastActivityDate="2020-02-17T10:09:31.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="512" PostTypeId="2" ParentId="507" CreationDate="2012-05-24T18:18:54.617" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Why are we not adding the contribution of water to the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions?  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; actually is &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; so you can write the equation as just dissociation of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; cancels out on both sides):&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HF &amp;lt;=&amp;gt; H+ + F-}.$$&lt;/span&gt;&#xA;Thus we do not take water into consideration.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Why are we assuming that the value &lt;span class=&quot;math-container&quot;&gt;$0.02x$&lt;/span&gt; is dissociated from &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; and not just &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;?  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because whenever we write an equation&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A -&amp;gt; B + C}$$&lt;/span&gt;&#xA;We write initial concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; while &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; are zero.&#xA;Final concentrations are written as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;: (Initial concentration) - (concentration dissociated)&lt;/li&gt;&#xA;&lt;li&gt;for &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;:  (concentration dissociated)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus re-writing the equations:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lccccc}&#xA;        &amp;amp; \ce{HF} &amp;amp; \ce{&amp;lt;=&amp;gt;} &amp;amp; \ce{H+} &amp;amp; + &amp;amp; \ce{F-} \\&#xA;  \text{Initial Amount}:&amp;amp; 0.2      &amp;amp;&amp;amp;    0  &amp;amp;&amp;amp;  0  \\&#xA;  \text{Final Amount}:  &amp;amp; 0.2-0.2x &amp;amp;&amp;amp;  0.2x &amp;amp;&amp;amp; 0.2x   \\&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="4945" LastEditDate="2020-02-17T10:05:59.153" LastActivityDate="2020-02-17T10:05:59.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="513" PostTypeId="1" AcceptedAnswerId="515" CreationDate="2012-05-24T18:22:13.180" Score="23" ViewCount="20036" Body="&lt;p&gt;In the title I am refering to silver spoons, which, when used to consume milk products (like yogurt), get a weird taste. It could be described as umami/meaty, a little bit like onions. I came to the conclusion that it must be the silver, because this taste is never observed when using steel spoons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess was: the silver (or maybe its oxide?) reacts with an ingredient of the milk product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is this documented/known process/reaction or is this some kind of misperception on my part? If the former is true, what exactly is happening? (By)-products?&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="30" LastEditDate="2012-05-24T19:02:44.500" LastActivityDate="2017-06-20T11:54:16.857" Title="Silver and milk (or milk products)" Tags="&lt;everyday-chemistry&gt;&lt;silver&gt;&lt;taste&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="514" PostTypeId="2" ParentId="456" CreationDate="2012-05-24T19:46:24.467" Score="19" Body="&lt;p&gt;The reaction coordinate is the progress of a reaction from reactants to products with various intermediates and transition states in between. It is an abstraction. It has no relation to time. Rather it is the progress of bond-forming and bond-breaking reaction steps. The free energy change of partially formed and partially broken bonds cannot be measured. The free energies of transition states (local maxima) cannot be directly measured. These states are transient and cannot be directly observed. For a simple reaction with few mechanistic steps, the activation energy (transition state energy) can be calculated from kinetics measurements. The free energy of intermediates can often be determined directly or indirectly if they are long-lived. Most of these diagrams are assembled from a few measured data points and either extrapolation or theoretical computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two reaction coordinate axes are presented, this often means that there are two separate pathways the reaction can take, and each pathway is presented on one of the two reaction coordinate axes. Perhaps the reaction involves two different steps, but those steps need not happen in the same order every time. As a generic example, consider nucleophilic substitution reactions in organic chemistry: A generic nucleophile ($\ce{Nu-}$) reacts with an organic molecule R-X, where X is a leaving group, and R is the rest of the molecule (which is unchanged or nearly unchanged). The overall reaction would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Nu- + R-X -&amp;gt; Nu-R + X-}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process involves two steps: forming the bond between Nu and R, and breaking the bond between R and X. These two steps can happen in three different orders: 1) the R-X bond breaks first and the R-Nu bond forms second (this is the dissociative or $\mathrm{S_{N}1}$ mechanism); 2) The R-Nu bond forms first and the R-X bond breaks second (this is the associative or $\mathrm{S_{N}Ac}$ mechanism); and 3) The R-X bond breaks and the R-Nu bond forms at the same time (this is the intermediate or $\mathrm{S_{N}2}$ mechanism). All three of these cases occur in reactions that are covered in any undergraduate organic chemistry text. These three processes would look like the following. The items in parentheses are transition states with partially formed/broken bonds ($\ce{\bond{....}}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissociative:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Step 1: $\ce{Nu- + R-X -&amp;gt; (Nu- + R\bond{....}X) -&amp;gt; Nu- + R+ + X-}$&lt;/li&gt;&#xA;&lt;li&gt;Step 2: $\ce{Nu- + R+ + X- -&amp;gt; (Nu\bond{....}R + X-) -&amp;gt; Nu-R + X-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Associative:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Step 1: $\ce{Nu- + R-X -&amp;gt; (Nu\bond{....}R-X) -&amp;gt; [Nu-R-X]-}$&lt;/li&gt;&#xA;&lt;li&gt;Step 2: $\ce{[Nu-R-X]- -&amp;gt; (Nu-R\bond{....}X) -&amp;gt; Nu-R + X-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Intermediate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Only step: $\ce{Nu- + R-X -&amp;gt; (Nu\bond{...}R\bond{....}X) -&amp;gt; Nu-R + X-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A reaction coordinate energy diagram for a substitution process could then be graphed over two different reaction coordinates: breaking the R-X bond and forming the R-Nu bond.  For ease of two dimensional representation, these are often presented as contour diagrams. Various trajectories could be taken through this space based on what real species R, Nu, and X represent. For example, if R+ is a stabilized cation, then the dissociative pathway becomes more likely. These plots are called More O'Ferrall-Jencks plots. &lt;a href=&quot;http://en.wikipedia.org/wiki/More_O%27Ferrall-Jencks_plot&quot; rel=&quot;noreferrer&quot;&gt;This link&lt;/a&gt; will take you to the wikipedia article, which is not a very good article. The references in that article to the papers by R. A. More O'Ferrall and W. P. Jencks are probably better places to go to find out more. Most advanced undergraduate or graduate level textbooks of physical organic chemistry, particularly &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0306434474&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Advanced Organic Chemistry, Part A: Structure and Mechanisms&lt;/em&gt; by Carey and Sundberg&lt;/a&gt;, will cover this topic, at least as it relates to organic chemistry. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="194" LastEditDate="2017-10-29T19:06:39.370" LastActivityDate="2017-10-29T19:06:39.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="515" PostTypeId="2" ParentId="513" CreationDate="2012-05-24T20:04:49.353" Score="12" Body="&lt;p&gt;You're right: the silver is reacting with sulfur compounds in the food to form a tarnish of silver sulfide. This is most commonly observed, in my experience, using silver teaspoons with boiled eggs, which are &lt;a href=&quot;http://healthyeatingclub.com/info/books-phds/books/foodfacts/html/data/data5g.html&quot; rel=&quot;noreferrer&quot;&gt;pretty rich in sulfur&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There are a number of reactions that can take place depending on the sulfur-containing species - the abstract from &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed077p328A&quot; rel=&quot;noreferrer&quot;&gt;this paper from the Journal of Chemical Education&lt;/a&gt; (unfortunately pay-walled) has a nice example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Silver and silver-plated objects react with sulfur and sulfur compounds to produce silver sulfide &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag2S}$&lt;/span&gt; or tarnish. Contact with materials that contain sulfur compounds, such as hard boiled eggs, mayonnaise, mustard, and rubber bands can cause tarnish. In the air, a silver object can tarnish owing to the reaction of silver with hydrogen sulfide (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt;). This is a gas found in the air as a result of some industrial processes and the decomposition of dead plants and animals. The reaction of silver with hydrogen sulfide to form tarnish is as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2 Ag(s) + H2S(g) -&amp;gt; Ag2S(s) + H2(g)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In Sterling silver specifically, you'll often also get copper sulfide formed.&lt;/p&gt;&#xA;&lt;p&gt;The paper also mentions that you can remove the tarnish abrasively, or through reacting the tarnish with aluminium (&lt;em&gt;via&lt;/em&gt; solution, electrochemically):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{3 Ag2S(s) + 2 Al(s) -&amp;gt; 6 Ag(s) + Al2S3(s)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But a more common approach used by many &lt;a href=&quot;http://www.amazon.co.uk/Johnson-Goddard-Silver-Dip-5Ltr/dp/B000TAP6UA/ref=sr_1_11?ie=UTF8&amp;amp;qid=1337889595&amp;amp;sr=8-11&quot; rel=&quot;noreferrer&quot;&gt;commercial products&lt;/a&gt; is to dissolve off the tarnish, usually in sodium thiosulfate or thiourea. This means you end up with slightly less silver on your item, but is otherwise pretty effective and very mild (it's &lt;a href=&quot;http://cool.conservation-us.org/waac/wn/wn11/wn11-1/wn11-103.html&quot; rel=&quot;noreferrer&quot;&gt;commonly used for museum pieces&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-04-08T21:48:56.007" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="516" PostTypeId="2" ParentId="270" CreationDate="2012-05-24T20:15:54.147" Score="9" Body="&lt;p&gt;I think your problem with this reaction lies in the easily confused term &lt;em&gt;syn&lt;/em&gt; and &lt;em&gt;cis&lt;/em&gt;. Both appear to mean that two functional groups are oriented in the same direction. On a cyclic molecule, &lt;em&gt;syn&lt;/em&gt; and &lt;em&gt;cis&lt;/em&gt; are the same. However, &lt;em&gt;cis&lt;/em&gt; is a stereochemical descriptor and &lt;em&gt;syn&lt;/em&gt; is a conformational descriptor. Simplifying this reaction to an acyclic case may help. Consider the meso isomer of  2,3-butanediol shown.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/7/78/Meso-2,3-butanediol.svg/200px-Meso-2,3-butanediol.svg.png&quot; alt=&quot;anti meso butanediol&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This isomer is drawn in the &lt;em&gt;anti&lt;/em&gt; conformation, which is probably the lowest energy for steric reasons. This isomer has a &lt;em&gt;syn&lt;/em&gt; conformation, which looks like the following. Note that the two -OH groups are not both oriented forward with solid wedges. That drawing implies a high energy eclipsed conformation.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G1vL7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1vL7.png&quot; alt=&quot;syn meso butanediol&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?Struct=C5341957&quot; rel=&quot;nofollow noreferrer&quot;&gt;nist.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now back to the reaction with HIO&lt;sub&gt;4&lt;/sub&gt;. In order to form the cyclic intermediate, the -OH groups need to be close in space. On an acyclic molecule, this requires a &lt;em&gt;syn&lt;/em&gt; conformation, which most acyclic molecules can achieve even if it is not the lowest energy conformation. In cyclic molecules, particularly cyclohexanes, the &lt;em&gt;syn&lt;/em&gt; conformation might not have the -OH groups close together. For &lt;em&gt;cis&lt;/em&gt;-1,2-cyclohexanediol, which is also &lt;em&gt;syn&lt;/em&gt; respective to the &amp;quot;plane&amp;quot; of the ring, to move the two -OH groups into the correct arrangement, it would need to wiggle into the higher energy boat or twist-boat conformations. However, &lt;em&gt;trans&lt;/em&gt;-1,2-cyclohexanediol, which is &lt;em&gt;anti&lt;/em&gt; has the two -OH groups oriented close together. In a sense, the &lt;em&gt;trans&lt;/em&gt; isomer has the two -OH groups oriented &lt;em&gt;syn&lt;/em&gt; with respect to the C-C bond that they share. Thus, &lt;em&gt;trans&lt;/em&gt;-1,2-cyclohexanediol should react more readily with HIO&lt;sub&gt;4&lt;/sub&gt; that &lt;em&gt;cis&lt;/em&gt;-1,2-cyclohexanediol, because the lowest energy conformer of &lt;em&gt;trans&lt;/em&gt;-1,2-cyclohexanediol has the close arrangement of two -OH groups. The &lt;em&gt;cis&lt;/em&gt; isomer would have to change conformation, which increases the energy input required, increasing the activation energy and slowing the reaction rate.&lt;/p&gt;&#xA;&lt;p&gt;The diaxial case (whether boat or chair), would definitely be unreactive.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="15751" LastEditDate="2020-12-19T17:20:29.040" LastActivityDate="2020-12-19T17:20:29.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="517" PostTypeId="2" ParentId="494" CreationDate="2012-05-24T20:32:46.497" Score="10" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/410/why-is-ceagcl-less-soluble-than-ceagno3&quot;&gt;This answer&lt;/a&gt; to a question about the solubility of silver nitrate and silver chloride will give you a good starting point to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve. There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, &lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lattice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this reference&lt;/a&gt;. The values for enthalpies of hydration and crystal lattice energies of sodium chloride and calcium fluoride can be found in standard chemistry texts or with an on-line search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have more questions after looking at this material, come back and ask for clarification.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2012-05-24T20:32:46.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="518" PostTypeId="1" AcceptedAnswerId="522" CreationDate="2012-05-24T21:43:42.177" Score="13" ViewCount="3435" Body="&lt;p&gt;I know that at extremely low temperatures (mK and lower), &lt;a href=&quot;http://isites.harvard.edu/fs/docs/icb.topic105883.files/deBroglie_wavelength/He2_Molecular_Diffraction.pdf&quot;&gt;Helium can form diatomic molecules&lt;/a&gt;. Do the other noble gasses also form molecules at extremely low temperatures?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="22" LastEditDate="2012-05-25T12:03:32.163" LastActivityDate="2015-09-11T06:05:37.513" Title="Do noble gasses besides Helium form diatomic molecules at low temperatures?" Tags="&lt;molecules&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="519" PostTypeId="1" AcceptedAnswerId="529" CreationDate="2012-05-25T04:53:54.440" Score="9" ViewCount="3546" Body="&lt;p&gt;I'm currently studying VCE BioChemistry, and we're studying the separation of DNA strings of different lengths via gel electrophoresis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This involves having 'clumps' of DNA at one end of a gel medium and applying an electric current, pushing the DNA strands across the gel. Different sized strands move at different speeds through the gel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, the strands stop moving such that there is a distribution of strands across the gel. What causes the strands to stop?&lt;br&gt;&#xA;Surely they'd continue until the end of the gel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If suppose the electric current could be stopped, thus causing the strands to stop moving, but how would one know when to stop the current?&lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="4945" LastEditDate="2016-06-29T13:11:36.317" LastActivityDate="2016-06-29T13:11:36.317" Title="What causes the DNA fragments to stop moving in gel electrophoresis?" Tags="&lt;electrochemistry&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="520" PostTypeId="2" ParentId="174" CreationDate="2012-05-25T05:44:52.690" Score="7" Body="&lt;p&gt;I keep hoping someone will jump in and answer this question to save me from thinking about it. Odds and ends from researching this question that might help someone else answer it.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Clusters of water and ammomia, such as $\ce{NH3(H2O)20}$ as well as various other ratios, are &lt;a href=&quot;http://www.engpapers.com/23762.html&quot; rel=&quot;noreferrer&quot;&gt;known&lt;/a&gt; and have been &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp104162k&quot; rel=&quot;noreferrer&quot;&gt;investigated&lt;/a&gt;. I didn't find any references that reported them in vapor phase in distillation columns.&lt;/li&gt;&#xA;&lt;li&gt;But, there is a &lt;a href=&quot;http://db.wdc-jp.com/mssj/search/pdf/200401/ms520001.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt; that reports that water clusters of varying sizes are found in steam during distillation, so it seems possible that water-ammonia complexes could exist during distillation since hydrogen bonding is strong between water and ammonia in solution. (I did see reports suggesting their existence as atmospheric pollutants.)&lt;/li&gt;&#xA;&lt;li&gt;Clusters involving both positive and negative ion complexes of ammonia and water have been reported. (Such as $\ce{(H2O)20NH4+}$ (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp104162k&quot; rel=&quot;noreferrer&quot;&gt;Ref&lt;/a&gt;) or $\ce{(H2O)p(NH3)q-}$ (&lt;a href=&quot;http://www.jhu.edu/chem/bowen/Publication%20PDF/Prediction%20and%20observation%20of%20a%20new,%20ground%20state,%20dipole-bound%20dimer%20anion%20the%20mixed%20water-ammonia%20system..pdf&quot; rel=&quot;noreferrer&quot;&gt;Ref&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;It's difficult for me to tell from the abstracts (or the papers) whether these species are important in regular laboratory conditions, since the researchers are normally looking at them by MS. &lt;/li&gt;&#xA;&lt;li&gt;A single $\ce{H2O}$ bonded with $\ce{NH4+}$ forms a very strong hydrogen bond 92.5 kJ/m.&lt;/li&gt;&#xA;&lt;li&gt;Why doesn't ammonia form an azeotrope with water? Of the common solvents that can form effective hydrogen bonds with water, only methanol and ammonia don't form azeotropes. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope_%28data%29&quot; rel=&quot;noreferrer&quot;&gt;ref&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The comments for the question mention stripping ammonia from a water in a digestate of a Biogas plant. Ammonia can be removed from waste streams by steam or air stripping after adding NaOH to suppress formation of ammonium ions. &lt;a href=&quot;http://nepis.epa.gov/Exe/ZyPURL.cgi?Dockey=30003U4A.txt&quot; rel=&quot;noreferrer&quot;&gt;Reports&lt;/a&gt; suggest that this is fairly efficient (&gt;99% removal). Perhaps the remaining ammonia in the effluent could come from the water-ammonia clusters that are not separated in the distillation. There also seem to be other options for treating the waste stream discussed on &lt;a href=&quot;http://www.finishing.com/336/95.shtml&quot; rel=&quot;noreferrer&quot;&gt;this forum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it, done thinking about this one (and not looking for the bounty, but it would be nice to have an answer).&lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-25T05:44:52.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="521" PostTypeId="2" ParentId="474" CreationDate="2012-05-25T11:59:13.420" Score="1" Body="&lt;p&gt;I wish I could answer your question outright, but this is outside my area of expertise. The best I can give you is a place to start. I expect any recent book on proteomics would cover this topic. For example &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471754021&quot; rel=&quot;nofollow&quot;&gt;Introduction to Proteomics: Principles and Applications (Methods of Biochemical Analysis)&lt;/a&gt; by Mishra and Blobel, Wiley 2010.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some promising papers with a little searching with SciFinder. Neither I nor my institution subscribe to &lt;em&gt;Protein Science&lt;/em&gt;, but the abstract looks promising. I have skimmed the article in &lt;em&gt;PNAS&lt;/em&gt;. Hopefully one of these articles has your answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zhou, &lt;em&gt;et al&lt;/em&gt;, &lt;a href=&quot;http://dx.doi.org/10.1002/pro.644&quot; rel=&quot;nofollow&quot;&gt;&quot;Revisiting the Ramachandran plot from a new angle&quot;&lt;/a&gt;, &lt;em&gt;Protein Science&lt;/em&gt; &lt;strong&gt;2011&lt;/strong&gt; 1166-1171.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Porter and Rose, &lt;a href=&quot;http://dx.doi.org/10.1073/pnas.1014674107&quot; rel=&quot;nofollow&quot;&gt;&quot;Redrawing the Ramachandran plot after inclusion of hydrogen-bonding constraints&quot;&lt;/a&gt;, &lt;em&gt;PNAS&lt;/em&gt;, *&lt;em&gt;2011&lt;/em&gt;, 108, 1, 109-112&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-25T11:59:13.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="522" PostTypeId="2" ParentId="518" CreationDate="2012-05-25T13:51:32.663" Score="11" Body="&lt;p&gt;Yeah, dimers of noble gas atoms are stable due to &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot;&gt;dispersion aka London forces&lt;/a&gt;. More generally, neutral, non-polar molecules can still form very weak bound states arising entirely from the induction and interaction of instantaneous dipole moments (that is, fluctuations in the electron density around an atom arising from the motion of electrons). My understanding is that physicists would explain this interaction as a special case of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Casimir_effect&quot;&gt;Casimir-Polder interaction&lt;/a&gt;. This interaction means that even noble gases are nonideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correctly predicting the negative interaction energy of a noble gas dimer is a pathological test of dispersion-corrected DFT functionals, as F'x's reference&lt;sup&gt;1&lt;/sup&gt; and the references therein indicate, and many methods fail qualitatively. I recall that Zhao and Truhlar used noble gas dimers to test the efficacy of their various metahybrid functionals&lt;sup&gt;2,3&lt;/sup&gt;. One of their papers&lt;sup&gt;2&lt;/sup&gt; makes reference to a table of binding energies and equilibrium distances derived&lt;sup&gt;4&lt;/sup&gt; by Wang and Ogilvy, giving for instance a $\ce{Ne\cdots{}Ne}$ binding energy of 0.35 kJ mol&lt;sup&gt;-1&lt;/sup&gt; (that's tiny) and less than half that for $\ce{He\cdots{}Ne}$. The $\ce{Ne\cdots{}Ne}$ equilibrium distance is given as 309 pm, which is extreme. As the paper&lt;sup&gt;5&lt;/sup&gt; you refer to indicates, however, the interaction energy for the $\ce{^4He\cdots{}^4He}$ dimer is much, &lt;em&gt;much&lt;/em&gt; smaller than either of these and has an equilibrium distance of 5.2 &lt;strong&gt;nm&lt;/strong&gt; (!!!), which is something else entirely.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Ruzsinszky, A.; Perdew, J.P.; Csonka, G.I.; &lt;em&gt;Binding Energy Curves from Nonempirical Density Functionals II. van der Waals Bonds in Rare-Gas and Alkaline-Earth Diatomics&lt;/em&gt;; J. Phys. Chem. A; &lt;strong&gt;2005&lt;/strong&gt;, 109, pp. 11015-11021&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; Zhao, Y.; Truhlar, D.G.; &lt;em&gt;Hybrid Meta Density Functional Theory Methods for Thermochemistry, Thermochemical Kinetics, and Noncovalent Interactions: The MPW1B95 and MPWB1K Models and Comparative Assessments for Hydrogen Bonding and van der Waals Interactions&lt;/em&gt;; J. Phys. Chem. A; &lt;strong&gt;2004&lt;/strong&gt;, 108, 6908-6918&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)&lt;/strong&gt; Zhao, Y.; Truhlar, D.G.; &lt;em&gt;Comparative DFT Study of van der Waals Complexes: Rare-Gas Dimers, Alkaline-Earth&#xA;Dimers, Zinc Dimer, and Zinc-Rare-Gas Dimers&lt;/em&gt;; J. Phys. Chem. A; &lt;strong&gt;2006&lt;/strong&gt;, 110, pp. 5121-5129&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(4)&lt;/strong&gt; Ogilvie, J.F.; Wang, F.Y.H.; &lt;em&gt;Potential-energy functions of diatomic molecules of the noble gases: II. Unlike nuclear species&lt;/em&gt;; J. Mol. Struct.; &lt;strong&gt;1993&lt;/strong&gt;, 291, pp. 313-322&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(5)&lt;/strong&gt; Grisenti, R.E.; Schöllkopf, W.; Toennies,  J.P.; Hegerfeldt, G.C.; Köhler, T.; Stoll, M.; &lt;em&gt;Determination of the Bond Length and Binding Energy of the Helium Dimer by Diffraction from a Transmission Grating&lt;/em&gt;; Phys. Rev. Lett.; &lt;strong&gt;2000&lt;/strong&gt;, 85, pp. 2284-2287&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-05-25T15:17:13.197" LastActivityDate="2012-05-25T15:17:13.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="523" PostTypeId="2" ParentId="519" CreationDate="2012-05-25T14:38:39.757" Score="7" Body="&lt;p&gt;As I understand it, the DNA strands don't stop, they just migrate at different rates and you have to stop the process (i.e. remove the electric field) at some point when you have a good pattern.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2012-05-25T14:38:39.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="524" PostTypeId="2" ParentId="194" CreationDate="2012-05-25T15:33:00.010" Score="9" Body="&lt;p&gt;This difference between living and non-living polymerizations is small in words but large in effects. In principle cationic and anionic polymerizations can be living. In practice, it is not as easy as it looks on paper. See below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A living polymerization is any chain (or addition) polymerization that is prevented from terminating, i.e. the ends of the chains are still reactive.  In a living polymerization, chain initiation has one initiation and propagation continues as long as there is monomer. Termination (almost) never happens.  If your polymerization stops because your monomer runs out, adding more monomer restarts the process and the existing chains get longer. The number of chains formed matches very closely with the number of equivalents of initiator, and the length of the chains depends on the ratio of monomer to initiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most chain polymerizations (radical, anionic, cationic, transition-metal insertion, and ring-opening) do not behave this way. There is a mechanism for termination that becomes increasingly kinetically favored as the concentration of free monomer decreases. Sadly, the wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_termination&quot;&gt;chain termination&lt;/a&gt; is pretty sad. Most chain polymerizations propagate and terminate quickly, with termination usually initiating the next chain. Thus, in a non-living polymerization, new chains are constantly being initiated and old chains are constantly terminating until their is no more monomer.  In these cases, adding additional monomer later would not accomplish anything because the chain ends are &quot;dead&quot;. Experimental evidence for this phenomenon includes the formations of more chains than there were equivalents of initiator. The length of the chain is more dependent on the relative rates of propagation and termination than on the initial ratio of monomer to initiator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For radical polymerization, termination mechanisms include fragmentation by &amp;beta;-hydrogen abstraction or by recombination. Both of these mechanisms are inherent in the radical nature of the polymerization, and so are not preventable by exclusion of other species. Radical polymerizations are made living by the addition of species that &quot;protect&quot; the radical at the end of the chain to prevent it from terminating. Examples include &lt;a href=&quot;http://en.wikipedia.org/wiki/Atom_transfer_radical_polymerization&quot;&gt;ATRP&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/RAFT&quot;&gt;RAFT&lt;/a&gt;. The wikipedia articles for both are pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cationic polymerization, termination mechanisms include &amp;beta;-hydrogen abstraction, chain transfer, and attack by nucleophiles like water. In principle, a cationic polymerization could be a living polymerization if great pains were taking to exclude all species which might induce termination. In practice this is doable, but difficult, as termination by chain transfer is always possible. See the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Living_cationic_polymerization&quot;&gt;living cationic polymerization&lt;/a&gt;. The article is not great, but suggests that cationic polymerizations done in nonpolar solutions (not in bulk) with nonnucleophilic Br&amp;oslash;nsted or Lewis acids at low temperatures can become living polymerization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For anionic polymerization, termination mechanisms include protonation by protic molecules, chain transfer, and reaction with electrophiles, which can include O&lt;sub&gt;2&lt;/sub&gt; in some reactions. Like cationic polymerizations, anionic polymerizations can be made living by rigorous exclusion of terminating species. Again, this is difficult, due to the possibility of chain transfer termination. However, with the correct choice of conditions, it is doable, as suggested by the wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Living_anionic_polymerization&quot;&gt;living anionic polymerizations&lt;/a&gt;. Another way to accomplish some anionic polymerizations in a living fashion is &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00355a039&quot;&gt;group transfer polymerization&lt;/a&gt;, even though it does not  follow a strictly anionic mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0130650560&quot;&gt;&lt;em&gt;Contemporary Polymer Chemistry&lt;/em&gt;&lt;/a&gt; by Allcock, Lampe, and Mark, Prentice Hall 2003 covers this topic well, but the book is a little intense to read. In particular, the book derives the kinetic expressions for various polymerization types in the living and non-living cases. I expect other textbooks focusing of the chemistry of polymers might also cover this topic in one fashion or another.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-25T15:33:00.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="525" PostTypeId="2" ParentId="249" CreationDate="2012-05-25T17:04:05.977" Score="6" Body="&lt;p&gt;Mad Scientist provides some good advice. I have a different approach that either renders some of those actions unnecessary or cancels them out through signal averaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One big issue for quantification is the signal to noise ratio. Doing many of the things in Mad Scientist's answer will increase the signal to noise ratio, but the easiest two changes are increasing the concentration and taking more scans. If you can increase the concentration in your sample, do it. I often would increase the concentration to the point where I would be able to collect publication quality 1D &lt;sup&gt;1&lt;/sup&gt;H NMR spectra in a single scan. If you collect just one scan, then the problems resulting from poor phase averaging and from delay times that are shorter than the relaxation time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are unable to change the concentration, then increasing the number of scans increases the signal to noise by the square root of the number of scans. 4 scans provides twice the singal to noise as 1 scan. 16 scans provides twice the signal to noise as 4 scans, and so on. When you opt to take multiple scans, then you must be worried about the things that Mad Scientist mentions. A phasing problem that aggregates over 100+ scans will not help you. Having too short a delay time such that all of your spins are not relaxed only gets worse over a large number of scans. In a recent project, I had to quantitatively integrate the 1D &lt;sup&gt;1&lt;/sup&gt;H NMR spectrum of a reaction mixture, looking for minor products and unreacted species (less than 5% of the total sample). I ended up taking a 3200 scan proton spectrum with long delay times. I was not popular with my fellow grad students, but I did get the data I needed. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-25T17:04:05.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="526" PostTypeId="1" AcceptedAnswerId="528" CreationDate="2012-05-25T21:38:27.817" Score="19" ViewCount="26802" Body="&lt;p&gt;This afternoon, I was having some trouble with an HSQC (Heteronuclear Single Quantum Coherence Spectroscopy) spectrum I was trying to acquire, and the spectroscopist recommended I try HMQC (Heteronuclear Multiple Quantum Coherence) spectra, as they are more robust. I did so and got a very nice spectrum in about half the time an equivalent HSQC would have taken, and didn't have to phase it, to boot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the downside? Why doesn't everyone use HMQC spectra instead of HSQC? I know they are rather different on a quantum-mechanical level, but as an end-user what, if any, is the difference in the data the two techniques give me, what are the advantages and disadvantages of each, and in what situations might I prefer one over the other?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="402" LastEditDate="2017-08-30T01:47:00.573" LastActivityDate="2017-08-30T01:47:11.083" Title="What are the differences between the HMQC and HSQC experiments?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="527" PostTypeId="1" CreationDate="2012-05-25T23:55:26.870" Score="8" ViewCount="473" Body="&lt;p&gt;What is the timescale of disulfide bond formation in a given protein? What influences it? Finally, how does one actually measure it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, the reduction and oxidation of bonds happens on the order of 10&lt;sup&gt;-8&lt;/sup&gt;s. However, it appears that the timescale of disulfide bond formation in macromolecules scales with the size of the macromolecule.&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2012-05-26T11:44:33.310" Title="What is the timescale of disulfide bond formation?" Tags="&lt;biochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="528" PostTypeId="2" ParentId="526" CreationDate="2012-05-26T06:38:02.210" Score="16" Body="&lt;p&gt;You get a better resolution in the indirect dimension with an HSQC compared to an HMQC. The HSQC signal is a singlet while the HMQC is usually a multiplet as you see certain couplings on it. So generally you should prefer the HSQC over the HMQC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage of the HMQC is that it uses significantly fewer pulses than the HSQC. This means that errors in the pulses don't accumulate as much as in the HSQC, you'll lose less signal if your pulses are wrong than in the HSQC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had trouble setting up an HSQC, that could mean that the probe wasn't tuned and matched properly for your sample, or that your pulse lengths were wrong. The proton pulse should be determined on each sample individually; the $\ce{^13C}$ and $\ce{^15N}$ pulses should be set up correctly already. If you're not using a probe with automatic tuning and matching you should also do that manually before the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot more HSQC and HMQC variants that are used, e.g. Sofast-HMQC or TROSY sequences that are used for larger molecules like proteins. Another interesting variant you should consider is the multiplicity-edited HSQC, where $\ce{CH2}$ signals will have the opposite sign from $\ce{CH}$ or $\ce{CH3}$ signals.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="402" LastEditDate="2017-08-30T01:47:11.083" LastActivityDate="2017-08-30T01:47:11.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="529" PostTypeId="2" ParentId="519" CreationDate="2012-05-26T06:47:06.390" Score="12" Body="&lt;p&gt;There are electrophoretic methods where the molecules stop automatically like &lt;a href=&quot;http://en.wikipedia.org/wiki/Isoelectric_focusing&quot;&gt;Isoelectric Focusing&lt;/a&gt;, but in the common agarose gel electrophoresis of DNA you mention this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DNA is negatively charged, it will move along the electric field you set up until you stop applying any current. It will happily move out of your gel if you let it run for too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see the progress of your gel some dyes are usually added to the sample like &lt;a href=&quot;http://en.wikipedia.org/wiki/Bromophenol_blue&quot;&gt;bromophenol blue&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Xylene_cyanol&quot;&gt;xylene cyanol&lt;/a&gt;. They don't interact with the DNA, they just provide you a reference on how far your gel has run. You still need to have a rough idea how your samples run compared to those dyes if you want to use the full length of your gel.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2012-05-26T06:47:06.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="530" PostTypeId="2" ParentId="527" CreationDate="2012-05-26T11:44:33.310" Score="7" Body="&lt;p&gt;I agree that the actual redox process is very fast. The rate limiting issue with proteins and other macromolecues involves the reactive groups finding each other. Thus, from the time of translation to the time of disulfide formation, the peptide chains need to fold properly. Perhaps other subunits need to be assembled and attached. Once everything is in the right place, the disulfide bond should form quickly provided there is an appropriate oxidant present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could measure the rate of the entire process by attaching reporters to positions on the protein - say a fluorophore and a quencher. When the process is complete, the quencher and the fluorophore are close in space, and fluorescence stops. This approach will measure the overall rate of the entire process. I am not sure if you can separate out the different components, e.g. specifically the S-S bond forming. I can't point to a specific paper where this is strategy is used, since I am away from my usual computer that likes my journal subscriptions. Hopefully I will be able to update my answer.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-26T11:44:33.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="531" PostTypeId="1" CreationDate="2012-05-26T17:30:30.810" Score="12" ViewCount="103903" Body="&lt;p&gt;In my textbook, for calculating the percentage dissociation of $\ce{HF}$ for the given equation:&#xA;$$\ce{HF + H2O &amp;lt;=&amp;gt; H3O+ + F-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Concentrations&#xA;$$[\ce{HF}] = 0.08~\mathrm{M}, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$&#xA;Equilibrium concentrations&#xA;$$[\ce{HF}] = 0.08~\mathrm{M} - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a &lt;a href=&quot;https://chemistry.stackexchange.com/q/507/160&quot;&gt;similar question&lt;/a&gt; earlier too but it is really hard for me to get my head around this concept.&lt;/p&gt;&#xA;" OwnerUserId="277" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2018-03-21T16:29:47.053" Title="How do I calculate the degree of dissociation in equilibrium?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="532" PostTypeId="2" ParentId="531" CreationDate="2012-05-26T18:27:54.843" Score="15" Body="&lt;p&gt;Let us first define the terms needed here.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Degree of dissociation (DOD)&lt;/strong&gt;&lt;br&gt;&#xA;Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.&lt;br&gt;&#xA;$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Number of moles dissociated&lt;/strong&gt;&lt;br&gt;&#xA;It is defined as the product of the &lt;em&gt;initial concentration of the reactant&lt;/em&gt; and the &lt;em&gt;degree of dissociation&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now suppose you have a reaction like this&lt;br&gt;&#xA;$$\ce{A-&amp;gt;B + C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial state of A is &lt;strong&gt;always&lt;/strong&gt; the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).&#xA;The final state of A is &lt;strong&gt;always&lt;/strong&gt; defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is &lt;strong&gt;just&lt;/strong&gt; $\text{amount of substance of A dissociated)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing our equation again,&#xA;\begin{array}{lcccc}&#xA; &amp;amp; \ce{A &amp;amp;-&amp;gt; &amp;amp;B &amp;amp;+ &amp;amp;C}\\&#xA;\text{Initial amount of substance}&amp;amp; a &amp;amp;&amp;amp; 0 &amp;amp;&amp;amp; 0\\&#xA;\text{Final amount of substance}&amp;amp;   a - a\cdot(\mathrm{DOD}) &amp;amp;&amp;amp; a\cdot(\mathrm{DOD}) &amp;amp;&amp;amp; a\cdot(\mathrm{DOD})\\&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="4945" LastEditDate="2016-10-04T04:52:21.097" LastActivityDate="2016-10-04T04:52:21.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="533" PostTypeId="2" ParentId="475" CreationDate="2012-05-26T19:15:39.100" Score="9" Body="&lt;p&gt;This question has a simple, but non-obvious answer. DMSO and water are miscible. Mixtures, and in particular homogeneous ones like that formed by water and DMSO, usually have lower melting (or more relevant in this case, freezing) points than either substance does alone. This is the reason salt melts ice: the salt solution has a lower freezing point than pure water. The physics behind this &lt;a href=&quot;http://en.wikipedia.org/wiki/Melting_point_depression&quot;&gt;melting point depression&lt;/a&gt; are complex, but a simple enough approximation involves disruption of the intermolecular forces holding the solid crystal lattice together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water freezes at 0 °C, and DMSO freezes at 19 °C. Mixtures of the two have lower freezing points. More importantly, mixtures of the two likely form less disruptive crystals when frozen. Water is well know for having a solid state that is less dense than its liquid state. Ice crystals take up more room than the liquid water they formed from. Single water crystals can grow quite large, especially as smaller crystals merge. Additionally, water crystals tend to form this hexagonal fractal pattern (snowflakes) which have &quot;pointy bits&quot;. DMSO, even at the 10% by weight customarily used, must prevent such large disruptive crystals from forming, especially when the everything is cooled to -80 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DMSO is used for this purpose instead of other water-miscible liquids such as ethanol because it has very low toxicity. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="30" LastEditDate="2012-05-26T21:48:03.477" LastActivityDate="2012-05-26T21:48:03.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="534" PostTypeId="1" AcceptedAnswerId="536" CreationDate="2012-05-26T21:17:01.987" Score="17" ViewCount="251" Body="&lt;p&gt;This is a question that has been bugging me for over 15 years!&#xA;In high school while studying elements our teacher showed us a clear white rock or element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was simply flat on both sides but when placed on a sheet of paper would transfer the letters from the page beneath it to the top surface of the rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember being amazed by this rock, but could not remember what it was.&#xA;Now I am working on a creative project and would love to see if I can incorporate the material in my design.&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorDisplayName="user95" LastEditDate="2012-07-07T17:36:22.147" LastActivityDate="2012-07-07T17:36:22.147" Title="Rock that transfers image" Tags="&lt;materials&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="535" PostTypeId="2" ParentId="534" CreationDate="2012-05-26T21:46:27.083" Score="8" Body="&lt;p&gt;I'm not sure I understand what you want from the description above “transfer the letters”, but there is a typical high-school demonstration that I can think of that closely matches this description: that of &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Birefringence&quot; rel=&quot;nofollow noreferrer&quot;&gt;birefringence&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;   &lt;img src=&quot;https://i.stack.imgur.com/AduDv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Calcite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calcite&lt;/a&gt; (featured above) was the first material with which this effect was observed, in 1669, and still an easily obtainable material for demonstrating it.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-26T21:46:27.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="536" PostTypeId="2" ParentId="534" CreationDate="2012-05-27T01:42:12.037" Score="18" Body="&lt;p&gt;I used F'x answer as a starting point, and I found the mineral. It's called &lt;a href=&quot;http://en.wikipedia.org/wiki/Ulexite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ulexite&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fibers of ulexite act as optical fibers, transmitting light along their lengths by internal reflection. When a piece of ulexite is cut with flat polished faces perpendicular to the orientation of the fibers, a good-quality specimen will display an image of whatever surface is adjacent to its other side (as shown in the photograph).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTOQY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YhK27.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YhK27m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZyenH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://2.bp.blogspot.com/-x0HDu2DcC0g/TaomMtYoGSI/AAAAAAAAAFI/AyBVU-uQX7w/s1600/Ulexite1-50%2525.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="281" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.310" LastActivityDate="2012-05-27T21:48:33.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="537" PostTypeId="1" AcceptedAnswerId="538" CreationDate="2012-05-28T02:04:45.637" Score="38" ViewCount="7551" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/472/what-are-the-mechanisms-for-capture-of-ceasiii-and-ceasv-by-magnet&quot;&gt;this question&lt;/a&gt;, I'm wondering why arsenous acid is frequently denoted &lt;span class=&quot;math-container&quot;&gt;$\ce{H3AsO3}$&lt;/span&gt;, as opposed to &lt;span class=&quot;math-container&quot;&gt;$\ce{As(OH)3}$&lt;/span&gt;, which would appear to more accurately reflect its connectivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[edit]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like such a doofus, forgetting about sulfuric acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;), phosphoric acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{H3PO4}$&lt;/span&gt;), and boric acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{B(OH)3}$&lt;/span&gt;), etc. Great discussion, people!&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="23561" LastEditDate="2019-03-19T04:32:13.677" LastActivityDate="2019-03-19T04:32:13.677" Title="Why is arsenous acid denoted H3AsO3?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="538" PostTypeId="2" ParentId="537" CreationDate="2012-05-28T03:54:23.870" Score="43" Body="&lt;p&gt;&lt;sup&gt;All quoted stuff copied from &lt;a href=&quot;http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;noreferrer&quot;&gt;IUPAC Red Book&lt;/a&gt;, IR-4 unless otherwise specified&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not only $\ce{H3AsO3}$ . What about $\ce{H3BO3,H3PO4,H2SO4,HClO4}$? Or indeed, any other oxyacid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, &lt;strong&gt;we have multiple ways of writing inline formulae of covalent compounds&lt;/strong&gt;. One is the boring way (&lt;strong&gt;molecular formula&lt;/strong&gt;), and one is the fun way (&lt;strong&gt;structural formula&lt;/strong&gt;) where the formula reveals a bit of the structure. $\ce{As(OH)3}$ reveals the structure, $\ce{H3AsO3}$ doesn't.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For compounds consisting of discrete molecules, the molecular formula, as opposed to the&#xA;  empirical formula, may be used to indicate the actual composition of the molecules. For the&#xA;  order of citation of symbols in molecular formulae, see Section IR-4.4.&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;A structural formula gives partial or complete information about the way in which the atoms&#xA;  in a molecule are connected and arranged in space. In simple cases, a line formula that is just&#xA;  a sequence of atomic symbols gives structural information provided the reader knows that&#xA;  the formula represents the order of the atoms in the linear structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, what are the rules for writing the molecular formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, initially these rules were basically that we just follow electronegativity--the most electropositive element is listed first and so on. &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry_2005#Sequencing_elements_-_the_.22electronegativity.22_list&quot; rel=&quot;noreferrer&quot;&gt;Looks like they slightly changed it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, there are actually two ordering standards--one is the &lt;em&gt;extremely&lt;/em&gt; boring alphabetical one, which I've never seen used (except in emperical formulae), and one is via electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, instead of following the absolute electronegativities, the IUPAC has assigned a &quot;pseudo-electronegativity&quot; order (my terminology). The rules are as follows &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry_2005#Sequencing_elements_-_the_.22electronegativity.22_list&quot; rel=&quot;noreferrer&quot;&gt;&lt;sup&gt;(source)&lt;/sup&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;for two elements in different groups - then the element in the higher numbered group has higher &quot;electronegativity&quot;&lt;/li&gt;&#xA;  &lt;li&gt;for two elements within the same group the element with the lower the atomic number has the higher &quot;electronegativity&quot;&lt;/li&gt;&#xA;  &lt;li&gt;Hydrogen is fitted in to be less electronegative than polonium and more electronegative than nitrogen. Hence the formulae of water and ammonia can be written H2O and NH3 respectively.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leads to the following &quot;pseudo-electronegativity&quot; order of elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VCSzW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCSzWl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(click to enlarge)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation behind this must probably be the erratic and hard-to-quantify nature of &quot;real&quot; enectronegativity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, we can apply these rules in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the compound is a binary compound (only two elements, &lt;strong&gt;not&lt;/strong&gt; two atoms), then we just follow the pseudo-electronegativity thingy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1)&#xA;  is most often used in binary species.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the compound is a coordination compound, we order &lt;em&gt;central&lt;/em&gt; atoms by pseudo-electronegativity, and &lt;em&gt;ligands&lt;/em&gt; by alphabetical order, &lt;em&gt;as written&lt;/em&gt; (eg $\ce{(en)}$ before $\ce{(dmg)}$, but $\ce{C2H4(NH2)2}$ after $\ce{(dmg)}$, even though $\ce{(en)}$ and $\ce{C2H4(NH2)2}$ are the same).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The order of citation of central atoms is based on electronegativity as described in&#xA;  Section IR-4.4.2.1. Ligands are cited alphabetically (Section IR-4.4.2.2) according to the&#xA;  first symbol of the ligand formula or ligand abbreviation (see Section IR-4.4.4) as written.&#xA;  Where possible, the ligand formula should be written in such a way that a/the donor atom&#xA;  symbol is closest to the symbol of the central atom to which it is attached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally (barring annoying extra rules for isotopes and hydration), &lt;strong&gt;we have what are called &quot;generalized salts&quot;&lt;/strong&gt;. This is for compounds which can be &lt;em&gt;treated&lt;/em&gt; as being made up of ions. The ions are ordered by electronegativity, and the elements within each ion are ordered alphabetically{*}.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the formula of a compound containing three or more elements is not naturally assigned&#xA;  using the preceding two sections, the compound can be treated as a generalized salt. This&#xA;  term is taken to mean any compound in which it is possible to identify at least one&#xA;  constituent which is a positive ion or can be classified as electropositive or more&#xA;  electropositive than the other constituents, and at least one constituent which is a negative&#xA;  ion or can be classified as electronegative or more electronegative than the rest of the&#xA;  constitutents. The ordering principle is then:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(i) all electropositive constituents precede all electronegative constituents;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(ii) within each of the two groups of constituents, alphabetical order is used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is the rule we need to write the formula for $\ce{H3AsO3}$. First, we need to clear up what is meant by &quot;&lt;em&gt;treated&lt;/em&gt; as being made up of ions&quot;. Well, for this, as @Terry noted above, we look for the weaker bond. In $\ce{NaOH}$, we have relative bond strengths $\ce{Na...O-H}$, whereas $\ce{H3AsO3}$ is $\ce{As-(O...H)3}$. Thus, in $\ce{NaOH}$, the &quot;ion&quot; breakup is $\ce{Na+, OH-}$, while in $\ce{H3AsO3}$, the ions are $\ce{H+,AsO3-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, arsenous acid is broken up into ions, of which $\ce{H+}$ is written first. Within each ion, we order alphabetically, so we get $\ce{AsO3}$, not $\ce{O3As}$. Putting them together, we get $\ce{H3AsO3}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*Despite being the IUPAC rule, it seems that, within ions, most people just follow the electronegativity rules again. &lt;a href=&quot;http://en.wikipedia.org/wiki/Selenous_acid&quot; rel=&quot;noreferrer&quot;&gt;Selenous acid&lt;/a&gt; seems to be commonly named $\ce{H2SeO3}$, when it &lt;em&gt;should&lt;/em&gt; be $\ce{H2O3Se}$. But this seems to be a minor offence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Deviation from alphabetical order of constituents in the same class is allowed to&#xA;  emphasize similarities between compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22" LastEditorUserId="30" LastEditDate="2012-05-28T09:41:41.193" LastActivityDate="2012-05-28T09:41:41.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="539" PostTypeId="1" AcceptedAnswerId="541" CreationDate="2012-05-28T15:14:49.023" Score="27" ViewCount="1573" Body="&lt;p&gt;I am currently reading the CRC Handbook of Phosphorus-31 Nuclear Magnetic Resonance Data by John C. Tebby (CRC Press, 1991) and on several figures (ex. pages 9 to 14) there is a caption that reads along these lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{^{31}P}$ NMR chemical shifts of three coordinate (λ3 σ3) phosphorus compounds. (Tables B to E.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does the (λ3 σ3) signify? I've not seen that notation before, and it is on the captions of some of the charts I need.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="16683" LastEditDate="2017-08-11T01:26:08.747" LastActivityDate="2017-08-11T01:51:46.483" Title="What does this lambda–sigma notation mean?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="540" PostTypeId="1" AcceptedAnswerId="571" CreationDate="2012-05-28T16:20:15.853" Score="8" ViewCount="9498" Body="&lt;p&gt;I'm attempting to oxidize a substance with potassium permanganate and I was wondering what organic solvents I have at my disposal. The compound is soluble in chloroform and ethanol, but obviously those would not do well with a powerful oxidizing agent. The phase-transfer catalyst I'm going to employ is tetra-n-butylammonium bromide.&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2019-06-20T09:07:40.757" Title="What organic solvents are suitable to use with potassium permanganate?" Tags="&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="541" PostTypeId="2" ParentId="539" CreationDate="2012-05-28T17:07:55.137" Score="24" Body="&lt;p&gt;The notation refers to the valency (number of valence electrons involved in bonds, i.e., either 3 or 5) and coordination number (number of substituents attached) in &lt;a href=&quot;http://en.wikipedia.org/wiki/Organophosphorus_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;organophosphorus&lt;/a&gt; compounds. For example, it could refer to the tautomeric forms of phosphonate esters:&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iq8yk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iq8yk.png&quot; alt=&quot;Tautomers of phosphonate esters&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the λ&lt;sup&gt;3&lt;/sup&gt;σ&lt;sup&gt;3&lt;/sup&gt; refers to a P with 3 bonding valence electrons (the lone pair doesn't count) and 3 substituents attached while λ&lt;sup&gt;5&lt;/sup&gt;σ&lt;sup&gt;4&lt;/sup&gt; refers to a P with all five of its valence electrons involved in bonds and 4 substituents attached. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reference&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Kraszewski, A.; Stawinski, J. H-Phosphonates: Versatile synthetic precursors to biologically active phosphorus compounds. &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2009,&lt;/strong&gt; &lt;em&gt;79&lt;/em&gt; (12), 2217–2227. &lt;a href=&quot;https://doi.org/10.1351/pac200779122217&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1351/pac200779122217&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="133" LastEditorUserId="16683" LastEditDate="2017-08-11T01:34:16.197" LastActivityDate="2017-08-11T01:34:16.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="542" PostTypeId="2" ParentId="464" CreationDate="2012-05-29T03:32:30.363" Score="9" Body="&lt;p&gt;Azane all by itself has an inversion barrier of about $6\ \mathrm{kcal/mol}$, which is  low compared to phosphine's approximately $30\ \mathrm{kcal/mol}$ barrier. Pyramidal nitrogen inversion requires a planar transition state. One can use a classical theory to model the thermal rate and we could write $$k \propto \mathrm e^{-E_\mathrm a/(RT)} $$ and using models like these we would predict that ammonia at $300\ \mathrm K$ would have a rate on the order of $10^8\ \mathrm{s^{-1}}$, instead what is observed is something on the order of $10^{10}\ \mathrm{s^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the potential energy surface has two minima equal in energy and a higher energy TS, two routes are possible: a thermal process going over the maxima or quantum-mechanical tunneling. Determining a tunneling frequency requires writing down vibrational wave functions, finding their linear combination and overlap. If we were to do all this for a particular vibration, we'd see that the tunneling frequency decreases exponentially with a dependence on increasing $\mu$, thickness of the barrier/shape, and the barrier height. As an example, $\ce{ND3}$ has approximately an order of magnitude slower rate of inversion than $\ce{NH3}$ at a sufficiently low temperature – this has been attributed to a decrease in its tunneling frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the difference in the energy barriers and corresponding rates depending on the atomic position in the periodic table? Perhaps the easiest/quickest explanation rests on considering what the TS requires for interconversion. Ideally the classical conversion requires passing through a trigonal planar structure with bond angles on the order of 120 degrees. If you consider the series of azane, phosphane, arsane then you'll see bond angles of 108, 94, 92 degrees respectively. In the absence of other effects, the barrier goes from smaller to larger. Indeed, asane-like molecules were the first of this series resolved, followed by phosphane-like. To the best of my knowledge the fastest have yet to be resolved, because rates on the order of $10^{-5}\ \mathrm{s^{-1}}$ are required at RT for a decent shot at isolating them. Naturally, this isn't the whole story, as I've previously alluded to above. Let me know if you would like to know more.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user7951" LastEditDate="2017-02-02T01:05:25.553" LastActivityDate="2017-02-02T01:05:25.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="543" PostTypeId="1" AcceptedAnswerId="545" CreationDate="2012-05-29T04:56:49.627" Score="23" ViewCount="5479" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-newtonian_fluid#Oobleck&quot;&gt;Oobleck&lt;/a&gt; is basically a mixture of cornstarch and water. It has the interesting property of going hard when pressure is applied. For example, it feels like a liquid when you swirl your fingers through it, but it becomes hard for a second if you punch it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common explanation for this is that the starch-water mixture is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Suspension_%28chemistry%29&quot;&gt;suspension&lt;/a&gt;, and the  starch particles &quot;lock&quot; when pressure is applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is the mechanism of this &quot;locking&quot;? &lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="84" LastEditDate="2012-05-29T04:59:52.697" LastActivityDate="2015-01-11T20:25:33.007" Title="How does oobleck work?" Tags="&lt;mixtures&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="544" PostTypeId="1" AcceptedAnswerId="548" CreationDate="2012-05-29T05:23:47.380" Score="8" ViewCount="4373" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Copper%28II%29_sulfate&quot;&gt;Copper sulphate&lt;/a&gt;, in its hydrated form, is crystalline, whereas the anhydrous form is amorphous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gypsum&quot;&gt;Gypsum&lt;/a&gt; has a similar story-- on heating the crystalline dihydrate we get an amorphous hemihydrate. (Gypsum in fact has two different crystalline forms of the dihydrate, and has an anhydrous form as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never really grasped how the water makes it crystalline. I suspect it has to do with the relative sizes of the compounds being re-compensated by ligands, but I'm not sure. How does the water affect/create the crystalline properties?&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2014-11-15T08:11:13.823" Title="How do copper sulphate and the like change crystal structure on adding water of hydration?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="545" PostTypeId="2" ParentId="543" CreationDate="2012-05-29T11:40:47.633" Score="13" Body="&lt;p&gt;Wikipedia already offers a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dilatant#Theories_behind_shear_thickening_behavior&quot;&gt;decent summary&lt;/a&gt; of two theories of shear thickening behavior - namely the generation of transient 'defects' in a high-density suspension that either resist further growth or are incompressible in the direction of shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A potential macroscale analogue of shear thickening is the dramatic increase in viscosity of large particle-air mixtures upon pneumatic removal of the interparticle fluid, as is exploited in &lt;a href=&quot;http://www.youtube.com/embed/SbqHERKdlK8&quot;&gt;jamming skin enabled locomotion&lt;/a&gt;. (paper &lt;a href=&quot;http://faculty2.ucmerced.edu/ebrown22/SMRBJ09_IEEE.pdf&quot;&gt;here&lt;/a&gt;) This should be roughly comparable to the transient decrease in particle spacing in a shear thickening fluid that is being compressed beyond its relaxation rate.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-29T11:40:47.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="546" PostTypeId="1" CreationDate="2012-05-29T15:20:02.703" Score="7" ViewCount="4222" Body="&lt;p&gt;Well, I have seen many comical acts where people balance a spoon (or a fork) on different parts of the face like on the tip of the nose, forehead etc.&lt;br&gt;&#xA;The do this by blowing moist air (see below) and immediately putting it on the face  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yEYRf.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i wanted to know is what is the force that temporarily binds the skin to the fork. I doubt it's friction or Van Der Walls force.&lt;br&gt;&#xA;Can someone please clarify this for me? Is it some sort of chemical bond?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some addition&lt;/strong&gt;&lt;br&gt;&#xA;Also another method of doing this was to rub the concave part of the spoon on the tip of the nose. This &lt;a href=&quot;http://www.wikihow.com/Hang-a-Spoon-from-Your-Nose&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; states how to do it properly. In the 2&lt;sup&gt;nd&lt;/sup&gt; point it states  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eventually you will feel a slight tendency for the spoon to resist on the downstroke.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this resisting tendency?&lt;/p&gt;&#xA;" OwnerUserId="124" LastEditorUserId="22" LastEditDate="2012-05-29T15:23:31.087" LastActivityDate="2017-08-17T05:03:09.863" Title="Hanging a spoon on the face" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="547" PostTypeId="2" ParentId="544" CreationDate="2012-05-29T16:53:37.370" Score="7" Body="&lt;p&gt;[Manishearth, I'm not sure I'm even answering the right question below. After reading your question more carefully, I think you were more interested in how water molecules help crystals form, rather than whether or not the anhydrous forms can form crystals also (versus being strictly amorphous). So, oops. Maybe some one else has a good answer on the water molecules part?]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the answer is an issue of process rather than of fundamentals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if you could find a non-aqueous medium that readily dissolves the anhydrous form of some salt, my very strong suspicion is that you could grow excellent clear crystals of it that would look almost nothing like the hydrous crystalline forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem for metallic-anion salts in particular is that finding a liquid that dissolves the salt without simultaneously forming strong coordination complexes with it is going to be tricky, since I'm guessing it's usually the coordination effect that help ionize the units of the material to make it soluble!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a wild guess: Someone, somewhere, has likely done research on how to grow crystals of substances such as anhydrous $\ce{CaSO4}$ (hmm, why does this phrase &quot;dead burnt plaster&quot; come to mind? :) or anhydrous $\ce{CuSO4}$ by using low-temperature molten solutions of... something or other? Not sure what!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what a delightful crystal that would be to have! It would be way out of the ordinary despite being so very... ordinary?... in one sense in its composition.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Apparently, &lt;a href=&quot;http://scripts.iucr.org/cgi-bin/paper?S0365110X58000955&quot;&gt;clear crystals of anhydrous $\ce{CuSO4}$&lt;/a&gt; do indeed exist, and are easier to create than I thought. You just add concentrated sulfuric acid [a] to ordinary copper sulphate pentahydrate solution and let it evaporate slowly. Since $\ce{H2SO4}$ is one of the most aggressive drying agents in existence, my guess is that the $\ce{CuSO4}$ just can't compete for the remaining water molecules and is forced to crystallize without hydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[a] Please, never ever attempt to use concentrated $\ce{H2SO4}$anywhere except in a real lab with protections and procedures fully in place. This is a compound that wants water so badly that it willy synthesize water right out of the hydrogen and oxygen bonded into proteins and carbohydrates, leaving only charred carbon. Ironically, the actual acidity of concentrated (vs hydrated) $\ce{H2SO4}$ is quite moderate, comparable to that of vinegar.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addenda:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Another common-exotic crystal may be lithium fluoride, $\ce{LiF}$, although the accuracy of that seems to depend on who you read. Unique how, you say? Even though lithium was an early result of the Big Bang, lithium and fluorine are both consumed rapidly by large hot stars, especially fluorine. Their existence on earth thus appears to depend on processes such as proton irradiation and possibly even intensive neutrino irradiation tht only occur &lt;em&gt;during&lt;/em&gt; the explosion of a supernova star, rather than being inherited from before the cataclysmic event. Pretty much all of the elements (except $\ce{H}$) here on Earth are star-forged of course, but the elements in $\ce{LiF}$ crystals thus may be remnants of the supernova event itself. That would... cool!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- And speaking of unique, wearable crystals: I've always wondered if there's a market out there for diamond crystals synthesized using carbon that was captured as carbon dioxide during the cremation of a loved one. If there is, feel free -- that is definitely not my bailiwick!&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="83" LastEditDate="2012-05-29T22:47:51.613" LastActivityDate="2012-05-29T22:47:51.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="548" PostTypeId="2" ParentId="544" CreationDate="2012-05-30T00:12:38.750" Score="6" Body="&lt;p&gt;I learned something in trying to answer this question since I've never quite understood what the water was doing in hydrated crystals either. The waters of hydration help to stabilize the crystalline form by holding the crystal together with hydrogen bonding. Water incorporated into the crystal allows the negative dipole of the water to partially diffuse the positive charge and the positive dipole of the water to partially diffuse the negative charge, and would help to minimize repulsive forces between ions of like charge &quot;close&quot; in the crystal. Speaking of $\ce{Cu(SO4)2 .5H2O}$...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Copper is surrounded by six oxygen atoms, provided by two different sulfate groups and four molecules of water. A fifth water resides elsewhere in the framework but does not bind directly to copper.....Water of crystallization is stabilized by electrostatic attractions, consequently hydrates are common for salts that contain +2 and +3 cations as well as -2 anions. &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_of_crystallization&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Water_of_crystallization&quot; rel=&quot;noreferrer&quot;&gt;Ref&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more &lt;a href=&quot;http://www.minsocam.org/ammin/AM55/AM55_1534.pdf&quot; rel=&quot;noreferrer&quot;&gt;interesting reference&lt;/a&gt; that takes some looking at to get the sense of is given below. $\ce{Fe2(SO4)3.xH2O}$, with iron with a 3+ charge, forms various hydrates with x = 6, 7, 9, 10. (Disclaimer....actually this is the crystal structure for coquimbite with has an occasional Al for Fe substitution thrown in, but the principle is the same)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJiax.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture, the sulfate ions are shown as tetrahedra and the $\ce{Fe^3+}$ ions are shown as octahedra (presumably because $\ce{Fe^3+}$ ions have 6-coordination sites). The crystal is formed from &quot;clusters of 6 sulfate tetrahedra and 3 iron octahedra which share only corners&quot; and there are two clusters per unit cell. Each of the clusters is linked to the others through hydrogen bond provided by the incorporated water, and parallel chains are held together only through hydrogen bonds. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The individual chain segments are linked through hydrogen bonds only. The geometrical arrangement of the chains gives rise to &quot;channels,&quot; ... which are occupied by water molecules linked to the chains by hydrogen bonds. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, you can see how removing the waters of hydration would cause the crystalline structure to break down and result in an amorphous substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information from the paper: Coquimbite has $\ce{.9H2O}$. The crystal structure shows that, of the 3 Fe found in a cluster, one Fe is associated with 4 oxygens from sulfate groups, the second Fe is surrounded by 3 oxygens from sulfate groups and 3 water molecules, and the third Fe (or Al) is surrounded by 6 water molecules, so the octahedral representation does make sense.  &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-05-31T05:17:34.437" LastActivityDate="2012-05-31T05:17:34.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="549" PostTypeId="1" AcceptedAnswerId="720" CreationDate="2012-05-30T05:12:52.573" Score="19" ViewCount="4176" Body="&lt;p&gt;I am confused about the justification for the corrections to the ideal gas law in the Van der Waals equation: $$p=\frac{nRT}{V-nb}-a\left(\frac{n}{V}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the equation attempts to correct for attractive and repulsive forces between molecules, and that at high volumes the corrections are negligible, at intermediate volumes the $V-nb$ correction dominates (since attractions are dominant and this shrinks the volume, as should occur for attractions), and at low volumes the $a(n/V)^2$ correction dominates, since repulsions are dominant.  However, the 7th edition of Atkins' and de Paula's &lt;em&gt;Physical Chemistry&lt;/em&gt; justifies part of this equation as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The repulsive interactions are taken into account by supposing that they cause the molecules to behave as small but impenetrable spheres.  The nonzero volume of the molecules implies that instead of moving in a volume $V$ they are restricted to a smaller volume $V-nb$.  This argument suggests that the perfect gas law $p=nRT/V$ should be replaced by $$p=\frac{nRT}{V-nb}$$ when repulsions are significant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am confused by this argument-- what exactly is the definition of 'volume' here?  They seem to be saying that it is the empty space around the gas molecules, but it seems to me that volume should be defined as the space 'taken up' by the gas.  Why should the space taken up by the particles themselves be subtracted?  This isn't done for solids or liquids as far as I know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These thoughts have also made me realize that I'm not quite sure what it means for a gas to 'take up' space.  Does anyone have a rigorous definition of gas volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The exchange with Chris in comments has raised further questions.  It now seems to me that the $V-nb$ correction actually accounts for repulsions rather than attractions.  I think I was incorrect in thinking that the $V-nb$ correction dominated at intermediate volumes and $a(n/V)^2$ at low volumes.  If $V-nb$ is for repulsions, then it should dominate at low volumes, but I can't tell from the equation which correction will actually be dominant.  Also I am now wondering whether it even makes sense to connect one correction with repulsions and one with attractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Follow-up questions for F'x:&lt;br&gt;&#xA;- I thought that 'density' meant mass/volume.  Is the use of it to represent the inverse of molar volume (as you have used it) common?&lt;br&gt;&#xA;- Where is the phase-transition in the red van der Waals curve?&lt;br&gt;&#xA;- I am still a little unclear on gas volume.  Are the 'volume available to the gas' and 'volume of the gas' the same?  You say it's the same as the shape of the container, but aren't we subtracting the volume of the actual gas particles from that?  The $V$ term represents the volume of the gas, correct?  And $V-nb$ represents the 'volume available to the gas'.&lt;/p&gt;&#xA;" OwnerUserId="289" LastEditorUserId="16683" LastEditDate="2016-09-30T09:29:47.503" LastActivityDate="2016-09-30T09:29:47.503" Title="What is a rigorous definition of gas volume, and how is the Van der Waals equation derived?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;&lt;equation-of-state&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="550" PostTypeId="1" AcceptedAnswerId="551" CreationDate="2012-05-30T16:04:49.573" Score="7" ViewCount="458" Body="&lt;p&gt;2D spectrum take up a large amount of hard disk space. At my last lab it was standard practice to delete the imaginary portions of the spectrum with &lt;code&gt;xfb n&lt;/code&gt; (We used TopSpin) and if needed later, to regenerate it from the Real part of the data with a Hilbert transform (Related, the TopSpin Commands and References, ~pg P-233 in version 2.1.2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question: If the raw data is still stored in the &lt;code&gt;ser&lt;/code&gt; file (TopSpin Commands and References v2.1.2, pg P-8) why can't you just reprocess the raw data? Is all the data not there? (I suspect this somewhat, as those files are much, much smaller then the processed files, and TopSpin processes 2D spectra as it acquires them, unlike 1D spectra)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second (highly related) question: Is there a downside to regenerating the imaginary data from the real data? Is it a perfect replication of the original data, or is it an approximation? Why do we ever bother to store this data if we can recreate it at any time? &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="22" LastEditDate="2012-05-31T00:30:51.397" LastActivityDate="2016-09-08T08:57:26.810" Title="Regenerating the imaginary data of a 2D NMR spectrum with a hilbert transform" Tags="&lt;computational-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="551" PostTypeId="2" ParentId="550" CreationDate="2012-05-30T18:54:41.403" Score="7" Body="&lt;p&gt;If you have the raw data (the FIDs in the &lt;code&gt;ser&lt;/code&gt; file) still available, there is no reason to use a Hilbert transformation, you can just process with &lt;code&gt;xfb&lt;/code&gt; instead of &lt;code&gt;xfb n&lt;/code&gt; again. &lt;code&gt;xfb n&lt;/code&gt; also doesn't really discard the imaginary part, it processes the spectrum from the raw data again, but doesn't save the imaginary processed data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;ser&lt;/code&gt; file contains all the recorded FIDs, that data is sufficient (together with the parameter files) to recreate the full spectrum. That your Topspin processes 2D spectra automatically is just some specific configuration, it doesn't do that by default.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand it, if you zero-fill your data at least once, all the information content of the imaginary part is then present in the real part of the processed spectrum. So I'd guess that under those conditions, the Hilbert transformation shouldn't have any downside, but I'm not totally sure about that.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2012-05-30T18:54:41.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="553" PostTypeId="2" ParentId="549" CreationDate="2012-05-31T08:33:10.307" Score="12" Body="&lt;p&gt;&lt;em&gt;I edited the first van der Waals equations in your question, because it was incorrect.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, the volume available to the gas is pretty much was you think: it's the space left for it to occupy, i.e. the volume delimited by the container. If you think of a gas tank, it's the interior volume of the tank. For systems of macroscopic dimensions, there is no real ambiguity as to its definition, which can be purely geometrical. For nanoscale systems, such as accessible volume in porous media, the definition is much harder: it is typically done by constructing a &lt;a href=&quot;http://goldbook.iupac.org/G02635.html&quot;&gt;Gibbs dividing surface&lt;/a&gt; between the &lt;a href=&quot;http://old.iupac.org/reports/2001/colloid_2001/manual_of_s_and_t/node18.html&quot;&gt;two phases considered&lt;/a&gt; (here, a solid and a gas).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second, importance of both terms in a real-life example. Let's write the van der Waals equation in terms of density, rather than volume and quantity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(P+a\rho^2\right)\left(\frac{1}{\rho}-b\right)=RT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a practical example of a gas with a large quadrupolar moment, CO&lt;sub&gt;2&lt;/sub&gt;. We have $a$ and $b$ parameters for it from &lt;a href=&quot;http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Thermal/vdWaalEquatOfState.html&quot;&gt;this source&lt;/a&gt;, and we can plot $\rho(P)$ at $T=300$ K:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xHKqC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(black: ideal gas; red: van der Waals gas; green: only $b$; blue: only $a$). We see that excluded-volume correction by itself is not much of a gain, but it's actually crucial once you include the attraction, because otherwise the phase transition (the cusp in the blue curve) occurs at much too low pressure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, regarding the derivation of the van der Waals equation, this is typically a topic of statistical thermodynamics and there are many such demonstrations. My favorite is the classical one by &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed025p347&quot;&gt;Hill&lt;/a&gt;, but it may not use the notations in modern textbooks. I'm also partial to that in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1891389157&quot;&gt;McQuarrie’s book&lt;/a&gt;. All statistical physics derivation introduce the excluded volume in a very natural fashion, but they show more variation on the way they arrive at the $a$ term (so of them find the exact analytical form of the equation, while in some derivations you find a series expansion whose first term is the van der Waals equation).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-05-31T08:33:10.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="554" PostTypeId="1" AcceptedAnswerId="557" CreationDate="2012-05-31T08:45:33.150" Score="22" ViewCount="9084" Body="&lt;p&gt;Most of the existing software I know (like VMD or Pymol) say that they can be used to create PDF files with 3D representations of molecules: actual 3D models, which you can rotate inside the Adobe Acrobat viewer. However, they all require the use of “Adobe Acrobat Pro 9 Extended” for its 3D capture feature. From what &lt;a href=&quot;http://www.adobe.com/fr/products/acrobatpro/faq.html#3d-geospatial&quot;&gt;I can read&lt;/a&gt;, this feature does not exist any more in Acrobat X and was transferred to a separate product, &lt;a href=&quot;http://www.tetra4d.com/&quot;&gt;Tetra4D&lt;/a&gt;, which doesn't work on Mac OS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I produce PDF files including 3D molecular models using free (or better, open source) tools?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5017" LastEditDate="2014-09-29T14:35:47.223" LastActivityDate="2014-09-29T14:35:47.223" Title="What open-source software can produce 3D PDF representation of molecules?" Tags="&lt;software&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="555" PostTypeId="1" AcceptedAnswerId="556" CreationDate="2012-05-31T08:49:39.867" Score="23" ViewCount="39778" Body="&lt;p&gt;It's a common cooking advice: if you want a somewhat softer caramel, &lt;a href=&quot;http://allrecipes.com/howto/sugar-cooking/&quot;&gt;add some lemon juice to the sugar&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To help prevent the caramel from crystallizing, you can add an acid to the sugar before you begin: add about half a tablespoon of lemon juice to each cup of sugar and mix it with your hands&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how does it work?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="275" LastEditDate="2012-06-28T02:40:01.900" LastActivityDate="2014-11-12T04:47:23.663" Title="How does adding lemon juice to sugar make better caramel?" Tags="&lt;everyday-chemistry&gt;&lt;acid-base&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="556" PostTypeId="2" ParentId="555" CreationDate="2012-05-31T11:15:19.447" Score="19" Body="&lt;p&gt;The citric acid in lemon juice hydrolyzes the sucrose molecules into glucose and fructose as you heat the sugar. The water in the lemon juice is also part of this process. This mixture of monosaccharides is much less crystalline in the same way that most mixtures are less crystalline than pure substances. Wikipedia has a good article on this process under the title &lt;a href=&quot;http://en.wikipedia.org/wiki/Invert_sugar&quot;&gt;&quot;Inverted sugar syrup&quot;&lt;/a&gt;. The syrup is &quot;inverted&quot; in the sense that the sign of its optical rotation has changed because sucrose has a large positive optical rotation while fructose has a large negative optical rotation. Glucose has a positive optical rotation, but of a lower magnitude than sucrose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is:&#xA;$$\ce{C12H22O11}(sucrose)+\ce{H2O -&amp;gt; C6H12O6}(glucose)+\ce{C6H12O6}(fructose)$$&#xA;The reaction can occur to a small degree by simply heating sugar with water in absence of acid catalysis. I assume the acid is necessary to get good conversion before you burn your sugar or vaporize all of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Honey&quot;&gt;Honey&lt;/a&gt; is a naturally occuring inverted sugar syrup of this type. Honey is resistant to crystallization because it is a mixture of gluclose and fructose. Honey is also slightly acidic.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-05-31T11:15:19.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="557" PostTypeId="2" ParentId="554" CreationDate="2012-05-31T15:03:11.743" Score="21" Body="&lt;h2&gt;Disclaimer&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This is &lt;em&gt;not&lt;/em&gt; for the faint-hearted.&lt;/li&gt;&#xA;&lt;li&gt;If you are not interested in using &lt;a href=&quot;http://en.wikipedia.org/wiki/LaTeX&quot; rel=&quot;noreferrer&quot;&gt;LaTeX&lt;/a&gt;, stop reading now, as this uses the &lt;code&gt;media15&lt;/code&gt; or &lt;code&gt;movie9&lt;/code&gt; packages to embed U3D in a pdfLaTeX generated PDF. This answer assumes basic familiarity with LaTeX. I have tested this with pdfLaTeX and it should work fine in XeLaTeX.&lt;/li&gt;&#xA;&lt;li&gt;One of the tools  I have used (DAZ Studio) is apparently 'free for a limited time'. It is also not available on Linux. FOSS replacement suggestions are welcome.&lt;/li&gt;&#xA;&lt;li&gt;I'm of the personal opinion that 3D in PDF is not as good as it sounds.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;What doesn't work&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;noreferrer&quot;&gt;VMD&lt;/a&gt; offers the ability to output as OBJ, which can then be processed with FOSS &lt;a href=&quot;http://meshlab.sourceforge.net/&quot; rel=&quot;noreferrer&quot;&gt;MeshLab&lt;/a&gt; into a U3D file that can be embedded into a PDF. This is sub-optimal for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no way to tell MeshLab that the VMD-generated mesh is supposed to be Gouraud-shaded. This means that the hard facets of the mesh will be glaringly obvious in the pdf, even if you use an unwieldy degree of subdivision.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Colours are not preserved. Your model will appear in the PDF with a uniform grey texture.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As such, MeshLab is &lt;em&gt;not suitable at this stage&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Preparing a model&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;VMD offers a number of decent export options to get our geometry out of .PDB or what have you and into a mesh format. I opted to use the VRML format to output a mesh of a ribbon representation of PDB entry &lt;a href=&quot;http://www.rcsb.org/pdb/explore/explore.do?structureId=4DM9&quot; rel=&quot;noreferrer&quot;&gt;4DM9&lt;/a&gt;. I then imported this into &lt;a href=&quot;http://www.blender.org/&quot; rel=&quot;noreferrer&quot;&gt;Blender&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/euW4y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blender has no reliable U3D export capabilities, however I can use it to assign Gouraud shading to all of the shapes and assign colours to them. I then use the blender export function to export the scene as a Wavefront .OBJ file, with attendant .MTL file.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Processing with DAZ Studio&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am indebted to the author of &lt;a href=&quot;http://artandlogic.blogspot.com.au/2008/08/blender-to-acrobat-3d.html&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; (John Nyquist), which indicated that &lt;a href=&quot;http://www.daz3d.com/&quot; rel=&quot;noreferrer&quot;&gt;DAZ Studio&lt;/a&gt; can convert Wavefront to U3D just fine, with preservation of textures. DAZ is a very specific program that seems to be heavily geared towards posing, clothing and rendering humanoid figures in the spirit of programs like Poser. However all we really need is the U3D export function and we can read in the structure just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, this is how I see myself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XyCgZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay in all seriousness, don't put a human figure in or on your molecule without good reason. Export as U3D and then we can embed in pdfLaTeX.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;pdfLaTeX and movie9&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As a proof of concept, I just followed the minimal example given in the &lt;a href=&quot;http://ctan.unsw.edu.au/macros/latex/contrib/media9/doc/media9.pdf&quot; rel=&quot;noreferrer&quot;&gt;media9 manual&lt;/a&gt; for getting a 3D model into a pdfLaTeX-generated PDF. Here it is, being viewed in Adobe Reader X (I know, I know!). Bear in mind that very few readers can actually display 3D PDF, which is one of the reasons why I feel that it is not such a good idea for chemical visualisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2T9l8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It starts off zoomed way too far in because I did not set up any options for the import (however you can find out all about that in the media9 or movie15 package manuals) and is a bit laggy on account of having over 1 million faces (I went way overboard with subdivision), however the important thing is that &lt;em&gt;it works&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Improvements welcome!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Obviously, this is a pretty &lt;a href=&quot;http://en.wikipedia.org/wiki/Rube_Goldberg_machine&quot; rel=&quot;noreferrer&quot;&gt;Rube Goldbergesque&lt;/a&gt; route to getting a molecule into a PDF. There probably exist easier ways which should definitely be suggested.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-31T15:03:11.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="558" PostTypeId="1" AcceptedAnswerId="567" CreationDate="2012-05-31T15:25:32.237" Score="50" ViewCount="9662" Body="&lt;p&gt;Especially when drinking water after the fact, mint can give a sharp cold sensation inside one's mouth.  What process causes the sensation to occur?&lt;/p&gt;&#xA;" OwnerUserId="170" LastActivityDate="2015-10-19T14:04:43.167" Title="Why does menthol (e.g. peppermint) feel cool to the tongue?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="559" PostTypeId="1" AcceptedAnswerId="564" CreationDate="2012-05-31T15:37:52.503" Score="7" ViewCount="3645" Body="&lt;p&gt;Here is my interpretation when asked to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By drawing arrows in the appropriate boxes, complete the outer electron structures for Cu and Cu&lt;sup&gt;2+&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTiMK.jpg&quot; alt=&quot;My electron box plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had no problem in drawing out the electron structure, but am a little confused by the orientation of my arrows.  I understand that each normal box has to have arrows in opposing directions to represent the opposite spins on the electrons, but the mark scheme defines which way round they want them to go:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQdGu.jpg&quot; alt=&quot;Markscheme extract&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;http://pdf.ocr.org.uk/download/ms_11/ocr_65225_ms_11_gce_uf333_jun.pdf?&quot; rel=&quot;noreferrer&quot;&gt;OCR GCE Chemistry B F334 June 2011 Markscheme pg. 11&lt;/a&gt;&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily I would have said that whether you go up or down first (or even consistently) is irrelevant and arbitrary, but the comment in the guidance &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;ALLOW &lt;em&gt;single&lt;/em&gt;&lt;/strong&gt; arrow in either direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;suggests to me that with the paired arrows, the order of directions matter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there any chemical reason for putting the downwards spin arrow first?  I would happily accept by convention as a reason but I just want to be sure that I'm not missing anything chemically relevant.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2012-06-01T10:56:14.693" Title="Is the order of orientation of electron box diagrams meaningful or arbitrary?" Tags="&lt;electronic-configuration&gt;&lt;structural-formula&gt;&lt;spin&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="560" PostTypeId="2" ParentId="554" CreationDate="2012-05-31T16:02:46.200" Score="10" Body="&lt;p&gt;If you are ready to invest a lot of time to learn &lt;a href=&quot;http://asymptote.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asymptote&lt;/a&gt; you can use it to create 3D PDF documents from &lt;code&gt;pdb&lt;/code&gt; files with it. &lt;a href=&quot;http://asymptote.sourceforge.net/gallery/100d.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is an example. It has the advantage that Asymptote has a good LaTeX integration and you can use the code directly in a LaTeX document. But it would be very messy to get the appearance right and I'd say it will only work for Ball-and-Stick models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;I found some other nice examples of crystal unit cells being depicted with Asymptote, which can serve as some kind of proof-of-concept on using Asymptote to depict (simple) molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://tex.stackexchange.com/questions/121169/labels-and-angles-in-crystal-structure-using-asymptote?lq=1&quot;&gt;Labels and angles in crystal structure using Asymptote&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2f0GK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://tex.stackexchange.com/questions/97271/drawing-3d-crystal-lattice-with-tikz-pstricks?lq=1&quot;&gt;Drawing 3D crystal lattice with tikz/pstricks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XIEo7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are from TeX.StackExchange, so if you want to try asymptote you might find some help there, when you run into problems.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:34:31.770" LastActivityDate="2013-07-08T21:14:47.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="561" PostTypeId="2" ParentId="559" CreationDate="2012-05-31T16:15:02.607" Score="5" Body="&lt;p&gt;I just cracked open five textbooks (Silberberg, Housecroft and Sharpe, West, McMurray, Simons) and they all adopt the $\uparrow\downarrow$ convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only conceivable reason to prefer any order (namely, alphabetic ordering of the spin orbitals $\alpha$ and $\beta$) would actually suggest &lt;em&gt;your&lt;/em&gt; chosen order. There is no chemically meaningful distinction between $\uparrow\downarrow$ and $\downarrow\uparrow$. I would &lt;em&gt;love&lt;/em&gt; to know the marker's rationale here.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-05-31T16:15:02.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="562" PostTypeId="2" ParentId="559" CreationDate="2012-05-31T17:31:04.047" Score="3" Body="&lt;p&gt;This is intended as an extension of RichardTerret's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally have used/been taught uparrow first (not _explicitly  taught, just that the teachers did it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I wanted to address the &quot;Allow single arrow in either direction&quot; issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you have misinterpreted this. Basically, it's allowing some deviation from Hund's rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, Hund's rule forbids stuff like $[\uparrow\downarrow][\uparrow][\downarrow] $, since the spins cancel out (Hund's rule valls for maximum total spin). The second two ought to be both up or both down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In early stages, this aspect of Hund's rule is not necessarily taught--Hund's rule is just portrayed as the thing that prevents pairing (from my experience). So, the paper setter may have decided to allow violations of the other aspect(if it is that) of Hund's rule, and thus the &quot;Allow single arrow in either direction&quot;. They mean that the directions of single arrows can be mixed up.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-05-31T17:31:04.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="563" PostTypeId="1" AcceptedAnswerId="565" CreationDate="2012-05-31T17:36:11.543" Score="9" ViewCount="3127" Body="&lt;p&gt;This is my first time taking a $\ce{{}^{13}C-NMR}$ spectrum and while I'm aware the signal to noise ratio is much lower compared to $\ce{{}^{1}H-NMR}$, I didn't realize it was quite this bad. I took a spectrum and I only see the $\ce{CDCl3}$ triplet. What concentration range is generally appropriate given say 256 scans (10 minutes on our machines?), or is it rare to see any good results after without drastically increasing the scan time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectrum for reference (taken on a 300 MHz Varian Mercury):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iz8lF.png&quot; alt=&quot;13C NMR Spectrum&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="4945" LastEditDate="2014-07-30T05:17:20.140" LastActivityDate="2014-07-30T05:17:20.140" Title="13C NMR spectrum only showing solvent" Tags="&lt;nmr-spectroscopy&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="564" PostTypeId="2" ParentId="559" CreationDate="2012-05-31T20:49:11.397" Score="11" Body="&lt;p&gt;&lt;strong&gt;Raise hell with the grader.&lt;/strong&gt; There's no reason whatsoever for this answer to be rejected. A &lt;a href=&quot;http://www.google.com/search?q=electron+box+diagram&amp;amp;btnG=Search+Books&amp;amp;tbm=bks&amp;amp;tbo=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google book search&lt;/a&gt; can substantiate your argument, and the first example from a textbook is actually in your order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;         &lt;img src=&quot;https://i.stack.imgur.com/9Wa7I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the guidance “allow single arrow in either direction” is not exclusive with “allow double arrows in either order”, and maybe only single arrows were explicitly mentionned because it was obvious to the writer that order in double arrows doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-06-01T06:01:08.307" LastActivityDate="2012-06-01T06:01:08.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="565" PostTypeId="2" ParentId="563" CreationDate="2012-05-31T21:15:22.737" Score="5" Body="&lt;p&gt;At 256 scans you should see something if your compound has reasonable concentration. The rule of thumb I was taught as an undergrad was ~10 mg of compound for a $^{13}$C scan in under a mL of solvent (400-500 uL if you are measuring precisely, modified by probe size and whatnot of course) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for an exact concentration based on MW? I'm not sure, I've never really worked it out, since the rule of thumb works well enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next question: Does your compound have protons on some of its carbons? Quaternary carbons will be much harder to see, so it is possible that you won't see them after 256 scans. &#xA;However yes, concentration is probably the problem. I made this mistake yesterday and found that after 256 scans I could see a tiny blot on the baseline. &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="121" LastEditDate="2012-06-04T15:49:52.460" LastActivityDate="2012-06-04T15:49:52.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="566" PostTypeId="1" AcceptedAnswerId="568" CreationDate="2012-05-31T22:50:08.773" Score="7" ViewCount="236" Body="&lt;p&gt;From what I understand, if you have a NMR spectra, you can determine the structure of a macromolecule but it is very difficult to determine what the NMR spectra of an expected structure would look like? Why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="121" LastEditDate="2012-06-01T20:50:17.490" LastActivityDate="2012-06-01T20:50:17.490" Title="Why is it so difficult to create mock NMR spectra of macromolecules?" Tags="&lt;nmr-spectroscopy&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="567" PostTypeId="2" ParentId="558" CreationDate="2012-05-31T23:39:06.363" Score="47" Body="&lt;p&gt;Menthol it self gives a cold feeling in the mouth because it is active at the same receptor (an ion channel) on the tongue that cold temperature triggers. Interestingly, although they act at the same receptor, they act at different sites, so that provides the intensified response when eating a mint and then drinking water. &lt;a href=&quot;http://boundlessthicket.blogspot.com/2012/04/that-cool-mint-feeling-when-i-was-in.html&quot;&gt;This reference&lt;/a&gt; gives an excellent detailed answer, with references to the original papers, which I'll  summarize here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Menthol acts at the TRPM8 protein which forms an ion channel that allows $\ce{Na+}$ and $\ce{Ca^2+}$ ions to flow into cells and this sends a signal saying &quot;cool&quot; to the brain. (As an aside, this protein monitors temperature across the body and not just on the tongue.) Cold temperatures actually change the confirmation of this protein, which allows the ions to flow more freely, and sends the signal to the brain. Menthol, on the other hand, stabilizes the open channel (allowing ions to flow even more freely) and also...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“menthol shifts the voltage dependence of channel activation to more negative values by slowing channel deactivation”.  This is very significant to my question because it supports a claim made by the first web page I visited which stated that menthol acts on the receptors, leaving them sensitized for when the second stimulus is applied (i.e. cold water) resulting in the enhanced sensation.  This mechanism of binding is very clearly different from the mechanism of cold affecting the TRP channels.  This is why the sensation is increased when both stimuli are applied, yet is not affected after addition stimulation from the same stimuli (i.e. eating another mint).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All in all, pretty cool. &lt;/p&gt;&#xA;" OwnerUserId="133" LastActivityDate="2012-05-31T23:39:06.363" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="568" PostTypeId="2" ParentId="566" CreationDate="2012-06-01T07:31:04.897" Score="8" Body="&lt;p&gt;Macromolecular NMR spectra are difficult to compute because macromolecular electronic structures are computationally expensive to solve with accuracy. Obtaining precise chemical shifts and splitting requires accurate treatment of electron density and geometry, especially for simulation of NOESY, which I understand is important in macromolecular structural elucidation. This requires good methodology and large basis sets.&lt;/p&gt;&#xA;&lt;p&gt;An interesting case study by Saielli and Bagno demonstrates the use of DFT (with cc-PVTZ basis) in predicting NMR spectra for two candidate structures and the correlating them with experimental data, however the mean error in chemical shifts is nontrivial&lt;sup&gt;1&lt;/sup&gt; (0.2 ppm in the best match). The same people do &lt;span class=&quot;math-container&quot;&gt;$^{1}\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^{13}C}$&lt;/span&gt; NMR predictions of a number of complex molecules&lt;sup&gt;2&lt;/sup&gt; (with intriguing discrepancies in shifts) however this is a far cry from predicting, say, a large protein &lt;span class=&quot;math-container&quot;&gt;$^{1}\ce{H}$&lt;/span&gt; NMR spectrum.&lt;/p&gt;&#xA;&lt;p&gt;I have seen EPR spectra of transition metal clusters simulated as abstract spin systems (for instance by &lt;a href=&quot;http://www.easyspin.org/&quot; rel=&quot;noreferrer&quot;&gt;EasySpin&lt;/a&gt;) however I presume that this is very qualitative.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;(1)&lt;/strong&gt; Saielli, G.; Bagno, A.; &lt;em&gt;Can two molecules have the same NMR spectrum? Hexacyclinol revisited.&lt;/em&gt;; Org. Lett.; &lt;strong&gt;(2009)&lt;/strong&gt; 11, pp. 1409-1412&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;(2)&lt;/strong&gt; Bagno, A.; Rastrelli, F.; Saielli, G.; &lt;em&gt;Toward the complete prediction of the &lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^{13}C}$&lt;/span&gt; NMR spectra of complex organic molecules by DFT methods. Application to natural substances.&lt;/em&gt;; Chem. Eur. J.; &lt;strong&gt;(2006)&lt;/strong&gt; 12, pp. 5514-5525&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-06-01T07:31:04.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="569" PostTypeId="1" AcceptedAnswerId="603" CreationDate="2012-06-01T08:47:46.733" Score="16" ViewCount="13157" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bead_test&quot;&gt;borax bead test&lt;/a&gt;, we take a crystal of borax and then heat it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A small loop is made in the end of a platinum or nichrome wire (as used in the flame test) and heated in a Bunsen flame until red hot. It is then dipped into powdered borax, and the adhering solid is held in the hottest part of the flame where it swells up as it loses its water of crystallization and then shrinks, forming a colourless, transparent glass-like bead (a mixture of sodium metaborate and boric anhydride)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whenever I've done the experiment, I've noticed that the process looks like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It heats up in a second, in this time there are no visible changes&lt;/li&gt;&#xA;&lt;li&gt;It starts expanding/blooming. Bubbles appear, and little pseudopodia(closest work I can think of) of borax flail about. This takes approx five seconds. At the end of this, the borax looks like whipped cream that's been...well.. &lt;em&gt;whipped&lt;/em&gt; . &lt;/li&gt;&#xA;&lt;li&gt;The thing suddenly contracts into a tiny bead.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, as mentioned in the quote, basically borax loses its water of crystallization, and then decomposes into $\ce{NaBO2 +B2O3}$, the small bead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flailing about is probably due to steam escaping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the reaction proceeds in such an interesting manner, I can't help but think that there's something else going on as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the details of the mechanism of the borax bead preparation reaction?&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2018-08-18T19:36:36.180" Title="When we heat a borax crystal, why does it first expand and then contract?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="570" PostTypeId="2" ParentId="559" CreationDate="2012-06-01T10:56:14.693" Score="5" Body="&lt;p&gt;The order definitely does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of diagram is for representing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot;&gt;quantum numbers&lt;/a&gt; of the electrons in the system. The 3d boxes mean the quantum numbers n=3, l=2 with ml =-2,-1,0,+1,+2 for the 5 boxes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a box has an up and a down arrow in it, this means there are 2 electrons with those quantum numbers, one with s=1/2, and one with 2=-1/2. It is completely meaningless which order your draw these in, the result is still the same two electrons with the same set of quantum numbers. Electrons are indistinguishable from each other. If you had a complex system with an exotic lepton such as a tau or muon then the order might matter, but in this case it certainly does not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only assume that the writer of the mark-scheme assumed that UP DOWN and DOWN UP being identical would be obvious to anyone, and so didn't bother to write it. If the answer to a maths question was x * y, most mark-schemes wouldn't bother to write y * x also acceptable, as its obvious to anyone with any subject knowledge.&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2012-06-01T10:56:14.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="571" PostTypeId="2" ParentId="540" CreationDate="2012-06-01T16:05:50.197" Score="7" Body="&lt;p&gt;It sounds like you need a solvent that 1) does not react with KMnO&lt;sub&gt;4&lt;/sub&gt; and 2) is immiscible with water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Janice DelMar said in her comment, some of the lower-chlorinated hydrocarbons (methylene chloride, 1,2-dichloroethane, 1,1,2-trichloroethane) should work. If you can live with a high boiling point solvent, you might consider chlorobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other solvents that fit these criteria include most alkanes (pentane, hexane, cyclohexane, petroleum ether, etc.), benzene (but not toluene), and diethyl ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick test to determine compatibility of your solvent with KMnO&lt;sub&gt;4&lt;/sub&gt; involves a TLC stain. Make (or borrow) a batch of &lt;a href=&quot;http://orgprepdaily.wordpress.com/2006/09/27/tlc-staining-solutions/&quot;&gt;KMnO&lt;sub&gt;4&lt;/sub&gt; stain as described here&lt;/a&gt;. Dip a TLC plate into your solvent and then dip it into the stain with minimal drying between. If the plate turns brown (MnO&lt;sub&gt;2&lt;/sub&gt;), then your solvent was oxidized and is no good. If the plate stays purple, then your solvent is resistant. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-01T16:05:50.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="572" PostTypeId="1" AcceptedAnswerId="582" CreationDate="2012-06-01T19:39:11.597" Score="9" ViewCount="162" Body="&lt;p&gt;Some NMR experiments can be halted partway through the acquisition and still give perfectly valid results: The traditional 1H experiment for example. Others, like a Saturation Transfer Difference Spectrum will have errors in the magnitude of the peaks in it if it is halted, due to the fact it uses groups of pulses and takes the difference of between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the standard proton-proton NOESY pulse sequence be halted halfway through and give a usable spectrum, or will this introduce artifacts?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="74" LastEditDate="2012-06-06T19:41:16.353" LastActivityDate="2012-06-06T19:41:16.353" Title="Can a NOESY NMR experiment be halted partway through and give meaningful results?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="573" PostTypeId="1" CreationDate="2012-06-01T21:36:01.317" Score="6" ViewCount="8970" Body="&lt;p&gt;I want to know which are the most stable furan resonance structures. I know that structure number 1 is the most stable, but which one comes after?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ulVgN.png&quot; alt=&quot;Resonance structures of furan&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="16683" LastEditDate="2017-02-27T21:00:31.623" LastActivityDate="2018-11-12T17:03:08.513" Title="What is the order of stability of the resonance structures of furan?" Tags="&lt;aromatic-compounds&gt;&lt;stability&gt;&lt;resonance&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="574" PostTypeId="1" AcceptedAnswerId="587" CreationDate="2012-06-01T21:42:04.283" Score="21" ViewCount="6571" Body="&lt;p&gt;I know that &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthocyanin&quot; rel=&quot;nofollow noreferrer&quot;&gt;anthocyanins&lt;/a&gt; are a class of compounds responsible for the purple colors found in flower petals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anthocyanins (also anthocyans; from Greek: ἀνθός (anthos) = flower + κυανός (kyanos) = blue) are water-soluble vacuolar pigments that may appear red, purple, or blue depending on the pH. (same reference as above)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The anthocyanins, when extracted from flower petals with water, can be used as acid/base indicators, producing red colors in acid and blue in neutral solutions (in basic solutions they turn green, yellow, and then become colorless). When I've done the extractions, some red petals produce a red color in acid and maintain the red color at higher pH than purple flowers (see plate below), but some red flowers produce colors that seem to follow the normal progression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkXOm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Row 2 from the top was a purple flower, row 3 a red flower and row 4 also a red flower. (pH values start at 2 on the left and increase by 1 pH unit as you move right by adding the water extracts to buffer samples in the wells.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there other classes of compounds responsible for the red and yellow colors found in flowers? Can the red colors that give a &quot;normal&quot; anthocyanin pH response arise from pigments held in an acidic environment in the plant?   (That may be a biology question, but I'm interested in the chemistry side of where the color comes from.)&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="402" LastEditDate="2017-01-26T20:03:03.127" LastActivityDate="2017-04-25T18:57:15.383" Title="Chemical compounds responsible for the colors in flowers?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="575" PostTypeId="1" CreationDate="2012-06-02T05:33:53.227" Score="31" ViewCount="153578" Body="&lt;p&gt;When purchasing chemicals from Sigma, Fisher, or wherever, there are often &lt;em&gt;-grade&lt;/em&gt;'s attached to their description like &lt;em&gt;reagent-grade&lt;/em&gt;, &lt;em&gt;technical-grade&lt;/em&gt;, &lt;em&gt;analytical-grade&lt;/em&gt;, or more niche-sounding &lt;em&gt;biotech-grade&lt;/em&gt;, &lt;em&gt;HPLC-grade&lt;/em&gt;, &lt;em&gt;DNA grade&lt;/em&gt; (DNase free perhaps?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sort of standard for what these actually mean or are they arbitrary, differing from supplier to supplier or even chemical to chemical?  Beyond the 'niche' grades, is there a common order of purity between the others or do they depend on the &lt;em&gt;types&lt;/em&gt; of impurities?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="9961" LastEditDate="2017-01-13T13:57:04.750" LastActivityDate="2019-02-15T22:24:35.167" Title="What do the different grades of chemicals mean?" Tags="&lt;experimental-chemistry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="576" PostTypeId="2" ParentId="575" CreationDate="2012-06-02T06:38:16.860" Score="33" Body="&lt;p&gt;Sigma-Aldrich gives a very useful &lt;a href=&quot;http://www.sigmaaldrich.com/technical-service-home/product-portfolio.html&quot; rel=&quot;noreferrer&quot;&gt;table&lt;/a&gt; outlining what the different purity levels are and suggested applications. I was less successful at finding equal documentation from some of the other suppliers, but the analysis for the purity of the chemicals they sell is in the catalog and on the bottle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, technical grade or laboratory grade are the lowest purity. &lt;a href=&quot;http://pubs.acs.org/reagents/index.html&quot; rel=&quot;noreferrer&quot;&gt;ACS Reagent grade&lt;/a&gt; means that the chemical conforms to specifications defined by the Committee on Analytical Reagents of the American Chemical Society (but Aldrich &lt;em&gt;&quot;ReagentPlus&quot;&lt;/em&gt; means &gt;95% pure). So, &quot;ACS Reagent grade&quot; chemicals should be comparable from different suppliers.  Analytical grade is generally the most pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some of the other grades, such as HPLC grade solvents, the issue is less about overall purity than about being free of substances that would interfere with a particular application. I've reproduced a piece of the table below to give you an idea of the information available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KX2Z0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KX2Z0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="41328" LastEditDate="2017-09-19T06:06:49.147" LastActivityDate="2017-09-19T06:06:49.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="577" PostTypeId="1" CreationDate="2012-06-02T14:46:19.420" Score="8" ViewCount="1439" Body="&lt;p&gt;I am trying to build a system of two connected chambers, one of which is also connected to a cuvette, so I can perform a spectrometry experiment with different concentrations of $\ce{KNO3}$ solution. In its simplest form, the cuvette and the first chamber would hold solution at the same concentration, while the second chamber would hold solution of higher concentration. I would perform one experiment, open a valve between the chambers, and wait for the concentration to equilibrate before performing the next experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advisor suggested this system and said that the equilibration time would likely be on the order of minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used this equation to calculate equilibration time between two chambers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t = \frac{\Delta m \cdot  l_{v}}{A_{v} \cdot D \cdot 10^6 \cdot \Delta c}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ = equilibration time in s&lt;br&gt;&#xA;$\Delta m$ = the difference in mass of $\ce{KNO3}$ between the chambers in g&lt;br&gt;&#xA;$l_{v}$ = the length of the valve in m&lt;br&gt;&#xA;$A_{v}$ = the orifice area of the valve in m^2&lt;br&gt;&#xA;$D$ = the mutual diffusion constant of $\ce{KNO3}$ and $\ce{H2O}$, which is on the order of 10^-9 m^2/s&lt;br&gt;&#xA;$\Delta c$ = the difference in concentration between the chambers in g/cm^3&lt;br&gt;&#xA;the factor of 10^6 is to convert between g/cm^3 and g/m^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cylindrical valve 1 cm long and 2 cm in radius (improbably short and wide) and chamber volumes 95 cm^3 and 400 cm^3, this gives an equilibration time of 4.7 days!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much more than I was expecting. I can reduce the sizes of the chambers to decrease the time demand, but I would prefer to keep them as large as I can, to increase the final concentration that can be achieved in the cuvette.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something obvious that I'm missing in this calculation? Is there a better way to do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation above was derived by equating the mass flux $j_{m}$ with its definition from Fick's first law (using $c$ as mass/volume rather than amount/volume):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;j_{m}=\frac{\Delta m}{A_{v}\cdot \Delta t}=D\cdot\frac{\Delta c}{\Delta x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that strictly the RHS should contain $\frac{\delta c}{\delta x}$ rather than $\frac{\Delta c}{\Delta x}$. Does this approximation make much difference? &lt;em&gt;For now&lt;/em&gt;, I am also making the fairly large assumptions that the volumes of solution in each chamber do not change, and that the density of the solution is independent of concentration. I think I can include these considerations later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an actual system that I need to build (well, ask our machine shop to build for me) and I need to have a good idea of the best valve sizes to use as they have a significant effect on the overall structure.&lt;/p&gt;&#xA;" OwnerUserId="300" LastEditorUserId="4231" LastEditDate="2014-10-31T18:46:06.590" LastActivityDate="2014-10-31T18:46:06.590" Title="How to calculate the equilibration time of two connected chambers of salt solution, initially at different concentrations?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="579" PostTypeId="1" CreationDate="2012-05-25T16:58:22.470" Score="19" ViewCount="3249" Body="&lt;p&gt;In a typical semiconducting material, we start with silicon (group IV) and introduce atoms of group V or group III depending on whether we are constructing an n- or p-type semiconductor respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it optimal to choose doping atoms one group away? Why not introduce atoms from groups VI or II, for instance?&lt;/p&gt;&#xA;" OwnerUserId="301" OwnerDisplayName="Simon S" LastEditorUserId="30836" LastEditDate="2020-05-03T08:15:53.783" LastActivityDate="2020-05-03T08:15:53.783" Title="Choice of doping elements in semiconductors" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;&lt;conductivity&gt;&lt;semiconductors&gt;&lt;electrical-properties&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="580" PostTypeId="2" ParentId="579" CreationDate="2012-05-26T02:44:24.457" Score="9" Body="&lt;p&gt;Look at the II-VI options from a chemical perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For silicon, Group III boron p-doping would be replaced by Group II beryllium doping, and Group V phosphorus n-doping would be replaced by Group VI sulfur doping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beryllium would tend to act like a Lewis base that donates a pair of electrons to some nearby silicon atom, forming a stronger and more localized bond than is possible with the odd electron count created by boron doping. Similarly, the sulfur would tend to act like a Lewis acid and form a localized sulfide bond with some nearby silicon atom, versus the odd electron count created by phosphorus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the main reason for picking elements immediately &quot;next door&quot; (off by one electron) is to avoid introducing paired electrons, which on average will bond more strongly and more locally via a Lewis base-acid process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there's enough complexity going on there that it's still possible that you could get II-VI or maybe even I-VII doping to work in the right situations.&lt;/p&gt;&#xA;" OwnerUserId="83" OwnerDisplayName="Terry Bollinger" LastActivityDate="2012-05-26T02:44:24.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="581" PostTypeId="2" ParentId="573" CreationDate="2012-06-02T18:51:51.610" Score="8" Body="&lt;p&gt;A quick post so you can get started on thinking about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, resonance is just a model. The electrons are delocalized all the time; the furan does not morph back and forth among the various structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a &quot;formal&quot; evaluation of which resonance forms have the most contribution to the structure by looking at &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_charge&quot;&gt;formal charges&lt;/a&gt;, and there are rules to help you decide, from the formal charges, which resonance forms are most important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking at reactivity, the property you're interested in may not be that the one that is &quot;more stable is less reactive&quot;. What you might look at instead, is the consequences of attack at the 2 and 3 position, look at the stability of the intermediates formed...a more stable intermediate would tend to drive the reaction along that path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at these ideas, feel free to come back with specific questions.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-06-02T19:17:32.173" LastActivityDate="2012-06-02T19:17:32.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="582" PostTypeId="2" ParentId="572" CreationDate="2012-06-02T19:00:28.323" Score="7" Body="&lt;p&gt;There are two big differences between the STD experiment and a NOESY that matter when aborting the experiment early. The STD is usually run as a 1D experiment, and it's using the phase cycle to substract the on- and off-resonance spectra from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you acquire a 1D spectrum with multiple scans, the scans are not saved individually but directly added up to one FID, at least on Bruker spectrometers. So if you abort a 1D experiment (you need to execute &lt;code&gt;tr&lt;/code&gt; to do that in most cases, as the FID is only written to disk after all scans are finished in most pulse sequences), you can't just throw away the last incomplete phasy cycle. And with experiments like the STD, your actual signal is the difference of two larger signals, so one incomplete substraction can ruin your spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 2D spectrum the increments for the indirect dimension are measured and written to disk sequentially, so you can throw away the last incomplete increment if you would need to. In most cases the incomplete increment will not be saved at all to disk, so you'll end up with one less data point in the indirect dimension, but no incomplete ones that were aborted in the middle of the phase cycle. This depends on how exactly the pulse program is written, but I suspect that most standard sequences handle it this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've stopped HSQC and NOESY spectra before they were finished pretty often, and never observed any artifacts that could be traced to the abortion of the experiment.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2012-06-02T19:00:28.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="583" PostTypeId="2" ParentId="563" CreationDate="2012-06-02T19:25:21.497" Score="7" Body="&lt;p&gt;There are several reasons why $\mathrm{^{13}C}$ experiments are far less sensitive than proton experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural abundance for $\mathrm{^{13}C}$ is only 1.1%, while $\mathrm{^{1}H}$ has a natural abundance of 99,9%. This difference alone means that if you don't label your sample with $\mathrm{^{13}C}$, the spectra will be around 90 times less sensitive than proton spectra due to the lower natural abundance alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{^{13}C}$ also has a four times lower gyromagnetic ration than $\mathrm{^{1}H}$, which also lowers the sensitivity. But the way the experiment is usually measured, you get magnetization transfer from directly-attached protons to the carbons, which increases the sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{^{13}C}$ also relaxes more slowly, which means that the time between experiments is longer and you can measure fewer scans than for proton experiments in the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor that is important is which kind of probe your spectrometer is using. For an inverse probe like e.g. a Bruker TXI, the inner coil is the proton coil, which means it's especially suited for proton experiments. For $\mathrm{^{13}C}$ experiments an observe probe like a BBO is more sensitive, as the carbon is measured on the inner coil. If you have multiple spectrometers, you should ask which one is better suited for $\mathrm{^{13}C}$ experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a very small amount of sample, a Shigemi tube allows you to measure in half the sample volume (200-250 $\mu$l) compared to standard tubes, you can achieve a higher concentration that way. They're pretty expensive though and they have to be matched to the solvent, so you would need a CDCl3 tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are various problems that could drastically reduce your sensitivity for $\mathrm{^{13}C}$, the most important ones if we disregard hardware problems, are proper tuning and matching for $\mathrm{^{13}C}$, and the correct pulse length for $\mathrm{^{13}C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've avoided your actual question, as the required concentration depends on quite a few factors. I'd recommend you to ask the person responsible for the spectrometers, as they should be able to give you a better estimate based on the actual hardware you have.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="121" LastEditDate="2012-06-04T17:38:26.803" LastActivityDate="2012-06-04T17:38:26.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="584" PostTypeId="1" CreationDate="2012-06-03T00:08:56.800" Score="7" ViewCount="1213" Body="&lt;p&gt;In one variation of the iodine clock demonstration, you add two colorless solutions together and apparently nothing happens. After about 30 seconds to a minute, you get an &quot;instantaneous&quot; change to a dark blue color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution &lt;strong&gt;A&lt;/strong&gt; is $\pu{0.006 M}$ $\ce{KIO3}$ and is stable over at least a year or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution &lt;strong&gt;B&lt;/strong&gt; is made with starch, $\pu{0.2 g}$ of sodium metabisulfite ($\ce{Na2S2O5}$), and $\pu{1 mL}$ of $\pu{1 M}$ $\ce{HCL}$ in $\pu{200 mL}$ of water. Starch can be from &quot;reagent grade starch&quot;, or starch from a packing peanut, or from laundry starch bought at the grocery store. The concentration of the starch ends up at about 0.05% w/v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution &lt;strong&gt;B&lt;/strong&gt; must be made up no more than 24 hours in advance or the reaction doesn't work: the blue color doesn't form or the timing of the reaction is way off, depending on how old the reagent is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens in solution &lt;strong&gt;B&lt;/strong&gt; that makes it unstable, and is there an easy way to stabilize it so that I could have the chemicals needed for the demo ready on the shelf?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions that happen in the demo are:&#xA;\begin{align}&#xA;\ce{S2O5^2- + H2O &amp;amp;-&amp;gt; 2 HSO3-} &amp;amp; \tag{1}\\&#xA;\ce{IO3- + 3 HSO3- &amp;amp;-&amp;gt; 3 SO4^2- + I- + 3 H+} &amp;amp;\text{(slow)} \tag{2}\\&#xA;\ce{IO3- + 8 I- + 6 H+ &amp;amp;-&amp;gt; 3 I3- + 3 H2O} &amp;amp;\text{(slow)} \tag{3}\\&#xA;\ce{I3- + HSO3- + H2O &amp;amp;-&amp;gt; 3 I- + SO4^2- + 3 H+} &amp;amp;\text{(fast)} \tag{4}\\&#xA;\ce{2 I3- + starch &amp;amp;-&amp;gt; blue complex + I-} &amp;amp;\text{(fast)} \tag{5}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see anything in these that tell me why the metabisulfite / $\ce{HCl}$ / starch reagent is unstable.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="41328" LastEditDate="2017-08-07T00:49:27.587" LastActivityDate="2017-08-07T00:49:27.587" Title="Why is a metabisulfite / starch / HCl solution unstable?" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="585" PostTypeId="2" ParentId="584" CreationDate="2012-06-03T17:08:04.420" Score="5" Body="&lt;p&gt;I can partially answer the question. I know what is happening to solution B. I don't know of a more stable version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One potential problem is the acid-catalyzed hydrolysis of the starch into glucose, which does &#xA;not form the blue complex. &#xA;$$\ce{nH2O + (C6H10O5)}_n \ce{-&amp;gt;nC6H12O6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the timing to be off, the metabisulfite must also be decomposing, either by reducing glucose or by reacting with dissolved oxygen. I'm not sure what the specific process is, nor do I know how to stop it.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-03T17:08:04.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="586" PostTypeId="2" ParentId="400" CreationDate="2012-06-03T19:35:17.483" Score="3" Body="&lt;p&gt;&lt;strong&gt;I have studied triple-point systems&lt;/strong&gt;, though not principally in hydrogen.  I have worked with phase equilibrium in dynamic (moving) water systems, looking at changes in density and surface tension.  Because of this, I have studied the phenomenon more broadly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a system of mass and energy to remains (is maintained) in phase-equilibrium (i.e. the triple point), the introduction of new forces – be they pressure, e-field, magnetic polarization, thermo, etc – can lead to an ongoing reaction cycle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I won't speculate too deeply about what that does in this H system, but will leave this paragraph for another specialist to answer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase-equilibrium point can make it easier to push a reaction toward a specific kind of change / characterization of the media.  Particularly, the closeness to the gas phase means the volatility of [part of] the media is being utilized to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bond in external media (doping) &lt;/li&gt;&#xA;&lt;li&gt;perform redox or some other electronic reaction &lt;/li&gt;&#xA;&lt;li&gt;liberate some media in a thermo exchange that crystalizes others &lt;/li&gt;&#xA;&lt;li&gt;change the crystallization morphology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pragmatically, I've read this is applied in reactor\feedstock scenarios, where the 'reaction' portion would in this case be held at the triple-point, whilst the feed/flow of media causes the outlet portion to take on a certain character, as the inlet media is prepped.  The principle to understand here is that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Permittivity#Explanation&quot; rel=&quot;nofollow&quot;&gt;reaction portion act more homogeneously, rather than local regions of differentiated materials.&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2012-06-03T19:35:17.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="587" PostTypeId="2" ParentId="574" CreationDate="2012-06-03T21:21:39.693" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;are there other classes of compounds responsible for the red and yellow colors found in flowers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interest in colors, specifically dyes, was a real motivator for serious chemical work. Some of the early efforts were aiming to mimic Tyrian purple, also known as Royal Purple, because in antiquity it was worth its weight in gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two classes of molecules come to mind other than anthocyanins in phytochemistry, such as the closely related &lt;a href=&quot;http://en.wikipedia.org/wiki/Flavones&quot; rel=&quot;noreferrer&quot;&gt;flavones&lt;/a&gt; (its root means yellow) and the distinct &lt;a href=&quot;http://en.wikipedia.org/wiki/Carotene&quot; rel=&quot;noreferrer&quot;&gt;carotenes&lt;/a&gt; (its root means carrot, but also responsible for flamingo color). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the red colors that give a &quot;normal&quot; anthocyanin pH response arise from pigments held in an acidic environment in the plant?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A real sharp organic chemist by the name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard_Willst%C3%A4tter&quot; rel=&quot;noreferrer&quot;&gt;Richard Willstatter&lt;/a&gt; proposed this  in the early 1900s. He was working on determining the structure of the pigment responsible for the blue color in &lt;a href=&quot;http://en.wikipedia.org/wiki/Cornflower&quot; rel=&quot;noreferrer&quot;&gt;cornflower&lt;/a&gt;. He extracted the same substance coming from a rose and proposed that all flower colors were due to a different pH in the sap. As your experiment clearly points out, however, alkaline conditions are typically required for the blue anion. A Japanese group argued that alkaline conditions in the sap were unlikely and that instead these different colors can be explained by the formation of metal-organic complexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't until 2005 that a definitive crystal structure was determined from the cornflower pigment called &lt;a href=&quot;http://en.wikipedia.org/wiki/Protocyanin&quot; rel=&quot;noreferrer&quot;&gt;protocyanin&lt;/a&gt;. It turns out that the blue color comes from a supermolecule containing one $\ce{Fe^{3+}}$, one $\ce{Mg^{2+}}$, two $\ce{Ca^{2+}}$, six anthocyanin and six flavone molecules packaged together in a complex structure. This color, as I vaguely recall, remains blue from pH 7 until pH 1! This agreed with some earlier evidence which suggested a complex species by the temperature dependence of the pigment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not easy studies to conduct and I'm not familiar with how in-depth red colors have been studied, but it is certainly possible a complex species is actually how the red pigments form &lt;em&gt;in vivo&lt;/em&gt;. Keeping in mind that plant vacuoles have a pH on the order of 5, we'll generally notice a loss in red above pH 3 in solution for a pure anthrocyanin in solution as hydration of the cation occurs. Stability of the color, and thus cation, obviously must be obtained either via copigmentation or with metals in a plant.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T19:54:53.667" LastActivityDate="2015-06-03T19:54:53.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="588" PostTypeId="2" ParentId="295" CreationDate="2012-06-03T22:31:32.583" Score="4" Body="&lt;p&gt;I've searched for this same data.  My interest regards electronic proporties of zeolite-class media.  &lt;a href=&quot;http://www.quora.com/Water-Purification/What-sources-exist-to-learn-about-materials-products-and-by-products-used-as-absorbing-flocking-agents-in-water-purification-and-metals-reclamation&quot; rel=&quot;nofollow&quot;&gt;I've written this same kind of question on Quora&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the reason this is rarely answered, and why there are no 'catalogs', is several-fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There exist a wide range of applications, based on the composition&#xA;of the water / etc. &lt;/li&gt;&#xA;&lt;li&gt;The treatment media usually varies in&#xA;composition, meaning that no directive for the use of a zeolite&#xA;product can be given by a sales rep (only on-staff engineers, who&#xA;know how to use it).&lt;/li&gt;&#xA;&lt;li&gt;Some uses create by-products that are regulated, creating possible liabilities for the seller by indicating a fitness-for-use.&lt;/li&gt;&#xA;&lt;li&gt;Regulations change; per year, per locale, etc.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think the groups that know about these are environmental engineering firms, whom have a stable of engineers that provide end-to-end services for their clientele.  Many of of these firms have Chemists on-staff, or a consultancy relationship that suits the same.  They know the chemical properties of their media, and need a zeolite / catalysts that can effect something &lt;em&gt;very&lt;/em&gt; specific.  As a result, they're out their hunting the journals for something that targets their needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard to out-sell that without already having a dog in the game.&lt;/p&gt;&#xA;" OwnerUserId="296" LastEditorUserId="30" LastEditDate="2012-06-04T08:34:29.767" LastActivityDate="2012-06-04T08:34:29.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="589" PostTypeId="1" CreationDate="2012-06-04T01:04:46.960" Score="21" ViewCount="10451" Body="&lt;p&gt;I've noticed that water obtained from melting ice cubes (by keeping them in the open) tastes different from water cooled to the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the taste goes away if you keep it in the open for a sufficient duration of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of these possible explanations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dissolved gases being released&lt;/li&gt;&#xA;&lt;li&gt;Something to do with the crystal structure of ice&lt;/li&gt;&#xA;&lt;li&gt;Something to do with dissolved minerals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Internet is giving me all sorts of varying answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what chemical differences exist between meltwater and cold water? &lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="9961" LastEditDate="2017-01-13T13:57:55.373" LastActivityDate="2018-03-03T05:49:48.610" Title="Why does meltwater taste different?" Tags="&lt;water&gt;&lt;home-experiment&gt;&lt;taste&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="590" PostTypeId="1" CreationDate="2012-06-04T10:14:13.070" Score="27" ViewCount="1296" Body="&lt;p&gt;I'm reading a paper&lt;sup&gt;1&lt;/sup&gt; by Sanakis, &lt;em&gt;et al.&lt;/em&gt; that characterises the magnetic coupling in the $\ce{Fe3S4}$ clusters present in bacterial ferredoxin II and beef heart aconitase as arising through something called 'antisymmetric exchange'. This looks like it may be a result of the apparent magnetic frustration of the system (three high-spin $\ce{Fe^{3+}}$ ions, all in essentially identical environments, all antiferromagnetically coupled with respect to their neighbours). The paper says that the clusters have a S=1/2 ground state, which is incompatible with a broken-symmetry model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Antisymmetric_Exchange&quot; rel=&quot;noreferrer&quot;&gt;antisymmetric exchange&lt;/a&gt; (all four sentences of it) is not particularly edifying. I'm hoping an expert can explain what antisymmetric exchange is and what causes it as I've not heard of it be before. The paper also notes that the exchange situation in the cluster gives rise to three slightly different $J_{ij}$ values (again, despite the symmetry of the ions) and terms this '&lt;em&gt;J-strain&lt;/em&gt;'. This makes me doubt that, for instance, 'antisymmetric' is being used as a synonym for 'antiferromagnetic'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Edit]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of the general connectivity of this kind of $\ce{Fe3S4}$ cluster (based on &lt;em&gt;D. Gigas&lt;/em&gt; FdII (PDB entry &lt;a href=&quot;http://www.rcsb.org/pdb/explore/explore.do?structureId=1F2G&quot; rel=&quot;noreferrer&quot;&gt;1F2G&lt;/a&gt;)). The cluster comprises a hemicubane. $\ce{S_{Cys}}$ denotes a cysteinyl sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PnwAl.jpg&quot; alt=&quot;Fe3S4 cluster&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) Sanakis, Y.; Macedo, A.L.; Moura, I.; Moura, J.J.G.; Papaefthymiou, V.; Münck, E.; &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja002658i&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Evidence for Antisymmetric Exchange in Cuboidal $\ce{[3Fe-4S]^+}$ Clusters&lt;/em&gt;&lt;/a&gt;; J. Am. Chem. Soc.; &lt;strong&gt;2000&lt;/strong&gt;, 122, pp. 11855-11863&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-06-04T10:48:15.057" LastActivityDate="2014-04-15T17:30:43.890" Title="What is antisymmetric exchange? What is J-strain? Where does it come from?" Tags="&lt;physical-chemistry&gt;&lt;spin&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="591" PostTypeId="2" ParentId="589" CreationDate="2012-06-04T17:18:57.943" Score="9" Body="&lt;p&gt;(not yet a fully grown answer: experiment still pending)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard that $\ce{CO2}$/$\ce{H2CO3}$/$\ce{HCO3-}$ is responsible for &quot;fresh&quot; taste of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@NickT: Chloramine has a very characteristic smell, so Manishearth could most probably tell us whether this was the smell/taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here are first questions for @Manishearth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;does water that has been boiled and then cooled to the same temperature taste similar to the taste you're asking for?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What about de-ionized water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, there's also that average-of-all-that-has-ever-been-in-the-freezer taste... &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-06-04T17:18:57.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="592" PostTypeId="1" CreationDate="2012-06-04T17:39:05.633" Score="51" ViewCount="1084" Body="&lt;p&gt;I have a PyMol session file I was using to create my figures.  It contains a protein and ligand broken into several groups of atoms with a fair bit of overlap.  There are also a few meshes built from a density map for a few figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I wasn't quite done with refinement, and now Refmac has given me shiny new PDB and MTZ files with subtly different coordinates and terms.  Is there a way to &quot;update&quot; my coordinates (despite the myriad groups, all the atoms still have the same name) with the new PDB?  With all the groups I made that are shown/hidden/colored differently for several scenes to produce all the figures, I'd rather not recreate it all manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm more pessimistic about the map/mesh, but I created them via a script so I can repeat that.  Should I also script the coordinate to figure stuff in the future if it's not finalized?&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="402" LastEditDate="2016-03-02T00:02:56.503" LastActivityDate="2021-01-21T03:43:11.910" Title="How can I update a map/mesh and coordinates in a PyMol session file?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="593" PostTypeId="1" AcceptedAnswerId="598" CreationDate="2012-06-04T20:02:25.213" Score="26" ViewCount="1589" Body="&lt;p&gt;I've been looking to use time-dependent density functional theory, but reading the literature I keep getting overwhelmed by how many different functionals there are.  How should I go about selecting which functional to use?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="84" LastEditDate="2012-06-05T02:44:53.640" LastActivityDate="2016-10-02T20:16:45.293" Title="How should I go about picking a functional for DFT calculations?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="594" PostTypeId="1" AcceptedAnswerId="1041" CreationDate="2012-06-05T01:32:31.257" Score="76" ViewCount="36501" Body="&lt;p&gt;Carbon is well known to form single, double, and triple $\ce{C-C}$ bonds in compounds. There is a &lt;a href=&quot;http://metallome.blogspot.com/2012/02/carbon-carbon-quadruple-bond.html&quot;&gt;recent report&lt;/a&gt; (2012) that carbon forms a quadruple bond in diatomic carbon, $\ce{C2}$. The excerpt below is taken from that report. The fourth bond seems pretty odd to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{C2}$ and its isoelectronic molecules $\ce{CN+}$, BN and $\ce{CB-}$ (each having eight valence electrons) are bound by a quadruple bond. The bonding comprises not only one σ- and two π-bonds, but also one weak ‘inverted’ bond, which can be characterized by the interaction of electrons in two outwardly pointing sp hybrid orbitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHKMf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Shaik, the existence of the fourth bond in $\ce{C2}$ suggests that it is not really diradical...&lt;br&gt;&#xA;  If $\ce{C2}$ were a diradical it would immediately form higher clusters. I think the fact that you can isolate $\ce{C2}$ tells you it has a barrier, small as it may be, to prevent that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem1.com/acad/webtext/chembond/cb08.html&quot;&gt;Molecular orbital theory for dicarbon&lt;/a&gt;, on the other hand, predicts a C-C double bond in $\ce{C2}$ with 2 pairs of electrons in $\pi$ bonding orbitals and a bond order of two. &quot;The bond dissociation energies (BDE) of $\ce{B2, C2}$, and $\ce{N2}$ show increasing BDE consistent with single, double, and triple bonds.&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diatomic_carbon&quot;&gt;Ref&lt;/a&gt;) So this model of the $\ce{C2}$ molecule seems quite reasonable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions, since this is most definitely not my area of expertise: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is dicarbon found naturally in any quantity and how stable is it? Is it easy to make in the lab? (The Wikipedia article reports it in stellar atmospheres, electric arcs, etc.)&lt;/li&gt;&#xA;&lt;li&gt;Is there good evidence for the presence of a quadruple bond in $\ce{C2}$ that wouldn't be equally well explained by double bonding? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="133" LastEditorUserId="16683" LastEditDate="2016-05-09T19:14:16.583" LastActivityDate="2020-06-13T14:57:09.473" Title="Bonding in diatomic C2, a carbon-carbon quadruple bond?" Tags="&lt;theoretical-chemistry&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;covalent-compounds&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="595" PostTypeId="5" CreationDate="2012-06-05T06:57:23.767" Score="0" Body="&lt;p&gt;Nuclear magnetic resonance (NMR) is a spectroscopic method used for the determination of the structures of molecules containing NMR-active nuclei, and may also be used as a method for studying mechanisms of chemical reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike methods such as IR spectroscopy or UV-vis spectroscopy, NMR is a quantitative technique that gives structural information such as the number of nuclei present, and the way in which they couple to one another, with a vast number of different NMR experiments available to probe various properties of the molecule of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though initially a technique for small molecule elucidation in organic chemistry, advances in technology have allowed the application of NMR to large biological molecules such as proteins, and gave rise to the related MRI (magnetic resonance imaging) which is widely used in medicine as a means of imaging the body. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Definition:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear%5C_magnetic%5C_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nuclear magnetic resonance (NMR) is a physical phenomenon in which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. This energy is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms; in practical applications, the frequency is similar to VHF and UHF television broadcasts (60–1000 MHz). NMR allows the observation of specific quantum mechanical magnetic properties of the atomic nucleus. Many scientific techniques exploit NMR phenomena to study molecular physics, crystals, and non-crystalline materials through nuclear magnetic resonance spectroscopy. NMR is also routinely used in advanced medical imaging techniques, such as in magnetic resonance imaging (MRI).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Applicability of the &lt;a href=&quot;/questions/tagged/nmr&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nmr&amp;#39;&quot; rel=&quot;tag&quot;&gt;nmr&lt;/a&gt; tag:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/nmr&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nmr&amp;#39;&quot; rel=&quot;tag&quot;&gt;nmr&lt;/a&gt; should be applied to all questions that deal with nuclear magnetic resonance, including…&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Theory questions about the theoretical chemistry behind the method, or about the way in which pulse sequences work &lt;/li&gt;&#xA;&lt;li&gt;Practical questions about the measurement of NMR spectra, choice of techniques, instrumentation&lt;/li&gt;&#xA;&lt;li&gt;Questions about the application of NMR, for instance in structure elucidation &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The majority of &lt;a href=&quot;/questions/tagged/nmr&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nmr&amp;#39;&quot; rel=&quot;tag&quot;&gt;nmr&lt;/a&gt; questions will need additional tags to fully describe them, for instance &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; if the question is asking about the NMR spectrum of a particular organic molecule or &lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;theoretical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt; if the question is asking about the physical basis for why the technique works. &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt; tag should also be applied to allow for easier searching, in the same way that a question might have tags for &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/carbonyl-compound&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;carbonyl-compound&amp;#39;&quot; rel=&quot;tag&quot;&gt;carbonyl-compound&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;/questions/tagged/analytical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;analytical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;analytical-chemistry&lt;/a&gt; tag may also be appropriate, but is perhaps best reserved for questions specifically regarding general analytical methods rather than for specific techniques.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related tags:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/nmr&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nmr&amp;#39;&quot; rel=&quot;tag&quot;&gt;nmr&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;theoretical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Further reading:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Many books/reviews on NMR have been published from theoretical and practical perspectives. A selection of the more approachable books aimed at chemists are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Keeler, J; &lt;em&gt;Understanding NMR Spectroscopy&lt;/em&gt;;S Wiley-Blackwell: Sussex, 2010.    &lt;/li&gt;&#xA;&lt;li&gt;Claridge,  T. D. W. ; &lt;em&gt;High Resolution NMR Techniques in Organic Chemistry&lt;/em&gt;; Elsevier: Oxford, 2009&lt;/li&gt;&#xA;&lt;li&gt;Levitt, M. H.; &lt;em&gt;Spin Dynamics: Basics of Nuclear Magnetic Resonance&lt;/em&gt;; Wiley: Sussex, 2008&lt;/li&gt;&#xA;&lt;li&gt;Macomber, R. S.; &lt;em&gt;A Complete Introduction to Modern NMR Spectroscopy&lt;/em&gt;; Wiley-Interscience: New York, 1998&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastEditorUserId="21328" LastEditDate="2018-01-25T05:06:28.613" LastActivityDate="2018-01-25T05:06:28.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="596" PostTypeId="4" CreationDate="2012-06-05T06:57:23.767" Score="0" Body="This tag should be applied to all questions about nuclear magnetic resonance, including the underpinning theory and the measurement or interpretation of spectra." OwnerUserId="22" LastEditorUserId="16683" LastEditDate="2017-08-21T11:43:14.770" LastActivityDate="2017-08-21T11:43:14.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="597" PostTypeId="1" AcceptedAnswerId="611" CreationDate="2012-06-05T13:21:48.333" Score="5" ViewCount="18391" Body="&lt;p&gt;I've seen a few videos of persons creating plastics from starch by adding an acid, glycerol, and water, however no explanation of the chemistry. &#xA;Does anyone know what $\ce{(C6H10O5)_{n} + C3H8O3 + C2H4O2 + H2O -&amp;gt; (C3H4O2)_{n} + (what)}$?&#xA;or does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have a basic understanding of organic chemistry, so sorry if the equations don't make sense. &lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastEditorUserId="189" LastEditDate="2012-06-05T19:17:20.027" LastActivityDate="2012-06-07T04:38:56.487" Title="The Chemistry behind creating Polylactic Acid (PLA)" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="598" PostTypeId="2" ParentId="593" CreationDate="2012-06-05T13:32:13.357" Score="17" Body="&lt;h2&gt;For Structures and Energies: Answer Hazy, Try Again Later&lt;/h2&gt;&#xA;&lt;p&gt;People can talk on this subject for literally days.&lt;/p&gt;&#xA;&lt;p&gt;It &lt;em&gt;mostly&lt;/em&gt; comes down to computational expense &lt;em&gt;vs&lt;/em&gt; accuracy - if you have a larger system, you can't afford the fancier functionals. The absolute highest accuracy state of the art functional-wise would probably be those using the Random-Phase Approximation (&lt;a href=&quot;http://prb.aps.org/abstract/PRB/v70/i24/e245115&quot; rel=&quot;noreferrer&quot;&gt;e.g.&lt;/a&gt;), and also the use of more complex &lt;a href=&quot;http://www.cecam.org/workshop-411.html&quot; rel=&quot;noreferrer&quot;&gt;range separation techniques&lt;/a&gt; (such as the Coulomb-Attenuated Method mentioned below).&lt;/p&gt;&#xA;&lt;p&gt;Having said that, a lot of people just use B3LYP for everything. B3LYP is parametrised for low-&lt;em&gt;Z&lt;/em&gt; main-group chemistry, and so does reasonably well there. It's a hybrid functional (i.e. it includes the exact Hartree-Fock form exchange operator), so it's not the cheapest. I tend to use the series of PBE, PBE0, TPSS - their only parameters are physical constants, so they should never give surprisingly bad results.&lt;/p&gt;&#xA;&lt;p&gt;A couple of basic guidelines, though (which should be taken as &amp;quot;as far as I know&amp;quot;s):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you have a delocalised system, you want exact exchange (so, a hybrid functional like PBE0, B3LYP, M06-2X, &lt;em&gt;etc&lt;/em&gt;). Without this, most functionals will over-delocalise the orbitals.&lt;/li&gt;&#xA;&lt;li&gt;If you have a system in which dispersion forces are important, worry. Then use something like &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;B97X-D which includes empirical dispersion corrections. Some packages can also apply &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.20495/abstract&quot; rel=&quot;noreferrer&quot;&gt;Grimme's empirical dispersion corrections&lt;/a&gt; to any functional.&lt;/li&gt;&#xA;&lt;li&gt;If you have a periodic system, you probably want to use screened exchange rather than full exact exchange - something like HSE (which tails off the exact exchange past a certain range) rather than PBE0 should be faster while not significantly less accurate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;For TDDFT: Probably CAM-B3LYP&lt;/h2&gt;&#xA;&lt;p&gt;Fortunately, this is a bit easier.&lt;/p&gt;&#xA;&lt;p&gt;Firstly, you want something that incorporates exact exchange (a hybrid functional), which quickly rules out many functionals. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ct600282k&quot; rel=&quot;noreferrer&quot;&gt;Pure GGAs and LDAs give qualitatively wrong answers for states involving charge transfer.&lt;/a&gt; Secondly, functionals that include long-range exchange corrections, such as CAM-B3LYP, &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2006/CP/B511865D&quot; rel=&quot;noreferrer&quot;&gt;also give better results.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, look out for &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ct200651r&quot; rel=&quot;noreferrer&quot;&gt;triplet instabilities&lt;/a&gt;, which can be a problem with the introduction of HF exchange.&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.dur.ac.uk/chemistry/staff/profile/?id=202&quot; rel=&quot;noreferrer&quot;&gt;David Tozer&lt;/a&gt; is my go-to author for TDDFT testing, though I would warn that despite the above answer, I've never used TDDFT in &amp;quot;production&amp;quot;.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(Now, bring on the functional flamewar.)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-10-02T20:16:45.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="599" PostTypeId="1" AcceptedAnswerId="620" CreationDate="2012-06-05T21:55:05.290" Score="7" ViewCount="1235" Body="&lt;p&gt;We're interested in purifying a protein that has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Azide&quot; rel=&quot;nofollow noreferrer&quot;&gt;azide&lt;/a&gt; moiety using ion exchange chromatography. Due to its unique structure, the moiety is a Zwitterion with a novel chemical behavior. My question is how will the molecule behave when I run it through an ion exchange column? Will it be considered positively charged, negatively charged, or neutral or is it much more complicated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/6c/Azide-2D.png/320px-Azide-2D.png&quot; alt=&quot;Azide functionality&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.337" LastActivityDate="2013-08-21T21:57:04.190" Title="What is the behavior of a Zwitterion during ion exchange chromatography?" Tags="&lt;ions&gt;&lt;experimental-chemistry&gt;&lt;purification&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="600" PostTypeId="1" AcceptedAnswerId="601" CreationDate="2012-06-06T00:24:15.123" Score="15" ViewCount="29974" Body="&lt;p&gt;In my teaching-lab experiments I've seen that nitric acid solutions are able to facilitate the dissolution of metals such as silver, even though they are more active than hydrogen. Does anyone know why nitric acid is so special? Why is it such a strong oxidizing agent?&lt;/p&gt;&#xA;" OwnerUserId="314" LastEditorUserId="4945" LastEditDate="2018-10-22T12:58:52.653" LastActivityDate="2018-10-22T22:41:16.577" Title="Why is nitric acid such a strong oxidizing agent?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="601" PostTypeId="2" ParentId="600" CreationDate="2012-06-06T01:17:23.913" Score="15" Body="&lt;p&gt;Because, unlike in other metal dissolution reactions, the &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; isn't reduced&amp;mdash;the &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt; is. The following data and balanced reactions are taken from &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitric_acid#Reactions_with_metals_and_strong_oxidizing_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align} &#xA;  \ce{NO3- + 4 H+ + 3 e- &amp;amp;-&amp;gt; NO  + 2 H2O} &amp;amp;&#xA;    E^\circ_\mathrm{red} &amp;amp;= \pu{0.96 V} \\&#xA;  \ce{NO3- + 2 H+ +   e- &amp;amp;-&amp;gt; NO2 +   H2O} &amp;amp;&#xA;    E^\circ_\mathrm{red} &amp;amp;= \pu{0.79 V} \\&#xA;  \ce{Ag+(aq) + e-       &amp;amp;-&amp;gt; Ag(s)} &amp;amp;&#xA;    E^\circ_\mathrm{red} &amp;amp;= \pu{0.799 V}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the standard reduction potential (SRP) of the &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; reaction &lt;em&gt;looks&lt;/em&gt; smaller than that of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; (I may be wrong, but by significant digits it can't be greater than &lt;span class=&quot;math-container&quot;&gt;$0.799$&lt;/span&gt;), one can conclude that it's the &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; reaction that's occurring here. And the &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; reaction has a large enough SRP to oxidise &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since nitrogen shows many oxidation states. So the simple reason for why &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; is so strong an oxidising agent (with respect to other acids) is that it has a different, &lt;em&gt;better&lt;/em&gt; path available to it to get reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the exact reduction path (i.e. final reduction products/oxidation state) depends upon the concentration of nitric acid&amp;mdash;so much that copper can be oxidised in three different ways. &lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="4945" LastEditDate="2018-10-22T13:04:43.417" LastActivityDate="2018-10-22T13:04:43.417" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="602" PostTypeId="2" ParentId="330" CreationDate="2012-06-06T02:40:23.073" Score="16" Body="&lt;p&gt;The reason for this is mainly because of a factor called &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleation&quot;&gt;Nucleation&lt;/a&gt;. Although it is not exactly obvious what is the source of the site for the nucleation to occur, what is clear is that there are sites present on ice-cream that are not present on ice alone. The carbon dioxide (CO$_2$) in the drink nucleates and forms bubbles (sometimes even a frothy foam) on the drink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several possibilities some of which are discussed in the excellent answer by Ashu. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I list some of them here starting with what I feel is the most likely. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Air in ice-cream&lt;/li&gt;&#xA;&lt;li&gt;Ice crystals in ice-cream&lt;/li&gt;&#xA;&lt;li&gt;Other chemicals found in the ice-cream&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="128" LastActivityDate="2012-06-06T02:40:23.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="603" PostTypeId="2" ParentId="569" CreationDate="2012-06-06T04:56:05.737" Score="10" Body="&lt;p&gt;I've never seen this reaction, but it sounds quite entertaining. And I learned a couple of new words in looking at the literature to explain this phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intumescence: The property of a material to swell when heated. Borax is intumescent.&lt;/li&gt;&#xA;&lt;li&gt;&quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Calcination&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calcination&lt;/a&gt; is a thermal treatment process in presence of air applied to ores and other solid materials to bring about a thermal decomposition, phase transition, or removal of a volatile fraction. The calcination process normally takes place at temperatures below the melting point of the product materials.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The mechanism for the calcination of borax quite complex and is covered in some detail in &lt;a href=&quot;http://journals.tubitak.gov.tr/chem/issues/kim-02-26-1/kim-26-1-10-0001-2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. Borax decahydrate (BDH), $\ce{Na2B4O7.10H2O}$ is easily dehydrated to Borax pentahydrate (BPH), $\ce{Na2B4O7.5H2O}$ without much visible change. The actual form of the borate ion in both BDH and BPH is $\ce{B4O5(OH)4^2-}$ with 2 of the water molecules actually incorporated as hydroxyl groups in the ion. (The other 2 hydroxyl groups come for the water hydrogens and 2 oxygens from the $\ce{B4O7^2-}$.) That means that the last two water molecules can only be removed by molecular decomposition. So in BPH, you actually have only 3 &quot;free&quot; (i.e., easy to remove) water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what happens. The borax decahydrate dehydrates smoothly with minimal density changes (no swelling) until you get to borax pentahydrate. At this point, the BPH dissolves  in its own water of dehydration and the &quot;blossoming&quot; you see is from the solution erupting through the partially dehydrated crystal surface as it's heated. The density of the borax changes dramatically as it goes from the pentahydrate (about $0.85\ \mathrm{g/cm^3}$) to the dihydrate  (about $0.14\ \mathrm{g/cm^3}$) which is why it ends up looking &quot;whipped&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the final transformation to a tiny bead, the remaining 2 waters of hydration (those tied up as hydroxide ions) are driven off as the $\ce{B4O5(OH)4^2-}$ decomposes to give metaborate, $\ce{BO2^2-}$ and boric anhydride, $\ce{B2O3}$. $$\ce{B4O5(OH)4^2- -&amp;gt; 2 BO2 + B2O3 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fascinating.&lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorDisplayName="user7951" LastEditDate="2018-08-18T19:36:36.180" LastActivityDate="2018-08-18T19:36:36.180" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="605" PostTypeId="1" CreationDate="2012-02-17T07:34:36.527" Score="23" ViewCount="341" Body="&lt;p&gt;I know how to measure the concentration of free thiols, which is reflective of the redox-potential of cellular compartment such as the cytoplasm or a lysosomes.       &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What other methods exist?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="57" OwnerDisplayName="bobthejoe" LastEditorUserId="15429" LastEditDate="2017-05-26T03:02:01.293" LastActivityDate="2017-05-26T03:02:01.293" Title="What is a good method to measure the redox potential of a cellular system?" Tags="&lt;biochemistry&gt;&lt;redox&gt;&lt;organosulfur-compounds&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="606" PostTypeId="1" AcceptedAnswerId="610" CreationDate="2012-06-06T15:25:24.117" Score="13" ViewCount="8150" Body="&lt;p&gt;I have some activated charcoal pads or small &quot;pillows&quot; that can be used to adsorb odors and chemical fumes from different materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instructions say to put them out in sunlight to reactivate them once they have been used for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism whereby sunlight is able to reactivate the charcoal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried googling but could not find a definitive answer. I ended up assuming it must be just heat that forces the gasses trapped in the charcoal to come loose and evaporate, thereby restoring some of the adsorption potential. Is that the mechanism or what is the sorcery going on here?&lt;/p&gt;&#xA;" OwnerUserId="319" LastEditorUserId="319" LastEditDate="2018-08-10T22:03:48.803" LastActivityDate="2019-03-19T18:48:43.257" Title="Activated Carbon - Desorption in sunlight?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="607" PostTypeId="2" ParentId="429" CreationDate="2012-06-06T16:35:21.397" Score="4" Body="&lt;p&gt;&lt;p&gt;Why do Magnesium and Lithium form &lt;em&gt;covalent&lt;/em&gt; organometallic compounds? I think this question touches a more general question in chemistry: Why do some metallic elements form 'covalent' bonds at all? And the other version of this question would be why sodium or potassium does not usually form covalent bonds. The answer for these questions would become long and winding very easily since it's going to touch on how electrons move and shift in forming a chemical bond, covalent or otherwise. &#xA;  &lt;/p&gt;&lt;p&gt;So I think a similar question that can have an easier answer would be: When does an element form 'covalent' bond with another element? Answer: If the electronegativities of the two elements differ by less than 1.9, the formed bond would be covalent. But then again, the answer is not absolute and serves only as a guideline. The case in point is Na-C bond, which has a delta(Eneg)=1.6 but is considered an ionic bond.&#xA;  &lt;/p&gt;&lt;p&gt;Back to the original question. Why do Magnesium and Lithium form &lt;em&gt;covalent&lt;/em&gt; organometallic compounds? A hand-waving answer would be that in forming a Li-C or Mg-C bond the electrons are not moved all the way from metal to carbon, but only partially to somewhere between the metal and carbon.&#xA;  &lt;p&gt;I hope these had touched the topic a little bit.&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2012-06-06T16:35:21.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="608" PostTypeId="1" AcceptedAnswerId="614" CreationDate="2012-06-06T19:15:55.000" Score="38" ViewCount="58708" Body="&lt;p&gt;What are the advantages and disadvantages of the various types of condensers commonly found in the laboratory? Obviously more intricate pieces of glassware are more costly but assuming they are all available why would one use a Dimroth condenser vs. a Friedrichs when refluxing something? I've heard that Graham condensers are to be avoided when refluxing due to the possibility of clogging, and yet it is still very common - when is it appropriate to use one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0N2T6.png&quot; alt=&quot;Different varieties of condensers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustration made with ChemDraw. Feel free to reference other designs but please attach a schematic for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit&lt;/em&gt;&lt;/strong&gt;: This question was motivated by an organic preparation which involved the bromination of an alkene in boiling water. Bromine has a boiling point of 58.8 &amp;deg;C and on top of that the reaction was exothermic. It was difficult to avoid the loss of Br&lt;sub&gt;2&lt;/sub&gt; gas with your typical Allihn condenser, but the increased cooling capacity provided by a Friedrichs returned Br&lt;sub&gt;2&lt;/sub&gt; back to the flask as it was produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while the Friedrichs forces vapors up a spiral path, the path itself is wide, and in my condenser at least, there was a bit of leeway for liquid to drop down the sides, helping to prevent blockages. I didn't try a Graham condenser but I imagine a much slower rate of addition would be supported by this condenser.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="74" LastEditDate="2012-06-13T14:18:16.443" LastActivityDate="2020-08-04T09:55:43.680" Title="Choosing the right condenser" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="609" PostTypeId="1" AcceptedAnswerId="612" CreationDate="2012-06-06T19:37:29.160" Score="8" ViewCount="191" Body="&lt;p&gt;I'm running a reaction that has ethylene gas as a byproduct. I have observed gas evolution as my reaction proceeded and I have other ways of verifying the identity of the main product, but I thought it might be cool if I could also test for the presence of ethylene.&lt;br&gt;&#xA;Are there any simple tests I can use for this purpose? Obviously it is not easy to detect since it's a fairly non-reactive gas, so it doesn't have to have 100% specificity.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="4945" LastEditDate="2014-07-30T05:12:48.400" LastActivityDate="2014-07-30T05:12:48.400" Title="How to characterize gaseous ethylene that forms as byproduct?" Tags="&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="610" PostTypeId="2" ParentId="606" CreationDate="2012-06-06T21:12:32.777" Score="11" Body="&lt;p&gt;Your guess is right: the mechanism of charcoal regeneration is undoubtably &lt;strong&gt;temperature-induced desorption&lt;/strong&gt;. Gas molecules are adsorbed on the large surface area of activated carbon by a mechanism of &lt;a href=&quot;http://en.wikipedia.org/wiki/Physisorption&quot;&gt;physisorption&lt;/a&gt; (or &lt;em&gt;physical sorption&lt;/em&gt;) caused by weak nonspecific interactions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot;&gt;van der Waals forces&lt;/a&gt;) between gas molecules and the carbon surface. Unlike &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemisorption&quot;&gt;chemisorption&lt;/a&gt;, this process is reversible and, since it involves only weak interactions, it can actually be reversed by relatively small increases of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia gives a &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot;&gt;specific example&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The activated carbon can adsorb a large amount of methanol vapours in ambient temperature and desorb it at a higher temperature (around 100 degrees Celsius). During the daytime, the sunshine irradiates the collector, so the collector is heated up and the methanol is desorbed from the activated carbon. In desorption, the liquid methanol adsorbed in the charcoal heats up and vaporizes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will surely notice the resemblance between this text and your case.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-06-06T21:12:32.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="611" PostTypeId="2" ParentId="597" CreationDate="2012-06-07T04:19:44.683" Score="14" Body="&lt;p&gt;Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some &lt;a href=&quot;http://www.smithsonianmag.com/science-nature/plastic.html?c=y&amp;amp;page=2&quot; rel=&quot;noreferrer&quot;&gt;controversy&lt;/a&gt; over the wisdom of this approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I've found a site that takes the approach you're basing your question on...but there are a couple of issues with the equation you've written. So, I'll answer this question in two parts. &lt;strong&gt;Plastic from starch&lt;/strong&gt; is what you saw in the video; &lt;strong&gt;PLA from starch&lt;/strong&gt; is what you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plastic from starch&lt;/strong&gt;. &lt;a href=&quot;http://green-plastics.net/discussion/54-student/131-qaa-help-with-cornstarch-pla-plastic-project&quot; rel=&quot;noreferrer&quot;&gt;The recipe&lt;/a&gt; for the plastics-from-starch uses starch, glycerol, acetic acid (the $\ce{C2H4O2}$ in your equation) and water as a solvent. (They do not add lactic acid, the $\ce{C3H8O3}$ in your equation, but lactic acid can be produced by hydrolysis of the starch.) With the starch-acetic acid-glycerol recipe, you do not actually make a PLA polymer, but a starch polymer that is somewhat like it instead. (&lt;a href=&quot;http://green-plastics.net/discussion/54-student/132-qaa-i-want-to-make-corn-plastic-utensils&quot; rel=&quot;noreferrer&quot;&gt;Ref&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, starch is a polymer that &lt;a href=&quot;http://djbreile.webs.com/reactionmechanism.htm&quot; rel=&quot;noreferrer&quot;&gt;looks like this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6OxNL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water, it tends to ball up into globular clusters due to hydrogen bonding. By adding acetic acid and heating the starch in water, you &quot;denature&quot; the starch and turn it into &quot;disordered chains&quot; that stretch out. When you let these disordered chains dry, they become entangled and form a flexible film, the plastic. In this recipe, the glycerol is included as a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasticizer&quot; rel=&quot;noreferrer&quot;&gt;plasticizer&lt;/a&gt;, so it is unchanged chemically in the reaction. The glycerol also helps to keep the starch stretched out by stabilizing it with hydrogen bonding. Plasticizers (as the name would indicate) are additives that increase the plasticity of a plastic and make it more durable or flexible. Changing the amount of glycerol in the recipe can have a big effect on the strength of the plastic film you make. This is the process you're seeing in the videos, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Making PLA from starch.&lt;/strong&gt; This is going to be a bit of a challenge to explain depending on your understanding of organic chemistry, but hopefully the pictures will help. When you make PLA commercially, you &lt;a href=&quot;http://www.springerlink.com/content/v14uxt2r823glaex/&quot; rel=&quot;noreferrer&quot;&gt;digest starch, usually with a bacteria&lt;/a&gt;, to turn it into lactic acid (see picture below for the structure of lactic acid). This lactic acid must be purified to make a high quality product. If you just combine a bunch of lactic acid molecules to make PLA, one water molecule would be lost each time they combine...sort of &lt;a href=&quot;http://www.materialsworldmodules.org/resources/polimarization/4-condensation.html&quot; rel=&quot;noreferrer&quot;&gt;like this&lt;/a&gt;. (This is an acid and alcohol combining, not two acids, but the idea is the same.) The water in the polymerization reaction leads to a poor quality plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/otkLQ.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in commercial synthetic schemes, two purified lactic acid molecules are reacted to form lactide...they come together in a controlled condensation reaction, with the loss of water. (&lt;a href=&quot;http://wiki.chemprime.chemeddl.org/index.php/BIODEGRADABLE_PLASTICS&quot; rel=&quot;noreferrer&quot;&gt;Ref for picture&lt;/a&gt;.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M1oaj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can use the lactide, i.e. &quot;lactic acids with the water removed&quot;, to do the polymerization to PLA. You add an initiator to open the lactic acid by breaking the bond between the $\ce{O-C}$ in  $\ce{O-C=O}$ and this gives you 2 lactic acid molecules hooked together...which react with another lactide molecule to add two more lactic acid molecules to the growing chain, and so on. And that's how the PLA forms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reaction sequence is strach-&gt; lactic acid-&gt; lactide + $\ce{H2O}$ -&gt; PLA &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably more than you wanted to know! &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-06-07T04:38:56.487" LastActivityDate="2012-06-07T04:38:56.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="612" PostTypeId="2" ParentId="609" CreationDate="2012-06-07T09:46:42.070" Score="9" Body="&lt;p&gt;If you want something simple and don't mind less than 100% certainty then you could bubble the gas through Bromine Water - in the presence of an alkene the bromine water would be decolourised from brown when reacting with the ethene.  Of course this does just indicate the presence of any alkene so isn't exactly specific! There is more about this on &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/alkenes/halogenation.html&quot;&gt;this website&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="118" LastActivityDate="2012-06-07T09:46:42.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="613" PostTypeId="1" CreationDate="2012-06-07T16:44:38.413" Score="14" ViewCount="3836" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kevlar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kevlar&lt;/a&gt; is an aramid fiber that is used in ballistics protection vests. From the structure (structure shown from Wikipedia article), you can see why it would work to dissipate the energy from a bullet since there is an extensive network of interactions between the linear chains. The wearer ends up bruised from the force, but unpunctured and avoids serious injury from handguns bullets.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4OPr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More modern &lt;a href=&quot;http://en.wikipedia.org/wiki/Ballistic_vest&quot; rel=&quot;nofollow noreferrer&quot;&gt;ballistics protection&lt;/a&gt; devices use ceramic plates (along with Kevlar) to give greater stab resistance and protection from higher velocity rifle rounds. My question is, what ceramics do they use? Do they have a &quot;flexible&quot; ceramic that will deform to absorb the energy and not break? Or do the plates work by breaking to absorb the energy and, thus, have to be replaced after each event? &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorDisplayName="user7951" LastEditDate="2016-12-30T01:22:27.077" LastActivityDate="2016-12-30T01:22:27.077" Title="Ceramics in &quot;bullet proof&quot; vests?" Tags="&lt;inorganic-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="614" PostTypeId="2" ParentId="608" CreationDate="2012-06-08T01:05:54.093" Score="26" Body="&lt;p&gt;@Mart 's comment impelled me to return to this question and correct my answer. I've deleted incorrect material and expanded the discussion to, hopefully, provide correct information. There is a good discussion (better than the reference previously cited) of the issue &lt;a href=&quot;http://www.sciencemadness.org/talk/viewthread.php?tid=14240&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.chem.wisc.edu/areas/organic/orglab/tech/reflux.htm&quot; rel=&quot;noreferrer&quot;&gt;Reflux&lt;/a&gt; is the process of boiling reactants while continually cooling the vapor returning it back to the flask as a liquid. It is used to heat a mixture for extended periods and at certain temperatures...A condenser is attached to the boiling flask, and cooling water is circulated to condense escaping vapors. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are refluxing a mixture, as you might in organic synthesis to increase the speed of the reaction by doing it at a higher temperature (i.e., the boiling point of the solvent), then any of the condensers that worked well enough to avoid the loss of solvent and avoid &quot;flooding&quot; would work equally well. When you're refluxing, you want the &quot;reflux ring&quot;, the place where the vapor is visibly condensing into a liquid, to be no more than 1/3 of the way up the reflux column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two different basic types of condensers shown, Graham-type condensers (the first 3) and coil condensers (the last two). In the coil condensers (the left condenser in the picture below), the water flows through the coil and the vapor moves up in the larger, outside  area of the condenser, condenses onto the cooled coils, then drips back into the pot. In a Graham-type condenser (the right condenser in the picture below), the water flows around a tube (whether straight or coiled) that contains the vapor/condensed liquid.(&lt;a href=&quot;http://www.sciencemadness.org/talk/viewthread.php?tid=14240&quot; rel=&quot;noreferrer&quot;&gt;picture source&lt;/a&gt;) The Graham-type condensers clog (or flood) more easily since they have a more restricted path for the liquid to return to the pot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PVM3p.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Graham-type condensers:&lt;/strong&gt; The Liebig condenser is simple, but has low cooling capacity and can be fairly easily clogged as the condensed liquid flows back into the flask and blocks the vapor that is trying to escape. The Allihn improves on this design by having a wider bore at the bottom and condensing the liquid on the &quot;bubbles&quot; where it can run down the sides and avoid blocking the vapor. (I've used this to good effect in refluxing many reactions.) The Graham condenser is the same basic design as the other two, but the condensation tube is coiled which provides more surface area for cooling...but also tends to send the condensed liquid right into the path of the vapor trying to move up. It is particularly prone to flooding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coil condensers&lt;/strong&gt;, such as the Dimroth and Freidrichs, have high capacity for cooling with fewer problems from flooding since the vapor condenses on the coils and drips back from the little prominence at the bottom of the coils into the center of the pot. The vapor has an easy time getting past the drops falling into the pot. If you can afford it, this seems like a good choice for most applications. The Freidrichs condensers, which incorporate a cold-finger with the spiral, are higher capacity, quite bulky and heavy. I have seen them used with rotovaps where you are taking a lot of solvent off quickly, but not with an ordinary reflux apparatus. This would be over-kill for a simple reflux reaction situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the incorrect information (for those of you who looked at this before) and hope this is helpful. &lt;/p&gt;&#xA;" OwnerUserId="133" LastEditorUserId="133" LastEditDate="2012-06-10T22:39:39.923" LastActivityDate="2012-06-10T22:39:39.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="615" PostTypeId="2" ParentId="441" CreationDate="2012-06-08T02:30:02.613" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;nofollow&quot;&gt;OpenBabel&lt;/a&gt; using SMARTS. &lt;a href=&quot;http://www.pymol.org/&quot; rel=&quot;nofollow&quot;&gt;pymol&lt;/a&gt; also does it.&lt;/p&gt;&#xA;" OwnerUserId="321" LastEditorUserId="30" LastEditDate="2012-06-11T10:05:36.280" LastActivityDate="2012-06-11T10:05:36.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="616" PostTypeId="2" ParentId="294" CreationDate="2012-06-08T02:36:47.860" Score="-2" Body="&lt;p&gt;You can use SMILES in Avogadro or openbabel and it produces 3D coordinates that&#xA;you can edit. And you can create Gaussian, NWChem, gamess... inputs.&lt;/p&gt;&#xA;" OwnerUserId="321" LastActivityDate="2012-06-08T02:36:47.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="617" PostTypeId="2" ParentId="613" CreationDate="2012-06-08T03:32:17.677" Score="12" Body="&lt;p&gt;As far as I can tell there are two material types used for this - pure ceramics or ceramic mixtures that are designed to break and be replaced, and composite metal + ceramic materials that are designed to be able to handle a couple of rounds without losing too much effectiveness. There are also techniques involving purely metal layers in overlapping structures with fibers of materials like Kevlar or ultra-high-molecular-weight polyethene (&lt;a href=&quot;http://en.wikipedia.org/wiki/UHMWPE&quot;&gt;UHMWPE&lt;/a&gt;) (of which Spectra is a brand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shopping around, it seems the best materials are boron and silicon carbides (&lt;em&gt;extremely&lt;/em&gt; hard materials, though not quite as hard as diamond), which can be formed into plates with multiple levels of structure by starting with hexagonal plates and forming larger plates from those. In addition to personal armour, the same materials and techniques are used for military vehicle (including aircraft) armour. In the ceramic/metal composite materials, you get ceramic plates formed up within a metal sheet, where a ceramic element will shatter to take most of the impact, while the rest will be absorbed by the metal matrix. The metal matrix then remains intact (but may be deformed) so that the armour is still useful but weakened around the area of impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good places to start from are the Wikipedia pages for &lt;a href=&quot;http://en.wikipedia.org/wiki/Chobham_armour&quot;&gt;Chobham armour&lt;/a&gt; (the common name for the tank armour that includes composite metal/ceramic matrix materials), and &lt;a href=&quot;http://en.wikipedia.org/wiki/Small_Arms_Protective_Insert&quot;&gt;Small Arms Protective Insert&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-06-08T03:32:17.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="618" PostTypeId="2" ParentId="577" CreationDate="2012-06-08T07:36:52.600" Score="5" Body="&lt;p&gt;In the future please specify where your equation came from, justify its use and specify all of your parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When only a concentration gradient is present, the diffusion coefficient, $D$, is typically described by Fick's first law of diffusion. To determine $D$ in these terms we must know the flux, $j$, and the gradient, $d c / dx $.  $c=$ concentration of our species divided by the volume and $x$ is the distance this occurs over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux, $j$, is the (flow quantity/time)/area or (flow rate)/area. Fluxes depending on potential gradients are molecular, but bulk flow flux is called convective. Here we have molecular flux present and the units of this are kmol/(m$^2 \cdot$ s). (It's true that diffusion creates its own convection but there is no bulk flow so we ignore this diffusion-induced convection.) Properly our flux is the product of a transport property by the gradient of the driving force, $ j = (D) ($Gradient of driving force$)$. Note that $D$ has units of m$^2$/s. For example, $D$ is $10^{-5}$, $10^{-9}$, $10^{-10}$ m$^2$/s for some gases, liquids and solids respectively at STP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's think back to basic physics. Say we have a feather at $h$ height here on Earth and we drop it. It will experience an acceleration due to gravity, $g$. Remember that gravity has units m/s$^2$. We can estimate the time it takes to travel distance $h$ crudely by dimensional analysis: knowing that $g$ has units of m/s$^2$ and $h$ has units of m then time is simply $\sqrt{\frac{h}{g}}$. In fact the equation that describes this is $h = \frac{1}{2}gt^2$ and so we actually needed to multiply our original answer by $\sqrt{2}$. In the same way we can crudely estimate the time by dimensional analysis on the diffusion coefficient, $D$ with units m$^2$/s and going a distance with units of m. So we need to take the distance$^2$ divided by $D$ to get seconds. Say ( $0.01$m )$^2$ divided by $10^{-9}$ m$^2$/s gives $10^5$ seconds which is about 1.1 days. This could easily be 4 days if  we needed to multiply by 3.7 which is certainly possible if your orifice is small. It is calculating these other things which turns out to be tricky. Although it's obvious &lt;strong&gt;the single easiest thing you can do is switch to a gas&lt;/strong&gt; when considering just diffusion because a gas would take &lt;em&gt;on the order of&lt;/em&gt; 10 seconds because $0.01^2/(10^{-5}) = 10$. &lt;strong&gt;You can also change the time by introducing another flux, such as by stirring the solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better answer requires differential equations. Are you sure you want to venture down that road?&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2012-06-08T17:30:44.017" LastActivityDate="2012-06-08T17:30:44.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="619" PostTypeId="7" CreationDate="2012-06-08T08:26:47.983" Score="0" Body="&lt;p&gt;&#xA;    Chemistry - Stack Exchange is for scientists, academics, teachers and students of &lt;b&gt;chemistry&lt;/b&gt;. &#xA;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions may be of any level, but should be of the following types:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Questions asking for explanation of a chemistry concept&lt;/li&gt;&#xA;&lt;li&gt;Questions relating to observed chemical phenomena&lt;/li&gt;&#xA;&lt;li&gt;Questions about experimental techniques and technology&lt;/li&gt;&#xA;&lt;li&gt;Questions about nomenclature, standards, et cetera pertaining to chemistry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some kinds of questions aren't allowed here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Do-my-homework questions:&lt;/strong&gt; Homework questions are OK, but they must follow &lt;a href=&quot;http://meta.chemistry.stackexchange.com/a/142/22&quot;&gt;these guidelines&lt;/a&gt;. Please don't ask &quot;do my homework for me&quot; type questions-- we only clear conceptual doubts in homework questions and will not do your work for you.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Pitches for your own personal theories or work:&lt;/strong&gt; We deal with mainstream chemistry here. Anything that couldn't be published in a reputable journal is probably not appropriate at this site.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Computational questions:&lt;/strong&gt; If your question is purely about numerical methods you are using in a simulation/etc, it is probably more appropriate at &lt;a href=&quot;http://scicomp.stackexchange.com/&quot;&gt;Computational Science&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;What notation and symbols are commonly used here?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Chemistry relies heavily on math. Many posts include mathematical notation, written using LaTeX commands and delimited by dollar signs. If you see dollar signs with what appears to be gibberish between them, it just means that MathJax, the math rendering we use, isn't working properly for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some basics for people new to LaTeX. Inline formulas are surrounded by single dollar signs (&lt;code&gt;$f(x) = ax^2 + bx + c$&lt;/code&gt; renders as  $f(x) = ax^2 + bx + c$), and block formulas by double dollar signs so that &lt;code&gt;$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$&lt;/code&gt; renders as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For chemical equations, one can use the &lt;code&gt;\ce{...}&lt;/code&gt; function from the &lt;a href=&quot;http://www.ctan.org/pkg/mhchem&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;mhchem&lt;/code&gt;&lt;/a&gt; package. This formats chemical formulae and reactions pretty well:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; $\ce{HCl}$ dissociates in water as follows:&#xA; $$\ce{H2O +HCl&amp;lt;=&amp;gt;H3O+ +Cl-}$$&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$ dissociates in water as follows:&#xA;$$\ce{H2O +HCl&amp;lt;=&amp;gt;H3O+ +Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://meta.chemistry.stackexchange.com/a/88/22&quot;&gt;More details on math formatting&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="22" LastEditDate="2013-04-10T17:09:33.227" LastActivityDate="2013-04-10T17:09:33.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="620" PostTypeId="2" ParentId="599" CreationDate="2012-06-08T15:07:34.350" Score="5" Body="&lt;p&gt;I would expect it to depend on the pH of the mobile phase. The charge of your stationary phase and whether you want the azide to elute will determine what charge you want on the azide. Given that, you should set the pH of the mobile phase either acidic or basic enough to ensure that the azide is fully charged in the direction that you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_chromatography#Separating_proteins&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proteins have numerous functional groups that can have both positive and negative charges. Ion exchange chromatography separates proteins according to their net charge, which is dependent on the composition of the mobile phase. By adjusting the pH or the ionic concentration of the mobile phase, various protein molecules can be separated. For example, if a protein has a net positive charge at pH 7, then it will bind to a column of  negatively-charged beads, whereas a negatively charged protein would not. By changing the pH so that the net charge on the protein is negative, it too will be eluted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If changing the pH is problematic, apparently increasing the ionic strength can help shield the azide's charge from the column's charge:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Elution by changing the ionic strength of the mobile phase is a more subtle effect - it works as ions from the mobile phase will interact with the immobilized ions in preference over those on the stationary phase. This &quot;shields&quot; the stationary phase from the protein, (and vice versa) and allows the protein to elute.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2013-08-21T21:57:04.190" LastActivityDate="2013-08-21T21:57:04.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="621" PostTypeId="1" AcceptedAnswerId="625" CreationDate="2012-06-08T16:26:46.480" Score="9" ViewCount="633" Body="&lt;p&gt;Is it possible (advisable?) to make a sealed tube by melting and twisting the end of a borosilicate test tube? One issue that I can think of is that the pressure might be too high if I then proceed to heat the tube. I thought that I could maybe connect the tube to a vacuum tube and heat it in the middle. Any experience with this?&lt;/p&gt;&#xA;" OwnerUserId="74" LastActivityDate="2012-06-10T06:30:40.827" Title="MacGyvering a sealed tube" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="622" PostTypeId="2" ParentId="425" CreationDate="2012-06-08T18:37:10.580" Score="4" Body="&lt;p&gt;Amines left out in the atmosphere will air-oxidize to give N-oxides, which typically have a yellow color. These N-oxides can decompose to give further byproducts. I would store your sparteine under nitrogen or argon atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit to give examples and mechanism]&#xA;The reaction proceeds through a radical chain mechanism with molecular oxygen. &lt;a href=&quot;http://www.springerlink.com/content/g162275744180331/&quot; rel=&quot;nofollow&quot;&gt;Here is one study of the mechanism&lt;/a&gt; [from first page of Google search for amines air-oxidation] . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Triethylenetetramine&quot; rel=&quot;nofollow&quot;&gt;wikipedia article for triethylenetetramine&lt;/a&gt;. &quot;This oily liquid is colourless but, like many amines, assumes a yellowish color due to impurities resulting from air-oxidation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It can be observed that arylamines develop a yellow or black color if left exposed to air for prolonged periods&quot;. &lt;a href=&quot;http://books.google.com/books?id=35xwBwjhe2MC&amp;amp;pg=PA99&amp;amp;lpg=PA99&amp;amp;dq=air+oxidation+amines&amp;amp;source=bl&amp;amp;ots=Y9jlI9SGmi&amp;amp;sig=0jeeyWwf1Ye2IEblo7_DHU08_cA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=NIr9T42mM-qi2gWqubS-Bg&amp;amp;ved=0CFkQ6AEwBw#v=onepage&amp;amp;q=air%20oxidation%20amines&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Ref&lt;/a&gt; Note - I realize that sparteine is not an aromatic amine, but the result is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to triethylamine or other amines in your lab, do a control experiment: leave one vial open to air, and leave the other under identical conditions except sealed under an inert atmosphere. A few days worth of observation will convince you of the role in oxygen in the yellowing of amines. &lt;/p&gt;&#xA;" OwnerUserId="324" LastEditorUserId="324" LastEditDate="2012-07-11T14:23:58.907" LastActivityDate="2012-07-11T14:23:58.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="623" PostTypeId="2" ParentId="563" CreationDate="2012-06-08T23:52:49.220" Score="1" Body="&lt;p&gt;You might start seeing signal if you set lb=1 (or, drastically, lb=2) and then type &quot;wft&quot; It won't give you a decent spectrum but you may at least see the beginnings of peaks. &#xA;Also, are you confident that this is a 1-H decoupled spectrum? If not, then your S/N is going to be very poor. &lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2012-06-08T23:52:49.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="624" PostTypeId="2" ParentId="621" CreationDate="2012-06-09T02:15:06.213" Score="4" Body="&lt;p&gt;Possible? Doubtless. Advisable? Depends what you're sealing, I'm sure. The pressure won't be a problem though, if anything it'll be a partial vacuum inside and you'd have to heat it near the tube's melting point to get it back to atmospheric pressure.  Consider that as you're heating the tube to seal it, you'll heat a good portion of the air inside as well, and once that cools it will be well below 1atm, only getting back there after heating to the melting point, the temperature at which it was sealed in.&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2012-06-09T02:15:06.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="625" PostTypeId="2" ParentId="621" CreationDate="2012-06-09T09:55:37.250" Score="12" Body="&lt;p&gt;Usual way to work with air sensitive compounds is to work on Schlenk-line (search Wiki). When you transfer your substance you can leave it in Schlenk tube, or if it's really sensitive and you need to keep out of the line for longer period of time you have in general three possibilities: (1) use glassware with J. Young valve (available even for NMR tubes); (2) store your standard Schlenk glass in glove box filled with inert gas and (3) the oldest is to melt your glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now back to your question. Usual way is to prepare glass with narrow place on it prior to doing experiment (that's were later you will melt it). By narrow I mean the hole inside should be ~2-3 mm (thinner glass, smaller hole). I used to use Schlenk lines, so on opened end I have standard 14 female cone that was connected to a vacuum line. I used to set vacuum ~0.1 Torr and with propane/butane/oxygen torch one can easily close it. The huge advantage of this old way is that you can prepare your glass even with 3mm thick walls that allows you to work with substances with very low boiling points (in this case to close the glass one need to freeze the stuff with liquid N2 or if not available at least cool it with dry ice bath). Of course later opening is quite tricky and usually also had to be done after freezing of chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last part if it's advisable. That one is as usually hard to answer. I have no idea how experienced you are. Judging by question not to much (sorry). So I wouldn't start with anything to reactive in the beginning. Generally try to play first with glass and torch and get a bit used to it. Do it on proper chemical glass (not just bottle glass).&#xA;If you are &quot;advanced guy&quot; and you want to make yourself f.e. PfBF2 borane you have to use this technique - any other gives just a miserable yield.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2012-06-09T09:55:37.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="626" PostTypeId="2" ParentId="621" CreationDate="2012-06-10T03:24:15.117" Score="3" Body="&lt;p&gt;It is indeed common, though you don't use standard borosilicate glass as I understand it. Also, it is a technique that needs special training to do right; it can be dangerous to both you and your compound if you don't get said training first. Ask around your local inorganic chem department, I'm sure someone can teach you to do it. &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="22" LastEditDate="2012-06-10T06:30:40.827" LastActivityDate="2012-06-10T06:30:40.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="627" PostTypeId="2" ParentId="466" CreationDate="2012-06-10T09:40:31.383" Score="7" Body="&lt;p&gt;The &lt;em&gt;Pmma&lt;/em&gt; is a minimal non-isomorphic supergroup of &lt;em&gt;Pmm2&lt;/em&gt;; or you can say that &lt;em&gt;Pmm2&lt;/em&gt; is one of maximal non-isomorphic sub-groups of &lt;em&gt;Pmma&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve the question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) fast and dirty: take &lt;em&gt;International Tables for Crystallography vol. A&lt;/em&gt; go to your group (in mine edition &lt;em&gt;Pmma&lt;/em&gt; starts on p.274). On second page describing space group you can find &lt;strong&gt;Maximal non-isomorphic subgroups&lt;/strong&gt;. The third listed is &lt;code&gt;[2] Pmm2 (25) 1; 2; 7; 8&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the operation in  #50 and #25 have a lot of similarities. No.1 you can ignore as it's [x,y,z] but 2, 7 and 8 for #50 are [-x+.5,-y,z], [x,-y,z] and [-x+.5,y,z]. As one can see the #25 has following general operators: (2) [-x,-y,z] (3) [x,-y,z] and (4) [-x,y,z].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The hard way: Mathematics is explain in the same volume in chapter &lt;em&gt;13.1. Isomorphic subgroups&lt;/em&gt;. Of course the knowledge of group theory and matrix calculus is to your advantage ;-)&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="327" LastEditDate="2012-06-10T20:18:16.827" LastActivityDate="2012-06-10T20:18:16.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="628" PostTypeId="2" ParentId="389" CreationDate="2012-06-10T10:22:47.867" Score="3" Body="&lt;p&gt;One of way to answer this question is to put it in some QM software. The system is small enough and has higher symmetry so computations shouldn't take to long, even on an older computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For co-planar geometry the energy calculated at B3LYP/6-31G(d,p) is –492.132206748 a.u.&#xA;The geometry with dihedral angle of 90° between $\ce{CO2}$ and $\ce{C3N2}$ is –492.116719625 a.u. That makes roughly 41 kJ/mol (10 kcal/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you really want to be sure if that's the right energy you should make frequency calculations on both structures to find out if they are minima or transition states. This can take a bit longer (I haven't done it) and I would expect that the better structure (with lower energy) should have no imaginary frequencies and the higher one should have one (TS). If by some accident both of them are minima than TS has probably a dihedral angel somewhere around 45°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is only computation. It gives as clues but not the exact answer. You can go for higher level of theory, add solvent modelling and... or you can take your compound and try to make temperature depending NMR spectra. With a bit of luck you can freeze your TS and find out how high is your barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;p.s.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be good if you mention in your post what tools (in this case computational) are available to you. I really would prefer to explain how to get the number instead of giving just &lt;em&gt;dry&lt;/em&gt; answer. &lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="4231" LastEditDate="2014-10-30T16:31:29.590" LastActivityDate="2014-10-30T16:31:29.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="629" PostTypeId="2" ParentId="441" CreationDate="2012-06-10T20:16:07.780" Score="4" Body="&lt;p&gt;Inorganic structures – that could be proper solid or molecular metalloorgnic. For the later one you can use xp (program written G.M.Shedrick - comes with Bruker diffractometers). If you have running license of Cambridge Crystal Data Base there is program called IsoStar, and I think that it should do the job. I haven't used it my self but apparently Olex2 can make aligning of fragments (this program is free so that's the place were you can start). I'm not quite sure but I think that old GROMOS code could also make a fit of orthogonal coordinates.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2012-06-10T20:16:07.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="630" PostTypeId="2" ParentId="76" CreationDate="2012-06-10T20:40:29.223" Score="6" Body="&lt;p&gt;A similar effect occurs in synthetic polymers. poly(N-isopropyl Acrylamide), (usually just p(NIPAM)) is a thermosensitive polymer. At room temperature, it dissolves easily in water. When you heat the solution past a certain temperature (the &quot;lower critical solution temperature,&quot; or LCST) the polymer chains undergo a conformational change and precipitate out of solution. When you cool the solution back down below the LCST, the polmer returns to its original conformation and re-dissolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp0690603&quot; rel=&quot;noreferrer&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jp0690603&lt;/a&gt; and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja0546424&quot; rel=&quot;noreferrer&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ja0546424&lt;/a&gt; are two papers from a group at Texas A&amp;amp;M that show that the Hofmeister series applies to these thermosensitive polymers in a similar manner. In this case, the ions lower the LCST. So even at room temperature, you could salt out the polymer with enough salt. The Hofmeister series indicates the efficacy of the ions in this salting out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those papers divide the ions in the Hofmeister series into &quot;chaotropes&quot; and &quot;kosmotropes,&quot; depending on whether the ion tends to break apart the natural structure of the water, or tends to stabilize the natural structure of the water. They claim that the chaotropes operate by changing the surface tension of the water and by direct anion binding to the polymer, and that the kosmotropes operate through surface tension and by changing the polarization of the water molecules. They correlate the magnitude of the effect of each ion on LCST with that ion's hydration entropy and surface tension increment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveats:&lt;/strong&gt; I'm only really comfortable with p(NIPAM) itself, and with the empirical fact of the salt effect. I don't know how well-understood these entropies of hydration or surface tension increments are, or if this would apply to salting-out of proteins.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2012-06-12T22:19:39.063" LastActivityDate="2012-06-12T22:19:39.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="631" PostTypeId="1" AcceptedAnswerId="632" CreationDate="2012-06-11T10:02:36.797" Score="5" ViewCount="125" Body="&lt;p&gt;I've read how putting a burning piece of sodium in chlorine would produce white $\ce{NaCl}$ fumes. However, I can't find out whether Na or Li can react with $\ce{Cl2}$ &lt;em&gt;without&lt;/em&gt; heating or burning.&lt;/p&gt;&#xA;" OwnerUserId="328" LastEditorUserId="9961" LastEditDate="2017-01-13T15:01:34.803" LastActivityDate="2017-01-13T15:01:34.803" Title="Do Na or Li need heat in order to react with Cl₂?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="632" PostTypeId="2" ParentId="631" CreationDate="2012-06-11T12:13:32.123" Score="5" Body="&lt;p&gt;From personal experience, Na(&lt;em&gt;s&lt;/em&gt;) and Cl&lt;sub&gt;2&lt;/sub&gt;(&lt;em&gt;g&lt;/em&gt;) react at room temperature without added energy. Li(&lt;em&gt;s&lt;/em&gt;) and Cl&lt;sub&gt;2&lt;/sub&gt;(&lt;em&gt;g&lt;/em&gt;) should do so also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following heat of formation data harvested from &lt;a href=&quot;http://webbook.nist.gov&quot;&gt;the NIST Chemistry Webbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7647145&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot;&gt;NaCl&lt;/a&gt;:&#xA;$$\ce{Na(s) + } \frac 1 2\ce{Cl2(g)-&amp;gt;NaCl(s)}\quad \Delta H_f^o=-411.12 \; kJ/mol $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7447418&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot;&gt;LiCl&lt;/a&gt;:&#xA;$$\ce{Li(s) + } \frac 1 2\ce{Cl2(g)-&amp;gt;LiCl(s)}\quad \Delta H_f^o=-408.27 \; kJ/mol $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These data suggest that both reactions should occur without added heat. However, the rate of reaction is not determinable from these data. The reaction between Li(&lt;em&gt;s&lt;/em&gt;) and Cl&lt;sub&gt;2&lt;/sub&gt;(&lt;em&gt;g&lt;/em&gt;) will likely be slower.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-11T12:13:32.123" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="633" PostTypeId="1" CreationDate="2012-06-11T13:10:31.330" Score="6" ViewCount="12101" Body="&lt;p&gt;Can we get chloroquine from 4-amino-7-chloroquinoline (7-chloroquinolin-4-amine) by adding 4-chloro-&lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-diethylpentan-1-amine $(\ce{CH3CHCl(CH2)3N(Et)2})$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2TluG.png&quot; alt=&quot;proposed chloroquine synthesis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="299" LastEditorUserId="418" LastEditDate="2016-11-21T07:07:17.963" LastActivityDate="2019-11-17T14:26:35.873" Title="Synthesis of chloroquine from 4-amino-7-chloroquinoline" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="634" PostTypeId="1" CreationDate="2012-06-11T16:40:51.667" Score="5" ViewCount="1786" Body="&lt;p&gt;I’m looking for a public database of bond energies of biochemical compounds. Does one exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, which is the standard computational methodology to calculate them? Any references would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="73" LastEditDate="2012-06-11T16:59:22.797" LastActivityDate="2013-01-28T05:42:05.063" Title="Bond energies database" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="635" PostTypeId="1" CreationDate="2012-06-11T18:19:19.220" Score="5" ViewCount="2573" Body="&lt;p&gt;I need a 0.01 N solution of $\ce{KIO3}$ to use as a standard in a Winkler titration.  The protocol that I am using specifies that this is 0.3567 g L&lt;sup&gt;-1&lt;/sup&gt;, which I have calculated to be 0.00167 M based on 214 g mol&lt;sup&gt;-1&lt;/sup&gt; for $\ce{KIO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to understand the calculations in the protocols that I am using, so how would I calculate that a 0.00167 M solution of $\ce{KIO3}$ is 0.01 N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (More detail on the reactions)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{KIO3}$ solution is added to an $\ce{NaOH} \cdot \ce{NaI}$ solution under acidic conditions for:  $\ce{IO3- + 5I- + 6H+ \rightarrow 3I2 + 3H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{I2}$ is then titrated to standardize a thiosulfate solution. &lt;/p&gt;&#xA;" OwnerUserId="115" LastEditorUserId="4231" LastEditDate="2015-03-04T16:01:11.700" LastActivityDate="2016-04-16T11:22:57.610" Title="How would I calculate the Normality of a 0.00167 molar solution of KIO₃?" Tags="&lt;solvents&gt;&lt;redox&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="639" PostTypeId="1" CreationDate="2012-06-12T16:01:50.727" Score="7" ViewCount="221" Body="&lt;p&gt;I have been looking all over the internet for forcefield parameters for a fluorescent probe called NBD (nitrobenzoxadiazole). &lt;a href=&quot;http://products.invitrogen.com/ivgn/product/N678&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an example of an NBD labeled fatty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have so far only found parameters for AMBER or &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0005273606003956&quot; rel=&quot;nofollow&quot;&gt;GROMACS&lt;/a&gt;. Also when following the articles from &lt;a href=&quot;http://www.springerlink.com/content/y7617857882117h3/?MUD=MP&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.mdpi.com/1420-3049/16/7/5437&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; reviews, there seem to be no CHARMM parameters anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What CHARMM forcefield parameters would you use for NBD?&lt;/li&gt;&#xA;&lt;li&gt;How could I transfer the AMBER or GROMACS parameters for use in CHARMM?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="309" LastEditorUserId="22" LastEditDate="2012-06-22T04:50:30.820" LastActivityDate="2013-01-23T20:34:16.993" Title="CHARMM parameter files for NBD (nitrobenzoxadiazole)" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="640" PostTypeId="2" ParentId="441" CreationDate="2012-06-12T20:48:52.390" Score="1" Body="&lt;p&gt;I'm sure &lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;nofollow&quot;&gt;VMD&lt;/a&gt; (although also geared towards protein alignment) is flexible enough to align any two sets of atoms. You will almost definitely have to learn some &lt;a href=&quot;http://www.tcl.tk/man/tcl/tutorial/tcltutorial.html&quot; rel=&quot;nofollow&quot;&gt;TCL&lt;/a&gt; scripting, but getting past the (perhaps steep) learning curve is well worth the effort. Besides VMD has &lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/current/docs.html&quot; rel=&quot;nofollow&quot;&gt;excellent documentation&lt;/a&gt;.&#xA; If you post a more specific problem I'll be happy to post a more specific answer:)&lt;/p&gt;&#xA;" OwnerUserId="309" LastActivityDate="2012-06-12T20:48:52.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="641" PostTypeId="1" AcceptedAnswerId="645" CreationDate="2012-06-13T19:54:56.220" Score="16" ViewCount="3780" Body="&lt;p&gt;I am looking for a liquid which is not miscible (or very close) with water or vegetable oil.  Ideally, I want to have a container which shows three layers (the three liquids) with clear boundary layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that I should find a non-polar compound of very high molecular weight - such that difference in the density of that compound and oil is enough to draw it down out of the oil.  However, I haven't found it yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer a compound which is non-toxic and non-flammable if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It feels like such a simple question, I must be missing something.  &lt;/p&gt;&#xA;" OwnerUserId="337" LastEditorUserId="73" LastEditDate="2012-06-14T20:03:03.683" LastActivityDate="2013-12-06T06:59:00.870" Title="Liquid which neither dissolves in water nor oil" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="642" PostTypeId="2" ParentId="87" CreationDate="2012-06-13T20:16:26.117" Score="11" Body="&lt;p&gt;I may have accidentally created copper cloride in addition to the copper acetate. I decided to just let the solution (made form Copper, vinegar, and salt) evaporate with a small fan speeding the evaporation for the last few weeks. The tallest copper acetate crystals started breaking the surface and then the surface of the liquid started to form a skin (pic 1) so I poured off the now green liquid into another jar. Copper acetate is blue, which is why the green solution seems to suggest that there is copper cloride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G7ZkK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Evaporation! Note the blue-green layer of copper carbonate below the dark copper acetate crystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1S2Al.jpg&quot; alt=&quot;Copper Chloride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top View&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/StH1J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green liquid on the left is what was left from evaporation (and I think is copper chloride) - next to another jar of copper acetate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SNUvD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper acetate crystals were covered in a thin layer of copper carbonate because last week I disturbed the solution and stirred up a cloud of copper carbonate accidantally. I rinsed them with vinegar and they cleaned nicely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7E42m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest crystal was just over 1 cm wide. These crystals were grown from a solution that started evaporating 5 years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IekAf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper carbonate goo in the bottom needed a use so I added more vinegar to it and it formed the copper acetate solution three pictures up (next to the green copper chloride).&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2012-06-13T23:15:42.997" LastActivityDate="2012-06-13T23:15:42.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="643" PostTypeId="2" ParentId="641" CreationDate="2012-06-14T07:19:47.100" Score="4" Body="&lt;p&gt;How about Mercury and Gallium? Mercury is rather toxic, but you can buy Gallium online and it's a liquid at 85 degrees Fahrenheit - so it might work if you left it in some sunlight when you wanted the Gallium to be liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't plan on opening it often or need to worry about it breaking, I'd try Mercury.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2012-06-14T07:19:47.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="644" PostTypeId="2" ParentId="635" CreationDate="2012-06-14T08:41:58.240" Score="2" Body="&lt;p&gt;Hmm.. I think that your data is wrong, I get that the normality:molarity ratio must be 5 in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ratio (I call it the $\eta$-factor) for a compound participating in a reaction is the net change of its oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, one iodine in $\ce{IO3-}$ (+5) becomes one iodine in $\ce{I2}$ (0). The net change in oxidation number is 5, so $\rm normality=5\times molarity$. Normality is like the reducing/oxidising &quot;power&quot; of a solution--or the concentration of electrons that it can furbish/absorb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your density, your $\mathrm{g~L^{-1}}$, or your value for normality is wrong here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that one needs to be careful with this reaction. It is comproportionation, where the same element is oxidising itself (Basically, two forms of the element in different oxidation states are reacting with each other). We can have some trouble when dealing with the &lt;em&gt;products&lt;/em&gt; of such a reaction, though dealing with the reactants is fairly simple, as shown above. I'm writing a detailed self-answered post on this later.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="16683" LastEditDate="2016-04-16T11:22:57.610" LastActivityDate="2016-04-16T11:22:57.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="645" PostTypeId="2" ParentId="641" CreationDate="2012-06-14T10:30:16.003" Score="14" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorocarbons&quot;&gt;Fluorocarbons&lt;/a&gt; are compounds that are hydrophobe and lipophobe. It's a special property of perfluorinated compounds: They are non-polar and thus hydrophobic but in addition as the high electronegativity of fluorine reduces the polarizability of the atom, fluorocarbons are only weakly susceptible to the fleeting dipoles that form the basis of the London dispersion force. As a result, fluorocarbons have low intermolecular attractive forces and are lipophobic. &#xA;I'm not sure about their toxicity, but this seems to be an issue, e.g. &lt;a href=&quot;http://www.nikwax-usa.com/en-us/aboutus/persistentflurocarbondanger.php&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2012-06-14T10:30:16.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="646" PostTypeId="1" CreationDate="2012-06-15T00:05:20.230" Score="0" ViewCount="471" Body="&lt;p&gt;A table in the book 'Chemistry NCEA Level 1' has 4 headings, the first of which is 'Volume (mL) ammonia solution provided (conc 2 g L&lt;sup&gt;-1&lt;/sup&gt;)'&#xA;Could someone please provide a comprehensive explanation of what the end part in brackets means.&#xA;This is for a chemistry newbie, as you may gather from the book title.&#xA;(this isn't really homework but I couldn't find a relevant tag. It's just someone studying chemistry for fun.)&lt;/p&gt;&#xA;" OwnerDisplayName="user338" LastEditorUserId="4945" LastEditDate="2018-12-11T12:33:37.687" LastActivityDate="2018-12-11T12:33:37.687" Title="Basic chemistry equation with ammonia" Tags="&lt;concentration&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2012-06-15T02:20:01.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="647" PostTypeId="2" ParentId="633" CreationDate="2012-06-15T17:10:20.617" Score="4" Body="&lt;p&gt;My intuition tells me that the nitrogen in the ring is more nucleophilic than the exocyclic amine. A quick-and-dirty electrostatic potential calculation using B88LYP/6-31G(d) on the PM3 geometry shows a large electron density (blue and violet) on the ring nitrogen and lower density (yellow) on the exocyclic amine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OpHpj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there conditions that might work? Yes. Should you try it, if you need that molecule? Yes. Will it be difficult? Yes.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-15T17:10:20.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="650" PostTypeId="1" AcceptedAnswerId="651" CreationDate="2012-06-16T09:28:29.200" Score="14" ViewCount="4753" Body="&lt;p&gt;For an upcoming exam, I have to describe the synthesis of 2-hydroxy-2-phenylacetic acid (&lt;a href=&quot;http://en.wikipedia.org/wiki/Mandelic_acid&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Mandelic acid&quot;&gt;mandelic acid&lt;/a&gt;) from benzaldehyde. The provided answers describe quite a long process, consisting of 8 stages. However, I've been suggested a much simpler method, consisting of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{C6H5CHO + HCN -&amp;gt; C6H5CH(OH)CN}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C6H5CH(OH)CN + HCl + 2H2O -&amp;gt; C6H5CH(OH)COOH + NH4Cl}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is even simpler than the 3-stage process described in the Wikipedia article. Could it be a viable answer?&lt;/p&gt;&#xA;" OwnerUserId="328" LastEditorUserId="16683" LastEditDate="2017-04-06T22:24:41.457" LastActivityDate="2017-04-06T22:24:41.457" Title="Synthesis of mandelic acid from benzaldehyde" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;&lt;nitriles&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="651" PostTypeId="2" ParentId="650" CreationDate="2012-06-16T10:04:43.087" Score="10" Body="&lt;p&gt;The direct reaction with HCN seems to be how the mandelic acid was synthesized in the 19th century. If you check how the reaction with sodium bisulfite is actually performed &lt;a href=&quot;http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv1p0336&quot; rel=&quot;noreferrer&quot;&gt;(Organic Syntheses, Coll. Vol. 1, p.336 (1941); Vol. 6, p.58 (1926).)&lt;/a&gt; you'll see that this is not more complicated than the direct HCN route as those first two steps are performed in one reaction step.&lt;/p&gt;&#xA;&lt;p&gt;From this preparation, several preparations via the HCN route are cited:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Naquet and Louguinine, Ann. 139, 299 (1866); Wallach, Ann. 193, 38&#xA;(1878); Spiegel, Ber. 14, 239 (1881); Ultee, Rec. trav. chim. 28, 254&#xA;(1909); Rule, J. Chem. Soc. 113, 12 (1918).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I looked at the first page of the &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jlac.18661390305/abstract&quot; rel=&quot;noreferrer&quot;&gt;oldest of those papers from Naquet and Louguinine from 1866&lt;/a&gt; and it states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bekanntlich bildet sich diese Säure bei der wechselseitigen Einwirkung des Benzoesäure-Aldehyds und der Cyanwasserstoffsäure bei Anwesenheit von verdünnter Chlorwasserstoffsäure gemäß der Gleichung:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C7H6O + CNH + 2 H2O + HCl -&amp;gt; NH4Cl + C8H8O3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Translated, this means that you can synthesize mandelic acid from benzaldehyde, HCN and HCl.&lt;/p&gt;&#xA;&lt;p&gt;I'm assuming you're not planning to actually synthesize it using this route yourself, I would guess that avoiding the use of HCN directly was one of the motivations behind the synthesis route using sodium bisulfite.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$%edit$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-03-11T23:23:38.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="652" PostTypeId="1" AcceptedAnswerId="653" CreationDate="2012-06-17T11:02:39.393" Score="18" ViewCount="834" Body="&lt;p&gt;Was there any false chemical element introduced in history of chemistry? I mean a substance that after a while proved that it was not an element.&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="61578" LastEditDate="2020-03-30T01:41:10.510" LastActivityDate="2020-11-17T18:29:22.760" Title="Any false chemical element in history of chemistry?" Tags="&lt;history-of-chemistry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="653" PostTypeId="2" ParentId="652" CreationDate="2012-06-17T13:33:37.223" Score="22" Body="&lt;p&gt;The example that comes to mind is &lt;a href=&quot;http://en.wikipedia.org/wiki/Didymium&quot;&gt;didymium&lt;/a&gt;, which turned out to be a mixture of praesodymium and neodymium. The term is still used, as far as I know, to refer to glass doped with a mixture of these lanthanides. This is discussed in one of the 'Chemistry in its element' &lt;a href=&quot;http://www.rsc.org/chemistryworld/podcast/interactive_periodic_table_transcripts/neodymium.asp&quot;&gt;podcast episodes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, wikipedia has a category called &lt;a href=&quot;http://en.wikipedia.org/wiki/Category%3aMisidentified_chemical_elements&quot;&gt;'Misidentified Chemical Elements'&lt;/a&gt;. Especially notable are illmenium, dianium and pelopium, which were all likely mixtures of niobium and tantalum, two rather similar elements.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-06-17T13:33:37.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="654" PostTypeId="1" AcceptedAnswerId="734" CreationDate="2012-06-18T09:57:30.713" Score="6" ViewCount="143" Body="&lt;p&gt;I know that perchloric acid could be potentially explosive and some organic solvents could form dangerous peroxides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to do an electrodeposition in diethyleneglycol solution containing $\ce{NH4Cl}$, $\ce{ZnCl2}$ and $\ce{FeCl2}$. Since in my case there are all elements to form $\ce{HClO4}$ and I am going to heat everything up to 60 degrees I want to be sure that it is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any danger of forming explosive reaction products during the electrodeposition?&lt;/p&gt;&#xA;" OwnerUserId="222" LastEditorUserId="84" LastEditDate="2012-06-18T10:06:39.403" LastActivityDate="2012-07-04T14:24:44.663" Title="Is there any danger of forming explosive reaction products during the electrodeposition in Diethylene glycol?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="656" PostTypeId="2" ParentId="87" CreationDate="2012-06-18T14:41:34.663" Score="7" Body="&lt;p&gt;Here are two more blue shades:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the bright blue of the hexa-hydrated $\ce{[Cu^{II}(H2O)6]^2+}$ (dissolve $\ce{CuSO4}$, maybe not household chemical, but used e.g. as fungicide in farming incuding organic farming), &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if you drop ammonia into this solution, you'll first see a white (blueish with the remaining hexahydrate) precipitate or clouds of the hydroxide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;which dissolves with more ammonia into the deep blue tetra ammine complex&#xA;$\ce{[Cu^{II}(NH3)4(H2O)2]^2+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-06-18T14:41:34.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="657" PostTypeId="1" CreationDate="2012-06-18T18:19:05.267" Score="8" ViewCount="426" Body="&lt;p&gt;I recently was told that once you find a good solvent mixture for separation on &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_layer_chromatography&quot; rel=&quot;nofollow&quot;&gt;thin layer chromatography&lt;/a&gt; (TLC), you should only use 10% of the more polar solvent (for silica) when you run your column. I tried it and it worked great! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I'd never heard that before, is that a standard rule of thumb, or did I just get lucky?&lt;/p&gt;&#xA;" OwnerUserId="345" LastEditorUserId="470" LastEditDate="2012-07-29T03:27:27.470" LastActivityDate="2017-07-21T22:02:36.303" Title="How to translate the mobile phase from thin layer chromatography to column?" Tags="&lt;chromatography&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="659" PostTypeId="1" CreationDate="2012-06-20T10:06:16.323" Score="5" ViewCount="261" Body="&lt;p&gt;&lt;strong&gt;Original question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at a Diopside-Hedenbergite solid solution, what is the difference between an ionic and molecular activity model? Why does the moleclar model yield higher values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post-Edit question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two activities calculated below seem to me like I'm just comparing some random values. What is the true activity of the Fe- or Mg-mineral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So far I got:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dipside and Hedenbergite are Pyroxenes. Pyroxenes have to metal sites (M1 and M2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;M^1M^2[Si_2O_6]&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between Hedenbergite and Diopside is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;Hedenbergit (Fe^{[M1]}Ca^{[M2]}[Si_2O_6]) \longleftrightarrow Diopside (Mg^{[M1]}Ca^{[M1]}[Si_2O_6])&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activity of an element is assumed to be equal to the concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a=\color{Red}{\gamma}x \\&#xA;a=x&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next step one has to calculate the concentrations on the M1 site because it's the site where the ion exchange happens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;x[Fe]=\frac{[Fe]}{[Fe]+[Mg]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;x[Mg]=\frac{[Mg]}{[Fe]+[Mg]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular mixing model apparently compares the activity of the Fe- and Mg-mineral on the M1 site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a[Mg]=x[Mg]&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a[Fe]=\frac{1-x[Mg]}{x[Fe]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the ionic mixing model compares the activities of all the cations of the minerals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a[Mg]=x[Mg]^{[M1]}x[Ca]^{[M2]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;a[Fe]=x[Fe]^{[M1]}x[Ca]^{[M2]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are there two activities? Shouldn't there be one true activity? What does the one activity mean and what does the other mean?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="351" LastEditorUserId="351" LastEditDate="2012-08-25T18:55:16.370" LastActivityDate="2012-08-25T19:24:40.407" Title="What is the difference between an ionic and molecular activity model? (Example: Diopside-Hedenbergite)" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="660" PostTypeId="1" AcceptedAnswerId="661" CreationDate="2012-06-20T10:31:54.430" Score="5" ViewCount="4264" Body="&lt;p&gt;I tried to figure out the MO-scheme of the tetragonal-bipyramidal complex &lt;em&gt;trans&lt;/em&gt;-$\ce{[Co(en)2(NCS)2]SCN}$ in which the isothiocyanate ligands are bound to the $\ce{Co^3+}$-Ion in $\eta^{1}$-mode (en = ethylenediamine). While undertaking this task I noticed that I don't know the MO-scheme of the isothiocyanate ion, which I need to determine whether it acts as a $\sigma$ and/or $\pi$ acceptor or donor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to construct the MO via a symmetry-based treatment but ran into the problem that the isothiocyanate ion - belonging to thet $C_{\infty v}$ point group - lacks symmetry elements perpendicular to its principal axis. Thus, I wasn't able to construct the bonding and antibonding sigma-orbitals with this method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could construct those $\sigma$-interactions by hand or use $\ce{CO2}$ as a model but I don't know whether this will give me the right results. I haven't found the MO-scheme in the internet so I would be happy if someone could advise me on how to construct a qualitative MO-scheme of $\ce{SCN-}$ or simply show it to me. I would also be very grateful for getting some information on how exactly the isothiocyanate ion acts as a ligand (which $\pi$-interactions and how strong are they, is there also $\sigma$-donation as with carbonyl ligands, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;I forgot to mention explicitly that the $\ce{SCN-}$ ion is bound to the metal center via its nitrogen atom.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="4945" LastEditDate="2014-06-12T08:44:02.303" LastActivityDate="2014-06-12T08:44:02.303" Title="MO-Scheme of SCN- and its bonding properties when used as a ligand" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;molecular-orbital-theory&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="661" PostTypeId="2" ParentId="660" CreationDate="2012-06-20T11:46:15.200" Score="6" Body="&lt;p&gt;One issue with SCN&lt;sup&gt;-&lt;/sup&gt;, is that it can bind either at the nitrogen or at the sulfur. The way you presented the formula, $\ce{[Co(en)2(NCS)2]SCN}$, suggested that the SCN&lt;sup&gt;-&lt;/sup&gt; ligand is bound via the nitrogen. If it were bound by the sulfur, the formula might be written  $\ce{[Co(en)2(SCN)2]SCN}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SCN&lt;sup&gt;-&lt;/sup&gt; should have a similar MO diagram to OCN&lt;sup&gt;-&lt;/sup&gt;, which is isoelectronic with CO&lt;sub&gt;2&lt;/sub&gt;, so your idea to compare to CO&lt;sub&gt;2&lt;/sub&gt; is not  a bad one. However, since SCN&lt;sup&gt;-&lt;/sup&gt; has a different symmetry group, you cannot trust that comparison too much. Your comparison to CO is also a good one, since SCN&lt;sup&gt;-&lt;/sup&gt; and CO have the same point group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you really need to do is determine the symmetry group for $\ce{[Co(en)2(NCS)2]SCN}$. Then, assign symmetry labels to the SCN&lt;sup&gt;-&lt;/sup&gt; ligands. The MO that is most important for SCN&lt;sup&gt;-&lt;/sup&gt; is, qualitatively, the &quot;lone pair&quot; on the atom pointing at the metal center, which can form a &quot;$\sigma$-bond&quot; with cobalt. I would only consider using those $\pi$ orbitals on SCN&lt;sup&gt;-&lt;/sup&gt; if the complex needs more electrons to get to 18. When you assign symmetry labels to the SCN&lt;sup&gt;-&lt;/sup&gt; ligands, consider them as a pair: &lt;em&gt;i.e.&lt;/em&gt; assign a symmetry label to the in-phase pair and then assign a symmetry label to the out-of-phase pair. The in-phase pair should probably have the same symmetry label as the cobalt $d_{z^2}$ orbital.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-20T11:46:15.200" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="662" PostTypeId="1" CreationDate="2012-06-20T13:46:22.960" Score="14" ViewCount="50874" Body="&lt;p&gt;I'm an electrical engineering and I hope you can give me some advice as I'm not much into chemistry.&#xA;I'm trying to start making my PCB (Printed Circuit Board) at home but I'd like to have some advice concerning how to handle and where to store these (potentially dangerous) chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need NaOH as a developer, hydrochloric acid + hydrogen peroxide as etchant (they should be a better alternative than Ferric Chloride) and isopropyl alcohol (to make soldering flux).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understood I need:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Glasses (to protect against vapors and splashes)&lt;/li&gt;&#xA;&lt;li&gt;Gloves (to protect when mixing the chemicals)&lt;/li&gt;&#xA;&lt;li&gt;Some kind of protective clothes (not sure if I need to, but obviously I'm not wearing nice clothes when doing etching)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have a few questions though:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do I need kind of mask / filter (to put on my nose) - are these vapors dangerous ?&lt;/li&gt;&#xA;&lt;li&gt;I read gloves should be stored in quite cold locations and not be exposed to sunlight. Does that apply to the chemicals listed above too ? I'd like to store them outside but maybe it's not safe ...&lt;/li&gt;&#xA;&lt;li&gt;Hydrochloric acid comes in 36% concentration but I think I'll need to dilute it to 1%-5% max. After it has been diluted it should be quite safe, isn't it ?&lt;/li&gt;&#xA;&lt;li&gt;Is it safe to store them inside a house (as long as it's well ventilated) ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="352" LastEditorUserId="7" LastEditDate="2012-06-20T16:51:28.450" LastActivityDate="2017-02-04T21:20:34.280" Title="Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="663" PostTypeId="2" ParentId="662" CreationDate="2012-06-20T17:27:14.027" Score="10" Body="&lt;p&gt;You probably don't need a mask - vapours from hydrochloric acid can be a little noxious, but if you're diluting it down and using it in a reasonably ventilated area, you shouldn't get much of that. Isopropyl alcohol vapour is &lt;a href=&quot;http://en.wikipedia.org/wiki/Isopropyl_alcohol#Toxicology&quot;&gt;toxic like most alcohols&lt;/a&gt;, but I don't think an ordinary mask will prevent that, and you'll be better off just making sure you have enough ventilation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard white cotton labcoats are inexpensive and can be quickly removed if you spill a nontrivial quantity of something on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide will probably come in an opaque container, but if not, it does break down in sunlight. Everything else is pretty stable to light, although you want to avoid storing the alcohol somewhere that's going to get too warm, because it evaporates readily and can pressurise the container more than is a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these materials are pretty safe to store indoors in unspecialised storage, although for commercial purposes or quantities you may need to take regulatory precautions that will obviously vary by locale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add&lt;/strong&gt;: The copper chloride that comes out should be regenerated as many times as possible, but when you eventually end up with too much of it, you'll want to already have thought about what you're going to do with it - it's pretty toxic and you can't (or, well, shouldn't) just wash it down the sink.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2012-06-21T16:03:49.280" LastActivityDate="2012-06-21T16:03:49.280" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="665" PostTypeId="2" ParentId="662" CreationDate="2012-06-20T19:16:12.097" Score="4" Body="&lt;p&gt;Storage considerations for chemicals are often noted on their Material Safety Data Sheet, or MSDS. All suppliers that I know of include a MSDS in the packaging when they ship chemicals. You can find lots of good sources of free MSDSs online through Google. Just google &quot;Hydrochloric acid MSDS&quot;. I would look for one that appears to come from a reputable source, such as a commercial supplier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, I have &lt;a href=&quot;http://fscimage.fishersci.com/msds/11155.htm&quot; rel=&quot;nofollow&quot;&gt;linked the MSDS for concentrated HCl from Fisher Scientific&lt;/a&gt;. Section 7 of this MSDS is titled &quot;Handling and Storage&quot;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-20T19:16:12.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="666" PostTypeId="1" AcceptedAnswerId="678" CreationDate="2012-06-20T20:35:19.687" Score="10" ViewCount="4844" Body="&lt;p&gt;I'm working with some very basic free-radical initiated polymerization reactions. For the most part, I just follow a recipe I'm given (by a member of my group or the broader literature), it works, and I don't worry too much about the chemistry of why it works. Clearly, that will only work for so long, and I'd like to better understand the chemistry of what I'm doing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So what general rules can I use to predict what kind of initiator will polymerize a given monomer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give a concrete example of what I mean: I know that some persulfate-based initiators will polymerize some acrylamide-based monomers. Can I generalize that to all persulfate initiators and all acrylamide monomers? What about methacrylamides? What about styrenes? Essentially, at what point have I changed the reactants so much that the chemistry doesn't work anymore?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odian's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0195124448&quot; rel=&quot;noreferrer&quot;&gt;Principles of Polymerization&lt;/a&gt; (4th ed) reviews some things that have been done, but doesn't discuss any general rules. Are there other good sources for this information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I am looking specifically for chemical reasons. Are some initiators better at reacting with some monomers? Are there any issues like side reactions I should take into account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT(2): from @Ben Norris's answer and the ensuing comments, it sounds like, as long as I can get the initiator to radicalize, and everything is soluble in the solvent, then I will be able to use any initiator with any monomer. Is that correct? I've found another textbook on this, &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0195124448&quot; rel=&quot;noreferrer&quot;&gt;Polymer Chemistry: An Introduction&lt;/a&gt; by Malcom Stevens, and that seems to support this. &lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2012-07-03T15:44:41.917" LastActivityDate="2018-10-24T18:49:03.820" Title="Are there any general chemical rules for choosing initiators for a polymerization reaction?" Tags="&lt;polymers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="667" PostTypeId="1" CreationDate="2012-06-20T20:40:16.843" Score="4" ViewCount="861" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/BET_theory&quot; rel=&quot;nofollow&quot;&gt;Brunauer–Emmett–Teller (BET) theory&lt;/a&gt; is comprehensible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is more geared towards what is physically and chemically happening within a BET device to get a reading?  How does a BET device actually go about measuring the physisorption or chemisorption of a material?  Why does this process tend to take multiple hours?  &lt;/p&gt;&#xA;" OwnerUserId="353" LastEditorUserId="4945" LastEditDate="2014-07-29T10:45:21.527" LastActivityDate="2014-07-29T10:45:21.527" Title="How does a Brunauer–Emmett–Teller device function?" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="668" PostTypeId="1" AcceptedAnswerId="670" CreationDate="2012-06-21T03:56:36.153" Score="12" ViewCount="16181" Body="&lt;p&gt;I know that when dissolved in water, hydrogen halides ($\ce{H-X}$), become acids. I know that $\ce{HF}$ forms a weak acid, but can it be generalized that all $\ce{HX}$ acids do so?&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="4231" LastEditDate="2015-02-21T22:26:57.660" LastActivityDate="2015-04-15T18:00:17.123" Title="Do hydrogen halides (HX) form strong or weak acids?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="669" PostTypeId="2" ParentId="668" CreationDate="2012-06-21T04:32:51.573" Score="2" Body="&lt;p&gt;No: HCl is a very strong acid, and HBr even more so. HF is the only weak acid. This is because it binds hydrogen much more strongly then the other halides. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2012-06-21T04:32:51.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="670" PostTypeId="2" ParentId="668" CreationDate="2012-06-21T05:22:11.803" Score="11" Body="&lt;p&gt;No, it can’t be generalised. Amongst the haloacids, only $\ce{HF}$ is weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acid strength can be determined by the stability of the conjugate base (for most acids, the conjugate base is what you get when you remove $\ce{H+}$ from the acid). In this case the conjugate bases are the halides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the size of the halide ion increases, the charge is dispersed over a larger volume, and thus the ion is more stable. In this manner, one can say that the order of acid strengths is $\ce{HF&amp;lt;HCl&amp;lt;HBr&amp;lt;HI}$ (I’m not considering the radioactive ones here; they fit in the same trend though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl, HBr, HI}$ are all strong acids. $\ce{HCl}$ has a $\mathrm{p}K_\text{a}$ of −7 (the smaller the $\mathrm{p}K_\text{a}$ is, the more acidic it is), $\ce{HBr}$ is approximately −9, and $\ce{HI}$ is also approx −9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\ce{HF}$ is 3.2. The $\ce{H-F}$ bond is pretty strong, so it has a smaller tendency to dissociate—making it a weak acid (not to say that it isn’t dangerous).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorDisplayName="user7951" LastEditDate="2015-04-15T18:00:17.123" LastActivityDate="2015-04-15T18:00:17.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="671" PostTypeId="1" AcceptedAnswerId="672" CreationDate="2012-06-21T11:12:39.000" Score="12" ViewCount="9726" Body="&lt;p&gt;In chemistry, the unit &quot;mole&quot; is used in many places, and is an important tool when working with quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not seen it used elsewhere--for example, we never count people or TVs with the unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the mole so useful in chemistry, and why isn't it used elsewhere?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="12860" LastEditDate="2015-10-30T17:26:28.380" LastActivityDate="2017-08-07T16:22:53.830" Title="The mole is used extensively in chemistry, why not elsewhere?" Tags="&lt;everyday-chemistry&gt;&lt;mole&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="672" PostTypeId="2" ParentId="671" CreationDate="2012-06-21T15:25:08.930" Score="21" Body="&lt;p&gt;Because the mole is not a convenient basis for people and TV sets (for instance, one mole of people would be approximately 86 trillion (short-scale) times the population of earth, or put another way the earth contains a human population concentration of approximately 12 femtomoles/planet), and because there's no connection back to atomic mass, which is what makes the mole useful in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was not part of your original question, but addresses Manishearth's modifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because atoms and molecules react in discrete units (cf. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_definite_proportions&quot; rel=&quot;noreferrer&quot;&gt;law of definite proportions&lt;/a&gt;), the &lt;em&gt;number&lt;/em&gt; of particles is more fundamental than the masses of those particles. The mole is useful in chemistry because it is defined such that the mass of Avogadro's number of particles (atoms or molecules) of a substance (the molar mass) is equal to the sum of the numerical values of the atomic masses of its constituent elements. The value of the mole is calibrated such that the molar mass of carbon-12 is exactly 12 grams per mole. On the periodic table you will note that the atomic masses given are non-integral (often markedly so, e.g. chlorine at 35.45 g/mol). This is because the atomic masses represent a weighted average of isotopic masses, and because the proton and neutron masses aren't quite equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mole makes stoichiometric calculations particularly convenient as we can measure out, for instance, 16 grams (one mole) of $\ce{^{16}O}$ and 2 grams (two moles) of $\ce{^1H}$ and obtain one mole of $\ce{H2O}$. In the case of water, oxygen makes up 89% of the molecule by mass, but is only one of the three atoms present in the molecule. The mole is necessary to transpose stoichiometries involving single atoms and molecules (with their tiny masses) into the everyday world of grams and litres.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2015-08-27T13:37:50.160" LastActivityDate="2015-08-27T13:37:50.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="673" PostTypeId="2" ParentId="671" CreationDate="2012-06-21T18:34:42.597" Score="11" Body="&lt;p&gt;The Si-unit &quot;mol&quot; is used in chemistry for three reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The quantity of atoms/molecules is the logical unit of chemistry (Reactions have to be balanced).&lt;/li&gt;&#xA;&lt;li&gt;Atoms/molecules can't be counted (It works for people and TV's), but have to be measured using mass or volume, and then converted into a quantity (e.g. mol/kg)&lt;/li&gt;&#xA;&lt;li&gt;The quantity of atoms/molecules involved in a typical experiment or measurement is very large, making a shorthand useful (Compare this to the Astronomical Unit which is a shorthand for a large number of km).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The mol (or a similar unit) isn't used elsewhere because:&#xA;Anything larger than chemicals can be measured with whichever measure seems to be the most logical. Chemistry is unique in requiring chemical quantities to make sense of the world but at the same time these quantities can not be determined directly.&lt;/p&gt;&#xA;" OwnerUserId="351" LastActivityDate="2012-06-21T18:34:42.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="674" PostTypeId="2" ParentId="662" CreationDate="2012-06-21T19:09:04.817" Score="9" Body="&lt;p&gt;A few additions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;shelf life&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H2O2}$ won't last forever, not even in the dark and cool&lt;br&gt;&#xA;edit: if it is stored cool and dark, and does not get contaminated, it will last for some years. However, the degradation is catalyzed by metal ions (famous: $\ce{Mn^2+}$) and bases ($\ce{OH-}$), so the problem is that once it gets contaminated (there is stabilizer added, but that of course won't last forever and cannot prevent every type of degradation), it will degrade rather fast.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the isopropanol must be kept tightly closed to prevent evaporation (the container it comes in is usually good), &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;but $\ce{NaOH}$ (pellets in tightly closed plastic container) and $\ce{HCl}$ (glass bottle with plastic cap) can be kept for years without problems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Storage of NaOH&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{NaOH}$ is deliquescent (very much hygroscopic). &lt;/li&gt;&#xA;&lt;li&gt;$\ce{NaOH}$ also reacts with $\ce{CO2}$ from air to $\ce{NaHCO3}$ and $\ce{Na2CO3}$. The corresponding reduction in base equivalents is not going to hurt you much, but&#xA;the carbonate can form a crust at the lid of $\ce{NaOH}$ bottles. You should not store $\ce{NaOH}$ in glass stoppered glass bottles as opening them becomes very hard and finally impossible with the carbonate crust. &lt;/li&gt;&#xA;&lt;li&gt;($\ce{NaOH}$ is also directly corrosive for glass, but not much)&lt;/li&gt;&#xA;&lt;li&gt;Overall, it is best to store the $\ce{NaOH}$ pellets in plastic bottles (they usually come stored this way) and prepare frequently small amounts of solution.&lt;/li&gt;&#xA;&lt;li&gt;You can also store the $\ce{NaOH}$ solution for a while in a tightly closed plastic bottle. &lt;/li&gt;&#xA;&lt;li&gt;How frequently you need to make fresh $\ce{NaOH}$ depends: If you use the $\ce{NaOH}$ in an open bath daily, it won't last as long as if you just need it every few weeks, and meanwhile it is stored away from fresh air. I believe your application is rather robust wrt. the real $\ce{NaOH}$ concentration, so you can probably reuse the $\ce{NaOH}$. On the other hand, preparing the solution is really fast &amp;amp; easy: take the water, count the needed number of $\ce{NaOH}$ pellets into it and let them dissolve.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorDisplayName="user15489" LastEditDate="2015-04-23T21:21:50.360" LastActivityDate="2015-04-23T21:21:50.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="675" PostTypeId="2" ParentId="662" CreationDate="2012-06-21T21:12:14.630" Score="7" Body="&lt;p&gt;Reactive chemicals should be stored in chemical storage cupboards to prevent fires. I've attached an example of the type of cupboard we use in my lab to give you an idea of what I am talking about. We use an under-fumehood model, but freestanding models are avalible. Acids and bases should not be stored together. MSDS sheets from the manufacture may specificity additional constraints on which chemicals can be stored alongside one another. Note that hydrogen peroxide is a strong oxidizing agent; Do not store with oxidizers is a common constraint. Consider contacting your local fire department or the chemistry department of a nearby university for advice on the safe storage of chemicals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note; While I'm not endorsing them, I know that Fisher Scientific provides a range of chemical storage cupboards of this type, and has information on which are good for acids, corrosive chemicals and flammable solvents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AQd3f.jpg&quot; alt=&quot;Example chemical storage cupboard&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2012-06-21T21:12:14.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="676" PostTypeId="2" ParentId="671" CreationDate="2012-06-21T21:38:44.070" Score="4" Body="&lt;p&gt;Apparently mol is huge unit and that makes in not very useful for &lt;em&gt;normal life&lt;/em&gt;, however there is another simple reason: there is no balance or cylinder that can measure mol. The unit has its logic in sense of reactions (equimolar etc.) but still every single experimental chemist has to start with weight or volume when he prepares his synthesis.&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="22" LastEditDate="2012-06-22T05:00:39.997" LastActivityDate="2012-06-22T05:00:39.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="677" PostTypeId="2" ParentId="662" CreationDate="2012-06-21T21:57:10.160" Score="5" Body="&lt;p&gt;For handling $\ce{NaOH}$ always use goggles (I know myself a case of a student that lost some 15 years ago her vision in the accident with $\ce{NaOH}$. In case of acid you'll get burn but with a bit of luck it will pass). In the case of even a diluted solution of a strong base, your eyes are gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing: check your gloves for the chemicals the cover. The usual latex gloves cannot protect against bases perfectly; nitrile gloves (usually bright blue) are much better in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Karen_Wetterhahn&quot; rel=&quot;nofollow&quot;&gt;Karen Wetterhahn&lt;/a&gt; had gloves on her hands when she worked with dimethylmercury solution. However, she didn't know that this compound goes through a latex in the blink of an eye. Of course, none of the compounds you have mentioned are that dangerous, but it's your health.  &lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorDisplayName="user15489" LastEditDate="2015-04-23T21:21:40.640" LastActivityDate="2015-04-23T21:21:40.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="678" PostTypeId="2" ParentId="666" CreationDate="2012-06-21T23:11:06.370" Score="10" Body="&lt;p&gt;It's been my experience that initiators are chosen for: 1) initiation temperature, 2) solubility, and 3) cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Initiation temperature - Most radical initiators are stable at room temperature; they need to be heated to dissociate. For example, the half-life of &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzoyl_Peroxide&quot;&gt;benzoyl peroxide&lt;/a&gt; at room temperature is very long, but it is about 30 minutes at 92 &amp;deg;C and 1 minute at 131 &amp;deg;C. However, some monomers boil or decompose at temperatures lower than those temperatures. AIBN &lt;a href=&quot;http://en.wikipedia.org/wiki/Azobisisobutyronitrile&quot;&gt;(azobisisobutyronitrile)&lt;/a&gt; decomposes above 65 &amp;deg;C, so less heat is needed. Both benzoyl peroxide and AIBN are suitable for bulk polymerization of most monomers, including acrylamides, acrylates, styrenes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Solubility - Benzoyl peroxide and AIBN are soluble in organic solvents, which include most monomers for bulk polymerizations. However, if you want to perform your polymerization in water, for example if you are doing an &lt;a href=&quot;http://en.wikipedia.org/wiki/Emulsion_polymerization&quot;&gt;emulsion polymerization&lt;/a&gt;, then you need a water-soluble initiator, like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxodisulfate&quot;&gt;peroxodisulfate salt&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Cost - Every initiator has a price, even if you are only using tiny amounts. AIBN may have a lower activation temperature than benzoyl peroxide, but AIBN is also more expensive. There are other more specialized inhibitors that are even more expensive. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-21T23:11:06.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="680" PostTypeId="1" AcceptedAnswerId="702" CreationDate="2012-06-24T00:44:47.633" Score="8" ViewCount="1021" Body="&lt;p&gt;Borax, $\ce{Na2B4O7}$, is often marketed as a &quot;laundry booster&quot; under the brand &quot;20 Mule Team Borax&quot;.  The unit crystal of borax can be seen below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4eRv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other laundry products in the past have added &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_perborate&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium perborate&lt;/a&gt; (which can be produced from borax, hydrogen peroxide, and sodium hydroxide) as a bleaching additive.  The &quot;peroxide&quot; portion can readily be seen below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zf2Bw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through what mechanism does the plain borax &quot;boost&quot; the detergent?  Would it be more effective to throw some weak peroxide and a small amount of base into the laundry to form the perborate instead of using plain borax? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="7475" LastEditDate="2015-10-30T16:59:19.477" LastActivityDate="2015-10-30T16:59:19.477" Title="Does borax do anything more for &quot;boosting&quot; detergent than adding active oxygen would?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="681" PostTypeId="1" AcceptedAnswerId="682" CreationDate="2012-06-24T13:13:29.740" Score="8" ViewCount="377" Body="&lt;p&gt;My symmetry lecture notes state that there are seven high-symmetry 3D point groups, which have more than one rotation axis of order greater than 2: $T$, $T_d$, $T_h$, $O$, $O_h$, $I$, $I_h$. I sometimes see them called the “cubic point groups”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is: most molecules (or molecular fragments) I have encoutered in these groups are always $T_d$ (tetrahedral) or $O_h$ (octahedral). What are examples of molecules in the other groups?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-07-21T17:28:33.690" Title="Are all cubic point groups encountered?" Tags="&lt;symmetry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="682" PostTypeId="2" ParentId="681" CreationDate="2012-06-24T14:27:57.680" Score="8" Body="&lt;p&gt;I refer you to the second link in my answer for &lt;a href=&quot;https://chemistry.stackexchange.com/a/327/84&quot;&gt;How does one recognized Td/Oh symmetry in molecules?&lt;/a&gt;, which (according to the abstract) describes some high symmetry species of unusual point groups $T$, $O$ and $I$. Additionally, the molecule &lt;a href=&quot;http://www.mazepath.com/uncleal/boojum_p.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[6.6]chiralane&lt;/a&gt; has $T$ symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckminsterfullerene is a well known $I_{h}$ symmetric molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't know of any $T_h$ molecules but I'm confident they can be devised.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2012-06-24T14:27:57.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="683" PostTypeId="1" AcceptedAnswerId="695" CreationDate="2012-06-24T15:53:16.840" Score="10" ViewCount="5144" Body="&lt;p&gt;Burning metals like sodium should not be extinguished with standard fire extinguishers, but they require either sand or a special class D fire extinguisher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there is a fire in the lab and the closest fire extinguisher is one of those class D extinguishers meant for metal fires, can those be used for other types of fires?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main case where there might not be enough time to get specific equipment is probably when people are on fire. Can a class D extinguisher be used in this case, if it's the closest available one?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="9961" LastEditDate="2017-01-13T14:05:35.743" LastActivityDate="2017-01-13T14:05:35.743" Title="Can class D fire extinguishers be used for other fire types than metal fires?" Tags="&lt;safety&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="685" PostTypeId="1" CreationDate="2012-06-24T19:57:21.753" Score="7" ViewCount="134" Body="&lt;p&gt;There are many food studies which determine the concentration of a substance in a food item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://web.archive.org/web/20120624082807/https://www.fda.gov/food/foodsafety/foodcontaminantsadulteration/chemicalcontaminants/acrylamide/ucm053549.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this study&lt;/a&gt; determines the concentration of acrylamide in solid foods like fries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By what experimental technique is this done?&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorDisplayName="user7951" LastEditDate="2017-06-12T09:27:22.270" LastActivityDate="2017-06-12T09:27:22.270" Title="How do you determine the amount of acrylamide in foods such as baked fries?" Tags="&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="687" PostTypeId="1" CreationDate="2012-06-25T02:42:55.250" Score="10" ViewCount="1234" Body="&lt;p&gt;I cannot find much guidance in the literature. Any ideas as to what the mechanism of this reaction might be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WX7Mw.png&quot; alt=&quot;http://malaria.ourexperiment.org/data/files/2141.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="363" LastEditorUserId="7475" LastEditDate="2015-12-21T00:14:02.003" LastActivityDate="2015-12-21T00:14:02.003" Title="Oxidative chlorination mechanism (sulfide to sulfonyl chloride)" Tags="&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;synthesis&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="688" PostTypeId="2" ParentId="685" CreationDate="2012-06-25T03:24:22.157" Score="6" Body="&lt;p&gt;Check out the &lt;a href=&quot;https://web.archive.org/web/20120428133428/http://www.fda.gov/Food/FoodSafety/FoodContaminantsAdulteration/ChemicalContaminants/Acrylamide/ucm053537.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;methodology page&lt;/a&gt; for the FDA section on acrylamide quantitation, which goes into considerable detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, they start by mashing up some fries, and then extract and purify a fraction containing the acrylamide through a number of steps leading up to &lt;a href=&quot;http://en.wikipedia.org/wiki/High-performance_liquid_chromatography&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid chromatography&lt;/a&gt;. They then use quantitative &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_spectrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass spectrometry&lt;/a&gt; to determine the amount of acrylamide present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They add a small known concentration of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopic_labeling&quot; rel=&quot;nofollow noreferrer&quot;&gt;isotopically labelled&lt;/a&gt; acrylamide (i.e. acrylamide with all the carbons as carbon-13, which has &lt;em&gt;very low natural abundance&lt;/em&gt; and results in a molecule that is &lt;em&gt;approximately 3 g/mol heavier&lt;/em&gt;). This shows up in the mass spectrum as an offset peak. Then they can integrate the normal acrylamide mass spectrum peak and the labelled peak and obtain the concentration from the relative peak areas.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="4945" LastEditDate="2017-06-12T05:53:24.640" LastActivityDate="2017-06-12T05:53:24.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="689" PostTypeId="1" CreationDate="2012-06-25T07:43:28.367" Score="10" ViewCount="196" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2627/why-are-solid-phase-organics-often-conjugated-to-hcl&quot;&gt;Why are solid phase organics often conjugated to HCl?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm wondering why a lot of drugs are in this hydrochloride form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cetirizine hydrochloride&lt;/li&gt;&#xA;&lt;li&gt;Methylphenidate hydrochloride&lt;/li&gt;&#xA;&lt;li&gt;Tramadol hydrochloride&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm really curious to know why this form is so common for the presentation of a drug.&lt;/p&gt;&#xA;" OwnerUserId="15" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-06-25T19:52:28.020" Title="Why are many drugs in the hydrochloride form?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ClosedDate="2012-11-27T22:55:31.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="690" PostTypeId="2" ParentId="2627" CreationDate="2012-06-25T08:04:20.157" Score="15" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrochloride&quot;&gt;Wikipedia page on &lt;em&gt;hydrochloride&lt;/em&gt;&lt;/a&gt; says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Converting insoluble amines into hydrochlorides is a common way to make them &lt;strong&gt;water- and acid-soluble&lt;/strong&gt;. This is particularly desirable for substances used in medications. Many pharmaceutical substances are prepared as hydrochlorides so that they may be quickly released in the gastrointestinal tract; the body usually absorbs a hydrochloride within fifteen or thirty minutes. Very often hydrochlorides of amines have longer shelf-lives than their respective free bases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-06-25T08:04:20.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="691" PostTypeId="1" AcceptedAnswerId="5801" CreationDate="2012-06-25T09:54:08.273" Score="7" ViewCount="323" Body="&lt;p&gt;I tried to fractionate proteins as a function of their size with an ultrafiltration membrane. I observed a higher concentration of chloride in my permeate than in my feed or retentate. The proteins do not pass through the membrane and the filtration has been done at pH = 7. I asked, how is it possible to have a higher concentration in the permeate than in the feed or retentate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this explanation but I'm not sure about it :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It may be explained by the Donnan effect. Negatively charged proteins are retained in the retentate. To compensate the positive charges from the cations that are on the other side of the membrane, they attract the chloride. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it a possible way to explain this issue?&lt;/p&gt;&#xA;" OwnerUserId="364" LastEditorUserId="2095" LastEditDate="2013-08-07T04:13:03.403" LastActivityDate="2013-08-07T04:13:03.403" Title="Negative retention for chloride in hydrolysate filtration" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;purification&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="693" PostTypeId="1" AcceptedAnswerId="707" CreationDate="2012-06-25T15:38:52.223" Score="7" ViewCount="581" Body="&lt;p&gt;Mixing solid copper, $5\%$ vinegar, and hydrogen peroxide ($\ce{H2O2}$) causes copper acetate to form. The process will occur very slowly without hydrogen peroxide. Adding $\ce{H2O2}$ speeds the formation of copper acetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I test the solution to see if all of the acetic acid has been converted into copper acetate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, provide answers that employ only household items or readily available devices.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="23561" LastEditDate="2016-12-21T02:54:04.707" LastActivityDate="2016-12-21T02:54:58.027" Title="How can one optimise the formation of copper acetate?" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="694" PostTypeId="2" ParentId="2627" CreationDate="2012-06-25T19:52:28.020" Score="6" Body="&lt;p&gt;The hydrochloride salts of most alkaloids are also more crystalline, and thus easier to purify by crystallization.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-06-25T19:52:28.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="695" PostTypeId="2" ParentId="683" CreationDate="2012-06-26T00:42:31.317" Score="4" Body="&lt;p&gt;Unfortunately, as far as I can tell, specific class D fire extinguishers aren't even suitable for all types of class D fire, much less other types of fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper powder variety in particular has limited effectiveness against fires that aren't lithium-based, and the thermoplastic additives in the sodium chloride-based extinguishers may also catch fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dry sodium carbonate-based class D extinguishers should be generally effective but suboptimal against many types of fire, but I'd say in general, even if you have to go a little further, you should go for the proper sort of extinguisher to minimise accidental horrible disasters.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-06-26T00:42:31.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="696" PostTypeId="1" AcceptedAnswerId="763" CreationDate="2012-06-26T11:44:39.630" Score="23" ViewCount="521" Body="&lt;p&gt;Synthesising $\ce {CH4}$ from air and water (in a non-biological process) has been proposed as one form of energy storage. What are the chemical constraints at play here? That is to say, &lt;strong&gt;what sort of catalysts might be used, and what are the performance / lifetime issues with them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {CO2}$ capture and recycling (CCR) has been suggested as one form of round-trip electricity storage, with a round trip being:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;electricity &lt;/li&gt;&#xA;&lt;li&gt;$\to$ something that can hang around for hours to months &lt;/li&gt;&#xA;&lt;li&gt;$\to$ electricity some time later&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The proposal goes something like this: capture the $\ce {CO2}$ produced from the combustion of $\ce {CH4}$ in air in a gas-turbine electricity plant, and store it. Then, when there's surplus electricity, use it to synthesise $\ce {CH4}$ from the $\ce {CO2}$ plus water (I'm guessing electrolysis of water and a Fischer-Trop / Sabatier type reaction). And next time there's a deficit of electricity, burn the $\ce {CH4}$ to generate electricity, and capture the $\ce {CO2}$, and go around again. So, for the sake of this question, I'm specifically interested in the &lt;strong&gt;non-biological&lt;/strong&gt; synthesis of methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's put aside for one moment the issue of energy efficiency, as that's effectively an economics question. (The question being &quot;is the round-trip efficiency less than the ratio of off-peak to peak electricity prices?&quot;) Let's just agree that round-trip energy efficiency is one of the factors that influences the economics, and leave it at that for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: what sort of catalysts might be used, and what are the performance / lifetime issues with them?&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="30836" LastEditDate="2020-04-21T20:15:29.327" LastActivityDate="2020-04-21T20:15:29.327" Title="Energy-efficiency aside, what are the chemical constraints on CO₂ capture and methanation?" Tags="&lt;physical-chemistry&gt;&lt;synthesis&gt;&lt;hydrocarbons&gt;&lt;catalysis&gt;&lt;green-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="697" PostTypeId="1" CreationDate="2012-02-28T22:42:48.937" Score="17" ViewCount="9661" Body="&lt;p&gt;PTFE and PVDF (durapore) are both used in protein binding filter membranes (Millipore specifically). Chemically speaking the two polymers differ quite significantly due to the additional fluorides and molecular weight. Their protein-binding properties also differ. I'm curious about how the chemical differences affect their properties as a membrane material.&lt;/p&gt;&#xA;&lt;p&gt;From what I understand, the durability between the two is quite different. The PTFE membranes are used for the filtration of organic solvents where as the PVDF membranes are generally for low protein binding. Furthermore the PTFE membranes typically require a polyethylene support.&lt;/p&gt;&#xA;&lt;p&gt;I don't believe that the degree of polymerization or the architecture; if we were dealing with a copolymer I expect differently. Based on my molecular intuition, I would imagine that this is completely based on the chemistry of repeat unit.&lt;/p&gt;&#xA;&lt;p&gt;My question concerns a few things. What makes the PTFE a better solvent resistant polymer? The other is how does the reduced number of fluroides result in reduced protein binding?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Truzo.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Teflon/PTFE&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CCErL.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h2&gt;PVDF&lt;/h2&gt;&#xA;" OwnerUserId="57" OwnerDisplayName="bobthejoe" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-12-14T12:24:31.537" Title="How does Polytetrafluoroethylene (Teflon) differ from Polyvinylidene fluoride (PVDF) as a protein binding membrane material?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="698" PostTypeId="1" AcceptedAnswerId="699" CreationDate="2012-06-26T18:01:17.667" Score="33" ViewCount="1294" Body="&lt;p&gt;I was reading up on a wonderful little chemical compound known as chlorine trifluoride (&lt;span class=&quot;math-container&quot;&gt;$\ce{ClF3}$&lt;/span&gt;). For a primer, check out Dr. Derek Lowe's blog post here: &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2008/02/26/sand_wont_save_you_this_time&quot; rel=&quot;noreferrer&quot;&gt;Sand Won't Save You This Time&lt;/a&gt;. The title of the post is quite telling; this compound is so reactive with other compounds - any other compound, really - that it will &lt;em&gt;set sand on fire&lt;/em&gt;, making this old standby for extinguishing lab fires useless and putting the chemical in very rare company indeed. Obviously, water is a huge no-no for chlorine trifluoride; it will exothermically react with the water to produce clouds of steam containing two nasty acids (hydrochloric and hydrofluoric) and a host of chlorate and fluoride compounds. Somewhere in the middle of all the heat, fluorine and organic-ish compounds, you could probably even get some fluorine oxides, which would just LOVE to rapidly reduce their oxidation state (an event normally accompanied by shrapnel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was intrigued; I went and looked up the &lt;a href=&quot;http://msds.lindeus.com/files/msds/WPS_LIND_082_NA_MSDS_FINAL_REV_9_10_10.pdf&quot; rel=&quot;noreferrer&quot;&gt;MSDS&lt;/a&gt; for this chemical to see what the instructions were for dealing with industrial spills of this stuff (apparently, there's a decent market for it in the semiconductor industry, and probably also for plastics to produce highly fluorinated polymers like Teflon). &quot;A good pair of running shoes&quot; can't be the only answer to a &lt;span class=&quot;math-container&quot;&gt;$\ce{ClF3}$&lt;/span&gt; fire. &lt;em&gt;(well, actually, it pretty much is; get everyone in a one mile radius and two miles downwind the hell away from the area)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here's the question. While this stuff has some serious &quot;synergy&quot; when it comes to being an oxidizing agent (it's a better oxidizer than pure oxygen, loads better than pure chlorine, better even than pure fluorine gas), the NFPA flammability rating of the chemical is 0. This, despite reports from those who originally studied it that the chemical is hypergolic with every known fuel substance, and most things you wouldn't consider fuel (like asbestos, sand, concrete, brick, earth, test engineers). The resulting reaction certainly looks like fire; intense heat, bright light, sparks, smoke, the works. So, if chlorine trifluoride doesn't &quot;burn&quot; by the NFPA's definition, what &lt;em&gt;is&lt;/em&gt; the NFPA's definition of &quot;burning&quot;?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2018-10-04T15:45:38.247" LastActivityDate="2018-10-04T15:45:38.247" Title="Flammability (NFPA) - how is it defined?" Tags="&lt;redox&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="699" PostTypeId="2" ParentId="698" CreationDate="2012-06-26T23:33:53.183" Score="28" Body="&lt;p&gt;You have found one of biggest flaws in the  &lt;a href=&quot;http://en.wikipedia.org/wiki/NFPA_704&quot;&gt;NFPA &quot;Fire Diamond&quot;&lt;/a&gt;. The NFPA rates the flammability of the compound &lt;strong&gt;as a fuel&lt;/strong&gt;. ClF&lt;sub&gt;3&lt;/sub&gt; is not a fuel. It is an oxidant. ClF&lt;sub&gt;3&lt;/sub&gt; is playing the role that oxygen normally does. It causes fuels (reductants) to burn by oxidizing them rapidly and exothermically. Oxidants are incapable of creating fire in absence of fuel regardless of temperature. Fires started by ClF&lt;sub&gt;3&lt;/sub&gt; continue to burn after being smothered because oxygen isn't necessary to continue the oxidation process. And since ClF&lt;sub&gt;3&lt;/sub&gt; is denser than air, it gets trapped under the sand. Many other powerful oxidants, like &lt;a href=&quot;http://www.msdshazcom.com/MSDS/A/ARCHIVE/WCD00008/WCD008DF.HTM&quot;&gt;fluorine&lt;/a&gt; and &lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9926440&quot;&gt;perchloric acid&lt;/a&gt;, have low flammability ratings. &lt;a href=&quot;http://www.praxair.com/praxair.nsf/AllContent/E88BE35D13EF81F685256A860081E837/%24File/p4638h.pdf&quot;&gt;Oxygen&lt;/a&gt; also has a flammability rating of zero. They do not burn; they cause other compounds to burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale for the flammability rating has to do with &lt;a href=&quot;http://en.wikipedia.org/wiki/Flash_point&quot;&gt;flash point&lt;/a&gt;, which is the temperature at which a fuel forms an ignitable mixture with air. Since ClF&lt;sub&gt;3&lt;/sub&gt; doesn't burn (isn't a fuel), and doesn't react with air (N&lt;sub&gt;2&lt;/sub&gt;, O&lt;sub&gt;2&lt;/sub&gt;, CO&lt;sub&gt;2&lt;/sub&gt;, Ar, etc, which are all inert towards oxidation), it does not have a measurable flash point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The danger posed by ClF&lt;sub&gt;3&lt;/sub&gt; is summed in the yellow (reactivity or stability) diamond and the white (physical and chemical hazards) diamond. Your &lt;a href=&quot;http://msds.lindeus.com/files/msds/WPS_LIND_082_NA_MSDS_FINAL_REV_9_10_10.pdf&quot;&gt;msds&lt;/a&gt; rates ClF&lt;sub&gt;3&lt;/sub&gt; as a 3 in reactivity/stability, which is appropriate. Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/NFPA_704&quot;&gt;NPFA rating article&lt;/a&gt;, even mentions chlorine trifluoride:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3 -  Capable of detonation or explosive decomposition but requires a strong initiating source, must be heated under confinement before initiation, reacts explosively with water, or will detonate if severely shocked (e.g. ammonium nitrate, chlorine trifluoride).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The white triangle will have appropriate hazard information. For ClF&lt;sub&gt;3&lt;/sub&gt;, it has the symbols &lt;strong&gt;OXY&lt;/strong&gt;, which means it is a strong oxidizer, and &lt;del&gt;W&lt;/del&gt;, which means it reacts with water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not found the specific reaction that ClF&lt;sub&gt;3&lt;/sub&gt; undergoes with sand (mostly SiO&lt;sub&gt;2&lt;/sub&gt;). I expect that it involves ${\ce{SiO2 -&amp;gt; SiF}}_6^{2-}$, except that this half reaction is not an oxidation.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-06-27T11:39:03.847" LastActivityDate="2012-06-27T11:39:03.847" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="702" PostTypeId="2" ParentId="680" CreationDate="2012-06-27T18:52:23.310" Score="4" Body="&lt;p&gt;Borax was used as a &quot;builder&quot; in washing formulations long before perborate was used as a bleach. &quot;Builder&quot; is the terminology for additives which reduce the influence of water hardness. Other examples of builders include trisodium phosphate, soda, polyphosphates (like Graham's salt) or today most often aluminosilicate ion exchangers (Phosphates are more or less forbidden nowadays because of eutrophication of rivers and seas).&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorDisplayName="user95" LastEditDate="2012-07-10T13:29:33.713" LastActivityDate="2012-07-10T13:29:33.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="703" PostTypeId="1" AcceptedAnswerId="755" CreationDate="2012-06-27T18:58:45.180" Score="7" ViewCount="895" Body="&lt;p&gt;I want to qualitatively test for the presence of oxygen within argon gas. I would like the test to be positive if approx $1\%$ or more is oxygen, but definitely negative if less than $0.1\%$. Anywhere in between is fine—A quantitative test is a bonus. I was thinking of heating iron filings in the gas to deduce it. I'm not sure what apparatus and setup  I would need since heating gas expands, I would like to use ultra low tech equipment. I'm thinking of using water or cooking oil displacement to hold the extra gas, but I don't know if argon would dissolve, or how much oxygen would dissolve. Preferably a nonflammable, certainly nontoxic liquids (or tests).&lt;/p&gt;&#xA;" OwnerUserId="371" LastEditorUserId="23561" LastEditDate="2016-12-21T02:52:34.143" LastActivityDate="2016-12-21T02:52:34.143" Title="Test presence of oxygen &gt; 1%" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="704" PostTypeId="1" AcceptedAnswerId="797" CreationDate="2012-06-27T21:00:01.530" Score="21" ViewCount="2598" Body="&lt;p&gt;I have adapted the lab experiment described in this &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed101195v&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Chem. Ed.&lt;/em&gt; article&lt;/a&gt; for the organic chemistry lab course I teach. In the past year, the first year of implementation, I have had 67 failures (100% failure rate) on the extraction part of this experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment involves the isolation/separation of the active components in Excedrin(R): acetylsalicylic acid, acetaminophen, and caffeine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pvGR8.jpg&quot; alt=&quot;Acetylsalicylic acid, acetaminophen, and caffeine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is in two parts. Part 1 involves analysis by TLC and separation by column chromatography. This part works beautifully. Part 2 involves separation by successive extractions with increasingly strong bases. Acetylsalicylic acid was extracted with 10% aqueous Na&lt;sub&gt;2&lt;/sub&gt;CO&lt;sub&gt;3&lt;/sub&gt;. Then, acetaminophen was extracted with 3 M NaOH. Finally, caffeine was recovered by evaporation of the remaining organic layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Part 2 We recovered lots of acetylsalicylic acid and caffeine, but no-one recovered any acetaminophen. Below is a list of things I tried in order to improve the results. None of them worked. I know from TLC analysis of the initial organic solution that acetaminophen was present. What happened to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I have tried:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Changing the concentration of both bases (up and down).&lt;/li&gt;&#xA;&lt;li&gt;Using a weaker base for the first extraction (like NaHCO&lt;sub&gt;3&lt;/sub&gt; or Na&lt;sub&gt;3&lt;/sub&gt;PO&lt;sub&gt;4&lt;/sub&gt;). &lt;/li&gt;&#xA;&lt;li&gt;Increasing the HCl concentration used to work up the second extraction. &lt;/li&gt;&#xA;&lt;li&gt;No acetaminophen was extracted with the weak base, nor was any left&#xA;behind after the second extraction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Where has my acetaminophen gone? &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="30" LastEditDate="2012-06-27T21:21:26.030" LastActivityDate="2014-02-21T22:49:53.127" Title="67 failures at extracting acetaminophen from Excedrin(R)" Tags="&lt;organic-chemistry&gt;&lt;teaching-lab&gt;&lt;extraction&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="705" PostTypeId="2" ParentId="704" CreationDate="2012-06-28T00:26:32.837" Score="5" Body="&lt;p&gt;Here's what I &lt;a href=&quot;http://en.wikipedia.org/wiki/Cold_water_extraction&quot;&gt;suggest&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mash up the pills in a mortar&lt;/li&gt;&#xA;&lt;li&gt;Dissolve the powder in cold distilled water, use a beaker&lt;/li&gt;&#xA;&lt;li&gt;Shake well&lt;/li&gt;&#xA;&lt;li&gt;Put the beaker in a freezer and wait until right before the water freezes to remove it&lt;/li&gt;&#xA;&lt;li&gt;Carefully decant the solution&lt;/li&gt;&#xA;&lt;li&gt;Check what's left both in the water and the powder residue&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly how much water is optimal for the separation, that you will have to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powder should contain acetaminophen.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-06-28T00:26:32.837" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="706" PostTypeId="1" CreationDate="2012-06-28T00:30:42.053" Score="6" ViewCount="1198" Body="&lt;p&gt;I am in high school, and I would like to do an experiment on this for a project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done quite a bit of research into the different methods I could use (see 2.4.2 &lt;a href=&quot;http://www.acgpubs.org/RNP/2011/Volume%205/Issue%201/17_RNP-1008-300.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) but I still cannot find enough information to formulate a hypothesis. It would be useful iff you could point me in the right direction. So, how does heat affect how much flavonoids are present in commercially available orange juice? (without pulp)&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="22" LastEditDate="2012-06-28T05:04:24.297" LastActivityDate="2013-01-24T00:02:55.460" Title="How does temperature affect the concentration of flavonoids in orange juice?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="707" PostTypeId="2" ParentId="693" CreationDate="2012-06-28T00:36:46.443" Score="6" Body="&lt;p&gt;Your question and title do not correspond, IMHO. I'll answer one of both:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vinegar is mostly water, with some acetic acid ($5\%$). A classic experiment, performed by a lot of kids at school, is to mix vinegar with baking soda. The baking soda neutralizes the acid and produces carbon dioxide. The carbon dioxide is what you're looking for: it's the sign that the reaction is taking place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pour a pinch of baking soda inside your solution. If it makes bubbles on contact, your solution still contains a lot of vinegar. If you don't see any bubbles, even when stirring, most of the acid has been consumed.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="23561" LastEditDate="2016-12-21T02:54:58.027" LastActivityDate="2016-12-21T02:54:58.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="708" PostTypeId="1" AcceptedAnswerId="5661" CreationDate="2012-06-28T08:06:30.327" Score="8" ViewCount="426" Body="&lt;p&gt;I am looking into implementation of Ewald summation techniques for a number of point charges in periodic 3D space (i.e. a molecular simulation box with periodic boundary conditions). The “mainstream” Ewald-type techniques I know are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;original Ewald scheme&lt;/li&gt;&#xA;&lt;li&gt;particle-mesh Ewald (PME)&lt;/li&gt;&#xA;&lt;li&gt;smooth particle-mesh Ewald (SPME)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, while I can find papers describing each of those, I have not been able to find a comprehensive article/white paper/technique note comparing their implementations and their CPU timings on a typical dense system (liquid, protein in solvent, whatever…). What resources do you know of on this topic?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-07-22T21:45:17.383" Title="Efficiency of various Ewald-type schemes for the summation of electrostatic interactions" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="709" PostTypeId="2" ParentId="696" CreationDate="2012-06-29T09:47:56.493" Score="4" Body="&lt;p&gt;Your question was about constraints, but see mostly the ways this could happen:&#xA;Methanogenic archae can metabolize Hydrogen and Carbondioxide to Methane, the input of electricity would be needed for Hydrolysis.&#xA;The $H_2$ path of methanogenesis is highly relevant in agricultural biogas plants, so you can probably get the right sort of archae from there or from grass eating animals. For anaerobic digestion in Wastewater treatment plant, the path via acetic acid seems to be far more prelevant, so that's maybe a worse place to look for your methane building workforce. One text on the subject: &lt;a href=&quot;http://www.zdnet.com/blog/green/h2o-co2-ch4-thanks-to-archaeans/3534&quot; rel=&quot;nofollow&quot;&gt;http://www.zdnet.com/blog/green/h2o-co2-ch4-thanks-to-archaeans/3534&lt;/a&gt; &#xA;The biological way of storing electricity as Methane is investigated by these guys, I think: &lt;a href=&quot;http://www.bioferm-energy.com/bf/de_de/company.html&quot; rel=&quot;nofollow&quot;&gt;http://www.bioferm-energy.com/bf/de_de/company.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I found zero info about how exactly the nearest commercial application ( &lt;a href=&quot;http://www.solar-fuel.net/&quot; rel=&quot;nofollow&quot;&gt;http://www.solar-fuel.net/&lt;/a&gt; have a pilot plant, the technology doens't handle load changes very well - pity, since that's the entire point) exactly works. I have to assume they use some Sabatier process.&lt;/p&gt;&#xA;" OwnerUserId="102" LastActivityDate="2012-06-29T09:47:56.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="710" PostTypeId="1" CreationDate="2012-06-29T10:44:52.307" Score="46" ViewCount="11229" Body="&lt;p&gt;I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, am I right to think that &quot;repulsion occurs when atoms are too close together&quot; comes from electrostatic interaction?&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="28918" LastEditDate="2016-04-19T03:10:01.467" LastActivityDate="2017-05-14T12:45:10.860" Title="Fundamental forces behind covalent bonding" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;covalent-compounds&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="37" ContentLicense="CC BY-SA 3.0" />
  <row Id="711" PostTypeId="2" ParentId="710" CreationDate="2012-06-29T17:39:32.880" Score="2" Body="&lt;p&gt;Both the attraction and repulsion are the result of the electromagnetic interaction.  At long distances, two atoms attract each other because of induced dipole-induced dipole interactions. When they get close enough together, the  exchange interaction acts on the non-valence electrons of the atoms and forces them into higher energy states. This makes the atoms repel each other at short distances.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="47" LastEditDate="2016-04-19T18:56:00.360" LastActivityDate="2016-04-19T18:56:00.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="712" PostTypeId="2" ParentId="710" CreationDate="2012-06-29T21:02:42.260" Score="10" Body="&lt;p&gt;Thinking about this like a physicist, there are four fundamental forces: the strong nuclear force, the weak nuclear force, the electromagnetic force, and gravity. The strong nuclear force holds the protons and neutrons in the nucleus together. The weak nuclear force causes beta decay. Those two might be considered chemistry. I don't, but some people do. Gravity is much too weak to have any effect on chemistry. So that leaves the electromagnetic force to control nearly all of chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a simple conceptual level, that's all there is. The nuclei are both positively charged, so they repel each other. The electrons are negatively charged, so they are attracted to their respective nuclei. When the electron clouds get close enough to interact with both nuclei, then they begin to pull the nuclei together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper explanation requires quantum mechanics. When the atoms are separated, you can use the Schrödinger equation with the electric potential from the nucleus. That gives you the electron orbitals for an atom all by itself. When the two atoms get close together, you use the electric potential for both nuclei in the Schrödinger equation. The solution is then the molecular orbital rather than the atomic orbitals. Because the Schrödinger equation is impossible to solve exactly for a molecule, chemists need an approximation. The usual approximation is to build the molecular orbital out of the atomic orbitals by adding and subtracting the atomic orbitals. This is where the ideas of $sp$-, $sp^2$-, and $sp^3$-hybridizaton, and $\pi$- and $\sigma$-bonding come from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information, most introductory college-level general chemistry texts should discus this. As an example, I pulled most of the above explanation from Zumdahl's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/061852844X&quot;&gt;Chemistry&lt;/a&gt;. In the 5th edition, this is in chapters 8 and 9 (the current edition appears to be the 7th). This is a much more important idea in organic chemistry, so those textbooks usually review it in the first one or two chapters. The organic chemistry book I have in front of me at the moment is McMurray's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0840054440&quot;&gt;Organic Chemistry&lt;/a&gt;. This is discussed in chapter 1 of the 3rd edition of that book (current edition is the 8th).&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-06-29T21:02:42.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="714" PostTypeId="1" AcceptedAnswerId="732" CreationDate="2012-06-30T20:55:43.890" Score="11" ViewCount="688" Body="&lt;p&gt;When looking at excited states of molecules, $\delta$ bonds are relatively common, but I've never come across a molecule with a $\delta$ bond in its ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there molecules with $\delta$ bonds in their ground states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about even higher angular momentum states ($\phi$ and $\gamma$ bonds)?&lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2017-01-13T14:02:47.030" Title="Are there any molecules with delta bonds in their ground states?" Tags="&lt;bond&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="715" PostTypeId="1" CreationDate="2012-06-30T21:17:53.863" Score="8" ViewCount="751" Body="&lt;p&gt;I've searched a lot for this, but I still cannot find any good explanations for how these two factors affect flavonoid concentrations. &lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2012-08-13T09:56:43.607" Title="How does brewing time and brewing temperature affect the flavonoid concentration of green tea?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="716" PostTypeId="2" ParentId="710" CreationDate="2012-07-01T02:24:59.713" Score="45" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The role of Pauli Exclusion in bonding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an unfortunate accident of history that because chemistry has a very convenient and predictive set of approximations for understanding bonding, some of the details of why those bonds exist can become a bit hard to discern. It's not that they aren't there -- they most emphatically are! -- but you often have to dig a bit deeper to find them. They are found in physics, in particular in the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;noreferrer&quot;&gt;Pauli exclusion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemistry as avoiding black holes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take your attraction question first. What causes that? Well, in one sense that question is easy: it's electrostatic attraction, the interplay of pulls between positively charged nuclei and negatively charged electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even in saying that, something is wrong. Here's the question that points that out: If nothing else was involved except electrostatic attraction, what would be the most stable configuration of two or more atoms with a mix of positive and negative charges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to that is a bit surprising. If the charges are balanced, the only stable, non-decaying answer for conventional (classical) particles is always the same: &quot;a very, very small black hole.&quot; Of course you could modify that a bit by assuming that the &lt;em&gt;strong&lt;/em&gt; force is for some reason stable, in which case the answer becomes &quot;a bigger atomic nucleus,&quot; one with &lt;em&gt;no&lt;/em&gt; electrons around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or maybe atoms as Get Fuzzy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point some of you reading this should be thinking loudly &quot;Now wait a minute! Electrons &lt;em&gt;don't&lt;/em&gt; behave like point particles in atoms, because quantum uncertainty makes them 'fuzz out' as they get close to the nucleus.&quot; And that is exactly correct -- I'm fond of quoting that point myself in other contexts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the issue here is a bit different, since even &quot;fuzzed out&quot; electrons provide a poor barrier for keeping other electrons away by electrostatic repulsion alone, precisely because their charge is so diffuse. The case of electrons that lack Pauli exclusion is nicely captured by &lt;a href=&quot;http://en.wikipedia.org/wiki/Richard_Feynman&quot; rel=&quot;noreferrer&quot;&gt;Richard Feynman&lt;/a&gt; in his &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Feynman_Lectures_on_Physics&quot; rel=&quot;noreferrer&quot;&gt;Lectures on Physics&lt;/a&gt;, in Volume III, Chapter 4, page 4-13, Figure 4-11 at the top of the page. The outcome Feynman describes is pretty boring, since atoms would remain simple, smoothly spherical, and about the same size as more and more protons and electrons get added in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Feynman does not get into atoms how such atoms would interact, there's a problem there too. Because the electron charges would be so diffuse in comparison to the nuclei, the atoms would pose no real barrier to each other until the nuclei themselves begin to repel each other. The result would be a very dense material that would have more in common with [neutronium[(&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutronium&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Neutronium&lt;/a&gt;) than with conventional matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now I'll just forge ahead with a more classical description, and capture the idea of the electron cloud simply by asserting that each electron is selfish and likes to capture as much &quot;address space&quot; (see below) as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Charge-only is boring!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while you can finagle with funny configurations of charges that might prevent the inevitable for a while by pitting positive against positive and negative against negative, positively charged nuclei and negatively charged electrons with nothing much else in play will always wind up in the same bad spot: either as very puny black holes, or as tiny boring atoms that lack anything resembling chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A universe full of nothing but various sizes of black holes or simple homogenous neutronium is not very interesting!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventing the collapse&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to understand atomic electrostatic attraction properly, you must &lt;em&gt;start&lt;/em&gt; with the inverse issue: What in the world is keeping these things from simply collapsing down to zero size -- that is, where is the &lt;em&gt;repulsion&lt;/em&gt; coming from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is your next question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, am I right to think that &quot;repulsion occurs when atoms are too close together&quot; comes from electrostatic interaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No; that is simply wrong. In the absence of &quot;something else,&quot; the charges will wiggle about and radiate until any temporary barrier posed by identical charges simply becomes irrelevant... meaning that once again you will wind up with those puny black holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What keeps atoms, bonds, and molecules stable is always something else entirely, a &quot;force&quot; that is not traditionally thought of as being a force at all, even thought it is unbelievably powerful and can prevent even two nearby opposite electrical charges from merging. The electrostatic force is enormously powerful at the tiny separation distances within atoms, so anything that can stop charged particles from merging is impressive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;repulsive force that is not a force&quot; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;noreferrer&quot;&gt;Pauli exclusion&lt;/a&gt; I mentioned earlier. A simple way to think of Pauli exclusion is that identical material particles (electrons, protons, and neutrons in particular) all insist on having completely unique &quot;addresses&quot; to tell them apart from other particles of the same type. For an electron this address includes: where the electron is located in space, how fast and in what direction it is moving (momentum), and one last item called spin, which can only have on of two values that are usually called &quot;up&quot; or &quot;down.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can force such material particles (called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermions&quot; rel=&quot;noreferrer&quot;&gt;fermions&lt;/a&gt;) into nearby addresses, but with the exception of that up-down spin part of the address, doing so always increases the energy of at least one of the electrons. That required increase in energy is a nutshell is why material objects &lt;a href=&quot;https://physics.stackexchange.com/a/23821/7670&quot;&gt;push back when you try to squeeze them&lt;/a&gt;. Squeezing them requires minutely reducing the available space of many of the electrons in the object, and those electrons respond by capturing the energy of the squeeze and using it to push right back at you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take that thought and bring it back to the question about where repulsion comes from when to atoms bond at a certain distance, &lt;em&gt;but no closer&lt;/em&gt;. They are the same mechanism!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, two atoms can &quot;touch&quot; (move so close, but no closer) only because they both have a lot of electrons that require separate space, velocity, and spin addresses. Push them together and they start hissing like cats from two households who have suddenly been forced to share the same house. (If you own multiple cats, you'll know exactly what I mean by that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens is that the overall set of plus-and-minus forces of the two atoms is trying really hard to crush &lt;em&gt;all&lt;/em&gt; of the charges down into a single very tiny black hole -- not into some stable state! It is only the hissing and spitting of the overcrowded and very unhappy electrons that keeps this event from happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orbitals as juggling acts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just how does that work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sort of a juggling act, frankly. Electrons are allowed to &quot;sort of&quot; occupy many different spots, speeds, and spins (mnemonic $s^3$, and &lt;em&gt;no&lt;/em&gt;, that is not standard, I'm just using it for convenience in this answer only) at the same time, due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;noreferrer&quot;&gt;quantum uncertainty&lt;/a&gt;. However, it's not necessary to get into that here beyond recognizing that every electron tries to occupy as much of its local $s^3$ address space as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Juggling between spots and speeds requires energy. So, since only so much energy is available, this is the part of the juggling act that gives atoms size and shapes. When all the jockeying around wraps up, the lowest energy situations keep the electrons stationed in various ways around the nucleus, not quite touching each other. We call those special solutions to the crowding problem &lt;em&gt;orbitals&lt;/em&gt;, and they are very convenient for understanding and estimating how atoms and molecules will combine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orbitals as specialized solutions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's still a good idea to keep in mind that orbitals are not exactly fundamental concepts, but rather outcomes of the much deeper interplay of Pauli exclusion with the unique masses, charges,  and configurations of nuclei and electrons. So, if you toss in some weird electron-like particle such as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon&quot; rel=&quot;noreferrer&quot;&gt;muon&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Positron&quot; rel=&quot;noreferrer&quot;&gt;positron&lt;/a&gt;, standard orbital models have to be modified significantly and applied only with great care. Standard orbitals can also get pretty weird just from having unusual geometries of fully conventional atomic nuclei, with the unusual dual hydrogen bonding found in boron hydrides such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Diborane&quot; rel=&quot;noreferrer&quot;&gt;diborane&lt;/a&gt; probably being the best example. Such bonding is odd if viewed in terms of conventional hydrogen bonds, but less so if viewed simply as the best possible &quot;electron juggle&quot; for these compact cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Jake! The bond!&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on to the part that I find delightful, something that underlies the whole concept of chemical bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that it takes energy to squeeze electrons together in terms of the main two parts of their &quot;addresses,&quot; the spots (locations) and speeds (momenta)? I also mentioned that spin is different in this way: the only energy cost for adding two electrons with different &lt;em&gt;spin&lt;/em&gt; addresses is that of conventional electrostatic repulsion. That is, there is no &quot;forcing them closer&quot; Pauli exclusion cost like you get for locations and velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might think &quot;but electrostatic repulsion is huge!&quot;, and you would be exactly correct. However, compared to the Pauli exclusion &quot;non-force force&quot; cost, the energy cost of this electrostatic repulsion is actually quite small -- so small that it can usually be ignored for small atoms. So when I say that Pauli exclusion is powerful, I &lt;em&gt;mean&lt;/em&gt; it, since it even makes the enormous repulsion of two electrons stuck inside the same tiny sector of a single atom look so insignificant that you can usually ignore its impact!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's secondary, because the real point is this: When two atoms approach each other closely, the electrons start fighting fierce energy-escalation battles that keep both atoms from collapsing all the way down into a black hole. But there is one exception to that energetic infighting: spin! For spin and spin alone, it become possible to get significantly closer to that final point-like collapse that all the charges &lt;em&gt;want&lt;/em&gt; to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin thus becomes a major &quot;hole&quot; -- the only such major hole -- in the ferocious armor of repulsion produced by Pauli exclusion. If you interpret atomic repulsion due to Pauli exclusion as the norm, then spin-pairing two electrons becomes another example of a &quot;force that is not a force,&quot; or a pseudo force. In this case, however, the result is a net &lt;em&gt;attraction&lt;/em&gt;. That is, spin-pairing allows two atoms (or an atom and an electron) to approach each other more closely that Pauli exclusion would otherwise permit. The result is a significant release of electrostatic attraction energy. That release of energy in turn creates a stable bond, since it cannot be broken unless that same energy is returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sharing (and stealing) is cheaper&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if two atoms (e.g. two hydrogen atoms) each have an outer orbital that contains only one electron, those two electrons can sort of look each other over and say, &quot;you know, if you spin downwards and I spin upwards, we could both share this space for almost no energy cost at all!&quot; And so they do, with a net release of energy, producing a covalent bond if the resulting spin-pair cancels out positive nuclear charges equally on both atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in some cases the &quot;attractive force&quot; of spin-pairing is so overwhelming greater for one of the two atoms that it can pretty much fully overcome (!) the powerful electrostatic attraction of the other atom for its own electron. When that happens, the electron is simply &lt;em&gt;ripped away&lt;/em&gt; from the other atom. We call that an ionic bond, and we act as it if it's no big deal. But it is truly an amazing thing, one that is possible only because the pseudo force of spin-pairing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line: Pseudo forces are important!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for having given such a long answer, but you happened to ask a question that cannot be answered correctly without adding in some version of Pauli &quot;repulsion&quot; and spin-pair &quot;attraction.&quot; For that matter, the size of an atom, the shape of its orbitals, and its ability to form bonds similarly all depend on pseudo forces.&lt;/p&gt;&#xA;" OwnerUserId="83" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2012-10-04T17:12:20.860" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="720" PostTypeId="2" ParentId="549" CreationDate="2012-07-02T02:37:58.397" Score="11" Body="&lt;p&gt;The van der Waals equation can't be derived from first principles.  It is an &lt;em&gt;ad-hoc&lt;/em&gt; formula.  There is a &quot;derivation&quot; in statistical mechanics from a partition function that is engineered to give the right answer.  It also cannot be derived from first principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gas is a collection of molecules that do not cohere strongly enough to form a liquid or a solid.  The volume of a gas is, as long as the molecules fit into it, the volume of the container holding the gas.  In other words, the volume of a gas is NOT a property of the gas directly, but a property of the container.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-07-02T02:37:58.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="722" PostTypeId="1" AcceptedAnswerId="723" CreationDate="2012-07-02T13:41:17.500" Score="16" ViewCount="2051" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How to name this cyclic compound with an O in it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was going to name it cyclobutane, but not all of the atoms making up the ring are carbons, one is an oxygen. So, then I decided to name it butane, but that doesn't make sense because there is a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DXF6v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DXF6v.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="7448" LastEditDate="2018-02-21T13:08:48.677" LastActivityDate="2018-02-21T13:08:48.677" Title="How to name this cyclic compound with an O in it?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="723" PostTypeId="2" ParentId="722" CreationDate="2012-07-02T14:53:39.897" Score="20" Body="&lt;p&gt;Well, the molecule you draw is &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahydrofuran&quot;&gt;&lt;strong&gt;tetrahydrofuran&lt;/strong&gt;&lt;/a&gt;, aka THF, but of course the only way you can guess that is if you already know the molecule. It's a very common organic solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you didn't know its common name, you could try to name it according to IUPAC nomenclature. It's an &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/orgnom/ethers/ethers-01.html&quot;&gt;ether&lt;/a&gt;, and it's cyclic, so it can be named as &lt;strong&gt;oxacyclopentane&lt;/strong&gt;. But IUPAC has specific names for cyclic ethers, which in this case is &lt;strong&gt;oxolane&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bmGvY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://books.google.com/books?id=rxRHzOS-3xoC&quot;&gt;this organic chemistry book&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-04-16T19:54:33.907" LastActivityDate="2013-04-16T19:54:33.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="724" PostTypeId="1" AcceptedAnswerId="726" CreationDate="2012-07-02T14:54:06.807" Score="8" ViewCount="13231" Body="&lt;p&gt;What are some qualitative guidelines one should know regarding the temperature dependencies (of the gas and maybe the radiation, if that makes sense) for an absorption spectroscopy measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does, for example, a rise in temperature influence the absorption lines?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="22" LastEditDate="2012-07-02T17:02:57.053" LastActivityDate="2012-07-02T18:42:57.177" Title="Temperature dependence in absorption spectroscopy" Tags="&lt;thermodynamics&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="725" PostTypeId="2" ParentId="724" CreationDate="2012-07-02T18:08:34.163" Score="4" Body="&lt;p&gt;Temperature will change the population of the different states of the molecule, particularly the rotational states (i.e. in temperatures in ranges that are easily reached in the lab/ some factor around 300 K). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will affect the intensity of the respective absorption lines in rotational (or rot.-vib.) spectra and therefore the hull shape of the P and Q branches will change&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just found example applets: &lt;a href=&quot;http://rkt.chem.ox.ac.uk/tutorials/rotation/rot_spectra.html&quot; rel=&quot;nofollow&quot;&gt;http://rkt.chem.ox.ac.uk/tutorials/rotation/rot_spectra.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-07-02T18:08:34.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="726" PostTypeId="2" ParentId="724" CreationDate="2012-07-02T18:42:57.177" Score="7" Body="&lt;p&gt;I am aware of one effect: &lt;a href=&quot;http://en.wikipedia.org/wiki/Doppler_broadening&quot;&gt;doppler broadening&lt;/a&gt;. When the temperature increases, the speed distribution spreads out. That spreads out the distribution of the velocity component in the direction of your light source. The light the particles see from the light source then get red- or blue-shifted relative to what actually comes out of the source. So the particle population ends up absorbing a broader range of wavelengths than it did at a lower temperature. The effect is  $\Delta\lambda/\lambda_0 \propto \sqrt{T/m}$&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-07-02T18:42:57.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="727" PostTypeId="1" AcceptedAnswerId="733" CreationDate="2012-07-02T21:39:07.597" Score="15" ViewCount="17398" Body="&lt;p&gt;I picked up a metal fence loop cap today and took it home just because I liked the shape.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mpDhB.jpg&quot; alt=&quot;fence loopcap&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(^ not mine)&#xA;I cleaned it, but some of the paint is chipping off. I'm only vaguely thinking about lead, and I'm not going to chew on it, so I don't want to buy a test kit. Is there a way I could test for the presence of lead ions with common/ideally household chemicals? I'd be willing to wait until I'm at my school lab again if I needed HCl or something.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-15T15:20:42.583" Title="Can I test for lead with household chemicals?" Tags="&lt;ions&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="728" PostTypeId="2" ParentId="727" CreationDate="2012-07-03T03:03:01.030" Score="10" Body="&lt;p&gt;If you’re testing for whether it’s entirely made of lead, you can do it with a simple density calculation rather than anything chemical – an easy typical way to do this is to immerse it in a known volume of water and see how much volume the object adds, then divide the weight by that. Lead is the one of the heaviest common metals, with a density of just over 11 g/cm³.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand it’s just lead &lt;em&gt;content&lt;/em&gt; you’re looking for, I think you can go for, coincidentally enough, HCl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB: I have not tested the following procedure(s).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead (II) chloride isn’t very soluble &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead(II)_chloride&quot; rel=&quot;nofollow&quot;&gt;(0.9 g/100 ml at 20 degrees Celsius)&lt;/a&gt;, so if you just add a small quantity of strong HCl to a small sample of the metal and wait a while (leave it overnight or longer if you can, the reaction’s supposedly quite slow with lead: &lt;a href=&quot;http://www.youtube.com/watch?v=ZywexSds-1c&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;cf&lt;/em&gt; this video of trying it quickly and nothing happening, even with strong acid and strong peroxide&lt;/a&gt;. You might get a white precipitate of lead chloride forming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing happens, try adding a little potassium iodide. Lead (II) iodide is over ten times less soluble than the chloride, so you might get a precipitate with that if you don’t with the chloride. Note that adding hydrogen peroxide in the initial step will speed up the reaction of the acid with the metal, but hydrogen peroxide will also decompose dangerously rapidly if you then add iodide later (see the &lt;a href=&quot;http://www.youtube.com/watch?v=4N0m95PExHY&quot; rel=&quot;nofollow&quot;&gt;‘elephant toothpaste’ demonstration&lt;/a&gt;, for example). Getting boiling acid everywhere is not something I recommend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also then do a flame test – adding methanol to a lead salt and igniting should give you a &lt;a href=&quot;http://www.youtube.com/watch?v=IzJIWv6O6Lc&quot; rel=&quot;nofollow&quot;&gt;somewhat ambiguous blue-ish flame&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorDisplayName="user7951" LastEditDate="2015-05-15T15:16:48.263" LastActivityDate="2015-05-15T15:16:48.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="729" PostTypeId="1" AcceptedAnswerId="730" CreationDate="2012-07-03T05:42:39.643" Score="42" ViewCount="87587" Body="&lt;p&gt;I know &lt;strong&gt;&lt;em&gt;phenols are more acidic as compared to alcohols&lt;/em&gt;&lt;/strong&gt;, but are they considered different from alcohols?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, you can study something as a subset, but are &lt;strong&gt;phenols&lt;/strong&gt; considered a subset of alcohols, or are they considered as completely different from alcohols? My confusion is due to the fact that both of them contain a hydroxyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ow6mK.png&quot; alt=&quot;Phenol&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/V6znn.png&quot; alt=&quot;Alcohol&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="388" LastEditorUserId="10185" LastEditorDisplayName="user95" LastEditDate="2016-05-24T22:10:42.567" LastActivityDate="2019-11-05T23:00:32.287" Title="Differences between phenols and alcohols" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="730" PostTypeId="2" ParentId="729" CreationDate="2012-07-03T08:10:38.030" Score="49" Body="&lt;p&gt;Nope. Alcohols consist of an -&lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; group bonded to a &lt;em&gt;saturated&lt;/em&gt; carbon (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized, no multiple bonds).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://goldbook.iupac.org/A00204.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC says:&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;alcohols&lt;/h1&gt;&#xA;&lt;p&gt;Compounds in which a hydroxy group, -&lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt;, is attached to a saturated carbon atom &lt;span class=&quot;math-container&quot;&gt;$\ce{R3COH}$&lt;/span&gt;. The term 'hydroxyl' refers to the radical species, &lt;span class=&quot;math-container&quot;&gt;$\ce{HO^.}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://goldbook.iupac.org/P04539.html&quot; rel=&quot;noreferrer&quot;&gt;and&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h1&gt;phenols&lt;/h1&gt;&#xA;&lt;p&gt;Compounds having one or more hydroxy groups attached to a benzene or other arene ring, &lt;em&gt;e.g.&lt;/em&gt;, 2-naphthol:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0BVVA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BVVA.png&quot; alt=&quot;2-naphthol structure&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://goldbook.iupac.org/SVG/P04539.png&quot; rel=&quot;noreferrer&quot;&gt;iupac.org&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A phenol consists of an -&lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; bonded to an unsaturated &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; carbon. Thus, it does not qualify as an alcohol. One &lt;em&gt;can&lt;/em&gt; classify it as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Enol&quot; rel=&quot;noreferrer&quot;&gt;enol&lt;/a&gt;, though.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Really, to me, the classification doesn't matter. Classifications are artificial, what is important is, how well the properties fit in the classification.&lt;/p&gt;&#xA;&lt;p&gt;Many of the alcohol properties depend upon:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its unsaturated nature: oxidation to ketone/aldehyde/acid&lt;/li&gt;&#xA;&lt;li&gt;The weaker &lt;span class=&quot;math-container&quot;&gt;$\ce{R-O}$&lt;/span&gt; bond and its ability to easily break and form an &lt;span class=&quot;math-container&quot;&gt;$\ce{R+}$&lt;/span&gt; cation (this makes it a good participant in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N1}$&lt;/span&gt; reactions)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Phenol can obviously not be oxidised at the &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; to a ketone/acid (though one &lt;em&gt;can&lt;/em&gt; do stuff to make it into a quinone). Phenylic carbocations are unstable, thus we don't get any &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N1}$&lt;/span&gt; reactions, and the &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph-O}$&lt;/span&gt; bond stays put.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, most of the reactions of phenol depend upon its&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aromatic phenyl ring: All the EAS reactions&lt;/li&gt;&#xA;&lt;li&gt;Weaker &lt;span class=&quot;math-container&quot;&gt;$\ce{O-H}$&lt;/span&gt; bond (&lt;em&gt;i.e.&lt;/em&gt;, acidic nature): Reimer-Tiemann reaction, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Thus phenols and alcohols don't have too many reactions in common. So, in this case, they have been classified in a sensible manner--if phenols were classified as alcohols, we would basically be clubbing two radically different classes of compounds under one umbrella.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-11-05T23:00:32.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="731" PostTypeId="2" ParentId="662" CreationDate="2012-07-03T10:33:18.960" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;Meh, personally I'm not worried about the vapours. You'll be sane and be doing this in an at least somewhat ventilated space, right? Don't do anything outright stupid, but also don't fear the substances so much that your defences cause you to trip and break a toe. $\ce{HCl}$ and $\ce{NaOH}$ aren't exactly harmless, but they're not dimethyl mercury either. There are loads of each of those ions in your body, but as another answer mentioned, $\ce{NaOH}$ doesn't play nice with eyes. (It also turns the oils in your skin into soap, but that's generally not a lifetime liability.)&lt;/li&gt;&#xA;&lt;li&gt;Like others have said, hydrogen peroxide won't like light. It develops oxygen, so storing it in a gas-tight container in conditions where it's likely to decompose, could cause trouble. I'll get back to this later.&lt;/li&gt;&#xA;&lt;li&gt;You might want to keep your acid undiluted. If you dilute it, etching can become slow.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;AFAICT, you want polyethylene containers throughout. I wouldn't go with glass as a dropped glass container doesn't simply harmlessly bounce off the tiled floor like a plastic container does. I guess a polypropylene ice cream box would work as an etching bath too, but IIRC polypropylene doesn't like the acid all that much (but I haven't tried, and it might well be good &lt;em&gt;enough&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By all accounts I can find, you really, really want to keep the used etchant - it'll become a cupric chloride solution (which you may have to regenerate occasionally either by aerating or adding hydrogen peroxide). That means you have yet another storage problem. This is no longer pure $\ce{H2O2}$, so you have to think about where the evolved oxygen goes. And if you let the evolved oxygen vent, hydrogen chloride will also vent, attacking any metal nearby. The electronics people have tales of ruined tools sharing a garage with etchant solution. I've resigned myself to keeping the etchant indoors (in my kitchen in fact! where the cost of damaged stuff is least) which is at least better-ventilated than my garage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And whatever you do, don't pour the beautiful blue cupric chloride down the drain. It'll eat your sink and bring forth chemical armageddon. Don't water your garden with it either, as it's a herbicide. I guess that leaves the options of making it someone else's problem (pay someone to dispose of it for you), or use it to electroplate something with copper, using an inert (not just more copper) anode. Plate faster than the solution etches.&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorDisplayName="user15489" LastEditDate="2015-04-23T21:21:34.620" LastActivityDate="2015-04-23T21:21:34.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="732" PostTypeId="2" ParentId="714" CreationDate="2012-07-03T14:07:24.467" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium%28II%29_acetate&quot;&gt;Chromium(II) acetate&lt;/a&gt; has a Cr-Cr quadruple bond - and hence a $\delta$ bond - in its ground state. This sort of &quot;paddlewheel&quot; configuration, with four carboxylic acids coordinating a pair of metal ions, is reasonably common (see also copper(II) acetate, for instance), although the extent to which the metals can be considered covalently bonded varies.&lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2012-07-03T14:07:24.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="733" PostTypeId="2" ParentId="727" CreationDate="2012-07-04T10:48:56.857" Score="14" Body="&lt;p&gt;General remark: get a bit of lead (metal or salt - whatever you look for) to have a positive control for your experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More non-chemical tests for lead (metal): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;lead is soft: you can cut/scrape it quite easily with a normal knife. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sheets of lead are easily bent, and they stay so. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For bigger pieces: Hammering on it would leave a rather big impression of the hammer. The sound would be dull.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it usually looks grayish, not silvery-metallic, but the cut will be silvery.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If it is about the paint:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;what colour is it? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;white lead: $\ce{2 PbCO3 \cdot Pb(OH)2}$ I'm more familiar with that in the context of paintings or painting of wooden things (window frames)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{PbSO4}$ also white. I think it has also been used for things like window frames.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;side thought: lead paints for wood may have been popular due to the fact that the toxicity to microorganisms helps very much with the anti-fouling effect. Anti-corrosion paints for metal need different properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;lead yellow: $\ce{PbCrO4}$ Wikipedia says that was the original pigment for school buses. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{2PbO \cdot PbO2}$(Mennige) red anti-corrosion pigment that was at least here in Germany the most important lead based priming paint. Only difficulty with recognizing by color: lots of non-toxic replacements are produced with similar colour to have the familiar look...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Very sensitive but but not very distinctive test$^*$ for $\ce{Pb^2+}$: the sulfide is black and very insoluble. Kitchen chemistry: put some egg on and even the surface of the metal should turn black. Possibly some heat is needed (I didn't test, though). Otherwise, use acidic $\ce{Na2S}$ solution from the school lab. If the $\ce{Na2S}$ solution doesn't give a black precipitate, there could in theory still be lead in the paint which is not accessible by the sulfide due to the binding material of the paint. In that case, however, the toxicity will be less of a problem, too (i.e. unless you dissolve the paint somehow, burn it, ...)&lt;br&gt;&#xA;$^*$ OTOH it is highly improbable that there are substantial amounts of silver in the paint...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;reductive borax bead should be gray, oxidative colourless.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Isolating Pb is quite easy, maybe you can do that in your school lab: precipitate $\ce{PbCl2}$ with &lt;em&gt;cold&lt;/em&gt; HCl, filter. The dissolve the chloride in hot water. You can then precipitate PbS (black) or $\ce{PbCrO4}$ (yellow).  Looking at the crystals under the microscope can help further. Qualitative inorganic textbooks often have microscope photos of different precipitates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorDisplayName="user95" LastEditDate="2012-07-09T12:17:51.697" LastActivityDate="2012-07-09T12:17:51.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="734" PostTypeId="2" ParentId="654" CreationDate="2012-07-04T14:24:44.663" Score="4" Body="&lt;p&gt;Diethylne glycol reacts readily with perchloricacid, however the oxidation and hazard potential of perchloric acid depends on its concentration in side the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If concentrations are high enough and more than 10 percent perchloric acid formation is expected with 100 percent reaction yield, then it can be explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually happens that as soon as diethylene gylcol meets with perchloric acid, diethylene glycol decomposes and some dangerous peroxides formed. These are very reactive to water, so if you can provide very dry and inert atmosphere it should be safe but you should give extra care to make sure that you do not have humidity and water both in your chemicals and in the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to do the experiment and use milimolar range chemicals then just do it, because you will not be able to form enough perchloricacid to get problems.&lt;/p&gt;&#xA;" OwnerUserId="222" LastActivityDate="2012-07-04T14:24:44.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="735" PostTypeId="1" AcceptedAnswerId="737" CreationDate="2012-07-04T15:08:57.030" Score="0" ViewCount="4485" Body="&lt;p&gt;Is $\ce{CCl_4}$ an electrolyte all by itself, or does something need to be added to it so that it becomes an electrolyte?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it &lt;em&gt;weak&lt;/em&gt; or &lt;em&gt;strong&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="380" LastEditorUserId="9961" LastEditorDisplayName="user95" LastEditDate="2017-01-13T13:59:19.653" LastActivityDate="2018-06-11T03:33:02.790" Title="Is carbon tetrachloride an electrolyte?" Tags="&lt;electrochemistry&gt;&lt;halides&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="737" PostTypeId="2" ParentId="735" CreationDate="2012-07-05T02:13:55.180" Score="13" Body="&lt;p&gt;No, $\ce{CCl4}$ is not an electrolye.  An electrolye is a material that causes ions (charged entities) to form in the solvent.  Carbon tetrachloride neither dissociates into ions nor induces ion formation in the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the $\ce{C-Cl}$ bond is rather strong and won't break under&#xA;normal solution conditions.  Nor does the molecule have any net dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="11367" LastEditDate="2017-01-13T14:08:51.187" LastActivityDate="2017-01-13T14:08:51.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="738" PostTypeId="1" AcceptedAnswerId="739" CreationDate="2012-07-05T09:32:23.807" Score="13" ViewCount="150374" Body="&lt;p&gt;I was told that an atom's atomic number is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of electrons or protons present in a &lt;strong&gt;neutral atom&lt;/strong&gt; is called atomic number. It is represented by &lt;code&gt;Z&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does neutral mean here? Why isn't it just &quot;..present in an atom...&quot;?&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="15410" LastEditDate="2019-07-30T03:31:33.373" LastActivityDate="2019-07-30T03:31:33.373" Title="What is a neutral atom?" Tags="&lt;atoms&gt;&lt;terminology&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="739" PostTypeId="2" ParentId="738" CreationDate="2012-07-05T10:48:03.233" Score="20" Body="&lt;p&gt;Electrons and protons are charged particles. The electrons have negative charge, while protons have positive charge. A neutral atom is an atom where the charges of the electrons and the protons balance. Luckily, one electron has the same charge (with opposite sign) as a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Carbon has 6 protons. The neutral Carbon atom has 6 electrons. The atomic number is 6 since there are 6 protons.&lt;/p&gt;&#xA;" OwnerUserId="396" LastEditorUserId="-1" LastEditDate="2015-01-28T12:45:56.073" LastActivityDate="2015-01-28T12:45:56.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="740" PostTypeId="2" ParentId="738" CreationDate="2012-07-05T10:50:28.153" Score="12" Body="&lt;p&gt;Here, a &quot;neutral atom&quot; is simply an atom that has no charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, an atom consists of protons, neutrons, and electrons. Protons are positively charged, electrons are negatively charged (with the same magnitude of charge per particle as a proton). Neutrons have no charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in a &quot;neutral atom&quot;, the number of protons must be equal to the number of electrons, otherwise it would not be neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the definition is saying that &quot;in a neutral atom, the atomic number is equal to both the number of protons , &lt;em&gt;and&lt;/em&gt; the number of electrons, because, well, they both are the same.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this with an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion&quot;&gt;ion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ion is an atom or molecule in which the total number of electrons is not equal to the total number of protons, giving it a net positive or negative electrical charge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, the atomic number cannot be equal to both--because they are unequal. In this case, a better definition would be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of protons present in an atom is it's atomic number. It is represented by $Z$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-07-05T10:50:28.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="741" PostTypeId="1" CreationDate="2012-07-05T14:53:52.693" Score="8" ViewCount="305" Body="&lt;p&gt;Why or when does non-thermal plasma become thermal? How do we know before hand if the ion temperature will differ from the electron temperature, on what parameters does it depend - where is the quantitative gap between the two options?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Has it to do with the heat release in ion-recombination reactions?)&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2014-04-12T00:08:17.957" Title="Difference between non-thermal and thermal ionized gases" Tags="&lt;ions&gt;&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;electrons&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="742" PostTypeId="1" CreationDate="2012-07-05T15:56:48.180" Score="7" ViewCount="955" Body="&lt;p&gt;Some recent research I'm dealing with has required me to look for tables of experimental pressure/volume/temperature values (equivalently, a table of compressibility/compression factors in terms of any two of pressure, volume, or temperature) for various substances. Thus far, I've managed to obtain compression factor tables for a short list of substances from &lt;em&gt;Perry's Chemical Engineers' Handbook&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, however, as to where I should be looking for more extensive tables (e.g. tables with finer sampling), as well as tables for substances that were not included in that handbook. What books/websites/journal articles should I be looking at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts at searching on the Internet have not been very fruitful; I would appreciate knowing what keywords should I be using with the usual search engines to get better results.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastActivityDate="2012-07-06T20:23:10.103" Title="Where can I obtain tables of P-V-T/compression factor data?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="743" PostTypeId="5" CreationDate="2012-07-05T19:32:44.433" Score="0" Body="&lt;p&gt;A branch of chemistry which uses computer simulations to assist in solving chemical problems by modeling and studying chemical systems, and chemical behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses various methods from theoretical chemistry (from approximations of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schrödinger equation&lt;/a&gt; to &lt;a href=&quot;https://en.wikipedia.org/wiki/Monte_Carlo_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monte-Carlo simulations&lt;/a&gt; to modeling with &lt;a href=&quot;https://en.wikipedia.org/wiki/Differential_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential equations&lt;/a&gt;), incorporated into efficient computer programs, to calculate the structures and properties of molecules and solids.&#xA;It is necessary because the quantum &lt;a href=&quot;https://en.wikipedia.org/wiki/Many-body_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;many-body problem&lt;/a&gt; cannot be solved analytically, much less in closed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While computational results normally complement the information obtained by chemical experiments, it can in some cases predict hitherto unobserved chemical phenomena. It is thus widely used in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Drug_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;design of new drugs&lt;/a&gt; and materials.&lt;/p&gt;&#xA;" OwnerUserId="40029" LastEditorUserId="9961" LastEditDate="2018-07-21T10:13:50.273" LastActivityDate="2018-07-21T10:13:50.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="744" PostTypeId="4" CreationDate="2012-07-05T19:32:44.433" Score="0" Body="A field which employs computers to model and study chemical systems and chemical behavior, using various models, from approximations of the Schrödinger equation to Monte-Carlo simulations, to modeling with differential equations." OwnerUserId="323" LastEditorUserId="9961" LastEditDate="2018-07-21T10:14:22.753" LastActivityDate="2018-07-21T10:14:22.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="745" PostTypeId="5" CreationDate="2012-07-05T19:41:56.773" Score="0" Body="&lt;p&gt;This tag can be applied to questions regarding the chemical processes behind common reactions observed in the day-to-day lives of people working outside laboratory or industrial settings.  Also acceptable are questions regarding the chemical composition of everyday &quot;household&quot; items, e.g. cleaning supplies, plastics, batteries, food and drink, and common fuels such as gasoline or natural gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry is central to virtually all aspects of modern living in industrialized nations. While the general definition of chemistry as &quot;the study of different types of atomic elements, their reactions and products&quot; makes it applicable to every aspect of the existence and functioning of the universe, the more limited common connotation of a &quot;chemical&quot; as being &quot;some created or extracted substance not typically seen in this form in nature&quot; typically limits the &quot;everyday&quot; aspect of chemistry to those types of &quot;artificial&quot; substances seen in the day-to-day lives of people not directly associated with the field of study. Even then, virtually every consumer item currently available for sale has been improved or even made possible through the targeted development of &quot;artificial&quot; chemical substances with desirable properties for the intended use of the end product.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="402" LastEditDate="2017-04-08T15:33:51.793" LastActivityDate="2017-04-08T15:33:51.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="746" PostTypeId="4" CreationDate="2012-07-05T19:41:56.773" Score="0" Body="Applications of chemistry to reagents, reactions and processes of everyday life. Use this tag for question on chemistry in an out-of-laboratory/class context. DO NOT use this tag merely because the question is about simple chemistry." OwnerUserId="323" LastEditorUserId="33991" LastEditDate="2017-06-25T10:24:56.830" LastActivityDate="2017-06-25T10:24:56.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="747" PostTypeId="5" CreationDate="2012-07-05T19:45:53.350" Score="0" Body="&lt;p&gt;This tag should be applied to questions concerning acid and base reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the IUPAC goldbook, an &lt;a href=&quot;http://goldbook.iupac.org/A00071.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;acid&lt;/a&gt; is a molecular entity or chemical species capable of donating a hydron (proton) (Brønsted acid) or capable of forming a covalent bond with an electron pair (Lewis acid).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Brønsted acid&lt;/strong&gt; (&lt;a href=&quot;http://goldbook.iupac.org/B00744.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;  A molecular entity capable of donating a hydron (proton) to a base, (i.e. a 'hydron donor') or the corresponding chemical species. For example: $\ce{H2O, H3O+, CH3CO2H, H2SO4, HSO4^{−}, HCl, CH3OH, NH3}$ .&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Lewis acid&lt;/strong&gt; (&lt;a href=&quot;http://goldbook.iupac.org/L03508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;  A molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:&lt;br&gt;&#xA;  $\ce{\underset{Lewis~acid}{(H3C)3B} + \underset{Lewis~base}{:NH3} -&amp;gt; \underset{Lewis~adduct}{(H3C)3\overset{\small{~~\ominus}}{B}-\overset{\small{\oplus}}{N}H3}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In conjunction to this, the definition of a &lt;a href=&quot;http://goldbook.iupac.org/B00601.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;base&lt;/a&gt; is a chemical species or molecular entity having an available pair of electrons capable of forming a covalent bond with a hydron (proton) (Brønsted base) or with the vacant orbital of some other species (Lewis base).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Brønsted base&lt;/strong&gt; (&lt;a href=&quot;http://goldbook.iupac.org/B00745.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;  A molecular entity capable of accepting a hydron (proton) from an acid (i.e. a 'hydron acceptor') or the corresponding chemical species. For example: $\ce{{}^{-}OH, H2O, CH3CO2^{−}, HSO4^{−}, SO4^{2−}, Cl^{−}}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Lewis base&lt;/strong&gt; (&lt;a href=&quot;http://goldbook.iupac.org/L03511.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;  A molecular entity (and the corresponding chemical species) able to provide a pair of electrons and thus capable of coordination to a Lewis acid, thereby producing a Lewis adduct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two kinds of molecules or chemical species are closely related as they form so called &lt;a href=&quot;http://goldbook.iupac.org/C01266.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjugated acid-base pair&lt;/a&gt;s in the Brønsted sense, or as mentioned previously, as a &lt;a href=&quot;http://goldbook.iupac.org/L03510.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis adduct&lt;/a&gt; in the alternative Description.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;conjugate acid–base pair&lt;/strong&gt;&lt;br&gt;&#xA;  The Brønsted acid $\ce{BH+}$ formed on protonation of a base $\ce{B}$ is called the conjugate acid of $\ce{B}$, and $\ce{B}$ is the conjugate base of $\ce{BH+}$. (The conjugate acid always carries one unit of positive charge more than the base, but the absolute charges of the species are immaterial to the definition.) For example: the Brønsted acid $\ce{HCl}$ and its conjugate base $\ce{Cl^{−}}$ constitute a conjugate acid–base pair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reactivity of acids and bases is dependent on the $\mathrm{p}\ce{H}$ of a solution and as a consequence, every reaction of acids and bases will change this property of a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution at $T = 25~^\circ\mathrm{C}$, acids usually have a $\mathrm{p}\ce{H}$ less than 7 and bases have a $\mathrm{p}\ce{H}$ greater than 7, this is because the neutrality of an aqueous solution is determined or governed by the &lt;a href=&quot;http://goldbook.iupac.org/A00531.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;autoprotolysis&lt;/a&gt; of water.&#xA;$$\ce{2H2O &amp;lt;=&amp;gt; H3+O + {}^{-}OH}$$ &#xA;Therefore neutrality of a solution is achieved when the activities of hydronium and hydroxide ions are equal, &#xA;$$a(\ce{H3+O})=a(\ce{{}^{-}OH}),$$ &#xA;or in simpler terms, when the concentrations of hydronium and hydroxide ions are equal, i.e. $$[\ce{H3+O}]=[\ce{{}^{-}OH}].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this nature these questions are usually related to &lt;a href=&quot;/questions/tagged/ph&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;ph&amp;#39;&quot; rel=&quot;tag&quot;&gt;ph&lt;/a&gt; and/or &lt;a href=&quot;/questions/tagged/aqueous-solution&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;aqueous-solution&amp;#39;&quot; rel=&quot;tag&quot;&gt;aqueous-solution&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fairly good answer, that covers acids and bases in general can be found in &lt;a href=&quot;http://chemistry.stackexchange.com/a/10528/4945&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="39539" LastEditDate="2017-05-16T06:10:10.320" LastActivityDate="2017-05-16T06:10:10.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="748" PostTypeId="4" CreationDate="2012-07-05T19:45:53.350" Score="0" Body="This tag should be applied to questions concerning acid and base reactions. An acid is capable of donating a hydron/ proton (Brønsted acid) or capable of forming a covalent bond with an electron pair (Lewis acid). A base on the other hand is a chemical species/ molecular entity having an available pair of electrons capable of forming a covalent bond with a hydron/ proton (Brønsted base) or with the vacant orbital of some other species (Lewis base)." OwnerUserId="323" LastEditorUserId="39539" LastEditDate="2017-05-16T06:10:26.323" LastActivityDate="2017-05-16T06:10:26.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="749" PostTypeId="2" ParentId="742" CreationDate="2012-07-06T00:53:17.673" Score="5" Body="&lt;p&gt;One place to look for simple fluids is &lt;a href=&quot;http://webbook.nist.gov/chemistry/fluid/&quot;&gt;http://webbook.nist.gov/chemistry/fluid/&lt;/a&gt;.  But there are only a limited number of compounds treated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be warned that the data in the NIST Webbook is smoothed.  Collecting data of this sort is very difficult.  Many different sets of observations are combined and fitted to an appropriate equation.  What is at the URL given above is computed from that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for overall compression factors, this will do very nicely.  If you want five figure precision in compression factors, well, this won't do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another place to check is the  which is a primary source of raw data.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-07-06T00:53:17.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="750" PostTypeId="1" AcceptedAnswerId="751" CreationDate="2012-07-06T10:21:47.283" Score="11" ViewCount="8307" Body="&lt;p&gt;I can't make heads or tails of the information on the net — although there seems to be plenty of it, it seems contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, calcium carbonate is used in antacids—yet there are warnings about ingestion and contact with skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to handle it in pure powder form. I'm mainly concerned about skin contact, should it be avoided? (Am I getting confused between two or more forms of calcium carbonate?&lt;/p&gt;&#xA;" OwnerUserId="371" LastEditorUserId="22" LastEditDate="2012-07-06T13:49:32.627" LastActivityDate="2018-07-05T19:08:38.317" Title="Calcium carbonate toxicity" Tags="&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="751" PostTypeId="2" ParentId="750" CreationDate="2012-07-06T13:46:38.460" Score="14" Body="&lt;p&gt;Looking at the &lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9927119&quot;&gt;Material Safety Data Sheet (MSDS)&lt;/a&gt;, most of the issues with it seem to stem from the physical consequences of it being in fine dry powder form. This means that if you get it on your skin, it can get into cracks and pores and easily dry out your skin to the point of cracking, and will cause irritation or worse if you get it in your eyes, for example. It's not &lt;em&gt;toxic&lt;/em&gt; if you get it on you, though -- rubbing a lump of limestone against your skin isn't going to do anything except abrade it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, if you're working with the powder, wear gloves. And safety glasses. And long sleeves. (See also: lab coat.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in doubt, follow the MSDS.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-07-06T13:46:38.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="753" PostTypeId="1" AcceptedAnswerId="754" CreationDate="2012-07-07T13:33:10.863" Score="14" ViewCount="22602" Body="&lt;p&gt;In Nuclear magnetic resonance (H-NMR) spectroscopy, the chemical shift of aliphatic hydrogen atoms are very much closer to the 1.0 ppm than that of aromatic hydrogen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromatic hydrogen atoms have a chemical shift value of about 7.0-9.0 ppm whereas the chemical shift value of aliphatic hydrogen atoms ranges between 2.0-3.5 ppm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="401" LastEditorUserId="7" LastEditDate="2012-07-07T13:54:23.230" LastActivityDate="2012-07-09T08:16:42.533" Title="Why do aromatic hydrogen atoms have a higher chemical shift value compared to aliphatic hydrogen atoms?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="754" PostTypeId="2" ParentId="753" CreationDate="2012-07-07T13:45:49.353" Score="11" Body="&lt;p&gt;The chemical shift gives you information about how well shielded the nuclei are from the magnetic field. A proton at higher chemical shift values is deshielded, so the aromatic protons are obviously less shielded than aliphatic protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One effect that causes deshielding is the presense of electronegative atoms that draw electrons away from other atoms and thereby deshield them. But that is not the cause of the aromatic chemical shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for that one is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromatic_ring_current&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring current that is induced in a magnetic field&lt;/a&gt;. The induced current creates a local magnetic field that has the same direction as the external field outside the aromatic ring where the attached protons are, leading to the deshielding of the nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSpU8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-07-09T08:16:42.533" LastActivityDate="2012-07-09T08:16:42.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="755" PostTypeId="2" ParentId="703" CreationDate="2012-07-07T15:42:27.207" Score="1" Body="&lt;p&gt;This should do it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Weather_p004.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Weather_p004.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="371" LastActivityDate="2012-07-07T15:42:27.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="756" PostTypeId="1" CreationDate="2012-07-08T10:31:14.987" Score="7" ViewCount="22002" Body="&lt;p&gt;I read in the local paper that common food products like turmeric can be adulterated. It continued to specify the adulterant which they called metanil yellow. On searching on the Net, I was able to find that metanil yellow is prepared using some basic raw materials like metanilic acid and diphenylamine. The paper continued to lay out procedures on how to find out whether the turmeric you buy from the local market is adulterated. To do this, one must first dissolve half a spoon full of turmeric powder in 20ml of lukewarm water. It then told to add a few drops of hydrochloric acid. Upon doing this, if the aqueous solution turned pink, violet or purple, it indicated the presence of metanil yellow. It continued to state the risks of consuming turmeric adulterated with metanil yellow. One of these risks was that it is a carcinogen. I'm an Indian, so turmeric is included in nearly all of the dishes I eat in a day, so I became very curious. My question is this:&lt;br&gt;&#xA;&lt;br&gt;&#xA;1.) What is the chemical formula for metanil yellow, how is it prepared (in layman's terms), and how does it react with hydrochloric acid to turn pink, purple or violet (a chemical equation would be helpful)?&#xA;&lt;br&gt;2.)What are the risks in consuming metanil yellow?&lt;/p&gt;&#xA;" OwnerUserId="406" LastEditorUserId="22" LastEditDate="2012-07-09T04:48:22.227" LastActivityDate="2017-06-14T04:58:44.330" Title="Adulteration of turmeric with Metanil yellow" Tags="&lt;organic-chemistry&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="757" PostTypeId="1" AcceptedAnswerId="762" CreationDate="2012-07-08T16:33:26.763" Score="9" ViewCount="4505" Body="&lt;p&gt;Why do peroxides tend to explode with little provocation? In general, what are typical products of their explosion?&lt;/p&gt;&#xA;" OwnerUserId="408" LastEditorUserId="19616" LastEditDate="2015-10-02T16:20:21.727" LastActivityDate="2015-10-02T16:20:21.727" Title="Why are peroxides explosive?" Tags="&lt;organic-chemistry&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="758" PostTypeId="1" AcceptedAnswerId="759" CreationDate="2012-07-08T22:41:41.727" Score="13" ViewCount="252" Body="&lt;p&gt;I remember that one of my old organic chemistry profs referred to mercury as being a &quot;fancy proton&quot;. I think it was in the context of oxymercuration reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might have been his rationale for offering this comparison?&lt;/p&gt;&#xA;" OwnerUserId="408" LastEditorUserId="73" LastEditDate="2012-07-09T00:32:17.787" LastActivityDate="2012-07-09T01:52:37.880" Title="Why is mercury a &quot;fancy proton&quot;?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="759" PostTypeId="2" ParentId="758" CreationDate="2012-07-09T01:52:37.880" Score="12" Body="&lt;p&gt;It is a large, soft, Lewis acid. One can use protic acid and water to promote the same reaction, but the product distribution can sometimes be different. &#xA; In the oxymercuration reaction Hg can form a pi complex with alkenes or alkynes, and these &quot;bridged&quot; cations will not undergo rearrangments to the extent that those produced by protic acids might. &#xA; Bottom line - they're both essentially performing the same function in this reaction - activating the pi bond towards attack - but the oxymercuration is milder. &lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2012-07-09T01:52:37.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="760" PostTypeId="2" ParentId="750" CreationDate="2012-07-09T02:00:39.670" Score="9" Body="&lt;p&gt;It's essentially chalk.  I'm not aware of any longitudinal studies of gymnasts' long term exposure to powdered calcium carbonate, but my sense is that the worst that can happen is your skin can get dry and irritated. If you have a part time job as a hand model, you will want to be especially careful to avoid skin contact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're in a lab, of course, it goes without saying that you should take necessary safety precautions such as proper eyewear and skin protection as well as a lab coat. &lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2012-07-09T02:00:39.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="762" PostTypeId="2" ParentId="757" CreationDate="2012-07-09T06:34:27.323" Score="7" Body="&lt;p&gt;I don't like to speak in generalities if I can help it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can talk, for instance, about hydrogen peroxide. One of the most familiar schemes is a disproportionation where,&#xA;$$2 \ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$$ &#xA;and as you study this particular species, you will see this reaction more and more often. Actually in solutions of hydrogen peroxide you can take a 30% concentrated solution and add some Mn and subsequently measure a significant increase in oxygen concentration over time. Although you should know the solution itself decomposes slowly in practice when exposed to light or due to trace impurities, but this pales in comparison to the case involving a catalyst. This reaction is, in part, why researchers believe a peroxide-type species is important for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photosystem_II#Oxygen-Evolving_Complex_.28OEC.29&quot;&gt;Oxygen Evolving Center&lt;/a&gt; of Photosystem II. Another fun demonstration of this process is often where foam is generated from a large graduation cylinder, see &lt;a href=&quot;http://chemed.chem.purdue.edu/demos/movies/large_movies/10.9.mov&quot;&gt;here&lt;/a&gt;, upon addition of KI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now 30% is weak, if instead you look at 80% solutions you will notice that the decomposition will evolve steam because it is so exothermic. In fact, the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Heinkel_He_176&quot;&gt;liquid-fueled rocket&lt;/a&gt; in 1939 used an 80% concentrated solution and decomposed it with calcium permanganate. Later engines utilized another important property of hydrogen peroxide for its decomposition until around 1980: the fact that it is a stronger oxidizer than chlorine/chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically we say the disproportionation is favorable since it has a fairly negative (~ -100 kJ/mol) enthalpy change and a considerable change in entropy ( ~ + 70 J/(mol K)). The basis of this is that it is highly favorable to form dioxygen and this is a large driving force of this reaction in terms of enthalpy and entropy. You'll also see this danger in species that are really close to forming dinitrogen too, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Diazomethane&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://pipeline.corante.com/archives/2008/04/30/how_not_to_do_it_diazomethane.php&quot;&gt;here&lt;/a&gt;. It only takes a catalyst for this down-hill process to take off. Which could be quite deadly if unintentionally concentrated and then introduced to a catalyst in a closed system. Perhaps the story is similar for other peroxides.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2012-07-09T06:39:45.757" LastActivityDate="2012-07-09T06:39:45.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="763" PostTypeId="2" ParentId="696" CreationDate="2012-07-09T08:36:16.627" Score="8" Body="&lt;p&gt;I'm only directly attacking your direct question, &quot;what sort of catalysts might be used, and what are the performance / lifetime issues with them?&quot; because the rest is ill-defined. I'll mention some electrochemical conversion aspects. I hope to illustrate why your question is problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several different products one could imagine reducing carbon dioxide to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;methane&lt;/li&gt;&#xA;&lt;li&gt;methanol&lt;/li&gt;&#xA;&lt;li&gt;formaldehyde&lt;/li&gt;&#xA;&lt;li&gt;carbon monoxide&lt;/li&gt;&#xA;&lt;li&gt;formic acid&lt;/li&gt;&#xA;&lt;li&gt;ethylene&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each will require different materials and conditions, but they all require energy for this to occur. Not to mention, these are all difficult reactions. Thermodynamically speaking carbon dioxide is at the bottom of this energy well and to get out of it requires spending energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach involves directly electrolyzing on a metal. Cu in aqueous solutions can give methane as a major product, but it also evolves hydrogen and many of the others I listed above. There are a couple of problems with this but the major problem is that it often requires large overpotentials at rates of interest for industrial purposes and this means more cost in terms of electrical input. Another major problem surrounding nearly all of these studies is the selectivity of a catalyst: in water at pH 7 evolving dihydrogen is thermodynamically more favorable than reducing carbon dioxide to everything except methane and methanol. Unfortunately these two are the most complex, as far as chemical transformations are concerned, to make. So the other products are much more likely to be produced along the way. This is why people have focused on making simpler things such as carbon monoxide or formic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this energy come from? Right now coal. So we would end up creating more carbon dioxide than we reduced to other species. It reminds me of how diesel-electric transmission works in freight trains: a dirty engine turns an electric generator that generates electricity that is used on motors instead of the motor directly driving the wheels; I think we can call this diesel-electric transmission an indirect drive whereas the latter is a direct drive. It turns out that making the former is far easier in engineering rather than the latter. The same is true here. It's easier to generate electricity somewhere else then reduce CO2 rather than have energy directly fed to reduce carbon dioxide. You can get efficiencies better by going the latter, but engineering/technical complexities increase immensely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a means to address this electrical burden via the direct-drive analogy, people have looked at these artificial photosynthesis approaches, i.e. using light to  drive these reactions or at least help ease the energy burden. This requires photocatalysts and often with these come semiconductors. Some semiconductors are naturally good at reducing carbon dioxide, such as CdTe, but this better performance for generating methanol is in turn related to an actual corrosion process of the semiconductor. Semiconductors are very complicated and expensive to manufacture well. They tend to corrode easily when they work well with sunlight. They require catalysts and this introduces huge compatibility issues and manufacturing complexities. We'll just leave the indirect approach at the state  the field is, complicated and incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the direct approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it's complicated by the fact that carbon dioxide is fundamentally limited in solubility at room temperature. Hence why other people prefer non-water systems. These unfortunately add other complexities...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually while typing this I came to the conclusion that I could write a much more succinct response to you rather than continuing to babble on: The best candidate will involve a transition metal, there is no doubt about that. It may end up being a molecular species, but there are so many possible structures that it is hard to say if we won't find a simpler heterogenous catalyst that is more capable first. By Murphy's law, it will be the most expensive material imaginable. But right now, considering everything, you can always look at all extensive work that has been done on copper, but realize it isn't very efficient or selective.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-07-09T08:36:16.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="765" PostTypeId="1" CreationDate="2012-07-09T12:01:13.267" Score="9" ViewCount="1970" Body="&lt;p&gt;Take a glass rod and rub vigorously the wall of the flask, the substance will crystallize out of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a fire polished stirring rod and etch (scratch) the glass of your beaker. The small pieces of glass that are etched off of the beaker serve as nuclei for crystal formation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But reaching back to my halcyon years of organic synthesis and crystallization, recalling vivid memories of scratching the sides of Erlenmeyer flasks with a glass rod to induce crystallization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a common lab technique. But does it have a name?&lt;/p&gt;&#xA;" OwnerUserId="409" LastEditorUserId="16683" LastEditDate="2016-04-05T00:45:31.907" LastActivityDate="2016-10-13T04:28:26.503" Title="What is the correct name for this lab technique in crystallization?" Tags="&lt;experimental-chemistry&gt;&lt;terminology&gt;&lt;ionic-compounds&gt;&lt;recrystallization&gt;&lt;teaching-lab&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="766" PostTypeId="2" ParentId="756" CreationDate="2012-07-09T14:29:17.797" Score="6" Body="&lt;p&gt;I hope to keep this short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metanil yellow is an &lt;em&gt;azo dye&lt;/em&gt; synthesized from the coupling of metanilic acid and diphenylamine, as described &lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;id=V0ErAAAAYAAJ&amp;amp;pg=PA548&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Here's a sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RKhmE.png&quot; alt=&quot;metanil yellow&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the thing with most azo dyes like metanil yellow is that when they are taken internally, liver enzymes or intestinal flora will usually reduce the azo dye to the components that were used in the diazo coupling. (See &lt;a href=&quot;http://dx.doi.org/10.3109/03602539109029761&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; for more details.) For metanil yellow, the metanilic acid is poorly absorbed from the intestinal tract (and is thus easily excreted), and your remaining worry is the diphenylamine, which is a suspected mutagen and carcinogen (apart from the usual toxicity associated with arylamines). (In short, the metanil yellow itself is not your worry, but the metabolite diphenylamine sure is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of contrast, the natural coloring agent in turmeric is the phenolic compound curcumin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GhAzO.png&quot; alt=&quot;curcumin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, both compounds are in fact &lt;em&gt;acid-base indicators&lt;/em&gt;, though they have different transition ranges; for metanil yellow, the pH transition range is 1.2-2.3 (red to yellow), while for curcumin, the transition range is 7.8-9.2 (yellow to red-brown). (Data taken from the &lt;a href=&quot;http://books.google.com/books?hl=en&amp;amp;id=z_TvAAAAMAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Handbook of Acid-Base Indicators&lt;/em&gt;&lt;/a&gt;.) I'm not quite sure where the purple/violet color would come from, but the key point is that natural turmeric would be reddish in an alkaline environment, while metanil yellow will be reddish in an acidic environment; however, vinegar (pH ~ 2.4) is not sufficiently acidic to display the color change for metanil yellow, and thus a stronger acid like hydrochloric acid is needed to display the color change.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastEditorUserId="11367" LastEditDate="2017-06-14T04:58:44.330" LastActivityDate="2017-06-14T04:58:44.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="767" PostTypeId="1" CreationDate="2012-07-09T16:27:20.947" Score="4" ViewCount="8642" Body="&lt;p&gt;Does ozone have a color or is it colorless? I have heard that ozone has a bluish color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain this observation scientifically? &lt;/p&gt;&#xA;" OwnerDisplayName="user411" LastEditorUserId="4231" LastEditDate="2015-02-21T22:27:21.220" LastActivityDate="2015-02-21T22:27:21.220" Title="Does ozone (O₃) gas have a color?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ClosedDate="2012-07-09T20:15:59.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="768" PostTypeId="2" ParentId="767" CreationDate="2012-07-09T17:10:10.860" Score="3" Body="&lt;p&gt;Ozone gas does indeed have a slight blue color. It is dark blue in a liquid form and purple-black in solid form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color comes from the fact that it absorbs all wavelengths of light other than those in the blue region. As the phase changes, the molecules get closer together, shifting the area of absorption, changing the color of the material.&lt;/p&gt;&#xA;" OwnerUserId="345" LastActivityDate="2012-07-09T17:10:10.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="769" PostTypeId="1" AcceptedAnswerId="770" CreationDate="2012-07-09T22:06:31.717" Score="6" ViewCount="2043" Body="&lt;p&gt;I got this question wrong on a test and I don't know why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: Which conformation is more stable?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qZYGP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: the chair on the left is more stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought that the ring conformation has zero ring strain so I'm confused. &lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="7" LastEditDate="2012-07-10T09:01:37.950" LastActivityDate="2016-01-12T12:22:56.487" Title="Which cyclohexane conformation is more stable?" Tags="&lt;organic-chemistry&gt;&lt;cyclohexane&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="770" PostTypeId="2" ParentId="769" CreationDate="2012-07-10T00:37:55.830" Score="9" Body="&lt;p&gt;By symmetry, the cyclohexane on the right requires each carbon atom to have a $\ce{C-C-C}$ angle of 120 degrees. From VSEPR theory, this is not optimal for tetrahedral carbon. A regular tetrahedron has angles between vertices of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron&quot;&gt;approximately 109.5 degrees&lt;/a&gt; and the equilibrium bond angles of a secondary carbon won't be too discrepant from this.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-07-10T00:37:55.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="771" PostTypeId="1" AcceptedAnswerId="772" CreationDate="2012-07-10T06:00:16.230" Score="20" ViewCount="2547" Body="&lt;p&gt;Before modern chemical techniques, say in the 19th century and earlier, how did (al)chemists determine whether a substance was an element or a compound? If it was a compound, how did they determine its molecular formula and structure?&lt;/p&gt;&#xA;" OwnerUserId="408" LastActivityDate="2017-04-08T22:18:37.013" Title="How were elements and molecular formulæ determined in the old days?" Tags="&lt;analytical-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="772" PostTypeId="2" ParentId="771" CreationDate="2012-07-10T12:16:10.680" Score="10" Body="&lt;p&gt;First thing, they couldn't tell elements and compounds apart. See Dalton's list of elements with their symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oy6n2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lime is right there--and it's not actually an element. Neither is &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryte&quot; rel=&quot;noreferrer&quot;&gt;Baryte(s)&lt;/a&gt;. Soda and Potash &lt;em&gt;probably&lt;/em&gt; refer to the oxides of Sodium and potassium, though terminology may have been different back then.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In those days, elements were things that could not be synthesized by addition reactions. If you have a reaction $\ce{A + B -&amp;gt;C}$, then $\ce{C}$ is a compound (can't say much about the other two). Alternatively, if the substance can be decomposed (via anaerobic heating/etc), then it's a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most probably they would stuff things into the list &quot;innocent until proven guilty&quot; (I don't know about this).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Molecular mass can be determined by osmotic pressure, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ebullioscopy&quot; rel=&quot;noreferrer&quot;&gt;ebullioscopy&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryoscopy&quot; rel=&quot;noreferrer&quot;&gt;Cryoscopy&lt;/a&gt;, etc. (Usually stuff delendant on colligative properties)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determination of molecular formula is much harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For organic compounds, it can be done by burning in a controlled environment. The ratios of oxides produced lets us know the relative quantities of all elements but oxygen. The quantity of oxygen can be determined by finding the decrease in Oxygen content of the chamber, and subtracting it from the Oxygen content of the oxides. The molecular formula can be found if you use the emperical formula and determine molecular mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inorganic compounds, they probably had various tests for ions/etc (like the silver nitrate test for halides). They also had flame tests (and the borax bead test), which could help identify certain ions. I'm not too sure here, though.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2012-07-10T14:00:15.427" LastActivityDate="2012-07-10T14:00:15.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="773" PostTypeId="2" ParentId="771" CreationDate="2012-07-10T20:52:05.507" Score="4" Body="&lt;p&gt;One very useful characterization tool would've been the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_fusion_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium fusion test&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is a general test for the detection of halogens, nitrogen and&#xA;  sulfur in an organic compound. These elements are covalently bonded&#xA;  to the organic compounds. In order to detect them, these have to be&#xA;  converted into their ionic forms. This is done by fusing the organic&#xA;  compound with sodium metal. The ionic compounds formed during the&#xA;  fusion are extracted in aqueous solution and can be detected by simple&#xA;  chemical tests. The extract is called sodium fusion extract or&#xA;  Lassaigne's extract.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When an organic compound is heated strongly with sodium, any halogens,&#xA;  nitrogen, and sulfur will be converted into inorganic sodium salts&#xA;  such as sodium halide (for halides), sodium cyanide (for nitrogen),&#xA;  sodium sulfide (for sulfur), and sodium thiocyanate (for sulfur and&#xA;  nitrogen).The nitrogen is confirmed with ferrous sulfate i.e. iron&#xA;  sulfate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example procedure for halogens: Melt a small pellet of sodium (first strip the oil off with hexanes) in a test tube using a Bunsen burner. When it gets red hot, remove from heat, add a small amount of sample (1 to 2 drops if liquid), allow it to react for a bit and then place back on flame for a few seconds. When adding the sample, I would make sure that the test tube is contained in another piece of glassware, should it break. After allowing the sample to cool, quench it with a few mL of water and then neutralize the mixture using dilute nitric acid. If there is sulfur or nitrogen present you should boil this solution for a minute. Now, add silver nitrate and you should get out a nice precipitate that varies slightly in colour depending on the halogen in question. There are milder procedures but this is the only one I've had experience with.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="402" LastEditDate="2017-04-08T22:18:37.013" LastActivityDate="2017-04-08T22:18:37.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="774" PostTypeId="2" ParentId="769" CreationDate="2012-07-11T11:53:36.043" Score="3" Body="&lt;p&gt;If you look at a Newman projection of &quot;flat&quot; cyclohexane, you'll see that along each C-C bond, the hydrogens and carbons are eclipsed. This leads to considerable torsional strain (imagine twisting a spring - it naturally wants to spring back) favoring relief of these unfavorable interactions. By contrast, in the chair form of cyclohexane, the Newman reveals that all of the hydrogens and carbons are staggered as you look along each C-C bond, which is a considerably more stable conformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the two forms is about 102 kJ/mol. For more references see here &lt;a href=&quot;http://www.ch.ic.ac.uk/local/organic/tutorial/cyclohexane/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ch.ic.ac.uk/local/organic/tutorial/cyclohexane/index.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2012-07-11T11:53:36.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="775" PostTypeId="2" ParentId="326" CreationDate="2012-07-11T19:47:19.330" Score="6" Body="&lt;p&gt;The quick way of doing that first step in the flowchart is to look for &lt;em&gt;too many principal $C_n$ axes&lt;/em&gt;. (Since the next step is to look for a principal axis anyway, this is a natural step to take.) In particular, you are looking for more than one axis of order &gt; 2. A threefold axis with three twofold axes perpendicular to it? Just $D_{3*}$ (where $* = d$, $h$, or nothing is yet to be determined, by using the flowsheet). But more than one threefold (or fourfold, or fivefold) axis? Has to be one of these special point groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next question, of course, is which one. Again the key is these &quot;excess&quot; principal axes. Lots of fivefold axes is the giveaway for icosahedral symmetry (to be sure, count six of them); lots of fourfold for octahedral (look for three); and neither of these means tetrahedral (you can look for four threefold axes to be sure, but you need to eliminate the other two since both of these also have threefold axes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice you can assume that icosahedral means $I_h$, octahedral $O_h$, and tetrahedral $T_d$. To be complete, however: for octahedral and icosahedral you need to find a centre of symmetry, otherwise it's just $I$ or $O$. For tetrahedral, if you have a centre of symmetry it's $T_h$; if you have mirror planes but no centre of symmetry it's $T_d$, and with neither it's $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially this is just the same answer as Richard's, but my point is just that there &lt;em&gt;are&lt;/em&gt; heuristics you can carry around in your head without having to check all 120 (or however many) symmetry elements. To take J.M.'s examples: adamantane has several (you don't really even have to count four) threefold axes but nothing higher: must be $T_d$. Buckyballs have a few fivefold axes: surely $I_h$.&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorUserId="390" LastEditDate="2012-07-12T07:26:47.017" LastActivityDate="2012-07-12T07:26:47.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="776" PostTypeId="1" CreationDate="2012-07-11T20:03:24.297" Score="4" ViewCount="115" Body="&lt;p&gt;I am on the lookout for, preferably, a cardboard (ish), that is both heat &amp;amp; grease resistant, for usage in microwave popcorn. As such, the material must be appropriate for direct contact with food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that a tray is placed on the bottom of a box, that will shield the body from any grease that will go liquid when heated (microwaved). Obviously, the most widely available material, and the cheaper the better is the quest that I have undertaken, and I hope that you folks can point me in the right direction, as I can't quite afford professional help for this project, as it stands.&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="73" LastEditDate="2012-07-11T23:41:58.313" LastActivityDate="2012-07-11T23:41:58.313" Title="Grease &amp; Thermal resistant cardboard for the foods industry?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="777" PostTypeId="2" ParentId="29" CreationDate="2012-07-12T00:22:31.907" Score="11" Body="&lt;p&gt;Almost certainly, yes. Applications of modern organic synthesis methods have drastically shortened the synthetic routes for such previously &quot;daunting&quot; targets such as strychnine &lt;a href=&quot;http://totallysynthetic.com/blog/?p=2585&quot;&gt;see Vanderwal&lt;/a&gt;, tetracyclines &lt;a href=&quot;http://www.chem.harvard.edu/groups/myers/page2/page4/page4.html&quot;&gt;Myers&lt;/a&gt;  and pyrroloindole alkaloids &lt;a href=&quot;http://totallysynthetic.com/blog/?p=1693&quot;&gt;Movassaghi&lt;/a&gt;, to name but a few - and recent efforts by &lt;a href=&quot;http://totallysynthetic.com/blog/?p=1730&quot;&gt;Baran&lt;/a&gt; and &lt;a href=&quot;http://www.scs.illinois.edu/white/index.php?p=publications&quot;&gt;White&lt;/a&gt; on C-H activation of alkanes have already yielded more efficient syntheses of many complex polycyclic terpenes. There is no reason to think that significant improvement could not occur for chlorophyll or B12. In fact, the absence of chlorophyll or Vitamin B12 from modern synthetic efforts represents a bit of a blind spot; it could well be that new chemistry could be discovered in the process. Although that 1,16 hydride shift by Eschenmoser will be hard to beat :-)&lt;/p&gt;&#xA;" OwnerUserId="324" LastActivityDate="2012-07-12T00:22:31.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="778" PostTypeId="1" CreationDate="2012-05-04T05:21:31.880" Score="9" ViewCount="4230" Body="&lt;p&gt;I'm looking for a way to take 3 separate liquids (liquids which function as paint -- they will all be different colors), put them together into a container, and then rapidly separate them back to distinct colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can handle all the electrical engineering aspects of my project, but this portion is beyond me. Does anyone have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;a quick method&quot;, I mean something that can be done easily without an excess of time or equipment. Ideally an way to make the paints intrinsically immiscible, but I need to be able to separate them &lt;em&gt;entirely&lt;/em&gt;, so that when they are poured together into a container, I can then separate and reuse them.&lt;/p&gt;&#xA;" OwnerUserId="416" OwnerDisplayName="Aerovistae" LastEditorUserId="41328" LastEditDate="2017-08-09T00:37:29.417" LastActivityDate="2017-08-09T00:37:29.417" Title="A quick method for separating liquids?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;mixtures&gt;" AnswerCount="6" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="779" PostTypeId="5" CreationDate="2012-07-12T05:14:54.530" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-07-12T05:14:54.530" LastActivityDate="2012-07-12T05:14:54.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="780" PostTypeId="4" CreationDate="2012-07-12T05:14:54.530" Score="0" Body="Questions on the harmful effects of various substances on the human body, or biological organisms in general." OwnerDisplayName="user95" LastEditorDisplayName="user95" LastEditDate="2012-07-12T05:48:29.847" LastActivityDate="2012-07-12T05:48:29.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="782" PostTypeId="2" ParentId="778" CreationDate="2012-07-12T12:33:12.167" Score="-1" Body="&lt;p&gt;i dont think their is any easy method(in a single step) for doing this kind of separation. if even one of the three liquids are immisible(not soluble) to others then you can try simple separation  using &lt;a href=&quot;http://www.gcsescience.com/e-separating-funnel.htm&quot; rel=&quot;nofollow&quot;&gt;funnel&lt;/a&gt; (most paints are &lt;a href=&quot;http://en.wikipedia.org/wiki/Emulsion&quot; rel=&quot;nofollow&quot;&gt;emulsions &lt;/a&gt;)and also try to learn about different kinds of &lt;a href=&quot;http://library.thinkquest.org/11430/research/techniques.htm&quot; rel=&quot;nofollow&quot;&gt;separation techniques&lt;/a&gt;. if possible find out the solubility of those paints(which are polar and nonpolar) and resubmit here.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditDate="2012-07-12T12:38:35.000" LastActivityDate="2012-07-12T12:38:35.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="783" PostTypeId="2" ParentId="714" CreationDate="2012-07-12T14:26:16.177" Score="6" Body="&lt;p&gt;Metal-metal quadruple bonds have seemingly been reported for several cases, quite often by F. A. Cotton. A short search revealed a more recent report on a species with a Cr-Cr quintuple bond: &lt;a href=&quot;http://www.sciencemag.org/content/310/5749/844&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthesis of a Stable Compound with Fivefold Bonding Between Two Chromium(I) Centers&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, I didn't bother to read the full article, it's just not the kind of chemistry I'm interested in. However, YMMV and you might find it helpful.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="9961" LastEditDate="2017-01-13T14:02:47.030" LastActivityDate="2017-01-13T14:02:47.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="784" PostTypeId="2" ParentId="778" CreationDate="2012-07-12T14:52:13.640" Score="0" Body="&lt;p&gt;The only thing that comes to my mind is laminar flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch &lt;a href=&quot;http://www.youtube.com/watch?v=p08_KlTKP50&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; and tell me if that's something like what you're look for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is, you're not really separating &quot;three liquids&quot;...&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-07-12T14:52:13.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="785" PostTypeId="1" AcceptedAnswerId="793" CreationDate="2012-07-12T17:53:30.053" Score="12" ViewCount="7799" Body="&lt;p&gt;I had a discussion with my supervisor today after I was a touch careless pumping down a flask with a liquid nitrogen cold trap attached. I checked, and there are &lt;a href=&quot;http://www.monash.edu.au/ohs/topics/hazard-alerts/vacuum-pumps-cold-traps.html&quot;&gt;reports of explosions&lt;/a&gt; after running air through a liquid nitrogen trap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, why does liquid oxygen not condense when you leave an open dewer of liquid nitrogen out? I mean, the air touching the surface of the nitrogen on top of the trap is the same as the air pumping through it. Is there too much nitrogen evaporating for the air to get close, or is the surface of the nitrogen too warm, or some other reasons?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="22" LastEditDate="2012-07-16T12:26:20.443" LastActivityDate="2015-07-03T16:15:27.843" Title="Why does oxygen not condense onto open dewers of liquid nitrogen?" Tags="&lt;physical-chemistry&gt;&lt;safety&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="786" PostTypeId="2" ParentId="785" CreationDate="2012-07-12T19:47:35.623" Score="7" Body="&lt;p&gt;The boiling point of nitrogen is 77 K, and the boiling point of oxygen is 90 K, so oxygen is liquid when liquid nitrogen is around. &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_oxygen&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; repeats the warning you mentioned:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Liquid nitrogen has a lower boiling point at −196 °C (77 K) than oxygen's −183 °C (90 K), and vessels containing liquid nitrogen can condense oxygen from air: when most of the nitrogen has evaporated from such a vessel there is a risk that liquid oxygen remaining can react violently with organic material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it's not because the surface of the nitrogen is too warm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the pressure due to the evaporating nitrogen is pushing air away from the opening. LN2 has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_nitrogen&quot; rel=&quot;nofollow&quot;&gt;expansion factor&lt;/a&gt; of 694, which you can compute by dividing the density of the liquid (0.808 g/mL) by the density of the gas at 20 °C (1.17 g/L). Of that expansion factor, a factor of 303 K/77 K = 3.94 comes from warming the gas from 77 K to 20 °C. The other factor of 176 comes from turning the liquid into a gas. That expansion is pushing air out of the headspace of the dewar.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T09:06:30.343" LastActivityDate="2015-05-01T09:06:30.343" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="787" PostTypeId="2" ParentId="778" CreationDate="2012-07-13T05:07:11.220" Score="1" Body="&lt;p&gt;You could use three immiscible liquids with different densities. Gravity will separate them for you. Here is a typical example of one of these experiments using water, cooking oil and corn syrup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sabrinaschool.wordpress.com/2008/02/08/liquid-density/&quot; rel=&quot;nofollow&quot;&gt;http://sabrinaschool.wordpress.com/2008/02/08/liquid-density/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For faster separation you could build a centrifuge.&lt;/p&gt;&#xA;" OwnerUserId="351" LastActivityDate="2012-07-13T05:07:11.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="788" PostTypeId="1" AcceptedAnswerId="789" CreationDate="2012-07-13T06:10:24.723" Score="16" ViewCount="13885" Body="&lt;p&gt;I have seen on a TV show that some  guy uses dry ice (solid $\ce{CO2}$) to make ice cream and on an another TV show, liquid nitrogen ($\ce{N2}$) is used with tequila.  Both are edible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that both gases are inert, but could these gases (perhaps in larger concentrations) be poisonous or create any side-effects in our body?  If these gases don't create any side effects, what mechanism(s) does the body use to remove these gases from the body?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="7448" LastEditDate="2015-10-30T17:14:36.313" LastActivityDate="2015-10-30T17:14:36.313" Title="What is the effect on humans of using dry ice and liquid nitrogen as food additives?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="789" PostTypeId="2" ParentId="788" CreationDate="2012-07-13T07:43:06.513" Score="14" Body="&lt;p&gt;Nitrogen has a boiling point of 77K (-196 °C), CO&lt;sub&gt;2&lt;/sub&gt; melts at 195K (-78 °C). You don't cool down the ice cream or whatever else you want to eat to those temperatures, that would be very unsafe. Both will evaporate or sublimate quickly when they are in contact with warmer material, so there won't be much left once you eat e.g. the ice cream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is composed of 78% N&lt;sub&gt;2&lt;/sub&gt; and 0.04% CO&lt;sub&gt;2&lt;/sub&gt;, nitrogen gas can only be dangerous when it replaces all the oxygen in the air. Carbon dioxide is toxic at around 7-10% concentration. Both conditions are not realistically achievable by the methods you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest danger I see would be ingesting a piece of dry ice shortly after it was put in a drink, which would be cold enough to hurt.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-10-13T08:05:39.983" LastActivityDate="2012-10-13T08:05:39.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="791" PostTypeId="2" ParentId="788" CreationDate="2012-07-13T15:18:14.457" Score="9" Body="&lt;p&gt;Liquid nitrogen is a dangerous substance to handle without proper knowledge. While food created using it will be perfectly safe (79% of the air you breathe is $N_2$, so no risk there), there are dangerous in the perpetration process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone in Germany lost both their arms while trying to use Liquid Nitrogen to produce food at home when they poured the left-over liquid nitrogen down the sink, causing an explosion which took their arm off. (See &lt;a href=&quot;http://news.sky.com/story/708871/blumenthal-style-chef-blows-off-his-hands&quot;&gt;Bluementhal Style Chef Blows Off His Hands&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CO_2$ is produced by your body in respiration and breathed out normally. It is also present in carbonated fizzy drinks and sparking wine. The only risk for dry-ice would be if you ate a lump of it, where you might suffer from the cold if it got stuck in your throat.&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="22" LastEditDate="2012-10-13T01:42:22.697" LastActivityDate="2012-10-13T01:42:22.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="793" PostTypeId="2" ParentId="785" CreationDate="2012-07-13T19:31:58.050" Score="5" Body="&lt;p&gt;If you look at the &lt;a href=&quot;http://encyclopedia.airliquide.com/images_encyclopedie/VaporPressureGraph/Oxygen_Vapor_Pressure.GIF&quot; rel=&quot;nofollow&quot;&gt;vapour pressure graph for oxygen&lt;/a&gt; you’ll see that at its partial pressure in air, it can only liquefy below about 77 K, which by coincidence is the temperature of boiling liquid nitrogen. The air immediately above the liquid nitrogen / air interface cannot be any colder than that 77 K (slightly less, because nitrogen is only 4/5ths of the air), so the 1/5th that is oxygen cannot actually condense – it &lt;em&gt;doesn’t get cold enough&lt;/em&gt;. Adding to that is the fact that the evaporating nitrogen further dilutes the oxygen that is present, further lowering the temperature that would be needed to condense oxygen at that pressure.&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T09:06:27.783" LastActivityDate="2015-05-01T09:06:27.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="794" PostTypeId="5" CreationDate="2012-07-13T22:03:39.300" Score="0" Body="&lt;p&gt;Physical chemistry is among &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt; one of the three &quot;classic&quot; branches of chemistry. It is closely related to &lt;a href=&quot;http://physics.stackexchange.com/&quot;&gt;physics&lt;/a&gt;, as it is the study of chemical systems and processes applying laws and concepts of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among its branches, there are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;theoretical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/thermodynamics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;thermodynamics&amp;#39;&quot; rel=&quot;tag&quot;&gt;thermodynamics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/kinetics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;kinetics&amp;#39;&quot; rel=&quot;tag&quot;&gt;kinetics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/electrochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;electrochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;electrochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical chemistry (Wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_physics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical physics (Wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.iupac.org/cms/wp-content/uploads/2015/07/Green-Book-PDF-Version-2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Green-Book-PDF-Version-2011.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2017-06-22T23:28:33.493" LastActivityDate="2017-06-22T23:28:33.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="795" PostTypeId="4" CreationDate="2012-07-13T22:03:39.300" Score="0" Body="The study of chemical systems using the laws and concepts of physics. This usually requires the techniques of thermodynamics, statistical mechanics and quantum mechanics." OwnerUserId="323" LastEditorUserId="323" LastEditDate="2012-07-13T23:08:13.637" LastActivityDate="2012-07-13T23:08:13.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="796" PostTypeId="2" ParentId="715" CreationDate="2012-07-14T09:20:12.347" Score="6" Body="&lt;p&gt;You might want to have a look at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Peterson et al., Major flavonoids in dry tea, &lt;em&gt;Journal of Food Composition and Analysis&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;18&lt;/em&gt;, 487--501, &lt;a href=&quot;http://dx.doi.org/10.1016/j.jfca.2004.05.006&quot;&gt;http://dx.doi.org/10.1016/j.jfca.2004.05.006&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the references therein, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Christiane Lakenbrink et al., Flavonoids and Other Polyphenols in Consumer Brews of Tea and Other Caffeinated Beverages, &lt;em&gt;J. Agric. Food Chem.&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, 48*, 2848--2852, &lt;a href=&quot;http://dx.doi.org/10.1021/jf9908042&quot;&gt;http://dx.doi.org/10.1021/jf9908042&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the abstract of the latter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] At brew times up to 2 min the composition of the brew solids was for each product practically independent of brew time, with flavonoids again comprising the major proportion (86−88%) of the total phenolics. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Longer extraction times (&gt; 10 min) or continuous (Soxhlet) extraction lead to partial decomposition of the flavonoids, most likely due to phenolic oxidation, cf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Won Jo Cheong et al., Determination of Catechin Compounds in Korean Green Tea Infusions under Various Extraction Conditions by High Performance Liquid Chromatography, &lt;em&gt;Bull. Korean. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;26&lt;/em&gt;, 747--754, &lt;a href=&quot;http://dx.doi.org/10.5012/bkcs.2005.26.5.747&quot;&gt;http://dx.doi.org/10.5012/bkcs.2005.26.5.747&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps a bit.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-07-14T09:20:12.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="797" PostTypeId="2" ParentId="704" CreationDate="2012-07-14T09:56:20.163" Score="7" Body="&lt;p&gt;Did you notice any change of colour (e.g. to brown) and/or the formation of gunk upon adding the 3M NaOH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd assume one would have to boil it, but is the following scenario conceivable under your conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;alkaline hydrolysis of the acetaminophen (paracetamol) to p-aminophenol&lt;/li&gt;&#xA;&lt;li&gt;aerial oxidation to the corresponding quinone imine&lt;/li&gt;&#xA;&lt;li&gt;partial polymerisation of the latter&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You have confirmed that acetaminophen was initially there. A a change in the constitution of Excedrine since the publication of the article can thus be excluded. Isn't a reaction (hydrolysis or whatever) under your extraction not the most likely explanation for the vanished acetaminophen?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-07-14T09:56:20.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="798" PostTypeId="1" CreationDate="2012-07-14T10:39:38.697" Score="19" ViewCount="862" Body="&lt;p&gt;I am looking at two phase diagrams, one has axes of $T$ and $\%\ce{MgClO4}$ in a $\ce{H2O-MgClO4}$ mix, and the other has axes of $T$ and relative humidity. The authors of the second phase diagram say &quot;The stability diagram is modiﬁed from [the earlier phase diagram] by converting salt concentration to relative humidity using the Pitzer model and parameters.&quot; How does changing the salt concentration affect the relative humidity, and how can I reproduce this calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am a geologist with minimal chemistry background).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LU1OU.png&quot; alt=&quot;first figure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top figure from Chevrier et al., GRL, 2009; bottom figure from Gough et al., EPSL, 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zpj4g.png&quot; alt=&quot;second figure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="422" LastEditorUserId="-1" LastEditDate="2012-07-19T09:27:18.437" LastActivityDate="2015-12-24T02:47:49.640" Title="Deliquescence/efflorescence: How can I convert between salt concentration and RH?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="799" PostTypeId="1" AcceptedAnswerId="803" CreationDate="2012-07-14T15:18:36.997" Score="2" ViewCount="16370" Body="&lt;p&gt;In some parts of the world food is stored with formalin so it looks fresh forever! This is astonishing, but true (&lt;a href=&quot;http://en.wikipedia.org/wiki/Formaldehyde#Contaminant_in_food&quot; rel=&quot;nofollow&quot;&gt;See refs here&lt;/a&gt;). As formalin is very bad for health and damages kidneys and liver very quickly, I am looking for a way to remove it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I buy fish, I know it has formalin in it. Every fruit I bought has formalin. I remember I bought 2 kilgrams of apples once. And it lasted for 1 month, looking fresh whole time. Then I threw it away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how can you remove formalin from food? (Especially fish and fruit). It's best if we can use a chemical found in the kitchen. If not, I'd think one that is found at a chemical store. &lt;/p&gt;&#xA;" OwnerUserId="423" LastEditorUserId="22" LastEditDate="2012-07-15T12:29:47.800" LastActivityDate="2013-07-17T05:49:50.407" Title="Remove formalin from food using chemical found in kitchen" Tags="&lt;food-chemistry&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="801" PostTypeId="1" CreationDate="2012-07-15T02:36:48.973" Score="8" ViewCount="12001" Body="&lt;p&gt;My course book says that kelvin is the unit of thermodynamic temperature, then what is the difference between temperature and thermodynamic temperature.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T11:09:06.793" LastActivityDate="2018-03-30T14:05:46.447" Title="What is the difference between temperature and thermodynamic temperature?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="802" PostTypeId="2" ParentId="801" CreationDate="2012-07-15T03:05:50.930" Score="11" Body="&lt;p&gt;Temperature is more of a relative notion--heat flows from higher temperature to lower temperature, but that's about it. The zero in a temperature scale is meaningless; it is an arbitrary convention. Usually, we measure temperature in Celsius, Fahrenheit, Rankine, etc (note that Kelvin measures temperature as well--but it also measures thermodynamic temperature)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thermodynamic&lt;/em&gt; temperature is such that zero temperature is absolute zero (unlike Celsius, where -273.15&amp;deg;C is absolute zero). It is related to the energy (specifically the kinetic energy) of the molecules of a substance-- for example, in a gas, kinetic energy per molecule is $\frac12k_BT$ (rotational energy, etc can be obtained if you know the type of molecule--but temperature is principally a measure of kinetic energy). It also has strong ties with entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We generally don't use &quot;temperature&quot; in thermodynamics, thermodynamic temperature is more convenient. Temperature is only useful in equations involving temperature difference--whereas equations involving the absolute value of temperature (Arrhenius equation in kinetics, entropy, Maxwell distribution equation, etc) demand thermodynamic temperature.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2012-07-16T11:11:42.377" LastActivityDate="2012-07-16T11:11:42.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="803" PostTypeId="2" ParentId="799" CreationDate="2012-07-15T13:42:52.967" Score="2" Body="&lt;p&gt;While the idea of filtering formaldehyde from food might sound naive, the motivation is understandable and the problem apparently real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found at least one publication on the topic: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T Yeasmin, *M S Reza, M N A Khan, F H Shikha and M Kamal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PRESENT STATUS OF MARKETING OF FORMALIN TREATED FISHES IN&#xA;DOMESTIC MARKETS AT MYMENSINGH DISTRICT IN BANGLADESH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Int. J. BioRes.&lt;/em&gt; &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;1&lt;/em&gt;, 21-24&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...]Study showed that formalin was not detected in any&#xA;  fish produced locally, but was detected in the imported ones of catla and rohu ranging from 0.5% to 1% which was&#xA;  sold in different markets of Mymensingh with comparatively lower price than those produced locally. The shelf life of&#xA;  the locally produced fish was much longer than those of imported fish which may be related to the loss of shelf life&#xA;  during its transport from importing countries. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Concerning the initial question:&#xA;I'm afraid but i don't see any chance to remove formaldehyde from food without turning it to another toxic cocktail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since formaldehyde was mostly used to increase the shelf time of imported fish, buying from markets predominantly selling local products might be a way--if available and affordable.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2012-07-15T15:12:16.287" LastActivityDate="2012-07-15T15:12:16.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="804" PostTypeId="2" ParentId="696" CreationDate="2012-07-15T16:54:06.363" Score="3" Body="&lt;p&gt;One of the earliest reports on the photochemical reduction of carbon dioxide was published about 30 years ago by Jean-Marie Lehn and Raymond Ziessel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photochemical generation of carbon monoxide and hydrogen by reduction of carbon dioxide and water under visible light irradiation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proc. Natl. Acad. Sci.&lt;/em&gt;, &lt;strong&gt;1982&lt;/strong&gt;, &lt;em&gt;79&lt;/em&gt;, 701-704&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They used Ru(bipy)&lt;sub&gt;2&lt;/sub&gt;Cl&lt;sub&gt;2&lt;/sub&gt; (the ruthenium complex absorbs the visible light) and CoCl&lt;sub&gt;2&lt;/sub&gt; as the catalytic system.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-07-15T16:54:06.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="805" PostTypeId="2" ParentId="290" CreationDate="2012-07-16T01:34:35.747" Score="7" Body="&lt;p&gt;I can't tell if the modifier &quot;formal&quot; applies only to &quot;definition&quot; or to &quot;algorithm&quot; also, in the first sentence of your question's body. In case any algorithm would do: this looks like a &lt;a href=&quot;https://stackoverflow.com/questions/261573/best-algorithm-for-detecting-cycles-in-a-directed-graph&quot;&gt;pretty standard instance of the cycle-detection problem&lt;/a&gt; that one learns in undergraduate computer science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only want to detect the presence of cycles, then the algorithm is essentially just to attempt a &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_sorting&quot; rel=&quot;nofollow noreferrer&quot;&gt;topological sorting&lt;/a&gt; of the graph; if the attempt fails, you have a cycle, and I guess that would mean that you have some sort of catalysis going on. (I didn't major in chemistry; please yell at me if that's &lt;em&gt;non sequitur&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so mere cycles are not enough - it seems you want cycles that have &quot;gain&quot;. How about a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycle_detection#Tortoise_and_hare&quot; rel=&quot;nofollow noreferrer&quot;&gt;tortoise and hare&lt;/a&gt; and have these multiply a state variable by the ratio of outgoing to incoming edges to reaction nodes (denoted by squares in your diagram)? If the hare catches up to the tortoise, that implies a cycle, hence catalysis; then if you compare their state, if the hare has accumulated a larger &quot;gain&quot; than the tortoise, I would say that that shows the presence of autocatalysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how this works if there are multiple, linked cycles. You could have one cycle in your graph that's autocatalytic, and another cycle linked to it that &quot;uses up&quot; its catalyst (something of a contradiction). For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B + C -&amp;gt; D + E&#xA;A + D + H -&amp;gt; 2C + F&#xA;2F + G -&amp;gt; H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is that still an autocatalytic system?&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorUserId="-1" LastEditDate="2017-05-23T12:41:54.053" LastActivityDate="2012-07-16T23:55:41.313" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="806" PostTypeId="1" AcceptedAnswerId="810" CreationDate="2012-07-16T08:46:25.980" Score="6" ViewCount="108" Body="&lt;p&gt;I'm a little confused as to what reactions are considered ionizations. Is there any requirement for the educts, for example? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references suggest that it's only considered an ionization if you produce additional charged atoms/molecules from the educts (and some processes are dubbed detachment, which intuitively look like ionizations to me). Then &quot;double ionization&quot; wouldn't be possible, strictly speaking. To what extend have electrons to be produced. At the moment I think that it's legal to call most processes that look like ionization ionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a remark in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; regarding the distiction from dissociation, but I don't really understand it.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="22" LastEditDate="2012-07-16T14:16:42.843" LastActivityDate="2012-07-16T14:16:42.843" Title="What is the exact nomenclature of ionization?" Tags="&lt;ions&gt;&lt;electrons&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="809" PostTypeId="2" ParentId="574" CreationDate="2012-07-16T09:23:44.307" Score="5" Body="&lt;p&gt;Janice, apart from the benzopyrylium (anthocyane) dyes you mentioned, there's quite some red to be found among the natural hydroxy-substituted naphthoquinones and anthraquinones, e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Purpurin [81-54-9] (1,2,4-Trihydroxyanthraquinone) from &lt;em&gt;Rubia tinctorum&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/59wLw.png&quot; alt=&quot;purpurin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Carminic acid [1260-17-9] from &lt;em&gt;Dactylopius occus&lt;/em&gt;, an insect living on opuntia&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alkannin [517-88-4] from &lt;em&gt;Alkanna tinctoria&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jtkkq.png&quot; alt=&quot;alkannin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ahh, I forgot to mention&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lycopene [502-65-8] from Tomato. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While others have already mentioned the carotenoids in general, lycopene is your favourite if you want it really red :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in lycopene in particular, don't bother to buy it from chemical suppliers. The lycopene concentration in tomato paste is that high (and the paste dirt cheap) that the extraction is the way to go. Colleagues of mine working in photosynthesis research did that all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I'm sorry, you asked for red dyes in flower petals, not natural red dyes in general. Otherwise I would have mentioned a member of another class of compounds we didn't cover so far: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Betanine [7659-95-2] from the red beet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, I won't ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="9961" LastEditDate="2017-04-25T18:57:15.383" LastActivityDate="2017-04-25T18:57:15.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="810" PostTypeId="2" ParentId="806" CreationDate="2012-07-16T14:10:48.627" Score="6" Body="&lt;p&gt;The reference for chemical terminology is the &lt;a href=&quot;http://goldbook.iupac.org/&quot;&gt;IUPAC Gold Book&lt;/a&gt;, or &lt;em&gt;IUPAC Compendium of Chemical Terminology&lt;/em&gt;. The &lt;a href=&quot;http://goldbook.iupac.org/I03183.html&quot;&gt;entry for &lt;strong&gt;ionization&lt;/strong&gt;&lt;/a&gt; is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The generation of one or more ions. It may occur, e.g. by loss of an electron from a neutral molecular entity, by the unimolecular heterolysis of such an entity into two or more ions, or by a heterolytic substitution reaction involving neutral molecules […] The loss of an electron from a singly, doubly, etc. charged cation is called second, third, etc. ionization. This terminology is used especially in mass spectroscopy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It includes examples that you might find instructive. It confirms your intuition that the term has a broad understanding, covering a wide variety of reactions.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-07-16T14:10:48.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="811" PostTypeId="1" CreationDate="2012-07-16T17:18:50.973" Score="7" ViewCount="1786" Body="&lt;p&gt;Is it possible to extract information on elemental ratios from X-ray photoelectron spectroscopy (XPS)?  For example, I am reading a paper and in it they say that the carbon to oxygen (C:O) ratios of the four samples are 2.8, 2.9, 5.1, and 23.3.  However, the authors do not provide the actual XPS spectra, not even in a Supporting Information-type addendum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a computational chemist and have no experience with XPS.  I do know that elements, and even particular hybridizations, give characteristic peaks at characteristic binding energies (in units of eV, for example).  This, I know, enables &lt;em&gt;qualitative&lt;/em&gt; analysis of compounds.  But is there a way to extract &lt;em&gt;quantitative&lt;/em&gt; information from XPS?  Suppose I want to determine the C:O ratio in a compound containing only C, O, and H.  Do I take the ratio of the C and O peak heights, or do I take the ratio of the area under the peaks corresponding to C and O?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know of any good review articles that would help me to understand the basics of XPS?&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2013-09-13T17:38:55.633" Title="Extracting elemental ratios from X-ray photoelectron spectroscopy (XPS)" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="813" PostTypeId="2" ParentId="811" CreationDate="2012-07-16T17:40:37.733" Score="3" Body="&lt;p&gt;I found some information in Skoog, &lt;em&gt;Principles of Instrumental Analysis&lt;/em&gt;, 6th ed.  On page 597:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Quantitative Applications.&lt;/strong&gt;  Once, XPS was not considered to be a very useful quantitative technique. However, there has been increasing&#xA;  use of XPS for determining the chemical composition of the surface&#xA;  region of solids.  If the solid is homogenous to a depth of several&#xA;  electron mean free paths, we can express the number of photoelectrons&#xA;  detected each second $I$ as $$I = n \phi \sigma \epsilon \eta A T \ell$$ where $n$ is the number density of atoms ($\text{atoms cm}^{-3}$) of the sample, $\phi$ is the flux of the incident X-ray&#xA;  beam ($\text{photons cm}^{-2} \text{ s}^{-1}$), $\sigma$ is the&#xA;  photoelectric cross section for the transition&#xA;  ($\text{cm}^{2}/\text{atom}$), $\epsilon$ is the angular efficiency&#xA;  factor for the instrument, $\eta$ is the efficiency of producing photo&#xA;  electrons ($\text{photoelectrons/photon}$), $A$ is the area of the&#xA;  sample from which photoelectrons are detected ($\text{cm}^{2}$), $T$&#xA;  is the efficiency of detection of the photoelectrons, and $\ell$ is&#xA;  the mean free path of the photoelectrons in the sample ($\text{cm}$).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For a given transition, the last six terms are constant, and we can&#xA;  write the atomic sensitivity factor $S$ as $$S = \sigma \epsilon \eta A T \ell$$  For a given spectrometer, a set of relative values of $S$&#xA;  can be developed for the elements of interest.  Note that the $I/S$ is&#xA;  directly porportional to the concentration $n$ on the surface.  The&#xA;  quantity $I$ is usually taken as the peak area, although peak heights&#xA;  are also used.  Often, for quantitative work, internal standards are&#xA;  used.  Relative precisions of about 5% are typical.  For the analysis&#xA;  of solids and liquids, it is necessary to assume that the surface&#xA;  composition of the sample is the same as its bulk composition.  For&#xA;  many applications this assumption can lead to significant errors. &#xA;  Detection of an element by XPS requires that it be present at a level&#xA;  of at least 0.1%.  Quantitative analysis can usually be performed if&#xA;  5% of the element is present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="271" LastActivityDate="2012-07-16T17:40:37.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="814" PostTypeId="1" AcceptedAnswerId="818" CreationDate="2012-07-16T21:35:35.183" Score="7" ViewCount="736" Body="&lt;p&gt;I'm researching cyanide for a biology project and noted that one of the largest sources of carbohydrates in the tropics is the cassava root -- which I previously researched as a food that contains a cyanogenic glycoside that is metabolized into hydrogen cyanide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the sources I reviewed claimed that, while boiling the cassava, the cyanogenic glycosides were broken down into components such hydrogen cyanide, which made the plant safer to consume. I have a limited understanding of chemistry and can't resolve why this is considered a safe way to cook the plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research indicates that inhalation of hydrogen cyanide is significantly more toxic than the oral route, so why would this method be considered beneficial? Not sure why this is considered safer, unless the information I've found is not correct. Any help in putting this together would be appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="432" LastEditorUserId="432" LastEditDate="2012-07-16T22:09:27.647" LastActivityDate="2015-02-21T22:27:10.093" Title="Cyanogenic glycosides and HCN" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;toxicity&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="815" PostTypeId="1" AcceptedAnswerId="819" CreationDate="2012-07-17T02:44:36.610" Score="5" ViewCount="13459" Body="&lt;p&gt;I have processed a si/polymer hybrid using a silane coupling agent which I need to determine the energy gap / band gap.  Would you please let me know the procedure to go about determining the band gap?&lt;/p&gt;&#xA;" OwnerUserId="433" LastEditorUserId="4356" LastEditDate="2015-07-16T23:04:38.117" LastActivityDate="2015-07-16T23:04:38.117" Title="How to find the band gap for a compound?" Tags="&lt;electrochemistry&gt;&lt;polymers&gt;&lt;solid-state-chemistry&gt;&lt;cyclic-voltammetry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="816" PostTypeId="1" CreationDate="2012-07-17T09:47:39.277" Score="7" ViewCount="607" Body="&lt;p&gt;So there is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Formation_enthalpy&quot;&gt;standard formation enthalpy&lt;/a&gt; $\Delta H^0_f$ and there is also formation entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there more (independent) quantities? What about the heat capacity? Is there a general theoretical structure where one can show one got &lt;em&gt;all&lt;/em&gt; the conserved quantities associated with a chemical reaction? It seems to me that one could this on a quite rigid structure and invesigate the properties of chemical processes from that point of view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to what extend are the procedures like using &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess%27s_law&quot;&gt;Hess's law&lt;/a&gt; just rules of thumb or what are some insights about he deeper structure where they come from.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="30" LastEditDate="2012-07-17T12:25:46.253" LastActivityDate="2012-07-31T12:04:08.673" Title="Algebra of conserved quantities for chemical reactions" Tags="&lt;theoretical-chemistry&gt;&lt;kinetics&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="818" PostTypeId="2" ParentId="814" CreationDate="2012-07-17T11:12:05.143" Score="2" Body="&lt;p&gt;OK, I was too lazy to look up LD50 and LC50 of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; for you ;)&lt;/p&gt;&#xA;&lt;p&gt;But if you have a chance, have a look at&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;G. Padmaja &amp;amp; Keith H. Steinkraus,&lt;br /&gt;&#xA;Cyanide detoxification in cassava for food and feed uses,&#xA;&lt;em&gt;Critical Reviews in Food Science and Nutrition&lt;/em&gt;, &lt;strong&gt;1995&lt;/strong&gt;, &lt;em&gt;35(4)&lt;/em&gt;, 299-339 &lt;a href=&quot;http://dx.doi.org/10.1080/10408399509527703&quot; rel=&quot;nofollow noreferrer&quot;&gt;[DOI]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Seemingly, the processing of cassava (manioc) involves several steps, such as drying in the sun, grating, watering and cooking, each of which either facilitates the enzymatic or thermal cleavage of the linamarin (i.e. the toxic compound) to glucose or supports the release of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In total, the processing minimizes the risk of exposition through inhalation. None of the steps is perfomed in confined space, watering helps to wash out the &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; which in water dissociates a bit (&lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; is a weak acid).&lt;/p&gt;&#xA;&lt;p&gt;Thus, digestion of the untreated roots is likely to be fatal, while the slow processing is apparently not.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Ok, I digged a bit and maybe you have you have access to this one:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bala Nambisan&lt;/p&gt;&#xA;&lt;p&gt;Strategies for elimination of cyanogens from cassava for reducing toxicity and improving food safety &lt;em&gt;Food and Chemical Toxicology&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt;, 690–693 &lt;a href=&quot;http://dx.doi.org/10.1016/j.fct.2010.10.035&quot; rel=&quot;nofollow noreferrer&quot;&gt;[DOI]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-21T22:27:10.093" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="819" PostTypeId="2" ParentId="815" CreationDate="2012-07-17T11:50:46.047" Score="11" Body="&lt;p&gt;There are two ways in which the band gap of a material: using optical spectroscopy and using electrochemistry. The way you go about analyzing the data to determine the band gap is independent of material, but the process of collecting the data is very specific to your material, and you should consult papers where others have made similar measurements on similar systems. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Optical spectroscopy. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You need to collect an absorbance spectrum of your material. Given what I can gather from your fabrication description, you will probably need to collect said spectrum in the solid state. The exact means of doing that vary somewhat by spectrometer model. Find someone who knows your spectrometer very well and ask them how to do it (or read the manual).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorbance process measures the energy difference between the ground state and the excited state as electromagnetic radiation is absorbed. Since there are a number of vibrational states associated with both the ground state and the excited state, the absorbance spectrum appears as a blob instead of a nice line. The &quot;true&quot; band gap, or $E_{00}$, the energy difference between the lowest vibrational states of the ground and excited states, is difficult to measure. However, the most commonly accepted approximation is the &lt;em&gt;onset of absorption&lt;/em&gt; from the low energy side. The onset value is the intersection of the extrapolations of the linear parts of the spectrum. See the image below. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/81MK4.png&quot; alt=&quot;onset of absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the wavelength (x) at that point, you can convert it into energy using the Planck relation:&#xA;$$E(\text{J})=\dfrac{hc}{\lambda(\text{m})}$$&#xA;If you want your energy in electronvolts, you can convert directly:&#xA;$$E(\text{eV})=\dfrac{1239.84187 \text{ eV nm}}{\lambda(\text{nm})}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Electrochemistry, specifically voltammetry&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Voltammetry gives you the ability to determine the peak oxidation potential, $V_{\text{ox}}$, and the peak reduction potential, $V_{\text{red}}$, of your material. Both &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_voltammetry&quot; rel=&quot;noreferrer&quot; title=&quot;cyclic voltammetry&quot;&gt;cyclic voltammetry&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_pulse_voltammetry&quot; rel=&quot;noreferrer&quot; title=&quot;differential pulse voltammetry&quot;&gt;differential pulse voltammetry&lt;/a&gt; can give you these data, and both should work on solids, though you may want to find someone who has experience in such things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption that we operate under when using electrochemistry to determine HOMO-LUMO or bad gap, is that when we oxidize a molecule or material, we take the first electron out of the HOMO (or valence band). When we reduce the molecule or material, we put an electron into the LUMO (or conductance band). Thus, $V_{\text{ox}}$ tells us something about the HOMO (valence band) and $V_{\text{red}}$ tells us something about the LUMO (conductance band). The energy gap (in electronvolts) is simply the difference between the two potentials (in volts) multiplied the charge on the electron $e=-1 \text{ eV/V}$.&#xA;$$E(\text{eV})=e(V_{\text{red}}-V_{\text{ox}})$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-07-17T11:50:46.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="820" PostTypeId="1" AcceptedAnswerId="821" CreationDate="2012-07-17T14:35:53.300" Score="29" ViewCount="16675" Body="&lt;p&gt;On &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Resistivity_of_various_materials&quot;&gt;this Wikipedia page&lt;/a&gt;, the electrical conductivity of various materials are given in the third column ($\sigma \text{ (S/m) at 20}^\circ \text{C}$).  I am interested in the entry for &lt;strong&gt;Carbon (graphite)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2$ to $3 \times 10^5 \text{ S/m} \perp \text{basal plane}$ &lt;/li&gt;&#xA;&lt;li&gt;$3.3 \times 10^2 \text{ S/m} \parallel \text{basal plane}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why is conductivity &lt;em&gt;lower&lt;/em&gt; parallel to the plane than in the direction perpendicular to the plane?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am surprised by the relative magnitudes of these $\sigma$ values because I thought that one of the amazing things about graphene -- an atomic layer of graphite -- was that its conductivity &lt;em&gt;in&lt;/em&gt; the plane (i.e., parallel to the basal plane) is very high.  I would expect a similar trend parallel to the basal plane in graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any suggestions for reading that I can do to understand this?&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="30836" LastEditDate="2020-04-30T14:44:55.730" LastActivityDate="2020-04-30T14:44:55.730" Title="Electrical conductivity of graphite" Tags="&lt;physical-chemistry&gt;&lt;conductivity&gt;&lt;carbon-allotropes&gt;&lt;electrical-properties&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="821" PostTypeId="2" ParentId="820" CreationDate="2012-07-17T14:59:10.453" Score="21" Body="&lt;p&gt;Good catch - you are absolutely correct. This is a mistake in the Wikipedia page, as a quick check with the reference listed there confirms. The values are the wrong way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing it now...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited to add:&lt;/strong&gt; this of course raises the question of &lt;em&gt;why&lt;/em&gt; it should be this way around. A somewhat hand-wavy answer is that, just like in benzene, the $p_z$ orbitals overlap within each graphene sheet so that the electron contributed by each carbon is delocalised in this plane (the other three are bound up in the three $\sigma$ bonds). So in a sense this is conjugation, but over a whole 2D plane rather than around a ring or along a single chain of atoms. On the other hand, the graphene sheets are rather further apart from one another than the covalently bonded atoms within a sheet (as you might expect since the sheets are only weakly bound through van der Waals interactions), and so the overlap is greatly reduced and conduction is much more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete answer would rely on a study of graphite's electronic band structure, which is an active research area and beyond the scope of this response!&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorUserId="390" LastEditDate="2012-07-17T17:29:56.677" LastActivityDate="2012-07-17T17:29:56.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="823" PostTypeId="2" ParentId="769" CreationDate="2012-07-18T10:39:25.263" Score="6" Body="&lt;p&gt;Pointing out the stable conformations of cyclohexane was a major part of Barton's &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1969/&quot;&gt;Nobel Prize&lt;/a&gt;. His original review is &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1956/QR/qr9561000044&quot;&gt;here&lt;/a&gt; (paywalled, but you can see the first page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic is discussed by Henry Rzepa on his blog starting with a little history:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like benzene, its fully saturated version cyclohexane represents an icon of organic chemistry. By 1890, the structure of planar benzene was pretty much understood, but organic chemistry was still struggling somewhat to fully embrace three rather than two dimensions. A grand-old-man of organic chemistry at the time, Adolf von Baeyer, believed that cyclohexane too was flat, and what he said went. So when a young upstart named Hermann Sachse suggested it was not flat, and furthermore could exist in two forms, which we now call chair and boat, no-one believed him. His was a trigonometric proof, deriving from the tetrahedral angle of 109.47 at carbon, and producing what he termed strainless rings.&#xA;  &lt;a href=&quot;http://www.ch.imperial.ac.uk/rzepa/blog/?p=1587&quot;&gt;henry rzepa blog&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some related calculations are available &lt;a href=&quot;http://www-jmg.ch.cam.ac.uk/software/QRC/example.html&quot;&gt;here&lt;/a&gt; which show the relative energy levels of the different conformers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these results are easy to visualise with simple molecular modelling kit, but you need space-filling models to really appreciate some of the subtle detail (eg why a twist-boat conformation is more stable than a boat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that, at least at room temperature, the conformers will interconvert rapidly as there is enough thermal energy to overcome the barriers between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, there is even a &lt;a href=&quot;http://www.youtube.com/watch?v=_6Qo3ABmECM&quot;&gt;Youtube&lt;/a&gt; visualisation showing some of this.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2012-07-18T10:39:25.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="824" PostTypeId="1" AcceptedAnswerId="826" CreationDate="2012-07-18T12:56:30.473" Score="11" ViewCount="29755" Body="&lt;p&gt;I know that standard latex and nitrile gloves don't stop common organic solvents such as DCM and acetone. However, I was reading a &lt;a href=&quot;http://www.reddit.com/r/chemistry/comments/wqdzl/worst_hand_model_ever/&quot; rel=&quot;noreferrer&quot;&gt;Reddit thread&lt;/a&gt; and there were comments that if you are coming into contact with solvents you are just wearing the wrong type of gloves. In a non-manufacturer specific fashion, what types of gloves will stop DCM and acetone? What will it slow down long enough for me to have a chance to finish what I'm doing and change my gloves in a safe manner? I am looking for a guide on what gloves to use. I just had DCM on my mind since nitrile doesn't slow it down, and it burns. And acetone, since it is hard to stop.&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="4945" LastEditDate="2015-02-10T07:02:43.137" LastActivityDate="2018-06-03T20:37:10.627" Title="What type(s) of gloves are effective against DCM and acetone?" Tags="&lt;solvents&gt;&lt;safety&gt;&lt;gloves&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="825" PostTypeId="2" ParentId="814" CreationDate="2012-07-18T13:11:08.747" Score="-1" Body="&lt;p&gt;And we can use google too @Greg !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The cyanide ion (CN-) is the toxic moiety in hydrogen cyanide. The toxicity of simple cyanide salts, such as potassium and sodium cyanide is, therefore, similar to that of hydrogen cyanide.&quot; and from &lt;a href=&quot;http://www.fas.org/nuke/guide/usa/doctrine/army/mmcch/Cyanide.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; &quot;Toxicities: Cyanide is the least toxic of the &quot;lethal&quot; chemical agents. The LCt50s of AC and CK by inhalation have been estimated to be 2500-5000 mg&quot;min/m3 for AC and about 11,000 mg&quot;min/m3 for CK. LD50s for hydrogen cyanide have been estimated to be 1.1 mg/kg for intravenous administration and 100 mg/kg after skin exposure. The oral LD50s for sodium and potassium cyanide are about 100 and 200 mg/kg respectively.&quot; AC - the term refers to the anion CN-, or to its acidic form, hydrocyanic acid (HCN).&#xA;Have a look at &lt;a href=&quot;http://www.hpa.org.uk/webc/HPAwebFile/HPAweb_C/1202487077231&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; absorption, metabolism and toxicity of cyanides&lt;/p&gt;&#xA;" OwnerUserId="417" LastActivityDate="2012-07-18T13:11:08.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="826" PostTypeId="2" ParentId="824" CreationDate="2012-07-18T13:16:01.850" Score="6" Body="&lt;p&gt;You can use tips from the site &lt;a href=&quot;http://www.aps.anl.gov/Safety_and_Training/User_Safety/gloveselection.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; - the table at the bottom or just google which glove can suit your needs.&#xA;As you can read from the link:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitrile gloves are Low cost, excellent physical properties, dexterity, but they are Poor vs. benzene, methylene chloride, trichloroethylene, many ketones. They are recommended for Oils, greases, aliphatic chemicals, xylene, perchloroethylene, trichloroethane; fair vs. toluene,&lt;/li&gt;&#xA;&lt;li&gt;According to Agronne for dealing with acetone you should use Natural Latex/Rubber or Butyl gloves.&lt;/li&gt;&#xA;&lt;li&gt;For DCM there are no good gloves - you could try using neoprenes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="417" LastEditorUserId="73" LastEditDate="2012-08-21T07:56:13.277" LastActivityDate="2012-08-21T07:56:13.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="827" PostTypeId="1" CreationDate="2012-07-19T14:30:27.070" Score="7" ViewCount="4538" Body="&lt;p&gt;I'm planning to do an experiment where I extract burnt toast and test how that affects the growth of mung beans. I know burnt food contains PAHs, but what are some other chemicals present and how would they affect plant growth?&lt;/p&gt;&#xA;" OwnerUserId="360" LastActivityDate="2016-10-24T02:36:28.137" Title="What chemicals are present in burnt toast?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="828" PostTypeId="1" CreationDate="2012-07-19T15:46:26.723" Score="6" ViewCount="1517" Body="&lt;p&gt;Is there a quick way to determine if a salt (or ionic compound in general) crystal should be birefringent from its formula without having to know/look up the crystal structure?  When I'm looking for protein crystals, if I know the condition has a lot of salt in it, but it shouldn't be birefringent (e.g. NaCl), it's a hint that birefringent crystals I do see &lt;em&gt;might&lt;/em&gt; be protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the only non-birefringent crystal type (e.g. not glasses) cubic?  Does it matter if it's primitive, face-centered, or body-centered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to looking for a generalization, are most (all?) salts with polyatomic ions birefringent?    All +1/-1 (NaCl, KBr, etc...) aren't?&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2012-07-19T21:30:06.420" Title="What salts are/are not birefringent?" Tags="&lt;crystal-structure&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="829" PostTypeId="2" ParentId="824" CreationDate="2012-07-19T16:39:15.880" Score="7" Body="&lt;p&gt;I just did a bit of an experiment. My lab uses &lt;a href=&quot;http://www.kcprofessional.com/us/product-details.asp?prd_id=55082&quot;&gt;Kimberly Clark nitrile gloves&lt;/a&gt;, and we have a huge jug of acetone. First, I put a glove on my hand, and poured acetone all over it. My hand felt cold, but not nearly as cold as it feels when I spill acetone on my bare hand. I allowed the acetone to evaporate from the glove (approximately 30 s), and removed the glove. There was absolutely no acetone on my hand or the inside of the glove. I then took a fresh glove, and poured enough acetone into it to fill the fingers with acetone. I held the opening shut for about 5 minutes, and poured out the acetone. During that time, the glove slowly felt thinner and easier to stretch. At the end, it was so thin that it tore with almost no effort. So for the particular pair of nitrile gloves and acetone, it looks like somewhere between 1 and 5 minutes' exposure will render the glove useless. &lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-07-19T16:39:15.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="830" PostTypeId="2" ParentId="593" CreationDate="2012-07-19T20:39:00.613" Score="7" Body="&lt;p&gt;This is a loaded question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some groups focus on precisely cooking a recipe from known ingredients to produce useful functionals, the &lt;a href=&quot;http://comp.chem.umn.edu/truhlar/docs/773.pdf&quot;&gt;Minnesota functionals&lt;/a&gt; being quite popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you will have to benchmark available functionals against your property and known analogues to your target of choice and ensure that you are capturing the relevant physics. This requires that you familiarize yourself with the respective assumptions in functional design, e.g., the homogeneous electron gas approximation.&lt;/p&gt;&#xA;" OwnerUserId="440" LastActivityDate="2012-07-19T20:39:00.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="831" PostTypeId="2" ParentId="828" CreationDate="2012-07-19T21:30:06.420" Score="4" Body="&lt;p&gt;Yes, the only crystal system that is guaranteed not to be birefringent is cubic, because all of the other crystal systems have at least one axis distinct from the others. What matters is the symmetry not the centring, so this is equally true for face-centred or body-centred as for primitive crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, however, there is no way to determine the crystal structure, or even the crystal system, simply from consideration of the formula - indeed, crystal structure prediction and prediction of polymorphism are active research fields. Counterexamples to your first generalisation include &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01103a006&quot; rel=&quot;nofollow&quot;&gt;ammonium chloride, bromide, and iodide&lt;/a&gt;; to your second include ammonium fluoride and silver iodide. &lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2012-07-19T21:30:06.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="833" PostTypeId="2" ParentId="162" CreationDate="2012-07-20T11:56:08.570" Score="-1" Body="&lt;p&gt;If you deal with cell culture its irrelevant; because the magnesium channels located at the surface of the cell will be forcing magnesium ions to flow into the cell, and if the balance between free and chelated magnesium ions is disturbed in favor of chelate – it will produce free magnesium ions due to the chelate constant.&lt;/p&gt;&#xA;" OwnerUserId="417" LastEditorUserId="30" LastEditDate="2012-08-25T07:42:58.570" LastActivityDate="2012-08-25T07:42:58.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="835" PostTypeId="1" AcceptedAnswerId="836" CreationDate="2012-07-20T20:34:24.583" Score="12" ViewCount="644" Body="&lt;p&gt;This question is sort of a sequel to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/822/how-can-capacitance-values-be-determined-from-a-cyclic-voltammogram&quot;&gt;my previous question about cyclic voltammetry (CV)&lt;/a&gt;.  One of the responses made reference to the fact that an ideal capacitor gives rise to a rectangular cyclic voltammogram.  Can you please help me understand why this is the case?  In other words, why does an ideal capacitor reach a constant current $I$ as soon as a voltage $V$ is applied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I indeed see nearly ideal CVs in many literature articles (CVs which are rather rectangular with rounded corners).  In other figures, though, I see relative deviation from &quot;rectangles with rounded corners,&quot; in that I see abrupt peaks, spikes, or valleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, below I have plotted two figures from &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0013468604010941&quot; rel=&quot;nofollow noreferrer&quot;&gt;Khomenko, &lt;em&gt;Electrochimica Acta&lt;/em&gt; &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;50&lt;/em&gt;, 2499-2506&lt;/a&gt;.  Just very roughly and &quot;hand wavy,&quot; what might be the qualitative reason for the &quot;rectangle with rounded corners&quot; behavior of Figure 8 (left) and the &quot;abrupt peaks&quot; behavior of Figure 4 (right)?  Could it be that the sample in Figure 8 (left) is relatively unreactive toward apllied potential, whereas the sample in Figure 4 (right) undergoes redox (Faradaic) reactions -- indicating the presence of so-called pseudocapacitance -- when an external potential is applied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M7ISm.gif&quot; alt=&quot;CV&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please know that I am not looking for an answer specific to the article to which I linked.  I am only asking this question in the context of basic, qualitative aspects of cyclic voltammetry.&lt;/p&gt;&#xA;" OwnerUserId="271" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2020-10-20T11:41:10.647" Title="Why does an ideal capacitor give rise to a rectangular cyclic voltammogram (CV)?" Tags="&lt;solid-state-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="836" PostTypeId="2" ParentId="835" CreationDate="2012-07-21T20:14:48.817" Score="8" Body="&lt;p&gt;The kind of analysis truly required is outside the scope of chemistry and would be best handled in the careful hands of an electrical engineer. I'll give a brief attempt that exploits an easy-to-use circuit simulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to first think of an equivalent circuit that we can use as a model to determine its behavior. I suggest the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZlrvB.png&quot; alt=&quot;equivalent circuit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where my variable is R1's value for its resistance, which I shall set to 0, 10 and 100 ohm. If we were to watch this happen in time then we would see the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zUvQ.png&quot; alt=&quot;voltage over time&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3hhaH.png&quot; alt=&quot;amperage over time&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quickly converting these to Current vs Voltage and running two other simulations at different resistances and we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/msQA2.jpg&quot; alt=&quot;CVs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These results are due to setting up differential equations and solving them appropriately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play with the circuit I made for yourself &lt;a href=&quot;https://www.circuitlab.com/circuit/5q6znb/cv-of-ideal-capacitor/&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="32" LastActivityDate="2012-07-21T20:14:48.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="837" PostTypeId="1" CreationDate="2012-07-22T18:23:28.347" Score="0" ViewCount="826" Body="&lt;p&gt;I'm confused about the octet rule.  As far as I understand, you need to complete the valence shell of the elements by sharing electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, carbon has 6 electrons which means it has 4 valence electrons, but because it has 6 electrons in total it exceeds the first shell with 4 electrons. The second shell has 4 orbitals with 2 electrons each, so to complete the second shell I need 8 valence electrons, but I had read somewhere that the octet rule holds true until down to the 4th period. How can that be when say, magnesium, has 12 total electrons and thus it exceeds the the second shell with 4 electrons but the third shell needs 18 valence electrons to be complete, not just 8? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I read that if the central atom has more then 8 electrons, then that's ok too? What happens if I have a choice of exceeding the 8 electrons by forming another bond with an adjacent atom or sticking those electrons on the opposite side of the adjacent atom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one strictly wrong/preferable? &lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="73" LastEditDate="2012-07-23T01:26:36.367" LastActivityDate="2017-12-09T00:23:13.540" Title="Lewis dot structures and exceeding the octet" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="839" PostTypeId="1" AcceptedAnswerId="841" CreationDate="2012-07-23T11:19:15.477" Score="6" ViewCount="1743" Body="&lt;p&gt;In nearly all default parameter sets for 2D NMR experiments on Bruker spectrometers, the window function is set to a pure squared cosine function for both dimensions (&lt;code&gt;QSINE&lt;/code&gt; with &lt;code&gt;SSB = 2&lt;/code&gt;). In contrast to that, 1D spectra are often processed with the &lt;code&gt;efp&lt;/code&gt; command, which uses an exponential window function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the squared cosine window function the default? Which advantages and disadvantages does it have compared to other common window functions?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-07-23T13:23:51.787" LastActivityDate="2012-09-28T01:06:12.703" Title="Why is a squared cosine window function commonly used for 2D spectra?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="840" PostTypeId="2" ParentId="837" CreationDate="2012-07-23T14:52:31.643" Score="3" Body="&lt;p&gt;For the first part of your question, you are correct that Mg has 12 electrons, which are arranged in shells in the pattern 2 - 8 - 2. But magnesium is a typical metal in that it tends to lose electrons to form a cation that then forms solids by ionic bonding. The octet rule predicts that the resulting cation will be stablest if its outermost shell contains eight electrons. And indeed the magnesium ion is $\mathrm{Mg}^{2+}$, two electrons less than the atom, with an electron configuration of 2 - 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if you had asked about a non-metal atom in the third period, say Cl? This has an electron configuration of 2 - 8 - 7, and tends to form a single covalent bond (e.g., to form $\mathrm{Cl}_2$ or $\mathrm{HCl}$), giving a configuration of 2 - 8 - 8 rather than 2 - 8 - 18. The reason for this is that the first eight valence electrons occupy the $s$ (2 electrons) and $p$ (6 electrons) subshells, while any further electrons in the third shell must go into the $d$ subshell. These orbitals are both higher in energy than those from the first two subshells (which is why, as you go through the periodic table, the $4s$ shell is filled before the $3d$) and extend less far from the nucleus, so that they are much less likely to overlap other atoms' orbitals to form a covalent bond. So it is often energetically preferable not to put electrons into this subshell if possible, which is why the octet rule can still apply to elements in this period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it is &lt;em&gt;possible&lt;/em&gt; to put electrons into this subshell, which is why it is possible for the central atom to have more than an octet in compounds like $\mathrm{SF_6}$. This leads into the second part of your question, which, as Manishearth suggested, looks like a duplicate of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/&quot;&gt;Can an atom have more than 8 valence electrons? If not, why is 8 the limit?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2012-07-23T14:52:31.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="841" PostTypeId="2" ParentId="839" CreationDate="2012-07-23T17:37:35.067" Score="5" Body="&lt;p&gt;The difference in window function is a result of the parameter being optimized for the respective experiment.  Exponential multiplication of the Free-Induction Decay (FID) weights the beginning of the FID where the amplitude is at it's greatest.  Weighting the beginning of the FID results increases the signal-to-noise ratio, since the noise should have the least impact with the strong signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2D NMR, we are usually more interested in resolution to aid in differentiating the various peaks and cross-peaks.  We can always take more scans to try and increase the signal strength, so resolution enhancement becomes the key objective.  A pure &lt;code&gt;sine&lt;/code&gt; function offers increased resolution at the loss of signal enhancement.  The shifted &lt;code&gt;sine&lt;/code&gt; function (or pure &lt;code&gt;cosine&lt;/code&gt; function with &lt;code&gt;SSB=2&lt;/code&gt;), provides a good mixture of the two objectives.  The enhancement doesn't weight the early parts of the FID as heavily as exponential weighting, but more than a &lt;code&gt;sine&lt;/code&gt; function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of &lt;code&gt;SSB=2&lt;/code&gt; for a &lt;code&gt;cosine&lt;/code&gt; function is used most often in 2D experiments where we need a usually more significant amount of signal enhancement, such as TOCSY, NOESY etc.  &lt;code&gt;sine&lt;/code&gt; enhancement is used most often for COSY, to reduce filtering out peaks from overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;strong&gt;UPDATE&lt;/strong&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer some further questions posted in the comments.  Using a &lt;code&gt;gaussian&lt;/code&gt; weighting for 2D-NMR data is not as preferable as &lt;code&gt;cosine&lt;/code&gt; or &lt;code&gt;sine&lt;/code&gt; weighting.  The &lt;code&gt;gaussian&lt;/code&gt; weighting has a much more narrow distribution, and rapidly approaches zero.  This can prematurely truncate the FID (by essentially weighting the later portion as zero), which introduces artificats into the resulting spectrum known as &quot;truncation wiggles&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring the &lt;code&gt;cosine&lt;/code&gt; (which remember is just the &lt;code&gt;sine&lt;/code&gt; function shifted) is done both to slow the increase and decrease of the weighting function over the course of the FID.  At the beginning of the FID, this allows for a greater amount of sensitivity to the early parts, while sacrificing a small amount of resolution.  Additionally, the slow decay also reduces the truncation artifacts discussed in the first question of the update.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="656" LastEditDate="2012-09-28T01:06:12.703" LastActivityDate="2012-09-28T01:06:12.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="842" PostTypeId="1" AcceptedAnswerId="846" CreationDate="2012-07-24T01:13:18.563" Score="5" ViewCount="1006" Body="&lt;p&gt;In a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/839/why-is-a-squared-cosine-window-function-commonly-used-for-2d-spectra&quot;&gt;recent question&lt;/a&gt; it was explained that Gaussian window functions are not used in 2D NMR spectra as they can introduce artifacts into the spectrum baseline by truncating the FID. Why then are they used with 1D spectra instead of sin or cosine functions?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2012-07-24T10:17:01.737" Title="Why is the Gaussian Window Function used to enhance resolution in 1D NMR spectra?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="843" PostTypeId="1" AcceptedAnswerId="32366" CreationDate="2012-07-24T04:49:35.883" Score="4" ViewCount="1170" Body="&lt;p&gt;Mixing solid copper, 5% vinegar, and hydrogen peroxide ($\ce{H2O2}$) causes copper acetate to form. The process will occur very slowly without hydrogen peroxide. Adding $\ce{H2O2}$ speeds the formation of copper acetate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does Hydrogen Peroxide promote the formation of Copper (II) Acetate? &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="9961" LastEditDate="2015-06-03T01:25:23.163" LastActivityDate="2015-06-03T01:25:23.163" Title="What role does H2O2 have in copper (II) acetate formation?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="844" PostTypeId="2" ParentId="843" CreationDate="2012-07-24T06:19:53.810" Score="7" Body="&lt;p&gt;Presumably hydrogen peroxide causes copper metal to donate 2 electrons to form $\ce{Cu(II)}$ -- which is soluble in water -- and we'd say the hydrogen peroxide is acting as an oxidizing agent. &lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorDisplayName="user15489" LastEditDate="2015-06-02T19:42:03.403" LastActivityDate="2015-06-02T19:42:03.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="845" PostTypeId="1" AcceptedAnswerId="848" CreationDate="2012-07-24T08:55:06.097" Score="8" ViewCount="10667" Body="&lt;p&gt;I always hear that some big corporations use &quot;diamond dust&quot; to filter water seven times to make &quot;diamond clear&quot; vodka. I tried googling something about using diamonds for filtering, but I could not find anything except corporate sites that tell us only the best things about &quot;diamond dust filtration technology&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a bogus claim? Because I can not think about diamonds as a viable filtering material.&lt;/p&gt;&#xA;" OwnerUserId="452" LastEditorUserId="470" LastEditDate="2012-07-29T03:27:16.253" LastActivityDate="2019-09-21T15:15:46.797" Title="Diamond filtering - is it real or bogus?" Tags="&lt;everyday-chemistry&gt;&lt;filtering&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="846" PostTypeId="2" ParentId="842" CreationDate="2012-07-24T09:47:24.987" Score="7" Body="&lt;p&gt;Under the conditions you measure a plain 1D proton spectrum, you usually never have any roblems with truncation of the FID. You have a relatively long relaxation delay, you don't gain anything by making your acquisition time much shorter than it can be. So a default setup might measure 64k or even 128k points, which is far, far larger than the values you use in 2D NMR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In two-dimensional spectra there are several aspects that severely limit the number of points you measure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Measuring more points in the indirect dimension takes time, if you want twice the points you have to measure twice as long. For that reason you only measure as much points in the indirect dimension as you have to, no more.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Processing can become a bit slow and take quite a bit of space when you increase the number of points in the direct dimension in a 2D spectrum. It's not a big problem with current hardware, but with a 3D (or higher) spectrum this gets much, much worse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you need to decouple during acquisition time, the acquisition time is severely limited by the probe hardware and heating effects of the decoupling.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So a typical 1D spectrum will have a lot more points measured, beyond the point where the FID has decayed below the noise. In such a case you won't get any truncation artifacts, as the FID was never truncated to begin with. You only need to worry about truncation artifacts if you actually truncate the FID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following image you can see&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lila: QSINE (SSB 4) 10k points in FID&lt;/li&gt;&#xA;&lt;li&gt;green: GM (LB -1, GB 0.5) 10k points in FID&lt;/li&gt;&#xA;&lt;li&gt;red:  QSINE (SSB 4) 2k points in FID&lt;/li&gt;&#xA;&lt;li&gt;blue: GM (LB -1, GB 0.5) 2k points in FID&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cDExz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can clearly see the truncation artifacts in the gaussian-processed spectra with 2k points, but already at 10k points they are not visible anymore. The quadratic cosine of the truncated FID doesn't look all that nice, but it suppresses the truncation artifacts much better than the gaussian window function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SSB 4 I used for this example is also a far more aggressive setting than I'd usually use. I chose it because it gives a relatively similar resolution as the gaussian I compared it to. Usually I'd use SSB 2 for most 2D spectra, which emphasizes signal/noise over resolution.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-07-24T10:17:01.737" LastActivityDate="2012-07-24T10:17:01.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="847" PostTypeId="1" CreationDate="2012-07-24T12:32:01.757" Score="12" ViewCount="215" Body="&lt;p&gt;I need to set a number of threaded bars into cement, to hold a new gatepost. When I cut the bar into shorter pieces, the metal near the cut became red hot, presumably hot enough to evaporate some or all of the zinc with which the entire rod was plated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to plate zinc back onto the ends, and have had mixed success so far. I dissolved some zinc oxide with hydrochloric acid and hooked it up to a little PV panel I have, and after a while it seems there is a light-coloured deposit on the cathode. The pencil-lead (6B) anode is no longer as smooth as it first was, but it doesn't seem to be losing (much) of its bulk. I assume it to be zinc, because... well what else can it be? It plated thick enough that I had to (ab)use a nut to re-cut the thread, leaving a light-coloured shiny surface that gives me more confidence that it is, in fact, zinc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: the edges of the threads seem to be taking more zinc than the valleys. In some cases the thread's valley doesn't show any evidence of plating. The fractal-like deposit on the edges is quite pretty actually, pity I can't take a (good) closeup photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything I can do (perhaps using some other common(ish) kitchen chemicals?) that will result in a more uniform layer of zinc? Is it worth trying lower / higher current or pulsed current? Am I overthinking things - is it even necessary for the whole stud to have a layer of zinc?&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorUserId="20847" LastEditDate="2016-02-15T04:23:41.417" LastActivityDate="2017-10-02T21:25:02.310" Title="How to plate zinc evenly on a threaded bar" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;&lt;electroplating&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="848" PostTypeId="2" ParentId="845" CreationDate="2012-07-24T14:19:07.443" Score="7" Body="&lt;p&gt;It appears that there is some validity to this technology.  Not surprisingly, the published journal literature available is basically non-existent, except for a handful of patent applications describing something along those lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The title of one journal publication &quot;Preparation of a filtering material using nanomaterials for removal of a heavy metals from wastewater&quot; uses a diamond nanopowder to aid in filtering out heavy metals from waste water.  The citation is &quot;Antonenko, L. P., Energotekhnologii i Resursosberezhenie, 6, 2011, 49-54&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there is a patent appplication from France (where at least one article I read cited as the technology originating from), title &quot;Apparatus for activation and mineralization of water for domestic use&quot;.  In this, they use a variety of powdered gemstones (diamond, ruby, sapphire) as well as precious metals (gold, silver) to filter the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, nano-particle scale materials are effective filtration agents, as most contaminants will not pass through such a small sized filtration device.  The use of some gemstones (certainly diamond) are beneficial for their relative chemical inertness, and certainly silver is well-known for it's antimicrobial activity, and synthesis of gold-nanoparticles or &lt;a href=&quot;http://dx.doi.org/10.1088/0957-4484/17/19/020&quot;&gt;gold-silver coatings&lt;/a&gt; are also actively under research&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2012-07-24T14:19:07.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="850" PostTypeId="2" ParentId="801" CreationDate="2012-07-24T14:55:03.483" Score="6" Body="&lt;p&gt;There is no difference between temperature and &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature#Thermodynamic_approach_to_temperature&quot; rel=&quot;noreferrer&quot;&gt;thermodynamic temperature&lt;/a&gt;. Wikipedia has a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature&quot; rel=&quot;noreferrer&quot;&gt;writeup&lt;/a&gt;. The book likely uses the adjective in order to differentiate from ill-defined conventional connotations. Temperature is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Statistical_ensemble_%28mathematical_physics%29&quot; rel=&quot;noreferrer&quot;&gt;ensemble&lt;/a&gt; property, e.g., an ideal gas has an &lt;em&gt;average&lt;/em&gt; energy of $\frac{1}{2}k_BT$. Outside of equilibrium, temperature is ill-defined.&lt;/p&gt;&#xA;" OwnerUserId="440" LastActivityDate="2012-07-24T14:55:03.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="851" PostTypeId="1" AcceptedAnswerId="852" CreationDate="2012-07-25T08:17:53.473" Score="8" ViewCount="4315" Body="&lt;p&gt;The most common drying agent I encountered when using a vacuum dessicator was phosphorus pentoxide in the form of Sicapent. But I've recently encountered people using KOH as a cheaper alternative (in a practical course for students, where such costs are amplified due to the high number of students).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the differences between KOH and phosphorus pentoxide as a general purpose drying agents to remove water? Is there a difference in effectiveness? Are there classes of substances where you can't use one of those drying agents?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2014-11-04T16:44:38.780" LastActivityDate="2018-03-20T08:50:13.647" Title="Differences between KOH and phosphorus pentoxide as drying agents?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="852" PostTypeId="2" ParentId="851" CreationDate="2012-07-25T11:39:44.193" Score="14" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.erowid.org/archive/rhodium/chemistry/equipment/dryingchemicals.html&quot;&gt;this page&lt;/a&gt;, which recreates information from &lt;em&gt;Purification of Laboratory Chemicals&lt;/em&gt;, by Perrin and Armarego. It provides a qualitative ranking of drying agents, listing phosphorous pentoxide higher than KOH. The two drying agents work by different means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;KOH&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anhydrous KOH is hygroscopic. It absorbs water from the atmosphere. It much better at doing so than more common hygroscopic drying agents like calcium chloride or calcium sulfate (drierite). A pellet of KOH left out in a beaker or dish open to the air will have absorbed so much water vapor within an hour that it will now be a solution of KOH (that will continue to be hygroscopic, at least until the solution is 50% water). Calcium chloride and calcium sulfate only continue to absorb water until they reach their maximum hydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorous pentoxide (really P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt;) is an anhydride of phosphoric acid. Phosphorous pentoxide reacts with water to form phosphoric acid: &#xA;$$\ce{P4O10 + 6H2O -&amp;gt; 4H3PO4}$$&#xA;Phosphoric acid is also hygroscopic and absorbs water to form a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt; is water reactive it is more difficult and dangerous to handle, especially for large groups of students. I might not use P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt; if the substance being dried is very acid sensitive. Eventually, the vacuum will start volatilizing the acid. I might not use KOH for compounds that are incredibly base sensitive, although it is unlikely that KOH will sublime. In an undergraduate lab with multiple students, KOH should be fine if you want to save money and hassle.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="22" LastEditDate="2012-07-25T13:46:00.310" LastActivityDate="2012-07-25T13:46:00.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="853" PostTypeId="1" AcceptedAnswerId="6713" CreationDate="2012-07-25T16:01:07.827" Score="8" ViewCount="1987" Body="&lt;p&gt;In certain applications some sorts of stainless steel &lt;code&gt;(V2A, 1.4301)&lt;/code&gt; corrode when exposed to $\small{\ce{Cl^-}}$ in water, or at least it is strongly advised to use other steels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions that were explained to me as being especially bad were: redox potential &gt;0, acidic environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these conditions, why do I need $\small{\ce{Cl^-}}$ to corrode steel? I would think that any electrolyte would have a bad effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to add 26.7:&#xA;Most stainless steel contains chrome, a layer of $\ce{CrO}$ is what's protecting the steel. Most articles mention that $\ce{Cl^-}$ damages this layer. More resistant steels use $\ce{Ni}$ (and are more expensive because of that). So, to narrow down the question, how does $\ce{Cl^-}$ attack the passive layer?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="1085" LastEditDate="2013-04-19T09:47:07.633" LastActivityDate="2013-10-30T23:40:33.647" Title="Why does Chloride attack CrO?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="854" PostTypeId="1" CreationDate="2012-07-26T07:56:16.660" Score="8" ViewCount="3393" Body="&lt;p&gt;Has anyone come across a scientific study of the S-Plasma Ions that Samsung claims will purify air from microbes, allergens and free radicals? I would prefer publicly available research findings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a quote from &lt;a href=&quot;http://www.samsungspi.com&quot; rel=&quot;nofollow&quot;&gt;http://www.samsungspi.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Samsung SPi generates Hydrogen atom ($\ce{H+}$) and Oxygen ion ($\ce{O_2-}$) which reacts together to form Hydroperoxyl radicals which the positively charged biological contaminants then reacts with three hydrogen elements of its protein structure before forming water vapor. Also, the Hydrogen atom ($\ce{H}$) will combine with $\ce{OH}$-radical to form a water vapor and neutralize harmful $\ce{OH}$-radical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It also matches the description found on the product itself. See &lt;a href=&quot;http://winniekepala.com/2011/03/ionizer-review-samsung-virus-doctor/&quot; rel=&quot;nofollow&quot;&gt;http://winniekepala.com/2011/03/ionizer-review-samsung-virus-doctor/&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a quote from the video on the same site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{H+}$ is born with the help of that plasma in this plus electrode. $\ce{H+}$ is hydrogen with a positive electric charge. Through the air, the $\ce{H+}$ moves to the minus electrode which generates electrons. In here, electron meets $\ce{H+}$ and active hydrogen like me are born.... The electron that is generated in minus electrode in the SPi ion device where [I] am born is naturally combined with oxygen. That is oxygen ion ($\ce{O2-}$). (Active hydrogen and oxygen ion together transform into $\ce{(H-O-O)-}$). $\ce{HOO}$($\ce{HOO-}$) ion envelops the positively charged airborne bio-contaminants and react with three hydrogen constituted of protein components, forming water vapor. This modifies the surface protein or breaks the cell membrane to remove bio-contaminants. $\ce{HOO}$($\ce{HOO-}$) ion reacts only with single cells such as virus or germ but does not react with the human body... $\ce{OH}$ radical is an active oxygen species and one of the highly reactive and strong oxidants. It can be harmful to human's body if exposed for a long time. (Active hydrogen combines with OH radical to form water vapor).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(It has numerous grammatical mistakes which I just copied verbatim).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what are the various compounds and radicals mentioned in the explanation, and how the reactions above take place. What is the plasma in the positive electrode, and what is the potential difference between the electrodes that allows the generation of the active hydrogen and oxygen ions? What are the real chemical states of the active hydrogen and the oxygen ions? How do they combine to form the $\ce{HOO-}$ ion? What is the $\ce{HOO-}$ ion and how is it stable enough to go around the room? How does the active hydrogen combine with the $\ce{OH}$ radical?&lt;/p&gt;&#xA;" OwnerUserId="458" LastEditorUserId="458" LastEditDate="2015-11-22T09:56:47.080" LastActivityDate="2020-02-04T18:54:55.413" Title="Samsung's S-Plasma Ion - scientific basis" Tags="&lt;ions&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="855" PostTypeId="1" AcceptedAnswerId="862" CreationDate="2012-07-26T09:12:48.123" Score="11" ViewCount="29525" Body="&lt;p&gt;What is the difference in purity betwen lab-grade and food-grade when talking about potentially consumable compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example let's take ascorbic acid powder: if it's marked as lab-grade, &lt;strong&gt;does that mean it is also fit for consumption?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm wondering whether lab-grade supersedes (and/or improves) a food-grade labeling, or if they are two totally different criteria? Can I say lab-grade &gt; food-grade? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can't say lab-grade &gt; food-grade, then what is the actual difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="319" LastEditorUserId="11367" LastEditDate="2017-01-14T14:54:23.120" LastActivityDate="2019-03-03T17:01:54.077" Title="Difference between lab-grade and food-grade purity?" Tags="&lt;terminology&gt;&lt;purification&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="856" PostTypeId="2" ParentId="855" CreationDate="2012-07-26T11:14:15.370" Score="9" Body="&lt;p&gt;Generally laboratory grade reagents have not been prepared with human consumption as a consideration, but with regard to their use as a chemical reagent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $95\%$ ethanol is mostly ethanol and $5\%$ water. This is fine for cleaning equipment, or using as a solvent for a TLC. If you want to use ethanol as a reagent, however, you want to use absolute ethanol (which is much more expensive). Absolute ethanol is NOT safe for human consumption though, as the water has been removed from the $95\%$ (which is the azeotrope, i.e. as far as you can get through distillation of water/ethanol alone) through a process which involves adding additives such as benzene. Benzene won't cause side reactions in the way that water would, and so absolute ethanol is much better for a lab reagent than $95\%$, but worse for consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, any chemical which is prepared for use as a reagent will contain impurities which don't generally cause side-reactions, while for consumption you want the impurities to be non-toxic.&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="11367" LastEditDate="2017-01-14T14:58:54.707" LastActivityDate="2017-01-14T14:58:54.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="857" PostTypeId="1" CreationDate="2012-07-26T14:38:16.260" Score="8" ViewCount="500" Body="&lt;p&gt;I will be using trimethyl tin chloride (1M in THF) in a few days for the first time. Looking at the MSDS, it looks like it is very toxic and has noxious vapors. Are there any more standard precautions for working with it other than the usual?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Check for proper airflow in the hood&lt;/li&gt;&#xA;&lt;li&gt;Wear lab coat, safety glasses, and double gloves&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="345" LastEditorUserId="73" LastEditDate="2012-07-26T22:11:06.757" LastActivityDate="2014-06-17T15:52:35.677" Title="How can I be safe and protect myself using trimethyl tin chloride?" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="858" PostTypeId="1" AcceptedAnswerId="4231" CreationDate="2012-07-26T16:18:15.760" Score="6" ViewCount="377" Body="&lt;p&gt;I have a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_table_file&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical Table files&lt;/a&gt; which include a &lt;a href=&quot;http://pastebin.com/cPg2p7GZ&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;M  ZZE&quot; tag in the properties block&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=85296601&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galactobuxin&lt;/a&gt;, I have:&lt;br&gt;&#xA;&lt;code&gt;M  ZZE  1  31  23&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=85292032&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubrobrassicin&lt;/a&gt; there is:&lt;br&gt;&#xA;&lt;code&gt;M  ZZE  3  28  23  36  23  48  23&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=4266213&quot; rel=&quot;nofollow noreferrer&quot;&gt;R-hexanoyl CoA&lt;/a&gt; there is:&lt;br&gt;&#xA;&lt;code&gt;M  ZZE  1  55  23&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical format for such tags is that it starts with an M, followed by two spaces. Then there is a three-letter code, in this case ZZE. Other codes include ISO (for isotope) or CHG (for charged atoms in the compound), then a number which indicates how many records on that line. This is followed by the actual data.&#xA;So&lt;br&gt;&#xA;&lt;code&gt;M  ZZE  3  28  23  36  23  48  23&lt;/code&gt;&lt;br&gt;&#xA;would probably mean that there are three sets of ZZE data, being &quot;28 23&quot; &quot;36 23&quot; and &quot;48 23&quot;. In every record I have the second number is 23. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ZZE tags seem to occur only for compounds containing ribose rings, and the first number is the index of one of the bonds inside the ribose ring. Specifically, of three three bonds adjacent bonds with a stereochemistry marker, it always refers to the one in the middle: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QwiAR.png&quot; alt=&quot;Rubrocrassin with ZZEs circled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the 23 is some property of the bond: so in rubrobrassicin the 28th, 36th and 48th bonds (there's one, and only one, ZZE record for each of the ribose rings in rubrobrassicin). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what the ZZE property in a chemical table file refers to, or what it indicates? I can't find any information on the format.&lt;/p&gt;&#xA;" OwnerUserId="459" LastEditorUserId="459" LastEditDate="2012-07-27T10:25:55.647" LastActivityDate="2013-03-26T03:55:22.820" Title="What does the ZZE tag in an Chemical Table file refer to?" Tags="&lt;organic-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="859" PostTypeId="1" AcceptedAnswerId="861" CreationDate="2012-07-26T22:07:43.263" Score="15" ViewCount="2750" Body="&lt;p&gt;I happened to stumble upon &lt;a href=&quot;http://www.nature.com/nchem/journal/v4/n5/full/nchem.1313.html&quot;&gt;this Nature Chemistry article&lt;/a&gt; today. It claims that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] use a 3D printer to initiate chemical reactions by printing the reagents directly into a 3D reactionware matrix, and so put reactionware design, construction and operation under digital control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I cannot obtain the full article to read, but even the abstract does not seem plausible in my opinion. My questions are...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;How can one &quot;print reagents&quot;? What is meant by this?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Most synthetic reactions require hours of refluxing and whatnot, not to mention isolation/purification afterwards. &lt;strong&gt;How can all of these techniques be performed with a printer?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Perhaps I'm not &quot;getting it&quot;, but the abstract of the article isn't providing me with the explanations and theory I'm looking for. &lt;em&gt;How can something like this be possible?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Symes, Mark D., Philip J. Kitson, Jun Yan, Craig J. Richmond, Geoffrey J. T. Cooper, Richard W. Bowman, Turlif Vilbrandt, and Leroy Cronin.&lt;/strong&gt; &lt;em&gt;“Integrated 3D-printed Reactionware for Chemical Synthesis and Analysis.”&lt;/em&gt; Nat Chem 4, no. 5 (May 2012): 349–354.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="43" LastActivityDate="2017-12-12T21:43:07.500" Title="Chemical synthesis via 3D printing?" Tags="&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="860" PostTypeId="1" CreationDate="2012-07-27T05:41:32.857" Score="10" ViewCount="1302" Body="&lt;p&gt;When a piece of cast iron is repeatedly covered in fat and exposed to heat it develops a hard, black, non-stick coating referred to as &quot;seasoning&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is this coating made of molecularly? How is it bonded to the cast iron substrate? What is the chemical process that transforms a liquid fat into this coating? And what if any characteristics of the fatty acids or triglycerides used have an effect on the final coating or how well the it bonds to the iron?&lt;/p&gt;&#xA;" OwnerUserId="461" LastActivityDate="2020-12-31T14:52:12.323" Title="What is the molecular structure of the coating on cast iron cookware known as seasoning?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="861" PostTypeId="2" ParentId="859" CreationDate="2012-07-27T07:42:28.753" Score="14" Body="&lt;p&gt;The authors use a &lt;strong&gt;3D printer&lt;/strong&gt; to generate the labware, and to place the reagents in its inlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we're not printing standard ink on paper but using a the &lt;em&gt;print head&lt;/em&gt; to exactly place very amounts of monomers at certain positions. These monomers cure (polymerize) upon heat or irradiation to a solid resin. By this, small reactors of arbitrary design (think in 3-neck reactors) with a volume of ~ 10 mL may be printed layer by layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, it is even possible to create reactors with &lt;em&gt;windows&lt;/em&gt; to allow for later reaction monitoring or print and assemble electrodes for spectroelectrochemistry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can one &quot;print reagents&quot;? What is meant by this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is all about precise positioning of the printhead! Thus same printhead used for the ink can be used to fill the inlets (think in dropping funnels of normal labware) of the reactor with componends whose reaction you want to monitor. Instead of dropping, material transfer is achieved by applying a vacuum to the reactor through a self-healing hole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most synthetic reactions require hours of refluxing and whatnot [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, a lot of reactions do proceed quite well at room temperature, think in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diels-Alder reactions of furanes and maleimides&lt;/li&gt;&#xA;&lt;li&gt;electrochemically triggered reactions&lt;/li&gt;&#xA;&lt;li&gt;photoinduced electron transfer reactions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;just to mention a few.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] not to mention isolation/purification afterwards. How can all of these techniques be performed with a printer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh, you got that one slightly wrong. The &quot;printer&quot; is just used to build the reaction sphere (the flask, if you like it) and to place the reagents in the right inlet of your vessel. The rest is &quot;just chemistry&quot; ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope i made myself clear enough - at least after the n-th edit ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the concept is pretty cool.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2012-07-27T13:38:31.967" LastActivityDate="2012-07-27T13:38:31.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="862" PostTypeId="2" ParentId="855" CreationDate="2012-07-27T08:26:07.933" Score="12" Body="&lt;p&gt;So-called food-grade stuff is meant to be consumed by humans. It's important that every single compound of a foodstuff is food-grade or at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Generally_recognized_as_safe&quot; rel=&quot;noreferrer&quot;&gt;GRAS&lt;/a&gt;. In the European Union the law giver has defined the state when a compound is good for consumption as &quot;save&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lab-grade chemicals often contain toxic materials, so when something is called lab-grade it is not save per definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: $\ce{NaOH}$ is widely used in preparation of foodstuff as well as in the lab. Now it's a fact that lab-grade (standard) $\ce{NaOH}$ (&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/s8045?lang=en&amp;amp;region=US&quot; rel=&quot;noreferrer&quot;&gt;Sigma Aldrich&lt;/a&gt;) contains a certain amount of heavy metals. On the other hand, food-grade $\ce{NaOH}$ will not contain any heavy metals (usually food-grade $\ce{NaOH}$ will have a purity of more than $99\%$ while standard $\ce{NaOH}$ pellets are sold with purities less than $97\%$). So using lab-grade $\ce{NaOH}$ in foodstuff is a crime (at least in the EU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my advice is: Do not use lab-grade materials for any food-related applications.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="11367" LastEditDate="2017-01-14T14:57:23.420" LastActivityDate="2017-01-14T14:57:23.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="863" PostTypeId="1" CreationDate="2012-07-27T18:44:07.340" Score="3" ViewCount="193" Body="&lt;p&gt;I'm just curious about this topic, as most portable, &quot;clean&quot; energy sources or carriers (in the sense of not being a primary energy source -fuel cell hydrogen, for example) are in essence electrochemical reactors, and most of the current work in the field is aimed at reducing irreversible losses at the electrodes, be it because of discharge overpotential, or polarization because of mass transfer along the membrane electrolyte; and of course, in making this structural elements cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a lot of proposed and existing waste treatment and recycling processes aimed at recovering rare and/or active metals (such as lithium from Li-ion batteries) involve at least one electrochemical unit op.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if nanotechnological advances, such as increasing the surface area of electrode materials or increasing the effecive diffusivity of relevant species in solid porous electrolytes, can reduce this irreversibilities, perhaps making use of cheaper raw materials than noble metals (i.e. platinum). I would also appreciate any references to current work in this applied field.&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="441" LastEditDate="2012-07-27T18:51:31.213" LastActivityDate="2017-01-13T00:38:09.287" Title="Can nanotechnology reduce electrode losses in electrochemical cells?" Tags="&lt;electrochemistry&gt;&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="864" PostTypeId="1" AcceptedAnswerId="865" CreationDate="2012-07-27T21:23:50.767" Score="9" ViewCount="9598" Body="&lt;p&gt;I did a short experiment in which I connected a $1.5\:\mathrm{V}$ battery with 2 paper clips into a bowl of water to which I added salt. After a while, I saw a yellowish-greenish cloud inside the water around one of the paper clips, and when I took it out, the part that was in the water was more black than the rest of the paper clip. After a while, some of it turned into a greenish color.&#xA;What caused the paper clip to change its color?&lt;/p&gt;&#xA;" OwnerUserId="466" LastEditorUserId="15410" LastEditDate="2015-05-20T01:38:06.977" LastActivityDate="2021-03-11T18:52:30.187" Title="Identity of a green substance formed during electrolysis reaction?" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;&lt;water&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="865" PostTypeId="2" ParentId="864" CreationDate="2012-07-27T22:30:06.077" Score="5" Body="&lt;p&gt;Without further details of your equipment, this might be a possible answer. I assume the electrolyte you added was table salt (NaCl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you saw as a yellowish cloud was gaseous chlorine which discharged on the clip you connected to the positive electrode (cathode) of your battery. The reason why you did see this, instead of seeing just colorless bubbles of oxygen, is because of something called &quot;overpotential&quot;. It's a joint property of the species in your cell and the material and surface termination of the electrode; in your case, the clip (made of wire steel) has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Overpotential&quot;&gt;lower overpotential&lt;/a&gt; for the discharge of chloride anions (from the salt you added) than oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elemental chlorine is a very reactive oxidant, and you had a thin chlorine film around your clip electrode. The reaction between chlorine and iron in aqueous solution most probably yielded &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_chloride&quot;&gt;iron (II) chloride tetrahydrate&lt;/a&gt; which is the reason you saw the greenish colour of the previously submerged fraction of the clip.&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="275" LastEditDate="2012-07-28T00:24:19.527" LastActivityDate="2012-07-28T00:24:19.527" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="869" PostTypeId="2" ParentId="827" CreationDate="2012-07-28T08:34:54.010" Score="7" Body="&lt;p&gt;First of all, one should have look a look on the ingredients of your toast. It mainly consist of starch, proteins (2% maybe), fat (a few %), water and some inorganic compounds such as NaCl. &#xA;Well Salt normally won´t interact with anything below 1400°C so no need to look at this. &#xA;Water just evaporates, i think we must not discuss that here. &#xA;Fat: that's more interesting because fat is known to degenerate into different compounds when heated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the definitely most interesting reactions will be present when one heats carbon hydrates together with protein: so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Maillard_reaction&quot;&gt;Maillard Reaction(s)&lt;/a&gt; those kind of reactions are still issues of ongoing research although this type of reactions have been discovered long ago. As a good starter on the various compound which can occur during Maillard Reaction you might read &lt;em&gt;Mottram&lt;/em&gt; or &lt;em&gt;Monnier&lt;/em&gt;. Especially &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrylamide&quot;&gt;acrylamide&lt;/a&gt; could be of interest when looking a toxicity. I don't have a certain paper in mind but i know that there has been research about toxicity of acrylamide in fish and other animal. So it's likely that there was some research about plants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at starch and protein separately you will encounter heat degeneration as well as &lt;a href=&quot;http://en.wikipedia.org/wiki/Caramelisation&quot;&gt;caramelisation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="462" LastEditorUserId="73" LastEditDate="2012-07-28T09:32:27.873" LastActivityDate="2012-07-28T09:32:27.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="871" PostTypeId="2" ParentId="657" CreationDate="2012-07-28T15:24:44.270" Score="3" Body="&lt;p&gt;I never heard about the &quot;10% rule&quot; and I never applied it. Transferring TLC (silica on aluminium plates, Merck) results to preparative column chromatography on similar silica usually worked without changes, although the situation is a bit different in the column: the upper part of the TLC plate is rather dry when letting it run, while the silica in the column is not. Note that I was never a friend of the dry packing method and always prefered the slurry method to prepare the columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I never thought about increasing the polarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad separation typically was a result of column overload, bad packing or cracking of the stationary phase, e.g., when the diethylether evaporates in the column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1 For citing the classic article on &lt;strong&gt;flash chromatography&lt;/strong&gt;! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can really recommend the technique and I used it ... more than once ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="73" LastEditDate="2012-07-29T03:28:15.940" LastActivityDate="2012-07-29T03:28:15.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="872" PostTypeId="1" CreationDate="2012-07-28T19:24:53.907" Score="12" ViewCount="13269" Body="&lt;p&gt;I plan on etching some PCBs with hydrochloric acid and hydrogen peroxide, which will therefore produce some kind of copper chloride (green color) which is highly toxic if released into the environnement (it may be useful as an herbicide, but I don't have any garden right now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While cupric acid (obtained from hydrochloric acid, hydrogen peroxide and some PCBs etching) may be bubbled with air/oxygen to re-activate it to etch again other PCBs I'm not sure this may what I want and I seek suggestions in case I need to dispose of this solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I heard (though none were actually on chemistry sites, except the 1st one):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Acid/Base neutralization: pour some baking soda to take care of the chloride&lt;/li&gt;&#xA;&lt;li&gt;Pour some alluminium balls to get back some of the copper (didn't really understand how this works)&lt;/li&gt;&#xA;&lt;li&gt;Copper electrolysis: basically get solid copper at the cathode, creating gaz at the anode and leaving (?) some residues that can be neutralized with baking soda (?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any insights on this? I think once the copper and chloride are taken care of, it shouldn't pose a huge problem to pour it down the drain or leave it outside to evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which metal should I choose for the anode / cathode ? In the first source it speaks about reactivity (and since copper is less reactive than hydrogen, copper is produced at the cathode). No idea about the metal I should use though.&lt;/li&gt;&#xA;&lt;li&gt;Would adding alluminium balls into the solution help make it more &quot;eco-friendly&quot; ?&lt;/li&gt;&#xA;&lt;li&gt;Any other suggestion is welcome&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mindset.co.za/resources//0000022572/0000029661/0000029601/default.htm&quot;&gt;Electrolysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hackaday.com/2011/03/03/simple-pcb-etchant-made-from-chemicals-you-can-put-in-your-mouth/&quot;&gt;How to etch PCBS - see comments&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20081009124104AAUjtpG&quot;&gt;Recovery of copper using electrolysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="352" LastActivityDate="2016-12-21T03:03:42.427" Title="Copper Chloride: neutralizing and disposal" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;purification&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="873" PostTypeId="2" ParentId="872" CreationDate="2012-07-28T20:17:53.623" Score="5" Body="&lt;p&gt;Regarding #2: Adding $\ce{Al}$ or $\ce{Mg}$ metal to $\ce{CuCl_2}$ solution reduces the $\ce{Cu_2^+}$ to $\ce{Cu}$, and you then have $\ce{Cu}$ metal (filter out and recover) and $\ce{AlCl_3}$ or $\ce{MgCl_2}$, respectively, in solution. Both solutions are safer in small quantities to flush down the drain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding #1: $\ce{2NaHCO3 + CuCl2 -&amp;gt; CuCO3 + 2NaCl + CO2 + H2O}$. $\ce{CuCO3}$ is insoluble, but should still be disposed of at an appropriate facility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, you get an insoluble copper compound (or metal) rather than a soluble one.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-07-25T02:03:00.187" LastActivityDate="2013-07-25T02:03:00.187" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="875" PostTypeId="2" ParentId="425" CreationDate="2012-07-29T22:47:10.700" Score="2" Body="&lt;p&gt;Tertiary amines are widely used as electron donors in photoinduced electron transfer (PET) reactions. Electron transfer to an electronically excited acceptor yields the radical anion of the acceptor and the radical cation of the amine. When the stereoelectronic requirements are met, the radical cation deprotonates to give an &amp;alpha;-aminoalkyl radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the deprotonation is observed for acyclic and thus flexible trialkylamines, but not for DABCO ([2.2.2]-diazabicyclooctane, lacking a sufficient orbital overlap between an &amp;alpha;-C-H bond and the (half-filled) orbital on the nitrogen. I'd say that the situation in sparteine is pretty similar to that in DABCO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With molecular oxygen, the formation of an &amp;alpha;-aminoalkyl radical as the first intermediate is nevertheless conceivable. The mechanism, however, is different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the triplet state character of ground state oxygen into account, i'd say the most likely step is the abstraction of an axial hydrogen atom at one of the &amp;alpha;-bridgehead carbon atoms next to a nitrogen atom.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-07-29T22:47:10.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="876" PostTypeId="1" CreationDate="2012-07-30T09:02:11.460" Score="6" ViewCount="162" Body="&lt;p&gt;Is radiation time the same thing as half life period? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if I'm talking about excited species in a gas, how would the gas interaction affect the radiation time, are there some formulas/approaches?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="4945" LastEditDate="2014-05-22T09:05:57.377" LastActivityDate="2014-05-22T09:05:57.377" Title="Radiation time vs. half life period" Tags="&lt;nuclear&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="877" PostTypeId="2" ParentId="876" CreationDate="2012-07-30T11:04:08.590" Score="4" Body="&lt;p&gt;In the context of photochemistry, &lt;em&gt;radiation time&lt;/em&gt; or &lt;em&gt;half life period&lt;/em&gt; are not used, as far as I know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only relevant technical term i'm aware of -in that context- is &lt;em&gt;half-life (time)&lt;/em&gt;, which describes the time needed for the concentration of an excited species to decrease to 50% of the initial value, &lt;em&gt;cf.&lt;/em&gt; S. E. Braslavsky, Glossary of terms used in photochemistry,&#xA;&lt;em&gt;Pure Appl. Chem.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;79&lt;/em&gt;, 293-465&#xA;[&lt;a href=&quot;http://dx.doi.org/10.1351/pac200779030293&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For longer-lived excited triplet states in the gas phase, radiationless deactivation upon collision with a ground-state molecules will play a role. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the lifetime &lt;em&gt;might&lt;/em&gt; be correlated with the &lt;em&gt;mean free path&lt;/em&gt; of the molecules.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="30" LastEditDate="2012-07-31T20:58:00.137" LastActivityDate="2012-07-31T20:58:00.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="878" PostTypeId="2" ParentId="872" CreationDate="2012-07-30T11:53:18.620" Score="3" Body="&lt;p&gt;Depending on where you are, there may be places where you can hand over hazardous waste for proper treatment. E.g. in Germany the communal waste collection centres take properly labelled hazardous waste. For private stuff this is paid for by the general waste fees, so no extra fees are charged. If you do this on a commercial basis (= large scale), you'll have to pay for it, of course (but in this case you shouldn't have to ask how to deal with this in a forum...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I'd recommend that you let (most of) the water in your solution evaporate so you get either a concentrated solution or the salt (chloride or carbonate) in order to reduce volume and mass of the hazardous waste.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-07-30T11:53:18.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="879" PostTypeId="1" CreationDate="2012-07-30T12:59:57.387" Score="5" ViewCount="83" Body="&lt;p&gt;The motivation for the question in the title is that I have some formula given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta T=\frac{3410\ \gamma_{O_2}}{2.5+\gamma_{O_2}},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma_{O_2}$ is the molar fraction of oxygen in a binary micture and the ratio of oxygen and rate gas heat capacities are supposed to be 7.5, if that helps. And there is another one with a factor $\frac{1-\eta}{\eta}$ multiplied, where $\eta$ is a singlet delta oxigen efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this formula come from?&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2012-07-30T12:59:57.387" Title="How can I compute the temperature change from the de-excitation of excited particle species?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="880" PostTypeId="2" ParentId="845" CreationDate="2012-07-30T19:39:15.470" Score="5" Body="&lt;p&gt;Altered properties (band gap, Stokes shift) of cristalline nanoparticles as compared to bulk have been demonstated for various semiconductors, namely CdS, CdSe, TiO&lt;sub&gt;2&lt;/sub&gt;, etc. They result from a significantly different surface-to-volume ratio and the fact that atoms at the crystal surfaces simply lack the proper number of neighbours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the crystal ends, the is just no other neighbour. Since there's &lt;em&gt;less than normal&lt;/em&gt; neighbours, centres at the surface are more reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the properties of the &lt;em&gt;diamond dust&lt;/em&gt; are concerned, i do however remain skeptic and to me, the whole thing mostly sounds like &lt;strong&gt;marketing lingo&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &lt;em&gt;diamond dust&lt;/em&gt; suggests something valuable: if the manufacturer uses something as precious as diamonds for filtering, the product &lt;em&gt;must&lt;/em&gt; be valuable too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would one actually want to use something with a rather inert diamond lattice of sp&lt;sup&gt;3&lt;/sup&gt;-carbon atoms instead of microporous activated carbon with well-known adsorption properties? &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2012-07-31T21:40:54.190" LastActivityDate="2012-07-31T21:40:54.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="881" PostTypeId="2" ParentId="816" CreationDate="2012-07-31T01:18:54.443" Score="5" Body="&lt;p&gt;What I think you are asking about is the number of thermodynamic parameters which characterize a chemical reaction. Those parameters are used to make &lt;strong&gt;quantitative predictions&lt;/strong&gt; of the final state of the reactive system (at least in the context I'm familiar with, i.e. chemical engineering); for example, the amount of heat exchanged with the reactor's environment, the composition of the mixture at equilibrium, it's material phase, overall density, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical thermodynamics, the way to address this problem is through the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot;&gt;potentials&lt;/a&gt; defined on the system of interest, and studying equilibrium as the extremum of one these. The chosen potential depends on the particular configuration of the system under study; and in all cases their relation with equilibrium is a consequence of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot;&gt;Second Law&lt;/a&gt;. For example, if the system is restricted to evolve at constant volume and entropy, the minimized potential is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Internal_energy&quot;&gt;internal energy&lt;/a&gt;; if it evolves at constant energy and volume (an isolated simple compressible system) it maximizes &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot;&gt;entropy&lt;/a&gt;; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical point of view of natural systems and industrial processes -which evolve at nearly constant temperature and pressure fixed by the environment- the involved minimized potential is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot;&gt;Gibbs free energy&lt;/a&gt;. That the minimization of $G$ yields an equilibrium state for a closed system at constant pressure and temperature is straightforward (it can be found in any physical chemistry textbook). Moreover, when we say here that the system at it's minum $G$ is at equilibrium, we are saying it's at &lt;em&gt;overall equilibrium&lt;/em&gt;: it's simultaneously at mechanical equilibrium (because of &lt;strong&gt;null &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_Law_for_Fluids&quot;&gt;pressure gradients&lt;/a&gt;&lt;/strong&gt;), thermal equilibrium (because of &lt;strong&gt;null &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier%27s_law#Fourier.27s_law&quot;&gt;temperature gradients&lt;/a&gt;&lt;/strong&gt;) and in chemical equilibrium (because of &lt;strong&gt;null &lt;a href=&quot;http://en.wikipedia.org/wiki/Ficks_law&quot;&gt;chemical potential gradients&lt;/a&gt;&lt;/strong&gt;); so there's no &quot;force&quot; (in a generalized sense) to drive any further process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to your question, which are the independent variables of a reactive system as such? Well, suppose that in your system you have $P$ chemical species which can be both reactives or products of $Q$ reactions taking place in a monophase. For every species, the balance on the number of moles of each one yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{i}=n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\xi_{j}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extent_of_reaction&quot;&gt;extent of the j-th reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we can define the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_potential&quot;&gt;chemical potential&lt;/a&gt; as the partial molar Gibbs free energy per species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{i}=\displaystyle\frac{\partial G}{\partial n_{i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, given that $G$ is an extensive property (or better phrased, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogeneous_function&quot;&gt;homogeneous function&lt;/a&gt; of first degree in mass) by Euler's Theorem it &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_molar_property&quot;&gt;follows&lt;/a&gt; that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G=\sum_{i=1}^P\mu_{i}n_{i}=\sum_{i=1}^P\mu_{i}\cdot (n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every chemical potential can be expressed as the sum of it's specific Gibbs free energy of it's &lt;strong&gt;pure form&lt;/strong&gt;, at the &lt;strong&gt;temperature&lt;/strong&gt; and &lt;strong&gt;pressure&lt;/strong&gt; of the mixture, plus a term which ponderates the influence of &lt;strong&gt;concentration&lt;/strong&gt; and &lt;strong&gt;chemical affinity&lt;/strong&gt; of each particular species towards the rest of the present compounds which constitute the mixture (for example, the &lt;em&gt;solvent matrix&lt;/em&gt;, if one species is so concentrated that it can be considered so). The latter term contains an adimensional parameter known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_%28chemistry%29&quot;&gt;&lt;em&gt;activity&lt;/em&gt;&lt;/a&gt; of the species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{i}=\mu_{0i}(P,T)+RTln(a_{i})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero index in the first term just means that it's referred to a pure system. This specific free energies can be related to the formation enthalpies and entropies which are tabulated through:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{0i}=h_{0i}(P,T)-T\cdot s_{0i}(P,T)\\\mu_{0i}=\left\{{h_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}dh_{0i}}\right\}-T\cdot \left\{{s_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}ds_{0i}}\right\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $h_{f}^0(P_{0},T_{0})$ and $s_{f}^0(P_{0},T_{0})$ are the standard formation enthalpy and entropy, respectively. The integrals evaluate the variation in enthalpy and entropy of the pure compounds due to a difference in temperature and pressure in reference to &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot;&gt;standard values&lt;/a&gt; (25 ºC and 1 bar). The evaluation of the integrals in terms of measurable properties must be done using &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell_relations&quot;&gt;Maxwell relations&lt;/a&gt; and definitions of potentials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dh_{0i}=\displaystyle\frac{\partial h_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial h_{0i}}{\partial {P}}dP=C_{p}dT+\frac{dP}{\rho}\\ds_{0i}=\displaystyle\frac{\partial s_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial s_{0i}}{\partial {P}}dP=Cp\frac{dT}{T}-\frac{\alpha}{\rho}dP$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the activities? These are non-linear functions of the composition, &lt;a href=&quot;http://es.scribd.com/doc/72373061/64/The-activity-coefficient-as-a-function-of-temperature-and-pressure&quot;&gt;temperature and pressure&lt;/a&gt; of the mixture, and usually that non-linearity is expressed through an &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_coefficient&quot;&gt;activity coefficient&lt;/a&gt; (it may be the case that further coefficients are introduced theoretically, but all they do is to &quot;partition&quot; the contributions to non-ideality into a product of these, and so here I talk about a unique coefficient without loss of generality). So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{i}=x_i\cdot\gamma_{i}(P,T,x_1,x_2,...,x_P)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i=\displaystyle\frac{n_i}{\sum_{k=1}^P n_k}=\displaystyle\frac{n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}}{\sum_{k=1}^P\left\{{n_{0k}+\sum_{j=1}^Q{\nu_{k}^j\cdot \xi_{j}}}\right\} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the (enormously complicated) expression for G turns out to be dependent of the variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G=G(P,T,h_{fi}^0,s_{fi}^0,Cp_{i},\rho_{i},\alpha_{i},n_{0i},\xi_j) \ \ \ \ \ \ i: 1 &amp;lt; i &amp;lt; P \ \ \textrm{and}\ \ \ j: 1 &amp;lt; j &amp;lt; Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are, a priori, the &quot;independent quantities&quot; involved in the overall reaction's equilibrium. Once you've fixed the temperature, pressure, the original amount of substances and get the intensive properties stated for all the chemical species present in the system along it's evolution, G becomes only a function of the extents of the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G=G(\xi_j) \ \ \ \ \ \  j: 1 &amp;lt; j &amp;lt; Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the equilibrium is reached when $G$ is at minimum; that's a problem solvable through &lt;a href=&quot;http://en.wikipedia.org/wiki/Hessian_matrix&quot;&gt;optimization&lt;/a&gt;. The values of the extent of every reaction at equilibrium is enough to calculate every other useful parameter of the system: it's composition, the heat involved, the change in overall volume, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;: the conserved quantities in a chemical reaction are physical constraints (conservation of total mass, balance of moles according to stecheometry, conservation of energy exchanged with the environment, strict increase in entropy, etc) but that's not enough information to uniquely evaluate the equilibrium parameters of a general reaction (or reactions). Most of the expressions I left in an undefined closed form in this answer (such as the activity coefficients, all of them functions of compositions) must be experimentally correlated, or modelled with some &lt;a href=&quot;http://en.wikipedia.org/wiki/UNIFAC&quot;&gt;theoretical formalisms&lt;/a&gt;, generally highly complex. The standard thermodynamic potentials might be tabulated though, as it's the case with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_databases_for_pure_substances&quot;&gt;intensive properties of pure substances&lt;/a&gt; (specific heats, thermal expansion coefficients, etc).&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="441" LastEditDate="2012-07-31T12:04:08.673" LastActivityDate="2012-07-31T12:04:08.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="882" PostTypeId="1" CreationDate="2012-07-31T17:56:21.537" Score="8" ViewCount="1696" Body="&lt;p&gt;I am an artist attempting to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Tollens%27_reagent&quot;&gt;Tollen's Reagent&lt;/a&gt; in a number of my projects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason I could not substitute Copper Nitrate for the Silver Nitrate in the reaction? Would the copper adhere to the glass? &lt;/p&gt;&#xA;" OwnerUserId="476" LastEditorUserId="73" LastEditDate="2012-08-01T07:28:07.910" LastActivityDate="2012-08-02T10:31:28.463" Title="Copper substituting silver in Tollen's reagent" Tags="&lt;inorganic-chemistry&gt;&lt;silver&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="883" PostTypeId="1" AcceptedAnswerId="898" CreationDate="2012-08-01T00:07:22.267" Score="8" ViewCount="262" Body="&lt;p&gt;Unfortunately, I don't have batteries to waste to test this myself, but I am curious... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have two identical batteries. AA Duracell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say you run one of the batteries through an LED until the LED starts emitting at half the intensity it was emitting at when you first started (effectively draining the battery to half voltage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets' say you touched the positive ends of both batteries for a reasonable amount of time, and then did the same for the negative ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the used battery would have a much lower voltage, raw electrical current would travel from the unused battery into the used battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this brings me to my question -- Would the batteries now effectively have the same amount of energy in them? I am not just talking voltage, but chemically too, or would the similarity in voltage on both batteries immediately disappear if either of them were used (because while, yes, either end of the battery does contain equal electrical charge to its counterpart, either end of each battery does NOT contain equal chemical potential)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify this, would touching two batteries as described above effectively drain the unused one, and charge the used one?&lt;/p&gt;&#xA;" OwnerUserId="477" LastEditorUserId="418" LastEditDate="2015-01-08T07:54:05.470" LastActivityDate="2015-01-08T07:54:05.470" Title="Can I use one alkaline battery to recharge another?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="884" PostTypeId="1" AcceptedAnswerId="885" CreationDate="2012-08-01T05:27:48.173" Score="12" ViewCount="2807" Body="&lt;p&gt;Michael Faraday melted sodium chloride in a silver spoon and ran an electric current through the molten salt, which released chlorine as a gas and left pure sodium behind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium chloride melts at $800\ \pu{^\circ C}$. How did Michael Faraday melt the salt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I know he invented the bunsen burner, but did he use that?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="23561" LastEditDate="2018-08-17T00:06:09.040" LastActivityDate="2018-08-17T00:06:09.040" Title="How did Michael Faraday melt Sodium Chloride?" Tags="&lt;experimental-chemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="885" PostTypeId="2" ParentId="884" CreationDate="2012-08-01T06:31:52.543" Score="11" Body="&lt;p&gt;The answer is given in Faraday’s own work, &lt;em&gt;Experimental Researches in Electricity, Volume 1&lt;/em&gt;, &lt;a href=&quot;http://www.gutenberg.org/files/14986/14986-h/14986-h.htm&quot; rel=&quot;noreferrer&quot;&gt;freely available&lt;/a&gt; thanks to The Project Gutenberg:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;399. &lt;strong&gt;Other substances, which could not be melted on glass, were fused by the lamp and blowpipe&lt;/strong&gt; on platina connected with one pole of the battery, and then a wire, connected with the other, dipped into them. In this way chloride of sodium, sulphate of soda, protoxide of lead, mixed carbonates of potash and soda, &amp;amp;c. &amp;amp;c., exhibited exactly the same phenomena as those already described: whilst liquid, they conducted and were decomposed; whilst solid, though very hot, they insulated the battery current even when four troughs were used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The text above shows that the lamp in question is the &lt;strong&gt;spirit lamp&lt;/strong&gt;, and the whole aparatus is &lt;a href=&quot;http://physics.kenyon.edu/EarlyApparatus/Thermodynamics/Blow_Lamp/Blow_Lamp.html&quot; rel=&quot;noreferrer&quot;&gt;described here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQDcn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of it is that an open flame might not have enough heat, so a conventional flame is simply contained into a more restricted space.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-01T06:31:52.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="886" PostTypeId="2" ParentId="882" CreationDate="2012-08-02T08:22:36.177" Score="4" Body="&lt;p&gt;For the particular reaction where you get a silver mirror with an aldehyde, I doubt it'll work for copper, due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduction_potential&quot; rel=&quot;nofollow&quot;&gt;reduction potentials&lt;/a&gt;. I'll take acetaldehyde $\ce{&amp;lt;-&amp;gt;}$ acetate as an example here, but I believe that the reduction potentials of most aldehydes will be similar. The nitrate is a spectator ion here, so we need not consider its presence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jesuitnola.org/upload/clark/refs/red_pot.htm&quot; rel=&quot;nofollow&quot;&gt;$$E^o_{\ce{Ag^+,Ag}}=+0.80$$&#xA;$$E^o_{\ce{Cu^{2+},Cu}}=+0.34$$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/redoxp.html&quot; rel=&quot;nofollow&quot;&gt;$$E^o_{\ce{CH3COO^{-},CH3CHO}}=+0.580$$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our reaction involves the metal ($\ce{Ag}$ or $\ce{Cu}$ being reduced, so it must have a higher reduction potential than the acetate/acetaldehyde pair. Of the two metals, only silver has a higher SRP, thus the reaction involving silver nitrate is spontaneous--whereas the copper nitrate reaction is  infeasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you could find some substrate other than an aldehyde which reacts with copper nitrate to achieve the effect you want--though I'll have to look up some stuff to figure this out. I &lt;em&gt;think&lt;/em&gt; the Tollen's reaction is a single step everyday redox reaction, which means that we just need to look for a stronger reducing agent which doesn't harm the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it looks like the standard process for plating nonconductors with metals is to give them a conducting layer by dipping them in something, and then using electroplating. Not sure about this, though.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-08-02T08:22:36.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="887" PostTypeId="2" ParentId="882" CreationDate="2012-08-02T09:07:46.493" Score="4" Body="&lt;p&gt;Manishearth points to the right direction, the crucial parameters to look at are the standard electrode potentials. One should however take into account that the relevant species that get reduced under the condition of the Tollens' reagent are not the aquo- but the amine complexes, namely $\ce{[Ag(NH3)2]+}$. and $\ce{[Cu(NH3)4]^{2+}}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately i don't have the data at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, a copper-based reagent comparable to the Tollens' reagent exists: &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fehling%27s_solution&quot; rel=&quot;nofollow&quot;&gt;Fehling's solution&lt;/a&gt;. However, in this case, Cu(II) is reduced to Cu(I) only but not to copper metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you just might give it a try and post the results :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2012-08-02T10:31:28.463" LastActivityDate="2012-08-02T10:31:28.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="888" PostTypeId="1" CreationDate="2012-08-02T22:05:46.167" Score="2" ViewCount="276" Body="&lt;p&gt;PEP (Phosphoenolpyruvic acid) is acidic but I need to make it neutral. I need a solution (pH 6-8) of PEP in a final 1 mL volume with a concentration of 100 mM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I make this?&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="735" LastEditDate="2012-10-12T12:55:27.443" LastActivityDate="2012-10-12T12:55:27.443" Title="How would I neutralize a PEP (Phosphoenolpyruvic) acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="893" PostTypeId="1" AcceptedAnswerId="897" CreationDate="2012-08-03T21:24:31.253" Score="57" ViewCount="219538" Body="&lt;p&gt;From school, I remember a very important rule: first you need to pour the water and &lt;em&gt;then&lt;/em&gt; the acid (when you need to mix them) not vice-versa. This is because otherwise the aсid becomes very hot and splashing may happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does it get hotter when water is poured into it? What reaction takes place?&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="402" LastEditDate="2019-05-17T06:42:01.900" LastActivityDate="2019-05-17T06:42:01.900" Title="Why add water first then acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="894" PostTypeId="2" ParentId="893" CreationDate="2012-08-03T23:02:23.120" Score="19" Body="&lt;p&gt;This happens with strong acids and bases. &lt;em&gt;Strong&lt;/em&gt; means the acid almost complete dissociates into ions; e.g. $\ce{HCl}$ to $\ce{H+}$ and $\ce{Cl-}$.  When the dissociation occurs, it releases a great deal of heat (i.e. &lt;em&gt;exothermic&lt;/em&gt;) because of the subsequent hydration of $\ce{H+}$ to $\ce{H3O+}$, as the other comments have indicated. The reason you add acid to water is that if you add water to acid, the first drop of water reacts immediately, and the heat might be high enough to boil the water instantly, which could spray acid out of the container.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="7475" LastEditDate="2015-12-04T21:08:11.837" LastActivityDate="2015-12-04T21:08:11.837" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="895" PostTypeId="1" AcceptedAnswerId="896" CreationDate="2012-08-04T03:45:01.367" Score="2" ViewCount="108" Body="&lt;p&gt;What would happen if I added $\ce{Na2CO3 \cdot n(H2O)}$ to a solution of $\ce{HCl}$? Could I consider the reaction the same as just $\ce{Na2CO3 + HCl -&amp;gt; NaCl + CO2 + H2O}$?&lt;/p&gt;&#xA;" OwnerUserId="484" LastEditorUserId="4231" LastEditDate="2015-02-22T23:22:30.507" LastActivityDate="2015-02-22T23:22:30.507" Title="Adding acid to hydrates" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="896" PostTypeId="2" ParentId="895" CreationDate="2012-08-04T12:12:34.453" Score="5" Body="&lt;p&gt;There is no difference in properties of salt hydrates and non-hydrates after they are dissolved in water, as they are hydrated to same state. So, yes, you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is difference in reaction of salt hydrates and non-hydrates outside water solutions. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-08-04T13:09:14.303" LastActivityDate="2012-08-04T13:09:14.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="897" PostTypeId="2" ParentId="893" CreationDate="2012-08-04T12:27:15.777" Score="54" Body="&lt;p&gt;This is mostly the case for sulfuric acid. Commercially available sulfuric acid is dense (~1.8 g/ml) and when water is added, it may not mix. In this case a layer of hot weak acid solution is formed, which boils and sprays around. When acid is poured into water, it flows down the flask and mixes much better, so no boiling occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this occurs is due to the large amount of energy released in the hydration reaction of sulfuric acid ions. Do not believe that heat comes from dissociation, as the dissociation of acids, bases, and salts always consumes energy. The energy is released from subsequent hydration, and the release may be high, especially if $\ce{H+}$ or $\ce{OH-}$ ions are hydrated. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="865" LastEditDate="2015-08-14T14:58:20.617" LastActivityDate="2015-08-14T14:58:20.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="898" PostTypeId="2" ParentId="883" CreationDate="2012-08-04T12:35:26.903" Score="3" Body="&lt;p&gt;First, some batteries are rechargeable, but some are not. However, even if we talk about rechargeable one, then&#xA;1) some degradation during recharge usually occurs&#xA;2) deep equality will require very long time to achieve&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-08-04T12:35:26.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="899" PostTypeId="1" CreationDate="2012-08-04T16:21:26.467" Score="3" ViewCount="1650" Body="&lt;p&gt;I am wondering how the cocoate ion (in natural soaps, in the form of sodium or potassium cocoate and from the saponification of coconut oil) differs structurally from the olivate ion (from the saponification of olive oil)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My family is attempting to implement the methods in solveeczema.org and believe that our children have detergent-reactive eczema and our baby is most sensitive.  In moving away from detergents and to natural soaps, we have started using soaps that result from the saponification of vegetable oils - mostly coconut oil, palm kernel oil, and olive oil.  We discovered that our other child (a toddler) seems to react to these soaps in a similar way that he would react to a cleaner containing much stronger/harsher surfactants (such as sodium lauryl sulfate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our toddler seems to get bad eczema flares using soaps that contain cocoate, but he has no reaction to soaps that contain 100% olivate.  I have no idea why this would be, since I would assume that both cocoate and olivate have the R-COO- structure.  Is cocoate much more hydrophilic than olivate?  Is the functional group different?  Is the cocoate ion a longer molecule that looks much more like a harsh detergent than a salt resulting from saponification of olive oil or tallow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that this question is related to two questions already asked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/416/is-soap-the-natural-form-of-sodium-lauryl-sulfate&quot;&gt;Is soap the natural form of Sodium Lauryl Sulfate?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  &lt;a href=&quot;https://parenting.stackexchange.com/questions/5047/how-to-test-whether-my-toddler-is-allergic-to-detergent/5049#comment7644_5049&quot;&gt;https://parenting.stackexchange.com/questions/5047/how-to-test-whether-my-toddler-is-allergic-to-detergent/5049#comment7644_5049&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="486" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2012-08-28T12:23:35.687" Title="What is the chemical structure of the cocoate ion vs. the olivate ion?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="900" PostTypeId="2" ParentId="899" CreationDate="2012-08-04T18:10:28.390" Score="7" Body="&lt;p&gt;Note 1. Both coconut oil and olive oil are natural oils. The soap, produced from them, is mixture of several compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference of olive and coconut oil may be easily found in wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Coconut_oil#Composition_and_comparison&quot;&gt;http://en.wikipedia.org/wiki/Coconut_oil#Composition_and_comparison&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, coconut oil contains mostly saturated acids esters, while olive contains mostly unsaturated ones. The unsaturated acid esters are responsible for olive oil ability to solidify on air. But here is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: In my knowledge, allergy is usually caused by carbohydrates or proteins. Fatty acids and their ions have too few binding sites to be recognized by immune system as something distinct from our own. The real source of reaction may be &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Minor components of initial oil. There are quite many of them and removing them completely is not easy. &lt;/li&gt;&#xA;&lt;li&gt;Other components of soap, such as perfume or pigment.&lt;/li&gt;&#xA;&lt;li&gt;Soap solution pH, that may differs from soap kind to kind, but this is unlikely. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastEditorUserId="275" LastEditDate="2012-08-05T10:45:44.080" LastActivityDate="2012-08-05T10:45:44.080" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="901" PostTypeId="1" AcceptedAnswerId="1124" CreationDate="2012-08-04T21:16:35.190" Score="17" ViewCount="3037" Body="&lt;p&gt;What are the different diagrams/tables used to organize the elements other than the Periodic Table? What are the advantages/disadvantages of each?&lt;/p&gt;&#xA;" OwnerUserId="489" LastEditorUserId="10256" LastEditDate="2015-04-15T20:16:33.590" LastActivityDate="2017-05-09T12:51:20.337" Title="What are the alternatives to the Periodic Table of the Elements?" Tags="&lt;periodic-table&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="903" PostTypeId="1" CreationDate="2012-08-06T13:11:50.873" Score="6" ViewCount="12583" Body="&lt;p&gt;Using Diffuse Reflectance Spectroscopy, I have obtained the band gap for $\small\ce{SiO2}$ (purchased from a chemical retailer) in the range of 3.25 to 4.75 eV.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if the values are in the right ballpark and, if not, would like to review the DRS procedure that I carried out.  I would also like to know if the band gap determined through DRS would have such a wide range.&lt;/p&gt;&#xA;" OwnerUserId="433" LastEditorUserId="73" LastEditDate="2012-08-06T13:41:26.893" LastActivityDate="2014-03-17T14:02:17.773" Title="What is the bandgap value of SiO2?" Tags="&lt;spectroscopy&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="904" PostTypeId="1" AcceptedAnswerId="906" CreationDate="2012-08-06T20:04:38.130" Score="8" ViewCount="1807" Body="&lt;p&gt;Most Lipids in a cell have ester bonds (made up of glycerol and fatty acids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WUK0d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned in my chemistry class that we can hydrolyze an ester by simply adding water to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOCH2CH3 + H2O &amp;lt;=&amp;gt;[H^+ (aq)] CH3COOH + EtOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why do lipids which  have an ester bond don't hydrolyze inside a cell which is 70 percent  water? While searching throught web I have found this &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/esters/hydrolysis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; which explains of different ways to hydrolyze an ester. In that it explains about how to hydrolyze esters using dilute acids like HCl. In our stomach HCl is present so logically HCl should destroy the ester in stomach lining (which are made up of cells) instantaneonusly, but this is not happening. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="24052" LastEditDate="2016-03-11T23:50:13.293" LastActivityDate="2016-03-11T23:50:13.293" Title="Why do lipids in a cell hydrolyze into glycerol and fatty acids?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="905" PostTypeId="1" CreationDate="2012-08-06T20:20:46.663" Score="16" ViewCount="440" Body="&lt;p&gt;&lt;em&gt;The previous question contained too much unnecessary information and was edited.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering about the &quot;hierarchy&quot; of wavefunctions. If one can combine atomic orbitals (AO) into molecular orbitals (MO) through the LCAO method, could one combine MOs to form a &lt;em&gt;supra&lt;/em&gt; wavefunction, or an LCMO?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some research and found a &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.560110612/abstract&quot;&gt;paper&lt;/a&gt; on an approach to finding the MO of molecules by effectively taking a LCMO of smaller molecules or constituents. While interesting, this is not what I'm referring to as a &quot;supra&quot; wavefunction, if there's such a thing.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-08-07T18:39:06.357" LastActivityDate="2012-08-10T13:42:22.657" Title="Hierarchy of electronic wavefunctions" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="906" PostTypeId="2" ParentId="904" CreationDate="2012-08-06T20:48:39.213" Score="7" Body="&lt;p&gt;Organisms control rates of chemical reactions in two ways. First, by controlling the conditions, and second, by controlling the location of reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conditions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrolysis of esters is very slow at normal physiological conditions: 37 &lt;sup&gt;o&lt;/sup&gt;C and pH 7.4. A mixture of water and simple esters like ethyl acetate remains stable for years. At higher or lower pH, the reaction becomes faster. As you mention, the HCl in the stomach would definitely increase the rate of hydrolysis. The reaction is also accelerated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Enzyme&quot;&gt;enzymes&lt;/a&gt;, and organisms use enzymes for all intentional ester hydrolysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Location&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reaction to occur, the water molecule and the acid catalyst needs to get to the ester. In &lt;a href=&quot;http://en.wikipedia.org/wiki/Cell_membrane#Lipids&quot;&gt;cell membranes&lt;/a&gt;, that ester functional group is not on the surface of the membrance; the ionic phosphate part is. The ester group is just below the surface, in the hydrophobic interior of the membrane. Neither water, nor acid can penetrate into this hydrophobic region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the stomach, there is an extra layer of protection: &lt;a href=&quot;http://en.wikipedia.org/wiki/Mucus#Digestive_system&quot;&gt;mucus&lt;/a&gt;, which prevents the stomach acid from reaching the cells.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-08-06T20:48:39.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="907" PostTypeId="2" ParentId="904" CreationDate="2012-08-07T04:32:32.710" Score="2" Body="&lt;p&gt;Stomach wall are made of mucous membrane, that secrets mucus like crazy. Mucous is made of glycoproteins -- sugar+protein complex, that makes diffusion slower, and protein structure make it acid hydrolysis slow. The mucous membrane cells maintains normal physiological pH, and while they are isolated from stomach acid by mucus they secret, they are fine, as hydrolysis is very slow without catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is wikipedia article on the question &lt;a href=&quot;http://en.wikipedia.org/wiki/Gastric_mucosal_barrier&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Gastric_mucosal_barrier&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barrier may be broken by some chemical agents or if mucus secreting cells are suppressed by illness. In latter case stomach damages itself routinely and peptic ulcer decease occurs. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-08-07T04:32:32.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="908" PostTypeId="5" CreationDate="2012-08-07T05:52:27.920" Score="0" Body="&lt;p&gt;In simple language, a home-experiment is defined as a small-scale experiment which can be undertaken with reagents readily available around the house, while observing standard laboratory safety principles and practices&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="5026" LastEditDate="2018-03-22T18:15:23.373" LastActivityDate="2018-03-22T18:15:23.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="909" PostTypeId="4" CreationDate="2012-08-07T05:52:27.920" Score="0" Body="For questions about a small-scale experiment which can be undertaken with reagents readily available around the house, while observing standard laboratory safety principles and practices. Do not confuse this tag with [everyday-chemistry] or [food-chemistry]." OwnerUserId="73" LastEditorUserId="5026" LastEditDate="2018-03-22T18:15:43.193" LastActivityDate="2018-03-22T18:15:43.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="910" PostTypeId="1" AcceptedAnswerId="943" CreationDate="2012-08-07T14:59:21.047" Score="3" ViewCount="38091" Body="&lt;p&gt;I've seen it many times... Why does soap become foam-like when it reacts with water (moisture) or when we touch it with wet hands.. What compound actually causes this physical change? And why does this physical change occur?&lt;/p&gt;&#xA;" OwnerDisplayName="user491" LastEditorUserId="4356" LastEditorDisplayName="user491" LastEditDate="2014-11-18T23:48:07.303" LastActivityDate="2016-11-02T13:26:34.057" Title="Reaction of soap with water" Tags="&lt;organic-chemistry&gt;&lt;surfactants&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="911" PostTypeId="1" AcceptedAnswerId="7584" CreationDate="2012-08-07T17:45:29.393" Score="5" ViewCount="2924" Body="&lt;p&gt;Can you tell me what is a burst nucleation process, and how can it be used for synthesising uniform nanoparticles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you also provide (if possible) any links to papers or materials regarding mechanism of nanoparticles growth?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="73" LastEditDate="2012-08-07T22:10:47.077" LastActivityDate="2013-12-31T13:17:35.987" Title="What is burst nucleation?" Tags="&lt;inorganic-chemistry&gt;&lt;nanotechnology&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-05-01T23:49:34.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="912" PostTypeId="2" ParentId="910" CreationDate="2012-08-07T19:19:32.680" Score="3" Body="&lt;p&gt;While this question should get a more elaborate answer, mine will at least give you leads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soaps act as &lt;a href=&quot;http://en.wikipedia.org/wiki/Surfactant&quot; rel=&quot;nofollow&quot;&gt;surfactants&lt;/a&gt;, that is, they are amphiphilic (both hydrophobic &lt;em&gt;and&lt;/em&gt; hydrophilic) molecules that lower the &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow&quot;&gt;surface tension&lt;/a&gt; of a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By lowering the surface tension of water, soap bubbles can form, the surface becoming more &quot;elastic&quot; - the molecules at the interface experience less inward/normal force than they would were surfactant not present. On the contrary, adding soap to a pond would prevent water striders from gliding on the surface of the pond.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-08-07T19:19:32.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="914" PostTypeId="2" ParentId="905" CreationDate="2012-08-07T23:48:57.933" Score="2" Body="&lt;p&gt;(atomic orbitals -&gt; molecular orbitals) -&gt; (&quot;Slater determinats&quot; -&gt; MCSCF wave functions) -&gt; mixed state N-electron density matrices -&gt; grand canonical ensemble (Fock space) density matrices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the first two are 1-electron objects, the next two are N-electron objects representing pure states at different level of approximations&lt;/p&gt;&#xA;" OwnerUserId="492" LastActivityDate="2012-08-07T23:48:57.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="915" PostTypeId="1" AcceptedAnswerId="938" CreationDate="2012-08-07T20:45:42.120" Score="43" ViewCount="78953" Body="&lt;p&gt;How can $\ce{CO2}$ be converted into carbon and oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 -&amp;gt; C + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 + ? -&amp;gt; C + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that plants are capable of transforming $\ce{CO2 + H2O}$ to glucose and oxygen via photosynthesis, but I'm interested in chemical or physical means rather than biological.&lt;/p&gt;&#xA;" OwnerUserId="498" OwnerDisplayName="Enrico Pallazzo" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-08-14T15:27:39.313" LastActivityDate="2018-08-14T15:31:15.673" Title="How can carbon dioxide be converted into carbon and oxygen?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;decomposition&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="916" PostTypeId="2" ParentId="915" CreationDate="2012-08-07T21:44:41.953" Score="3" Body="&lt;p&gt;$\ce{CO2}$ into $\ce{CO + O2}$ is easier but you need a catalyst and about 1500C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rsc.org/chemistryworld/news/2008/january/03010801.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rsc.org/chemistryworld/news/2008/january/03010801.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short:&lt;br&gt;&#xA;Fe3O2 is heated to 1500C driving off oxygen. The resulting FeO is moved to CO2 chamber where it absorbs oxygen from the CO2. Result is CO and cooled Fe3O2. Cycle repeats.&lt;/p&gt;&#xA;" OwnerDisplayName="Martin Beckett" LastEditorUserId="5078" LastEditorDisplayName="user15489" LastEditDate="2018-04-25T05:10:35.590" LastActivityDate="2018-04-25T05:10:35.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="917" PostTypeId="1" AcceptedAnswerId="6800" CreationDate="2012-08-08T05:16:37.123" Score="2" ViewCount="2808" Body="&lt;p&gt;For the purposes of x-ray absorption, I'm trying to discover the average chemical composition of a typical PCB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research so far has indicated that the most common type of PCB is designated FR-4.  It is composed of fiberglass and epoxy resin.  The resin often contain Bromine to add fire resistance.  Let's ignore any printed copper for the mean time.  So we are looking at H, C, O, Si, Br, and probably some other bits and pieces.  But what are the percentage of these elements?&lt;/p&gt;&#xA;" OwnerUserId="494" LastActivityDate="2014-11-17T07:35:50.690" Title="Composition of a typical printed circuit board (PCB)?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="921" PostTypeId="1" CreationDate="2012-08-08T11:00:02.413" Score="14" ViewCount="5320" Body="&lt;p&gt;What is the order of dipole moments for &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3, SiO2, P2O5,}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7}$&lt;/span&gt;? This question is from a book and the answer given is &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7 &amp;lt; SO3 &amp;lt; P2O5 &amp;lt; SiO2}$&lt;/span&gt;. Can someone explain why? When I drew the Lewis structures and applied VSEPR theory, it seemed that all the given compounds have zero dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="497" LastEditorUserId="17277" LastEditDate="2019-07-19T22:04:19.120" LastActivityDate="2021-04-12T17:33:00.337" Title="Order of dipole moments" Tags="&lt;inorganic-chemistry&gt;&lt;lewis-structure&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="922" PostTypeId="1" CreationDate="2012-08-08T11:01:31.513" Score="4" ViewCount="189" Body="&lt;p&gt;Consider a chemical reaction where two different particles form another one &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O + O_2 -&amp;gt; O_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it confusing how this can be an exothermic process. How to picture the release of $+\Delta H$ to the surroundings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And am I right in viewing this energy value only as the net energy coming from an integration of the particle potentials as they get closer along the reaction coordinate?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="4231" LastEditDate="2015-02-21T22:27:55.010" LastActivityDate="2015-02-21T22:27:55.010" Title="What is the energy package released to the outside world?" Tags="&lt;energy&gt;&lt;enthalpy&gt;&lt;reaction-coordinate&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="923" PostTypeId="2" ParentId="905" CreationDate="2012-08-08T11:45:34.640" Score="12" Body="&lt;p&gt;I've been reading up on this, and will take a shot at the answer. It is going to be an interesting answer, and probably not what you expect. There has been an interesting conversation going on in the &lt;em&gt;Journal of Chemical Education&lt;/em&gt; in response to &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed100155c#citing&quot;&gt;this letter&lt;/a&gt; advocating the retirement of the localized hybrid orbitals because they &quot;aren't real&quot; and cannot be used to predict &quot;all properties&quot; of molecules. Rebuttals included comments that delocalized molecular orbitals also fail to predict &quot;all properties&quot;. However, the final word in the discussion came from &lt;a href=&quot;http://pubs.acs.org/doi/full/10.1021/ed200615j&quot;&gt;this letter&lt;/a&gt;, which reminds us what orbitals are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;Orbitals&lt;/a&gt; are not real. The electrons that populate orbitals are real. Orbitals are mathematical constructs created to describe the curious wave-like probabilistic quantum mechanical properties of electrons. When we talk about atomic orbitals, we generally mean the 1-electron hydrogenic atomic orbitals (HAOs), which are mathematically represented as a wavefunction ($\Psi$) and graphically represented by the probability density function ($\Psi ^ * \Psi$) at either 90% or 95% level. Despite valiant and massive efforts, wavefunctions for multi-electron atoms cannot be solved exactly. They are approximated as a parameterized HAOs. When we talk about the atomic orbitals of carbon, we usually mean the HAOs that closely approximate what we think the AOs of carbon should be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules and their orbitals are even more troublesome. The LCAO approach creates a bunch of nice 1-electron wavefunctions out of the set of HAOs. These methods give rise to both the canonical delocalized molecules orbitals (CMOs), and the localized valence bond orbitals (VBOs). As the authors of that letter point out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term canonical implies that CMOs are not the only possible MOs, and there are many orbitals that generate an identical “chemical reality”.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The approximate polyelectron wavefunction of a molecule (at least in HF theory) is the Slater determinant of a series of 1-electron CMOs $\psi _i$:&#xA;$$  \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ... $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting property of Slater determinants is that unitary operations (like rotations) on the set of CMOs return a different set of MOs, which produce the same expectation values for the polyelectron wave function as original set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ...~~ \underrightarrow{unitary~~transformation} ~~\psi _a \overline{\psi} _a \psi _b \overline{\psi} _b \psi _c \overline{\psi} _c \psi _d \overline{\psi} _d ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, it seems to me, that a linear combination of MOs that are themselves linear combinations of AOs would still be a linear combination of AOs, just a different one than you started with. If that is true, then all you have is a different set of MOs than you started with, and according to HF theory, that should give you the same expectation values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a set of &quot;LCMOs&quot; are no more or less valid than the original set of LCAOs. They likely do not produce any additional insights into the structure of the molecule. However, the approach might be computationally more efficient when trying to determine the MOs of very large molecules and polymers.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-08-08T11:45:34.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="925" PostTypeId="2" ParentId="905" CreationDate="2012-08-08T15:11:58.593" Score="5" Body="&lt;p&gt;As a supplement to Ben Norris's answer, I thought I'd add the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you use a basis to construct an orbital, your original basis functions can already be represented as linear combinations of your whole set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi(x,y,z,s)=\chi(\mathbf{x})= c_1\varphi_1 + c_2\varphi_2+c_3\varphi_3\dots \\&#xA;\varphi_1 = 1\varphi_1 + 0\varphi_2 + 0 \varphi_3\dots$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\chi$ is an MO and $\varphi$ is a basis function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, altering the coefficients or linearly combining MOs doesn't increase the complexity of your result:&#xA;$\chi_a + \chi_b = (c_{a,1} + c_{b,1})\varphi_1 + (c_{a,2}+c_{b,2})\varphi_2 + (c_{a,3}+c_{b,3})\varphi_3\dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you construct a wave function from basis functions, the usual method is to use a Slater determinant - the determinant of a matrix as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Phi(\mathbf{x}_1,\mathbf{x}_2,\dots\mathbf{x}_N) = \frac{1}{\sqrt{N!}}\left|&#xA;   \begin{matrix} \chi_1(\mathbf{x}_1) &amp;amp; \chi_2(\mathbf{x}_1) &amp;amp; \cdots &amp;amp; \chi_N(\mathbf{x}_1) \\&#xA;                      \chi_1(\mathbf{x}_2) &amp;amp; \chi_2(\mathbf{x}_2) &amp;amp; \cdots &amp;amp; \chi_N(\mathbf{x}_2) \\&#xA;                      \vdots &amp;amp; \vdots &amp;amp; \ddots &amp;amp; \vdots \\&#xA;                      \chi_1(\mathbf{x}_N) &amp;amp; \chi_2(\mathbf{x}_N) &amp;amp; \cdots &amp;amp; \chi_N(\mathbf{x}_N)&#xA;   \end{matrix} \right|$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater_determinant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you an expression involving products of your MOs rather than just linear sums, increasing the complexity. This is kind of awkward looking for the n-function case, so instead, for a 3 MO case:&#xA;$\Phi(\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3) = \frac{1}{\sqrt{3!}} ( -\chi_3(\mathbf{x}_1)\chi_{2}(\mathbf{x}_2)\chi_1(\mathbf{x}_3) + \chi_2(\mathbf{x}_1)\chi_{3}(\mathbf{x}_2)\chi_3(\mathbf{x}_1) + \chi_3(\mathbf{x}_1)\chi_{1}(\mathbf{x}_2)\chi_2(\mathbf{x}_3) + \chi_1(\mathbf{x}_1)\chi_{1}(\mathbf{x}_2)\chi_3(\mathbf{x}_3) + \chi_2(\mathbf{x}_1)\chi_{1}(\mathbf{x}_2)\chi_3(\mathbf{x}_3) ) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, just as you can use AOs to build MOs, you can either use that alone as your wave function, or you can use linear combinations of them to obtain a &lt;em&gt;multiconfigurational&lt;/em&gt; wave function - the MCSCF methods mentioned in jjj's answer:&#xA;$\Psi = c_1\Phi_1 + c_2\Phi_2 + c_3\Phi_3 \dots$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is the sort of thing you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2012-08-10T13:42:22.657" LastActivityDate="2012-08-10T13:42:22.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="926" PostTypeId="2" ParentId="922" CreationDate="2012-08-08T16:41:49.803" Score="2" Body="&lt;p&gt;Essentially yes - if you consider the two particles colliding into a potential well, the kinetic energy associated with each will (depending on the angular specifics) be apportioned into rotational, vibrational, and even occasionally electronic excitations. So, the two particles collide, and are then vibrating and/or rotating about each other with all the translational energy converted in the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-reaction collisions with other ambient particles allow relaxation into lower-energy states, transmitting energy to those particles. After some collisions the new species formed is in equilibrium, transferring between low-energy rotational and vibrational states as it collides further, and the change in potential energy has been dissipated as rotational, vibrational, and translational energy in other particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I guess I could really use some sort of animation for this.)&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-08-08T16:41:49.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="927" PostTypeId="1" AcceptedAnswerId="928" CreationDate="2012-08-08T20:11:48.177" Score="1" ViewCount="3322" Body="&lt;p&gt;I wonder, what software packs / environments are available for chemical reaction simulation. The goal is to learn or just rehearse some basic knowledge in chemistry. I suppose there should be some to buy but are there any open source projects dedicated to this?   &lt;/p&gt;&#xA;" OwnerUserId="50" LastActivityDate="2012-08-12T01:40:51.177" Title="Software for chemical reaction simulation" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2012-08-12T01:41:02.860" ContentLicense="CC BY-SA 3.0" />
  <row Id="928" PostTypeId="2" ParentId="927" CreationDate="2012-08-08T20:34:39.587" Score="6" Body="&lt;p&gt;If you mean finding saddle point, then you should consider Gamess and Quantum Espresso packages: free, stable, mature packages, useful for molecule simulations and estimating reaction barriers. They, however, have to be manually compiled and require some skill to setup, as they are meant to be used in POSIX environment (various UNIX clones), not windows, and does not have user friendly result viewer or job builder/scheduler. It is not a problem for experienced linux user, though...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently using quantum espresso and I found it useful for my cases.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-08-09T04:58:25.820" LastActivityDate="2012-08-09T04:58:25.820" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="929" PostTypeId="2" ParentId="922" CreationDate="2012-08-08T20:37:04.080" Score="2" Body="&lt;p&gt;The &quot;packet&quot; of energy exchanged fall into one of three forms: heat, work, or electromagnetic radiation. The first two are kinetic energy phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding on &lt;a href=&quot;https://chemistry.stackexchange.com/questions/922/what-is-the-energy-package-released-to-the-outside-world/926#926&quot;&gt;Aesin's answer&lt;/a&gt; about potential and kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions that are exothermic convert potential energy into kinetic energy. The potential energy is the energy stored in the chemical bonds (electrons), and the kinetic energy is the molecular motion (rotation, translation, and vibration). Consider your &quot;system&quot; to be the electrons and the &quot;surroundings&quot; to be the nuclei of the atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to thermochemistry data from the NIST chemistry webbook for &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C10028156&amp;amp;Units=SI&amp;amp;Mask=1#Thermo-Gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;ozone&lt;/a&gt; and &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C17778802&amp;amp;Units=SI&amp;amp;Mask=1#Thermo-Gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomic oxygen&lt;/a&gt;, this reaction is exothermic (the heat of formation of molecular oxygen is by definition zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H ^0_{rxn}=\sum  \Delta H ^0_{f}(products)-\sum  \Delta H ^0_{f}(reactants)   $$&#xA;$$\Delta H ^0_{rxn}= \Delta H ^0_{f}(\text{O}_3 )- \Delta H ^0_{f}(\text{O}) $$&#xA;$$\Delta H ^0_{rxn}= 142.67~~\text{kJ/mol}-249.18~~\text{kJ/mol}=-106.51~~\text{kJ/mol} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the potential energy of the electrons in ozone (total bond order of 3) is lower than the potential energy of a molecule of oxygen and an atom of oxygen (total bond order of 2). That extra potential energy is needed to keep O and O&lt;sub&gt;2&lt;/sub&gt; apart. When that energy is no longer needed, it has to become some other kind of energy (First Law of Thermodynamics: Conservation of Energy). In all cases of exothermic reactions, the extra potential energy becomes kinetic energy. If this kinetic energy increase occurs without a change in volume, then the temperature increases, and we say that &quot;heat&quot; has been exchanged. If the volume is allowed to change, then the increase in kinetic energy is converted to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, for an endothermic reaction, kinetic energy (work or heat) is converted into potential energy (less stable electronic arrangements). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electromagnetic radiation is a little different. Instead of increasing the kinetic energy of the nuclei, an exothermic reaction can release a photon of light. In the case of the ozone reaction, the photon would have a wavelenth of 1.12 micrometers, in the near-IR.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2012-08-08T20:37:04.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="930" PostTypeId="1" AcceptedAnswerId="931" CreationDate="2012-08-09T05:21:23.343" Score="3" ViewCount="1477" Body="&lt;p&gt;I was drawing out all the isomers of $\ce{C5H8}$ and came upon two structures of cyclopropane attached with double bonds. The group is neither a methylene nor an ethyl... what is it? What are the names of these two molecules?&lt;/p&gt;&#xA;" OwnerDisplayName="user499" LastEditorDisplayName="user499" LastEditDate="2014-04-22T00:55:55.830" LastActivityDate="2014-04-22T00:55:55.830" Title="What is the name of cyclopropane with a double bond?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="931" PostTypeId="2" ParentId="930" CreationDate="2012-08-09T07:46:44.197" Score="9" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/apQGD.png&quot; alt=&quot;ethylidenecyclopropane&quot;&gt; Ethylidenecyclopropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DQAgJ.png&quot; alt=&quot;vinylcyclopropane&quot;&gt; Vinylcyclopropane (or Ethenylcyclopropane)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-08-09T07:46:44.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="932" PostTypeId="1" CreationDate="2012-08-09T14:15:15.153" Score="2" ViewCount="2111" Body="&lt;p&gt;I'm calculating molar changes in thermodynamic properties due to reactions between gasses (assumed to be ideal gases).  I can calculate $\Delta H$ easily enough, because it's just $\sum_i \nu_i\Delta_f H^\circ_i$, with $\nu_i$ the stoichiometric coefficients. $\Delta G$ (at standard pressure) can be calculated from $\sum_i \nu_i\mu_i$, with $\mu_i = \Delta_f G^\circ_i + RT\log[i]$. Since $\Delta G = \Delta H - T\Delta S$, I can calculate $\Delta S$ as $\frac{1}{T}(\Delta H - \Delta G)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good. But I'm inexperienced in working with tabulated quantities (I'm a more of a theoretical physicist than a chemist) and keep making mistakes with signs and units. So I figured that as a sanity check I would calculate $\Delta S$ directly, using tabulated values of $S^\circ$. But then I realised I don't know how to do this, and it seems that none of the textbooks on my desk explain it either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based mostly on intuition, it seems like it should be $\Delta S = \sum_i\nu_i(S^\circ_i + R\log[i])$, with the $R\log[i]$ term having something to do with the entropy of mixing. For the example reaction I chose, this gave something with the approximately correct magnitude but the wrong sign (-15.6 instead of 14.2, which is what I get by calculating it from the Gibbs energy and the enthalpy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what is the correct way to calculate the entropy change due to an ideal gas reaction if, for some reason, you only have access to the concentrations and the standard entropies of the reactants?&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2012-08-10T12:48:08.167" Title="How to calculate entropy change for a gas reaction using $S^\circ$?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="933" PostTypeId="2" ParentId="915" CreationDate="2012-08-09T15:00:10.350" Score="7" Body="&lt;p&gt;Since the carbon atom is linked to two oxygen atoms via double bonds, usually more amount of energy must be supplied in order to separate it. About 94 kcal of energy is required per mol of $\ce{CO2}$ (about 44 g). This energy input could come from any source, but the major source of conversion is through photosynthesis using solar energy which is very well known by the famous equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6CO2 + 6H2O -&amp;gt; C6H12O6 + 6O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in fact, there is a machine built by Sandia researchers known as Counter-Rotating-Ring Receiver Reactor Recuperator (CR5), which uses solar power to convert carbon dioxide and water to carbon monoxide, water, hydrogen and oxygen at a temperature of about 1500 °C using a solar concentrator. Iron oxide acts as an extractor of oxygen from $\ce{CO2}$ forming $\ce{CO}$. But, the main aim of this method is to produce fuel and not &lt;strong&gt;carbon&lt;/strong&gt;. But it’ll take at least 15–20 years to come into usage because, only the prototype of this machine has been invented and tested.&lt;/p&gt;&#xA;" OwnerDisplayName="user491" LastEditorUserId="24052" LastEditDate="2016-03-11T23:04:14.413" LastActivityDate="2016-03-11T23:04:14.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="935" PostTypeId="2" ParentId="910" CreationDate="2012-08-09T20:03:29.133" Score="1" Body="&lt;p&gt;The soap dissolves in the water. Dissolution is not a chemical change because the same chemicals are present, they just interact with each other differently. Like CHM said, soaps act as surfactants and the difference in surface tension improves the survivability of bubbles, thus creating a &quot;foam&quot;.&lt;/p&gt;&#xA;" OwnerUserId="482" LastActivityDate="2012-08-09T20:03:29.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="936" PostTypeId="2" ParentId="778" CreationDate="2012-08-09T20:49:11.240" Score="3" Body="&lt;p&gt;A stable 3-phase liquid system can be created, as Spießbürger mentioned with three mutually immiscible liquids. The best one that I know of is the organic-aqueous-fluorous 3-phase system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorocarbon&quot; rel=&quot;nofollow&quot;&gt;Fluorocarbons&lt;/a&gt; (hydrocarbons in which every hydrogen atom is replaced by a fluorine atom), are interestingly immiscible in both water and common organic solvents. This property has added a dimension to chemical separations using &lt;a href=&quot;http://fluorous.com/news/2010/03/technical-newsletters/822/&quot; rel=&quot;nofollow&quot;&gt;fluorous phase extractions&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have three paints, one oil-based, one water-based, and one fluorous-based, then you could separate them easily using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Separatory_funnel&quot; rel=&quot;nofollow&quot;&gt;separatory funnel&lt;/a&gt; or similar device. You would have to make sure that the pigments in each paint was only significantly soluble in that solvent, which would take experimentation. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-08-09T20:49:11.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="937" PostTypeId="1" AcceptedAnswerId="944" CreationDate="2012-08-10T04:49:21.997" Score="5" ViewCount="2477" Body="&lt;p&gt;I'm working on a program that needs to determine if a bond between two or more elements will result in a stable state.  I understand at a high-level how to fill electron subshells using the Aufbau principle, but I also read that in some cases, electrons will jump from a lower energy shell or orbital to a higher one in order to maintain a stable state (&lt;a href=&quot;http://en.wikibooks.org/wiki/General_Chemistry/Filling_Electron_Shells&quot; rel=&quot;nofollow noreferrer&quot;&gt;Filling Electron Shells&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I want to determine if Hydrogen and Nitrogen will form a stable bond I would fill the shells for Nitrogen like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W7HZf.png&quot; alt=&quot;Nitrogen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When adding Hydrogen, would the two electrons in the 2s shell jump to the 2p shell because with the one electron from Hydrogen the HN would then have a full 2p shell? Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tUfdz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this not a stable state? Would it require a large amount of energy to excite or promote those two 2s electrons to the 2p shell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond (on there own w/o adding a large amount of energy to the system) using their valence electron configuration like this, or if there are too many exceptions, making it not a simple programming task to estimate this.&lt;/p&gt;&#xA;" OwnerUserId="501" LastEditorUserId="501" LastEditDate="2012-08-10T14:50:25.773" LastActivityDate="2012-08-10T15:13:51.190" Title="How to determine stable electron states in ionic and covalent bonds?" Tags="&lt;electrons&gt;&lt;bond&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="938" PostTypeId="2" ParentId="915" CreationDate="2012-08-10T08:45:37.547" Score="32" Body="&lt;p&gt;In my opinion, the catalytic, solar-driven conversion of carbon dioxide to methanol, formic acid, etc. is much more interesting and promising, but since Enrico asked for the conversion of carbon dioxide to carbon itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The group around &lt;strong&gt;Yutaka Tamaura&lt;/strong&gt; was/is active in this field. In one of their earlier publications,&lt;sup&gt;[1]&lt;/sup&gt; they heated magnetite ($\ce{Fe3O4}$) at 290 °C for 4 hours in a stream of hydrogen to yield a material which turned out to be stable at room temperature &lt;strong&gt;under nitrogen&lt;/strong&gt;. This material, $\ce{Fe_{3+\delta}O4}$ $(\delta=0.127)$, i.e. the metastable cation-excess magnetite is able to incorporate oxygen in the form of $\ce{O^2-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under a $\ce{CO2}$ atmosphere, the oxygen-deficient material converted to &quot;ordinary&quot; $\ce{Fe3O4}$ with &lt;strong&gt;carbon deposited on the surface&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This remarkable reaction however is &lt;strong&gt;not catalytic&lt;/strong&gt;, but a short recherche showed that the authors have published a tad more in this field. Maybe somebody else finds a a report on a catalytic conversion among their publications. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tamaura, Y.; Tahata, M. Complete reduction of carbon dioxide to carbon using cation-excess magnetite. &lt;em&gt;Nature&lt;/em&gt; &lt;strong&gt;1990,&lt;/strong&gt; &lt;em&gt;346&lt;/em&gt; (6281), 255–256. &lt;a href=&quot;https://doi.org/10.1038/346255a0&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1038/346255a0&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2018-08-14T15:31:15.673" LastActivityDate="2018-08-14T15:31:15.673" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="939" PostTypeId="2" ParentId="915" CreationDate="2012-08-10T10:37:49.850" Score="9" Body="&lt;p&gt;Electrolysis of carbonates in anhydrous environment can produce either $\ce{CO + O_2}$ or $\ce{C + O_2}$, &lt;sup&gt;[ref]&lt;/sup&gt; depending on conditions. A base, remaining in electrolyzed liquid then can capture carbon dioxide from other sources and be recirculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; L. Massot, P. Chamelot, F. Bouyer, P. Taxil; Electrodeposition of carbon films from molten alkaline fluoride media. &lt;em&gt;Electrochimica Acta&lt;/em&gt;, &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;47&lt;/em&gt; (12), 1949-1957. &lt;a href=&quot;https://doi.org/10.1016/S0013-4686(02)00047-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1016/S0013-4686(02)00047-6&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-04-25T05:52:08.267" LastActivityDate="2018-04-25T05:52:08.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="940" PostTypeId="2" ParentId="905" CreationDate="2012-08-10T10:55:26.923" Score="3" Body="&lt;p&gt;I am writing this as a comment, since all the replies are fine, but I cannot write comments. I hope though that the answer is not fully redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum Mechanics is linear. In the space of a particle you can write the wave function as a linear combination of whatever functions. So you can write MOs (molecular orbitals) as LC (linear combination) of AOs (atomic orbitals) or AOs as LC of MOs. The distinction between these set of basis is physical, not mathematical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should realise that the electronic wave function of a molecule is not an orbital: it &quot;dwells&quot; in a 3N space (not including spin) where N is the number of electrons, and it must by antisymmetrized. Again, since Quantum Mechanics is linear, the LC of wave functions of 3N electrons is an electronic wave function of 3N electrons. You can always approximate one in terms of others that are simpler and indeed this is the basis of all methods beyond Hartree-Fock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the LC of MOs is another MO of a single electron in the molecule, not the wave function of a &quot;molecule&quot; or &quot;super molecule&quot;. The LC process does not move you away from the initial space where the function is defined.&lt;/p&gt;&#xA;" OwnerUserId="504" LastActivityDate="2012-08-10T10:55:26.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="942" PostTypeId="2" ParentId="932" CreationDate="2012-08-10T12:48:08.167" Score="3" Body="&lt;p&gt;Just to make sure that I get your question correctly; the process in question is a chemical reaction that takes place at a temperature and pressure close enough to the reference temperature and pressure (those used for the tabulated values) that any changes to the tabulated values are neglible? (I assume this, as this is what you have done in the enthalpy calculation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, it is probably easier that you would expect - you can use (almost) the same approach as you used to calculate $\Delta H$ as both entropy and enthalpy are state functions. That is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_{rx} S = \sum_i \nu_i S^\circ_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S^\circ_i$ is the tabulated value for each of the components. This is described on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess_law#Extension_to_entropy_and_free_energy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature deviates sufficiently from the reference temperature used then you should adjust the entropy change accordingly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{T_1 \rightarrow T_2} = \int_{T_1}^{T_2}\frac{C_V}{T}dT=C_V\ln(T_2/T_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the last equality holds if $C_V$ can be assumed to stay constant in the interval $T_1 \rightarrow T_2$. A similar adjustment can be made if the pressure, volume etc. changes during the course of the reaction. Again, as the entropy is a state function, it is enough to adjust the total change in entropy for the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps! Please let me know if I am not being sufficiently clear, and I'll try to elaborate things a bit.&lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2012-08-10T12:48:08.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="943" PostTypeId="2" ParentId="910" CreationDate="2012-08-10T14:22:06.540" Score="11" Body="&lt;p&gt;The simplest kinds of soap are just a fatty acid chain with something more ionic on the end - something like the ubiquitous sodium laureth sulphate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXAcz.png&quot; alt=&quot;Sodium laureth sulphate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it's in bar form, it behaves a bit like a solid block of plastic, because chemically speaking, those long chains of carbon are quite a lot like a common plastic such as polythene. These chains attract each other and repel and exclude water, keeping the block together. The sulphate group on the end is hydrophilic, though - it's chemically stable surrounded by water. This means that when you add water, it infiltrates and teases apart the structure of strands, getting to the ionic ends on each chain and forming a &quot;solvation shell&quot; of waters around each one. There's no real chemical reaction here - as you say, it's a physical process, which is why rubbing the soap helps - you're applying shear to the structure and helping to separate the strands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once they're separated from the bulk, the soap forms tiny bubbles called &lt;a href=&quot;http://en.wikipedia.org/wiki/Micelle&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;micelles&quot;&lt;/a&gt;, which have the chains pointed inwards towards each other and the ionic sulphate groups meeting the water at the outside.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-08-10T14:22:06.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="944" PostTypeId="2" ParentId="937" CreationDate="2012-08-10T15:13:51.190" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should try to find good textbook/course about Molecular Orbital theory. As I recall, good university-level textbooks on general/inorganic chemistry dive into this aspect and make analysis for dimers of second row elements. This heuristics, however, are of limited usability: they can be used to explain differences in family of compounds with similar structure, but are virtually useless for predicting geometry in complicated cases. And yet, this is the best you can get without diving into quantum chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is quickly found link on the subject. &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html#valence&quot; rel=&quot;nofollow&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html#valence&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-08-10T15:13:51.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="949" PostTypeId="1" AcceptedAnswerId="950" CreationDate="2012-08-12T18:59:56.650" Score="1" ViewCount="3359" Body="&lt;p&gt;As I was doing my Ap Chem homework, sometimes the question would use Iron(III) and sometimes it would use Ferric.  What's the difference in usage?  Also, what's the difference between Iron(II) and Ferrous?&lt;/p&gt;&#xA;" OwnerUserId="510" LastActivityDate="2012-08-12T19:36:10.443" Title="Iron nomenclature differences between Iron(III) and Ferric" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2012-08-12T20:03:38.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="950" PostTypeId="2" ParentId="949" CreationDate="2012-08-12T19:23:35.030" Score="3" Body="&lt;p&gt;There is no difference, in both the examples you cite.  The forms Iron(II) and Iron(III) are preferred today, versus the older forms ferric and ferrous, respectively, because they more explicitly give the oxidation states of Fe, +2 and +3 in this example.  &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2012-08-12T19:23:35.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="951" PostTypeId="2" ParentId="949" CreationDate="2012-08-12T19:36:10.443" Score="3" Body="&lt;p&gt;The difference in usage is simply a choice.  The usage of the Roman numerals to indicate oxidation state is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Stock_nomenclature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stock nomenclature&lt;/a&gt;, devised by Alfred Stock in 1919.&lt;br /&gt;&#xA;The other method, using the &lt;em&gt;-ic&lt;/em&gt; and &lt;em&gt;-ous&lt;/em&gt; suffixes is derived from Latin.  The &lt;em&gt;-ic&lt;/em&gt; suffix is used for the higher oxidation state of the metal, while the &lt;em&gt;-ous&lt;/em&gt; suffix is used for the lower oxidation state of the metal.&lt;/p&gt;&#xA;&lt;p&gt;However, the use of the Latin suffixes is no longer recommended for indication of oxidation state, but it may be used for metals exhibiting not more than two valences.&lt;/p&gt;&#xA;&lt;p&gt;From IUPAC:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The following systems are in use but not recommended:&lt;/p&gt;&#xA;&lt;p&gt;The system of indicating valence by means of the suffixes -ous and -ic added to the root of the name of the cation may be retained for elements exhibiting not more than two valences.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="446" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-08-12T19:36:10.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="952" PostTypeId="1" AcceptedAnswerId="953" CreationDate="2012-08-12T19:36:13.437" Score="3" ViewCount="4943" Body="&lt;p&gt;What's the difference between bisulfate and &quot;something&quot; hydrogen sulfate?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has there historically been a preference for one over the other?&lt;/p&gt;&#xA;" OwnerUserId="510" LastEditorUserId="73" LastEditDate="2012-08-12T20:08:55.843" LastActivityDate="2012-08-17T18:20:14.937" Title="Bisulfate vs Hydrogen sulfate" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="953" PostTypeId="2" ParentId="952" CreationDate="2012-08-12T19:43:40.947" Score="4" Body="&lt;p&gt;Nope, no rule, and again there is no difference, like your question about the names for iron oxidation states.  The &quot;bi&quot; form is the older form, and the &quot;hydrogen&quot; form is now preferred, again because it more explicitly describes the form of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some additional detail was requested.  When I say &quot;preferred&quot;, I mean the name sanctioned by the IUPAC (International Union of Pure &amp;amp; Applied Chemistry), also known as the &lt;em&gt;systematic&lt;/em&gt; name, so called because the rules to form the name are &quot;systematic&quot;; i.e. consistent and clear. Older names are called &lt;em&gt;common&lt;/em&gt; names, which were in used before the IUPAC.  So &quot;bisulfate&quot; is a common anion name, while &quot;hydrogen sulfate&quot; is systematic.  &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2012-08-17T18:20:14.937" LastActivityDate="2012-08-17T18:20:14.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="956" PostTypeId="2" ParentId="888" CreationDate="2012-08-13T07:09:12.027" Score="4" Body="&lt;p&gt;For our work, we also need to make 100 mM PEP and we neutralize it using KOH. Traditionally, using Na+ based salts isn't ideal for biological assays so K+ salts are better for our applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you're adding the PEP to a different solution, you can buffer the reaction solution. While the Phosphate does have a good amount of buffering capacity, effective concentrations of the molecule will be low so you ultimately won't be adding much.&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2012-08-13T07:09:12.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="957" PostTypeId="2" ParentId="554" CreationDate="2012-08-13T07:14:36.760" Score="5" Body="&lt;p&gt;There is online converter from PubChem CID to 3D PDF/U3D&#xA;at &lt;a href=&quot;http://xemistry.com/&quot; rel=&quot;noreferrer&quot;&gt;http://xemistry.com/&lt;/a&gt; (http://85.214.192.197/pdf3d to be exact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to Asymptote - go to its forum and ask for help.&#xA;3D PDF export is under development, so early adopters (AKA guinea pigs) are welcome.&#xA;Among other things OBJ import is in the works.&#xA;The apparent lack of demand is one of the reasons FOSS support for 3D PDF is slow to emerge - and at lest asking for it in appropriate place is the minimal effort I can think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can inspire the developers of your favorite tools to take code from Asymptote or Intel U3D Sample library to produce 3D PDF directly or develop standalone converters.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2012-08-13T07:14:36.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="958" PostTypeId="1" AcceptedAnswerId="961" CreationDate="2012-08-13T13:41:40.623" Score="5" ViewCount="1501" Body="&lt;h2&gt;Problem:&lt;/h2&gt;&#xA;&lt;p&gt;On my boat I have a marine toilet. It is operated with salt water.&lt;/p&gt;&#xA;&lt;p&gt;The problem is, on the interior of the outflow tubing it accumulates some sort of hard grime. I am looking for a good way to dissolve this grime as it is impossible to replace the tubing.&lt;/p&gt;&#xA;&lt;p&gt;From experience it is known that citric acid work - but not very well. Also it leaves a &amp;quot;sandy&amp;quot; residue that is difficult to flush.&lt;/p&gt;&#xA;&lt;p&gt;Does anybody have an idea of what this &amp;quot;grime&amp;quot; consists of - and thus how best to dissolve it?&lt;/p&gt;&#xA;&lt;h2&gt;Materials:&lt;/h2&gt;&#xA;&lt;p&gt;On hand I have:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;citric acid&lt;/li&gt;&#xA;&lt;li&gt;56% phosphoric acid&lt;/li&gt;&#xA;&lt;li&gt;23% HCl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Materials:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tubing: PVC&lt;/li&gt;&#xA;&lt;li&gt;toilet itself: porcelain&lt;/li&gt;&#xA;&lt;li&gt;pump: ABS or polypropylene with selected parts in acetal resin, with 316 passivated stainless steel fastenings, brass weights and neoprene seals and gaskets&lt;/li&gt;&#xA;&lt;li&gt;valve: grey, harder plastic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I know this quite vague but am hoping somebody can help anyway.&lt;/p&gt;&#xA;" OwnerUserId="514" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-10-30T16:58:41.083" Title="Good acid for cleaning marine toilets" Tags="&lt;everyday-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="961" PostTypeId="2" ParentId="958" CreationDate="2012-08-13T18:01:57.923" Score="3" Body="&lt;p&gt;To my knowledge, marine water contains great amount of $\ce{Ca}$ and $\ce{Mg}$ ions, that gives insoluble carbonates and phosphates. So, most probably, you have mix of carbonates, phosphates and, possibly but unlikely, sulfates of this metals and possibly some organic salts, like oxalates. Carbonates are readily dissolved by $\ce{HCl}$ wich in weak solutions will not dissolve copper, but will slowly dissolve steel. It is more interesting to use $\ce{Na2H2edta}$ . It dissolves $\ce{CaCO_3}$ easily and, to my knowledge, in hight concentration dissolves $\ce{Ca_3(PO_4)_2}$ as well, but not sure in the latter.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="7475" LastEditDate="2015-10-30T16:56:15.957" LastActivityDate="2015-10-30T16:56:15.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="962" PostTypeId="2" ParentId="958" CreationDate="2012-08-13T20:20:49.993" Score="5" Body="&lt;p&gt;A cheaper and easier to find solution than EDTA (permeakra's answer) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid#Cleaning_and_chelating_agent&quot;&gt;citric acid&lt;/a&gt;. Citric acid reacts with metal carbonates (bicarbonates, phosphates, chlorides, etc.) to produce the soluble citrates. The citrate ion keeps the metals in solution by &lt;a href=&quot;http://en.wikipedia.org/wiki/Chelating&quot;&gt;chelation&lt;/a&gt;, the same process that &lt;a href=&quot;http://en.wikipedia.org/wiki/Edta&quot;&gt;EDTA (ethylene diamine tetra acetate)&lt;/a&gt; does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citric acid can be purchased in the grocery store or convenience store as one of the principle components for any sour powdered drink mix (like &lt;a href=&quot;http://en.wikipedia.org/wiki/Tang_%28drink%29&quot;&gt;Tang&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use any substance or cleaner designed to remove hard water deposits or lime scale, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_Lime_Rust&quot;&gt;CLR&lt;/a&gt;, which contains lactic and gluconic acids as chelating agents.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-08-13T20:20:49.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="964" PostTypeId="1" CreationDate="2012-08-14T12:49:09.260" Score="7" ViewCount="1893" Body="&lt;p&gt;From &lt;a href=&quot;http://www.newscientist.com/article/dn18977-innovation-methane-capture-gives-more-bang-for-the-buck.html&quot; rel=&quot;nofollow&quot;&gt;an article in the New Scientist in 2010&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since pre-industrial times, methane levels in the atmosphere have risen from around 715 parts per billion to nearly 1800 ppb, claims Folberth. &quot;If methane could be taken back to pre-industrial levels it would bring about half a degree of cooling and would be equivalent to removing one-third of anthropogenic carbon dioxide,&quot; says Folberth. Furthermore, the cooling would take place within 10 years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Klaus Lackner and others have developed technology such as the &lt;a href=&quot;http://dx.doi.org/10.1021/es201180v&quot; rel=&quot;nofollow&quot;&gt;moisture-swing sorbent&lt;/a&gt; to capture ambient $\ce{CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where might be the chemical options for a material that would absorb or adsorb ambient methane?&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="7475" LastEditDate="2015-10-16T16:49:28.660" LastActivityDate="2016-10-10T15:06:36.967" Title="What chemicals could absorb/adsorb ambient atmospheric methane?" Tags="&lt;environmental-chemistry&gt;&lt;absorption&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="965" PostTypeId="2" ParentId="964" CreationDate="2012-08-14T13:35:43.107" Score="3" Body="&lt;p&gt;There is none. Alkanes are very hard to activate. They require either free radicals, or coordinated unsaturated sites either in the molecule or on surface of solid. But in both cases, the atmosphere contains molecules that will react more readily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the average life-time of methane in atmosphere is quite small, less than 10 years.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="-1" LastEditDate="2016-10-10T15:06:36.967" LastActivityDate="2016-10-10T15:06:36.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="966" PostTypeId="1" CreationDate="2012-08-14T14:43:23.983" Score="4" ViewCount="783" Body="&lt;p&gt;I'm given a flow rate of a gas mixture of $3\ \mathrm{Nm^3/h}$ in &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_cubic_foot&quot; rel=&quot;nofollow&quot;&gt;standard cubic meters&lt;/a&gt; through a cylinder of length $50\ \mathrm{cm}$ and cross section $70\ \mathrm{cm^2}$, can I compute how long a particle will stay inside of the cylinder (which would equal the time the particle can react inside the cylinder)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what if the temperature changes during the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flow rate would be given in volume per time, I had an idea to compute this, but the use of standard cubic meters is baffling to me.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorDisplayName="user7951" LastEditDate="2016-10-09T11:01:51.870" LastActivityDate="2016-10-09T11:01:51.870" Title="How to compute the residence time for given flow rate in standard cubic meters?" Tags="&lt;experimental-chemistry&gt;&lt;units&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="967" PostTypeId="1" AcceptedAnswerId="972" CreationDate="2012-08-14T15:17:21.100" Score="3" ViewCount="5475" Body="&lt;p&gt;I watched a video on the construction of &lt;a href=&quot;http://en.wikipedia.org/wiki/Burj_Khalifa&quot; rel=&quot;nofollow&quot;&gt;Burj Khalifa&lt;/a&gt;, Dubai. The construction engineers answered that they're using a special type of protection known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathodic_protection&quot; rel=&quot;nofollow&quot;&gt;Cathodic protection&lt;/a&gt; to protect from the corrosion due to local ground water. They told a compound named &quot;&lt;a href=&quot;http://www.ritchiewiki.com/wiki/index.php/Burj_Dubai#Preparing_the_Foundation&quot; rel=&quot;nofollow&quot;&gt;Polymer Slurry&lt;/a&gt;&quot; (What's that?). They're using viscous Polymer Slurry as their basic coat to dump the high density &amp;amp; low permeability concrete in order to create a &quot;GOOD FOUNDATION&quot;. What is it made of? (Simply any polymer?) And what does it do? Does it actually repel water? Many organic compounds repel water... Then Why this?&lt;/p&gt;&#xA;" OwnerDisplayName="user491" LastEditorUserId="5078" LastEditorDisplayName="user491" LastEditDate="2015-05-31T05:20:33.797" LastActivityDate="2017-04-10T16:20:03.033" Title="Polymer Slurry used in boring foundation holes?" Tags="&lt;polymers&gt;&lt;viscosity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="969" PostTypeId="1" AcceptedAnswerId="970" CreationDate="2012-08-14T20:15:58.677" Score="6" ViewCount="1205" Body="&lt;p&gt;I am currently reading &lt;a href=&quot;http://pubman.mpdl.mpg.de/pubman/item/escidoc:39376:7/component/escidoc:52433/sengbusch_357_PDFA.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this German paper&lt;/a&gt; on the dissolution of phosphates by acids and chelants.&lt;/p&gt;&#xA;&lt;p&gt;The authors do a &amp;quot;fast-test&amp;quot; of the dissolving power of different acids which goes as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For each test, increasing quantities (10-50ml) of 0.1 n acids are&#xA;added to 100 mg of tricalcium phosphate and agitated for 1 hour. The&#xA;residue is filtrated and weighed.&lt;/p&gt;&#xA;&lt;p&gt;For 0.1 n hydrochloric acid 20ml suffice for complete dissolution of&#xA;100mg tricalcium phosphate. Citric acid [...] shows &amp;gt;90% dissolution&#xA;with 40ml.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not being a chemist I am having trouble interpreting this. What quantity is effectively being controlled by specifying 0.1 n acid? The paper is from the 60s. I thus suspect that n stands for normality but I am not sure.&lt;/p&gt;&#xA;&lt;p&gt;What exactly is being measured?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The speed at which different acids dissolve the material? &lt;em&gt;(Why then vary the volume of acid?)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;The quantity of acid needed to dissolve a given of material? &lt;em&gt;(Measured in what exactly? What does the volume effectively stand for? Moles?)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Something else?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Edit 1&lt;/h2&gt;&#xA;&lt;p&gt;J M's answer makes this much clearer. I still have some problems though:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Regarding normality: Does this ensure that the same concentration of protons is in each solution? I assume this is not the same as &amp;quot;constant PH&amp;quot; as some acids may be strong acids while others may be weak?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If even more citric acid was added, there would still remain 10% residue? Why did they not try with 50ml citric acid? With this test there are two variables that change: volume and residue. Does this not make it difficult to compare the results of different acids? I would understand a test like this: how much acid is needed to dissolve the material 100%. Then one could compare easily.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Regarding weak acids: Once one of the available protons from the citric acid has reacted, should not another previously non-dissociated acid molecule dissociate (to keep the dissociation proportion in equilibrium) and keep everything going to dissolve the material without any residue? Given infinite time, why do not 30ml citric acid also dissolve 10 mg of 100 mg of tricalcium phosphate?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am still not quite sure however what the point of this test is. Should it not be reasonably easy to calculate this using the reaction equations and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of the acids? (See reaction equation below.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thought experiment: If not the acid quantity was the limiting factor but the amount of solution what would limit the reaction: i.e. infinite supply of acid molecules (high acid concentration) and infinite supply of phosphates but finite volume for the solution to take up the reaction product: Would the reaction be limited by the solubility of the reaction products? I assume, in this case a strong acid would be advantageous, right?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Reaction Equation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I understand the papers I read correctly, one can dissolve 15 moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca_3(PO_4)_2}$&lt;/span&gt; with 60 moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; or 20 moles of citric acid (disregarding the fact that it is a weak acid?!):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{ 15 Ca_3(PO_4)_2 + 60HCl -&amp;gt; 15Ca(H_2PO_4)_2 + 30 CaCl_2 }$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{ 15 Ca_3(PO_4)_2 + 20 CitH_3 -&amp;gt; 10 Ca(Cit)_2 + 25 Ca(H_2PO_4)_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="514" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-11-08T03:14:07.937" Title="Comparison of acids to dissolve phosphates" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="970" PostTypeId="2" ParentId="969" CreationDate="2012-08-14T20:37:32.480" Score="8" Body="&lt;p&gt;N, or Normality is a way of standardizing the # of acid equivalencies a particular acid has across multiple acids.  Acids that have only 1 acidic hydrogen for example $\text{HCl}$, the Normality will equal Molarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normality is defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Normality = \frac{c_i}{f_{eq}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $c_i$ is the molar concentration of the acid, and $f_{eq}$ is the equivalency factor.  In the cases of something like $\text{H}_2\text{SO}_4$, the equivalency factor is 0.5 because it is a diprotic (two proton) acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This test also is showing the effect of weak vs strong acids.  $\text{HCl}$ is a strong acid, so it dissociates essentially 100%, while citric is a weak acid.  A weak acid exists in equilibrium between the dissociated acid species, and the neutral species.  This equilibrium is defined by the acid-dissociation constant $K_a$, or it's logarithmic version $pK_a$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For citric acid though, it is &lt;strong&gt;triprotic&lt;/strong&gt;, meaning there are 3 acidic hydrogens, so it's $f_{eq}$ value is $1/3=0.3\overline{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your numbered questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  The speed is being measured to a certain point.  However, the chance is that all of these reactions are complete at this point.  The hour timepoint is to normalize everything to the same time frame.  But as indicated, the reactions are likely complete at this point.  What is really being measured is your second point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  Absolutely, the quantity of acid needed to dissolve the material is being tested.  As you use more and more volume of acid, you are providing larger amounts of acid to the reaction.  This helps it dissolve more of the powder (as there is more acid now present), and also increase the rate of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They standardize the reaction to normality because if they used molarity, 1 mole of citric acid has 3 acid equivalents, while $\text{HCl}$ has only 1 acid equivalent.&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="446" LastEditDate="2012-08-14T21:05:38.357" LastActivityDate="2012-08-14T21:05:38.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="971" PostTypeId="2" ParentId="969" CreationDate="2012-08-14T20:54:06.133" Score="3" Body="&lt;p&gt;Indeed, &lt;em&gt;n&lt;/em&gt; (pretty weird, chemists would use the capital &lt;em&gt;N&lt;/em&gt;) is used for normality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their &lt;em&gt;fast test&lt;/em&gt;, the authors try to dissolve 100 mg (0.322 mmol) of $\ce{Ca3(PO4)2}$ with different acids and buffers at room temperature while shaking for 1 hr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, they state that 20 mL of 0.1N HCl is sufficient to dissolve all the triphosphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the other solvents (solution according to STUBY, etc.), where solution was incomplete, they determine the weight of the dried residue (i.e. how much solid material was left) and try to correlate it with the volume of the solvent used (cf. Fig. 1). From the volume used and the concentration of the citric acid in the stock solution the absolute amount can be calculated.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2012-08-14T20:54:06.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="972" PostTypeId="2" ParentId="967" CreationDate="2012-08-14T22:46:12.700" Score="6" Body="&lt;p&gt;After reading the &lt;a href=&quot;http://www.ritchiewiki.com/wiki/index.php/Burj_Dubai#Preparing_the_Foundation&quot; rel=&quot;nofollow&quot;&gt;link provided in the question&lt;/a&gt;, and doing a little googling, of &quot;polymer slurry&quot; and &quot;foundation&quot;, I found something useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ashland.com/products/sc-50-fluidized-polymer-slurry&quot; rel=&quot;nofollow&quot;&gt;Polymer slurry&lt;/a&gt; seems to be made of one or more hydrophilic &lt;a href=&quot;http://en.wikipedia.org/wiki/Superabsorbent_polymer&quot; rel=&quot;nofollow&quot;&gt;superabsorbent&lt;/a&gt; polymers, such as polyethylene glycol, polyvinyl alcohol, or carboxymethylcellulose that produce a thick viscous slurry when mixed with water. It does not repel water so much as the opposite. It is so hydrophilic that it absorbs water before the water can reach the concrete or the steel. Then this slurry becomes thicker the more water it absorbs so that the rate of diffusion of water through the slurry decreases. Ultimately, this action creates a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogel#Hydrogels&quot; rel=&quot;nofollow&quot;&gt;hydrogel&lt;/a&gt;, which may be quite strong, despite being mostly water, due to extensive hydrogen bond crosslinks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other applications of polymer slurry seem to be in the drilling industry to modulate the viscosity of drilling fluids.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5078" LastEditDate="2015-05-31T04:44:43.477" LastActivityDate="2015-05-31T04:44:43.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="973" PostTypeId="2" ParentId="966" CreationDate="2012-08-15T15:57:59.893" Score="4" Body="&lt;p&gt;The N stands for &quot;norm&quot; and should be avoided (in fact, the German Wiki page lists the Nm³ as wrong abbreviation) as it can create confusion.(qed) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your flow should be written as &quot;3 m³/h under standard condtions&quot;. &#xA;If you can guess German, have a look at the German Wiki page. It is much better than the English one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard conditions vary a bit between different norms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = 0 °C or 15 °C or 20 °C &#xA;p = 1.01325 bar (sometimes 1 bar)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas is supposed to be dry.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-08-15T15:57:59.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="974" PostTypeId="1" AcceptedAnswerId="975" CreationDate="2012-08-15T16:38:41.143" Score="41" ViewCount="4811" Body="&lt;p&gt;While revisiting some of my old notes about the Miller-Urey experiment, I stumbled across the &quot;equation&quot;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electricity + $\ce{CH4~/~ NH3~/~H2O~/~CO}$ = Amino Acids&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This got me thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventionally, why are molecules like $\ce{CH4}$ and $\ce{NH3}$'s molecular formula written differently (in H placement) than others like $\ce{H2O}$ and $\ce{HF}$? Is there a particular reason, or did it just happen to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I realize the $\ce{H4C}$ or $\ce{H3N}$ make perfect sense, especially when drawing out molecules. It's just that these are &lt;em&gt;rarely&lt;/em&gt; used in literature.&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="4231" LastEditDate="2015-02-21T22:27:49.277" LastActivityDate="2015-02-21T22:27:49.277" Title="Why is methane's molecular formula conventionally &quot;CH₄&quot;, while water is &quot;H₂O&quot; (among others)?" Tags="&lt;nomenclature&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="975" PostTypeId="2" ParentId="974" CreationDate="2012-08-15T17:24:10.997" Score="39" Body="&lt;p&gt;The choice of whether to write binary hydrides with the hydrogen first or second depends on whether that hydrogen is considered acidic, with water marking the delineation point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By convention, if a binary hydride is more acidic than water, then they are written in the form $\ce{H}_{\text{n}}\ce{X}$. If a binary hydride is less acidic than water, they are written as $\ce{YH}_{\text{n}}$. It so happens that this changeover neatly divides the periodic table. Binary hydrides from groups 16 and 17 (VIA and VIIA) are written with the hydrogen atoms first: $\ce{H2O,~H2S,~H2Se,~H2Te,~HF,~HCl,~HBr,~HI}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary hydrides from groups 1 (IA) through 15 (VIA) are all written with the hydrogen atoms last, regardless of whether the hydride is ionic $(\ce{LiH,~NaH,~KH,~CaH2})$ or molecular $(\ce{BH3,~CH4,~SiH4,~NH3,~PH3})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This notation is in keeping with that for more complex compounds. If the compound is an acid, the acidic hydrogen atoms are put at the from of the formula, while nonacidic hydrogen atoms are placed in the main part of the formula. For example, the hydrogen atoms in phosphoric acid $(\ce{H3PO4})$ are considered acidic and the hydrogen atoms in octane $(\ce{C8H18})$ are not. Acetic acid $(\ce{HC2H3O2})$ has one hydrogen atom that is considered acidic and three that are not, this the formula $\ce{HC2H3O2}$ is considered more helpful than $\ce{C2H4O2}$, although both are &quot;correct&quot; by different sets of rules.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-08-15T17:24:10.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="976" PostTypeId="1" AcceptedAnswerId="977" CreationDate="2012-08-15T17:53:52.333" Score="6" ViewCount="12830" Body="&lt;p&gt;Can I pipette concentrated sulfuric acid using a plastic pipette tip? Will either the plastic or the acid be affected?&lt;/p&gt;&#xA;" OwnerUserId="520" LastEditorUserId="22" LastEditDate="2012-08-16T03:44:53.747" LastActivityDate="2012-08-16T03:44:53.747" Title="Pipetting strong acid with plastic material" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="977" PostTypeId="2" ParentId="976" CreationDate="2012-08-15T19:14:20.193" Score="10" Body="&lt;p&gt;This depends on type of plastic. You definitely can if plastic is perfluoropolyetylene (Teflon). You probably can if plastic is polyethylene or polypropylene, but both will degrade in long term (polypropylene will do it faster). You should not, if plastic is PET or other polyester, or polystyrene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probably most reliable to find which plastic you dealing with and search for 'chemical resistance chart' for this plastic.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-08-15T19:14:20.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="978" PostTypeId="1" CreationDate="2012-08-16T18:01:05.133" Score="0" ViewCount="408" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hexamethylene_triperoxide_diamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hexamethylene triperoxide diamine&lt;/a&gt;, &quot;Hex&quot; for short, is an explosive organic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the products of this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_6H_{12}N_2O_6 -&amp;gt; ? + Energy}$$ &lt;/p&gt;&#xA;" OwnerUserId="522" LastEditorUserId="16683" LastEditDate="2017-03-22T18:53:11.513" LastActivityDate="2017-03-22T18:53:11.513" Title="Products of HMTD decomposition" Tags="&lt;organic-chemistry&gt;&lt;reference-request&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="980" PostTypeId="2" ParentId="978" CreationDate="2012-08-17T02:56:14.073" Score="6" Body="&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://energetics.chm.uri.edu/system/files/Oxley_ThermochimActa_2002.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; provide two reactions for the thermal decomposition of HMTD, depending on reaction conditions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(at or below 150°C, air or vacuum)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_6H_{12}N_2O_6 -&amp;gt; 3CO_2 + N(CH_3)_3 + NH_3}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(160°C to 180°C, vacuum)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_6H_{12}N_2O_6 -&amp;gt; 3CO + N(CH_3)_3 + 1/2 N_2 + 3/2 H_2O + 3/4 O_2}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Oxley, J.C., J.L. Smith, H. Chen, and Eugene Cioffi.&lt;/strong&gt; &lt;em&gt;“Decomposition of Multi-peroxidic Compounds: Part II. Hexamethylene Triperoxide Diamine (HMTD).”&lt;/em&gt; Thermochimica Acta 388, no. 1–2 (June 18, 2002): 215–225.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="43" LastEditorUserId="35555" LastEditDate="2017-03-22T18:47:58.043" LastActivityDate="2017-03-22T18:47:58.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="983" PostTypeId="1" CreationDate="2012-08-17T09:49:05.710" Score="13" ViewCount="486" Body="&lt;p&gt;I'm a non-chemist attempting to put together a chemistry paper. Up to now I've got my thermodynamic data from various random pdf files found on the web, together with Wikipedia. Now that the paper is nearly finished I need to replace these with values from a single, respectable source. But I've realised I don't know how to find such a source. I have a few chemistry textbooks on my desk, which tend to have tables in the back, but they're not terribly comprehensive and none of them have all the data I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a standard resource for thermodynamic tables? I particularly need the Gibbs energy and enthalpies of formation for gases found in the atmosphere, such as $\ce{O2}$, $\ce{CO2}$, $\ce{CH4}$ and $\ce{H2O(vapour)}$?  In an ideal world this would either be something I can access online or a book that's likely to be found in my institute's rather small library.&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2012-08-20T15:19:17.193" Title="standard reference for thermodynamic properties?" Tags="&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="984" PostTypeId="1" AcceptedAnswerId="1039" CreationDate="2012-08-17T11:26:26.717" Score="8" ViewCount="331" Body="&lt;p&gt;Is there a way to make a liquid, such as a liquid crystal material, transition from cloudy or clear and vice-versa using only electric current?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cloudy, I mean, similar to the effect that cholesterol or &lt;a href=&quot;http://en.wikipedia.org/wiki/4-Cyano-4%27-pentylbiphenyl&quot;&gt;5CB&lt;/a&gt; exhibits, thermally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, where could this be obtained, and how quickly does it switch between states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how else might such an effect be achieved using electric current? I've read about the Kerr effect, but it didn't change opacity so much as refractive index, and it also appeared to use a great deal of electricity. I'm thinking something within the single volts.&lt;/p&gt;&#xA;" OwnerUserId="525" LastActivityDate="2012-09-26T22:08:33.720" Title="Is there a liquid crystal or similar material that gets cloudy when an electric field is applied, but is otherwise clear?" Tags="&lt;optical-properties&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="985" PostTypeId="1" CreationDate="2012-08-17T12:11:51.317" Score="15" ViewCount="1643" Body="&lt;p&gt;Battery leaks was an issue in the 80s and 90s, and since quality and innovation is constantly improving, I thought battery leak might be slowly going away.  But I found that even nowadays, name brand alkaline batteries still leak after a while, even if originally concealed in a tight package of the metal shell.  Is it true that the chemical in an alkaline battery inherently expands in volume over time, and make the alkaline battery leak issue unavoidable?&lt;/p&gt;&#xA;" OwnerUserId="527" LastEditorUserId="7448" LastEditDate="2015-10-30T16:56:36.923" LastActivityDate="2015-10-30T16:56:36.923" Title="Does the chemical in an alkaline battery make battery leaks unavoidable?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="986" PostTypeId="2" ParentId="983" CreationDate="2012-08-17T13:15:35.767" Score="11" Body="&lt;p&gt;A reference that I've used quite a lot (as a student) is &lt;a href=&quot;http://eu.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001348.html&quot;&gt;SI Chemical Data&lt;/a&gt;, which I guess is quite common. It should contain the data you ask for, among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for an online source, the &lt;a href=&quot;http://kinetics.nist.gov/janaf/&quot;&gt;NIST-JANAF Thermochemical Tables&lt;/a&gt; or &lt;a href=&quot;http://webbook.nist.gov/chemistry/form-ser.html&quot;&gt;NIST Chemistry WebBook&lt;/a&gt; should be good places to start. I think both should be respectable enough. The Chemistry WebBook might be a bit more comprehensive, but I suspect that much of the data is from the same source anyway, since NIST is behind both services. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you can use one or the other!&lt;/p&gt;&#xA;" OwnerUserId="475" LastEditorUserId="475" LastEditDate="2012-08-17T13:38:25.770" LastActivityDate="2012-08-17T13:38:25.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="987" PostTypeId="2" ParentId="983" CreationDate="2012-08-17T15:31:26.393" Score="6" Body="&lt;p&gt;What you'll want to look at is the &lt;a href=&quot;http://www.springermaterials.com/&quot;&gt;Landolt-Börnstein&lt;/a&gt; series of handbooks, particularly Group IV (Physical Chemistry). Volume 19 deals with the &quot;Thermodynamic Properties of Inorganic Materials&quot;, which will have the Gibbs energy and formation enthalpy included among the tabulated properties.&lt;/p&gt;&#xA;" OwnerDisplayName="user95" LastActivityDate="2012-08-17T15:31:26.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="988" PostTypeId="2" ParentId="983" CreationDate="2012-08-17T22:06:54.593" Score="2" Body="&lt;p&gt;I think the &lt;a href=&quot;http://www.amazon.de/Handbook-Chemistry-Physics-93rd-Edition/dp/1439880492/ref=sr_1_2?ie=UTF8&amp;amp;qid=1345240538&amp;amp;sr=8-2&quot; rel=&quot;nofollow&quot;&gt;CRC Handbook of Chemistry and Physics&lt;/a&gt; is a very good source to look for all sorts of physical or chemical data and it can be considered as a kind of standard reference. You can also find an Internet version of the book which would be a bit more &quot;lightweight&quot; than the printed book with its more than 2600 pages.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2012-08-17T22:06:54.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="989" PostTypeId="1" AcceptedAnswerId="990" CreationDate="2012-08-18T20:00:28.223" Score="10" ViewCount="4746" Body="&lt;p&gt;I saw &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/32ZGo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the Internet and I'm curious as to what compound is represented by it.&lt;/p&gt;&#xA;" OwnerUserId="528" LastEditorUserId="10185" LastEditDate="2015-10-02T16:20:42.033" LastActivityDate="2015-10-02T16:20:42.033" Title="Interpreting a formula tattoo" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="990" PostTypeId="2" ParentId="989" CreationDate="2012-08-18T20:24:48.473" Score="15" Body="&lt;p&gt;The molecule seen in the tattoo is a protein-based mammalian hormone called &lt;strong&gt;oxytocin&lt;/strong&gt;, commonly called &quot;the love hormone&quot; because it is involved in several aspects of sexual reproduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TXfG8.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/5/55/Oxytocin_with_labels.png/800px-Oxytocin_with_labels.png&quot; rel=&quot;noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2012-08-18T20:24:48.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="993" PostTypeId="1" AcceptedAnswerId="1120" CreationDate="2012-08-20T05:05:58.740" Score="5" ViewCount="1018" Body="&lt;p&gt;I have a few cases of brand-new, nitrile gloves of a type and vendor that &lt;em&gt;used&lt;/em&gt; to be good.  But apparently, the vendor has moved manufacturing to a different country and outsourced customer service to Hell. I can't return the gloves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gloves all have a chemical smell, resembling a burnt transistor with a hint of burnt rubber. If worn for a while, they irritate the skin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this model of glove used to be outstanding, and the cases were not cheap, I'm hoping there's some way to salvage the gloves.  I've tried repeated washings in mild detergent, or just airing them out for 4 weeks. No change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical(s) is/are this likely to be?  How can I tell?&lt;br&gt;&#xA;Is there a way to neutralize this/these compound(s) without otherwise destroying the gloves?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="537" LastEditDate="2012-08-20T09:00:41.487" LastActivityDate="2016-09-27T06:24:39.710" Title="How can I identify and neutralize the chemical leaching out of new nitrile gloves?" Tags="&lt;safety&gt;&lt;solvents&gt;&lt;gloves&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="995" PostTypeId="1" AcceptedAnswerId="996" CreationDate="2012-08-20T12:33:16.403" Score="5" ViewCount="570" Body="&lt;p&gt;I know almost nothing beyond 10th grade chemistry. I'm trying to name a web application, and was wondering what the products are of a reaction between the element Chromium and the gas Ozone (ozonolysis of chromium?).&lt;/p&gt;&#xA;" OwnerUserId="538" LastEditorUserId="16683" LastEditDate="2017-04-06T22:25:38.883" LastActivityDate="2017-04-06T22:27:03.220" Title="What does the reaction of chromium with ozone form?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="996" PostTypeId="2" ParentId="995" CreationDate="2012-08-20T15:04:31.973" Score="10" Body="&lt;p&gt;Ozone is a strong oxidizing agent, even stronger than molecular oxygen. But it is still a donor of oxygen atoms. Thus, depending on conditions, you'll have different chromium oxides. The most stable and common chromium oxide is chromium (III) oxide $\ce{Cr2O3}$. It has numerous nicknames. Other oxides are much harder to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substances named ozonides do exist and in two different flavours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inorganic ozonides are salts of big, inoxidizable cations, like $\ce{Cs}$ and have formulas like $\ce{KO3}$. I'm not sure which cations have ozonides that have been already produced, but to my knowledge the only reasonably easy-to-produce ozonides are $\ce{KO3}$, $\ce{RbO3}$ and $\ce{CsO3}$ by reaction of the corresponding hydroxides with ozone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another type of ozonides is organic ozonides that are quite weird heterocyclic compound, which are product of reaction between alkenes and ozone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a name based on word games, you may consider that chromium is a derivative of an ancient word for 'color' and ozone, as a strong oxidizer, has bleaching and antiseptic properties.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2017-04-06T22:27:03.220" LastActivityDate="2017-04-06T22:27:03.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="997" PostTypeId="2" ParentId="983" CreationDate="2012-08-20T15:19:17.193" Score="6" Body="&lt;p&gt;I too usually refer to the CRC Handbook. Both of my copies (86th ed, 2005-06 and 46th ed, 1965-66) have a section on thermodynamic properties of materials. The CRC Handbook works for the most of the general data I need to look up. There is some field-specific data I need that isn't in the CRC Handbook; I use the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471166286&quot;&gt;Polymer Handbook&lt;/a&gt; for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of actually citing the data in your manuscript, I consider it to be best practice to cite an original research paper if possible. The CRC Handbook and Polymer Handbook cite every datum they contain (sometimes with more than one source). Ideally, you would look up that citation, verify both the datum and the method, and then cite that source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally consider a citation to the reference book itself to be sub-optimal, but acceptable. NIST is a slightly different beast, as they often are the original source of the data they compile. They usually supply a preferred format for citations, so I'm a little surprised that it's not obviously placed on the NIST-JANAF page that @Kjetil Sonerud linked to. Google led me to a &lt;a href=&quot;http://www.nist.gov/manuscript-publication-search.cfm?pub_id=903913&quot;&gt;citation page&lt;/a&gt; that you might find useful, in addition to the other page Kjetil linked. EDIT: The page @J.M. linked to in his comment is exactly what I was thinking of. I can't figure out if that's the same source as the NIST-JANAF tables.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-08-20T15:19:17.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="998" PostTypeId="1" CreationDate="2012-08-21T01:10:15.753" Score="7" ViewCount="3153" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following possesses the highest energy electron?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{Br-}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{Ca^2+}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{Cr+}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{As}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't bromine and arsenic have equally high energy electrons (i.e. in the $\ce{4p}$ subshell)? My book says that arsenic is the correct answer. But I don't see why $\ce{Br^{–}}$ is wrong.&lt;/p&gt;&#xA;" OwnerUserId="545" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-24T10:49:50.913" LastActivityDate="2018-02-24T10:49:50.913" Title="Which of the following possesses the highest energy electron?" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="999" PostTypeId="2" ParentId="998" CreationDate="2012-08-21T05:07:40.343" Score="5" Body="&lt;p&gt;The energy of orbital changes via row in periodic table and falls to the end of row. Because of it in orbitals for particles from the end of row have lesser energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept, however, seems... questionable.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-08-21T05:07:40.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1000" PostTypeId="1" AcceptedAnswerId="1001" CreationDate="2012-08-21T05:49:39.237" Score="16" ViewCount="20014" Body="&lt;p&gt;While performing an inventory of my school's chemical storage closet today, my fellow teacher and I found some sodium tucked away in the back corner of our flammables cabinet.  One container appears to have leaked the oil the sodium was being stored in, and we aren't sure about the second container.  We don't know how long it's been there, but neither of us is going to use it, and, honestly, we don't want it.  How do we dispose of this?  Can someone with a little more chemistry know-how point me in the right direction?  Are we stuck with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event that the administration won't let us dispose of the sodium, can you tell me how to remedy the storage situation?  We don't feel that there is an adequate amount of oil in the container for proper storage of the remaining sodium, and if we have to keep it I want to make sure it is being stored correctly.&lt;/p&gt;&#xA;" OwnerUserId="548" LastActivityDate="2018-04-15T21:21:35.833" Title="How to properly dispose of sodium...or store it properly" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1001" PostTypeId="2" ParentId="1000" CreationDate="2012-08-21T06:43:41.463" Score="20" Body="&lt;p&gt;If you handle sodium, the first thing you need is a way to extinguish it if it catches on fire. You should have a bucket of sand ready to use, and additionally a class D fire extinguisher, if available. I'm assuming you know this, but don't let the sodium come into contact with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You always want the sodium stored in an inert liquid, usually mineral oil is used for that. Organic solvents like n-hexane are also used for short time periods. You should refill the mineral oil immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way to neutralize sodium is to put small pieces in an organic solvent like n-hexane and slowly add either isopropyl alcohol or ethanol to it. Isopropyl alcohol reacts slower, and is therefore a bit safer. When all the sodium is dissolved, you can add water to make sure there is no elementary sodium left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How dangerous this is depends a lot on the scale, it's easy to do if you are only getting rid of a bit of sodium you haven't used in your reaction. I have no experience with neutralizing large amounts of sodium, this method should still work but it might take a long time if you are careful as you'll have to do it in small batches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest, the fact that you asked this question here on the internet indicates to me that you should at least let someone more experienced supervise the procedure.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2012-08-21T06:43:41.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1002" PostTypeId="2" ParentId="750" CreationDate="2012-08-21T13:40:21.280" Score="3" Body="&lt;p&gt;So far the skin questions have been answered nicely. Here's the ingestion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;used in antacids—yet there are warnings about ingestion &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;half of the problem is the acid &lt;em&gt;neutralizing&lt;/em&gt; capacity: your stomach may like to keep up its acidic pH. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other half is of course how that neutralization happens:&lt;br&gt;&#xA;$\ce{CaCO3 + 2 H+ \rightarrow Ca^2+ + H2O + CO2 \uparrow}$&lt;br&gt;&#xA;You'll produce lots of carbondioxide gas in your stomach&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-08-21T13:40:21.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1005" PostTypeId="1" AcceptedAnswerId="1008" CreationDate="2012-08-22T01:58:43.567" Score="2" ViewCount="1980" Body="&lt;p&gt;I'm working through &lt;em&gt;Chemistry - Principles and Practice, 3rd Edition&lt;/em&gt; and I have few complaints so I don't think it's the text. At one point, I was asked to draw the Lewis structure, with formal charge, of SO&lt;sub&gt;3&lt;/sub&gt;. The central sulfur atom has six covalent bonds with the surrounding oxygen atoms. Based on the calculation for formal charge, you'd expect one of +3, but there is none. How am I to know when there is or isn't a formal charge, and why?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="323" LastEditDate="2012-08-22T03:00:55.333" LastActivityDate="2012-08-22T14:57:17.410" Title="Why don't atoms with expanded valence shells have a formal charge?" Tags="&lt;bond&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1006" PostTypeId="2" ParentId="984" CreationDate="2012-08-22T03:58:18.593" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrochromism&quot; rel=&quot;nofollow&quot;&gt;Electrochromic materials&lt;/a&gt; are produced for this purpose and are typically sold as “smart glass”.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="30" LastEditDate="2012-08-24T08:51:32.170" LastActivityDate="2012-08-24T08:51:32.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1008" PostTypeId="2" ParentId="1005" CreationDate="2012-08-22T11:17:08.987" Score="3" Body="&lt;p&gt;Formal charge is the difference between the number of valence electrons that a neutral unbonded atom has and the number of valence electrons assigned to that atom when after it forms bonds in a molecule. It looks like you are making your mistake in assigning the proper number of electrons to the sulfur atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons assigned to an atom are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All nonbonding (lone pair) electrons on that atom in the Lewis Structure.&lt;/li&gt;&#xA;&lt;li&gt;One half of all bonding electrons (one electron for each single bond, two electrons for each double bond, etc.). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8LbS.png&quot; alt=&quot;Lewis Structure of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider water, $\ce{H2O}$. The oxygen atom has two bonds and two lone pairs. Oxygen originally had 6 valence electrons. We assign 6 electrons to it (4 nonbonding and 2 bonding). The formal charge is $6-6=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T1Coa.png&quot; alt=&quot;Lewis structure of hydroxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the hydroxide ion, $\ce{OH-}$. The oxygen in hydroxide has 7 electron assigned to it: 6 nonbonding and 1 bonding. The formal charge is $6-7=-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r1En1.png&quot; alt=&quot;Lewis Structure of nitrate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the nitrogen atom in the nitrate ion, $\ce{NO3-}$. Nitrogen originally has 5 electrons. In the nitrate ion we assign 4 electrons to that nitrogen (1 for each single bond and 2 for the double bond. There are no lone pairs). The formal charge is $5-4=+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfQ7S.png&quot; alt=&quot;Lewis Structure of phosphate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final example, let's look at the P in phosphate, $\ce{PO4^3-} $. Phosphorous original had 5 electrons, like nitrogen. We assign 5 electrons to phosphorous in phosphate (1 for each P-O bond and two for the P=O bond). The formal charge is $5-5-=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many electrons are assigned to the sulfur atom in $\ce{SO3}$? From the question it sounds like you have the correct Lewis Structure. It looks like you are making your mistake in assigning the proper number of electrons based on the number of bonds.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-08-22T14:57:17.410" LastActivityDate="2012-08-22T14:57:17.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1009" PostTypeId="1" AcceptedAnswerId="1021" CreationDate="2012-08-22T12:50:08.390" Score="3" ViewCount="10781" Body="&lt;p&gt;I am denaturing a protein using organic solvent and acid (49:49:2% water:methanol:acetic acid), but I want to maintain the disulfide bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry knowledge isn't good but disulfides are broken under reducing conditions and the reaction also uses free protons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I8KKE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wanted to check if acids could break disulfide bonds without reducing conditions.&lt;/p&gt;&#xA;" OwnerUserId="552" LastEditorUserId="402" LastEditDate="2017-04-01T02:13:25.620" LastActivityDate="2017-04-01T02:13:25.620" Title="Can acidic conditions break disulfide bonds" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1010" PostTypeId="1" CreationDate="2012-08-22T15:23:01.753" Score="6" ViewCount="106" Body="&lt;p&gt;I know that the answer to this question will vary by country, but some aspects must be similar amongst those countries that have authorities overseeing chemical and industrial hygiene. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Msds#National_and_international_requirements&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on the topic is not so good. My understanding for the US is that a MSDS must be correct and complete to the best knowledge of the preparer of the sheet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question was inspired by the commercial for &lt;a href=&quot;http://www.scotch-brite.com/wps/portal/3M/en_US/Scotch-BriteBrand/Scotch-Brite/Products/Product-Catalog/~/Scotch-Brite-Products?N=4294582256%205889599&amp;amp;loc=en_US&amp;amp;plmlblid=1319209952481&amp;amp;rt=r3&quot; rel=&quot;nofollow&quot;&gt;this product&lt;/a&gt; from 3M. You can view the commercial by clicking on the videos link at the bottom of the site. Never mind how much nonsense their claim of &quot;botanically pure ingredients&quot; is, I was intrigued by the MSDS link at the top of the page. I wanted to know what the &quot;botanically pure&quot; substances were. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two versions of this product, with different scents. The MSDS for each is linked: &lt;a href=&quot;http://multimedia.3m.com/mws/mediawebserver?mwsId=SSSSSuUn_zu8l00xMY_BM8215v70k17zHvu9lxtD7SSSSSS--&quot; rel=&quot;nofollow&quot;&gt;Breeze&lt;/a&gt; and &lt;a href=&quot;http://multimedia.3m.com/mws/mediawebserver?mwsId=SSSSSuUn_zu8l00xMY_9O829Nv70k17zHvu9lxtD7SSSSSS--&quot; rel=&quot;nofollow&quot;&gt;Lemongrass&lt;/a&gt;. Section 2: Ingredients lists the same four items for both products:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water 65-80%&lt;/li&gt;&#xA;&lt;li&gt;Cloth 20-30%&lt;/li&gt;&#xA;&lt;li&gt;Sodium lauryl sulfate &amp;lt;3% (surfactant, not a naturally occurring substance)&lt;/li&gt;&#xA;&lt;li&gt;Thyme Oil &amp;lt;0.1% (antibacterial, at least thymol is, and botanical in origin, but certainly not a pure substance)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;No odorant is listed in the ingredients, but Section 3 clearly indicated they have different smells. 3M knows what odorant was added, and if legal requirements exist for completeness, then 3M is earning a finger-shaking and a slap on the wrist. So, again, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there are legal requirement to the correctness and completeness (to the best of the preparer's knowledge) for a MSDS? If not, then why bother with MSDS at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2014-06-12T08:28:30.087" LastActivityDate="2014-06-12T08:28:30.087" Title="What legal guidelines exist for completeness of a Materials Safety Data Sheet?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1011" PostTypeId="2" ParentId="1010" CreationDate="2012-08-22T16:24:28.870" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.osha.gov/html/faq-hazcom.html#faq9&quot;&gt;OSHA's FAQ&lt;/a&gt; says that chemicals that make up less than 1% (0.1% if it is a carcinogen) of a product do not need to be listed on the MSDS. So it looks like 3M could be in compliance if the oderants are present in trace amounts (in that case, I have no idea why they feel the need to include the thyme oil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only relevant thing OSHA seems to say about &lt;a href=&quot;http://www.osha.gov/html/faq-hazcom.html#faq8&quot;&gt;consumer products&lt;/a&gt; specifically is that employers are not required to provide an MSDS for consumer products used in the workplace, as long as the consumer product is used in the same manner as a consumer would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.cpsc.gov/&quot;&gt;Consumer Product Safety Commission&lt;/a&gt; specifically disclaims jurisdiction over chemical products, and the &lt;a href=&quot;http://www.chemsafety.gov/&quot;&gt;Chemical Safety Board&lt;/a&gt; doesn't appear to have anything related to consumer chemical products.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-08-22T16:24:28.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1012" PostTypeId="1" AcceptedAnswerId="20330" CreationDate="2012-08-23T07:50:28.937" Score="21" ViewCount="1982" Body="&lt;p&gt;Are there any heuristics to get the relative absorbtion of left and right circularly polarized light by a molecule from its molecular structure? Is it even possible to predict which polarization is selectively absorbed?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="16683" LastEditDate="2015-12-24T04:49:23.780" LastActivityDate="2015-12-24T04:49:23.780" Title="Is there a simple way to get the circular dichroism of a molecule from its structure?" Tags="&lt;physical-chemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1013" PostTypeId="1" CreationDate="2012-08-23T15:29:38.760" Score="2" ViewCount="298" Body="&lt;p&gt;I'm trying to get the spectral analysis data of the periodic table of elements in the &lt;strong&gt;time domain format&lt;/strong&gt; or find a site that has the data in the &lt;strong&gt;time domain format&lt;/strong&gt;.  All the sites&#xA;I know have the data in the frequency domain which is great but &lt;strong&gt;I need to do my own fft analysis starting from the time domain because the frequency domain does not include the phase angle data, which I also need to use for my calculations/animations&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where I can get this data or can someone send me the spectral analysis data for sodium and potassium in the time domain? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="554" LastEditorUserId="39240" LastEditDate="2017-03-23T05:56:36.373" LastActivityDate="2017-03-23T05:56:36.373" Title="Spectral analysis of periodic table of elements in time domain" Tags="&lt;computational-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1014" PostTypeId="1" CreationDate="2012-08-23T16:13:33.703" Score="16" ViewCount="4036" Body="&lt;p&gt;Specifically, I'm curious if there are any programs which come close to &lt;a href=&quot;http://gaussian.com/&quot;&gt;Gaussian&lt;/a&gt; in breadth but also take advantage of things such as GPU processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_modeling_on_GPU&quot;&gt;page on wikipedia&lt;/a&gt; that goes through several options but I'd like to get feedback from actual users.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorUserId="30" LastEditDate="2012-08-24T07:17:59.640" LastActivityDate="2015-10-29T09:25:15.343" Title="Modern alternatives to Gaussian" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" ClosedDate="2015-10-29T15:52:25.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="1016" PostTypeId="2" ParentId="1014" CreationDate="2012-08-24T20:44:58.153" Score="7" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;&quot;Specifically, I'm curious if there are any programs which come close to Gaussian in breadth&quot;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not really, as far as I can tell. Gaussian is a massive work by many, many people over many, many years, and the reason it's still one of the most popular packages, as far as I can tell, is that it does &lt;em&gt;most things&lt;/em&gt; (notably excluding non-GTOs), even if it's not necessarily the fastest or the best at the things it does in every case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;&quot;but also take advantage of things such as GPU processing.&quot;&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double no. As LordStryker says above, GPU programming is pretty new, and most of the resources I've seen for it are C-based, while a significant chunk of scientific development still takes place in Fortran. I suspect library-based encapsulation of GPU-running numerical routines may improve this over time (I mean, by introducing people to the idea of getting speedup by using the GPUs more easily), but I wouldn't hold my breath. It'll be interesting to see what hybrid chips with multiple types of core, or the Xeon Phis will be able to do, though.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-08-24T20:44:58.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1017" PostTypeId="2" ParentId="1014" CreationDate="2012-08-24T23:25:56.977" Score="3" Body="&lt;p&gt;I've always heard &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;nofollow&quot;&gt;GAMESS&lt;/a&gt; was the canonical alternative to Gaussian.  I don't know much about either, though, aside from the fact that GAMESS doesn't use GPU processing either and is free as in beer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how these programs work on a low level, but matrix-vector operations usually don't get much of a speed boost from using GPUs. If these sorts of packages need lots of matrix-vector or vector-vector operations, GPU processing would probably not speed things up very much. &lt;/p&gt;&#xA;" OwnerUserId="47" LastActivityDate="2012-08-24T23:25:56.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1018" PostTypeId="2" ParentId="851" CreationDate="2012-08-25T00:51:53.353" Score="2" Body="&lt;p&gt;$\ce{P2O5}$ is an acidic oxide while $\ce{KOH}$ is a base. So if you want to dry substances which are acidic you must use $\ce{P2O5}$, while if you want to dry a substance that is basic in nature, you have to use $\ce{KOH}$.&lt;/p&gt;&#xA;" OwnerUserId="519" LastEditorUserId="17368" LastEditDate="2016-12-23T08:25:15.527" LastActivityDate="2016-12-23T08:25:15.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1019" PostTypeId="2" ParentId="851" CreationDate="2012-08-25T06:59:11.730" Score="3" Body="&lt;p&gt;Usage of drying agent is ruled by four factors: cost, effectiveness, reactions it can provoke in substance being dried, level of discomfort in working process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{P2O5}$ is much more effective, $\ce{KOH}$ is cheaper and easier to handle. $\ce{P2O5}$ is also acidic and $\ce{KOH}$ is basic. There are substances that die in acidic environment (acetales) and those that die in basic environment (halogen-anhydrides). &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="17368" LastEditDate="2016-12-23T08:24:31.570" LastActivityDate="2016-12-23T08:24:31.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1020" PostTypeId="1" CreationDate="2012-08-25T15:03:35.000" Score="8" ViewCount="1312" Body="&lt;p&gt;I'd like to preface this with a declaration of my understanding that this is not necessarily possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew of a chemical clock reaction where the period before the first change (the induction period?) is very long (between 3 hours and a day) and the change is à la Landolt or Old Nassau reactions (i.e they cross a concentration threshold to create sudden color change).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Question&lt;/strong&gt;:&#xA;Is there any remote possibility that a reaction like this can happen in pigments or dyes once they are absorbed by a cloth? &lt;/p&gt;&#xA;" OwnerUserId="562" LastEditorUserId="16683" LastEditDate="2017-04-06T22:28:09.347" LastActivityDate="2017-04-06T22:28:09.347" Title="Very slow chemical clock reaction" Tags="&lt;kinetics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1021" PostTypeId="2" ParentId="1009" CreationDate="2012-08-25T16:38:37.763" Score="5" Body="&lt;p&gt;No, acidic conditions (i.e. low pH) shouldn't be enough by itself to break a disulfide bond. The main reaction that breaks a S–S bond is &lt;a href=&quot;http://en.wikipedia.org/wiki/Disulfide_bond#Scission_of_disulfides&quot; rel=&quot;noreferrer&quot;&gt;cleavage by reduction&lt;/a&gt;. This is in a way similar to peroxides, which are stable even at very low pH (see e.g. mixtures of &lt;a href=&quot;http://en.wikipedia.org/wiki/Piranha_solution&quot; rel=&quot;noreferrer&quot;&gt;sulfuric acid and hydrogen peroxide&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-25T16:38:37.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1022" PostTypeId="2" ParentId="659" CreationDate="2012-08-25T19:24:40.407" Score="2" Body="&lt;p&gt;The &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;activity&lt;/a&gt;&lt;/em&gt; of a chemical species is a key concept in thermodynamics of solutions. While the thermodynamic potential is the “true thermodynamic variable”, i.e. the one that plays a primary role in statistical mechanics, chemists do tend to like activity (which is univocally related to chemical potential) because it is close to another quantity which has clear physical meaning: &lt;em&gt;concentration&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, relating the activity (or chemical potential) of a species to its concentration in a mixture is impossible to do in a general way. Models have been developed to provide approximations, however, the best known being the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_solution&quot; rel=&quot;nofollow&quot;&gt;ideal solution&lt;/a&gt;&lt;/em&gt;, where activity equals molarity (concentration). Your “molecular mixing model” appears to be a model of ideal solid solution. However, one cannot expect this approximation to hold in case of strongly non-ideal solutions, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_%28chemistry%29#Ionic_solutions&quot; rel=&quot;nofollow&quot;&gt;ionic solutions&lt;/a&gt; are prominent in that category. The “ionic mixing model” you describe thus appears to be another such approximation, more suited to these ions in a solid solution. I have never encountered it before, but I found a &lt;a href=&quot;http://www.sci.uidaho.edu/geol555/PDF/lectures_pdf/GEOL555_Topic_6.pdf&quot; rel=&quot;nofollow&quot;&gt;series of lecture slides&lt;/a&gt; where it seems to be very neatly introduced (see around slide 31).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I cannot fully address the question, I hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-25T19:24:40.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1023" PostTypeId="2" ParentId="1012" CreationDate="2012-08-26T07:06:38.490" Score="6" Body="&lt;p&gt;Predicting the circular dichroism of a molecule ab initio is possible, but pretty difficult. The only way I know of doing that is advanced quantum chemistry calculations, for example in the time-dependent density functional theory (TD-DFT) approach. You can find an example of its use on a small organic molecule in &lt;a href=&quot;http://dx.doi.org/10.1002/chir.20388&quot; rel=&quot;noreferrer&quot;&gt;this publication&lt;/a&gt;, and a mention of this on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vibrational_circular_dichroism&quot; rel=&quot;noreferrer&quot;&gt;VCD&lt;/a&gt; page of Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding heuristics, I don't think there are any… a fully characterization of the spectrum by quantum chemistry calculations seem unavoidable due to the complexity of the problem.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-26T07:06:38.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1024" PostTypeId="2" ParentId="340" CreationDate="2012-08-26T12:58:10.673" Score="4" Body="&lt;p&gt;Toilet bowls are made of ceramics which can contain feldspar. Toilet bowls also often have a feldspar glaze. There is a study that disolved Cr(VI) can be removed by feldspar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Studies on the removal of Cr(VI) from waste-water by feldspar&lt;/strong&gt;&#xA;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jctb.280530204/abstract&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/jctb.280530204/abstract&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IMA currently recognizes 90 Cr bearing minerals. 16 minerals if you only count those that contain also Si. Feldspar can be either K, Na or Ca-Feldspar. Of those 16 minerals 6 contain Na, 2 K and 5 Ca. Reading those papers might even narrow the choices down to a couple of minerals that can form in standard-toilet-bowl-conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mineral database that lets you select minerals by their chemistry:&#xA;&lt;a href=&quot;http://rruff.info/ima/&quot; rel=&quot;nofollow&quot;&gt;http://rruff.info/ima/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that unless you are willing to donate your toilet bowl to science that the answer won't get any more precise than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Cu would be the favorite choice for a blue color mineral (Azurite, ...). There is also a study about the removal of copper by feldspar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental studies of the interaction of aqueous metal cations with mineral substrates: Lead, cadmium, and copper with perthitic feldspar, muscovite, and biotite&#xA;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0016703797001178&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/S0016703797001178&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again there are 603 Cu-minerals. Cu+Si=32 minerals. With the common Feldspar cations your left with about 10 each. I don't think the minerals could be recognized visually. The favorite method would be x-ray diffraction of a powder sample of the blue stuff.&lt;/p&gt;&#xA;" OwnerUserId="351" LastEditorUserId="351" LastEditDate="2012-08-26T20:21:08.533" LastActivityDate="2012-08-26T20:21:08.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1025" PostTypeId="2" ParentId="835" CreationDate="2012-08-26T20:31:11.777" Score="2" Body="&lt;p&gt;This is veering into physics, but the reason for your initial observation - that an ideal capacitor has a square cyclic voltammogram, as shown in Chris's diagram - is that the charge on an ideal capacitor is proportional to the voltage across it, with the constant of proportionality being the capacitance: $q = CV$. Differentiating with respect to time, we see that the current across the capacitor is $I = \frac{\mathrm{d}q}{\mathrm{d}t} = C\frac{\mathrm{d}V}{\mathrm{d}t}$. Now when you put a sawtooth wave across the capacitor, $\frac{\mathrm{d}V}{\mathrm{d}t}$ flips back and forth between a positive and a negative value, hence so does $I$, giving the characteristic square shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge flowing for some other reason - such as a redox reaction - will indeed give the peaks shown in the second diagram.&lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2012-08-26T20:31:11.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1026" PostTypeId="1" CreationDate="2012-08-26T21:22:43.353" Score="5" ViewCount="208" Body="&lt;p&gt;I'm wondering how to properly perturb a quantum mechanical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to express the return to equilibrium of such a perturbed state, where the perturbation is in the Hamiltonian. The unperturbed Hamiltonian is $H_{\text{u}}$ and the perturbed Hamiltonian is $H_{\text{p}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can compute the eigensystem for both $H$ and thus find the relevant energy spectra and eigenfunctions, and can add the time-dependence factor $e^{-iEt}$, but I do not know how to generate the final system from its component, isolated systems. I would guess an inner product to be of use here, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What theory/mathematical methods do I need to learn to perform such simulations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Something I want to model is a molecule in a polarized state, that is, with an asymmetric electron distribution.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2012-08-27T04:04:12.697" LastActivityDate="2012-08-27T10:11:30.500" Title="Modeling perturbations of a quantum mechanical system" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1028" PostTypeId="1" AcceptedAnswerId="1029" CreationDate="2012-08-26T22:07:42.750" Score="7" ViewCount="2335" Body="&lt;p&gt;When an electron is added to cesium, why is energy released?&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2019-07-30T00:53:09.297" Title="Why does cesium have a positive electron affinity?" Tags="&lt;electron-affinity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1029" PostTypeId="2" ParentId="1028" CreationDate="2012-08-26T22:22:07.733" Score="2" Body="&lt;p&gt;&lt;strong&gt;All elements have a positive (first) &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity&quot; rel=&quot;nofollow&quot;&gt;electron affinity&lt;/a&gt;, except noble gases&lt;/strong&gt;, for which it has not been measured conclusively, but might be either slightly positive or slightly negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can more clearly see the reason why electron affinity is positive if you think of the opposite process: detaching an electron from the negative ion X&lt;sup&gt;–&lt;/sup&gt; to form neutral atom X. Because the anion X&lt;sup&gt;–&lt;/sup&gt; is stable, it takes energy to take away the electron.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-26T22:22:07.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1030" PostTypeId="1" AcceptedAnswerId="1037" CreationDate="2012-08-27T03:47:59.360" Score="17" ViewCount="34524" Body="&lt;p&gt;I was looking at the &lt;a href=&quot;http://www.webanswers.com/education/high-school/what-is-the-chemical-symbol-of-sulfuric-acid-23fae0&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical structure of $\ce{H2SO4}$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/izuvo.png&quot; alt=&quot;H2SO4 chemical structure&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/c1Mup.jpg&quot; alt=&quot;H2SO4 ball-and-stick model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I would have expected this molecule to be square planar in accordance with $p^2d^2$ or $sp^2d$ hybridization, but instead it is shown to be in a tetrahedral geometry consistent with $sp^3$ hybridization. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that an alternative way of asking this question is: What is the hybridization of the sulfur atom in $\ce{H2SO4}$?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="11367" LastEditDate="2016-11-20T04:35:24.333" LastActivityDate="2020-05-10T23:48:36.420" Title="What is the chemical structure / hybridization of sulfuric acid?" Tags="&lt;inorganic-chemistry&gt;&lt;structural-formula&gt;&lt;molecular-orbital-theory&gt;&lt;vsepr-theory&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1031" PostTypeId="1" AcceptedAnswerId="1032" CreationDate="2012-08-27T04:02:52.097" Score="10" ViewCount="8684" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Compound 1 was hydrothermally synthesized under autogenous pressure&lt;/em&gt;&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've seen this term used a few times, but with no real explanation of what it means forthcoming. I suspect it refers to the synthesis being performed at equilibrium pressure inside a sealed container (the aforementioned synthesis is performed in an autoclave at 160 C). Is my intuition correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[1]&lt;/strong&gt; Sun, et al., CrystEngComm, &lt;strong&gt;(2012)&lt;/strong&gt;, 14, 3982-3988&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-08-27T07:59:11.417" Title="What is 'autogenous pressure'?" Tags="&lt;physical-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1032" PostTypeId="2" ParentId="1031" CreationDate="2012-08-27T07:59:11.417" Score="4" Body="&lt;p&gt;&lt;em&gt;Autogenous pressure&lt;/em&gt; synthesis is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvothermal_synthesis&quot; rel=&quot;nofollow&quot;&gt;solvothermal synthesis&lt;/a&gt; (also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrothermal_synthesis&quot; rel=&quot;nofollow&quot;&gt;“hydrothermal synthesis”&lt;/a&gt; if the solvent is water). In most publications I could find, the two terms (“autogenous pressure” and “—thermal synthesis”) are used together, making it clear what it is: you seal your solution in a stainless steel autoclave, and heat it up. The pressure thus generated is called &lt;em&gt;&lt;a href=&quot;http://www.uio.no/studier/emner/matnat/kjemi/KJM5100/h07/undervisningsmateriale/15KJM5100_2007_solvothermal_a.pdf&quot; rel=&quot;nofollow&quot;&gt;autogenous pressure&lt;/a&gt;&lt;/em&gt; because you don't apply it externally, but it simply derives from the closed nature of the container and the $(\rho,P,T)$ relationship for the fluid (its equation of state).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-27T07:59:11.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1033" PostTypeId="2" ParentId="1026" CreationDate="2012-08-27T10:11:30.500" Score="5" Body="&lt;p&gt;I don't understand exactly what is your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start in an eigenstate of $H_p$, say $\varphi_j$ and then your Hamiltonian changes like $H(t) = H_u + f(t) (H_p - H_u)$ where $f(0) = 1$ and $f(T) = 0$ -$T$ being the final time of the perturbation- then in order to follow the evolution of $\psi(t)$ all you need to do is to solve the time dependent Schroedinger equation &#xA;$$i\hbar \partial \psi(t) / \partial t = H(t) \psi(t)$$ &#xA;where $\psi(0) = \varphi_j$. There are many numerical procedures to do so. For instance if you are interested, as it seems, in following the dynamics in the basis of the eigenstates of $H_u$, then you expand $\psi(t) = \sum_k a_k(t)&#xA;\phi_k$, where $\phi_k$ are eigenstates of $H_u$, and you solve a set of ordinary differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $f(t)$ is very fast ($T &amp;lt;&amp;lt; (\Delta E)^{-1}$, where $\Delta E$ is a characteristic energy of the system, say the energy difference between adjacent eigenstates) then you can use the sudden approximation; if $f(t)$ is very slow you can use the adiabatic approximation, if $H_p - H_u$ is small you can use first order perturbation theory, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case it is important to realise that in Quantum Mechanics there is &lt;strong&gt;no relaxation&lt;/strong&gt; unless you couple your initial state to a continuum and there is &lt;strong&gt;no return to equilibrium&lt;/strong&gt; unless you do the same (couple to continuum) and work with the proper formalism that allows you to define the (thermal) equilibrium, which are the density matrices, unless by equilibrium you are referring to the initial state. Again, in ordinary Quantum Mechanics $\psi(t)$ will oscillate and will not relax to the initial state $\varphi_j$ (or $\phi_0$); at most, it will return to the initial state as part of an oscillation, called revival.&lt;/p&gt;&#xA;" OwnerUserId="567" LastActivityDate="2012-08-27T10:11:30.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1034" PostTypeId="2" ParentId="1028" CreationDate="2012-08-27T11:19:58.460" Score="1" Body="&lt;p&gt;I guess&#xA;Isaac’s main problem arises from the usual chemical &#xA;experience dealing about ions in water. Then of &#xA;course Cs- is not a real reaction product. &#xA;Electron affinities deal about reactions of atoms&#xA;and electrons in vacuum! &#xA;The cesium atom &quot;well&quot; to plunge in is not very deep,&#xA;but better than nothing and there is no better &#xA;alternative for the electron  in a vacuum. &#xA;The true reason is the atomic orbital energy for &#xA;an additional electron on cesium, not easy to calculate. &lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="30" LastEditDate="2012-08-27T11:46:20.337" LastActivityDate="2012-08-27T11:46:20.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1035" PostTypeId="2" ParentId="860" CreationDate="2012-08-27T11:36:07.047" Score="3" Body="&lt;p&gt;Some hints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if you mean that &quot;what exactly&quot; in the original sense of the words,&#xA;then the answer is: nobody knows and will ever know. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are content with some more vague answer, then &#xA;&lt;strong&gt;look at the pyrolysis chemistry&lt;/strong&gt; occuring from 150 °C &#xA;up with fats, proteins and starch/sugars, and all that &#xA;at once. This is a vast field of chemistry, and all &#xA;is gives to a trained chemist is a idea of the structures &#xA;of that coating. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bonds to iron will not occur, first because on cast iron and &#xA;on steel pans there is always a layer of iron oxide, &#xA;on cast iron  you find a glaze formed by fusion of the&#xA;molds finish layer (often containing graphite and clay) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditDate="2012-08-27T12:27:26.493" LastActivityDate="2012-08-27T12:27:26.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1036" PostTypeId="2" ParentId="238" CreationDate="2012-08-27T12:03:08.610" Score="-1" Body="&lt;p&gt;The following is to clarify the degeneracy issue debated in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothesis: Electron orbitals are quantum wave functions (the truth value is left to reader).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text below is an example adabted from Liboff: introductory quantum mechanics, pages 119 - 121.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is in degenerate state of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi = \frac{3\phi_2+4\phi_9}{\sqrt{25}}     (5.17)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probability of finding energy $E_n$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(E_n) = \langle\phi_n|\phi_n\rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(E_2) = \langle\phi_2|\phi_2\rangle = \frac{9}{25}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(E_9) = \langle\phi_9|\phi_9\rangle = \frac{16}{25}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(E_(2+9)) = \langle\phi_2|\phi_9\rangle = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(E_(a+b)) = \langle\phi_a|\phi_b\rangle = 0$, if $a \neq b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is a direct quotation: &quot;In an ensemble of 2500 identical one-dimensional boxes, each containing an identical particle in the same state $\phi(x,0)$ given by (5.17), measurement of $E$ at $t=0$ finds about 900 particles to have energy $E_2 = 4 E_1$ and 1600 particles to have energy $E_9 = 81E_1$.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the above states that the particle can be only in one state, not in many states and not in any linear combination of states. The &lt;strong&gt;ensemble average&lt;/strong&gt; can have a linear combination of states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, adapt this to orbitals. Electron can be only in one orbital. On average, electrons of bulk material are on linear combination of orbitals.&lt;/p&gt;&#xA;" OwnerUserId="144" LastEditorUserId="144" LastEditDate="2012-08-27T12:08:29.490" LastActivityDate="2012-08-27T12:08:29.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1037" PostTypeId="2" ParentId="1030" CreationDate="2012-08-27T13:11:37.050" Score="11" Body="&lt;p&gt;It is much easier to explain it on simpler example – $\ce{O3}$ molecule. It has structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, central atom has hybridization state $sp^2$. One bond here is normal covalent bond and another bond is dative: an electron pair is donated onto vacant orbital of $\ce{O}$ atom with all electrons paired. in $\ce{H2SO4}$ molecule to bonds are simple covalent ($\ce{S-OH}$ ones) and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved brocken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little more interesting example is $\ce{XeF2}$ molecule, where three-atom four-electron bond $\ce{F-Xe-F}$ are formed, that can be think of as reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take course of quantum chemistry, it should be enough. However, I'll recommend to search for &quot;MO LCAO model&quot;: it is quite simple and very useful. It is also often used in advanced chemistry books and articles.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="15410" LastEditDate="2016-11-19T22:16:36.440" LastActivityDate="2016-11-19T22:16:36.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1039" PostTypeId="2" ParentId="984" CreationDate="2012-08-27T17:52:54.463" Score="2" Body="&lt;p&gt;I would expect this to exist. One major class of stimuli-responsive polymers is thermosensitive polymers. These have a sudden drop in solubility with an increase in temperature above a certain point. Depending on the particular polymer and the concentration, this transition can be accompanied by a clouding of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://dx.doi.org/10.1016/j.progpolymsci.2009.10.008&quot; rel=&quot;nofollow&quot;&gt;review&lt;/a&gt; of stimuli-responsive polymers from Brent Sumerlin's group contains a section about electric field sensitive polymers. These are usually in the context of the field causing a mechanical response from the polymer. Sumerlin's review cites &lt;a href=&quot;http://dx.doi.org/10.1016/S0022-2860%2800%2900564-0&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; from a Hungarian group discussing polymer gels that change shape under electric field. You would want the electric field to affect the solubility of the polymer. Sumerlin's review doesn't appear to directly mention anything exactly like that, but I would expect it to be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not as familiar with ionic liquids, but &lt;a href=&quot;http://dx.doi.org/10.1039/C1SM05027C&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; uses an electric field to turn a hydrophobic ionic liquid hydrophilic. That's what happens with the thermosensitive polymers, and more what I was thinking of.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-08-27T17:52:54.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1040" PostTypeId="2" ParentId="921" CreationDate="2012-08-27T19:00:02.093" Score="8" Body="&lt;p&gt;I wondered about that for a long time, but Ben Norris and BigGenius have the truth of it: the question is about the bond dipole moment, otherwise the answer would make sense. Here's why.&lt;/p&gt;&#xA;&lt;h2&gt;Molecular dipoles&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_trioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sulfur trioxide&lt;/a&gt; has planar trigonal geometry, and its molecular dipole is zero. Similarly, molecular &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; has the same geometry as &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;, i.e. linear, and thus its molecular dipole is zero.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7}$&lt;/span&gt; is bent at its central oxygen atom:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vVxaa.png&quot; alt=&quot;Cl2O7&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;and thus has a non-zero dipole moment. While &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt; generally occurs as its “dimer” &lt;span class=&quot;math-container&quot;&gt;$\ce{P4O10}$&lt;/span&gt;, it is possible to write a molecular structure for &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt; which then strongly resembles that of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7}$&lt;/span&gt;, being bent at the central O atom. It thus also has non-zero molecular dipole.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, ordering between &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt;: because the P is less electronegative than Cl, the partial charges borne by the P atoms will be larger than those of the Cl atoms, and thus I expect the molecular dipole of &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt; to be greater than that of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2O7}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Bond dipoles&lt;/h2&gt;&#xA;&lt;p&gt;If the question is talking about the &lt;span class=&quot;math-container&quot;&gt;$\ce{X–O}$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$\ce{X=O}$&lt;/span&gt;) bond dipole in the series where X = S, Si, P or Cl, it becomes easier to understand. For one thing, because you're asked about &lt;span class=&quot;math-container&quot;&gt;$\ce{X–O}$&lt;/span&gt; bonds where X spans a row of the periodic table, which is typical of this sort of problems. The answer would then be that bond dipole moment get bigger when the electronegativity difference of X and O becomes larger (the sharing of electrons gets more asymmetric). This leads to the answer you gave.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2021-04-12T17:33:00.337" LastActivityDate="2021-04-12T17:33:00.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1041" PostTypeId="2" ParentId="594" CreationDate="2012-08-27T19:29:10.517" Score="40" Body="&lt;p&gt;Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)&lt;/p&gt;&#xA;&lt;p&gt;A few of the things worthy of note are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The bond energy found by the authors for this fourth bond is &lt;span class=&quot;math-container&quot;&gt;$\pu{13.2 kcal/mol}$&lt;/span&gt;, &lt;em&gt;i.e.&lt;/em&gt; about &lt;span class=&quot;math-container&quot;&gt;$\pu{55 kJ/mol}$&lt;/span&gt;. This is very weak for a covalent bond. You can compare it to &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_bond#Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;other values here&lt;/a&gt;, or to the energies of the first three bonds in triple-bonded carbon, which are respectively &lt;span class=&quot;math-container&quot;&gt;$348, 266$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\pu{225 kJ/mol}$&lt;/span&gt;. This fourth bond is actually even weaker than the strongest of hydrogen bonds (&lt;span class=&quot;math-container&quot;&gt;$\ce{F\bond{...}H–F}$&lt;/span&gt;, at &lt;span class=&quot;math-container&quot;&gt;$\pu{160 kJ/mol}$&lt;/span&gt;). Another point of view on this article could thus be: &lt;em&gt;“valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the &lt;em&gt;&lt;a href=&quot;https://doi.org/10.1038/nchem.1263&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature Chemistry&lt;/a&gt;&lt;/em&gt; paper) which have found a bond order between 3 and 4 for molecular dicarbon.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is &lt;span class=&quot;math-container&quot;&gt;$\pu{6.32 eV}$&lt;/span&gt;, &lt;em&gt;i.e.&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{609 kJ/mol}$&lt;/span&gt;. This latter value is much more in line with typical double bonds, &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_bond#Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;reported at an average of &lt;span class=&quot;math-container&quot;&gt;$\pu{614 kJ/mol}$&lt;/span&gt;&lt;/a&gt;. This is still a bit of a misery to me…&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastEditorUserId="61578" LastEditDate="2020-06-13T14:09:52.537" LastActivityDate="2020-06-13T14:09:52.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1045" PostTypeId="1" AcceptedAnswerId="1046" CreationDate="2012-08-28T11:05:28.187" Score="7" ViewCount="37758" Body="&lt;p&gt;I'm trying to find the oxidation number of $ \ce N$ in $ \ce{(NH4)2SO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is supposed to be $-3$ but I can't figure out the logic behind this.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="5026" LastEditDate="2018-02-19T03:36:17.157" LastActivityDate="2018-02-19T03:36:17.157" Title="Oxidation number for nitrogen in ammonium sulfate" Tags="&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2018-02-19T13:45:40.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="1046" PostTypeId="2" ParentId="1045" CreationDate="2012-08-28T11:14:05.353" Score="11" Body="&lt;p&gt;Ammonium sulfate is an ionic compound, composed of two ammonium ions ($\ce{NH_{4}^{+}}$) and one sulfate ion ($\ce{SO_{4}^{2-}}$). As such, you can neglect the sulfate entirely from your consideration of the oxidation state of the nitrogen, as well as treating each ammonium ion as a separate entity. Given the total charge on the ammonium cation, it should be easy to see why the nitrogen has the specified oxidation state.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-08-28T11:14:05.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1048" PostTypeId="1" AcceptedAnswerId="1049" CreationDate="2012-08-28T14:26:18.317" Score="8" ViewCount="174" Body="&lt;p&gt;I am reading a paper[1] that contains the line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most of the $\ce{H2}$ in the Earth's mantle escaped to space early in the Earth's history; consequently, the overwhelming majority of the abiotic geochemical reactions are based on acid/base chemistry, i.e., transfers of protons without electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The connection between the loss of $\ce{H2}$ and acid/base chemistry is not clear to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would the loss of $\ce{H2}$ to space make acid/base chemistry the default?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Falkowski, P. G., T. Fenchel, and E. F. Delong. (2008) The microbial engines that drive Earth's biogeochemical cycles. Science 320:1034-1038.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="115" LastEditorUserId="5591" LastEditDate="2015-11-13T19:02:47.870" LastActivityDate="2015-11-13T19:02:47.870" Title="Why would the loss of H₂ to space mean that the Earth's geochemistry would be based on acid/base reactions?" Tags="&lt;acid-base&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1049" PostTypeId="2" ParentId="1048" CreationDate="2012-08-28T14:57:02.613" Score="3" Body="&lt;p&gt;The statement you cite contains a reference to &lt;a href=&quot;http://www.atmos.washington.edu/~davidc/papers_mine/Catling2005-EPSL.pdf&quot; rel=&quot;nofollow&quot;&gt;“How Earth’s atmosphere evolved to an oxic state: A status report”&lt;/a&gt;, D. C. Catling and M. W. Claire, &lt;em&gt;Earth and Planetary Science Letters&lt;/em&gt; 237 (2005) 1–20. This paper in turns give some more details on this assertion, which I'll summarize below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all turns around the question of how reducing the prebiotic atmosphere was. The paper I cited above contains this quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;before life existed the atmosphere contained mainly $\ce{N2}$, with negligible $\ce{O2}$, probably more $\ce{CO2}$ than today, and only minor levels of reducing gases such as $\ce{H2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in the absence of widely available gases such as oxidative $\ce{O2}$ or reducing $\ce{H2}$, oxygen was mainly present as $\ce{H2O}$ and $\ce{CO2}$. This means redox reactions are not big players in the overall chemical scheme of the periodic atmosphere, and acide-base reactions dominate (by default).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-25T15:48:03.540" LastActivityDate="2015-02-25T15:48:03.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1050" PostTypeId="1" CreationDate="2012-08-28T15:55:30.213" Score="1" ViewCount="105" Body="&lt;p&gt;Let us consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Method X fails at describing intermolecular non-covalent interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Method X excels at describing molecules of type Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Method X &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;em&gt;&lt;/strong&gt;?&lt;/em&gt;&lt;strong&gt;&lt;em&gt;_&lt;/em&gt;__&lt;/strong&gt; at describing molecules of type Y_1 containing intramolecular non-covalent interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an existing general consensus on number 3? I haven't found an example in the literature yet which would highlight a clear answer to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT for clarification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: An non-covalent intermolecular interaction would be the interaction of two stacked benzene molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: A molecule of type Y could be like an amino acid. This molecule has well defined, strong covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 3: An organic molecule with a non-covalent interaction occurring between two functional groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... regarding #3, would we expect the method to do well at characterizing a molecule in Example 3, would we expect the method to fail (as it did in Example 1) or is there no general consensus on the matter which is essentially the same as saying, &quot;Just have to try it and see!&quot;?&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2012-08-28T18:25:57.843" LastActivityDate="2012-08-29T02:13:36.790" Title="Methods: Does failing at intermolecular properties mean the same for intramolecular properties?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1051" PostTypeId="1" AcceptedAnswerId="1053" CreationDate="2012-08-28T18:49:36.753" Score="3" ViewCount="6296" Body="&lt;p&gt;When I was a teenager, I played with a chemistry set. I tried adding potassium permanganate, alum, and sodium hydrogen sulphite (not sodium hydrogen sulphate) to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon adding the third ingredient, the water would rapidly turn clear, and a brown precipitate would form. After filtering this out, I allowed the water to evaporate, which left a white crystalline deposit. All three ingredients were definitely needed in order for the reaction to happen, which is why I remember it, since none of the other reactions I knew about needed three reactants. (This was a long time ago, so there's a very slight possibility I've mis-remembered the ingredients, but I'm 99% sure they're right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time, neither my teachers nor my parents could help me work out what kind of reaction it might be, so out of sheer curiosity, I thought I'd ask here. I'm no chemist, so simple explanations would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="160" LastEditDate="2012-08-30T12:04:07.383" LastActivityDate="2016-10-13T16:29:57.817" Title="Reaction between potassium permanganate, alum, and sodium hydrogen sulfite" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1052" PostTypeId="2" ParentId="998" CreationDate="2012-08-28T19:21:09.427" Score="10" Body="&lt;p&gt;My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - &lt;em&gt;if&lt;/em&gt; we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Webelements gives the electron affinity of $\ce{Br}$ (&lt;em&gt;i.e.&lt;/em&gt;, the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, &lt;strong&gt;experimentally, the bromide ion has the highest energy electron&lt;/strong&gt; (which therefore requires the lowest energy to remove it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorDisplayName="user15489" LastEditDate="2015-06-13T08:16:59.793" LastActivityDate="2015-06-13T08:16:59.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1053" PostTypeId="2" ParentId="1051" CreationDate="2012-08-28T19:45:43.440" Score="5" Body="&lt;p&gt;The fact that the solution turned clear is a pretty good guarantee that this was a redox reaction: the permanganate oxidised some other species, and in turn was reduced, probably to manganese dioxide, the brown precipitate you noticed. Unfortunately, neither of the other reactants you mention are reducing agents!&lt;/p&gt;&#xA;&lt;p&gt;Do you think it's possible that the third reagent was actually sodium hydrogen sulphite, NaHSO&lt;span class=&quot;math-container&quot;&gt;$_3$&lt;/span&gt;? If so, this would make a lot more sense: the (hydrogen)sulfite ion would have been oxidised to (hydrogen)sulfate. This reaction requires an acidic medium, and the reason the alum (potassium aluminium sulfate) was necessary was to lower the pH.&lt;/p&gt;&#xA;&lt;p&gt;These individual steps have the following equations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Al&lt;span class=&quot;math-container&quot;&gt;$^{3+}$&lt;/span&gt; + 4 H&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;O &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; [Al(OH)&lt;span class=&quot;math-container&quot;&gt;$_4$&lt;/span&gt;]&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt; + 4 H&lt;span class=&quot;math-container&quot;&gt;$^+$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;MnO&lt;span class=&quot;math-container&quot;&gt;$_4^-$&lt;/span&gt; + 4 H&lt;span class=&quot;math-container&quot;&gt;$^+$&lt;/span&gt; + 3 &lt;span class=&quot;math-container&quot;&gt;$e^-$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; MnO&lt;span class=&quot;math-container&quot;&gt;$_2\downarrow$&lt;/span&gt; + 2 H&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;O&lt;/p&gt;&#xA;&lt;p&gt;HSO&lt;span class=&quot;math-container&quot;&gt;$_3^-$&lt;/span&gt; + H&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;O &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; HSO&lt;span class=&quot;math-container&quot;&gt;$_4^-$&lt;/span&gt; + 2 H&lt;span class=&quot;math-container&quot;&gt;$^+$&lt;/span&gt; + 2 &lt;span class=&quot;math-container&quot;&gt;$e^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(note that the last two are half-equations -- they can’t occur on their own, but I’ve written them separately for clarity) giving an overall reaction of&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Al&lt;span class=&quot;math-container&quot;&gt;$^{3+}$&lt;/span&gt;  + 2 MnO&lt;span class=&quot;math-container&quot;&gt;$_4^-$&lt;/span&gt; + 3 HSO&lt;span class=&quot;math-container&quot;&gt;$_3^-$&lt;/span&gt; + 3 H&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;O &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; [Al(OH)&lt;span class=&quot;math-container&quot;&gt;$_4$&lt;/span&gt;]&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt; + 2 MnO&lt;span class=&quot;math-container&quot;&gt;$_2\downarrow$&lt;/span&gt; + 3 HSO&lt;span class=&quot;math-container&quot;&gt;$_4^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The white crystalline deposit in this case would just be alum again, more or less. You have some sodium in your solution now, from the sodium hydrogensulfite, so you'd get some sodium aluminium sulfate as well, but the only anion in your solution will still be (hydrogen)sulfate, and my guess is that with aluminium this will crystallise out as the sulfate.&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-08-28T19:45:43.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1054" PostTypeId="2" ParentId="1050" CreationDate="2012-08-29T02:13:36.790" Score="5" Body="&lt;p&gt;I think the distinction that you are trying to draw is (as you have hinted) better expressed in terms of covalent and noncovalent interactions, rather than inter- versus intramolecular interactions. A technique that fails to correctly describe noncovalent interactions will fail whether or not these interactions are between two moieties on the same molecule, or on different molecules. The usefulness of a technique that fails noncovalent interactions is contingent on whether or not those interactions are significant to the particular properties of a molecule that you are investigating.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-08-29T02:13:36.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1055" PostTypeId="1" AcceptedAnswerId="1058" CreationDate="2012-08-29T03:38:39.170" Score="0" ViewCount="565" Body="&lt;p&gt;I was wondering what the product will be in the following reaction:&#xA;$$\ce{H3COOCH3 -&amp;gt;[\ce{H+}]} \:?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about $\ce{CH3COOH}$  being a product but I do not know the mechanism.&lt;/p&gt;&#xA;" OwnerDisplayName="user570" LastEditorUserId="7448" LastEditDate="2015-10-16T16:57:02.153" LastActivityDate="2015-10-16T16:57:02.153" Title="Product/(s) of the following reaction" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1056" PostTypeId="1" AcceptedAnswerId="1057" CreationDate="2012-08-29T07:01:38.123" Score="13" ViewCount="15243" Body="&lt;p&gt;In my self-study, I recently came across the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Choose the solute of each pair that would be more soluble in hexane ($\ce{C6H14}$). Explain your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) $\ce{CH3(CH2)10OH}$ or $\ce{CH3(CH2)2OH}$ ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undecanol is more soluble in hexane because it is apparently less polar than propanol. Further Internet searches revealed that alcohols decrease in polarity as the chain length increases (assuming a basis in an alkane; don't know if this rule generalizes which I guess is a sub-question) but it is still not clear why. None of the sources I found explained this. The way I imagine things, the carbon-hydrogen bonds should add a bunch of zero vectors to the the O–H vector that both molecules share, giving the same polarity in both cases. But this is apparently wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is undecanol less polar than propanol then? Does it have something to do with a more advanced bonding theory?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="30" LastEditDate="2012-08-29T08:37:00.627" LastActivityDate="2012-08-29T11:06:30.850" Title="Why are alcohols with longer chains less polar?" Tags="&lt;solvents&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1057" PostTypeId="2" ParentId="1056" CreationDate="2012-08-29T08:36:16.383" Score="13" Body="&lt;p&gt;If you're looking at the overall polarity of the solvent, you're not only consider the value of the individual dipole moment on a single molecule, but also their “density”, i.e. the number of these dipoles per unit of volume. You are right that the dipole moment of an individual linear-chain alcohol will, in first approximation, be independent of the chain length. However, in a given volume, you'll fit fewer of those long molecules, and the polarity of the overall solvent is thus lower.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-29T08:36:16.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1058" PostTypeId="2" ParentId="1055" CreationDate="2012-08-29T08:45:53.140" Score="3" Body="&lt;p&gt;&lt;strong&gt;If you're sure that the initial formula is $\ce{H3COOCH3}$&lt;/strong&gt; (and that's a big &lt;strong&gt;if&lt;/strong&gt;), then I disagree with Richard's answer in comments. &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C690028&amp;amp;Units=SI&quot; rel=&quot;nofollow&quot;&gt;$\ce{H3COOCH3}$ is dimethyl peroxide&lt;/a&gt;. While peroxides are not the most stable of molecules, and decompose rapidly (often by oxidizing other compounds), they do not spontaneously transform in acid conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the formula were $\ce{H3C–COOCH3}$ (note the added C), then yes, it's be &lt;a href=&quot;http://www.chemguide.co.uk/physical/catalysis/hydrolyse.html&quot; rel=&quot;nofollow&quot;&gt;acid-catalysed ester hydrolysis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-29T08:45:53.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1059" PostTypeId="1" AcceptedAnswerId="1061" CreationDate="2012-08-29T10:49:19.617" Score="12" ViewCount="1012" Body="&lt;p&gt;Could a filter be used to isolate salt from sea water? Could we say that the water molecules are smaller than the molecules of salt dissolved in the water. I have no relation to chemistry or physics, so pardon the question if it's too naive.&lt;/p&gt;&#xA;" OwnerUserId="572" LastEditorUserId="21964" LastEditDate="2016-07-27T22:36:46.270" LastActivityDate="2020-08-24T08:22:06.597" Title="Isolating salt from sea water using filters" Tags="&lt;water&gt;&lt;purification&gt;&lt;green-chemistry&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1060" PostTypeId="2" ParentId="1056" CreationDate="2012-08-29T11:06:30.850" Score="2" Body="&lt;p&gt;In the context of a single molecule, the distance between the poles is inversely proportional to the polarity. The farther apart they are, the less polar the molecule is if they have the same electronegative potentials. &lt;a href=&quot;http://www.elmhurst.edu/~chm/vchembook/214organicchain.html&quot; rel=&quot;nofollow&quot;&gt;This Link&lt;/a&gt; explains it well.&lt;/p&gt;&#xA;" OwnerUserId="91" LastActivityDate="2012-08-29T11:06:30.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1061" PostTypeId="2" ParentId="1059" CreationDate="2012-08-29T11:07:25.420" Score="13" Body="&lt;p&gt;Yes, this is possible and is a process called &lt;a href=&quot;http://en.wikipedia.org/wiki/Reverse_osmosis&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;reverse osmosis&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt;. It is based on a membrane which lets water pass but not the salt ions (and other water contaminants). From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The membranes used for reverse osmosis have a dense layer in the polymer matrix -- either the skin of an asymmetric membrane or an interfacially polymerized layer within a thin-film-composite membrane -- where the separation occurs. In most cases, the membrane is designed to allow only water to pass through this dense layer, while preventing the passage of solutes (such as salt ions). This process requires that a high pressure be exerted on the high concentration side of the membrane, usually 2–17 bar (30–250 psi) for fresh and brackish water, and 40–82 bar (600–1200 psi) for seawater&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Water desalination features prominently among applications of reverse osmosis:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The largest and most important application of reverse osmosis is to the separation of pure water from seawater and brackish waters; seawater or brackish water is pressurized against one surface of the membrane, causing transport of salt-depleted water across the membrane and emergence of potable drinking water from the low-pressure side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hLMPv.png&quot; alt=&quot;Reverse osmosis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-29T11:07:25.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1062" PostTypeId="1" AcceptedAnswerId="1063" CreationDate="2012-08-29T15:35:48.883" Score="4" ViewCount="2951" Body="&lt;p&gt;What does the moment of inertia for a molecule mean? How is it useful? Most importantly, how would I go about solving it for a triatomic molecule?&lt;/p&gt;&#xA;" OwnerUserId="573" LastEditorUserId="470" LastEditDate="2012-09-13T18:12:35.457" LastActivityDate="2012-09-13T18:12:35.457" Title="Moment of inertia for a triatomic molecule" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1063" PostTypeId="2" ParentId="1062" CreationDate="2012-08-29T18:24:12.260" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_of_inertia&quot;&gt;moment of inertia&lt;/a&gt; is defined for a molecule as it would be for any other solid: you consider the nuclei as points with mass, ignore the mass of the electrons, and you can then calculate either the scalar moment of inertia or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_of_inertia#Moment_of_inertia_tensor&quot;&gt;tensor of the same name&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia of a molecule is involved in the quantification of its rotational movements. Of the top of my head, the two areas of chemistry where this plays a big role are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rotational_spectroscopy&quot;&gt;rotational spectroscopy&lt;/a&gt;: the rotation of a molecule is quantized, and the value that the angular momentum can take are related to the moment of inertia of the molecule.&lt;/li&gt;&#xA;&lt;li&gt;the rotational entropy of a molecule is linked through statistical mechanics to its inertia tensor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-29T18:24:12.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1065" PostTypeId="1" AcceptedAnswerId="1067" CreationDate="2012-08-30T16:15:48.893" Score="6" ViewCount="14125" Body="&lt;p&gt;I tried to find info on tin toxicity but there is not much.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_poisoning&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; here says that it is not very poisonous; but there could be cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it dangerous to say have a coffee boiled in a tin container? I know that coffee has some acidity, so it can probably react with tin. &lt;/p&gt;&#xA;" OwnerUserId="577" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T15:16:30.767" LastActivityDate="2020-07-27T03:18:27.107" Title="Is tin and tinware toxic?" Tags="&lt;inorganic-chemistry&gt;&lt;toxicity&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1066" PostTypeId="2" ParentId="876" CreationDate="2012-08-31T07:42:45.813" Score="3" Body="&lt;p&gt;In the context of radioactivity (which might be what you meant by “radiation”), the &lt;a href=&quot;http://goldbook.iupac.org/H02717.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;half-life&lt;/em&gt; of a radionuclide&lt;/a&gt; is a well-defined term and commonly used. &lt;em&gt;“Radiation time”&lt;/em&gt;, on the other hand, is not used to describe radionuclides.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-31T07:42:45.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1067" PostTypeId="2" ParentId="1065" CreationDate="2012-08-31T10:11:33.003" Score="12" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_poisoning&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; you quoted clearly says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The low toxicity is relevant to the widespread use of tin in dinnerware and canned food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moreover, tin cookware, which used to be very common in earlier centuries, is still sold for outdoor cooking purposes. It is thus without a doubt deemed safe to use under current health regulations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a side note, the pH of coffee is around 5, which is only weakly acidic (same as unpolluted rain, and less acidic than tomato juice or beer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.floridacoastalcooking.com/2012/03/how-important-is-ph-in-body.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yiIhf.gif&quot; alt=&quot;Food pH scale&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-08-31T10:11:33.003" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1069" PostTypeId="1" AcceptedAnswerId="1071" CreationDate="2012-09-02T11:39:26.323" Score="1" ViewCount="2073" Body="&lt;p&gt;If we know the properties of two elements e.g Oxygen and Hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the following properties&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sound speed&lt;/li&gt;&#xA;&lt;li&gt;boiling point&lt;/li&gt;&#xA;&lt;li&gt;melting point&lt;/li&gt;&#xA;&lt;li&gt;refractive index &lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;given that information, is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments? i.e deduce its boiling point&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="275" LastEditDate="2012-09-02T16:21:12.613" LastActivityDate="2012-09-14T10:33:09.693" Title="deduce chemical properties of a compound from elements" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1070" PostTypeId="2" ParentId="1069" CreationDate="2012-09-02T12:58:55.657" Score="1" Body="&lt;p&gt;No. Calculating the physical properties of materials is very, very difficult and can't simply be done from the elemental composition. &lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2012-09-02T12:58:55.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1071" PostTypeId="2" ParentId="1069" CreationDate="2012-09-02T16:20:56.917" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Not yet, assuming you mean extrapolating such information from the molecule's structure&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical and chemical properties are hard enough to extrapolate from a compound's structure, let along its composition. The entire field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Theoretical_chemistry&quot; rel=&quot;nofollow&quot;&gt;theoretical chemistry&lt;/a&gt; has been working on predicting the structure-property relationship for several decades. Great strides have been made, but accurate physical property prediction has not been achieved (though I would love to be proven wrong). In most cases, very accurate computation of properties of all but the smallest molecules takes more computing time than you would need to do the experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question implies an interest in deriving such properties from the formula. As Canageek said, this is not possible, given the phenomenon of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomerism&quot; rel=&quot;nofollow&quot;&gt;isomerism&lt;/a&gt;. One example I just used with my class is that there are 22 known molecules sharing the formula $\ce{C6H12}$, which is a relatively simple formula. Two of the more common are&lt;a href=&quot;http://en.wikipedia.org/wiki/1-hexene&quot; rel=&quot;nofollow&quot;&gt;1-hexene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclohexane&quot; rel=&quot;nofollow&quot;&gt;cyclohexane&lt;/a&gt;, which you can see have very different structures and properties (chemical and physical).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-09-14T10:33:09.693" LastActivityDate="2012-09-14T10:33:09.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1072" PostTypeId="1" CreationDate="2012-09-02T16:39:22.637" Score="7" ViewCount="721" Body="&lt;p&gt;Are there theoretical reasons for or against the existence of materials that will &lt;em&gt;not&lt;/em&gt; become solid at some fixed, sufficiently large temperature when enough pressure is applied? Or is this even the normal state of affairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual phase diagrams typically only give a finite portion of the thermodynamical state space, probably as more cannot be experimentally tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But theory might tell us what happens, one way or another. &lt;/p&gt;&#xA;" OwnerUserId="258" LastActivityDate="2012-10-03T01:46:50.670" Title="What is known about the phase of matter at very high temperature and enough pressure?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1073" PostTypeId="1" AcceptedAnswerId="1076" CreationDate="2012-09-02T17:13:49.813" Score="6" ViewCount="161" Body="&lt;p&gt;So we are told that a unimolecular elementary reaction has a rate law of $k[\text{A}]$ where a termolecular reaction with three unique reagents, $A$, $B$ and $C$ has a rate law of $k[\text{A}][\text{B}][\text{C}]$. Now, other things being equal, and assuming for the sake of argument that the initial concentrations are all greater than 1M, this means that the termolecular elementary reaction is &lt;em&gt;more likely&lt;/em&gt; at the outset. But that doesn't make much sense. There may be some statistical mechanics principle that's lost on me but it seems more likely that A bumps into A than, A, B and C come together all at once. So I don't quite follow this reasoning. Is it the case that $k$ is generally quite smaller in these termolecular elementary reactions?&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2012-09-02T20:50:22.533" Title="Why do collisions in elementary reactions of higher-orders appear to be more likely?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1074" PostTypeId="1" AcceptedAnswerId="1099" CreationDate="2012-09-02T17:53:15.553" Score="2" ViewCount="184" Body="&lt;p&gt;I wonder if the  words on my T-shirt are removable? If yes, what kind of (chemical or physical) ways are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They look like paints on the cloth, as the letter &quot;C&quot; has become blurred in the way shown in the picture over the time when I was wearing it. But I admit that I don't know anything about how words are printed on clothes generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MvinJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="19616" LastEditDate="2015-10-30T17:56:18.233" LastActivityDate="2015-10-30T17:56:18.233" Title="Are words on my T-shirt removable?" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-10-30T19:35:32.967" ContentLicense="CC BY-SA 3.0" />
  <row Id="1075" PostTypeId="2" ParentId="1073" CreationDate="2012-09-02T17:57:20.637" Score="4" Body="&lt;p&gt;That is exactly the case.  My professor is famous (well within our group at least) for often saying &quot;A rate is a rate constant times a concentration&quot;  You can flood your reaction with a high concentration of reactants, but if the rate constant is 0, nothing happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a termolecular reaction (which usually unlikely), the rate constant is going to be very small.&lt;/p&gt;&#xA;" OwnerUserId="446" LastActivityDate="2012-09-02T17:57:20.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1076" PostTypeId="2" ParentId="1073" CreationDate="2012-09-02T20:50:22.533" Score="5" Body="&lt;p&gt;I second J M’s answer: the rate does not depend only on concentrations, but the rate constant is key. However, I will add two more comments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, you cannot compare values of rate constants for reactions of different orders… they don't have the same dimension (unit). So, you cannot say things like “rate constants of third order reactions are typically smaller than those of first-order reactions”.&lt;/li&gt;&#xA;&lt;li&gt;However, you can say (and it is true) that &lt;em&gt;reactions rates&lt;/em&gt; of high-order elementary reactions tend to be small, because of the low probability of the molecules crossing path.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-02T20:50:22.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1077" PostTypeId="2" ParentId="1030" CreationDate="2012-09-02T22:08:05.067" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess that an alternative way of asking this question is what is the&#xA;  hybridization of H2SO4?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A better way to ask this might be: &quot;What is the hybridization of the sulfur atom in Hydrogen Sulfate?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sulfur atom has a bond with four other atoms in this molecule. Because that the electrons try to stay as far away from each other as possible according to Valence Shell Electron Pair Repulsion Theory (VSEPR), the other atoms will seperate as far as possible, which results in bonds that are 109.5 degrees apart from one another. If this were a square planar arrangement, the bonds would only be 90 degrees apart, which is not as satisfactory as the tetrahedral arrangement would provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my book, if the number of effective pairs is 4, the arrangement is tetrahedral, and therefore the hybridization required is $sp^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&#xA;Zumdahl, Zumdahl, Chemistry: An Atoms First Approach&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ISBN-13: 978-0840065322&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This certainly answers the question, but why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has to do with molecular orbitals, and how they hold only a certain number of electrons, and they prefer to spread themselves out evenly because in the quantum world, electrons tend to prefer the lowest possible energy state. Once the sigma bond is occupied, which is the first orbital referred to as the s orbital, the electrons end up occupying the p orbitals, and when electrons are shared through these orbitals they are called pi bonds, they actually form a whole new orbital, a molecular orbital, which allows the sharing of electrons between two atoms. Let me refer you to &lt;a href=&quot;http://www.khanacademy.org/science/organic-chemistry/v/sp3-hybridized-orbitals-and-sigma-bonds&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; website to help visually show you what is happening.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-03T02:39:16.260" LastActivityDate="2012-09-03T02:39:16.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1078" PostTypeId="2" ParentId="1074" CreationDate="2012-09-02T22:25:25.543" Score="1" Body="&lt;p&gt;As you say, it appears to be a kind of paint? So perhaps if you get a razor or something sharp you could try scraping and it will probably come off eventually, that would be one physical way. On the other hand if you want to go crazy try putting some paint thinner on there and see what happens, but don't blame me if you ruin your shirt experimenting on it, I have never used paint thinner before but I am guessing it consists of a non-polar compound and dissolves similar compounds (paint) so that it might be washed off with water. This is all hypothesis of mine, I have no idea what will happen except for what a razor can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I believe that &lt;a href=&quot;http://www.youtube.com/watch?v=NQevyIy8hzs&quot; rel=&quot;nofollow&quot;&gt;Mr. Bean&lt;/a&gt; may have already demonstrated this experiment for us.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-02T22:33:34.190" LastActivityDate="2012-09-02T22:33:34.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1079" PostTypeId="1" AcceptedAnswerId="1081" CreationDate="2012-09-02T22:46:19.760" Score="1" ViewCount="859" Body="&lt;p&gt;For yellow rings on shirts caused by skin grease or sweat:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some recommended to use an &quot;Oxyclean&quot; stain remover or stain remover stick&#xA;for cleaning.&lt;/li&gt;&#xA;&lt;li&gt;Some recommended to use dishwashing detergent to brush and wash the&#xA;stains, saying &quot;dish washing liquid is designed to remove grease, and&#xA;why buy another product (that doesn't work very well -- I've tried)&#xA;when you already have one that does?&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From a chemistry perspective, do dishwashing detergent and stain remover powder/stick have similar chemical composition and a similar ability to remove collar yellow rings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original dirty collar&#xA;&lt;img src=&quot;https://i.stack.imgur.com/a3r6a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After dishwashing detergent and laundry detergent&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3mJRp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="865" LastEditDate="2015-10-30T16:58:02.290" LastActivityDate="2015-10-30T16:58:02.290" Title="Do dishwashing detergent and stain remover powder/stick have the similar ability to remove collar rings?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1080" PostTypeId="2" ParentId="1065" CreationDate="2012-09-02T22:54:57.723" Score="3" Body="&lt;p&gt;I don't believe it is the metal itself that should concern you, but when it reacts with something in your food it might lose its metallic bond and therefore allow you to ingest tiny amounts of tin via a molecule that includes tin itself. This is perhaps what can potentially happen when food is cooked inside a tin container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had a conversation about this last year in general-chemistry when the OCHEM instructor was talking about using metals as containers for dishes during recitation for our lab on cell potential. She was saying that the acids in food might create undesirable compounds when reacting with the metal containers, but that was probably a guess. She was saying something about how pop cans, for example, are made of aluminum, but the insides are lined with a plastic to prevent any undesired reactions (presumably). But then again she also went on to say that the FDA would not allow plastic containers (2 Litre pop bottles for example) until they finally passed off the plastic as a &quot;food additive.&quot; Interesting stuff...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, I always say that too much of anything is bad for you, and eating food from tin containers might be one of those ways of getting tin toxicity, but this is all speculation here so take it with a grain of salt, or a grain of tin whichever you prefer.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="402" LastEditDate="2017-08-24T22:45:42.327" LastActivityDate="2017-08-24T22:45:42.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1081" PostTypeId="2" ParentId="1079" CreationDate="2012-09-02T23:11:24.410" Score="2" Body="&lt;p&gt;Detergents seem to consist of both polar and non-polar compounds. Probably because they need the non-polar part of the mixture to dissolve the grease, and the polar compound part of the mixture to get washed off with the water so there is not much left over on your shirt. The ideal solution would probably depend on just how bad the stain is (how much grease needs to be dissolved). If your home solutions did not work the first time, try to look for signs if it did anything at all, and it might not be something that will remove all the grease the first go around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no experience using detergents or removing stains, but I am going off a lot of speculation here as to how the process might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;Like dissolves like&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this article if there are lots of Mg, Ca, or Iron ions in your water (likely) then it could reduce the effectiveness of soaps. A good question to ask yourself is whether or not your water contains many ions such as these. &lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-02T23:21:07.860" LastActivityDate="2012-09-02T23:21:07.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1083" PostTypeId="1" AcceptedAnswerId="1085" CreationDate="2012-09-03T04:13:34.443" Score="8" ViewCount="3232" Body="&lt;p&gt;Take an ice cube for example. Heat is applied in a closed container until it is vaporized completely. Will the molecule's size be larger (on average)? Is there a substance that you know of that has odd behaviour such as changing molecular size due to temperature change? Obviously there would be more pressure in the case of $H_2O$. Have you heard of any substances that are exceptions to this rule? I would love to hear about it if you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this question hasn't quite boggled your mind or you are having trouble thinking of an exception, let's say that an average diamond at about room temperature (something with entropy, ie: not a perfect lattice) and we say that this is a single macro-molecule because every carbon atom present in the diamond shares a bond with another carbon in the diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretend we heat said diamond up a bit or let it sit for awhile. Are covalent bonds being broken and reformed as time goes on? If we heated it up enough without breaking the lattice completely, would the diamond itself, the molecule as a whole, increase in size at all? Or would the bonds just be broken more often and reformed, and the size change would be so negligible that it could hardly be considered an increase in volume?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="9961" LastEditDate="2015-10-23T17:15:17.970" LastActivityDate="2016-10-11T15:03:42.960" Title="Is there any scenario in which the size of a molecule increases due to an increase in temperature?" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1085" PostTypeId="2" ParentId="1083" CreationDate="2012-09-03T07:57:50.073" Score="12" Body="&lt;p&gt;First, let me remark to “the size of a molecule” is not particularly well-defined. I assume in the following that you are interested about bond lengths (which are averages over time of distances between bonded atoms), as seems to be the case in your example. In the simplest case, you can consider an isolated diatomic molecule (think: gas phase N&lt;sub&gt;2&lt;/sub&gt;), and we suppose we're not going to temperature high enough to ionize it or otherwise break its bond. Because the interatomic potential is asymmetric:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mHj1k.png&quot; alt=&quot;Morse potential&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see that, &lt;strong&gt;as the temperature increases&lt;/strong&gt; and the system is allowed to explore more of the potential energy curve (climb higher up the well), &lt;strong&gt;the average bond length will increase&lt;/strong&gt; and not stay constant. So, for a single bond in an isolated molecule, it is expected that the bond lenghtens with temperature.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, one case where the overall “size expansion” can be clearly studied is that of crystalline materials. In solids, the evolution of volume as a function of temperature is named &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;thermal expansion&lt;/strong&gt;&lt;/a&gt;, and is characterized by the parameter of the same name, defined as: $$\large \alpha = \frac{1}{V} \frac{\partial V}{\partial T}$$ Because bonds grow longer with increasing temperature, &lt;strong&gt;most materials display a positive thermal expansion, i.e. they expand with temperature&lt;/strong&gt;. This is, for example, the case for diamond, with $\alpha \approx +10^{-6} \text K^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some materials display the oppositive behavior, and shrink upon temperature increase. This phenomenon is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_thermal_expansion&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;negative thermal expansion&lt;/strong&gt;&lt;/a&gt; (NTE), and is rather uncommon.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-03T07:57:50.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1087" PostTypeId="1" CreationDate="2012-09-03T21:17:58.017" Score="19" ViewCount="509" Body="&lt;p&gt;I was drinking some tea, and I randomly decided to pour some into the toilet in an unbroken stream; can particles travel &quot;upstream&quot; quickly enough to reach the mug of tea? How fast can diffusion occur in this system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I pour it into highly concentrated toxic waste?  Given that even trace amounts could make it toxic, should I not drink my tea?&lt;/p&gt;&#xA;" OwnerUserId="588" LastEditorUserId="4945" LastEditDate="2014-10-23T10:12:34.723" LastActivityDate="2020-01-14T01:56:31.753" Title="Can molecules diffuse upstream when emptying a cup of tea?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1088" PostTypeId="1" AcceptedAnswerId="1089" CreationDate="2012-09-04T09:04:58.240" Score="5" ViewCount="1053" Body="&lt;p&gt;The equation for the reaction between a &lt;a href=&quot;http://en.wikipedia.org/wiki/Permanganate&quot; rel=&quot;nofollow&quot;&gt;permanganate&lt;/a&gt; and iron(II) ions in &lt;strong&gt;acidic&lt;/strong&gt; solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{MnO}_4^- + \text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + \text{Mn}^{2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book, as well as a certain &lt;a href=&quot;http://www.webqc.org/balance.php&quot; rel=&quot;nofollow&quot;&gt;equation balancer&lt;/a&gt;, shows that the answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5\text{Fe}^{2+} + \text{MnO}_4^-  8\text H^+ \longrightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text H_2\text O$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They check the charges to ensure that it is indeed balanced. My book says that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(2+) + (1-) + 8(1+) = 17+$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5(3+) + (2+) + 0 = 17+$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And voila, they must be therefore balanced. However, there must be more than one way to balance such a reaction, like many problems. Is there a different, perhaps a more interesting, method than the half-reaction method for balancing this reaction and others?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="1085" LastEditDate="2013-01-12T07:32:50.890" LastActivityDate="2013-08-15T17:30:18.363" Title="Is there an alternative to the half-reaction method to solving red-ox reactions?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1089" PostTypeId="2" ParentId="1088" CreationDate="2012-09-04T09:54:23.200" Score="8" Body="&lt;p&gt;There is one other way that I have used, using a &lt;strong&gt;linear system&lt;/strong&gt; to solve this reaction. Although it can be used for other more casual reactions too, such as the combustion of ethanol for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, in this special case (because the reaction is relatively easy to solve already) we see that there must be $H_2O$ molecules on the product side of the equation, because there is no oxygen on the products already, but there is oxygen in the reactants. So start by assuming the products include $H_2O$ with a coefficient of at least 4 (because there is 4 oxygen in permanganate ion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is no hydrogen anywhere in the equation, so we can assume that to balance the water (which has at least 4 moles) there must be some hydrogen ions added as reactants to balance the reactants side (it is acidic conditions in this case). As it turns out, since there must be at least 4 moles of water, then there must be &lt;strong&gt;at least&lt;/strong&gt; 8 moles of hydrogen ions to satisfy the convention of having all mole coefficients in the lowest terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can start building the linear system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we obviously do not know the coefficients yet, we will assign variables to them. Let &lt;code&gt;a&lt;/code&gt; &lt;code&gt;b&lt;/code&gt; &lt;code&gt;c&lt;/code&gt; &lt;code&gt;d&lt;/code&gt; &lt;code&gt;e&lt;/code&gt; and &lt;code&gt;f&lt;/code&gt; be the molar coefficients of each respective molecule in the following reaction we have established so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a&lt;/code&gt;$H^+$ + &lt;code&gt;b&lt;/code&gt;$MnO_4^-$ + &lt;code&gt;c&lt;/code&gt;$Fe^{2+}$ ==&gt; &lt;code&gt;d&lt;/code&gt;$Fe^{3+}$ + &lt;code&gt;e&lt;/code&gt;$Mn^{2+}$ + &lt;code&gt;f&lt;/code&gt;$H_2O$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do we build the linear system for this reaction? We might do it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the hydrogen atom, simply because it comes first when reading it from left to right. We write this equation to represent the hydrogen atoms in our reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a = 2f&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is as simple as it looks, there is one mole of hydrogen atoms per mole of hydrogen ions, and there is two moles of hydrogen atoms per mole of water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can keep doing this for the rest of the elements in our reaction, until we have &lt;strong&gt;three&lt;/strong&gt; more equations along with the one for hydrogen, because there are three more different elements to account for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;b = e&lt;/code&gt; This one is pretty straightforward... This accounts for Manganese atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4b = f&lt;/code&gt; Oxygen atoms are now accounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;c = d&lt;/code&gt; Also straightforward, now iron atoms will be accounted for as well. &lt;strong&gt;Notice that &lt;code&gt;c&lt;/code&gt; and &lt;code&gt;d&lt;/code&gt; are lone variables, they do not show up with the other equations, but at least they must be equivalent&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that we established some mathematical axioms when we said that there must be at least four moles of water molecules to satisfy the one mole of permanganate ion. An axiom is just something that we assume to be true, such as &lt;code&gt;f = 4&lt;/code&gt; in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the equation we established for hydrogen atoms, the first equation &lt;code&gt;a = 2f&lt;/code&gt;. Since we are assuming that &lt;code&gt;f = 4&lt;/code&gt; we can therefore safely assume that &lt;code&gt;a = 8&lt;/code&gt; without going against our original axiom. Also if we assume that &lt;code&gt;f = 4&lt;/code&gt; we can also go ahead and solve the equation to account for oxygen atoms &lt;code&gt;4b = f&lt;/code&gt; and solving it gives us &lt;code&gt;b = 1&lt;/code&gt; and the rest is pretty straight forward. If &lt;code&gt;b = 1&lt;/code&gt; and &lt;code&gt;b = e&lt;/code&gt; then &lt;code&gt;e = 1&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given what we have found out so far, lets take a look at our equations again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;a = 8&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;b = 1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;c = d&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;e = 1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;f = 4&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;8&lt;/code&gt;$H^+$ + &lt;code&gt;1&lt;/code&gt;$MnO_4^-$ + &lt;code&gt;c&lt;/code&gt;$Fe^{2+}$ ==&gt; &lt;code&gt;d&lt;/code&gt;$Fe^{3+}$ + &lt;code&gt;1&lt;/code&gt;$Mn^{2+}$ + &lt;code&gt;4&lt;/code&gt;$H_2O$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So now we can look at the charges to figure out the correct coefficients for iron(II) and iron(III) ions, which should be the same of course (&lt;code&gt;c = d&lt;/code&gt;). As it stands currently (ignoring the iron ions), there is a total charge of &lt;code&gt;7+&lt;/code&gt; on the reactants side, and &lt;code&gt;2+&lt;/code&gt; on the products side. These should be the same for the equation to be balanced, so we can start by creating another axiom and saying that &lt;code&gt;c &amp;gt; 1&lt;/code&gt;. If there are &lt;code&gt;2&lt;/code&gt; moles iron then our charges would be &lt;code&gt;11&lt;/code&gt; on the reactants and &lt;code&gt;8&lt;/code&gt; on the products. Incrementing it to &lt;code&gt;3&lt;/code&gt; would increment the charge on the reactants side by &lt;code&gt;2+&lt;/code&gt; while the product's total charge would increment &lt;code&gt;3&lt;/code&gt; due to the charge of the ions themselves. Eventually things are balanced when the moles for iron reach &lt;code&gt;5&lt;/code&gt; making both sides have a total charge of &lt;code&gt;17&lt;/code&gt; and everything is nicely balanced.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-04T10:03:33.597" LastActivityDate="2012-09-04T10:03:33.597" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1090" PostTypeId="1" CreationDate="2012-09-05T03:04:31.857" Score="9" ViewCount="463" Body="&lt;p&gt;For multi-electron atoms for which the Hamiltonian (including the spin-orbit coupling) reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H=\sum_i T_i-Z\sum_i V_i+\sum_i V_i^{\text{s.o.}}+\sum_{i&amp;gt;j}V_{ij}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $T_i$ are the kinetic energy of the individual electrons, $V_i$ are the nuclear Coulomb attraction, $V_i^{\text{s.o.}}$ is the spin-orbit coupling, and $V_{ij}$ are interelectron Coulomb repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_atomic.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_atomic.pdf&lt;/a&gt; (page 103), using Wigner-Eckert, they are able to diagonalize the spin-orbit in terms of the &lt;em&gt;total&lt;/em&gt; orbital and spin angular momenta of all the electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it says it is beyond the scope of the course, I would still like to know how to do it.  I am pretty familiar with the Wigner-Eckert theorem in its basic form (that the matrix elements of a tensor-operator can be written in terms of a C.G. coefficient times a reduced matrix element).  Thanks.&lt;/p&gt;&#xA;" OwnerUserId="592" LastEditorUserId="592" LastEditDate="2017-02-20T17:35:49.707" LastActivityDate="2017-02-20T17:35:49.707" Title="multielectron spin-orbit coupling and energy level" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1091" PostTypeId="1" CreationDate="2012-09-05T04:00:42.693" Score="7" ViewCount="233" Body="&lt;p&gt;On the physics.nist.gov site: &lt;a href=&quot;http://physics.nist.gov/Pubs/AtSpec/node06.html&quot;&gt;http://physics.nist.gov/Pubs/AtSpec/node06.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table titled &quot;Atomic structural hierarchy in LS coupling and names for the groups of all transitions between structural entities&quot; contains the hierarchy of atomic LS coupling.  But there is a rather interesting footnote in the table: &lt;em&gt;The letter $\gamma$ represents any additional quantum numbers, such as ancestral terms, necessary to specify a particular term.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's &lt;em&gt;ancestral term&lt;/em&gt;?  Is there an example of this in the literature?&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2015-01-12T21:51:54.703" Title="What is the 'ancestral term' in the LS coupling scheme?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1092" PostTypeId="1" CreationDate="2012-09-05T18:30:45.667" Score="21" ViewCount="226864" Body="&lt;p&gt;I have read conflicting advice on various sites about the appropriate methods for cleaning plastic and rubber items and would like to ask people who &lt;strong&gt;&lt;em&gt;actually know what's going on&lt;/em&gt;&lt;/strong&gt; what they would recommend.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize some of my assumptions may be false:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am (perhaps wrongly) assuming that most plastics would have similar properties as it relates to cleaning with alcohol or soaps/detergents and that you don't need me to give you the actual chemical structure of a specific polymer.&lt;/li&gt;&#xA;&lt;li&gt;I am (perhaps wrongly) assuming that most soaps and detergents would act similarly on either a plastic or a rubber item.  If this is false please give your understanding of what an &quot;average&quot; soap or detergent would do.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;am not&lt;/em&gt; assuming that plastics and rubber will respond the same to either alcohol or soaps and detergents.  If they are different, please note which will have what effect.&#xA;I &lt;em&gt;am not&lt;/em&gt; assuming that synthetic and real rubber will act the same.  If they are different and you know the answer for both, please mention the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am &lt;em&gt;hoping&lt;/em&gt; that it's okay to ask &quot;applied&quot; questions here.  This seems like the best of all stackexchange.com sites, not just because none of the less theoretical sites are deal with this sort of question, but also because it really boils down to a matter of chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;: by &quot;alcohol&quot; I mean rubbing alcohol, or isopropanol.&lt;/p&gt;&#xA;" OwnerUserId="593" LastEditorUserId="402" LastEditDate="2017-04-02T20:33:03.323" LastActivityDate="2017-04-02T20:34:47.633" Title="Will alcohol or soap damage plastic or rubber?" Tags="&lt;everyday-chemistry&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1093" PostTypeId="2" ParentId="1092" CreationDate="2012-09-05T21:58:02.200" Score="3" Body="&lt;p&gt;&lt;strong&gt;My advice: Look up manuals where applicable, or experiment with methods if you cannot find good answers, hopefully, you can figure out what exact material it is you are cleaning.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can tell you for sure is that I have a small 'plastic' refrigerator which, by quickly looking up the model No. and finding &lt;a href=&quot;http://www.eofficedirect.com/PSNFSR3770S/Counter-Height-Office-Refrigerator-with-Crisper-3-7-cu-ft-Black-Stainless-St.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; order website, and the &lt;a href=&quot;https://www.google.com/#hl=en&amp;amp;sclient=psy-ab&amp;amp;q=SR-3770S+manual&amp;amp;oq=SR-3770S+manual&amp;amp;gs_l=hp.3..0j0i5i30l2.12733.14616.1.14725.7.7.0.0.0.0.125.556.6j1.7.0.les;..0.0...1c.5USeEuzokmE&amp;amp;pbx=1&amp;amp;bav=on.2,or.r_gc.r_pw.r_qf.&amp;amp;fp=e726f0c03c60e912&amp;amp;biw=939&amp;amp;bih=596&quot; rel=&quot;nofollow noreferrer&quot;&gt;manual&lt;/a&gt;, it is made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyurethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polyurethane&lt;/a&gt;. And the manual indicates under &quot;cleaning and maintenance&quot; that I should &lt;strong&gt;&quot;wash the&lt;/strong&gt; inside &lt;strong&gt;(polyurethane) material with a solution of warm water and two tablespoons baking soda per quart water.&quot;&lt;/strong&gt; And, specifically, &lt;strong&gt;do not use &quot;soaps, detergents, scouring powder, spray cleaners, or the like because it may cause odors&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps about.com has answers you wish, but I prefer the good old owners manual for the cold hard truth. By the way, I went ahead and used dish soap and cold water from my hose (I am a very lazy person). It smells slightly.. goofy, and that was a couple days ago. So I think they know what they are talking about. Note that the manual says nothing about Polyurethane!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps clarify something or another.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="402" LastEditDate="2017-04-02T20:34:47.633" LastActivityDate="2017-04-02T20:34:47.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1094" PostTypeId="2" ParentId="1030" CreationDate="2012-09-06T00:46:08.713" Score="6" Body="&lt;p&gt;Permeakra's explanation is spot on but I didn't really get a good understanding of the result until I visited the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sulfate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central hallmark of the question and the confusion was why a ten electron configuration would adopt a sp3 configuration more reflective of an octet. This was a source of confusion that goes back to Gilbert Lewis who originally proposed the below structure fitting of a sp3 hybridization (model 2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GtTql.png&quot; alt=&quot;Two different models&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linus Pauling steps in to propose that two of the d orbitals play a role suggesting that there should be a sp3d2 hybridization. The issue then arises with the role of Pi bonds and how they fit in the structure with the proposal that the occupied p orbitals overlap with the empty d orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately from computational calculations it turns out that the sulfur is indeed charged and there is very little pi character. As concluded in &lt;a href=&quot;http://www.springerlink.com/content/d21kpda6jgjd7vlr/?MUD=MP&quot; rel=&quot;nofollow noreferrer&quot;&gt;How relevant are S=O and P=O Double Bonds for the Description of the Acid Molecules H2SO3, H2SO4, and H3PO4, respectively?&lt;/a&gt;, model 1 with S=O bonds is complete crap even though this is how we teach highschool and college chemistry.&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="9961" LastEditDate="2016-11-19T21:59:22.530" LastActivityDate="2016-11-19T21:59:22.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1095" PostTypeId="2" ParentId="1092" CreationDate="2012-09-06T01:14:44.713" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Will soap damage plastic or rubber?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Probably not.&lt;/strong&gt; Most plastics and rubbers are resistant to aqueous solutions and ionic compounds (including most surfactants and soaps).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will alcohol damage plastic or rubber?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maybe.&lt;/strong&gt; There are two problems, in that neither &lt;em&gt;alcohol&lt;/em&gt; nor &lt;em&gt;plastic&lt;/em&gt; refer to specific compounds, but are rather generic terms encompassing hundreds of compounds each. The number of possible combinations is endless. As Leonardo suggests, the best approach is either to look in the manual or experiment in a small inconspicuous part of the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, assuming that by &lt;em&gt;alcohol&lt;/em&gt; you either mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethanol&quot; rel=&quot;noreferrer&quot;&gt;ethanol&lt;/a&gt;, which is the drinking kind of alcohol, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Isopropanol&quot; rel=&quot;noreferrer&quot;&gt;isopropanol&lt;/a&gt;, which is the alcohol in many cleaning products, then I can give you some advice based on common plastic types. Both are similar enough in their solvent properties that we can generalize.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohol will damage some plastics, but not all. I will list my answers by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Resin_code&quot; rel=&quot;noreferrer&quot;&gt;resin code&lt;/a&gt;, or &quot;recycling symbol&quot;, found on most plastic items:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Poly(ethylene terephthalate), PET or PETE - PET is not very soluble in ethanol or isopropanol, but prolonged exposure may cause &lt;a href=&quot;http://en.wikipedia.org/wiki/Crazing&quot; rel=&quot;noreferrer&quot;&gt;crazing&lt;/a&gt; or stiffening due to the dissolution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasticizer&quot; rel=&quot;noreferrer&quot;&gt;plasticizers&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;High-Density polyethylene, HDPE - HDPE is resistant to most things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Poly(vinyl chloride), PVC - PVC is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Low-Density polyethylene, LDPE - LDPE is resistant to most things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Polypropylene, PP - PP is resistant to most things.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Polystyrene PS - PS is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This stands for &quot;other&quot;, but the most common is polycarbonate, which is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rubber&lt;/strong&gt; - most rubbers are probably resistant to alcohols, but prolonged exposure will cause loss of elasticity&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="402" LastEditDate="2017-04-02T20:34:21.537" LastActivityDate="2017-04-02T20:34:21.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1096" PostTypeId="2" ParentId="1087" CreationDate="2012-09-06T03:49:58.093" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://blogs.discovermagazine.com/discoblog/2011/05/17/small-particles-can-flow-up-waterfalls-say-tea-drinking-physicists/&quot; rel=&quot;noreferrer&quot;&gt;was it those crazy tea fanatic Cuban scientists again?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particles can &lt;strong&gt;&lt;em&gt;supposedly&lt;/em&gt;&lt;/strong&gt; travel &quot;upstream&quot; if you were to say, pour it into your toilet as closely as your hand can get without a hazmat team busting down your door. But I would be more concerned about the sneaky particles that instead diffuse through the air itself, and not the stream you are considering...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the more highly concentrated solution Y is, and the more pure solution X is, the more likely it is for solute in solution Y to end up in solution X. Higher concentrations flow into lower concentrations. And the closer the two solutions are during the decanting, the easier it is for this counter-current to form. But it is assumed that at more than 1 centimeter height, a counter-current should not form according to that article.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-06T21:20:23.763" LastActivityDate="2012-09-06T21:20:23.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1098" PostTypeId="1" CreationDate="2012-09-06T15:42:56.230" Score="4" ViewCount="5137" Body="&lt;p&gt;In one of my chem classes I had to do a lab that involved finding the mass and volume of a substance.  I did three trials and for my mass I got 9.3g, 9.3g, and 9.4 g.  For my volume, I got 9.7ml, 9.9ml, and 9.9 ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am being asked to find the inherent instrument error calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For volume, I used a graduate cylinder which has &lt;code&gt;1 % vol/reading&lt;/code&gt; as an inherent error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mass I used a one place electronic balance which has error of &lt;code&gt;0.1g/reading&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the error in mass and the percent error in mass?&lt;/p&gt;&#xA;" OwnerUserId="595" LastEditorUserId="73" LastEditDate="2012-09-22T17:48:37.363" LastActivityDate="2012-09-22T17:48:37.363" Title="Inherent error in mass and volume?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1099" PostTypeId="2" ParentId="1074" CreationDate="2012-09-06T15:43:26.443" Score="1" Body="&lt;p&gt;It would depend on the type of dye/ink and the type of fabric.  The are several different types of dye removers available.  Dharma Trading Company makes &quot;Dharma's Dyehouse Color Remover&quot; which is Thiourea dioxide (a.k.a. Thiox, Spectralite).&lt;/p&gt;&#xA;" OwnerUserId="596" LastEditorUserId="30" LastEditDate="2012-10-04T21:07:53.010" LastActivityDate="2012-10-04T21:07:53.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1100" PostTypeId="1" AcceptedAnswerId="1101" CreationDate="2012-09-06T18:39:47.390" Score="7" ViewCount="1483" Body="&lt;p&gt;Is it possible to work out without post-Bohrian mathematics?&lt;/p&gt;&#xA;" OwnerUserId="597" LastActivityDate="2012-09-06T20:33:24.400" Title="What determines the electronegativity of an element?" Tags="&lt;electrochemistry&gt;&lt;electronic-configuration&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1101" PostTypeId="2" ParentId="1100" CreationDate="2012-09-06T20:33:24.400" Score="5" Body="&lt;p&gt;There are various slightly different electronegativity scales.  The two most common are the Pauling scale and the Mulliken scale.  Of the two the Mulliken is the simplest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mulliken scale is based on two quantities, both measurable.  One is the ionization energy $(E_i)$, the energy required to remove an electron from a neutral atom.  The other term is the electron affinity $(E_a)$.  That is the energy released when the neutral atom gains an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mulliken electronegativity $\chi$ is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\chi = (E_a + E_i)/2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is that simple!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details and explanation see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Electronegativity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-09-06T20:33:24.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1102" PostTypeId="1" AcceptedAnswerId="1105" CreationDate="2012-09-07T01:28:41.833" Score="8" ViewCount="61842" Body="&lt;p&gt;I've been struggling with this problem that I can't seem to figure out. I think I know how to solve it but I think there's missing information.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The combustion of 1 mole of glucose $\ce{C6H12O6}$ releases $\pu{2.82\times10^3 kJ}$ of heat. If $\pu{1.25 g}$ of glucose are burnt in a calorimeter containing $\pu{0.95 kg}$ of water and the temperature of the entire system raises from $\pu{20.10 ^\circ C}$ to $\pu{23.25 ^\circ C}$. What is the heat capacity of the calorimeter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I need the specific heat of glucose (which I haven't found yet), but also I don't know why they give me the heat released by 1 mole of glucose.  I need to know the method of how to solve such problems.&lt;/p&gt;&#xA;" OwnerUserId="599" LastEditorUserId="5026" LastEditDate="2018-02-25T08:13:48.647" LastActivityDate="2018-02-25T08:15:46.333" Title="Calculating the heat capacity of a calorimeter" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1103" PostTypeId="1" CreationDate="2012-09-07T01:46:35.893" Score="5" ViewCount="1260" Body="&lt;p&gt;Why is it that oxidizers (specifically $\ce{KClO3}$) are at risk of causing a fire when they come into contact with the rubber stopper used to contain them in their container, but can be slowly heated if they do not contact the rubber stopper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, why is it that oxidizers are incompatible with rubber?&lt;/p&gt;&#xA;" OwnerUserId="67" LastEditorUserId="4231" LastEditDate="2015-02-22T23:20:11.607" LastActivityDate="2016-11-17T12:26:11.093" Title="Why must an oxidizer not touch rubber?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1104" PostTypeId="1" AcceptedAnswerId="1148" CreationDate="2012-09-07T03:19:54.030" Score="3" ViewCount="3025" Body="&lt;p&gt;I have some caffeine tablets and I want to extract the pure caffeine from the rest of the ingredients (caffeine 200 mg. Inactive ingredients in order are corn starch, dextrose, dicalcium phosphate, D&amp;amp;C yellow #10 aluminum lake, magnesium stearate, microcrystalline cellulose, silica gel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the result would be aqueous solution but any method which works would be great.&lt;/p&gt;&#xA;" OwnerUserId="600" LastEditorUserId="73" LastEditDate="2012-09-08T09:24:42.520" LastActivityDate="2012-09-12T22:11:40.270" Title="How to extract pure caffeine from tablets?" Tags="&lt;home-experiment&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1105" PostTypeId="2" ParentId="1102" CreationDate="2012-09-07T07:07:43.937" Score="4" Body="&lt;p&gt;$12.5\ \mathrm{kJ}$ of heat was absorbed by the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this by using the mcat formula and the specific heat capacity of water (4.18 J/(g °C)):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q=m\cdot c\cdot\Delta T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = 950\ \mathrm{g} \times (4.18\ \mathrm{J\cdot g^{-1}\cdot {^\circ C^{-1}}}) \times (23.25\ \mathrm{^\circ C} - 20.10\ \mathrm{^\circ C}) = 12508.7\ \mathrm{J}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to use this whole formula for solving the calorimeter's specific heat capacity, you would need to know the mass of the calorimeter as well, which is not given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your book is probably asking is for what is called the &quot;calorimeter constant&quot;. This is given in units of $\pu{J/^\circ C}$ &lt;strong&gt;notice that it does not include mass.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: Sometimes &quot;the calorimeter's specific heat capcity&quot; is used instead of referring to the calorimeter constant, but in this case we cannot find a value which will include mass in the units, so I think it is more clear to use the term &quot;calorimeter constant.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can determine the constant by this formula: $$Q_\text{cal} = C_\text{cal} \times \Delta T_\text{cal}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q_\text{cal}$ is the energy absorbed, $C$ is the constant and $\Delta T$ is the same as the change in temperature of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may calculate $Q_\text{cal}$ by using this formula: $$Q_\text{cal} = -(Q_\text{water} + Q_\text{glucose})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also help to think of $Q_\text{water}$ = $Q_\text{surroundings}$ and $Q_\text{glucose}$ = $Q_\text{system}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $Q_\text{glucose}$ I did: (glucose has lost energy, it is negative value)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-2820\ \mathrm{kJ} \times 0.007\ \mathrm{mol}$ and $Q_\text{water}$ is simply the $12508.7\ \mathrm{J}$ positive because $\Delta T$ is positive for the surroundings (the system/glucose lost energy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_\text{cal} = -(12508.7\ \mathrm{J} + (-19740\ \mathrm{J}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my final answer is then: $2.3\times10^3\ \mathrm{J/{^\circ C}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important that heat capacities are positive, think about what it would mean if this were a negative value.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the laboratory, it is necessary to do a calculation such as this one before using a calorimeter for anything. Normally it can be done by heating a piece of nickel or something, recording the temperature of the metal and the water, and then dropping the metal into the calorimeter to find the final temperatures, and then calculate the calorimeter constant. You can then proceed with further experimentation using that calorimeter, but only after this constant has been found can you find the specific heat capacity of other materials.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="5026" LastEditDate="2018-02-25T08:15:46.333" LastActivityDate="2018-02-25T08:15:46.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1106" PostTypeId="2" ParentId="1103" CreationDate="2012-09-07T09:07:34.040" Score="0" Body="&lt;p&gt;$\ce{KClO3}$ and $\ce{KMnO4}$ are special kind of oxidizers, that donate oxygen very, very easy. Mixtures of this oxidizers with some reducers can self-combust on friction. Rubber stopper on regular friction produce fine powder of rubber and oxidize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for it is a special property of this oxidizers, that give out oxygen of surprisingly low heating.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user7951" LastEditDate="2016-11-17T12:26:11.093" LastActivityDate="2016-11-17T12:26:11.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1107" PostTypeId="1" AcceptedAnswerId="1110" CreationDate="2012-09-07T13:45:36.343" Score="26" ViewCount="32592" Body="&lt;p&gt;Water ($\ce{H2O}$) is a dipole. The reason why is simply because it is not symmetrical, there are more electrons on the oxygen side than on the hydrogen side, and the electronegativity of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why isn't $\ce{H2O}$ symmetrical like $\ce{CO2}$? Why isn't $\ce{H2O}$ non-polar like $\ce{CO2}$? Does it have anything to do with the orbitals?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="194" LastEditDate="2018-01-29T04:51:06.753" LastActivityDate="2019-09-15T14:08:01.680" Title="Why is water a dipole?" Tags="&lt;water&gt;&lt;molecular-structure&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1108" PostTypeId="2" ParentId="1098" CreationDate="2012-09-07T13:53:44.020" Score="1" Body="&lt;p&gt;Is the &lt;strong&gt;percent&lt;/strong&gt; error formula you are looking for something like the formula &lt;a href=&quot;http://www.percenterrorcalculator.com/index.php&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; website service offers? Your lab instructor should have given you the formula, perhaps you missed that. Do you have a reliable lab partner you can ask? The error in mass would probably be something to do with how far each measurement deviates from the average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for example I used a 250 ml Erlenmeyer flask which had marks at only 25ml increments. I would then say that I could make a 'guess' as to what the volume is at if I were trying to measure something less than 25ml or between two marks. But the error for that instrument would then be somewhere around +/- 12.5ml error. My general-chem teacher was fine with that, maybe you should ask yours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that the convention is to expect at least 2% error in the last significant digit if not explicitly defined. When taking a measurement with a device like a graduated cylinder, you always measure down to the most precise &quot;known&quot; value, and then the last significant digit is your own judgement.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-07T13:59:51.233" LastActivityDate="2012-09-07T13:59:51.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1109" PostTypeId="2" ParentId="1107" CreationDate="2012-09-07T14:29:06.160" Score="7" Body="&lt;p&gt;Fundamentally, dipolar simply means that a region of the molecule has a center of positive charge and a center of negative charge. We can say that water has a dipole moment because it is not a 'balanced' molecule, like $CO_2$ is. I understand what you are really asking though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it likely has something to do with orbitals. You can look into VSEPR Theory for an explanation of why water has a bent shape. It might have to do with both reasons you mentioned, the electronegative properties, and also that oxygen just has more electrons in general (and therefore needs to put them somewhere that has a low amount of energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to VSEPR, the electron's try to separate themselves out as far as possible, that is one explanation for why the molecule has a bent shape. The electrons orbiting oxygen will inevitably occupy what they call the p orbitals. These extend outward a greater distance, allowing the electrons to occupy a space that is further away from the other electrons, and therefore requires less energy to do so (because electrons repel each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model seems slightly counter-intuitive though, and requires a bit of imagination and understanding of the theory itself, if not a great deal quantum physics. But, like any good model, it tends to predict molecular properties and shapes, and this is why it is taught in schools.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="22" LastEditDate="2012-09-07T16:33:37.773" LastActivityDate="2012-09-07T16:33:37.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1110" PostTypeId="2" ParentId="1107" CreationDate="2012-09-07T16:44:04.033" Score="47" Body="&lt;p&gt;Yep, it has to do with the orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is linear, so even though the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; is &quot;bent&quot;, which means that the individual dipole moments of the bond are at an angle to each other. They add up to give a net dipole moment (shown with grey in the diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8tEfA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colors indicate electron density, red is more dense/blue is less dense. Dipole moment is from low density to high density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so why do these molecules have differing shapes? This is where orbitals come in. I'll try to explain as much as I can without going into orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon has an outer shell electronic configuration as &lt;span class=&quot;math-container&quot;&gt;$2s^22p^2$&lt;/span&gt;. Out of these &lt;em&gt;four&lt;/em&gt; electrons, two are used in &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds, and two in &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds. If you don't know what those are, just look at it like this for now: A set of bonds between two atoms will have one and only one &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond, with the rest of them &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds. So, any single bond is made up of just a &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond, a double bond is made up of one &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; and one &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;, and a triple bond is made up of one &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; and &lt;em&gt;two&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds. What these types of bonds actually are can be explained if you know what an orbital is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what VSEPR says is that the geometry of the molecule is only decided by the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds and lone pairs on the central atom. You count up the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds and lone pairs (lets say they add up to &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;), and decide the geometry based on that. The geometry is the most stable configuration of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; hybrid orbitals. In simple terms, if we took &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; balloons and tied them together, the directions the balloons point in help us correspond to where the bonds and lone pairs lie:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tSD1M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;, we have two &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds and two &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds (as each double bond has one of each type). Each bond takes up one electron from carbon, so we have no leftover electrons for forming any lone pair. Since we have two &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds and 0 lone pairs, &lt;span class=&quot;math-container&quot;&gt;$x=2$&lt;/span&gt;, giving us the structure given by the first set of balloons, which is linear. And &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is indeed linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5fsG2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take water. The central atom (Oxygen) has a valence configuration of &lt;span class=&quot;math-container&quot;&gt;$2s^22p^4$&lt;/span&gt;, that is, 6 electrons. In water, since we have two single bonds, we have one &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond each (and no &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds). So we have total two &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this leaves us with &lt;span class=&quot;math-container&quot;&gt;$6-2=4$&lt;/span&gt; unpaired valence electrons. These form two &quot;lone pairs&quot; (pairs of electrons which do not bond). With two lone pairs and two &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bonds, &lt;span class=&quot;math-container&quot;&gt;$x=4$&lt;/span&gt;. This gives us a tetrahedral structure (third in the balloon diagram). Two of the four points in the tetrahedron are occupied by the lone pairs, and two by bonds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iXOCx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the angle 104.5 is not the same as the angle in perfect tetrahedra, 109.25--this is due to the lone pairs repelling each other)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally, we have the following &quot;bent&quot; structure for water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xC0NU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the structure, as shown above, it is very easy to check if the molecule has a dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2019-09-15T14:08:01.680" LastActivityDate="2019-09-15T14:08:01.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1111" PostTypeId="1" AcceptedAnswerId="1118" CreationDate="2012-09-08T01:14:44.573" Score="7" ViewCount="979" Body="&lt;p&gt;I am reading &lt;em&gt;Ionic Equilibrium&lt;/em&gt; by James N. Butler.  Using silver(I) chloride in excess aqueous chloride as an example it speaks of several forms:  $\ce{AgCl (s)}$ i.e. the ionic crystalline salt, $\ce{Ag+(aq)}$ and $\ce{Cl^{-}(aq)}$ the aqueous ions from dissociation, and finally $\ce{AgCl(aq)}$, which Butler purports to be &quot;dissolved covalent silver chloride molecules as distinct from solid salt or dissolved ions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this &quot;molecule&quot; of $\ce{AgCl}$ really exist in solution?  If so, is there a way to quantify the concentrations of this covalent species?  Equilibrium solubility product constants seem to give the concentration of the aqueous ions only.&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="4231" LastEditDate="2015-02-23T19:51:32.940" LastActivityDate="2021-04-28T13:30:00.953" Title="Does &quot;covalent&quot; AgCl (aq) exist and if so, how to quantify its concentration?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1112" PostTypeId="2" ParentId="1103" CreationDate="2012-09-08T01:52:57.787" Score="4" Body="&lt;p&gt;A lot of rubbers possess &lt;a href=&quot;https://en.wikipedia.org/wiki/Elastomer&quot; rel=&quot;nofollow&quot;&gt;C=C double bonds&lt;/a&gt;. Alkenes are very easily oxidized by KMnO&lt;sub&gt;4&lt;/sub&gt; or KClO&lt;sub&gt;3&lt;/sub&gt;, therefore in the case where you're not aware of the chemical nature of the rubber, it's best not to expose it to strong oxidizers.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-09-08T01:52:57.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1113" PostTypeId="1" AcceptedAnswerId="1115" CreationDate="2012-09-08T05:55:53.443" Score="5" ViewCount="179" Body="&lt;p&gt;I am currently studying the basics of simulation and want to try out some experiments. It is a known fact that gold nanoparticles can bind to chemical groups which contains &lt;a href=&quot;http://en.wikipedia.org/wiki/Thiol&quot; rel=&quot;nofollow&quot;&gt;thiol&lt;/a&gt; (R–SH group). I want to simulate the binding of a thiol group to a gold nanoparticle. How can I do this? Where do I have to start? Which software can be used for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="470" LastEditDate="2012-09-13T18:12:53.133" LastActivityDate="2012-09-13T18:12:53.133" Title="Simulation of a particle and chemical molecule" Tags="&lt;computational-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1114" PostTypeId="1" AcceptedAnswerId="1123" CreationDate="2012-09-08T06:00:47.787" Score="8" ViewCount="4977" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When you are diluting air in the system and/or properly evacuating it, you must consider water temperature. Methane is completely dissolved in water at 42 degrees Fahrenheit (5.5 C) but can be completely released (as a gas) at 58 degrees F (14.5 C). &lt;sup&gt;&lt;a href=&quot;http://www.watertechonline.com/drinking-water/article/the-fine-points-of-methane-removal&quot; rel=&quot;nofollow&quot;&gt;(source)&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would seem to imply that the enthalpy of solution for methane in water is negative (i.e. the process is exothermic). But, intuitively, it seems that you would have to input energy to force non-polar and polar to mingle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain why this might be the case?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="143" LastEditDate="2013-09-03T03:38:34.180" LastActivityDate="2013-09-03T03:38:34.180" Title="Methane seems to dissolve more readily in cold water. Why?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;solubility&gt;&lt;enthalpy&gt;&lt;water&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1115" PostTypeId="2" ParentId="1113" CreationDate="2012-09-08T07:52:06.523" Score="7" Body="&lt;p&gt;Density Functional Theory is worth investigating if you're looking for a fairly efficient way of investigating these interactions at an electronic level. The term 'nanoparticle' is wonderfully vague as it encompasses two to three orders of size magnitude - the size of the nanoparticle will inform the specific choice of method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small (as in, only a couple of atoms) nanoparticles are amenable to computation with atom-centered basis sets in programs such as &lt;a href=&quot;http://www.gaussian.com/&quot;&gt;Gaussian&lt;/a&gt;, &lt;a href=&quot;http://scm.com/&quot;&gt;ADF&lt;/a&gt;, &lt;a href=&quot;http://www.thch.uni-bonn.de/tc/orca/&quot;&gt;ORCA&lt;/a&gt;, &lt;a href=&quot;http://classic.chem.msu.su/gran/gamess/index.html&quot;&gt;GAMESS&lt;/a&gt; and so on. The latter two may be acquired free for academic use. If your particle/ligand complex is highly symmetric and/or you have supercomputer time you might be able to work with a moderately large particle. I recall a paper abstract where researchers did TDDFT on a 120 atom silver nanoparticle in ADF (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp802707r&quot;&gt;here it is!&lt;/a&gt;), aided by high symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large systems, where it's a) computationally unreasonable to model the whole particle and b) to a first-order approximation the ligand-particle interaction can be treated as occurring on an infinite plane or edge, may be more amenable to periodic DFT using plane-wave basis sets (PW-DFT). There are many codes available, ranging from free to very expensive. Off the top of my head, &lt;a href=&quot;http://www.vasp.at/&quot;&gt;VASP&lt;/a&gt;, &lt;a href=&quot;http://icmab.cat/leem/siesta/&quot;&gt;SIESTA&lt;/a&gt;, &lt;a href=&quot;http://www.quantum-espresso.org/&quot;&gt;Quantum Espresso&lt;/a&gt;, &lt;a href=&quot;http://www.abinit.org/&quot;&gt;Abinit&lt;/a&gt; and &lt;a href=&quot;http://accelrys.com/products/materials-studio/quantum-and-catalysis-software.html&quot;&gt;CASTEP&lt;/a&gt; are fairly popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also an atom-centered periodic franken-version of ADF called &lt;a href=&quot;http://www.scm.com/Products/BAND.html&quot;&gt;ADF-BAND&lt;/a&gt; which can model 1D to 3D periodic systems, which is worth mentioning. It is not particularly well documented, however.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-09-08T07:52:06.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1116" PostTypeId="1" CreationDate="2012-09-08T08:09:48.797" Score="1" ViewCount="5743" Body="&lt;p&gt;The question was posed as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fertilizers supply the essential elements, nitrogen, phosphorus and potassium for plant growth. A bag of fertilizer contains 500g of ammonium sulfate, $ \small \ce{(NH4)2 SO4}$, and 500g of potassium nitrate, $\small \ce{KNO3}$. Calculate the percentage by mass of nitrogen in the bag of fertilizer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This came in a test of mine and I got the answer wrong, now I did a redo and here's my solution:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(42 * 2)/227  * 100 = 37% &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this answer correct?&lt;/p&gt;&#xA;" OwnerUserId="604" LastEditorUserId="73" LastEditDate="2012-09-21T05:07:30.493" LastActivityDate="2012-09-21T05:07:30.493" Title="Percent composition of nitrogen in fertilizer" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1117" PostTypeId="2" ParentId="1116" CreationDate="2012-09-08T09:17:16.727" Score="2" Body="&lt;p&gt;I got 17.542% by mass nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An explanation for beginners, a refresher for the rest of us:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make sure anyone reading knows what molar mass means: The mass of one &lt;strong&gt;mole&lt;/strong&gt; of something, in our case the mass of a mole of nitrogen, or the mass of a mole of ammonium sulfate, etc. A &lt;strong&gt;mole&lt;/strong&gt; is just a way of saying a really big number which is actually $6.022 * 10^{23}$. It is a word we use instead of saying that big number, like when we say a &lt;strong&gt;dozen&lt;/strong&gt; of eggs, but we know it means $12$ eggs. The same concept applies to a mole of eggs (that's a lot of eggs). To calculate the molar mass you need to add the atomic masses of each atom present in the molecule. If it is water, then we must count hydrogen's atomic mass twice (two hydrogen atoms per molecule), oxygen only once, since the formula is $H_2O$. Atomic mass can be found on any respectable periodic table, be sure not to confuse the atomic mass with the atomic number! The atomic mass always represents the weight in grams of a mole of that element. This convention makes our calculations much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately we need to know how many $mol$ of nitrogen exist in this bag of fertilizer, so we can find out the mass of nitrogen in said bag (atomic mass can be used both ways because it is a ratio of grams per mole). To find that we need to do this process for both of the compounds separately because they have different molar masses. If there are two $chickens$ per $hen$ and we have $4_{mols}$ of $hens$ we should be able to tell that we have $8_{mols}$ of $chickens$. The same concept is used to determine how many $mols$ of nitrogen we have, based on how many nitrogens we expect per molecule of each substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how that calculation might look in dimensional analysis in case it surprises you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$81_{grams} * \frac{1_{mol}}{12_{grams}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice units of grams cancel each other out, and the remainder is in moles. This is how you would find the number of moles of carbon there exist in a pure sample of 81 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice we used the ratio $12$ $\frac{grams}{mol}$ which is carbon's atomic mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend writing out the entire problem using dimensional analysis so that you can be absolutely sure you have the answer correct for whatever problem you are solving. You will know this is true when all the units cancel correctly.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To solve the problem&lt;/strong&gt; it may be good to take a step back from the calculator and consider what we know. What is the molar mass of these two substances? It should be the summation of the atomic masses of all the atoms in the molecule. Just like a water balloon's mass is not only the mass of the water, but also the rubber. Similarly the mass of water is not just the mass of the hydrogen, but the oxygen as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is $500_{grams}$ of ammonium sulfate, but if we are trying to figure out the moles based on this information, we will need to do some dimensional analysis. Use the &lt;strong&gt;molar mass&lt;/strong&gt; of ammonium sulfate which has units $grams/mol$ to multiply this ratio we found by the total mass of ammonium sulfate $(500_{grams})$ in such a way that the $_{grams}$ cancel each other, leaving $mol$ units. This is how many moles of that &lt;strong&gt;molecule&lt;/strong&gt; exist in a 500g sample. We need to know this because the relationship of nitrogen per molecule is not a gram to gram relationship. You could of course, go another route and figure out the percent by mass of nitrogen per &lt;strong&gt;molecule&lt;/strong&gt; and your final calculation should be equivalent. You should get about $21$% nitrogen by mass in a molecule of ammonium sulfate, and therefore 21% of the mass of that substance should equal the mass of nitrogen in that substance. It will be different for the other substance in the bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the total moles of nitrogen, convert back into grams of nitrogen, make sure your units cancel. From there you know the mass of nitrogen in the fertilizer bag, and the total mass of the fertilizer bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find the answer: what percent mass is the nitrogen atoms compared to the total mass of the fertilizer bag (all the atoms/molecules)?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-09T07:27:38.057" LastActivityDate="2012-09-09T07:27:38.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1118" PostTypeId="2" ParentId="1111" CreationDate="2012-09-08T17:29:51.203" Score="7" Body="&lt;p&gt;Some amount of &lt;span class=&quot;math-container&quot;&gt;$[\ce{AgCl(H2O)}]$&lt;/span&gt; particles should exist in solution from pure thermodynamic point of view. See, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; ion can coordinate two ligands, and in solution both &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; are present. However, it should be really hard to detect trace amounts of this particle, as &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; is almost insoluble. In case one can get concentrated salt solutions, similar particles may exist and hints for their existence may be got from conductivity measurements and osmotic pressure. Sometimes a direct study using NMR is possible, which can give some structural information if used properly (see 2D-NMR methods). As I remember, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(CH_3COO)2}$&lt;/span&gt; hydrate is slightly soluble in water but undergoes little to no dissociation.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="69830" LastEditDate="2021-04-28T13:30:00.953" LastActivityDate="2021-04-28T13:30:00.953" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1120" PostTypeId="2" ParentId="993" CreationDate="2012-09-09T04:54:12.857" Score="5" Body="&lt;p&gt;Pack a glove into a glass container, let it sit for a few hours or days, get the container to your institute's mass spectroscopy and gas chromatography departments, and have them have a go at the gas phase in the container. I think that might be your best bet at finding out what is leaking out of these gloves. An IR or UV/Vis of the gas phase might also yield results, as Spießbürger suggested above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, if you got an analytics department at your university/institute, why not go to them, they might be able to have someone do a B.Sc. or M.Sc. thesis on that. Even if you end up trashing the gloves, at least then it won't have been a complete waste.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2012-09-09T04:54:12.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1122" PostTypeId="1" AcceptedAnswerId="1127" CreationDate="2012-09-09T10:55:34.127" Score="5" ViewCount="1224" Body="&lt;p&gt;Basically, does the lanthanide contraction also occur with the 4d, 6d groups as well? I am at a lack of understanding in this topic, I am just wondering what specific elements does this jump in a lower atomic radii occur. It seems to happen due to the d orbitals being occupied? I am making that judgement based on this graph.&lt;br /&gt;&#xA;I am hoping someone can explain, as one orbital seems to make all the difference.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TE4Tm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TE4Tm.jpg&quot; alt=&quot;graph&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-11-22T08:49:33.837" Title="Is there a similar characteristic of the lanthanide contraction in the other transition metal groups?" Tags="&lt;physical-chemistry&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1123" PostTypeId="2" ParentId="1114" CreationDate="2012-09-09T15:30:45.340" Score="3" Body="&lt;p&gt;Never mind, I think I found the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the gas phase, attractions between solute particles are negligible, while solvent-solute attractions may reduce the energy of solute particles, resulting in an exothermic process that favors lower solvent temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record.&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2012-09-09T15:30:45.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1124" PostTypeId="2" ParentId="901" CreationDate="2012-09-10T04:09:02.807" Score="12" Body="&lt;p&gt;One periodic table that apparently physicists find useful is ADOMAH periodic table, for its usefulness is finding electron configuration of a particular element. More information at this &lt;a href=&quot;http://www.perfectperiodictable.com/userguide&quot; rel=&quot;nofollow noreferrer&quot;&gt;guide&lt;/a&gt;. A disadvantage to this table, along with any new table configuration you have not experienced, will be the lack of knowledge in the trends on your new table. There are many spiral looking tables, but the trends such as the change in atomic radii, or electronegative characteristics, might not be as obvious, and you might have to find a guide or relearn old tricks to methods you would learn in your first year of chemistry. At least the ADOMAH Table is built in such a way that these trends should still be recognizable, although clearly not the same as the normal table you find in any high-school science classroom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of particular interest to me are dynamic periodic tables. These are tables that are basically computer applications which will change according to what you are looking for. One outstanding example is the widely popular &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ptable.com&lt;/a&gt; which will maintain all the common trends you probably know already, and dynamically show you various states of the elements when changing the temperature. I learned from this table that rhenium has the highest known boiling temperature, while I thought it was tungsten (although tungsten has the highest melting temperature of the transition metals). And better yet, carbon has the highest melting temperature of all elements, very cool table! The disadvantage is obviously that you cannot carry it around in your notebook without some flavor of a mobile computer and/or internet connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure there are many other great tables still to be found, but I hope you have learned something new from my post.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="402" LastEditDate="2017-05-09T12:51:20.337" LastActivityDate="2017-05-09T12:51:20.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1125" PostTypeId="1" AcceptedAnswerId="1126" CreationDate="2012-09-10T06:12:19.763" Score="1" ViewCount="619" Body="&lt;p&gt;Why is it referred to as &quot;carbonation&quot; and we drink &quot;carbonated&quot; beverages when carbonate is $\ce{CO3}$ while $\ce{CO2}$ (carbonite?) is present in carbonation? &lt;/p&gt;&#xA;" OwnerUserId="609" LastEditorUserId="865" LastEditDate="2015-10-30T17:12:37.233" LastActivityDate="2015-10-30T17:12:37.233" Title="Why is it called &quot;carbonation&quot;?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1126" PostTypeId="2" ParentId="1125" CreationDate="2012-09-10T07:31:52.980" Score="6" Body="&lt;p&gt;Nope. Carbonate is $\ce{CO}_3^\color{red}{2-}$ (an ion), not $\ce{CO3}$. Carbonite is $\ce{CO}_2^\color{red}{2-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbonation involves dissolving $\ce{CO2}$ gas in water. It turns out that $\ce{CO2}$ reacts (maybe not the best term to use here*) with water via the following reaction $$\ce{CO2 + H2O -&amp;gt; CO3^{2-} + 2H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have effectively &lt;em&gt;carbonated&lt;/em&gt; the soft drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you depressurize the bottle by opening it, the reverse reaction occurs and you get carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Such dissociation is normal and integral to dissolution of polar solutes in water, so generally it is considered as &lt;em&gt;part&lt;/em&gt; of the dissolution&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-09-10T07:31:52.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1127" PostTypeId="2" ParentId="1122" CreationDate="2012-09-10T17:28:02.247" Score="6" Body="&lt;p&gt;Hope I interpret your question correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decrease of atomic radii occurs in every period (from left to right). Atomic radii of lower periods are in general larger (Increase from top to bottom). This simple behavior only holds true for elements that only have p-block and s-block electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first anomaly of this behavior is the &lt;strong&gt;d-block contraction&lt;/strong&gt;. This can be seen in the 4p, 5p, 6p and 7p elements (Elements to the right of Ga, In, Tl). For example the atomic radius increases about 60 % from C to Si. The difference of Si and Ge with the added p-block contraction is just 20 %.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;lanthanoide contraction&lt;/strong&gt; occurs when the f-block is added. For this reason the 4d and 5d elements are very similar in size. For example Hf (5d, 160 pm) is even smaller than Zr (4d, 159 pm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the &lt;strong&gt;actinoide contraction&lt;/strong&gt; should have the same effect on 6d elements. But I'm not sure if realiable numbers for the atomic radius exist for those elements.&lt;/p&gt;&#xA;" OwnerUserId="351" LastActivityDate="2012-09-10T17:28:02.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1128" PostTypeId="1" AcceptedAnswerId="1250" CreationDate="2012-09-11T02:03:04.233" Score="2" ViewCount="4088" Body="&lt;p&gt;I am using an old book and they almost use both terms interchangeably sometimes. Do they mean the same thing? Similiarly for &quot;steam point&quot; and &quot;boiling point&quot;&lt;/p&gt;&#xA;" OwnerUserId="611" LastActivityDate="2012-09-28T13:25:13.327" Title="Is &quot;ice point&quot; the same as &quot;freezing point&quot;?" Tags="&lt;thermodynamics&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1129" PostTypeId="1" CreationDate="2012-09-11T02:35:28.017" Score="3" ViewCount="3278" Body="&lt;p&gt;$$\ce{Ba(NO3)2(aq) + Na2CO3(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above reaction both the substances are in water, however water plays no role in the reaction. Why is it that although water is present in the reaction, it does not play a reactant role? To clarify, why isn't the above written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ba(NO3)2(aq) + Na2CO3(aq) + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does water act as a bystander and not a reactant?&lt;/p&gt;&#xA;" OwnerDisplayName="user612" LastEditorUserId="4231" LastEditDate="2015-02-22T23:20:23.840" LastActivityDate="2016-07-14T21:56:48.353" Title="Why doesn't water play a role in reactions in aqueous reactions?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1130" PostTypeId="1" CreationDate="2012-09-11T02:44:07.223" Score="-4" ViewCount="3256" Body="&lt;p&gt;I checked the Wikipedia page, but it did not seem to provide a very intuitive explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What distinguishes a molecule such that it behaves like a polyatomic ion? I understand that it is a covalent bond, but what distinguishes it from other covalent bonds that allow it to act like one element. Why does it have a &lt;em&gt;charge&lt;/em&gt;, unlike many other compounds (ionic and covalent)?&lt;/p&gt;&#xA;" OwnerDisplayName="user612" LastActivityDate="2012-09-11T03:30:35.900" Title="What makes a molecule a polyatomic ion?" Tags="&lt;inorganic-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1131" PostTypeId="1" CreationDate="2012-09-11T02:53:38.393" Score="1" ViewCount="479" Body="&lt;p&gt;We were taught that there are primarily four kind of reactions, synthesis, combustion, single displacement, and double displacement. When the reactants are given to you, the reaction can be predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reactants are AB + CD, it will be a double displacement (two pairs of compounds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reactants are AB + C, it will be a single displacement (one pair and one compound/element)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reactants are A + B, it will be synthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the &quot;format&quot; of the reactants determine the type of reaction? Why can't, for example, AB + CD synthesize to create ABCD? Or AB + C synthesize to create ABC? Why is it assumed that if the reactants are in a specific &quot;format&quot; a specific reaction will occur?&lt;/p&gt;&#xA;" OwnerDisplayName="user612" LastEditorUserId="470" LastEditDate="2012-09-13T18:12:21.197" LastActivityDate="2012-09-13T18:12:21.197" Title="What causes reactions to react the way they do?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1133" PostTypeId="2" ParentId="1130" CreationDate="2012-09-11T03:30:35.900" Score="2" Body="&lt;p&gt;A polyatomic ion is simply a molecule that has gained or lost one or more electrons, thus acquiring a net charge. This can occur due to oxidation or reduction of the molecule, by association of an atomic ion with a neutral molecule or by autoionisation of a neutral species into a anion/cation pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example of the latter two processes is the autoionisation of water, which to a first-order approximation may be considered as $\ce{2H_{2}O &amp;lt;=&amp;gt; H_{3}O^{+} + OH^{-}}$. This is coincidentally an example (one of many) of an aqueous reaction involving covalency with a water molecule, addressing in part one of your other &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1129/why-doesnt-water-play-a-role-in-reactions-in-aqueous-reactions&quot;&gt;questions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There also exist polyatomic ions known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Zwitterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;zwitterions&lt;/a&gt; (from German 'zwitter', meaning 'hermaphrodite' or 'hybrid') which possess localised opposite charges on different atoms, but are neutral overall.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2012-09-11T03:30:35.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1134" PostTypeId="2" ParentId="1129" CreationDate="2012-09-11T05:13:14.217" Score="1" Body="&lt;p&gt;It does play a role--it's all hidden in the $_\ce{(aq)}$. Whenever an ionic compound is dropped in water, it dissociates into the ions. This dissociation is facilitated by the water, as it solvates the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that a certain amount of energy is required to dissociate an ionic compound (lattice energy). If a pure compound was dissociated, there would be quite a bit of energy required to do this, and you would end up with something unstable (the molecule would probably just recombine.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, water, being polar, solvates the ions. What happens is that water molecules aggregate around the ion, and the net charge gets effectively dispersed. This leads to a gain in stability (and a decrease in potential energy known as the &quot;hydration energy&quot;). Ionic compounds with enough hydration energy (specifically, when the absolute value of hydration energy&gt;lattice energy) are said to be &quot;soluble&quot;, and dissociate readily in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the actual reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2CO3 -&amp;gt;[\ce{H2O}] 2Na+_{(aq)} + CO3^{2-}_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the $\ce{_{(aq)}}$ signifies solvation. There will be a similar one for barium nitrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when one mentions $\ce{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}$, it is actually $\ce{2Na+_{(aq)} + CO3^{2-}_{(aq)} + Ba^{2+}_{(aq)} + 2NO3^{-}_{(aq)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens is that the barium and carbonate ions react to form insoluble barium carbonate. This precipitates out, and we get the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\underbrace{Ba(NO3)2_{(aq)} + Na2CO3_{(aq)}}_{Mixture~of~ions} -&amp;gt;[\ce{H2O}] \underbrace{BaCO3}_{Insoluble~solid} v + \underbrace{NaNO3_{(aq)}}_{Mixture~of~ions}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water isn't reacting here, it is more of a catalyst. Without it, the reaction wouldn't occur (unless you melt the compounds--molten ionic compounds behave similar to their aqueous counterparts).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="9961" LastEditDate="2016-07-14T21:56:48.353" LastActivityDate="2016-07-14T21:56:48.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1135" PostTypeId="2" ParentId="1129" CreationDate="2012-09-11T05:20:48.653" Score="4" Body="&lt;p&gt;Water is far from a bystander in aqueous solutions. It has a number of features that greatly facilitate chemical reactions, as well as being an important ligand (the &lt;em&gt;aqua&lt;/em&gt; ligand) of transition metal complexes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Here are some (unbalanced) examples:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water has a rather large dielectric constant and as such is very effective at shielding charges from one another, facilitating the charge separations inherent in the dissolution of ionic compounds. This is why you can easily dissolve sodium chloride in water, but not in hexane. The nature of this shielding is often articulated as several concentric 'solvation shells', where 'shells' of water molecules cluster around an ion with their dipoles approximately aligned towards or away from the ion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's an &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation_shell&quot; rel=&quot;nofollow noreferrer&quot;&gt;image from Wikipedia&lt;/a&gt; qualitatively depicting the first solvation shell surrounding a sodium ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PJn6v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;More generally, the large dipole moment of water allows it to partake in hydrogen bonding, where water will weakly associate with other polar molecules or functional groups. This extends to water forming (short lived, highly fluxional) clusters with itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Water is also happy to accept (forming hydronium, $\ce{H3O+}$) and donate (forming hydroxide, $\ce{OH-}$) protons in solution. So much so that it will actually do this with itself, in a process called autoionisation. This process makes water molecules significant in a wide range of acid/base and redox equilibria.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="84" LastEditorUserId="73" LastEditDate="2012-09-28T02:26:48.203" LastActivityDate="2012-09-28T02:26:48.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1136" PostTypeId="2" ParentId="1131" CreationDate="2012-09-11T06:01:05.403" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the reactants are AB + CD, it will be a double displacement (2 pairs of compounds)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example of this might be two salts that dissociate into their respective ions in a solvent, and then the cation and anions that were not previously paired find their way to each other, if they form a salt that is not easily dissociated in this solvent, we have a new ionic compound (salt) which precipitates out as a visible solid in the mixture container. This might not always be the case, when we write &lt;strong&gt;net ionic equations&lt;/strong&gt; we leave out anything that was not involved in creating the precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have the net ionic equation: $\ce{A^{+}_{(aq)} + B^{-}_{(aq)} -&amp;gt; AB_{(s)}}$ occurring when the solutions mix, then $\ce{AB}_{(s)}$ is said to be the precipitate. Notice that we leave out the other ions C and D because they were observers in the chemical equation, perhaps their attraction to each other is not strong enough to overcome the solvent. We should not always be able to tell whether this will happen or not unless we refer to experiments in which this substance had been seen to precipitate out before, perhaps even measuring its solubility in some experiments, the extent to which it will dissociate can also be calculated from experimentation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't, for example, AB + CD synthesize to create ABCD?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are trying to determine if a covalent bond is compatible, one method is to use the octet rule and see what combinations might occur. For example, you have reactants of silicon and fluorine. What will form? Looking at the periodic table to see the electrons they contribute, draw a lewis structure for fluorine and silicon. Silicon contributes 4 electrons, and 4 fluorine atoms contribute the rest to give all elements full octets. Silicon tetrafluoride seems to be the most stable using this model. This is one method, but often times reactions occur the opposite way too. However, perhaps for some reason it is more likely for the &lt;strong&gt;reactants&lt;/strong&gt; to be present in a given reaction, even after it seems to have stopped on a macroscopic scale (even thought it really has not stopped activity), we need to come up with more things to explain why this behaviour might happen. So as scientists, we devise new models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a less than obvious and bigger topic than you might realise. A general chemistry textbook might have a chapter called &lt;strong&gt;chemical kinetics&lt;/strong&gt; or &lt;strong&gt;reaction mechanisms&lt;/strong&gt; which describe this ten times better than I can. Some things that will help you in understanding some models of reactions and their mechanisms are things like &lt;strong&gt;rate laws&lt;/strong&gt; and how to determine rate constants in the laboratory, or how to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy&quot;&gt;&lt;strong&gt;activation energy&lt;/strong&gt;&lt;/a&gt; proposed by Svante Arrhenius in 1880's which basically means that two chemicals need to have at least a certain amount of energy (the activation energy) to overcome their repulsive behaviors and other limiting factors (the energy required to actually break apart a molecule's covalent bonds). For a collision of two particles to produce a different particle, there are other things that determine if a new molecule is formed from a collision of two molecules. First, like I said they must overcome repulsions with &lt;strong&gt;kinetic energy&lt;/strong&gt; but they often need to be &lt;strong&gt;aligned&lt;/strong&gt; spatially for a certain molecule to form in some proposed mechanisms. A mechanism is a proposition that an in-between step occurs in the reaction, but it is very unlikely that two hydrogen molecules will smash into a carbon dioxide molecule and create methane and oxygen. We must propose intermediate steps to make it sound more probable. although this would represent the reverse reaction of combusting methane, and will likely never occur because the potential energy of the system would increase, and we know how much the universe hates that kind of behaviour, if &lt;strong&gt;not&lt;/strong&gt; then read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some concepts that will also help you understand &quot;why reactions happen&quot; are some laws like the 2nd law of thermodynamics which states that the entropy of the universe is always increasing. This explains why a reaction in which a reaction of type that you describe $\ce{ABC + D -&amp;gt; ABCD}$ is actually quite possible. The problem is that the reverse reaction &lt;strong&gt;also&lt;/strong&gt; takes place with a higher probability, these molecules might actually break apart more than they combine. This &lt;strong&gt;two-way&lt;/strong&gt; speciality is why this reaction will likely be written as such: $\ce{ABC + D &amp;lt;=&amp;gt; ABCD}$ indicating that it does possibly happen, but can go both ways, and most often goes backwards. The reason for this is that there is more &lt;strong&gt;entropy&lt;/strong&gt; (aka: disorder) when there are more &lt;strong&gt;moles&lt;/strong&gt; of molecules present. In your examples the reactions are unlikely because the reactants ($\ce{ABC}$ and $\ce{D}$) are twice the number of moles than the products ($\ce{ABCD}$). This is a limiting factor for the reaction to commonly take place. It is useful to note that this reaction &lt;strong&gt;might&lt;/strong&gt; occur once in awhile, but it is more often that molecules fluttering about will break apart than create larger and more complex molecules. Indeed, it is more likely that this reaction will be reversed and proposed as a &lt;strong&gt;mechanism of a larger process&lt;/strong&gt;, if at all, where D breaks off and combines with something else, lets say E and ABC still remains along with a very unbreakable compound called ED which has a triple bond that will likely not be broken at the given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets talk more about &lt;strong&gt;entropy&lt;/strong&gt;: You have a snowball you just rolled up in your hand (it is composed of many flakes of ice). Think of each ice particle that makes up the snowball as an atom in a larger molecule (the snowball). Chances are when this snowball starts moving (like molecules tend to do) it will run into something, and when that happens it will become many different smaller flakes and chunks (atoms and smaller molecules). This is because the universe prefers disorder and chaos. If you threw your clothes freshly out of the dryer into your closet, will they miraculously fold themselves neatley and in order? No, because that would make them more orderly and less chaotic, instead they will flail about in the air and land in a disorganized manner, this then makes sense by the phenomenon of the second law of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look back on your equations now, why can't $\ce{AB + CD -&amp;gt; ABCD }$ occur? Hopefully this is more obvious to you now. There is more disorder in the reactants than the products. This reaction might still occur, but normally balanced chemical equations represent the &lt;strong&gt;net&lt;/strong&gt; affect that tends to occur. Perhaps occasionally someone will win at the game BINGO, but that does not necessarily mean that the same person will keep on winning at BINGO, more likely it will be someone else in the room, and then someone else, etc. With that BINGO analogy, the person who won at bingo played again, and lost. The same way ABCD might have been created, but then almost immediately was destroyed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical engineers might desire to create your molecule $\ce{ABCD}$ and to do this what they might do is add a large amount of $\ce{AB}$ molecules into the reaction chamber. Due to the overwhelming number of these molecules, the reaction is more likely to take place. Not everything in the container will eventually be $\ce{ABCD}$ molecules, but there will be more of them that there was before. These kinds of reactions are actually called &lt;strong&gt;equilibriums&lt;/strong&gt; meaning that the reaction can occur &lt;strong&gt;both ways&lt;/strong&gt;. We write the equilibrium reaction as $\ce{AB + CD &amp;lt;=&amp;gt; ABCD}$ to explicitly show this. The quotient of reactants to products will not be 1:1. Sometimes the rate constant is very large for a reaction implying that it will go to completion, in which case we simply write the reaction in a single direction as you often see with chemical equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is far as I am willing to explain because I am merely repeating snippets of what any chemistry book will explain in a chapter on chemical kinetics, hopefully my repetitive explanation will help you get a better idea of what is actually going on here, but there is still a lot more that you could learn or I will bother to explain.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Food for thought: In the equilibrium reaction $\ce{AB + CD &amp;lt;=&amp;gt; ABCD}$, A chemical engineer decreases the pressure of the reaction chamber, will the reaction shift more in favor of the reactants or the products? These are some questions that chemistry books will ask you after chemical kinetics are discussed. But I do not expect you to be able to answer it yet.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-11T15:04:59.000" LastActivityDate="2012-09-11T15:04:59.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1137" PostTypeId="1" AcceptedAnswerId="1138" CreationDate="2012-09-11T06:03:50.797" Score="21" ViewCount="23388" Body="&lt;p&gt;I haven’t quite reached the point where I can read a full-fledged text on chemical kinetics and thermodynamics yet, so bear with me, please. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m wondering why a value like $K_\text{eq} = \frac{[\ce{NO}]^2[\ce{O2}]}{[\ce{NO2}]^2}$ wouldn't have units of M?&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorDisplayName="user7951" LastEditDate="2015-04-20T12:56:17.257" LastActivityDate="2020-01-15T13:28:22.683" Title="Why are equilibrium constants unitless?" Tags="&lt;kinetics&gt;&lt;equilibrium&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1138" PostTypeId="2" ParentId="1137" CreationDate="2012-09-11T06:24:02.083" Score="14" Body="&lt;p&gt;I goofed up the first time I tried to answer this question, erroneously applying dimensional analysis to your equilibrium expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that Silberberg&lt;sup&gt;[1]&lt;/sup&gt; gives a good explanation of why $K_\text{eq}$ is dimensionless, which is often glossed over as the terms of the equilibrium expression are generally taught as concentrations. In actual fact, the terms are ratios of the concentration or activity of each species with a reference concentration (1 $\mathrm{mol\cdot{L^{-1}}}$ for solutions.) For example, a concentration of 2 $\mathrm{mol\cdot{L^{-1}}}$ divided by a reference of 1 $\mathrm{mol\cdot{L^{-1}}}$ yields a ratio of 2, with no units. As each term has no units, so too does $K_\text{eq}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Silberberg, M.E.; Chemistry – The Molecular Nature of Matter and Change 3e; 2003, p. 719&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorDisplayName="user7951" LastEditDate="2015-04-20T12:57:30.833" LastActivityDate="2015-04-20T12:57:30.833" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1139" PostTypeId="2" ParentId="1137" CreationDate="2012-09-11T09:05:12.467" Score="0" Body="&lt;h1&gt;Note&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;This answer is incorrect, but because it is a common pitfall I'll leave it to show. It is wrong 1) because the OP meant &lt;span class=&quot;math-container&quot;&gt;$\lvert \ce{NO} \rvert ^2$&lt;/span&gt; in the numerator not &lt;span class=&quot;math-container&quot;&gt;$\lvert \ce{NO2} \rvert$&lt;/span&gt; as I took it (edited) and 2) because the &lt;span class=&quot;math-container&quot;&gt;$K_{eq}$&lt;/span&gt; formula is incorrect. I will post a better answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The concise answer: Because the units cancel. Here I'll use &lt;span class=&quot;math-container&quot;&gt;$\lvert \rvert$&lt;/span&gt; = concentration = moles/liter = m/L&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert^2} (?) units$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (?) units$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert m/L \lvert O_2 \rvert m/L}{\lvert {NO}_2 \rvert m/L \lvert {NO}_2 \rvert m/L}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2/L^2}{m^2/L^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} \frac{m^2}{m^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{eq} = \frac{\lvert NO_2 \rvert \lvert O_2 \rvert}{\lvert {NO}_2 \rvert \lvert {NO}_2 \rvert} (unitless)$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-09-11T10:42:35.627" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1140" PostTypeId="2" ParentId="1137" CreationDate="2012-09-11T10:54:20.913" Score="16" Body="&lt;p&gt;&lt;a href=&quot;http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf&quot; rel=&quot;noreferrer&quot;&gt;This source&lt;/a&gt; explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of &lt;span class=&quot;math-container&quot;&gt;$K_{eq}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct.  A more correct version is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where the &amp;quot;ss&amp;quot; subscripts refer to the concentration of that species in the standard state.  (By this definition, Keq is always unitless.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It then goes on to state:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.)  We will NEVER use this &amp;quot;correct&amp;quot; version of the equation in this class (well, never except in one problem on this week’s problem set...), and &lt;span class=&quot;math-container&quot;&gt;$K_{eq}$&lt;/span&gt; for a reaction with unequal numbers of reactants and products is ALWAYS given with units,&#xA;even in published papers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-09-11T10:54:20.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1141" PostTypeId="1" AcceptedAnswerId="1142" CreationDate="2012-09-11T16:38:48.123" Score="16" ViewCount="20990" Body="&lt;p&gt;Hand sanitizer is a gel in its bottle, but when I pour it on my palm, it turns into a free flowing liquid. Why does that happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't turn to a liquid when I pour it on my floor or my table. It doesn't seem to be related to heat, as I tried it also with a hot surface.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="12860" LastEditDate="2015-10-30T17:26:26.867" LastActivityDate="2019-06-28T22:39:56.103" Title="Why does hand sanitizer turn into a liquid after coming out of the bottle?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;alcohols&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1142" PostTypeId="2" ParentId="1141" CreationDate="2012-09-11T17:34:17.367" Score="8" Body="&lt;p&gt;A little research and perhaps a small experiment might help you figure something like this out on your own. First, have you tried &lt;a href=&quot;http://en.wikipedia.org/wiki/Hand_sanitizer&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;? A small mention of research can go a long way. Describing what product you are interested in never hurts either. Is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyacrylic_acid&quot; rel=&quot;noreferrer&quot;&gt;polyacrylic acid&lt;/a&gt;, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersant&quot; rel=&quot;noreferrer&quot;&gt;dispersant&lt;/a&gt; in this substance? That material can absorb water effectively, so here is what I would do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a very small amount of water and smear it on the same surface that it does not &quot;turn into a free flowing liquid&quot; and see if it does with the addition of water. If it works, you might conclude that your alcohol-based rubs use a suspension material that when contacted with a little water, such as from your palms, will cause the gel to have just enough water absorbed inside to decrease its &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity&quot; rel=&quot;noreferrer&quot;&gt;viscosity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you will like to share the results of your experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: OP seems to be observing a gel demonstrating &lt;a href=&quot;http://en.wikipedia.org/wiki/Thixotropy&quot; rel=&quot;noreferrer&quot;&gt;Thixotropy&lt;/a&gt;, this I have determined from a &lt;a href=&quot;https://www.google.com/search?q=hand%20sanitizer%20turns%20from%20gel%20to%20liquid&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&amp;amp;channel=fflb#hl=en&amp;amp;client=firefox-a&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;channel=fflb&amp;amp;sclient=psy-ab&amp;amp;q=hand%20sanitizer%20goes%20from%20gel%20to%20liquid&amp;amp;oq=hand%20sanitizer%20goes%20from%20gel%20to%20liquid&amp;amp;gs_l=serp.3...68397.72606.1.72690.38.20.0.8.8.0.463.3112.5j9j4j0j1.19.0.les;..0.0...1c.1._Q53iJCYm2w&amp;amp;pbx=1&amp;amp;bav=on.2,or.r_gc.r_pw.r_qf.&amp;amp;fp=fcff299af63d19f1&amp;amp;biw=1024&amp;amp;bih=650&quot; rel=&quot;noreferrer&quot;&gt;quick google search&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-16T14:17:21.040" LastActivityDate="2012-09-16T14:17:21.040" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1143" PostTypeId="1" CreationDate="2012-09-11T17:52:09.897" Score="3" ViewCount="1217" Body="&lt;p&gt;How do I find the change in Gibbs free energy of the reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 + \frac{1}{2}O2 -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when it is at $\pu{75^\circ C}$ and $\pu{1 atm}$? Do I calculate the Gibbs free energy of each species and then do &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum\Delta G{\mathrm{(products)}} - \sum\Delta G{\mathrm{(reactants)}}?$$&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="16683" LastEditDate="2017-08-06T07:40:17.137" LastActivityDate="2017-08-06T07:40:17.137" Title="How do I find change in Gibbs free energy of water formation?" Tags="&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1144" PostTypeId="2" ParentId="1114" CreationDate="2012-09-11T20:17:44.317" Score="14" Body="&lt;p&gt;The enthalpy of solution going from the ideal gas state to the solution at infinite dilution is exothermic.  However, more importantly, the Gibbs energy of solution is positive meaning that dissolution of methane into water is not a favored process.  Most of this comes from the relatively large negative entropy change associated with the dissolution at room temperature.  This positive Gibbs energy is manifested in the very small solubilities observed and gives rise to the whole concept of something being &quot;hydrophobic.&quot;  The fact that &lt;em&gt;any&lt;/em&gt; methane dissolves in water is due to equilibrium effects based on a finite positive Gibbs energy.  Since methane is dissolved to a greater extent at lower temperatures must mean that the Gibbs energy is less positive at the lower temperatures.&lt;/p&gt;&#xA;" OwnerUserId="42" LastActivityDate="2012-09-11T20:17:44.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1145" PostTypeId="2" ParentId="1137" CreationDate="2012-09-12T00:41:10.197" Score="3" Body="&lt;p&gt;Dimensional analysis is useless. The correct answer is the one already given involving activities, which are dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activities are defined as ratios. For example a pressure activity (there are many kinds) is defined in terms of the ratio of the actual pressure of a gas divided by the reference pressure, often 1 atm or 1 bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the present example, the activity is the ratio of the molality divided by the reference molality of 1 molal. This assumes ideal solutions, which is good enough if the solutions are dilute. If the solution is not ideal one has to correct the molality for nonideality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of these complications, detailed discussions of activities are usually left for a course in physical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while we use square brackets and molarities, we have to understand that we are really dealing with activities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the activity of a pure liquid or solid is 1, which is why $[\ce{H2O}]$, for example, is dropped from equilibrium calculations.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorDisplayName="user7951" LastEditDate="2015-04-20T13:00:06.133" LastActivityDate="2015-04-20T13:00:06.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1146" PostTypeId="2" ParentId="1143" CreationDate="2012-09-12T04:49:13.277" Score="1" Body="&lt;p&gt;&lt;strong&gt;Warning: this is only a way to estimate and not necessarily good practice.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never actually did one like this, but I believe that one must use the equation $\Delta G$&amp;deg; $= \Delta H$&amp;deg;$ - T\Delta S$&amp;deg; with the assumptions that the information you will need to look up for $\Delta H$&amp;deg; and $\Delta S$&amp;deg; are temperature independent. This will only allow you to estimate $\Delta G$&amp;deg; and remember to convert to kelvin when plugging in the temp.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-12T13:55:57.900" LastActivityDate="2012-09-12T13:55:57.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1147" PostTypeId="1" AcceptedAnswerId="1161" CreationDate="2012-09-12T09:27:30.513" Score="5" ViewCount="917" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The focus of my question here is this: In a laboratory there is a Bunsen burner, a hot plate, hydrochloric acid, and concentrated ammonia. What would you mention about safety precautions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As students of chemistry and science, we often need to write detailed conclusions about our laboratory experiments. This eventually becomes second-nature, but to have an an idea of which of the safety measures taken to include are often useful to ensure ourselves that we have not left anything out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the following are some of the the most important components of any well written conclusion in a lab entry, something that your lab instructor will read and grade you on. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Purpose&lt;/strong&gt;: Explain the goal and purpose of the experiment in a clear and concise manner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Findings&lt;/strong&gt;: Presents a reasonable interpretation of, and logical explanation for all findings pertaining to the problem and stated purpose.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Discussion&lt;/strong&gt;: Discusses possible sources of error in detail, including their effect on the results and ways of avoiding them in the future.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Referencing&lt;/strong&gt; experimental findings and explaining the known/expected results we were looking for. Mentioning and discussing reasons for trends, if any.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular my instructors last year were often interested in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Safety&lt;/strong&gt;: This is often what I always got marked down for. I would explain that hydrochloric acid is a caustic substance and should be treated with care to not get on your tissue by wearing clothing that does not expose skin and closed toe shoes, and to always carry out the experiment with safety goggles securely fastened. It never seems to be enough for them, even if I mention eye flush and chemical shower in case of emergencies. &lt;em&gt;Should I mention to not snort or freebase it? /end sarcasm&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have I missed any safety components that should be included in a &quot;complete&quot; conclusion for the average experiment? What else might I have left out that instructors tend to look for?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="73" LastEditDate="2012-09-13T06:09:07.833" LastActivityDate="2012-09-13T16:10:56.800" Title="What components of safety should be included in a chemistry laboratory experiment conclusion?" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1148" PostTypeId="2" ParentId="1104" CreationDate="2012-09-12T10:03:58.643" Score="4" Body="&lt;p&gt;The best option would be an organic extraction with dichloromethane. All the other ingredients (except possibly D&amp;amp;C yellow #10 aluminum lake) will be insoluble and therefore unaffected. Filter, let the dichloromethane evaporate and you should have your caffeine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I agree with J M. This caffeine IS NOT EDIBLE. You could easily become intoxicated by the small amounts of dichloromethane that will remain. Also be careful when handling the dichloromethane. Vapors are also toxic and if the liquid is heated or shaken in a closed recipient it might explode. Use caution.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="-1" LastEditDate="2012-09-12T22:11:40.270" LastActivityDate="2012-09-12T22:11:40.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1149" PostTypeId="2" ParentId="493" CreationDate="2012-09-12T10:49:06.543" Score="8" Body="&lt;p&gt;The reaction you propose would be formally equivalent to the protection of aldehydes with 1,3-propanedithiol to form the correspondent &lt;a href=&quot;http://en.wikipedia.org/wiki/Dithiane&quot;&gt;dithiane&lt;/a&gt;. There are many organic reactions in which aldehydes participate this way, but all of them have something in common: the other species(lets call it donor) must have a free pair of electrons to attack the carbon atom in the aldehyde. In case this pair is not present, but the donor has an acidic hydrogen, we could use a strong base to remove it and form in situ the pair of electrons. (See &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldol_reaction&quot;&gt;aldol reactions&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the problem: diamonds have no free pairs of electrons. And terminal hydrogens in a diamond lattice are by no means acidic, hence can't be removed by any base to the best of my knowledge. So there's no way a diamond will react with aldehydes.&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2012-09-12T10:49:06.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1151" PostTypeId="1" CreationDate="2012-09-12T13:54:21.987" Score="1" ViewCount="2669" Body="&lt;p&gt;I don't know how to obtain the &quot;exact charge of the ions&quot; that will be formed by each atom mentioned. I also have to figure out whether they are positive or negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen, for example, can form the ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H -&amp;gt; H+ + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if I were to try to get the exact charge for say, boron using only the periodic table, how would I do so?&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="7448" LastEditDate="2016-04-02T13:17:35.367" LastActivityDate="2016-04-02T13:17:35.367" Title="Determining the exact charge of ions that will be formed by an atom using the periodic table" Tags="&lt;ions&gt;&lt;electrons&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1152" PostTypeId="2" ParentId="1151" CreationDate="2012-09-12T14:22:45.837" Score="4" Body="&lt;p&gt;Each atom tends to imitate the noble gases which have a full valence shell. Looking at your table you can see that boron is 3 spaces above helium, but 5 spaces behind neon. Boron would &quot;prefer&quot; to have a configuration like the noble gases, and because losing 3 electrons seems easier than gaining 5, boron, as boring as he is, decides to do the obvious thing and three electrons walk the plank... Now he has 5 protons and only 2 electrons therefore imitating helium's electron configuration. This means boron has a charge of $+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation states are common on most periodic tables, which show you the result of less than obvious behavior of atoms when this trend does not always hold. In the case of hydrogen, it is common than it will bond with another hydrogen to imitate helium, otherwise it will tend to form a $+1$ ion.&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2012-09-12T14:22:45.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1153" PostTypeId="5" CreationDate="2012-09-12T22:47:54.463" Score="0" Body="&lt;p&gt;For questions which might pertain in particular to students of chemistry, researchers, instructors and assistants in the academic world.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2012-09-13T13:59:23.963" LastActivityDate="2012-09-13T13:59:23.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1154" PostTypeId="4" CreationDate="2012-09-12T22:47:54.463" Score="0" Body="For questions which might pertain in particular to students of chemistry, researchers, instructors and assistants in the academic world." OwnerUserId="584" LastEditorUserId="73" LastEditDate="2012-09-13T13:59:23.963" LastActivityDate="2012-09-13T13:59:23.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1155" PostTypeId="1" CreationDate="2012-09-13T02:45:17.487" Score="5" ViewCount="2150" Body="&lt;p&gt;I've been looking on the net for a reference which shows a phase diagram for elements at least from $\ce{H}$ up to $\ce{Ne}$. Specifically, I'm looking which of these elements can be solid or liquid in a vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't really found anything, except some anecdotical comments here and there that hydrogen and helium evaporate in a vacuum at any temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions as to where these data might be published?&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorDisplayName="user7951" LastEditDate="2016-02-23T19:26:37.787" LastActivityDate="2016-02-23T19:26:37.787" Title="Phase diagram for first 10 elements" Tags="&lt;phase&gt;&lt;elements&gt;&lt;periodic-table&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1156" PostTypeId="1" AcceptedAnswerId="1176" CreationDate="2012-09-13T03:01:25.000" Score="6" ViewCount="591" Body="&lt;p&gt;Let's say I truncate the virial equation after the third term for use as my equation of state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${P\over\rho RT}=1+B\rho+C\rho^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tabulated values for $B(T)$ and $C(T)$.  I know $P$ and $T$ and want to solve for $\rho$.  The equation is cubic in $\rho$ meaning I could potentially have three real solutions.  I suspect that since the virial equation is intended to calculate the densities of vapors or supercritical gasses at relatively low pressures, that the lowest real root is the &quot;real world&quot; value.  Is this correct?  Do the other real values, if they exist, have any meaning?  Can this be generalized to higher order virial expansions?&lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="42" LastEditDate="2012-09-13T03:07:27.600" LastActivityDate="2012-09-16T06:00:58.713" Title="What meaning do multiple real solutions of a truncated virial equation have, if any?" Tags="&lt;equation-of-state&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1157" PostTypeId="1" CreationDate="2012-09-13T05:30:37.483" Score="3" ViewCount="123" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some barium chloride is added to a solution that contains both &lt;span class=&quot;math-container&quot;&gt;$\ce{K2SO4}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.050 M}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na3PO4}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.020 M}$&lt;/span&gt;).&lt;br&gt;&#xA;  (a) Which begins to precipitate first: the barium sulfate or the barium phosphate?&lt;br&gt;&#xA;  (b) The concentration of the first anion species to precipitate, either the sulfate or phosphate, decreases as the precipitate forms. What is the concentration of the first species when the second begins to precipitate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've figured out the first part of this problem, that barium phosphate is the first to precipitate (due to lower &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt;), but I can't seem to figure out the other part. I even read about selective precipitation and I can't get the answer that was in the back of the book (namely, &lt;span class=&quot;math-container&quot;&gt;$\pu{1.1 \times 10^{-6} M}$&lt;/span&gt;).&lt;br&gt;&#xA;Granted, sometimes I'm pretty there are errors in the books I use to teach myself, but this is not one of those instances where I am so sure.&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="4945" LastEditDate="2019-03-14T15:16:19.673" LastActivityDate="2019-03-14T15:16:19.673" Title="How much of the anion species is left over after the first precipitation?" Tags="&lt;equilibrium&gt;&lt;precipitation&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1158" PostTypeId="2" ParentId="1155" CreationDate="2012-09-13T07:38:59.387" Score="3" Body="&lt;p&gt;&lt;strong&gt;All elements exist in as a solid&lt;/strong&gt; in at least some pressure and temperature conditions, i.e. in some region of their phase diagram. In fact, all elements present a solid form at zero pressure (or atmospheric pressure) if you cool them enough, i.e. when they reach 0 K. All…  &lt;a href=&quot;http://en.wikipedia.org/wiki/Helium#Solid_and_liquid_phases&quot; rel=&quot;nofollow&quot;&gt;except helium&lt;/a&gt;, which requires some pressure to solidify.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-13T07:38:59.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1159" PostTypeId="1" CreationDate="2012-09-13T13:21:45.473" Score="2" ViewCount="112" Body="&lt;p&gt;I wish to grow algae in a 40% $\ce{CO2}$ (water under air with 40% $\ce{CO2}$ content) aqueous environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple DIY methods are available, but they are generally unreliable (&lt;a href=&quot;http://www.theplantedtankfaq.com/CO2_FAQ.html#&quot; rel=&quot;nofollow&quot;&gt;For more info on DIY read here&lt;/a&gt;) because they rely on a &lt;a href=&quot;http://www.theaquatools.com/co2-calculations&quot; rel=&quot;nofollow&quot;&gt;pH/KH chart&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atmospheric $\ce{CO2}$ levels are measured based on &lt;a href=&quot;http://www.esrl.noaa.gov/gmd/ccgg/about/co2_measurements.html&quot; rel=&quot;nofollow&quot;&gt;IR absorbency&lt;/a&gt;. However, such measurements are difficult and &quot;complicated by the temperature signal generated by carbon dioxide as well as by the &lt;a href=&quot;http://airs.jpl.nasa.gov/story_archive/Measuring_CO2_from_Space/History_CO2_Measurements/&quot; rel=&quot;nofollow&quot;&gt;signatures of ozone and water vapor&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reliable way to measure the $\ce{CO2}$ concentration that is relatively inexpensive (under $1,000)? Is the &lt;a href=&quot;http://www.aquaticplantenthusiasts.com/co2/4-drop-checker-revealed.html&quot; rel=&quot;nofollow&quot;&gt;drop checker method&lt;/a&gt; reliable?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4231" LastEditDate="2015-02-23T19:51:41.233" LastActivityDate="2015-02-23T19:51:41.233" Title="Maintaining a 40% CO₂ aqueous environment?" Tags="&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1161" PostTypeId="2" ParentId="1147" CreationDate="2012-09-13T16:10:56.800" Score="3" Body="&lt;p&gt;Perhaps your instructors want more specific information about the hazard itself, rather than generic safety advice.  Some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) HCl: Strong acid, highly corrosive (nitpick: caustic is usually reserved for strong bases, corrosive for acids), emits corrosive and toxic fumes (not all acids do)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 29% NH4OH: Caustic, emits strong fumes (lachrymator--makes your eyes water), work in a fume hood, outdoors or in a well-ventilated area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Bunsen burner: No open flames in the presence of flammable vapors or liquids with high vapor pressure (much bigger hazard than just burning yourself)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Hotplate: Hotplate looks the same whether hot or cold, so assume it is hot, take care with flammable liquids with low flash points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these are just examples, but I tried to tailor the answer to the item in question.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2012-09-13T16:10:56.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1162" PostTypeId="2" ParentId="1155" CreationDate="2012-09-13T16:54:32.773" Score="4" Body="&lt;p&gt;If you want to know the temperature at atmospheric pressure below which an element is a solid, just look up the melting point of that element.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for evaporation of solids, ALL materials evaporate.  That is, all materials exhibit a positive non-zero vapor pressure.  That vapor pressure will vary with temperature.  When the vapor pressure hits 1 atm, you are at the boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For phase diagrams of the elements simply Google for them.  One promising source is &lt;a href=&quot;http://www.osti.gov/bridge/purl.cover.jsp?purl=/4010212-0BbwUC/4010212.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="73" LastEditDate="2012-09-25T18:44:21.107" LastActivityDate="2012-09-25T18:44:21.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1163" PostTypeId="2" ParentId="1155" CreationDate="2012-09-13T17:53:56.463" Score="4" Body="&lt;p&gt;from thermodynamical point of view at absolute zero pressure each and every compound will evaporate, given enough time. So, it is better to stick with finite (i.e. small, but non-zero) pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, phase diagrams usually has 'triple point' (at which solid, liquid and gas coexists) with three areas - lower-right one, where gas is stable, left one, where solid is stable, and upper, where liquid is stable, with barrier between liquid and gas fading at some point. This means, that liquid is doubtful to exist in vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, you can hope for slowly evaporating solids, which are ones with crystal structure hold by metallic or covalent bond. Evaporation speed should drop with boiling point raise, which drops in the row of $\ce{C}$ (3915 K), $\ce{B}$ (3927 K), $\ce{Be}$ (2725 K), $\ce{Li}$ (1342 K). $\ce{Be}$ is sometimes used as construction material in satellite construction, so it should work fine, but is very toxic. Carbon should work fine as well, as to my knowledge was used in thermonuclear fusion program (where both vacuum and hight temperature stability are required) , but is hard to work with. $\ce{Li}$ and $\ce{B}$ are out of consideration except VERY special application, as $\ce{Li}$ is highly corrosive and flammable and $\ce{B}$ is almost as hard to work with as diamond.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T04:34:44.303" LastActivityDate="2015-06-11T04:34:44.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1164" PostTypeId="1" CreationDate="2012-09-13T22:50:52.707" Score="7" ViewCount="432" Body="&lt;p&gt;Both $\ce{H2}$ or $\ce{O2}$ are stable molecules. However when given energy they heat up and form $\ce{H2O}$. My teacher explained that the increased energy breaks the bonds, allowing new ones to form. However, why don't the old forms reform instead? Why are new bonds created?&lt;/p&gt;&#xA;" OwnerDisplayName="user612" LastEditorUserId="4231" LastEditDate="2015-02-22T23:20:39.127" LastActivityDate="2015-02-22T23:20:39.127" Title="Why do H₂ and O₂ create H₂O?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1165" PostTypeId="2" ParentId="1156" CreationDate="2012-09-14T01:16:04.200" Score="3" Body="&lt;p&gt;The virial equation is set up for the gas phase only.  The virial coefficients $B$ and $C$ are found from gas phase measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get three solutions, I suspect that it is the largest one that you want.  I'd do a trial solution with something like water vapor to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher order virial equations are problematic since it is exceptionally difficult to determine higher virial coefficients from experimental data.  Higher order virial equations do have theoretical interest, but that's another story.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-09-14T01:16:04.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1167" PostTypeId="2" ParentId="1164" CreationDate="2012-09-14T08:14:57.010" Score="10" Body="&lt;p&gt;Are you familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;activation energy&lt;/a&gt;?  Consider a hole at the top of a large hill.  If a ball or some water was in that hole it would seek the lowest point it could find.  Clearly, the bottom of the hill would be preferable to the bottom of the hole at the top of the hill, however, in order to get there it must first be brought up over the top of the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oik1Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture, the $H_2$ and $O_2$ are on the right side of the curve.  Just like water in the above analogy, they'd like to move to the lower area, but first must be brought up over the bump.  So some energy must be added to accomplish this, and then they end up in the much lower energy position on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once over the bump, they stay as $H_2O$ because that is the state they'd like to be in at any time, but were prevented from being in before by the activation energy bump.&lt;/p&gt;&#xA;" OwnerUserId="632" LastActivityDate="2012-09-14T08:14:57.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1168" PostTypeId="2" ParentId="1164" CreationDate="2012-09-14T10:57:45.033" Score="5" Body="&lt;p&gt;Adding on to DaleSwanson's correct answer: The ball-and-hill analogy is just saying that a ball will roll down a hill. This is because if the path is available it will find a way to the bottom of the hill due to there being less potential energy in the ball down at the bottom than up at the peak. Think of the activation energy as the little 'push' needed to get the ball moving off the edge of the cliff, once it is nudged off the side it will release almost all the potential energy that was released through kinetic energy when it hits the ground (small amounts of potential energy will also be released in other forms, i.e: friction of the air and ground).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To answer your question, I ask you this: you accidently knock a pen off your desk, it required so little energy it was almost unnoticeable. Would it require more energy for you to reach down and put the pen back on your desk? What position is &quot;more stable&quot; for the pen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and oxygen are stable, yes, they have full valence shells and are satisfied. Just like a ball sitting on the top of the hill is stable, until you give it a nudge. Now think of the top of the hill as the hydrogen and oxygen molecules. Also, think of the bottom of the hill as the water molecule. Take a look at the &lt;strong&gt;balanced&lt;/strong&gt; chemical equation now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 H2 + O_2 -&amp;gt; 2H_2_O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;strong&gt;less&lt;/strong&gt; energy at the bottom of the hill, the same way that there is &lt;strong&gt;less&lt;/strong&gt; energy in the covalent bonds that make up the products. There is &lt;strong&gt;more energy&lt;/strong&gt; stored in the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very possible to calculate the energies of the bonds and compare the energies of the reactants to the products in the balanced chemical equation by looking up heat of formation's for these species, and calculate the change in energy from going to reactants to products. That is, we can calculate how much energy is released when two moles of $\ce{H_2O}$ are created from this reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we can calculate how much energy a ball will release when falling from the cliff. The release in energy to the surroundings during a chemical reaction is said to be an exothermic process, whereas an endothermic process would take away energy from the surroundings, such as vaporizing water molecules into steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using the analogy you have learned, would it be reasonable to say that an equally small nudge that caused the ball to fall would be enough to cause the ball to rise up back onto the peak of the hill? You tell me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better picture of the path from reactants to products would be different than a smooth 2d hill picture. Sure, the change in energy is not as outrageous as I portray it, and perhaps in some miraculously rare situation some invaders from another country find the ball at the bottom of the hill, they put it in a catapult and launch it at their enemies at the top of the hill. This is extremely rare and very situational. In other words there is almost no chance it will happen. As it turns out, you are thinking very rationally when you ask why the reverse process does not happen! The problem is you have to think on the molecular scale where molecules are colliding into each other. It is very likely that a $H_2$ will collide with the $O_2$ and one side of the oxygen breaks into the bond of the hydrogen and you now have $H_2O + O$. But how would seperate $H_2 + O_2$ form from two $H_2O$ colliding, the answer is that it is just extremely unlikely, it would more likely create $H_2O_2 + H_2$ &lt;strong&gt;if at all!&lt;/strong&gt; There are other steps in-between that would need to account for going backwards in the reaction. You can think of this as a separate path &lt;strong&gt;around&lt;/strong&gt; the hill. There are more ways back up the hill in other words, but they may lead to dead-ends!&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-14T11:21:21.203" LastActivityDate="2012-09-14T11:21:21.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1169" PostTypeId="1" AcceptedAnswerId="1170" CreationDate="2012-09-14T21:16:22.573" Score="3" ViewCount="1312" Body="&lt;p&gt;Are there ways to compute what chemicals kill a give bacteria but not another given bacteria ?&#xA;Does that help much to make medication ?&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="4945" LastEditDate="2014-10-24T10:40:42.080" LastActivityDate="2014-10-24T10:40:42.080" Title="Are there ways to predict which chemicals will kill certain bacteria?" Tags="&lt;biochemistry&gt;&lt;medicinal-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1170" PostTypeId="2" ParentId="1169" CreationDate="2012-09-14T23:26:32.290" Score="10" Body="&lt;p&gt;The basic idea is sound, and is used to some degree; however, there are significant difficulties inherent in &quot;tailoring&quot; a chemical to selectively target certain organisms and not others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three things that a chemical can do that make it toxic to cellular life:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Rip some vital chemical to shreds.&lt;/strong&gt; Chlorine and other light halogens (fluorine, iodine) are &lt;em&gt;huge&lt;/em&gt; here, because they have such high electron affinity that they can tear apart the otherwise relatively stable bonds within most organics, destroying that molecule. The most effective ones don't form any stable compounds; fluorine, for instance, is very effective at shredding organic compounds because it isn't happy just barging in on any one organic molecule and making itself at home; it'll deprotonate this one, split that one in half, before finally binding tightly and insolubly to calcium (robbing your body of that essential mineral).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bind to a vital chemical and not let go.&lt;/strong&gt; Carbon monoxide is a deadly danger to most aerobic organisms because the hemoglobin (or related compound) that loosely binds oxygen for transport to cells will bind more stably with CO, so the chemical can't transport oxygen and can't get rid of the CO. A lot of more selective poisons attach themselves to enzymes, not necessarily destroying them, but preventing the enzyme doing the work it's supposed to. Cyanide is poisonous specifically because it works like (well, it is) a nitrile group, of which there are very many in a lot of inorganic biochemicals, and specifically it targets a key enzyme involved in cell respiration and programmed cell death.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mimic a chemical's interaction with a key receptor, causing it to go haywire.&lt;/strong&gt; Many other &quot;selective&quot; poisons and antibiotics work this way; they fit into the receptor sites that a neurotransmitter or enzyme would, triggering some behavior of the cell, and by flooding the cellular environment with this chemical you cause the underlying programmed reaction to go haywire (or stop completely). A lot of insecticides and pesticides do this; they increase or shut down cell metabolism, or in creatures with a nervous system they mimic or trigger the release of massive amounts of neurotransmitters, basically causing cardiac or respiratory arrest because any normal message to control these actions is completely overridden.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem with all this comes in identifying which bacteria have a particular enzyme, receptor or other targeted chemical process, and thus would be susceptible to poisoning by the chemical. In addition, the chemical must not be toxic to human biochemistry (otherwise, no matter how beneficial it might be as a disinfecting agent, it will never see approval as an antibiotic medicine). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antibiotics must have selective toxicity; while they do not affect eukaryotic cells (such as the ones that make up most plants and animals, thus allowing the chemical to be taken internally or directly applied), they have a much more pronounced effect on bacterial and archaeal cells. Usually, the chemical targets something common to a large class of bacteria (such as the thick peptidoglycan mesh forming the cell wall of Gram-positive bacteria), or most/all bacterial cells (such as the S30 robosomal subunit, key to bacterial protein synthesis and thus found in nearly all bacteria, but completely absent from eukaryote cells which use a different mechanism). However, even within an antibiotic's &quot;target class&quot;, there are bacteria that are naturally resistant (perhaps they don't accept the chemical through the cell membrane due to some variation unique to the genus) or develop genetic resistance through overuse or misuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, bacteria are fickle; most are at least harmless if not beneficial in certain areas (E. coli, for instance, produces the essential vitamin K&lt;sub&gt;2&lt;/sub&gt; in your intestine), but lethal in others (ingestion of e-coli or introduction through an open cut can cause a fatal infection). Staph is relatively harmless on the skin, but fatal in the spine. Antibiotics don't stop and question bacteria; they kill any bacteria that comes across them, and so overuse of antibiotics to kill one harmful bacteria will kill the bacteria in beneficial places as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, not every infection is bacterial. Fungi and protozoa produce a significant portion of food-borne illnesses and poisonings, as well as other types of infection. These life forms are eukaryotes and unaffected by most antibacterial agents (in fact, penicillin was discovered when a fungus contaminated a bacterial culture; the fungus &lt;em&gt;produces&lt;/em&gt; the antibiotic chemical as a natural defense).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-01-31T20:30:08.247" LastActivityDate="2013-01-31T20:30:08.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1171" PostTypeId="1" AcceptedAnswerId="1173" CreationDate="2012-09-14T23:53:17.770" Score="6" ViewCount="607" Body="&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/88K9F.png&quot; alt=&quot;picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the first row of the iCe-table, $[\ce{H3O+}]$ is said to be 0. But water autoionizes to some extent, so it shouldn't &lt;em&gt;really&lt;/em&gt; be 0. I accept that, because the pH of this solution is 2.5, the autoionization effect is negligible but are there instances where this can't be ignored (e.g. where the pH is close to 7?)? Academic, industrial (or hobby) examples would all be fine.&lt;/p&gt;&#xA;" OwnerUserId="550" LastEditorUserId="12860" LastEditDate="2015-11-06T19:18:15.770" LastActivityDate="2015-11-06T19:18:15.770" Title="How useful is the assumption that water doesn't auto-ionize in the following?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1173" PostTypeId="2" ParentId="1171" CreationDate="2012-09-15T06:19:31.593" Score="8" Body="&lt;p&gt;First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem.&#xA;You can write the full system of equations describing the aqueous equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).&lt;/li&gt;&#xA;&lt;li&gt;electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$&lt;/li&gt;&#xA;&lt;li&gt;water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$&lt;/li&gt;&#xA;&lt;li&gt;acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, &lt;strong&gt;this is indeed a good approximation.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorDisplayName="user7951" LastEditDate="2015-04-15T18:15:24.747" LastActivityDate="2015-04-15T18:15:24.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1175" PostTypeId="2" ParentId="374" CreationDate="2012-09-16T03:31:51.707" Score="1" Body="&lt;p&gt;Here iron becomes +1 due to presence of the ligand $\ce{NO}$ +1. The compound $\ce{[Fe(H2O)6]^{2+}}$ is formed when $\ce{FeSO4}$ dissolves in water during the brown ring test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we add an aqueous solution of nitrate ions in $\ce{FeSO4}$ solution $\ce{Fe^{(II)}}$ are converted to $\ce{Fe^{(III)}}$ and $\ce{NO3-}$ are converted to $\ce{NO}$.&lt;br&gt;&#xA; These $\ce{NO}$ being an odd electron compound can use  three of its electrons among which it donates two electrons to $\ce{Fe^{(II)}}$ while one it shares with incompletely filled $\ce{d}$ orbital and displace one water ligand bonded with secondary valency to $\ce{Fe}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way $\ce{Fe^{(II)}}$ is converted to $\ce{Fe^{(I)}}$ and not $\ce{Fe^{(III)}}$ In these regard we get the colour brown instead of yellow for $\ce{Fe^{(III)}}$ or green for $\ce{Fe^{(II)}}$&lt;/p&gt;&#xA;" OwnerUserId="637" LastEditorUserId="4945" LastEditDate="2015-01-05T03:00:13.777" LastActivityDate="2015-01-05T03:00:13.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1176" PostTypeId="2" ParentId="1156" CreationDate="2012-09-16T06:00:58.713" Score="3" Body="&lt;p&gt;I don’t have much practical experience with virial equations, though I teach it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendations would be to identify the correct root based on physical knowledge of the system: first, the root has to be real and its value of $\rho$ has to be positive. This part is obvious. The second thing we know is that the virial equation represents a deviation from  the ideal gas law. So, I would take &lt;strong&gt;the root that is closest to ideal behavior&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to confirm it with an example, so I investigate water at 298 K (based on virial coefficients given in K.M. Benjamin et al., &lt;em&gt;J. Phys. Chem. C&lt;/em&gt; &lt;strong&gt;2007&lt;/strong&gt;, 111, 16021-16027). Here's a plot hastily made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TDUeb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least in these conditions, this rule works.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-16T06:00:58.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1177" PostTypeId="1" AcceptedAnswerId="1182" CreationDate="2012-09-16T08:35:51.023" Score="5" ViewCount="496" Body="&lt;p&gt;I received the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Complete the missing values for the four quantum numbers:&#xA;  $n=?$, $\ell = 2 , m_\ell = 0$ , $m_s= ? $ .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problem is that I think $n$ can be any number greater then 2, and $m_s$ can be $\frac{1}{2} $ or $-\frac{1}{2}$, so there seems to be no unique solution to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="638" LastEditorUserId="73" LastEditDate="2012-09-16T19:45:02.423" LastActivityDate="2012-09-16T19:45:02.423" Title="Does this set of quantum numbers have a unique solution?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1178" PostTypeId="1" AcceptedAnswerId="1180" CreationDate="2012-09-16T09:29:57.380" Score="2" ViewCount="486" Body="&lt;p&gt;As far as I know, when writing Lewis structures for molecules, we need to consider the octet rule and the valence electrons each atom has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if we take for example, the molecule $ Cl O_2^- $ , we have the following Lewis structure: (as it appears &lt;a href=&quot;http://www.youtube.com/watch?v=j1klCSe62-I&amp;amp;feature=channel&amp;amp;list=UL&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; ) &#xA;O - Cl - O with 6 valence electrons surrounding each oxygen atom, and 4 valence electrons surrounding the chlorine atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is:&#xA;The oxygen atom has 6 valence electrons. By forming a chemical bond with the chlorine atom, it shares one electron, which leaves it with a total of 5 non-bonded electrons. So, how can it be that in the Lewis structure of this molecule, it has 6 non-bonded electrons and one shared electron? (or a total of 7 electrons! ) &lt;/p&gt;&#xA;" OwnerUserId="638" LastEditorUserId="73" LastEditDate="2012-09-18T17:34:17.443" LastActivityDate="2012-09-18T17:34:17.443" Title="Lewis structures of molecules containing oxygen" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1179" PostTypeId="1" AcceptedAnswerId="1181" CreationDate="2012-09-16T09:41:57.377" Score="4" ViewCount="1006" Body="&lt;p&gt;I need to find the Lewis structure of the ion &lt;span class=&quot;math-container&quot;&gt;$\ce{SCH^-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the Lewis structure should be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{S=C-H}$&lt;/span&gt;, with one lone pair on the carbon and two lone pairs surrounding the sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="638" LastEditorUserId="4945" LastEditDate="2020-02-22T12:15:49.987" LastActivityDate="2020-02-22T20:31:26.527" Title="What is the correct lewis structure of SCH⁻?" Tags="&lt;organic-chemistry&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1180" PostTypeId="2" ParentId="1178" CreationDate="2012-09-16T11:21:43.480" Score="5" Body="&lt;p&gt;If you consider the $\ce{ClO2-}$ as coming together from neutral atoms (plus the extra electron), all the bonding &lt;em&gt;electrons&lt;/em&gt; &quot;come from&quot; the chlorine. This is commonly called a '&lt;a href=&quot;http://en.wikipedia.org/wiki/Dative_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;dative covalent bond&lt;/a&gt;' (or just 'dative bond' for short), and is sometimes drawn like this, instead of just a line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JbU8K.png&quot; alt=&quot;Dative Bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this image suggests, you can think of it as an atom with a lone pair bonding with an atom with no spare valence electron.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-09-16T11:21:43.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1181" PostTypeId="2" ParentId="1179" CreationDate="2012-09-16T13:52:48.043" Score="8" Body="&lt;p&gt;This model seems reasonable, as you can see from the formal charges carbon has the -1 value while the rest are 0. One thing that you have forgot is to add brackets around the structure to show that it is a ion, include its overall charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[\overset{..}{\underset{..}{S}}=\overset{..}{C}-H]-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, I suppose a structure of $\ce{[\overset{..}{C}#\overset{..}{S}-H]-}$ would work with the same formal charge, except that sulfur has the expanded octet, which is possible but I am not sure if that is more reasonable or not, it probably is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a very odd ion, to say the least, probably very unstable. Perhaps you have mistaken the hydrogen with nitrogen? It would be more reasonable to be asked to draw the lewis structure for $\ce{[SCN]-}$, thiocyanate ion. I am not even know how to pronounce or name the $\ce{[SCH]-}$ion you are required to draw, or if it even exists. Maybe that is why you need to draw it though.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="41328" LastEditDate="2017-10-31T04:44:10.307" LastActivityDate="2017-10-31T04:44:10.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1182" PostTypeId="2" ParentId="1177" CreationDate="2012-09-16T15:06:03.303" Score="5" Body="&lt;p&gt;You haven't gotten anything wrong, you are entirely correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of quantum numbers can be used to identify the specific electron in the sub-atomic world of an atom, and what path it occupies in three dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ell = 2$ refers to the $d$ orbital. Now how many electrons can a d orbital contain? It has 5 &quot;slots&quot; because we can go from -2, -1, 0, 1, 2. These represent the orientations, each one holding up to two electrons at different spins (up or down, -1/2 or +1/2). We therefore know that $\ell = 2$ refers to a possible 10 different unique sets of quantum numbers, because each set could represent a specific electron, of which there could be up to 10 in the d orbitals, only 6 for p orbitals, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;principal quantum number&lt;/strong&gt; ($n$): Has only integral values (1, 2, 3, etc) starting at 1. Think of this as the &lt;strong&gt;power level&lt;/strong&gt;. An electron in a 3s, 3p, 3d, etc orbital is in the third &lt;strong&gt;power level&lt;/strong&gt;. It represents the &lt;strong&gt;numbers&lt;/strong&gt; before the letter in the electron configuration such as 1s$^2$2s$^2$2p$^6$ so in other words, it represents what level the electron is on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;angular momentum quantum number&lt;/strong&gt; ($\ell$): goes only from 0 to $n-1$, also uses integral values like $n$. Think of this as the &lt;strong&gt;letter&lt;/strong&gt; after the number in the electron configuration such has 1s$^2$2s$^2$2p$^6$ so in other words it represents which orbital we are talking about on that power level. 0 represents $s$, 1 for $p$, 2 for $d$, 3 for $f$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;magnetic quantum number&lt;/strong&gt; ($m_\ell$): Goes from $-\ell$ to $\ell$ including 0. Think of this value as representing a particular orientation of an orbital. The number of orientations will vary depending on the orbital. s orbitals are spherical and can only have one orientation, s orbitals are referred to by saying $\ell = 0$. If we said $\ell = 1$ we would be talking about the p orbital, which has -1, 0, and 1 orientations (3 total). This can be thought of as the x, y, and z axis. $d$ orbitals have 5 possibilities, $f$ has 7, etc. I like to refer to each of these orientations as a &quot;slot&quot; which can hold two electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value for $m_s$ is used to identify the spin either up or down, but not both.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-09-16T18:18:40.713" LastActivityDate="2012-09-16T18:18:40.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1185" PostTypeId="1" AcceptedAnswerId="1188" CreationDate="2012-09-17T03:12:48.683" Score="8" ViewCount="395" Body="&lt;p&gt;I'm reading an &lt;a href=&quot;https://www.netl.doe.gov/technologies/carbon_seq/refshelf/project%20portfolio/2009/Overview/Selected%20CS%20Program%20Papers/2008%20Jan%20Advances%20in%20CO2%20Capture%20IJGGC.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; about carbon sequestration, and I came across a strange phrase:  &quot;The flue gas is dilute in $\ce{CO2}$&quot;.  I can't quite ascertain its meaning from the context.  I think it means that the $\ce{CO2}$ is in low concentration in the flue gas, but I'm no chemistry expert.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have a technical meaning in chemstry when a gas is dilute in another substance?  &lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="4231" LastEditDate="2015-02-23T19:52:14.017" LastActivityDate="2015-02-23T19:52:14.017" Title="What does it mean when &quot;a gas is dilute in CO₂&quot;?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1186" PostTypeId="1" AcceptedAnswerId="1194" CreationDate="2012-09-17T04:02:01.620" Score="7" ViewCount="730" Body="&lt;p&gt;I'm reading &lt;a href=&quot;https://www.netl.doe.gov/technologies/carbon_seq/refshelf/project%20portfolio/2009/Overview/Selected%20CS%20Program%20Papers/2008%20Jan%20Advances%20in%20CO2%20Capture%20IJGGC.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article about carbon capture from the combustion of fossil fuels.  In this article, it says that amines are used as a chemical absorbent to capture the $\ce{CO2}$ from the exhaust of the combustion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I have read that N-Methyl diethanolamine (MDEA) is used (a &lt;a href=&quot;http://training.swpartnership.org/CCSclass/CCS_Module_files/co2_capture_sept_2012.pdf&quot; rel=&quot;nofollow&quot;&gt; reference &lt;/a&gt; lists the formula as $\ce{CH3N (C2H4OH)2}$).  I am curious if anyone can explain what happens during the reaction with $\ce{CO2}$ and how it absorbs the $\ce{CO2}$ from the flue gas.  &lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="10185" LastEditDate="2015-09-13T15:31:55.620" LastActivityDate="2015-09-13T15:33:22.870" Title="How do amines react with flue gases to absorb CO₂?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1187" PostTypeId="1" AcceptedAnswerId="5803" CreationDate="2012-09-17T04:11:43.123" Score="5" ViewCount="589" Body="&lt;p&gt;I'm trying to understand pre-combusiton carbon capture processes as described in &lt;a href=&quot;http://training.swpartnership.org/CCSclass/CCS_Module_files/co2_capture_sept_2012.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; presentation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told that solvents such as chilled methanol ($\ce{CH3OH}$) and dimethyl ethers of polyethylene glycol ($\ce{C_{2n}H_{4n + 2}O_{n + 1}}$) can dissolve $\ce{CO2}$ from a gasified coal fuel (before combustion) under the action of increased pressure.  I'm struggling in understanding how the increased pressure induces the dissolution of $\ce{CO2}$ from the gasified coal.  I'm also a bit rusty with my chemistry, by saying $\ce{CO2}$ is dissolved by the solvents, what is really going on, chemically?&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="4231" LastEditDate="2015-03-02T23:13:02.917" LastActivityDate="2015-03-02T23:13:08.260" Title="Understanding dissolution of CO₂ in Selexol and Rectisol" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;solvents&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1188" PostTypeId="2" ParentId="1185" CreationDate="2012-09-17T08:47:52.783" Score="9" Body="&lt;p&gt;In this case, it's comparing post-air-combustion $\ce{CO2}$ capture with pre-combustion capture, and with post oxy-combustion capture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the two latter cases, the  $\ce{CO2}$ stream is very highly concentrated.  However, with air-combustion, the flue gases contain a lot of nitrogen and NOx; hence the $\ce{CO2}$ is diluted in the flue gas, compared to the other two methods, where it's concentrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as to Leonardo's supplementary question in the comments, about why the pre-combustion  stream has higher concentrations of $\ce{CO2}$ than the post-air combustion stream:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the comparison here isn't between two points in the same process. It's between analagous points in two completely different processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With post-combustion capture, the $\ce{CO2}$ is stripped out after the hydrocarbons (coal, gas, oil) have been burnt in air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas with pre-combustion capture, the hydrocarbon goes through a gasifier/shift process, resulting in a stream of $\ce{H2}$ mixed with $\ce{CO2}$. And at this point, the $\ce{CO2}$ is stripped out, and the $\ce{H2}$ is burnt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's why the $\ce{CO2}$ concentration is higher in the pre-combustion capture process, than in the post air-combustion capture process.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2012-09-17T12:20:11.737" LastActivityDate="2012-09-17T12:20:11.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1189" PostTypeId="1" AcceptedAnswerId="1191" CreationDate="2012-09-17T10:35:26.803" Score="2" ViewCount="1427" Body="&lt;p&gt;I need to calculate the angles between CCC, CCH, HCH in the following molecules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ethene ($\ce{H2CCH2}$), &lt;/li&gt;&#xA;&lt;li&gt;propa-1,2-diene ($\ce{H2CCCH2}$), and &lt;/li&gt;&#xA;&lt;li&gt;buta-1,2,3-triene ($\ce{H2CCCCH2}$). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I'm not mistaken, the Lewis structures of all of these molecules have the carbons in the middle, connected by double bonds to each other, and the hydrogen atoms connected to the carbons in the edge. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H           H&#xA; \         /&#xA;  C = C = C&#xA; /         \&#xA;H           H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The carbons at the edge are $\mathrm{sp^2}$ and the middle carbon is always $\mathrm{sp}$ . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the angle CCC is always 180°. But how can I determine the angles CCH, HCH? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, how can I determine if the hydrogen atoms in the three molecules lie on the same plane?&lt;/p&gt;&#xA;" OwnerUserId="638" LastEditorDisplayName="user7951" LastEditDate="2016-11-17T13:18:50.470" LastActivityDate="2016-11-17T13:18:50.470" Title="Lewis structures of H₂(C)xCH₂ molecules" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1191" PostTypeId="2" ParentId="1189" CreationDate="2012-09-17T16:52:32.297" Score="6" Body="&lt;p&gt;What are the bond angles in a normal trigonal planar arrangement? Treat the center atom as in H &lt;strong&gt;C&lt;/strong&gt; H as a trigonal planar arrangement with the middle (carbon) atom in the center with a third atom attached on the left over position (another carbon). What angle maximizes their separation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other question about the plane of the molecule is a bit more interesting. The pi bonds between the carbons ensure that the molecule is not rotating in space. To maximize the distance between the terminal hydrogen atoms, one set should be aligned up and down, while the other two on the opposite side should be rotated 90&amp;deg; in space, otherwise they would ellipse one another. In other words, one pair is horizontal while the opposite is vertical for example, depending how you want to imagine it.&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2012-09-17T16:52:32.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1192" PostTypeId="2" ParentId="1186" CreationDate="2012-09-18T16:20:16.377" Score="3" Body="&lt;p&gt;All amines are proton bases. $\ce{CO2}$ in water containing solutions acts as a weak acid. An overall reaction is $\ce{NR3 + H2O + CO2 -&amp;gt; [NHR3][HCO3]}$ and is possible for all amines. Aminoalcohols are cheap as they are produced from easily available oxirane (ethylene oxide) and have a high boiling point, so for them this reaction can be reversed by heating, with the regeneration of working solution. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="10185" LastEditDate="2015-09-13T15:33:22.870" LastActivityDate="2015-09-13T15:33:22.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1193" PostTypeId="1" AcceptedAnswerId="1277" CreationDate="2012-09-18T19:06:59.890" Score="5" ViewCount="1957" Body="&lt;p&gt;I am planning to install something on a boat that will be in salt water, and my question is, will the nickel plated brass fastener not corrode if I connect it to stainless steel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked it up on the chart and it says that there should be less then &lt;code&gt;0.15 V&lt;/code&gt; difference in the &quot;Anodic Index&quot; for harsh environments and while brass and stainless is within the &lt;code&gt;0.15&lt;/code&gt;, nickel is just outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The environment is not entirely harsh as well, as these will be used to fasten just the windows, so while there could be water there, mostly it will be fresh water from rain and the occasional spray from salt water. &lt;/p&gt;&#xA;" OwnerUserId="577" LastEditorUserId="73" LastEditDate="2012-09-18T19:21:31.560" LastActivityDate="2012-10-02T05:17:45.010" Title="Galvanic corrosion between nickel brass nut and stainless bolt" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1194" PostTypeId="2" ParentId="1186" CreationDate="2012-09-18T19:27:46.310" Score="7" Body="&lt;p&gt;In addition to the reaction in permeakra's answer, some amines may react with $\ce{CO2}$ in absence of water also. This type of reaction is a Lewis acid-base reaction forming a zwitterionic complex: $\ce{R3N + CO2 -&amp;gt; R3NCO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CAPWK.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MDEA adds some stability to this complex. The alcohol groups can stabilize the carboxylate group through hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G2JY8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-09-18T19:27:46.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1195" PostTypeId="1" CreationDate="2012-09-19T07:02:48.503" Score="2" ViewCount="154" Body="&lt;p&gt;Reading up on the theory of cyclic voltammetry, I note that the current at the inversion potential (that is, the point at which the applied potential goes negative) in every reversible CV trace I have seen is nonzero. My intuition says that this is related to the finite speed of enlargement of the depletion zone around the working electrode and suspect that given sufficient time this current will drop away to zero as the bulk solution homogenises. Is this correct? If not, what accounts for the current?&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-09-20T03:16:56.640" LastActivityDate="2012-09-20T03:16:56.640" Title="Will the current at the inversion potential of a CV trace asymptotically approach zero?" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1196" PostTypeId="1" AcceptedAnswerId="1198" CreationDate="2012-09-19T13:25:43.757" Score="22" ViewCount="5396" Body="&lt;p&gt;Atoms that have eight electrons in their outer shell are extremely stable. It can't be because both the $s$ and the $p$ orbitals are full, because then an atom with 13 or 18 valence electrons would be extremely stable. ($d$ has 10, and 5 is also stable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that atoms with eight electrons in the outer shell are extremely stable?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="10212" LastEditDate="2015-10-14T15:55:07.490" LastActivityDate="2021-04-29T15:08:26.737" Title="Why are atoms with eight electrons in the outer shell extremely stable?" Tags="&lt;orbitals&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1197" PostTypeId="1" CreationDate="2012-09-19T13:26:11.603" Score="9" ViewCount="11401" Body="&lt;p&gt;Nitric acid obtained in laboratory is slightly yellow due to the dissolution of nitrogen dioxide which is produced due to thermal decomposition of a portion of $\ce{HNO3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {4HNO3  -&amp;gt; 2H2O + 4NO2 + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its says here that if $\ce{CO2}$ is bubbled through the acid, it turns colourless- because it drives out $\ce{NO2}$ from warm acid which is further oxidized to nitric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean by &lt;code&gt;drives out&lt;/code&gt; ? How can carbon dioxide drive out nitrogen dioxide that's dissolved in the acid? And also, the oxidation of 'what' is being talked about here?&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="4231" LastEditDate="2015-02-26T16:22:57.847" LastActivityDate="2015-02-26T16:22:57.847" Title="How is the yellow colour of nitric acid removed?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1198" PostTypeId="2" ParentId="1196" CreationDate="2012-09-19T20:19:33.577" Score="18" Body="&lt;p&gt;First, this isn't quite true.  It is true for the first row of the periodic chart (from lithium to neon). It is almost true for the second row (from sodium to argon.  But there are exceptions there.  Beyond that it really isn't true at all for the elements beyond the first two columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the increased stability for the first two rows lies in quantum mechanics.  Classically we can note that there are no $d$ electrons there.  Another ways of looking at it from a classical point of view is that the early elements are too small to allow too many other atoms or groups of atoms around them.  That tends to go away as you go down the periodic chart and the atoms get &quot;fatter&quot;.   A typical example is &lt;em&gt;chloroplatinic acid&lt;/em&gt; which has six chlorines around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most transition metals also can have more than four groups around them as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that this isn't an exceptionally useful explanation.  As I said, the answer really lies in quantum mechanics.  In looking up &quot;molecular orbital theory&quot;,  one reference can be found &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="7475" LastEditDate="2015-10-14T16:30:36.800" LastActivityDate="2015-10-14T16:30:36.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1200" PostTypeId="1" AcceptedAnswerId="1201" CreationDate="2012-09-20T03:33:18.020" Score="5" ViewCount="343" Body="&lt;p&gt;Going through the Bohr's model and his assumptions, I came across with this formula to find the energy of the n-th level of any atom: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = - \frac{Z k_e e^2}{2r_n} = -\frac{Z^2(k_e e^2)^2m_e}{2\hbar n^2} \approx \frac{-13.6Z^2}{n^2}~\pu{eV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say we take as an example the atom of hydrogen, the energy of the $n = 1$ and only level is ($\pu{-13.6 eV}$). Now, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (not really a reliable source) this can be interpreted as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An electron in the lowest energy level of hydrogen ($n = 1$) therefore has about $\pu{13.6 eV}$ less energy than a motionless electron infinitely far from the nucleus. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't really understand why there is a &quot;minus&quot; in the result. Why does it say that it has less energy than a motionless electron when he, himself, suggested that the electron's acceleration does not result in radiation and energy loss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand this formula only worked for the atom of hydrogen (that's why I took it as an example) and due to this problem and that's why the Sommerfeld model was postulated.&lt;/p&gt;&#xA;" OwnerUserId="599" LastEditorUserId="41328" LastEditDate="2017-08-07T00:32:05.933" LastActivityDate="2017-08-07T00:32:05.933" Title="Energy of the n-th level for an atom" Tags="&lt;quantum-chemistry&gt;&lt;energy&gt;&lt;electrons&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1201" PostTypeId="2" ParentId="1200" CreationDate="2012-09-20T06:00:30.380" Score="8" Body="&lt;p&gt;It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is.  The total energy is still zero. The &quot;falling&quot; from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy.  When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until &quot;crashes&quot; - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.)  The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6.  This is  called binding energy, and is negative by the convention physicists and chemists have followed since long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, that formula is only for any atom with exactly one electron, and Z not too big.  With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.&lt;/p&gt;&#xA;" OwnerUserId="651" LastActivityDate="2012-09-20T06:00:30.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1203" PostTypeId="1" AcceptedAnswerId="1205" CreationDate="2012-09-20T07:10:41.240" Score="5" ViewCount="257" Body="&lt;p&gt;So many atoms can be surrounded by oxygen atoms to make anions - chromate, phosphate, etc.   In particular there's Tellurate $\ce{TeO4^{-2}}$, Selenate $\ce{SeO4^{-2}}$ and Sulfate $\ce{SO4^{-2}}$ - why not go one step more and have &quot;Oxygate&quot; $\ce{OO4^{-2}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that such a thing, if it could be created at all, would break up into an molecular oxygen and ozone $\ce{O2 + O3}$ with the electrons doing whatever they please.  But sulfate and the heavier VI central atoms don't break up.  I've never encountered a good clear explanation about this.   &lt;/p&gt;&#xA;" OwnerUserId="651" LastEditorUserId="4231" LastEditDate="2014-11-04T16:39:22.040" LastActivityDate="2014-11-04T16:39:22.040" Title="Why not &quot;Oxygate&quot; OO4(2-)?" Tags="&lt;molecules&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1205" PostTypeId="2" ParentId="1203" CreationDate="2012-09-20T18:04:00.380" Score="7" Body="&lt;p&gt;It's not necessary that an element shares some chemical property with all the others in its group. If it &lt;em&gt;did&lt;/em&gt; share all chemical properties with the rest of the group, then we could just as well call the elements (something akin to) isotopes of each other. You'd have a pretty boring periodic table then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening here is that the most striking difference between elements in a group is at play. That difference being the number of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our electron shells are broken into &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow noreferrer&quot;&gt;orbitals&lt;/a&gt;&quot;. Every orbital can hold 2 electrons. $s$ orbitals come in sets of one, $p$ in sets of 3, $d$ in sets of 5, etc. Not every shell is made up of all orbitals--the first shell has only one $s$ orbital, the second shell has $s$ and $p$, and so on. Electrons fill into these orbitals by a triad of rules (which I won't go into here)-- the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aufbau principle&lt;/a&gt;, the Pauli exclusion principle (which is implicit in my statement that each orbital holds two electrons), and Hund's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an interesting thing is that bonding can only occur with orbitals with close energies. Thus, if the $4s$ orbital is involved in bonding, the $1s$ orbital will not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK. Now, let's first look at Sulfur's outer electron configuration. There are 2 electrons in a $3s$ orbital, 4 in the $3p$ orbitals (so there is enough space for 2 more electrons here), and you have some &quot;vacant space&quot; in the $3d$ orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sulfate ion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BnvQW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sulfur is &quot;accepting&quot; 6 electrons in this bond. There's some nifty mixing of orbitals that goes on, but basically, those electrons need some place in sulfur's outer shell to hole up--two electrons per orbital as usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have 6 extra electrons, we need 3 orbitals. We have one &quot;left over&quot; $p$ orbital, and 5 $d$ orbitals, so we have enough space to accommodate those electrons and form a bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK. Now, let's try the same thing for Oxygen. It has, in its outer shell, a full $2s$ orbital, 2 full $2p$ orbitals, leaving an &quot;extra&quot; $2p$ orbital empty. As mentioned before, the second shell only has $s$ and $p$ orbitals, so there obviously aren't any $d$ orbitals. You've probably noticed the issue here-- we have six electrons and only one extra orbital to stuff them in. This isn't possible, so oxygen never forms $\ce{OO4-}$. The other chalcogens($\ce{Se, Te}$, etc) all have $d$ orbitals in their outer shells, so they have no trouble forming such ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I've grossly oversimplified the bonding process, deliberately leaving out a few crucial aspects (which would otherwise take extremely long to explain). If you want to know about the mechanism behind bonding, you may want to read up on orbitals first, then VSEPR theory, and then MO theory.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-09-20T18:04:00.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1208" PostTypeId="1" AcceptedAnswerId="1209" CreationDate="2012-09-21T02:59:42.670" Score="4" ViewCount="343" Body="&lt;p&gt;At high temperatures (almost) everything burns to ash. Is ash 'fireproof'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is further combustion of ash possible for example by combining with hydrogen, oxygen, etc?&lt;/p&gt;&#xA;" OwnerUserId="656" OwnerDisplayName="user41631" LastEditorUserId="656" LastEditDate="2012-10-24T05:38:44.210" LastActivityDate="2012-10-24T05:38:44.210" Title="Is ash fireproof?" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1209" PostTypeId="2" ParentId="1208" CreationDate="2012-09-21T03:41:48.777" Score="5" Body="&lt;p&gt;Short answer no, ash can be burnt. It may contain unburned particles that did not receive enough air (oxygen) to burn completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although one may consider ash to be product of complete burning. In which case it won't burn easily. But given the right conditions it can burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact even water can burn, there is quite a bit of research in burning aluminum with water (especially for under water applications like submarines), this produces aluminum oxide and hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="655" OwnerDisplayName="mythealias" LastEditorUserId="655" LastEditDate="2012-09-21T04:59:42.970" LastActivityDate="2012-09-21T04:59:42.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1212" PostTypeId="5" CreationDate="2012-09-22T01:58:30.023" Score="0" Body="&lt;p&gt;Please see the guidelines on &lt;a href=&quot;http://meta.chemistry.stackexchange.com/questions/3066/how-to-ask-and-answer-nomenclature-questions/3067#3067&quot;&gt;how to ask and answer nomenclature questions&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://doi.org/10.1515/pac-2014-0718&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brief guide to the nomenclature of inorganic chemistry (IUPAC Technical Report 2015)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://doi.org/10.1515/pac-2019-0104&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brief guide to the nomenclature of organic chemistry (IUPAC Technical Report 2020)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="73" LastEditorUserId="101286" LastEditorDisplayName="user7951" LastEditDate="2021-02-28T11:29:45.643" LastActivityDate="2021-02-28T11:29:45.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1213" PostTypeId="4" CreationDate="2012-09-22T01:58:30.023" Score="0" Body="This tag should be used for questions about the naming of compounds and elements, usually using systematic names according to IUPAC recommendations, but also using retained traditional names." OwnerUserId="73" LastEditorUserId="865" LastEditDate="2016-11-02T21:26:52.553" LastActivityDate="2016-11-02T21:26:52.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1214" PostTypeId="1" CreationDate="2012-09-22T02:12:42.830" Score="4" ViewCount="418" Body="&lt;p&gt;Hot tub shock is potassium peroxymonosulfate. When added to a hot tub, it creates bubbles. It is said to oxidize organic debris in the water. What is the reaction? &lt;/p&gt;&#xA;" OwnerUserId="658" LastEditorUserId="22" LastEditDate="2012-09-23T07:12:16.890" LastActivityDate="2012-09-23T18:09:50.017" Title="How does hot tub &quot;shock&quot; work?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1215" PostTypeId="1" AcceptedAnswerId="1225" CreationDate="2012-09-22T06:31:23.133" Score="9" ViewCount="2860" Body="&lt;p&gt;All metals form basic oxides. Metallic oxides are basic in character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polonium is a metal, so it should form basic oxides but it forms acidic oxides. Why are its oxides acidic in character?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="652" LastEditDate="2012-09-24T06:55:49.380" LastActivityDate="2015-04-22T13:22:07.833" Title="Why does polonium form acidic oxides?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1217" PostTypeId="1" AcceptedAnswerId="1223" CreationDate="2012-09-22T07:02:31.043" Score="1" ViewCount="625" Body="&lt;p&gt;Is there a way to extract caffeine from coffee completely?&lt;/p&gt;&#xA;" OwnerUserId="656" LastEditorUserId="30" LastEditDate="2012-09-22T07:09:26.963" LastActivityDate="2012-09-23T17:54:14.333" Title="Can caffeine be separated from coffee?" Tags="&lt;safety&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1219" PostTypeId="2" ParentId="1217" CreationDate="2012-09-22T09:32:28.360" Score="7" Body="&lt;p&gt;Yes, this is how decaffeinated coffee is made. If I recall correctly, this is done commercially by dissolving the caffeine in dichloromethane (DCM), or more recently in supercritical carbon dioxide. This switch is probably over concerns regarding chronic industrial exposure to DCM, which is mildly toxic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From personal experience, I have extracted caffeine from instant coffee using a separatory funnel with a system of water and dichloromethane. Purification by sublimation finished the job.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-09-22T09:32:28.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1220" PostTypeId="1" AcceptedAnswerId="1232" CreationDate="2012-09-22T20:33:29.807" Score="3" ViewCount="992" Body="&lt;p&gt;As I am looking at titration curves, a few things stand out. For now, I have a two part question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3 is several orders of magnitude than the number of particles involved in the jump from 6 to 7?&lt;/li&gt;&#xA;&lt;li&gt;When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle? That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium. This should retard the titration to some degree. Is that the case?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="550" LastEditorUserId="550" LastEditDate="2012-09-22T20:45:33.797" LastActivityDate="2012-09-25T10:07:15.417" Title="Understanding the rate of change in titration curves" Tags="&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1221" PostTypeId="1" CreationDate="2012-09-23T02:10:19.710" Score="4" ViewCount="3328" Body="&lt;p&gt;I have butene being burned in a fuel rich engine and I can't seem to balance this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C4H8 + $a$(O2 + 3.76 N2) &#xA;-&amp;gt; $b$ CO2 + $c$ H2O + $d$ C4H8 + $e$ (3.76 N2)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try write equations for the elements, I get 5 equations and 4 unknowns which can't be solved. Here are my equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\text{Carbon}   &amp;amp;:&amp;amp; 4     &amp;amp;= b  + 4d  \\&#xA;\text{Oxygen}   &amp;amp;:&amp;amp; 2a    &amp;amp;= 2b + c   \\&#xA;\text{Nitrogen} &amp;amp;:&amp;amp; 3.76a &amp;amp;= 3.76e    &amp;amp;  a &amp;amp;= e \\&#xA;\text{Hydrogen} &amp;amp;:&amp;amp; 2a    &amp;amp;= 2c + 8d  \\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this makes a different but it says that water in the combustion products is removed before the dry analysis of the exhaust. The dry analysis has a composition of &lt;span class=&quot;math-container&quot;&gt;$14.95\% \ce{CO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0.75\% \ce{C4H8}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0\% \ce{CO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0\% \ce{H2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0\% \ce{O2}$&lt;/span&gt;, and balance &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; (I don't know what that means).&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="4945" LastEditDate="2020-01-28T18:35:27.767" LastActivityDate="2020-01-28T18:39:02.433" Title="How do I balance the combustion reaction of butene?" Tags="&lt;energy&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="1222" PostTypeId="1" AcceptedAnswerId="1305" CreationDate="2012-09-23T12:51:16.527" Score="13" ViewCount="423" Body="&lt;p&gt;I created the L-Phenylalanine based Evans auxiliary bearing a benzyl group with help of a molecular model set and got into some questions. The structural formula will do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PJ3M5.png&quot; alt=&quot;formula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I suppose a low rotational barrier, the phenyl group can be aligned off (a) and towards (b) the alpha position of the &lt;em&gt;N&lt;/em&gt;-propionyl group. I suppose maximum sterical repulsion for attacking electrophiles for (b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenylalanine and valine based auxiliaries (Bn or &lt;em&gt;i&lt;/em&gt;-Pr) are the most commonly involved types of Evans auxiliaries. But why do I get good stereoselectivities for this Bn auxiliary? There are no corresponding carbon-carbon rotational freedoms in the valine species.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="21296" LastEditDate="2017-09-24T10:44:18.277" LastActivityDate="2017-09-24T10:44:18.277" Title="The Evans auxiliary: rotational freedom of the benzyl group - stereoselectivity still guaranteed?" Tags="&lt;organic-chemistry&gt;&lt;stereoselectivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1223" PostTypeId="2" ParentId="1217" CreationDate="2012-09-23T17:54:14.333" Score="2" Body="&lt;p&gt;Regarding &quot;completely&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is &lt;strong&gt;no&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for all intents and purposes, &lt;strong&gt;yes&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By submitting coffee to &lt;a href=&quot;https://en.wikipedia.org/wiki/Partition_coefficient&quot; rel=&quot;nofollow&quot;&gt;several small volume extractions&lt;/a&gt;, one can drastically reduce the concentration of caffeine present in the grains. This method of extraction (multiple small volume aliquots) is more effective than a single large volume extraction. The link I shared provides a good discussion.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-09-23T17:54:14.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1224" PostTypeId="2" ParentId="1214" CreationDate="2012-09-23T18:09:50.017" Score="5" Body="&lt;p&gt;There are several properties you can infer from the compound's name.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The compound is ionic, which the structure &lt;code&gt;cation [...]-ate&lt;/code&gt; informs us. In this case, we're dealing with the potassium salt of peroxymonosulfuric acid. This should therefore dissolve pretty well in water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The compound will be an oxidizer, in the conditions present in your bathtub. &lt;code&gt;peroxy&lt;/code&gt; means the compound contains an $\ce{O-O}$ bond, which breaks up homolytically (one electron per oxygen) to give radicals (quite reactive species, electronically unhappy).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As per both hints, the compound is expected to be pretty effective at &quot;eating&quot; through organic deposits, by oxidizing them, rendering some soluble. It should also decrease pH, but not so drastically as it is already the conjugate base of sulfuric acid (which can let go of two protons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the bubbling: it most probably is $\ce{CO2}$ coming from the oxidation of the organic matter.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2012-09-23T18:09:50.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1225" PostTypeId="2" ParentId="1215" CreationDate="2012-09-24T11:06:10.610" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All metals form basic oxides. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's true. I cannot remember metal, that does not form basic oxide&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metallic oxides are basic in character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessary. Polonium is not the only one metal, forming acidic oxides. Mn, Cr and some others do it as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxide is considered acidic, when it forms hydroxide, that is acidic (well, it is simplification, but let's stick with it for a while). Hydroxide is acidic, when it can lose proton, giving stable anion. Acidity of hydroxide and metallicity of element formed it does not connect any way. Acidity depends of ability of remaining part of molecule to stabilize negative charge, which usually correlate with $\ce{O/OH}$ ratio and electronegativity of central atom. So, $\ce{H6TeO6}$ is quite weak acid and $\ce{H2CrO4}$ or $\ce{HMnO4}$ are quite quite strong acids. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user7951" LastEditDate="2015-04-22T13:22:07.833" LastActivityDate="2015-04-22T13:22:07.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1226" PostTypeId="1" CreationDate="2012-09-24T12:54:18.773" Score="9" ViewCount="3183" Body="&lt;p&gt;Specifically, if I wanted to purchase a large amounts of helium, what are my options for purchasing helium in a state that is suitable for long term storage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's my understanding that if I were to buy helium in bulk it would be shipped cold, as a liquid.  Keeping helium in liquid form for 10-20+ years would involve a lot of serious refrigeration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other reasonable options for bulk, long term storage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not seem like much of a &quot;chemistry&quot; question but since this is used in chemical sciences, I assume this is a need that some may be aware and may have a good answer since this is an elemental material that some may use on a very common basis.&lt;/p&gt;&#xA;" OwnerUserId="665" LastEditorUserId="73" LastEditDate="2012-09-24T13:24:25.600" LastActivityDate="2013-02-15T22:46:15.560" Title="Options for long term storage of helium?" Tags="&lt;safety&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1227" PostTypeId="1" AcceptedAnswerId="1230" CreationDate="2012-09-24T14:11:31.767" Score="4" ViewCount="10917" Body="&lt;p&gt;What is the boiling point of vodka?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found the boiling point of ethanol, ~173 (degrees) F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am unable to find the boiling point of vodka. I found some information on it, but I do not trust the source. &lt;/p&gt;&#xA;" OwnerUserId="666" LastEditorUserId="73" LastEditDate="2012-09-24T23:27:14.967" LastActivityDate="2014-12-20T21:42:56.373" Title="Boiling point of vodka" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1229" PostTypeId="1" AcceptedAnswerId="1231" CreationDate="2012-09-24T17:18:42.697" Score="5" ViewCount="472" Body="&lt;p&gt;What is the shape of the $\ce{C2H2}$ molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that is has a triple covalent bond between the 2 carbon atoms and 2 other bonding pairs between the hydrogen and carbon atoms leads me to believe that it could be linear shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure if the triple bond counts as a single electron bonding pair or 3. If it counts as 3 bonding pairs then the molecule would have 5 overall, and therefore be a trigonal bipyramidal shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it octohedral or linear? Pretty sure it's linear &lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="4231" LastEditDate="2015-02-23T19:52:26.203" LastActivityDate="2015-02-23T19:52:26.203" Title="Shape of C₂H₂ molecule" Tags="&lt;molecules&gt;&lt;covalent-compounds&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1230" PostTypeId="2" ParentId="1227" CreationDate="2012-09-24T18:18:05.040" Score="11" Body="&lt;p&gt;Vodka is a solution of water and ethanol.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water boils at 100 °C&lt;/li&gt;&#xA;&lt;li&gt;Ethanol boils at 78.37 °C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A mixture of both liquids will start boiling close to the boiling point of ethanol (otherwise distillation wouldn't work). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest boiling temperature is a mixture of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;95.63 mass-% ethanol&lt;/li&gt;&#xA;&lt;li&gt;4.37 mass-% water&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which boils at 78.2 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more about this effect at: &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;nofollow&quot;&gt;Azeotrope&lt;/a&gt; There is also a picture of a phase diagram there.&lt;/p&gt;&#xA;" OwnerUserId="351" LastEditorUserId="351" LastEditDate="2014-12-20T21:42:56.373" LastActivityDate="2014-12-20T21:42:56.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1231" PostTypeId="2" ParentId="1229" CreationDate="2012-09-24T21:16:09.530" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot;&gt;VSEPR theory&lt;/a&gt;, each carbon has &lt;strong&gt;2&lt;/strong&gt; electron &quot;pairs&quot;.  These electron pairs are not what you would typically think of as electron pairs because VSEPR theory considers double bonds and triple bonds to still be a single &quot;pair&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considering each carbon atom, the shape is $AX_2E_0$ ($A$ is the carbon atom, $X$ is a bonding pair of electrons, and $E$ is a non-bonding pair (lone pair) of electrons).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$AX_2E_0$ corresponds directly to a &lt;em&gt;linear&lt;/em&gt; molecule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer your question, when trying to determine the shape of a molecule, consider double and triple bonds as only a single electron bonding pair.  &lt;/p&gt;&#xA;" OwnerUserId="12" LastActivityDate="2012-09-24T21:16:09.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1232" PostTypeId="2" ParentId="1220" CreationDate="2012-09-25T10:07:15.417" Score="0" Body="&lt;p&gt;This answer from a member of Chemical Forums who seems to have been quite useful to other posters in the past given his or her &quot;karma&quot; would appear to confirm that the shape of titration curves owes in part to their logarithmic nature: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=6140.0&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalforums.com/index.php?topic=6140.0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2012-09-25T10:07:15.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1233" PostTypeId="1" AcceptedAnswerId="1296" CreationDate="2012-09-25T10:31:35.030" Score="3" ViewCount="3335" Body="&lt;p&gt;I am trying to determine the electron dot structure of an alkyne like &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that hydrocarbons with at least one carbon-carbon triple bond are called alkynes, and their general formula is &lt;span class=&quot;math-container&quot;&gt;$\ce{C_{n}H_{2n-2}}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is number of carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what will the electron dot structure of &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; look like?&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="4945" LastEditDate="2018-12-11T13:34:26.703" LastActivityDate="2018-12-11T13:37:00.060" Title="What is the electron dot structure of an alkyne?" Tags="&lt;hydrocarbons&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2015-06-19T19:23:15.190" ContentLicense="CC BY-SA 4.0" />
  <row Id="1236" PostTypeId="1" CreationDate="2012-09-26T05:20:04.487" Score="1" ViewCount="183" Body="&lt;p&gt;I am taking a course on fuel cells and we were talking about the activation energy needed to start a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that the answer to this question is that the anode potential should be increased to generate current so less energy is needed to begin the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? Or am I way off track.&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2017-05-01T06:45:47.947" Title="Should the anode potential be increased or decreased to generate current?" Tags="&lt;homework&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1237" PostTypeId="2" ParentId="1215" CreationDate="2012-09-26T20:16:17.933" Score="5" Body="&lt;p&gt;Another way of looking at this is to understand that the structure is always&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{M-O-H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{M}$ is the metal atom.  If the $\ce{O-H}$ bond is weaker than the $\ce{M-O}$ bond, the material acts as&#xA;an acid.  Otherwise it acts as a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines the relative strength is a bit complex.  Suffice it to say that it depends on oxidation state of the $\ce{M}$ atom as well as the constituents in the rest of the molecule and the environment all play a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some materials are &lt;em&gt;amphoteric&lt;/em&gt;.  That is they can split either way.  A typical&#xA;example is aluminum hydroxide, $\ce{Al(OH)_3}$, which has three $\ce{O-H}$ bonds.&#xA;It is usually a base, but under the right conditions it breaks the other way and&#xA;is a weak acid called &lt;em&gt;aluminic acid&lt;/em&gt;.  The formula for that is usually written as $\ce{H3AlO_3}$, but the structure is the same.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="30" LastEditDate="2012-09-26T21:11:16.550" LastActivityDate="2012-09-26T21:11:16.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1238" PostTypeId="1" AcceptedAnswerId="1249" CreationDate="2012-09-26T21:42:08.187" Score="9" ViewCount="9922" Body="&lt;p&gt;Are there chemical reactions that could cool down an average sized room by a noticeable amount (say 5 &amp;deg;C)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to investigate if it is possible to have a situation where I can mix 2 reagents at room temperature and pressure and in open air then they should react and become colder than room temperature without evaporation of some type, with an eye to making a noticeable drop in the room temperature.&lt;/p&gt;&#xA;" OwnerUserId="677" LastEditorUserId="402" LastEditDate="2017-04-01T01:16:44.370" LastActivityDate="2018-01-30T04:42:49.607" Title="Chemical reactions with a room-scale cooling effect" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1239" PostTypeId="1" CreationDate="2012-09-26T22:26:44.433" Score="4" ViewCount="60986" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/974/why-is-methanes-molecular-formula-conventionally-ch4-while-water-is-h20-a&quot;&gt;Why is methane&amp;#39;s molecular formula conventionally &amp;ldquo;CH4&amp;rdquo;, while water is &amp;ldquo;H20&amp;rdquo; (among others)?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;If I have a formula containing 4 hydrogen atoms and 1 carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be written $\small\ce{CH4}$ or $\small\ce{H4C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, how do I know the order to put elements in, in a chemical formula?&lt;/p&gt;&#xA;" OwnerUserId="678" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2018-01-18T21:26:28.757" Title="Order of elements in a formula?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2012-09-28T14:25:13.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="1240" PostTypeId="1" AcceptedAnswerId="1242" CreationDate="2012-09-26T22:30:56.613" Score="6" ViewCount="18836" Body="&lt;p&gt;Is there a way to calculate how fast water will cool if you know the water's starting temperature and temperature of the environment that you put it in, such as a freezer? &lt;/p&gt;&#xA;" OwnerUserId="679" LastEditorUserId="296" LastEditDate="2012-12-27T14:22:32.327" LastActivityDate="2012-12-27T14:22:32.327" Title="Is there a way to calculate how fast water will cool?" Tags="&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1241" PostTypeId="2" ParentId="1239" CreationDate="2012-09-26T22:35:20.377" Score="6" Body="&lt;p&gt;There's something called Hill System. This is the system of writing chemical formulas.&#xA;In this system the carbon atoms are first, then hydrogen atoms and then other in alphabetical order. When the formula contains no carbon or hydrogen atoms, then all the elements are sorted alphabetically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Hill_system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it should be: $\ce{CH4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Hill system is not the only method of ordering elements in a chemical formula. We can instead choose to list the elements based on their electronegativity. ManishEarth explains this in his answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/537/why-is-arsenous-acid-denoted-h3aso3/538#538&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="676" LastEditorUserId="9961" LastEditDate="2018-01-18T21:26:28.757" LastActivityDate="2018-01-18T21:26:28.757" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1242" PostTypeId="2" ParentId="1240" CreationDate="2012-09-27T03:28:38.300" Score="6" Body="&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_conduction#Fourier.27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fourier's law&lt;/a&gt;. The rate of heat flux (transfer of thermal energy per unit time per unit surface area)  is proportional to the difference in temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FIUn6.png&quot; alt=&quot;Heat eqn&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a change in temperature is directly proportional to the change in energy, you can form this into a fairly simple first-order differential equation. If we take one side (air in a room), at a temperature we'll call $T_{0}$ as a cold reservoir, we can say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d(cT_{1})}{dt}=c\frac{dT_{1}}{dt}=-kA\frac{T_{1}-T_{0}}{w}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T_{1}$ is the temperature of the water, $c$ is the water's specific heat, $A$ is the surface area, $k$ is the thermal conductivity of the material between them (i.e. styrofoam cup, say it has a styrofoam lid too so it surrounds the water completely), and $w$ is the thickness of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will notice that this equation leads to an exponential decay in the temperature difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply plug in the appropriate constants and solve this equation for any given $\Delta T$ and you can find how long it takes to get that close (it will never quite reach zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'm a bit rusty, so please forgive any notational errors&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="107" LastActivityDate="2012-09-27T03:28:38.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1243" PostTypeId="2" ParentId="1155" CreationDate="2012-09-27T06:52:40.010" Score="7" Body="&lt;p&gt;This may be what &lt;s&gt;suffices&lt;/s&gt; you are looking for :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.qdusa.com/sitedocs/Quantum_Design_Periodic_Table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Table of Elements w. Selected Phase Transition Temperatures&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vRU0u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This may get you started:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://nsdl.library.cornell.edu/websites/nsdlnetwork/content/am-poster/773/matdl-materials-digital-library-pathway.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MatDL - The Materials Digital Library Synopsis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nist.gov/srd/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST SRD&lt;/a&gt; ( I would write them)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://lectureonline.cl.msu.edu/~mmp/period/phase.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Java Applet: Phase Transition Temperatures&lt;/a&gt; (untested)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is a very good question, which I came across as well. You could just google-image search &lt;code&gt;phase diagram&lt;/code&gt; + &lt;code&gt;ELEMENT&lt;/code&gt; to copy and past the most appealing charts into a table in your editor of choice.&#xA;If the intend is &lt;em&gt;personal use&lt;/em&gt;, this might be the fastest route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;wolfram alpha&lt;/strong&gt; should allow you to look up the state of the element at a given pressure as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sodium phase at 0 bar&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Helium itself is a particularly interesting subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps for a start...&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditDate="2015-06-11T04:27:58.417" LastActivityDate="2015-06-11T04:27:58.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1244" PostTypeId="1" CreationDate="2012-09-27T12:23:49.513" Score="2" ViewCount="187" Body="&lt;p&gt;In crystallization, crystals are formed from liquids by cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can crystals be formed from a gas by cooling?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="73" LastEditDate="2012-09-27T19:31:07.187" LastActivityDate="2012-09-27T19:31:07.187" Title="How are crystals formed from a gas?" Tags="&lt;experimental-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1245" PostTypeId="2" ParentId="1244" CreationDate="2012-09-27T12:32:00.857" Score="2" Body="&lt;p&gt;It's called deposition or desublimation. It's for example how snow is created in clouds or how frost forms on leaf. When leaf is very cold, then vapour surrounding leaf looses so much energy that it changes into solid state.&lt;/p&gt;&#xA;" OwnerUserId="676" LastActivityDate="2012-09-27T12:32:00.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1246" PostTypeId="1" AcceptedAnswerId="1247" CreationDate="2012-09-27T19:42:10.523" Score="-1" ViewCount="4900" Body="&lt;p&gt;The thing I am having trouble with: why does $\small\ce{PCl4}$ form a +1 ion and $\small\ce{Cl}$ a -1 ion?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="73" LastEditDate="2012-09-27T20:31:39.500" LastActivityDate="2015-03-22T07:14:31.257" Title="How does Phosphorus Pentachloride bond?" Tags="&lt;ions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1247" PostTypeId="2" ParentId="1246" CreationDate="2012-09-27T20:38:14.297" Score="3" Body="&lt;p&gt;There is no ionic bonding that occurs, the phosphorus is able to make a hybrid orbital out of its $\small3s$ and $\small3p$ electrons, along with an empty $\small d$ orbital to form 5 $sp^3d$ molecular orbitals, of equal energy, which yields 5 spots for bonding electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a known exception to the octet rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dboy4.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(graciously borrowed from &lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/covalent.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2012-09-27T20:44:29.097" LastActivityDate="2012-09-27T20:44:29.097" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1248" PostTypeId="1" AcceptedAnswerId="1292" CreationDate="2012-09-27T21:48:26.563" Score="4" ViewCount="5759" Body="&lt;p&gt;I was wondering if there is a way to detect for presence or residue of bleach &lt;em&gt;using household items&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious about either chlorine-based or peroxide-based. It can be any sort of reaction that shows the presence of it. Well, as long as the reaction doesn't produce a lot of heat or is somehow toxic.&lt;/p&gt;&#xA;" OwnerUserId="681" LastEditorUserId="73" LastEditDate="2012-09-27T21:51:28.197" LastActivityDate="2012-10-04T21:08:28.970" Title="Is there a way to detect bleach residue on a surface?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1249" PostTypeId="2" ParentId="1238" CreationDate="2012-09-28T06:53:23.650" Score="3" Body="&lt;p&gt;The reaction between ammonium thiocyanate and barium hydroxide octahydrate is endothermic. It absorbs heat from the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=MyAzjSdc3Fc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=MyAzjSdc3Fc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="656" LastActivityDate="2012-09-28T06:53:23.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1250" PostTypeId="2" ParentId="1128" CreationDate="2012-09-28T13:25:13.327" Score="1" Body="&lt;p&gt;Ice point is the freezing point of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freezing point is the point where liquid turns to solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that in the case of ice point is substance is always water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that there is a slight discrepancy of 0.01 K between the triple point of water, melting point, freezing point and/or ice point. And technically, Ice point is the point where water and Ice are in equillibrium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See e.g. &lt;a href=&quot;http://www.thefreedictionary.com/freezing+point&quot; rel=&quot;nofollow&quot;&gt;http://www.thefreedictionary.com/freezing+point&lt;/a&gt;, &lt;a href=&quot;http://www.thefreedictionary.com/ice+point&quot; rel=&quot;nofollow&quot;&gt;http://www.thefreedictionary.com/ice+point&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="144" LastActivityDate="2012-09-28T13:25:13.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1251" PostTypeId="1" CreationDate="2012-09-28T16:47:29.847" Score="8" ViewCount="714" Body="&lt;p&gt;Here's the molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/60D9o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/60D9o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've tried: 6-methylcyclohex-2-en-ol and 4-methylcyclohexen-3-ol but both are wrong. I'm rather confused as to where I should start: the alcohol or the double bond?&lt;/p&gt;&#xA;" OwnerUserId="684" LastEditorUserId="5026" LastEditDate="2018-02-25T07:35:28.150" LastActivityDate="2018-02-25T07:35:28.150" Title="Proper IUPAC name of a substituted cyclohexenol" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1252" PostTypeId="1" AcceptedAnswerId="1295" CreationDate="2012-09-28T18:28:54.677" Score="8" ViewCount="4734" Body="&lt;p&gt;I am tasked with writing a molecular equation, an overall ionic equation, and a net ionic equation for the reaction between an aqueous solution of phosphoric acid, $\ce{H3PO4{(aq)}}$ and an aqueous solution of sodium hydroxide. The products are sodium phosphate and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already figured out the balanced molecular equation of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3PO4 {(aq)} + 3 NaOH {(aq)} -&amp;gt; Na3PO4 {(aq)} + 3 H2O {(l)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am uncertain how to write the proper ionic equations. I know that the ions are separated, but how do you format this properly when writing with multiple molecules. If this is based on the disassociation of the chemicals, how do you determine which ions disassociate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have arrived at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3H+ {(aq)} + PO4^3- {(aq)} + 3 Na+ {(aq)} + 3 OH- -&amp;gt; 3 Na+ {(aq)} + 3 H2O &#xA;{(l)} + PO4^3- {(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I still do not know why this is the best answer. Why does the $\ce{NaOH}$ show separated in the equation while the $\ce{PO4}$ does not? How would you know which ones separate by simply looking at the problem?&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="5026" LastEditDate="2018-05-16T10:02:27.153" LastActivityDate="2018-05-16T10:03:06.620" Title="Which species dissociate completely in ionic equations?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1253" PostTypeId="2" ParentId="1251" CreationDate="2012-09-28T19:29:45.327" Score="2" Body="&lt;p&gt;The functional group with highest precedence is the alcohol, so it will be the suffix and bears number 1. The main chain is cyclohexene, and you have two ways of numbering it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5-methylcyclohex-2-en-1-ol&lt;/li&gt;&#xA;&lt;li&gt;2-methylcyclohex-5-en-1-ol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, IUPAC rules state that you must give double bonds at the lowest possible numbering, even before considering prefix groups (rule 6.1 to 6.4 of &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Basic_principles&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page&lt;/a&gt;). So, it has to be &lt;strong&gt;5-methylcyclohex-2-en-1-ol&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-28T19:29:45.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1254" PostTypeId="1" AcceptedAnswerId="1259" CreationDate="2012-09-28T20:12:56.510" Score="21" ViewCount="5337" Body="&lt;p&gt;I've always wondered what the chemistry behind fire is. What are the basic chemical reactions behind a simple wood fire, and how do they manifest into this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="685" LastEditorUserId="22" LastEditDate="2012-09-29T05:21:02.027" LastActivityDate="2018-05-03T12:02:07.360" Title="What are the chemical reactions behind fire?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1255" PostTypeId="1" AcceptedAnswerId="1270" CreationDate="2012-09-28T20:15:43.933" Score="3" ViewCount="621" Body="&lt;p&gt;I've seen it in a couple of videos, it's used to electrolyze water in 2 different containers at the same time to make a comparison and thus is very handy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it called (besides just a hydrolizer)? &#xA;&lt;img src=&quot;https://i.stack.imgur.com/buWWA.jpg&quot; alt=&quot;http://i1274.photobucket.com/albums/y432/mocDOTtekcubotohp/hydrolizer_zps38a63656.jpg&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="686" LastEditorUserId="73" LastEditDate="2012-09-28T23:13:01.343" LastActivityDate="2012-10-01T00:59:27.267" Title="What's the name of this electrolysis device?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1256" PostTypeId="1" AcceptedAnswerId="1263" CreationDate="2012-09-29T04:25:44.317" Score="9" ViewCount="133" Body="&lt;p&gt;I've been looking over the lists of materials used together to make metallic glass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it purely coincidental that if you round up/down any decimals from the atomic weight of elements used to construct metallic glass, you're always very close to +-1 (if not at) a prime number for the atomic weight of each element?&lt;/p&gt;&#xA;" OwnerUserId="687" LastEditorUserId="23561" LastEditDate="2019-04-08T13:05:18.837" LastActivityDate="2019-04-08T13:05:18.837" Title="Materials used in metallic glass" Tags="&lt;metal&gt;&lt;materials&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1258" PostTypeId="1" CreationDate="2012-09-29T09:06:53.063" Score="0" ViewCount="2040" Body="&lt;p&gt;Can someone explain me how can I use the periodic table in order to find the specific heat capacities of some elements? (for example, of Aluminium )&lt;/p&gt;&#xA;&lt;p&gt;The question I have encountered is as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Given the specific heat capacities of the following liquids, use the periodic table in order to find the specific heat capacities of the metals &lt;span class=&quot;math-container&quot;&gt;$Al, Fe, Cu, Au $&lt;/span&gt; .&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ H_2 O - 4.18 J  /g \cdot C $&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$ C_2 H_5 OH - 2.46 $&lt;/span&gt; , &lt;span class=&quot;math-container&quot;&gt;$CCl_4 - 0.861 $&lt;/span&gt; , &lt;span class=&quot;math-container&quot;&gt;$CCl_2 F_2 - 0.598 $&lt;/span&gt; .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="638" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2012-09-29T15:05:43.367" Title="Specific heat capacity and the periodic table" Tags="&lt;homework&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1259" PostTypeId="2" ParentId="1254" CreationDate="2012-09-29T12:59:20.620" Score="18" Body="&lt;p&gt;Even though fire is one of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Classical_element#Classical_elements_in_Greece&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greek classical elements&lt;/a&gt;, it is the only one that is not matter in our current understanding. What we experience as fire is the energy (in the form of light and heat) given off by the exothermic combustion of (usually) organic materials, like wood. The large amounts of thermal energy released by combustion often cause the gases in the fire to &lt;a href=&quot;http://en.wikipedia.org/wiki/Incandescence&quot; rel=&quot;nofollow noreferrer&quot;&gt;incandesce&lt;/a&gt;, that is, some of the kinetic energy produced by the fire is converted into electromagnetic radiation that we see (visible) and feel (infrared). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical reaction for combustion is pretty simple. Take, for example, the combustion of n-octane $\ce{C8H18}$, which produces more than 5 MJ of energy per mole during combustion (from &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C111659&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot; rel=&quot;nofollow noreferrer&quot;&gt;webbook.nist.gov&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C8H18 +25O2-&amp;gt;16CO2 +18H2O} \space\space \Delta_cH^o=-5430 \text{ kJ/mol} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combustion of wood is more complex. The majority of the organic mass of dry dead wood is lignin and cellulose. &lt;a href=&quot;http://en.wikipedia.org/wiki/Lignin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lignin&lt;/a&gt; is a highly-cross-linked copolymer of p-coumaryl alcohol, $\ce{C9H10O2}$, &lt;a href=&quot;http://en.wikipedia.org/wiki/Coniferyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;coniferyl alcohol&lt;/a&gt;, $\ce{C10H12O3}$, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sinapyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;sinapyl alcohol&lt;/a&gt;, $\ce{C11H14O4}$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cellulose&lt;/a&gt; is a linear polymer of &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;nofollow noreferrer&quot;&gt;glucose&lt;/a&gt;, with formula $\ce{(C6H10O5)}_n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different species will have different ratios of lignin to cellulose, different cross-link densities in the lignin, and different ratios of coumaryl to coniferyl to sinapyl alcohols. The formula for lignin could be expressed then as $\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the combustion reaction for cellulose is:&#xA;$$\ce{(C6H10O5)}_n +6n\ce{O2-&amp;gt;}+6n\ce{CO2}+5n\ce{H2O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since lignin is more complex, its combustion equation is more complex:&#xA;$$2[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+(21x+23y+25z)\ce{O2}\\ \ce{-&amp;gt;}(18x+20y+22z)\ce{CO2}+(10x+12y+14z)\ce{H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And assuming the variable composition of wood as a ratio (A:B) of cellulose to lignin, the overall combustion reaction becomes the following monstrosity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A\ce{(C6H10O5)}_n +2B[\ce{(C9H10O2)}_x \cdot \ce{(C10H12O3)}_y \cdot \ce{(C11H14O4)}_z]+[6An+B(21x+23y+25z)]\ce{O2}\\ \ce{-&amp;gt;}[6An+B (18x+20y+22z)]\ce{CO2}+[5An+B(10x+12y+14z)]\ce{H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the number of variables ($A, B, n, x, y, z$), the $\Delta_cH^o$ for this reaction is difficult to determine (but not impossible, assuming we know some thermochemical reference data). Whatever its value, it is exothermic enough to induce incandescence. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2018-05-03T12:02:07.360" LastActivityDate="2018-05-03T12:02:07.360" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1260" PostTypeId="1" AcceptedAnswerId="1261" CreationDate="2012-09-29T14:47:51.737" Score="5" ViewCount="1053" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A container of volume &lt;span class=&quot;math-container&quot;&gt;$\pu{4.0 L}$&lt;/span&gt; contains nitrogen at &lt;span class=&quot;math-container&quot;&gt;$\pu{25 °C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{803 kPa}.$&lt;/span&gt; This container is connected by a valve to another container of volume &lt;span class=&quot;math-container&quot;&gt;$\pu{10.0 L}$&lt;/span&gt; that contains argon at &lt;span class=&quot;math-container&quot;&gt;$\pu{25 °C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{47.2 kPa}.$&lt;/span&gt; We now open the valve so that the two gases are mixing.&lt;/p&gt;&#xA;&lt;p&gt;Calculate the partial pressure of each gas after the mixing and the total pressure of the mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using the Dalton's law of partial pressures, isn't the partial pressure of each gas exactly &lt;span class=&quot;math-container&quot;&gt;$\pu{803 kPa}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{47.2 kPa}$&lt;/span&gt; and the total pressure &lt;span class=&quot;math-container&quot;&gt;$\pu{803 kPa} + \pu{47.2 kPa}?$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="638" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-03-21T06:50:31.903" Title="Using Dalton's law of partial pressures" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1261" PostTypeId="2" ParentId="1260" CreationDate="2012-09-29T14:59:31.997" Score="5" Body="&lt;p&gt;Dalton's law of partial pressures means that the total pressure in the final volume $V=V_1+V_2$ is equal to the sum of the partial pressures P(N&lt;sub&gt;2&lt;/sub&gt;) + P(Ar) &lt;em&gt;in this volume&lt;/em&gt;. However, the partial pressure of N&lt;sub&gt;2&lt;/sub&gt; in the total volume $V$ is not the same as the pressure of N&lt;sub&gt;2&lt;/sub&gt; in its initial volume $V_1$. You have to account for the expansion. This probably means you also need to use the &lt;em&gt;ideal gas law&lt;/em&gt; for that.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-09-29T20:31:51.730" LastActivityDate="2012-09-29T20:31:51.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1262" PostTypeId="2" ParentId="1258" CreationDate="2012-09-29T15:05:43.367" Score="2" Body="&lt;p&gt;Either you have a periodic table that features the specific heat capacity of elements in their standard state at STP (standard temperature and pressure), in which case you merely have to look it up, or you don't, and there's nothing you can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific heat capacity of materials, even elemental solids, are physical properties resulting from complex behavior of the systems, and cannot be determined by simple laws or equations. They have to be determined experimentally, or by &lt;a href=&quot;http://en.wikipedia.org/wiki/Ab_initio_quantum_chemistry_methods&quot; rel=&quot;nofollow&quot;&gt;first principles calculations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS: water, methanol, carbon tetrachloride and dichlorodifluoromethane have got nothing to do with it.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-09-29T15:05:43.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1263" PostTypeId="2" ParentId="1256" CreationDate="2012-09-29T15:20:36.443" Score="7" Body="&lt;p&gt;&lt;strong&gt;Yes, it’s coincidental&lt;/strong&gt;. There is no expectation that prime (or near-prime) atomic weight means anything for the element itself. In fact, your affirmation is not even true: on the list given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_metal&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the first two examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Au_75 Si_25}$, your assertion is true&lt;/li&gt;&#xA;&lt;li&gt;“an alloy of 77.5% palladium, 6% copper, and 16.5% silicon”, and your assertion isn’t met for copper&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's go further. For a given element, how likely is it that its atomic weight, rounded, is within ±1 of a prime number? Well, atomic weights are relatively low integers, and prime numbers are plenty numerous among small integers, so it is relatively frequent. The answer can be computed with Mathematica:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;weights = {#, Round@ElementData[#, &quot;AtomicWeight&quot;]} &amp;amp; /@ ElementData[All];&#xA;f[i_] := PrimeQ[i] || PrimeQ[i - 1] || PrimeQ[i + 1]&#xA;res = {f[#[[2]]], #[[1]]} &amp;amp; /@ weights&#xA;Select[res, ! #[[1]] &amp;amp;][[All, 2]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;74 out of the first 118 elements turn out to satisfy your “near-prime atomic weight” condition&lt;/strong&gt;, which is 63%. This seems high, but it's not: if you do the same check, not only for atomic weights, but for all integers below 200, 61% of them are near-prime:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Total@Boole@Select[f /@ Range[1, 200], # &amp;amp;]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-09-29T20:37:08.823" LastActivityDate="2012-09-29T20:37:08.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1264" PostTypeId="2" ParentId="1255" CreationDate="2012-09-30T03:47:04.590" Score="2" Body="&lt;p&gt;Based on &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/elan.201100346/abstract&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, in which the authors describe a system that can do 25 cells simultaneously, the term 'Parallel Electrolyser' would seem appropriate.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-09-30T03:47:04.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1266" PostTypeId="2" ParentId="1254" CreationDate="2012-09-30T11:56:25.217" Score="6" Body="&lt;p&gt;The chemical reactions of a &quot;typical fire&quot; using wood as the inflammable matter are extremely complex and not known to their full extend. The wood isn't actually burning but gasified (mostly superficial) components of the wood, which yield a plasma in an environment that provides sufficient excitation energy. Accelerants are often used to this end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is aggravated by the many radical species which are produced in chain reactions and the temperature gradient of the flame which favors different species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The form of fire depends on many parameters, gravity included...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search for &quot;&lt;em&gt;estimating the number of chemical species in a wood fire&lt;/em&gt;&quot; brings up interesting literature which may serve you as a starting point, in addition to Ben Norris's answer.&lt;/p&gt;&#xA;" OwnerUserId="660" LastActivityDate="2012-09-30T11:56:25.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1267" PostTypeId="1" CreationDate="2012-09-30T19:29:04.843" Score="3" ViewCount="638" Body="&lt;p&gt;I am trying to build a homemade thermos. It needs to be able to conserve the heat in 125 ml of water for 15 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, unlike a normal thermos, it also needs to be able to be heated. What homemade substances can I use that are not flammable, yet provide a relatively high degree of insulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that my thermos will be heated with a flame from a candle for 25 minutes.&lt;/p&gt;&#xA;" OwnerDisplayName="user690" LastEditorDisplayName="user7951" LastEditDate="2016-10-09T14:25:49.183" LastActivityDate="2016-10-09T14:25:49.183" Title="Are there any nonflammable thermal insulators which can be made at home?" Tags="&lt;heat&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1269" PostTypeId="1" AcceptedAnswerId="1286" CreationDate="2012-10-01T00:20:17.217" Score="7" ViewCount="332" Body="&lt;p&gt;In one of the solar system documentaries it was mentioned there are moons in the solar system with lakes of frozen methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming humans or robots or whatever could make it there, would there be a way to extract energy from frozen methane without using oxygen? Otherwise, it would seem that the oxygen would have to be carried there as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to recycle the oxygen efficiently enough to burn methane again and again? &lt;/p&gt;&#xA;" OwnerUserId="692" LastEditorUserId="73" LastEditDate="2012-10-01T08:59:45.887" LastActivityDate="2012-11-30T08:29:19.167" Title="Is it possible to get energy from frozen methane without oxygen?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1270" PostTypeId="2" ParentId="1255" CreationDate="2012-10-01T00:59:27.267" Score="1" Body="&lt;p&gt;It's a precipitator, also known as TDS indicator.&lt;/p&gt;&#xA;" OwnerUserId="686" LastActivityDate="2012-10-01T00:59:27.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1271" PostTypeId="1" CreationDate="2012-10-01T02:29:11.510" Score="7" ViewCount="174" Body="&lt;p&gt;Are the half-reactions used in analyzing redox reactions a real thing or are they a conceptual shortcut?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to, however briefly, run a half-reaction by itself without the corresponding other half?  For example, reduce metal atoms to metal ions without the corresponding oxidation half-reaction?  &lt;/p&gt;&#xA;" OwnerUserId="693" LastActivityDate="2012-10-01T07:22:59.450" Title="Are redox half-reactions &quot;real&quot;?" Tags="&lt;ions&gt;&lt;redox&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1272" PostTypeId="2" ParentId="1271" CreationDate="2012-10-01T04:40:11.337" Score="7" Body="&lt;p&gt;Electrolysis/Galvanic cell is basically redox reaction split in half-reactions, running on different electrodes. For example, hydrogen fuel cell has two electrodes. On one hydrogen donates electrons to electrode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_2 = 2H^+ + 2e^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on another, oxygen drains electrons from electrode&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_2 + 4 e^- + 4 H^+ = 2 H_2O $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons travel from one electron to another via external electricity consumer. If electrons are forced to travel by external force, the reaction may be reversed. This ideas are utilized in electric chemical batteries, accumulators, fuel cells, industrial electrolyzers and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, half-reactions are separated between different electrode. Running half-reaction by itself without other half elsewhere is problematic at best, because of electroneitrality violation. You'll need extreme voltage source and even then it will stop quickly.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-10-01T07:22:59.450" LastActivityDate="2012-10-01T07:22:59.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1273" PostTypeId="1" CreationDate="2012-10-01T16:00:44.510" Score="10" ViewCount="3019" Body="&lt;p&gt;I am reading up on entropy in a textbook and I got confused by this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that to calculate the entropy for an irreversible process using heat flow, one must imagine the reversible process in which the initial and final states are the same as for the irreversible process. Why is this so and why do we need to think of it in the reversible way?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="30" LastEditDate="2012-10-02T21:08:18.850" LastActivityDate="2015-10-25T03:48:21.247" Title="Calculating entropy: why consider a reversible path?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1274" PostTypeId="1" CreationDate="2012-10-01T16:38:28.857" Score="4" ViewCount="207" Body="&lt;p&gt;It's been a long, long time since I've had to worry about electron excitation levels and band gaps. Please forgive (and correct!) my terminology and misconceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've become interested in volumetric displays. In particular I'm looking at a pair of lasers scanning a volume of gas such that the intersection of the lasers would supply enough energy to cause the gas to emit visible light. Since this is the start of my research I would like to investigate a number of gasses in order to determine a number of gasses which might be suitable for this application. I have, however, forgotten how to determine the amount of energy needed to cause an outer shell electron to emit a photon, and then trying to back-calculate the required wavelength(s) and power requirements of the laser emitters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some information on how to identify the required information for a given gas and how to calculate the energy required to cause visible light emission, and also information on how the wavelength of laser light impacts the ability of the gas to absorb the energy with the intended effect of emitting visible light. Links to introductory texts are always helpful, as are answers which help me get the terminology correct so that I can improve my searches for this information.&lt;/p&gt;&#xA;" OwnerUserId="1041" LastEditorUserId="418" LastEditDate="2014-03-07T19:17:23.563" LastActivityDate="2014-03-07T19:18:44.083" Title="Calculating laser wavelength/power to cause emission of light in a gas?" Tags="&lt;energy&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1275" PostTypeId="2" ParentId="1273" CreationDate="2012-10-01T16:41:02.797" Score="9" Body="&lt;p&gt;There are two parts to this answer.  The first is that the entropy change is defined only for reversible processes.  The second is that the entropy is a state function.  Many, but not all, the functions in thermodynamics (including the internal energy) are state functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A state function is one that is independent of path.  That is, if one goes from state $A$ to state $B$, the change in internal energy is independent of how the change takes place.  The same is true for the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the entropy change in going from state $A$ to state $B$ is independent of how the change takes place.  But we can calculate the change ONLY for reversible changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we mentally find a reversible path from state $A$ to state $B$ and calculate the entropy change.  Since entropy is a state function, that must also be the change for a non-reversible path.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-10-01T16:41:02.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1277" PostTypeId="2" ParentId="1193" CreationDate="2012-10-02T05:17:45.010" Score="3" Body="&lt;p&gt;While I cannot comment in a technical level, I can speak from a background in aerospace experience (in my family) and lifetime of building things that live outdoors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In time, dissimilar metal corrosion always sets in. Most stainless steels are inhibiting, not a 'way out' of the problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that, I think the ambient salts will kills you on this in the marine environment.  If you have a regular maintenance plan, then perhaps you can re-service the bolts yearly, as needed.&lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2012-10-02T05:17:45.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1278" PostTypeId="1" AcceptedAnswerId="1279" CreationDate="2012-10-02T10:33:35.863" Score="8" ViewCount="260" Body="&lt;p&gt;I would like to print some molecules rather than building them using a kit. What file formats work?&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2012-10-02T11:48:20.557" Title="What chemical file formats are compatible with 3D printing?" Tags="&lt;organic-chemistry&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1279" PostTypeId="2" ParentId="1278" CreationDate="2012-10-02T11:48:20.557" Score="10" Body="&lt;p&gt;&lt;strong&gt;None of the chemical file formats are typically accepted by 3D printing services.&lt;/strong&gt; The reason is that chemical file formats are fundamentally different from those describing 3D objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest case, a chemical file format describes a list of atoms and their positions (plus possible metadata). An example is the XYZ format, which stores exactly that. Other “chemical objects” which can also be described in some chemical file formats include bonds, protein structure, crystalline structure information (unit cell, symmetry, occupancy of each atomic position, …), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D file formats, on the other hand, describe a series of objects. In the simplest case (e.g., the STL format or the Wavefront .obj file format), the object itself is not stored, but an unstructured triangulated surface delimiting the object. In those, you can't describe a sphere, but only a tesselation of the sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need a conversion step to create a 3D representation of your chemical structure. Chemical file formats don't typically contain representation information: you need to pick the one you want (CPK, liccorice, protein cartoon, …) and use software to create a 3D scene. &lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot;&gt;VMD&lt;/a&gt;, for example, allows export to STL format, which is suitable for 3D printing.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-02T11:48:20.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1280" PostTypeId="1" AcceptedAnswerId="1282" CreationDate="2012-10-02T15:54:56.367" Score="10" ViewCount="132" Body="&lt;p&gt;I'm seeking a &lt;strong&gt;lay&lt;/strong&gt; explanation for how the work of Dr Marc Taillefer that won the &lt;a href=&quot;http://www.euchems.eu/news/single-news/article/marc-taillefer-wins-the-european-sustainable-chemistry-award.html&quot; rel=&quot;nofollow&quot;&gt;2012 European Sustainable Chemistry Award&lt;/a&gt;, contributes to &lt;strong&gt;sustainable&lt;/strong&gt; chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://www.euchems.eu/fileadmin/user_upload/awards/ESCA2012_Taillefer.pdf&quot; rel=&quot;nofollow&quot;&gt;press release&lt;/a&gt;, Dr. Taillefer &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is being recognised for his seminal contribution to the field of&#xA;  homogeneously catalysed coupling reactions leading to C—C, C—N, C—O, C—P bonds. His team at&#xA;  the Institut Charles Gerhardt, ICG (Montpellier, France) is investigating for a decade the&#xA;  environmentally sustainable conversion of small molecules into more valuable substances&#xA;  catalysed by copper and iron molecular complexes. This renaissance of “Ullmann type arylations” is now often used at the academic or industrial level and avoids the use of more expensive catalysts based on palladium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The objectives  of the award are (to quote from &lt;a href=&quot;http://www.euchems.eu/news/single-news/article/matthias-beller-winner-of-the-first-european-sustainable-chemistry-award.html&quot; rel=&quot;nofollow&quot;&gt;this press release&lt;/a&gt;) to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Recognise individuals or small research groups which make an outstanding contribution to sustainable development by applying green and sustainable chemistry.&lt;/li&gt;&#xA;  &lt;li&gt;Promote innovation in chemistry and chemicals that will deliver clear improvements in the sustainable production and use of chemicals and chemical products.&lt;/li&gt;&#xA;  &lt;li&gt;Demonstrate that chemistry and chemicals can play a central role in delivering society’s needs, while minimizing and solving environmental problems.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25" LastEditorUserId="24093" LastEditDate="2016-07-18T11:47:14.330" LastActivityDate="2016-07-18T11:47:14.330" Title="How does the work that won the 2012 Sustainable Chemistry Award contribute to sustainable chemistry?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1281" PostTypeId="1" CreationDate="2012-10-02T16:57:08.410" Score="4" ViewCount="2738" Body="&lt;p&gt;Generally, noble gases are very unreactive. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that it has to do with electron shielding prospective electrons that could join the atom or leave and the energies associated with both actions, but why are 5 electrons in the p orbital (to take Fluorine as an example) so much worse in shielding the nuclear charge than 6 (Neon)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What IS electron shielding, anyway? I understand it in the classical lie-to-children way, but not satisfactorily.&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="30836" LastEditDate="2020-02-25T10:37:08.460" LastActivityDate="2020-02-25T10:37:08.460" Title="Why are noble gases stable?" Tags="&lt;energy&gt;&lt;stability&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-04-14T13:06:57.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="1282" PostTypeId="2" ParentId="1280" CreationDate="2012-10-02T18:12:51.023" Score="12" Body="&lt;p&gt;His work is about developping new catalysts based on copper and iron, to &lt;strong&gt;replace to traditional catalysts based on palladium&lt;/strong&gt;. Copper and iron are both very common elements in nature, while &lt;strong&gt;palladium is considered a high supply risk&lt;/strong&gt; (see the &lt;a href=&quot;http://www.bgs.ac.uk/mineralsuk/statistics/riskList.html&quot;&gt;2012 British Geological Survey risk list&lt;/a&gt; for details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new &lt;a href=&quot;http://www.rsc.org/periodic-table/&quot;&gt;RSC Visual Elements Periodic Table&lt;/a&gt; can be used to check this kind of information.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-02T18:12:51.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1283" PostTypeId="1" AcceptedAnswerId="1284" CreationDate="2012-10-02T22:57:18.417" Score="7" ViewCount="9653" Body="&lt;p&gt;I know like dissolves like, but I am still unclear.  9-fluorenone has a polar $\ce{C—O}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because 9-fluorenone can only form London forces with other 9-fluorenone molecules and hexane can break these forces?&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="4231" LastEditDate="2015-02-24T16:25:39.007" LastActivityDate="2015-02-24T16:25:42.063" Title="Why does 9-fluorenone dissolve in hexane?" Tags="&lt;solvents&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1284" PostTypeId="2" ParentId="1283" CreationDate="2012-10-03T00:15:10.973" Score="5" Body="&lt;p&gt;Firstly, as a general rule, organic compounds with more then 6 carbon have higher London forces then polar forces.  This is a very general rule and doesn't work in all circumstances. Thus, this molecule is not as polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, cyclic compounds have much weaker London forces then molecules that have straight chains. Thus, this makes the London forces not so high(relative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, to have something dissolve in hexane it must have weak intermolecular forces, but not weaker then hexane itself, as hexane only has to suspend the solute molecules in solution and prevent the solute from clumping up and falling to the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, quite simply, what happens is that the hexane molecules surround the 9-Fluorenone molecules, which diffuse through the solution.&lt;/p&gt;&#xA;" OwnerUserId="698" LastEditorUserId="73" LastEditDate="2012-10-05T17:08:39.307" LastActivityDate="2012-10-05T17:08:39.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1285" PostTypeId="2" ParentId="1281" CreationDate="2012-10-03T01:00:19.917" Score="1" Body="&lt;p&gt;It's all about the energy levels of electrons. Electrons exist in different energy levels and each energy level can only be occupied by a specific number of electrons (google quantum theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Neon has an electron configuration of $1s^2 2s^2 1p^6$, which simply means that it has a full $1s$ orbital (the maximum is 2 electrons as denoted by the superscript), and has a full valence shell which is the $2s$ and $1p$ orbitals (which is the valence shell for molecules in row 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for neon to lose an electron (to bond) it would have to take an electron out of a full orbital ($1p$), which is what makes molecules reactive (a non-full valence shell), and would just take the electron right back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of Ne$^+$ as more electronegative (wanting electrons) than elemental Fluorine, so it will pull an electron from almost anything and go back to its elemental form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this answer is not satisfactory then I can explain further, without diving head first into quantum theory.&lt;/p&gt;&#xA;" OwnerUserId="698" LastEditorUserId="107" LastEditDate="2012-10-03T04:20:02.307" LastActivityDate="2012-10-03T04:20:02.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1286" PostTypeId="2" ParentId="1269" CreationDate="2012-10-03T01:10:43.203" Score="6" Body="&lt;p&gt;The planet will have gaseous oxygen if it does have an atmosphere, and if that is the case then with enough energy(like a spark) atomized hydrocarbons could be lit for energy. What is atomizing you say? Well its just breaking a liquid apart into little particles, making it easier for the oxygen to react with said methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for recycling the oxygen from the combustion reaction, it will not be possible to do in an energy efficient way, CO2 is too stable and the bonds are too strong.&lt;/p&gt;&#xA;" OwnerUserId="698" LastActivityDate="2012-10-03T01:10:43.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1287" PostTypeId="2" ParentId="1252" CreationDate="2012-10-03T01:26:25.927" Score="1" Body="&lt;p&gt;Firstly what you have here is an acid base problem. The way you can tell when you have an acid base problem is when your aqueous reactants produce water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly you know that anything with an (aq) subscript are ions in solution. AKA they are telling you which molecules are disassociated in water.&lt;/p&gt;&#xA;" OwnerUserId="698" LastActivityDate="2012-10-03T01:26:25.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1288" PostTypeId="2" ParentId="1072" CreationDate="2012-10-03T01:46:50.670" Score="1" Body="&lt;p&gt;There is a point in which it is too hot for a molecule to remain bonded and will break apart, no matter how much pressure is applied. Now with more pressure the electrons and nuclei will start free flowing in what is known as a plasma. And add a little more heat and fission happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the reason they don't show you any more of the curve because your molecule under enough pressure and heat is no longer a molecule.&lt;/p&gt;&#xA;" OwnerUserId="698" LastActivityDate="2012-10-03T01:46:50.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1289" PostTypeId="1" AcceptedAnswerId="1290" CreationDate="2012-10-03T10:22:36.430" Score="27" ViewCount="96352" Body="&lt;p&gt;The case of polar solvents is clear to me - we get an attraction between opposite charges. However, how do non-polar substances dissolve in non-polar solvents? How could it be explained on a molecular level?&lt;/p&gt;&#xA;" OwnerUserId="222" LastActivityDate="2015-08-16T16:43:26.620" Title="How do non-polar substances dissolve in non-polar solvents?" Tags="&lt;solvents&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1290" PostTypeId="2" ParentId="1289" CreationDate="2012-10-03T11:56:06.860" Score="29" Body="&lt;p&gt;The electron density distribution in molecules (including nonpolar ones) is not static. Therefor, as a function of time, the electron density is not uniform. Occasionally, randomly, the electron density in a molecule will shift to produce a spontaneous dipole: part of the molecule now has more electron density ($\delta^-$) and part of the molecule now has less electron density ($\delta^+$). This spontaneous dipole is transient. The electron density will shift back to negate it and then shift again to create a new spontaneous dipole. In a vacuum, this behavior would be a curiosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of other molecules, these spontaneous dipoles have a propagating effect. If molecule A develops a spontaneous dipole, then the electron density in neighboring molecule B will by respond by developing a spontaneous (but opposite) dipole. The $\delta^+$ region of B will be close to the $\delta^-$ region of A. This new dipole in molecule B is an induced dipole. This induced dipole in B will then induce a new induced dipole in another neighboring molecule C. By this time the transient dipole in A is already fading, perhaps to be replaced by a new induced dipole from another molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These random, transient, but continuously propagating dipoles have attractive forces associated with them. These forces are named London dispersion forces after the physicist who proposed them. The magnitude of these forces scales with increasing surface area of the molecule. Thus, larger molecules will overall have stronger London dispersion forces than smaller molecules. Linear rod-like molecules will have stronger forces than spherical molecules (spheres having the smallest surface area to volume ratio of the 3-dimensional solids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex attractive forces also arise from nonpolar molecules. Benzene (and other aromatic molecules) can &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi_stacking&quot;&gt;pi stack&lt;/a&gt;, which relies on the strong permanent electric &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadrupole&quot;&gt;quadrupole&lt;/a&gt;. Benzene also has a (weak but) permanent magnetic dipole due to its ring current, so some component of the attraction between benzene molecules may be magnetic and not coulombic. Or, the two attractions may be one and the same, as benzene's electric quadrupole may result from its having a magnetic dipole. As his part of the discussion is devolving into physics, I will end here. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-03T11:56:06.860" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1291" PostTypeId="2" ParentId="1289" CreationDate="2012-10-03T15:19:42.027" Score="19" Body="&lt;p&gt;I think @Ben Norris's answer is great, but I wanted to add one other reason that non-polar molecules will dissolve in non-polar solvents, even in the absence of interactions: entropy. If there are no intermolecular interactions at all (ie, the solvent molecules ignore each other and the solute molecules, and the solute molecules ignore each other and the solvent molecules), then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_of_mixing&quot;&gt;entropy of mixing&lt;/a&gt; will be positive, and mixing will be favored. As a result, species always want to mix; it takes unfavorable conditions to cause non-solubility.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2012-10-03T15:19:42.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1292" PostTypeId="2" ParentId="1248" CreationDate="2012-10-03T22:19:01.360" Score="3" Body="&lt;p&gt;What immediately comes to mind is something rather obvious, don't use ammonia on any surface that could possibly be contaminated with bleach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search revealed a &lt;a href=&quot;http://www.youtube.com/watch?v=tOhJr-iLsGo&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; showing red cabbage juice used as an indicator.  I do not believe the limit of detection would be anything to write home about.  However, this got me thinking about a homemade pH paper.  This would allow you to dampen the surface with a very small amount of neutral water, apply the paper, and possibly get an accurate reading on whether something acidic or basic is present on the spot you tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, someone else has already put together a procedure on how to make these strips, &lt;a href=&quot;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Chem_p041.shtml&quot; rel=&quot;nofollow&quot;&gt;found here&lt;/a&gt;.  With another source showing the expected colors using this &lt;a href=&quot;http://chemistry.about.com/od/acidsbase1/a/red-cabbage-ph-indicator.htm&quot; rel=&quot;nofollow&quot;&gt;indicator&lt;/a&gt;.  Since bleach aka sodium hypochlorite is a base, I would expect the color of the test strips to be in the blue to green range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could also be a cool home science project.  The strips will be more effective at detection for lower concentrations of bleach than a liquid solution of the indicator. (pH is concentration dependent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I linked many articles to fill in some relevant information but I was limited to 2 links.  If you have questions please feel free to ask!&lt;/p&gt;&#xA;" OwnerUserId="700" LastEditorUserId="30" LastEditDate="2012-10-04T21:08:28.970" LastActivityDate="2012-10-04T21:08:28.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1293" PostTypeId="1" AcceptedAnswerId="1317" CreationDate="2012-10-04T01:18:12.560" Score="1" ViewCount="1614" Body="&lt;p&gt;Applying pressure can result in chemical changes, does this hold true for an avocado? &lt;/p&gt;&#xA;" OwnerUserId="701" LastActivityDate="2014-02-01T15:33:49.613" Title="Is crushing an avocado a chemical or physical change?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" ClosedDate="2015-10-31T06:08:09.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="1294" PostTypeId="2" ParentId="1293" CreationDate="2012-10-04T02:59:29.200" Score="9" Body="&lt;p&gt;I don't think so.  A chemical change implies changing molecular-level structure usually accompanied by transferring electrons.  When you squish an avocado you're only rearranging the fats, carbohydrates, and fiber in it - I don't think a chemical reaction occurs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you left it out, and it turned brown (oxidized), that would be a chemical change! &lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="702" LastEditDate="2013-03-05T18:25:06.960" LastActivityDate="2013-03-05T18:25:06.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1295" PostTypeId="2" ParentId="1252" CreationDate="2012-10-04T03:17:18.540" Score="5" Body="&lt;p&gt;The second equation you’ve given appears to be the right overall ionic equation. Phosphate is a weak acid, $\ce{NaOH}$ is a strong base. We expect NaOH to dissociate at any pH, where the speciation of phosphate ($\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$) depends on pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A “net ionic equation” doesn’t include spectator ions. Spectator ions exist on both sides of the arrow and can be “canceled like” (think as in algebra). For any acid base reaction, it’s really easy to remember!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H+ {(aq)} + OH- {(aq)} &amp;lt;=&amp;gt; H2O {(l)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly don’t forget the state: $\ce{OH {(aq)}-}$. I was taught to right them as subscripts, but I’ve seen then as regular text too: $\ce{OH- (aq)}$.&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="5026" LastEditDate="2018-05-16T10:03:06.620" LastActivityDate="2018-05-16T10:03:06.620" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1296" PostTypeId="2" ParentId="1233" CreationDate="2012-10-04T03:26:34.410" Score="4" Body="&lt;p&gt;Carbon needs four electrons to fill its octet.  In a &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; covalent bond carbon shares one electron with hydrogen.  A triple consists of one &amp;sigma; covalent bond and two &amp;pi; covalent bonds.  The carbons share two electrons in each of these bonds; in total six electrons are shared between them.  You can draw this as a dot structure:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H:C:::C:H}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorUserId="4945" LastEditDate="2018-12-11T13:37:00.060" LastActivityDate="2018-12-11T13:37:00.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1297" PostTypeId="2" ParentId="1267" CreationDate="2012-10-04T04:51:51.710" Score="3" Body="&lt;p&gt;Consider using a ceramic container.  You could adjust the density, thickness, and size to match your insulating requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;home made' is loose, here, since you would need to acquire oven bake-able ceramics, or a similar product.  (unless you have a kiln lying around).&lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2012-10-04T04:51:51.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1298" PostTypeId="2" ParentId="1293" CreationDate="2012-10-04T07:30:22.857" Score="6" Body="&lt;p&gt;I second Phillip’s answer, which is right in practice, with a small additional detail: &lt;strong&gt;it depends on the pressure&lt;/strong&gt; applied. If you apply pressure by hand, you are not likely to create or destroy chemicals bonds, and the changes will be physical. &lt;strong&gt;Applying pressure can change chemical structure&lt;/strong&gt;, as in the case of solid phase transitions or pressure-induced reactions, but the pressure has to be much larger than that: &lt;strong&gt;typically of the order of gigapascals&lt;/strong&gt; (1 GPa is rougly 10,000 atmospheres).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-04T07:30:22.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1300" PostTypeId="1" AcceptedAnswerId="1301" CreationDate="2012-10-05T02:52:23.747" Score="10" ViewCount="1400" Body="&lt;p&gt;For drawing MO diagram for heteronuclear diatomic molecule, how do I know which is lower in energy than the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, given a heteronuclear diatomic molecule $\ce{NO}$, which one is lower in energy?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="12860" LastEditDate="2015-10-30T04:11:06.280" LastActivityDate="2015-10-30T04:11:06.280" Title="Molecular orbitals of heteronuclear diatomic molecules" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1301" PostTypeId="2" ParentId="1300" CreationDate="2012-10-05T11:45:57.973" Score="7" Body="&lt;p&gt;The rule as generally stated in textbooks is that &lt;strong&gt;“atomic orbitals of the more electronegative element are lower in energy than the corresponding orbitals of the less electronegative element”&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, MO diagrams are only simple for elements of the second row of the periodic table ($\ce{Li}$ through $\ce{Ne}$). Involving heavier atoms makes it harder to guess at molecular orbital diagrams, and there is need for quantum chemistry calculations. Thus, the rule becomes: the further to the right your element is, the lower its energy levels are. The reason for this is that the charge of the nucleus increases with its atomic number, and thus lowers the energy levels.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-24T16:27:19.433" LastActivityDate="2015-02-24T16:27:19.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1302" PostTypeId="1" CreationDate="2012-10-05T13:18:00.713" Score="18" ViewCount="41729" Body="&lt;p&gt;I saw in a textbook that for carbonate ion, there are 3 resonance forms and the bond order is 1 and 1/3. So in general, how should we calculate the bond order for resonance structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any definite way to do this?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="12860" LastEditDate="2015-10-30T04:14:20.583" LastActivityDate="2020-08-13T13:16:19.160" Title="Bond order for carbonate ion for resonance" Tags="&lt;inorganic-chemistry&gt;&lt;resonance&gt;&lt;molecular-structure&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1303" PostTypeId="2" ParentId="1302" CreationDate="2012-10-05T14:23:07.453" Score="14" Body="&lt;p&gt;There definitely is an easy way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the easiest examples that comes to mind is benzene.  If we look at the structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hQvU6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction.  If we draw the other resonance structure these bonds shift by 1 atom to look like the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ztLH5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them.  For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The &quot;$2$&quot; on the bottom of the equation is the number of resonance structure being considered.  Thus, for benzene the bond order is $1.5$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.  Since there are 3 structures the answer is $(2+1+1)/3 = 1.33333$ or $4/3$.&lt;/p&gt;&#xA;" OwnerUserId="700" LastEditorUserId="7475" LastEditDate="2015-10-30T12:53:06.243" LastActivityDate="2015-10-30T12:53:06.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1305" PostTypeId="2" ParentId="1222" CreationDate="2012-10-05T16:59:34.997" Score="4" Body="&lt;p&gt;The comment left by CHM is very helpful.  You need to consider what this molecule actually looks like in 3D.  That may help your understanding greatly, but to be honest there is considerably more going on here than you would expect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple pieces of information would be very helpful for clarifying this question.  The first is what type of a metal are you using for this synthesis step.  The second is what is the outcome.  Believe it or not, based on those two pieces of information the mechanisms for this reaction can be explained entirely differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the answer basically breaks down into two possibilities I will attempt to explain both.  One of the most helpful articles I came across in my search was entitled &quot;Transition State Models for Probing Stereoinduction in Evans Chiral Auxiliary-Based Asymmetric Aldol Reactions&quot; &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja101913k&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;132&lt;/em&gt; (35), pp 12319–12330.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this article they go into depth about how the electronic and steric effects balance out to produce the almost complete stereocontrol throughout the course of this reaction.  The scheme below is what I find the most helpful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/347BC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the selectivity is varied based on whether the auxiliary is chelating to the titanium or not.  This is why the metal lewis acid used in these reactions is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note not shown here is the rational behind the non chelating setup.  The opposing dipoles on &quot;y&quot; and aldehyde oxygen in the Zimmerman-Traxler model help to hold the six member ring in place while the aldol reaction occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scheme shown is incredibly useful for visualizing this transformation and I would like to compliment the authors for their efforts in this publication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="700" LastActivityDate="2012-10-05T16:59:34.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1306" PostTypeId="2" ParentId="1283" CreationDate="2012-10-05T18:08:41.587" Score="3" Body="&lt;p&gt;I would argue differently from Mike. 9-fluorenone has one polar $\ce{C=O}$ bond, but it also has two aromatic rings. Dispersion interactions are strong for such aromatic rings, and the point can be made that this factor will drive the solvation of 9-fluorenone in hexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is confirmed by the fact that &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorenone&quot; rel=&quot;nofollow&quot;&gt;9-fluorenone&lt;/a&gt; is insoluble in water.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-24T16:25:42.063" LastActivityDate="2015-02-24T16:25:42.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1307" PostTypeId="1" AcceptedAnswerId="1308" CreationDate="2012-10-05T20:09:35.620" Score="2" ViewCount="233" Body="&lt;p&gt;I'd like to illustrate my question with an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine nickel reacts with fluoride and you start with $0.766\ \mathrm{g}$ of nickel, an unknown amount of fluoride and you end with $1.261\ \mathrm{g}$ of nickelfluoride. The mass of fluoride in the end must then be $1.261\ \mathrm{g} - 0.766\ \mathrm{g} = 0.495\ \mathrm{g}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the end there's $0.013$ moles of nickel and $0.026$ moles of fluoride. Now the question is, what is the empirical formula for nickel fluoride? ($\small\ce{NiF2}$). I see that for every mole of nickel you need $2$ moles of fluoride but then why is the empirical formula $\small\ce{Ni2F4}$ incorrect?&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="18002" LastEditDate="2015-11-15T01:29:38.423" LastActivityDate="2015-11-15T01:29:38.423" Title="Find empirical formula when individual masses are known" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1308" PostTypeId="2" ParentId="1307" CreationDate="2012-10-05T20:30:51.280" Score="6" Body="&lt;p&gt;An &lt;strong&gt;empirical formula&lt;/strong&gt; is, according to &lt;a href=&quot;http://goldbook.iupac.org/E02063.html&quot;&gt;IUPAC’s &lt;em&gt;Gold Book&lt;/em&gt; definition&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Formed by juxtaposition of the atomic symbols with their appropriate subscripts to give the &lt;strong&gt;simplest possible formula expressing the composition&lt;/strong&gt; of a compound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Empirical_formula&quot;&gt;Wikipedia&lt;/a&gt; goes even further:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the simplest positive integer ratio of atoms of each element present in a compound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An empirical formula makes no reference to isomerism, structure, or &lt;strong&gt;absolute number of atoms&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, NiF&lt;sub&gt;2&lt;/sub&gt; is an empirical formula, while Ni&lt;sub&gt;2&lt;/sub&gt;F&lt;sub&gt;4&lt;/sub&gt; simply isn't. Another example is given by, again, Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH&lt;sub&gt;2&lt;/sub&gt;O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Their molecular formulas (which account for the absolute number of atoms in a molecule) differ, though: formaldehyde CH&lt;sub&gt;2&lt;/sub&gt;O, acetic acid C&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;2&lt;/sub&gt;, glucose C&lt;sub&gt;6&lt;/sub&gt;H&lt;sub&gt;12&lt;/sub&gt;O&lt;sub&gt;6&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-05T20:30:51.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1309" PostTypeId="1" AcceptedAnswerId="2382" CreationDate="2012-10-05T23:18:00.850" Score="4" ViewCount="3915" Body="&lt;p&gt;I recently learned that attempts to compare the spectra of two isomers of $\ce{C_3H_3^+}$ were frustrated by the difficulty of separating the two species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes these isomers difficult to separate?&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="4231" LastEditDate="2015-02-23T19:52:35.630" LastActivityDate="2015-02-23T19:52:35.630" Title="Why are isomers difficult to separate?" Tags="&lt;carbocation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1310" PostTypeId="2" ParentId="1309" CreationDate="2012-10-06T02:33:16.650" Score="5" Body="&lt;p&gt;Molecules which are made up of exactly the same elements but differ in atomic arrangement are called isomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the two isomers have the same mass and charge, any seperation techniques which rely on a difference in mass and/or density would be very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass spectrometry is an analytical technique in which components of different mass are separated and detected, typically by ionizing (charging), accelerating through an electric field to a given velocity and finally, 'bending' through a magnetic field. The final position of the is detected as a function of its charge-to-mass ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any two isomers of $\small\ce{(C3H3)^+}$ have the same charge-to-mass ratio and therefore would be very difficult to separate and detect with mass spectrometry.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="73" LastEditDate="2012-10-06T10:35:49.193" LastActivityDate="2012-10-06T10:35:49.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1312" PostTypeId="5" CreationDate="2012-10-06T16:26:16.197" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-10-06T16:26:16.197" LastActivityDate="2012-10-06T16:26:16.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1313" PostTypeId="4" CreationDate="2012-10-06T16:26:16.197" Score="0" Body="Questions regarding chemical and lab safety. This includes questions about proper lab procedure, safe chemical handling, and such." OwnerUserId="323" LastEditorUserId="5137" LastEditDate="2014-06-02T23:15:29.553" LastActivityDate="2014-06-02T23:15:29.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1314" PostTypeId="2" ParentId="1293" CreationDate="2012-10-06T16:56:47.293" Score="1" Body="&lt;p&gt;My 5 cents.&#xA;In the beginning it's only physical. However if you crush it a lot of things change. The most obvious is that oxygen can penetrate it much easier than before. There are a lot of substances that can be readily oxidated (&lt;em&gt;f.e.&lt;/em&gt; unsaturated compounds). You gain bigger surface so the cells on surface will dry. The nice equilibrium in which this fruit exist is changed. I can't explain it on the level of biochemistry but as every living thing avocado must have some mechanisms that protect it from decomposing. Some of them can't work properly if there is not enough moisture and to much O2.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2012-10-06T16:56:47.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1315" PostTypeId="1" AcceptedAnswerId="1316" CreationDate="2012-10-06T17:17:30.833" Score="2" ViewCount="1510" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which element is reduced in this reaction:&#xA;  $$\ce{2KMnO_4 +3NaSO_3 +H_2O -&amp;gt;2MnO_2 +3NaSO_4 +2KOH}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out which elements lose their electrons, but how am I supposed to know when elements are neutral and when they are ionized? I have tried the following, but I don't believe it to be correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2K^+Mn^+O_4^{-2} +3Na^+SO_3^{-2} +H_2O -&amp;gt;2Mn^{+2}O_2^{-2} +3Na^{+}SO_4{-2} +2K^+OH^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning being that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Oxygen atoms are always doubly anionized &lt;/li&gt;&#xA;&lt;li&gt;Sulfate and sulfite are also doubly ionized&lt;/li&gt;&#xA;&lt;li&gt;Sodium and potassium are always singly cationized&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Based on these rules I tried to make the bonds neutral, but my assumptions must be incorrect. Any help would be greatly appreciated&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorUserId="5026" LastEditDate="2018-02-20T16:14:58.493" LastActivityDate="2018-02-20T16:14:58.493" Title="Which element is being reduced in this reaction?" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1316" PostTypeId="2" ParentId="1315" CreationDate="2012-10-06T19:54:42.930" Score="4" Body="&lt;p&gt;Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in KMnO&lt;sub&gt;4&lt;/sub&gt;, Mn must be (+VII). In MnO&lt;sub&gt;2&lt;/sub&gt;, it is (+IV). Thus, Mn is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sulfur: in NaSO&lt;sub&gt;3&lt;/sub&gt;, it must be be (+V). In NaSO&lt;sub&gt;4&lt;/sub&gt;, it must be (+VII). Thus, sulfur is oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; (big &lt;em&gt;however&lt;/em&gt;, in fact): the sulfate anion is not SO&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;–&lt;/sup&gt;, but SO&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;2–&lt;/sup&gt;. Similarly, sulfite is SO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;2–&lt;/sup&gt;, not SO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;–&lt;/sup&gt;. So, your equation should be instead:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -&amp;gt; 2MnO_2 +3Na_2SO_4 +2KOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means that sulfur is oxidized from (+IV) to (+VI).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2012-10-07T18:52:40.807" LastActivityDate="2012-10-07T18:52:40.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1317" PostTypeId="2" ParentId="1293" CreationDate="2012-10-06T22:43:04.330" Score="3" Body="&lt;p&gt;Your title and question are a bit different.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is crushing an avocado a chemical or physical change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Physical. Crushing (or any other physical manipulation of) any substance is, by definition, a physical change.  However, those physical changes can affect the conditions for a chemical change, potentially causing a noticeable change in the rate of said reaction. You somewhat acknowledge this in your question, though I'm not sure it's quite intentional:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Applying pressure can &lt;em&gt;result in&lt;/em&gt; chemical changes, does this hold true for an avocado? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under reasonable conditions, I'm fairly sure that nothing in an avocado is reactive enough to undergo a chemical change because of any conditions the physical crushing altered. Except that it will increase the surface area and speed up oxidization.  Given enough of a crushing pressure (e.g. equivalent to the pressure put on coal to create diamond) I'm sure some chemical reaction or another will occur.&lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditDate="2014-02-01T15:33:49.613" LastActivityDate="2014-02-01T15:33:49.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1320" PostTypeId="1" AcceptedAnswerId="1352" CreationDate="2012-10-07T19:02:44.260" Score="4" ViewCount="423" Body="&lt;p&gt;For example, in phosphorus pentachloride five orbitals hybridize, but why not 6, as in the hydrated Al ion? I have read that the reason six orbitals hybridize is that a maximum of six oxygens can stably fit around it. Why is it not the same case for phosphorus?&#xA;Secondly, why does water $sp^3$ hybridize even though it has enough lone pairs in its p orbital to bond with 2 oxygens. Surely it would take more energy to pull up the lower $s$ orbital to bond than to allow the $p$'s, already elevated to that energy, to do the bonding?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="597" LastEditDate="2012-10-18T16:26:16.140" LastActivityDate="2012-10-18T16:26:16.140" Title="In orbital hybridisation, what determines the number of orbitals that hybridise to be at the same energy level?" Tags="&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1321" PostTypeId="1" AcceptedAnswerId="1327" CreationDate="2012-10-07T20:14:59.360" Score="6" ViewCount="772" Body="&lt;p&gt;I was wondering, can I separate the $\ce{NaClO}$ from the $\ce{H2O}$ in bleach. The decomposition point for $\ce{NaClO}$ is $374.14\ \mathrm{K}$, so I'm risking decomposing the compound if I do simple distillation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I do?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS: I don't have a vacuum pump.&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="5456" LastEditDate="2014-10-09T17:31:24.347" LastActivityDate="2014-10-09T17:31:24.347" Title="Separating NaClO from its aqueous form in bleach" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1322" PostTypeId="1" AcceptedAnswerId="1326" CreationDate="2012-10-07T20:21:06.627" Score="3" ViewCount="213" Body="&lt;p&gt;Imagine two atoms, and only consider the Van der Waals force. The electron cloud will jitter due to its quantum mechanical nature- some of these jitters forming dipoles, some not. However, on average they form dipoles. Why? &#xA;This question is equivalent to asking (although it is useful here to ask in another way): why is the potential energy of a dipole lower than a non-dipole? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom (I assume that a dipole is formed like this $e^-..ion...........e^-..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge): given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter?&#xA;Am I falling into a rut because I am not considering the wave nature of electrons?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="1362" LastEditDate="2014-11-02T14:20:17.407" LastActivityDate="2014-11-02T14:20:17.407" Title="When considering Van der Waals forces, why do dipoles form?" Tags="&lt;energy&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1323" PostTypeId="2" ParentId="1320" CreationDate="2012-10-07T21:55:58.777" Score="1" Body="&lt;p&gt;&lt;em&gt;This is only partly an answer to your question. I think teaching hybridization in depth is a bad pedagogical choice, so I am not the right person to explain the inner workings of this theory.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't spend too much time try to get the details of the how's and why's of hybridization…  There are arguments over how useful it is as a concept in theoretical chemistry, and it certainly has severe limitations. For example, the picture derived from hybridization for the water molecule is simply &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation#Explanation_of_the_shape_of_water&quot; rel=&quot;nofollow&quot;&gt;wrong&lt;/a&gt;: it predicts two equivalent lone pairs on the oxygen atom, and they are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just think in terms of molecular orbitals!&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-07T21:55:58.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1324" PostTypeId="1" AcceptedAnswerId="1325" CreationDate="2012-10-07T22:31:36.117" Score="8" ViewCount="347" Body="&lt;p&gt;I have a clogged drain from my shower. Commercial products like Savo or WC Net totally don't work. Is there any chemical substance which can be bought without permission or which can be made easily to clear the drain? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Czech Republic, it's part of European Union, just for your information about the law system here (can be helpful to decide what's legal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea is sodium hydroxide, is it powerful enough?&lt;/p&gt;&#xA;" OwnerUserId="718" LastEditorUserId="73" LastEditDate="2012-10-07T23:46:39.237" LastActivityDate="2016-06-28T14:18:25.040" Title="Clogged drain, commercial products are ineffective" Tags="&lt;everyday-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1325" PostTypeId="2" ParentId="1324" CreationDate="2012-10-07T23:07:37.017" Score="10" Body="&lt;p&gt;I don't know what the products you mention contain, but you basically have four chemical weapons, and two physical weapons against a clogged drain. Choosing depends on the nature of the clog, which is hard to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry first:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Acidity: sulfuric acid. ’nough said.&lt;/li&gt;&#xA;&lt;li&gt;Basicity (causticity): lye, a.k.a. “caustic soda”, a.k.a. sodium hydroxide.&lt;/li&gt;&#xA;&lt;li&gt;Oxidation: bleach or peroxide.&lt;/li&gt;&#xA;&lt;li&gt;Enzymes: take time to act, but can be very powerful. This you should find at an eco-friendly store or supermarket aisle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, on to physics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hot water: if the pipe is clogged by solidified fats, boiling water can melt them and unclog. Hardly typical for a shower drain, but very often useful for kitchen sink.&lt;/li&gt;&#xA;&lt;li&gt;Last, but not least: mechanical action. Either a drain snake, or a plunger. Most often, the clog is a localized problem, and by breaking it into smaller parts, you will restore the flow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a typical shower drain, hair and other organic matter is most likely the issue, so I would choose drain snake first, then sodium hydroxide, then enzymes.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-07T23:07:37.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1326" PostTypeId="2" ParentId="1322" CreationDate="2012-10-07T23:30:29.210" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;on average they form dipoles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite. On average they form the nice (electrically neutral) &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;electron orbitals&lt;/a&gt; we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why is the potential energy of a dipole lower than a non-dipole&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing. The  &quot;movement&quot; of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic.  And strictly speaking, you can't tell electrons apart anyway. There is no &quot;which electron,&quot; at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't think of those as being electrons, just the statistical average of the negative charge distribution. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive. &lt;/p&gt;&#xA;" OwnerUserId="107" LastEditorUserId="107" LastEditDate="2012-10-07T23:54:58.190" LastActivityDate="2012-10-07T23:54:58.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1327" PostTypeId="2" ParentId="1321" CreationDate="2012-10-08T01:14:26.780" Score="4" Body="&lt;p&gt;Sodium hypochlorite, $\ce{NaClO}$, is unstable and can't be separated from the water in which you usually find it without it decomposing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the short answer.  To learn more about it check Wikipedia for sodium hypochlorite.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="22" LastEditDate="2012-10-08T06:21:42.127" LastActivityDate="2012-10-08T06:21:42.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1329" PostTypeId="1" AcceptedAnswerId="1330" CreationDate="2012-10-08T08:09:10.643" Score="9" ViewCount="2881" Body="&lt;p&gt;I have purchased an &quot;unknown&quot; white coloured powder from the market which can frost glass objects. Here are some properties of this powder&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;White in colour&lt;/li&gt;&#xA;&lt;li&gt;When mixed with HCL or water, it results in the soluion becoming ice cold, the reaction is endothermic.&lt;/li&gt;&#xA;&lt;li&gt;When kept for a while (10 minutes) the white powder settles at the bottom of the plastic vessel and the acid floats above it.&lt;/li&gt;&#xA;&lt;li&gt;When stirred, the entire solution becomes milky&lt;/li&gt;&#xA;&lt;li&gt;When glass is dipped in this solution for 90 seconds, it becomes frosted/translucent.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since I am not sure what this chemical is, I cannot determine the safest method to dispose it. The person from whom I brought this told me to put it down the drain, but I wouldnt choose to do that before I know if it will not effect the water table near my locality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point out what could this powder be, and the best way to dispose it off?&lt;/p&gt;&#xA;" OwnerUserId="721" LastEditorUserId="23561" LastEditDate="2019-04-08T16:54:46.630" LastActivityDate="2019-04-08T16:54:46.630" Title="Identifying a glass frosting chemical" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;identification&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1330" PostTypeId="2" ParentId="1329" CreationDate="2012-10-08T09:22:55.947" Score="6" Body="&lt;p&gt;The most common glass-etching powders are &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_bifluoride&quot; rel=&quot;noreferrer&quot;&gt;solid ammonium bifluoride&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_bifluoride&quot; rel=&quot;noreferrer&quot;&gt;solid potassium bifluoride&lt;/a&gt;. See for example the &lt;a href=&quot;http://www.etchon.co.in/pdf/MSDS%20Acid-Frost.pdf&quot; rel=&quot;noreferrer&quot;&gt;Material Safety Data Sheet&lt;/a&gt; for “EtchON Acid-Frost Glass Frosting Powder”, or &lt;a href=&quot;http://www.hillbrothers.com/msds/pdf/n/ammonium-bifluoride.pdf&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;. This is not a definite identification, but it fits your description, and those two are the most commonly available such chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disposal is a bit tricky. &lt;strong&gt;Definitely don't put it down the drain!&lt;/strong&gt; The first MSDS linked above says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ecological Information:&lt;br&gt;&#xA;  Noxious for the water inhabitants. Avoid disposal into rivers, sewer system and land penetration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second gives a hint at disposing of it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Disposal Instructions:&lt;br&gt;&#xA;  All wastes must be handled in accordance with local, state and federal regulations. Material can be converted to a less hazardous material by weak reducing agents followed by neutralization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, contact your local waste management authorities, and ask them how to proceed. &lt;strong&gt;Do not attempt it neutralize it by yourself unless (i) you know what you are doing, and (ii) you have the appropriate hardware&lt;/strong&gt; (fume hood, e.g.).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-08T09:22:55.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1332" PostTypeId="1" CreationDate="2012-10-08T12:43:14.250" Score="2" ViewCount="47" Body="&lt;p&gt;Dendrites that form on the electrodes of cells shorten the cell's lifetime, and result in reduced capacity before the end of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dendrites could be detached from the electrode, would they then dissolve back into the electrolyte, or would they remain as solids suspended in it? How does this effect battery lifetime and capacity?&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2012-10-08T12:43:14.250" Title="Would the detaching of dendrites from electrodes in a cell, result in the crystal dissolving back into the electrolyte?" Tags="&lt;electrochemistry&gt;&lt;crystal-structure&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1333" PostTypeId="2" ParentId="1322" CreationDate="2012-10-08T12:53:48.480" Score="3" Body="&lt;p&gt;''What determines which electrons are going to move to the middle of the atoms to form a dipole,...?''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons are indistinguishable particles, so the question doesn't make sense. Electrons are excitations of the electron field (a quantum version of like wavelets on a lake), and their formation somewhere is just by concentrating the field at some place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some molecules such as benzene, some electrons don't even clump at all near particular locations but are strongly delocalized.&lt;/p&gt;&#xA;" OwnerUserId="258" LastActivityDate="2012-10-08T12:53:48.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1334" PostTypeId="1" CreationDate="2012-10-08T13:19:19.587" Score="16" ViewCount="536" Body="&lt;p&gt;What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over on the Skeptics StackExchange, a question has come up about &lt;a href=&quot;https://skeptics.stackexchange.com/q/11184/6236&quot;&gt;charging Lithium-ion cells to just 80% of their maximum, in order to extend their lifetime&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a sound basis in chemistry for such a recommendation, and if so, what is it?&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:44.273" LastActivityDate="2013-05-20T23:39:21.363" Title="What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1335" PostTypeId="1" AcceptedAnswerId="1336" CreationDate="2012-10-08T13:31:26.763" Score="2" ViewCount="111150" Body="&lt;p&gt;I know how to calculate the specific heat, but not molar specific heat. What is molar specific heat and how do you calculate it?&lt;/p&gt;&#xA;" OwnerUserId="723" LastEditorUserId="13339" LastEditDate="2015-06-14T14:54:17.777" LastActivityDate="2018-03-16T17:42:09.883" Title="How do you calculate molar specific heat?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1336" PostTypeId="2" ParentId="1335" CreationDate="2012-10-08T14:28:00.080" Score="3" Body="&lt;p&gt;Specific heat has the units of $\mathrm{J/(K\cdot kg)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar specific heat is in units of $\mathrm{J/(K\cdot mol)}$, and is the amount of heat needed (in joules) to raise the temperature of $1$ mole of something, by $1$ kelvin (assuming no phase changes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the conversion factor you need, from dimension analysis, will have unit $\pu{kg/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pu{kg/mol}$ is the SI unit for molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the specific heat by the molar mass to get the molar specific heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example,&lt;/strong&gt; the molar mass of water is $\approx \pu{0.018 kg/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat of water is $\approx 4186\ \mathrm{J/(K\cdot kg)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the molar specific heat of water is $4186\ \mathrm{J/(K\cdot kg)} \times \pu{0.018 kg/mol} \approx 75\ \mathrm{J/(K\cdot mol)}$&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorDisplayName="user7951" LastEditDate="2018-03-16T17:42:09.883" LastActivityDate="2018-03-16T17:42:09.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1337" PostTypeId="1" AcceptedAnswerId="1339" CreationDate="2012-10-08T14:35:18.210" Score="8" ViewCount="12854" Body="&lt;p&gt;While drinking milk (or better 'while seeing the milk I'm gonna drink') a question came up to my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the milk sometimes flake, even if not in contact with some other substances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the correct term to use is &lt;em&gt;flocculation&lt;/em&gt;, at least this is the word that raises up when talking about cheese production. I just learned the word &lt;em&gt;casein&lt;/em&gt;. However, my question is not about cheese production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is also not about flaking milk when poured into a cup of coffee which seems to be dependent on the acidity of the coffee. But &lt;em&gt;casein&lt;/em&gt; is again the keyword.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer to my actual question will also contain the word &lt;em&gt;casein&lt;/em&gt;, but here's the scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I open a milk carton for the first time. This package doesn't have any hole and the shelf life isn't reached yet. I pour the milk into a clean cup and then observing some flakes at top of the milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I was told as a child, when curiously asking, flaking is an indicator that the milk is getting acid. And in that case the keyword &lt;em&gt;casein&lt;/em&gt; would fit again, but there's still one thing why I'm not persuaded: the milk was quite tasty and my stomach also didn't complain about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what exactly has happened to my milk when I see some flakes on top of it? Is the milk indeed spoiled and I actually shouldn't have drunk it (and just was lucky) or is there anything else which can lead to flakes?&lt;/p&gt;&#xA;" OwnerUserId="724" LastEditorUserId="724" LastEditDate="2012-10-09T19:38:01.523" LastActivityDate="2016-06-15T18:14:22.520" Title="Why does milk flake?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;food-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1338" PostTypeId="2" ParentId="1324" CreationDate="2012-10-08T14:51:23.297" Score="7" Body="&lt;p&gt;Not a chemistry answer: but a very successful one, in my experience: &lt;strong&gt;get a Ropump&lt;/strong&gt;. It's a cross between a plunger and a bicycle pump. It makes drain-unclogging a pleasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical action is just so much more successful for this task, than chemical.  Because water is extremely resistant to being compressed, the Ropump's action, in turning water into a liquid drain rod and forcing it down the drain, has a pretty high success rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pull plunger out, while the far end sits in a reservoir of water. This fills the body of the pump with water. You then fit the far end over the blocked drain, and push the plunger in, forcing the water down the drain, and displacing the blockage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you don't know what the chemistry is at the blockage. So it's easier to anticipate the unintended consequences of a mechanical, rather than chemical, intervention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QZpzd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2012-10-08T14:51:23.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1339" PostTypeId="2" ParentId="1337" CreationDate="2012-10-08T14:57:35.500" Score="8" Body="&lt;p&gt;I think the right word for it is &lt;strong&gt;curdle&lt;/strong&gt;. Have you read, for example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Curd&quot;&gt;Wikipedia article on &lt;em&gt;curd&lt;/em&gt;&lt;/a&gt;? It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;“Milk that has been left to sour (raw milk alone or pasteurized milk with added lactic acid bacteria or yeast) will also naturally produce curds, and sour milk cheese is produced this way”&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or &lt;a href=&quot;http://en.wikipedia.org/wiki/Milk#Spoilage_and_fermented_milk_products&quot;&gt;this excerpt of the “Milk” page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When raw milk is left standing for a while, it turns &quot;sour&quot;. This is the result of fermentation, where lactic acid bacteria ferment the lactose in the milk into lactic acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;milk naturally contains bacteria which turn it acid&lt;/strong&gt;, and hence it curdles spontaneously. You see the beginning of this process.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An invaluable source of information on this topic is McGee’s &lt;a href=&quot;http://www.amazon.co.uk/McGee-Food-Cooking-Encyclopedia-Kitchen/dp/0340831499&quot;&gt;&lt;em&gt;On food and cooking&lt;/em&gt;&lt;/a&gt;. Great reading if you like chemistry and food.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-08T14:57:35.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1340" PostTypeId="1" AcceptedAnswerId="1349" CreationDate="2012-10-08T17:03:08.450" Score="7" ViewCount="326" Body="&lt;p&gt;Could anyone tell me how I would go about calculating the pH of 100.0 mL of 0.100 M $\ce{H2NNH2}$ ($K_\text{b} = 3.0 \times 10^{-6}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, how would this pH change when 20.0 mL of 0.200 M $\ce{HNO3}$ is added to the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not looking for only an answer, but rather the general method to solve these types of problems&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just to include what I’ve tried:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first thought that (for the initial solution) $\mathrm{pH} = 14 - \log(\sqrt{K_\text{b} [\ce{H2NNH2}]})$, but that doesn’t seem to be the case.&lt;/p&gt;&#xA;" OwnerUserId="725" LastEditorUserId="402" LastEditDate="2015-11-06T21:15:37.893" LastActivityDate="2015-11-06T21:15:37.893" Title="How can I determine the pH of solution?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1341" PostTypeId="1" AcceptedAnswerId="1343" CreationDate="2012-10-09T07:14:37.523" Score="4" ViewCount="1995" Body="&lt;p&gt;We are asked to sort these four acids by increasing Brønsted acidity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OyNHz.png&quot; alt=&quot;sort these four acids by increasing Brønsted acidity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, I am confused why $\ce{H2F+}$ is the strongest acid of them all. I understand that stronger acids will have weaker bonds, and the $\ce{F-H}$ bonds should be the most electronegative of them all, leading me to believe that they are the strongest bonds. So then why are these bonds the most easily broken? What have I forgotten here? Is the correct order really from weakest to strongest acid as follows? $\ce{CH3OH}$, $\ce{(CH3)2OH+}$, $\ce{CH3SH2+}$, $\ce{H2F+}$&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="194" LastEditDate="2018-02-19T04:47:18.757" LastActivityDate="2018-02-19T04:47:18.757" Title="Is this the correct relative Brønsted acidity in these four acids?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1342" PostTypeId="1" CreationDate="2012-10-09T07:45:31.100" Score="6" ViewCount="3678" Body="&lt;p&gt;&lt;a href=&quot;http://www.newscientist.com/article/mg20928014.800-power-of-cool-liquid-air-to-store-clean-energy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid nitrogen&lt;/a&gt; seems to be attracting a bit of attention at the moment as a medium of energy storage, both for electricity grid applications and for transport.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://web.archive.org/web/20121010203742/http://www.highview-power.com/wordpress/?page_id=1320f&quot; rel=&quot;nofollow noreferrer&quot;&gt;Highview (via the Internet Archive)&lt;/a&gt; are doing round-trip electricity storage via liquid nitrogen.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p8WId.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://web.archive.org/web/20120427110723/http://www.dearmanengine.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dearman Engine Company (via the Internet Archive)&lt;/a&gt; are developing a &amp;quot;liquid-air&amp;quot; vehicle engine.&lt;/p&gt;&#xA;&lt;p&gt;And the UK’s Institution of Mechanical Engineers has formed a &lt;a href=&quot;https://web.archive.org/web/20121014225430/http://www.playitback.org/details.aspx?v=377&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid Air working group (via the Internet Archive)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What is the energy density of liquid nitrogen, per unit volume and per unit weight, assuming we’ll be bringing it up to 20 °C, 1 bar? How does this compare to the energy densities of isooctane and of Li-ion batteries?&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="4945" LastEditorDisplayName="user7951" LastEditDate="2021-02-10T22:15:41.603" LastActivityDate="2021-02-10T22:15:41.603" Title="What's the energy density of liquid nitrogen?" Tags="&lt;energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="1343" PostTypeId="2" ParentId="1341" CreationDate="2012-10-09T08:09:33.763" Score="6" Body="&lt;p&gt;Fluorine being the most electronegative element in the periodic table, it doesn’t like bearing a positive charge. Thus $\ce{H2F+}$, which has the positive charge on the fluorine nucleus, is not very stable and willing to expel the extra proton. So, it is indeed a strong acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2F+}$ is formed, e.g., by autoionization of liquid $\ce{HF}$, but only because that reaction also involves creating a $\ce{F-}$ ion (associated with another molecule of $\ce{HF}$ to form $\ce{HF2-}$), which is &lt;em&gt;very&lt;/em&gt; favorable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3 HF &amp;lt;=&amp;gt; H2F+ + HF2-}$$&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2016-10-09T14:06:17.990" LastActivityDate="2016-10-09T14:06:17.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1344" PostTypeId="1" CreationDate="2012-10-09T16:26:31.480" Score="4" ViewCount="2670" Body="&lt;p&gt;Can we make a solution containing only positive ions by using electrodialysis but instead of using 2 kinds of membrane (one blocks the positive ions and one blocks the negative ions) we can use only one kind (the one which blocks the positive ions) so the positive ions wont reach the electrode and wont be electrolysed while the negative ions will be electrolysed leaving the positive ions in the solution &lt;/p&gt;&#xA;" OwnerUserId="815" OwnerDisplayName="Abdelrahman Esmat" LastActivityDate="2012-10-10T12:47:14.683" Title="Can we separate positive ions from negative ions in a solution?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1345" PostTypeId="2" ParentId="1344" CreationDate="2012-10-09T18:15:22.517" Score="3" Body="&lt;p&gt;While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: &lt;strong&gt;you cannot separate positive and negative charges in a solution&lt;/strong&gt; at a macroscopic level. See for example this excerpt from &lt;a href=&quot;http://books.google.com/books?id=IgBv7Gdyvr8C&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Ionic Transport Processes:in Electrochemistry and Membrane Science&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;       &lt;img src=&quot;https://i.stack.imgur.com/wRoOG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-09T18:15:22.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1346" PostTypeId="1" CreationDate="2012-10-09T19:39:57.887" Score="7" ViewCount="1256" Body="&lt;p&gt;If I buy a loaf of freshly-made bread and leave it out at room temperature for a day or two, when I reach into the bag the air inside is much warmer than the rest of the room. Why?&lt;/p&gt;&#xA;" OwnerUserId="730" LastEditorDisplayName="user7951" LastEditDate="2016-12-30T01:20:54.667" LastActivityDate="2016-12-30T01:20:54.667" Title="Why is bread warm at room temperature?" Tags="&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1347" PostTypeId="1" CreationDate="2012-10-10T04:32:26.740" Score="3" ViewCount="102" Body="&lt;p&gt;For example, consider a pot of water with a drop of oil floating on it, with surfactants present in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the surfactants in the water spread evenly across all three interfaces (water/metal, water/air, and water/oil)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there material-specific surfactants, such that each of the three interfaces might attract different surfactants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, are there also surfactants that would tend to spread across all three interfaces evenly?&lt;/p&gt;&#xA;" OwnerUserId="731" LastActivityDate="2012-10-18T12:26:38.367" Title="How do surfactants spread in the case of multiple materials?" Tags="&lt;surface-chemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1348" PostTypeId="2" ParentId="1238" CreationDate="2012-10-10T10:43:52.420" Score="5" Body="&lt;p&gt;One way to do it is, as user41631 says, with an endothermic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is to use phase-change materials, from which you remove the thermal insulation, to allow the phase-change to take place. Now, all materials change their phases, at some combinations of temperature and pressure. The phrase &quot;phase-change material&quot; is used not to classify materials, but to identify the paricular use of a material - that is, the material is being used because one of its phase changes happens at a temperature and pressure that makes it useful for a specific process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such materials are being looked at for the temperature regulation of buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a material with a melting point of, say, 18&amp;deg;C, and a high latent heat of fusion. Freeze it, insulate it, and take it into a room that's at 22&amp;deg;C. While the material melts, it will absorb heat from the air, cooling the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle has been applied in building material design, to provide smoothing of temperature variations. See, for example, Schossig et al's &quot;Micro-encapsulated phase-change materials integrated into construction materials&quot;, &lt;a href=&quot;http://dx.doi.org/10.1016/j.solmat.2005.01.017&quot; rel=&quot;nofollow noreferrer&quot;&gt;[DOI]&lt;/a&gt;, from which the following graph is taken, showing the smoothing of temperatures over a week in August , when a phase-change material (PCM) is used (&lt;code&gt;T_Wall_PCM&lt;/code&gt;) - this is compared to an identical room without phase-change materials (&lt;code&gt;T_Wall_REF&lt;/code&gt;) - as you can see, the PCM reduces the amount of time that the room spends above 26&amp;deg;C, providing more thermal comfort for people in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9aPnS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="73" LastEditDate="2012-10-10T11:18:11.563" LastActivityDate="2012-10-10T11:18:11.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1349" PostTypeId="2" ParentId="1340" CreationDate="2012-10-10T12:27:15.540" Score="6" Body="&lt;p&gt;Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the important thing: The mass action law uses the concentrations of our components in the &lt;strong&gt;equilibrium&lt;/strong&gt;! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.&lt;/p&gt;&#xA;" OwnerUserId="732" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T17:35:08.253" LastActivityDate="2015-05-11T17:35:08.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1350" PostTypeId="2" ParentId="1344" CreationDate="2012-10-10T12:47:14.683" Score="1" Body="&lt;p&gt;For water and on macroscopic level we can't, as there is non-zero water dissociation constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropping solvent dissociation from consideration, we should consider molar charge. Faraday's number is molar charge and it is ~10^5 Coulomb/mol. So, if you can store really much charge (at least some Coulombs) in your solution, you can. But can anybody? It is really hard.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-10-10T12:47:14.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1352" PostTypeId="2" ParentId="1320" CreationDate="2012-10-10T18:59:32.457" Score="5" Body="&lt;p&gt;The difficulty in talking about hybridization lies in the fact that it tends to confuse atomic and molecular orbitals, or (equivalently?) one- and many-electron wave functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing atomic orbitals generally only becomes possible when we do a mathematical approach which treats more than one electron at once. That, however, is mathematically rather complex, and the resulting wave functions are molecular orbitals, which are entirely different from atomic orbitals. Molecular orbitals are constructed as combinations of atomic orbitals, which means that it is possible to e.g. add &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt; wave functions together to obtain something that looks like &lt;em&gt;sp&lt;/em&gt; hybrids. But, and this is important, there are actually many different possibly ways for such combinations, and none of them should really be considered physical reality; and furthermore, this &quot;mixing together&quot; has to follow very strict mathematical rules which in the end affects the entire set of molecular orbitals at once. Describing molecular orbitals in terms of hybrid orbitals (or something similar) is possible if one does it thoroughly, in terms of a full and correct mathematical treatment. Still, even when applying the concept in this way, it is somewhat arbitrary, since it represents only one of many possible solutions (and depending on the problem, not even the best one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simple and qualitative terms often introduced in chemistry (especially organic chemistry), it is a very crude concept which should be regarded with great caution and especially not as something that would be some sort of reality. Hybridization is a mathematical procedure for many-electron wave functions; nothing more, nothing less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more technical terms: The role of one- and many-electron wave functions in atoms is crucial here. Atomic orbitals are assumed to be hydrogen-like wave functions; in the one-electron atom, the orbitals are $\ell$-degenerate, so linear combinations of wave functions for different orbitals would mathematically still be solutions to the Schrödinger equation. In many-electron atoms, the degeneracy breaks down, so combining different orbital wave functions would be invalid, since they belong to different eigenvalues. We do, however, employ a many-electron Hamiltonian such as the Fock operator in this case, and unless I am mistaken, the electron-electron interactions lift this restriction. So, by some unitary transformation of the many-electron wave function of a single atom, I guess one might arrive at something like hybrid orbitals. Then again, this does not reflect physical reality.&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-10T18:59:32.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1353" PostTypeId="2" ParentId="1302" CreationDate="2012-10-10T19:53:37.660" Score="4" Body="&lt;p&gt;Only in simple cases, it is possible to make reliable estimates of the relative weight of resonance structures from the Lewis structure alone. If you want a somewhat quantitative yet simple approach to bond orders in delocalized systems, have a look at the &lt;a href=&quot;http://wiki.chemeddl.org/mediawiki/index.php/H%C3%BCckel_MO_Theory&quot; rel=&quot;nofollow&quot;&gt;Hückel method&lt;/a&gt;. It is a quantum-mechanical theory for determining the energies of $\pi$ orbitals, and it defines a method to determine a measure of the $\pi$ bond order between two atoms. If you want to go beyond that, you will have to look further into quantum chemistry and different schemes for bond localization and population analysis. In many cases there is no one definite answer, but rather, your results will depend heavily on the quantum-chemical calculations and assumptions for the bond-order analysis itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are probably also ways to tackle the problem from an experimental side, although I'm not all sure about this; I'm thinking about perhaps determining electron distributions through NMR chemical shifts, or bond strengths through vibrational spectroscopy.)&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-10T19:53:37.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1354" PostTypeId="1" AcceptedAnswerId="1357" CreationDate="2012-10-11T00:26:30.373" Score="12" ViewCount="15313" Body="&lt;p&gt;I was wondering, how can I clean my RBF after doing a distillation? Cleaning it with a simple pipe-cleaner or similar cleaner is a headache, due to the &lt;em&gt;roundness&lt;/em&gt; of the flask. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more effective method?&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="73" LastEditDate="2012-10-11T00:34:38.023" LastActivityDate="2018-04-09T08:57:37.190" Title="Cleaning a round-bottom flask, especially after distillation" Tags="&lt;equipment&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1355" PostTypeId="2" ParentId="1354" CreationDate="2012-10-11T03:53:14.797" Score="1" Body="&lt;p&gt;Make up a basebath out of conc KOH in isopropyl alcohol (Or whatever recipe you prefer). Leave it in there for an hour or two, then rinse with distilled water. Most of the time it will come out sparkling clean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful: Saturated base baths should be handled with care, rubber gloves (dish gloves) over lab gloves are a very, very good idea.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2012-10-11T03:53:14.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1356" PostTypeId="2" ParentId="1354" CreationDate="2012-10-11T04:37:25.507" Score="1" Body="&lt;p&gt;The simplest way is using some cleaning solution. The nature of solution depends on nature of remnants. In my practice, I used&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H_2SO_4(conc) + K_2Cr_2O_7$ &lt;/li&gt;&#xA;&lt;li&gt;$H_2SO_4(conc) + H_2O_2$ (porous glass filters mostly)&lt;/li&gt;&#xA;&lt;li&gt;$KOH + iPrOH$&lt;/li&gt;&#xA;&lt;li&gt;$KOH + KMnO_4 + H_2O$&lt;/li&gt;&#xA;&lt;li&gt;$H_2C_2O_4 + H_2O$ (cleaning $MnO_2$ and $FeOOH$)&lt;/li&gt;&#xA;&lt;li&gt;$HCl$(2%).&lt;/li&gt;&#xA;&lt;li&gt;$HCl$ (conc.)&lt;/li&gt;&#xA;&lt;li&gt;Organic solvents (toluene, alcohol, hexane, dichloromethane)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, the nature of cleaning solution depends on what you are cleaning after. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="73" LastEditDate="2012-10-11T08:46:35.707" LastActivityDate="2012-10-11T08:46:35.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1357" PostTypeId="2" ParentId="1354" CreationDate="2012-10-11T10:53:40.323" Score="17" Body="&lt;p&gt;@Permeakra suggests some good cleaning solutions. I have rearranged them (adding a few) in a suggested order from mildest to most evil. The specific thing you want to use will vary based on what you want to clean. If an initial rinse isn't good enough, you may want to let it soak for a few hours. Always take safety precautions (don't add acid washes to basic compounds, for example) and where the appropriate personal protection equipment (PPE).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Detergent + water + &quot;elbow grease&quot; - there is a lab grade soap out there called &quot;microwash&quot; that is pretty good, and I also tell my students that things are more soluble in elbow grease than water.&lt;/li&gt;&#xA;&lt;li&gt;Acetone - Acetone will dissolve most organics, and it is less harmful to you and the environment than other organics.&lt;/li&gt;&#xA;&lt;li&gt;Dilute acid (HCl or H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt;) - good for some &quot;baked&quot;-on stuff or for inorganic compounds. Watch out for vigorous acid-base reactions.&lt;/li&gt;&#xA;&lt;li&gt;Dilute nitric acid (HNO&lt;sub&gt;3&lt;/sub&gt;) - since most nitrate salts are soluble, nitric acid does a better job with resistant inorganics. It is also an oxidant and will convert resistance organics to more soluble compounds. Also watch for acid-base reactions.&lt;/li&gt;&#xA;&lt;li&gt;Other organic solvents - Use whatever will dissolve your crud, but use proper personal protective equipment (PPE), fume hoods, and waste disposal. &lt;/li&gt;&#xA;&lt;li&gt;&quot;Base bath&quot; (saturated KOH in isopropanol) - This solution eats a fine layer off the glass, which also removes the crud stuck to it. Rinse first with water or neutralize any acid present to avoid vigorous reaction with the bath. Immerse the flask into the bath for a few hours to overnight and it should come out clean. Base bath will dissolve silicone grease. Leave it in too long and it will come out etched. Use heavy butyl gloves.&lt;/li&gt;&#xA;&lt;li&gt;Concentrated HCl - more powerful than dilute acids, but more dangerous. Neutralize any bases present to avoid vigorous reaction. Use proper PPE.&lt;/li&gt;&#xA;&lt;li&gt;KOH + KMnO&lt;sub&gt;4&lt;/sub&gt; + H&lt;sub&gt;2&lt;/sub&gt;O - this solution is basic (eats a fine layer of glass) and oxidative (converts many things, like organics, to a more soluble form). Neutralize any acids present and wear appropriate PPE. &lt;/li&gt;&#xA;&lt;li&gt;Concentrated H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; - has the added bonus of reacting with some insoluble organics, like polyaromatics, to form soluble sulfonates. H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; is more dangerous than HCl. Follow the same precautions, and be more careful. I would use use butyl gloves instead of nitrile or latex. It will cause serious chemical burns.&lt;/li&gt;&#xA;&lt;li&gt;&quot;No-Chromix&quot; - a non-chromium alternative to chromerge (below), no-chromix is a proprietary cleaner that is added to sufluric acid. The main ingredient is ammonium persulfate (NH&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;S&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;8&lt;/sub&gt;. You can make a reasonable knock-off by dissolving 2 g of (NH&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;S&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;8&lt;/sub&gt; in 100 mL of water and adding to 2.5 L of sulfuric acid. Feel free to scale down, as this stuff has no shelf-life to speak of.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The following five items have dangerous reputations, and for good reasons. Use them as last resorts. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Concentrated HNO&lt;sub&gt;3&lt;/sub&gt; - strongly acidic and powerfully oxidizing. Dissolves many resistant things, including burned-on crud, but may ignite low molecular weight organics (like solvents). Rinse the flask first with water to remove bases, then with acetone to remove organics, and then with water to remove acetone before adding HNO&lt;sub&gt;3&lt;/sub&gt;. Use butyl gloves. &lt;/li&gt;&#xA;&lt;li&gt;Dilute HF (no more than 5%) - HF eats glass also, and it does so more quickly than base baths. However, the the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrofluoric_acid#Health_and_safety&quot;&gt;hazards&lt;/a&gt; associated with HF are severe. HF burns are not like other acid burns. The HF continues to bore into you and the damage might not be evident until later. &lt;strong&gt;DO NOT USE UNLESS YOU HAVE THE BURN ANTIDOTE HANDY.&lt;/strong&gt; The burn antidote&#xA;is &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_gluconate#Treatment_of_HF_burns&quot;&gt;calcium gluconate&lt;/a&gt;, which neutralizes the acid and more important sequesters the fluoride ions. It is available as a gel or cream. You should still go overboard on PPE.&lt;/li&gt;&#xA;&lt;li&gt;Aqua Regia: 1 part Concentrated HNO&lt;sub&gt;3&lt;/sub&gt; to 3 parts concentrated HCl. This is more powerful than just concentrated HNO&lt;sub&gt;3&lt;/sub&gt;, since HCl is a stronger acid. It also decomposes to form NO&lt;sub&gt;2&lt;/sub&gt; and Cl&lt;sub&gt;2&lt;/sub&gt;, both more powerful oxidants than Concentrated HNO&lt;sub&gt;3&lt;/sub&gt; alone. Aqua regia is famous as the only substance that can dissolve gold. It will also dissolve all manner of inorganics and organics, including silicone and hydrocarbon grease. Treat aqua regia with the respect deserved of its two components and keep it in the fume hood. Do not seal the vessels since it gives off gas. Aqua regia has no shelf life, but it remains a strong acid and will need neutralized before disposal. Note that aqua regia will not dissolve metallic rhodium, iridium, osmium, and ruthenium (all less active than gold). &lt;/li&gt;&#xA;&lt;li&gt;Piranha solution: H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; + H&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;2&lt;/sub&gt; - this strongly acidic strongly oxidizing solution will eat off a lot of metal, organic, and inorganic residues by oxidizing then to more soluble forms. As a strong acid oxidant, it shares the same incompatibility warnings that concentrated nitric acid has. Piranha solution is good for glassware that needs to remain calibrated (volumetrics) or uniform (NMR tubes) since it does not eat glass. Piranha solution does not have a long shelf life.&lt;/li&gt;&#xA;&lt;li&gt;Chromerge H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; + K&lt;sub&gt;2&lt;/sub&gt;Cr&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;7&lt;/sub&gt; - forms chromic acid, a powerful oxidant. This stuff is considered toxic waste in most jurisdictions, meaning it will be expensive to get rid of. You can store it, however, if you store it in teflon. Do not use on NMR tubes, since chromium is paramagnetic. Treat chromerge with a respect due to the sum of everything above it on this list.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7475" LastEditDate="2015-12-12T23:39:34.693" LastActivityDate="2015-12-12T23:39:34.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1360" PostTypeId="2" ParentId="1293" CreationDate="2012-10-11T20:55:32.693" Score="2" Body="&lt;p&gt;The actual act of crushing the avocado (applying simple pressure) is a physical change.  While the act of applying pressure may allow oxidation to happen faster, the application of pressure is still physical.  The type of pressure you would apply to an avocado under normal circumstances would be insufficient to result in chemical changes.  Chemical changes result in the breaking or formation of molecular bonds which are simply not possible by normal kitchen appliances.  Pressure and heat are related which could cause the enzymes and proteins within the avocado to denature, but the act of applying the pressure is not directly responsible for any chemical change in this case, and the application would only be a physical change.  If there were chemical changes, these changes would be caused by factors other than the pressure itself.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2012-10-11T20:55:32.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1361" PostTypeId="2" ParentId="1342" CreationDate="2012-10-11T21:13:00.953" Score="10" Body="&lt;p&gt;Just skimmed over the links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Highview&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using ambient heat to warm it, the process &lt;strong&gt;recovers around 50 per cent of the electricity that is fed in&lt;/strong&gt;, says Highview's chief executive Gareth Brett. The efficiency &lt;strong&gt;rises to around 70 per cent&lt;/strong&gt; if you harness waste heat from a nearby industrial or power plant to heat the cryogen to a higher than ambient temperature, which increases the turbine's force, he says.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Batteries under development .. have efficiencies of around 80 to 90 per cent&lt;/strong&gt;, but cost around \$4000 per kilowatt of generating capacity. Cryogenic storage would cost just \$1000 per kilowatt because it requires fewer expensive materials, claims Brett.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's some evidence that people who know about gas and liquid air are interested in the technology: &lt;a href=&quot;http://www.messergroup.com/de/Presse/wpresse/050712_cooperation-highview/index.html&quot;&gt;strategic partnership between Messer and Highview&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually Highview has a presentation on their web-site that contains a lot of information. Note that they propose to run a 4-stage process against heat (ca. 100 °C), not ambient temp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some values for energy storage efficiency comparison found after a very quick search:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rechargeable_battery#Table_of_rechargeable_battery_types&quot;&gt;Batteries (Wiki)&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://de.wikipedia.org/wiki/Akkumulator#Energiedichte_und_Wirkungsgrad&quot;&gt;German Wikipedia has lower efficiency values&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_conversion_efficiency#Example_of_energy_conversion_efficiency&quot;&gt;Energy conversion table (Wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pump-stored hydroelectricity: 70/75 - 80% (&lt;a href=&quot;http://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity&quot;&gt;Wiki&lt;/a&gt; + &lt;a href=&quot;http://de.wikipedia.org/wiki/Pumpspeicherkraftwerk#Wirkungsgrad&quot;&gt;de.Wiki&lt;/a&gt;)&lt;br&gt;&#xA;(Note the difference between technology that is actually in use, so efficiencies that are achieved in practical use are known and prototypes/scientific studies about achievable efficiency.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dearman&lt;/strong&gt; claims 70% efficiency for the engine. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Unfortunately they don't say % of what (remember the low-temp heater manufacturers claiming &gt; 100% efficiency because they used achievable efficiency with high temp exhaust gas as their denominator?).  &lt;/li&gt;&#xA;&lt;li&gt;Neither do they say something about the power at which they have this efficiency (ideal thermodynamic efficiency is at P $\rightarrow$ 0 ($t_{cycle} \rightarrow \infty$))&lt;/li&gt;&#xA;&lt;li&gt;Neither can we assume that they start their calculation taking into account the efficiency of producing liquid nitrogen. &lt;/li&gt;&#xA;&lt;li&gt;Nor storage loss.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;=&gt; so far, Highview's stationary plant concept looks more sound.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's have a look what thermodynamics says:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;ideal efficiency&lt;br&gt;&#xA;boiling T of liquid air: 77 K&lt;br&gt;&#xA;ambient T: assume 290 K =&gt; $\eta_{ideal} = 1 - \frac{77}{290}$ = 73% &#xA;waste heat: assume 370 K =&gt; $\eta_{ideal} = 1 - \frac{77}{370}$ = 79% &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(in other words, Dearman claims close-to-ideal efficiency)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;energy density of liquid $\ce{N2}$: 620 kJ/l = 712 kJ/kg (from presentation at Highview)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;The Linde process is said to need approx. 500 kWh / t = 1750 kJ/kg of air&lt;br&gt;&#xA;(that would be ≈40% efficiency; no real source, just something someone answered in a &lt;a href=&quot;http://www.wer-weiss-was.de/theme241/article4856757.html&quot;&gt;German-language chemistry forum&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ffe.de/die-themen/speicher-und-netze/94-ln-kuehlung-in-aegypten&quot;&gt;Here's short report&lt;/a&gt; that gives ≈20% efficiency for producing liquid air in Egypt. They also give storage loss (ca. 0.17 \%/d)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, the 50% (achieved by Highview) to 70/80% (short term goal Highview against waste heat, quote for Dearman's engine) mean that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;either the liquefaction was not part of the efficiency calculation,&lt;/li&gt;&#xA;&lt;li&gt;or they have a super efficient air liquefaction that is about 2 - 4 times as efficient as the two numbers I found.&lt;br&gt;&#xA;(Anyone around here who knows? Linde, Messer, AL?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Update: just had a closer look at the Highview presentation. Their 50 - 80% are cycle efficiency based on &quot;Available enthalpy for cold cycle ~300kJ/kg&quot;, so less than half of the energy density they give. Calculating with 1750 kJ/kg to produce the kg liquid air, we're at 17% (close to the &quot;Egyptian&quot; 20% efficiency for producing liquid air), of which they recover 0.5 - 0.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion: electric -&gt; liquid air -&gt; electric cycle with an overall efficiency in the order of magnitude of 10%&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-10-11T21:13:00.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1362" PostTypeId="5" CreationDate="2012-10-12T01:38:41.690" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-10-12T01:38:41.690" LastActivityDate="2012-10-12T01:38:41.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1363" PostTypeId="4" CreationDate="2012-10-12T01:38:41.690" Score="0" Body="Halocarbons have one or more carbon atoms that are linked with Halogen atoms like Fluorine, Chlorine, Bromine or Iodine." OwnerUserId="670" LastEditorUserId="670" LastEditDate="2012-10-12T03:58:00.957" LastActivityDate="2012-10-12T03:58:00.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1364" PostTypeId="1" AcceptedAnswerId="1366" CreationDate="2012-10-12T01:50:14.390" Score="3" ViewCount="23310" Body="&lt;p&gt;While writing up a lab for biology, I realized that the dialysis tubing for sale &lt;a href=&quot;http://www.sigmaaldrich.com/labware/labware-products.html?TablePage=21942948&quot; rel=&quot;nofollow&quot;&gt;online&lt;/a&gt; seemed to retain large molecules (12-14,000 Da) but didn't specify an upper bound. I learned that the tubing is impermeable to &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=5988&quot; rel=&quot;nofollow&quot;&gt;Sucrose&lt;/a&gt;, molecular weight 342 Da, because it's too big. This doesn't make sense to me; the site says the tubing &quot;retains &gt;12,000 Da molecules&quot;, but Sucrose is clearly smaller than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either A) Sucrose does indeed pass through dialysis tubing and I misheard/mis-learned or B) my class used tubing with much smaller pores.&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="30" LastEditDate="2012-10-12T07:55:35.653" LastActivityDate="2015-10-28T23:04:48.150" Title="Why is dialysis tubing impermeable to sucrose when it should retain &gt;12,000 Da?" Tags="&lt;osmosis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1366" PostTypeId="2" ParentId="1364" CreationDate="2012-10-12T07:55:27.273" Score="2" Body="&lt;p&gt;There exist many types of semi-permeable membranes (the ones used for dialysis tubing), with various pore sizes. One of the very common lab experiment on the topic of osmosis is using a sucrose solution (sucrose is cheap) and small-pores membrane, such that water and small ions (typically Na&lt;sup&gt;+&lt;/sup&gt; and Cl&lt;sup&gt;–&lt;/sup&gt;) can pass, but not sucrose. The one which you link to simply happens to be a variety with much wider pores.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-12T07:55:27.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1367" PostTypeId="2" ParentId="1364" CreationDate="2012-10-12T10:37:20.823" Score="1" Body="&lt;p&gt;Dialysis tubing (and most semi-permeable membranes) operate based on differences of &lt;strong&gt;size&lt;/strong&gt; and not molecular weight. While the size of a molecule or ion does increase as its molecular weight increases, there is some nonintuitive size-mass discrepancies that can occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these discrepancies are well known to polymer chemists. Most of the ways that we attempt to measure the &lt;em&gt;average molecular weight&lt;/em&gt; of a polymer sample are actually measures of &lt;em&gt;average molecular size&lt;/em&gt;. The shape that a molecule adopts in solution greatly impacts its apparent size. A polymer of MW in the 40 kDa range that adopts a rigid-rod conformation in solution will usually have a larger apparent size than a polymer of MW in the 100 kDa range that folds up into a tight globular conformation (spheres have a smaller surface area to volume ratio than cylinders).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor in the &quot;size&quot; of a molecule is the number of solvent molecules that get dragged along with it. If the molecule is dissolved in a good solvent (for example sucrose in water), it will always have a number os solvent molecules associated with it. The solvodynamic radius of a molecule may be much greater that the molecule's actual size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If indeed you had dialysis tubing rated to retain only greater than 12,000 Da (and like @F'x, I'm not sure that you did) it is still possible for sucrose to not pass through if the sucrose is trying to drag too many water molecules (from hydrogen bonding) with it. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-12T10:37:20.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1368" PostTypeId="2" ParentId="788" CreationDate="2012-10-12T20:41:22.333" Score="6" Body="&lt;p&gt;To add to the answer from Nick about the dangers and from Georg about the use of liquid nitrogen for flair, a woman recently had her stomach removed to save her life after she consumed a liquid nitrogen alcoholic beverage.  You can read about it at on the &lt;a href=&quot;http://ca.shine.yahoo.com/blogs/shine-on/teen-girl-loses-stomach-drinking-liquid-nitrogen-cocktail-180639813.html&quot;&gt;Yahoo! News&lt;/a&gt; web site.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must be careful with these things and even dry ice can be dangerous enough to cause burns.&lt;/p&gt;&#xA;" OwnerUserId="618" LastEditorUserId="73" LastEditDate="2012-10-13T11:34:04.830" LastActivityDate="2012-10-13T11:34:04.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1369" PostTypeId="1" CreationDate="2012-10-13T00:49:24.923" Score="2" ViewCount="116" Body="&lt;p&gt;Here's a good link as an introduction to DRIE:&lt;br&gt;&#xA;&lt;a href=&quot;http://www-bsac.eecs.berkeley.edu/scripts/show_pdf_publication.php?pdfID=1102529945.&quot; rel=&quot;nofollow&quot;&gt;BSAC: DRIE Process Development&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{C4F8}$ (refer Pg. 9 on the link) passivates the entire top layer. But when $\ce{SF6}$ ion bombardment occurs, only the region not covered by the mask is etched. Why does the passivation layer protect just the mask and not the substrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional reference: &lt;a href=&quot;http://www-bsac.eecs.berkeley.edu/projects/ee245/Lectures/lecturepdfs/Lecture3.SurfaceMicromachiningI.pdf&quot; rel=&quot;nofollow&quot;&gt;Surface Micromachining Techniques and Materials.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="736" LastEditorUserId="4231" LastEditDate="2015-02-23T19:52:57.383" LastActivityDate="2015-02-23T19:52:57.383" Title="Why does the passivation layer in DRIE protect just the mask and not the substrate?" Tags="&lt;ions&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1370" PostTypeId="1" AcceptedAnswerId="1377" CreationDate="2012-10-13T04:45:56.680" Score="0" ViewCount="102" Body="&lt;p&gt;I'm gonna get kicked all the way to the Moon, but whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing some nitroglycerin. I know the process, since the reaction itself is pretty basic. I know that it is an electrophilic substitution and that it releases a hell lot of energy. I've handled that with an ice bath and by doing the respective stoichiometry. I add around 5 ml of glycerol every 10 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of that works great, until I have to do the wash. I've learned how to do washing and extractions from MIT's OCW. Nitroglycerin is insoluble in water, so I require water to wash the mixture and pull out the acid. &amp;lt;- There's the problem. When I'm doing the wash, the temperature rises and a runway reaction begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice, besides the usual: &quot;Just… &lt;em&gt;don't&lt;/em&gt;&quot;?&lt;/p&gt;&#xA;" OwnerUserId="715" LastActivityDate="2012-10-13T18:22:53.820" Title="Acid-base reaction is screwing up my synthesis" Tags="&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1371" PostTypeId="1" AcceptedAnswerId="1372" CreationDate="2012-10-13T06:24:54.530" Score="2" ViewCount="10028" Body="&lt;p&gt;Since there is no liquid state of carbon dioxide, what will carbon dioxide turn into when exposed to high pressure?&lt;/p&gt;&#xA;" OwnerUserId="737" LastActivityDate="2014-03-10T17:22:17.440" Title="If gases turn into liquids under pressure, what does carbon dioxide turn into?" Tags="&lt;gas-laws&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1372" PostTypeId="2" ParentId="1371" CreationDate="2012-10-13T07:34:56.250" Score="9" Body="&lt;p&gt;There is a liquid state for carbon dioxide. Borrowing the $\ce{CO2}$ &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=File:Carbon_dioxide_pressure-temperature_phase_diagram.svg&amp;amp;page=1&quot; rel=&quot;noreferrer&quot;&gt;phase diagram&lt;/a&gt; from Wikipedia, we can see that $\ce{CO2}$ will condense at a few atmospheres, dependent on temperature. At still higher pressures, the liquid will solidify. Below the triple point temperature, the gas will transition directly to solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SvG64.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-10-13T07:34:56.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1373" PostTypeId="1" AcceptedAnswerId="1374" CreationDate="2012-10-13T08:49:46.253" Score="5" ViewCount="993" Body="&lt;p&gt;Why do molecules like methoxyethane, ethoxyethane and so on have lower boiling points than isomeric alcohols? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has something to do with the structure, but I don't exactly know what it is.&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="7448" LastEditDate="2015-10-02T16:18:00.863" LastActivityDate="2015-10-02T16:18:00.863" Title="The effect of structure on the boiling points of organic molecules" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;isomers&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1374" PostTypeId="2" ParentId="1373" CreationDate="2012-10-13T12:22:44.740" Score="6" Body="&lt;p&gt;Yes, structure is important in this respect. Most importantly, the alcohols are able to form hydrogen bonds among each other, which is not possible for the ethers. Since evaporation always means breaking intermolecular interactions in the liquid phase, any additional stabilizing interaction raises the boiling point.&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-13T12:22:44.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1375" PostTypeId="1" AcceptedAnswerId="1376" CreationDate="2012-10-13T12:57:52.030" Score="3" ViewCount="612" Body="&lt;p&gt;What are the chemical reactions responsible for the breakdown of valine in the body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know oxidative deamination is part of the process but how is valine further broken down from an keto acid to isobutyrate?&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorUserId="22" LastEditDate="2012-10-18T05:32:57.483" LastActivityDate="2012-10-18T05:32:57.483" Title="Breakdown of valine" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1376" PostTypeId="2" ParentId="1375" CreationDate="2012-10-13T17:44:34.270" Score="2" Body="&lt;p&gt;From Wikipedia’s article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Propionic_acidemia&quot; rel=&quot;nofollow noreferrer&quot;&gt;propionic acidemia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Enzyme propionyl-CoA carboxylase] is required for the normal breakdown of the essential amino acids valine, isoleucine, threonine, and methionine, as well as certain odd-chained fatty-acids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In healthy individuals, the enzyme propionyl CoA carboxylase converts propionyl CoA to methylmalonyl CoA. This is one step in the process of converting certain amino acids and fats into sugar for energy. Individuals with PA cannot perform this conversion because the enzyme propionyl CoA carboxylase is nonfunctional. The essential amino acids isoleucine, valine, threonine, and methionine and odd-chain fatty acids are simply converted to propionyl CoA, before the process stops, leading to a buildup of propionyl CoA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find some more or less detailed descriptions of valine’s degradation pathways by Googling:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EP1Vh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rc28n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-13T17:44:34.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1377" PostTypeId="2" ParentId="1370" CreationDate="2012-10-13T18:22:53.820" Score="1" Body="&lt;p&gt;Just do a decantation. Simply discard the acid layer and then do the wash.&lt;/p&gt;&#xA;" OwnerUserId="715" LastActivityDate="2012-10-13T18:22:53.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2376" PostTypeId="1" CreationDate="2012-10-14T01:10:22.590" Score="2" ViewCount="89" Body="&lt;p&gt;How does dichlorotriazenetrione supply chlorine to a swimming pool and how does it oxidize organic material?&lt;/p&gt;&#xA;" OwnerUserId="658" LastActivityDate="2014-05-01T17:13:37.900" Title="Mechanism of oxidation by dichlorotriazenetrione" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2377" PostTypeId="1" CreationDate="2012-10-14T03:11:34.623" Score="5" ViewCount="2064" Body="&lt;p&gt;I'm thinking of sucrose, which I &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=5988&quot; rel=&quot;nofollow&quot;&gt;learned&lt;/a&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Reducing_sugar&quot; rel=&quot;nofollow&quot;&gt;nonreducing&lt;/a&gt;, what I take to mean it doesn't accept/take electrons from the surrounding solution. Does this also mean that it doesn't ionize and thus has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_factor&quot; rel=&quot;nofollow&quot;&gt;Van't Hoff factor&lt;/a&gt; &lt;em&gt;( i )&lt;/em&gt; of 1?&lt;/p&gt;&#xA;" OwnerUserId="354" LastEditorUserId="30" LastEditDate="2012-10-14T12:16:22.530" LastActivityDate="2012-10-14T17:23:39.203" Title="If a sugar is nonreducing, does that mean it doesn't ionize in water?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2378" PostTypeId="2" ParentId="1309" CreationDate="2012-10-14T03:49:07.243" Score="1" Body="&lt;p&gt;Firstly, the reasons above—identical density and molar mass—are very true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that at least for some enantiomers they &lt;a href=&quot;http://en.wikipedia.org/wiki/Racemize&quot; rel=&quot;nofollow&quot;&gt;racemize&lt;/a&gt;, or readily and spontaneously convert between both forms. This means that even if you start with a pure sample of one form, you will eventually end up with both. A famous example of this is with the banned drug &lt;a href=&quot;http://en.wikipedia.org/wiki/Thalidomide&quot; rel=&quot;nofollow&quot;&gt;Thalidomide&lt;/a&gt;, made in the 50s to treat the morning sickness of pregnant women. The R-form worked fine. Unfortunately, its S-form caused severe birth defects. If someone could figure out how to separate the two without it racemizing (there is some work on similarly structured molecules, I'm sure), they would have done so.&lt;/p&gt;&#xA;" OwnerUserId="354" LastActivityDate="2012-10-14T03:49:07.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2379" PostTypeId="2" ParentId="2377" CreationDate="2012-10-14T06:22:01.417" Score="5" Body="&lt;p&gt;I believe the confusion lies with the term &lt;a href=&quot;http://en.wikipedia.org/wiki/Reducing_sugar&quot; rel=&quot;nofollow&quot;&gt;Reducing sugar&lt;/a&gt;. From my perspective sucrose is not an ionic compound, it is covalently bonded. I believe that this should indicate to you that it does not &quot;ionize&quot; or the term that I was taught &quot;dissociate&quot; since it has nothing to dissociate from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The van 't Hoff factor is the ratio between the actual concentration&#xA;  of particles produced when the substance is dissolved, and the&#xA;  concentration of a substance as calculated from its mass.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since it does not &quot;dissociate,&quot; I would say that the answer is no, it does not &quot;ionize,&quot; and therefore has a Van't Hoff factor of 1 when mixed with pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like anything with electrons, they can be lost. This takes place in our bodies through more complicated reactions than I understand completely. I believe however, according to wikipedia, that sucrose has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycosidic_bond&quot; rel=&quot;nofollow&quot;&gt;Glycosidic bond&lt;/a&gt; which when broken, will separate sucrose into smaller sugars, if this takes place in your solution then I believe it will have a Van't Hoff factor &gt; 1.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-10-14T17:23:39.203" LastActivityDate="2012-10-14T17:23:39.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2380" PostTypeId="2" ParentId="2377" CreationDate="2012-10-14T13:43:17.527" Score="2" Body="&lt;p&gt;There's more than one way to &lt;s&gt;skin a cat&lt;/s&gt; accept an electron. Actually, in the case of a reducing sugar, it donates an electron. When it does so, it is oxidised. But that doesn't mean that it ionizes in water!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you keep a reducing sugar in the presence of an oxidising agent, the electron is transferred over to the agent, reducing it. The sugar may form an ion, or it may form a carboxylic acid (depending on the medium). The electron is fuelling the reduction half reaction of the oxidising agent we have used. If we do get a carboxylic acid, there is a chance for it to ionize(again, depends upon the medium) and we &lt;em&gt;may&lt;/em&gt; get a Van 't Hoff factor $&amp;gt;1$ {*}. But remember that this is in the presence of an oxidising agent. If we just dunk a sugar in water, it won't be able to reduce anything, and will be chemically inert, giving a Van 't Hoff factor of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*Also, the reaction with the oxidising agent may provide additional ions, increasing the Van 't Hoff factor for the oxidising agent&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-10-14T13:43:17.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2381" PostTypeId="2" ParentId="1346" CreationDate="2012-10-14T17:15:38.150" Score="8" Body="&lt;p&gt;After reading some more stuff I actually think that crystallization of starch might be the answer to this question. There is a lot of literature about this and it might very well be that it can be sufficiently exothermic to warm the small volume of air in a bag.&lt;/p&gt;&#xA;" OwnerUserId="351" LastActivityDate="2012-10-14T17:15:38.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2382" PostTypeId="2" ParentId="1309" CreationDate="2012-10-14T17:55:02.273" Score="4" Body="&lt;p&gt;It would be important to know what kind of &quot;separation&quot; you are talking about -- spectral separation, which would just mean suppression on the spectroscopic response from undesired species, or physical separation, where you actually remove those species from your sample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that not only may your sample contain different isomers, but these may actually interconvert through rearrangement reactions after ionization. This is a very common phenomenon in mass spectrometry, for example. If you want to single out the spectral signals of a certain isomer in a bulk sample, the simplest method would be to take the spectra of all unwanted components (from the literature or, preferably, separate measurements) and subtract them from your &quot;mixed&quot; spectrum. You would need information about the composition of your sample, though, or at least some indicator which tells you how much and what you have to subtract. Physically separating isomers may also be possible, e.g. if one of the isomers can be selectively ionized by laser irradiation, then separated from the remaining non-ionized sample. Such techniques are actually somewhat common in physical chemistry (not so much in everyday analytics, I think), although their applicability does depend heavily on the compounds and isomers under investigation. And there certainly are cases where different isomers are just too difficult to separate, or interconvert so readily that one structure virtually don't exist without the other.&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-14T17:55:02.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2383" PostTypeId="1" AcceptedAnswerId="2385" CreationDate="2012-10-14T18:56:53.723" Score="5" ViewCount="1605" Body="&lt;p&gt;I have drawn the figures and determined intuitively that the larger molecule's $\ce{C-C}$ would break first. I made this assumption that ethane has the stronger $\ce{C-C}$ bond and I am correct, however I do not understand the underlying reasons why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both molecules we have single bonds, but in ethane we have more space for the hydrogen atoms to fill. This was my initial reasoning, that the larger molecule is less compact and will have longer $\ce{C-C}$ bonds. Am I correct in my reasoning, or does it have something to do with the radical stabilities? My book is saying &quot;radical stability increases along the series from primary to secondary to tertiary; consequently, the energy required to create them decreases.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ethane we would have primary carbons because they bond to only one other carbon. In 2,2-dimethylpropane the central carbon is bonded to four other carbons, making it quaternary. Is this the underlying reason, and if so then what is the reason for this trend?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="10185" LastEditDate="2015-10-02T16:16:44.353" LastActivityDate="2015-10-02T16:16:44.353" Title="Which C-C bond would break first, the one in ethane, or 2,2-dimethylpropane?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2384" PostTypeId="1" CreationDate="2012-10-14T19:07:20.733" Score="25" ViewCount="1307" Body="&lt;p&gt;Obviously smoke is very complex, but apparently phenol compounds are the main components of wood smoke that produce the characteristic odor.  (One of more significant of these compounds seems to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Syringol&quot;&gt;syringol&lt;/a&gt;.)  The human nose seems to be pretty sensitive to these compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smoke seems to be very easily absorbed by various fabric-like materials, which then release the compounds back into the air over long periods of time, making the odor quite persistent.  I'd like to know what is going on with these compounds and the fabric-like materials to cause this behavior.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="7448" LastEditDate="2015-12-05T22:58:59.077" LastActivityDate="2015-12-05T22:58:59.077" Title="Why are smoke odors persistent in fabrics?" Tags="&lt;everyday-chemistry&gt;&lt;aromatic-compounds&gt;&lt;polymers&gt;&lt;smell&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2385" PostTypeId="2" ParentId="2383" CreationDate="2012-10-14T21:04:13.603" Score="8" Body="&lt;p&gt;It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inductive_effect&quot;&gt;inductive effect&lt;/a&gt;. Figuratively speaking, carbon side-chains attached to an central atom &quot;push&quot; electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density is donated towards the site of the unpaired electron, which is the case in the &lt;em&gt;tert&lt;/em&gt;-butyl radical formed from dimethylpropane, but not in the methyl radical from ethane.&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-14T21:04:13.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2386" PostTypeId="1" AcceptedAnswerId="2389" CreationDate="2012-10-15T10:36:29.357" Score="1" ViewCount="3344" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A backpacker wants to carry enough fuel to heat $\pu{3.0 kg}$ of water from $\pu{20 ^\circ C}$ to $\pu{100.0 ^\circ C}$. If the fuel he carries produces $\pu{36 kJ}$ of heat per gram when it burns, how much fuel should he carry? (For the sake of simplicity, assume that the transfer of heat is $100\%$ efficient.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My hunch is that it involves the specific heat equation $Q_\mathrm p = m  c_\mathrm p  \Delta T$ where $Q$ is heat, $m$ is mass, $c_\mathrm p$ is the specific heat capacity (in this case $\pu{4.184 J/(g ^\circ C)}$) under constant pressure and $\Delta T$, but twice I calculated the wrong answer, and would like a more intuitive understanding of the problem.&lt;/p&gt;&#xA;" OwnerUserId="361" LastEditorUserId="5026" LastEditDate="2018-02-21T14:57:50.347" LastActivityDate="2018-02-21T14:57:50.347" Title="Should I use specific heat to calculate how much fuel is required?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2387" PostTypeId="1" CreationDate="2012-10-15T17:59:02.643" Score="3" ViewCount="4412" Body="&lt;p&gt;At $\pu{25^\circ C}$ I used the free energy table for $G$ for the reaction and got $K_\mathrm{sp}= 10^{-9.02}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now need to find out what the solubility is at $10$, $20$ and $\pu{30^\circ C}$. Could you also show what equation you used and how to work out the solubility by one of these other temperatures?&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="41328" LastEditDate="2017-08-06T06:39:52.560" LastActivityDate="2017-08-06T06:39:52.560" Title="How to calculate the solubility of CaF2 at different temperatures?" Tags="&lt;solubility&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2389" PostTypeId="2" ParentId="2386" CreationDate="2012-10-15T22:46:41.103" Score="1" Body="&lt;p&gt;$$\Delta H = mc_p\Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Delta H$ is either energy absorbed or released at constant pressure.&#xA;Assuming this specific problem refers to $3\ \mathrm{kg}$ of water:&#xA;$$&#xA;\Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that the energy required is: $1\,004\,160$ joules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram.&#xA;$$&#xA;m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorDisplayName="user7951" LastEditDate="2015-09-13T19:30:32.183" LastActivityDate="2015-09-13T19:30:32.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2390" PostTypeId="2" ParentId="2387" CreationDate="2012-10-16T07:27:24.847" Score="1" Body="&lt;p&gt;It depends what you have access to… but if we suppose that you have access to decent thermodynamic tables, here's a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know how solubility is linked to dissolution reaction (Gibbs) free energy. Now, how does the free energy depend on temperature? In other words, what is $\displaystyle\frac{\partial G}{\partial T}$?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-16T07:27:24.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2391" PostTypeId="1" AcceptedAnswerId="2392" CreationDate="2012-10-16T08:46:51.417" Score="2" ViewCount="457" Body="&lt;p&gt;There are lots of software packages like Schr­­­ödinger Maestro or Argus Lab, which are used for simulations of docking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always been fascinated as to how such software works. What is the fundamental principle working &quot;under the hood&quot; of this kind of sofware? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I wanted to make a simple piece of software which, when I input the atoms and atom number (e.g., &lt;code&gt;C1H4&lt;/code&gt;) can be used for finding different energy levels (or other properties) in a molecule such as methane ($\small\ce{CH4}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does one have to start in attempting to make such a package, given a considerable knowledge of programming? What theory is applicable?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="10185" LastEditDate="2015-09-28T10:26:15.607" LastActivityDate="2015-09-28T10:26:15.607" Title="How does simulation software work at a fundamental level?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2392" PostTypeId="2" ParentId="2391" CreationDate="2012-10-16T10:57:37.427" Score="8" Body="&lt;p&gt;Well, that's a rather diffuse question. From what I have seen, the software packages you mention do Molecular Mechanics (MM), which is a treatment of molecules with classical physics. For each atom in a given molecule, the forces acting upon it due to chemical bonding with other atoms is calculated using a set of empirically determined parameters (a force field), and then an estimate of the potential energy is provided. But given that this question is tagged with &quot;Computational Chemistry&quot;, I also want to mention quantum chemistry, i.e. the application of quantum mechanics to atoms and molecules. This field is both far more interesting and far more difficult than the classical MM treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first shot would be the Wikipedia article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Computational_chemistry#Electronic_structure&quot;&gt;Computational Chemistry&lt;/a&gt; which already points you to other relevant topics, i.e. the article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_mechanics&quot;&gt;Molecular Mechanics&lt;/a&gt;, which is quite concise about the topic (although I don't know how exhaustive it is in this regard, since I don't deal with MM in my research). For quantum chemistry, the necessary knowledge is more spread-out; relevant Wikipedia articles would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Eigenproblem&quot;&gt;Eigenproblems&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot;&gt;Schrödinger equation&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Ab_initio_quantum_chemistry_methods&quot;&gt;Ab initio methods&lt;/a&gt;. I don't know any literature about MM, but if you want to look deeper into quantum chemistry, I would recommend the following two books (and lots and lots of time and patience for reading them):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Attila Szabo, Neil Ostlund: &quot;Modern Quantum Chemistry&quot;. Provides a good mathematical introduction and thorough treatments of the Hartree-Fock and Moller-Plesset methods. It's cheap, also.&lt;/li&gt;&#xA;&lt;li&gt;Ira Levine: &quot;Quantum Chemistry&quot;. Covers a wider range of methods, but is in my opinion a bit more difficult to consume than Szabo/Ostlund.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-16T10:57:37.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2394" PostTypeId="2" ParentId="2384" CreationDate="2012-10-16T20:04:04.830" Score="5" Body="&lt;p&gt;This is just a guess, but the surface area of 'fabric' is very large.  The smaller&#xA;the threads and the more strands there are, the larger the surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many organic materials are adsorbed (i.e. stuck to the surface of) onto organic materials.  The larger the surface area, the more will be adsorbed.  Since the adsorbtion &quot;bond&quot; is weak, molecules will break loose from the surface over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget, there can be a lot of molecules in even a little bit of smoke.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-10-16T20:04:04.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2395" PostTypeId="1" AcceptedAnswerId="67898" CreationDate="2012-10-17T00:40:14.797" Score="10" ViewCount="1782" Body="&lt;p&gt;In more powerful toothpastes, like Crest Pro-Health, stannous fluoride is used in place of sodium fluoride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin%28II%29_fluoride&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stannous fluoride converts the calcium mineral apatite into fluorapatite, which makes tooth enamel more resistant to bacteria generated acid attacks. In toothpastes containing calcium minerals, sodium fluoride becomes ineffective over time, while stannous fluoride remains effective in strengthening tooth enamel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the stannous fluoride more effective in this reaction simply due to having two fluoride ions per mole, or is there something more complicated in the mechanism of the reaction in which the tin spectator ion is beneficial? &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="41328" LastEditDate="2020-04-01T07:13:43.853" LastActivityDate="2020-04-02T17:00:10.513" Title="Why is tin(II) fluoride more effective in turning apatite into fluorapatite?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2396" PostTypeId="1" AcceptedAnswerId="2399" CreationDate="2012-10-18T05:09:01.347" Score="6" ViewCount="3516" Body="&lt;p&gt;I am a bit unclear on the definition of salinity.  I have always thought of it as simply total dissolved ions.&#xA;Wikipedia seems to agree &quot;saltiness or dissolved salt content&quot; but many other sources seem to suggest that only certain ions contribute.&lt;br&gt;&#xA;So is there a list of specific anions and cations that contribute to salinity and is there an underlying logic involved in their selection or is it simply arbitrary? Is the definition dependent on the solute in question?&lt;/p&gt;&#xA;" OwnerUserId="755" LastActivityDate="2012-10-19T00:12:02.897" Title="What is the exact definition of Salinity?" Tags="&lt;ions&gt;&lt;solvents&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2397" PostTypeId="1" CreationDate="2012-04-11T20:27:24.553" Score="13" ViewCount="1187" Body="&lt;p&gt;Can a polypropylene 55 gallon drum previously filled with chemicals used in the chrome plating process (possibly Nickel Sulfate) ever really be &quot;cleaned out&quot; for re-use? Can polypropylene ever absorb and retain such chemicals, or are these drums practical for this use because they do not?&lt;/p&gt;&#xA;" OwnerDisplayName="Brian" LastEditorUserId="4945" LastEditDate="2014-06-12T08:27:48.463" LastActivityDate="2017-04-28T12:57:27.677" Title="Polypropylene 55 gallon drum" Tags="&lt;safety&gt;&lt;polymers&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2398" PostTypeId="2" ParentId="1347" CreationDate="2012-10-18T12:26:38.367" Score="3" Body="&lt;p&gt;A surfactant is a material that has two regions, one that tends to be attracted to its environment, and another that is repelled by the environment. The amount to which they will be attracted to one interface or another depends upon whatever would minimize the total free energy of the system. Thus, unless the interactions for water/oil, water/air, and water/surface were equivalent, you should see that the surfactant preferentially moves to the interface where it will have the biggest impact on the free energy.  &lt;/p&gt;&#xA;" OwnerUserId="758" LastActivityDate="2012-10-18T12:26:38.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2399" PostTypeId="2" ParentId="2396" CreationDate="2012-10-18T13:50:30.900" Score="3" Body="&lt;p&gt;The term 'salinity' is measure of 'saltiness', particularly used in reference to sea water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact scientific definition has evolved over time to incorporate our ever growing understanding of physics and chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1889 the International Council for the Exploration of the Sea (ICES) used a measure of salinity based on the total dissolved solids of salt (g/kg) found in seawater after evaporation and drying at 480$^{o}$C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By taking samples of ocean water from around the world and at different depths, oceanographers have determined that whilst the concentration of salts varies, the relative ionic composition remains essentially constant. The average salinity of the world's ocean water is usually stated as 35 g/kg, with the relative ionic composition of this &quot;average sea water&quot; made up as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major cations present in 1 kg of sea water are Na$^{+}$ (10.7653 g), Mg$^{2+}$ (1.2942 g), Ca$^{2+}$ (0.4117 g), K$^{+}$ (0.3991 g) and Sr$^{2+}$ (0.0079 g).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major anions present in 1kg of seawater are Cl$^{-}$ (19.3534 g), SO$_{4}$$^{2-}$ (2.7124 g), HCO$_{3}$$^{-}$ (0.1178 g), CO$_{3}$$^{2-}$ (0.0122 g), Br$^{-}$ (0.0674 g), F$^{-}$ (0.0013 g), B(OH)$_{4}$$^{-}$ (0.0066 g).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of determining salinity from evaporation &amp;amp; drying can be quite difficult, particularly on board ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1963, the ICES adopted a new definition based on &quot;chlorinity&quot; measurement (using titration). The &quot;average sea water&quot; with ionic composition given above was determined to have chlorinity of 19.374%. By measuring the relative chlorinity of a sample of sea water, the salinity could be calculated using the formula S = 1.80655 Cl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1978, with the advent of modern electronics, the United Nations Educational, Scientific and Cultural Organization (UNESCO), International Council for the Exploration of the Sea (ICES), Scientific Committee on Oceanic Research (SCOR) and International Association for the Physical Sciences of the Oceans (IAPSO) adopted a method based on electrical conductivity measurements compensated for temperature. The Practical Salinity Scale of 1978 (PSS78) defines S (a dimensionless quantity) from conductivity and temperature measurements, in the range 2 $\leq$ S $\leq$ 42, by the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = 0.0080 - 0.1692$K_{15}$$^{1/2}$ + 25.3851$K_{15}$ + 14.0941$K_{15}$$^{3/2}$ - 7.0261$K_{15}$$^{2}$ + 2.7081$K_{15}$$^{5/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $K_{15}$ is the ratio of conductivity of the sea water sample to that of potassium chloride (KCl) solution of mass fraction 32.4356x10$^{-3}$, at 15$^{o}$C and atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A formula is also provided for compensation of the conductivity measurement at different temperatures (see &lt;a href=&quot;http://unesdoc.unesco.org/images/0004/000479/047932eb.pdf&quot; rel=&quot;nofollow&quot;&gt;http://unesdoc.unesco.org/images/0004/000479/047932eb.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2010, the Intergovernmental Oceanographic Commission released &quot;the international thermodynamic equation of seawater (TEOS-10)&quot; in which it recommended a return to the use of &quot;Absolute Salinity&quot; to take into account variations in the composition of sea water which can occur for example in regions where rivers run into the ocean or where some other regional variability results in &quot;anomalous&quot; sea water composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rational for this approach is based on understanding the effect each chemical component of seawater has on its thermodynamic properties such as density, pressure, temperature and Gibbs function. Various measures of salinity are defined and their relationships with thermodynamic properties described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Added-Mass Salinity, which incorporates the Reference Composition Salinity of &quot;standard sea water&quot; (the Practical Salinity Scale) with an adjustment for the Absolute Salinity Anomaly (mass fraction of material which must be added to standard seawater to achieve the concentration levels in the sample) is sometimes difficult to determine in-situ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern techniques for determining Absolute Salinity involve taking various other measurements such as conductivity, temperature, pressure and density by incorporating measurements from sensitive electronic and optical instruments with data-driven computer algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Density Salinity, which is the salinity argument in the thermodynamic expression for density which gives the sample's actual measured density, can be used where accurate measurements of density are available, such as a vibrating-tube densimeter or optical refractometer. When combined with Practical Salinity, temperature and pressure measurements, this can provide an accurate and easily determined measure of the amount of dissolved material in seawater (within a certain range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composition anomalies can be estimated by computer algorithms which take into account longitude, latitude and depth data. The accuracy of such algorithms is dependent upon the collection of accurate data from field measurements and laboratories across the world and is therefore an evolving process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For more details, see: &lt;a href=&quot;http://www.teos-10.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.teos-10.org/&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-10-19T00:12:02.897" LastActivityDate="2012-10-19T00:12:02.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2400" PostTypeId="2" ParentId="2397" CreationDate="2012-10-18T17:27:35.377" Score="9" Body="&lt;p&gt;Polypropylene is hydrocarbon and does not adsorb polar and ionic compounds. So, several treatments with mild hydrochloric acid followed with several treatments with clean water should work. However, I wouldn't store drinking water there without a proven test. For example, leave water there for several weeks and then perform analysis of the water for metal ions. If all concentrations are within safe levels, it should be alright. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="42741" LastEditDate="2017-04-28T12:57:27.677" LastActivityDate="2017-04-28T12:57:27.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2401" PostTypeId="1" CreationDate="2012-10-19T02:53:37.447" Score="1" ViewCount="6393" Body="&lt;p&gt;Is there an exothermic reaction which produces 1 to 2 minutes of heat at around 175 degrees Fahrenheit, no toxic gasses and is considered to be more or less safe?&lt;/p&gt;&#xA;" OwnerUserId="761" LastEditorUserId="865" LastEditDate="2015-10-30T17:13:07.233" LastActivityDate="2016-06-09T21:24:56.227" Title="Looking for a safe exothermic reaction" Tags="&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-06-10T13:11:07.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="2402" PostTypeId="2" ParentId="2401" CreationDate="2012-10-19T11:54:33.910" Score="1" Body="&lt;p&gt;Simplest thing that comes to my mind is dissolving sodium or potassium hydroxide in water, and/or neutralizing it. Both steps are exothermic, and also rather safe (given that you don't just blindly throw things together). Should become hot enough, I'd guess.&lt;/p&gt;&#xA;" OwnerUserId="732" LastActivityDate="2012-10-19T11:54:33.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2403" PostTypeId="2" ParentId="2401" CreationDate="2012-10-19T12:58:06.487" Score="4" Body="&lt;p&gt;Crystallisation of supersaturated sodium acetate is a good one, as &lt;a href=&quot;https://chemistry.stackexchange.com/a/469/84&quot;&gt;seen in my answer here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is the reaction of water with calcium oxide, which is commonly seen in self-heating coffee and soup cans. This reaction can get very hot, though.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2012-10-19T12:58:06.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2404" PostTypeId="1" CreationDate="2012-10-19T14:02:30.373" Score="17" ViewCount="650" Body="&lt;p&gt;A &lt;a href=&quot;http://www.independent.co.uk/news/uk/home-news/exclusive-pioneering-scientists-turn-fresh-air-into-petrol-in-massive-boost-in-fight-against-energy-crisis-8217382.html&quot;&gt;recent news report&lt;/a&gt; in the UK claimed a breakthrough in making a petrol equivalent from carbon dioxide and water:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A small British company has produced the first &quot;petrol from air&quot; using a revolutionary technology that promises to solve the energy crisis as well as helping to curb global warming by removing carbon dioxide from the atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It sounds like the news reports are over-hyping it a little (and I'm not sure it is that new), so what is the real chemistry behind it? What are the key reactions and how much input energy is required?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="22" LastEditDate="2013-06-06T16:21:05.157" LastActivityDate="2013-06-18T13:04:30.853" Title="What is the chemistry behind this &quot;petrol from air&quot; technology?" Tags="&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;surface-chemistry&gt;&lt;catalysis&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2405" PostTypeId="1" CreationDate="2012-10-19T15:25:35.670" Score="3" ViewCount="707" Body="&lt;p&gt;If water is mixed with lard oil and heated (creating some &lt;em&gt;super-critical&lt;/em&gt; water), how does this affect the volatility of the mixture in comparison if it is a pure mix of either water, or lard oil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when a mixture of lard oil and water is compressed and then heated to over 800 degrees Fahrenheit almost instantaneously (up to 1200 degrees Fahrenheit)? If it is in a well sealed copper container, would it be volatile enough to burst through the container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm no scientist or chemist, so use of laymen's terms, if possible, would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="763" LastEditorUserId="73" LastEditDate="2012-10-19T23:08:02.783" LastActivityDate="2012-10-19T23:08:02.783" Title="What would happen to a Water/Lard Mix under pressure and high heat?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2406" PostTypeId="2" ParentId="2405" CreationDate="2012-10-19T15:47:40.033" Score="4" Body="&lt;p&gt;I'm not sure what &lt;em&gt;aspect&lt;/em&gt; you are asking about exactly… but &lt;strong&gt;all supercritical liquids are completely miscible with each other&lt;/strong&gt;, so you would get one supercritical phase with the mixture of the two… supposing the oil doesn't decompose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://www.cookingforengineers.com/article/50/Smoke-Points-of-Various-Fats&quot; rel=&quot;nofollow&quot;&gt;the smoke point of lard is 182°C&lt;/a&gt; while the critical temperature of water is 374°C, so what most probably happens is that your lard will burn, &lt;strong&gt;breaking down to glycerol, acrolein and free fatty acids&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-19T15:47:40.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2408" PostTypeId="1" CreationDate="2012-10-19T20:27:15.440" Score="7" ViewCount="1432" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/OLED&quot;&gt;OLED displays&lt;/a&gt; have been emerging as competitors for LCD displays for some time and are now common in mobile phones (see comparison of a Google Nexus 1 and an iPhone &lt;a href=&quot;http://www.displaymate.com/Nexus_One_ShootOut.htm&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displays use small molecules or polymers that emit light when a current is passed through them (like a semiconductor LED). But what is the structure of the emitters?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="23561" LastEditDate="2018-07-02T16:56:21.060" LastActivityDate="2018-07-02T16:56:21.060" Title="What sort of molecules or polymers are used in OLED displays?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;semiconductors&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2409" PostTypeId="1" AcceptedAnswerId="2410" CreationDate="2012-10-20T00:28:14.583" Score="9" ViewCount="2072" Body="&lt;p&gt;I seek to put a layer of fluorine on a rotationally molded cross-linked polyethylene (XPE) plastic gasoline tank. I sell tanks and they will be more eco-friendly if I can accomplish this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an electrical engineer and my last experience with chemistry was &lt;a href=&quot;http://www.coursehero.com/file/1963024/CHEM200SyllabusS09/&quot; rel=&quot;nofollow&quot;&gt;Chem 200 lecture and lab at SDSU&lt;/a&gt;, a one semester intensive &quot;all of chemistry for engineers&quot; kind of course. I asked a chemistry professor at my alma mater about this, who said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Xenon difluoride is a powerful fluorinating agent, and is soluble&#xA;  (according to Wikipedia) in anhydrous methyl cyanide and several&#xA;  fluorinated solvents.  Offhand, this looks to me like it would do the&#xA;  job, but it's not a friendly compound (toxic, possibly corrosive).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The concept is already in place all around us: any time you buy a plastic jug with harsh chemicals in it, the jug has been fluorinated. So far, the majority of my learning is confirmation that yes, indeed, fluorinating polyethylene (PE) surfaces with $\ce{XeF2}$ leads to decreased gas permeability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a bit of Googling and bought a couple of great textbooks, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000WEF3QK&quot; rel=&quot;nofollow&quot;&gt;Fluoropolymers 1; Synthesis&lt;/a&gt;&lt;/em&gt; and &lt;em&gt;&lt;a href=&quot;http://books.google.com/books?id=08y8kcvRS6AC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Fluoropolymers 2; Properties&lt;/a&gt;&lt;/em&gt;. &lt;em&gt;Fluoropolymers 1&lt;/em&gt; seems to have great stuff in a couple sections. Perusing &lt;em&gt;Fluoropolymers 2&lt;/em&gt;, I haven't seen anything jump out at me yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two methods I have read about:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Involves an oven, steady high temperature, and plasma discharge. &lt;/li&gt;&#xA;&lt;li&gt;&quot;Surface Fluorination of Polymers Using Xenon Difluoride&quot;, or Gaseous $\ce{XeF2}$ and PE are placed in a &quot;stainless steel reactor&quot; and left alone.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Seeing the options, the 2nd seems the way to go, and this seems to be the epic equation concerning my quest quoted from &lt;em&gt;Fluoropolymers 1&lt;/em&gt; page 231, section 15.3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comparative accessibility and desirable properties of $\ce{XeF2}$ enable&#xA;  its use for the synthesis of fluoroorganic compounds under “mild”&#xA;  conditions. As a fluorinating agent $\ce{XeF2}$ can be used to effect&#xA;  electrophilic fluorination of various organic compounds with good&#xA;  yield and it is a useful reagent for generating radical cations.&#xA;  ... In many cases XeF2 is an attractive alternative to the&#xA;  conventional Balz–Schiemann reaction for the preparation of&#xA;  fluorinated aromatic compounds, e.g., fluorobenzenes:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{RC6H6 -&amp;gt;C[XeF2] Rc6H5F + Xe + HF -&amp;gt;C[XeF2] RC6H4F2 }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Xenon difluoride reacts with carbon–carbon single, double, and triple bonds&#xA;  giving addition fluoroorganic compounds, e.g.,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{R-CH=CH2 -&amp;gt;C[XeF2] RCH2-CHF2 + Xe}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{RCH2-CH3 -&amp;gt;C[XeF2] RCHF-CH3 + Xe + HF -&amp;gt;C[XeF2] RCF2-CH3 + Xe + HF}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this quote from page 236 gives me great hope:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another argument in favor of the $\ce{XeF2}$ method is the evidence that since 1995 it&#xA;  has been used by a manufacturing company that had no previous experience with&#xA;  direct fluorination or fluorine at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;... for I'm in the same position now!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I actually perform a fluorination surface treatment my XPE products?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where / how do I go about getting $\ce{XeF2}$? And if it is soluble in a &quot;fluorinated solvent&quot; like my prof said, is there a way I can flush this through the interior of my tanks in a controlled manner that would accomplish some degree of fluorination?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edited to add 10-22-12:&lt;/em&gt; I called a few companies today. Everyone is pointing me to pay Fluoro-Seal to do this. They are the only company providing this service in the US. I'll need to ship my products to them, wait a couple weeks, and pay $15~45 each tank. I'd still like some kind of bath or wash to do this. This is why no-one making replacement tanks has them emissions legal... all this triples my cost.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edited to add 12-30-13:&lt;/em&gt; Fluoro-Seal has given up on fluorinating rotationally molded PE. I'm pushing a plastics manufacturer to develop their existing acetel product to work with rotational molding. Their solution for injection and blow molding is in production and impervious to fuels (they have parts in nearly all automotive fuel systems in the US right now). So there's hope for a solution; however it seems impossible to fluorinate PE.&lt;/p&gt;&#xA;" OwnerUserId="344" LastEditorUserId="24052" LastEditDate="2016-03-16T01:09:42.977" LastActivityDate="2018-07-02T20:15:06.660" Title="How to fluorinate polyethylene?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2410" PostTypeId="2" ParentId="2409" CreationDate="2012-10-20T12:29:54.197" Score="3" Body="&lt;p&gt;There is a simpler way to get fluorinated polyethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyethylene is polymerized ethylene $\ce{CH2=CH2}$. As part of the polymerization, the double bond is no longer there. &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polyethylene&lt;/a&gt; is essentially a very long saturated hydrocarbon:  $\ce{(-CH2-CH2 -)}_n.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdYvZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another polymer already exists that replaces all the hydrogen atoms with fluorine atoms: &lt;a href=&quot;http://en.wikipedia.org/wiki/Teflon&quot; rel=&quot;nofollow noreferrer&quot;&gt;poly(tetrafluoroethylene)&lt;/a&gt; (PTFE), made from tetrafluoroethylene $\ce{CF2=CF2}$. PTFE is a long saturated ferfluorocarbon: $\ce{(-CF2-CF2 -)}_n.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yKzi1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most well known trademarked brand of PTFE is Teflon. Get your coated with or made out of Teflon/PTFE. It will be easier, cheaper, and safer in the long run. If XeF&lt;sub&gt;2&lt;/sub&gt; will fluorinate polyethylene (which is reasonably inert), imagine what it could do to yyou.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-20T12:29:54.197" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2411" PostTypeId="1" AcceptedAnswerId="2412" CreationDate="2012-10-20T15:04:40.257" Score="11" ViewCount="22167" Body="&lt;p&gt;What is the meaning of $\ce{(en)}$ on this two coordination compounds: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Co(en)_2(NO_2)_2]Cl}$ &#xA;&lt;br&gt;&#xA;$\ce{[Cd(en)_2(CN)_2]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a type of compound? If yes, what is the meaning?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="7448" LastEditDate="2015-07-03T11:01:32.363" LastActivityDate="2015-07-03T11:05:19.707" Title="What is the meaning of '(en)'?" Tags="&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2412" PostTypeId="2" ParentId="2411" CreationDate="2012-10-20T16:44:24.697" Score="13" Body="&lt;p&gt;It's an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethylenediamine&quot; rel=&quot;noreferrer&quot;&gt;ethylenediamine&lt;/a&gt; ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit - elaborating as per your request:&lt;/strong&gt; Unfortunately, you don't provide any information as to how much you already know about &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_complex&quot; rel=&quot;noreferrer&quot;&gt;complex chemistry&lt;/a&gt;, so I'm going back to the roots here. If you already know what a complex is, skip this paragraph. As you probably know, atoms and molecules may bind to each other, forming new compounds. We all know covalent bonds in which two atoms move close together and mutually share their electrons, and ionic bonds wherein one of the partaking partners is more egocentric in terms of electrons, &quot;stealing&quot; it from the other but staying in the vicinity nontheless. What these types have in common is that each atom or molecule provides an electron for the formation of the bond. However, it may also happen that both electrons come fron one partner alone; an electron-rich molecule or ion approaches another atom or ion, &quot;donating&quot; his electrons without really giving them up. This is the situation in coordination chemistry: A &quot;central atom&quot; or ion -- usually, but not necessarily a transition metal -- is surrounded by a varying number of &quot;ligands&quot; which donate electron pairs to the central atom, all while these electron pairs remain &quot;located&quot; at their respective ligands (&lt;em&gt;very dangerous expression, I know ...&lt;/em&gt;). The compound that is formed is called a complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ligands are atoms or molecules which can provide lone pairs for coordination bonds, i.e. electrons in non-bonding, doubly filled orbitals. In the above example of ethylenediamine, the binding centers of the molecule are the nitrogen atoms. The en ligand is an example of a multidentate (or chelate) ligand which may form more than one coordinating bond per molecule, while the $\ce{NO2^-}$ and $\ce{CN-}$ are monodentate ligands. The chloride anions in your first compound does not bind to the copper ion, but rather stay outside its &quot;coordination sphere&quot; to balance the positive charge of the cobalt complex (which you might then interpret as an ionic bonding between the complex cation and the chlorine anion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there are, of course, many quirks to this. Complexes with multiple central atoms, different theories about bonding, bridging or radical ligands, charge-transfer complexes, etc. This is a very wide and important field of chemistry, after all; nothing you could exhaustively explain in one or two paragraphs.&lt;/p&gt;&#xA;" OwnerUserId="732" LastEditorUserId="7448" LastEditDate="2015-07-03T11:03:03.017" LastActivityDate="2015-07-03T11:03:03.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2415" PostTypeId="1" AcceptedAnswerId="2416" CreationDate="2012-10-21T02:44:20.027" Score="14" ViewCount="3416" Body="&lt;p&gt;Thought experiment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like the way a piece of litmus paper (yes, it is laboratory equipment and &lt;strong&gt;easily available to the layman&lt;/strong&gt; at that!) may be used to test whether a solution is an acid, or alkali.&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="9961" LastEditDate="2018-11-02T23:39:26.690" LastActivityDate="2020-03-15T17:52:50.393" Title="Is there a simple field test for heavy water?" Tags="&lt;water&gt;&lt;home-experiment&gt;&lt;isotope&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2416" PostTypeId="2" ParentId="2415" CreationDate="2012-10-21T03:40:59.030" Score="21" Body="&lt;p&gt;There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-10-21T03:40:59.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2417" PostTypeId="1" CreationDate="2012-10-21T15:37:31.980" Score="5" ViewCount="16359" Body="&lt;p&gt;Which would have one monochlorination possibility? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2,2-dimethylpropane&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t846j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;strong&gt;2-methylpropane&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Hov5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems I'm missing something because I thought both would only have one monochlorination possibility. (unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl?)&lt;/p&gt;&#xA;" OwnerUserId="766" LastEditorUserId="10185" LastEditDate="2015-10-02T16:17:30.290" LastActivityDate="2019-10-02T15:34:49.110" Title="Which has one monochlorination possibility?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2418" PostTypeId="2" ParentId="2415" CreationDate="2012-10-21T18:58:03.100" Score="4" Body="&lt;p&gt;If you don't have access to a freezer, you could allow sun light to pass through&#xA;your glass vessels and then through a prism. The heavy water will absorb less orange red light and the difference might be observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5YMBf.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of &lt;a href=&quot;http://publiclaboratory.org/tool/spectrometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; might help (not complicated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I did not want to post this as an answer because I'm not sure if it's       realistic but I can not see were to add a comment.&lt;/p&gt;&#xA;" OwnerUserId="755" LastEditorUserId="755" LastEditDate="2012-10-21T19:31:37.007" LastActivityDate="2012-10-21T19:31:37.007" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2419" PostTypeId="2" ParentId="2415" CreationDate="2012-10-21T23:15:14.280" Score="5" Body="&lt;p&gt;Mass. Heavy water is noticeably heavier then light water. Simply weigh an equal amount of each water and see which weights more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light water weights 1 g/mL (&lt;a href=&quot;http://webbook.nist.gov/cgi/fluid.cgi?Action=Load&amp;amp;ID=C7732185&amp;amp;Type=SatP&amp;amp;Digits=5&amp;amp;THigh=26&amp;amp;TLow=24&amp;amp;TInc=1&amp;amp;RefState=DEF&amp;amp;TUnit=C&amp;amp;PUnit=atm&amp;amp;DUnit=g/ml&amp;amp;HUnit=kJ/mol&amp;amp;WUnit=m/s&amp;amp;VisUnit=uPa%2as&amp;amp;STUnit=N/m&quot; rel=&quot;noreferrer&quot;&gt;0.99700 g/mL at 25 °C&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/151882?lang=en&amp;amp;region=CA&quot; rel=&quot;noreferrer&quot;&gt;Heavy water weights 1.107 g/mL at 25 °C&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have 100 mL you will see an 11 g difference, assuming pure samples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The Science and Technology museum in Ottawa actually has samples you can lift to feel this).&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2012-10-21T23:15:14.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2421" PostTypeId="1" AcceptedAnswerId="2425" CreationDate="2012-10-22T03:41:54.777" Score="9" ViewCount="943" Body="&lt;p&gt;I am reading some articles about the electronic properties of graphene and while reading it I came across graphene armchair and zigzag structures, which have different bond structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two different orientations (armchair and zigzag) are due to the arrangement of carbon atoms in the periphery (&lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot;&gt;wiki&lt;/a&gt;), but I am unable to understand what armchair(5,5) and zigzag(5,0) mean. &lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="4945" LastEditDate="2015-01-06T06:58:41.127" LastActivityDate="2015-01-06T06:58:41.127" Title="What do graphene &quot;armchair(5,5)&quot; and &quot;zigzag(5,0)&quot; refer to?" Tags="&lt;everyday-chemistry&gt;&lt;nomenclature&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2422" PostTypeId="1" CreationDate="2012-10-22T05:58:00.067" Score="5" ViewCount="580" Body="&lt;p&gt;I am not sure where to start with this problem. I know that $\small q = mc(T_f-T_i)$, but that seems like it would not help here. I think I need to balance the equation, but I am not sure how the rate factors into the problem. Anyway, if anyone could help me find the relevant equations/method for solving this problem, it would be greatly appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural gas mixture is burned in a furnace at a power-generating station at a rate of 13.0 mol per minute.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the fuel consists of 9.3 mole $\small\ce{CH4}$, 3.1 mole $\small\ce{C2H6}$, 0.40 mol $\small\ce{C3H8}$, and 0.20 mole $\small\ce{C4H10}$, what mass of $\small\ce{CO2} (g)$ is produced per minute?  &lt;/li&gt;&#xA;&lt;li&gt;How much heat is released per minute?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="770" LastEditorUserId="73" LastEditDate="2012-10-22T08:55:21.203" LastActivityDate="2012-10-30T17:55:23.637" Title="Find mass of CO₂ and heat released per minute in a combustion reaction" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2423" PostTypeId="1" CreationDate="2012-10-22T10:20:28.047" Score="8" ViewCount="1332" Body="&lt;p&gt;To simulate the action of gastric $\ce{HCl}$, we have titrated 100 ml of a meal with 0.1 N $\ce{HCl}$. To plot the data, use (R-code)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;vol = read.table(&quot;http://menne-biomed.de/uni/volph.txt&quot;,sep=&quot;\t&quot;,header=TRUE)&#xA;plot(vol$vol,vol$pH,pch=16,type=&quot;l&quot;,main=&quot;Titration of 100 ml meal with 0.1 N HCl&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am trying to determine, for example, how much 1N $\ce{HCl}$ I need to add to get a pH of 1.5; if I understand correctly, I have to fit the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Hasselbalch equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a statistics guy, the numerics is no problem, I just do not understand how to get the terms in that equation together with my data.&lt;/p&gt;&#xA;" OwnerUserId="771" LastEditorUserId="4231" LastEditDate="2015-02-24T20:49:49.960" LastActivityDate="2015-02-24T20:49:49.960" Title="Fitting titration curve" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2424" PostTypeId="1" CreationDate="2012-10-22T13:22:33.887" Score="4" ViewCount="48" Body="&lt;p&gt;Are there any compounds that reduce water's sensitivity to extreme heat change and/or volatility to high heat (over 800 degrees Fahrenheit), whether organic or inorganic?&lt;/p&gt;&#xA;" OwnerUserId="763" LastEditorUserId="73" LastEditDate="2012-10-22T23:10:54.190" LastActivityDate="2012-10-22T23:10:54.190" Title="Compounds that vary water's superheated and supercritical limits" Tags="&lt;physical-chemistry&gt;&lt;water&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2014-03-19T02:43:01.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="2425" PostTypeId="2" ParentId="2421" CreationDate="2012-10-22T17:52:05.227" Score="6" Body="&lt;p&gt;The third picture, subpic (d) &lt;a href=&quot;http://www.ifw-dresden.de/institutes/iff/research/Carbon/molecular-nanostructures/graphene&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; shows you the geometric armchair(red) and zigzag(blue) structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fpupV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's derived from  structure description of &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_nanotube#Single-walled&quot; rel=&quot;noreferrer&quot;&gt;Carbon nanotubes&lt;/a&gt; similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_reconstruction&quot; rel=&quot;noreferrer&quot;&gt;surface reconstruction&lt;/a&gt;/&lt;a href=&quot;http://www.chem.qmul.ac.uk/surfaces/scc/scat6_1.htm&quot; rel=&quot;noreferrer&quot;&gt;overlayer structures&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The integers n and m denote the number of unit vectors along two directions in the honeycomb crystal lattice of graphene. If m = 0, the nanotubes are called zigzag nanotubes, and if n = m, the nanotubes are called armchair nanotubes. Otherwise, they are called chiral. The diameter of an ideal nanotube can be calculated from its (n,m) indices&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope you have some background, otherwise here is a &lt;a href=&quot;https://vimeo.com/1086112&quot; rel=&quot;noreferrer&quot;&gt;nice visualisation&lt;/a&gt;, or look the lecture linked above. It's important to understand what the unit cell of a 3D crystal or 2D surface is.&lt;/p&gt;&#xA;" OwnerUserId="165" LastEditorUserId="30" LastEditDate="2012-10-22T18:39:35.997" LastActivityDate="2012-10-22T18:39:35.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2426" PostTypeId="1" AcceptedAnswerId="2427" CreationDate="2012-10-22T20:23:15.703" Score="9" ViewCount="18783" Body="&lt;p&gt;I'm here to share with you something that totally confuses me, as I can't see the logic behind it, and my teacher doesn't know either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a set of bonds that's trigonal pyramidal, with a lone pair of electrons and three bonds. The angle between the bonds is 107 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the part where I get confused is if, say, we have one double and two single bonds (I know this makes the atom have 10 electrons, but just ignore that. It's not the point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in my brain, it would be logical that the double bonds, containing four electrons, would repel more than a single bond, containing just two. However, according to what I've been taught, it doesn't. What is that?&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="30" LastEditDate="2012-10-22T20:35:04.683" LastActivityDate="2012-10-22T20:42:26.800" Title="Why do single, double and triple bonds repel each other equal amounts?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;intermolecular-forces&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2427" PostTypeId="2" ParentId="2426" CreationDate="2012-10-22T20:42:26.800" Score="5" Body="&lt;p&gt;The way you have learnt to predict molecular geometries is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;noreferrer&quot;&gt;VSEPR theory&lt;/a&gt;. As all theories, it makes assumptions and has a certain number of limitations. The issue you are asking about is one such limitation.&lt;/p&gt;&#xA;&lt;p&gt;When predicting the geometry of molecules with VSEPR, one should treat multiple bonds as a single entity. However, the geometry obtained is just a first approximation, or “idealized” geometry. In fact, multiple bonds induce distortions from this ideal geometry, pretty much like lone pairs do (I assume you know that already. If not, well, check the above-linked &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; for “lone pair”.) Citing from &lt;a href=&quot;http://wps.prenhall.com/wps/media/objects/3311/3391094/blb0902.html&quot; rel=&quot;noreferrer&quot;&gt;this online course&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Multiple bonds contain a higher electronic-charge density than do single bonds, so multiple bonds also represent larger electron domains (&amp;quot;fatter balloons&amp;quot;). Consider the structure of phosgene, Cl&lt;sub&gt;2&lt;/sub&gt;CO, which is shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7E9R.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;From the Lewis structure of phosgene, we might expect a trigonal planar geometry with 120°-bond angles. However, the double bond seems to act much like a nonbonding pair of electrons, reducing the ClCCl bond angle from 120° to 111°. In general, electron domains for multiple bonds exert a greater repulsive force on adjacent electron domains than do single bonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So: &lt;strong&gt;yes, lone pairs and multiple bonds give rise to deviations from the ideal VSEPR geometries&lt;/strong&gt;. It's nice that you figured it out on your own!&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-10-22T20:42:26.800" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2428" PostTypeId="1" AcceptedAnswerId="2430" CreationDate="2012-10-22T21:12:09.343" Score="14" ViewCount="777" Body="&lt;p&gt;Most solids objects turn black after they are burnt.&#xA;But some turn white.&#xA;Why is that ? &lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="4945" LastEditDate="2014-10-24T09:51:57.187" LastActivityDate="2014-10-24T09:54:09.557" Title="Why do some materials turn black and others white after combustion?" Tags="&lt;everyday-chemistry&gt;&lt;combustion&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2429" PostTypeId="1" AcceptedAnswerId="7873" CreationDate="2012-10-22T21:43:40.813" Score="6" ViewCount="1651" Body="&lt;p&gt;In the old days computer monitors and televisions were built from &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathode_ray_tube&quot;&gt;cathode ray tubes&lt;/a&gt; (CRTs) where the image was generated by an electron beam hitting phosphors which emitted light approximating the three primary colours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What were the phosphors made from and what were the key reasons why the particular substances were chosen?&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-08-02T12:32:43.310" Title="What chemical compounds were used in CRT phosphors?" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2430" PostTypeId="2" ParentId="2428" CreationDate="2012-10-22T22:50:40.953" Score="9" Body="&lt;p&gt;The objects which turn black are usually carbon based/organic compounds. When burned/combusted they follow one of the following unbalanced reactions:&#xA;\begin{align}&#xA;\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &amp;amp;\ce{-&amp;gt;} \ce C_\text{(solid)} + \ce{H_2O}_\text{(liquid)} + {\ce{CO_2}}_\text{(gas)}\\&#xA;\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &amp;amp;\ce{-&amp;gt;}  \ce{H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)}\\&#xA;\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &amp;amp;\ce{-&amp;gt;}  \ce {H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)} + \ce{CO}_\text{(gas)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that some of the reactions will produce solid carbon, which is the black thing you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, inorganic compounds begin to decompose into their ions and/or release their  &quot;trapped&quot; water molecules. As you'll notice, most ionic compounds are white solid salts.&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="4945" LastEditDate="2014-10-24T09:54:09.557" LastActivityDate="2014-10-24T09:54:09.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2431" PostTypeId="2" ParentId="2415" CreationDate="2012-10-23T01:42:14.913" Score="2" Body="&lt;p&gt;(Completely untested since, sadly, I have no access to heavy water.) Slowly pour the contents of one glass into the other. The index of refraction of heavy water is 1.328, that of light water is 1.333. Not much, but maybe just enough to see the inhomogeneities. If the decanted glass's contents sink, it had the heavy water. Otherwise, light water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you have even a cheap kitchen scale, that would probably be the better test, since the densities differ by about 10%, as opposed to the index of refraction, which differs by far less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, just add a drop of food colouring to one of the glasses. Or a drop of milk. Just to help out with the contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the test with just a small sample from each glass if you want to avoid contaminating the expensive heavy water.&lt;/p&gt;&#xA;" OwnerUserId="389" LastActivityDate="2012-10-23T01:42:14.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2432" PostTypeId="1" AcceptedAnswerId="2434" CreationDate="2012-10-23T04:50:19.070" Score="5" ViewCount="214" Body="&lt;p&gt;I am in the process of constructing a kitchen waste digester (vegetable/fruits) from which, hopefully, over time adequate methane may be collected to burn. The digester won't be complete for several months yet - a few minutes each week or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if a &lt;a href=&quot;http://datasheetreference.com/component/content/article/16/87-ignition-coil-driver.html&quot; rel=&quot;noreferrer&quot;&gt;high frequency high-voltage field&lt;/a&gt; were to be applied to the methane (taking due care to avoid sparks) would the methane decompose into carbon, and hydrogen? A google search seems to indicate that methane may decompose, or form a plasma; I understand there may not be a simple/straight answer for this - but a general direction would be nice to have.&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="73" LastEditDate="2012-10-23T06:46:50.610" LastActivityDate="2012-10-23T13:28:46.613" Title="Thought Experiment: Applying a high frequency high voltage to methane" Tags="&lt;organic-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2433" PostTypeId="2" ParentId="2423" CreationDate="2012-10-23T12:21:10.933" Score="4" Body="&lt;p&gt;The best thing would be if you went back to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_action_law&quot; rel=&quot;nofollow&quot;&gt;mass action law&lt;/a&gt;, considered &lt;em&gt;all&lt;/em&gt; reactions that take place ($\ce{Ca}$ buffer!) and derived a formula for your specific problem. The Henderson-Hasselbalch equation actually does exactly that for a one-component acid/base system with a single protonation stage. Remember that the concentrations that occur in the mass-action law are those at chemical equilibrium, not the initial ones; i.e. for any component $C$, the eq. concentration is $[C] = [C]_0 - \Delta_C$. The problem at hand is then to find how the changes in concentration depend on the added volume of titer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just took the Henderson-Hasselbalch equation (equating the change in concentrations with the added concentration of $\ce{HCl}$), you do not include the $\ce{Ca}$ buffer, which will probably play an role in this. (You get a $pK_a$ of around 0.7, which I guess is a bit on the strong side ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing I just noticed: You apparently added 100 mL of titer to 100 mL of sample. That's kind of bad, because you'll also have to consider how the change of overall volume affects your concentrations (effectively the pH) during titration.&lt;/p&gt;&#xA;" OwnerUserId="732" LastEditorUserId="4231" LastEditDate="2015-02-24T20:49:44.077" LastActivityDate="2015-02-24T20:49:44.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2434" PostTypeId="2" ParentId="2432" CreationDate="2012-10-23T13:28:46.613" Score="2" Body="&lt;p&gt;To decompose methane molecule you have to apply energy about several electron-volts directly to molecule. Since microwave radiation wavelength measured in millimeters, to really decompose methane you'll need ridiculous voltage, much more them mere kilovolts needed to ignite electrical discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And btw, electrical discharge is the easiest way to supply the energy needed for molecule to break.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-10-23T13:28:46.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2435" PostTypeId="1" AcceptedAnswerId="2436" CreationDate="2012-10-23T15:07:01.623" Score="3" ViewCount="275" Body="&lt;p&gt;I have seen in some textbooks that such a reaction is listed under reduction of alkenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the double bond is getting reduced to single bond and one hydrogen atom is added to each of the carbons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I classify this reaction as an &lt;em&gt;addition reaction&lt;/em&gt; or a &lt;em&gt;reduction reaction&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="778" LastEditorUserId="402" LastEditDate="2017-04-01T00:17:46.237" LastActivityDate="2017-04-01T00:17:46.237" Title="Is alkene reaction with hydrogen to form an alkane an addition reaction?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2436" PostTypeId="2" ParentId="2435" CreationDate="2012-10-23T15:20:57.303" Score="3" Body="&lt;p&gt;Yes, the &lt;em&gt;hydrogenation&lt;/em&gt; reaction of alkenes is typically classified as an &lt;em&gt;addition&lt;/em&gt; reaction. See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Catalytic_Hydrogenation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkene#Hydrogenation&quot; rel=&quot;nofollow&quot;&gt;there&lt;/a&gt;, for example.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-23T15:20:57.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2437" PostTypeId="2" ParentId="1114" CreationDate="2012-10-23T16:29:01.443" Score="2" Body="&lt;p&gt;Almost all gasses are more soluble in cold water than in hot.  To test this for carbon dioxide open a bottle of warm soda.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-10-23T16:29:01.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2438" PostTypeId="1" AcceptedAnswerId="2440" CreationDate="2012-10-24T02:36:47.680" Score="1" ViewCount="1459" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We know:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$&lt;/li&gt;&#xA;  &lt;li&gt;The mass of water inside it: $1450\ \mathrm g$&lt;/li&gt;&#xA;  &lt;li&gt;The change in temperature: $2.62\ \mathrm K$&lt;/li&gt;&#xA;  &lt;li&gt;The end product is $\ce{K2O}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is &lt;em&gt;extravagant&lt;/em&gt;; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="5026" LastEditDate="2018-02-28T13:55:39.730" LastActivityDate="2018-02-28T13:57:27.067" Title="Determining enthalpy and internal energy of formation for potassium superoxide" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2439" PostTypeId="1" CreationDate="2012-10-24T03:50:56.327" Score="5" ViewCount="988" Body="&lt;p&gt;I am living in my dorms at the moment and apparently it is an accepted thing that the water in the building is brown and yellow. According to my RA, this is a characteristic of the town I live in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only &quot;clean&quot; water we get is a water filter at the zeroth floor of our building where transparent water comes out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that the water that comes out of the tap, the kitchen, the shower, and the toilet bowl is light yellowish-brown water. I haven't done enough research as the cause of this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling water doesn't seem to change the colour either (is this even possible?). I for one cannot live like this, so I am asking the community here if they know a household method to remove the discoloration from the water.&lt;/p&gt;&#xA;" OwnerUserId="611" LastEditorUserId="21964" LastEditDate="2016-08-17T17:38:57.890" LastActivityDate="2016-08-17T17:38:57.890" Title="Is there a household method to decolorize potentially contaminated water?" Tags="&lt;everyday-chemistry&gt;&lt;water-treatment&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2440" PostTypeId="2" ParentId="2438" CreationDate="2012-10-24T09:14:27.513" Score="2" Body="&lt;p&gt;Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. &quot;Standard state&quot; doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="5026" LastEditDate="2018-02-28T13:57:27.067" LastActivityDate="2018-02-28T13:57:27.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2441" PostTypeId="1" AcceptedAnswerId="2442" CreationDate="2012-10-24T14:32:54.620" Score="3" ViewCount="7009" Body="&lt;p&gt;Chemically speaking, what are the biggest differences between saturated fatty acids and unsaturated fatty acids? What are the effects of those differences on the geometry of the molecule? And what is the effect of that on our body?&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="10185" LastEditDate="2015-10-02T16:20:07.127" LastActivityDate="2015-10-02T16:20:07.127" Title="What is the difference between saturated and unsaturated fatty acids?" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2442" PostTypeId="2" ParentId="2441" CreationDate="2012-10-24T15:56:58.007" Score="3" Body="&lt;p&gt;Saturated fatty acids are &lt;a href=&quot;http://en.wikipedia.org/wiki/Fatty_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;fatty acids&lt;/a&gt; with no double bonds and unsaturated fatty acids are fatty acid with one or more double bonds. This is the biggest difference between a saturated and unsaturated fatty acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nGg9v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is two types of structural configuration for unsaturated fatty acids due to the presence of double bond; cis (kink/bend in fatty acid struture) and trans(No kink/bend in fatty acid struture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iWQhv.png&quot; alt=&quot;enter image description here&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;cis fatty acid (wikipedia)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A cis configuration means that adjacent hydrogen atoms are on the same side of the double bond. The rigidity of the double bond&#xA;  freezes its conformation and, in the case of the cis isomer, causes&#xA;  the chain to bend and restricts the conformational freedom of the&#xA;  fatty acid. The more double bonds the chain has in the cis&#xA;  configuration, the less flexibility it has. When a chain has many cis&#xA;  bonds, it becomes quite curved in its most accessible conformations.&#xA;  For example, oleic acid, with one double bond, has a &quot;kink&quot; in it,&#xA;  whereas linoleic acid, with two double bonds, has a more pronounced&#xA;  bend. Alpha-linolenic acid, with three double bonds, favors a hooked&#xA;  shape. The effect of this is that, in restricted environments, such as&#xA;  when fatty acids are part of a phospholipid in a lipid bilayer, or&#xA;  triglycerides in lipid droplets, cis bonds limit the ability of fatty&#xA;  acids to be closely packed, and therefore could affect the melting&#xA;  temperature of the membrane or of the fat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iodine value&lt;/a&gt; test is used to determine degree of unsaturation in a unsaturated fatty acids; that is to measure number of double bonds present in a unsaturated fatty acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.fitday.com/fitness-articles/nutrition/fats/saturated-vs-unsaturated-fatty-acids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; will explain the health benifits of saturated and unsaturated fatty acids&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditDate="2012-10-24T16:02:38.373" LastActivityDate="2012-10-24T16:02:38.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2443" PostTypeId="1" CreationDate="2012-10-24T16:46:54.773" Score="7" ViewCount="3022" Body="&lt;p&gt;I'm having a hard time deducing the structure of the product formed when dimedone and benzaldehyde react in the presence of catalytic piperidine. I think that a Knoevenagel reaction will get me to this product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VMiur.gif&quot; alt=&quot;2-[hydroxy(phenyl)methyl]-5,5-dimethylcyclohexane-1,3-dione&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the answer says that the product has a molar mass of 368 g/mol. The product above is only 246 g/mol. What else can have occurred?&lt;/p&gt;&#xA;" OwnerUserId="266" LastEditorUserId="16683" LastEditDate="2017-11-28T23:26:47.787" LastActivityDate="2017-11-28T23:26:47.787" Title="What is the product formed in the reaction of dimedone and benzaldehyde?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;aldol-reaction&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2444" PostTypeId="2" ParentId="2443" CreationDate="2012-10-24T17:20:12.843" Score="1" Body="&lt;p&gt;This  image  should  help  you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;       &lt;a href=&quot;http://www.hindawi.com/journals/chem/2013/652805/sch1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7AYIc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="780" LastEditorUserId="30" LastEditDate="2012-10-25T07:30:46.020" LastActivityDate="2012-10-25T07:30:46.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2445" PostTypeId="1" CreationDate="2012-10-24T19:02:43.113" Score="12" ViewCount="3514" Body="&lt;p&gt;Surfactants decrease surface tension between two fluids.  What do you call a substance that increases the surface tension between two fluids?  How is this effect acheived?&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="22" LastEditDate="2012-11-02T11:27:04.550" LastActivityDate="2018-03-21T15:15:07.580" Title="What is the opposite of a surfactant?" Tags="&lt;physical-chemistry&gt;&lt;surfactants&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2446" PostTypeId="1" AcceptedAnswerId="2453" CreationDate="2012-10-25T11:57:17.773" Score="8" ViewCount="2796" Body="&lt;p&gt;Is there a standard symbol for molar density, i.e. $n/V$, where $n$ is the total number of moles of every species in a system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that there is standard term &quot;molar volume&quot;, which is the reciprocal of what I'm asking for, but my equations are going to look pretty silly with $1/V_m$ written everywhere. As a simple example, I want to say that if my system consists of a mixture of $\ce{A}$ and $\ce{B}$ with the reaction $\ce{A &amp;lt;-&amp;gt; B}$, the total concentration of $\ce{A}$ and $\ce{B}$ will be constant and we can write&#xA;$$&#xA;\ce{[B]} = c - \ce{[A]},&#xA;$$&#xA;where $c$ is the thing I'm looking for the symbol for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't find &quot;molar density&quot; in the IUPAC green book, but I don't know my way around it very well, and if it has a different name it might be hiding somewhere I didn't think to look.&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2015-03-10T14:05:45.913" Title="Is there a standard symbol for molar density?" Tags="&lt;nomenclature&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2447" PostTypeId="2" ParentId="2446" CreationDate="2012-10-25T21:23:25.683" Score="3" Body="&lt;p&gt;I don't know of a standard symbol for &lt;em&gt;molar&lt;/em&gt; density, but rho ($\rho$) is a standard symbol for density in general, so if there's no better symbol, I'd go for something like $\rho_m$. As long as you explicitly define it somewhere, it should be okay.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-10-25T21:23:25.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2448" PostTypeId="2" ParentId="2417" CreationDate="2012-10-26T13:07:28.967" Score="3" Body="&lt;p&gt;&lt;em&gt;“unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl”&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why wouldn't you be? I think that's the key to the answer.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-26T13:07:28.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2449" PostTypeId="1" CreationDate="2012-10-26T19:52:54.093" Score="8" ViewCount="6554" Body="&lt;p&gt;An exothermic reaction is apparently one that releases energy to it's environment. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Explosion&quot;&gt;Wikipedia&lt;/a&gt;, an explosion is a rapid increase in volume &amp;amp; energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that an explosion is basically an exothermic reaction?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="73" LastEditDate="2012-10-26T21:10:59.620" LastActivityDate="2016-10-23T00:07:33.610" Title="Is an explosion merely an exothermic reaction?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2450" PostTypeId="2" ParentId="2449" CreationDate="2012-10-26T23:20:31.727" Score="16" Body="&lt;p&gt;The key parts to the definition of an explosion are &quot;rapid&quot; and &quot;increase in volume&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exothermic reactions do not necessarily meet these requirements. Exothermic reactions need only release energy to the surroundings by converting chemical potential energy into kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the combustion of methane is very exothermic, with a standard enthalpy change of -890 kJ/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH4(g) + 2O2(g)-&amp;gt;CO2(g) +2H2O(g)} \space  \space \space \space \Delta_c H^o=-890 \text{ kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, at room temperature this reaction is very slow. Mixtures of oxygen and methane do not spontaneously ignite without a spark source. Also, there is no change in pressure with this reaction, since there are 3 moles of gas on both sides of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider nitroglycerin. Yes, it's combustion is significantly exothermic. More importantly, the combustion happens rapidly, needs only the heat of friction or the energy of impact to ignite, and produces significantly more gas than it consumes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4C3H5O9}(l) \ce{+1O2(g)-&amp;gt;12CO2(g) +10H2O(g) +6N2(g)} \space  \space \space \space \Delta_c H^o=-5660 \text{ kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All thermodynamic data were mined from the &lt;a href=&quot;http://webbook.nist.gov/chemistry/name-ser.html&quot;&gt;NIST Chemistry Webbook&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-26T23:20:31.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2451" PostTypeId="2" ParentId="2446" CreationDate="2012-10-26T23:33:17.973" Score="3" Body="&lt;p&gt;I agree with Aesin, but I have another comment to make.  The term &quot;molar density&quot; has a confused meaning.  That is because the density of one mole of a substance is identical to the density of two moles of the substance, or any number of moles of the substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;molar&quot; is most often used to make an extensive quantity into an intensive one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extensive quantity is one that doubles if you take two identical copies of a system and put them together.  Volume is extensive, temperature is not, it is intensive.  Most importantly, &lt;strong&gt;density&lt;/strong&gt; is already intensive, so that &quot;molar density&quot; is essentially devoid of any meaning other than that the question states, the inverse of the molar volume.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-10-26T23:33:17.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2452" PostTypeId="2" ParentId="2422" CreationDate="2012-10-26T23:36:54.177" Score="5" Body="&lt;p&gt;You know the amounts of the four gases burned in a minute.  Just write combustion reactions for each of these and work out the heat produced by each reaction &lt;strong&gt;as written&lt;/strong&gt;. That's your answer to the second part.  Adding up the carbon dioxide produced in each of the four combustion reactions is the answer to the first part.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="675" LastEditDate="2012-10-30T17:55:23.637" LastActivityDate="2012-10-30T17:55:23.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2453" PostTypeId="2" ParentId="2446" CreationDate="2012-10-27T11:15:02.317" Score="6" Body="&lt;p&gt;To my knowledge, n/V is known as molar concentration and uses symbol $C$ and it seems okay for me to use $C_t$ i.e. total concentration to mark sum of concentrations of all particles in system. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-10-27T11:15:02.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2454" PostTypeId="1" CreationDate="2012-10-27T11:40:50.287" Score="12" ViewCount="738" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Octane_rating&quot;&gt;octane rating&lt;/a&gt; of petrol (gasoline) is a measure of the how much compression a fuel air mixture can have before detonating (which in petrol engines is a bad thing as you want the mix to combust only when the spark plug arcs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the key chemical characteristics that determine this rating? What is the molecular-level explanation? And how do the various non-hydrocarbon additives work to modify the result?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="30836" LastEditDate="2019-04-18T08:45:22.667" LastActivityDate="2019-04-18T08:45:22.667" Title="What are the key chemical characteristics that determine the octane rating of fuel?" Tags="&lt;reaction-mechanism&gt;&lt;combustion&gt;&lt;fuel&gt;&lt;petrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2455" PostTypeId="2" ParentId="2454" CreationDate="2012-10-27T13:06:50.400" Score="8" Body="&lt;p&gt;Combustion&#xA; is (usually) chain radical reaction with heavy branching. More &#xA;branching means faster combustion and less branching is slower &#xA;combustion. And detonation is wave of super-fast combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you can suppress branching or deaden some branches, you can &#xA;increase durability of air-fuel mixture. Actually, this means, that some&#xA; radical trap is needed. The oldest trap was tetraethyllead, but it is &#xA;currently dropped (lead oxides as fine dust is not a health friendly &#xA;substance). Non-saturated hydrocarbons of different types also can trap &#xA;active radicals, transforming them into more stable allyl radicals, so &#xA;fuel companies tries to keep as much of them in fuel as possible. Some &#xA;other compounds, like ferrocene, also can be used to increase octane &#xA;rating.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="25206" LastEditDate="2016-03-27T14:54:17.483" LastActivityDate="2016-03-27T14:54:17.483" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2457" PostTypeId="1" CreationDate="2012-10-27T22:13:12.970" Score="6" ViewCount="8951" Body="&lt;p&gt;Scientists are certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium. How? and why?&lt;/p&gt;&#xA;" OwnerUserId="773" LastEditorUserId="73" LastEditDate="2012-10-28T03:07:53.297" LastActivityDate="2020-04-04T06:30:38.627" Title="How can scientists be certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium?" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2458" PostTypeId="2" ParentId="2457" CreationDate="2012-10-27T22:40:47.750" Score="14" Body="&lt;p&gt;You are talking about the first 103 elements.  An element is &lt;em&gt;by definition&lt;/em&gt; identified with the number of protons in its nucleus, the so-called &lt;em&gt;atomic number&lt;/em&gt;.  Any atom with 6 protons is carbon, it cannot be anything else.  So there cannot be any missing elements out to 103, as all of these elements have been confirm discovered by accepted scientific methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here is a link to a CHEM Study film showing how transuranium elements (i.e. atomic number &gt; 92 and man-made) can be separated from one another and verified as distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://archive.org/details/transuranium_elements&quot;&gt;http://archive.org/details/transuranium_elements&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2012-10-27T22:49:19.737" LastActivityDate="2012-10-27T22:49:19.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2459" PostTypeId="1" CreationDate="2012-10-27T22:46:07.703" Score="11" ViewCount="15990" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the thermite reaction&#xA;  $$\ce{2Al(s) + Fe2O3(s) -&amp;gt; 2Fe(l) + Al2O3(s)}$$&#xA;  what is the maximum temperature attainable by the products? Assume reactants at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$ and that all heat capacities remain constant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started by using Hess's law to find the enthalpy change of the reaction. I couldn't find a heat of formation for liquid iron so I assume I have to find that out. I just wrote out the equation&#xA;$$\ce{Fe(s) + heat + enthalpy of fusion -&amp;gt; Fe(s)}$$&#xA;and figured the heat of formation for liquid iron at $25\ \mathrm{^\circ C}$ is around $-302.71\ \mathrm{kJ/mol}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the max temperature I used &#xA;$$\Delta H_\mathrm r=\Delta T\cdot C_\mathrm p(\text{products})$$&#xA;and came out with an equation like &#xA;$$-14445\ \mathrm{kJ/mol}=0.147\ \mathrm{kJ/(mol\ K)}\cdot T_\text{max}$$&#xA;This gives me a negative temperature, so I assume it is wrong. Even if I were to take the absolute value, the answer seems awfully high. I was expecting something around a few thousand degrees not near ten thousand. &lt;/p&gt;&#xA;" OwnerUserId="788" LastEditorUserId="5026" LastEditDate="2018-02-25T08:02:38.993" LastActivityDate="2018-02-25T08:02:38.993" Title="Maximum temperature attainable by thermite reaction" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2460" PostTypeId="1" CreationDate="2012-10-28T15:34:07.627" Score="7" ViewCount="5997" Body="&lt;p&gt;We use Scherrer's equation to calculate crystalline size of a particle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d = \frac{Bλ}{β\cos{θ}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But which theta should we take from the XRD pattern for this calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the following plot shows the diffraction pattern of nano (above) and bulk (below) silicon and it has two diffraction peaks at &lt;span class=&quot;math-container&quot;&gt;$2θ \approx 69^\circ$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2θ \approx 76^\circ$&lt;/span&gt;. Which theta should we take for calculating crystalline size of nano silicon and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fg9X0.png&quot; alt=&quot;Diffractograms of nano and bulk silicon&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="41328" LastEditDate="2019-07-18T14:23:38.437" LastActivityDate="2019-07-19T20:48:48.203" Title="Which theta should be taken from x-ray diffraction pattern for calculating Scherrer's equation?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2461" PostTypeId="1" AcceptedAnswerId="2468" CreationDate="2012-10-28T15:56:41.187" Score="9" ViewCount="1431" Body="&lt;p&gt;Is the half-full rule and full rule followed in the 6th and 7th periods? (Note: Half-full rules is Hund's rule)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; What is the correct electron configuration?&#xA;$$\ce{W = [Xe] 6s^2 4f^{14} 5d^4}$$ doesn't use half full rule, or&#xA;$$\ce{W = [Xe] 6s^1 4f^{14} 5d^5}$$ uses half full rule.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="5026" LastEditDate="2018-03-31T12:50:12.957" LastActivityDate="2018-03-31T12:53:13.657" Title="Is the half-full rule and full rule followed in the 6th and 7th periods?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2462" PostTypeId="1" AcceptedAnswerId="2463" CreationDate="2012-10-28T16:19:08.100" Score="7" ViewCount="9907" Body="&lt;p&gt;I drew the two chair conformers of  $ \beta $-D-glucose:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dwy92.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is more stable?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's the one with all of the OH groups in the equatorial position because of less steric hindrance, but I also know that the anomeric effect makes the axial one favored as well.  So I'm unsure which is more stable.&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="21296" LastEditDate="2017-01-02T16:39:39.003" LastActivityDate="2017-11-05T10:29:22.157" Title="Which conformer of beta D-glucose is more stable?" Tags="&lt;carbohydrates&gt;&lt;conformers&gt;&lt;stereoelectronics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2463" PostTypeId="2" ParentId="2462" CreationDate="2012-10-28T22:11:36.803" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I also know that the anomeric effect makes the axial one favored as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The anomeric effect does stabilize an axial group, but &lt;em&gt;only&lt;/em&gt; at the anomeric position. Other axial groups are not similarly stabilized. Thus $\alpha$-D-glucose is more stable than you think it should be based on steric considerations alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, which is the more stable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - very pretty chairs. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-28T22:11:36.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2464" PostTypeId="2" ParentId="2459" CreationDate="2012-10-28T22:39:36.233" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This gives me a negative temperature, so I assume it is wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remember that the energy given off as an exothermic reaction progresses is absorbed by the surroundings. The chemical potential energy stored in aluminum and rust is converted into kinetic energy. You are interested in $\Delta H_{surr}$, which should be $-\Delta H_{sys}$. The &quot;system&quot; in this case is the electrons of the materials engaged in various bonding patterns, and the &quot;surroundings&quot; are the bulk materials.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was expecting something around a few thousand degrees not near ten thousand. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would agree. Both aluminum oxide and iron melt and then boil below 10000 &lt;sup&gt;o&lt;/sup&gt;C. Are you sure you are determining the $\Delta _f H^o$ of $\ce{Fe}(l)$ correctly?  Melting is an endothermic process (absorbs heat from the surroundings).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-28T22:39:36.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2465" PostTypeId="2" ParentId="2459" CreationDate="2012-10-28T22:50:37.907" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermite&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; the maximum temperature is approximately 2500 °C, limited by the boiling point of aluminum, which is 2519 °C.  The article provides some discussion of how the temperature is limited, and how to calculate it using Hesse's law.&lt;/p&gt;&#xA;" OwnerUserId="789" LastEditorUserId="30" LastEditDate="2012-10-30T21:30:24.950" LastActivityDate="2012-10-30T21:30:24.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2466" PostTypeId="1" CreationDate="2012-10-28T23:38:00.297" Score="9" ViewCount="5510" Body="&lt;p&gt;I recently made self-assembled monolayers (SAMs) of alkanethiols on a copper substrate. After scrubbing off the $\ce{Cu2O}$ top layer and rubbing the surface with ether and ethanol, I let my copper plates stand in a 5% (w/w) $\ce{NaOH}$ solution for 5 minutes, which I then throughly rinsed off with deionized water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if I could skip this step. I'm not sure what it is needed for. My knowledge of chemistry tells me that $\ce{Cu(0)}$ should not react with sodium hydroxide. I'm also positive that any grease or other contaminants were washed off in the previous steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="4231" LastEditDate="2015-02-24T20:45:55.737" LastActivityDate="2015-02-24T20:45:55.737" Title="Treatment of copper (0) with sodium hydroxide" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2467" PostTypeId="2" ParentId="2459" CreationDate="2012-10-29T07:49:53.450" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am coming out with an equation like -14445 kJ/mol = 0.147 kJ/(mol K) * maxT&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is wrong. $C_p$ is temperature dependent, so you have to use a more sophisticated equation with integrals. Moreover, parts for heating liquid and solid phases will be different.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-10-29T07:49:53.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2468" PostTypeId="2" ParentId="2461" CreationDate="2012-10-29T10:25:03.030" Score="3" Body="&lt;p&gt;&lt;strong&gt;Short answer: no!&lt;/strong&gt; The first known exception to occur in the periodic table is in period 5:  niobium’s ground state electronic configuration is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Nb: [Kr] 5s^1 4d^4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let’s look at Wikipedia’s &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_configuration#Other_exceptions_to_Madelung.27s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of exceptions to Madelung’s rule&lt;/a&gt;. In period 4, the only exceptions are &lt;strong&gt;$\ce{Cr}$&lt;/strong&gt; and &lt;strong&gt;$\ce{Cu}$&lt;/strong&gt;, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel#Electron_configuration_dispute&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ni&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In period 5, the exceptions to Madelung’s rule are classified in three groups:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;$\ce{Mo}$&lt;/strong&gt; and &lt;strong&gt;$\ce{Ag}$&lt;/strong&gt;: they are the analogues of &lt;strong&gt;Cu&lt;/strong&gt; and &lt;strong&gt;Cr&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;$\ce{Pd}$&lt;/strong&gt; is $\ce{5s^0 4d^10}$: a different type of application of the explanation that &lt;em&gt;“fully filled shells are particularly stable”&lt;/em&gt;, because in this case &lt;em&gt;two&lt;/em&gt; electrons from the s shell were pulled into the d shell. Not what you usually learn.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;$\ce{Nb}$&lt;/strong&gt; is $\ce{5s^1 4d^4}$,  and &lt;strong&gt;$\ce{Ru}$&lt;/strong&gt; is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5026" LastEditDate="2018-03-31T12:53:13.657" LastActivityDate="2018-03-31T12:53:13.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2469" PostTypeId="1" AcceptedAnswerId="5243" CreationDate="2012-10-29T15:11:32.923" Score="61" ViewCount="51579" Body="&lt;p&gt;As cited in an answer to &lt;a href=&quot;https://chemistry.stackexchange.com/q/2461/30&quot;&gt;this question&lt;/a&gt;, the ground state electronic configuration of niobium is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Nb: [Kr] 5s^1 4d^4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is that so? What factors stabilize this configuration, compared to the obvious $\ce{5s^2 4d^3}$ (Aufbau principle), or the otherwise possible $\ce{5s^0 4d^5}$ (half-filled shell)?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5026" LastEditDate="2018-03-31T12:47:12.580" LastActivityDate="2019-08-22T15:31:51.407" Title="How can one explain niobium’s weird electronic configuration?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="2470" PostTypeId="2" ParentId="2460" CreationDate="2012-10-29T19:16:37.507" Score="4" Body="&lt;p&gt;Scherrer’s formula is a simple equation for the simple case of a single peak broadened only due to crystalline particle size. Application of the formula to other, more complex cases, gives only an estimate. Quoting &lt;a href=&quot;http://en.wikipedia.org/wiki/Shape_factor_%28X-ray_diffraction%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is important to realize that &lt;strong&gt;the Scherrer formula provides a lower bound on the particle size&lt;/strong&gt;. The reason for this is that a variety of factors can contribute to the width of a diffraction peak; besides crystallite size, the most important of these are usually inhomogeneous strain and instrumental effects. If all of these other contributions to the peak width were zero, then the peak width would be determined solely by the crystallite size and the Scherrer formula would apply. If the other contributions to the width are non-zero, then the crystallite size can be larger than that predicted by the Scherrer formula, with the &quot;extra&quot; peak width coming from the other factors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scherrer formula gives an estimate&lt;/strong&gt;, an order of magnitude. Here, calculate the particle sizes for the two peaks, and they should be close. The differences are covered by the large uncertainty of your resulting value.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-29T19:16:37.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2471" PostTypeId="2" ParentId="2466" CreationDate="2012-10-29T19:24:05.597" Score="4" Body="&lt;p&gt;$\ce{Cu(0)}$ will not be attacked by $\ce{NaOH}$, and will not decompose in this aqueous environment at room temperature. So, this is not to create some reaction with the copper. Two uses I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{NaOH}$ increases the solubility of $\ce{Cu2O}$, so maybe this is a way of getting rid of the last remaining copper oxide.&lt;/li&gt;&#xA;&lt;li&gt;To bind the thiols with copper, you'd better be in alkaline conditions rather than acidic conditions, right? If so, then maybe this is the way of making sure you don't have any acidic groups or molecules remaining adsorbed on the copper surface, to improve bonding at the next step.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-24T20:45:52.407" LastActivityDate="2015-02-24T20:45:52.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2472" PostTypeId="1" AcceptedAnswerId="2475" CreationDate="2012-10-29T19:31:53.287" Score="6" ViewCount="500" Body="&lt;p&gt;I am currently exploring the synthesis of xRNA, which contains benzo-expanded purine and pyrimidines. I have found &lt;a href=&quot;http://pubmedcentralcanada.ca/pmcc/articles/PMC3039074/&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; which deals with exactly this subject, and provides the following synthetic pathway for the xA ribonucleotide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Osaid.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the first 5 synthetic steps are omitted from the paper, and are likely to be found in one of the following three papers, none of which are publicly available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/14746479&quot; rel=&quot;noreferrer&quot;&gt;1.&lt;/a&gt; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/15651812&quot; rel=&quot;noreferrer&quot;&gt;2.&lt;/a&gt; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/819479&quot; rel=&quot;noreferrer&quot;&gt;3.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I am interested in &lt;strong&gt;any synthetic pathway which will lead me from the first to the second molecule shown above&lt;/strong&gt;. I understand the presented material, but as an undergraduate cell biologist, I am a little rusty on my synthetic organic chem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The first thing that came to mind was the condensation of a biscarbonyl electrophile and bisamine nucleophile like in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrimidine#Organic_synthesis&quot; rel=&quot;noreferrer&quot;&gt;pyrimidine synthesis&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2012-10-30T07:23:36.123" Title="Adenine precursor synthesis" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2473" PostTypeId="1" CreationDate="2012-10-29T19:52:50.937" Score="-6" ViewCount="171" Body="&lt;p&gt;The beginning of the picture, from the word 'Acrylonitril' to the word 'structuurformule:', means the following: &lt;em&gt;Acrylotitrile reacts with 1,4-butanediamine. When you use a lot of acrylonitrile, under certain circumstances you almost exclusively get the following:&lt;/em&gt; Then you can see the first of the 2 structural formulas. The text between the 2 structural formulas, says &lt;em&gt;This 'bond' can react with hydrogen. You'll get addition. When this addition finished, you get a bond with the following structural formula:&lt;/em&gt; And then you see the picture. The final blob of text: &lt;em&gt;We can make polymer molecules which keep branching on and on because of reactions with hydrogen and acrylonitrile.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final question is: &lt;em&gt;We start with 1 molecule of 1,4-butanediamine. Explain how many molecules of hydrogen are needed to make a polymer molecule with 16 $\ce{NH2}$ groups, assume that all reactions are complete.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer should be 56, but I have no clue why, can someone explain? I have half of 56, I just did 14x2, because you get 2 $\ce{H2}$ molecules per added acrylonitrile, and you have to get 14 acrylonitrile molecules to get to 16 $\ce{NH2}$ groups, so that would equal 2x14=28.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJR0v.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="194" LastEditDate="2018-02-05T02:40:45.590" LastActivityDate="2018-02-05T02:40:45.590" Title="Number of hydrogen molecules necessary in acetonitrile/butanediamine polymerisation" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-25T02:43:27.477" ContentLicense="CC BY-SA 3.0" />
  <row Id="2474" PostTypeId="1" AcceptedAnswerId="4358" CreationDate="2012-10-29T22:57:58.560" Score="6" ViewCount="3295" Body="&lt;p&gt;When I prepare sodium perborate by mixing sodium hydroxide with metaborate ($\ce{NaBO}$) and hydrogen peroxide and cooling it I applied a test to it by acidifying it with $\ce{HCl}$ and adding $\ce{KMnO4}$. A precipitate appeared and it turned black. What is the reason? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested another sample of sodium perborate by acidifying it with $\ce{NH4Cl}$ and after that KI and this turned white. Again, why?&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="4231" LastEditDate="2015-02-26T16:22:01.527" LastActivityDate="2018-10-16T22:02:14.557" Title="Tests on sodium peroxoborate" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2475" PostTypeId="2" ParentId="2472" CreationDate="2012-10-30T07:23:36.123" Score="4" Body="&lt;p&gt;Well, you are right, the synthesis of your second compound is described in DOI:&lt;a href=&quot;http://dx.doi.org/10.1021/ja038384r&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1021/ja038384r&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3G4Jq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6301.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-30T07:23:36.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2476" PostTypeId="1" AcceptedAnswerId="2477" CreationDate="2012-10-30T10:39:17.503" Score="15" ViewCount="1115" Body="&lt;p&gt;I noted, flipping through the paperback Alfa Aesar catalogue today, that they sell &lt;a href=&quot;http://www.alfa.com/en/GP100w.pgm?DSSTK=042369&quot;&gt;deuterium-depleted water&lt;/a&gt;. Under the usage category, they list 'NMR'. I'm trying to work out what specific use this water has. Unfortunately, on the &lt;a href=&quot;http://stevekochscience.blogspot.com.au/2011/03/i-am-maximally-skeptical-that-there.html&quot;&gt;basis of very limited clinical evidence&lt;/a&gt; of improved cancer outcomes I've been bombarded mostly with companies recommending deuterium depleted water for drinking, which trips my snake oil sensor and is something I doubt Alfa has in mind (at US$1930 per liter, and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Drinking_water#Requirements&quot;&gt;US RDI of 3.7 L/day for an adult male&lt;/a&gt;, that would seem to be a comically expensive drinking habit.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect biophysics experiments are a small part of the market for this water, but that there's some other NMR-related use I can't think of, possibly calibration related. What is it?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-14T11:03:52.347" Title="What is deuterium-depleted water actually used for?" Tags="&lt;nmr-spectroscopy&gt;&lt;water&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2477" PostTypeId="2" ParentId="2476" CreationDate="2012-10-30T12:57:16.053" Score="13" Body="&lt;p&gt;When you perform &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_NMR&quot;&gt;&lt;sup&gt;1&lt;/sup&gt;H-NMR spectroscopy&lt;/a&gt; in solution, there are many cases where you want to use deuterated solvent, so that signals coming from solvent hydrogen nuclei don’t interfere with the signal from your target molecule. Hence the frequent use of deuterated water, deuterated acetone, deuterated methanol, and deuterated chloroform. Aprotic solvents are also in common use, including carbon tetrachloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is reversed when you want to perform &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deuterium_NMR&quot;&gt;&lt;sup&gt;2&lt;/sup&gt;D-NMR&lt;/a&gt;: you then need deuterium-free solvents, so their signal doesn’t mess with the molecule you're interested in&lt;/strong&gt;. Hence, you will be interested in buying deuterium-depleted water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you're wondering about real use cases for solution deuterium NMR, you can browse the literature for articles, there are plenty. One recent exemple is &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/bi300281k&quot;&gt;this one&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-30T12:57:16.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2478" PostTypeId="2" ParentId="2473" CreationDate="2012-10-30T16:54:17.890" Score="9" Body="&lt;p&gt;This question appears to be about the preparation of polyacrylonitrile dendrimers. Each addition of acrylonitrile completely reacts away the NH&lt;sub&gt;2&lt;/sub&gt; groups previously present:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;assume that all reactions are complete. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your answer of 28 assumes that the reactions could be controllable to allow partial consumption of NH&lt;sub&gt;2&lt;/sub&gt; groups, which requires less addition of acrylonitrile and fewer equivalents of H&lt;sub&gt;2&lt;/sub&gt;. &#xA;You need 16 NH&lt;sub&gt;2&lt;/sub&gt; groups, which is 8 on each side. This is the third generation dendrimer: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Generation 0 is butanediamine (2 NH&lt;sub&gt;2&lt;/sub&gt; groups). Let $G_n$ represent the interior region of the $n^{th}$ generation. &#xA;$$\ce{NH2CH2CH2CH2NH2} \equiv \ce{NH2}G_0\ce{NH2}$$&lt;/li&gt;&#xA;&lt;li&gt;Generation 1 involves reaction of 4 acrylonitriles followed by 8 molecules of hydrogen to make 4 NH&lt;sub&gt;2&lt;/sub&gt; groups.&#xA;$$\ce{NH2}G_0\ce{NH2 +4CH2=CHCN-&amp;gt;(NCCH2CH2)2N}G_0\ce{N(CH2CH2CN)2} $$&#xA;$$\ce{(NCCH2CH2)2N}G_0\ce{N(CH2CH2CN)2 +8H2-&amp;gt; (NH2CH2CH2CH2)2N}G_0\ce{N(CH2CH2CH2NH2)2} \equiv \ce{(NH2)2}G_1\ce{(NH2)2} $$&lt;/li&gt;&#xA;&lt;li&gt;Generation 2 involves reaction of 8 acrylonitriles followed by 16 molecules of hydrogen to make 8 NH&lt;sub&gt;2&lt;/sub&gt; groups.&#xA;$$ \ce{(NH2)2}G_1\ce{(NH2)2 +8CH2=CHCN-&amp;gt; [(NCCH2CH2)2N]2}G_1\ce{[N(CH2CH2CN)2]2} $$&#xA;$$\ce{[(NCCH2CH2)2N]2}G_1\ce{[N(CH2CH2CN)2]2 +16H2 -&amp;gt; [(NH2CH2CH2CH2)2N]2}G_1\ce{[N(CH2CH2CH2NH2)2]2} \equiv \ce{(NH2)4}G_2\ce{(NH2)4}$$&lt;/li&gt;&#xA;&lt;li&gt;Generation 3 involves reaction of 16 acrylonitriles followed by 32 molecules of hydrogen to make 8 NH&lt;sub&gt;2&lt;/sub&gt; groups.&#xA;$$ \ce{(NH2)4}G_2\ce{(NH2)4 +16CH2=CHCN-&amp;gt; [(NCCH2CH2)2N]4}G_2\ce{[N(CH2CH2CN)2]4} $$&#xA;$$\ce{[(NCCH2CH2)2N]4}G_2\ce{[N(CH2CH2CN)2]4 +32H2 -&amp;gt; [(NH2CH2CH2CH2)2N]4}G_2\ce{[N(CH2CH2CH2NH2)2]4} \equiv \ce{(NH2)8}G_3\ce{(NH2)8}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total number of H&lt;sub&gt;2&lt;/sub&gt; molecules = 8 + 16 + 32 = 56 starting at $G_0$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-10-30T16:54:17.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2479" PostTypeId="1" CreationDate="2012-10-30T17:13:43.770" Score="15" ViewCount="1987" Body="&lt;p&gt;I know two distinct recipes for cleaning silver:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.&lt;/li&gt;&#xA;&lt;li&gt;Gently clean silverware with a fine chalk powder (obtained from a pharmacy).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd be happy to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?&lt;/li&gt;&#xA;&lt;li&gt;How toxic (if at all) are the byproducts?&lt;/li&gt;&#xA;&lt;li&gt;How much silver is lost in such a procedure?&lt;/li&gt;&#xA;&lt;li&gt;Are these procedures &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Idempotent&quot;&gt;idempotent&lt;/a&gt;&quot;? I mean, if I apply one of these procedures to an already cleaned item, will it consume some more silver, or will it remain intact? In other words, do the reactions consume just tarnish or do they consume pure silver too?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="793" LastEditorUserId="73" LastEditDate="2012-11-01T00:41:55.090" LastActivityDate="2018-08-22T08:53:07.247" Title="What happens during cleaning silverware?" Tags="&lt;everyday-chemistry&gt;&lt;purification&gt;&lt;silver&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2480" PostTypeId="2" ParentId="2479" CreationDate="2012-10-30T20:12:55.270" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am familiar with this process. I am not familiar with the other. Perhaps someone else can provide an answer for it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This process cleans silver by a redox process. Aluminum is a more &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;active metal&lt;/a&gt; than silver. This means that aluminum metal, Al(s), will reduce silver salts (salts of Ag&lt;sup&gt;+&lt;/sup&gt;). The most common type of tarnish on silver is silver oxide, Ag&lt;sub&gt;2&lt;/sub&gt;O.&#xA;The reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag2O(s) + 2Al(s) -&amp;gt; 3Ag(s) + Al2O3(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The club soda - containing sodium bicarbonate - plays two roles. First it acts as a supporting electrolyte so that electrons can be conducted through the solution. Second, the carbonic acid in it dissolves the &lt;a href=&quot;http://en.wikipedia.org/wiki/Passivation_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;passivation layer&lt;/a&gt; of aluminum oxide that protects the aluminum foil. This produces aluminum carbonate, which is unstable in water and decomposes into amphoteric aluminum hydroxide, which is soluble through the formation of a number of complex ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al2O3(s) + 3H2CO3(aq) -&amp;gt; Al2(CO3)2(s) + 3H2O(l)-&amp;gt; 2Al(OH)3 (aq) + 3CO2(g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net reaction is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -&amp;gt; 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The byproducts are carbon dioxide and aluminum hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How toxic (if all) are the byproducts?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not very. &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_hydroxide#Potential_adverse_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aluminum salts&lt;/a&gt; have very low health impact on you or the environment. Additionally, you produce only very small amounts of aluminum salts this way.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much silver is lost in such a procedure?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;None (Almost). The process coverts solid silver oxide into solid silver metal, both of which are negligibly soluble in club soda (or any other aqueous solution). The $K_{sp}$ of AgOH (which is what the surface of Ag&lt;sub&gt;2&lt;/sub&gt; looks like when exposed to water) is $1.52\times10^{-8}$. All of the action happens at the surface of both metals. The only common silver salts of appreciable solubility in aqueous solutions are silver nitrate and silver sulfate. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this procedure &quot;idempotent&quot;? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes. This procedure converts silver oxide into metallic silver. It has no effect on metallic silver: metallic silver does not react with most acids; it cannot be reduced farther by aluminum metal; and it is not soluble in water or most aqueous solutions.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5078" LastEditDate="2018-08-22T08:53:07.247" LastActivityDate="2018-08-22T08:53:07.247" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2481" PostTypeId="1" CreationDate="2012-10-30T20:52:09.687" Score="4" ViewCount="130" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit&quot;&gt;Can an atom have more than 8 valence electrons? If not, why is 8 the limit?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I was watching this youtube clip ( &lt;a href=&quot;http://www.youtube.com/watch?v=XY1XLgHUAIQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.youtube.com/watch?v=XY1XLgHUAIQ&lt;/a&gt; ) in which a lewis diagram was drawn for a sulfite ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur ends up with four bonds as well as a lone pair of electrons. This gives sulfur 10 electrons. Isn't this too many?&lt;/p&gt;&#xA;" OwnerUserId="775" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-10-30T20:52:09.687" Title="Is it okay for a sulfite ion to have 10 electrons around the sulfur?" Tags="&lt;bond&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2012-10-30T21:27:24.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="2482" PostTypeId="2" ParentId="2445" CreationDate="2012-10-30T21:05:07.323" Score="8" Body="&lt;p&gt;Briefly, the cause for surface tension is the difference in the energetic state between a molecule at the surface compared to a molecule inside the liquid phase: the molecule inside has lower energetic state as it has more inermolecular interactions (or, to get to the surface, a molecule needs to break intermolecular bonds, which needs energy). This is more pronounced with higher polarity or when ions (charges) are around as the interactions are then stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can increase the polarity of a (polar) solvent to get higher surface tension, by dissolving &lt;em&gt;salts&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2012-10-30T21:05:07.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2484" PostTypeId="2" ParentId="1045" CreationDate="2012-10-31T01:33:05.630" Score="3" Body="&lt;p&gt;To easily determine the oxidation number on an individual atom in an ion it is necessary to write an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First determine what the oxidation number of your known atoms are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we know the oxidation number for H is +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then set this value equal to the overall net charge of the ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it is +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our equation now looks like this:  1(4) = 1, You use the multiplier of 4 to indicate that the ammonium ion has 4 hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next substitute a variable in the equation for the missing oxidation number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1(4) + N = +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1(4) + N = +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N + 4 = +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = +1 - 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the oxidation number for Nitrogen is -3.&lt;/p&gt;&#xA;" OwnerUserId="795" LastActivityDate="2012-10-31T01:33:05.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2485" PostTypeId="2" ParentId="2439" CreationDate="2012-10-31T11:27:48.270" Score="1" Body="&lt;p&gt;A simple test you can do is to boil a large amount of the water and see if a sediment is left in the vessel. This will tell you certain things about you contaminant.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it particulate? We expect so because a filter is removing the contaminant.&lt;/li&gt;&#xA;&lt;li&gt;Is it's boiling point below that of water?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you get a residue what is it's state? I'm expecting a solid, I dont anticipate a coloured liquid that is miscible in water with a boiling point higher than water.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does it burn? How does it burn? Colour of flame? Further residue?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd also try adding a colourless surfactant to the water, does it remain coloured?&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2012-10-31T11:27:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2486" PostTypeId="1" AcceptedAnswerId="2487" CreationDate="2012-10-31T15:33:36.210" Score="1" ViewCount="313" Body="&lt;p&gt;In a $\pu{Zn-Cu}$ voltaic cell, the electrolyte is sodium chloride solution. The question was: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this cell, which of the following undergoes reduction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer was copper, but why is the exact answer &quot;a hydrogen ion&quot;?&lt;/p&gt;&#xA;" OwnerUserId="797" LastEditorUserId="5026" LastEditDate="2018-02-20T15:48:51.383" LastActivityDate="2018-02-20T15:48:51.383" Title="What undergoes reduction in a zinc-copper voltaic cell?" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2487" PostTypeId="2" ParentId="2486" CreationDate="2012-10-31T15:59:13.040" Score="2" Body="&lt;p&gt;Copper cannot undergo reduction because, in the list of components of the cell you listed, it is only present in its most reduced state, Cu(0). Thus, it cannot be copper that is reduced. See &lt;a href=&quot;http://scienceline.ucsb.edu/getkey.php?key=2466&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for additional hints about your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have been confused by the fact that, in the historical (and widely taught) &lt;a href=&quot;http://en.wikipedia.org/wiki/Daniell_cell&quot; rel=&quot;nofollow&quot;&gt;Daniell cell&lt;/a&gt;, copper is indeed reduced: it is present as Cu(+II) in a solution of copper sulfate.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-10-31T15:59:13.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2488" PostTypeId="1" AcceptedAnswerId="5261" CreationDate="2012-10-31T22:36:10.613" Score="6" ViewCount="4819" Body="&lt;p&gt;I drew the conformer (below) and I'm wondering if this is the most stable chair conformation.&lt;img src=&quot;https://i.stack.imgur.com/fBhzx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three of the bonds are equatorial so it looks like the most stable to me.&lt;/p&gt;&#xA;" OwnerUserId="384" LastActivityDate="2013-06-09T15:10:30.403" Title="Is this the lowest chair conformer of Beta D-galactose" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2490" PostTypeId="1" AcceptedAnswerId="2491" CreationDate="2012-11-01T02:44:23.990" Score="10" ViewCount="1378" Body="&lt;p&gt;If we have a benzene with a single substituent, such as chlorobenzene, what do we call the carbon atom to which the group is attached? I'm familiar with the ortho- meta- para- nomenclature, but the last one escapes me.&lt;/p&gt;&#xA;" OwnerUserId="798" LastEditorUserId="22" LastEditDate="2012-11-02T07:45:22.450" LastActivityDate="2012-11-02T07:45:22.450" Title="What is the name of the &quot;primary&quot; carbon in a benzene derivative?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2491" PostTypeId="2" ParentId="2490" CreationDate="2012-11-01T07:26:28.077" Score="7" Body="&lt;p&gt;It's &lt;em&gt;ipso&lt;/em&gt;-.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Arene_substitution_patterns#Ipso.2C_meso.2C_and_peri_substitution&quot;&gt;Wikipedia page&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;IUPAC &lt;em&gt;&lt;a href=&quot;http://goldbook.iupac.org/index.html&quot;&gt;Gold Book&lt;/a&gt;&lt;/em&gt; &lt;a href=&quot;http://goldbook.iupac.org/I03251.html&quot;&gt;definition of the &lt;em&gt;ipso&lt;/em&gt;-attack&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="84" LastEditorUserId="30" LastEditDate="2012-11-01T10:45:01.897" LastActivityDate="2012-11-01T10:45:01.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2492" PostTypeId="1" AcceptedAnswerId="2494" CreationDate="2012-11-01T10:39:09.543" Score="15" ViewCount="1071" Body="&lt;p&gt;For some DFT calculations I performed with localized (Gaussian) basis sets, I wonder how the basis sets I used are properly named. The names I could gather are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H, C and O: 6-311G(d,p)&lt;/li&gt;&#xA;&lt;li&gt;Al: 8-511G* &lt;/li&gt;&#xA;&lt;li&gt;Ga: 86-4111d41G&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(They all come from &lt;a href=&quot;http://www.crystal.unito.it/Basis_Sets/Ptable.html&quot;&gt;this website&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am not sure that the above names are consistent. Specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Shouldn't I write 6-311G** instead of 6-311G(d,p), for consistency with 8-511G*?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;86-4111d41G looks weird to me, but I don't know how it could otherwise be named. My knowledge of basis sets naming conventions didn't really cover anything beyond the third period of the periodic table!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-30T00:01:26.543" Title="Naming of quantum chemistry basis sets" Tags="&lt;computational-chemistry&gt;&lt;basis-set&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2493" PostTypeId="1" AcceptedAnswerId="2497" CreationDate="2012-11-02T00:32:19.440" Score="12" ViewCount="13921" Body="&lt;p&gt;When carbon combusts with oxygen, is this considered a redox reaction since the oxygen atoms gain electrons and the carbon atoms lose them?&lt;/p&gt;&#xA;" OwnerUserId="799" LastEditorUserId="73" LastEditDate="2012-11-02T03:46:47.397" LastActivityDate="2019-08-19T01:23:47.333" Title="Is combustion considered a redox reaction?" Tags="&lt;redox&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2494" PostTypeId="2" ParentId="2492" CreationDate="2012-11-02T12:09:58.127" Score="8" Body="&lt;p&gt;Regarding 6-311G**:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The * notation (one for polarization functions on the heavy atoms and ** for polarization functions on the heavy and hydrogen atoms) is an older notation belonging to the split-valence basis sets of John Pople. Many people still use this notation today but if you want to be more precise and more descriptive, the (d,p) notation is what you would want to use.  In fact, we only ever use the explicit (d,p) type notation and never use the *'s in my lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;X-YZg&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split-valence basis sets have the aforementioned structure for us to quickly and easily determine how that basis set was constructed. The &lt;em&gt;X&lt;/em&gt; simply is the number of primitive Gaussians that describe the core electron shells of an atom. Most of the time you will be freezing core in your computations so you split the core and valence regions of your atoms and assign various basis functions to each one.  The &lt;em&gt;Y&lt;/em&gt; and &lt;em&gt;Z&lt;/em&gt; components mean you have two sets of primitive Gaussian functions which describe your valence region. Because you have &lt;strong&gt;two&lt;/strong&gt; sets here, you can denote this basis set as a &lt;em&gt;double-zeta&lt;/em&gt;. In your case, a 6-311G** is in fact a &lt;em&gt;triple-zeta&lt;/em&gt; quality basis set, which is better than a &lt;em&gt;double-zeta&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of interest, a 6-31+G* includes a + symbol. This simply means that diffuse functions would be added to the heavy atoms.  A ++ symbol would extend these diffuse functions to the hydrogen atoms. Of course we would never want to use the '*' notation so I should have written this as 6-31+G(d,p).  Of course polarization functions can come in sets as well such as 6-31+G(2df,2pd), and other combinations exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, &lt;strong&gt;when using split-valence basis sets such as 6-311G*&lt;/strong&gt;, &lt;strong&gt;you must be very careful to check the manual and see what function space the program is going to use by default&lt;/strong&gt;.  I will use Gaussian 09 as an example.  By default, Gaussian 09 uses spherical harmonics as its default... EXCEPT FOR A FEW DIFFERENT TYPES OF BASIS SETS including the family of 6-31G* type basis sets which uses the CARTESIAN SPACE BY DEFAULT.  This means that rather than the 5D, 7F stuff that you may be used to using on a regular basis, invoking the 6-31G* type basis set will automatically use the 6D, 10F function set which inevitably changes the size of your basis set which means that you may have data that is not readily comparable to other things.  I always opt out for the 5D, 7F type functions.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2013-10-30T00:01:26.543" LastActivityDate="2013-10-30T00:01:26.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2495" PostTypeId="1" AcceptedAnswerId="2496" CreationDate="2012-11-02T18:15:06.720" Score="17" ViewCount="71966" Body="&lt;p&gt;I am studying different means of collecting water for consumption. From resources available on Internet, I understand that distilled water is not advised for regular consumption because of the lack of minerals and salts, while rainwater can be consumed if not overloaded with unwanted chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me is that I know water to be pouring down from clouds, which is water vapor turning into droplets through condensation. To me, this looks to be exactly the same as the process of distillation, i.e. evaporation and condensation of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is my assumption about rainwater formation wrong? What is the difference between rainwater and distilled water? What causes this difference to happen?&lt;/p&gt;&#xA;" OwnerUserId="801" LastActivityDate="2015-12-22T19:37:03.747" Title="What is the difference between distilled water and rainwater?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2496" PostTypeId="2" ParentId="2495" CreationDate="2012-11-02T22:16:37.977" Score="11" Body="&lt;p&gt;Rain water passes through the atmosphere and picks up whatever is present there, including particulates such as dust, trace amounts of atmospheric gases, nitrates and nitrites, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rain water is relatively low in mineral content.  In practice, people drink rainwater from streams and wells, where it has had ample opportunity to pick up minerals from soil and rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=rain+water+composition&quot;&gt;https://www.google.com/search?q=rain+water+composition&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="803" LastActivityDate="2012-11-02T22:16:37.977" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2497" PostTypeId="2" ParentId="2493" CreationDate="2012-11-02T23:23:37.097" Score="8" Body="&lt;p&gt;In general, yes.  If the reaction involves oxygen going from the oxidation state of zero in $\ce{O_2}$ to an oxidation state of -2, then there is oxidation and the reaction is a redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-02T23:23:37.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2498" PostTypeId="1" AcceptedAnswerId="2503" CreationDate="2012-11-03T05:35:19.907" Score="11" ViewCount="6166" Body="&lt;p&gt;What alloys can be used to make a metal pencil like &lt;a href=&quot;http://www.wired.com/gadgetlab/2010/09/neither-pen-nor-pencil-write-endlessly-in-metal/&quot;&gt;this one&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HqANd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="804" LastEditorUserId="30" LastEditDate="2012-11-04T10:24:50.870" LastActivityDate="2017-12-12T20:42:19.593" Title="Alloys for inkless metal pen" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2499" PostTypeId="1" AcceptedAnswerId="2557" CreationDate="2012-11-03T05:43:53.620" Score="8" ViewCount="12015" Body="&lt;p&gt;Would a high voltage ( of the order of a few score kV ) drive electrolysis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, does it require a large current and low voltage? Alternatively, does electrolysis require both large voltage &amp;amp; large current?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="22" LastEditDate="2012-11-03T17:12:57.307" LastActivityDate="2017-03-06T04:50:20.580" Title="Does potential difference or current drive electrolysis?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2500" PostTypeId="2" ParentId="2499" CreationDate="2012-11-03T10:11:01.010" Score="5" Body="&lt;p&gt;Current is, by definition, a flow of charged particles, so it doesn't drive anything.  Electrical potential is a store of energy (you can think of a voltage source as an ideal &quot;battery&quot; of sorts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One volt can produce $\pu{1 A}$ of current through a $\pu{1 \Omega}$ resistor.  One ampere of current is equal to a coulomb per second.  One &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb&quot; rel=&quot;nofollow noreferrer&quot;&gt;coulomb&lt;/a&gt; is $6.24150965(16)×10^{18}$ electrons.  Were you to use enough voltage that oxidation at the anode released enough electrons to produce a massive current, you'd likely deplete your anode material before you ever achieved it. &lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="11367" LastEditDate="2017-03-06T04:50:20.580" LastActivityDate="2017-03-06T04:50:20.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2501" PostTypeId="2" ParentId="2404" CreationDate="2012-11-03T13:58:18.847" Score="14" Body="&lt;p&gt;Step one: trap $\ce{CO2}$. This can be done using specially selected amines, like triethanolamine. Aminosilicones (a common chemical used in hair conditioner), can also do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step two: release trapped $\ce{CO2}$. In case of triethanolamine you'll have to boil mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step three: mix $\ce{CO2}$ with hydrogen (hydrogen is produced by electrolysis of water). Volume proportion should be somewhere 1:3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step four: threat the mixture with proper catalyst. This step may be divided into two stages: reverse &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_gas_shift_reaction&quot; rel=&quot;noreferrer&quot;&gt;water-shift reaction&lt;/a&gt;  and some variant of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fischer%E2%80%93Tropsch_process&quot; rel=&quot;noreferrer&quot;&gt;Fischer–Tropsch process&lt;/a&gt; with some type of &lt;a href=&quot;http://dx.doi.org/10.1016/S0166-9834%2800%2982447-1&quot; rel=&quot;noreferrer&quot;&gt;alcohol synthesis&lt;/a&gt; as very interesting candidate. There are &lt;a href=&quot;http://www.google.com/patents/US7858667&quot; rel=&quot;noreferrer&quot;&gt;some&lt;/a&gt; multi-functional catalysts that can perform both stages at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another variation on step 4 is the Sabatier process, which involves generating methane from $\ce{CO2}$ and hydrogen.  Typically the catalyst is Rainy nickel (powder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required energy in form of electrical current is probably from 1.5 to 3 times of heat of combustion of resulting fuel, which is in turn at least twice more that can be converted into actual work again.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="275" LastEditDate="2013-06-18T13:04:30.853" LastActivityDate="2013-06-18T13:04:30.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2502" PostTypeId="1" AcceptedAnswerId="2504" CreationDate="2012-11-03T21:52:22.340" Score="0" ViewCount="632" Body="&lt;p&gt;If I am given the molarity of two substances and I know how they react, how can I write the balanced chemical reaction? i.e. experimentally I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pu{1.0 M}$ $\ce{NH_3(aq)}$ mixed with $\pu{0.1 M}$ $\ce{BaCl_2(aq)}$  yielded &lt;strong&gt;no&lt;/strong&gt; precipitate when $\pu{1ml}$ of each were mixed together. My unbalanced chemical equation so far looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N^{-3}H3+(aq) + (1/10)Ba^{+2}Cl2-(aq) -&amp;gt; N^{-3}Ba^{+2} + Cl^{-2}H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My logic being: since the solutions are administered in equal volumes, their molarity determines the molar coefficients. Right off the bat I can tell that $\ce{NH}$ is &lt;strong&gt;not&lt;/strong&gt; soluble, but the other compound I think is, which would contradict my observations. Either way, I am incredibly lost.&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorUserId="5026" LastEditDate="2018-02-25T02:49:19.763" LastActivityDate="2018-02-25T02:49:19.763" Title="Writing the chemical equation given molarity" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2503" PostTypeId="2" ParentId="2498" CreationDate="2012-11-04T00:02:04.773" Score="7" Body="&lt;p&gt;&quot;Silverpoint&quot; nibs have a long history in art and significantly predate modern drawing tools such as the lead pencil (made of graphite and clay but possibly called &lt;em&gt;lead&lt;/em&gt; as lead metal was one possible material in metal pens). Silverpoint writing and drawing doesn't smudge or fade and many examples have survived almost unchanged over hundreds of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a useful site with a good history of the original artistic uses of silverpoint pens &lt;a href=&quot;http://www.silverpointweb.com&quot;&gt;here&lt;/a&gt;. According to this site a wide variety of metals will work. Historically which was used depended on status:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wealthy and royal persons who needed to keep records (business inventories and so forth) employed these scribes, and the stylus material served as notice to all of the status of the employer - &lt;strong&gt;gold&lt;/strong&gt; and &lt;strong&gt;silver&lt;/strong&gt; for the upper classes, &lt;strong&gt;lead&lt;/strong&gt; for somewhat lesser beings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems tht many metals will work. The dictionary of Art is quoted as saying:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metalpoint. Drawing instrument (the forerunner of the pencil) made from a small, pointed metal tip, usually of lead, silver, copper or gold, encased in a wooden holder. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current preferred metal for artistic use seems to be fairly pure silver. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Best results are obtained from a thicker point that is &quot;pure&quot; silver, or &quot;.999 fine.&quot; This is much softer than sterling, so it will deposit more metal on the drawing surface&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.silverpointweb.com/catalog.html&quot;&gt;silverpoint.com&lt;/a&gt; web site sells ready made points of both silver and gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't obvious what metal or alloy is used in the pens mentioned in the Wired story, but it isn't silver. It apparently contains at least some lead as the pencils come with warnings not to let children chew them!&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2012-11-04T00:02:04.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2504" PostTypeId="2" ParentId="2502" CreationDate="2012-11-04T02:00:48.040" Score="3" Body="&lt;p&gt;Have you considered the possibility that your reaction is wrong.  where, for example, is the barium on the product side.  I'll give you one hint here:  barium chloride dissociates in water to barium and chloride ions.  Ammonia reacts with water to produce some ammonium ion and the same amount of hydroxide ion.  However, barium hydroxide is totally ionized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you take it from there?&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-04T02:00:48.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2506" PostTypeId="1" AcceptedAnswerId="2507" CreationDate="2012-11-05T13:12:20.780" Score="8" ViewCount="25586" Body="&lt;p&gt;How can acetic acid be distilled by freezing when its freezing point is above the freezing point of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples on the web show the water portion of the vinegar as ice and the acid as a liquid but Wikipedia shows that freezing point of acetic acid is 16.6&amp;deg;C.&lt;/p&gt;&#xA;" OwnerUserId="810" LastEditorUserId="275" LastEditDate="2012-11-05T22:07:15.440" LastActivityDate="2013-09-10T11:53:32.203" Title="Acetic acid freezing distilation" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2507" PostTypeId="2" ParentId="2506" CreationDate="2012-11-05T22:06:57.983" Score="12" Body="&lt;p&gt;The procedure you are describing sounds like &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_freezing&quot; rel=&quot;noreferrer&quot;&gt;fractional freezing&lt;/a&gt;. Acetic acid is frequently purified by this method, hence 100% acetic acid is often called &quot;glacial&quot; acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As an example of how this works, let's consider a mixture of acetic acid and water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acetic acid has a freezing point of 16.6 &lt;sup&gt;o&lt;/sup&gt;C. Water has a freezing point of 0  &lt;sup&gt;o&lt;/sup&gt;. They form a eutectic at -26.7 &lt;sup&gt;o&lt;/sup&gt;C (although I do not know the composition). The phase diagram probably looks something like (but is not) the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebvLZ.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Phase Legend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A = Liquid solution&lt;/li&gt;&#xA;&lt;li&gt;B = Solid solution of acetic acid in water PLUS liquid solution&lt;/li&gt;&#xA;&lt;li&gt;C = Solid solution of water in acetic acid PLUS liquid solution&lt;/li&gt;&#xA;&lt;li&gt;D = Solid solution of acetic acid in water &lt;/li&gt;&#xA;&lt;li&gt;E = Solid solution of water in acetic acid &lt;/li&gt;&#xA;&lt;li&gt;F = Solid solution of acetic acid in water PLUS Solid solution of water in acetic acid &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A mixture of acetic acid and water that is approximately 72% acetic acid is cooled until it begins to freeze (solid blue vertical line), which happens around 2 &lt;sup&gt;o&lt;/sup&gt;C. The phase changes from A (liquid solution) to C (Solid solution of water in acetic acid PLUS liquid solution). The solid solution is enriched in acetic acid compared to the original liquid solution. The composition of the remaining supernatant liquid solution is enriched in water compared to the original liquid solution. The composition of the solid solution can be determined by drawing a horizontal (red) line to the phase boundary between C and E and reading down to 91% acetic acid. The composition of the remaining supernatant liquid solution can be determined by drawing a horizontal (green) line to the phase boundary between A and C and reading down to 63% acetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supernatant phase is removed and discarded, and the solid phase is allowed to melt. The process is repeated. Each successive freezing yields a new solid solution that is enriched in acetic acid, until the purity is as high as you want it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that since water and acetic acid have a eutectic point, if we has started on the left side of the eutectic point, the phases would have been reverse. The solid would be enriched in water and the supernatant would have been enriched in acetic acid. The supernatant would be remove and kept, and the solid would be discarded. The supernatant would then be cooled and fractionally frozen again to further enrich the supernatant in acetic acid until the eutectic point is reached (at which point the composition of the supernatant is equivalent to the composition of the solid phase). If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-11-05T22:06:57.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2509" PostTypeId="1" AcceptedAnswerId="2513" CreationDate="2012-11-06T01:44:20.920" Score="23" ViewCount="233107" Body="&lt;p&gt;How do I type a simple chemical equation in Microsoft Word? I can do subscripts, but long arrows are more difficult. I can't get them to align with the text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I can't figure out how to put a delta above the arrow for heat. I have tried the Chemistry add-on from Microsoft, but that does not seem to help with equations. &lt;/p&gt;&#xA;" OwnerUserId="812" LastEditorUserId="402" LastEditDate="2016-10-30T04:41:22.280" LastActivityDate="2018-09-23T13:34:06.520" Title="How do I type a simple chemical equation in Microsoft Word?" Tags="&lt;software&gt;&lt;notation&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2510" PostTypeId="2" ParentId="2509" CreationDate="2012-11-06T06:25:54.750" Score="3" Body="&lt;p&gt;Frequently the best solution is to draw the equation in a chemical drawing software (there are several freeware options) and paste it in as a graphic in Word. &lt;/p&gt;&#xA;" OwnerUserId="806" LastEditorUserId="402" LastEditDate="2017-04-02T20:16:14.103" LastActivityDate="2017-04-02T20:16:14.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2511" PostTypeId="1" AcceptedAnswerId="2527" CreationDate="2012-11-06T06:39:22.960" Score="24" ViewCount="4348" Body="&lt;p&gt;I knew while learning about electrolysis that if the ionic compound is molten it becomes free moving ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, what will happen if I continued heating till it reaches the boiling point so that the ionic compound evaporates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it still be free moving ions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, shouldn't the result be more efficient at electrolysis than in the liquid state due to increased mobility of ions? If not, why?&lt;/p&gt;&#xA;" OwnerUserId="815" LastEditorUserId="22" LastEditDate="2012-11-19T07:12:25.327" LastActivityDate="2020-04-20T01:58:05.457" Title="Will gaseous ionic compounds be free moving ions?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;ionic-compounds&gt;&lt;boiling-point&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2513" PostTypeId="2" ParentId="2509" CreationDate="2012-11-06T11:33:13.983" Score="30" Body="&lt;p&gt;If you are using MS Word 2007 or newer, use the equation feature. It is designed for math but works okay for chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to the insert tab. (For shortcut you can press Alt+= sight together)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBF2v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on the equation button on the far right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ix52E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type in your equation. Use the buttons in the ribbon to do superscripts and subscripts. Alternatively you can use _ for subscript and ^ for superscript.  The default is to have letters italicized (as variables), so you will want to fix that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRCce.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also shortcut commands to render most the common things you want. For example, underscore _ creates a subscript and a caret ^ creates a superscript &lt;a href=&quot;https://www.youtube.com/watch?v=SdzgM9GacCY&amp;amp;t=21s&quot; rel=&quot;noreferrer&quot;&gt;Shortcut for typing subscript and superscript in MS Word 2007|2010|2013|2016 and office 365&lt;/a&gt;  . You have access to a wide range of arrows from a pull-down menu, but -&gt; will give you a simple right arrow (although it is not very long). This feature on Word will also accept some (but not all) &lt;a href=&quot;http://tex.stackexchange.com&quot;&gt;tex&lt;/a&gt; commands for formatting equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a long arrow, click on the operator but and choose the arrow with the word &quot;yields&quot; written over it under common operator structures. For up arrow and down arrow showing gas liberation and precipitation use \uparrow or \downarrow followed by space &lt;a href=&quot;https://cpgupta1201.wordpress.com/2018/09/21/quickest-way-to-type-chemical-reaction-chemical-equation-reaction-arrows-in-word-2007-above/&quot; rel=&quot;noreferrer&quot;&gt;Shortcut for typing arrows of chemical equation in Word 2007 and above&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o0yC6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on the word &quot;yields&quot; and replace it with as many spaces as you need to create an arrow of whatever length you want. Shortcut for other types of arrows is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NjD8m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, finish your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fgTro.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to type above or below arrow just type &quot;\above(text above arrow goes here)[space]&quot;.Similarly tying below arrow just type \below(test below goes here)[space]&quot;. &lt;a href=&quot;https://www.youtube.com/watch?v=IXML84P3P-c&amp;amp;t=29s&quot; rel=&quot;noreferrer&quot;&gt;How to type chemical equation and arrows in Word 2007 and above&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For older versions of MS Word, go to the insert menu and click on the equation, which launches the Equation Editor Program (you can also find this program on your computer by searching for eqnedt.exe), which gives you the same ability to create equations.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="68230" LastEditDate="2018-09-23T13:34:06.520" LastActivityDate="2018-09-23T13:34:06.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2516" PostTypeId="1" CreationDate="2012-11-07T08:55:49.553" Score="10" ViewCount="1041" Body="&lt;p&gt;I recently heard a talk from a physicist about experiments on simple two-atom molecules in vacuum that showed that certain behaviour of the molecules was explained by quantum entanglement. Now this experiment was about creating holes in the inner electron shells of the molecule, so pretty much in the realm of physics and not chemistry. It was also performed in vacuum, and the lecturer himself said that in solution such an effect would likely be destroyed by frequent collisions in the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are any examples of chemical behaviour of molecules that can only be explained by quantum entanglement. Does it ever play a role on the scale of chemistry, or is this effect just not relevant on anything beyond the small moleculs in vacuum stage physicists tend to use?&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2020-01-24T16:08:29.040" Title="Can quantum entanglement affect the chemistry of molecules?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2517" PostTypeId="2" ParentId="2516" CreationDate="2012-11-07T09:16:14.133" Score="7" Body="&lt;p&gt;I've not really read anything on it, but a short while ago research on the role of entanglement in excitonic transfer in the photosynthetic light harvesting apparatus caused quite a stir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an &lt;a href=&quot;http://arxiv.org/pdf/0905.3787&quot;&gt;ArXiV preprint of a paper&lt;/a&gt; on entanglement in an antenna protein from &lt;em&gt;C. tepidum&lt;/em&gt;. It was &lt;a href=&quot;http://www.nature.com/nphys/journal/v6/n6/abs/nphys1652.html&quot;&gt;published&lt;/a&gt; in Nature Physics. The authors note the extraordinary result of entanglement being observed in an inherently quite 'noisy' system.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-11-07T09:16:14.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2520" PostTypeId="2" ParentId="2516" CreationDate="2012-11-07T21:41:38.177" Score="1" Body="&lt;p&gt;I've hear that &lt;a href=&quot;http://en.wikipedia.org/wiki/Superatom&quot; rel=&quot;nofollow&quot;&gt;superatomic&lt;/a&gt; systems are theorized to involved strange quantum behaviour (compared with other molecular systems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-classical behaviour is also believed to be the root of unusual carbon / graphene behaviour, &lt;a href=&quot;http://www.nature.com/nature/journal/v401/n6754/abs/401680a0.html&quot; rel=&quot;nofollow&quot;&gt;such as in C60&lt;/a&gt;, and other higher-order materials / chemistry.&lt;/p&gt;&#xA;" OwnerUserId="296" LastEditorUserId="296" LastEditDate="2012-11-07T21:47:44.597" LastActivityDate="2012-11-07T21:47:44.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2521" PostTypeId="2" ParentId="2511" CreationDate="2012-11-07T22:03:11.670" Score="9" Body="&lt;p&gt;I believe that gaseous ionic compounds such as NaCl are in fact diatomic NaCl.  One can't create a plasma by simply boiling ordinary compounds.  The forces between ions are way too strong.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-07T22:03:11.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2523" PostTypeId="2" ParentId="2511" CreationDate="2012-11-08T04:38:00.133" Score="3" Body="&lt;p&gt;In liquid state, NaCl become free moving ions but the overall system is stable because of the attractions between the oppositely charged ions.&#xA;I believe that gaseous NaCl would exists as very polar covalent or ionic molecules with their overall affect being electrically neutral (no plasma).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a purely intuited answer and should be taken with a grain of salt&#xA;as I could find no numbers to back it up.&#xA;If someone could change this into a comment for me I would be much obliged.&lt;/p&gt;&#xA;" OwnerUserId="755" LastEditorUserId="755" LastEditDate="2012-11-08T05:32:46.947" LastActivityDate="2012-11-08T05:32:46.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2524" PostTypeId="1" AcceptedAnswerId="2528" CreationDate="2012-11-08T04:44:23.343" Score="0" ViewCount="528" Body="&lt;p&gt;I want to know what is the phenomenon and explanation behind this corrosion. What is the reaction?&lt;/p&gt;&#xA;" OwnerUserId="652" LastEditorUserId="5026" LastEditDate="2018-03-04T02:49:28.350" LastActivityDate="2018-03-04T02:49:28.350" Title="Why do lithium and sodium corrode so easily?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2018-03-04T19:09:18.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="2526" PostTypeId="1" CreationDate="2012-11-08T07:25:21.170" Score="5" ViewCount="1453" Body="&lt;p&gt;Dissolution reactions are normally endothermic, and crystallization reactions are normally exothermic, but there are some exceptions -- such as the dissolution of calcium chloride. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="826" LastEditorUserId="22" LastEditDate="2012-11-08T08:17:01.423" LastActivityDate="2012-12-09T13:47:31.920" Title="Why can dissolution reactions (e.g. dissolution of calcium chloride) be exothermic?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2527" PostTypeId="2" ParentId="2511" CreationDate="2012-11-08T09:18:48.230" Score="23" Body="&lt;p&gt;Usually not. Boiling point rarely exceeds 4-5 thousand kelvin. A typical ionic bound energy is about 5 eV. 1 eV is roughly 11 thousand kelvin, so ions in low-temperature vapors exist as molecules. When the temperature becomes enough to break ionic molecules, it is enough to strip one or two electrons from atoms, so high-temperature vapors will be plasma with electrons and positive ions. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="-1" LastEditDate="2020-04-20T01:58:05.457" LastActivityDate="2020-04-20T01:58:05.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2528" PostTypeId="2" ParentId="2524" CreationDate="2012-11-08T11:38:36.560" Score="1" Body="&lt;p&gt;$$\ce{2Li + 2H2O -&amp;gt; 2LiOH + H2}$$&#xA;$$\ce{2LiOH + CO2 -&amp;gt; Li2CO3 + H2O}$$ &#xA;$$\ce{4Li + O2 -&amp;gt; 2Li2O}$$ &#xA;$$\ce{Li2O + CO2 -&amp;gt; Li2CO3}$$&#xA;$$\ce{6Li + N2 -&amp;gt; 2Li3N}$$&#xA;Lithium and sodium are reacting with the gases in the air; a destructive physical process.&lt;/p&gt;&#xA;" OwnerUserId="802" LastEditorUserId="22" LastEditDate="2012-11-08T14:09:53.360" LastActivityDate="2012-11-08T14:09:53.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2529" PostTypeId="1" CreationDate="2012-11-08T15:23:55.700" Score="3" ViewCount="307" Body="&lt;p&gt;I am reading up on boron in my notes and there is a certain paragraph that I do not understand&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tendency of $\ce{B}$ to form $\ce{B-B}$ bonds is less pronounced than that of&#xA;  carbon, its periodic table neighbor on the right, to form $\ce{C-C}$ bond,&#xA;  but it is much greater than that of the nearest element in the group,&#xA;  $\ce{Al}$, to form $\ce{Al-Al}$ bonds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone can explained to me why is $\ce{B-B}$ bonds can be formed in the first place?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="10185" LastEditDate="2015-02-27T09:11:30.717" LastActivityDate="2020-12-24T12:09:00.030" Title="Boron-Boron bonds" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2530" PostTypeId="1" AcceptedAnswerId="2531" CreationDate="2012-11-09T00:18:49.527" Score="11" ViewCount="478" Body="&lt;p&gt;I remember my Physical Chemistry Professor saying that very tiny negative Kelvin temperatures have been achieved on the quantum level. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="829" LastEditorUserId="19616" LastEditDate="2015-10-23T16:55:37.553" LastActivityDate="2015-10-23T16:55:37.553" Title="Negative Kelvin Temperature" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2531" PostTypeId="2" ParentId="2530" CreationDate="2012-11-09T00:49:48.737" Score="20" Body="&lt;p&gt;Yes.  But it is a bit of a joke of a sort.  In any system with energy levels, the lower energy levels never have a lower population than the higher ones.  In fact there is an equilibrium arrangement of the populations of such levels characterized by the temperature.  The higher the temperature, the more evenly populated the levels are.  There is even a formula for calculating the populations in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy levels of some systems depends on a magnetic field.  You can let the populations of the energy levels come to equilibrium at some temperature $T$ and then quickly invert the magnetic field.  What was a low energy level now becomes a high one and vice versa.  You have an upside down distribution of populations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon enough that upside down population decays to a right side up population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while the distribution is &quot;upside down&quot; one finds that it is given by the same formula used for an ordinary distribution, but with a negative temperature.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-09T00:49:48.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2532" PostTypeId="2" ParentId="2526" CreationDate="2012-11-09T00:53:20.207" Score="4" Body="&lt;p&gt;The endothermicity (taking up heat) or exothermicity (giving off heat) depends on a balance of factors including the entropy change, the degree with which the ions become associated with water, and the possibility of actual chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most notorious example is sulfuric acid, which reacts so strongly with water that it can actually boil the water, and spatter the acid, which is why one always adds the acid to the water(water as solvent) and precautionary measures (gloves and proper clothing) are needed to handle the reaction.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="73" LastEditDate="2012-12-09T13:47:31.920" LastActivityDate="2012-12-09T13:47:31.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2535" PostTypeId="2" ParentId="2509" CreationDate="2012-11-09T22:38:21.030" Score="19" Body="&lt;p&gt;Even though your question has already been answered (and this is not an alternative answer), but if you're open to it, switching from Word to LaTeX with the &lt;a href=&quot;http://www.math.washington.edu/tex-archive/macros/latex/contrib/chemmacros/chemmacros_en.pdf&quot;&gt;&lt;code&gt;chemmacros&lt;/code&gt;&lt;/a&gt; package (PDF) will benefit you greatly in the long run.&lt;/p&gt;&#xA;" OwnerUserId="82" LastActivityDate="2012-11-09T22:38:21.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2537" PostTypeId="2" ParentId="2397" CreationDate="2012-11-11T03:46:01.360" Score="8" Body="&lt;p&gt;Many of the chemicals used in metal plating are extremely poisonous (cyanide), cause severe allergies (Ni), and are carcinogenic (Cr(VI)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd consider it as extremely unprofessional (besides violating many laws and regulations), to re-use anything that has been in contact with these materials for any purpose that requires a 'clean' container, regardless of the theoretical possibility of cleaning everything off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use such a drum for food or drinking water would be criminal negligence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd only consider it ok if it was thoroughly cleaned (my myself) and only used by myself as a trashcan in my machine shop outside of my home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would even disallow to use it as a trashcan in the machine shop at work. Why risk it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if PP does not absorb ions in principle, there is always the real world:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The guy who does the cleaning gets lazy and thinks you are an idiot requiring him to clean something off he doesn't see and only rinses the drum with water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The plating bath contained organics that made the ions stick to the drum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some stuff precipitated from the plating bath and stuck to the bottom of the drum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;During the cleaning process cleaned and contaminated drums get mixed up.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some plating solution dried in a crack, or under the rim, or wherever you can't clean in off.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cleaning is itself a hazardous action that creates a lot of contaminated wastewater.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't know where the drums might end up. Of course the risk of using them as a trashcan in the machine shop is minuscule, agreed. But what if a machinist takes them home to his home shop? What if his kids use them to play with? I'd have been delighted as a kid to have a 55-gallon drum as a toy. Maybe the machinist's wife will take them to store food.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are hundreds of accidents where contaminated containers have been improperly cleaned and re-purposed. Please don't do it, even if it is actually safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the drums can be cleaned, imagine anything happening, like some kid being admitted to the ER with a Nickel allergy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd be in a tough position proving that no one got injured from your unprofessionally cleaned drums. The people who used the drums in the first place will have all the paperwork to 'prove' that it wasn't them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw. we had an accident with a metal plating bath that was not properly cleaned and injured a technician who had do do some modifications to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might not hurt anyone, but it's better not to do it.&lt;/p&gt;&#xA;" OwnerUserId="804" LastActivityDate="2012-11-11T03:46:01.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2538" PostTypeId="1" AcceptedAnswerId="2539" CreationDate="2012-11-11T04:47:30.193" Score="6" ViewCount="142" Body="&lt;p&gt;While doing computational chemistry for both organic and inorganic compounds, an energy optimization of that compound is done before starting a study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we do drug docking studies using aspirin, first we would make aspirin's chemical structure, and then we would optimize it. So what does &quot;energy optimization&quot; mean in the context of organic and inorganic compounds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is energy optimization different for organic and inorganic materials (e.g.,aspirin or gold nanoparticles, respectively)?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="73" LastEditDate="2012-11-11T12:13:34.303" LastActivityDate="2012-11-11T13:18:31.947" Title="What is in silico energy optimization?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2539" PostTypeId="2" ParentId="2538" CreationDate="2012-11-11T09:50:10.993" Score="11" Body="&lt;p&gt;Generally, energy optimization is done as iterative process with following steps repeating &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;calculate forces of nuclei&lt;/li&gt;&#xA;&lt;li&gt;move nuclei according forces known so final structure were less tense.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Most focus in science is focused on how first step is made, as choice of the second step does not affect final structure except pathological cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference is usually tied to computational capabilities of 'silica' available. Complexity of quantum mechanical computations usually grows as $O(n^3)$, $O(n^4)$ or even $O(e^n)$ where n is number of atoms of same type in system. So, the bigger system is, the 'worse' approximation is used. Biological systems are usually so large, that any sort of quantum mechanical computation is not an option at all, so &quot;molecular mechanics/dynamics&quot; approach is used, with empiric formulas of forces based on relative positions of atoms&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-11-11T13:18:31.947" LastActivityDate="2012-11-11T13:18:31.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2541" PostTypeId="1" AcceptedAnswerId="2545" CreationDate="2012-11-11T22:46:36.390" Score="7" ViewCount="3206" Body="&lt;p&gt;When we calculate the electrons and orbital numbers in $\ce{CH4}$ $sp^3$ hybridization we have taken 3 electrons from $2p$ of $\ce{C}$ and 1 electron from $2s$ of $\ce{C}$ but in $\ce{C2H4}$ we have $2sp^2$ (2 electrons from $2p$ and 1 from $2s$).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why do $\ce{C}$'s atomic orbitals make $2sp^2$ in $\ce{C2H4}$ and $sp^3$ in $\ce{CH4}$?&lt;/li&gt;&#xA;&lt;li&gt;How we associate the numbers of electrons from $2p$ and $2s$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="792" LastEditorUserId="22" LastEditDate="2012-11-12T03:55:34.930" LastActivityDate="2012-11-13T01:15:05.360" Title="Hybridization sp3 and sp2" Tags="&lt;hybridization&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2542" PostTypeId="1" CreationDate="2012-11-12T01:11:59.257" Score="8" ViewCount="621" Body="&lt;p&gt;What do you call the proton in a C&lt;b&gt;H&lt;/b&gt;-OH configuration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When discussing an NMR spectrum, I can talk about the hydroxyl proton, but it seems clumsy to refer to &quot;the proton on the hydroxyl carbon&quot;.&lt;/p&gt;&#xA;" OwnerUserId="798" LastEditorUserId="7" LastEditDate="2012-11-13T07:50:04.820" LastActivityDate="2012-11-13T07:50:04.820" Title="Name of a proton bonded to a hydroxyl carbon" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2543" PostTypeId="2" ParentId="2542" CreationDate="2012-11-12T03:58:07.420" Score="9" Body="&lt;p&gt;I would call it the $\alpha$-proton to the $\ce{-OH}$. The alpha proton is the proton bonded to the alpha carbon (i.e., the hydroxyl carbon).&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-11-12T03:58:07.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2545" PostTypeId="2" ParentId="2541" CreationDate="2012-11-12T13:30:46.167" Score="7" Body="&lt;p&gt;This can be explained in terms of orbital overlap and therefore bond strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a single carbon in an ethene molecule with two sp3 hybridised carbon atoms. The orbitals on the carbon repel each other and arrange themselves into a tetrahedron. Two of the four sp3 hybrids are used to form sigma bonds with hydrogen atoms. The carbon is then left to share its remaining two sp3 hybridized orbitals with its neighbouring carbon. Draw this system and you can see that two sigma bonds can be formed, though they are &quot;bent&quot; as the sp3 orbitals on the carbon are at a 109.5 degree angle to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a single sp2 hybridised carbon in an ethene molecules with two sp2 hybridised carbon atoms. Again two orbitals are used to bond with hydrogen. This time we have an sp2 orbital pointing directly at the neighbouring carbon with which to form a strong sigma bond. The remaining p orbital which is orthogonal to the C-C sp bond can also forms a C-C pi bonding interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw these orbital systems and compare.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sp2 hybridised system constructs a strong linear sigma bond.&lt;/li&gt;&#xA;&lt;li&gt;The sp2 hybridised system better shields the nuclei of the bonding atoms.&lt;/li&gt;&#xA;&lt;li&gt;The sp2 hybridised system provides more orbital overlap, leading to stronger bonding and lower free energy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="784" LastActivityDate="2012-11-12T13:30:46.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2546" PostTypeId="2" ParentId="2541" CreationDate="2012-11-13T01:15:05.360" Score="5" Body="&lt;p&gt;I agree with Horba's answer.  However the usual case is that the bond structure of a given molecule is already known and &lt;em&gt;then&lt;/em&gt; the hybridization is chosen to fit the observed structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $\ce{CH4}$ and $\ce{C2H4}$.  It is clear that in methane the central carbon has four bonds and hence will need hybridization that gives four bonds.  Thus we get $sp^3$.  Ethene, on the other hand, has two carbons each of which needs three bonds.  So we get $sp^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, hybridization is not a predictive scheme, it is an explanatory scheme.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-13T01:15:05.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2547" PostTypeId="1" AcceptedAnswerId="2549" CreationDate="2012-11-13T12:26:06.927" Score="24" ViewCount="32873" Body="&lt;p&gt;I notice that salt solutions of $\ce{NaCl}$ and $\ce{KCl}$ are colourless while those of $\ce{CuSO4}$ and $\ce{FeSO4}$ are coloured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got as far as figuring that it has to do with the transition metal ions, but I can't explain why the salt solution of $\ce{ZnSO4}$ is colourless even though zinc itself is one of the transition metals.&lt;/p&gt;&#xA;" OwnerUserId="401" LastEditorUserId="73" LastEditDate="2012-11-13T15:10:47.200" LastActivityDate="2015-03-14T17:21:38.863" Title="Why are some salt solutions coloured?" Tags="&lt;color&gt;&lt;transition-metals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2548" PostTypeId="1" AcceptedAnswerId="2550" CreationDate="2012-11-13T12:53:41.137" Score="9" ViewCount="178" Body="&lt;p&gt;A Koppers-Totzek reactor (there seems to be only &lt;a href=&quot;http://de.wikipedia.org/wiki/Koppers-Totzek-Reaktor&quot;&gt;a German Wikipedia page&lt;/a&gt;) is an entrained flow gasifier that blows coal dust, oxygen and water (as steam, of course) into a burning chamber where the coal burns at about 1600 °C. The output contains around 60% $\ce{CO}$ and 33% $\ce{H2}$, plus some $\ce{CO2}$ and (very little) $\ce{CH4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When burning coal, I would expect $\ce{CO2}$ and $\ce{H2O}$ as output, is the reactor simply providing too little oxygen, or is there something about the process that produces the output observed?&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditDate="2012-11-13T15:09:13.250" LastActivityDate="2012-11-14T13:15:50.127" Title="In a Koppers-Totzek reactor, why do I get CO and H₂ instead of CO₂ and H₂O?" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2549" PostTypeId="2" ParentId="2547" CreationDate="2012-11-13T13:11:38.750" Score="25" Body="&lt;p&gt;You're right--it's got to do with them being transition metals (usually). Transition metal ions form &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_complex&quot; rel=&quot;noreferrer&quot;&gt;coordination complexes&lt;/a&gt;. Their empty $d$ orbitals accept lone pairs from other molecules (called &quot;ligands&quot;) and form larger molecules (though we don't call them that--we call them &quot;complexes&quot;). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper%28II%29_sulfate&quot; rel=&quot;noreferrer&quot;&gt;Copper Sulphate)&lt;/a&gt;. Thus, metals with no empty $d$ orbitals of comparable energy become colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPOjk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Octahedral_molecular_geometry&quot; rel=&quot;noreferrer&quot;&gt;octahedral&lt;/a&gt; complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're &quot;mixing&quot; these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called &quot;d-orbital splitting&quot; {*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KVSgn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, electrons may &quot;jump&quot; between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is &lt;em&gt;before&lt;/em&gt; the ligand bonds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*For &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedral_molecular_geometry&quot; rel=&quot;noreferrer&quot;&gt;tetrahedral complexes&lt;/a&gt;, the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-06-22T17:58:09.680" LastActivityDate="2013-06-22T17:58:09.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2550" PostTypeId="2" ParentId="2548" CreationDate="2012-11-13T17:47:31.313" Score="3" Body="&lt;p&gt;The reaction occurs in two stages: in first $\ce{C}$ oxidized to $\ce{CO2}$ and then $\ce{CO2}$ is reduced to $\ce{CO}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of having carbon react with oxygen and water vapor is called steam-oxygen conversion. It is used, because in reality, due to the high reaction temperatures, the allothermic principle is hard to provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C + O2-&amp;gt;CO2}$$&#xA;$$\ce{C + 2H2O-&amp;gt;CO2 + 2H2}$$&#xA;$$\ce{C + H2O-&amp;gt;CO2 + H2}$$&#xA;$$\ce{C + CO2-&amp;gt;2CO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium concentrations of $\ce{O2}$ and $\ce{H2O}$ in the gasification products at all temperatures are negligible. Steam-oxygen gasification products are a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before carrying out the reaction of the shift&#xA;$$\ce{CO + H2O-&amp;gt;CO2 + H2}$$ from a mixture of $\ce{CO2}$, $\ce{CO}$ and $\ce{H2}$ must be separated $\ce{CO2}$. To this must be expended work division (again burn some $\ce{C}$).&lt;/p&gt;&#xA;" OwnerUserId="802" LastEditorUserId="22" LastEditDate="2012-11-14T13:15:50.127" LastActivityDate="2012-11-14T13:15:50.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2551" PostTypeId="1" AcceptedAnswerId="2552" CreationDate="2012-11-14T03:38:38.020" Score="8" ViewCount="164" Body="&lt;p&gt;If we were trying to figure out the time scale for a gas-phase reaction between two hydrogen atoms in a molecular cloud (which has density $~\pu{10^4cm^-3}$), apparently the reaction would happen on a time scale proportional to the inverse of the density multiplied by $10^{15}$ years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the cloud not being dense and the probability that a collision will surpass the activation energy is small, is the time elongated because you need to induce a dipole moment between two hydrogen atoms to actually bind them together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently read somewhere that even if the reactants would go into the unbound state of $\ce{H2}$ they can't emit a photon that would let it go into the bound state of $\ce{H2}$. I guess they are talking about quantum mechanics, but it is not obvious to me what the bound and unbound states of molecular hydrogen are. (If someone could clarify that too)&lt;/p&gt;&#xA;" OwnerUserId="845" LastEditorUserId="5026" LastEditDate="2018-02-25T07:55:36.727" LastActivityDate="2018-02-25T08:00:10.247" Title="Why does hydrogen form on such a long time scale?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2552" PostTypeId="2" ParentId="2551" CreationDate="2012-11-14T16:49:51.987" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the time-scale so long?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time-scale is so long because the density is not just low, it's virtually nonexistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle density of any gas can be determined at any temperature and pressure using the ideal gas law ($PV=nRT$). Let's use air as an example. At $\pu{25 o^\circ C}$ ($\pu{298 K}$), and 1 atmosphere, the particle density of air ($n/V$) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{n}V =\frac{RT}P $$&#xA;$$ \frac{n}V =\frac{(\pu{0.08026 L atm (K mol)^{-1})(\pu{298  K})}}{{\pu{1 atm}}} = \pu{23.9 mol L^-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to molecules per $\pu{cm^3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$23.9 \frac{\mathrm{mol}}{\mathrm{L}} \times 6.022\times10^{23}\frac{\text{molecules}}{\text{mol}}\times\frac{1 \text{ L}}{1000 \text{ cm}^3}=1.44\times10^{22}\frac{\text{molecules}}{\text{cm}^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of gasses we are used to dealing with is 18 orders of magnitude higher than the density you quoted. If we have a gas of hydrogen atoms at air density, then the timescale for formation of H&lt;sub&gt;2&lt;/sub&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 10^{15} \frac{\text{years}\times\text{cm}^3}{\text{particles}} \times  \frac{1 \text{ cm}^3}{1.44\times10^{22}\text{ molecules}}=6.94\times 10^{-8}\text{ years} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6.94\times 10^{-8}\text{ years}= 2.54\times 10^{-5}\text{ days}=6.08\times 10^{-4} \text{ hours}=2.19 \text{ seconds}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &quot;normal&quot; densities, the timescale is much shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of your question deals with the quantum mechanics of the activation barrier to this reaction, which I cannot do justice to at this time.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-02-25T08:00:10.247" LastActivityDate="2018-02-25T08:00:10.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2554" PostTypeId="1" AcceptedAnswerId="2555" CreationDate="2012-11-15T17:29:53.150" Score="9" ViewCount="1132" Body="&lt;p&gt;I understand why smaller particles have more velocity, but I don't understand what velocity has to do with rate of effusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is thus:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Pressure is the number of impacts of particles in a given period of time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If He and Ar are both in balloons at, say, 1.5 atm, both gasses have the same average Kinetic energy but He is moving faster (because it's smaller). Simple enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If both gasses are subject to the same pressure, they have the same number of impacts over a given area over a given period in time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This should mean that each gas has an equal number of particles approaching a hole over a given period of time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a hole is big enough for both particles to fit through, the rate of effusion should be the same, since the same number of particles are approaching the exit hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For an analogy to explain my thinking ... if you have two lanes of cars going through a checkpoint, one has 30 cars per minute at 50 miles per hour and one has 30 cars per minute at 30 miles per hour, the resulting number of cars through the checkpoint should still be 30 cars per minute in each lane, regardless of their velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book (and professor, and Wikipedia...) say that it is the higher velocity of He particles which cause the faster rate of effusion; however, from my reasoning, that doesn't make sense except with small enough holes where the size of the particles would have an effect, i.e. the holes are very small in size and He could fit through but Ar could not, or there are simply &lt;em&gt;more&lt;/em&gt; holes that He could fit through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Graham's law is referring to pinholes, which should be large enough for either atom to fit through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does velocity affect rate of effusion?&lt;/p&gt;&#xA;" OwnerUserId="853" LastEditorUserId="16683" LastEditDate="2016-09-30T09:49:25.013" LastActivityDate="2018-08-19T03:01:28.180" Title="Why does gas particle velocity affect rate of effusion?" Tags="&lt;physical-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2555" PostTypeId="2" ParentId="2554" CreationDate="2012-11-15T19:05:27.073" Score="10" Body="&lt;p&gt;Your point 1 is mistaken (or incomplete)&#xA;$$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is assuming all particles hit perpendicularly (you can always modify this for all colissions by taking the average component of velocity in the perpendicular direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, pressure is proportional to number of impacts, &lt;strong&gt;mass, and velocity&lt;/strong&gt;. Pressure is &lt;em&gt;not simply &quot;number of impacts&quot;, rather it is a combination of number of impacts with other stuff&lt;/em&gt;. Since the energies are equal, we can say that $v\propto m^{-\frac12}$, and we get $P\propto n m^{\frac12}$. P is also the same, so $n\propto\frac1{\sqrt m}\propto v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, when avg KE and pressure are the same, rate of diffusion is inversely proportional to square root of mass.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-11-15T19:05:27.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2556" PostTypeId="1" AcceptedAnswerId="2558" CreationDate="2012-11-15T19:48:37.917" Score="3" ViewCount="1180" Body="&lt;p&gt;Sodium Perborate is a well known bleaching agent in detergents. What makes Sodium Perborate an effective bleaching agent and how is this related to its oxidation chemistry?&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="81" LastEditDate="2012-11-16T22:15:17.203" LastActivityDate="2012-11-16T22:15:17.203" Title="Sodium Perborate as a detergent ingredient" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2557" PostTypeId="2" ParentId="2499" CreationDate="2012-11-15T22:45:06.683" Score="8" Body="&lt;p&gt;For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work produced by electrical energy, $W = V\cdot I\cdot t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E^\circ_\text{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\Delta G^\circ$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O(aq) -&amp;gt;[\text{elect}] H+(aq) + OH- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4OH- (aq) -&amp;gt; 2H2O + O2 + 4e- }$ (anode half-cell reaction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+(aq) + 2e- -&amp;gt; H2(g)}$ (cathode half-cell reaction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall cell reaction is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2O(aq) -&amp;gt; 2H2(g) + O2(g)}$ ; $\Delta G^\circ=237.2\ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum (theoretical) voltage required is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in practice, a slightly larger voltage of 1.48 V is required since the  enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T21:42:31.713" LastActivityDate="2015-04-24T21:42:31.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2558" PostTypeId="2" ParentId="2556" CreationDate="2012-11-16T11:01:09.957" Score="4" Body="&lt;p&gt;Connecting with water, sodium perborate releases hydrogen peroxide.&#xA;$$\ce{NaBO2*H2O2*3H2O-&amp;gt;Na+ + BO2- + H2O2 + 3H2O}$$&#xA;$$\ce{H2O2-&amp;gt;H+ + HOO-}$$&#xA;$$\ce{HOO- -&amp;gt;HO- + O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolution of sodium percarbonate in water at hand washing and the washing machine washing, leads to the formation of the perhydroxy anion. This anion is an effective bleach, but it is formed in sufficient quantities only at temperatures above 60 ° C.&#xA;From the perhydroxy anion obtain atomic oxygen, which has a strong oxidizing capacity.&lt;/p&gt;&#xA;" OwnerUserId="802" LastEditorUserId="802" LastEditDate="2012-11-16T12:01:26.897" LastActivityDate="2012-11-16T12:01:26.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2560" PostTypeId="1" CreationDate="2012-11-16T16:30:04.457" Score="4" ViewCount="839" Body="&lt;p&gt;I prepared a compound &quot;$\ce{SO2}$-Quinolclathrate&quot; by dissolving hydroquinone and sodium sulfite with hydrochloric acid in warm water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, I tested a little sample of the product on a flame and it vaporised, following which I smelled the gas. I am supposed to define this gas and my prediction is that this could be $\ce{SO2}$(with orange flame).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clathrate compounds include a trapping system with guest molecules (held &quot;guest&quot; in cages) and the host (hydroquinone here). &#xA;$\ce{SO2}$ is guest in the cages of hydroquinone and the gas is held due to crystal lattice forces. Does this mean there are no bonds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the reason for $\ce{SO2}$ vaporizing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the product was added to starch/iodine solution to find out if it is soluble in that solution and it is so. What can be the reason of this solubility. &lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="1322" LastEditDate="2013-04-17T04:45:10.817" LastActivityDate="2014-04-11T23:03:29.003" Title="Clathrate Compounds with Quinone" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2561" PostTypeId="1" AcceptedAnswerId="2562" CreationDate="2012-11-16T18:12:55.957" Score="4" ViewCount="919" Body="&lt;p&gt;The apparatus that I am using is a glass container, with two electrodes,two glass cylinders, water, and a battery. In the glass container will be filled with water and two electrodes kept inside the glass cylinder and then dipped in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two electrodes are then connected with the positive and negative terminals of the battery and the container is sealed.  As the current is started the water will be separated into hydrogen and oxygen. After extraction of the gas, the two gases are passed separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this process, will the hydrogen gas start burning immediately as soon as produced? If yes, then what to do to make it pass without burning?  The separated gas is passed through glass pipes, will the extraction work properly? &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="296" LastEditDate="2012-12-27T14:22:15.947" LastActivityDate="2015-06-30T16:50:52.343" Title="Extraction of hydrogen and oxygen from water" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;electrons&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2562" PostTypeId="2" ParentId="2561" CreationDate="2012-11-17T05:24:16.363" Score="6" Body="&lt;p&gt;No, it won’t burn immediately. As your electrodes are generally separated by some distance, you should be able to isolate the produced oxygen and hydrogen gas by catching the respective bubble streams under a submerged test tube or the like. Alone, neither oxygen nor hydrogen will burn, however combining the oxygen and hydrogen gas will produce an &lt;strong&gt;explosive mixture&lt;/strong&gt;. Mixing the hydrogen with air will also produce an &lt;strong&gt;explosive mixture&lt;/strong&gt; as the atmosphere is about 22 % oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the hydrogen to burn, you need to supply the necessary activation energy, which can be supplied by a spark or flame. After this, the reaction will be self-sustaining due to its exothermicity. There are in fact entirely self-contained apparatus which can be purchased from scientific teaching shops that can electrolytically split water and then combust the hydrogen and oxygen using a piezoelectric ignition system, the ensuing tiny explosion causing water to spurt out of a small hole. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen#Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia mentions&lt;/a&gt; that the autoignition temperature of hydrogen in air is 500 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a small scale, this reaction is quite safe, however you should &lt;strong&gt;always&lt;/strong&gt; take safety precautions such as wearing &lt;strong&gt;protective eyewear&lt;/strong&gt; and having &lt;strong&gt;fire-suppression equipment&lt;/strong&gt; at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s an illustration of a reasonable water-splitting setup. Remember that your water should contain an electrolyte such as dissolved salt. Keep in mind that if your potential difference is high enough you will start to produce small quantities of &lt;strong&gt;chlorine gas&lt;/strong&gt; and aqueous &lt;strong&gt;sodium hydroxide&lt;/strong&gt;, both of which are hazardous in high concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XfICX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should draw attention to the fact that the $\ce{H2}$ headspace is about twice the volume of the $\ce{O2}$ headspace. This comes down to the stoichiometry of the reaction, and also means that the $\ce{H2}$ volume will increase at twice the rate of $\ce{O2}$. The wires I’ve illustrated connect to the terminals of a battery or lab PSU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: Here’s a &lt;a href=&quot;http://zamcopter.web.fc2.com/electrolysis/electrolysis.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice guide&lt;/a&gt; on how to construct and safely operate a water-splitting/combusting apparatus using household equipment, aimed at secondary school students.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorDisplayName="user7951" LastEditDate="2015-06-30T16:50:52.343" LastActivityDate="2015-06-30T16:50:52.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2563" PostTypeId="1" AcceptedAnswerId="2572" CreationDate="2012-11-17T14:00:12.067" Score="5" ViewCount="2537" Body="&lt;p&gt;I'm doing a basic chemistry course, and we are currently learning how to compute $\text{pH}$ from the acid dissociation constant (using $\left[\text{H}^{+}_{(\text{aq})}\right]=\sqrt{K_{a}\left[\text{HA}_{(\text{aq})}\right]}$) along with computing the $\text{pH}$ of strong bases by assuming full dissociation into $\text{OH}^{-}$ ions, and then using the ionic product of water to calculate the concentration  of protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have been given is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the pH of the solution obtained when $14.9\text{ cm}^{3}$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide solution has been added to $25.0\text{ cm}^{3}$ solution of methanoic acid of concentration $0.100\text{ mol dm}^{-1}$; $K_{a}=1.60\times 10^{-4}\text{ mol dm}^{-3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how I should go about solving this problem, I can calculate concentrations of hydrogen and hydroxide ions for each of the solutions but I'm unsure how to combine them (as presumably some of the $\text{OH}^{-}$ ions will react with the $\text{H}^{+}$ ions to form $\text{H}_{2}\text{O}$?)&lt;/p&gt;&#xA;" OwnerUserId="857" LastEditorUserId="46306" LastEditDate="2018-03-05T01:06:52.430" LastActivityDate="2018-03-05T01:06:52.430" Title="Computation of pH when an acid and base are mixed in solution" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2564" PostTypeId="1" AcceptedAnswerId="2567" CreationDate="2012-11-17T14:26:40.893" Score="25" ViewCount="5873" Body="&lt;p&gt;What is the IUPAC nomenclature for alkanes with large numbers of $\ce{C}$ atoms (more than 200)? For example, what is the IUPAC accepted name for $\ce{C_{205}H_{412}}$?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="11367" LastEditDate="2015-09-06T18:32:22.957" LastActivityDate="2016-09-09T07:48:42.943" Title="What is the IUPAC nomenclature of alkane with large numbers of C atoms (more than 200)?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2565" PostTypeId="1" CreationDate="2012-11-17T22:26:52.217" Score="4" ViewCount="62" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Possible Duplicate:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions&quot;&gt;Will gaseous ionic compounds be free moving ions?&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I know it sounds extremely expensive, but what if we could heat the ionic compound to evaporate it and achieve the lattice energy and break them into free ions, wouldn't the electrolysis be faster, as the ions can move faster freely than in liquid state?&lt;/p&gt;&#xA;" OwnerUserId="815" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2012-11-18T01:53:51.847" Title="Can the electrolytes be in gaseous state?" Tags="&lt;ions&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2012-11-19T07:13:53.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="2566" PostTypeId="2" ParentId="2564" CreationDate="2012-11-18T12:38:13.577" Score="4" Body="&lt;p&gt;Presuming it is linear, then it should follow the trend of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Higher_alkanes&quot; rel=&quot;nofollow&quot;&gt;higher alkanes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C10}:  \ce{CH3(CH2)8CH3}$ - the name is decane (&lt;em&gt;deca&lt;/em&gt; means &lt;em&gt;ten&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C20}:  \ce{CH3(CH2)18CH3}$ - the name is icosane (&lt;em&gt;icos&lt;/em&gt; means &lt;em&gt;twenty&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C25}:  \ce{CH3(CH2)23CH3}$ - the name is pentacosane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After icosane, the higher alkanes follow the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C30}:  \ce{CH3(CH2)28CH3}$ - the name is triacontane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C40}:  \ce{CH3(CH2)38CH3}$ - the name is tetracontane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C50}:  \ce{CH3(CH2)48CH3}$ - the name is pentacontane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C100}:  \ce{CH3(CH2)98CH3}$ - the name would be decacontane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C200}:  \ce{CH3(CH2)198CH3}$ - the name would probably be icosacontane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for $\ce{C205}:  \ce{CH3(CH2)203CH3}$ - the name might be &lt;strong&gt;pentacosacontane&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-11-18T12:38:13.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2567" PostTypeId="2" ParentId="2564" CreationDate="2012-11-18T14:33:56.807" Score="27" Body="&lt;p&gt;The definitive answer is, of course, in the rules established by the IUPAC’s “Commission on nomenclature in organic chemistry”. The reference you are looking for is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Extension of Rules A-1.1 and A-2.5 Concerning Numerical Terms Used in Organic Chemical Nomenclature”, &lt;em&gt;Pure Appl. Chem.&lt;/em&gt;, &lt;strong&gt;1986&lt;/strong&gt;, &lt;em&gt;58&lt;/em&gt;, 1693-1696&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can be found &lt;a href=&quot;http://www.iupac.org/publications/pac/1986/pdf/5812x1693.pdf&quot; rel=&quot;noreferrer&quot;&gt;here as a PDF&lt;/a&gt; and &lt;a href=&quot;http://www.chem.qmul.ac.uk/iupac/misc/numb.html&quot; rel=&quot;noreferrer&quot;&gt;here as an HTML version&lt;/a&gt;. In particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NT-1.1 - The fundamental numerical terms for use in hydrocarbon names or as multiplying prefixes for simple features are given in the following list:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{array}{|c:c|c:c|c:c|c:c|}\hline&#xA;    1&amp;amp;\rm \text{mono- or hen-} &amp;amp;10 &amp;amp;\rm \text{deca-}&amp;amp;100&amp;amp;\rm \text{hecta-}&amp;amp; 1000&amp;amp;\rm \text{kilia-} \\\hline&#xA;    2&amp;amp;\rm \text{di- or do-}&amp;amp;20&amp;amp;\rm \text{icosa-}&amp;amp;200&amp;amp;\rm \text{dicta-}&amp;amp;2000&amp;amp;\rm \text{dilia-}\\\hline&#xA;    3&amp;amp;\rm \text{tri-}&amp;amp; 30&amp;amp;\rm \text{triaconta-}&amp;amp;300&amp;amp;\rm \text{tricta-}&amp;amp;3000&amp;amp;\rm \text{trilia-} \\\hline&#xA;    4&amp;amp;\rm \text{tetra-}&amp;amp;40&amp;amp;\rm \text{tetraconta-}&amp;amp;400&amp;amp;\rm \text{tetracta-}&amp;amp;4000&amp;amp;\rm \text{tetralia-} \\\hline&#xA;    5&amp;amp;\rm \text{penta-}&amp;amp;50&amp;amp;\rm \text{pentaconta-}&amp;amp;500&amp;amp;\rm \text{pentacta-}&amp;amp;5000&amp;amp;\rm \text{pentalia-}\\\hline&#xA;    6&amp;amp;\rm \text{hexa-}&amp;amp;60&amp;amp;\rm \text{hexaconta-}&amp;amp;600&amp;amp;\rm \text{hexacta-}&amp;amp;6000&amp;amp;\rm \text{hexalia-}\\\hline&#xA;    7&amp;amp;\rm \text{hepta-}&amp;amp;70&amp;amp;\rm \text{heptaconta-}&amp;amp;700&amp;amp;\rm \text{heptacta-}&amp;amp;7000&amp;amp;\rm \text{heptalia-}\\\hline&#xA;    8&amp;amp;\rm \text{octa-}&amp;amp;80&amp;amp;\rm \text{octaconta-}&amp;amp;800&amp;amp;\rm \text{ octacta-}&amp;amp;8000&amp;amp;\rm \text{octalia-}\\\hline&#xA;    9&amp;amp;\rm \text{nona-}&amp;amp;90&amp;amp;\rm \text{nonaconta-}&amp;amp;900&amp;amp;\rm \text{nonacta-}&amp;amp;9000&amp;amp;\rm \text{nonalia-}\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, $\ce{C_205H_412}$ is the molecular formula of pentadictane. $\ce{C_7547H_15096}$ is heptatetracontapentactaheptaliane.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2016-09-09T07:48:42.943" LastActivityDate="2016-09-09T07:48:42.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2568" PostTypeId="1" AcceptedAnswerId="2582" CreationDate="2012-11-18T14:55:37.513" Score="8" ViewCount="219" Body="&lt;p&gt;People said that wine is good for health. So I take a cup (less than 100 mL) of wine every evening. (Yes, wine is cheap where I live.) So to finish a bottle of wine (750 mL), it takes me a week.&#xA;(Of course I put it in the refrigerator.) I have to say that it doesn't taste good after 3 days (but still OK).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know if it is not safe to do so. Maybe the wine begins to produce something not good for health after a few days when it is opened. So, my question is: what chemical species are formed after opening a wine bottle, making it go bad? Are they only bad for the taste, or also bad unhealthy?&lt;/p&gt;&#xA;" OwnerUserId="860" LastEditorUserId="7475" LastEditDate="2015-10-30T16:48:06.863" LastActivityDate="2015-10-30T16:48:06.863" Title="Is it not safe for health to finish a bottle of wine during a week" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-10-30T19:33:21.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="2569" PostTypeId="1" CreationDate="2012-11-18T15:29:54.027" Score="4" ViewCount="996" Body="&lt;p&gt;Usually two separate p orbitals would have '4 lobes' while in a $\ce{sp^2}$ hybrid those 2 p orbitals would only have '3 lobes'. why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm referring to diagrams such as the one shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZFYn.gif&quot; alt=&quot;http://i.stack.imgur.com/Cb74T.gif&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user781" LastEditorUserId="4231" LastEditDate="2015-02-24T20:49:33.717" LastActivityDate="2015-07-06T20:04:18.570" Title="In sp² hybrid orbital, why does the p orbitals only have 3 sides instead of 4?" Tags="&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2570" PostTypeId="2" ParentId="2569" CreationDate="2012-11-18T18:55:37.840" Score="16" Body="&lt;p&gt;First, an $sp^2$ orbital &lt;em&gt;does&lt;/em&gt; have two lobes. One is large, and one is tiny. Secondly, $sp^2$ means &quot;a 'mixture' of &lt;strong&gt;one $s$&lt;/strong&gt; and two $p$ orbitals&quot;. Like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nv1wK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you plug in three orbitals, and you get out three more orbitals. When drawn in place, you get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCAC3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a total of six lobes here, but they are distributed amongst three $sp^2$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of lobes need not stay constant--it's the number of orbitals that does, since each orbital corresponds to two electrons, and there's no way for new electrons to be added/removed (yet). What's happening is that the wavefunctions(orbitals) of the electrons are &quot;mixing up&quot; to distribute the electrons in a way that stimulates bonding (one per orbital). The lobes have a significance, but it's much deeper.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="7448" LastEditDate="2015-07-06T20:04:18.570" LastActivityDate="2015-07-06T20:04:18.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2571" PostTypeId="1" AcceptedAnswerId="4631" CreationDate="2012-11-18T20:04:45.687" Score="6" ViewCount="54366" Body="&lt;p&gt;If someone were to pour about 5 ml of 94% pure ethanol on a flat surface and at room temperature, approximately how much time would it take for it to evaporate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?&lt;/p&gt;&#xA;" OwnerUserId="861" LastEditorUserId="4231" LastEditDate="2015-02-23T16:12:38.487" LastActivityDate="2017-08-29T03:41:55.187" Title="Time required for a thin layer of ethanol to evaporate?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2572" PostTypeId="2" ParentId="2563" CreationDate="2012-11-19T01:04:13.603" Score="5" Body="&lt;p&gt;You are headed in the right direction.  Pretend that the $\ce{H+}$ ions and the $\ce{OH-}$ ions neutralize each other 1 for 1.  That will leave you with only one of those ions left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water ionization won't affect your problem as that ionization is repressed by the excess ion present in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should be able to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-19T01:04:13.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2574" PostTypeId="2" ParentId="151" CreationDate="2012-11-19T06:29:51.830" Score="6" Body="&lt;p&gt;We know that the electronic configuration of chromium is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 3d^5 4s^1}$&lt;/span&gt;. It is because promotion occurs in case of &lt;span class=&quot;math-container&quot;&gt;$\ce{3d}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{4s}$&lt;/span&gt; orbitals &amp;mdash; in other words, the electron is shifted from a lower energy level to a higher one (also known as excitation). Promotional energy and pairing energy both are endergonic so the process which requires less energy would preferably take place.&lt;/p&gt;&#xA;" OwnerUserId="863" LastEditorUserId="4945" LastEditDate="2019-01-30T12:46:16.103" LastActivityDate="2019-01-30T12:46:16.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2575" PostTypeId="1" CreationDate="2012-11-19T12:42:36.050" Score="10" ViewCount="12574" Body="&lt;p&gt;According to Pauli’s exclusion principle, an $s$ orbital contains at most two electrons with the opposite spin (up and down). Why can't an $s$ orbital contain a third electron whose state is the linear combination of spin up and down?&lt;/p&gt;&#xA;" OwnerDisplayName="user864" LastEditorUserId="4945" LastEditDate="2014-10-23T10:22:49.783" LastActivityDate="2017-10-13T11:03:45.750" Title="Why can an s orbital only hold two electrons?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2576" PostTypeId="1" AcceptedAnswerId="2579" CreationDate="2012-11-19T20:08:08.303" Score="9" ViewCount="60124" Body="&lt;p&gt;I am trying to figure out the configuration for this chiral carbon to answer a homework question, however they arbitrarily include a &quot;D&quot; group as shown in the image. My guess is that it is some higher priority group than the carbon, and lower priority than the halide, making the configuration (S). Is this correct? What does the D group represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3KnSa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2018-04-27T01:16:11.910" Title="What does the &quot;D&quot; group represent on a chiral center?" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2577" PostTypeId="1" AcceptedAnswerId="2578" CreationDate="2012-11-19T20:16:18.667" Score="2" ViewCount="488" Body="&lt;p&gt;If I have a complete silicon nitride ball bearing unit, is it safe to hold without any protection? If it cracks, or disintegrates is it also safe?&lt;/p&gt;&#xA;" OwnerUserId="691" LastEditorUserId="5026" LastEditDate="2018-07-03T02:37:14.950" LastActivityDate="2018-07-03T02:37:14.950" Title="Is it safe to hold silicon nitride ceramic without protection?" Tags="&lt;safety&gt;&lt;ceramics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2578" PostTypeId="2" ParentId="2577" CreationDate="2012-11-19T20:24:45.897" Score="2" Body="&lt;p&gt;I have done a minimal amount of research to try and help you understand this material. The credibility of this research is very debatable, as it basically involves me typing in silicon nitride into YouTube. That being said, understand that silicon nitride is first of all extremely strong, &lt;a href=&quot;https://www.youtube.com/watch?v=EAY6hL4exVc&quot; rel=&quot;nofollow&quot;&gt;inert&lt;/a&gt;, according to that link, and being considered as a contemporary 'ideal' material for use in the human body. Inert just means it will not react which is generally safe and certainly not caustic to life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know however, that if this material is present in powder form I would highly recommend a breathing mask at the very least as it is usually inadvisable to inhale solid particles. In the scenario that it shatters it could potentially be sharp. However, if it is merely a &lt;strong&gt;bearing&lt;/strong&gt; we are talking about, then, you probably have very little to worry about other than contaminating your materials if you are going to use the part in high performance devices, and you therefore would want to wear clean gloves so as not to get dust on the bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I advise you to search you-tube and other sources for people handling the material, but more than likely it would be less beneficial to the bearing or your project to handle the material without gloves than it would be to your health.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2012-11-20T01:33:23.987" LastActivityDate="2012-11-20T01:33:23.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2579" PostTypeId="2" ParentId="2576" CreationDate="2012-11-19T20:41:19.073" Score="16" Body="&lt;p&gt;I assume it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Deuterium&quot;&gt;deuterium&lt;/a&gt;, the isotope of hydrogen that has a single neutron as well as a proton in the nucleus. As such, it has the same atomic number as hydrogen, which means you &lt;a href=&quot;http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules#Isotopes&quot;&gt;break the tie on atomic mass&lt;/a&gt; --  deuterium is higher priority than hydrogen but lower than everything else.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-11-19T20:41:19.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2580" PostTypeId="1" CreationDate="2012-11-19T21:39:06.400" Score="3" ViewCount="598" Body="&lt;p&gt;I am trying to create a small 10 to 20cm cube that can conserve the heat in 100 to 200ml of 60-90c water in a small Pyrex glass for about an hour. The problem is that I can not use commercial insulators like foam, bubble wrap, and the like. I have searched Google for ideas, but all the results are commercial. I also can not heat or cool the device beforehand, and I have to have a 2cm hole in the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried cotton balls and a packed cotton balls with felt mixture, and cotton balls worked the best.&lt;/p&gt;&#xA;" OwnerUserId="867" LastActivityDate="2012-11-22T06:21:54.960" Title="Non-Commercial Thermal Insulation" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2581" PostTypeId="2" ParentId="2575" CreationDate="2012-11-20T02:29:27.467" Score="2" Body="&lt;p&gt;Because we can have only one electron per quantum state.  Spin up and spin down are two different states.  A linear combination of the two is not a new independent state.  It is obviously formed from the spin up and spin down states.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-20T02:29:27.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2582" PostTypeId="2" ParentId="2568" CreationDate="2012-11-20T03:48:56.583" Score="5" Body="&lt;p&gt;The bad taste left after drinking wine can be attributed a number of agents found in wine, particularly alcoholic wines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcoholic wines typically contain 8-15% ethanol. When ethanol mixes with oxygen in air it gets oxidised, gradually producing acetic acid (vinegar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the reaction of ethanol with urea (naturally present in wine) to form ethyl carbomate (urethane) can produce a saline, bitter taste, although probably not toxic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ethanol in wine is produced through the fermentation of carbohydrates by yeast. This process also results in the production of acetaldehydes which contribute to its complex set of flavours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethanol itself however is tasteless. It is however toxic and a &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_IARC_Group_1_carcinogens&quot;&gt;class 1 carinogen&lt;/a&gt; according to the International Agency for Research on Cancer (IARC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When consumed, ethanol is &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/10563665&quot;&gt;oxidised by alcohol dehydrogenase (ADH) enzymes in the body to produce acetaldehyde&lt;/a&gt;, which is 10-30 times more toxic than alcohol and &lt;a href=&quot;http://monographs.iarc.fr/ENG/Publications/internrep/08-001.pdf&quot;&gt;possibly carcinogenic in humans&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recent scientific research on cancer seems to indicate even low levels of alcohol consumption (1 cup per day) 'increases the risk of cancer of oral cavity and pharynx, esophagus and female breast' Source: &lt;a href=&quot;http://annonc.oxfordjournals.org/content/early/2012/08/21/annonc.mds337.full&quot;&gt;http://annonc.oxfordjournals.org/content/early/2012/08/21/annonc.mds337.full&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a bad taste may not be the worst thing alcohol leaves in one's mouth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to the health effects of alcohol, please refer to the &lt;a href=&quot;http://www.who.int/substance_abuse/publications/globalstatusreportalcohol2004_healtheffects.pdf&quot;&gt;World Health Organisation report&lt;/a&gt; which is based on data collected scientifically from around the world over many years.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2012-11-20T03:48:56.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2583" PostTypeId="2" ParentId="2577" CreationDate="2012-11-20T05:15:38.120" Score="5" Body="&lt;p&gt;I have a few silicon nitride balls in my drawer. They look pretty cool, that's why I keep them and touch them all of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXbdH.jpg&quot; alt=&quot;http://i.stack.imgur.com/WnjnP.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a very hard, glasslike material. It won't be easy to crack. There is nothing coming off the surface, ball bearings have extremely smooth surfaces (nm roughness), if anything would rub off, they would not be usable as ball bearings. Besides that, I don't think it is toxic. It is an extremely inert material, much more inert than steel, does not dissolve in much. It is even used in prosthetic parts that go into the body. I assume it needs a lot of certification for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you break it (which won't be easy), you'll have some sharp pieces that might cut you, that's all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only risk would be from a ball bearing that was ground to a very fine dust.&lt;/p&gt;&#xA;" OwnerUserId="872" LastEditorUserId="275" LastEditDate="2012-11-20T11:35:45.097" LastActivityDate="2012-11-20T11:35:45.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2584" PostTypeId="2" ParentId="2580" CreationDate="2012-11-20T10:35:00.190" Score="2" Body="&lt;p&gt;Often the key with stopping the conduction of heat is to provide as many phase transitions as possible. Most thermal insulators achieve this by having voids of Air which is achieved in a variety of ways, bubble wrap, fibre glass etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &quot;non comercial&quot; I assume they mean stuff you have lying around the lab?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To attempt this I'd layer blue roll / cotton wool / glass wool with tin foil. The more layers the better.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2012-11-20T10:35:00.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2586" PostTypeId="1" AcceptedAnswerId="2587" CreationDate="2012-11-20T21:57:25.360" Score="15" ViewCount="2940" Body="&lt;p&gt;When two atomic orbitals make one molecular orbital there must be an antibonding orbital also. Why should they make also an antibonding molecular orbital, such as the $2\mathrm b_2$ orbital in the scheme below? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrcJl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="16683" LastEditDate="2018-08-27T17:40:32.257" LastActivityDate="2020-07-20T16:06:03.460" Title="Why are antibonding molecular orbitals formed?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2587" PostTypeId="2" ParentId="2586" CreationDate="2012-11-20T23:03:58.320" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why should they make also an antibonding molecular orbital?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Orbital symmetry must be conserved. Thus, when you mix orbitals, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2569/in-sp2-hybrid-orbital-why-does-the-p-orbitals-only-have-3-sides-instead-of-4/2570#2570&quot;&gt;the number of orbitals must also be conserved&lt;/a&gt;. As a consequence, when two orbitals are operated on to produce new orbitals, two new orbitals must be created. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it conceptually this way: When two orbitals that have the same symmetry and appropriately matched energy come together, a bonding orbital is formed. This bonding orbital contains all of the &quot;good orbital stuff&quot; from that pair of orbitals. But, because there were two orbitals originally, there has to be &quot;orbital stuff&quot; left over. This leftover &quot;orbital stuff&quot; forms the antibonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the screwy part is that &lt;a href=&quot;https://chemistry.stackexchange.com/questions/905/hierarchy-of-electronic-wavefunctions/923#923&quot;&gt;orbitals are not real&lt;/a&gt;. They are mathematical constructs to describe the wave-like probabilistic properties of electrons. If there are no electrons in the orbital, then the orbital is not really &quot;there&quot;. So, while an antibonding orbital is formed, it is not real unless it has electrons in it. In the case of water above, the antibonding orbitals are not populated, so they don't exist. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2012-11-20T23:03:58.320" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2588" PostTypeId="1" AcceptedAnswerId="2592" CreationDate="2012-11-21T06:15:14.130" Score="6" ViewCount="23560" Body="&lt;p&gt;I am reading up on how galvanic cell works and I realised that the flow of electrons is from Zn to Cu. But Zn is more electronegative compared to Cu according to periodic table trends&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read it somewhere that when comparing transition metals, it may be different, but it was not  further elaborated. Can someone explain this to me how Cu is more electronegative than Zn?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="73" LastEditDate="2012-11-21T23:17:33.060" LastActivityDate="2017-11-01T11:07:52.380" Title="Electronegativity of zinc vs copper in galvanic cell" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2589" PostTypeId="1" CreationDate="2012-11-21T15:38:26.370" Score="6" ViewCount="244" Body="&lt;p&gt;I'm having a hard time finding data about an experiment that I know has been done a bunch of times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a sheet of single crystal rutile $\ce{TiO_{2}}$ (110) that I want to reduce to various $\ce{TiO_{2-x}}$ forms where the x is what I want to vary. I want to try to avoid conversion to the other Magnéli phases if possible $(\ce{Ti_{n}O_{2n-1}})$. To do this reduction, I have a high temp oven that I can flow with 95% $\ce{N_2}$ and 5% $\ce{H_2}$ or simply evacuate. I can set the flow rate to cubic centimeters per second. I need a chart that will tell me what temperature, flow rate, and duration I need to convert my sheets surface to a desired reduced value (i.e., I want a specific x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then want to reoxidize the surface by using lower temperatures in a furnace that I can fill with oxygen at different pressures. I need a chart that will tell me what temp, partial pressure of $\ce{O_2}$, and duration I need to re-Oxidize the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal here is not to simply get the surface, but to test other properties. I was hoping this step (finding the proper charts) would have been fairly straightforward, but my literature searches haven't turned up quite what I need. I was hoping someone here can point me to a resource that has this data.&lt;/p&gt;&#xA;" OwnerUserId="874" LastEditorUserId="1389" LastEditDate="2013-05-09T08:24:11.947" LastActivityDate="2014-01-21T21:22:39.423" Title="Reduction and re-oxidation of TiO₂" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2590" PostTypeId="1" CreationDate="2012-11-21T16:12:48.817" Score="6" ViewCount="6068" Body="&lt;p&gt;I was looking up silica gel beads on Wikipedia and noticed that some (and from experience most) are doped with a colouring agent to indicate that they have absorbed all that they can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Silica_gel#Moisture_indicators&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; says that usually cobalt(II) chloride or methyl violet is used, both of which are toxic. But it mentions nothing about what quantity these become toxic? I assumed the quantity in a normal silica gel bag (that come in shoes for example) isn't enough to be harmful, otherwise why would they be using it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it mentions that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Crystalline silica dust can cause silicosis but synthetic amorphous silica gel is indurated, and so does not cause silicosis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are the silica gel beads synthetic amorphous, i.e. do not cause silicosis?&lt;/p&gt;&#xA;" OwnerUserId="877" LastEditorUserId="41328" LastEditDate="2019-04-14T18:57:20.957" LastActivityDate="2019-04-14T18:57:20.957" Title="How dangerous is cobalt(II) chloride used in silica gel balls?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2591" PostTypeId="2" ParentId="2580" CreationDate="2012-11-21T17:25:23.490" Score="0" Body="&lt;p&gt;Take a carton box in which the glass fits easily. Wrap the outside of the glass with a thin layer of plastic (sandwich bag). Now, buy some gap filler (PU spray foam, aerosol) to fill the rest of the space. The PU foam has good filling and isolation properties. Once it is cured (24h), you can take out the glass and use it when necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please read the safely instructions on the label !! Good luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: the justification for buying the PU foam, lies in the that you already spent some time trying to reach your goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;em&gt;background info:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It concerns polyurethane chemistry. Here in Europe this type of canistered foam is widely used on the building site. Its purpose is predominantly to fill cavities and act as an isolation material. It's made out of an isocyante (R-NCO) and a polyol (R-(OH)n) forming a urethane bond (R-C=O-O-N-H-R). Together with a propellant (hydrocarbons), it is pressed out of a canister. The left-over isocyantes will react with ambient water, to form poly-urea bonds. This embodies the curing mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for me to suggest this particular product is because it fills easily and shapes any form. It will cure within a couple of hour. Quit often this material is chosen to prepare surprises (packages with surprises in them....) &lt;/p&gt;&#xA;" OwnerUserId="878" LastEditorUserId="878" LastEditDate="2012-11-22T06:21:54.960" LastActivityDate="2012-11-22T06:21:54.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2592" PostTypeId="2" ParentId="2588" CreationDate="2012-11-22T02:16:02.533" Score="8" Body="&lt;p&gt;There are two common confusions in your question. I hope I can clear them up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrons are not flowing from zinc to copper. Not exactly. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the Zinc-Copper galvanic cell, the net reaction is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-&amp;gt; Zn}^{2+} (aq)\ce{ + Cu(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc is being oxidized:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(s)-&amp;gt;Zn}^{2+}(aq) \ce{+2e-} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is being reduced: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu}^{2+}(aq)\ce{+2e- -&amp;gt;Cu(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electronegativity has nothing to do with electrochemistry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electronegativity&lt;/a&gt; is the attraction an atom has for a pair of electrons in a covalent bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrochemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrochemistry&lt;/a&gt; is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this &lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS34.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST publication&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For zinc:&#xA;$$\ce{Zn-&amp;gt;Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{  kJ/mol}$$&#xA;$$\ce{Zn+-&amp;gt;Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{  kJ/mol}$$&#xA;$$\ce{Zn-&amp;gt;Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For copper:&#xA;$$\ce{Cu-&amp;gt;Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{  kJ/mol}$$&#xA;$$\ce{Cu+-&amp;gt;Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{  kJ/mol}$$&#xA;$$\ce{Cu-&amp;gt;Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{  kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is zinc oxidized and copper reduced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn-&amp;gt;Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$&#xA;$$\ce{Cu}^{2+} \ce{+ 2e- -&amp;gt; Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{  kJ/mol}$$&#xA;$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-&amp;gt; Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -11 \text{  kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G=-nFE$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn&lt;sup&gt;2+&lt;/sup&gt;. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu&lt;sup&gt;2+&lt;/sup&gt;. Zinc reduces the Cu&lt;sup&gt;2+&lt;/sup&gt; ions and copper metal plates onto the zinc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_standard_electrode_potentials&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, which mines them from a number of legitimate sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SHE:&#xA;$$\ce{H2(1 atm) + 2e- -&amp;gt; 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$&#xA;For zinc:&#xA;$$\ce{Zn-&amp;gt;Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$&#xA;For copper:&#xA;$$\ce{Cu-&amp;gt;Cu}^{2+} \ce{+ 2e-}  \ \ \ E^o =+0.34 \text{ V}$$&#xA;To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive:&#xA;$$\ce{Zn-&amp;gt;Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$&#xA;$$\ce{Cu}^{2+} \ce{+ 2e- -&amp;gt; Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$&#xA;$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-&amp;gt; Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$&#xA;This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn&lt;sup&gt;2+&lt;/sup&gt;]&gt;&gt;[Cu&lt;sup&gt;2+&lt;/sup&gt;], then the reverse reaction could be spontaneous.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="40029" LastEditDate="2017-11-01T11:07:52.380" LastActivityDate="2017-11-01T11:07:52.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2593" PostTypeId="1" AcceptedAnswerId="2594" CreationDate="2012-11-22T20:47:00.960" Score="7" ViewCount="28033" Body="&lt;p&gt;Simple question, the average neutral copper atom has 2 valance electrons, and sulfur 6. Wouldn't the two electrons jump from copper and bind to sulfide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The empirical formula is: $\ce{Cu + S -&amp;gt; Cu2S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So apparently two copper atoms bind with every one sulfur atom. Does that mean that the copper atoms each gives one electron away and leaves them with 1 valance electron each?&lt;/p&gt;&#xA;" OwnerUserId="882" LastEditorUserId="7475" LastEditDate="2016-07-06T18:09:46.600" LastActivityDate="2020-02-05T09:46:24.677" Title="If copper has 2 valance electrons and sulfur 6, why don't they bind in pairs?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2594" PostTypeId="2" ParentId="2593" CreationDate="2012-11-22T21:32:34.083" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the average neutral copper atom has 2 valance electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope. Only one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BkwrL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the transition metals, the &lt;em&gt;previous&lt;/em&gt; shell is being filled. Generally, the valence shell can't have more than 8 electrons, but lower shells can. So, you have the two outer electrons filled in in Group 1 and 2, and the rest are filled in in the inner shell. With copper, we have 17 electrons in the inner shell, and 2 on the outer shell by this rule. But--it is sometimes more favorable for the inner shell to be &quot;complete&quot; with 18 electrons, leaving the outer shell with only one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much better explained if you understand what &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbitals&quot; rel=&quot;noreferrer&quot;&gt;atomic orbitals&lt;/a&gt; are. Think of these as as &quot;subshells&quot;. In copper, we have two possible shell configurations: 10 electrons in $3d$ and one in $4s$, or 9 in $3d$, and 2 in $4s$. The numbers signify the shell, and the letters signify the orbital type. There are $5$ d orbitals per shell (not all shells have $d$ orbitals, though), and 1 $s$ orbital. Each orbital holds two electrons. Here, we have the trade-off between having a fully filled $3d$ orbital set and having a fully filled $4s$ orbital. Fully filled orbital sets lead to stability. Anyway, since both choices lead to a fully filled orbital set, a lot of other external factors decide which electronic configuration copper uses. So you have both $\ce{Cu^+}$ and $\ce{Cu^{2+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, &lt;a href=&quot;http://en.wikipedia.org/wiki/Covellite&quot; rel=&quot;noreferrer&quot;&gt;$\ce{CuS}$ exists as well&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-11-22T21:32:34.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2598" PostTypeId="1" CreationDate="2012-11-23T12:49:54.240" Score="6" ViewCount="4449" Body="&lt;p&gt;In my lecture notes for inorganic chemistry, it is stated that magnesium has high affinity for oxygen but is rather stable at room temperature in air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain why this is the case?&lt;/p&gt;&#xA;" OwnerDisplayName="user646" LastEditorUserId="73" LastEditDate="2012-11-24T01:06:32.903" LastActivityDate="2017-04-16T04:27:17.527" Title="Reaction of magnesium with the surrounding air" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2599" PostTypeId="1" CreationDate="2012-11-23T15:28:44.020" Score="2" ViewCount="56" Body="&lt;p&gt;Is there anything special about (S)-3-hydroxypalmitoyl-CoA comparing to those with different number of carbons, e.g. (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemist at all; I am looking at &lt;a href=&quot;http://www.ebi.ac.uk/chebi&quot; rel=&quot;nofollow&quot;&gt;the ChEBI ontology&lt;/a&gt; and wondering why (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA are descendants of hydroxy fatty acyl-CoA, while (S)-3-hydroxypalmitoyl-CoA is not (it is only a descendant of long-chain fatty acyl-CoA).&lt;/p&gt;&#xA;" OwnerUserId="886" LastEditorUserId="4945" LastEditDate="2014-06-12T08:50:03.683" LastActivityDate="2014-06-12T08:50:03.683" Title="Is (S)-3-hydroxypalmitoyl-CoA a hydroxy fatty acyl-CoA?" Tags="&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2601" PostTypeId="1" AcceptedAnswerId="2602" CreationDate="2012-11-23T20:28:29.310" Score="5" ViewCount="1369" Body="&lt;p&gt;Data on &lt;a href=&quot;http://periodictable.com/Elements/060/data.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; conveys that Neodymium is paramagnetic. Yet the Neodymium magnets out there make extremely strong magnets. I'm guessing these are not pure neodymium magnets but some kind of compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to predict the change in magnetic property of a compound given the knowledge of it's participating elements?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="73" LastEditDate="2012-11-24T01:08:14.777" LastActivityDate="2016-03-13T19:56:28.640" Title="How is the magnetic property of an element affected when the element is part of a compound?" Tags="&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2602" PostTypeId="2" ParentId="2601" CreationDate="2012-11-23T21:17:14.173" Score="4" Body="&lt;p&gt;Magnetism appears in many very different forms that have very different origins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All elements show some &lt;strong&gt;diamagnetism&lt;/strong&gt;, even though often overshadowed by other, stronger, forms of magnetism, because it is a property of the bound electrons.&#xA;The diamagnetic contribution caused by the inner electrons that are not part of a chemical bond will therefore remain unchanged if the compound bonds chemically. The diamagnetic contribution from the electrons participating in the bond will of course change, so even for the most simple form of magnetism, it is not possible to predict the diamagnetic behavior of a compound from that of the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Paramagnetic&lt;/strong&gt; materials have a permanent magnetic moment from uncompensated magnetic moments, typically electron spins (there are exceptions where a paramagnet is formed from larger units, like magnetic particles). &#xA;In a very simplified way, you will find unpaired electrons typically on the outer levels of elements, where they are influenced and participate in chemical bonds. This is even the case for the strongly localized d and f electrons that are characteristic for the strongest paramagnets like $\ce{Gd}$.&#xA;So clearly, paramagnetism is influenced by the chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both diamagnetism and paramagnetism are extremely weak effects that normally don't play a role in everyday life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the 'real' forms of magnetism, &lt;strong&gt;ferromagnetism&lt;/strong&gt; (ferri- antiferro and ferri etc.): All of these types of magnetism are &lt;strong&gt;collective phenomena&lt;/strong&gt;, i.e. they are the result of the internal interactions of a large system, through conduction electrons (itinerant), &lt;a href=&quot;http://en.wikipedia.org/wiki/Superexchange&quot; rel=&quot;nofollow&quot;&gt;superexchange&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_exchange&quot; rel=&quot;nofollow&quot;&gt;double exchange&lt;/a&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the collective nature, it is not possible to predict the behavior of a compound from that of the physical parameters of the elements. (Of course, you can simulate the magnetic behavior of more or less any material, but I think that was not the question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could not call $\ce{Fe}$ itself a ferromagnet, the ferromagnetism is a property of a huge number of $\ce{Fe}$ atoms that are acting collectively. This ensemble is very sensitive to changes, often even a small temperature change (e.g. at 20°C for $\ce{Gd}$) will completely change the magnetic properties. A chemical reaction or even simple alloying, will completely change the magnetic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The macroscopic magnetic properties of a material (susceptibilitiy, coercivity, remanence, saturation magnetization, frequency dependence of parameters etc.) do not allow to predict the properties of a compound that contains this material. So you can't predict the coercivity of  $\ce{NdFeB}$ given the coercivities of  $\ce{Nd}$, $\ce{Fe}$ and $\ce B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to simulate the complete electronic structure of a compound with &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;Density Functional Theory&lt;/a&gt;. This simulation will tell you the magnetic parameters of the material, including the ordering temperatures in the case of ferromagnetic materials, to a varying degree of accuracy. It is a very computationally intensive approach, to compute the properties of a new material is a research project, requires some ad-hoc assumptions and will need to be experimentally verified as it is not yet 100% reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This computational approach does not make any use of the macroscopic physical properties of the elements in a compound, it directly solves the many body problem of electrons and nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, it is not possible to predict the change in magnetic property of a compound given the knowledge of the magnetic properties of its participating elements, but you can simulate the electronic structure of a compound, and that will tell you how the compound will behave magnetically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;em&gt;Neodymium&lt;/em&gt; magnets aren't made out of $\ce{Nd}$ metal. They are made from sintered $\ce{Nd2Fe14B}$. $\ce{Nd}$ is only a small fraction of the entire material. This is a very complex material. The exceptionally high coercivity, i.e. the resistance against a change in the magnetization, is the result of high magnetocrystalline anisotropy of $\ce{Nd2Fe14B}$ and the interfaces between the grains that pin the magnetization. This is something you would not think about given just the magnetic parameters of $\ce{Nd}$, $\ce{Fe}$ and $\ce{B}$.&lt;/p&gt;&#xA;" OwnerUserId="804" LastEditorUserId="24052" LastEditDate="2016-03-13T19:56:28.640" LastActivityDate="2016-03-13T19:56:28.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2604" PostTypeId="1" CreationDate="2012-11-23T23:31:42.383" Score="2" ViewCount="5800" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Cosmeceutical&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cosmeceuticals refers to the combination of cosmetics and&#xA;  pharmaceuticals. Cosmeceuticals are cosmetic products with&#xA;  &lt;strong&gt;biologically active ingredients&lt;/strong&gt; purporting to have medical or&#xA;  drug-like benefits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering what  ingredients being &quot;biologically active&quot; means? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cosmetics that are not cosmeceuticals, are their ingredients not &quot;biologically active&quot;?&lt;/p&gt;&#xA;" OwnerUserId="167" LastEditorUserId="7475" LastEditDate="2017-01-20T00:19:23.973" LastActivityDate="2017-01-20T00:19:41.930" Title="What does ingredients being &quot;biologically active&quot; mean?" Tags="&lt;terminology&gt;&lt;medicinal-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2605" PostTypeId="2" ParentId="2598" CreationDate="2012-11-24T01:58:20.690" Score="7" Body="&lt;p&gt;I believe that magnesium exposed to air quickly ends up with a coating of magnesium oxide on its surface.  This oxide coating is quite uniform and does not allow any further oxygen to reach the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several metals, but NOT iron, have this property of a protective oxide coating.  In iron the coating forms a spongey surface easily penetrated by further oxygen.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-11-24T01:58:20.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2606" PostTypeId="1" AcceptedAnswerId="24402" CreationDate="2012-11-24T02:04:40.097" Score="38" ViewCount="8676" Body="&lt;p&gt;There are lots of videos on YouTube showing sodium, potassium, etc. exploding when dropped into water (&lt;a href=&quot;http://www.youtube.com/watch?v=JD85OUkEKKw&quot;&gt;this&lt;/a&gt;, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that when an alkali metal is exposed to water, a violent exothermic reaction occurs where a hydroxide and hydrogen gas are produced, but why and how does the sample of metal end up detonating and fragmenting?  Physically speaking, how can a block of metal seemingly blast apart from the inside when the reaction occurs on the surface of the sample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkali_metal#Reaction_with_water_.28alkali_metal_hydroxides.29&quot;&gt;Wikipedia article on alkali metals&lt;/a&gt; explains this, but I still don't seem to understand how this would result in an explosion.  Diagrams would be very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key of the question is how does the release of hydrogen gas and energy result in an explosion?  Is there thermal runaway, and is the metal physically destabilized in some way during the reaction?&lt;/p&gt;&#xA;" OwnerUserId="888" LastEditorUserId="19616" LastEditDate="2015-10-23T17:11:09.147" LastActivityDate="2020-07-07T16:57:57.817" Title="What exactly is happening when sodium, potassium, or other alkali metal explodes in water?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;explosives&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2608" PostTypeId="2" ParentId="706" CreationDate="2012-11-24T12:33:21.883" Score="3" Body="&lt;p&gt;Orange juice is often subjected to temperature variation during processing, pasteurization, storage and refrigeration at 4$^{\circ}$C. &lt;a href=&quot;http://europepmc.org/abstract/AGR/IND43937669&quot; rel=&quot;nofollow&quot;&gt;According to some research&lt;/a&gt;, there appears to be a dramatic decrease in the concentration of certain flavonoids at storage temperatures especially above 20$^{\circ}$C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variety of flavonoids found in orange juice includes didymin, eriocitrin, hespertin, narirutin, poncirin, apigenin and neoeriocitrin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effective analytical methods used to &lt;a href=&quot;http://www.mdpi.com/1420-3049/12/8/1641/pdf&quot; rel=&quot;nofollow&quot;&gt;identify and quantify individual flavonoids&lt;/a&gt; include gas chromatography mass spectroscopy (GCMS) and high-performance liquid chromatography (HPLC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst chromatography is the most effective method of identifying specific flavonoids, you may be interested to know that ultra-violet spectrophotometry &lt;a href=&quot;http://www.cabdirect.org/abstracts/20093220760.html&quot; rel=&quot;nofollow&quot;&gt;has been used to measure total flavonoid&lt;/a&gt; content before &amp;amp; after treatment.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-11-24T23:25:35.983" LastActivityDate="2012-11-24T23:25:35.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2609" PostTypeId="2" ParentId="2606" CreationDate="2012-11-24T20:38:19.877" Score="12" Body="&lt;p&gt;&lt;strong&gt;EDIT/Retraction:&lt;/strong&gt; This answer was written prior to the publication of the research described in &lt;a href=&quot;https://chemistry.stackexchange.com/a/24402/39&quot;&gt;the accepted answer&lt;/a&gt;, and is now known to be incorrect. &lt;strong&gt;&lt;em&gt;Science!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I assume this is due to mesoscopic defects -- if you think about a piece of metal on a molecular scale, it's not going to be presenting a totally flat sheet of atoms at its surface. Hydrogen gas forming within cracks in the metal can force those cracks to open wider, and this then exposes more surface to react, and produces more cracks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process will result in the force of the generated hydrogen ripping the bulk metal apart if it happens quickly enough. If the heat generated from the reactions then also starts boiling water in those same cracks and defects, that'll only add to the effect.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-16T14:22:33.210" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2610" PostTypeId="2" ParentId="2408" CreationDate="2012-11-24T23:47:02.763" Score="7" Body="&lt;p&gt;A typical organic light emitting diode (OLED) structure consisting of a glass substrate, a indium titanium oxide (ITO) anode, and an organic bilayer diffused into a metallic cathode, is given below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/II2HD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organic bilayer consists of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a &lt;strong&gt;hole transport layer (HTL)&lt;/strong&gt;, typically of N, N'-diphenyl-N, N'-bis(3-methylphenyl) 1, 1'-biphenyl-4, 4' diamine (TPD) or N,N′-di(naphthalene-1-yl)-N,N′- diphenylbenzidine (NPB)&lt;/li&gt;&#xA;&lt;li&gt;an &lt;strong&gt;emitting/electron transport layer (EML/ETL)&lt;/strong&gt;, typically of tris(8- hydroxyquinoline) aluminum (Alq3)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://espace.library.uq.edu.au/eserv.php?pid=UQ:8598&amp;amp;dsID=SMS.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://espace.library.uq.edu.au/eserv.php?pid=UQ:8598&amp;amp;dsID=SMS.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2012-11-24T23:47:02.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2612" PostTypeId="1" AcceptedAnswerId="2690" CreationDate="2012-11-25T05:25:46.703" Score="5" ViewCount="299" Body="&lt;p&gt;I'm working on some combustion problems using &lt;a href=&quot;http://code.google.com/p/cantera/&quot; rel=&quot;nofollow&quot;&gt;Cantera&lt;/a&gt; and I can't find references for the thermodynamic or transport properties for many of my chemical species. I'm looking for NASA-7 or NASA-9 curve-fits for the thermodynamics, although if there is another equation format that exists, I will make it work.  And I also need properties for transport that Cantera calls &quot;well depth&quot;, &quot;diameter&quot;, and &quot;rotational relaxation&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I am looking for: HClO&lt;sub&gt;4&lt;/sub&gt;, HCl (transport information), ClO&lt;sub&gt;3&lt;/sub&gt;, ClO&lt;sub&gt;2&lt;/sub&gt; (transport information), ClO (transport information), Cl&lt;sub&gt;2&lt;/sub&gt; (transport information), Cl (transport information), and ClOH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I suppose the real question is &lt;strong&gt;where can I find thermodynamic and transport properties for gaseous species with chlorine?&lt;/strong&gt; Finding data on hydrocarbons and air is simple, and I found data on liquid refrigerants but nothing on gaseous chlorine species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I checked NIST and JANAF and neither had the information. I can find constant values, but running a very detailed reaction mechanism with calorically perfect equation of state and power or Sutherland law transport isn't worth the computational cost.&lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="785" LastEditDate="2012-11-29T06:24:11.357" LastActivityDate="2012-12-04T06:53:32.263" Title="Source for thermodynamic and/or transport properties" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2617" PostTypeId="1" CreationDate="2012-11-25T11:41:58.113" Score="2" ViewCount="1089" Body="&lt;p&gt;Can silica gel beads (which I believe are amorphous) when crushed to dust cause silicosis?&lt;/p&gt;&#xA;" OwnerUserId="892" LastEditorUserId="4945" LastEditDate="2014-05-22T09:09:53.813" LastActivityDate="2014-05-22T09:09:53.813" Title="Can silica gel beads cause silicosis if crushed?" Tags="&lt;inorganic-chemistry&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2618" PostTypeId="2" ParentId="2606" CreationDate="2012-11-25T14:28:36.140" Score="7" Body="&lt;p&gt;I don't think any clever effects need to be invoked to explain this phenomenon. The simple fact that the heat of the reaction is always enough to melt the blobs of metal involved is enough. Once you have a mobile molten drop of metal (which is often on fire), very little energy is required to cause the drop to split up so the violence of the continuing reaction is often quite sufficient to cause the drop to split and disperse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily verify that drops of liquids don't need much energy to cause them to disperse and, if you were still allowed to play with mercury, you could also prove this applies to liquid &lt;em&gt;metals&lt;/em&gt;. Drops of mercury will easily split into smaller drops when small amounts of energy are added (e.g. letting a drop fall a short distance onto a surface). These smaller drops are a lot harder to contain than their parent (which is one reason why mercury contamination is so prevalent and hard to remove in old labs). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short, exothermal reaction + liquid metal = readily dispersed drops of burning metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS&#xA;While what I say in this answer is broadly correct, the recent research (see accepted answer) add additional explanatory power to the detail of the reaction. The reaction is &lt;em&gt;more&lt;/em&gt; violent that my simple mechanism would explain and the high-speed camera footage demonstrates that more factors were at wok than most people expected.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2016-04-18T00:06:14.073" LastActivityDate="2016-04-18T00:06:14.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2619" PostTypeId="1" AcceptedAnswerId="2630" CreationDate="2012-11-25T17:37:10.277" Score="5" ViewCount="196" Body="&lt;p&gt;I am looking for introduction to modeling of chemical reactions. I think there is the base approach, where concentrations of chemical species are given, plus ratios of each possible reaction / outcome. Is there a paper lightly explaining internal mechanics of such reactions, and their modeling via system of differential equations?&lt;/p&gt;&#xA;" OwnerUserId="893" LastEditorUserId="30836" LastEditDate="2017-10-13T15:02:09.330" LastActivityDate="2017-10-13T15:02:09.330" Title="Introduction to modeling chemical reactions" Tags="&lt;kinetics&gt;&lt;reference-request&gt;&lt;books&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2620" PostTypeId="1" AcceptedAnswerId="2622" CreationDate="2012-11-25T20:46:51.473" Score="3" ViewCount="759" Body="&lt;p&gt;I prepared an isotonic solution of table salt in water following this recipe: 1 leveled coffee spoon of salt in 0,5L of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9g/L would be an isotonic solution, a leveled coffee spoon is commonly given as 5g - close enough. I use water from the tap, which was freshly boiled in a water cooker. The water is not chlorinated, I don't know the hardness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even given hours of time, a significant amount of salt doesn't go into solution - visible as turbidity/precipitation. How can this be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the solution to flush my child's nose, so I don't want to use too much salt.&lt;/p&gt;&#xA;" OwnerUserId="895" LastEditorUserId="73" LastEditDate="2012-11-26T01:07:07.197" LastActivityDate="2012-11-26T07:28:35.033" Title="Why is my NaCl solution seemingly saturated, when I followed the recipe for an isotonic solution?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2622" PostTypeId="2" ParentId="2620" CreationDate="2012-11-26T05:03:11.117" Score="3" Body="&lt;p&gt;Salt added to food could be sea salt, common table salt or iodized table salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common Table salt typically contains anti-caking agents (desiccants) such as magnesium carbonate, calcium silicate, dicalcium phosphate, tricalcium phosphate, calcium carbonate and sodium alumina silicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodised table salt may contains added potassium iodide for people who may otherwise have low consumption of this nutrient which is essential for proper thyroid function. Since iodide will slowly oxidize in moist air, stabilizing agents such as sodium thiosulphate or dextrose are also added. Some will also contain a base such as sodium bicarbonate, disodium phosphate or sodium pyrophosphate to also inhibit oxidation of iodide. There may also be added sequestering agents which bond to trace metal ions present as impurities to prevent them oxidizing the iodide.$^{[1]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of magnesium and calcium will add &quot;hardness&quot; to the water which could cause precipitates such as calcium carbonate and magnesium carbonate to form.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-11-26T07:28:35.033" LastActivityDate="2012-11-26T07:28:35.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2623" PostTypeId="2" ParentId="2417" CreationDate="2012-11-26T06:36:30.587" Score="6" Body="&lt;p&gt;Replacing Hydrogen is the only way how chlorination (or halogenation) works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in 2,2-dimethylpropane the central carbon (carbon-number 2) is quaternary, meaning all four of its valencies are satisfied by carbon atoms, which implies that there are no hydrogen atoms which can be substituted by chlorine (or any halogen), whereas all the remaining 4 carbon atoms are identical, so substitution on any one of those will result in the same compound (1-chloro-2,2-dimethylpropane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas in 2-methylpropane(commonly known as isobutane) the carbon-number 2 is only tertiary meaning only 3 of its valencies are satisfied by carbon-atoms and it has one hydrogen atom on it. So on substitution of this hydrogen atom by chlorine the product is 2-chloro-2-methylpropane and on substitution of any one of the terminal (last) carbon atoms would all result in 1-chloro-2-methylpropane. That is two different products.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="10185" LastEditDate="2015-10-02T16:18:31.977" LastActivityDate="2015-10-02T16:18:31.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2624" PostTypeId="2" ParentId="2617" CreationDate="2012-11-26T06:55:57.377" Score="3" Body="&lt;p&gt;No, it won't. Silica beads dust may cause irritation to the skin and eyes. Unlike silica crystalline,  synthetic amorphous silica gel is  indurated, and so does not cause silicosis.&lt;/p&gt;&#xA;" OwnerUserId="815" LastActivityDate="2012-11-26T06:55:57.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2625" PostTypeId="2" ParentId="2604" CreationDate="2012-11-26T13:11:25.913" Score="4" Body="&lt;p&gt;Chemists usually marks as biologically active compounds, that have clear effect on living being in small amounts. For example, methane is not biologically active, but potassium cyanide clearly is. Of course, the borderline is quite blurry and can move if context changes.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-11-26T13:11:25.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2627" PostTypeId="1" AcceptedAnswerId="2629" CreationDate="2012-11-26T19:37:55.423" Score="17" ViewCount="3891" Body="&lt;p&gt;Many uncharged organic compounds are sold in solid phase with HCl conjugate - e.g. the antihistamine &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Diphenhydramine_HCl&amp;amp;redirect=no&quot;&gt;diphenhydramine.HCl&lt;/a&gt; or the derivitizing agent &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/fluka/76735?lang=en&amp;amp;region=US&quot;&gt;pentafluorobenzyl hydroxylamine.HCl&lt;/a&gt;, among many others. Assuming the compound will be dissolved at relatively low concentrations in a buffered medium (for instance when a drug enters the bloodstream) the HCl shouldn't have much chemical effect. Why is it there - does it aid crystallization, or perhaps it's leftover from synthesis and is difficult to separate?&lt;/p&gt;&#xA;" OwnerUserId="36" LastActivityDate="2014-09-17T03:58:43.577" Title="Why are solid phase organics often conjugated to HCl?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2628" PostTypeId="1" AcceptedAnswerId="2648" CreationDate="2012-11-26T21:00:52.283" Score="2" ViewCount="104" Body="&lt;p&gt;I'm a gold smith and I want to start experimenting with non-standard metal-alloys. Now high school is a long time ago for me so while I think I've got this right, I'm just not as sure as I should be when working with precious metals costing 40 dollar a gram. So could anyone confirm that what I'm doing is right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two metals&#xA;Metal X, with an atomic-weight of 20&#xA;Metal Y, with an atomic-weight of 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to create an intermetallic alloy of $\ce{XY2}$, where one atom of X is combined with two atoms of Y and I'd want 10 grams of that alloy. I would alloy (20*10)/26=&lt;strong&gt;7.692g&lt;/strong&gt; of metal X and (6*10)/26=&lt;strong&gt;2.308g&lt;/strong&gt; of metal Y, correct?&lt;/p&gt;&#xA;" OwnerUserId="900" LastEditorUserId="73" LastEditDate="2012-11-27T21:10:06.770" LastActivityDate="2012-11-29T00:54:46.453" Title="Alloy-component weight based on desired molecule" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2629" PostTypeId="2" ParentId="2627" CreationDate="2012-11-26T21:14:55.533" Score="20" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it aid crystallization?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the drugs that you see as &quot;hydrochlorides&quot; are &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaloid&quot; rel=&quot;noreferrer&quot;&gt;alkaloids&lt;/a&gt; - a class of basic natural products containing one or more amine functional groups. The dimethylamino group in the upper right of the structure of &lt;a href=&quot;http://en.wikipedia.org/wiki/Diphenhydramine&quot; rel=&quot;noreferrer&quot;&gt;diphenhydramine&lt;/a&gt; (below) is a prime example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/clWPD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these compounds are not crystalline (at best amphorous, at worst waxy or liquid). These compounds are expensive to purify, since the inexpensive method of &lt;a href=&quot;http://en.wikipedia.org/wiki/Recrystallization_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;recrystallization&lt;/a&gt; is not an option. &lt;a href=&quot;http://en.wikipedia.org/wiki/Column_chromatography&quot; rel=&quot;noreferrer&quot;&gt;Chromatography&lt;/a&gt; is an expensive and wasteful process on the industrial scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reacting alkaloids with HCl (or other acids - for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Dextromethorphan&quot; rel=&quot;noreferrer&quot;&gt;dextromethorphan&lt;/a&gt; HBr) usually produces a crystalline compound. These crystalline compounds are easier to purify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yKwTm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a second more important benefit associated with the hydrochloride salts - they have higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Bioavailability&quot; rel=&quot;noreferrer&quot;&gt;bioavailability&lt;/a&gt; than the freebases. The freebase is nonpolar and poorly soluble in water. The hydrochloride salt has an ionic group, which increases water solubility.  &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-11-26T22:07:31.237" LastActivityDate="2012-11-26T22:07:31.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2630" PostTypeId="2" ParentId="2619" CreationDate="2012-11-26T22:25:59.513" Score="5" Body="&lt;p&gt;The topic you are asking about is called &lt;strong&gt;reaction dynamics&lt;/strong&gt;, and is closely related to &lt;strong&gt;chemical kinetics&lt;/strong&gt; (the former is more focussed on atomistic events, reactive and nonreactive collisions, while the later usually describes the mathematical treatment of the kinetic laws governing species concentrations). It was the topic of the &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1986/index.html&quot; rel=&quot;noreferrer&quot;&gt;1986 Nobel Prize in Chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is covered by most &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198792859&quot; rel=&quot;noreferrer&quot;&gt;physical chemistry&lt;/a&gt; textbooks, and more in depth by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0137371233&quot; rel=&quot;noreferrer&quot;&gt;specialized textbooks&lt;/a&gt; on that matter. It is sometimes taught together with basics of statistical mechanics or statistical thermodynamics. Check your favorite (or local) university for a list of reading material on this topic!&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-11-26T22:25:59.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2631" PostTypeId="1" CreationDate="2012-11-26T23:27:49.940" Score="5" ViewCount="3284" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Liquid_oxygen&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; on liquid oxygen tells us&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Liquid oxygen is also a very powerful oxidizing agent [...], if soaked in liquid oxygen, some materials such as coal briquettes, carbon black, etc., can detonate unpredictably from sources of ignition such as flames, sparks or impact from light blows.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have heard similar warnings about LOX in texts about amateur and semi-professional rocketry, but I know nobody who has ever handled stuff like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, say I spill some LOX and soak paper lying around (that maybe has been warmed by a desk lamp), will that burst into a fireball?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe more to the point, how do I assess such risks short of conducting foolhardy experiments? Is there a way to estimate the maximum effect a given amount of spillage on a given material may have?&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2012-11-30T09:49:21.037" Title="In practice, how dangerous is liquid oxygen as an oxidizing agent?" Tags="&lt;safety&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2632" PostTypeId="2" ParentId="2599" CreationDate="2012-11-27T03:43:28.400" Score="1" Body="&lt;p&gt;This just looks like an arbitrary distinction they made in the ontology between 'short chain' acyl-CoAs and 'long chain'. As you can see, the 3-hydroxypalmitoyl chain is somewhat longer (16C) than the others. It stands to reason that the creators of the ontology have decided to subdivide hydroxy fatty acyl-CoAs into those with shorter hydroxyl fatty chains:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hydroxyoctanoyl:   8C&lt;/li&gt;&#xA;&lt;li&gt;hydroxydecanoyl:  10C&lt;/li&gt;&#xA;&lt;li&gt;hydroxylauroyl:   12C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and long:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hydroxypalmitoyl: 16C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As to where the dividing line is, the hydroxytetradecanoyl-CoA (14C) homologue is not 'long chain', whilst there is no entry for hydroxypentadecanoyl-CoA. The octadecanoyl homologue belongs to its own class (of 3-hydroxyoctadecanoyl-CoAs, of which there are only two conceivable enantiomeric members), and there does not appear to be icosanoyl homologue in the database. There may be other members of the 'long chain' family if you search around, who knows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it appears to be arbitrary.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-11-27T03:43:28.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2633" PostTypeId="1" AcceptedAnswerId="2640" CreationDate="2012-11-27T19:56:19.917" Score="16" ViewCount="7306" Body="&lt;p&gt;I've seen people arguing for some questionable &quot;miracle cure-all&quot; alternative medicines not scientifically, but by their own logic. A common claim is, that if a compound looks very similar to an important compound found in our body, it must be very beneficial. By looks very similar I mean the drawings of their structures look similar, as in, similar as images. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this reasoning alone to be unscientific. (correct me if I'm wrong). I would like to explain it to people, but it's hard to do if they lack even basic understandings in chemistry, and have claims like &quot;I think it might happen to be like this, so it's a proof that it's like this&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm looking for a counter-example. Are there famous and well-known (organic) compounds which look similar on a structure drawing, but have vastly different properties, for example one is either beneficial or a necessary component of our body, and the other is highly toxic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I'm sorry for the possible unclear terminology, English is not my first language and I didn't find any better term to say &quot;structure drawing&quot; for &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/b/be/Heme_b.svg/543px-Heme_b.svg.png&quot;&gt;something like this&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="10185" LastEditDate="2015-10-02T16:04:40.697" LastActivityDate="2015-10-02T16:04:40.697" Title="Are there notable organic compunds which look very similar but have very different properties?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;structural-biology&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2634" PostTypeId="2" ParentId="2633" CreationDate="2012-11-27T20:55:09.057" Score="9" Body="&lt;p&gt;Recall that carbon in an organic molecule has 4 bonds in a tetrahedral shape.  If each of those bonds connects to a different group, then there are two configurations of the bonds which have the same formula but a mirror image geometric relationship in space.  Such a carbon (or molecule) is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29&quot;&gt;chiral&lt;/a&gt;.  Chiral molecules have the same formula but different spacial structures and thus are different molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Carvone&quot;&gt;carvone&lt;/a&gt; has two enantiomers (mirror images):  R-(-)-carvone smells like spearmint, while S-(+)-carvone, smells like caraway.  This is because different olfactory receptors in the nose fit the molecules differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.periodicvideos.com/videos/mv_carvone.htm&quot;&gt;video about carvone&lt;/a&gt; and chirality from the good folks at &lt;a href=&quot;http://www.periodicvideos.com/index.htm&quot;&gt;Periodic Videos&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2012-11-27T21:28:28.113" LastActivityDate="2012-11-27T21:28:28.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2635" PostTypeId="2" ParentId="2633" CreationDate="2012-11-27T21:36:36.787" Score="17" Body="&lt;p&gt;One example of disastrously different effects of very similar molecules is &lt;a href=&quot;http://en.wikipedia.org/wiki/Thalidomide#cite_note-Stephens-56&quot; rel=&quot;noreferrer&quot;&gt;thalidomide&lt;/a&gt;. There are two isomers, only differing in their stereochemistry. The (R)-isomer is responsible for the sedating effect it was initially used for. The (S)-isomer lead to horrifying birth defects in the children of woman that took that drug. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xU5PK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is somewhat more complicated, as the isomers are racemized quickly &lt;em&gt;in vivo&lt;/em&gt;, so even if you administer only the (S)-isomer, it will still be converted to the (R)-isomer in the body.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2012-11-27T21:49:59.973" LastActivityDate="2012-11-27T21:49:59.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2636" PostTypeId="1" AcceptedAnswerId="2643" CreationDate="2012-11-27T21:38:27.970" Score="-1" ViewCount="857" Body="&lt;p&gt;I'm working with Poly(allylamine hydrochloride), and I found that the chemical formula is&#xA;$$\ce{(C_3 H_7 N)_{$n$}. HCl}$$&#xA;But I need a diagram like below, but more explicit. I mean with the carbons, where they are exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sKuUj.png&quot; alt=&quot;structural diagram of a polyamine&quot;&gt; (&lt;a href=&quot;http://www.sigmaaldrich.com/medium/structureimages/96/mfcd00084396.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.74263.html?rid=ad9aab54-f204-4c1a-97ea-ba67ddb5927f&quot; rel=&quot;nofollow noreferrer&quot;&gt;web page&lt;/a&gt;, with a lot information of my molecule. What I was looking for is that kind of diagram (the one with 3D aspect). That is the structure of the allylamine hydrochloride, now I want to know what is the structure of the poly(allylamine hydrochloride).&lt;/p&gt;&#xA;" OwnerUserId="903" LastEditorUserId="4945" LastEditDate="2015-07-01T03:11:27.683" LastActivityDate="2015-07-01T03:11:27.683" Title="What is the specific structure of Poly(allylamine hydrochloride) (C3H7N)n HCl?" Tags="&lt;structural-formula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2637" PostTypeId="1" AcceptedAnswerId="2639" CreationDate="2012-11-27T21:59:25.110" Score="6" ViewCount="1105" Body="&lt;p&gt;There were ninety original elements on the Periodic Table. Over time, new ones were founded and put on. However, they were all either the original elements experimented on. For example, &lt;a href=&quot;http://www.lenntech.com/periodic/elements/pu.htm&quot;&gt;Plutonium was discovered by deuteron bombardment of uranium&lt;/a&gt;. Therefore, I would like to know: &lt;strong&gt;How were the rest of the elements discovered?&lt;/strong&gt; It would also be nice if an example could be given. Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user904" LastActivityDate="2012-11-27T23:37:07.727" Title="Where did the rest of the elements come from?" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2638" PostTypeId="1" AcceptedAnswerId="4053" CreationDate="2012-11-27T22:01:54.723" Score="2" ViewCount="3284" Body="&lt;p&gt;I'm running &lt;a href=&quot;http://en.wikipedia.org/wiki/Gel_permeation_chromatography&quot; rel=&quot;nofollow&quot;&gt;gel permeation chromatography&lt;/a&gt; (GPC) of a polar, uncharged polymer (acrylamide-based) in a polar solvent (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethylformamide&quot; rel=&quot;nofollow&quot;&gt;dimethylformamide&lt;/a&gt;). The literature recommends adding a salt to the eluent; for my particular experiment, I'm using 0.05 M LiBr (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/pola.22338/abstract&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;). Other salts are reported (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ma902378k&quot; rel=&quot;nofollow&quot;&gt;Sodium Sulfate&lt;/a&gt;, for example) in the literature, both for this eluent/polymer mixture, and for similar polymers and eluents. Why is this done? Most papers simply report the eluent and the additives they used, with no mention of their reasoning. One paper just identifies the eluent, with no mention of any added salts; given the prevalence of the use of salts, I'm not sure if this is an oversight on their part, or if they just didn't use a salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were working with a polyelectrolyte, I would know that adding a salt would screen the charges along the polymer, reduce the $A_2$ effect, and make my data much easier to interpret. &lt;a href=&quot;http://www.waters.com/waters/nav.htm?cid=10167847&amp;amp;locale=en_US&quot; rel=&quot;nofollow&quot;&gt;One GPC column manufacturer&lt;/a&gt; claims (under &quot;What additives are important...&quot;) that dipole-dipole interactions can have a similar effect, but I'm not not sure how much to trust that claim. In addition to confirming or denying that this is the reason, I would also appreciate pointers to any relevant literature.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2012-11-29T22:49:28.323" LastActivityDate="2013-01-31T20:08:04.700" Title="Why should I add lithium bromide to my GPC eluent?" Tags="&lt;experimental-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2639" PostTypeId="2" ParentId="2637" CreationDate="2012-11-27T23:37:07.727" Score="10" Body="&lt;p&gt;By the &quot;original periodic table&quot; I assume you're talking about the &quot;elements ocurring in measurable quantities in nature&quot;.--everything till Uranium sans Technetium and Promethium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of preparation of these elements is pretty much the same -- bombardment. Just the projectile and substrate differ.For example, the (accidental) neutron bombardment of Molybdenum in old cyclotron parts led to the discovery of &lt;a href=&quot;http://en.wikipedia.org/wiki/Technetium&quot;&gt;Technetium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other sources as well. Analyzing fission products can get you new elements--that's how &lt;a href=&quot;http://en.wikipedia.org/wiki/Promethium#Discovery_and_synthesis_of_promethium_metal&quot;&gt;Promethium&lt;/a&gt; was discovered. The process of fission creates smaller nuclei--but there is a shower of neutrons created as well (which can bombard everything in sight). Most of the discoveries have been due to the &quot;shower of neutrinos&quot;, because &lt;em&gt;generally&lt;/em&gt; all &quot;smaller nuclei&quot; are of already-known elements. Promethium was synthesized by the &quot;formation of smaller nuclei&quot;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since fission occurs in nature (Usually it's just a little bit, though we sometimes have &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_nuclear_fission_reactor&quot;&gt;natural fission reactors&lt;/a&gt;), we actually have all the elements up till californium existing in minute quantities in nature &lt;sup&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table#Further_development&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2012-11-27T23:37:07.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2640" PostTypeId="2" ParentId="2633" CreationDate="2012-11-28T03:45:38.417" Score="17" Body="&lt;p&gt;trb456 and Mad Scientist have raised the point that enantiomeric forms of a bioactive molecule may have markedly different activities, however there's more to it than that. A molecule that has a similar shape and surface charge distribution to another biologically important molecule is liable to bind to the same protein receptor sites, but in different ways. This may be a good or a bad thing and is a crucial observation of modern drug design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the potent hallucinogen &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethyltryptamine&quot; rel=&quot;noreferrer&quot;&gt;DMT&lt;/a&gt; (all image credits go to wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBab7.png&quot; alt=&quot;DMT&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the fairly pedestrian neurotransmitter &lt;a href=&quot;http://en.wikipedia.org/wiki/Serotonin&quot; rel=&quot;noreferrer&quot;&gt;serotonin&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BBhaP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are strikingly similar in the presence of both an indole and ethylamine group in both molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the ur-example of two extremely similar molecules with tremendously different effects in the body are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Estradiol&quot; rel=&quot;noreferrer&quot;&gt;Estradiol&lt;/a&gt; (female sex hormone):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/isI52.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Testosterone&quot; rel=&quot;noreferrer&quot;&gt;Testosterone&lt;/a&gt; (male sex hormone):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/he8Ls.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The reasoning that a molecule that looks similar to a biologically important species must be beneficial is flawed.&lt;/strong&gt; A molecule that looks similar to a biologically important species may just as easily strongly bind to and jam up the active site of something &lt;em&gt;really important&lt;/em&gt;. This is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_monoxide_poisoning#Myoglobin&quot; rel=&quot;noreferrer&quot;&gt;carbon monoxide&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sarin&quot; rel=&quot;noreferrer&quot;&gt;sarin&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Disulfiram&quot; rel=&quot;noreferrer&quot;&gt;disulfiram&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Roundup_%28herbicide%29&quot; rel=&quot;noreferrer&quot;&gt;glyphosate&lt;/a&gt; work.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-11-28T03:45:38.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2643" PostTypeId="2" ParentId="2636" CreationDate="2012-11-28T09:22:56.770" Score="3" Body="&lt;p&gt;My intuition is telling me that, based on your empirical formula and the image you provided, all three carbons are at the bends in the line-notation as is conventional in bond-line notation. The structure inside the brackets is repeated virtually indefinitely, as shown with the $n$ at the bottom right of the image. I have circled the alleged carbons in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This adds up if you count the number of Hydrogen atoms bonded to these carbon I believe it will add up to 7 as your formula suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure in the link that you provided in your comment does not show us a structure of the hybridization that is, well very good to be honest. For one thing it is not a picture of the polymer, instead it is just a small molecule with a double bond, and the Hydrogen on the Carbon connected to the Nitrogen appear to be in a funky alignment, they should face away from the flat shape the Jmol applet is showing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the structure in the polymer, we should be able to determine from the bond-line notation that if each carbon is $sp^3$ hybridized, then as the bends suggest we have a molecule which bends roughly 109.5&amp;deg; between all of the Carbon-R bonds. This can be visualized as a tetrahedral shape where the Carbon atom sits in the center of the geometric shape and the atoms that it bonds to are at the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;visualize hybridizations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given that the Nitrogen atom is $sp^3$ hybridized, since it has 3 covalent bonds (and one lone pair that is left off for clarity), it should have bond angles of ~108&amp;deg; between itself and either of the two Hydrogen or the one Carbon atom allowing a 'trigonal pyramidal' shape. Not very easy to imagine after staring at this picture, which is why we use model kits to help us visualize the molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nu0Rl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="275" LastEditDate="2012-11-28T11:21:38.293" LastActivityDate="2012-11-28T11:21:38.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2644" PostTypeId="1" AcceptedAnswerId="2646" CreationDate="2012-11-28T16:12:32.833" Score="31" ViewCount="14099" Body="&lt;p&gt;I was learning about galvanic cells and I had a problem understanding why electrons do not travel through the electrolyte solutions themselves, instead preferring to travel through metals. Can electrons travel through an electrolytic solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a galvanic cell, without the wiring between the two electrodes and instead we only have the salt bridge. Won't the electrons (although it is very unfavorable to happen) travel through the salt bridge and the solutions? In other terms will the transformation of the electrodes occur? &lt;/p&gt;&#xA;" OwnerUserId="909" LastEditorUserId="22" LastEditDate="2012-11-28T17:20:40.160" LastActivityDate="2019-03-27T17:01:23.047" Title="Why don't the electrons move through the electrolyte (instead of the circuit) in a galvanic cell?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;electrons&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="2646" PostTypeId="2" ParentId="2644" CreationDate="2012-11-28T19:30:35.107" Score="12" Body="&lt;p&gt;Not in water. Free electron in water is really unfavorable, so no significant concentration of them can be generated chemically, and it almost immediately reduces water itself to hydrogen (but I heard rumors about generation of solvated electrons in water in very special experiment with short half-life)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In liquid $\ce{NH_3}$, however, solvated electrons can occur, so self-discharge of galvanic cells with $\ce{NH_3}$-based electrolyte may occur through travel of electrons via electrolyte. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user5764" LastEditDate="2016-05-24T10:57:03.573" LastActivityDate="2016-05-24T10:57:03.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2647" PostTypeId="1" CreationDate="2012-11-28T23:06:15.883" Score="-3" ViewCount="544" Body="&lt;p&gt;I have a question about electrical potential in galvanic cells ,suppose that we have only half a cell wich means that we have only one Ox/Red couple (one metal and the other as ionic solution ...)clearly nothing will happen ,suppose that we put a wire between the metal and the ground .The ground has no voltage while the Red/OX couple has a potential does It mean that the oxidation will happen ? and the electrons will go through the ground ?If the oxidation happens with any kind of OX/RED couple ,and if we take a couple wich the deltaG of the oxidation reaction is very high (so It's very unlikely to happen all by Itself ) won't we be creating a free source of energy ? Thank you !&lt;/p&gt;&#xA;" OwnerUserId="909" LastActivityDate="2015-05-18T08:53:07.183" Title="Can we create a galvanic cell with only half a cell connected to the ground?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-09-27T22:23:12.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="2648" PostTypeId="2" ParentId="2628" CreationDate="2012-11-29T00:54:46.453" Score="3" Body="&lt;p&gt;Yes, you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.692 g / (20 g/mol) = 0.3846 mol X (or 2.316 * 10^23 atoms X)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.308 g / (3 g/mol) = 0.7693 mol Y (or 4.633 * 10^23 atoms Y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles are just a count of atoms, so if you have two moles of Y per mole of X, you have two Y atoms per X atom.&lt;/p&gt;&#xA;" OwnerUserId="911" LastActivityDate="2012-11-29T00:54:46.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2649" PostTypeId="2" ParentId="2633" CreationDate="2012-11-29T11:49:26.353" Score="1" Body="&lt;p&gt;1-propanol and 2-propanol are isomers but have different boiling points: 370 K and 356K. As I remember my undergraduate studies correctly, my distillation was to separate a mixture of these two alcohols. The metabolism of the two alcohols are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further readings:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propan-1-ol&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Propan-1-ol&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/2-propanol&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/2-propanol&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="396" LastActivityDate="2012-11-29T11:49:26.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2651" PostTypeId="2" ParentId="2633" CreationDate="2012-11-29T13:20:08.027" Score="4" Body="&lt;p&gt;In biochemistry branch searching for new drugs, it was common to search for new active substances using chemical similarity. It does have reasoning behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most drugs acts, binding themselves to some protein site, in the place, where normal ligand of the protein will bind. However, chemical similarity is not enough to be a drug. Sure, a substance must bind to given protein strongly, and imitating typical protein ligand is wonderful for it. But to be a drug, substance must differ from natural substance to modify protein behavior, usually blocking its activity or activating in heavily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of it, drugs are often searched in modified natural protein ligands or their structural analogs, but usually modified. For example, there is well known group of &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_blocker&quot; rel=&quot;nofollow&quot;&gt;beta-blockers&lt;/a&gt;, that mimics &lt;a href=&quot;http://en.wikipedia.org/wiki/Norepinephrine&quot; rel=&quot;nofollow&quot;&gt;noradrenaline&lt;/a&gt;. However, they have to indestructible &lt;em&gt;in vivo&lt;/em&gt; and bind strongly, preferably to some, but not all types of &lt;a href=&quot;http://en.wikipedia.org/wiki/Adrenergic_receptor&quot; rel=&quot;nofollow&quot;&gt;Adrenergic receptor&lt;/a&gt;, so all of them has structural twist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, theoretically it is possible to find a drug in chemically unrelated structures, if they exhibit required groups in right positions or can bind to target protein in some other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question, the most infamous example is previously reported thalidomide. However, it is not the only one. For example, most aminoacides has to &lt;a href=&quot;http://en.wikipedia.org/wiki/Enantiomer&quot; rel=&quot;nofollow&quot;&gt;enantiomers&lt;/a&gt;, but only one of them is usable for our bodies&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get basic understanding of the subject I recommend to dwell the wiki starting from &lt;a href=&quot;http://en.wikipedia.org/wiki/Drug_design&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-11-29T13:28:06.850" LastActivityDate="2012-11-29T13:28:06.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2652" PostTypeId="2" ParentId="2631" CreationDate="2012-11-29T15:52:07.850" Score="7" Body="&lt;p&gt;I'll have to acknowledge my question might be too vague for a straightforward answer, as can be seen by the various comments. The correct answer, as given by &lt;a href=&quot;https://chemistry.stackexchange.com/users/675/paul-j-gans&quot;&gt;Paul J. Gans&lt;/a&gt;, probably is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[V]ery. I know of no way to quantify this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, using the comments as a starting point, I read some more around the topic, and I would like to expand on the comments to the question and to summarize what I found out.&lt;/p&gt;&#xA;&lt;p&gt;Since there's no unit to measure &amp;quot;dangerous&amp;quot; with, we'll first have to get more specific. In the context above, we can offer the following dimensions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ease with which LOX + X is set on fire or detonated, the easier, the more dangerous&lt;/li&gt;&#xA;&lt;li&gt;predictability of ignition, the less predictable, the more dangerous&lt;/li&gt;&#xA;&lt;li&gt;energy output of ignition, the more energy the more dangerous&lt;/li&gt;&#xA;&lt;li&gt;speed of reaction, the faster the worse&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The core problem with all of the above is that the LOX is not a fuel and will, on its own, do nothing at all. So before we can even speculate on possible reactions, we will have to narrow down the fuel options our spilled LOX might encounter. When we also want to account for oxygen-enriched atmosphere by boiled off LOX, a danger in itself, even without spilled, as &lt;a href=&quot;https://chemistry.stackexchange.com/users/81/matt-black&quot;&gt;matt_black&lt;/a&gt; correctly points out, the fuel can be almost anything burnable in the room.&lt;/p&gt;&#xA;&lt;p&gt;That pretty much takes care of 2), predictability, which is low, so keeping the whole setup safe is difficult. -&amp;gt; Danger: high&lt;/p&gt;&#xA;&lt;p&gt;Ease of ignition depends on the fuel, and the number we're after is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;activation energy (Wikipedia)&lt;/a&gt;. I found various Internet pages mentioning LOX spill accidents that turned tarmac into an explosive that can be set off by walking or driving over it, although I have not found an actual report of such an accident. So apparently, there are common materials that combine with LOX to a mixture requiring a very low activation energy. -&amp;gt; Danger: high.&lt;/p&gt;&#xA;&lt;p&gt;Energy content and speed of reaction also depend on the fuel, but the fact that it's being used as rocket propellant and that LOX was used to build explosive charges in mining (so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxyliquit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxyliquits (Wikipedia)&lt;/a&gt;, via &lt;a href=&quot;https://chemistry.stackexchange.com/users/485/permeakra&quot;&gt;permeakra&lt;/a&gt;'s comment) means that we can expect plenty of both for a number of materials. The article also emphasizes the unpredictability, even of things that are &lt;em&gt;meant to&lt;/em&gt; explode. -&amp;gt; Danger: high.&lt;/p&gt;&#xA;&lt;p&gt;Finally, various safety guidelines inspire vivid imaginations of some scenarios:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The whole floor of an office is known to have caught fire when oxygen vapors contacted a lighted cigarette butt. -- &lt;a href=&quot;http://www.air-separation-plants.com/precautions-for-oxygen-plants.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saftey Precautions for Oxygen Plants&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If liquid oxygen spills on asphalt or other surfaces contaminated with combustibles, do not walk on or roll equipment over the area of the spill. &lt;a href=&quot;http://www.ehs.ufl.edu/programs/lab/cryogens/oxygen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid Oxygen, Environmental Health and Safety&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="589" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2012-11-30T09:49:21.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2653" PostTypeId="1" AcceptedAnswerId="2658" CreationDate="2012-11-29T17:44:32.873" Score="3" ViewCount="117" Body="&lt;p&gt;Recently someone is claiming that the laminar wood panels used for building our new primary school contains a &quot;carcinogenic and dangerous compound&quot;, namely phenolphthalein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any evidence of it being used as that on the net.&#xA;Could you confirm my findings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="914" LastActivityDate="2012-11-30T05:43:59.873" Title="Is phenolphthalein used in laminar wood construction panels?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2655" PostTypeId="1" CreationDate="2012-11-29T23:08:19.337" Score="-2" ViewCount="6369" Body="&lt;p&gt;What would be the reaction mechanism for the reaction between epoxide and $\ce{H2SO4}$?&lt;/p&gt;&#xA;" OwnerUserId="916" LastEditorUserId="5026" LastEditDate="2018-03-13T13:18:31.210" LastActivityDate="2018-03-13T13:18:31.210" Title="SN1 like reaction acid-cataylzed hydrolysis of epoxide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-03-14T01:03:43.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="2656" PostTypeId="1" AcceptedAnswerId="2657" CreationDate="2012-11-30T02:15:28.013" Score="8" ViewCount="288" Body="&lt;p&gt;My chemistry is not good enough to solve this problem, and it's driving me crazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my job, a sales agent approached me to be their representative in my area for their product.  It's a product that when mixed with diesel or gasoline, it reduces carbon emissions in the vehicle to almost 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this seems impossible to me.  The burning of hydrocarbons necessarily creates $\ce{CO}$ or $\ce{CO2}$, and the $\ce{CO}$ can be converted to $\ce{CO2}$ in the catalytic converter by oxidizing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the test results, $\ce{CO}$ dropped to almost 0, and $\ce{CO2}$ stayed the same or went down (also my own calculations show the change in $\ce{CO2}$ was not statistically significant with the small sample size - 3 before and 3 after tests on 3 cars).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen the test results (yes, they could have been faked, but they filmed the testing process).  My buddy also personally tested it on his car and got the same results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anecdotal evidence shows that the car's mileage is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me to somewhere to try and figure out what is going on here? It sounds too good to be true, and as such, there is no way I'm going to recommend my company represent this product until I know what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the mean test results (ppm).  Some of them have high variance.  The only statistically significant changes are $\ce{CO}$ reduction in all cars, and all changes in the first car.  The $\ce{NOx}$, $\ce{CO2}$, and $\ce{CxHy}$ changes in car2 and car3 are not significant.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        Before                          After&#xA;        CO    CO2    CxHy    NOx        CO    CO2    CxHy    NOx&#xA;Car1    3491  9.4    0.39    46.7       33.3  7.8    0.07    9.7&#xA;Car2    164   11.5   0.13    2.7        12.9  11.4   0.37    0.7&#xA;Car3    2868  11.1   0.18    14         123   10.9   0.06    39.3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm suspicious this may just be ethanol, but the manufacturer swears it isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advertised ratio is 1 oz to 12.5 gallons of fuel which is less than 0.1% (I believe they used double that in the tests).  Too low to be ethanol anyway, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow-up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have confirmed that I was originally mistaken in that all measurements are ppm.  I received a new copy of the original results, and $\ce{CxHy}$ and $\ce{CO2}$ are indeed measured in percent, rather than ppm.  It was incorrectly rewritten in the report.  The numbers are the same, though.&lt;/p&gt;&#xA;" OwnerUserId="917" LastEditorUserId="39539" LastEditDate="2017-02-27T02:57:48.657" LastActivityDate="2017-02-27T02:57:48.657" Title="How is this mystery chemical reducing carbon emissions?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;&lt;environmental-chemistry&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2657" PostTypeId="2" ParentId="2656" CreationDate="2012-11-30T05:03:00.767" Score="10" Body="&lt;p&gt;With reference to the discussion in the question comments, this product does not appear to appreciably reduce carbon emissions in total. It does however appear to reduce carbon monoxide emissions. As carbon monoxide is a product of incomplete combustion, it is plausible that the additive is helping the fuel burn more completely. Broadly speaking, you would hope that the majority of the carbon in your exhaust gas would come out as carbon dioxide, which is 'merely' a greenhouse gas and not truly nasty like CO or PAHs. Prior to the advent of catalytic converters, carbon monoxide poisoning by vehicle exhaust was a much greater risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that &lt;em&gt;conservation of mass&lt;/em&gt; tells us some powerful things about what to expect from an engine - namely that the carbon that goes in to the engine as a component of fuel must come out as exhaust gasses, particulates, polycyclic aromatic hydrocarbons etc., or else be sequestered inside the engine (though the latter is achievable by putting a banana into the exhaust pipe, it is generally done as a prank).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction of carbon emissions in general is achieved through obtaining more work per mass of fuel burnt, which, when coupled with the mass and other design features of the car confers the fuel efficiency. Unfortunately, the data provided don't tell us anything about that. The more complete burning of the fuel, evinced by the reduction of CO concentration in the exhaust gas does imply some increase in efficiency, however CO is a very small component of exhaust gas, indicating alongside the low concentration of unburnt hydrocarbons that the fuel is already burning virtually to completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the specific claims made by the manufacturer of the additive, I will not make any judgment of good or bad faith, however my distilled conclusions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The product &lt;strong&gt;does not dramatically impact&lt;/strong&gt; &lt;em&gt;total carbon&lt;/em&gt; emissions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The product &lt;strong&gt;appears to appreciably reduce&lt;/strong&gt; &lt;em&gt;carbon monoxide&lt;/em&gt; emissions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-11-30T05:03:00.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2658" PostTypeId="2" ParentId="2653" CreationDate="2012-11-30T05:28:33.193" Score="3" Body="&lt;p&gt;Could it be that he meant &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol_formaldehyde_resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;phenol-formaldehyde resin&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These resins are often used to bond wood chips. They contain unreacted formaldehyde and the resin decomposes slowly, giving off more formaldehyde, which is volatile and very toxic. This is a known health concern. Many companies offer indoor air testing for formaldehyde, you can even buy dubious tests to do it yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My bet would be on such a resin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are indeed &lt;a href=&quot;http://www.google.com/patents/EP0617070A2&quot; rel=&quot;nofollow noreferrer&quot;&gt;some resins that use phenolphtalein&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenolphtalein might be cancerogenic, even though it was used as a laxative &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/18784766&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phenolphthalein was commonly used as a laxative for most of the 20th century. The use of phenolphthalein in laxatives has decreased since 1997 when the United States Food and Drug Administration (FDA) proposed to withdraw its classification as an over-the-counter drug (21 CFR, Part 310). Phenolphthalein has been previously evaluated in 2-year carcinogenicity studies by the National Toxicology Program (1996). The major route of human exposure to phenolphthalein is via ingestion, dermal contact, and inhalation of contaminated air originating from process units manufacturing the compound. &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any case, phenolphtalein is a powder that is not very volatile and would not pose a big risk. &#xA;Commonly the formaldehyde resin is used for particleboard, I have never heard about the phenolphthalein resin being used for this purpose (but this is far from my area of work)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is a simple test. Phenolphtalein changes color in a basic solution and looks like this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WORql.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolve the material in ethanol and add NaOH (drain cleaner), phenolphtalein will turn fuchsia.&lt;/p&gt;&#xA;" OwnerUserId="804" LastEditorUserId="22" LastEditDate="2012-11-30T05:43:59.873" LastActivityDate="2012-11-30T05:43:59.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2659" PostTypeId="2" ParentId="845" CreationDate="2012-11-30T06:59:51.620" Score="1" Body="&lt;p&gt;Yes, diamond filters are used. Usually in the filtration of ethanol and water (vodka). All drinkable ethanol must be filtered through specialized carbon filters. So in the thinking that diamonds are formed from compressed carbon, therefore a diamond filter is the extreme sports of filtration. Whether it's better? I don't know, but it is a great marketing tool. &lt;/p&gt;&#xA;" OwnerUserId="918" LastEditorUserId="470" LastEditDate="2012-12-30T12:23:16.020" LastActivityDate="2012-12-30T12:23:16.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2660" PostTypeId="1" CreationDate="2012-11-30T07:11:06.830" Score="17" ViewCount="64079" Body="&lt;p&gt;Why is the electronic configuration of platinum &#xA;$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or &#xA;$\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="59123" LastEditDate="2018-08-08T11:31:16.437" LastActivityDate="2021-01-15T23:30:30.193" Title="Anomalous electronic configuration of platinum" Tags="&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2662" PostTypeId="2" ParentId="89" CreationDate="2012-11-30T08:15:29.120" Score="3" Body="&lt;p&gt;The data provided by @F'x are very helpful towards answering the question. But the relative stabilization of the bases when hydrogen bonded to each other is germane to the question. Because the role of stacking is still being worked out, I don't think this question has a simple answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since wobble pairs are observed, but I haven't seen them outside a context where conventional base pairs are also present in the same molecule, I'm not even sure their formation is exergonic.&lt;/p&gt;&#xA;" OwnerUserId="912" LastActivityDate="2012-11-30T08:15:29.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2663" PostTypeId="2" ParentId="1269" CreationDate="2012-11-30T08:29:19.167" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to recycle the oxygen efficiently enough to burn methane again and again?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't think so. Sounds a little like perpetual motion. Assuming you're burning the methane into $\mathrm{CO_2}$ and $\mathrm{H_2O}$, you would need a large supply of something that would regenerate the $\mathrm{O_2}$. The water's a much easier target -- you could electrolyze it, for example, but you'd only recover 1/2 the oxygen each cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of many things to act as chemical $\mathrm{X}$ in:&#xA;$\mathrm{CO_2}+\mathrm{X}\rightarrow{}\mathrm{CX}+\mathrm{O_2}$. That would be a killer reaction for the &quot;carbon capture and storage&quot; crowd -- we could just process our carbon emissions into oblivion! Goodbye greenhouse effect! Would sure help terraform Mars!&lt;/p&gt;&#xA;" OwnerUserId="912" LastActivityDate="2012-11-30T08:29:19.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2664" PostTypeId="2" ParentId="2655" CreationDate="2012-11-30T12:41:49.817" Score="1" Body="&lt;p&gt;Note that your question could do with some more data. Here is my answer based on what I &lt;em&gt;think&lt;/em&gt; you mean:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You ought to get a glycol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AUZrz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a proton from the acidic medium attacks the neucleophile, $\ce{O}$. This creates a positively charged trivalent oxygen. The electrons in one of the two $\ce{C-O}$ bonds transfer to the $\ce{O+}$, breaking the bond and effectively transferring the positive charge to the carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, note that the carbocation thus formed may &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;rearrange&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have the nucleophilic attack. An oxygen atom (specifically the lone pair of an oxygen) from a nearby water molecule attacks the carbocation, forming another trivalent positively charged oxygen. This is deprotonated, giving the final glycol product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the same mechanism,  showing movement of electrons (pairs):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oVx0V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-01-31T03:53:11.083" LastActivityDate="2013-01-31T03:53:11.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2665" PostTypeId="1" AcceptedAnswerId="2704" CreationDate="2012-11-30T13:21:39.123" Score="8" ViewCount="2354" Body="&lt;h2&gt;Rephrasing:&lt;/h2&gt;&#xA;&lt;p&gt;The following reactions are a small example of my big system&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(1)\  \ce{A + B &amp;lt;=&amp;gt; AB}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$(2)\  \ce{AB -&amp;gt; CB}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$(3)\ \ce{A -&amp;gt; C}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$(4)\ \ce{C + B &amp;lt;=&amp;gt; CB}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\frac{[AB]}{[A][B]}} = 100 = K_e$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\frac{[CB]}{[C][B]}} = 100 = K_e$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I know the rate constants for reactions (2) and (3), and I know that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[A] + [AB] + [C] + [CB] = T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with T =  some constant.&lt;/p&gt;&#xA;&lt;p&gt;Now, although &lt;strong&gt;there is a fast equilibrium between&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A + B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{AB}$&lt;/span&gt;, and between &lt;span class=&quot;math-container&quot;&gt;$\ce{C + B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CB}$&lt;/span&gt;, &lt;strong&gt;I want to know how the concentration of each form changes with time.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;But I dont know how to write down the differential equations of the system and include the equilibrium constant. Could someone please help me out? My only problem is how to include &lt;span class=&quot;math-container&quot;&gt;$K_e$&lt;/span&gt; in the system of ordinary differential equations.&lt;/p&gt;&#xA;" OwnerUserId="921" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-01-05T01:36:43.190" Title="Simultaneous equilibria" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2666" PostTypeId="1" AcceptedAnswerId="2670" CreationDate="2012-11-30T21:12:31.897" Score="12" ViewCount="34493" Body="&lt;p&gt;Most cheap items (from China mostly) have a strong tangy odor to the plastics (or some resin, but I can mostly smell it on plastics). The smell is always consistent over a broad range or products and I get it for at least a decade. So I do not believe this is a one-off thing and I bet most people have experienced it. The only variation is intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On some really bad products that smell impregnates other things and I failed to remove it with soap, baking soda, acid, scrubbing...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious to know what exactly is that smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently it got some attention already, but there's no consensus, and a lot of crywolf(?) e.g. &lt;a href=&quot;http://dynamics.org/Altenberg/CURRENT_AFFAIRS/CHINA_PLASTIC/&quot; rel=&quot;noreferrer&quot;&gt;Campaign to Halt the Import of &#xA;Chemical-Emitting &#xA;Smelly Plastic from China&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="922" LastEditorUserId="7448" LastEditDate="2015-09-21T22:09:17.130" LastActivityDate="2017-07-07T14:48:14.990" Title="What's that tangy smell on cheap plastic?" Tags="&lt;everyday-chemistry&gt;&lt;plastics&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2667" PostTypeId="1" AcceptedAnswerId="2668" CreationDate="2012-11-30T21:26:47.327" Score="2" ViewCount="11796" Body="&lt;p&gt;I am doing my chemistry homework and I am supposed to write the products when strong acids react with water, but I am unsure of the physical state for hydronium ions. Should it be liquid (l) or aqueous (aq)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HI(aq) + H2O(l) -&amp;gt; ?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3O+(aq) + I^{-}(aq) }$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or should it be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3O+(l) + I^{-}(aq) }$?&lt;/p&gt;&#xA;" OwnerUserId="799" LastEditorUserId="22" LastEditDate="2012-12-01T09:14:29.587" LastActivityDate="2012-12-01T09:14:29.587" Title="Physical state of hydronium" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2668" PostTypeId="2" ParentId="2667" CreationDate="2012-12-01T02:40:46.010" Score="7" Body="&lt;p&gt;This is a bit touchy.  The problem is that a bare proton, $\ce{H^+}$, won't exist in water.  It attracts electrons too strongly.  Once this was recognized, chemists started to write the bare proton as being bound to a water molecule, thus $\ce{H3O^+}$.  However, studies also showed that water molecules are typically linked together in clusters containing a fair number of water molecules held together by hydrogen bonds (Google for this if you are not familiar with it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus today's version really is $\ce{H+_aq}$ but that looks awful, so it is as you note $\ce{H3O+_aq}$ is more often used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point is that how you write it isn't really important.  It is the realization that the bare proton attaches itself to any nearby source of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you should handle your homework depends on what your teacher does in class.  I'd follow that example.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-01T02:40:46.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2669" PostTypeId="1" AcceptedAnswerId="2691" CreationDate="2012-12-01T15:14:31.990" Score="9" ViewCount="2227" Body="&lt;p&gt;In essence what I'm asking is if the applications of Electrochemically activated water (ECA water) as a bactericidal-disinfectant is real and profitable (especially this last part). The principle at first seems very dull just take salt water through an electrolysis process and you get anolytes and catholytes, with disinfectant proprieties. Why is that only now is it viable?&lt;/p&gt;&#xA;" OwnerUserId="924" LastActivityDate="2015-08-15T14:19:28.790" Title="Is electrochemically activated water a real technology?" Tags="&lt;electrochemistry&gt;&lt;water&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2670" PostTypeId="2" ParentId="2666" CreationDate="2012-12-01T15:25:03.310" Score="13" Body="&lt;p&gt;Without a sample and a gaschromatograph it is hard to say what exactly it can be but... there are at least three well-known sources of odour in plastics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Some residual of the monomer that makes up the plastic (that is a polymer).&lt;/li&gt;&#xA;&lt;li&gt;Some residual of some other substance used during the manufatcturing process (catalyzer, co-polymers, modifiers and so on). For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyethylene_terephthalate&quot;&gt;PET&lt;/a&gt;can sometime contain small traces of &lt;a href=&quot;https://en.wikipedia.org/wiki/Terephthalate&quot;&gt;terephthalic acid&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Some odourizing substance that is added to the plastic just to cover others smells.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Usually, tangy smells are a symptom of some kind of  &lt;a href=&quot;https://en.wikipedia.org/wiki/Ester&quot;&gt;ester&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="925" LastEditorUserId="22" LastEditDate="2012-12-01T16:43:25.830" LastActivityDate="2012-12-01T16:43:25.830" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2671" PostTypeId="2" ParentId="2669" CreationDate="2012-12-01T15:35:03.437" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysed_water&quot;&gt;wikipedia page&lt;/a&gt; says all:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The electrolysis of such salt solutions produces a solution of sodium hypochlorite, which is the most common ingredient in store-bought household bleach. The resulting water is a known surfactant (soap) and sanitizer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is: the effect of electrolysis is equivalent to adding a (very) small amount of bleach. There is not real reason to prefer this energy-consuming process to the tradionat bleach.&lt;/p&gt;&#xA;" OwnerUserId="925" LastEditorUserId="22" LastEditDate="2012-12-01T16:44:02.397" LastActivityDate="2012-12-01T16:44:02.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2672" PostTypeId="1" AcceptedAnswerId="2674" CreationDate="2012-12-02T01:14:17.447" Score="7" ViewCount="7394" Body="&lt;p&gt;I am asked to identify the organic/inorganic products of an S&lt;sub&gt;N&lt;/sub&gt;1 reaction between 3-iodopentane and ethanol. I have identified correctly that the substrate is the 3-iodopentane, the leaving group is the iodide, and that the nucleophile is the ethanol. My problem is that I do not understand what the products will be, and I am unsure why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided that the products should be 3-pentanol and ethyl iodide (or iodoethane) because iodide is a good leaving group, and the oxygen takes two electrons and leaves ethane to bond with the pentane group where the iodide left off. Now it makes sense that the ethane and iodide react to create ethyl iodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I am wrong in this case, and I just can not put my finger on why. What is the error in my thinking and what is the concept that I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that my major organic product will be the ether compound which I am having trouble naming, and the other compounds are a protonated ethanol plus the iodide. The most confusing part now is the protonated ethanol and the iodide, why not just create the ethyl iodide?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="5026" LastEditDate="2018-02-27T08:20:52.843" LastActivityDate="2018-02-27T08:23:23.717" Title="What are the products in this SN1 reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2673" PostTypeId="1" AcceptedAnswerId="2675" CreationDate="2012-12-02T02:45:17.673" Score="1" ViewCount="1954" Body="&lt;p&gt;Before I can even begin to contemplate the products of a given &lt;a href=&quot;http://en.wikipedia.org/wiki/SN1_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;S&lt;sub&gt;N&lt;/sub&gt;1 reaction&lt;/a&gt;, I need to be able to determine what my leaving group will be, which should be a species with a strong conjugate acid, and as I understand it the species should be relatively stable once it has left the substrate. Please correct me if I am making any mistakes in my understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydroxide ion will be a terrible leaving group, since its conjugate acid is water, and hydroxide is not very stable by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when I am asked this question &quot;Draw the major organic product(s) for this S&lt;sub&gt;N&lt;/sub&gt;1 reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMk6J.png&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know what will be the best leaving group? Hydroxide is not very good, so perhaps $\ce{CH_3OH}$ will work better? How do I know it is not $\ce{CH_3CH_2OH}$ instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this only one decent option, and I am over thinking this?&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="7475" LastEditDate="2015-09-28T09:37:10.873" LastActivityDate="2015-09-28T09:37:10.873" Title="How should I identify my leaving group in SN1 reactions if there might be more than one option?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2674" PostTypeId="2" ParentId="2672" CreationDate="2012-12-02T05:17:11.567" Score="7" Body="&lt;p&gt;OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's S&lt;sub&gt;N&lt;/sub&gt;1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will then be deprotanated by something acting as a base. There's probably some water around, but I suppose saying protonated ethanol and iodide is fine. This gives the final product of: $\ce{(CH3CH2)2CH-O-CH2CH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ethyl iodide to form, one of two things would need to happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;S&lt;sub&gt;N&lt;/sub&gt;1 mechanism: $\ce{OH-}$ would have to spontaneously leave forming a primary carbocation. It is SUPER DISFAVORED, so much so that it straight up won't happen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;S&lt;sub&gt;N&lt;/sub&gt;2 mechanism: $\ce{I-}$ would have to act as a nucleophile AND $\ce{OH-}$ would have to leave. Iodide is an OK nucleophile, but OH- is a pretty bad leaving group. Both things considered this pathway is unlikely.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This was a pretty weak question by your (I am assuming) professor. This reaction would be horribly slow, and probably would barely go. Now, if they used sodium ethoxide instead of ethanol you get a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Williamson_ether_synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Williamson ether synthesis&lt;/a&gt; and form the same ether and NaI, much more favored thermodynamically.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just realized you probably meant why doesn't $\ce{I-}$ perform an S&lt;sub&gt;N&lt;/sub&gt;2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming there is any water around (there is pretty much always water around in the real world (especially with ethanol), unless remarkably heroic measures are taken to avoid it), the protonated ethanol will quickly react with it and form hydronium, ethanol and $\ce{I-}$. (Protonated ethanol has a $\mathrm pK_\mathrm a$ of about -2.4 and hydronium has a $\mathrm pK_\mathrm a$ of about -1.7.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming there is not water around, you are right it will probably react and form small amounts of ethyl iodide, however this will generate water which will stop this from continuing for the above reason.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could also write the mechanism where the iodide deprotonates to form the ether (and avoid this whole fuss). BUT $\ce{HI}$ has a $\mathrm pK_\mathrm a\approx10$ where as oxonium proton $\mathrm pK_\mathrm a$s are usually between -7 and -2, so this would NOT be an accurate way to write the mechanism (in my opinion), but some professors get a little hand wavy here and allow it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know those are kind of sketchy workaround reasons, but honestly it won't be formed in any real quantity. There is probably another (probably better) argument from a thermodynamic stand point, but I don't know the bond energies off hand.&lt;/p&gt;&#xA;" OwnerUserId="926" LastEditorUserId="5026" LastEditDate="2018-02-27T08:23:23.717" LastActivityDate="2018-02-27T08:23:23.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2675" PostTypeId="2" ParentId="2673" CreationDate="2012-12-02T11:25:52.440" Score="9" Body="&lt;p&gt;This case (and similar cases) you need to think about the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.&lt;/li&gt;&#xA;&lt;li&gt;What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-02T11:25:52.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2676" PostTypeId="1" CreationDate="2012-12-02T14:20:10.173" Score="8" ViewCount="1515" Body="&lt;p&gt;$\ce{R-OH + HBr -&amp;gt; R-Br + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $\ce{R+}$ carbocation is formed during this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{R}$ is any alkyl group instead of methyl or ethyl, can the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_and_beta_carbon&quot; rel=&quot;noreferrer&quot;&gt;beta hydrogen&lt;/a&gt; to the hydroxy group be removed to form a double bond, similar to a dehydration??&lt;/p&gt;&#xA;" OwnerUserId="928" LastEditorUserId="897" LastEditDate="2013-01-02T03:17:09.173" LastActivityDate="2019-06-29T07:25:15.690" Title="Preparation of alkyl halide from alcohols" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2677" PostTypeId="1" CreationDate="2012-12-02T16:30:15.030" Score="8" ViewCount="558" Body="&lt;p&gt;I just came across this reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2S2O3 + 2HCl -&amp;gt; 2NaCl + SO2 + S + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it was possible to predict the products of this reaction if I didn't know it (or any inorganic reaction for that matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we use arrow-pushing (as we do in organic chemistry) to get the products of an inorganic reaction? Can we devise a mechanism to find the products? It seems unrealistic to me how anyone could learn the many reaction of inorganic chemistry.&lt;/p&gt;&#xA;" OwnerUserId="509" LastEditorUserId="9961" LastEditDate="2016-03-13T16:02:09.780" LastActivityDate="2016-03-13T16:02:09.780" Title="Thiosulphate acidic decomposition" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2678" PostTypeId="2" ParentId="2676" CreationDate="2012-12-02T17:16:29.440" Score="5" Body="&lt;p&gt;It can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Direct S&lt;sub&gt;N&lt;/sub&gt;1 Substitution&lt;/strong&gt; &#xA;$$\ce{(CH3)2CH-OH + HBr -&amp;gt; (CH3)2CH-OH2+ + Br-}$$&#xA;$$\ce{(CH3)2CH-OH2+ -&amp;gt; (CH3)2CH+ + H2O}$$&#xA;$$\ce{(CH3)2CH+ + Br- -&amp;gt; (CH3)2CH-Br}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Substitution by elimination-addition&lt;/strong&gt;&#xA;$$\ce{(CH3)2CH-OH + HBr -&amp;gt; (CH3)2CH-OH2+ + Br-}$$&#xA;$$\ce{(CH3)2CH-OH2+ -&amp;gt; (CH3)2CH+ + H2O}$$&#xA;$$\ce{(CH3)2CH+ + H2O &amp;lt;=&amp;gt; CH3-CH=CH2 + H3O+}$$&#xA;$$\ce{(CH3)2CH+ + Br- -&amp;gt; (CH3)2CH-Br}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-02T17:16:29.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2679" PostTypeId="1" AcceptedAnswerId="111065" CreationDate="2012-12-02T18:23:50.710" Score="11" ViewCount="14335" Body="&lt;p&gt;The other day I was dipping out Neapolitan(strawberry, vanilla, &amp;amp; chocolate) ice cream and noticed that the chocolate ice cream tended to melt faster than that of the vanilla and strawberry.  I couldn't remember in which order I dipped the ice cream so I decided to do a controlled experiment with it the next time I dipped it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took three bowls of the same type/shape and dipped out the strawberry first, then vanilla, and chocolate last.  Even dipping the chocolate ice cream last showed that the chocolate ice cream would melt faster than the strawberry or vanilla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither the strawberry nor chocolate use artificial coloring, though the vanilla states that it does.  Is there a reason as to why chocolate ice cream melts faster?  Is it because of the darker coloring that attracts heat or perhaps the low melting temperature of chocolate in general?&lt;/p&gt;&#xA;" OwnerUserId="929" LastActivityDate="2019-03-16T10:12:47.197" Title="Why does chocolate ice cream melt faster than others?" Tags="&lt;food-chemistry&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2680" PostTypeId="1" AcceptedAnswerId="2681" CreationDate="2012-12-02T19:22:51.827" Score="0" ViewCount="7417" Body="&lt;p&gt;List in order of increasing radius: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Rb, K, C5, Kr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this one I got C, Kr, K, Rb&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Ar, Cs, Si, Al&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I got: Ar, Si, Al, Cs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if this is correct? I don't have the solution for these in my book :S&lt;/p&gt;&#xA;" OwnerUserId="930" LastEditorUserId="30" LastEditDate="2012-12-03T15:03:00.173" LastActivityDate="2012-12-05T04:17:59.043" Title="List in order of increasing radius" Tags="&lt;homework&gt;&lt;atoms&gt;&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2681" PostTypeId="2" ParentId="2680" CreationDate="2012-12-02T19:45:45.197" Score="1" Body="&lt;p&gt;A) Rb:265 pm&#xA;   K: 243 pm&#xA;   C: 67 pm&#xA;   Kr: 88 pm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Ar: 71 pm&#xA;   Cs:298 pm&#xA;   Si: 111 pm&#xA;   Al: 118 pm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Pm is picometer =1.0 × 10^-12 meter&lt;/p&gt;&#xA;" OwnerUserId="815" LastActivityDate="2012-12-02T19:45:45.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2682" PostTypeId="1" AcceptedAnswerId="2683" CreationDate="2012-12-02T23:12:01.747" Score="6" ViewCount="2317" Body="&lt;p&gt;Considering the Sn1 reaction of (S)-2,3-dimethyl-3-pentanol, knowing that this molecule has a chiral carbon, should I always assume that both enantiomers are present, and therefore the products of a reaction will always have $2^n$ products for $n$ stereocenters on the substrate, &lt;strong&gt;assuming that each stereocenter has equal chance of reacting&lt;/strong&gt;? Is there two products of this Sn1 reaction due to the fact that the molecule is chiral, and therefore its enantiomer is present and has equal chance of reacting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hIoFi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="22" LastEditDate="2012-12-02T23:19:13.040" LastActivityDate="2012-12-03T02:18:52.733" Title="If I have a chiral substrate, should I always assume that both enantiomers are present?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;optical-properties&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2683" PostTypeId="2" ParentId="2682" CreationDate="2012-12-03T02:18:52.733" Score="3" Body="&lt;p&gt;In this S&lt;sub&gt;N&lt;/sub&gt;1 reaction there would be two products even if only one enantiomer reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enantiomerically pure (S)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALSO, enantiomerically pure (R)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this kind of phenomenon happens, there must be an achiral intermediate that can react to produce either enantiomer with equal propensity. In this case, what is the structure (geometry and everything) of the carbocation formed when the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2673/how-should-i-identify-my-leaving-group-in-sn1-reactions-if-there-might-be-more-t&quot;&gt;leaving group leaves&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2012-12-03T02:18:52.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2684" PostTypeId="2" ParentId="2660" CreationDate="2012-12-03T14:40:09.527" Score="0" Body="&lt;p&gt;At this point in the periodic table, s (outer) and d almost have same energy. So, it is better having 4 fully filled and 2 half filled orbitals, than having 5 fully filled and 1 empty ( this empty 6s creates a problem as it is not energetically favored ), since in the first case all of them are stable whereas, in the second the empty 6s is not. &#xA;The initial configuration is : [Xe] 4f&lt;sup&gt;14&lt;/sup&gt; 5d&lt;sup&gt;8&lt;/sup&gt; 6s&lt;sup&gt;2&lt;/sup&gt;, then one electron is transfered from 6s to 5d, so that all orbitals become stable, either through full filling or half filling, which is better then having one empty and unstable.&#xA;This makes it : [Xe] 4f&lt;sup&gt;14&lt;/sup&gt; 5d&lt;sup&gt;9&lt;/sup&gt; 6s&lt;sup&gt;1&lt;/sup&gt;.&#xA;It cannot be the other two because, in both of, one orbital is empty and unstable.&lt;/p&gt;&#xA;" OwnerUserId="509" LastEditorUserId="735" LastEditDate="2012-12-04T17:01:40.780" LastActivityDate="2012-12-04T17:01:40.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2686" PostTypeId="1" AcceptedAnswerId="2692" CreationDate="2012-12-03T21:12:53.580" Score="3" ViewCount="1858" Body="&lt;p&gt;As the question title says, I would like to know what the physical and chemical properties are. If possible, please include examples. Since it seems to be related, I would also like to know what physical and chemical changes are. I believe that things such as crushing a can is a physical change, but I am not sure what would constitute a chemical change &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1293/is-crushing-an-avocado-a-chemical-or-physical-change&quot;&gt;(perhaps doing something that rearranges the molecules?)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I originally wanted to ask if cooking an egg constitutes as a chemical or physical change, but I thought that that would be closed as too vague. If possible, please explain which one it is too. Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user904" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2019-01-06T11:35:25.497" Title="What are physical properties and chemical properties?" Tags="&lt;everyday-chemistry&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2687" PostTypeId="2" ParentId="2686" CreationDate="2012-12-03T21:27:35.180" Score="-1" Body="&lt;p&gt;Physical changes are reversible, whereas chemical changes are irreversible. For example, boiling water is a physical change because you can reverse it by cooling it down, but setting fire to wood or boiling an egg cannot be undone.&lt;/p&gt;&#xA;" OwnerUserId="878" LastEditorDisplayName="user904" LastEditDate="2012-12-04T00:23:57.153" LastActivityDate="2012-12-04T00:23:57.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2688" PostTypeId="1" CreationDate="2012-12-04T00:11:23.590" Score="4" ViewCount="2163" Body="&lt;p&gt;I am fully aware that the pKa is the pH at half the equivalence point, but are there some programs or methods I could use to accurately estimate the pKa without manually drawing lines on the graph?&lt;/p&gt;&#xA;" OwnerUserId="935" LastActivityDate="2012-12-04T01:09:46.647" Title="How can I find pKa from a titration curve?" Tags="&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2690" PostTypeId="2" ParentId="2612" CreationDate="2012-12-04T06:53:32.263" Score="3" Body="&lt;p&gt;I have a CHEMKIN files with HCL:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HCL                1   344.7       3.339     0.000     0.000     1.000  !(singh)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;344.7 should be $\varepsilon/k$ in K and 3.339 should be $\sigma$ in Angstroms.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;! Species: HCL              CAS Number: 7647-01-0&#xA;! Name:    Hydrogen Chloride&#xA;! Source:  SNLL fit to 1986 JANAF tables, 4/21/89&#xA;! H0(298K) =      -22.0600 (Kcal/mole),  S0(298K) =       44.6500 (cal/mole-K)&#xA;HCL                42189CL  1H   1          G   300.000  5000.000 1000.00      1&#xA; 2.75533500e+00 1.47358100e-03-4.97125400e-07 8.10865800e-11-5.07206300e-15    2&#xA;-1.19180600e+04 6.51511600e+00 3.33853400e+00 1.26820700e-03-3.66691700e-06    3&#xA; 4.70399200e-09-1.83601100e-12-1.21315100e+04 3.19355500e+00                   4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also, Poling [1] has some data on HCl, Cl$_2$, ClNO. You could estimate the other properties using group contributions to define critical properties and then corresponding state principle methods (like Chung's [2]) to get your temperature/pressure dependencies. You can then make your own curvefit for your conditions of interest. Not as good as curvefits from experimental data but it should get you in the ballpark...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] B. E. Poling, J. M. Prausnitz, and J. P. O’Connell. The Properties of Gases and Liquids, Fifth Edition. McGraw-Hill, 2001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] T. H. Chung, M. Ajlan, L. L. Lee, and K. E. Starling. Generalized multiparameter corresponding state correlation for polyatomic, polar fluid transport properties. 27:671–679, 1988.&lt;/p&gt;&#xA;" OwnerUserId="936" LastActivityDate="2012-12-04T06:53:32.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2691" PostTypeId="2" ParentId="2669" CreationDate="2012-12-04T07:03:06.050" Score="10" Body="&lt;p&gt;Far from 'only now' becoming viable, the production of disinfectant from electrolysis of salt water (brine) is not only 'real' it is &lt;a href=&quot;http://www.ehow.com/about_6382595_chlorine-bleach-made_.html&quot;&gt;how bleach sold on supermarket shelves is produced!&lt;/a&gt; Industrially, this is called the  &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloralkali_process&quot;&gt;'chor-alkali process'&lt;/a&gt; and is well established for over 100 years. Both chlorine and sodium hypochlorite are &lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/21c/chemicals_in_our_lives/manufacture_chemicalsrev5.shtml&quot;&gt;manufactured industrially by electrolysis of brine&lt;/a&gt;. Where gaseous chlorine presents a significant hazard however, liquid sodium hypochlorite is offered as an alternative, however it has a limited shelf-life and degrades rapidly with temperature. Recent innovations in terms of design and manufacture of ion-exchange membranes has opened up new possibilities for small-scale on-site production of socium hypochlorite which is especially useful for remote communities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the question of economic 'viability' or profitability, like any question of economics, it depends upon the available technology and specific circumstances. For example, large power stations typically have the need to disinfect huge volumes of water which passes through their cooling towers. Since they literally 'generate electricity at cost', it is often more economic for such &lt;a href=&quot;http://www.electrochemsci.org/papers/vol7/7053929.pdf&quot;&gt;power stations to produce disinfectant on-site using eletrolysis&lt;/a&gt; than to truck in volumes of bleach manufactured at a chemical plant. Of course, this does not necessarily mean its 'economically viable' for you to make your own bleach at home right now, although &lt;a href=&quot;http://www.siemens.com/press/en/pressrelease/?press=/en/pressrelease/2012/industry/industry-automation/iia2012042822.htm&quot;&gt;this may change in the future&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.environmental-expert.com/Files/5306/articles/13867/501.pdf&quot;&gt;on-site generation of hypochlorite for disinfection&lt;/a&gt; may not only be useful for cooling towers on power stations but also for treating wastewater &lt;a href=&quot;http://www.water.siemens.com/SiteCollectionDocuments/Product_Lines/Electrocatalytic_Products/Brochures/IndustrialWaterWorld-Article_April2006.pdf&quot;&gt;on board ships&lt;/a&gt; and in in remote locations. Since the chemical is produced 'on demand', there is no need for the extra waste created by packaging, shipping and storage of large volumes of hazardous chemicals.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-12-04T07:57:51.117" LastActivityDate="2012-12-04T07:57:51.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2692" PostTypeId="2" ParentId="2686" CreationDate="2012-12-04T07:10:02.633" Score="5" Body="&lt;p&gt;This is a tricky question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding the egg:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-12-04T07:20:40.907" LastActivityDate="2012-12-04T07:20:40.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2693" PostTypeId="2" ParentId="2686" CreationDate="2012-12-04T11:37:45.580" Score="2" Body="&lt;p&gt;Paul's comment and Richard's answer are correct that that chemical and physical change span a continuum. However, at the introductory level, there is a difference between physical and chemical &lt;strong&gt;properties&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical changes do not involved the conversion of one substance into another. They can be reversible (ice melting and freezing at 0 °C) or essentially irreversible (free expansion of a gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical changes are those that involve the conversion of one substance into another. They can be reversible (acid-base equilibria) or essentially irreversible under normal conditions (combustion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical Properties&lt;/strong&gt; are those properties of a substance or mixture than seem to match with physical descriptions of objects and with the physical changes, especially changes of state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intensive physical properties include such characteristic data as densities, specific heat capacities, boiling points, freezing points, &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henry's Law&lt;/a&gt; constants, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extensive physical properties vary with the sample (are not characteristic to the substance): temperature, pressure, volume, mass, size, shape, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemical Properties&lt;/strong&gt; are those properties which, as Richard mentions, seem to have to do with electronic transitions (aka chemical changes). These include all of the known reaction, reactivity, thermochemistry, kinetics, and equillibrium data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, (and non-intuitively) most of the properties of substances and mixture that interact with our five senses are also chemical properties: hardness, color, taste, smell, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the following properties (and others like them) are also chemical properties: toxicity, flammability, conductivity, nutritional value, etc.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorDisplayName="user7951" LastEditDate="2019-01-06T11:35:25.497" LastActivityDate="2019-01-06T11:35:25.497" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2694" PostTypeId="2" ParentId="2677" CreationDate="2012-12-05T02:00:19.087" Score="4" Body="&lt;p&gt;The general answer to this question is no, one cannot simply predict the products of any reaction.  How do we know that one of the products is not $\ce{S2O6^{2-}}$ or perhaps even $\ce{S2O5Cl2^{2-}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best that can be done without any guidance from experiment is to compute the thermodynamically favored products.  One does this (at standard temperature and pressure) by writing down all plausible products and then computing the Gibbs free energy ($\Delta G$) for the reactions.  The one with the smallest $\Delta G$ is the thermodynamically preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But be warned:  there are mechanistic and kinetic factors to take into account.  The thermodynamically favored reaction may, as a practical matter, not take place.  The standard example is diamond --&gt; graphite, which is thermodynamically allowed, but don't hold your breath waiting for it to happen.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-05T02:00:19.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2695" PostTypeId="1" CreationDate="2012-12-05T02:09:54.973" Score="1" ViewCount="980" Body="&lt;p&gt;Can someone draw the reaction that leads to the generation of HOCl when &lt;a href=&quot;http://en.wikipedia.org/wiki/Dichloroisocyanuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;dichloroisocyanuric acid&lt;/a&gt; (structure shown below) is dissolved in water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpkyc.png&quot; alt=&quot;1,3-dichloro-1,3,5-triazinane-2,4,6-trione&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="658" LastEditorDisplayName="user7951" LastEditDate="2016-10-09T17:41:26.517" LastActivityDate="2016-10-09T17:41:26.517" Title="How does 1,3-dichloro-1,3,5-triazinane-2,4,6-trione (dichloroisocyanuric acid) generate hypochlorous acid when dissolved in water?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2696" PostTypeId="2" ParentId="2680" CreationDate="2012-12-05T04:17:59.043" Score="2" Body="&lt;p&gt;There is a trend that follows an arrow from the lower left of the periodic table to the upper right.  As the atomic number increases along a period, the radius for each element decreases.  Here is a useful image of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nyxec.jpg&quot; alt=&quot;Atomic Radii&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, with the exception of the noble gases, you can draw an arrow from the lower left to the upper right.  The justification for what you see here is that as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principal_quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;principal quantum number&lt;/a&gt; increases, the probability that the electrons are more distance from the center of the atom also increases.  The higher the principal quantum number, the higher the energy of the electron and therefore its probable distance from the center of the atom.  Therefore, as you move down a group, the radius increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another factor is that as the number of protons increase in an atom, so does the positive charge which happens to pull the electrons toward the center of the atom.  So, there is the impact of the protons causing a decrease in radius (probability of the electron being closer to the center) and the probability that the electrons are in an orbital that is farther from the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition metals, however, have &lt;a href=&quot;http://en.wikipedia.org/wiki/D_orbital&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;d&lt;/em&gt; orbital electrons&lt;/a&gt;.  These electrons enter into a shell that is closer to the nucleus and the change is less noticeable.  That is part of the reason why they do not seem to change as much as the other atoms, according to my understanding.&lt;/p&gt;&#xA;" OwnerUserId="618" LastActivityDate="2012-12-05T04:17:59.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2697" PostTypeId="1" AcceptedAnswerId="2712" CreationDate="2012-12-05T18:18:49.807" Score="5" ViewCount="904" Body="&lt;p&gt;What is the significance of enclosing marks in chemical names in determining the uniqueness of a substance (organic and inorganic)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a simple change in the type of enclosing mark used indicate that we are dealing with a different substance or does it just mean that the names were written slightly differently (or incorrectly)? If I decided that I was always going to use parentheses and never use brackets when naming chemicals would it stop me from being able to differentiate between two substances based on their chemical name or would it just be non-standard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_61.htm&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; makes me think that switching from a parentheses to a bracket has a different meaning but that a brace could be used in place of brackets just because it makes the name look nicer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help illustrate what I mean, the following chemical names differ only by the use of curly braces instead of brackets. Are these two different substances or are they the same substance named slightly differently?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Poly(oxy-1,2-ethanediyl),.alpha.-[4-[(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo]phenyl]-.ome&#xA;  Poly(oxy-1,2-ethanediyl),.alpha.-{4-{(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo}phenyl}-.ome&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The &lt;a href=&quot;http://www.cas.org/File%20Library/Training/STN/User%20Docs/indexguideapp.pdf&quot; rel=&quot;nofollow&quot;&gt;Naming and Indexing of Chemical Substances for Chemical Abstracts&lt;/a&gt; talks a lot about how punctuation is used to distinguish between different types of substances. It talks a lot about brackets and parentheses in these discussions but almost ignores curly braces. The publication backs up my theory that the differentiation between parentheses and brackets can be important but that curly braces are not. If I am correct then the two chemical names that I listed above are two different names for the same substance. If I can get verification on this or even opinions from chemists, that would be helpful as I am far from a chemist.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="941" LastEditDate="2012-12-06T17:15:16.133" LastActivityDate="2012-12-06T20:44:01.937" Title="Are different enclosing mark types in chemical names needed to differentiate between substances?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2698" PostTypeId="1" AcceptedAnswerId="2703" CreationDate="2012-12-05T19:39:06.627" Score="7" ViewCount="80029" Body="&lt;p&gt;I've been studying the roles of hydroxide &amp;amp; hydronium in acids and bases, and it was mentioned that a hydroxide ion (OH&lt;sup&gt;-&lt;/sup&gt;) has a negative charge. Can someone give me a layman's explanation of what causes this charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since hydrogen and oxygen are sharing a covalent bond, is the charge negative because an electron is donated from oxygen, thus giving hydrogen 2 electrons?&lt;/p&gt;&#xA;" OwnerUserId="943" LastEditorUserId="755" LastEditDate="2012-12-06T16:10:21.580" LastActivityDate="2021-02-19T12:17:47.637" Title="Why does the hydroxide ion have a negative charge?" Tags="&lt;acid-base&gt;&lt;hydrogen&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2699" PostTypeId="1" AcceptedAnswerId="2701" CreationDate="2012-12-05T19:55:49.567" Score="4" ViewCount="180" Body="&lt;p&gt;What is the correct name for a reaction like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{ 4MnO2 -&amp;gt;[500\ ^\circ \text{C}] 2Mn2O3 + O2 ^}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a synthesis, or a Decomposition? &#xA;I tried to find it out with &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chemical_reaction&lt;/a&gt; but i did not get the link between the formula and the nomenclature as it seems to be both.&lt;/p&gt;&#xA;" OwnerUserId="944" LastEditorUserId="275" LastEditDate="2013-03-15T11:03:43.667" LastActivityDate="2013-03-15T11:03:43.667" Title="Correct nomenclature for reaction types" Tags="&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2701" PostTypeId="2" ParentId="2699" CreationDate="2012-12-05T22:15:59.420" Score="6" Body="&lt;p&gt;This is a decomposition reaction.  Synthesis requires a minimum of two reactants.  Decomposition is where a reactant breaks down into two or more products.  Here, one reactant (MnO2) goes to two products.  It seems like it may not be decomposition because MnO2 is becoming MnO3, but it is still one reactant to two products.&lt;/p&gt;&#xA;" OwnerUserId="875" LastActivityDate="2012-12-05T22:15:59.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2702" PostTypeId="2" ParentId="2698" CreationDate="2012-12-05T22:52:33.303" Score="0" Body="&lt;p&gt;Hydrogen exists as $\ce{H2}$, while oxygen exists as $\ce{O2}$.  They are both diatomic elements, meaning that their stable form is a pair.  There are several others...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Hydrogen ($\ce H$) has a positive charge +1.  This has nothing to do with it having more protons than electrons, as it has one of each.  Think of it as 'willing' to donate one electron, or it has +1 electrons... see, it is on the plus side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Oxygen ($\ce O$) has a negative charge  of -2. Think of it as 'wanting' to borrow two electrons, or it has -2 electrons... see, it is on the plus side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, put one $\ce H^+$ with one $\ce {O}$ and the $\ce H$ will donate, the $\ce O$ will borrow and the $\ce H$ will be 0, and the O will be -1.  Now the combination has a -1 charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Hydrogen is sharing one electron with the oxygen, filling its outermost shell with 2, but the Oxygen is needing 8 in its outermost shell, and is only at 7, so the compound is at -1, or 'wanting' one more electron to make it stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add one more $\ce H^+$ the $\ce{O}$ will be satisfied, and you will get $\ce {H2O}$, a very stable compound we know as water.  &lt;/p&gt;&#xA;" OwnerUserId="875" LastEditorUserId="30" LastEditDate="2013-10-22T16:59:33.573" LastActivityDate="2013-10-22T16:59:33.573" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2703" PostTypeId="2" ParentId="2698" CreationDate="2012-12-05T23:52:25.537" Score="14" Body="&lt;p&gt;A water molecule is charge neutral because there is the same number of positive &#xA;charges as there are negative charges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram, called a Lewis structure, the dots represent electrons while the lines or dashes represent a covalent bond of two electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t4ReA.gif&quot; alt=&quot;h2o lewis structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water ionizes one of the hydrogen atoms absconds with itself and leaves&#xA;it's electron behind, giving us the hydroxide ion. The extra electron gives hydroxide a net charge of -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AtlaK.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brackets indicate that this is an ion, charge is denoted at top right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go deeper down the rabbit hole on this one I recommend reading up on &lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;noreferrer&quot;&gt;the Octet rule&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;Electronegativity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="755" LastActivityDate="2012-12-05T23:52:25.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2704" PostTypeId="2" ParentId="2665" CreationDate="2012-12-06T00:38:39.520" Score="7" Body="&lt;p&gt;Ok.  You have five equations, two are differential equations and three are algebraic.  And you have the explicit assumption that the equilibria are fast enough so that the two equilibrium equations are always satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have five unknowns, A, B, C, AB, and BC.  One method of solution is to use the algebraic equations to eliminate three of the variables.  This should leave you with two differential equations with two unknowns.  If you are very lucky (I've not worked out the algebra) you will get two equations that can be integrated independently.  More likely you will have two simultaneous differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case the result is apt to be a mess.  So I'd cheat.  If this is a textbook problem or homework problem, there is probably a way to combine the equations to get simple enough expressions.  Otherwise, you may have serious troubles solving the set.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-06T00:38:39.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2706" PostTypeId="1" AcceptedAnswerId="2709" CreationDate="2012-12-06T10:03:22.583" Score="12" ViewCount="7952" Body="&lt;p&gt;Is it possible to synthesise ethers from carboxylic acids, and if so, how? By ether I mean an ether in general ($\ce{R-O-R'}$), not just specific examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I theorised one could perform Fischer esterification, $\ce{ROH +R'COOH -&amp;gt; RCOOR'}$, and then perform Wolff-Kishner reduction on the carbonyl group. The problem with this theory is that it assumes that the single-bonded oxygen between the $\ce{R}$ and $\ce{R'}$ doesn't influence the reduction process. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="16683" LastEditDate="2017-04-08T22:18:00.877" LastActivityDate="2017-04-08T22:18:00.877" Title="Is it possible to synthesise ethers from carboxylic acids?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2707" PostTypeId="1" AcceptedAnswerId="2708" CreationDate="2012-12-06T10:11:44.923" Score="9" ViewCount="424" Body="&lt;p&gt;When hydrocarbons burn with little oxygen, carbon is produced. What form do these particulates of carbon come in? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a simple molecule like $\ce{S8}$, or a small fragment of a larger structure, like graphite? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this form more stable than others?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="4231" LastEditDate="2015-02-24T16:40:49.720" LastActivityDate="2015-02-24T16:40:49.720" Title="Forms of carbon in incomplete combustion" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2708" PostTypeId="2" ParentId="2707" CreationDate="2012-12-06T12:41:15.667" Score="6" Body="&lt;p&gt;It heavily depends on reaction parameters (oxygen/oxidants availability, temperature, pressure, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a free-air incomplete combustion at a (relatively) high temperature, you get mainly carbon-based molecules of various kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a internal combustion engine, that usually works at high temperature and high pressure, you can also get nanoparticles of various kind, including graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a free-air, mixed, relatively cold combustion, the original small molecules, graphite and other carbon-bases compounds can further react with other molecules giving you larger molecules and/or can stick to larger particles forming nano or micro-particles of various kind. Predicting/modelling this kind of reactions can be very complicated.&lt;/p&gt;&#xA;" OwnerUserId="925" LastActivityDate="2012-12-06T12:41:15.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2709" PostTypeId="2" ParentId="2706" CreationDate="2012-12-06T13:05:23.653" Score="12" Body="&lt;p&gt;This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a Google search for &quot;reduce ester to ether&quot; brought up the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/COgY7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction appears to be complete within 3 hours with moderate to good (but not stellar) yields. The abstract and key figures are available for free at &lt;a href=&quot;http://www.organic-chemistry.org/abstracts/lit1/800.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.organic-chemistry.org&lt;/a&gt;. &lt;a href=&quot;http://dx.doi.org/10.1021/jo070814z&quot; rel=&quot;nofollow noreferrer&quot;&gt;The paper&lt;/a&gt; is published in the &lt;em&gt;Journal of Organic Chemistry&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that $\ce{Et3SiH}$ is the terminal reductant, but most of the work is being done by indium. The combination of $\ce{Et3SiH}$ and $\ce{InBr3}$ produces the active reductant $\ce{HInBr3}$, which provides the equivalent of $\ce{H.}$. The second $\ce{H.}$ comes from $\ce{Et3SiH}$. The proposed mechanism from the paper is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JEc7z.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other mild sequential one-electron reductions with a source of $\ce{H.}$ would also likely work. The two-electron reductions, like Wolff-Kishner ($\ce{NH2NH2 +NaOH}$) and hydride-transfer (e.g. $\ce{LiAlH4}$), are all heavy-handed and nucleophilic with the tendency to take apart or over-reduce the ester.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-06T13:05:23.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2710" PostTypeId="1" CreationDate="2012-12-06T16:41:51.077" Score="28" ViewCount="592" Body="&lt;p&gt;If I understand correctly, $\ce{H2}$ in the presence of $\ce{Pd}$ readily dissociates as it dissolves into the metal. With the dissociation energy for the $\ce{H—H}$ bond being so large, how is this possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought that the $\ce{H}$ atoms were falling to a lower energy level in the $\ce{Pd}$ that was somehow only available if they were monatomic, but if that were true wouldn't some heat be generated by the absorption? &lt;/p&gt;&#xA;" OwnerUserId="755" LastEditorUserId="4231" LastEditDate="2015-02-25T15:40:25.540" LastActivityDate="2015-02-25T15:40:25.540" Title="How does Palladium dissociate H₂ so easily?" Tags="&lt;reaction-mechanism&gt;&lt;hydrogen&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2711" PostTypeId="1" CreationDate="2012-12-06T20:15:12.303" Score="4" ViewCount="462" Body="&lt;p&gt;What are the biggest environmental issues with using methane, $\ce{CH4}$, (which is found as the major energy source in natural gas, shale gas, biogas, and synthetic natural gas), as a combustible fuel?&lt;/p&gt;&#xA;" OwnerUserId="523" LastEditorUserId="30836" LastEditDate="2019-04-13T22:43:32.647" LastActivityDate="2019-04-13T22:43:32.647" Title="What are the major environmental drawbacks with using methane as a combustable fuel?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;&lt;hydrocarbons&gt;&lt;fuel&gt;&lt;green-chemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2712" PostTypeId="2" ParentId="2697" CreationDate="2012-12-06T20:44:01.937" Score="3" Body="&lt;p&gt;There is no real difference between the types of bracket - indeed, in online searching in the CA Registry file on STN &lt;em&gt;all&lt;/em&gt; brackets are replaced by round () brackets. However, there are conventions with regards to square brackets in display in some names and these should be kept.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2012-12-06T20:44:01.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2713" PostTypeId="2" ParentId="2711" CreationDate="2012-12-06T21:32:29.703" Score="8" Body="&lt;p&gt;Methane has many lovely properties. It's a very dense source of energy. It is plentiful, it seemed to be cheap to extract (until we worked out the real cost), and similarly seemed to be cheap to transport long distances. It can be stored for many months in vast quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a greenhouse gas - a powerful one at that. On a 100-year horizon, it has about 20 times the global warming potential of $\ce{CO2}$. That needn't be a problem, if natural-gas extraction had no leakages, and distribution systems had no leakages.  But they do have leakages, which means the supply chain worsens global warming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when methane is burnt, $\ce{CO2}$ is released into the atmosphere, worsening global warming even further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've learnt to handle methane's other main problems: that it's colourless and odourless, so can just accumulate unnoticed waiting to explode, if you let it; and that it's also dangerous if handled carelessly in any other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also generates $NO_x$ when burnt in air, and they are local as well as global pollutants. It is possible to have flue-scrubbers to remove the $NO_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean that that methane is permanently bad.  It can be created in a carbon-neutral fashion (from biological sources, or synthesised from atmospheric  $\ce{CO2}$), and if leakages are minimised, offers a potential useful form of energy storage. (note that in these forms, it would be an energy &amp;amp; storage vector, not a net energy source, so its economic value would be very different - but that's economics, not chemistry)&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2012-12-06T21:47:36.973" LastActivityDate="2012-12-06T21:47:36.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2714" PostTypeId="1" AcceptedAnswerId="2716" CreationDate="2012-12-06T22:34:20.440" Score="13" ViewCount="561" Body="&lt;p&gt;I understand that oxidation numbers are a method for keeping track of electrons in a reaction and how they are generally assigned (electrons in a bond are assigned to the more electronegative atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this method work? Why does assigning these arbitrary numbers based on electronegativity work in predicting and balancing reactions? &lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="73" LastEditDate="2012-12-09T00:50:25.757" LastActivityDate="2016-01-22T17:40:43.197" Title="Why do oxidation numbers work?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2716" PostTypeId="2" ParentId="2714" CreationDate="2012-12-07T01:26:40.020" Score="12" Body="&lt;p&gt;First of all they are not really arbitrary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the main point, regarding balancing the equations is that: You can keep track of the total number of electrons per species, which is a well defined value. Like, for one permanganate ($\ce{MnO4-}$) you have 25+4*8+1 electrons, for one $\ce{Mn(II)}$ you have 23 electrons, for one water you have 10 electrons, for one $\ce{H+}$ you have 0 electrons. So for this half-equation, you must have 5 electrons on the left side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{8H+ + MnO4- + 5e- -&amp;gt; Mn(II) + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or you can say that, here I assume all H atoms are +, all oxygens are -2, and for consistency, Mn in $\ce{MnO4-}$ is +7. I can dismiss H and O's since all are same, and all change is because of Mn, which goes from +7 to +2 and difference is 5 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do this like that as well: O is -4 everywhere, so H's in $\ce{H2O}$ should be +2 and they were +1 on the left side. They donated 8 electrons. Mn in $\ce{MnO4-}$ should be +15 and it is +2 on the right side, which means it accepted 13 electrons. And the difference is 5 electrons, again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottomline: It is all about consistency. If all elements in each and every species sum up to the charge of this species, whatever you say their formal charges are, it will work -because this just another way of keeping track of total electron count.  So it is better to treat most of them constant like, O is -2, H is +1 almost always. &lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="946" LastEditDate="2016-01-22T17:40:43.197" LastActivityDate="2016-01-22T17:40:43.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2717" PostTypeId="5" CreationDate="2012-12-07T01:43:01.413" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-12-07T01:43:01.413" LastActivityDate="2012-12-07T01:43:01.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2718" PostTypeId="4" CreationDate="2012-12-07T01:43:01.413" Score="0" Body="Questions regarding the chemical properties of hydrogen and its behavior in reactions and compounds. " OwnerUserId="755" LastEditorUserId="22" LastEditDate="2012-12-07T01:45:00.010" LastActivityDate="2012-12-07T01:45:00.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2719" PostTypeId="2" ParentId="2711" CreationDate="2012-12-07T07:52:24.133" Score="2" Body="&lt;p&gt;From what point of view? Environmental? Industrial? Practical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Europe (I live in Italy) we use methane a lot, both for automotive and for house heating/cooking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the environmental point ov view, $\ce{CH4}$ is one of the best choices (together with LPG). When burnt (both in thermodynamics engines and in free-flames) it releases $\ce{H2O}$ and $\ce{CO2}$ &lt;em&gt;only&lt;/em&gt;, even when burn at low temperatures and pressures (gasoline would produce a lot of pollutants in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an industrial point of view it is even better. There is a large abundance of $\ce{CH4}$ underground. We import a huge amount of it from Russia and north-Africa at a quite low price. Even better: $\ce{CH4}$ is easy to produce from other sources. For example, here in my area we use $\ce{CH4}$ (and LPG) produced (as a by-product) from petrolchemical industry and oil refineries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical point of view, it is relatively easy to manage. In my house (as in 99% of European houses) we use $\ce{CH4}$ both for heating and for cooking. We just have a high/low pressure pipeline connected to the national gas network. Nothing actually more complicated that the usual drinkable water supply pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drove $\ce{CH4}$- and LPG-powered automobiles for years in my life. $\ce{CH4}$ forces you to have heavy, high-pressure bottles in you car (usually in the hood) but still can give a car that reach 90% of the usual performance and costs you something like 50% of a normal gasoline-powered car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LPG is even better. My wife has a LPG car in this period. The LPG bottle is relatively  light (50% of the high-pressure bottle used for $\ce{CH4}$) and cheap (30% less than a $\ce{CH4}$ system). Her car reach 90% of the gasoline-powered equivalent car and she spend something like 60% of a corresponding gasoline-powered car. It is even competitive with a corresponding diesel car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The version of $\ce{CH4}$ (and LPG) that you can buy for automotive and in-house uses in Europe is &lt;em&gt;always&lt;/em&gt; added with a specific odorizer that makes very easy to spot any leakage. Our gas sensors are aimed at spotting this specific odorizer so using $\ce{CH4}$ and LPG is actually quite safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, $\ce{CH4}$ is still a fuel gas. It still requires high-pressure bottle and high-pressure pipeline so it still is a dangerous product and still have to be handle with care. For this reason, for example, you cannot refill your car by yourself and you have to wait for the service station staff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way, every $\ce{CH4}$ pipeline used for in-house cooking or heating must be verified and certified once per year by a certified technician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In EU and IT, there are plans to produce $\ce{CH4}$ and LPG from many new, sometime weird, sources (organic waste, biomasses, etc.) so we expect to see an even larger diffusion of $\ce{CH4}$ in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that many of those sources are totally renewable so there isn't any &lt;em&gt;new&lt;/em&gt; $\ce{CO2}$ added to earth atmosphere. There is just a circulation of existing $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="925" LastEditorDisplayName="user10259" LastEditDate="2015-05-21T23:19:21.157" LastActivityDate="2015-05-21T23:19:21.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2720" PostTypeId="1" CreationDate="2012-12-07T09:38:51.657" Score="2" ViewCount="104" Body="&lt;p&gt;The output from an anerobic kitchen waste digester is mostly $\ce{CH4}$, and $\ce{CO2}$.&#xA;If the methane is collected over water; algae/moss may use the $\ce{CO2}$ to grow. Yet oxygen is released ... which, together with methane, is a potent environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the oxygen, in this closed jar filled with water+algae+moss, recombine to form $\ce{CO2}$ and $\ce{H2O}$? Is there a means to prevent methane combining with oxygen released by the algae?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="4231" LastEditDate="2015-02-25T15:44:40.687" LastActivityDate="2015-02-25T15:44:40.687" Title="Is it possible to prevent methane combining with oxygen?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2722" PostTypeId="2" ParentId="2695" CreationDate="2012-12-07T13:34:34.717" Score="2" Body="&lt;p&gt;$\ce{N-Cl}$ bond is polarized toward nitrogen: $\ce{N^--Cl+}$ This means, that any chloramine can produce $\ce{HOCl}$ in water by mean of hydrolysis of $\ce{N-Cl}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R2NCl + H2O -&amp;gt; R2NH + HOCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism is more subtle matter. Basically, it should be one of $S_N$ mechanisms, but it is hard to choose right one here ($S_N1$ or $S_N2$). The central atom in mechanism here should be $\ce{Cl}$, and $\ce{R2N-}$ is leaving group. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user7951" LastEditDate="2016-10-09T17:21:44.217" LastActivityDate="2016-10-09T17:21:44.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2723" PostTypeId="2" ParentId="2720" CreationDate="2012-12-07T14:01:00.397" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the oxygen, in this closed jar filled with water+algae+moss, recombine to form $\ce{CO2}$ and $\ce{H2O}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very slowly, you will not see this. However, it is much worse, as explosive composition of $\ce{CH4}$ and $\ce{O2}$ will form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industrial applications it is possible to separate $\ce{CO2}$ from $\ce{CH4}$ and them move $\ce{CO2}$ into separate chamber with algae. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4231" LastEditDate="2015-02-25T15:44:38.177" LastActivityDate="2015-02-25T15:44:38.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2724" PostTypeId="1" CreationDate="2012-12-07T14:21:31.877" Score="10" ViewCount="2059" Body="&lt;p&gt;I wonder why crystals of Naphthalene are needle-shaped after sublimation. Googling the fact, I could only confirm that it is true but failed to find the reason. Could someone please shed some light as to why this happens?&lt;/p&gt;&#xA;" OwnerUserId="8235" LastEditorUserId="8235" LastEditDate="2012-12-07T15:24:46.293" LastActivityDate="2015-08-21T20:06:00.063" Title="Why does Naphthalene turn into needle-shaped crystals by sublimation?" Tags="&lt;organic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2725" PostTypeId="2" ParentId="2695" CreationDate="2012-12-07T14:34:22.390" Score="2" Body="&lt;p&gt;To build on permekra's answer, I have drawn a mechanism. I think that is would be S&lt;sub&gt;N&lt;/sub&gt;2 (due to the unlikeliness of $\ce{Cl+}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rI4uK.png&quot; alt=&quot;dichlor hydrolysis mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water attacks a chlorine atom as a nucleophile, and the chloroisocynurate anion leaves resulting in $\ce{H2OCl+}$. The chloroisocyanurate anion can act as a leaving group due to its resonance stabilization (see below). Then, the chloroisocyanurate anion depropotonates $\ce{H2OCl+}$ to form choroisocyanuric acid and hypochlorous acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sfitO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since there is another chlorine atom, this substitution reaction occurs again to form cyanuric acid $\ce{(NCOH)3}$ and another equivalent of $\ce{HOCl}$. The net reaction for the hydrolysis is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ C3HCl2N3O2 + 2H2O -&amp;gt; C3H3N3O3 + 2HOCl}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For curiosity - the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_dichloroisocyanurate&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium salt&lt;/a&gt; of dichloroisocyanuric acid is the active ingredient in the household cleaning product &lt;a href=&quot;http://en.wikipedia.org/wiki/Comet_%28cleanser%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comet&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-07T14:34:22.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2726" PostTypeId="1" CreationDate="2012-12-08T03:14:35.463" Score="4" ViewCount="5405" Body="&lt;p&gt;When hypochlorous acid is dissolved in water, is it the chlorine atom or the oxygen that serves as the anti microbial agent?&lt;/p&gt;&#xA;" OwnerUserId="658" LastActivityDate="2012-12-16T08:40:24.677" Title="Hypochlorous acid mechanism" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2727" PostTypeId="1" CreationDate="2012-12-08T03:31:53.797" Score="7" ViewCount="272" Body="&lt;p&gt;Has diatomic oxygen ever been observed with a 3− charge, i.e. $\ce{O_2^{3-}}$? Have any ab intio calculations been done on this ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to summarize some of the answers below and add my two cents, it seems that no studies have reported observing this ion experimentally, either in isolation or as part of a condensed phase. Ab initio calculations suggest that $\ce{O_2^{3-}}$ is unstable in isolation, meaning that the loss of an electron is exothermic. However, first-principles calculations have predicted that a molecular oxygen trianion can exist as part of a solid; see &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevB.85.035210&quot; rel=&quot;nofollow&quot;&gt;Phys. Rev. B 85, 035210 (2012)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="4945" LastEditDate="2015-03-05T11:30:13.363" LastActivityDate="2015-03-05T11:30:13.363" Title="Has the molecular diatomic oxygen trianion been observed or predicted?" Tags="&lt;ions&gt;&lt;molecules&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2728" PostTypeId="1" AcceptedAnswerId="2749" CreationDate="2012-12-08T12:16:45.367" Score="6" ViewCount="9153" Body="&lt;p&gt;How do you know if a polymer is a thermoset or a thermoplastic? For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;hydroxybutanoic acid &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2-amino-3- hydroxybutanoic acid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I am not sure if these are correct in English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know what thermosets and thermoplastics are, and their properties, but I don't know how to see if polymers of a certain molecule is a thermoset or a thermoplastic by looking at the structural formula. &lt;/p&gt;&#xA;" OwnerUserId="738" LastActivityDate="2016-06-03T02:10:23.243" Title="A thermoset or a thermoplastic?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2730" PostTypeId="2" ParentId="2727" CreationDate="2012-12-08T15:30:32.903" Score="2" Body="&lt;p&gt;I don't know of (and a quick search couldn't reveal) any studies on the $\ce O_2^{3-}$ ion. Note that in this ion, oxygen wouldn't have an oxidation state of –3, but –1.5.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2012-12-08T15:30:32.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2731" PostTypeId="2" ParentId="1221" CreationDate="2012-12-08T15:31:22.670" Score="2" Body="&lt;p&gt;The balanced equation for the combustion of butene is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C4H8 + 6O2 -&amp;gt; 4CO2 + 4H2O}.$$&lt;/span&gt;&#xA;Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount &amp;ndash; well technically it is included to make question look complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C4H8}$&lt;/span&gt; in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room, etc. &amp;ndash; which is why you have 4 equations, 5 unknowns: d is free to vary.  But you are given the dry analysis to fix its value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the equations given, you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  b &amp;amp;= c = 4(1-d)\\&#xA;  a &amp;amp;= e = b + c/2 = 6(1-d) = 3b/2&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And dry analysis says ratio of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;:&lt;span class=&quot;math-container&quot;&gt;$\ce{C4H8}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$14.95:0.75 = 19.93 (b/d)$&lt;/span&gt; after the combustion. And this is the last equation you need.&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="4945" LastEditDate="2020-01-28T18:39:02.433" LastActivityDate="2020-01-28T18:39:02.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2732" PostTypeId="2" ParentId="2714" CreationDate="2012-12-08T19:52:13.030" Score="7" Body="&lt;p&gt;There are several ways to keep track of things.  Oxidation number is one, formal charge is another.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With oxidation numbers, the electrons in a molecule or ion are assigned to the more electronegative atom in a bond.  Thus in $\ce{H2O}$ oxygen is the most electronegative so the oxidation number of hydrogen is $+1$ (it has &quot;lost&quot; an electron) and oxygen is $-2$ (it has &quot;gained&quot; two electrons.  Using this scheme in a consistent manner one can balance equations quite nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are times when the formal charge is more important.  In this scheme one always splits the electrons in any bond between the two atoms.  Thus in water all the atoms have a formal charge of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can think of these two schemes as follows:  oxidation numbers pretend that all bonds are in effect totally ionic with the electrons going to the more electronegative atom.  Formal charge pretends that all bonds are covalent and the electrons in a bond are shared equally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course neither one is totally correct, but either one can be used as an &quot;electron accounting&quot; scheme.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-08T19:52:13.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2733" PostTypeId="1" CreationDate="2012-12-09T08:57:09.463" Score="-1" ViewCount="790" Body="&lt;p&gt;I don't understand how to find the volume in this problem..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flask containing 63.2 mL of neon gas was collected over water at 21C and 788 mm Hg. What is this volume of gas at STP? (WVP at 21C = 18.6 mm Hg)&lt;/p&gt;&#xA;" OwnerUserId="713" LastEditorUserId="4945" LastEditDate="2018-12-11T12:34:01.740" LastActivityDate="2018-12-11T12:34:01.740" Title="Homework - Ideal gas law" Tags="&lt;gas-laws&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2012-12-10T04:37:19.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="2734" PostTypeId="1" AcceptedAnswerId="2735" CreationDate="2012-12-09T16:45:39.660" Score="4" ViewCount="6101" Body="&lt;p&gt;Why water can't mix with oil or cooking oil (both saturated and unsaturated)?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="296" LastEditDate="2012-12-27T14:23:28.477" LastActivityDate="2015-05-20T18:46:53.080" Title="Why water can't mix with oil or cooking oil?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2735" PostTypeId="2" ParentId="2734" CreationDate="2012-12-09T22:22:30.053" Score="8" Body="&lt;p&gt;The basic answer is that water molecules attract each other and so clump together forcing almost all of the oil out of the clump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attraction exists because water is a polar molecule.  That is, it has a positive end and an negative end.  Thus the molecules tend to clump together just as a bag of small magnets tend to clump together.  Oils, on the other hand, are not polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle can be expanded.  In general polar materials tend not to dissolve in oils or, if they are liquids, dissolve in them.  The rule in chemistry is &quot;like dissolves like&quot;.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-09T22:22:30.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2736" PostTypeId="2" ParentId="2724" CreationDate="2012-12-09T22:32:37.667" Score="6" Body="&lt;p&gt;Every solid material has a crystal structure.  Sublimation tends to produce good crystals because the solid can grow slowly.  In bulk form naphthalene crystals are broken and compressed and so bulk solid naphthalene looks quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact same situation applies to snowflakes.  They are flat hexagonal structures when seen individually.  Crunch them into a snowball and they lose that pretty structure.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-09T22:32:37.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2737" PostTypeId="2" ParentId="2734" CreationDate="2012-12-10T03:08:58.857" Score="8" Body="&lt;p&gt;The textbook answer for why water can't mix with oil is to use the rule 'like dissolves like'. But a rule in science is none other than summarizing a few collected facts. Answering a question with a rule is like pointing a single fact - why water can't mix with oil - to a collection of facts instead. This easily invites a follow-up question, why like dissolves like. Besides, this rule is vague and superficial; it does not address the facts that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ice is not dissolved in the same matter, i.e., water at 0 °C, &lt;/li&gt;&#xA;&lt;li&gt;cooking oil has a polarity comparable to that of water (as measured in Debye units),  &lt;/li&gt;&#xA;&lt;li&gt;oil molecules do not clump together as strong as water, then how could they prevent water molecules, which are attracted more to oil than oil itself, from coming in? or &lt;/li&gt;&#xA;&lt;li&gt;stuff like calcium fluoride is very polar but insoluble.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, IMO, it's better explained in terms of energy. When water clumps with water and oil does the same, the total energy is lower than when they are mixed. The system is more stable if unmixed. But watch for a delicate point in this argument. Energy can be lowered when water molecule is moved into oil molecules. How do we account for it? The lowered energy is not enough to compensate the energy needed to pull that water molecule out of its pool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the difference in energy is not the only factor of insolubility. Temperature plays a role too. When temperature is raised, the difference between dissolved and undissolved states becomes less steep, relatively speaking. When thermal energy chips in, the two substances are more miscible.&lt;/p&gt;&#xA;" OwnerUserId="320" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T18:46:53.080" LastActivityDate="2015-05-20T18:46:53.080" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2738" PostTypeId="1" AcceptedAnswerId="2752" CreationDate="2012-12-10T03:46:40.033" Score="8" ViewCount="1073" Body="&lt;p&gt;Like evaporation happens only on the surface of a liquid, and boiling/vapourisation takes place throughout the body, how does sublimation occur? Why?&lt;/p&gt;&#xA;" OwnerUserId="961" LastEditorUserId="22" LastEditDate="2012-12-10T04:19:26.750" LastActivityDate="2012-12-11T19:24:59.297" Title="Does sublimation happen only on the surface or throughout the body?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2739" PostTypeId="1" AcceptedAnswerId="2803" CreationDate="2012-12-10T03:49:48.350" Score="5" ViewCount="1203" Body="&lt;p&gt;How would you go about synthesising a fatty acid amide, preferably by means of a high-yield reaction. Please provide an explanation that I, as a grade 12 chemistry graduate, could understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fatty acids&lt;/li&gt;&#xA;&lt;li&gt;Carboxylic acid&lt;/li&gt;&#xA;&lt;li&gt;Equilibrium reactions &amp;amp; constants&lt;/li&gt;&#xA;&lt;li&gt;Amide groups&lt;/li&gt;&#xA;&lt;li&gt;Amines&lt;/li&gt;&#xA;&lt;li&gt;Alkanols&lt;/li&gt;&#xA;&lt;li&gt;Esterification&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you need to use terms I may be unfamiliar with, it would be nice if you could explain them or provide a link to an explanation. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="7448" LastEditDate="2015-10-16T17:54:03.263" LastActivityDate="2016-12-12T17:48:09.157" Title="How do you Synthesise a Fatty Acid Amide?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2742" PostTypeId="1" AcceptedAnswerId="2746" CreationDate="2012-12-10T11:46:42.713" Score="2" ViewCount="237" Body="&lt;p&gt;How does one synthesise 2-monoglycerides from triglycerides with additional ingredients if necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct would be to add a catalyst for the reaction of $triglyceride \rightarrow 3 FA + glycerol$ (FA=fatty acid; for instance a catalyst could be sodium hydroxide) and then add additional glycerol so as to cause the reaction: $FA + glycerol \rightarrow 2$-$monoglyceride $ to occur but I just wanted to ask if this is true because to me this seems too simple to be true. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2012-12-16T10:36:13.757" LastActivityDate="2012-12-16T10:36:13.757" Title="How does one synthesise 2-Monoglycerides from Triglycerides?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2743" PostTypeId="1" CreationDate="2012-12-10T16:53:13.607" Score="6" ViewCount="383" Body="&lt;p&gt;My teacher in my physics course attributed this effect to phonons, more &lt;a href=&quot;https://physics.stackexchange.com/questions/45910/why-do-phonons-cause-excellent-heat-conduction-in-diamonds&quot;&gt;here&lt;/a&gt;, but I am skeptical about this argument, it feels like he is overlooking the whole question -- what about with Silica that has similar structure to diamond with covalent bonds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend attributed this to the small mass of atoms and large Young constant, considering the system to be made of springs -- $v_{sound}=\sqrt{\frac{Y}{\rho}}$. He suggested to think of heat as sound to use the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silica has less strong bonds to diamond so perhaps this is a reason why diamondis a good conductor of hear and silica is not. Air is good insulator so does it mean that carbon atoms are so closely packed and silicon atoms not? Is this the reason for the difference in conductance?&lt;/p&gt;&#xA;" OwnerUserId="966" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2013-12-05T11:52:19.593" Title="Classical reason for good heat conductance in diamond?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2744" PostTypeId="1" AcceptedAnswerId="2745" CreationDate="2012-12-11T06:07:37.897" Score="2" ViewCount="2641" Body="&lt;p&gt;I am trying to understand the synthesis of nitrocellulose.&#xA;A mixture of sulfuric acid and nitric acid is used to nitrate cellulose. The sulfuric acid acts as a catalyst, producing $\ce{NO2^{+}}$ ions from the nitric acid which bind at certain sites on the cellulose chain to nitrate it. So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions. The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions. If so, wouldn't any sufficiently strong acid, such as hydrochloric acid, do just as well. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As long as the K&lt;sub&gt;a&lt;/sub&gt; for that acid is higher than nitric acid's, it would work the same right? &lt;/li&gt;&#xA;&lt;li&gt;Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway? &lt;/li&gt;&#xA;&lt;li&gt;Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="799" LastEditorUserId="4231" LastEditDate="2015-02-24T17:31:46.500" LastActivityDate="2015-02-24T17:31:46.500" Title="Synthesis of nitrocellulose" Tags="&lt;acid-base&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2745" PostTypeId="2" ParentId="2744" CreationDate="2012-12-11T12:09:39.073" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electrophilic nitration reactions like the production of nitrocellulose require sulfuric acid because of the dehydrative properties of sulfuric acid. Sulfuric acid is a non-nucleophilic acid - all it can do is mediate proton transfers and dehydrate things (sulfuric acid has a large negative enthalpy change of mixing with water). Thus,$\ce{NO2+}$ is free to react with whatever you want it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1)\ \ \  \ce{H2SO4 + HNO3 &amp;lt;=&amp;gt; HSO4- + H2NO3+} $$&#xA;$$2)\ \ \ \ce{H2NO3+ &amp;lt;=&amp;gt; H2O + NO2+} $$&#xA;$$\text{overall)    }\ \ \ce{HNO3 + H2SO4 &amp;lt;=&amp;gt; H2O + NO2+ +HSO4-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrochloric acid and nitric acid produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Aqua_regia&quot; rel=&quot;nofollow&quot;&gt;aqua regia&lt;/a&gt;. Hydrochloric acid is a nucleophilic acid, and thus the decomposition products after proton transfer to nitric acid are different. The chloride ion is a good nucleophile, and reacts with the nitronium ion, eventually producing $\ce{NO2}(g)$ and $\ce{Cl2}(g)$. There is a big difference between $\ce{NO2+}$ and $\ce{NO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1) \ \ \ \ce{HCl + HNO3 &amp;lt;=&amp;gt; Cl- + H2NO3+}$$&#xA;$$2) \ \ \ \ce{H2NO3+ &amp;lt;=&amp;gt; H2O + NO2+} $$&#xA;$$3) \ \ \ \ce{2NO2+ + 2Cl- &amp;lt;=&amp;gt; 2NO2 + Cl2}$$&#xA;$$\text{overall)    }\ \ \ce{2HCl + 2HNO3 &amp;lt;=&amp;gt; 2H2O + 2NO2 + Cl2}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. However, starting with nitric acid shifts the equilibrium towards nitronium a little.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway? Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe, but not enough to be useful. Nitric acid is not available pure. It is only stable to a maximum concentration of 70-75% by weight (remainder water). The other reason that sulfuric acid must be used is that it is available &quot;pure&quot; (or at least 98% by weight) and it is so hydrophilic so as to sequester the water. Hydrochloric acid has a max concentration of 35-40% by weight remainder water. All the extra water gets in the way and shifts the equilibrium in the opposite direction (the reaction produces water).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-12-11T16:53:34.127" LastActivityDate="2012-12-11T16:53:34.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2746" PostTypeId="2" ParentId="2742" CreationDate="2012-12-11T15:17:15.090" Score="4" Body="&lt;p&gt;One way I have learnt is by taking a page out of the human body's book and using pancreatic lipase, found in pancreatic juice, to catalyse the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$triglyceride+ 2H_2 O \rightleftharpoons   2$-$monoglyceride+2 FAA$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where FAA stands for, 'Fatty acid anion.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone has a better answer I'll readily accept it over this one, this is just one method.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2012-12-14T00:46:07.487" LastActivityDate="2012-12-14T00:46:07.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2747" PostTypeId="5" CreationDate="2012-12-11T16:10:11.517" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2012-12-11T16:10:11.517" LastActivityDate="2012-12-11T16:10:11.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2748" PostTypeId="4" CreationDate="2012-12-11T16:10:11.517" Score="0" Body="For questions seeking answers deduced from or composed of theories regarding the chemical substances and about models and methods of theoretical chemistry." OwnerUserId="876" LastEditorUserId="9961" LastEditDate="2015-11-29T00:34:09.403" LastActivityDate="2015-11-29T00:34:09.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2749" PostTypeId="2" ParentId="2728" CreationDate="2012-12-11T16:33:04.563" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some background, since most readers would probably need it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermoplastic&quot; rel=&quot;nofollow&quot;&gt;thermoplastic&lt;/a&gt; is a polymer that becomes pliable/moldable upon heating, and then returns to a solid (keeping its new shape) when cooled. Thermoplastics are generally high molecular weight polymers with very few or no crosslinks. Polymer chains are held together by intermolecular forces (most often van der waals) and physical entanglement, both of which can be overcome with addition of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermosetting_plastic&quot; rel=&quot;nofollow&quot;&gt;thermoset&lt;/a&gt; is a polymer that is irreversibly cured, usually through covalent crosslinking (during or after polymerization). The covalent crosslinks hold the chains together in such a way that the chains cannot be separated by the addition of (small amounts of) kinetic energy. Thermosets tend to dramatically fail after the addition of lots of heat, following which they do not return to their original state. It is possible for the cumulative forces from many strong non-covalent interactions (ionic, hydrogen-bonding, and even pi stacking) to make a polymer behave like a thermoset. For example, proteins are thermosets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you tell if a monomer will produce a thermoplastic or themoset based on its structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt; All polymers produced from only mono- (&lt;a href=&quot;http://en.wikipedia.org/wiki/Addition_polymerization&quot; rel=&quot;nofollow&quot;&gt;addition polymerizations&lt;/a&gt;) or bi- (&lt;a href=&quot;http://en.wikipedia.org/wiki/Step-growth_polymerization&quot; rel=&quot;nofollow&quot;&gt;condensation polymerizations&lt;/a&gt;) functional monomers will always produce thermoplastics. All of the hydroxybutanoic acids (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma-Hydroxybutyric_acid&quot; rel=&quot;nofollow&quot;&gt;$\gamma$-hydroxybutanoic acid&lt;/a&gt;) are difunctional monomers for condensation polymerizations. Their polymers will be thermoplastics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unless...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of tri- or tetra- functional monomers produces crosslinks, which always produces thermosets. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycerol&quot; rel=&quot;nofollow&quot;&gt;glycerol&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Malic_acid&quot; rel=&quot;nofollow&quot;&gt;malic acid&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Lysine&quot; rel=&quot;nofollow&quot;&gt;lysine&lt;/a&gt; could be trifunctional monomers in condensation polymerizations. 2-Amino-3-hydroxybutanoic acid (&lt;a href=&quot;http://en.wikipedia.org/wiki/2-amino-3-hydroxybutanoic_acid&quot; rel=&quot;nofollow&quot;&gt;threonine&lt;/a&gt;) is a trifunctional monomer. It has three functional groups (alcohol, amine, carboxylic acid), that can all react in a condensation polymerization. Any synthetic polymer containing threonine is likely to contain crosslinks and be a thermoset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thermoplastic polymers can be converted into thermosets by curing them afterwards. Some curing is thermal, some is irradiative (espccially using UV), and some is chemical. A prime example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyisoprene&quot; rel=&quot;nofollow&quot;&gt;polyisoprene&lt;/a&gt; (natural latex rubber), which is a thermoplastic (with a very low melting point). It is also an elastomer - it will stretch and return to its original shape (although melting it erases the &quot;memory&quot;). Rubber is converted into a thermoset by &lt;a href=&quot;http://en.wikipedia.org/wiki/Vulcanization&quot; rel=&quot;nofollow&quot;&gt;vulcanization&lt;/a&gt; to make it more useful while still having the elastomeric properties. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-11T16:33:04.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2752" PostTypeId="2" ParentId="2738" CreationDate="2012-12-11T19:24:59.297" Score="8" Body="&lt;p&gt;The situation is akin to boiling.  If a molecule at the surface frees itself of its attractions to its neighbors, it can move into the vapor phase easily.  Same thing with evaporation at the surface of a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a molecule evaporate from the bulk liquid?  The answer is no.  Imagine such a thing happening.  The escaping molecule(s) form a bubble inside the liquid.  The pressure inside the bubble is the vapor pressure of the liquid at that temperature.  But that temperature is less than the boiling temperature so the vapor pressure is less than the pressure external to the liquid.  The result is that the bubble collapses.  The boiling temperature is the lowest temperature at which the bubble pressure is equal to the external pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think of sublimation in the bulk of a solid.  Here the &quot;bubble&quot; has to have enough pressure to break apart the solid.  It generally won't develop that much pressure.  Indeed, one can watch &quot;dry ice&quot; sublime well above its freezing point and nothing happens inside the bulk of the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the general answer is no, sublimation does not take place in the bulk of a solid.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-11T19:24:59.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2753" PostTypeId="1" AcceptedAnswerId="2756" CreationDate="2012-12-11T23:10:51.213" Score="4" ViewCount="992" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the following gasses are to mix, what is the partial pressure of neon?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZ8Kb.png&quot; alt=&quot;http://i.stack.imgur.com/8Pn7w.png&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is &lt;strong&gt;0.1&lt;/strong&gt;. I cannot figure out how my professor derived this equation. From the ideal gas law, all I can arrive to is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_x = \frac{P_tV_t}{V_x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives me an answer of 4.4, which is obviously incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've determined the equation he used is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_x = \frac{P_xV_x}{V_t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the correct answer of &lt;strong&gt;0.1&lt;/strong&gt;. How did he arrive to this answer?&lt;/p&gt;&#xA;" OwnerUserId="971" LastEditorUserId="5026" LastEditDate="2018-02-24T03:06:45.470" LastActivityDate="2018-02-24T03:09:10.680" Title="How to derive partial gas equation?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ClosedDate="2020-04-27T06:03:30.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="2755" PostTypeId="2" ParentId="2753" CreationDate="2012-12-11T23:40:20.627" Score="3" Body="&lt;p&gt;The total pressure of the system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. just the sum of the $PV$s divided by the sum of the $V$s.  Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures.  So this at least suggest that the individual partial pressures are near $0.1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you figure out how to parcel out the individual pressures from here?&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2012-12-11T23:40:20.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2756" PostTypeId="2" ParentId="2753" CreationDate="2012-12-12T01:28:35.267" Score="3" Body="&lt;p&gt;Prepare for an amazing feat of algebra! I sincerely hope there is a shorter way to the endpoint. (there is! See the end of my answer.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate partial pressure, $P_{\ce{He}}$, we need the total pressure $P_t$ and the fraction of the gas that is He $X_{\ce{He}}$: $$P_{\ce{He}}=X_{\ce{He}} P_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $PV=nRT$. $T$ is not given, so assume it is constant. $n$ is unknown, but the system is closed, so $n_x$ for each gas is constant. Thus we have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=\text{ constant}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total Pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is tempting to write variations of $$P_1 V_1 = P_2 V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for each gas, &lt;strong&gt;both $P$ and $V$ are changing&lt;/strong&gt;, so we need to consider their products: $$(PV)_1 = (PV)_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as trb456 suggests, $$P_t V_t = (PV)_t \implies P_t =\frac{\sum{(PV)_i}}{\sum{V_i}}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fraction of He&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fraction of the mixture that is helium is determined by the ratio $\dfrac{n_{\ce{He}} }{n_t}$. Since $n=PV/RT$, and $R$ and $T$ are constant, we can write: $$X_{\ce{He}}=\frac{n_{\ce{He}}}{\sum{n_i}}=\frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At last!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$P_{\ce{He}}=X_{\text{He}} P_t = \left( \frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}\right) \left( \frac{\sum{(PV)_i}}{\sum{V_i}}\right)$$&#xA;$$P_{\ce{He}_2}=\frac{(PV)_{\ce{He}_1}}{V_t}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or, as it struck me as I finished:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Helium expands to fill the total volume. Now we can use $P_1 V_1 = P_2 V_2$ and completely ignore the other gasses (there is a lot empty space for the He atoms to fit in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\ce{He}_2}=\frac{P_{\ce{He}_1}V_{\ce{He}_1}}{V_t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is very similar to your second equation, except the subscripts are added to denote initial and final pressure of helium. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-02-24T03:09:10.680" LastActivityDate="2018-02-24T03:09:10.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2758" PostTypeId="1" AcceptedAnswerId="2763" CreationDate="2012-12-12T15:31:08.360" Score="1" ViewCount="2333" Body="&lt;p&gt;So during school today, I am running an electrolysis experiment with water and sodium bicarbonate copper leads on both the anode and cathode.  I am noticing the usual $\ce{NaO}$ on the anode, and $\ce{H2C2}$ on the cathode.  However, on the anode, the copper is obviously oxidizing, and changing the hue of the water to a bright green-blue.  How can this be calculated into the chemical equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is this equation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + NaC2 + energy  -&amp;gt;  NaO + H2C2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(with the order of anode then cathode)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am aware that this is the best possible situation, but in general, is this what is supposed to be happening?&lt;/p&gt;&#xA;" OwnerUserId="154" LastEditorUserId="4231" LastEditDate="2015-02-24T17:32:04.210" LastActivityDate="2015-02-24T17:32:04.210" Title="Oxidization of Copper In Electrolysis" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2759" PostTypeId="1" AcceptedAnswerId="2761" CreationDate="2012-12-12T15:53:21.763" Score="10" ViewCount="10373" Body="&lt;p&gt;I am trying to make rocket fuel using hydrazine, hydrogen peroxide and home products. I want to put an action figure in orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to synthesize hydrazine with bleach and ammonia, using &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; as a catalyst?  &lt;/p&gt;&#xA;" OwnerUserId="898" LastEditorUserId="30836" LastEditDate="2019-04-21T17:03:59.433" LastActivityDate="2019-12-02T19:35:26.003" Title="Synthesis of hydrazine with bleach and ammonia" Tags="&lt;everyday-chemistry&gt;&lt;synthesis&gt;&lt;home-experiment&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="2760" PostTypeId="2" ParentId="30" CreationDate="2012-12-12T15:59:28.727" Score="6" Body="&lt;p&gt;The two above answers are academic/scholastic. I'll give an intuitive one. &#xA;When impurity is in a solid, it usually (not always, as said in Answer 1) weakens the connections/forces between molecules, and hence makes it more vulnerable to heat (read lower melting point). A solid is like an army in rank and file. When you put a civilian in the army, it doesn't matter whether that civilian is as strong as Arnold Schwarzenegger, the formation will be messed to some extent. &#xA;&lt;br&gt;However, weaker intermolecular forces do not mean that solid will becomes softer. It may become harder and more brittle. An example is that copper is changed to a harder bronze if tin is added. &lt;/p&gt;&#xA;" OwnerUserId="320" LastEditorUserId="320" LastEditDate="2012-12-12T17:20:37.630" LastActivityDate="2012-12-12T17:20:37.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2761" PostTypeId="2" ParentId="2759" CreationDate="2012-12-12T16:40:04.163" Score="20" Body="&lt;p&gt;I sympathize with your desire to launch an action figure. However, the chemistry you propose is really dangerous. You would be better off with black-powder based rockets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any way to synthesize hydrazine with bleach and ammonia?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do not do this.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is not &quot;yes&quot; or &quot;no&quot;. My answer is &lt;strong&gt;Don't&lt;/strong&gt;. Do not try this without proper safety equipment and training, and then think twice or three times before doing so, and then have an ambulance standing by. Certainly don't do it at home, or by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular reaction is capable of producing &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrazine&quot;&gt;hydrazine&lt;/a&gt;. However, it also producing a whole host of additional stuff, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloramine&quot;&gt;chloramine&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxylamine&quot;&gt;hydroxylamine&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/NCl3&quot;&gt;nitrogen trichloride&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_chloride&quot;&gt;hydrogen chloride&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine&quot;&gt;chlorine gas&lt;/a&gt;, all of which are either gases or volatile liquids, and all of which are poisonous. Hydrazine is poisonous. Nitrogen trichloride is a shock-sensitive explosive on the order of a sneeze can set it off. The reactions are as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloramine:&#xA;$$\ce{NH3 + HOCl -&amp;gt; Cl-NH2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dichloramine: &#xA;$$\ce{Cl-NH2 + HOCl -&amp;gt; NHCl2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen trichloride:&#xA;$$\ce{NHCl2 + HOCl -&amp;gt; NCl3 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrazine and HCl:&#xA;$$\ce{NH3 + NH2Cl -&amp;gt; H2N-NH2 + HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydroxylamine and HCl&#xA;$$\ce{NH2Cl + H2O -&amp;gt; NH2OH + HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorine gas:&#xA;$$\ce{HCl + HOCl -&amp;gt; H2O + Cl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any organic impurities in the mix can lead to other deadly stuff being produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is not controllable to produce a single specific product. Everything it produces is poisonous and/or explosive. I will recommend against, and I will not help you find a recipe.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2012-12-12T16:59:50.293" LastActivityDate="2012-12-12T16:59:50.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2763" PostTypeId="2" ParentId="2758" CreationDate="2012-12-12T21:05:37.227" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sodium compounds:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Hydrocarbons&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetylene&quot; rel=&quot;nofollow&quot;&gt;It is a gas.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na2C2 + 2H2O -&amp;gt; 2NaOH + H2C2 ^}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What actually happened?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{2Cu + 2OH- -&amp;gt; Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cu -&amp;gt; Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{4OH- -&amp;gt;O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{2H2O -&amp;gt; O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{2H2O + 2e- -&amp;gt; H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cu^{2+} + e- -&amp;gt; Cu+} \ \ \ E^o=+0.16 \ \text{V}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, &lt;strong&gt;but not initially.&lt;/strong&gt; Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -&amp;gt; H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -&amp;gt; H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="154" LastEditDate="2012-12-13T14:54:27.607" LastActivityDate="2012-12-13T14:54:27.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2765" PostTypeId="1" CreationDate="2012-12-13T01:57:13.463" Score="20" ViewCount="75648" Body="&lt;p&gt;Let's say you have a solution that is vinegar and it has salt (&lt;strong&gt;NaCl&lt;/strong&gt;) dissolved in it. Then you place old dull pennies in it. Then you look at it 5 minutes later and the pennies are clean and looking like new again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand acid can etch the copper compound but why is the salt necessary for this process and what exactly does it do?&lt;/p&gt;&#xA;" OwnerUserId="977" LastEditorUserId="23561" LastEditDate="2019-04-10T16:47:28.280" LastActivityDate="2019-04-10T16:47:50.457" Title="Why is salt needed when using vinegar to clean pennies?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="2766" PostTypeId="1" CreationDate="2012-12-13T11:16:54.780" Score="3" ViewCount="9938" Body="&lt;p&gt;I want to put an action figure in orbit but I need to know how to make a secure yet not too  expensive rocket fuel engine. The design of the engine must very primitive and not too complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read some information about hydrazine and hydrogenperoxide but it is too dangerous to use and make at a noncommercial scale. Although I studied some chemistry in highschool, I don't have so much knowledge about this matter. So far my best bet is blackpowder based fuel engine. What do you suggest? &lt;/p&gt;&#xA;" OwnerUserId="898" LastEditorUserId="7448" LastEditDate="2015-10-30T17:17:00.653" LastActivityDate="2016-04-18T07:28:48.627" Title="Homemade Rocket Fuel and Engine" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;home-experiment&gt;&lt;explosives&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ClosedDate="2020-05-28T02:03:08.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="2767" PostTypeId="2" ParentId="2766" CreationDate="2012-12-13T11:43:52.377" Score="9" Body="&lt;p&gt;Note: I'm not a modeler myself, but tested several recipes of rocket candy and read rocket modeler forum for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you have to think if you will be satisfied with solid propellant or want a liquid fuel engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building home-made liquid fuel rocket engine is very hard. One need to find pumps, make combustion chamber from something very heat-proof and reasonably stable chemically. Anyway, some successful reports may be found on the internetz. Note, however, that if you can choose safe liquid fuel, like kerosene or some alcohol, &lt;strong&gt;there is no safe oxidant&lt;/strong&gt; for liquid-fueled rocket engine. The closest thing to safe oxidant I can imagine is nitric acid, that require work on fresh air or in specially designed room and tends to form explosives on contact with many things, like milk protein (yep!), some of which are volatile&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid-fuel model propellants are much easier to prepare and use, there are numerous articles in the internetz about such fuels. The most common one is rocket candy, prepared from potassium nitrate and sugar. I tried it to see how it works (I didn't build any rockets), but instead of sugar I used sorbitol. Rocket candy based on sorbitol can be melted and then can be casted in any form. &lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_candy&quot;&gt;wiki article&lt;/a&gt; contains both info and links to sites with well-testes recipes by successful modelers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible option is making missile model based on some classical jet. There is quite simple pulse jet and there are some books about home-made turbine jet designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I recommend to search nearby rocket modeling club and find supervisor.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2012-12-13T11:53:43.997" LastActivityDate="2012-12-13T11:53:43.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2768" PostTypeId="1" CreationDate="2012-12-13T13:08:47.757" Score="16" ViewCount="4908" Body="&lt;p&gt;Why are the energy levels for vibrational energy in a molecule discrete as opposed to continuous?&#xA;I don't understand how a vibration can't have continuous amounts of energy that depend on the frequency of any infrared photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone combine this question onto my stack exchange account: olly_price, because I can't sign in on this computer.&lt;/p&gt;&#xA;" OwnerUserId="978" LastEditorUserId="323" LastEditDate="2012-12-13T14:13:09.637" LastActivityDate="2013-01-07T16:57:08.640" Title="Why are there quantised energy levels in the vibrational energy of a molecule?" Tags="&lt;quantum-chemistry&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2769" PostTypeId="1" AcceptedAnswerId="2771" CreationDate="2012-12-14T02:27:16.827" Score="11" ViewCount="1478" Body="&lt;p&gt;So I understand molecular orbitals and how to do VSEPR models, but I seem to be struggling with understanding localized bonding theory and how to do hybridizations.  After drawing the initial lewis structure then molecular orbital, I do not know where to go in terms of whether something has a $\pi$ or $\sigma$ bond.&lt;br&gt;&#xA;Also, how do I know when an atom has a carbon attached as an extra sigma bond or not? I understand how to know if it is $\mathrm{sp}$, $\mathrm{sp}^2$, etc. but I do not know what that means.&lt;/p&gt;&#xA;" OwnerUserId="982" LastEditorUserId="4945" LastEditDate="2014-11-17T06:17:51.383" LastActivityDate="2019-02-24T08:47:59.390" Title="How does localized bonding theory and hybridisation work?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2770" PostTypeId="2" ParentId="2768" CreationDate="2012-12-14T03:59:53.593" Score="14" Body="&lt;p&gt;&quot;Why&quot; is a good question, and one that science has yet to fully answer. We generally have a good understanding of &quot;how&quot; things work at the subatomic level, based on over a century of observation followed by theory and math backed up by experimentation. However, the really basic, naively simple questions, like &quot;where do particles get the charges, spins and masses that define them&quot; are the questions scientists are asking now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the discrete quantum energy states theorized by quantum theory, those are the result of a lot of math that explains what we see much to well to be coincidence. Basically, an electron can be thought of as orbiting its nucleus, but it's really more like a vibration. Similar to a vibration, there are frequencies above the &quot;fundamental&quot; that produce &quot;harmonics&quot;; standing waves with wavelengths of a perfect integer fraction of the fundamental, and perfect integer multiples of frequency, which are self-stabilizing. Electrons can only exist in these discrete energy states because if they didn't, they'd &quot;break orbit&quot; and go zooming off as free radicals, or have an unstable or unenergetic path that sends them crashing into the nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's weird is that that doesn't ever happen as far as we know; not only &lt;em&gt;do&lt;/em&gt; electrons exist in stable states, they &lt;em&gt;only&lt;/em&gt; exist in these states, and we can't get them to exist in any energy state between two discrete states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Planck first posited these quantum energy states existed while attempting to solve a conundrum of classical physics. Back to the vibration analogy, consider a guitar string. By damping the string at precise points on the string, we can induce clear harmonics. however, the higher the harmonic, the harder you have to pluck to produce a wave of the same amplitude. As the wavelengths get smaller, the energy you have to put into them to obtain a constant amplitude (and thus the energy inherent in the wave; Newton's third law) increases exponentially, diverging to infinity as the wavelengths approach zero. Now, this was obviously wrong; observed visible spectra of a glowing piece of metal as it is heated clearly shows a &quot;peak&quot; in the frequency of light produced, and at frequencies above this peak, amplitude of emitted light drops off sharply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planck solved this &quot;ultraviolet catastrophe&quot; by describing what was going on as similar to this string having a number of springs of different length attached to it, with weights on the other end of each spring. As the string vibrated, it in turn induced vibrations in the springs (known technically as &quot;harmonic oscillators&quot;), and as the frequency increases, it approaches, matches, and then falls away from the harmonic frequencies of the springs, increasing their relative amplitudes and then making them fall off as the vibrations begin to cancel out, while simultaneously increasing the energy imparted to a different string whose harmonic was being excited. The relative harmonic frequencies of the springs are what determine the colors of light seen, and the total energy of the string is divided among the springs, keeping the same total energy as frequency increases instead of diverging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the math work, Planck had to describe the springs as each receiving an integer portion of the energy of the entire system, and described these energy levels as proportional to the length of the spring. The proportionality is defined by a slope known today as Planck's Constant. Thus, Planck asserted that all energy existed in discrete amounts, which he called &quot;quanta&quot;, hence &quot;quantum theory&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this worked, it seemed to do so by convenience, until Einstein then further developed photon theory: light was made of of photons, &quot;packets&quot; of energy that could be quantified in proportion to the photon's frequency, based on Planck's Constant. This in turn explained puzzling properties of the &quot;photoelectric effect&quot;, namely that the threshold where electrons began to be ejected from metal when exposed to light (forming an electrical charge) was dependent not on intensity of light, but on the frequency of the light. Einstein's photons explained that, because an electron was likely to only be hit by a single photon of light, no matter how many there were, the energy imparted to the electron could only come from that single photon, and as a photon's energy increases with frequency, there is a minimum frequency required to impart enough energy with a single photon to eject an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planck's constant, and the highly definite nature of how atoms interact with photons, continues to show up in math that accurately describes things we observe about matter, such as how materials emit very specific wavelenghs of light while absorbing all others. That was explained by the Bohr model of the atom, which was the first model to introduce quantum theory to atomic structure. The Bohr model was based partially on a mathematical model first proposed by Johann Balmer and then improved on by Johannes Rydberg, showing that the emission spectra of hydrogen was predictable by an equation with integer solutions. Bohr showed that the energy states of electrons, which can be altered by absorption of photons, must exist in certain integral states, defined by an equation derived from Rydberg's but that also includes, you guessed it, Planck's Constant. The famous &quot;double-slit experiment&quot; showing that light is both wave and particle, and the resulting mathematical models by Louis de Broglie showing the movement of electrons as a series of stable standing waves of vibration in the electromagnetic field of the nucleus (electrons, in fact, are similar to photons in that they are as much a wave as a particle), reproduced the hypothetical energy states of Bohr's model by a different method, and in math, when you get independent confirmation of your results by a completely different mathematical model, you know you're on to something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that's the theory to date (well, actually I stopped about 100 years' worth of research ago, but them's the basics of the discovery of discrete quanta). What we are still trying to explain is why these quanta actually exist. A spaceship can have an infinite number of stable orbits around the Earth, and there are countless stable orbits of planets and planetoids around the Sun; why then does an electron stabilize so deterministically in these discrete states? Our best answer is that the electron's wave motion is only stable in these discrete states; otherwise the wave would interfere with itself or others and cancel out. However, we have never seen this &quot;turbulence&quot; happen and can't seem to force it to happen; it seems to be a fundamental quality of matter.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-01-07T16:57:08.640" LastActivityDate="2013-01-07T16:57:08.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2771" PostTypeId="2" ParentId="2769" CreationDate="2012-12-14T13:21:51.700" Score="9" Body="&lt;p&gt;First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\text{sp}^2$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning. It's just like vector cross product in Physics (for example the torque on a rotating wheel acts along the axis, i mean what does that even mean physically). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly regarding how to know if an atom has a $\pi$ or a $\sigma$ bond: observing many structures(Lewis structures) and recognizing certain trends is the way out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some rules of the thumb that will help you draw structures are as follows:&lt;br&gt;&#xA;1. Identify the central atom as the one that can make the most bonds.&lt;br&gt;&#xA;2. Hydrogen makes only one bond. If Oxygen is present in the same compound, the Hydrogen is     &lt;strong&gt;generally&lt;/strong&gt; attached to the Oxygen forming a $\ce{O-H}$ group.&lt;br&gt;&#xA;3. Second period elements cannot expand their octets(i.e have more than 8 electrons in valence shell)&lt;br&gt;&#xA;4. &lt;strong&gt;Mostly&lt;/strong&gt; second period elements can make $\pi$ bonds.(That really solves part of your query). Although in some rare cases third period elements may also make $\pi$  bonds.&lt;br&gt;&#xA;5. Never neglect the lone pairs(or non-bonding electrons) of the central atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that these are only &lt;em&gt;thumb rules&lt;/em&gt; and may be violated sometimes. &lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="897" LastEditDate="2012-12-23T15:33:57.040" LastActivityDate="2012-12-23T15:33:57.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2773" PostTypeId="2" ParentId="2765" CreationDate="2012-12-15T14:11:27.493" Score="-3" Body="&lt;p&gt;Copper(II) oxide is basic and reacts with acids to form a copper salt and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A strong acid, like HCl, reacts faster than a weak acid - like acetic acid (which is the main acid component of vinegar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one adds NaCl to vinegar, you get HCl:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOH +NaCl&amp;lt;=&amp;gt;CH3COONa +HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper(II) chloride dissolves in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the salt is added to speed up the process.&lt;/p&gt;&#xA;" OwnerUserId="963" LastActivityDate="2012-12-15T14:11:27.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2774" PostTypeId="1" AcceptedAnswerId="2791" CreationDate="2012-12-15T14:14:41.127" Score="16" ViewCount="1942" Body="&lt;p&gt;During a reaction if addition on one of the $\pi$ bonds of an allene (containing even number of $\pi$ bonds like $\ce {H2C=C=CH2}$) takes place then a carbocation is formed. To decide the position of the positive charge, we look at the stability of the carbocation. Example:&#xA;$$\ce {H2C=C=CH2 -&amp;gt;[\ce{H3O+}]} \quad?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above reaction the proton ($\small{\ce{H+}}$) attacks the electron rich $\pi$ bond (say the left one) and can form one of the following carbocation:&#xA;$$\ce{H2C+\bond{-}CH=CH2}\quad\quad \text{or }\quad\quad \ce{CH3\bond{-}C+=CH2}$$&#xA;$$ (\text{I})\quad\qquad\qquad\qquad (\text{II})$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now the major product (acetone) is formed as a result of the carbocation on the right (II), which apparently is less stable than the one on the left (I) as the positive charge in I is in resonance/conjugation with the $\pi$ bond. But our teacher said that since in allenes the overlapping p-orbitals (say $\ce{p_z}$) of one $\pi$ bond are perpendicular to the overlapping p-orbitals of the other (say $\ce{p_y}$; which makes $\ce{p_x}$ in the direction of internuclear axis $x$) and hence the positive charge cannot resonate with the $\pi$ bond (formed by overlap of $\ce{p_z}$) as the positive charge is effectively an empty p-orbital (the $\ce{p_y}$) and hence the carbocation (II) is &lt;strong&gt;relatively&lt;/strong&gt; more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once addition has taken place the $\ce{C_1-C_2}$ $\pi$ bond no longer exists and the $\sigma$ bond is free to rotate so the empty p-orbital/positive charge can freely rotate and become coplanar with the p-orbitals of the $\pi$ bond. &lt;strong&gt;But why does this not happen?&lt;/strong&gt;&#xA;The reason cannot be steric hindrance as the two groups on $\ce{C_1}$ are the smallest, $\ce{H}$.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="5026" LastEditDate="2018-04-21T15:18:32.603" LastActivityDate="2018-05-08T13:18:06.093" Title="Resonance in carbocation derived from allenes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;resonance&gt;&lt;carbocation&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2775" PostTypeId="1" AcceptedAnswerId="2777" CreationDate="2012-12-16T02:37:16.427" Score="0" ViewCount="735" Body="&lt;p&gt;I need to do some plastic molding design. I want to know the names/types of plastic which are commonly/widely used in industries (please see the pictures below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZUEiN.jpg&quot; alt=&quot;Picture1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tM1oi.jpg&quot; alt=&quot;Picture2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="897" LastEditDate="2012-12-16T04:12:18.817" LastActivityDate="2016-03-02T01:31:05.040" Title="What are the common names of this two types of plastic?" Tags="&lt;everyday-chemistry&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2776" PostTypeId="1" CreationDate="2012-12-16T02:56:36.323" Score="2" ViewCount="185" Body="&lt;p&gt;I need to create a container which hold ~150 to 500 ml water. My plan is to create something like a inverse cone. I will apply pressure (~20psi) on the top of the cone and I want to create a hole at and only at the tip of the cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to make the tip of the cone thinner, so that it will  be easier to be break. However, I am new to plastic. Would anyone please give me some suggestion of material that I should use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2014-09-24T02:50:15.627" Title="Which plastic can be partially broken by pressure easily?" Tags="&lt;plastics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2777" PostTypeId="2" ParentId="2775" CreationDate="2012-12-16T04:03:25.470" Score="4" Body="&lt;p&gt;The most common types of plastics in use today are polyethylene (PE) and polystyrene (PS). Each of these comes in several variants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to tell just from a photograph exactly what plastics are used in the products shown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, various additives are often added to plastic to enhance certain properties such as high-impact, chemical resistance, recyclability and flexibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The type of plastic is sometimes marked on the item itself to help with recycling, typically in the form of an extruded imprint, or you could try to contact the manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both PE and PS are suitable for moulding and extrusion. You can also do some simple physical and/or chemical testing to determine what the plastics contain, although this will involve destroying at least a part of the item. Low density polyethylene (LDPE) typically has a lower melting point $(105-130^{o}C)$ than polystyrene $(200-240^{o}C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LDPE is also softer and slightly rubbery whereas polystyrene is typically more rigid &amp;amp; crackly when crushed. Without physically touching the items to get a proper feel for what they are made of, as a first guess, I would say that the colourful containers such as plastic bins and laundry baskets in your first photo look like some sort of polyethylene, whereas the clear plastic packaging in the second photo looks more like polystyrene.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-12-16T04:47:43.813" LastActivityDate="2012-12-16T04:47:43.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2778" PostTypeId="2" ParentId="2726" CreationDate="2012-12-16T07:13:07.363" Score="3" Body="&lt;p&gt;The chemical disinfection (inactivation of bacteria) in fresh water can occur through a number of mechanisms, including oxidation of cell walls, inactivation of key enzymes and disruption of nucleic acids, thereby rendering them non-functional. The precise mechanism of inactivation depends upon the nature of micro-organism (bacteria, spores, viruses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When chlorine was first used as a disinfectant in the USA in 1908 (slightly earlier in Europe), its germicidal power was commonly attributed to the liberation of 'nascent oxygen' from hypochlorous acid $(HOCl)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HOCl\rightarrow HCl+\frac{1}{2}O_{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequent investigations (Chang, 1944, Green and Stumpf, 1946) debunked this theory and showed that chlorine reacts irreversibly with the enzymatic system of bacteria, thereby killing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang demonstrated by showing that hydrogen peroxide or potassium permanganate which release considerable amounts of nascent oxygen, nevertheless exhibit only weak germicidal activity. Chang was able to demonstrate that there is no liberation of oxygen involved in the inactivation of bacteria by chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, oxygen gas $(O_{2})$ is a much weaker oxidizing agent than hypochlorous acid, as shown in the table below. Hypochlorous acid is likely to be effective as both an oxidizing agent as well as direct chlorination of the protoplasm and reaction with lipoproteins to form toxic chloro compounts that interfere with cell division (Chang, 1944).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EVX7U.png&quot; alt=&quot;Oxidation power&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypochlorous acid $(HOCl)$ dissociates in water to $H^{+}$ and hypochlorite ion $(OCl^{-})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HOCl\leftrightarrow H^{+} + OCl^{-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because hypochlorous acid $(HOCl)$ is uncharged, it is better able to penetrate cell walls than other chlorine species and is about &lt;strong&gt;80 times&lt;/strong&gt; more effective than hypochlorite $(OCl^{-})$ at chlorination disinfection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;microbial disinfection due to total chlorine&lt;/em&gt; ($HOCl + OCl^{-}$) therefore depends upon &lt;em&gt;pH&lt;/em&gt; and &lt;em&gt;temperature&lt;/em&gt;. Whilst low pH and temperature favour disinfection by chlorine as $HOCl$, rate of diffusion through the cell membrane occurs faster at high temperature. The metabolic activity of the cell,rate of reaction of chlorine with enzymes and hence rate of inactivation of the cell is faster at higher temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Fion.png&quot; alt=&quot;HOCl and OCl vs pH&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disinfection by redox reaction using ozone gas $(O_{3})$ (a fast and powerful oxisizing agent) is sometimes used as an alternative to chlorination to reduces or eliminate the toxicity caused by residual chlorine, especially where there is concern over organic matter in the water which may produce trichloromethanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang, S.L., Destruction of microorganisms, J. Am. Water Works Assoc., 36 (1944), pp. 1192–1206&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Green, D.E. and Stumpf, P.K., The mode of action of chlorine, J. Am. Water Work Assoc., 38 (1946), pp. 1301–1305&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water Quality Control Handbook, Second Edition, by E. Roberts Alley (2007, McGraw Hill).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com.au/books?id=mGVbIoW2lNAC&amp;amp;printsec=frontcover&amp;amp;dq=Handbook%20of%20Chlorination%20and%20Alternative%20Disinfection&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=T3TNUPL8JojvmAW3w4D4AQ&amp;amp;ved=0CDgQ6AEwAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;White's Handbook of Chlorination and Alternative Disinfectants&lt;/a&gt;, 5th Edition, by Black &amp;amp; Veatch Corp (2010, John Wiley &amp;amp; Sons)&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2012-12-16T08:40:24.677" LastActivityDate="2012-12-16T08:40:24.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2779" PostTypeId="1" CreationDate="2012-11-13T19:38:03.087" Score="12" ViewCount="4959" Body="&lt;p&gt;Which materials or compounds expand in response to cold temperatures?&#xA;This material or compound should expand at a slow (but constant rate), from about 0–20 months when in the following temperature range −25 °C to −10 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the compound should not react with plastic or be toxic to humans or the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain &lt;em&gt;why&lt;/em&gt; the compound does this as well.&lt;/p&gt;&#xA;" OwnerDisplayName="natasa" LastEditorUserId="402" LastEditDate="2019-05-19T06:18:51.730" LastActivityDate="2019-05-19T13:07:42.140" Title="Which materials/compounds expand in response to cold temperatures?" Tags="&lt;physical-chemistry&gt;&lt;materials&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2780" PostTypeId="2" ParentId="2779" CreationDate="2012-11-15T17:27:37.383" Score="8" Body="&lt;p&gt;Perhaps one of the most studied materials to exhibit &lt;a href=&quot;http://en.wikipedia.org/wiki/Negative_thermal_expansion&quot; rel=&quot;nofollow noreferrer&quot;&gt;negative thermal expansion&lt;/a&gt; is cubic zirconium tungstate &lt;span class=&quot;math-container&quot;&gt;$(\ce{ZrW2O8}).$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="Mark Rovetta" LastEditorUserId="41328" LastEditDate="2019-05-19T13:07:42.140" LastActivityDate="2019-05-19T13:07:42.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2781" PostTypeId="5" CreationDate="2012-12-16T09:45:30.510" Score="0" Body="&lt;p&gt;Carbocations (a portmanteau of carbon and cation) are one of the three main reactive intermediates in organic reactions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation&quot; rel=&quot;nofollow&quot;&gt;carbocations&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbanion&quot; rel=&quot;nofollow&quot;&gt;carbanions&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Radical_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;radicals&lt;/a&gt;). They are species which have a positive charge on a carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positively-charged carbon in a carbocation only has six electrons in its valence shell, and hence is &lt;em&gt;electron-deficient&lt;/em&gt; or &lt;em&gt;electrophilic&lt;/em&gt;, making it extremely susceptible towards nucleophilic attack. Carbocations generally show $\mathrm{sp^2}$ hybridisation and are therefore trigonal planar about the positive carbon.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="16683" LastEditDate="2016-04-17T21:09:30.427" LastActivityDate="2016-04-17T21:09:30.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2782" PostTypeId="4" CreationDate="2012-12-16T09:45:30.510" Score="0" Body="Carbocations are species bearing a positive charge on carbon. They are intermediates generally formed during organic reactions, which can be stabilised by various electronic effects. Less stable carbocations are capable of undergoing rearrangements to form more stable carbocations in the course of a reaction." OwnerUserId="897" LastEditorUserId="16683" LastEditDate="2016-04-17T21:09:39.773" LastActivityDate="2016-04-17T21:09:39.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2784" PostTypeId="1" AcceptedAnswerId="2785" CreationDate="2012-12-16T12:52:14.837" Score="10" ViewCount="16193" Body="&lt;p&gt;Relative atomic masses of atoms of all chemical elements are numbers without units, being the value of proportion compared to  $\frac{1}{12}^\text{th}$ the mass of the carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the relative atomic mass of carbon is never 12! Instead, it is 12.01 or, more accurately,  12.011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="747" LastEditorUserId="4945" LastEditDate="2014-10-23T10:24:05.937" LastActivityDate="2020-10-10T12:19:47.653" Title="Why is the relative atomic mass of carbon not exactly 12?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2785" PostTypeId="2" ParentId="2784" CreationDate="2012-12-16T14:41:48.657" Score="16" Body="&lt;p&gt;Simply because the atomic mass is defined as 1/12 of the mass of &lt;sup&gt;12&lt;/sup&gt;C. Others isotopes of carbon (&lt;sup&gt;13&lt;/sup&gt;C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2012-12-16T14:41:48.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2786" PostTypeId="1" AcceptedAnswerId="2790" CreationDate="2012-12-17T05:16:35.613" Score="18" ViewCount="2566" Body="&lt;p&gt;I have been researching for a Physics/Chemistry exam and thought; what will the future periods in the periodic table (periods 8 and above) would entail? Each block contains its own properties that allow for accurate predictions of elements in these blocks. Each block has very unique properties and benefits society in major ways. The g-block is the theoretical block thought to exist in period-8 though no elements have been produced in this period for any sustained time. My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the predicted properties of the G-block if any?  Who and what methods have they been predicted by?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know if there are any ways of &quot;stabilizing&quot; these elements so that they take more time to decay.&lt;/p&gt;&#xA;" OwnerUserId="994" OwnerDisplayName="Bored915" LastEditorUserId="22" LastEditDate="2012-12-17T05:39:14.513" LastActivityDate="2012-12-28T21:43:12.690" Title="What are the g-block's predicted properties?" Tags="&lt;atoms&gt;&lt;periodic-table&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2789" PostTypeId="2" ParentId="2727" CreationDate="2012-12-17T20:43:18.457" Score="-2" Body="&lt;p&gt;Yes, it is, in $\ce{H3O2}$. This is from an a CCB bond. However, this is highly unlikely.&lt;/p&gt;&#xA;" OwnerUserId="997" LastEditorUserId="22" LastEditDate="2012-12-17T22:01:05.003" LastActivityDate="2012-12-17T22:01:05.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2790" PostTypeId="2" ParentId="2786" CreationDate="2012-12-18T03:27:27.677" Score="11" Body="&lt;p&gt;First of all, from the perspective of chemistry, it is the group (column) that matters more than the period (row), and subshells act at the period level.  Second, the elements beyond the actinides have virtually no significant chemical research owing to their extreme radioactivity.  So there may not be many predicted properties for the g-block beyond the simplistic theoretical assumptions about groups, which are much less relevant for a hyper transitional period like the g-block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for you question about potential stability, Glenn Seaborg made some predictions here, which you can read that this Wikipedia page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot;&gt;Island of Stability&lt;/a&gt;.  In general, though, these elements are not predicted to be stable.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2012-12-18T03:27:27.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2791" PostTypeId="2" ParentId="2774" CreationDate="2012-12-18T11:21:33.497" Score="8" Body="&lt;p&gt;The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : &lt;a href=&quot;http://journals.sagepub.com/doi/abs/10.1255/ejms.12&quot; rel=&quot;nofollow noreferrer&quot;&gt;G. van der Rest &lt;em&gt;et al. Eur. Mass Spectrom.&lt;/em&gt; &lt;strong&gt;3&lt;/strong&gt; 323 (1997)&lt;/a&gt; and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00409a004&quot; rel=&quot;nofollow noreferrer&quot;&gt;K. Raghavashari &lt;em&gt;et al. J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;103&lt;/strong&gt; 5649 (1981)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relative stability of the allyl and 2-propenyl cations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both articles make it clear that the allyl cation (I) is the most stable structure for the $\ce{C_{3}H_{5}^{+}}$ carbocations. It is around $\pu{21 kJ mol-1}$ more stable than its 2-propenyl (II) isomer. The other possible isomers are all higher in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Protonation of allene&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the first reference above, the protonation of allene on an extremal carbon to lead to the allyl cation requires to pass a high energy barrier, whereas no high barrier is present to lead to the 2-propenyl cation. Figure 4 gives the structure of the interaction complex between $\ce{H_{3}O^{+}}$ and allene, showing clearly that the incoming proton is initially in interaction with the $\pi$ orbital on one side of the allene. Their is obviously no conjugation contribution between this protonation structure and the other $\pi$ orbital, since the proton arrival plane is perpendicular to the other $\pi$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this position, formation of the 2-propenyl cation (II) requires only a minimal energy barrier, requiring only a bending of the $\ce{C-C-C}$ bond and bending of the $\ce{CH_{2}}$ group. Once this is done, the $\ce{CH_{3}}$ group can rotate freely, but one should notice that it does not involve the $\ce{sp^{2}}$ vacant orbital located on the central carbon. This orbital will always remain orthogonal to the remaining $\pi$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this position, one could also consider direct protonation of the central carbon, followed by a rotation of the $\ce{CH_{2}}$ group since no conjugation is present at this point. But here, one should consider that stabilization by conjugation can only proceed &lt;em&gt;after&lt;/em&gt; formation of a (non stabilized) primary carbocation. These species are much less stable than secondary carbocations (on the order of $\pu{75 kJ mol-1}$), thus leading to the presence of a large barrier for the direct formation of an allyl cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A last pathway that one should consider is a 1,2 hydride transfer between (II) and (I). Although hydride transfers are generally easy between carbocations, this transfer proceeds through a three-center system which can be considered a proton attached to a $\pi$ orbital. Thus, this leads us back to the initial problem, which is that no conjugation can serve as a driving force leading to the allyl cation. And only once the unstable primary carbocation is formed can a rotation of the $\ce{CH_{2}}$ group lead to formation of the allyl cation.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2018-05-08T13:18:06.093" LastActivityDate="2018-05-08T13:18:06.093" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="2793" PostTypeId="1" AcceptedAnswerId="2794" CreationDate="2012-12-18T15:25:04.910" Score="19" ViewCount="14145" Body="&lt;p&gt;Sorry for the FAQ, but I really don't know where to find a full spreadsheet for element properties. I mean a table that has one row for each element, and the columns are its properties, such as name, atomic number, density, melting point, etc. I know that there are many sites or references for that, but I need it in tabular format, because I want to analyze data, make charts, etc.&lt;/p&gt;&#xA;" OwnerUserId="989" LastEditorUserId="73" LastEditDate="2014-08-20T07:44:00.693" LastActivityDate="2020-01-08T16:36:26.607" Title="Where can I find a downloadable spreadsheet of element properties?" Tags="&lt;periodic-table&gt;&lt;reference-request&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2794" PostTypeId="2" ParentId="2793" CreationDate="2012-12-19T11:22:09.970" Score="11" Body="&lt;p&gt;Well, here's a CSV file that I parsed out of the JSON data provided by &lt;a href=&quot;http://www.pnathan.com/#tab2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paul Nathan's website&lt;/a&gt;, which in turn was produced from &lt;a href=&quot;http://www.frantz.fi/software/gperiodic.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;gPeriodic&lt;/a&gt; data in response to &lt;a href=&quot;https://stackoverflow.com/questions/4697829/finding-easily-parseable-chemical-element-data&quot;&gt;this question&lt;/a&gt;. gPeriodic is FOSS, so I can only assume that the element data contained within is good to share, though I have no idea of its provenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parser uses the python json module to read the data, which I then wrote into comma separated values, which should be readable by any halfway competent spreadsheet program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is: &lt;a href=&quot;http://pastebin.com/raw.php?i=CKwm136x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pastebin Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;save it as &lt;strong&gt;elementdata.csv&lt;/strong&gt; and you're good to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some gotchas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This reproduces the gPeriodic data, warts and all. I make no guarantees regarding its correctness, recency, etc. I just munged the data.&lt;/li&gt;&#xA;&lt;li&gt;Some of the data is augmented with tildes, notes about temperature/polymorph/state etc. I've left these as they are but you may need to trim them if you want to plot them as numeric values.&lt;/li&gt;&#xA;&lt;li&gt;Argon had an atomic radius of '2-', whatever that means. I cut it.&lt;/li&gt;&#xA;&lt;li&gt;Excel is &lt;em&gt;joke software&lt;/em&gt; and habitually interprets numbers wrapped in parentheses as negatives, because apparently some accountants decided that surrounding numbers in parens is a more sensible option than using a minus sign. Needless to say, this is wrong. I've addressed the problem by wrapping the affected numbers in angle brackets.&lt;/li&gt;&#xA;&lt;li&gt;There are no ionic radii in the source data. At all. Not my fault.&lt;/li&gt;&#xA;&lt;li&gt;The columns inherit the units of the source data. I've left the units out on purpose because I wanted each column to have no spaces in the name for ease of processing in R or whatever. Furthermore, the units in the source file have some inconvenient characters from an encoding perspective.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tried plotting Z versus covalent radius in R from this data - looks alright:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EROe4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the data points are missing, some are not read by R because they are wrapped in &amp;lt;&gt; - pre-process to your heart's content.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2017-05-23T12:41:54.053" LastActivityDate="2012-12-19T11:22:09.970" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2795" PostTypeId="1" AcceptedAnswerId="4848" CreationDate="2012-12-19T19:13:11.890" Score="25" ViewCount="16293" Body="&lt;p&gt;I'm wondering why radium appears to buck the general trend that first ionisation energies decrease as you move down a group in the periodic table: barium (the group 2 element preceding it) has a first ionisation energy of &lt;span class=&quot;math-container&quot;&gt;$\pu{502.9 kJ/mol}$&lt;/span&gt;, whereas radium has a slightly higher first I.E. of &lt;span class=&quot;math-container&quot;&gt;$\pu{509.3 kJ/mol}$&lt;/span&gt; (from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, although my textbook agrees). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any explanation for this at present? (I imagine that quantum mechanics may be involved in some way, but I'm not entirely sure how)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are most likely other examples of the trend being broken, but this is the only one I've come across so far and I'm curious to know why this is the case.&lt;/p&gt;&#xA;" OwnerUserId="753" LastEditorUserId="61578" LastEditDate="2019-06-29T08:38:02.577" LastActivityDate="2019-06-29T08:46:28.500" Title="Why does radium have a higher first ionisation energy than barium?" Tags="&lt;physical-chemistry&gt;&lt;periodic-table&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2796" PostTypeId="2" ParentId="2768" CreationDate="2012-12-20T00:49:49.713" Score="3" Body="&lt;p&gt;There is a general result in quantum mechanics that &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_level&quot; rel=&quot;nofollow&quot;&gt;a bound system has quantized energy states&lt;/a&gt;. Atoms in a molecule are clearly bound to each other, so they have quantized energy states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More quantitatively, one easy model for the vibrational energy of molecules is to model the atoms as a set of masses, and the bonds as a set of springs connecting the masses. To see where this comes from, back up for a second to classical mechanics. The classical harmonic oscillator has the equation $\ddot x + \omega^2x = 0$, with $\omega = \sqrt{k/m}$. $k$ is the stiffness constant for the spring, and $m$ is the mass, but physicists usually keep everything in terms of $\omega$. This represents a mass on a spring, attached to an immovable wall. The energy of that oscillator is $E = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$, where $A$ is the amplitude of the oscillation. $A$ is determined by the initial conditions, and can be any continuous value. So the energy spectrum is continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To translate this to the quantum mechanical simple harmonic oscillator, we write the Schroedinger equation as $(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2x^2)\psi = E\psi$. The solutions to this equation are the &lt;em&gt;discrete&lt;/em&gt; set of wave functions $\psi_n(x)$ with energies $E_n = \hbar\omega(n + \frac{1}{2})$. Again, this is a mass $m$ attached to an immovable wall via a spring. If you replace the wall with another (movable) mass of $M$, the only complication is to replace the $m$ dependence in $\omega$ with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduced_mass&quot; rel=&quot;nofollow&quot;&gt;reduced mass&lt;/a&gt; $\mu = \frac{mM}{m+M}$. In the case of molecules, the mass terms $m$ and $M$ have the obvious meaning of the mass of the relevant atom. The spring stiffness $k$ corresponds to the bond energy (though I'm not sure of the exact relationship). In practice, the energy is what you actually measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules with more than two atoms are more complicated. There is one characteristic frequency $\omega_i$ (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_mode&quot; rel=&quot;nofollow&quot;&gt;normal modes&lt;/a&gt; for each degree of freedom in the molecule. Actually solving for them involves solving a set of coupled equations. As a simple example, consider carbon dioxide, and restrict the motion of the atoms to be only along the bonds; there are two bonds, so there are two normal modes. One normal mode corresponds to the carbon atom remaining stationary while the two hydrogen atoms move oscillate. The two &quot;springs&quot; are in phase with each other: they are compressed at the same time, and extended at the same time. The other normal mode corresponds to the two hydrogen atoms remaining stationary while the carbon atom oscillates. In this normal mode, the two bonds are out of phase: one is compressed when the other is extended, and vice versa. Any motion along the bonds can be written as a sum of these two normal modes, and they are the only vibrational energies the molecule can absorb. More generally, if you allow the atoms free range of motion in space, there are nine degrees of freedom, so there are nine normal modes. Three correspond to motion of the center of mass of the atom in space. Two are rotations of the molecule as a whole around the center oxygen (it &quot;can't&quot; rotate around its axis). None of these contribute to the vibrational spectrum. Two are the springs I described above, and are called the stretching modes. The final two are due to the oxygen atom oscillating out of the line connecting the three atoms, and are called the bending modes. &lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2012-12-22T02:44:59.930" LastActivityDate="2012-12-22T02:44:59.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2797" PostTypeId="1" CreationDate="2012-12-20T01:23:02.643" Score="7" ViewCount="1298" Body="&lt;p&gt;The LSPR properties of nanoparticles depend on a number of factors, including shape, size, material, crystal structure, dielectric environment, etc. However, it has been consistently shown that shape and size have the greatest effect on the optical properties of nanoparticles. Exactly WHY, however, is this true? Does anyone have an explanation at the molecular level?&lt;/p&gt;&#xA;" OwnerUserId="1003" LastActivityDate="2012-12-22T01:57:23.427" Title="Why do shape and size have the greatest effect on the optical properties of nanoparticles?" Tags="&lt;materials&gt;&lt;nanotechnology&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2799" PostTypeId="2" ParentId="2797" CreationDate="2012-12-20T17:58:26.527" Score="4" Body="&lt;p&gt;As Surface Plasmon Resonance phenomenon is mainly seen in metal nanoparticles, I don't know whether this answer will help you? The size dependent optical phenomenon of semiconductor nanoparticles (e.g., $\ce{CdS}$) can be explained using the &lt;a href=&quot;http://inbt.jhu.edu/igert/jc_Semiconductor_Quantum_Dots.ppt&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective mass approximation model&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3GSL.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;where r is radius of the particle (the rest details of the above equation can be find in the link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the energy gap is inversely proportional to the radius of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ioGki.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the radius of the particle  decreases, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_dot&quot; rel=&quot;nofollow noreferrer&quot;&gt;band gap increases so it will emit high energy wavelength&lt;/a&gt; (i.e., wavelength in blue region) and when the particle size is large the energy gap will be less so it will emit light in red region (lesser energy).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="73" LastEditDate="2012-12-22T01:57:23.427" LastActivityDate="2012-12-22T01:57:23.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2800" PostTypeId="1" CreationDate="2012-12-20T18:21:48.840" Score="7" ViewCount="662" Body="&lt;p&gt;If I used chloralkali electrolysis of brine to produce hydrogen, and can't use the other products (or not all of them), what route(s) can I take to revert them back to brine? I figured, since I have to put in energy to take the brine apart, it should have some desire to reform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the $\ce{H2}$ taken away, the remaining products are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NaOH (aq) + Cl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just mix them, I get sodium chlorate or hypochlorite. Can I process those further?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $\ce{H2O}$ and $\ce{NaCl}$ have a negative $\Delta H_\mathrm{f}^\circ$ ($\pu{-285.8 kJ/mol}$ and $\pu{-407.0 kJ/mol}$ respectively), and $\ce{Cl2}$ is already in its standard state, so we would need to overcome the $\Delta H_\mathrm{f}^\circ$ of the $\ce{NaOH}$, which is negative as well ($\pu{-470.1 kJ/mol}$). Can I use ion filters to some benefit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Routes that allow retrieval of some of the energy spent for the electrolysis are favored (i.e. the more exothermic, the better).&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="16683" LastEditDate="2017-04-06T22:29:44.287" LastActivityDate="2017-04-06T22:30:16.987" Title="How can I dispose of chlorine and sodium hydroxide as table salt and water?" Tags="&lt;electrolysis&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2801" PostTypeId="2" ParentId="2797" CreationDate="2012-12-20T18:25:20.847" Score="2" Body="&lt;p&gt;Optical properties of nanoparticles come from electronic spectra of one excited electron trapped within. The energy levels in the spectra are derived from lengths of waves, that resonates with energy well within particle. Of course, this lengths depends on size and form of the resonator. Note: this IS oversimplification, but it allows to grab the concept.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-12-20T18:25:20.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2802" PostTypeId="2" ParentId="2743" CreationDate="2012-12-22T02:41:50.323" Score="2" Body="&lt;p&gt;I think that there is no really good classical reason.  Heat conductivity in a solid depends on bonding in that solid and bonding is inherently a quantum mechanical phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, recall that graphite is a good conductor of heat (and electricity) in two dimensions and a fairly good insulator (heatwise and electrically) in the third.  This is due to the layered structure in graphite and that is due to the bonding in graphite.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2012-12-22T02:41:50.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2803" PostTypeId="2" ParentId="2739" CreationDate="2012-12-22T05:22:31.647" Score="4" Body="&lt;h2&gt;Schotten-Baumann Reaction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I found the solution to my question and it is the Schotten-Baumann reaction which takes an acyl chloride (which can be synthesised from a carboxylic acid via the methods detailed in the next section) and an amine to form an amide, a hydrogen chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QtQrq.gif&quot; alt=&quot;enter image description here&quot;&gt; (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Net reaction: $ \ce{RCOCl+NH_2 R' \rightarrow RCONHR' + HCl}$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Synthesis of Acyl Chlorides&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To create acyl chlorides from carboxylic acids the following three reactions will suffice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \ce{RCOOH (l) + PCl_5 (s) \rightarrow RCOCl (aq) + POCl_3 (aq) + HCl (g) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \ce{3 RCOOH (l)+ PCl_3 (s) \rightarrow 3 RCOCl (aq) + H_2 PO_3 (aq)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \ce{RCOOH (l) + SOCl_2 (s) \rightarrow RCOCl (aq) + SO_2 (g) + HCl (g)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reference List (APA style)&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Schotten-Baumann Reaction Schotten-Baumann Conditions&lt;/em&gt; (n.d.). Retrieved December 22 2012 from &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/schotten-baumann-reaction.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.organic-chemistry.org/namedreactions/schotten-baumann-reaction.shtm&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Clarke, J. (2004). &lt;em&gt;CONVERTING CARBOXYLIC ACIDS INTO ACYL CHLORIDES (ACID CHLORIDES)&lt;/em&gt;. Retrieved December 22 2012 from &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/acids/pcl5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemguide.co.uk/organicprops/acids/pcl5.html&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastEditorDisplayName="user15489" LastEditDate="2015-05-08T06:15:44.707" LastActivityDate="2015-05-08T06:15:44.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2804" PostTypeId="2" ParentId="2800" CreationDate="2012-12-22T10:50:17.123" Score="5" Body="&lt;p&gt;It looks as if the simplest way would be to mix the products, and let the mixture warm up, so we get sodium chlorate in water. We can then have that crystalize and heat it until it disintegrates into $\ce{NaCl}$ and $\ce{O2}$. That would need no additional substances and have a clean end result. The only remaining question would be the energy budget, but the forming of the chlorate is exothermic and additional heat might be available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, we might find something useful to oxidize with the $\ce{NaClO3}$.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="16683" LastEditDate="2017-04-06T22:30:16.987" LastActivityDate="2017-04-06T22:30:16.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2805" PostTypeId="2" ParentId="2793" CreationDate="2012-12-22T12:12:08.300" Score="6" Body="&lt;p&gt;Fine, &lt;strong&gt;&lt;a href=&quot;http://pastebin.com/raw.php?i=Wj9xgEYd&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's&lt;/a&gt;&lt;/strong&gt; my data taken from the reference mentioned in my above comment.&#xA;I cannot vouch for it's accuracy.&lt;br&gt;&#xA;This is in the &lt;code&gt;.csv&lt;/code&gt; file format for its wide use as enlightened by &lt;strong&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/users/84/richard-terrett&quot;&gt;Richard Terrett&lt;/a&gt;&lt;/strong&gt;. You can combine  certain ranges from this csv/spreadsheet to the one prepared by Richard, because it contains certain properties not contained in Richard's source.&lt;br&gt;&#xA;Also there are certain properties in Richard's file which are absent/improper in mine so this file is best used in conjunction with Richard's file.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-12-22T12:12:08.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2806" PostTypeId="1" AcceptedAnswerId="2810" CreationDate="2012-12-22T17:22:05.810" Score="5" ViewCount="1927" Body="&lt;p&gt;I have been intrigued by one of the principles of chemistry lately- electrolysis. In my knowledge, electrolysis is a quite dangerous operation as the decomposition of water produces hydrogen and oxygen gas, both which are flammable. However, its potent to be volatile isn't what I am concerned with; rather, it is the production of these gases sufficiently and cheaply that I have been wondering about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, what is the cheapest way available right now to build an efficient electrolysis apparatus that could be used for a relatively long amounts of time? In addition to this, what are the side effects with electrolysis of water? In addition to this, could solar panels be used in the process of electrolysis? I want this electrolysis apparatus to be as independent from the grid as possible! &lt;/p&gt;&#xA;" OwnerUserId="884" LastEditorUserId="39539" LastEditDate="2017-04-09T01:55:29.557" LastActivityDate="2017-04-09T01:55:29.557" Title="How can I produce hydrogen gas efficiently and cheaply?" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2807" PostTypeId="2" ParentId="2743" CreationDate="2012-12-22T18:14:39.073" Score="1" Body="&lt;p&gt;I think the classical explanation for good heat-conductance in diamond is the spring-analogy. But it does not explain all elements such as metals where electron-gas in valence shell contribute to the good heat-conductance. In order to understand this deeper, you need to study the statistics outlined below. The phonon gas occurs between the bonds in diamond while in the electron gas with metals of blocks I/II.&lt;/p&gt;&#xA;&lt;p&gt;Let's think about classical features such as heat capacity and heat-conductance. As said, they are different in nature depending on the location in periodic table.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I/II blocks in periodic table&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let's consider the first block and second block. The good heat-conductance in metals such as Li, Na and Mg is due to the electron-gas that forms fermi-balls according to Fermi-Dirac statistics. When you experience the heat-change, phonos are the interactive quantitative pieces of energy.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hvbrL.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Only the electrons in the highlighted region contribute meaningfully to the specific heat capacity. The source of this picture is the page 166 &lt;a href=&quot;https://noppa.aalto.fi/noppa/kurssi/tfy-0.2062/materiaali/Tfy-0_2062_opetusmoniste_2012_-_luku_4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, a course booklet in 2062 Physics course in Aalto University.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;In order to understand this, you must understand Fermi-Dirac statistics and Fourier Transforms by which you can get the location-frequency-spectrums (in Finnish paikkataajuus spektri): you have distribution of particles and then you use that distribution to deduce the frequency plot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;IV block elements such as Si and C&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now the heat conductance here with structures such as diamond or Silica cannot be due to electron-gas because there is no moving electrons in diamond or silica. The very good heat conductance in diamond is due to mechanical vibration and very strong bonds. Silica has much weaker bonds than carbon atoms with four covalent bonds. Young's modulus is analogical to the mechanical Hooke's constant. Einstein and Debyen explained this so that atoms act like massless springs (page 155 of the earlier lecture slide). Now you can calculate the internal energy with the help of Bose-Einstein statistics.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KeyrZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;This plot shows the heat capacities as a function of temperature. The parameter &lt;span class=&quot;math-container&quot;&gt;$\theta_D$&lt;/span&gt; is the Debye temperature, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the gas-constant and dots show real temperatures. The source is page 158 of the earlier lecture slides.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now in Chemistry, people often use the Maxwell-Boltzman statistics. In Quantum mechanics, they say that Maxwell-Boltzman is like an approximation -- it only work well with very low-temperature or very high temperature -- otherwise it is a bad approximation even though useful in practice but not correct. The statistics Fermi-Dirac and Bose-Einstein are the more realistic description of nature. Depending on the spin-number of atom, you choose the statistics. If your spin is some integer (or zero with modulo 1), you have a boson. If your spin has half (or half with modulo 1), you have a fermion. Examples of fermions contain electron, neutron, proton -- and they all have spin of half.&lt;/p&gt;&#xA;&lt;p&gt;Let's consider Rubidium 86 (atomic number 37 so 37 electrons, 37 protons and 49 neutrons) to its isotope 87 (37 electrons, 37 protons and 50 neutrons). Because 86's spin is odd and 87's spin is even, 86 is fermion while 87 is boson. Now use Boson-Einstein statistics with 87 and Fermi-Dirac with 86.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;P.s. You could mathematically approach this problem this way:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;1. calculate amount of particles &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;2. derivate &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; with respect to &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; energy&lt;/em&gt;&lt;br /&gt;&#xA;&lt;em&gt;3. find the point when the derivative &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial N}{\partial E}$&lt;/span&gt; is zero i.e. &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial N}{\partial E}=0$&lt;/span&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="966" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-12-05T11:52:19.593" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2808" PostTypeId="1" AcceptedAnswerId="2809" CreationDate="2012-12-22T20:17:35.473" Score="1" ViewCount="1516" Body="&lt;p&gt;I'm looking for a software which can solve chemical equations. For example something like this (given with result):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \ce{CH4 + 2O2 &amp;amp; → CO2 + 2H2O} \\&#xA;    \ce{C3H6 + HCl &amp;amp; → C3H7Cl}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there are such software? I'm studying chemistry as a hobby and want to check if I'm doing the reactions correct.&lt;/p&gt;&#xA;" OwnerDisplayName="user822" LastActivityDate="2012-12-22T20:21:08.843" Title="Chemical software for solving reactions" Tags="&lt;reaction-mechanism&gt;&lt;software&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2012-12-22T21:29:28.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="2809" PostTypeId="2" ParentId="2808" CreationDate="2012-12-22T20:21:08.843" Score="2" Body="&lt;p&gt;No, there is no such thing as chemical reaction are not that easily predictable. You can get result right in some cases, but it is impossible to count all factors.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2012-12-22T20:21:08.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2810" PostTypeId="2" ParentId="2806" CreationDate="2012-12-22T20:34:38.693" Score="9" Body="&lt;p&gt;Electrolysis of water involves the splitting (lysis) of water molecules using electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above equation, we see that the production of hydrogen will occur twice as fast as oxygen (on a volume basis). The electrode materials must be inert (non-reactive with water or its constituents). Platinum is good but very expensive, so titanium and/or nickel alloy catalysts are sometimes used. The current flows into the anode where oxygen gas is produced, whereas hydrogen gas is produced at the cathode. Generally, the higher the current, the higher the rate of production of hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is not a spontaneous reaction, energy is required from an external source in order to drive it. This is provided by a direct current (DC) electrical power supply, such as a battery or solar cell. The amount of energy required can be determined from the Gibbs energy of formation of water, which is -237.2 kJ/mol (at 298.15 K or 25 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical minimum voltage needed for the reaction to occur is the standard electrode potential of oxygen, which is +1.23V. In practice, a slightly higher voltage (about 1.48V) is needed because of an over-potential of about 0.25V which is used up as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure (demineralised) water can be expensive and has very low electrical conductivity, so some impurities in the water are desirable. Salt (sodium chloride) dissolved in the water will increase its conductivity dramatically, thereby reducing the current necessary to produce a produce hydrogen at a given rate. However, the chlorine gas can be produced at the anode (instead of oxygen) which is highly corrosive and hazardous. Minerals such as magnesium or calcium (present in sea water) will form insoluble precipitants on the cathode which could build up a 'passive' layer, reducing the rate of hydrogen production. Some acids or bases such as sulfuric acid or sodium hydroxide are therefore sometimes preferred.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="402" LastEditDate="2017-04-08T22:18:00.290" LastActivityDate="2017-04-08T22:18:00.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2811" PostTypeId="1" AcceptedAnswerId="2812" CreationDate="2012-12-23T00:14:44.020" Score="1" ViewCount="1151" Body="&lt;p&gt;Could you please explain how the bent molecular geometry of water is due to hydrogen bonding? I was under the impression that it was because of the lone pairs (VSEPR Theory).&lt;/p&gt;&#xA;" OwnerUserId="743" LastEditorUserId="743" LastEditDate="2012-12-23T06:42:21.437" LastActivityDate="2012-12-23T17:58:33.903" Title="Bent molecular geometry of water" Tags="&lt;bond&gt;&lt;water&gt;&lt;structural-formula&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2812" PostTypeId="2" ParentId="2811" CreationDate="2012-12-23T03:31:11.930" Score="3" Body="&lt;p&gt;Your understanding is correct. The usual explanation given in introductory classes is &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow&quot;&gt;VESPR&lt;/a&gt; theory. The oxygen atom has four bond-like &quot;things&quot; coming out of it. Two of those are the covalent bonds with the hydrogen bonds. The other two are the two lone pairs. VESPR theory calls this a &lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_number&quot; rel=&quot;nofollow&quot;&gt;steric number&lt;/a&gt; of four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slightly more advanced explanation comes from &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow&quot;&gt;molecular orbital theory&lt;/a&gt;. Oxygen's valence electrons are all $2\text{p}$ electrons; hydrogen's electron is $1\text{s}$. The only way to mix those orbitals to get enough bonding orbitals is by using two $\text{sp}^3$ hybrid orbitals to generate the two bonds. The $\text{sp}^3$ orbitals are arranged in a tetragonal geometry.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="897" LastEditDate="2012-12-23T17:58:33.903" LastActivityDate="2012-12-23T17:58:33.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2813" PostTypeId="1" AcceptedAnswerId="2814" CreationDate="2012-12-23T04:39:59.900" Score="4" ViewCount="4376" Body="&lt;p&gt;To be more specific, could the carbon dioxide fumes produced in the reaction be harmful to a small child? I don't have a strong background in chemistry, and I was prompted to ask this question when it was suggested by a friend that a mixture of 2 cups of vinegar and baking soda is commonly used to euthanize rodents. I'm aware rodents are more sensitive than humans, but I was concerned that a small child could inadvertantly breathe enough fumes to cause harm. I think it would be a great experiment to introduce children to chemistry if not unsafe. Thanks for any advice or guidance.&lt;/p&gt;&#xA;" OwnerUserId="432" LastEditorUserId="323" LastEditDate="2012-12-23T07:47:11.527" LastActivityDate="2018-02-16T18:23:14.830" Title="Is the vinegar and baking soda experiment safe for children?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2814" PostTypeId="2" ParentId="2813" CreationDate="2012-12-23T05:54:09.720" Score="8" Body="&lt;p&gt;A basic safety rule in chemistry is &quot;don't breathe fumes&quot;. Maximum waft a sample of evolved gas to the nose with your hand. Given a controlled environment, adequate size vessels and small quantities and adequate ventilation, it could be a good introduction to chemistry. However - start in the way you should continue - safety glasses, safe operating procedures etc.. BTW, red cabbage makes a good indicator for this.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2012-12-23T05:54:09.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2815" PostTypeId="1" CreationDate="2012-12-23T17:18:53.350" Score="5" ViewCount="907" Body="&lt;p&gt;In multiple sources (&lt;a href=&quot;http://www.phy.cuhk.edu.hk/course/surfacesci/mod3/m3_s4.pdf&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;), I found the information that the kinetic energy $E_k$ measured using (Ultraviolet / X-ray) Photoelectron Spectroscopy is given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_k = h\nu - B.E._F - \phi_{spec}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h\nu$ is the energy carried by the incident photon, $B.E._F$ the energy difference between the energy level of the electron and the Fermi energy and $\phi_{spec}$ the work function &lt;strong&gt;of the spectrometer&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understood the measurement process, the actually measured quantity in Photoelectron Spectroscopy is the kinetic energy $E_k$ of the electron. This energy is measured by the radius of the electron in a homogeneous electric field. Thus, the kinetic energy of the electron after the emission would $E_k = h\nu - B.E._F - \phi_{sample}$, this independent of the work function of the spectrometer. So where does this difference come from?&lt;/p&gt;&#xA;" OwnerUserId="1011" LastEditorUserId="189" LastEditDate="2014-10-17T10:52:40.063" LastActivityDate="2014-10-17T18:16:25.323" Title="Kinetic energy of electrons emitted in Photoelectron Spectroscopy" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;spectroscopy&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2818" PostTypeId="1" CreationDate="2012-12-24T10:32:27.817" Score="6" ViewCount="1516" Body="&lt;p&gt;I have a first-order reaction, for the binding of two molecules, of the form: $\text{k}_\text{f} = \ce{k_1 (M*s)^{-1}}$.  What happens at infinite concentration?  How is this upperbound typically handled for approximations in chemistry?  Do we need to measure additional kinetics parameters in this regime?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has any specific knowledge, I'm particularly concerned with DNA hybridization, where we have $\text{k}_1 \approx 10^6$.  Will we simply converge on the elementary step time for the formation of a base $\text{k}_+ \approx 10^6$ to $10^7$ (Craig et al., 1971; Porschke &amp;amp; Eigen, 1971)?  It seems reasonable to me that as $\text{C} \to \infty$ we'd have $\text{k}_\text{f} \to \text{k}_+$, which, I'd guess is the point where the rate limiting step is the time for the molecule to sample different rotamer/etc. configurations.  Perhaps it isn't an accident that these orders of magnitude match up?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, should we think about the entropy of complex formation in the same way in the limit of infinite concentration?  My guess for this would be &quot;yes&quot; since the entropic penalty of holding two molecules together, derived from studies on monoatomic gases, should be concentration independent as long as diffusion/mixing is allowed.  In practice, however, we'll probably have a gel at these higher concentrations.&lt;/p&gt;&#xA;" OwnerUserId="1015" LastEditorUserId="897" LastEditDate="2012-12-25T00:00:18.457" LastActivityDate="2013-04-13T10:19:51.367" Title="A first-order reaction at infinite concentration" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2819" PostTypeId="1" AcceptedAnswerId="2848" CreationDate="2012-12-24T13:28:09.467" Score="13" ViewCount="2513" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_of_unsaturation&quot; rel=&quot;noreferrer&quot;&gt;Degree of Unsaturation&lt;/a&gt; (DU) of an organic compound tells us about the total number of $\pi$ bonds (both double bonds and triple bonds) and number of rings/cyclic chains in the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degree of unsaturation forms an integral part of the four tools used to identify the structure of an organic compound (some of the others being structure after hydration and ozonolysis (&lt;em&gt;there's one more I don't remember&lt;/em&gt;)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states the formula of degree of unsaturation as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\text{Rings + $\pi$-bonds} = C - \frac{H}2 - \frac{X}2 + \frac{N}2 + 1$$&#xA;  where $C$ = number of carbons, $H$ = number of hydrogens, $X$ = number of halogens and $N$ = number of nitrogens&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A more general formula is stated as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$DU = 1 + \frac{1}{2}\sum{n_i(v_i - 2)}$$&#xA;  where $n_i$ is the number of atoms with valence $v_i$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I'm interested is the underlying concept from which this formula is derived.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="41328" LastEditDate="2017-08-07T20:00:20.383" LastActivityDate="2017-08-07T20:00:20.383" Title="Derivation of formula for the degree of unsaturation" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2820" PostTypeId="1" CreationDate="2012-12-24T17:35:08.393" Score="7" ViewCount="1325" Body="&lt;p&gt;What could be the possible basis set that can be used for a metal base macrocyclic complex in gaussian09? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using transition metals such as $\ce{Ni}$, $\ce{Co}$, etc.  Also, I have tried using DFT and HF theory for optimization but I am not getting appropriate results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am basically looking for binding of metal with the ligand in these coordinate complexes on basis of which i can use them for a particular application.&#xA;The basis set that I am using right now for DFT is &lt;code&gt;Opt=(Cartesian,maxcycles=500)  b3lyp/gen  nosymm   pseudo=read SCF=(conver=5,vshift=400,maxcycles=1000)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Earlier I was using a specific basis set for transition metals and now I am using &lt;code&gt;lanl2&lt;/code&gt; as a single basis set for the DFT calculation. I am also using &lt;code&gt;hf cal&lt;/code&gt; in which the &lt;code&gt;sto3g&lt;/code&gt; basis set is used but iam getting two consistent errors in optimization which are convergence failure and number of step exceed error processed link 9999.&lt;/p&gt;&#xA;" OwnerUserId="1017" LastEditorUserId="22" LastEditDate="2012-12-27T19:30:40.847" LastActivityDate="2020-03-25T21:06:44.243" Title="Basis sets that can be used for macrocyclic complexes" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2823" PostTypeId="2" ParentId="2820" CreationDate="2012-12-25T02:00:01.347" Score="3" Body="&lt;p&gt;In order to check on the availability of basis sets you may want to look at the &lt;a href=&quot;https://www.basissetexchange.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basis Set Exchange&lt;/a&gt;. After determining which basis set suits your problem (this will likely involve some literature search, Nickel, DFT, basis set, in &lt;a href=&quot;http://isiknowledge.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Web of Science&lt;/a&gt; should do the trick), you can enter basis sets into Gaussian using the &lt;a href=&quot;http://gaussian.com/gen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gen keyword&lt;/a&gt;. BSE will export any selected basis sets in Gaussian format, so that shouldn't be a problem.&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="41556" LastEditDate="2020-03-25T21:06:44.243" LastActivityDate="2020-03-25T21:06:44.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2825" PostTypeId="1" AcceptedAnswerId="9284" CreationDate="2012-12-25T11:45:55.450" Score="5" ViewCount="6646" Body="&lt;p&gt;I expected ethene ($\ce{C2H4}$) to undergo electrophilic substitution due to $\ce{HCN}$ but it doesn't happen, as I am told by my textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't figure out the reason, as I found out that $\ce{HCN}$ does react with $\ce{CH3CH2Br}$ by nucleophilic substitution.  Upon ionisation, the proton of $\ce{HCN}$ is an electrophile,while the cyanide group is a nucleophile.  Correct?&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="22" LastEditDate="2012-12-30T12:34:15.427" LastActivityDate="2014-03-18T09:35:58.610" Title="Why cannot HCN react with ethene?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2827" PostTypeId="1" CreationDate="2012-12-27T15:33:56.090" Score="7" ViewCount="20013" Body="&lt;p&gt;Both sulfur and oxygen have 6 electrons in their outermost shell. Since both only need two electrons to stabilize, why doesn't just one atom of oxygen react with one sulfur atom, forming sulfur monoxide SO instead of sulfur dioxide $\ce{SO2}$?&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="402" LastEditDate="2017-04-01T00:12:50.997" LastActivityDate="2017-04-01T00:12:50.997" Title="Why do sulfur and oxygen form sulfur dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2828" PostTypeId="2" ParentId="2827" CreationDate="2012-12-27T15:42:52.273" Score="-5" Body="&lt;p&gt;Its just like answering formation of $\ce{CCl4}$ against $\ce{CCl2 }$&#xA;for $\ce{SO}$ and $\ce{SO2}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is highly unstable&lt;/li&gt;&#xA;&lt;li&gt;It becomes more stable after forming a double bond with O ( maybe resonance )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7YIL9.png&quot; alt=&quot;enter image description here&quot;&gt; ,  &lt;img src=&quot;https://i.stack.imgur.com/QjfIk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has an electronic configuration of $\ce{[Ne]} \: 3s^2\: 3p^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has 4 bonds due to uncoupling of p electrons (a little energy is spent), but due to the formation of four bonds, a large amount of energy is released.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="22" LastEditDate="2012-12-27T15:54:57.050" LastActivityDate="2012-12-27T15:54:57.050" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2829" PostTypeId="1" AcceptedAnswerId="2831" CreationDate="2012-12-27T17:39:38.870" Score="4" ViewCount="191" Body="&lt;p&gt;The conversion of sulfur dioxide to sulfur trioxide in the contact process for the manufacture of sulfuric acid requires the following condition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Temperature&lt;/strong&gt;: 450 degrees&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Pressure&lt;/strong&gt;: 1-2 atm, although High pressure would fasten the reaction, creating those pressures is not economical&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Vanadium Pentoxide ($\ce{V2O5}$) as &lt;strong&gt;catalyst&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;and &lt;strong&gt;excess of oxygen&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{2SO2 + O2 -&amp;gt; 2SO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook says that high pressure favors the rate of the reaction. The reason it states is that the volume of product is less than that of the reactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain how the volume of the product being less causes high pressure to be required? &lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="402" LastEditDate="2017-04-08T22:17:38.100" LastActivityDate="2017-04-08T22:17:38.100" Title="Why does the conversion of SO₂ to SO₃ require high pressure?" Tags="&lt;gas-laws&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2830" PostTypeId="2" ParentId="409" CreationDate="2012-12-28T00:50:35.550" Score="12" Body="&lt;p&gt;I had success with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Glass bottle, ideally with flat sides.  If you have a round bottle you may not be able to see the ferrofluid due to diffraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A solution of ~25% distilled water (a.k.a. deionized water) and ~75% of store bought isopropyl alcohol at 91% concentration.  That means that ~70% of my solution was actually isopropyl alcohol.  I eyeballed the ratio between the water and the isopropyl alcohol, those numbers are likely to not be exactly correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure that I followed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add some isopropyl alcohol to the glass container.  Then add some (or all) of your ferrofluid to the container.  Then add some water to bring your isopropyl alcohol concentration to about 70%.  Next, use a magnet to settle the ferrofluid to the bottom and then remove the magnet.  If your ferrofluid has little bits that are floating, then you need to add more water.  If your ferrofluid looks too stringy then I would add more alcohol.  When I was done the ferrofluid had zero staining, like you see in the videos online, except for in a few spots.  I think for better results I would need a higher quality glass container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;notes: Don't put too much ferrofluid in your container, otherwise you will have an uninteresting blob.  I recommend using gloves.  To clean the ferrofluid stuff off my hands I used warm water, laundry detergent, and lots of scrubbing with a towel.  If you mess up, you can use a magnet to hold your ferrofluid in place while you pour off some of your suspension fluid.  In addition, you can use a straw and your thumb to pipet some of the ferrofluid, however some will get stuck in the straw, so I wouldn't do that unless you have a bunch of ferrofluid.  Finally, I only finished like 30 minutes ago, so I hope it will all look good tomorrow.  If something bad happens I will make another post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;notes, part 2: If all you want is a cool desk toy, then I recommend just buying one of the pre-made ferrofluid toys online. (I saw some for ~$30).  You will get a lot more ferrofluid for your money if you buy just the ferrofluid, however making the toy is a bit of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GOOD LUCK!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:  no combinations of the following worked for me: acetic acid, hydrogen peroxide, pep plastic.&lt;/p&gt;&#xA;" OwnerUserId="1024" LastEditorUserId="1024" LastEditDate="2012-12-28T18:02:00.473" LastActivityDate="2012-12-28T18:02:00.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2831" PostTypeId="2" ParentId="2829" CreationDate="2012-12-28T03:14:14.140" Score="6" Body="&lt;p&gt;As &lt;a href=&quot;https://chemistry.stackexchange.com/users/189/philipp&quot;&gt;Philipp&lt;/a&gt; stated in the comments it is Le Chatelier's principle in play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle in layman's language states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you try to bring about any physical change in a system the system will try to but not necessarily succeed in cancelling the change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reaction you have given is an equilibrium reaction where all the compounds are in gaseous state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you count the total number of moles on the left-hand side it will be $\ce{2SO2 + O2}$&#xA;i.e. $\ce{2 + 1}$ that is $\text{3}$.&#xA;And on the right-hand side there are only $\text{2}$ moles of $\ce{SO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ultimately if you consider the reaction to proceed in forward direction &lt;strong&gt;3&lt;/strong&gt; moles of gas are converting into &lt;strong&gt;2&lt;/strong&gt; moles of gas and hence the volume &lt;em&gt;decreases&lt;/em&gt; in &lt;em&gt;forward&lt;/em&gt; direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think according to Gas Law(or logically) an increase in pressure generally corresponds to a decrease in volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use Le Chatelier's principle to say that if I increase the pressure on the the system ,in order the system &#xA;to counteract to the change will decrease it's own volume and it has no other way of achieving that but to proceed in the forward direction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But if the query arises that high pressure should favour any and every reaction take a look at the example below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PCl5 &amp;lt;=&amp;gt; PCl3 + Cl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the gaseous moles on the left-hand side is $\ce{1}$ and on the right-hand side $\ce{2}$. So increasing pressure makes the reaction proceed in the backwards direction as there are less number of gaseous moles on the left-hand side.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, in higher level this idea is what is called $\Delta\text{n}_\text{g}$ i.e the difference($\Delta$) of gaseous($_g$) moles($\ce{n}$).  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\Delta\text{n}_\text{g} \quad = \text{gaseous moles on right-hand side - gaseous moles on left-hand side}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If $\Delta\text{n}_\text{g}$ is positive, then on increasing pressure reaction proceeds backwards and vice-cersa&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2012-12-28T12:34:12.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2832" PostTypeId="1" AcceptedAnswerId="2835" CreationDate="2012-12-28T04:47:08.067" Score="46" ViewCount="48953" Body="&lt;p&gt;Why do the names of most chemical elements end with -um or -ium for both primordial and synthetic elements?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="16683" LastEditDate="2017-02-04T22:28:44.147" LastActivityDate="2017-03-21T14:52:14.883" Title="Why do the names of most chemical elements end with -um or -ium?" Tags="&lt;nomenclature&gt;&lt;elements&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2833" PostTypeId="1" AcceptedAnswerId="2839" CreationDate="2012-12-28T05:56:19.007" Score="2" ViewCount="5438" Body="&lt;p&gt;In the section about fuels, my textbook mentions that the boiling and freezing-points of alkanes (which are non-polar) increase with an increase in molecular mass due to London dispersion forces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand what these are and why they become relevant when talking about non-polar molecules, but I can't figure out how they are connected to the boiling and freezing point of the matter. How do these London dispersion forces influence the boiling and freezing point?&lt;/p&gt;&#xA;" OwnerUserId="996" LastEditorUserId="22" LastEditDate="2012-12-28T13:33:24.183" LastActivityDate="2012-12-28T13:48:46.370" Title="Why does the boiling and freezing-point of non-polar molecules rise with increasing molecular mass?" Tags="&lt;physical-chemistry&gt;&lt;boiling-point&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2834" PostTypeId="2" ParentId="2832" CreationDate="2012-12-28T10:13:15.847" Score="17" Body="&lt;p&gt;The -ium suffix is a Latin suffix which forms abstract nouns, thus it is used to form chemical elements' name from its naming origin, such as minerals (calcium from calx) or person names (gadolinium).&lt;/p&gt;&#xA;" OwnerUserId="1026" LastActivityDate="2012-12-28T10:13:15.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2835" PostTypeId="2" ParentId="2832" CreationDate="2012-12-28T10:35:32.067" Score="35" Body="&lt;p&gt;To expand on @BelieveInvis's answer -- in the early 19th century, when the Royal Society was really in the swing of things, the dominant language of scholarship was still Latin. Since Latin didn't have words for the new metallic elements, new words were coined from the existing terms for the substances and given Latinate endings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://oed.com/view/Entry/100366?redirectedFrom=-ium&amp;amp;&quot;&gt;OED's entry on &lt;em&gt;-ium&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;The Latin names of metals were in -um, e.g. aurum, argentum, ferrum; the names of sodium, potassium, and magnesium, derived from soda, potassa or potash, and magnesia, were given by Davy in 1807, with the derivative form -ium; and although some of the later metals have received names in -um, the general form is in -ium, as in cadmium, iridium, lithium, osmium, palladium, rhodium, titanium, uranium; in conformity with which aluminum has been altered to aluminium.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I think after that, other elements were simply given the suffix to fit the generally useful naming scheme, and then, metal names which were already in common use kept their common language names (&lt;em&gt;e.g.&lt;/em&gt; gold as opposed to aurum) simply by force of usage.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2012-12-28T10:35:32.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2836" PostTypeId="1" AcceptedAnswerId="2840" CreationDate="2012-12-28T11:49:26.410" Score="8" ViewCount="871" Body="&lt;p&gt;I'm partnered with a jewelry production company whose plating baths require $\ce{KOH}$. We're located in Indonesia, where import fees are absurdly high, and reliable suppliers are rare. A local shop sells $\ce{KOH}$ at \$5/kg, but whatever impurities are present in their mix cause enough problems to spoil the entire bath. (Expensive mistake when doing gold plating.) The only alternative we can find is a Merck redistributor, who sells $\ce{KOH}$ at close to \$100/kg, which is ridiculous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a relatively simple process we could learn (of the DIY variety) for removing impurities, turning non-lab-grade $\ce{KOH}$ into lab-grade $\ce{KOH}$?&lt;/p&gt;&#xA;" OwnerUserId="1027" LastEditorDisplayName="user7951" LastEditDate="2017-07-16T11:58:33.837" LastActivityDate="2017-07-16T11:58:33.837" Title="What is the process for purifying KOH?" Tags="&lt;inorganic-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2838" PostTypeId="1" CreationDate="2012-12-28T13:42:38.320" Score="17" ViewCount="16201" Body="&lt;p&gt;I'm wondering about how does the litmus solution work as a pH indicator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you put drops of litmus solution into an colorless acid (e.g. $\ce{HCl}$), it turns red. But then if you add colorless base (e.g. $\ce{NaOH}$) to the mixture, it turns blue. How come?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that the litmus solution doesn't change color after it's changed color once already after a chemical change with an acid or a base. In this case, I thought that the mixture wouldn't turn blue after adding the $\ce{NaOH}$ into the mixture. But it seems that it still works well after used once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come?&lt;/p&gt;&#xA;" OwnerUserId="747" LastEditorUserId="4231" LastEditDate="2015-02-25T15:56:42.993" LastActivityDate="2020-02-24T15:01:50.930" Title="How does the litmus pH indicator work?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2839" PostTypeId="2" ParentId="2833" CreationDate="2012-12-28T13:48:46.370" Score="4" Body="&lt;p&gt;In general, the larger the molecule, the larger the accessible surface area of the molecule and the more surface polarisable electron density. There are more atoms that can participate in dispersion interactions and the effect stacks up. To melt or boil a substance you need to overcome the intermolecular forces holding the molecules together and dispersion forces -  whilst weak even with respect to hydrogen bonding - can add up to large interaction energies. One of the more interesting examples of strong dispersion interactions is the gecko's foot, which permits the reptile to easily stick to sheer surfaces due to the exceptional surface area of the footpad, which is covered in many many tiny hairlike protrusions.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-12-28T13:48:46.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2840" PostTypeId="2" ParentId="2836" CreationDate="2012-12-28T13:50:13.677" Score="4" Body="&lt;p&gt;According to Armarego, &lt;em&gt;et al.&lt;/em&gt; (&lt;em&gt;Purification of Laboratory Chemicals, 6th edition&lt;/em&gt;), they only provide a procedure to purify a solution of $\ce{KOH}$, likely owing to its very high solubility.  Further, the only impurity of note they cite is carbonate (from standing in contact with air), so if this is not your impurity then this procedure will not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is your impurity, they recommend adding a saturated solution of a barium salt, either $\ce{BaCl}$ or $\ce{Ba(OH)2}$, shaking, and then setting aside to allow the insoluble $\ce{BaCO3}$ to settle.  It sounds like you are working on large scale, so it is important to recall that the procedure just described should be done in several small batches to work best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not certain this is what you are looking for, but I hope it's a start.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="4231" LastEditDate="2015-02-25T15:53:32.383" LastActivityDate="2015-02-25T15:53:32.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2841" PostTypeId="2" ParentId="2838" CreationDate="2012-12-28T14:36:30.240" Score="19" Body="&lt;p&gt;This explanation is true for most organic indicators.  Litmus is a complex mixture of organic molecules.  The molecule responsible for the color change (the &lt;em&gt;chromophore&lt;/em&gt;), is &lt;em&gt;7-hydroxyphenoxazone&lt;/em&gt; (image taken from Wikipedia, public domain):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsVu2.png&quot; alt=&quot;7-hydroxyphenoxazone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See that $\ce{OH}$ group at the lower left?  That's where the so-called &quot;acidic proton&quot; is.  This molecule is actually a weak acid, meaning that in the presence of a strong base, the $\ce{OH}$ group can lose a proton, which reacts with an hydroxide ion ($\ce{OH^{-}}$) to form a water.  The 7-hydroxyphenoxazone is now a negatively charged ion: it looks the same except it is missing the $\ce{H}$ on the $\ce{OH}$ group, and it leaves behind an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To think about the equation that describes what's going on, pretend that the entire 7-hydroxyphenoxazone molecule is represented by the formula $HLit$, where $\ce{H}$ is the hydrogen on the $\ce{OH}$ group and $Lit$ is everything else (this is common procedure: only the hydrogen matters in the acid-base equation).  Then the acid-base equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{OH^{-} + HLit &amp;lt;=&amp;gt;H2O + Lit^{-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more strong base is added, more hydroxide ion is present, driving the equilibrium to the right and producing more $Lit^-$ ion.  And now, the color change: $HLit$ is red, but $Lit^-$ is bue.  So the more acidic your solution, the more the equilibrium is to the left and thus there is more $HLit$ than $Lit^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Litmus is pretty good for checking rough acidity or basicity, because its midpoint pH is close to 7 (neutral).  Litmus is as red as it gets at a pH of 4.5, and is as blue as it gets at a pH of 8.3, so the midpoint is about 6.4, a bit more acidic than pure water.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="4231" LastEditDate="2015-02-25T15:56:47.400" LastActivityDate="2015-02-25T15:56:47.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2842" PostTypeId="1" AcceptedAnswerId="2844" CreationDate="2012-12-28T16:04:56.693" Score="10" ViewCount="22855" Body="&lt;p&gt;Is diamond an allotrope or polymorph of carbon? When is something considered an allotrope vs a polymorph?&lt;/p&gt;&#xA;" OwnerUserId="1028" LastEditorUserId="10185" LastEditDate="2015-05-18T12:27:40.817" LastActivityDate="2020-11-11T09:25:02.393" Title="Is diamond an allotrope or polymorph of carbon?" Tags="&lt;organic-chemistry&gt;&lt;crystal-structure&gt;&lt;carbon-allotropes&gt;&lt;allotropes&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2843" PostTypeId="1" CreationDate="2012-12-28T16:31:12.477" Score="7" ViewCount="1053" Body="&lt;p&gt;I wish to know the trends in the periodic table and why that trend exists, for the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Atomic size&lt;/li&gt;&#xA;&lt;li&gt;Ionization energy&lt;/li&gt;&#xA;&lt;li&gt;Metallic character&lt;/li&gt;&#xA;&lt;li&gt;Positive ion size&lt;/li&gt;&#xA;&lt;li&gt;Negative ion size&lt;/li&gt;&#xA;&lt;li&gt;Electronegativity&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My progress (if my reasoning is wrong for any of these please correct me):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Atomic size - I memorized from my class lecture that the greatest atomic size is towards the bottom-left of the periodic table. I think the reason for this is because as you go down the groups (IA, IIA, etc.) you add more electrons. Adding more electrons means the atom will logically grow bigger. I am not sure why the answer is not bottom-right, though. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ionization energy - This is the energy required to remove an electron from an atom. If there are more electrons then the electron will be further from the nucleus and by Coulomb's law, since the distance increases, the force attraction that keeps the electron in the nucleus's orbit is weak. Thus, the ionization energy is the highest toward the top-left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have no clue what to think for metallic character.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Positive ion size I know the answer is that it increases towards the bottom left, but I don't know why.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Negative ion size I know the answer is that it increases towards the bottom left, but I don't know why.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronegativity I am pretty sure is highest at F which is 4.0 and is lowest towards the bottom left. So the trend increases towards the top-right.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1029" LastEditorUserId="5026" LastEditDate="2018-06-24T11:51:48.430" LastActivityDate="2018-06-24T11:51:48.430" Title="What trends exist in the Periodic Table?" Tags="&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-07-17T14:07:20.187" ContentLicense="CC BY-SA 4.0" />
  <row Id="2844" PostTypeId="2" ParentId="2842" CreationDate="2012-12-28T16:34:02.303" Score="9" Body="&lt;p&gt;Allotropy is a special case of polymorphism. Polymorphism is the phenomenon of a substance exhibiting different crystal structures. Allotropism is the same phenomenon limited to the subset of all substances that contains only the chemical elements. According to the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymorphism_%28materials_science%29&quot; rel=&quot;noreferrer&quot;&gt;polymorphism&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Polymorphism ... is the ability of a solid material to exist in more than one form or crystal structure. Polymorphism can potentially be found in any crystalline material ...  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropy&quot; rel=&quot;noreferrer&quot;&gt;allotropy&lt;/a&gt; article defines allotropy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allotropy or allotropism (coined from Greek &quot;other&quot; + &quot;form&quot;) is the property of some chemical elements to exist in two or more different forms ... The term allotropy is used for elements only, not for compounds. The more general term, used for any crystalline material, is polymorphism. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, diamond and graphite represent two different allotropes of carbon, which also makes them different polymorphs of carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_diagram_of_water#Phase_diagram&quot; rel=&quot;noreferrer&quot;&gt;Water&lt;/a&gt; is well-known for having multiple solid phases. These phases are polymorphs, but, because water is not an element, they are not allotropes. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2012-12-28T16:34:02.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2845" PostTypeId="2" ParentId="2843" CreationDate="2012-12-28T21:31:06.190" Score="12" Body="&lt;p&gt;&lt;ol&gt;&#xA;&lt;li&gt;You are correct on atomic size being due to number of electrons&#xA;and their &lt;em&gt;shells&lt;/em&gt; and the reason why atomic size decreases from&#xA;left to right is due to the number of the protons being greater&#xA;on the right than on the left of the periodic table with same number of shells due to shell theory. &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_radius&quot;&gt;(Reference)&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The concept you are forgetting is&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_electron&quot;&gt;Valence Electrons&lt;/a&gt;.&#xA;Atoms are stabilized most often when there outer most shell is&#xA;filled with 8 valence electrons (except for hydrogen and helium with&#xA;0 or 2). The elements on the left side of the periodic table lose&#xA;electrons easily since they have a very large atomic radius allowing&#xA;for lower ionization energy. on the right side of the periodic&#xA;table(nitrogen, oxygen , and halogens) have a smaller atomic radius&#xA;creating a stronger pull making it harder to lose an electron thus&#xA;they will steal an electron to become stable. &lt;/li&gt;&#xA;&lt;li&gt;The best way to summarize this is to quote Chemistry.about.com &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are trends in&#xA;  metallic character as you move across and down the periodic table.&#xA;  Metallic character decreases as you move across a period in the&#xA;  periodic table from left to right. This occurs as atoms more readily&#xA;  accept electrons to fill a valence shell than lose them to remove&#xA;  the unfilled shell. Metallic character increases as you move down an&#xA;  element group in the periodic table. This is because electrons&#xA;  become easier to lose as the atomic radius increases, where there is&#xA;  less attraction between the nucleus and the valence electrons&#xA;  because of the increased distance between them.&quot;&lt;br&gt;&#xA;  &lt;a href=&quot;http://chemistry.about.com/od/periodicitytrends/a/Metallic-Character.htm&quot;&gt;(Reference)&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Positive ion size decreases due to the lack of electrons which allows for the protons to pull stronger on fewer electrons.&lt;/li&gt;&#xA;  &lt;li&gt;The inverse applies for negative ion size. Due to the fact that there are more electrons, the size will increase since they will occupy more space.&lt;/li&gt;&#xA;  &lt;li&gt;Electronegativity is the pull a elements has on electrons. This determines how it bonds with other elements i.e.(whether it shares or looses electrons). If a element has over a 1.7 Electronegativity difference then these the element with the smaller electronegativity will give away several electrons 1-4. Elements on the left side of the periodic table have the smallest Electronegativity due to their atomic size and inability to gain enough electrons to be stable ions. Elements from left to right in a period show increasing elctronegativity, and electrons form up to down in a group will have a lower electronegativity. Also fluorine has a 3.98 elctronegativity. &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20080924041039AAyrZDW&quot;&gt;(Reference)&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;/ol&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="994" LastEditorUserId="897" LastEditDate="2012-12-29T08:15:29.693" LastActivityDate="2012-12-29T08:15:29.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2847" PostTypeId="1" CreationDate="2012-12-30T05:12:40.000" Score="5" ViewCount="241" Body="&lt;p&gt;I am trying to generate a phase diagram of pressure versus temperature in a NPT ensemble using molecular dynamics method. I am using the Lennard Jones potential with 108 particles and Berendsen thermostat and barostat to fix the parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to observe phase transition I plot the radial distribution function (RDF) and also density versus temperature graph for various fixed temperature. However it is hard to differentiate liquid and gaseous phase from RDF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another peculiar thing is that there should be a discontinuity in density as the transition temperature is reached, which I observe only for solid liquid transition but not for the liquid vapor transition. The change in density should be huge for liquid vapor transition. Any idea on what possibly goes wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameters:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Periodic boundary  &#xA;cut off radius = 2.5  &#xA;Truncated and shifted Lennard Jones potential  &#xA;number of particle = 108  &#xA;number of steps = 2000  &#xA;timestep = 0.004  &#xA;Berendsen barostat of scale coeefficient of 0.1 is used to scale the size of simulation box&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The density versus temperature graph for pressure of 1.0MPa:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TeY77.jpg&quot; alt=&quot;http://i.stack.imgur.com/x0d0p.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid liquid transition can be seen but not the liquid vapor transition.&lt;/p&gt;&#xA;" OwnerUserId="1032" LastEditorUserId="73" LastEditDate="2012-12-30T05:18:48.127" LastActivityDate="2013-01-30T16:10:04.080" Title="Lennard Jones liquid vapor phase transition" Tags="&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2848" PostTypeId="2" ParentId="2819" CreationDate="2012-12-30T05:25:31.880" Score="13" Body="&lt;p&gt;Well, that is simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Carbon (C) makes 4 bonds, a Hydrogen (H) or a halogen (X) makes 1 bond and Nitrogen (N) makes three bonds. So in a molecule with $c$ Carbons, $h$ Hydrogens, $x$ halogens and $n$ Nitrogens, we will have $4c+h+x+3n$ &quot;&lt;em&gt;bonding capacity&lt;/em&gt;&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there are no rings, no double/triple bonds, we must have $c+h+x+n-1$ bonds in between these atoms [think like how many dashes you can draw between $m$ $\ce{O}$'s: $\ce{O-O-O}$ two dashes for three $\ce{O}$'s apparently].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each bond uses one &quot;bonding capacity&quot; from each of the two atoms its binding, $c+h+x+n-1$ bonds are using $2(c+h+x+n-1)$ bonding capacity. So the remaining capacity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(4c+h+x+3n)-2(c+h+x+n-1)=2c-h-x+n+2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is larger than zero, it means our molecule is unsaturated, i.e. still has a capacity to make bonds. And our assumption that &quot;there are no rings, no double/triple bonds&quot; is wrong. This bonding capacity is used in forming new bonds either in rings or in pi-bonds.  Since each bond uses two capacity one from each side we should have $c-h/2-x/2+n/2+1$ &quot;unsaturated&quot; bonds.&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="24052" LastEditDate="2016-03-11T22:09:59.363" LastActivityDate="2016-03-11T22:09:59.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2849" PostTypeId="1" CreationDate="2012-12-30T06:57:51.700" Score="1" ViewCount="729" Body="&lt;p&gt;What is the oxidation state of Calcium found in human teeth?&lt;/p&gt;&#xA;" OwnerUserId="1033" LastEditorUserId="4945" LastEditDate="2014-06-12T08:39:42.543" LastActivityDate="2014-06-12T08:39:42.543" Title="What is the oxidation state of Calcium in teeth?" Tags="&lt;biochemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2850" PostTypeId="1" AcceptedAnswerId="2860" CreationDate="2012-12-30T10:19:34.263" Score="4" ViewCount="196" Body="&lt;p&gt;A paper says the following,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dichloroglyoxime was prepared by chlorination of glyoxime in 10 % HCl at 0 deg C and crystallized from toluene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When they say &quot;crystallized from toluene&quot; do they unambiguously mean recrystallized from boiling toluene at standard pressure? Thermally these compounds are unstable and instead of a white nice product I've ended up with a purple one much to my dismay and I'm not sure what actually causes the purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had problems with this synthesis because the $^1$H NMR always shows more than one oxime environment. &lt;/p&gt;&#xA;" OwnerUserId="1034" LastEditorUserId="-1" LastEditDate="2014-12-03T01:44:40.053" LastActivityDate="2014-12-03T01:44:40.053" Title="Ambiguity in synthesis (crystallization)" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2851" PostTypeId="1" AcceptedAnswerId="2858" CreationDate="2012-12-31T01:51:17.047" Score="6" ViewCount="583" Body="&lt;p&gt;I can understand redox equations of the following form, I break these down into half equations and combine them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4^- + C2O4^{2-} -&amp;gt; Mn^{2+} + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I have issue is with the following (unbalanced) equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 + NO -&amp;gt; NH_3 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to show balanced half equations and the final combined equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see the following changes in oxidation states&lt;br&gt;&#xA;reduction: $$\ce{ H -&amp;gt; H^{1+} + e^{1-} }$$ (for both the $\ce{H}$ in water and ammonia)&lt;br&gt;&#xA;oxidation: $$\ce{ N^{2+} + 5e- -&amp;gt; N^{3-} }$$   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My issue is then with balancing these and combining them, I am not sure if the half equation involving $\ce{H}$ should have $\ce{H2O}$ as the product or the $\ce{H3}$ from $\ce{NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How should I go about breaking this into half equations?  &lt;/li&gt;&#xA;&lt;li&gt;Is it ok for both the half-equations to have molecules of the same compound in the product?   &lt;/li&gt;&#xA;&lt;li&gt;Should I ever have water as the product when trying to show a half equation (ie $\ce{H -&amp;gt; H_2O}$)?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="5026" LastEditDate="2018-02-25T03:16:01.413" LastActivityDate="2018-02-25T03:18:28.110" Title="Balancing and combining the half equations in redox" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2853" PostTypeId="2" ParentId="2849" CreationDate="2012-12-31T15:06:18.543" Score="3" Body="&lt;p&gt;As a group II metal, you will be hard pressed to find calcium in the 1+ oxidation state (a quick google search yields an isolated report of calcium(I) chloride). Calcium's willingness to oxidise is evinced by its vigorous reaction with water. As you have noted, the calcium in tooth enamel is hydroxyapatite, often with varying degrees of fluorination. Which brings me to your next question, which hopefully you or a mod can slipstream into the main body -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would definitely warn against eating chalk without good reason as it will evolve $\ce{CO2}$ in your tummy and raise your stomach pH, and is associated with a variety of negative health outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Milk-alkali_syndrome&quot; rel=&quot;nofollow&quot;&gt;Milk-Alkali Syndrome&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypercalcemia&quot; rel=&quot;nofollow&quot;&gt;Hypercalcaemia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Antacid#Side_effects&quot; rel=&quot;nofollow&quot;&gt;Antacid side effects&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As always, all things in moderation. On a personal note eating chalk sounds like exactly the thing certain sectors of the 'alternative health' crowd would be into (namely, those who assert that the body is 'too acidic' (there's another crowd who claim the opposite and will recommend, say, drinking $\ce{HCl}$ (through a glass straw, natch)), ignoring for a moment the fact that the human body is pretty good at maintaining its pH and will up and have a cardiac arrest or something if the pH drifts more than a bit)&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2012-12-31T15:23:33.737" LastActivityDate="2012-12-31T15:23:33.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2854" PostTypeId="2" ParentId="2847" CreationDate="2012-12-31T15:57:59.833" Score="2" Body="&lt;p&gt;I think your transition, if present, may be drowned out by noise, as your RDF is only produced by 108 particles. As the Lennard-Jones interaction is quite simple and you are using cutoffs (good idea!), is it possible to increase the number of particles in your simulation somewhat, or else take the mean of some large number of trials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, it does not take much imagination to see a continuous PV plot from about 0.7 onwards, which might mean you have something analogous to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supercritical_fluid&quot; rel=&quot;nofollow&quot;&gt;supercritical fluid&lt;/a&gt;. Maybe try a number of trials at different pressures?&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2012-12-31T15:57:59.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2855" PostTypeId="1" CreationDate="2013-01-01T01:51:33.260" Score="1" ViewCount="72" Body="&lt;p&gt;Can someone please explain what photodissociation is and if it is more efficient than electrolysis while splitting water into its components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="884" LastActivityDate="2013-01-15T17:44:08.833" Title="Clarification on Photodissociation" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2858" PostTypeId="2" ParentId="2851" CreationDate="2013-01-01T06:48:10.957" Score="6" Body="&lt;p&gt;&lt;strong&gt;Assuming this reaction is taking place in aqueous phase&lt;/strong&gt;, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.&lt;br&gt;&#xA;I will be writing the equation as it should be after each step.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 + NO -&amp;gt; NH3 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 1: Ionise the required compounds and remove spectator ions&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.*&#xA;$$\small\ce{H2 + NO -&amp;gt; NH3 + H+ + OH-}$$&#xA;*-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 2: Split into Oxidation and reduction halves&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -&amp;gt; NH3}$  (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -&gt; -3) the other element, Hydrogen must get oxidised (0 -&gt; +1).&#xA;So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -&amp;gt; H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;\ce{H2 -&amp;gt; H+ + OH-} &amp;amp; \ce{NO -&amp;gt; NH3}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 3: Balance only those atoms undergoing redox&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;\ce{H2 -&amp;gt; H+ + OH-} &amp;amp; \ce{NO -&amp;gt; NH3} \\&#xA;\text{(because Hydrogen is } &amp;amp; \text{(because only Nitrogen } \\&#xA;\text{undergoing redox but} &amp;amp; \text{is undergoing redox,} \\&#xA;\text{Oxygen is not.)} &amp;amp; \text{it's already balanced)} \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;\ce{H2 + H2O -&amp;gt; H+ + OH-} &amp;amp; \ce{NO -&amp;gt; NH3 + H2O}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;\ce{H2 + H2O -&amp;gt; H+ + OH- + 2H+} &amp;amp; \ce{NO + 5H+ -&amp;gt; NH3 + H2O} \\&#xA;\ce{H2 + H2O -&amp;gt; [H+ + OH- ] + 2H+} &amp;amp; \ce{NO + 5H+ -&amp;gt; NH3 + H2O} \\&#xA;\ce{H2 + H2O -&amp;gt; \qquad H2O \qquad + 2H+} &amp;amp; \ce{NO + 5H+ -&amp;gt; NH3 + H2O} \\&#xA;\hline&#xA;\ce{H2 + \qquad-&amp;gt; \qquad \qquad \qquad \quad 2H+} &amp;amp; \ce{NO + 5H+ -&amp;gt; NH3 + H2O}\\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This step must follow &lt;strong&gt;Step 5a&lt;/strong&gt; and means nothing on it's own.&#xA;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;\ce{H2 \qquad -&amp;gt; 2H+} &amp;amp; \ce{NO + 5H+ \qquad \qquad -&amp;gt; NH3 + H2O \qquad \qquad} \\&#xA;\ce{H2 + 2OH- -&amp;gt; 2H+ + 2OH-} &amp;amp; \ce{NO + 5H+ + 5OH- -&amp;gt; NH3 + H2O + 5OH-} \\&#xA;\ce{H2 + 2OH- -&amp;gt; [2H+ + 2OH- ]} &amp;amp; \ce{NO + [5H+ + 5OH- ]-&amp;gt; NH3 + H2O + 5OH-} \\&#xA;\ce{H2 + 2OH- -&amp;gt; \quad \quad 2H2O} &amp;amp; \ce{NO + \qquad 5H2O \qquad -&amp;gt; NH3 + H2O + 5OH-} \\&#xA;\hline&#xA;\ce{H2 + 2OH- -&amp;gt; \quad \quad 2H2O} &amp;amp; \ce{NO + \qquad 4H2O \qquad -&amp;gt; NH3 + 5OH- \quad \qquad  }\\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Till the above steps mass has been balanced, now charge balancing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Step 6: Balance charge in each side of each half by adding electrons&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half}  \\&#xA;\hline&#xA;(acidic)\quad \ce{H2  -&amp;gt; 2H+ + 2e-} &amp;amp; \ce{NO + 5H+ + 5e- -&amp;gt; NH3 + H2O} \\&#xA;\hline&#xA;(basic) \quad \ce{H2 + 2OH- -&amp;gt; 2H2O + 2e-} &amp;amp; \ce{NO + 4H2O + 5e- -&amp;gt; NH3 + 5OH-} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;rarr; Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;rarr; Also verify if net change in oxidation state is equal to the number of electrons appearing in each half.&#xA;eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons.&#xA;Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Step 7: Make 'net electrons produced' equal to 'net electrons consumed'&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|r}&#xA;\text{Oxidation Half} &amp;amp; \text{Reduction half} \\&#xA;\hline&#xA;(acidic) \ce{5H2 -&amp;gt; 10H+ + 10e-} &amp;amp; \ce{2NO + 10H+ + 10e- -&amp;gt; 2NH3 + 2H2O}\\&#xA;\hline&#xA;(basic) \ce{5H2 + 10OH- -&amp;gt; 10H2O + 10e-} &amp;amp; \ce{2NO + 8H2O + 10e- -&amp;gt; 2NH3 + 10OH-} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally add the two halves and cancel out the common cmopounds and you get the balanced equation&#xA;&lt;strong&gt;Acidic Medium&lt;/strong&gt;&#xA;$$\begin{matrix}&#xA;\ce{&amp;amp;5H2 + &amp;amp;2NO + &amp;amp;10H+ + &amp;amp;10e- &amp;amp;-&amp;gt; &amp;amp;10H+ + &amp;amp;10e- + &amp;amp;2NH3 + &amp;amp;2H2O} \\&#xA;\ce{&amp;amp;5H2 + &amp;amp;2NO &amp;amp; &amp;amp; &amp;amp;-&amp;gt; &amp;amp; &amp;amp; &amp;amp;2NH3 + &amp;amp;2H2O}\\&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic Medium&lt;/strong&gt;&#xA;$$\begin{matrix}&#xA;\ce{&amp;amp;5H2 + &amp;amp;10OH- + &amp;amp;2NO + &amp;amp;8H2O + &amp;amp;10e- &amp;amp;-&amp;gt; &amp;amp;10H2O + &amp;amp;10e- + &amp;amp;2NH3- + &amp;amp;10OH-} \\&#xA;\ce{&amp;amp;5H2 + &amp;amp; &amp;amp;2NO &amp;amp; &amp;amp; &amp;amp;-&amp;gt; &amp;amp;2H2O + &amp;amp; &amp;amp;2NH3}\\&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -&amp;gt; 2H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform &lt;strong&gt;Step 5b&lt;/strong&gt; which is less tedious.&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="897" LastEditDate="2013-12-05T06:47:19.520" LastActivityDate="2013-12-05T06:47:19.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2859" PostTypeId="1" CreationDate="2013-01-01T16:20:13.907" Score="14" ViewCount="81780" Body="&lt;p&gt;What makes Coke and other soft drinks acidic?  These drinks are carbonated, so a freshly-opened can should have a lot of dissolved carbonic acid, but the ingredients also lists phosphoric acid.  Would flat Coke be significantly less acidic than carbonated Coke?  &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="19383" LastEditDate="2015-10-08T21:13:55.370" LastActivityDate="2015-10-08T21:13:55.370" Title="What makes Coke acidic?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2860" PostTypeId="2" ParentId="2850" CreationDate="2013-01-01T16:33:07.087" Score="3" Body="&lt;p&gt;No.  This could also mean applying a vacuum to the toluene-dichloroglyoxime solution, and letting the solvent evaporate-away at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2013-01-01T16:33:07.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2861" PostTypeId="1" AcceptedAnswerId="2864" CreationDate="2013-01-02T01:17:00.487" Score="4" ViewCount="4518" Body="&lt;p&gt;So, a simple question: what makes neon a gas; what makes water liquid; what makes plutonium-238 a solid at room temperature? Of course, the phase is determined by the excitation of the molecules, but what determines that excitation?&lt;/p&gt;&#xA;" OwnerUserId="1042" LastActivityDate="2018-10-23T01:31:44.320" Title="What defines matter's phase at room temperature?" Tags="&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2862" PostTypeId="2" ParentId="2859" CreationDate="2013-01-02T07:18:06.257" Score="16" Body="&lt;p&gt;If you look at the list of ingredients, you will find that Coca-Cola and many other brands contain phosphoric acid (food grade - don't worry). This is probably the dominant cause of acidity. Citric acid is another common component. Of course, the carbon dioxide is also acidic, but weakly, so there should be very little difference between carbonated and flat.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2013-01-02T07:18:06.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2863" PostTypeId="2" ParentId="2842" CreationDate="2013-01-02T07:39:15.383" Score="3" Body="&lt;p&gt;Diamond is an &lt;strong&gt;Allotrope&lt;/strong&gt; of Carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As cleared by &lt;strong&gt;@Ben Norris&lt;/strong&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Allotropy&lt;/em&gt; is property of any &lt;strong&gt;element&lt;/strong&gt; to exist in two or more different forms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the term &lt;em&gt;Polymorphism&lt;/em&gt; meant the ability of a &lt;strong&gt;solid material&lt;/strong&gt; to exist in more than one form or crystal structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In diamond, each carbon (an &lt;strong&gt;element&lt;/strong&gt;)  is bonded to four Carbon atoms forming a rigid 3-dimensional structure.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2013-01-02T07:39:15.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2864" PostTypeId="2" ParentId="2861" CreationDate="2013-01-02T16:35:31.563" Score="8" Body="&lt;p&gt;It's all in the intermolecular forces (and the air pressure). Intermolecular forces are the forces of attraction and repulsion between molecules to neighboring particles. To understand how they determine physical properties like enthalpies of vaporization, fusion, etc..., you need to think about what states of matter are and how they affect the molecules. When you heat up water to a temperature (average kinetic energy) of 90 °C, the energy within the molecules isn't strong enough to overcome the intermolecular bonds (and the atmospheric pressure above the sample) and become a gas. However, it does have enough energy to overcome the bonds to a certain extent and thus can remain in a liquid state instead of in a solid state. This principle applies to any substance. In order for there to be a change in state of matter, the energy within the molecules has to overcome the intermolecular bonds and atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned several times above, another factor is the atmospheric pressure. A common experiment is to put water in a vacuum and watch it boil, something many first-time-experimenters wouldn't expect. The reason water is able to vaporize at room temperature is because it doesn't have the atmosphere to prevent it from vaporizing and therefore, in order to reach an equilibrium, the water is able to vaporize with much less energy than normally required at STP (standard temperature/pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a direct response to your question about neon, the reason that the noble gases are usually gases unless at extreme temperatures has to do with the weak intermolecular forces they have (van der Waals forces, or more specifically, london dispersion forces). Because the noble gases don't usually bond &lt;em&gt;intramolecularly&lt;/em&gt; with anything else, including themselves, and have full energy levels, they have extremely weak intermolecular forces allowing them to almost always be in the gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've listed some helpful links below if you want to take your research to the next step.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Intermolecular_force&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wps.prenhall.com/esm_hillpetrucci_genchem_4/16/4219/1080299.cw/index.html&quot; rel=&quot;noreferrer&quot;&gt;http://wps.prenhall.com/esm_hillpetrucci_genchem_4/16/4219/1080299.cw/index.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=A30lguyG0WM&quot; rel=&quot;noreferrer&quot;&gt;http://www.youtube.com/watch?v=A30lguyG0WM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html&quot; rel=&quot;noreferrer&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Van_der_Waals_force&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1044" LastEditorUserId="1044" LastEditDate="2013-01-03T00:49:53.010" LastActivityDate="2013-01-03T00:49:53.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2865" PostTypeId="2" ParentId="2859" CreationDate="2013-01-02T18:34:47.613" Score="0" Body="&lt;p&gt;Mostly Carbonic acid, probably. Because it is the only species in coke, that can escape in the form of gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just shake a coke, or heat it and see how much of the acidity is gone. It is the part that is due to $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="946" LastActivityDate="2013-01-02T18:34:47.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2867" PostTypeId="1" CreationDate="2013-01-03T04:40:11.300" Score="1" ViewCount="755" Body="&lt;p&gt;Anhydrous magnesium chloride will absorb ammonia to the tune of ($\ce{Mg(NH3)6Cl2}$).&lt;br&gt;&#xA;Incidentally, magnesium chloride is usually a hexahydrate ($\ce{MgCl2(H2O)6}$) at STP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, could magnesium chloride be induced to absorb both ammonia and water at the same time? Since both of the absorbed molecules in question are close in shape, size, structure and since six seems to be the occupancy limit for either one (at STP), would they be able to coexsist?&#xA;For example, could I expose some magnesium chloride dihydrate to ammonia and get&#xA;($\ce{Mg(NH3)4Cl2(H2O)2}$)?&lt;/p&gt;&#xA;" OwnerUserId="755" LastActivityDate="2013-01-03T04:40:11.300" Title="Can magnesium chloride absorb water and ammonia at the same time?" Tags="&lt;solubility&gt;&lt;molecules&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2868" PostTypeId="1" AcceptedAnswerId="2875" CreationDate="2013-01-03T08:47:57.427" Score="6" ViewCount="2941" Body="&lt;h3&gt;Background:&lt;/h3&gt;&#xA;&lt;p&gt;I was a chemistry major and have recently started watching this series of videos by &lt;a href=&quot;http://www.youtube.com/user/NurdRage?feature=watch&quot; rel=&quot;noreferrer&quot;&gt;NurdRage&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I learned that &lt;a href=&quot;http://www.youtube.com/watch?v=APxL87X92t4&quot; rel=&quot;noreferrer&quot;&gt;Platinum dissolves in Aqua Regia&lt;/a&gt; but &lt;a href=&quot;http://www.youtube.com/watch?v=H7Ng4sOVkns&quot; rel=&quot;noreferrer&quot;&gt;Ruthenium is only attacked by high temperature halogens and some alkalis&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Question:&lt;/h3&gt;&#xA;&lt;p&gt;Would a Ruthenium and Platinum alloy be resistant to both acid and alkali?&lt;/p&gt;&#xA;&lt;p&gt;But more importantly, is there any alloy or pure metal that is completely chemically inert to every strong oxidizing reagent? Is there any research about the most chemically inert metal alloys?&lt;/p&gt;&#xA;" OwnerUserId="479" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-04-14T04:42:33.113" Title="Completely chemically inert metal alloy?" Tags="&lt;metal&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2869" PostTypeId="2" ParentId="2868" CreationDate="2013-01-03T14:53:43.507" Score="2" Body="&lt;p&gt;According to the Wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorine#Chemical_reactivity&quot; rel=&quot;nofollow&quot;&gt;chemical reactivity of fluorine&lt;/a&gt;, fluorine (a very strong oxidizing agent) reacts with &lt;em&gt;every&lt;/em&gt; element except helium and neon.  So it looks like no metal has the property you want.  The reason helium and neon are so unreactive is that their outer electron shells are completely filled, with no hidden subshell available to accept an electron.  No metal (or any other element, for that matter) has this property.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-01-03T14:53:43.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2870" PostTypeId="1" AcceptedAnswerId="2873" CreationDate="2013-01-03T13:23:04.600" Score="14" ViewCount="30613" Body="&lt;p&gt;I found as I was preparing to teach &quot;Introduction to thermodynamics&quot; that there are varying definitions or examples of pure substances. For instance the text book I have to use &lt;strong&gt;&quot;Thermodynamics an engineering approach, Cengel and Boles&quot;&lt;/strong&gt; defines a pure substance as one that has the same chemical composition throughout. Air in the gaseous form is cited as an example. Air in its liquefied form is NOT since it seperates into it's various components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.dummies.com/how-to/content/how-to-distinguish-pure-substances-and-mixtures.html&quot;&gt;Dummies book on Chemistry&lt;/a&gt; suggests that sugar can be considered a pure substance as it has the same chemical composition throughout. And water can be considered a pure substance for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would a homogeneous mixture of sugar and water be considered a pure substance? After all, a homogeneous mixture has &quot;identical properties&quot; throughout it's phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemsite.lsrhs.net/Intro/Pure_vs_mixtures.html&quot;&gt;Yet another definition&lt;/a&gt; of a pure substance suggests that any material made of the same kind of atoms is a pure substance. So individual elements in the periodic table and stuff made &quot;purely&quot; of them are pure substances. Other stuff is just homogeneous or heterogeneous mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are pure substances (irrespective of whether we view them as elements, compounds or mixtures) just very difficult to break down into their individual components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would the definition of a pure substance change depending on who you speak to (chemists, phycisists, thermodynamicists)?&lt;/p&gt;&#xA;" OwnerUserId="1051" OwnerDisplayName="drN" LastActivityDate="2015-07-13T01:40:10.443" Title="What is a pure substance?" Tags="&lt;thermodynamics&gt;&lt;physical-chemistry&gt;" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2871" PostTypeId="2" ParentId="2870" CreationDate="2013-01-03T14:57:47.817" Score="1" Body="&lt;p&gt;In the lab, we refer to a pure substance as any substance with no impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say: pure water is 100 % $\ce{H2O}$ with nothing else in it; pure acetone is 100 % acetone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do however frequently use the term with respect to known mixtures. One can say we have pure clay despite the clay containing several KNOWN components because there are (practically) no impurities that have not been identified or would upset our experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be careful describing a mixture as pure, however, unless you know everyone is on the same page. Pure tap water sounds absurd but pure air, oil, product, or rubber doesn’t.&lt;/p&gt;&#xA;" OwnerDisplayName="kaine" LastEditorDisplayName="user7951" LastEditDate="2015-06-19T15:20:39.030" LastActivityDate="2015-06-19T15:20:39.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2873" PostTypeId="2" ParentId="2870" CreationDate="2013-01-03T20:09:05.573" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Thermodynamics an engineering approach, Cengel and Boles&quot; defines a pure substance as one that has the same chemical composition throughout.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the correct definition of a pure substance. However, air, regardless of phase, is not a pure substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All matter is categorized as either a &quot;pure&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_substance&quot;&gt;substance&lt;/a&gt; or a mixture. The word &quot;pure&quot; in front of &quot;substance&quot; is unnecessary, since the definition of substance implies purity. A &lt;strong&gt;substance&lt;/strong&gt; is a sample that is chemically uniform in composition. In other words, all particles of that sample are the same chemically (whether they are atoms or molecules or ionic lattices or what have you). Substances cannot be separated into simpler components by any physical process (i.e. without breaking chemical bonds). Mixtures are samples that contain two or more substances. All mixtures can be physically separated, although the means to do so may be neither easy nor practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Air#Composition&quot;&gt;Air&lt;/a&gt; is a &lt;strong&gt;mixture&lt;/strong&gt; of nitrogen ($\ce{N2}$), oxygen ($\ce{O2}$), carbon dioxide ($\ce{CO2}$), argon ($\ce{Ar}$), water vapor, and traces of other gases. Each of these gases separately is a substance. Air can be easily (but not in the comfort of your own home) separated into its constituent parts. Cooling air will cause the constituent components to condense out of it at progressively lower temperature: first water vapor, and then $\ce{CO2}$ (195 K), followed by $\ce{Kr}$ (120 K), $\ce{O2}$ and $\ce{CH4}$ (90 K), $\ce{Ar}$ (87 K), $\ce{N2}$, $\ce{Ne}$ (27 K), $\ce{H2}$ (20 K), and $\ce{He}$ (4 K). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As each substance condenses out of air it can be physically separated from the remaining gas, reducing the complexity of the mixture. Each substance, if returned to the gaseous state separate from the original mixture is physically and chemically distinct from the original mixture (or whatever is left of it). Such physical separations do not need to be easy to make air a mixture, they need to be &lt;strong&gt;possible&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pure substance does not exhibit this behavior. When pure water is cooled, it undergoes a phase change to ice, which is physically separable from liquid water. However, the solid and liquid samples are still water. You have not reduced the complexity of the sample, just changed its form. If you allow the separated solid water to melt, it will be physically and chemically indistinguishable from the original water sample. Pure water cannot be separated into other chemically distinct substances without performing a chemical reaction, which means breaking and forming bonds between atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of matter that we interact with in our everyday lives is comprised of mixtures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We encounter a few pure substances in the kitchen. Table sugar is a single compound (sucrose). So is table salt (sodium chloride), baking soda (sodium bicarbonate), and cream of tartar (potassium bitartrate). Water is a pure substance, so long as it has been purified. Most water that we encounter has other substances dissolved in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold (as long as it is 24K) is a pure elemental substance. So is diamond (carbon), graphite (also carbon), and the neon in neon lights. All minerals are pure substances, including asbestos. Glass is a pure substance. Most precious stones are pure substances. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-01-05T14:41:36.977" LastActivityDate="2013-01-05T14:41:36.977" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2874" PostTypeId="2" ParentId="2849" CreationDate="2013-01-03T20:49:24.003" Score="3" Body="&lt;p&gt;The answer to the question asked is the +2 oxidation state.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-01-03T20:49:24.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2875" PostTypeId="2" ParentId="2868" CreationDate="2013-01-04T01:30:57.413" Score="3" Body="&lt;p&gt;Due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;electronegativities&lt;/a&gt; no alloy would be thermodynamically stable in the presence of fluorine (F$_2$). Even in an alloy the constituent metal atoms are either &quot;sharing&quot; electrons or receiving/donating electrons. Fluorine is the element with the &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow&quot;&gt;highest known electronegativity&lt;/a&gt; and would &quot;pry&quot; the electrons from the electron-accepting metal atoms.&lt;/p&gt;&#xA;" OwnerUserId="440" LastActivityDate="2013-01-04T01:30:57.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2876" PostTypeId="1" AcceptedAnswerId="2878" CreationDate="2013-01-04T02:24:40.510" Score="11" ViewCount="27904" Body="&lt;p&gt;I love Kleenex Cool Touch tissues.  I'm just wondering what makes them actually feel cool.  For some reason it makes me think of refrigerant but I'm sure that's not in them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the combination of common ingredients that somehow gives it a moist feel, making it feel cooler than normal tissues?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the tissues safe to put to your face, if you are potentially inhaling the ingredient(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ingredients:  Stearyl heptanoate, stearyl alcohol, polyethylene, mineral oil, coconut oil, aloe&lt;/p&gt;&#xA;" OwnerUserId="1052" LastEditorUserId="73" LastEditDate="2013-01-06T00:45:25.247" LastActivityDate="2017-06-25T00:56:37.020" Title="What makes Kleenex Cool Touch tissues cool to the touch, and is it safe?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="16" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2877" PostTypeId="2" ParentId="2870" CreationDate="2013-01-04T09:44:12.573" Score="7" Body="&lt;p&gt;Simply, a pure substance is a substance which is composed by molecules or directly by atoms of only one kind. Diamond is a pure substance because it is directly made up of carbon atoms. Salt water is not because it contains sodium chloride and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is almost no 100% pure substance in this world (Exceptions exist, as there can be only one or two molecule and you can call that a substance), thus we normally recognize relatively pure substances as pure substances.&lt;/p&gt;&#xA;" OwnerUserId="1054" LastEditorUserId="73" LastEditDate="2013-01-06T00:49:43.973" LastActivityDate="2013-01-06T00:49:43.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2878" PostTypeId="2" ParentId="2876" CreationDate="2013-01-04T10:12:43.793" Score="5" Body="&lt;p&gt;My first guess as to the nature of this behaviour is similar to water. Hypothermia is a realistic risk if you are drenched in water. This is because water can absorb large amounts of energy (it has a large specific heat capacity) to only rise in temperature slightly. Water itself evaporates into steam spontaneously, if some water is spilled then it will evaporate readily, regardless of the temperature. Likewise, the solids in the tissue are probably coming closer to melting temperature, which I believe is higher than your body temperature, but quickly absorbing a lot of energy from your body in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something similar to properties of water are probably characteristic to the tissue. Leave the tissue in the open in a cool room, along with a glass of water. After awhile, touching the glass of water and the tissue should invoke the same sense of coolness. This is because your body is at a higher temperature than the items in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is likely another property of matter besides specific heat capacity that is a measurement of how &lt;strong&gt;quickly&lt;/strong&gt; heat is transferred, this is likely to play a role as well. I do not know what this property is called and it was never covered in my classes so far, but take metal for example. Metal has a very small specific heat capacity, it can not hold much energy. However, touching a metal object invokes a sense of coolness. I believe this is because heat is transferred very quickly through metal, although it warms very quickly due to its limited heat capacity. If you put a metal cup and a porcelain cup on the stove, the metal will burn you much quicker if you grab it than grabbing the porcelain/ceramic cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were very interested and inquistive you could test the specific heat of the tissue through experimentation. Also, try holding the napkin for a few seconds to try and find out how &lt;strong&gt;quickly&lt;/strong&gt; the tissue warms to the touch. Chances are, it warms relatively fast and can absorb a lot of energy, giving you the impression that it is 'cooler' than most objects, to which marketers will be overwrought with the joy of hearing.&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2013-01-04T10:12:43.793" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2879" PostTypeId="1" CreationDate="2013-01-04T22:43:47.240" Score="22" ViewCount="220384" Body="&lt;p&gt;I am looking for the precise definitions, as I am very confused as to what they are exactly because although I mostly understand what they mean, I have encountered some conflicting definitions that confused me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it stands, this is what I understand them to mean: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Compound: two or more different atoms bonded together.&lt;/li&gt;&#xA;  &lt;li&gt;Mixture: two or more different atoms together but not joined.&lt;/li&gt;&#xA;  &lt;li&gt;Molecule: two particles (same or different) bonded together.&lt;/li&gt;&#xA;  &lt;li&gt;Element: only 1 type of atom; this definition is applied to things both bonded and not to itself.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know whether these terms apply microscopically, macroscopically, or both. For example, I think 1 $\ce{H2O}$ molecule is a compound but is a bathtub of them called a mixture (as it contains more than 1 atom type), a compound, or both? That is, to be a compound, do all the atoms in the compound have to be bonded physically together? Generally, to what extent can these four terms overlap?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would very much like a systematic way of thinking about these terms.&lt;/p&gt;&#xA;" OwnerUserId="1057" LastEditorUserId="11367" LastEditDate="2017-04-03T02:20:16.820" LastActivityDate="2018-07-26T18:02:31.513" Title="What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?" Tags="&lt;terminology&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2880" PostTypeId="2" ParentId="2879" CreationDate="2013-01-04T23:40:43.120" Score="11" Body="&lt;p&gt;In a sample of water, you'll find the &lt;strong&gt;compound&lt;/strong&gt; $\ce{H2O}$. In a sample of hydrogen gas, you'll find the &lt;strong&gt;element&lt;/strong&gt; hydrogen, though it'll take the form of the &lt;strong&gt;molecule&lt;/strong&gt; $\ce{H2}$. In a mixture, you may find different &lt;strong&gt;molecules&lt;/strong&gt;, &lt;strong&gt;compounds&lt;/strong&gt;, and &lt;strong&gt;elements&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, you may find it helpful to look at the following classification of matter.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wsfz0.png&quot; alt=&quot;Matter tree&quot;&gt;&lt;br&gt;&#xA;The left side of the tree denotes chemical bonds/changes. The right is physical. (i.e., you can get a heterogeneous mixture to convert to a homogeneous mixture by changing physical things about the mixture like adding in another element. You can't do the same to elements and compounds.)&lt;/p&gt;&#xA;" OwnerUserId="1044" LastEditorUserId="41556" LastEditDate="2017-04-03T01:17:20.860" LastActivityDate="2017-04-03T01:17:20.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2881" PostTypeId="1" CreationDate="2013-01-05T13:40:29.220" Score="42" ViewCount="127388" Body="&lt;p&gt;What is the difference between thermodynamic and kinetic stability? I'd like a basic explanation, but not too simple. For example, methane does not burn until lit -- why?&lt;/p&gt;&#xA;" OwnerUserId="1058" LastEditorUserId="22" LastEditDate="2013-01-05T19:16:43.073" LastActivityDate="2019-07-27T18:47:01.933" Title="Difference between thermodynamic and kinetic stability" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;stability&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="2882" PostTypeId="2" ParentId="2881" CreationDate="2013-01-06T03:20:14.077" Score="9" Body="&lt;p&gt;There is often a &quot;reaction barrier&quot; that needs to be overcome.  Thus one must supply energy to get the reaction started, but once started it becomes self-sustaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can think of it this way:  one has to break bonds in a reaction before one can make bonds.  Energy is needed to break the first bonds.  Then making new bonds can supply enough energy to keep the reaction going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm talking about thermodynamically allowed reactions.  If a reaction is not thermodynamically allowed, then even if some bonds are broken by adding energy, the reaction will not spontaneously continue.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-01-06T03:20:14.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2883" PostTypeId="1" CreationDate="2013-01-06T08:07:47.320" Score="11" ViewCount="70955" Body="&lt;p&gt;For example, Magnesium has a higher charge density compared to Sodium, therefore its metallic lattice is stronger and it has a higher melting point. Does that mean that Magnesium has more protons and a smaller atomic radius or is it about its charge (2+), that makes its lattice stronger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does charge density show?&lt;/p&gt;&#xA;" OwnerUserId="1058" LastEditorUserId="22" LastEditDate="2013-01-06T11:12:00.850" LastActivityDate="2017-03-19T19:47:47.387" Title="What is charge density?" Tags="&lt;bond&gt;&lt;metal&gt;&lt;periodic-trends&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2884" PostTypeId="2" ParentId="2883" CreationDate="2013-01-06T12:10:26.920" Score="7" Body="&lt;p&gt;If it was about the number of protons in the nucleus, then Mercury would hardly be liquid at room temperature, would it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, it is about the density of &lt;em&gt;free electrons&lt;/em&gt;, i.e. the density of the &quot;electron gas&quot; that makes up the &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bonding&quot;&gt;metallic bonding&lt;/a&gt;. Obviously, this density has to be the negative of the ions' charge, so you might equivalently say it's about magnesium's (2+) charge, though that doesn't make much sense physically unless you also consider the conduction electrons, as the ions themselves of course &lt;em&gt;repulse&lt;/em&gt; each other.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2013-01-06T14:16:50.643" LastActivityDate="2013-01-06T14:16:50.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2885" PostTypeId="2" ParentId="2881" CreationDate="2013-01-06T13:04:46.373" Score="22" Body="&lt;p&gt;Kinetics deals with all things that impact the rate at which a reaction occurs. One implication of that is that reaction kinetics are partially a function of the stability of the intermediate species (and transition states) that the pathway of a chemical reaction necessarily has to pass through en route to the formation of the final product. If those intermediates are highly unstable (i.e., highly energetic) relative to the initial reactants, then the initial barrier of activation energy that must be surpassed to set the reaction in motion will be relatively high, and the reaction will tend to proceed slowly (if at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamic stability, on the other hand, is strictly a function of the change in free energy (&amp;Delta;G), which is a state function, meaning that its value is determined exclusively by the difference between the initial state and the final state (or the free energy of the initial products and the final products). In other words, thermodynamic stability is totally independent of the pathway between reactants and products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it's true that thermodynamic stability is the ultimate arbiter of the spontaneity of a reaction, if the kinetics render the reaction too slow then, in practice, the reaction may not actually occur at all (under some given set of conditions). The classic textbook example is the conversion of diamond to graphite, which is thermodynamically favorable because the free energy of graphite is lower, but doesn't occur under ordinary conditions because the kinetics of the reaction (in the form of the immense activation energy required) are extremely unfavorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have some set of reactants, A + B, and there are two different pathways through which A + B can react, both of which lead to final products that are lower in free energy than the reactants A + B. Pathway #1 has much higher activation energy than Pathway #2, but the final products at the end of Pathway #1 are lower in free energy (i.e., more thermodynamically stable) than those at the end of Pathway #2. This means that Pathway #1 is thermodynamically favored, but Pathway #2 is kinetically favored (that is, it poses a lower barrier in terms of activation energy and the reaction will proceed more rapidly via this pathway). The consequence of this is that the kinetic pathway, Pathway #2, will be the favored one, unless environmental conditions (temperature, pressure, concentrations of catalysts, etc.) are tuned in a such a way that the higher activation energy of Pathway #1 can be surpassed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, thermodynamic stability (in terms of free energy difference between reactants and products) will determine whether a given reaction could be theoretically spontaneous, but kinetic factors will decide whether the reaction occurs in practice under the given environmental conditions, as well as which pathway for a given reaction takes primacy. &lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-01-06T13:04:46.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2888" PostTypeId="2" ParentId="605" CreationDate="2013-01-07T05:32:32.313" Score="5" Body="&lt;p&gt;Electrochemistry;  using microelectrodes.  So, basically, place one microelectrode inside the cell, and another (the reference electrode) just outside the cell, and then use a potentiostat to measure the potential between them. &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrophysiology#Intracellular_recording&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reference.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="15429" LastEditDate="2017-05-21T21:00:39.807" LastActivityDate="2017-05-21T21:00:39.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2889" PostTypeId="1" AcceptedAnswerId="2890" CreationDate="2013-01-07T10:57:11.933" Score="25" ViewCount="7229" Body="&lt;p&gt;In the documentary about Chernobyl, it was mentioned that Uranium should not  come in contact with water, otherwise an explosion occurs. What is the reason for that? What kind of reaction makes it explode?&lt;/p&gt;&#xA;" OwnerUserId="1068" LastEditorUserId="19616" LastEditDate="2015-10-23T16:46:06.663" LastActivityDate="2021-05-12T22:19:48.290" Title="Why shouldn't Uranium come in contact with water?" Tags="&lt;physical-chemistry&gt;&lt;reactivity&gt;&lt;explosives&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2890" PostTypeId="2" ParentId="2889" CreationDate="2013-01-07T11:46:43.367" Score="24" Body="&lt;p&gt;Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible temperature of the reactor failure. This is as far as I know why the failure mode is called a 'meltdown', because the corium melts and pools in the bottom of the reactor vessel. In a molten pool, the fissile material may achieve uncontrolled criticality and just get hotter and hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature is high enough the corium may melt through the bottom of the reactor. This is really bad, because at least hypothetically the lava may melt down to the water table and cause a steam explosion. This is quite a bit like what happens if you try to extinguish an oil fire by throwing water on it (something you should &lt;strong&gt;never do&lt;/strong&gt;) - the water violently boils and expands and sprays burning oil everywhere. However in this case the burning oil is &lt;strong&gt;radioactive lava.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a chemical reaction, per se, but rather a case of extremly vigorous heating of water and associated violent expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remember the Fukushima disaster, at least one of the buildings was destroyed in a hydrogen explosion. As I understand it, in the event of a meltdown, the fissile material can get hot enough to effect the thermolysis of water. (The &lt;a href=&quot;http://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster&quot;&gt;wikipedia article&lt;/a&gt; of the event seems to confirm this) Inside a sealed building, the water will develop a headspace filled with hydrogen and oxygen gas. This mixture is very explosive.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2013-01-07T11:59:56.833" LastActivityDate="2013-01-07T11:59:56.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2891" PostTypeId="2" ParentId="2720" CreationDate="2013-01-07T15:51:15.297" Score="2" Body="&lt;p&gt;I doubt that the conditions in an Anaerobic digester (organic acids, ammonia (!), etc) allow algae to grow. If they do, the oxygen they set free will create oxic conditions and stop the AD - Methanogens are obligate anaerobe bacteria.&#xA;In nature, Methanogens and Algae inhabit different habitats - methanogens in the mud at the bottom of a swamp, algae (maybe) in the water above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Processes for separating $\ce{CO2}$ from $\ce{CH4}$ usually work on the $\ce{CO2}$ (absorption in amines or water under high pressure), so it would be hard to pull the $\ce{CH4}$ from the headspace of your vessel - but given the biological points, this is a non-issue.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="4231" LastEditDate="2015-02-25T15:44:35.207" LastActivityDate="2015-02-25T15:44:35.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2892" PostTypeId="2" ParentId="605" CreationDate="2013-01-07T20:13:47.320" Score="8" Body="&lt;p&gt;There are many complexes of d-elements, that can reversibly lose/gain one electron, changing their color in process. They are known as Redox Indicators. Some of them are stable enough to be used &lt;em&gt;in vivo&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="73" LastEditDate="2013-01-08T00:46:26.963" LastActivityDate="2013-01-08T00:46:26.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2893" PostTypeId="2" ParentId="2889" CreationDate="2013-01-08T00:49:45.577" Score="13" Body="&lt;p&gt;Another possible explanation: water &lt;a href=&quot;http://www.researchgate.net/post/Can_anyone_suggest_a_good_neutron_reflecting_material&quot; rel=&quot;noreferrer&quot;&gt;can act as a neutron reflector&lt;/a&gt;, at least under some conditions.  Therefore, it might be possible for a subcritical mass of uranium to become supercritical if immersed in water, as some of the emitted neutrons could be reflected back into the uranium.  This could lead to a criticality accident, with large amount of ionizing radiation emitted and possibly some type of explosion.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: I found a reference to this in Richard Feynman's essay &lt;a href=&quot;http://calteches.library.caltech.edu/34/3/FeynmanLosAlamos.htm&quot; rel=&quot;noreferrer&quot;&gt;Los Alamos From Below&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So [Emilio Segrè] finally went down [to the Oak Ridge uranium enrichment plant] to see what they were doing, and as he was walking through he saw them wheeling a tank carboy of water, green water—which is uranium nitrate solution.&lt;/p&gt;&#xA;&lt;p&gt;He says, “Uh, you're going to handle it like that when it's purified too? Is that what you're going to do?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;They said, “Sure—why not?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Won't it explode?&amp;quot; he says.&lt;/p&gt;&#xA;&lt;p&gt;Huh! Explode?&lt;/p&gt;&#xA;&lt;p&gt;And so the Army said, “You see! We shouldn't have let any information get to them! Now they are all upset.”&lt;/p&gt;&#xA;&lt;p&gt;Well, it turned out that the Army had realized how much stuff we needed to make a bomb—20 kilograms or whatever it was - and they realized that this much material, purified, would never be in the plant, so there was no danger. But they did not know that the neutrons were enormously more effective when they are slowed down in water. And so in water it takes less than a tenth - no, a hundredth - as much material to make a reaction that makes radioactivity. It kills people around and so on. So, it was very dangerous, and they had not paid any attention to the safety at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is more about water being a neutron &lt;em&gt;moderator&lt;/em&gt; than a &lt;em&gt;reflector&lt;/em&gt;, but the idea is the same - uranium in water could lead to a criticality accident.&lt;/p&gt;&#xA;" OwnerUserId="1075" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-01-08T06:22:46.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2894" PostTypeId="1" CreationDate="2013-01-08T01:19:32.103" Score="5" ViewCount="66" Body="&lt;p&gt;Using a molecular viewer like PyMOL, I understand that its often incorrect to simply look at positive and negative residues and use those residues to calculate the charge density and instead, you should be using something like APBS to calculate the electrostatic potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct way of going about this?&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="73" LastEditDate="2013-01-08T07:56:57.477" LastActivityDate="2013-01-08T14:38:29.753" Title="What is the appropriate method to visualize electrostatic potentials?" Tags="&lt;structural-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2895" PostTypeId="2" ParentId="2889" CreationDate="2013-01-08T06:03:59.653" Score="3" Body="&lt;p&gt;It appears that uranium metal reacts with water &lt;a href=&quot;http://www.pnl.gov/main/publications/external/technical_reports/PNNL-17815.pdf&quot; rel=&quot;nofollow&quot;&gt;Uranium Metal Reaction Behavior in&#xA;Water, Sludge, and Grout Matrices&lt;/a&gt; - when molten, steam will be generated by the heat of the molten metal &lt;em&gt;and&lt;/em&gt; by the exotherm. Not a healthy situation in an enclosed environment.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2013-01-08T06:03:59.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2896" PostTypeId="2" ParentId="2894" CreationDate="2013-01-08T14:38:29.753" Score="4" Body="&lt;p&gt;One way to do it is to perform a quantum chemistry calculation of the electrostatic field generated by the molecule. Another is to determine (or assume) partial atomic charges for each of the atoms of the system, and use a Poisson solver (as I believe APBS is) to calculate the electrostatic field generated by this point charge distribution.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-01-08T14:38:29.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2897" PostTypeId="1" CreationDate="2013-01-08T18:30:09.160" Score="1" ViewCount="1015" Body="&lt;p&gt;I am struggling on this question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of neutralization of $\ce{NaOH}$ and $\ce{HCl}$ is $-57.1\:\mathrm{kJ}$, calculate the numbers of moles of $\ce{HCl}$ and if $x$ moles of $\ce{HCl}$ were used how many $\:\mathrm{kJ}$ would be given out if one mole results in $57.1 \:\mathrm{kJ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already got the moles as $0.05$ by doing $n=\frac{v}{1000}  c$ where I had $50\:\mathrm{mL}$ of $2\:\mathrm{M}$ $\ce{HCl}$&lt;/p&gt;&#xA;" OwnerUserId="1076" LastEditorUserId="865" LastEditDate="2021-05-22T21:16:33.540" LastActivityDate="2021-05-22T21:16:33.540" Title="Calculating heat capacity of a calorimeter" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2898" PostTypeId="2" ParentId="2879" CreationDate="2013-01-09T00:08:36.187" Score="18" Body="&lt;p&gt;I can't really provide a systematic approach, but I can attempt to clarify (as a student myself).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Elements&lt;/strong&gt; are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isotopes&lt;/a&gt; which are physical variations among atoms of the same element).&lt;br&gt;&#xA;A definite (I suppose, systematic) way to distinguish elements is that their corresponding atoms have different numbers of electrons orbiting the nucleus.&lt;br&gt;&#xA;The term is also used to describe a collective of the same atom (element). A bar of gold (Gold being an 'element'; a chemical class) is said to be an element itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Molecules&lt;/strong&gt; are a group of atoms &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;covalently&lt;/a&gt; bonded to each other (which can be considered a 'direct connection', if you will). The molecule can consist of atoms of the same element, or atoms of different elements.&lt;br&gt;&#xA;If you joined to molecules via covalent bonding, you'd have created a new molecule (it's nothing special)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Compounds&lt;/strong&gt; are (in the chemical sense of the term) made of two or more elements. This means a lot of molecules (like H2O) are considered compounds (though H2 is not). 'Compound' is often used interchangeably with the word &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_substance&quot; rel=&quot;nofollow noreferrer&quot;&gt;substance&lt;/a&gt;, which describes a collective of molecules (water is a substance. It is a large quantity of neighboring H2O molecules).&lt;br&gt;&#xA;A substance (and a compound, when a compound is used to describe a quantity of molecules) only consists of the same molecule. In water, all the molecules are H2O. If there were also some H2 molecules among them, it would no longer be considered a compound/substance (but instead; a mixture).&lt;br&gt;&#xA;These molecules are not directly bonded to each other (or they'd be one big molecule), they are just in close* proximity to one another and are &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;attracted&lt;/a&gt; to one another, keeping them in proximity. Even when the substance is a solid, like ice, they are not directly connected, the molecules are just slower (less energetic) and closer together (and in some special cases, like ice, stronger, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Bonding&lt;/a&gt;. Don't be confused, though it is referred to as 'bonding', there is no direct covalent bond, just an attraction).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Mixtures&lt;/strong&gt; are collectives of different molecules or atoms. The molecules and atoms are like that in a substance; they're not bonded to each other, but are attracted, and in a mixture specifically, the molecules/atoms are not all the same. Juice, for instance, is a mixture. It contains some Vitamin C molecules, some water molecules, some sugar molecules, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a horribly drawn diagram to help clarify the distinctions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7EJs0.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflecting on this, here is what in your understanding needs specific attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound: &lt;strike&gt;two or more different atoms bonded together&lt;/strike&gt; &lt;strong&gt;A molecule or a group of identical molecules.&lt;/strong&gt;&lt;br&gt;&#xA;Mixture: two or more different atoms &lt;strong&gt;or molecules&lt;/strong&gt; together but not joined &lt;strong&gt;(they're not covalently bonded)&lt;/strong&gt;.&lt;br&gt;&#xA;Molecule: two or more atoms(same or different) bonded together.  &lt;/p&gt;&#xA;" OwnerUserId="97" LastEditorUserId="5026" LastEditDate="2018-07-26T18:02:31.513" LastActivityDate="2018-07-26T18:02:31.513" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="2899" PostTypeId="2" ParentId="2897" CreationDate="2013-01-09T18:54:20.850" Score="1" Body="&lt;p&gt;First, is this a question from your book, and do you have the correct answer to check your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know how to use dimensional analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with 50 mL. You have 2 Molarity of some subtance, how do you write that as a quotient and what are the units of the denominator and numerator? You will also need to know how many mL are in a liter and you should get the correct moles of that substance. It is very important that you write this down on paper correctly and make sure your units will cancel leaving only moles. It is hard to be wrong when your units cancel correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you mean the heat of neutralization is $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more information you give us, the more help you will get.&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorDisplayName="user7951" LastEditDate="2016-10-31T22:23:50.270" LastActivityDate="2016-10-31T22:23:50.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2900" PostTypeId="1" CreationDate="2013-01-10T16:25:49.000" Score="3" ViewCount="292" Body="&lt;p&gt;Does Hund's rule allow both of the following scenarios?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Filling each orbital with a single electron, so that a sub-shell, at first, only electrons with a negative spin&lt;/li&gt;&#xA;&lt;li&gt;Filling each orbital with a single electron, such that a sub-shell, at first, only electrons with a positive spin?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I assume two H atoms whose electrons have different spins would form $\ce{H-H}$ and atoms where the spins are the same would form $\ce{H:H}$. Is that correct?&lt;/p&gt;&#xA;" OwnerDisplayName="Alexander Wilms" LastEditorUserId="4231" LastEditDate="2015-02-25T15:49:29.940" LastActivityDate="2015-02-25T15:49:29.940" Title="Hund's rule &amp; different H₂ molecules" Tags="&lt;spin&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2901" PostTypeId="7" CreationDate="2013-01-10T17:44:14.130" Score="0" Body="&lt;p&gt;&lt;strong&gt;Chemistry&lt;/strong&gt; is a question and answer site for scientists, academics, teachers and students in the field of chemistry. It's built and run &lt;em&gt;by you&lt;/em&gt; as part of the &lt;a href=&quot;http://stackexchange.com&quot;&gt;Stack Exchange&lt;/a&gt; network of Q&amp;amp;A sites. With your help, we're working together to build a library of detailed answers to every question about chemistry.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="16683" LastEditDate="2019-05-12T23:39:17.870" LastActivityDate="2019-05-12T23:39:17.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2902" PostTypeId="1" AcceptedAnswerId="4428" CreationDate="2013-01-11T00:22:37.597" Score="2" ViewCount="470" Body="&lt;p&gt;In a simple reaction between butane and chlorine, it is more likely for the secondary carbons of the butane to lose a hydrogen and gain a chlorine. The carbocation is a more stable intermediate in that situation compared to a primary carbon losing the hydrogen. So if the desired synthesis is 1-chlorobutane as opposed to 2-chlorobutane then it seems reasonable that a chemist might use stratification. The reason that I believe it is less desirable to use temperature as a method of separation is because that, at least in this example, both compounds will move between the gas and liquid phase randomly providing the possibility of still having mixed samples. Is this true, and will stratification work more reliably if purification is the desired result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I never really learned why mono-chlorinated products form. Will there not also be compounds that have been chlorinated twice? Or will, over time due to entropy, all compounds be mono-chlorinated assuming that the reagents are in a ratio of 1 chlorine atom per molecule of butane?&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2013-03-20T11:00:10.313" Title="Is it more reasonable to use stratification to separate similar compounds rather than using temperature?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2903" PostTypeId="1" CreationDate="2013-01-11T19:07:47.290" Score="3" ViewCount="142" Body="&lt;p&gt;I'm reading about electrodeposition of conductive polymers in a chemistry textbook, and the author mentions 'trace-crossing' in the 'potentiodynamic response'. From the context I can figure out that 'potentiodynamic response' means cyclic voltammogram.  But what is the 'trace crossing'?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="22" LastEditDate="2013-01-12T14:16:41.953" LastActivityDate="2013-01-15T17:41:13.847" Title="What is a 'Trace Crossing'?" Tags="&lt;electrochemistry&gt;&lt;polymers&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2904" PostTypeId="1" AcceptedAnswerId="2906" CreationDate="2013-01-11T22:06:59.363" Score="5" ViewCount="1293" Body="&lt;p&gt;First question, and this is one I've searched for in vain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of setting up cooling baths, I've come across the term &quot;lagged&quot; bath.  I cannot extract what this means from any available context.  I assume it has something to do with time passing (i.e. a lag), but I'm not certain.  If I had to guess, I'd assume it means allowing the cooling agent to come to equilibrium temperature, but again, not certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any direct technical experience that can explain this term would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2020-08-13T03:20:24.393" Title="What is a &quot;lagged bath&quot;?" Tags="&lt;experimental-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2906" PostTypeId="2" ParentId="2904" CreationDate="2013-01-12T00:11:36.287" Score="9" Body="&lt;p&gt;I'm not 100% certain on this, but my best guess on this would be referring to 'lagged' as in 'lagging', as in insulation material. So, a lagged bath is an insulated bath.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-01-12T00:11:36.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2907" PostTypeId="2" ParentId="1030" CreationDate="2013-01-12T01:12:25.903" Score="3" Body="&lt;p&gt;In the end there is no way to predict the structure of a given molecule.  Yes, with experience one can guess correctly in many cases, but not in all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way all this works is to &lt;strong&gt;first&lt;/strong&gt; find the structure and then &lt;strong&gt;second&lt;/strong&gt; work out the bonding scheme that gives this structure.  In principle one can do an &lt;em&gt;ab initio&lt;/em&gt; quantum mechanical calculation, but that's not really practical for molecules with a fair number of electrons such as $\ce{(NH3)2SO4}$.  So most often an approximation procedure is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two simple approximations.  One is molecular orbitals, the other is the LCAO (linear combination of atomic orbitals) approximation.  They describe the same thing but use different language.  The choice of which to use is up to the person wanting the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of all this, try predicting the bonding and structure of $\ce{H3PO4}$ without knowing the answer.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-01-12T01:12:25.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2908" PostTypeId="1" CreationDate="2013-01-12T10:06:05.007" Score="2" ViewCount="72" Body="&lt;p&gt;I have 12 experimental tables with OD(substrate concentration, time), where OD is optical density and one table with $\mu$ (substrate), where $\mu$ is growth speed. OD and $\mu$ are functions.&#xA;The experiment is &quot;growth of proteins on a methanol substrate&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With passage of time, the concentration of substrate and oxygen is decreasing (substrate to zero, and $\ce{O2}$ to 10% from 100% and the process is terminated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With substrate decrease the number of particles in bioreactor is growthing. I must determine the parameters $(s, \ce{O2})$ when the growth speed is $\rm max$ to $\mathrm{max }- x\%$. $x$ may be $10\%-15\%$ of $\rm max$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a model to determine the functions $\mu(s)$, $\mu(\ce{O2})$ and $\mu(s, \ce{O2})$ or the max value of miu(s,o2), where $\mu$ = growth speed, $s$ = substrate concentration and $\ce{O2}$ = oxygen concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must simulate the function $\mu(s,\ce{O2})$, to observe the maximization surface(sized by $s$ and $\ce{O2}$) through variation of parameters $s$ and $\ce{O2}$. The surface formed by $s$ and $\ce{O2}$ when $\mu$ values are between $\mathrm{max}(\mu) - 10\%$ and $\mathrm{max}(\mu)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of bioreactor is $15\:\mathrm{L}$ and the density inside is $\approx 1$.&lt;/p&gt;&#xA;" OwnerUserId="1086" LastEditorUserId="22" LastEditDate="2013-01-12T12:37:52.603" LastActivityDate="2013-01-12T12:37:52.603" Title="Determine max of 2 parameter functions" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2909" PostTypeId="7" CreationDate="2013-01-12T13:32:00.203" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Chemistry concepts&lt;/li&gt;&#xA;&lt;li&gt;Observed chemical phenomena&lt;/li&gt;&#xA;&lt;li&gt;Experimental techniques and technology&lt;/li&gt;&#xA;&lt;li&gt;Nomenclature, standards, etc. pertaining to chemistry&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="847" LastEditDate="2013-01-17T00:08:24.267" LastActivityDate="2013-01-17T00:08:24.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2910" PostTypeId="7" CreationDate="2013-01-12T13:34:20.667" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Homework (read &lt;a href=&quot;http://meta.chemistry.stackexchange.com/a/142/22&quot;&gt;the homework policy&lt;/a&gt; if you have such a question)&lt;/li&gt;&#xA;&lt;li&gt;Personal medical questions&lt;/li&gt;&#xA;&lt;li&gt;Legal questions relating to chemical substances and techniques &lt;/li&gt;&#xA;&lt;li&gt;Non-mainstream chemistry&lt;/li&gt;&#xA;&lt;li&gt;Anything not directly related to chemistry&lt;/li&gt;&#xA;&lt;li&gt;Questions that are primarily opinion-based&lt;/li&gt;&#xA;&lt;li&gt;Questions with too many possible answers or that would require an extremely long answer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="16683" LastEditDate="2020-01-18T19:13:02.993" LastActivityDate="2020-01-18T19:13:02.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2911" PostTypeId="1" CreationDate="2013-01-13T15:24:06.103" Score="0" ViewCount="228" Body="&lt;p&gt;Can you tell me what's the nomenclature of $\ce{C H_2 O_2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is its IUPAC name?&lt;/p&gt;&#xA;" OwnerUserId="1091" LastEditorUserId="73" LastEditDate="2013-01-14T00:45:05.870" LastActivityDate="2013-01-14T04:32:53.740" Title="What is the proper nomenclature for CH₂O₂?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2913" PostTypeId="1" AcceptedAnswerId="2915" CreationDate="2013-01-13T18:11:37.503" Score="4" ViewCount="594" Body="&lt;p&gt;Alanin and dyes contain nitrogen, is being colorful a special feature in nitrogenous substances?&lt;/p&gt;&#xA;" OwnerUserId="1092" LastEditorUserId="73" LastEditDate="2013-01-13T22:04:25.067" LastActivityDate="2013-01-13T22:06:13.230" Title="Why are nitrogenous hydrocarbons sometimes colorful?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-09T19:15:48.717" ContentLicense="CC BY-SA 3.0" />
  <row Id="2914" PostTypeId="2" ParentId="2911" CreationDate="2013-01-13T18:48:52.700" Score="2" Body="&lt;p&gt;The compound you provided can be written more clearly as $\ce{HCOOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IUPAC name : Formic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Systematic name :Methanoic Acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other names:&#xA;Aminic acid, Formylic acid, Hydrogen carboxylic acid, Hydroxymethanone, Hydroxy(oxo)methane, Metacarbonoic acid, Oxocarbinic acid, Oxomethanol&lt;/p&gt;&#xA;" OwnerUserId="1092" LastEditorUserId="84" LastEditDate="2013-01-14T04:32:53.740" LastActivityDate="2013-01-14T04:32:53.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2915" PostTypeId="2" ParentId="2913" CreationDate="2013-01-13T21:31:45.503" Score="5" Body="&lt;p&gt;Introduction from &lt;a href=&quot;http://people.chem.umass.edu/samal/269/color.pdf&quot; rel=&quot;noreferrer&quot;&gt;Color Introduction&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;What causes a compound to appear as being colored? The answer to this question lies in the way that electromagnetic energy interacts with matter. The portion of the electromagnetic spectrum that is of interest to this discussion is the visible portion, or what we call light. This visible light includes all the colors that we perceive: violet, blue, green, yellow, orange, red, and the various combinations of these. Although light can interact with matter in several different ways, this discussion will limit the modes of interaction to absorption and reflection.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;In most cases color comes from the excitation of electrons due to an absorption of energy performed by the chemical. But - there is still ... :)&#xA;&amp;quot;(...) The vast majority of simple inorganic (e.g. sodium chloride) and organic compounds (e.g. ethanol) are colorless. Transition metal compounds are often colored because of transitions of electrons between d-orbitals of different energy.&#xA;Organic compounds tend to be colored when there is extensive conjugation, causing the energy gap between the HOMO and LUMO to decrease, bringing the absorption band from the UV to the visible region. Similarly, color is due to the energy absorbed by the compound, when an electron transitions from the HOMO to the LUMO.(...)&amp;quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Color_of_chemicals&quot; rel=&quot;noreferrer&quot;&gt;Wiki - color of chemicals&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can also assume that if a compound has double or triple conjugated (2 or more) bonds it can be colorful.&lt;/p&gt;&#xA;&lt;p&gt;To answer your question: in alanine there is almost no chance to make double bond in amine group (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{=NH}$&lt;/span&gt;), but nitrogen group (&lt;span class=&quot;math-container&quot;&gt;$\ce{O-N=O}$&lt;/span&gt;) has a capability to conjunct with other double bonds (can be also delocalized electrons or pi electrons - e.g., nitrobenzene)&lt;/p&gt;&#xA;" OwnerUserId="417" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-01-13T22:06:13.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2916" PostTypeId="1" AcceptedAnswerId="2917" CreationDate="2013-01-14T01:23:15.883" Score="12" ViewCount="1882" Body="&lt;p&gt;Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart? E.g., is it possible to change dextrodopa to levodopa?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4325" LastEditDate="2018-01-26T01:50:44.350" LastActivityDate="2018-01-26T01:50:44.350" Title="Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2917" PostTypeId="2" ParentId="2916" CreationDate="2013-01-14T02:27:53.950" Score="10" Body="&lt;p&gt;Yes,this process is called stereoconversion. Both SN1 and SN2 reactions can invert the chirality of a carbon center. In the case of the SN1 mechanism, departure of a leaving group generates a prochiral intermediate which is then open to nucleophilic attack on either face. If both faces are equally acessible to the nucleophile this will result in racemisation of the molecule. In the case of the SN2 reaction the nucleophile will enter a trigonal bipyramidal intermediate with the molecule. Assuming the nucleophile has the same stereochemical precedence as the leaving group, the stereochemistry will be inverted.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-14T02:27:53.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3917" PostTypeId="1" AcceptedAnswerId="3924" CreationDate="2013-01-14T08:42:29.857" Score="6" ViewCount="1819" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mQbFP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me, the answer should be (A) because that's a Wurtz reaction, but the answer given in the book is (B).&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorDisplayName="user7951" LastEditDate="2018-07-11T20:16:18.367" LastActivityDate="2018-07-11T20:16:18.367" Title="What will be the major product of of the Wurtz reaction of 3,6-dibromocyclohexa-1,4-diene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3918" PostTypeId="1" AcceptedAnswerId="3931" CreationDate="2013-01-14T08:51:32.570" Score="14" ViewCount="1757" Body="&lt;p&gt;I'm tutoring few Class 12 students for their chemistry exams. I'm stuck with few problems. I don't know, if the key is wrong or I am wrong. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7AyT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me product (A) must be the cyclic diene, as given in option (A). Now, I don't know, what happens when you heat cyclic diene with selenium. But I do know that,  $\ce{SeO2}$  oxidizes allylic positions of alkenes into alcohols. But, none of the options given are fitting into that argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer given in the book is option (B).&#xA;How can that be explained?&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="4945" LastEditDate="2018-02-20T08:31:30.563" LastActivityDate="2019-05-12T07:59:33.003" Title="What will be the product when 3,4-dibromocyclohexene is reacted with sodium followed by heating with selenium?" Tags="&lt;organic-chemistry&gt;&lt;elimination&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3919" PostTypeId="1" AcceptedAnswerId="3926" CreationDate="2013-01-14T09:24:54.057" Score="3" ViewCount="1553" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/slJEL.png&quot; alt=&quot;Reductive ozonolysis of 1-Methoxy-4-methyl-1,4-cyclohexadiene&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, both the left side and the right side double bonds undergo cleavage and hence the product must be, two molecules: Keto aldehyde and Ester aldehyde. But since, such product is not given in options, let us assume that only one bond gets oxidized. But which one? Which bond is the most reactive? I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer given in the book:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(B)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How come?&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="83758" LastEditDate="2020-05-13T06:31:09.450" LastActivityDate="2020-05-13T06:31:09.450" Title="What will be the Ozonolysis product?" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3920" PostTypeId="1" AcceptedAnswerId="3928" CreationDate="2013-01-14T09:33:07.583" Score="7" ViewCount="4360" Body="&lt;p&gt;Alkynes are more reactive towards catalytic hydrogenation than that of alkenes, because they are more unsaturated. This is the correct reasoning, right?&lt;/p&gt;&#xA;&lt;p&gt;But why are carbonyl compounds more reactive towards catalytic hydrogenation than alkynes? The &lt;span class=&quot;math-container&quot;&gt;$\ce{C=O}$&lt;/span&gt; in carbonyl group is polar, but how does that affect reactivity?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Vrgjk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Answer: Option (D), &lt;span class=&quot;math-container&quot;&gt;$c&amp;gt;a&amp;gt;b$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="385" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-05T02:48:11.757" Title="Why are carbonyl compunds more reactive towards catalytic hydrogenation than alkynes and alkenes?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3921" PostTypeId="1" AcceptedAnswerId="3927" CreationDate="2013-01-14T09:49:55.750" Score="9" ViewCount="328" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qtkjx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qtkjx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{A}$ can be?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;concentrated sulphuric acid&lt;/li&gt;&#xA;  &lt;li&gt;alcoholic potassium hydroxide&lt;/li&gt;&#xA;  &lt;li&gt;triethylamine&lt;/li&gt;&#xA;  &lt;li&gt;potassium &lt;em&gt;tert&lt;/em&gt;-butoxide&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My comments:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option (A):&lt;/strong&gt; Conc. $\ce{H2SO4}$, is an acid, it gives $\ce{H+}$ ions which must act like electrophile and add to the double bond, which must produce a carbocation and then, $\ce{HSO4^-}$ ion must act like a nucleophile and form a bond with the carbocation. Hence, the reaction must be addition of $\ce{H+}$ and $\ce{HSO4^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Options (B),(C),(D):&lt;/strong&gt; are bases, with basicity order: (D) &gt; (B) &gt; (C). But what would they do to an alkene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer given in the book is A. How is this so?&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="5026" LastEditDate="2018-02-25T07:53:57.140" LastActivityDate="2018-02-25T07:53:57.140" Title="What would be the mechanism of the given reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3922" PostTypeId="2" ParentId="3917" CreationDate="2013-01-14T11:00:01.413" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;https://wikivividly.com/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon. This is because the activation energy required for the SN2 reaction in the second step becomes significantly high, so the alternate elimination mechanism is favoured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't see the Wurtz reaction happening to make (A) in this case with the two double bonds, there isn't enough flexibility to get the radical to attack across the ring. Also, (B) is making an aromatic ring, which is normally the most stable product when given the option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'll go with (B) for this.&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="5026" LastEditDate="2018-07-11T03:29:39.700" LastActivityDate="2018-07-11T03:29:39.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3924" PostTypeId="2" ParentId="3917" CreationDate="2013-01-14T12:04:28.983" Score="7" Body="&lt;p&gt;The answer is (B) because (B) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzene&quot;&gt;benzene&lt;/a&gt;, which is the prototypical &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromatic_hydrocarbon&quot;&gt;aromatic hydrocarbon&lt;/a&gt;. The aromaticity of (B) increases its thermodynamic stability relative to (A), which is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Dewar_benzene&quot;&gt;Dewar benzene&lt;/a&gt; (mistakenly attributed to James Dewar), which is notoriously difficult to synthesis and isolate. It reverts back to benzene with a half-life of two days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, if a pathway exists for a reaction to produce an aromatic product, then the reaction likely will do so. In this particular case, the diradical intermediate that might be produced as part of the Wurtz reaction's mechanism, is the triplet excited state of benzene.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-01-14T12:04:28.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3925" PostTypeId="2" ParentId="3919" CreationDate="2013-01-14T12:48:34.827" Score="2" Body="&lt;p&gt;Any reaction in general will try to attain maximum possible stability.&lt;br&gt;&#xA;The reactivity of a molecule or a group is inversely proportional to its stability.&lt;br&gt;&#xA;The general order of reactivity of carbonyl compounds is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\small\ce{Aldehyde &amp;gt; Ketone &amp;gt; Acid halide &amp;gt; Anhydride &amp;gt; Ester &amp;gt; Carboxylic Acid &amp;gt; Acid Amide}$$ &#xA;&lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=4778.0&quot; rel=&quot;nofollow&quot;&gt;(Reference)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the left bond breaks it results in a ketone which is more reactive(hence less stable) as compared to the product formed by the cleavage of the right bond which is an ester.&lt;br&gt;&#xA;Hence the bond on the right would break to form an ester.&lt;/p&gt;&#xA;" OwnerUserId="897" LastActivityDate="2013-01-14T12:48:34.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3926" PostTypeId="2" ParentId="3919" CreationDate="2013-01-14T13:02:57.977" Score="5" Body="&lt;p&gt;There are two types of control, that governs reaction path: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;kinetic control (path with least activation energy) and &lt;/li&gt;&#xA;&lt;li&gt;thermodynamic control (path with least energy of products). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given temperature favors kinetic control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozonolysis starts as 2+3 electrocyclic addition (with later regroup of the product), similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction&quot; rel=&quot;nofollow&quot;&gt;Diels-Alder reaction&lt;/a&gt; . Please, read about this reaction in good organic chemistry textbook, especially about what controls geometry of products, it will help to understand what I'm going to say next. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozone has low energy of LUMO, so alkene part here provides HOMO, that is still lower then Ozone's LUMO. Oxygen from $OMe$ group donates some electronic density to $C-C$ double bound, so this bond becomes more active, then opposite one.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="385" LastEditDate="2013-01-18T20:42:47.547" LastActivityDate="2013-01-18T20:42:47.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3927" PostTypeId="2" ParentId="3921" CreationDate="2013-01-15T09:07:00.540" Score="16" Body="&lt;p&gt;Bottom line: $\ce{HSO4-}$ is a really bad nucleophile, and it doesn't enter in the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bnydy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnydy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize after scanning that I'm missing the &quot;+ H&lt;sup&gt;+&lt;/sup&gt;&quot; alongside the final product.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Question for extra credit: after the second step, why is the third step (carbocation migration) favorable rather than directly reforming the double bond on the right-hand side ring?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="55803" LastEditDate="2018-01-26T11:15:45.670" LastActivityDate="2018-01-26T11:15:45.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3928" PostTypeId="2" ParentId="3920" CreationDate="2013-01-15T09:16:30.083" Score="6" Body="&lt;p&gt;That is not true in general: the stronger nature of the C=O bonds (larger bond energy) means that they are not as easily hydrogenated as C=C bonds. However, your compound (c) is not a regular carbonyl group, but an acyl chloride, which is a quite reactive species and thus very easily hydrogenated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bit of reading, see &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid2.htm#react2&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; the section on “Catalytic Hydrogenation”:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds.&lt;/strong&gt; Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. &lt;strong&gt;The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions&lt;/strong&gt;, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. &lt;strong&gt;This reduction stops at the aldehyde stage&lt;/strong&gt;, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. &lt;strong&gt;The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastEditorUserId="385" LastEditDate="2013-01-18T20:42:43.337" LastActivityDate="2013-01-18T20:42:43.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3930" PostTypeId="1" CreationDate="2013-01-15T14:03:08.140" Score="1" ViewCount="133" Body="&lt;p&gt;I am going to stick a strain gauge to the side of an expanding piezocrystal using a basic 2 part epoxy glue. I am eventually also going to be sticking other things to it. Does anyone know a safe way to remove the glue without damaging the piezostack? The manual says specifically not to use acetone.&lt;/p&gt;&#xA;" OwnerUserId="1098" LastActivityDate="2013-01-15T14:03:08.140" Title="Removing Epoxy Glue from a ceramic piezostack" Tags="&lt;solvents&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3931" PostTypeId="2" ParentId="3918" CreationDate="2013-01-15T14:04:56.100" Score="14" Body="&lt;p&gt;Assuming your Product A is correct then product B should be benzene. Selenium act as a mild oxidizing agent at temperature 220 °C to 330 °C. In the above reaction selenium oxidizes cyclohexadiene to form benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a reaction in which 1-methyl-1,3-cyclohexadiene is converted into toluene.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/E9NEz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can also go to this &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/aroma_ringe.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/entfernen_wasserstoff/aromatisierung_6_ringe.vscml.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link for more info&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorDisplayName="user7951" LastEditDate="2019-05-12T07:59:33.003" LastActivityDate="2019-05-12T07:59:33.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3932" PostTypeId="2" ParentId="2855" CreationDate="2013-01-15T16:07:44.213" Score="2" Body="&lt;p&gt;Photodissociation is a unimolecular reaction triggered by absorption of light. You start with one molecule, a bond (or more) breaks and you end up with two (or more) products. In water you can get $\ce{OH}$ and $\ce{H}$ (etc), usually in excited electronic states. You can not get the products of the electrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water electrolysis is a more complex reaction. The final balance is: &#xA;$\ce{2H2O -&amp;gt; 2H2 + O2}$,&#xA;so you break four bonds and make three bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both reactions are not spontaneous (in the electronic ground state of water). The net balance is not energetically favorable (you start with more energy in bonds than what you end up). However, the way you add this energy is also different. In photodissociation the energy is given by light; in electrolysis the energy is given by forcing an electron current.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="22" LastEditDate="2013-01-15T17:44:08.833" LastActivityDate="2013-01-15T17:44:08.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3933" PostTypeId="2" ParentId="2903" CreationDate="2013-01-15T16:37:54.437" Score="3" Body="&lt;p&gt;'Trace crossing' is in the literal sense, where 'trace' is another word for 'line'. 'Trace crossing' basically means that the line in the voltammogram crosses itself.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="22" LastEditDate="2013-01-15T17:41:13.847" LastActivityDate="2013-01-15T17:41:13.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3934" PostTypeId="2" ParentId="2469" CreationDate="2013-01-15T17:00:22.103" Score="11" Body="&lt;p&gt;Unfortunately, the Aufbau rule cannot predict all electron configuration as it doesn't take into account electron-electron interactions. In the end the Aufbau is only a rule of thumb. Electronic levels have to be found using quantum calculations taking into account electron-electron interactions (not to mention spin orbit coupling). Therefore there is no simple or rational explanation for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=239512&quot;&gt;http://www.physicsforums.com/showthread.php?t=239512&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_configuration#Aufbau_principle_and_Madelung_rule&quot;&gt;Other exceptions to Madelung's rule paragraph&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1055" LastEditorUserId="22" LastEditDate="2013-01-15T17:45:45.497" LastActivityDate="2013-01-15T17:45:45.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3936" PostTypeId="1" CreationDate="2013-01-15T21:17:54.900" Score="2" ViewCount="164" Body="&lt;p&gt;The ionic radius of iodine in the high temperature form can be determined once the unit cell is known, but how can I calculate the radius of $\ce{I}$ given that the cubic lattice parameter is 6.10 Å. &lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="69830" LastEditDate="2021-05-11T05:36:04.387" LastActivityDate="2021-05-11T05:36:04.387" Title="Ionic Radius of Iodine in High Temperature Form" Tags="&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3937" PostTypeId="1" CreationDate="2013-01-15T22:39:35.197" Score="9" ViewCount="5786" Body="&lt;p&gt;I have three questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To find the parent chain of a saturated hydrocarbon, we have to find the longest chain and then if we have a choice, we choose the one with the more substituents. But what if we still had a choice? What would we do next?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, for numbering the chain, the last step is to number the chain such that the substituent which comes alphabetically first gets the lower number. But what if we again had a choice? I'm not sure here, but I think we can't possibly have another choice. Can anyone confirm this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If an alkyl substituent is attached to the parent chain by a double or triple bond, how would we name it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;: Here are some examples which demonstrate the above 'tie-breakers'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;For question 1:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJL4e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (green + blue) and (green + magenta) chains are the longest and both have two substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4WRcU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (green + blue) and (green + yellow) chains contain the maximum double bonds and are also the longest. They both have one substituent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;For question 2:&lt;/em&gt;&lt;/strong&gt; As I said, I'm not sure if there can possibly be a 'tie-breaker' here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;For question 3:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTIb8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longest chain is the green one. I'm not sure how to name the substituent attached by the double bond.&lt;/p&gt;&#xA;" OwnerUserId="1200" LastEditorUserId="1200" LastEditDate="2013-01-17T23:14:26.690" LastActivityDate="2016-08-07T16:01:43.093" Title="&quot;Tie-breaker&quot; rules for IUPAC nomenclature of organic compounds" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3938" PostTypeId="2" ParentId="3936" CreationDate="2013-01-16T03:29:35.530" Score="5" Body="&lt;p&gt;Considering by &lt;code&gt;cubic lattice&lt;/code&gt; you mean a Simple Cubic structure, you could do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/qLTB2.jpg&quot; rel=&quot;noreferrer&quot;&gt;Reference image1&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemistry.umeche.maine.edu/~amar/spring2012/bur25542_1219.jpg&quot; rel=&quot;noreferrer&quot;&gt;Reference image2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(could not put it here because of copyright)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2). &#xA;$$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$&#xA;$$\therefore\ a \ = \ 2\times r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\therefore\ce{Length\ of\ face\ diagonal&amp;amp;\ =\ radius\ of\ top\ left\ atom +\\&#xA;                                        &amp;amp;\quad\ diameter\ of\ atom\ at\ face-center + \\&#xA;                                        &amp;amp;\quad\ radius\ of\ bottom\ left\ atom}\\&#xA;\therefore\ce{Length\ of\ face\ diagonal&amp;amp;\ =\ r\ + ( 2\times r ) + r}\\&#xA;\therefore\ce{\sqrt{2}\times a&amp;amp;\ =\ 4\times r}\\&#xA;\therefore\ce{a&amp;amp;\ = 2\sqrt{2}\times r}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Similarly for bcc structure the atoms touch each other along the body-diagonal &#xA;$$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="897" LastEditDate="2013-01-16T09:36:27.743" LastActivityDate="2013-01-16T09:36:27.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3940" PostTypeId="2" ParentId="3937" CreationDate="2013-01-16T22:26:50.993" Score="2" Body="&lt;p&gt;I think in case of &lt;strong&gt;#1&lt;/strong&gt;, you should keep the chain which contains the overall less-complex substituents (which would be green and magenta). Stating it into a rule I guess it would mean comparing the longest substituent of both alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question &lt;strong&gt;#2&lt;/strong&gt;, you should chose the numbering that gives an overall lower numbering result, even if you had a choice based on the alphabetically first substituent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea for question &lt;strong&gt;#3&lt;/strong&gt;. My &lt;em&gt;instinct&lt;/em&gt; tells me that naming that =C-CH3 as a substituent feels &lt;em&gt;wrong&lt;/em&gt; and I should choose the shorter chain even if its &lt;em&gt;against the rules&lt;/em&gt;. But there's nothing scientific about that.&lt;/p&gt;&#xA;" OwnerUserId="1110" LastActivityDate="2013-01-16T22:26:50.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3941" PostTypeId="2" ParentId="2779" CreationDate="2013-01-17T01:22:20.737" Score="8" Body="&lt;p&gt;In old time printing using type, the alloy used for the type was designed to have a zero coefficient of thermal expansion over a wide range so that they type did not change shape when cast at a high temperature and then cooled off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very likely that slight modifications of that formula would yield an alloy with either a positive or negative thermal coefficient.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-01-17T01:22:20.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3942" PostTypeId="1" AcceptedAnswerId="3943" CreationDate="2013-01-17T03:35:46.637" Score="13" ViewCount="51896" Body="&lt;p&gt;By a primary amine I mean a compound where the hydroxyl group of the corresponding alcohol is replaced with an amine group. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="418" LastEditDate="2016-11-17T18:11:46.430" LastActivityDate="2018-02-07T19:31:33.683" Title="How do you create primary amines from alcohols?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;&lt;amines&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3943" PostTypeId="2" ParentId="3942" CreationDate="2013-01-17T05:53:57.123" Score="4" Body="&lt;p&gt;For  &lt;strong&gt;aliphatic compounds&lt;/strong&gt; you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Amine#Synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;alkylation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-OH + NH3 -&amp;gt; R-NH2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;aromatic compounds&lt;/strong&gt; like phenol &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{matrix}&#xA;\ce{C6H6O &amp;amp;\quad +&amp;amp; Zn&amp;amp; \ -&amp;gt;&amp;amp; C_6 H_6\\&#xA;Phenol&amp;amp;  &amp;amp; (dust\ form)&amp;amp; \ &amp;amp; Benzene&amp;amp;}&#xA;\end{matrix}&#xA;$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc dust reduces the OH group in phenol&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\ce{C6H6&amp;amp; + &amp;amp;HNO3&amp;amp; -&amp;gt;[H2SO4]&amp;amp; C6H5NO2 + &amp;amp;H_2 O&amp;amp;\\&#xA;Benzene&amp;amp; &amp;amp; (high\  concentration)&amp;amp;&amp;amp; Nitrobenzene&amp;amp;  water&amp;amp;&#xA;}&#xA;\end{matrix}$$&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aromatic_nitration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nitration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\ce{C6H5NO2&amp;amp; +&amp;amp; Sn&amp;amp; +&amp;amp; \qquad HCl_{(aq)} -&amp;gt;&amp;amp; C6H5NH2&amp;amp;\\&#xA;Nitrobenzene&amp;amp; &amp;amp; Tin&amp;amp; &amp;amp;Hydrochloric &amp;amp; Aniline&amp;amp;\\&#xA;&amp;amp;&amp;amp;&amp;amp;&amp;amp;acid}&#xA;\end{matrix}$$ &#xA;$ $ &lt;br&gt; Tin and hydrochloric acid act as reducing agents &lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="418" LastEditDate="2016-11-17T18:11:33.797" LastActivityDate="2016-11-17T18:11:33.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3944" PostTypeId="2" ParentId="3942" CreationDate="2013-01-17T16:50:15.870" Score="2" Body="&lt;p&gt;In addition to the method provided by &lt;a href=&quot;https://chemistry.stackexchange.com/users/139/t3st&quot;&gt;t3st&lt;/a&gt; you could also try the following for &#xA;$$\begin{matrix}&#xA;\ce{Ph-OH&amp;amp; +&amp;amp; Zn&amp;amp; -&amp;gt;&amp;amp; C6H6 \\&#xA;Phenol&amp;amp; &amp;amp; Zinc&amp;amp; &amp;amp; Benzene&amp;amp;}&#xA;\end{matrix}$$&#xA;(Just like the first step of t3st 's answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Then you could do &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_halogenation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrophilic halogenation&lt;/a&gt; of benzene using $\small\ce{AlCl3\ or\ FeBr3}$ as catalyst.&#xA;$$\begin{matrix}&#xA;\ce{C6H6&amp;amp; -&amp;gt;[\ce{Cl2/AlCl3}]&amp;amp; Ph-Cl\\&#xA;Benzene&amp;amp; &amp;amp; Aryl\ halide&amp;amp;}&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally you could react aryl-halide with soda-amide($\small\ce{NaNH2}$) to substitute $\ce{Cl}$ with $\ce{NH2}$&lt;sup&gt;&lt;sup&gt;[citation needed]&lt;/sup&gt;&lt;/sup&gt;&#xA;$$\begin{matrix}&#xA;\ce{Ph-Cl&amp;amp; -&amp;gt;[\ce{NaNH2}]&amp;amp; Ph-NH2&amp;amp;\\&#xA;Aryl halide&amp;amp; &amp;amp; Aniline&amp;amp;}&#xA;\end{matrix}$$&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2013-01-17T16:50:15.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3945" PostTypeId="1" AcceptedAnswerId="3953" CreationDate="2013-01-17T23:50:15.917" Score="4" ViewCount="815" Body="&lt;p&gt;Why is the Electronegativity difference for atoms in bonding uncertain while determining what the compound will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the IB(International Baccalaureate) they say that the Electronegativity difference required for Ionic bonds is 1.7 though compounds like Hydrogen Fluoride are covalent even though they have a Electronegativity difference of 1.9. If the IB scale of above 1.7 being ionic and below 1.7 being covalent then why are there many exceptions to this rule? Also, why does Electronegativity not change, since having more electrons in atom would probably cause repulsion of new electrons?&#xA;If you search for what the Electronegative difference to predict bonding and were to look at Yahoo.answers you will see 3 different answers to how to predict these reactions at 1.7,1.8 or 1.9 electronegative difference &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20070125052137AAFZZGH&quot; rel=&quot;nofollow&quot;&gt;Yahoo question&lt;/a&gt;. Another exception is &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_carbide&quot; rel=&quot;nofollow&quot;&gt;Calcium Carbide&lt;/a&gt; which forms a ionic bond at 1.5 electronegative difference.&lt;/p&gt;&#xA;" OwnerUserId="994" LastEditorUserId="994" LastEditDate="2013-01-19T23:57:53.590" LastActivityDate="2020-04-06T16:04:16.010" Title="Electronegativity" Tags="&lt;electrochemistry&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3946" PostTypeId="2" ParentId="3937" CreationDate="2013-01-18T02:54:05.933" Score="9" Body="&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;For Question 1&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Example 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YuIVj.png&quot; alt=&quot;Example 1&quot;&gt;&lt;br&gt;&#xA;You should select the (&lt;code&gt;green + blue&lt;/code&gt;) chain because, no matter which among the &lt;code&gt;blue&lt;/code&gt; and &lt;code&gt;yellow&lt;/code&gt; you pick, the numbering would not start from the &lt;code&gt;green&lt;/code&gt; chain rather the yellow or blue.&lt;br&gt;&#xA;Now if you pick the &lt;code&gt;blue&lt;/code&gt; chain the substituent ($\ce{methyl}$) would get number &lt;strong&gt;4&lt;/strong&gt;, whereas if you picked the green chain the substituent($\ce{isopropyl\ or\ methylethyl}$) would get the number &lt;strong&gt;5&lt;/strong&gt;, so in accordance to &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/orgnom/main/difference.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the First Point of Difference Rule&lt;/a&gt; you should select the &lt;code&gt;green + blue&lt;/code&gt; chain.&lt;br&gt;&#xA;Also Worth noticing is the fact that no matter which chain you pick, the other chain which becomes the substituent would always get the number &lt;strong&gt;7&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So locants &lt;strong&gt;(4,7)&lt;/strong&gt; are better than &lt;strong&gt;(5,7)&lt;/strong&gt;.&#xA;$$\begin{matrix}&#xA;\text{Better name:}&amp;amp; \text{7-(2-isopropylhexyl)-4-methylheptadecane}&amp;amp;\\&#xA;\text{Rather than:}&amp;amp; \text{5-isopropyl-7-(3-methylhexyl)heptadecane}&amp;amp;&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Example 2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C15MY.png&quot; alt=&quot;Example 2&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case according to the above mentioned rule, you should check the position of the double bond, now since there is a double bond already in the &lt;code&gt;green&lt;/code&gt; chain at the first position it would get number &lt;strong&gt;1&lt;/strong&gt; and the numbering would start from there.&lt;br&gt;&#xA;If you choose the &lt;code&gt;green + blue&lt;/code&gt; chain the other double bond would get number &lt;strong&gt;5&lt;/strong&gt; and for the &lt;code&gt;green + yellow&lt;/code&gt; chain the other double bond would get number &lt;strong&gt;6&lt;/strong&gt;. So you should go for the &lt;code&gt;green + blue&lt;/code&gt; chain in this case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So locants &lt;strong&gt;(1,5)&lt;/strong&gt; are better than &lt;strong&gt;(1,6)&lt;/strong&gt; with reference to double bonds.&#xA;$$\begin{matrix}&#xA;\text{Better name:}&amp;amp; \text{(5}\textit{Z}\text{)-5-ethenylhepta-1,5-diene}\\&#xA;\text{Rather than:}&amp;amp; \text{(3}\textit{Z}\text{)-3-ethylidenehepta-1,6-diene}&#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;For Question 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If the chain has to be numbered and two substituents have the same locant the n you see the alphabetical order of the names of the groups/substituents and if they are also the same it only means that they are the same substituent and hence are written together with there locants separated by a &lt;code&gt;(comma),&lt;/code&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in you would name the below compound as &#xA;$\text{2,4-dichloropentane}$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uEA2h.gif&quot; alt=&quot;Example 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;For Question 3&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Example 1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIjd9.png&quot; alt=&quot;Example 4&quot;&gt;&#xA;The substituent in this example would be named $\ce{ethylidene}$. &#xA;$$&#xA;\begin{matrix}&#xA;\text{Final name:}&amp;amp; \text{5-ethylidenenona-1,8-diene}&#xA;\end{matrix}&#xA;$$&#xA;Similar substituents can be named as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|}&#xA;\hline&#xA;\ce{Substituent&amp;amp; Saturated\ form&amp;amp; Name\ of&amp;amp; Name\ of \\&#xA;&amp;amp;&amp;amp; saturated\ form&amp;amp; substituent \\&#xA;\hline&#xA;R=CH2&amp;amp; R-CH3&amp;amp; Methyl&amp;amp; Methylidene\\&#xA;\hline&#xA;R=CH-CH3&amp;amp; R-CH2-CH3&amp;amp; Ethyl&amp;amp; Ethylidene\\&#xA;\hline&#xA;R-CH=CH2&amp;amp; R-CH2-CH3&amp;amp; Ethyl&amp;amp; Ethenyl\\&#xA;\hline&#xA;R=CH-CH2-CH3&amp;amp; R-CH2-CH2-CH3&amp;amp; Propyl&amp;amp; Propylidene\\}&#xA;\hline&#xA;\begin{align}&#xA;\ce{R=C&amp;amp;-CH3\\&#xA;|&amp;amp; \\&#xA;C&amp;amp;H3}&#xA;\end{align}&amp;amp;   &#xA;\begin{array}{rl}&#xA;\ce{R-C&amp;amp;-CH3\\&#xA;|\ &amp;amp;\\&#xA;C&amp;amp;H3}&#xA;\end{array}&amp;amp; &#xA;\ce{Isopropyl&amp;amp; Isopropylidene}\\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorDisplayName="user7951" LastEditDate="2016-04-07T18:09:49.247" LastActivityDate="2016-04-07T18:09:49.247" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3947" PostTypeId="1" AcceptedAnswerId="3955" CreationDate="2013-01-18T05:54:26.430" Score="7" ViewCount="1630" Body="&lt;p&gt;Is chemical energy basically stored energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to define &quot;chemical energy&quot; while I know many physical formulas for energy e.g. potential, movement and other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is chemical energy an energy that is &quot;conserved&quot; and if so, why is chemical energy not just potential energy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If chemical energy is an &quot;own&quot; form of energy, why does it not have a specific equation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Did we have chemical energy before the big bang or when did it start?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you have chemical energy at or close enough to absolute zero?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much of the total energy in the universe is chemical energy?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the point of many different definitions of energy? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="22" LastEditDate="2013-01-18T07:51:23.443" LastActivityDate="2013-01-19T04:27:48.843" Title="What is the nature of chemical energy?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2017-07-19T20:48:16.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="3948" PostTypeId="2" ParentId="2529" CreationDate="2013-01-18T08:24:29.870" Score="5" Body="&lt;p&gt;As mentioned in the comments, &lt;em&gt;why not&lt;/em&gt;? A boron-boron bond will form as long as there is an increase in stability of the system. Forming a bond requires energy, but there is a lot of energy released once it gets formed. The feasibility of a bond depends which energy is greater.&lt;/p&gt;&#xA;&lt;p&gt;One example of where a boron-boron bond occurs is in &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_diboride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aluminium diboride&lt;/a&gt; / &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_diboride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnesium diboride&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hw0Ja.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, you have graphite-like &amp;quot;sheets&amp;quot; of boron (pink), with aluminium layers between them.&lt;/p&gt;&#xA;&lt;p&gt;This will probably be quite stable, due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Delocalized_electron&quot; rel=&quot;nofollow noreferrer&quot;&gt;delocalization&lt;/a&gt;. The singly-filled aluminium &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;-orbitals will overlap with the empty &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbitals of Boron, and will delocalize across the graphite-like sheet (I cannot confirm this, though). This will lead to a large increase in stability.&lt;/p&gt;&#xA;&lt;p&gt;A few more compounds with &lt;span class=&quot;math-container&quot;&gt;$\ce{B-B}$&lt;/span&gt; bonds are &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanthanum_hexaboride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lanthanum hexaboride&lt;/a&gt; and various &lt;a href=&quot;http://en.wikipedia.org/wiki/Yttrium_borides&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yttrium borides&lt;/a&gt;. These are probably very stable due to the crystal structure (though delocalization may be a factor as well)&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="17175" LastEditDate="2020-12-24T12:09:00.030" LastActivityDate="2020-12-24T12:09:00.030" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="3949" PostTypeId="1" AcceptedAnswerId="3960" CreationDate="2013-01-18T09:03:54.123" Score="13" ViewCount="23662" Body="&lt;p&gt;Watching some movie/TV this question came to mind, purely theoretical of course, if one wanted to dissolve a human corpse, like getting rid of a body after killing someone, is it better to use an acid or a base? In the movies (as well the popular perception for some reason) they always use acid. The movie I was watching they had a bathtub full of acid in which they throw in the body. But I also remember my chem teacher telling me (a long long time ago) that if she had a choice of an acid or a base in her eye, she would take acid any day because a base will dissolve organic tissue faster. So assuming that the acid and base have the same level of strength (say a pH of 1 and 13 respectively) which is better for getting rid of a corpse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1115" LastEditorUserId="16683" LastEditDate="2017-04-08T22:33:07.167" LastActivityDate="2017-04-08T22:33:07.167" Title="Dissolving Organic Tissues" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;chemical-biology&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ClosedDate="2017-04-09T15:07:30.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="3950" PostTypeId="2" ParentId="2900" CreationDate="2013-01-18T10:24:14.287" Score="4" Body="&lt;p&gt;For H$_2$ there are only two electrons and thus you can only have singlet (electrons with opposite spin projection) and triplet states (electrons with equal spin projection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin wave function of the singlet state is a linear combination (a superposition state) of the first electron with spin projection $\alpha$ and the second electron with spin projection $\beta$ and the opposite situation. Both spin states in the quantum superposition have opposite phase: &#xA;$$|\uparrow\downarrow\rangle -|\downarrow\uparrow \rangle$$&#xA;If the electrons are in the same molecular orbital then (by the Pauli principle) only the singlet state exists. If they are filling $\pi$ molecular orbitals then they could be in different degenerate orbitals (with angular momentum projection on the molecular axis $M_z = \pm \hbar$) and thus both singlet and triplet terms would exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three degenerate states belonging to the triplet term, that is, there are three wave functions that have equal energy. There spin components are:&#xA;$$|\uparrow \uparrow \rangle$$&#xA;$$|\downarrow \downarrow \rangle$$&#xA;$$|\uparrow \downarrow\rangle + | \downarrow\uparrow \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that it doesn't matter if you fill one molecular orbital with one electron in spin $\alpha$ and a different molecular orbital with the electron with $\alpha$ spin or you do so with both $\beta$ and $\beta$ spins. However you can not put both electrons with the same spin projection in the same molecular orbital. That is against the Pauli principle, that is, against the anti-symmetry of the electronic wave function.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2013-01-18T10:24:14.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3953" PostTypeId="2" ParentId="3945" CreationDate="2013-01-18T16:29:38.263" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are several methodics of assgning Electronegativity, each giving slightly different results. There is no strict border between ionic and covalent bound, and it usually have some weigh of covalent bonding too. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronegativity of atom in molecule changes in response to its neighborhood&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hydrogen without electrons will be naked proton. Because of its size it produces so strong electrostatic field, that it ruptures electronic shell of other atoms. So, hydrogen never forms positive ions, it is &lt;strong&gt;always&lt;/strong&gt; attached to some other atom(s) covalently. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastEditorUserId="22" LastEditDate="2013-01-18T17:22:00.087" LastActivityDate="2013-01-18T17:22:00.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3954" PostTypeId="1" AcceptedAnswerId="3956" CreationDate="2013-01-18T20:22:07.753" Score="2" ViewCount="2904" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch18/ch18-3-1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MECHANISM OF α-HALOGENATION OF ALDEHYDES AND KETONES&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBzCX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the third step. Bromide ion is picking up a base from protonated acetone. Doesn't that step mean that protonated acetone is stronger acid than Hydrobromic acid? Is that even possible?&lt;/p&gt;&#xA;" OwnerUserId="385" LastActivityDate="2013-01-18T21:01:10.933" Title="Can bromide ion ever act like a base?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3955" PostTypeId="2" ParentId="3947" CreationDate="2013-01-18T20:52:45.327" Score="7" Body="&lt;p&gt;Chemical energy is really nothing but electric potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial difference is that chemical energy involves energy that can be extracted by breaking/making bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we say that the reaction $\ce{2H2 + O2-&amp;gt;H2O}$ is accompanied by a release of energy. This energy is the difference of potential energies of the bonds in the reactants and products. Which potential energies? The bond potential energies. Bonds are, at the most basic level, formed due to attraction and repulsion of electrons and nuclei (coupled with some quantum mechanical effects). This carries some potential energy. The change in this potential energy achieved by breaking the bond is the &quot;bond energy&quot;, and forms part of your &quot;chemical energy&quot;. Some other parts are lattice energy, hydration energy etc, which are (in the end), due to electromagnetic forces/electromagnetic potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah, chemical energy is just a &quot;subset&quot; of potential energy,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If chemical energy is an &quot;own&quot; form of energy, why does it not have a specific equation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since chemical energy is an amalgamation of various types of electric potential energy in atoms, it cannot be given a formula (though you can calculate it for some compounds if you know some standard values). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is chemical energy an energy that is &quot;conserved&quot; and if so, why is chemical energy not just potential energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemical energy isn't really conserved anywhere unless there is no change to the chemical composition of the system. However, in a chemical reaction, most of the time any changes in energy are compensated for by absorption/release of heat. But it is just potential energy, as mentioned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Did we have chemical energy before the big bang or when did it start?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaningless question, there was no &quot;before&quot; the big bang. Also, energy really isn't something that you should classify into compartmentalized types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, since atoms formed a few minutes after the Big Bang, there wasn't anything to facilitate the existence of any chemical energy -- so the first occurrence of chemical energy was then.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you have chemical energy at or close enough to absolute zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yep. Bonds exist at absolute zero. Note that we can never reach absolute zero, but bonds exist both close to and hypothetically &quot;at&quot; absolute zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much of the total energy in the universe is chemical energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't answer that, and I doubt it can be answered to any reasonable approximation. I think the percentage would be rather low -- our universe is mainly stars (besides being empty space). The average star doesn't have many chemical bonds--it contains hydrogen/helium plasma, and plasma can't exactly form bonds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the point of many different definitions of energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In certain situations, you know that there is no method to facilitate conversion of one form of energy into another. In such a case, one can separately conserve that form of energy. So the forms of energy help in analyzing/solving a situation, but not much more.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-01-19T04:27:48.843" LastActivityDate="2013-01-19T04:27:48.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3956" PostTypeId="2" ParentId="3954" CreationDate="2013-01-18T21:01:10.933" Score="3" Body="&lt;p&gt;Come on, you can figure out this one on your own! How do you compare acidities? &lt;a href=&quot;http://cactus.dixie.edu/smblack/chem2310/summary_pages/pKa_chart.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's a helpful hint&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Armed with that information, I have to add something. I believe the mechanism as drawn above is not universal: it sure depends on the conditions and solvent used. Some &lt;a href=&quot;http://faculty.swosu.edu/william.kelly/pdf/ch22.pdf&quot; rel=&quot;nofollow&quot;&gt;other proposed mechanisms&lt;/a&gt; involve the enol, rather than the protonated ketone, which might be more plausible in mild conditions. In water, anyway, HBr is fully dissociated and it shouldn't be “reformed” like that. The proton will return to be solvated in the solution.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-01-18T21:01:10.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3957" PostTypeId="1" AcceptedAnswerId="3964" CreationDate="2013-01-18T21:12:38.150" Score="6" ViewCount="255" Body="&lt;p&gt;I was thinking back to my chemistry major days and remembering the bunch of cool spectra we produced by various means (HNMR, IR, UV/Vis) and I remember all the cool 'self-interference' like couplets, triplets doublets, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is basically motivated by the complexity of spectra, and the reductionist instinct to 'break it down' into simpler components. The FFT or DFT is known to separate out some periodic components of signals, as are SVD or tensor decompositions known to isolate independent components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am completely not an expert in any of this but I am interested to hear from any coal-face chemists if they regularly use some 'spectra decomposition' method to aid in interpretation or analysis. I remember being taught how to read various peaks in spectra and analyze them, but I can't help thinking, for some very complex molecules, surely it gets pretty hard. And what if, we don't actually know the structure beforehand? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given that Fourier transform is used in NMR, my question is : is there ever a need to take a spectrum and do a computer analysis of it that isn't just about matching peaks to a database of known signals? What other nifty things can be done with a spectrum once it is obtained?&lt;/p&gt;&#xA;" OwnerUserId="479" LastEditorUserId="479" LastEditDate="2013-01-19T12:07:51.417" LastActivityDate="2013-01-19T12:33:13.533" Title="Fourier transform for spectroscopy spectra?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3958" PostTypeId="1" CreationDate="2013-01-18T22:05:53.220" Score="5" ViewCount="7478" Body="&lt;p&gt;I've noticed that when adding lemon juice to carbonated water, large 'soapy' bubbles appear on the surface. Since I'm adding another acid to a solution of carbonic acid, I would maybe expect a faster release of $\ce{CO2}$, but the bubbles I'm seeing do not resemble the usual fizzy drink bubbles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on?&lt;/p&gt;&#xA;" OwnerUserId="1099" LastEditorUserId="73" LastEditDate="2013-01-19T00:31:21.180" LastActivityDate="2020-05-28T16:44:41.183" Title="Does citric acid react with carbonated water?" Tags="&lt;everyday-chemistry&gt;&lt;acid-base&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3959" PostTypeId="1" CreationDate="2013-01-18T23:27:00.173" Score="21" ViewCount="6599" Body="&lt;p&gt;I've read entire &lt;em&gt;Chapter 14: Organometallic Compounds&lt;/em&gt; of  Francis Carey's &quot;Organic Chemistry&quot; but I still didn't get an answer to my question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote from the book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of their basicity &lt;em&gt;organolithium compounds&lt;/em&gt; and &lt;em&gt;Grignard reagents&lt;/em&gt; cannot be prepared or used in the presence of any material that bears a &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; hydroxyl group. Nor are these reagents compatible with &lt;span class=&quot;math-container&quot;&gt;$\ce{-NH}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{-SH}$&lt;/span&gt; groups, which can also convert an organolithium or organomagnesium compound to a hydrocarbon by proton transfer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Fine. No arguments with that. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Terminal alkynes also &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/addyne1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;give alkanes with Organolithium and Organomagnesium compounds&lt;/a&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Must be true if he says so. Now coming to the point,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Ammonia and Amines have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; around 36.&lt;/li&gt;&#xA;  &lt;li&gt;Terminal alkynes have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of 25.&lt;/li&gt;&#xA;  &lt;li&gt;Aldehyde &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 17$&lt;/span&gt;, Ketone &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 19$&lt;/span&gt; and an Ester &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 25$&lt;/span&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Water &amp;amp; Alcohols have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of 15–18.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;organolithium compounds&lt;/em&gt; and &lt;em&gt;Grignard reagents&lt;/em&gt; act like bases with alcohols, terminal alkynes, and amines/ammonia, why do they act like nucleophlies and give addition reaction with carbonyl compounds? Why don't they act as bases and give alkanes instead? What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="61578" LastEditDate="2019-04-20T02:34:00.663" LastActivityDate="2019-04-20T02:34:00.663" Title="Why do Organolithium or Grignard reagents act as nucleophiles and not as bases with aldehydes and ketones" Tags="&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;&lt;amines&gt;&lt;organometallic-compounds&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="3960" PostTypeId="2" ParentId="3949" CreationDate="2013-01-19T04:52:33.553" Score="11" Body="&lt;p&gt;I would say that acids are better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Bases and acids dissolve flesh pretty easily (I'm thinking of $\ce{NaOH}$ and $\ce{HCl}$ here, weaker acids/bases -- not so much). Most of our body is made up of proteins, which are overall &lt;em&gt;pretty much&lt;/em&gt; neutral (generally). Both acids and bases can dissolve protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there is one major component of our body which is easier to dissolve with acid. Bones. The major component of bone is $\ce{CaCO3}$. The reaction $\ce{2HCl +CaCO3 -&amp;gt; H2O + CO2 ^ + CaCl2}$ is much more efficient/faster than $\ce{2NaOH + CaCO3 -&amp;gt; Na2CO3 + Ca(OH)2}$. Why is this? Well, for one, in the acid reaction, a gas is being released. This speeds up the reaction, by &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot;&gt;Le Châtelier's principle&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, NaOH &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hydroxide#Tissue_digestion&quot;&gt;is not able to dissolve bone completely&lt;/a&gt;, you are left with a fragile shell. Either way, it will be much slower than $\ce{HCl}$ while dissolving bone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://the7thwreck.wordpress.com/2005/12/25/why-not-to-use-quicklime-to-dispose-of-a-corpse/&quot;&gt;Here's&lt;/a&gt; an interesting article (no claims on validity) of why lime shouldn't be used to dissolve a corpse.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While (as far as I can tell) acids may be the better way to dissolve a body, I feel that filmmakers use acids because the term is known to most of the general public (unlike &quot;bases&quot;, which isn't)&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-01-19T04:52:33.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3961" PostTypeId="1" CreationDate="2013-01-19T10:23:08.020" Score="8" ViewCount="158" Body="&lt;p&gt;How can one find whether or not there is enough oxygen for breathing present inside a (dry) well of diameter 1 m at a depth of 50 m?                    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people in Nepal are killed while cleaning wells because of a lack of oxygen.&lt;/p&gt;&#xA;" OwnerUserId="1123" LastEditorUserId="16683" LastEditDate="2017-03-28T12:53:39.597" LastActivityDate="2017-03-28T12:53:39.597" Title="How can I find whether there is oxygen present inside a well of diameter 1 m at 50 m's deep?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3962" PostTypeId="2" ParentId="3961" CreationDate="2013-01-19T11:35:07.930" Score="7" Body="&lt;p&gt;The ideal solution would be a single gas personal detector, such as &lt;a href=&quot;http://catalog.enmet.com/compare/l-detector-for-co-h-sub-2-sub-s-o-sub-2-sub-or-l-2/l-detector-for-co-h-sub-2-sub-s-o-sub-2-sub-or-lel?forward=1&amp;amp;itemids=1002+1003+1004+1005+&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; (it's only the first example I found, I don't know this brand, but there are many equivalent apparatuses from other vendors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While digital personal detectors are expensive ($200+ range), there are simpler systems which also provide some basic safety (better than nothing). The &lt;a href=&quot;http://en.wikipedia.org/wiki/Davy_lamp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Davy lamp&lt;/a&gt; is perhaps the best known example of these, and it was a huge breakthrough in mine safety when it was popularized after 1815.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsGIl.jpg&quot; alt=&quot;Davy lamp&quot;&gt;&lt;br&gt;&#xA;(Source: &lt;a href=&quot;http://lindal-in-furness.co.uk/MinersLamps/MinersLamp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lindal-in-furness.co.uk/MinersLamps/MinersLamp.jpg&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2017-03-28T12:41:53.907" LastActivityDate="2017-03-28T12:41:53.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3963" PostTypeId="2" ParentId="3958" CreationDate="2013-01-19T12:02:02.363" Score="3" Body="&lt;p&gt;The chemistry you're describing the the one you expect: CO&lt;sub&gt;2&lt;/sub&gt; is formed, and released from the liquid. However, your question is more about &lt;em&gt;kinetics&lt;/em&gt; than about the final product. We can assume that slow release of gas leads to larger bubbles, because nucleation rate for the bubble formation is lower. Now, in fizzy drink, there is a large quantity of carbon dioxide accommodated by the liquid due to pressurization. In your case, it will depend on the concentration of your lemon juice (natural lemon juice has pH ~ 2), and more importantly on that of the carbonated water: I just tried with a few grams of sodium bicarbonate dissolved in a glass of water in my kitchen, and the kinetics and bubble size really depends on initial carbonate concentration. (I haven't been able to make pictures or video, because I had to handle my kids who were very fast drawn to this impromptu experiment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I second Richard's comment: there are many additional chemicals in lemon juice, some of which surely act as surfactants… which changes bubble formation and kinetics.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-01-19T12:02:02.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3964" PostTypeId="2" ParentId="3957" CreationDate="2013-01-19T12:33:13.533" Score="5" Body="&lt;p&gt;As Mad Scientist said, many spectroscopy methods are already based on Fourier transform of a time-dependent observable. (This is not true of the basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrophotometry&quot; rel=&quot;noreferrer&quot;&gt;spectrophotometry&lt;/a&gt; measurements, where you directly measure the transmittance as a function of frequency.) NMR is a method based on Fourier transform, as is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform_infrared_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;FTIR&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding NMR signal processing, it's a very wide field, which includes interesting research ranging from academic interest to software development. So, it's hard to answer your question other than saying “yes, of course, further processing of NMR signals is sometime possible”. One thing you have to realize is: while you have probably learnt &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_NMR&quot; rel=&quot;noreferrer&quot;&gt;proton NMR&lt;/a&gt; during your studies, this is only one NMR technique, and the simplest of them all. &lt;strong&gt;There are plenty of different NMR techniques&lt;/strong&gt;, depending on the type of excitation and sequence of RF pulses, many of them being multidimensional. Each method can use a variety of signal processing techniques.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-01-19T12:33:13.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3965" PostTypeId="1" AcceptedAnswerId="9311" CreationDate="2013-01-19T13:20:25.870" Score="4" ViewCount="145" Body="&lt;p&gt;I would like to know how you go about synthesising glycosides of flavonoids and stilbenoids from their respective sugars and flavonoids/stilbenoids. I read &lt;a href=&quot;http://www.netsci-journal.com/97v1/97005/005p20.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.netsci-journal.com/97v1/97005/005p20.htm&lt;/a&gt; but I found it too specific (as it only applied to aliphatic compounds) and required an enzymatic catalyst that is impossible to predict. I would like a general formula that if it does contain an enzymatic catalyst the enzyme is either constant or at least usually predictable.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="418" LastEditDate="2014-03-19T09:12:56.697" LastActivityDate="2014-06-17T10:55:46.057" Title="How do you Synthesise Glycosides of flavonoids and stilbenoids?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3966" PostTypeId="1" AcceptedAnswerId="3967" CreationDate="2013-01-19T14:16:04.433" Score="3" ViewCount="369" Body="&lt;p&gt;How would you go about synthesising amino sugars from sugars? e.g. How would you synthesise the compound below from D-glucose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GySO2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would adding ammonia ($\ce{NH3}$) to it work?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4325" LastEditDate="2015-02-25T16:26:12.850" LastActivityDate="2015-02-25T16:26:12.850" Title="How would you go about synthesising amino sugars from sugars?" Tags="&lt;synthesis&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3967" PostTypeId="2" ParentId="3966" CreationDate="2013-01-19T14:41:15.000" Score="3" Body="&lt;p&gt;First, the anwer is &lt;em&gt;“Would glucose and ammonia work?”&lt;/em&gt; is: &lt;strong&gt;no, definitely not&lt;/strong&gt;. The $\ce{OH}$ group is a poor leaving group, and alcohol + ammonia definitely won't yield an amine. Moreover, in your case, you have many hydroxyl groups and you only want one transformed into an amine, which is tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how the synthesis of the above compound can be achieved, but I know one way to help you start a literature search: &lt;strong&gt;your expected compound is called an &lt;em&gt;amino sugar&lt;/em&gt;&lt;/strong&gt;. In this particular case, it's the &lt;strong&gt;6-amino-&lt;em&gt;D&lt;/em&gt;-glucose&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-25T15:51:26.500" LastActivityDate="2015-02-25T15:51:26.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3968" PostTypeId="1" AcceptedAnswerId="3982" CreationDate="2013-01-19T14:55:58.003" Score="3" ViewCount="230" Body="&lt;p&gt;If it is possible, how would you estimate a compound's water solubility from data like its partition coefficient, $ \log P $, its dissociation constant, $pK_a $, its molar mass, etc.?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-01-19T16:05:04.760" LastActivityDate="2013-02-21T17:25:22.383" Title="Is it possible to estimate a substance's water solubility from data like the substance's partition coefficient, dissociation constants, etc.?" Tags="&lt;solvents&gt;&lt;theoretical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3969" PostTypeId="1" AcceptedAnswerId="4241" CreationDate="2013-01-19T15:22:24.580" Score="1" ViewCount="204" Body="&lt;p&gt;The chemical structure of the compounds in question are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZBy9N.png&quot; alt=&quot;Hyperforin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 1: Hyperforin&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xThuE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 2: Tetrahydrohyperforin&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MY3mu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 3: Octahydrohyperforin&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would heating hyperforin in the presence of water manage to saturate the required bonds and thus allow one to synthesise these compounds?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="22" LastEditDate="2013-02-25T05:21:58.227" LastActivityDate="2013-02-25T05:21:58.227" Title="How would you synthesise tetrahydrohyperforin and octahydrohyperforin from hyperforin?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3971" PostTypeId="2" ParentId="2706" CreationDate="2013-01-20T04:27:44.040" Score="-1" Body="&lt;p&gt;If you want &lt;strong&gt;symmetrical ethers&lt;/strong&gt; you could do the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce carboxylic acid to alcohol using &lt;strong&gt;Lithium aluminium hydride&lt;/strong&gt; and then perform bimolecular dehydration on it using &lt;strong&gt;concentrated sulfuric&lt;/strong&gt; acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\ce{R-COOH&amp;amp; -&amp;gt;[\large\ce{LiAlH4}]&amp;amp; R-OH\\&#xA;Carboxylic\  acid&amp;amp; &amp;amp; Alcohol\\&#xA;2\ R-OH&amp;amp; -&amp;gt;[\large\ce{conc. H2SO4}]&amp;amp; R-O-R\\&#xA;Alcohol&amp;amp; &amp;amp; Symmetrical\\ &amp;amp; &amp;amp; Ether} &#xA;\end{matrix}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;asymmetrical ethers&lt;/strong&gt; yo could d o the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduce carboxylic acid of both the constituent alkyl groups of ether using above method.&lt;br&gt;&#xA;React one of the alcohols with Sodium metal to form Sodium alkoxide, and then make the alkoxide react with the other alcohol to form ether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{matrix}&#xA;\ce{R-COOH&amp;amp; -&amp;gt;[\large\ce{LiAlH4}]&amp;amp; R-OH&amp;amp; (reduction)\\&#xA;R'-COOH&amp;amp; -&amp;gt;[\large\ce{LiAlH4}]&amp;amp; R'-OH&amp;amp; (reduction)\\&#xA;\hline&#xA;\\&#xA;R-OH&amp;amp; -&amp;gt;[\ce{Na}]&amp;amp; R-O- Na+&amp;amp; (generation\ of\ nucleophile)\\&#xA;Alcohol&amp;amp; &amp;amp; Alkoxide&amp;amp;\\&#xA;\hline&#xA;\\&#xA;R'-OH\ +\ R-O- Na+&amp;amp; -&amp;gt;&amp;amp; R-O-R'&amp;amp; (attack\ of\ nucleophile)\\ &#xA;Alcohol\qquad Alkoxide&amp;amp; &amp;amp; Asymmetric\ Ether&amp;amp; &#xA;}&#xA;\end{matrix}$$&#xA;&lt;sub&gt;NOTE: This is only from a theoretical point of view&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="897" LastEditorUserId="402" LastEditDate="2017-04-08T22:16:56.840" LastActivityDate="2017-04-08T22:16:56.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3972" PostTypeId="1" AcceptedAnswerId="3978" CreationDate="2013-01-20T08:18:46.537" Score="20" ViewCount="1668" Body="&lt;p&gt;I'm tutoring few students for Chemistry. During the course, I many times use the phrase &quot;Out of these possibilities this one is stable. So it is formed&quot;. I get questions like, &quot;How do the molecules know?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly I don't know. How do the molecules know the future? How do they proceed to form &quot;Most Stable&quot; compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross Post:&lt;/strong&gt; &lt;a href=&quot;https://physics.stackexchange.com/questions/51894/how-do-the-molecules-of-reacting-compounds-proceed-to-form-most-stable-molecul&quot;&gt;https://physics.stackexchange.com/questions/51894/how-do-the-molecules-of-reacting-compounds-proceed-to-form-most-stable-molecul&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="385" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-10-02T16:29:23.570" Title="How do the molecules know the future? How do they proceed to form &quot;Most Stable&quot; compound?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;molecules&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3974" PostTypeId="2" ParentId="3972" CreationDate="2013-01-20T18:20:22.333" Score="13" Body="&lt;p&gt;Clearly, molecules are not in any sense volitional agents. They're simply subject to the basic and immutable laws of thermodynamics, the second of which states that any isolated system always spontaneously develops toward a state of equilibrium in which entropy (which can be informally described as &quot;disorder&quot; or &quot;randomness,&quot; meaning thermal energy distributed in such a way that it cannot be used to perform useful work) is maximized. Increased stability is really another term for the minimization of free energy, i.e., energy available to perform useful work. All irreversible spontaneous chemical reactions are ones in which the free energy change is negative, either because the chemical reagents involved experience an increase in entropy, or because the thermal energy released by the reaction sufficiently increases the entropy of the surroundings. The basic, simplified equation for describing this situation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\Delta G\ =\ \Delta H\ -\ T\Delta S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\small\ce{\Delta G}$ is the change in free energy, $\small\ce{\Delta H}$ is enthalpy (change in heat energy), $\small\ce{T}$ is absolute temperature, and $\small\ce{\Delta S}$ is the change in entropy. If $\small\ce{\Delta G}$ is negative for a particular reaction at a given temperature, the reaction will proceed spontaneously. In essence, the entire universe is evolving toward a state in which entropy is maximized and free energy is minimized, and only chemical reactions which contribute to that phenomenon can spontaneously take place.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="897" LastEditDate="2013-01-21T03:29:54.033" LastActivityDate="2013-01-21T03:29:54.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3975" PostTypeId="1" AcceptedAnswerId="3976" CreationDate="2013-01-20T21:29:39.097" Score="8" ViewCount="8062" Body="&lt;p&gt;I'm new to Chemistry and in my textbook, it describes the &lt;em&gt;Law of Definite Proportions&lt;/em&gt; (aka Proust's Law) and then goes on to describe the Law of Multiple Proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example they give is carbon monoxide and carbon dioxide, where the mass ratio of oxygen to carbon in carbon dioxide is $2.67:1$ and the mass ratio of oxygen to carbon in carbon monoxide is $1.33:1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the above example, and I understand that when you compare them in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2.67~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon dioxide)$/ 1.33~\mathrm{g}$ (proportion of oxygen to $1~\mathrm{g}$ carbon in carbon monoxide) $= 2.00$, a small whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{Mass oxygen to 1 g carbon in carbon dioxide}}{\text{Mass oxygen to 1 g carbon in carbon monoxide}} = \frac{2.67~\mathrm{g}}{1.33~\mathrm{g}} = 2.00$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook doesn't describe what a small whole number is, or what it means … just that, in this example, $2.00$ is a small whole number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the significance of the $2.00$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that carbon dioxide has double the oxygen. Is this where the $2.00$ plays a role?&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="7475" LastEditDate="2016-12-14T00:30:24.793" LastActivityDate="2019-09-04T21:18:26.357" Title="Law of Multiple Proportions: What is the significance of small whole numbers?" Tags="&lt;stoichiometry&gt;&lt;molecules&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3976" PostTypeId="2" ParentId="3975" CreationDate="2013-01-20T22:37:54.017" Score="10" Body="&lt;p&gt;There's no universally accepted hard cut-off, as far as I'm aware, which determines the range of numbers that would constitute &quot;small&quot; mass ratios. Generally, all textbook examples select compounds that yield single-digit mass ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The significance of the ratios being small (as well as their constancy and the limited number of different mass ratios for binary compounds actually found in nature) is that they allow for reliable conclusions to be made about the stoichiometric proportions in which elements combine based on crude empirical data alone, as well as leading to conjectures about the number of possible configurations, bond arrangements, and oxidation states yielding said combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law is valid for a large variety of binary compounds simply by virtue of the fact that there is typically a limited number of permutations in which two given elements can be combined to yield a stable compound. For example, C and O have only two, N and O have only four, metals have a limited number of oxidation states, therefore giving rise to relatively few possibly formulas for binary ionic compounds, etc. The law breaks down for cases in which two elements can combine to form long chains or complex molecules (e.g., large hydrocarbons and [mostly hypothetical] silicon analogues), when the mass ratios can sometimes become quite large.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="402" LastEditDate="2018-12-13T03:41:19.610" LastActivityDate="2018-12-13T03:41:19.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3977" PostTypeId="1" CreationDate="2013-01-21T09:47:10.300" Score="3" ViewCount="86" Body="&lt;p&gt;I'm looking into building an absorption chiller as a decidedly non-professional DIY project. For safety reasons and because temps. &gt; 0°C are sufficient, I won't be using ammonia but some saline solution as absorber and water as cooling agent.&#xA;The obvious choice for the salt would be LiBr, as this is used in comercial installations. However, I'm not sure I can procure it (chemicals suppliers don't sell to private persons in germany) and it's not cheap (~ 30€/100g).&#xA;What other salts are strongly hydroscopic and commonly availible?&#xA;Alternativly, where can I find a concise list with the physical properties of many common (=household) salts?&lt;/p&gt;&#xA;" OwnerUserId="1069" LastEditorUserId="1069" LastEditDate="2013-01-22T12:00:51.823" LastActivityDate="2013-01-23T15:31:55.517" Title="What commonly availible salts are strong dessicants?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3978" PostTypeId="2" ParentId="3972" CreationDate="2013-01-22T01:34:00.980" Score="17" Body="&lt;p&gt;Chemical reactions occur because the thermal energy in a substance makes molecules bump into each other, and each time they bump into each other there is some chance for a reaction to take place.  This is an oversimplified model, but it suffices to get some intuition for what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecules of course do not &quot;know&quot; that some configuration is more stable than others.  What happens is that the molecules react randomly, again and again; they take a sort of random walk through the available configurations.  Every now and then, a molecule will purely randomly find its way into the stable state.  Once it's there, it's &quot;stuck&quot; - it has a much smaller chance of getting out of the stable state.  That's more or less what it means for a state to be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over time, more and more molecules will find their way into the stable state and get stuck, so the result looks as if the molecules &quot;knew&quot; which state to go to and just went straight there.  But actually they only got there through the process of randomly bouncing around the configuration space, plus the stable state acting as a trap that molecules can enter but not (as easily) leave.&lt;/p&gt;&#xA;" OwnerUserId="1075" LastActivityDate="2013-01-22T01:34:00.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3979" PostTypeId="2" ParentId="3975" CreationDate="2013-01-22T01:48:45.383" Score="11" Body="&lt;p&gt;The law of multiple proportions is largely obsolete these days because we all believe that atoms exist.  Prior to about 1900 this and similar laws were used to show that, for example, the ratio of oxygen in $\ce{CO}$ to oxygen in $\ce{CO2}$ is 2:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that's trivially obvious when we write the formulas, but before we believed in atoms the &quot;small whole number&quot; ratios were used as an example of an experimental situation (the 2:1 ratio) could be explained by atoms (i.e. quantizing atomic masses.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, the law causes more confusion than it cures.  Atoms are here to stay.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="402" LastEditDate="2017-04-02T19:55:55.287" LastActivityDate="2017-04-02T19:55:55.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3982" PostTypeId="2" ParentId="3968" CreationDate="2013-01-22T17:20:09.940" Score="1" Body="&lt;p&gt;It could certainly be &lt;em&gt;estimated&lt;/em&gt; from those properties, the structure, etc.  A lot of companies have developed proprietary algorithms for things like that:  solubility, diffusion coefficient, polarity.... (For example of a company, see &quot;ChemAxon&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water solubility for many chemicals is already known, though.  Search for a chemical &lt;a href=&quot;https://www.chemspider.com/&quot; rel=&quot;nofollow&quot;&gt;ChemSpider&lt;/a&gt;, and look under the &quot;Properties&quot; tag.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They've even got an API, in case you're interested in using water solubility in an app.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2013-01-22T17:20:09.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3983" PostTypeId="1" AcceptedAnswerId="3984" CreationDate="2013-01-22T18:30:23.070" Score="2" ViewCount="1381" Body="&lt;p&gt;When an electron absorbs energy and is in a higher energy orbit (I guess the atom would be in an unstable state), when the electron releases this energy, would all of this energy be released at the same time, or could it release energy one &quot;level&quot; at a time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example if a Hydrogen atom absorbed a lot of energy and its electron went to some level n = 5, could it slowly make its way back down to level 1?&lt;/p&gt;&#xA;" OwnerUserId="1119" LastActivityDate="2016-07-17T19:25:02.450" Title="Does an &quot;excited atom&quot; lose energy gradually?" Tags="&lt;electrons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3984" PostTypeId="2" ParentId="3983" CreationDate="2013-01-22T19:37:25.303" Score="6" Body="&lt;p&gt;Yes, it certainly can. In fact, it can do either, traverse gradually across excitation states down to its ground state, or jump from its current state to its ground state, or to any lower state for that matter. So essentially, as long as the target state is equal or lower in energy than the current one the electron can jump to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those jumps have different probabilities though. Determining which process is more likely to occur, and what the probabilities are is a matter of quantum mechanic calculations. Have a look here for example: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/fermi.html&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/fermi.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-01-22T19:37:25.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3985" PostTypeId="1" AcceptedAnswerId="3987" CreationDate="2013-01-22T20:46:01.453" Score="2" ViewCount="104" Body="&lt;p&gt;Why is there a max radius for an orbit of an electron around a nucleus ?&#xA;I had a course in electromagnetism but I do not get this.&lt;/p&gt;&#xA;" OwnerUserId="633" LastActivityDate="2013-01-23T03:14:18.567" Title="Why is there a max radius for an orbit of an electron around a nucleus?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3986" PostTypeId="1" CreationDate="2013-01-23T01:29:41.753" Score="5" ViewCount="165" Body="&lt;p&gt;What is needed to  measure distance between two electrodes imersed in seawater ? &#xA;I have  tried measuring the voltage across the path (a suposed resistance) with a constant current source (about 10uA) passing the water to GND of circuit. (the resistance is suposed proportional to distance of electrodes) The probes are 100Hz reversing switched to avoid polarization. And I am using AG/ AGCl pellets as electrodoes. But until now I could not have stable results.&#xA;What may be wrong ?  Could someone help ?&#xA;Thanks &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The AgAgCl sinterized disks was choose because it is very used as reference electrode for seawater potential measurements (cathodic protection) with very good results and low maintenance, but I really dont know if it is the the best for my application. The suface area is about 1 cm2 then we are expecting about 20 Ohms per cm of distance in seawater. With a 10uA as (AC square wave) current source  between electrodes voltage will range from about few milivolts up to 1 volt at 50m.  As you see the potential is bellow of water electrolisys. What you mean &quot;with a lot of corrosion protection&quot; for Stainless Steel plates ? &#xA;I understand that salinity, temperature etc could vary in the media but these factors are not expected  to change so quickly and I also intend to use a known distance sensors for reference as well averaging the measures.&#xA;I intend use this sistem in a dynamic environment where the reletive speed between sensors or the flow may be about 10 cm / sec. &#xA;What you think about the method ? There are a better one ?&lt;/p&gt;&#xA;" OwnerUserId="1130" LastEditorUserId="22" LastEditDate="2013-01-23T15:01:55.860" LastActivityDate="2013-01-24T23:37:04.700" Title="Measuring distance in a conductive liquid" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3987" PostTypeId="2" ParentId="3985" CreationDate="2013-01-23T03:14:18.567" Score="3" Body="&lt;p&gt;Perhaps you can clarify your question a bit, but if you are referring to the size of the radial distribution of electron density, this increases with increasing principal quantum number, up to the ionisation limit, at which point the spacing between energy levels trends towards 0 and the interaction energy trends towards zero. After this, the electron is for all intents and purposes unbound. This is illustrated for the hydrogen atom in the Lyman series (image borrowed from &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyman_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j4DeL.png&quot; alt=&quot;Lyman series, from wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the largest transition $n=\infty \rightarrow n=1$ is 13.6 eV, or half a Hartree (this is the energy of a photon with a wavelength of 912 Ångström). This is the binding energy of the hydrogen atom. Even the Lyman gamma line - representing a $n=4 \rightarrow n=1$ transition - is quite close to the ionisation limit. As n increases, from a strictly electrostatic point of view the orbital becomes far less stable, and this is reflected in the general negative trend of ionisation energies across the periodic table, however the complete picture is a bit more complex than that because electrons can interact with electrons as well as the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is that as n increases, the energy required lose an electron entirely becomes smaller and smaller, and this fixes a practical limit on where we can say the electron is actually bound. This has a caveat though - cold gasses can form unusually long-lived &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_atom&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg states&lt;/a&gt; with &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;ridiculous&lt;/a&gt; principal quantum numbers with correspondingly enormous orbital radii (up to micrometers!*) and semiclassical electron orbits. The reasons for the stability of these states is outside of my knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;**To paraphrase, Alisa Bokulich points out in 'Re-examining the Quantum-classical Relation' that Rydberg atoms are of comparable size to grains of (fine) sand, which is pretty neat*&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-23T03:14:18.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3988" PostTypeId="1" CreationDate="2013-01-23T04:59:14.813" Score="2" ViewCount="801" Body="&lt;p&gt;I just &lt;a href=&quot;http://mattson.creighton.edu/H2-Uptake.pdf&quot; rel=&quot;nofollow&quot;&gt;read&lt;/a&gt; that saturated palladium hydride, $\ce{PdH}$&lt;sub&gt;0.7&lt;/sub&gt;, when exposed to oxygen will generate heat and water on it's surface, $\ce{4 H}$&lt;sub&gt;(Pd)&lt;/sub&gt; $\ce{+O2}$&lt;sub&gt;(g)&lt;/sub&gt; $\ce{\to 2 H2O}$&lt;sub&gt;(g, l)&lt;/sub&gt;. &#xA;What is going on? &lt;/p&gt;&#xA;" OwnerUserId="755" LastActivityDate="2016-03-13T19:57:59.443" Title="Palladium hydride forming water and heat at room temperature" Tags="&lt;reaction-mechanism&gt;&lt;hydrogen&gt;&lt;surface-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3989" PostTypeId="2" ParentId="3988" CreationDate="2013-01-23T04:59:14.813" Score="1" Body="&lt;p&gt;I'm guessing that when the $\ce{H}$ atoms leave the palladium are they still monatomic for an instant, but without the $\Delta H^0_f = 436 kJ/mol$ extra energy that free  $\ce{H}$ atoms would normally have, as bypassed by the $\ce Pd$ somehow. Most of the $\ce{H}$ atoms simply fall in step with one another, reforming $\ce{H2}$ without a peep. But some latch onto nearby $\ce{O2}$ molecules, disassociating them and forming $\ce{H2O}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat generated is the difference between heat required to dissociate the $\ce{O2}$ and the heat generated by the formation of $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 \to 2 O}$   $\Delta H^o_f = 247.5 kJ/mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ H2 + 1/2 O2 \to  H2O}$ $\Delta H^o_f = -285.83 kJ/mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$∴ \ce{2H_{(Pd)} + 1/2O2_{(g)} -&amp;gt; H2O_{(g, l)}} \quad \Delta H^o_{\mathrm{f}} = -38.33 \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being as this is only$\ 5.68 kJ/mol$ short of the heat required to vaporise $\ce{H2O}$, it explains why some of the product would be gas and some liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a complete guess, am I hilariously wrong here?&lt;/p&gt;&#xA;" OwnerUserId="755" LastEditorUserId="24052" LastEditDate="2016-03-13T19:57:59.443" LastActivityDate="2016-03-13T19:57:59.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3990" PostTypeId="1" AcceptedAnswerId="3991" CreationDate="2013-01-23T05:25:30.623" Score="3" ViewCount="641" Body="&lt;p&gt;So I have been freezing ice a lot recently, and I will come out and say it, as it turns out it is, albeit unlikely, it is sometimes possible for ice to &quot;explode.&quot; I was not so aware of this before, but from now on I consume my ice beverages with deliberate care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first action I have taken is to make sure no bubbles are visibly present before freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about how one might reduce these frozen bubbles when making ice with ice trays and a freezer (the bubbles are sometimes a few millimeters across). I was thinking perhaps the bubbles are more likely to form if there are more dissolved minerals and other compounds commonly found in tap water. I think it is time to review osmosis to see if anything will shed light on this idea. My first instinct was to distill water and freeze it to see if there would be less bubbles, but since there will always be dissolved gases even in a freezer it does not follow that this will greatly affect anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the case that at lower temperatures water will not be able to hold as many dissolved gases? If that is the case, then it might be worthwhile to lower the temperature of ice in the fridge or to a temperature just above freezing and allow the solution to stabilize, and then put it in the freezer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your insight is appreciated, thank you.&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2013-01-23T09:04:31.647" Title="How might one reduce the concentration of bubbles in their ice, from ice cube trays?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;osmosis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3991" PostTypeId="2" ParentId="3990" CreationDate="2013-01-23T06:11:45.877" Score="5" Body="&lt;p&gt;IIRC, the solubility of virtually all gases in water decreases with temperature. This is the reason why heating water will result in bubble nucleation, and this can be exploited to produce very clear ice by boiling water, perhaps several times, before putting it into the trays.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-23T06:11:45.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3992" PostTypeId="2" ParentId="3990" CreationDate="2013-01-23T08:26:48.523" Score="3" Body="&lt;p&gt;To minimize the air bubbles in ice, either 1) hold the water under a vacuum or 2) 'degas' the water with an ultrasonic cleaner before freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Holding the water under a vacuum will pull the dissolved gases out of solution.  The will diffuse to the air-water surface, and then be vacuumed away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) An ultrasonic cleaner will vibrate the liquid at ultrasonic frequencies.  Vibrating water-gas mixtures at the right frequency will drive out dissolved gases in much the same way that an earthquake will drive ground water out of water-saturated soil (ie liquifaction).  Many such ultrasonic cleaners even have a 'degas' setting for just such a purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also:&lt;br&gt;&#xA;* Water will freeze as soon as it dips below 0 degrees C (assuming it's pure water, it's not being agitated, and it's at standard atmospheric pressure).  So, cooling the ice at a lower temperature freezer will make the ice form quicker.  Thermodynamically then, it's creating a greater temperature difference between the air-water interface and the center of the water, so the heat flow out of the ice will be greater, and water will reach 0 degrees C sooner.&#xA;* Gases do not &lt;em&gt;immediately&lt;/em&gt; dissolve into water.  Assuming you had water with absolutely no dissolved gas it takes some time for the gas atoms/molecules at the air-water interface to dissolve into the center of the water.  -- So I guess this means, assuming you degas the water, forming the ice in a lower temperature freezer will help, too.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditDate="2013-01-23T09:04:31.647" LastActivityDate="2013-01-23T09:04:31.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3993" PostTypeId="2" ParentId="3986" CreationDate="2013-01-23T08:58:08.710" Score="3" Body="&lt;p&gt;(Initial, preliminary answer, here.)&#xA;There are several possible reasons for your unstable results:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Possibility 1:  Wrong electrodes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ag|AgCl pellets seem like an odd choice of electrodes for this application.  What are the reasons for using them as electrodes?  What is their surface area?  Their complex shape means the electrical current maybe isn't uniform between the two electrodes.  Also, the fact that you're drawing current through an electrode with an AgCl coating may mean the coating is building-up/breaking-down (which could mess up your measurements) ...  In your current application, it seems like you'd want to use two, big, parallel stainless steel plates (with a lot of corrosion protection!) as your electrodes.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Possibility 2:  Changing salinity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Depending on the local conditions, the conductivity of the seawater may be changing.  For example, if you are by a river delta, the tides may be washing fresh water between your plates.  Or, if the electrodes are at the water surface in really sunny location, day-time evaporation may increase salinity ...  If either of those are the case, install a second set of plates with a &lt;em&gt;known, fixed electrode spacing&lt;/em&gt; to measure salinity, and use that to modify the results from the original set of electrodes.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Possibility 3:  Convection Currents&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is the seawater between the two plates moving?  If so, how fast?  Does the speed change over time?  All of these factors could contribute to unstable electrochemical readings, too ... If so, you may add mesh or baffles to create more-consistent flow conditions. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Follow-up answer, here)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, you're method sounds alright from an electrochemical perspective.  Cathodic protection is fine for minimizing corrosion.  Having &quot;known distance sensors&quot; will avoid problems with changing salinity, etc (as you mentioned).  &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditDate="2013-01-24T23:37:04.700" LastActivityDate="2013-01-24T23:37:04.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3995" PostTypeId="2" ParentId="3977" CreationDate="2013-01-23T15:31:55.517" Score="1" Body="&lt;p&gt;Have you looked into silica gel or zeolites? Both are non-toxic, and thus likely obtainable by private persons. Zeolites are used in self-refrigerating beer kegs.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-01-23T15:31:55.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3996" PostTypeId="1" CreationDate="2013-01-23T17:00:39.433" Score="3" ViewCount="5836" Body="&lt;p&gt;On our chemistry exam, a question asks &quot;Which force is present between Fe particles? a) Van der Waals b) Metallic&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first it seemed obvious that it was metallic; however, upon closer reading, it seemed like a trick question. First of all, there's no such thing as a metallic force. In addition, I'm not sure if Van der Waals forces exist for metallic bonds. If they do exist, then the answer was definitely &quot;Van der Waals&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, would Van der Waals/London dispersion forces exist between particles of iron?&lt;/p&gt;&#xA;" OwnerUserId="1131" LastActivityDate="2016-09-04T13:01:45.357" Title="Do metallic bonds contain London dispersion forces?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3997" PostTypeId="1" AcceptedAnswerId="3999" CreationDate="2013-01-23T18:39:48.417" Score="2" ViewCount="2510" Body="&lt;p&gt;Are there any simple at-home experiments to test the relative viscosity of alcohol to water?&#xA;If they are the same or practically the same, are there any household liquids that are less viscous than water? If so, how can I demonstrate this to first-graders?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2013-01-23T19:06:17.453" Title="How to compare viscosity of alcohol to water?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3998" PostTypeId="2" ParentId="3996" CreationDate="2013-01-23T18:45:21.893" Score="3" Body="&lt;p&gt;If by &quot;Fe particles&quot; you mean distinct particles (as opposed to the atoms inside a continuous Fe crystal structure), then the answer would be VdW/London forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;This is the only attractive intermolecular force present between neutral atoms&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which, in a way, can be applied to particles or particle surfaces as well, as long as they're not charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas, if you are talking about a continuous iron crystal structure, and by particles you mean atoms within that structure, then you have a metal-type bond between them caused by delocalization of electrons into the conductive band. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bond&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for further details.&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="605" LastEditDate="2013-02-24T21:31:30.080" LastActivityDate="2013-02-24T21:31:30.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3999" PostTypeId="2" ParentId="3997" CreationDate="2013-01-23T19:06:17.453" Score="3" Body="&lt;p&gt;A common experiment would be to have two concentric cylinders. The outer one can rotate almost freely (with some slight and constant retardation/breaking applied), while the inner one is driven by a motor, &lt;a href=&quot;http://hmf.enseeiht.fr/travaux/CD0203/travaux/optmfn/gpfmho/02-03/grp7/Couette_geometry.gif&quot; rel=&quot;nofollow&quot;&gt;like so&lt;/a&gt;. Between them is a layer of the liquid that you wish to determine the viscosity of. The speed of the outer cylinder is a measure for the viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since you want to explain this to first-graders, you might wanna start with something more obvious than the difference in viscosity between ethanol and water. How about the difference in viscosity between water and honey? As a demonstration experiment, you could use a funnel in which you pour an equal amount of either liquid, and then measure the time it takes until it has passed through the funnel. This should be more tangible and easier to understand. You could also prepare beakers with water, honey and maybe peanut butter, and have your students run a spatula through all of them. That way they can experience the difference in force that is necessary to move the spatula first hand.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-01-23T19:06:17.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4000" PostTypeId="1" CreationDate="2013-01-23T19:51:34.073" Score="2" ViewCount="358" Body="&lt;p&gt;I want to study what happens in a internal combustion engine in which fossil fuels are mixed along with other chemical additives.&lt;/p&gt;&#xA;&lt;h3&gt;Example:&lt;/h3&gt;&#xA;&lt;p&gt;Acetone drastically reduces surface tension.&lt;br /&gt;&#xA;Most fuel molecules are sluggish when it comes to their natural frequency. But acetone has inherent molecular vibration that &amp;quot;stirs up&amp;quot; the fuel molecules, to break the surface tension. &#xA;This results in a more complete vaporization with the various other factors remaining the same. More complete vaporization means less wasted fuel, hence the increased gas mileage from the increased thermal efficiency.&lt;/p&gt;&#xA;&lt;p&gt;Along with this I also want to study the emission of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2,\  O2\,\ H2O}$&lt;/span&gt; and more.&lt;/p&gt;&#xA;&lt;p&gt;Is it possible to carry out the above phenomena/process in a virtual simulator ?&lt;/p&gt;&#xA;" OwnerUserId="1132" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-01-16T11:04:41.967" Title="Software for chemical reaction simulation in ICE" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;computational-chemistry&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4002" PostTypeId="2" ParentId="639" CreationDate="2013-01-23T20:34:16.993" Score="3" Body="&lt;p&gt;Converting between force fields is always an iffy affair, as different force fields were parametrized with different reference sets, and possibly even with different intent in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two options here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can extract the raw parameters from the particular GROMACS or AMBER force field used here, and try to translate them into the parameter formalism CHARMM uses. Generally speaking, force fields consist of a set of parameters for each atom type, and a set of equations these parameters are used in to calculate forces and geometries. Those equations represent aspects of the atom's or molecule's properties that have an influence on the calculated forces or the molecule geometry, like bonding interactions, non-bonding interactions, angle contraints, bond length constraints, and so on. You need to find the corresponding equations in CHARMM's force field framework, and then translate the parameters from AMBER or GROMACS so that the force field is the same in CHARMM.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alternatively, you can look at the various CHARMM force fields and try to find one that works with NBD and any other molecules you got in your simulation. The first article you cited details where the various general parameters they used were from and how their simulation in GROMACS was set up. You should be able to define the same simulation parameters in CHARMM and rerun their simulation there. You can then use their published data to benchark your own simulation against theirs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-01-23T20:34:16.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4003" PostTypeId="1" CreationDate="2013-01-23T20:57:24.643" Score="9" ViewCount="681" Body="&lt;p&gt;Any suggestions on diagramming the chemical structure for &lt;a href=&quot;http://en.wikipedia.org/wiki/Cummingtonite&quot; rel=&quot;nofollow noreferrer&quot;&gt;cummingtonite&lt;/a&gt;? &#xA;$$\ce{(Mg,Fe)_7 Si_8 O_{22} (OH)_2}$$&#xA;As I understand it, the $\ce{(Mg,Fe)_7}$ says the molecule has seven atoms that could be either Magnesium or Iron, with probably more Magnesium than Iron, meaning the structure isn't precisely defined, i.e. a mineral not a specific molecule.  Is there a suitable way to diagram this that looks nice but not excessively complex? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, I'm rather ignorant about chemistry, but started screen printing subtlety goofy neck ties.  I'm envisioning doing essentially &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B009KNBL3I&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; but vertically and adding a chemical structure diagram.  I'll do the final pattern in &lt;a href=&quot;https://tex.stackexchange.com/questions/52722/can-you-make-chemical-structure-diagrams-in-latex&quot;&gt;latex&lt;/a&gt; if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s.  If I were making ties professionally, maybe I'd diagram an assortment of different cummingtonite molecules as a background for the formula.  I'm not doing that, way beyond my screen printing ability.  Just curious though, how does one know the probability distribution for the Magnesium and Iron?  Just varies by sample perhaps? &lt;/p&gt;&#xA;" OwnerUserId="1134" LastEditorUserId="-1" LastEditDate="2017-04-13T12:34:31.770" LastActivityDate="2015-02-25T15:37:00.180" Title="Chemical diagram for cummingtonite?" Tags="&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4004" PostTypeId="1" CreationDate="2013-01-23T22:57:35.600" Score="-4" ViewCount="3433" Body="&lt;p&gt;It says that the mass of Carbon and Hydrogen are 8:45. We are talking about an alkane here.&#xA;General formula is C&lt;sub&gt;n&lt;/sub&gt;H&lt;sub&gt;2n+2&lt;/sub&gt;. What is the particular formula?&lt;/p&gt;&#xA;" OwnerUserId="811" LastActivityDate="2013-01-24T02:25:33.553" Title="How to find the formula of this alkane?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2013-01-24T07:44:39.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="4005" PostTypeId="2" ParentId="4004" CreationDate="2013-01-24T02:25:33.553" Score="4" Body="&lt;p&gt;I presume that you mean the ratio by mass of hydrogen to carbon is 8:45. Depending on how many significant figures you select for your carbon and hydrogen masses, as well as the tolerable degree of error in your calculation, you may have several reasonable answers. If we take the mass of one mole of carbon to be 12g, and one mole of hydrogen to be 1g, then the following equality should hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;12n/(2n+2) = 45/8&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows naturally from the general formula for alkanes. Solving the equation, we find $n = 15$, yielding the chemical formula C&lt;sub&gt;15&lt;/sub&gt;H&lt;sub&gt;32&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-01-24T02:25:33.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4006" PostTypeId="2" ParentId="4003" CreationDate="2013-01-24T15:34:43.773" Score="10" Body="&lt;p&gt;Crystalline structures of inorganic materials aren't usually represented by diagrams, but rather by 3D renderings. First, you need to find a crystallographic structure file, which are usually in CIF format. A Google search leads to &lt;a href=&quot;http://www.crystallography.net/information_card.php?cif=9000141&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt; for cummingtonite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you need some molecular visualization software. Here, I use &lt;a href=&quot;http://www.crystalmaker.com&quot; rel=&quot;noreferrer&quot;&gt;CrystalMaker&lt;/a&gt;, which can create nice photorealistic 3D images, and has a simple tool to record animations. The result is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CiEXm.gif&quot; alt=&quot;Cummingtonite?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\ce{Fe}$ is pink, $\ce{Mg}$ is yellow, $\ce{Si}$ is blue; both $\ce{Mg}$ and $\ce{Si}$ are at the center of the tetrahedra, and oxygen atoms correspond to all the tetrahedra edges)&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-02-25T15:36:57.313" LastActivityDate="2015-02-25T15:36:57.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4007" PostTypeId="2" ParentId="4000" CreationDate="2013-01-24T19:35:22.597" Score="1" Body="&lt;p&gt;I am not sure about the phenomena you are describing but simulators are just tools with built-in models and thermodynamic data together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need is, for example, to find some kinetic (combustion kinetic) that it has in its equation a parameter regarding the acetone concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to find at the literature some of these kinetic equations, if you are student you can access with your university credentials into sciencedirect.com papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, check arxiv.org or google ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have these, then you can use simulators to reproduce your &quot;special&quot; kinetic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to describe better your system behaviour (including graphs) and try to fit them suing a correlation or a polynomial fit. Or ask the mathematicians, they could recognise models just seeing the patterns and give you an equation type that you can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this paper as an example;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.combustion.org.uk/ECM_2007/ecm2007_papers/1-20.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.combustion.org.uk/ECM_2007/ecm2007_papers/1-20.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1138" LastEditorUserId="1138" LastEditDate="2013-01-24T19:41:47.373" LastActivityDate="2013-01-24T19:41:47.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4008" PostTypeId="1" AcceptedAnswerId="4013" CreationDate="2013-01-25T06:50:51.057" Score="3" ViewCount="1520" Body="&lt;p&gt;How do you add a benzene group to cyclic compounds like piperidine, piperazine and phenyl derivatives?&lt;/p&gt;&#xA;&lt;h3&gt;Piperidine&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDXPC.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Piperazine&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/37jKD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-01-25T13:18:39.240" Title="How do you Benzylate (add a benzene group to) cyclic compounds?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4009" PostTypeId="1" AcceptedAnswerId="4012" CreationDate="2013-01-25T11:42:45.977" Score="4" ViewCount="206" Body="&lt;p&gt;To my knowledge, metalic ions only precipitate from a solution as salts &amp;amp; and a process that could possibly produce monocristalline metal would be very interesting indutrially - I've never heard of one, so I don't think it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: is it not conceivable to dissolve a metal in an acid (or maybe base), and then (later) neutralize the acid, without resulting ions that form a salt with the metal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I arrived at this question while thinking if there could be a biological, and thus wet-chemical, pathway from ores to pure metals (in metallic form). The last step in a such a process would be the precipitation/crystallization as a metal.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2013-01-25T15:55:53.457" Title="Is it possible to crystallize a solid ingot of metal from an (acidic) aquaous solution?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4010" PostTypeId="1" CreationDate="2013-01-25T12:09:51.620" Score="2" ViewCount="1059" Body="&lt;p&gt;So, I have 1 litre of acetate buffer, and 0.1 mol of oxonium ions/$\ce{H3O+}$ is added. The task is to calculate the pH of this solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I was thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOH + H2O &amp;lt;=&amp;gt; CH3COO- + H3O+}$$&#xA;$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \mathrm{p}K_\mathrm{a} + \log\frac{n(\ce{Ac-})}{n(\ce{HAc})}$$&#xA;$$\mathrm{pH} = 4.74 + \log\frac{n(\ce{Ac-}) - 0.1\ \mathrm{mol}}{n(\ce{HAc}) + 0.1\ \mathrm{mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But neither $c(\ce{Ac-})$ nor $c(\ce{HAc})$ (which should, in theorey, be equal) is given. The only thing I know (or I think I know) is this: $$\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \frac{1}{1} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it possible to calculate the pH from that? Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1140" LastEditorDisplayName="user7951" LastEditDate="2015-07-04T19:18:02.910" LastActivityDate="2018-01-25T22:59:09.103" Title="How can one calculate the pH of a solution?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4011" PostTypeId="2" ParentId="4009" CreationDate="2013-01-25T12:23:28.873" Score="3" Body="&lt;p&gt;I think you've just described the general concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrowinning&quot; rel=&quot;nofollow&quot;&gt;Electrowinning?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a large number of ways to recover metals from solutions or molten salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Producing monocrystalline metals is a different matter - crystal grains in metals tend to be microscopic, but there are techniques that permit growing large crystals of metals such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Czochralski_process&quot; rel=&quot;nofollow&quot;&gt;Czochralski process&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Zone_melting&quot; rel=&quot;nofollow&quot;&gt;zone melting&lt;/a&gt;. Monocrystalline metals sometimes have attractive mechanical properties not achievable in polycrystalline metals - for example, Callister Materials Science and Engineering 6e p. 225 has a nice photo of a single-crystal jet turbine blade that is highly resistant to creep failure with respect to conventional blades.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-25T12:23:28.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4012" PostTypeId="2" ParentId="4009" CreationDate="2013-01-25T13:08:55.780" Score="3" Body="&lt;p&gt;The easiest way to produce big crystals of metal from water solution is reactions like this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe_{(s)} + CuSO_4_{(aq)}\to Cu_{(s)} + FeSO_4_{(aq)}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;performed at very low concentration of the salt. The lesser know method is usage of reducer, like hydrazine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuSO_4_{(aq)} + N_2H_4_{(aq)} \to N_2_{(g)} + H_2_{(g)} + H_2SO_4_{(aq)} + Cu_{(s)}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of reactions usually produces fine powder (when I tried it, the resulting powder oxidezed with air in less then half of hour), but it is possible to produce visible crystals this way, working in gel with reagent transfer by diffusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For industrially valuable path from ores to metals it not absolutely necessary to get metal directly as result of the process. For example, iron oxalate when heated slightly decomposes, releasing metallic iron&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{FeC_2O_4_{(s)} \to Fe_{(s}) + 2 CO_2_{(g)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process gives very fine powder, like in previous example.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="1085" LastEditDate="2013-01-25T15:55:53.457" LastActivityDate="2013-01-25T15:55:53.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4013" PostTypeId="2" ParentId="4008" CreationDate="2013-01-25T13:18:39.240" Score="5" Body="&lt;p&gt;Brenton, the term &quot;benzene group&quot; is a bit misleading; it seems that you'd like to add a benzyl group, i.e. $\ce{C6H5-CH2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, these reaction are carried out using freshly distilled benzyl halides. In the case of the diamines, monoalkylation is usually achieved using the bis-hydrochloride of the amine and a stochiometric amount of base, thus keeping one nitrogen atom protected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, that the Australian regulations, kindly mentioned by Richard Terrett, are likewise true in the EC: 1-benzylpiperazine, an infamous psychoactive party drug, is a controlled substance!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-01-25T13:18:39.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4014" PostTypeId="2" ParentId="633" CreationDate="2013-01-25T15:42:17.640" Score="-4" Body="&lt;p&gt;yes, it is quite possible as aromatic anines nitrogen being moore nucleophilic further the secondary chloro reacts easily, the reaction may either be SN1 or SN2. I have done couple of reaction of this type, if you need help feel free to contact me. &lt;/p&gt;&#xA;" OwnerUserId="1141" LastActivityDate="2013-01-25T15:42:17.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4015" PostTypeId="1" AcceptedAnswerId="4022" CreationDate="2013-01-25T19:24:56.770" Score="9" ViewCount="3429" Body="&lt;p&gt;Formaldehyde has the formula $\ce{CH2O}$, and the ratio of atoms in a simple carb is $\ce{1C:2H:1O}$. This fits the formula of formaldehyde. When I researched this, I found some sources saying that formaldehyde is the simplest carb, but other sources saying it is glycolaldehyde ($\ce{C2H4O2}$). Is formaldehyde a carbohydrate?&lt;/p&gt;&#xA;" OwnerUserId="1142" LastEditorDisplayName="user7951" LastEditDate="2018-01-01T10:59:13.673" LastActivityDate="2019-02-19T22:42:33.480" Title="Is formaldehyde a carbohydrate?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4016" PostTypeId="2" ParentId="633" CreationDate="2013-01-25T19:46:01.090" Score="4" Body="&lt;p&gt;It's a pretty untypical approach to chloroquine and you didn't tell how you would synthesize 7-chloro-3-aminoquinoline and the secondary chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a good chance that the planned reaction will furnish a quarternary ammonium salt as a side product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UWP6y.png&quot; alt=&quot;formation of a quartenary ammonium salt&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction will not necessary stop at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ia addition, there's a good chance to bis-alkylate the $\ce{-NH2}$ group of the quinoline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is not very promising and any attempt to optimize this reaction is fighting problems you would not have if you had decided for a more reasonable and well-established alternative in the first!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would be the synthesis of chloroquine from &lt;strong&gt;4,7-dichloroquinoline&lt;/strong&gt; and &lt;strong&gt;4-diethylamino-1-methylbutylamine&lt;/strong&gt;, where both starting materials are available through well-known procedures.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-19T19:40:54.033" LastActivityDate="2015-05-19T19:40:54.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4017" PostTypeId="1" AcceptedAnswerId="4023" CreationDate="2013-01-25T22:01:58.197" Score="4" ViewCount="80" Body="&lt;p&gt;I was hoping there'd be someone who studies pesticides here. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are they water soluble? &lt;/li&gt;&#xA;&lt;li&gt;Do they get absorbed in to the fruit? &lt;/li&gt;&#xA;&lt;li&gt;How well would you have to wash the fruit to fully remove the residue?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1144" LastEditorUserId="73" LastEditDate="2013-01-26T03:59:21.857" LastActivityDate="2013-01-26T08:06:11.560" Title="Are pesticides used on fruit easily washed off?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4018" PostTypeId="1" CreationDate="2013-01-25T23:28:21.627" Score="4" ViewCount="25858" Body="&lt;p&gt;I'm not an expert. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple words how do they differ:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-H-O}$ and $\ce{H-O-H}$ molecules?&lt;/p&gt;&#xA;" OwnerUserId="1145" LastEditorUserId="4945" LastEditDate="2015-01-14T04:21:21.013" LastActivityDate="2015-06-09T19:04:35.897" Title="Differences between H-H-O and H-O-H" Tags="&lt;molecules&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4019" PostTypeId="2" ParentId="4018" CreationDate="2013-01-26T01:23:36.597" Score="7" Body="&lt;p&gt;There is no molecule in existence with the structure &lt;code&gt;H-H-O&lt;/code&gt;, for the simple reason that hydrogen possesses only one orbital and is therefore chemically incapable of forming more than one bond or maintaining more than two electrons in its orbit. Therefore, the formula $\ce{HHO}$ is either a very idiosyncratic way of denoting a molecule of water (normally written $\ce{H2O}$ and occasionally $\ce{HOH}$, in order to emphasize its structure, i.e., &lt;code&gt;H-O-H&lt;/code&gt;), or it refers to oxyhydrogen, which isn't actually a molecule at all, but rather a mixture of hydrogen and oxygen gases (the molecules $\ce{H2}$ and $\ce{O2}$, respectively) used as a fuel.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-01-26T01:23:36.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4020" PostTypeId="2" ParentId="4018" CreationDate="2013-01-26T01:24:05.413" Score="20" Body="&lt;p&gt;$\ce{H_{2}O}$ is the water molecule, two hydrogen atoms attached to a central oxygen atom, $\mathrm{C}_{2v}$ symmetry, thermodynamically minimal structure of these atoms, Adam's ale, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HHO is a poorly defined term often bandied around by 'water powered car'... &lt;em&gt;enthusiasts&lt;/em&gt;. I'm not sure that it's supposed to represent a molecule so much as a state of hydrogen and oxygen that conventional chemistry has no concept of.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://skeptics.stackexchange.com/questions/8067/man-using-water-as-fuel-for-a-welding-machine-ho-hho&quot;&gt;Here's a post on skeptics.SE about it&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rationalwiki.org/wiki/Water_powered_car#Theory_of_HHO&quot; rel=&quot;nofollow noreferrer&quot;&gt;And here's an acerbic article on rationalwiki that mentions 'HHO'&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think we're supposed to believe that it's a gas of hydrogen and hydroxyl radicals, or of atomic hydrogen and oxygen, or protons and hydroxide ions, or something, that is metastable at room temperature and that also happens to violate the first law of thermodynamics when produced and combusted. Why this glaringly obvious anomaly in quantum electrodynamics and thermodynamics has never been observed anywhere in nature before is anyone's guess (&lt;em&gt;probably part of the conspiracy&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These guys should write up a few papers, get published in Nature, collect the Nobel chemistry, physics and peace prizes...&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:44.273" LastActivityDate="2013-01-26T01:24:05.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4021" PostTypeId="1" CreationDate="2013-01-26T02:09:45.883" Score="5" ViewCount="1995" Body="&lt;p&gt;I was trying to isolate elemental bismuth from pepto bismol (bismuth subsalicylate) using the process seen &lt;a href=&quot;http://www.youtube.com/watch?v=Hp1fYtYEfoQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; but after I added hydrochloric acid, instead of getting a pink foamy substance, I got a white liquid with some white foam. The only reason I can think of, for why this would happen, is that the pepto bismol was expired by a month. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I realized the reaction didn't work, I added a homemade universal indicator (red cabbage - phenolphthalein - preserved with a little isopropyl alcohol) and started adding sodium hydroxide to neutralise any remaining acid. When I started adding the sodium hydroxide, the solution started heating up and then changed color fairly rapidly from pink to a dull grey green which didn't change when I tried adding more acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shortly after the color change, something started to precipitate out. I filtered off the precipitate and got a clear orange liquid. I then added a little hydrochloric acid to the orange liquid and something immediately started precipitating out. The remaining liquid then turned clear and smells slightly like mint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only guess for what I have made is bismuth hydroxide (the first precipitate) but I have no idea what the second precipitate is. &lt;a href=&quot;https://imgur.com/a/qGyYA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here are pictures of the final precipitates.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1146" LastEditorUserId="5026" LastEditDate="2018-02-23T13:56:22.547" LastActivityDate="2018-02-23T13:56:22.547" Title="How to extract elemental bismuth from pepto bismol?" Tags="&lt;inorganic-chemistry&gt;&lt;purification&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4022" PostTypeId="2" ParentId="4015" CreationDate="2013-01-26T04:07:50.920" Score="7" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.iupac.org/publications/pac/1996/pdf/6810x1919.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC definition&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;‘carbohydrate’ includes monosaccharides, oligosaccharides and polysaccharides as well as substances derived from monosaccharides by reduction of the carbonyl group (alditols), by oxidation of one or more terminal groups to carboxylic acids, or by replacement of one or more hydroxy group(s) by a hydrogen atom, an amino group, a thiol group or similar heteroatomic groups. It also includes derivatives of these compounds [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But according to &lt;a href=&quot;http://en.wikibooks.org/wiki/Principles_of_Biochemistry/The_Carbohydrates:_Monosaccharides,_Disaccharides_and_Polysaccharides&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikibooks&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the term is generally understood in the biochemistry sense, which excludes compounds with only one or two carbons. Natural saccharides are generally built of simple carbohydrates called monosaccharides with general formula &lt;span class=&quot;math-container&quot;&gt;$\ce{(CH2O)_n}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is three or more [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because formaldehyde and glycolaldehyde (&lt;a href=&quot;http://en.wikipedia.org/wiki/Glycolaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;not a true sugar&lt;/a&gt;) has one and two carbon respectively in its structures we can exclude both as a carbohydrate.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="41328" LastEditDate="2019-02-19T22:41:38.703" LastActivityDate="2019-02-19T22:41:38.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4023" PostTypeId="2" ParentId="4017" CreationDate="2013-01-26T08:06:11.560" Score="5" Body="&lt;p&gt;Pesticides never were my main field of research, so you might want to take the following with a grain of salt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most pesticides i'm aware of are insoluble in water, but you might want to cross-check that for particular substances in the huge &lt;a href=&quot;http://ec.europa.eu/sanco_pesticides/public/index.cfm&quot; rel=&quot;noreferrer&quot;&gt;Pesticide database of the EC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This insolubility seems reasonable, otherwise the substances would get washed off after the next rain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your second question, the answer unfortunately is yes. To my knowledge, different pesticides are typically applied at the various state of plant growth. In order to suppress weeds, one might spray fields in an early, pre-emergent state. Later, fungicides or other substances might be applied during blossom. In order to protect the bees, it is however not recommended to apply insecticides during that time. Reasonable and minimized use of pesticides will keep the load in the fruit close to the limits of detection or beyond. Depending on the origin of the fruit, this might however not be the norm. In these cases, high concentrations of pesticides, even those banned in your home country, are detected both in the fruit and its products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with the question what we can do as consumers. If the fruit is highly contaminated, we can't do anything. Therefore, it might be reasonable to avoid products from countries infamous for breaking environment protection rules and go &quot;organic&quot; instead. Honestly, i never peel apples, but i wipe the surface with a piece of cloth. But i avoid industrial sorts, buy from local farmers, or pick them from the ground of &lt;a href=&quot;http://en.wikipedia.org/wiki/Orchard&quot; rel=&quot;noreferrer&quot;&gt;meadow orchards&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that this contribution, although more personal and less scientific, was helpful.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-01-26T08:06:11.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4024" PostTypeId="1" AcceptedAnswerId="4026" CreationDate="2013-01-26T11:07:21.390" Score="2" ViewCount="3165" Body="&lt;p&gt;I'm looking for more information on the composition of a &lt;a href=&quot;http://www.rubbermaid.com/Category/Pages/ProductDetail.aspx?Prod_ID=RP091418&quot; rel=&quot;nofollow&quot;&gt;Rubbermaid Roughneck container.&lt;/a&gt;  The description says &quot;Roughneck containers are made with a durable polyethylene material for a lifetime of durability and strength.&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia entry&lt;/a&gt; indicates that polyethylene is the most common type of plastic with chemical composition $\ce{(C2H4)_{n}H2}$. But if this is so how are these containers able to flex instead of break like other plastic containers as well as tolerate greater temperature changes? I have a project that requires a custom shaped material of similar attributes. How feasible would it be to make my own mold at home? The shape I want to mold is very simple: a very thin rectangle about 35 width and 23 length. Are the chemicals involved dangerous to work with?&lt;/p&gt;&#xA;" OwnerUserId="1022" LastEditorUserId="84" LastEditDate="2013-01-26T13:46:30.973" LastActivityDate="2013-01-26T14:28:46.863" Title="What is the chemical composition of a Rubbermaid Roughneck container and is it feasible for me to make my own molds?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4025" PostTypeId="1" AcceptedAnswerId="4034" CreationDate="2013-01-26T11:41:39.417" Score="2" ViewCount="2154" Body="&lt;p&gt;I'm going to use an example (I could have used a different one);&#xA;On the document it is about the study of the chemical balance of an acid-basic solution.&lt;br&gt;&#xA;The chemists have written the chemical reaction equation to make a table of the evolution of pH during the reaction.&lt;br&gt;&#xA;If I understand with this equation I'll be able to understand others.&#xA;$\ce{CH3COOH + H2O -&amp;gt;CH3COO- + H3O+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I already understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is the equation of methanoic acid with water&lt;/li&gt;&#xA;&lt;li&gt;I understood why the general formula of methanoic acid is $\ce{CH3COOH}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I want to know is how do we know that  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH3COOH}$ changes into $\ce{CH3COO-}$ ?   &lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2O}$ changes into $\ce{H3O+}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, how does one find the products when one knows the reagents?&#xA;Does reagents molecules always correspond the same ions?&lt;/p&gt;&#xA;" OwnerUserId="1147" LastEditorUserId="1147" LastEditDate="2013-01-27T19:21:24.230" LastActivityDate="2013-01-29T23:32:26.570" Title="How can one find products when one knows the reagents?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;solvents&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4026" PostTypeId="2" ParentId="4024" CreationDate="2013-01-26T14:28:46.863" Score="5" Body="&lt;p&gt;This is a pretty broad question and might be better split into several, however a general comments on polymers, such as poly(ethylene), is that the properties of polymers can be tweaked quite a bit by control over the mean molecular mass and dispersity (probability distribution of molecular mass) of the polymer chains, as well as the addition of all kinds of stabilisers and plasticisers. Plasticisers especially may dramatically alter the flexibility of plastics. The exact composition of the Rubbermaid Roughneck is unlikely to be forthcoming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molding of thermoplastics by injection molding or casting is carried out by melting down plastic pellets and squirting the melt into or onto a mold. A mold release agent is often used to help get the cooled plastic out of the mold. The safety of the plastic depends on its specific composition (read the material safety data sheet for the pellets) and how hot it is (don't get molten plastic on yourself, wear protective clothing etc.). I don't know anything about the safety of mold release agents except for the fact that miniature model kits typically still have the agent on the plastic when purchased. This suggests they are fairly safe, though their presence may interfere with paint adhesion which is why pros scrub kits down before painting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometric simplicity of your custom part, however, suggests that you could cut it from a plastic sheet, obviating molding and the potentially expensive equipment that goes with it.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-01-26T14:28:46.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4027" PostTypeId="1" AcceptedAnswerId="4029" CreationDate="2013-01-26T16:41:28.297" Score="29" ViewCount="21760" Body="&lt;p&gt;I've heard that, even though according to Molecular Orbital Theory there is no chance of having nobel gases bonded to each other, it is &lt;strong&gt;not totally impossible&lt;/strong&gt;. For example, under extreme conditions, Ar2 can be synthesised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering whether a carbon can make 5 bonds if the required conditions are provided.&lt;/p&gt;&#xA;" OwnerUserId="1060" LastActivityDate="2019-01-27T10:33:05.560" Title="Carbon with 5 bonds?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4029" PostTypeId="2" ParentId="4027" CreationDate="2013-01-26T19:26:42.913" Score="25" Body="&lt;p&gt;Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe6C}$&lt;/span&gt; fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Fe}$&lt;/span&gt; bonds will be still 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is different if we consider exited states. Indeed, it is possible for hight excited state of carbon to be able to form 5 or 6 bonds. The resulting structure, however, will die quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noble gases are able to form molecules in excited states, or, to be precise, their molecules are stable only in excited state. This kind of molecules is known as excimer (excited dimer).&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="41328" LastEditDate="2019-01-27T10:33:05.560" LastActivityDate="2019-01-27T10:33:05.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4030" PostTypeId="2" ParentId="3959" CreationDate="2013-01-26T20:25:01.143" Score="19" Body="&lt;p&gt;In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ether molecules. When reacts, it coordinates first with most nucleophilic atom. In case of alcohol it is oxygen of $\ce{OH}$ fragment, and of course it  reacts with closest easily migrating atom: hydrogen, producing alkane. In case of ketone it coordinates with $\ce{CO}$ group, and alkyl group attacks closest electrophile : carbon from $\ce{CO}$ group, whose electrophility was just boosted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If right compound, effectively bounding to metal atom, is added and no other way of reactions is readily available, both reagents can work as bases. For example, organolithium compounds can replace hydrogen atoms in alkynes, alkenes and arenes as you noted earlier. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="9961" LastEditDate="2016-05-11T21:52:05.503" LastActivityDate="2016-05-11T21:52:05.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4031" PostTypeId="1" AcceptedAnswerId="5712" CreationDate="2013-01-27T04:30:26.360" Score="4" ViewCount="151" Body="&lt;p&gt;Say you have two compounds with their own unique chemical structures and you know the environment (e.g. temperature, pressure, presence of enzymes) could you then, using this information, approximate the rate of reaction? And if so, how would you? Please provide a mathematical answer, not just use some software. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-01-27T07:45:39.287" LastActivityDate="2013-07-28T09:05:48.783" Title="Could you estimate the rate of reaction from the chemical structure of the reactants? If so, how?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4032" PostTypeId="2" ParentId="4027" CreationDate="2013-01-27T06:49:35.597" Score="17" Body="&lt;p&gt;Carbon can make 5 bonds on methanium, $\ce{CH5+}$. It is a kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonium_ion&quot;&gt;carbonium ion&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Superacid&quot;&gt;superacid&lt;/a&gt;. Methanium can be produced in the laboratory as a dilute and low-temperature gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction:&#xA;$$\ce{CH4 + H+ -&amp;gt; CH5+}$$&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="4945" LastEditDate="2015-04-30T05:30:57.023" LastActivityDate="2015-04-30T05:30:57.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4033" PostTypeId="1" CreationDate="2013-01-27T10:21:35.440" Score="7" ViewCount="1376" Body="&lt;p&gt;Is it possible to achieve an optical mirror coating to a flexible plastic part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If vacuum metalization is possible, would the reflective coating layer be durable enough to withstand the bending of a flexible plastic such as polyurethane or a thermoplastic elastomer, without cracking over time?&lt;/p&gt;&#xA;" OwnerUserId="1150" LastEditorUserId="73" LastEditDate="2013-01-27T17:41:52.567" LastActivityDate="2016-11-08T23:52:02.687" Title="Can flexible plastics have reflective coating (vacuum metalized)?" Tags="&lt;materials&gt;&lt;plastics&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4034" PostTypeId="2" ParentId="4025" CreationDate="2013-01-27T19:17:48.853" Score="4" Body="&lt;p&gt;If I went into specifics, this answer would span a few thousand books. Instead, I'll give an overview:&lt;/p&gt;&#xA;&lt;p&gt;There are many ways of determining the products given the reagents:&lt;/p&gt;&#xA;&lt;h1&gt;Actually reacting the reactants&lt;/h1&gt;&#xA;&lt;p&gt;One can, of course, actually make the reaction take place. Once you've done that, you need to identify the contents of the resulting &amp;quot;soup&amp;quot;. A few methods for this  (You may have to &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_separation_techniques&quot; rel=&quot;nofollow noreferrer&quot;&gt;separate the products&lt;/a&gt; first) :&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;NMR Spectroscopy&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Here, a nifty technique is used to determine the magnetic properties of a compound. From these properties, one can make reasonable guesses at the substructures existing in a molecule, and check against a list of possible products. This is a very powerful technique, and usually gives good results.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;Qualitative analysis&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Here, we use various chemical reagents on the products, and analyse these products, making conclusions about the nature of the original products.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Microscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microscopy&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Here, using an &lt;a href=&quot;http://en.wikipedia.org/wiki/Scanning_tunneling_microscope&quot; rel=&quot;nofollow noreferrer&quot;&gt;STM&lt;/a&gt;, one figures out the structure of the molecule. This isn't used that often, but is an interesting technique.&lt;/p&gt;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://en.wikipedia.org/wiki/Analytical_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;many other such techniques&lt;/a&gt;, with varying accuracies and applicabilities.&lt;/p&gt;&#xA;&lt;h1&gt;Predicting the products in the absence of experiment&lt;/h1&gt;&#xA;&lt;p&gt;A lot of chemical research is into improving its (already vast) predictive power. The end-game of most sciences, after all, &lt;em&gt;is&lt;/em&gt; to be able to predict stuff. So, in many cases, we can just &lt;em&gt;predict&lt;/em&gt; how the reaction proceeds, without actually making the reaction occur in a lab.&lt;/p&gt;&#xA;&lt;h2&gt;Figuring out a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction mechanism&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;This generally occurs in organic reactions. Here, you break the reaction into elementary steps. Each step can be explained by looking at electronic attraction/repulsion, as well as the increase in stability (which can generally be estimated by taking into account the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;inductive&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Mesomeric_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;mesomeric&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperconjugation&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;steric&lt;/a&gt; effects, though other things may effect the overall stability as well). From these steps, one gets a proposed mechanism, which may be verified.&lt;/p&gt;&#xA;&lt;h2&gt;Knowing solubilities and other numerical values&lt;/h2&gt;&#xA;&lt;p&gt;In ionic/inorganic reactions, the path a reaction takes generally hinges on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility&quot; rel=&quot;nofollow noreferrer&quot;&gt;solubility&lt;/a&gt; of its final products. This is sometimes the case for other parameters as well. Knowing various numerical values lets one compare two possible reaction paths&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Computational_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Computational chemistry&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Here, one uses the equations of quantum mechanics (&lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schrodinger's equation&lt;/a&gt; and a few of its cronies), and solves them for the given problem using &lt;a href=&quot;http://en.wikipedia.org/wiki/Numerical_methods&quot; rel=&quot;nofollow noreferrer&quot;&gt;numerical methods&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The particular reaction you have given&lt;/h1&gt;&#xA;&lt;p&gt;This one can be easily explained by resonance. In &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt;, the &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; is loosely bonded, as the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=O}$&lt;/span&gt; bond resonates with the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; bond. This makes it easy for the &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; to float away as &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, leaving behind a pretty stable &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COO-}$&lt;/span&gt; (made stable by resonance of the negatively charged lone pair over the oxygen atoms).&lt;/p&gt;&#xA;&lt;p&gt;So far, we know that &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH -&amp;gt; CH3COO- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we know that it is behaving like an acid. It is extremely well known that water forms a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydronium&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;) ion by self-dissociation, and this is accelerated by making the medium acidic (All free &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;s find the nearest &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; molecule and form &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;). &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; is stable due to resonance, again.&lt;/p&gt;&#xA;&lt;p&gt;So we now know that &lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + H2O -&amp;gt; H3O+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Put the two reactions together and you get &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + H2O -&amp;gt; H3O+ + CH3COO-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Though, in this case, the historical method of finding out the reaction may have been different -- acetic acid has been well known and studied for a very long time.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-01-29T23:32:26.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4035" PostTypeId="2" ParentId="2795" CreationDate="2013-01-28T05:12:36.667" Score="3" Body="&lt;p&gt;Elements 57+ have electrons in f orbitals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radium has a full 4f sublevel these electrons as less effective at shielding than d electrons (f&#xA;" OwnerUserId="1153" LastActivityDate="2013-01-28T05:12:36.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4036" PostTypeId="2" ParentId="634" CreationDate="2013-01-28T05:42:05.063" Score="3" Body="&lt;p&gt;The Comprehensive Handbook of Chemical Bond Energies might be able to help.  It walks through bonds in different functional groups.  It can be found at &lt;a href=&quot;http://www.ebah.com.br/content/ABAAABH2cAD/comprehensive-handbook-of-chemical-bond-energies-0849373662&quot; rel=&quot;nofollow&quot;&gt;http://www.ebah.com.br/content/ABAAABH2cAD/comprehensive-handbook-of-chemical-bond-energies-0849373662&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2013-01-28T05:42:05.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4040" PostTypeId="1" AcceptedAnswerId="4042" CreationDate="2013-01-28T14:42:53.363" Score="5" ViewCount="1903" Body="&lt;p&gt;By &lt;strong&gt;chemical database software&lt;/strong&gt; I mean software with similar functionalities to &lt;em&gt;&lt;a href=&quot;http://www.chemaxon.com/products/instant-jchem/&quot; rel=&quot;noreferrer&quot;&gt;Instant JChem&lt;/a&gt;&lt;/em&gt; in that they have a large database of different compounds and have the basic capabilities you, or at least I expect from a chemistry software. Namely: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can approximate the chemical and physical characteristics of the compounds in question.&lt;/li&gt;&#xA;&lt;li&gt;Can approximate products of reactants and their corresponding proportions. &lt;/li&gt;&#xA;&lt;li&gt;Can show the 2d and 3d structure of compounds in the database.&lt;/li&gt;&#xA;&lt;li&gt;Can connect to online chemical databases like PubChem, ChemSpider, NCBI, etc.&lt;/li&gt;&#xA;&lt;li&gt;Has a molecule editor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would also prefer it if it is free of charge. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-01-28T16:50:45.990" LastActivityDate="2016-07-11T21:39:37.243" Title="Which Chemical Database Software Are Available?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4041" PostTypeId="2" ParentId="4027" CreationDate="2013-01-28T15:11:29.390" Score="7" Body="&lt;p&gt;There are stable crystal structures where one could notionally see carbon atoms with more than 4 'bonds'. One example that comes to mind is a carborane unit; e.g. see&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carborane&quot;&gt;http://en.wikipedia.org/wiki/Carborane&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example. But of course, such 'bonds' are a notional thing really and don't reflect the true nature of the underlying electronic structure.&lt;/p&gt;&#xA;" OwnerUserId="1155" LastActivityDate="2013-01-28T15:11:29.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4042" PostTypeId="2" ParentId="4040" CreationDate="2013-01-28T16:28:24.327" Score="3" Body="&lt;p&gt;A widely used database in academia and industry is ChemACX.The only one of your requirements that I don't believe it does is the product prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemACX/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemACX/Default.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2013-01-28T16:28:24.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4043" PostTypeId="2" ParentId="2825" CreationDate="2013-01-28T16:51:58.430" Score="-1" Body="&lt;p&gt;Summarizing Phillip and Siddahrtha's comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under normal conditions ethene normally undergoes addition since the double bond is nucleophilic.  In order to have a substitution reaction on ethene H $^{-}$, would need to be the leaving group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_2H_4 + HBr \rightarrow CHCHBr + H^{-} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not energetically favorable under normal conditions.  Under high pressure and with a catalyst the reaction would be possible.&lt;/p&gt;&#xA;" OwnerUserId="1153" LastActivityDate="2013-01-28T16:51:58.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4044" PostTypeId="2" ParentId="4010" CreationDate="2013-01-28T17:59:23.830" Score="1" Body="&lt;p&gt;I'm tring to write the methodology&lt;br&gt;&#xA;If anything is wrong please point it out to me i'll be happy to correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $\small\ce{pH} $ of a solution you can:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the conjugated base of your acid  (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Write the equation of the solution&lt;br&gt;&#xA;Now you want to prepare an aqueous solution. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the concentration of $\small\ce{[CH_3COOH]}$ &#xA;you use the formula to determine the concentration in solution of acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;   \ce{[CH_3COOH]\ =\ C\ -\ [CH3COO- ]}&#xA;   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then you write that:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$&#xA;as both dissociate from the same acid as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOH\ -&amp;gt;\ CH3COO- \ +\ H+}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the $\small\ce{pH}$ concentration is given by:&lt;br&gt;&#xA;$$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$pH$ here you are! (without any unit)&lt;/p&gt;&#xA;" OwnerUserId="1147" LastEditorUserId="1147" LastEditDate="2013-01-29T17:31:10.277" LastActivityDate="2013-01-29T17:31:10.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4045" PostTypeId="1" CreationDate="2013-01-28T18:34:10.843" Score="1" ViewCount="1640" Body="&lt;p&gt;Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets? I know for sure it has to do with the van der Waals force and the fact that the molecules are huge, but why? Why does a big molecule break down before the van der Waals forces break?&lt;/p&gt;&#xA;" OwnerUserId="738" LastActivityDate="2014-11-27T09:13:10.887" Title="Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4046" PostTypeId="1" CreationDate="2013-01-28T22:26:41.700" Score="8" ViewCount="32813" Body="&lt;p&gt;For some reason it has been very difficult to find the right answer for this question as there's so much conflicting information to be found online. Some mention a reaction where carbonic acid is made, others mention carbonates and hydroxide end product but nothing seems to be precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can the reaction be as simple as this (a double displacement reaction) where magnesium bicarbonate and sodium chloride are the end products:&#xA;$$&#xA;\ce{2NaHCO3 + MgCl2 -&amp;gt;  2NaCl + Mg(HCO3)2}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1158" LastEditorUserId="897" LastEditDate="2013-01-29T04:16:30.090" LastActivityDate="2019-07-26T01:17:45.710" Title="What is the reaction between Magnesium chloride and Sodium bicarbonate?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4047" PostTypeId="1" CreationDate="2013-01-29T14:06:05.103" Score="2" ViewCount="1293" Body="&lt;p&gt;I would like to know What is the best method for predicting the water solubility (and in other solvents) of a compound given its molecular structure at different pHs. What I know now is based on the seminal paper from Eisenberg et al.;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Eisenberg, D. &amp;amp; McLachlan, A. D. Solvation energy in protein folding and binding. Nature 319, 199–203 (1986).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where partition coefficient can be estimated depending on each atom type and its exposed surface to the solvent (SASA). This seems to work well for many substances, but I have the problem that I need to know the solubility dependence of some substances against pH, and I the only thing I found now as &lt;a href=&quot;http://www.chemaxon.com/products/marvin/&quot; rel=&quot;nofollow&quot;&gt;Chemaxon&lt;/a&gt; tools, but their results are not very accurate. It seems that the best option might be quantum mechanical calculations and tools such as &lt;a href=&quot;http://www.cosmologic.de/index.php&quot; rel=&quot;nofollow&quot;&gt;COSMO-RS&lt;/a&gt;. At the moment we can not pay its license.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, I wonder what approaches do you suggest for calculating solubility of small chemical compounds depending on pH. Any method/software/server would be fine for us.&lt;/p&gt;&#xA;" OwnerUserId="256" LastEditorUserId="256" LastEditDate="2013-02-27T16:53:48.837" LastActivityDate="2013-02-27T16:53:48.837" Title="Prediction of solubility (from chemical structure) at different pHs" Tags="&lt;solubility&gt;&lt;software&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2019-12-03T19:18:00.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="4048" PostTypeId="1" AcceptedAnswerId="4075" CreationDate="2013-01-29T21:05:02.140" Score="3" ViewCount="422" Body="&lt;p&gt;What makes electrolysis or electroplating take so long ? What stage takes longer when ions are attracted to electrodes or when they gain or lose electrons at electrode.&#xA;And if we are talking about an electrolyte composed of positive ions and electrons , which is faster, electrons entering the circuit at positive electrode or gaining electrons by positive ions at negative electrode?&lt;/p&gt;&#xA;" OwnerUserId="815" LastActivityDate="2013-02-04T03:29:55.297" Title="Why does electrolysis take so much time?" Tags="&lt;electrochemistry&gt;&lt;ions&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4049" PostTypeId="2" ParentId="2710" CreationDate="2013-01-30T03:24:12.863" Score="15" Body="&lt;p&gt;$\ce{Pd}$ can dissociate $\ce{H2}$ because the resulting $\ce{Pd-H}$ bonds are more stable than the starting $\ce{H2}$. &#xA;But the reason why $\ce{Pd}$ is so good at dissociating $\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of $\ce{H2}$ on a $\ce{Pd}$ surface (and on $\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the $\ce{H2}$ in an excited state to go over a barrier and create $\ce{Pd-H}$ bonds.&#xA;On $\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite $\ce{H2}$ to dissociate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/andrew3/Public/Papers/1995/Hammer/1995_Nature_Why%20gold%20is%20the%20noblest_Hammer.pdf&quot;&gt;Here is a good theoretical article&lt;/a&gt; (with a great title): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows the different barriers to dissociation of $\ce{H2}$ on $\ce{Pt}$, $\ce{Ni}$, $\ce{Cu}$ and $\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).&lt;/p&gt;&#xA;" OwnerUserId="1162" LastEditorUserId="22" LastEditDate="2013-01-30T04:30:07.247" LastActivityDate="2013-01-30T04:30:07.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4051" PostTypeId="2" ParentId="4046" CreationDate="2013-01-30T17:41:43.140" Score="1" Body="&lt;p&gt;I thought I would start out in saying that I think this is a random, but great question. This is because as you said there's so little information out there regarding this sort of reaction so it was an interesting little puzzle for me to figure out for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;guess&lt;/strong&gt; would be that the reaction you detailed would &lt;strong&gt;not&lt;/strong&gt; occur. This double replacement reaction would occur in &lt;strong&gt;aqueous solution&lt;/strong&gt; as otherwise the product of magnesium bicarbonate probably wouldn't have a chance of forming since at least one other alkali earth metal bicarbonate salt, calcium bicarbonate, does not exist outside of aqueous solution. I tell you this piece of information instead of whether magnesium bicarbonate exists outside of aqueous solution because I just don't know for certain whether it does. Evidence for this theory that it doesn't exist out of aqueous solution includes (in addition to the point raised above about calcium bicarbonate):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Magnesium bicarbonate's predicted partition coefficient is -0.809 according to &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.92335.html&quot; rel=&quot;nofollow&quot;&gt;chemspider&lt;/a&gt; which is identical to the predicted partition coefficient for calcium bicarbonate which makes it &lt;strong&gt;very&lt;/strong&gt; likely that it is highly water soluble. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since it occurs in aqueous the only driving force to the reaction would be the production of an insoluble salt, in your reaction, however, there is &lt;strong&gt;no&lt;/strong&gt; insoluble products that results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I suspect an alternate reaction would occur. Namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NaHCO_3 + MgCl_2 -&amp;gt; 2NaCl + MgO + H_2 O + 2 CO_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this reaction would lead to three insoluble products and thus would invoke a greater drive for the reaction.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-01-30T17:52:45.410" LastActivityDate="2013-01-30T17:52:45.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4052" PostTypeId="1" AcceptedAnswerId="4066" CreationDate="2013-01-31T01:51:23.270" Score="10" ViewCount="5704" Body="&lt;p&gt;I have a chemistry homework problem that asks whether $Br_2$ or $O_2$ is more soluble in water in room temperature. However, I can't seem to find a way to figure it out, short of looking it up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a guiding rule for such figures, such as relating to molar mass?&lt;/p&gt;&#xA;" OwnerUserId="1169" LastEditorUserId="22" LastEditDate="2013-01-31T04:34:58.203" LastActivityDate="2013-02-02T16:09:18.223" Title="How to determine which gas is more soluble" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4053" PostTypeId="2" ParentId="2638" CreationDate="2013-01-31T20:08:04.700" Score="1" Body="&lt;p&gt;The answer to my question appears to be in a 1977 paper by Nancy D Hann (DOI: &lt;a href=&quot;http://dx.doi.org/10.1002/pol.1977.170150604&quot; rel=&quot;nofollow&quot;&gt;10.1002/pol.1977.170150604&lt;/a&gt;; full citation below). In it, she compares GPC of polyurethane in DMF with and without added LiBr. The chromatograms with no added LiBr show multiple peaks. The chromatagrams with added LiBr show just a single peak. The addition of LiBr also shifts the peak to later elution volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hann explains these results by the polymer having a strong polar interaction with the pure solvent. This causes a higher effective hydrodynamic radius, resulting in the earlier elution time. The polar interactions also case aggregation of the polymer, giving multiple peaks. Adding a salt breaks up the polar interactions, giving just a single, later, peak.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hann, N. D. (1977). &quot;Effects of lithium bromide on the gel-permeation chromatography of polyester-based polyurethanes in dimethylformamide.&quot; Journal of Polymer Science: Polymer Chemistry Edition 15(6): 1331-1339.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2013-01-31T20:08:04.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4054" PostTypeId="1" AcceptedAnswerId="4057" CreationDate="2013-01-31T20:10:28.870" Score="11" ViewCount="4956" Body="&lt;p&gt;The two broad ways a halogen can leave a haloalkane are &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;nucleophilic substitution&lt;/a&gt; and elimination reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that &lt;a href=&quot;http://en.wikipedia.org/wiki/Elimination_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;elimination reactions&lt;/a&gt; (that form a double bond as a $H^+$ is also removed) do not occur when the haloalkane is dissolved in water; here nucleophilic substitution (anion replaces halogen only) is favoured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? I guess it's not just water - what are the general conditions (&lt;em&gt;including&lt;/em&gt; type of solvent) that prefer nucleophilic substitution over elimination, and vice-versa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand there are different variants of both elimination and substitution ($S_N1, S_N2, E1,E2$); are any of them more prone to occur than others of the same type in water?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="16683" LastEditDate="2017-04-06T22:31:16.393" LastActivityDate="2017-04-06T22:34:59.367" Title="Why does water favour nucleophilic substitution over elimination?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;halides&gt;&lt;solvents&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4057" PostTypeId="2" ParentId="4054" CreationDate="2013-02-01T05:27:07.263" Score="9" Body="&lt;p&gt;The nature of the substrate (in particular the degree of substitution of the α-carbon) is of principal importance in determining which mechanism predominates. Generally, $\mathrm{S_N1}$ and $\mathrm{E1}$ are both less common than $\mathrm{S_N2}$ and $\mathrm{E2}$, partially as a consequence of their first-order kinetics and the stability constraints imposed by the charged intermediate species the mechanisms generate. $\mathrm{S_N1}$ and $\mathrm{E1}$ both require tertiary, allylic, or benzylic α-carbons in order to be viable (although secondary carbons can occasionally react by these pathways as well, but slowly, and only if conditions disfavor competing reactions), mainly because the carbocation intermediate that forms during the rate-determining step can be stabilized by the neighboring electron-donating alkyl groups via hyperconjugation (or via resonance in the case of carbons in allylic or benzylic positions). Tertiary carbons also disfavor $\mathrm{S_N2}$ because the approach of the nucleophile is sterically hindered by the β-alkyl groups. Under those conditions, $\mathrm{E1}$ and $\mathrm{S_N1}$ reactions would be in competition. $\mathrm{E1}$ would be favored if there were no strong nucleophiles present, as well as by the application of heat. Obviously, $\mathrm{S_N2}$ reactions are favored by the presence of strong nucleophiles (ideally, ones that don't also act as strong bases) and a primary α-carbon substrate. $\mathrm{E2}$ reactions are favored over $\mathrm{S_N2}$ if strong bases, especially non-nucleophilic, are present. $\mathrm{E2}$ also prevails when either the substrate and/or the base is sterically hindered, since abstraction of a proton is vastly less impacted by steric factors than is the approach of a nucleophile attacking a carbon with simultaneous loss of the leaving group. Finally, $\mathrm{E2}$ reactions impose an additional restriction in that the β-hydrogen and the leaving group must be anti-periplanar to one another, in order for simultaneous loss of the leaving group and formation of the π-bond to occur in a coordinated fashion. If the molecule can't adopt such a conformation (for example, it may be rotationally locked due to the presence of π-bonds, or lack the kinetic energy necessary for a ring flip, etc.), then $\mathrm{E2}$ is categorically ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Parenthetically, I should add that elimination reactions, in general, require higher temperatures than substitution reactions. The naive explanation for this is based on the equation for the free energy change of a reaction: $\Delta G = \Delta H\ -\ T\Delta S$. Elimination reactions, at the most simplified level, involve two reactant molecules [namely, a base and the substrate], and yield three products [the protonated base, the substrate molecule that has undergone elimination, and the substrate's leaving group]. Consequently, elimination reactions are entropically favorable [i.e., the $T\Delta S$ term is positive]. However, elimination reactions are also endothermic [i.e., $\Delta H$ is positive], because heat energy is required to break the two &amp;sigma;-bonds of the abstracted proton and the leaving group, while the resulting products form only one &amp;sigma;- and one π-bond, the latter of which is less stable. The implication of this is that, in order for the overall change in free energy, $\Delta G$, to be negative [a requisite condition for spontaneity of the reaction], $T\Delta S$ must be greater than $\Delta H$, so the temperature needs to be sufficiently high if the reaction is to take place.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solvent is of only slightly less importance. Polar protic solvents (water, alcohols, weak carboxylic acids, etc.) have partial-positive hydrogens, which serve to stabilize negatively charged species. Consequently, $\mathrm{E2}$ and $\mathrm{S_N2}$ mechanisms, which are aided by having negatively charged bases or nucleophiles in high concentration, are somewhat dampened by protic solvents, allowing otherwise disfavored $\mathrm{E1}$ and $\mathrm{S_N1}$ pathways to compete. Polar aprotic solvents (THF, DMF, DMSO, chloroform, ethers, ketones, etc.) cannot stabilize negative charges as effectively, making them the preferred choice when $\mathrm{E2}$ or $\mathrm{S_N2}$ mechanisms are desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to your specific question about haloalkanes. I've explained above why various combinations of conditions might favor one class of reaction over another. In water, specifically, a few distinct factors, beyond the general principles outlined above, might be relevant:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Due to the leveling effect, no base stronger than hydroxide can exist in water at any appreciable concentration (that is, any base much stronger than hydroxide will simply deprotonate the water molecules, forming hydroxide and the base's conjugate acid). However, strong, non-basic nucleophiles are not subject to this effect, and though they may be solvated and weakened by hydrogen bonding, this effect would be less drastic than the protonation that strong bases would undergo.&lt;/li&gt;&#xA;&lt;li&gt;Hydroxide is actually a much stronger base in many other solvents than it is in aqueous solution, which is chiefly the result of it being very thoroughly solvated and stabilized by hydrogen bonding. Hence, competing nucleophiles become stronger by comparison, while stronger bases simply end up being consumed, producing more hydroxide.&lt;/li&gt;&#xA;&lt;li&gt;Because elimination reactions require heat, substitution reactions will be heavily favored and immediately take place unless the reaction vessel is heated prior to mixing of the reagents. The problem is compounded by the fact that, with hydroxide as the nucleophile, an alcohol would form, which would be stabilized by hydrogen bonding with water molecules. Further, because haloalkanes are poorly soluble in water, and immiscible liquids in the same container will boil at lower temperatures than they would unmixed, it could become very difficult in practice to achieve the temperatures necessary to favor elimination reactions while also maintaining the agitation necessary to keep the haloalkane and the aqueous phase from completely separating.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="305" LastEditorUserId="16683" LastEditDate="2017-04-06T22:34:59.367" LastActivityDate="2017-04-06T22:34:59.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4058" PostTypeId="1" AcceptedAnswerId="4059" CreationDate="2013-02-01T08:08:56.860" Score="16" ViewCount="232739" Body="&lt;p&gt;I'm sure most of us have heard that saturated fats are solid at room temperature, and unsaturated fats are liquid at room temperature. I'm wondering how this relates to their chemical structure -- saturated fats contain only single bonds between carbons, yet to qualify as an unsaturated fat a C=C double bond must exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a double bond is stronger than a single bond, and the length of the C=C double bond is shorter than that of the single bond, why is it that the fat containing a double bond is a liquid and saturated fats are solids at room temperature? Seems like the double bond would inhibit movement and the resulting substance would be less like olive oil and more like butter. &lt;/p&gt;&#xA;" OwnerUserId="1176" OwnerDisplayName="Alexandria" LastEditorUserId="402" LastEditDate="2017-04-02T20:10:26.233" LastActivityDate="2017-04-02T20:10:26.233" Title="Saturated vs unsaturated fats - Structure in relation to room temperature state?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;&lt;fats&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4059" PostTypeId="2" ParentId="4058" CreationDate="2013-02-01T10:11:16.250" Score="14" Body="&lt;p&gt;In the solid state, the individual triacylglycerol molecules are interacting with each other primarily through Van der Waals interaction. These weak bonds between molecules are broken at the solid-liquid transition. The amount of energy needed to disrupt these interactions (which determines the melting point of the fat or oil) is determined by the energy associated with all of these bonds added together. In a saturated fat, the acyl chains are able to align perfectly right along their length, maximizing intermolecular interactions. This effect is reflected in the fact that the melting temperature of a pure triacylglycerol increases as the chain length increases.&lt;/p&gt;&#xA;&lt;p&gt;You can see this effect clearly in the melting temperatures of individual fatty acids. (C18:0 means an 18 carbon molecule with zero double bonds in the acyl chain):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;C18:0 (stearic acid) 70°C&lt;/p&gt;&#xA;&lt;p&gt;C16:0 (palmitic acid) 63°C&lt;/p&gt;&#xA;&lt;p&gt;C14:0 (myristic acid) 58°C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the addition of a single -CH&lt;sub&gt;2&lt;/sub&gt;- group in the acyl chain increases melting temperature by a few degrees.&lt;/p&gt;&#xA;&lt;p&gt;When a cis double bond is introduced into the acyl chain this creates a kink in the structure. Because of this, the acyl chains cannot align completely along their length - they don't pack together as well. Because of this, the sum of the energy associated with intermolecular Van der Waals interactions is reduced. Again this is seen clearly in the melting temperatures of fatty acids:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;stearic acid C18:0 70°C&lt;/p&gt;&#xA;&lt;p&gt;oleic acid C18:1 16°C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As you can see from these numbers, the effect of introducing a double bond is large compared to the chain length effect.&lt;/p&gt;&#xA;&lt;p&gt;A typical fat or oil will, of course, be a mixture of different triacylglycerols, but the underlying principle is the same.&lt;/p&gt;&#xA;" OwnerDisplayName="Alan Boyd" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-01-22T10:15:53.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4060" PostTypeId="1" CreationDate="2013-02-01T17:08:52.953" Score="4" ViewCount="1744" Body="&lt;p&gt;I'm not detecting ammonia in my samples while using a NOESY pulse program for a 1H-NMR experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for ammonia to be detected by a 1D-1H-NOESY experiment?&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="73" LastEditDate="2013-02-02T00:38:04.967" LastActivityDate="2013-02-04T09:23:53.470" Title="Can ammonia be detected by a 1D-1H-NOESY experiment?" Tags="&lt;nmr-spectroscopy&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4061" PostTypeId="2" ParentId="4060" CreationDate="2013-02-01T17:45:39.137" Score="6" Body="&lt;p&gt;In the &lt;sup&gt;1&lt;/sup&gt;H,&lt;sup&gt;1&lt;/sup&gt;H-NOESY experiment you are using the nuclear Overhauser effect to transfer magnetization from one proton to another proton in close proximity (&amp;lt; 5 Å).  This only works within a single molecule, unless you have a reasonably stable complex you can't observe signals in the NOESY between protons of different molecules in your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia has three protons which are at least chemically equivalent, their signals will appear at the same chemical shift. In a 1D NOESY you irradiate at the frequency of a specific proton and then observe the transferred magnetization at other frequencies. You can't observe magnetization transfer between protons at the same frequency that way, as your irradiation will destroy any signal at this specific frequency.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can ignore the explanation above, as the NOESY experiment you are using is different. In metabolomics the NOESY 1D is used for water suppression, this is a different experiment than the 1D NOESY I explained above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common variant of this I found also uses presaturation to suppress water. This presaturation will also affect all signals that exchange quickly with the solvent, and the protons of ammonia are exchanging with water. So I would expect the ammonia signal to be far weaker or even not visible in such an experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to observe ammonia, you have to use a water suppression method that doesn't affect exchanging protons.  &lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2013-02-04T09:23:53.470" LastActivityDate="2013-02-04T09:23:53.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4062" PostTypeId="2" ParentId="4058" CreationDate="2013-02-01T20:27:19.170" Score="20" Body="&lt;p&gt;This has to do with the chemical structure of the fatty acids. Single bonds create a straight molecule, which can be easily packed together to form a solid, while the double bond introduces a kink in the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fy6xf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fy6xf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fats, which are mostly from animal sources, have all single bonds between the carbons in their fatty acid tails, thus all the carbons are also bonded to the maximum number of hydrogens possible. Since the fatty acids in these triglycerides contain the maximum possible amount of hydrogens, these would be called &lt;strong&gt;saturated fats&lt;/strong&gt;. The hydrocarbon chains in these fatty acids are, thus, fairly straight and can pack closely together, making these fats solid at room temperature. Oils, mostly from plant sources, have some double bonds between some of the carbons in the hydrocarbon tail, causing bends or “kinks” in the shape of the molecules. Because some of the carbons share double bonds, they’re not bonded to as many hydrogens as they could if they weren’t double bonded to each other. Therefore these oils are called &lt;strong&gt;unsaturated fats&lt;/strong&gt;. Because of the kinks in the hydrocarbon tails, unsaturated fats can’t pack as closely together, making them liquid at room temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://biology.clc.uc.edu/courses/bio104/lipids.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The source website&lt;/a&gt; is no longer accessible, but biochemistry textbooks will state the same thing.&lt;/p&gt;&#xA;" OwnerUserId="1175" LastEditorUserId="402" LastEditDate="2017-01-22T10:16:24.677" LastActivityDate="2017-01-22T10:16:24.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4063" PostTypeId="1" CreationDate="2013-02-02T04:22:34.363" Score="13" ViewCount="15985" Body="&lt;p&gt;The definition of an acid is a compound with a hydrogen cation and a non-metal anion. Water is a hydroxide bonded to a hydrogen atom. This hydroxide has a net negative charge, too, since the negative oxygen's charge is stronger than the positive hydrogen's. Water has a pH of 7, though, meaning that it's neutral. Why isn't water acidic&lt;/p&gt;&#xA;" OwnerUserId="1142" LastEditorUserId="-1" LastEditDate="2017-01-26T18:10:24.053" LastActivityDate="2017-01-26T18:10:24.053" Title="Why isn't water acidic?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;ph&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4064" PostTypeId="2" ParentId="4063" CreationDate="2013-02-02T06:06:11.140" Score="20" Body="&lt;p&gt;If you look at what happens in the equilibrium dissociation of water, both a hydronium ion (Arrhenius acid) and a hydroxide ion (Arrhenius base) are formed. Since these two ions are  formed in equal quantities ($\ce{10^{-7} mol/L}$ each in pure water), the water will be neither acidic nor basic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O_{(l)} + H2O_{(l)} &amp;lt;=&amp;gt; H3O+_{(aq)} + OH^{-}_{(aq)}}$$  (from wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia can explain more about acids and their definitions if you are interested: &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid&quot;&gt;http://en.wikipedia.org/wiki/Acid&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="897" LastEditDate="2013-02-02T09:04:47.467" LastActivityDate="2013-02-02T09:04:47.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4065" PostTypeId="2" ParentId="4060" CreationDate="2013-02-02T08:18:39.147" Score="3" Body="&lt;p&gt;First off, NOESY is a tool that gives information more useful for structure determination and confirmation (as Mad Scientist has explained). A simple and quick 1H NMR experiment should be enough to quantify your compounds assuming you have sufficient peak resolution (often a problem with multiple compound quantification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you know that you cannot detect the ammonia? Where do you expect to find its peak (I believe it is a triplet found around 0 ppm, depending on solvent/conditions)?. Remember that the chemical shift may be affected by temperature, concentration, or hydrogen bonding, so it may not be where you expect to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect your problem could also be with the deuterated solvent you use to prepare samples. For example, if you are using D2O, you will not see any OH, NH (ammonia), or SH protons as they will rapidly exchange with deuterium from the solvent, effectively making your ammonia &quot;invisible&quot;. You may try to dissolve your compounds in deuterated solvents that do not exchange as readily (CDCl3, DMSO-d6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility which I already mentioned is your peak resolution. If you have multiple compounds in this sample at various concentrations then you will likely have many peaks where the ammonia signal could &quot;hide&quot; under, so be careful there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found by experience that when trying to quantify mixtures in 1H NMR (especially as part of a routine) it is extremely valuable to first collect a reference spectrum of each individual compound (when possible) using identical preparation techniques and experimental conditions in order to make peak identification and quantification of mixtures a breeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps you sort out your problem!&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2013-02-02T08:18:39.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4066" PostTypeId="2" ParentId="4052" CreationDate="2013-02-02T16:09:18.223" Score="6" Body="&lt;p&gt;You could answer this question by looking at the possible interactions between a gas molecule and water molecules. The only I can see are van der Waals interactions (because these gas molecules cannot participate in hydrogen bonding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O_2}$ and $\ce{Br_2}$ have no permanent dipole so the only interactions are Debye forces (interaction between water permanent dipole and a gas molecule induced dipole) and London forces (between instantaneous dipoles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The induced/instantaneous dipoles strength increases with the size of the atom because the bigger the electron cloud the easier it is to modify its shape.It happens because the electrons far away from the nucleus are less attracted by it and can move around a bit more easily than electrons very close to the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason why van der waals forces are stronger for bigger atoms.&#xA;This would explain why $\ce{Br_2}$ has a better solubility than $\ce{O_2}$ in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London interactions are very important if you look at liquefaction temperature of gases. For instance, it explains why $\ce{Ar}$ becomes a liquid below 87K while $\ce{He}$ become a liquid only below 4K. &lt;/p&gt;&#xA;" OwnerUserId="1162" LastActivityDate="2013-02-02T16:09:18.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4067" PostTypeId="1" CreationDate="2013-02-02T20:42:00.830" Score="4" ViewCount="3213" Body="&lt;p&gt;I came across some hairy math that's used for analytical centrifugation, mostly for things like determining the sedimentation rate for liquids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I don't understand is how someone determines how long a test tube should be inside a centrifuge? Is it based off of others experimental data or is there a fixed way of doing it? &lt;/p&gt;&#xA;" OwnerUserId="1181" LastEditorUserId="73" LastEditDate="2013-02-03T02:41:22.800" LastActivityDate="2013-05-24T05:28:55.433" Title="How is the time for a test tube in a centrifuge determined?" Tags="&lt;biochemistry&gt;&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4068" PostTypeId="2" ParentId="4046" CreationDate="2013-02-03T01:29:07.143" Score="5" Body="&lt;p&gt;The chemicals must be in water, since magnesium bicarbonate decomposes when dried. So I will assume that the mixing is done in water.  To start with, both salts are strong electrolytes and so they immediately dissociate to give $\ce{Na+}$ ions,&#xA;$\ce{HCO3-}$ ions, $\ce{Mg^{2+}}$ ions, and $\ce{Cl-}$ ions.  As Brenton Home has pointed out, none of the possible compounds formed from these are insoluble and there is no gas formation due to the decomposition of the bicarbonate because the resulting solution is not acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is that there is no reaction.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="675" LastEditDate="2013-02-03T01:46:06.063" LastActivityDate="2013-02-03T01:46:06.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4069" PostTypeId="1" AcceptedAnswerId="4070" CreationDate="2013-02-03T02:22:39.977" Score="4" ViewCount="163" Body="&lt;p&gt;The question states the following  equation:&#xA; $$ \ce{C(s) + 2H2(g) -&amp;gt; CO2(g) + 2H2O}$$&#xA; And then asks you to calculate the enthalpy change for it, given enthalpies of combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stated values are $C(s)$ -393,  $\ce{H2}(g)$ -286,  and $\ce{CH4}(g)$ -890&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to why oxygen atoms have suddenly appeared on the products side. The only thing I have been able to think is that it refers to the enthalpy of formation of methane, and that the answer would be -393 + (2*-286) + 890. But it still doesn't make sense to me considering the equation provided. I understand the products are the products expected from combustion, but no oxygen has been added on the reactants side.&lt;/p&gt;&#xA;" OwnerUserId="1182" LastEditorUserId="1362" LastEditDate="2014-11-04T08:02:24.960" LastActivityDate="2014-11-04T08:02:24.960" Title="How can oxygen atoms suddenly appear on the product side of an equation?" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4070" PostTypeId="2" ParentId="4069" CreationDate="2013-02-03T08:19:19.083" Score="1" Body="&lt;p&gt;They probably want you to calculate the enthalpy change required to &lt;em&gt;convert&lt;/em&gt; the reactants into the products. Another way of looking at it is the enthaply &lt;em&gt;difference&lt;/em&gt; between the products and reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, $\ce{O2(g)}$ has 0 enthalpy of formation in standard state. (so do $\ce{C(s)}$ and $\ce{H2(g)}$, but we don't need that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, adding $\ce{2O2}$ to the reactants side won't change $\Delta H^o=H^o_{products}-H^o_{reactants}$. So calculating $\Delta H^o$ for the given &quot;reaction&quot; is the same as calculating it for $$\ce{C(s) + 2H2(g) +2O2 -&amp;gt; CO2(g) + 2H2O}$$. And $\Delta S^o$ for this reaction can be calculated very easily using enthalpies of combustion. You don't even need to use the given value for methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while the question is wrong in calling it a &lt;em&gt;reaction&lt;/em&gt;, it is still asking us to do something reasonable -- that is, calculate $$\Delta H^o=H^o_{\ce{CO2(g) + 2H2O}} - H^o_{\ce{C(s) + 2H2(g)}}$$. Since we can subtract $H^o_{\ce{2O2(g)}}=0$ easily, the problem can be solved by taking a combustion path.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-02-03T08:19:19.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4071" PostTypeId="1" AcceptedAnswerId="4072" CreationDate="2013-02-03T08:41:21.767" Score="1" ViewCount="123" Body="&lt;p&gt;When we run a current through water (pure or not), $\ce{O_2}$ forms at the anode while $\ce{H_2}$ forms at the cathode. If I understand correctly, as water is already slightly ionized, $\ce{H}$'s snap together (and so do $\ce{O}$'s) during the electron exchange. Is it possible to apply enough heat or enough pressure to water to prevent $\ce{H}$'s and $\ce{O}$'s from snapping together and so we could thus form bubbles of ionized hydrogen and ionized oxygen? &lt;/p&gt;&#xA;" OwnerUserId="1183" LastEditorUserId="22" LastEditDate="2013-02-03T09:27:53.297" LastActivityDate="2013-02-03T09:58:33.170" Title="Is it possible to form bubbles of ionized hydrogen and ionized oxygen under the right pressure/temperature conditions during electrolysis of water?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4072" PostTypeId="2" ParentId="4071" CreationDate="2013-02-03T09:46:29.910" Score="3" Body="&lt;p&gt;The electrolysis will not form any ionized oxygen or ionized hydrogen, but instead proceeds through radical species. Some experiments have attempted to show the presence of $H^{\bullet}$ and $OH^{\bullet}$ radicals at the anode / cathode respectively (see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100494a024&quot; rel=&quot;nofollow&quot;&gt;P.H. Kasai, D. McLeod &lt;em&gt;J. Am. Chem. Soc&lt;/em&gt; (1978)&lt;/a&gt;). Provided that the surface concentration is sufficient, these radical species will very quickly recombine to lead to $H_{2}$ and $O_{2}$. The surface of the electrode is very likely to influence the process, as the radicals species are likely to interact with it and not be released in high concentration in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other factors are also influential, and the radical species might also react with other constituents of the solution (see for instance &lt;a href=&quot;http://www.plosone.org/article/info%3adoi/10.1371/journal.pone.0046392&quot; rel=&quot;nofollow&quot;&gt;T. Mokudai,  K. Nakamura,  T. Kanno,  Y. Niwano, &lt;em&gt;PLOS One&lt;/em&gt; (2012)&lt;/a&gt; on the formation of hydroxyl radical vs formation of hypochlorous acid through electrolysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating the solution is likely to increase radical concentration, but the temperature should be extremely high to prevent radical recombination within the gas. As for pressure, my guess would be that a reduced pressure is likely to improve radical isolation by displacement of the solvation equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2013-02-03T09:58:33.170" LastActivityDate="2013-02-03T09:58:33.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4074" PostTypeId="1" AcceptedAnswerId="4244" CreationDate="2013-02-03T15:38:54.767" Score="2" ViewCount="258" Body="&lt;p&gt;I am trying to run a molecular dynamics simulation with AMBER forcefield. However, it's the first time I use is, and while I know the form of the various intra- and intermolecular energy terms, some details are not clear to me. So, for the AMBER forcefield:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What pairs are excluded in the summation of intramolecular Lennard-Jones interactions?&lt;/strong&gt; Typical choices are “excluding bonded atoms”, “excluding 1-2 and 1-3 pairs”, “excluding 1-2 and 1-3, applying a factor 1/2 to 1-4 pairs”, and “excluding 1-2, 1-3 and 1-4 pairs”.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;For torsion potentials, is the potential scaled by the number of such torsions?&lt;/strong&gt; (as is done in the DREIDING forcefield, to ensure a constant torsional barrier for the central bond) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-02-25T19:42:31.980" Title="Questions about intramolecular interactions in AMBER forcefield" Tags="&lt;theoretical-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4075" PostTypeId="2" ParentId="4048" CreationDate="2013-02-04T02:36:54.540" Score="0" Body="&lt;p&gt;It depends: The rate of electrolysis is limited &lt;strong&gt;either&lt;/strong&gt; by the electric source &lt;strong&gt;or&lt;/strong&gt; by the ion concentration. I am inexperienced with electrolysis but I imagine that quality may increase as the the process slows, much like crystals grow with more imperfections when the rate of crystallization is fast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as an electrolyte solution composed of positive ions and electrons. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Activity coefficients of single ions cannot be measured&#xA;  experimentally because an electrolyte solution must contain both&#xA;  positively charged ions and negatively charged ions.&quot; -&lt;a href=&quot;http://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_theory&quot; rel=&quot;nofollow&quot;&gt;Debye–Hückel&#xA;  theory&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electroplating systems tend to use an electrode source that replenishes the solvent ions plated onto the target electrode. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2013-02-04T03:29:55.297" LastActivityDate="2013-02-04T03:29:55.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4076" PostTypeId="1" AcceptedAnswerId="4077" CreationDate="2013-02-04T19:51:16.443" Score="11" ViewCount="5623" Body="&lt;p&gt;For example, take $\ce{O2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation state of oxygen is -2, yet once its in a molecule its oxidation state becomes zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this so?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorDisplayName="user7951" LastEditDate="2017-07-16T11:26:18.977" LastActivityDate="2021-05-21T16:04:52.273" Title="Why do single elements in a molecule have an oxidation state of zero?" Tags="&lt;molecules&gt;&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4077" PostTypeId="2" ParentId="4076" CreationDate="2013-02-04T20:08:34.840" Score="20" Body="&lt;p&gt;Oxidation states are something of an &quot;accounting fiction&quot;, used to describe where electrons are spending most of their time in a molecule.  This is turn is dictated by electronegativity: the more electronegative, the more the element attracts electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you say oxygen's oxidation state is -2, this is only with respect to other elements of lower electronegativity. Now in the case of oxygen, the only other element with higher electonegativity is fluorine. For example, the molecule Oxygen Difluoride ($\ce{OF2}$) has oxygen in the +2 oxidation state because fluorine is more electronegative!  For any other element in combination with oxygen, however, we &lt;em&gt;assume&lt;/em&gt; the electrons spend their time with the oxygen filling its outer valence shell, thus the -2 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But two oxygens in a molecule of gas have the &lt;em&gt;same&lt;/em&gt; electronegativity, and there is no way to say that the electrons want to spend more time with one or the other oxygen.  Therefore, the &quot;average&quot; electron state for an oxygen molecule is as for the individual atoms, and so the oxidation state is 0.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-02-06T03:10:31.107" LastActivityDate="2013-02-06T03:10:31.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4078" PostTypeId="1" AcceptedAnswerId="4079" CreationDate="2013-02-04T21:58:20.433" Score="11" ViewCount="5027" Body="&lt;p&gt;Below is a $\ce{^1H}$ Nuclear Magnetic Resonance (NMR) spectrum for benzyl alcohol. The hydrogen in the alcohol group gives rise to a unique signal. The two hydrogens in $\ce{CH2}$ give rise to a unique signal with around twice the signal area. What I don't understand is why all hydrogens in the phenyl group are equivalent. My guess would have been there would be 3 signals from the phenyl ring. One pair of $\ce{H}$ that are equally distant from the methanol, another pair that is also equally distant, and last, a single hydrogen that is the furthest away from the methanol. Clearly this is not the case, and the 5 hydrogen give rise to a single signal only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? What is a good way to think about equivalent and non-equivalent atoms in NMR spectrometry? I would rather avoid rules like the one that says &quot;if you can substitute the hydrogen with another atom and it gives the same compound, the hydrogens are equivalent&quot;. I would rather like an intuitive way to think about it, if it's possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zYlGV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="4945" LastEditDate="2015-02-26T06:23:31.030" LastActivityDate="2020-02-20T22:43:59.367" Title="Why are all the phenyl protons in benzyl alcohol equivalent in the ¹H-NMR spectrum?" Tags="&lt;nmr-spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4079" PostTypeId="2" ParentId="4078" CreationDate="2013-02-04T22:16:41.947" Score="15" Body="&lt;p&gt;You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are so close in chemical shift that they overlap and result in this multiplet.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2015-09-08T08:01:36.330" LastActivityDate="2015-09-08T08:01:36.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4080" PostTypeId="1" CreationDate="2013-02-05T05:00:49.590" Score="2" ViewCount="270" Body="&lt;p&gt;I have posted two reactions below. The top reaction is well known and occurs readily and near quantitatively (e.g. in protein modification), while the bottom reaction to my knowledge does not exist in the literature, but would be valuable to me. I would like to ask the experienced organic chemists out there if you think &lt;strong&gt;2&lt;/strong&gt; and &lt;strong&gt;4&lt;/strong&gt; will have similar reactivities with &lt;strong&gt;1&lt;/strong&gt;, or whether the phosphonate moeity will cause trouble or  reduce reactiveness (resonance stabilization or side reactions?).&lt;/p&gt;&#xA;&lt;p&gt;Forgot to add it to the image, but I normally run the top reaction in dichloromethane at a little above 0 C, with &lt;a href=&quot;http://en.wikipedia.org/wiki/N-Hydroxysuccinimide&quot; rel=&quot;nofollow noreferrer&quot;&gt;N-hydroxysuccinimide&lt;/a&gt; produced as a by-product.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hffnz.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Note: I currently don't have the time/resources to test this, but plan to in the future if I think it will work.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-10-16T17:45:24.160" Title="Primary amine versus phosphoramidate reactivity with succinate esters" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;&lt;amines&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4081" PostTypeId="1" CreationDate="2013-02-05T08:19:11.477" Score="0" ViewCount="39512" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following statements best characterizes the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) Both the strong acid and the strong base are completely dissociated in solution whereas the weak acid dissociation equilibrium constant is small enough that there is a measurable amount of the un-dissociated acid present in aqueous solution.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;B) The equivalence point of a titration of a strong acid with a strong base may be observed with an indicator whereas the equivalence point of a titration of a weak acid with a strong base can only be observed with a pH meter.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;C) The titration of a strong acid with a strong base is a fast reaction whereas the titration of a weak acid with a strong base is generally a very slow reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gcsescience.com/aa7.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; seems to indicate that choice C is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For B, all I know is that the pH at the equivalence point does not equal 7.00 (pH &gt; 7.00) for the weak acid titration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For A, I know the first part of that sentence is true. Strong acids and bases are characterized by completely dissociating whereas weak acids do not dissociate to the same extent.  But is it true that there's a &quot;measurable amount&quot; of the un-dissociated acid present in aqueous solutions?&lt;/p&gt;&#xA;" OwnerUserId="1188" LastEditorUserId="4945" LastEditDate="2017-05-19T05:27:25.643" LastActivityDate="2021-05-05T17:03:39.400" Title="What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4082" PostTypeId="2" ParentId="4081" CreationDate="2013-02-05T08:39:26.327" Score="7" Body="&lt;p&gt;The answer is A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a weak acid: acetic acid. It has a $K_a$ of $1.8\cdot  10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of $1~\mathrm{M}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;K_a &amp;amp;= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\&#xA;1.8 \cdot 10^{-5} &amp;amp;= \frac{x \cdot x}{1~\mathrm{M} - x}\\&#xA;x &amp;amp;= 4.233 \cdot 10^{-3}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, at equilibrium, $[\ce{CH3COOH}] = 1~\mathrm{M}- 4.233 \cdot 10^{-3} = 0.995~\mathrm{M}$. As you can see, by far the largest part is not dissociated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is not true, you can still observe the equivalence point. Have a look at the following picture, that shows the titration curve for weak / strong acid titrated with strong base. When choosing an indicator for colorimetric titration select one so that the pH jump at the equivalence point contains the interval $\mathrm{p}K_a\pm1$. Phenolphthalein has a $\mathrm{p}K_a \approx 9$, so to decide if it is a suitable indicator you would check if the pH jumps from 8 to 10 at the equivalence point. It does:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.chemicool.com/img1/graphics/titration-strong-weak.gif&quot; alt=&quot;Titration curve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is not true either. If it was a very slow reaction, you could not titrate accurately, as you would have to wait a long time between every drop of $\ce{NaOH}$.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.383" LastActivityDate="2016-03-18T21:38:25.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4083" PostTypeId="1" AcceptedAnswerId="4084" CreationDate="2013-02-05T19:19:26.800" Score="0" ViewCount="515" Body="&lt;p&gt;I'm having serious problems in the electron configuration of iron dioxide $\ce{Fe(OH)2}$ because all I could come up with was $\ce{Fe}$ double bonded with each oxygen and single bonded with each hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything seems good except for the iron that's bonded with 5 atoms (which I think is weird). Could someone you give me the right electron configuration (or geometry)?&lt;/p&gt;&#xA;" OwnerUserId="913" LastEditorUserId="73" LastEditDate="2013-02-06T03:46:36.553" LastActivityDate="2017-04-08T22:15:40.717" Title="Electronic and geometric configuration of Fe(OH)₂" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4084" PostTypeId="2" ParentId="4083" CreationDate="2013-02-05T20:03:09.597" Score="2" Body="&lt;p&gt;I apologize if there's something to this that I don't know about, but my understanding of it is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\ce{Fe(OH)2}$, also called iron(II)hydroxide (iron-two-hydroxide). The reason it's not called dihydroxide, is that the number (II) describes the oxidation state of the iron, so you can easily calculate how many hydroxyl-groups (which are always -1) it needs to have an overall charge of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In iron(II)hydroxide you don't have covalent bonds (double bonds, single bonds)... You have ionic bonds. This particular compound consists of a $\ce{1Fe}^{2+}\ce{+2OH-}$. They don't SHARE electrons, they are simply held together by their charges!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds group together in lattices. They don't have &quot;geometric&quot; configurations like sulfur dioxide for example. Instead, they are clustered together like in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJFS5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJFS5.gif&quot; alt=&quot;ion lattice&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative dots would be the $\ce{OH-}$ and the positive dots would be $\ce{Fe}^{2+}$. Then, of course, you would have two negatives for each positive. Ion bonds are the strongest type of bonds there is.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="402" LastEditDate="2017-04-08T22:15:40.717" LastActivityDate="2017-04-08T22:15:40.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4085" PostTypeId="1" AcceptedAnswerId="14806" CreationDate="2013-02-05T23:44:30.037" Score="3" ViewCount="220" Body="&lt;p&gt;We made a solution of Tetrakis Copper(I) hexafluorophosphate and the usual color is yellowish. As it oxidizes to Cu(II), it should turn blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, our current mixture has an off color that is pink. Looking up alternative copper colors, there will be copper rose. What would cause that color?&lt;/p&gt;&#xA;" OwnerUserId="57" LastActivityDate="2014-08-01T22:13:34.267" Title="What causes Copper Rose?" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;color&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4086" PostTypeId="2" ParentId="3988" CreationDate="2013-02-05T23:54:01.250" Score="3" Body="&lt;p&gt;I'm not an expert, but typically there are a series of surface reactions where the reactive species are all adsorbed on the surface. So atomic hydrogen wouldn't &quot;leave&quot; the palladium before it reacts, it just diffuses from an internal crystal site to a surface site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the oxygen would a&lt;b&gt;d&lt;/b&gt;sorb onto the Pd surface, I'll assume dissociatively. So now there is atomic O and H diffusing around the surface sites of the Pd. These can react and form adsorbed OH, which can further react and form adsorbed H2O, which can then desorb and give you your water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your enthalpy calculation is off, the equation for water already includes the fact that an oxygen molecule must break apart to form it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you include the hydrogen gas reaction with Pd in your picture, all you're doing is turning hydrogen gas and oxygen gas into water, using Pd as a catalyst. So the total enthalpy change is just -285.83kJ/mol. You don't have enough information to determine how much heat is released (or absorbed possibly) in the hydrogen absorption reaction and how much is released in the water production reaction, but the total of the two will be -285.83kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="1189" LastActivityDate="2013-02-05T23:54:01.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4087" PostTypeId="1" AcceptedAnswerId="4089" CreationDate="2013-02-06T00:31:54.313" Score="3" ViewCount="17745" Body="&lt;p&gt;How can one determine the oxidation number of nitrogen in $\ce{KCN}$?&lt;/p&gt;&#xA;" OwnerUserId="1190" LastEditorUserId="4231" LastEditDate="2015-02-26T16:37:02.637" LastActivityDate="2019-07-17T11:13:50.203" Title="Oxidation number of nitrogen in KCN" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4089" PostTypeId="2" ParentId="4087" CreationDate="2013-02-06T01:57:17.397" Score="2" Body="&lt;p&gt;KCN is a ionic compound consisting of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;. You can know that the cyanide ion will have a negative charge, because a potassium ion always has a positive. Also the elements in the cyanide ion are more electronegative - that means they will have the negative oxidation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know that since potassium has an oxidation state of +1, the cyanide ion has an overall oxidation state of -1. As stated above, ions have oxidation states corresponding to their charge. But you want to know what the oxidation state of the NITROGEN atom has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are rules you can use for polyatomic ions (ions with multiple atoms in it), such as &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;. Nitrogen is more electronegative than carbon. Thus it will have the negative oxidation state (think: it's better at grabbing electrons, which are negatively charged). It is 3 columns away from being a noble gas, thus its oxidation state it -3. The overall oxidation state of the ion is -1, thus carbon must have an oxidation state of +2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up: K = +1, C = +2, N = -3. If you're trying to balance an equation though, you don't need to look at C and N individually. You just look at the &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt; as a whole and the &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I know &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt; was an ion? Since potassium is a metal, and only forms ionic bonds. C and N do not form ionic bonds, they bond covalently, thus they act as a single ion in ionic compounds. Does that make sense? Other than that it's just something you learn to recognize with time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read rules for determining oxidation states here:&#xA;&lt;a href=&quot;http://www.tmcleod.org/cgi-bin/chem1/wiki.cgi?action=browse&amp;amp;diff=3&amp;amp;id=OxidationNumber&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.tmcleod.org/cgi-bin/chem1/wiki.cgi?action=browse&amp;amp;diff=3&amp;amp;id=OxidationNumber&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a comment to this answer explaining what causes electronegativity. A general rule of thumb is that electronegativity increases as you go towards higher group number and low period number. You can study it for yourself in the picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kJCxO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJCxO.jpg&quot; alt=&quot;Electronegativity in periodic table&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://chesterhowe360.wikispaces.com/file/view/electronegativity-chart.jpg/192579142/electronegativity-chart.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikispaces.com&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="15751" LastEditDate="2019-07-17T11:13:50.203" LastActivityDate="2019-07-17T11:13:50.203" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4090" PostTypeId="1" AcceptedAnswerId="4097" CreationDate="2013-02-06T15:10:21.527" Score="7" ViewCount="2156" Body="&lt;p&gt;When you study the electronegativity of the elements, the general trend is that it rises with increasing group number, and decreasing period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly this is because the attractive forces of the nucleus increase quicker than the shielding effect of the electrons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this shielding effect simply a repellant force, or is there another phenomenon at play?&lt;/li&gt;&#xA;&lt;li&gt;Why does the shielding effect not increase as rapidly?&lt;/li&gt;&#xA;&lt;li&gt;Why does the difference between the attractive forces and the shielding effect become less pronounced with increasing period?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YpziJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YpziJ.jpg&quot; alt=&quot;Electronegativity periodic table&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://chesterhowe360.wikispaces.com/file/view/electronegativity-chart.jpg/192579142/electronegativity-chart.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikispaces.com&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="15751" LastEditDate="2019-07-13T14:31:59.797" LastActivityDate="2019-07-13T14:31:59.797" Title="Why does attractive forces of nucleus increase more than shielding across groups?" Tags="&lt;periodic-trends&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4091" PostTypeId="1" AcceptedAnswerId="5783" CreationDate="2013-02-06T21:20:39.520" Score="4" ViewCount="108" Body="&lt;p&gt;How can you estimate how much of a substance with a known partition coefficient and other physical/chemical data regarding the substance will dissolve in a given lipid? In particular I would like to know the answer to this question in the context of the blood-brain barrier (BBB). &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2015-05-02T18:22:28.470" LastActivityDate="2015-05-02T18:22:28.470" Title="How can you estimate how much of a substance will dissolve in a given lipid?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4092" PostTypeId="1" AcceptedAnswerId="4093" CreationDate="2013-02-06T21:30:57.513" Score="7" ViewCount="22688" Body="&lt;p&gt;How does a substance with an endothermic heat of solution dissolve?&lt;/p&gt;&#xA;" OwnerUserId="1196" LastEditorUserId="1160" LastEditDate="2013-02-07T00:50:29.010" LastActivityDate="2013-02-07T00:50:29.010" Title="Why does a substance with an endothermic heat of solution dissolve?" Tags="&lt;heat&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4093" PostTypeId="2" ParentId="4092" CreationDate="2013-02-06T22:37:39.917" Score="10" Body="&lt;p&gt;Dissolution happens in three steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solute-solute attractions must be broken (consumes energy, endotherm), solvent-solvent attractions must be broken (also endotherm), and finally solute-solvent attractions form. This results in a lower energy state and is exotherm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolution will be endotherm if it takes more energy to break the mentioned attractions, than is released in the last step. You are asking then, why is this reaction spontaneous? In other words, why does it occur? To find out we need chemical thermodynamics. $$\Delta G=\Delta H - T\Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G$ = Change in Gibb's free energy for a reaction. $\Delta G&amp;lt;0$ = reaction is spontaneous. Otherwise it is NOT spontaneous (ie. will not occur on its own).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H$ = Change in enthalpy. $\Delta H &amp;lt;0$ = reaction is exotherm. If larger than 0, the reaction is endotherm. Already now we can see that an endotherm reaction will make $\Delta G$ for the reaction more positive, and thus the reaction is less likely to be spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ = temperature in K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S$ = Change in entropy. $\Delta S &amp;gt; 0$ = reaction leads to a state of higher entropy. Entropy describes disorder in a system. Dissolution of a solute in a solvent will always lead to a state of higher disorder, since we go from having all the solute concentrated in a cluster (more ordered) to being spread evenly throughout the solution (more disordered). We can see from the equation that if the reaction leads to higher entropy, then higher temperatures increase likeliness of $\Delta G &amp;lt; 0$, and the reaction will be spontaneous. On the other hand, if we're going to a more ordered state, lower temperatures increase likelihood for a spontaneous reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So how do we apply this to dissolution?&lt;/strong&gt;&#xA;You know that the dissolution reaction is endotherm. That means that $\Delta H &amp;gt; 0$ for this particular reaction. Yet we know that the reaction is spontaneous, since you report that it does indeed occur! This means $\Delta G &amp;lt; 0$. The only way to make that happen is for $\Delta S$ to be positive. So the reaction must go to a more disordered state. Since it's a dissolution, we know that it does. The solute goes from being concentrated in a cluster (ordered) to being spread throughout the solvent (disordered). Pay attention to where temperature is in the equation. It is quite easy to see, that for a dissolution reaction where  $\Delta S &amp;gt; 0$, you can increase the chance of the reaction being spontaneous by increasing temperature!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line&lt;/strong&gt;&#xA;Your dissolution reaction is endotherm, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless! If it wasn't, heating the system could make it so.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2013-02-06T22:50:38.003" LastActivityDate="2013-02-06T22:50:38.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4094" PostTypeId="1" AcceptedAnswerId="4095" CreationDate="2013-02-07T05:16:21.740" Score="13" ViewCount="136506" Body="&lt;p&gt;I used my water kettle (metal ones, not the plastic ones) to boil my water. Usually, if I did not consume all the water in the kettle and the water was cold, I usually topup extra water to the kettle and turn it on to boil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after many weeks of boiling in this way, I notice that there are white powder outside the kettle. (I notice that when the water is boil, some water was splashed to the outside of the kettle and I left those water splashed out to dry as time goes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the white powder actually called fluoride? How do I tested it to verify if the white powder chemical is safe or not?  &lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-11-09T02:38:33.830" Title="White powder observed after boiling water in electric kettle for many weeks" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4095" PostTypeId="2" ParentId="4094" CreationDate="2013-02-07T06:36:54.523" Score="19" Body="&lt;p&gt;It's likely that your home is supplied with &quot;hard&quot; water (that is, water with a high mineral content). These dissolved minerals typically include calcium or magnesium ions (from dissolved calcium bicarbonate and magnesium bicarbonate, respectively). Heating or boiling your &quot;hard&quot; water in a kettle will soften it by precipitating the calcium carbonate as its solubility decreases with increasing temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4PlKK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (calcium carbonate) could be the white powder you have observed after evaporation. Hard water should not pose any health risks, and in fact can add some extra calcium and magnesium to your diet. If it's still concerning to you, there are hundreds of products available that can soften your water before you use it, just google &quot;water softener&quot; and you'll be overwhelmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_water&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Hard_water&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="1178" LastEditDate="2013-02-07T06:50:14.063" LastActivityDate="2013-02-07T06:50:14.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4096" PostTypeId="2" ParentId="4087" CreationDate="2013-02-07T20:04:05.333" Score="0" Body="&lt;p&gt;The oxidation state of $\ce{N}$ is effectively zero, while the oxidation state of $\ce{C}$ is -1. You may rationalize this by looking at the molecular orbitals (MOs) of the cyanide ion. Carbon and nitrogen are sp-hybridized and triply bonded. As the MOs are filled, the bond MOs are filled first followed by the lone pair MOs, one for N and one for $\ce{C}$. Both lone pairs are much lower in energy than the $\sigma^\ast$, $\pi^\ast$, or $lp^\ast$ MOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you may follow the 8-electron rule and use the fact, that carbon and nitrogen are covalently bonded. This also gives a -1 oxidation state for carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behavior is also consistent with various reactions for cyanide, which acts as a nucleophile binding with the carbon instead of nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="4231" LastEditDate="2015-02-26T16:37:45.347" LastActivityDate="2015-02-26T16:37:45.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4097" PostTypeId="2" ParentId="4090" CreationDate="2013-02-07T20:42:04.083" Score="3" Body="&lt;h2&gt;Question (1)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To at least a very good approximation, the shielding effect is merely due to a repulsive force.  Like much of chemistry at the atomic level, electrons experience a balance of counteracting forces due to the nucleus and other electrons.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question (2)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is usually explained in the following way.  As one goes across a period from left to right, the charge on the nucleus increases at each step, increasing the attractive potential for electrons (which you already noted).  In addition, an electron is added to the &lt;em&gt;valence&lt;/em&gt; shell for each step.  Since the added electrons are in the same shell as the preexisting valence electrons, they do not contribute as strongly to shielding as one might initially expect.  Electrons that are in lower shells contribute to shielding more effectively since they are &quot;more likely&quot; to be between a valence shell electron and the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question (3)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One way to justify the increasing effectiveness of shielding in later periods is an extension of the answer above to question (2).  There it was stated that valence shell electrons do not effectively shield other valence shell electrons since they are at the same level.  An natural extension of this would be that the further apart two electronic shells are, the more the lower shell screens the higher shell.  Compared to the valence electrons in Pb, the 1s electrons are essentially right up against the nucleus and, from the view of Pb's 6p electrons, should almost completely neutralize the charge of two protons in the nucleus (i.e. very effective shielding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of looking at this is that as the principle quantum number (n) increases, the orbitals become less penetrating.  This means they spend less time near the nucleus (where the shielding is weak) and more time further from the nucleus (where shielding is strong).&lt;/p&gt;&#xA;" OwnerUserId="1201" LastEditorUserId="1201" LastEditDate="2013-02-11T18:06:54.617" LastActivityDate="2013-02-11T18:06:54.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4098" PostTypeId="1" AcceptedAnswerId="4100" CreationDate="2013-02-07T23:35:14.570" Score="3" ViewCount="182" Body="&lt;p&gt;&lt;strong&gt;How would I synthesise LY-293,284?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MBP2i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Figure 1: LY-293,284&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have satisfied the conditions at the page &lt;a href=&quot;https://chemistry.meta.stackexchange.com/questions/205/should-we-allow-synthesis-questions-subject-to-the-six-constraints-listed-below&quot;&gt;Should we allow Synthesis questions subject to the six constraints listed below?&lt;/a&gt; so please don't just write off this question and close it. The constraints are satisfied as followed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Starter material(s): &lt;strong&gt;5-acetylindole&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aMvRt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Figure 2: 5-acetylindole&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I have completed a &lt;strong&gt;Google search&lt;/strong&gt; to check for &lt;strong&gt;LY-293,284 synthesis&lt;/strong&gt; pathways by typing &quot;LY-293,284 synthesis&quot; and there were no applicable results. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would imagine the synthesis would occur via the following pathways: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Phenylation&lt;/strong&gt; of the &lt;strong&gt;5-acetylindole&lt;/strong&gt; to &lt;strong&gt;phenyl-5-acetylindole&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Oxidation&lt;/strong&gt; of the resulting &lt;strong&gt;phenyl-5-acetylindole&lt;/strong&gt; with &lt;strong&gt;oxidising agents&lt;/strong&gt; such as potassium nitrate?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Di-n-propylamination&lt;/strong&gt; of the resulting &lt;strong&gt;5-acetylindolylphenol&lt;/strong&gt; with &lt;strong&gt;di-n-propylamine&lt;/strong&gt; to &lt;strong&gt;LY-293,284&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The synthesis I'm hoping is &lt;em&gt;simple&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;LY-293,284&lt;/strong&gt; is a &lt;strong&gt;5-HT&lt;sub&gt;1A&lt;/sub&gt;&lt;/strong&gt; receptor &lt;strong&gt;full agonist&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="876" LastEditorUserId="-1" LastEditDate="2017-03-16T15:50:39.383" LastActivityDate="2019-06-27T13:41:29.880" Title="How would you Synthesise LY-293,284?" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4099" PostTypeId="1" AcceptedAnswerId="5819" CreationDate="2013-02-08T04:34:49.723" Score="2" ViewCount="406" Body="&lt;p&gt;There was some spirited debate between the members of my chemistry lab today about the theoretical consequences of ingesting a large amount of the detergent Triton X-100 at a 100% concentration. The &lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=US&amp;amp;language=en&amp;amp;productNumber=93420&amp;amp;brand=FLUKA&amp;amp;PageToGoToURL=http://www.sigmaaldrich.com/catalog/product/fluka/93420?lang=en&quot; rel=&quot;nofollow&quot;&gt;MSDS&lt;/a&gt; provided by Sigma Aldrich indicates that it is &quot;harmful if swallowed,&quot; with a LD50 of 1800mg/kg in rats, but does not provide further detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that it is capable of &lt;a href=&quot;http://en.wikipedia.org/wiki/Triton_X-100#Uses&quot; rel=&quot;nofollow&quot;&gt;decellularizing animal-derived tissues&lt;/a&gt;, would it literally &quot;wash&quot; one's extracellular matrix of its cells after degrading the lipid bilayers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Triton X-100 is incapable of doing so, are there more powerful detergents that would pose a genuine threat if ingested?&lt;/p&gt;&#xA;" OwnerUserId="1202" LastEditorUserId="2095" LastEditDate="2013-08-08T01:08:43.707" LastActivityDate="2013-08-08T01:08:43.707" Title="Impact of detergent on internal living tissue?" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;&lt;safety&gt;&lt;toxicity&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4100" PostTypeId="2" ParentId="4098" CreationDate="2013-02-08T07:19:35.933" Score="5" Body="&lt;p&gt;If you look at Eli Lilly patents and other literature, there doesn't seem to be a &lt;em&gt;simple&lt;/em&gt; synthesis... I've put together an example of how it could be done. Although there are many ways you could go depending on starting materials, this way is based primarily on literature reactions with the exception of the last 3 steps which are Eli Lilly patents. I stopped after about 16 steps as I think that's more than enough to demonstrate how complex these designer molecule syntheses can be. I also left out reagents/reaction conditions just to show the general pathway, and to save on eye strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/agTeJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really the simplicity (or complexity) will depend on where you are able to start on the synthetic pathway (e.g. what starting materials are available to you).&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2013-02-08T07:19:35.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4101" PostTypeId="1" CreationDate="2013-02-08T07:27:29.277" Score="2" ViewCount="175" Body="&lt;p&gt;My friend spilt a &quot;level 4 carcinogen&quot; on their lap.  What is it and how dangerous is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They had washed if off after.&lt;/p&gt;&#xA;" OwnerUserId="1203" LastEditorUserId="22" LastEditDate="2013-02-08T11:17:15.553" LastActivityDate="2013-02-08T11:17:15.553" Title="What is a level 4 carcinogen?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4102" PostTypeId="2" ParentId="4101" CreationDate="2013-02-08T07:50:25.180" Score="3" Body="&lt;p&gt;There are &lt;a href=&quot;http://toxipedia.org/display/toxipedia/Hazard+Rankings#HazardRankings-CancerInformation&quot; rel=&quot;nofollow&quot;&gt;several different ways to rate carcinogens&lt;/a&gt;.  According to &lt;a href=&quot;http://www.mathesongas.com/pdfs/msds/MAT19990.pdf&quot; rel=&quot;nofollow&quot;&gt;its MSDS&lt;/a&gt;, Pyridine belongs to &lt;strong&gt;Group 3, A3:  Known to cause cancer in animals, inadequate evidence in humans&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was just contact with the skin, your friend should wash it off with soap and water immediately for 15 minutes &lt;em&gt;(this is again from the MSDS)&lt;/em&gt;.  Short term exposure can cause &lt;em&gt;&quot;irritation (possibly severe), allergic reactions, sensitivity to light.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if any irritation occurs, it is probably not serious, but &lt;strong&gt;I would recommend seeing a doctor anyways.&lt;/strong&gt;  I doubt there is anything to worry about, but cancer is not something to mess around with.&lt;/p&gt;&#xA;" OwnerUserId="1204" LastActivityDate="2013-02-08T07:50:25.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4103" PostTypeId="1" AcceptedAnswerId="4109" CreationDate="2013-02-08T08:21:05.763" Score="14" ViewCount="372" Body="&lt;p&gt;A very long time ago I performed an experiment in my kitchen, part of which was to remove all water from a solution. Unfortunately I don't remember what was in it, but the point is that as the solution got more concentrated, it gradually stopped boiling, and would instead remain completely still for seconds, until a huge bubble would form instantly at the bottom of the test tube, making a pop sound and shooting some of the solution out of the tube. The effect got stronger and stronger until it became too strong for me to feel safe to continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was unable to get an explanation of this effect out of my school teachers at the time. What I expected was that as the solution got more concentrated, the dissolved compound would start to precipitate, until eventually no water is left. Why did the popping occur instead?&lt;/p&gt;&#xA;" OwnerUserId="1205" LastEditorUserId="4945" LastEditDate="2014-05-13T09:57:14.447" LastActivityDate="2014-05-13T09:57:14.447" Title="Why do some solutions eventually stop boiling and start &quot;popping&quot; as they get more concentrated?" Tags="&lt;water&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4104" PostTypeId="2" ParentId="4094" CreationDate="2013-02-08T09:31:55.213" Score="2" Body="&lt;p&gt;I live a place where the water is extremely hard, and I regularly remove calcium carbonate from my electric heater. The method is quite simple: fill the kettle with water, add citric acid (as solid) and heat. You will notice a lot of gas which is carbondioxide (CaCO3 + 2H+ -&gt; Ca2+ + H2O + CO2). You can use any acid but citric acid is cheap.&lt;/p&gt;&#xA;" OwnerUserId="396" LastActivityDate="2013-02-08T09:31:55.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4105" PostTypeId="1" CreationDate="2013-02-08T16:59:21.547" Score="2" ViewCount="1911" Body="&lt;p&gt;For docking calculations at different values of pH, I am working with flavonoids and would like to know about any software/server that can prepare ligands depending on the pH. Up to now I read about Maestro (Schrodinger) but it is too expensive for us. I wonder if you could suggest other alternatives.&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2015-03-06T19:06:35.810" Title="Software/server for calculation of protonation states of small molecules depending on pH" Tags="&lt;software&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4106" PostTypeId="1" AcceptedAnswerId="4127" CreationDate="2013-02-08T21:32:01.040" Score="3" ViewCount="8100" Body="&lt;p&gt;When talking about gaseous fuel / oxidizer mixture what does one mean by oxidizer if there is no oxygen involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to think of oxidizer as the more electronegative specie, because it is pulling the electron? Or is this definition only valid for solutions?&#xA;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidizer&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; is not clear whether the definition applies to gaseous reactants or only valid for solutions.&lt;/p&gt;&#xA;" OwnerUserId="655" LastEditorUserId="1160" LastEditDate="2013-02-11T12:03:48.150" LastActivityDate="2013-02-11T14:30:38.720" Title="What is an oxidizer?" Tags="&lt;redox&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4107" PostTypeId="2" ParentId="4106" CreationDate="2013-02-08T21:56:04.583" Score="1" Body="&lt;p&gt;Any electron-withdrawing reactant is an oxidizer, whether in gas or liquid phase (solid too). They're called this simply because they act like oxygen, oxygen being the most common oxidizing agent on earth. &lt;/p&gt;&#xA;" OwnerUserId="1207" LastActivityDate="2013-02-08T21:56:04.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4108" PostTypeId="2" ParentId="2897" CreationDate="2013-02-08T22:15:48.647" Score="0" Body="&lt;p&gt;First lets work out moles of $\ce{HCl}$. The concentration is $2\ \mathrm M$, or in other words, 2 moles per liter. To figure out how many moles you have then, you need to figure out how many liters you have. $50\ \mathrm{mL} = 5\times10^{-2}\ \mathrm L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying concentration and volume, you are saying $\frac{\mathrm{mol}}{\mathrm L}\cdot\mathrm L=\mathrm{mol}$. So you have $2\ \mathrm M\times5\times10^{-2}\ \mathrm L=0.1\ \mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then return to some information you've been given. The heat of neutralization is: $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$. So to get the total energy in kJ, you can see by the unit that you have to multiply by the number of moles. That cancels the units like we did with the volume: $$-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\times0.1\ \mathrm{mol}=-5.17\ \mathrm{kJ}$$&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorDisplayName="user7951" LastEditDate="2016-10-31T22:23:10.250" LastActivityDate="2016-10-31T22:23:10.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4109" PostTypeId="2" ParentId="4103" CreationDate="2013-02-09T01:15:11.397" Score="11" Body="&lt;p&gt;What you witnessed was probably something similar to &lt;a href=&quot;http://en.wikipedia.org/wiki/Superheating&quot;&gt;superheating&lt;/a&gt;/&lt;a href=&quot;http://en.wikipedia.org/wiki/Bumping_%28chemistry%29&quot;&gt;bumping&lt;/a&gt;. Essentially, when a solution is heated too rapidly, large pockets of vapor can form beneath the surface of the liquid that have an internal temperature higher than the expected boiling temperature of the liquid. In those situations, even small physical disturbances and/or further addition of heat can provoke the bubbles to burst explosively as the vapor is released. Usually, boiling chips (made of inert, inorganic materials like aluminum oxide, silicates, carbonates, etc.) are added to the reaction vessel in order to prevent that occurrence. The boiling chips provide an uneven and porous surface around which smaller and more numerous vapor nuclei can form, rather than the large bubbles that can form in smooth containers which have relatively few nucleation sites. Furthermore, the chips themselves provide additional surface area for the radiation of heat, allowing more uniform heat distribution. The likelihood of bumping can also be reduced by using wider reaction vessels, which allow superheated bubbles to escape without disturbing all of the surrounding liquid above them.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-02-09T01:23:17.510" LastActivityDate="2013-02-09T01:23:17.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4110" PostTypeId="1" AcceptedAnswerId="4115" CreationDate="2013-02-09T01:21:53.900" Score="6" ViewCount="485" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fentanyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fentanyl&lt;/a&gt; is a potent anilidopiperidine opioid. Where did its INN name come from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TpBHT.png&quot; alt=&quot;Fentanyl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Figure 1: Fentanyl&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the INN name of paracetamol came from its chemical name para-acetylaminophenol (&lt;strong&gt;par-acet-am-ol&lt;/strong&gt;). Is there a meaningful origin of the INN name of fentanyl?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="30836" LastEditDate="2019-04-21T15:56:15.410" LastActivityDate="2019-04-21T15:56:15.410" Title="Where did the INN name of fentanyl come from?" Tags="&lt;nomenclature&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4111" PostTypeId="5" CreationDate="2013-02-09T01:24:14.680" Score="0" Body="&lt;p&gt;International Nonproprietary Name (INN) is the official nonproprietary or generic name given to a pharmaceutical substance, as designated by the World Health Organization (WHO).&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-02-09T04:16:49.290" LastActivityDate="2013-02-09T04:16:49.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4112" PostTypeId="4" CreationDate="2013-02-09T01:24:14.680" Score="0" Body="International Nonproprietary Name (INN) is the official nonproprietary or generic name given to a pharmaceutical substance, as designated by the World Health Organization (WHO)." OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-02-09T04:16:44.330" LastActivityDate="2013-02-09T04:16:44.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4113" PostTypeId="1" CreationDate="2013-02-09T03:39:43.597" Score="3" ViewCount="72" Body="&lt;p&gt;How can long chain alcohols(not thiols) be functionalised on the surface of silver or gold nanoparticles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read in an article that Polyvinyl alcohol is used as a capping agent for silver nanoparticle synthesis, but what is the mechanism for that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of bonds form between nanoparticle and the long chain alcohol(covalent or ionic)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any relationship between chain length of alcohols(or thiols) to the functionalisation of nanoparticles?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="73" LastEditDate="2013-02-09T04:20:27.707" LastActivityDate="2013-02-25T04:51:03.690" Title="What is the mechanism of functionalisation of nanoparticle using long chain alcohols" Tags="&lt;synthesis&gt;&lt;reaction-mechanism&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4114" PostTypeId="1" AcceptedAnswerId="4116" CreationDate="2013-02-09T09:37:53.767" Score="3" ViewCount="132" Body="&lt;p&gt;My teacher gave me the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What are the possible products of these reactions? Then, what is the mechanism of those reaction (elimination (E1 or E2) or subtitution (S&lt;sub&gt;N&lt;/sub&gt;1  or S&lt;sub&gt;N&lt;/sub&gt;2)? Explain in as much detail as possible.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce {CH3Cl + OH- -&amp;gt;} ?$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce {(CH3)3CCl + OH- -&amp;gt;}?$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I suggest that the first reaction will produce methanol and chloride:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce {CH3Cl + OH- -&amp;gt; CH3OH + Cl-}$$&lt;/span&gt;&#xA;and it is a S&lt;sub&gt;N&lt;/sub&gt;2 reaction, because &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; attacking from behind and this makes &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; leave the chloromethane. It becomes &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; because it brings one electron from the carbon atom.&lt;/p&gt;&#xA;&lt;p&gt;Then I suggest the second reaction is a S&lt;sub&gt;N&lt;/sub&gt;1 reaction because there are some barriers behind the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; (the barrier is &lt;span class=&quot;math-container&quot;&gt;$\ce {(CH3)3}$&lt;/span&gt;). But I don't know the product that can be formed.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="95133" LastEditDate="2021-04-12T06:11:16.280" LastActivityDate="2021-04-12T06:13:52.257" Title="Reaction of haloalkane with hydroxide ions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4115" PostTypeId="2" ParentId="4110" CreationDate="2013-02-09T20:16:18.617" Score="5" Body="&lt;p&gt;My guess: the name may be based on two of the immediate precursors to the drug, N-Phenethyl-4-piperidone (FENT, from Phen + ethyl?) and aniline (ANYL):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwCTj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitely not as straight-forward as your para-acetylaminophenol example.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2013-02-09T20:16:18.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4116" PostTypeId="2" ParentId="4114" CreationDate="2013-02-09T21:12:16.963" Score="4" Body="&lt;p&gt;You are on the right track with the hydroxide substituting the chloride, so you can predict the alcohol product in each reaction. In order to determine whether the reactions will be S&lt;sub&gt;N&lt;/sub&gt;1 or S&lt;sub&gt;N&lt;/sub&gt;2, we must look at the mechanism of each reaction:&lt;/p&gt;&#xA;&lt;p&gt;S&lt;sub&gt;N&lt;/sub&gt;1&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WpHMe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;S&lt;sub&gt;N&lt;/sub&gt;2&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WEW7b.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia:Nucleophilic_Substitution&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So we can see that in an S&lt;sub&gt;N&lt;/sub&gt;1 reaction, a &lt;strong&gt;stable&lt;/strong&gt; carbocation intermediate must be formed, while in an S&lt;sub&gt;N&lt;/sub&gt;2 reaction there is no intermediate. We can determine the stability of our carbocation intermediate by looking at the following stability diagram:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uxuT0.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia:Carbocation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From the diagram, you can see that carbocation stability increases with the number of alkyl groups attached to the ionized carbon. If the carbocation is relatively stable, the reaction will likely be S&lt;sub&gt;N&lt;/sub&gt;1, if the carbocation is relatively unstable, the reaction will likely be S&lt;sub&gt;N&lt;/sub&gt;2.&lt;/p&gt;&#xA;&lt;p&gt;Another important idea, which you have mentioned, is that a central carbon surrounded by bulky groups will be more difficult to attack due to steric hindrance (thus not likely an S&lt;sub&gt;N&lt;/sub&gt;2 reaction), and becomes more easily attacked when dissociated to a carbocation.&lt;/p&gt;&#xA;&lt;p&gt;You can use these concepts to find/explain the answers to your question.&lt;/p&gt;&#xA;" OwnerUserId="1178" LastEditorUserId="95133" LastEditDate="2021-04-12T06:13:52.257" LastActivityDate="2021-04-12T06:13:52.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4117" PostTypeId="1" CreationDate="2013-02-10T01:03:05.450" Score="3" ViewCount="110" Body="&lt;p&gt;Ammonia has a heat of formation of $-46 \:\mathrm{kJ/mol}$ (which means that it's formation from the elements releases heat). When mixed with iodine at room temperature, it makes nitrogen triodide (heat of formation +146 complexed with ammonia, +287 pure). With the slightest touch it decomposes explosively back into nitrogen and iodine. Thus the net reaction for both steps seems to be ammonia decomposing into its elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine seems to be catalyzing its decomposition. However, such a highly endothermic reaction seems unlikely (evaporating water is only $40.65 \:\mathrm{kJ/mol}$). Also, ammonia's formation is spontaneous ($-26 \:\mathrm{kJ/mol}$) despite the entropy cost with $1 \:\mathrm{atm}$ reactants. This shouldn't go backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where do we get the energy to make this unstable compound? Iodine is a weak halogen, it barely reacts with hydrogen. Could oxygen drag this reaction forward by stealing hydrogen from ammonia?&lt;/p&gt;&#xA;" OwnerUserId="1211" LastEditorUserId="22" LastEditDate="2013-02-11T08:05:28.430" LastActivityDate="2013-02-11T08:05:28.430" Title="Nitrogen triodide energetics" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4120" PostTypeId="2" ParentId="4117" CreationDate="2013-02-10T08:32:08.560" Score="3" Body="&lt;p&gt;I'm not sure what your question is. The reaction from NI&lt;sub&gt;3&lt;/sub&gt; is largely exothermic (Wikipedia gives –290 kJ/mol as the reaction enthalpy), which is consistent with its explosiveness. The reaction from NI&lt;sub&gt;3&lt;/sub&gt; · NH&lt;sub&gt;3&lt;/sub&gt; is also exothermic: you quote the heat of formation of the NI&lt;sub&gt;3&lt;/sub&gt; · NH&lt;sub&gt;3&lt;/sub&gt; as being +146 kJ/mol, and the heat of formation of product NH&lt;sub&gt;4&lt;/sub&gt;I is &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C12027064&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot; rel=&quot;nofollow&quot;&gt;–202 kJ/mol&lt;/a&gt;. Both are largely favored entropically because they release gas.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-02-10T08:32:08.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4121" PostTypeId="1" CreationDate="2013-02-10T15:03:20.730" Score="1" ViewCount="186" Body="&lt;p&gt;I'm trying to do homework and I have the answers to the problems but I can't read the explanation because it's illegible and I can't find how it's done in my notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the simplest example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ C + O_2 \to CO_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the answer is 44/12 but I don't understand where the 12 comes from at all. How is a question like this approached? (Didn't know what to search in google to get an explanation).&lt;/p&gt;&#xA;" OwnerUserId="1214" LastEditorUserId="865" LastEditDate="2021-05-22T21:11:29.113" LastActivityDate="2021-05-22T21:11:29.113" Title="How to calculate weight of CO₂ in one kg of fuel" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4122" PostTypeId="2" ParentId="4121" CreationDate="2013-02-10T19:31:54.840" Score="7" Body="&lt;p&gt;You have 1 mol $\ce{C}$ reacting with 1 mol $\ce{O_2}$ to give 1 mol of $\ce{CO_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need now is something to relate weight and mol. This is called molar mass, notated with the letter M, and has the unit g/mol. You'd just look that up on the internet, but it's calculated from the weight of the atoms in each molecule. If a molecule weighs 32.2 units, then it'd have a molar mass of 32.2 g/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations are pretty straight forward. How many moles of fuel per kilo? $1000g / M(\ce{C})$, where M, as noted, indicates molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that 1 mol carbon turned into 1 mol carbondioxide. So the number you just calculated is thus also the moles of carbondioxide. You need to find out how many grams those x mol carbondioxide corresponds to: $n(\ce{CO_2}) * M(\ce{CO_2}) = m(\ce{CO_2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not confident in this, pay attention to the units. n = mol; M = g/mol; m = g. Thus multiplying M by n yields a figure in grams. Is that clear?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-02-10T19:31:54.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4123" PostTypeId="1" AcceptedAnswerId="4150" CreationDate="2013-02-10T22:43:41.340" Score="-1" ViewCount="9842" Body="&lt;p&gt;I just don't understand how to do these and would really appreciate your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Ammonium dichromate decomposes in a reaction when heated. Calculate the heat transferred for the decomposition of 53.0 g of ammonium dichromate according to &#xA;$\ce{(NH4)2Cr2O7(s) -&amp;gt; Cr2O3(s) + N2(g) + 4H2O(g)}$ ΔH° = -315 kJ&#xA;Do I convert grams to moles and then multiply by ΔH° ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The standard heat of formation $\Delta H_f$ for $\ce{C2H4}$ is +52.3 kJ/mol. If $\ce{C2H4}$ (ethene) reacts with $\ce{H2}$(g) to produce $\ce{C2H6}$ (g) (ethane) ΔH = -137 kJ. What is the heat formation of ethane?&lt;/p&gt;&#xA;" OwnerUserId="1217" LastEditorUserId="73" LastEditDate="2013-02-10T23:59:08.850" LastActivityDate="2013-02-13T20:44:00.823" Title="Enthalpy Calculations, Heat of Formation/Combustion" Tags="&lt;combustion&gt;&lt;enthalpy&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4124" PostTypeId="1" AcceptedAnswerId="4125" CreationDate="2013-02-11T03:55:34.767" Score="5" ViewCount="2328" Body="&lt;p&gt;Most materials become more dense as they cool, but ice floats in water, meaning that it's less dense. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="1142" LastActivityDate="2018-09-02T12:21:37.033" Title="Why does water become less dense when it cools" Tags="&lt;water&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4125" PostTypeId="2" ParentId="4124" CreationDate="2013-02-11T04:21:54.273" Score="10" Body="&lt;p&gt;Under normal conditions (meaning your regular everyday freezer, or a cold day) water forms a crystal lattice below its melting point (as shown below). This structure is held together by hydrogen bonds, which are no longer overcome by the internal energy of the molecules. The hexagonal arrangement of the water molecules in this structure leads to a lower density due to the spaces left in the center of each hexagon (you can now fit fewer molecules in the same volume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DIjxn.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_Ih&quot; rel=&quot;noreferrer&quot; title=&quot;From Wikipedia&quot;&gt;From Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other solid phases of water (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ice#Phases&quot; rel=&quot;noreferrer&quot;&gt;at least 16 on wikipedia&lt;/a&gt;), some which have densities higher than normal liquid water, though most are not naturally formed (at least on Earth).&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2013-02-11T04:21:54.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4126" PostTypeId="1" AcceptedAnswerId="4129" CreationDate="2013-02-11T05:46:55.623" Score="9" ViewCount="9876" Body="&lt;p&gt;Metals tend to lose electrons to obtain the stable noble gas configuration of 8 valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do they want to obtain this configuration, and how does the strength of their &quot;desire&quot; to obtain this configuration compare with the &quot;desire&quot; to maintain neutral charge.  If the answer depends on the chemical, I'm happy for you to provide some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1219" LastEditorUserId="1160" LastEditDate="2013-02-11T07:17:45.923" LastActivityDate="2014-11-20T01:41:22.867" Title="Why do metals tend to lose electrons, as opposed to maintaining electric neutrality?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4127" PostTypeId="2" ParentId="4106" CreationDate="2013-02-11T07:36:33.847" Score="5" Body="&lt;p&gt;&lt;strong&gt;Oxidation:&lt;/strong&gt; losing electron(s), &lt;strong&gt;Oxidizer / oxidizing agent:&lt;/strong&gt; a chemical that can oxidize another reagent. &lt;strong&gt;Reduction:&lt;/strong&gt; gaining electron(s) - think of it as reduction of charge! &lt;strong&gt;Reducing agent:&lt;/strong&gt; A chemical that can reduce another reagent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And oxidation can never occur without a reduction, meaning if you use an oxidizing agent to oxidize a substance, when the oxidizing agent is in turn being reduced itself. It gains the electrons of the oxidized substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe how well a substance functions as an oxidizer, we use &lt;strong&gt;reduction potentials&lt;/strong&gt;, measured in volts! A substance with a large, positive reduction potential is easy to reduce, which in turn means it is a good oxidizing agent. Similarly, if a substance has a large, negative reduction potential, it means that it is hard to reduce this substance. This is because it's good at reducing other substances itself - it's a reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out if a redox reaction is spontaneous (if it can potentially proceed on its own), we use standard reduction potentials, &lt;strong&gt;$E_0$&lt;/strong&gt;. This is for concentrations of 1 M and a temperature of 25 degrees celcius. You look these up in your textbook or online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An example: Is the following reaction spontaneous?&lt;/strong&gt;&#xA;$$\ce{Cu^{2+}(aq) + 2Ag(s) -&amp;gt; Cu(s) + 2Ag+(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out we first find the standard reduction potentials for each participating species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^{2+}(aq) + 2e- -&amp;gt; Cu(s)}, E_0=+0.34V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ag^{+}(aq) + e- -&amp;gt; Ag(s)}, E_0=+0.80V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This describes how much they want to be reduced (GAIN electrons). The silver ion wants it the most! However, notice that in our reaction we're asking for it to LOSE electrons (be oxidized). THEREFORE: No, the reaction is not spontaneous under the aforementioned standard conditions. However, the REVERSE reaction is spontaneous. To calculate the potential for our reaction, we simply say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_0 = (+0.34V) - (+0.80V) = -0.46V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction potential for the substance that we want to gain electrons (reduced), minus the potential for the substance we want to lose electrons (be oxidized). The fact that the result is negative, is what tells us the reaction will not proceed - and that it will in fact proceed in the OPPOSITE direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take notice that these reduction potentials are not given the unit Volt for no reason! They are true electric potentials. You can think of electric potentials as &quot;electron pressures&quot;. The pole with the highest electron pressure will be (-), because electrons are negatively charged, and the other will be (+) relative to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the reduction potential (the willingness to take electrons) for the silver ion is higher than that of the copper ion. Silver wants to take up electrons more than copper, meaning that electrons will travel from the copper to the silver. you could just as well say that the copper wants to get rid of its electrons more than the silver - copper has the higher electron pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What about electronegativity?&lt;/strong&gt;&#xA;Electronegativity does indeed correlate somewhat with reduction potentials, at least for simple reactions involving pure elements. And you can see that Ag does indeed have a higher electronegativity than Cu, which makes sense. Electronegativity doesn't take into account the oxidation states of different species etc., and quickly it becomes a better idea to just leave electronegativity out of the picture when dealing with redox reactions.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2013-02-11T14:30:38.720" LastActivityDate="2013-02-11T14:30:38.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4128" PostTypeId="2" ParentId="4106" CreationDate="2013-02-11T13:57:51.497" Score="3" Body="&lt;p&gt;Brian's answer is very good and thorough, but there is one rather important empirical fact you need to consider with reduction potentials.  While the difference in $\ce{E_0}$ values can tell you if a reaction &lt;em&gt;can&lt;/em&gt; occur, it cannot tell you if a reaction &lt;em&gt;will&lt;/em&gt; occur.  There are other factors, like reaction rate and activation energy, that can interfere with what $\ce{E_0}$ values indicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two terrific CHEM Study videos that demonstrate these points.  The first, &lt;a href=&quot;http://www.youtube.com/watch?v=NFjc57a2u4M&quot; rel=&quot;nofollow&quot; title=&quot;Bromine: Element from the Sea&quot;&gt;Bromine: Element from the Sea&lt;/a&gt;, shows a procedure for oxidizing bromide ion in seawater to elemental bromine. At about the 9:30 mark, they discuss looking up $\ce{E_0}$ values to find an oxidizer for bromine.  They first try oxygen, which indicates a spontaneous reaction by $\ce{E_0}$, but in fact the reaction does not occur, likely because the rate is too slow.  They then try chlorine, which works to oxidize the bromine.  In this example, the difference in $\ce{E_0}$ values is higher between $\ce{Br_2}$ and $\ce{Cl_2}$ than between $\ce{Br_2}$ and $\ce{O_2}$, so you might reasonably conclude that a higher difference in $\ce{E_0}$ values indicates a faster reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there is more to it, as the second video shows.  The CHEM Study video on &lt;a href=&quot;http://www.youtube.com/watch?v=cGTU55jf0cY&quot; rel=&quot;nofollow&quot; title=&quot;Nitric Acid&quot;&gt;Nitric Acid&lt;/a&gt; shows that nitric acid--$\ce{HNO_3}$--can be used as a strong oxidizing agent, owing to nitrogen in the $\ce{+5}$ state.  At around the 10:30 mark, they discuss the potential reduction products that can be produced from nitric acid (mostly nitrogen-oxygen gases) by consulting $\ce{E_0}$ values. The highest potential is for nitrogen gas, $\ce{N_2}$.  But when an experiment to oxidize metals is conducted, poisonous nitrogen dioxide $\ce{(NO_2)}$ is the product, not $\ce{N_2}$, even though $\ce{NO_2}$ has a smaller $\ce{E_0}$ value.  The answer is that reduction to $\ce{N_2}$ requires a higher activation energy, and at room temperature this reaction mostly does not occur, but reduction to $\ce{NO_2}$ dominates.  Not realizing this possibility could literally be the difference between life $\ce{(N_2)}$ and death $\ce{(NO_2)}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always in science, experimentation is the ultimate determinant of what works. Theory such as reduction potentials helps to determine what &lt;em&gt;could&lt;/em&gt; work, but it is only the beginning of the process.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-02-11T13:57:51.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4129" PostTypeId="2" ParentId="4126" CreationDate="2013-02-11T17:19:45.207" Score="6" Body="&lt;p&gt;Firstly, atoms &quot;want&quot; to achieve the noble gas configuration of 8 valence electrons because it is the most stable form. All that means is that it doesn't tend to react under normal conditions that we experience on Earth, therefore it will stay in that configuration for quite a while and are less likely to react. There is a more complex quantum physical answer for that but you'll have to go elsewhere for than. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main force that keeps electrons in atoms is the electrical attraction between the electrons and the protons in the nucleus and so, if it is more energetically favourable to lose that electron in order to form a bond, then that is what will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focusing on the Alkali metals as an example, as you move down the group, they get more and more reactive. This is because of two main reasons that are a result of the electrons being further away from the nucleus:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Because they're further away, the attraction between the protons and the outer most electron is less&lt;/li&gt;&#xA;&lt;li&gt;Secondly, taking Rubidium as an example, it has 37 electrons and 37 protons. From the perspective of the outer-most electron, there are 36 electrons repelling it, and 37 protons attracting it, therefore acting as a net charge of 1. However, if you take into account the first point, the repulsion of the closer electrons is stronger than the attraction of the protons so it could even be less than one&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1220" LastActivityDate="2013-02-11T17:19:45.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4130" PostTypeId="1" AcceptedAnswerId="4134" CreationDate="2013-02-11T17:45:41.013" Score="7" ViewCount="852" Body="&lt;p&gt;I understand Le Chatelier's principle and how every change to an equilibrium system causes an opposing reaction from the system. I also understand how, when pressure is increased, the equilibrium shifts to the side with the lowest number of moles of gas and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does this happen? How is it more energetically favourable - or is it something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are these explanations of why the ration changes for the other factors right:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Temperature - equilibrium moves to the endothermic side of the reaction as temperature increases because the activation enthalpy of the exothermic reaction increases as the overall energy of the system is less &lt;/li&gt;&#xA;&lt;li&gt;Concentration - ratio of products/reactants changes immediately after one part of the reaction's concentration increases, it then approaches the same ratio (assuming one whole side of the reaction is added)&lt;/li&gt;&#xA;&lt;li&gt;Catalysts do nothing more than decrease the time it takes to reach equilibrium as they catalyse both sides of a reversible reaction and reduce the activation enthalpy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1220" LastActivityDate="2014-08-30T18:36:23.830" Title="Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4131" PostTypeId="1" CreationDate="2013-02-11T17:54:06.717" Score="-1" ViewCount="935" Body="&lt;p&gt;If I had an ionic molecule which needed $\pu{4 eV}$ to break the ionic bond, a $\pu{7 eV}$ photon is shot at it. If it is absorbed by the molecule and breaks,  where does the rest of the energy go?&lt;/p&gt;&#xA;" OwnerUserId="1221" LastEditorUserId="41328" LastEditDate="2017-08-15T22:10:19.710" LastActivityDate="2017-08-15T22:10:49.557" Title="Effect of light on ionic compounds" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;&lt;photochemistry&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4132" PostTypeId="1" AcceptedAnswerId="4133" CreationDate="2013-02-11T18:58:40.423" Score="2" ViewCount="8215" Body="&lt;p&gt;During the lecture in my chemistry class, my teacher posed the following problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Arrange in terms of increasing vapor pressure:&lt;/p&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5m}$&lt;/span&gt; sugar&lt;/p&gt;&#xA;&lt;p&gt;b) &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5m} ~ \ce{NaCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;c) &lt;span class=&quot;math-container&quot;&gt;$\pu{2m}$&lt;/span&gt; ethylene glycol&lt;/p&gt;&#xA;&lt;p&gt;d) &lt;span class=&quot;math-container&quot;&gt;$ \pu{2m} ~\ce{ LiBr}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer turned out to be: d, c, b, a. However, I do not understand why this is the case. I know the formula for vapor pressure is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta P = -XP^\circ $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which means that the pressure of the solvent is proportional to the mole fraction of the solute. I'm still not grasping why the aforementioned answer is correct though. Can anyone provide me with an explanation?&lt;/p&gt;&#xA;" OwnerUserId="1222" LastEditorUserId="69830" LastEditDate="2021-05-19T06:08:22.433" LastActivityDate="2021-05-19T06:12:44.130" Title="How to arrange substances in order of increasing vapor pressure?" Tags="&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4133" PostTypeId="2" ParentId="4132" CreationDate="2013-02-11T20:02:49.683" Score="2" Body="&lt;p&gt;To solve this problem, use Raoult's law. In Raoult's law, the vapor pressure of a fluid is lowered proportionally with the mole fraction of solute particles in the fluid.&lt;/p&gt;&#xA;&lt;p&gt;You know the formula, but you only need it if you want to calculate the &lt;strong&gt;actual&lt;/strong&gt; resulting vapor pressure. We can know which is bigger like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5M }$&lt;/span&gt; sugar. Sugar does not ionize, thus there are &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 M }$&lt;/span&gt; particles in the solvent.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;B.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5M \ce{NaCl}}$&lt;/span&gt;. Sodium chloride dissociates to two ions. Thus there are &lt;span class=&quot;math-container&quot;&gt;$0.5 \times \pu{2 M}$&lt;/span&gt; particles in the solvent.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;C.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{2M}$&lt;/span&gt; ethylene glycol. Again, this doesn't dissociate, so there are &lt;span class=&quot;math-container&quot;&gt;$\pu{2M}$&lt;/span&gt; particles.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;D.&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{2M \ce{LiBr}}$&lt;/span&gt;. Dissociates, resulting in &lt;span class=&quot;math-container&quot;&gt;$2 \times \pu{2 M}$&lt;/span&gt; particles.&lt;/p&gt;&#xA;&lt;p&gt;Thus, listing the concentration of particles in the solvent, you see where the answer comes from:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;D:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{4 M}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;C:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;B:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 M}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="69830" LastEditDate="2021-05-19T06:12:44.130" LastActivityDate="2021-05-19T06:12:44.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4134" PostTypeId="2" ParentId="4130" CreationDate="2013-02-11T20:44:31.940" Score="6" Body="&lt;p&gt;With gasses, what you're doing by changing the pressure is you change the partial pressures or the reactants. As long as there's the same moles of gas on either side, the equilibrium is unaffected, but if there's an un unequal number, the reaction quotient is changed. The same would happen if you added water to an aqueous reaction. You can play with the numbers yourself, I'll give you an example to use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2(g) + 3H2(g) &amp;lt;-&amp;gt;2NH3(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use the reaction quotient with partial pressures, but it's more clear if we use the one with concentrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c\ce{ = \frac{[NH3]^2}{[N2][H2]^3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $c=\frac{n}{V}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c\ce{ = \frac{\frac{n(NH3)^2}{V^2}}{\frac{n(N2)}{V}\times \frac{n(H2)^3}{V^3}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take notice of how this fraction depends on volume! So it's really just the system reacting to attempt to reach equilibrium again (making it so that K = Q).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for temperature. My understanding is that it's not to do with activation energy. It IS related to the enthalpy of the reaction though, and your understanding of what a temperature change means for a particular reaction is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentration is the same deal as I've talked about in the beginning of this post, and incidentally it matches your own understanding of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've got catalysts right too. Have a look at the following diagram:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OYbW8.png&quot; alt=&quot;Activation energy with and without catalyst&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-02-11T20:44:31.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4135" PostTypeId="1" AcceptedAnswerId="64379" CreationDate="2013-02-12T03:02:28.480" Score="46" ViewCount="44009" Body="&lt;p&gt;A lot of the organometallics are rather... interesting compounds to work with. The most famous (among those who care, anyway) is &lt;em&gt;tert&lt;/em&gt;-butyllithium or t-BuLi. It is the textbook example of a pyrophoric substance, demonstrated to pretty much every chemistry major as an air-sensitive chemical requiring special handling (Syringe and cannula transfers, gas-tight septa, argon/nitrogen blankets, that kind of thing). Despite the inherent &lt;a href=&quot;https://www.google.com/search?q=Sheharbano%20Sangji&quot; rel=&quot;nofollow noreferrer&quot;&gt;hazards&lt;/a&gt; of the stuff, it's widely used in the industry (hence widespread notoriety) to add butyl groups to an organic molecule (most organometallics are useful for this kind of carbon-carbon bond formation), and you can buy it by the - airtight - gallon canister.&lt;/p&gt;&#xA;&lt;p&gt;There are, surprisingly (except to most organic chemists), even more dangerous compounds in the average organic chemistry lab. Along the pyrophoric line, many of the multi-methyl-metallics are &lt;em&gt;violently&lt;/em&gt; pyrophoric (even compared to BuLi), including trimethylaluminum, dimethylzinc, and dimethylmagnesium. All of these are also extremely poisonous (anything these compounds can do to your wonder-drug-in-progress, they can also do to various key structures in your own body), with dubious honors reserved for &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethylmercury&quot; rel=&quot;nofollow noreferrer&quot;&gt;dimethylmercury&lt;/a&gt;. Because many of these, especially the organoalkaline compounds, react violently with water, they also automatically rate at least a 3 on the &amp;quot;reactivity&amp;quot; scale.&lt;/p&gt;&#xA;&lt;p&gt;When leads me to wonder, because I wonder about these things from the safety of my office chair in a completely unrelated field; just how bad can it get? Specifically, &lt;strong&gt;is there any compound with an accepted use in laboratory or industrial application that is nasty enough to max out the &lt;em&gt;entire&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/NFPA_704&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFPA-704 diamond&lt;/a&gt;?&lt;/strong&gt; The closest I can find is trimethylaluminum, at a 3-4-3 (heath-fire-reactivity). T-BuLi is a 3-3-4. I can't find NFPA data on straight diazomethane (which may be because nobody in their right mind ever works with the stuff in its pure gaseous form; it's always used in a &lt;em&gt;dilute&lt;/em&gt; diethyl ether solution, and even then is never sold or shipped that way) but it would probably be a finalist, as the gas is acutely toxic, autoignites at room temperature, &lt;em&gt;and&lt;/em&gt; detonates on standing (something for everyone!). I'm thinking that these two inorganic families - light alkyl-metallics and organo-polyazides - would be the most likely candidates to produce a compound so toxic, so flammable, and so readily reactive, yet so interesting to chemistry, that the NFPA would see fit to rate it, and would give it highest honors.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="1782" LastEditDate="2021-04-29T16:21:48.153" LastActivityDate="2021-04-29T16:21:48.153" Title="Is there any substance that's a 4-4-4 on the NFPA diamond?" Tags="&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="5" CommentCount="8" FavoriteCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="4136" PostTypeId="1" CreationDate="2013-02-12T04:14:28.737" Score="6" ViewCount="457" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I originally posted &lt;a href=&quot;https://math.stackexchange.com/q/300382/24293&quot;&gt;a similar question on math.stackexchange.com&lt;/a&gt;, but I believe I might have a better chance of having my question answered here after giving it some time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was thinking about the activation energy of chemical reactions, and I was wondering if there exists a bifurcation node somewhere in the transition state.   I'm wondering about this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yo3Y0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what's bugging me about it. I don't have an actual equation to the curve, so I can't necessarily do a quantitative analysis. Regardless, I can at least make conjectures about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, my instincts are telling me, provided there truly is a bifurcation node, it would be stable at the peak, which sounds odd. From the mathematical definition, stability indicating that any small perturbations from the bifurcation point will &quot;dampen&quot; out in time, rather than flying off into infinity. In context though, that doesn't sound correct. The reaction has to go one way or another. The peak, representing the transition state, is said to be unstable chemically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is with the parameter of time. It has to go forward. In order to analyze it, we would more than likely have to nondimensionalize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just some of my current thoughts on it. My question is regarding the mathematics of it. I'm just debating whether there is a bifurcation point (which I believe there is), and if so whether it is stable, unstable, or halfstable. Basically, I'm not sure if my reasoning is correct right now. Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Let me also add that instability does not necessarily mean that a phase point will fly off into infinity. Stability and instability can be be local. In other words, an unstable point can fly into another equilibrium point.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to add a lot more detail to this post as I've done some more research and thinking on this topic. First off, my assumption was correct - in fact, transition state theory is, in part, based on potential energy surface concepts. Hence, saddle-points (bifurcations), are incorporated into the theory. Second, time is not necessarily the independent variable. Typically, the graph of potential energy vs. I.V. where the I.V. is usually a coordinate of some sorts - for instance, bond lengths. That said, I believe that modifies my question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain, what I am curious about is the stability of the transient point as a function of time. Right now, my graph does not reflect that, and therefore my conclusions are not necessarily correct. The slope would be, in terms of instantaneous slope, $\frac{dV}{dq}$. But that's not what I want. Rather, I'm looking for $\frac{dV}{dt}$. Without showing the derivation, one can come up with an equivalency: $\frac{dV}{dt} = -(\frac{dV}{dx})^2$. What winds up happening, now, is that the local maxima in the picture corresponds to an unstable fixed point. That makes sense. Any thoughts on this?&lt;/p&gt;&#xA;" OwnerUserId="1224" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:09.340" LastActivityDate="2013-02-13T03:26:00.173" Title="Is there a bifurcation node in the potential energy as a function of reaction progress equation of chemical reactions?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4137" PostTypeId="1" AcceptedAnswerId="4139" CreationDate="2013-02-12T05:09:11.590" Score="4" ViewCount="3236" Body="&lt;p&gt;I’m currently trying to learn some chemistry and have run into some trouble understanding some of the basics. The confusion stems from a passage in my textbook, so I will list the brief passage and then ask my question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to predicting molecular formulas, Avogadro’s hypothesis gives correct results for the relative atomic masses of the elements. Analysis by chemists during the 18th century had revealed that 1g of hydrogen combines fully with 8g of oxygen to make 9g of water. If Dalton’s formula for water, HO, were correct, then an atom of oxygen would have to weigh eight times as much as an atom of hydrogen; that is, Dalton’s assumption requires the relative atomic mass of oxygen to be 8 on a scale where the relative atomic mass of hydrogen is set at 1. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far so good, I believe that oxygen has a relative mass of 8 because there are 8g of oxygen in the original reaction and likewise with the 1g of hydrogen, but this next part is where it starts to get confusing. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From Avogadro’s hypothesis, however, each water molecule contains twice as many atoms of hydrogen as oxygen, so to achieve the experimental mass relationship, each oxygen atom must have twice as large a relative atomic mass. This gives a relative atomic mass of 16 for oxygen, a result consistent with modern measurements. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alright, so basically what I’m confused about is how they arrived at a relative atomic mass of 16 given that there is 8g of oxygen and 1g of H in the original experiment. What exactly are they observing in order to establish the relative atomic mass of oxygen as 16. Any help understanding this and relative mass in general would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1225" LastActivityDate="2013-02-12T18:42:49.807" Title="Basic Understanding of Relative Atomic Mass" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4138" PostTypeId="1" AcceptedAnswerId="4141" CreationDate="2013-02-12T07:35:59.157" Score="12" ViewCount="4721" Body="&lt;p&gt;Here are the MOs (more accurately, the breakdown of the MOs) for conjugated hexatriene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ff8Fu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when first asked to draw the MOs myself, I drew this one for $\psi_3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Snikc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It still has two nodes, and is still symmetric. I see nothing wrong with it, except that it doesn't match the &quot;correct&quot; MOs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for $\psi^*_4$, I can draw&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bdGil.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, as I understand it, MOs are formed from various combinations of atomic orbitals, subject to a few rules regarding symmetry. From what I know, the two MOs I've given above should be valid MOs for hexatriene, or at least components of an MO of hexatriene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't these two MOs valid? Also, is there a methodical way to obtain the sequence of MOs in the general case(of a conjugated polyene)?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-02-12T07:46:56.950" LastActivityDate="2016-02-13T14:24:14.810" Title="Why are these molecular orbitals invalid for hexatriene?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4139" PostTypeId="2" ParentId="4137" CreationDate="2013-02-12T11:00:40.250" Score="3" Body="&lt;p&gt;It's because there's two hydrogen atoms in water, not one. In order to show you, let's assume we're making HO instead of H2O. This would clearly only require half the amount of hydrogen, but still the same amount of oxygen. Thus we have .5 grams of hydrogen combining with 8 grams of oxygen in a 1:1 relationship. Thus the oxygen weighs 16 times more than the hydrogen. Do you understand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In H2O they combine in a 2:1 relationship. We have 1 gram of hydrogen combining with 8 grams of oxygen, but because there's 2 hydrogen for every oxygen, we actually need TWICE as many oxygen atoms to have the same amount of atoms. In other words, there will be the same number of atoms in 1g of hydrogen as there is in 16g of oxygen.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-02-12T11:00:40.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4140" PostTypeId="2" ParentId="4137" CreationDate="2013-02-12T11:13:11.463" Score="3" Body="&lt;p&gt;Basically, the error here is Dalton's in assuming that each mole of water contains an equal number of oxygen and hydrogen atoms. It doesn't; the molecular formula of water is H2O - each mole of water contains 2 moles of hydrogen (2*1g = 2g) and one mole of oxygen (16g). Water has a relative molecular mass of 18g. For every mole of oxygen there are &lt;em&gt;two&lt;/em&gt; moles of hydrogen:&#xA; $$\ce{2H2 + O2 -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this particular case, &lt;em&gt;one&lt;/em&gt; mole of hydrogen (1g) and &lt;em&gt;half&lt;/em&gt; a mole of oxygen (0.5*16 = 8g) are reacting to produce &lt;em&gt;half&lt;/em&gt; a mole of water(9g). To get the original relative &lt;em&gt;molar&lt;/em&gt; mass of oxygen they would have to double the mass of oxygen used - 8g - which would mean the result of 16g. This then gives a relative atomic mass of 16. Oxygen contains more isotopes (variants of an element with different relative atomic masses) than just oxygen-16 (such as oxygen-17 or oxygen-18), but to the level of accuracy we're working with it doesn't matter too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, oxygen &lt;em&gt;does&lt;/em&gt; have a relative atomic mass of 16. Ignoring isotopes, it could be worth looking at the (probably simplified) structure of the nuclei of oxygen and hydrogen . Certainly at this level, the relative mass of protons and neutrons is taken to be &quot;1&quot; (the two particles do have very slightly different masses, but this makes things easier). A 'typical' hydrogen nucleus just contains a proton with around 0 neutrons, so it has a relative atomic mass of 1 whereas a 'typical' oxygen nucleus contains 8 protons  and around 8 neutrons; it has a relative atomic mass of 16. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as mentioned before, different isotopes containing different numbers of neutrons do exist. Relative atomic mass is just a weighted mean of these (which is why copper can have a relative atomic mass of 63.5 or chlorine can have one of 35.5), but the model above, whilst being a massive simplification and better for relative &lt;em&gt;isotopic&lt;/em&gt; mass, might help you visualise the difference better.&lt;/p&gt;&#xA;" OwnerUserId="753" LastEditorUserId="753" LastEditDate="2013-02-12T18:42:49.807" LastActivityDate="2013-02-12T18:42:49.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4141" PostTypeId="2" ParentId="4138" CreationDate="2013-02-12T13:11:59.047" Score="6" Body="&lt;p&gt;&lt;strong&gt;The general method to find molecular orbitals of π electrons in conjugated hydrocarbons in first approximation is called &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;noreferrer&quot;&gt;Hückel’s method&lt;/a&gt;&lt;/strong&gt;. It is an undergrad-level topic, and you should find lots of explanations on the internet or in undergrad physical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave function you propose is actually equal to $(-\psi_1 +\psi_3 + \psi_5^*)$   (with normalization factor not indicated). Thus, it is not an eigenstate of the system, and not a molecular orbital. But… unless you know Hückel's method, I am not sure there is an intuitive reason to discard this possibility: it has the right number of nodes (i.e. +/– inversions) for $\psi_3$.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-02-12T13:47:17.020" LastActivityDate="2013-02-12T13:47:17.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4142" PostTypeId="2" ParentId="4105" CreationDate="2013-02-12T16:28:29.747" Score="2" Body="&lt;p&gt;For this, you can try to use the LigpKa database server.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-02-12T16:28:29.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4143" PostTypeId="1" AcceptedAnswerId="4146" CreationDate="2013-02-12T19:40:50.123" Score="9" ViewCount="4406" Body="&lt;p&gt;Can someone help me identify the steps for the following reaction? I'd like to keep track of everything that happens. I can't figure out what happens with oxidation in this context, and thus can't get started.&lt;/p&gt;&#xA;&lt;p&gt;It's 2-methylpropan-1-ol oxidized with potassium dichromate(VI) in aqueous sulfuric acid to yield 2-methylpropanoic acid (also known as isobutyric acid).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J9vHn.png&quot; alt=&quot;Reaction schematic&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-01-14T08:39:30.927" Title="Reaction mechanism for oxidation of primary alcohol to carboxylic acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4144" PostTypeId="1" CreationDate="2013-02-13T03:19:06.993" Score="6" ViewCount="2983" Body="&lt;p&gt;I ran an IR spectrum ($\ce{KBr}$ pellet) and found a sharp absorbance at $2360 \text{cm}^{-1}$, which was unexpected. As far as I can tell, this indicates either a nitrile or $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there is no nitrile in my product. Would you expect atmospheric $\ce{CO2}$ to show up to a noticeable degree in a typical IR spectrum?&lt;/p&gt;&#xA;" OwnerUserId="1239" LastEditorUserId="73" LastEditDate="2013-02-13T03:32:41.540" LastActivityDate="2013-02-13T07:50:30.750" Title="How much CO₂ should one expect in an IR spectrum?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4145" PostTypeId="2" ParentId="4144" CreationDate="2013-02-13T04:38:04.550" Score="5" Body="&lt;p&gt;From what I've read (and personal experience with IRs I've run), a relatively intense $\ce{CO2}$ asymmetric bond stretching signal at approximately $\ce{2350cm^{-1}}$ is an extremely common background artifact in IR spectra (along with the broad, weak signals downfield of $\ce{3000cm^{-1}}$ attributable to $\ce{O-H}$ bonds from ambient moisture). Depending on your software, it's often possible to run a &quot;dry&quot; IR reading with no sample present in order to establish a baseline spectrum of the surrounding air, which can then be checked against by your software (so that overlapping signals from the baseline air spectrum are deleted from the spectrum of your sample). Alternatively, you can take measures to dry the optics (I've once used $\ce{N2}$ for that purpose; alternatively, air directed through $\ce{CaCl2}$ drying tubes could possibly be used, but I can't attest to the efficacy of that). That manual drying procedure takes some effort, and might be futile in any case depending on the conditions of the lab. My one experience with that approach was disappointing.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-02-13T04:38:04.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4146" PostTypeId="2" ParentId="4143" CreationDate="2013-02-13T04:59:21.550" Score="7" Body="&lt;p&gt;A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid &lt;em&gt;in situ&lt;/em&gt;, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl &amp;pi;-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="402" LastEditDate="2017-01-14T08:39:30.927" LastActivityDate="2017-01-14T08:39:30.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4147" PostTypeId="2" ParentId="4144" CreationDate="2013-02-13T06:42:34.340" Score="4" Body="&lt;p&gt;It's rather common in my experience. If you background substraction doesn't work perfectly, you'll get a noticeable $\ce{CO2}$ signal. It can be either negative or positive due to the incomplete substraction of the background signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the spectrum multiple times, this signal should vary in intensity as the $\ce{CO2}$ concentration varies slightly. Though this is probably exaggerated in the situation where I observed it, as we have an open IR spectrometer and often multiple students exhaling $\ce{CO2}$ around it.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2013-02-13T07:50:30.750" LastActivityDate="2013-02-13T07:50:30.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4149" PostTypeId="1" CreationDate="2013-02-13T13:06:06.487" Score="8" ViewCount="817" Body="&lt;p&gt;I am currently in the process of connecting transition state (TS) structures to other minima in order to determine a reaction pathway.  Now I understand the basic concepts of this method.  You start with your TS, run an IRC in both the Forward and Reverse directions, then you optimize the final structure in the successful IRC run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what constitutes connectivity between a TS and a minimum energy structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) I run an IRC and I take the last resulting structure and optimize it. The optimization takes 4 steps and the geometry barely changed.  Is this structure connected to the TS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) I run an IRC and I take the last resulting structure and optimize it.  The optimization takes 20 steps and the geometry changed fairly appreciably, however, connectivity of the atoms remains the same.  Is this structure connected to the TS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) I have a TS where a water is being formed next to some other molecule.  I run the IRC and following the pathway shows you that the water forms and is being moved away from the molecule.  The IRC successfully finishes so I take the last structure and optimize it.  The optimization brings the water back towards the molecule to some particular spot where H-bonding occurs with the molecule.  Is this structure connected to the TS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These examples were purposely laid out in order to determine at what point TS-&gt;Minima are or are not connected.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2013-02-13T15:13:17.077" LastActivityDate="2017-01-23T15:56:51.713" Title="Intrinsic Reaction Coordinate - What does 'Connectivity' really mean?" Tags="&lt;reaction-mechanism&gt;&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;reaction-coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4150" PostTypeId="2" ParentId="4123" CreationDate="2013-02-13T20:44:00.823" Score="1" Body="&lt;p&gt;&lt;strong&gt;Exercise 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decomposition of ammonium dichromate has a standard enthalpy change of $-315 \:\mathrm{kJ/mol}$. So you have a figure that relates the heat to moles. You need to find out how many moles correspond to 53.0 g of the ammonium dichromate. We use molar mass for this. You can calculate molar mass by simply summing the weights of all atoms in the molecule, but I use &lt;a href=&quot;http://www.webqc.org/mmcalc.php&quot; rel=&quot;nofollow&quot;&gt;this tool&lt;/a&gt; to do it. $$\ce{M}_{\ce{(NH4)2Cr2O7}}=252.06 \:\mathrm{\frac{g}{mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\:\mathrm{g/\frac{g}{mol}=mol}$$&#xA;$$\Downarrow$$&#xA;$$\frac{53.0\:\mathrm{g}}{252.06 \:\mathrm{\frac{g}{mol}}}=0.21\:\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good. Our figure of enthalpy change is per mole, thus you simply multiply the two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\:\mathrm{\frac{kJ}{mol}*mol=kJ}$$&#xA;$$\Downarrow$$&#xA;$$\Delta H^\Theta-315 \:\mathrm{\frac{kJ}{mol}}*0.21\:\mathrm{mol}=-66.15\:\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exercise 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this exercise you need to apply Hess' law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H^\Theta_{reaction}=\sum \Delta H^\Theta _{f(products)}-\sum \Delta H^\Theta _{f(reactants)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we actually need a figure for hydrogen. You look that up on the internet for example, and see that  $\Delta H^\Theta _f \ce{(H2)}=0 \:\mathrm{\frac{kJ}{mol}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now use Hess' law. We have the enthalpy of the reaction, and we have the formation enthalpies of the reactants. So we rearrange the equation to find the formation enthalpy for the single product. Notice we leave out hydrogen only because it's 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H^\Theta_{reaction}=\Delta H^\Theta _{f(ethane)}-\Delta H^\Theta _{f(ethene)}$$&#xA;$$\Downarrow$$&#xA;$$\Delta H^\Theta _{f(ethane)} = \Delta H^\Theta_{reaction}+\Delta H^\Theta _{f(ethene)}$$&#xA;$$\Delta H^\Theta _{f(ethane)} = -137\:\mathrm{\frac{kJ}{mol}} + 53.3\:\mathrm{\frac{kJ}{mol}}=-83.7\:\mathrm{\frac{kJ}{mol}} $$&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-02-13T20:44:00.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4151" PostTypeId="1" AcceptedAnswerId="4153" CreationDate="2013-02-14T02:38:07.730" Score="3" ViewCount="2305" Body="&lt;p&gt;I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms. 1° carbocations are unstable to the point of not having been observed in solution, ever. 2° are more stable, and 3° carbocations are the most stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that higher stability carbocations do also require less activation energy in their formation. Why is that? Why exactly are they more stable? And is this another way of saying they are also less reactive? Does this in turn mean they require a better nucleophile compared to less stable carbocations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small side question. How should I think about resonance structures? You don't view it as changing between the forms, spending more time being the more stable form, but rather as a seamless &lt;em&gt;blend&lt;/em&gt; of the forms, with more characteristics of the more stable form, right?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2015-01-01T15:31:06.707" Title="Why does carbocation stability increase in the order 1°, 2°, 3°?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4152" PostTypeId="1" AcceptedAnswerId="4166" CreationDate="2013-02-14T02:49:03.623" Score="17" ViewCount="9307" Body="&lt;p&gt;I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms, and I have a few questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book has a couple of tables, one lists a bunch of substances grouped as good, moderate and poor nucleophiles. Among the good are $\ce{Br-}$ and $\ce{I-}$. So a couple of halogens. I'm figuring chlorine is in that group too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it has a table showing ability to function as leaving group. Iodine, bromine and chlorine anions are at the top of that list too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: How can these be quite attracted to the positive charge AND quite easy to remove at the same time?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-03-18T16:45:29.173" Title="How can a group be both a good nucleophile and a good leaving group?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4153" PostTypeId="2" ParentId="4151" CreationDate="2013-02-14T02:53:20.807" Score="7" Body="&lt;p&gt;One word: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot;&gt;hyperconjugation&lt;/a&gt;! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.&lt;/p&gt;&#xA;" OwnerUserId="1242" LastActivityDate="2013-02-14T02:53:20.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4154" PostTypeId="1" AcceptedAnswerId="4155" CreationDate="2013-02-14T11:10:07.750" Score="13" ViewCount="519" Body="&lt;p&gt;Why aren't aromatic compounds with 3, 4 or 5 carbon atoms taught at high school level chemistry? We are instead directly taught about aromatic compounds with 6 carbons/benzene rings. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My core question is: is because the lower members are tough to understand? Do they have any strange properties?&lt;/p&gt;&#xA;" OwnerUserId="1243" LastEditorUserId="22" LastEditDate="2013-02-14T12:52:00.403" LastActivityDate="2013-02-16T12:59:31.823" Title="Why aren't lower aromatic organic compounds taught?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4155" PostTypeId="2" ParentId="4154" CreationDate="2013-02-14T12:18:03.277" Score="18" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule&quot;&gt;Hueckel's rule&lt;/a&gt; (i.e. that aromatic compounds have $4n+2$ $\pi$ electrons) places limits on aromaticity in cycles with less than 6 carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there are only 3 cases to look at, let's enumerate them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5 carbons&lt;/strong&gt;: cyclopentadiene ($\ce{C5H6}$) is not aromatic, but the cyclopentadienide anion ($\ce{[C5H5]^{-}}$) is. This species is well known in organometallic chemistry, due to its propensity to act as an $\eta^5$ ligand and form sandwich complexes with transition metals. The most famous of these is no doubt ferrocene, where two cyclopentadienide ligands sandwich an $\ce{Fe^{2+}}$ ion. This ion is didactically problematic and in my experience not covered in high school or early undergraduate curricula because most of its interest lies in its sandwich bonding mode with metal ions, which cannot be easily reconciled with the simplified VB/VSEPR/hybridisation theory used to teach introductory organic chemistry. The same goes for the under-loved tropylium cation ($\ce{[C7H7]^+}$), which has &lt;strong&gt;7 carbons&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4 carbons&lt;/strong&gt;: cyclobutadiene is antiaromatic, and as such only exists at very low temperatures. If I recall correctly, it possesses two unusually long bonds and two unusually short bonds, quite at odds with the bond equivalency found in benzene. This molecule would have to lose 2 e- to become aromatic. It does so happen that the cyclobutadienyl ligand is known in $\eta^{4}$ metal complexes such as $\ce{(C4H4)Fe(CO)3}$ &lt;strong&gt;(1)&lt;/strong&gt; - I was not aware of these complexes until an expert in the field pointed them out to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3 carbons&lt;/strong&gt;: I was not sure about this, but apparently the &lt;a href=&quot;http://www.quirkyscience.com/aromatic-cyclopropenyl-cation/&quot;&gt;cyclopropenyl cation is a thing&lt;/a&gt; and is aromatic. Cotton &lt;strong&gt;(2)&lt;/strong&gt; also mentions this being known as a $\eta^3$ ligand. I suspect that it is not taught because it's &lt;em&gt;too weird&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In conclusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller aromatic rings are principally known as polyhaptic ligands to metals due to their non-neutrality. This places them outside of the comfort zone of the theories of molecular structure and bonding used in introductory organic chemistry, and as such they are usually left until later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Cotton, F.A., Wilkinson, G.; &lt;em&gt;Advanced Inorganic Chemistry&lt;/em&gt;; Wiley Interscience, 5e, (&lt;strong&gt;1988&lt;/strong&gt;), p. 1170&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;em&gt;Ibid&lt;/em&gt;, p. 1168&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2013-02-15T05:57:51.997" LastActivityDate="2013-02-15T05:57:51.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4156" PostTypeId="1" AcceptedAnswerId="4161" CreationDate="2013-02-14T13:18:49.550" Score="15" ViewCount="821" Body="&lt;p&gt;I’m trying to get a deeper understanding of how mass spectroscopy works. Most tutorials and textbooks I’ve encountered omit certain details about the process and I’m hoping someone out there who understands the process can fill in the gaps. I’ll start by explaining the way I currently understand the process and then list where I start to get confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 First a particular molecule is bombarded with a beam of electrons (ionization step). This step works to free an electron from the molecule under analysis which generates a cation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: It’s important to remember that the only reason why this molecule was ionized is to make use of a magnetic field to bend its direction. Only ions will be affected by this magnetic field which allows us to pinpoint the molecule in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Next, these ions are accelerated by an electric field toward a magnetic field that bends the moving ions by a certain amount.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 The ions that bend will travel around a tube by a certain degree and hit a detector. The amount of bending will tell us about the mass of a particular molecule. Heavier molecules will bend less and lighter molecules will bend more. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any help understanding this concept would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1225" LastActivityDate="2013-02-15T16:21:00.320" Title="Understanding how mass spectroscopy works" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4157" PostTypeId="2" ParentId="4138" CreationDate="2013-02-14T13:30:43.253" Score="7" Body="&lt;p&gt;I read up a bit about Huckel theory. There's a &lt;a href=&quot;http://www.chem.qmul.ac.uk/software/download/mo/6.pdf&quot; rel=&quot;noreferrer&quot;&gt;pdf&lt;/a&gt; that explains this particularly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one point, the following formula is derived at one point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_s\propto\sin\left(\frac{\pi k s}{n+1}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives the coefficient (sans a positive normalisation constant) of the atomic orbital of the $s$th atom in the $k$th MO of a conjugated linear or monocyclic system on $n$ atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick and dirty method has also been provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vgfck.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vgfckl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, one can get a general idea of how the MO should look, and then one can get the exact +/- sequence by considering the number of nodes. Note that this is really the same thing as applying the $c_s$ formula given above, just in a graphical manner.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-02-14T13:30:43.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4158" PostTypeId="1" CreationDate="2013-02-14T19:05:15.887" Score="3" ViewCount="2211" Body="&lt;p&gt;Part of the total reaction is mentioned here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/demonstrationsexperiments/ht/instantfire.htm&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.about.com/od/demonstrationsexperiments/ht/instantfire.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2KClO3(s)} + \text{heat} \rightarrow 2\ce{KCl(s)} + 3\ce{O2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is a mystery. (And where does the heat come from?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I regret to say that I have delayed acquisition of organic chemistry principles that might have answered this question already but can't help but wonder &lt;em&gt;why&lt;/em&gt;, right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that once the role of sulfuric acid is cleared up, the reaction with sucrose not described above will explain itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And no I am going to run around burning things down with this knowledge.)&lt;/p&gt;&#xA;" OwnerUserId="550" LastActivityDate="2017-04-08T22:15:14.217" Title="How does sulfuric acid catalyze the reaction of potassium chlorate and sucrose?" Tags="&lt;combustion&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4159" PostTypeId="2" ParentId="4149" CreationDate="2013-02-14T19:53:55.860" Score="6" Body="&lt;p&gt;Various criteria may be applied in order to ascertain the connectivity of structures. A common solution is a path generated by requiring it to be a Newton path on the potential energy surface (E):&#xA;$$&#xA;(\nabla\nabla^\ast E(x))^{-1}\nabla E(x) = -\frac{dx}{dt}&#xA;$$&#xA;Two points $x_0$ and $x_1$ are connected only if $\exists x:\mathbb{R}\to\Omega,\tau_0, \tau_1\in \mathbb{R}$ such that $x(t)$ is a Newton path and $x(\tau_i)=x_i$, i.e., starting and ending structures, where $\Omega$ is your space of structures (usually centered 3D coordinates). $x$ is then a parameterization of that path on $[\tau_0,\tau_1]\subset\mathbb{R}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes it is additionally required that there is at most 1 maximum on $E(x)$ between $\tau_0$ and $\tau_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be interested in reading &lt;a href=&quot;http://cims.nyu.edu/~eve2/string.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;. In a literature search, you may want to read up on the work by Vanden-Eijnden, Bofill, Quapp, and Henkelman. This topic is too rich to discuss in this forum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the specific examples, assuming the starting point is in the convex attraction basin of the nearest optimum and the used optimizer converges in such instances, the optimum is connected in the above sense. Most optimizers used in geometry optimizations are some variant of (quasi-) Newton-Raphson optimizer, so this criterion is given. There are several reasons why the IRC may not converge to the same optimum as the subsequently used geometry optimizer, including changes of basis sets, convergence tolerances, or in case of quasi-Newton methods, the Hessian guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the optimizer brings the water closer to the other molecule is no indication of lack of connectedness. Instead the IRC run did not end at an optimum. It is actually expected that the energy of two (neutral) molecules decreases as they approach each other from infinity even in the complete basis set limit (van der Waals forces). You just found the vdW/BSSE-bound pre/post-complex.&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="440" LastEditDate="2017-01-23T15:56:51.713" LastActivityDate="2017-01-23T15:56:51.713" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4160" PostTypeId="2" ParentId="4158" CreationDate="2013-02-14T22:44:10.503" Score="5" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; the sulfuric acid isn't really a catalyst here -- I think you use the heat from the reaction between sulfuric acid and sucrose (which will start happily at room temperature) to start the reaction between the potassium chlorate and the sucrose (which usually won't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ETA:&lt;/strong&gt; Once you've done that, it's pretty much a straight-up messy combustion reaction, only with the oxygen being provided from a solid source (thermal decomposition of chlorate to chloride and oxygen as the reaction you mentioned shows) rather than anything mechanistically clean. Compare, for example, this fairly typical demonstration just using hot chlorate and a sugar source: &lt;a href=&quot;http://lecturedemos.chem.umass.edu/chemReactions5_5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxidation of Sugar or Gummi Bear with Potassium Chlorate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="402" LastEditDate="2017-04-08T22:15:14.217" LastActivityDate="2017-04-08T22:15:14.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4161" PostTypeId="2" ParentId="4156" CreationDate="2013-02-15T16:21:00.320" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electric fields accelerate charged particles (linear acceleration). Magnetic fields deflect them (angular acceleration). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charged particles respond to &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_fields#Magnetic_field_and_electric_currents&quot;&gt;electric fields and magnetic fields differently&lt;/a&gt;. Electric fields accelerate charged particles linearly in the direction of the flux. For example, if an electric field exists between two plates, one with a buildup of positive charge and the other with a buildup of negative charge, then positively charged particles will accelerate from the positive part toward the negative plate. The acceleration of the particle ($\vec{a}$) depends on the strength of the field ($\vec{E}$), the mass of the particle ($m$), and the charge on the particle ($q$). Since the acceleration is a scalar multiple of the electric field, they have the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}=q\vec{E}$$&#xA;$$\vec{F}=m\vec{a}$$&#xA;$$q\vec{E}=m\vec{a}$$&#xA;$$\vec{a}=\frac{q}{m}\vec{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields are orthogonal to electric fields. Magnetic fields alter the angular acceleration of charged particles (i.e., they deflect them). Thus, charged particles have a circular trajectory in a magnetic field. The radius of the circular path ($r$) is dependent on the strength of the magnetic field ($\vec{B}$), the charge of the particle ($q$), the mass of the particle ($m$), and the velocity of particle ($\vec{v}$). The radius of curvature is proportion to the mass to charge ration ($m/q$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\vec{F}|=m\frac{|\vec{v}|^2}{r}$$&#xA;$$|\vec{F}|=|q\vec{v}||\vec{B}|$$&#xA;$$m\frac{|\vec{v}|^2}{r}=|q\vec{v}||\vec{B}|$$&#xA;$$r=\frac{m}{q} \frac{|\vec{v}|}{|\vec{B}|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration of the particle depends on the velocity, magnetic field, radius, and mass. Since acceleration in a magnetic field is a scalar multiple of the cross product of velocity and field strength, the direction of acceleration is always changing.&#xA;$$\vec{F}=q(\vec{v}\times\vec{B})$$&#xA;$$m\vec{a}=q(\vec{v}\times\vec{B})$$&#xA;$$\vec{a}=\frac{q}{m}(\vec{v}\times\vec{B})$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that we know the relationship between radius of curvature and mass-to-charge ratio, this question is easy. The detector counts the number of particles that hit it. That is all that it can do. That is all that it needs to do. The hard work of separating the ions by m/q is done by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_analyzer#Mass_selection&quot;&gt;mass analyzer&lt;/a&gt;, which is the magnetic field in curved tube. The magnetic field scans (quickly) over a range of $\vec{B}$ to alter the curvature of the ions by their m/q. Since the tube has a fixed curvature, only the ions with the correct m/q will pass to the detector. All others will hit the walls of the tube since their radius of deflection would be too large or too small. Thus, since the instrument controls the strength of the magnetic field, and physics determines what m/q will have the correct deflection at that field strength, there is no need to calibrate the detector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex mass analyzers exist, for example the quadrupole mass analyzer. Instead of the typical bipolar magnetic field, it generates a quadrupolar magnetic field. Ions are thrown all over the place to hit a spherical detector. The physics is more complex, but the trajectory once again depends on m/q, so the detector only has to count hits in certain spots. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-02-15T16:21:00.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4162" PostTypeId="2" ParentId="1226" CreationDate="2013-02-15T22:46:15.560" Score="3" Body="&lt;p&gt;In my experience, for small quantities in laboratories helium usually comes compressed in cylinders. Except possibly in the case that you require extremely high purity, I don't see why helium couldn't be stored for long periods in this form, it would be relatively safe and would avoid the need for refrigeration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what sort of quantities you are thinking about, but it seems that you can get big cylinders. For example the largest cyclinder size available from BOC gases, a British company, is 194.7 $m^3$ filled to a pressure of 300 $bar$ (&lt;a href=&quot;http://www.boconline.co.uk/en/products-and-supply/speciality-gas/pure-gases/pure-gases.html&quot; rel=&quot;nofollow&quot;&gt;source: BOC website&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1247" LastActivityDate="2013-02-15T22:46:15.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4164" PostTypeId="1" CreationDate="2013-02-16T14:52:39.273" Score="5" ViewCount="15207" Body="&lt;p&gt;Why do some people say that hydrogen should be above lithium in the periodic table and others argue it should be above fluorine?&lt;/p&gt;&#xA;" OwnerUserId="1248" LastEditorUserId="402" LastEditDate="2017-07-16T23:52:15.833" LastActivityDate="2017-07-16T23:52:15.833" Title="Where hydrogen belongs in the Periodic Table" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4166" PostTypeId="2" ParentId="4152" CreationDate="2013-02-16T15:47:43.130" Score="11" Body="&lt;p&gt;&lt;sup&gt; I'm not entirely sure of this, comments appreciated&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -&gt; more stable while solvated -&gt; better leaving group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, nucleophilic tendency is basically &quot;how willing to sacrifice a lone pair is the group?&quot;.  It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it would be via the &quot;accessibility&quot; of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to &quot;give up&quot; the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the &lt;em&gt;reverse&lt;/em&gt; process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to &quot;extract&quot; it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more &quot;accessible&quot;.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-02-16T15:47:43.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4168" PostTypeId="1" CreationDate="2013-02-16T21:10:26.967" Score="2" ViewCount="96" Body="&lt;p&gt;Alfatoxin and benzene are carconogenic, and this is the major concern for low, chronic doses. They are regulated to the few parts per billion level in food. However, according to &lt;a href=&quot;http://toxnet.nlm.nih.gov/cpdb/pdfs/ChemicalTable.pdf&quot; rel=&quot;nofollow&quot;&gt;http://toxnet.nlm.nih.gov/cpdb/pdfs/ChemicalTable.pdf&lt;/a&gt;,&#xA;aflatoxin is &lt;strong&gt;three to five&lt;/strong&gt; orders of magnitude more carcinogenic, as evident from the lower doses needed to induce tumors in rodents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the permissible thresholds similar?&lt;/p&gt;&#xA;" OwnerUserId="1211" LastEditorUserId="73" LastEditDate="2013-02-17T03:46:35.067" LastActivityDate="2015-06-16T10:21:41.450" Title="Aflatoxin versus Benzene, permissible levels" Tags="&lt;biochemistry&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4169" PostTypeId="2" ParentId="4164" CreationDate="2013-02-17T02:19:41.240" Score="9" Body="&lt;p&gt;This comes down to the fact that hydrogen has a $1s$ valence shell, capable of holding only two electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense, hydrogen is like the halogens, in that it can achieve the electronic configuration of a noble gas (namely, helium, which has a full $1s$ level) by gaining an electron. This reduction yields the hydride ion, $\ce{H-}$. It should be noted that hydride does not have the same stability as the halogen anions (e.g. hydrogen has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29&quot;&gt;electron affinity&lt;/a&gt; of 0.75 eV vs. 3.40 eV for fluorine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another sense, hydrogen is like the alkali metals, as it only contains a single electron in its valence shell. Whilst the alkali metals can attain noble gas configurations by losing their valence electrons, it's a bit abstract to talk about hydrogen doing the same by losing its single electron to generate a proton ($\ce{H^{+}}$)(1). Nevertheless there is strong analogy between the electronic structure of hydrogen and the alkali metals which motivates hydrogen being often placed in this group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Periodic table makers sometimes get away with duplicating hydrogen above the alkali metals &lt;em&gt;and&lt;/em&gt; the halogens, or putting it in no group in particular.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="84" LastEditDate="2013-02-17T02:30:34.167" LastActivityDate="2013-02-17T02:30:34.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4171" PostTypeId="1" AcceptedAnswerId="4183" CreationDate="2013-02-17T08:50:30.393" Score="11" ViewCount="477" Body="&lt;p&gt;Please correct me if I’m wrong, but I’ve read that the reason why we use significant figures is to avoid making the result of a calculation more accurate than the starting values prior to the calculation. For example, if we were seeking the ratio of two weights, lets call them A and B, using a cheap scale and an expensive scale we could get the following varied results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheap Scale &#xA;$$A = 34 \space oz$$&#xA;$$B = 23 \space oz$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expensive Scale &#xA;$$A =  34.0000 \space oz$$&#xA;$$B = 23.0000 \space oz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the ratio $\frac{A}{B}$ using a calculator we get the following value:&#xA;$$\frac{34}{23} = 1.47826$$&#xA;Now this is a correct ratio depending on the scale used (ie the cheap scale could not have produced such an accurate measurement given the starting values)&#xA;This leads to the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are the zeros behind a decimal point ignored if they are just place holders?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the value 0.000023 oz only has two significant figures, but isn’t that value something only a very fine tuned scale could measure (keeping in line with the previous weight example). So why are these zeros ignored?&lt;/p&gt;&#xA;" OwnerUserId="1225" LastEditorUserId="22102" LastEditDate="2017-03-12T01:50:51.143" LastActivityDate="2017-03-12T01:50:51.143" Title="Significant Figures Interpretation" Tags="&lt;terminology&gt;&lt;significant-figures&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4172" PostTypeId="1" CreationDate="2013-02-17T10:37:37.523" Score="1" ViewCount="87" Body="&lt;p&gt;When the electrons that are excited by the energy of burning drop back into lower energy levels they emit light at the wavelength of colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When metals are burnt, metal oxides(some coloured) are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When metals are burnt under high temperature, will they turn to plasma state? If so, should the metal ions be responsible for the colours and sparks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is the reason for the beautiful colours of fireworks? &lt;/p&gt;&#xA;" OwnerUserId="1251" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:07:46.127" LastActivityDate="2015-10-23T17:07:46.127" Title="The colour of fireworks" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4173" PostTypeId="1" CreationDate="2013-02-17T11:28:20.067" Score="1" ViewCount="1211" Body="&lt;p&gt;In the reaction of combustion of gunpowder,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sulfur is used to speed up the reaction and acts like a catalyst in that it increases the rate of reaction. However, unlike a catalyst, it is used up in the reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that transition metals are often used as catalysts but why can sulfur act like a catalyst?&lt;/p&gt;&#xA;" OwnerUserId="1251" LastEditorUserId="402" LastEditDate="2016-12-04T21:40:50.393" LastActivityDate="2017-04-08T22:14:08.090" Title="Why can sulfur act like a catalyst?" Tags="&lt;inorganic-chemistry&gt;&lt;catalysis&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4174" PostTypeId="1" CreationDate="2013-02-17T13:11:45.950" Score="3" ViewCount="2873" Body="&lt;p&gt;I need to control oxygen in a erlenmeyer flask in a bioprocess.  I need to remove overall oxygen in a erlenmeyer flask, and after I need to add oxygen in controlled amounts to a culture in a shaken flask.  Someone knows how can I do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for advance for your ideas.&lt;/p&gt;&#xA;" OwnerUserId="1252" LastEditorUserId="1252" LastEditDate="2013-02-17T13:18:51.747" LastActivityDate="2013-02-20T14:51:11.520" Title="How can I control the oxygen in an erlenmeyer flask in a bioprocess?" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4175" PostTypeId="2" ParentId="4174" CreationDate="2013-02-17T13:46:55.747" Score="5" Body="&lt;p&gt;Controlling oxygen in a flask can be achieved by bubbling a purge gas of some description through it. This will entrain and remove oxygen from the solution as well as creating an oxygen-depleted headspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a regulated nitrogen supply, a rubber stopper with a hole in it, an open-ended pasteur pipette and a length of hose, you can bubble nitrogen into the solution to purge oxygen. This requires the vessel to be open somewhere else to equalise pressure, which is achievable with a multi-neck flask but not with a standard Erlenmeyer flask. You can get around this with a stopper with two holes. Continual positive pressure should keep the external atmosphere out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBXqs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how vigorously you want to shake your flask, this may or may not be reasonable. If you are using a stirrer bar you will obviously want to give the pipette sufficient clearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as controlling addition of oxygen, this would seem to be a bit trickier as you are probably not going to be providing continual positive pressure. You might be able to get around this, depending on the specifics of what you are doing, by pumping in oxygen in the same way and using a balloon or a gas syringe or something to deal with the overpressure.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-02-17T13:46:55.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4176" PostTypeId="2" ParentId="4173" CreationDate="2013-02-17T14:15:22.990" Score="3" Body="&lt;p&gt;It's not a catalyst here. It speeds up the reaction, which is why it acts &lt;em&gt;like&lt;/em&gt; a catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the reaction without sulfur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6 KNO3 + C7H8O -&amp;gt; 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here it is with &lt;em&gt;sulfur&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4 KNO3 + C7H4O + 2 S —&amp;gt; 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Gunpowder#Combustion_characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur decreases the ignition temperature, as well as probably contributing a bit to the rate. Here, it is effectively turning $\ce{CO2}$ and $\ce{K2CO3}$ to $\ce{CO}$ and $\ce{K2S}$. If you take into account &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Le Chatelier's&lt;/a&gt; principle, the conversion of the products of the basic reactants speeds it up. (The decrease in concentrations/activities of the products due to their conversion means that the equilibrium shifts in the forward direction to compensate for it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is speeding up the reaction and making it easier to carry out. Not a catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note: There's no reason why any element/compound can't act as a catalyst. Transition metals are generally catalysts in a certain type of reaction, and they generally catalyze the reaction by adsorbing the gases. Catalysts can work in many other ways—generally, they form intermediates in the reaction and present an alternative path via the intermediate. ($\ce{V2O5}$, for example, is a catalyst in many reactions involving organic compounds since it forms cyclic intermediates with various groups).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="402" LastEditDate="2017-04-08T22:14:08.090" LastActivityDate="2017-04-08T22:14:08.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4177" PostTypeId="1" CreationDate="2013-02-17T15:30:01.700" Score="6" ViewCount="2864" Body="&lt;p&gt;In organic chemistry, people draw 2p orbitals like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5x5Ka.png&quot; alt=&quot;2p static&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then they explain how the orbitals combine to non-bonding (π*) or bonding (π) molecular orbitals, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTWVg.png&quot; alt=&quot;orbitals bonding&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depending on whether the orbitals overlap out-of-phase (blue + white) or in-phase (blue + blue, or white + white).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in reality, the orbitals are time-dependent solutions of the time-dependent Schrödinger equation, so they are not really like the first image above, but more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aYPWQ.gif&quot; alt=&quot;2p animated&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what does it mean, when the orbitals overlap in-phase or out-of-phase? As far as I understand, the phase difference of the orbitals of two random atoms could be anything from 0 to 2π, and the cases of perfect overlap (0 phase difference) or perfect anti-overlap (π, or half-cycle, phase difference) should be quite improbable.&lt;/p&gt;&#xA;" OwnerUserId="1253" LastEditorUserId="1253" LastEditDate="2013-06-28T18:33:57.197" LastActivityDate="2013-06-28T18:33:57.197" Title="Bonding and anti-bonding orbitals in the light of time-dependent Schrödinger equation?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4178" PostTypeId="2" ParentId="4172" CreationDate="2013-02-18T06:06:47.947" Score="1" Body="&lt;p&gt;The colours of fireworks are caused by the electrons emitting energy when dropping orbital levels. Metal oxides are solids, mainly dull colours (even the yellows and reds are not in the same class as firework colours), and could not cause bright light in the air. The temperature of fireworks is far below that needed to form a plasma.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2013-02-18T06:06:47.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4181" PostTypeId="1" CreationDate="2013-02-18T11:36:42.193" Score="1" ViewCount="790" Body="&lt;p&gt;I'm currently studying ion exchange resins and generally getting a bit stuck on some points of information which I do not fully understand / are not very well explained within the literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that resins having varying selectivity or affinity for a particular ion and selectivity generally increases with increased valance number, atomic weight, reduced hydrated radius, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen some information that selectivity reversal will occur when the solution has high TDS e.g., sulphate will have a far higher affinity than chloride for a strong base anion resin within a solution with low TDS; however, at high TDS the affinity of these ions for the resin will be reversed - chloride will have a greater affinity than sulphate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea whether this is correct and what the mechanism causing this change is?&lt;/p&gt;&#xA;" OwnerUserId="1257" LastEditorUserId="73" LastEditDate="2013-02-19T00:24:54.100" LastActivityDate="2013-02-19T00:24:54.100" Title="Selectivity Coefficients of Ion Exchange Resins - Selectivity Reversal with TDS / Ionic Strength" Tags="&lt;reaction-mechanism&gt;&lt;ions&gt;&lt;water&gt;&lt;equilibrium&gt;&lt;surface-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4182" PostTypeId="2" ParentId="4171" CreationDate="2013-02-18T19:00:15.897" Score="1" Body="&lt;p&gt;In your example 0.000023 oz does indeed have two significant figures. It is indeed probably something only a very fine tuned scale could measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider that you add that mass to another sample that you measured with the much less accurate scale, let's say the other scale measured that other weight as 23 oz. Both have two significant figures, but considering the weight read by the inaccurate scale, you have no legitimate idea whether it is really &lt;em&gt;23.28125 oz&lt;/em&gt; or &lt;em&gt;22.87284 oz&lt;/em&gt;, or some other number for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do know is that it probably falls within some range of values, we usually look at the last significant digit and if the uncertainty is not mentioned we can assume it is +/- 2%. That means the value 1.0 oz might be somewhere between 0.98 oz and 1.02 oz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now when we add the two masses together, we do not want decimals since 23 oz has none. The result is 23.000023 oz. Which is 23 oz knowing that we can only go down to the ones place. The reason is that you have no clue whether it is &lt;em&gt;23.38125 oz&lt;/em&gt; + .000023 oz, or &lt;em&gt;22.87284 oz&lt;/em&gt; + .000023 oz, or any other values for that matter. The fact is that these zero's are &lt;strong&gt;not&lt;/strong&gt; ignored, they are giving us the magnitude of the number. But when we do something with them in conjunction with some other &lt;strong&gt;uncertain&lt;/strong&gt; numbers, we cannot be sure of the result to high certainty. One rotten apple spoils the whole barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in this scenario described above, even if we said 0.000023 oz has 6 significant figures. The other number 23 oz still has only two. Therefore we must go with 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we divide or multiply we count the sig figs. 23 oz has 2 sig figs, so our result of multiplying should have 2 as well. So how many times greater is the inaccurate weight (23 oz) compared to the accurate weight (0.000023 oz). 23 oz / 0.000023 oz ~= 1000000 times greater. Notice that the number 1000000 has only 1 significant figure. We want two significant figures so we write. $1.0 * 10^6$&lt;/p&gt;&#xA;" OwnerUserId="584" LastEditorUserId="584" LastEditDate="2013-02-18T19:07:48.560" LastActivityDate="2013-02-18T19:07:48.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4183" PostTypeId="2" ParentId="4171" CreationDate="2013-02-18T19:12:14.047" Score="12" Body="&lt;p&gt;This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but isn’t that value something only a very fine tuned scale could measure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessarily. The cost of an instrument is much more closely related to the &lt;em&gt;relative&lt;/em&gt; precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2013-02-18T19:12:14.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4184" PostTypeId="2" ParentId="4174" CreationDate="2013-02-18T19:25:00.043" Score="2" Body="&lt;p&gt;For clarity, I'm going to separate this into two parts: purging the oxygen from the flask, and filling the flask with a particular mixture of gas(es). It's the same idea in both cases, but the first is much easier than the second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Purging the flask&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is usually done with Nitrogen, as it is one of the cheapest inert gases. You'll take a line from your nitrogen tank, pass it into your flask, and open it. You will need to have a second opening in your flask so pressure doesn't build up, as Richard Terrett describes in his answer. As long as you have the nitrogen flowing, there will be enough positive pressure in the flask that oxygen won't get in from the room. The headspace in your flask should purge quickly, but your culture will take longer. Agitation, either by shaking the flask or by stirring the culture, will make that go much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Establishing a given atmosphere&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really the same process as above: instead of pure nitrogen, you pump in the mixture of nitrogen and oxygen you want. The system will eventually equilibrate to the mixture you're pumping in. The hard part is getting the mixture you want. When I did this in the past, we used digital mass flow controllers (DMFC's). We would flow oxygen through one DMFC and nitrogen through another. The output of the two DMFC's would mix together before going into the flask. Once you have it set up, it's just a matter of setting the two DMFC's to the mixture you want. The big difficulty with that is that DMFC's are expensive. I don't know how much ours were, but they were made by &lt;a href=&quot;http://www.bronkhorst.com/&quot; rel=&quot;nofollow&quot;&gt;Bronkhorst&lt;/a&gt;, and it looks like that type of DMFC starts at around $2000 each. The nice thing about using DMFC's is that you can control the oxygen composition to within less than 1% by mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how precisely you need to control the mixture, you might be able to get away with a simple flow meter. Again, you would run each gas into its own flow meter, and use that to control the output of each gas. Then you set the two flow rates so that the composition and the total flow rate are what you want them to be. I don't know how precisely you would be able to control the composition, but it might be enough for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the atmosphere you're running into the culture, you can either run the purge line into the culture itself, or just into the headspace. I have done it both ways; running it into the culture works faster, but might not be feasible depending on your setup. In one case I worked on, the culture was only a few millimeters thick, and there just wasn't room to stick a needle from the purge line into it. In that case, just pumping gas into the headspace and agitating the culture was sufficient because of how shallow the culture was. In my more recent work, I have a much deeper volume of fluid to purge. In this case, running the purge line only into the headspace probably would be too slow for my purposes, but it's pretty easy to run a needle from the purge line into the fluid without impeding anything else.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2013-02-20T14:51:11.520" LastActivityDate="2013-02-20T14:51:11.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4185" PostTypeId="2" ParentId="4164" CreationDate="2013-02-19T01:03:02.130" Score="3" Body="&lt;p&gt;The simplest explanation is that hydrogen has a valence shell that can only accept two electrons. Hydrogen is thus in the unique position of being just as likely to want to gain an electron as to lose the one it has; either way, it reaches a stable valence state. For this reason, it defines the &quot;center point&quot; of the Pauling electronegativity scale; elements with a lower number are &quot;electropositive&quot; (they prefer to get rid of their valence electrons), while elements with a higher number are &quot;electronegative&quot; (they want more valence electrons than they have).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elsewhere in the periodic table, alkali metals which have only one or two valence electrons want to get rid of them, to get back to a complete valence shell underneath. They'd need to gain 6, 7 or even more electrons to stabilize their existing valence shells, and that's unlikely (they don't have enough electrons to share, and their nucleus size won't support an additional 7 electrons at that energy level). Halogens, on the other side of the table, have a nearly-complete valence shell, and just need one electron to get there, so they'll take it from anything that'll give it up (and in the case of some of the more vigorous halogens, like fluorine, they'll even take it from things that &lt;em&gt;won't&lt;/em&gt; give them up, like other halogens and even noble gases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen, however, can do either one. It can lose its electron and become a naked proton, the H&lt;sup&gt;+&lt;/sup&gt; cation, or it can gain an electron and become a H&lt;sup&gt;-&lt;/sup&gt; anion. Which of these it is more likely to do depends on what it's reacting with. With alkalis, it'll happily accept the extra proton and form a hydride, such as LiH. With halogens, it'll happily give up the proton and form a hydrohalic acid like HF, HCl or HBr. With other elements, it'll share, generally forming &quot;polar covalent&quot; bonds where the other atom in the bond attracts the electrons more closely. Hydrogen's very &quot;zen&quot; about the whole thing; give one up, take one in, share, it's all good.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-06T17:58:15.773" LastActivityDate="2013-05-06T17:58:15.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4187" PostTypeId="1" AcceptedAnswerId="4207" CreationDate="2013-02-19T03:36:05.510" Score="4" ViewCount="1929" Body="&lt;p&gt;I am working on reaction predictions for the AP Chemistry and I need some help with this problem. On a practice quiz I tried to use the rule of &quot;Metal + Water = Metal Oxide&quot; so I answered with $\ce{2Li + O^{2-} -&amp;gt; Li2OH}$. I got zero points for this answer meaning both reactants and products were wrong. What is the correct answer and how would you get to the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="1262" LastEditorUserId="1262" LastEditDate="2013-02-19T04:24:17.543" LastActivityDate="2016-10-11T18:33:53.027" Title="What is the product when Lithium solid is dropped in water?" Tags="&lt;reaction-mechanism&gt;&lt;water&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4188" PostTypeId="1" CreationDate="2013-02-19T16:25:29.247" Score="15" ViewCount="86548" Body="&lt;p&gt;I do not like body wash always with strong favor. I like hand wash soap to wash my body. Are they difference at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the comments, so that I add the following information&#xA;I am really sorry that I posted the question using my phone at my office parking lot right before I went to work this morning, so not much detail was added&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my limited knowledge, Body wash and hard wash soaps are detergent, they both are OK to use on skin. If that is the case, why body wash soap is body wash soap? What I mean is that from the chemistry point of view, what makes body wash soap difference from hand wash soap?&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="7448" LastEditDate="2015-10-30T17:22:20.567" LastActivityDate="2019-11-17T00:44:17.437" Title="What is the difference between hand wash and body wash soap?" Tags="&lt;everyday-chemistry&gt;&lt;surfactants&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4189" PostTypeId="1" AcceptedAnswerId="4225" CreationDate="2013-02-19T16:25:39.250" Score="6" ViewCount="126" Body="&lt;p&gt;I am unfortunately not very good at chemistry, so I am having trouble figuring out where to start on this project.  I need to have some way of get some amide functional groups stuck on a substrate (most likely some glass beads or slides) so I can try to show some interactions of the amide with a molecule I am interested in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any literature that shows a good way to functionalize silica with amides? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help with this, I really appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I found some of &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/440140?lang=en&amp;amp;region=US&quot;&gt;this stuff ((3-Aminopropyl)triethoxysilane)&lt;/a&gt; in our lab.  Could I just put a glass slide in it and leave it over night?&#xA;Then maybe come back and wash acetyl chloride over it as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Schotten-Baumann_reaction&quot;&gt;this reaction?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2013-02-23T21:59:04.750" Title="Covalently bonding amide to glass" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4190" PostTypeId="1" AcceptedAnswerId="4192" CreationDate="2013-02-19T19:00:04.207" Score="20" ViewCount="580" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/va1So.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above are some spectral emission lines for hydrogen, helium and neon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Schroedinger equation, it's possible to derive the colours that hydrogen will emit when light is fired at it. A little more complicated (although beyond my rudimentary knowledge of quantum mechanics), the same can be done for helium, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course as we progress to molecules, even the relatively small $C_5H_{11}Br$, it's (I'm told) &lt;em&gt;very hard&lt;/em&gt; to predict the colours emitted from first principles, and even then incorrect assumptions must be made (like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation&quot; rel=&quot;noreferrer&quot;&gt;Born-Oppenheimer approximation&lt;/a&gt;), so upon progressing to larger molecules, we find that these simplifications diverge from experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: with the advent of computers, are we able to (in practice, not theoretically in a few decades) accurately compute the colours given off by a molecule  by numerically solving the laws of quantum mechanics for complicated systems? Much like nonlinear differential equations can now be 'solved' with computers without any simplifying assumptions (like $\sin(x) \approx x$), is it in humanity's capacity today to approximate colours from first principles, so that computing time is the only restraint? Note the difference between an incorrect model of reality (Born-Oppenheimer), and one that converges to the correct one (even if slowly).&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="9961" LastEditDate="2016-11-20T22:41:18.937" LastActivityDate="2016-11-20T22:41:18.937" Title="Is it possible to compute the colour of *any* molecule?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4191" PostTypeId="1" CreationDate="2013-02-19T21:07:06.950" Score="4" ViewCount="387" Body="&lt;p&gt;I need to find the activity &lt;span class=&quot;math-container&quot;&gt;$a_\pm$&lt;/span&gt; for a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0120 M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na3PO4}$&lt;/span&gt;, and have calculated the activity coefficient, &lt;span class=&quot;math-container&quot;&gt;$\gamma_\pm = 0.389$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$$a_\pm = \frac{m_\pm}{m_0}\gamma_\pm,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I must be plugging in the wrong numbers. What does the top &lt;span class=&quot;math-container&quot;&gt;$m_\pm$&lt;/span&gt; mean? And isn't the bottom &lt;span class=&quot;math-container&quot;&gt;$m_0$&lt;/span&gt; simply &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1266" LastEditorUserId="16683" LastEditDate="2020-06-06T03:14:18.280" LastActivityDate="2020-06-06T03:14:18.280" Title="What does m± represent in the equation for mean activity of a salt solution?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4192" PostTypeId="2" ParentId="4190" CreationDate="2013-02-20T02:05:16.127" Score="16" Body="&lt;p&gt;It is possible to compute electronic spectra with decent accuracy. Time dependent density functional theory (TDDFT) permits fairly rapid calculation of UV/vis spectra at a quantum mechanical level of theory. Whilst DFT is not truly a first principles method, it can furnish electronic spectra accurate enough to be useful, depending on care taken in selecting appropriate functionals, accounting for solvatochromic effects, etc. Obviously, DFT uses the Born-Oppenheimer approximation, but the beauty of the approximation is that it's actually very close to reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, for instance, &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ct700187z&quot;&gt;this paper&lt;/a&gt; by Jacquemin et al., where they calculate $\lambda_{max}$ for a wide range of dyes using TDDFT and get a MAE of 22 nm using CAM-B3LYP (a long-range hybrid functional), reducing that to 10 nm with an empirical calibration curve. The paper only really cares about accurately reproducing the location of $\lambda_{max}$, however &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ct2000974&quot;&gt;this paper&lt;/a&gt; by Quartarolo and Russo computes complete spectra with fairly reasonable accuracy. Not perfect, but the general shape of the spectra are reproduced adequately. Part of the divergence between the experimental and computed spectra probably comes down to higher order broadening effects not reflected in the simple Gaussian convolution of the computed data, but I'm not a spectroscopist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional: as for computing the colour of &lt;strong&gt;&lt;em&gt;any&lt;/em&gt;&lt;/strong&gt; molecule, well, that does depend heavily on the size of your molecule and any mitigating factors such as symmetry elements. Most QM techniques scale fairly viciously with the number of electrons in your molecule, and whilst you can always reduce the size of your basis set and heap on less and less conservative approximations, accuracy will suffer.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-02-20T02:05:16.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4193" PostTypeId="1" CreationDate="2013-02-20T10:10:08.803" Score="4" ViewCount="305" Body="&lt;p&gt;I have absorbance values for blood plasma. Absorbance was measured in range of 380 – 600 nm every 5 nm ending up with absorbance values for 45 wave lengths.&#xA;I was browsing through literature, and  I came  across chromacity coordinates (XYZ) and this seems to be a very convenient way of summarizing data about color in just two data points. &#xA;I have been reading some literature about different colorimetric systems, but since my background is in genetics, it didn’t make much sense for me. &#xA;Therefore, I would like to ask you if it is possible to calculate these chromacity coordinates using my absorbance values and if so does anyone now some kind of software for this or exact formulas on how to do it? &lt;/p&gt;&#xA;" OwnerUserId="1268" LastActivityDate="2013-12-25T01:40:26.060" Title="Is it possible to calculate chromacity coordinates using absorbance values?" Tags="&lt;color&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4194" PostTypeId="1" CreationDate="2013-02-20T12:16:09.580" Score="-1" ViewCount="170" Body="&lt;p&gt;Is there a positive radical Aurous, $\ce{{Au}^{x+}}$? If yes what is the value of x?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have Auric as $\ce{{Au}^{3+}}$.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="4231" LastEditDate="2015-02-26T16:32:30.027" LastActivityDate="2015-02-26T16:32:30.027" Title="Is there a positive radical Aurous, Au (x+)?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4195" PostTypeId="2" ParentId="4194" CreationDate="2013-02-20T12:29:39.497" Score="4" Body="&lt;p&gt;&lt;em&gt;Aurous&lt;/em&gt; would appear to denote Gold(I), that is, $x=+1$, as &lt;a href=&quot;http://gold.atomistry.com/aurous_chloride.html&quot; rel=&quot;nofollow&quot;&gt;detailed on this site&lt;/a&gt;, which gives a profile of Gold(I) chemistry. This atom is not a radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google is your friend.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-02-20T12:29:39.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4198" PostTypeId="2" ParentId="531" CreationDate="2013-02-21T04:22:27.800" Score="-1" Body="&lt;p&gt;x is subtracted from reactants , because the reaction is forward, so it will change in to ions, or we can say the concentration of reactants will decrease therefore we write minus x, and the acid you mention here is weak acid so it will not dissociate completely, so we donot know directly how much it will dissociate thus we use a variable x ,to find the value of x  we make equation ,eg c-x=x+x then kc =x2/c-x ,then we solve for x , and get the amount which dissociated,&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2013-02-21T04:22:27.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4199" PostTypeId="1" CreationDate="2013-02-21T13:58:31.897" Score="3" ViewCount="2662" Body="&lt;p&gt;My textbook states that for most reactions, the reaction rate decreases with time. I wonder what reactions will have increasing reaction rates with time. I know that adding a catalyst will increase the reaction rate but the final reaction rate will finally be zero. So, what reactions will their reaction rates increase with time?&lt;/p&gt;&#xA;" OwnerUserId="1251" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:53:10.350" LastActivityDate="2015-10-23T16:53:10.350" Title="What reactions will their reaction rates increase with time?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4202" PostTypeId="2" ParentId="4199" CreationDate="2013-02-21T15:12:43.693" Score="0" Body="&lt;p&gt;There are several, but the rate can only increase in time under certain limits, that is, assuming certain approximations. Obviously at &quot;final time&quot; (for instance, when the equilibrium is reached) the rate will not be increasing continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, in order to have rates increasing with time, you need to have negative orders of reaction with respect to some reactant, that is, the velocity must depend with the inverse of the population (or less likely, other exponent). This is typical of chain reactions, the most typical case being explosive reactions.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2013-02-21T15:12:43.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4203" PostTypeId="2" ParentId="4177" CreationDate="2013-02-21T15:42:14.023" Score="6" Body="&lt;p&gt;As F'x has indicated, the sum of atomic orbitals $\phi_k$ (AO) that form the molecular orbital $\psi_j$ (MO)&#xA;$$ \psi_j = \sum_k c_{jk} \phi_k $$&#xA;is just a solution (an approximate solution) to a self-consistent variational procedure to solve the &lt;strong&gt;time-independent&lt;/strong&gt; Schrödinger equation, obtained after choosing a particularly well suited basis to represent the solution. They approximately solve this equation:&#xA;$${\sf F}\psi_j = \epsilon_j \psi_j$$&#xA;where ${\sf F}$ is the Fock operator or if you wish, an effective one-electron energy operator and $\epsilon_j$ is an effective &quot;one-electron&quot; energy that gives a good approximation to the ionization potential from that orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have to construct your overall n-electron wave function $\Psi({\bf x}_1,\ldots,{\bf x}_n)$, which in the simplest case (the one that follows from the minimal molecular orbital theory) is a Slater determinant. You can notice that for this wave function the coefficients $c_{jk}$ are not unique. Following the general properties of determinants, by a similarity transformation of the basis you can get a new set of coefficients $c'_{jk}$ that will not solve the Fock equation, but after which you obtain the same overall electronic wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In everything that I have said so far, there are no time-dependent phases: we only have one &quot;real&quot; (observable) energy, an eigenvalue of the time-independent Schrödinger equations for the electronic Hamiltonian&#xA;$$ {\sf H}_{elec} \Psi({\bf x}_1,\ldots,{\bf x}_n) = E \Psi({\bf x}_1,\ldots,{\bf x}_n)$$&#xA;for a fixed geometry. Now is where you can put the time-dependent Schrödinger equation into work. In principle you could create electronic superposition states combining $\Psi({\bf x}_1,\ldots,{\bf x}_n)$ of different energy. These are the ones that would evolve in a manner similar to what you have assumed, with time-dependent evolving phases.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2013-02-21T15:42:14.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4204" PostTypeId="1" CreationDate="2013-02-21T16:28:57.520" Score="20" ViewCount="5401" Body="&lt;p&gt;Why doesn't Nitrogen monoxide dimerize even though there is an odd electron present whereas nitrogen dioxide does (because of the odd electron on nitrogen)?&lt;/p&gt;&#xA;" OwnerUserId="1272" LastEditorUserId="402" LastEditDate="2016-10-12T02:08:33.703" LastActivityDate="2018-08-03T04:31:16.000" Title="Why doesn't nitrogen monoxide dimerize?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4205" PostTypeId="1" AcceptedAnswerId="4233" CreationDate="2013-02-21T18:49:43.497" Score="8" ViewCount="22012" Body="&lt;p&gt;I'm going to have an online machine shop build a part for me using a CAD drawing that I specify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, these parts are made of aluminium, but this manufacturer also mills parts out of these plastic materials:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Polypropylene (PP)&lt;/li&gt;&#xA;&lt;li&gt;Polystyrene (PS)&lt;/li&gt;&#xA;&lt;li&gt;High impact polystyrene (HIPS)&lt;/li&gt;&#xA;&lt;li&gt;Acrylonitrile butadiene styrene (ABS)&lt;/li&gt;&#xA;&lt;li&gt;Polyethylene terephthalate (PET)&lt;/li&gt;&#xA;&lt;li&gt;Polyester (PES)&lt;/li&gt;&#xA;&lt;li&gt;Polyamides (PA) (Nylons)&lt;/li&gt;&#xA;&lt;li&gt;Poly(vinyl chloride) (PVC)&lt;/li&gt;&#xA;&lt;li&gt;Polyurethanes (PU)&lt;/li&gt;&#xA;&lt;li&gt;Polycarbonate (PC)&lt;/li&gt;&#xA;&lt;li&gt;Polyethylene (PE)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Polypropylene, I know, is the plastic used in water bottles, so I know that is not what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fabricated part will be used for the clutch cover on a motorcycle transmission and will have a raised section that needs to be tapped to allow attaching a gear shifter to, like in the aluminium version below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tOcvX.png&quot; alt=&quot;Clutch Cover with Shift Mount&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like something rigid and strong. Plexiglass comes to mind, but there could be a better alternative that I am unaware of, as long as I am not getting into something exotic that would cost more than aluminium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could suggest a good plastic for me, which would need to be oil resistant and withstand typical engine transmission temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X7D2T.jpg&quot; alt=&quot;Location on Transmission&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1274" LastEditorUserId="4945" LastEditDate="2018-10-31T12:47:18.570" LastActivityDate="2018-10-31T12:47:18.570" Title="Which kind of plastic material would be suitable for a clutch cover on a motorcycle transition?" Tags="&lt;polymers&gt;&lt;materials&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4207" PostTypeId="2" ParentId="4187" CreationDate="2013-02-22T03:27:02.603" Score="5" Body="&lt;p&gt;I think the major error you made is applying what you call the &quot;Metal + Water = Metal Oxide&quot; rule.  This may apply to &lt;em&gt;some&lt;/em&gt; metals, but not all.  Specifically, Lithium is a Group 1 metal, the so-called Alkali Metals.  This means that Group 1 metals tend not to form oxides with water, but rather form &lt;em&gt;hydroxides&lt;/em&gt; with water.  That is, the metal ions bond with hydroxide ion, $\ce{OH-}$, which has a -1 charge.  If you have ever seen this reaction (lots of YouTube videos if you have not done the reaction yourself), you will see that lithium in water produces lots of gas.  And in some videos, you will see that this gas can be set on fire!  Given the elements involved, that's a good indication that the gas is hydrogen, a very flammable gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the most likely equation for lithium in water forms an hydroxide of lithium along with hydrogen gas.  Is that enough information for you to propose a different equation than what you answered?&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-02-22T03:27:02.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4208" PostTypeId="2" ParentId="2469" CreationDate="2013-02-22T04:00:26.370" Score="14" Body="&lt;p&gt;The question of anomalous electronic configurations, meaning $\mathrm{s^1}$ or $\mathrm{s^0}$ in one case (Pd) is very badly explained in textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the anomalous configuration of Cr ($\mathrm{3d^5~4s^1}$) is typically explained as being due to &quot;half-filled subshell stability&quot;.  This is wrong for several reasons. First of all there is nothing especially stable about half-filled subshells. Secondly it does not explain the fact that many second transition series elements show anomalous configurations even though they do not possess half-filled subshells. The possession of a half-filled subshell (henceforth abbreviated as hfss) is neither necessary nor sufficient for there to be an anomalous configuration. There are atoms that have hfss but do not have anomalous configurations and there are atoms that have anomalous configurations but do not have hfss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion is, the use of the hfss explanation is &lt;em&gt;ad hoc&lt;/em&gt; and should be avoided. It so happens that Cr and just one other atom have both hfss AND show an anomalous configuration. But it cannot be generalized.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="16683" LastEditDate="2016-09-09T07:13:21.767" LastActivityDate="2016-09-09T07:13:21.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4210" PostTypeId="2" ParentId="4204" CreationDate="2013-02-22T13:16:12.300" Score="7" Body="&lt;p&gt;Nitrogen monoxide's dimer exists, but it is very unstable. NO consists of dimeres only at low temperatures (about −163 °C). Solid NO seems to consist mostly of dimers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems that unpaired electrons on antibonding orbitals can only pi-overlap. In NO the electronic density is shifted towards oxygen atom, and the &quot;petals&quot; of pi-orbitals don't have enough electronic density to support stable bond (some call it a half-bond). So the dimer bond is weak (enthalpy of formation is about 17 kilojoules)&lt;/p&gt;&#xA;" OwnerUserId="1233" LastEditorUserId="16683" LastEditDate="2016-10-12T10:35:45.487" LastActivityDate="2016-10-12T10:35:45.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4211" PostTypeId="1" CreationDate="2013-02-22T13:48:55.727" Score="2" ViewCount="50" Body="&lt;p&gt;Is there an enzyme which is used as a drug (let it be called version $A$) and a mutated version of it ($M$), where the activity of $M$ was found to be higher than $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references on such a case would be greatly acknowledged.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2014-07-31T16:05:40.550" Title="Which mutated enzymes act as better drugs than their non-mutated form?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4212" PostTypeId="1" AcceptedAnswerId="4238" CreationDate="2013-02-22T16:07:21.777" Score="6" ViewCount="15274" Body="&lt;p&gt;So I was wading through several textbooks yesterday reviewing the concepts of buffers. I found them lot more challenging than I used to. One thing in particular disturbed me: The books and internet all said &quot;Yeah, buffers work pretty great in as long as the  $\rm pH \approx pK_a\pm 1$.&quot; But I am not one to leave these things alone. I finally found the &lt;strong&gt;Quantitative&lt;/strong&gt; explanation I was looking for over at &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution#Buffer_capacity&quot; rel=&quot;nofollow noreferrer&quot;&gt;buffer capacity&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the graph inside the link says the buffering capacity becomes great when you solution becomes very acidic or very basic, as well. And, of course, looking at titration curves there seem to be a few &quot;buffer&quot; zones e.g. where the solution is highly acidic or highly basic, the $\rm pH$ doesn't change as much after a certain point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wj6Do.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, obviously there is something wrong with my reasoning, since nobody calls those areas buffer zones. Is it because it is not useful to have such a highly acid/basic solution? Or is it just because &lt;em&gt;technically&lt;/em&gt;, there is no buffering, just small $\rm pH$ changes? Or maybe something else entirely...&lt;/p&gt;&#xA;" OwnerUserId="1279" LastEditorUserId="4945" LastEditDate="2014-07-30T03:46:57.177" LastActivityDate="2018-01-21T21:20:33.790" Title="Can you create a buffer with a strong acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4213" PostTypeId="1" AcceptedAnswerId="4214" CreationDate="2013-02-22T17:02:35.103" Score="4" ViewCount="3575" Body="&lt;p&gt;As we know on fusing two hydrogen atoms we can convert it to helium and by breaking them we get two $\ce{H}$ atoms. So why can't we convert mercury to gold by removing 1 proton per atom?&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="7475" LastEditDate="2016-05-05T12:09:30.100" LastActivityDate="2016-10-11T21:34:02.687" Title="Why can't we use fusion to make ourselves rich?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4214" PostTypeId="2" ParentId="4213" CreationDate="2013-02-22T18:53:16.067" Score="10" Body="&lt;p&gt;While it is possible to synthesize gold from mercury-196 (exists as 0.15% of natural mercury) the resultant gold will be radioactive and will not be gold for long!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is actually the spallation neutron source which creates gold from liquid mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Synthesis_of_precious_metals#Gold&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1280" LastEditorUserId="33610" LastEditDate="2016-10-11T21:23:57.783" LastActivityDate="2016-10-11T21:23:57.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4215" PostTypeId="2" ParentId="4213" CreationDate="2013-02-22T18:58:44.703" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't we use fusion to make ourselves rich?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because at the current stage of technological progress making gold from other elements by fusion in large quantities is so costly (energy-wise and therefore money-wise), that it is just extraordinarily unprofitable to do that.&lt;/p&gt;&#xA;" OwnerUserId="1233" LastEditorUserId="1233" LastEditDate="2013-02-22T19:10:04.197" LastActivityDate="2013-02-22T19:10:04.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4216" PostTypeId="1" AcceptedAnswerId="4217" CreationDate="2013-02-22T20:25:44.993" Score="6" ViewCount="216" Body="&lt;p&gt;I lit a mosquito coil and used it to melt down small down candle wax which had fallen on the floor. The coil then produced some dense fumes. I directed the fumes to the flame of a candle. As soon the fumes reached the candle, my coil caught fire again and started burning. I mean the flame actually traveled the path of the fumes and re lit my mosquito coil. I just wanted to know what this phenomenon is called? Why does this happen? &lt;/p&gt;&#xA;" OwnerUserId="1281" LastEditorUserId="73" LastEditDate="2013-02-24T22:25:49.840" LastActivityDate="2013-02-24T22:25:49.840" Title="Phenomenon where fumes can relight an extinguished flame" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4217" PostTypeId="2" ParentId="4216" CreationDate="2013-02-22T20:41:37.853" Score="4" Body="&lt;p&gt;It looks like you accidentally created a combustion mixture.  I checked the Material Safety Data Sheet (MSDS) for several mosquito coils, and in addition to the insecticide, one of them also contained the oxidizer potassium nitrate $\ce{(KNO3)}$.  The burning candle likely had some vaporized paraffin (fuel) in the flame, so when you directed the $\ce{KNO3}$ vapor into the flame, you had heat + fuel + oxidizer = combustion!  You literally set that vapor on fire.  I'd be careful with that: although the MSDS indicated a small amount of $\ce{KNO3}$ in the coil, you could still be injured if you were not prepared.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-02-22T20:41:37.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4219" PostTypeId="2" ParentId="4187" CreationDate="2013-02-22T23:41:31.030" Score="4" Body="&lt;p&gt;I think you here looking for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{aM(s) + bH2O(l) -&amp;gt; M_a(OH)_b(s) + $\tfrac b2$\,H2(g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if you remove the water from the medium you are left with the $\ce{LiOH}$. The thing is, the lithium breaks up the water and molecular hydrogen is released. Nevertheless, this ionic compound ($\ce{LiOH}$) is soluble in water, hence the aqueous forms.&lt;/p&gt;&#xA;" OwnerUserId="1282" LastEditorDisplayName="user7951" LastEditDate="2016-10-11T18:33:53.027" LastActivityDate="2016-10-11T18:33:53.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4220" PostTypeId="1" CreationDate="2013-02-23T00:02:05.307" Score="4" ViewCount="110" Body="&lt;p&gt;How do they find out exactly what reactions they can use to produce some substance they want to produce? Is there some reaction database that helps them find a way to make the target substance out of available chemicals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say they want to produce ethanol. There are multiple ways to do this: they could use the fact that yeast can transform sugars into ethanol and carbon dioxyde and use distillation to produce concentrated ethanol, or use ethylene hydration to get the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: how could they find those ways? How do they know what will be produced by a given reaction, or inversely, what reaction will yield a given substance? Some reactions are well known and were just observed and studied, but what about the obscure ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1283" LastActivityDate="2013-02-23T04:36:08.120" Title="How do chemists find a procedure to produce a given substance?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4222" PostTypeId="2" ParentId="4191" CreationDate="2013-02-23T00:56:38.077" Score="3" Body="&lt;p&gt;The top &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mean ionic molality of the solution you have.  You must know the concentration since you need it to compute the activity coefficient.  The bottom &lt;span class=&quot;math-container&quot;&gt;$m_0$&lt;/span&gt; is the standard molality which in general is &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt;.  It is there to make the activity dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required formula for &lt;span class=&quot;math-container&quot;&gt;$m_\pm$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$m_-$&lt;/span&gt; is the molality of the negative ions and &lt;span class=&quot;math-container&quot;&gt;$m_+$&lt;/span&gt; is the molality of the positive ions. Also &lt;span class=&quot;math-container&quot;&gt;$\nu_-$&lt;/span&gt; is the stoichiometric coefficient of the negative ions and &lt;span class=&quot;math-container&quot;&gt;$\nu_+$&lt;/span&gt; is the same for the positive ions. Also &lt;span class=&quot;math-container&quot;&gt;$\nu = \nu_- + \nu_+$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in your specific case, where you have &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0120 M}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na3PO4}$&lt;/span&gt;, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\nu_+ &amp;amp;= 3 &amp;amp; \nu_- &amp;amp;= 1 &amp;amp; \nu &amp;amp;= 4 \\&#xA;m_+ &amp;amp;= \pu{0.0360 M} &amp;amp; m_- &amp;amp;= \pu{0.0120 M}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;m_\pm &amp;amp;= \left[(\pu{0.0360 M})^3(\pu{0.0120 M})^1\right]^{1/4} \\&#xA;&amp;amp;= \pu{0.0274 M}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="16683" LastEditDate="2020-06-06T03:13:28.903" LastActivityDate="2020-06-06T03:13:28.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4223" PostTypeId="2" ParentId="4220" CreationDate="2013-02-23T04:36:08.120" Score="5" Body="&lt;p&gt;Chemists use many sources of information to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of being trained as a chemists is learning, in general terms, what kinds of reactions are known and what their best uses are. That training can go 5-10 years beyond a bachelors degree and is very intensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond what you can keep in your head, one of the most important information sources is the &quot;primary literature&quot;, consisting of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Patents&lt;/li&gt;&#xA;&lt;li&gt;Journal publications&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As an example, see this &lt;a href=&quot;http://www.google.com/patents/US7449313&quot;&gt;patent about ethanol manufacture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, see this &lt;a href=&quot;https://www.beilstein-journals.org/bjoc/single/articleFullText.htm?vt=f&amp;amp;publicId=1860-5397-9-40&quot;&gt;very recent journal publication dealing a new catalytic reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having all of this information out there is a start, but it's a vast collection spread out over many publishers and 100 years or more. To find what you need, it's essential to have a good indexing system. This allows you to search large databases of reactions and molecules for references that exactly match, or which are close to what you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last part - being able to reason by analogy - is critical to being able to get the most out of the primary literature. Maybe I don't find an exact match for the exact reaction I want to do. But I find something pretty close. Here's where my training and reasoning ability come in to either decide to adapt what I've found or keep looking for something better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of one of these search systems is &lt;a href=&quot;http://www.cas.org/&quot;&gt;Chemical Abstracts Service&lt;/a&gt;, but there are others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a professional chemist, you're expected to keep up with all of the main journals in your field so you stay up to date. You do this by reading articles, talking to people, giving talks, and going to conferences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another category of information consisting of 'reviews'. These are articles that summarize a large area of chemistry - for example, recent large-scale methods for the production of ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discovery/invention of new reactions is a very active area of current chemical research.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastActivityDate="2013-02-23T04:36:08.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4224" PostTypeId="2" ParentId="4204" CreationDate="2013-02-23T07:01:43.730" Score="16" Body="&lt;p&gt;Highstaker's answer is almost correct.  The antibonding orbitals that are occupied in NO are in fact pi symmetry, but when the dimer forms that is no longer relevant.  It is a sigma bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enthalpy of the newly formed sigma bond in the dimer is weak because the net gain in bond is off set by the loss of a very odd set of single-electron resonance forms available for NO monomer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\Delta G= -17 \:\mathrm{kJ/mol}$ , and that dimerization is entropically disfavored, when the total free energy is considered there is no gain since entropic effects are on the order of $10-30 \:\mathrm{kJ/mol}$.  Thus any small gain in enthalphy is offset by the loss of entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always look at these things through Gibbs free energy and not just enthalpy.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="22" LastEditDate="2013-02-24T12:12:56.920" LastActivityDate="2013-02-24T12:12:56.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4225" PostTypeId="2" ParentId="4189" CreationDate="2013-02-23T07:11:53.630" Score="4" Body="&lt;p&gt;Ethanol solution, soaked over night of APTEOS will work.  You won't get great coverage, but it will be likely good enough for what you want to do.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subsequent chemistry is a bit more touchy.  The problem is acid byproduct which forms ammonium salt with the amine functions that is difficult to remove.  This leaves latent acid catalyst at your surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are better served forming the amide from your APTEOS and coating with that product.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2013-02-23T21:59:04.750" LastActivityDate="2013-02-23T21:59:04.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4226" PostTypeId="2" ParentId="4213" CreationDate="2013-02-23T22:07:38.333" Score="6" Body="&lt;p&gt;The premise of this question is false.  We don't fuse two hydrogen atoms to make Helium, we fuse two deuterium (Hydrogen-2) to make helium, and it is definitely not going to split back to two deuterium atoms under any circumstances, it is too expensive energetically.  This process is called fission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highstaker's answer is partially correct, and Meditate's answer is almost correct.  The process Mediate describes is radioactive decay, not fusion, or even fission.  Even using fusion processes, 'atom smashing' tends to lead to radioactive products because of the high energy (Hightaker comments) required to overcome electrostatic repulsion.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2016-05-05T14:46:10.457" LastActivityDate="2016-05-05T14:46:10.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4227" PostTypeId="2" ParentId="2488" CreationDate="2013-02-23T22:24:33.293" Score="0" Body="&lt;p&gt;The conformation shown loses the anomeric effect, worth roughly 17 kcal/mol, and about 0.5 kcamol  for the axial OH.  This is opposed by the penalty for two axial OH and an axial CH2OH.  The penalty for these interaction sis harder to estimate, but it probably amounts to 5 kcal/mol.  Thus, on the basis of sterics alone, I expect you have drawn the least stable isomer.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonding is probably not a factor because in both conformations, the same number of hydrogen bonds is available.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-23T22:24:33.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4228" PostTypeId="2" ParentId="4131" CreationDate="2013-02-23T22:35:58.230" Score="2" Body="&lt;p&gt;Breaking an (single) ionic bond is a nonsensical concept.  Ionic bonding it the aggregation combination of many electrostatic interactions with a bulk solid.  To liberate an ion from an ionic solid you have to break many bonds, not just one.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-23T22:35:58.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4231" PostTypeId="2" ParentId="858" CreationDate="2013-02-24T02:11:28.553" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;The current version of the CTFile spec&lt;/a&gt; can be downloaded for free from Accelrys. Based on the files you have, you'll want to look at the &quot;V2000&quot; section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: I believe [M  ZZE] is a vendor-specific property key, possibly introduced by ACD Labs, given it's the package cited in the header on line 2. The CTFile format spec for V2000 files says nothing about an &quot;M  ZZE&quot; property key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CTFile spec says that property keys that are not recognized by parsing software are to be ignored. This makes vendor-specific extensions feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be on the right track in your interpretation. The Galactobuxin entry has this line in the properties block:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;M  ZZE  1  31  23&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Breaking this down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{M  ZZE} is the mystery properties key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1} is the number of data entries on the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{31} is the 1-based index of either an atom or bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atom 31 is one of the rather boring methoxy groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, bond 31 is a real problem child:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V6aCS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This file is abusing the meaning of 'wedge bonds' by trying to use them to indicate chair conformation perspective, rather that what they're intended for - flat stereoprojection. My guess is that the file is trying to annotate this bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question left is - what does {23} refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond 23 is unremarkable. Atom 23 is one of the origins of a wonky wedge bond. But it's not clear why it should be associated with bond 31.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that the software that created this file knew it was abusing wedge bonds and wanted to include a flag on bond 31 so that software reading the file wouldn't misinterpret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you're seeing the same {23} in every file, this could be a code to specially process bond 31's stereo flag.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastActivityDate="2013-02-24T02:11:28.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4233" PostTypeId="2" ParentId="4205" CreationDate="2013-02-24T11:38:20.820" Score="8" Body="&lt;p&gt;Most plastics consist of linear molecules with little inter-molecular interactions, so most plastics melts too easy. The only ones you mentioned having a chance are polyuretanes and polyamides. They have (CO)(NH) fragments, that forms strong intermolecular hydrogen bonds. So, they are ones of the strongest polymeric materials &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I will recommend aluminum anyway as it is &lt;strong&gt;much&lt;/strong&gt; harder to screw things with aluminum. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-02-24T11:38:20.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4235" PostTypeId="2" ParentId="4188" CreationDate="2013-02-24T13:42:32.230" Score="2" Body="&lt;p&gt;Basically, they are the same thing. But, by adding some ingredients and changing the pH, soaps can act like mild or more harsh detergents. There are a lot of different types of soaps in the market; like ones especially made to treat baby skin, for face skin, body, hands, etc. &lt;/p&gt;&#xA;" OwnerUserId="1288" LastActivityDate="2013-02-24T13:42:32.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4236" PostTypeId="1" CreationDate="2013-02-24T17:26:54.190" Score="0" ViewCount="279" Body="&lt;p&gt;Is 2,3-dimethylprop-1-ene a possible organic structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to approach this, what would the correct name of this molecule be?&lt;/p&gt;&#xA;" OwnerUserId="1289" LastEditorDisplayName="user7951" LastEditDate="2016-11-22T16:28:19.927" LastActivityDate="2016-11-22T16:28:19.927" Title="Is 2,3-dimethylprop-1-ene a possible organic structure?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4237" PostTypeId="2" ParentId="4236" CreationDate="2013-02-24T20:49:49.920" Score="12" Body="&lt;p&gt;2-methylbut-1-ene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For future reference: an alkyl substituent at the 3 position of a propyl group probably indicates you did not select the longest carbon chain to name the parent molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your answer was not 'formally' correct.  That being said there was only possible structure to be drawn from the name you used, so correct or otherwise, it is effective communication.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorDisplayName="user7951" LastEditDate="2015-03-28T20:02:42.107" LastActivityDate="2015-03-28T20:02:42.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4238" PostTypeId="2" ParentId="4212" CreationDate="2013-02-25T00:57:16.567" Score="3" Body="&lt;p&gt;I think the issue with buffer systems is that they are resistant to changes from strong acids/bases &lt;em&gt;in their normal state&lt;/em&gt;.  For example, biological systems often are buffered so that they have some resistance to changes in pH, which can wreck havoc on biochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall the basic buffer equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA &amp;lt;-&amp;gt; H+ + A-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{HA}$ is a &lt;em&gt;weak&lt;/em&gt; acid, and there is an excess of $\ce{A-}$, the conjugate base. Because $\ce{HA}$ is weak, it is a relative poor proton donor, but $\ce{A-}$ is a relatively good proton acceptor.  So if you now add a strong acid, the $\ce{H+}$ it donates reacts with the $\ce{A-}$ to produce $\ce{HA}$, shifting the equilibrium to the &lt;em&gt;left&lt;/em&gt;, rather than to the right as one might expect.  Or perhaps, preventing as much shifting to the right as might be expected. This gets to your issue of buffering capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This line of reasoning does not work with strong acids/bases because they donate/accept protons with gusto, and thus are not resistant to shifting the equilibrium.  You would need to be &lt;em&gt;way&lt;/em&gt; at the extremes of pH for there to be a small change to pH, as your graph shows.  In contrast, buffers are resistant to pH change in their normal state.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-02-25T00:57:16.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4239" PostTypeId="2" ParentId="4113" CreationDate="2013-02-25T04:51:03.690" Score="2" Body="&lt;p&gt;The longer the alykylchains, the more likely you can form self-assembled monolayers (SAMs), which will hold themselves together by covalent interactions between chains.  The nanoparticle analog would be miceles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, alcohols are a fairly weak ligand for silver, so I'm not sure this is feasible.  SAMs are known to form on silver with carboxylate ligands.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-25T04:51:03.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4240" PostTypeId="2" ParentId="2827" CreationDate="2013-02-25T04:59:04.633" Score="1" Body="&lt;p&gt;One atom of oxygen is a fictional species outside of a plasma.  Oxygen is naturally diatomic, and the process by which it oxidizes sulfur is radical-based, leading to both oxygen atoms reacting.  The fact that SO2 forms instead of SO is a result of the [redundant answer] thermodynamics favor SO2 over SO.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that simple fact here is a quantum mechanical rationalization: Sulfur can expand its octet, which means in SO2 it has 12 electrons 'around' it.  This allows resonances forms where O- has a significant contribution to resonance, while S still has at least 8 electrons around it.  In SO no such resonance forms exist, meaning the highly electronegative oxygen must share its electrons with with sulfuer to avoid exposing a six-eletron sulfur species.  Hence SO2 should be a better structure than SO overall.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-25T04:59:04.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4241" PostTypeId="2" ParentId="3969" CreationDate="2013-02-25T05:09:18.400" Score="3" Body="&lt;p&gt;That route won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a hard base reduction that does not attack the enone, or you need to protect the enone to make Tetrahydrohyperforin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalytic hydrogenation that is selective against carbonyl reduction is necessary to make Octahydrohyperforin.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-25T05:09:18.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4242" PostTypeId="1" CreationDate="2013-02-25T05:31:26.493" Score="8" ViewCount="252" Body="&lt;p&gt;Is there a chemical that has the bond structure like below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IMY1P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 5 atoms linking together.&#xA;Two of them have 4 bonds. (They are different atoms.)&#xA;Another two of them have 3 bonds. (They can be either same or different atoms.)&#xA;The last atom has only 2 bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested SMILES for unknown $\ce{X}$ elements is &lt;code&gt;[X]2[X]14[X]3[X]1[X]234&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there such kinds of chemicals?&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="11616" LastEditDate="2017-06-30T13:57:32.940" LastActivityDate="2017-07-01T00:24:06.507" Title="Are there any chemicals with this structural skeleton?" Tags="&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4243" PostTypeId="1" AcceptedAnswerId="4245" CreationDate="2013-02-25T07:24:42.103" Score="3" ViewCount="186" Body="&lt;p&gt;I need some help in a electrophilic substitution reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bfvGQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the major product formed in this reaction and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that&#xA;(1) Activator wins over deactivator.&#xA;(2) A powerful wins over a weak activator.&#xA;(3) When a substituent possesses a lone pair, it it usually extends the product determining influence.&#xA;(4) Steric factor always plays a key role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO2}$ is a powerful deactivator, so I think the middle ring will not undergo substitution. I can't decide which of the other two rings will undergo substitution and at what position.&lt;/p&gt;&#xA;" OwnerUserId="1291" LastEditorUserId="51553" LastEditDate="2018-03-21T19:12:48.180" LastActivityDate="2018-03-21T19:12:48.180" Title="Electrophilic Substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4244" PostTypeId="2" ParentId="4074" CreationDate="2013-02-25T19:42:31.980" Score="2" Body="&lt;p&gt;As for your first question, &lt;a href=&quot;http://www.quimica.urv.es/~bo/MOLMOD/General/Forcefields/AMBER.html#Special&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; states that 1-4-interactions are scaled by a factor of 0.5, which I suppose implies 1-2 and 1-3 interactions being excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your second question is yes, as can be seen &lt;a href=&quot;http://www.scm.com/Doc/Doc2010/ADF/ADF-QMMM/page44.html&quot; rel=&quot;nofollow&quot;&gt;here (section TORSIONS)&lt;/a&gt;. The potential is divided (and thus scaled) by N&lt;sub&gt;tors&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-02-25T19:42:31.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4245" PostTypeId="2" ParentId="4243" CreationDate="2013-02-25T19:50:50.170" Score="3" Body="&lt;p&gt;I no longer am a practicing chemist, so I have no access to a structure drawing program; I'll do my best with words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accept the initial premise that the NO2 is strongly deactivating so only the 'end' rings will be attacked.  As we know o,p- substituion have functionalyl equivalent resonance forms, so we only have to consider 4 possible products: p-, m-substitutuion on the 3-ring and similar on the 4-ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing all four of these structures and the reasonanc forms indicates that the p-,4-ring susbtitution and the m-,3-ring susbtitution are poor choices because of high energy intermediates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final question to resolve is whether or not the m-,4-ring is better than the p-,3-ring.  Based on the intermediates, this is a even, but the nitro group is in conjugation with the 4-ring, whle only in cross-conjugation with the 3-ring  You can see this by drawing a rsonance form with a carbocation at the para position, and rearranging the bonds via resonance all the way to the nitro group.  This can't be done with the with the 3-ring, meaning the nitro group deactivates the 4-ring more than the 3-ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the product is likely p-substitution on the 3-ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advanced:&#xA;The resonance for arguments require all three structures be coplanar.  This is not the case; biphenyls have some torsion and twisting because o-hydrogens bump.  However, there is still enough 'flatness' and orbital overlap to make some use of the resonance arguments presented above, albeit the magnitude of better and worse are smaller than they would be for more classical EAS problems.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-02-25T19:50:50.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4246" PostTypeId="2" ParentId="2627" CreationDate="2013-02-25T19:59:00.487" Score="4" Body="&lt;p&gt;I think another argument for the use of HCl here, as opposed to HF, HI, or HBr, is simply that chloride is physiologically inert in concentrations found in medical drugs, as opposed to the other halogenides. In other words, it won't poison you.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-02-25T19:59:00.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4248" PostTypeId="1" CreationDate="2013-02-26T19:17:45.633" Score="4" ViewCount="123" Body="&lt;p&gt;This is a question that has bothered me for a while.  In the past many years, no company has been able to develop a material that can transmit light nearly as well as fiber optic glass, or at least not as cheaply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing a bit of research, I found that although there is fiber optic plastic, the quality is considerably poor when compared to the standard glass medium.  I would assume that developing a highly-transparent, high-quality plastic that is cheap to manufacture would be considered the holy grail of data communication (not to mention, quite profitable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that in the many, many years that fiber optic data transmission has been around, no one has been able to develop such a form of plastic?  In fact, I've never even heard of such research making headlines on the more common tech news sites.&lt;/p&gt;&#xA;" OwnerUserId="665" LastEditorUserId="73" LastEditDate="2013-02-27T01:49:23.900" LastActivityDate="2013-02-27T17:59:24.793" Title="What barriers prevent plastic from becoming a fiber optic, data transmission medium?" Tags="&lt;plastics&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4250" PostTypeId="1" AcceptedAnswerId="4257" CreationDate="2013-02-27T04:26:36.677" Score="1" ViewCount="158" Body="&lt;p&gt;I know sulfuric acid is an important catalyzer and is used in various processes. My question is, how do I recover the remaining sulfuric acid? It will be impure, and I don't know how to do the &quot;standard&quot; procedure (is there one?)&lt;/p&gt;&#xA;" OwnerUserId="1282" LastEditorUserId="1282" LastEditDate="2013-02-27T04:36:04.873" LastActivityDate="2013-04-28T19:11:03.497" Title="Sulfuric acid recovery" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4252" PostTypeId="1" AcceptedAnswerId="4298" CreationDate="2013-02-27T11:03:28.240" Score="3" ViewCount="530" Body="&lt;p&gt;Given some small chemical compounds whose solubility varies with pH, and in the case of compounds that can form complexes, I have seen that the complexation constant varies with pH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there might be two main reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Chemical structure of the host system is affected by pH (for instance, protonation) and therefore it affects complexation constant;   &lt;/li&gt;&#xA;&lt;li&gt;It might be related to the small chemical compound, since its solubility might vary according to pH. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I guess that in this case there are many chemical/physical equilibria involved, in this case mainly phase solubility and protonation for both species. I wonder what is the best approach for describing this, perhaps one can use different equilibrium constants for the mentioned equilibria, and from there derive a global constant that depends on pH. How do these two factors (protonation and phase/solvation) influence the value of the complexation constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a particular case, I am mostly interested in complexation of flavonoids with cyclodextrins.&lt;/p&gt;&#xA;" OwnerUserId="256" LastEditorUserId="41328" LastEditDate="2017-08-06T06:34:44.280" LastActivityDate="2017-08-06T06:34:44.280" Title="Influence of solubility on the complexation constants at different pH values" Tags="&lt;thermodynamics&gt;&lt;solubility&gt;&lt;theoretical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4253" PostTypeId="1" CreationDate="2013-02-27T14:19:40.890" Score="2" ViewCount="401" Body="&lt;p&gt;My teacher gave us the equations for zero-order, first-order and second-order reactions as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [A]_t = -akt + [A]_0 $$&#xA;$$ \ln [A]_t=-akt+\ln [A]_0$$ &#xA;$$\frac1{[A]_t}=akt+\frac1{[A]_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a is the coefficient of $A$ in the chemical reaction, but everywhere I've seen, the $a$ is not there. I have the Central Science 12th edition and there's no $a$ either, is my teacher right? Because that value does affect the answer.&lt;/p&gt;&#xA;" OwnerUserId="599" LastActivityDate="2017-10-13T16:20:43.133" Title="Integrated rate laws" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4256" PostTypeId="2" ParentId="4248" CreationDate="2013-02-27T17:42:54.243" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, I've never even heard of such research making headlines on&#xA;  the more common tech news sites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tasks like that are called deveopment, not research, normally. &#xA;And no industrial company will talk much about such work &#xA;if its not sucessful. &#xA;In fact there are such organic fibers for minor purposes (eg the Laserjets &#xA;II and III used such a fiber to send a synchronising signal from the &#xA;laser scanning unit down to the control board, for about 10 inches.&#xA;30 years ago one could buy such fibres for experiments at radio shack) &#xA;If You read some literature on optical fibres, You will learn that &#xA;low scattering is the grail. And plastics scatter a lot, for &#xA;very basic reasons of polymer structure. Try to look throug a plate of &#xA;acrylic glass &quot;edge on&quot;, and You will understand.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditDate="2013-02-27T17:59:24.793" LastActivityDate="2013-02-27T17:59:24.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4257" PostTypeId="2" ParentId="4250" CreationDate="2013-02-27T17:53:06.450" Score="1" Body="&lt;p&gt;In laboratory, nobody cares. It's too cheap to care about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industry, other catalysts are usually used (heterogeneous ones).&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-02-27T17:53:06.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4258" PostTypeId="2" ParentId="4253" CreationDate="2013-02-27T21:06:54.007" Score="5" Body="&lt;p&gt;From the reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2A -&amp;gt; B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we have $a = 2$, which leads to the differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d[A]}{dt} = - k [A]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which then leads to the integrated rate law you've cited.  The $2$ is nowhere as a coefficient, it is related to the order of the equation, and determines which integral/form should be used.  Similar conventions also apply for 1st and zeroth order kinetics.  I disagree with your instructor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be stressed that these rate laws apply to elementary reactions, not complex reactions.  Experimental measurement may differ from that predicted using an elementary approach, which then leads to the conclusion there are other steps involved.  In the case of OP's comment regarding oxidation of carbon monoxide with NO2, elementary reactions would predict a rate proportional to $$k [\ce{NO2}][\ce{CO}]$$ if the reaction is the result of a simple bimolecular collision. The fact that a different rate is observed indicates this process is more elaborate.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="30836" LastEditDate="2017-10-13T16:20:43.133" LastActivityDate="2017-10-13T16:20:43.133" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4259" PostTypeId="1" CreationDate="2013-02-28T16:59:06.520" Score="2" ViewCount="155" Body="&lt;p&gt;There is a process named solar furnace pyrolysis which is used to reduce the oxides of various elements by concentrating solar energy. Oxygen is liberated but an oxide slag is left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SiO2 -&amp;gt;[pyrolysis] SiO + 1/2O2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the element can be extracted from the oxide produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am making a project for NASA on space settlements in solar furnace pyrolysis. Can solar furnace pyrolysis be used for obtaining both metals as well as oxygen from the metal containing oxides left after pyrolysis? Can it be done by electrolysis? If possible, please provide with a link with your answer.&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="4945" LastEditDate="2019-04-18T09:44:22.703" LastActivityDate="2019-04-18T12:26:10.807" Title="Metal extraction by pyrolysis" Tags="&lt;metal&gt;&lt;extraction&gt;&lt;pyrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4260" PostTypeId="2" ParentId="4259" CreationDate="2013-03-01T05:01:39.130" Score="5" Body="&lt;p&gt;The answer appears to be forthcoming in the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_monoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;silicon monoxide&lt;/a&gt; and references therein — to quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{(SiO)_{n}}$&lt;/span&gt; irreversibly disproportionates into &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Si}$&lt;/span&gt; in a few hours between 400 and 800 °C and very rapidly between 1000 and 1440 °C, although the reaction does not go to completion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article implies that &lt;span class=&quot;math-container&quot;&gt;$\ce{(SiO)_{n}}$&lt;/span&gt; generates a passivating layer of &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; in contact with dioxygen, so presumably the slag would be encapsulated in a thin layer of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, silicon is not generally considered to be a metal, but rather a &lt;em&gt;semimetal&lt;/em&gt; or &lt;em&gt;metalloid&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="41328" LastEditDate="2019-04-18T12:26:10.807" LastActivityDate="2019-04-18T12:26:10.807" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4261" PostTypeId="1" AcceptedAnswerId="4272" CreationDate="2013-03-01T05:08:14.277" Score="5" ViewCount="153" Body="&lt;p&gt;I have recently read many papers where neural networks (NNs) are trained to predict chemical properties (starting from the structure of small chemical compounds) for compounds rather close to the ones with which they were trained. In some cases the predictions are rather good. But I see that apart from the predictive capability, people use NNs as &quot;black boxes&quot; in the sense that they do not tell you or give you chemical insight about why that compound has that value for that predicted property. At least that is my impression now. So my question here is, do you know cases where useful chemical information can be extracted from NN predictions?&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2013-03-04T18:36:39.407" Title="Obtain chemical insight from neural network results" Tags="&lt;computational-chemistry&gt;&lt;structural-biology&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4262" PostTypeId="1" CreationDate="2013-03-01T16:13:15.427" Score="4" ViewCount="520" Body="&lt;p&gt;I've watched over this Khan video titled &quot;Reactions in Equilibrium&quot; and am still confused on the interpretation of equilibrium. &lt;a href=&quot;http://www.youtube.com/watch?v=CGJjjm7-DJ4&amp;amp;list=PL166048DD75B05C0D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reactions in Equilirbium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N2(g) + 3H2(g) &amp;lt;=&amp;gt;2NH3(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever I see an equilibrium like the one above, the amounts of each molecule are only what happens at the start of the reaction, correct? In the video, it was made out as if the starting coefficients next to each molecule were what was needed to reach equilibrium. However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of $\ce{NH3}$ would not necessarily be 2. He does this in other videos too. He starts off with something such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{aA + bB &amp;lt;=&amp;gt; cC + dD}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and immediately plugs those into an equilibrium equation. I guess what he is getting at is that we figure out the final concentrations necessary to reach equilibrium but I'm not positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Towards the end of the video, arbitrary (I guess) concentration amounts of each molecule were plugged into the formula to output the matching equilibrium constant. He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3 H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, let's say the concentrations I have for a particular reaction were not the concentrations at equilibrium. If I still plugged them into the equation, that would tell me not only if I were at equilibrium (if I looked up the constant) but also if the forward or backward reactions were favored right?&lt;/p&gt;&#xA;" OwnerUserId="1302" LastEditorUserId="323" LastEditDate="2013-03-01T20:03:48.750" LastActivityDate="2013-05-14T10:57:40.833" Title="Interpreting equilibrium equations correctly" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4263" PostTypeId="2" ParentId="2679" CreationDate="2013-03-01T16:19:33.487" Score="3" Body="&lt;p&gt;This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-03-01T16:19:33.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4264" PostTypeId="2" ParentId="4261" CreationDate="2013-03-01T16:58:55.520" Score="4" Body="&lt;p&gt;Since you are interested in Artificial Neural Networks that also explain their decisions/predictions you may be interested in &lt;a href=&quot;http://www.math.chalmers.se/~olleh/TNN.pdf&quot; rel=&quot;nofollow&quot;&gt;transparent neural networks&lt;/a&gt;, although they have not been used to my knowledge on chemical information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other interesting approaches include &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_logic_programming#cite_note-1&quot; rel=&quot;nofollow&quot;&gt;inductive logic programming&lt;/a&gt; which have among other things been applied to &lt;a href=&quot;http://www.biomedcentral.com/1471-2105/13/162/abstract&quot; rel=&quot;nofollow&quot;&gt;predicting binding of proteins to hexose&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="440" LastEditDate="2013-03-04T18:36:39.407" LastActivityDate="2013-03-04T18:36:39.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4267" PostTypeId="1" AcceptedAnswerId="4269" CreationDate="2013-03-01T23:00:22.860" Score="13" ViewCount="6105" Body="&lt;p&gt;I know that some chemical substances are used as explosives. Amongst the most famous nitroglycerin in &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamite&quot; rel=&quot;nofollow&quot;&gt;dynamite&lt;/a&gt; and cyclonite in &lt;a href=&quot;http://en.wikipedia.org/wiki/C-4_%28explosive%29&quot; rel=&quot;nofollow&quot;&gt;C-4&lt;/a&gt;. Which chemical properties (I suspect that physical properties have less of an effect) make a substance a good explosive? If we know, for example, the formula and its geometry, is it easy to predict how explosive it can be?&lt;/p&gt;&#xA;" OwnerUserId="50" LastEditorUserId="10185" LastEditDate="2015-06-17T11:25:19.827" LastActivityDate="2015-06-17T11:25:19.827" Title="Which chemical properties make a substance explosive?" Tags="&lt;reaction-mechanism&gt;&lt;theoretical-chemistry&gt;&lt;explosives&gt;&lt;nitro-compounds&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4268" PostTypeId="2" ParentId="4267" CreationDate="2013-03-02T06:44:07.667" Score="10" Body="&lt;p&gt;I'm going to assume by &lt;em&gt;'good explosive'&lt;/em&gt; you mean &lt;em&gt;'high energy to mass ratio'&lt;/em&gt;, which by no means implies anything about whether the explosive is &lt;em&gt;'good'&lt;/em&gt; in the &lt;em&gt;'won't suddenly explode when thought about'&lt;/em&gt; sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not intended to be a full answer as it doesn't touch on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Acetone_peroxide&quot;&gt;terrifying explosivity&lt;/a&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/Diethyl_ether_peroxide&quot;&gt;organic peroxides&lt;/a&gt;, however it's a start. The presence of nitro ($\ce{R-NO2}$) or azide ($\ce{R-N=N=N}$) groups in a molecule is strongly suggestive of explosivity. This has a lot to do with the large negative free energy change of formation of dinitrogen and (I guess) oxides of nitrogen upon decomposition of these molecules. $\ce{N2}$ has one of the strongest bonds known and releasing a lot of $\ce{N2_{(g)}}$ entails a large increase in entropy. Inorganic nitrates and azides are also likely to be explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state of the art in explosive research seems to involve cramming as many consecutive nitrogen atoms as possible into a molecule, as showcased on Derek Lowe's blog:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pipeline.corante.com/archives/2010/10/14/whoa_time_to_clean_the_fishtank_uh_root_canal_appointment_look_at_the_time.php&quot;&gt;$\ce{C4H4N8}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pipeline.corante.com/archives/2013/01/09/things_i_wont_work_with_azidoazide_azides_more_or_less.php&quot;&gt;$\ce{C2N14}$ (what?!)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can take a &lt;em&gt;highly strained&lt;/em&gt; polyhedrane of some description and stick nitro groups all over it. The ur example of this sort of philosophy is &lt;a href=&quot;https://en.wikipedia.org/wiki/Octanitrocubane&quot;&gt;octanitrocubane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0166128003008972&quot;&gt;tetranitrotetrahedrane&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer: Explosives are deadly and people who prepare them are either highly trained experts with exquisite safety infrastructure, or morons. Nothing in my answer should be taken by any person as an endorsement to attempt to synthesise anything described.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-03-02T06:44:07.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4269" PostTypeId="2" ParentId="4267" CreationDate="2013-03-02T09:58:34.990" Score="8" Body="&lt;p&gt;There are, actually, several types of explosives by use: initiating (primary) explosives, secondary explosives and blasting agents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initiating explosive must be sensitive to shock/friction, but not too sensitive. The most famous primary explosive is mercury fulminate, but others are known. To act as primary explosive, substance must have hi-energetic unstable groups, like peroxides, azides and similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondary explosives are substances, that are not sensitive to shock/friction and requires initiating blast to explode. Unlike primary explosives, they can be manufactured as hi-density blocks with high speed of detonation. It is exactly what is needed for military applications. The most common type of secondary explosives are (substituted) polynitroarenes. Most well-known examples are TNT and picric acid. It is very beneficial to have zero oxygen balance for such explosives, i.e. to explode without extra oxygen produced and without extra reducing gases produced. TNT, for example, has negative oxygen balance, producing much carbon on explosion, and dinitromethane has positive oxygen balance, producing extra oxygen on explosion. A less stable class of secondary explosives is nitroamines, with hexo/octo-gen as most well-known example. Anyway, main variables here are relative stability and high density paired with medium-high energy of explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tertiary explosives or blasting agents are explosives, used to move hight amounts of materials. They are usually used in big amounts in mining/engineering and must be cost-effective and produce much gases in explosion. The most common typeы of blasting agents are oxyliquits and compositions, based on ammonia nitrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also special kinds of explosives/pyrotechnic mixtures. For example, a mixture used in car airbags must produce much gases very quickly, but said gases must have low temperature. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we know, for example, the formula and its geometry - is it easy to predict how explosive it can be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, not that easy. Di/Trinitroarenes are usually explosive, but their stability and performance may vary greatly and depends on crystal structure, density and form (crystals/fine dust/puffy things). &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2013-03-02T15:32:21.317" LastActivityDate="2013-03-02T15:32:21.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4270" PostTypeId="1" AcceptedAnswerId="4326" CreationDate="2013-03-02T10:26:07.023" Score="16" ViewCount="7997" Body="&lt;p&gt;This is not the typical chemistry question on this website, but I think it's an important practical question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I got a gas string trimmer, the woman who worked at the gardening shop told me that E10 (fuel commonly sold in the USA - 10% Ethanol, 90% gas) left sitting in a gas tank for months would be harmful. So I wonder, is this true? And if so, what about for cars and other engines (boats, tractors, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effects my life because I was considering at some point getting a plug-in hybrid car because during most weeks I would never have to use or buy fuel. However, could I be damaging this car by leaving the same fuel inside the tank for months?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various claims about ethanol either helping or hurting things, but I have not seen any supported by science. What is the chemical basis of these claims? A layman translation is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="30836" LastEditDate="2020-04-06T11:53:57.667" LastActivityDate="2020-04-06T11:53:57.667" Title="Is there a point at which Ethanol (E10) fuel becomes harmful to gas tanks or engines if not used?" Tags="&lt;everyday-chemistry&gt;&lt;alcohols&gt;&lt;combustion&gt;&lt;materials&gt;&lt;fuel&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4271" PostTypeId="1" AcceptedAnswerId="5426" CreationDate="2013-03-02T11:34:39.290" Score="17" ViewCount="443" Body="&lt;p&gt;Quasicrystals are materials that have long-range atomic order but lack the translational symmetry of conventional crystals. All quasicrystalline tilings and packings I have read about thus far, whilst lacking translational symmetry, have rotation axes and reflection planes. I'm wondering if anyone is aware of experimental evidence of quasicrystals which lack reflection symmetry (and thus possess implicit chirality), or can provide a theoretical rationale for why such structures can or cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, a cursory google search turns up a paper interpreting a viral capsid as having chiral quasicrystalline order - Whilst I don't have access to the paper, I specifically exclude this sort of intepretation from consideration as I'm talking about structures with potentially infinite spatial extent, which viruses do not have.&lt;/p&gt;&#xA;" OwnerUserId="84" LastEditorUserId="30836" LastEditDate="2020-04-06T08:55:05.067" LastActivityDate="2020-04-06T08:55:05.067" Title="Are chiral quasicrystals possible?" Tags="&lt;theoretical-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;chirality&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4272" PostTypeId="2" ParentId="4261" CreationDate="2013-03-02T12:00:24.253" Score="4" Body="&lt;p&gt;I did some work on interpreting neural network QSAR models - I won't claim that they explain everything and has a number of limitations (linearizes network connections, applicable only to feed forward networks). But maybe it will be useful&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ci050022a&quot; rel=&quot;nofollow&quot;&gt;Interpreting Computational Neural Network QSAR Models:  A Measure of Descriptor Importance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ci050110v&quot; rel=&quot;nofollow&quot;&gt;Interpreting Computational Neural Network Quantitative Structure−Activity Relationship Models:  A Detailed Interpretation of the Weights and Biases&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the descriptors going into the network are opaque (as in abstract mathematical properties such as many topological descriptors) then I don't think any interpretation method is going to be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="1306" LastActivityDate="2013-03-02T12:00:24.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4273" PostTypeId="2" ParentId="4262" CreationDate="2013-03-02T19:45:18.590" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of NH3 would not necessarily be 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true. I have not watched the video, but if he is suggesting that the stoichiometric coefficients define the ratios of the equilibrium concentrations of the compounds, then he is wrong. Stoichiometric coefficients do play an important role in equilibrium expressions, but not that one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is how the equilibrium constant (and more broadly the reaction quotient) is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your generic reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{aA + bB &amp;lt;=&amp;gt; cC + dD}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction quotient $Q$ is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. Since $\ce{a}$ equivalents of $\ce{A}$ are in the reactants, $[\ce{A}]$ is multiplied $\ce{a}$ times for $[\ce{A}]^{\ce{a}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c = \frac{\prod\lbrace[\text{Products}]_i\rbrace}{\prod\lbrace[\text{Reactants}]_i\rbrace} = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure version of $Q$ replaces concentrations with partial pressures $P_i$.&#xA;$$Q_p = \frac{\prod\lbrace (P_{\text{products}})_i\rbrace}{\prod\lbrace( P_{\text{reactants}})_i\rbrace} = \frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant ($K_c$ for concentrations and $K_p$ for pressures) is defined as the reaction quotient at equilibrium concentrations/pressures. At equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c = K_c = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_p = K_p =\frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values of $K_c$ and $K_p$ are determined by &lt;strong&gt;experimentally measuring&lt;/strong&gt; the concentrations or pressures of $\ce{A}$, $\ce{B}$, $\ce{C}$, and $\ce{D}$ and plugging them into the expressions above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the specific reaction of nitrogen and hydrogen to form ammonia, the equilibrium constant (at a given temperature) is determined by &lt;strong&gt;measuring&lt;/strong&gt; the equilibrium concentrations/pressures and plugging them into the expressions below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3H2(g) + 2N2(g) &amp;lt;=&amp;gt; 2NH3 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p = \frac{P_{\ce{NH3}}^2}{P_{\ce{H2}}^3 P_{\ce{N2}}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know the value of $K_c$ or $K_p$, we can then use it and some partial concentration/pressure data to predict the equilibrium concentrations/pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 300 K, the $K_c$ of this process (&lt;a href=&quot;http://en.wikipedia.org/wiki/Haber_process#Reaction_rate_and_equilibrium&quot;&gt;the Haber Reaction&lt;/a&gt;) is $4.34\times10^{-3} \text{ M}^{-3}$. If we started with $[\ce{H2}]=3.0 \text{ M}$, $[\ce{N2}]=2.0 \text{ M}$, and $[\ce{NH3}]=2.0 \text{ M}$, we can calculate the equilibrium concentrations of all three species when the dust has settled. First we calculate $Q_c$ and compare to $K_c$. If $Q_c$ is greater than $K_c$, there is too much product, and the reaction will shift toward reactants. If $Q_c$ is less than $K_c$, there is too much reactant, and the reaction will shift toward products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2} = \frac{(2.0 \text{ M})^2}{(3.0 \text{ M})^3 (2.0 \text{ M})^2}= \frac{1}{27 \text{ M}^3}=3.70\times10^{-2} \text{ M}^{-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 3.70\times10^{-2} &amp;gt; 4.34\times10^{-3} \space \therefore \space Q_c &amp;gt; K_c  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this combination, the reaction shifts toward reactants. $[\ce{H2}]$ and $[\ce{N2}]$ increase, and $[\ce{NH3}]$ decreases, all by a factor of $x$ multiplied by the stoichiometric coefficients of each species. The new concentrations are $[\ce{H2}]=3.0+3x \text{ M}$, $[\ce{N2}]=2.0+2x \text{ M}$, and $[\ce{NH3}]=2.0-2x \text{ M}$. The equilibrium expression is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c 4.34\times 10^{-3} = \frac{(2-2x)^2}{(3+3x)^3 (2+2x)^2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I chose the starting concentrations to be those values implied by the stoichiometry, I ended up (coincidentally) with a reaction near equilibrium. Thus, I cannot make assumptions about the value of $x$ relative to the concentrations, so no simplification is possible. To avoid mistakes, I will let &lt;a href=&quot;http://www.wolframalpha.com/input/?i=4.34e-3%20=%20%28%282-2x%29%5E2%29/%28%28%283%2b3x%29%5E3%29%28%282%2b2x%29%5E2%29%29&quot;&gt;WolframAlpha handle the algebra&lt;/a&gt; and solve for $x=3.18\times 10^{-1}$. The equilibrium concentrations are thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H2}]_{eq} = 3+3(3.18\times 10^{-1})=3.95 \text{ M}$$&#xA;$$[\ce{N2}]_{eq}= 2+2(3.18\times 10^{-1})=2.64 \text{ M}$$&#xA;$$[\ce{NH3}]_{eq} = 2-2(3.18\times 10^{-1})=1.36 \text{ M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if we had started at different initial concentrations, we would end with different equilibrium concentrations. The maths are easier if you start with zero of one or two species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could also go on about the other way to determine the equilibrium constant (i.e. from the relationship $\Delta G^o=-RT\ln{K}$), but I have gone on long enough.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-02T19:45:18.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4274" PostTypeId="1" AcceptedAnswerId="4275" CreationDate="2013-03-02T20:39:13.410" Score="7" ViewCount="11703" Body="&lt;p&gt;I am looking for some kind of material that can be used to easily latch something to someone's finger. This is for a first year engineering project. The idea is to use this material to latch small motors on someone's fingers in the same way that the motor in &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/6/62/Piezoelectric_pickup1.jpg&quot;&gt;this picture&lt;/a&gt; is attached. The person should be able to move their fingers around without the motor falling off and the motor should be removable with a slightly strong pull. Also the material should create as small a barrier between the finger and motor as possible so that when the motor vibrates, it can easily be felt by the finger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin than some sort of adhesive or fluid glue, but I'm open to hearing all ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there polymer materials out there (or particular functional groups) that match this description?&lt;/p&gt;&#xA;" OwnerUserId="1019" LastActivityDate="2020-09-04T03:21:51.643" Title="Easily removable material that sticks to skin" Tags="&lt;polymers&gt;&lt;materials&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4275" PostTypeId="2" ParentId="4274" CreationDate="2013-03-02T20:51:07.987" Score="5" Body="&lt;p&gt;I have an engineering solution for you that bypasses the need to find a chemistry solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You want an &lt;a href=&quot;http://en.wikipedia.org/wiki/Adhesive_bandage&quot; rel=&quot;noreferrer&quot;&gt;adhesive bandage&lt;/a&gt;. Adhesive bandages are thin, stick to skin, and are easy to remove. In my experience, the less expensive generic brands are both thinner and stickier, so go with a generic brand. If you are worried about the thickness, do not use the central part with the embedded gauze pad. Only use the wings. Use a strong glue to fix your motor to the back of the adhesive wing.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-02T20:51:07.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4276" PostTypeId="1" AcceptedAnswerId="4589" CreationDate="2013-03-02T22:50:07.540" Score="3" ViewCount="1031" Body="&lt;p&gt;I have been looking into certain approximations for band gap energies in quantum dots and I found a paper &lt;a href=&quot;https://journals.aps.org/prb/abstract/10.1103/PhysRevB.72.125325&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;. However, I noticed that the equations which they obtain are not useful to the situation which I would like to use them in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to do is gather information from other articles that show the absorption and emission spectra and understand the size of the quantum dots if the sizes are not given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at several papers where $\ce{ZnSe}$ QDs are absorbing and emitting around $\pu{\sim 300 .. 450 nm}$ with sizes from $\pu{2 .. 6 nm}$ in diameter.  However, if you use the LDA+C equation for $\ce{ZnSe}$ quantum dots (QDs) ($E_g = 2.54/d^{1.54}$) with diameters of $\pu{\sim 2 nm}$, you get band gaps around $\pu{0.9 eV}$ or $\pu{\sim 1370 nm}$, not $\pu{\sim 400 nm}$ as you would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I am missing here?  Is this a $E_{11}$ transition band gap energy and I am seeing like a $E_{66}$ absorption to $E_{55}$ fluorescence energy emission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_dot#Quantum_confinement_in_semiconductors&quot; rel=&quot;nofollow noreferrer&quot;&gt;exciton binding energies which may take up a large portion of the total exciton energy&lt;/a&gt;. Are these binding energies important to this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, is there a better way to search for and obtain some literature on a simple correlation for these type of energy transitions?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Li, J.; Wang, L.-W. &lt;em&gt;Phys. Rev. B&lt;/em&gt; &lt;strong&gt;2005&lt;/strong&gt;, 72 (12), 125325. DOI &lt;a href=&quot;https://journals.aps.org/prb/abstract/10.1103/PhysRevB.72.125325&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1103/PhysRevB.72.125325&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="41328" LastEditDate="2017-08-06T06:29:37.017" LastActivityDate="2017-08-06T06:30:37.190" Title="Quantum dot band gap energy equation" Tags="&lt;quantum-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4277" PostTypeId="1" AcceptedAnswerId="4295" CreationDate="2013-03-03T04:50:17.200" Score="10" ViewCount="8657" Body="&lt;p&gt;There is a debate about &quot;how reducing&quot; the atmosphere of the early Earth was, and this question is my attempt to grasp toward an understanding of what that that phrase really means.  As a bit of context, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Miller%E2%80%93Urey_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Miller–Urey experiment&lt;/a&gt; produces a wide variety of complex organic molecules, but people say it requires a very reducing atmosphere to work, and it's disputed whether the early Earth's atmosphere was reducing enough for analogous processes to occur. Ultimately I would like to understand what it is about reducing atmospheres that allows complex organic molecules to be created so easily, but the first step is to understand what it really means to say one mixture of gases is &quot;more reducing&quot; than another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything can be more reducing than something else, it seems like $\ce{CH4}$ gas should be more reducing than $\ce{O2}$, so I'm trying to understand the extent to which this can be quantified. However, I'm currently self-learning chemistry, and I'm having difficulty seeing connections between some of the concepts. (So apologies in advance for the rambling nature of this question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seemed like one place to start might be to draw the net reaction of methane oxidation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8GMEH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if you add up the total oxidation state of every atom, you get zero - the carbon gets oxidised, the oxygen gets reduced, and the hydrogen stays the same. This must always be the case for any reaction (right?) since electrons are conserved. At first this made me think that maybe you can't really say that one thing is objectively &quot;more reducing&quot; than another. Though of course you can still talk about its tendency to oxidise or reduce &lt;em&gt;carbon&lt;/em&gt;, and maybe that's what people are talking about when they say an atmosphere is &quot;reducing&quot; or &quot;oxidising&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I came across the concept of redox potential, which does seem to be about things being objectively more reducing or oxidising than others. The problem is, the discussions of the concept that I can find online seem to be in terms of electrons &quot;being transferred&quot; from one molecule to another. But in the reaction above, as far as I can see, electrons aren't being transferred from the $\ce{O2}$ to the $\ce{CH4}$ but instead both molecules are being destroyed and replaced with completely different ones. So I'm having trouble seeing whether, or how, the concept of redox potential can be applied in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this confusion, I guess my specific questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When people talk about an atmosphere being &quot;more reducing&quot; than another, are they referring to redox potential or something else, such as the degree to which the atmosphere reduces or oxidises carbon? If it's the latter, is it something that can be quantified, or is it more of a subjective judgement?; and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the relationship between the concepts of the redox potential of a substance on the one hand, and the oxidation states of atoms on the other? Can the concept of redox potential be applied to gases like $\ce{CH4}$ and $\ce{O2}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="148" LastEditorDisplayName="user7951" LastEditDate="2018-07-01T16:30:04.667" LastActivityDate="2018-07-01T16:30:04.667" Title="What does the phrase &quot;reducing atmosphere&quot; mean in quantitative terms?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4278" PostTypeId="1" AcceptedAnswerId="4861" CreationDate="2013-03-03T06:49:26.347" Score="5" ViewCount="176" Body="&lt;p&gt;What happens to volatile molecules that reach the main olfactory epithelium in the nasal cavity after they bind and the neural stimulus fades?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent do such factors as receptor kinetics and diffusion (or any others) direct where these molecules end up?&lt;/p&gt;&#xA;" OwnerUserId="1309" LastEditorUserId="73" LastEditDate="2013-03-03T19:44:38.363" LastActivityDate="2013-05-06T21:29:19.737" Title="What are the factors controlling the fate of post-adhesed volatile molecules of the olfactory epithelium?" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4279" PostTypeId="1" CreationDate="2013-03-03T08:09:14.427" Score="2" ViewCount="312" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does it work? I've read that its about the similarity of the energy levels of $3d$ and $3p$, but shouldn't the electron shell $4s$ be filled before $3d$ if filled? So what is done with the $3d$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'd also like to know in what cases will atoms of elements expand their valence electrons? Which blocks or groups or periods do this?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="73" LastEditDate="2013-03-03T20:07:36.347" LastActivityDate="2014-01-06T06:15:46.320" Title="How does expansion of electron shells work?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4280" PostTypeId="2" ParentId="4277" CreationDate="2013-03-03T12:37:22.370" Score="4" Body="&lt;p&gt;I'm not certain mine will be the best answer, but I may be able to get this started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is yes, redox potentials are the key to determining how reducing or oxidizing a reaction is.  What your simple balanced redox equation (which is correct) does not show are the various half reactions that produce the complete reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big oxidizing half reaction in your equation is the classic reduction of oxygen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2O2 +8H+ + 8e- -&amp;gt;2H2O}, E_0=+1.23V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already balanced this half reaction for your combustion equation.  The positive E0 value of 1.23 volts says this reaction tends to occur spontaneously (given some activation energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the other half reaction, the oxidation of carbon in methane, is likely MUCH more complex, because it involves many steps.  Here's where my answer my not be completely correct.  Below is one &lt;em&gt;possible&lt;/em&gt; sequence of half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH4 +H2O -&amp;gt; CH3OH + 2H+ + 2e-}, E_0=-0.50V$&#xA;$\ce{CH3OH -&amp;gt; HCHO + 2H+ + 2e-}, E_0=-0.13V$&#xA;$\ce{HCHO +H2O -&amp;gt; HCOOH + 2H+ + 2e-}, E_0=+0.03V$&#xA;$\ce{HCOOH -&amp;gt; CO2 + 2H+ + 2e-}, E_0=+0.11V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice first that the half reaction oxidizing methane to methanol has a negative potential of -0.50 volts and so is not spontaneous; it needs an oxidizer like oxygen to get started.  In the sequence of half-reactions I list, the potentials get progressively bigger, so that by the time we are oxidizing formic acid to carbon dioxide, the half reaction is slightly spontaneous (positive potential).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this shows that you were correct that the reaction involves the creation and destruction of various compounds, but the sequence (the &lt;em&gt;mechanism&lt;/em&gt;) depends on redox reactions which DO transfer electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the reason my answer may not be best is that the mechanism I propose may be wrong.  While all these half reactions do exist and produce a spontaneous total, the total potential sums to 0.74 volts.  At least one source I read suggested that the actual potential of methane oxidation to carbon dioxide is 0.92 volts.  So there may be some other sequence of half reactions, which I could not track down, which is a better mechanism than the one I propose.  Still, I think my answer gets the principles right.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-03-03T12:54:09.557" LastActivityDate="2013-03-03T12:54:09.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4281" PostTypeId="1" AcceptedAnswerId="5648" CreationDate="2013-03-03T13:49:07.607" Score="4" ViewCount="256" Body="&lt;p&gt;We installed an artesian well and we had some surprises. We asked a few places around and they all told me that we needed 4 different systems to solve our water problem. I am ready to accept this fact but before having our basement transformed to a factory I just wanted to confirm that this is the only solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the result of my analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Barium  $6.9 \:\mathrm{mg/L}$    (2nd analysis revealed same results)  &lt;/li&gt;&#xA;&lt;li&gt;Apparent Color 17.5 UCA&lt;/li&gt;&#xA;&lt;li&gt;Iron  $0.47 \:\mathrm{mg/L}$&lt;/li&gt;&#xA;&lt;li&gt;Total sulfur $0.07 \:\mathrm{mg/L}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also the water is hard ($210 \:\mathrm{mg} \ce{CaCO3}/\rm L$) which is not recommended&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, do I really need 4 different systems to treat my water?&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="9961" LastEditDate="2016-07-25T17:10:49.510" LastActivityDate="2016-07-25T17:10:49.510" Title="Do I really need 4 different water treatment systems with this water analysis?" Tags="&lt;water-treatment&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4282" PostTypeId="1" CreationDate="2013-03-03T15:15:38.540" Score="7" ViewCount="5167" Body="&lt;p&gt;Do you know any tools that generate electrostatic potential 2d maps of the molecules as an image?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastEditorUserId="1322" LastEditDate="2013-04-16T18:02:15.407" LastActivityDate="2017-04-09T04:54:06.437" Title="Tool that generates electrostatic maps of molecules?" Tags="&lt;computational-chemistry&gt;&lt;molecules&gt;&lt;software&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4283" PostTypeId="1" AcceptedAnswerId="4284" CreationDate="2013-03-03T18:57:35.843" Score="6" ViewCount="1315" Body="&lt;p&gt;I'm looking to experiment with distillation to have a basis for understanding the theory slightly better, and get a better feel for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping for some suggestions as to what I could try distilling, such as purifying drinking alcohol. Any single idea would be appreciated, best answer would be awarded to an idea where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The resulting purity would be straight forward to approximately test&lt;/li&gt;&#xA;&lt;li&gt;No part of the distillation presents a significant hazard (e.g. toxic fumes, high volatility)&lt;/li&gt;&#xA;&lt;li&gt;No particularly specialist or expensive equipment is required (e.g. extreme temperatures)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think that pretty much covers it. I'm just looking for one or two ideas which I can experiment with besides alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="371" LastActivityDate="2014-01-29T00:42:24.180" Title="What household substances could be distilled" Tags="&lt;home-experiment&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4284" PostTypeId="2" ParentId="4283" CreationDate="2013-03-03T20:40:25.840" Score="4" Body="&lt;p&gt;This is tricky. Besides water, nearly everything has some hazard associated with it.  And even water distillation involves high temperature (100C).  Be careful distilling alcohol: the azeotrope of distilled alcohol is 95% ethanol which is VERY flammable! Plus, technically speaking, distilling alcohol may be illegal where you live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you could do is distill water with dissolved substances, most likely table salt (NaCl).  This would help you test the principle that distillation is a technique for &lt;em&gt;separating&lt;/em&gt; substances. In this case, distilling salt water should yield pure water. A more visual example would be brewed coffee or tea, which is a suspension and should separate into water and crud!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after that, I'd be careful! For example, concentrating acetic acid (white vinegar) is dangerous, as concentrated acetic acid is both corrosive and has choking fumes.  I would not even consider trying something that is not food grade, given your requirements.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-03-03T21:41:14.290" LastActivityDate="2013-03-03T21:41:14.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4285" PostTypeId="1" AcceptedAnswerId="4286" CreationDate="2013-03-03T22:10:04.300" Score="6" ViewCount="1376" Body="&lt;p&gt;I'm looking for ways to manipulate molten glass with (e.g. bending a heated glass tube). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/eng99/eng99295.htm&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; mentions that oxides tend to adhere, hence molten glass to heated stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What materials could I use to manipulate hot glass? Perhaps I could coat the glass or instrument doing the manipulation with some sort of powder or liquid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass is soda lime glass which has a transition temperature of 550 degrees C, so the temperature it needs to cope with will be a fair bit above that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="371" LastEditorUserId="23561" LastEditDate="2019-04-07T01:07:27.593" LastActivityDate="2019-04-07T01:07:27.593" Title="Materials that resist adhesion with molten glass even when hot" Tags="&lt;surface-chemistry&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4286" PostTypeId="2" ParentId="4285" CreationDate="2013-03-04T01:16:01.377" Score="5" Body="&lt;p&gt;Professional glass blowers use graphite paddles to &quot;push&quot; the glass around.  But to bend a heated glass tube you need to know that glass is a very poor conductor of heat.  Thus it is possible to grasp the heated tube six or so inches from the heated spot and just bend the glass.  This works well for small bore tubing.  Just be careful putting the bent tube down.  Putting it on a high heat conductivity surface will result in the glass breaking.  In the old days asbestos pads were used, but that's a big no-no now.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-03-04T01:16:01.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4287" PostTypeId="1" CreationDate="2013-03-04T03:26:27.490" Score="2" ViewCount="61" Body="&lt;p&gt;Could anyone offer a suggestion as to how an economy-minded chemist ought to implement a DIY-ed positive displacement alternative to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh4.googleusercontent.com/-MjSfgQg6rlE/UTQaF9zsdrI/AAAAAAAAAPI/U3ZKipb3Ah4/w700-h708-p-o-k/photo.JPG&quot; alt=&quot;failure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1309" LastEditorUserId="15477" LastEditDate="2015-05-23T17:36:49.163" LastActivityDate="2015-05-23T17:36:49.163" Title="Is there a way to cheaply transfer (plunge by pushing) a viscous liquid in to another container without plunger-to-tube discharge?" Tags="&lt;titration&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4288" PostTypeId="1" AcceptedAnswerId="4290" CreationDate="2013-03-04T05:39:15.483" Score="5" ViewCount="2670" Body="&lt;p&gt;If I start with a small amount of a substance, and I dilute that substance with billions of gallons of water, what is the scientific explanation for there being no (actually zero) remaining molecules of that original substance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is related to the practice of homeopathic dilutions. I was reading some articles which stated that homeopathic dilutions don't work because at most homeopathic levels of dilution, there are very few or no molecules or the original substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm confused about is how there can no longer be any molecules of the original substance (or even fewer molecules) of an original substance when diluted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are X number of molecules in a solution, and Y liters of water are poured into the solution for dilution, don't the X number of molecules remain? How can a molecule just &quot;disappear&quot;? How can &lt;em&gt;even&lt;/em&gt; one molecule just &quot;disappear&quot;? It makes more sense if these molecules are undergoing a chemical reaction, but if two kind of molecules (the substance, and water) are not undergoing any chemical reaction, then how can one molecule of either kind simply vanish?&lt;/p&gt;&#xA;" OwnerUserId="750" LastEditorUserId="19616" LastEditDate="2015-10-30T17:15:32.317" LastActivityDate="2019-12-06T02:42:29.690" Title="How can a substance be diluted to nothing?" Tags="&lt;everyday-chemistry&gt;&lt;concentration&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4289" PostTypeId="1" CreationDate="2013-03-04T05:47:27.877" Score="0" ViewCount="2350" Body="&lt;p&gt;My textbook said $\ce{NO2+}$ and $\ce{NO2-}$ both have resonance structures, now I am a little bit confused since I cannot figure out the resonance structure for $\ce{NO2+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it?&lt;/p&gt;&#xA;" OwnerUserId="1310" LastEditorUserId="4231" LastEditDate="2015-03-08T17:03:51.600" LastActivityDate="2015-03-08T17:03:51.600" Title="Does resonance apply to NO₂⁺?" Tags="&lt;inorganic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4290" PostTypeId="2" ParentId="4288" CreationDate="2013-03-04T08:13:45.730" Score="18" Body="&lt;p&gt;You correctly point out that the number of molecules in a solution is finite and constant, however the volumetric concentration (that is, how many molecules per litre) changes upon dilution. If, for instance, you take one liter of a 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro_constant&quot;&gt;mol/L&lt;/a&gt; solution of ethanol in water ($\approx{6.02\times{10}^{23}}$ ethanol molecules per liter) and add 9 liters of distilled water, you now have a 1 mole of ethanol in 10 liters, giving a concentration of 0.1 mol/L. If you were to draw out 1 liter of this and count the number of ethanol molecules in that liter, you would find approximately $\approx{6.02\times{10}^{22}}$ - a tenfold reduction. The rest of the ethanol is retained in the other 9 liters of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalently, if you were to measure out 100 mL of 1 mol/L ethanol solution and then make it up to 1 liter by adding distilled water, you would also get a 0.1 mol/L solution. This is the kind of thing taken to absurd extremes in homoeopathy, where tiny quantities of solution are repeatedly diluted, with some shaking. Obviously, they don't literally add billions of liters of water to their solutions, but rather pipette out a tiny amount of solution and discard the remainder, adding water to make up the desired volume. This process is repeated many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite number of molecules present in a solution leads to the concept of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Homeopathic_dilutions#The_molar_limit&quot;&gt;dilution limit&lt;/a&gt;, which is where you have carried out sufficient dilutions that statistically speaking, your chances of finding even a single molecule of the dissolved ingredient become vanishingly small and that by extrapolation, you would have to drink approximately a zillion liters of the preparation to get a single molecule of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did all those other molecules go? They didn't simply disappear - they were poured down the drain during the dilution process, which curiously is the same place money goes when spent on homoeopathy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: &lt;a href=&quot;http://www.theodoregray.com/periodictable/Samples/094.5/index.s15.html&quot;&gt;homoeopathic plutonium&lt;/a&gt;! (courtesy of Theodore Gray's periodic table)&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-03-04T08:13:45.730" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="4293" PostTypeId="2" ParentId="4288" CreationDate="2013-03-04T13:03:33.260" Score="6" Body="&lt;p&gt;They don't vanish. It's just that if there are X molecules and Z bottles with &lt;code&gt;Z &amp;gt; X&lt;/code&gt;, part of the bottles will not have even one molecule.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-03-04T13:03:33.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4294" PostTypeId="2" ParentId="854" CreationDate="2013-03-04T15:35:52.083" Score="5" Body="&lt;p&gt;if you can take a look at the main web site, then it seems they make $\ce{HO2}$ and not $\ce{HO2-}$. This a peroxy radical which exists at trace levels in the ambient atmosphere, usually from reactions of light with oxides of nitrogen and volatile organic compounds. Its really reactive and I don't know for sure how long it will last in the air of the room. It will however react with $\ce{OH}$ radical and make water and $\ce{O2}$. $\ce{HO2 + OH -&amp;gt; O2 + H2O}$. $\ce{HO2}$ will also react with $\ce{NO}$ to form $\ce{NO2}$ and then depending if there is sunlight could make ozone. One of the issues for the machine will be that $\ce{HO2}$ will need to made at some point and therefore one of the first molecules it will encounter will be itself and therefore make $\ce{H2O2}$ (hydrogen peroxide). I think getting one into a lab and testing it would be a start.&lt;/p&gt;&#xA;" OwnerUserId="1318" LastEditorUserId="275" LastEditDate="2013-03-05T02:15:47.430" LastActivityDate="2013-03-05T02:15:47.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4295" PostTypeId="2" ParentId="4277" CreationDate="2013-03-04T23:17:51.970" Score="10" Body="&lt;p&gt;I’m an environmental chemist and we frequently refer to ‘oxidizing’ or ‘reducing’ conditions in the environment. I’ll use soils as an example. The ‘oxic zone’ is the area of soil that oxygen can reach by diffusion and is seen as being under ‘oxidizing’ conditions because the reduction of $\ce{O2}$ is the dominant redox reaction that occurs in that zone. Elements that have multiple oxidation states tend be in the higher oxidation states. For example, sulfur will exist primarily as sulfate ($\ce{SO4^2-}$). As you get further away from the Earth’s surface, the oxygen is depleted (by microbes) and the redox conditions change: the dominant reduction reaction changes. Usually in this order: $\ce{O2}$, nitrate, manganese(III,IV), iron(III), sulfate and finally $\ce{H2}$. When the conditions become so reducing that the microbes are using $\ce{H2}$ as an electron acceptor, they making methane via a process called methanogenesis. This order arises because of reduction potentials. Microbes get the most energy for the least effort from reducing $\ce{O2}$, so they do that until it runs out, which is when they switch to the next best thing available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately what it means to be a ‘reducing environment’ is that if you added a compound with a higher reduction potential then the dominant redox pair, it would become reduced. For example, if you could put $\ce{SO4^2-}$ in a methanogenic environment, it would be reduced to $\ce{H2S}$ very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve read that the composition of Earth’s atmosphere in pre-life earth was very different from today. If I recall correctly, there was no $\ce{O2}$ as it was formed by microbes later on. Without $\ce{O2}$ in Earth’s atmosphere, it makes sense that conditions would be very reducing based on an analogy to the soil redox zones.&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorDisplayName="user7951" LastEditDate="2015-05-08T18:39:20.617" LastActivityDate="2015-05-08T18:39:20.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4298" PostTypeId="2" ParentId="4252" CreationDate="2013-03-05T00:00:16.113" Score="2" Body="&lt;p&gt;It is possible to calculate out all of the complexation and speciation reactions using the concepts of Aquatic Chemistry. I’m a PhD student in environmental chemistry and I TA a course in Aquatic Chemistry. Usually, when pH matters it’s because the protonation state controls the chemistry and can be described using equilibrium constants. I bet that your solubility issue is related to the deprotonated form being an anion and thus more soluble in water than the protonated, neutral form (very common for organic compounds). The protonation state can be calculated from the $\mathrm{p}K_\text{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{L- + H+ &amp;lt;=&amp;gt; HL}\\&#xA;K_\text{a}=10^{-\mathrm{p}K_\text{a}}=\dfrac{[\ce{L-}][\ce{H+}]}{[\ce{HL}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $K_\text{a}$ is the equilibrium constant, and L is your compound. I would guess that your flavanoids and cyclodextrins might have multiple $\mathrm{p}K_\text{a}$s, you can write an expression like this for each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complexation reactions can be written in a similar way. Let’s say that $\ce{Y}$ is the molecule your $\ce{L}$ molecule complexes with, and it is neutral when protonated ($\ce{HY}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{L- + HY &amp;lt;=&amp;gt; HLY-}\\&#xA;K=\dfrac{[\ce{L-}][\ce{HY}]}{[\ce{HLY-}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $K$ is the equilibrium constant for this complexation reaction. Keep in mind that which species (protonation state) is in the expression matters, the value for the equilibrium constant is only valid for those species. Above, I’ve written that the deprotonated form of L ($\ce{L-}$) complexes with the protonated form of Y ($\ce{HY}$). This is where pH comes in. pH controls how much of L is deprotonated and how much of Y is deprotonated and able to participate in complexation, and $K$ controls how much of the available $\ce{L-}$ and $\ce{HY}$ form the complex. Complexes themselves can also have $\mathrm{p}K_\text{a}$s the one above ($\ce{HLY-}$) could probably be protonated to a neutral form $\ce{H2LY}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to guess that the mathematics gets a little complex at this point. It is possible to do these calculations by hand, but there are computer programs that can handle them. The most easily accessible one is Visual MinTEQ. The other hard part is having the data for the equilibrium constants, there may be data for your system, but you’ll have to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to summarize. Complexation constants don’t change with pH (they are, in fact, constant) it’s just that the speciation matters, and speciation is controlled by pH. The ‘global constant’ you refer to is pH, not something else.&lt;/p&gt;&#xA;" OwnerUserId="702" LastEditorDisplayName="user7951" LastEditDate="2015-04-16T13:38:28.390" LastActivityDate="2015-04-16T13:38:28.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4299" PostTypeId="1" AcceptedAnswerId="4301" CreationDate="2013-03-05T03:26:17.780" Score="3" ViewCount="143" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This linear polyatomic ion containing three atoms of carbon has a negative four charge and is only found bonded with lithium and magnesium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could anyone identify this for me? It's from a quiz bowl clue that doesn't have the answer with it.&lt;/p&gt;&#xA;" OwnerUserId="1321" LastEditorUserId="41328" LastEditDate="2019-11-03T18:29:45.110" LastActivityDate="2021-05-11T05:41:43.057" Title="Identifying a quadruply charged anion containing three carbon atoms" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4301" PostTypeId="2" ParentId="4299" CreationDate="2013-03-05T12:28:44.167" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sesquicarbide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick &lt;a href=&quot;https://www.google.com/search?q=li4c3&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot; rel=&quot;noreferrer&quot;&gt;Google search&lt;/a&gt; suggests that $\ce{Li4C3}$ is a real compound, but based on what I saw in the links, it probably is not ionic in the same way that lithium acetylide $\ce{Li2C2}$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg2C3}$ appears to be the formula for &lt;a href=&quot;https://www.google.com/search?q=mg2c3&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot; rel=&quot;noreferrer&quot;&gt;magnesium carbide&lt;/a&gt;, or at least &lt;em&gt;a magnesium carbide&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbide&quot; rel=&quot;noreferrer&quot;&gt;carbide&lt;/a&gt; is a binary compound of carbon and a less electronegative element, usually a metal or metalloid. Carbide is also a generic name for mono- and polyatomic anions containing only carbon. The following ions would all carbide ions: $\ce{C^{4-}, C2^2-, C2^{4-}, C2^{6-}, C3^{4-}, C3^{6-1}, C3^{8-}}$, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The special name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbide#Salt-like_.28saline.29_carbides&quot; rel=&quot;noreferrer&quot;&gt;three of these&lt;/a&gt; carbide ions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C^{4-}}$ - Methide&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C2^{2-}}$ - Acetylide&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C3^{4-}}$ - Sesquicarbide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-05T12:28:44.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4302" PostTypeId="1" CreationDate="2013-03-05T15:43:59.680" Score="3" ViewCount="897" Body="&lt;p&gt;I'm trying to get a set of FTIR or (FTIR like range) x/y numerical data points for a couple of elements,  namely Silver, Platinum, Palladium, Gold and Sodium.  I've looked on line and found images only of the FTIR but I need the numerical x/y data points with the largest wavelength range.  I tried NIST but they just show the NMR data.  Is there a site where I can get just these or can someone send these/post these for me?  And is it possible to get an FTIR of Sodium just by itself or does it need to be combined with another element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note I know metals (due to covalent bonds) don't work well with FTIR but I'm trying to get / see if it's possible to get a large wavelength range analysis of the items above like in FTIR.  Sorry for using the FTIR term I don't know the proper term to use at the moment.&lt;/p&gt;&#xA;" OwnerUserId="554" LastEditorUserId="554" LastEditDate="2013-03-05T16:33:46.107" LastActivityDate="2014-04-07T16:58:42.117" Title="large wavelength range and list of x/y numerical data points of elements" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;nanotechnology&gt;&lt;resonance&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4303" PostTypeId="2" ParentId="4279" CreationDate="2013-03-05T15:52:02.543" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You're right! The 4s subshell is filled before 3d. Electrons fill the shell   with least energy first and the 4s shell has a lower energy than the 3d subshell.&#xA;Use &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow&quot;&gt;the Aufbau Principle&lt;/a&gt; to find the shell with least energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it says that shells with a lower n+l (n and l are the principal and azimuthal quantum numbers) value are filled first. &#xA;If two shells have the same n+l value, then the shell with lower n is filled first.&#xA;There are a few exceptions to this rule - Copper, Chromium, Palladium, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason: n+l gives the number of nodes in an orbital. The more the number of nodes, the less stable the orbital and so the higher the energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Valence shells are filled in the s and p blocks. The inner shells are filled in the d and f blocks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2013-03-05T21:09:36.587" LastActivityDate="2013-03-05T21:09:36.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4305" PostTypeId="1" AcceptedAnswerId="8468" CreationDate="2013-03-05T16:39:59.673" Score="51" ViewCount="15696" Body="&lt;p&gt;Not to be confused with &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2655/sn1-like-reaction-acid-cataylzed-hydrolysis-of-epoxide&quot;&gt;what is the mechanism of acid-catalyzed ring opening of epoxides&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the correct order of regioselectivity of acid-catalyzed ring-opening of epoxides: $3^\circ$ &gt; $2^\circ$ &gt; $1^\circ$ or $3^\circ$ &gt; $1^\circ$ &gt; $2^\circ$? I am getting ready to teach epoxide ring-opening reactions, and I noticed that my textbook has something different to say about the regioselectivity of acid-catalyzed ring-opening than what I learned. My textbook does not agree with 15 other introductory texts I own, but it does agree with one. None of my Advanced Organic Chemistry texts discuss this reaction at all. Thus, I have no ready references to go read. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  My textbook is in the minority, and it is a first edition. Is it wrong, or do the other 15 texts (including some venerable ones) oversimplify the matter? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What I learned&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the acid-catalyzed ring-opening of epoxides follows a mechanism with S&lt;sub&gt;N&lt;/sub&gt;2 features (inversion of stereochemistry, no carbocation rearrangements), the mechanism is &lt;strong&gt;not&lt;/strong&gt; strictly a S&lt;sub&gt;N&lt;/sub&gt;2 mechanism. The transitition state has more progress toward the C-LG bond breaking than an S&lt;sub&gt;N&lt;/sub&gt;2, but more progress toward the C-Nu bond forming than S&lt;sub&gt;N&lt;/sub&gt;1. There is significantly more $\delta ^+$ character on the carbon than in S&lt;sub&gt;N&lt;/sub&gt;2, but not as much as in S&lt;sub&gt;N&lt;/sub&gt;1. The transition states of the three are compared below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xNojp.png&quot; alt=&quot;Epoxy_TS.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/More_O%27Ferrall-Jencks_plot&quot; rel=&quot;noreferrer&quot;&gt;More O’Ferrall-Jencks diagram&lt;/a&gt;, the acid-catalyzed ring-opening of epoxides would follow a pathway between the idealized S&lt;sub&gt;N&lt;/sub&gt;2 and S&lt;sub&gt;N&lt;/sub&gt;1 pathways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CB5Qb.png&quot; alt=&quot;Epox_MOFJ.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the significant $\delta ^+$ character on the carbon, the reaction displays regioselectivity inspired by carbocation stability (even though the carbocation does not form): the nucleophile preferentially attacks at the more hindered position (or the position that would produce the more stable carbocation if one formed). If a choice between a primary and a secondary carbon is presented, the nucleophile preferentially attacks at the secondary position. If a choice between a primary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position. If a choice between a secondary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SmNb8.png&quot; alt=&quot;Epoxy_regio1.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall order of regioselectiveity is $3^\circ$ &gt; $2^\circ$ &gt; $1^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What my textbook says&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My text agrees that the mechanism is somewhere in between the S&lt;sub&gt;N&lt;/sub&gt;2 and S&lt;sub&gt;N&lt;/sub&gt;1 mechanisms, but goes on to say that because it is in between, electronic factors (S&lt;sub&gt;N&lt;/sub&gt;1)  do not always dominate. Steric factors (S&lt;sub&gt;N&lt;/sub&gt;2) are also important. My text says that in the comparison between primary and secondary, primary wins for steric factors. In other words, the difference between the increased stabilization of the $\delta ^+$ on secondary positions over primary positions is not large enough to overcome the decreased steric access at secondary positions. For the comparison of primary and tertiary, tertiary wins. The increased electronic stabilization at the tertiary position &lt;em&gt;is&lt;/em&gt; enough to overcome the decreased steric access at the tertiary position. The comparison between secondary and tertiary is not directly made, but since $3^\circ$ &gt; $1^\circ$ and $1^\circ$ &gt; $2^\circ$, it is implied that $3^\circ$ &gt; $2^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/19PBa.png&quot; alt=&quot;Epoxy_regio2.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this pattern is true, then other cyclic &quot;onium&quot; ions (like the bromonium ion and the mercurinium ion) should also behave this way. They don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical of introductory texts, no references are provided. A Google search did not yield satisfactory results and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Epoxide#Reactions&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on epoxides&lt;/a&gt; is less than helpful.&#xA;Since I 15 other introductory texts on my bookshelf, I consulted all of them on this reaction. The following is a summary of my findings. Only two of the texts (the one I am using and one other) describe the regioselectivity as $3^\circ$ &gt; $1^\circ$ &gt; $2^\circ$. All of the other books support the other pattern, including Morrison and Boyd (which lends credence to the pattern that I learned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Books that have $3^\circ$ &gt; $2^\circ$ &gt; $1^\circ$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Brown, Foote, Iverson, and Anslyn&lt;/li&gt;&#xA;&lt;li&gt;Hornsback&lt;/li&gt;&#xA;&lt;li&gt;Ege&lt;/li&gt;&#xA;&lt;li&gt;Wade&lt;/li&gt;&#xA;&lt;li&gt;Bruice&lt;/li&gt;&#xA;&lt;li&gt;Smith&lt;/li&gt;&#xA;&lt;li&gt;Fessenden and Fessenden&lt;/li&gt;&#xA;&lt;li&gt;Volhardt and Schore&lt;/li&gt;&#xA;&lt;li&gt;Solomons and Fryhle&lt;/li&gt;&#xA;&lt;li&gt;Jones&lt;/li&gt;&#xA;&lt;li&gt;Baker and Engel&lt;/li&gt;&#xA;&lt;li&gt;Ouellette and Rawn&lt;/li&gt;&#xA;&lt;li&gt;Carey&lt;/li&gt;&#xA;&lt;li&gt;Morrison and Boyd&lt;/li&gt;&#xA;&lt;li&gt;Streightweiser and Heathcock&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Books that have $3^\circ$ &gt; $1^\circ$ &gt; $2^\circ$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Klein (the text I am using)&lt;/li&gt;&#xA;&lt;li&gt;McMurray&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also surveyed my various Advanced Organic texts (March, Smith, Carey and Sundberg, Wyatt and Warren, Lowry and Richardson, etc.). Interestingly, none of them even mention acid-catalyzed ring-opening of epoxides (either by Brønsted or Lewis acids). I suspect that these omissions mean that this reaction 1) has difficult to predict regioselectivity (despite the predominance of introductory books that suggest otherwise), and thus 2) is synthetically useless. If #2 is true, then why is this reaction in introductory organic texts?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="23561" LastEditDate="2018-09-02T00:40:07.250" LastActivityDate="2018-09-02T00:40:07.250" Title="Regioselectivity of acid-catalyzed ring-opening of epoxides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;erratum&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="4306" PostTypeId="2" ParentId="4302" CreationDate="2013-03-05T17:39:55.853" Score="0" Body="&lt;p&gt;I don't know where you can get the X/Y data, but you can try to digitize the images you have found using something like a &lt;a href=&quot;http://plotdigitizer.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;plot digitizer&lt;/a&gt;. I know it's not the best solution, but in bad circumstances it can work.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2013-03-05T17:39:55.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4307" PostTypeId="2" ParentId="4305" CreationDate="2013-03-05T18:18:12.323" Score="16" Body="&lt;p&gt;Extending the pattern of epoxides to bromonium ions and mercurinium ions is hazardous at best.  Below the second row (the carbon row) bond lengths get much longer, and acid strengths are much lower than compared to sulfuric acid in methanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general acid-catalyzed reactions involving epoxides where selectivity is a concern will have carbocation intermediates, but the are not required to do such.  But this potential is most likely to scare off a potential synth-jock simply because it is effectively a lower throughput at whatever step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pedagogical value of teaching epoxides at the undergraduate level is 4-fold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It allows discussion of the physical chemistry of strained&#xA;ring-systems &lt;/li&gt;&#xA;&lt;li&gt;Synthetic Organic Chemistry is the business of making and breaking carbon-carbon bonds.  Via Grignard reagent, epoxides allow a two-carbon extension of a carbon skeleton, which is fairly simple to understand and at the point where the topic is introduced, often the first example. &lt;/li&gt;&#xA;&lt;li&gt;It can serve as an introduction to polymer chemistry. &lt;/li&gt;&#xA;&lt;li&gt;In the example you've presented, it forces students to evaluate the relative strengths of competing phenomena, which is the sine qua non of organic chemistry as an educational tool.  In this case to explicitly address the question: Does the effect 2&lt;sup&gt;o&lt;/sup&gt; carbocation stabilization for gain outweigh the steric hindrance?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the instructor or students are obsessed with the 'right' answer rather than an intelligently described answer (which may also be right), epoxides are less useful as a teaching topic.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="402" LastEditDate="2017-04-08T22:13:11.010" LastActivityDate="2017-04-08T22:13:11.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4308" PostTypeId="2" ParentId="4289" CreationDate="2013-03-05T18:30:06.500" Score="5" Body="&lt;p&gt;In the following below, please interpret formal charges as marked below the atom.&#xA;$$\ce{O=N=O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O\equiv N-O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O-N \equiv O} $$&#xA;$$\ \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \ +\ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ +\ \ + $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triply charged species are valid resonance forms, but very likely are small contributors.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="275" LastEditDate="2013-03-06T00:43:47.417" LastActivityDate="2013-03-06T00:43:47.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4309" PostTypeId="2" ParentId="4279" CreationDate="2013-03-05T18:32:09.993" Score="-1" Body="&lt;p&gt;Regarding #2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything below the carbon row &lt;em&gt;can&lt;/em&gt; expand their octets.  The question makes little sense in any other context.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-03-05T18:32:09.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4310" PostTypeId="1" AcceptedAnswerId="4311" CreationDate="2013-03-05T21:13:18.877" Score="6" ViewCount="167219" Body="&lt;p&gt;I've already searched online some information about this equation, but everywhere I read, there's also sulfuric acid, which we did not use in the laboratory. We only used potassium permanganate and oxalic acid and water, there was no sulfuric acid. The purpose of the laboratory was to calculate experimentally the rate of the reaction at different concentrations, but in order to do so, I need the balanced equation. Can anyone help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the oxalic acid was dihydrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone is wondering, one of the experiments was as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we took $1.5~\pu{mL}$ of $\ce{H2C2O4}$ and added $1.5~\pu{mL}$ of $\ce{H2O}$, and finally, we added $1~\pu{mL}$ of $\ce{KMnO4}$ and then we shook the test tube until the colour changed from purple to ochre. Because from what I read, they said the $\ce{H2SO4}$ was necessary so the reaction takes place, but in the experiment, the reaction did occur and we didn't use it.&lt;/p&gt;&#xA;" OwnerUserId="599" LastEditorUserId="4945" LastEditDate="2017-01-14T10:08:44.520" LastActivityDate="2020-03-27T18:39:07.817" Title="What is the reaction between oxalic acid and potassium permanganate?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4311" PostTypeId="2" ParentId="4310" CreationDate="2013-03-05T21:39:31.047" Score="9" Body="&lt;p&gt;We had a titration of oxalic acid vs. potassium permanganate for an experiment in which we used concentrated $\ce{H2SO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the reaction requires an acidic medium i.e. $\ce{H+}$ is involved as a reactant... I suppose the $\ce{H+}$ released by the oxalic acid would be enough since you are studying the rate of the reaction though I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced ionic equation is:&#xA;$\ce{2MnO4- + 5H2C2O4 + 6H+ -&amp;gt; 2 Mn^2+ + 10CO2 + 8H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, $\ce{H+}$ is involved as a reactant and that is why you might have read that conc.$\ce{H2SO4}$ is necessary.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="402" LastEditDate="2017-04-02T20:29:39.173" LastActivityDate="2017-04-02T20:29:39.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4312" PostTypeId="1" CreationDate="2013-03-05T22:52:14.873" Score="3" ViewCount="3123" Body="&lt;p&gt;Why is $\ce{NaCl}$ more soluble in water than $\ce{NaOH}$? Well, this is a question from Chemical Equilibrium and the question hints for explanation in terms of Le Chatelier's principle.&lt;/p&gt;&#xA;" OwnerUserId="1273" LastEditorUserId="69830" LastEditDate="2021-05-09T06:47:09.430" LastActivityDate="2021-05-09T06:47:09.430" Title="Solubility of compounds in water" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4313" PostTypeId="1" CreationDate="2013-03-06T10:11:35.373" Score="7" ViewCount="16591" Body="&lt;p&gt;A very &lt;a href=&quot;http://www.chemistry.ccsu.edu/glagovich/teaching/316/qualanal/tests/ceric.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple test&lt;/a&gt; for the presence of alcohols in the lab involves adding ceric(IV) ammonium nitrate solution which gives a pinkish red colour in the presence of alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I looked up the structure of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ceric_ammonium_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;this compound&lt;/a&gt; it turns out to be quite interesting. So a couple of questions about lanthanides came up in my mind (a Google search only comes up with very complicated research papers):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Wouldn't this compound be highly unstable since there 12 donor atoms (forming an icosahedral structure) surrounding it (but it is quite stable)?&lt;/li&gt;&#xA;&lt;li&gt;Is the formation of complexes in lanthanides and actinides similar to that of transition metals? Do $\ce{f}$ orbitals undergo crystal field splitting just like $\ce{d}$ orbitals?&lt;/li&gt;&#xA;&lt;li&gt;Are higher coordination numbers like 12 more common (than octahedral and tetrahedral structures) among lanthanides and actinides because they have more orbitals for bonding, or is this compound an exception?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="41328" LastEditDate="2017-10-06T20:13:05.233" LastActivityDate="2019-12-01T07:17:02.287" Title="Complexes formed by lanthanides and actinides" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4315" PostTypeId="1" CreationDate="2013-03-06T18:18:31.653" Score="6" ViewCount="107447" Body="&lt;p&gt;During the formation of a molecule of water ($\ce{H2O}$), what kind of bond occurs between the elements? I'm studying this subject on my own and from some sources I saw that there is a covalent bond and others showed an hydrogen bond, responsible for the non-evaporation of water at $-90\ ^{\circ} \rm{C}$. Can you guys explain to me exactly how this bonds behave? Both of them occurs to the molecules of water?&lt;/p&gt;&#xA;" OwnerUserId="1325" LastEditorUserId="9961" LastEditDate="2015-10-02T16:51:46.993" LastActivityDate="2019-01-15T19:23:31.860" Title="Covalent Bond/Hydrogen Bond" Tags="&lt;bond&gt;&lt;water&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4316" PostTypeId="2" ParentId="4315" CreationDate="2013-03-06T19:20:17.270" Score="1" Body="&lt;p&gt;Water is a polar covalent molecule. Take a look at their electronegativity values, Hydrogen is 2.20 and Oxygen is 3.44. This is a difference of 1.24 meaning that it is Polar Covalent. A Covalent bond is present when there is a difference 1.7 +/- 0.5. Also there is Hydrogen Bonding in water also, this is because the hyrdogens will have a slightly psotove charge and the oxygen will have a slightly negative charge, which causes hydrogens in other water molecules to become attracted to the oxygens. I hope this helps&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="1326" LastEditDate="2013-03-07T18:18:51.160" LastActivityDate="2013-03-07T18:18:51.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4317" PostTypeId="2" ParentId="4312" CreationDate="2013-03-06T19:35:47.913" Score="-4" Body="&lt;p&gt;The Water molecules move around the $\ce{Na}$ and $\ce{Cl}$ and break the salt apart into ions. The polarity of the water molecule allows this by the more negative oxygen being attracted to the positive Na ions and the positive hydrogen being attracted to the negative chlorine ions. In the case of NaOH, I am assume that bond between $\ce{Na+}$ and $\ce{OH-}$ is stronger than the bond in $\ce{NaCl}$. Also I don't see a case for the use of an equilibrium in this case because $\ce{NaOH}$ almost completely dissociates into its ions $\ce{Na+~~ OH-}$. &lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="4231" LastEditDate="2014-11-01T15:55:43.947" LastActivityDate="2014-11-01T15:55:43.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4318" PostTypeId="1" CreationDate="2013-03-06T20:35:49.097" Score="11" ViewCount="422" Body="&lt;p&gt;I'm looking for the dipole transition matrix elements for electronic transitions between the ground and excited states of the carbon dioxide cation, $\text{CO}_2^+$, that is, objects like&#xA;$$\langle\text{CO}_2^+ \,\tilde{\text{A}}\,\Pi_\text{u}|\mathbf{r}|\text{CO}_2^+ \,\tilde{\text{X}}\,\Pi_\text{g}\rangle.$$&#xA;I feel this is a standard enough quantity that it ought to be easy to find in some appropriate database of chemical data. However, I've searched for quite some time and I can't even find something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should I be looking?&lt;/p&gt;&#xA;" OwnerUserId="1327" LastActivityDate="2013-06-09T03:37:24.497" Title="Database for molecular dipole transition moments?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4319" PostTypeId="2" ParentId="4312" CreationDate="2013-03-06T23:14:46.907" Score="-4" Body="&lt;p&gt;Pure water, pH 7, has a concentration of roughly .1 ppm (100 ppb) of hydrogen cations (protons), in the form of hydronium ($\ce{H3O+}$). This is a natural state of equilibrium; the solution wants neither more nor fewer protons (if it did, it would take them or give them up, and so it would not be &quot;neutral&quot;; that's a bit circular, but it's a fundamental property of water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium hydroxide, $\ce{NaOH}$, is a strong base; it is a &quot;proton acceptor&quot;, wanting desperately to share one of its lone pairs with a hydrogen cation. The sodium, which gave up the valence electron to the hydroxide, doesn't want it back, so the hydroxide effectively owns the electron. But, hydroxide isn't extremely electronegative, like say, the halogens; it doesn't want to own the electron outright and would rather share. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen atoms, for their part, either want two electrons or none, and they aren't picky about which; again, this leads to a situation in which hydrogen would rather share. Hydrogen is the textbook definition of a neutral element in terms of electronegativity, and because of this it will readily form either ionic compounds with elements at the extremes, or covalent compounds with elements closer to the center. This makes combinations of hydrogen and halogens very acidic, because the hydrogen wants to get away from the electron-greedy halogen and form a more stable, equal covalent bond; the compound becomes a &quot;proton donor&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduce $\ce{NaOH}$ to water, and the hydroxide ion will display its affinity for the protons of the hydronium. The water, however, was in equilibrium, and so the removal of the cations to form new water molecules, and the resulting release of un-paired sodium ions, disturbs this equilibrium. According to Le Chatelier's principle, this disruption is resisted with an opposing reaction (in this case the re-creation of sodium hydroxide from the sodium deprotonating water to form more hydroxide, releasing or re-releasing hydrogen cations). Because of this love-hate relationship between all of the component ions, the dissassociation of $\ce{NaOH}$ into water is resisted, but because of the extremely high solubility of Group I elements including sodium, it does eventually happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium chloride, however, is a salt. It is quite literally the byproduct of an acid and base getting together to form water. The chlorine and sodium ions, abandoned by their respective proton and hydroxide when those two formed the much more stable polar covalent water molecule, form a marriage of convenience between them. The sodium, lacking the electron that the hydroxide now shares with the proton in the water molecule, is attracted to the chlorine, which had as an acid stripped a hydrogen of its electron to form a proton. The chlorine is very happy to &quot;own&quot; the extra electron that produces this charge difference, the sodium is equally happy to be rid of it, and as such, there is no &quot;shock&quot; to the system where the introduced components try to do something that disrupts the equilibrium of the water. Thus, there is no resistance to the natural tendency of the polar water molecules to dissociate the sodium and chloride ions (which happens readily, because the sodium and chloride are such opposites in terms of electron affinity that only their charge difference holds them together, and so any other charge difference, even a partial one, can pull them apart).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="4231" LastEditDate="2014-11-01T15:57:36.670" LastActivityDate="2014-11-01T15:57:36.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4320" PostTypeId="1" AcceptedAnswerId="4323" CreationDate="2013-03-06T23:31:18.190" Score="5" ViewCount="138" Body="&lt;p&gt;As a surprise gift to a friend I would like to discover a new theoretical mineral and have it named after them. I have some understanding that not all minerals have been discovered, and some minerals exist in theory but have not been found on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a computer search that looks for new minerals? Would it be difficult to write one, if one does not exist?&lt;/p&gt;&#xA;" OwnerUserId="1328" LastEditorUserId="84" LastEditDate="2013-03-07T02:36:30.457" LastActivityDate="2013-03-07T06:44:28.467" Title="Are there computer searches for new theoretical minerals in a similar vein to Folding@Home or SETI@Home?" Tags="&lt;inorganic-chemistry&gt;&lt;computational-chemistry&gt;&lt;nomenclature&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4321" PostTypeId="2" ParentId="4315" CreationDate="2013-03-07T01:38:16.030" Score="1" Body="&lt;p&gt;&#xA;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In H2O molecule, two water molecules are bonded by a Hydrogen bond but the bond between two H - O bonds within a water molecule are covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer this diagram : &lt;a href=&quot;http://goo.gl/84w9T&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/84w9T&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt; The dotted lines represent a hydrogen bond and the solid lines represent a covalent bond.&lt;/p&gt;&#xA;" OwnerUserId="1331" LastEditorUserId="1331" LastEditDate="2013-03-07T01:54:08.323" LastActivityDate="2013-03-07T01:54:08.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4322" PostTypeId="1" CreationDate="2013-03-07T05:54:25.800" Score="5" ViewCount="1727" Body="&lt;p&gt;I'm looking for a table showing which size of bar to use for a given volume to stir.&#xA;Or a formula if there is one.&lt;/p&gt;&#xA;" OwnerUserId="686" LastEditorDisplayName="user7951" LastEditDate="2015-08-17T17:09:18.143" LastActivityDate="2015-08-17T17:09:18.143" Title="Stir bar size and volume to stir" Tags="&lt;equipment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4323" PostTypeId="2" ParentId="4320" CreationDate="2013-03-07T06:44:28.467" Score="5" Body="&lt;p&gt;By Ross Snider's request, I have moved my comment into an answer - whilst it does not address computational searches for minerals, it touches on the complexities of actually getting a mineral officially named:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Mineralogical Association Commission on New Minerals, Nomenclature and Classification seems to &lt;a href=&quot;http://pubsites.uws.edu.au/ima-cnmnc/cnmmn98.pdf&quot; rel=&quot;noreferrer&quot;&gt;require a physical specimen for consideration for official naming of a new mineral&lt;/a&gt;. Under current rules, anthropogenic minerals are also excluded, so if you came up with a hypothetical mineral it seems you would have to also find a naturally occurring specimen to get it named.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add to this, I am not aware of any software for systematically searching for plausible minerals, however computational packages for the simulation of crystalline structures certainly exist. Determining the electronic structure of a crystal - which is what is really required to make conclusions about geometry, stability and so on - generally involves the use of plane wave QM codes such as &lt;a href=&quot;http://www.vasp.at/index.php/about-vasp/59-about-vasp&quot; rel=&quot;noreferrer&quot;&gt;VASP&lt;/a&gt;, &lt;a href=&quot;http://www.castep.org/&quot; rel=&quot;noreferrer&quot;&gt;CASTEP&lt;/a&gt; or &lt;a href=&quot;http://www.abinit.org/&quot; rel=&quot;noreferrer&quot;&gt;ABINIT&lt;/a&gt;, the latter of which is free. The feasibility of mineralogical calculations using these methods is strongly dependent on how crystalline the mineral is and how ideal it is (from a perspective of not being heavily defected or substituted) as they work within the constraints of a periodic unit cell or supercell.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-03-07T06:44:28.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4324" PostTypeId="1" CreationDate="2013-03-07T16:31:32.237" Score="4" ViewCount="204" Body="&lt;p&gt;I'm writing a script and would like to use a &quot;Gödel numbering&quot; to search for specific reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any used procedure to assign an symbol (integer, possibly) to a chemical reaction? E.g. &#xA;$$CH_4+4O_2=CO_2+2H_2O\ \ \ \Longrightarrow \ \ \  121231412$$&lt;/p&gt;&#xA;" OwnerUserId="145" LastActivityDate="2013-03-08T15:04:21.253" Title="Unique calling code for a chemical reaction" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4325" PostTypeId="1" CreationDate="2013-03-07T18:55:15.813" Score="3" ViewCount="1232" Body="&lt;p&gt;Is there any kind of program that is able to calculate the end result of a specific chemical reaction? Say I wanted to know what happens when you mix $\ce{HClO + NaOH}$, is there anything that could give me the answer to any question like this? Wolfram alpha works pretty good, but it doesn't give me an answer if it isn't a balanced equation, and I don't know what that means. $\ce{HClO}$ is an acid and $\ce{NaOH}$ is a base, so there has to be some reaction, balanced or not.&lt;/p&gt;&#xA;" OwnerUserId="412" LastEditorUserId="4231" LastEditDate="2015-02-26T17:49:08.227" LastActivityDate="2015-02-26T17:49:08.227" Title="Chemical reaction calculator" Tags="&lt;everyday-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4326" PostTypeId="2" ParentId="4270" CreationDate="2013-03-08T01:51:49.467" Score="17" Body="&lt;p&gt;Well first off, pure ethanol is hygroscopic; it attracts water, to the point that it will pull it out of the air. Ethanol and gasoline will mix, but ethanol, gasoline and water will not; the ethanol-water mixture will come out of solution and settle on the bottom of your tank. Add a little oxygen to the mix, and you get rust. However, the more common side effect of this is more immediate; turn the car on, and the fuel pump will draw the water from the bottom of the tank into the engine, where it will promptly kill it, and require a costly dry-out process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down here in the south, where it's hot and humid most of the year, you hear a news story or two about this every year, usually when a gas station didn't properly purge its storage tank on a regular basis. The regular addition of thousands of gallons of fresh ethanol fuel &quot;recharges&quot; this hygroscopic process, causing a rather large pool of water to form at the bottom of the tank which the station is supposed to drain regularly. If they don't, and you fill up when the store's tank is just a &lt;em&gt;little&lt;/em&gt; too low, or just a &lt;em&gt;little&lt;/em&gt; too soon after the tanker's refilled it, the pump will draw a water-gas emulsion, or even straight water, into your tank, and the water will settle to the bottom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual fix for long-term storage of ethanol fuel is to add a stabilizer, which will contain a high concentration of isopropanol. The main thing that will do is form an azeotropic mixture with water, meaning that its components will evaporate at the same rate, so when the isopropanol evaporates, it takes the water with it. This same mixture, in sufficient concentration, will also burn, so while it's not the greatest thing for your engine it can at least be flushed out the normal way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, gasoline, as you probably well know, isn't a pure substance; it was originally what was left over after the refinery had extracted the &quot;useful&quot; heavier compounds (kerosene, diesel fuel, lubricating oil, paraffin waxes) out of crude oil by fractional distillation. Nowadays it's much more carefully purpose-made by thermal decomposition of heavier alkanes, but it's still an amalgam of relatively lightweight carbon compounds (usually between 6-12 carbons, with the dividing line for &quot;octane rating&quot; drawn between heptane and octane), plus additives and pollutants. It contains three components of interest; sulfur (a naturally occurring pollutant and difficult to remove completely; &quot;summer blend&quot; gas in the U.S. is more expensive because they have to get rid of more of it to prevent smog and acid rain), oxygen (modern fuels are oxygenated to make them more clean-burning), and ethanol (added to make the fuel burn cleaner and be more sustainable to produce than neat gasoline).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of water (pulled in by the ethanol) and oxygen, sulfur naturally forms sulfuric acid by a variation of the &quot;contact process&quot;; it oxidizes easily to sulfur dioxide, then more slowly to sulfur trioxide, which when dissolved in water becomes sulfuric acid. This reaction, specifically the formation of sulfur trioxide, is sped up in industrial processes by heat and by contact with catalyst metals such as vanadium oxide (hence &quot;contact process&quot;), but it will also happen on its own without any special treatment (i.e. acid rain). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sulfuric acid will do a lot of corrosion damage, reacting with the tank's sheet metal producing hydrates of iron sulfate. When burned, those sulfates decompose into iron oxides and sulfur dioxide again, which do more damage to your exhaust system on the way out (rust is a catalyst for more rust, and we've discussed sulfur dioxide's contribution to the party in detail). In addition, sulfuric acid is even more hygroscopic than ethanol; at concentrations in water as low as 10%, it'll attract more water. Lastly, acids and alcohols form esters, in our case ethyl sulfate, and esters and alcohols form ethers, here diethyl ether (with the sulfuric acid and water as a byproduct). These are reversible and ultimately circular reactions, which will form an equilibrium of roughly equal concentrations of the intermediates. A dehydration reaction to produce ethylene is also possible in very high acid-to-water concentrations of sulfuric acid (possible when filling up in the winter, when it's dry and the gas has more sulfur). Ethylene will form a number of intermediate products with just about anything in that gas tank, some better for your engine than others (ethylbenzene is an antiknock compound; methyl ethyl ketone is a solvent and a pollutant byproduct of combustion engines).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-10T15:24:50.540" LastActivityDate="2013-05-10T15:24:50.540" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4327" PostTypeId="2" ParentId="4324" CreationDate="2013-03-08T02:42:57.563" Score="3" Body="&lt;p&gt;I don't know about Godel numbering specifically (though this is interesting, my being a math guy at heart), but there are existing textual systems for encoding molecules (not reactions that I know of).  From these, you may be able to encode reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples are &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Chemical_Identifier&quot; rel=&quot;nofollow&quot;&gt;InChI&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplified_molecular-input_line-entry_system&quot; rel=&quot;nofollow&quot;&gt;SMILES&lt;/a&gt;.  These are Wikipedia links.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-03-08T02:42:57.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4329" PostTypeId="2" ParentId="4324" CreationDate="2013-03-08T12:02:49.250" Score="3" Body="&lt;p&gt;My suggestion is to dabble in combining SMILES representations of the reactants and products and a stoichiometric matrix in a consistent fashion and then using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hashing_function&quot; rel=&quot;nofollow&quot;&gt;hashing function&lt;/a&gt; (Now that I think of it, I guess Godel numbering is an example of hashing) to assign the reaction a unique identifier. You need to use a representation of the reactants and products that unambiguously describes the molecular graph (including stereochemistry) so that stereo- and/or constitutional isomers cannot give rise to the same identifiers, and so that different orderings of atoms in a string describing identical molecules do not give rise to differing identifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist a number of reaction ontologies that can be used to categorise and describe chemical reactions. &lt;a href=&quot;http://www.genome.jp/kegg/reaction/&quot; rel=&quot;nofollow&quot;&gt;KEGG RCLASS and RPAIR&lt;/a&gt; is one such example which gives rise to numerical identifiers, however I'm not totally clear on how it works and it appears to be tailored to metabolomics. It is however a good example of the kind of thing people are doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ci0502698&quot; rel=&quot;nofollow&quot;&gt;CMLReact format&lt;/a&gt; is an XML format for representing chemical reactions that may be suitable for generating short identifiers from through hashing, however being XML it has enough cruft in it that chemically insignificant changes to a CMLReact file may give rise to differing derived identifiers. Open documentation of the format is sadly not terribly forthcoming on the internet however there is an &lt;a href=&quot;http://openbabel.org/wiki/CML_React&quot; rel=&quot;nofollow&quot;&gt;library for reading and writing CMLReact in OpenBabel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-03-08T12:02:49.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4331" PostTypeId="1" CreationDate="2013-03-08T13:01:55.473" Score="7" ViewCount="387" Body="&lt;p&gt;I have two identical graphite strips of dimensions 3.5 cm by 1 cm. Both are 3/4 submerged in aqueous $\ce{NaCl}$ solution (0.34 mol/dm$^3$, 2% wt) separated by a distance of 7 cm. I have attached crocodile clips to each strip and a voltmeter is used to record the potential difference between the two graphite strips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. There is a potential difference that ranges between 1 mV and 100 mV; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. This potential difference can change quite dramatically within that range over time periods of an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any chemical or physical reason for these potential differences?&lt;/p&gt;&#xA;" OwnerUserId="1337" LastEditorUserId="865" LastEditDate="2015-08-02T19:08:45.980" LastActivityDate="2015-08-03T10:43:16.813" Title="Varying potential diffrence between two graphite strips in NaCl solution" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;surface-chemistry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4332" PostTypeId="1" CreationDate="2013-03-08T13:32:44.263" Score="12" ViewCount="30246" Body="&lt;p&gt;Cu has an anomalous electron configuration. $\ce{Cu ~=~ 1s^2~2s^2~2p^6~3s^2~3p^6~4s^1~3d^{10}}$, it does not follow the usual pattern. In this case, the 3d subshell is filled before the 4s, which usually happens in the reverse order ($\ce{1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^{9}})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, can you tell by an element's position on the periodic table (Group #, Row #, Block, etc.) that it will have an anomalous electron configuration? Do I have to memorize which elements have this property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am noting that Cu is in the 1B Group of Transition metals which implies that it has one valence electron, this is consistent with its configuration. will this be true for all cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: $\ce{Fe ~=~ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$ and is a transition metal as well (Group 8). This implies it must have 8 valence electrons. But following the definition of valence electrons (The number of electrons in outermost shell), $n = 4$ is the outermost shell and I get a conundrum. Obviously if I add $\ce{4s + 3d}$ electrons everything is good. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorUserId="5026" LastEditDate="2018-03-08T10:28:31.677" LastActivityDate="2018-03-08T10:29:16.803" Title="Anomalous electronic configurations" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4334" PostTypeId="2" ParentId="4325" CreationDate="2013-03-08T14:56:03.977" Score="4" Body="&lt;p&gt;This is a fundamentally very difficult question. Your example of of $\ce{HClO + NaOH}$ hides several assumptions, e.g., time frame and aqueous solution, which really means some number of $\ce{HClO}$, some number of $\ce{NaOH}$, and presumably a huge number of $\ce{H2O}$. When mixing all three components a wide variety of reactions take place, e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Disproportionation&quot; rel=&quot;nofollow&quot;&gt;disproportionation&lt;/a&gt; of water ($\ce{H2O + H2O -&amp;gt; H3O^{+} + OH^{-}}$), solvation of $\ce{NaOH}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a full exploration of all possible reactions and their barriers have to be assessed in order to make a prediction. For instance, $\ce{HClO + H2O -&amp;gt; HCl + O2H2}$ is a possible reaction, or $\ce{NaOH + 2 HClO -&amp;gt; NaCl + H2O2}$ even in the absence of extra water. The number of possible recombinations of the input materials generally increases exponentially with the number of atoms (e.g., the number of possible isomers of an n-alkane).&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="4231" LastEditDate="2015-02-26T17:49:05.967" LastActivityDate="2015-02-26T17:49:05.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4335" PostTypeId="2" ParentId="4324" CreationDate="2013-03-08T15:04:21.253" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.daylight.com/dayhtml/doc/theory/theory.smarts.html&quot; rel=&quot;nofollow&quot;&gt;SMARTS&lt;/a&gt; are a canonical way of representing chemical reactions (and other pattern-matching problems). This is a formal language so you could assign a &lt;a href=&quot;http://en.wikipedia.org/wiki/Goedel_number&quot; rel=&quot;nofollow&quot;&gt;Goedel numbering&lt;/a&gt; to the SMARTS. If you need a library to handle SMARTS/SMILES etc., take a look at &lt;a href=&quot;http://www.openbabel.org&quot; rel=&quot;nofollow&quot;&gt;OpenBabel&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="440" LastActivityDate="2013-03-08T15:04:21.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4336" PostTypeId="2" ParentId="4332" CreationDate="2013-03-08T15:33:14.620" Score="2" Body="&lt;p&gt;First of all the configuration of $\ce{Cu}$ you see has anomalous electron configuration because completely filled or half filled sub shells are more stable than any other configuration and you can see the configuration at last it is $\ce{4s^13d^{10}}$ and the basic configuration is $\ce{4s^23d^{10}}$ so you can see that $\ce{4s}$ is half filled and $\ce{3d}$ is completely filled. It is said this configuration is more stable due to symmetry. And it has no connection to their position in periodic table, and you only have to remember the concept and be careful while writing the configuration to notice this type of conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As much I know they both are to be taken as outer most shell because $\ce{4s}$ is a lower energy cell as compared to $\ce{3d}$ and it filled first. I don't know the details about this, I hope some one will give a better answer to the second part as I also wish to know about that.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="5026" LastEditDate="2018-03-08T10:29:16.803" LastActivityDate="2018-03-08T10:29:16.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4337" PostTypeId="2" ParentId="4332" CreationDate="2013-03-08T16:29:00.350" Score="7" Body="&lt;p&gt;Many related questions have been asked on this topic before. I suggest you read these previous posts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations&quot;&gt;Why do elements in columns 6 and 11 assume &amp;#39;abnormal&amp;#39; electron configurations?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2469/how-can-one-explain-niobiums-weird-electronic-configuration&quot;&gt;How can one explain Niobium’s weird electronic configuration?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aufbau Principle&lt;/a&gt; is a general rule of thumb to find the electronic configuration of an element but usually many exceptions arise in the d and f block elements due to electron-electron interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of valence electrons does get a little fuzzy when you consider $d$ block elements since the energies of the $\ce{3d}$ and $\ce{4s}$ subshells are quite similar so I think its better to avoid it. $\ce{d}$ block elements can show a staggeringly large number of oxidation states (for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese#Chemical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;manganese&lt;/a&gt; which shows almost all oxidation states from 0 to +7). This can't be explained using the concept of valence electrons and the octet rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe}$ itself can lose 2 electrons to form $\ce{Fe^{2+}}$ or 3 electrons to form $\ce{Fe^{3+}}$ so now again you can see that you can't explain this by saying that $\ce{Fe}$ has 8 valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other exceptional configurations: (there are quite a few):&#xA;niobium, platinum, gold, cerium, gadolinium, etc.  &lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-08T10:28:56.807" LastActivityDate="2018-03-08T10:28:56.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4338" PostTypeId="2" ParentId="4322" CreationDate="2013-03-08T17:55:27.643" Score="2" Body="&lt;p&gt;No chart I'm aware of.  However, an informal and incomplete survey indicates most chemists to this by trial and error.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For pipe-type stir bars in flat bottom flasks/beakers, and you want a 'fast' stir, generally you want one roughly two thirds the size of your magnet arm to minimum the motor moving ahead of the magnet spin and 'flinging' it around.  Obviously this is not possible if your flask is much smaller than the motor.  If your stir bar is too close to the size of your vessel, you have 'catching' issues. with the sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For egg-type stir bars in round bottom flasks, the concavity of the surface will help, but not eliminate 'flinging'.If you need a fast stir, you need a bigger stir bar.  Personally, I've never needed larger than 3/4&quot; because of my use preference, (also do not typically use roundbottoms larger than 1L).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, however, remember stir bars are for mild to moderate stirring.  If you need vigorous, you are better served with a mechanical stirrer.  Additionally, if you are using very large (5L+) flasks, mechanical stirring is best.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-03-08T17:55:27.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4339" PostTypeId="2" ParentId="4315" CreationDate="2013-03-08T21:50:46.397" Score="7" Body="&lt;p&gt;Covalent bonds are the bonds between the atoms within the same water molecule. Hydrogen bonds are the bonds between two water molecules. All molecules have covalent bonds, but only some molecules have hydrogen bonds. As an example, water has hydrogen bonds, but carbon dioxide does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the requirements for hydrogen bonding is that the molecule must be polar. Water molecules are polar because of two effects. The first is that oxygen is more electronegative than hydrogen, so each of the OH bonds will be polar. The electrons shared in each of those bonds will spend more time near the oxygen than near the hydrogen. The other effect is the geometry of the molecule: the oxygen atom has two free electron pairs in addition to the two bonds, so it assumes a tetrahedral arrangement. The $\ce{O-H}$ bonds are two corners of that tetrahedron, and the free electron pairs are the other two corners. This arrangement means that, if you only look at the atoms, the molecule has a bent geometry. Because of that, the polarity of the two bonds partially adds together. By contrast, $\ce{CO2}$ has a linear geometry, and the two polar bonds cancel each other out, leaving the molecule non-polar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the water molecule is polar, with a lot of excess positive charge around the hydrogen atoms, and a lot of negative charge on the side of the oxygen atom away from the hydrogen atoms. Hydrogen bond is the name given to the electrostatic interaction between the positive charge on a hydrogen atom and the negative charge on the oxygen atom of a neighboring molecule. The covalent bond is the electrostatic interaction between two atoms in the same molecule. Covalent bonds are much stronger than hydrogen bonds: the $\ce{O-H}$ has a strength of 467 kJ/mol, while the hydrogen bond is usually between 4 to 40 kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2013-03-08T21:50:46.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4340" PostTypeId="1" CreationDate="2013-03-08T22:00:39.743" Score="9" ViewCount="333" Body="&lt;p&gt;The electron affinity of a neutral oxygen atom is −142 kJ (it releases this energy). The electron affinity for the now double anion $\ce{O^2-}$ is 710 kJ (work must be done on the atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why the large jump in electron affinity? We are taught that the electrons do not interact much with each other in the way of shielding. I think this to be incorrect, because otherwise, the effective nuclear charge on all of the valence electrons should be about the same, including the new ones added to ionize the atom. This would make the electron affinity about the same for each electron. Plus, doesn’t the oxygen atom want to complete its shell? So why is it putting up a fight?&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorUserId="194" LastEditDate="2017-10-29T19:47:54.833" LastActivityDate="2017-10-29T20:53:03.453" Title="Why does oxygen not like to be a double anion?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;&lt;electronic-configuration&gt;&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4341" PostTypeId="2" ParentId="4340" CreationDate="2013-03-08T22:08:22.657" Score="8" Body="&lt;p&gt;The addition of an electron to any negatively charged ion will need to overcome the repulsive force of the negative ion. This is generally less favorable than adding an electron to a neutral atom, and the electron affinity will be positive. Even though the additional electron will complete the octet, as in the case of an  ion where the addition of electron will complete the  subshell, the electron affinity is a positive value.&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorDisplayName="user15489" LastEditDate="2015-04-29T18:28:10.347" LastActivityDate="2015-04-29T18:28:10.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4342" PostTypeId="2" ParentId="4312" CreationDate="2013-03-09T04:36:52.917" Score="6" Body="&lt;p&gt;It is not. $\ce{NaOH}$ is more soluble ($&amp;gt;100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($&amp;lt;30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="7475" LastEditDate="2015-10-30T18:40:58.830" LastActivityDate="2015-10-30T18:40:58.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4343" PostTypeId="2" ParentId="4313" CreationDate="2013-03-09T04:46:50.433" Score="3" Body="&lt;p&gt;1) No, because it is not donor-acceptor relationship, but electrostatic one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) No, f-orbitals of lanthanide and actinides are unavailable for valence interactions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) No, they are common because amount of ligands in the case is determined by geometry, and not electronic structure. Lanthanide cations are big.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-03-09T04:46:50.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4344" PostTypeId="1" CreationDate="2013-03-09T06:41:56.593" Score="8" ViewCount="14185" Body="&lt;p&gt;$\ce{CaF2}$ is insoluble in water, but $\ce{CaCl2}$, $\ce{CaBr2}$, $\ce{CaI2}$ are soluble. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="22" LastEditDate="2013-03-09T15:40:56.720" LastActivityDate="2018-10-16T13:26:35.357" Title="Solubility of Calcium Fluoride in water" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4345" PostTypeId="1" AcceptedAnswerId="4352" CreationDate="2013-03-09T06:57:49.347" Score="4" ViewCount="8641" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suggest an explanation for the coordination number in the $\ce{CsCl}$ lattice being different from those in $\ce{NaCl}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standard answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Caesium ion is larger than sodium ion, hence it can be surrounded by greater number of chloride ions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The question didn’t specify whether it is asking for the coordination number of $\ce{Cs+}$ or $\ce{Cl-}$... Is it because the coordination numbers of both ions are the same? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The given answer sounds like each $\ce{Cl-}$ ion will be surrounded by less $\ce{Cs+}$ ions, how is it possible that the ratio of $\ce{Cs:Cl}$ is 1:1 (as from its empirical formula)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1340" LastEditorUserId="4945" LastEditDate="2015-12-14T08:26:10.227" LastActivityDate="2015-12-14T08:26:10.227" Title="How to rationalise the coordination number of CsCl versus NaCl?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;crystal-structure&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4346" PostTypeId="1" CreationDate="2013-03-09T08:29:08.970" Score="6" ViewCount="1336" Body="&lt;p&gt;A couple months ago, one of my room mates accidentally used a cup of salt to make &lt;a href=&quot;http://allrecipes.com.au/recipe/9877/the-perfect-apple-crumble.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;apple crumble&lt;/a&gt;. She asked if there was any way to separate it, probably because I had recently reduced the acidity of a tomato sauce by adding baking soda. I said &amp;quot;probably not&amp;quot;, but ever since then I have wondered if there is a way to get the salt out.&lt;/p&gt;&#xA;&lt;p&gt;At this point, I would settle for a difficult method that requires a lab and ruins the dish, as long as it is still harmless to eat and retains most of it's nutritional value.&lt;/p&gt;&#xA;&lt;h3&gt;Here are some of the ideas I have thought of:&lt;/h3&gt;&#xA;&lt;p&gt;First, scrape off the oat layer, because it is the only hard solid and the salt wasn't mixed with it.&lt;/p&gt;&#xA;&lt;p&gt;Seperate it by mass:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The apple crumble could be puréed and spun in a centrifuge to separate the salty water, which could be poured off and replaced with sugar water.&lt;/li&gt;&#xA;&lt;li&gt;It could be atomized and sprayed sideways into a cold vacuum chamber, and then if the salt fell in a consistent radius around the nozzle, we could scrape up the salt, mix the rest back together and eat it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Alter the the properties of the salt:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a way to encourage the salt to form large crystals, even when it is a relatively minor component of a complicated mixture? If so, those crystals could be filtered out of puréed apple crumble with a coarse filter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Crazy biochemistry:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there proteins or enzymes that could break the salt apart and bind the sodium to something oily that would float up by itself, leaving the chlorine to evaporate? I saw a documentary about bull sharks which mentioned that they have a gland that can absorb salt against the salinity gradient, so that the shark can expel it's salt and move up into fresh water.&lt;/li&gt;&#xA;&lt;li&gt;Are there bacteria that require lots of salt to survive, and by using it, convert it into something that doesn't taste salty, and wouldn't release tons of sodium in my stomach?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1341" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-26T12:33:08.820" Title="Is it possible to separate salt from food, or at least convert it into something that doesn't taste salty?" Tags="&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;&lt;food-chemistry&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4347" PostTypeId="2" ParentId="4346" CreationDate="2013-03-09T08:58:16.033" Score="7" Body="&lt;p&gt;Even if there is a method to extract salt from food, specially from apple crumble, it’ll probably ruin it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way I can think of doing this is dissolving the apple crumble in water and blending it. Filter it and then boil the filtrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that every solid ingredient but salt is organic, when all the water boils off you should be left with salt and burnt carbohydrates, since NaCl boiling point is way above 1000 °C and a stove will reach 400 °C tops. Now, if you ever burnt sugar you know the caramel sticks to the pan, so if you’re lucky just add water again to separate the salt crystals from the burnt carbs and transfer it to a different recipient. Boil the water to get the salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can add the burnt carbs back to the apple crumble but I don’t think it’ll taste like apple crumble again.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T12:33:08.820" LastActivityDate="2015-04-26T12:33:08.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4348" PostTypeId="2" ParentId="4313" CreationDate="2013-03-09T09:46:16.457" Score="7" Body="&lt;p&gt;After a lot of digging around, I finally found something about this in J.D. Lee's &lt;em&gt;Concise Inorganic Chemistry.&lt;/em&gt; I thought I'd post whatever I found over here anyway since it's very interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer to question 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer to question 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher coordination numbers are apparently extremely common among f-block elements (strange right?). &#xA;Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;7 (Capped trigonal prismatic)&lt;/strong&gt; &lt;a href=&quot;http://www.chemicalbook.com/ProdSupplierGWCB3344275_EN.htm&quot; rel=&quot;nofollow&quot;&gt;yttrium acetylacetonate hydrate&lt;/a&gt; (Yttrium seems to be grouped with the f-block elements due to similar properties) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;8 (Square antiprismatic and Dodecahedral)&lt;/strong&gt; cerium acetylacetonate and holmium tropolonate&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;9 (Tri-capped trigonal prismatic)&lt;/strong&gt; nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;10 (very complex) and 12 (icosahedral)&lt;/strong&gt; are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="7475" LastEditDate="2016-10-15T12:29:19.590" LastActivityDate="2016-10-15T12:29:19.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4349" PostTypeId="1" AcceptedAnswerId="4350" CreationDate="2013-03-09T15:08:39.593" Score="0" ViewCount="66" Body="&lt;p&gt;I don't think a combination also can be called a &quot;state&quot; i.e. that the chemical state of vapour or solid would be just a combination. When I read Wikipedia about &quot;chemical state&quot; it refers to just a combination. But AFAIK, an entity that has just combination does not have a state. So how do we know what is a state and what is just a combination?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastEditorUserId="22" LastEditDate="2013-03-09T15:22:49.677" LastActivityDate="2013-03-09T15:35:14.357" Title="What is the difference between a chemical state and just a combination?" Tags="&lt;terminology&gt;&lt;solid-state-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4350" PostTypeId="2" ParentId="4349" CreationDate="2013-03-09T15:35:14.357" Score="2" Body="&lt;p&gt;You seem to be confused with &quot;state of matter&quot;/&quot;phase&quot; and &quot;chemical state&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical state encompasses all the macroscopic parameters necessary to fully define a given system. For example, &quot;pure $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$&quot; is an example of chemical state. A more complicated example would be something like: &quot;10 moles of liquid $\ce{O2}$ at $-100^{\mathrm o}\:\mathrm{C}$, mixed with 1 mole of gaseous $\ce{O2}$ at $1\:\mathrm{atm}$ and $100^{\mathrm o}\:\mathrm{C}$ and 2 moles of gaseous $\ce{N2}$ at $1\:\mathrm{atm}$ and $200^{\mathrm o}\:\mathrm{C}$&quot; (Yes, this isn't at equilibrium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the phase or sate of matter only applies to a homogeneous system. A gas is in gas phase, but a system consisting of a gas and a solid doesn't have a phase/state of matter. Note that if a gas/solid is dissolved in liquid, then the phase is liquid since solutions are homogeneous.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-03-09T15:35:14.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4351" PostTypeId="2" ParentId="4331" CreationDate="2013-03-09T15:35:31.607" Score="1" Body="&lt;p&gt;What type of voltmeter are you using? to verify the answer to my question, you need to determine 1) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Accuracy_and_precision&quot; rel=&quot;nofollow&quot;&gt;precision&lt;/a&gt; and 2) the &lt;a href=&quot;http://en.wikipedia.org/wiki/Detection_limit&quot; rel=&quot;nofollow&quot;&gt;detection limit&lt;/a&gt; of the instrument .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision is a measure of the error in a typical measurement. The smaller the standard deviation in repeat measurements, the higher the precision. To get a guess at the precision, prepare a system that should generate a known potential difference in the same range as your current measurements (like 100 mV). If the variation in that value is similar to what you are seeing, then you are recording the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noise_%28electronics%29&quot; rel=&quot;nofollow&quot;&gt;noise&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detection limit is the lowest value your instrument can reliably measure. Generally, you cannot reliably measure values smaller than the twice the precision of your instrument. If you determine a standard deviation of 50 mV when determining precision, then 100 mV is your the detection limit. If this is true, you are recording the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noise_%28electronics%29&quot; rel=&quot;nofollow&quot;&gt;noise&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision and detection limit of an instrument depend on quality (price), but also on purpose. A voltmeter designed to be used by an electrician will have high precision in the range 1-200 V, but low precision below 1 V (because that range is not as important). A sensitive voltmeter used for electrochemical measurements may have high precision between 1 uV and 1 V, and behave miserably outside that range.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-09T15:35:31.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4352" PostTypeId="2" ParentId="4345" CreationDate="2013-03-09T16:09:12.020" Score="6" Body="&lt;p&gt;Since both compounds are of the form $\ce{AB}$ (and not $\ce{AB2}$ or $\ce{A2B3}$/etc), the coordination numbers will be equal for the ions. Similarly, for $\ce{AB2}$, the coordination number for $\ce{A}$ will be twice that of $\ce{B}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that each ion is connected to more ions of the opposite type in $\ce{CsCl}$. Each $\ce{Cs+}$ ion is connected directly to 8 $\ce{Cl-}$ ions, &lt;em&gt;and vice versa&lt;/em&gt;. The net effect is that their ratio stays the same, they just have more connections.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Look at it this way: There are four men and four women in a room. First, each man holds hands with only one woman and vice versa. Now, we tell each man to hold the hands of two women, and we tell each women to do the opposite. Now, we shift it up to 3 hands (with the reasonable approximation that humans have an infinite supply of hands), and finally 4. Try this on a piece of paper -- you'll see that the man:woman ratio doesn't need to change as long as the &quot;handholding number&quot; (coordination number) for each person is the same.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taking a look at the actual lattices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{CsCl}$ (body-centered cubic):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3apTw.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/lqE2B.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each $\ce{Cs+}$ ion touches 8 $\ce{Cl-}$ ions and vice versa. Also note that the $\ce{Cs:Cl}$ ratio is indeed $1:1$. For each $\ce{Cs+}$ ion, there is a corresponding $\ce{Cl-}$ ion on its NE diagonal tilting into the plane of paper, and for each $\ce{Cl-}$ ion there is a corresponding $\ce{Cs+}$ ion on the SW diagonal tilting out of the plane of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's $\ce{NaCl}$ (face centered cubic, close packing):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDr9G.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/xCvfWm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here each $\ce{Na+}$ ion touches 6 $\ce{Cl-}$ ions and vice versa. We can define a similar pairing as before -- each $\ce{Na+}$ ion has a corresponding $\ce{Cl-}$ ion to its east, and each $\ce{Cl-}$ ion has a corresponding $\ce{Na+}$ ion to its west.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-03-09T16:09:12.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4353" PostTypeId="1" AcceptedAnswerId="4361" CreationDate="2013-03-09T17:38:41.583" Score="4" ViewCount="1267" Body="&lt;p&gt;In this video, Steve Spangler says the substance he is using (a polyethylene oxide): &lt;a href=&quot;http://www.youtube.com/watch?v=m-88M75_PCI&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=m-88M75_PCI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He mentions that one molecule would go to the moon and back twice! I have a hard time believing this, but I guess it must be true.. Is it true that polyethylene molecules can be that long?&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="1344" LastEditDate="2013-03-10T14:43:56.980" LastActivityDate="2013-03-22T11:18:29.380" Title="How long can polyethylene molecules be?" Tags="&lt;polymers&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4354" PostTypeId="1" AcceptedAnswerId="4427" CreationDate="2013-03-09T18:15:37.450" Score="0" ViewCount="600" Body="&lt;p&gt;In the ebeam lithography with PMMA resists isopropanol is often used in a mixture 3:1 (IPA:MIBK) as a developer solution. To my knowledge the developing process is usually stopped by putting the sample in distilled water and drying it with a nitrogen airgun. But I also see collegeaus who use the 3:1 mixture for developing and then &quot;stop&quot; the development in pure IPA. I actually think they develop further, as I know some people also use pure IPA instead of 3:1 mixture for developing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing, I know there many different recipes in lithography, but this IPA thing doesn't make sense to me, using IPA as a stopper means to me the developing process is strengthened further and the resist isotropically resolved, which you normally want to avoid with kind of written sharp edged rectangular structures written by ebeam&lt;/p&gt;&#xA;" OwnerUserId="1307" LastActivityDate="2013-03-19T19:07:28.667" Title="Lithography: Is isopropanol a developer and/or a stopper?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4355" PostTypeId="1" CreationDate="2013-03-09T20:08:44.617" Score="5" ViewCount="2186" Body="&lt;p&gt;I’m dealing with electrochemistry problems.&#xA;For the following reaction&#xA;$$\ce{2H+ +2e- &amp;lt;=&amp;gt;H2}\quad E^\circ=0\ \mathrm{V}$$&#xA;Does it make sense to say &#xA;$$\Delta G^\circ=-nFE^\circ=0$$&#xA;Which would mean the reaction is in equilibrium?&lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T14:17:30.920" LastActivityDate="2015-04-19T14:17:30.920" Title="Delta G for Standard Hydrogen Electrode reaction" Tags="&lt;electrochemistry&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4356" PostTypeId="1" AcceptedAnswerId="4600" CreationDate="2013-03-09T20:17:38.420" Score="9" ViewCount="6011" Body="&lt;p&gt;Why is it that dry erase markers allow permanent markings to come off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.youtube.com/watch?v=GD-MJ-nF55M&quot; rel=&quot;noreferrer&quot;&gt;demonstration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="7448" LastEditDate="2015-10-30T17:31:26.783" LastActivityDate="2017-04-21T20:25:11.167" Title="Why does permanent marker writing come off a white board when coated with white board marker?" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4357" PostTypeId="2" ParentId="4344" CreationDate="2013-03-09T21:43:55.437" Score="8" Body="&lt;p&gt;Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluoride is very electronegative, and thus the bonding forces between $\ce{Ca^2+}$ and each fluoride anions are very strong, they are tightly held together in the crystal form. A few water molecules do not have enough strength to get in between and dissolve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. So weaker the bond strength, greater the solubility – at least that’s the expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, $\ce{CaCl2}$ is more soluble than $\ce{CaF2}$, and $\ce{CaBr2}$ is even more – as expected. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. It might be true, though, since there are a variety of factors that may or may not contribute to it’s solubility in water, such as it’s crystal structure and the particular characteristics of iodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. I couldn’t find anything on the solubility of $\ce{CaI2}$ in my CRC Handbook of Chemistry and Physics either, so I can’t provide you with a precise answer on why it’s less soluble than $\ce{CaBr2}$ (if in fact it is), but I hope you can now understand some of the factors that make $\ce{CaF2}$ the least soluble compound on that list.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorDisplayName="user7951" LastEditDate="2015-05-04T17:34:02.387" LastActivityDate="2015-05-04T17:34:02.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4358" PostTypeId="2" ParentId="2474" CreationDate="2013-03-10T10:01:25.633" Score="2" Body="&lt;p&gt;In your first test, the $\ce{KMnO4}$ probably reacted with the $\ce{H2O2}$ excess remaining after your preparation. The ionic equation for the reaction is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2MnO4- + 3H2O2 -&amp;gt; 2MnO2 + 2H2O +3O2 + 2OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{MnO2}$ formed is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;insoluble black precipitate&lt;/a&gt;. Although there is an $\ce{OH-}$, the reaction apparently occurs in neutral and faintly acidic medium as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of your second test doesn't seem to be very clear. Sodium perborate apparently reacts very quickly with any water so if you added any water during the second test, it must have reacted to form oxygen gas bubbles (which appear white).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="17368" LastEditDate="2016-12-23T08:43:35.847" LastActivityDate="2016-12-23T08:43:35.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4359" PostTypeId="2" ParentId="4355" CreationDate="2013-03-10T10:08:12.430" Score="12" Body="&lt;p&gt;You cannot apply the $\Delta{G}$ equation to a single electrode potential. It can be applied to a cell though so if the hydrogen electrode is connected to another electrode (say copper dipped in copper sulfate solution) then you can find the free energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's better to remember the formula as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta{G} = -nFE_{cell}^o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{cell}^o = E_{cathode}^o - E_{anode}^o$ (using all standard reduction potentials)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way you won't accidentally apply it anywhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, for $\ce{\Delta G}$ to be zero you would have to take Hydrogen electrodes for cathode as well as anode.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A More Complicated Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just in case you're curious about the reasons behind this (Since I was also very confused about this when I first learned Electrochemistry. It's ok if you don't understand this),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very difficult to measure the &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_electrode_potential&quot;&gt;absolute potential&lt;/a&gt; of a single electrode and so the standard reduction potentials you are using are actually based on a &lt;strong&gt;reference system&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually very similar to the system of atomic masses. The atomic masses you use in calculations are not the actual masses of the elements in grams but they are measured in &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_mass_unit&quot;&gt;unified atomic mass units&lt;/a&gt;. One unified atomic mass unit is defined as 1/12th the mass of a carbon-12 atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;strong&gt;Note&lt;/strong&gt;: Now of course, after the mole was made a fundamental SI unit, the unified atomic mass unit has become equivalent to $1 g/mol$. This is just for the sake of explanation.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you come across a situation where you have to use the mass of the atom (maybe some nuclear chemistry calculations) you can't use those normal atomic masses but you would have to convert it into kg (One unified atomic mass unit is about $1.6\times10^{-27}$ kg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the standard reduction potentials are actually measured &lt;strong&gt;with reference&lt;/strong&gt; to a standard electrode (Whose reduction potential is &lt;strong&gt;arbitrarily&lt;/strong&gt; taken to be zero) which is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_hydrogen_electrode&quot;&gt;standard hydrogen electrode&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This standard reduction potential of an electrode can be measured easily by connecting the electrode to a standard hydrogen electrode (SHE) and then measuring the potential difference using a voltmeter and applying the $E_{cell}^o$ equation. (If $H^+$ is being reduced to $H_2$, then $E_{cathode}^o$ is zero otherwise $E_{anode}^o$ is zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this just means that when you say $E^o$ of the SHE is zero, you are actually saying that &quot;the potential of the SHE with respect to itself is zero&quot; which is quite obvious. That doesn't mean that the potential of the electrode itself is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why you cannot apply the first equation in this case. But this does not apply to cells. If we assume the absolute potential of the SHE to be some $x$, then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{abs}^o = E^o + x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&#xA;$E^o_{cell} = E_{abs}(cathode) - E_{abs}(anode)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$= E^o_{cathode} + x - [E^o_{anode} + x]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^o_{cell} = E_{cathode}^o - E_{anode}^o$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute potential of the SHE cancels out in the difference.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2013-03-12T05:37:04.850" LastActivityDate="2013-03-12T05:37:04.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4360" PostTypeId="1" AcceptedAnswerId="4442" CreationDate="2013-03-10T14:00:43.873" Score="4" ViewCount="3081" Body="&lt;p&gt;When we plot molecular orbital diagrams we use a linear combination of atomic orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find the energies of particular atomic orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture below, the energies of 2s and 2p orbitals of Oxygen are lower than Carbon's 2s and 2p orbitals of Carbon, what is the logic behind this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Enfns.png&quot; alt=&quot;Molecular orbital diagram of CO2 from wikipedia&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="222" LastActivityDate="2017-04-23T05:38:31.327" Title="Energies of atomic orbitals on molecular orbital diagrams" Tags="&lt;energy&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4361" PostTypeId="2" ParentId="4353" CreationDate="2013-03-10T18:14:42.483" Score="8" Body="&lt;p&gt;Short answer? Possibly, but these ones aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the other thing mentioned in the video is that the specific material he's using has a molecular weight of approximately four million. If we assume it's ordinary poly-ethylene oxide, we can get the rough number of $\ce{(CH2)2O}$ subunits by just dividing by the molecular weight of the subunit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$(12 \times 2) + (1 \times 2) + 16 = 42$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{4000000}{42} \approx 95000$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, if we assume that each $\ce{C-O-C}$ back-bone unit is stretched out like a di-methyl ether molecule, a quick check on a fairly standard Ghemical force field gives me a subunit length of approximately 3.7 Å.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$3.7 \times 10^{-10} \times 95000 = 0.00003515 \text{m}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, these molecules should be about 0.04 millimetres long. Not exactly astronomical distances, but impressively long for a single molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;theory&lt;/em&gt;, I don't think there's anything stopping you making a molecule as long as they claim, but these ones certainly aren't. I think also that the longest (synthetic) polymers in common use are of similar molecular weights and thus similar lengths: ultra-high molecular weight polyethene seems to top out around 4.5 million for commercial applications, from a quick search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biological molecules frequently get longer than this, though, especially DNA: the average human chromosome &lt;a href=&quot;http://hypertextbook.com/facts/1998/StevenChen.shtml&quot; rel=&quot;noreferrer&quot;&gt;apparently contains a single DNA molecule approximately 5 centimetres long.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man explaining in the video may have erroneously given the figure for the total length of all the molecules in that beaker.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2013-03-22T11:18:29.380" LastActivityDate="2013-03-22T11:18:29.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4363" PostTypeId="2" ParentId="316" CreationDate="2013-03-10T21:38:56.183" Score="21" Body="&lt;p&gt;This is a complex issue, particularly because people often like to think in terms of an indepedent-particle picture (i.e. the aufbau filling up orbitals), even though the exact many-body wavefunction has strong electron-electron correlations. So let me rephrase your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between the KS eigenfunctions and the exact many-body wavefunction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, as you say, the KS eigenfunctions strictly speaking have no physical meaning (as far as we know). However, the KS eigenfunctions do give a useful qualitative (and sometimes quantitative) picture. The reason for this is that the KS eigenfunctions are a pretty good approximation to something in many-body perturbation theory called the quasiparticle wavefunction. The quasiparticle wavefunction is a well-defined physical property of a system that essentially tells you if you add (or remove) an electron with a certain amount of energy, where it will go. For example, see &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevB.74.045102&quot;&gt;Phys. Rev. B 74, 045102 (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there examples of when the KS eigenfunctions don't give a good desription of the quasiparticle wavefunctions? Well there are certainly many situations where the approximations we typically use in DFT (such as the local density approximation) lead to serious problems. However, I don't know of any examples where someone has shown that the exact KS eigenfunctions (i.e. those obtained with the true exchange-correlation functional) don't agree at least qualitatively with the quasiparticle wavefunctions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, everything I have said above applies equally well to the Hartree-Fock wavefunctions. In fact, there is a solid mathematical basis for interpreting the HF wavefunctions as an approximation to the quasiparticle wavefunctions. See Chapter 4 of Fetter's &lt;em&gt;Quantum Theory of Many-Particle Systems&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the KS eigenvalues? Strictly speaking, in general they do not correspond to ionization energies (or any other physically useful quantity). The one exception is the highest occupied eigenvalue, which is exactly equal to the ionization energy of the system. Janak's theorem tells us that the other eigenvalues are related to the derivative of the energy with respect to the occupancy of that eigenfunction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_i=\frac{dE}{dn_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevB.18.7165&quot;&gt;Phys. Rev. B 18, 7165 (1978)&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevB.56.16021&quot;&gt;Phys. Rev. B 56, 16021 (1997)&lt;/a&gt;. It turns out that empirically these eigenvalues are nonetheless pretty good approximations to the true  energy levels of the system with some caveats. In particular, the band gaps of solids are systematically underestimated.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-03-10T21:38:56.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4364" PostTypeId="1" CreationDate="2013-03-10T21:44:49.783" Score="5" ViewCount="1160" Body="&lt;p&gt;In theory you should be able to turn $\ce{CO2}$ to $\ce{CO}$ and then to $\ce{C}$ and connect all the C-atoms together to form graphene. &lt;br/&gt;&#xA;That should be possible, because plants do something similar, they take $\ce{H2O}$ and $\ce{CO2}$ and link the C-atoms together as $\ce{C6H12O6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to produce graphene out of air? Here is a related article from the RCS: &quot;&lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2012/January/graphene-oxide-carbon-dioxide-reactions.asp&quot; rel=&quot;nofollow&quot;&gt;Conjuring graphene oxide from thin air&lt;/a&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="1347" LastEditorUserId="7448" LastEditDate="2014-12-31T19:17:18.697" LastActivityDate="2014-12-31T19:17:18.697" Title="Is it possible to produce graphene out of air?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;green-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4365" PostTypeId="2" ParentId="4364" CreationDate="2013-03-11T02:56:15.683" Score="1" Body="&lt;p&gt;There are a few standard ways of making graphene (e.g. exfoliating graphite or chemical vapor deposition), but making it out of air isn't one of these ways (though I suppose the ability to do so would fix a few problems...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one has ever been able to do anything like this.  I suppose it wouldn't be a thermodynamically favorable process, and the concentration of CO2 in air isn't particularly high.&lt;/p&gt;&#xA;" OwnerUserId="1348" LastActivityDate="2013-03-11T02:56:15.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4366" PostTypeId="2" ParentId="4364" CreationDate="2013-03-11T16:09:03.820" Score="5" Body="&lt;p&gt;A graphene sheet is easily obtained from a block of graphite using adhesive tape. Since graphite itself is essentially a pure grade of coal, it is the result of biological processes in vegetation some millions of years ago. As is pointed out in answers to &lt;a href=&quot;https://biology.stackexchange.com/questions/2398/do-trees-really-get-a-large-share-of-their-mass-from-the-carbon-in-the-air&quot;&gt;this Biology.StackExchange question&lt;/a&gt;, the majority of a plant's carbon comes from the air via photosynthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a sense, we already do that, but we just use the huge stores of graphite from millions of years of high pressure processing beneath our feet rather than start from the beginning.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:14.547" LastActivityDate="2013-03-11T16:09:03.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4367" PostTypeId="1" CreationDate="2013-03-11T19:46:38.310" Score="2" ViewCount="14094" Body="&lt;p&gt;I've bought a white powder. The label on the package states that the powder contains sodium chloride, potassium chloride and potassium citrate. How can I make sure that the powder does indeed contain potassium chloride? I'm mainly interested in the presence of potassium ions.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastEditorUserId="865" LastEditDate="2015-10-30T16:58:22.673" LastActivityDate="2017-01-27T00:01:20.410" Title="Detection of potassium ions" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4368" PostTypeId="1" CreationDate="2013-03-11T23:48:36.117" Score="1" ViewCount="196" Body="&lt;p&gt;I know the reaction proceeds as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pyruvate accepts the electrons from NADH to form lactate. What would be the intermediate in this reaction? Would it be NADH or NAD+?&lt;/p&gt;&#xA;" OwnerUserId="1354" LastActivityDate="2013-06-09T13:38:28.387" Title="What metabolite intermediate will accumulate in a mammalian cell-free extracts capable of glycolysis if lactate dehydrogenase is inhibited?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4369" PostTypeId="1" AcceptedAnswerId="4370" CreationDate="2013-03-12T13:04:31.690" Score="11" ViewCount="2073" Body="&lt;p&gt;For the papal election, white and black smoke will be used to signal success or failure in the voting process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, the ballots are burnt and chemicals are added to create the color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what chemicals are used to get black or white smoke?&lt;/p&gt;&#xA;" OwnerUserId="1355" LastEditorUserId="16683" LastEditDate="2015-10-02T16:55:51.010" LastActivityDate="2016-07-30T08:30:22.737" Title="What chemicals are going to be used to make the smoke black or white?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4370" PostTypeId="2" ParentId="4369" CreationDate="2013-03-12T15:52:03.153" Score="6" Body="&lt;p&gt;I think &lt;a href=&quot;http://en.wikipedia.org/wiki/Titanium_tetrachloride&quot; rel=&quot;noreferrer&quot;&gt;Titanium Tetrachloride&lt;/a&gt; would be a good option for white smoke (I've heard it's used quite often in smoke bombs). But it releases quite a lot of &lt;span class=&quot;math-container&quot;&gt;$HCl$&lt;/span&gt; gas and so it might not be very safe.&lt;/p&gt;&#xA;&lt;p&gt;A quick Google search comes up with a lot of answers to your question though. This site even answers your question in the same context (The elections going on at the Vatican):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://thehappyscientist.com/science-experiment/black-and-white-smoke&quot; rel=&quot;noreferrer&quot;&gt;http://thehappyscientist.com/science-experiment/black-and-white-smoke&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Apparently you can make white smoke using Magnesium powder as well. (The white colour is due to fine particles of Magnesium Oxide &lt;span class=&quot;math-container&quot;&gt;$MgO$&lt;/span&gt; )&lt;/p&gt;&#xA;&lt;p&gt;It seems the Vatican has experimented using many sources before (Military flares, Smoke bombs and so on) though apparently they didn't turn out too well...&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Colored_smoke&quot; rel=&quot;noreferrer&quot;&gt;Coloured smoke&lt;/a&gt; gives another possibility. You can add Titanium Dioxide (&lt;span class=&quot;math-container&quot;&gt;$TiO_2$&lt;/span&gt;) to a mixture of Potassium Chlorate (&lt;span class=&quot;math-container&quot;&gt;$KClO_3$&lt;/span&gt;) and Lactose for white smoke. &lt;span class=&quot;math-container&quot;&gt;$KClO_3$&lt;/span&gt; is a powerful oxidizing agent which reacts with Lactose (which is a reducing sugar) to create an explosion. &lt;span class=&quot;math-container&quot;&gt;$TiO_2$&lt;/span&gt; is safe and is used as a pigment in most white paints and as a food colouring as well.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The official recipes have been revealed by the Vatican press office. See &lt;a href=&quot;http://www.nytimes.com/2013/03/13/science/vatican-reveals-recipes-for-conclave-smoke.html?_r=0&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; story on the New York Times website.&lt;/p&gt;&#xA;&lt;p&gt;The previous edit was partially correct. They use a mixture of &lt;span class=&quot;math-container&quot;&gt;$KClO_3$&lt;/span&gt;, lactose and pine rosin for white smoke. For black smoke, they use Potassium perchlorate (&lt;span class=&quot;math-container&quot;&gt;$KClO_4$&lt;/span&gt;), anthracene (an aromatic compound made of three fused benzene rings) and sulfur.&lt;/p&gt;&#xA;&lt;p&gt;The process (quoted from the site):&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The chemicals are electrically ignited in a special stove first used for the conclave of 2005, the statement said. The stove sits in the Sistine Chapel next to an older stove in which the ballots are burned; the colored smoke and the smoke from the ballots mix and travel up a long copper flue to the chapel roof, where the smoke is visible from St. Peter’s Square. A resistance wire is used to preheat the flue so it draws properly, and the flue has a fan as a backup to ensure that no smoke enters the chapel.&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-03-13T09:06:21.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4371" PostTypeId="2" ParentId="369" CreationDate="2013-03-12T16:07:35.520" Score="2" Body="&lt;p&gt;Some density functional theory packages, such as &lt;a href=&quot;http://www.quantum-espresso.org/&quot; rel=&quot;nofollow&quot;&gt;Quantum Espresso&lt;/a&gt;, can calculate XPS core level shifts using the pseudopotential formalism. See &lt;a href=&quot;http://dx.doi.org/10.1103%2FPhysRevB.74.045430&quot; rel=&quot;nofollow&quot;&gt;Phys. Rev. B 74, 045430 (2006)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-03-12T16:07:35.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4372" PostTypeId="1" CreationDate="2013-03-12T22:05:42.107" Score="7" ViewCount="1554" Body="&lt;p&gt;According to the biogenic theory of hydrocarbon formation, hydrocarbons come from pressurized carbohydrates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an explanation of what happens to the oxygen in the process. How could the oxygen break away from the hydroxyl groups of molecules like glucose to form hydrocarbons?&lt;/p&gt;&#xA;" OwnerUserId="1358" LastEditorUserId="148" LastEditDate="2013-03-13T12:59:55.433" LastActivityDate="2013-03-14T21:12:58.870" Title="What happens to the oxygen when carbohydrates turn into hydrocarbons?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;geochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4373" PostTypeId="2" ParentId="4372" CreationDate="2013-03-13T04:09:23.270" Score="2" Body="&lt;p&gt;This will largely depend upon the reagents that perform the reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using $\ce{HI}$  (or a strong acid and heat or $\ce{H2}$ and $\ce{Pd/C}$) the reaction goes as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H12O6 -&amp;gt;[H^+] C6H14 + H2O}$ (organic products only) &lt;/p&gt;&#xA;" OwnerUserId="1359" LastEditorUserId="22" LastEditDate="2013-03-13T04:52:09.233" LastActivityDate="2013-03-13T04:52:09.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4374" PostTypeId="2" ParentId="4372" CreationDate="2013-03-13T11:28:49.677" Score="6" Body="&lt;p&gt;The process likely involves &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrolysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyrolysis&lt;/a&gt;, a generic term for the thermal degradation pathways of organic compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of carbohydrates, like glucose (shown below), the most common function groups are alcohols (R-OH). Alcohols undergo &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehydration_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;dehydration&lt;/a&gt; in the presence of strong acids, but the reaction can also be promoted by extreme heat in the absence of oxygen. The presence of oxygen leads to combustion (bad!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LZmvp.png&quot; alt=&quot;Glucose&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a dehydration elimination reaction, the OH on one carbon and the H on a neighboring carbon leave to form water. This usually follows an &lt;a href=&quot;http://en.wikipedia.org/wiki/Elimination_reaction#E1_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;E1-like mechanism&lt;/a&gt;, involving a carbocation intermediate. In the geochemical version, the necessary acid could come from nearby slightly acidic mineral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protonation of the alcohol:&#xA;$$\ce{(CH3)2CH-OH + H+ -&amp;gt; (CH3)2CH-OH2+}$$&#xA;Loss of leaving group:&#xA;$$\ce{(CH3)2-CH-OH2+ -&amp;gt; (CH3)2CH+ + H2O}$$&#xA;Loss of proton (regeneration of acid):&#xA;$$\ce{CH3-CH+-CH2-{\bf{H}} -&amp;gt; CH3-CH=CH2 + {\bf{H+}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall:&#xA;$$\ce{(CH3)2CH-OH -&amp;gt;[\ce{H+}][\Delta] CH3-CH=CH2 + H2O }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of carbohydrates, they tend to dehydrate to carbonyl compounds, which can then decompose further into hydrocarbons by reductive and radical pathways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, fructose is converted into furan derivatives, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxymethylfurfural&quot; rel=&quot;nofollow noreferrer&quot;&gt;5-hydroxymethylfurfural&lt;/a&gt;, by a series of dehydrations and rearrangements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12O6 -&amp;gt;[\ce{H+}][\Delta] C6H6O3 + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hr9e2.png&quot; alt=&quot;HMF synthesis from fructose&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above was created by Wikipedia user &lt;a href=&quot;http://en.wikipedia.org/wiki/User:V8rik&quot; rel=&quot;nofollow noreferrer&quot;&gt;V8rik&lt;/a&gt; and released under the &lt;a href=&quot;http://commons.wikimedia.org/wiki/Commons:GNU_Free_Documentation_License&quot; rel=&quot;nofollow noreferrer&quot;&gt;GNU Free Documentation License&lt;/a&gt; and the &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Creative Commons Attribution-Share Alike 3.0 Unported&lt;/a&gt; License.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-13T11:28:49.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4375" PostTypeId="2" ParentId="4177" CreationDate="2013-03-13T15:41:36.193" Score="2" Body="&lt;p&gt;It is important to note that the phases of the MOs are irrelevant. Assuming that the Hamiltonian can be solved with a Slater determinant $\Psi_0=|\phi_1(1)\dots\phi_N(N)|$ of MOs $\phi_i$ with eigenvalue $E:= \sum \epsilon_i$, where $\epsilon_i$ are the eigenvalues of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fock_operator&quot; rel=&quot;nofollow&quot;&gt;Fock operator&lt;/a&gt;, then your time-dependent solution will be&#xA;$$ \Psi(t) = e^{-\frac{Et}{\Im \hbar}} \Psi_0.$$&#xA;Now you see that no phase can be uniquely attributed to any one MO. Canonically the MOs are chosen to be real-valued.&lt;/p&gt;&#xA;" OwnerUserId="440" LastActivityDate="2013-03-13T15:41:36.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4376" PostTypeId="2" ParentId="4367" CreationDate="2013-03-13T16:33:24.917" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;What do you want?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you're just looking for a source of potassium ions you could just buy $\ce{KCl}$ or some other compound, they're relatively cheap if not pro analysis. If you actually want to test for $\ce{K+}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What else do you know about the product?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tests are completely different for different amounts of $\ce{K+}$. Does the vendor specifies the amount of each compound on the product's label?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How sure do you need to be?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you need to do a QUALITATIVE analysis or a QUANTITATIVE analysis? If you need to be absolutely sure about the concentration then you need to guess it's ridiculously low - you'll probably need at least AAS (Atomic Absorption Spectroscopy) equipment, which can't be accessed by anyone nor find anywhere, but you can order an analysis. Alternatively, you can use an electrode - keep reading .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some ways of testing that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course you'll need to dissolve the powder in at least distilled water to do all the following tests, and do a blank with the same water - the more pure the water the more accurate your results.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For big amounts of $\ce{K+}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The flame test suggested by Paul J. Gans on the comments above could work, although I never actually did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is doing the following titration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{KCl(aq) + AgNO3(aq) -&amp;gt; AgCl(s) + KNO3(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AgNO3}$ can be found at any store that sells chemicals. We used 0.1 M for small amounts of $\ce{Cl-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are actually testing for is $\ce{Cl-}$ ions, but if the amount of K = the amount of $\ce{Cl}$ the job is done. Now, since the powder contains potassium citrate this is just a qualitative test, although it can be very accurate for small concentrations if properly executed (see &lt;a href=&quot;http://www.outreach.canterbury.ac.nz/chemistry/documents/chloride_mohr.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mohr&lt;/a&gt; method). Silver chloride is a white solid that is barely visible in small amounts, and a similar  titration's end point is just a few drops (Mohr method), and there are a lot of possible interference so you should not expect precision unless you do some pro work.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For small amounts of $\ce{K+}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the best way of doing this is using a potassium-selective electrode. I never used one of those, but other ion-selective electrodes work great on water analysis, detection levels were a few mg/L on our old ones!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_selective_electrode&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Potassium_selective_electrode&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, you can use AAS.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="39539" LastEditDate="2017-01-27T00:01:20.410" LastActivityDate="2017-01-27T00:01:20.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4377" PostTypeId="1" AcceptedAnswerId="4480" CreationDate="2013-03-13T21:26:57.743" Score="8" ViewCount="276" Body="&lt;p&gt;Last week I was discussing with a friend how we thought the stripeless cleaning of windows is achieved when using a cleaner like &lt;a href=&quot;http://www.windex.com/en-US/Products/Pages/original-glass-cleaner.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Windex Glass Cleaner&lt;/a&gt; as opposed to just using hot water, in which case you do get stripes left behind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first we thought that the stripeless nature was caused by surfactants in the cleaner that allow the (polar) dirt on the windows to be collected in the water through micellation. However, when we use normal cleaner not specifically designed to clean glass/windows, we also get the stripes even though we are sure that there are also surfactants in the normal cleaner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what is the difference between the glass cleaner and the regular cleaner that makes sure that windows become stripeless clean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought we considered but couldn't really work out was that the stripe formation would be something caused by a combination of the marangoni effect and evaporation, similar to &lt;a href=&quot;http://www.nature.com/nature/journal/v389/n6653/full/389827a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the coffee-stain effect&lt;/a&gt; which can be overcome by specific tuning of the surfactants as shown &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp3009628&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain how the stripeless cleaning works? Can fluid dynamics explain what happens or is it really something chemical going on?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When searching I also ran into &lt;a href=&quot;http://www.baudoin.nl/content_en/pag3_03.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;a solution&lt;/a&gt; that (at least the manufacturers claim) will also give stripeless clean windows. The interesting thing is that they use pure $\ce{H2O}$ so without any surfactants but also without any dissolved salts like regular tap water has.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cross-post&lt;/strong&gt;: &lt;a href=&quot;https://physics.stackexchange.com/q/56800/7433&quot;&gt;https://physics.stackexchange.com/q/56800/7433&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="5026" LastEditDate="2018-05-15T01:51:55.360" LastActivityDate="2018-05-15T01:51:55.360" Title="Stripeless cleaning of windows" Tags="&lt;everyday-chemistry&gt;&lt;surface-chemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4378" PostTypeId="1" CreationDate="2013-03-13T22:52:18.593" Score="7" ViewCount="3162" Body="&lt;p&gt;I'm doing an experiment where I'm measuring the effect of changing the $\ce{Cu^{2+}}$ ion concentration on the voltage in a copper-zinc galvanic cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be doing at least five trials for each of the five concentrations I'm testing for (from $0.2-1.0\:\mathrm{M}$), so in total that is 25 trials. However, do I need a new piece of filter paper as a salt bridge for each trial, or can I reuse the same piece of filter paper and dip it in a $\ce{KCl}$ solution each time? Do I really need to make 25 separate pieces of filter paper?&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="22" LastEditDate="2013-03-14T09:38:46.200" LastActivityDate="2016-08-19T18:44:19.200" Title="What is the procedure of designing a salt bridge for a galvanic cell?" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4381" PostTypeId="1" CreationDate="2013-03-14T17:28:30.233" Score="7" ViewCount="6018" Body="&lt;p&gt;I'm looking for a wet-etchant to remove aluminum oxide created by anodizing, but not etch the unoxidized metal (with greater than 10:1 selectivity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about chromium(IV) oxide and fluorine-based etchants, but those are problematic for safety and disposal. Are there other etchants? It can be an acid or a base.&lt;/p&gt;&#xA;" OwnerUserId="1365" LastEditorUserId="23561" LastEditDate="2018-10-19T01:17:42.133" LastActivityDate="2018-10-19T01:17:42.133" Title="Selective etchant for Al2O3 (Does not etch Al, Cu, Zn, Sn)" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;applied-chemistry&gt;&lt;chemical-engineering&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4383" PostTypeId="2" ParentId="4372" CreationDate="2013-03-14T18:11:47.740" Score="2" Body="&lt;p&gt;As others answers have pointed out, the main pathways for reducing carbohidrates to hydrocarbons involves the dehydration elimination of the alcohol groups. This reaction is catalized by acidic conditions and can be carried out with linear alcohols (i.e. like ethanol, resulting in an olefin) in laboratory conditions, at temperatures higher than 160 °C and atmospheric pressure. The products are unsaturated compounds and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the geochemical conditions don't allow a strongly acidic medium, the pressures and temperatures attained should be enough to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehydration_reaction&quot; rel=&quot;nofollow&quot;&gt;pyrolitically dehydrate &lt;/a&gt; sugars and cellulosic material which makes up most of the cell walls of vegetal biomass. Nevertheless, pyrolitic conversion of mainly-cellulosic biomass in sedimentary basins ussually doesn't produce aliphatic hydrocarbons. Instead, the carbon skeletons are further dehydrogenated and compressed forming &lt;a href=&quot;http://en.wikipedia.org/wiki/Coal&quot; rel=&quot;nofollow&quot;&gt;coal deposits&lt;/a&gt; containing minor aromatic derivatives like phenols and aniline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main source of hydrocarbon deposits is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Decarboxylation&quot; rel=&quot;nofollow&quot;&gt;pyrolitic decarboxylation&lt;/a&gt; of fatty acids, in turn produced by the hydrolisis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lipids&quot; rel=&quot;nofollow&quot;&gt;lipids&lt;/a&gt;. Those lipids mostly come from microphytes (microalgae) which comprise most of the phytoplancton in the seas, and which have been accumulated at the bottom of ancient oceans. For example, the rich sedimentary basin of modern Middle East was once the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tethys_Ocean&quot; rel=&quot;nofollow&quot;&gt;Tethys Ocean&lt;/a&gt; seabed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in essence, both in the dehydration of carbohidrates and in the decarboxylation of fatty acids, the oxygen goes away in the form of water or carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="441" LastEditorUserId="441" LastEditDate="2013-03-14T21:12:58.870" LastActivityDate="2013-03-14T21:12:58.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4384" PostTypeId="2" ParentId="4378" CreationDate="2013-03-14T18:26:34.060" Score="2" Body="&lt;p&gt;If you wish to do 25 trials with 5 trials per concentration, I would suggest the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can use the same filter paper as long as the $\ce{Cu^{2+}}$ concentration remains constant. However, since you are changing the concentration 5 times, you will need 5 pieces of filter paper.&lt;/li&gt;&#xA;&lt;li&gt;After every trial, thoroughly wash the filter paper with distilled water. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I would say you need at least 5 pieces, but you would not need to go all the way to 25 pieces. Just make sure to follow step 2 after &lt;em&gt;each&lt;/em&gt; trial.&lt;/p&gt;&#xA;" OwnerUserId="1048" LastEditorUserId="275" LastEditDate="2013-03-15T11:08:39.573" LastActivityDate="2013-03-15T11:08:39.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4385" PostTypeId="2" ParentId="2699" CreationDate="2013-03-14T18:34:23.293" Score="2" Body="&lt;p&gt;This reaction is definitely decomposition. Decomposition reactions involve taking one reactant and destabilizing its bonds in order to produce 2 or more products. In this case, the 500 degree Celsius condition (indicated above the reaction arrow) indicates that heat energy is being added to decompose the molecule of &lt;code&gt;MnO2&lt;/code&gt;. This is more commonly called &quot;thermal decomposition&quot; (Check out this link: &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_decomposition&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thermal_decomposition&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1048" LastActivityDate="2013-03-14T18:34:23.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4386" PostTypeId="2" ParentId="1227" CreationDate="2013-03-14T19:28:48.933" Score="0" Body="&lt;p&gt;Wait, &lt;a href=&quot;http://en.wikipedia.org/wiki/Vodka&quot; rel=&quot;nofollow&quot;&gt;vodka&lt;/a&gt; is quoted at 40% ethanol V/V (so mole fraction about 1/5) and not azeotropic! It should be about 83 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/1/1e/Vapor-Liquid_Equilibrium_Mixture_of_Ethanol_and_Water.png&quot; rel=&quot;nofollow&quot;&gt;Graph of vapour-liquid equilibrium for mixture of ethanol and water&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1366" LastActivityDate="2013-03-14T19:28:48.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4387" PostTypeId="1" CreationDate="2013-03-15T12:35:24.950" Score="1" ViewCount="693" Body="&lt;p&gt;What is the best method for calculating partial charges and protonation/deprotonation states of a small molecule such as phenol or caffeine, depending on the pH? For instance, it is well known that above pH=10 or so phenol is deprotonated, but there is data about its protonation around pH=2. For more complex molecules such as caffeine, similar things happen. So I am more interested in calculating partial charges of protonated and deprotonated forms than on pKas (sometimes, experimental data is available).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I head about Gaussian and the like, but I also wonder about software or methods that are easy to use.&lt;/p&gt;&#xA;" OwnerUserId="256" LastActivityDate="2015-09-13T08:13:55.743" Title="Calculation of partial charges and protonation/deprotonation states depending on pH" Tags="&lt;theoretical-chemistry&gt;&lt;ph&gt;&lt;formal-charge&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4388" PostTypeId="2" ParentId="1197" CreationDate="2013-03-15T17:20:29.750" Score="4" Body="&lt;p&gt;I think you can explain this using the equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO_2(g) &amp;lt;=&amp;gt; NO_2(dissolved)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you pass $CO_2$ through the warm acid, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_pressure&quot; rel=&quot;nofollow&quot;&gt;partial pressure&lt;/a&gt; of $CO_2$ ($P_{CO_2}$) above the gas increases (Since partial pressure generally represents the amount of gas above the solution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;total pressure $P = P_{NO_2} + P_{CO_2}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dalton%27s_law&quot; rel=&quot;nofollow&quot;&gt;Dalton's Law&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And $P$ remains constant at atmospheric pressure (assuming that the system is not closed).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$P_{NO_2}$ will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier's Principle&lt;/a&gt;, the equilibrium proceeds backward to compensate the decrease in $P_{NO_2}$ and thus more $NO_2$ gas is released. The gas flows quickly along with the bubbled gas and this prevents the forward reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like quite a roundabout method but it works. I'm sure there could be a much simpler explanation though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation mentioned is most probably the disproportion reaction of $NO_2$ when it comes in contact with water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3NO_2 + H_2O -&amp;gt; 2HNO_3 + NO}$&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-03-15T17:20:29.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4390" PostTypeId="1" AcceptedAnswerId="4392" CreationDate="2013-03-16T10:15:51.490" Score="21" ViewCount="11384" Body="&lt;p&gt;I was given the first structure, and then drew the other 5 resonance structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xbna3.jpg&quot; alt=&quot;resonance structures of 1-(cyclopenta-2,4-dien-1-ylidene)-N,N-dimethylmethanamine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same number of electrons, and no valence orbitals exceeding capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is that the first structure contributes more. There's no electron deficiency on any of them, so we can disregard that. There &lt;em&gt;is&lt;/em&gt; charge separation on the last 5 though, and in combination with this, the negative charge is on the least electronegative atoms, thus making the last 5 structures noticably higher in energy, and thus they contribute less to the true electronic image of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I'm correct so far; &lt;strong&gt;here's my question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first structure, nitrogen is sp³ hybridised, but on all others it's clearly sp² hybridised. So what does that mean? It's somewhere in between, but closer to sp³-hybridised?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="10185" LastEditDate="2015-09-13T15:35:09.817" LastActivityDate="2020-05-15T05:24:34.840" Title="How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;&lt;hybridization&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4391" PostTypeId="2" ParentId="2902" CreationDate="2013-03-16T12:30:17.433" Score="3" Body="&lt;p&gt;The answer to your second question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halogenation of saturated hydrocarbons occurs by &lt;strong&gt;free radical substitution&lt;/strong&gt; in the presence of UV light or heat and not by formation of carbocations. There's a very clear representation of the reaction mechanism on &lt;a href=&quot;http://en.wikipedia.org/wiki/Free_radical_halogenation&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since free radical intermediates are involved, this is a chain reaction and so what you said is correct, a large number of substituted products can be formed depending on the reaction conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example&lt;/strong&gt;: If you consider the simple reaction between Methane and Chlorine in the presence of sunlight, Chloromethane, Dichloromethane, Chloroform, Carbon tetrachloride and a small amount of Ethane (by combination of Methyl radicals) are formed depending on the amount of chlorine available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these products will be formed even if the reactants are in a 1:1 ratio but they will be present in low concentration since there is less chlorine available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of higher alkanes like Butane and Pentane, the secondary carbons are substituted faster since the secondary free radical intermediates are more stable (reasons are the same as that for secondary carbocations). &lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-03-16T12:30:17.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4392" PostTypeId="2" ParentId="4390" CreationDate="2013-03-16T13:04:43.320" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the first structure, nitrogen is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridised, but on all others it's clearly &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridised. So what does that mean? It's somewhere in between, but closer to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt;-hybridised?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized so that the the lone pair is in a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-like orbital to ensure appropriate symmetry for &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-overlap. Put another way, if an atom is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized in one resonance structure, then it is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized in all of them. Atoms that are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt;-hybridized have differing geometries, which is not permitted in the resonance phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized (not the other way around). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; orbitals formed from &lt;a href=&quot;http://en.wikipedia.org/wiki/LCAO&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear combinations&lt;/a&gt; of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugated_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjugated systems&lt;/a&gt;, which is not very good but will give you the idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My reasoning is that the first structure contributes more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromaticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;aromatic&lt;/a&gt; like the surprisingly stable &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopentadienyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclopentadienyl carbanion&lt;/a&gt;. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="61578" LastEditDate="2020-05-15T05:24:34.840" LastActivityDate="2020-05-15T05:24:34.840" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="4393" PostTypeId="1" AcceptedAnswerId="4394" CreationDate="2013-03-16T13:05:10.700" Score="3" ViewCount="19766" Body="&lt;p&gt;In the text that I'm reading it is stated that :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the molecular mass increases in any homologous series, a gradation in physical properties is seen. This is because the melting points and boiling points increase with increasing molecular mass. Other physical properties such as solubility in a particular solvent also show a similar gradation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is regarding the last line; what kind of gradation in solubility is observed? Increase or decrease?&lt;/p&gt;&#xA;" OwnerUserId="1376" LastEditorUserId="9961" LastEditDate="2015-10-16T17:03:22.977" LastActivityDate="2018-02-25T13:44:37.703" Title="Changes in physical properties in homologous series : Solubility" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4394" PostTypeId="2" ParentId="4393" CreationDate="2013-03-16T13:34:58.697" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is regarding the last line, what kind of gradation in solubiltiy is observed? Increase or decrease?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility is not one single property, but a collection of related properties. Solubility in water is different from solubility in ethanol, which is different from solubility in benzene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause of the gradation of (all of these) properties is a systematic change in intermolecular forces. The number and strength of these forces increase as molecular mass increases, which increases the boiling points and melting points. For solubility, we care about the interplay between the solute and the solvent, so we also need to know the &lt;em&gt;type&lt;/em&gt; of intermolecular force (polar vs. nonpolar) in addition to the total strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the homologous series of alcohols below, I have listed their water solubilities (mined from UC Davis's &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Alcohols/Properties_of_Alcohols#Hydrogen_Bonding_and_Solubility&quot; rel=&quot;nofollow&quot;&gt;ChemWiki&lt;/a&gt; and from Wikipedia articles). Notice that as the number of $\ce{CH2}$ units increases, the water solubility decreases. This decrease occurs because, while the strength of nonpolar intermolecular forces (London-Dispersion) increases with each added $\ce{CH2}$ until the nonpolar interactions overpower the polar interactions (dipole-dipole and hydrogen-bonding) of the alcohol functional group. Water is a polar solvent, and with each added $\ce{CH2}$, the alcohol molecules begin to have stronger attraction to themselves than to water, and the solubility decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solubility of the homologous series of linear n-alkanols in water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Methanol $\ce{CH3OH}$ - miscible (infinite)&lt;/li&gt;&#xA;&lt;li&gt;Ethanol $\ce{CH3CH2OH}$ - miscible (infinite)&lt;/li&gt;&#xA;&lt;li&gt;n-propanol $\ce{CH3CH2CH2OH}$ - miscible (infinite)&lt;/li&gt;&#xA;&lt;li&gt;n-butanol $\ce{CH3CH2CH2CH2OH}$ - 80 g / L&lt;/li&gt;&#xA;&lt;li&gt;n-pentanol $\ce{CH3CH2CH2CH2CH2OH}$ - 22 g / L&lt;/li&gt;&#xA;&lt;li&gt;n-hexanol$\ce{CH3CH2CH2CH2CH2CH2OH}$ - 5.9 g / L&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At the same time, the solubility of this homologous series of alcohols increases in nonpolar solvents like hexane and benzene. Those data are not readily available, but a quick experiment will verify it. The increased amount of nonpolar interactions in the alcohols increases their affinity with nonpolar solvents.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-16T13:34:58.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4395" PostTypeId="1" AcceptedAnswerId="4396" CreationDate="2013-03-16T14:59:41.900" Score="9" ViewCount="948" Body="&lt;p&gt;I'm doing exercises on hybridisation, and I was given this molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2sGV9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering about this (electron deficient) oxygen atom. My intuition says it should be ${sp^2}$ like the answer says, but honestly I only see that it's $sp$ hybridised. Is it because you ask what it would look like if it was not electron deficient?&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="4325" LastEditDate="2015-03-01T15:14:42.897" LastActivityDate="2015-03-01T15:14:42.897" Title="Why is this atom sp² hybridised?" Tags="&lt;hybridization&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ClosedDate="2016-11-24T09:23:57.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="4396" PostTypeId="2" ParentId="4395" CreationDate="2013-03-16T17:37:46.003" Score="7" Body="&lt;p&gt;This would become much clearer if you show the lone pair on the oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair. &lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-03-16T17:37:46.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4398" PostTypeId="1" AcceptedAnswerId="4400" CreationDate="2013-03-17T05:32:06.983" Score="1" ViewCount="2955" Body="&lt;p&gt;Should chemical formula of ethanol be written as $\ce{CH3CH2OH}$ or $\ce{C2H5OH}$? And whats the logic behind it? I have seen the former being used more often, so why is ethane represented by $\ce{C2H6}$ than $\ce{CH3CH3}$?&lt;/p&gt;&#xA;" OwnerUserId="1376" LastEditorUserId="7448" LastEditDate="2015-10-16T17:09:18.620" LastActivityDate="2018-10-31T09:58:16.683" Title="Chemical formula of ethanol and other carbon compounds" Tags="&lt;organic-chemistry&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4399" PostTypeId="1" AcceptedAnswerId="4401" CreationDate="2013-03-17T05:37:32.333" Score="55" ViewCount="778386" Body="&lt;p&gt;I'm learning how to apply the VSEPR theory to Lewis structures and in my homework, I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn.&lt;/p&gt;&#xA;&lt;p&gt;I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electron regions, and figured out the shape of each molecule. I'm being asked to figure out the hybridization of the central atom of various molecules.&lt;/p&gt;&#xA;&lt;p&gt;I found a sample question with all the answers filled out:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized.&lt;/p&gt;&#xA;&lt;p&gt;Where does this come from? I understand how to figure out the standard orbitals for an atom, but I'm lost with hybridization.&lt;/p&gt;&#xA;&lt;p&gt;My textbook uses &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt; as an example.&#xA;Carbon has &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2s^2 \,2p^2}$&lt;/span&gt;, but in this molecule, it has four &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt;. I understand the purpose of four (there are four hydrogens), but where did the &amp;quot;3&amp;quot; in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; come from?&lt;/p&gt;&#xA;&lt;p&gt;How would I figure out something more complicated like &lt;span class=&quot;math-container&quot;&gt;$\ce{H2CO}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="95133" LastEditDate="2021-05-17T05:52:46.140" LastActivityDate="2021-05-17T05:52:46.140" Title="How do I figure out the hybridization of a particular atom in a molecule?" Tags="&lt;hybridization&gt;&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="33" ContentLicense="CC BY-SA 4.0" />
  <row Id="4400" PostTypeId="2" ParentId="4398" CreationDate="2013-03-17T07:34:32.913" Score="4" Body="&lt;p&gt;$\ce{CH3CH2OH}$, and $\ce{C2H5OH}$ or $\ce{EtOH}$ (Ethyl OH) tells you specifically what are the bonds in the molecule. $\ce{CH3H6O}$, which would be the equivalent of writing ethane like $\ce{C2H6}$, on the other hand, tells you nothing, since it can be both ethanol or dimethyl ether. Bigger formulas can represent thousends of different compounds if written like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why $\ce{C2H6}$ is used instead of $\ce{CH3CH3}$ is because it can only represent ethane - it's a very simple compound with no isomers, so you can write it both ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write ethanol both ways you suggested, although I've never seen $\ce{C2H5}$ instead of $\ce{Et}$ anywhere but in examples. What you shouldn't do is write it like $\ce{C2H6O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were asking what is the logic of $\ce{CH3CH2OH}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it's telling you is that the compound it is representing [ethanol] consists of a carbon atom bonded to 3 hydrogen atoms and to another carbon atom, which is bonded to 2 hydrogen atoms and 1 oxygen atom, which is finally bonded to 1 hydrogen atom, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/047Hr.png&quot; alt=&quot;CH3 CH2 OH&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="4231" LastEditDate="2015-02-26T16:28:21.593" LastActivityDate="2015-02-26T16:28:21.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4401" PostTypeId="2" ParentId="4399" CreationDate="2013-03-17T10:39:15.370" Score="45" Body="&lt;p&gt;If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.&lt;/li&gt;&#xA;&lt;li&gt;Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane. &lt;/li&gt;&#xA;&lt;li&gt;Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.&lt;/li&gt;&#xA;&lt;li&gt;Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals.&lt;/li&gt;&#xA;&lt;li&gt;Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NH3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For $\ce{H2CO}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement). &lt;/li&gt;&#xA;&lt;li&gt;Determine the number of electron domains on the central atom.&lt;/li&gt;&#xA;&lt;li&gt;Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization. &lt;/li&gt;&#xA;&lt;li&gt;Practice until you can do this quickly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-20T06:58:22.927" LastActivityDate="2018-03-20T06:58:22.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4403" PostTypeId="1" CreationDate="2013-03-17T17:43:41.847" Score="2" ViewCount="2749" Body="&lt;p&gt;The melting and boiling point of Lithium as compared to other members of the Alkali group metals is high. But, Lithium does not occur in the free state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what does it mean that it has high melting and boiling point? How are its BP and MP measured if it isn't in the free state?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="22" LastEditDate="2013-03-17T19:26:04.777" LastActivityDate="2016-06-06T19:39:54.023" Title="Boiling and Melting point of Lithium (or any other element that does not occure in free state)" Tags="&lt;physical-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4404" PostTypeId="1" CreationDate="2013-03-17T17:48:48.760" Score="0" ViewCount="161" Body="&lt;p&gt;How can I reconcile all of the chemical properties, physical properties and peculiar behaviour of metals in inorganic chemistry for the purposes of studying these systematically? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very hard to learn all of these for each group.&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="73" LastEditDate="2013-10-20T12:07:09.913" LastActivityDate="2014-08-06T20:02:40.687" Title="Reconciling multiple inorganic properties for the purposes of learning these elements" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4405" PostTypeId="2" ParentId="4399" CreationDate="2013-03-17T17:55:45.563" Score="21" Body="&lt;p&gt;You can find the hybridization of an atom by finding its steric number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;steric number&lt;/strong&gt; = the number of atoms bonded to the atom + the number of lone pairs the atom has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steric number is 4, the atom is $\mathrm{sp^3}$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steric number is 3, the atom is $\mathrm{sp^2}$  hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steric number is 2, the atom is $\mathrm{sp}$  hybridized.&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="15410" LastEditDate="2016-05-09T01:55:18.273" LastActivityDate="2016-05-09T01:55:18.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4406" PostTypeId="2" ParentId="4395" CreationDate="2013-03-17T18:01:02.587" Score="3" Body="&lt;p&gt;You can find the hybridization of the atom by finding its steric number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;steric number = no of atoms bonded (to the atom you are finding the hyb. of) + lone pairs with that atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if Steric no comes to be 4, there is $\ce{sp^3}$ hybridization in the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if steric no comes to be 3, there is $\ce{sp^2}$ hybridization in the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if steric no comes to be 2, there is $\ce{sp}$ hybridization in the atom.&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="4231" LastEditDate="2015-03-01T14:35:22.810" LastActivityDate="2015-03-01T14:35:22.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4407" PostTypeId="2" ParentId="4403" CreationDate="2013-03-17T19:32:30.470" Score="8" Body="&lt;p&gt;Your premise here is wrong, pure metal Lithium exists and can be handled without much difficulty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, materials which &lt;em&gt;are&lt;/em&gt; extremely reactive can always be kept in an inert gas when you want to measure their physical properties. Of course, &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2013/01/09/things_i_wont_work_with_azidoazide_azides_more_or_less&quot; rel=&quot;nofollow&quot;&gt;there are compounds which can don't need an excuse to explode&lt;/a&gt; (and  will explode in a vacuum/inert gas as well), in which case it is harder to find out BP/MPs (one can put them at low pressure and then extrapolate, but that's not as accurate).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="11367" LastEditDate="2016-06-06T19:39:54.023" LastActivityDate="2016-06-06T19:39:54.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4409" PostTypeId="1" AcceptedAnswerId="4415" CreationDate="2013-03-17T19:50:45.920" Score="4" ViewCount="285" Body="&lt;p&gt;Will borosilicate glass absorb 808nm laser light? If it instead refracts, then what is the refractive index? Also, will borosilicate reflect lower frequencies of light like those used in non contact temp. measurements?&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastEditorUserId="23561" LastEditDate="2019-04-09T02:02:19.347" LastActivityDate="2019-04-09T16:29:06.917" Title="Will borosilicate glass absorb 808nm laser light?" Tags="&lt;equipment&gt;&lt;glass&gt;&lt;light&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4410" PostTypeId="1" AcceptedAnswerId="4419" CreationDate="2013-03-18T05:34:55.297" Score="5" ViewCount="263" Body="&lt;p&gt;I was wondering if there was a difference between $K_{eq}$ and $K_c$. I think they both refer to the Equilibrium Constant. If I'm wrong, could you please tell me the difference between the two? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="636" LastEditorUserId="22" LastEditDate="2013-03-18T08:10:39.787" LastActivityDate="2013-03-18T13:22:01.997" Title="$K_{eq} $ vs $K_c$" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4411" PostTypeId="1" CreationDate="2013-03-18T06:01:59.213" Score="10" ViewCount="36789" Body="&lt;p&gt;Is astatine a non-metal or a metalloid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metalloid is an element which exhibits some of the properties of a metal as well as those of a non metal.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="16683" LastEditDate="2017-07-26T12:48:33.527" LastActivityDate="2017-07-26T12:48:33.527" Title="Is astatine a non-metal or a metalloid?" Tags="&lt;elements&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4412" PostTypeId="5" CreationDate="2013-03-18T07:51:44.853" Score="0" Body="&lt;p&gt;Electrochemistry is the scientific discipline studying the relationship between chemistry and electricity. It includes all technologies and techniques such as work on electrolysis, corrosion, batteries, fuel cells, accumulators, and electroplating.&lt;/p&gt;&#xA;&lt;p&gt;In general, electrochemistry studies the relationship between chemical reactions and electricity. Electricity is the movement of electrons moving together in one direction. The chemical reaction that is happening in an electrochemical reaction will make electrons move from one side to another. These reactions are called oxidation-reduction reactions, meaning that electrons move between atoms. Some atoms are being oxidized, meaning that they lose an electron, and other atoms are being reduced, meaning that they gain an electron.&lt;/p&gt;&#xA;" OwnerUserId="102302" LastEditorUserId="102302" LastEditDate="2021-04-21T22:09:34.080" LastActivityDate="2021-04-21T22:09:34.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4413" PostTypeId="4" CreationDate="2013-03-18T07:51:44.853" Score="0" Body="The branch of chemistry that deals with the study of redox reactions and how they can be applied to generate electricity (in electrochemical cells) and to carry out non-spontaneous reactions using electricity (electrolysis)." OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2013-03-18T07:58:43.993" LastActivityDate="2013-03-18T07:58:43.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4414" PostTypeId="1" AcceptedAnswerId="4434" CreationDate="2013-03-18T07:58:20.147" Score="2" ViewCount="121" Body="&lt;p&gt;This is a screenshot of  a portion of a metal organic framework (MOF).  These are two oxygens of two &lt;a href=&quot;http://en.wikipedia.org/wiki/N-Methyl-2-pyrrolidone&quot; rel=&quot;nofollow noreferrer&quot;&gt;NMP&lt;/a&gt; solvent molecules.  What I am wondering is why they are so close and overlapping like this.  I checked the CIF file, and occupancy for all atoms in this structure is 1, so it doesn't seem to be the case that there is only one NMP molecule in two equally likely positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain this?  Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JRGkm.png&quot; alt=&quot;Overlapping oxygens&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="130" LastActivityDate="2013-03-20T13:10:25.967" Title="Overlapping oxygens in crystal structure -- what's going on?" Tags="&lt;crystal-structure&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4415" PostTypeId="2" ParentId="4409" CreationDate="2013-03-18T09:27:23.430" Score="6" Body="&lt;p&gt;Borosilicate glasses are a family that has many members with differing properties. You can look up the provider for your glass, and check whether it is transparent in the near infrared. I think it would be in most cases (see &lt;a href=&quot;http://www.edmundoptics.com/optics/windows-diffusers/ultraviolet-uv-infrared-ir-windows/borofloat-borosilicate-windows/1913&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for one example).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-03-18T09:27:23.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4416" PostTypeId="1" AcceptedAnswerId="4418" CreationDate="2013-03-18T09:29:04.853" Score="14" ViewCount="52723" Body="&lt;p&gt;Ketones are relatively much harder to oxidise, but they do undergo oxidation reactions at extreme temperatures. Our teacher also told us about Popoff's rule which states that, on oxidation of unsymmetrical ketones, the $\ce{&amp;gt;C=O}$ group remains with the smaller alkyl group. Could you please give me a hint as to how this happens, and the probable reaction mechanism for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please give the complete balanced equation when butan-2-one is oxidized to a carboxylic acid.&lt;/p&gt;&#xA;" OwnerUserId="487" LastEditorUserId="5026" LastEditDate="2018-04-02T16:07:51.170" LastActivityDate="2018-04-02T16:11:36.083" Title="What is the mechanism for oxidation of ketones by the Popoff's rule?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;organic-oxidation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="4417" PostTypeId="2" ParentId="4411" CreationDate="2013-03-18T12:10:38.310" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is Astatine a non metal or a metalloid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unknown&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Astatine&quot;&gt;astatine&lt;/a&gt;, the longest-lived isotope has a half-life of 8.1 hours. The element is so radioactive that large samples would melt and vaporize under the heat generated by its own radioactivity. Such behavior makes it difficult to assess whether astatine has the appropriate traits to be a metalloid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metalloid_%28comparison_of_properties_with_those_of_metals_and_nonmetals%29&quot;&gt;Metalloids&lt;/a&gt; tend to be lustrous, brittle, semiconductors with other properties intermediate between metals and nonmetals. Nonmetals are dull/matte, brittle, and poor conductors of electricity and heat. All of these properties require a visible solid (or liquid) sample to assess, and thus astatine is impossible to assess and classify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astatine is usually placed with the metalloids or the nonmetals by different authors based on its position in periodic table, but in truth, astatine is so unstable in its elemental form, we have no idea which one it is.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-18T12:10:38.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4418" PostTypeId="2" ParentId="4416" CreationDate="2013-03-18T12:38:11.020" Score="25" Body="&lt;p&gt;I am uncertain about the &quot;Popoff's&quot; rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&quot; rel=&quot;noreferrer&quot;&gt;Baeyer-Villager oxidation&lt;/a&gt; is an oxidation of ketones to esters using a peracid in the presence of a mild base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COCH3 + RCO3H-&amp;gt;[\ce{Na2HPO4}] CH3CO-OCH3 + RCOOH}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism involves the fragmentation of one of the $\ce{C-CO}$ bonds. Unsymmetrical ketones fragment in a predictable pattern, but not always that the carbonyl remains with the smaller group. The fragment that would be a more stable carbocation (even though carbocations are not formed in this reaction) is the one to move. The migratory aptitude is tertiary alkyl &gt; cyclohexyl &gt; secondary alkyl, aryl &gt; H &gt; primary alkyl &gt; methyl . For exampl, with 2-methyl-3-octanone, the isopropyl group moves because it is attached via a $2^\circ$ carbon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)2CHCOCH2CH2CH2CH2CH3 + RCO-O-OH -&amp;gt;[\ce{Na2HPO4}] (CH3)2CH{\bf O}COCH2CH2CH2CH2CH3 + RCOOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid chews ketones apart into two carboxylic acids. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kmno4#Organic_chemistry&quot; rel=&quot;noreferrer&quot;&gt;Concentrated $\ce{KMnO4}$ in acid&lt;/a&gt; also does this. These reactions are a little bit harder to find information on, since they tend to be considered uncontrollable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction goes through a series of oxidations from ketone to $\alpha$-hydroxyketone to cleaved acids through several enol intermediates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{RCH2COCH2R &amp;lt;=&amp;gt;[\ce{HNO3}] RCH=C(OH)CH2R -&amp;gt;[\ce{HNO3}] RCH(OH)COCH2R}$$&#xA;$$\ce{RCH(OH)COCH2R &amp;lt;=&amp;gt;[\ce{HNO3}] RC(OH)=C(OH)CH2R -&amp;gt;[\ce{HNO3}] RCO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ketone is unsymmetrical, there is no guarantee that is will cleave predictably. For example, 2-butanone could cleave into propanoic acid and carbon dioxide or two equivalents of acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2COCH3 -&amp;gt;[\ce{HNO3}] n(CH3CH2COOH + CO2) + (1-n)(2CH3COOH)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A balanced equation for the formation of acetic acid would look like:&#xA;$$\ce{CH3CH2COCH3 + 3NO3^- +  -&amp;gt; 2 CH3CO2H + 3NO2^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A balanced equation for the formation of propanoic acid and carbon dioxide would look like:&#xA;$$\ce{CH3CH2COCH3 + 4NO3^- +  -&amp;gt; CH3CH2COOH + CO2 + 4NO2^- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The permanganate reactions are tougher to balance, since permanganate is a three electron oxidant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I wrote this answer, I had never heard of Popoff's rule. No textbook I own mentions this rule, and a Google search about it brings up this question as the top hit (and similar questions at other sites like ask.yahoo.com as the other hits). I now know that Aleksandr Popov published a &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jlac.18721620112/abstract;jsessionid=DFFB556A4039E246998CD60367611F5E.f01t04?systemMessage=Wiley%20Online%20Library%20will%20be%20disrupted%20Saturday,%2015%20March%20from%2010:00-12:00%20GMT%20%2806:00-08:00%20EDT%29%20for%20essential%20maintenance&quot; rel=&quot;noreferrer&quot;&gt;paper in Liebigs Annalen&lt;/a&gt; in 1872 describing a variation of the reaction that would become the Baeyer-Village oxidation. This article is behind a paywall for me, and the first page preview confirms I would not be able to make much of it. I only know a small amount of German and the PDF sadly looks like a low quality copy of a copy of a copy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from the title &quot;Die Oxydation der Ketone als Mittel zur Bestimmung der Constitution der fetten Säuren und der Alkohole&quot; I can parse what the paper was about. Roughly this paper was about &quot;The oxidization of ketones - a means for determining the constitution of the fatty acids and alcohols&quot;. Thus, his method involved the oxidation to the ester and hydrolysis of said ester in one step. I don't know what reagents Popov used, but I'm pretty sure it was not a peroxyacid (since this reagent will not hydrolyze the ester). I have &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6269/bayer-villiger-reaction-migratory-aptitude/6272#6272&quot;&gt;previously answered a question&lt;/a&gt; about migratory aptitude in Baeyer-Villager reactions. Since I cannot read Popov's paper, nor can I find any authoritative resource on his rule (on- or offline), I have to assume that the known migratory aptitude for the Baeyer-Village reaction is the same as Popov's rule. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditorDisplayName="user1395" LastEditDate="2018-04-02T16:11:36.083" LastActivityDate="2018-04-02T16:11:36.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4419" PostTypeId="2" ParentId="4410" CreationDate="2013-03-18T13:22:01.997" Score="7" Body="&lt;p&gt;$K_{eq}$ is the generic equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_c$ is the equilibrium constant for a reaction explicitly in solution, for which the mass action expression can be written using concentrations only.&#xA;$$\ce{aA + bB &amp;lt;=&amp;gt; cC + dD}$$&#xA;$$K_c = Q_c (\text{at equilibrium}) = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_p$ is the equilibrium constant for a reaction explicitly in the gas state, for which the mass action expression can be written using partial pressures only:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p = Q_p (\text{at equilibrium}) = \frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is the generalizd equilibrium constant, and is otherwise used when reactions are in mixed states. $K$ should perhaps be written with mole fractions $\chi_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = Q (\text{at equilibrium}) = \frac{\chi_{\ce{C}}^c \chi^d_{\ce{D}}}{\chi^a_{\ce{A}} \chi^b_{\ce{B}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since concentration ($[i]=\frac{n_i}{V}$) is related to pressure from the ideal gas law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n}{V}=\frac{P}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_p$ amd $K_c$ are related by powers of $RT$, depending on the stoichiometric coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}=\frac{(\frac{P_{\ce{C}}}{RT})^c (\frac{P_{\ce{D}}}{RT})^d}{(\frac{P_{\ce{A}}}{RT})^a (\frac{P_{\ce{B}}}{RT})^b}= \frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}} (RT)^{a+b-c-d}=K_p (RT)^{a+b-c-d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar relations exist between mole fraction ($\chi_i=\frac{n_i}{n_T}$) and concentration ($\chi_i=\frac{c_i}{c_T}$) or pressure ($\chi_i=\frac{P_i}{P_T}$), and $K_c$ or $K_p$ can similarly be converted to $K$ or vice versa.&#xA;$$K= \frac{\chi_{\ce{C}}^c \chi^d_{\ce{D}}}{\chi^a_{\ce{A}} \chi^b_{\ce{B}}} = \frac{(\frac{P_{\ce{C}}}{P_T})^c (\frac{P_{\ce{D}}}{P_T})^d}{(\frac{P_{\ce{A}}}{P_T})^a (\frac{P_{\ce{B}}}{P_T})^b} =\frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}} (P_T)^{a+b-c-d}=K_p (P_T)^{a+b-c-d}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-18T13:22:01.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4421" PostTypeId="1" AcceptedAnswerId="4425" CreationDate="2013-03-18T20:10:12.217" Score="8" ViewCount="5681" Body="&lt;p&gt;Titanium is expensive, although it is not rare and titanium dioxide is a ubiquitous component, it's part of such mundane articles as toothpaste, paper or plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for that is that the processes that extract titanium from titanium dioxide (which occurs naturally and just needs some inexpensive purification) are quite costly. The established industrial method for this is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kroll_process&quot;&gt;Kroll process&lt;/a&gt;. It works by first chlorinating the titanium to $TiCl_4$ and then moving the chlorine over to some $Mg$, yielding $MgCl_2$ and pure $Ti$. Then follows a costly remelting process under vacuum to further purify the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recycling titanium scrap is even harder, and removing the other components from the alloys is not entirely possible, so recycling only yields titanium based alloys, not the pure stuff (which is not necessarily a problem since it's very rarely used in its pure form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What specific chemical properties of titanium make it so hard to purify? For example, the chlorine process at the beginning of the Kroll process takes place with coke in a fluidized bed reactor, very similar to the reduction of $Fe$ in a blast furnace, but if the carbon takes care of the reduction, why is that not sufficient? Why is the chlorine route needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research into titanium production dates back to the late 1880s, and the Kroll process, patented in 1940, made it feasible for mass production, but remains costly and has not been improved upon since, despite much effort. How would a chemist look for suitable reactants during such research?&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2013-12-04T14:07:26.080" Title="Why is it hard to reduce titanium dioxide?" Tags="&lt;redox&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4422" PostTypeId="1" CreationDate="2013-03-18T20:56:35.170" Score="-4" ViewCount="3559" Body="&lt;p&gt;I was trying to know what is the probability for the carbon atoms to rotate .I know that in single bond it is 0 % and I think it is 100 % in double bond but   how about the triple bond ?&lt;/p&gt;&#xA;" OwnerUserId="1380" LastEditorUserId="7448" LastEditDate="2015-11-15T15:17:24.037" LastActivityDate="2015-11-15T15:17:24.037" Title="Carbon atoms in triple bonds" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-23T18:27:15.860" ContentLicense="CC BY-SA 3.0" />
  <row Id="4424" PostTypeId="2" ParentId="4422" CreationDate="2013-03-19T12:27:28.237" Score="12" Body="&lt;p&gt;Actually, it's the opposite. Most Carbon-Carbon single bonds can rotate without much difficulty. Ethane, for example, has a (potential energy of conformer vs dihedral angle) graph like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SqZcb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy barrier isn't that much here, and ethane probably rotates without much effort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propane and all will have higher energy barriers, and will rotate less often. Even cyclic compounds like cyclohexane undergo &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_flip&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring flipping&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, not all single bonds can easily rotate a full turn. Systems like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yAiWQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have a hard time rotating about the central bond due to steric repulsion of the groups on the biphenyl system (which makes the two compounds shown above noninterchangeable). These systems can only rotate a small angle, and rarely go 360&amp;deg; (it may be possible to force this to happen by providing the system with energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, there are systems where it's obvious that there cannot be much rotation. For example, in bicyclo[2.2.2]octane, most bonds will rotate very little, and it is impossible for any bond to rotate a full 360&amp;deg; without breaking any bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qbqdY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, single bonds can rotate because they are &lt;a href=&quot;http://en.wikipedia.org/wiki/Sigma_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;sigma bonds&lt;/a&gt;. A sigma bond is formed by overlap of one lobe of an orbital from each of the two atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9YQPh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is ethane)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing preventing the carbons from rotating here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, double/triple bonds have pi bonds (in which two lobes each overlap) as well as sigma bonds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZAFmR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is ethene)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, it is impossible to rotate without breaking the pi bond. Note that this has some importance in how our eyes see -- one of the mechanisms in eyesight is a double bond in rhodopsin being broken (by incident light) into a single bond, rotating, and reforming (which leads to some nerve impulse).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A way to  visualise this is by holding a pencil between your fists. You can easily rotate your hands. Now, hold three pencils (a bit spaced out) between your fists -- you will no longer be able to rotate your hands.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, to answer your question: The probability/average time depends on the compound and the surroundings. For double/triple bonds, it's more or less 0%.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-03-19T12:27:28.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4425" PostTypeId="2" ParentId="4421" CreationDate="2013-03-19T16:09:22.937" Score="6" Body="&lt;p&gt;Reasons why Titanium is hard to extract and purify:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. &lt;strong&gt;For example&lt;/strong&gt;: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider these reactions which could take place in the fluidized bed reactor:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -&amp;gt; TiCl_4(g) + 2CO(g)}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Ti(s) + C(s) -&amp;gt; TiC(s)}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.&lt;br&gt;&#xA;And the equation for change in Gibbs' Energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta{G} = \Delta{H} - T\Delta{S}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}&amp;gt;&amp;gt;\Delta{H}$ so we can neglect $\Delta{H}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Notes: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-03-19T16:09:22.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4427" PostTypeId="2" ParentId="4354" CreationDate="2013-03-19T19:07:28.667" Score="1" Body="&lt;p&gt;Methyl isobutyl ketone is the solvent for PMMA - the alcohol is just the carrier. MIBK has low solubility in water, so if you stopped with water, you would probably leave more MIBK residue in your PMMA.&lt;/p&gt;&#xA;" OwnerUserId="1365" LastActivityDate="2013-03-19T19:07:28.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4428" PostTypeId="2" ParentId="2902" CreationDate="2013-03-20T11:00:10.313" Score="2" Body="&lt;p&gt;Density stratification will only work if the substances are immiscible (mutually insoluble). For example, water and hexane could be separated this way because they are immiscible. Water will settle to the bottom (density = 1.0 g/mL) and hexane will float on top (0.65 g/mL). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) cannot be separated this way because they are miscible with each other. They form a homogenous solution when mixed. One of the properties of homogeneous solutions is that they do not separate into their components without a transfer of heat to or from the solution. You do not see bottles of soda separating into their components on the shelf because they are homogenous mixtures. You do see bottles of salad dressing separate because they are not homogeneous mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the various monochlorinated, dichlorinated, trichlorinated and so forth derivatives of butane have different densities, they are likely all miscible in each other, which means the homogeneous mixture will not separate by density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for an alternative to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_distillation&quot; rel=&quot;nofollow&quot;&gt;fractional distillation&lt;/a&gt;, you could try &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_freezing&quot; rel=&quot;nofollow&quot;&gt;fractional freezing&lt;/a&gt;, in which you cool down the mixture, and the substances crystallize out according to the order of their freezing points. To achieve high purity, you would need to repeat the procedure several times for each substance. A good fractional distillation apparatus already has the repeated distillations built in, so it would probably be the fastest technique.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-03-20T11:00:10.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4429" PostTypeId="1" AcceptedAnswerId="4435" CreationDate="2013-03-20T11:26:57.917" Score="7" ViewCount="38215" Body="&lt;p&gt;Why are intramolecular hydrogen bonds weaker than intermolecular hydrogen bonds?&lt;/p&gt;&#xA;" OwnerUserId="1378" LastEditorUserId="402" LastEditDate="2016-11-16T00:14:43.207" LastActivityDate="2018-04-05T19:32:41.057" Title="Strength of intramolecular vs intermolecular hydrogen bonds" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4430" PostTypeId="5" CreationDate="2013-03-20T12:56:22.680" Score="0" Body="&lt;p&gt;The periodic table is a systematic arrangement of the known chemical elements in order of atomic number. The elements are arranged in groups on the basis of similarity in chemical and physical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fMhFd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fMhFd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is the standard periodic table used by most scientists around the world. However, alternatives exist, ranging from useful ones which order the elements based on electron configuration, to artistic designs which are purely aesthetic in their appeal.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-09-18T20:36:37.697" LastActivityDate="2015-09-18T20:36:37.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4431" PostTypeId="4" CreationDate="2013-03-20T12:56:22.680" Score="0" Body="This tag should be applied to questions about the layout, history and interpretation of the periodic table, not to questions relating to specific elements or trends within the table. For these, the tags 'elements' or 'periodic-trends' should be used where appropriate." OwnerUserId="1322" LastEditorUserId="10185" LastEditDate="2015-09-18T20:36:33.907" LastActivityDate="2015-09-18T20:36:33.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4432" PostTypeId="5" CreationDate="2013-03-20T13:06:14.953" Score="0" Body="&lt;p&gt;It is an exothermic process by which a substance is oxidized (by an oxidizing agent such as oxygen) to release a large amount of energy in the form of heat and often light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion is a reaction with complicated radical chain mechanisms, depending on substrates - combustible substance and oxidiser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustible materials include most of the organic compounds (for example petrol and ethanol) and many inorganic ones (like ammonia and boranes), as also many elementary substances, like carbon, sulfur or more reactive metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is most popular oxidizer for combustion, but also chlorine, fluorine, and compounds of these elements can be used.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="5026" LastEditDate="2018-03-19T14:43:02.090" LastActivityDate="2018-03-19T14:43:02.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4433" PostTypeId="4" CreationDate="2013-03-20T13:06:14.953" Score="0" Body="For questions about the combustion of an organic molecule or inorganic substance. Do not use this tag if your question is about the [stoichiometry] of a combustion reaction, or for the calculations &#xD;&#xA;of [thermodynamics] functions (like enthalpy change)." OwnerUserId="1322" LastEditorUserId="5026" LastEditDate="2018-03-19T14:42:58.053" LastActivityDate="2018-03-19T14:42:58.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4434" PostTypeId="2" ParentId="4414" CreationDate="2013-03-20T13:10:25.967" Score="3" Body="&lt;p&gt;It's a bit hard to answer without having coordinates file (*.res or *.xyz), but the simplest answer that I would expect is that these oxygen atoms belong to two different molecules that are disordered over the same position (i.e. they have only partial occupation on the site that in sum is 1). It's very common for &lt;em&gt;f.e.&lt;/em&gt; thf coordinated to alkali-metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have shelx input for the structure look for the line that defines one of oxygens and see if 6th parameter is 11.0&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O1    5   -0.180348    0.245831    0.298215    11.00000    0.02208    0.02526 = &#xA;                                               ^^  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This means fixed (10) occupancy of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have cif-file look for the atom loop &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;loop_ &#xA; _atom_site_label &#xA; _atom_site_type_symbol &#xA; _atom_site_fract_x &#xA; _atom_site_fract_y &#xA; _atom_site_fract_z &#xA; _atom_site_U_iso_or_equiv &#xA; _atom_site_adp_type &#xA; _atom_site_occupancy &#xA; _atom_site_symmetry_multiplicity &#xA; _atom_site_calc_flag &#xA; _atom_site_refinement_flags &#xA; _atom_site_disorder_assembly &#xA; _atom_site_disorder_group &#xA;O1 O -0.18035(6) 0.24583(6) 0.29822(5) 0.02794(8) Uani 1 1 d . . .  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;8th parameter is occupancy (_atom_site_occupancy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These both examples were for not disordered atoms. If they are disordered then it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shelx-ins/res&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O1    4    0.297097   -0.198418   -0.064164    10.50000    0.34896    0.34935 =&#xA;                                                  ^ occup. 1/2  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;FVAR       0.22777   0.51340   0.51266  &#xA;...&#xA;O1    4    0.551971    0.536866    0.295322    21.00000    0.03100    0.03031 =&#xA;                                                ^ occup. bound to 2nd free variable&#xA;                                                  here 51%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;cif-file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O1_20 O 0.6673(9) 0.3140(7) 0.1724(3) 0.0434(14) Uani 0.513(3) 1 d PDU C 1&#xA;                                                      ^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-03-20T13:10:25.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4435" PostTypeId="2" ParentId="4429" CreationDate="2013-03-20T13:36:18.200" Score="3" Body="&lt;p&gt;I don't think there is much of a difference between the strengths of intramolecular or intermolecular hydrogen bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the strength of interactions &lt;strong&gt;between&lt;/strong&gt; molecules depends on the type of hydrogen bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a typical example: ortho-Nitrophenol has a lower melting and boiling point than para-Nitrophenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/40hC0.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interactions &lt;strong&gt;between&lt;/strong&gt; molecules are much stronger when there are &lt;strong&gt;intermolecular&lt;/strong&gt; hydrogen bonds (as in para-Nitrophenol) because the bonds are formed between molecules. This results in more stability and lower melting and boiling points (which might appear to be because of stronger bonds) because the molecules prefer to be closer to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;intramolecular&lt;/strong&gt; hydrogen bonds are present (as in the case of ortho-Nitrophenol) the interactions &lt;strong&gt;within&lt;/strong&gt; individual molecules are greater and so there is less attraction between molecules. This causes less stability and higher melting and boiling points because the molecules are not really attracted to each other much and not because the hydrogen bonds are weaker.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-03-20T13:36:18.200" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4436" PostTypeId="5" CreationDate="2013-03-20T16:11:25.427" Score="0" Body="&lt;p&gt;Coordination compounds are those that contain a dative or coordinative bond, usually formed by one atom donating a free lone pair to another, electron-deficient atom. In most cases, the electron deficient atom — a Lewis acid — is a (transition) metal atom while the donating atom is often a non-metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every metal can more or less be considered to form coordination compounds in solution; for some, e.g. zinc, these are poorly defined and easily transition into each other. For others, e.g. aluminum or chromium, the ligand sphere is very stable and ligand exchanges require substantial activation energy. This is usually due to d-orbital splitting observed in complexes: While for a lone metal ion in a vacuum all five d-orbitals are degenerate if a set of ligands approaches said ion, the orbitals split up into sets depending on where the ligands approach from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large number of complexes is octahedral in structure. For those, the $\mathrm d_{xy}$, $\mathrm d_{xz}$ and $\mathrm d_{yz}$ (usually termed $\mathrm{t_{2g}}$ due to the way they transform in $O_\mathrm{h}$) remain unaltered at first approximation while the $\mathrm{e_g}$ orbitals $\mathrm d_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ — which point towards the ligands — are destabilised. This simple approximation is called &lt;em&gt;crystal field approximation.&lt;/em&gt; For tetrahedral complexes, the $\mathrm{t_1}$ orbitals are destabilised and $\mathrm{e}$ orbitals slightly stabilised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple crystal field model provides no quantitative approximation for the energy difference between the orbitals and thus for the light wavelength absorbed. If the energy difference is large enough, it may be thermodynamically favourable to keep all spins paired up in the lower energy levels generating a &lt;em&gt;low-spin complex.&lt;/em&gt; If not, a &lt;em&gt;high-spin complex&lt;/em&gt; is formed where the first five d-electrons are always filled in with parallel spins. Useful starting points for deciding whether a complex is high-spin or low spin is the &lt;em&gt;spectrochemical series&lt;/em&gt; which orders ligands by the field split they typically cause. Halides are one extreme, usually termed &lt;em&gt;weak field&lt;/em&gt; ligands since they almost always form high-spin complexes. Oxygen and nitrogen donors are generally medium field. The &lt;em&gt;strong field&lt;/em&gt; end of the spectrum is formed by carbonyl and cyanide ligands which often exclusively form low-spin complexes such as $\ce{[Fe(CN)6]^4-}$. Note that the specialised tag &lt;a href=&quot;/questions/tagged/carbonyl-complexes&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;carbonyl-complexes&amp;#39;&quot; rel=&quot;tag&quot;&gt;carbonyl-complexes&lt;/a&gt; exists for complexes containing entirely or predominantly $\ce{CO}$ as ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, complexes are not perfect octahedra but distorted. This is often due to the &lt;em&gt;Jahn-Teller effect:&lt;/em&gt; If the ligands on the $z$ axis are moved away slightly, all orbitals with a $z$ component are stabilised thus generating discreet energy levels rather than degenerate ones allowing for stabilisation of the ion’s state. Square planar complexes, which often occur for $\mathrm{d^8}$ systems, can be thought of as extreme Jahn-Teller cases where a pair of ligands has been completely removed from the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many advanced organic reactions rely on catalysis by transition metal complexes, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;palladium-catalysed cross-couplings&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;metathesis reactions (Grubbs and Shrock; alkene and alkyne metathesis)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;homogenous hydrogenations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the Wacker process&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and many more&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions on the catalysts may be tagged with &lt;a href=&quot;/questions/tagged/coordination-compounds&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;coordination-compounds&amp;#39;&quot; rel=&quot;tag&quot;&gt;coordination-compounds&lt;/a&gt; if the structure or specific properties of the catalytic complex are part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many essential enzymes such as cytochrome C oxidase also have a catalytic metal domain and many other proteins also contain metal centres such as haemoglobin. If the properties of said metal centre are the focus of the question, this tag should be used.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="402" LastEditDate="2017-04-09T16:22:22.803" LastActivityDate="2017-04-09T16:22:22.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4437" PostTypeId="4" CreationDate="2013-03-20T16:11:25.427" Score="0" Body="This tag is for questions concerning coordination compounds including but not limited to ligand properties, metal properties, orbital splitting, micro- and macroscopic properties of entire complexes etc. For complexes where carbon monoxide is the only ligand, use the [carbonyl-complexes] tag instead. For organic catalysts or proteins, the tag is applicable if the question is about the metal’s direct coordination sphere." OwnerUserId="1322" LastEditorUserId="15751" LastEditDate="2020-03-02T10:38:59.923" LastActivityDate="2020-03-02T10:38:59.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4438" PostTypeId="1" AcceptedAnswerId="4444" CreationDate="2013-03-20T18:21:33.270" Score="7" ViewCount="297" Body="&lt;p&gt;What method was used to solve the quantum three-body problem?  I heard it was solved and was not sure about the status of the problem.&lt;/p&gt;&#xA;" OwnerUserId="1386" LastEditorUserId="16683" LastEditDate="2016-11-12T20:01:52.123" LastActivityDate="2016-11-12T20:01:52.123" Title="How was the quantum three-body problem solved?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4439" PostTypeId="1" AcceptedAnswerId="4449" CreationDate="2013-03-20T20:09:16.690" Score="8" ViewCount="44216" Body="&lt;p&gt;A question in my general chemistry textbook gives me a list of formulas for molecules and ions. It then asks me to determine, using molecular orbital theory, whether or not each would exist in a relatively stable form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I tell this? The chapter covers molecular shapes, valence bond theory, and molecular orbital theory which all builds on what was taught in the previous chapter (mainly on Lewis structures.) I can't tell what concepts to use in telling what will be stable and what won't.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2016-03-01T01:26:58.577" LastActivityDate="2017-12-14T13:48:28.010" Title="Molecular orbital theory &amp; predicting the stability of a molecule?" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4440" PostTypeId="1" AcceptedAnswerId="4464" CreationDate="2013-03-20T22:20:14.953" Score="18" ViewCount="16371" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{NaOH &amp;amp;-&amp;gt; Na+ + OH-} &amp;amp; \Delta H^\circ &amp;amp;= \pu{-44.51 kJ/mol}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="743" LastEditorUserId="4945" LastEditDate="2019-10-30T09:54:08.233" LastActivityDate="2019-10-30T09:59:03.873" Title="Why is heating increasing the solubility of sodium hydroxide?" Tags="&lt;solubility&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="4441" PostTypeId="2" ParentId="4440" CreationDate="2013-03-20T23:38:03.430" Score="-2" Body="&lt;p&gt;According to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = -RT \ln K$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K is dependent on Temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle is explicitly a mass balance issue, and thus is not affected by temperature, although that distinction may be pedantic.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2013-03-21T23:51:44.300" LastActivityDate="2013-03-21T23:51:44.300" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4442" PostTypeId="2" ParentId="4360" CreationDate="2013-03-20T23:43:48.013" Score="5" Body="&lt;p&gt;Ben Norris's comment is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An estimate of the atomic energies for orbitals may be gleaned from ionization energies, for example here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://periodictable.com/Properties/A/IonizationEnergies.html&quot;&gt;http://periodictable.com/Properties/A/IonizationEnergies.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the energies need to be calculated quantum mechanically.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-03-20T23:43:48.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4443" PostTypeId="2" ParentId="4422" CreationDate="2013-03-20T23:46:30.227" Score="5" Body="&lt;p&gt;Manish's Answer is appropriate for single and double bonds, however:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotation about C-C triple bonds is a nonsensical concept.  There is no observable change in geometry.  It is impossible to determine whether or not any rotation happens.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-03-20T23:46:30.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4444" PostTypeId="2" ParentId="4438" CreationDate="2013-03-20T23:50:07.760" Score="7" Body="&lt;p&gt;It never was solved, analytically (which means an exact mathematical solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was solved approximately, using perturbation theories.  In the case of the helium atom, an estimate of the energy electron-electron repulsion is added to the calculation.  The appropriateness of the perturbation added is based on the comparison of the result to experimentally determined values.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="16683" LastEditDate="2016-11-12T20:01:44.003" LastActivityDate="2016-11-12T20:01:44.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4445" PostTypeId="1" CreationDate="2013-03-21T00:28:57.677" Score="2" ViewCount="2210" Body="&lt;p&gt;What would happen if air heated to 1200 degree's Fahrenheit in a sealed environment was injected with colder air? &#xA;No specific pressure defined yet just a general idea of potential interaction between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;My apologies, I meant within a sealed environment like an engine cylinder. I know that the hot air would naturally expand and push a piston but how would the injection of cold air alter this, if at all?&lt;br&gt;&#xA;For example purposes we can assume the cylinder is iron while the piston is aluminum which is common design today. &lt;/p&gt;&#xA;" OwnerUserId="1387" LastEditorUserId="4945" LastEditDate="2014-07-29T10:04:26.063" LastActivityDate="2015-09-16T15:26:33.510" Title="Hot and cold air interaction in a sealed environment" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4446" PostTypeId="1" CreationDate="2013-03-21T00:40:19.650" Score="2" ViewCount="23215" Body="&lt;p&gt;First off, I know $\ce{SiO2}$ is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Network_covalent_bonding&quot; rel=&quot;nofollow noreferrer&quot;&gt;network solid&lt;/a&gt;. I am looking at its intramolecular forces between the $\ce{Si}$ and the $\ce{O}$ molecule. On the one hand, it seems to me that this should be a polar covalent bond, but on the other, both electrons are given to the $\ce O $ by $\ce{Si}$ which makes it coordinate covalent.&lt;/p&gt;&#xA;" OwnerUserId="1388" LastEditorUserId="16035" LastEditDate="2017-06-17T01:54:51.723" LastActivityDate="2017-06-17T01:54:51.723" Title="Does SiO₂ have polar covalent bonds or coordinate covalent bonds?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4447" PostTypeId="2" ParentId="4446" CreationDate="2013-03-21T01:38:42.747" Score="3" Body="&lt;p&gt;The bonds are polar covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Given' is a strange term in this context and generally only of relevance for ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinate covalent bonds imply the donation of two electrons to a bond with another atom.  In the case of $\ce{ SiO2 }$, The structure is actually tetrahedral, with an oxygen between and two silicon atoms, and where each oxygen forms two bonds and each silicon forms four bonds.  In these bonds, each of the four valence electrons for silicon is shared singly with a single electron from one of the oxygens.  Thus, in each bond, one of the electrons is from silicon and one from oxygen, which is why we don't normally consider this a coordinate covalent or dative bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of those examples where the empirical formula, $\ce{ SiO2 }$, is highly misleading.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1085" LastEditDate="2013-03-21T11:46:42.170" LastActivityDate="2013-03-21T11:46:42.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4448" PostTypeId="2" ParentId="4445" CreationDate="2013-03-21T01:42:08.220" Score="-1" Body="&lt;p&gt;If the volume is constant (sealed container that is not a rubber ballon, for example), the overall temperature would cool.  And the net pressure would be a weighted average of the two pressures if the volume was constant.  Without actual amounts it cannot be determined if the pressure would be higher or lower than the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the container were a rubber balloon, it is not possible to determine the final volume (even as bigger or smaller) without information about relative amounts and the temperature of the 'colder air'.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-03-21T01:42:08.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4449" PostTypeId="2" ParentId="4439" CreationDate="2013-03-21T02:00:14.080" Score="14" Body="&lt;p&gt;You need to mix the orbitals, populate them with the electrons and see if you have net bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg:  H + H two 1s orbitals mix to form sigma and sigma*.  Two electrons total, both occupy the sigma orbital, two more electrons in bonding than antibonding orbitals, the compound is stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZAbo.png&quot; alt=&quot;H2 MO diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg:  He + He; same mixing as above.  Four electrons, two in the sigma, two in the sigma*.  Since there are as many bonding electrons as as antibonding, there is no net bond.  He2 is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xAhQr.png&quot; alt=&quot;He MO Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg:  He + H; same mixing as above.  Three electrons, two in sigma, one in sigma*.  One more electron in bonding than antibonding.  He-H forms a very weak bond.  Please note the diagram is for He2+ but the He-H is very similar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K2Nfs.png&quot; alt=&quot;He2+ MO diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg: Li + H; Li has 1s + 2s, while H has 1s.  This mix to form a sigma orbital from H1s+Li2s, a sigma* orbital and H1s-Li2s, and a non bonding orbital from Li1s (lower in energy than the sigma).  Four total electrons.  Nonbonding sigma is occupied, and then the sigma orbital is occupied.  Net effect: Li-H forms a stable bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xAhQr.png&quot; alt=&quot;LiH MO diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might have noticed that the first three diagrams had identical structure, even if their energy levels were different.  This is a property of quantum mechanical symmetry, and solving the problem only requires knowing a few diagram types and filling the electrons in.  Consider the second row.  The diagram is as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9fnSg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You simply count the valence electrons and fill in the diagram.  Then you subtract the number of bonding orbital electrons from antibonding, and if you have more, you will have a bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of webpages are very good for further explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source of the diagrams:&#xA;&lt;a href=&quot;http://www.chem1.com/acad/webtext/chembond/cb08.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chem1.com/acad/webtext/chembond/cb08.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&#xA;&lt;a href=&quot;http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2014-03-18T17:34:54.307" LastActivityDate="2014-03-18T17:34:54.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4450" PostTypeId="1" CreationDate="2013-03-21T10:35:17.863" Score="6" ViewCount="672" Body="&lt;p&gt;We can predict whether a reaction can occur spontaneously by calculating the change in Gibbs' free energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering whether there is any general measure of the thermodynamic stability of any substance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I look at a random substance and using a table of thermodynamic values predict whether this substance is stable or not within a range of given conditions (pressure, temperature, etc.).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-04-08T17:59:08.150" Title="Measuring Thermodynamic Stability" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4451" PostTypeId="1" CreationDate="2013-03-21T16:16:19.720" Score="4" ViewCount="47832" Body="&lt;p&gt;I am an engineer and I understand the difference between baking power and baking soda as baking powder &lt;em&gt;is&lt;/em&gt; baking soda (alkali) with tartaric acid (acidic) and an inert starch (why? Is this a catalyst in the presence of humidity and temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why use &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_bicarbonate#As_a_cleaning_agent&quot; rel=&quot;nofollow&quot;&gt;baking soda to deodorize a carpet&lt;/a&gt; and &lt;strong&gt;not&lt;/strong&gt; baking powder? Is it because baking powder has both an acid and base and thus one would neutralize the other and would render the baking powder useless for adsorbing smells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My carpet would be at a temperature of about $25~\mathrm{^\circ C}$. I am guessing this temperature and a relative humidity of $50{-}60~\%$  will not catalyze a violent reaction with either baking powder or baking soda?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input from chemists would be useful! &#xA;I don't want to use either on my synthetic carpet in my &lt;strong&gt;rented&lt;/strong&gt; house to experiment before I get enough input from people with know-how!&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="7475" LastEditDate="2015-10-30T17:05:08.830" LastActivityDate="2015-10-30T17:05:08.830" Title="Deodorizing using baking powder instead of baking soda" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4455" PostTypeId="1" CreationDate="2013-03-22T11:53:36.523" Score="1" ViewCount="4392" Body="&lt;p&gt;1 mol of $\ce{F2}$ is turned into $\ce{2F}$ (atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the specific enthalpy of atomization, or two times that?&lt;/p&gt;&#xA;" OwnerUserId="1393" LastEditorUserId="4231" LastEditDate="2015-02-26T17:49:45.280" LastActivityDate="2018-03-26T19:07:06.683" Title="What defines the specific enthalpy of atomization?" Tags="&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2018-03-26T18:35:20.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="4456" PostTypeId="1" AcceptedAnswerId="72153" CreationDate="2013-03-22T12:10:22.743" Score="4" ViewCount="270" Body="&lt;p&gt;Is there a combination of an acid and a base that when mixed results in a solution that is more basic or more acidic than the initial base or acid, respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, do such combinations have a name?&lt;/p&gt;&#xA;" OwnerUserId="1394" LastEditorUserId="194" LastEditDate="2017-04-08T02:52:38.260" LastActivityDate="2018-03-04T18:17:25.420" Title="Is it possible for an acid and base to react to form an even stronger base or acid?" Tags="&lt;acid-base&gt;&lt;solutions&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4457" PostTypeId="2" ParentId="4456" CreationDate="2013-03-22T13:36:23.187" Score="0" Body="&lt;p&gt;NO, adding acid in base is addition of $\ce{H+}$ ions in $\ce{OH-}$ ions which will certainly reduce the concentration of $\ce{OH-}$ ions in solution and solution will become less basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 + \log_{10}[\ce{OH-}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:(after further question in comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatsoever, you say addition of $\ce{H+}$ would be followed by a decrease in concentration of $\ce{OH-}$ due to release of water. In the case of amphoteric oxides, they behave depending on the nature of the medium; in basic pH it will act as an acid. A polyprotic acid will neutralise the base faster and even in case or organic acids the process will be slow (being weak acids).&lt;/p&gt;&#xA;" OwnerDisplayName="user1395" LastEditorUserId="16683" LastEditorDisplayName="user1395" LastEditDate="2017-04-07T12:33:16.020" LastActivityDate="2017-04-07T12:33:16.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4458" PostTypeId="2" ParentId="4455" CreationDate="2013-03-22T13:53:11.587" Score="-2" Body="&lt;p&gt;Enthalpy of atomisation is energy required to break 1 mole of E-E bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg: $\ce{1 F2 -&amp;gt; 2 F}$ .............................$\Delta H_{atomisation}$&lt;/p&gt;&#xA;" OwnerDisplayName="user1395" LastEditorUserId="4231" LastEditorDisplayName="user1395" LastEditDate="2015-02-26T17:49:43.383" LastActivityDate="2015-02-26T17:49:43.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4459" PostTypeId="1" AcceptedAnswerId="4470" CreationDate="2013-03-22T14:44:56.157" Score="2" ViewCount="616" Body="&lt;p&gt;I am looking for hydrophilic transparant materials which are solid at room temperature and that don't dissolve in water. I need these for an experiment in which I condense droplets on a plate made out of/or coated with the material&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just thinking `chemically' I imagine that any polymer with a lot of hydroxyl groups will be hydrophilic (like Polyvinylalcohol), but PVA is water soluble. Polyethylenetereftalate (PET) is reasonable but still has a contact angle around 70 degrees &lt;a href=&quot;http://www.accudynetest.com/polytable_03.html?sortby=contact_angle&quot; rel=&quot;nofollow&quot;&gt;(common polymer contact angles)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know transparant materials with contact angles closer to 0 that don't dissolve in water? An additional `demand' is that the material does not break easily (like glass)&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-03-23T08:58:20.610" LastActivityDate="2013-03-23T19:14:14.973" Title="Solid hydrophilic transparant materials that don't dissolve in water" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;polymers&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4461" PostTypeId="1" CreationDate="2013-03-22T18:35:41.677" Score="4" ViewCount="167" Body="&lt;p&gt;In acidic solution, what does $\ce{ClO3-}$ reduce into and, most importantly, why? It's $$\ce{6e- +ClO_{3}^- +6H^+ -&amp;gt; Cl^- + 3 H_2O}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with the gift of foresight, why was the equation for its oxidation not instead $$\ce {2e^- +ClO_{3}^- +2H^+ \rightarrow Cl^- + O_2 + H_2O}?$$&#xA;Obviously it balances and charge is conserved. However, the latter equation would seem more likely: less charges have to be pulled apart (so less work has to be done to move these charges), and the bond energy of bonds in a water molecule is almost twice that in an oxygen molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is there a flaw in my logic and does there exist a rough set of rules to determine how things will split up under oxidation or reduction?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="4945" LastEditDate="2014-08-29T03:47:56.277" LastActivityDate="2014-08-29T03:47:56.277" Title="In acidic solution, what does ClO₃⁻ reduce into?" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4462" PostTypeId="1" AcceptedAnswerId="4476" CreationDate="2013-03-23T05:02:33.483" Score="0" ViewCount="7074" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The density of a gas &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times that of hydrogen. In that case, what is the molecular weight of gas &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well what I've done up to now is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular mass of hydrogen is &lt;span class=&quot;math-container&quot;&gt;$M(\ce{H2})=2$&lt;/span&gt;, the density of hydrogen is &lt;span class=&quot;math-container&quot;&gt;$\rho(\ce{H2})= \pu{0.089 kg/m3}$&lt;/span&gt;.&#xA;Assuming that equal volumes &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; of both gases are taken, the mass of hydrogen &lt;span class=&quot;math-container&quot;&gt;$m(\ce{H2})= (0.089 \cdot v) \pu{kg}$&lt;/span&gt; and the mass of the unknown gas &lt;span class=&quot;math-container&quot;&gt;$m(\ce{X}) = (0.89 \cdot v) \pu{kg}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this hasn't really gotten me anywhere. How will it be done? &lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="4945" LastEditDate="2019-01-22T20:03:30.630" LastActivityDate="2019-01-22T20:03:30.630" Title="How to find the molecular weight of a given gas based on its density?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;hydrogen&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4463" PostTypeId="2" ParentId="4462" CreationDate="2013-03-23T05:26:48.733" Score="1" Body="&lt;p&gt;Assuming equal volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, equal temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, and equal pressure conditions &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, we can apply Avogadro's Law:&#xA; An equal volume of two gasses contain an equal number of molecules at the same temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can proceed: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \rho &amp;amp;= \frac{\text{Mass}}{\text{Volume}}\\&#xA;       &amp;amp;= \frac{n\times M (\text{molar  mass})}{\text{Volume}}\\&#xA;\implies\frac{\rho_1}{\rho_2}&amp;amp;=\frac{M_1}{M_2}\\&#xA;\implies M_{\ce{X}}&amp;amp;=10\times M_{\ce{H2}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1395" LastEditorUserId="4945" LastEditDate="2019-01-22T20:02:14.500" LastActivityDate="2019-01-22T20:02:14.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4464" PostTypeId="2" ParentId="4440" CreationDate="2013-03-23T07:25:08.343" Score="9" Body="&lt;p&gt;The question to the &lt;em&gt;Newton - Ask a Scientist&lt;/em&gt; program by Danna C Griffiths (to be found via &lt;a href=&quot;https://web.archive.org/web/20150228170949/http://www.newton.dep.anl.gov/askasci/chem99/chem99198.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Internet Archive&lt;/a&gt;) has an answer for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I understood from the explanation on that site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; in a saturated solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you dissolve &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; in pure water, the process is highly exothermic, but as the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; to dissolve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="4945" LastEditDate="2019-10-30T09:59:03.873" LastActivityDate="2019-10-30T09:59:03.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4465" PostTypeId="1" AcceptedAnswerId="4847" CreationDate="2013-03-23T09:38:25.603" Score="7" ViewCount="26115" Body="&lt;p&gt;I was reading this other question on the site: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4451/deodorizing-using-baking-powder-instead-of-baking-soda&quot;&gt;Deodorizing using baking powder instead of baking soda&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, baking soda is used quite often as a cleaning and deodorizing agent. From what I know, it's cleaning effects are because it can denature proteins and saponify oils (convert oils to soap) since it is a weak base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you take a look the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Aroma_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;aroma compounds&lt;/a&gt;, the only compounds that could possibly be affected by a weakly basic substance are esters (which would undergo hydrolysis). Most of the compounds are polyunsaturated aldehydes, ketones and alcohols which are not affected by (weak) bases. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: I assume here that compounds which cause bad odors would have a similar structure to compounds that are fragrant since they both have to activate whatever receptors there are in the nose. Also I have read that usually the same compound can cause both a bad and a good smell depending on the concentration of the compound, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Indole&quot; rel=&quot;nofollow noreferrer&quot;&gt;indole&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So I ask: Does $\ce{NaHCO3}$ really have any deodorizing effect and if so, what is the mechanism (reactions involved) behind it. Also would it be possible to achieve the same effect with any random alkaline substance ($\ce{NaOH}$ maybe, although not very practical).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2019-05-19T10:30:23.347" Title="How does sodium bicarbonate act as a deodorizing agent?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4468" PostTypeId="1" AcceptedAnswerId="4576" CreationDate="2013-03-23T17:10:37.130" Score="5" ViewCount="8013" Body="&lt;p&gt;Given a sample that is probably modafinil plus common fillers, but could be fake, what reagent tests would best determine whether it contains modafinil? Assume a known-good sample for comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Modafinil&quot; rel=&quot;noreferrer&quot;&gt;Modafinil&lt;/a&gt; is a sulfur-containing atypical stimulant which has become common in some circles, but which, due to off-label use and the high price of the US name brand, is mostly obtained by mail order from slightly-sketchy websites in India and the UK. There are rumors that some fraction of the modafinil imported this way is fake, but this is only rumor because the effects vary to begin with, and most people don't seem willing to pay for laboratories to do proper testing.)&lt;/p&gt;&#xA;" OwnerUserId="1397" LastActivityDate="2014-02-21T07:59:35.117" Title="Presumptive test for modafinil?" Tags="&lt;experimental-chemistry&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4469" PostTypeId="2" ParentId="4456" CreationDate="2013-03-23T18:06:07.733" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{6HF + Al(OH)3 (s) -&amp;gt; H3[AlF6] + 3H2O}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{6HCN + Fe(OH)2 (s) -&amp;gt; H4[Fe(CN)6] + 2 H2O}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{TlOH + HOCl + H2O -&amp;gt; Tl(OH)3(s) + HCl}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;May be one of these will work. Even if they don't, you should get the idea where to search.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2017-04-07T12:34:35.990" LastActivityDate="2017-04-07T12:34:35.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4470" PostTypeId="2" ParentId="4459" CreationDate="2013-03-23T19:14:14.973" Score="2" Body="&lt;p&gt;Several commercially available titanium dioxide coatings are available, commonly used to kill bacteria or aid UV stability (I think). &lt;a href=&quot;http://en.wikipedia.org/wiki/Superhydrophilicity&quot; rel=&quot;nofollow&quot;&gt;When exposed to light, the titanium dioxide has a contact angle of almost 0 degrees.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to deposit a thin layer on something like a PVA or polycarbonate sheet. The commercial coatings are &lt;a href=&quot;http://www.oxititan.com/store/&quot; rel=&quot;nofollow&quot;&gt;somewhat expensive&lt;/a&gt;, though, as far as I can tell, and the common process for doing it yourself involves &lt;a href=&quot;http://www.solideas.com/solrcell/coatglas.html&quot; rel=&quot;nofollow&quot;&gt;heating the surface in a mid-temperature flame for 10-15 minutes&lt;/a&gt; which excludes pretty much all plastics I know of that would otherwise be suitable.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-03-23T19:14:14.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4471" PostTypeId="2" ParentId="4199" CreationDate="2013-03-23T19:42:45.060" Score="2" Body="&lt;p&gt;Reactions that are &lt;a href=&quot;http://en.wikipedia.org/wiki/Autocatalysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;autocatalytic&lt;/a&gt; can have their rate increase as a function of conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example is the decomposition of cellulose acetate by hydrolysis, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Vinegar_syndrome#Decay_and_the_.22vinegar_syndrome.22&quot; rel=&quot;nofollow noreferrer&quot;&gt;vinegar syndrome&lt;/a&gt;. The reaction is acid catalyzed, and it produces acid. In theory, the reaction gets faster. In practice, diffusion and limiting reagents many prevent a rate acceleration. Acetylsalicylic acid decomposes similarly. In the following equation, $\ce{R}$, could be the cellulose backbone or the salicylyl group. The acetyl group $\ce{-COCH3}$ is represented by $\ce{Ac}$. Thus acetic acid is $\ce{HOAc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-OAc + H2O -&amp;gt;[\ce{H+ catalyst}] R-OH + HOAc}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the alpha halogenation of ketones or aldehydes, which is acid catalyzed and acid-producing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-CO-CH3 + Br2-&amp;gt;[\ce{H+ catalyst}] CH3-CO-CH2Br + HBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A historically interesting example is the allotropic decomposition of tin, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_pest&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;tin pest&quot;&lt;/a&gt;, in which tin metal transforms to a brittle form in cold weather. The book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1585423319&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Napoleon's Buttons&lt;/em&gt;&lt;/a&gt;, which is about the influence of chemistry (wittingly or otherwise) on history, opens with the conjecture that tin pest is what really stopped Napoleon's invasion of Russia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, as &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4475/at-equilibrium-the-mas-of-reactants-and-products-are-equal-does-that-mean-react/4479#4479&quot;&gt;this answer reminds us&lt;/a&gt;, the rate of the &lt;em&gt;reverse&lt;/em&gt; reaction increases to a maximum during the establishment of dynamic equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2013-03-25T11:38:53.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4472" PostTypeId="1" CreationDate="2013-03-24T08:51:45.490" Score="-2" ViewCount="22425" Body="&lt;p&gt;How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols? What rules can be applied to make these choices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which one of the following compounds  has the highest solubility in water?&lt;br&gt;&#xA;  Which one of the following compounds  has the lowest boiling point?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;ethane, $\ce{CH3CH3}$&lt;/li&gt;&#xA;  &lt;li&gt;dimethylether, $\ce{CH3-O-CH3}$&lt;/li&gt;&#xA;  &lt;li&gt;acetaldehyde, $\ce{CH3CHO}$ &lt;/li&gt;&#xA;  &lt;li&gt;methanol, $\ce{CH3OH}$&lt;/li&gt;&#xA;  &lt;li&gt;chloromethane, $\ce{CH3Cl}$&lt;/li&gt;&#xA;  &lt;li&gt;methane, $\ce{CH4}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1399" LastEditorUserId="4945" LastEditDate="2018-08-22T11:37:23.363" LastActivityDate="2018-08-22T11:37:23.363" Title="How to determine the order of boiling points and solubilities for (chloro-)alkanes, ethers, aldehydes, and alcohols?" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4473" PostTypeId="1" AcceptedAnswerId="4491" CreationDate="2013-03-24T10:58:05.083" Score="0" ViewCount="13510" Body="&lt;p&gt;How do you find the pH of a solution with a known pK&lt;sub&gt;a&lt;/sub&gt; and Molarity?&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2015-04-02T12:53:19.817" Title="How do you find the pH of a solution with a known pKa and Molarity?" Tags="&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4475" PostTypeId="1" AcceptedAnswerId="4479" CreationDate="2013-03-24T11:59:08.770" Score="2" ViewCount="18160" Body="&lt;p&gt;This one is troubling me, and I even got it wrong in my exam:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please explain this one. &lt;/p&gt;&#xA;" OwnerUserId="1400" LastEditorUserId="7475" LastEditDate="2016-01-25T15:51:42.757" LastActivityDate="2016-01-25T15:57:35.827" Title="At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops?" Tags="&lt;equilibrium&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4476" PostTypeId="2" ParentId="4462" CreationDate="2013-03-24T12:04:40.980" Score="2" Body="&lt;p&gt;Indeed you can use Avogadro's law here, but I guess it is useful to give a little bit more insight in Avogadro's Law. What you can do is apply the ideal gas law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p V = n R T $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the amount of substance of gas, which you can also write as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ n=\frac{m}{M},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the total mass of the gas and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the molecular mass.&#xA;You can rewrite the ideal gas law now to:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$p M = \frac{m}{V} R T,$$&lt;/span&gt;&#xA;where you should recognize that &lt;span class=&quot;math-container&quot;&gt;$\frac{m}{V}=\rho$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas constant &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is (as the name shows) a constant so you can write&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  R &amp;amp;=\frac{p M}{\rho T} &amp;amp;&#xA;  \rightarrow            &amp;amp;&amp;amp;&#xA;  \frac{p_1 M_1}{\rho_1 T_1}&amp;amp;=\frac{p_2 M_2}{\rho_2 T_2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therefore, for equal pressure and temperature you will find that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2},$$&lt;/span&gt;&#xA;which means that if the density of gas 2 (the unknown gas &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt;) is &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times as high, the molecular mass should be as well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What is interesting to add is that this assumes the ideal gas law to be applicable. If, for example, your gases would be at high pressure (e.g. &lt;span class=&quot;math-container&quot;&gt;$\pu{200 bar}$&lt;/span&gt;) this is typically not a good assumption so the molecular mass that you would find for a &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times as dense gas will not be equal to &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times the molecular mass of hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="4945" LastEditDate="2019-01-22T20:02:47.897" LastActivityDate="2019-01-22T20:02:47.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4477" PostTypeId="2" ParentId="546" CreationDate="2013-03-24T12:32:19.413" Score="5" Body="&lt;p&gt;I would say the answer on &lt;a href=&quot;http://www.flyingcircusofphysics.com/News/NewsDetail.aspx?NewsID=51&quot; rel=&quot;noreferrer&quot;&gt;this website&lt;/a&gt; is pretty near perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, you need 2 things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enough friction between your nose and the spoon&lt;/li&gt;&#xA;&lt;li&gt;The center of mass of the spoon should be on a line vertically downwards THROUGH the tip of your nose as illustrated in this image from the same website:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AKXMw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction is highest if both your skin and the spoon are free of greasy `lubricants'. Breathing on the spoon will add more friction by depositing a very thin (only a few atomic layers) film of water which, on this scale, behaves solid-like due to ordering effects &lt;a href=&quot;http://polymer.bu.edu/hes/articles/hcks10.pdf&quot; rel=&quot;noreferrer&quot;&gt;as shown in this paper&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-03-24T12:32:19.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4479" PostTypeId="2" ParentId="4475" CreationDate="2013-03-25T08:25:13.637" Score="3" Body="&lt;p&gt;At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the first order reaction $\ce{A &amp;lt;=&amp;gt;[k_f][k_b] B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpHJq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-03-25T08:25:13.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4480" PostTypeId="2" ParentId="4377" CreationDate="2013-03-25T09:48:45.263" Score="3" Body="&lt;p&gt;After quick digging I found, that glass cleaner solutions usually contain alcohols and surfactants. This additions reduce surface tension and allows the solution to interact with glass more easily, distributing the solution over the entire surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white stripes visible after water dried are usually solid salts. If the water does not contain any salts, but only compounds, that can quickly evaporate, it will not leave stripes. But it is not addition, but absence of additions that do the trick. For example, most bleacher/toilet cleaners contain hypoclorites, salts of unstalble $\ce{HOCl}$ acid and leave $\ce{NaCl}$ after evaporation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When tap water is allowed to dry, it usually leaves white stripes, but they are not visible on most surfaces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only ideas I can get why the salts don't deposit if using a glass cleaner is that salts are removed (and if you have a cleaner not meant to be diluted, the water in it may be deionized) or to trap salt ions into organic ligands so resulted salts were not crystalline (possible, but I do not like the idea). &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="61578" LastEditDate="2018-05-14T23:12:32.737" LastActivityDate="2018-05-14T23:12:32.737" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="4481" PostTypeId="1" CreationDate="2013-03-25T11:19:30.727" Score="3" ViewCount="2449" Body="&lt;p&gt;Can someone explain the following types of molecules and predict their structure and shape? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will help my understanding if you could provide an example.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AX3E,AX5 and AX6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;note that A = central atom and X = bonding atoms.&lt;/p&gt;&#xA;" OwnerUserId="1403" LastEditorUserId="73" LastEditDate="2013-04-02T04:00:56.403" LastActivityDate="2013-05-24T01:53:43.890" Title="Predicting molecular geometry from a formula" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4482" PostTypeId="1" CreationDate="2013-03-25T12:19:44.443" Score="6" ViewCount="12237" Body="&lt;p&gt;According to my notes, octahedral site stabilization energy (OSSE), sterical effects and electronic effects in stability create a confusion in this theory. However, we are able to solve this problem with few rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tetrahedral splitting energy is nearly equal to 4/9 octahedral splitting energy. &lt;/li&gt;&#xA;&lt;li&gt;As the OSSE increases, the $O_\mathrm{h}$ geometries will be more favored. &#xA;These are useful to decide to whether the structure is spinel or inverse spinel.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I got $\ce{ZnCr2O4}$ ($\ce{Zn^2+}$, $\ce{2Cr^3+}$, $\ce{4O^2-}$):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Zn^2+}$; $\mathrm{d^{10}}$ ion with CFSE = $0$ and thus, there will be no effect on $O_\mathrm{h}$ holes.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cr^3+}$; $\mathrm{d^3}$ ion with CFSE = $\pu{-8 Dq}$ and OSSE = $\pu{8.45 Dq}$ so, this is thought to occupy $O_\mathrm{h}$ holes. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{Zn^2+}$ can only occupy $T_\mathrm{d}$ holes, then it suits the rules for being normal spinel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if we got a group II metal with a non-zero CFSE, then what would the result be?&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="41328" LastEditDate="2017-08-07T02:52:41.923" LastActivityDate="2017-08-07T02:52:41.923" Title="Normal Spinel and Inverse Spinel" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;coordination-compounds&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4483" PostTypeId="1" CreationDate="2013-03-25T21:28:22.973" Score="2" ViewCount="1582" Body="&lt;p&gt;How, chemically, would you remove protruding phenyl groups? I mean for instance in the molecule phenethylamine there's a phenyl group sticking out (or protruding) how would you go about removing this phenyl group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDX0Z.png&quot; width=&quot;255&quot; alt=&quot;2-phenylethan-1-amine to ethanamine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorDisplayName="user7951" LastEditDate="2016-10-12T10:26:45.373" LastActivityDate="2016-10-12T10:26:45.373" Title="How, chemically, would you remove protruding phenyl groups?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4484" PostTypeId="1" CreationDate="2013-03-25T23:04:32.953" Score="2" ViewCount="238" Body="&lt;p&gt;How would you change the chirality of a compound such that (S,R) becomes (S,S)? That is how would you synthesise a compound from its diastereomer?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2013-03-26T00:07:07.360" LastActivityDate="2013-03-26T00:07:07.360" Title="How would you change the chirality of a compound such that (S,R) becomes (S,S)?" Tags="&lt;chirality&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2014-05-03T23:30:32.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="4486" PostTypeId="1" AcceptedAnswerId="4487" CreationDate="2013-03-26T08:36:23.760" Score="12" ViewCount="758" Body="&lt;p&gt;I. e., what if we just burn trash completely or dissolve it in acid, get the resulting paste and extract pure chemical elements out of it by applying some physics and/or chemistry to it until no more can be extracted? Just curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be perfectly clear: my main concern is whether there &lt;em&gt;are&lt;/em&gt; any means to extract any useful element from this bulk of burned/dissolved/blender-ed trash. For example, we burn with Thermite the whole contents of some waste disposal site, until it's only multicolored ash there. Or pour the HSbF6 over it. From there, can we get the pure carbon out of the resulting mass? Pure aluminium? Pure iron? Gases, maybe, like oxygen?&lt;/p&gt;&#xA;" OwnerUserId="1404" LastEditorUserId="5354" LastEditDate="2014-07-26T05:33:30.157" LastActivityDate="2014-07-26T05:33:30.157" Title="Can we process waste just by decomposing it to homogenous mass and extracting the elements from it?" Tags="&lt;elements&gt;&lt;extraction&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4487" PostTypeId="2" ParentId="4486" CreationDate="2013-03-26T12:11:04.143" Score="11" Body="&lt;p&gt;Sure, this absolutely can be done.  In particular, you cannot destroy elements using conventional means like burning, so all the initial elements are still in the mass.  The question is whether or not this is a good way to extract what you want.  There may be better chemical techniques to do the extraction that are short of burning.  The branch of chemistry which deals with the optimal conversion of elements and chemicals to other useful substances is &lt;em&gt;chemical engineering&lt;/em&gt;, and figuring out which process is &quot;best&quot; (e.g. fastest, cheapest, highest yield) is crucial for industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a very funny story related to this idea on the Periodic Table of Videos.  A scientist was trusted with the entire UK supply of plutonium (10mg, a very small amount) during World War II.  He accidentally spilled it on to his lab bench.  He then took a saw, sawed a large circle out of bench, burned that wood, and recovered nearly all the plutonium!  So yes, this works.  Here's the video, and this story begins around 15:30: &lt;a href=&quot;http://www.periodicvideos.com/videos/094.htm&quot;&gt;Plutonium Video&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As to your update, there are lots of methods available, but it depends greatly on the starting conditions and materials.  For example, burning is a great way to extract carbon: burning hydrocarbons or carbohydrates burns off the hydrogen or water (respectively) and leaves mostly pure carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But extracting metals might take more work.  Often, metals exist as ionic compounds.  For example, silver nitrate, $\ce{AgNO3}$, contains siver $\ce{(Ag)}$ in the +1 state.  While it might be possible to dissociate the silver with high enough temperature, it is far easier to extract metallic silver by &lt;em&gt;reducing&lt;/em&gt; silver in the presence of a more reactive metal.  &quot;More reactive&quot; simply means a metal that will more easily give up an electron to the less reactive sliver, changing the silver ion $\ce{Ag+}$ to metallic $\ce{Ag}$.  Silver nitrate is highly soluble in water, and if you simply drop in a piece of metallic copper $\ce{(Cu)}$ metallic silver will slowly come out of solution and plate the copper with silver, while copper dissolves and creates copper nitrate.  The equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2AgNO3 + Cu-&amp;gt;2Ag + Cu(NO3)2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the crucial part is the ionic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Ag+ + Cu-&amp;gt;2Ag + Cu^2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows that the silver ion is &lt;em&gt;reduced&lt;/em&gt; to silver, and the copper metal is &lt;em&gt;oxidized&lt;/em&gt; to copper ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, Periodic Videos has the footage of this exact reaction: &lt;a href=&quot;http://www.periodicvideos.com/videos/047.htm&quot;&gt;Silver Video&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I picked this example precisely because of this footage, but it's not the best example because silver is so low in reactivity that it exists metallically in nature (i.e. you can find lumps of silver if you are lucky!)  But many metals are so reactive that it takes a huge amount of energy to extract them.  Examples are sodium or potassium, which are so reactive they explode in the presence of water (go search YouTube and have a field day!).  To extract them as metals commercially usually requires &lt;em&gt;electrolysis&lt;/em&gt;, which uses electricity to force electron transfer which would not otherwise occur spontaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on and on with other methods.  There is no universal extraction method, at least not yet, and this is why we have chemcial engineers to figure out better and better procedures to use.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-03-26T13:55:15.037" LastActivityDate="2013-03-26T13:55:15.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4488" PostTypeId="1" CreationDate="2013-03-26T15:31:43.230" Score="2" ViewCount="1059" Body="&lt;p&gt;I am in mathematics, but I am writing a paper on molecular vibrations (because you can use representation theory to handle this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to prove the selection-rule, but I also have to give a presentation of my intermediate results. Can someone give a little (maybe trivial) explanation (in one or maximal two small sentences) of what the selection-rule does? (for example for the IR-spectrum). This is for my presentation if someone doesn't know something about the mathematical and chemical background.&lt;/p&gt;&#xA;" OwnerUserId="1407" LastEditorUserId="73" LastEditDate="2013-04-02T04:05:53.770" LastActivityDate="2013-04-05T16:54:45.587" Title="Explanation of the Selection rule in light of IR spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4489" PostTypeId="1" AcceptedAnswerId="4495" CreationDate="2013-03-26T16:15:53.540" Score="3" ViewCount="1038" Body="&lt;p&gt;I am interested in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Biological_half-life&quot; rel=&quot;nofollow&quot; title=&quot;Biological Half Life&quot;&gt;Biological Half Life&lt;/a&gt; of substances. In the previous wikipedia link, I've already found the half lives of common pharmaceutical products. I can't seem to find similar data for recreational substances, does anyone know these half lives?&lt;/p&gt;&#xA;" OwnerUserId="1408" LastActivityDate="2016-10-27T08:22:59.090" Title="Recreational drugs by half life" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4490" PostTypeId="2" ParentId="4481" CreationDate="2013-03-26T16:20:37.477" Score="7" Body="&lt;p&gt;It will depend on the number of lone electron pairs as well.&lt;/p&gt;&#xA;&lt;p&gt;For example:&#xA;&lt;span class=&quot;math-container&quot;&gt;$BCl_3$&lt;/span&gt; will be triangular planar as it has no lone electron pairs.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$NH_3$&lt;/span&gt; however will be triangular based-pyramidal because there are 3 bonded pairs of electrons and one lone electron pair.&lt;/p&gt;&#xA;&lt;p&gt;Basically, it comes down to the number of electron pairs (both bonded and unbonded) around the central atom.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Two electron pairs - Linear&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/03Ngn.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Three electron pairs - Triangular planar&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cI9NW.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Four electron pairs - Tetrahedral&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5b1Po.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Five electron pairs - Triangular based bi-pyramidal (it's like triangular planar, but with one on top and one below, or two triangular pyramids with their base stuck together)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M8vvn.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Six electron pairs - Octahedral (Square based bi-pyramidal)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j209J.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Those geometries include the lone electron pairs. For the molecular geometry, you will then need to exclude the electron pair.&lt;/p&gt;&#xA;&lt;p&gt;E.g. &lt;span class=&quot;math-container&quot;&gt;$H_2O$&lt;/span&gt; : central atom O. O has 4 electron pairs: 2 bonded, 2 lone. The shape, including lone electron pairs is Tetrahedral. Ignore the lone pairs now, you will be left with a bent molecular geometry.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nt9Yw.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Blue is oxygen;&lt;br&gt;&#xA;Yellow is hydrogen;&lt;br&gt;&#xA;White/Grey are lone/unbonded electron pairs.&lt;/p&gt;&#xA;&lt;p&gt;I hope this makes sense. :)&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-05-22T19:24:50.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4491" PostTypeId="2" ParentId="4473" CreationDate="2013-03-26T16:28:09.610" Score="3" Body="&lt;p&gt;Do you know the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to find a concentration first, however, on the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA &amp;lt;=&amp;gt; H+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solve for the dissociated hydrogen ion, say $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{a} = \frac{[x][x]}{[\ce{HA}]-x} = \frac{x^2}{[\ce{HA}]-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is: if $x$ of $\ce{HA}$ dissociated, then we would get $x$ of $\ce{H}$, $x$ of $\ce{A}$ and $[\ce{HA}]-x$ left of $[\ce{HA}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T12:53:19.817" LastActivityDate="2015-04-02T12:53:19.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4493" PostTypeId="1" AcceptedAnswerId="4504" CreationDate="2013-03-27T00:47:37.850" Score="1" ViewCount="54" Body="&lt;p&gt;I got a forcefield (OPLS-AA in this case) that lists lennard-jones parameters for all homonuclear van der waals interactions. Is there a way to deduct the heteronuclear vdw-interaction parameters from those?&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-03-29T00:14:19.043" Title="Can I deduct heteronuclear van der waals parameters from homonuclear parameters in a forcefield?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4494" PostTypeId="1" AcceptedAnswerId="4497" CreationDate="2013-03-27T10:49:12.957" Score="0" ViewCount="335" Body="&lt;p&gt;According to Wikipedia caesium’s density is $1.90\ \mathrm{g/cm^3}$ at $\Theta =20\ \mathrm{^\circ C}$. How does this change when $T$ changes? E.g. will it expand when melting? &lt;/p&gt;&#xA;" OwnerUserId="1412" LastEditorDisplayName="user7951" LastEditDate="2015-07-03T11:35:17.610" LastActivityDate="2015-07-03T11:35:17.610" Title="How does caesium’s density change with temperature?" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4495" PostTypeId="2" ParentId="4489" CreationDate="2013-03-27T14:00:11.213" Score="2" Body="&lt;p&gt;Wikipedia reports biological half-lives for several recreational drugs I just tried (cocaine, THC, MDMA). In each case, I went to the page for the particular substance, and Wikipedia keeps an information box at the top of the article containing the pharmacokinetic data. THC's half-life has a citation to the scientific literature, while the other two I checked do not. I think it might be slightly harder in general to get good data for the purely recreational drugs. I would expect drugs without an accepted medical benefit to be less well-studied. To find what's there, though, it looks like you will be able to look up each active ingrediate individually. Depending on what you need, Wikipedia might be okay. If it's not, follow the Wikipedia citation back to the literature if it's there, or try PubMed.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2013-03-27T14:00:11.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4496" PostTypeId="2" ParentId="4494" CreationDate="2013-03-27T18:13:57.467" Score="0" Body="&lt;p&gt;I think I found what I was looking for on &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. They give the thermal expansion coefficient for Caesium as $97 µm\cdot m^{−1}\cdot K^{−1}$. &lt;/p&gt;&#xA;" OwnerUserId="1412" LastActivityDate="2013-03-27T18:13:57.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4497" PostTypeId="2" ParentId="4494" CreationDate="2013-03-27T18:16:42.467" Score="5" Body="&lt;p&gt;Looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page&lt;/a&gt; you can see that the density is $\rho=1.93$ kg/l at room temperature and $\rho=1.843 kg/l$ at it's melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional information can be found in &lt;a href=&quot;http://dx.doi.org/10.1007/s10740-006-0026-y&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, where they experimentally explore the density of liquid Cesium in the vicinity of its melting point. In the range of 302 to 375 K to be exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From their experimental data they deduce the following approximate equation for the density as a function of the temperature:&#xA;$$\rho= 1829.12- 0.61483 \left(T-T_{melt}\right) $$&#xA;where $T_{melt}=301.6$ K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the density reduces with temperature. Plotting the information that we have from the paper and from wikipedia yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h4k7K.png&quot; alt=&quot;rho_T_plot for Cesium&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jump at the melting point is from the mismatch between the paper and wikipedia. In the paper they specify specifically that the density is for the liquid case, on wikipedia it might (I'm not sure) be the density for the solid case at the melting point. If that is the case then the jump in density from solid to liquid is: $1843-1829=14 $kg/m$^3$&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-03-27T18:16:42.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4498" PostTypeId="2" ParentId="4494" CreationDate="2013-03-27T21:29:28.367" Score="3" Body="&lt;p&gt;Since I cannot comment yet, the only other way I know to say something here is to post an answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your question about the units, the thermal expansion coefficient is a percentage. From wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_L = \dfrac{1}{L}\dfrac{dL}{dT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a change &lt;code&gt;dT&lt;/code&gt; of temperature, the metal will expand by a percentage of $\dfrac{dL}{L}$.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-03-27T21:29:28.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4499" PostTypeId="1" AcceptedAnswerId="8338" CreationDate="2013-03-28T00:06:15.327" Score="2" ViewCount="1738" Body="&lt;p&gt;I require Congo red indicator for an experiment I wish to perform involving dipping a cellulose sponge into an acid and a base.  I first want to dye the cellulose sponge with Congo red, and observe the color change of the sponge as I place it in acids and bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congo red apparently binds to cellulose and thus is suitable for this experiment.  Unfortunately the laboratory doesn't have this indicator at the moment.  Are there any other indicators that could be used for this effect instead?&lt;/p&gt;&#xA;" OwnerUserId="1219" LastEditorUserId="12860" LastEditDate="2015-11-13T19:20:16.417" LastActivityDate="2015-11-13T19:20:16.417" Title="Alternatives to Congo Red?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4500" PostTypeId="2" ParentId="151" CreationDate="2013-03-28T06:09:12.617" Score="24" Body="&lt;p&gt;&lt;em&gt;This is just a confirmation to Aesin's answer...&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Say, we take copper. The expected electronic configuration (as we blindly fill the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals along the period) is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]}\mathrm{3d^9 4s^2}$&lt;/span&gt;, whereas the real configuration is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]}\mathrm{3d^{10} 4s^1}$&lt;/span&gt;. There is a famous interpretation for this, that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbitals are more stable when half-filled and completely-filled. That's a complete myth. There are very few pages explaining this myth, like &lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&quot; rel=&quot;noreferrer&quot;&gt;chemguide.co.uk&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As we fill the electrons starting from &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d^1}$&lt;/span&gt;, we'd be stuck at chromium and also at copper. While filling chromium and copper, it has been observed that the energies of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt; orbitals are fairly close to each other. The increasing nuclear charge (as we go along the period) and the size and shape of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbital should be a reason. This similarity makes the energy for pairing up electrons in d-orbital very less than that of pairing up in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital (&lt;em&gt;i.e.&lt;/em&gt;) the energy difference between these orbitals is much less than the pairing energy required to fill the electrons in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt; orbital. Moreover, the energy for the configuration &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d^5 4s^1}$&lt;/span&gt; is much less than that of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d^4 4s^2}$&lt;/span&gt;. Since we usually fill electrons in the order of increasing energy, the next electron (in case of manganes) goes into the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt;-orbital.&lt;/p&gt;&#xA;&lt;p&gt;The same reason for effective nuclear charge makes the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt;-orbitals somewhat lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt;-orbitals and hence, the unusual configuration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From a paper of Richard Hartshorn  and  Richard Rendle of the University of Canterbury (&lt;a href=&quot;http://www.chemteach.ac.nz/documents/chemed09/misc/Half%20full%20shells%20etc.doc&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;*.doc&lt;/code&gt; file&lt;/a&gt;), which supports that this is quite &lt;em&gt;untrue&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the case of chromium, this means that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s^1 3d^5}$&lt;/span&gt; will be lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s^2 3d^4}$&lt;/span&gt;, because in the second case you have to &amp;quot;pay&amp;quot; the electron pairing energy. Since this pairing energy is larger than any difference in the energies of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt; orbitals, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. Effectively this is Hund's rule applying not just to strictly degenerate orbitals (orbitals with the same energy), but to all orbitals that are (significantly) closer in energy than the electron pairing energy.&lt;/p&gt;&#xA;&lt;p&gt;In the case of copper, the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt;-orbital has dropped in energy below that of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt;, so that it is better to have the paired electrons in the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; and the unpaired one in the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;. The reason why the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3d}$&lt;/span&gt; is lower than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4s}$&lt;/span&gt; is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; compared with the earlier transition metals, and also means that orbitals in inner shells are more stabilised with respect to those further out for copper than for earlier elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1414" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-12-17T14:04:40.820" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4501" PostTypeId="1" CreationDate="2013-03-28T08:59:04.540" Score="2" ViewCount="3391" Body="&lt;p&gt;I am currently taking a general chemistry course and during a lab they had us do the following procedure:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place &lt;span class=&quot;math-container&quot;&gt;$\pu{2 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$5\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ \ce{NaClO} $&lt;/span&gt; solution in a test tube.&lt;/li&gt;&#xA;&lt;li&gt;Add &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; drops of cyclohexane.&lt;/li&gt;&#xA;&lt;li&gt;Add ~&lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; drops &lt;span class=&quot;math-container&quot;&gt;$\pu{6 M}\,\ce{HCl}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;In another test tube, add a pinch of solid &lt;span class=&quot;math-container&quot;&gt;$\ce{KBr}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Use a dropping pipette to withdraw the chlorine/cyclohexane layer from above and add it to the test tube containing &lt;span class=&quot;math-container&quot;&gt;$\ce{KBr}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;When I did this, the mixture turned dark red and bubbled with a hint of purple to it. In my lab report, my teacher wants me to write the net ionic equation(s) of this reaction, but I have no idea how to do that.&lt;/p&gt;&#xA;&lt;p&gt;How would you proceed?&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="95133" LastEditDate="2020-09-11T20:38:10.003" LastActivityDate="2020-09-11T20:47:24.523" Title="Writing the net reaction for reaction of sodium hypochlorite with hydrochloric acid and potassium bromide" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4502" PostTypeId="1" CreationDate="2013-03-28T17:08:23.193" Score="0" ViewCount="189" Body="&lt;p&gt;I am not sure how sharpness is measured, but I do know that certain materials can be designated sharper than others, for example &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1273673/&quot; rel=&quot;nofollow&quot;&gt;obsidian is said to be the sharpest material&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The prismatic glass blade is infinitely sharper than a honed steel edge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my question is, will a honed edge of high-tin bronze be sharper than a honed edge of low-tin bronze, and if so why?&lt;/p&gt;&#xA;" OwnerUserId="1416" LastActivityDate="2013-03-30T12:39:01.980" Title="Can high-tin bronze be made sharper than low-tin bronze?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4504" PostTypeId="2" ParentId="4493" CreationDate="2013-03-29T00:07:32.497" Score="2" Body="&lt;p&gt;Found the answer myself. In case someone else stumbles over this, the combining rules for Lennard-Jones parameters of van der Waals interactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance parameter: &#xA;$r_{XY}=\frac{1}{2}(r_{XX}+r_{YY})$ or $\sigma_{XY}=\frac{1}{2}(\sigma_{XX}+\sigma_{YY})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depth parameter: &#xA;$\varepsilon_{XY}=\sqrt{\varepsilon_{XX}\varepsilon_{YY}}$&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="605" LastEditDate="2013-03-29T00:14:19.043" LastActivityDate="2013-03-29T00:14:19.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4505" PostTypeId="2" ParentId="697" CreationDate="2013-03-29T00:13:50.800" Score="5" Body="&lt;p&gt;Essentially you are asking why about the different surface properties of PTFE vs. PVDF.  Generally PTFE is more 'non stick' than PVDF, which is more non-stick than polyethene (PE).  PTFE is much more chemically resistant to corrosion than PVDF or PE.  You are correct to look at the chemical structure.  The more Fluorine atoms on the backbone, the more proteins will not be able to chemically latch onto the surface.  Measuring things like contact angle will confirm this.  I don't know how much the pore structure of the membranes affect things, but from a chemical standpoint, there are clear differences.&lt;/p&gt;&#xA;" OwnerUserId="1419" LastActivityDate="2013-03-29T00:13:50.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4506" PostTypeId="1" AcceptedAnswerId="4508" CreationDate="2013-03-29T03:33:19.107" Score="6" ViewCount="558" Body="&lt;p&gt;When people talk about oxides like $\ce{SiO2}$ or $\ce{Al2O3}$, they use expressions like stoichiometric or under-stoichiometric. I understand that this refers to the relative composition of materials, but I don't really know what it exactly means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for $\ce{SiO2}$, does under-stoichiometric mean that it isn't exactly $\ce{SiO2}$, but more like $\ce{SiO_{1.8}}$? Does the term usually refer to the oxygen composition?&lt;/p&gt;&#xA;" OwnerUserId="1420" LastEditorUserId="1420" LastEditDate="2014-05-06T16:12:57.950" LastActivityDate="2014-05-06T16:12:57.950" Title="What does under-stoichiometric or stoichiometric mean?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4507" PostTypeId="1" CreationDate="2013-03-29T04:18:42.917" Score="3" ViewCount="592" Body="&lt;p&gt;I'm working with an abstract model of chemical reaction networks. In this model, some of the reactions have the form $\ce{A + B &amp;lt;=&amp;gt; C}$ and others have the form $\ce{A + B&amp;lt;=&amp;gt;C + D}$. I'm wondering if there is some existing terminology I can use to refer to these two cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the terms &quot;unimolecular&quot; and &quot;bimolecular&quot; refer to the number of reactants, but in this case the important thing is the number of products. The ideal thing would be if there's a general term describing whether a reaction has the same number of products as it does reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, since the species I'm modelling are abstract polymers, I can use the terms &quot;cleavage&quot; and &quot;ligation&quot; to refer to the $\ce{A + B &amp;lt;=&amp;gt; C}$ case, depending on the direction in which the reaction proceeds. If there is a corresponding term for $\ce{A + B&amp;lt;=&amp;gt;C + D}$ type reactions in polymers, that would also be useful to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose at a stretch I could call it an exchange or a substitution reaction, but (as far as I know) that tends to imply that the thing being substituted or exchanged is a single atom, rather than a whole section of polymer. (In my case the reaction might be something along the lines of $\ce{aaabb + ccdddd -&amp;gt; aaadddd + ccbb}$, where $\ce{a}$, $\ce{b}$, $\ce{c}$ and $\ce{d}$ are different types of monomer.)&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="4945" LastEditDate="2015-01-16T12:09:38.793" LastActivityDate="2015-01-16T12:09:38.793" Title="Is there a specific term for reactions of the form A + B → C + D?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4508" PostTypeId="2" ParentId="4506" CreationDate="2013-03-29T07:30:04.727" Score="10" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, the convention is to refer to the oxygen atoms when talking about over- and under-stoichiometric in metaloxides. Normally they also indicate this explicitly by writing $\ce {SiO}_x$ (like in &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/pssa.200521443/abstract&quot;&gt;the article here&lt;/a&gt; for example) or $\ce {Al_2O}_x$. Being explicit about it is a good habit because it avoids ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1085" LastEditDate="2013-03-30T15:59:37.873" LastActivityDate="2013-03-30T15:59:37.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4509" PostTypeId="2" ParentId="4507" CreationDate="2013-03-29T07:44:56.160" Score="5" Body="&lt;p&gt;There are 4 '&lt;a href=&quot;http://www2.ucdsb.on.ca/tiss/stretton/CHEM1/stoich2.html&quot; rel=&quot;noreferrer&quot;&gt;base-types&lt;/a&gt;' of chemical reactions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Synthesis reactions ($\ce{A + B -&amp;gt; AB}$)&lt;/li&gt;&#xA;&lt;li&gt;Decomposition reactions ($\ce{AB -&amp;gt; A + B}$)&lt;/li&gt;&#xA;&lt;li&gt;Substitution reactions ($\ce{A + BC -&amp;gt; B + AC}$)&lt;/li&gt;&#xA;&lt;li&gt;Metathesis reactions ($\ce{AB + CD -&amp;gt; AD + CB}$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given that you talk about polymers that in one case combine to form a new (single) polymer: $\ce{A + B &amp;lt;=&amp;gt; C}$ you could call this a &lt;strong&gt;synthesis&lt;/strong&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You other example, $\ce{A + B &amp;lt;=&amp;gt; C + D}$, is an exchange of parts of the polymers so that would be a &lt;strong&gt;metathesis&lt;/strong&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For polymers there are a couple of additional options to name the types of reactions, but these depend a lot on the specific chemistry (e.g. &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/108802/chemical-reaction/277196/Polymerization-reactions&quot; rel=&quot;noreferrer&quot;&gt;hydrolysis&lt;/a&gt; if water is one of the reactants) so that is probably difficult if you have 'abstract' polymers.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-03-29T07:44:56.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4510" PostTypeId="2" ParentId="4501" CreationDate="2013-03-30T05:31:36.247" Score="2" Body="&lt;p&gt;A ionic equation of a reaction isn't much different from a regular reaction equation, except that you write the ions separately. As per usual, reactants go on the left, products go on the right.&lt;/p&gt;&#xA;&lt;p&gt;As an example, let's let hydrochloric acid react with sodium carbonate in aqueous solution. First you figure out what ions your reactants dissociate into. For &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; that would be &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;, and for &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3}$&lt;/span&gt; that's &lt;span class=&quot;math-container&quot;&gt;$\ce{2 Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt;, and lastly you got water, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;. That's the left side of this example equation. Which in sum would look like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ +Cl- +2 Na+ +CO3^2- + H2O -&amp;gt;?}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Next you need to figure out what ions react and in what way. In this example, first, &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; react to form &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;, and after that, &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt; react to form &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;. You would write the two parts of the reaction like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + Cl- + 2 Na+ + CO3^2- + H2O -&amp;gt; Cl- +2Na+ + CO3^2- + H3O+}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Cl- + 2 Na+ + CO3^2- + H3O+ -&amp;gt; Cl- +2Na+ + CO2 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now for the last bit, stoichiometry. Stoichiometry simply means making sure the numbers of atoms left and right are the same, and the ratio of atoms needed to make the reaction work is right. The first line works as is, but in the second line you'll notice that we're using a single positive charge in &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; to react with two negative charges in &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt;. Something's fishy there, as charges need to match too. And indeed, we need two &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; for one &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt;. After we changed that in the second line, we need to alter the first line too to match, like so:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + 2Cl- + 2 Na+ + CO3^2- + 2H2O -&amp;gt; 2Cl- +2Na+ + CO3^2- + 2H3O+}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Cl- + 2 Na+ + CO3^2- + 2H3O+ -&amp;gt; 2Cl- +2Na+ + CO2 + 3H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;That's what your teacher wants you to do for the reaction you cited.&lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="95133" LastEditDate="2020-09-11T20:47:24.523" LastActivityDate="2020-09-11T20:47:24.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4511" PostTypeId="2" ParentId="4501" CreationDate="2013-03-30T12:24:34.493" Score="4" Body="&lt;p&gt;I'd like to add some things to what Tanith Rosenbaun said.&lt;/p&gt;&#xA;&lt;p&gt;First, the point of &lt;em&gt;net&lt;/em&gt; ionic equations is not just to write out the equations in terms of ions, but to focus on only those ions that experience a change in charge or valence.  Ions which don't change in charge don't really participate in the reaction and called &lt;em&gt;spectator ions&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Here's the last equation from Rosenbaum's answer:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H3O+ + 2Cl- + 2Na+ + CO3^2- -&amp;gt; 2H2O + 2Cl- + 2Na+ + CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice that the sodium and chloride ions are unchanged in this reaction.  So we &lt;em&gt;drop&lt;/em&gt; them, and just write out the ions that change:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H3O+ + CO3^2- -&amp;gt; 2H2O + CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is the &lt;em&gt;net&lt;/em&gt; ionic equation: it focuses precisely on what is interesting, which in this case is the neutralization reaction.&lt;/p&gt;&#xA;&lt;p&gt;Now let's go back to your problem:  People have been asking what you know about it.  You seem to know that you are transferring a solution of chlorine dissolved in cyclohexane.  You got that chlorine from the following reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2HCl + NaClO -&amp;gt; Cl2 + NaCl + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and the chlorine preferentially dissolves in the cyclohexane.  Now it's important to understand that the cyclohexane doesn't play any role in the second part of your experiment, it's just how you transport the chlorine.  (And it's why you need to be careful with acids around bleach: chlorine gas is poisonous!)&lt;/p&gt;&#xA;&lt;p&gt;The key question is: what happens when you put elemental chlorine in the presence of dissolved potassium bromide?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2 + KBr -&amp;gt; ?}$&lt;/span&gt;  (Note: this is not balanced yet!)&lt;/p&gt;&#xA;&lt;p&gt;In particular, do you know what you produced when that solution turned dark red?  Hint: this is a very important property that halogens like chlorine and bromine have towards each other, which is indeed seen best &lt;em&gt;via&lt;/em&gt; a net ionic equation.  I'll happily provide the answer if you've tried some stuff and are still stuck.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="95133" LastEditDate="2020-09-11T20:38:46.017" LastActivityDate="2020-09-11T20:38:46.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4512" PostTypeId="2" ParentId="4502" CreationDate="2013-03-30T12:39:01.980" Score="1" Body="&lt;p&gt;Unfortunately I don't have a solid reference for this, but what I remember from my undergrad courses is that the limit to sharpening is basically determined by the porosity and tensile strength of the metal you are trying to sharpen. At some point you will just run into the microscopic roughness of the metal itself and that is where sharpening ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low- and high-tin bronze are pretty rough classifications so I couldn't tell you which will result in a lower porosity or higher strength, but if I would have to guess than I think low-tin bronze can be made sharper, because &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper&quot; rel=&quot;nofollow&quot;&gt;copper&lt;/a&gt; has a higher strength than &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin&quot; rel=&quot;nofollow&quot;&gt;tin&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if you can get your hands on &lt;a href=&quot;http://openarchaeology.info/bibliography/distinguishing-metal-steel-and-low-tin-bronze-stone-flint-and-obsidian-tool-cut-marks&quot; rel=&quot;nofollow&quot;&gt;this book chapter&lt;/a&gt;, you can find out a more exact answer.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-03-30T12:39:01.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4513" PostTypeId="1" AcceptedAnswerId="4514" CreationDate="2013-03-30T13:19:16.083" Score="1" ViewCount="67690" Body="&lt;p&gt;$\ce{Mg + O2 → MgO}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium, oxygen and magnesium oxide form the above equation. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How?&lt;/li&gt;&#xA;&lt;li&gt;What do I need to learn to do this? atomic numbers/ valencies?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1426" LastEditorUserId="4231" LastEditDate="2015-02-26T17:49:54.000" LastActivityDate="2015-02-26T17:49:54.000" Title="How does the reaction Mg + O₂ → MgO occur?" Tags="&lt;reaction-mechanism&gt;&lt;molecules&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-27T15:42:38.287" ContentLicense="CC BY-SA 3.0" />
  <row Id="4514" PostTypeId="2" ParentId="4513" CreationDate="2013-03-30T13:35:12.377" Score="2" Body="&lt;p&gt;First of all, do you know that chemical equations are means to express chemical reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, this equation represents the reaction of magnesium with oxygen. Magnesium and Oxygen are the reactants and Magnesium Oxide is the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrow indicates the direction of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To write basic chemical equations, you only need to know your reactants, your products and how to write the symbols for them. Usually, you can 'guess' what the products can be from generic patterns or mechanism, but I don't think you will be there yet (or required to remember them) soon if you just got started on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the molecular formula or ionic formula in this case, you need to know the valencies of the elements constituting that compound. For instance here, Magnesium has valency 2, Oxygen has valency 2, so $\ce{MgO}$ is the formula for Magnesium Oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Sodium Oxide, Sodium has valency 1, Oxygen has valency 2, so $\ce{Na_2O}$ is the formula for Sodium Oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron (III) Oxide becomes: $\ce{Fe_2O_3}$ (Iron valency = 3; Oxygen valency = 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium Chloride becomes $\ce{MgCl_2}$ (Magnesium valency = 2; Chlorine valency = 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this answer your question or were you looking for something different?&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-03-30T13:35:12.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4515" PostTypeId="1" AcceptedAnswerId="36880" CreationDate="2013-03-30T14:55:44.630" Score="3" ViewCount="11725" Body="&lt;p&gt;I suppose this question can be extrapolated to food in general, but it is the case that I would like to know if there are any methods that would help me to keep the aloe gel fresh, i.e. avoiding it to rot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mostly squeeze the gel and pour it into a small jar. So far I have tried the following methods, being some of them obvious, but would appreciate corroboration anyway:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sterilize the container and the tools, e.g. boiling or steaming.&lt;/li&gt;&#xA;&lt;li&gt;Keep it in the fridge.&lt;/li&gt;&#xA;&lt;li&gt;Stir with C and E vitamin powder to avoid oxidation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know little about chemistry, but have heard about pasteurization and some other methods that I couldn't afford at home so, even if you mention those, I will be mainly interested in homemade methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am forced to choose the tags for the post but I am not sure if they are the ones that best suit the question. Please, let me know if I should change them.&lt;/p&gt;&#xA;" OwnerUserId="1428" LastEditorUserId="1428" LastEditDate="2013-03-31T10:27:58.540" LastActivityDate="2018-12-02T05:08:51.127" Title="Homemade preservative for natural aloe-vera gel?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4516" PostTypeId="1" CreationDate="2013-03-30T21:24:22.493" Score="6" ViewCount="5465" Body="&lt;p&gt;A friend of mine gave me a bottle of home-made liquor which is a concoction of pure alcohol and sweetened milk. I accepted his gift happily and then I put the bottle in my cocktail cabinet. After an hour, we happened to hear a terrible noise of a some sort of explosion. We were indeed right - the bottle had exploded with actually no reason! Could someone please explain this strange phenomenon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure whether this is important but he claims he had been keeping it in his fridge, although the difference of temperatures (in the fridge and in the cocktail cabinet) does not seem terribly significant to me. Also, the bottle was almost full meaning there was little room for the air in the bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please find attached the full list of ingredients:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.5 litre of condensed, unsweetened milk,&lt;/li&gt;&#xA;&lt;li&gt;0.5 litre of condensed, sweetened milk,&lt;/li&gt;&#xA;&lt;li&gt;0.25 litre of pure alcohol,&lt;/li&gt;&#xA;&lt;li&gt;few sweets.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1429" LastEditorUserId="7475" LastEditDate="2015-10-30T16:38:34.870" LastActivityDate="2015-10-30T16:38:34.870" Title="Spontaneous explosion of a bottle containing alcohol and milk" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4517" PostTypeId="1" AcceptedAnswerId="8039" CreationDate="2013-03-31T03:20:31.730" Score="44" ViewCount="7750" Body="&lt;p&gt;In molecular orbital theory, the fact that a bonding and antibonding molecular orbital pair have different energies is accompanied by the fact that the energy by which the bonding is lowered is less than the energy by which antibonding is raised, i.e. the stabilizing energy of each bonding interaction is less than the destabilising energy of antibonding. How is that possible if their sum has to equal the energies of the combining atomic orbitals and conservation of energy has to hold true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Antibonding is more antibonding than bonding is bonding.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, the fact that $\ce{He2}$ molecule is not formed can be explained from its MO diagram, which shows that the number of electrons in antibonding and bonding molecular orbitals is the same, and since the destabilizing energy of the antibonding MO is greater than the stabilising energy of bonding MO, the molecule is not formed. This is the common line of reasoning you find at most places.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2016-09-09T07:17:47.857" LastActivityDate="2018-12-27T12:46:25.397" Title="How can antibonding orbitals be more antibonding than bonding orbitals are bonding?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4518" PostTypeId="1" CreationDate="2013-03-31T03:29:29.123" Score="1" ViewCount="209" Body="&lt;p&gt;Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?&#xA;if a system contains for example an inflammable gas dispersed with another combustible gas such that the molecular speed are insufficient to cause the reaction. but as the molecular speeds keep fluctuating, at some point, it may cross the energy barrier and may start the reaction which will eventually spread evenly but till then the temperature there would be high and thus the system in inequilibrium.&#xA;isn't this also an example that although the free energy is minimum and entropy maximum, some useful work would've been done if the temperature gradient had been utilised?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-04-07T14:00:12.117" LastActivityDate="2013-07-11T21:57:17.337" Title="Spontaneous inequilibrium in a gas in an isolated system?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4519" PostTypeId="2" ParentId="4517" CreationDate="2013-03-31T04:39:20.053" Score="5" Body="&lt;p&gt;The sum of the energies over all orbitals is constant. The lower energy in a bonded state arises from the fact that not all MOs are occupied by electrons. Because the energy of the electrons is lowered, the total system energy is lowered by bonding. See &lt;a href=&quot;http://www.dlt.ncssm.edu/tiger/diagrams/bonding/BondingEnergy.gif&quot; rel=&quot;noreferrer&quot;&gt;http://www.dlt.ncssm.edu/tiger/diagrams/bonding/BondingEnergy.gif&lt;/a&gt; for an illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;This phenomenon is due to the nuclear-nuclear repulsion of the two (or more) atoms that are binding. See Ball, D.W. &lt;em&gt;Physical Chemistry&lt;/em&gt; p. 408 &lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="1389" LastEditDate="2013-03-31T07:53:06.893" LastActivityDate="2013-03-31T07:53:06.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4522" PostTypeId="1" CreationDate="2013-03-31T15:08:45.813" Score="3" ViewCount="813" Body="&lt;p&gt;I'm asked to find the Lewis Structure  of sulfuric acid, $SO_4H^-$. After struggling for a while, I sumply looked it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3zWnq.png&quot; alt=&quot;Source: Wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be? Sulfur has 6 electrons in its valence shell, so it should form bonds so as to collect 2 more electrons and satisfy the octet rule. So, surely it should form 2 bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's involved in no less than &lt;em&gt;six&lt;/em&gt; bonds here. Huh?&lt;/p&gt;&#xA;" OwnerUserId="1433" LastActivityDate="2013-03-31T15:41:12.700" Title="Number of bonds in a Lewis Structure" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-04-01T14:29:41.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="4523" PostTypeId="2" ParentId="4522" CreationDate="2013-03-31T15:41:12.700" Score="2" Body="&lt;p&gt;The octet rule only applies firmly to elements below main energy level 3 (in fact, there are rare instances in which even that isn't entirely accurate, but molecular orbtal theory needs to be invoked to explain those situations). At main energy level 3, which Sulfur occupies, unfilled d orbitals become available, and those d orbitals serve to accommodate the additional electrons in Sulfur's bonds. In reality, Sulfur is a much larger atom than Oxygen, and the &amp;pi; orbital overlap between the two elements is fairly poor, giving the bonds significant single-bond character. You might therefore see resonance structures in which single bonds are drawn instead, with formal positive charges assigned to Sulfur and formal negatives assigned to Oxygen.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-03-31T15:41:12.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4524" PostTypeId="2" ParentId="4450" CreationDate="2013-03-31T16:33:42.273" Score="2" Body="&lt;p&gt;I might not be sure on this, but the instantaneous partial molar gibbs free energy change at a particular P and T (chemical potential) shows the chemical equivalent of potential energy and thus the graph of it versus the conditions gives us an idea of the &quot;driving force&quot; of the reaction which is maximum if the rate of change of chemical potential is maximum and hence the substance under consideration has the greatest tendency to react or rather to move from one phase to another, i.e equivalent to say that it is less stable.&lt;br&gt;&#xA;Partial molar gibbs free energy change is nothing but the change in gibbs free energy per a small  change in phase i.e for a small part of reactants to be converted into products in a reaction under consideration. If this is high, slope of the graph of G versus Moles is high the reaction has a greater tendency to move forward   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;well unfortunately there is no know-it-all thermodynamic function ( atleast that i know of!) because, similar to the variations in stability caused by different pressure and temperature, number of moles and other possible phases in the system also affect the stability. Hence you can only talk about the resistance of of a substance to a particular reaction or a change of phase (in the thermodynamic sense).&lt;br&gt;&#xA; For example, methane is quite stable in presence of intense sunlight alone, but if fluorine or chlorine is introduced into the system under consideration, there is explosive conversion of methane into methyl halides and thus the stability is out of question. therefore you would have to compute the potential changes for a mixture with all properties defined before you can comment on its stability.&lt;br&gt;&#xA; But as is given in most texts, an almost improper way of understanding stability, is through seeing how negative the heat of formation is but it only gives details as to how stable it is in comparison only to the reactants.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-04-01T17:45:10.617" LastActivityDate="2013-04-01T17:45:10.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4525" PostTypeId="1" AcceptedAnswerId="4537" CreationDate="2013-03-31T17:44:31.103" Score="4" ViewCount="191" Body="&lt;p&gt;The question title is a bit odd, I couldn't find a plainer way of saying it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given the concentration and conductivity of a solution of a particular acid, let's just call it $\ce{AH}$. I then have to deduce that acid's acidity constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first instinct is to write out the expression for the solution's conductivity... but then I realize I don't know how to do that. I have the coefficients for all the relevant ions (I'm not sure what the technical term is, but essentially I mean the lambdas in the second equation &lt;a href=&quot;http://en.wikipedia.org/wiki/Conductivity_%28electrolytic%29#Strong_electrolytes&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), in particular, the ions $\ce{H_3O^+}$, $\ce{OH^-}$ and the conjugate base $\ce{A^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I had the concentrations of those ions in the solution, I could solve the problem! But hang on... I have no idea how to figure them out. I know the concentration of the acid in the solution. Suppose the acid has just been added to the water. Then it won't have started to disassociate yet, so the water will be exactly as it was before we added the acid: $[\ce{H_3O^+}] = [\ce{OH^-}] = 10^{-7}$. But what if originally there was a lot more acid, but it's had time to react with the water? Then the water will have been filled with an amount of $\ce{A^-}$ dependent on how long ago the acid was dissolved in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I assume the acid is at equilibrium? The question doesn't specify that.&lt;/p&gt;&#xA;" OwnerUserId="1433" LastEditorUserId="24052" LastEditDate="2017-03-15T16:12:16.677" LastActivityDate="2017-03-15T16:12:16.677" Title="How am I to interpret the phrase &quot;a solution of acid&quot;?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4526" PostTypeId="1" AcceptedAnswerId="30999" CreationDate="2013-03-31T17:44:50.110" Score="3" ViewCount="924" Body="&lt;p&gt;If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems that since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs.&lt;/p&gt;&#xA;" OwnerUserId="1372" LastEditorUserId="1372" LastEditDate="2013-04-01T21:19:46.867" LastActivityDate="2015-05-07T10:55:33.727" Title="Minimizing Gibbs free energy with Helmholtz free energy expressions" Tags="&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4527" PostTypeId="1" AcceptedAnswerId="4538" CreationDate="2013-04-01T06:06:36.220" Score="4" ViewCount="192" Body="&lt;p&gt;I know this is simple, compared to most questions asked here, but I notice this several times a week and I'm curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice if I break an egg and drop it into the frying pan, and the pan is still warming up, the egg sticks to the pan.  But if I wait until the pan is quite hot, and I break an egg and drop it in, the egg barely sticks to the surface at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the heat make a non-stick surface work better then when it is merely warm or is still warming up?&lt;/p&gt;&#xA;" OwnerUserId="1436" LastEditorUserId="9961" LastEditDate="2015-10-30T17:01:37.873" LastActivityDate="2015-10-30T17:01:37.873" Title="Why do frying pans stick less with heat?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4528" PostTypeId="1" CreationDate="2013-04-01T07:24:59.443" Score="2" ViewCount="11030" Body="&lt;p&gt;At standard conditions(at 1atm pressure and unit activity(1molal or 1molar concentration)of all dissolved compounds),the electrode potential is equal to the standard electrode potential by the Nernst equation. $$ \Delta E = \Delta E^0 $$&#xA;At the same time,while the electrochemical reaction is at equilibrium,the electrode potential of the cell, $$ \Delta E =0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this must mean,that when ANY reaction is at equilibrium under the standard conditions, $$ \Delta E = \Delta E^0 = 0 $$&#xA;Which is quite surprising.Is this correct?please explain.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="1018" LastEditDate="2013-04-01T08:44:21.730" LastActivityDate="2015-04-28T19:03:19.637" Title="Standard electrode potential of a galvanic half cell is zero at equilibrium at standard conditions?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4529" PostTypeId="1" AcceptedAnswerId="4530" CreationDate="2013-04-01T10:51:18.047" Score="5" ViewCount="3205" Body="&lt;p&gt;I'm familiar with the periodic table / periodic system, but I wonder why it's called &quot;periodic&quot; since there seems not much periodic about (there seems to be little or no predictability of which elements are stable) and from physics I know Z-value of elements and that is not periodic either. So what is a &quot;period&quot; in this context?&lt;/p&gt;&#xA;" OwnerUserId="1106" LastActivityDate="2020-01-07T12:48:26.707" Title="What is a &quot;period&quot; of the periodic table?" Tags="&lt;periodic-table&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4530" PostTypeId="2" ParentId="4529" CreationDate="2013-04-01T11:07:38.540" Score="6" Body="&lt;p&gt;I agree that the &quot;periods&quot; in the periodic table are not mathematically regular.  The simplest definition is that a &lt;em&gt;period&lt;/em&gt; begins when a new &lt;em&gt;s&lt;/em&gt;-subshell starts to fill.  Recall that in the &lt;em&gt;s, p, d, f&lt;/em&gt; subshells there are 2, 8, 10, and 14 electrons, respectively, so the periods have to get bigger over time and thus cannot be regular.  As to the lack of predictability of stability, that may be true, but the periodic table is pretty good at predicting the &lt;em&gt;chemistry&lt;/em&gt; of elements by collecting together similar elements into &lt;em&gt;groups&lt;/em&gt; (the columns).  This similarity is mostly expressed by the &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt; subshells, which is where a lot of the reactivity occurs.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="-1" LastEditDate="2020-01-07T12:48:00.053" LastActivityDate="2020-01-07T12:48:00.053" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4531" PostTypeId="2" ParentId="4364" CreationDate="2013-04-01T11:20:28.963" Score="2" Body="&lt;p&gt;Yes, any compound of carbon that can be synthesised, can be synthesised from carbon that comes from captured atmospheric $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We discussed related subjects recently on the new &lt;a href=&quot;https://sustainability.stackexchange.com/a/715/48&quot;&gt;Sustainability Stack Exchange, on the options for plastics without oil&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least &lt;a href=&quot;http://www.earth.columbia.edu/articles/view/2523&quot; rel=&quot;nofollow noreferrer&quot;&gt;three&lt;/a&gt; &lt;a href=&quot;http://www.keith.seas.harvard.edu/AirCapture.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;research&lt;/a&gt; &lt;a href=&quot;https://www.llnl.gov/news/aroundthelab/2012/May/ATL-052412_catalyst.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;outfits&lt;/a&gt; working on the atmospheric extraction of $\ce{CO2}$. Now, inevitably, capturing it and turning it back into carbon is going to take more energy than gets released from burning coal or gas in the first place to release the $\ce{CO2}$, but that doesn't make the chemistry impossible (and it isn't a deal-breaker economically either, because energy costs vary hugely by time and space.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as &lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2012/January/graphene-oxide-carbon-dioxide-reactions.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;the article from the UK's Royal Chemistry Society from Jan 2012&lt;/a&gt; that you linked to, highlights, this is already being done:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Graphene oxide can be easily reduced to graphene but is also useful in its own right ... Lee and colleagues produced graphene oxide through a two-step process: first, they fixed the $\ce{CO2}$ to ammonia borane ($\ce{NH3BH3}$) and then they heated the resulting solid to create a honeycomb structure of graphene oxide. In the first reaction, they placed $\ce{NH3BH3}$ into a stainless steel cell and then added $\ce{CO2}$ until the cell reached a target pressure. They then gradually heated the cell to 100&amp;deg;C over a few hours. At a pressure of about 30atm, the mass of the $\ce{NH3BH3}$ doubled thank to the addition of carbon dioxide, producing $\ce{OCH3}$, $\ce{HCOO}$ and aliphatic groups. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's the original paper on the research: &lt;a href=&quot;http://dx.doi.org/10.1021/jp210295e&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of Graphene Oxide Nanocomposites from Carbon Dioxide Using Ammonia Borane, Junshe Zhang et al, 2012&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:27.820" LastActivityDate="2013-04-01T11:20:28.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4532" PostTypeId="2" ParentId="4527" CreationDate="2013-04-01T11:42:22.167" Score="0" Body="&lt;p&gt;The solid egg is not sticky like the liquid. If the pan is hotter it is solid on reaching the pan surface, so it doesn't stick.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-04-01T11:42:22.167" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4533" PostTypeId="2" ParentId="4516" CreationDate="2013-04-01T12:35:29.750" Score="0" Body="&lt;p&gt;I am guessing there was only a small gap between the cap and the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was not a fermentation with that much alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be a question of gas being less soluble in liquid at a higher temperature (with the gas being CO2) causing a pressure rise.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-04-01T12:35:29.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4534" PostTypeId="1" CreationDate="2013-04-01T13:34:42.933" Score="-3" ViewCount="4523" Body="&lt;p&gt;which chemical element has the lowest boiling point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it Helium(He) or is it Hydrogen(H)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is lighter than Helium.&lt;/p&gt;&#xA;" OwnerUserId="1269" LastActivityDate="2013-04-01T14:59:37.330" Title="which chemical element has the lowest boiling point?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2013-04-01T16:13:43.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="4535" PostTypeId="2" ParentId="4528" CreationDate="2013-04-01T14:37:06.427" Score="4" Body="&lt;p&gt;There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what  concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\Delta E$ however, consider the Nernst equation, &#xA;$$E = E^\circ - (RT/nF)\ln Q$$&#xA;Change in $E$&#xA;$$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$&#xA;therefore&#xA;$$\Delta E = -(RT/nF)\Delta\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero). &lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorDisplayName="user7951" LastEditDate="2015-04-28T19:03:19.637" LastActivityDate="2015-04-28T19:03:19.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4536" PostTypeId="2" ParentId="4534" CreationDate="2013-04-01T14:59:37.330" Score="4" Body="&lt;p&gt;At its present form, this question is quite bad. I think you should've had a look at Wiki before asking this question. The boiling point of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen&quot; rel=&quot;nofollow&quot;&gt;Hydrogen&lt;/a&gt; is around 20 K, while for &lt;a href=&quot;http://en.wikipedia.org/wiki/Helium&quot; rel=&quot;nofollow&quot;&gt;Helium&lt;/a&gt;, it's around 4 K. I don't memorize these. But, I had a dirty look at Wiki. If you want a reason for &lt;strong&gt;why&lt;/strong&gt; it has...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium has a stablest $1s$ orbital, which has somewhat lower energy. And, this makes it chemically inert towards other elements and to other Helium atoms, which is also the reason for its lowest boiling point.&lt;/p&gt;&#xA;" OwnerUserId="1414" LastActivityDate="2013-04-01T14:59:37.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4537" PostTypeId="2" ParentId="4525" CreationDate="2013-04-01T16:27:22.307" Score="4" Body="&lt;p&gt;The acidity constant is defined as the equilibrium constant for the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA &amp;lt;=&amp;gt; H^+ + A^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="402" LastEditDate="2017-03-15T10:23:22.283" LastActivityDate="2017-03-15T10:23:22.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4538" PostTypeId="2" ParentId="4527" CreationDate="2013-04-01T17:05:22.290" Score="5" Body="&lt;p&gt;Heat creates steam bubbles, which hold the egg away from the pan until it is somewhat cooked - see &lt;a href=&quot;http://en.wikipedia.org/wiki/Leidenfrost_effect&quot;&gt;Leidenfrost effect&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1365" LastActivityDate="2013-04-01T17:05:22.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4539" PostTypeId="2" ParentId="4516" CreationDate="2013-04-02T00:17:38.763" Score="5" Body="&lt;p&gt;The bottle can't have exploded for &quot;no actual reason&quot;; there may have been no &lt;em&gt;apparent&lt;/em&gt; reason, but everything does indeed happen for a reason, at least at this level. But this isn't English.SE...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My bet is on a combination of things. As with many things in everyday chemistry, the substances involved are never pure enough to pinpoint one exact cause. Case in point, milk is an emulsion containing a significant percentage of fats and proteins. So, possible causes include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hygroscopy and acidity of the alcohol causing the milk to curdle, by taking the water from the milk to form an alcohol solution that the complex milk proteins and fats precipitate out of due to lower solubility. This may produce a &quot;greater than the sum&quot; volume of product by leaving these proteins and fats in a porous gel (basically cottage cheese, though usually a stronger acid than ethanol is used in cheesemaking).&lt;/li&gt;&#xA;&lt;li&gt;Esterification. Milk, as stated, contains fats including fatty acids. Add alcohol, and the fatty acids react with the alcohol to produce esters (neutral fats). The byproduct is water, formed from the acid stripping the alcohol's hydroxyl group. Again, the products of this reaction can take up more volume than the reactants.&lt;/li&gt;&#xA;&lt;li&gt;Biological reduction. I actually do not think the concentration of alcohol in this solution would be antiseptic; if the alcohol is pure then we're only talking about a 20% solution by volume, and that would have to be a lab-distilled product like Everclear; &quot;natural&quot; distillation in open air only gets you to about 150 proof, and using even 100-proof vodka gives this mixture only a 10% concentration by volume of ethanol. According to at least &lt;a href=&quot;http://www.wisegeek.org/why-is-alcohol-a-good-antiseptic.htm&quot; rel=&quot;nofollow&quot;&gt;one source&lt;/a&gt;, ethanol needs to be concentrated to at least a 65% solution by volume to be effective as an antiseptic. So, to a bacteria's point of view, the alcohol is an overworked police officer; it'll slow down bacterial growth, but there's &lt;em&gt;way&lt;/em&gt; too much fat and sugar here for the alcohol to be much of a deterrent. Consumption of organic fuel by cellular metabolism, as a rule, produces CO&lt;sub&gt;2&lt;/sub&gt; gas, and in a closed container at room temperature, well, you get the picture.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-04-11T16:42:46.770" LastActivityDate="2013-04-11T16:42:46.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4540" PostTypeId="1" CreationDate="2013-04-02T05:14:09.823" Score="5" ViewCount="990" Body="&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/oIJlVbr.png&quot; alt=&quot;enter link description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no clue what the substituent may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of the unsubstituted benzaldehyde is 105 g/mol (subtracting one H, where the substituent will be). So the substituent is around 30–33 g/mol. Also I'm having trouble identifying the molecular ion. Is it the really small 138 peak or the 137 peak?&lt;/p&gt;&#xA;" OwnerUserId="1438" LastEditorDisplayName="user7951" LastEditDate="2018-12-06T12:11:12.013" LastActivityDate="2018-12-06T12:11:12.013" Title="What is the substituent on this benzaldehyde? (MS attached)" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4544" PostTypeId="2" ParentId="4540" CreationDate="2013-04-02T19:39:32.200" Score="4" Body="&lt;p&gt;The molecular ion $\ce{[M]^{\bullet +}}$ peak is likely at 136 m/z. The peak at 137 m/z is due to the 1.1% natural abundance of &lt;a href=&quot;http://en.wikipedia.org/wiki/13C&quot; rel=&quot;nofollow&quot;&gt;$\ce{^{13}C}$&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak at 135 m/z is the $\ce{[M-1]^{+}}$ peak typical of aldehydes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ [R-C(=O)-H]^{\bullet +} -&amp;gt; [R-C+=O &amp;lt;-&amp;gt; R-C\equiv O+] +H ^{\bullet} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another typical fragmentation of aldehydes is the loss of the CHO for $\ce{[M-29]^{+}}$, which you see at 107 m/z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ [R-C(=O)-H]^{\bullet +} -&amp;gt; R+ + [HC=O] ^{\bullet} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the main contributor to the $\ce{[M+1]^{\bullet +}}$ peak is $^{13}\ce{C}$ we can estimate the number of carbon atoms ($C$) in the molecule from the ratio of intensities of the $\ce{[M]^{\bullet +}}$ to the $\ce{[M +1]^{\bullet +}}$ peak:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{I_{M+1}}{I_{M}}=\frac{1.1}{98.9}C$$&#xA;$$C=\frac{98.9I_{M+1}}{1.1I_{M}}=89.9\frac{I_{M+1}}{I_{M}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensities appear to be $I_M \approx75$ and $I_{M+1} \approx 7$.&#xA;$$\therefore C=89.9\frac{7}{75}=8.4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number of carbon atoms needs to be an integer, round down to $C=8$. We can now work backward to get a formula. Eight carbon atoms have a mass of $8*12=96$ g/mol, leaving 40 g/mol to be the rest of the formula. We know one is an oxygen (aldehyde) (16 g/mol), leaving 24 g/mol. The most number of hydrogen atoms that can be associated with eight carbon atoms is 18 (2n+2), so there has to be some other atom, maybe an oxygen or nitrogen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave rest to you:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Maybe there is a second oxygen? &lt;/li&gt;&#xA;&lt;li&gt;How many hydrogens do you have? &lt;/li&gt;&#xA;&lt;li&gt;What is the final formula?&lt;/li&gt;&#xA;&lt;li&gt;How is it different from benzaldehyde?&lt;/li&gt;&#xA;&lt;li&gt;How could those extra atoms be placed on the ring? (there are many possibilities) &lt;/li&gt;&#xA;&lt;li&gt;Do you have any other evidence that suggests which of the possibilties from #5 is correct (IR, NMR, etc.)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-02T19:39:32.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4547" PostTypeId="1" CreationDate="2013-04-03T13:25:10.700" Score="8" ViewCount="26618" Body="&lt;p&gt;How to find the valencies of elements by using its distribution of electrons?&#xA;Please explain the method in simple words. Do you have to study the valencies or is there a simple way of remembering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: There is supposed to be some method of getting the valency after knowing the distribution of electrons. For example the element sodium, $\ce{Na}$ has a distribution of electrons: 2, 8, 1. What is it's valency?  &lt;/p&gt;&#xA;" OwnerUserId="1426" LastEditorUserId="13339" LastEditDate="2015-06-14T14:15:47.780" LastActivityDate="2017-04-21T23:41:48.027" Title="How to get the valencies of elements?" Tags="&lt;atoms&gt;&lt;molecules&gt;&lt;electrons&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4548" PostTypeId="2" ParentId="4547" CreationDate="2013-04-03T15:12:46.553" Score="8" Body="&lt;p&gt;Look at it this way: The atom needs to have a complete octet in its outermost shell, and will give away/accept electrons to achieve this goal. Sodium, for example, can give away one electron (+1 valency) or accept 7 (-7). Generally accepting more than 4 electrons is hard, so you don't see the latter happening. Similarly, oxygen has $2,8,6$ -- it can accept 2 electrons (-2) or give up 6 (+6, never observed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/&quot;&gt;the octet rule is not universally applicable&lt;/a&gt;. Elements can have multiple valencies, not all satisfying the octet rule. To understand this you need to understand orbital theory first (which will be hard to explain in the scope of one post).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2013-04-03T15:12:46.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4550" PostTypeId="2" ParentId="4529" CreationDate="2013-04-03T20:04:45.677" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So what is a &quot;period&quot; in this context?&#xA;  To answer this directly, &lt;strong&gt;a period is a row of the table.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As user467 indicates, the repeating periods reflect the filling-up of atomic orbitals. Because the valence electrons (those in the outer shell of an atom) determine many of the chemical properties, the periodic structure places atoms with similar valences in similar groups (vertical columns.)&lt;/p&gt;&#xA;" OwnerUserId="912" LastEditorUserId="-1" LastEditDate="2020-01-07T12:48:26.707" LastActivityDate="2020-01-07T12:48:26.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4552" PostTypeId="1" CreationDate="2013-04-04T00:56:52.380" Score="2" ViewCount="149" Body="&lt;p&gt;I need to 'adhese' paper to plastic, but I need the adhesive to be flammable and quick burning, but without producing any ash or too many toxic fumes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is an absurd question, but does anything like this exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much&lt;/p&gt;&#xA;" OwnerUserId="1443" LastActivityDate="2013-10-31T14:07:15.017" Title="I'm looking for a substance that can be used as an adhesive, but that will burn without producing any ash" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4554" PostTypeId="1" CreationDate="2013-04-04T03:30:39.263" Score="9" ViewCount="93214" Body="&lt;p&gt;Why does the melting point get lower  going down the Alkali Metal Group with increase in atomic number?&lt;/p&gt;&#xA;" OwnerUserId="1269" LastEditorUserId="22" LastEditDate="2013-04-04T07:50:54.927" LastActivityDate="2016-02-10T13:03:46.967" Title="Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?" Tags="&lt;physical-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4555" PostTypeId="2" ParentId="242" CreationDate="2013-04-04T05:25:07.690" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, I ask, what are the trends leading to the brittleness of a metal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Brittleness&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brittle&lt;/a&gt; materials absorb very little energy before fracturing. The only trend that can be linked to this is the strength of bonding in the metal (Other factors like crystal structure and purity do not have any trends and so there are a lot of exceptions). This also means that the trend in brittleness is not very regular since it is complex and depends on too many other factors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why are these trends there? (usually trends can be explained in terms of simpler things like atomic size/sheilding/etc)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since brittleness cannot be explained by a very simple trend, I will explain what reasons I have found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stronger the bonding in the metal, the more difficult it is to change its structure and thus it tends to break instead of deforming. Here's a graph (melting point vs. Group number) that gives a general measure of strength of metallic bonding in all the transition elements of the first three series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wzX1C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the graph, we predict that Tungsten (Period 6 Group VIB) should the most brittle and this is true for impure tungsten. Again, this successfully predicts that Silver, Gold and Copper (Group IB) are least brittle (and so most malleable and ductile). Also Lanthanum (Period 6 Group IIIB) is predicted to be quite malleable and this is also true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this also explains the high malleability of alkali metals (the graph is not very clear at that point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be much more exceptions to this trend when examined closely so take this only as an approximation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Crystal structures also play an important role in brittleness but they cannot be explained by a trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purity of a metal tends to increase malleability and decrease brittleness because there are less points at which a fracture can take place.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="1322" LastEditDate="2013-04-04T05:32:53.020" LastActivityDate="2013-04-04T05:32:53.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4556" PostTypeId="2" ParentId="4552" CreationDate="2013-04-04T12:12:42.143" Score="2" Body="&lt;p&gt;Nitrocellulose(flash paper). Using acetone or something you get a gel. Caution recommended. &lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-04-04T12:12:42.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4558" PostTypeId="1" AcceptedAnswerId="4560" CreationDate="2013-04-04T14:46:46.000" Score="8" ViewCount="11296" Body="&lt;p&gt;I am considering making a sheet of graphene using the LightScribe technique and I have read that I need graphite oxide. However when I look up graphite oxide for sale, graphene oxide pops up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two oxides similar? What is the difference? &lt;/p&gt;&#xA;" OwnerUserId="1446" LastEditorUserId="4945" LastEditDate="2015-01-06T06:58:56.867" LastActivityDate="2015-01-06T06:58:56.867" Title="What is the difference between graphite oxide and graphene oxide?" Tags="&lt;terminology&gt;&lt;nanotechnology&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4559" PostTypeId="2" ParentId="4554" CreationDate="2013-04-04T17:09:37.187" Score="14" Body="&lt;p&gt;You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-04-04T17:09:37.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4560" PostTypeId="2" ParentId="4558" CreationDate="2013-04-04T17:27:28.577" Score="9" Body="&lt;p&gt;In short, graphite is several graphene sheets piled one above another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphene is made up of one single sheet of carbon atoms arranged in hexagonal pattern (like a honeycomb), and graphite are several such sheets, each sheet linked to another by weak intermolecular forces, which gives the graphite its lubricative properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Similarly, graphene oxide is a layer of graphene with oxygen atoms, $OH$ groups and $COOH$ groups attached to it, while graphite oxide are layers of graphene oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the presence of $COOH$ groups, graphite oxide can disperse in basic solutions (or even water!) to give graphene oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CO_2$ is simply carbon dioxide, one carbon atom doubly bonded two two oxygen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, graphite oxide is like a huge field of carbon with the groups I mentioned earlier while carbon dioxide is like a small ball.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2014-07-19T21:23:46.963" LastActivityDate="2014-07-19T21:23:46.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4563" PostTypeId="1" CreationDate="2013-04-05T09:28:20.153" Score="3" ViewCount="1992" Body="&lt;p&gt;Is there a technical difference between the terms &quot;autocatalysis&quot; and &quot;branching chain reaction&quot;? They both seem to refer to something very similar, namely sequences of reactions in which one of the intermediates is also one of the products, leading to the possibility of exponential growth. Is one of the terms more specific than the other, or are they generally understood to mean the same thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There do seem to be some differences in how the terms are used. I've never seen the formose cycle, for example, referred to as a chain reaction, which does seem to imply that the term &quot;chain reaction&quot; means something more specific. If this is the case, what is the defining difference between the two terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of the formose cycle (from the &lt;a href=&quot;http://de.wikipedia.org/wiki/Formose&quot; rel=&quot;nofollow noreferrer&quot;&gt;German Wikipedia entry&lt;/a&gt;, because it has a much better diagram than the English one) - is there a specific reason why it would not be referred to as a chain reaction?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UhZna.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="148" LastEditDate="2013-04-06T12:48:50.593" LastActivityDate="2013-04-06T12:48:50.593" Title="Difference between the terms &quot;autocatalysis&quot; and &quot;(branching) chain reaction&quot;" Tags="&lt;theoretical-chemistry&gt;&lt;kinetics&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4564" PostTypeId="2" ParentId="4563" CreationDate="2013-04-05T11:16:33.293" Score="4" Body="&lt;p&gt;An &lt;a href=&quot;http://en.wikipedia.org/wiki/Autocatalysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;autocatalytic reaction&lt;/a&gt; is a reaction where the product catalyzes the reaction. An autocatalytic reaction takes the following form (or more complex variations of it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A + B -&amp;gt; 2B}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An autocatalytic reaction must produce the exact species $\ce{B}$ that cataylzes the reaction. More about autocatalytic reactions, including examples, can be found at &lt;a href=&quot;https://chemistry.stackexchange.com/questions/290/definitions-of-autocatalysis?rq=1&quot;&gt;this question&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4199/what-reactions-will-their-reaction-rates-increase-with-time/4471#4471&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_reaction#Chemical_chain_reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;chain reaction&lt;/a&gt; is one where every step of the mechanism produces another activated intermediate of some kind that can continue to react under the same conditions. Chain reactions continue until they terminate by means specific to the type of chain reaction. Chain reactions may appear to be autocatalytic, but there is a net consumption of both reagents, and the end product does not tend to cataylze the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B}* \ce{ -&amp;gt; C}$$&#xA;$$\ce{C + B -&amp;gt; D + B}*$$&#xA;Net:&#xA;$$\ce{ A + B -&amp;gt; D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the generic example above, the second step produces more $\ce{B}*$, which can reenter the first step. Overall, however, the reaction is not autocatalytic, since no more $\ce{B}$ is being produced and the concentration of $\ce{B}*$ is likely unchanging (&lt;a href=&quot;http://en.wikipedia.org/wiki/Steady_state_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;steady state&lt;/a&gt;). All chain reactions have this feature. While the reactive species may appear on both the reactant side and the product side, there is no net change in the number of reactive species present in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic introductory chain reaction is the radical halogenation of an alkane like methane. This reaction's mechanism is divided into three regimes: 1) Initiation (generation of radicals where there were none previously), 2) propagation (the chain part), and 3) termination (in which there is a net reduction in radicals, usually by recombination). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initiation:&#xA;$$\ce{Cl2 -&amp;gt;[hv] 2Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propagation:&#xA;$$1) \ \ \ \ \ \ce{Cl* + CH4 -&amp;gt; HCl + CH3*}$$&#xA;$$2) \ \ \ \ \ \ce{CH3* + Cl2 -&amp;gt; CH3Cl + Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Termination:&#xA;$$\ce{CH3* + Cl* -&amp;gt; CH3Cl}$$&#xA;$$\ce{2CH3*  -&amp;gt; CH3CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above example is a cyclic chain reaction. Chain reactions can be linear, for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain-growth_polymerisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;chain-growth polymerization&lt;/a&gt;. In this kind of chain reaction, two species, where one is the activated end of a polymer chain $\ce{P}*$ and the other is a new monomer $\ce{M}$, react to form a new activated chain end $\ce{P}'*$ that is one monomer unit longer. $\ce{P}'*$ can then reenter the reaction and react with another $\ce{M}$. Again, this reaction is not autocatalytic. The number of activated chain ends is not increasing (or decreasing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{P}* +\ce{M -&amp;gt; P'}*$$&#xA;$$\ce{P'}* +\ce{M -&amp;gt; P''}*$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activated chain ends, $\ce{P}*$, can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Radical_polymerization&quot; rel=&quot;nofollow noreferrer&quot;&gt;radicals&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Anionic_polymerization&quot; rel=&quot;nofollow noreferrer&quot;&gt;anions&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cationic_addition_polymerization&quot; rel=&quot;nofollow noreferrer&quot;&gt;cations&lt;/a&gt;, the products of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring-opening_polymerization&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring-opening&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_polymerization&quot; rel=&quot;nofollow noreferrer&quot;&gt;organometallic complexes&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2013-04-05T11:16:33.293" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4565" PostTypeId="2" ParentId="4302" CreationDate="2013-04-05T11:46:33.913" Score="8" Body="&lt;p&gt;I don't know about the IR spectra of metals. Certainly metals will have vibrational spectra, but I'm sure where to find it (if NIST does not have).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I can tell you how to get x/y data from NIST&lt;/strong&gt;, if they have the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C65850&amp;amp;Units=SI&amp;amp;Type=IR-SPEC&amp;amp;Index=2#IR-SPEC&quot; rel=&quot;noreferrer&quot;&gt;benzoic acid&lt;/a&gt;. At the bottom of the page, there is a link to &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?JCAMP=C65850&amp;amp;Index=2&amp;amp;Type=IR&quot; rel=&quot;noreferrer&quot;&gt;download the sepctrum in JCAMP-DX format&lt;/a&gt;. The .JDX file format is really a space-delimited text file containing x/y data with some markup at the front for the JCAMP plugin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bsRvw.png&quot; alt=&quot;JCAMP-DX file in Notepad - save as .TXT&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See all that lovely x/y data? Force the .JDX file open in your favorite text editor, like notepad. Do not use a word processor - they will add additional markup. Do not use a spreadsheet program &lt;strong&gt;yet&lt;/strong&gt;. Spreadsheets can open .JDX files, but they won't always choose the most helpful delineation. Once open in notebook, resave the file as a text (ascii) file. Make sure you type in the .txt to change the extension (alternately, you can just change the extension to .txt using any other means you may feel comfortable with). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... open your favorite spreadsheet program, and use it to open your shiny new text file. You may have to tell the program to look for &quot;all files&quot; (MS Excel is bad about this). You should get some dialog/options on how to open the file. Choose &quot;Delimited&quot; if its an option, and when it comes time to choose your delimiters, make sure your choice includes &quot;space&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eKSOU.png&quot; alt=&quot;Choosing the space delimiter in MS Excel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should have a file with the x/y data. The first column is wavenumber (at least for IR) and the second is transmission ($\frac{I_{\text{out}}}{I_{\text{in}}}$). Some of these files have more than one column of y-data. When I graphed each column from the benzoic acid file, I got the same spectrum (almost).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wmnjr.png&quot; alt=&quot;x/y data and IR spectrum in Excel&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-05T11:46:33.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4567" PostTypeId="2" ParentId="4302" CreationDate="2013-04-05T15:41:55.100" Score="3" Body="&lt;p&gt;The absorption spectrum of any material is related to the dielectric constant/refractive index.  Since there are no sharp resonance peaks for metals in the infrared, the absorption spectrum is pretty boring.  However, there are many sources of data for metals.  The most comprehensive set of data is the Handbook of Optical Constants of Solids by Edward D. Palik.  It is a four volume set that should be available in most university libraries.  The numbers for the most common metals can be downloaded in csv form from &lt;a href=&quot;http://refractiveindex.info/&quot; rel=&quot;nofollow&quot;&gt;http://refractiveindex.info/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the refractive index in hand, the imaginary part of the index can be related to the extinction coefficient as $\alpha=\frac{\omega}{c}k$ where $\omega$ is the angular frequency of light, $c$ is the speed of light, and k is the imaginary part of the index.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2013-04-05T15:41:55.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4568" PostTypeId="2" ParentId="4488" CreationDate="2013-04-05T16:54:45.587" Score="3" Body="&lt;p&gt;The operator for interaction with infrared light is the permanent dipole of the molecule, $\mu$.  I don't know anything about representation theory, but in group theory, you decompose the molecule by its symmetry elements to find the eigenmodes.  The infrared active modes are those modes in which the permanent dipole changes over the course of a vibration.  An excellent reference is Symmetry and Spectroscopy by Harris and Bertolucci which should be available in your library.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2013-04-05T16:54:45.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4569" PostTypeId="1" AcceptedAnswerId="4570" CreationDate="2013-04-05T19:09:12.410" Score="0" ViewCount="549" Body="&lt;p&gt;I'm studying for my final exam, and these are the questions I got wrong on the mid term, could anyone help correct me? I know people are against giving answers for homework, so I must add that is now exam period for Universities, so this is only for my further understanding!&#xA;Thanks so much! &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My attempts will be in this format&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1)The pH of a 0.2M unknown base X OH is 8.15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)Calculate [OH-]&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;10^(-8.15) = 7E-9M&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;b)What is the % disassociation&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(7E-9/0.2M) x 100 = 3.5E-6%&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;c)What is the Kb of the base?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This one I don't even know where to begin..&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;2)Calculate the pH of the following solutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 0.10M KCN, Ka (HCN) = 6.2E-10 (Hint: KCN is the salt of a weak acid and a strong base)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;x = sqrt(Ka*c)&#xA;x = sqrt(6.2E-10 * 0.1) = 7.8E-6&#xA;-log(7.8E-6) = pH = 5.1&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="1450" LastEditorUserId="1450" LastEditDate="2013-04-05T19:41:50.653" LastActivityDate="2014-12-16T19:43:22.793" Title="Acids / Bases - Need some clarity for final exam" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2013-04-06T06:46:21.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="4570" PostTypeId="2" ParentId="4569" CreationDate="2013-04-05T19:35:05.830" Score="0" Body="&lt;p&gt;&lt;strong&gt;1a)&lt;/strong&gt; First use $pOH = 14 - pH$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use: $pOH = \log([OH^-])$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt; Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c)&lt;/strong&gt; Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Dissolving $KCN$ in water gives: &#xA;$\ce{KCN \rightarrow K^+ + CN^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following equilibrium reaction is thus formed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2014-12-16T19:43:22.793" LastActivityDate="2014-12-16T19:43:22.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4571" PostTypeId="1" CreationDate="2013-04-06T00:15:55.973" Score="3" ViewCount="3015" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For two reactions,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\ce{A -&amp;gt;[$k_1$] B} \tag{1} \\&#xA;\ce{A -&amp;gt;[$k_2$] C} \tag{2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;you have determined that the activation energies are &lt;span class=&quot;math-container&quot;&gt;$\pu{71.9 kJ/mol}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{142.8 kJ/mol}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt;. If the &lt;em&gt;rate constants&lt;/em&gt; are equal at a temperature of &lt;span class=&quot;math-container&quot;&gt;$\pu{321 K}$&lt;/span&gt;, at what temperature will &lt;span class=&quot;math-container&quot;&gt;$k_1/k_2$&lt;/span&gt; be equal to &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know I have to use the Arrhenius equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k=A\exp\left(-\frac{E_\mathrm A}{RT}\right),$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I've tried solving for &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; after filling in my data into the formula as ratios, ie &lt;span class=&quot;math-container&quot;&gt;$k_1/k_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{A}^{(1)}/E_\mathrm{A}^{(2)}$&lt;/span&gt;, etc, but I can't seem to get a coherent answer. Any guidance on figuring this out would be much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1451" LastEditorUserId="16683" LastEditDate="2020-05-30T17:19:32.730" LastActivityDate="2020-05-30T17:32:43.557" Title="Using Arrhenius equation to calculate the ratio of two reaction rates" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4572" PostTypeId="1" AcceptedAnswerId="4574" CreationDate="2013-04-06T02:55:52.943" Score="3" ViewCount="30753" Body="&lt;p&gt;Of all the elements in periodic table, which one has smallest nearest neighbor distance? I tried searching on net, but could not find any reference. My earlier guess was one which goes into FCC with smallest unit cell dimension should be one. But I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an FCC, if unit cell dimension is a, then nearest neighbor distance will be a/sqrt(2) because nearest atom to an atom on the edge of the cube will be face atom. But this is for FCC. For SC of size a, nearest atom will be at a distance of a. So, for different elements, there will be different nearest neighbor distance based on its unit cell geometry and unit cell length. I want to know which combination of these result in shortest nearest neighbor distance. &lt;/p&gt;&#xA;" OwnerUserId="542" LastEditorUserId="542" LastEditDate="2013-04-06T07:39:28.587" LastActivityDate="2013-08-09T07:48:47.523" Title="Smallest nearest neighbor distance among elements" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4573" PostTypeId="2" ParentId="4571" CreationDate="2013-04-06T07:24:41.333" Score="5" Body="&lt;p&gt;To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy#Temperature_independence_and_the_relation_to_the_Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, although more precisely, there is some temperature variation). However, the activation energies for &lt;em&gt;both&lt;/em&gt; reactions need not be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider both reactions at &lt;span class=&quot;math-container&quot;&gt;$\pu{321 K}$&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\ln k_1 &amp;amp;= \ln k_2 \tag{1}\\&#xA;\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &amp;amp;= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt]&#xA;\ln A_1 - \ln A_2 &amp;amp;= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt]&#xA;&amp;amp;= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt]&#xA;&amp;amp;= -26.56 \tag{4}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, denote the desired temperature at which &lt;span class=&quot;math-container&quot;&gt;$k_1  = 2k_2$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$T^*$&lt;/span&gt;. At this new temperature, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\ln k_1 &amp;amp;= \ln 2 + \ln k_2 \tag{5}\\&#xA;\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &amp;amp;= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt]&#xA;\ln A_1 - \ln A_2 - \ln 2 &amp;amp;= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can reuse the previously calculated value for &lt;span class=&quot;math-container&quot;&gt;$\ln A_1 - \ln A_2$&lt;/span&gt;, and we can also plug in the same activation energies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;T^* &amp;amp;= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt]&#xA;&amp;amp;= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt]&#xA;&amp;amp;= \pu{313 K}. \tag{10}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="16683" LastEditDate="2020-05-30T17:32:43.557" LastActivityDate="2020-05-30T17:32:43.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4574" PostTypeId="2" ParentId="4572" CreationDate="2013-04-06T08:57:45.637" Score="5" Body="&lt;p&gt;Both &lt;a href=&quot;http://www.virginia.edu/bohr/mse209/chapter3.htm&quot; rel=&quot;noreferrer&quot;&gt;FCC and HCP packings&lt;/a&gt; have nearest neighbours as close as possible (in nature at least, the simple cube is even closer but doesn't exist). Their nearest neighbour distances in terms of $a/R$ are $2\sqrt{2}$. Here $a$ is the length of a side of the unit cell and $R$ is the radius of the atom the cell consists of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are looking for the smallest nearest neighbour distance this means that you are looking for the smallest $a$ in an FCC or HPC packing. You can translate this into: you are looking for the atom which forms an FCC or HPC packing and has the smallest value for $2\sqrt{2}R$ so for $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't know all the options but for example $\ce{Cu, Au, Ag}$ form FCC structures and $\ce{Zn}$ forms an HCP structure (see e.g. &lt;a href=&quot;http://che.uri.edu/course/che333/Structure.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). You can look up the &lt;a href=&quot;http://www.sciencegeek.net/tables/AtomicRadius.pdf&quot; rel=&quot;noreferrer&quot;&gt;atomic radii&lt;/a&gt; and find that Zinc has the smallest: $122$ pm and will thus have a nearest neighbour distance in the lattice of $345$ pm.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-04-06T08:57:45.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4575" PostTypeId="2" ParentId="4468" CreationDate="2013-04-06T09:30:00.850" Score="2" Body="&lt;p&gt;&lt;strong&gt;Complex structure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.drugs-forum.com/forum/showwiki.php?title=Modafinil&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modafinil&lt;/a&gt; is a pretty complex molecule and judging by its structure (see below) I think it will be extremely difficult to find a single reagent test that will identify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8rSKj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post-consumption identification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly it is much easier to find after consumption, either by the &lt;a href=&quot;http://www.drugs-forum.com/forum/showthread.php?t=161837&quot; rel=&quot;nofollow noreferrer&quot;&gt;strong sulfur smell of your urine&lt;/a&gt; or through &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/15993026&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromatography techniques&lt;/a&gt; (also &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/15834965&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;'Sketchy' tests&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are apparently some simple tests you can try like &lt;a href=&quot;http://www.drugs-forum.com/forum/showthread.php?t=161837&quot; rel=&quot;nofollow noreferrer&quot;&gt;dissolving it in hot water and vinegar&lt;/a&gt; (which should happen without formation of gas), but these tests are far from fool-proof and might even cause you to think something is a fake while it is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Combination of physical/chemical tests as a 'solution'&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing to do if you really want 'kitchen-level' tests is to test for a whole range of chemical/physical properties. For example &lt;a href=&quot;http://www.drugbank.ca/drugs/DB00745&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt; has some information on the pH, water solubility and melting point which are fairly easy to test. Passing the tests will certainly not guarantee anything, but at least it will take away a significant part of the doubts that the product is fake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something to consider when designing these tests is that there will always be fillers &lt;a href=&quot;http://www.rxlist.com/provigil-drug.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;(e.g. lactose monohydrate, microcrystalline cellulose, pregelatinized starch, croscarmellose sodium, povidone, and magnesium stearate)&lt;/a&gt; in the sample so you have to find ways to correct for that.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-04-06T09:30:00.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4576" PostTypeId="2" ParentId="4468" CreationDate="2013-04-06T17:27:44.893" Score="1" Body="&lt;p&gt;Honestly, I don't think the average kitchen is equipped to identify (with any accuracy) the presence of this molecule in pills that likely vary in impurities, fillers, etc. depending on the website you buy them from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, I don't think it's unreasonable that someone could develop an inexpensive testing kit that could at least give a &quot;positive&quot; or &quot;negative&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the molecule I would focus on is not the amide or sulfoxide, but the diphenyl group attached to the sulfoxide. These conjugated rings will absorb UV (literature suggests lambda max around 220-230 nm). So you have a molecule that absorbs UV and is hydrophobic, and you need to separate it from the other pill contents for analysis. One way to do this would be with a mini-&lt;a href=&quot;http://en.wikipedia.org/wiki/Thin_layer_chromatography&quot; rel=&quot;nofollow&quot;&gt;TLC&lt;/a&gt; kit, where the Modafinil could be detected as dark spots under UV light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example Kit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Small vial + cap &lt;/li&gt;&#xA;&lt;li&gt;Organic solvent (this would be determined by your&#xA;own testing) in dropper bottle &lt;/li&gt;&#xA;&lt;li&gt;TLC plate(s) &lt;/li&gt;&#xA;&lt;li&gt;Small glass capillary (for spotting plates) &lt;/li&gt;&#xA;&lt;li&gt;UV-LED Light &lt;/li&gt;&#xA;&lt;li&gt;Cool UV-protective glasses&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Example Procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Crush pill, add to vial&lt;/li&gt;&#xA;&lt;li&gt;Add a few drops of organic solvent (provided by you in dropper bottle)&lt;/li&gt;&#xA;&lt;li&gt;Cap vial, shake well (trying to extract Modafinil into organic phase)&lt;/li&gt;&#xA;&lt;li&gt;Spot TLC plate, elute with appropriate solvent&lt;/li&gt;&#xA;&lt;li&gt;Compare Rf of spot to the Rf of Modafinil standard (alternatively, the presence or absence of the Modafinil spot may be enough to provide a positive or negative result).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are the person developing such a kit, calibration would need to be done using a known &quot;good&quot; sample of the Modafinil. I would be careful to watch for other UV absorbing impurities that may be found in the pills (e.g. reagents from synthesis). You could confirm these results using an HPLC with UV detector or NMR (or whatever other appropriate techniques you have access to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all else fails, do as the previous answer says and smell your pee!&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2013-04-06T17:27:44.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4578" PostTypeId="1" CreationDate="2013-04-07T03:22:53.720" Score="0" ViewCount="692" Body="&lt;p&gt;I'm wondering if it's possible to balance a chemical equation with an incorrect mole ratio.  I have the unbalanced equation $\ce{AgNO3 + Cu -&amp;gt; Ag + Cu(NO3)2}$, I have a $2:3$ ratio of $\ce{Cu}$ and $\ce{Ag}$ that I must plug in.  Is this possible, because it doesn't look like it. &lt;/p&gt;&#xA;" OwnerUserId="1456" LastEditorUserId="5026" LastEditDate="2018-02-23T02:07:02.883" LastActivityDate="2018-02-23T02:07:02.883" Title="Balancing a chemical equation, but with an incorrect mole ratio" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4580" PostTypeId="2" ParentId="4578" CreationDate="2013-04-07T03:32:26.413" Score="2" Body="&lt;p&gt;The balanced reaction is $$\ce{2AgNO3 +Cu -&amp;gt; 2Ag +Cu(NO3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, it's not necessary that all the reactants are depleted in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the concept of a &quot;limiting reagent&quot; -- the reagent that gets depleted first and drives the extent of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the limiting reagent is $\ce{AgNO3}$, and you will have half a unit of $\ce{Cu}$ left over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, the limiting reagent need not be the one in the least quantity. For example, for $\ce{2A +B -&amp;gt;C +D}$ , if $\ce{A:B}$ is 3:2, then $\ce{A}$ is still the limiting reagent as it will get depleted first.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-04-07T03:32:26.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4581" PostTypeId="2" ParentId="4578" CreationDate="2013-04-07T07:48:30.763" Score="1" Body="&lt;p&gt;The balanced reaction is this:&#xA;$$\ce{2AgNO3 +Cu -&amp;gt; 2Ag +Cu(NO3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is normally what you would write and then just mention that your reactants ratio is different so you have a &lt;em&gt;limiting species&lt;/em&gt;. However, if you really want the 3:2 ratio inside your the reaction equation you could write this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3AgNO3 + 2Cu -&amp;gt; 3 Ag + 3/2 Cu(NO3)2 + Cu}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have just added equal amounts of Cu and Ag on the left and right and explicitly put in the copper that is left over. Of course by standard convention (leaving no fractions in the reaction equation) you would actually have to write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{6AgNO3 + 4Cu -&amp;gt; 6 Ag + 3 Cu(NO3)2 + 2Cu}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have a reactants ratio $\ce{Ag:Cu}=3:2$ and you can see that this will result in left over copper, because the silvernitrate is the limiting species.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-04-07T07:48:30.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4585" PostTypeId="1" CreationDate="2013-04-07T17:25:24.030" Score="5" ViewCount="23224" Body="&lt;p&gt;After you react rotundone ($\ce{C_{15}H_{22}O}$) with water it loses all of the double bonds (except the carboxylic bond, $\ce{C=O}$) and gets two of the functional group $\ce{-OH}$ (alcohol). Two compounds are formed, one (A) where both the alcohol groups are tertiary and another (B) where one is tertiary and the other is primary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asked to state and explain the colour change of both A and B when they react separately with acidified potassium dichromate(VI) solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can see that compound A would have no colour change because the $\ce{-OH}$ groups are tertiary so the potassium dichromate won't react or oxidise them.&#xA;I can also see that there will be a colour change on compound B because one of the $\ce{-OH}$ groups is primary, so the potassium dichromate will oxidise and react with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm expected to know that when compound B reacts, it changes from orange to green. How would I know that rotundone is orange in the first place? And what makes it go green?&#xA;I know that I'm expected to use intuition as we definitely aren't required to know the colour of rotundone. &lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="4231" LastEditDate="2015-03-01T14:32:31.250" LastActivityDate="2016-05-09T16:26:49.497" Title="How can I know the colour change of this reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4586" PostTypeId="1" CreationDate="2013-04-07T17:35:08.063" Score="4" ViewCount="1692" Body="&lt;p&gt;We have a home carbonator for makeing seltzer. It blows $\ce{CO2}$ under high pressure into water. Could I use it to carbonate corn oil? Is $\ce{CO2}$ soluble in a no aqueous solution? &lt;/p&gt;&#xA;" OwnerUserId="658" LastEditorUserId="275" LastEditDate="2013-04-07T19:03:26.543" LastActivityDate="2015-07-02T22:28:55.477" Title="Can you carbonate a non aqueous liquid?" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4588" PostTypeId="1" AcceptedAnswerId="6301" CreationDate="2013-04-07T17:53:25.027" Score="3" ViewCount="1170" Body="&lt;p&gt;This question probably applies to other heavier metals as well.  The only rationalization I can figure out is that lead in the +2 oxidation state (which is most common) is a borderline soft acid and prefers to take electrons from a soft sulfur instead of trying to react with a hard oxygen.&lt;/p&gt;&#xA;&lt;p&gt;Is this the right direction?  Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="838" LastEditorUserId="69830" LastEditDate="2021-05-13T13:17:38.000" LastActivityDate="2021-05-13T13:17:38.000" Title="Why is lead sulfide found in nature, whereas lead oxide is less common?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;hsab&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4589" PostTypeId="2" ParentId="4276" CreationDate="2013-04-07T19:12:41.313" Score="2" Body="&lt;p&gt;I believe the problem was that the LDA+C energies quoted in the paper were &lt;em&gt;extra, additive&lt;/em&gt; energies which were supposed to be added to the main band gap energy of the bulk material. In this case that was something like $\pu{2.7 eV}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the quantum dots will have higher band gaps the smaller they get. As the quantum dots get larger, in a sense, they approach a bulk material size, hence their band gap energy gets smaller and smaller and approaches the bulk material band gap energy limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps anyone who happens upon a similar problem.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="41328" LastEditDate="2017-08-06T06:30:37.190" LastActivityDate="2017-08-06T06:30:37.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4591" PostTypeId="2" ParentId="4586" CreationDate="2013-04-07T20:34:04.240" Score="-1" Body="&lt;p&gt;$\ce{CO2}$ is also soluble in various organic liquids, but your mileage may vary&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From an excellent comment by Greg E. that he didn't opt to turn into an answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CO2 is a non-polar molecule, so its solubility will be partially a function of the polarity of the solvent in question. Furthermore, solubility of gases is very sensitive to temperature. Solvation of a gas is entropically disfavored (lower degrees of freedom for the gas molecules), so at some point a further increase in temperature will take primacy over the heat of solution (enthalpy change due to solvation) and any further increase in temperature will actually result in more CO2 coming out of solution. Additionally, higher temperature will increase the vapor pressure of solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1461" LastEditorUserId="73" LastEditDate="2015-07-02T22:28:55.477" LastActivityDate="2015-07-02T22:28:55.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4593" PostTypeId="1" CreationDate="2013-04-08T02:26:42.220" Score="4" ViewCount="37078" Body="&lt;p&gt;A buffer is made from $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ benzoic acid, $K_a = 6.3\cdot10^{-5}$, and $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ sodium benzoate.&lt;br&gt;&#xA;a) Calculate the pH of this buffer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this do I simply use $-\log(6.3\cdot 10^{-5})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)Calculate the new pH when $0.010~\mathrm{M}$ $\ce{HCl}$ is added to $100~\mathrm{mL}$ of this buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one I do not understand.&lt;/p&gt;&#xA;" OwnerUserId="1450" LastEditorUserId="4945" LastEditDate="2014-10-31T02:17:50.887" LastActivityDate="2014-10-31T02:20:27.617" Title="How to calculate the pH of a buffer after HCl was added?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4594" PostTypeId="1" CreationDate="2013-04-08T02:49:30.110" Score="3" ViewCount="652" Body="&lt;p&gt;Why are the oxides usually insulating?  It would be nice to get an explanation with chemistry terms (e.g., electronegativity, electron affinity, etc...) and a different explanation involving the language of band structure (i.e. energy gaps).&lt;/p&gt;&#xA;" OwnerUserId="1348" LastActivityDate="2014-11-09T06:23:31.243" Title="Why are the oxide materials insulating?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4595" PostTypeId="2" ParentId="4593" CreationDate="2013-04-08T03:23:11.977" Score="7" Body="&lt;p&gt;For (a), the Henderson-Hasselbalch equation, $$\ce{pH} = \mathrm{p}K_a + \log([\ce{A^{-}}]/[\ce{HA}]),$$ comes in handy. Because your molarities and volumes of the acid and its conjugate base are equal, this indeed reduces to simply $\ce{pH} = -\log(6.3 \cdot 10^{-5})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b), the volume of $\ce{HCl}$ added is required, as the concentration of the solution alone is not sufficient information. The standard practice is to assume that $\ce{HCl}$ (being a strong acid) reacts fully with the conjugate base in your buffer solution to produce an equal amount of the conjugate acid (i.e., if $x$ moles of $\ce{A^{-}}$ are consumed by $\ce{HCl}$, $x$ moles of conjugate acid $\ce{HA}$ are produced). Therefore, you can use the Henderson-Hasselbalch equation to recalculate the $\ce{pH}$, subtracting the moles of $\ce{HCl}$ added from your conjugate base, and adding that some number of moles to your conjugate acid.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="4945" LastEditDate="2014-10-31T02:20:27.617" LastActivityDate="2014-10-31T02:20:27.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4596" PostTypeId="2" ParentId="4585" CreationDate="2013-04-08T04:48:43.183" Score="5" Body="&lt;p&gt;Potassium dichromate (VI) is the species that turns from orange and to green when reduced. Most of the organic substances I came across for testing purposes are colourless, so I would assume that compound B would be as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is just like you are expected to know the colour change of potassium manganate (VII) from purple to colourless when it is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those two oxidants were used frequently in my labs, and I'm pretty sure you'll be using them often as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question! :)&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-04-08T04:48:43.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4597" PostTypeId="1" CreationDate="2013-04-08T05:10:56.213" Score="6" ViewCount="10853" Body="&lt;p&gt;I read in many MSDS that Isopropyl alcohol tends to form peroxides (by reacting to air OR light) which may explode when they reach a given concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;a href=&quot;http://nj.gov/health/eoh/rtkweb/documents/fs/1076.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nj.gov/health/eoh/rtkweb/documents/fs/1076.pdf&lt;/a&gt; or &lt;a href=&quot;http://www.hvchemical.com/msds/isal-70.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.hvchemical.com/msds/isal-70.htm&lt;/a&gt; as well as many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I own a few liters of isopropyl alcohol (a few bottles) and I'm getting concerned.&#xA;It came in standard transparent (polypropylene I suppose) bottles with a plastic bag wrapped around it (I suppose to prevent contact with air). I only keep one bottle ready to use, all the others are as when I bought them (with plastic bag wrapped around it), stored in a relatively cool (&amp;lt;= 25°C) dark area of my house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at home and I couldn't find any fire resistant storage cabinet in my area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not sure if I understand the datasheet correctly, but as far as I understood the problem lies with the opened bottle (which I open and close regularly) and not with the air-tight never opened other bottles, right?&#xA;Therefore a simple solution may be to say: after one year I empty the opened bottle, trash it away and buy another one. Should this be enough?&lt;/p&gt;&#xA;" OwnerUserId="352" LastEditorDisplayName="user467" LastEditDate="2015-06-23T00:35:43.247" LastActivityDate="2015-06-23T00:35:43.247" Title="Isopropyl alcohol forming peroxides" Tags="&lt;safety&gt;&lt;distillation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4598" PostTypeId="1" CreationDate="2013-04-08T05:54:46.640" Score="0" ViewCount="1480" Body="&lt;p&gt;The 2 Hydrogens near the carbonyl (the lower 2 $\ce{H}$s are not bonded to the nitrogen) should be the most deshielded so I think they appear as the rightmost peak in the NMR.  Are these 2 $\ce{H}$s a doublet of doublets?  I think this because they are split by 2 different $\ce{H}$s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RjRjr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="4231" LastEditDate="2015-03-01T14:32:17.467" LastActivityDate="2015-03-01T14:32:17.467" Title="What is the multiplicity of this signal in the ¹H NMR of glutamine in D₂O" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4600" PostTypeId="2" ParentId="4356" CreationDate="2013-04-08T12:12:53.493" Score="10" Body="&lt;p&gt;The solvent of the marker dissolves the permanent ink and mixes it with the whiteboard ink, leaving a dark dust that doesn't stick to the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link in Wikipedia doesn't name common ethyl alcohol (denatured or not) as being a solvent for old marks. That is what I have used.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="4489" LastEditDate="2014-02-10T18:22:51.650" LastActivityDate="2014-02-10T18:22:51.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4602" PostTypeId="2" ParentId="4598" CreationDate="2013-04-08T13:35:19.083" Score="2" Body="&lt;p&gt;You're thinking about deshielding the wrong way around, the most deshielded signal appears at the highers ppm value, so to the left. In this case it would be the signal at around 3.8 ppm, as the signal at 4.7 ppm is residual water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two signals at the right are multiplets, if you take a look they're obviously caused by rather complicated coupling patterns that you can't easily resolve. The reason for that is that the protons you see in this spectrum are not magnetically equivalent, even the CH2 protons are not magnetically equivalent. This leads to the complicated couplings you can observe here.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-04-08T13:35:19.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4603" PostTypeId="2" ParentId="4450" CreationDate="2013-04-08T17:59:08.150" Score="3" Body="&lt;p&gt;Thermodynamics isn't quite that general.  It will tell you about the possibility of going from one well-defined state to another well-defined state, but as has been said, there is no way to do it in general.  What you are asking is akin to &quot;is it hard to get there from here?&quot; without specifying where both here and there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to specify enough variables such as temperature or composition to unambiguously determine both the initial and final states.  How many variables do you need?  The Gibbs phase rule provides the answer.  Once that has been done, your question can be answered.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-04-08T17:59:08.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4604" PostTypeId="2" ParentId="4598" CreationDate="2013-04-08T19:25:44.843" Score="1" Body="&lt;p&gt;The signal at ~3.8ppm is looks like a triplet[t], C-alpha&#xA;The signal at ~2.5ppm looks like a triplet of doublets [td], C-gamma.&#xA;The signal at ~2.2ppm looks like a doublet of doublet of doublets [ddd], c-beta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2.2 protons are assigned 2.2 because they are the farthest from carbonyls and thus the least deshielded by these functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ddd signal indicates clearly that both protons on C-beta are non-equivalent, because you need 2 3-bond couplings to make a ddd when combined with the 2-bond geminal coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the C-beta protons are non-equivalent, the triplet is actually a superimposed doublet of doublets, and the td is actually a similarly superimposed double of doublets [the superimposed-apparent triplet pair and then...] of doublets.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-04-08T19:25:44.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4605" PostTypeId="2" ParentId="4597" CreationDate="2013-04-08T20:11:43.587" Score="2" Body="&lt;p&gt;Not quite enough.  If you ever see crystals in the bottle, year old or otherwise, evacuate and call the bomb squad.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-04-08T20:11:43.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4606" PostTypeId="1" AcceptedAnswerId="4608" CreationDate="2013-04-08T20:26:21.293" Score="0" ViewCount="269" Body="&lt;p&gt;I am using the &lt;a href=&quot;http://www.ebi.ac.uk/chebi/searchId.do?chebiId=CHEBI:17303&quot; rel=&quot;nofollow&quot;&gt;Morphine&lt;/a&gt; &lt;a href=&quot;https://www.google.com/url?url=http://en.wikipedia.org/wiki/Chemical_table_file#Molfile&amp;amp;rct=j&amp;amp;q=mol%20file%20explanation%20type%20of%20bond&amp;amp;usg=AFQjCNGDSqUybnNKqToBALc-O0K8TRaj0A&amp;amp;sa=X&amp;amp;ei=SiVjUf3gGYLp2QWGlIHgDQ&amp;amp;ved=0CDcQygQwAA&quot; rel=&quot;nofollow&quot;&gt;mol&lt;/a&gt; file and I'm attempting to create a matrix which contains a table of the atoms and how they bond similar to a ball and stick format.&#xA;In the mol file bond block listed bellow what do the type place holders mean?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  7  1  1  0  0  0  0&#xA;  1  2  1  0  0  0  0&#xA;  2 12  1  0  0  0  0&#xA;  1  3  2  0  0  0  0&#xA;  3 14  1  0  0  0  0&#xA;  2  4  2  0  0  0  0&#xA;  4 18  1  0  0  0  0&#xA;  3  5  1  0  0  0  0&#xA;  4  6  1  0  0  0  0&#xA;  5  6  2  0  0  0  0&#xA;  7  8  1  0  0  0  0&#xA;  7  9  1  0  0  0  0&#xA;  7 22  1  1  0  0  0&#xA;  8 10  1  0  0  0  0&#xA;  8 11  1  0  0  0  0&#xA;  8 20  1  1  0  0  0&#xA;  9 12  1  6  0  0  0&#xA;  9 13  1  0  0  0  0&#xA; 10 15  1  1  0  0  0&#xA; 10 14  1  0  0  0  0&#xA; 11 16  2  0  0  0  0&#xA; 13 17  1  6  0  0  0&#xA; 13 16  1  0  0  0  0&#xA; 15 19  1  0  0  0  0&#xA; 21 15  1  0  0  0  0&#xA; 22 21  1  0  0  0  0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the format I'm trying to create but not all the atoms have the full amount of bonds. What is wrong?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C   7   2   3   &#xA;C   1   12  4   12&#xA;C   1   14  5   5&#xA;C   2   18  6   18&#xA;C   3   6   3   6&#xA;C   4   5   4   5&#xA;C   1   8   9   22&#xA;C   7   10  11  20&#xA;C   7   12  13  12&#xA;C   8   15  14  15&#xA;C   8   16      16&#xA;O   2   9   2   9&#xA;C   9   17  16  16&#xA;C   3   10      10&#xA;N   10  19  21  &#xA;C   11  13  11  13&#xA;O   13          &#xA;O   4   4       &#xA;C   15          &#xA;H   8           &#xA;C   15  22  22  22&#xA;C   7   21  21  21&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerDisplayName="user1464" LastEditorUserId="4945" LastEditDate="2014-06-12T08:53:00.027" LastActivityDate="2014-06-12T08:53:00.027" Title="Mol file of Morphine" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4607" PostTypeId="2" ParentId="4597" CreationDate="2013-04-08T22:04:03.370" Score="9" Body="&lt;p&gt;I think this may be an overreaction.  First, MSDS are designed for workers working with industrial quantities in industrial conditions.  Second, the danger comes from concentration of the peroxide from evaporation or, more typically, distillation.  Small quantities of peroxides in solution are not usually an explosion hazard.  Third, I have never seen isopropanol listed in lists of hazardous peroxide prone chemicals as used by lab safety departments (ethers are the typical risk here).  The closest I have seen is sec-butanol, which is the secondary alcohol from butanol, and I've only seen this listed once, in a list that seemed much more cautious than some other lists.  Isopropanol is the secondary alcohol from propanol.  Given your storage, cool and dark, I think you are fine.  Remember that 91% isopropanol is sold in drugstores, though these may contain stabilizers and you larger bottles may not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But having said all this, Lighthart's brief answer is correct though probably unlikely.  If you DO see any crystals, don't even try to open the bottle (crystals in the threads are the most common source of explosion).  But again, seems unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are still concerned, there are peroxide test strips you can buy that are very easy to use and can tell you what concentration of peroxides are present.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-04-08T22:04:03.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4608" PostTypeId="2" ParentId="4606" CreationDate="2013-04-08T22:04:56.023" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;http://www.chemaxon.com/marvin/help/formats/mol-csmol-doc.html&quot; rel=&quot;nofollow&quot;&gt;bond block&lt;/a&gt; has the number of the atoms followed by the number of bonds mol files support up to three bonds. Not all the atoms are supposed to have all their bonds filled. &lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastActivityDate="2013-04-08T22:04:56.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4609" PostTypeId="1" AcceptedAnswerId="4659" CreationDate="2013-04-09T02:21:50.903" Score="7" ViewCount="911" Body="&lt;p&gt;Are there any common elemental (single-step) reactions with three reactants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything I've found either says that they're very rare, or can be broken down into two bimolecular steps with combined reaction rates far greater than the termolecular step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What reactions &lt;em&gt;require&lt;/em&gt; an indivisible termolecular step?&lt;/p&gt;&#xA;" OwnerUserId="1467" LastActivityDate="2013-04-12T20:24:11.177" Title="Common Elemental Termolecular Reactions" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4610" PostTypeId="1" CreationDate="2013-04-09T03:41:10.543" Score="2" ViewCount="14877" Body="&lt;p&gt;As we all know, the freezing point of absolute sulfuric acid is 10 °C. The average lead-acid battery (sulfuric) uses ~30% sulfuric acid, and has a freezing point much lower than 10 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What constituents contribute to this effect? What is the diluent used in these batteries? What would the freezing point be?&lt;/p&gt;&#xA;" OwnerUserId="1468" LastEditorUserId="41328" LastEditDate="2019-01-30T20:41:18.940" LastActivityDate="2019-01-30T20:59:19.150" Title="What is the freezing point of sulfuric battery acid?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="4611" PostTypeId="2" ParentId="4597" CreationDate="2013-04-09T10:36:33.477" Score="7" Body="&lt;p&gt;Isopropanol is not one of the notorious peroxide formers, but it will eventually do so in the presence of oxygen (with or without light - the process is autocatalytic). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a graduate student TA in an organic chemistry class, isopropanol was a distillation &quot;unknown&quot;. However, the &quot;unknown&quot; bottle had probably been in existence for 20 years and was periodically refilled. Some of the isopropanol had been around in the presence of air for a long time. While the bottle never developed the crystals typical of ether peroxides, peroxides did form. Our proof was the explosion of three distillations using the stuff. Fortunately no one was injured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a qualitative test for the presence of peroxides. Buy yourself some &lt;a href=&quot;http://www.fishersci.com/ecomm/servlet/itemdetail?itemdetail=%27item%27&amp;amp;storeId=10652&amp;amp;productId=7160192&amp;amp;catalogId=29104&amp;amp;matchedCatNo=50300309&amp;amp;fromSearch=1&amp;amp;searchKey=potassium+iodide+starch+paper&amp;amp;highlightProductsItemsFlag=Y&amp;amp;endecaSearchQuery=%23store%3DScientific%23nav%3D0%23rpp%3D25%23offSet%3D0%23keyWord%3Dpotassium%2Biodide%2Bstarch%2Bpaper&amp;amp;xrefPartType=From&amp;amp;savings=0.0&amp;amp;xrefEvent=1365503004087_0&amp;amp;searchType=PROD&quot;&gt;KI Starch test paper&lt;/a&gt;. KI reacts with oxidants to produce $\ce{KI3}$, which binds with starch to form a blue-purple complex. This chemistry is also the basis of the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_clock&quot;&gt;iodine clock reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ROOH + 2H+ + 3I- -&amp;gt; ROH + H2O + I3-}$$&#xA;$$\ce{I3- + starch -&amp;gt; [I3-starch]^-} \ \ \text{(blue)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-09T10:36:33.477" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4612" PostTypeId="1" AcceptedAnswerId="4615" CreationDate="2013-04-09T12:09:25.217" Score="0" ViewCount="1946" Body="&lt;p&gt;I can't find the $K_{sp}$ value for Manganese Phosphate ($\ce{Mn_3(PO_4)_2}$) anywhere online. Has it been measured? What is it?&lt;/p&gt;&#xA;" OwnerUserId="1169" LastEditorUserId="4231" LastEditDate="2015-03-01T14:32:11.880" LastActivityDate="2015-03-01T14:32:11.880" Title="Ksp of Manganese Phosphate" Tags="&lt;solubility&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4613" PostTypeId="1" CreationDate="2013-04-09T12:33:35.337" Score="3" ViewCount="498" Body="&lt;p&gt;Let us represent molecules by graphs (atoms are vertices and bonds are edges).&#xA;Let $A$ be the adjacency matrix of the molecular graph. $A_{ij} = b$ where $b$ is either binary: &lt;em&gt;0&lt;/em&gt; when atom &lt;em&gt;i&lt;/em&gt; and &lt;em&gt;j&lt;/em&gt; are not bonded else &lt;em&gt;1&lt;/em&gt; (Repr. 1). Or, further we may take &lt;em&gt;b&lt;/em&gt; from the set $\{0,1,2,3\}$ depending on whether the no bond (0) or single(1) double(2) or triple (3) bond is present (Rper. 2). Here were are not distinguishing the atoms - for our purose they represent a node in a graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider only non-isomers and molecules with at least 10 atoms. How different will be two molecules with same graphs (with respect to Repr. 1 or Repr. 2) in terms of their properties or particularly about drug-like properties. My interest is to know how faithfully the &lt;a href=&quot;http://en.wikipedia.org/wiki/Topological_index&quot; rel=&quot;nofollow&quot;&gt;topological indices&lt;/a&gt; characterizes the molecular activity. Specific or general examples of pairs of  non-isomer molecules with same graph but very different property could be useful. It would be helpful if I can find enough examples to convince myself about the usefulness or otherwise of topological indices which are calculated from the grpahical representation outlined above. Topological indices are popular in drug discovery research, but I am not very much convinced about their utility since most the results published are not statistically significant (use of very small data size - from 15 to some hundreds molecules, among other weakness) nor rigorously validated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="4945" LastEditDate="2014-06-12T08:53:59.180" LastActivityDate="2019-11-05T06:51:31.240" Title="How different are molecule with same connectivity" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4615" PostTypeId="2" ParentId="4612" CreationDate="2013-04-09T15:58:32.813" Score="1" Body="&lt;p&gt;Well, the &lt;a href=&quot;http://fs1.agrian.com/pdfs/Tech-Flo_Mn-15_0-7-0_%2882295%29_MSDS.pdf&quot; rel=&quot;nofollow&quot;&gt;MSDS entry&lt;/a&gt; tells me that the solubility is 50%. Converting, we get 0.5 g per gram of water, or 0.0005 g per liter of water. This is $s=1.4\times 10^{-6}$ moles per liter of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{sp}=(3s)^3\times(2s)^2=6.14\times 10^{-27} \rm M^5 $&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-04-09T15:58:32.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4616" PostTypeId="1" CreationDate="2013-04-09T17:40:38.403" Score="7" ViewCount="266" Body="&lt;p&gt;Has anyone seen a version of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_clock_reaction&quot; rel=&quot;noreferrer&quot;&gt;iodine clock&lt;/a&gt; reaction where instead of the system oscillating between clear and blue it shows blue bands travelling down the vessel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because someone not unknown to this site has made a version of the experiment with such moving bands and I was wondering if it was worth sending a write up of it to the Journal of Chemical Education or some similar journal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clarification:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if I was unclear, I am asking the phenomenon has been published before or if it is unpublished but already well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the BZ reaction can produce all sorts of patterns, but I have never seen travelling bands in the iodine clock. A quick Google has found no mention of the phenomenon. I am not asking &quot;is it worth publishing&quot;, which I believe is against site rules anyway. I'm just asking about prior art.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="1470" LastEditDate="2013-04-10T05:50:56.147" LastActivityDate="2013-07-09T06:56:51.123" Title="Iodine clock with moving bands" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4617" PostTypeId="2" ParentId="4135" CreationDate="2013-04-09T18:56:52.643" Score="24" Body="&lt;p&gt;Answering my own question based on the comments, &lt;em&gt;tert&lt;/em&gt;-butyl-hydroperoxide is at least one such chemical. As stated on &lt;a href=&quot;http://cameochemicals.noaa.gov/chemical/2692&quot;&gt;this MSDS from a government website&lt;/a&gt;, it's a 4-4-4, with additional special warning of being a strong oxidizer. The only thing that it does not do that could make the 704 diamond any worse is react strongly with water. It is in fact water soluble, though marginally, preferring to float on top (and therefore traditional water-based fire suppression is ineffective, but foam/CO2 will work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone else can find a chemical that, in a form that is used in the lab or industrially, is a 4-4-4 that is a strong oxidizer &lt;em&gt;and&lt;/em&gt; reacts strongly with water, that's pretty much &quot;as bad as it gets&quot; and they'll get the check.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2015-09-02T18:03:38.393" LastActivityDate="2015-09-02T18:03:38.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4618" PostTypeId="2" ParentId="4616" CreationDate="2013-04-10T01:18:23.067" Score="3" Body="&lt;p&gt;Whilst not specifically an iodine clock reaction, &lt;a href=&quot;http://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Belousov-Zhabotinski&lt;/a&gt; reactions are oscillatory and can produce moving &lt;a href=&quot;http://www.youtube.com/watch?v=3JAqrRnKFHo&quot; rel=&quot;nofollow noreferrer&quot;&gt;bands, spirals and concentric circles of colour&lt;/a&gt;. This works best in thin films (limiting convection) as the top commenter on the linked video points out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a great &lt;a href=&quot;http://hantz.web.elte.hu/cikkfile/hantzth.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;thesis by Peter Hantz&lt;/a&gt; which looks at these kinds of reactions in detail (the first bit is in french, the rest is in english) and specifically covers Liesegang banding, where an oscillating chemical reaction in a mobile phase causes bands of precipitation in a gel column. As Hantz points out on page 22, the gel is required to prevent convection from destroying the bands, however the viscosity of the gel also freezes the precipitate in place as it forms. This would seem to be close to what you are interested in, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of Liesegang bands in test tubes from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yeo0e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-04-10T01:18:23.067" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4619" PostTypeId="2" ParentId="4610" CreationDate="2013-04-10T02:16:40.653" Score="3" Body="&lt;p&gt;The general effect you're talking about is &lt;em&gt;freezing point depression&lt;/em&gt;, one of the basic colligative properties. There are several flavors of lead-acid batteries, so there are different additives that can modulate the freezing point.  The actual composition of the electrolyte &lt;strong&gt;varies with the charge state of the battery,&lt;/strong&gt; so that means the freezing point also varies with how charged it is. I think all of the lead-acid batteries use water as a diluent (it will always be present as it's a species in the electrochemical equilibrium.) From the &lt;em&gt;Wikipedia&lt;/em&gt; article on &quot;lead-acid battery&quot;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Due to the freezing-point depression of the electrolyte, as the battery discharges and the concentration of sulfuric acid decreases, the electrolyte is more likely to freeze during winter weather.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="912" LastActivityDate="2013-04-10T02:16:40.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4620" PostTypeId="1" AcceptedAnswerId="4623" CreationDate="2013-04-10T04:18:41.167" Score="3" ViewCount="60818" Body="&lt;p&gt;Elements on the left side of the periodic table tend to be solid and metallic, elements on the right side of the periodic table are nonmetal and tend to be gases at room temperature, and the semi-metals are in between.  Why is that?  Why are the elements on the right side nonmetallic and in a gas phase?&lt;/p&gt;&#xA;" OwnerUserId="1348" LastActivityDate="2019-05-28T04:43:22.270" Title="Why are elements on the right side of the periodic table nonmetallic and gas at room temperature?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4622" PostTypeId="1" CreationDate="2013-04-10T05:23:43.997" Score="4" ViewCount="143" Body="&lt;p&gt;After reading a section on this particular oscillatory reaction, something started bugging me about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let me lay out the basis of my question.  From what I've researched, Lengyel and Epstein proposed a model that captures the overall behavior of the system. There are a few different versions of the model in regard to the dimensions, but this seems to be the most popular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MA+I_2 \rightarrow IMA + I^- + H^+;\hspace{1mm} \frac{d[I_2]}{dt} = -\frac{k_{1a}[MA][I_2]}{k_{1b}+[I_2]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO_2 + I^2 \rightarrow ClO_2^+ + \frac{1}{2}I_2; \hspace{1mm}\frac{d[CLO_2]}{dt} = -k_2[CLO_2][I^-]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO_2^- + 4I^- + 4H^+ \rightarrow Cl^- + 2I_2 + 2H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\frac{d[CLO_2^-]}{dt} = -k_{3a}[ClO_2^-][I^-][H^+]-k_{3b}[CLO_2^-][I_2]\frac{[I^-]}{u+[I^-]^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I am much more familiar with the mathematics than the chemistry.  &lt;strong&gt;First, are these rate laws empirical or were they derived from physical laws?&lt;/strong&gt; Pardon me if this is a naive question, but I am not as familiar with chemistry than the other hard sciences. In basic cases, rate laws are a straight multiplication of the concentrations of the present species. However, that neglects situations involving, for example, autocatalycity.  My suspicion is that it is a mixture of processes that I'm not familiar with and simple observation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, these reactions don't violate the laws of thermodynamics. &lt;strong&gt;My assumption is that the overall free energy of the system continues to decline - meaning that the system eventually reaches equilibrium.  Unfortunately, I've been at a loss to find good sources on oscillatory chemical reactions.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1224" LastEditorUserId="9961" LastEditDate="2017-04-18T20:27:15.350" LastActivityDate="2017-04-18T20:27:15.350" Title="Where do the rate laws for the Chlorine Dioxide-Iodine-Malonic Acid reaction come from?" Tags="&lt;redox&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4623" PostTypeId="2" ParentId="4620" CreationDate="2013-04-10T05:45:40.570" Score="5" Body="&lt;p&gt;The elements on the left-side of the periodic table are relatively electron deficient (i.e., they have few valence electrons), and due to their comparatively low effective nuclear charges (the net positive charge of the protons minus the shielding core electrons below the valence level), their electrostatic hold on these electrons are weak. Consequently, the electrons of metal atoms tend to be highly delocalized, not forming localized covalent bonds or possessing tightly held lone pairs. In metals, the electrons are effectively shared between many atoms, loosely bound in what is often described as an electron &quot;sea,&quot; and the atoms of metals arrange to achieve maximally efficient packing in a crystal lattice, through which the electrons (particularly the most extensively delocalized electrons in the s and p orbitals at the highest main energy level) move freely. The tight packing of the crystal lattice accounts for the density and hardness of metals, while the extensive delocalization of electrons partially explains their malleability and conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements further right on the period table, on the other hand, have higher effective nuclear charges and stabilize electrons more effectively, leading to localized covalent bonding and the formation of molecules (vs. the ionic bonding and crystal lattices of metals). Electrostatic intermolecular forces of attraction are generally weaker than ionic bonding or delocalized metallic bonding, so non-metals and molecular compounds are more likely to exist in the liquid or gaseous phase at comparatively low temperatures than metals or ionic compounds.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-04-10T05:52:19.343" LastActivityDate="2013-04-10T05:52:19.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4624" PostTypeId="1" AcceptedAnswerId="4625" CreationDate="2013-04-10T07:44:01.457" Score="24" ViewCount="12639" Body="&lt;p&gt;I just read &lt;a href=&quot;http://what-if.xkcd.com/40/&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; that mentioned that &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride&quot; rel=&quot;noreferrer&quot;&gt;dioxygen difluoride&lt;/a&gt; is very dangerous. The terms it uses are &quot;awful&quot;, &quot;violently hideous&quot; and &quot;deeply alarming&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I couldn't get a handle on exactly what it would do that warrants those descriptions.&lt;/p&gt;&#xA;" OwnerUserId="1475" LastEditorUserId="9961" LastEditDate="2019-01-22T21:32:05.927" LastActivityDate="2019-01-22T21:32:05.927" Title="Please explain why dioxygen difluoride is so dangerous" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;safety&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="4625" PostTypeId="2" ParentId="4624" CreationDate="2013-04-10T08:51:34.677" Score="21" Body="&lt;p&gt;$\ce {O_2F_2}$ doesn't spontaneously combust. It is a supporter of combustion, which means that it's basically a better version of oxygen when it comes to supporting fires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, when placed in contact with $\ce{O_2F_2}$, &lt;em&gt;other&lt;/em&gt; materials spontaneously combust. Here's an analogy: Substances such as nitroglycerin and TNT are like a person with a short temper. They are easy to combust. Substances like &lt;a href=&quot;https://chemistry.stackexchange.com/questions/698/flammability-nfpa-how-is-it-defined&quot;&gt;$\ce{ClF3}$&lt;/a&gt; and $\ce{O_2F_2}$, on the other hand, are like a person with a knack of making others lose their temper. Mild-mannered on his own, but he can make people around him combust with relative ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It still can be prepared, though, because it becomes a lot more mellow at low temperatures. Also, it won't act on stuff which is already a good oxidizing agent (oxygen, ozone, fluorine, sulfuric acid for example), or at least it won't act so vigorously. Coupling these two together, one can probably prepare it in a low temperature environment inside a container lined with a relatively inert material.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(h/t to &lt;a href=&quot;https://chemistry.stackexchange.com/users/367/keiths&quot;&gt;KeithS&lt;/a&gt; for pointing this out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\ce{O2F2}$ has problems on its own as well. It decomposes when on its own to $\ce{OF2}$. Generally, when $\ce{O2F2}$ is oxidising something, it is really the $\ce{OF2}$ which does this (though $\ce{O2F2}$ is strong enough to do the same as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After burning something, $\ce{OF2}$ leaves behind some dangerous residues as well, like $\ce{HF}$ (which will eat through most of the material that was saved from burning)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are more details in @KeithS' answer)&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2013-04-11T05:24:29.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4626" PostTypeId="1" CreationDate="2013-04-10T17:26:31.790" Score="1" ViewCount="124" Body="&lt;p&gt;Ice will melt when heat is applied; paper will catch fire. In trying to figure out why — what the difference is between things that melt and things that catch — I found &lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/mats05/mats05170.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;Burn, Char, Melt&quot;&lt;/a&gt; by Roberto Gregorius, who says it depends on&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;whether the substance easily forms compounds with some oxidant [usually oxygen]. This term &quot;easily&quot; can be taken to mean that the energy required to start the reaction (called the energy of activation) is low enough so that the reaction starts (as in striking a match), or the compounds produced in the burning are low enough in energy that there is a tendency to form it. Of these two, energy of activation is usually the controlling factor in deciding if a reaction will take place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All right, so what determines a substance's energy of activation with air? (&lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_of_activation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; isn't helping.) In case it isn't obvious, I'm uneducated in chemistry, and an answer tailored accordingly, to the extent possible, would be ideal.&lt;/p&gt;&#xA;" OwnerUserId="1477" LastEditorUserId="1477" LastEditDate="2013-04-11T17:16:12.613" LastActivityDate="2015-09-04T15:26:33.760" Title="What determines a substance's energy of activation with air?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;combustion&gt;&lt;phase&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4627" PostTypeId="2" ParentId="4594" CreationDate="2013-04-10T17:57:12.887" Score="1" Body="&lt;p&gt;To have conductuing valence zone requires to have infinite amount of ovarlapping valence orbitals with lesser amount of electrons. This is not usually possible for oxides, because they contain direct contacts O-O, but not M-M. O-atoms in oxides has completely filled electronic shell. Pure metals has electronic shells with much less electrons then available orbitals so they can allow electrons to move freely. This is the difference.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-10T06:03:16.680" LastActivityDate="2014-09-10T06:03:16.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4628" PostTypeId="2" ParentId="4624" CreationDate="2013-04-10T19:55:16.743" Score="33" Body="&lt;p&gt;XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his &quot;Things I Won't Work With&quot; list, and the article is found &lt;a href=&quot;http://pipeline.corante.com/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride.php&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous oxidizer; in plain English, it starts roaring fires with almost anything it touches, even with things that simply will not burn in any other circumstances, like sand, concrete, brick, iron, copper, even asbestos and fiberglass. Both the oxygen and the fluorine are pretty highly electronegative (fluorine is &lt;em&gt;the&lt;/em&gt; most electronegative element in the table, and oxygen's #2), and so they will readily react with anything even slightly more willing to donate an electron; carbon, silicon, metals, alkalis, most other gases, etc. When that happens, the high-energy bonds of this molecule (it can only be formed at 700*C, and that thermal energy remains &quot;locked&quot; in the structure of the molecule) are rearranged into structures that require less energy, and the excess energy is released. The resulting rapid, aggressive, highly exothermic reaction of dioxygen difluoride with just about any other material would not be described as anything less than an explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, even in the absence of something to oxidize, the oxygen would &lt;em&gt;much&lt;/em&gt; rather exist in its &quot;native&quot; diatomic gas state. Again, there's a lot of energy tied up in the molecular bonds of this chemical, and the lower the energy required to form a bond, the more stable the molecule. Therefore, even at extremely cold temperatures (-160*C) in complete isolation, dioxygen difluoride decomposes to oxygen difluoride ($\ce{OF_2}$) and $\ce{O_2}$ gas. At room temperature, this decomposition is uncontrollable (no-one's gotten it anywhere close to RT without it coming apart), and even at extremely cold temperatures, the heat produced by slow decomposition can cause what's called a &quot;thermal runaway&quot;; the heat produced by one molecule decomposing is absorbed by nearby molecules causing them to decompose, releasing more heat, and so on. When things rapidly decompose into more stable gaseous components, the event is normally accompanied by a loud noise and shrapnel, formed from the remains of whatever had been holding the substance and anything else nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the products formed by the self-decomposition of FOOF, oxygen gas is a pretty strong oxidizer, but it's the least of our worries here. The other product, oxygen difluoride, is almost as strong an oxidizer as dioxygen difluoride, which is only stronger because it's oxygen difluoride &lt;em&gt;plus&lt;/em&gt; more oxygen. The &quot;shared&quot; electron pairs forming the bonds between the oxygen and each fluorine atom are much more closely attracted to the fluorine, as opposed to a structure like water where the oxygen is the more attractive. This gives the oxygen a formally positive oxidation state of +2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By its very nature, oxygen &lt;em&gt;yearns&lt;/em&gt; to flip that sign and achieve its &quot;natural&quot; -2 ox state by bonding with something less electronegative (which, as previously mentioned, is almost everything else in the periodic table besides fluorine). However, unlike the other oxygen atom in the FOOF molecule which flew the coop, this second oxygen can't escape, because the two fluorine atoms won't let it go. The fluorines prefer to bond to it rather than to each other... until something else, &lt;em&gt;anything&lt;/em&gt; else, is introduced that's even more attractive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{OF_2}$, like its FOOF parent, reacts very strongly and exothermically with almost anything, especially water. It will literally burn ice, rapidly; that's one of the reactions guaranteed to produce a powerful explosion. The most stable ultimate products of that reaction are hydrogen fluoride gas and more oxygen, and when you call HF gas a &quot;stable&quot; product of any reaction you are speaking in very relative terms; a release of HF gas into the air is one of those &quot;drop everything and &lt;em&gt;run&lt;/em&gt;&quot; types of industrial accidents. HF gas dissolves in water, including the mucus of your lungs, to produce hydrofluoric acid, a persistent, toxic, corrosive acid that will shred just about any organic chemical it comes across from hemoglobin to proteins to DNA, and you &lt;em&gt;really&lt;/em&gt; want to keep those kinds of things intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, there are a lot of things in an inorganic chemistry lab that have a Medusa-like ability to kill at a glance, but fluorine deserves special notice, in part because of &lt;a href=&quot;http://www.lateralscience.co.uk/Fluorine/Fluorine.html&quot;&gt;the number of notable chemists who were poisoned or blown up in the early heady days of isolating it&lt;/a&gt;. When you combine it with other elements on the right hand side of the periodic table in &quot;lightweight&quot; combinations (between two and about ten atoms per molecule), your day is about to get &lt;em&gt;reeeeealy&lt;/em&gt; interesting; see the work of A.G. Streng, who was tasked not only with producing large quantities of dioxygen difluoride, but also with reacting it with things that normal chemists would keep very far away from it (and in fact &quot;normal chemists&quot; stay very far away from FOOF itself for the obvious reasons). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorine only starts becoming more stable and easier to work with as an ionic salt, such as sodium or calcium fluoride (in which form it naturally occurs as the mineral fluorite; still poisonous in sufficient doses but much less of an explosion/reactivity hazard), or when you start throwing in elements of the carbon group (the carbon-fluorine bond is among the strongest known to chemistry, and fluorocarbons have a variety of uses as propellants, refrigerants and polymers, including Teflon), or heavier chalcogens like sulfur (sulfur hexafluoride, $\ce{SF_6}$, has virtually no reaction chemistry and is favored as a dielectric and to blanket air-sensitive materials, though other sulfur-fluoride compounds like sulfur tetrafluoride and disulfur decafluoride are highly toxic).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-12-30T19:08:18.947" LastActivityDate="2013-12-30T19:08:18.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4630" PostTypeId="1" CreationDate="2013-04-11T04:25:16.973" Score="1" ViewCount="354" Body="&lt;p&gt;There are some fundamental charactaristics of liquid crystal. One of them is that all liquid crystals are organic compounds and at one side of their long molecular foundation there exists a polar group such as -CN, -NO2, -NH2, -OR etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A strong dipole effect in molecules is created by the existance of polar group; so the molecular axies are parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually I don't understand the characteristic perfectly because of having confusion at several terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the polar polar groups?&lt;/li&gt;&#xA;&lt;li&gt;What is the strong dipole effect?&lt;/li&gt;&#xA;&lt;li&gt;What is molecular axis?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I especially want to know  about molecular axis with an extant explanation.&lt;/p&gt;&#xA;" OwnerUserId="1480" LastEditorUserId="22" LastEditDate="2013-04-14T19:18:43.350" LastActivityDate="2013-04-14T19:18:43.350" Title="Polar group strong dipole effect molecular axis" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;molecules&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4631" PostTypeId="2" ParentId="2571" CreationDate="2013-04-11T11:29:04.410" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In others words, it is to know approximately how much time it would take for a small amount (about 5 ml) of pure alcohol to evaporate if poured over a surface such as a table. Would it take about 15 seconds/30 seconds/2 minutes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It be simpler and faster to do the experiment than to try to predict the time needed.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic-molecular_theory&quot; rel=&quot;noreferrer&quot;&gt;kinetic molecular theory&lt;/a&gt; explains why liquids evaporate at temperatures lower than their boiling points. At any temperature, the molecules in a liquid have a range of kinetic energies described by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_distribution&quot; rel=&quot;noreferrer&quot;&gt;Boltzmann distribution&lt;/a&gt;. Some percentage of the molecules have enough kinetic energy to escape into the gaseous phase. These molecules contribute to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;noreferrer&quot;&gt;vapor pressure&lt;/a&gt; of the liquid. As the temperature increases, more molecules are above the gas threshold and the vapor pressure increases. As the vapor pressure increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a closed system, &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor-liquid_equilibrium&quot; rel=&quot;noreferrer&quot;&gt;equilibrium&lt;/a&gt; would be established with an unchanging ratio of liquid to vapor. Molecules would be exchanged between the liquid and the vapor, but their relative amounts would remain constant. The surface you describe is not closed - the vapor molecules can wander away by diffusion or convection. Equilibrium is disrupted, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;noreferrer&quot;&gt;Le Châtelier's Principle&lt;/a&gt; tells us that the equilibrium shifts to compensate. As more ethanol molecules escape, more vaporize to replace them until there are no molecules remaining in the liquid. In the following equation, $K$ is the equilibrium constant, $p$ is the partial pressure of ethanol vapor (the vapor pressure) and [$\ce{C2H6O}$] is the concentration of ethanol in the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H6O (l) &amp;lt;=&amp;gt; C2H6O (g)} $$&#xA;$$K=\frac{p_{\ce{C6H6O}}}{[\ce{C2H6O}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After establishment of equilibrium (fast), the rate-determining step of the evaporation is probably the diffusion of the gaseous ethanol molecules away. The average kinetic energy of a gas particle can be expressed as a function of mass ($m$, in kg) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_mean_square&quot; rel=&quot;noreferrer&quot;&gt;root mean square&lt;/a&gt; velocity ($v^2_\text{rms}$) and separately as a function of temperature ($T$, in kelvin) times the Boltzmann constant ($k_\text{B} = 1.38 \times 10^{-23} \frac{\text{J}}{\text{K}}$). We can derive a formula for rms velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{E_\text{k}}=\frac{1}{2}mv^2_\text{rms}$$&#xA;$$\overline{E_\text{k}}=\frac{3}{2}k_\text{B} T$$&#xA;$$v^2_\text{rms}=\frac{3k_\text{B} T}{m}$$&#xA;$$v_\text{rms} = \sqrt{\frac{3k_\text{B} T}{m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could calculate the velocity of ethanol particles escaping. If you set an arbitrary distance (one meter is probably fine), then you can calculate a time it takes one particle to travel that distance (on average). If we know the equilibrium constant, we can determine how much vapor is above the liquid and then calculate the amount of time it takes to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we know the equilibrium constant? It varies with temperature! The value $\Delta G^\circ_\text{vap}$ in the equation below is the free energy change of vaporization of ethanol in the standard thermodynamic state. $R$ is the ideal gas constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\mathrm{e}^{-\frac{\Delta G^\circ_\text{vap}}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model above ignores complicating factors like &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_free_path&quot; rel=&quot;noreferrer&quot;&gt;mean free path&lt;/a&gt;, the fact that evaporation is endothermic (meaning the liquid cools as it evaporates and the vapor pressure decreases with time), the temperature and heat capacity of the surface govern how much kinetic energy is available to the liquid to begin with, and that any amount of air current in the vicinity will move the vapor away significantly faster than diffusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete model includes consideration of the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Constants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the Boltzmann constant&lt;/li&gt;&#xA;&lt;li&gt;The ideal gas constant&lt;/li&gt;&#xA;&lt;li&gt;the free energy change of vaporization of ethanol (not actually constant, but it varies only slightly over the temperature range)&lt;/li&gt;&#xA;&lt;li&gt;the vapor pressure of ethanol as a function of temperature&lt;/li&gt;&#xA;&lt;li&gt;the mass of an ethanol molecule&lt;/li&gt;&#xA;&lt;li&gt;the heat capacity of the surface&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;variables&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;temperature of the air&lt;/li&gt;&#xA;&lt;li&gt;volume of ethanol&lt;/li&gt;&#xA;&lt;li&gt;temperature of the surface&lt;/li&gt;&#xA;&lt;li&gt;atmospheric pressure (needed for mean free path corrections)&lt;/li&gt;&#xA;&lt;li&gt;velocity of air currents&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in principle, you could do it. Even so, the best answer cannot be had without some serious calculus. In practice, it would be faster to do the experiment (if often is). We often forget that science is empirical. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorDisplayName="user7951" LastEditDate="2015-02-23T16:48:30.597" LastActivityDate="2015-02-23T16:48:30.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4632" PostTypeId="1" AcceptedAnswerId="4633" CreationDate="2013-04-11T11:41:56.597" Score="2" ViewCount="216" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the products (more than one) for the hydrogenation of the cycloalkene ring in (3⁠&lt;em&gt;R&lt;/em&gt;,4⁠&lt;em&gt;R&lt;/em&gt;,5⁠&lt;em&gt;S&lt;/em&gt;)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid. Reagents are molecular hydrogen with Pt catalyst. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My problem here is that the question implies there are several products for hydrogenation of the cycloalkene. I only see one product, where each end of the alkene obtains a hydrogen, and the pi bond is gone. How can a hydrogenation of this alkene result in more than one product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/vbPWYPr.png&quot; alt=&quot;Why is there more than one product?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="5026" LastEditDate="2018-07-10T09:27:37.487" LastActivityDate="2018-07-10T09:27:37.487" Title="Products for the hydrogenation of the cycloalkene in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid" Tags="&lt;organic-chemistry&gt;&lt;organic-reduction&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4633" PostTypeId="2" ParentId="4632" CreationDate="2013-04-11T15:27:04.947" Score="5" Body="&lt;p&gt;Hydrogenation of the alkene introduces a stereo centre, thus the products are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HBRIx.png&quot; alt=&quot;The products!&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-04-11T15:27:04.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4634" PostTypeId="1" AcceptedAnswerId="4682" CreationDate="2013-04-11T15:41:43.603" Score="2" ViewCount="3854" Body="&lt;p&gt;In ICP-OES (&lt;a href=&quot;https://en.wikipedia.org/wiki/ICP-OES&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;), you introduce a sample into a very hot plasma and then run a spectral analysis on the resulting light emissions. Liquid samples are sprayed into the flame. From the Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A peristaltic pump delivers an aqueous or organic sample into an analytical nebulizer where it is changed into mist and introduced directly inside the plasma flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From that I can imagine that you would probably dissolve solid samples (the technique is widely used in metallurgy and mineralogy) in something suitable to obtain a liquid sample. If that's correct, I'm wondering how the feeding apparatus is constructed to withstand possibly aggresive solvents, like nitric acid or aqua regia. The injection tube is apparently made from fused silica, I guess that's a probable material for the feed as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how about gases? Are they analysed with ICP-OES as well? Are they simply pumped into the injection tube? For example, can we just introduce a stream of air into the plasma and get the emission spectrum for 78% nitrogen, 21% oxygen (I guess the argon will be hidden because the plasma gas is argon)? I guess since every sample will be beyond vaporized anyway by the intense heat, it can't make much difference, but are gas samples too &quot;thin&quot;, i. e. do they contain too little atoms per volume for useful analysis?&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2013-04-16T05:13:11.407" Title="How are samples prepared for analysis with ICP-OES? Can it analyse gases?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4635" PostTypeId="2" ParentId="4626" CreationDate="2013-04-11T17:00:56.433" Score="3" Body="&lt;p&gt;Energy of activation is pretty much an experimental thing. You can try to explain obtained results, but predicting exact values without numerical analysis is not really possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, given a compound to react with air, you first devise a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_mechanism&quot; rel=&quot;nofollow&quot;&gt;reaction mechanism&lt;/a&gt; -- how the reaction proceeds. Now, using some rules, you determine the highest energy &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state&quot; rel=&quot;nofollow&quot;&gt;transition state&lt;/a&gt;. Finally, you try to approximate the energy difference between this and the reactants by comparing with known quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the rules that are used to compare energies and energy differences: There are many of them, and out of the scope of this answer (indeed, one would be hard pressed to fit them all on a single bookshelf). Basically, one uses general rules that compare stability. For example, a positively-charged carbon is more stable when it has more alkyl groups attached to it (the reason usually given for this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;nofollow&quot;&gt;hyperconjugation&lt;/a&gt;). A carbon conjugated with a double bond is considered even more stable, due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance&quot; rel=&quot;nofollow&quot;&gt;resonance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding &lt;em&gt;paper&lt;/em&gt;, paper is made of molecular fibres which are effectively infinitely long. To melt paper, one would have to disentangle these (as molecules need to be much farther apart and with much less attraction in liquid form), and even then one would be left with a very, very viscous liquid (which is not too different from a solid). So paper would be extremely hard to melt, if not impossible.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-04-11T17:00:56.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4636" PostTypeId="1" AcceptedAnswerId="4645" CreationDate="2013-04-11T17:48:56.383" Score="18" ViewCount="932" Body="&lt;p&gt;For more fun with Dr. Derek Lowe, see this for a primer: &lt;a href=&quot;http://pipeline.corante.com/archives/2005/03/03/things_i_wont_work_with_carbon_diselenide.php&quot;&gt;Things I Won't Work With - Carbon Diselenide&lt;/a&gt;. The short of it is that we carbon-based life forms generally like oxygen. However, move just one row down in the group, to sulfur, and things start going horribly wrong. Sulfur, and the thiol/mercaptan analogs of otherwise relatively benign hydrocarbons and alcohols, gives us the thrilling &lt;em&gt;eaudeur&lt;/em&gt; of rotten eggs, burning tires, skunks, farts, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Move one row further, to selenium, and abandon all hope ye who enter here; the tamest common example of selenium in our daily lives is garlic, an instantly recognizable scent, especially when your body starts excreting the selenium-containing essential oil back out of your pores. It's supposed to be the last word in mosquito repellent; down here in Texas the little biters apparently have a taste for Italian. Selenium analogs of thiols often defy chemists' descriptions of their olfactory power; one of Dr. Lowe's other selenium-based posts, on &lt;a href=&quot;http://pipeline.corante.com/archives/2012/05/15/things_i_wont_work_with_selenophenol.php&quot;&gt;selenophenol&lt;/a&gt; (the selenium analog of thiophenol, which is bad enough), puts it thusly: &quot;Imagine 6 skunks wrapped in rubber innertubes and the whole thing is set ablaze; that might approach the metaphysical stench of this material&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This poses the rather interesting question; &lt;em&gt;why&lt;/em&gt; do things change so suddenly with this group? I'm willing to believe that all of them have similar chemical properties, and we as humans and other oxygen-respiring life forms differentiate oxygen from the rest of the group because oxygen's good for us while the others are bad. But, that borders on the simplistic, and poses more questions, like why we work so much better on oxygen than sulfur (there are a few bacterial species hanging out near deep-sea vents that just can't get enough of those sulfur gases, so obviously there's sulfur-based &quot;organic&quot; metabolism just as there is oxygen-based). I guess what I'm looking for is an explanation of why two chalcogens, both obeying the octet rule as Periods 2 and 3 tend to, are treated so differently by our sensibilities.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="7448" LastEditDate="2015-10-17T20:40:01.087" LastActivityDate="2015-10-17T20:40:01.087" Title="Why do chalcogens (Group VI) stink so badly?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4637" PostTypeId="5" CreationDate="2013-04-11T18:44:00.903" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-04-11T18:44:00.903" LastActivityDate="2013-04-11T18:44:00.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4638" PostTypeId="4" CreationDate="2013-04-11T18:44:00.903" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-04-11T18:44:00.903" LastActivityDate="2013-04-11T18:44:00.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4639" PostTypeId="1" AcceptedAnswerId="4643" CreationDate="2013-04-11T19:43:09.103" Score="2" ViewCount="3492" Body="&lt;p&gt;I'm in chemistry and I still have a hard time comprehending how to balance a chemical equation. &#xA;What are the main steps and general things to know?&#xA;thank you.&lt;/p&gt;&#xA;" OwnerUserId="1484" LastEditorUserId="16683" LastEditDate="2018-12-13T01:50:26.597" LastActivityDate="2018-12-13T01:50:26.597" Title="Balancing a chemical equation" Tags="&lt;stoichiometry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ClosedDate="2018-12-13T01:50:12.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="4640" PostTypeId="5" CreationDate="2013-04-11T20:16:03.440" Score="0" Body="&lt;p&gt;Water, chemical formula H&lt;sub&gt;2&lt;/sub&gt;O, is a common chemical compound on Earth, covering over two-thirds of the surface area of the planet (in either frozen or liquid form) and comprising 0.02% of the total mass of the Earth. The majority of lifeforms on the Earth depend on water and are primarily composed of it, and many natural mineral compounds contain water molecules in their basic structure. As such, water is a key component of many physical and chemical processes of interest to modern science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is naturally produced by the oxidation of pure hydrogen gas by pure oxygen gas. This reaction, like many oxidation reactions, produces heat. It is also a byproduct of the reaction of most strong acids, containing an H&lt;sup&gt;+&lt;/sup&gt; ion, and strong bases, containing a hydroxide (OH&lt;sup&gt;-&lt;/sup&gt;) ion. The remaining components of the acid and base generally produce a salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the arrangement of the electrons of an oxygen atom, a water molecule has a &quot;bent&quot; structure, with the hydrogen atoms at a 105° angle to the oxygen and each other, instead of a linear 180° arrangement. Because of this, a partial positive charge can be observed on the side of the molecule containing the two hydrogens, and a partial negative charge directly opposite. These partial charges make each water molecule, in effect, a tiny magnet, which gives water several interesting properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is a polar solvent; the molecular structure and partial polar charges can break weaker chemical bonds, such as hydrogen bonds and some ionic bonds. This causes materials which have these bonds, such as many salts including table salt, to readily dissolve in water. This property makes water crucial for most biological processes, which use the water to carry nutrients and electrolytes dissolved within it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting property caused by its molecular shape is the tendency to form a crystalline structure when frozen which is less dense (occupying more space) than the same mass of liquid water. This causes ice to float, but also causes problems when living organic tissues, which contain water, are frozen; the water expands, bursting and destroying the cells, damaging the tissue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These properties, paired with water's abundance on Earth, cause many other compounds to incorporate water into their structures. Many minerals, composed of various metallic and nonmetallic solid elements, are &quot;hydrated&quot;; their molecular arrangement incorporates water molecules in the structure. The water can often be released by heating the mineral, which produces a material with different properties. The production of Portland cement, for instance, involves &quot;dehydrating&quot; limestone, creating anhydrous calcium carbonate powder. Re-adding water to the cement during the formation of concrete in effect &quot;reconstitutes&quot; the limestone.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-04-07T01:15:21.653" LastActivityDate="2017-04-07T01:15:21.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4641" PostTypeId="4" CreationDate="2013-04-11T20:16:03.440" Score="0" Body="For questions relating to water (H₂O), specifically its structure, properties, and uses in chemistry. For questions not about water but rather about solutions in water, use tags [aqueous-solution] or [solubility] instead." OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-05-09T11:48:54.050" LastActivityDate="2018-05-09T11:48:54.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4642" PostTypeId="1" AcceptedAnswerId="9022" CreationDate="2013-04-11T21:27:59.097" Score="2" ViewCount="397" Body="&lt;p&gt;I have heard that a &lt;a href=&quot;http://io9.com/5987086/meet-the-scientific-accident-that-could-change-the-world&quot; rel=&quot;nofollow&quot;&gt;super capacitor can be made with a dvd burner&lt;/a&gt;. What materials do I need to to get and what software for my DVD burner do I need to make this amazing device?&lt;/p&gt;&#xA;" OwnerUserId="1485" LastEditorUserId="7448" LastEditDate="2015-04-02T18:30:03.043" LastActivityDate="2015-04-02T18:30:03.043" Title="How can I make a graphene supercapacitor?" Tags="&lt;electrochemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4643" PostTypeId="2" ParentId="4639" CreationDate="2013-04-11T23:03:38.807" Score="2" Body="&lt;p&gt;OK, here goes a semester of college chemistry in one post...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, remember the law of conservation of matter; in chemical reactions, matter is neither created nor destroyed. Therefore, every atom of every element on the left side of the arrow has to appear on the right side, with all their protons and neutrons intact. No strictly chemical reaction will cause a carbon to become a nitrogen by gaining a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;can&lt;/em&gt; change in a reaction is the number of electrons each atom &quot;owns&quot;. For that, remember the concept of valence shells; the number of electrons in the outermost &quot;shell&quot; of its orbitals. Most elements do not have a stable valence shell, and so will want to react in ways that increase their stability. For elements not in the &quot;transition metal&quot; groups, this is pretty straightforward; elements to the left of the transition metals generally want to lose electrons to empty their valence shell (acquiring the stable configuration of the next lightest noble gas), while elements to the right of the transition metals want to share or gain electrons to fill their valence shell, acquiring the configuration of the next &lt;em&gt;heaviest&lt;/em&gt; noble gas. The transition metals are &quot;stable&quot; in various valence configurations based on their more complex &lt;em&gt;d&lt;/em&gt;-orbital valence structure (the &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt; orbitals exposed by the first two and last six columns of the table are more predictable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When atoms outright gain or lose electrons, they become ions, and gain a charge based on the difference between the protons and electrons in their configuration. When they gain them, they become negative anions, and then they lose them, positive cations. Cations are then attracted to anions, and these charges must balance; a +2 cation such as Mg&lt;sup&gt;2+&lt;/sup&gt; or Ca&lt;sup&gt;2+&lt;/sup&gt; requires one -2 anion (carbonate, sulfate, hydrogen-phosphate) or two -1 anions (chloride, fluoride, bicarbonate). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alkali metals, in the 1 and 2 groups, usually want to lose their one or two valence electrons; the heavier the alkali metal, the more the atom wants to lose the electrons, with cesium being the most reactive alkali metal that has a non-radioactive isotope. The elements in the halogen group (fluorine, chlorine, bromine, iodine) like to gain electrons, especially from the alkalis. The &lt;em&gt;lighter&lt;/em&gt; the compound, the more it wants to gain an electron, with fluorine being the most aggressive electron receptor in the table (to the point where it will try to take electrons from other elements that don't want to give them up, like other halogens and noble gases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Columns between the halogens and transition metals (from left to right, the boron group or triels, the carbon group or tetragens, nitrogen group or pnictogens, and oxygen group or chalcogens) are more of the &quot;sharing&quot; type; they'll take electrons outright from the alkalis, and if they &lt;em&gt;really&lt;/em&gt; have to they'll give them up to a halogen like fluorine or chlorine, but they prefer a &quot;give-and-take&quot; relationship, and form shared-orbital &quot;covalent bonds&quot; with most other elements of these groups and with the transition metals. How strong these bonds are depend on what elements are being bonded and how many electrons each is sharing with its neighbor. A double or triple bond is very strong, while certain other combinations are strong for other reasons (bonds between carbon and the &quot;Big 3&quot; highly electronegative elements, namely oxygen, fluorine and chlorine, are commonly very strong). You can compute a balanced bond structure using valence electrons, much like calculating ionic charges; oxygen needs two valence electrons and so it will form no more than two bonds with other elements. When two atoms in these columns get together, expect to see a few multiple bonds; oxygen and nitrogen love forming double and triple bonds when they can, and carbon will oblige with these two and a few others, but typically prefers single bonds with other carbons and most other nonmetals (allowing it to form those long chains that create the backbone of life).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining what the reaction is, keep in mind that there has to be a driving force that causes an equilibrium shift. This can be a number of things, but basically, the reactants have to want to react and produce a product more than they want to stay where they are, and also more than any other component of their environment wants them not to react. Compounds generally want to react when the bonds of the newly-formed compounds are of lower energy and therefore more stable; such reactions are exothermic. Compounds can be forced to react when additional energy is added to the system (usually in the form of heat, sometimes with electricity to get the electrons jumping).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds also generally want to react when they're attracted so strongly that nothing can keep them apart, so learn your solubility rules. Sodium chloride and aluminum bromide will both sit in solution together quite happily as dissociated ions; you can call some of it sodium bromide and some of it aluminum chloride, but because all four possible products are soluble and no electrons change hands, the solution is already in equilibrium and so there's no reaction. However, calcium chloride and sodium carbonate will react because one of the possible cation/anion combinations, calcium carbonate, is insoluble in water; these two will meet, bond, and precipitate out, leaving sodium chloride in the water.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-04-30T18:07:44.670" LastActivityDate="2013-04-30T18:07:44.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4644" PostTypeId="2" ParentId="4624" CreationDate="2013-04-12T05:14:59.730" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is a very strong oxidizer. That means it can make a lot of materials &quot;burn&quot;, i. e. undergo a reaction (oxidation) that releases energy, often in violent ways, because there's so much energy released in such a short time that it has trouble dissipating in any other way than exploding with vigour. This includes a lot of materials that you wouldn't normally consider burnable. Derek Lowe has another article (&lt;a href=&quot;http://pipeline.corante.com/archives/2008/02/26/sand_wont_save_you_this_time.php&quot; rel=&quot;noreferrer&quot;&gt;&quot;Sand won't save you this time&quot;&lt;/a&gt;, about a similarly hideous substance that's also a fluorine compound), which has some very illustrative examples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It contains fluorine, which forms many highly toxic compounds that you don't want to have flying around hot and steaming. Even on itself (as $F_2$), it is a gas that's far from harmless.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you like firsthand stories about compounds like this, and some insight on why people might want to experiment with them at all (although imho, the &lt;em&gt;real&lt;/em&gt; reason is they are simply lunatics), I can recommend the book &lt;a href=&quot;http://www.bookride.com/2007/03/ignition-informal-history-of-liquid.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Ignition!&quot;&lt;/a&gt; about rocket fuel research. It's out of print, but PDFs float around the web.&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditDate="2013-04-12T05:36:34.670" LastActivityDate="2013-04-12T05:36:34.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4645" PostTypeId="2" ParentId="4636" CreationDate="2013-04-12T06:48:09.527" Score="11" Body="&lt;p&gt;I think that there are two reasons: the sulfur is quite common in biochemistry and at the same time a lot of sulfur compounds are toxic. The bio-degradation of sulfur compounds mostly leads to some sulfides which very strongly bind transition metals, &lt;em&gt;i.e.&lt;/em&gt; they can destroy/block Metalloenzymes. If you add some millions years of evolution it's a v.good feature to smell them (that would also work for small phosphines and amines). Your chances for food-poisoning are lower, chance for survival higher (Long Live &quot;The selfish gene&quot;). &lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-04-12T06:48:09.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4646" PostTypeId="1" AcceptedAnswerId="4648" CreationDate="2013-04-12T06:59:18.127" Score="4" ViewCount="8916" Body="&lt;p&gt;I would like to know a general procedure for the synthesising cis isomers from their trans counterparts and vice versa. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2013-04-12T16:03:15.670" Title="How do you change the cis-trans isomerism of compounds?" Tags="&lt;synthesis&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4647" PostTypeId="2" ParentId="4639" CreationDate="2013-04-12T08:02:23.257" Score="3" Body="&lt;p&gt;Basically, balanced chemical reactions must 1) have the same number of each element on each side, and 2) have the same net charge on each side. Consider the reaction $H_2 + O_2 → H_2O$. It is unbalanced because there are two oxygen atoms on the left side but only one on the right. If we assign variable coefficients to each reactant and product, $a H_2 + b O_2 → c H_2O$, we can set up a system of equations by multiplying the variables by the number of atoms in each species. For oxygen, $2b = c$ and for hydrogen, $2a = 2c$. While the system has infinitely many solutions, we take the lowest integer solution. Thus, the balanced equation is $2 H_2 + O_2 → 2 H_2O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical reactions involving charges are a bit more complicated. A good explanation can be found &lt;a href=&quot;http://www.chemteam.info/Redox/Redox.html&quot; rel=&quot;nofollow&quot;&gt;this tutorial&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-12T08:02:23.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4648" PostTypeId="2" ParentId="4646" CreationDate="2013-04-12T08:51:34.460" Score="4" Body="&lt;p&gt;In general, any setup that gives enough energy to the compound can isomerize it. This can be done by heating it, irradiating it with visible or UV light, or by using a catalyst (reducing the energy required for transition rather than supplying more energy). Excitation of electrons from the $\pi$ to the $\pi^*$ (bonding orbital to anti-bonding orbital) allow free rotation about the $\sigma$ bond. Relaxation back to the $\pi$ orbital locks the configuration. The specific procedures depend on the compound of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, photoisomerization (light-induced isomerization) is the principle on which sight is based.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-12T08:51:34.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4649" PostTypeId="1" CreationDate="2013-04-12T11:57:44.600" Score="34" ViewCount="104367" Body="&lt;p&gt;How can one predict whether a given complex ion will be square planar or tetrahedral when its coordination number is 4 using crystal field theory ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to theoretically predict this?&lt;/p&gt;&#xA;" OwnerUserId="1486" LastEditorUserId="22" LastEditDate="2013-04-12T12:52:47.283" LastActivityDate="2019-01-26T02:39:41.493" Title="Tetrahedral or Square Planar" Tags="&lt;coordination-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4651" PostTypeId="2" ParentId="4649" CreationDate="2013-04-12T13:34:46.333" Score="20" Body="&lt;p&gt;This is a really quick and dirty method since I don't know much about the calculations involved in crystal field theory. This method is based on some general observations I've made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, if your complex metal ion is not in group 8 ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) or it doesn't have a $d^8$ configuration (like $\ce{Rh(I)}$ or $\ce{Au(III)}$), it is usually tetrahedral (Assuming you already know that the coordination number is 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your metal ion is in group 8 or has a $d^8$ configuration, look at the crystal field splitting diagram. Square planar complexes have a four tiered diagram (i.e. four different sets of orbitals with different energies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mkc9g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has a two tiered crystal field splitting diagram then it is tetrahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7JOGx.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this assumes you have the crystal field splitting diagram of the complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how you obtain these diagrams (the calculations involved), I don't know exactly how it's done for specific molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I mentioned before, this is just a very basic way to distinguish between the two geometries. I'm sure the calculations involved to obtain these diagrams would be really complicated but would make more sense in some cases if you're looking for the actual reasons behind why some complexes are square planar while others are tetrahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicting the structure of a complex depends on how much info you're given. If you only have the number of ligands in the complex or the molecular formula of the complex, then you can predict the structure based on this list &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_complex#Geometry&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, steric factors and electronic factors are also important in determining the structure and your prediction would be more accurate if you know what ligand(s) you're dealing with (for steric effects) and the electronic configuration and oxidation state of the central metal ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about how exactly these factors influence the structure but in general, you only have to be worried about these if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the ligands are bulky or have some other special features&lt;/li&gt;&#xA;&lt;li&gt;certain special electronic configurations (I've read something about certain electronic configurations causing distortions in the structure especially in copper(II) complexes)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1322" LastEditorDisplayName="user15489" LastEditDate="2015-05-23T23:35:09.980" LastActivityDate="2015-05-23T23:35:09.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4652" PostTypeId="5" CreationDate="2013-04-12T15:33:48.710" Score="0" Body="&lt;p&gt;A chemical compound is &quot;soluble&quot;, simply, if it will dissolve in a specified solvent and, all other things being equal, remain in this solution indefinitely. A solid compound that will not dissolve in a particular solvent is &quot;insoluble&quot; in that solvent; a liquid which will not mix with a solvent to form a homogeneous solution is &quot;immiscible&quot; with that solvent. Solvents are important in chemistry as a basic environment in which to mix reactants; solvents can catalyze or deter reactions, as compared to simply combining the non-dissolved reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound &quot;dissolves&quot; when the attraction between molecules of the compound (known as the &quot;solute&quot;) are broken by the introduction of the solvent, freeing individual molecules (or in the case of ionic salts, individual ions) which then disperse and remain mixed with the solvent's molecules. Compounds that, at least at some level, retain these bonds between ions or molecules do not dissolve; they may form tiny, even invisible particles, but at the molecular level there are clear regions dominated by either solvent or solute. Eventually, the compound will separate and settle either to the top or bottom of the container depending on the relative weight and density of the compound compared to the solvent. These &quot;heterogeneous&quot; mixtures are typically called &quot;suspensions&quot; when mixing a solid with water, or &quot;emulsions&quot; when mixing an oil with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common solvent for which solubility is data is widely available is water, as it is a fundamental, abundant, non-toxic polar solvent that can be used for a variety of reactions. Other common solvents used in conditions near room temperature and pressure include hydrocarbons, such as relatively light alkanes and aromatics, and their derivatives such as alcohols, aldehydes, ketones, esters, and ethers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvents are typically classified as either polar or nonpolar, but polarity can be described using many criteria and is relative. A polar solvent has a molecular structure that produces a partial positive and negative charge at different atoms of the molecule. Water is one such molecule because its highly polar bonds create charge separation. Polar solvents typically work by &quot;pulling apart&quot; weaker bonds, such as some hydrogen and ionic bonds found in organic compounds and salts, respectively. It is not successful in solvating all of these, however; oils and waxes, for example, are defined in part by their immiscibility in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nonpolar solvent has no localized partial charge and usually works instead by breaking and reforming hydrogen bonds between molecules of the solute. Most of these are organic hydrocarbon derivatives, with alcohols such as methanol, ethanol, and phenol being a primary subcategory and ketones such as acetone and methyl ethyl ketone commonly seen in everyday use. These typically work well to dissolve other hydrocarbons, and are also used as solvents for certain compounds which react strongly with water.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="39539" LastEditDate="2017-05-17T07:20:17.163" LastActivityDate="2017-05-17T07:20:17.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4653" PostTypeId="4" CreationDate="2013-04-12T15:33:48.710" Score="0" Body="For questions related to the solubility of compounds in various solvents, especially in water. Do not confuse with [solutions] or [aqueous-solution]." OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-03-19T13:17:50.243" LastActivityDate="2018-03-19T13:17:50.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4654" PostTypeId="2" ParentId="4646" CreationDate="2013-04-12T16:03:15.670" Score="0" Body="&lt;p&gt;As a particular example of what buckminst is talking about, one common undergraduate organic chemistry laboratory experiment is the one where a tube full of dimethyl maleate is isomerized to dimethyl fumarate, with the help of sunlight and a teeny amount of bromine. Since solid dimethyl fumarate has low solubility in liquid dimethyl maleate, this makes for a nice visual demonstration. See e.g. &lt;a href=&quot;http://dx.doi.org/10.1021/ed018p477&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1039/TF9656101138&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user1489" LastActivityDate="2013-04-12T16:03:15.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4655" PostTypeId="1" CreationDate="2013-04-12T16:26:29.883" Score="2" ViewCount="23446" Body="&lt;p&gt;Please explain the physical meaning it carries ? And how is it that it can increase with dilution ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why is it called Molar ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is equivalent conductivity ?&lt;/p&gt;&#xA;" OwnerUserId="1490" LastActivityDate="2015-09-22T03:19:49.210" Title="What is Molar Conductivity of an electrolytic solution" Tags="&lt;electrochemistry&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4656" PostTypeId="1" CreationDate="2013-04-12T17:01:12.407" Score="2" ViewCount="614" Body="&lt;p&gt;Referring to &lt;a href=&quot;http://www.askiitians.com/iit-jee-chemistry/physical-chemistry/salt-of-weak-acid-and-weak-base.aspx&quot; rel=&quot;nofollow&quot;&gt;this,&lt;/a&gt; when &quot;hydrolysis of a salt&quot; is its reaction with water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it that both the cation and anion of the salt are reacting in an equal amount with water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To state this specifically in reference to the link provided, how is it that there is a unique $h$ (degree of hydrolysis)?&lt;/p&gt;&#xA;" OwnerUserId="1492" LastEditorUserId="87609" LastEditDate="2020-06-29T17:30:49.447" LastActivityDate="2020-06-29T17:30:49.447" Title="Salt Hydrolysis" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;salt&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4657" PostTypeId="2" ParentId="4649" CreationDate="2013-04-12T20:06:40.073" Score="8" Body="&lt;p&gt;There are all sorts of approximations available, but the short of it is that in general one cannot predict square planar vs tetrahedral without either doing experiments or doing some very detailed calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, ligand field theories and the like are after the fact justifications of observations rather than predictions of observations.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-04-12T20:06:40.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4658" PostTypeId="2" ParentId="4655" CreationDate="2013-04-12T20:10:39.007" Score="6" Body="&lt;p&gt;I will answer your question about the definition of molar conductivity and why it increases with decreasing concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jerry pointed out, you can probably learn what you need about equivalent conductivity using a &lt;a href=&quot;https://www.google.com/search?q=equivalent%20conductivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google search&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;molar&lt;/strong&gt; conductivity of an electrolyte solution is the conductivity of the solution divided by the &lt;strong&gt;molar&lt;/strong&gt; concentration of the dissolved electrolyte. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductivity ($\kappa$, in $\text{S/cm}$) is proportional to the square root of the concentration of ions ($\sqrt{[\text{ions}]}$, withy $\ce{[ions]}$ in $\text{mol/L}$). In the following equation, A is an empirical proportionality constant with units needed to convert from $\text{M}^{1/2}$ to $\text{S/cm}$, namely $\frac{\text{S/cm}}{\text{M}^{1/2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa = A\sqrt{\ce{[ions]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ion concentration is related to the electrolyte concentration ($C$) by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;van't Hoff factor&lt;/a&gt; ($i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{ions}]=iC$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore \ \kappa = A\sqrt{iC}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar conductivity $\kappa_M$ is the conductivity divided by the molar concentration of electrolyte ($C$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\kappa_M=\frac{\kappa}{C}=\frac{A\sqrt{iC}}{C}=A\frac{\sqrt{i}}{\sqrt{C}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $C$ decreases, so does $\sqrt{C}$, and thus $\frac{1}{\sqrt{C}}$ increases, while $A$ remains constant. The van't Hoff factor changes a little with concentration, but not dramatically. ($A$ probably also varies a little with concentration. It certainly varies with temperature.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have a series of solutions of sodium chloride, concentration decreases faster than conductivity, so conductivity divided by concentration increases. Note that the conductivity data are fabricated with an arbitrary value of $A$. I could not find the specific value of $A$ for sodium chloride. The van't Hoff Factors come from the &quot;Colligative Properties&quot; Chapter of &lt;em&gt;Chemistry: Atoms First&lt;/em&gt; by Burdge and Overby, McGraw Hill, 2012.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSVcu.png&quot; alt=&quot;Molar Conductivity of Sodium Chloride Solutions&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-12T20:10:39.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4659" PostTypeId="2" ParentId="4609" CreationDate="2013-04-12T20:24:11.177" Score="5" Body="&lt;p&gt;Termolecular reactions almost never happen.  The chance of three molecules &quot;colliding&quot; at the same instant is so small as to be ignorable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there are &quot;almost&quot; termolecular reactions when one molecule hits another and &quot;sticks&quot; to it, and then a third molecule hits the two that are stuck together.  This most often happens when one of the molecules is part of a solid.  But this is really two sequential bimolecular reactions.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-04-12T20:24:11.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4660" PostTypeId="2" ParentId="4518" CreationDate="2013-04-12T20:29:39.877" Score="1" Body="&lt;p&gt;The short answer to the question &quot;Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?&quot; is NO.  Once the system is in equilibrium it will remain there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, as is asked in the second part of the question, to have some reaction be triggered at a higher temperature reached by fluctuation, then as has been commented, the answer is that the system was never in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should never forget that there is no &quot;absolute&quot; starting temperature for a reaction.  If the reaction is physically possible, it will take place, but the &lt;em&gt;rate&lt;/em&gt; of that reaction can be horribly slow.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-04-12T20:29:39.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4662" PostTypeId="1" AcceptedAnswerId="4663" CreationDate="2013-04-13T00:17:51.790" Score="3" ViewCount="195" Body="&lt;p&gt;When blood / breath alcohol levels are taken to determine how intoxicated a person is, does it depend on their tolerance ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. If you had two people of identical age, weight, fitness, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;person A, a non drinker, drank 1 measure of alcohol over a period and felt quite &quot;drunk&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;while person B, a frequent drinker drank the same measure of alcohol over the same period, but because of their built up tolerance felt and acted considerably more sober.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would they give the same blood, or breath alcohol readings, despite the effects on the person being different ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a very poorly formatted question from a chemistry point of view, i know absolutely nothing, nor have interacted with any of the sciences since school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also not sure whether this comes under chemistry or biology, if it needs changing to the biology stack exchange site please advise&lt;/p&gt;&#xA;" OwnerUserId="1494" LastEditorUserId="1322" LastEditDate="2013-04-13T05:30:51.073" LastActivityDate="2013-04-13T05:30:51.073" Title="Does tolerance to alcohol affect blood/breath alcohol levels?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4663" PostTypeId="2" ParentId="4662" CreationDate="2013-04-13T04:46:55.567" Score="3" Body="&lt;p&gt;Typical roadside 'breath-testing' measures metabolites produced in the metabolism of alcohol (primarily, they measure ethanal, oxidised ethanol). The level measured level of intoxication therefore definitely differs from person to person, and even day-to-day for everyone. Some people have higher metabolic rates, and will metabolise the alochol faster. Some people have greater amounts of alcohol dehydrogenase (enzyme that breaks down alcohol) than others. It follows, that a larger person will likely be able to consume more alcohol compared to a smaller person (compare the concentration of 1 g of salt in a small glass, compared to a large glass)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there isn't really any fudge room for arguing against the tests. And most police-forces will back up breath testing with far more rigorous blood tests.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2013-04-13T04:46:55.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4664" PostTypeId="2" ParentId="4586" CreationDate="2013-04-13T07:30:22.000" Score="3" Body="&lt;p&gt;$\ce{CO2}$, being non-polar (due to the 180 degree bond angles),  will probably be more soluble in solvents that are non-polar as well, although the quadrupole moment might cause higher solubility in polar solvents as well (thanks to @georg for this addition). Also temperature will play a significant role, the higher the temperature the lower the solubility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These qualitative remarks however only take you so far. If you really want to know the solubility in a specific liquid you will have to look it up or, if that is not possible, test it yourself. In &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/00908310500434481&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; they test the solubility of $\ce{CO2}$ in different types of crude oil and make predictions based on the structure of the oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find anything specific for corn oil, although the reverse process of dissolving &lt;a href=&quot;http://dx.doi.org/10.2202/1556-3758.2186&quot; rel=&quot;nofollow&quot;&gt;palm oil in supercritical $\ce{CO2}$&lt;/a&gt; might set you on the right track to find what you want to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a fun addition: don't ever try to &lt;a href=&quot;http://www.comedycentral.nl/nieuws/26808?video=0acc5664e58786e2079a&quot; rel=&quot;nofollow&quot;&gt;carbonate milk as these guys do&lt;/a&gt; - it will get messy! (it is in Dutch, but you can still watch what happens). Click &lt;a href=&quot;http://youtu.be/E30rtHVG6Cc&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; an English video with similar footage&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-04-16T14:05:31.873" LastActivityDate="2013-04-16T14:05:31.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4665" PostTypeId="2" ParentId="4067" CreationDate="2013-04-13T09:52:42.990" Score="2" Body="&lt;p&gt;&lt;strong&gt;The short answer: check out equation 5 below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will explain the whole story in terms of sedimenting particles, but you can easily read this as sedimenting droplets as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Necessary condition for sedimentation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want particles to sediment the first thing to check is whether their density is higher than the density of the liquid, if not: don't even bother. This can be easily shown on the basis of a force balance with buoyancy and gravity acting on the particles:&#xA;$$\tag{1} F_{net}=(\rho_{p}-\rho_{l})V_{p}g $$&#xA;where subscripts $p$ and $l$ stand for particle and liquid respectively.&#xA;What I should also mention here is that I neglect Brownian motion which can, for very small $V_p$ become a significant effect that avoids sedimentation even when the density of the particles is above that of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, let's assume the particles are big enough to ignore Brownian motion. If we have a centrifuge it is typical to specify the centrifugal acceleration in terms of $a=Ng$ where $N$ is a multiplication factor dependent on the rotation speed of your centrifuge and the distance of the test tube to the center of rotation ($a=\frac{v^2}{r}$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Centrifugal_force&quot; rel=&quot;nofollow&quot;&gt;the wiki for details&lt;/a&gt;). The net force acting on the particles in a centrifuge is thus:&#xA;$$\tag{2} F_{net}=(\rho_{p}-\rho_{l})V_{p}Ng $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Settling velocity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the initial acceleration of the particles there will quickly be a force balance between the net force introduced by the centrifuge and the drag force on the particle. For a perfectly spherical particle that doesn't settle to fast this is simply the Stokes drag:&#xA;$$\tag{3}F_{drag}=6\pi\mu_l R_p v_s$$&#xA;where $\mu_l$ is the liquid viscosity, $R_p$ is the particle radius and $v_s$ is the settling velocity of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balancing the drag with the centripetal force we obtain:&#xA;$$\tag{4} v_s=\frac{2(\rho_p-\rho_l)R_p^2Ng}{9\mu} $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to determine the settling time we need the liquid height in the test tube $h$ so we find $$\tag{5} t_s=\frac{h}{v_s} =  \frac{9\mu h}{2(\rho_p-\rho_l)R_p^2Ng } $$&#xA;Note that with respect to the settling height we take the worst case scenario here because we want all particles (also those close to the surface of the liquid) to settle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 5 nicely shows the influence of various parameters on the time a test tube has to stay in the centrifuge. Smaller particles means longer times, lower density difference also and obviously higher centrifugal acceleration means shorter times. Plugging in some typical values for $1 \mu$m particles in water $(\Delta\rho=100$kg/m$^3)$ and a strong centrifuge ($N=10^4$) we find that we need $O(10 s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Discussion of the simplifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there are quite some assumptions in the derivation above. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Brownian motion might not be negligible therefore either increasing the settling time, or avoiding settling altogether.&lt;/li&gt;&#xA;&lt;li&gt;Particles need to accelerate to their settling velocity first, this will also increase the settling time a bit.&lt;/li&gt;&#xA;&lt;li&gt;In general particles will not be spherical. If, for example, the particles are more rod-shaped then the orientation will determine whether the settling time increases or decreases with respect to the spherical particle prediction. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For many types of particles a &lt;a href=&quot;http://en.wikipedia.org/wiki/Svedberg&quot; rel=&quot;nofollow&quot;&gt;Svedberg coefficient&lt;/a&gt; is tabulated in literature which will allow you to calculate the settling time without having to assume sphericity of the particles. In &lt;a href=&quot;http://www.monzir-pal.net/Bioseparation/Lectures/L12.pdf&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt; they indicate the use of Svedberg coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, these simplifications can mean that the prediction of equation 5 is of by a factor 10 (or maybe even 100 if you're unlucky) therefore in most practical situations you would use this equation as a starting point for some trial-and-error experiments. You can also use it to determine what will happen to the settling time when you e.g. run a new experiment with a different liquid or particle type.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-04-24T05:28:05.460" LastActivityDate="2013-04-24T05:28:05.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4666" PostTypeId="2" ParentId="2818" CreationDate="2013-04-13T10:19:51.367" Score="4" Body="&lt;p&gt;When you go to the limit of infinite concentration (obviously you can never get there), a reaction cannot stay first order, because you will always run into some other rate-limiting mechanism thus making the reaction zeroth order in the highly concentrated species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an example, for the binding of 2 molecules if both molecular species are 'infinitely' concentrated, the rate will not be limited by molecules 'finding' each other (&lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation&quot; rel=&quot;nofollow&quot;&gt;as is normally the case for first order reactions&lt;/a&gt;), but it will be limited by the finite time binding process itself takes.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-04-13T10:19:51.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4667" PostTypeId="1" AcceptedAnswerId="40342" CreationDate="2013-04-13T14:16:11.683" Score="30" ViewCount="238158" Body="&lt;p&gt;Why do transition metals element  make colored compounds both in solid form and in solution? Is it related to their electrons?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="402" LastEditDate="2015-11-07T06:45:28.477" LastActivityDate="2017-04-02T20:38:29.527" Title="Why do transition elements make colored compounds?" Tags="&lt;transition-metals&gt;&lt;color&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="4668" PostTypeId="2" ParentId="4667" CreationDate="2013-04-13T15:24:59.067" Score="17" Body="&lt;p&gt;The partially full d-orbitals in transition metals have energy splittings that happen to lie in the visible range.  Depending on the arrangement of substituents (known as ligands) that attach to them, the electron energies split according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_field_theory#Explaining_the_colors_of_transition_metal_complexes&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;crystal field splitting&quot;&gt;crystal field theory&lt;/a&gt;.  Similar splitting in the s or p orbitals produce gaps in the ultraviolet, and any visible light goes right through, so we don't see any color.  In transition metals, however, visible light excites the electrons from a lower d orbital to a higher one and only letting some light through.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastEditorUserId="402" LastEditDate="2017-04-02T20:37:05.197" LastActivityDate="2017-04-02T20:37:05.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4669" PostTypeId="1" CreationDate="2013-04-14T03:41:01.287" Score="2" ViewCount="9007" Body="&lt;p&gt;Recently I saw &lt;a href=&quot;http://youtu.be/x-GQk8Q96PY&quot; rel=&quot;nofollow&quot;&gt;a video&lt;/a&gt; in YouTube about how to make ice in seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this video, salt is added to a glass of water and then a straw is closed from one end and the other side is lit using a lighter and after which it is stirred for 5 seconds and removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as it is removed the ice is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the chemical reaction that forms ice so quickly?  &lt;/p&gt;&#xA;" OwnerUserId="832" LastEditorUserId="22" LastEditDate="2013-04-15T13:51:54.517" LastActivityDate="2017-08-05T18:35:39.877" Title="What are the factors behind making ice faster?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4670" PostTypeId="2" ParentId="2868" CreationDate="2013-04-14T04:42:33.113" Score="1" Body="&lt;p&gt;My best guess would be a combination Pt, Ru and Ni.&lt;/p&gt;&#xA;" OwnerUserId="1496" LastActivityDate="2013-04-14T04:42:33.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4671" PostTypeId="2" ParentId="4451" CreationDate="2013-04-14T04:52:45.930" Score="1" Body="&lt;p&gt;Baking soda is an effective deodorant because of its basicity. Many foul-smelling compounds are acidic. Neutralizing the compound gives it an ionic character which reduces the vapor pressure (meaning it is less available for the nose to smell it) and makes it much more water soluble (it can be washed away). Odorous compounds like fatty acids (butyric acid is BO) and thiols  have much higher $pK_a$s (~4.8 and 7-10, respectively) than tartaric acid ($pK_a$ ~3) means that they would remain un-neutralized with baking powder. It would not be an effective  deodorizer.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-14T04:52:45.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4673" PostTypeId="1" AcceptedAnswerId="4674" CreationDate="2013-04-14T21:17:56.610" Score="14" ViewCount="3905" Body="&lt;p&gt;I am reading a paper (&lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2007/JM/b706139k&quot; rel=&quot;noreferrer&quot;&gt;Gutel et al., &lt;em&gt;J. Mater. Chem.&lt;/em&gt; &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;17&lt;/em&gt;, 3290-3292&lt;/a&gt;) and I have come across the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In organic solvents, the decomposition under mild conditions of&#xA;  ($\eta^4$-1,5-cyclooctadiene)($\eta^6$-1,3,5-cyclooctatriene)ruthenium(0),&#xA;  Ru(COD)(COT), under a dihydrogen atmosphere gives ruthenium&#xA;  nanoparticles (RuNPs), by releasing cyclooctane (COA).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\text{Ru(COD)(COT)} + \text{H}_2 \rightarrow \text{RuNPs} + \text{COA}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do the $\eta^4$ and $\eta^6$ notations represent?  Do they somehow specify the coordination &quot;connectivity&quot; of the Ru(0) and the COD and COT coordinating ligands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, I have drawn the (two-dimensional) structures of COD and COT.  How would I draw the connectivity of Ru(COD)(COT), or is it not specified in the given name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSTuB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2013-04-15T01:55:11.590" Title="What does the eta notation mean in the naming of a transition metal complex?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4674" PostTypeId="2" ParentId="4673" CreationDate="2013-04-15T01:55:11.590" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What do the $\eta^4$ and $\eta^6$ notations represent? Do they somehow specify the coordination &quot;connectivity&quot; of the Ru(0) and the COD and COT coordinating ligands?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;They do.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\eta$-notation denotes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hapticity&quot; rel=&quot;noreferrer&quot;&gt;hapticity&lt;/a&gt; of the ligand, which is the number of atoms in the ligand bonded to the metal center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQdxk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your image, shown above, cyclooctadiene (COD), has two alkene functional groups. A transition metal can coordinate to just one alkene, which would be $\eta^2$, or to both alkenes, which would be $\eta^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclooctatriene (COT) has three alkenes. A metal center can be coordinated at one ($\eta^2$), two ($\eta^4$), or all three ($\eta^6$) alkenes.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-15T01:55:11.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4675" PostTypeId="2" ParentId="4558" CreationDate="2013-04-15T05:00:52.733" Score="6" Body="&lt;p&gt;I'm hardly an authority on graphene structures (only been reading the literature for about a year), but let me try to contribute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it seems that research into graphene and its derivatives is progressing at such a pace that it can be hard to standardize notation. I have actually previously had a small discussion on the internet regarding the difference between graphene oxide and graphite oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's talk about graphene itself (for the sake of completeness). Originally the term meant a honeycomb sheet with single atom thickness, however it seems to have slightly evolved in usage to mean &quot;graphite thin enough to show peculiar properties&quot;. According to Geim and Novoselov (The rise of graphene, 2007), there is a distinction to be made between single-layer graphene, bilayer graphene and few-layer graphene (3-10 layers) based on their electronic structure. Anything above 10 sheets thick already is electronically sufficiently similar to bulk graphite that it can be called as such. However, there also seems to be another term on the rise for stacks over 10 layers thick, with a still small but not well-defined upper limit, called graphite nanoplatelets (or confusingly, graphene nanoplatelets, depending on who you ask).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding graphene/graphite oxide, I follow the interpretation suggested by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0008622311001965&quot; rel=&quot;nofollow&quot;&gt;Ruoff's group (2011)&lt;/a&gt;. As far as I understand, when dry or dissolved in a highly concentrated, non-basic solution with little stirring, the individual oxidized sheets tend to stay stacked, and one may speak of graph*ite* oxide. If the oxidized sheets are in a basic, dilute, and/or vigorously stirred or sonicated solution, or if the sheets are supported on a substrate and proven to be composed of few layers, then they are unstacked (exfoliated) and one can speak of graph*ene* oxide. Theoretically, there is no composition difference between graphite and graphene oxide, since it's merely a matter of aggregation (The chemistry of graphene oxide, 2010). The difference would be like that between bulk gold and colloidal gold; the composition is the same, but they &lt;strong&gt;are&lt;/strong&gt; chemically different. I have been informed that it is also common to refer to the &lt;em&gt;stacked&lt;/em&gt; oxidized sheets as graphene oxide, and that &lt;em&gt;separate&lt;/em&gt; sheets are called exfoliated graphene oxide. I don't agree very much with that standard, though, since it doesn't make too much sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things increasingly complicated, it seems that what most people considered to be graphene/graphite oxide is actually &lt;em&gt;neither&lt;/em&gt;, but a mixture of oxidized graphitic sheets with highly oxidized polycyclic molecules, collectively called tannic acids or oxidized debris, given they appear as side-products during the process of vigorous oxidation of graphite &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.201007520/abstract&quot; rel=&quot;nofollow&quot;&gt;(Rourke et al., 2011)&lt;/a&gt;. These debris can make up as much as 30% of the dry weight of graphite oxide prepared via Hummers and Offeman graphite oxidation, and they have also been reported as side products of the oxidation of CNTs with HNO$_{3}$. It could well be that many properties previously attributed to graphene/graphite oxide are in fact strongly altered by the presence of these debris, which stay adsorbed to the graphene sheets via $\pi - \pi$ interactions and hydrogen bonding. For example, graphite oxide is only appreciably soluble in polar solvents such as water, DMSO and NMP because the debris act as a surfactant of sorts, stabilizing the suspension. When removed, the purified graphene oxide sheets are actually quite insoluble in most solvents, much like the non-oxidized sheets. Also, perhaps the existence of previously undetected debris is part of the reason why there are still debates on the proper model for the structure of graphene oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, graphene and graphite oxide, while not exactly the same, are terms used interchangeably, and it's possible to go from one to the other reversibly by adding/removing solvent (as long as no concentrated base is added, or dilute base with heating). &quot;Graphene&quot; is a trendier term though, so that's usually what oxidized graphite will be advertised as. Though in actuality, due to the presence of debris, without purification it is really neither!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Ever since I wrote this answer, it seems that efforts to begin standardizing the nomenclature of graphene-related carbon nanostructures &lt;a href=&quot;http://www.materialstoday.com/carbon/features/all-in-the-graphene-family-a-recommended-nomenclat/&quot; rel=&quot;nofollow&quot;&gt;are underway&lt;/a&gt; (free access). Quoting from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Graphene oxide (GO)&lt;/strong&gt; – chemically modified graphene prepared by oxidation and exfoliation that is accompanied by extensive oxidative modification of the basal plane. Graphene oxide is a monolayer material with a high oxygen content, typically characterized by C/O atomic ratios less than 3.0 and typically closer to 2.0.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Graphite oxide&lt;/strong&gt; – a bulk solid made by oxidation of graphite through processes that functionalize the basal planes and increase the interlayer spacing. Graphite oxide can be exfoliated in solution to form (monolayer) graphene oxide or partially exfoliated to form few-layer graphene oxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that the definitions proposed in the article will not necessarily take hold, and even if they do the transition will take some time. Be wary of nomenclature mismatches during this period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in my post I mention the presence of oxidized debris in graphite oxide. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cm401922e&quot; rel=&quot;nofollow&quot;&gt;Research is still ongoing&lt;/a&gt; to determine their nature and properties.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-03-21T00:45:34.067" LastActivityDate="2014-03-21T00:45:34.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4676" PostTypeId="1" AcceptedAnswerId="4695" CreationDate="2013-04-15T06:04:29.370" Score="6" ViewCount="2976" Body="&lt;p&gt;We all know that an $\ce{NO2}$ group attached to a benzene is a deactivator, and it directs and substituents that would be adding to the meta position to the $\ce{NO2}$.  For an alkyl group attached to a benzene it is an activator, and it directs to the positions ortho/para to the alkyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found any information on how a $\ce{-CH2-OH}$ group would affect the reactivity of the benzene ring shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pMgvv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1500" LastEditorUserId="1362" LastEditDate="2013-04-15T13:09:43.137" LastActivityDate="2013-04-16T18:56:56.873" Title="What is the influence of an alkyl-hydroxy group on a benzene ring?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4677" PostTypeId="2" ParentId="4669" CreationDate="2013-04-15T09:53:29.587" Score="2" Body="&lt;p&gt;I am not sure about this specific video (although it indeed appears fake), but the trick in general is NOT FAKE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works by &lt;a href=&quot;http://en.wikipedia.org/wiki/Supercooling&quot; rel=&quot;nofollow&quot;&gt;undercooling&lt;/a&gt; (also called supercooling) the water, which essentially means that you cool the water (far) below its freezing point but avoid ice formation. A common way to do this is to pressurize the water. This will lower the freezing point so it allows you to cool to say -20$^\circ$C, if you then change the pressure back to normal the freezing will only start when triggered by a disturbance, like a tap, shake or added ice crystals (like in the movie below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding salt is another way of lowering the freezing temperature and achieving a similar result. Also with beer it is possible to achieve undercooling. Just put a closed bottle in the freezer for $\approx 2$ hours then open it and tap the top to see the entire beer freeze up. In this case the freezing occurs because your tapping removes (part of) the dissolved gases from the water thus elevating the freezing point back to normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decent (visual) explanation is given &lt;a href=&quot;http://www.squidoo.com/Instant_water_freezing&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-04-16T11:42:39.490" LastActivityDate="2013-04-16T11:42:39.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4679" PostTypeId="1" AcceptedAnswerId="4704" CreationDate="2013-04-15T13:17:16.587" Score="0" ViewCount="97" Body="&lt;p&gt;What exactly happens? Does the butylithium chain add to the carbon where the bromine is attached? or do the 2 undergo halogen metal exchange?&lt;/p&gt;&#xA;" OwnerUserId="1500" LastActivityDate="2013-04-17T23:55:24.133" Title="Reaction between n-butylithium and either an aromatic bromide or an alkylbromide" Tags="&lt;organic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4680" PostTypeId="1" AcceptedAnswerId="4702" CreationDate="2013-04-15T14:44:09.583" Score="6" ViewCount="505" Body="&lt;p&gt;I'm asking as I fear that the 2-chloropropane might react with the pyridine ring to form k-isopropylpridine-4-carbohydrazidine where k=2,3,5. Or that no reaction will take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cl0ho.png&quot; alt=&quot;Isoniazid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 1: Isoniazid&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8p8Kl.png&quot; alt=&quot;Iproniazid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 2: Iproniazid&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/89vxF.png&quot; alt=&quot;Isopropyl chloride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 3: 2-chloropropane&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="1362" LastEditDate="2013-04-15T19:51:20.680" LastActivityDate="2016-10-23T00:02:13.323" Title="Can iproniazid be prepared by reacting isoniazid with 2-chloropropane?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4681" PostTypeId="2" ParentId="162" CreationDate="2013-04-15T17:42:40.470" Score="2" Body="&lt;p&gt;If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. &lt;a href=&quot;http://probes.invitrogen.com/media/pis/mp01290.pdf&quot; rel=&quot;nofollow&quot;&gt;This write-up&lt;/a&gt; describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and &lt;a href=&quot;http://www.sc.ehu.es/powgep99/dcytp/teoricos/txema/tesia/node54.html&quot; rel=&quot;nofollow&quot;&gt;forms a complex with magnesium which breaks planarity&lt;/a&gt;, which would certainly change the fluorescence spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would only take a day to get accurate quantitative results if you have access to a fluorometer.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-15T17:42:40.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4682" PostTypeId="2" ParentId="4634" CreationDate="2013-04-15T18:17:04.460" Score="4" Body="&lt;p&gt;&lt;strong&gt;About the measurement of solids&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Injections tubes are indeed often made out of fused quartz. Dealing with aggressive solvents is actually not that much of an issue because of the extremely low amounts of material that can be detected with ICP-OES. Therefore, you can simply dilute your aggressive solvent before use. You can look for example at this &lt;a href=&quot;http://www.chem.agilent.com/Library/applications/icpes-6.pdf&quot; rel=&quot;nofollow&quot;&gt;research note&lt;/a&gt; where they dissolve a sample in $\ce{HCl}$ and $\ce{HNO3}$ and then dilute it before use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About the measurement of gases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is possible to measure gases directly, but it is not recommended and some adaptations to the normal procedure are needed. In this &lt;a href=&quot;http://dx.doi.org/10.1016/j.wasman.2012.05.015&quot; rel=&quot;nofollow&quot;&gt;research paper&lt;/a&gt; they investigated aerosol analysis by ICP. The main issue was keeping sampling lines at elevated temperatures to allow proper calibration. A secondary problem arose because their target analyte, metallic compound in flue gases, which are volatile at (or close to) room temperature, are typically highly unstable or explosive. Therefore, they recommend dissolving areosols in a liquid solution of water and some organic solvent (1-methoxy-2-propanol in their case) and use this as feed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the measurement of 'normal' gases it seems that the problems they encounter will not occur, because you don't need elevated temperatures in your calibration. So in principle that means that you can indeed measure air. It should be noted that in some machines a shear flow of dried air or nitrogen is used to cut the plasma (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductively_coupled_plasma_atomic_emission_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;) which will disallow measurement of air, as this Perkin Elmer instrument (overview table in &lt;a href=&quot;http://www3.nd.edu/~cest/news/inorgmetals.pdf&quot; rel=&quot;nofollow&quot;&gt;slide 7 of this presentation&lt;/a&gt;). Some other instruments, like the one in the &lt;a href=&quot;http://dx.doi.org/10.1016/j.wasman.2012.05.015&quot; rel=&quot;nofollow&quot;&gt;research paper&lt;/a&gt; I referenced, do not use such a shear flow and will thus be able to detect air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas samples are not at all 'too thin', because the normal nebulized feed that goes into the plasma is similarly dilute.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1389" LastEditDate="2013-04-16T05:13:11.407" LastActivityDate="2013-04-16T05:13:11.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4683" PostTypeId="2" ParentId="4472" CreationDate="2013-04-16T05:39:04.493" Score="3" Body="&lt;p&gt;&lt;strong&gt;Solubility in water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two most important (and somewhat related) factors are the polarity of the molecule and the presence of hydrogen bonding groups. As a rule of thumb, the higher the polarity the more soluble a molecule will be in water (which is highly polar as well). Also, the more hydrogen bonding groups, the more soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these two rules of thumb you should be able to order the compounds you listed. I'll leave it to you to actually do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Boiling point&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website &lt;em&gt;masterorganicchemistry.com&lt;/em&gt;[1] gives a fairly good summary of 3 important factors that determine the boiling point of organic compounds. They are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The relative strength of the four intermolecular forces is: ionic &gt; hydrogen bonding &gt; dipole dipole &gt; van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present, but generally stronger forces means a higher boiling point.&lt;/li&gt;&#xA;  &lt;li&gt;Boiling points increase as the number of carbons is increased.&lt;/li&gt;&#xA;  &lt;li&gt;Branching decreases boiling point.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With these 3 'rules' you should be able to figure out the ordering of your listed compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/10/25/3-trends-that-affect-boiling-points/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Master Organic Chemistry: 3 Trends That Affect Boiling Points&lt;/a&gt; or via &lt;a href=&quot;https://web.archive.org/web/20171015041640/https://www.masterorganicchemistry.com/2010/10/25/3-trends-that-affect-boiling-points/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Internet Archive&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="4945" LastEditDate="2018-08-22T11:33:54.313" LastActivityDate="2018-08-22T11:33:54.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4684" PostTypeId="2" ParentId="2395" CreationDate="2013-04-16T05:41:49.847" Score="2" Body="&lt;p&gt;$\ce{SnF2}$ appears to have a predominantly covalent structure while $\ce{NaF}$ is ionic. So I think $\ce{SnF2}$ might release fluoride ions slower (more reluctantly) than $\ce{NaF}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since toothpaste contains $\ce{Ca}$ containing minerals as well (as mentioned in the Wikipedia page), highly insoluble $\ce{CaF2}$ is formed easily when more fluoride ions are present. This is bad since $\ce{CaF2}$ will be inactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So since $\ce{SnF2}$ doesn't release fluoride ions as much (unless it comes into contact with water – it undergoes hydrolysis fast. Again this is good since more fluoride will be available only while you are brushing your teeth.), it has a longer shelf life than $\ce{NaF}$ I guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that it doesn't matter what fluoride you use as long as its covalent. However I suspect tin is used because it happens to be one of the covalent fluorides relatively harmless to us (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_poisoning&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all just a guess so take it with a grain of salt... I don't have many sources to back it up (except Wikipedia of course!). Do point out any mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Georg has mentioned in the comments, toothpastes have a significant amount of water. Apparently toothpastes containing stannous fluoride contain a strong complexing agent which prevents oxidation of the $\ce{Sn^2+}$ ion (I don't have access to the paper but I read the abstract &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/8021041?dopt=Abstract&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the abstract mentions that $\ce{Sn^2+}$ has a antibacterial effect (most probably due to it being a heavy metal ion) and this is another reason why it is used (in contrast to $\ce{Na+}$ which has almost no antibacterial effects).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorDisplayName="user7951" LastEditDate="2015-07-04T20:23:02.240" LastActivityDate="2015-07-04T20:23:02.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4685" PostTypeId="1" CreationDate="2013-04-16T08:05:55.023" Score="1" ViewCount="675" Body="&lt;p&gt;Now in an electrochemical cell , we make use of the fact that whenever there is a potential drop , electrons will flow from lower to higher potential as for a potential drop , there must be an electric field .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose there are two single electrodes , what creates the electrostatic potential on each of them in the first place ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just to confirm , a redox reaction is used just that we can maintain a steady flow ?And so why is it that we always put in that electrolyte that has one of the ions same as the electrode element ?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="1322" LastEditDate="2013-04-17T12:10:18.863" LastActivityDate="2019-10-08T03:38:25.937" Title="Electrochemical Cell" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4686" PostTypeId="1" CreationDate="2013-04-16T10:55:34.723" Score="2" ViewCount="5019" Body="&lt;p&gt;How does a SHE electrode work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the $H_2$ gas get absorbed on the platinum electrode? How is it reduced? How are electrons transferred from the adsorbed gas to the solution? How is $H^+$ liberated or how does the $H^+$ in the solution take electrons from the adsorbed $H_2$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If platinum doesn't adsorb the $H_2$ gas, what is its use? How does it provide a surface for reduction/oxidation of hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in a concentration cell, why is it, in evaluating Q(reaction quotient)&#xA;you take pressure of a gas numerically and put it directly as a concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in evaluating log(Q) in nernst equation, Q isn't always unitless. But inside log, you can put only a unitless number, isn't that so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in concentration cells, if [c1] &gt; [c2], then c1 acts as cathode. In Gases, considering Pressure=k[C], where k is a constant fixed at a particular temperature. then in gas concentration cell opposite happens if pressure 1 &gt; pressure 2 then pressure 1 part acts as anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is P? is it pressure of the gas above the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In net, how does a gas electrode work?&lt;/p&gt;&#xA;" OwnerUserId="1505" LastEditorUserId="1389" LastEditDate="2013-04-16T19:55:18.317" LastActivityDate="2014-04-22T10:43:10.940" Title="Gas Electrode Working" Tags="&lt;electrochemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4688" PostTypeId="2" ParentId="4669" CreationDate="2013-04-16T12:27:35.270" Score="4" Body="&lt;p&gt;The simplest explanation would be that it isn't (pure) water at all. It is most likely a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supersaturation&quot; rel=&quot;nofollow noreferrer&quot;&gt;supersaturated solution&lt;/a&gt; of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium acetate&lt;/a&gt; - &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COONa}$&lt;/span&gt; (or sometimes &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_thiosulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium thiosulfate&lt;/a&gt; - &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2S2O3}$&lt;/span&gt;) in water.&lt;/p&gt;&#xA;&lt;p&gt;A supersaturated solution is a solution that contains more solute than normally soluble at those conditions. So as soon as a rough surface is provided for &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleation&quot; rel=&quot;nofollow noreferrer&quot;&gt;nucleation&lt;/a&gt; (formation of crystals), the solute starts crystallizing and comes out of solution as a big crystal.&lt;/p&gt;&#xA;&lt;p&gt;The rough surface used over here is most probably the rough end of the burnt straw. This produces enough disturbance to cause crystallization. Usually an already prepared crystal of pure sodium acetate is used for nucleation but only if you want high purity crystals.&lt;/p&gt;&#xA;&lt;p&gt;This is a pretty common trick (also the original video is from trick life -- as mentioned on the youtube description --) and most people who don't know about this phenomenon are usually easily tricked (then again, the &amp;quot;ice&amp;quot; formed would actually be hot since it is an exothermic reaction but you can't feel it through a video right :) ).&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.wikihow.com/Make-Hot-Ice&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; and &lt;a href=&quot;http://chemistry.about.com/od/homeexperiments/a/make-hot-ice-sodium-acetate.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; if you want more info.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-08-05T18:35:39.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4689" PostTypeId="1" CreationDate="2013-04-16T12:29:23.903" Score="1" ViewCount="508" Body="&lt;p&gt;I obtained some Rf values via TLC by adding two or three drops methanolic triethylamine to an eluent mixture of let's say DCM/MeOH = 9:1. I couldn't find an example of how a mixture like that is written down in short in publications. I think that calculating the absolute content of dissolved ammonia wouldn't be suitable.&lt;/p&gt;&#xA;" OwnerUserId="661" LastEditorUserId="9961" LastEditDate="2015-10-16T17:24:32.793" LastActivityDate="2015-10-16T17:24:32.793" Title="How to write down the composition of eluent mixtures for TLC" Tags="&lt;organic-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4690" PostTypeId="1" AcceptedAnswerId="4691" CreationDate="2013-04-16T12:50:03.337" Score="2" ViewCount="84" Body="&lt;p&gt;In &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071624422&quot; rel=&quot;nofollow&quot;&gt;Goodman and Gilman's The Pharmacological Basis of Therapeutics 12th Edition&lt;/a&gt; pg. 406, two drugs, nor&lt;sub&gt;1&lt;/sub&gt;-citalopram and nor&lt;sub&gt;2&lt;/sub&gt;-citalopram are referred to but I have no idea what they are. Could somebody tell me a more commonly used name of these drugs or their IUPAC names if common names do not exist. Also please show me what exactly the nor&lt;sub&gt;1&lt;/sub&gt; and nor&lt;sub&gt;2&lt;/sub&gt; refers to. I'm guessing there's methyl groups missing in them at different positions because I know that norpethidine (or normeperidine) refers to the N-demethylated metabolite of pethidine (or meperidine).&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="84" LastEditDate="2013-09-01T05:48:59.807" LastActivityDate="2013-09-01T05:48:59.807" Title="What compounds do Nor1 and Nor2 citalopram refer to?" Tags="&lt;nomenclature&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4691" PostTypeId="2" ParentId="4690" CreationDate="2013-04-16T13:12:05.463" Score="4" Body="&lt;p&gt;Those names are referring to &lt;a href=&quot;http://en.wikipedia.org/wiki/Desmethylcitalopram&quot; rel=&quot;nofollow&quot;&gt;desmethylcitalopram&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Didesmethylcitalopram&quot; rel=&quot;nofollow&quot;&gt;didesmethylcitalopram&lt;/a&gt;, which, as you correctly predicted, are the &lt;em&gt;N&lt;/em&gt;-demethylated and &lt;em&gt;N&lt;/em&gt;-didemethylated metabolites of citalopram. Per IUPAC conventions, the nor- prefix is used to refer to any structural derivative of some parent compound in which a single carbon and (possibly) its hydrogens have been replaced (typically by a hydrogen, but not exclusively). See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nor-&quot; rel=&quot;nofollow&quot;&gt;relevant Wikipedia page&lt;/a&gt; on the prefix for more detail.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-04-16T13:12:05.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4692" PostTypeId="1" AcceptedAnswerId="4693" CreationDate="2013-04-16T16:59:58.603" Score="1" ViewCount="8276" Body="&lt;p&gt;What is the result when glucose is mixed with excess acetic anhydride and excess pyridine?&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="865" LastEditDate="2015-10-02T16:37:46.223" LastActivityDate="2017-04-07T12:35:37.493" Title="Result of glucose with acetic anhydride and pyridine?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4693" PostTypeId="2" ParentId="4692" CreationDate="2013-04-16T17:26:21.263" Score="2" Body="&lt;p&gt;Most probably it forms glucose pentaacetate. The acetic anhydride esterifies with all the alcohol groups on the glucose ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ziUQx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-04-16T17:26:21.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4694" PostTypeId="2" ParentId="4282" CreationDate="2013-04-16T18:39:09.820" Score="6" Body="&lt;p&gt;Most computational chemistry software packages geared toward students/education should be able to do this. Examples include &lt;a href=&quot;http://scigress.com/&quot; rel=&quot;noreferrer&quot;&gt;Scigress, formerly CAChe&lt;/a&gt;, &lt;a href=&quot;http://wavefun.com/products/spartan.html&quot; rel=&quot;noreferrer&quot;&gt;Spartan&lt;/a&gt;, which has a 'lite/student' version that can probably do it, etc. Generally, any modeling software, for example, &lt;a href=&quot;http://gaussian.com/&quot; rel=&quot;noreferrer&quot;&gt;Gaussian&lt;/a&gt; or &lt;a href=&quot;http://www.msg.ameslab.gov/GAMESS/&quot; rel=&quot;noreferrer&quot;&gt;GAMESS&lt;/a&gt; should generate a file with the electrostatic potential map provided you give it the correct instructions. The output files of these methods can be visualized using many types of software (like Scigress and Spartan) and open source software like &lt;a href=&quot;http://avogadro.openmolecules.net/wiki/Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for an inexpensive solution, GAMESS is free for research/education and Avogadro is open source.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-16T18:39:09.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4695" PostTypeId="2" ParentId="4676" CreationDate="2013-04-16T18:56:56.873" Score="6" Body="&lt;p&gt;According to Hansch, Leo, and Hoekman (Exploring QSAR: Hydrophobic, Electronic and Steric Constants 1995) the Hammet sigma value for $-CH_2-OH$ is 0 (zero) in both either the meta or para position.  Positive values indicate deactivators or electron withdrawing (+.778 for nitro in para position) and negative values are activators or electron donating (-0.268 for methoxy group).  The Hammett relationship gives a way to compare electronic effects of substituents on a benzene ring.  Further info on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_equation&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="702" LastActivityDate="2013-04-16T18:56:56.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4696" PostTypeId="5" CreationDate="2013-04-16T18:57:19.983" Score="0" Body="&lt;p&gt;Trends which are observed in the properties of elements as you move along the periodic table in a given direction. Some of these properties include &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomic radius&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionization energy&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_affinity&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron affinity&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronegativity&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Valence_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;number of valence electrons&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Metal&quot; rel=&quot;nofollow noreferrer&quot;&gt;metallic properties&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="42741" LastEditDate="2017-06-18T08:08:04.160" LastActivityDate="2017-06-18T08:08:04.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4697" PostTypeId="4" CreationDate="2013-04-16T18:57:19.983" Score="0" Body="Trends which are observed in the properties of elements as you move along the periodic table in a given direction. " OwnerUserId="1322" LastEditorUserId="22" LastEditDate="2013-04-16T19:09:03.080" LastActivityDate="2013-04-16T19:09:03.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4698" PostTypeId="2" ParentId="4686" CreationDate="2013-04-16T19:21:50.707" Score="3" Body="&lt;p&gt;&lt;strong&gt;Standard Hydrogen Electrode&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A SHE consists of four parts: a platinum lead wire, a platinum foil contact, an acid solution, and a hydrogen gas flow. The word standard refers to the fact that all concentrations and pressures are at standard state, thus the acid is 1M and the gas flow is 1 atm. At these conditions, the electrode has a potential of 0 V (this is how potential is defined). If there are any deviations from these values, the electrode is no longer considered a SHE. The potential can then be calculated using the Nernst equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrode is one half of an electrochemical circuit. The other half cell determines whether the SHE will behave as the cathode or the anode. If it is the anode, the electron flow comes from the external system to the platinum contact. 2 electrons reduce 2 acidic hydrogens from the solution to form one molecule of hydrogen, as in $2H^+ + 2e^- → H_2$. The molecular hydrogen then bubbles up above the solution. If the electrode is cathodic, hydrogen gas is absorbed on the platinum contact plate and oxidized to two acidic hydrogens which remain in solution. The electrons then flow to the external circuit. Electrons do not flow from hydrogen gas to the acid, or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reaction Quotient&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction quotient is indeed unitless. While it is common to use concentrations and pressures to calculate Q, it is more properly calculated by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;activity&lt;/a&gt; of each species, which are also unitless. The activity is the apparent concentration or pressure of a species (can be viewed as a correction to concentration or pressure). Because in most chemical applications activity is very close numerically to concentration, using concentrations or pressures directly for calculations is a close approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NOHqm.png&quot; alt=&quot;Diagram of a Standard Hydrogen Electrode in a electrochemical circuit&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="1389" LastEditDate="2013-04-16T19:34:18.500" LastActivityDate="2013-04-16T19:34:18.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4702" PostTypeId="2" ParentId="4680" CreationDate="2013-04-17T02:54:13.350" Score="2" Body="&lt;p&gt;Alkylation of the pyridine ring under normal conditions is very unlikely, so that's not a major concern. The nitrogen is electron withdrawing, hence electrophilic aromatic substitution reactions are much more difficult to achieve on pyridine than benzene, particurarly absent catalysis. Additionally, the &amp;alpha;-amino ketone substituent already on the ring is further deactivating (since it contains a &amp;pi; bond conjugated to the ring and a highly electronegative atom, i.e., oxygen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue is that pyridine is liable to form quaternary ammonium salts when reacted with alkyl halides. See for example the substances &lt;a href=&quot;http://en.wikipedia.org/wiki/Paraquat&quot; rel=&quot;nofollow&quot;&gt;paraquat&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diquat&quot; rel=&quot;nofollow&quot;&gt;diquat&lt;/a&gt;, and the &lt;a href=&quot;http://www.elsevierdirect.com/brochures/CompHeterocyclic/PDFs/00602.pdf&quot; rel=&quot;nofollow&quot;&gt;following paper on pyridine chemistry&lt;/a&gt;. While the quaternization of H-bearing ammonium cations is easily reversible (given that they can be deprotonated), the same can't be said for the nitrogen of the pyridine ring. Generally, the only methods I'm aware of for dealkylation of quaternary ammonium salts involve hydride donors and strong reducing agents, such as lithium aluminum hydride. Certainly, the ketone group would need to be protected in that situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue of avoiding polyalkylation, as well as selecting which of the two amines in your substituent react, in view of the fact that the basicity and nucleophilicty of amines increases with the number of alkyl substituents. Sterics, however, might preclude that from becoming a problem in this instance.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-04-17T04:02:04.737" LastActivityDate="2013-04-17T04:02:04.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4703" PostTypeId="2" ParentId="151" CreationDate="2013-04-17T03:57:10.930" Score="16" Body="&lt;p&gt;Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann.&lt;sup&gt;[1]&lt;/sup&gt; He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to him, there is a purely quantum mechanical energy term proportional to &lt;span class=&quot;math-container&quot;&gt;$$\frac{n_{↑}(n_{↑}-1)}{2} + \frac{n_{↓}(n_{↓}-1)}{2}, $$&lt;/span&gt;&#xA; where &lt;span class=&quot;math-container&quot;&gt;$n_{↑}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n_{↓}$&lt;/span&gt; represent the number of spin-up and spin-down electrons in a subshell. This term decreases the potential energy of the atom, and it can be shown that the difference in exchange energy between two consecutive subshell populations (such as &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}^{3}/\mathrm{p}^{4}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{f}^{9}/\mathrm{f}^{10}$&lt;/span&gt;, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{1}\mathrm{d}^{5}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{1}\mathrm{d}^{10}$&lt;/span&gt; configurations relative to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{2}\mathrm{d}^{4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{2}\mathrm{d}^{9}$&lt;/span&gt;, even though there would be a slight increase of electron density in the more compact &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; subshells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. So, which is (more) true? Or is neither?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cann, P. Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2000,&lt;/strong&gt; &lt;em&gt;77&lt;/em&gt; (8), 1056 &lt;a href=&quot;https://doi.org/10.1021/ed077p1056&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed077p1056&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="4945" LastEditDate="2019-01-30T12:36:44.893" LastActivityDate="2019-01-30T12:36:44.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4704" PostTypeId="2" ParentId="4679" CreationDate="2013-04-17T06:21:35.010" Score="0" Body="&lt;p&gt;It's quite a long time that I did myself, but I used BuLi to transmetalate. &lt;em&gt;I.e.&lt;/em&gt; $\ce{MesBr + BuLi -&amp;gt; MesLi + BuBr}$&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-04-17T06:21:35.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4705" PostTypeId="1" CreationDate="2013-04-17T08:28:09.190" Score="1" ViewCount="138" Body="&lt;p&gt;A very common analytical method for detection of more than 70 elements is ICP-OES (&lt;a href=&quot;https://en.wikipedia.org/wiki/ICP-OES&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;). But apparently, it's not suited for some substances (cited from &lt;a href=&quot;http://www.wfu.edu/chemistry/courses/jonesbt/334/icpreprint.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, section 6.2, page 12):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The elements that are&#xA;  not usually determined by ICP/OES fall into three basic&#xA;  categories. The ﬁrst category includes those elements&#xA;  that occur either as trace contaminants in the argon gas&#xA;  used in the ICP/OES ($C$ from $CO_2$), constituents of the&#xA;  sample solvent ($C$, $O$, $H$), or as contaminants from the&#xA;  environment or atmosphere ($N$ for example). The second&#xA;  category encompasses those elements that require high&#xA;  excitation energy, such as the halogens. These elements&#xA;  could be determined with poor LOD, however. The third&#xA;  category is the family of short-lived radioactive elements&#xA;  that are commonly determined by $\gamma$-ray spectrometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, apparently, when I'm working with, say carbon-based fuels, and I want to determine $C$ contents, ICP-OES would not be the method of choice, at least not the standard variety. So, what else would be the typical choice? Or are there modifications to ICP-OES, such as using very pure $Ar$, or $He$, as the plasma gas and sealing the plasma chamber from the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="589" LastEditorUserId="589" LastEditDate="2013-04-17T08:43:29.403" LastActivityDate="2013-04-17T15:40:20.110" Title="What analytical methods are used to quantify C, O, N, and H in a sample?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4706" PostTypeId="1" AcceptedAnswerId="4733" CreationDate="2013-04-17T08:42:11.353" Score="4" ViewCount="7543" Body="&lt;p&gt;ICP-OES is a very common analytical method (Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/ICP-OES&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), and I have heard a number of times (from chemists, too) that it is used for quantitative analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from the description of the process, I'm not certain if it really does give quantitative information, how it does so, and whether it can do so on its own, or whether there are additional methods run in combination (like with gas chromatography and mass spectrometry) to achieve that result.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2019-07-24T11:58:01.077" Title="(How) Does ICP-OES determine the quantity of detected elements, or just their occurance in the sample?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4708" PostTypeId="1" AcceptedAnswerId="4714" CreationDate="2013-04-17T08:57:25.367" Score="7" ViewCount="1792" Body="&lt;p&gt;I am a high school student . While studying the atomic structure a question raised in my mind. We know that an atom can have the highest limit of 7 bohr energy shells or principle energy shells ; K,L,M,N,O,P,Q respectively. My question is, can there exist any element that consists more than 7 energy shells having more electrons? Or , can there exist any element that has an 8th energy shell of R shell? &lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2018-03-19T05:52:25.693" Title="Can there exist any element which has an 8th principle shell?" Tags="&lt;electrons&gt;&lt;orbitals&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4709" PostTypeId="1" AcceptedAnswerId="4710" CreationDate="2013-04-17T09:41:04.390" Score="1" ViewCount="1255" Body="&lt;p&gt;I came across the following part in my text book. Bohr's atomic theory explained the stability of an atom by the idea of energy shells. He said that while orbiting the nucleus through a particular orbit or energy shell the energy of electrons neither increases nor decreases. The electron does not carry energies while orbiting. The shells are charged and the energy of electrons orbiting through the shell equals to the energy of the shell and as the shells are stable therefore the energy of electrons are also stable . Now, the energy increases from shell to shell like(K,L,M,N,O,P,Q)[ascending order]. So,the energy of an electron in K shell should be less than that of an electron in Q shell. But we know that the energy of an electron is constant and is equal to (4.8*10^-10) esu. Then how can this is possible that an electron with a certain energy does not carry any energy while orbiting and it gains the same energy of its orbit while orbiting? &#xA;Note that there may be some mistakes in my informations as there may be many false informations printed in my book but still I am curious about the question. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1509" LastActivityDate="2015-07-09T17:08:45.783" Title="Confusion about the charge of electrons while orbiting through bohr's energy shells" Tags="&lt;electrons&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4710" PostTypeId="2" ParentId="4709" CreationDate="2013-04-17T11:41:47.170" Score="3" Body="&lt;p&gt;The main confusion I can see in your question is charge vs. energy. &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy&quot; rel=&quot;nofollow&quot;&gt;Energy&lt;/a&gt; is the capacity to do &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;work&lt;/a&gt;. Energy comes in many forms. &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;nofollow&quot;&gt;Kinetic energy&lt;/a&gt; is the energy of motion. &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;nofollow&quot;&gt;Potential energy&lt;/a&gt; is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has &lt;strong&gt;unchanging&lt;/strong&gt; energy. The only way for the electron to change its energy is to &lt;em&gt;move to a different&lt;/em&gt; shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model_of_the_atom#Electron_energy_levels&quot; rel=&quot;nofollow&quot;&gt;total energy&lt;/a&gt; of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model_of_the_atom#Rydberg_formula&quot; rel=&quot;nofollow&quot;&gt;Rydberg constant&lt;/a&gt; by $R_H=R_E/hc$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_n=-\frac{Z^2 R_E}{n^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants  $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant&quot; rel=&quot;nofollow&quot;&gt;fine-structure&lt;/a&gt; constant. Eventually, $R_E$ simplifies down to a relativistic energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_charge&quot; rel=&quot;nofollow&quot;&gt;Electric charge&lt;/a&gt; is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Coloumb_law&quot; rel=&quot;nofollow&quot;&gt;Coulomb's Law&lt;/a&gt;. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, charge$\ \neq\ $energy.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-17T11:41:47.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4711" PostTypeId="2" ParentId="785" CreationDate="2013-04-17T11:47:37.640" Score="1" Body="&lt;p&gt;Vapor pressure is property of liquid and partial pressure is property of a gas in a mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal boiling point for the pure component in liquid phase is the temperature at which vapor pressure exerted by the liquid becomes equal to 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a gas mixture, as long as partial pressure of a component remains less than its liquid phase vapor pressure at the temperature of the mixture, the component will not condense. Condensation starts only from the point when $PP=VP$ and starting from the case where $PP&amp;gt;VP$, condensation continues only till $PP=VP$. When $PP&amp;lt;VP$, condensation stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air at 1 bar total pressure will have 0.21 bar partial pressure of $\ce{O2}$ and say it is at -183C (the normal boiling point of $\ce{O2}$). Vapor pressure of liquid $\ce{O2}$ at this temperature is 1.01 bar, but partial pressure is 0.21 bar  which is less than 1.01 bar so $\ce{O2}$ should not condense. But if Air is at 5 bar total pressure, this will have $0.21\times 5=1.05$ bar partial pressure of $\ce{O2}$ and this air is at -183C then $\ce{O2}$ will condense till its partial pressure has gone below 1.01 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vapor pressure is independent of composition in liquid phase but dependent on temperature. As temperature increases, the vapor pressure of a liquid component increases. Vapor pressure does not depend on gas phase mixture properties. &#xA;At a given temperature, the component having the higher vapour pressure in liquid phase is the more volatile components and will have lower boiling point.&lt;/p&gt;&#xA;" OwnerUserId="1510" LastEditorUserId="22" LastEditDate="2013-04-18T06:33:29.420" LastActivityDate="2013-04-18T06:33:29.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4712" PostTypeId="2" ParentId="4705" CreationDate="2013-04-17T11:55:37.870" Score="2" Body="&lt;p&gt;For N:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total N = N from NO2 + N from NO3 + Total Kjeldahl N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods from &lt;a href=&quot;http://www.standardmethods.org/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid Chromatography: NO2, NO3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific electrodes: NO3, Kjeldahl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectrophotometry: NO2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This for water including wastewater. I don't know the aplicability for other samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also ICP-MS is worth &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductively_coupled_plasma_mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;researching&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="1300" LastEditDate="2013-04-17T12:06:00.400" LastActivityDate="2013-04-17T12:06:00.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4713" PostTypeId="2" ParentId="2590" CreationDate="2013-04-17T11:58:52.920" Score="4" Body="&lt;p&gt;Cobalt poisoning is apparently very rare. The only situation in which cobalt poisoning is a serious problem is when it is exposed directly to your bloodstream (Recently it has been an issue with people who have undergone certain hip implants -- see &lt;a href=&quot;http://www.nlm.nih.gov/medlineplus/ency/article/002495.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; page).&lt;/p&gt;&#xA;&lt;p&gt;The amount of Cobalt chloride in silica gel balls would be absolutely harmless. According to most sources, you have to be exposed to significant amounts of cobalt (at least more than 20g for an average adult) for a long period of time for symptoms to appear.&lt;/p&gt;&#xA;&lt;p&gt;Note: Cobalt is actually necessary for our body (in small amounts) in the form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vitamin_B12&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vitamin B12&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-04-17T11:58:52.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4714" PostTypeId="2" ParentId="4708" CreationDate="2013-04-17T12:10:51.100" Score="8" Body="&lt;p&gt;That is certainly possible, and in fact very likely. Nuclear physics predicts an island of stable elements around an atomic number of 120 &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;$^{[1]}$&lt;/a&gt;, with the two heavier of the three elements predicted to be most stable having an $\ce{8s^2}$&lt;a href=&quot;http://en.wikipedia.org/wiki/Unbinilium&quot; rel=&quot;nofollow noreferrer&quot;&gt;$^{[2]}$&lt;/a&gt; and an $\ce{8s^2 8p^1}$&lt;a href=&quot;http://en.wikipedia.org/wiki/Unbihexium&quot; rel=&quot;nofollow noreferrer&quot;&gt;$^{[3]}$&lt;/a&gt; configuration, respectively. &lt;/p&gt;&#xA;" OwnerUserId="605" LastEditorUserId="5026" LastEditDate="2018-03-19T05:52:25.693" LastActivityDate="2018-03-19T05:52:25.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4715" PostTypeId="2" ParentId="4706" CreationDate="2013-04-17T13:01:47.343" Score="3" Body="&lt;p&gt;Quantitative results are obtained by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Measuring the emited light of known samples (standards) and making a curve of emission for elements of interest.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Comparing the emission of the unknown samples and calculating the concentration of analytes with the curve.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using samples fortified with standards and duplicates. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using statistics to measure accuracy, precision, etc. of the results, and in general evaluate the quality of the results. Most of this (if not all) is done by the software that controls the equipment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-04-17T13:01:47.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4716" PostTypeId="1" CreationDate="2013-04-17T15:10:59.743" Score="2" ViewCount="736" Body="&lt;p&gt;Can someone please explain the bonding in a bridged metal carbonyl (like diiron nonacarbonyl)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot understand what kind of bonds (sigma/pi) exist between the bridging carbonyls and the metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also why aren't the metals charged in such a case?&#xA;What is the bond order in such a case?&lt;/p&gt;&#xA;" OwnerUserId="1511" LastEditorUserId="22" LastEditDate="2013-04-17T16:31:13.863" LastActivityDate="2013-06-23T12:44:44.973" Title="Bonding in a bridged metal carbonyl" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4717" PostTypeId="1" CreationDate="2013-04-17T15:31:00.193" Score="9" ViewCount="514" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Mendeleev%27s_predicted_elements#Later_predictions&quot;&gt;this Wikipedia page&lt;/a&gt;, Mendeleev originally believed that the inert gases belonged in Group 0 (to the left of the alkali metals). Thus, helium would be placed in the second period to the left of lithium and argon would be placed in the third period to the left of sodium. Apparently this made sense to him at the time, since atomic theory was in its infancy (and Mendeleev didn't like it) and quantum mechanics was only a twinkle in the eyes of Planck, Bohr, and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this arrangement, there is an open spot above helium and to the left of hydrogen for an ultralight inert gas he called &quot;proto-helium&quot;. We now know that hydrogen must be the lightest known element, meaning Mendeleev was wrong about the placement of the inert gases. However, host of the other holes in Mendeleev's table were filled with real elements having properties nearly identical to what Mendeleev predicted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mendeleev predicted that proto-helium would have a miniscule mass (ca. $ 5.3\times10^{-11}$ amu) and high (near relativistic) velocity. Mendeleev believed that this near massless particle would be able to permeate all matter, rarely interacting chemically. He thought that they might be responsible for radioactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino&quot;&gt;neutrino&lt;/a&gt; is massless (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutrino#Mass&quot;&gt;nearly so&lt;/a&gt;), and travels at the speed of light (if massless) or just under (if it has mass). Otherwise, it ignores the electromagnetic force, and so has no chemical interaction. Neutrinos are often produced during radioactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant references point back to Mendeleev's original writings (in Russian), which I cannot access, so I cannot verify if he made a prediction about &quot;proto-helium&quot;. Did he predict this particle? Has anyone previously made the connection between Mendeleev's proto-helium and neutrinos?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-18T01:24:41.133" Title="Did Mendeleev predict the existence of neutrinos?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4718" PostTypeId="2" ParentId="4705" CreationDate="2013-04-17T15:40:20.110" Score="1" Body="&lt;p&gt;The analytical method preferred for CHNO (and also S, X, etc) determination is &lt;a href=&quot;http://en.wikipedia.org/wiki/Elemental_analysis&quot; rel=&quot;nofollow&quot;&gt;elemental analysis&lt;/a&gt;, usually by combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if a molecule has the generic formula $\ce{C_wH_xN_yO_z}$, combustion would produce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_wH_xN_yO_z + (w +\frac{x}{4}+\frac{y}{4}+\frac{z}{2})O2 -&amp;gt; wCO2 + \frac{x}{2}H2O +\frac{y}{2}N2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining the masses of $\ce{CO2}$, $\ce{H2O}$, and $\ce{N2}$ given off by the combustion is the first step in determining how much carbon, hydrogen, and nitrogen was in the original sample. The remaining mass is oxygen. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-17T15:40:20.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4720" PostTypeId="1" CreationDate="2013-04-17T20:50:31.560" Score="7" ViewCount="7555" Body="&lt;p&gt;In para-aminobenzoic acid, how can I know whether the amino group loses a proton or not when $\mathrm{pH}$ exceeds $8.5$? I have only found two values for $\mathrm{p}K_\mathrm{a}$ but I never found info about the pKa (in this case it would be the third $\mathrm{p}K_\mathrm{a}$ at $\mathrm{pH}&amp;gt;8.5$) of the amino group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I guess at basic $\mathrm{pH}$, negative charge over $\ce{NH2}$ would be stabilized by aromatic ring and $\ce{COO-}$.&lt;/p&gt;&#xA;" OwnerUserId="256" LastEditorUserId="7475" LastEditDate="2015-11-13T19:03:57.923" LastActivityDate="2020-05-22T10:09:45.883" Title="In aminobenzoic acid, do we have three different pKa's?" Tags="&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4722" PostTypeId="1" AcceptedAnswerId="4729" CreationDate="2013-04-17T21:12:41.880" Score="2" ViewCount="13860" Body="&lt;p&gt;When is it okay to use $Q=mc\Delta T$, is the equation only for calorimetry questions? When a question talks about heat how do I know if i'm measuring $\Delta H$ or Q?&lt;/p&gt;&#xA;" OwnerUserId="1512" LastEditorUserId="305" LastEditDate="2013-04-18T04:30:36.820" LastActivityDate="2013-04-18T14:06:46.003" Title="When is it okay to use $Q=mc\Delta T$, is this equation only for calorimetry questions?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4723" PostTypeId="2" ParentId="4720" CreationDate="2013-04-17T23:49:30.217" Score="2" Body="&lt;p&gt;The third &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; would refer to the deprotonation of the amino group of the &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2-Ph-COO-}$&lt;/span&gt; (singly deprotonated) species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be very difficult to deprotonate since the molecule is already negatively charged.  Deprotonation would make a doubly charged species which is normally unfavorable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to your assertion, the &lt;span class=&quot;math-container&quot;&gt;$\ce{COO-}$&lt;/span&gt; provides no stabilization for the negative charge on amine, and is likely to modestly inhibit delocalization into the benzene ring as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very strong bases &lt;span class=&quot;math-container&quot;&gt;$(\ce{BuLi}$&lt;/span&gt; or similar) &lt;em&gt;might&lt;/em&gt; be able to deprotonate this compound.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="41328" LastEditDate="2020-05-22T10:09:45.883" LastActivityDate="2020-05-22T10:09:45.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4724" PostTypeId="2" ParentId="4717" CreationDate="2013-04-17T23:53:04.983" Score="0" Body="&lt;p&gt;Neutrinos are chargeless, which is a concept that does not fit nicely in Mendeleev's periodic table; it seems improbable he would predict this particle.  The closest bet would be positronium.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-04-17T23:53:04.983" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4725" PostTypeId="2" ParentId="4679" CreationDate="2013-04-17T23:55:24.133" Score="0" Body="&lt;p&gt;Coupling BuLi with an alkylbromide to extend an alkyl chain is an extremely minor produt.  Transmetalation or elimination products will dominate.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2013-04-17T23:55:24.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4726" PostTypeId="2" ParentId="4720" CreationDate="2013-04-18T00:21:26.113" Score="7" Body="&lt;p&gt;The two acidity constants in 4-aminobenzoic acid are due to the loss of &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt; from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt; turning it into the 4-aminobenzoate anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tSnv.png&quot; alt=&quot;Like so.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two ionizations have a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of around 5, while the next ionization quite likely has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; well below ~25 (&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/pkatable/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;compare with other substituted anilines&lt;/a&gt;), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="41328" LastEditDate="2020-05-22T10:08:35.367" LastActivityDate="2020-05-22T10:08:35.367" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="4727" PostTypeId="2" ParentId="4717" CreationDate="2013-04-18T01:15:22.657" Score="10" Body="&lt;p&gt;While it's an interesting thought, I don't think it's very useful to suggest Mendeleev predicted neutrinos. It's far more sensible for it to be a curious coincidence. His suggestions of ghost-like properties similar to neutrinos (very hard to detect, tiny mass, little interaction with matter, etc) are more easily explained as an attempt to justify (properly) why it was so hard to detect: if it had a much larger mass, someone should've weighed it by then; if it interacted more, someone should've managed to capture it physically or chemically, and so on. It's not uncommon for theoretical predictions to temporarily &quot;take refuge&quot; just beyond the limit of observability (for example, proton decay), with varying degrees of justification. In this case, Mendeleev's prediction was wrong in its original terms, and happened to almost match something else entirely because neutrinos really &lt;em&gt;are&lt;/em&gt; hard to detect! Also, the connection with radioactivity could very likely have been made merely due to good timing with its discovery. Since no one had a good explanation, why not forward your own!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as a sidenote, neutrinos definitely have mass, otherwise they could not suffer the observed flavour oscillation.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-04-18T01:15:22.657" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4728" PostTypeId="2" ParentId="4722" CreationDate="2013-04-18T01:16:33.297" Score="3" Body="&lt;p&gt;You can use $q=mc\Delta T$ for any process in which heat is transferred, not just calorimetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, when a question talks about heat transfer, that means $q$. If the question meant enthalpy change, the question would have used the word &quot;enthalpy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under constant pressure conditions, the enthalpy change is equal to the heat transfer. Enthalpy equals internal energy plus $PV$. Thus, change in enthalpy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H=U+PV$$&#xA;$$\Delta H = \Delta U + P\Delta V + V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the change in internal energy is equal to the sum of heat transfer and work (and work is $w=-P\Delta V$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = q -P\Delta V + P\Delta V + V\Delta P = q+ V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant pressure, $q=q_p$ and $\Delta P = 0$, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = q_p = mc_p \Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constant pressure is common in the lab when the reaction vessel is open to the air. At constant pressure, if you are measuring $q$, you are also measuring $\Delta H$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-18T01:16:33.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4729" PostTypeId="2" ParentId="4722" CreationDate="2013-04-18T01:24:21.477" Score="4" Body="&lt;p&gt;The equation $Q = mc \Delta T$ is appropriate for calorimetry at a constant volume. $Q$ here refers to the net heat transfer of the isolated system. Change in enthalpy, $\Delta H$, on the other hand, is really the total change in energy of the system, that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \Delta U + P\Delta V + V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is pressure, $V$ is volume, and $U$ is the internal energy of the system, which is the sum of the system's net heat transfer ($Q$) and the work done on the system ($W$): $\Delta U = Q + W$. If we substitute the definition of $\Delta U$ into our equation for enthalpy, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = Q + W + P\Delta V + V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because any mechanical pressure-volume work performed on a closed system is equal to $-P\Delta V$, the equation further reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = Q + V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it becomes apparent that $Q$ and $\Delta H$ are only equal if pressure is constant (i.e., $\Delta P = 0$). The upshot of all this is that you should not assume $\Delta H$ and $Q$ are equal unless you're explicitly instructed to assume constant volume and pressure. When a problem refers to heat, it's almost certainly referring to heat transfer, $Q$, and not enthalpy change ($\Delta H$).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-04-18T14:06:46.003" LastActivityDate="2013-04-18T14:06:46.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4730" PostTypeId="2" ParentId="4685" CreationDate="2013-04-18T05:31:59.723" Score="4" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Creating a Galvanic Cell&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify, let's forget galvanic cells for a moment. Instead, just picture an oxidation/reduction reaction in aqueous solution.  If we had a solution of $\ce{Cu^+}$ and $\ce{Co^{3+}}$, the metal ions would exchange an electron, and we would end up with $\ce{Cu^{2+}}$ and $\ce{Co^{2+}}$. This happens because of the relative stability of the two ions; the electron is at a lower energy with the cobalt than with the copper (at these oxidation states). Now, if we separate the two metals into two different solutions, insert an electrode into each, and connect the two with a conductive wire, we can still transfer an electron. The potential between the electrodes comes from the same source as the mixture: the chemical potential between the two ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A salt bridge is also necessary. The salt bridge balances charge in the two solution. Otherwise, we would have a net positive charge in the copper solution and a net negative charge in the cobalt solution. This would create a potential difference and the reaction would eventually and stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Galvanic Cell using Non-ionic Species&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galvanic cells do not have to use species that are ionic in the oxidized and reduced states (or either, for that matter). Consider the cell pictured below. In this example, zinc metal is being oxidized to $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$ is being reduced to copper metal. In other words, the zinc electrode is getting dissolved, and the copper electrode is growing. The anodic salt solution is used to create what is called a poised electrode. If a zinc electrode were inserted into pure water, according to the Nernst equation, the system would have an infinite potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal E = \mathcal E^0 - \frac {RT} {nF} \ln 0 \rightarrow -\infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as even a few molecules per liter are in solution (which would happen very quickly) the potential drops to around 50 V, and would continue to drop as more zinc was oxidized (although there are certainly deviations from the Nernst equation in an observed system). The salt solution puts the system into a steady state. In terms of the cathode, the metal does not matter. Anelectrode made from any conductive material would behave the same as a copper electrode in a solution of $\ce{Cu^{2+}}$; copper would plate onto the electrode as it is reduced. The system potential is not altered because the activity of a pure solid is unity. It is normally just convention (and for learning purposes) that the electrode material is the same as the oxidant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xgOuh.png&quot; alt=&quot;Zn|Zn2+||Cu2+|Cu&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-18T05:31:59.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4731" PostTypeId="1" CreationDate="2013-04-18T06:37:50.250" Score="2" ViewCount="350" Body="&lt;p&gt;I am designing a fairly simple distillation for dichloromethane.&#xA;Being an engineer I feel more comfortable working with precise dimensions of both the separate glass vessels (condensers, evaporators, flask, etc) and of the complete apparatus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking to find a CAD library that has various glassware blocks. &#xA;This way I can visually assemble my set and be able to show a drawing to my supervisor. &lt;/p&gt;&#xA;" OwnerUserId="1513" LastActivityDate="2013-04-18T06:37:50.250" Title="Glassware Test Instruments CAD blocks library" Tags="&lt;analytical-chemistry&gt;&lt;equipment&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4732" PostTypeId="1" CreationDate="2013-04-18T14:12:47.657" Score="8" ViewCount="11344" Body="&lt;p&gt;I have this iron pan that got rusty from not being properly dried. Scrubbing it I was able to get rid of most of the rust, but there's still some I just can't remove. I thought I could chemically remove it, but I myself am a bit rusty on the inorganic chemistry subject, so I couldn't think of any way of doing it without leaving any unwanted residue on the pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rust is the brownish-red type, which according to Wikipedia consists of $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Most rust-removal &quot;recipes&quot; I found involve all sorts of acids, but I can't tell what the reaction(s) would be with $\ce {Fe2O3·nH2O}$ so I can't tell if there would be any residues left behind that I could not remove. Anyone with any experience on the matter?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I contacted the manufacturer and the pan is hand-made, just Iron and natural coating (fat).&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="7475" LastEditDate="2015-10-30T16:43:31.747" LastActivityDate="2015-10-30T16:43:31.747" Title="Chemically removing rust without leaving any unwanted residues" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4733" PostTypeId="2" ParentId="4706" CreationDate="2013-04-18T16:24:55.370" Score="6" Body="&lt;p&gt;Although &lt;a href=&quot;https://chemistry.stackexchange.com/users/1300/f-p&quot;&gt;fp&lt;/a&gt;'s answer is absolutely correct and I see that you have something more than high school chemistry, I'd like to give a more elaborated, yet simplistic answer, if you or anyone else reading this has any interest on the basic principles of an ICP-OES - you only need high school chemistry to understand it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;I'm not certain if it really does give quantitative information&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It does, and very &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accuracy_and_precision&quot; rel=&quot;nofollow noreferrer&quot;&gt;accurately&lt;/a&gt;&quot;. Detection limits &lt;a href=&quot;http://perkinelmer.com/PDFs/Downloads/BRO_WorldLeaderAAICPMSICPMS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be in ppb&lt;/a&gt;, that is, it can detect 0.0000001g (0.1 µg) of something in a 1L sample. That's really, really impressive if you stop to think about it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;how it does so&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Wizardry! Well, almost. Read on and you'll see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all the sample must be prepared, you can't simply toss something in it. Depending on what your sample is, it must be mixed with a known amount of this or that chemical (you'll see later why). This answers your next question: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;and whether it can do so on its own&lt;/code&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although the actual analysis and preparation can indeed be done purely by the machine in some cases (&lt;a href=&quot;http://www.youtube.com/watch?v=fB3FL7p795A&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;'s a video), and there are methods such as &lt;em&gt;laser ablation&lt;/em&gt; which make it possible to almost just toss something in the ICP, the know-how of a trained analyst is still require to tell the machine what to do, to interpret the results it outputs, and to tell what chemicals to mix in the preparation process, if there is one at all. Equipments such as gas chromatographs can (and might need to) be coupled to assist in the sample preparation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to &lt;code&gt;how it does so&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you introduce the prepared sample into the machine (it basically sucks it) some processing is done to it, and at some point it reaches the atomizer - the plasma torch. The atomizer does what it's name suggests: It turns stuff into separate atoms (by heating them to &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductively_coupled_plasma#Applications&quot; rel=&quot;nofollow noreferrer&quot;&gt;absurd temperatures&lt;/a&gt;). Here's a picture I took from an ICP atomizer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJyO3.jpg&quot; alt=&quot;ICP atomizer&quot;&gt;&#xA;&lt;em&gt;Note: Not 100% of the sample is atomized. In fact, only a small percentage is.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it's turned on it looks like the picture you saw at Wikipedia. You can see by the size of the parts that only a few milliliters of sample - if not less - is atomized at a time. Now that it got our sample atomized, it heats the sample up just a bit more, so the electrons get exited and go from their current level of energy to a higher one. While the sample passes through the equipment it gets cooler, so the electrons get back to the lower energy level, and by doing so, the atom emits light, hence the name optical EMISSION spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AHVxD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AHVxD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/55/Bohr-atom-PAR.svg/310px-Bohr-atom-PAR.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikimedia.org&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what elements are emitting light, and at what wavelength? The principle of every spectroscopy analysis is that each element emits and absorbs specific, known, predictable wavelengths ranges, and only when heated to specific temperatures (in the case of emission). Because the temperature was set to a specific one for the analyte and a specific set of chemicals were mixed with the sample, only the atoms of the element we want to quantify will emit light at a certain element-specific wavelength range. We set the equipment to detect only that wavelength range and ignore others and that's how we know how much of that element there is in the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you kept track of the proportions you might be thinking: &quot;if it detects ppb amounts in only a few ml of sample where only a few % of it is atomized and only certain elements will emit only a certain wavelength, how the hell can it detect that little light?&quot;. That's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Photomultiplier&quot; rel=&quot;nofollow noreferrer&quot;&gt;photomultiplier&lt;/a&gt;'s job. It's a fantastic piece of equipment and you should definitively read about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After it detects the light emitted by the analyte in the sample, it compares it to a &lt;a href=&quot;http://wiki.answers.com/Q/In_analytical_chemistry_what_is_mean_by_blank&quot; rel=&quot;nofollow noreferrer&quot;&gt;blank&lt;/a&gt;, does some calculations and displays the results in the form of a graph. Here's a screenshot of an ICP-OES software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eUt7E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, this is a simplistic explanation, most of the process isn't that simple in reality. So as you can see, it's basically witchcraft.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="15751" LastEditDate="2019-02-18T13:14:09.007" LastActivityDate="2019-02-18T13:14:09.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4735" PostTypeId="2" ParentId="4732" CreationDate="2013-04-18T19:33:37.760" Score="1" Body="&lt;p&gt;If you're worried about residues, just use a common kitchen item that is acidic. Vinegar is just acetic acid. That should do the trick.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-18T19:33:37.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4736" PostTypeId="2" ParentId="2870" CreationDate="2013-04-18T19:36:57.880" Score="4" Body="&lt;p&gt;It is quite possible for a material consisting entirely of one element to be &quot;impure&quot; - for example, a mixture of carbon nanotubes, graphite and buckyballs could hardly be called &quot;pure&quot; despite being made of carbon only. For some applications, even a mixture of different types or lengths of carbon nanotube isn't &quot;pure&quot; enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you can buy &quot;pure&quot; orange juice in the shops that is neither entirely one element nor entirely one molecule. In fact, &quot;pure (from concentrate)&quot; orange juice isn't even entirely made up of stuff that was squeezed out of an orange. So &quot;purity&quot; is in effect a matter of how you define it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For thermodynamics purposes, &quot;purity&quot; is a kind of order - if this order is lost by mixing two pure substances together, it takes work to un-mix them. Conversely, with a carefully designed machine it is possible to harness the entropic &quot;force&quot; in favour of mixing to generate useful work.&#xA;ATPase, the enzyme which performs the final stage of converting food energy into ATP, the internal energy &quot;currency&quot; used by chemical processes in cells, is the most widespread example (although in this case it was evolved rather than designed). As a crude summary, it uses the energy released by mixing acid and water to drive a rotor, which as it turns forces ADP and phosphate together making ATP. (The acid is generated in earlier stages of the process.)&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2013-04-18T19:36:57.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4737" PostTypeId="1" AcceptedAnswerId="4739" CreationDate="2013-04-18T21:08:21.220" Score="5" ViewCount="2516" Body="&lt;p&gt;I have a decanter which I use fairly often to pour out wines. I typically rinse it with hot water after the last wine has come out, but I cannot prevent it from forming 'residual' wine stains over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time I wanted to clean it, I looked on the internet for advice on cleaning. There are tons of it (&lt;a href=&quot;https://www.google.nl/search?um=1&amp;amp;hl=nl&amp;amp;q=Stained%20Wine%20Decanter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stained wine decanter&lt;/a&gt;), but I noticed that there are quite a lot of things around that don't work (or don't work well) and I figured that the question is basically plain chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my question is, can somebody explain how I can best clean the wine stains from my decanter? Please, also explain the chemistry behind the method you propose so I know why it should work.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="402" LastEditDate="2017-04-10T02:23:20.603" LastActivityDate="2017-04-10T02:23:20.603" Title="Removing wine stains from a decanter" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4738" PostTypeId="2" ParentId="4732" CreationDate="2013-04-18T23:33:03.787" Score="8" Body="&lt;p&gt;The common DIY rust remover is cola of some sort -- most colas contain a significant quantity of phosphoric acid, which reacts with the iron oxide to form iron (III) phosphate which you can then scrub off. Any remaining iron phosphate should come off with time and/or more water, and is harmless in domestic quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphoric_acid#Rust_removal&quot;&gt;Wikipedia:Phosphoric acid - Rust Removal&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-04-18T23:33:03.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4739" PostTypeId="2" ParentId="4737" CreationDate="2013-04-18T23:44:48.720" Score="8" Body="&lt;p&gt;It's a bit of a blunt instrument, but I'd tend to just go straight for hydrogen peroxide if it's available where you are. It breaks up all sorts of organic dyes via oxidation, and it slowly decomposes by itself, especially in sunlight, so you can just rinse (&lt;em&gt;thoroughly&lt;/em&gt;) and leave it to dry in the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;NB&lt;/strong&gt;: I have never owned a wine decanter, and take no responsibility if your decanter dissolves, explodes, or spontaneously gains the ability to sing operatically.)&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-04-18T23:44:48.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4741" PostTypeId="1" AcceptedAnswerId="4742" CreationDate="2013-04-19T08:18:38.610" Score="0" ViewCount="5987" Body="&lt;p&gt;The dissociation of formic acid ($\ce{HCOOH}$) is:&#xA;$$&#xA;\ce{HCOOH -&amp;gt; H+ + COOH-}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a charge on $\ce{COOH^{-1}}$&#xA; though?  Is it achieved through:&#xA;$$1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does carbon exist as a 4+ ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a somewhat unrelated question, can an acid donate multiple $\ce{H+}$ ions (or can a base receive multiple $\ce{H+}$ ions)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="1520" LastEditorUserId="4231" LastEditDate="2015-03-01T14:31:47.573" LastActivityDate="2015-03-01T14:31:47.573" Title="Why is there a charge on COOH⁻¹?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;ions&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4742" PostTypeId="2" ParentId="4741" CreationDate="2013-04-19T10:03:36.200" Score="7" Body="&lt;p&gt;Not all bonds are ionic bonds. While saying that &quot;Carbon is 4+ valency and is a $\ce{C^{4+}}$ ion&quot; works for carbides like $\ce{Be2C}$, it will not work for a covalent compound like $\ce{HCOOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of ionic bonds, here we have &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;covalent bonds&lt;/a&gt;. Whereas in an ionic bond one atom donates an electron to the other, in a covalent bond, two atoms each contribute some electrons and share them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with $\ce{O2}$, each atom shares two electrons with the other. Neither &quot;give up&quot; any electrons, they are just shared and contribute to the octets of both atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lqHQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formate ($\ce{HCOO-}$) ion is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central carbon shares one electron with a hydrogen, and two electrons with an oxygen. It also shares one electron with a negatively charged oxygen. This completes the octets (or doublets in the case of hydrogen) of all four atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ZNJJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With electrons drawn on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G246z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that while the &lt;em&gt;charges&lt;/em&gt; are not of the $1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})$ form, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxidation numbers&lt;/a&gt; are.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="275" LastEditDate="2013-04-19T11:40:53.140" LastActivityDate="2013-04-19T11:40:53.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4743" PostTypeId="1" CreationDate="2013-04-19T11:49:36.860" Score="1" ViewCount="572" Body="&lt;p&gt;I am conducting a &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/nl2045414&quot; rel=&quot;nofollow&quot;&gt;bio-chemistry related experiment&lt;/a&gt; and I have been unable to understand a step which is commonly performed. The procedure that is confusing me is outlined in the &lt;a href=&quot;http://pubs.acs.org/doi/suppl/10.1021/nl2045414&quot; rel=&quot;nofollow&quot;&gt;supporting information&lt;/a&gt;, beginning on page S2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aim in this step is to apply a PEG (Polyethylene glycol) silane layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After immersing ITO (Indium Tin Oxide) slides in a PEG-concentration, the slides are incubated for 18 hours at a temperature of 60 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me why the incubation is performed? And why is it for 18 hours at 60 degrees Celsius?&lt;/p&gt;&#xA;" OwnerUserId="1521" LastEditorUserId="323" LastEditDate="2013-04-19T18:06:31.530" LastActivityDate="2013-04-19T18:06:31.530" Title="Why do we need to incubate this system for 18 hours at 60 degrees Celsius?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4744" PostTypeId="2" ParentId="4689" CreationDate="2013-04-19T11:52:47.017" Score="2" Body="&lt;p&gt;Describe what you did. The key is make sure your communication allows your experiment to be reproduced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... R&lt;sub&gt;f&lt;/sub&gt; 0.34 (10 mL of 9:1 DCM/MeOH + 3 drops of 1.0 M Et&lt;sub&gt;3&lt;/sub&gt;N in MeOH).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you do all of your TLC with the additional of triethylamine, then you can include a sentence in the general methods about this procedure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All TLC plates were developed in a 9:1 mixture of DCM and MeOH that was made alkaline by the addition of three drops of 1.0 M Et&lt;sub&gt;3&lt;/sub&gt;N in MeOH for every 10 mL original solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The procedure you describe is common, especially in the elution of basic molecules (since silica is slightly acidic). You might find examples of how to state this sort of procedure in the synthesis of amines and alkaloid natural products. Ultimately, in a publication, you should err on the side of thoroughness for your experimental methods. If it's too long, a reviewer of the editor will tell you.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-04-20T13:14:51.303" LastActivityDate="2013-04-20T13:14:51.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4745" PostTypeId="2" ParentId="4743" CreationDate="2013-04-19T12:12:01.070" Score="2" Body="&lt;p&gt;Until you post the original source procedure, my answer is going to be generic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When I am repeating an experiment I found in the literature, why do I need to do [insert step that seems unnecessary, difficult, or just plain long]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer is, &lt;strong&gt;because it works.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to incubate the slides at 60 &lt;sup&gt;o&lt;/sup&gt;C for 18 hours because that's what the author of the paper found to be necessary to guarantee that the slides were properly PEGylated. It's possible that the author of your procedure is also using a procedure that he or she found in the literature, and the the rationale for the specific incubation procedure will require some digging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longer answer is, &lt;strong&gt;the reaction probably needs some incubation to run to completion.&lt;/strong&gt; The reaction between the PEG and the ITO surface is not instantaneous. If you want the coating to be good, you need to give it time to cure, and mild heating can accelerate the process. While 18 hours seems like a long time, it would likely take interminably longer at room temperature. Could you heat at a higher temperature for less time? Probably, but I would wait until you are certain you can make the existing procedure work before messing with it. That way you have some successfully coated slides to compare to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 18 hour period has one of two origins. The second is more likely.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The developer of the procedure maximized the efficiency of the process: He/she carefully experimented with the conditions until he/she discovered the exact time/temperature combination that maximized the efficiency of the reaction - getting the best coating in the shortest time.&lt;/li&gt;&#xA;&lt;li&gt;The developer of the procedure maximized the convenience of the process: He/she determined 1) that the coated slide needed to cure for several hours (3 or 4) at 60 &lt;sup&gt;o&lt;/sup&gt;C,  2) that higher temperatures were detrimental to the quality and integrity of the coating, and 3) prolonged heating at 60 &lt;sup&gt;o&lt;/sup&gt;C is not detrimental. In this case, the researcher puts the slides into the oven in the afternoon to cure overnight and picks up the cured slides the next morning. 4 PM on Monday afternoon to 10 AM on Tuesday morning is 18 hours. As a synthetic chemist, I often wrote down '18 hours', when I really meant 'overnight'. The reaction may have been done at 6 hours or 8 hours, but 18 hours is more convenient. I set up the procedure one afternoon a few hours before I left for the day so that I could make sure it was stable. I worked it up the next morning. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-19T12:12:01.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4747" PostTypeId="1" CreationDate="2013-04-19T16:10:09.907" Score="12" ViewCount="396" Body="&lt;p&gt;My daughter is a 3rd grader, but she reads at 8th grade level, and she is 2 grade levels ahead in math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She reads everything she gets her hands on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She took some chemistry activities, and she loved them. (&quot;Mad Science&quot;, &quot;Ooo, goo and stinky too&quot;,...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She knows that matter is made of atoms, and that atoms are made of protons and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to find some chemistry books that would be appropriate for her. A good book would be one that starts from beginning, help her connect together the stuff she already knows, and then take her further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reason is that I don't want school to be too easy for her. I want to nurture her curiosity, love of challenge, and motivation to study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I don't want to push her even further ahead in the subjects that she has to take in emementary or middle school anyway -- because that would make our dealing with the school system even more complicated... So, I think Chemistry and Physics are good subjects.&lt;/p&gt;&#xA;" OwnerUserId="1523" LastEditorUserId="1523" LastEditDate="2013-04-23T15:30:14.077" LastActivityDate="2013-04-23T15:30:14.077" Title="Books for a very advanced 3rd grader (elementary school)" Tags="&lt;books&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2016-11-02T17:43:20.233" ContentLicense="CC BY-SA 3.0" />
  <row Id="4748" PostTypeId="2" ParentId="4747" CreationDate="2013-04-19T16:21:57.573" Score="8" Body="&lt;p&gt;One of the best is &lt;a href=&quot;http://chemistry.about.com/b/2012/10/01/download-the-golden-book-of-chemistry-experiments.htm&quot;&gt;The Golden Book of Chemistry Experiments&lt;/a&gt;.  By today's standards, this book is considered &quot;dangerous&quot;, but keep in mind that it was written for teenagers in the 1960s.  In any case, parental supervision should always be required in chemistry, and if you carefully research the experiments before doing them you should be fine.  It takes you all the way through technique, building apparatus, and safety.  The only problem is obtaining chemicals: the sources cited therein are hopelessly outdated (i.e. your will NOT find most of this stuff in drugstores anymore), and you will need to use educational supply companies to obtain small quantities of chemicals.  Have fun!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-04-19T16:21:57.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4750" PostTypeId="1" AcceptedAnswerId="4753" CreationDate="2013-04-19T19:49:22.020" Score="5" ViewCount="5588" Body="&lt;p&gt;Why do sodium halides react so differently with sulfuric acid? &#xA;\begin{align}&#xA;  \ce{NaF + H2SO4 &amp;amp;-&amp;gt; NaHSO4 + HF} &#xA;    \tag{1a}\label{NaF}\\&#xA;  \ce{NaCl + H2SO4 &amp;amp;-&amp;gt; NaHSO4 + HCl} &#xA;    \tag{1b}\label{NaCl}\\&#xA;  \ce{2 NaBr + H2SO4 + 2H+ &amp;amp;-&amp;gt; Br2 + SO2 + 2H2O + 2Na+} &#xA;    \tag2\label{NaBr}\\&#xA;  \ce{8 NaI + H2SO4 + 8H+ &amp;amp;-&amp;gt; 4 I2 + H2S + 4 H2O + 8Na+}&#xA;    \tag3\label{NaI}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conventional explanation:&lt;/strong&gt; $\ce{NaI}$ is a strong enough reducing agent to reduce the sulfur, and $\ce{NaBr}$ is a little less strong, so sulfur is not reduced as much. Equivalently, $\ce{H2SO4}$ is not as strong an oxidising agent to reduce $\ce{NaF}$ and $\ce{NaCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My ensuing question&lt;/strong&gt;: Why is reaction \eqref{NaI} more favourable than \eqref{NaBr} and $(\ref{NaF},\ref{NaCl})$ for sodium iodide, so that it reacts with sulfuric acid like it does. What factors are involved: what is more stable, for instance, (or there is lower Gibbs free energy or something else) about reducing sulfur as much as possible (as $\ce{NaI}$ 'chooses' \eqref{NaI} over $(\ref{NaF},\ref{NaCl},\ref{NaBr})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto for sodium bromide. I understand why \eqref{NaI} is impossible, but why is \eqref{NaBr} chosen over $(\ref{NaF},\ref{NaCl})$?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="4945" LastEditDate="2018-09-06T11:29:48.717" LastActivityDate="2018-09-06T11:29:48.717" Title="Why do sodium halides react so differently with sulfuric acid?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;redox&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4751" PostTypeId="1" CreationDate="2013-04-19T20:01:44.827" Score="2" ViewCount="522" Body="&lt;p&gt;The acid in question has &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; about 1.29. Density of 98% w/w solution is &lt;span class=&quot;math-container&quot;&gt;$d=\pu{1.323 g/cm^3}$&lt;/span&gt;. I would like to find the number of undissociated acid molecules (&lt;span class=&quot;math-container&quot;&gt;$\ce{HX}$&lt;/span&gt;), water molecules (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;), dissociated acid molecules (&lt;span class=&quot;math-container&quot;&gt;$\ce{X-}$&lt;/span&gt;), and hydronium ions (&lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The molar mass of the acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{HX}$&lt;/span&gt;) is &lt;span class=&quot;math-container&quot;&gt;$\pu{125.1 g/mol}$&lt;/span&gt;. Thus 98% w/w solution corresponds to &lt;span class=&quot;math-container&quot;&gt;$\pu{10.57 M (mol/L)}$&lt;/span&gt;. If I'm not missing sth, the solution contains 1.5 mole of water and 10.57 mole of acid. What I'm not sure is how these numbers convert into the number of ions/molecules which are, let's say in a volume of &lt;span class=&quot;math-container&quot;&gt;$1~\pu{\mu L}$&lt;/span&gt; (one microliter).&lt;/p&gt;&#xA;&lt;p&gt;Though my attempt would be:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Firstly I have to determine &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of the 98% (w/w) solution in question; I did some rough calculations and got &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; = 0.509&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, I need to establish the ratio &lt;span class=&quot;math-container&quot;&gt;$\dfrac{[\ce{X-}]}{[\ce{HX}]}$&lt;/span&gt;; this can be done using Henderson-Hasselbach equation; I got 0.166&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus, for let's say 1000 &lt;span class=&quot;math-container&quot;&gt;$\ce{HX}$&lt;/span&gt; molecules and 172 &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; molecules (this stems from 10.57:1.5 molar ratio), I have at equilibrium: 834 &lt;span class=&quot;math-container&quot;&gt;$\ce{HX}$&lt;/span&gt;, 166 &lt;span class=&quot;math-container&quot;&gt;$\ce{X-}$&lt;/span&gt;, 166 &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; and 6 &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; molecules.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The solution, seems to be correct. 834 &lt;span class=&quot;math-container&quot;&gt;$\ce{HX}$&lt;/span&gt;, 166 &lt;span class=&quot;math-container&quot;&gt;$\ce{X-}$&lt;/span&gt;, 166 &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; and 6 &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; molecules should have a total mass of &lt;span class=&quot;math-container&quot;&gt;$2.129\times10^{-19} \pu{g}$&lt;/span&gt;. Since the density is known, such a group of molecules should occupy a volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{160.9 nm^3}$&lt;/span&gt; which is perfectly reasonable.&lt;/p&gt;&#xA;" OwnerUserId="1524" LastEditorUserId="83758" LastEditDate="2020-06-18T17:57:03.077" LastActivityDate="2020-06-18T17:57:03.077" Title="How many of different types of chemical species are in a certain volume of aqueous solution of an acid" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4752" PostTypeId="1" AcceptedAnswerId="4770" CreationDate="2013-04-19T21:46:43.453" Score="4" ViewCount="1845" Body="&lt;p&gt;I am reading a paper (&lt;a href=&quot;http://link.aip.org/link/doi/10.1063/1.1587119&quot; rel=&quot;nofollow&quot;&gt;Shirts, M. R.; and Pitera, J. W. and Swope, W. C. and Pande, V. S. &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;2003&lt;/strong&gt;, &lt;em&gt;119&lt;/em&gt;, 5740-5761&lt;/a&gt;).  Near the beginning of the paper (page 5745), the authors state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Experimental free energies of hydration for weakly soluble solutes are&#xA;  determined from concentration measurements made on two phase systems,&#xA;  where one phase consists of a vapor with a partial pressure $P_s$ of&#xA;  some solute molecule of type $s$, and the other phase consists of an&#xA;  aqueous solution with a number density concentration for solute&#xA;  molecules of $\rho_s^{\ell}$.  When such a two phase system is at&#xA;  equilibrium with respect to transfer of molecules of type $s$ between&#xA;  the phases, the solvation free energy is given by $$\Delta G_{\text{solv}} = kT\ln(P_s / (\rho_s^{\ell} kT)).$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does this mean in practice?&lt;/strong&gt;  For example, suppose that I would like to measure the hydration free energy $\Delta G_{\text{solv}}$ of a small molecule (e.g., the amino acid arginine).  That is to say, I would like to measure the &lt;em&gt;solvation&lt;/em&gt; free energy of arginine in &lt;em&gt;water&lt;/em&gt;.  (Since arginine has a charged side chain at neutral pH, I would expect $\Delta G_{\text{solv}}$ to be negative, because arginine is likely stabilized by water, which is polar.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the expression for $\Delta G_{\text{solv}}$ above imply that I just need an aqueous solution of arginine?  Then would I need to somehow measure the partial vapor pressure $P_s$ of the arginine vapor which exists above the arginine aqueous solution?  How is such a partial vapor pressure vapor measurement done in practice?  Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="271" LastActivityDate="2013-04-22T09:03:02.120" Title="How is hydration free energy of a solute measured experimentally?" Tags="&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4753" PostTypeId="2" ParentId="4750" CreationDate="2013-04-19T23:36:30.827" Score="10" Body="&lt;p&gt;A factor you didn't touch upon which may have some weight in determining the reactions is the acidities of the hydrogen halides formed by reactions of the first type (which is fundamentally a coarse qualitative analysis of the free energy of reaction). In aqueous solutions, the order of acidity is $\ce{HF} \ll \ce{HCl} &amp;lt; \ce{HBr} &amp;lt; \ce{HI}$. While concentrated sulfuric acid is obviously not the same as water, perhaps it is fair to say the relative acidities of the hydrogen halides are approximately the same, given they are both polar protic solvents with very high dielectric constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you go down the family, the acid formed in reactions of the first type become stronger, and there is less thermodynamic drive to follow such a reaction pathway. $\ce{HI}$ is somewhat close in acidity to $\ce{H2SO4}$ since neither conjugate base is very coordinating. Meanwhile, as you say, the reducing power of the halides increases, and since sulfur is in a high oxidation number in $\ce{H2SO4}$, redox reaction pathways become more attractive. A detailed analysis of free energies of reactions should agree with the change in the mechanism since it is unlikely these reactions are kinetically controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Another neat consequence of the acidity strength argument is that it explains why the reactions stop at the bisulfate anion instead of going all the way to the halide sulfates; $\ce{HSO4-}$ is a much, much weaker acid than $\ce{H2SO4}$, so there is no free energy to lose out of consuming it to produce a much stronger acid (in the case of $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ at least). $\ce{HF}$ is similar in acidity to $\ce{HSO4-}$ (in water), so it's not as readily explainable in such a qualitative analysis.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-08T22:09:26.443" LastActivityDate="2017-04-08T22:09:26.443" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4754" PostTypeId="1" CreationDate="2013-04-20T11:28:21.160" Score="6" ViewCount="14382" Body="&lt;p&gt;Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does having &quot;no d-orbital&quot; mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="402" LastEditDate="2016-12-04T21:01:54.033" LastActivityDate="2017-04-08T22:08:24.103" Title="What is meant by &quot;no d-orbital&quot;?" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4755" PostTypeId="1" CreationDate="2013-04-20T12:00:30.427" Score="3" ViewCount="6211" Body="&lt;p&gt;Sulfur is a diamagnetic element, meaning that it has no unpaired electrons. But its electronic configuration for valence shell is just like oxygen because they are in the same group i.e ns2, np4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that why is sulfur diamagnetic? If there are 4 electrons in p-orbital of its valence shell then shouldn't it have 2 unpaired electrons?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="402" LastEditDate="2016-12-04T17:20:53.290" LastActivityDate="2016-12-04T22:13:00.240" Title="Diamagnetism in sulfur" Tags="&lt;physical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4756" PostTypeId="2" ParentId="4754" CreationDate="2013-04-20T12:56:00.777" Score="2" Body="&lt;p&gt;Oxygen has a d-orbital as well. Just that there are $3s$ and $3p$ orbitals in between, so extra electrons will get filled there first. The $3d$ orbital has too high energy for it to be of any use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, hybridization almost always (the exception is with coordination compounds) happens for orbitals with the same principal quantum number (shell number). So the $3s$ and $3p$ orbitals are not of any use for accomodating extra electrons/bonds.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-04-20T12:56:00.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4757" PostTypeId="2" ParentId="4754" CreationDate="2013-04-20T12:59:17.330" Score="10" Body="&lt;p&gt;There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are &lt;strong&gt;vacant&lt;/strong&gt;), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is &lt;strong&gt;accessible&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-08T22:08:24.103" LastActivityDate="2017-04-08T22:08:24.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4758" PostTypeId="2" ParentId="4755" CreationDate="2013-04-20T13:12:43.037" Score="5" Body="&lt;p&gt;You seem to be confusing the &lt;em&gt;elements&lt;/em&gt; with the pure &lt;em&gt;simple substances&lt;/em&gt; of the elements. In atomic terms, both the oxygen atom and the sulfur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the &lt;em&gt;substances&lt;/em&gt; oxygen ($\ce{O2}$) and sulfur (generally intended as $\ce{S8}$) are not the same as the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simple picture of Lewis structures, both substances have all atoms forming two bonds, completing the octet rule and therefore without unpaired electrons. This would imply that both are diamagnetic. However, simple valence bond theory is inadequate to explain the bonding in oxygen. Simple molecular orbital theory suggests that there are in fact &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen_MO_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;two unpaired electrons&lt;/a&gt; in the $\ce{O2}$ molecule. This gives rise to paramagnetism in the oxygen molecule, also.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2016-12-04T22:13:00.240" LastActivityDate="2016-12-04T22:13:00.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4759" PostTypeId="2" ParentId="4751" CreationDate="2013-04-20T14:54:01.540" Score="3" Body="&lt;p&gt;The following is an approximate calculation, since among other things it makes the approximation that the activities of each substance are equal to their molarities, which is not true in such concentrated solutions. Also I take some shortcuts to make the calculation quicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a volume of $1\ L$. The mass would be $1323\ g$, 98% of which ($1296.54\ g$) is due to the acid, and 2% ($26.46\ g$) due to water. This initially translates into $10.36\ mol$ of acid, and $1.47\ mol$ of water. However, some of the water and acid are consumed to create hydronium and conjugate base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the dissociation of the acid, you may want to apply $K_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[HA]}$ (the Henderson-Hasselbach equation is derived from this by taking the log on both sides, so they're equivalent), but this is actually a very bad approximation, since you need to take into account the concentration of water in non-dilute solutions. We'll do it anyway and I worked it out to be $[H_{3}O^{+}]=[A^-]=0.704\ mol.L^{-1}$. Since the volume we're working with is $1\ L$, that means $n_{H_{3}O^{+}} = n_{A^-} = 0.704\ mol$. Subtracting those amounts from the initial amounts of water and acid, we reach $$ n_{H_{2}O}=0.77\ mol$$ $$ n_{HA}= 9.66\ mol$$ $$n_{H_{3}O^{+}}=n_{A^-}=0.704\ mol$$&#xA;To find the actual number of each entity, just multiply by the Avogadro number $6.022\times 10^{23}$ and you have the amount of each in a litre of solution.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-04-20T14:54:01.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4760" PostTypeId="1" CreationDate="2013-04-20T15:55:56.903" Score="50" ViewCount="34173" Body="&lt;p&gt;Both $\ce{SF6}$ and $\ce{SH6}$ and $\ce{SF4}$ and $\ce{SH4}$ have the same central atom and the same hybridization, but my teacher specifically mentioned that $\ce{SH6}$ and $\ce{SH4}$ don't exist. I've looked everywhere but I can't figure out why? I'd appreciate some insight into the problem.&lt;/p&gt;&#xA;" OwnerUserId="1526" LastEditorUserId="16683" LastEditDate="2017-07-02T06:57:25.677" LastActivityDate="2021-01-21T11:41:46.917" Title="Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;halides&gt;&lt;hybridization&gt;" AnswerCount="8" CommentCount="1" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="4761" PostTypeId="2" ParentId="4760" CreationDate="2013-04-20T16:12:36.370" Score="17" Body="&lt;p&gt;This is caused by the molecule $\ce{SF6}$ being &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypervalent&lt;/a&gt;, which means that the main element (in this case sulfur) has more then 8 valence electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why this can happen is extremely complex and, to be honest, I am not even sure whether it is a fully solved issue. I do know that the effect is related to the &lt;a href=&quot;http://alpha.chem.umb.edu/chemistry/Seminar/06-09%20WQE/InorgI.Carter.pdf&quot; rel=&quot;noreferrer&quot;&gt;electronegativity of the ligands&lt;/a&gt;, which is very high for $\ce{O}$, $\ce{F}$ and $\ce{Cl}$ atoms and somewhat lower for $\ce{H}$-atoms. This might explain why it doesn't happen for hydrogen, but this is just speculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An incredibly extensive explanation of hypervalency (and related phenomena ) can be found in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit&quot;&gt;this post&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2013-04-20T16:22:40.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4762" PostTypeId="2" ParentId="4760" CreationDate="2013-04-20T16:27:17.150" Score="12" Body="&lt;p&gt;Hypervalency is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalency#History_and_controversy&quot; rel=&quot;noreferrer&quot;&gt;bit of a touchy subject&lt;/a&gt; since there are strong opposing proponents for both the &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;noreferrer&quot;&gt;hybridization camp&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_four-electron_bond&quot; rel=&quot;noreferrer&quot;&gt;3-center-4-electron bond camp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not directly related to your question's compounds, I shall point you to a fun book by Errol Lewars on the computational analysis of uncommon molecules: &lt;a href=&quot;http://books.google.com.br/books?id=IoFzgBSSCwEC&amp;amp;lpg=PA53&amp;amp;pg=PA53#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Modeling Marvels&lt;/a&gt;. While some of it isn't displayed on Google Books, a good amount of the relevant chapter is there. Indeed it even mentions the theoretical higher hydrides of the second-period elements.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-08T22:06:10.830" LastActivityDate="2017-04-08T22:06:10.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4763" PostTypeId="1" CreationDate="2013-04-20T18:42:34.987" Score="2" ViewCount="1320" Body="&lt;p&gt;With reference to crystal field splitting in octahedral complexes, my textbook shows the following table of distribution of $d$-orbital electrons in octahedral complexes, based on the energy difference of $t_{2g}$ and $e_g$ orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/Hhepzmy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do ALL the $d$ orbital configurations (i.e. pertaining to all examples given) form octahedral complexes? I doubt it because not all of them have two vacant $d$ orbitals, especially in the case of weak field ligands where Hund's rule is never violated.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="1322" LastEditDate="2013-04-21T08:18:07.410" LastActivityDate="2013-05-21T08:28:39.677" Title="Which d-orbital configurations form an octahedral complex?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4764" PostTypeId="2" ParentId="2870" CreationDate="2013-04-20T20:10:33.673" Score="0" Body="&lt;p&gt;Basically, a pure substance in this context means a substance that is always available. As in the expression of rate law, if a solvent is always available,say water. Then rate won't depend on the concentration of water , as it is always available. Hence its activity can be considered as 1. &#xA;Certain textbooks refer to pure substances in physical chemistry in this sense.&lt;/p&gt;&#xA;" OwnerUserId="1528" LastActivityDate="2013-04-20T20:10:33.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4765" PostTypeId="2" ParentId="4763" CreationDate="2013-04-21T06:13:17.857" Score="3" Body="&lt;p&gt;A metal ion doesn't actually need vacant d orbitals to form a coordination complex according to Crystal field theory. All the ions mentioned in the table form complexes with water (which is a weak ligand) - see &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_aquo_complex#Stoichiometry_and_structure&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_field_theory#Overview_of_crystal_field_theory_analysis&quot; rel=&quot;nofollow&quot;&gt;Crystal field theory&lt;/a&gt; assumes that all bonding between ligand and metal ion in a complex is electrostatic (ionic). That means there is no exchange of electrons between the metal ion and ligand. Instead, the negatively charged (or polar in case of neutral ligands like $NH_3$) ligand is attracted to the positively charged metal ion by purely electrostatic forces. This is why vacant d orbitals are not required for bonding.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2013-04-21T06:13:17.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4766" PostTypeId="1" CreationDate="2013-04-21T09:55:33.417" Score="23" ViewCount="66397" Body="&lt;p&gt;The melting and boiling points of transition elements increases from scandium ($1530~\mathrm{^\circ C}$) to vanadium ($1917~\mathrm{^\circ C}$). They increase because as we go across the group, we have more unpaired (free) electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at chromium ($1890~\mathrm{^\circ C}$) however, the melting point decreases even though it has more unpaired electrons than the previous atoms. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="9961" LastEditorDisplayName="user15489" LastEditDate="2016-03-07T12:13:21.753" LastActivityDate="2019-12-13T12:37:07.110" Title="Melting and boiling points of transition elements" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;periodic-trends&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4767" PostTypeId="2" ParentId="4754" CreationDate="2013-04-21T11:49:18.737" Score="5" Body="&lt;p&gt;As Nicolau hints in his answer, orbitals are not regions of space around an atomic nucleus. They are mathematical constructs formulated as &lt;a href=&quot;http://en.wikipedia.org/wiki/Wavefunction&quot; rel=&quot;noreferrer&quot;&gt;wavefunctions&lt;/a&gt; describing the properties of electrons, including their energy, angular momentum, and probabilistic distribution in space. The pictures of &quot;orbitals&quot; that you are used to seeing are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_density_function&quot; rel=&quot;noreferrer&quot;&gt;probability density functions&lt;/a&gt; (the absolute value of the wavefunction squared). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the only way we can know what a probability density function looks like or what the energy of orbital would be is to pretend there is an electron in the orbital and then do the math. So, orbitals without electrons in them might as well not exist. As Nicolau and ManishEarth point out for oxygen, the orbitals in the 3d subshell are basically inaccessible to the electrons in oxygen. Thus those orbitals do not exist. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-21T11:49:18.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4768" PostTypeId="2" ParentId="4732" CreationDate="2013-04-21T16:25:13.477" Score="7" Body="&lt;p&gt;Oxalic acid or citric acids works just fine and are not too toxic (I tested oxalic acid myself, it works great! much better, then most mineral acids) . It is preferable to not use strong mineral acids as they will eat some iron too (in fact, they have problems dissolving iron (III) oxide in acceptable time), and oxalic and citric acids are not strong, but gives complexes with iron ions and dissolves rust just fine. Acetic acid cannot give complexes, so I wouldn't bet on it.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2013-04-21T16:25:13.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4769" PostTypeId="1" AcceptedAnswerId="4784" CreationDate="2013-04-22T01:49:52.567" Score="4" ViewCount="40910" Body="&lt;p&gt;I have the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2 O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine how many electrons are transferred in this reaction? More specifically, I need to know the value of $v_e$ which I am supposed to plug into the Nernst equation, which my textbook explains as the coefficient of the number of electrons transferred in each half-reaction. How does one go about splitting this into two half-reactions?&lt;/p&gt;&#xA;" OwnerUserId="1535" LastEditorUserId="4231" LastEditDate="2015-03-01T14:32:01.960" LastActivityDate="2015-03-01T14:32:01.960" Title="How many electrons are transferred in this redox reaction?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4770" PostTypeId="2" ParentId="4752" CreationDate="2013-04-22T09:03:02.120" Score="3" Body="&lt;p&gt;Your interpretation is correct. Theoretically, you will need only an aqueous solution and a fixed volume of gas, and then be able to measure equilibrium concentrations. Although, you may not observe very much vapor pressure without using elevated temperatures (as you noted, the zwitterion is very soluble). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a species $x$ is in equilibrium between two states, $$x_{state1} \rightleftharpoons x_{state2}$$ the equilibrium constant is given by the ratio of concentration of $x$ in each state: $$K = \frac {[x_{state2}]} {[x_{state1}]}.$$ The equilibrium constant relates to Gibbs free energy by the following equation: $$\Delta G = -RT \ln(K)=-RT\ln\left(\frac{x_{state2}}{x_{state1}}\right)=RT\ln\left(\frac{x_{state1}}{x_{state2}}\right).$$ Substituting for an aqueous/vapor phase system yields the cited equation. The literature uses Boltzmann's constant, $k$, in place of the gas constant $R$ because they are using units of molecules instead of moles. (The extra $kT$ term within the logarithm comes from the conversion of pressure to number density concentration using the ideal gas law: $\frac{n}{V}=\frac{P}{RT}$, again using molecular units instead of moles will require $k$ instead of $R$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Measurement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining the &quot;best&quot; method for gas phase measurement depends on what instrumentation you have available to you and how accurate you need/want to your results. Mass spectrometry would be ideal if it employs a soft ionization technique. If the molecular ion doesn't fraction, you could use the raw counts compared to an internal standard to calculate the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't necessarily need to measure the vapor pressure directly, either. If you have a method for measuring the aqueous concentration, you could calculate the vapor pressure based on decrease in solution concentration: $$\Delta G=RT \ln \left(\frac{[arg_{aq}]_i-[arg_{aq}]_f}{[arg_{aq}]_f}\right)=RT \ln \left(\frac{[arg_{aq}]_i}{[arg_{aq}]_f}-1\right)$$ where $[arg_{aq}]_i$ is the initial aqueous arginine concentration (total arginine) and $[arg_{aq}]_f$ is the final concentration (equilibrium amount).  HPLC (or even &lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/7643/1/IJPAP%2048%284%29%20251-255.pdf&quot; rel=&quot;nofollow&quot;&gt;UV-Vis&lt;/a&gt; if that is all you have available) would be a good method, but you would have to either make a calibration curve or use an internal standard (your buffer may work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to &lt;em&gt;J. Phys. Chem.&lt;/em&gt; and/or &lt;em&gt;Biochemistry&lt;/em&gt;, &lt;a href=&quot;http://www.scfbio-iitd.res.in/doc/SolvationThermodynamics.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; cites articles (ref. 29, 82) that have experimental data for free energies of hydration for amino acids (all but arginine). They may be helpful to consult for methods.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-22T09:03:02.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4771" PostTypeId="2" ParentId="4751" CreationDate="2013-04-22T12:10:31.870" Score="2" Body="&lt;p&gt;While searching for a correct answer to my own question I performed a simple simulation using &lt;a href=&quot;http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/&quot; rel=&quot;nofollow&quot;&gt;PHREEQC&lt;/a&gt;. It is critical to realize that the calculation procedure outlined by Nicolau in his answer to my question works properly only for dilute solutions. The highly concentrated solutions have high ionic strength ($IS$) which makes Debye–Hückel theory inadequate. Thus the activity coefficients need to be find in a different way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is claimed that &lt;a href=&quot;http://en.wikipedia.org/wiki/Pitzer_equation&quot; rel=&quot;nofollow&quot;&gt;Pitzer equations&lt;/a&gt; can handle solutions with $IS &amp;gt; 2$. This morning I got accustomed with basic functionality of PHREEQC program (it has amazing possibilities, though it seems to have extremely steep learning curve..). At the moment it seems that it is not possible to perform calculations for alkyl phosphates, which I had been originally interested in, so I decided to determine population of chemical moieties for phosphoric acid. In the calculations described below, I basically start with 14.616 mole of $\ce{H3PO4}$ dissolved in $1 kg$ of water. Then I trigger evaporation to finally get 85% wt solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input file for PHREEQC is as follows (these are my first calculations in this program so I'd appreciate if someone more experienced could comment on the correctness of the input):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SOLUTION 1&#xA;  pH 2.0 charge&#xA;  pe 0&#xA;  P 14616  # 85% wt solution has 14.62 moles of H3PO4&#xA;  water 1.0 # start with 1 kg of water&#xA;REACTION 1&#xA;  H2O -1.0&#xA;  41.520 moles in 10 steps # end with 252g of water&#xA;PITZER&#xA;  -macinnes   true&#xA;  -use_etheta true&#xA;  -redox      true&#xA;DEBUG&#xA;  -iterations 200&#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The results include calculated $pH$ ($0.197$), ionic strength ($1.363$), and molality ($58$). The really interesting, however, are final activities calculated for such a condensed solution of $\ce{H3PO4}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{H3O+}$: $0.635$&lt;br&gt;&#xA;  $\ce{H2O}$:  $0.346$&lt;br&gt;&#xA;  $\ce{H3PO4}$: $56.64$&lt;br&gt;&#xA;  $\ce{H2PO4-}$: $0.635$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using concentrations instead of activities in solving the equation for acid dissociation constant as shown by Nicolau, lead to $\ce{H3O+}$ of $0.324$ and $\ce{H3PO4}$ of $14.756$.&lt;/p&gt;&#xA;" OwnerUserId="1524" LastEditorUserId="1524" LastEditDate="2013-04-22T16:25:17.770" LastActivityDate="2013-04-22T16:25:17.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4772" PostTypeId="1" AcceptedAnswerId="4780" CreationDate="2013-04-22T12:30:05.077" Score="1" ViewCount="281" Body="&lt;p&gt;I was trying to figure out the different peaks in a XPS spectrum for Molybdenum, but I could not understand them. About XPS, I know that this technique is based on the binding energy of the electrons of different elements&lt;/p&gt;&#xA;" OwnerUserId="1111" LastActivityDate="2013-05-23T05:28:50.797" Title="What does the signals in a XPS spectrum mean?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4774" PostTypeId="1" AcceptedAnswerId="4778" CreationDate="2013-04-22T19:02:48.627" Score="14" ViewCount="11934" Body="&lt;p&gt;If electrolysis splits water, then that means that $\ce{H2O}$ is split into $\ce{H}$ and $\ce{OH}$ or $\ce{O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come that if a water molecule is split at e.g. the negative pole (anode), only the $\ce{O}$ bubbles up. What happens to the $\ce{H}$ and/or $\ce{OH}$? Do they react with the pole and stay there?&lt;/p&gt;&#xA;" OwnerUserId="1538" LastEditorUserId="1538" LastEditDate="2013-04-23T07:21:04.457" LastActivityDate="2019-12-06T17:54:25.360" Title="If electrolysis splits water, why does only either hydrogen or oxygen (but not both) bubble up on one pole?" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4775" PostTypeId="5" CreationDate="2013-04-22T19:19:33.700" Score="0" Body="&lt;p&gt;An &lt;em&gt;orbital&lt;/em&gt; is a one-electron wavefunction, which describes the state of an electron in an atom or molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The form of the familiar atomic orbitals is derived by solving the time-independent Schrodinger equation for the hydrogen atom; this leads to solutions with three quantum numbers ($n$, $l$, and $m_l$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ is the principal quantum number, and is generally associated with the idea of an electron shell; one shell comprises electrons with the same value of $n$. The allowed values of $n$ are the positive integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The azimuthal quantum number, $l$, describes the angular momentum of an electron in such an orbital; s-orbitals have $l = 0$, for p-orbitals $l = 1$, and so on. Therefore, orbitals that share the same values of $n$ and $l$ form a subshell. $l$ is allowed to take integer values between 0 and $(n - 1)$ inclusive; therefore, for the first shell, $(n - 1) = 0$ and only s-orbitals are allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the magnetic quantum number, $m_l$, conventionally dictates the projection of the angular momentum onto the $z$-axis. $m_l$ can take integer values between $-l$ and $l$ inclusive; therefore, for a set of p-orbitals with $l = 1$, there are three different orbitals with $m_l=-1,0,+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the fourth quantum number, $m_s$, describes the spin of the electron. It can take two different values, +1/2 and -1/2, and these are associated with &quot;spin up&quot; and &quot;spin down&quot; states respectively. Spin is not accounted for in the Schrodinger equation; in non-relativistic quantum mechanics the existence of spin is a postulate, and it can only be derived from first principles using the relativistic Dirac equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neglecting finer corrections (such as the Lamb shift), the energy of an orbital in a hydrogen-like atom is determined solely by $n$, and is proportional to $-1/n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many-electron atoms, the total electronic wavefunction is assumed to be an antisymmetrized product of one-electron wavefunctions (the &lt;em&gt;orbital approximation&lt;/em&gt;). This allows the idea of an electronic configuration, where each electron is assigned to an orbital. In these species, the $l$-degeneracy is removed, and in general the orbital energies increase in the sequence: $\pu{1s &amp;lt; 2s &amp;lt; 2p &amp;lt; 3s &amp;lt; 3p &amp;lt; 4s &amp;lt; 3d}\ldots$ According to the Pauli exclusion principle, no two electrons can have the same four quantum numbers. Therefore, each &lt;em&gt;spatial orbital&lt;/em&gt;, described by the first three quantum numbers, can only accommodate two electrons, one with spin up and one with spin down. The electronic configuration is then obtained via the Aufbau principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitals are not always obtained by solution of the Schrodinger equation for a single electron. So-called &quot;hybrid orbitals&quot;, such as sp&lt;sup&gt;3&lt;/sup&gt; orbitals, are obtained via a linear combination of the atomic orbitals. In all molecules (apart from $\ce{H2+}$), the Schrodinger equation cannot be solved analytically, so molecular orbitals are generated in different ways, most commonly via linear combination of atomic orbitals. In the Hartree-Fock method, the orbitals are obtained iteratively from the series of one-electron Fock equations. In density functional theory, the electronic wavefunction is not the central quantity, but it still proves useful to model the electron density using the fictitious Kohn-Sham orbitals.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-03-21T17:50:52.943" LastActivityDate="2018-03-21T17:50:52.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4776" PostTypeId="4" CreationDate="2013-04-22T19:19:33.700" Score="0" Body="An orbital is a one-electron wavefunction, usually derived by solving the Schrodinger equation. This tag applies to questions about all forms of orbitals; additionally, questions about the construction and properties of molecular orbitals should be tagged with [molecular-orbital-theory]." OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-03-22T06:26:05.230" LastActivityDate="2018-03-22T06:26:05.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4777" PostTypeId="1" AcceptedAnswerId="4779" CreationDate="2013-04-22T19:38:16.850" Score="3" ViewCount="16725" Body="&lt;p&gt;I would like to ask if there is a source where I can find all the isomers for the alkanes, or other organic molecules. With other words, a list where each molecule of that group is stated in ascending order, and beside there is a value for the no. of isomers it can form. &lt;/p&gt;&#xA;" OwnerUserId="1539" LastActivityDate="2013-04-23T16:37:44.247" Title="Source of the no. of isomers in alkanes, and other organic molecules" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-10-02T16:51:33.367" ContentLicense="CC BY-SA 3.0" />
  <row Id="4778" PostTypeId="2" ParentId="4774" CreationDate="2013-04-22T21:05:46.393" Score="24" Body="&lt;p&gt;Water, as you may know, has a dualist nature between covalent and ionic bonding; oxygen is the second most electronegative element in the periodic table, while hydrogen's simplistic construction makes it very zen about how it forms bonds (it defines the center point of most electronegativity scales). While the bond between the hydrogens and oxygen of a water molecule are termed covalent, the electrophilic nature of the oxygen makes the bonds border on the ionic, and so water is just as easily thought of as $\ce{2H^+ + O^2-}$, or in more stable terms as $\ce{H^+ + OH^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrolysis, this ionic character of water is exploited in a series of &quot;half-reactions&quot;; at each node, a gas is formed by adding or removing electrons from water to liberate one of its two components, and when that gas precipitates out, what's left remains dissolved in the water as charged ions, which are repelled by the node that created it and attracted to the other node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When electrons are added to water at the cathode, they are attracted to the partial positive charge of the &quot;hydrogen side&quot; of water, and can then form an orbit around a hydrogen. That liberates the hydrogen; it no longer needs the electron it's sharing with the oxygen to balance its charge, but it still needs a second electron to stabilize its &lt;em&gt;1s&lt;/em&gt; shell, and the best bet is to pair with another hydrogen to form hydrogen gas, which precipitates out. That leaves soluble hydroxide ions in the water, which are repelled by the cathode (and each other, so these ions wouldn't form any compounds) and move toward the anode. It's highly unlikely that both hydrogen atoms would be liberated from a single oxygen, and even if it did, the oxygen, with extra electrons, couldn't bond with another negative-charge oxygen to form gas; it would be much more likely to deprotonate another water molecule forming two hydroxides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4H2O + 4e^- -&amp;gt; 2H2 (g) (^) + 4OH^- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode, the positive potential causes water to give up electrons, liberating the hydrogens again, this time as naked protons, which are attracted to water's partial negative pole to form hydronium ($\ce{H_3O^+}$ or simply $\ce{H^+ (aq)}$). The now chargeless but bond-seeking oxygen is left to pair up as oxygen gas, which precipitates out. Hydrogen peroxide is a possible intermediate, but very short-lived, as this peroxide will either find another peroxide and pair up the extra oxygens, or it will give up its hydrogens as protons under the positive voltage potential the same way water would, and become oxygen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H_2O - 4e^- \to O_2(g)\uparrow + 4H^+ (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... Then, somewhere in the middle, the hydronium and hydroxide meet, and as these proton donors and proton acceptors will do, they reform water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4H^+(aq) + 4OH^- (aq) \to 4H_2O }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that only one gas precipitates out at each node; the other atoms are trapped as soluble ions, either protons/hydronium or hydroxide, which cannot form other compounds because their like charges repel them from each other. Instead, they remain dissolved as local acidic and basic solutions. If the potential charge in the nodes is high enough, dipping a litmus strip at each end of the water bath (or dropping in a few mL of pH indicator dye) will yield an acidic reading at the anode and a basic reading at the cathode.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="24052" LastEditDate="2016-07-20T23:16:34.690" LastActivityDate="2016-07-20T23:16:34.690" CommentCount="0" CommunityOwnedDate="2013-07-29T22:21:36.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="4779" PostTypeId="2" ParentId="4777" CreationDate="2013-04-22T22:43:10.067" Score="2" Body="&lt;p&gt;Not counting stereoisomers and enantiomers, the number of isomers of the first ten alkanes is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Methane: 1 &lt;/li&gt;&#xA;&lt;li&gt;Ethane: 1 &lt;/li&gt;&#xA;&lt;li&gt;Propane: 1 &lt;/li&gt;&#xA;&lt;li&gt;Butane: 2 &lt;/li&gt;&#xA;&lt;li&gt;Pentane: 3 &lt;/li&gt;&#xA;&lt;li&gt;Hexane: 5&lt;/li&gt;&#xA;&lt;li&gt;Heptane: 9 &lt;/li&gt;&#xA;&lt;li&gt;Octane: 18 &lt;/li&gt;&#xA;&lt;li&gt;Nonane: 35 &lt;/li&gt;&#xA;&lt;li&gt;Decane: 75&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf&quot; rel=&quot;nofollow&quot;&gt;http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf&lt;/a&gt; and Wikipedia articles for these alkanes. The PDF has the number of isomers for the first 30 n-alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of isomers possible with an n-carbon alkane can be calculated by a recursive combinatorial function; however, as the linked paper states, this math isn't for the weak. The number of structural isomers of a 30-carbon alkane is over 4 billion, and a 50-carbon alkane has over 10&lt;sup&gt;31&lt;/sup&gt; possible isomers. Even organic chemists and petrochemists usually don't bother to identify any one isomer beyond about the pentane mark. Some of the more symmetrical isomers are interesting on their own, since a few of the more compact ones (like the n-methyl types) have markedly different behaviors such as boiling point from longer isomers (for example, tetramethylbutane, an isomer of octane notable for its relatively narrow liquid window). The 2,2-dimethylpropane isomer of pentane is visually interesting because of its isohedral geometry, and likewise the tetraethylmethane/diethylpentane isomer might find interest among a few supremacist groups but is otherwise just another 9-carbon alkane.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-04-23T16:29:21.770" LastActivityDate="2013-04-23T16:29:21.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4780" PostTypeId="2" ParentId="4772" CreationDate="2013-04-23T05:13:40.593" Score="3" Body="&lt;p&gt;you are quite right, XPS does measure the 'binding energy' of electrons. More specifically, XPS bombards a surface with X-rays and when those X-rays have sufficient energy they are able to sufficiently excite electrons to be emitted from the sample. Each peak in an XPS spectrum corresponds to the energy necessary to emit a given electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what would you expect to see for Mo?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Outer lying electrons (for Mo, we are talking about electrons in the&#xA;4p and 4s orbitals) will emit at a lower energy. You should observe&#xA;low-energy peaks that correspond to these emissions (approx. 30 - 60&#xA;eV).  &lt;/li&gt;&#xA;&lt;li&gt;you should see a cluster of peaks at approx. 230 eV that correspond to the d-orbital electrons.&lt;/li&gt;&#xA;&lt;li&gt;Finally, you might see discrete peaks at very high energy, approx. 2600 - 2800 eV that    correspond to emission from the 2s and 2p orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, outer orbitals emit at lower energy, and inner orbitals emit at higher energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2013-04-23T05:13:40.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4781" PostTypeId="1" AcceptedAnswerId="4782" CreationDate="2013-04-23T07:37:37.387" Score="7" ViewCount="521" Body="&lt;p&gt;I'm probably missing a simple component of this reaction, but for some reason I not seeing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are two steps that compose a radical process of the Briggs-Rauscher reaction (that are autocatalytic):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$IO_3^- + HIO_2+H^+ \rightarrow 2IO_2 \cdot +H_2O$ &lt;br /&gt;&#xA;$IO_2 \cdot + Mn^{2+} + H_2O \rightarrow HIO_2 + Mn(OH)^{2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is autocatalytic in the sense that 2 $HIO_2$ is produced for each one consumed.  However, I'm just not seeing that for some reason.  Like I said, it's probably a simple thing I overlooked. So, I'm looking for an explanation for what's going on here.&lt;/p&gt;&#xA;" OwnerUserId="1224" LastActivityDate="2013-04-23T10:37:29.950" Title="How is this an autocatalytic reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4782" PostTypeId="2" ParentId="4781" CreationDate="2013-04-23T10:37:29.950" Score="9" Body="&lt;p&gt;In your first reaction, $\ce{HIO2}$ and $\ce{HIO3}$ react to form $\ce{2IO2.}$ and $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second reaction, $\ce{IO2\cdot}$ is being reduced back to $\ce{HIO2}$ by $\ce{Mn^{2+}}$. Since the first reaction produced $\ce{2IO2.}$, this second reaction needs to happen twice for each time the first reaction occurs. The pair of reactions is really:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ IO3- + HIO2 + H+ -&amp;gt; 2IO2. +H2O }$$&#xA;$$2\left(\ce{IO2. +Mn^{2+} +H2O-&amp;gt; HIO2 + Mn(OH)^{2+}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you combine these reactions into an overall reaction, the instances of $\ce{IO2.}$ cancel and you are left with a net gain of $\ce{HIO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{IO3- + H+ +2Mn^{2+} + H2O-&amp;gt; HIO2 + 2Mn(OH)^{2+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the net reaction does not hint at the autocatalytic behavior. To emphasize that the reaction is autocatalytic, you could write the overall reaction in a not quite simplest form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{IO3- + HIO2 +  H+ +2Mn^{2+} + H2O-&amp;gt; 2HIO2 + 2Mn(OH)^{2+}}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-23T10:37:29.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4783" PostTypeId="2" ParentId="4777" CreationDate="2013-04-23T16:37:44.247" Score="0" Body="&lt;p&gt;KeithS's answer is good for the saturated alkanes (generic formula $\ce{C_{n}H_{2n +2}}$). For other organic molecules...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introducing unsaturation or heteroatoms makes it even more complicated. There are two &quot;possible&quot; structures for $\ce{C3H6}$ - &lt;a href=&quot;http://en.wikipedia.org/wiki/Propene&quot; rel=&quot;nofollow&quot;&gt;propene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopropane&quot; rel=&quot;nofollow&quot;&gt;cyclopropane&lt;/a&gt; and three for $\ce{C3H4}$ - &lt;a href=&quot;http://en.wikipedia.org/wiki/Propyne&quot; rel=&quot;nofollow&quot;&gt;propyne&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Allene&quot; rel=&quot;nofollow&quot;&gt;allene&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/cyclopropene&quot; rel=&quot;nofollow&quot;&gt;cyclopropene&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $\ce{C4H10}$ has two isomers, &lt;a href=&quot;http://en.wikipedia.org/wiki/Butane&quot; rel=&quot;nofollow&quot;&gt;butane&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Isobutane&quot; rel=&quot;nofollow&quot;&gt;isobutane&lt;/a&gt;, $\ce{C4H10O}$ has seven: &lt;a href=&quot;http://en.wikipedia.org/wiki/1-butanol&quot; rel=&quot;nofollow&quot;&gt;1-butanol&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Butanol&quot; rel=&quot;nofollow&quot;&gt;2-butanol&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Isobutanol&quot; rel=&quot;nofollow&quot;&gt;isobutanol&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tert-Butanol&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;tert&lt;/em&gt;-butanol&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Deithyl_ether&quot; rel=&quot;nofollow&quot;&gt;diethyl ether&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_propyl_ether&quot; rel=&quot;nofollow&quot;&gt;methyl propyl ether&lt;/a&gt;, and &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.11228.html&quot; rel=&quot;nofollow&quot;&gt;methyl isopropyl ether&lt;/a&gt;. For $\ce{C4H8O}$, there are at least 23 valid structures, not including stereoisomers, although not all of them are stable molecules (enols, I am looking at you!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond maybe 5 or so heavy atoms, the whole effort becomes futile as the number of isomers increases exponentially. Instead, try Googling the formula you care about. ChemSpider will list all known/reported compounds with a given formula. For $\ce{C4H8O}$, there are 42 &lt;a href=&quot;http://www.chemspider.com/Search.aspx?rid=371e2471-8194-4fec-b7ec-3cbf0521b8d9&quot; rel=&quot;nofollow&quot;&gt;entries including stereoisomers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-23T16:37:44.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4784" PostTypeId="2" ParentId="4769" CreationDate="2013-04-23T19:51:07.383" Score="5" Body="&lt;p&gt;The carbon atom must have an oxidation state (O.S.) of -4 on the left hand side since each hydrogen bonded to it must have an O.S. of +1. On the right hand side, the carbon atom has an O.S. of +4 since each oxygen bonded to it has an O.S. of -2. The carbon's oxidation state increased by 8. This is a hint that 8 electrons were transferred. To verify, examine the oxygen atoms. On the left, we only have free oxygen, therefore the O.S. is 0. On the right, we have 4 total oxygen atoms, each with an O.S. of -2. Thus our net change in oxidation state is 0 - 8 = -8. We can almost conclude that all these electrons were transferred from carbon, but we must check the oxidation states of the hydrogen atoms. On both sides, each H atom has an O.S. of +1, so there is no net O.S. change. Thus we can confirm that 8 electrons were indeed transferred in this reaction.&lt;/p&gt;&#xA;" OwnerUserId="1535" LastActivityDate="2013-04-23T19:51:07.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4785" PostTypeId="5" CreationDate="2013-04-23T23:42:49.757" Score="0" Body="&lt;p&gt;Spectroscopy is the study of the interactions between matter and radiated energy. Experimentally, the results are returned as a function of wavelength versus amplitude (intensity), where the spectral lines represent the resonance frequency of the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different types of spectroscopy. Common spectroscopy techniques include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infrared&lt;/a&gt; (IR) &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ultraviolet%E2%80%93visible_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultraviolet-Visual&lt;/a&gt; (UV-Vis) &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;X-ray&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nuclear Magnetic Resonance&lt;/a&gt; (NMR) &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Raman_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raman Spectroscopy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectroscopic techniques are often used in chemistry to characterize a compound and determine its constituents. In biochemistry, spectroscopy is often used to study the conformers and physical structure of proteins and other biological molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A subset of these topics is &lt;a href=&quot;https://en.wikipedia.org/wiki/Time-resolved_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;time-resolved&lt;/em&gt; spectroscopy&lt;/a&gt;, which uses ultra-fast laser pulses (on the pico or femtosecond timescale) to analyze transient species as they are forming. Time-averaged spectroscopy, such as an FT-IR encountered in an organic lab, provides an average IR spectrum of all the molecules in your sample, whereas &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultrafast_laser_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;ultra-fast spectroscopy&lt;/a&gt; can provide information on the structures of intermediates and even very short-lived transition states as a reaction proceeds along its reaction coordinate. Whilst transient UV/vis and IR spectroscopy is common, ultra-fast NMR is not possible due to the relative longevity of the beta state. &lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="9961" LastEditDate="2018-07-20T07:04:31.850" LastActivityDate="2018-07-20T07:04:31.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4786" PostTypeId="4" CreationDate="2013-04-23T23:42:49.757" Score="0" Body="Use this tag for questions about spectroscopic methods in practice and in theory, especially where more specific tags for the individual method (such as [nmr-spectroscopy] or [ir-spectroscopy]) are not available." OwnerUserId="1495" LastEditorUserId="15751" LastEditDate="2020-03-22T20:52:30.310" LastActivityDate="2020-03-22T20:52:30.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4787" PostTypeId="5" CreationDate="2013-04-24T02:36:54.523" Score="0" Body="&lt;p&gt;Quantum mechanics is the discipline of physics that focuses on physical phenomenon that occur at an atomic or sub-atomic scale. In particular quantum mechanics attempts to describe the interaction of energy and matter, which can be described using both particle-like and wave-like interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum theories were first postulated in the early 1900s in an effort to describe atomic-level phenomenon that could not be suitably described by classical mechanics. The 1920s saw critical pioneers such as Heisenberg and Schroedinger formulate theories that would underpin this science. Today, quantum theory is used in computing and many other areas of science and has contributed to our philosophical understanding of life and the universe.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2013-04-24T03:20:20.477" LastActivityDate="2013-04-24T03:20:20.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4788" PostTypeId="4" CreationDate="2013-04-24T02:36:54.523" Score="0" Body="Quantum mechanics is the discipline of physics that focuses on physical phenomenon that occur at an atomic or sub-atomic scale. In particular quantum mechanics attempts to describe the interaction of energy and matter, which can be described using both particle-like and wave-like interactions." OwnerUserId="1495" LastEditorUserId="1495" LastEditDate="2013-04-24T03:20:30.837" LastActivityDate="2013-04-24T03:20:30.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4790" PostTypeId="1" AcceptedAnswerId="4800" CreationDate="2013-04-24T08:00:31.803" Score="3" ViewCount="1447" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the standard enthalpy of formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3(aq)} = \pu{-99 kJ/mol}$&lt;/span&gt; and the standard enthalpy of formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+(aq)} = \pu{+105 kJ/mol}$&lt;/span&gt;, what's the standard heat of formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-(aq)}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I know this if the standard enthalpy of reaction is not known? This is self test 10.1 in Atkins, PC.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="83758" LastEditDate="2020-03-24T14:14:41.403" LastActivityDate="2020-03-24T14:14:41.403" Title="What is the standard heat of formation of the nitrate ion, based on the standard enthalpy of formation of silver nitrate?" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4792" PostTypeId="1" CreationDate="2013-04-24T12:18:22.393" Score="3" ViewCount="1105" Body="&lt;p&gt;Starting point for &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt;&lt;/em&gt; (AIM) is the electron density $\rho(r)$. It could be determined experimentally (X-ray diffractions). I want to know how can we estimate them &lt;em&gt;ab initio&lt;/em&gt;. In &lt;strong&gt;AIM&lt;/strong&gt;&#xA;it is often emphasized that the focus is on density rather than a wave function for the later is just a mathematical function while the former is experimentally observable. Are there ways to &lt;strong&gt;estimate electron density without first calculating wave function&lt;/strong&gt; (which will obviously lead to density function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest is to use AIM in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantitative_structure%E2%80%93activity_relationship&quot; rel=&quot;nofollow&quot;&gt;quantitative structure-activity relationship&lt;/a&gt;. I will typically deal with molecules of around 50 atoms. Usually I will already &lt;strong&gt;have the graph (connectivity) of the molecule (eg .SDF file, &lt;em&gt;smile&lt;/em&gt; notation etc), how can we incorporate this to speed up electron density calculation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if brief comment on the computational cost is also included in the answer.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1586" LastEditDate="2013-05-05T11:25:02.130" LastActivityDate="2013-05-05T11:25:02.130" Title="Approximation of electron density of a small molecule" Tags="&lt;computational-chemistry&gt;&lt;molecules&gt;&lt;quantum-chemistry&gt;&lt;qtaim&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4793" PostTypeId="2" ParentId="4716" CreationDate="2013-04-24T12:24:54.217" Score="2" Body="&lt;p&gt;I don't like to reinvent wheel -- if you go to Wikipedia's  &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_carbonyl&quot; rel=&quot;nofollow&quot;&gt;Metal carbonyl&lt;/a&gt; and follow just two 1st literature position, you'll find your answer concerning bridging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For MO calculations and a general scheme applicable to systems featuring “dative” MM bonds and non-compensating, semi-bridging carbonyl ligands, see: R. D. Barr, T. B. Marder, A. G.&#xA;Orpen, I. D.Williams, J. Chem. Soc. Chem. Commun. 1984, 112–114&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-04-24T12:24:54.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4794" PostTypeId="1" CreationDate="2013-04-24T16:53:05.837" Score="4" ViewCount="1540" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Write a program that determines the molecular weight of a chemical compound based on the number of hydrogen, carbon, and oxygen atoms in the compound. You should use the following atomic weights:&#xA;  \begin{array}{|c|c|c|c|}&#xA;\hline&#xA;\text{Atom} &amp;amp; \text{Weight}~\pu{gm/mole} \\\hline&#xA;\ce{H} &amp;amp; 1.0079 \\\hline&#xA;\ce{C} &amp;amp; 12.011\\\hline&#xA;\ce{O} &amp;amp; 15.9994\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don’t understand the relation between molecular weight and the number of $\ce{H}$, $\ce{C}$, and $\ce{O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The professor has said &quot;there is no need to be able to handle chemical compound names like $\ce{H2O}$ or $\ce{C2H5OH}$. (I plan to discuss how to handle these later in class) Instead, the user should just be asked to specify the number of hydrogen, carbon, and oxygen atoms in the compound.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help to proceed.&lt;/p&gt;&#xA;" OwnerUserId="1542" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-22T10:35:46.880" LastActivityDate="2018-02-22T10:35:46.880" Title="I have to write a Python program relating with molecular weight. Any insight?" Tags="&lt;physical-chemistry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4795" PostTypeId="2" ParentId="4794" CreationDate="2013-04-24T17:23:27.910" Score="13" Body="&lt;p&gt;It's pretty simple actually. The molecular weight of a compound is the sum of the atomic weights of all its atoms. The same numbers apply to a single atom as for a mole of that atom, but at the single-molecule level the unit of measure is the atomic mass unit or amu (aka the dalton or Da). The number of grams per mole of an element and the number of daltons that a single atom of the element weighs are intrinsically linked due to the definition of what a mole of something is (the amount of a pure chemical substance that weighs the same amount in grams as the mass of one atom or molecule of the substance in daltons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, methane (CH&lt;sub&gt;4&lt;/sub&gt;), for instance, has a molecular weight of 1.0079 x 4 + 12.011 = 16.0426 Da. A mole of this gas would weigh 16.0426 grams (which is - interesting science fact - lighter by almost half than that of N&lt;sub&gt;2&lt;/sub&gt; or O&lt;sub&gt;2&lt;/sub&gt; gases making up most of the atmosphere, so a balloon of it will float). So, your Python script would simply prompt for the number of hydrogens (accepting integer values only), carbons (ditto) and oxygens (ditto), and plug them into the expression:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;total_mass = num_hyd * 1.0079 + num_carb * 12.011 + num_oxy * 15.9994&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;... before returning &lt;code&gt;total_mass&lt;/code&gt; as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, the energy contained as enthalpy in the bonds between atoms contributes to a molecule's mass by general relativity and the first law of thermodynamics. In practice, the amount of energy inherent in even the least stable bonds (which produce the most net energy when they decompose to form more stable ones) is a rounding error when you run it through E=mc&lt;sup&gt;2&lt;/sup&gt;, and so chemists don't bother. Nuclear physicists, who actually change the atoms themselves, &lt;em&gt;do&lt;/em&gt; care, because the equation works in reverse for them; they get &lt;em&gt;lots&lt;/em&gt; of energy from very small changes in the atomic masses of the reactants vs the products.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-04-24T18:00:48.367" LastActivityDate="2013-04-24T18:00:48.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4796" PostTypeId="2" ParentId="4792" CreationDate="2013-04-24T19:30:48.540" Score="2" Body="&lt;p&gt;You are looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;density functional theory&lt;/a&gt;. The theory is somewhat involved and you may be interested in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3527303723&quot; rel=&quot;nofollow&quot;&gt;A Chemist's Guide to Density Functional Theory&lt;/a&gt; or similar text book. A full description of the variety of ways to compute the (ground state) electron density is beyond the scope of an answer and better suited to a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0195092767&quot; rel=&quot;nofollow&quot;&gt;textbook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general gist is to optimize the energy with respect to the electron density numerically.&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="440" LastEditDate="2013-04-29T17:07:30.260" LastActivityDate="2013-04-29T17:07:30.260" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4797" PostTypeId="1" CreationDate="2013-04-24T20:02:59.017" Score="2" ViewCount="93" Body="&lt;p&gt;$$\ce{HCO3- (aq) + H2O(l) &amp;lt;=&amp;gt; H2CO3(aq) +OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand why this makes an Arrhenius base and why it doesn't form:&#xA;$$\ce{HCO3- (aq) + H2O(l) &amp;lt;=&amp;gt; CO3^2- (aq) +H3O+ (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="1543" LastEditorUserId="69830" LastEditorDisplayName="user7951" LastEditDate="2021-05-21T07:43:52.227" LastActivityDate="2021-05-21T07:43:52.227" Title="How do you determine which entity gains or loses an H+ in an equation?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4798" PostTypeId="1" AcceptedAnswerId="4809" CreationDate="2013-04-24T20:06:08.273" Score="1" ViewCount="206" Body="&lt;p&gt;I was told in this question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4789/synthesizability-and-energy-minimization-of-a-structure/4791#4791&quot;&gt;https://chemistry.stackexchange.com/questions/4789/synthesizability-and-energy-minimization-of-a-structure/4791#4791&lt;/a&gt; to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Count the] &quot;number of accessible functional groups (alkenes, amines, alcohols, esters, etc) which make it easier to prepare, 5) number of inaccessible functional groups (aromatic rings, polycyclic systems, quaternary carbons, complex or long alkyl groups),&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I googled accessible functional groups and inaccessible functional groups no pages had a longer list. I found a &lt;a href=&quot;http://openbabel.svn.sourceforge.net/viewvc/openbabel/openbabel/trunk/data/SMARTS_InteLigand.txt?view=markup&quot; rel=&quot;nofollow noreferrer&quot;&gt;SMARTS&lt;/a&gt; file which lists some common functional groups. Could someone help me determine which are readily synthesizable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorUserId="-1" LastEditorDisplayName="user1464" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2013-04-25T11:58:43.287" Title="synthesizable functional groups" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2015-10-09T19:15:37.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="4799" PostTypeId="2" ParentId="4797" CreationDate="2013-04-24T20:36:23.410" Score="6" Body="&lt;p&gt;Both of these equations are valid, as $\ce{HCO3-}$ is amphoteric (i.e. both an acid and a base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, one has to consider that the $\mathrm{p}K_\text{a}$ value for $\ce{H2CO3}$ is 6.37 and that for $\ce{HCO3-}$ is 10.32. So if we started from &quot;pure&quot; $\ce{HCO3-}$ (let's assume sodium bicarbonate) and add it to water, what would happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{p}K_\text{b}$ for $\ce{HCO3-}$ is of $14 - 6.37 = 7.63$, it is larger than the $\mathrm{p}K_\text{a}$ for the acidic function (10.32). Therefore although both a weak base and a weak acid, it is a stronger base than acid, increasing the pH when added in water.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorDisplayName="user7951" LastEditDate="2015-04-20T20:49:33.890" LastActivityDate="2015-04-20T20:49:33.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4800" PostTypeId="2" ParentId="4790" CreationDate="2013-04-24T21:24:39.920" Score="2" Body="&lt;p&gt;Silver nitrate, &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; is a strong electrolyte. Based on the information given, I expect you are meant to assume that it dissociates completely when dissolved in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{AgNO3(aq) -&amp;gt; Ag+(aq) + NO3- (aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can make the assumption that &lt;span class=&quot;math-container&quot;&gt;$$\ce{AgNO3(aq)} \equiv \ce{Ag+(aq) + NO3- (aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, we can state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_\mathrm{f} H (\ce{AgNO3(aq)}) = \Delta_\mathrm{f} H (\ce{Ag+(aq)})+\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\implies \pu{-99kJ/mol} = \pu{+105kJ/mol} +\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus &lt;span class=&quot;math-container&quot;&gt;$$\Delta_\mathrm{f} H (\ce{NO3- (aq)})=\pu{-204kJ/mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="83758" LastEditDate="2020-03-24T14:14:12.923" LastActivityDate="2020-03-24T14:14:12.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4801" PostTypeId="1" CreationDate="2013-04-24T21:53:21.003" Score="3" ViewCount="165" Body="&lt;p&gt;Metals contain free moving electrons, in both the solid and liquid state. Does this happen when it is in vapor form?&lt;/p&gt;&#xA;" OwnerUserId="1544" LastEditorUserId="22" LastEditDate="2013-04-24T22:53:22.560" LastActivityDate="2013-04-25T04:11:45.447" Title="Are there any free electrons in metal vapor?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4803" PostTypeId="1" AcceptedAnswerId="4806" CreationDate="2013-04-25T03:31:29.817" Score="3" ViewCount="1988" Body="&lt;p&gt;I recently conducted an experiment where I electrolyzed a solution of sodium chloride. I was using a copper coin as the anode, and I observed a pale blue precipitate in the cup when I was done with the experiment. I then deduced the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chloride ions get oxidized at the anode: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Cl- -&amp;gt; Cl2 + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chlorine gas then reacts with the copper anode to form copper(II) chloride which goes into solution as it is completely soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode, the hydrogen in the water gets reduced to form hydrogen gas: &#xA;$$\ce{2H2O + 2e- -&amp;gt; 2H2 + 2OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the above half reactions exist from &lt;a href=&quot;http://apcentral.collegeboard.com/apc/members/repository/ap04_chem_standard_9793.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the list of standard reduction potentials&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the copper(II) ions react with the hydroxide ions being formed to produce copper(II) hydroxide, which is both insoluble and pale blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://www.ibchem.com/IB/ibnotes/full/red_htm/19.3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt; mentions that the hydronium ions in the water get reduced at the cathode: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H+ + 2e- -&amp;gt; H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? In my understanding, the concentration of hydronium ions ($\pu{1e-7 M}$) would be much too low, and where would the hydroxide ions come from to form copper(II) hydroxide?&lt;/p&gt;&#xA;" OwnerUserId="743" LastEditorUserId="41328" LastEditDate="2017-08-06T05:40:23.380" LastActivityDate="2017-08-06T05:40:23.380" Title="Electrolysis: Water to Hydrogen gas?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4804" PostTypeId="2" ParentId="4801" CreationDate="2013-04-25T04:11:45.447" Score="3" Body="&lt;p&gt;Interesting question. There are free electrons and ions in a metal's gas phase, but even at rather high temperatures, they are relatively rare. The formation of free electrons is given by the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce {M_{(g)} → M^+_{(g)} + e^{-}_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an endothermic reaction for all atoms and molecules, and the energy involved receives a special name: the (first) ionization energy. Comparing &lt;a href=&quot;http://dept.astro.lsa.umich.edu/~cowley/ionen.htm&quot; rel=&quot;nofollow&quot;&gt;a chart of ionization energies&lt;/a&gt; for the elements, one finds that almost all elements require over $ 500\ kJ/mol$ (or about $5\ eV/atom$) to ionize. Even at $5750\ K$, the temperature of the surface of the Sun, the average thermal energy of a particle is about only about $0.5\ eV$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a large part of the electrons that did manage to free themselves from their parent neutral atoms would actually be consumed by other neutral atoms, in a different reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce {M_{(g)}  + e^{-}_{(g)} → M^{-}_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is slightly exothermic for most elements and molecules, and the energy associated with it also receives a special name: the (first) electron affinity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the equilibrium concentrations of each species at a given temperature is possible but is a bit of work. However, for most metals, you can expect to find significantly less than one free electron per ten thousand neutral atoms at $5750\ K$, and the relative amount of free electrons decreases exponentially with temperature.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-04-25T04:11:45.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4805" PostTypeId="1" AcceptedAnswerId="4807" CreationDate="2013-04-25T04:47:06.883" Score="4" ViewCount="573" Body="&lt;p&gt;I was reviewing some analytical chemistry and stumbled upon a section that explained the imperfection of using a salt bridge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It said that the using dissimilar ions is a problem because in, for example, the case of KCl the K+ and Cl- have different mobility and so you get regions that are rich in K+ and others rich in Cl-. This means that some specific parts of solution (even though it's small) to be not electrically neutral i.e. electrically charged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I imagine in theory at least it must be possible to somehow extract these charge-rich areas and put them into a beaker. So what is all this I've been learning about solutions having to be electrically neutral?&lt;/p&gt;&#xA;" OwnerUserId="1546" LastEditorUserId="22" LastEditDate="2013-04-25T05:58:33.240" LastActivityDate="2013-04-25T06:03:54.177" Title="Why do aqueous solutions always have to be electrically neutral?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4806" PostTypeId="2" ParentId="4803" CreationDate="2013-04-25T05:04:18.127" Score="4" Body="&lt;p&gt;In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O &amp;lt;=&amp;gt; OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$  as you keep using them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to question my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="1546" LastEditorUserId="1546" LastEditDate="2014-06-22T00:15:53.180" LastActivityDate="2014-06-22T00:15:53.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4807" PostTypeId="2" ParentId="4805" CreationDate="2013-04-25T06:03:54.177" Score="4" Body="&lt;p&gt;Because the electrostatic force is much stronger than you can imagine. Yes, you can get charge-rich areas in a cell. However, the amount of charge held here will be negligible. A rather small amount of charge can overcome the field near the plates and prevent more charge from approaching. The amount is quite small; it's not something that you would notice (and it won't give you a static shock&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. Everyday static shocks are stronger by orders of magnitude because they deal with a lot of charge accumulated on the &lt;em&gt;surface&lt;/em&gt; of an &lt;em&gt;insulator&lt;/em&gt;, while here it is a relatively small amount of charge dispersed through the &lt;em&gt;volume&lt;/em&gt; of a &lt;em&gt;conductor&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-04-25T06:03:54.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4809" PostTypeId="2" ParentId="4798" CreationDate="2013-04-25T11:47:08.743" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What you are asking amounts to the collected knowledge, wisdom, and intuition of the organic chemistry community. Your best bet will be to find an organic chemist, preferably one with a graduate degree, to work with you on this project. Said organic chemist can apply his/her knowledge/wisdom/intuition to each functional group in less time than it would take you to look up the answer. Doing so frees you to work on the algorithm you describe in your other question. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The beginning of the answer to this question takes me two semesters to get through with my introductory organic chemistry students, and that is just for the &lt;strong&gt;common&lt;/strong&gt; functional groups. After covering the basics of organic structure and reactivity, most texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;common&lt;/strong&gt; functional groups covered in most introductory sequences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;alkanes&lt;/li&gt;&#xA;&lt;li&gt;cycloalkanes&lt;/li&gt;&#xA;&lt;li&gt;alkyl halides&lt;/li&gt;&#xA;&lt;li&gt;alkenes&lt;/li&gt;&#xA;&lt;li&gt;conjugated dienes and other unsaturated systems&lt;/li&gt;&#xA;&lt;li&gt;alkynes&lt;/li&gt;&#xA;&lt;li&gt;aromatic ring systems&lt;/li&gt;&#xA;&lt;li&gt;alcohols&lt;/li&gt;&#xA;&lt;li&gt;glycols and other diols&lt;/li&gt;&#xA;&lt;li&gt;phenols&lt;/li&gt;&#xA;&lt;li&gt;ethers&lt;/li&gt;&#xA;&lt;li&gt;epoxides&lt;/li&gt;&#xA;&lt;li&gt;aldehydes&lt;/li&gt;&#xA;&lt;li&gt;ketones&lt;/li&gt;&#xA;&lt;li&gt;carboxylic acids&lt;/li&gt;&#xA;&lt;li&gt;esters&lt;/li&gt;&#xA;&lt;li&gt;anhydrides&lt;/li&gt;&#xA;&lt;li&gt;acyl halides&lt;/li&gt;&#xA;&lt;li&gt;amides&lt;/li&gt;&#xA;&lt;li&gt;amines&lt;/li&gt;&#xA;&lt;li&gt;nitriles&lt;/li&gt;&#xA;&lt;li&gt;thiols&lt;/li&gt;&#xA;&lt;li&gt;sulfides&lt;/li&gt;&#xA;&lt;li&gt;acetals&lt;/li&gt;&#xA;&lt;li&gt;imines&lt;/li&gt;&#xA;&lt;li&gt;enamines&lt;/li&gt;&#xA;&lt;li&gt;heterocycles&lt;/li&gt;&#xA;&lt;li&gt;nitro compounds&lt;/li&gt;&#xA;&lt;li&gt;diazo compounds&lt;/li&gt;&#xA;&lt;li&gt;carbohydrates&lt;/li&gt;&#xA;&lt;li&gt;amino acids&lt;/li&gt;&#xA;&lt;li&gt;nucleic acids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to know about &lt;strong&gt;every&lt;/strong&gt; functional group, you will need an advanced, probably graduate level, synthesis course. However, even I have not heard of some of the designations on your list, like semithiocarbazide. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some good ideas on where to start, if you choose not to consult with an organic chemist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Check an undergraduate level organic chemistry book out of your local library and get some sense of the subject. In your case, it does not look like you need to know how the reactions work, just that they exist, so you can use the text as a reference source. &lt;/li&gt;&#xA;&lt;li&gt;Sign up for a free &lt;a href=&quot;https://www.coursera.org/courses?search=organic%20chemistry&quot; rel=&quot;nofollow&quot;&gt;MOOC in organic chemistry&lt;/a&gt;. Again, you just want to know that the reactions exist, so you can pick and choose the content from the course.&lt;/li&gt;&#xA;&lt;li&gt;Acquire a copy of one of the seminal books in organic chemistry, &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470462590&quot; rel=&quot;nofollow&quot;&gt;March's Advanced Organic Chemistry&lt;/a&gt;&lt;/em&gt;. This book is not for the fainthearted but it gives good treatment to the functional groups, including the less common ones. &lt;/li&gt;&#xA;&lt;li&gt;Find (do not buy - these books are not worth the price tag) a copy of either &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471190314&quot; rel=&quot;nofollow&quot;&gt;Comprehensive Organic Transformations: A Guide to Functional Group Preparation&lt;/a&gt;&lt;/em&gt; by Richard Larock (which is now 14 years old) or the set of &lt;em&gt;&lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_ss_i_0_30?url=node=13585&amp;amp;field-keywords=comprehensive%20organic%20functional%20group%20transformations&amp;amp;sprefix=comprehensive%20organic%20function,null,274&amp;amp;rh=n:283155,n:75,n:13570,n:13585,k%3acomprehensive%20organic%20functional%20group%20transformations&quot; rel=&quot;nofollow&quot;&gt;Comprehensive Organic Functional Group Transformations&lt;/a&gt;&lt;/em&gt;, which appears to be many volumes (now 9-10 years old).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-04-25T11:58:43.287" LastActivityDate="2013-04-25T11:58:43.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4810" PostTypeId="1" CreationDate="2013-04-25T12:04:50.193" Score="6" ViewCount="840" Body="&lt;p&gt;So I understand, electrons exist in orbitals, mainly s, p, d, f and that when they bond with one another it will form sigma and pi bonds and that whether it bonds in a certain orientation a molecular bonding orbital or an anti bonding orbital will form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I DON'T understand is that when an element bonds it's energy levels will ALWAYS form a series of energy levels that contain bonding and anti bonding orbitals so when they come together, the electrons are always orientated such that these orbitals will form? I also don't really understand what happens when these two atoms meet (the elements that are bonding) how are the orbitals interacting with one another? And what are the orbitals involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who would like to answer the question feel free to loosely address these questions and refer to fundamental concepts or explain bonding through orbitals briefly.&lt;/p&gt;&#xA;" OwnerUserId="1548" LastActivityDate="2013-05-04T04:55:28.490" Title="Bonding and Antibonding" Tags="&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4813" PostTypeId="2" ParentId="4810" CreationDate="2013-04-25T15:33:31.917" Score="4" Body="&lt;p&gt;Orbitals are defined by a set of wavefunctions. Hence, when two orbitals overlap they can either interfere constructively (bonding) or destructively(anti bonding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take two pz orbitals orbitals overlapping with each other. You must understand that the p orbitals look like a dumbbell. One side of the dumbbell is + and the other side is -.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If - meets - we have constructive overlap. If - meets + we have destructive overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former gives a bonding orbital ($\sigma$) and the latter an antibonding orbital ($\sigma*$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the molecule it is often energetically favourable to be in a bonding state hence bonding orbitals are lower in energy to their anti-bonding counterparts.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2013-04-25T15:33:31.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4814" PostTypeId="1" CreationDate="2013-04-25T20:47:21.750" Score="2" ViewCount="1269" Body="&lt;p&gt;Say I have an &lt;a href=&quot;http://en.wikipedia.org/wiki/Effervescence&quot; rel=&quot;nofollow&quot;&gt;effervescent&lt;/a&gt; tablet and I put in water.  I thought that if I break the tablet in two halves it would dissolve quicker than with the whole table (without breaking), but it actually takes the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the scientific reason of this?&lt;/p&gt;&#xA;" OwnerUserId="1550" LastEditorUserId="1362" LastEditDate="2013-04-25T22:36:35.543" LastActivityDate="2016-10-12T15:30:15.880" Title="Time to effervescent an effervescent tablet" Tags="&lt;aqueous-solution&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4815" PostTypeId="2" ParentId="4766" CreationDate="2013-04-25T22:11:10.550" Score="10" Body="&lt;p&gt;To make an argument based on electron configuration, one could see that the orbital occupation does affect the melting point:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Element     Z   Electron Config.    Melting Point / K&#xA;-----------------------------------------------------&#xA;Vanadium    23  [Ar] 4s2 3d3        2183&#xA;Chromium    24  [Ar] 4s1 3d5        2180&#xA;Manganese   25  [Ar] 4s2 3d5        1519&#xA;Iron        26  [Ar] 4s2 3d6        1811&#xA;---&#xA;Copper      29  [Ar] 4s1 3d10       1358&#xA;Zinc        30  [Ar] 4s2 3d10        693&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is apparent in both Mn and Zn: the filled s-orbitals and (half-)filled d-orbitals impact the bonding character of the elements. This is also apparent in the 2nd and 3rd row transition metals, though to a lesser extent due to more diffuse orbitals, greater shielding of the valence electrons and relativistic effects from heavier nuclei. If you look at @NicolauSakerNeto's graph of the melting points, notice that there is still a dip in the graph when the d-orbitals are (half-)filled (i.e., for elements Tc, Cd, Os and Hg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, melting point boils (haha!) down to bonding:  more unpaired electrons lead to the creation of stronger bonds and thus the higher melting points.&lt;/p&gt;&#xA;" OwnerUserId="1299" LastActivityDate="2013-04-25T22:11:10.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4816" PostTypeId="2" ParentId="4814" CreationDate="2013-04-25T23:59:41.927" Score="1" Body="&lt;p&gt;Technically, it should dissolve faster, just like you thought! It will dissolve faster because there is a greater surface area : solution ratio (i.e. the broken edges, which used to be 'in the middle' of the tablet, are now exposed to the water). In general, the more surface area exposed, the faster it will dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why aren't you observing a noticeable increase in speed? How big is your tablet? When you break it in half, are the 'broken edges' small (in surface area) compared to the rest of the tablet? Just for fun, try cutting the tablet horizontally through the middle (like you would cut a bread roll). This should significantly increase the surface area and the speed to dissolve.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2013-04-25T23:59:41.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4818" PostTypeId="1" CreationDate="2013-04-26T08:08:50.940" Score="0" ViewCount="974" Body="&lt;p&gt;I am trying to transform a lipid with a tertiary alcohol into a carboxylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I want to convert to diplopterol (see below) into something homologous to cholesteryl hemisuccinate (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any suggestions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Diplopterol:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LRRVV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cholesteryl hemisuccinate:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3EA3h.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1552" LastEditorUserId="1362" LastEditDate="2013-04-26T12:33:41.640" LastActivityDate="2019-02-20T18:54:31.803" Title="How to transform a tertiary alcohol to a carboxyl?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4819" PostTypeId="2" ParentId="4814" CreationDate="2013-04-26T09:24:35.363" Score="2" Body="&lt;p&gt;&lt;strong&gt;Dissolution in general&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolution is normally a function of several aspects of the solid and liquid and one of them is the interfacial area between the solid and the liquid. A common relation for the dissolution rate which is not always exact, but shows all trends/dependencies very well is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dissolution_%28chemistry%29#Rate_of_dissolution&quot; rel=&quot;nofollow&quot;&gt;following&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm dm}{\mathrm dt}=A \frac{D}{d}\left(C_\mathrm s-C_\mathrm b\right) $$&#xA;where $A$ is the interfacial area, $D$ is the diffusion coefficient of your solid molecules in the liquid, $d$ is the thickness of the solute boundary layer and $C_\mathrm s$ and $C_\mathrm b$ are the concentrations of the solid molecules at surface and in the bulk (this term is the driving force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dissolution of a whole or broken tablet&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a simple calculation for the whole vs. the broken tablet let us do the math.  Assuming that we have a disc-like tablet the surface area is 2 times the frontal area $+$ the area of the sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A=2 A_\mathrm f+A_\mathrm s $$&#xA;&lt;em&gt;For the whole tablet&lt;/em&gt;&#xA;$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For the broken tablet&lt;/em&gt;&#xA;$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t  + 2 h D_\mathrm t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I assume that the tablet is cleanly broken in the exact middle of the disc. So you can already see that the surface area is bigger for the broken tablet, but how much will depend on the specifics of the tablet. Rewriting the area we can see how the different aspects of the broken tablet will influence the area:&#xA;$$A=\pi D_\mathrm t \left[\frac{D_\mathrm t}{2}+h\left(1+\frac{2}{\pi}\right)\right] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation it is clear that if $D_\mathrm t \gg h$ the height term doesn't do much and therefore it will not matter much that the tablet is broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a final check: let's assume some typical numbers now. A typical effervescent tablet is about $D_\mathrm t=2\ \mathrm{cm}$ in diameter and $h=3\ \mathrm{mm}$ thick. This will result in the following surface areas for the broken and the whole tablet: $A_\text{whole}=8.2\ \mathrm{cm^2}$ and $A_\text{broken}=9.4\ \mathrm{cm^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the difference in dissolution rate will only be about $20\ \%$ which is probably hard to notice due to other effects that disturb the comparison. One such effect is that $d$ depends on the amount of stirring which will be different already when you drop the tablets in the glass in a slightly different way (which is unavoidable, because you have 1 broken and 1 whole tablet). &lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorDisplayName="user7951" LastEditDate="2016-10-12T15:30:15.880" LastActivityDate="2016-10-12T15:30:15.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4820" PostTypeId="1" AcceptedAnswerId="4821" CreationDate="2013-04-26T12:07:35.163" Score="3" ViewCount="1681" Body="&lt;p&gt;The energy barrier for activation appears in the derivation of a bimolecular reaction through collision or transition state theory, but it seems that from looking at diagram such as this one &lt;img src=&quot;https://i.stack.imgur.com/NCDX1.jpg&quot; alt=&quot;enter image description here&quot;&gt; it seems like the barrier to activation would be related to the energy required to break bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1372" LastActivityDate="2013-11-04T18:39:48.403" Title="Is the activation barrier of a reaction related to the free energy or heat of formation of its reactants?" Tags="&lt;enthalpy&gt;&lt;free-energy&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4821" PostTypeId="2" ParentId="4820" CreationDate="2013-04-26T13:45:03.840" Score="3" Body="&lt;p&gt;They are certainly related, but not necessarily the same. If all bond-breaking and forming would be sequential than the activation energy would equal the bond-breaking energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible in some reaction mechanisms that the forming bonds 'donate' energy to the break-up and therefore the overall activation energy will be lower. &lt;a href=&quot;http://bouman.chem.georgetown.edu/S02/lect6/lect6.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;An example is the reaction of hydrogeniodide&lt;/a&gt; to hydrogen and iodine show in the figure below. Instead of a full dissocation, two molecules of $\ce{HI}$ line up and the breakup occurs simultaneously with the formation of the new bonds thus allowing a lower energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YDdy5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing you can safely say is that the sum of breaking energies is an upper bound for the activation energy, because single atom level is 'as bad as it gets' (excluding nuclear reactions). &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-04-26T13:45:03.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4822" PostTypeId="1" AcceptedAnswerId="4838" CreationDate="2013-04-26T14:04:01.323" Score="4" ViewCount="3349" Body="&lt;p&gt;Why is it that the extent of racemisation is greater in an $S_N 1$ reaction caried out using a polar protic solvent but the racemisation is not as extensive if an aprotic solvent is used although the carbocation intermediate is common in both?&lt;br&gt;&#xA;eg:- 2-Bromobutane is solvolysed with $\ce{H2O}$ and $\ce{CH3OH}$ but the racemisation is 98% in water and 73% in methanol&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="10185" LastEditDate="2015-10-02T16:23:26.250" LastActivityDate="2015-10-02T16:23:26.250" Title="Why does the extent of racemisation in an SN1 reaction depend on the solvent used?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;solvents&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4823" PostTypeId="1" AcceptedAnswerId="4833" CreationDate="2013-04-26T15:16:41.047" Score="6" ViewCount="2426" Body="&lt;p&gt;I know this is broad so I will narrow:&#xA;Why does molten &lt;a href=&quot;http://en.wikipedia.org/wiki/Polylactic_acid&quot; rel=&quot;nofollow&quot;&gt;PLA&lt;/a&gt; stick to Aluminum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because I'm looking for a way to prevent/reduce the friction of the plastic inside the aluminum barrel of a 3D Printer. I believe that I can treat the surface of the aluminum with a High Temp enamel however I need to be sure of the physics of why the plastic sticks in the first place. Is it the surface finish, or is the molecular attractiveness? If it is the finish, then I can make it smooth. If it is static attraction then I can charge the barrel etc.&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastEditorUserId="1362" LastEditDate="2013-04-27T10:45:10.333" LastActivityDate="2018-03-06T15:07:30.637" Title="Why does molten plastic (PLA) stick to metals (AL)?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4824" PostTypeId="1" AcceptedAnswerId="4830" CreationDate="2013-04-26T18:54:14.263" Score="3" ViewCount="434" Body="&lt;p&gt;I know that equilibrium constant depends on temperature and pressure. But according to Le Chatelier's principle when we change pressure of reagents the equilibrium is the same. Why? &lt;/p&gt;&#xA;" OwnerUserId="1553" LastEditorUserId="275" LastEditDate="2013-04-27T12:23:24.960" LastActivityDate="2018-05-19T07:03:38.203" Title="Equilibrium constant" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4825" PostTypeId="1" AcceptedAnswerId="4826" CreationDate="2013-04-26T19:28:20.760" Score="2" ViewCount="2288" Body="&lt;p&gt;I am doing a chemistry lab report on figuring out the volume of air trapped inside a bottle.  Here is the procedure: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fill a larger container or tub about half full with tap water&lt;/li&gt;&#xA;&lt;li&gt;Fill a 2 L plastic soft drink bottle with tap water.  Put the cap on loosely&lt;/li&gt;&#xA;&lt;li&gt;Carefully turn the bottle upside down without spilling any water&lt;/li&gt;&#xA;&lt;li&gt;Put the bottle into the large container of water so that the mouth of the bottle is underwater.  Remove the cap underwater&lt;/li&gt;&#xA;&lt;li&gt;Feed the flexible tubing under the water so that one end goes inside the bottle&lt;/li&gt;&#xA;&lt;li&gt;Put your straw into the other end of the tubing.  Do not share straws&lt;/li&gt;&#xA;&lt;li&gt;When it is your turn, exhale into the straw to collect the air of one normal breath&lt;/li&gt;&#xA;&lt;li&gt;With a marking pen, mark the volume of air on the soft drink bottle&lt;/li&gt;&#xA;&lt;li&gt;Figure out the volume of the air trapped inside the bottle.  Record this volume&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How might I do number 9? My proposed procedure is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;With a marking pen, mark the new volume of air on the soft drink&#xA;bottle&lt;/li&gt;&#xA;&lt;li&gt;Record this volume, which will be smaller than the volume found in&#xA;step 8 as some of the water will have been displaced by the breath&lt;/li&gt;&#xA;&lt;li&gt;Subtract the new volume of air from the old volume of air to find&#xA;the volume of the air trapped inside the bottle&lt;/li&gt;&#xA;&lt;li&gt;Record this volume&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1456" LastEditorUserId="865" LastEditDate="2015-10-30T17:11:37.820" LastActivityDate="2015-10-30T17:11:37.820" Title="Figuring out the volume of air trapped inside a bottle of water" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4826" PostTypeId="2" ParentId="4825" CreationDate="2013-04-26T19:40:53.897" Score="4" Body="&lt;p&gt;Displacement! Let the container that you have bottle in be filled to the brim inside a bigger container. Then when you breathe into the bottle, the amount of water that is displaced will be the same as the volume of the air.&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastActivityDate="2013-04-26T19:40:53.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4828" PostTypeId="1" AcceptedAnswerId="4836" CreationDate="2013-04-27T01:14:01.017" Score="6" ViewCount="33887" Body="&lt;p&gt;I found a site with this formula if you will, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Formal Charge = [Number of valence electrons on atom] – [non-bonded electrons + number of bonds]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think I am using it correctly in finding the formal charges of each atom in $\ce{SO_4^{2-}}$. $\ce{S}=2$, $\ce{O}=-1$ and the other $\ce{O}=0$ and if you add them together I don't get the overall charge of -2.&lt;br&gt;&#xA; How do you work out formal charges? &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="46306" LastEditDate="2018-03-01T08:47:24.960" LastActivityDate="2018-03-01T08:47:24.960" Title="How to calculate the formal charge of sulfate ion?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4829" PostTypeId="2" ParentId="4766" CreationDate="2013-04-27T04:44:13.427" Score="13" Body="&lt;p&gt;I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a backdrop to this answer, I have plotted the &lt;a href=&quot;https://i.stack.imgur.com/dEEuW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;melting points&lt;/a&gt; and &lt;a href=&quot;https://i.stack.imgur.com/JtS0Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;boiling points&lt;/a&gt; of the transition metals, so make sure to stare at these for a while before reading on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the physicists' &quot;electron sea&quot; point of view of metal bonding, the higher the ionic charge the metal atom can support, the higher the element's melting and boiling points. This explains why group 1 metals such as sodium have quite low melting/boiling points, since the metal would be composed of electrons delocalized in a &lt;span class=&quot;math-container&quot;&gt;$\ce{M}^+$&lt;/span&gt; lattice. Going towards group 2 and group 3 elements, one can expect to find a &lt;span class=&quot;math-container&quot;&gt;$\ce{M}^{2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{M}^{3+}$&lt;/span&gt; lattice, and so on. However, this does not mean that that, for example, metallic osmium is expected to be made out of &lt;span class=&quot;math-container&quot;&gt;$\ce{Os}^{8+}$&lt;/span&gt; cations surrounded by a dense electron sea; as each successive &quot;ionization&quot; of the metal atom is performed, it becomes more unlikely that the next electron can become delocalized, since it has to fight against a higher effective nuclear charge relative to the electron before it. Therefore, we can expect some &quot;optimal average charge&quot; for the metal ions in the pure metal. Since melting and boiling points tend to increase towards groups 7 and 8, we can safely assume that this optimal average charge is at least greater than &lt;span class=&quot;math-container&quot;&gt;$+3$&lt;/span&gt;, and it is likely higher for the heavier elements since they have significantly higher melting/boiling points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can somehow figure out how easily electrons can be removed from their parent atoms to create the electron sea, we may have an argument to explain the trends and apparent discontinuities. First, notice that the &lt;span class=&quot;math-container&quot;&gt;$4^{\ce{th}}$&lt;/span&gt; period metals are in general a little bit underwhelming compared their &lt;span class=&quot;math-container&quot;&gt;$5^{\ce{th}}$&lt;/span&gt; and 6&lt;span class=&quot;math-container&quot;&gt;$^{\ce{th}}$&lt;/span&gt; period counterparts when looking at the boiling points. This can be attributed to the fact that &lt;em&gt;the &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt; orbitals are &lt;strong&gt;anomalously compact&lt;/strong&gt; relative to higher &lt;span class=&quot;math-container&quot;&gt;$nd$&lt;/span&gt; orbitals&lt;/em&gt;. This happens because all first occurrences of a given sublevel (i.e. &lt;span class=&quot;math-container&quot;&gt;$1s$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2p$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$4f$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$5g$&lt;/span&gt;, etc.) are composed of orbitals whose wavefunctions have &lt;em&gt;no radial node&lt;/em&gt;. Since the wavefunction's value is monotonically increasing up to its maximum value, it tapers off quickly. Thus, when a 4&lt;span class=&quot;math-container&quot;&gt;$^{\ce{th}}$&lt;/span&gt; period metal atom tries to delocalize its &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt; electrons, it meets a slight increase in resistance due to its stronger interaction with the nucleus, meaning that the optimal average charge of the ions is slightly lower and leading to lesser interatomic interactions holding the metal together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the sudden dips, most visible for &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr,\, Mn}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Tc}.$&lt;/span&gt; As has been suggested, one is tempted to take into consideration the fact that they all have &lt;span class=&quot;math-container&quot;&gt;$d^5$&lt;/span&gt; configurations, which represent a half-filled &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; subshell. It can be argued that &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed077p1056&quot; rel=&quot;nofollow noreferrer&quot;&gt;they are particularly stable due to exchange energy&lt;/a&gt;, though Cann suggests it is not the most adequate explanation and instead forwards the &quot;parallel spin avoidance factor&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In half-filled subshell configurations, there is a maximum in the effective nuclear charge felt by the electrons (compared to the previous elements with no doubly-occupied orbitals) combined with relatively low interorbital repulsions due to the Pauli exclusion principle. The element that follows a half-filled subshell must now put an electron in a previously occupied orbital, creating stronger, intraorbital electron repulsion. These effects combine to suggest that &lt;a href=&quot;https://i.stack.imgur.com/ov63p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;removing an electron from&lt;/a&gt; or &lt;a href=&quot;https://i.stack.imgur.com/o2hCR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;forcing an electron into&lt;/a&gt; half-filled configurations is particularly difficult (notice that &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr}$&lt;/span&gt; doesn't follow &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Tc},$&lt;/span&gt; more on that soon). If this is the case, then the delocalization of the third electron (which would create &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}^{3+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Tc}^{3+}$&lt;/span&gt; ions) and all those beyond it would be somewhat suppressed. This would effectively reduce the optimal average charge of the ions in the metal, and decrease the strength of interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr}$&lt;/span&gt; doesn't show the same electronic behaviour as &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Tc},$&lt;/span&gt; even though it has a &lt;span class=&quot;math-container&quot;&gt;$d^5$&lt;/span&gt; configuration, and yet it is still anomalous. What gives? Well, according to Greenwood and Earnshaw (Chemistry of the Elements, 1997), the low boiling point of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr}$&lt;/span&gt; is due to a combination of increased effective nuclear charge and the anomalously small size of &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt; orbitals. When going from &lt;span class=&quot;math-container&quot;&gt;$\ce{V}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr},$&lt;/span&gt; the increase in nuclear charge is enough to pull the already small &lt;span class=&quot;math-container&quot;&gt;$3d$&lt;/span&gt; orbitals quite close to the nucleus, disfavouring electron delocalization to the extent that a sudden drop appears. Mo has the same electron configuration as &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr},$&lt;/span&gt; yet it does not show nearly as large a dip because its &lt;span class=&quot;math-container&quot;&gt;$4d$&lt;/span&gt; orbitals are already significantly more diffuse, and aren't so easily suppressed from delocalizing. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditorDisplayName="user5764" LastEditDate="2019-12-13T12:37:07.110" LastActivityDate="2019-12-13T12:37:07.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4830" PostTypeId="2" ParentId="4824" CreationDate="2013-04-27T08:23:17.353" Score="1" Body="&lt;p&gt;Good question here - the equilibrium constant is all a little bit abstract, and therefore tough for some of us chemists, at least initially :) I have always thought about the concept of 'equilibrium' in terms of a balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a balloon, and you've blown the balloon up until it is as full as can be. At this stage the balloon is in one of two 'equilibrium' states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;balloonDeflated $\rightleftharpoons$  balloonInflated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, while blowing up the balloon, you and the balloon are involved in a delicate dance. You want the balloon to be the biggest balloon in the world! But, the balloon doesn't want to be too fat. You and the balloon dance a delicate equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while you try to blow the balloon up as big as possible, the balloon says: 'No!', until it finally explodes in your face :( Too much pressure from one side. Blown out the equilibrium, and lost the battle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Temperature:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've huffed and puffed just as much as you can - (or as much as you can be bothered) - and the balloon is just the right size, determined by the effort you can be bothered putting into it. The balloon size is now in equilibrium with your effort. This is the same with temperature, you can push a reaction faster and further with higher temperatures (more puff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equilibrium State&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, once you and the balloon have agreed on the optimal size, and you have tied a knot in the balloon, you have reached a happy equilibrium. Yay! Now the fun can begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will happen if you squeeze on one side of the balloon?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;roundShape $\rightarrow$ distortedShape&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this analogy, you have added more reagent to the left hand side of the equilibrium. But, by definition, this is no longer an equilibrium, there is more on the left than the right. So, the balloon &lt;strong&gt;bulges&lt;/strong&gt; out to the right, to even up the pressure (i.e. you get more right hand reactants produced to counter the influx of left hand reagent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, the concept of the '&lt;em&gt;pressure&lt;/em&gt;' of reagents is a reference to the concentration of reagents on either side of the equilibrium. A greater concentration on the left, causes a flow to the right and &lt;em&gt;vice versa&lt;/em&gt;. Note that this is distinct from the concept of pressure in the normal terms (&lt;em&gt;pounds per square&lt;/em&gt; or &lt;em&gt;force&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this whole process is fluid, a change to one side, affects the other side. Temperature / environmental pressure will affect the initial equilibrium state (your effort / lung power and the potential size of the balloon will affect the initial equilibrium state) and after this, the pressures imposed on either sides of the equilibrium will be countered by a shift to the low-pressure side.&lt;/p&gt;&#xA;" OwnerUserId="1495" LastEditorUserId="5026" LastEditDate="2018-05-19T07:03:38.203" LastActivityDate="2018-05-19T07:03:38.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4831" PostTypeId="1" CreationDate="2013-04-27T09:24:29.353" Score="1" ViewCount="35" Body="&lt;p&gt;Certain reduction reactions require electrons as reducing agents. Is the overlapping of orbitals is necessary for electron transfer? I mean if a element $\ce{X}$ is heated and it gives out thermionic electrons then can an oxidizing agent $\ce{Y}$ acquire it and get reduced? If so then will it acquire fast moving electrons or slow moving electrons? What will happen if the inert gas is replaced by $\ce{CO2}$ in an cathode ray tube? &lt;/p&gt;&#xA;" OwnerUserId="1556" LastEditorUserId="275" LastEditDate="2013-04-27T12:09:58.157" LastActivityDate="2013-04-27T12:09:58.157" Title="Reduction by thermionic electrons" Tags="&lt;redox&gt;&lt;electrons&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4832" PostTypeId="2" ParentId="4188" CreationDate="2013-04-27T09:32:18.200" Score="1" Body="&lt;p&gt;In common composition they are the same but there may be a little difference due to the antibacterial materials added and the smell.&lt;/p&gt;&#xA;" OwnerUserId="1556" LastActivityDate="2013-04-27T09:32:18.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4833" PostTypeId="2" ParentId="4823" CreationDate="2013-04-27T12:07:04.443" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium&quot;&gt;Aluminum&lt;/a&gt; reacts with oxygen in the air to produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_oxide&quot;&gt;aluminum oxide&lt;/a&gt;, which forms a fine coating on any exposed aluminum surface protecting against further oxidation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ 4Al + 3O2 -&amp;gt; 2Al2O3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum oxide is polar, amphoteric, and hygroscopic: it absorbs moisture from the atmosphere. The surface of aluminum oxide is likely coated with $\ce{Al-OH}$ groups. The polar aluminum oxide and aluminum hydroxide surface is attracted to the polar ester functional groups in PLA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum surfaces can be rendered nonpolar by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0166662292800627&quot;&gt;silanization&lt;/a&gt;. This article may be behind a paywall for you. I apologize. I could not find a good open access reference. The Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Silanization&quot;&gt;silanization&lt;/a&gt; is pretty terrible. The surface is rendered nonpolar, which may decrease the interaction with PLA. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-27T12:07:04.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4834" PostTypeId="2" ParentId="4824" CreationDate="2013-04-27T13:04:13.260" Score="3" Body="&lt;p&gt;The &lt;strong&gt;position&lt;/strong&gt; of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium &lt;strong&gt;constant&lt;/strong&gt; depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: consider a hypothetical reaction in which two molecules of gas react to form a third. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A(g) + B(g)&amp;lt;=&amp;gt;C(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{P_C}{P_A P_B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_{C}}{P_{A} P_{B}}=2$$&#xA;$$P_A + P_B + P_C = 7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q.&#xA;$$P_A =1\times \frac{2}{7}=\frac{2}{7}$$&#xA;$$P_B =2\times \frac{2}{7}=\frac{4}{7}$$&#xA;$$P_A =4\times \frac{2}{7}=\frac{8}{7}$$&#xA;$$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $C&amp;gt;K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than goof up the algebra, I let &lt;a href=&quot;http://www.wolframalpha.com/input/?i=2=%28%288/7%29-x%29/%28%28%282/7%29%2bx%29%28%284/7%29%2bx%29%29&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; do it. The solution for $x$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$&#xA;$$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$&#xA;$$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of equilibrium is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the temperature changes the values of $Q$ which define the equilibrium state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=e^{-\Delta G^\circ/RT}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-27T13:04:13.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4836" PostTypeId="2" ParentId="4828" CreationDate="2013-04-27T19:49:32.270" Score="8" Body="&lt;p&gt;Your formula is correct. If $V$ is the number of valence electrons originally assigned to an atom, $N$ is the number of nonbonding electrons, and $B$ is the number of bonds ($\frac{1}{2}$ of the number of bonding electrons to be more precise), then the formal charge $FC$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$FC = V - (B+N)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfate ion has two valid structures that you could draw, one with sulfur having a formal charge of zero and one with sulfur having a formal charge of +2. The structures below (from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfate_ion&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;, released into the public domain) follow the annoying but allowable convention of replacing lone pairs with bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0fKp9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the structure on the left (#1) is correct, then the sulfur atom ($V=6$) has six bonds ($B=6$) and no lone pairs ($N=0$). Two oxygen atoms (V=6) have two bonds ($B=2$) and two lone pairs ($N=4$), while the other two oxygen atoms have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom and two oxygen atoms have $FC=0$ and the remaining two oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$FC_{\ce{S}}=6-(6+0)=0$$&#xA;$$FC_{\ce{O_{1,2}}}=6-(2+4)=0$$&#xA;$$FC_{\ce{O_{3,4}}}=6-(1+6)=-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the structure on the right (#2) is correct, then the sulfur atom ($V=6$) has four bonds ($B=4$) and no lone pairs ($N=0$). All four oxygen atoms (V=6) have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom has $FC=+2$ and the oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.&#xA;$$FC_{\ce{S}}=6-(4+0)=+2$$&#xA;$$FC_{\ce{O}}=6-(1+6)=-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, wait! A more fundamental question might be &quot;How can the sulfate ion have two very different structures that puts different formal charges on the sulfur and oxygen atoms?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;noreferrer&quot;&gt;Formal charge&lt;/a&gt; is a nice bookkeeping mechanism, but has no experimentally valid relationship to the actual charge on any atom in most molecules or ions. Formal charge only equals actual charge on monoatomic species. Formal charge falls into the category of models that we use in chemistry that are 1) helpful, 2) produce the correct answer when used correctly, and 3) completely bogus.  Other models in this category include oxidation number, VSEPR, resonance, and electronegativity. The experimental evidence suggests the real structure of sulfate combines features of both structure #1 and structure #2, but would be challenging to draw using the formalisms we have adopted:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All four $\ce{S-O}$ bonds are equal in length (#2).&lt;/li&gt;&#xA;&lt;li&gt;The length of the $\ce{S-O}$ bonds is shorter than a normal $\ce{S-O}$ single bond and longer than a normal $\ce{S-O}$ double (#1).&lt;/li&gt;&#xA;&lt;li&gt;The sulfur atom has a partial positive charge (#2) (note that partial charges, unlike formal charges, have some experimental basis).&lt;/li&gt;&#xA;&lt;li&gt;The four oxygen atoms have equivalent partial negative charges (#2).&lt;/li&gt;&#xA;&lt;li&gt;The partial negative charges on the oxygen atoms add to more than $-2$ but not anywhere near $-4$ (#1).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-04-28T23:14:42.053" LastActivityDate="2013-04-28T23:14:42.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4837" PostTypeId="2" ParentId="4828" CreationDate="2013-04-27T20:17:27.147" Score="7" Body="&lt;p&gt;Ben Norris' answer is excellent. I'll collaborate with a more visual procedure, so it may provide a different point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I've learned it, oxidation numbers can be determined by figuring out where the bonding electrons would be assuming bonds are 100% ionic, while formal charges can be determined by figuring out where the electrons would be if the bond were 100% covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of applying an equation directly, draw the Lewis structure of the compound. Now cut all covalent bonds homolytically, i.e. evenly distributing the two electrons between the bonding atoms. Now count the number of electrons attached to each atom, and subtract the number of valence electrons the free atom has. The number you obtain will be an &quot;electron excess&quot;, which is the opposite of the formal charge, so just multiply by -1. Here's your example worked out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/12O36.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-04-27T20:17:27.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4838" PostTypeId="2" ParentId="4822" CreationDate="2013-04-27T20:21:00.440" Score="4" Body="&lt;p&gt;Secondary haloalkanes can undergo substitution by both the $S_N 1$ and $S_N 2$ routes. In general these two mechanisms compete with each other. The following diagram is a &lt;a href=&quot;http://en.wikipedia.org/wiki/More_O%27Ferrall-Jencks_plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;More-O'Ferrall-Jencks Plot&lt;/a&gt; showing the possible mechanism space for a nucleophilic substitution reaction. The reactants are in the upper left. The products are in the lower right. The horizontal axis represents the extent to which the C-LG bond has broken. The vertical axis represents the degree to which the C-Nu bond has formed. The $S_N 1$ pathway is represented by traveling along the perimeter of the plot first to the right and then down. The $S_N 2$ pathway is actually the red diagonal line represented both processes happening simultaneously. There is a third pathway (down and then right), representing nucleophilic attack and then loss of leaving groups. $S_N$ reactions at $sp^2$ centers often follow this route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LSmSL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very few reactions stick completely to one of these three extreme paths. For $sp^3$ centers, the mechanism sticks to the upper right of the diagonal depending on a variety of factors like the polarity of the solvent, the nucleophilicity of the nucleophile, the strength of the C-LG bond, steric hindrance at the subsitution center, and how stable the carbocation would be if it formed. The arrows on the plot above suggest the directions that the transition state would move if the nucleophilic attack step became more favored (stronger nucleophile, less polar solvent, less steric hindrance). The end position is not possible for this particular reaction, but such is the result of reusing and image from somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you can tell how close the actual reaction pathway sticks to one of the idealized pathways. Or, you can think of these numbers as the relative likelihood that a substrate will undergo solvolysis by $S_N 1$ or $S_N 2$ mechanisms in that solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, methanol $\ce{CH3OH}$ is less polar than water $\ce{H2O}$, so the $S_N 1$ pathway is less favorable in methanol than in water. However, the $S_N 1$ pathway in methanol is still way more favorable than the $S_N 2$ pathway in methanol (by 73:27).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-05-01T11:37:34.503" LastActivityDate="2013-05-01T11:37:34.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4841" PostTypeId="1" CreationDate="2013-04-28T01:41:30.703" Score="1" ViewCount="148" Body="&lt;p&gt;How do I find partial pressures if I am only given their percentages? Where do I begin?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Assume $\ce{O}=21\%$ , $\ce{N}=78\%$ , $\ce{Ar}=1\%$ by volume. How would I calculate the partial pressures if total pressure is $\pu{1 atm}$. &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="5026" LastEditDate="2018-02-28T02:48:03.137" LastActivityDate="2018-02-28T02:48:03.137" Title="How to find partial pressures of gases only given their percentages?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4842" PostTypeId="1" AcceptedAnswerId="4845" CreationDate="2013-04-28T04:00:18.147" Score="1" ViewCount="1882" Body="&lt;p&gt;A binary compound of lead and oxygen contains $90.66 \% \ce{Pb}$. What is the empirical formula for the compound? How do I use the percentage to get to the normal formula, then how do I change the normal formula to empirical?&lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="5026" LastEditDate="2018-02-22T02:13:01.243" LastActivityDate="2018-02-22T02:13:01.243" Title="Empirical formula when given percentages?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4843" PostTypeId="2" ParentId="4841" CreationDate="2013-04-28T05:48:43.257" Score="3" Body="&lt;p&gt;&lt;strong&gt;Pressures of gasses are additive.&lt;/strong&gt; For example, 1 atm X plus 1 atm Y makes 2 atm total pressure. Because all the components of a mixed gas are at the same temperature and pressure, the fraction of each component by volume is also the same fraction of the total pressure.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-28T05:48:43.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4844" PostTypeId="2" ParentId="4842" CreationDate="2013-04-28T10:36:08.027" Score="0" Body="&lt;p&gt;I'm not going to this question for you. The point of this forum is to educate and not to do your homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you 100 g of compound XY where X and Y are elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the percentage of X is 70 and of Y is 30.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there must be 70 g and 30 g of X and Y respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert this to moles you need to do :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{70}{molar\;mass\;of\;X}$ and similarly for Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then make use of the definition of the empirical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE : To obtain the molecular formula you need the molecular mass.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2013-04-28T10:36:08.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4845" PostTypeId="2" ParentId="4842" CreationDate="2013-04-28T10:51:22.293" Score="1" Body="&lt;p&gt;You will get the empirical formula straight from this data. In order to convert to a molecular formula, you need more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of another example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A binary compound of tin and chlorine contains 62.89% tin by mass. Determine the empirical formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our first goal is to figure out the ratio of &lt;em&gt;moles&lt;/em&gt; of tin to chlorine in this compound. Once we know the ratio of moles, we can determine the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do we know? We know what percentage of tin is in this compound by mass. What is the percentage of chlorine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get to moles, we need at least a mass of the sample. We were not given that. Since percent by mass is a ratio of masses, it turns out we can choose any mass of sample we want and get the same formula (Try it yourself!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that we started with 100.0 g of this stuff, then how many grams of tin would you have? How many grams of chlorine would you have? 100.0 g is a nice starting mass since you are working with percents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as Ari's answer says, you need to convert mass of tin and mass of chlorine into moles of tin and moles of chlorine by dividing each mass by the atomic molar mass of each element (look it up on your periodic table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you are in moles, determine the lowest whole number ratio of the two elements by dividing each number of moles by the smaller number of moles. If your numbers are close to whole numbers, you can round. For example, round 1.97 to 2. If your numbers are not close to whole numbers, you may need to do a little more arithmetic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you get a number like 1.48, this is close to 1.5 which is 3/2. Multiply both numbers by 3.&lt;/li&gt;&#xA;&lt;li&gt;If you get numbers like 1.34 (close to 4/3) or 1.67 (close to 5/3) multiply by 3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now you have the empirical formula. As Ari's answer says, you need a molar mass of the species to get a molecular formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for my example is $\ce{SnCl2}$. Work through the steps and see if you can get to this answer. Then repeat for your lead and oxygen compound.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-04-28T10:51:22.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4846" PostTypeId="2" ParentId="2795" CreationDate="2013-04-28T11:41:16.597" Score="10" Body="&lt;p&gt;This behavior can be attributed to the same phenomenon as that which causes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanthanide_contraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;lanthanide contraction&lt;/a&gt;. The electrons in &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; subshells are very poor at &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuclear shielding&lt;/a&gt;, so the &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; electrons in the next higher shell are closer (on average) to the nucleus than you might expect. If these electrons are closer to the nucleus, then the atom exhibits a smaller radius than expected and these &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; electrons are more difficult to remove. According to the Wikipedia article on the lanthanide contraction, there are both quantum mechanical and relativistic causes for the lanthanide contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behavior also occurs in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hafnium&quot; rel=&quot;nofollow noreferrer&quot;&gt;hafnium&lt;/a&gt; (72), which is right after the lanthanides. Hafnium's first ionization energy is &lt;span class=&quot;math-container&quot;&gt;$\pu{658.5 kJ/mol}$&lt;/span&gt;, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Zirconium&quot; rel=&quot;nofollow noreferrer&quot;&gt;zirconium&lt;/a&gt; (40, right above hafnium) is &lt;span class=&quot;math-container&quot;&gt;$\pu{640.1 kJ/mol}$&lt;/span&gt;. Barium is before the lanthanides and radium is after them. Any pair so split will probably exhibit this new trend in ionization potentials. Go and check it out!&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="61578" LastEditDate="2019-06-29T08:46:28.500" LastActivityDate="2019-06-29T08:46:28.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4847" PostTypeId="2" ParentId="4465" CreationDate="2013-04-28T11:48:08.520" Score="4" Body="&lt;p&gt;The key factor is that baking soda is &lt;a href=&quot;http://en.wikipedia.org/wiki/Amphoteric&quot; rel=&quot;nofollow noreferrer&quot;&gt;amphoteric&lt;/a&gt;. This means that it can react both with strong acids and with strong bases which are typically smelly compounds &lt;a href=&quot;http://oconto.uwex.edu/files/2011/02/Baking-Soda.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(sour milk, rotten fish&lt;/a&gt;, fatty acids in sweat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation on the third page of article by Lamb [&lt;a href=&quot;https://doi.org/10.1038/jid.1946.13&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] gives a nice summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The chemical explanation of why bicarbonate of soda should act as a deodorant&#xA;  is of interest. Sweat (3) has an acid reaction, &lt;span class=&quot;math-container&quot;&gt;$\ce{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$5.2\text{–}6.75$&lt;/span&gt; which is due to the fatty acids or acid phosphates of sodium and potassium which it contains. It also contains &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt;, alkaline salts, organic acids and urea. Minute amounts of unstable fats and oils are secreted in sweat and hydrolysed to their corresponding fatty acids and glycerin. Free fatty acids may be secreted. Many of these acids are volatile and have a characteristic disagreeable odor. These include butyric acid, formic acid, caproic acid and valerie acid. The following are among the possible explanations of the manner in which sodium bicarbonate acts as a deodorant in the axilla.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;It forms a sodium salt with butyric, caproic and valerie acids. These sodium salts are comparatively mild in odor in comparison to the volatile &quot;rancid-smelling&quot; fatty acids.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;By changing the &lt;span class=&quot;math-container&quot;&gt;$\ce{pH}$&lt;/span&gt; of the axillae, the bacteria which decompose the fats are unable to act. However, as a result of this the staphylococcic flora which prefer higher alkalinity might thrive and axillary furunculosis might result.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To give an example. The reaction of baking soda with &lt;a href=&quot;http://en.wikipedia.org/wiki/Butyric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;butyric acid&lt;/a&gt; looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C3H7COOH + NaHCO3 -&amp;gt; C3H7COONa +H2O + CO2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while butyric acid is extremely smelly (an intense vomit smell), the salt hardly smells at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this explanation, I think &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; should obviously also work well for deodorizing in case the smell is caused by acids. The problem would be that smells caused by basic compounds are left untouched.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lamb, J. H. Sodium Bicarbonate: An Excellent Deodorant. &lt;em&gt;Journal of Investigative Dermatology&lt;/em&gt; &lt;strong&gt;1946&lt;/strong&gt;, 7 (3), 131–133. &lt;a href=&quot;https://doi.org/10.1038/jid.1946.13&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1038/jid.1946.13&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="41328" LastEditDate="2019-05-19T10:30:23.347" LastActivityDate="2019-05-19T10:30:23.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4848" PostTypeId="2" ParentId="2795" CreationDate="2013-04-28T14:37:56.140" Score="19" Body="&lt;p&gt;I think it's also important to mention &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev-physchem-032511-143755&quot; rel=&quot;nofollow noreferrer&quot;&gt;relativistic effects&lt;/a&gt; here.  They already start becoming quite visible after &lt;span class=&quot;math-container&quot;&gt;$Z=70$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ra}$&lt;/span&gt; lies a good bit after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very heavy atoms, the electrons of the &lt;span class=&quot;math-container&quot;&gt;$\ce{1s}$&lt;/span&gt; orbital (actually, all orbitals with some electron density close to the nucleus, but the &lt;span class=&quot;math-container&quot;&gt;$\ce{1s}$&lt;/span&gt; orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, (&lt;span class=&quot;math-container&quot;&gt;$p_{\text{relativistic}}=\gamma p_{\text{classical}}$&lt;/span&gt;) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a &lt;strong&gt;relativistic contraction&lt;/strong&gt; of the &lt;span class=&quot;math-container&quot;&gt;$\ce{1s}$&lt;/span&gt; orbital (and other orbitals with electron density close to the nucleus, especially &lt;span class=&quot;math-container&quot;&gt;$\ce{ns}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{np}$&lt;/span&gt; orbitals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all &lt;span class=&quot;math-container&quot;&gt;$\ce{ns}$&lt;/span&gt; orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the &lt;span class=&quot;math-container&quot;&gt;$\ce{7s}$&lt;/span&gt; valence electrons in &lt;span class=&quot;math-container&quot;&gt;$\ce{Ra}$&lt;/span&gt; are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the first (and second) ionization energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ra}$&lt;/span&gt; becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium (&lt;span class=&quot;math-container&quot;&gt;$Z=120$&lt;/span&gt;) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to &lt;span class=&quot;math-container&quot;&gt;$\ce{Ra}$&lt;/span&gt;. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="61578" LastEditDate="2019-06-29T08:43:29.970" LastActivityDate="2019-06-29T08:43:29.970" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="4849" PostTypeId="1" CreationDate="2013-04-29T18:16:14.207" Score="-2" ViewCount="2235" Body="&lt;p&gt;So I did for my AS Level exam today a titration experiment, seeing how much sulfuric acid of (I think) about $0.100 - 0.200 \text{ mol dm}^{-3}$, which went in the burette, is needed to react with 0.1M sodium hydroxide, in the conical flask. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We lose a mark for every $0.10\text{ cm}^3$ we are off of the actual titre value that should be obtained by a professional, i.e. our chemistry teacher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I obtained an average titre of $24.28\text{ cm}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what the titre should be from experience or memory in the lab of this experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this isn't a constructive question, so feel free to downvote - I just really want to know so that I'm taken off of the edge I'm sitting on!&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="275" LastEditDate="2013-05-01T11:29:48.787" LastActivityDate="2013-05-01T11:29:48.787" Title="What is the titre of a sodium hydroxide-sulfuric acid titration experiment?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4850" PostTypeId="2" ParentId="4849" CreationDate="2013-04-29T19:01:37.257" Score="2" Body="&lt;p&gt;There isn't really an answer to your question. The problem is, there is many possible ways in which that lab might have been set up and there's no telling which one you had to sit, not to say that the concentration of sulfuric acid given to you might have been &lt;em&gt;any&lt;/em&gt; concentration between 0.100 and 0.200 M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you do AS Chemistry, you should also be able to calculate the volume range you should be able to get. With 0.1 M sulfuric acid and 0.1 sodium hydroxide, you expect to use a titre of 12.5 mL of acid against 25 mL of base, to 6.25 mL of 0.2 M acid against 25 mL of 0.1 M base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your titre value seems so off by calculation, I guess that the concentration of the acid could have been about 0.5 M instead, but really, there is no telling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See a sample marking scheme (I took one from &lt;a href=&quot;http://papers.xtremepapers.com/CIE/Cambridge%20International%20O%20Level/Chemistry%20%285070%29/5070_s12_ms_32.pdf&quot; rel=&quot;nofollow&quot;&gt;June 2010&lt;/a&gt; and the syllabus doesn't seem to have changed for this May/June sitting):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(a) Accuracy [8]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For the two best titres give:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;4 marks for value within 0.2 cm^3 of supervisor&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;2 marks for value within 0.3 cm^3 of supervisor&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;1 marks for value within 0.4 cm^3 of supervisor&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Concordance [3]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Give:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;3 marks if all the ticked values are within 0.2 cm^3&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;2 marks if all the ticked values are within 0.3 cm^3&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;1 marks if all the ticked values are within 0.4 cm^3&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Average [1]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Give 1 mark if the candidate calculates a correct average (error not greater than 0.05) of all his ticked values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second and third mark allocations are really easy to get. As long as you get your readings close to each other and calculate the average titre correctly, you will get 4 marks. I'm not sure where you got this &quot;you lose 1 mark for every 0.1 mL off the supervisor's titre value&quot;. As you can see, if you get 2 best titre values withinn 0.2 mL of the supervisors, you get full marks.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-04-29T19:01:37.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4852" PostTypeId="2" ParentId="2647" CreationDate="2013-04-29T19:26:26.740" Score="4" Body="&lt;p&gt;The answer to your question is difficult to explain and understand without a correct understanding of charge and the redox reaction. Another way of asking this question is: Why does current not flow if I connect two batteries using only one wire from the positive terminal of one battery to the negative terminal of the other. The common answer you will get is that you need a complete circuit, but that avoids the deeper question of why you need a circuit. To understand it a little better, you could think of a cell as two parts, one tries to pull electrons from its metal terminal and another tries to push electrons into its terminal. However, neither can do so unless the oppositely charged ions that are left behind can be exchanged between the two metals inside the battery. This means that while the cathode does have a negative charge, it will not let the electrons go unless there is a negative ion to pair up with the positive ion that gets left behind inside the cell. This negative ion comes from positive anode ions leaving the solution and attaching to the anode, but more negative ions only get created if the anode receives the electrons needed to attract the positive anode ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, the one Ox/Red half might be the cathode. In that case it will not have a potential at all since there is nothing but other electrons to balance the charge of the positive ions created at the cathode as the electrons leave it for the wire. The ground connection acts only as a conductor of the electrons. There is no reason for an electron to leave behind a positive ion just to push other electrons through a wire to ground and recombine with a positive ion. That would require energy since the wire has some resistance. The reason the battery works is because a chemical reaction takes place that releases energy. In other words, the positive and negative ions from each half cell would rather be combined together to reach a lower energy state. This can only occur if the positive metal ions that detach from the cathode as it releases electrons can combine with negative ions in the electrolyte. However, the negative ions in the electrolyte must be replenished and that can only occur if other positive metal ions in the electrolyte are pulled out of solution by the anode, which can occur only if the anode receives electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may know, the cells' anode and cathode are typically made of two different materials such as zinc and copper, and the electrolyte is a solution made of a salt containing positive ions of the anode metal such as copper sulfate. As the cell discharges, negative zinc ions are leaving the cathode and dissolving into the solution, while positive copper ions are leaving the solution and attaching to anode. One terminal dissolves while the other grows. As negatively charged electrons move from the zinc cathode to the copper anode via the circuit outside the battery, positively charged zinc ions now in the solution are soaking up or balancing the negative charge created in the electrolyte as the positively charged copper ions leave the solution and attach to the anode that received the extra electrons via the wire from the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason all of this takes place is because the process results in a lower overall energy state after it occurs. In other words, the zinc atoms would rather be combined with $\ce{SO4^2-}$ ions than with each other, and the copper ions would rather be combined with each other at the anode than with the $\ce{SO4^2-}$ in the solution. The only way this can happen is if electrons are moved from the cathode to the anode and the easiest way is via the wire and circuit outside the battery. Without the electron flow, the chemical reaction between the two half cells is inhibited. Conversely, without the chemical energy potential of the reaction between the two half cells, there is no reason for electrons to flow.&lt;/p&gt;&#xA;" OwnerUserId="1563" LastEditorDisplayName="user7951" LastEditDate="2015-05-18T08:53:07.183" LastActivityDate="2015-05-18T08:53:07.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4854" PostTypeId="1" AcceptedAnswerId="4855" CreationDate="2013-04-29T22:58:35.613" Score="7" ViewCount="15792" Body="&lt;p&gt;My textbook, &lt;em&gt;Organic Chemistry&lt;/em&gt; (6th edition) by Brown, Foote, Iverson, and Anslyn, states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alkynes in which the triple bond is between carbons 1 and 2 are commonly referred to as &lt;strong&gt;terminal alkynes&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is not this definition incomplete? I can imagine scenarios where we have a terminal triple bond that does not reside between carbons 1 and 2 according to IUPAC rules. E.g., $\ce{CH2=CH-C#CH}$. Here, the terminal triple bond is between carbons 3 and 4 (according to IUPAC rules, doubles bonds take priority when numbering).&lt;/p&gt;&#xA;" OwnerUserId="1565" LastEditorUserId="24052" LastEditDate="2016-03-11T22:20:12.693" LastActivityDate="2016-08-31T15:20:30.973" Title="Correct definition of terminal alkynes?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4855" PostTypeId="2" ParentId="4854" CreationDate="2013-04-30T00:42:40.863" Score="8" Body="&lt;p&gt;You are correct about the absolute numbering of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=CH-C#CH}$&lt;/span&gt;. By the IUPAC priority rules, the alkene has the lower absolute numbering.&lt;/p&gt;&#xA;&lt;p&gt;However, numbering can also be &lt;em&gt;relative&lt;/em&gt;. In your case, the relative numbering 1,2 refer to either of the following:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;relative position from the end of the chain, where the chain end is #1&lt;/li&gt;&#xA;&lt;li&gt;relative position from the functional group you care about (where that group is #1)&lt;/li&gt;&#xA;&lt;li&gt;suggesting that a reaction occurs on neighboring positions in a molecule.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;That being said, the definition given by Brown &lt;em&gt;et al.&lt;/em&gt; is imprecise. A better definition is one of the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A terminal alkyne is an alkyne in which one (or both) of the two alkyne carbons is at the end of a chain.&lt;/p&gt;&#xA;&lt;p&gt;A terminal alkyne is an alkyne &lt;span class=&quot;math-container&quot;&gt;$\ce{RC#CR}$&lt;/span&gt; in which one or both &lt;span class=&quot;math-container&quot;&gt;$\ce{R}=\ce{H}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-04-30T00:42:40.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4856" PostTypeId="2" ParentId="4854" CreationDate="2013-04-30T00:43:11.310" Score="4" Body="&lt;p&gt;You're right; the definition in your book is simplistic. A terminal alkyne would be better defined as &quot;an alkyne with the formula $\ce{RC_2H}$&quot;, differentiating from an internal alkyne, which would have the formula $\ce{RC_2R'}$ and the added stipulation that neither R nor R' can be a hydrogen. Equivalently stated as in Wikipedia, an internal alkyne has carbon substitutents on all acetylenic carbons, and a terminal alkyne is therefore not an internal alkyne because at least one acetylenic carbon has a hydrogen bonded to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example formula you cite is vinylacetylene, systematic name butenyne or 1-buten-3-yne (here's your IUPAC nomenclature rules favoring double bonds at work; otherwise it'd be ethynylbutene or similar). It is indeed an example of a terminal alkyne, having the exploded formula (C&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;3&lt;/sub&gt;)C&lt;sub&gt;2&lt;/sub&gt;H.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2013-04-30T00:43:11.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4857" PostTypeId="1" AcceptedAnswerId="4859" CreationDate="2013-04-30T01:19:18.963" Score="11" ViewCount="2966" Body="&lt;p&gt;Maybe this is better in English.SE, but I think I'm more likely to get better responses here in the &quot;industry&quot; site. &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4854/correct-definition-of-terminal-alkynes&quot;&gt;This post&lt;/a&gt; got me thinking; how do you pronounce &quot;alkyne&quot; (a hydrocarbon with a carbon-carbon &lt;em&gt;triple&lt;/em&gt; bond) differently enough from &quot;alkene&quot; (a hydrocarbon with a carbon-carbon &lt;em&gt;double&lt;/em&gt; bond) to differentiate the two in spoken conversation? They're obviously spelled differently and so can be separated in the literature, but I can't find a satisfactory pronunciation for either that isn't &quot;al-&lt;em&gt;keen&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2013-05-01T02:35:27.917" Title="Pronunciation of &quot;alkene&quot; vs. &quot;alkyne&quot;" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4858" PostTypeId="2" ParentId="4857" CreationDate="2013-04-30T02:05:08.647" Score="1" Body="&lt;p&gt;I've always heard it pronounced with a long I sound: &quot;al-kine.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-04-30T02:05:08.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4859" PostTypeId="2" ParentId="4857" CreationDate="2013-04-30T02:16:53.777" Score="12" Body="&lt;p&gt;English isn't my first language, and in truth I'm always doubtful about the pronunciation of words in chemistry. Sometimes it's possible to find videos where the words are mentioned, for example &lt;a href=&quot;https://www.youtube.com/watch?feature=player_detailpage&amp;amp;v=aaS9IyfB6wo#t=37s&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.youtube.com/watch?v=HbDWBeRJboE&amp;amp;feature=player_detailpage&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.youtube.com/watch?v=mqvn9WOg5Z8&amp;amp;feature=player_detailpage&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.youtube.com/watch?v=l6m7zOto9oU&quot;&gt;here&lt;/a&gt; (several examples of alkenes and alkynes being pronounced in this last one). That's what I use, al-keen and al-kyne (ky as in kite or sky). &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2013-04-30T02:23:44.130" LastActivityDate="2013-04-30T02:23:44.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4860" PostTypeId="2" ParentId="4639" CreationDate="2013-04-30T02:40:50.150" Score="0" Body="&lt;p&gt;Balancing a chemical equations is fundamentals of chemistry. There is one recommended way to balance the equations is that you balance the Oxygen in the equation first, do not use fractions only natural numbers are allowed.&lt;/p&gt;&#xA;" OwnerUserId="1556" LastActivityDate="2013-04-30T02:40:50.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4861" PostTypeId="2" ParentId="4278" CreationDate="2013-04-30T02:46:39.970" Score="5" Body="&lt;p&gt;Simply, the odorant particles/molecules never actually enter the cell through endocytosis. Instead, these olfactory receptors, belonging to the G-protein-coupled receptor family, are transmembrane receptors whose structure passes through the cellular wall (several times actually) to functional complexes on the other side, and they use simple and reversible ligand binding as their primary activation mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odorant particles are dissolved in the mucous lining the nasal cavity, and in this form come into contact with the receptors in the cell membrane of the epithelial cells. These bind as ligand complexes to receptor sites, triggering a change in the receptor's structure which in turn causes the coupled G-protein to release its bound guanosine diphosphate, and instead bind guanosine triphosphate from the interior cell environment. When that happens, the G-protein is &quot;activated&quot;; it decouples from the receptor, splits in half into Ga and Gbg subunits, and those two halves bind to effector enzymes which trigger the cell signalling. As a result of the protein decoupling, the charge balance of the receptor changes, releasing the complexed ligand from its binding to the receptor, back into the extracellular environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reverse all this, an enzyme called RGS (Regulator of G-protein Signalling) binds to the Ga subunit, triggering hydrogenation of the GTP molecule it still has bound into GDP, causing the protein's subunits to recombine and recouple to the receptor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The released ligand, meanwhile, is washed away again by the mucous flow from the Bowman's glands. It may complex with other receptors along the way, but eventually it makes its way past the nasal membrane to the pharynx at the back of the throat, where the mucous is swallowed and the odorant broken down in the stomach. This is the final fate of virtually all odorant agonists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A select few, however, bind more strongly, and are not released by the structure changes of the receptor. When this happens, the receptor is stuck in the &quot;on&quot; position. One of two things then happens; either the receptor continues to activate, causing neural impulses which your brain eventually ignores (odor fatigue), or mechanisms inside the cell notice the faulty receptor, bring it into the cell in a process similar to endocytosis, and attempt to break down its components and fix the damage to reuse it. More often than not, the bound chemical ends up poisoning the cell by binding to something more vital than an ordinary G-protein receptor, eventually triggering programmed cell death. That's fine; the basal cells behind the surface of the olfactory epithelium divide fast enough to replenish every cell in the nasal membrane about once every two days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Wikipedia - &lt;a href=&quot;http://en.wikipedia.org/wiki/Olfactory_epithelium&quot; rel=&quot;nofollow&quot;&gt;Olfactory epithelium&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Olfactory_receptor&quot; rel=&quot;nofollow&quot;&gt;Olfactory receptor&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/G_protein-coupled_receptor&quot; rel=&quot;nofollow&quot;&gt;G protein-coupled receptor&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-06T21:29:19.737" LastActivityDate="2013-05-06T21:29:19.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4862" PostTypeId="1" AcceptedAnswerId="4868" CreationDate="2013-04-30T14:14:53.790" Score="2" ViewCount="37131" Body="&lt;p&gt;JB Weld, a brand name of steel-reinforced epoxy, makes claims of being the &quot;world's strongest bond&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the maximum pressure it can withstand, and what is it about the interaction between the steel and the epoxy that makes it so strong?&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="73" LastEditDate="2013-05-02T02:44:41.493" LastActivityDate="2019-09-30T15:21:12.170" Title="What is the maximum pressure that JB Weld can withstand?" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4863" PostTypeId="2" ParentId="4857" CreationDate="2013-05-01T02:35:27.917" Score="0" Body="&lt;p&gt;Perhaps a different way of thinking about the molecule: Very rarely have I heard people using these nouns. Certainly in an organic chemistry world, it is much more likely to talk about the functional group, so you would actually talk about the c-c double bond, or the c-c triple bond, or the carboxyl group... You get the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this doesn't really address the pronounciation issue, but think of it this way:&#xA;An alkene IS an alkene at the point where there is a c=c bond, the rest of the molecule is kinda unimportant, and it is really this functional group that makes it chemically interesting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So talk explicitly about the bond of interest, not the whole molecule (at least when you are talking about certain properties you want to exploit that relate directly to this functional group).&lt;/p&gt;&#xA;" OwnerUserId="1495" LastActivityDate="2013-05-01T02:35:27.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4864" PostTypeId="2" ParentId="4818" CreationDate="2013-05-01T14:05:24.483" Score="1" Body="&lt;p&gt;I am not sure that what you claim to want to do is what you actually want to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cholesterol hemisuccinate is prepared from cholesterol, but not by converting the alcohol functional group into a carboxyl group. The alcohol oxygen becomes one of the oxygen atoms in the ester, but not the carboxyl oxygen. Cholesterol hemisuccinate can be prepared in the lab probably from cholesterol and succinic anhydride in the presence of a non-nucleophilic base (like pyridine or triethyl amine), which can act as a proton shuttle. An acidic workup will be needed to remove the base and protonate the carboxylic acid. The scheme below shows this reaction (sans acidic workup). The alcohol oxygen is highlighted in red.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oiS2H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An identical reaction could be done with your molecule, diplopterol. Dissolve both (1 eq) diplopterol and succinic anhydride (1.1 eq.) in dichloromethane or diethyl ether (or any low polarity aprotric solvent you like). Cool on an ice bath. With stirring, add pyridine (1.1 eq.) dropwise. Let the reaction warm to room temperature, probably about 2 hours. The product will likely precipitate as the pyridinium salt. Check the progress by TLC or some other means. The product will probably be more polar than the reactant. Work up by adding dilute HCl (probably 0.1 M). Extract the aqueous layer with your organic solvent of choice. Combine the organic layers. Wash with brine. Dry over magnesium or sodium sulfate. Remove the solvent by evaporation (rotary being preferred), and purify by chromatography if necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if the product does precipitate at the pyridium salt, collect the salt by filtration, wash the precipitate with ether and then with cold ethanol (will wash away the reactants). Then, add the precipitate to the 0.1 M HCl and extract with ether if necessary. Further purification is probably not needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q21XN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-01T14:05:24.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4865" PostTypeId="1" AcceptedAnswerId="4866" CreationDate="2013-05-01T16:22:34.080" Score="1" ViewCount="5624" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_grease&quot; rel=&quot;nofollow&quot;&gt;Thermal Grease&lt;/a&gt; is used to remove the air gaps between CPU and heatsink. However, computing aside, what would determine the max temp that the grease can handle before it burns or breaks down?&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastActivityDate="2013-05-01T21:19:11.427" Title="How hot can Thermal Grease get?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4866" PostTypeId="2" ParentId="4865" CreationDate="2013-05-01T21:19:11.427" Score="2" Body="&lt;p&gt;Since most thermal greases are ceramic and/or metal suspended in a silicone or organic oil, the temperature the grease remains stable at will likely primarily be determined by the properties of that oil -- either its ease of evaporation, or decomposition temperature (or, I suppose, its flashpoint, but that seems unlikely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Ben Norris says in the comment above, your best bet is to contact the manufacturer or check their data sheets: for common CPU uses, most manufacturers will give you stable and extreme temperature ranges which are likely to be fairly conservative &lt;a href=&quot;http://www.arcticsilver.com/as5.htm&quot; rel=&quot;nofollow&quot;&gt;(for example, here)&lt;/a&gt;, and they may be able to help with more esoteric data if you're using it for other purposes.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-05-01T21:19:11.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4867" PostTypeId="1" CreationDate="2013-05-02T08:07:18.660" Score="2" ViewCount="2400" Body="&lt;p&gt;I'm no chemist, but much of my work revolves around biochar.  I've always understood it with the following rough approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH2O (dry plant matter) + heat -&amp;gt; black carbon + H2 + CO + CH4 + hydrocarbons}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the above a decent approximation?  In what ways is it misleading (as all approximations always are)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A colleague recently told me that methane is only formed in this reaction when there is water vapor present -- he said that it mixes with CO at high temperature to form the methane.  Is this true?  Will one not get any methane from oven-dried biomass that is exposed to heat in the absence of oxygen?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1570" LastEditorUserId="30836" LastEditDate="2020-04-05T21:02:25.710" LastActivityDate="2020-04-05T21:02:25.710" Title="Better understanding the pyrolysis reaction" Tags="&lt;organic-chemistry&gt;&lt;heat&gt;&lt;hydrocarbons&gt;&lt;pyrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4868" PostTypeId="2" ParentId="4862" CreationDate="2013-05-02T11:47:33.273" Score="4" Body="&lt;p&gt;This is only a partial answer, addressing the maximum pressure a JB weld can withstand. I am still thinking about the reason for the strength so I will leave it to the 'real' chemists among you for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.repairproducts.co.uk/page42.htm&quot; rel=&quot;nofollow&quot;&gt;Repair Products&lt;/a&gt; the JB weld in fully hardened state has the following properties (in psi):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tensile Strength:   3960 &#xA;Adhesion:           1800 &#xA;Flex Strength:      7320 &#xA;Tensile Lap Shear:  1040 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow&quot;&gt;Tensile strength&lt;/a&gt; is simply the pressure needed to break the material by pulling it apart. &lt;a href=&quot;http://en.wikipedia.org/wiki/Flexural_strength&quot; rel=&quot;nofollow&quot;&gt;Flex strength&lt;/a&gt; is the pressure needed to bend the material which, in case of a perfectly homogeneous sample would equal the tensile strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adhesion is the pressure needed to pull the bond from the surface when pulling perpendicular to the surface (parallel to the bond). The &lt;a href=&quot;http://info.admet.com/specifications/bid/43089/How-to-Perform-an-Adhesive-Lap-Joint-Shear-Strength-Test-ASTM-D1002&quot; rel=&quot;nofollow&quot;&gt;tensile lap shear&lt;/a&gt; is also an 'adhesion', but measured when pulling parallel to the surface (perpendicular to the bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the latter is the lowest it means that this would be the maximum pressure the JB weld could withstand, but only if it is applied exactly perpendicular to the bond. If the direction is in a different direction than you will get some mixture between the adhesion and the tensile lap shear pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is clear from these numbers is that the bond itself will not break but rather it will detach from the materials it is keeping together.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-05-02T11:47:33.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4869" PostTypeId="1" CreationDate="2013-05-02T12:39:51.880" Score="1" ViewCount="578" Body="&lt;p&gt;For example, $\ce{CH3CHBrCH3}$ reacts with $\ce{:NH_3}$ to make $\ce{CH3CH(NH3^+)CH3 +:Br-}$,$$\ce{CH3CHBrCH3 \rightarrow CH3CH(NH3^+)CH3 +:Br-}$$then reacting again with $\ce{:NH_3}$  to make $\ce{CH3CH(NH2)CH3 +:Br^- +NH4^+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH_3 + CH3CH(NH3^+)CH3 +:Br- \rightarrow CH3CH(NH2)CH3 +:Br^- +NH4^+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: why does the last step not produce $\ce{CH3CH(NH2)CH3 +HBr}$? That is, why is Ammonia a better nucleophile than the halide ion, despite having a lower partial negative charge (the halide has a $-1$ charge)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: a suggestion made by a teacher is that my proposed step &lt;em&gt;does&lt;/em&gt; in fact happen, it's just that it occurs as an intermediary to the last step.  Apparently, it's very favourable to have a Hydrogen Halide and Ammonia acid-base reaction. Is he correct, and if so, why is this?&lt;/p&gt;&#xA;" OwnerUserId="1571" LastEditorUserId="10185" LastEditDate="2015-10-09T16:46:44.753" LastActivityDate="2015-10-09T16:46:44.753" Title="Why does nucleophilic substitution of a haloalkane with ammonia produce ammonium, not a hydrogen halide?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4871" PostTypeId="1" CreationDate="2013-05-02T21:37:28.520" Score="0" ViewCount="377" Body="&lt;p&gt;I'm studying for my chemistry final and having a hard time with this question on one of the practice exams. Here's the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $20$ degreees Celsius, Henry's Law constant for Ar is half of that for Kr. If a gaseous Ar/Kr mixture containing 20% molar fraction of Kr is placed over water at $20$ degrees Celcius, what is the ratio of concentrations of Ar to Kr in water?&lt;/p&gt;&#xA;" OwnerUserId="1551" LastEditorUserId="84" LastEditDate="2013-08-31T07:51:20.673" LastActivityDate="2013-08-31T07:51:20.673" Title="Ratio of concentrations" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4872" PostTypeId="2" ParentId="4869" CreationDate="2013-05-03T01:15:01.833" Score="6" Body="&lt;p&gt;The answer to this is quite similar to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4750/reactions-of-sodium-halides-with-sulphuric-acid/4753#4753&quot;&gt;one I gave recently&lt;/a&gt; for another set of reactions with hydrogen halides as possible products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proposed last step is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH(NH3^+)CH3 +Br- \rightarrow CH3CH(NH2)CH3 +HBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, you have a weak acid $\ce{(CH3CH(NH3^+)CH3)}$ and a very weak base $\ce{(Br^{-})}$ reacting with each other to turn into a weak base $\ce{(CH3CH(NH2)CH3)}$ and a very strong acid $\ce{(HBr^)}$. This is highly unfavourable from a thermodynamic sense, because you're trying to inject a lot of chemical potential into the molecules at no cost. Though your teacher suggests it might happen, I wouldn't be very comfortable saying that, if only for the fact that there's another, easy option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the previous argument is slightly circular, so it's best to explain &lt;em&gt;why&lt;/em&gt; $\ce{HBr}$ is such a strong acid / $\ce{Br^-}$ is such a weak base. As you mention, $\ce{Br^-}$ has a negative charge, and one might expect that to mean it is a better base. However, that affirmation is only useful as a rule of thumb when comparing substances of different structure, being strictly valid solely when looking at a single sequences of related structures (such as $\ce{PO_4^{3-}}$/$\ce{HPO_4^{2-}}$/$\ce{H_2PO_4^{-}}$). The strength of a base is determined by several factors, such as charge localization, bond strength between base and a proton, steric hindrance, and strength of interaction of the protonated/deprotonated species with the solvent (if it exists). There are &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_sponge&quot; rel=&quot;nofollow noreferrer&quot;&gt;very strong neutral bases&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;very weak negatively charged bases ($\ce{SbF_6^-}$)&lt;/a&gt; known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br^-}$ is a poor base because, while it has a net negative charge, that extra electron is bound to a somewhat electronegative atom, thereby stabilizing the charge. Not only that, but the bromide anion is quite large, so the charge is &quot;spread out&quot; over a significant volume and would attract a proton less, in some sense. Meanwhile, though $\ce{NH_3}$ has no extra charge, and even though the central nitrogen atom is more electronegative than bromine, the non-bonding pair of valence electrons in $\ce{NH_3}$ is highly localized, especially due to the fact that second period p-block elements are anomalously small. This makes $\ce{NH_3}$ an attractive target for a proton.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2013-05-03T01:20:33.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4873" PostTypeId="2" ParentId="4331" CreationDate="2013-05-03T04:19:05.960" Score="0" Body="&lt;p&gt;It might be asymmetry potential. &#xA;Asymmetry potentials are due to differences between the electrodes themselves, and are not due &#xA;to differences in potential within the NaCl solution.&lt;/p&gt;&#xA;" OwnerUserId="1575" LastEditorUserId="22" LastEditDate="2013-05-03T15:38:06.617" LastActivityDate="2013-05-03T15:38:06.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4874" PostTypeId="2" ParentId="4871" CreationDate="2013-05-03T04:20:46.790" Score="2" Body="&lt;p&gt;Henry's Law is mathematically expressed by $$p=k_Hc$$ where $p$ is the partial pressure above the liquid, $k_H$ is the Henry constant for the specific gas at the given temperature, and $c$ is the concentration of gas in liquid. If you set up one equation for each gas and divide the two, you get $$\frac{p_{Ar}}{p_{Kr}}=\frac{k_{H, Ar}}{k_{H, Kr}}\frac{c_{Ar}}{c_{Kr}}.$$ You are given both the ratio of partial pressures and the ratio of the Henry's Law constants; the ration of concentrations should be easily calculated from that point.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-05-03T04:20:46.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4875" PostTypeId="1" AcceptedAnswerId="4876" CreationDate="2013-05-03T10:00:06.497" Score="6" ViewCount="355" Body="&lt;p&gt;Why won't $\ce{Ni^2+}$ form a complex with $\ce{Cl-}$, while the ions $\ce{Cu^2+}$ and   $\ce{Co^2+}$ form the complexes $\ce{[CuCl4]^2-}$ and $\ce{[CoCl4]^2-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the HSAB theory, $\ce{Cl-}$ is considered a hard/borderline base. So I figured the strongest complex would be the one in which the center ion is the hardest acid. The $\ce{Co^2+}$ ion is the largest and the least electronegative. So it should be the softest acid (although they are all considered borderline). It forms a complex with $\ce{Cl-}$. $\ce{Ni^2+}$ should be the hardest acid, and yet it does not form a complex with $\ce{Cl-}$ at all. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www2.chemistry.msu.edu/courses/CEM812/HSAB_Theory.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.chemistry.msu.edu/courses/CEM812/HSAB_Theory.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;R. Pearson, Chemical hardness and density functional theory, J. Chem. Sci., Vol. 117, No. 5, September 2005, pp. 369–377&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1577" LastEditorUserId="9961" LastEditDate="2015-10-02T23:10:35.827" LastActivityDate="2015-10-02T23:10:35.827" Title="Strongest chloride complex as predicted by HSAB theory" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;hsab&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4876" PostTypeId="2" ParentId="4875" CreationDate="2013-05-03T10:58:19.377" Score="10" Body="&lt;p&gt;The link to your handout explains why $\ce{Fe^{3+}}$ (and it is $\ce{Fe^{3+}}$ in $\ce{FeCl4^{-}}$) is a harder acid than $\ce{Co^{3+}}$ or $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hard acids (in context, HA) are characterized by (s,f blocks, &lt;strong&gt;left side of d block&lt;/strong&gt; in higher OS's)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Iron is the furthest to the left of the four metals you mention. Thus, it has fewer electrons in its $d$ shell to balance the positive charge:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Fe^{3+}}$ is $\ce{[Ar]}3d^5$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Co^{3+}}$ is $\ce{[Ar]}3d^6$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Ni^{2+}}$ is $\ce{[Ar]}3d^8$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cu^{2+}}$ is $\ce{[Ar]}3d^9$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Low electronegativity (χ) of the acidic atom. A value in the range 0.7-1.6 is typical of hard acids;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Iron has the lowest &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements&quot;&gt;electronegativity&lt;/a&gt; of the four metals:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Fe}$ 1.83&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Co}$ 1.88 &lt;/li&gt;&#xA;&lt;li&gt;$\ce{Ni}$ 1.91 &lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cu}$ 1.90&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Iron&lt;/strong&gt; is the hardest Lewis acid of the metal cations you list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for $\ce{Ni^{2+}}$, it &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel%28II%29_chloride#Coordination_chemistry&quot;&gt;does form a complex&lt;/a&gt; $\ce{NiCl4^{2-}}$ with $\ce{Cl-}$. Here is a &lt;a href=&quot;https://www.google.com/search?q=nicl4&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla%3aen-US%3aofficial&amp;amp;client=firefox-a&quot;&gt;Google search&lt;/a&gt; for it.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-03T10:58:19.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4877" PostTypeId="1" AcceptedAnswerId="4915" CreationDate="2013-05-03T12:08:02.953" Score="95" ViewCount="260560" Body="&lt;p&gt;In the TV show &quot;Breaking Bad&quot;, Walter White frequently gets rid of people who get in his way by submerging them in a plastic container full of hydrofluoric acid. This, at least in the TV show, completely dissolves the body leaving nothing but a red sludge behind at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it actually possible to dispose of a body with hydrofluoric acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hydrofluoric acid wouldn't work, are there any acids corrosive enough to achieve the stated effect from the show?&lt;/p&gt;&#xA;" OwnerDisplayName="user1579" LastEditorUserId="7448" LastEditDate="2015-08-30T10:03:14.083" LastActivityDate="2017-04-08T22:25:51.550" Title="Is it actually possible to dispose of a body with hydrofluoric acid?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="4879" PostTypeId="2" ParentId="4877" CreationDate="2013-05-03T19:27:20.837" Score="8" Body="&lt;p&gt;This subject is touched upon &lt;a href=&quot;http://chemistry.about.com/b/2008/02/12/hydrofluoric-acid-breaking-bad.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and explained in detail on &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_hydrolysis_%28death_custom%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, it is impractical to dispose of a body using an &lt;em&gt;acid&lt;/em&gt;, though there exists a process called &lt;strong&gt;alkaline hydrolysis&lt;/strong&gt;, which, as its name implies, makes use of a base to degrade the materials comprising the human body to a liquid state within a matter of hours.&lt;/p&gt;&#xA;" OwnerUserId="1202" LastEditorUserId="402" LastEditDate="2017-04-08T22:24:57.727" LastActivityDate="2017-04-08T22:24:57.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4881" PostTypeId="1" AcceptedAnswerId="4882" CreationDate="2013-05-03T23:31:36.457" Score="12" ViewCount="632" Body="&lt;p&gt;&lt;a href=&quot;http://xkcd.com/1162/&quot; rel=&quot;nofollow noreferrer&quot;&gt;XKCD #1162&lt;/a&gt; piqued my curiosity. Obviously, there's nothing we generally use as a chemical fuel that can match the energy density of uranium fission. But I wondered what it would take, hypothetically, to match it mole-for-mole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;a href=&quot;http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;bond enthalpy charts&lt;/a&gt;, I see that the carbon-oxygen triple bond has the highest enthalpy of formation (the energy required to break it and form something else, and by extension the energy produced when this bond is formed, minus the energy required to break the previous ones) at &lt;span class=&quot;math-container&quot;&gt;$\pu{1072 kJ mol-1}$&lt;/span&gt;. However, the carbon-oxygen single bond has an EoF of &lt;span class=&quot;math-container&quot;&gt;$\pu{358 kJ mol-1}$&lt;/span&gt;, and other bonds likely to be involved (&lt;span class=&quot;math-container&quot;&gt;$\ce{H-C}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{413 kJ mol-1}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{H-O}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{463 kJ mol-1}$&lt;/span&gt;, etc.) have higher bond enthalpies, meaning the formation of carbon monoxide by combustion of an organic molecule is very likely a losing proposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen, on the other hand, has an infamously high bond energy ratio; breaking a single nitrogen bond costs only &lt;span class=&quot;math-container&quot;&gt;$\pu{163 kJ mol-1}$&lt;/span&gt;, while forming a triple bond releases &lt;span class=&quot;math-container&quot;&gt;$\pu{941 kJ mol-1}$&lt;/span&gt;. Nitrogen single bonds, especially in high percentages of the total molecular mass, are notoriously explosive; the azide anion, covalently bonded, forms the backbone of the eminent &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2011/02/25/things_i_wont_work_with_chlorine_azide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Lowe's &lt;em&gt;Things I Won't Work With&lt;/em&gt;&lt;/a&gt;. The nitro group (&lt;span class=&quot;math-container&quot;&gt;$\ce{RNO2}$&lt;/span&gt;), well-known in explosives, can be made even more potent by bonding it to another nitrogen to form a nitramine, and these are central to the well-known explosive cyclotrimethylenetrinitramine (better known in pure form as RDX, and with added plasticizers as C-4) and its big brother, HMX (cyclotetramethylenetetranitranime).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's consider a theoretical molecule composed of only single-bonded nitrogen (eek). &lt;span class=&quot;math-container&quot;&gt;$\ce{N4}$&lt;/span&gt;, tetranitrogen, has been synthesized, but in a form theorized to be more like two pairs of double-bonded nitrogens that are joined by two single bonds (kind of hard to tell, since it only exists at very high temperature and for less than a microsecond at a time). Let's instead bond each nitrogen to each other one, forming an isohedron. When that structure reverts to diatomic nitrogen (and it will, oh, you betcha), we can expect a heat of detonation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$4\cdot\pu{163 kJ mol-1} - 2\cdot\pu{941 kJ mol-1} = \pu{-1230 kJ mol-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a pretty aggressive exotherm; RDX has a detonation energy of 1118 kJ/mol, and its molecular mass is four times that of our theoretical &quot;isohedral tetranitrogen&quot;. RDX is as heavy as it is for stability; high molecular weights sacrifice pure energy density for a structure that won't detonate given the slightest excuse (or even no excuse at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, bring in the champion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^{235}_{92}U + ^1_0n -&amp;gt; ^{142}_{56}Ba + ^{91}_{36}Kr + 3 ^1_0n}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in the atomic chaos of a nuclear reaction, a lot of other products are possible, but this is one of the most common). The masses of the reactants total &lt;span class=&quot;math-container&quot;&gt;$\pu{236.0526 Da}$&lt;/span&gt;, and the masses of the products, &lt;span class=&quot;math-container&quot;&gt;$\pu{235.8658 Da}$&lt;/span&gt;, for a delta of mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1868 Da}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{3.102e-28 kg}$&lt;/span&gt;). &lt;span class=&quot;math-container&quot;&gt;$E = mc^2$&lt;/span&gt;, so the energy produced by one uranium atom splitting this way is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E = \pu{3.102e-28 kg}\cdot(\pu{3.00e8 m s-1})^2 = \pu{2.79e-12 J}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which when multiplied by Avogadro's number and converted to kilojoules, gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{2.79e-12 J}\cdot\pu{6.022e23 mol-1} = \pu{1.68e8 kJ mol-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 168 &lt;em&gt;million&lt;/em&gt; kilojoules per mole available by fissioning uranium-235.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that double the nitrogen equals double the heat of reaction, in order to have the same heat of decomposition, mole-for-mole, our nitrogen molecule would instead have to be an isometric &quot;sheet&quot; of single-bonded nitrogen, or a geodesic dome (buckyazide?), consisting of &lt;span class=&quot;math-container&quot;&gt;$546340$&lt;/span&gt; nitrogen atoms single-bonded to each other (a concept that, if ever seriously proposed, should have any student of chemistry falling out of their chair either laughing or horrified). One mole of this material would weigh &lt;span class=&quot;math-container&quot;&gt;$\pu{14.0067 g mol-1}\cdot 546340 = \pu{7652.42 kg mol-1} \approx \pu{8.4 t}$&lt;/span&gt;. One mole of U-235 weighs &lt;span class=&quot;math-container&quot;&gt;$\pu{235 g}$&lt;/span&gt;, just a little more than half a pound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long live the king.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;(thank you for reading this far)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the math look right? I've done it three times and come up with three different answers for the heat of detonation of my hypothetical isohetral tetranitrogen, and thus the size and weight of the resulting molecule, possibly based on the use of different enthalpy figures (I've seen the single bond energy of formation of nitrogen to be as little as &lt;span class=&quot;math-container&quot;&gt;$\pu{158 kJ mol-1}$&lt;/span&gt;, which produces a heat of detonation of &lt;span class=&quot;math-container&quot;&gt;$\pu{1250 kJ mol-1}$&lt;/span&gt;, meaning, all other things being equal, a required molecule of only &lt;span class=&quot;math-container&quot;&gt;$537600$&lt;/span&gt; nitrogen atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can anyone come up with something that would be more energetic to go up against the champ? There are molecules that have been successfully created in the lab, such as octanitrocubane, that are more powerful than RDX while being usably shock-insensitive, but are prohibitively expensive to manufacture.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="41328" LastEditDate="2019-01-28T11:27:12.917" LastActivityDate="2019-01-28T11:27:12.917" Title="A theoretical molecule that could match uranium mole-for-mole" Tags="&lt;bond&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4882" PostTypeId="2" ParentId="4881" CreationDate="2013-05-04T01:53:54.107" Score="15" Body="&lt;p&gt;Getting a large exotherm per mole of molecules is a bit of a bad cheat if you're allowed to make molecules arbitrarily big! Energy per mass is the only honest way to do it (or energy per mole of atoms would work too I guess). However, there really is no chance to chemically match a nuclear exotherm; nuclear energies are simply in a class of their own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically all chemical energy is merely a function of electric* repulsion and attraction on distance scales from around a picometre (being very generous, as that is quite small compared to the dimensions of any valence orbital) to perhaps a few micrometres (to include long-distance ion interactions). However, nuclear energies are based on electric attractions and repulsions on distances scales of a few &lt;em&gt;femtometres&lt;/em&gt; or less. This makes quite a difference when you remember that electrostatic potentials go as &lt;span class=&quot;math-container&quot;&gt;$1/r^n$&lt;/span&gt;! Not only that, but the nucleus also plays &lt;em&gt;a whole other ball game&lt;/em&gt;, displaying strong nuclear potentials. It is commonly mentioned that, apples to apples, the strong nuclear force is about &lt;a href=&quot;http://scienceworld.wolfram.com/physics/FundamentalForces.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;three orders of magnitude stronger&lt;/a&gt; than the electrostatic force. Therefore, no chemical reaction involving structures stable at STP will ever reach the feet of nuclear energies. There is probably no &quot;reasonable&quot; suggestion like yours that could do better by a factor of &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;, much less a factor of &lt;span class=&quot;math-container&quot;&gt;$10^5$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can we do about it? Well, if interaction distances being too large is part of the problem, one way to cheat is to simply add pressure, an &lt;em&gt;unreasonable&lt;/em&gt; amount of pressure. This effectively brings atoms closer and closer together, so you climb the &lt;span class=&quot;math-container&quot;&gt;$1/r^n$&lt;/span&gt; potential for each atom. Chemistry gets pretty weird at ultrahigh pressures. For one, lots of things (perhaps everything?) seem to like turning into a metal. With all those orbitals crammed up against each other, eventually bonds overlap extensively and you get bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going, and the bands start to overlap, until there's no gap for electron excitation, and room temperature thermal energy can freely knock electrons all over the place. You talk of making huge polynitrogen molecules. Well, &lt;a href=&quot;http://iopscience.iop.org/1367-2630/15/1/013010&quot; rel=&quot;nofollow noreferrer&quot;&gt;metallic nitrogen&lt;/a&gt; would be the natural limit of that. Apparently even at pressures of the order of a terapascal, metallic nitrogen has &lt;a href=&quot;http://arxiv.org/abs/1210.4358&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N-N}$&lt;/span&gt; single bonds around &lt;span class=&quot;math-container&quot;&gt;$\pu{110 pm}$&lt;/span&gt;&lt;/a&gt;, which is only a little smaller than &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrazine#Molecular_structure_and_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;STP &lt;span class=&quot;math-container&quot;&gt;$\ce{N-N}$&lt;/span&gt; bonds&lt;/a&gt;, at around &lt;span class=&quot;math-container&quot;&gt;$\pu{145 pm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this would already create quite a bang if pressure were suddenly relieved. But how much pressure would it take to get down to N-N distances of &lt;span class=&quot;math-container&quot;&gt;$\pu{10 pm}$&lt;/span&gt;? And &lt;span class=&quot;math-container&quot;&gt;$\pu{1 pm}$&lt;/span&gt;? How much energy would be pent up per gram of whatever that material is? It's a crazy suggestion, but it takes beyond crazy to get electrons to carry energy comparable to nucleons. Keep going and at some point you'll arguably leave even the theoretical domain of chemistry and reach astrophysically relevant &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter#Electron_degeneracy&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron degenerate matter&lt;/a&gt;, the stuff in the core of relatively light dead stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go even further, and eventually electronic potentials &lt;em&gt;finally&lt;/em&gt; become comparable to nuclear ones, and so it actually becomes favourable for nuclei to gobble electrons up. &lt;a href=&quot;https://en.wikipedia.org/wiki/Degenerate_matter#Neutron_degeneracy&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is the matter inside a neutron star&lt;/a&gt;. The fact that it takes the mass of a Sun in a sphere a few kilometres wide to press electron energies into the realm of nuclear ones should be a sobering thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Magnetism is rarely, if ever, the dominant source of electromagnetic potential energy in chemical systems. Magnetic charges do not exist (except maybe for &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_monopole&quot; rel=&quot;nofollow noreferrer&quot;&gt;monopoles&lt;/a&gt;, but those aren't chemically relevant!), so all magnetism is induced by electric charges in motion (well, spin magnetism is a little more complicated). As far as I understand, which isn't much, magnetism induced from electric charges takes a &lt;span class=&quot;math-container&quot;&gt;$1/c^2$&lt;/span&gt; hit in the equations of electromagnetism, and that's hard to recover from! &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="41328" LastEditDate="2019-01-28T11:26:48.607" LastActivityDate="2019-01-28T11:26:48.607" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="4883" PostTypeId="2" ParentId="4792" CreationDate="2013-05-04T04:17:35.813" Score="2" Body="&lt;p&gt;As far as obtaining the density without first computing the wave function--there's almost no way to do it with any accuracy for anything &quot;realistic.&quot;  The easiest way to obtain the electron density is to perform an ab initio electronic structure theory calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you can use the &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;nofollow&quot;&gt;GAMESS&lt;/a&gt; quantum chemistry program to perform a Hartree-Fock or DFT calculation, and then have it write the resulting wave function to a AIMPAC input deck (WFN file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aimpac/&quot; rel=&quot;nofollow&quot;&gt;AIMPAC&lt;/a&gt; program is the original version of QTAIM, written by the late Richard Bader and his research group. There are more modern versions, such as Todd Keith's &lt;a href=&quot;http://aim.tkgristmill.com&quot; rel=&quot;nofollow&quot;&gt;AIMALL&lt;/a&gt; which is certainly more powerful and user-friendly than the original AIMPAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably perform some calculation on fifty or so atoms, assuming a typical organic molecule. It all depends on how fast your computer(s) is and how many basis functions you use.  Constructing molecular graphs with AIM is quite fast.  Where things get expensive is when you start calculating properties of atomic basins, as these are high-accuracy integrations over &quot;irregular&quot; (but well-defined) regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to ruin your day, but the connectivities in your SDF file are probably of little value.  There are no such things as &quot;single&quot;/&quot;double&quot;/&quot;triple&quot; bonds, at least in the context of QTAIM.  Either they are bonded or they are not. Bonding appears when the interaction between two atoms lowers the total energy of the system. QTAIM will give you that information unambiguously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the answers are surprising, but it is guaranteed that they are more &quot;correct&quot; than what we unfortunately call &quot;chemical intuition&quot; and/or simple proximity of two atoms.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-04T04:33:05.937" LastActivityDate="2013-05-04T04:33:05.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4884" PostTypeId="2" ParentId="4810" CreationDate="2013-05-04T04:55:28.490" Score="4" Body="&lt;p&gt;A proper answer to your question would require at least a semester of Physical Chemistry. I would like to touch on a couple of issues (very loosely speaking):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We know from quantum mechanics that electrons have certain properties, which are couched in something called a wave function.  Mathematically, wave functions may be complex (i.e. have real and imaginary components)  It is impossible to &quot;see&quot; complex quantities. So there must be something else.&lt;/li&gt;&#xA;&lt;li&gt;In fact, the only thing that we can &quot;see&quot; is the electron density, and the electron density of an atom is spherical.  The electron density is the square (actually complex modulus) of the wave function, and so a possible set of (perhaps imaginary) wave functions are the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;Spherical Harmonics&lt;/a&gt;&quot; (s, p, d, ... fuctions)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only system that has an analytic solution is the Schroedinger equation for a hydrogen-like atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens if we &quot;cram&quot; two or more atoms together to make a molecule?  Is there an analytic solution for molecules that gives sigma, pi, etc.?  Nope.  But we can try to make an approximation, called the linear combination of Atomic Orbitals (s,p,d...) to give Molecular Orbitals (LCAO-MO)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sigma and pi bonds (and anti bonds) are a &quot;basis&quot; or a representation of molecules that we construct from things we know everything about, the atomic orbitals of a hydrogen-like atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a coarse analogy, when we watch TV, the pixels on the screen are a &quot;basis set&quot; which, individually, can be turned red green blue.  Their combination gives rise to the image of whatever you think you see on the screen (State of the Union address, football game, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple fact is: two atoms in separation look like spheres.  If they are pushed together, then their electron density forms a dumbbell shape.  You can not look at the electron density and pick out sigma and pi features.  There is no such thing as &quot;hybridization&quot; to form sp, sp2, sp3...   These are convenient &quot;lies&quot; (linear combinations of primitive atomic orbitals) which sometimes get in the way of something so very simple:   Electrons of Atom A are attracted to Protons in Electron B and vice versa. (This is a slight bend of the truth, because electrons are shared between both.  But it is possible to identify atomic basins, please see Quantum Theory of Atoms in Molecules)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to follow the conventional explanation: electrons &quot;go&quot; into bonding orbitals (e.g. sigma bonds) first because sigma bonds usually look more like s atomic orbitals.  s atomic orbitals are primarily centered upon the nucleus, and since electrons and protons have an attractive interaction, that is the lowest energy configuration.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-04T04:55:28.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4885" PostTypeId="2" ParentId="4760" CreationDate="2013-05-04T05:08:45.463" Score="15" Body="&lt;p&gt;&lt;em&gt;The reason&lt;/em&gt; that they do not exist (or at least are not the most stable form) is because the decomposition reaction is exothermic.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}\ce{&#xA;(1) &amp;amp;&amp;amp; SF6 &amp;amp;-&amp;gt; SF4 + F2\\&#xA;(2) &amp;amp;&amp;amp; SH6 &amp;amp;-&amp;gt; SH2 + 2 H2&#xA;}\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Reaction &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; is much more exothermic than &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;, and it can be argued in two ways: either there is something very bad about the reactants (&lt;span class=&quot;math-container&quot;&gt;$\ce{SX6}$&lt;/span&gt;) or something very good about the products.&lt;/p&gt;&#xA;&lt;p&gt;If you look at the bond strength of &lt;span class=&quot;math-container&quot;&gt;$\ce{F-F}$&lt;/span&gt; vs. the bond strengths of &lt;span class=&quot;math-container&quot;&gt;$\ce{H-H}$&lt;/span&gt; (well-known quantities) you will see that &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; has a much stronger bond than &lt;span class=&quot;math-container&quot;&gt;$\ce{F-F}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;There is every reason that &lt;span class=&quot;math-container&quot;&gt;$\ce{SH6}$&lt;/span&gt; should exist if you are looking at how many &amp;quot;slots&amp;quot; sulfur has.  If it has six slots for fluorine, then surely it could accommodate six hydrogens.&lt;/p&gt;&#xA;&lt;p&gt;I don't really want to get into it, but I tend to think that &amp;quot;hypervalency&amp;quot; is borderline quackery, preoccupation with non-explanation.  Either the atoms want to be there because their interaction is favourable electrostatically, or they would rather be in some other configuration.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="16683" LastEditDate="2021-01-21T11:41:46.917" LastActivityDate="2021-01-21T11:41:46.917" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="4886" PostTypeId="1" CreationDate="2013-05-04T05:16:23.390" Score="3" ViewCount="2106" Body="&lt;p&gt;What sort of changes in the properties of organic or drug molecules can be anticipated if you substitute some or all of the oxygen atoms with sulfur atoms, and vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest in this question is for chemical graph theory perspective. There are numerous parameters calculated on molecular graphs (in trying to relate to chemical or biological activity) which are indifferent to the above substitution. I would like to make an appropriate adjustment if the effects are really significant. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="402" LastEditDate="2017-04-08T22:06:31.260" LastActivityDate="2017-04-08T22:06:53.860" Title="What is the effect of substituting oxygen with sulfur in drug molecules?" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;&lt;medicinal-chemistry&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-04-08T23:50:53.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="4887" PostTypeId="2" ParentId="444" CreationDate="2013-05-04T05:20:35.323" Score="33" Body="&lt;p&gt;There is a big difference between a &quot;rule&quot; and a law of nature. The &quot;octet rule&quot; is a turn-of-the-last-century concept that somehow managed to get into introductory chemistry books and never got kicked out with the advent of modern quantum mechanics. (Circumstantial proof: it is impossible to identify individual electrons to label them &quot;valence&quot; or &quot;not valence&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you won't find any answer based on physical evidence as to why/why not a rule that is not based on physical evidence will hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms take their spatial configuration because it happens to be an electrostatically-favorable circumstance, not because electrons avail themselves like &quot;slots&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="16683" LastEditDate="2017-08-14T15:38:01.740" LastActivityDate="2017-08-14T15:38:01.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4888" PostTypeId="2" ParentId="4395" CreationDate="2013-05-04T05:35:50.130" Score="1" Body="&lt;p&gt;+10 to kaliaden's comment.  The &quot;reason&quot; is because linear molecules are $\ce{sp}$, trigonal-like molecules are $\ce{sp^2}$, and tetrahedral are $\ce{sp^3}$.  Hybridization rules are completely bogus, and basically you know the answer by first observing the molecule experimentally, and post-hoc hammering it into these resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is nothing but a bunch of memorization. It is interesting to note that if you made $\ce{O}$ into $\ce{S}$, then it _may_not_ have the same shape.  (Actually, it probably does, but where it is different, we sound off the d-orbital participation mumbo-jumbo)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="4231" LastEditDate="2015-03-01T14:35:19.113" LastActivityDate="2015-03-01T14:35:19.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4889" PostTypeId="2" ParentId="4315" CreationDate="2013-05-04T05:52:18.930" Score="7" Body="&lt;p&gt;Just want to point out that there is a subtlety to the word &quot;water&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water can mean the molecule, $\ce{H2O}$, and those bonds (between central O and H's) may be covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water also means the bulk water, many molecules of water.  The interactions between two or more molecules is what dictates the bulk properties of a liquid, such as evaporation and freezing.  Here we would talk about hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a picture of bonding of one configuration of &quot;water&quot; in &quot;water.&quot;  This is computed from electronic structure theory, and the Quantum Theory of Atoms in Molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P4h40.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-04T14:24:03.583" LastActivityDate="2013-05-04T14:24:03.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4890" PostTypeId="2" ParentId="4267" CreationDate="2013-05-04T06:02:45.187" Score="3" Body="&lt;p&gt;It is extremely difficult to look at a molecule and know whether it is explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, generally speaking, highly exothermic chemical reactions are the &quot;explosive&quot; ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A --&gt; B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the reaction exothermic?  Either A is very unstable, or B and C are very stable.  Any molecule that has lots of very stable B's and C's is going to be very explosive. Examples of very stable B and C could be carbon soot, water, and also NO2.  (Actually there are lots of them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to keep A stable enough (metastable) to use it for some practical purpose. So, this somewhat limits how &quot;unstable&quot; A can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said another way, I would design an explosive A to be an effective delivery mechanism of B's and C's just waiting to be formed.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-04T06:07:49.427" LastActivityDate="2013-05-04T06:07:49.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4891" PostTypeId="2" ParentId="594" CreationDate="2013-05-04T06:20:38.227" Score="19" Body="&lt;p&gt;The real issue is that no one has ever taken a picture (i.e. electron density) of genuine, unambigious, cases of a single, double, triple, quadruple??? bonds. And they never will, because these concepts are not based on quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two atoms reside next to each other, and if they have a favorable electrostatic interaction, then a certain type of topology arises in their electron density.  (q.v. Quantum Theory of Atoms in Molecules)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might as well say that every &quot;bond&quot; is a single-bond, or, equivalently, and infinity-bond.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of articles are bogus, as they can not be confirmed experimentally.  They got lucky with the reviewers, and/or an editor who knows their readership is just dying to hear news of a quadruple bond, having heard for so many years that triple is the highest you can go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, what are people looking for? Four &quot;ropes&quot; that link between the two carbon atoms?  Where is the unambiguous, unbiased dividing line between bond energies of a single/double/triple/quadruple bond?&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-04T06:25:48.213" LastActivityDate="2013-05-04T06:25:48.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4892" PostTypeId="1" AcceptedAnswerId="4896" CreationDate="2013-05-04T07:14:27.073" Score="15" ViewCount="7389" Body="&lt;p&gt;I was reading &lt;a href=&quot;https://chemistry.stackexchange.com/questions/594/bonding-in-cec2-a-carbon-carbon-quadruple-bond&quot;&gt;this question and the answer by user EricBrown&lt;/a&gt; to it, and this got me thinking about covalent chemical bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I was always taught is that a single bond contains 2 paired electrons, a double bond 4 etc (&lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Lewis dots drawings&lt;/a&gt; always show this nicely). In his answer Eric mentions that the whole definition of single, double etc bonds is ambiguous because there haven't been measurements of the electron density. This notion strikes me as odd, because a simple googling gives me articles like this: How similar is a molecule to another? An electron density measure of similarity between two molecular structures &lt;a href=&quot;https://doi.org/10.1002/qua.560170612&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Int. J. Quantum Chem.&lt;/em&gt; &lt;strong&gt;1980,&lt;/strong&gt; &lt;em&gt;17&lt;/em&gt; (6), 1185–1189&lt;/a&gt;. But it did make me think about the way the different types of bonds are defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read up a bit on the &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aim/aim_0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theory of atoms in molecules&lt;/a&gt; and indeed this theory only specifies 1 bond type: the line with maximum electron density between two nuclei. According to the theory this line is unique for a given pair of nuclei in a given molecule. I could imagine that a molecule which has a triple bond (6 bonding electrons in the 'classical' theoretical sense) would have a relatively high electron density along this maximum line as compared to e.g. a single bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: if we would be able to measure the electron density in a molecule (are we?) couldn't we use this to define a relation between the value of the maximum electron density and the notion of a single, double or triple bond? Or perhaps use the spread in electron density between the nuclei for this?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="5026" LastEditDate="2018-07-18T02:49:40.167" LastActivityDate="2018-07-18T02:49:40.167" Title="Electron density definition of single, double, triple bonds?" Tags="&lt;molecules&gt;&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;covalent-compounds&gt;&lt;qtaim&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="4893" PostTypeId="2" ParentId="230" CreationDate="2013-05-04T12:07:10.780" Score="3" Body="&lt;p&gt;The answer is that there is one (or perhaps two) types of bonds. A &lt;em&gt;bond&lt;/em&gt; occurs when two atoms are attracted in a net-electrostatically favorable way.  (Of course, the electrons and protons are subject to their quantum nature)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why two? In the &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aim/aim_0.html&quot; rel=&quot;nofollow&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt;, the &quot;procedure&quot; is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ascertain whether a &lt;a href=&quot;http://aim.tkgristmill.com&quot; rel=&quot;nofollow&quot;&gt;bond path&lt;/a&gt; exists between two atoms (atomic basins).  This is a yes/no answer -- the is a bond or no bond -- because it is electrostatically favorable or unfavorable.&lt;/li&gt;&#xA;&lt;li&gt;If a bond path does exist, then the sign of the Laplacian of the electron density at the bond critical point tells you whether the interaction is &quot;covalent&quot; or &quot;not covalent&quot; though these terms are not strictly applicable. +/- yeilds only TWO possible types of bonds. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;BTW: The Laplacian of any field is a measure of whether the electron density at a particular point is a &quot;sink&quot; or a &quot;source.&quot; The reality is: there is an entire, continuous spectrum of values of the Laplacian. None of them present themselves as being called &quot;dipole&quot; or &quot;van der Waals&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, there is one language of bonding, and that is couched in Quantum Physics.  You will never hear of van der Waals forces in a physics [quantum electrodynamics] class, because there is no such thing.  There is no van der Waals term in the Schroedinger equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between chemical and physical bonds as taught in class is simple:  chemistry as a whole has not shaken the pre-quantum revolution description of a bond.  It is disastrously complicated for students and professionals in the field, as there is a never ending squabble as to whether something is a &quot;dipole-dipole&quot; or &quot;dipole-induced dipole&quot; or &quot;three center,  two electron&quot;  But these are all arguments built on shaky scaffolding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is that the rules of physical bonding are almost &quot;boring.&quot;  The second fact is that most bond descriptions in chemistry have only certain elements of reality. To the credit of chemists, they need to quickly and efficiently describe certain oft-appearing motifs in bonding--without resorting to a quantum chemistry calculation--and they have done a pretty good job.  It's just that these descriptions are somewhat ambigious and always open to interpretation, hence not &lt;em&gt;purely&lt;/em&gt; quantum mechanical.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-04T12:44:54.410" LastActivityDate="2013-05-04T12:44:54.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4894" PostTypeId="1" CreationDate="2013-05-04T13:08:10.697" Score="5" ViewCount="208" Body="&lt;p&gt;So I'm carrying out the distillation of some essential oil and I have to remove the water content in it. The usual method is using diethyl ether but I'm looking for a greener method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do?&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="7475" LastEditDate="2015-10-30T16:46:03.857" LastActivityDate="2015-10-30T16:46:03.857" Title="Drying a distillate liquid" Tags="&lt;everyday-chemistry&gt;&lt;purification&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4895" PostTypeId="1" CreationDate="2013-05-04T15:55:07.443" Score="0" ViewCount="2527" Body="&lt;p&gt;how do you calculate the wavelength of light? For chemsitry lab, we calibrated the scale reading of Hg, for example, to be 4.91 cm which is violet and the given wavelength says 436.  For Hg, it says scale reading of 5.51 cm for green and wavelength of 546. But for overhead fluorescent, I took the scale reading of 4.93 for violet but don't know how to calculated the wavelength. Any help with steps shown would be great!&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2013-08-04T20:57:52.290" Title="Wavelength of light" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4896" PostTypeId="2" ParentId="4892" CreationDate="2013-05-04T17:57:35.023" Score="8" Body="&lt;p&gt;Below are plots of the electron density for staggered ethane, ethene, and ethyne, computed from electronic structure theory.  For ethene, which is planar, imagine that the hydogens are jutting forward and backward into the paper, so that the contour plots should contain any putative $\pi$ bonds. It is the orthogonal plane that still contains the two carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of points that I wish to make:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You are correct (in this case for sure) that there is some association between stronger bonds and higher electron density at the C-C bond critical point $(\star)$.  This was noted by Bader and co-workers.&lt;/li&gt;&#xA;&lt;li&gt;There is no evidence (to my eyes) that there are any side-wise $p$ orbital interactions forming $\pi$ bonds. These look like single $\sigma$ bonds!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is, where do we draw the dividing line in the correlation between electron density at the bond critical point, and what constitutes single, double, and triple bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxbMn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Responding to @michaelm: I might have expected this, if for example, ethene were to have a &quot;double bond&quot; between the carbons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Av2Bd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-06T15:40:32.937" LastActivityDate="2013-05-06T15:40:32.937" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="4897" PostTypeId="2" ParentId="285" CreationDate="2013-05-04T18:51:30.363" Score="5" Body="&lt;p&gt;(Fairness in reporting: I am a hard-core Quantum Theory of Atoms in Molecules partisan)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NBO is one of many attempts the apply rotations among molecular orbitals, each of which can spread over the entire molecule, (i.e. the wave function) so that they maximally-resemble localized atomic orbitals or &quot;hybrid atomic orbitals&quot;. Then one may be able to talk about &quot;pieces&quot; of a overall state wave function, and draw connections to the historical descriptions of bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the whole process of wave function localization/transformation is puzzling, because it leaves the observable--the electron density--just the way that it was, at least it ought to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that its motivation is good-intentioned.  However, Schroedinger himself cautioned strongly against attempts to analyze the wave function directly. For example, since the wave function may be complex, what does one do with a localized wave function that has real and imaginary components?  Why not just analyze the electron density (which is real and observable) directly?&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-06T16:17:51.073" LastActivityDate="2013-05-06T16:17:51.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4898" PostTypeId="2" ParentId="228" CreationDate="2013-05-04T19:02:23.467" Score="22" Body="&lt;p&gt;Here is a plot of the Quantum Theory of Atoms in Molecules answer to you question. I have shown the bond paths of $\ce{B2H6}$.  Indeed, they are &quot;banana-like&quot; but interestingly they are curved inward, unlike the case of cyclopropane which are curved outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hybridization does not exist. Also, I'm not sure if there is a point of ascribing &quot;number of electrons&quot;--like they are aliquots--to any bonding interaction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice that I have drawn the bond paths between the B's and the four similar hydrogens as solid (covalent), and the set of bond paths along the &quot;bridge&quot; as dotted (not covalent). This is because the sign of the Laplacians of the electron density at their respective bonc critical points (yellow spheres) are opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z70pJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-06T17:50:08.090" LastActivityDate="2013-05-06T17:50:08.090" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4899" PostTypeId="1" AcceptedAnswerId="4912" CreationDate="2013-05-04T19:32:55.680" Score="5" ViewCount="280" Body="&lt;p&gt;Consider this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3(s) &amp;lt;=&amp;gt; CaO(s) + CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for this, $K_c=\left[\ce{CO2(g)}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given temperature, $K_c$ will remain fixed. That means, the  amount of $\ce{CO2}$ released will be same, irrespective of amount of $\ce{CaCO3}$ taken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, that means, at the same temperature, irrespective of  whether we start with 1 g of $\ce{CaCO3}$, or 10 kg of $\ce{CaCO3}$, the amount of $\ce{CO2}$ at equilibrium will be same(!) because, according to the $K_c$ expression, it should be constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is against my intuition, because, it seems that if we start with 10 kg of $\ce{CaCO3}$, we should end up with a much larger amount of $\ce{CO2}$ at equilibrium than if we started with just 1 g of the carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I being mistaken somewhere, or is this is what will actually happen? I am confused.&lt;/p&gt;&#xA;" OwnerUserId="487" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T15:12:05.117" LastActivityDate="2015-06-17T15:12:05.117" Title="Equilibrium constant expression for heterogeneous equilibrium" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4901" PostTypeId="1" AcceptedAnswerId="7563" CreationDate="2013-05-04T23:15:20.743" Score="3" ViewCount="283" Body="&lt;p&gt;Does cysteine break down in filtered water?  Should this made more acidic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I mix cysteine powder in juice (something acidic maybe), does it remain at cysteine?  It seems that it can oxidize into cystine and lose some of its qualities. &lt;/p&gt;&#xA;" OwnerUserId="1590" LastEditorUserId="4325" LastEditDate="2015-05-18T19:05:27.440" LastActivityDate="2015-05-18T19:05:27.440" Title="Best way to dissolve and consume cysteine HCl monohydrate powder" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4903" PostTypeId="1" CreationDate="2013-05-05T07:32:15.073" Score="3" ViewCount="3135" Body="&lt;p&gt;Why is that in an external magnetic field(uniform) the degeneracy of d,f orbitals is lost but the degeneracy of p orbitals remain intact if the main cause of losing degeneracy is the difference in energy of differently oriented orbitals in the magnetic field?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-06-06T00:32:08.917" Title="Degeneracy of orbitals?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4904" PostTypeId="1" CreationDate="2013-05-05T11:37:35.627" Score="6" ViewCount="6338" Body="&lt;p&gt;I am not able to find any examples of the inductive effect operating through multiple bonds. Why can't electron displacement take place in multiple bonds?&lt;/p&gt;&#xA;" OwnerUserId="1614" LastEditorUserId="10185" LastEditDate="2015-10-02T16:26:47.023" LastActivityDate="2020-10-29T07:00:07.517" Title="Why Inductive effect operates only through single bonds?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4906" PostTypeId="1" CreationDate="2013-05-05T18:47:25.147" Score="1" ViewCount="266" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A balloon is half filled with water at sea level ($P = \pu{1 atm}$, $T = \pu{298K}$) such that the partial pressure of water is $\pu{24 mmHg}$.  It is then submerged to a point where $P = \pu{2 atm}, T = \pu{298K}$.  What will be the vapor pressure of water at that point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first instinct is to find the mole fraction of water in the balloon.  Using Dalton's law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{total} = P_{\ce{H2O}} + P_{\mathrm{atm}} = 24 + 760 = \pu{784 mmHg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{\ce{H2O}} = X_{\ce{H2O}}\,P_{total}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $X_{\ce{H2O}}$, I get $24/784$.  Then I use that to find the new partial pressure at the new depth, $(24/784)\cdot 1520$.  But the answer seems to be that the partial pressure of water remains constant. Can someone explain why the partial pressure will be constant?&lt;/p&gt;&#xA;" OwnerUserId="1593" LastEditorUserId="35555" LastEditDate="2018-02-20T16:08:18.477" LastActivityDate="2018-02-20T16:08:18.477" Title="Partial pressure of water in a submerged balloon" Tags="&lt;physical-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4907" PostTypeId="1" CreationDate="2013-05-06T01:07:43.850" Score="4" ViewCount="147" Body="&lt;p&gt;I'm just wondering if anybody here knows of a Consumer-Report-like publication for lab equipment. Something that would give the specifics, as, for example, the flow rate of a distilling condenser, etc.&lt;/p&gt;&#xA;" OwnerUserId="1595" LastEditorUserId="1362" LastEditDate="2013-05-06T07:37:09.537" LastActivityDate="2013-05-06T07:37:09.537" Title="Consumer Report for Lab Equipment" Tags="&lt;materials&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4908" PostTypeId="1" CreationDate="2013-05-06T09:32:27.497" Score="3" ViewCount="307" Body="&lt;p&gt;If we are given a molecular graph i.e. al the atoms involved and their connectivity, how can we make reasonable prediction if the atom would lie in the surface or in the inner part. On way might be check it degree (how many atoms it is bonded to), higher the valency higher it surface accessibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will most of the time have small molecules (drug molecules) which have in average 50 atoms.&#xA;My broader interest is to find out which atoms directly affect the activity of the molecule more than the other. Probably, atoms lying at surface have more role to play. Any other suggestions aimed at this identification would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1586" LastEditDate="2013-05-06T13:47:04.010" LastActivityDate="2014-11-19T06:01:09.990" Title="Prediction of surface atoms in molecule from its graph" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;surface-chemistry&gt;&lt;medicinal-chemistry&gt;&lt;qtaim&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4910" PostTypeId="2" ParentId="4908" CreationDate="2013-05-06T12:39:47.657" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aim/aim_0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt; (QTAIM) provides a fairly unambiguous answer. First, I would like demonstrate the concept of an Atomic Basin. A region of space belongs to a given atom if the steepest ascent path through the electron density terminates at that atom.  Consider a plane that contains the nuclei of the water molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YMYtL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the atomic basins could also be represented in 3D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MUPUA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These atomic basins are the QTAIM definitions of ATOMS in MOLECULES. Note that there are gradient paths for each atom that extend to infinity, hence each atom could be considered an &quot;external atom.&quot;  (This is a quite trivial case, of course.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider formamide in this configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pvhOg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic basins of form amide look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X5owr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the carbon atom is fairly &quot;inaccessible&quot; in the plane of the molecule, but it does have some extent &quot;peeking&quot; out of the plane of the molecule (here, the maroon-colored basin, sorry for change-up in color):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FKpdl.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a completely general concept, and so can be extended to any scenario. Here is a molecular graph of thiol on a silver cluster (meant to mimic a surface) (note the bond paths unambiguously show which atoms are bonded to what, even for C-Ag, S-Ag, H-Ag interactions where these &quot;cooked-up&quot; rules of hybridized bonding and &quot;valency&quot; are totally out of their league):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DdDbu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting the atomic basins is just an exercise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x2HbJ.png&quot; alt=&quot;p&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer: you can define &quot;interior atoms&quot; as QTAIM atomic basins that do not have steepest ascent trajectories starting at infinity that terminate at that atom, such as an interior silver atom in the above example. If you are conceptually separating a system into a substrate and an adsorbate, then you could say &quot;exterior&quot; atoms of the substrate include the ones that have bond paths connecting the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must caution that as two or more atoms approach each other, their atomic basins change their shape.  (Of course they do!)  And so one must consider the possibility that atomic basins of &quot;exterior&quot; atoms move out of the way and lead to exposure of what was, in isolation, an interior atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS:  the atomic basins of peripheral atoms actually extend to infinity.  Here, I have applied cutoffs so that one can actually plot the molecule in a finite region of space.]&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-06T13:54:36.527" LastActivityDate="2013-05-06T13:54:36.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4911" PostTypeId="1" CreationDate="2013-05-06T15:12:32.513" Score="3" ViewCount="348" Body="&lt;p&gt;We know the Michaelis Menten derivation for the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{E + S &amp;lt;=&amp;gt; ES -&amp;gt; E + P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what if the reaction took place in a different scenario whereby:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{E + S &amp;lt;=&amp;gt; ES1 -&amp;gt; ES2 -&amp;gt; E + P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the derived Michaelis Menten equation be now?&lt;/p&gt;&#xA;" OwnerUserId="1597" LastEditorUserId="7475" LastEditDate="2017-10-14T02:14:21.800" LastActivityDate="2017-10-14T02:14:21.800" Title="Michaelis Menten derivation for 2 enzyme substrates" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4912" PostTypeId="2" ParentId="4899" CreationDate="2013-05-06T15:40:54.903" Score="7" Body="&lt;p&gt;First, as kaliaden commented, the true equilibrium constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K(T,p) = a(\textrm{CaO,s}) a(\textrm{CO$_2$,g}) / a(\textrm{CaCO$_3$,s}) \approx&#xA;p_\textrm{CO$_2$} / p^\circ \equiv K_p(T)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ are the activities, $p$ the partial pressure and $p^\circ$ is one bar.&#xA;Thus $K$ basically only depends on temperature (unless you are interested in brutally large pressures, and we safely disregard the sublimation of CaCO$_3$ and the effect of the air). Obviously you can obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c(T) \approx K_p(T)/ RT = [\textrm{CO$_2$}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your question is: how is it that this constant does not depend on the amount of matter that we have? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is that as long as there is enough matter to reach the equilibrium (to saturate the air with carbon dioxide vapor) it doesn't matter how much CaCO$_3$ you start with. The reaction will proceed until reaching an equilibrium characterized by $K(T) \approx p_\textrm{CO$_2$} / p^\circ$. Obviously if you don't have enough CaCO$_3$ (for instance, a few micrograms) then everything will react and you will not reach the equilibrium. &lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditDate="2013-05-08T16:07:57.523" LastActivityDate="2013-05-08T16:07:57.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4913" PostTypeId="2" ParentId="4886" CreationDate="2013-05-06T16:56:28.200" Score="4" Body="&lt;p&gt;In asking this question - &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4636/why-do-chalcogens-group-vi-stink-so-badly&quot;&gt;Why do chalcogens (Group VI) stink so badly?&lt;/a&gt; - I received a couple answers and comments along the lines of &quot;it's not quite so simple&quot;. The same would generally apply here; there isn't one change in behavior, or even a finite set of behavioral changes, that could be predicted by an oxygen-to-sulfur exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a basic example, compare methanol ($\ce{CH_3(OH)}$) to methyl mercaptan, ($\ce{CH_3(SH)}$). Methanol is a colorless liquid at RT with a sweetish odor. Methyl mercaptan, with a higher molecular mass due to the sulfur switch, is nevertheless a gas at RT and smells like rotting cabbage. Methanol metabolizes into formic acid and formaldehyde which in sufficient doses causes nerve damage, specifically to the optic nerve. Methyl mercaptan is pretty much its own metabolite; while also toxic in sufficient doses, it's readily filtered out by the liver and kidneys, as it's produced by bacterial breakdown of many of the foods we eat (especially highly-fibrous vegetables, legumes etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most common occurrence of sulfur in otherwise organic molecules is a sulfhydryl group (R-SH), forming a thiol. The alcohol-to-thiol switch is relatively common in human biochemistry and as such is also of interest in pharmacology. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thiol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on thiols&lt;/a&gt; lists the following basic changes between an alcohol and its thiol analog:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;More easily oxidized, producing disulfides, esters, ethers, etc&lt;/li&gt;&#xA;&lt;li&gt;As thiols, more acidic than alcohols&lt;/li&gt;&#xA;&lt;li&gt;Contrasting this, thiols dehydrogenate easily to form thiolate bases&lt;/li&gt;&#xA;&lt;li&gt;Thiolates complex strongly to most metals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, you can expect the sulfur analog of pretty much any organic functional group (alcohols-&gt;thiols, aldehydes-&gt;thials, ketones-&gt;thioketones, esters-&gt;thioesters, ethers-&gt;thioethers, etc) to be more reactive. However, exactly how the sulfur switch will affect the actual reactions the compound will undergo is best studied case by case, as there is no one way that a particular sulfur-bearing functional group will behave when introduced into the complex chemical stew of human biology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One near-constant; sulfur in small doses &lt;em&gt;can&lt;/em&gt; smell pleasant, because sulfur is an essential not-so-trace element in humans (oranges and grapefruit, for instance, have some volatile sulfur compounds that contribute to their smell and taste). Sulfur in high concentrations stinks, because for the same reasons it is essential in biochemistry, too much of it is toxic (which is a pretty good illustration of the First Rule of Toxicology - &lt;em&gt;every&lt;/em&gt; substance, in sufficient dosage, is toxic).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-04-08T22:06:53.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4914" PostTypeId="2" ParentId="4895" CreationDate="2013-05-06T17:39:49.263" Score="2" Body="&lt;p&gt;Well, the equipment appears to be giving you the wavelength; however, the scale seems to be skewed (off-calibration) so you have to correct for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale readings you give appear to be off a few orders of magnitude; the wavelengths you list as &quot;given wavelengths&quot; are in nanometers. Wavelengths of 4-5 &lt;em&gt;centimeters&lt;/em&gt; as your measurements suggest would place the emission spectra in the &quot;ELF&quot; radio frequency range at the extreme low end of detectable electromagnetic spectra, and &lt;em&gt;well&lt;/em&gt; out of the range of visible light. So, just for the sake of argument, let's say you interpreted the measurements incorrectly, and these numbers should be in the hundreds and represent nanometers of wavelength (or, as written, they represent tenths of a micrometer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that assumption, the numbers are still off a bit, indicating the instrument is slightly out of calibration. If the instrument itself isn't adjustable to correct the values,  &quot;calibrating&quot; it boils down to a percentage error calculation that is then applied to further readings: (experimental-theoretical)/theoretical * 100%. For your violet wavelength, you got 491nm but expected 436. That is (491-436)/436 * 100% = 12.6% error. The green wavelength reading was 551 when you expected 546; those two are much closer (0.915% error), which indicates that the equipment you're using has some chromatic aberration. That's understandable in an instrument designed to &quot;fracture&quot; light into a prismatic spectrum; other light-sensing equipment, like cameras, generally try to minimize it so all the wavelengths emanating from one point on the subject pass through the lens to expose one point on the film or photocell. But I digress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your violet spectral line reading off of an ordinary fluorescent lamp is 493nm. Given the percentage error measures on this end of the scale, this reading is about 12.6% too high; so, we divide by 112.6% to get the actual measured wavelength of 437.8. That measurement, against the known emission spectra of mercury vapor, has a percentage error of only (437.8-436)/436 * 100% = 0.42%, which for a measurement made with classroom lab equipment and other less-than-ideal conditions is pretty darn close.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-06T19:58:37.323" LastActivityDate="2013-05-06T19:58:37.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4915" PostTypeId="2" ParentId="4877" CreationDate="2013-05-06T18:31:40.587" Score="91" Body="&lt;p&gt;Hydrofluoric acid is toxic and corrosive, but actually isn't that strong of an acid compared to other hydrohalic acids; the fluorine has a very good orbital overlap with hydrogen and is also not very polarizable, therefore it resists donating its proton, unlike other hydrohalic acids which are good proton donators. It will break down some tissues, but it will take a relatively long time and won't turn the entire body into stuff that can be rinsed down the drain. Hydrochloric acid is a much stronger acid, and as it has several uses from pH-balancing pool water to preparing concrete surfaces, it's available by the gallon from any hardware store. However, it isn't very good at dissolving bodies either; while it will eventually work by breaking down the connective tissues, it will make a huge stink and take several days to dissolve certain types of tissues and bones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard body-dissolving chemical is lye aka sodium hydroxide. The main source is drain clog remover because most drain clogs are formed by hair and other bio-gunk that accumulates naturally when humans shower, exfoliate etc. It works, even though the body's overall chemistry is slightly to the basic side of neutral (about 7.35-7.4) because the hydroxide anion is a strong proton acceptor. That means that it strips hydrogen atoms off of organic molecules to form water (alkaline hydrolysis, aka saponification), and as a result, those organic molecules are turned into simpler molecules with lower melting points (triglycerides are turned into fatty acids, saturated fats are dehydrogenated to form unsaturated fats, alkanes become alcohols, etc). Sodium hydroxide is also a ready source of the sodium ion; sodium salts are &lt;em&gt;always&lt;/em&gt; water-soluble (at least I can't think of a single one that isn't). The resulting compounds are thus either liquids or water-soluble alcohols and salts, which flush down the drain. What's left is the brittle, insoluble calcium &quot;shell&quot; of the skeleton; if hydrolyzed by sodium hydroxide, the resulting calcium hydroxide (&quot;slaked lime&quot;) won't dissolve completely but is relatively easy to clean up.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-01-22T09:56:54.963" LastActivityDate="2017-01-22T09:56:54.963" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4920" PostTypeId="1" AcceptedAnswerId="4925" CreationDate="2013-05-07T00:36:02.803" Score="1" ViewCount="263" Body="&lt;p&gt;I'm looking for a reducing sugar such as glucose or maltose (any reducing sugar will do) for some home chemistry experiments. Is there a readily available reducing sugar that I might source without contacting a chemical company? A simple &quot;synthesis&quot; of glucose from sucrose would work as well. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorDisplayName="user7951" LastEditDate="2016-11-26T19:22:24.140" LastActivityDate="2016-11-26T19:22:24.140" Title="How To Obtain A Reducing Sugar At Home?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4921" PostTypeId="1" CreationDate="2013-05-07T01:22:18.423" Score="2" ViewCount="154" Body="&lt;p&gt;In &lt;a href=&quot;http://www.rsc.org/suppdata/ce/c1/c1ce05243h/c1ce05243h.pdf&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; I am trying to understand, the following occurs: (summarized below)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Synthesis of octahedral $\ce{Cu_2O}$ seeds:In a typical synthesis, 2.9946 g of $\ce{Cu(CH_3COO)_2}$ was dissolved in deionized water (20 mL) using a beaker under a constant stirring at 70°C for 2 min. A dark precipitate was produced when a sodium hydroxide solution (9 M, 10 mL) was added dropwise to the above solution. After being stirred for 5 min, D-(+)-glucose powder (0.3 g) was added into the dark precursor with a constant stirring for another 60 min at 70°C,and then was allowed to cool to room temperature naturally. Afterward, the obtained products were centrifuged at 5000 rpm for 1min (XIANYI TG16-WS centrifuge). The precipitates were centrifuged twice more in deionized water and anhydrous ethanol, respectively. And finally were dried at 70°C for 12 hours in a vacuum oven. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My understanding of this paragraph is that copper acetate reacted with sodium hydroxide to form copper hydroxide and sodium acetate. A reducing sugar reduces the copper hydroxide to $\ce{Cu_2O}$. The ethanol dissolves the sodium acetate to leave only $\ce{Cu_2O}$ (octahedral) crystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next morphology of $\ce{Cu_2O}$ crystals, &quot;polyhedral M-pyramid $\ce{Cu_2O}$ architecture,&quot; was grown through essentially the same process, except using seed crystals from the first batch of octahedral crystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do octahedral seed crystals give rise polyhedral ones?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4231" LastEditorDisplayName="user4076" LastEditDate="2015-03-08T17:05:21.900" LastActivityDate="2015-03-08T17:05:21.900" Title="How do octahedral seed crystals yield polyhedral Cu₂O crystals?" Tags="&lt;crystal-structure&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4923" PostTypeId="1" AcceptedAnswerId="4927" CreationDate="2013-05-07T09:52:20.100" Score="2" ViewCount="621" Body="&lt;p&gt;I read a paper about &lt;a href=&quot;http://dx.doi.org/10.1039/C39930001831&quot; rel=&quot;nofollow&quot;&gt;enzyme-catalysed synthesis of octanamide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is octanamide used for?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="1362" LastEditDate="2013-05-07T16:47:21.217" LastActivityDate="2013-05-08T15:39:00.007" Title="What is octanamide used for?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4924" PostTypeId="1" CreationDate="2013-05-07T14:08:39.240" Score="5" ViewCount="22034" Body="&lt;p&gt;How would we use our $\mathrm{p}K_{\mathrm{a}}$ of the acid component of the buffer to calculate $\large \frac{[\ce{A-}]}{[\ce{HA}]}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$$ rearranging gives me $$\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}$$&#xA;if for example the $\mathrm{pH}=4.5$, $\mathrm{p}K_{\mathrm{a}}=3.74$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to say&#xA;$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{19}{250}$$&#xA;since &#xA;$$&#xA;\begin{align}&#xA;\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&amp;amp;=\mathrm{pH} - \mathrm{p}K_{\mathrm{a}}F\\&#xA;\implies\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)&amp;amp;=4.5- 3.74\\&#xA;\implies\frac{[\ce{A-}]}{[\ce{HA}]}&amp;amp;=\frac{\operatorname{e}^{0.76}}{10}&#xA;\end{align}&#xA;$$&#xA;if $\dfrac{[\ce{A-}]}{[\ce{HA}]}=\dfrac{V_\text{b}}{V_\text{a}}= \dfrac{19}{250}$&#xA;how would I use this to find the volumes of acid and base needed to make my buffer?&lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="5026" LastEditDate="2018-02-28T16:08:11.707" LastActivityDate="2018-02-28T16:08:11.707" Title="Calculating volumes of acid/base required to make buffer using pH and pKa" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4925" PostTypeId="2" ParentId="4920" CreationDate="2013-05-07T14:49:26.750" Score="5" Body="&lt;p&gt;As permeakra mentions, glucose can be purchased pretty easily. It might be labeled as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dextrose&quot; rel=&quot;nofollow noreferrer&quot;&gt;dextrose&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another commonly available is &lt;a href=&quot;http://en.wikipedia.org/wiki/Maltose&quot; rel=&quot;nofollow noreferrer&quot;&gt;maltose&lt;/a&gt; (or malt sugar). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sucrose&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sucrose&lt;/a&gt; (or table sugar) is not a reducing sugar, but one of its constituents (glucose) is. Glucose can be made from sucrose by heating with a little bit of acid to make &lt;a href=&quot;http://en.wikipedia.org/wiki/Invert_sugar&quot; rel=&quot;nofollow noreferrer&quot;&gt;invert sugar syrup&lt;/a&gt;. Invert sugar is approximately a 1:1 ratio of glucose to fructose, and only the glucose in this mixture is reducing. An example procedure can be found here:&lt;a href=&quot;https://chemistry.stackexchange.com/questions/555/how-does-adding-lemon-juice-to-sugar-make-better-caramel/556#556&quot;&gt;How does adding lemon juice to sugar make better caramel?&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Honey&quot; rel=&quot;nofollow noreferrer&quot;&gt;Honey&lt;/a&gt; is a naturally occurring invert sugar. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2013-05-07T14:49:26.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4927" PostTypeId="2" ParentId="4923" CreationDate="2013-05-07T16:47:50.373" Score="4" Body="&lt;p&gt;Well, one thing is as plastics. Nylon is a linear polyamide; it's traditionally produced from alkyl-diamines and a dicarboxylic acid (to donate the acyl group of the amides, as well as a second alkyl chain joined directly to the carboxyl), and the typical product is a poly-hexanamide, but nylon-8,x variants are also possible, as are reactions involving a more direct route of welding the monomers together and arriving at the same place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Google-fu turns up a number of possible end products for which octanamide aka caprylamide would be a useful intermediate. By itself, it's an emulsifier and common solvent in organic synthesis. Other alkyl amides including derivatives of octanamide, often with a few additional alkyl or amino groups welded on, are used in drugs, inks/dyes, fragrances, and skin care items. N,N-dimethyloctanamide and N,N-diethyloctanamide appear to be common ingredients in deodorants and sunblocks, though I'm not sure if they're an actual fragrance or a carrier/stabilizer (Edit: they seem to be emulsifiers and stabilizers for essential oils of the fragrances in water-based products).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-08T15:39:00.007" LastActivityDate="2013-05-08T15:39:00.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4928" PostTypeId="5" CreationDate="2013-05-07T17:25:57.677" Score="0" Body="&lt;p&gt;This tag should be applied to questions relating to the theory, design, techniques or performance of chemical experiments. This includes but is not limited to theoretical structures, expected products, thermal calculations, and empirical analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional tags should be applied to narrow down the field of research.&lt;br&gt;&#xA;Among others, this includes, but is not limited &lt;a href=&quot;/questions/tagged/glassware&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;glassware&amp;#39;&quot; rel=&quot;tag&quot;&gt;glassware&lt;/a&gt; &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt; &lt;a href=&quot;/questions/tagged/physical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;physical-chemistry&lt;/a&gt; &lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt; &lt;a href=&quot;/questions/tagged/synthesis&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis&lt;/a&gt;...&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-07-29T09:31:03.970" LastActivityDate="2014-07-29T09:31:03.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4929" PostTypeId="4" CreationDate="2013-05-07T17:25:57.677" Score="0" Body="This tag should be applied to questions relating to the theory, design, techniques or performance of chemical experiments. This includes but is not limited to theoretical structures, expected products, thermal calculations, and empirical analysis. Additional tags should be applied to narrow down the field of research." OwnerUserId="367" LastEditorUserId="4945" LastEditDate="2014-07-29T09:30:57.670" LastActivityDate="2014-07-29T09:30:57.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4930" PostTypeId="5" CreationDate="2013-05-07T17:44:40.700" Score="0" Body="&lt;p&gt;Water is colloquially known as the &quot;universal solvent&quot;; its polar nature and high availability make it central to many key biological processes and &quot;everyday&quot; chemical reactions. As such, many reagents used in the laboratory, especially in elementary chemistry, are typically dissolved in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of simple formula, water displays highly complex behavior on molecular level - hydrogen bonding and autodissociation create various ionic clusters. Adding both ionic and molecular solutes further complicates attempts to characterise solutions, as water is a strongly solvating most of them, using a wide range of interactions.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="9961" LastEditDate="2020-04-23T18:30:56.140" LastActivityDate="2020-04-23T18:30:56.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4931" PostTypeId="4" CreationDate="2013-05-07T17:44:40.700" Score="0" Body="For questions relating to compounds dissolved in water and resulting reactions and acid-base equilibria." OwnerUserId="367" LastEditorUserId="9961" LastEditDate="2020-04-23T19:17:10.327" LastActivityDate="2020-04-23T19:17:10.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4932" PostTypeId="1" AcceptedAnswerId="4939" CreationDate="2013-05-07T18:12:36.657" Score="4" ViewCount="1709" Body="&lt;p&gt;I read about perchloride in a talk and after trying to search for its chemical formula, the search results imply that perchloride is actually named perchlorate. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="710" LastEditorUserId="83758" LastEditDate="2020-03-21T10:09:18.213" LastActivityDate="2020-03-21T12:16:25.907" Title="Is perchloride the same as perchlorate ion?" Tags="&lt;ions&gt;&lt;oxidation-state&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4933" PostTypeId="5" CreationDate="2013-05-07T18:17:20.210" Score="0" Body="&lt;p&gt;Energy, in physics, is defined as &quot;an indirectly observed quantity that is a prerequisite for performing work&quot;. Exactly what it is is an open question of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry, energy most often manifests as thermal energy or heat. Chemical bonds store a given amount of energy and also require the addition of more energy from the outside environment to break these bonds. This energy that is required to break the bonds between two atoms of a compound and form another bond is known as its &quot;enthalpy of formation&quot; or &quot;heat of formation&quot;, and is typically measured either in kilojoules per mole (the Joule is the SI unit favored in physical sciences) or kilocalories per mole (more easily obtained by thermal measurement). The units can be easily converted; 1kcal/mol = 4.184kJ/mol. Chemical bonds with lower enthalpies of formation are characterized as less stable; they have more energy inherent in the bond and so will more easily break that bond to form another. Bonds with higher enthalpies of formation are more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical reactions that break less stable bonds to form more stable ones release the additional energy, typically in the form of heat. These reactions are known as &quot;exothermic&quot;. By contrast, chemical reactions that break more stable bonds to form less stable ones require the addition of energy from the environment in which the reaction takes place; these are known as &quot;endothermic&quot;. The total energy production or consumption inherent in a chemical reaction can be calculated by computing the sum difference of the enthalpies of the existing bonds broken and the new bonds formed. The resulting quantity, typically also in kJ/mole or kcal/mole, is the &quot;heat of reaction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical energy can also manifest itself as kinetic or electrical energy. Reactions involving the transfer of electrons and/or protons, inducing a charge imbalance in one or more of the reactants, can reach equilibrium by transferring electrons, creating an electrical current. Highly exothermic reactions, especially those that create gaseous products, can also produce kinetic energy as the newly-formed, highly-compressed gases rapidly expand to equalize pressure with the surrounding environment.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-04-08T14:52:21.410" LastActivityDate="2017-04-08T14:52:21.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4934" PostTypeId="4" CreationDate="2013-05-07T18:17:20.210" Score="0" Body="For questions relating to the energy required for or produced by reactions, including questions of endothermicity/exothermicity, bond enthalpy, etc." OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-04-08T14:52:07.420" LastActivityDate="2017-04-08T14:52:07.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4935" PostTypeId="2" ParentId="4932" CreationDate="2013-05-07T19:17:23.410" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;per- -ide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Refers, as far as I know, to double anions. For example: &lt;span class=&quot;math-container&quot;&gt;$\ce{HgCl2, H2O2}$&lt;/span&gt; are named &quot;Mercury dichloride&quot; and &quot;hydrogen peroxide&quot; respectively. I believe that &quot;per- -ide&quot; has been dropped for most cations (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{HgCl2}$&lt;/span&gt; was previously called &quot;mercury perchloride&quot;). &quot;Peroxide&quot; is the only one I've still see a lot around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;per- -ate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Refers, as far as I know, to anions derived from 'per- -ic' acids. Perchloric acid is &lt;span class=&quot;math-container&quot;&gt;$\ce{HClO4}$&lt;/span&gt; and perchlorate is &lt;span class=&quot;math-container&quot;&gt;$\ce{ClO4-}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system of naming isn't so common now. We generally refer to 'chlorate' or 'chloride' and indicate the oxidation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{NaClO4}$&lt;/span&gt; is now often referred to as &quot;sodium chlorate (VII)&quot; instead of &quot;sodium perchlorate&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look a bit into the &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry#System&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="83758" LastEditDate="2020-03-21T10:09:39.540" LastActivityDate="2020-03-21T10:09:39.540" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4936" PostTypeId="2" ParentId="4924" CreationDate="2013-05-07T19:21:42.640" Score="3" Body="&lt;p&gt;First of all, note that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{10}\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)=4.5- 3.74$$&#xA;$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)\neq\dfrac{e^{0.76}}{10}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)= 10^{0.76}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, the volume will depend on the concentration of the acid and of the salt (if it's a solution) that you are provided with. This ratio will give you the relative concentrations required to get a buffer of the pH 4.5.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-07T19:21:42.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4937" PostTypeId="5" CreationDate="2013-05-07T19:28:39.003" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Plastic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plastic&lt;/a&gt;, as a noun, is defined as &quot;any one of a wide range of synthetic or semi-synthetic organic solid compounds which are moldable&quot;. The word was originally an adjective, from Latin and Greek roots for &quot;moldable&quot;. Another definition commonly describing modern plastics is &quot;a class of synthetic organic resins&quot;, a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;resin&lt;/a&gt;&quot; in turn generally defined as an organic substance originating as a thick liquid that hardens into a translucent solid, for instance, amber (a natural resin of fossilized tree sap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of plastics are organic &quot;polymers&quot;; long chains of repeating linear or aromatic compounds known as monomers. The exact method and bonding of the monomers differs between plastics, as do the general properties of the resulting resin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastics are typically separated into two categories; thermoset resins and thermoplastics. The primary difference is in their behavior when heated; thermoset resins, once hardened (typically by the evaporation of a solvent), remain solid and rigid even under high heat, while thermoplastics will melt and can be remolded when heated. Common examples of plastics of both types include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermoset:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol_formaldehyde_resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phenol formaldehyde resin&lt;/a&gt; - used commonly in gaming devices, such as billiard balls, dice, dominoes etc. Its density and hardness make it desirable over lighter and softer thermoplastics which can deform with use.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Epoxy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Epoxy resin&lt;/a&gt; - typically used as a resin suspension for other materials, such as paper/cloth (forming &lt;a href=&quot;http://en.wikipedia.org/wiki/Micarta&quot; rel=&quot;nofollow noreferrer&quot;&gt;Micarta&lt;/a&gt;), carbon fiber weave (forming &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_fiber&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon fiber composite&lt;/a&gt;), and silicates (forming &lt;a href=&quot;http://en.wikipedia.org/wiki/Fiberglass&quot; rel=&quot;nofollow noreferrer&quot;&gt;fiberglass&lt;/a&gt;, in turn a common substrate for the production of printed circuit boards)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Melamine_resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Melamine resin&lt;/a&gt; - formed from polymerization of melamine (cyanamide trimer) and formaldehyde, this lightweight thermoset is commonly used in dinnerware as a more durable substitute for clay ceramics, as well as being a key component in Formica, some laminate flooring, and other chemically-resistant hard surfaces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thermoplastics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethyleneterephthalate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poly-ethylene terephthalate&lt;/a&gt; (PET) - known as polyester when spun into fibers, it is also seen in various vacuum-molded products such as food packaging, cold drink cups, blister and bubble packaging, soda bottles, etc. Its use with food is slightly controversial as the phthalate compounds used as a plasticizer can leach out when the plastic is heated, and certain of these compounds are known to disrupt human hormones.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polycarbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polycarbonate&lt;/a&gt; - optically clear and rigid, it is commonly used in applications requiring impact and pressure resistance, such as in safety eyewear, headlight enclosures, waterproof camera housings, etc. Most formulations use &lt;a href=&quot;http://en.wikipedia.org/wiki/Bisphenol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bisphenol-A&lt;/a&gt; as the monomer, joined by ester linkages, and when the plastic degrades with age, heat, and contact with water, it can release this chemical, which is also a known endocrine disruptor. As such, most food-based uses for BPA polycarbonate plastics, especially use by infants, have been replaced with other formulations.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyvinyl_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poly-vinyl chloride&lt;/a&gt; (PVC) - A versatile plastic produced from the vinyl chloride monomer, the &quot;raw&quot; polymer is very brittle and degrades easily, so it is almost always combined with additional stabilizers and plasticizers such as phthalates. Its use in several applications, particularly in food packaging and water/wastewater handling, has raised controversy due to the unstable nature of the polymer-phthalate matrix, leading to release of the endocrine-disrupting phthalates into the food or water.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polyethylene&lt;/a&gt; - A common &quot;food-grade&quot; plastic, this polymer is naturally flexible without the use of phthalate plasticizers, and so is favored for applications involving food storage, especially those requiring heating in the packaging. Two variants are commonly seen; high-density, forming a more rigid plastic commonly used for milk jugs and condiment bottles, and low-density, often milky in appearance and used for squeeze bottles.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polypropylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polypropylene&lt;/a&gt; - Another common food-grade plastic that doesn't require a plasticiser, this plastic is commonly seen in reusable food storage containers, in bags for prepared fresh vegetables and snack foods, and for protective covers for paper products.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polystyrene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polystyrene&lt;/a&gt; - In rigid form, it's transparent (but can be colored) and somewhat brittle, and is commonly used for CD &quot;jewel cases&quot; and small moldable parts such as for scale models and some toys and gaming pieces. In foamed form, it's used for disposable food packaging, packing peanuts, etc. Because it doesn't biodegrade and was originally produced using CFCs, the use of polystyrene foam in disposable containers is highly controversial, and at one point it had been largely abandoned in favor of other options. It remains popular in a non-CFC formulation for packaging hot foods and foods that would dissolve or soak through paper products, like raw meat.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="367" LastEditorUserId="402" LastEditDate="2017-05-04T00:56:43.667" LastActivityDate="2017-05-04T00:56:43.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4938" PostTypeId="4" CreationDate="2013-05-07T19:28:39.003" Score="0" Body="For questions relating to the structures, properties, production methods and potential hazards of polymer materials commonly known as plastics" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-07T22:17:44.250" LastActivityDate="2013-05-07T22:17:44.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4939" PostTypeId="2" ParentId="4932" CreationDate="2013-05-07T21:41:41.280" Score="6" Body="&lt;p&gt;Short answer: Yes, sometimes &quot;perchlorate&quot; can be called &quot;perchloride&quot; in older literature, but because of the classical definition of a &quot;perchloride&quot;, it's so confusing and technically wrong that the $\ce{ClO_4^-}$ anion should be called either &quot;perchlorate&quot; or &quot;chlorate(VII)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In older naming systems, pre-IUPAC, a &quot;per-ide&quot; is quite simply &quot;any substance containing an unusually high proportion of the named element&quot;. So, a &quot;&lt;a href=&quot;http://en.wiktionary.org/wiki/perchloride&quot; rel=&quot;nofollow&quot;&gt;perchloride&lt;/a&gt;&quot; is any substance with more chlorine than is &quot;normal&quot; for that mixture of elements. This name was typically given to metal halide salts where the metal has multiple possible oxidation states; the highest oxidation state of the metal forms the metal &quot;per-X-ide&quot;. for instance, titanium perchloride aka titanium tetrachloride, $\ce{TiCl_4}$ (IUPAC name titanium(IV) chloride), as differentiated from the trichloride and dichloride salts of titanium, and iron perchloride aka ferric chloride (that name's rarely used as iron only has two stable chlorides, and so the traditional &quot;ferric&quot; and &quot;ferrous&quot; descriptors work fine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the perchlorate anion isn't actually a &quot;perchloride&quot; by that definition; it's more a &quot;peroxide&quot;, as there's only one chlorine (which is relatively &quot;normal&quot;) that has been &quot;superoxidized&quot; by four oxygens (not normal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This use of &quot;perchloride&quot; as a synonym for &quot;perchlorate&quot; is more the result of a bad simplification of the traditional naming system, where salts of &quot;-ic&quot; acids are &quot;-ides&quot;. While the rule follows for the hydrohalic acids (HF, HCl, HBr, HI) and most oxoacids (which, when they end in &quot;-ous&quot;, form &quot;-ites&quot;, as in hypofluorous acid, chlorous acid and sulfurous acid), the highly oxygenated oxoacids like chloric, perchloric, sulfuric, nitric, phosphoric and boric acids form &quot;-ates&quot;, not &quot;-ides&quot;, to mirror the low-oxygen &quot;-ous&quot; acids and their &quot;-ite&quot; anions/salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When not talking about the perchlorate anion, you sometimes hear about organic molecules that have had all their hydrogens replaced by halogens. Perchlorocarbons and perfluorocarbons are the chlorine and fluorine derivatives of hydrocarbons (for instance, the dry-cleaning chemical perchloroethylene, &quot;Perc&quot;, and the perfluoroethylene monomer of polytetrafluoroethylene, aka Teflon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Peroxide&quot; has found a home in the IUPAC system as the official name of the $\ce{(O_2)^{2-}}$ anion, composed of two single-bonded oxygens (often as the related &quot;hydroperoxide&quot;, which is an OOH functional group bound to a carbon or to hydrogen). However, because each chlorine anion in a &quot;perchloride&quot; is typically bound to whatever the chlorine is oxidizing, the multiple chlorides are not a single anion and so aren't named as such in IUPAC.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-03T16:08:14.103" LastActivityDate="2013-09-03T16:08:14.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4940" PostTypeId="1" AcceptedAnswerId="4941" CreationDate="2013-05-08T07:15:21.107" Score="17" ViewCount="21712" Body="&lt;p&gt;I've read that when a carboxylic acid reacts with &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; the corresponding alcohol is formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vjhRt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vjhRt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I try to think of the mechanism, I get stuck here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; produces &lt;span class=&quot;math-container&quot;&gt;$\ce{H-}$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$\ce{H-}$&lt;/span&gt; is a strong base it should immediately abstract a proton from the carboxylic acid to give the corresponding carboxylate ion (just like in the reaction of carboxylic acids with Grignard reagents), instead of undergoing nucleophilic addition to give the alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I resolve this?&lt;/p&gt;&#xA;" OwnerUserId="1604" LastEditorDisplayName="user7951" LastEditDate="2020-01-03T16:52:52.447" LastActivityDate="2020-01-03T16:52:52.447" Title="Reduction of carboxylic acids by LiAlH4" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="4941" PostTypeId="2" ParentId="4940" CreationDate="2013-05-08T07:44:21.343" Score="9" Body="&lt;p&gt;The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Od3ym.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/INc9j.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the pictures &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu/Page/vsc/en/ch/12/oc/c_acid/organoli/organoli.vscml.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-08T07:44:21.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4942" PostTypeId="1" AcceptedAnswerId="4944" CreationDate="2013-05-08T17:47:03.693" Score="2" ViewCount="844" Body="&lt;p&gt;Consider a one dimensional predominance diagram for a simple acid-base reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{AH &amp;lt;=&amp;gt; A^- + H^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With equilibrium constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_a = \frac{[A^-]_{eq}[H^+]_{eq}}{[AH]_{eq}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, $[H^+] &amp;gt; K_a$, then $[AH]&amp;gt;[A^-]$. At first I thought I understood why this was so: the $[H^+]$ cancels out on both sides of the inequality, and then you bring $[AH]$ to the other side. Only now I realized that you can't actually cancel $[H^+]$ because on one side of the inequality it's actually $[H^+]_{eq}$, not the current amount of protons but the amount at equilibrium. So now I'm lost. What justifies this rule for determining predominance?&lt;/p&gt;&#xA;" OwnerUserId="1433" LastEditorUserId="22" LastEditDate="2013-05-09T08:24:58.810" LastActivityDate="2015-12-22T22:18:28.770" Title="The reasoning behind predominance diagrams" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4943" PostTypeId="2" ParentId="4942" CreationDate="2013-05-08T20:51:44.493" Score="0" Body="&lt;p&gt;If you rearrange the equation to $$K_a[HA] = [A^-][H^+]$$ the relationship becomes more obvious. If $[H^+]&amp;gt;K_a$, then $[HA]&amp;gt;[A^-]$ in order to keep the two products equal. otherwise $[H^+]&amp;gt;K_a$ and $[A^-]&amp;gt;[HA]$, then the relationship would be an inequality: $$[A^-][H^+] &amp;gt; K_a[HA].$$&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-05-08T20:51:44.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4944" PostTypeId="2" ParentId="4942" CreationDate="2013-05-08T21:03:24.163" Score="3" Body="&lt;p&gt;I think you might be making a fundamental mistake when you differentiate between the &quot;current amount of protons&quot; and the &quot;amount at equilibrium.&quot; All dissociation constants are based on equilibrium concentrations, and predominance diagrams refer to the concentrations present only after equilibrium has already been established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the concrete example of the dissociation of ammonium: $\ce{NH4+ &amp;lt;=&amp;gt; NH3 + H+}$. The equilibrium constant expression for this chemical process is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_\text{a} = \frac{\ce{[NH3][H+]}}{\ce{[NH4+]}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{p}K_\text{a}$ of this process has been experimentally determined to be approximately $9.25$. Therefore, the following equation should hold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\log(K_\text{a}) = -\log\frac{\ce{[NH3][H+]}}{\ce{[NH4+]}} = \mathrm{p}K_\text{a} = 9.25&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a given solution has a measured $\mathrm{pH}$ precisely equal to the $\mathrm{p}K_\text{a}$ of the above process, then what we have is the following (keeping in mind that $\ce{[H+]} = 10^{-\mathrm{pH}}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\log\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = -9.25&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the anti-log of both sides of the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = 10^{-9.25}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then dividing both sides of the equation by $10^{-9.25}$, we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, the concentrations of ammonium and ammonia in the solution are perfectly equal when the $\mathrm{pH}$ and $\mathrm{p}K_\text{a}$ are equal. If the $\mathrm{pH}$ is higher (that is to say, more basic), then you'll find that your conjugate base (in this case, $\ce{NH3}$) will predominate by factor of $10^{\mathrm{pH} - \mathrm{p}K_\text{a}}$. Conversely, if the $\mathrm{p}K_\text{a}$ is higher, then the conjugate acid ($\ce{NH4+}$) will predominate by a factor of $10^{\mathrm{p}K_\text{a} - \mathrm{pH}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer to use an inequality and some generic conjugate acid/base pair $\ce{HA/A- }$, then the logic goes as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathrm{pH} &amp;gt; \mathrm{p}K_\text{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{pH} &amp;gt; \mathrm{p}K_\text{a}\\&#xA;-\log\ce{[H+]} &amp;gt; -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\&#xA;\ce{[H+]} &amp;lt; \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\&#xA;1 &amp;lt; \frac{\ce{[A- ]}}{\ce{[HA]}}\\&#xA;\ce{[HA]} &amp;lt; \ce{[A- ]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathrm{pH} = \mathrm{p}K_\text{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{pH} = \mathrm{p}K_\text{a}\\&#xA;-\log\ce{[H+]} = -\log{\frac{\ce{[H+][A- ]}}{\ce{[HA]}}}\\&#xA;\ce{[H+]} = \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\&#xA;1 = \frac{\ce{[A- ]}}{\ce{[HA]}}\\&#xA;\ce{[HA]} = \ce{[A- ]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\mathrm{pH} &amp;lt; \mathrm{p}K_\text{a}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{pH} &amp;lt; \mathrm{p}K_\text{a}\\&#xA;-\log\ce{[H+]} &amp;lt; -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\&#xA;\ce{[H+]} &amp;gt; \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\&#xA;1 &amp;gt; \frac{\ce{[A- ]}}{\ce{[HA]}}\\&#xA;\ce{[HA]} &amp;gt; \ce{[A- ]}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="24052" LastEditDate="2015-12-22T22:18:28.770" LastActivityDate="2015-12-22T22:18:28.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4945" PostTypeId="1" AcceptedAnswerId="4947" CreationDate="2013-05-09T00:36:10.270" Score="6" ViewCount="3650" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Las_Palmas_De_Gran_Canaria&quot; rel=&quot;nofollow&quot;&gt;Las Palmas de Gran Canaria&lt;/a&gt;, tap water is chlorinated. In an Internet forum, someone   &lt;a href=&quot;http://www.nexusboard.net/showthread.php?siteid=6689&amp;amp;threadid=305174&amp;amp;showpage=2&amp;amp;nx=967fe3d9949c0d609ccf3c64bef92e8b#pid3356533&quot; rel=&quot;nofollow&quot;&gt;claims&lt;/a&gt; that in 2008 the water had a chlorine concentration of 0,5 - 1mg/l. On the &lt;a href=&quot;http://www.emalsa.es&quot; rel=&quot;nofollow&quot;&gt;official site&lt;/a&gt; of Emalsa, the water provider, I couldn't find any information about chlorine concentration. I tried contacting them online, but the contact form is broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another forum, I read that ascorbic acid (vitamin C) can be used to neutralize the chlorine: My taste buds say it works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much ascorbate/l do I need to neutralize the above amount of chlorine?&lt;/p&gt;&#xA;" OwnerUserId="1607" LastEditorUserId="21964" LastEditDate="2016-08-04T22:49:51.730" LastActivityDate="2016-08-04T22:49:51.730" Title="How much ascorbate to neutralize one liter of chlorinated tap water?" Tags="&lt;water&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4946" PostTypeId="1" CreationDate="2013-05-09T05:54:44.533" Score="4" ViewCount="253" Body="&lt;p&gt;In my introductory chemistry course, we learned that for reversible processes, entropy is defined as $ΔS = q/T$. However, we did not spend that long on this before moving on to other things like Gibbs free energy. My questions are: &#xA;&lt;br&gt;1. What exactly is a reversible process?&#xA;&lt;br&gt;2. Why must we use things like standard molar entropies to calculate $\Delta S$ instead of just using $\Delta H$ according to this equation?&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="1389" LastEditDate="2013-05-09T21:26:05.220" LastActivityDate="2013-05-10T02:37:56.820" Title="Entropy, enthalpy, and reversibility" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4947" PostTypeId="2" ParentId="4945" CreationDate="2013-05-09T07:52:55.253" Score="5" Body="&lt;p&gt;This is just a very short explanatory note on &lt;a href=&quot;http://www.fs.fed.us/t-d/pubs/html/05231301/05231301.html&quot; rel=&quot;noreferrer&quot;&gt;the article suggested by Greg E.&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction between ascorbic acid and hypochlorous acid is as follows:&#xA;$$ \ce{C5H5O5CH2OH + HOCl → C5H3O5CH2OH + HCl + H2O}$$&#xA;so for 1 mole of hypochlorous acid you need 1 mole of ascorbic acid. This can be recalculated into a mass ratio by the use of their molecular masses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(\ce{C5H5O5CH2OH})=176\;\text{g/mole}$$&#xA;$$M(\ce{HOCl})=52.5\;\text{g/mole}$$&#xA;From this we can see that if the molar ratio is $1:1$, the mass ratio is $176:52.5$ which is $3.3:1$ somewhat higher than the article claims. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not exactly sure of the reason, it could be that there is some side reaction of ascorbic acid that also removes some additional hypochlorous acid although I wouldn't know which. Perhaps the article is talking about the amount of ascorbic acid you need to get a concentration which you don't taste anymore?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-05-09T07:52:55.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4948" PostTypeId="1" AcceptedAnswerId="4962" CreationDate="2013-05-09T08:41:01.643" Score="1" ViewCount="2037" Body="&lt;p&gt;I am currently researching as too which is better for maintaining the chemical balance in a swimming pool. I have obviously tried the standard Google search for &lt;code&gt;Salt Water or Chlorinated Pools&lt;/code&gt; and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What chemical reasons are there for choosing a salt water pool over a chlorinated pool?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Salt water meaning the chlorine is generated and chlorinated meaning that you add the tablets/pucks etc.&lt;/p&gt;&#xA;" OwnerUserId="1610" LastActivityDate="2013-05-10T15:09:40.043" Title="Chemical reasons for choosing Salt Water or Chlorinated pool?" Tags="&lt;everyday-chemistry&gt;&lt;homework&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4949" PostTypeId="1" CreationDate="2013-05-09T19:08:25.047" Score="1" ViewCount="498" Body="&lt;p&gt;To me it seems like diatomic elements like $\ce{F2}$, $\ce{Cl2}$, etc. occur more prevalently than diatomic molecules with both, like $\ce{ClF}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the case, and if so, what causes diatomic elements to be favored?&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="1362" LastEditDate="2013-05-10T07:58:10.400" LastActivityDate="2014-11-19T11:09:06.543" Title="Do halogens bond more commonly with each other or themselves?" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4950" PostTypeId="2" ParentId="853" CreationDate="2013-05-09T21:36:31.187" Score="3" Body="&lt;p&gt;Perhaps chromium can form a chloride-containing complex e.g. $ \left[ \ce{Cr \cdot 2 Cl \cdot 4 H_2O}\right] $. (I know copper forms an analogous complex, but only at high concentrations of $ \ce{Cl^-} $.) The possibility of forming this complex would make chromium oxide more soluble in water in the presence of chloride than in its absence.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastEditorUserId="22" LastEditDate="2013-05-09T21:39:37.743" LastActivityDate="2013-05-09T21:39:37.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4951" PostTypeId="2" ParentId="4949" CreationDate="2013-05-10T02:20:16.903" Score="2" Body="&lt;p&gt;There's a lot of circumstantial evidence that heterodiatomics are more stable than homodiatomics. It is usually ascribed to the observation that atoms of different &quot;electronegativity&quot; tend to have stronger bonds than atoms of the same electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your particular example, consider the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 + F2 -&amp;gt; 2 ClF}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the tables handy, but I calculate with electronic structure theory that this reaction is &lt;em&gt;exothermic&lt;/em&gt; by about 25 kcal/mol. That's quite a bit! (The equilibrium lies far to the right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would suggest that a mixture of Chlorine and Fluorine, heated up to very high temperature and then cooled down, would tend to reform as ClF.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="4945" LastEditDate="2014-11-19T11:09:06.543" LastActivityDate="2014-11-19T11:09:06.543" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4952" PostTypeId="2" ParentId="4946" CreationDate="2013-05-10T02:37:56.820" Score="3" Body="&lt;p&gt;Your text will best explain reversibility.  The short answer is that a reversible process is one whose direction can be reversed by a microscopic change in external conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a gas in a cylinder expanding against a piston will do so reversibly as long as the external pressure and the internal pressure are essentially identical.  Of course for expansion the internal pressure must be infinitesimally greater than the external pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, entropy is really only defined for systems at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the equation for the entropy above is not correct except at constant temperature.  In reality it is $dS = dq/T$, which generally must be integrated in order to determine the entropy by this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, there is some confusion here. $\Delta H$ is equal to $q$ only at constant pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there are lots of ways to compute the entropy change in a process, but this space is too small to go into them.  They should be in your textbook.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-10T02:37:56.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4953" PostTypeId="2" ParentId="4267" CreationDate="2013-05-10T02:48:29.670" Score="3" Body="&lt;p&gt;I would add that explosive reactions are extremely &lt;strong&gt;rapid&lt;/strong&gt; reactions.  It is hard to judge rapidity of a reaction just by knowing the structure of a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A given substance can qualify as an explosive under special conditions.  For example we cook with methane gas.  It burns nicely and gives us no problems.  But mix it with the proper amount of oxygen and toss in a match and there goes the neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper amount is often a range of concentrations, wide or narrow depending on the molecules involved.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-10T02:48:29.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4954" PostTypeId="1" AcceptedAnswerId="4957" CreationDate="2013-05-10T08:11:46.087" Score="4" ViewCount="1058" Body="&lt;p&gt;Does there exist a formula I can use to calculate alcohol content of a solution after heating it for some time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following use-case:&#xA;I am making a pecan pie. I have a thick sugary solution. I am not sure how much the makeup of the solution matters. Just assume its more viscous than water. I add 75 ml of 40%v  alcohol and I heat it at a temperature high enough to cause boiling for 20 minutes. How much alcohol remains at the end of 20 mins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there are a lot of variables here that would make it impossible to give a precise answer. I would be interested in an answer that solves an &quot;easier&quot; version of this problem, for example: having 1 liter of water, adding to it 100ml of 100%v alcohol (Ethanol), boiled for 100 minutes, how much alcohol is left?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How quickly does alcohol boil off? Is it dependent on is concentration within a solution? is there a formula for that?  &lt;/p&gt;&#xA;" OwnerUserId="1615" LastEditorUserId="1615" LastEditDate="2013-05-10T14:06:28.643" LastActivityDate="2013-05-10T14:06:28.643" Title="Calculate Ethanol content after heating for N minutes" Tags="&lt;home-experiment&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4955" PostTypeId="1" AcceptedAnswerId="4956" CreationDate="2013-05-10T09:44:51.147" Score="3" ViewCount="867" Body="&lt;p&gt;Why don't we need to consider the $\ce{H+}$ ions of water when finding the $\ce{pH}$ of $0.01~\mathrm{M}$ of $\ce{HCl}$ solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{pH}$ value of something is depending on the molarity of $\ce{H+}$ ions. But the teachers says the calculation is just $-\log(0.01)$, don't we need to consider the $\ce{H+}$ ions of water? &lt;/p&gt;&#xA;" OwnerUserId="1617" LastEditorUserId="4945" LastEditDate="2014-10-28T07:15:36.467" LastActivityDate="2014-10-28T07:15:36.467" Title="Why do we neglect the protons from water's autoprotolysis?" Tags="&lt;ph&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4956" PostTypeId="2" ParentId="4955" CreationDate="2013-05-10T10:40:10.047" Score="9" Body="&lt;p&gt;You could consider them, but what would be the result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.01 M$ of $\ce{HCl}$ with $1\times10^{-7} M$ of $\ce{H^+}$ from water gives you $0.0100001 M$ of $H^+$. Find the pH of this concentration of $\ce{H^+}$ and compare with the pH you would obtain if you used $0.01 M$ directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You shouldn't see that the resulting pH differ significantly.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2013-05-10T14:22:36.203" LastActivityDate="2013-05-10T14:22:36.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4957" PostTypeId="2" ParentId="4954" CreationDate="2013-05-10T11:12:33.923" Score="2" Body="&lt;p&gt;For a pure substance, to be able to calculate how much has boiled off after $N$ minutes you will need the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_of_vaporization&quot; rel=&quot;nofollow noreferrer&quot;&gt;enthalpy of vaporization&lt;/a&gt;, $L$ (in kJ/kg) and the heat input, $Q$ (in kJ) which will give you:&#xA;$$\tag{1} m=\frac{\epsilon \, Q}{L}$$&#xA;where $m$ is the mass that has boiled off and $\epsilon$ is the fraction of the heat input that actually goes into the liquid. $1-\epsilon$ would be the fraction of heat lost to the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The very crude story&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extremely crude calculation would be to take the mixture enthalpy of vaporization and assume that only ethanol vaporizes such that the fraction of ethanol can be directly related to the total mass loss. &lt;em&gt;Note that this calculation will give you a lower bound for the time needed to get rid of the ethanol&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the mixture enthalpy of vaporization by averaging the enthalpies of vaporization of water and ethanol:&#xA;$$\tag{2} L_{mixture}=x_{\ce{EtOH}} L_{\ce{EtOH}}+x_{\ce{H2O}}L_{\ce{H2O}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to figure out the heat input. A precise calculation is impossible here, but we can use some reasonable estimates. According to an estimate on this (dutch) website when we are cooking we use about $164$ l/h of methane. Assuming that that is for cooking with 2 pots this translates to a gas flow rate of $\dot{V}=1.4$ l/min for a single pot. With the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_of_combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat of combustion of methane&lt;/a&gt;, $\Delta H_c$ and the density $\rho$ we can then calculate the amount of heat released per minute:&#xA;$$\tag{3} \dot{Q}=\rho \dot{V} \Delta H_c $$&#xA;plugging in the values for methane we find $\approx 50$kJ/min&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enthalpy of vaporization of the mixture will constantly change, because the composition of the mixture constantly changes over time during the boiling process. Therefore, instead of simply using Eq. 3 in Eq. 1 to find $\dot{m}$ and multiplying with $N$ we will need to do an integration:&#xA;$$\tag{4} m_{boil-off}=\int_0^N \dot{m}\; dt $$&#xA;where $\dot{m}=\frac{\epsilon \, \dot{Q}}{L(t)}$. Now this is a bit problematic because $L(t)$ depends on $m_{boil-off}$, but it can be solved numerically (I feel like this should also be possible analytically, but can't figure it out right now, anybody?!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example question (with $\epsilon=0.5$) leads to the following curve for the fraction of ethanol in your mixture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WziqP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the ethanol is gone after about 7 minutes. To do this calculation for your pecan pie mixture you will need to know the heats of vaporization of all your ingredients and the mass of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The more precise story&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more precise story is that not only the ethanol, but also part of the water will vaporize therefore slowing down the vaporization rate of ethanol. If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid. In that case you can use the theory available for distillation columns and something called the &lt;a href=&quot;http://rowiescribbles.blogspot.nl/2008/11/ponchon-savarit-method.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ponchon-Savarit Method&lt;/a&gt; (the complete diagram for water-ethanol can be found &lt;a href=&quot;http://demonstrations.wolfram.com/PonchonSavaritDiagramForAnEthanolWaterBinaryMixture/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), which I will not recap here, because it is basically a book chapters worth of undergrad chemical engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equilibrium vapor assumption is probably not a very good one due to air flows over the pot which keep pushing the system out of equilibrium. For the evaporation of ethanol this is good news by the way, because it will speed up the process a bit. Also note that with a true equilibrium you would run into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;nofollow noreferrer&quot;&gt;azeotrope&lt;/a&gt; of ethanol-water mixtures.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-05-10T11:12:33.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4958" PostTypeId="1" AcceptedAnswerId="4963" CreationDate="2013-05-10T13:24:20.077" Score="1" ViewCount="648" Body="&lt;p&gt;I would like to know what are the main problems currently studied in Chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, it seems that there is far less vulgarisation in chemistry than the other fields, and it's hard to find an overview of the field accessible to a layman (compared to math or physics for instance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only found this wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_chemistry&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt;, but I have no idea how relevant it is.&lt;/p&gt;&#xA;" OwnerUserId="1618" LastActivityDate="2013-05-10T15:52:37.593" Title="What are the main axis of research in Chemistry?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2013-05-10T16:39:46.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="4959" PostTypeId="2" ParentId="4958" CreationDate="2013-05-10T13:52:54.560" Score="0" Body="&lt;p&gt;Chemistry is the study of the transformation of matter. People are studying every single aspect of how this happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get a simple answer for this question.  For example, if you were to ask, &quot;what are the important aspects of chemistry research?&quot;  You might get back drug design, nanotechnology, etc. etc.  But chemical engineers are cranking out new paints and lacquers (notice that almost everything in our human world is covered by some sort of lacquer.)  Food chemists are working in genetics and flavorings.  Petroleum chemists are working on better cracking and new products.  The list goes on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Importance is in the eye of the beholder.  Quantification of importance is impossible.  If you want to know what is deemed newsworthy by, e.g. the American Chemical Society, pick up the past issue of Chemical &amp;amp; Engineering news and take a look.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-10T13:52:54.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4960" PostTypeId="1" CreationDate="2013-05-10T14:04:09.063" Score="1" ViewCount="195" Body="&lt;p&gt;Electron affinity of chlorine = -349 kJ per mole. And, first ionization energy of sodium = 496 kJ per mole while energy required for combining= 147 kJ per mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it a contradiction against genereal physics and G.N. Lewis' &quot;electronic theory of valency&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1619" LastEditorUserId="22" LastEditDate="2013-05-12T14:43:40.413" LastActivityDate="2013-05-12T20:22:20.623" Title="Lewis model and modern physics" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4961" PostTypeId="1" CreationDate="2013-05-10T14:11:21.113" Score="3" ViewCount="284" Body="&lt;p&gt;I want to understand the degradation mechanism of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxymethylfurfural&quot; rel=&quot;nofollow&quot;&gt;hydroxymethylfurfural (HMF)&lt;/a&gt;. According to the Wikipedia article, HMF is degraded in honey. Someone said that the degradation proceeds via &lt;strong&gt;2&lt;/strong&gt; possible routes and that the products are 2,5-dioxo-3-hexenal and &lt;a href=&quot;http://en.wikipedia.org/wiki/Levulinic_acid&quot; rel=&quot;nofollow&quot;&gt;4-oxopentanoic acid&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism by which these compounds form? If you add a picture of the mechanism, it will be better.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="10185" LastEditDate="2015-10-02T16:25:24.607" LastActivityDate="2015-10-02T16:25:24.607" Title="Degradation mechanism of hydroxymethylfurfural to 2,5-dioxo-3-hexenal and 4-oxopentanoic acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4962" PostTypeId="2" ParentId="4948" CreationDate="2013-05-10T15:09:40.043" Score="2" Body="&lt;p&gt;A salt water pool still has hypochlorous acid $(\ce{HOCl})$ as the active component so essentially it CAN be the same thing as a chlorinated pool. For a chlorinated pool the $\ce{HOCl}$ is directly added, whereas for a salt water pool the salt water is electrolysed to create $\ce{Cl2}$ which then reacts with water to form $\ce{HOCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is indeed often a difference and the explanation is partly chemistry and partly chemical engineering. &lt;a href=&quot;http://articles.chicagotribune.com/2009-02-05/entertainment/0902030478_1_chlorine-gas-pools-saline&quot; rel=&quot;nofollow&quot;&gt;In terms of chemistry&lt;/a&gt;  the overabundance of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloramine&quot; rel=&quot;nofollow&quot;&gt;chloramines&lt;/a&gt; $(\ce{NH2Cl})$ in chlorinated pools over salt water pools makes the difference. When hypochlorous acid, the active component in both, comes into contact with ammonia (from sweat, saliva etc) you get the formation of these chloramines:&#xA;$$\ce{NH3 + HOCl → NH2Cl + H2O} $$&#xA;Chloramines are the major reason for irritated eyes and skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the difference between salt water and chlorinated pools has a chemical engineering reason. Salt water pools have a constant generation of chlorine as opposed to a 'batch' delivery in the case of chlorinated pools. Therefore the control over the concentration of hypochlorous acid in the water is better, which also allows better control over the concentration of chloramines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to sum up: salt water pools allow higher control over the concentration of chloramines and therefore might be better, however, if this control is not properly applied, it could also even be worse!&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-05-10T15:09:40.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4963" PostTypeId="2" ParentId="4958" CreationDate="2013-05-10T15:52:37.593" Score="9" Body="&lt;p&gt;Even though I voted to close this question as not constructive, these are among the most important unsolved problems in chemistry:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Liquid Water - We still know so little about the molecular level interactions of water molecules with themselves and other species. The great extent of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot;&gt;hydrogen bonding&lt;/a&gt; networks in liquid water make it hard to model. Just how &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_cluster&quot;&gt;many water molecules&lt;/a&gt; are solvating a proton in acid? Why are organic reactions between hydrophobic molecules &lt;a href=&quot;http://en.wikipedia.org/wiki/On_water_reactions&quot;&gt;accelerated in the presence of water&lt;/a&gt;? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Molecular origins of life, &lt;strong&gt;especially homochirality&lt;/strong&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot;&gt;formose reaction&lt;/a&gt; is believed to be the origin of carbohydrates. Amino acids, lipids, and nucleic acids have similar hypothetical origins. However, how did these molecules come together &lt;a href=&quot;http://en.wikipedia.org/wiki/Abiogenesis&quot;&gt;to create living organisms&lt;/a&gt;? Why are most (not all) biomolecules present as one &lt;a href=&quot;http://en.wikipedia.org/wiki/Homochirality&quot;&gt;enantiomer and not the other&lt;/a&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Protein folding. Can we predict the three-dimensional shape of a protein simply from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Protein_folding&quot;&gt;order of the amino acids&lt;/a&gt;? If so, then we could design proteins to do whatever we wanted.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;True designer proteins. Once we solve the folding problem, then we can design proteins with active sites of specific geometries and specific amino acids in those active sites to do &lt;a href=&quot;http://en.wikipedia.org/wiki/Protein_design&quot;&gt;whatever reaction we need&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;High Temperature superconductors. Although &lt;a href=&quot;http://en.wikipedia.org/wiki/Room-temperature_superconductor&quot;&gt;room temperature superconductors&lt;/a&gt; would be nice, inexpensive superconductors that operate at liquid nitrogen temperature (77 K) instead of liquid helium (5 K) temperatures would make NMR, MRI, and a host of other important techniques much cheaper.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Energy Crisis. We need a chemical solution to replace fossil fuels as the world's primary energy source. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_economy&quot;&gt;Hydrogen&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuel_cell&quot;&gt;fuel cells&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Biomass&quot;&gt;biomass&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_cell&quot;&gt;solar power&lt;/a&gt; all have serious drawbacks currently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replacement of petrochemical feedstocks. The petroleum industry remains the primary source of organic compounds. What will we do when the petroleum runs out?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bonding theory. Modern bonding theories have developed into bloated mathematical formalisms that have removed any resemblance of intuition from the study of the structure of molecules. Can a unified modern bonding theory be developed that retains only the best features of &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot;&gt;valence bond&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot;&gt;molecular orbital&lt;/a&gt; theories?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Improved ideality and efficiency of synthesis. Can the synthesis of compounds be improved so as to remove wasteful steps (like adding and removing protecting groups) or complex purifications to better adhere to the principles of &lt;a href=&quot;http://en.wikipedia.org/wiki/Green_chemistry&quot;&gt;Green Chemistry&lt;/a&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Structure-Property Relationship. All established relations between structure and activity is empirical, especially when the desired activity is biochemical. Can we design the best drug with the fewest side effects on the first try? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I find some of the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_chemistry&quot;&gt;unsolved problems&lt;/a&gt; on the to be too narrow (stability of norbonyl cation - this is a subset of the structure-activity problem) or trivial (origin of rotation barrier in ethane - the problem is important but the answer is useless).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-10T15:52:37.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4964" PostTypeId="1" CreationDate="2013-05-11T05:11:46.400" Score="1" ViewCount="1239" Body="&lt;p&gt;when searching in the internet about why are alkenes more polar than alkanes (but still they are non-polar), they are always saying that the double bond is more polarizable , i understood the whole story about how the sp2 attract electrons from the sp3 and forming london forces, but they always said that vinylic bonds are slightly polar but i never understood why.&lt;/p&gt;&#xA;" OwnerUserId="815" LastActivityDate="2013-05-12T10:36:07.003" Title="Why are the vinylic bonds in alkenes slightly polar?" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4965" PostTypeId="1" CreationDate="2013-05-11T05:30:48.283" Score="1" ViewCount="3437" Body="&lt;p&gt;This is for determination of copper in a penny Lab. After making standard solutions we made a solution with a penny, dissolving it in nitric acid and adding ammonium hydroxide to form a more absorbtive copper complex. We didn't control the amount of ammonium hydroxide used in our samples.&#xA;How can I test whether or not it has an effect, specifically?&lt;/p&gt;&#xA;" OwnerUserId="1587" LastActivityDate="2013-05-11T13:24:17.077" Title="How do I know if concentration of Ammonia affected my absorbency measurements?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4966" PostTypeId="2" ParentId="4955" CreationDate="2013-05-11T06:19:46.130" Score="3" Body="&lt;p&gt;Do note, however, that the self-ionization of water technically must be taken into account. Though very often justifiable, it is an &lt;em&gt;approximation&lt;/em&gt; to leave it out. Ask yourself, what is the pH of a $10^{-8}\  mol.L^{-1}$ solution of $\ce{HCl}$? It doesn't make sense to add a strong acid to pure water, however dilute, and obtain a basic pH! If you don't rely on the approximate formula, you will find that such a solution has a pH of about 6.98.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.chem.ubc.ca/courseware/pH/section16/index.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a good source to get you going. The more proper way of solving for $\ce{[H^+]_{eq}}$ is based on the solution of a polynomial found from the combination of the charge balance equation and chemical entity balance equations along with equilibria constant definitions, and it is detailed &lt;a href=&quot;http://www2.chem.ubc.ca/courseware/pH/section16/weak.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Though used specifically to analyse the behaviour of a weak acid solution, the same equations can be used for the strong acid solution, with the added approximation that $\ce{[HA]_{eq}} \approx  0$ (i.e. $\ce{c_{HA}} \approx \ce{[A^{-}]}$).&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-05-11T06:19:46.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4967" PostTypeId="1" AcceptedAnswerId="4968" CreationDate="2013-05-11T10:02:54.960" Score="5" ViewCount="1637" Body="&lt;p&gt;The justification or rather intuition behind Markovnikov's rule is the stability of intermediate carbocation. The more substituted the carbocation the more stable the product. But based on this ideas how can we satisfactorily explain the following reaction?&lt;br&gt;&#xA;$$\ce{CH2=CHCl + HCl -&amp;gt; CH3-CHCl2}$$&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="83758" LastEditDate="2020-05-08T10:17:57.867" LastActivityDate="2020-05-08T10:17:57.867" Title="Markovnikov's addition to halogenated alkene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4968" PostTypeId="2" ParentId="4967" CreationDate="2013-05-11T13:04:13.497" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Markovnikov%27s_rule&quot;&gt;Markovnikov's rule&lt;/a&gt; is usually stated in terms of subsitution on the alkene, for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{HCl}$ end up on the carbon atom that already has two hydrogen atoms. The chlorine atom ends up on the other carbon atom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH2=CHCl +} {\bf \ce{H-Cl}} \ce{-&amp;gt; {\bf H}-CH2-CHCl-{\bf Cl} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better version of the rule has to do with the stability of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation&quot;&gt;carbocation&lt;/a&gt; intermediate (and can be generalized to electrophiles other than $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An alkene reacts with electrophiles in such a way as the electrophile attaches to the carbon atom which will result in the more stable carbocation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The order of stability for simple carbocations is $3^\circ &amp;gt;2^\circ &amp;gt;1^\circ &amp;gt;\text{methyl}$, however there can always be more complex carbocations (de)stabilized by heteroatoms and/or resonance. The original wording of Markovnikov's rule may fail in these more complex cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the addition of $\ce{HX}$ to $\ce{CH3O-CH=C(CH3)2}$. In this case the major product (before it decomposes into the aldehyde) is $\ce{CH3O-CHCl-CH(CH3)2}$. Two possible carbocations form, and the one where the proton attaches at the more substituted position is the one that is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First carbocation, $2^\circ$, but stabilized by resonance (&lt;strong&gt;more stable&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-O-CH+ -CH(CH3)2 &amp;lt;-&amp;gt; CH3-O+ =CH-CH(CH3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second carbocation, $3^\circ$, but not stabilized by resonance (&lt;strong&gt;less stable&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-O-CH2-C+(CH3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first carbocation goes on to form the product. Can you extend this to what happens to your molecule?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-11T13:04:13.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4969" PostTypeId="2" ParentId="4965" CreationDate="2013-05-11T13:24:17.077" Score="3" Body="&lt;p&gt;The concentration of ammonia will matter, but only up to a certain point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you dissolve copper in nitric acid, you form copper nitrate, which is very soluble, but as you say, not very absorptive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu(s) +2HNO3(aq) -&amp;gt; Cu(NO3)2(aq)+H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems with copper nitrate is that it dissociates readily in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu(NO3)2(aq) -&amp;gt; Cu^{2+}(aq) + 2NO3^{-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper(II) ions then coordinate with varying amounts of water molecules. Each balue of $\ce{x}$ in the equation below is a slightly different aqua complex with slightly different optical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^{2+}(aq) +xH2O-&amp;gt; Cu(H2O)_{x}^{2+}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of ammonia changes the situation. Ammonia binds strongly to copper to preferentially form the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schweizer%27s_reagent&quot; rel=&quot;nofollow&quot;&gt;tetraamminecopper(II) complex&lt;/a&gt;, technically $\ce{Cu(NH3)4(H2O)2^{2+}}$, which is highly absorptive. Since this formation is highly favored, you only need to add enough ammonia to provide at least 4 stoichiometric equivalents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ammonia solution you added was labeled &quot;Ammonium Hydroxide&quot; (as you say in your question), then you know how much ammonia you added. Your instructor just might not have told you how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonium hydroxide exists only by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_hydroxide#Basicity_of_ammonia_in_water&quot; rel=&quot;nofollow&quot;&gt;ionization of ammonia&lt;/a&gt; in water (which is very small except at high concentrations). Labeling a solution as &quot;Ammonium Hydroxide&quot; has come to mean a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_hydroxide#Saturated_solutions&quot; rel=&quot;nofollow&quot;&gt;saturated solution of ammonia in water&lt;/a&gt;&quot;, which is 35% ammonia by mass (approx. 18 molar). Now you can calculate the amount of ammonia that you added (as long as you know the volume of solution that you added).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To determine the effect of amount of ammonia,&lt;/strong&gt; you can do the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dilute the original ammonium hydroxide solution. For example, dilute 25 mL of the solution to 50 mL with water to get an ammonia solution of about 9 molar. Then repeat the penny experiment and add exactly the same volume of the new ammonia solution. This guarantees that the overall copper concentration remains constant even when the ammonia concentration changes. Then measure the absorbance. Then dilute the ammonia solution again (either dilute the new solution or dilute the original solution by a different amount). Rinse and repeat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-11T13:24:17.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4970" PostTypeId="1" AcceptedAnswerId="4973" CreationDate="2013-05-11T14:14:39.360" Score="1" ViewCount="2549" Body="&lt;p&gt;Quick run down, I was given pH of buffer 4.5 and then pKa of acid 4.79&#xA;I then had to work out volume of acid and volume of base using $\text{pH=pKa}+\log_{10}{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$. Using that I calculated the volumes of acid and base required to make 15 ml of the buffer $\ce{CH3COOH,CH3COONa}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{\ce{HA}}=9.7\ \text{ml}$ and $V_{\ce{A-}}=5.3\ \text{ml}$ (both acid and base used to make buffer are of 0.1 M conc, probably come in handy later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I am now asked to calculate the amounts of acid and base in 15ml of my buffer&#xA;they give me&#xA;Acid: $n_{\ce{HA}} = [\ce{HA}]\cdot V_{\ce{HA}}=0.1\cdot V_{\ce{HA}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base: $n_{\ce{A-}} = [\ce{HA}]\cdot V_{\ce{A-}}=0.1\cdot V_{\ce{A-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which I don't really get?&#xA;I then have to find out the amount of acid was added, some context is that this is n experiment I conducted using a pH probe I added three drops and recorded the pH for a total of 60 drops the final pH reading is 3.52 of an initial reading of 4.38, would I use the simple $\text{pH} = -\log_10[\ce{H+}]$?&#xA;They have given me $[\ce{H^+}]\cdot V=0.1\text{ M}\cdot\dfrac{1\text{ ml}}{1000\text{ L}}=...\text{mol}$ which I also don't really know what to do with it. &#xA;After that I calculate the new concentrations of acid and conjugate base after all added H+ reacts&#xA;initial and final.&#xA;you don't have to use my example just explain EVERY SINGLE STEP and guide me through the process :) much appreciated!  I know I am asking a lot in one go but this all correlates with one another, if anyone can help thanks a lot! &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="865" LastEditDate="2021-05-22T21:31:55.993" LastActivityDate="2021-05-22T21:31:55.993" Title="Buffers calculating amounts of acid &amp; base" Tags="&lt;acid-base&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4971" PostTypeId="2" ParentId="4760" CreationDate="2013-05-11T19:50:03.207" Score="11" Body="&lt;p&gt;As michielm said, it is because of electronegativity. The bond $\ce{S-F}$ is strongly polarized toward the fluorine (~more electrons are near fluorine), while the $\ce{S-H}$ bond is polarized toward the sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{SH6}$ molecule there would be very high electron density around sulfur. This would increase both the electrostatic repulsion of electron and also the kinetic energy connected with occupation of higher atomic orbitals (this is hand-waving argument).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More rigorously it can be explained by a linear combination of atomic orbitals (LCAO), Hartree-Fock method (HF) or density functional theory (DFT). The increased electrostatic repulsion corresponds to Hartree term in HF or DFT. But even more important is the kinetic energy due to Pauli exclusion principle and requirement of orthogonalization of orbitals of each electron (resp. spin pairs). The orthogonalization requires a certain number of nodes in the wavefunction. Each such node increases the orbital energy (similar to states of particle in the box).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the LCAO point of view this can be viewed as the contribution of higher atomic orbitals (d-orbitals, for example) to the bonding molecular orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{SF6}$ the contribution of sulfur atomic orbitals to the bonding states is lower (because most of the electrons are localized on fluorines). Because of that the energy of bonding molecular orbitals is not increased much by either Coulomb repulsion or by kinetic energy of higher atomic orbitals of sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough argument can be given also using &lt;a href=&quot;https://en.wikipedia.org/wiki/Thomas%E2%80%93Fermi_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thomas-Fermi model&lt;/a&gt; where the kinetic energy of electrons is proportional to $\rho^{5/3}$. However, Thomas-Fermi model is not appropriate for molecules.&lt;/p&gt;&#xA;" OwnerUserId="1624" LastEditorUserId="41328" LastEditDate="2017-10-08T08:25:48.633" LastActivityDate="2017-10-08T08:25:48.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4972" PostTypeId="1" CreationDate="2013-05-12T04:53:09.190" Score="7" ViewCount="16098" Body="&lt;p&gt;Whenever I read about coordination compounds in my textbooks, I always find a discussion about spin-only magnetic moment which is given by $\sqrt{n(n+2)}\cdot\mu_\mathrm{B}$, where $n$ is the number of unpaired electrons and $\mu_\mathrm{B}$ the Bohr magneton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how to calculate the total magnetic moment of a given atom (spin only + orbital). For example let's take $\ce{Ni^2+}$. The spin only magnetic moment comes out to be $\sqrt{8}$ BM. Now how do I add the orbital magnetic moment? I've seen a bunch of expressions for orbital magnetic moment in terms on quantum numbers $n$ and $l$, but I am unable to understand which one to add.&lt;/p&gt;&#xA;" OwnerUserId="1604" LastEditorUserId="16683" LastEditDate="2016-10-21T23:43:10.920" LastActivityDate="2017-07-25T11:27:04.387" Title="Total magnetic moment of atom" Tags="&lt;orbitals&gt;&lt;magnetism&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4973" PostTypeId="2" ParentId="4970" CreationDate="2013-05-12T09:03:35.183" Score="1" Body="&lt;p&gt;I'm going to address only the first part since that was in the original question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_a$ means the number of moles of the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equal to the concentration of the acid multiplied by the volume of the acid, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_a = [\ce{HA}].V_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-12T09:03:35.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4974" PostTypeId="2" ParentId="4964" CreationDate="2013-05-12T10:36:07.003" Score="2" Body="&lt;p&gt;Maybe because of resonance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{2}C=CH_{2}$    $\leftrightarrow$    $H_{2}C^{-}-C^{+}H_{2}$&lt;/p&gt;&#xA;" OwnerUserId="1604" LastActivityDate="2013-05-12T10:36:07.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4975" PostTypeId="2" ParentId="4960" CreationDate="2013-05-12T13:01:17.407" Score="4" Body="&lt;p&gt;The &lt;strong&gt;Born-Haber Cycle&lt;/strong&gt; might help you visualizing how those values are used. Notice the values used bellow are a bit different from the ones you got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mQwOU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electron affinity&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The amount of energy released when an electron is added to a neutral&#xA;  atom or molecule to form a negative ion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ionization energy&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The amount of energy required to remove an electron from the atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Enthalpy of formation (ΔH)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The change of enthalpy that accompanies the formation of 1 mole of the&#xA;  compound from its elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2013-05-12T13:01:17.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4976" PostTypeId="2" ParentId="4960" CreationDate="2013-05-12T20:22:20.623" Score="2" Body="&lt;p&gt;Your question assumes that there is a total transfer of an electron from sodium to chlorine when some kind of sodium chloride compound is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in compound formation there is &lt;em&gt;never&lt;/em&gt; a total transfer of an electron from one atom to another.  In other words while there are &lt;em&gt;pure&lt;/em&gt; covalent compounds (electrons completely shared) there are no &lt;em&gt;pure&lt;/em&gt; ionic bonds (electron totally transferred.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, as the other answers have indicated, the nature of the final compound also has an influence.  In solid sodium chloride a network of shared electrons is formed linking many sodium and chloride atoms together.  In gaseous sodium chloride, $\ce{NaCl}$ there are only the two atoms present.  In sodium chloride solutions the situation is even more complex.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-12T20:22:20.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4977" PostTypeId="1" AcceptedAnswerId="4981" CreationDate="2013-05-12T21:04:21.937" Score="11" ViewCount="7947" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Consider the following reversible reaction.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Cr2O7^2-(aq) + H2O(l) &amp;lt;=&amp;gt; 2 CrO4^2-(aq) + 2 H+(aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What will happen to the position of equilibrium and the value of &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; when more &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions are added at constant temperature?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{lcc}&#xA;\hline&#xA;  &amp;amp; \text{Position of equilibrium} &amp;amp; \text{Value of}~K_c \\&#xA;\hline&#xA;\text{A.} &amp;amp; \text{shifts to the left} &amp;amp; \text{decreases} \\&#xA;\text{B.} &amp;amp; \text{shifts to the right} &amp;amp; \text{increases} \\&#xA;\text{C.} &amp;amp; \text{shifts to the right} &amp;amp; \text{does not change} \\&#xA;\text{D.} &amp;amp; \text{shifts to the left} &amp;amp; \text{does not change} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer given is D.&lt;/p&gt;&#xA;&lt;p&gt;I understand that when more &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions are added, the reaction shifts in reverse. So I've understood that this is known as having the equilibrium shift to the left, toward the reactants.&lt;/p&gt;&#xA;&lt;p&gt;What confuses me is why &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; doesn't change. If the equilibrium shifts to the left, I suppose that must mean more reactants are formed in relation to the products, so &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; would decrease.&lt;/p&gt;&#xA;&lt;p&gt;But the answer key says that &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; remains unchanged. This is coherent with what I've been taught, but I don't understand how the equilibrium can change yet have &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; remain constant.&lt;/p&gt;&#xA;&lt;p&gt;I'm just afraid I've got some of the theory mixed up, so I wanted to get some clarity. Could someone explain this please?&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-07-03T06:48:29.773" Title="How can the equilibrium shift, while Kc remains constant?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4978" PostTypeId="1" CreationDate="2013-05-12T21:15:52.587" Score="0" ViewCount="101" Body="&lt;p&gt;I would just like some reputable textbooks and or publication companies that have extensive research on plastics, as, for example, dealing with life cycles of plastics, degradation factors, effect on life, polymer types, etc.&lt;/p&gt;&#xA;" OwnerUserId="1309" LastActivityDate="2013-07-12T12:57:29.030" Title="Plastic: Reputable Publications" Tags="&lt;biochemistry&gt;&lt;polymers&gt;&lt;plastics&gt;&lt;reference-request&gt;&lt;books&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4979" PostTypeId="1" AcceptedAnswerId="5113" CreationDate="2013-05-13T07:27:13.923" Score="12" ViewCount="16966" Body="&lt;p&gt;I keep trying to figure out how current is conducted through an electrolyte but all I can find are BS half answers. They say the ions conduct, but the specifics are poorly explained or absent. I understand that if you, for example, put sodium chloride in water it will disassociate into positive sodium ions and negative chlorine ions, and that an electric field would cause the ions to move, allowing a current to flow. But what happens once the charges get to the electrodes? Obviously, they can't just stay there in elemental form stably. I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas. If this is all correct, it seems one would end up with sodium and hydroxide ions. Does it still conduct then? If so, how so? And don't you have a solution of sodium hydroxide? It seems this process necessitates that the water be split. Is there any salt or other compound which would act similarly but not result in electrolysis (just conducting)?&lt;/p&gt;&#xA;" OwnerUserId="799" LastActivityDate="2015-05-15T06:40:22.307" Title="Conducting current in electrolytes" Tags="&lt;electrochemistry&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4980" PostTypeId="2" ParentId="4979" CreationDate="2013-05-13T10:37:50.993" Score="5" Body="&lt;p&gt;Actually if you look at electroplating you would see that the metal ions once give an electron at the cathode do turn into elemental metals and plate the electrode. This is how you electroplate things. You can't conduct forever as eventually you will deplete the ions in the solution, much like a battery will run out of charge if it is driving a current rather than the reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use molten salt, you will get sodium at one electrode and chlorine gas at the other. If you use a solution you will either get O2 and H2 at the two electrodes, or the ions precipitating out depending which reaction is more favourable.&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2013-05-13T10:37:50.993" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4981" PostTypeId="2" ParentId="4977" CreationDate="2013-05-13T10:43:27.170" Score="13" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; is related to the ratio of reactants to products &lt;strong&gt;at equilibrium&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction is currently at equilibrium, and you add more products then the reaction is now &lt;strong&gt;out&lt;/strong&gt; of equilibrium and the reverse reaction will happen until it is back in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like the wording of &quot;equilibrium shifts to the left&quot; myself, I would say that reverse reactions occurs to restore equilibrium. But &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; doesn't change based on reactants/products concentrations since its the ratio at equilibrium. &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; &lt;strong&gt;will&lt;/strong&gt; normally depend on temperature though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the options you have, &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="41328" LastEditDate="2019-07-03T06:48:29.773" LastActivityDate="2019-07-03T06:48:29.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="4982" PostTypeId="2" ParentId="4977" CreationDate="2013-05-13T11:51:22.333" Score="12" Body="&lt;p&gt;To add to Nick's great answer, I believe you are confusing $K_c$, the equilibrium constant, with $Q_c$, the reaction quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction quotient is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, under any conditions or relative amounts of the various species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_\mathrm c =\dfrac{[\ce{CrO4^{2-}}]^2[\ce{H+}]}{[\ce{Cr2O7^{2-}}][\ce{H2O}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant describes which combinations of reactants and products constitute equilibrium, and there exists an infinite set of such combinations at a given temperature. However, they all must have $Q_\mathrm c = K_\mathrm c$. The difference in the formula for $K_\mathrm c$ below (from the formula for $Q$ above) is the use of equilibrium concentrations. &#xA;$$K_\mathrm c= Q_\mathrm {c,eq} =\frac{[\ce{CrO4^{2-}}]^2_{\text{eq}}[\ce{H+}]_{\text{eq}}}{[\ce{Cr2O7^{2-}}]_{\text{eq}}[\ce{H2O}]_{\text{eq}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an example, where $K_\mathrm c=1$. There are infinitely many concentration combinations that we could have that would work. Three of them are shown below.&#xA;$$1. \ \ \ [\ce{CrO4^{2-}}]=[\ce{H+}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{1 M}$$&#xA;$$2. \ \ \ [\ce{CrO4^{2-}}]=\pu{4 M};\ \ \ \ [\ce{Cr2O7^{2-}}]=[\ce{H2O}]=\pu{2 M}; \ \ \ \ [\ce{H+}]=\pu{1 M}$$&#xA;$$3. \ \ \ [\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=\pu{1 M};\ \ \ \ [\ce{H2O}]=[\ce{H+}]=\pu{0.987 M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take the first case, where all concentrations are equal, and increase the concentration of $\ce{H+}$ by $\pu{0.5 M}$, now we have the following:&#xA;$$[\ce{CrO4^{2-}}]=[\ce{Cr2O7^{2-}}]=[\ce{H2O}]=1\ M; \ \ \ \ [\ce{H+}]=\pu{1.5 M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we calculate $Q_c$, we find that $Q_c=1.5$. Thus, the system is &lt;em&gt;no longer at equilibrium&lt;/em&gt; because $Q_c\neq K_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c=\frac{(\pu{1 M})^2(\pu{1.5 M})}{(\pu{1 M})(\pu{1 M})}=1.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will shift towards the reactants (since $Q_\mathrm c&amp;gt;K_\mathrm c$) to reestablish equilibrium. A little algebra, and we can figure out just how much. The product concentrations will decrease and the reactant concentrations will increase, each by  $nx$, where $x$ is the concentration change and $n$ is the stoichiometric coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{CrO4^{2-}}]_{\text{eq}}=1-2x\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.5-x\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1+x\ M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm c=\frac{(1-2x\ M)^2(1.5-x\ M)}{(1+x\ M)(1+x\ M)}=1$$&#xA;$$K_\mathrm c=\frac{(1-4x+4x^2)(1.5-x)}{(1+2x+x^2)}=\frac{(1.5-7x+10x^2-16x^3)}{(1+2x+x^2)}=1$$&#xA;$$(1+2x+x^2)&amp;gt;0\ \ \text{for}\ \ x&amp;gt;0$$&#xA;$$1.5-7x+10x^2-16x^3=1+2x+x^2$$&#xA;$$0=16x^3-9x^2+9x-0.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above expression has three roots, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=0=16x%5E3-9x%5E2%2b9x-0.5&quot; rel=&quot;noreferrer&quot;&gt;two of which are imaginary&lt;/a&gt;. The real root is $x=0.0586353$. Thus, the new equilibrium concentrations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{CrO4^{2-}}]_{\text{eq}}=0.883\ M; \ \ \ \ [\ce{H+}]_{\text{eq}}=1.441\ M\ \ \ \ [\ce{Cr2O7^{2-}}]_{\text{eq}}=[\ce{H2O}]_{\text{eq}}=1.059\ M$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-01T02:14:28.337" LastActivityDate="2018-03-01T02:14:28.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4983" PostTypeId="2" ParentId="4978" CreationDate="2013-05-13T12:01:30.930" Score="1" Body="&lt;p&gt;A quick search in my university's library for &lt;code&gt;Life cycle polymer&lt;/code&gt; brings me two reputable looking books:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1855737396&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Green composites: Polymer composites and the environment&lt;/em&gt;&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1107000238&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Fundamentals of Materials for Energy and Environmental Sustainability&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://scholar.google.com/scholar?hl=en&amp;amp;q=lifecycle%20analysis%20of%20plastic&amp;amp;btnG=&amp;amp;as_sdt=1,21&amp;amp;as_sdtp=&quot; rel=&quot;nofollow&quot;&gt;Google Scholar search&lt;/a&gt; for &lt;code&gt;life cycle analysis of plastic&lt;/code&gt; yields several good looking hits. A &lt;a href=&quot;http://scholar.google.com/scholar?q=life%20cycle%20analysis%20of%20polymers&amp;amp;btnG=&amp;amp;hl=en&amp;amp;as_sdt=0,21&quot; rel=&quot;nofollow&quot;&gt;similar result&lt;/a&gt; for &lt;code&gt;life cycle analysis of polymers&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of information is out there, and there is a lot of it. My examples hopefully will inspire you to spend the effort (and it's not that much effort) to look for it. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-13T12:01:30.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4985" PostTypeId="1" CreationDate="2013-05-13T23:24:34.083" Score="1" ViewCount="790" Body="&lt;p&gt;I am trying to draw the structure of this complex &lt;span class=&quot;math-container&quot;&gt;$\ce{[Co(NH3)5NO2]^{2+}},$&lt;/span&gt; however I am not sure if the linkage isomers of this complex like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oasrg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oasrg.png&quot; alt=&quot;Linkage isomers of the complex&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="792" LastEditorUserId="41328" LastEditDate="2019-05-31T12:12:35.590" LastActivityDate="2019-05-31T12:12:35.590" Title="What Could Formations of Linkage Isomers of [Co(NH₃)₅NO₂]²⁺ be like?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4986" PostTypeId="2" ParentId="4985" CreationDate="2013-05-13T23:24:34.083" Score="3" Body="&lt;p&gt;This Co-ordination compound of were studied by Jorgensen and Werner who observed that were two compounds of the same chemical formula but of different colors. (We can see these two geometries in the image). &#xA;A red form of low stability converted readily to a yellow form the $\ce{M-NO2}$ nitro isomer, based on comparison with compounds with similar colors. This conclusion was later confirmed, and kinetic and $\ce{^{18}O}$ labeling experiments showed that conversion of one form to the other is strictly intramolecular, not a result of dissociation of the $\ce{NO2^-}$ ion followed by reattachment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;&lt;em&gt;Miessler, Tarr,Inorg. Chem., 2nd Edition, 1998&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="4231" LastEditDate="2015-03-01T14:31:16.597" LastActivityDate="2015-03-01T14:31:16.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4987" PostTypeId="1" CreationDate="2013-05-14T02:32:19.550" Score="5" ViewCount="2239" Body="&lt;p&gt;I am using DCM ( dichloromethane) in my research, as a solvent. At the end of the day I have to separate the salute and measure it. &#xA;Up until now, I have been using a constant temperature oven, set a about 65 degree Celsius (evaporation temp of DCM). The exhaust DCM is expelled into the atmosphere and the dried salute is then cooled down and measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to replace the oven with a rotary evaporator. However I have an argument with my advisor. He claims that I require two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aspirator /vacuum pump&lt;/li&gt;&#xA;&lt;li&gt;Draft Chamber&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While I agree that a vacuum pump can be of use, I do not require my solvent to boil, therefore I do not need vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the draft chamber, I completely fail to see its implication.&#xA;I believe that since the rotary evaporator is a completely closed system, there is no way DCM vapor could leak, rendering the draft chamber useless.&lt;/p&gt;&#xA;" OwnerUserId="1513" LastActivityDate="2020-03-02T06:29:46.170" Title="Does a rotary evaporator need a draft chamber and a vacuum pump?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4988" PostTypeId="2" ParentId="4987" CreationDate="2013-05-14T02:57:13.630" Score="5" Body="&lt;p&gt;I think you are taking too dogmatic a position overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Vacuum also increases the rate of evaporation, without boiling.  This is just practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Your claim that your system is &quot;completely closed&quot; is over confident. What if you are wrong and you have a leak?  Now you are free to dismiss this, but this is the safety issue your adviser is getting at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, it would be easy to incorporate your practice with industry best practice.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-05-14T02:57:13.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4989" PostTypeId="1" CreationDate="2013-05-14T03:32:19.270" Score="11" ViewCount="1367" Body="&lt;p&gt;We know that the Greeks were the first scientists (or better, philosophers) to introduce the idea that matter is made up of little “chunks”, and they named these chunks “atoms”. This idea that matter comes in discrete small pieces was termed wrong and right over centuries when John Dalton, in the 19th century, constructed a theoretical framework in which he could explain in an elegant way the work of scientists such as Joseph Priestley and Antoine Lavoisier on qualitative and quantitative properties of matter and its transformations.&#xA;The framework which John Dalton used to better understand the work of other scientists had in its core the “language” of atoms, that is, he used the concept of atoms to explain, for example, why reactions only occurs if we have quantities “in proportion”. Note that the work of John Dalton doesn't proof that atoms are real, it only shows that thinking of matter in terms of atoms is a suitable way to understand the results coming out of experiments.&#xA;Scientists only proved that atoms are real at the beginning of the 20th century, and that discovery gave birth to Quantum Mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now comes my question:&lt;/strong&gt;&#xA;When did scientists prove that molecules “are real”? In other words, what is the history of the discovery of (the shape of) molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists have been playing around with toy models of particles for a long time but only a short time ago IBM researchers managed to “take a picture” of an actual molecule, and its configuration matches exactly the models we've been using for a long time, for example this pentacene molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qyzswm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When did we first learn that molecules have a “geometry”? That is, an exact arrangement in space?&lt;/p&gt;&#xA;" OwnerUserId="1631" LastEditorUserId="22" LastEditDate="2013-05-18T18:22:40.830" LastActivityDate="2014-09-28T13:36:39.940" Title="When did the atom Theory Become indisputable?" Tags="&lt;molecules&gt;&lt;bond&gt;&lt;history-of-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4990" PostTypeId="2" ParentId="4987" CreationDate="2013-05-14T07:55:04.960" Score="4" Body="&lt;p&gt;I've never seen a rotary evaporator without a vacuum pump. They generally use a water bath and are severely limited in the temperature they can achieve due to that. It is also much easier to adjust the pressure to get the desired evaporation rate than to try that by changing the temperature. And in your case, a rotary evaporator without a vacuum pump would have to be an open system, so DCM could easily get into the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotary evaporators are a closed system, and you would generally notice any big leaks because you won't be able to achieve a good vacuum. For common solvents I see no huge advantage in putting them into a fume hood, and they often aren't because the space in the hood is rather precious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But DCM is potentially carcinogenic, so I can certainly see where your advisor is coming from and I would also prefer to put it into a fume hood. You also handle the open flasks when attaching and detaching your sample and the solvent waste to the rotary evaporator, doing that under a fume hood is a good idea.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-05-14T07:55:04.960" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4992" PostTypeId="1" CreationDate="2013-05-14T08:30:08.803" Score="6" ViewCount="42159" Body="&lt;p&gt;In polysachharides like sucrose or any other disachhaharides, how is it that we assign whether the glycosidic bond is alpha/beta ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the images here: &lt;a href=&quot;http://forums.studentdoctor.net/showthread.php?t=898328&quot; rel=&quot;nofollow&quot;&gt;http://forums.studentdoctor.net/showthread.php?t=898328&lt;/a&gt; (kindly refer to the images of the original question, both of them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one decide which glycosidic bonding is alpha and which is beta ? What if one of the monosachharides is alpha and one is beta constituting the disachharides ? How do you know that in the final acetal formation which sachharide acted as nucleophile and which acted as electrophile after seeing the disaccharide , take for example sucrose , glucose/sucrose (?) which acted as nucleophile any why ?&lt;/p&gt;&#xA;" OwnerUserId="1634" LastEditorUserId="7448" LastEditDate="2015-10-09T15:57:25.437" LastActivityDate="2016-03-05T19:33:33.950" Title="How do we assign alpha/beta status in polysaccharides?" Tags="&lt;biochemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4993" PostTypeId="1" CreationDate="2013-05-14T08:44:01.833" Score="2" ViewCount="1583" Body="&lt;p&gt;In &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198792816&quot; rel=&quot;nofollow&quot;&gt;Price &amp;amp; Dwek&lt;/a&gt;, I find a small section discussing entropy of activation for three examples&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1: \ce{Ce^{4+}} + \ce{EDTA}^{4-} \rightarrow (\ce{CeEDTA})\\&#xA;2, \text{S}_N\text{1 reaction}: \ce{AB} \rightarrow [A^+...B^-]^\ddagger\\&#xA;3, \text{S}_N\text{2 reaction}: \ce{AB} + \ce{C^-} \rightarrow [\ce{C...A...B}^-]^\ddagger&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the signs of the entropy of activtion $\Delta S ^\ddagger$ in 1-3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1), the argument is that the formation of the complex releases water (because the complex is neutral), and so the entropy change is positive ($S_{final} &amp;gt; S_{initial}$ and $\Delta S = S_{final} - S_{initial}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3) (yes 3), the argument is that two molecules combine to form one, the solvation being less strongly affected compared to 1) because the charges are lower, and so $\Delta S &amp;lt; 0$ (two molecules have more degrees of translational freedom than one and on combination the average volume decreases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for 2), I dont get it that $\Delta S &amp;gt; 0$. They only say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Separation of ions in transition state, hence a positive $\Delta S^\ddagger$.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain what this means? I would have guessed that the charges in the transition state require ordering of the solvation shell around the complex and so $S_{final} &amp;lt; S_{initial}$. &lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-05-14T08:44:01.833" Title="Entropy of activation" Tags="&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4997" PostTypeId="1" AcceptedAnswerId="4998" CreationDate="2013-05-14T12:42:12.223" Score="1" ViewCount="28654" Body="&lt;p&gt;You don't need to work out anything for me. I am just giving an example so you can explain something to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;$$\ce{ClF3(g) &amp;lt;=&amp;gt; ClF(g) + F2(g)}\\ \ \ K_c=8.77\times10^{-14} \ \ @\  279\ \text{K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $[\ce{ClF}]$ at equilibrium if initially $[\ce{ClF3}]=2.50 \ M$ and $[\ce{F2}]= 1.00\ M$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get everything that just happened above. Thats how I would do it but then, my lecturer said the $x$ then becomes negligible for some reason something about the CIF. and now the equation becomes:&#xA;$K_c= \dfrac{(x)(1.00)}{2.50}$ and so $x=2.2\times10^{-13}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the $x$ and why? Also, what exactly is $K_c$ in chemistry what does it represent? I know how to work out questions but I don't know why. &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="22102" LastEditDate="2017-03-12T01:54:02.900" LastActivityDate="2017-03-12T01:54:02.900" Title="Why x in equilibrium calculations can be &quot;ignored&quot;" Tags="&lt;equilibrium&gt;&lt;theoretical-chemistry&gt;&lt;significant-figures&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4998" PostTypeId="2" ParentId="4997" CreationDate="2013-05-14T15:40:34.540" Score="8" Body="&lt;p&gt;In problems like this, the maths can sometimes be simplified by making assumptions so that you do not need to do difficult algebra. You could always do the harder algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a valid assumption is that when the equilibrium constant is very small in comparison to the concentrations given, then the change in those concentrations (which is represent by $x$) will also be very small and can be neglected. I'll work through your example the long way to show you why this assumption is valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working through the algebra exactly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(x)(1.00+x)=(8.77\times10^{-14})(2.50-x)$$&#xA;$$1.00x-x^2=2.19_{25}\times 10^{-13}-8.77\times10^{-14}x$$&#xA;$$0=x^2+(-8.77\times10^{-14}-1)x+2.19_{25}\times 10^{-13}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_formula#Derivation_of_the_quadratic_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;quadratic formula&lt;/a&gt; gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\dfrac{(1+8.77\times10^{-14}\pm \sqrt{(-1-8.77\times10^{-14})^2 +4(2.19_{25}\times 10^{-13})}}{2}=[1,\ 2.19\times10^{-13}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first root $x = 1$ is nonsense. The second is the root you were given (when rounded to the correct number of sig figs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $x$ is so small compared to the initial concentrations of $\ce{ClF3}$ and $\ce{F2}$ that the concentrations of these species &lt;strong&gt;do not change within the magnitude of correct use of significant figures&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{ClF3}]_{eq}=2.50-2.1925\times10^{-13}=2.50 \ \text{to 2 decimal places}$$&#xA;$$[\ce{F2}]_{eq}=1.00+2.1925\times10^{-13}=1.00 \ \text{to 2 decimal places}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maths above are equivalent to the assumption that since $x&amp;lt;&amp;lt;1.00$ and $x&amp;lt;&amp;lt;1.00$, $1.00+x\approx1.00$ and $2.50-x\approx2.50$. Whether you do it at the beginning or at the end, you get the same result: $x=2.19\times10^{-13}$. Making the simplification at the beginning is operationally easier and avoid potential algebra and arithmetic errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption is not valid if the initial concentrations are close in value to $x$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="22102" LastEditDate="2017-03-12T01:05:34.223" LastActivityDate="2017-03-12T01:05:34.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4999" PostTypeId="1" CreationDate="2013-05-14T19:31:52.660" Score="6" ViewCount="42244" Body="&lt;p&gt;(This example is purely hypothetical.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the reaction $$\ce{H2(g) + O2(g) &amp;lt;=&amp;gt; H2O2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $T = 500\ \mathrm{K}$. The reaction reaches equilibrium at the following concentrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[H2]} = \ce{[O2]} = 5 \times 10^{-3}\ \mathrm{mol\ dm^{-3}}$$&#xA;$$\ce{[H2O2]} = 4\times10^{-5}\ \mathrm{mol\ dm^{-3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $$K_c = \left(\frac{\ce{[H2O2]}}{\ce{[H2][O2]}}\right) = 1.6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the total concentration of the reactants is $250$ times higher than the concentration of the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, as $K_c &amp;gt; 1$, per definition, the products are favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems counterintuitive to me, and my Chemistry teacher couldn't really explain it to me, so I was hoping someone here could explain why, even when the reactants are so much more plentiful, the products are considered to be favored.&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="16683" LastEditDate="2016-10-16T15:25:01.427" LastActivityDate="2016-10-16T15:25:33.547" Title="What does it mean for a reaction to favor the reactants/products?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5000" PostTypeId="1" CreationDate="2013-05-15T01:19:26.917" Score="1" ViewCount="8467" Body="&lt;p&gt;Why does Butane burn blue? From my (very) basic understanding of chemistry, carbon and hydrogen do not burn blue, their radiation spectra are different-so why should the combination burn blue?&#xA;Please elaborate as much as possible, about bonds, orbitals, electrons, and the like!&lt;/p&gt;&#xA;" OwnerDisplayName="user1638" LastEditorUserId="4945" LastEditDate="2015-10-23T16:56:17.903" LastActivityDate="2015-10-23T16:56:17.903" Title="Butane Burning Color" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5001" PostTypeId="2" ParentId="4999" CreationDate="2013-05-15T01:59:33.193" Score="5" Body="&lt;p&gt;Fundamentally, though it may seem intuitive, adding concentrations on either side of the reaction and comparing them to see which is larger isn't a measure of favourability, since $K_c$ is defined by the multiplication of the concentrations.  It's like taking two rectangles and trying to measure which has a larger area by comparing their perimeters; it doesn't work in general unless you specify some additional constraint(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under your definition of favourability by addition, you're weighing reagents and products unfairly, in some sense, because in the expression of $K_c$ there are two reagents and one product. This is a bit (if indirectly) reminiscent of trying to compare salt solubilities by comparing their $K_\mathrm{sp}$ directly instead of actually calculating their solubilities; it doesn't always work, because $K_\mathrm{sp}$ of salts are weighted by how many ions are released when dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not strictly true that the products would be favoured by your measure for any $K_c &amp;gt; 1$ even if the number of reagents and products were equal - just add one reagent in heavy excess. For example, consider the hypothetical reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B -&amp;gt; C + D} \qquad \qquad  K_c=\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}=1000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution for equilibrium concentrations is $[\ce{A}]=1\ \mathrm{M}$, $[\ce{B}]=10^{-5}\ \mathrm{M}$ and $[\ce{C}]=[\ce{D}]=10^{-1}\ \mathrm{M}$, and here $[\ce{A}]+[\ce{B}]&amp;gt;[\ce{C}]+[\ce{D}]$. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="16683" LastEditDate="2016-10-16T15:25:15.693" LastActivityDate="2016-10-16T15:25:15.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5002" PostTypeId="1" AcceptedAnswerId="5004" CreationDate="2013-05-15T06:26:24.470" Score="3" ViewCount="2341" Body="&lt;p&gt;If a reaction is exothermic it means that the energy state of the products is lower than that of the reactants so this will be the state that 'nature' will naturally strive for. Often you see that stability of molecules is argued in this way (e.g. in one of the answers &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4760/why-do-compounds-like-cesf6-and-cesf4-exist-but-cesh6-and-cesh&quot;&gt;here&lt;/a&gt;), saying that the higher the exothermicity of the decomposition reaction, the more unstable the molecule and thus the more unlikely that it will exist in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether this reasoning is correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am doubting this, is that the reaction enthalpy $\Delta H_r$ is not the only parameter of importance in the energetics of chemical reactions: the activation energy $\Delta E_a$ can also play a major role. To clarify my point, take a look at the diagram for 2 hypothetical reactions I have made. Reaction 1 has a low activation energy and a low exothermicity, whereas 2 has a high activation energy and a high exothermicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging solely by exothermicity you would conclude that the reactants of reaction 2 are more unstable, but if you look at the activation energies for the reactions than I would argue that the reactants of reaction 1 are more unstable because they more easily react to form their products. Of course having a high activation energy would at best make the reactants meta-stable, but metastability could last for decades as well given the right conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UrIQU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my general question is: can reaction exothermicity be used to say something about the stability of molecules and what is the role of activation energy in this story?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2013-05-15T11:49:49.663" Title="What does reaction exothermicity say about the stability of molecules" Tags="&lt;reaction-mechanism&gt;&lt;theoretical-chemistry&gt;&lt;energy&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5004" PostTypeId="2" ParentId="5002" CreationDate="2013-05-15T11:49:49.663" Score="3" Body="&lt;p&gt;Good question. There exists two types of stability - thermodynamic stability (determined by $\Delta G_{reaction}$, or $\Delta H_{reaction}$ approximately) and kinetic stability (determined by $\Delta G ^{\ddagger}$, or $E_{activation}$ approximately). The reactant compound in your first reaction is &lt;em&gt;thermodynamically&lt;/em&gt; more stable than the second, since it sits at a lower potential energy, but it is &lt;em&gt;kinetically&lt;/em&gt; less stable, since there is a smaller barrier for decomposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually context is sufficient to determine what kind of stability is being talked about, though in general I think stability is more often mentioned in the thermodynamic sense than the kinetic, since thermodynamic stability is related to the &lt;em&gt;reaction&lt;/em&gt; free energy (a sum of its enthalpy change and a factor containing its entropy change) and is more commonly tabulated and more readily measured than &lt;em&gt;activation&lt;/em&gt; free energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, all explosives are quite thermodynamically unstable (otherwise they wouldn't be very good at exploding!), but we still speak of &quot;stable explosives&quot;, such as TNT or RDX and &quot;unstable explosives&quot;, such as $\ce{NI_3}$ or $\ce{Hg(CNO)_2}$. Clearly kinetic stability is what is taken into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In constrast, when studying acidity and basicity of substances, it's common to justify reactions based on the &quot;stability&quot; of the acids/bases/conjugate species, but in this case, we're talking about thermodynamic stability, as it defines the acidity/basicity constants.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-05-15T11:49:49.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5005" PostTypeId="1" AcceptedAnswerId="5014" CreationDate="2013-05-15T12:27:48.767" Score="1" ViewCount="15626" Body="&lt;p&gt;I missed my lesson on periodicity so had to teach myself, and have always forgotten to ask my teacher to explain to me why these trends are seen, which, unfortunately, the textbooks don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Density&lt;/strong&gt;: the density of an element tends to increase across a period up to group 3, and then starts to decrease and become very low at group 5 or so. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Melting and boiling point&lt;/strong&gt;: they tend to rise up to about group 3 or 4, before falling rapidly after that. I would suggest that this is because group 4 elements tend to have giant covalent structures, so also a higher boiling point, whereas on the right hand side of a period they are simple molecular structures, so have a low boiling point. Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atomic size&lt;/strong&gt;: atomic size always decreases along a period from left to right. My suggestion is that this is because there are more electrons and more protons as the group increases, so there becomes a stronger attraction between them, pulling the electrons closer to the nucleus. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only one I have no idea about really is density, and the other two I just need clarification on.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2019-10-20T16:32:56.477" Title="Why is periodicity seen in these certain properties?" Tags="&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5006" PostTypeId="1" CreationDate="2013-05-15T13:16:34.933" Score="6" ViewCount="3451" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a bottle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Salicylaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;salicylaldehyde&lt;/a&gt;. And you want to make &lt;a href=&quot;http://en.wikipedia.org/wiki/Salicylic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;salicylic acid&lt;/a&gt; from salicylaldehyde. You can use all reagents and catalyst to make salicylic acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think salicylic acid can be synthesized from salicylaldehyde since salicylaldehyde has a structure similar to salicylic acid. I'm thinking that it might be possible to use $\ce{KMnO4}$, but I'm afraid the hydroxyl group attached to the benzene would be oxidized too.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="46306" LastEditDate="2018-03-01T05:41:30.880" LastActivityDate="2018-03-01T05:41:30.880" Title="Can salicylic acid be synthesized from salicylaldehyde?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5008" PostTypeId="2" ParentId="4999" CreationDate="2013-05-15T14:59:06.533" Score="7" Body="&lt;p&gt;You may not have had kinetics (reaction rates) yet, but I find it really useful in understanding equilibria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kinetics of a reaction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules in solution (gas or liquid) move randomly. Because they need to bump into each other to react, the probability of reaction increases as concentrations of reactants increase. The rate of the reaction is then proportional to the concentrations of reactants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B \rightarrow C}$$&#xA;$$\mathrm{rate} = k_\mathrm{forward}\ce{[A][B]}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the reverse reaction has a rate dependent on concentration of product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{rate} = k_\mathrm{reverse}\ce{[C]}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From Kinetics to Equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, the forward rate is equal to the reverse rate, and concentrations of both reactants and products remain unchanged. $$k_\mathrm{forward}\ce{[A][B]}=k_\mathrm{reverse}\ce{[C]}$$ We can rearrange to get: $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}} = \ce{\frac{[C]}{[A][B]}}.$$ We can substitute the right side of the equation to be the equilibrium constant, and we get $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}}=K_c$$ Thus, when we say that $K_c&amp;gt;1$ we are also saying that $k_\mathrm{forward} &amp;gt; k_\mathrm{reverse}$ and inversely if $K_c&amp;lt;1$ then $k_\mathrm{reverse} &amp;gt; k_\mathrm{forward}.$ To favor either the reactants or the products in equilibrium is to say the &lt;em&gt;formation&lt;/em&gt; of either the reactants or products is favored, as indicated by the rate constants.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="16683" LastEditDate="2016-10-16T15:25:33.547" LastActivityDate="2016-10-16T15:25:33.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5009" PostTypeId="2" ParentId="5006" CreationDate="2013-05-15T16:04:24.463" Score="1" Body="&lt;p&gt;Salicylic acid can be synthesized from salicylaldehyde using $\ce {KMnO4}$ without oxidising the hydroxy group ($\ce {-OH}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydroxy group that attached to the benzene isn't oxidized, because the C atom that bonded with the hydroxy has already use 2 electrons for sigma bonding and 1 electron for pi bonding. The electron for pi bonding is used for resonance on benzene cycle, so it can't be used for bond with O atom to make oxo group ($\ce {=O}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the electron of pi bonding is used for bonding with O atom to make oxo group, the benzene will change to a cyclohexadiene.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2013-05-15T16:04:24.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5011" PostTypeId="1" CreationDate="2013-05-15T17:25:48.793" Score="0" ViewCount="4138" Body="&lt;p&gt;Below you can see the chemical reaction I'm working with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH_3CH_2OH(l) + 3O_2(g) -&amp;gt; 2CO_2 (g) + 3H_2O(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got so far:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/88gwA.png&quot; alt=&quot;http://i.stack.imgur.com/3zgNk.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the volume of the product of carbon dioxide gas (the pressure is 1 bar, and the temperature is 20 deg Celsius).&lt;/p&gt;&#xA;" OwnerUserId="1642" LastEditorUserId="194" LastEditDate="2018-02-19T04:48:28.753" LastActivityDate="2018-02-19T04:48:28.753" Title="Find the volume of the carbon dioxide product" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5012" PostTypeId="2" ParentId="5011" CreationDate="2013-05-15T17:56:35.103" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You need to determine the limiting reactant. Use the equation and the number of moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_3CH_2OH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In the reaction, you need 1 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_3CH_2OH}$&lt;/span&gt; and 3 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt; to get 2 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt;. If you have less than 1 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_3CH_2OH}$&lt;/span&gt; but 3 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt;, you cannot get 2 mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt;. In such a case, &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_3CH_2OH}$&lt;/span&gt; is the limiting reactant. The amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt; will thus be lower than 2 mol.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the number of moles of the limiting reactant, you should be able to calculate the number of moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt; produced. Use proportions!&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last step will involve using the ideal gas equation, by assuming that &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt; behaves like an ideal gas. &lt;span class=&quot;math-container&quot;&gt;$$PV=nRT$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You have the pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, the temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; (the number of moles of the &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is a constant. Find the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-05-15T17:56:35.103" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5014" PostTypeId="2" ParentId="5005" CreationDate="2013-05-15T18:25:38.640" Score="1" Body="&lt;p&gt;For 2. I would add, on top of what you already have, that the elements start as metallic elements, hence we have metallic bonding. Across the period, the number of electrons increase, meaning there are more delocalised electrons among the metal cations so that electrostatic attraction increases along the period. For period 2, that's the trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For period 3 however, it is a bit more tricky because of the transition metal elements and I think Nicolau explains it better than I could in &lt;a href=&quot;https://chemistry.stackexchange.com/a/4829/1409&quot;&gt;this answer&lt;/a&gt;. But I don't think that you will be required to know those for your level, but it's still interesting for your own knowledge. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3. I would add that the number of electron shells containing electrons remains constant. If it increased instead, the atomic radius would increase as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you could use 2. and 3. to explain 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we observe that the atomic mass increases along the period, and that the atomic radius decreases. For giant structures, this generally means that the density increases ($\rho = \dfrac{m}{V}$ and while $m$ increases, $V$ is decreasing slightly, so $\rho$ increases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we hit the simple molecular structures, the density sharply falls again, because of the average distance between the atoms of the element has drastically risen, so the volume increases. With $m$ increasing gradually and a sharp rise in $V$, you get a sharp decrease in $\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These observations change down the groups though. For instance, Astatine is a solid, so you won't expect a huge fall in density across period 6 around group 5. But for the purpose of your syllabus coverage (I'm assuming High School), those should be enough.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2013-05-16T06:48:55.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5015" PostTypeId="1" CreationDate="2013-05-15T21:14:09.157" Score="6" ViewCount="285" Body="&lt;p&gt;The title more or less contains my question - why does the Förster resonance energy transfer (FRET) (photon?)-based energy transfer mechanism, which folks sometimes use as a spectroscopic ruler to measure the inter-particle spacing between dye pairs, exhibit $1/r^6$ scaling for energy transfer efficiency?  Since we're in 3-space when performing a FRET measurement, I could understand some $1/r^3$ contribution due to us having an expanding sphere of radius $r$, but where does the rest of the distance-dependent efficiency dropoff come from?&lt;/p&gt;&#xA;" OwnerUserId="1644" LastEditorUserId="16683" LastEditDate="2017-01-08T12:46:44.997" LastActivityDate="2017-01-09T11:25:44.420" Title="Why does the Förster resonance energy transfer scale as 1/r⁶?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5017" PostTypeId="2" ParentId="4989" CreationDate="2013-05-16T01:03:02.513" Score="0" Body="&lt;p&gt;This is hard to answer.  The concept of atoms and molecules was very fruitful in the 19th century.  It could be used to prove that there were only three isomers of chlorobenzene, just for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the turn of the 20th century and the working out of kinetic theory, molecules became more &quot;real&quot;.  But it was not until late in the 20th century that molecules were first &quot;seen&quot; using a scanning electron microscope.  You can look that up in Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I personally feel it is wrong to attribute the notion of atoms to the Greeks.  Things only become &quot;discovered&quot; when the idea catches hold and informs ongoing research.  That would make Dalton the &quot;discoverer&quot; of atoms.  Just my view, others may differ.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-16T01:03:02.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5018" PostTypeId="1" CreationDate="2013-05-16T02:21:07.817" Score="0" ViewCount="4570" Body="&lt;p&gt;How does the balanced reaction look for these compounds? I found that phosphoric acid instead of hydrogen phosphate creates a much easier problem, but the teacher said hydrogen phosphate and I can't seem to balance it.&lt;/p&gt;&#xA;" OwnerUserId="1645" LastEditorUserId="16683" LastEditDate="2017-04-07T12:40:00.263" LastActivityDate="2017-04-07T12:40:00.263" Title="Calcium carbonate and hydrogen phosphate producing calcium phosphate, carbon dioxide, and water" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5023" PostTypeId="1" CreationDate="2013-05-16T19:29:13.090" Score="11" ViewCount="91307" Body="&lt;p&gt;I'm not quite sure I understand this. My question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calcium chloride is a salt used widely to melt ice on sidewalks and roads.  Explain why one mole of $\ce{CaCl2}$ would be more effective than one mole of $\ce{NaCl}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So from everything I've learned, $\ce{NaCl}$ would be more effective for the &lt;strong&gt;melting&lt;/strong&gt;, because calcium chloride dissociates into three ions and sodium chloride dissociates into two, so that would make the boiling point of water with calcium chloride higher.  Therefore, it shouldn't be used to &lt;strong&gt;melt&lt;/strong&gt; ice over sodium chloride, as it would make the water have a higher boiling point so wouldn't melt it as well!&lt;/p&gt;&#xA;" OwnerUserId="1456" LastEditorUserId="5026" LastEditDate="2018-02-25T07:37:48.120" LastActivityDate="2018-02-25T07:37:48.120" Title="Why calcium chloride is used to melt ice over sodium chloride" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5025" PostTypeId="1" AcceptedAnswerId="9407" CreationDate="2013-05-16T23:58:55.747" Score="2" ViewCount="3848" Body="&lt;p&gt;I made a solution of iron acetate by placing steel wool in one litre of acetic acid (vinegar $5\%$) and adding a few tablespoons of hydrogen peroxide of questionable age. The solution turned a nice brown almost instantly upon shaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q5Z1Dm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I left it overnight and it lost its colour while forming a white precipitate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ioFbWm.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added some more peroxide (new, good peroxide this time), and the moment the peroxide hit the clear an colorless solution it looked like dark red-brown food-coloring! Unfortunately I could only get out my camera fast enough to make the following photo: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZMNNm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I squeezed the air out of the steel wool before the above picture, which caused it to sink (noted this so you don't think something of its sinking). Added one-fourth cup hydrogen peroxide and shook:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DgCpJm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a flash from the top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hc3ram.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Held up to the light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mSf38m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result seems to be ferric acetate (III), since &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferric acetate (II)&lt;/a&gt; is light brown in color (according to the Wikipedia link)--possibly like the light brown in the first/top photo. &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ferric acetate (III)&lt;/a&gt; is supposed to be a red-brown color (again according to the Wikipedia link). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the clear an colourless solution in the second picture that I am not certain as to its composition. Any assistance will be appreciated in identifying that solution. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would hypothesize that the white precipitate is either acetate or carbonate (more likely carbonate). I would guess that the clear solution occurred as a result of the solution interacting with either the steel wool or the air. I took the steel wool out after making the final dark red solution and it has been 2 days now without it diminishing in colour, so it is likely not the air that was interacting with the solution. Perhaps, if I placed the steel wool back in the solution, it might go clear again. I'll post again in a day or two.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="5026" LastEditDate="2018-02-23T13:54:18.580" LastActivityDate="2018-02-23T13:54:18.580" Title="What is this colorless ferric solution?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5026" PostTypeId="2" ParentId="4989" CreationDate="2013-05-17T01:02:52.203" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short Answer&lt;/strong&gt;: Henry Moseley provided the first experimental evidence for the nuclear model of atoms through &lt;a href=&quot;http://en.wikipedia.org/wiki/Moseley%27s_law&quot; rel=&quot;nofollow&quot;&gt;X-ray observations&lt;/a&gt; in 1913. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proving something is a rather nebulous task when there is no judge or jury and consensus is arrived at gradually over a century rather than instantaneously. Many scientists would consider the atom &quot;proved&quot; decades apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro was the first to distinguish atoms from molecules, which alleviated much confusion regarding atoms and allowed quantitative description/experimental evidence as relating to Advogadro's number (a mole). That was during the early decades of the 1800s (19th century)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various scientists became convinced at various times over the next century. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Max Plank for an example. Around 1898 Plank became persuaded that the molecular approach was the only way forward for physics. He had formerly ascribed to the &quot;pea soup&quot; model where there were not discreet atoms. It was the molecular approach that allowed Plank to discover energy level quantization that is the foundation for quantum mechanics. E=hv&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_structure&quot; rel=&quot;nofollow&quot;&gt;Lewis dot structures&lt;/a&gt; were some of the earliest electronic approaches to molecular geometry (circa 1910), but long before that there was an understanding of basic molecular geometries as demonstrated in the correct postulation of Benzene's hexagonal ring structure in 1865. Carbon was known to form 4 bonds with other carbons in the late 1800s. Linus Pauling was the first to calculate bond angles based on quantum mechanics in 1931.&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2013-05-18T16:59:41.580" LastActivityDate="2013-05-18T16:59:41.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5027" PostTypeId="1" AcceptedAnswerId="10440" CreationDate="2013-05-17T02:45:26.730" Score="3" ViewCount="1353" Body="&lt;p&gt;Ferric Acetate (II) and (III) are &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mordant&quot; rel=&quot;nofollow&quot;&gt;mordants&lt;/a&gt;,&quot; which intensify wood staining in direct proportion to the tannin concentration in the wood. Some wood stainers add tea to increase the tannin content and thereby make the stain darker. How does Ferric Acetate act as a mordant in this case of &lt;a href=&quot;http://www.wwgoa.com/ebonizing-wood/&quot; rel=&quot;nofollow&quot;&gt;ebonizing&lt;/a&gt; (&lt;a href=&quot;https://web.archive.org/web/20130519014340/http://www.wwgoa.com/articles/projects/ebonizing-wood/&quot; rel=&quot;nofollow&quot;&gt;web archive&lt;/a&gt;) wood?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4945" LastEditDate="2015-01-13T06:01:40.047" LastActivityDate="2015-01-13T06:01:40.047" Title="How Does The Mordant, Ferric Acetate, Interact With Tannins When Ebonizing Wood?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;color&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5028" PostTypeId="1" AcceptedAnswerId="5029" CreationDate="2013-05-17T06:48:25.473" Score="5" ViewCount="153" Body="&lt;p&gt;In &lt;a href=&quot;http://accessmedicine.com/resourceTOC.aspx?resourceID=651&quot; rel=&quot;nofollow noreferrer&quot;&gt;Goodman &amp;amp; Gilman's The Pharmacological Basis of Therapeutics, 12e&lt;/a&gt; R,S-citalopram is referred to. The problem is that from what I am aware of citalopram only has one stereocentre, and hence I am forced to ask were the author having a stroke when they added the comma and really meant the racemic mixture of R and S stereoisomers?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wAe6w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1: Screenshot of the Relevant Page in Goodman and Gilman&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ugL9L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 2: Citalopram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The latter has the stereocentral carbon highlighted)&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="84" LastEditDate="2013-09-01T05:48:29.833" LastActivityDate="2013-09-01T05:48:29.833" Title="In Goodman and Gilman what does R,S-citalopram refer to?" Tags="&lt;nomenclature&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5029" PostTypeId="2" ParentId="5028" CreationDate="2013-05-17T08:30:29.993" Score="5" Body="&lt;p&gt;It seems fairly clear that the authors intended to refer to the racemic mixture there. You're right that a comma would normally be used to separate the locants of distinct stereocenters when multiple ones are present, but then the individual chiral carbons would have to be numbered, which obviously isn't the case here.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-05-17T08:30:29.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5030" PostTypeId="1" AcceptedAnswerId="5419" CreationDate="2013-05-17T10:39:19.920" Score="4" ViewCount="1057" Body="&lt;p&gt;My guess at why diesel produces less carbon monoxide is that more heat is required for combustion of diesel fuel, so there is less incomplete combustion. Therefore rather than carbon monoxide being formed and released, the ability to completely combust means that carbon dioxide is often formed instead. Correct me if I'm wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a risk of diesel fuel is that more carbon particulates are produced. But surely this is just a contradiction, because carbon particulates are a result of a lack of oxygen supply and incomplete combustion. Surely if there is enough complete combustion to form carbon dioxide over carbon monoxide, then there is a high enough oxygen supply and temperature to stop carbon particulates forming?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="30836" LastEditDate="2020-04-05T21:01:18.620" LastActivityDate="2020-04-05T21:01:18.620" Title="How can diesel fuel produce less carbon monoxide but more carbon particulates?" Tags="&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5033" PostTypeId="5" CreationDate="2013-05-17T15:34:11.790" Score="0" Body="" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-05-17T15:34:11.790" LastActivityDate="2013-05-17T15:34:11.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5034" PostTypeId="4" CreationDate="2013-05-17T15:34:11.790" Score="0" Body="For questions regarding the separation process consisting in the separation of a substance from a matrix. This tag can be used for generic extraction of elements from ore or more specific types of extraction such as: liquid-liquid extraction, and solid phase extraction." OwnerUserId="22" LastEditorUserId="9961" LastEditDate="2017-05-04T00:17:19.380" LastActivityDate="2017-05-04T00:17:19.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5035" PostTypeId="1" AcceptedAnswerId="5036" CreationDate="2013-05-17T15:34:31.347" Score="6" ViewCount="13493" Body="&lt;p&gt;Let's say I have a large quantity of rust and want to extract the oxygen from it in a way that allows me to collect the oxygen in a container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical process can I use to do this and what equipment does it require?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's preferable if the process is doable at home so I can demonstrate the process but it's not mandatory.&lt;/p&gt;&#xA;" OwnerDisplayName="user1616" LastActivityDate="2020-12-18T13:50:24.620" Title="Chemical methods to extract Oxygen from Rust" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5036" PostTypeId="2" ParentId="5035" CreationDate="2013-05-17T19:48:11.713" Score="11" Body="&lt;p&gt;Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why &quot;oxidation&quot; has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor (&quot;oxidizer&quot;). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest &quot;household&quot; method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's &lt;em&gt;much&lt;/em&gt; higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;thermite reaction&quot; between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are &quot;household chemicals&quot; to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the &lt;em&gt;extremely&lt;/em&gt; relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a &quot;drop everything and &lt;em&gt;run&lt;/em&gt;&quot; type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2015-05-04T16:16:56.490" LastActivityDate="2015-05-04T16:16:56.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5037" PostTypeId="1" CreationDate="2013-05-18T11:13:07.557" Score="2" ViewCount="182" Body="&lt;p&gt;The binding of proteins to an ITO surface can be prevented by coating it with PEG groups (polyethylene glycol). PEG-silane is used to create the coating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can somebody help with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does PEG-silane bind to ITO (indium tin oxide)? and how? (which bond etc.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does PEG repel proteins? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there better coatings to use to repel proteins?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1651" LastEditorDisplayName="user65670" LastEditDate="2018-07-02T08:02:54.220" LastActivityDate="2018-07-02T08:02:54.220" Title="PEG (polyethylene glycol) and ITO (indium tin oxide)" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5038" PostTypeId="1" CreationDate="2013-05-18T16:49:28.107" Score="7" ViewCount="671" Body="&lt;p&gt;By chance, we received some monolayer &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;noreferrer&quot;&gt;graphene&lt;/a&gt; sheets (20 cm x 20 cm) and mixed coper-graphene wires for free at our university. I would like to prepare some very-easy-to-do experiments for the chemistry students that show the awesome properties of graphene, and which could be carried out with very simple lab equipment or even home experiments. What are some good experiments we could do? &lt;/p&gt;&#xA;" OwnerUserId="256" LastEditorUserId="7448" LastEditDate="2015-04-02T18:29:23.637" LastActivityDate="2015-04-02T18:29:23.637" Title="Easy-to-do experiments that clearly show outstanding properties of graphene" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5039" PostTypeId="1" CreationDate="2013-05-18T17:46:19.777" Score="0" ViewCount="1606" Body="&lt;p&gt;Those are the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(H2N)2CO + 3 H2O -&amp;gt; CO2 + 2 NH4+ + 2OH-}$$&#xA;$$\ce{CH3COCH2COCH3 + OH- -&amp;gt; acac- + H2O}$$&#xA;$$\ce{CrCl3.6H2O -&amp;gt; Cr3+ + 3Cl- + 6H2O}$$&#xA;$$\ce{Cr3+ + 3acac- -&amp;gt; Cr(acac)}3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get from where comes the $\ce{C5H8O2}$ in reaction #2! &lt;/p&gt;&#xA;" OwnerUserId="1653" LastEditorUserId="4231" LastEditDate="2015-03-03T21:14:57.560" LastActivityDate="2015-03-03T21:14:57.560" Title="Where did the acac come from in Cr(acac)₃ synthesis by CrCl₃ reagent?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5040" PostTypeId="2" ParentId="5039" CreationDate="2013-05-18T19:36:58.600" Score="3" Body="&lt;p&gt;$\ce{CH3COCH2COCH3}$ is the condensed formula of pentane-2,4-dione, commonly known as acetylacetone. The $\ce{acac^-}$ is just a shorthand for the acetylacetonate ligand, i.e., $\ce{C5H7O2^-}$, which is just the deprotonated form of the acetylacetone used in reaction two.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-05-18T19:36:58.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5041" PostTypeId="1" CreationDate="2013-05-18T20:02:07.710" Score="0" ViewCount="918" Body="&lt;p&gt;I'd like to calculate Bond Dissociation Energy and Bond Energy for ATP to a similar molecule. What free (at least for academic use) computer programs which natively support and document BDE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't mind, could you provide a tutorial for using those programs to perform the calculations?&lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorUserId="116" LastEditDate="2013-05-22T00:52:53.067" LastActivityDate="2017-10-05T05:20:40.563" Title="Calculating bond disassociation energy and bond energy for ATP" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-05-24T03:32:32.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="5042" PostTypeId="1" AcceptedAnswerId="5043" CreationDate="2013-05-18T20:59:34.767" Score="2" ViewCount="272" Body="&lt;p&gt;My question is if any energy is required to form bonds, for instance when there is a phase change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am correct, energy might be required in the beginning, to make the reaction start and then release a bigger amount of energy than it was putted in. However, in terms of molecular behaviour in a phase change, I think there should not be any energy required to make the molecules to form bonds, if they are placed in a system which has a lower temperature than the molecules. Would they not lose the kinetic energy after a specific time?&lt;/p&gt;&#xA;" OwnerUserId="1539" LastActivityDate="2013-05-18T21:18:56.277" Title="Is energy required to form bonds [phase change]" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5043" PostTypeId="2" ParentId="5042" CreationDate="2013-05-18T21:18:56.277" Score="2" Body="&lt;p&gt;Bond formation always lowers the energy of the system (or bond formation is a consequence of lower energy, take your pick.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, you may have to add energy, because presumably certain bonds must break in order to rearrange the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding energy won't guarantee formation of much higher energy isomers, because the atoms have a large amount of kinetic energy, and so they can just as easily turn around and go back whence they came.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the kinetic energy can be dissipated by solvent, collisions with inert gasses, etc. and then you may end up with a measurable quantity of the higher energy species.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-18T21:18:56.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5044" PostTypeId="2" ParentId="5041" CreationDate="2013-05-18T21:34:10.397" Score="10" Body="&lt;p&gt;An example program is GAMESS.  It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do.  I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE:&#xA;\begin{align}&#xA;\ce{A-B &amp;amp;-&amp;gt; A+ + B-}   \tag{ionic}\\&#xA;\ce{A-B &amp;amp;-&amp;gt; A. + .B}  \tag{homolytic}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most electronic structure theory programs assume gas-phase, absence of solvent.  We know that homolytic dissociation is favored in the gas phase, as charge separation into ions is unfavorable in a Coulombic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the run-of-the-mill electronic structure theory calculations (e.g., HF, DFT, MP2, CCSD, etc. etc.) will not break the atoms apart in a homolytic manner.  They tend to break apart atoms in the ionic fashion.  But this is a problem, as ions are abhorrent in the gas phase!  Basically, simple electronic structure theory methods over-estimate the bond strength, as the product of the dissociation (ionic) is erroneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must use multiconfigurational methods to get it right (e.g. CASSCF, MCSCF). But these are very, very difficult unless you have had a lot of training.  I did them for years, and they were still hard for me.  But they will give you $\ce{A. + .B}$ as a product!  That is the gas-phase BDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to study closed-shell, closed-shell dissociations without much problem, using the HF and DFT methods.  Examples include dimers, host-guest interactions where the products look like good-old fashioned molecules that can stand by themselves. (Think water dimers, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, beware of Solvent Effects (favoring ionic dissociation $\ce{A-B -&amp;gt; A+ + B-}$) and solvation energy.  In fact, the solvation energy is why $\ce{Na-Cl}$ breaks apart into $\ce{Na+}$ and $\ce{Cl-}$.  In the gas phase, it would be $\ce{Na.}$  and $\ce{Cl.}$ radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, BDEs are very difficult things to compute to qualitative accuracy. Quantitative accuracy is even harder.  You may want to have an expert look over your procedure and results before you trust them completely.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="4945" LastEditDate="2017-10-05T05:20:40.563" LastActivityDate="2017-10-05T05:20:40.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5046" PostTypeId="2" ParentId="5038" CreationDate="2013-05-19T01:42:29.067" Score="4" Body="&lt;p&gt;You may want to post this in the Physics forum, as many of the outstanding properties of graphene are expected to be physical.  See this Periodic Video: &lt;a href=&quot;http://www.periodicvideos.com/videos/mv_graphene.htm&quot; rel=&quot;nofollow&quot;&gt;Graphene&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, two things to try.  First, graphene should conduct electricity, like graphite but not like carbon.  Second, graphene may dissolve in nonpolar solvents like carbon tetrachloride, similar to buckyballs, to produce colorful solutions.  See this Periodic Video: &lt;a href=&quot;http://www.periodicvideos.com/videos/mv_buckyball.htm&quot; rel=&quot;nofollow&quot;&gt;BuckyBall&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both these experiments should be checked in advance: I'm extrapolating from other experiments, but have not done these myself.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-05-19T01:42:29.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5047" PostTypeId="1" CreationDate="2013-05-19T03:00:24.760" Score="1" ViewCount="2439" Body="&lt;p&gt;I am a confused (and somewhat anal retentive) non-chemist who is trying to understand what pH is.  I have come to understand it is a measure of acidity, but I have not yet figured out what that means to a precision that I find satisfying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse me if these are dumb questions, but they will help me understand the definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does &lt;em&gt;everything&lt;/em&gt; have a pH?  What is the scale that pH works upon?  Does a proton have a pH?  A quark?  Does the sun have a pH?  Does the Universe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1655" LastActivityDate="2013-06-12T17:39:49.280" Title="On the scale of stuff with a pH" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5048" PostTypeId="1" AcceptedAnswerId="5070" CreationDate="2013-05-19T05:13:58.313" Score="4" ViewCount="14592" Body="&lt;p&gt;I'd like know if this is right. The reaction between $\ce{NaCl}$ and $\ce{H_2SO_4}$ should yield $\ce{HCl}$ and sodium sulfate. This is a double substitution, true? No oxidation took place here. The thing is, when I tried the reaction in the lab, the resulting solution (gas bubbled into water) seemed to be sulfuric acid. Simply enough, the solution didn't react with aluminum. I presume that $\ce{SO_2}$ was formed from the oxidation the sulfur (it a little bit like it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Shouldn't the substitution take place instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\ce{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl}\\&#xA;\ce{NaCl + NaHSO_4 \rightarrow Na_2SO_4 + HCl}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\ce{2\space NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2\space HCl }&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right?&lt;/p&gt;&#xA;" OwnerUserId="1282" LastEditorUserId="1282" LastEditDate="2014-12-06T21:14:49.763" LastActivityDate="2019-10-15T08:29:49.347" Title="HCl synthesis from sulfuric acid and sodium salt" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5049" PostTypeId="2" ParentId="5047" CreationDate="2013-05-19T07:09:08.457" Score="5" Body="&lt;p&gt;Did you not carry out some research?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you can find information about it in &lt;a href=&quot;https://en.wikipedia.org/wiki/PH#pH&quot;&gt;Wikipedia&lt;/a&gt;, or even dictionaries like &lt;a href=&quot;http://www.merriam-webster.com/dictionary/ph&quot;&gt;Merriam-Webster&lt;/a&gt; or &lt;a href=&quot;http://dictionary.reference.com/browse/Ph&quot;&gt;Dictionary Online&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that pH is calculated through:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pH = -\log(a_{\ce{H+}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or which can most of the time be stated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pH = -\log(\ce{[H+]})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{\ce{H+}}$ is the hydrogen ion activity, the 'effective concentration' of hydrogen ions. Without active hydrogen ions, you don't have any pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find that the pH scale is from 0 to 14, but you can also have circumstances where you have negative pH and pH above 14. It depends on the extent of dissociation of an acid or base to produce active hydrogen ions, or hydroxide ions in the case of bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I mentioned hydroxide ions, it might also be worth nothing that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$14 = pH + pOH$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that basically, a pH of 1 is equivalent to a pOH of 13.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-19T07:09:08.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5050" PostTypeId="2" ParentId="5025" CreationDate="2013-05-19T12:03:30.437" Score="5" Body="&lt;p&gt;I don't believe you have any ferric in the colorless picture since you have a metallic iron which would reduce the ferric to a ferrous. I believe you have a ferrous (Iron(II)) acetate characterized as an off-white precipitant. &lt;/p&gt;&#xA;" OwnerUserId="1658" LastActivityDate="2013-05-19T12:03:30.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5051" PostTypeId="2" ParentId="4906" CreationDate="2013-05-19T14:51:05.123" Score="1" Body="&lt;p&gt;Excellent question. On the one hand, the vapor pressure of water (or any other volatile liquid) is a function of the temperature alone, but on the other, if you assume that the mole fraction is constant you find that the vapor pressure of the water changes even though the temperature is constant. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The short answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The water vapor pressure stays constant because as the pressure in the balloon increases, the mole fraction of the water reduces proportionally to maintain the equilibrium it has established.&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The longer answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's dig into the former approach I mentioned in the introduction first. We can begin with &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation&quot; rel=&quot;nofollow&quot;&gt;the Clausius-Clapeyron relationship&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Clausius-Clapeyron relationship (often is simplified to):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln \left(P_{vapor}\right) = -\frac{L}{R}\left(\frac{1}{T}\right) + C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ and $C$ are constants and $R$ is the ideal gas constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this model, we would expect the vapor pressure of the water to remain constant and therefore the mole fraction of water in the gas phase to decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dalton's law tells us: $$X_{water}=\frac{n_{water}}{n_{tot}} = \frac{P_{water}}{P_{tot}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this expression we can see that, as the total pressure of the gas increases and the vapor pressure of the water stays constant, the mole fraction of water in the gas phase must go down. If we assume, as a first approximation, that the total number of gas molecules in the gas phase is constant, then a reduction in mole fraction can only be accomplished by some of the molecules of water in the gas phase being reabsorbed into the liquid water phase. This consequence is in agreement with &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;the Le Chatelier principle&lt;/a&gt; prediction. A decrease in volume available to the water molecules will shift the equilibrium below to the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_2O\,(l) \rightleftharpoons H_2O\,(g)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for such a rich question! How interesting that there are so many topics it relates too! Excellent!&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other thoughts&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When we calculate the initial pressure of the other gases in the volume, we should use:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$P_{tot} = P_{water} + P_{all\,other\,gas}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{tot} = P_{external} = 760\,mm\,Hg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{all\,other\,gas} = P_{tot} - P_{water} = 760\,mm\,Hg - 24\,mm\,Hg = 736\,mm \,Hg$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The dissolved gas in the water in the balloon will also increase at higher pressures. How will this affect our calculations?&lt;/li&gt;&#xA;&lt;li&gt;We have not considered energy concepts explicitly in this argument about the pressures being equal (in the first bullet above), but we could have considered the elastic energy stored in the bonds of the balloon. How could we include this to get a better estimate of the pressure of the other gases in the balloon?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="1659" LastEditDate="2013-05-19T17:51:00.910" LastActivityDate="2013-05-19T17:51:00.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5052" PostTypeId="2" ParentId="5047" CreationDate="2013-05-19T18:22:13.987" Score="3" Body="&lt;p&gt;pH scales are just a special case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Inorganic_nonaqueous_solvent#Nonaqueous_acid-base_chemistry&quot; rel=&quot;nofollow&quot;&gt;solvonium/solvate scales&lt;/a&gt;, which are &lt;em&gt;a way to keep track of the concentration of solvonium/solvate ions&lt;/em&gt;. They can be defined for any (usually protic) liquid solvent in macroscopic amounts (the limit of many molecules). Particles don't have pH, and I'm not sure it's possible to define such scales in a useful manner for solids, gasses, plasmas or mixtures of solvents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at some cases. &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow&quot;&gt;Water is capable of self-ionization&lt;/a&gt;, according to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H_2O_{(l)} -&amp;gt; H_3O^+_{(aq)} + OH^{-}_{(aq})} \ \ \ \ \ K=\frac{a_{H_3O^+}a_{OH^{-}}}{a_{H_2O}^2}\approx \frac{[H_3O^+][OH^-]}{[H_2O]^2}=3.2\times10^{-18}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, where the equilibrium constant $K$ is given at 25°C (it varies with temperature). If we consider relatively dilute solutions, then $[H_2O]\approx55.5\ mol.L^{-1}$ and one can obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[H_3O^+][OH^-]=K_w^{25°C}=1\times10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_w$ is called the &lt;em&gt;ionic product&lt;/em&gt; of water. Taking the logarithm of both sides and multiplying everything by $-1$, we find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-log[H_3O^+]-log[OH^-]=-log\ K_w^{25°C}\rightarrow pH+pOH=pK_w^{25°C}=14$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(At the end I switched $\ce{H_3O^+}$ for $\ce{H^+}$ since they are essentially the same thing. $\ce{H_3O^+}$ is nothing but $\ce{H^+}$ solvated in water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $p$ in $pH$ is nothing special; it is merely shorthand for the operator $-log()$, and it can be applied to any measure, though it's not very common to do so. Though at 25°C, the sum of pH and pOH is constant (equal to 14), there is nothing immediately forbidding either value being negative. They just can't be very negative because the approximations in the first two equations breaks down as such solutions are no longer dilute. Also, &lt;a href=&quot;http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html&quot; rel=&quot;nofollow&quot;&gt;the sum of pH and pOH is only 14 at 1 bar for a temperature of 25°C&lt;/a&gt;. At 0°C, $K_w^{0°C}=1.1\times10^{-15}$, while at 100°C, $K_w^{100°C}=5.1\times10^{-13}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for other solvents. Water isn't the only substance to self-ionize; another example is is liquid ammonia ($\ce{NH_3}$). Here, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NH_3\ _{(l)} -&amp;gt; NH_4^+\ _{(am)} + NH_2^{-}\ _{(am})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ionic product can also be defined for ammonia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[NH_4^+][NH_2^-]=K_{am}^{-34°C}\approx 10^{-30}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the ionic product, we see that ammonia has a smaller tendency to self-ionize compared to water. This is due to several factors, such as the smaller electronegativity of the nitrogen atom and the lower dielectric constant of ammonia, which destabilize the ions formed. In this solvent, the neutral point is around pH 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all this, what's the point of pH? As was mentioned at the start, it's a way to keep track of the concentration (or activity rather) of $\ce{H^+}$ in solvents, and it turns out this species is particularly important to chemists, not only because it's so common and a part of water. Bonding/dissociation/charge transfer happens due to the interaction of electrostatic potentials in molecules which push and pull electrons. Usually, a handful of fundamental units of charge are distributed in volume of a few atoms. However, if you stop to think about it, $\ce{H^+}$ has no electron cloud and is actually a lone proton, characterized by a charge of $+e$ concentrated &lt;em&gt;in the volume of a nucleus&lt;/em&gt;. If you recall, an atomic nucleus has about a trillionth the volume of an atom's electron cloud. This means $\ce{H^+}$ has a &lt;em&gt;phenomenally high&lt;/em&gt; charge density capable of generating a massive electric potential, and is therefore extremely reactive. Though helium and neon are the most inert substances we know, $\ce{H^+}$ will &lt;a href=&quot;https://en.wikipedia.org/wiki/Proton_affinity&quot; rel=&quot;nofollow&quot;&gt;easily react exothermically with even them&lt;/a&gt; to produce $\ce{HeH^+}$ and $\ce{NeH^+}$. In fact, $\ce{H^+}$ cannot exist as a true lone ion in a condensed phase, and it will demand some electron density from any substance. A very large amount of acid-base theory was created solely to understand more about the reactivity of this unique species.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-05-19T18:22:13.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5055" PostTypeId="1" CreationDate="2013-05-20T00:09:34.247" Score="1" ViewCount="8901" Body="&lt;p&gt;I want to remove a layer of 2 mm thick carbon black on a copper bowl on the outside from hanging over wood and coal fire. &#xA;I have tried different solvents, but the carbon seems to be very resistive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which formula solves carbon black without solving copper? The bowl is made from pure copper. It will melt at roughly 1000 degree Celsius and has a high heat capacity.&lt;/p&gt;&#xA;" OwnerUserId="944" LastEditorUserId="944" LastEditDate="2014-04-20T11:37:16.573" LastActivityDate="2018-06-24T15:17:37.433" Title="Solving carbon black" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;solvents&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5056" PostTypeId="1" CreationDate="2013-05-20T09:15:02.707" Score="2" ViewCount="74" Body="&lt;p&gt;So, I need a method of estimating the influence of a surface (rate constant would do for a start), the idea is to estimate just Reactant + Surface -&gt; Product in a turbulent liquid flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am aware of the book &quot;The properties of gases and liquids&quot; by Poling and Prausnitz which gives me two terms for estimating diffusion in a liquid, one general formula from 1975 and a formula specific to hydrocarbons (applicable in my case) from 1982, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tyn and Calus, 1975:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(AB) = \frac{7.4\,10^{-4} (\phi M(B))^{1/2}T}{\eta(B) V(A)^{0.6}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hayduk and Minhas, 1982:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(AB) = \frac{13.3\,10^{-8}T^{1.47} \eta(B)^{\epsilon}}{V(A)^{0.71}}$$&#xA;where $\epsilon = \frac{10.2}{V(A)}-0.791$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other variables are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M(B)$ = molecular weight of solvent B &lt;/li&gt;&#xA;&lt;li&gt;$T$ = temperature  &lt;/li&gt;&#xA;&lt;li&gt;$\eta(B)$ = viscosity of solvent B &lt;/li&gt;&#xA;&lt;li&gt;$V(A)$ = molar volume of solute A &lt;/li&gt;&#xA;&lt;li&gt;$\phi$ = association factor of solvent B (1 if unassociated)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given that in  my special case I am looking at a liquid in a turbulent flow, I can use the kinetic theory of gases to estimate a surface collision rate - and include surface efficiency using the work on Baldwin and Howarth from 1982.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cylindrical pipe with the solute tending to zero, an upper limit for my rate constant for a surface reaction would be:&#xA;$k(s) = \frac{8D}{r^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is one big problem with this: The diffusion estimate is not for a turbulent flow (I might be able to get get a diffusion term from Fluent to get a more accurate estimate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, at best this would give me an upper limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, do you know of any better more accurate methods of estimating (at least limits) a rate constant for surface reactions? (Or at least the collision rate with the surface.)&lt;/p&gt;&#xA;" OwnerUserId="1665" LastEditorUserId="16683" LastEditDate="2016-09-30T09:24:36.283" LastActivityDate="2016-09-30T09:24:36.283" Title="Estimating Surface Reactions - Liquid, Turbulent Flow" Tags="&lt;kinetics&gt;&lt;surface-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5057" PostTypeId="1" AcceptedAnswerId="5075" CreationDate="2013-05-20T09:20:19.530" Score="6" ViewCount="396" Body="&lt;p&gt;Independent from my general kinetics question here - &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5056/estimating-surface-reactions-liquid-turbulent-flow&quot;&gt;Estimating Surface Reactions - Liquid, Turbulent Flow&lt;/a&gt; - I am only interested in responses regarding Gaussian in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if it is feasible to predict or model the influence of a surface (metal) on a reaction between two hydrocarbons (aromatic and aldehyde for example) in Gaussian.&#xA;If yes, how would I go about it? (Building a small metal grid and working with it is very computationally intensive - it is the only approach I can think off...) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have access to Gaussian at my university - however I do not have access to Linda, so no shared nodes.)&lt;/p&gt;&#xA;" OwnerUserId="1665" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-06-02T10:54:57.497" Title="Gaussian and Surface Reactions - Doable?" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5059" PostTypeId="1" CreationDate="2013-05-20T16:13:51.143" Score="5" ViewCount="525" Body="&lt;p&gt;I just started with Molecular dynamics simulations. My background is computer science but I'm learning about scientific computing and thus my jump into MD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, I was asked to calculate the percentage of each conformer in vacuum at a certain temperature of the Methylcyclohexane molecule. Assuming that entropies and molar volumes are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me out to some website or something where I can see how to calculate this?&lt;/p&gt;&#xA;" OwnerUserId="1668" LastActivityDate="2013-05-20T22:02:06.080" Title="Percentage of a conformer in vacuum" Tags="&lt;molecules&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5060" PostTypeId="1" AcceptedAnswerId="5069" CreationDate="2013-05-20T16:21:37.853" Score="1" ViewCount="486" Body="&lt;p&gt;I am trying to create the molecule of the methylcyclohexane in the equatorial and axial conformers. I read &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/teach/cyclohex.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this site&lt;/a&gt; about the chair conformation. My molecule looks like the following picture (created in Molden)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2vMRB.png&quot; alt=&quot;Methylcyclohexane axial conformer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point out if this is actually good? I am quite doubtful about the &quot;mid&quot; carbons, should they be push to the center of the molecule like in the drawings in the website?&lt;/p&gt;&#xA;" OwnerUserId="1668" LastActivityDate="2018-12-26T13:47:27.097" Title="Methylcyclohexane chair conformers" Tags="&lt;molecules&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5061" PostTypeId="1" AcceptedAnswerId="5067" CreationDate="2013-05-20T16:42:06.627" Score="3" ViewCount="755" Body="&lt;p&gt;Carbon can be turned to diamond artificially, but while watching a video I learned that a diamond can be made from the ashes of a man, how is this possible?&lt;/p&gt;&#xA;" OwnerUserId="815" LastEditorUserId="9961" LastEditDate="2016-10-20T00:04:31.743" LastActivityDate="2018-07-03T08:43:08.533" Title="How to make artificial diamond?" Tags="&lt;carbon-allotropes&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5063" PostTypeId="1" CreationDate="2013-05-20T17:30:11.593" Score="6" ViewCount="617" Body="&lt;p&gt;In a carbonyl or $\ce{-OH}$ group, it's clear the oxygen acts as a hydrogen bond acceptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do metals act as hydrogen bond acceptors or donors and if so, how can I tell as what it's acting?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorDisplayName="user15489" LastEditDate="2015-05-08T19:13:42.017" LastActivityDate="2015-05-08T20:09:57.387" Title="Are metals hydrogen bond acceptors or donors?" Tags="&lt;metal&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5065" PostTypeId="2" ParentId="5023" CreationDate="2013-05-20T18:32:46.907" Score="13" Body="&lt;p&gt;That one mole of $\ce{CaCl2}$ is more effective in melting ice than one mole of $\ce{NaCl}$ is explained by the van 't Hoff factor. The freezing point depression of a solution is calculated by&#xA;$$\Delta T=K_fbi$$&#xA;where $K_f$ is a cryogenic constant which is specific to the solvent, $b$ is the molal concentration of the solute, and $i$ is the van 't Hoff factor, which indicates number of solute particles. In this case, it is the number of ions produced upon dissociation. Because $K_f$ and $b$ are constant, and $i=3$ for $\ce{CaCl2 \rightarrow Ca^2+ + 2 Cl-}$ and $i = 2$ for $\ce{NaCl \rightarrow Na+ + Cl-}$ the freezing point depression is greater for $\ce{CaCl2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason $\ce{CaCl2}$ is widely used as melting agent may have more to do with keeping the concrete intact. Concrete is composed of many calcium-containing species. When water flows over the concrete, the calcium can be leeched out, and the concrete becomes brittle. The higher the calcium content in the water, the less leeching occurs, and the concrete remains structurally sound.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="5387" LastEditDate="2015-04-28T16:31:05.097" LastActivityDate="2015-04-28T16:31:05.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5066" PostTypeId="2" ParentId="5063" CreationDate="2013-05-20T19:55:15.330" Score="1" Body="&lt;p&gt;Only very electronegative atoms— $\ce{O}$, $\ce{N}$ and $\ce{F}$— participate in hydrogen bonding. The donor is the group with the covalently-bonded hydrogen participating in hydrogen-bonding and the acceptor is the atom not covalently bonded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TjAPp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\ce{C}$ atom may also act as a hydrogen bond donor if there are highly electronegative atoms attached to it such as in the case of chloroform $\ce{CHCl3}$.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="1804" LastEditDate="2013-06-20T16:33:46.953" LastActivityDate="2013-06-20T16:33:46.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5067" PostTypeId="2" ParentId="5061" CreationDate="2013-05-20T20:42:29.407" Score="9" Body="&lt;p&gt;First off, take a look at the Wikipedia article: &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_diamond&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthetic diamond&lt;/a&gt;. Basically, there are two ways to artificially generate the heat and pressure needed. First is the &quot;conventional&quot; high-temperature, high-pressure method, which uses a press to put a sample of pure carbon graphite under pressure of about 5 gigapascals (1GPa = about 10,000 atmospheres or about 145,000 psi) and about $\pu{1500 ^\circ C}$. This combination encourages the flat hexagonal lattice structure of the graphite to morph into the trihexagonal 3-D lattice of diamond. Second is the chemical vapor deposition method, in which a substrate of diamond crystal is exposed to sublimated carbon plasma (temperatures exceeding $\pu{3000 ^\circ C}$) in a hard vacuum (less than .0001 torr). The sublimated carbon attaches to the crystal substrate, &quot;growing&quot; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these can be used to create so-called &quot;memorial diamonds&quot; from the cremated remains of a human. To do that, first the ash is chemically filtered to separate the carbon from &quot;pollutants&quot; naturally occurring in the body that don't combust into gaseous substances during cremation, such as calcium, phosphorous, iron etc. Then, this &quot;amorphous carbon&quot;, much like any other carbon, can be used in HTHP or CVD processes, along with additional carbon as may be required to grow the diamond to the proper size. Usually, the HTHP process is used, as it makes the best use of the available carbon (CVD requires sublimating a lot of carbon, more than would be left behind in the ash, and most of which ends up on the chamber walls and not the substrate), and requires less energy (it's easier to maintain mechanical pressure once applied than to maintain temperature).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="5026" LastEditDate="2018-07-03T08:43:08.533" LastActivityDate="2018-07-03T08:43:08.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5068" PostTypeId="2" ParentId="5059" CreationDate="2013-05-20T22:02:06.080" Score="8" Body="&lt;p&gt;Page 11699 of &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja011357l&quot;&gt;this paper&lt;/a&gt; has one approach to what you want: basically, you compute the relative Gibbs free energies of the possible stable conformers, assume they can interconvert, and use a Boltzmann distribution to work out what the percentages are.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, we selected a reference structure $R$ for a given species (neutral or protonated). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Next, for every structure $M$ other than the reference structure $R$ we determined the&#xA;  equilibrium constant $K_M$ $$K_M = \frac{[M]}{[R]}$$ from the difference in Gibbs free energies for $M$ and $R$. The fraction of $M$ in the equilibrated sample is given by:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$x_M = K_M/\left(1 + K_1 + K_2 + ...\right)$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where the sum in the denominator goes through all structures for a&#xA;  given species. The fraction of $R$ in the sample is: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$x_R = 1/\left(1 + K_1 + K_2 + ...\right)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't include it in the paper, but the way to calculate an equilibrium constant $K$ from a difference in Gibbs free energy $\Delta G$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G = RT \ln K $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_constant&quot;&gt;molar gas constant&lt;/a&gt;, and $T$ is the temperature (in Kelvin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part is then working out what your stable conformers are, generally using some form of either guesswork or conformation sampling algorithm. They use a genetic algorithm of some sort in the paper I linked to, but given that methylcyclohexane isn't that complex a molecule, you might want to just take a look at an &lt;a href=&quot;http://www.ch.imperial.ac.uk/rzepa/blog/?p=7926&quot;&gt;energy diagram and discussion of conformers of cyclohexane&lt;/a&gt; for some hints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're getting deep into molecular dynamics, you could also use some sort of algorithm designed to use dynamic simulation to sample a wide area on a potential surface, like metadynamics or umbrella sampling, to find your conformers, and probably calculate your free energies at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a static method to calculate the free energies instead, though -- the method you use to do this will depend on what sort of accuracy you're after, and may vary from a classical forcefield technique through Density Functional Theory to coupled-cluster theory (in ascending order of computational expense).&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-05-20T22:02:06.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5069" PostTypeId="2" ParentId="5060" CreationDate="2013-05-20T22:25:26.863" Score="7" Body="&lt;p&gt;It's a little hard to tell from this angle, but that looks like a pretty sensible representation of the axial methyl geometry given on the website you linked to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what you mean by &quot;push to the center of the molecule&quot;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWNB5.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ETA&lt;/strong&gt;: These might help a little more: this isn't exactly the most accurate geometry, but it's a pretty good representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/g6Yz7JGm.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/ifuZCkbm.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/vy3SO2lm.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="15751" LastEditDate="2018-12-26T13:47:27.097" LastActivityDate="2018-12-26T13:47:27.097" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5070" PostTypeId="2" ParentId="5048" CreationDate="2013-05-20T23:36:20.537" Score="5" Body="&lt;p&gt;Ideally, that's the reaction. Realistically, there are a few speed bumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a &quot;water-free&quot; combination of the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-03T16:12:06.150" LastActivityDate="2013-09-03T16:12:06.150" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5071" PostTypeId="2" ParentId="1334" CreationDate="2013-05-20T23:39:21.363" Score="9" Body="&lt;p&gt;Most commercially-available lithium-ion batteries use graphite as the anode (negative electrode, where the lithium ions go when the battery is charged) and $Li Co O_2$ as the cathode (positive electrode, where the lithium ions go when the battery is discharged). Both graphite and lithium cobalt oxide have a structure composed of layers stacked on top of one another; the lithium ions fit in between these layers in a process called intercalation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lithium sitting between the layers props the layers apart. If too much of the lithium is removed from either the cathode (over-charging) or the anode (over-discharging) then the layers can collapse back together and it becomes hard to put the lithium back in. Normally batteries have control circuitry to prevent the cells from over-charging or over-discharging but this circuitry is not always perfect, particularly for smaller, simpler battery packs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, AFAIK the circuitry is already set to limit charging and discharging to well below the theoretical maximum capacity, so I would not recommend limiting the charge any further. Perhaps unscrupulous vendors will cut the amount of active material inside their batteries while keeping the rated capacity the same and therefore run the cells closer to their theoretical maximum, in which case keeping charge and discharge within their rated limits would be a good idea, but it should not be necessary for high-quality batteries. &lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2013-05-20T23:39:21.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5072" PostTypeId="2" ParentId="5047" CreationDate="2013-05-21T00:20:47.880" Score="1" Body="&lt;p&gt;pH stands for &quot;potential hydrogen&quot;, originally &quot;power of hydrogen&quot;. In simplest terms, it is a logarithmic scale (based on powers of 10, similar to the Richter scale for earthquakes, the decibel scale for sound and signal levels, and others that seek to measure very large differences with smaller numbers) of how many free protons there are in an aqueous solution of a substance, which is in turn a measure of that solution's acidity or basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number is the absolute value of the exponent of this measurement, so if, for instance, a substance has a pH of 7, that means there's one free proton per 10 million water molecules (1/10,000,000 = 10&lt;sup&gt;-7&lt;/sup&gt;). A pH of one, or less, means there's at least one proton for every 10 water molecules, while a pH of 14 means there is one proton or fewer per 100 &lt;em&gt;trillion&lt;/em&gt; water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons, or H&lt;sup&gt;+&lt;/sup&gt; ions, are attracted to the negative charge region of a water molecule. Water's bent molecular structure, with the two hydrogens closer to one side and oxygen's extra lone pairs on the other, effectively makes each molecule a tiny magnet, which is the foundation for many important characteristics of water, like why ice floats and why some things dissolve in water while others don't. Anyway, the hydrogen loosely &quot;binds&quot; to one of those lone pairs, forming hydronium (H&lt;sub&gt;3&lt;/sub&gt;O&lt;sup&gt;+&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural equilibrium of one hydronium ion per 10 million molecules. At this point, pure water (again being a polar solvent) will actually &lt;em&gt;dissolve itself&lt;/em&gt; to a small degree, producing one hydronium and one hydroxide per 10 million. At this pH of 7, the opposite measure, pOH (power of hydroxide) is also 7, balancing the reactivity of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below a pH of 7, the positively-charged hydronium atoms come to dominate this equilibrium, decreasing the number of hydroxide ions and looking for a substance to which to donate its protons to reach a balance again. This solution, which is now a &quot;proton donor&quot;, is one of the classical definitions of an acid (the Lowry-Bronsted definition). Ideally, it will donate its proton to a substance that needs one as badly as the acid needs to get rid of it, but if the acid has enough hydronium, it will stick those hydrogens wherever they'll fit into any other substance's molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above pH 7, the negatively-charged hydroxide ions reduce the hydronium and dominate equilibrium. The solution is now a &quot;base&quot;, a &quot;proton acceptor&quot;, and is looking for something from which to take protons. The ideal substance to do so is a proton donor - an acid - but similar to an acid, a base, if strong enough, will take hydrogen atoms from anywhere they're not nailed down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, not all substances which are acidic have a free proton to donate (and so not all of them increase the amount of hydronium ions in water), and not all substances which are basic have a hydroxide anion, nor do they accept free protons into their structure. Therefore, there is another definition, the Lewis definition, stating that an acid is a substance that accepts a &quot;lone pair&quot; of electrons from another substance (a pair of electrons not forming a bond between atoms of the molecule) to equalize its own molecular charge, and a base is a substance that readily donates an electron pair. Regardless of the definition or the exact mechanism, these acids and bases have similar observed properties, and so even if explicit exchange of hydronium atoms isn't involved, the substances still indicate a pH using various testing methods. This prompted the change in name from &quot;power of hydrogen&quot; to &quot;potential hydrogen&quot;, in effect making the measurement an &quot;equivalent&quot;, applying to both Lewis acids and bases and Bronsted-Lowry acids and bases.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-06-12T17:39:49.280" LastActivityDate="2013-06-12T17:39:49.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5074" PostTypeId="1" CreationDate="2013-05-21T02:58:51.887" Score="4" ViewCount="163" Body="&lt;p&gt;Does anyone know what causes things like xanthan gum to make water into a slimy substance on the molecular level?&lt;/p&gt;&#xA;" OwnerUserId="1669" LastEditorUserId="7448" LastEditDate="2015-07-09T09:57:54.320" LastActivityDate="2015-07-09T09:57:54.320" Title="How do xanthan gum and other hydrocolloids work?" Tags="&lt;water&gt;&lt;solutions&gt;&lt;colloids&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5075" PostTypeId="2" ParentId="5057" CreationDate="2013-05-21T07:28:07.567" Score="7" Body="&lt;p&gt;You did go to &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_pbc.htm&quot; rel=&quot;noreferrer&quot;&gt;Gaussian's Documentation on the subject&lt;/a&gt;, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, it should work.  In practice, you should consider using massive firepower, and the plane-wave codes (NWChem, abinit, VASP, etc. etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I question whether any of these codes &lt;em&gt;properly&lt;/em&gt; model bond breaking on a surface, i.e.  metal-catalyzed reactions.  Big simulations don't guarantee correct wave functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out that surface reactions with plane waves are also difficult. At least they used to be.  There are/were issues with Ewald summation having to be done in 3D, and so the surface normal dimension used to have to be HUGE to avoid image effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a tough business.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-21T07:28:07.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5077" PostTypeId="1" CreationDate="2013-05-21T22:01:44.227" Score="4" ViewCount="3194" Body="&lt;p&gt;As I understand it, the boiling point elevation in a particular solvent (in this case water) is a function of a particular constant of that solvent, and the molality of a dissolved substance. In turn, the maximum molality is a function of the solubility of the compound, as well as the van 't' Hoff factor. Thus it would follow that the highest possible elevation of water's boiling point would be the compound with the greatest ratio of solubility to van 't' Hoff factor. So the question remains: which compound is that?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2013-05-22T00:57:04.917" Title="What is the highest the boiling point of water can be raised?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5078" PostTypeId="1" CreationDate="2013-05-21T23:42:48.950" Score="5" ViewCount="1958" Body="&lt;p&gt;Why does the graph of $\ln[\ce A]$ vs $t$ produce a straight line for the first order rate law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also why does the graph of $t$ vs $\frac{1}{B}$ produce a straight line for the second order rate law?&lt;/p&gt;&#xA;" OwnerUserId="1473" LastEditorDisplayName="user7951" LastEditDate="2015-09-30T21:31:21.877" LastActivityDate="2017-10-13T16:06:38.493" Title="Integrated rate law graphs" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5079" PostTypeId="2" ParentId="5077" CreationDate="2013-05-22T00:57:04.917" Score="4" Body="&lt;p&gt;The &quot;particular constant&quot; $k$ you refer to in the question isn't an arbitrary constant.  This &quot;law&quot; is usually written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T = km = \frac{M_1 RT_b^2}{1000 \Delta H_1} m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k is the &quot;constant you refer to; $M_1$ is the molecular weight of the solvent, $R$ is the gas constant, $T_b$ is the boiling point of the pure solvent, the constant 1000 makes the formula applicable only to water, and $\Delta H_1$ is the heat of vaporization of the solvent, here assumed to be water.  $m$ is the molality of the solute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is derived in courses in physical chemistry and is itself a simplification to allow slide rule simplification (this is true for a lot of formulas in chemistry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you want to do to make $\Delta T$ as large as possible is to find a solvent with a large molecular weight and a high boiling point and a low heat of vaporization.  The constant 1000, which is the number of grams of solvent in a liter of solvent will also have to be adjusted.  And of course $m$ can be played with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think anyone has ever worked out what works best for a large $\Delta T$ since boiling point elevation, once used for molecular weight approximation, isn't used for that purpose any more -- and thermometers are a lot more sensitive and accurate than they used to be.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-22T00:57:04.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5080" PostTypeId="2" ParentId="5078" CreationDate="2013-05-22T01:08:01.943" Score="9" Body="&lt;p&gt;The short answer is that the equations you describe as graphs are the solutions of the &lt;em&gt;differential equations&lt;/em&gt; that define the rate laws in question.  If you don't understand what a differential equation is, any sensible answer is going to be problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you do know what a differential equation is, we proceed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a first order rate law, the differential equation is something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the concentration and time to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the differential equation for the second order rate law is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the concentration and time to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again integrate to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all comes from the following Wikipedia page, which has a lot more details: &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rate equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="30836" LastEditDate="2017-10-13T16:06:38.493" LastActivityDate="2017-10-13T16:06:38.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5081" PostTypeId="1" AcceptedAnswerId="5082" CreationDate="2013-05-22T07:50:04.713" Score="4" ViewCount="11860" Body="&lt;p&gt;I saw ingeniously simple &quot;cooler box&quot; yesterday - a double-walled plastic box with some liquid between the walls. You freeze it, and it keeps temperature near the freezing point of that liquid, until the liquid melts, while you can store dry foods inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the efficiency of such a box will depend on volume of the liquid, to a lesser degree - how far below its melting point it was frozen, and on latent heat of its melting. (and insulation of course, but that's an entirely different matter.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I suspect the liquid used in the box I saw was just water with a little coloring added, I wonder if I could do better than that. Is there a substance that can be reasonably obtained by a DIY tinkerer, that has a latent heat of melting significantly higher than water, and a melting point at normal pressure somewhere &quot;within reason&quot; - achievable by normal kitchen freezer, and not above &quot;food inside will spoil&quot;? (bonus points for it not being overly dangerous...) Or maybe is there some way to increase latent heat of melting water ice by adding something to it?&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2017-02-06T21:25:56.777" Title="Substance of higher latent heat of melting than water, similar melting point" Tags="&lt;melting-point&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5082" PostTypeId="2" ParentId="5081" CreationDate="2013-05-22T14:11:35.077" Score="2" Body="&lt;p&gt;Indeed, it will be hard to find a substitute for water. You might want to look into how &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_pack&quot; rel=&quot;nofollow&quot;&gt;ice packs&lt;/a&gt; work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the things that are added to water in such packs include substances to prevent bacterial growth and to prevent the water to become fully liquid. The water thus become a gel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gel packs may be made by adding hydroxyethyl cellulose (Cellusize) or vinyl-coated silica gel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One factor that you might consider, is that if there is more mass (for example, dissolving something in the water) will provide more substance for the any heat to go to, so that the temperature rise will be slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, is hydrogen bonds, which are very present in water molecules and which explains its high specific heats (latent heat and heat capacity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, adding in stuff to the water will tend to disrupt the hydrogen bonds. I haven't tried it, nor can find something to support my next sentence, but I would expect ammonia to be a possible additive. It has a higher specific heat capacity than water and dissolves easily as well since it does not disrupt hydrogen bonds that much due to its molecular structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glucose is also a potential additive since it doesn't disrupt the hydrogen bonds much, but its specific heat is apparently a bit lower than ammonia. I found a &lt;a href=&quot;http://sugartech.co.za/heatcapacity/&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; where I could input some parameters and it outputs the heat capacity of the sugar water and the outputs often say that the specific heat capacity is lower than that of ammonia alone (from the same &lt;a href=&quot;http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; that michielm mentioned, ammonia has a specific heat capacity of a bit less than 4.6 at $0^oC$).&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-22T14:11:35.077" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5084" PostTypeId="1" CreationDate="2013-05-22T17:29:37.683" Score="10" ViewCount="5704" Body="&lt;p&gt;I've always heard that different isotopes of the same element have exactly the same chemical  properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for example, I've read that some compounds which have hydrogen atoms interchanged by deuterium become odorless (this is one of the arguments for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vibration_theory_of_olfaction#Isotope_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;vibrational theory of olfaction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 2001 study by Haffenden et al. showed humans able to distinguish benzaldehyde from its deuterated version. In addition, tests with animals have shown fish and insects able to distinguish isotopes by smell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, are there any situations in which  neutrons do matter (besides nuclear decays)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 2013/10/28&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another curious effect is that water is slightly blue, due to &lt;a href=&quot;http://www.webexhibits.org/causesofcolor/5B.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;vibration&lt;/a&gt;. But heavy water is not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heavy water is colorless because all of its corresponding vibrational transitions are shifted to lower energy by the increase in isotope mass. For example the H2O band at 760 nm is shifted to approximately 1000 nm in D2O.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/REJQH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Left: tube if filled with (light) water. Right: empty tube.)&lt;/p&gt;&#xA;" OwnerUserId="1680" LastEditorUserId="9961" LastEditDate="2018-09-24T19:43:25.843" LastActivityDate="2018-09-24T19:43:25.843" Title="Chemical properties of isotopes" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;isotopologues&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5086" PostTypeId="2" ParentId="5084" CreationDate="2013-05-22T18:40:39.003" Score="7" Body="&lt;p&gt;I think isotopic analogues have mostly the same &lt;em&gt;bulk&lt;/em&gt; physical properties.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There are some striking differences, e.g. water and heavy water (D2O) have perceptibly different boiling points.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are many spectroscopies where isotopic substitution has a dramatic effect on the spectrum: including vibrational spectroscopies (IR/Raman), and nuclear spin (NMR, EPR, Mossbauer?).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rates of reaction can be substantially altered by using isotopic substitution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I think there is some confusion here, as far as smell.  It is when you change chiral centers from $\ce{R -&amp;gt; S}$ or $\ce{S-&amp;gt;R}$, thereby generating enantiomers and/or diastereomers, that you begin to affect things like smell.  It is profound in, e.g. menthol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diastereomers can have drastically different properties.  Enantiomers may interact differently with olfactory sensors, because our bodies (amino acids, sugars) have an intrinsic chirality.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-22T18:51:46.607" LastActivityDate="2013-05-22T18:51:46.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5088" PostTypeId="2" ParentId="4481" CreationDate="2013-05-22T21:12:14.853" Score="0" Body="&lt;p&gt;If you convert your equation to &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_table_file&quot; rel=&quot;nofollow&quot;&gt;mol format&lt;/a&gt; or another format supported by &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;ved=0CC8QFjAA&amp;amp;url=http://openbabel.org/&amp;amp;ei=vTOdUYX_GYnO9gS_8oGoAw&amp;amp;usg=AFQjCNH6r0ZBYNM7Tvtq7g21uw5NeBY21A&amp;amp;sig2=gJtRlqMQPJbEaL7lCZjY_A&amp;amp;bvm=bv.46751780,d.dmQ&quot; rel=&quot;nofollow&quot;&gt;OpenBabel&lt;/a&gt; you can use its tools to add 3D structure information to the files. Also if you need quality conformations you could use tools &lt;a href=&quot;http://www.eyesopen.com/omega%E2%80%8E&quot; rel=&quot;nofollow&quot;&gt;OpenEye's Omega&lt;/a&gt; or as I prefer &lt;a href=&quot;http://users.abo.fi/mivainio/balloon/faq.htm&quot; rel=&quot;nofollow&quot;&gt;Balloon&lt;/a&gt; which can quickly give biologically significant structures. Balloon works on many different atom types and most smiles Omega is parameterised for a select few smiles. &lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorDisplayName="user1464" LastEditDate="2013-05-24T01:53:43.890" LastActivityDate="2013-05-24T01:53:43.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5090" PostTypeId="2" ParentId="5084" CreationDate="2013-05-23T05:12:48.647" Score="2" Body="&lt;p&gt;My first idea when I read this question was: the masses are different, so everything that is affected by the mass of the ion or molecule must be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffusion and dissociation also come to mind. I did a quick search and found this: &lt;a href=&quot;http://www.mendeley.com/research/secondary-deuterium-isotope-effects-acidity-glycine/&quot; rel=&quot;nofollow&quot;&gt;http://www.mendeley.com/research/secondary-deuterium-isotope-effects-acidity-glycine/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1682" LastEditorUserId="1974" LastEditDate="2014-03-28T15:48:09.637" LastActivityDate="2014-03-28T15:48:09.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5091" PostTypeId="1" CreationDate="2013-05-23T08:41:55.807" Score="2" ViewCount="194" Body="&lt;p&gt;Can someone explain me whether a protein (like an antibody) would bind to carbon? The carbon arises due to electron beam induced deposition of Polyethylglycol (PEG). The carbon molecules that arise are either due to the PEG that dissociates or due to molecules, that are present in the microscope chamber as contamination, that fall apart. I am thus not sure about what the carbon molecule is specifically. You can look in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/nl2045414&quot; rel=&quot;nofollow&quot;&gt;this experiment&lt;/a&gt; to see more of the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why? (what bond?)&lt;/p&gt;&#xA;" OwnerUserId="1684" LastEditorUserId="275" LastEditDate="2013-05-23T10:38:15.063" LastActivityDate="2014-07-29T01:57:13.573" Title="Do proteins bind to carbon? If so how?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5092" PostTypeId="1" CreationDate="2013-05-23T10:37:29.860" Score="2" ViewCount="397" Body="&lt;p&gt;I am using a rotary evaporator to separate DCM (dichloromethane) from some nasty tars.&#xA;Before the evaporation is completed, I stop the process and collect the remaining slurry.&#xA;The slurry is then positioned in a constant temperature oven, for a secondary evaporation. After cooling in a desiccator, the tar is measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that some of the tars remain deposited on the flask's wall. I can easily remove them with a soft brush, however this is a problem for 2 reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Some of the tar is lost, retained on the brush itself ( I require a very high precision since I work with samples of 100 or so mg)&lt;/li&gt;&#xA;&lt;li&gt;The amount of DCM for secondary evaporation is increased (I am trying to reduce that amount as much as possible)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a round bottomed flask that has a TEFLON like coating on the &lt;strong&gt;inside&lt;/strong&gt;, that can prevent the tar deposition on the walls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seem to be a lot of products that are referred as &quot; Flask with Plastic Safety Coating&quot;. On which side is the coating applied (I assume is the outside)&lt;/p&gt;&#xA;" OwnerUserId="1513" LastEditorUserId="1513" LastEditDate="2013-05-23T11:05:45.720" LastActivityDate="2014-03-21T15:29:03.217" Title="Round Bottom Flask with Inner Coating" Tags="&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5093" PostTypeId="2" ParentId="5092" CreationDate="2013-05-23T12:47:25.847" Score="2" Body="&lt;p&gt;I've seen a couple of companies making flasks &lt;em&gt;entirely&lt;/em&gt; from PTFE or PFA, which might be suitable. They may not hold up to heating with internal vacuum though, for example, check out the notes &lt;a href=&quot;http://www.brandtech.com/prodpage.asp?prodid=V107797&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the typical safety coatings are on the outside to hold the glass together on fracture.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-05-23T12:47:25.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5095" PostTypeId="1" CreationDate="2013-05-23T14:54:40.280" Score="6" ViewCount="171" Body="&lt;p&gt;Aoyagi et al. (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00077a044&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. Am. Chem. Soc., Vol. 115, No. 24, p 11393&lt;/a&gt;) reported that the addition of 2-lithio-1,3-dithiane to the alpha-chiral (&lt;em&gt;S&lt;/em&gt;)-2-Acetylpiperidine gives the (2 &lt;em&gt;S&lt;/em&gt;,3 &lt;em&gt;S&lt;/em&gt;)-configured product. It is stated that the diastereoselectivity of this 1,2-addition (&lt;em&gt;dr&lt;/em&gt; = 6.2) is predicted by Cram's model involving complexation of the Li-ion by the piperidine-N. I tried to reproduce the diastereoselectivity in the following scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs9aO.png&quot; alt=&quot;http://abload.de/img/cram-ruleseej1y.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see I come up with the wrong product by applying Cram's rule with a O,N-complexed Li-ion (right side of the scheme). By applying the non-chelate Cram rule (left side; biggest substituent oriented anti to the carbonyl, attack via smallest group) I get the correctly configured (2 &lt;em&gt;S&lt;/em&gt;,3 &lt;em&gt;S&lt;/em&gt;)-product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the mistake?&lt;/p&gt;&#xA;" OwnerUserId="1685" LastEditorUserId="1685" LastEditDate="2013-05-24T07:41:31.263" LastActivityDate="2015-01-17T17:17:00.807" Title="acyclically stereocontrolled reaction: organolithium and 2-chiral compound" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5096" PostTypeId="2" ParentId="5084" CreationDate="2013-05-23T21:28:16.977" Score="12" Body="&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: nuclear neutrons greatly affect chemistry involving hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is the lightest element, and deuterium is just less than twice as heavy. This makes a pretty massive (pun intended) difference to any vibrational modes involving moving hydrogen/deuterium. This means that, for example, the deuterated equivalents of weak acids are less acidic, and any reaction involving the loss of a proton will be significantly slower with the deuterated version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since helium is pretty unreactive, and by the time you're at lithium an extra neutron is only increasing the mass by about 14%, the effects on reaction energetics are much, much smaller for anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest looking up the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot;&gt;Kinetic Isotope Effect&lt;/a&gt;&quot; for more information and more complex effects the mass change on substituting isotopes can cause to chemical activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Eric Brown said in his answer, different isotopes may also have very different spectroscopic properties if the spectroscopic technique depends either on vibrational or nuclear properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a minor curiosity, I'm given to understand this also means that heavy water is mildly hazardous to your health, since deuterated water and deuterated analogues of various vital chemicals have a nasty tendency to react mostly irreversibly or very slowly with vital enzymes that are &quot;tuned&quot; for hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-05-23T21:28:16.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5097" PostTypeId="2" ParentId="4656" CreationDate="2013-05-24T00:15:18.420" Score="1" Body="&lt;p&gt;I think @ManishEarth has it right.  The actual phenomenon involves not only the hydrolysis of the salt, but the autoionization of water.  To keep things simple the autoionization of water in this case is often omitted in elementary textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the hydrolysis of the anion and the cation are not normally identical and the resulting solution is slightly acidic or basic, depending on the nature of the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, for each such reaction there is a definite amount of hydrolysis (which you call $h$ in the question).  This constant will vary somewhat with temperature.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-05-24T00:15:18.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5098" PostTypeId="1" CreationDate="2013-05-24T04:31:18.370" Score="5" ViewCount="294" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.&#xA;&lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Some ideas&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I can think of a few ways to start this proof:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: &lt;strong&gt;&lt;em&gt;A&lt;/em&gt;&lt;/strong&gt;(0, 0, 0), &lt;strong&gt;&lt;em&gt;B&lt;/em&gt;&lt;/strong&gt;(1, 1, 0), &lt;strong&gt;&lt;em&gt;C&lt;/em&gt;&lt;/strong&gt;(1, 0, 1) and &lt;strong&gt;&lt;em&gt;D&lt;/em&gt;&lt;/strong&gt;(0, 1, 1), and a central point &lt;strong&gt;&lt;em&gt;O&lt;/em&gt;&lt;/strong&gt;($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\&#xA;g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\&#xA;g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\&#xA;g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\&#xA;\end{array}\right.$$&#xA;$$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i&amp;lt;j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right] $$&#xA;$$\nabla f=-\lambda \nabla g$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="1659" LastEditDate="2013-05-24T05:25:28.367" LastActivityDate="2013-05-24T05:25:28.367" Title="Tetrahedron is minimum energy conformation for atom with 4 electron domains in VSEPR theory" Tags="&lt;inorganic-chemistry&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5099" PostTypeId="2" ParentId="5098" CreationDate="2013-05-24T04:31:18.370" Score="2" Body="&lt;h1&gt;Sketch of a Simpler Solution&lt;/h1&gt;&#xA;&lt;p&gt;There is one remaining way that is simpler than all of the above, and it has to do with circumscribing a tetrahedron with a cube.&lt;/p&gt;&#xA;&lt;p&gt;I will leave it to the reader to work on this most beautiful version. The properties of the cube ensure that the distances are maximized, and some more geometry will show that the inscribed shape is indeed a regular tetrahedron. For a hint, check out &lt;a href=&quot;http://bit.ly/10WV6lt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this excellent answer to this question on the math stack exchange site&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1uLqx.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-05-24T04:55:41.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5100" PostTypeId="1" CreationDate="2013-05-24T05:26:12.367" Score="5" ViewCount="223" Body="&lt;p&gt;I'm reasonably unfamiliar with chemical notation, and recently I've come across something I'm not sure how to interpret.  In &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp036735i&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, the authors refer to the &quot;surface groups&quot; denoted by $\ce{#Ti-F}$ and $\ce{#Ti-OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0142961203006835&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; refers to surface functional groups $\ce{#Si-O-}$ and $\ce{#Si-OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my (limited) knowledge, &quot;$\ce{#}$&quot; denotes a triple bond.  If this is the case, what am I to take from this?  Are they referring to a (triple) dangling bond, or is this some notation I'm not familiar with to denote surface species?  The first page of the first article does state that &quot;$\ce{#Ti-X}$&quot; refers to surface species, but I'm finding it difficult to believe the notation was chosen arbitrarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any explanation.  As I've said, I am a novice with chemistry and chemical notation.&lt;/p&gt;&#xA;" OwnerUserId="1690" LastEditorUserId="16683" LastEditDate="2015-10-23T16:58:34.273" LastActivityDate="2015-10-23T16:58:34.273" Title="Surface Functional Groups/Notation" Tags="&lt;surface-chemistry&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5101" PostTypeId="1" AcceptedAnswerId="5556" CreationDate="2013-05-24T05:44:04.687" Score="16" ViewCount="7921" Body="&lt;p&gt;I was handling alum one day and accidentally I touched toilet soap after that. They reacted with each other and formed a rubberlike sticky substance (like dried glue on palm). I am curious what was the chemical reaction and the substance which was formed. I re-tried the same thing with detergent and got similar results.&lt;/p&gt;&#xA;" OwnerUserId="1691" LastEditorUserId="22" LastEditDate="2013-05-24T13:45:43.483" LastActivityDate="2018-04-19T16:11:48.840" Title="What happens when you mix alum with soap?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5103" PostTypeId="1" AcceptedAnswerId="5105" CreationDate="2013-05-24T10:07:12.217" Score="15" ViewCount="18520" Body="&lt;p&gt;I aim for two things with this post – on the one hand to produce a manual for other people to use, but also to verify what I obtained via searching, asking a more experienced user as well as receiving a reply from Gaussian support.&lt;br&gt;&#xA;I will assume that the user is familiar with the general basic idea of the format of a Gaussian input file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to find a transition state for an electrophilic aromatic substitution reaction in decane as  a solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal initial method to obtain a transition state is a QST2 calculation which requires a geometry for the reactants as well as the products. If it fails, a QST3 calculation which also provides a guess for the geometry of the transition state may also be attempted or as a last resort, a transition state may be manually built and optimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally the user should have optimized geometries for the reactants as well as the product(s) in a reaction, however it is possible to build the geometry “on the spot”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example I provide the input geometries for m-toludine and acetone: &#xA;In this case the oxygen atom  has been shifted out of the way – however the hydrogen has not been (should be though) tweaked on the m-toluidine. If bonds are broken, it is advisable to lengthen the bonds slightly as for example singe bonds will be longer than double bonds.&#xA;In the product, the initial molecules have been “rebounded” for the product.&#xA;“Imaginary bonds”, bonds from the product in the reactants or bonds from the reactants in the product are shows as dashed lines).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%nprocshared=4&#xA;%rwf=m-toluidine-acetone-2.rwf&#xA;%nosave&#xA;%mem=750MB&#xA;%chk=m-toluidine-acetone-2.chk&#xA;#p opt=(calcall,qst2) b3lyp/6-31++g(d,p) scrf=(solvent=n-decane) nosymm&#xA;geom=connectivity&#xA;&#xA;Title Card Required&#xA;&#xA;0 1&#xA; C                 -0.66738150    0.29867781   -0.01472477&#xA; C                  0.72777850    0.29867781   -0.01472477&#xA; C                  1.42531650    1.50642881   -0.01472477&#xA; C                  0.72766250    2.71493781   -0.01592377&#xA; C                 -0.66716250    2.71485981   -0.01640277&#xA; C                 -1.36476350    1.50665381   -0.01540677&#xA; H                 -1.10194426   -0.62306930    0.00047259&#xA; H                  1.27728650   -0.65383519   -0.01340977&#xA; H                  1.27786250    3.66708081   -0.01598277&#xA; H                 -2.46436750    1.50683681   -0.01558677&#xA; N                 -1.40248531    3.98773009   -0.01767660&#xA; H                 -1.16147100    4.51273462    0.79858460&#xA; H                 -1.16060228    4.51160224   -0.83440824&#xA; C                  2.96531624    1.50654085   -0.01383691&#xA; H                  3.32245021    2.00928088   -0.88825546&#xA; H                  3.32144259    2.01265639    0.85904400&#xA; H                  3.32205530    0.49776307   -0.01168239&#xA; C                 -2.94426822   -1.38737908   -0.21119110&#xA; C                 -2.43095250   -2.83931123   -0.21119110&#xA; H                 -2.78762331   -3.34370870    0.66246165&#xA; H                 -2.78762737   -3.34371014   -1.08484136&#xA; H                 -1.36095250   -2.83932442   -0.21119359&#xA; C                 -4.48426822   -1.38736010   -0.21119110&#xA; H                 -4.84094078   -1.89192796    0.66236254&#xA; H                 -4.84092213   -0.37854994   -0.21099580&#xA; H                 -4.84094128   -1.89158925   -1.08494009&#xA; O                 -2.63797974   -0.83944966    0.94169704&#xA;&#xA; 1 2 1.5 6 1.5 7 1.0 18 0.5&#xA; 2 3 1.5 8 1.0&#xA; 3 4 1.5 14 1.0&#xA; 4 5 1.5 9 1.0&#xA; 5 6 1.5 11 1.0&#xA; 6 10 1.0&#xA; 7 27 0.5&#xA; 8&#xA; 9&#xA; 10&#xA; 11 12 1.0 13 1.0&#xA; 12&#xA; 13&#xA; 14 15 1.0 16 1.0 17 1.0&#xA; 15&#xA; 16&#xA; 17&#xA; 18 19 1.0 23 1.0 27 2.0&#xA; 19 20 1.0 21 1.0 22 1.0&#xA; 20&#xA; 21&#xA; 22&#xA; 23 24 1.0 25 1.0 26 1.0&#xA; 24&#xA; 25&#xA; 26&#xA; 27&#xA;&#xA;Title Card Required&#xA;&#xA;0 1&#xA; C                 -1.10949687    0.47711117    1.88331619&#xA; C                  0.28566313    0.47711117    1.88331619&#xA; C                  0.98320113    1.68486217    1.88331619&#xA; C                  0.28554713    2.89337117    1.88211719&#xA; C                 -1.10927787    2.89329317    1.88163819&#xA; C                 -1.80687887    1.68508717    1.88263419&#xA; H                 -1.73208670   -0.29950945    3.46519163&#xA; H                  0.83517113   -0.47540183    1.88463119&#xA; H                  0.83574713    3.84551417    1.88205819&#xA; H                 -2.90648287    1.68527017    1.88245419&#xA; N                 -1.84460068    4.16616345    1.88036435&#xA; H                 -1.60358637    4.69116797    2.69662556&#xA; H                 -1.60271765    4.69003560    1.06363271&#xA; C                  2.52320087    1.68497420    1.88420404&#xA; H                  2.88033484    2.18771423    1.00978550&#xA; H                  2.87932722    2.19108974    2.75708496&#xA; H                  2.87993993    0.67619643    1.88635857&#xA; C                 -2.60557153   -0.76810572    1.76628847&#xA; C                 -2.09225582   -2.22003788    1.76628847&#xA; H                 -2.44892663   -2.72443535    2.63994122&#xA; H                 -2.44893069   -2.72443679    0.89263821&#xA; H                 -1.02225582   -2.22005106    1.76628598&#xA; C                 -4.14557153   -0.76808675    1.76628847&#xA; H                 -4.50224410   -1.27265460    2.63984211&#xA; H                 -4.50222545    0.24072341    1.76648378&#xA; H                 -4.50224460   -1.27231589    0.89253948&#xA; O                 -2.55774834   -0.63242367    2.73721431&#xA;&#xA; 1 2 1.5 6 1.5 7 0.5 18 1.0&#xA; 2 3 1.5 8 1.0&#xA; 3 4 1.5 14 1.0&#xA; 4 5 1.5 9 1.0&#xA; 5 6 1.5 11 1.0&#xA; 6 10 1.0&#xA; 7 27 1.0&#xA; 8&#xA; 9&#xA; 10&#xA; 11 12 1.0 13 1.0&#xA; 12&#xA; 13&#xA; 14 15 1.0 16 1.0 17 1.0&#xA; 15&#xA; 16&#xA; 17&#xA; 18 19 1.0 23 1.0 27 1.0&#xA; 19 20 1.0 21 1.0 22 1.0&#xA; 20&#xA; 21&#xA; 22&#xA; 23 24 1.0 25 1.0 26 1.0&#xA; 24&#xA; 25&#xA; 26&#xA; 27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The calculation can then be started.&#xA;Once it has successfully finished, it is advisable to verify that the molecule exhibits one imaginary frequency only and that the motion of the atoms or molecules under that frequency agree with the expected reaction. If this is the case, one should follow up with an IRC calculation to follow the reaction path to verify one has indeed found the desired transition state.&#xA;(The geometry below is the transition state from the calculation above, set up for an IRC calculation.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%nprocshared=4&#xA;%rwf=m-toluidine-acetone-2.rwf&#xA;%nosave&#xA;%mem=750MB&#xA;%chk=m-toluidine-acetone-2.chk&#xA;#p irc=(maxpoints=200,recalc=5,calcfc) b3lyp/6-31++g(d,p)&#xA;scrf=(solvent=n-decane) nosymm geom=connectivity&#xA;&#xA;Title Card Required&#xA;&#xA;0 1&#xA; C                 -1.20983500    0.26428000    0.39208600&#xA; C                  0.14314000    0.04829600    0.07611800&#xA; C                  0.97119600    1.12420400   -0.26335000&#xA; C                  0.42483200    2.41568200   -0.28932800&#xA; C                 -0.92232100    2.65929200    0.02889700&#xA; C                 -1.73598700    1.56297000    0.36139500&#xA; H                 -1.80370000   -0.35728900    1.39005500&#xA; H                  0.56171000   -0.95292700    0.09719200&#xA; H                  1.06141800    3.25561300   -0.56193000&#xA; H                 -2.77794400    1.73200600    0.62392300&#xA; N                 -1.45199400    3.94909800   -0.04588200&#xA; H                 -2.29620600    4.10853800    0.48772900&#xA; H                 -0.78926200    4.70638200    0.05450600&#xA; C                  2.43332600    0.91594500   -0.58932400&#xA; H                  2.68752500    1.33957100   -1.56739900&#xA; H                  3.07670000    1.40280600    0.15294900&#xA; H                  2.68804800   -0.14719300   -0.60548500&#xA; C                 -2.49137900   -1.22149000    0.43327000&#xA; C                 -1.66737300   -2.42548800    0.01309500&#xA; H                 -2.34686200   -3.28416400    0.08629100&#xA; H                 -1.29025700   -2.36043500   -1.01081300&#xA; H                 -0.84776200   -2.60357900    0.71204300&#xA; C                 -3.57286200   -0.83131000   -0.55877300&#xA; H                 -4.30815500   -1.64705100   -0.52976700&#xA; H                 -4.08297000    0.08110500   -0.24873300&#xA; H                 -3.20072400   -0.72720100   -1.58114900&#xA; O                 -2.83178700   -1.17441600    1.73871300&#xA;&#xA; 1 2 1.5 6 1.5&#xA; 2 3 1.5 8 1.0&#xA; 3 4 1.5 14 1.0&#xA; 4 5 1.5 9 1.0&#xA; 5 6 1.5 11 1.0&#xA; 6 10 1.0&#xA; 7&#xA; 8&#xA; 9&#xA; 10&#xA; 11 12 1.0 13 1.0&#xA; 12&#xA; 13&#xA; 14 15 1.0 16 1.0 17 1.0&#xA; 15&#xA; 16&#xA; 17&#xA; 18 19 1.0 23 1.0 27 1.0&#xA; 19 20 1.0 21 1.0 22 1.0&#xA; 20&#xA; 21&#xA; 22&#xA; 23 24 1.0 25 1.0 26 1.0&#xA; 24&#xA; 25&#xA; 26&#xA; 27&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am aware that the initial reactants guess is not great in this calculation as it dates back to me starting to work with Gaussian.&lt;/li&gt;&#xA;&lt;li&gt;Is there anything I am missing that could have a significant impact on the obtained transition state (and the resulting energy barrier for the reaction)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1665" LastEditorUserId="4945" LastEditDate="2017-11-28T05:15:07.507" LastActivityDate="2017-11-28T05:34:36.353" Title="How to find a transition state for an electrophilic addition with Gaussian and map the reaction pathway?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="5104" PostTypeId="1" AcceptedAnswerId="5106" CreationDate="2013-05-24T14:12:38.163" Score="2" ViewCount="2644" Body="&lt;p&gt;I ran molecular dynamics simulation with a molecule of methylcyclohexane soaked into a water box. On the other hand I ran Monte Carlo with the molecule alone (no soaking). I performed this for the equatorial and axial conformers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial conformer with MD stayed axial (the methyl group). The equatorial with MD stayed equatorial. So far no visual change. Is there a way to compute the average energy in VMD for these systems (I have around 250 frames).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the axial conformer alone with MC was transformed into equatorial. While the equatorial stayed the same (visually speaking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand what's happening here. What I read is that always the &lt;strong&gt;equatorial&lt;/strong&gt; configuration is more stable as it has less steric hindrance (BTW I am computer scientist learning this), so this explains the interconversion in Monte Carlo simulation from axial to equatorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I would expect also that in MD the axial conformer turn into equatorial but it didn't, what I believe here is that the water molecules are having some &quot;Solvation effect&quot; on the methylcyclohexane that prevents the interconversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to know if my thought are good and if you could give some references to get more info about solvation effects.&lt;/p&gt;&#xA;" OwnerUserId="1668" LastActivityDate="2013-05-24T16:36:44.547" Title="Solvation effect on cyclohexane" Tags="&lt;solvents&gt;&lt;conformers&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5105" PostTypeId="2" ParentId="5103" CreationDate="2013-05-24T16:34:22.707" Score="17" Body="&lt;h2&gt;Mapping a Reaction Pathway: A Simple Guide&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are various ways to go about determining a reaction pathway using electronic structure theory.  This post is to serve as a brief guide, highlighting a few useful techniques that may make this process more efficient.  I will restrict this guide to a simple reaction mentioned below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preliminaries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a reaction of interest such as A + B -&gt; C.  In terms of relative energies we could define the electronic energies of an isolated A and an isolated B (i.e. (A)+(B)) as the separated reactants (SR) energy.  Furthermore, we could define the electronic energy of C as the product (prod) energy.  So now we have determined two electronic energies that can be set relative to each other in order to determine the deltaE of reaction which I will call dE which is simply (E(prod)-E(SR)).  However, when it comes to reaction pathways, we are also interested in the activation energy (Ea) of the reaction.  This means, for our basic reaction example, that there is one other electronic energy which we must consider, which leads us to the transition state (TS).  Resolving this structure will be discussed from this point on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Technique #1 - Finding a TS Using the QST2 &amp;amp; QST3 Methods&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The QST2 and QST3 methods allow the user to specify two minimum energy structures lying on each 'side' of a TS, and have an algorithm determine a TS which connects the two.  In our example, SR and Prod are these two structures.  Now, for our simple reaction example, using QST2 is unwise as our SR is simply two molecules, A and B, separated at infinite distance.  This is not a meaningful geometry in terms of a QST method.  Our Prod structure would work quite well, however.  The QST methods work well if both of your input geometries represent coherent molecular systems.  So if you were resolving a more complicated reaction pathway with one or more intermediates, you could use the QST method to determine TS structures connecting an intermediate to another intermediate or an intermediate to a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between QST2 and QST3 is simply that the latter allows the user to input a TS guess.  If you have fully characterized both the low-lying minimum energy structures which you wish to find a connecting TS structure, you may use your own intuition and guess what the TS may or should look like.  This can help speed up the process of locating a TS with QST3.  However, if you do not know what the TS would look like, you can revert to using QST2 and not provide a TS guess.  In my experience with these methods, it is usually more efficient to provide the program with your own user-defined internal coordinates rather than allowing the program to create its own. If you are lucky enough to have this type of job complete successfully, you will want to optimize the resulting geometry and perform a frequency analysis to determine the nature of the stationary point.  Since we are looking for a TS, we would have to end up with one imaginary mode.  If you don't get this, you need to go back try finding another TS.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;QST2 only requires two fully characterized minimum energy structures.  These must be coherent chemical systems.&lt;/li&gt;&#xA;&lt;li&gt;QST3 allows the user to provide a TS guess&lt;/li&gt;&#xA;&lt;li&gt;Consider using your own internal coordinates rather than allowing the program to create its own&lt;/li&gt;&#xA;&lt;li&gt;Fully characterize the resulting structure by performing an optimization and a corresponding harmonic vibrational frequency analysis.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Technique #2 - Finding a TS via Scanning&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have already stated that SR is simply (A)+(B) where A and B are separated at infinite distance.  We can look at our Prod and gain some insight as to how A and B come together to form Prod.  So rather than leaving our TS guess up to some program to decide, we can be smart about the way we go about finding it by using our own, well-informed intuition.  Perhaps by observing SR and Prod we make a reasonable guess that the distance between some atom on molecule A and some atom on molecule B decreases as we go from SR to Prod.  Looking at the fully characterized Prod, we know the final distance between these two atoms.  In SR, this distance is infinite.  What we do is we take our Prod and scan across this distance, partially optimizing the geometry as we go.  So if the distance between the two atoms is 1 Angstrom in the Prod, we would gradually increase this distance in increments of say 0.2 Angstroms a total of, perhaps, 10 times.  Our final distance would then be 3.0 Angstroms of separation.  From this can, we can look at the electronic energies and, if we guessed a good coordinate to scan over, we should have a smooth energy curve.  From this we can yank the geometry with the highest electronic energy from the scan and use this as our guess to the TS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your guess TS from the scan and insert it into a TS optimization job.  With a bit of luck, your guess will be close to a real TS.  Again, perform a frequency analysis and determine the nature of the stationary point.  We must end up with one imaginary mode for our structure to be a TS.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Scanning requires a good guess as to which coordinate to scan across; you will have to specify your own internal coordinates&lt;/li&gt;&#xA;&lt;li&gt;The maximum energy structure from your scan is used for a TS optimization&lt;/li&gt;&#xA;&lt;li&gt;This manual approach maximizes user intuition&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Verifying the TS - The IRC Computation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because we've now characterized a TS does not mean that the TS connects, in our case, SR and Prod.  We must perform an intrinsic reaction coordinate (IRC) computation.  Simply, we give a program our fully characterized TS and we follow the reaction path in the forward and reverse directions.  Be sure to give your IRC calculation a large number of steps.  IRC calculations require force constants on your TS.  Since you've already fully characterized your TS via a frequency analysis, you already have the force constants.  Just feed them into your IRC job, eliminating the need to recalculate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once your IRC calculation has finished, you must optimize the final structures from both the forward and reverse directions.  The resulting structures should have the same energy and same geometry as the structures you used to find the TS.  Of course in our simple reaction case, you would only be comparing the Prod as SR is molecules A and B at infinite distance.  For this, just ensure that the IRC proceeded in such a way that A and B are separating from each other across the entire IRC pathway.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perform an IRC on the TS you have characterized in both the forward and reverse directions&lt;/li&gt;&#xA;&lt;li&gt;Optimize the final structures from the IRC computation and compare to the fully characterized structures you used to initially find your TS.  If they match both energetically and geometrically, then you have found a plausible TS which connects both structures.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2013-05-24T16:49:38.307" LastActivityDate="2013-05-24T16:49:38.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5106" PostTypeId="2" ParentId="5104" CreationDate="2013-05-24T16:36:44.547" Score="3" Body="&lt;p&gt;The difference in behavior between the two methods is likely how much &quot;wiggle&quot; the molecule is allowed to have, or how much additional kinetic energy you allow the molecule to have when searching for the minimum energy conformation. The equatorial conformation is the global minimum conformation, but the axial conformation is a local minimum. Additionally, between the two conformations, there is another minimum called the twist-boat. You can read about them on this &lt;a href=&quot;http://en.wikipedia.org/wiki/Twist_boat&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; or you could check out an introductory organic chemistry book from a library. Molecular dynamics tend to stop when &lt;strong&gt;a minimum&lt;/strong&gt; is found. In other words, if a small alteration to the geometry increases the potential energy, the calculation stops. Monte Carlo calculations on molecules are designed to examine the evolution of conformation with time, and so geometry moves out of energy minima are allowed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsubstituted cyclohexane has four different conformations: Chair (relative energy 0 kJ/mol), half chair (+45 kJ/mol), twist boat (+23 kJ/mol), and boat (+30 kJ/mol). A molecular dynamics calculation may not allow large enough steps to get over the activation barrier to get out of a chair conformation. The following energy diagram shows the conformational space of cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdJ8j.png&quot; alt=&quot;conformations of cyclohexane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy difference between equatorial and axial methylcyclohexane is 7.6 kJ/mol (1.6 kcal/mol &lt;a href=&quot;http://en.wikipedia.org/wiki/A_value#Table_of_A-Values&quot; rel=&quot;nofollow noreferrer&quot;&gt;on this table&lt;/a&gt;), which is small compared to the difference between the chair and the half-chair conformation. The energy diagram below approximates the energy steps between equatorial and axial methylcyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QKZ75.png&quot; alt=&quot;conformations of methylcyclohexane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any computational method that allows potential energy steps large enough to get to the half-chair will certainly isomerize the axial conformer to the equatorial conformer. Presence or absence of solvent molecules in the simulation may be irrelevant. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-24T16:36:44.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5107" PostTypeId="1" CreationDate="2013-05-25T08:50:18.670" Score="3" ViewCount="1022" Body="&lt;p&gt;When hex-1-ene is treated with bromotrichloromethane in the presence of a peroxide initiator (e.g. dibenzoyl peroxide), what is the regioselectivity of the addition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the radical addition of HBr to an alkene is considered anti-Markovnikov, but is this reaction considered a Markovnikov or anti-Markovnikov addition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q8UmU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8UmU.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user1694" LastEditorUserId="16683" LastEditDate="2018-10-18T23:33:00.857" LastActivityDate="2018-10-19T00:02:19.217" Title="Regioselectivity in Kharasch addition of bromotrichloromethane" Tags="&lt;organic-chemistry&gt;&lt;radicals&gt;&lt;regioselectivity&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5108" PostTypeId="1" CreationDate="2013-05-25T13:40:27.227" Score="5" ViewCount="599" Body="&lt;p&gt;I've learnt in Chemistry that for example in Chromium, the 4s subshell only has one electron in favor of making the 3d subshell half-full. Also, for Copper the same happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other circumstances does this happen under? Do ions obey the same pattern, e.g. Mn+?&lt;/p&gt;&#xA;" OwnerUserId="1344" LastActivityDate="2016-08-09T01:05:24.510" Title="In which cases does the 3d subshell take precedence over the 4s subshell?" Tags="&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5109" PostTypeId="1" CreationDate="2013-05-25T15:54:42.003" Score="2" ViewCount="124" Body="&lt;p&gt;I've got the following from a &lt;a href=&quot;http://megglobal.com/commitment.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;supposed new energy-storage technologist&lt;/a&gt;. It's supposed to be a fancy new cheap, high-powered, highly scalable battery. I want to know if this is known chemistry, and if there's anything genuinely revolutionary about it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Basic materials in MEG's E-SOURCE™ power cells are lead metal, absorptive glass mat paper, and sulfuric acid solution. Materials unique to the MEG advanced power cell are S2 glass fiber yarn; sodium sulfate; Teflon emulsion, glass-filled polypropylene; bituminous, petroleum-based wax; and metal-free nano-oxides. &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The lead is apparently in the form of lead oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The claim is that this is a radical new cheap energy-storage system - but someone makes that claim for yet another technology every week or two. Is there anything in the recent literature that would give credibility to this one?&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="402" LastEditDate="2017-04-08T22:02:13.463" LastActivityDate="2017-04-08T22:02:13.463" Title="Is there anything in the literature that supports revolutionary claims for this &quot;new&quot; sulfuric-acid battery?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5110" PostTypeId="1" AcceptedAnswerId="5111" CreationDate="2013-05-25T11:02:39.493" Score="8" ViewCount="10343" Body="&lt;p&gt;Increased vibration causes the atoms in an object to spread out farther away. Thus, with added heat, and object goes form a solid to a liquid to a gas (and then to plasma, with enough heat). Why is it then, that when a heat something like cake batter, it solidifies?&lt;/p&gt;&#xA;" OwnerDisplayName="Hmir" LastEditorUserId="1362" LastEditDate="2013-05-26T12:18:01.963" LastActivityDate="2017-09-11T18:16:56.987" Title="Why does heat solidify cake batter?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;food-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5111" PostTypeId="2" ParentId="5110" CreationDate="2013-05-25T11:19:29.507" Score="8" Body="&lt;p&gt;It is caused by &lt;a href=&quot;http://www.sugar.org/sugar-basics/sugar-in-bakery-foods.html&quot; rel=&quot;noreferrer&quot;&gt;physical and chemical reactions&lt;/a&gt; in the cake batter of two types:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) When heated and mixed with water, starches present in the flour will undergo a process called &lt;a href=&quot;http://en.wikipedia.org/wiki/Starch_gelatinization&quot; rel=&quot;noreferrer&quot;&gt;gelatinization&lt;/a&gt; in which water is absorbed in the structure. This will cause the batter to 'set' from a liquid into a solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Additionally, reactions of the proteins from the egg that you added can also work to 'set' the batter. At first the proteins will unfold (denature) after which they start to form bigger proteins through cross-linking reactions that turn the liquid into a solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The baking soda in the batter turns into sodium carbonate, water and carbon dioxide. The latter will make your cake nice and fluffy.&lt;/p&gt;&#xA;" OwnerUserId="1362" OwnerDisplayName="michielm" LastEditorUserId="1362" LastEditDate="2013-05-26T12:13:09.353" LastActivityDate="2013-05-26T12:13:09.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5113" PostTypeId="2" ParentId="4979" CreationDate="2013-05-25T19:50:33.327" Score="8" Body="&lt;h2&gt;An approach using a classical model of electron flow&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You can think of an electron flowing through a medium as analogous to a pinball bouncing around a pinball machine. The electric (potential) energy difference driving the displacement of the electron is analogous to the gravitational energy difference in the pinball machine. Now consider two pinball machines with differing numbers of obstacles to a falling pinball. The time it takes for a pinball to reach the bottom in each machine would be different by virtue of these obstacles (on average, taking more time with more obstacles). Similarly, the flow of electrons, &lt;em&gt;I&lt;/em&gt;, in the presence of an electric potential, &lt;em&gt;V&lt;/em&gt;, is higher for media with less resistance, &lt;em&gt;R&lt;/em&gt; , to electron flow. We can use this model (&lt;a href=&quot;http://en.wikipedia.org/wiki/Drude_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drude&lt;/a&gt; see below) to understand &lt;a href=&quot;http://en.wikipedia.org/wiki/Ohm%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ohm's law&lt;/a&gt;: $$I = \frac{V}{R}$$ The observation that electrons flow more readily through electrolyte solutions than non-electrolyte solutions indicates that solutions with charge carriers reduce the resistance of the media. A better understanding of why this is true at a molecular level will require a more in depth look. Notice, though, that we now have answers to your good questions! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;What happens once the charges get to the electrodes?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sodium stays in solution as an ion, but the hydrogen ion (from the hydrolysis of water) can be reduced to form hydrogen gas at the cathode, and the chlorine ion can be oxidized to form chlorine gas at the anode. Electrons can be passed through the solution as solvated electrons under some circumstances, but this is unlikely due to the larger energy barrier to this process (around &lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2010/March/07031001&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 eV&lt;/a&gt;). With sufficient energy, these electrons (called solvated electrons) flow through metal and solutions. The resistance to electron flow in metals is much less than in electrolyte solutions, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;Would it cause the water to split into hydrogen gas and hydroxide ions?&quot;&lt;/em&gt; and &lt;em&gt;&quot;I would expect the sodium, once given an electron, to be looking for a way to give that electron back.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any additional redox reaction would require electric potential. You mentioned the splitting of water or the formation of sodium metal. This would not occur according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;the standard electrode potential&lt;/a&gt; unless the electric potential provided is sufficiently high to overcome the &quot;uphill&quot; energy barrier and as you mention &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrolysis&lt;/a&gt; would be involved.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Drude Model&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In 1900, Paul Drude used the following schematic of a lattice of metal ions sitting in a sea of electrons to derive a relationship between electron flow and electric potential applied.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IiDsO.png&quot; alt=&quot;&amp;lt;Insert image&amp;gt;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Drude Model electrons (shown here in blue) constantly bounce between heavier, stationary crystal ions (shown in red). -From wikipedia entry on &lt;a href=&quot;http://en.wikipedia.org/wiki/Drude_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drude model&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A metal's electrical resistance, &lt;em&gt;R&lt;/em&gt;, (as seen in Ohm's law above) should take the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = \left(\frac{m}{nq^2\tau}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,  $m$ is mass of electron; $n$ is number density of electrons; $q$ is charge of electron and $\tau$ is the mean free path. While this classical model does not explain the resistances for all substances perfectly and quantum mechanics can be used to do a better job, it does suggest that if the mean free path for electron flow is increased and the assumptions of the model are valid, the electrical resistance of the media should be reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to understand why there is lower electrical resistance (higher electrical conductance) in aqueous solutions of electrolytes is that there is less water in the way of the Drude ion lattice! In other words, in the limit that the solution is a molten salt, we approach a Drudian model. As you add more and more molecules of water to this model, you are reducing the mean free path of the electron and increasing the resistance of the solution. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Notes&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The fact that an electrolyte solution is comprised of positive &lt;em&gt;and negative&lt;/em&gt; ions does not diminish the utility of Drude's model: electrons are equally attracted (and speed up) moving toward a (metal) cation as they are attracted (and slow down) moving away from a metal cation. The situation is reversed but the effect is the same for a nonmetal anion. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For more evidence of the utility of note 1., read about the additive nature of the molar electrical conductance (inverse of resistance) of ions (cations and anions): &lt;a href=&quot;http://en.wikipedia.org/wiki/Conductivity_%28electrolytic%29#Strong_electrolytes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kohlrausch's Law of independent migration of ions&lt;/a&gt;. The additive nature suggests that both cations and anions improve the conductance of an electrolyte solution as the note above suggests.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="1659" LastEditDate="2014-07-27T21:11:23.007" LastActivityDate="2014-07-27T21:11:23.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5116" PostTypeId="1" AcceptedAnswerId="5118" CreationDate="2013-05-25T20:30:53.387" Score="5" ViewCount="336" Body="&lt;p&gt;Usually textbooks on CI takes the SCF Hartree-Fock orbitals as a starting point. Can one use the configuration interaction method starting directly from the single-particle Slater determinants avoiding the iterative procedure. This question concerns mainly small molecules like the hydrogen molecule. Do I get a high inaccuracy in this case? If yes, why? Could this inaccuracy be decreased taking into account more configurations?&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2021-03-21T22:02:17.190" LastActivityDate="2021-03-21T22:02:17.190" Title="CI without Hartree-Fock SCF" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5118" PostTypeId="2" ParentId="5116" CreationDate="2013-05-25T22:41:44.423" Score="6" Body="&lt;p&gt;If you were to perform Full CI (within a finite basis) then I don't think that it matters whether you start with SCF orbitals or some other choice.  This solution &lt;em&gt;will&lt;/em&gt; have the lowest conceivable energy no matter what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are doing truncated (less than full) CI, then you are very much at the mercy of your starting orbitals. In general, one would want to start with orbitals that resemble those of the actual state that you are trying to describe via CI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, your precious excitations (singles and doubles) may be consumed correcting the &quot;slop&quot; that you provided, rather than moving toward the full CI limit.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-25T22:41:44.423" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5119" PostTypeId="1" CreationDate="2013-05-25T23:43:30.727" Score="6" ViewCount="6890" Body="&lt;p&gt;I read this at this webpage: &lt;a href=&quot;http://www.whfoods.com/genpage.php?dbid=90&amp;amp;tname=nutrient#impactcookingstorageprocessing&quot;&gt;http://www.whfoods.com/genpage.php?dbid=90&amp;amp;tname=nutrient#impactcookingstorageprocessing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How do cooking, storage, or processing affect potassium?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Potassium losses from cooking of high-potassium foods can be&#xA;  significant. In the case of spinach for example, potassium levels have&#xA;  been shown to drop from 6.9 to 3.0 grams in 3 and 1/2 ounces of&#xA;  spinach after blanching for several minutes (a loss of about 56%).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand this. How can potassium disappear after cooking ? &lt;/p&gt;&#xA;" OwnerUserId="1699" LastActivityDate="2015-04-03T16:05:30.587" Title="How can potassium disappear after cooking?" Tags="&lt;biochemistry&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5120" PostTypeId="2" ParentId="5119" CreationDate="2013-05-25T23:58:32.033" Score="3" Body="&lt;p&gt;The potassium is leeched out into the water used to cook it, and unfortunately poured down the drain -- vitamins and all.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-25T23:58:32.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5121" PostTypeId="2" ParentId="5119" CreationDate="2013-05-26T07:57:12.073" Score="8" Body="&lt;p&gt;As Eric has mentioned in his answer, the potassium present in foods will leach out into water when you cook them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium is present in food in the form of potassium &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion&quot; rel=&quot;nofollow&quot;&gt;ions&lt;/a&gt; $(\ce{K+})$. These $\ce{K+}$ ions happen to be extremely soluble in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When food is boiled, the cells break down and release their contents more easily which makes the $\ce{K+}$ ions free to move into the water. The $\ce{K+}$ ions remain in the water since they are so highly soluble in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens to all water-soluble nutrients like the B vitamins, vitamin C, sodium, etc. Some ways to avoid the loss of all these nutrients would be to steam such foods instead of boiling them since the food wouldn’t come into contact with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Side note:&lt;/strong&gt; The same problem occurs when you deep fry foods containing fat-soluble nutrients like vitamins A and E. These leach out easily into oil and so all those nutrients would also be lost in the process.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T16:05:30.587" LastActivityDate="2015-04-03T16:05:30.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5122" PostTypeId="1" CreationDate="2013-05-26T16:48:47.533" Score="1" ViewCount="131962" Body="&lt;p&gt;For example, iron and tin(II) sulfate will react to produce iron(II) sulfate and tin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe + SnSO4 -&amp;gt; FeSO4 + Sn}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the following half equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe + SO4^{2-} -&amp;gt; FeSO4 + 2e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Sn^{2+} + 2e- -&amp;gt; Sn}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but where does the &quot;2&quot; come from? Why can't it be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe + SO4^{5-} -&amp;gt; FeSO4 + 5e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Sn^{5+} + 5e- -&amp;gt; Sn}$&lt;/p&gt;&#xA;" OwnerUserId="1702" LastEditorUserId="22" LastEditDate="2013-06-29T23:23:05.943" LastActivityDate="2021-02-12T23:32:55.880" Title="How is it determined how many electrons are transferred in redox reactions?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5123" PostTypeId="1" AcceptedAnswerId="6098" CreationDate="2013-05-26T21:43:35.510" Score="1" ViewCount="180" Body="&lt;p&gt;I would like to know if anybody here knows of a method to detect the presence of ovalbumin--or any unique, egg-related molecule, in a baked good. Here I am anticipating that the &quot;unique egg-related molecule&quot; might alter under high temperatures, and so the suspect molecule for detection may or may not be any molecule which is present in a fresh, uncooked egg. I'm thinking of something that can be brushed on to a pastry to evolve a color change. Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1309" LastEditorUserId="1309" LastEditDate="2013-08-28T21:40:15.610" LastActivityDate="2013-09-01T11:55:58.003" Title="Egg-Detection in Pastries: An Analysis in Heat-Alterated Molecule Identification" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;molecules&gt;&lt;color&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5125" PostTypeId="1" AcceptedAnswerId="5126" CreationDate="2013-05-27T04:10:36.190" Score="11" ViewCount="44259" Body="&lt;p&gt;In chemistry, we recently did a lab where we observed the different ways of reading pH of several different substances. We used things such as red cabbage juice, litmus paper, pH paper and a pH probe. One of the substances we tested was water. When we used the pH probe, it came back with a reading around 8.0. When we tested the distilled water, it came back with a reading very close to 7.0. So what is different between distilled and tap water that causes the tap water to be basic?&lt;/p&gt;&#xA;" OwnerUserId="1703" LastEditorUserId="21964" LastEditDate="2016-08-02T01:55:46.913" LastActivityDate="2018-07-11T13:22:05.713" Title="What chemicals are in tap water that gives it a slightly basic pH level" Tags="&lt;water&gt;&lt;ph&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5126" PostTypeId="2" ParentId="5125" CreationDate="2013-05-27T06:06:41.153" Score="12" Body="&lt;p&gt;The tap water is likely &quot;hard,&quot; i.e., contains some dissolved mineral salts, most likely (primarily) calcium and magnesium carbonates and bicarbonates. The anions of these salts are slightly basic, consuming $\ce{H^+}$ ions and thereby raising the pH. Carbonates (with the obvious exception of alkali metal salts) are only sparingly soluble in water, while bicarbonates are typically very soluble. A couple of other subtle factors may affect the pH:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lower atmospheric pressure encourages the evaporation of dissolved carbon dioxide from water. Since dissolved carbon dioxide exists in equilibrium with carbonic acid, it would tend to lower water pH. If you live in an area with comparatively low atmospheric pressure, your tap water might have a lower concentration of $\ce{CO_2}$, which would favor higher pH.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At higher temperatures, bicarbonate decomposes to water, carbonate, and carbon dioxide. Carbonates are relatively insoluble, and the consumption of bicarbonate should drive forward the dissociation of any dissolved carbonic acid, so a higher temperature should have the effect of lowering pH.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-05-27T06:06:41.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5128" PostTypeId="1" CreationDate="2013-05-28T11:05:39.190" Score="5" ViewCount="13316" Body="&lt;p&gt;I need to make a saturated solution of $\ce{Na2CO3}$ and according to the available data I should add $22\ \text{g}$ in $100\ \text{ml}$ of water at $25\ ^\circ\text{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to make $300\ \text{ml}$ solution, which would require $66\ \text{g}$. But, even after adding $110\ \text{g}$ of the salt, no turbidity appears which might indicate saturation. So I believe that the solution is still not saturated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I kept the flask overnight, almost the entire salt precipitated out in large crystalline chunks. I am a bit confused as to what the problem might be. Has anyone faced this kind of trouble?&lt;/p&gt;&#xA;" OwnerUserId="1707" LastEditorDisplayName="user7951" LastEditDate="2015-06-16T18:21:50.817" LastActivityDate="2015-06-16T18:21:50.817" Title="Making a saturated solution of Na₂CO₃" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5129" PostTypeId="2" ParentId="5128" CreationDate="2013-05-28T14:24:04.510" Score="3" Body="&lt;p&gt;One possibility that occurs to me is that some proportion of the carbonate anions underwent an acid-base reaction with the water to form bicarbonate, which is substantially less soluble (96g/L for $\ce{NaHCO3}$ vs. 215g/L for $\ce{Na2CO3}$ at 20&amp;deg;C according to the respective Wikipedia pages, meaning the carbonate salt is nearly twice as soluble in terms of molar ratios), and therefore would precipitate out. The acid-base equilibrium would be driven forward by the formation of said precipitate over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why no precipitate formed initially, even when the solution was supersaturated, I'm not sure. Normally, preparation of a supersaturated solution would require heating to increase solubility. Then, upon cooling, crystals will precipitate out if given nucleation sites around which to form and/or the solution is physically disturbed. The solvation of sodium carbonate is an exothermic process, so the process would raise the temperature of the solution, but I doubt that's sufficient to explain the unexpectedly great solubility that you observed.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-05-28T14:32:36.007" LastActivityDate="2013-05-28T14:32:36.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5130" PostTypeId="1" CreationDate="2013-05-28T22:16:19.093" Score="2" ViewCount="1531" Body="&lt;p&gt;In my textbook it says, that a composed substance is a substance made of more than 1 type of atom. And It also says, in another chapter it says that a mixture is the grouping of more than one substance without any chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between a mixture and a composed substance?&lt;/p&gt;&#xA;" OwnerUserId="1711" LastEditorUserId="4945" LastEditDate="2014-11-19T10:13:50.107" LastActivityDate="2014-11-19T14:03:41.160" Title="What is the difference between a mixture and a composed substance?" Tags="&lt;periodic-table&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5131" PostTypeId="1" CreationDate="2013-05-28T22:23:39.500" Score="2" ViewCount="361" Body="&lt;p&gt;I would like to know what is the difference between a chemical element and a type of atom? Can  anyone give the definition, please!&lt;/p&gt;&#xA;" OwnerUserId="1711" LastActivityDate="2013-05-29T22:36:28.447" Title="What is a chemical element?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5132" PostTypeId="2" ParentId="5131" CreationDate="2013-05-28T23:54:43.257" Score="3" Body="&lt;p&gt;From a scientific standpoint, there is no difference.  Elements=Atoms=the 100 or so basic building blocks of chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is purely conjecture, but I suspect that the term/concept &quot;elements&quot; pre-date &quot;atoms&quot; (at least atoms as we know them today).  I could see &quot;elements&quot; referring to the macroscopic description, e.g. a jar of carbon (graphite), but it may not have been realized back-in-the-day that these were aggregations of many carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Atom&quot; strikes me as quantum scale, and &quot;Element&quot; strikes me as something on the macroscopic scale, perhaps hearkening back to the pre-atomic days.  But really, there is no difference.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-28T23:54:43.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5133" PostTypeId="2" ParentId="4188" CreationDate="2013-05-29T00:15:30.110" Score="15" Body="&lt;p&gt;(My sister-in-law works at Unilever, and she makes these soaps)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all detergents.  The primary differences are harshness (hand soaps can be harsher, as hands are often more soiled), lathering, concentration, perfuming/scents, and marketing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Emphasis on the marketing.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-29T13:01:57.487" LastActivityDate="2013-05-29T13:01:57.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5134" PostTypeId="1" CreationDate="2013-05-29T01:02:30.887" Score="21" ViewCount="33866" Body="&lt;p&gt;Shouldn't iron oxide be able to form without water? It &lt;em&gt;is&lt;/em&gt; just iron and oxygen. I don't really understand what the dot followed by the $\ce{H2O}$ means either. I was reading on wikipedia, but I have a rather terrible understanding of chemistry. &#xA;What purpose does water serve in rust formation, and what does it mean for a molecule to be hydrated-how is that different then just being surrounded by water molecules?&lt;/p&gt;&#xA;" OwnerDisplayName="user1638" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:22.413" LastActivityDate="2020-03-03T15:14:11.587" Title="Why can't rust form without water?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5135" PostTypeId="1" AcceptedAnswerId="5146" CreationDate="2013-05-29T01:49:53.660" Score="5" ViewCount="3972" Body="&lt;p&gt;I hear about things like dioxin, but I really don't see what's dangerous about plastics-shouldn't you be ble to just breathe it out, or pass it out of your system? How would it cause cancer?&lt;/p&gt;&#xA;" OwnerDisplayName="user1638" LastEditorUserId="305" LastEditDate="2013-05-29T02:48:56.680" LastActivityDate="2015-06-26T19:04:55.817" Title="Chemically Speaking, What's Bad About Inhaling Plastics?" Tags="&lt;everyday-chemistry&gt;&lt;plastics&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5136" PostTypeId="2" ParentId="5134" CreationDate="2013-05-29T04:27:55.507" Score="1" Body="&lt;p&gt;Water is the catalyst in the process of rusting, therefore you cannot get rust without water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;a href=&quot;http://www.chemicalformula.org/rust&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalformula.org/rust&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The overall chemical equation for the formation of rust is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Iron + water + oxygen --&gt; rust&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{4 Fe(s) + 6 H2O(l) + 3 O2(g) -&amp;gt; 4 Fe(OH)3(s)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Iron(III) hydroxide, $\ce{Fe(OH)3}$ then dehydrates to produce $\ce{Fe2O3.nH2O(s)}$&#xA;  or rust&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1713" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:18.917" LastActivityDate="2015-03-06T23:22:18.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5137" PostTypeId="2" ParentId="5134" CreationDate="2013-05-29T07:55:39.500" Score="23" Body="&lt;p&gt;We could explain it from an electrolytic point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron has a certain tendency to dissolve in water according to the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe (s) -&amp;gt; Fe^2+ (aq) + 2e^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The release of electrons causes a small current to flow in the metal (Remember that metals conduct electricity). This turns the point where iron dissolves into an anode, and the region around this area becomes rich in electrons, a cathode. Now, we have to remember that water dissociates to small extents according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O (l) &amp;lt;=&amp;gt; H^+ (aq) + OH^- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was an acid, the $\ce{H^+}$ would have quickly taken all the electrons liberated from the dissolution of iron earlier, but the concentration of hydrogen ions is not large enough in water, so that we get another reaction taking place at the cathode [The hydrogen ions actually 'gather up' near to the surface but cannot do anything, creating a thin protective layer of hydrogen ions around the cathode, but not strong enough to prevent further reaction]. This reaction involves oxygen and other water molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2O (l) + O2 (g) + 4e^- -&amp;gt; 4OH^- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is the $\ce{OH^-}$ that reacts with the earlier liberated iron (or ferrous) ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe^2+ (aq) + 2OH^- (aq)-&amp;gt;Fe(OH)2 (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now more oxygen will react to oxidise the iron (II) hydroxide...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe(OH)2 (aq) \xrightarrow[O] Fe(OH)3 (aq) \xrightarrow[O] Fe2O3 (s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably can recognise the last product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the $\ce{Fe^2+}$ are consumed, more iron will get dissolved (Le Chatelier's Principle) and keep the whole process going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a little picture to go along with this :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0r8Ka.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2013-05-29T08:14:56.753" LastActivityDate="2013-05-29T08:14:56.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5138" PostTypeId="1" CreationDate="2013-05-29T08:58:36.480" Score="1" ViewCount="737" Body="&lt;p&gt;Could someone explain to me where the additional energy goes when applied to an object in the state of melting (such as water) so that the temperature of the object does not increase, and could someone please explain to me why Helium has a specific latent heat.&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2013-12-25T15:34:22.070" Title="Specific latent heat of fusion of a noble gas" Tags="&lt;heat&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5139" PostTypeId="2" ParentId="5130" CreationDate="2013-05-29T11:26:23.030" Score="3" Body="&lt;p&gt;From the definitions in your textbook, it seems that a mixture is a combination of two or more substances that--though combined--maintain their chemical identifiability. Examples of mixtures might be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water/Alcohol Mixtures&lt;/li&gt;&#xA;&lt;li&gt;Salt/Pepper Mixture&lt;/li&gt;&#xA;&lt;li&gt;In pharmacy, pills, pastes/creams, and suppositories are mixtures&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(there are many example of mixtures)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consituent compounds might be separated with somewhat mild techniques, including dissolving in certain solvents, distillation, and crystallization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composed Substances (according to your textbook) are mixtures of two or more species that under go some chemical reaction. The species involved could lose most of their identifiability, becoming completely different compound(s).  Loosely speaking, chemical reactions can not be reversed by simple separation techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course, there is some arbitrariness here: particularly in what comprises &quot;most&quot; of their identifiability and what defines &quot;simple&quot; separation techniques.  Also, mixtures are actually held together with bonds, just like in composed substances.  They may be weaker, but even &quot;weakness&quot; of the interaction is a subjective concept.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-29T11:42:55.223" LastActivityDate="2013-05-29T11:42:55.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5140" PostTypeId="2" ParentId="5138" CreationDate="2013-05-29T12:11:16.783" Score="1" Body="&lt;p&gt;When you heat a substance (not at its melting point or boiling point), the energy supplied goes to increase the kinetic energy of the atoms/molecules in the substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the same heat is applied to the substance at its melting point (or boiling point), the heat goes to increase the potential energy of the atoms/molecules in the substance. You could refer to the potential energy as the energy used to break the inter-particle bond. In ice, there are relatively strong forces of intermolecular attraction, which makes the ice hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, temperature is directly related to the kinetic energy of the atoms/molecules of a substance, and since the kinetic energy of the atoms/molecules does not increase during a change of state, the temperature remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least, this is what happens at a macroscopic level and in the long term. But if you heat ice at one particular point (at particle level and for a short period of time), that point will melt faster and the temperature will tend to rise there, but this heat energy will then be distributed to the other parts of the ice still cold for melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only when all the ice has molten than the heat supplied will contribute once again to the increase in kinetic energy, that is, temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the last part, every substance that undergoes a change of state has latent heat.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-05-29T12:11:16.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5141" PostTypeId="1" AcceptedAnswerId="5144" CreationDate="2013-05-29T13:27:11.953" Score="7" ViewCount="34077" Body="&lt;p&gt;What do you mean by a pseudohalogen/pseudohalide and how do you tell whether a molecule/ion is a pseudohalogen/pseudohalide ?&lt;/p&gt;&#xA;" OwnerUserId="1715" LastEditorUserId="84" LastEditDate="2013-08-29T02:46:06.020" LastActivityDate="2019-03-24T04:26:02.447" Title="what is a pseudohalogen/pseudohalide?" Tags="&lt;inorganic-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5142" PostTypeId="1" CreationDate="2013-05-29T08:13:53.883" Score="3" ViewCount="71" Body="&lt;p&gt;To fill with drinkable water, I have lots of empty 19lt carboy bottles. Since they came from different people and companies, I have to check their healthiness to decide wheter they can be sterilise for reuse or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to detect if a 19lt carboy bottle was used previously for gas, lpg, diesel etc. or is there any other harmful chemical (such as urine) in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be the cheapest and trustable solution for it? Could I use gas sensors or infrared imaging or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="1786" OwnerDisplayName="Pertinaks" LastActivityDate="2013-06-10T09:14:11.717" Title="How to detect harmful gas occurence in carboy bottle" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5143" PostTypeId="1" AcceptedAnswerId="5148" CreationDate="2013-05-29T15:07:03.097" Score="6" ViewCount="326" Body="&lt;p&gt;Could you suggest, please, a manual/textbook/tutorial for implementing FCI (full configuration interaction) method on computers. I am interesting in algorithms mainly. I would like to write my own code, so I want to study how one can provide convenient indexing for configurations and what tricks are usually used to decrease computation time.  &lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2013-10-20T21:27:38.800" Title="FCI implementation" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5144" PostTypeId="2" ParentId="5141" CreationDate="2013-05-29T15:21:26.047" Score="9" Body="&lt;p&gt;Pseudohalides are &quot;fake halides.&quot;   Their chemistry bears some resemblance to true halides ($\ce{F-}$, $\ce{Cl-}$, $\ce{Br-}$, $\ce{I-}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of pseudohalides include $\ce{CN-}$ and $\ce{N3-}$. These are generally weak Lewis bases that bear a formal negative 1 charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, it is rather subjective as to what can be labeled as a pseudohalide. In general, it is in the eye of the beholder, and for a certain set of chemical circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, there is no &quot;dividing line&quot; for being a pseudohalide. If we assume that halides undergo certain (characteristic) types of chemistry, and the species X in question also generally does those same types of chemistry with a similar mechanism, then you may be able to label X a &quot;pseudohalide.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that within the true halides themselves, they can have very different characteristics, and so there is a bit of a moving target for what constitutes a pseudohalide.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-29T15:21:26.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5145" PostTypeId="2" ParentId="5141" CreationDate="2013-05-29T15:22:49.027" Score="3" Body="&lt;p&gt;A pseudohalogen is an anion with similar properties to a halogen, but that isn't a halogen. The true halogens are of course the group one column to the left of the noble gases; fluorine, chlorine, bromine, iodine, and astatine, as well as the theoretical ununseptium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pseudohalogens are generally -1 compound anions with stable structure (typically featuring multiple bonds) that form compounds with similar reaction behaviors and properties as halides, and can be used as general substitutes in reactions with similar mechanisms. These properties include being weak bases by themselves that form strong acids when protonated, forming neutral salts in combination with metals, and oxidizing electropositive elements. Common pseudohalogens include cyanide/isocyanide, cyanate/isocyanate, azide, their sulfur analogs such as thiocyanate, and complexes such as tetracarbonyls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, pseudohalogens are not always identical. Take for instance sodium chloride, and the pseudohalogen analog sodium azide. While the former is relatively harmless table salt, consumed in tens of grams every day, the latter is so generally toxic to life that biochem labs use it as a reference poison (and to make sure whatever nasties they've been growing &lt;em&gt;in vitro&lt;/em&gt; for an experiment are good and dead before disposing of the medium). Similarly, while the human body needs a decent dose of chlorine anion as an electrolyte, and smaller amounts of iodine and even bromine are essential trace elements, most of the pseudohalogens are quite toxic, including cyanide, azide and carbonyl complexes. Thiocyanates and isothiocyanates are essential functional groups and anions in human biochemistry, so not overtly toxic like azide, but - First Lesson in Toxicology - too much of anything is bad.&lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2013-05-29T15:22:49.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5146" PostTypeId="2" ParentId="5135" CreationDate="2013-05-29T15:48:17.710" Score="10" Body="&lt;p&gt;The primary dangers from plastics themselves come when the monomer, and any compound put in to make the polymer matrix flexible, separate from the polymer and are released (leached) into whatever the plastic is containing. There are several such compounds which are known to be problematic to human biochemistry:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Phthalates&lt;/strong&gt; - These are aromatic diesters, with a structure incorporating a phenyl ring with two ester linkages, typically to alkanes like ethyl or butyl groups. Phthalates are used in certain plastics to link the monomers, such as in poly-ethyl terephthalate (PET aka polyester), and in others as a plasticizer to make the normally rigid plastic flexible (PVC). Phthalates mimic certain hormones in the human body and so can cause issues with growth and maturation, as well as feminizing characteristics in males (reduced body hair, weight gain, moobs). PET is OK for packaging and storage of cold foods (it is, for instance, the plastic used for soda bottles), but you don't want to heat it because that will degrade the plastic to release the phthalate.&lt;/p&gt;&#xA;&lt;p&gt;The ones used in PVC are especially problematic because they're not well-bonded; instead of being an integral piece of the polymer chain, they form weak hydrogen bonds between the polymer chains (thus the flexibility, even in relatively thick layers of the plastic), and so these phthalates easily leach out into water or &amp;quot;off-gas&amp;quot; into the air, even without heat. These phthalates are a big part of the &amp;quot;plastic smell&amp;quot; of new plastic products, and of water stored in them; the more flexible, and the longer the plastic has been kept enclosed, the more intense the smell. There was a big stink (NPI) in the early 90s when PVC was being used for water supply lines in home construction, and then they found all these problems with the stuff and a lot of homebuilders and chemical companies went bust paying for all this piping to be replaced with copper.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bisphenols&lt;/strong&gt; - These are variations of 2,2-bis-(dihydroxyphenyl)propane (which is the infamous Bisphenol A or BPA). Bisphenol A is the monomer for most of the original formulations of high-impact polycarbonate, such as Lexan. Again, when heated, the polymer breaks down and leaches the water-soluble monomer, which is structurally similar in many ways to phthalate and again has hormone-mimicking, endocrine-disrupting properties, which is why newer polycarbonates intended for food storage are made with other bisphenol monomers (most of which have similar endocrine properties, but because they aren't Bisphenol A they're still legal for now and allow the manufacturer to label the product &amp;quot;BPA-free&amp;quot;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Vinyl Chloride&lt;/strong&gt; - The monomer of poly-vinyl chloride or PVC, this is typically a hazard in production facilities for PVC and when old, aged PVC begins to break down. The vinyl chloride monomer is a known carcinogen, which is the other reason PVC is no longer used for food packaging or water supply, and is being phased out from wastewater lines in favor of nylon.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;As far as answering the question &amp;quot;shouldn't we just be able to pass it out of our system&amp;quot;, the same could naively be asked of a whole host of chemicals that are bad for us. The biggest problem with that statement is the First Law of Toxicology; &lt;em&gt;anything&lt;/em&gt;, in sufficient amounts, is harmful, even if it's absolutely essential for life in the &lt;em&gt;right&lt;/em&gt; amounts. Selenium is an essential trace mineral in human biochemistry, required in certain protein synthesis mechanisms, but the LD50 blood concentration is about 3 mg/kg, and harmful effects such as stunted growth begin to be observed at concentrations within foodstuffs of as low as 4 to 5 ppm. Heck, you can die from drinking too much water, faster than you could from lack of it (it causes a condition called hyponatremia where the added water dilutes electrolytes to the point the body can no longer function).&lt;/p&gt;&#xA;&lt;p&gt;What causes many chemicals to be harmful is in fact our body's own affinity for them as a beneficial chemical. Either the chemical &lt;em&gt;is&lt;/em&gt; beneficial to us, or it &lt;em&gt;looks&lt;/em&gt; like one that is, and the chemical is normally so rare in our diet and environment that our bodies grab all of it that they find, without an &amp;quot;off switch&amp;quot; to regulate intake.&lt;/p&gt;&#xA;&lt;p&gt;Which leads to the second problem with that statement; sometimes our bodies &lt;em&gt;don't&lt;/em&gt; efficiently metabolize and dispose of chemicals. Vitamin A, for instance, is an essential nutrient, and our body absorbs all it can, and gets rid of it only very slowly. As a result, it &amp;quot;bioaccumulates&amp;quot;, and eventually becomes toxic. That's why beta carotene is the preferred &amp;quot;form&amp;quot; of Vitamin A in supplements; it's in fact an intermediate that our bodies can use to synthesize the vitamin as needed, but being water-soluble unlike actual Vitamin A, any unused excess of beta carotene is simply flushed out via the kidneys.&lt;/p&gt;&#xA;&lt;p&gt;That process in itself can cause problems; the latest round of drug recalls the lawyers are running commercials on are for drugs that did exactly what they were supposed to, but were then metabolized into carcinogenic compounds that concentrated in the urine and caused various cancers of the excretory system.&lt;/p&gt;&#xA;&lt;p&gt;And the last problem, tying back into the case in point, is that even if the body selectively absorbs it, efficiently processes it, and it isn't &lt;em&gt;acutely&lt;/em&gt; toxic in any form or intermediate along the way, as long as there's an input of the chemical into your system, you'll always have some of it. Plastics are ubiquitous in consumer products, and especially food prep and packaging. As a result, virtually every resident of the industrialized world has a detectable amount of artificial phthalates in their blood, because it is simply impossible &lt;em&gt;not&lt;/em&gt; to come into contact with them. Even USDA-certified organic foods have phthalates in them, because the cows eat food that's been stored or served in plastic bins and it shows up in their milk and meat, or the vegetable was fertilized with compost stored in a plastic bin or watered from a PVC irrigation system (even the farmer's market items from someone's backyard were likely grown using water collected in a plastic rain catchment and/or dispensed from a hose or a drip irrigator system that leaches plasticizer). What you store it in or prepare it with once it gets to you is immaterial; the damage is done.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-26T19:04:55.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5147" PostTypeId="2" ParentId="5134" CreationDate="2013-05-29T18:46:20.220" Score="8" Body="&lt;p&gt;The dot means that the compound is a &lt;em&gt;hydrate&lt;/em&gt; of the parent chemical; the parent compound forms a weakly-bonded structure with one or more water molecules per parent molecule, because either the hydrogen atoms of the water, or the partial positively-charged region of the water molecule caused by its &quot;bent&quot; molecular shape, is attracted to partially negative regions of the parent molecule. It's not a &quot;full&quot; traditional ionic bond but it works in somewhat of a similar way. In solid form, this mixture of the parent compound and water molecules forms a crystal structure like ice (it doesn't have to be a well-ordered crystal like you typically get with salts), with properties that differ from a form of the substance that has more, less or no water in its structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An everyday example of the difference between a dehydrated and hydrated compound is Portland cement, used in concrete. In its raw form, it's &quot;dehydrated&quot; calcium carbonate along with a few other components, made by heating limestone until the water is released from the rock's structure. The remaining solid forms a fine powder that can't hold much of a shape. Add the water back in while mixing concrete, and the water molecules are re-incorporated into a solid structure with the calcium carbonate to essentially reform a rock in the desired shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as why rust only forms in the presence of water, water's &quot;polar&quot; nature is at work here as well. Being a polar solvent, it has some affinity for electrons, which are easily given up by the iron as a transition metal, and are attracted to the hydrogen atoms in the water. If a hydrogen atom successfully &quot;captures&quot; an electron, it balances its own electric charge, and is &quot;liberated&quot; from the bond it has with the oxygen in the water molecule, instead pairing with another liberated hydrogen to form a diatomic gas. Now, this hydrogen gas molecule won't get far, because water usually has some &quot;dissolved&quot; oxygen gas in it, and there's more oxygen just beyond the water, waiting to re-oxidize this hydrogen gas back into water. The net result is a positively-charged iron atom (typically in the +2 or +3 oxidation states, having given up 2 or 3 electrons to the water, respectively), negatively-charged hydroxide ions ($\ce{OH-}$), and water. These form a mixture of iron hydroxides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2(Fe - 2e^{-}) + 4(H_2O + e^+) + O_2 \\ \to 2Fe^{2+} + 4OH^- + 2H_2 + O_2 \\ \to 2Fe(OH)_2 + 2H_2O}\\$$&#xA;$$\ce{4(Fe - 3e^{-}) + 12(H_2O + e^+) + 3O_2 \\ \to 4Fe^{3+} + 12OH^- + 6H_2 + 3O_2 \\ \to 4Fe(OH)_3 + 6H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That second equation typically occurs &quot;stepwise&quot;, as you'll notice that's quite a lot of electrons and ions floating around at once. Typically, iron(II) hydroxide is formed first, and the iron will then readily oxidize further to its +3 state, either taking on an additional hydroxide to form iron(III) hydroxide, or liberating a hydrogen that will oxidize to form water, resulting in the monohydrate of iron(III) oxide-hydroxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4(Fe(OH)2 - e^{-}) + 2H_2O + O_2 \to 4FeO(OH)*H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron(III) hydroxide will readily rearrange itself into much the same state (it's not so much decomposition, as the water stays weakly bonded to the oxide-hydroxide to form the monohydrate):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(OH)3 \to FeO(OH)*H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, two of these iron oxide-hydroxide molecules are generally considered to &quot;share&quot; an oxygen from one of the hydroxide ions between them, with the hydrogen instead bonding to the oxygen of the other hydroxide, reforming water as a hydrate (possibly with some of it being released):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2FeO(OH)*H_2O \to Fe_2O_3*xH_2O + yH_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various hydrate structures of this iron oxide exist, which result in different colors of the compound from dark brown to deep red to red-orange. Other oxides are also possible by more direct action of the oxygen on the iron itself, such as iron(II) oxide ($\ce{FeO}$). Heating rust releases the water and forces the iron oxides to rearrange, which can also produce the third oxide of iron, iron (II,III) oxide (known in mineral form as magnetite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that the water participates in the overall reaction without really being consumed by it to form the product; it's destroyed and then re-formed in equal amounts, creating intermediates in the process. As such, water is a catalyst for these reactions. You'll also hopefully notice that this &quot;ideal&quot; reaction chain is just one of the possibilities; &quot;rust&quot; is an ill-defined term in chemistry, with the everyday compound consisting of a mixture of various hydrates of the three iron oxides, all of which are produced depending on subtle localized differences in the temperature and ratios of water, oxygen and iron.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:13.860" LastActivityDate="2015-03-06T23:22:13.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5148" PostTypeId="2" ParentId="5143" CreationDate="2013-05-29T18:55:21.320" Score="5" Body="&lt;p&gt;MOLPRO has a full CI module (by Peter Knowles), but it also costs quite a bit of money.  You can also do full CI in GAMESS (I think Ruedenberg's implementation), and perhaps any MCSCF/CASSCF program that is efficient enough to include all orbitals in the active space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I regret to inform you: you may be in for a lot of work if try to write your own code.  :-(  This is an area of very deep, highly mathematical and numerical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as far as a manual/textbook:  You may want to try Molecular-Electronic Structure Theory, especially Chapter 11 and 12.  There is just a crazy amount of work that you will have to do prior to CI, including the coding of the various four-center integrals for electron repulsion, kinetic energy, and overlap (earlier chapters).&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-05-30T00:52:10.520" LastActivityDate="2013-05-30T00:52:10.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5149" PostTypeId="2" ParentId="5131" CreationDate="2013-05-29T19:39:49.580" Score="3" Body="&lt;p&gt;At the chemical level, an &quot;element&quot; is &quot;a type of atom&quot;; the later is literally the simple definition of the former (with &quot;type&quot; being further clarified as a specific number of protons found in the nuclei of all atoms of that element). So for most purposes, the two are one and the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is sometimes a distinction made in chemistry literature between a substance in its &quot;molecular&quot; form as opposed to its &quot;elemental&quot; or &quot;atomic&quot; form. &quot;Molecular nitrogen&quot; refers to the pure form of natural gaseous nitrogen, N&lt;sub&gt;2&lt;/sub&gt;. &quot;Elemental nitrogen&quot; or &quot;Atomic nitrogen&quot; refers not only to the pure substance, but in its unbonded, monoatomic form. Most non-noble gases don't naturally exist this way, and in fact a few, like nitrogen, are far more stable in diatomic gas form than any other. However, when making stoichiometric calculations, it's an important distinction; a mole of &quot;molecular&quot; nitrogen is double the mass of a mole of &quot;atomic&quot; nitrogen, because each nitrogen molecule is two nitrogen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further confusing things is that depending on who you talk to, the &quot;elemental&quot; form of a substance could be its &quot;natural pure&quot; form, the form the pure substance exists in at room temperature and pressure, or its &quot;atomic&quot; form as single unbonded atoms. If there is a difference, as there is with gases, then the term &quot;elemental&quot; can be very confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And perhaps most confusing of all is that the definition of &quot;pure&quot; a lab chemist might use is different from the definition that a nuclear physicist would use, because of the existence of isotopes. While atoms of a specific element must have a constant number of protons in order to be that element, many elements can exist for a measurable length of time with varying numbers of neutrons. Some have multiple &quot;stable&quot; isotopes (which will exist indefinitely in their current state without decay); chlorine, for example, is stable with either 35 or 37 total nucleons. Usually (but not always), the chemistry of various isotopes is unaffected; where it matters, in the molar mass of reagents, an average of the &quot;naturally-occurring stable isotopes&quot;, weighted by percentage in naturally occurring form, is used and is close enough for chemists' purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what a chemist would call &quot;pure chlorine&quot; would be unusable for a nuclear physicist researching molten salt reactors, because of the differing behaviors of Chlorine-35 and Chlorine-37 when bombarded with neutrons. Chlorine-35 can capture a neutron to become Chlorine-36, which then degrades (with a rather long half-life of 300,000 years give or take a few thousand) into Argon-36 and Sulfur-36, neither of which are chlorine and therefore don't produce the halide salt that is necessary for the chemistry of the reactor design, and sulfur being highly corrosive to the alloys used for the reactor vessel and coolant loops.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-05-29T22:36:28.447" LastActivityDate="2013-05-29T22:36:28.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5150" PostTypeId="1" AcceptedAnswerId="5154" CreationDate="2013-05-29T21:49:33.907" Score="1" ViewCount="129" Body="&lt;p&gt;I understand(I think) the mass and density aspect, i.e. the more protons you have, the more the element weighs, also the denser the atom is. What about everything else(color, for example)? Elements are not just masses and densities.&lt;/p&gt;&#xA;" OwnerUserId="1720" LastActivityDate="2013-05-30T00:45:53.077" Title="How do the differences in the number of protons result in such great differences in elemental properties?" Tags="&lt;atoms&gt;&lt;periodic-table&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5151" PostTypeId="1" AcceptedAnswerId="5153" CreationDate="2013-05-29T23:15:59.300" Score="5" ViewCount="2979" Body="&lt;p&gt;I'm attempting to write a small program that will calculate the values for the Ideal Gas Law, specifically for molecular Hydrogen (H2) in space.  Eventually, this will grow to be a sort of &quot;simulator&quot;, but right now I seem to be having problems with remembering my college chemistry and physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of the form that I created:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nOlAq.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my program (written in C#), I have the following as initial values for those variables:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        P = ?&#xA;        v = ?&#xA;        n = 2.0158 (weight in grams of H2)&#xA;        T = 2.725 (average temp in space)&#xA;        R = 1.67E-27 (1.67 * 10^-27)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My problem is that, how does one go about trying to solve for P or v?  Do you assign an arbitrary number for what you suppose the volume of the gas would be in space, and then calculate P based on that?  Or is it the other way around?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra, yet entirely unneeded info:&lt;/strong&gt;&#xA;This was brought about because my 7 year old daughter asked me how stars were made.  I gave her the explanation that anyone who's got even a remote interest in it gives to their kids, but thought it would be pretty cool to make a little program that would give her a visual.  Here was my explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;we think that Hydrogen stuck together to make even larger clouds of&#xA;  Hydrogen, which increased the total mass of the gas cloud, increasing&#xA;  the gravitational pull on other atoms/molecules of nearby gas, etc. &#xA;  This eventually got to the point that the gases started to heat up as&#xA;  they were squished together by their own mass, but that increase in&#xA;  temperature created an outward push (pressure), which got into a&#xA;  tug-of-war with gravity acting on the mass of the gas itself.  And&#xA;  later, we got a star.  That's how most of us non-scientists think it&#xA;  happens, anyway!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1721" LastEditorUserId="7448" LastEditDate="2015-11-14T12:58:26.780" LastActivityDate="2015-11-14T12:58:26.780" Title="Finding Pv in Pv=nRT for molecular hydrogen" Tags="&lt;physical-chemistry&gt;&lt;hydrogen&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5152" PostTypeId="2" ParentId="5150" CreationDate="2013-05-29T23:33:55.027" Score="0" Body="&lt;p&gt;It's not the protons, it's the &lt;em&gt;electrons&lt;/em&gt;.  Electrons are the main determinant of chemical bonding, and it is the bonding of differing atoms that creates the huge variety of chemical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the color of molecules often is driven by the &lt;em&gt;transition elements&lt;/em&gt;, starting at atomic number 21.  These metals have the ability to obtain many different &lt;em&gt;oxidation states&lt;/em&gt;; i.e. ways to share elections with other atoms that create a certain implied positive charge, and those differing positive charges produce different colors because of the different wavelengths of light that the molecule can absorb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really nice video on this comes from the old Chem Study series.  The specific topic is the element vanadium, but they also discuss transition elements generally, complete with a big period table of actual substances.  Here is a link: &lt;a href=&quot;http://www.youtube.com/watch?v=s7w2egRycqs&quot; rel=&quot;nofollow&quot;&gt;Chem Study Vanadium&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-05-29T23:33:55.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5153" PostTypeId="2" ParentId="5151" CreationDate="2013-05-29T23:56:46.317" Score="3" Body="&lt;p&gt;You can solve for either $P$ or $V$.  Choose one, say $P$, and then divide that into the right hand side $nRT$ to get $V$; i.e. $V=\frac{nRT}{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More seriously, you need to check your units!  You seem to be using Kelvins for $T$, which is right, and $n$ is just a count in moles.  But I do not recognize your value of $R$.  The SI value for $R$ is $8.3144621(75)$, in units of joules per Kelvin-mole.  Joules, in turn, are Pascal-meters$^3$. Personally, I think most people find Pascals too small a unit for pressure.  Far more practical would be Torr-liters, and then $R=62.36367(11)$.  You can examine lots of different unit combinations for $R$ at the ideal gas constant page on Wiki: &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_constant&quot; rel=&quot;nofollow&quot;&gt;Ideal Gas Constant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-05-29T23:56:46.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5154" PostTypeId="2" ParentId="5150" CreationDate="2013-05-30T00:45:53.077" Score="4" Body="&lt;p&gt;Unfortunately, I have to disagree with trb456's answer.  It is both the protons and the electrons that make up chemistry. (Neutrons are considered massive &quot;spectators&quot; though they are what keep protons (all + charged) together)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonds are formed due to the attraction of electrons and protons. (cf. Slater, Bader, et al.) (The precise shape and nature of the electron density is additionally determined by constraints such as Pauli Exclusion and electron-electron repulsion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you think about it, there is a very complex interplay upon mixing electrons up with protons. More positive charge in the nucleus tends to keep electrons drawn tight, but in some ways that is counter-balanced by electron-electron repulsion--and the fact that electrons are attacted to other nuclei as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although we ascribe color (&quot;electronic transitions&quot;) and reactivity to electrons, this is only part of the story. The protons are also there, and they provide some governance (via Coulombic interaction) over what electrons are permitted to do.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-30T00:45:53.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5155" PostTypeId="1" AcceptedAnswerId="5157" CreationDate="2013-05-30T01:10:19.147" Score="2" ViewCount="2089" Body="&lt;p&gt;If a reaction requires a &quot;strong&quot; acid - what does this mean? Does it mean any acid of a specific concentration or pH? Are only certain acids classed as &quot;strong&quot;? I have read on Wikipedia that it needs to ionise completely in an aqueous solution - in this case, is a strong acid (for example, HCl) still classed as a strong acid if it is in a 0.00001M solution? &lt;/p&gt;&#xA;" OwnerUserId="842" LastEditorUserId="9961" LastEditDate="2016-06-07T15:58:26.697" LastActivityDate="2016-06-07T15:58:26.697" Title="What is a &quot;Strong&quot; Acid?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5156" PostTypeId="2" ParentId="5151" CreationDate="2013-05-30T01:15:27.577" Score="2" Body="&lt;p&gt;You need one of p or V to solve for the other. If you have a spherical body, then $V=\frac{4}{3}\pi r^3$ and plug in what you want the radius of your star to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related note, the ideal gas law can't be applied to stars because IGL assumes the particles are infinitely spread such that gravity between particles is negligible - which is obviously not true for a star!&lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-05-30T01:15:27.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5157" PostTypeId="2" ParentId="5155" CreationDate="2013-05-30T01:22:42.517" Score="6" Body="&lt;p&gt;The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$.&#xA;$$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$&#xA;This value will either be very small (&amp;lt;&amp;lt;1) or very large (&gt;&gt;1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into  $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.&lt;/p&gt;&#xA;" OwnerUserId="1719" LastEditorUserId="-1" LastEditDate="2015-09-16T08:40:34.837" LastActivityDate="2015-09-16T08:40:34.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5158" PostTypeId="1" CreationDate="2013-05-30T06:14:55.287" Score="7" ViewCount="594" Body="&lt;p&gt;How would I grow my own and build functional solar cells? I've read and researched that they're mostly a crystal structure of something like silicon. I would assume I'd be able to grow my own crystals and sandwich this in a process of some sort with a &lt;code&gt;p-type&lt;/code&gt; and &lt;code&gt;n-type&lt;/code&gt;, etc.&lt;/p&gt;&#xA;" OwnerUserId="1722" LastEditorUserId="22" LastEditDate="2013-06-07T03:32:54.447" LastActivityDate="2016-09-27T05:21:22.617" Title="Create my own solar cells?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;energy&gt;&lt;crystal-structure&gt;&lt;green-chemistry&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5159" PostTypeId="1" AcceptedAnswerId="5164" CreationDate="2013-05-30T08:40:37.360" Score="0" ViewCount="1019" Body="&lt;p&gt;Right, how do you systematically name them? Also, do you number them from the double bond end or the Carboxyl end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ie would it be something like.  2-methyl Hept-3-enoic acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scuse the bad question and I made up the example, sorry! I'm really confused!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="1723" LastEditorUserId="275" LastEditDate="2013-05-30T15:22:59.247" LastActivityDate="2013-05-30T15:22:59.247" Title="Nomenclature of alkenoic acids - which group has priority: the alkene or the acid?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5160" PostTypeId="2" ParentId="5123" CreationDate="2013-05-30T09:06:07.420" Score="1" Body="&lt;p&gt;Egg white consists of mostly Ovalbumin. It is a protein, and so will be hard to distinguish from other proteins in a baked product. Egg white changes structure to a solid polymer on cooking. You could dissolve it in a mixture of detergent and hair perm, returning the Ovalbumin to its uncooked state, and then run it through a column to identify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might also be biological assays you could run using enzymes/antibodies which would only interact with Ovalbumin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there are any chemical means to identify a particular cooked protein easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask on Biology exchange?&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2013-05-30T09:06:07.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5161" PostTypeId="2" ParentId="5158" CreationDate="2013-05-30T09:47:26.750" Score="5" Body="&lt;p&gt;You have to study existed architectures. First, it is necessary to distinguish between photo-conductive and photovoltaic elements. In photovoltaic elements one needs a mechanism of separation of positive and negative charges formed under an irradiation of the semiconductor. Even a simple p-n junction can do this. But, this is all about efficiency. The solar cell has to absorb as much light as possible without reflections, and, at the same time, the charge transport inside should be without losses that is impossible in thick structures. Also we have to care about the absorption spectrum. To absorb in all spectrum of the solar radiation, people make cascade structures with different materials absorbing in different ranges of spectrum.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2013-05-30T11:12:44.020" LastActivityDate="2013-05-30T11:12:44.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5162" PostTypeId="2" ParentId="4517" CreationDate="2013-05-30T12:31:23.250" Score="4" Body="&lt;p&gt;Usually, when atoms form bounds they release energy (heating, radiation etc). So, the energy conservation law holds, but this kind of sum rule does not work. Vice versa, one has to spend energy to get dissociation of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2018-12-08T12:18:47.393" LastActivityDate="2018-12-08T12:18:47.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5163" PostTypeId="1" CreationDate="2013-05-30T12:56:05.977" Score="5" ViewCount="2616" Body="&lt;p&gt;I have a sdf file for around 50 small-molecules for which I am doing structure-activity study. I would like to draw 2d graphs of all the molecules such that the image would fit into a single journal page. For example, in this &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/19757815&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt;, page number 3, they have listed 2d molecular structure. I have attached the image file from the same paper. I have software tools like MarvinSketch which give me 2d or 3d graph for a given sdf file. But I want the integrated image like shown below suitable for publication. &lt;img src=&quot;https://i.stack.imgur.com/TjDvW.png&quot; alt=&quot;List of 2d molecular graphs&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-12-12T10:25:07.413" Title="Open Source tools to draw a set of 2d molecular graphs from sdf file" Tags="&lt;software&gt;&lt;medicinal-chemistry&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5164" PostTypeId="2" ParentId="5159" CreationDate="2013-05-30T12:56:44.840" Score="3" Body="&lt;p&gt;The carboxylic acid moiety takes precedence over the alkene moiety, so the molecule has to be numbered starting from the carboxylic acid carbon. Hence, &lt;a href=&quot;http://en.wikipedia.org/wiki/Methacrylic_acid&quot; rel=&quot;nofollow&quot;&gt;methacrylic acid&lt;/a&gt;, for example, has the systematic IUPAC name of 2-methylpropenoic acid.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-05-30T12:56:44.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5165" PostTypeId="2" ParentId="5122" CreationDate="2013-05-30T14:04:51.243" Score="9" Body="&lt;p&gt;In short, the species you mention ($\ce{Sn^{5+}}$ and $\ce{SO_4^{5-}}$) do not exist in any chemical system at anywhere near ordinary conditions. The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions. Hence in your reaction, $\ce{Sn}$ atoms start with oxidation number +2 and end with 0 (therefore gaining two electrons) and $\ce{Fe}$ atoms start with oxidation number 0 and end with +2 (therefore losing two electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example of electron counting, here's the reaction between $\ce{Ba}$ and $\ce{Al^3+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: $\ce{Ba^0 -&amp;gt; Ba^{2+} + 2e^{-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: $\ce{Al^{3+} + 3e^{-} -&amp;gt; Al^0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To balance the equations, the oxidation half-reaction must be multiplied by 3 and the reduction half-reaction must be multiplied by two. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: $\ce{3Ba^0 -&amp;gt; 3Ba^{2+} + 6e^{-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: $\ce{2Al^{3+} + 6e^{-} -&amp;gt; 2Al^0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Global: $\ce{3Ba^0 + 2Al^{3+} -&amp;gt; 3Ba^{2+} + 2Al^0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the global reaction, &lt;em&gt;six&lt;/em&gt; electrons are involved.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-05-30T14:04:51.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5167" PostTypeId="2" ParentId="5163" CreationDate="2013-05-30T17:44:46.697" Score="3" Body="&lt;p&gt;Check out this &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecule_editor&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; which lists molecule editors, many of which are licensed under GNU GPL. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inexpensive (50-60 USD) program for drawing chemical structures is &lt;a href=&quot;http://www.chemdoodle.com/&quot; rel=&quot;nofollow&quot;&gt;Chemdoodle&lt;/a&gt;. Despite not being free, you get some user friendliness and features (like reaction building) for the cost.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-05-30T17:44:46.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5168" PostTypeId="2" ParentId="5163" CreationDate="2013-05-30T19:36:22.230" Score="7" Body="&lt;p&gt;This looks like a combination of individual SDF images and some sort of Table editor. A program that can take your PNGs rendered from SDF files and create this table layout is $\LaTeX$, or perhaps plain $\TeX$ or (my favorite) ConTeXt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a number of constraints:  a) a free program that b) makes intricate tables and c) handles SDF images. You probably want it to be easy to use, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any single open source program that will simultaneously satisfy these conditions. There probably isn't one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for sure that Mathematica can take care of b) and c).  You can get it for $100 if you are academic.  But then you have to learn it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to make this is probably to render all of your images as graphics, and then lay everything out in Microsoft Excel or Word Table Editor, changing the column widths and grid lines to suit you.  (But this is Microsoft Office, and is not open source, so again, b) and c) )&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-05-30T19:36:22.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5170" PostTypeId="1" CreationDate="2013-05-31T03:39:26.483" Score="3" ViewCount="134" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Using a state-of-the-art atomic force microscope, the scientists have taken the first atom-by-atom pictures, including images of the chemical bonds between atoms, clearly depicting how a molecule's structure changed during a reaction. Until now, scientists have only been able to infer this type of information from spectroscopic analysis.  &amp;mdash; &lt;a href=&quot;http://www.sciencedaily.com/releases/2013/05/130530142007.htm&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there any papers how this thing actually works?&lt;/p&gt;&#xA;" OwnerUserId="1727" LastEditorUserId="7448" LastEditDate="2015-11-14T12:51:45.747" LastActivityDate="2015-11-14T12:51:45.747" Title="How did Scientists Capture The First Images of Molecules Before and After Reaction?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5171" PostTypeId="1" AcceptedAnswerId="5199" CreationDate="2013-05-31T05:35:58.100" Score="0" ViewCount="151" Body="&lt;p&gt;I do not understand how equation(3) occurs.(taken from the book Internal Combustion Engines by John B.Heywood)&#xA;$$ $$&#xA;Consider a reactive mixture of ideal gases. The reactant species $M_a,M_b$ etc.and the product species $M_l,M_m$ etc. are related by the general reaction whose stoichiometry is given by:&#xA;$$\upsilon _aM_a+\upsilon _bM_b+...=\upsilon _lM_l+\upsilon _mM_m+...$$&#xA;$$ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be otherwise written as:&#xA;$$ \sum_{i}\upsilon _iM_i=0$$&#xA;where the $\upsilon _i$ are the stoichiometric coefficients and by convention are positive for the product species and negative for the reactant species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\delta n_a$ of $M_a$ react with $\delta n_b$ of $M_b$,etc. and produce $\delta n_l$ of $M_l$,$\delta n_m$ of $M_m$,etc.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;These amounts are in proportion,given by the equation (3):&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $$\delta n_i=\upsilon _i\delta n$$              &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;em&gt;Does $\delta n$ here signify 'extent of reaction'?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;em&gt;I would eventually like to use the number of moles of each species in expressing the chemical potential,with the gibbs free energy already known. But,the part I don't understant is equation 3. What does the author mean by proportional? Could you give a example?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs free energy is given as:&#xA;$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\delta n_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which by equation(3) can be re-written as(WHY??):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\nu _i\delta n$$&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="1728" LastEditDate="2013-05-31T05:50:54.013" LastActivityDate="2013-06-02T05:48:36.717" Title="chemical equilibrium reaction for burned gases in engine" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5172" PostTypeId="2" ParentId="5170" CreationDate="2013-05-31T06:12:29.157" Score="5" Body="&lt;p&gt;There are a ton of papers on how AFM works, but I'll have a go at explaining it anyhow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic Force Microscopy (AFM) is a venerable technology for imaging nanostructures down to individual molecules by physically contacting atoms on surfaces. The whole process is very similar to a finger reading Braille.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to physically contact an atom? It means that the AFM works by detecting some kind of repulsion between an atom and a very, very sharp probe due to their respective electron clouds repelling. The forces between surfaces on the nanoscale are a complicated mixture of attraction due to Casimir-Polder/Van der Waals interactions, dipole-dipole interactions, electrostatic repulsion, Pauli exclusion, &lt;em&gt;etc. etc.&lt;/em&gt;, but the key point is that if two objects get too close together they will experience a rapidly increasing repulsive interaction, which the AFM picks up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This repulsion is detected through the deflection of a cantilever (a thin, long, elastic beam of material). The probe is mounted on the end of this cantilever, and the cantilever bends when the probe is pushed up against a substrate. The deflection of this cantilever is measured by bouncing a laser beam off the end, the change in position of the reflected beam indicating the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AFM probe has an atomically sharp tip that can localise single atoms on a surface, and the tip is moved around using exquisitely precise actuators so that the tip can be scanned back and forth over a surface, picking up the bumps and valleys of the sample that indicate the positions of atoms. I've seen ads for AFM probes that actually have flared tips - these can obtain even more 3D information about nanostructures as they can hook into nanostructures from the side and measure the texture and relief parallel to the surface.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-05-31T06:12:29.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5173" PostTypeId="1" CreationDate="2013-05-31T13:07:04.360" Score="2" ViewCount="1558" Body="&lt;p&gt;I am using Avogadro for visualizing some cluster systems. Is there a way to hide the bonds of the system, ie. only showing atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1730" LastEditorUserId="275" LastEditDate="2013-05-31T18:36:17.363" LastActivityDate="2014-09-12T00:43:46.680" Title="Avogadro - How can bonds be hidden and only show atoms?" Tags="&lt;bond&gt;&lt;software&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5174" PostTypeId="1" CreationDate="2013-05-31T15:14:49.200" Score="9" ViewCount="6982" Body="&lt;p&gt;Let's take Hematoxylin, which is a pH indicator and a stain for histology, and therefore a chromophore, but does not fluoresce:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwAPP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take Eosin, which stains tissue pinkish, but is also fluorescent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQYo2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know which features of the molecules decide if it is a chromophore or a fluorophore. As far as I can see it (I'm a biologist with some training in organic chemistry), both have conjugated systems, although the system in Eosin spans a lot of more bonds. Is it really just the size of the conjugated system that decides this, e.g., if Hematoxylin had a larger conjugated system, would it also be fluorescent?&lt;/p&gt;&#xA;" OwnerUserId="1731" LastActivityDate="2013-11-13T02:16:41.027" Title="What's the physicochemical difference between a chromophore and a fluorophore?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5175" PostTypeId="1" CreationDate="2013-05-31T16:23:25.900" Score="3" ViewCount="3017" Body="&lt;p&gt;I am trying to run single point calculations with Gaussian 09 for a Mulliken population analysis on large graphene/graphite (940-1411 atoms) sheets/clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was having difficulty with memory issues (single CPU license) running &lt;code&gt;DFTBA&lt;/code&gt; for single point calculations. I have stepped my convergence (beginning with &lt;code&gt;conver=1&lt;/code&gt;, and increasing by reading in my previous checkpoint files).  I found my results oscillating and failing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to try &lt;code&gt;scf=(qc,conver=5,maxcycle=500)&lt;/code&gt;.  This worked, which concerned me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It not only worked, but it worked much faster than my failed &lt;code&gt;sp&lt;/code&gt; calculations.  I know Gaussian's old default convergence for single point calculations was N=4.  As my calculation was done in less than two hours compared to the days it was taking for &lt;code&gt;DIIS&lt;/code&gt;, I have become concerned at the accuracy of the QC using &lt;code&gt;conver=5&lt;/code&gt;.  I cannot find anything indicating my approach is bad, but also I cannot find anything syaing my approach should yield reasonable results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please tell me if this approach is reasonably accurate?&lt;/p&gt;&#xA;" OwnerUserId="1732" LastEditorUserId="4945" LastEditDate="2017-01-19T09:15:39.313" LastActivityDate="2017-01-19T09:15:39.313" Title="What is the accuracy of scf=qc when convergence criteria is lowered?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5176" PostTypeId="1" CreationDate="2013-05-31T20:16:27.450" Score="2" ViewCount="1047" Body="&lt;p&gt;I was wondering if there is a decent and non-complex way to predict the outcome of mixing two chemicals together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that molecular dynamics is used in computational chemistry to understand some of these types of problems, but in many cases I have noticed that they seem to be restricted to classical physics and therefore cannot accurately predict any real reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point in asking is such that I can have a broader understanding of this; However, at the moment, I am interested in the difference between this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://imgur.com/WkcrgxI.jpeg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the same reaction while replacing the triflic acid ($\ce{CF3SO3H}$) with chlorosulfonic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be any differences if these two superacids were switched?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper from which I obtained this reaction from is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Klumpp, Douglas A., Rendy Rendy, Yun Zhang, Alma Gomez, and Aaron McElrea. &quot;Dicationic Intermediates Involving Protonated Amides: Dual Modes of Reactivity Including the Acylation of Arenes.&quot; Organic Letters 6, no. 11 (2004): 1789-1792.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="7448" LastEditDate="2015-11-14T12:44:12.153" LastActivityDate="2016-06-14T20:20:55.890" Title="What tools are used to predict the products of a chemical reaction?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2016-06-15T05:12:32.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="5179" PostTypeId="2" ParentId="5175" CreationDate="2013-06-01T01:53:59.857" Score="5" Body="&lt;p&gt;Something that I have discovered is that when Gaussian (or other good ab initio software) fails to converge, there is something sketchy about what one is trying to calculate. Sometimes DFT calculations just aren't appropriate, especially in cases that require multiconfigurational SCF for even a zero-th order description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, given the sheet-like character, I suspect the problem is that the molecular orbitals are so close together, and so similar, that there are serious linear dependency problems that lead to all sorts of havoc in the linear algebra routines and convergence accelerators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are trying to simulate an infinite sheet, then I suggest that you try plane wave codes like those found in solid state chemistry (Gaussian's PBC code, NWChem, abinit, etc.)  These sorts of calculations have gotten easier in recent years, but sometimes they require serious horsepower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really does seem to be a problem with SCF convergence, and this could be because you are approaching some kind of band structure limit. I have always used QC as an absolute last resort when I &lt;em&gt;must&lt;/em&gt; do something wacky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One other thing, you have &quot;capped&quot; your peripheral benzenoid rings with hydrogens?)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-01T01:53:59.857" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5180" PostTypeId="2" ParentId="5176" CreationDate="2013-06-01T02:14:30.420" Score="5" Body="&lt;p&gt;I don't know for sure, but I suspect these two reagents would lead to the same product.  They are both superacids, and extremely harsh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever is a better Bronsted acid (proton donor) in benzene probably leads to a faster reaction, or better conversion.  But who knows?  Maybe the reactions are so acidic (two plus charges in a small organic molecule!) that they are just about the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some superacid expert lurking on chemistry.se who knows this answer, but I think the only way to know is to do the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to write the corresponding author of the paper. They may already know the answer. Or, they might be flattered and may find the question interesting enough to go to the lab and do the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't think computational chemistry will help here.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-01T02:14:30.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5181" PostTypeId="2" ParentId="5173" CreationDate="2013-06-01T09:05:10.847" Score="4" Body="&lt;p&gt;Not in the current version, as far as I know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need it, you could either make a &lt;a href=&quot;http://sourceforge.net/p/avogadro/feature-requests/&quot; rel=&quot;nofollow&quot;&gt;feature request on the Sourceforge page&lt;/a&gt;, or you could do something with extensions, such as making a new version of the ball and stick renderer removing the code that renders bonds, or making an extension that adds a button to delete all the bonds.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-06-01T09:05:10.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5182" PostTypeId="1" CreationDate="2013-06-01T09:17:38.957" Score="1" ViewCount="755" Body="&lt;p&gt;Does $\ce{Ni((en)2(Cl)2)}$ show cis-trans form ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if No &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not ?&lt;/p&gt;&#xA;" OwnerUserId="1735" LastActivityDate="2013-06-26T17:16:30.197" Title="Geometrical Isomerism of Metal Complex" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5183" PostTypeId="1" AcceptedAnswerId="5193" CreationDate="2013-06-01T09:22:16.300" Score="2" ViewCount="22265" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I add equal amounts of lemon juice and sugar into water, does the ph-value remain unchanged, as both cancel each other out?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question arose in my head as I learned that pouring milk into coffee increases the ph-level of the beverage making it taste less sour (You might want to check out the following video at 5:43: &lt;a href=&quot;https://www.youtube.com/watch?v=eTYsQUqodeE&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=eTYsQUqodeE&lt;/a&gt;; it's in German but you will get the overall point, I'm sure).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I add lemon into my tea, it tastes more acidic, so the ph-value would be more acidic right? If I add additionally some sugar, the tea tastes not sour anymore. So, I wonder if sugar is able to make a acidic drink more alkaline as it tastes less acidic...Or am I comparing apple with pears? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, if my question sounds a bit unscientific...)&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="1736" LastEditDate="2013-06-02T06:39:15.773" LastActivityDate="2015-11-12T14:08:12.477" Title="The influence of lemon juice &amp; sugar on the ph-value" Tags="&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5185" PostTypeId="2" ParentId="5174" CreationDate="2013-06-01T12:03:14.847" Score="1" Body="&lt;p&gt;The physical chemistry of fluorescence is a little bit complicated, and so you should consult a text book for &quot;the reason&quot; that things fluoresce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the things that I have noticed is that fluorescent compounds often have several &quot;heavy halogens&quot; Br and I attached to some poly-aromatic hydrocarbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: this is not to say that the halogens are the fluorescing portion of the molecule (it's probably almost the whole molecule, conjugation + halogen), and it does not mean that all compounds that contain Br and I will be fluorescent. It also does not mean that Br and I are required for fluorescence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that a lot of fluorescent molecules were discovered by chance, or even brute-force changing substituents until the molecule exhibits the desired fluorescence.  (It's really too hard to predict with qualitative theory, and electronic structure theory descriptions of fluorescence are also very difficult.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-01T12:33:53.043" LastActivityDate="2013-06-01T12:33:53.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5186" PostTypeId="1" CreationDate="2013-06-01T15:10:57.893" Score="8" ViewCount="251" Body="&lt;p&gt;I found betaine in my face wash. I think it is there because it has a hydrophobic (non-polar) end that can pick up grease, and a hydrophilic (polar) end that means it is attracted to polar solvents, such as water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why are betaines used over other detergents? What are the factors that contribute to the which detergents are used in face wash (or indeed shampoo/shower gels)?&lt;/p&gt;&#xA;" OwnerUserId="1738" LastEditorUserId="1738" LastEditDate="2015-10-31T08:22:55.167" LastActivityDate="2015-10-31T08:22:55.167" Title="How do chemists decide on which detergent to use?" Tags="&lt;everyday-chemistry&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5189" PostTypeId="1" AcceptedAnswerId="33551" CreationDate="2013-06-01T19:06:39.930" Score="17" ViewCount="19320" Body="&lt;p&gt;I'm currently in high school and I did two titrations. The first one is a titration of $10.0~\mathrm{mL}$ $0.1~\mathrm{M}$ $\ce{HCl}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSJPL.jpg&quot; alt=&quot;first titration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue curve is what I measured and the red curve is what the curve would look like in theory, with these data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second titration is a titration of $10.0~\mathrm{mL}$ $0.2~\mathrm{M}$ $\ce{CH3COOH}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VWO3Y.jpg&quot; alt=&quot;second titration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, my measurements are in blue and the red curve is the theoretical model. I noticed that (looking at the second curve in specific) the measured pH value is most of the time lower than in the theoretical model, and close to the equivalence point, the graph is not as steep. Why is this? Not a valid reason would be that the pH sensor does not measure the right value instantly, because near the equivalence point, only one drop of $\ce{NaOH}$ was added about every two seconds. Do you have a more reasonable explanation?&lt;/p&gt;&#xA;" OwnerUserId="1740" LastEditorUserId="4945" LastEditDate="2014-07-29T11:33:54.013" LastActivityDate="2015-07-02T15:30:09.013" Title="Why is a theoretical titration curve higher than a practical one?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5190" PostTypeId="1" AcceptedAnswerId="5191" CreationDate="2013-06-01T20:09:36.570" Score="0" ViewCount="146" Body="&lt;p&gt;I came across a compound $\ce{[FeO4]^{2-}}$ but $\ce{Fe}$ has +6 charge according to my calculations. How this is possible ? Or it is possible but my periodic table is kind of non-detailed one. (On the table I got $d$-metals possible charges).  &lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="73" LastEditDate="2013-06-02T04:36:37.893" LastActivityDate="2013-06-02T04:36:37.893" Title="Transition Metals' Charges" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5191" PostTypeId="2" ParentId="5190" CreationDate="2013-06-01T20:26:10.343" Score="3" Body="&lt;p&gt;If you write out the electron configuration of Fe, you will find it has $3\text{d}^6 4\text{s}^2$ in its valence level, meaning it can theoretically take on any oxidation state from +1 to +8 (and even some negative states, for that matter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice it is extremely uncommon to find any oxidation states other than +2 or +3 for iron - which explains why a simple periodic table containing only the most common naturally occurring states will omit them - but they are possible, just not very stable or long-lived. The polyatomic ion you have observed is called &lt;a href=&quot;http://enwp.org/Ferrate&quot; rel=&quot;nofollow&quot;&gt;ferrate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-06-01T20:26:10.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5192" PostTypeId="2" ParentId="5186" CreationDate="2013-06-01T20:39:57.943" Score="4" Body="&lt;p&gt;If it's polar, then it &lt;em&gt;won't&lt;/em&gt; pick up grease because lipids (fats and oils) are generally &lt;em&gt;non&lt;/em&gt;-polar. I don't know specifically how a detergent is chosen, but soaps are chemically the salts of a fatty acid. When choosing a detergent, you want a molecule with a polar head with a long, nonpolar tail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that water is polar and polar substances dissolve polar substances, and nonpolar dissolve nonpolar. The detergent in a sense forms a &quot;bridge&quot; between the water and the nonpolar substances (dirt, grease, oil, etc.) which would not dissolve in pure water. The detergent molecules form little &quot;cells&quot; (my choice of this word is not a coincidence - this is also how your cell membranes work) with the polar heads facing out and the nonpolar tails mingling with nonpolar dirt molecules inside. This allows the dirt be washed away by water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why &lt;em&gt;certain&lt;/em&gt; detergents are chosen, it is probably a balance struck between safety, effectiveness, cost to mass-produce, possibly environmental considerations associated with production, etc. &lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-06-01T20:39:57.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5193" PostTypeId="2" ParentId="5183" CreationDate="2013-06-01T21:39:39.080" Score="2" Body="&lt;p&gt;You're right about the lemon juice and the milk but wrong about the sugar. Lemon juice is weakly acidic (hence its sour taste) and milk is slightly basic (or alkaline, the terms are used interchangeably). A sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have background in chemistry, then let's not talk about pH - I thought pH was confusing when I learned it because &quot;lowering the pH&quot; makes a solution more acidic, and &quot;raising the pH&quot; makes it more basic! (Ironically, the pH scale was originally devised to make the concepts of acidity and alkalinity more accessible to the general public)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your observations about tea probably have nothing to do with acidity or alkalinity since the foods you've described are not significantly acidic or alkaline (except lemon juice). Tea is slightly bitter because of tannins in the leaves. You've evolved to think of things that taste bitter as unappetizing because they often do not have nutritional value and may even be poisonous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sugar (and sweet things in general), on the other hand, is very tasty because you've evolved to know that consuming sugars (carbohydrates) will give you energy. Milk also contains sugars, as well as lipids (fats) which give it its creamy taste, which you've also evolved to find appealing because they give you energy - more than simple sugars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The creamy taste of milk or sweet taste of sugar are usually enough to cover up the bitter taste of tannins (if the tea is brewed right, that is, but that's an issue for another discussion) and make your tea tasty. &lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-06-01T21:39:39.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5194" PostTypeId="1" AcceptedAnswerId="5195" CreationDate="2013-06-01T22:52:53.097" Score="4" ViewCount="39021" Body="&lt;p&gt;I am having trouble understanding how valence electrons are counted in Helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My references are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://funnyexam.com/answers/unmoderated/89095-self-doubt-the-most-common&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; where it appears that Helium has 0 valence electrons because the shells are finished&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_electron&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; where it appears that Helium has two valence electrons (which is what I thought)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How many valence electrons does helium have? &lt;/p&gt;&#xA;" OwnerUserId="1741" LastEditorUserId="7448" LastEditDate="2015-11-14T12:37:20.543" LastActivityDate="2016-01-09T07:15:02.040" Title="Valence electrons in helium?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5195" PostTypeId="2" ParentId="5194" CreationDate="2013-06-01T23:10:39.020" Score="10" Body="&lt;p&gt;Helium has two electrons in total, and according to the aufbau principle, it adopts the electronic configuration $\mathrm{1s^2}$. This means it has two electrons in s orbitals with a principal quantum number of 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last (and only) level of helium's electronic configuration is $\mathrm{1s^2}$, and therefore $\ce{He}$ has 2 valence electrons.&lt;/p&gt;&#xA;" OwnerUserId="792" LastEditorUserId="16683" LastEditDate="2016-01-09T07:15:02.040" LastActivityDate="2016-01-09T07:15:02.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5197" PostTypeId="1" CreationDate="2013-06-02T03:06:16.090" Score="2" ViewCount="203" Body="&lt;p&gt;I know that analytic forms of the Helium wavefunctions are not known.  However, are there empirical expressions for the low-lying states of Helium?  I'd like to use them to calculate some transition matrix elements for fun.&lt;/p&gt;&#xA;" OwnerUserId="592" LastActivityDate="2013-12-26T06:41:29.923" Title="Helium empirical wavefunctions" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5198" PostTypeId="1" AcceptedAnswerId="5201" CreationDate="2013-06-02T04:54:23.743" Score="4" ViewCount="1112" Body="&lt;p&gt;My chemistry text book says that &quot;Atomic Absorption Spectrometry (AAS) can be used to determine the concentration of metal ions, but other methods must be used for anions. One convenient method is to use an ion-selective electrode (ISE).&quot; What is the reason? Why can't ASP be used for determining the concentration of anions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author then argues that &quot;Lanthanum fluoride crystal doped with europium&quot; is the element used for measuring the concentration of fluorine in water because &quot;fluoride ions bind to the crystal&quot;. What is the reason behind choosing &quot;Lanthanum doped with europium&quot;?&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="13101" LastEditDate="2015-03-14T19:12:22.063" LastActivityDate="2015-03-14T19:12:22.063" Title="Why is Lanthanum a &quot;fluoride ion sensitive electrode&quot; and hence used for measuring &quot;flouride concentration&quot;?" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5199" PostTypeId="2" ParentId="5171" CreationDate="2013-06-02T05:48:36.717" Score="0" Body="&lt;p&gt;It's simpler than it might look. As a random example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3A + 2B-&amp;gt;2C + 4D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ &quot;due to the stoichiometry&quot;. In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for &quot;how many times the reaction happened&quot; ($\delta n&amp;gt;0$ if left to right, $\delta n&amp;lt;0$ if right to left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-06-02T05:48:36.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5201" PostTypeId="2" ParentId="5198" CreationDate="2013-06-03T00:49:59.893" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;What is the reason behind choosing &quot;Lanthanum doped with europium&quot;?&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, you said it yourself, &lt;code&gt;because &quot;fluoride ions bind to the crystal&quot;&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a great explanation from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the lanthanum fluoride electrode, the sensing element is a crystal of lanthanum fluoride LaF3, doped with europium fluoride EuF2 to create lattice vacancies. Such a crystal is an ionic conductor by virtue of the mobility of fluoride ions which jump between lattice vacancies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rough sketch I made to help you visualizing this. This is a simplistic drawing and the actual crystalline structure of the compound is 3-dimensional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ejYDb.jpg&quot; alt=&quot;Tip of the electrode&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must stress this: Both the structure and the proportions are probably nowhere near the ones in my drawing, it is only for didactic purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;An electrochemical cell may be constructed using such a crystal as a membrane separating two fluoride solutions. This cell acts as a concentration cell with transference where the fluoride transport number is 1. As transference of charge through the crystal is almost exclusively due to fluoride, the electrode is highly specific to fluoride. The only ion which significantly interferes is hydroxide (OH-). Generally such &quot;alkaline error&quot; can be avoided by buffering the sample to a pH below 7.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had the pH around 5.5. We also used TISAB for the preparation of water samples. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Total_ionic_strength_adjustment_buffer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on TISAB&lt;/a&gt; gives an excellent explanation on it's usage in fluoride ion analysis.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;Why ASP cannot be used for determining the concentration of anions?&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, I cannot give you an authoritative answer on that. I'll ask a teacher of mine this week and edit the answer with an update - if someone reading this knows it feel free to improve this answer or add a new one. In the case of fluoride it's probably because the Lanthanum electrode has pretty decent detection limits (it does). It's a lot cheaper to use it instead of AAS.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="1342" LastEditDate="2013-06-03T00:56:45.770" LastActivityDate="2013-06-03T00:56:45.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5202" PostTypeId="1" CreationDate="2013-06-03T08:01:19.903" Score="5" ViewCount="3423" Body="&lt;p&gt;Which has hotter temperature, blue flame or yellow flame? &lt;/p&gt;&#xA;" OwnerUserId="1750" LastEditorUserId="19616" LastEditDate="2015-10-30T17:44:30.253" LastActivityDate="2015-10-30T17:44:30.253" Title="Flame temperature for cooking food" Tags="&lt;everyday-chemistry&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5203" PostTypeId="2" ParentId="4364" CreationDate="2013-06-03T14:43:10.303" Score="1" Body="&lt;p&gt;Graphene can be made from $\ce{CO2}$ by a process of special electrolysis known as SMD. This is a patent pending process which produces graphene in bulk from $\ce{CO2}$ with very little energy. The inventor Michael Nunnerley came across this by accident when he was working on a new system of methane synthesis from $\ce{CO2}$ and $\ce{H2O}$.&lt;/p&gt;&#xA;" OwnerUserId="1751" LastEditorUserId="323" LastEditDate="2013-06-03T18:15:05.813" LastActivityDate="2013-06-03T18:15:05.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5204" PostTypeId="1" AcceptedAnswerId="5237" CreationDate="2013-06-03T17:07:09.487" Score="1" ViewCount="6278" Body="&lt;p&gt;As the title suggests – does there exist a liquid with a lower specific heat then water but with a higher thermal conductivity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of this in the context of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sous-vide&quot; rel=&quot;nofollow&quot;&gt;sous vide cooking&lt;/a&gt;: Ideally, the liquid involved would transfer heat to the food quickly (hence high thermal conductivity), but would also respond to quick changes in temperature (hence the low specific heat), reducing the risk / time of heater element overshoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there such liquids? (preferable ones that stay liquid between 20 and 85 degrees Celsius).&lt;/p&gt;&#xA;" OwnerUserId="1754" LastEditorDisplayName="user7951" LastEditDate="2015-11-13T19:06:19.213" LastActivityDate="2015-11-13T19:06:19.213" Title="Liquid with low specific heat and high thermal conductivity (relative to water)?" Tags="&lt;thermodynamics&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5205" PostTypeId="1" AcceptedAnswerId="5211" CreationDate="2013-06-03T17:22:26.090" Score="1" ViewCount="228" Body="&lt;p&gt;I ask because I'm trying to write an equilibrium equation for this reaction and I'm wondering whether an extra 2 protons will be required after one is taken to attack the $\ce{OH}$ group and turn it into $\ce{H_2O}$ in order to convert the freed ammonia (with one proton missing) into a conjugate acid. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4231" LastEditDate="2015-03-01T14:30:51.040" LastActivityDate="2015-03-01T14:30:51.040" Title="How many protons are required to satisfy hydroxyamine's (NH₂OH) lust for protons?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5206" PostTypeId="2" ParentId="5202" CreationDate="2013-06-03T17:53:06.430" Score="7" Body="&lt;p&gt;Blue flame has a hotter temperature. In physics there is a relationship called Wien's displacement law, which states&#xA;$$\lambda_{max}T=2.898$$&#xA;where $\lambda_{max}$ is the peak wavelength of light emitted (which is the wavelength you see) and T is temperature in kelvins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relationship can be rewritten $$T=\frac{2.898}{\lambda_{max}}$$, or just $T\propto \frac 1 \lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a table we find that the wavelength of blue light is about 450-500 nm and yellow light is 570-590 nm. Since temperature is inversely proportional to wavelength, the light with the smaller wavelength has the higher temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Now you might quibble that Wien's law is for idealized blackbodies but it can be used to model most things which emit light because of their temperature regardless.)&lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-06-03T17:53:06.430" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5207" PostTypeId="1" AcceptedAnswerId="5208" CreationDate="2013-06-03T18:24:43.443" Score="2" ViewCount="1528" Body="&lt;ul&gt;&#xA;&lt;li&gt;What is the general name for acids that can only lose one hydrogen ion? (i.e. $\ce{HCl}$)&lt;/li&gt;&#xA;&lt;li&gt;What is the general name for acids that can lose $n$ hydrogen ions? (i.e. $\ce{H_3PO_4}$, which can lose 3)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answers to these questions are probably mentioned a million times on the internet, but I could not find them.&lt;/p&gt;&#xA;" OwnerUserId="1740" LastEditorUserId="9961" LastEditDate="2015-11-13T19:50:42.453" LastActivityDate="2015-11-13T19:50:42.453" Title="Acids that donate just one proton" Tags="&lt;acid-base&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5208" PostTypeId="2" ParentId="5207" CreationDate="2013-06-03T18:29:35.297" Score="5" Body="&lt;p&gt;An acid that can donate one hydrogen ion is called monoprotic (i.e., $\ce{HCl}$). An acid that can donate two is called diprotic (i.e., $\ce{H_2SO_4}$). An acid that can donate 3 is called triprotic ($\ce{H_3PO_4}$). I don't know of any which can donate four.&lt;/p&gt;&#xA;" OwnerUserId="1719" LastEditorUserId="1740" LastEditDate="2013-06-03T18:44:59.233" LastActivityDate="2013-06-03T18:44:59.233" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5210" PostTypeId="1" AcceptedAnswerId="5222" CreationDate="2013-06-03T20:33:00.257" Score="2" ViewCount="328" Body="&lt;p&gt;I'm not a chemist but I'm working on an eco-friendly project that is supposed to recycle $\ce{CO2}$ from the air and convert it to $\ce{O2}$ or at least a usable fuel source ,the system can use the solar system as an energy source (I know that such system isn't efficient) but for the country in which the project is applicable it would be a great economical benefit (over direct sun power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-What is the most appropriate chemical reaction to achieve such goal ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reaction of course shouldn't emit toxic material(the whole purpose of the project is to benefit the environment, plus the cost of any filter or disposal mechanism isn't affordable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there any way to construct a circular system (&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_dioxide_scrubber#Regenerative_carbon_dioxide_removal_system&quot; rel=&quot;nofollow&quot;&gt;Regenerative carbon dioxide removal system &lt;/a&gt;)&#xA;Wikipedia's article doesn't elaborate further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help tag this question(this is my first time on chemistry.stachexchange)&lt;/p&gt;&#xA;" OwnerUserId="1753" LastEditorUserId="4231" LastEditDate="2015-03-01T14:31:07.430" LastActivityDate="2015-03-01T14:31:07.430" Title="Air recycling system(CO₂--&gt;O₂) or (CO₂--&gt;fuel)?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5211" PostTypeId="2" ParentId="5205" CreationDate="2013-06-03T21:11:56.090" Score="1" Body="&lt;p&gt;Well, hydroxylamine is technically amphoteric, meaning it can act as either a Bronsted acid or base. The acidic proton is presumably the one on the oxygen (although the resulting salts are apparently explosively unstable), while the nitrogen is the proton acceptor. I wouldn't expect the oxygen to be basic (which, unless I've misunderstood, seems to be what your question implies), at least not in any measurable way or for any appreciable length of time, given hydroxylamine's instability and tendency to spontaneously decompose in myriad ways. When treated with acid, it forms hydroxylammonium salts, which disproportionate upon heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;http://books.google.com/books?id=Mtth5g59dEIC&amp;amp;pg=PA657#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; was my primary reference; it contains a concise but informative summary of the reactions hydroxylamine undergoes.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-06-03T21:11:56.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5214" PostTypeId="1" AcceptedAnswerId="5217" CreationDate="2013-06-04T05:25:07.047" Score="0" ViewCount="203" Body="&lt;p&gt;Here is a question from my practice problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dMSNJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the answer is A, but I don't know why.  Is it because A has the most moles of gas on the left? thanks&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="1362" LastEditDate="2013-06-04T05:26:17.047" LastActivityDate="2013-06-05T10:32:47.157" Title="How does pressure shift equlibrium" Tags="&lt;homework&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5215" PostTypeId="1" AcceptedAnswerId="5216" CreationDate="2013-06-04T07:48:32.910" Score="5" ViewCount="1748" Body="&lt;p&gt;My question pertains generally to any form of spectroscopy, but I'll focus on UV spectroscopy. I am trying to get information on carrying out spectroscopy, but most of it is theoretical. Wikipedia is completely useless. Textbooks generally go into more detail about the underlying sub atomic processes being carried out. However, they do not provide too much details about how spectroscopy works in practice. So my question is this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will something like a fingerprint, or a speck of dust, or a label on the beaker (not even sure what material the beaker would be made of), or even air for that matter interfere with the absorption spectrum?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1759" LastActivityDate="2013-06-04T08:27:56.993" Title="Can dust or fingerprints interfere in ultra-violet spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5216" PostTypeId="2" ParentId="5215" CreationDate="2013-06-04T08:27:56.993" Score="4" Body="&lt;p&gt;You can deal with this through background correction of your spectra, which is a very common practice in spectrometry and implemented in any good spectrometry software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the most general sense, background correction consists of recording the spectrum of the spectrometer path &lt;em&gt;without&lt;/em&gt; the analyte, and then subtracting that &lt;em&gt;background spectrum&lt;/em&gt; from the spectrum of the analyte. In that sense, background correction is a form of difference spectroscopy. The spectrum you obtain from this ideally contains only the peaks attributable to the analyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example is subtracting the absorption spectrum of a solvent in a cuvette from an analyte. You would run a cuvette with solvent but without the analyte and use that as your background, followed by a cuvette with the analyte. Subtracting the first spectrum from the second would yield the signal attributable only to the analyte, assuming that the cuvettes, solvents and atmosphere between the two runs are &lt;em&gt;negligibly different&lt;/em&gt;.&#xA;Obviously, there are a lot of potential confounds here that need to be accounted for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Cuvettes have to be made of the same material, have the same dimensions (path length is important!) Cuvettes and such are often designed to be maximally transparent to radiation in the region of the spectrum of interest to maximise sensitivity. This is why they sell uber-expensive quartz cuvettes for UV/vis.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solvents/substrates have to be of identical composition between runs. You also don't want a solvent or substrate that absorbs strongly in the spectral region of interest for similar reasons to the above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cuvettes have to be homogeneously clean - if one is dirty, greasy or scuffed and scatters light more than the other, it may ruin the results.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The atmosphere has to be the same between runs - I've heard stories about quantitative IR spectroscopy of $\ce{C=O}$ stretches being thrown out by an increase in $\ce{CO2}$ due to people breathing (how rude!). You can understand that researchers who look at atmospheric $\ce{CO2}$ concentration need very careful sampling and analysis protocols for reasons similar to this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-06-04T08:27:56.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5217" PostTypeId="2" ParentId="5214" CreationDate="2013-06-04T08:54:22.503" Score="0" Body="&lt;p&gt;Shifts in equilibrium position of chemical reactions can be addressed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier's Principle&lt;/a&gt;. In short, the response of a system will be such that it counteracts the change that you are applying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a closed system, an increase in pressure at equal temperature will result in a decrease in volume according to the ideal gas law:&#xA;$$\frac{P}{RT}=\frac{n}{V} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This change will cause an upward shift in partial pressures which the system will counteract. In this case the equilibrium will shift to the right for reaction A, because this will lower the partial pressures. Indeed this means that for an increase in pressure the reaction will shift to the side with the lower number of moles. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-06-04T08:54:22.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5220" PostTypeId="1" AcceptedAnswerId="5848" CreationDate="2013-06-05T05:44:50.350" Score="7" ViewCount="74864" Body="&lt;p&gt;How can we chill water in a plastic bottle without a freezer or refrigerator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the office and I have two plastic bottles of water (brought from home). I'm in the office for 3 to 4 hours and this means the bottles I brought from home will get hot (and won't be drinkable in a hot season). Now I want to make it cool so that I can drink chilled water, without access to ice with salt or a refrigerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="1762" LastEditorUserId="2095" LastEditDate="2013-08-10T03:46:45.110" LastActivityDate="2015-08-27T04:00:42.060" Title="How do I chill water without a refrigerator?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5221" PostTypeId="2" ParentId="5214" CreationDate="2013-06-05T10:32:47.157" Score="2" Body="&lt;p&gt;I think the most intuitive way to think about this is rates of reaction. A reaction involving 2 molecules in the gas phase only happens when they collide. At higher pressures there will be more collisions between molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the reaction with the most molecules will be sped up more, so if a reaction has 2 molecules reacting to give 1 molecule of product, the forward reaction will be sped up more than the reverse reaction by increasing the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the kinetic calculations properly will require all the mechanistic steps involved. The general principle is that that at equilibrium, both reactions are happening at the same rate, but when you increase the pressure you will speed up the reaction which has more moles of gas, resulting in that reaction happening faster than the other reaction, shifting the equilibrium until the rates of the two reactions are balanced again.&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2013-06-05T10:32:47.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5222" PostTypeId="2" ParentId="5210" CreationDate="2013-06-05T13:43:55.777" Score="6" Body="&lt;p&gt;This is quite a hot topic at the moment - the current buzz-phrase is &lt;strong&gt;&lt;a href=&quot;https://duckduckgo.com/?q=power+to+gas&quot; rel=&quot;noreferrer&quot;&gt;power to gas&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are quite a few conference papers and technical papers, and at least one dedicated conference, for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;strong&gt;it's too soon for there to be a known &quot;best&quot; process.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ecogeek.org/biofuels/3817-audis-carbon-neutral-e-gas&quot; rel=&quot;noreferrer&quot;&gt;Audi&lt;/a&gt; and others are using a modified &lt;a href=&quot;http://www.pennenergy.com/articles/pennenergy/2010/03/the-sabatier-reaction.html&quot; rel=&quot;noreferrer&quot;&gt;Sabatier process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 + 4 H2 -&amp;gt; CH4 + 2 H2O }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others are looking at a modified &lt;a href=&quot;http://www.netl.doe.gov/technologies/coalpower/gasification/gasifipedia/5-support/5-11_ftsynthesis.html&quot; rel=&quot;noreferrer&quot;&gt;Fischer-Tropsch&lt;/a&gt; process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(2n +1) H2  +  n CO -&amp;gt; C_{n} H_{2n +2}  +  n H2O}$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SR96v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(image courtesy of the &lt;a href=&quot;http://www.rwe.com/web/cms/en/1846074/rwe/innovation/projects-technologies/energy-storage/project-power-to-gas/&quot; rel=&quot;noreferrer&quot;&gt;RWE power to Gas Project&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, you've got to prise the Oxygen off the Carbon, which is an energy-intensive process; and then attach the Hydrogen, which is not. But the Hydrogen typically comes from electrolysis of water, which &lt;strong&gt;is&lt;/strong&gt; energy-intensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only are the processes being prototyped high-energy, but they're high-exergy too, requiring very high heat and / or electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's unlikely that this is a better option than using solar power for solar-thermal or photovoltaics, unless the application is long-distance travel, which would require an energy source with a very high energy density. In which case, there are several renewable-fuel synthesisers being prototyped at the moment, but it's too early to tell which is the best.&lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2013-06-05T13:43:55.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5224" PostTypeId="1" AcceptedAnswerId="5236" CreationDate="2013-06-05T16:28:24.780" Score="13" ViewCount="9929" Body="&lt;p&gt;I have been curious about this question for a while. If you want to warm up a large amount of water, is it feasible to do this by adding a substance that has an exothermic reaction with the water? What is the relevant chemistry one has to know to work out how much the water would be heated up by?&lt;/p&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="7475" LastEditDate="2015-11-13T21:09:09.207" LastActivityDate="2020-03-11T18:02:01.860" Title="Can you heat water with additives?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5227" PostTypeId="1" CreationDate="2013-06-05T20:02:49.367" Score="4" ViewCount="2862" Body="&lt;p&gt;For example, calcium carbide (CaC$_2$) has covalent C‒C bonds and ionic Ca$^{2+}$‒ C$_2^{2-}$ bonds.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2016-08-22T17:07:54.543" Title="Is there a word for a compound that has both ionic and covalent bonds?" Tags="&lt;bond&gt;&lt;nomenclature&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5228" PostTypeId="2" ParentId="5224" CreationDate="2013-06-05T20:16:30.563" Score="13" Body="&lt;p&gt;You’re looking for materials that either react with water with a large reaction enthalpy $\Delta H_\mathrm{r}$, or have a large enthalpy of solution $\Delta H_\mathrm{sol}$ (which really amounts to the same thing but with a slightly different scope).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this as a general method, though, is that water has a &lt;em&gt;really&lt;/em&gt; high heat capacity: take a look at this &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities&quot; rel=&quot;nofollow noreferrer&quot;&gt;table of heat capacities&lt;/a&gt;, which gives it as approximately 4.2 joules per gram per kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common method of producing self-heating coffee cans is to combine anhydrous calcium chloride with water in a closed vessel surrounding the can. &lt;a href=&quot;http://www.tetrachemicalseurope.com/applications/chemical_heating.aqf&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to a vendor of calcium chloride&lt;/a&gt;, 100 % anhydrous $\ce{CaCl2}$ has a solution enthalpy of 737 joules per gram when dissolved in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming an approximate density of water of 1 gram per millilitre, that means to heat a litre of water by 10 K, you’d need about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_{\ce{CaCl2}} = \frac{m_\text{water} c_p \Delta T}{\Delta H_\text{sol}} = \frac{1000\ \mathrm{g} \times 4.2\ \mathrm{J\ g^{-1} K^{-1}} \times 10\ \mathrm{K}}{737\ \mathrm{J\ g^{-1}}} = 57\ \mathrm{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;em&gt;that’s&lt;/em&gt; feasible, but you may have a different idea of a large amount of water. And, let’s not forget, you're still adding stuff to the water, so it may not be suitable for what you wanted it for anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notable reactions that can heat water include reacting it with lithium metal (~220 J/g), or dissolving sulfuric acid (~90 J/g). You’re looking for materials that don’t hold together particularly strongly (&lt;em&gt;e.g.&lt;/em&gt; if they’re a crystal, they have a relatively weak lattice), because you have to break those bonds to dissolve/react it, and also bind to water &lt;em&gt;very&lt;/em&gt; strongly, because that’s what gives you the exothermicity. Because of that, ionic materials and materials that form at least doubly charged ions when reacting with water are likely to be better candidates.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="402" LastEditDate="2017-04-08T22:01:09.653" LastActivityDate="2017-04-08T22:01:09.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5229" PostTypeId="1" AcceptedAnswerId="9453" CreationDate="2013-06-05T20:20:03.643" Score="2" ViewCount="5445" Body="&lt;p&gt;In reading a COA (Certificate of Analysis), I came across an Assay line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The Assay is 100.5% where &quot;Lower Limit&quot; is 98.5 and &quot;Upper Limit&quot; is 101.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This product is L-Cysteine Hydrochloride Monohydrate.  How can the Assay be over 100%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found this statement: &quot;If the substance under test is purer than the standard, the assay comes more than 100 pct.&quot;  If that is the case, then the question becomes, where can I find out about this standard?&lt;/p&gt;&#xA;" OwnerUserId="1590" LastEditorUserId="73" LastEditDate="2013-11-28T05:53:27.853" LastActivityDate="2014-03-25T01:49:30.747" Title="Assay over 100% (L-Cysteine Hydrochloride Monohydrate)" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5230" PostTypeId="2" ParentId="5220" CreationDate="2013-06-05T20:30:02.130" Score="1" Body="&lt;p&gt;It depends on what you can use:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Leave the bottle in the cold air from the air conditioner,&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Freeze it at night and wrap in bubble plastic,&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a peltier cooler.&lt;/p&gt;&#xA;&lt;p&gt;This last one is expensive and not too good, but I've used the others with good results.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I think this is Physics more than chemistry.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-06-05T20:30:02.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5231" PostTypeId="2" ParentId="5227" CreationDate="2013-06-06T01:21:10.833" Score="5" Body="&lt;p&gt;There may be a strong case to call this a &quot;salt&quot;, like other ionic compounds.  Very often, the anion (negative ion) of a salt is an ion comprised of different atoms with covalent bonds.  For example, $\ce{ SO_4^{2-} }$ is the &lt;em&gt;sulfate anion&lt;/em&gt;, and the bonds between $\ce O$ and $\ce S$ are covalent to some degree.  So $\ce{ Na_2SO_4 }$ is an ionic compound under this definition.  Thus, any compound with some component that is strongly ionic can be called an ionic compound; i.e. a salt, regardless of the bonding structure of the components.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="1085" LastEditDate="2013-06-07T15:56:35.457" LastActivityDate="2013-06-07T15:56:35.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5232" PostTypeId="2" ParentId="4131" CreationDate="2013-06-06T04:56:20.660" Score="2" Body="&lt;p&gt;The energy can go to several different places:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The electrons may get some of the energy. In this case the dissociated atoms will be in excited states.  &lt;/li&gt;&#xA;&lt;li&gt;A photon might be emitted.   &lt;/li&gt;&#xA;&lt;li&gt;The atoms may have some kinetic energy left over after they are separated.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="955" LastEditorUserId="41328" LastEditDate="2017-08-15T22:10:49.557" LastActivityDate="2017-08-15T22:10:49.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5233" PostTypeId="5" CreationDate="2013-06-06T16:14:58.517" Score="0" Body="" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2013-06-06T16:14:58.517" LastActivityDate="2013-06-06T16:14:58.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5234" PostTypeId="4" CreationDate="2013-06-06T16:14:58.517" Score="0" Body="For questions dealing with sustainable living, including questions about better fuels, pollution reduction, CO2 capture and recycling (CCR), the chemistry of photovoltaics, photochemical splitting of water, catalysts for hydrolysis of water and other routes to hydrogen generation." OwnerUserId="22" LastEditorUserId="25" LastEditDate="2013-06-07T04:09:14.377" LastActivityDate="2013-06-07T04:09:14.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5235" PostTypeId="1" CreationDate="2013-06-06T23:07:16.637" Score="3" ViewCount="939" Body="&lt;p&gt;I know that elements (ex. Ag) have spectral/emission/absorption lines, but do chemical compounds (ex. NaCl) have spectral lines as well? Ones that when seen you can tell that that compound (NaCl) is made up of certain elements (Na and Cl).&lt;/p&gt;&#xA;" OwnerUserId="1767" LastActivityDate="2015-10-23T07:53:53.317" Title="Do chemical compounds have spectral lines, or only elements?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5236" PostTypeId="2" ParentId="5224" CreationDate="2013-06-06T23:09:03.653" Score="16" Body="&lt;p&gt;From the comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do you want to do with the hot water? Swim in it. I was thinking&#xA;  of thousands of liters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be constantly adding more chemicals to it. At some point, you'd have a pool of chemicals instead of a pool of water. Not to mention that the chemicals would probably end up being more expensive than the electricity or the gas bill to heat up the pool to the same temperature the conventional way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice way of heating up pools is using &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_water_heating&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar water heating systems&lt;/a&gt;, and coupling them with the conventional heating systems, if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... Is it feasible? I'd say it isn't. But since you said this was more of a thought experiment... The only scenario I can imagine where the best solution to heat up a pool is to add chemicals to it is: You have a small pool, very little time to heat it up, you want to heat it up to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sauna&quot; rel=&quot;nofollow noreferrer&quot;&gt;sauna-like temperatures&lt;/a&gt;, and although you'll only use it for a couple of hours you are willing to spend a lot of money to do so. In this case, I'd add some water to the pool, and then add the following reagents to it, making the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NaOH + HCl -&amp;gt; NaCl + H2O} \\ \Delta H = -56\,200\ \mathrm{\frac{J}{mol}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the math, but I assure you it heats up pretty quickly. We used to neutralize concentrated acid in ~10 L plastic buckets, and if we &quot;accidentally&quot; neutralized too much at a time the bucket would start to heat up to a point you couldn't even touch it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NaCl is kitchen salt, so no issues with toxicity if you do the math correctly. Salt water also makes you float better, which I think is a nice quality in a pool. There would be salt on the bottom, but I think that's nice too, makes it feel like the sea.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorDisplayName="user7951" LastEditDate="2019-01-13T19:50:29.273" LastActivityDate="2019-01-13T19:50:29.273" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5237" PostTypeId="2" ParentId="5204" CreationDate="2013-06-07T09:47:16.233" Score="1" Body="&lt;p&gt;Most heat transfer through fluids is actually convection rather than conduction, so you need something that expands a lot with heat and has a low viscosity. Mineral oils or silicone are used as heat transfer fluids when temperatures above 100 °C are needed, but they are much more viscous than water.&#xA;However, for your system I guess that heat capacity is actually the most important number, as convection will be sufficient to ensure an equal temperature throughout the system (unless you actively impede it) even with a viscosity slightly higher than that of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eyeballing some of the tables at &lt;a href=&quot;http://www.engineeringtoolbox.com&quot; rel=&quot;nofollow&quot;&gt;http://www.engineeringtoolbox.com&lt;/a&gt; I’d suggest trying ethanol – its heat capacity is around half that of water, its viscosity is not much higher and it stays liquid over the desired range. Plus of course, unlike some of the more exotic solvents, it’s not a disaster if some of it does leak in to your food. If the food isn’t actually going to be eaten (just for demonstration purposes) then you could perhaps try toluene or petrol (gasoline).&lt;/p&gt;&#xA;" OwnerUserId="1001" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T11:35:30.030" LastActivityDate="2015-04-26T11:35:30.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5238" PostTypeId="2" ParentId="5235" CreationDate="2013-06-07T13:20:45.930" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4706/how-does-icp-oes-determine-the-quantity-of-detected-elements-or-just-their-oc/4715&quot;&gt;icp&lt;/a&gt; analysis the sample is heated to a temperature in which you don't have compounds and you detect only elements by their emissions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For compounds there is absortion or fluorescence in which the sample is at near room temperature mostly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course this too general, and the technique to use depends on what you are looking for and where because each technique has its strengths and weakness.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2013-06-07T13:20:45.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5239" PostTypeId="1" AcceptedAnswerId="5240" CreationDate="2013-06-07T14:06:28.627" Score="12" ViewCount="36914" Body="&lt;p&gt;Why do molecules like alkanes with higher relative molecular&#xA;mass ($M_\mathrm r$) have stronger intermolecular forces? For example, methane ($\ce{CH4}$) has a weaker intermolecular force than pentane ($\ce{C5H12}$). Pentane has a higher $M_\mathrm r$ than methane and therefore pentane possesses stronger intermolecular force than methane. These are members of the same family in the homologous series, known as alkanes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is why does pentane have stronger intermolecular force than methane? Why is there a relationship between relative molecular mass ($M_\mathrm r$) and inter-molecular forces. The stronger the intermolecular force the higher the boiling point and the melting point?&lt;/p&gt;&#xA;" OwnerUserId="856" LastEditorUserId="402" LastEditDate="2017-04-18T18:57:15.610" LastActivityDate="2018-01-13T12:42:37.013" Title="Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5240" PostTypeId="2" ParentId="5239" CreationDate="2013-06-07T14:52:59.327" Score="10" Body="&lt;p&gt;There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't have a real internal dipole--all $\ce{C-C}$ bonds are non-polar and the electronegativity of the $\ce{C-H}$ bond is close enough that those bonds are also considered non-polar). This matters because polar interactions are a relatively strong IMF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a detour into e- distributions. Typically, electrons have the highest probability of being evenly distributed about the atoms in a molecule, proportional to the electronegativity of those atoms. That being said, that is only the most probable state, and so, of course, there are times when the e- are NOT evenly distributed. Remember how alkanes normally aren't polar? Well, when the electrons get unbalanced in the molecule, surprise! An &lt;em&gt;Instantaneous Dipole&lt;/em&gt; is temporarily introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? Well, when an alkane is next to another alkane that is subject to an instantaneous dipole, the electrostatic forces from the quasi-polarized alkane cause the other alkane to also attain a lopsided distribution of electrons, causing it to gain a temporary polarity, and this effect both attracts these molecules while also propagating throughout the surrounding molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now that you have a little understanding of what LDF's or instantaneous dipoles are, why the trend with molecular weight? The short answer is that larger molecules have more e-, and are affected by LDF's more than smaller molecules with less e-. With more electrons and more atoms for those electrons to be distributed over, there is a greater chance for frequent or large instantaneous dipoles, increasing the overall Van der Waals forces.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorDisplayName="user7951" LastEditDate="2018-01-13T12:42:37.013" LastActivityDate="2018-01-13T12:42:37.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5242" PostTypeId="2" ParentId="444" CreationDate="2013-06-07T16:55:26.733" Score="52" Body="&lt;p&gt;Something worth adding to this discussion that I'm surprised hasn't been mentioned about such &quot;hypervalent&quot; molecules like $\ce{SF6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my professors at university informed me that the common explanation (that the empty d-orbitals are empty and are thus accessible) is actually most likely incorrect. This is an old-model explanation that is out-of-date, but is for some reason continuously taught in schools. A quote from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on orbital hybridisation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1990, Magnusson published a seminal work definitively excluding the role of d-orbital hybridization in bonding in hypervalent compounds of second-row elements.&lt;br&gt;&#xA;  (&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1990,&lt;/strong&gt; &lt;em&gt;112&lt;/em&gt; (22), 7940–7951. &lt;a href=&quot;https://doi.org/10.1021/ja00178a014&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/ja00178a014&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you actually look at the numbers, the energy associated with those orbitals is significantly higher than the bonding energy found experimentally within molecules like $\ce{SF6}$, meaning that it is highly unlikely that the d-orbitals are involved at all in this type of molecular structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us stuck, in fact, with the octet rule. Since $\ce{S}$ cannot reach into its d-orbitals, it cannot have more than 8 electrons in its valence (see other discussions on this page for definitions of valence etc, but by the most basic definition, yes, only 8). The common explanation is the idea of a 3-centered 4-electron bond, which is essentially the idea that sulfur and two fluorines 180 degrees apart share only 4 electrons between their molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of comprehending this is to consider a pair of resonance structures where sulfur is bonded covalently to one $\ce{F}$ and ionically to the other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{F^{-}\bond{...}^{+}S-F  &amp;lt;-&amp;gt; F-S+\bond{...}F-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you average these two structures out, you will notice that sulfur maintains a positive charge and that each fluoride has a sort of &quot;half&quot; charge. Also, note that sulfur only has &lt;em&gt;two electrons associated with it&lt;/em&gt; in both structures, meaning that it has successfully bonded to two fluorines while only accumulating two electrons. The reason they have to be 180 degrees apart is due to the geometry of the molecular orbitals, which is beyond the scope of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, just to review, we've bonded to two fluorines to the sulfur accumulating two electrons and 1 positive charge on sulfur. If we bonded the remaining four fluorides from $\ce{SF6}$ in the normal covalent way, we'd still end up with 10 electrons around sulfur. So, by utilizing another 3-center-4 electron bond pair, we achieve 8 electrons (filling both the s and p valence orbitals) as well as a $+2$ charge on the sulfur and a $-2$ charge distributed around the four fluorines involved in the 3c4e bonding. (Of course, all of the fluorines have to be equivalent, so that charge will actually be distributed around all of the fluorines if you consider all of the resonance structures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There actually is a lot of evidence to support this style of bonding, the simplest of which is observed by looking at bond lengths in molecules such as $\ce{ClF3}$ (T-shapes geometry), where the two fluorines 180 degrees apart from each other have a slightly longer bond length to chlorine than the other fluorines do, indicating a weakened amount of covalency in those two $\ce{Cl-F}$ bonds (a result of averaging out a covalent and ionic bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in the details of the molecular orbitals involved, you may wish to read &lt;a href=&quot;https://chemistry.stackexchange.com/a/49844/16683&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR Hypervalency doesn't really exist, and having more than $\ce{8 e-}$ in non-transition metals is much harder than you would think.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="16683" LastEditDate="2017-08-14T15:45:57.890" LastActivityDate="2017-08-14T15:45:57.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5243" PostTypeId="2" ParentId="2469" CreationDate="2013-06-07T18:23:51.813" Score="41" Body="&lt;p&gt;There is an explanation to this that can be generalized, which dips &lt;em&gt;a little&lt;/em&gt; into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d:   ↿ ↿ ↿ _ _        ↿ ↿ ↿ ↿ _        ↿ ↿ ↿ ↿ ↿     ^&#xA;                 OR               OR               | &#xA;s:   ⥮               ↿                _            Energy gap (E)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The top row is for the d-orbitals, which are higher in energy, and the bottom row is for the s-orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s-orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principle and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Repulsion energy:&lt;/strong&gt; Pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exchange energy:&lt;/strong&gt; This is a little tricky, and probably the main reason this isn't taught until later in your chemistry education. Basically (due to quantum chemistry which I won't bore you with), there is a &lt;em&gt;beneficial energy&lt;/em&gt; associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy. (This is very simplified, but really &quot;stabilizing energy&quot; is nothing more than negative energy. I hope your thermodynamics is in good shape!) The thing with exchange (or K) energy is that you get one for &lt;strong&gt;every&lt;/strong&gt; pair, so in the case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;↿ ↿ ↿&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from say a p-subshell, you would get 3 K, for &lt;strong&gt;each&lt;/strong&gt; pair, while from this example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;⥮ ↿ ↿ ↿ ↿&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from a $\ce{d^6}$, you would get 10 K (for each unique pair, and none for the opposite spin e-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This K is quantifiable as well (and like the repulsion energy is unique for each atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the combination of these two energies &lt;strong&gt;when compared to the band gap&lt;/strong&gt; determines the state of the electron configuration. Using the example we started with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;d:   ↿ ↿ ↿ _ _         ↿ ↿ ↿ ↿ _         ↿ ↿ ↿ ↿ ↿    ^&#xA;s:   ⥮           OR   ↿           OR   _            | &#xA;PE:  3K + 1C          6K + 0C          10K + 0C     Energy gap (E)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can see from the example that shoving 1 e- up from the s to the d-subshell results in a loss of 1C (losing positive or &quot;destabilizing&quot; repulsive energy) and gaining 3K (gaining negative or &quot;stabilizing&quot; exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e. 3K - 1C &gt; E) then the electron will indeed be found in the d shell in $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, lets look at perhaps exciting the second s e- up to the d-subshell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d-subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that for $\ce{Nb}$: 4K + 0C &amp;lt; E  (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="5026" LastEditDate="2018-03-31T12:37:44.407" LastActivityDate="2018-03-31T12:37:44.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5245" PostTypeId="2" ParentId="579" CreationDate="2013-06-07T21:06:56.540" Score="3" Body="&lt;p&gt;This is a great question, and one that I also have and would like to see a better explanation for the answer, but I can offer what I know (I would comment, but I do not have that usability at the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add a dopant to say, Si, you are essentially shortening the distance between the valence and the conduction band by adding either an accessible conduction (empty) band just above the original valence band (in the case of p-type like B) or are making the original conduction band more accessible by adding a higher energy valence band (in the case of n-type like N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guessed answer to your question is one of two things: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;either the group III/V elements are optimal in shortening this band gap, and that II produces excess conduction too high and VI excess valence too low, or...&lt;/li&gt;&#xA;&lt;li&gt;the change in valence of the dopant too greatly affects the lattice stabilization of the Si crystal. Essentially by causing too many localized vacancies/defects that the lattice strain negatively affects the overall conductivity of the Si.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-07T21:06:56.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5246" PostTypeId="1" CreationDate="2013-06-08T00:05:30.890" Score="0" ViewCount="13257" Body="&lt;p&gt;As title says, why is methyl group more electron-donating than tert-butyl group? The context behind this is stabilization of conjugate base. (&lt;a href=&quot;http://www.khanacademy.org/science/organic-chemistry/organic-structures/acid-base-review/v/stabilization-of-conjugate-base-iii&quot; rel=&quot;nofollow&quot;&gt;http://www.khanacademy.org/science/organic-chemistry/organic-structures/acid-base-review/v/stabilization-of-conjugate-base-iii&lt;/a&gt; look at 10:51)&lt;/p&gt;&#xA;" OwnerUserId="1769" LastActivityDate="2013-08-07T17:51:51.777" Title="Why is methyl group more electron-donating than tert-butyl group?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5247" PostTypeId="1" CreationDate="2013-06-08T05:25:50.367" Score="2" ViewCount="6928" Body="&lt;p&gt;We had a test and and we had a question asking the number of valence electrons in an azide ion, $\ce{N3^{-}}$.  I find out that the answer is 16 but I could not understand how. &#xA; I saw the structure. It says that the end nitrogen have a -1 charge and the middle nitrogen has a +1 charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From where does the one of the end nitrogen get the electron assuming the other end nitrogen gets it from the middle nitrogen?&lt;/p&gt;&#xA;" OwnerUserId="1763" LastEditorUserId="13339" LastEditDate="2015-05-18T07:38:30.110" LastActivityDate="2019-07-21T14:01:37.650" Title="How many valence electrons does the azide ion have?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5249" PostTypeId="2" ParentId="5246" CreationDate="2013-06-08T11:11:17.643" Score="4" Body="&lt;p&gt;Methyl group is &lt;strong&gt;not&lt;/strong&gt; more electron donating than tert-butyl group, as the video provided in the link suggests (11:20 of the video), which is true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_effect&quot; rel=&quot;nofollow&quot;&gt;Inductive effect&lt;/a&gt; is a permanent effect which results in displacement of electrons along the chain of atoms in the molecule (here, carbon atoms).  The reason for this displacement can be worked out from the electronegativity difference between the atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the tert-butyl group.  In a C-H bond, the electron density is directed towards &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon%E2%80%93hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;carbon&lt;/a&gt;.  Methyl group, which has three hydrogen atoms attached to a carbon, pushes the electron density towards $\ce{C2}$ of tert-butyl acetic acid.  So do the two other methyl groups.  The amount of partial negative charge accumulated on $\ce{C2}$ of tert-butyl group (has 3 methyl groups) is more than that on methyl group (only has 3 hydrogens) of acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more negative charge on $\ce{C2}$ in tert-butyl acetic acid destabilizes the conjugate base, compared to a simple acetate ion.  Hence acetic acid has less pKa than tert-butyl acetic acid.&lt;/p&gt;&#xA;" OwnerUserId="1734" LastEditorUserId="1878" LastEditDate="2013-08-07T17:51:51.777" LastActivityDate="2013-08-07T17:51:51.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5250" PostTypeId="2" ParentId="2563" CreationDate="2013-06-08T12:52:43.397" Score="0" Body="&lt;p&gt;The trick is: after the addition, a buffer solution is formed.  Find the concentrations of the salt and acid and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson's equation&lt;/a&gt; to arrive at the answer. &lt;/p&gt;&#xA;" OwnerUserId="1734" LastActivityDate="2013-06-08T12:52:43.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5251" PostTypeId="1" AcceptedAnswerId="5252" CreationDate="2013-06-08T14:01:07.200" Score="3" ViewCount="11739" Body="&lt;p&gt;$\ce{[Cr(H2O)4Cl2]Cl}$, Tetraaquadichlorochromium(III) chloride&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a book, the (III) indicates the oxidation state of chromium, but it doesn't explain why. Cr has 6 valence electrons. How does it get to +3?&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="4945" LastEditDate="2014-07-31T15:35:27.243" LastActivityDate="2016-07-23T16:51:06.537" Title="Oxidation State in Tetraaquadichlorochromium(III) chloride" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5252" PostTypeId="2" ParentId="5251" CreationDate="2013-06-08T15:08:12.417" Score="4" Body="&lt;p&gt;I think the easiest way to see this is to count the formal charges on the ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four waters do not have net charge.  The remaining ligands--the three Cl's--are likely $\ce{Cl-}$, and so their total charge contribution is $-3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way the complex can be neutral is if the chromium has a $+3$ charge:  hence Chromium (III).&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-08T15:08:12.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5253" PostTypeId="2" ParentId="5247" CreationDate="2013-06-08T15:14:12.037" Score="4" Body="&lt;p&gt;Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence.  Therefore, a single elemental nitrogen has five valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form.  So, the total number of valence electrons is $15 + 1 = 16$.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-08T20:05:21.580" LastActivityDate="2013-06-08T20:05:21.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5254" PostTypeId="2" ParentId="5189" CreationDate="2013-06-09T01:03:17.553" Score="7" Body="&lt;p&gt;Nicolau Saker Neto's last comment is probably what's going on here.  All of your titrations have too low a pH compared to theoretical, and so are less basic than expected.  This suggests that your concentration of $\ce{NaOH}$ is lower than you think, and the fact that $\ce{NaOH}$ is hydroscopic is a good explanation for this (i.e. you have a more dilute solution than you think).  Neto continues to be correct in that the usual procedure to correct this is to standardize your titration solution against a standard that is known to be more reliable with respect to producing accurate concentrations.  $\ce{KHPh}$ is indeed one of these standards, and is used after drying in a drying oven.  So the full procedure is: prepare $\ce{NaOH}$ solution as accurately as possible, do the same with $\ce{KHPh}$, titrate $\ce{NaOH}$ against $\ce{KHPh}$, and record the resulting concentration of $\ce{NaOH}$ accurately for use in future titrations.  Also, $\ce{NaOH}$ loses strength over time with exposure to the air, so you have to re-standardize over time.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-06-09T01:03:17.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5255" PostTypeId="2" ParentId="4318" CreationDate="2013-06-09T03:37:24.497" Score="5" Body="&lt;p&gt;You can find some experimental information on dipole transition moments in the NIST Chemistry Webbook in the section on constants of diatomic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=B3000043&amp;amp;Units=SI&amp;amp;Mask=1000&quot; rel=&quot;noreferrer&quot;&gt;http://webbook.nist.gov/cgi/cbook.cgi?ID=B3000043&amp;amp;Units=SI&amp;amp;Mask=1000&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-06-09T03:37:24.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5256" PostTypeId="1" CreationDate="2013-06-09T11:11:07.673" Score="6" ViewCount="13624" Body="&lt;p&gt;Are there substances that cannot undergo combustion? If so, what are they?&lt;/p&gt;&#xA;" OwnerUserId="1773" LastActivityDate="2016-07-31T00:35:50.533" Title="Are there substances that cannot undergo combustion?" Tags="&lt;combustion&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5257" PostTypeId="2" ParentId="5251" CreationDate="2013-06-09T13:14:53.327" Score="1" Body="&lt;p&gt;Eric is correct about the easiest method to determine the most likely oxidation state of Cr. As for the rest of your question, the oxidation state of Cr in this compound means that it has three electrons less than in its neutral state, i.e. three valence electrons instead of six. This +3 oxidation state is the most stable one for chromium, although many other states are also possible. Note that a wide range of oxidation states is characteristic of d-block elements such as chromium.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2013-06-09T13:14:53.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5259" PostTypeId="2" ParentId="4368" CreationDate="2013-06-09T13:38:28.387" Score="1" Body="&lt;p&gt;As you probably know, lactate dehydrogenase catalyzes the interconversion of pyruvate and lactate, with electron transfer via NADH/NAD⁺. If lactate dehydrogenase is inhibited, the fate of pyruvate in the cell extract will depend on what cell structures as wells as enzymes and other compounds are available. In other words, the answer to your question will likely be strongly dependent on the organism, as well as the particular circumstances of the experiment you are performing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you a speculative answer: Assuming that the metabolic pathways of the mammalian cell still function and that oxygen is available to accept electrons in respiration, pyruvate could be metabolized through the tricarboxylic acid cycle, forming energy-rich compounds and carbon dioxide. If other fermentative pathways are available, pyruvate could be funnelled into those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, predicting what compunds would accumulate would require more information than that which is given in your question.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2013-06-09T13:38:28.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5260" PostTypeId="2" ParentId="5256" CreationDate="2013-06-09T14:24:12.723" Score="-4" Body="&lt;p&gt;If you define &lt;em&gt;combustion&lt;/em&gt; as a chemical process that consumes $\ce{O2}$ and produces $\ce{H2O}$, then the answer is: any $\ce{X}$ that does not contain hydrogen is not combustible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{X + O2 -&amp;gt; Y + H2O}$&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-09T14:24:12.723" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="5261" PostTypeId="2" ParentId="2488" CreationDate="2013-06-09T15:10:30.403" Score="2" Body="&lt;p&gt;The chair conformation that you have drawn (&lt;sup&gt;4&lt;/sup&gt;C&lt;sub&gt;1&lt;/sub&gt;) is likely to be the most stable one, as it minimizes the number of heavy axial groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the chair conformation of a hexose, it is generally easiest to draw it and compare it with β-D-glucose, where all heavy groups are equatorial and the conformation is &lt;sup&gt;4&lt;/sup&gt;C&lt;sub&gt;1&lt;/sub&gt;. If the number of heavy axial groups becomes smaller when the conformation is changed to &lt;sup&gt;1&lt;/sup&gt;C&lt;sub&gt;4&lt;/sub&gt; (all equatorial groups in &lt;sup&gt;4&lt;/sup&gt;C&lt;sub&gt;1&lt;/sub&gt; become axial and vice versa), then it is likely that the conformation is &lt;sup&gt;1&lt;/sup&gt;C&lt;sub&gt;4&lt;/sub&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2013-06-09T15:10:30.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5262" PostTypeId="2" ParentId="4911" CreationDate="2013-06-09T15:42:56.537" Score="3" Body="&lt;p&gt;I will propose an outline for determining an expression for the rate of substrate conversion in the given case:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $$\ce{E + S &amp;lt;=&amp;gt;[k_1][k_{-1}] ES_1-&amp;gt;[k_2] ES_2 -&amp;gt;[k_3] E + P}$$&lt;/li&gt;&#xA;&lt;li&gt;Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$.&lt;/li&gt;&#xA;&lt;li&gt;Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES&lt;sub&gt;1&lt;/sub&gt; and ES&lt;sub&gt;2&lt;/sub&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Make the &lt;em&gt;pseudo-steady-state assumption&lt;/em&gt; (PSSA): assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e. $\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$.&lt;/li&gt;&#xA;&lt;li&gt;Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2013-06-09T15:42:56.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5263" PostTypeId="1" CreationDate="2013-06-09T16:52:09.107" Score="5" ViewCount="256" Body="&lt;p&gt;I'm interested in measuring the stable isotope ratios of some bird feathers. I'm aware that a mass spectrometer is required to measure stable isotope ratios. Is it possible to measure stable isotope ratios at home, using your own 'DIY' spectrometer?&lt;/p&gt;&#xA;" OwnerUserId="1776" LastEditorUserId="4945" LastEditDate="2014-11-19T09:08:57.507" LastActivityDate="2019-11-30T07:59:10.800" Title="Can I perform a stable isotope analysis at home?" Tags="&lt;home-experiment&gt;&lt;isotope&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5266" PostTypeId="2" ParentId="5263" CreationDate="2013-06-09T23:47:50.567" Score="6" Body="&lt;p&gt;I am assuming that you wish to measure isotopes of carbon.  It is unknown to me how you will extract sample from a bird feather that is suitable for measurement in a spectrometer, and does not taint the isotope ratios. The spectrometers that can distinguish between isotopes are generally quite sensitive and complicated. So the literal answer to your question is &quot;NO, you can not do it at home with a DIY spectrometer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you might be able to purchase a spectrometer. You might be able to differentiate carbon isotopes in carbonyl compounds with infrared spectroscopy. (But don't count on it, unless you are somehow making a gaseous product such as $\ce{CO}$ or $\ce{CO2}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way of measuring carbon isotopes is via mass spectrometry. You can probably get a crude measurement with a machine that costs $20,000 (on eBay) but then you must get it calibrated and keep it supplied with carrier gas, etc.  In short, it will bankrupt you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;User permeakra has the best idea: send a sample to a lab.  But, I am not sure of any lab will allow you to send samples through their &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_14_dating#Measurements_and_scales&quot; rel=&quot;noreferrer&quot;&gt;spectrometer&lt;/a&gt;, not knowing anything about your reputation or the volume of business that they might expect to get from you.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-10T01:01:46.617" LastActivityDate="2013-06-10T01:01:46.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5267" PostTypeId="1" CreationDate="2013-06-10T02:07:33.867" Score="8" ViewCount="24746" Body="&lt;p&gt;Just now I got some room temperature water and added some ice to cool it, and it just came to me, like the scientist I wish I was ... why doesn't frozen (freezing point) ice added to higher temperature liquid compound force the water molecules to solidify?&lt;/p&gt;&#xA;" OwnerDisplayName="user1779" LastEditorUserId="21964" LastEditDate="2016-02-25T20:11:12.953" LastActivityDate="2021-04-22T17:18:20.830" Title="Why won't water freeze if you put ice in it, despite ice being frozen?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5268" PostTypeId="1" CreationDate="2013-06-10T02:14:43.083" Score="-3" ViewCount="714" Body="&lt;p&gt;Is there a device that can measure the air in the room in lbs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate the anseer&lt;/p&gt;&#xA;" OwnerDisplayName="user1779" LastActivityDate="2013-06-10T02:41:53.060" Title="Why can't you see air but can feel it" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2013-06-10T19:45:53.007" ContentLicense="CC BY-SA 3.0" />
  <row Id="5269" PostTypeId="2" ParentId="5267" CreationDate="2013-06-10T02:28:13.330" Score="1" Body="&lt;p&gt;At room temperature, water is a liquid. So the ice will gradually turn into liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask your question in a different way:  if you add water to ice, why doesn't the ice turn into liquid?  It does, at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add water to ice, at freezing temperature (Antarctica), the water turns to ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is mostly due to the surrounding air temperature.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-10T02:28:13.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5270" PostTypeId="2" ParentId="5268" CreationDate="2013-06-10T02:36:13.397" Score="2" Body="&lt;p&gt;A scale.  Consider an experiment where a container of &quot;air&quot; is submerged into liquid helium (just about the coldest stuff known.)  This will turn the air into solid $\ce{N2}$ and solid $\ce{O2}$ which can then be weighed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you knew the area of the room, then you would know the &quot;pounds per square inch&quot; or &quot;p.s.i.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the above answer, I think, is a little cheeky.  I think it is usually done via the ideal gas law:  $ P V = n R T$, and solving for $n$, the number of moles of gas. Then through Avogadro's number, one can deduce the number of molecules.  The molecules (e.g. $\ce{N2}$) have a known molecular weight, which can be expressed in pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A barometer, a thermometer, and a ruler are the only three devices one would need to get the number of pounds of air. &lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-10T02:41:53.060" LastActivityDate="2013-06-10T02:41:53.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5271" PostTypeId="1" AcceptedAnswerId="5272" CreationDate="2013-06-10T06:41:07.763" Score="0" ViewCount="7472" Body="&lt;p&gt;The volume of a mole of gas is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{22.4 l}$&lt;/li&gt;&#xA;&lt;li&gt;directly proportional to pressure and Kelvin temperature.&lt;/li&gt;&#xA;&lt;li&gt;directly proportional to pressure, inversely proportional to Kelvin temperature. &lt;/li&gt;&#xA;&lt;li&gt;inversely proportional to pressure, directly proportional to Kelvin temperature.&lt;/li&gt;&#xA;&lt;li&gt;inversely proportional to pressure and to Kelvin temperature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I said the answer is 3 because the ideal gas law is $PV=nRT$ and &quot;proportional to&quot; is the same as saying which ones are on opposite sides. So $n$ and $P$ are on opposite sides and temp is on the same side as $n$.  But I'm wrong and the answer is not 3, but I don't know why.&lt;/p&gt;&#xA;" OwnerUserId="384" LastEditorUserId="5026" LastEditDate="2018-02-25T02:46:19.563" LastActivityDate="2018-02-25T02:47:22.820" Title="How does the volume of a mole of ideal gas relate to pressure and temperature?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5272" PostTypeId="2" ParentId="5271" CreationDate="2013-06-10T06:53:34.513" Score="2" Body="&lt;p&gt;Your starting point of reasoning is good, but you misread the equation. The question is about the volume of 1 mole of gas so let's rewrite the ideal gas law to give this relation:&#xA;$$P V = n R T \rightarrow \frac{V}{n}=\frac{R T}{P} $$&#xA;Now we have the volume of a mole of gas $\frac{V}{n}$ on the left hand side so we can see the proportionalities on the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of a mole of gas is inversely proportional to the pressure and directly proportional to the temperature, i.e. &lt;strong&gt;4&lt;/strong&gt; in your list of answers&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="5026" LastEditDate="2018-02-25T02:46:53.717" LastActivityDate="2018-02-25T02:46:53.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5273" PostTypeId="2" ParentId="5267" CreationDate="2013-06-10T06:57:41.967" Score="22" Body="&lt;p&gt;&lt;code&gt;Why won't water freeze if you put ice in it?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will, even at room temperature. You just need a big enough, cold enough ice cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second law of thermodynamics we know that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Heat cannot spontaneously flow from a colder location to a hotter location.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Isolated_system&quot;&gt;isolated system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorDisplayName="user7951" LastEditDate="2015-04-25T18:03:48.493" LastActivityDate="2015-04-25T18:03:48.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5274" PostTypeId="2" ParentId="5271" CreationDate="2013-06-10T07:20:38.183" Score="0" Body="&lt;p&gt;According to Boyle's law, at constant temperature, pressure is inversely proportional to volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Charles law, at constant pressure, volume is directly proportional to temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the above and the Avagadro's law, we get the ideal gas equation.&#xA;Hence, the correct answer is 4.&lt;/p&gt;&#xA;" OwnerUserId="1763" LastEditorUserId="5026" LastEditDate="2018-02-25T02:47:22.820" LastActivityDate="2018-02-25T02:47:22.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5275" PostTypeId="2" ParentId="5220" CreationDate="2013-06-10T09:10:16.470" Score="5" Body="&lt;p&gt;The old system of keeping things cool utilized the evaporation of water, which takes its latent heat from the item you want to keep cool (milk, butter etc. normally). Either using a cloth soaked in water (and dipping into water at its base) or a coarse earthenware vessel, it worked.&lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2013-06-10T09:10:16.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5276" PostTypeId="2" ParentId="5142" CreationDate="2013-06-10T09:14:11.717" Score="1" Body="&lt;p&gt;Gas sensors are probably too specific. Infrared imaging is going to have to take account of the carboy material itself. However, sampling the air and performing a gc/ms analysis could work - basically its headspace analysis. &lt;/p&gt;&#xA;" OwnerUserId="806" LastActivityDate="2013-06-10T09:14:11.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5277" PostTypeId="1" CreationDate="2013-06-10T16:47:48.350" Score="3" ViewCount="1328" Body="&lt;p&gt;This might be a slightly daft question as I'm a physicist rather than a chemist, but I have a slight problem in using Henry's law which I think I can circumvent using the ideal gas law – I'd be grateful if anyone can confirm or deny my logic! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an equation that yields the volume of oxygen gas per unit mass of tissue. We'll call this quantity $C$ with S.I units of $\mathrm{m^3\: kg^{-1}}$. We also have the partial pressure $p$ measured in $\mathrm{mmHg}$ and we wish to relate the two; several texts have suggested using Henry's law of the form $C = kp$ where $k$ is Henry's constant for oxygen. However, this causes a problem, as I can find no form of Henry's law with the correct units to allow this conversions (units of $\text{vol} / ({\text{pressure} \times \text{mass}})$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is does such a series of constants exist and does anyone know where?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interim, I decided to try to circumvent this problem by using the ideal gas law, which I believe holds at the relatively low pressures in very shallow water. Knowing one mole of oxygen gas has a mass of $32\:\mathrm{g}$, and occupies a volume of $22.4\:\mathrm{l}$ at standard temperature and pressure of $760\:\mathrm{mmHg}$, I re-wrote the expression for the mass of $\ce{O_2}$ in a given volume; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\ce{O_2}} = \frac{0.024\,p}{(0.032)760RT} = \frac{0.7\,p}{760RT}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ideal gas constant, and a temperature of $310.15\:\mathrm{K}$, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\ce{O_2}} = 3.572 \times 10^{-7} \cdot p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I define $M_{\ce{O_2}}$ as the unitless mass fraction of $m_{\ce{O_2}}$ per unit mass, then knowing the density of oxygen gas is around $1.331\:\mathrm{kg/m^3}$, I can divide this into $M_{\ce{O_2}}$ to get the expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C = 2.6835 \times 10^{-7} \cdot p$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the correct units. My second question is then is this valid, and if not, why not?&lt;/p&gt;&#xA;" OwnerUserId="1781" LastEditorUserId="61" LastEditDate="2013-06-10T23:04:41.220" LastActivityDate="2013-08-10T06:43:10.460" Title="Forms of Henry's law versus ideal gas law" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5278" PostTypeId="2" ParentId="4972" CreationDate="2013-06-10T20:57:50.747" Score="6" Body="&lt;p&gt;Firstly I recommend you to understand the concept of &quot;Magnetic Susceptibility&quot;. My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Ni&lt;sup&gt;2+&lt;/sup&gt;, its electronic structure is 4s&lt;sup&gt;0&lt;/sup&gt;3d&lt;sup&gt;8&lt;/sup&gt;. This is a d&lt;sup&gt;8&lt;/sup&gt; ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you should know gyromagnetic ratio which is the ratio of magnetic dipole moment to angular momentum of the system. (We got a system where quantum mechanics may be obeyed). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, considering orbitals and electrons that fill these orbitals we got two magnetic moment here. Spin magnetic moment(sum of it is S) and angular magnetic moment(sum of it is L).&#xA;I don't know properly write all calculations with codes. So I attached an image. &lt;img src=&quot;https://i.stack.imgur.com/rLcrQ.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="792" LastActivityDate="2013-06-10T20:57:50.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5279" PostTypeId="2" ParentId="5277" CreationDate="2013-06-10T21:18:44.057" Score="1" Body="&lt;p&gt;I wrote out an answer to this but realized I was wrong. But I've noticed a number of things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, your form of the ideal gas law has the Molar Mass inverted. If you make the substitution that $n=m/M$, then the rearrangement reads $m = MVp/RT$, you have the $M$ on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, You are selecting a Volume of 1 Liter, so your calculation will solve for a relation of mass and pressure @ 1L, which I don't think you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly (?), I'm guessing that the 760 in the denominator is there so you can use a gas constant in terms of atm? Or did I miss something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So those are my observations about your attempt with the ideal gas law. I also am concerned over the bit with the mass fraction, as I think you converted a tissue mass into an $\ce {O2}$ volume using $\ce {O2}$'s density, which would be wrong if that was the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking again at Henry's law in the $p=kC$, you can see that in order to use the standard k units available, you would need to convert the $C$ you have from $L/kg$ to $mol/L$. You can convert $L$ to $mol$ using the ratio you used earlier, and you end up with $C$ in terms of $mol/kg_{tissue}$. The trick here is that you need the density of the tissue (not of $\ce {O2}$) to convert this into the volume you're looking for in order to use the equation. I think haha.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="2095" LastEditDate="2013-08-10T06:43:10.460" LastActivityDate="2013-08-10T06:43:10.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5286" PostTypeId="1" AcceptedAnswerId="5309" CreationDate="2013-06-11T08:15:30.380" Score="5" ViewCount="4176" Body="&lt;p&gt;Recently the water cooler was replace with a new filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I refill my water kettle with the water from the water cooler, there are many bubbles and the taste was a bit like soap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical cause such effects? (Is it related to the new filter?)&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="1362" LastEditDate="2013-06-11T19:51:37.997" LastActivityDate="2013-07-29T22:01:03.220" Title="What is the substance in the new water cooler filter that give bubbles and taste a bit like soap" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5287" PostTypeId="1" CreationDate="2013-06-11T14:42:48.747" Score="3" ViewCount="811" Body="&lt;p&gt;There is a poor quality photo in the diagram of a manometer below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UoLdT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture is too poor for me to understand what they mean by &quot;glass pointer marks calibrated volume&quot;.  I would be interested if anyone (a) knows what this is - preferably with a manufacturer's number so I can inquire further or (b) has a higher quality image of the part.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-06-23T02:39:54.397" Title="Parts of a manometer" Tags="&lt;equipment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5288" PostTypeId="1" AcceptedAnswerId="5459" CreationDate="2013-06-11T14:53:39.673" Score="0" ViewCount="267" Body="&lt;p&gt;I was wondering if there is any other way to crystallize materials from a solution other than using supersaturation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="1787" LastActivityDate="2013-06-26T16:59:26.713" Title="Crystallization from a solution" Tags="&lt;organic-chemistry&gt;&lt;purification&gt;&lt;recrystallization&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5289" PostTypeId="1" AcceptedAnswerId="5294" CreationDate="2013-06-11T18:37:37.513" Score="8" ViewCount="1245" Body="&lt;p&gt;I'm attempting to extract the diels alder reaction into a generic reaction and am trying to gain a better understanding of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a conjugated diene react with another conjugated diene via the diels alder reaction? And is the reaction limited to only dienes or can it be expanded to polyenes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone happen to have a link as to the details on hetero diels alder reactions and the structures that can undergo such reactions and limitations on chemical species that can react via this reaction.&lt;/p&gt;&#xA;" OwnerUserId="1788" LastEditorUserId="305" LastEditDate="2013-06-12T02:06:11.013" LastActivityDate="2021-05-24T13:48:40.357" Title="A few questions about the conditions of the Diels Alder Reactions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5290" PostTypeId="2" ParentId="5289" CreationDate="2013-06-11T19:50:02.963" Score="1" Body="&lt;p&gt;I'm speaking from memory, so this could be a little loose, but I believe dienes react just as well with other dienes via the diels alder reaction as with simple alkenes (of course, only attacking one of the conjugated double bonds). As for polyenes, I believe the same is true, but I'm not certain if there are added complexities there, and it would probably have to be evaluated on a case by case basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the disclaimer is that it's possible that the configuration of double bonds may or may not lock in the polyene into a planar or partially planar structure, in which case depending on how many other double bonds and their locations in one or both of the reagents could interfere with the traditional diers alder reaction. But as to your question, yes, you can apply the reaction to a general case as far as I know. &lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-11T19:50:02.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5291" PostTypeId="2" ParentId="5288" CreationDate="2013-06-11T20:07:09.120" Score="1" Body="&lt;p&gt;Not really. By definition, if some solute is in a solution below the saturation point at that temperature, it's going to remain in solution. The only way to get it out is to lower the temperature or evaporate some of the solvent to increase the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't care especially about the integrity of the chemical, you could always precipitate it with some other reagent, but then of course your crystals would not be of the chemical you started with, but perhaps could be purified through some other means...&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-11T20:07:09.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5293" PostTypeId="1" AcceptedAnswerId="5295" CreationDate="2013-06-11T21:33:44.907" Score="8" ViewCount="11096" Body="&lt;p&gt;In an esterification reaction between a carboxylic acid and an alcohol, a carboxylate ester and water is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the following diagram online, and my textbook has a similar one as well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eH9uy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In forming the water in an esterification of a carboxylic acid, why does the hydrogen ion come from the alcohol and the hydroxide come from the carboxylic acid and not the other way around?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it might be because carbon's bond to the OH&lt;sup&gt;-&lt;/sup&gt; is weaker than the R bond to the H&lt;sup&gt;+&lt;/sup&gt; ion and requires less energy to remove those two ions from their respective compounds.&lt;/p&gt;&#xA;" OwnerUserId="1790" LastEditorUserId="4945" LastEditDate="2014-07-17T10:40:31.503" LastActivityDate="2019-10-29T09:09:20.203" Title="Esterification of a carboxylic acid and the production of water" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5294" PostTypeId="2" ParentId="5289" CreationDate="2013-06-11T22:31:19.220" Score="7" Body="&lt;p&gt;Yes, conjugated dienes can react with other conjugated dienes in Diels-Alder reactions. A well known example of this is the dimerization of cyclopentadiene. All the standard caveats, restrictions, and other considerations still apply, namely:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The diene must not be rotationally locked in such a way that it cannot adopt an s-cis/synperiplanar conformation. As a logical corollary, cyclic dienes that are already locked into the correct s-cis conformation often have lower activation barriers.&lt;/li&gt;&#xA;&lt;li&gt;The kinetics of the reaction are accelerated by electron-withdrawing substituents on the dienophile and electron-donating substituents on the diene.&lt;/li&gt;&#xA;&lt;li&gt;The stereochemistry of the major product is highly substrate-dependent. A preference for the &lt;em&gt;endo&lt;/em&gt; isomer exists in cases where the dienophile has unsaturated substituents. In those cases, the transition state is stabilized by an overlap of the $\pi$-electron system between said unsaturated substituents and the $\pi$ orbitals of the newly forming double-bond at the back of the diene. (Just as a side note, Carey &amp;amp; Sundberg refer to this as the &quot;Alder rule&quot; in &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt;, but I've never seen that term used anywhere else.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In principle, there's no reason why two conjugated $\pi$-bonds of a polyene cannot react similarly, provided they satisfy the orbital symmetry and conformational requirements and the reaction kinetics are favorable. Even aromatic molecules can undergo Diels-Alder reactions under the right set of conditions (see, for example, the Diels-Alder reaction between benzene and furan via a benzyne intermediate). There are also numerous instances of well-documented Diels-Alder reactions involving heteroatomic dienophiles, alkynes as dienophiles, polymerization reactions, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get your hands on the aforementioned Carey &amp;amp; Sundberg book, they discuss the Diels-Alder reaction at length (including relatively rigorous explanations of the stereospecifity and reaction kinetics via molecular orbital theory). If you're looking for a comprehensive reference, I doubt there's anything superior to the venerable &lt;em&gt;March's Advanced Organic Chemistry&lt;/em&gt;. There's a chapter on Diels-Alder reactions (along with virtually every other organic chemistry reaction known to science), as well as a massive list of accumulated primary literature references if further detail is required.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-06-12T00:07:14.270" LastActivityDate="2013-06-12T00:07:14.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5295" PostTypeId="2" ParentId="5293" CreationDate="2013-06-11T23:31:18.350" Score="6" Body="&lt;p&gt;The mechanism of the reaction doesn't involve the direct exchange of ions in the way that, e.g., a metathesis/double replacement reaction might. Take a careful look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aFischer_esterification_mechanism.svg&quot;&gt;reaction mechanism from the Wikipedia page&lt;/a&gt;. Notice that one of the intermediates (the product after step three) is a tetrahedral complex with two hydroxyl ($\ce{-OH}$) groups. One of those hydroxyls is indeed the original hydroxyl of the carboxylic acid moiety, but the other one is actually formed from the protonated oxygen of the carbonyl. Either one of them could, with equal likelihood, be protonated in a subsequent step to provide the water molecule that serves as a leaving group. In the image that you linked, it's convenient to circle those specific atoms, but that shouldn't lead you to believe that the mechanism of the reaction is in fact a simple exchange of ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question that you didn't ask directly, but perhaps implied, is: why is protonation of the alcohol by the carboxylic acid, followed by nucleophilic attack by the carboxylate on the carbon of the protonated alcohol, not a viable mechanism? First, a protonated alcohol is more acidic than a carboxylic acid by six or seven orders of magnitude, which means that the initial proton transfer would be slow and unfavorable. Second, the resulting ester is apt to be hydrolyzed eventually (albeit slowly without catalysis, but then the whole reaction proceeds very slowly), yielding the original carboxylic acid and alcohol.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-06-11T23:31:18.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5296" PostTypeId="1" AcceptedAnswerId="5297" CreationDate="2013-06-12T00:00:00.067" Score="2" ViewCount="10195" Body="&lt;p&gt;I need help on identifying precipitation, acid-base, and redox reactions. I know what each one is in theory:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Precipitation: Creating a solid&lt;/li&gt;&#xA;&lt;li&gt;Acid-Base: neutralization&lt;/li&gt;&#xA;&lt;li&gt;Redox: exchange of electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But given a chemical reaction like:&#xA;HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to understand how to classify them. Are there any tell-tale signs to look out for when determining. For example, does a precipitation reaction ALWAYS produce a &lt;code&gt;(s)&lt;/code&gt; as a product and therefore any reaction with a &lt;code&gt;(s)&lt;/code&gt; on the product side is a precipitation reaction?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="11367" LastEditDate="2017-01-14T14:49:03.407" LastActivityDate="2017-01-14T14:49:03.407" Title="Identifying Reaction Types" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5297" PostTypeId="2" ParentId="5296" CreationDate="2013-06-12T00:38:14.037" Score="4" Body="&lt;p&gt;It's true that a precipitation reaction will always yield at least one solid product, but not every reaction that yields a solid is, strictly speaking, properly categorized as a precipitation reaction. Precipitation reactions are ones in which at least one of the reactants is in the aqueous phase (i.e., dissolved in water), and a solid forms on the product side which was not present on the reactant side. If all the reactants are solid, then a solid product forming should not be called a precipitate. To be called a precipitate, an insoluble product must form from within solution, either from a solid and solute interacting, or from an interaction strictly between solutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of precipitation reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -&amp;gt; AgCl_{(s)} + NaNO3_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuCl2_{(aq)} + Zn_{(s)} -&amp;gt; ZnCl2_{(aq)} + Cu_{(s)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that at least one of the reactants was in the aqueous phase, and a new solid chemical species is produced on the product side. $\ce{AgCl_{(s)}}$ and $\ce{Cu_{(s)}}$, respectively, are the precipitates. Incidentally, the second reaction is also an example of a redox reaction, since it involved a change in the oxidation states of the reactants (here, zinc is oxidized while copper is reduced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Identifying acid-base and redox reactions is a lengthier topic, so I suggest you consult a textbook or any of the numerous online resources.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-06-12T00:38:14.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5299" PostTypeId="1" AcceptedAnswerId="15696" CreationDate="2013-06-12T05:05:54.373" Score="9" ViewCount="82954" Body="&lt;p&gt;The boiling point of water is $\pu{100^\circ C}$. However, we find that even at room temperature, water can evaporate into gas.&#xA;Therefore, water vapor can exist at temperatures of, say, $\pu{50^\circ C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what temperature, can you say for certain that all water vapor will turn into liquid?&#xA;In other words, what is the condensation point of water?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="4945" LastEditDate="2017-11-01T15:45:15.460" LastActivityDate="2018-05-16T16:37:43.533" Title="What is the condensation point of water?" Tags="&lt;water&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5300" PostTypeId="2" ParentId="5299" CreationDate="2013-06-12T19:31:37.883" Score="5" Body="&lt;p&gt;From a chemical engineering perspective (where we do a lot with steam), the answer is dependent on if you're in a closed system or not. If I put some water in a closed container, it would evaporate only enough such that the gaseous water would reach the vapor pressure at that specific temperature. All temperatures above $\pu{0 K}$ have a non-zero vapor pressure, so you could say that water evaporates at all temperatures, at least for a while, and then it will be in equilibrium with the liquid (or solid) state. However, if you are not in a closed system (say, outside with a cup of water), then the water will continue to evaporate to reach the vapor pressure, but the gas will continue to escape, so the water will never stop evaporating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite, and the answer to your question, is also true. If you have circumstance when the surrounding pressure due to water vapor is higher than the vapor pressure of water at that temperature (like a rain cloud cooling down), then the water will start to precipitate. Of course, like all equilibrium, even if the current net phase shift is from gas towards liquid water, some of the liquid will still be evaporating - it's just that more gas will be condensing.&lt;/p&gt;&#xA;" OwnerUserId="518" LastEditorUserId="4945" LastEditDate="2017-11-01T15:49:01.407" LastActivityDate="2017-11-01T15:49:01.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5301" PostTypeId="1" AcceptedAnswerId="5303" CreationDate="2013-06-12T20:41:10.057" Score="19" ViewCount="8617" Body="&lt;p&gt;I've been searching around the internet for a while and I know that Chlorine, Bromine and Iodine are used as disinfectants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the property of the halogens that make them suitable for killing microbes? Is it just their toxicity?&lt;/p&gt;&#xA;" OwnerUserId="1220" LastEditorUserId="305" LastEditDate="2013-06-13T06:37:25.777" LastActivityDate="2015-06-01T19:12:02.903" Title="Why are the halogens good disinfectants?" Tags="&lt;biochemistry&gt;&lt;toxicity&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5302" PostTypeId="1" CreationDate="2013-06-13T02:22:42.460" Score="1" ViewCount="9934" Body="&lt;p&gt;This is a quote from my Chemistry textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitric acid ionises according to the following equation:&#xA;  $$\ce{HNO3(l) + H2O(l) -&amp;gt; H3O+(aq) + NO3- (aq)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the formula for nitric acid is $\ce{HNO3}$. But I cannot understand what is the product of this equation. What is $\ce{H3O+(aq) + NO3- (aq)}$? Is it a diluted acid with less concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another quote (from my textbook) that confuses me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitric acid and sulfuric acid are not quite as strong as hydrochloric acid. Chemical and spectroscopic studies have shown that concentrated solutions of &lt;strong&gt;both these acids do contain some unionised molecules.&lt;/strong&gt; For our purpose, however, their ionisation is &lt;strong&gt;effectively complete&lt;/strong&gt;. This is particularly true in dilute solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot understand the part that is in the bold. What are these unionised molecules? I know that these acids are considered &lt;strong&gt;strong acids&lt;/strong&gt; which &lt;strong&gt;completely ionise&lt;/strong&gt; in water solution. If that is the case, then why does the text indicate that there are &lt;strong&gt;some unionised molecules&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="5026" LastEditDate="2018-03-12T06:19:38.450" LastActivityDate="2018-03-12T06:19:38.450" Title="Which one is nitric Acid, HNO3 by itself, or when it is included in H2O?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5303" PostTypeId="2" ParentId="5301" CreationDate="2013-06-13T02:25:42.430" Score="17" Body="&lt;p&gt;The halogens, particularly in their diatomic free states and within various oxoacids, are strong oxidizing agents by virtue of their high electronegativities, electron affinities, and reduction potentials. The polarizability of the heavier halogens also makes them almost uniquely versatile as both good leaving groups and strong nucleophiles, depending on conditions. Additionally, the relatively low bond dissociation energies of the halogens contribute to their aptitude to react by free-radical mechanisms (coupled with the exothermicity that usually accompanies the propagation of those reactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their diatomic forms, they readily add across the $\pi$-bonds of alkenes and alkynes (which are ubiquitous and essential structural features of myriad types of biomolecules), and they can oxidize numerous other functional groups under the right conditions and in combination with other substances. Various oxoacids that incorporate halogens are often extremely strong oxidizers as well (household bleach, for example, is an aqueous solution of sodium hypochlorite, which is mild by comparison to certain other halogen-containing oxoacids). In those reactions, they typically serve to add oxygen to molecules in place of, e.g., hydrogen, while themselves serving as leaving groups (or parts of leaving groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The antiseptic properties of halogens (and halogen-containing oxoacids) are directly consequent to those chemical properties. At least &lt;a href=&quot;http://books.google.com/books?id=tFPW4D70BmgC&amp;amp;pg=PA149&amp;amp;lpg=PA149#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;one source&lt;/a&gt; suggests that elemental iodine, for example, reacts by mechanisms similar to the ones described above to disrupt key structures of bacterial enzymes, proteins, and lipid membranes: it oxidizes the thiol groups of cysteine or methionine to disulfides, interfering with protein synthesis; it reacts with unsaturated lipids to alter their structure; etc. These are biological equivalents of the very same reactions that are routinely carried out in organic chemistry labs. If you want more details/examples, they should be easy to find with a web search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the very same awesome chemical reactivity that makes them so indispensable in organic synthesis, also means that they are highly reactive towards (and hence, usually harmful to) all manner of organisms.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-06-13T02:25:42.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5304" PostTypeId="2" ParentId="5302" CreationDate="2013-06-13T02:55:14.847" Score="3" Body="&lt;p&gt;In your first reaction, $\ce{H2O}$ acts as a base to abstract an $\ce{H+}$ from the nitric acid. The resulting $\ce{H3O+}$, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;nofollow&quot;&gt;hydronium&lt;/a&gt;, is the conjugate acid, while the $\ce{NO3-}$, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrate&quot; rel=&quot;nofollow&quot;&gt;nitrate&lt;/a&gt;, is the conjugate base (this is the molecule of nitric acid, but stripped of its $\ce{H+}$). You could view this as a dilution, in the sense that the resulting solution is less acidic than highly concentrated nitric acid would be, but the process of dilution of acid by mixing with water inevitably involves an acid-base reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your second question, the unionized molecules are the neutral water and nitric acid on the reactant side, while the ions are the products resulting from the acid-base reaction. Nitric acid is relatively strong, meaning it tends to ionize &lt;em&gt;almost completely&lt;/em&gt; in aqueous solution. Additionally, the less acidic the solution already is prior to addition of the acid, the more extensively the added acid will dissociate into ions (to understand this, please read up on acid-base equilibria and Le Chatelier's principle, as the topics are too lengthy and complex to cover adequately here). This is why the author emphasizes that the dissociation is more complete in dilute solutions. Statistically, however, some proportion of the nitric acid molecules will inevitably not dissociate into ions. The stronger an acid, the greater the extent of dissociation, but there is always some statistical chance that a proportion of the molecules will remain unreacted. For strong acids, however, the overwhelming preponderance of the molecules will dissociate, while only a tiny fraction remain unreacted. Hence, for many (if not most) purposes, it's reasonable when dealing with strong acids to proceed &lt;em&gt;as if&lt;/em&gt; the dissociation is truly complete, since the unreacted portion is typically negligible in quantity.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-06-13T02:55:14.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5305" PostTypeId="2" ParentId="5302" CreationDate="2013-06-13T02:59:17.047" Score="-1" Body="&lt;p&gt;The product is simply the ionized form of $\text{HNO}_3$; if you prefer it is equally correct to write $$\text{HNO}_3(aq)\rightarrow \text{H}^+(aq)+\text{NO}_3^-(aq)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that technically nitric acid is $\text{HNO}_3(aq)$and not $\text{HNO}_3(l)$ (which would presumably be the salt hydrogen nitrate in liquid form) As hydrogen nitrate will simply decompose under standard conditions if not dissolved so $\text{HNO}_3$ is sufficient to refer to nitric acid; however, it is understood to refer to $\text{HNO}_3$ dissolved in water, not just $\text{HNO}_3$ alone. &lt;/p&gt;&#xA;" OwnerUserId="1719" LastActivityDate="2013-06-13T02:59:17.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5308" PostTypeId="1" AcceptedAnswerId="5312" CreationDate="2013-06-13T15:51:43.277" Score="4" ViewCount="1742" Body="&lt;p&gt;I am planning an experiment for a chemistry class about corrosion. I am going to show how different prevention methods affects the corrosion of iron and aluminium parts using bleach as the oxidizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the methods I plan on is protecting the parts using cathodic protection, the iron nail will be protected by aluminium but I don't know a good, easily available metal to protect the  aluminium stud. Any ideas? &lt;/p&gt;&#xA;" OwnerUserId="1796" LastEditorUserId="73" LastEditDate="2013-12-15T07:12:26.137" LastActivityDate="2017-01-15T05:47:44.290" Title="Cathodic protection of aluminium" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5309" PostTypeId="2" ParentId="5286" CreationDate="2013-06-13T18:04:55.227" Score="4" Body="&lt;p&gt;The most likely cause of a &quot;soapy&quot; taste and sudsing from filtered water is a reverse osmosis filter, for which the usual &lt;a href=&quot;http://www.wateranywhere.com/Water_Filtration_Products/index.php?cPath=386_389&amp;amp;osCsid=c9ffc85fb9c8ea376d6e14dd028eb446&quot; rel=&quot;nofollow&quot;&gt;membrane preservative&lt;/a&gt; used to keep the filter from degrading before its first use is plain old glycerin, a common ingredient in and byproduct of traditional soap (though most cleansing products nowadays, even those marketed as soaps, are technically anionic surfactant &quot;detergents&quot;, based on alkyl-sulfates). The glycerin is hygroscopic, meaning it will attract and hold water on the membrane surface to keep it from drying out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An activated carbon filter sometimes has a similar preservative to improve the material's shelf life, or else a binder that allows the manufacturer to mold a block of the carbon rather than loading loose powder into filter housings, which reduces carbon dust in the factory (a &lt;em&gt;major&lt;/em&gt; fire and health hazard of these types of plants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, the manufacturer usually recommends flushing any new filter with between one and five gallons of source water to wash out these substances, along with any loose filter media or other byproducts of manufacturing. This obviously didn't happen in your case. The glycerin in the water won't kill you; it's an FDA-approved food additive used as a sweetener and &quot;plasticizer&quot; (it's used in cake frosting for instance, to avoid it hardening after it sets). In sufficient doses it's an irritant to the mucous membranes and the GI tract, which can cause nausea and diarrhea, but you'd have to eat almost a kilo of the pure stuff for it to become dangerously toxic.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-07-29T22:01:03.220" LastActivityDate="2013-07-29T22:01:03.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5310" PostTypeId="2" ParentId="5308" CreationDate="2013-06-13T18:10:58.257" Score="0" Body="&lt;p&gt;Of the metals lower in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow&quot;&gt;Standard Electrode Potential list&lt;/a&gt;, Magnesium is probably easiest to come by and least problematic to handle.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-06-13T18:10:58.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5311" PostTypeId="2" ParentId="5267" CreationDate="2013-06-13T19:06:54.127" Score="5" Body="&lt;p&gt;Water, like most substances, requires more energy gain or loss to undergo a state change (from solid to liquid, liquid to gas) than to be warmed or cooled by a degree of temperature without changing state. Liquid water requires about 4.2 Joules per milliliter per degree Celsius to raise or lower its temperature. However, the Latent Heat of Fusion (L&lt;sub&gt;f&lt;/sub&gt;) of water (the energy that must be added to ice to melt it or removed from water to solidify it at the freezing point) is 334 Joules per milliliter.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, for ice to freeze the surrounding water, the ice must be cold enough that it can afford to gain enough energy to invoke this state change, without melting itself. Now, solid ice has a lower volumetric heat capacity; it only takes about 1.9 J/(mL * K) to change its temperature. So, for an ice cube, let's say 10mL in volume, to freeze 1mL of the surrounding water (at 0°C) without melting itself, the ice must be able to absorb 334J of energy without melting. That means the ice must be -16.7°C, or about 2°F.&lt;/p&gt;&#xA;&lt;p&gt;That's well within the range of your average freezer (which in the US is normally kept around 0°F to ensure everything stays good and frozen), but that back-of-the-envelope calculation assumes the energy is gained evenly throughout the entire ice cube. This doesn't happen; it's common knowledge that ice melts from the outside in, and that's because it's a relatively poor conductor of heat compared to substances like metals, which sometimes seem to instantly liquefy when they reach their melting point. For the ice to be able to cool that kind of water volume without melting, therefore, it must be much colder; cold enough that the ice can afford to gain enough heat just in a few millimeters of its outer surface layer to freeze the water.&lt;/p&gt;&#xA;&lt;p&gt;Regardless of this, if you have a fridge with in-door ice and water, and you fill a cup with the ice and then pour the water over it, the ice will clump; the combination of two cubes in close proximity, a small volume of already-cold liquid water in a thin layer between them, and the ice being colder than freezing, results in the ice freezing the water, joining the two cubes. This happens throughout the ice. If you go with crushed ice, which has smaller pieces but more surface area and a larger volume of frozen water to total water, the effect is even more pronounced. But, eventually, the overall thermal energy of the room adds more energy into the glass of water than the ice can absorb without melting. So, it melts.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2021-04-22T17:18:20.830" LastActivityDate="2021-04-22T17:18:20.830" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5312" PostTypeId="2" ParentId="5308" CreationDate="2013-06-14T00:31:11.813" Score="2" Body="&lt;p&gt;You might want to look at &lt;a href=&quot;http://www.npl.co.uk/upload/pdf/cathodic_protection_in_practise.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; which has a nice general discussion about cathodic protection, with an abbreviated activity series for the metals of interest in cathodic protection.  Typically, aluminum, magnesium and zinc are used as sacrificial anodes for cathodic protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking from a thermodynamic standpoint, one would suspect that magnesium would be needed to protect aluminum.  However &lt;a href=&quot;http://www.duroboat.com/experience/AluminumBoatsandCorrossion.htm&quot; rel=&quot;nofollow&quot;&gt;aluminum boat manufacturers&lt;/a&gt; state that zinc can be used for protecting aluminum boats in salt water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm completely unfamiliar with the die casting industry, but they use Al/Zn alloys which, if affordable, may provide an interesting pedagogical component to your corrosion activity (namely, what do scientists do when the properties of the elements on the periodic table don't quite fit what we have in mind).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-06-14T00:31:11.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5313" PostTypeId="1" AcceptedAnswerId="5341" CreationDate="2013-06-14T03:53:30.607" Score="31" ViewCount="19217" Body="&lt;p&gt;This question is meant for a simple unstabilised ylide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of the Wittig reaction, as given on &lt;a href=&quot;http://www.chemtube3d.com/AsymReagent-ZselectivewittigReaction.html&quot; rel=&quot;noreferrer&quot;&gt;ChemTube3d&lt;/a&gt;, involves a concerted formation of the oxaphosphetane (this is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/65233/&quot;&gt;generally favoured&lt;/a&gt; over the traditional stepwise mechanism with betaine formation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wNnDn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the puckered transition state has been clearly shown but involves the direct formation of the oxaphosphetane (without any betaine intermediate). In this transition state, the two possibilities are for the methyl groups to be either &lt;em&gt;cis&lt;/em&gt; or &lt;em&gt;trans&lt;/em&gt; to each other. Surely the transition state with the methyl groups trans to each other will be more stable. Doesn't this support formation of the trans product?&lt;/p&gt;&#xA;" OwnerUserId="1804" LastEditorUserId="16683" LastEditDate="2017-09-23T15:49:50.813" LastActivityDate="2017-09-23T15:49:50.813" Title="Why does the unstabilised Wittig reaction selectively form cis alkenes?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;wittig-reactions&gt;&lt;stereoselectivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5315" PostTypeId="1" AcceptedAnswerId="5344" CreationDate="2013-06-14T06:11:12.220" Score="4" ViewCount="1079" Body="&lt;p&gt;This question is about delaying/postponing chemical reactions in general, but I will explain my point based on an example. Most of you will know the famous vinegar and baking soda experiment which allows you to make a volcano or a &lt;a href=&quot;http://www.youtube.com/watch?v=lwiosotOlWU&quot; rel=&quot;nofollow&quot;&gt;rocket&lt;/a&gt;. What I have noticed when doing this experiment (in particular the rocket experiment) is that if you just put soda in powder form in the bottle and pour in vinegar, a lot of the formed gas escapes before you can put the cork on thus resulting in a less spectacular launch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the video I linked above they wrap the soda in paper to allow some time between the addition of vinegar and the start of the reaction. This is a nice trick, but in many cases you don't want any paper inside your 'reaction' vessel. My question is: what are my options to chemically postpone the start of the reaction?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-06-16T23:49:42.760" Title="How to chemically postpone a chemical reaction" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5316" PostTypeId="1" AcceptedAnswerId="5332" CreationDate="2013-06-14T11:07:44.327" Score="2" ViewCount="262" Body="&lt;p&gt;I'm currently working with the band structure of graphene and graphene-like compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;α-&lt;a href=&quot;http://en.wikipedia.org/wiki/Boron_nitride&quot; rel=&quot;nofollow&quot;&gt;Boron Nitride&lt;/a&gt; (hexagonal, single layer) has many interesting properties. I was wondering if the next period analogue, a hexagonal single-layered Aluminium Phosphide existed. Wikipedia lists &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_phosphide&quot; rel=&quot;nofollow&quot;&gt;this compound&lt;/a&gt;, but it seems to only be found with a multilayered sphalerite structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does an allotrope of AlP exist (in a stable state at normal conditions) where the structure is similar to graphene or silicene? If not, why?&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="10185" LastEditDate="2015-05-18T12:28:19.683" LastActivityDate="2015-05-18T12:28:19.683" Title="Does single-layered Aluminium Phosphide exist?" Tags="&lt;inorganic-chemistry&gt;&lt;allotropes&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5317" PostTypeId="2" ParentId="5315" CreationDate="2013-06-14T13:28:32.967" Score="1" Body="&lt;p&gt;From what I remember, the powder was kept separate.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The powder was put on a piece of tissue paper.&lt;/li&gt;&#xA;&lt;li&gt;Put the vinegar in the body of the rocket.&lt;/li&gt;&#xA;&lt;li&gt;Put the cork in with the paper wrapped powder.&lt;/li&gt;&#xA;&lt;li&gt;On putting it in place the vinegar gets to the paper/powder.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This brings me back... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could work on the cork to put the powder in it without paper or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you could make a paste and put it in a hole on the cork.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="73" LastEditDate="2013-06-14T23:07:35.057" LastActivityDate="2013-06-14T23:07:35.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5318" PostTypeId="2" ParentId="4378" CreationDate="2013-06-14T15:15:55.693" Score="2" Body="&lt;p&gt;You might also want to consider using &lt;a href=&quot;http://www.uwplatt.edu/chemep/chem/chemscape/labdocs/catofp/measurea/echem/video/01183315.mov&quot; rel=&quot;nofollow&quot;&gt;string&lt;/a&gt; as a disposable salt bridge.  The general procedure would be to cut 6-8 inches of string, twist it and soak in the KCl solution.  &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-06-14T15:15:55.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5321" PostTypeId="1" AcceptedAnswerId="5322" CreationDate="2013-06-15T01:20:31.263" Score="10" ViewCount="2154" Body="&lt;p&gt;I really can't understand why is the mole a unit of measurement and not just a constant. I mean, we use units to express things that we can measure but that are not countable, that are continuous - and not discrete. However, the number of atoms (or whatever) of something is clearly countable, discrete. It is also a huge number, which would request for a constant, a number to multiply to: Avogadro's constant is that number, like mega or kilo in SI for other units. We could still write 14 mole, where mole is the Avogadro constant (exactly the way it is), but every book I read so far states that the mole is a unit of measurement, official in the S.I., and so does &lt;a href=&quot;http://en.wikipedia.org/wiki/International_System_of_Units&quot;&gt;Wikipedia&lt;/a&gt;. It just makes no sense to me that it is considered that way, but there is obviously a reason why... What is it?&lt;/p&gt;&#xA;" OwnerUserId="1809" LastEditorUserId="1362" LastEditDate="2013-06-15T06:36:47.547" LastActivityDate="2016-09-20T10:39:20.273" Title="Why is the mole a unit of measurement?" Tags="&lt;mole&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5322" PostTypeId="2" ParentId="5321" CreationDate="2013-06-15T02:22:28.670" Score="9" Body="&lt;p&gt;I think you are confusing &lt;code&gt;unit of measurement&lt;/code&gt; with &lt;code&gt;fundamental unit&lt;/code&gt;.  A unit of measurement is a &lt;a href=&quot;https://mathematica.stackexchange.com/questions/18683/applying-textrecognize-on-alpha-numerical-table&quot;&gt;definite magnitude of a physical quantity&lt;/a&gt;.  Consider a very common grouping unit, the &lt;code&gt;dozen&lt;/code&gt; which is conceptually the same as a &lt;code&gt;mole&lt;/code&gt; and is a little easier for us to grasp since it is a commonly used small number.  As far as I know, the only requirements for a unit of measurement to be valid is general consensus.  If you and I agree that we will measure distance in units of 'Chemistry Textbook Thickness', then it is a valid unit of measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;code&gt;fundamendal unit&lt;/code&gt; is a part of a particular set of units upon which all &lt;a href=&quot;http://en.wikipedia.org/wiki/Fundamental_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;other units can be derived&lt;/a&gt;.  In this case, the &lt;code&gt;mole&lt;/code&gt; is related to the fundamental unit of &lt;code&gt;mass&lt;/code&gt; and is defined as an amount of items equal to the amount of carbon atoms in 12 grams of $^{12}C$.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2017-04-13T12:56:10.427" LastActivityDate="2013-06-15T02:22:28.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5323" PostTypeId="1" CreationDate="2013-06-15T03:05:44.703" Score="1" ViewCount="2421" Body="&lt;p&gt;I would like to know if formic acid (concentrated to 85%) is stable (no chemical reaction) when frozen.&#xA;Is it a good solution to store it for many years?&lt;/p&gt;&#xA;" OwnerUserId="1810" LastEditorUserId="10185" LastEditDate="2015-03-25T13:17:30.967" LastActivityDate="2015-03-25T13:17:30.967" Title="is it possible to store formic acid longer if frozen?" Tags="&lt;acid-base&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5324" PostTypeId="1" AcceptedAnswerId="5334" CreationDate="2013-06-15T06:03:06.713" Score="6" ViewCount="3801" Body="&lt;p&gt;Why doesn't cyclohexene have a cis-trans isomer whereas cyclodecene has one?&#xA;Is strain a factor?&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorUserId="30" LastEditDate="2013-07-16T14:47:53.843" LastActivityDate="2016-03-05T19:51:22.973" Title="Geometrical isomerism in cycloalkenes" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5326" PostTypeId="1" AcceptedAnswerId="5335" CreationDate="2013-06-15T14:04:07.473" Score="-1" ViewCount="1205" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which ions do not react with each other in solution?&lt;/p&gt;&#xA;&lt;p&gt;A. &lt;span class=&quot;math-container&quot;&gt;$Ba^{2+},SO_4^{2-},NO_3^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;B. &lt;span class=&quot;math-container&quot;&gt;$Cu^{2+},SO_4^{2-},OH^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;C. &lt;span class=&quot;math-container&quot;&gt;$Ag^+,Cl^-,H^+$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;D. &lt;span class=&quot;math-container&quot;&gt;$K^+,Fe^{3+},Br^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have no idea why the answer is &lt;strong&gt;D&lt;/strong&gt;. Please help.&lt;/p&gt;&#xA;" OwnerUserId="1617" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-10-28T11:14:45.453" Title="Which ions do not react with each other in a solution?" Tags="&lt;ions&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2014-10-28T13:20:52.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="5327" PostTypeId="1" AcceptedAnswerId="5329" CreationDate="2013-06-15T16:02:56.967" Score="2" ViewCount="171" Body="&lt;p&gt;I am getting confused about this the more problems I do on it. My understanding was that only strong acids and strong bases will react to produce water and a salt. Something like this:&lt;br&gt;&lt;br&gt;&#xA;Molecular Eq: $\ce{HCl(aq) + NaOH(aq) \rightarrow H2O(l) + NaCl}$&lt;br&gt;&#xA;Net Ionic Eq: $\ce{H+(aq) +OH- (aq) \rightarrow H_2O(l)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Makes perfect sense, both the acid and base break apart and combine with each other. But then I ran across something like this:&lt;br&gt;&#xA;Molecular Eq: $\ce{2 CH3CO2H (aq) + Ba(OH)2 (aq) → Ba(CH3CO2)2 (aq) + 2 H2O (l)}$&lt;br&gt;&#xA;Net Ionic Eq: $\ce{2CH3CO2H (aq) + 2OH- (aq) → 2CH3CO2- (aq) + 2H2O (l)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net ionic eq tells me that $\ce{CH3CO2H}$ will not break apart, but the molecular equation tells me that they will break apart and form a salt with $\ce{Ba}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm so confused by this. Why are they giving two opposing pieces of information? Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="7448" LastEditDate="2015-07-12T19:33:00.887" LastActivityDate="2017-04-06T16:05:57.443" Title="How to solve this contradiction?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5329" PostTypeId="2" ParentId="5327" CreationDate="2013-06-15T19:23:48.903" Score="8" Body="&lt;p&gt;The idea behind this is that a weak acid like $\ce{CH_3CO_2H}$ will break apart, but won't break apart completely. If you have some $\ce{CH_3CO_2H}$ in water, most of it will remain as $\ce{CH_3CO_2H}$, so in the net ionic equation it is written whole, but some will separate into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $\ce{CH_3CO_2H -&amp;gt; CH_3CO_2^{-}  + H^+}$goes both ways: $\ce{CH_3CO_2H}$ decomposes into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, and $\ce{CH_3CO_2^-}$ reacts with $\ce{H^+}$ for form $\ce{CH_3CO_2H}$. Since the forward reaction happens at a rate proportional to the concentration of $\ce{CH_3CO_2H}$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, eventually the ratio $\frac{[\ce{CH_3CO_2^-}][\ce{H^+}]}{[\ce{CH_3CO_2H}]}$ becomes constant, and we call this constant the $K_a$ of the $\ce{CH_3CO_2H}$. For $\ce{CH_3CO_2H}, K_a=1.7\times 10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now its time to connect all this back to your question. When you add $\ce{Ba(OH)_2}$ to a solution of $\ce{CH_3CO_2H}$, the $\ce{OH^-}$ ions present from the dissociation of $\ce{Ba(OH)_2}$ react with the $\ce{H^+}$ ions from the (partial) dissociation of $\ce{CH_3CO_2H}$. This lowers the concentration of $\ce{H^+}$, and so more $\ce{CH_3CO_2H}$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $\ce{Ba(OH)_2}$, $\ce{CH_3CO_2H}$ dissociates almost completely, even though, on its own, $\ce{CH_3CO_2H}$ dissociates only to a very small extent.&lt;/p&gt;&#xA;" OwnerUserId="1815" LastEditorUserId="24052" LastEditDate="2016-03-13T20:04:21.900" LastActivityDate="2016-03-13T20:04:21.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5330" PostTypeId="1" CreationDate="2013-06-15T20:36:06.117" Score="2" ViewCount="120" Body="&lt;p&gt;I used to live near a volcano where the groundwater contains naturally occurring arsenic compound. Since $\ce{As2O3}$, $\ce{As2S3}$, and $\ce{As4S4}$ are predominantly occurring in drinking water, what are the processes for removing them?&lt;/p&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="5016" LastEditDate="2016-07-16T14:46:11.727" LastActivityDate="2016-07-16T14:46:11.727" Title="How are arsenic compounds removed from drinking water?" Tags="&lt;inorganic-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5331" PostTypeId="2" ParentId="5330" CreationDate="2013-06-15T23:55:55.007" Score="2" Body="&lt;p&gt;The US Environmental Protection Agency (EPA) has a page dedicated to arsenic in drinking water.  &lt;a href=&quot;http://water.epa.gov/drink/contaminants/basicinformation/arsenic.cfm#How%20will%20arsenic%20be%20removed%20from%20my%20drinking%20water?&quot; rel=&quot;nofollow&quot;&gt;It contains a section that suggests ways to remove it from water&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I have no idea if the arsenic removal methods on the linked page actually work.  You should consult someone trained in the science of water purification for definitive information.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-16T00:07:27.960" LastActivityDate="2013-06-16T00:07:27.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5332" PostTypeId="2" ParentId="5316" CreationDate="2013-06-16T00:14:20.690" Score="2" Body="&lt;p&gt;I suspect that it does not.  As far as I know, all trivalent phosphorous is highly pyramidal--or at least not planar. This would rule out any graphene-like structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning that I have seen is based on hybridization. In the second row (third period) the 3s and 3p atomic orbitals are well separated, implying that s and p can not mix and that the outer valence is unhybridized. If bonding is with unhybridized p's, then bonding would occur at right angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Kutzelnigg, W. Angew. Chem., Int. Ed. Engl. 1984, 23, 272.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Brown, E. Organometallics 2000, 19, 2208-2214.)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="30" LastEditDate="2013-11-14T08:46:04.260" LastActivityDate="2013-11-14T08:46:04.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5333" PostTypeId="2" ParentId="5326" CreationDate="2013-06-16T00:22:57.020" Score="0" Body="&lt;p&gt;The question (and its choices) are bogus. One can make species (and solutions) of all of these combinations of cations with anions.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-16T00:22:57.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5334" PostTypeId="2" ParentId="5324" CreationDate="2013-06-16T00:29:42.997" Score="6" Body="&lt;p&gt;You simply can't make a transoid molecule of cyclohexene because there is not enough &quot;length&quot; in the carbon chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it is &quot;strain&quot; per se--but a putative transoid cyclohexene would have extraordinarily exotic bonding arrangement--thus being very high in energy.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-06-16T00:35:07.003" LastActivityDate="2013-06-16T00:35:07.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5335" PostTypeId="2" ParentId="5326" CreationDate="2013-06-16T00:35:41.410" Score="3" Body="&lt;p&gt;I believe the question is slightly misleading, non of the ions &lt;em&gt;react&lt;/em&gt; however some do &lt;em&gt;precipitate&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BaSO&lt;sub&gt;4&lt;/sub&gt;, Cu(OH)&lt;sub&gt;2&lt;/sub&gt; , and AgCl will precipitate in A, B, and C respectively.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-06-16T00:35:41.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5337" PostTypeId="1" CreationDate="2013-06-16T07:07:48.893" Score="4" ViewCount="277" Body="&lt;p&gt;Imagine I'm performing a primary amine to primary amine (i.e. PA to PA) crosslinking reaction at $37~^{\circ}\mathrm{C}$ in phosphate-buffered saline (pH $\approx 7.4$) with glutaraldehyde drawn from a $4 \%$ stock solution stored at $\approx 4~{^\circ}\mathrm{C}$.  I've heard that glutaraldehyde, like other aldehydes, can self-polymerize via &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldol_condensation&quot; rel=&quot;nofollow&quot;&gt;aldol condensation&lt;/a&gt; under certain conditions.  What might be a good approximation for the number of PA-PA crosslinks that consist of a length $N=1$ chain of glutaraldehyde?&lt;/p&gt;&#xA;" OwnerUserId="1817" LastEditorUserId="9961" LastEditDate="2015-10-16T18:04:04.210" LastActivityDate="2015-10-16T18:04:04.210" Title="What is the average chain length of a glutaraldehyde crosslink between primary amines?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;amines&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5338" PostTypeId="2" ParentId="4680" CreationDate="2013-06-16T07:43:26.913" Score="4" Body="&lt;p&gt;The $\ce{NH_2}$ group in given compounds is very similar to regular amine fragment and is more active than acylated $\ce{NH}$. However, alkylated amines are more active than non-substituted ones. So, I'd expect a very significant amount of double-alkylated product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid purification issues I recommend to consider synthesis of acetone hydrazone followed by reduction with $\ce{Na[BH_3(CN)]}$ or similiar low-active hydrogen donor (hi-active ones will reduce amide fragment)&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="876" LastEditDate="2015-05-01T19:07:53.353" LastActivityDate="2015-05-01T19:07:53.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5339" PostTypeId="1" CreationDate="2013-06-16T08:29:04.430" Score="7" ViewCount="164" Body="&lt;p&gt;In &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/nl2045414?prevSearch=Painting%2bwith%2bbiomolecules%2bat%2bthe%2bnanoscale%253A%2bbiofunctionalization&amp;amp;searchHistoryKey=&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; on page 1985 they say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fluorescence analysis confirmed that enhanced protein binding at higher electron dose depends only to a small extent on the slight increase in the lateral dimensions of the nanodeposits but to a much larger degree on their height (Figure 4E)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oHkw2m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The amount of adsorbed IgG-Cy3 is largely independent of the actual area of the nominal 100 nm islands but scales with the height of the nanodeposits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand this yet: why do I see that the fluorescence is proportional to area in this figure?&lt;/p&gt;&#xA;" OwnerUserId="1818" LastEditorUserId="194" LastEditDate="2018-03-02T13:32:31.387" LastActivityDate="2018-03-02T13:32:31.387" Title="Interpretation of Figure - biofunctionalization" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;fluorescence&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5341" PostTypeId="2" ParentId="5313" CreationDate="2013-06-16T15:18:29.647" Score="26" Body="&lt;p&gt;To my knowledge the mechanism of the Wittig reaction isn't fully resolved yet. But maybe I can give you some ideas about why the Wittig reaction with &lt;strong&gt;unstabilized&lt;/strong&gt; ylides is (&lt;em&gt;Z&lt;/em&gt;)-selective (well, with the exception of the &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/schlosser-modification.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schlosser modification&lt;/a&gt;) instead of (&lt;em&gt;E&lt;/em&gt;)-selective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the excellent book by &lt;a href=&quot;https://global.oup.com/academic/product/organic-chemistry-9780199270293&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clayden, Warren, and Greeves&lt;/a&gt;, there is a section beginning p. 690 that describes the oxaphosphetane formation as a concerted, antarafacial [2+2]-cycloaddition reaction (similar to your &quot;puckered transition state&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the phosphonium ylide and the carbonyl compound approach each other at right angles. The substituents are arranged in order to minimise steric repulsions in the transition state. In particular, the phenyl group (Ph) points away from the PPh&lt;sub&gt;3&lt;/sub&gt; group, and then the R group on the ylide points away from the Ph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QfHgD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QfHgD.jpg&quot; alt=&quot;Stereoselectivity of Wittig reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of the oxaphosphetane is irreversible and kinetically controlled. Hence, even though the &lt;em&gt;trans&lt;/em&gt; oxaphosphetane is more stable, it is not formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(By the way, the book also has some justification why the Wittig reaction of &lt;strong&gt;stabilized&lt;/strong&gt; ylides is E-selective.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the OP asked for it I will try to give some justification on why the ylide and the carbonyl group approach one another perpendicularly. For a proper description one would need the frontier orbitals of both compounds and then reason about the orbital interactions that bring about the [2+2] cycloaddition. Since I don't have those I will try to give some oversimplified description that should convey the general principle at work here. In the approximate orbital pictures I'm going to use I'll leave out the substituents for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin with, if one imagines the orbital interactions between the ylide and the carbonyl group, it is clear that the most important (primary) interaction for the reaction is the one between the HOMO of the ylide (nucleophile) and the LUMO of the carbonyl group (electrophile). What happens when both groups approach one another head-on or at right angles is shown in the following picture (for the perpendicular approach a top-view is used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dsQy8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dsQy8.png&quot; alt=&quot;Comparison of suprafacial and antarafacial [2+2] cycloadditions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases there are as many bonding interactions (black) as there are antibonding interactions (red). These situations are overall non-bonding and cannot lead to the desired product. This is the reason why [2+2] cycloadditions are usually thermally forbidden (by symmetry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some compounds have additional orbitals that: (1) are similar in energy to the frontier orbitals drawn in the picture, and (2) possess the right symmetry to interact with the other orbitals. These &quot;secondary orbital interactions&quot; are weaker than the HOMO-LUMO interaction, but might provide enough bonding to tip the balance from an overall non-bonding to a bonding situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does this translate to the situation of the Wittig reaction? Since O is more electronegative than C, the HOMO of the carbonyl group will have a larger orbital coefficient at O, while the LUMO will have a larger orbital coefficient at C. The situation is similar with the ylide. Since C is more electronegative than P, the HOMO of the ylide group will have a larger orbital coefficient at C. The ylide group is the nucleophile so the primary interaction will be between its HOMO and the LUMO of the electrophilic carbonyl group. Since the orbitals interact best where the coefficients are large, this primary interaction will be biased towards C–C bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is now, is there a suitable secondary interaction? One can argue that P has a lot of orbitals (valence shell extension) and that the LUMO of the ylide might be some empty phosphorus orbital (or at least an orbital with a big coefficient on $\ce{P}$) - this could be a p-orbital perpendicular to the ylide's C=P π-orbitals, or a d-orbital, or something like that. This phosphorus orbital can then interact with the carbonyl group's HOMO, which has its biggest coefficient on O. So this bonding interaction will be biased towards P–O bonding. The situation is shown in the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XW1n1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XW1n1.png&quot; alt=&quot;Secondary orbital interactions in cycloaddition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biases in the primary and secondary orbital interactions will lead to some distortions from the pure perpendicular approach which will become more pronounced the nearer the two species get to each other. For example, the ylide might rotate a bit so that its C and P atoms are closer to the carbonyl's C and O atoms respectively, in order to maximize the bonding overlap. And of course the perpendicular approach of ylide and carbonyl group will proceed in such a way that their respective biggest substituents are as far away from each other as possible (see diagram above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this quite lengthy answer I want to make clear that this description of mine is only a guess. The real reaction path is still a matter of debate and there are some compounds that rather react via a radical or ionic pathway. But I think it gives a good explanation for most of the observed behaviour and hope it helped the understanding. A more exact frontier orbital description of another thermal [2+2] cycloaddition, namely that of ketene with ethene, can be found in the textbook by &lt;a href=&quot;http://www.springer.com/gp/book/9783642036507&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brueckner&lt;/a&gt; on p 653.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2017-08-24T09:53:35.323" LastActivityDate="2017-08-24T09:53:35.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5342" PostTypeId="1" AcceptedAnswerId="5346" CreationDate="2013-06-16T16:46:41.530" Score="5" ViewCount="3056" Body="&lt;p&gt;For a titration, if you want to know the concentration of an acid you use a strong base such as $\ce{NaOH}$. My question is; why you can't use a weak base for this purpose? Is it because the calculation would be tougher? Is it because you'd need more of the base?&lt;/p&gt;&#xA;" OwnerUserId="1822" LastEditorUserId="4231" LastEditDate="2015-03-02T18:19:56.277" LastActivityDate="2019-03-07T08:48:35.790" Title="Why do you need a strong base for a titration" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5343" PostTypeId="2" ParentId="5321" CreationDate="2013-06-16T23:46:56.340" Score="3" Body="&lt;p&gt;Let me add to bobthechemist's answer.  The mole counts quantity.  The fundamental unit involved is Avogadro's Number, approximately $6\times 10^{23}$.  So $6\times 10^{23}$ units of anything is a mole of that thing, whether it is atoms of hydrogen or sand particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind reread bobthechemist's answer.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-06-16T23:46:56.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5344" PostTypeId="2" ParentId="5315" CreationDate="2013-06-16T23:49:42.760" Score="4" Body="&lt;p&gt;In general the answer to your question is no, you cannot postpone the reaction once the reactants are actually in contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction rate usually decreases with temperature.  Thus it is possible, but not certain, that if you freeze a reaction mixture down to liquid helium temperatures the reaction would stop.  This isn't really practical in general.&lt;/p&gt;&#xA;" OwnerUserId="675" LastActivityDate="2013-06-16T23:49:42.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5345" PostTypeId="2" ParentId="5342" CreationDate="2013-06-16T23:59:12.737" Score="0" Body="&lt;p&gt;Because a strong base should be able to titrate nearly &lt;em&gt;any&lt;/em&gt; acid, strong or weak. Recall that &lt;em&gt;strong/weak&lt;/em&gt; does &lt;strong&gt;not&lt;/strong&gt; mean &lt;em&gt;concentrated/dilute&lt;/em&gt;!  It means that, for an acid, it is easy/hard for the acid to donate a proton to the base's hydroxide ion.  If you use a weak base to titrate a weak acid, you may not be able to accomplish anything if the base is too weak, because the weak acid doesn't want to donate the proton, and the weak base doesn't want to accept it.  Even if you use a strong base, it may be the case that your acid is so weak that even your strong base is not strong enough, but this is rare.  This is where you use the so called $\ce{pK_a}$ values of the acid and base to determine if you can get a reaction, which is the titration.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-06-16T23:59:12.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5346" PostTypeId="2" ParentId="5342" CreationDate="2013-06-17T02:37:14.837" Score="7" Body="&lt;p&gt;There is no reason why you &lt;strong&gt;cannot&lt;/strong&gt; use a weak base to titrate a strong acid.  The calculation is probably considered a bit more complex but I think the real reason is a practical one.  For the sake of simplicity, I swapped your question around just a bit, and the figure below shows the titration of a strong base (blue) weak base (reddish) and very weak base (dashed line) with a strong acid (say, for example 1 M &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x43JL.png&quot; alt=&quot;simulated titration curves&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case the concentration of the acid is the same (1.0 M) and the concentration of the base is 0.100 M.  The volume of the base titrated is 100 mL.  For the standard strong acid/strong base scenario, one finds a well defined equivalence point.  When you transition to a weak base, however, there is additional information presented in the titration curve, namely the pKb of the base, which is a measure of its strength.  The pKb of the base can be obtained by finding the pH at the half-way mark between no acid added and the equivalence point (and if you know a bit of calculus, this is the point of minimum slope prior to the equivalence point). I've marked the pKb of the reddish titration curve with a dot, and the pKb is 9.  As the base gets weaker (less willing to accept a proton) the region before the equivalence point becomes less well defined.  In an extreme example, I have shown a &lt;strong&gt;simulated&lt;/strong&gt; titration curve of acetic acid (a very poor base indeed!) with a strong acid.  Notice how the equivalence point in this curve would be very hard to pick out if we didn't know where we were looking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the bottom line is this: the calculations aren't harder (one uses the Henderson–Hasselbalch equation and this is a standard exercise in a first year analytical chemistry class) but the &lt;strong&gt;data&lt;/strong&gt; one obtains from the experiment are harder to analyze.  And in practice, we &lt;strong&gt;do&lt;/strong&gt; perform weak/strong acid/base titrations when we are interested in determining the strength of the acid or base under consideration.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorDisplayName="user7951" LastEditDate="2019-03-07T08:48:35.790" LastActivityDate="2019-03-07T08:48:35.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5348" PostTypeId="1" CreationDate="2013-06-09T09:46:18.477" Score="5" ViewCount="2590" Body="&lt;p&gt;If you look online you can find various graphs of the absorption spectrum of water. The graph below for examples comes from &lt;a href=&quot;http://backreaction.blogspot.co.uk/2009/01/water-is-blue-because-water-is-blue.html&quot; rel=&quot;noreferrer&quot;&gt;this blog post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TL3o1.png&quot; alt=&quot;absorption spectrum of light&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell these graphs come from empirical measurements. Is there a theoretical model that can derive the same or similar results?  Of course if it was easy to implement in software that would be ideal.&lt;/p&gt;&#xA;" OwnerUserId="1764" OwnerDisplayName="marshall" LastActivityDate="2013-06-17T03:28:10.107" Title="How do you calculate the absorption spectrum of water?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5349" PostTypeId="2" ParentId="5348" CreationDate="2013-06-09T15:43:55.403" Score="4" Body="&lt;p&gt;The easy answer is no. It is possible to calculate absorption spectra from molecular structures using quantum mechanical calculation programs (for instance DFT using Gaussian) starting from the Schrödinger equation. These programs only work for a relatively small&#xA;number of electrons though and therefore usually assume gas phase including a small correction for the surroundings. Liquid phase water, where the molecules are very strongly coupled will be extremely difficult to calculate since it is almost impossible to create a &quot;box&quot; of relevant molecules. I don't have SciFinder or something equivalent, but a search on Google only gave articles calculating X-ray spectra. &lt;/p&gt;&#xA;" OwnerDisplayName="Groda.eu" LastActivityDate="2013-06-09T15:43:55.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5350" PostTypeId="2" ParentId="4867" CreationDate="2013-06-17T07:29:59.297" Score="2" Body="&lt;p&gt;The presumption here is that the biomass is purely sugars: glucose- and xylose-based starches and cellulose.  The essential components of lignin, which make up about 20% of the biomass, is phenols and propane. You may need to rebalance based on those additional components.&lt;/p&gt;&#xA;" OwnerUserId="1824" LastActivityDate="2013-06-17T07:29:59.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5352" PostTypeId="1" AcceptedAnswerId="5363" CreationDate="2013-06-17T10:28:42.503" Score="6" ViewCount="3394" Body="&lt;p&gt;I would like to measure the pH value of dough with a pH-Meter, to ensure a pH value of 4.1 (max). Because of the high viscosity I would thin down the dough with demineralized water for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I expect more accurate measurements&lt;/li&gt;&#xA;&lt;li&gt;This reduces pollution of the measurement device.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as I understand this will affect the measured value, so I would thin down the dough in a defined ratio of 1:10.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it correct that the thinning with demineralized water changes the pH value?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ratio 1:10 should cause an offset of 1 in ph value, right? In which direction? Do I have to add 1.0  to or subtract 1 from the measurement to estimate the real value?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Edit #1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As it seems to be important: We are talking about weak acids produced by fermentation of rye flour (or wheat flour) using sourdough (lactose bacteria + yeast). Some samplings are taken during the fermentation process to check that the pH value is low enough (value of 4.1 or lower).&#xA;I'm less interested in the -details- of the theoretical background, more in a practical solution for the issue. The actual real value isn't that important (I need to check, if the pH limit is reached), but it would be interesting how to measure it as there might be future experiments where the value matters...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of additional chemicals except (demineralized) water should be avoided for simplicity. Of course the use of tab water would be preferable, but I guess this becomes more complicated because of unknown water hardness and varying ion concentrations.&lt;/p&gt;&#xA;" OwnerUserId="1826" LastEditorUserId="1826" LastEditDate="2013-06-18T16:14:46.827" LastActivityDate="2015-04-11T10:01:06.743" Title="How can I measure the pH of high-viscosity fluids like dough?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;&lt;ph&gt;&lt;food-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5353" PostTypeId="1" AcceptedAnswerId="5354" CreationDate="2013-06-17T10:56:49.427" Score="4" ViewCount="1676" Body="&lt;p&gt;I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.&lt;/p&gt;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$\Delta G = \Delta H − T\Delta S$&lt;/span&gt;, but I don't know how to put them together to show why the equilibrium would shift.&lt;/p&gt;&#xA;&lt;p&gt;To summarise, I already know what happens when, and I'm interested in &lt;strong&gt;why&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1220" LastEditorUserId="4945" LastEditDate="2021-05-27T22:46:27.083" LastActivityDate="2021-05-27T22:54:16.473" Title="Why does changing the temperature shift the equilibrium?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5354" PostTypeId="2" ParentId="5353" CreationDate="2013-06-17T12:53:09.370" Score="10" Body="&lt;p&gt;I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.&lt;/p&gt;&#xA;&lt;p&gt;If you increase the temperature, what happens to the rates of the forward and reverse reactions?&lt;/p&gt;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arrhenius equation&lt;/a&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$k = A\exp\left(\frac{-E_\mathrm{a}}{RT}\right),$$&lt;/span&gt;&#xA;you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{a}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.&lt;/p&gt;&#xA;" OwnerUserId="244" LastEditorUserId="4945" LastEditDate="2021-05-27T22:49:14.907" LastActivityDate="2021-05-27T22:49:14.907" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5355" PostTypeId="1" CreationDate="2013-06-16T20:04:55.387" Score="1" ViewCount="38" Body="&lt;p&gt;I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\Delta G = \Delta H - T \Delta S$ but I don't know how to put them together to show why the equilibrium would shift.&lt;/p&gt;&#xA;" OwnerUserId="1220" OwnerDisplayName="Kian" LastActivityDate="2013-06-17T17:04:57.720" Title="Chemical Equilibrium - Why does changing the temperature shift the equilibrium?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2013-06-17T17:05:20.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="5356" PostTypeId="2" ParentId="5353" CreationDate="2013-06-17T11:42:29.497" Score="11" Body="&lt;p&gt;Suppose you have some reaction&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A &amp;lt;=&amp;gt; B}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The equilibrium constant for the reaction is &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and the Gibbs free energy change is &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt;. The equilibrium constant is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K = \frac{[\ce{B}]}{[\ce{A}]}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Increasing value of &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; shifts the equilibrium towards the right, i.e. more &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt;, and reducing the value shifts it to the left, i.e. more &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;. The equilibrium constant and the Gibbs free energy are related by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ K = \exp\left(-\frac{\Delta G}{RT}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Putting in the expression you give for &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; gives us&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA; K &amp;amp;= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\&#xA;   &amp;amp;= \exp\left(-\frac{\Delta H}{RT}\right)\, &#xA;      \exp\left(\frac{\Delta S}{R}\right).&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we assume the variation of &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta S$&lt;/span&gt; is small we can ignore the entropy term because it's a constant, and we get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the reaction is endothermic &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease &lt;span class=&quot;math-container&quot;&gt;$\Delta H/(RT)$&lt;/span&gt; and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.&lt;/p&gt;&#xA;&lt;p&gt;To understand why this happens consider what happens when we make the temperature very high.  If you do this the value of &lt;span class=&quot;math-container&quot;&gt;$\Delta H/(RT)$&lt;/span&gt; approaches zero and the exponential approaches unity, i.e. the concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.&lt;/p&gt;&#xA;" OwnerUserId="1470" OwnerDisplayName="John Rennie" LastEditorUserId="4945" LastEditDate="2021-05-27T22:54:16.473" LastActivityDate="2021-05-27T22:54:16.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5357" PostTypeId="1" CreationDate="2013-06-17T17:22:56.937" Score="5" ViewCount="15782" Body="&lt;p&gt;To get the oxidation state of carbon (denoted $x$) in octane, $\ce{C8H18}$, I used the fact that hydrogen has an oxidation state of $+1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8x + 18\cdot 1 = 0 \qquad \Longrightarrow \qquad x = -\frac{18}{8} = -2.25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in that way, I get an oxidation number that is fractional, or not an integer. Is that normal?&lt;/p&gt;&#xA;" OwnerUserId="1829" LastEditorUserId="16683" LastEditDate="2018-09-03T10:36:07.767" LastActivityDate="2019-01-23T10:48:42.860" Title="Are fractional oxidation states possible?" Tags="&lt;oxidation-state&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5358" PostTypeId="2" ParentId="5357" CreationDate="2013-06-17T19:08:44.783" Score="20" Body="&lt;p&gt;The &quot;oxidation number&quot; is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons &quot;owned&quot; by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="5026" LastEditDate="2018-03-08T10:37:44.980" LastActivityDate="2018-03-08T10:37:44.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5359" PostTypeId="2" ParentId="5357" CreationDate="2013-06-17T20:24:53.333" Score="9" Body="&lt;p&gt;In addition to bob's answer, it's worth noting that oxidation number assumes that bonding is happening mostly ionically, which isn't really the case with alkanes where the electronegativities of Carbon and Hydrogen are very similar.&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-17T20:24:53.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5361" PostTypeId="2" ParentId="5352" CreationDate="2013-06-18T01:02:40.690" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, dilution changes pH. pH is defined as the negative log of $\ce{H+}$ concentration, so changing the concentration will change pH.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Not necessarily.&lt;/strong&gt; For dilute acids and bases, the auto-ionization of water becomes a factor in calculating pH, and will change by less than one pH unit. For more concentrated acids (pH&amp;lt;5 or pH&gt;9 for initial solution), one pH unit is correct. The pH will change toward neutral (pH=7) upon dilution, so it depends on initial pH whether you must subtract or add to make a correction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-06-18T01:02:40.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5363" PostTypeId="2" ParentId="5352" CreationDate="2013-06-18T13:03:12.943" Score="4" Body="&lt;p&gt;In addition to buckminst’s answer, for weak acids, which are not fully ionized, the pH of the solution depends on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;ratio of the concentrations&lt;/a&gt; of the conjugate base $\ce{A-}$ to the neutral acid $\ce{HA}$, especially in the pH regime close to the $\mathrm{p}K_\text{a}$ of the acid (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_region&quot; rel=&quot;nofollow&quot;&gt;buffer region&lt;/a&gt;). Diluting the solution will change the concentrations, but in the same way. The ratio will not change. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; can be derived from the mass action expression for $K_\text{a}$ of a weak acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA &amp;lt;=&amp;gt;H+ + A-}$$&#xA;$$K_\text{a}=\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}$$&#xA;$$\text{p}K_\text{a}=-\log{K_\text{a}}=-\log \left(\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}\right)=-\log[\ce{H+}]-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$&#xA;$$\text{p}K_\text{a}=\text{pH}-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$&#xA;$$\text{pH}=\text{p}K_\text{a}+\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the volume increases by a factor of 10, then $[\ce{A-}]$ and $[\ce{HA}]$ both decrease by a factor of 10, and $\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}$ remains constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the acids in dough are likely acetic or lactic from fermentation; carbonic or tartaric from leavening; or citric or ascorbic from preservatives – all of which are weak acids – diluting the dough might not change the pH.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorDisplayName="user7951" LastEditDate="2015-04-11T10:01:06.743" LastActivityDate="2015-04-11T10:01:06.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5364" PostTypeId="1" AcceptedAnswerId="5366" CreationDate="2013-06-18T14:20:39.773" Score="6" ViewCount="30438" Body="&lt;p&gt;When we are titrating acid/base using ph meter we add distilled water to immerse the ph electrode.&#xA;Won't this affect the concentration of the acid/base: I mean isn't  this dillution. &#xA;Won't  this affect our calcultion of the concentration at the end.&lt;/p&gt;&#xA;" OwnerUserId="1836" LastActivityDate="2013-06-18T15:08:38.303" Title="Effect of dillution on titration" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5366" PostTypeId="2" ParentId="5364" CreationDate="2013-06-18T15:08:38.303" Score="5" Body="&lt;p&gt;What's important to note in this situation is that the number of moles of the titrand isn't changing. This matters because all acid/base reactions go to completion, so every drop of titrant that falls into the solution will react with the titrand until there are no moles of titrand left &lt;em&gt;regardless of concentration&lt;/em&gt;, at which point you've reached the equivalence point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding a little water here or there doesn't really change the pH at the equivalence point, because ideally there are no moles of acid or base present in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this doesn't mess up your measurement for the concentration is because you measured the volume of titrand before the titration, and then recorded the volume of the known titrant needed to reach the equivalence point. From this information, you can calculate the moles of titrant needed to reach equivalence, which in most cases equals the number of moles of titrand present, and with the original volume of titrand noted previously, you can calculate the original concentration of the solution the sample was taken from. voila!&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-18T15:08:38.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5369" PostTypeId="1" AcceptedAnswerId="5441" CreationDate="2013-06-18T18:05:41.427" Score="10" ViewCount="1744" Body="&lt;p&gt;What is the fundamental argument behind explaining why Ta metal is so inert and does not usually participate in corrosion? I know that it forms a protective oxide, but nothing explains why in an acidic solution it is resistant to nearly everything (aside from HF). I want to know if it has to do with a bond strength of the metal or the molecular orbitals.&#xA;An in-depth citation is required.&lt;/p&gt;&#xA;" OwnerUserId="1034" LastEditorUserId="16683" LastEditDate="2017-04-08T22:47:41.863" LastActivityDate="2017-04-08T22:47:41.863" Title="Why is tantalum so unreactive?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5370" PostTypeId="1" AcceptedAnswerId="5371" CreationDate="2013-06-07T03:14:50.340" Score="0" ViewCount="51" Body="&lt;p&gt;I have a question that explains the weights and abundances of various hydrogen isotopes, and the question concludes by asking to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the atomic mass of hydrogen $\mathrm{A}_r(\mathrm{H})$ (with the $r$ in subscript).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer simply summed the mass of each isotope multiplied by its natural abundance. However, the text never mentioned the meaning of $\mathrm{A}_r(\mathrm{H})$ and I couldn't find anything online. &lt;/p&gt;&#xA;" OwnerUserId="1986" OwnerDisplayName="jake9115" LastActivityDate="2013-06-18T20:46:14.803" Title="What does $\mathrm{A}_r(\mathrm{H})$ represent when calculating the average natural weight of an elemental atom?" Tags="&lt;notation&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5371" PostTypeId="2" ParentId="5370" CreationDate="2013-06-07T03:31:21.493" Score="2" Body="&lt;p&gt;The relative atomic mass $A_r$ is 'the ratio of the average mass of atoms of an element ... to 1/12 of the mass of an atom of carbon-12' (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Relative_atomic_mass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Relative_atomic_mass&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic mass $m_a$ is the mass of a particular isotope of an atom, and if it is in 'u' (unified atomic mass units) it is relative to 1/12 of the mass of an atom of carbon-12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to get $A_r$ you take a weighted average of $m_a$ in 'u' for each isotope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the standard atomic weight is the relative atomic mass of an element using natural abundance on Earth (&lt;a href=&quot;http://en.wikipedia.org/wiki/Relative_atomic_mass&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Relative_atomic_mass&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1841" OwnerDisplayName="Ramashalanka" LastActivityDate="2013-06-07T03:38:17.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5376" PostTypeId="1" CreationDate="2013-06-19T14:55:17.240" Score="22" ViewCount="464" Body="&lt;p&gt;Molecular imaging, using STM and AFM technologies, appear so far to visually prove most of what we already know about chemical structures, such as VSEPR theory. For instance, here are the ideal structures and AFM of a couple of cyclization reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvdfN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've also taken single-atom imaging to pretty scary levels. Behold a field emission electron microscope image of the &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt; orbitals of carbon, long relegated to conceptual mathematical images in textbooks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vut4Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, have there been any surprises to be had from these images, where something we thought was structured one way was in fact radically different? You can see in the AFM image that some of the bond angles deviate from the ideal regular shapes a little, and it may also be that the electrons in the pi bonds, instead of being regularly distributed, seem to prefer the &quot;far&quot; end of the phenyl rings, but the predicted structures are uncannily accurate. Again, with the orbital imaging, the s orbital isn't &lt;em&gt;perfectly&lt;/em&gt; spherical and we don't see the full p-orbital shell, just two &quot;lobes&quot;, but again, they're powerful evidence that our theories are right so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this ever not been the case? Have any of these recent images turned chemistry on its head by showing something totally unexpected?&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-06-19T15:15:01.507" LastActivityDate="2016-04-08T14:33:08.837" Title="Molecular Imaging - Any surprises to be had?" Tags="&lt;physical-chemistry&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5377" PostTypeId="1" CreationDate="2013-06-19T15:05:11.313" Score="6" ViewCount="735" Body="&lt;p&gt;I ran into an issue trying to run dftb.org parameter files on G09 and found questions asked on this but not answers.  I thought I would post the solution here to help anyone else who may run into this problem.&lt;/p&gt;&#xA;" OwnerUserId="1732" LastEditorUserId="1732" LastEditDate="2013-06-20T19:30:42.657" LastActivityDate="2013-12-13T17:19:59.837" Title="Having problems running DFTB using dftb.org parameter files on Gauss 09 (G09)?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5378" PostTypeId="1" CreationDate="2013-06-19T16:29:56.773" Score="5" ViewCount="13233" Body="&lt;p&gt;What will happen to the melting point temperature of ice if some common salt is added to it? How to justfy the answer?&lt;/p&gt;&#xA;" OwnerUserId="1846" LastEditorUserId="31775" LastEditDate="2018-02-19T16:24:04.513" LastActivityDate="2018-02-19T16:24:04.513" Title="What is the effect of adding salt to ice?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;phase&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5379" PostTypeId="2" ParentId="5378" CreationDate="2013-06-19T20:04:01.083" Score="6" Body="&lt;p&gt;From &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml&quot; rel=&quot;noreferrer&quot;&gt;http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that ice, a crystalline solid, forms via weak partial-polar forces between water molecules, known as hydrogen bonding. As liquid water cools, it loses the energy that keeps these forces from attracting and capturing neighboring water molecules, and so these bonds form in a crystalline structure. Exactly what that structure is depends on the temperature and pressure at which the ice forms; your everyday freezer ice as well as most of the stuff outside in January is Ice I&lt;sub&gt;h&lt;/sub&gt;, which is a very spacious hexagonal arrangement that makes ice less dense than liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that even within what looks like solid ice, individual molecules &quot;melt&quot; and &quot;freeze&quot; all the time as they transiently gain enough energy to break free of the lattice, and then lose it and are recaptured. When the rate at which molecules melt and freeze are equal, the ice is stable. This happens at 0.01*C; below this temperature, more water will freeze, while above this temperature, more water will melt. Right at 0.01*C, water remains water and ice remains ice (the other half of this coexistence, besides temperature, is that water like most substances requires more energy gain or loss to change its phase, known as the Latent Heat of Vaporization, and for water that's 334J/mL, 80 times what's needed to cool liquid water from 2*C to 1*C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introduce salt, and these dissolved ions interfere with this hydrogen bonding within the ice, preventing it from happening at the normal freezing point of water. Now the melting molecules don't refreeze, because their partial polar charges are more strongly attracted to the full charges of the sodium and chloride ions than to other water molecules. That upsets the balance; more water is melting than freezing at the normal freezing temperature, and so the ice melts. It takes more energy loss within this system before the water is too unenergetic to stay liquid and settles into crystal, releasing the salt, which then finds other liquid water at the boundary layer between liquid and ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates a vicious cycle; as the salt concentrates within the remaining liquid water, its freezing point is lowered even further as the sodium and chlorine, which won't resolidify as salt crystals until there's just too little water to remain dissolved, will become even more attracted to the remaining water, even further lowering the remaining liquid's freezing point as the water must free itself from even more ions before it can be captured by the ice crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, an aqueous salt solution, in nearly any concentration, will lower the boiling point of the solution below 0*C, but the minimum freezing point is at -21.1*C, which is the freezing point of a fully-saturated salt solution. Below this temperature, salt that loses its water to freezing can't stay dissolved, and precipitates out as a separate crystal solid suspended within the ice.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-06-19T20:09:16.373" LastActivityDate="2013-06-19T20:09:16.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5381" PostTypeId="2" ParentId="5377" CreationDate="2013-06-20T12:49:04.627" Score="4" Body="&lt;p&gt;First, at least for C.01, dftb=readparameters, in the route section, does not work dftb=read does but is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skip a line after molecular descritption and include your parameter files there. Make sure you include all pair interaction files, not just bonded atoms.  It should look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@C:/G09W/C-C_mio.skf&#xA;@C:/G09W/C-H_mio.skf&#xA;@C:/G09W/H-C_mio.skf&#xA;@C:/G09W/H-H_mio.skf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I included the mio because we are running more than one parameter set, not a requirement just a name)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real glitch comes from my final hint.  The DFTB parameters generally have more than 500 points, but the first line in the code is 500 (not 519, or 520, etc).  G09 Rev. D.01 doesn't care about this, B.01 and C.01 do.  You simply need to change 500 ro reflect the correct amount of points and add in the correct order for your atomic numbers (this part is explained on Gauss' website) and you are good to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this helps anyone trying to run DFTB on G09 for the first time, and helps save some time.&lt;/p&gt;&#xA;" OwnerUserId="1732" LastActivityDate="2013-06-20T12:49:04.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5382" PostTypeId="1" AcceptedAnswerId="5383" CreationDate="2013-06-20T13:30:39.037" Score="8" ViewCount="191003" Body="&lt;p&gt;I don't understand how the freezing point of a substance is the same temperature as the melting point of the same substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if liquid water freezes at 0 °C how can ice also melts at 0 °C?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="-1" LastEditDate="2019-09-13T05:44:04.873" LastActivityDate="2019-09-13T05:44:04.873" Title="How can melting point equal freezing point?" Tags="&lt;energy&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="5383" PostTypeId="2" ParentId="5382" CreationDate="2013-06-20T13:55:08.333" Score="11" Body="&lt;p&gt;Because melting point and freezing point describe the same &lt;em&gt;transition of matter&lt;/em&gt;, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may not realize is that while water is freezing or melting, its temperature is &lt;strong&gt;not&lt;/strong&gt; changing!  It is stuck on $0\ \mathrm{^\circ C}$ during the entire melting or freezing process.  It is easier to see this for &lt;em&gt;boiling points&lt;/em&gt;.  if you put a thermometer in water and heat it, the temperature will rise until it reaches $100\ \mathrm{^\circ C}$, and then it starts boiling.  And while it boils, it will &lt;em&gt;stay&lt;/em&gt; at $100\ \mathrm{^\circ C}$!  All the way until the water has all boiled away.  Now if you could somehow trap the steam (gaseous water) and keep heating it, the steam could have a temperature higher than $100\ \mathrm{^\circ C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to sum this all up, when matter is transitioning from solid to liquid (melting) or liquid to solid (freezing), its temperature is fixed at the melting/freezing point, which is the &lt;strong&gt;same&lt;/strong&gt; temperature.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="42741" LastEditDate="2017-05-20T03:52:55.733" LastActivityDate="2017-05-20T03:52:55.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5384" PostTypeId="2" ParentId="5382" CreationDate="2013-06-20T14:01:46.450" Score="9" Body="&lt;p&gt;I think this is an interesting question where the confusion is mostly due to semantics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider an ice cube in your freezer (a typical kitchen freezer has a temperature of -10 C).  When that ice cube is taken out of the freezer and placed in a warm kitchen, heat from the surroundings (air, counter top, etc.) is transferred to the ice cube.  We observe that the temperature of the ice cube increases.  The ice cube stays a cube because the energy of the intermolecular forces that keep the water molecules together is greater than the heat energy added to the ice cube so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the melting point, however, there &lt;em&gt;is&lt;/em&gt; enough thermal energy to start breaking those intermolecular forces.  What we observe is that the temperature does not rise, but bonds are breaking and the solid starts to melt.  Once all the solid melts, the temperature of the (now liquid) water can increase when thermal energy is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar explanation can be used for the reverse process (freezing water) only in this case the thermal energy is being transferred from the water to the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we come to your question, how can melting point = freezing point.  This &quot;point&quot; is &lt;em&gt;the temperature at which the solid and liquid forms of the molecule are in equilibrium.&lt;/em&gt;  When we use a term like &lt;strong&gt;melts&lt;/strong&gt; oftentimes we mean &lt;strong&gt;melts completely&lt;/strong&gt;.  In this case, the temperature of the liquid would be just &lt;strike&gt;beyond&lt;/strike&gt; &lt;em&gt;above&lt;/em&gt; the melting point.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="7475" LastEditDate="2015-10-27T19:26:01.270" LastActivityDate="2015-10-27T19:26:01.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5385" PostTypeId="1" AcceptedAnswerId="5387" CreationDate="2013-06-20T15:05:06.037" Score="4" ViewCount="3418" Body="&lt;p&gt;So this experiment goes like this :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You take an ice-cube&lt;/li&gt;&#xA;&lt;li&gt;You put a thread on top face of it&lt;/li&gt;&#xA;&lt;li&gt;You sprinkle some common salt on the ice-cube wherever its touching the thread&lt;/li&gt;&#xA;&lt;li&gt;You wait for sometime (approx 1 min)&lt;/li&gt;&#xA;&lt;li&gt;Voila! now you can lift the thread and ice-cube will stick to the end of the thread so you can lift the whole thing altogether&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How do you explain this in chemistry?&lt;/p&gt;&#xA;" OwnerUserId="1821" LastActivityDate="2017-12-11T21:54:18.850" Title="How do you explain the thread sticking to an ice-cube when salt is sprinkled over it?" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5386" PostTypeId="1" CreationDate="2013-06-20T16:28:32.367" Score="1" ViewCount="7542" Body="&lt;p&gt;There are alkalinizing water ionizers on the market that lower the ORP of water. Does heating or cooling that water have an impact on its ORP?&lt;/p&gt;&#xA;" OwnerUserId="1852" LastActivityDate="2013-07-17T18:40:43.070" Title="What impact does heating or cooling have on the oxidation reduction potential of water?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5387" PostTypeId="2" ParentId="5385" CreationDate="2013-06-20T17:33:27.723" Score="12" Body="&lt;p&gt;I conjecture that there are several things happening.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Step 3:  Addition of salt will melt the ice at its immediate point of contact. The string will naturally be surrounded by the highly concentrated brine (salt water).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Step 4:  This is enough time for the brine to get diluted, noting that there is an infinite reservoir of potential water by way of the ice cube.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Step 5:  Once the salt gets sufficiently diluted, then the comparatively large ice cube provides the necessary cold for the (now) very dilute salt water to reform as ice.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Recalling that in Step 3 the salt water surrounded the string, then ice naturally forms around the string in Step 5, hence you can pick up the ice cube with the string.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-20T17:33:27.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5388" PostTypeId="1" CreationDate="2013-06-20T18:13:37.400" Score="1" ViewCount="2543" Body="&lt;p&gt;I read somewhere that a silane group will react with hydroxyl groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know why this would happen? &lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorUserId="73" LastEditDate="2013-06-21T04:04:27.607" LastActivityDate="2017-08-15T22:07:56.490" Title="Why does a silane group react with hydroxyl groups?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5389" PostTypeId="1" AcceptedAnswerId="5401" CreationDate="2013-06-21T06:51:57.767" Score="7" ViewCount="25066" Body="&lt;p&gt;This is balanced equation of fermentation of glucose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H12O6(aq) -&amp;gt; 2C2H5OH(aq) + 2CO2(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the result of glucose fermentation is ethanol (&lt;span class=&quot;math-container&quot;&gt;$\ce{C2H6O}$&lt;/span&gt;). What represented in the aforementioned equation is slightly different. I would like to know what &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5OH}$&lt;/span&gt; is? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the OH functional group doing here? Does it make ethanol acidic? Does it mean that ethanol is diluted with water?&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="402" LastEditDate="2019-10-31T02:01:22.913" LastActivityDate="2019-10-31T02:01:22.913" Title="Why is ethanol written as C2H5OH not C2H6O?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5390" PostTypeId="2" ParentId="5389" CreationDate="2013-06-21T06:58:25.420" Score="1" Body="&lt;p&gt;I confused compound formula with the actual formula of ethanol. Compound formula is $\ce{C2H6O}$. But the actual molecular structure is $\ce{CH3CH2OH}$. The oxygen is attached to hydrogen which it in turn is attached to Carbon.&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="1743" LastEditDate="2013-06-22T02:17:16.393" LastActivityDate="2013-06-22T02:17:16.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5391" PostTypeId="1" CreationDate="2013-06-21T07:06:04.300" Score="4" ViewCount="10166" Body="&lt;p&gt;Methyl orange, bromothymol blue and phenolphthalein indicators were mixed together to form a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4bzh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over what pH range would the solution be yellow? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot understand the question. If two or more different indicators are mixed, &lt;strong&gt;do they still work as indicators?&lt;/strong&gt; Because each their colours will blend with each other and result to formation of a new colour, how can we know what the color will be?&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="73" LastEditDate="2013-06-21T12:30:32.733" LastActivityDate="2019-02-21T11:17:06.013" Title="How do indicators work when they mixed with each other?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5393" PostTypeId="1" CreationDate="2013-06-21T10:05:10.177" Score="-1" ViewCount="10997" Body="&lt;p&gt;Assuming every other conditions are the same, what would be the comparison of energy required to break single bond, double bond and triple bond of a carbon atom in an organic molecule?&lt;/p&gt;&#xA;" OwnerUserId="1857" LastActivityDate="2015-03-17T09:13:52.650" Title="Energy required to break single bond vs double bond vs triple bond of a carbon atom in an organic molecule" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5394" PostTypeId="1" CreationDate="2013-06-21T10:14:23.280" Score="10" ViewCount="113363" Body="&lt;p&gt;Is the melting and boiling point of ionic bond usually higher than covalent bond? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that compounds with ionic bonds are usually solid at room temperature, so I want other answers than this. (and this question can also be about why compounds with ionic bonds are usually solid in room temperature.)&lt;/p&gt;&#xA;" OwnerUserId="1857" LastEditorUserId="9961" LastEditDate="2017-12-06T00:59:53.217" LastActivityDate="2017-12-06T00:59:53.217" Title="Is melting/boiling point of ionically bonded substance higher than of covalently bound?" Tags="&lt;bond&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5395" PostTypeId="2" ParentId="5391" CreationDate="2013-06-21T14:06:38.353" Score="3" Body="&lt;p&gt;In general, the chemistry of one indicator does not usually affect another as far as I know, but the colors will mix as you would expect. It gets tricky here because the question shows the colored ranges but leaves a blank in the transition pH ranges, but I would guess the answer they're looking for is from 3-7.5&lt;/p&gt;&#xA;" OwnerUserId="518" LastActivityDate="2013-06-21T14:06:38.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5396" PostTypeId="2" ParentId="5393" CreationDate="2013-06-21T14:39:30.253" Score="2" Body="&lt;p&gt;365 kJ/mol for the single bond, 598 kJ/mol for the double bond, and 813 kJ/mol for the triple bond. These are usually listed as &lt;em&gt;bond enthalpy&lt;/em&gt; or &lt;em&gt;heat of formation&lt;/em&gt; in scientific table books. If you don't find the one you need you can calculate it in steps according to Hess' Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the energy required to break one bond as opposed to a mol of bonds, you divide the number I gave by Avogadro's number, which is the number of atoms or molecules in a mol of a compund or element.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2013-06-21T14:39:30.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5397" PostTypeId="2" ParentId="5394" CreationDate="2013-06-21T17:25:09.090" Score="3" Body="&lt;p&gt;One set of comparisons is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it take so much energy to turn sodium chloride molten, whereas candle wax melts virtually upon touch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Ionic compounds&quot; tend to form crystals.  You may have seen pictures of a crystal lattice of sodium chloride:  cubic packing with a sodium being surrounded by many chlorides.  It takes a lot of energy to break up all of those bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of wax (also a solid at room temperature), there are not these strong, &quot;ionic&quot; interactions between molecules. Things are more &quot;layered&quot; and disordered rather than crystalline. (Though there is definitely inter-molecular bonding Otherwise it would never be solid!!!)&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2013-06-21T17:25:09.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5398" PostTypeId="2" ParentId="5388" CreationDate="2013-06-21T18:05:38.613" Score="6" Body="&lt;p&gt;It is well known that silicon forms very strong bonds with oxygen (think sand, $\ce{SiO2}$). Hence, any time one can convert a $\ce{Si-H}$ bond into $\ce{Si-OH}$, it will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that silanes are extremely air sensitive due to the moisture content.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="41328" LastEditDate="2017-08-15T22:07:56.490" LastActivityDate="2017-08-15T22:07:56.490" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5399" PostTypeId="2" ParentId="5391" CreationDate="2013-06-21T22:52:31.037" Score="1" Body="&lt;p&gt;The mixture of indicators should still work because most indicators are weak acids, and the presence of other weak acids does not interfere with any given weak acid.  If you had an indicator that happened to be a weak base, you would likely have a problem, but that is not the case for your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a nice video from the CHEM Study series on exactly this topic, including a thorough discussion on mixing the indicators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=2mwjg9i00Qs&quot; rel=&quot;nofollow&quot;&gt;Acid Base Indicators CHEM Study&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the specific answer to your assigned question, I think @Gavin Kramar is spot on.  I'm just adding this answer for some additional information.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-06-21T22:52:31.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5400" PostTypeId="2" ParentId="5385" CreationDate="2013-06-22T02:00:48.013" Score="5" Body="&lt;p&gt;@Eric Brown has the result right qualitatively.  We can add some more quantitative and chemical information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: When salt is added, the freezing point of the water is decreased at that point, so the ice melts to water.  This is a &lt;em&gt;colligative&lt;/em&gt; property; i.e. has only to do with amount of substance, and not the particular substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4: As the melting water dilutes, its phase shifts back towards pure water, which has a higher freezing point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5: At some point, the phase shifts back to freezing at 0 °C, and the water re-freezes given the ambient ice at 0 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Wikipedia article shows you the phase diagram corresponding to my description:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:WatNaCl.png&quot; rel=&quot;nofollow&quot;&gt;NaCl phase diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, my answer merely augments your already selected correct answer, but I thought this might be useful additional information.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user7951" LastEditDate="2015-10-03T18:44:29.627" LastActivityDate="2015-10-03T18:44:29.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5401" PostTypeId="2" ParentId="5389" CreationDate="2013-06-22T02:13:58.413" Score="9" Body="&lt;p&gt;It is better to write &lt;span class=&quot;math-container&quot;&gt;$\ce{C_2H_6O}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$\ce{C_2H_5OH}$&lt;/span&gt;, because there is more than one compound with the same overall formula, and emphasising the &lt;em&gt;hydroxyl group&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt;, which is what characterizes an &lt;em&gt;alcohol&lt;/em&gt; makes it &lt;em&gt;ethanol&lt;/em&gt; and nothing else.  So when you ask, &quot;What is OH functional group doing here&quot;, it is exactly that the product is an alcohol, characterized by a hydroxyl group. And yes, alcohols &lt;em&gt;are&lt;/em&gt; acidic in the right situations, given that their hydroxyl groups do produce acidic hydrogen ions, and yes, that is why it's so soluble in water. &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="402" LastEditDate="2019-10-31T02:00:53.767" LastActivityDate="2019-10-31T02:00:53.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5403" PostTypeId="1" CreationDate="2013-06-22T04:26:11.877" Score="2" ViewCount="431" Body="&lt;p&gt;We've got a mixture of carbonate salts. It's a buffer, used in fish tanks. How might we go about identifying the components and their concentrations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect it'd be hard to separate some of this. Is this correct? If so, why?&lt;/p&gt;&#xA;" OwnerUserId="603" LastEditorUserId="22" LastEditDate="2013-06-22T07:34:08.453" LastActivityDate="2013-06-22T07:52:43.173" Title="Analysis of a mixture of carbonate salts" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5406" PostTypeId="1" CreationDate="2013-06-22T12:52:45.587" Score="2" ViewCount="1158" Body="&lt;p&gt;A slightly deranged curiosity: if we had a 3D printer able to “print” anything starting from “cartridges” of pure chemical elements, how many of the 118+ possible cartridges (or of the 98ish for non-artificial elements) would we actually need to print most - or all - foodstuff?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(OK, one could have phrased the same question in a more mundane way, I agree.)&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2017-04-08T22:00:37.443" Title="How many elements in food?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5409" PostTypeId="1" AcceptedAnswerId="5413" CreationDate="2013-06-22T20:57:00.290" Score="5" ViewCount="5409" Body="&lt;p&gt;Can natural gas appliances run on Hydrogen? If so, does the appliance have to be modified and how significant of a modification would it need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2021-04-27T14:07:16.313" Title="Can natural gas appliances run on Hydrogen?" Tags="&lt;energy&gt;&lt;combustion&gt;&lt;hydrogen&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5410" PostTypeId="2" ParentId="5406" CreationDate="2013-06-22T21:29:21.447" Score="5" Body="&lt;p&gt;For the record, I am not a biochemist, a nutritional expert, or any sort of person who is required to know this for my profession.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, as far as I know, to get the vast majority of nutritional food substances, you'll need to be able to produce at least:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sugars and starches&lt;/li&gt;&#xA;&lt;li&gt;lipids&lt;/li&gt;&#xA;&lt;li&gt;amino acids&lt;/li&gt;&#xA;&lt;li&gt;cellulose&lt;/li&gt;&#xA;&lt;li&gt;various other non-polymeric organic molecules of various levels of complexity&lt;/li&gt;&#xA;&lt;li&gt;salts&lt;/li&gt;&#xA;&lt;li&gt;proteins&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These together will give you a reasonable approximation of food which will be most of what you need to stay alive, without necessarily going into the complexity of actually producing a better analogue of dead tissue &quot;under the hood&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sugars and starches are entirely composed of carbon, hydrogen, and oxygen. The same goes for lipids, cellulose, and the majority of the organic components you'll need. For amino acids, and the other organic bits, you'll need to add nitrogen and sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need phosphorus for DNA, RNA, ADP, and ATP, as well as a whole bunch of other signalling and respiration-based molecules, and bone material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need sodium and potassium ions which are needed in living tissue for nerve signalling, and you also need calcium ions to make bones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many plant materials seem to contain small amounts of boron, though the biological role of it isn't concrete as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the proteins are where the metallic awkwardness comes in. A lot of proteins have a reaction site containing a metal, metal-sulfur, metal-porphyrin, or another metal-organic molecule complex, which can come in various metallic flavours. For example, haemoglobin requires an iron-porphyrin complex, while chlorophyll requires magnesium-porphyrin, and a lot of respiration-based proteins seem to involve iron-sulfur complexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly confident you could make reasonable analogues of most food with C, H, O, N, S, P, Na, K, Cl, and Ca, with relatively small additional quantities of B, Fe, and Mg. You might suffer after a while from micronutrient deficiencies, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'll probably find that I've forgotten something important somewhere.)&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="402" LastEditDate="2017-04-08T22:00:37.443" LastActivityDate="2017-04-08T22:00:37.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5411" PostTypeId="2" ParentId="5409" CreationDate="2013-06-22T22:10:47.413" Score="5" Body="&lt;p&gt;I'm not entirely sure of this, feel free to correct me.&lt;/p&gt;&#xA;&lt;p&gt;This probably depends on the type of appliance. I'll try two, the internal combustion engine and the stove.&lt;/p&gt;&#xA;&lt;h2&gt;Internal combustion engine&lt;/h2&gt;&#xA;&lt;p&gt;The nice thing about an internal combustion engine is that it takes in gas and oxygen, isolates it, and &lt;em&gt;then&lt;/em&gt; ignites it. Besides, it uses a spark to ignite it. Hydrogen is more explosive than natural gas, so the isolation is quite useful here. And the reaction with air is more easily triggered by a spark rather than by heat.&lt;/p&gt;&#xA;&lt;p&gt;On modification needed to make it work would be to recalibrate the apertures of the oxygen and hydrogen feeds such that the gas mixture is sucked in in a 1:2 ratio.&lt;/p&gt;&#xA;&lt;p&gt;Secondly, there needs to be a cleaner way of disposing the water formed. Internal combustion engines assume that the exhaust is gas and particulate matter -- which can be compressed and expelled entirely, and isn't adhesive. Water can't be compressed. This makes it easier to expel, but some of it may adhere to the walls. As water, oxygen, and electricity are being supplied, this may quickly rust the walls of the cylinder. So there needs to be a way to protect the cylinder  from rusting. Also, the engine will just stop working if the spark plugs get shorted by water. There needs to be a way to prevent that from happening, too.&lt;/p&gt;&#xA;&lt;p&gt;Actually, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_internal_combustion_engine_vehicle&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen internal combustion engine&lt;/a&gt; exists, but it looks like they got there by starting from scratch, not modifying a regular internal combustion engine.&lt;/p&gt;&#xA;&lt;h2&gt;The stove&lt;/h2&gt;&#xA;&lt;p&gt;The only issue I see here is that the aperture on a normal stove is too large. Oxygen may enter and end up exploding the whole setup.&lt;/p&gt;&#xA;&lt;p&gt;In such a case, having one pipe (with a very small bore) for each individual flame may work better. This would probably have to be lit automatically -- I don't see how one can safely light a set of such pipes without there being an individual knob for each pipe.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Nowadays, the more efficient way of using hydrogen for fuel is in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuel_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;fuel cell&lt;/a&gt;. This way, expensive hydrogen doesn't get wasted, and is instead reused (the energy input comes from other sources)&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-06-22T22:16:50.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5412" PostTypeId="2" ParentId="5369" CreationDate="2013-06-22T23:31:19.770" Score="4" Body="&lt;p&gt;I think, there are no secret properties of tantalum. The primer corrosion resistant is thin layer of Tantalum pentoxide $(\ce{Ta2O5})$. It's more case of a reaction of an oxide with fluorides. Here some related quotes from the book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444516042&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry of Tantalum and Niobium Fluoride Compounds&lt;/a&gt; by Anatoly Agulyansky:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As opposed to the oxides, fluoride compounds are characterized by the&#xA;  formation of mostly ionic bonds. This peculiarity is related to fluorine's high&#xA;  electronegativity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxygen and fluorine ions have very similar ionic radii. &lt;strong&gt;This steric similarity&#xA;  ensures relatively easy substitution of the oxygen in the compounds, by&#xA;  fluorine ions&lt;/strong&gt;. Such substitution opens up huge possibilities for the synthesis of&#xA;  oxyfluoride compounds with desired crystal structure and properties.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Fluorine-oxygen substitution, yielding oxyfluoride compounds, can be&#xA;  performed without changing the initial oxide crystal structure, provided the&#xA;  changes in the anionic sublattice are compensated by appropriate substitutions&#xA;  in the cationic sublattice. In these particular cases, the properties of the&#xA;  compounds will change due to variations in the character of the chemical&#xA;  bonds. On the other hand, fluorine-oxygen substitution can also cause cardinal&#xA;  changes in the initial crystal structure. Hence, compounds with desired&#xA;  structures and specific properties can be obtained only through controlled&#xA;  fluorine-oxygen substitution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1502" LastEditorUserId="402" LastEditDate="2017-04-08T22:33:22.247" LastActivityDate="2017-04-08T22:33:22.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5413" PostTypeId="2" ParentId="5409" CreationDate="2013-06-23T00:24:59.773" Score="6" Body="&lt;p&gt;Pure hydrogen, as opposed to hydrocarbons, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_damage&quot;&gt;degrades metals&lt;/a&gt;.  $\ce {H_2}$ molecules are tiny, so they wedge themselves into the relatively large metallic crystal structures and form hydrides.  This is the entire reason nickel-metal hydride (NiMH) batteries work - the hydrogen is stored in an iron-hydride matrix. An appliance that might last 20 or 30 years burning natural gas might not make it past ten years burning $\ce{H_2}$, and a failure may take the form of a dangerous leak.&lt;/p&gt;&#xA;" OwnerUserId="1824" LastEditorUserId="22" LastEditDate="2013-06-26T09:07:40.513" LastActivityDate="2013-06-26T09:07:40.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5414" PostTypeId="1" CreationDate="2013-06-23T05:57:49.203" Score="1" ViewCount="9633" Body="&lt;p&gt;Why is methoxide strong base? In case of halogens, I get why one is weak base, and one is strong base, but in the case of ones with oxygen, this seems to be hard to figure out.&lt;/p&gt;&#xA;" OwnerUserId="1857" LastEditorUserId="73" LastEditDate="2013-06-23T08:53:51.167" LastActivityDate="2016-07-14T16:03:23.970" Title="Why is methoxide a strong base?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5415" PostTypeId="1" CreationDate="2013-06-23T08:48:04.263" Score="2" ViewCount="1309" Body="&lt;p&gt;The process of racemization of an optically compound is entropy driven and hence nature prefers a racemic mixture to an optically active one, which is consistent with the idea of the intuitive understanding of entropy as the degree of randomness or disorder. Then, how can the entropy change of a process of conversion of an optically pure compound to its racemic mixture be computed theoretically or atleast estimated accurately enough so that it can be compared for two different reaction outcomes or the spontaneity of a reaction be estimated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, which has higher entropy, a racemic mixture or a mixture of all possible stereoisomers of a compound including the meso form? In the later case, there is more randomness and hence should possess greater entropy. Can we treat a mixture of stereoisomers as a mixture of two different compounds and then use their mole fractions (or activities) to calculate the entropy of mixing which in case of gases is $-nR(x_1\ln {x_1} +x_2\ln{x_2})$.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-09-20T10:21:16.613" LastActivityDate="2013-09-20T10:21:16.613" Title="Entropy and racemization?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;stereochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5418" PostTypeId="1" CreationDate="2013-06-23T15:57:25.537" Score="2" ViewCount="1856" Body="&lt;p&gt;I have searched the internet and I found that reusing PET no.1 is dangerous. Why it is dangerous and what is the equation of the reaction between it and water tap ? And why it is not danger to when it is used for the first time in water bottles but it is danger when reusing it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links I have found useful: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://greenliving.nationalgeographic.com/plastic-drink-bottles-safe-reuse-drinking-water-3110.html&quot; rel=&quot;nofollow&quot;&gt;http://greenliving.nationalgeographic.com/plastic-drink-bottles-safe-reuse-drinking-water-3110.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.plasticsinfo.org/Main-Menu/MicrowaveFood/Need-to-Know/Plastic-Bev-Bottles/The-Safety-of-Polyethylene-Terephthalate-PET.html&quot; rel=&quot;nofollow&quot;&gt;http://www.plasticsinfo.org/Main-Menu/MicrowaveFood/Need-to-Know/Plastic-Bev-Bottles/The-Safety-of-Polyethylene-Terephthalate-PET.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1868" LastActivityDate="2014-12-10T20:43:03.553" Title="Polyethylene terephthalate reaction with tap water" Tags="&lt;plastics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5419" PostTypeId="2" ParentId="5030" CreationDate="2013-06-23T16:28:16.653" Score="4" Body="&lt;p&gt;It seems that the mechanism of particulate formation is not yet well known. &lt;a href=&quot;http://www.dieselnet.com/tech/dpm.php&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; (preview) states some basics about the composition and formation of Diesel Particulate Matter (DPM): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diesel particulates form a very complex aerosol system. Despite considerable amount of basic research, neither the formation of PM in the engine cylinder, nor its physical and chemical properties or human health effects are fully understood.&#xA;  [...]&#xA;  The basic fractions of DPM are elemental carbon, heavy hydrocarbons derived from the fuel and lubricating oil, and hydrated sulfuric acid derived from the fuel sulfur. DPM contains a large portion of the polynuclear aromatic hydrocarbons (PAH) found in diesel exhaust&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, it seems that the particulates are formed both during combustion and from gas during dilution. During combustion, carbonaceous agglomorates (soot), metallic ash from lubricating oil, and precursors to particle formation during dilution (sulfur oxides and partially burned hydrocarbons from fuel and lubricating oil) are formed. During dilution, materials such as sulfuric acid and other sulfates, as well as heavy hydrocarbons and hydrocarbon derivatives, are for the most part adsorbed onto the carbonaceous agglomorates formed during combustion. Some material may also nucleate by itself, but most nucleation occurs on the agglomorates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aqmd.gov/tao/ultrafine_presentations/Session1_1_Kittleson.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF with explanation of ultrafine particle formation mechanisms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, i found it difficult to find any reason why this is such a big problem with diesel and not with gasoline. Perhaps the composition of the diesel is to blame, not temperature or oxygen supply.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditDate="2013-09-11T11:03:34.063" LastActivityDate="2013-09-11T11:03:34.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5420" PostTypeId="1" CreationDate="2013-06-23T21:44:14.740" Score="1" ViewCount="1763" Body="&lt;p&gt;I have a can of spray for self-protection which has expired way back in 2008. It hasn't been used at all, and it doesn't seem a good idea to just throw it away without disabling it somehow first. I'm looking for ways to disarm it with minimum risk to myself and others. Obviously, I can just spray it away, but I'd rather not if I can avoid it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The active ingredient is &lt;a href=&quot;http://en.wikipedia.org/wiki/CS_gas&quot; rel=&quot;nofollow&quot;&gt;2-chlorobenzalmalononitrile&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not water-soluble, but I was thinking of some combination of cooking oil and alkaline agent (baking soda, for example) to neutralize it. I'm still doing research on it, and I won't be doing anything rash just yet :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any particular advice how to neutralize this spray?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If jurisdiction is somehow important, assume European Union.&lt;/p&gt;&#xA;" OwnerUserId="1870" LastEditorUserId="22" LastEditDate="2013-06-24T10:51:37.093" LastActivityDate="2013-06-24T11:41:55.943" Title="How to disarm and correctly dispose of tear gas spray?" Tags="&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5421" PostTypeId="1" AcceptedAnswerId="5429" CreationDate="2013-06-23T22:05:53.850" Score="2" ViewCount="10662" Body="&lt;p&gt;Can a spark ignite hydrogen? Also, what about the tip of a cigarette? I've heard that hydrogen is very explosive. Thanks&lt;/p&gt;&#xA;" OwnerUserId="1865" LastActivityDate="2021-03-26T15:20:39.197" Title="Can a spark ignite hydrogen?" Tags="&lt;hydrogen&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5422" PostTypeId="2" ParentId="5418" CreationDate="2013-06-24T01:34:07.483" Score="3" Body="&lt;p&gt;It is probably not a chemical reaction that's the problem, it's that the container should be cleaned before it's refilled, and an unclean container might or will have bacteria.  While you're thinking 'refilled' might be in a matter if minutes, someone labeling the bottle is thinking about it sitting on a shelf with the cap off for months before reuse.  At that point there could be any number of unhealthy contaminants.&lt;/p&gt;&#xA;" OwnerUserId="1824" LastActivityDate="2013-06-24T01:34:07.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5425" PostTypeId="2" ParentId="5158" CreationDate="2013-06-24T04:35:08.650" Score="3" Body="&lt;p&gt;Making pure silicon is somewhat like making aluminum, you heat SiO2 (sand, or more specifically, quartz) to a liquid state and hydrolyze with two carbon electrodes. One electrode combines the oxygen with the carbon in the electrode, releasing CO2, the silicon and other impurities remain behind.  When this cools, you have foundry grade silicon, used mostly as a steel additive.  Since this involves extremely high temperatures it's unlikely you'll be doing this in your garage (although it is, literally, possible).  As an alternative, some 'thin film' panels are printed, usually on a plastic backing, you would have to find the printer and the proprietary ink.  Once you had bought a $100,000 printer and very expensive 'ink', you're ready to roll.&lt;/p&gt;&#xA;" OwnerUserId="1824" LastActivityDate="2013-06-24T04:35:08.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5426" PostTypeId="2" ParentId="4271" CreationDate="2013-06-24T06:59:37.257" Score="8" Body="&lt;p&gt;Some recent developments shown in article &lt;a href=&quot;http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=6458107&quot; rel=&quot;noreferrer&quot;&gt;Transversely quasicrystallographic 3D photonic chiral lattices: Polarization-controllable complex photonic band gap structures&lt;/a&gt;, where autors claimed experimantal results on produsing 3d chiral photonic quasicrystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pFVe0.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Figure. Computer simulation of the light intensity distribution of&#xA;  the interference pattern for transversely quasicrystallographic 3D&#xA;  chiral photonic lattices respectively with rotational symmetries&#xA;  (a)5-fold (d) 8-fold (g) 12-fold. (b), (e) and (h) are their respective&#xA;  x-y planes. (c), (f) and (i) are their respective x-z planes&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;&quot;So the fabricated class of 3D PQC chiral lattices integrate both these wonder features on asingle platform in view of their possible applications in new generation integrated optical photonic devices with tunable polarization band gaps as well as frequency band gaps&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I also found fascinating that some researchers, use combination of structures like simple lattice and chiral lattice. This presentation &lt;a href=&quot;http://www.chemie.tu-darmstadt.de/media/ak_schneider/fsp_nanomaterialien/ringvorlesung/vortrag_prof_wegener.pdf&quot; rel=&quot;noreferrer&quot;&gt;Photonic Crystals and Metamaterials&lt;/a&gt;&#xA;has a lot of pictures, which could be good &quot;food for thoughts&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zAe5C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this was helpful, I think your area of research is very interesting.&lt;/p&gt;&#xA;" OwnerUserId="1502" LastEditorUserId="1502" LastEditDate="2013-06-24T12:03:54.023" LastActivityDate="2013-06-24T12:03:54.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5427" PostTypeId="2" ParentId="5420" CreationDate="2013-06-24T11:41:55.943" Score="4" Body="&lt;p&gt;&lt;strong&gt;Do not open, do not try to neutralize it yourself!&lt;/strong&gt; Bring it to &lt;strong&gt;Recycling center&lt;/strong&gt; (&lt;em&gt;DE&lt;/em&gt;: Wertstoffhof). If your city is 100k+ there should be one. That's the place where you bringing your old furniture, the rubbish after renovations, and so on. They have a part for disposal of &lt;strong&gt;hazardous waste&lt;/strong&gt; (&lt;em&gt;DE&lt;/em&gt;: Sondermüll). What usually comes in there are paints, old car batteries, old pesticides, &lt;em&gt;etc...&lt;/em&gt;. Another possibility - if you have some friend at chemistry school of some local university, give it him. Each Chemistry faculty has some &quot;disposal&quot; facility.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-06-24T11:41:55.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5428" PostTypeId="1" CreationDate="2013-06-24T14:51:13.547" Score="0" ViewCount="4749" Body="&lt;p&gt;Is water neutral in nature or amphoteric in nature? I understand it as an neutral compound in complex compounds, but I've been told that it is amphoteric, too.&lt;/p&gt;&#xA;" OwnerUserId="1874" LastEditorUserId="73" LastEditDate="2013-06-24T15:37:25.857" LastActivityDate="2018-03-04T02:55:17.030" Title="What is the nature of water?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;water&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5429" PostTypeId="2" ParentId="5421" CreationDate="2013-06-24T15:37:13.007" Score="6" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Can a spark ignite hydrogen? Also, what about the tip of a cigarette?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, if the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2(g)}$&lt;/span&gt; concentration in oxygen or air is greater than about 4% not only will it burn, but most likely explode. You should avoid all heat sources when dealing with it, including direct sunlight. Just so you know, hydrogen safety datasheets include words such as &amp;quot;extremely flammable&amp;quot; and &amp;quot;DANGER&amp;quot; in red capital letters.&lt;/p&gt;&#xA;&lt;p&gt;Besides its high flammability, another thing that makes it so dangerous is that you can barely see its flames when there's a high oxygen concentration, as seen below. It also reacts violently (A.K.A. explodes) with most oxidizing compounds, so you should be careful where you are storing it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;A rocket engine at full thrust. There's a barely visible bluish flame below it, that's the hydrogen-oxygen mix used as fuel.&lt;/em&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7vs5S.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="2133" LastEditDate="2021-03-26T15:20:39.197" LastActivityDate="2021-03-26T15:20:39.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5431" PostTypeId="2" ParentId="5428" CreationDate="2013-06-24T23:24:49.957" Score="6" Body="&lt;p&gt;$\ce{H2O}$ is a both a neutral molecule and an amphoteric molecule &amp;mdash; the two are not mutually exclusive terms. An amphoteric molecule is simply one that can act as either an acid or a base, while a neutral molecule is one in which the total number of protons is equal to the number of electrons, such that there is zero net charge. Water can act as a Brønsted acid (i.e., an $\ce{H+}$ donor), as per the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + NH2- -&amp;gt; OH- + NH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it can also act as a Brønsted base (i.e., an $\ce{H+}$ acceptor):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + HCl -&amp;gt; H3O+ + Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When proton transfer is involved, as in the Brønsted definition of acids and bases, the more precise term is &lt;em&gt;amphiprotic&lt;/em&gt;. However, water can also act as both a Lewis acid (an electron pair acceptor) and Lewis base (an electron pair donor). Notably, as a Lewis base, it is not limited merely to proton abstraction, but can act as a nucleophile with any number of electrophilic molecules.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-04T02:55:17.030" LastActivityDate="2018-03-04T02:55:17.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5432" PostTypeId="1" CreationDate="2013-06-25T03:32:55.213" Score="2" ViewCount="549" Body="&lt;p&gt;I have a bottle of baby powder that is 15% Zinc oxide and the rest is talc. Is there an over-the-counter way of purifying zinc oxide from it? I'm trying to make some sunscreen from it.&lt;/p&gt;&#xA;" OwnerUserId="1876" LastEditorUserId="22" LastEditDate="2013-06-25T12:07:42.610" LastActivityDate="2013-06-25T15:51:24.750" Title="Extracting Zinc (oxide) from baby powder" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5435" PostTypeId="2" ParentId="5432" CreationDate="2013-06-25T14:16:58.137" Score="4" Body="&lt;p&gt;This might be not worth it, because you will need special equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Description totally theoretical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixture consist of $Mg_3Si_4O_{10}(OH)_2$ - tale powder, $ZnO$ - zink oxide powder, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Starch&quot; rel=&quot;nofollow&quot;&gt;starch&lt;/a&gt;. 81% to 15% to 4% it's most common variation with 15% zink oxide, there also 10% mixtures, but process will be the same.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You need to get rid of starch. Best choice to convert it to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dextrin&quot; rel=&quot;nofollow&quot;&gt;dextrine&lt;/a&gt;, because starch in not fully soluble in water. You need to heat it in oven on a foil for about 2 hours, with temperature $200^{\circ}C$. It will have no effect for other components, zync oxide might colorize to orange, but when heat is gone it will be white again. Dextrine will be yellow color and it's fully soluble in water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mix new mixture with water, and pour it out. Now you have only $Mg_3Si_4O_{10}(OH)_2$ and $ZnO$ in you mixture. Both of them not soluble in water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then you need to obtain or build equipment for &lt;a href=&quot;http://en.wikipedia.org/wiki/Sublimation_%28phase_transition%29&quot; rel=&quot;nofollow&quot;&gt;sublimation&lt;/a&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Sublimation_apparatus&quot; rel=&quot;nofollow&quot;&gt;sublimation apparatus&lt;/a&gt;. Because compounds have different melting point &lt;a href=&quot;http://www.wolframalpha.com/input/?i=talc&quot; rel=&quot;nofollow&quot;&gt;Talc&lt;/a&gt; - $800^{\circ}C$, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=zink%20oxide&quot; rel=&quot;nofollow&quot;&gt;zync oxide&lt;/a&gt; - $1975^{\circ}C$, you can get them one by one. You will need to freeze and dry your mixture first.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat over $800^{\circ}C$ to obtain talc, left over structure will be zinc oxide.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about sublimation from start, with starch inside, but this method &lt;strong&gt;in theory&lt;/strong&gt; should work.&lt;/p&gt;&#xA;" OwnerUserId="1502" LastEditorUserId="1502" LastEditDate="2013-06-25T15:51:24.750" LastActivityDate="2013-06-25T15:51:24.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5437" PostTypeId="1" CreationDate="2013-06-25T16:36:28.513" Score="-11" ViewCount="34186" Body="&lt;p&gt;I am taking an Organic Chemistry class in my college and I was wondering what the Molecular Formula for CH3-C≡C-CH2-CH3 was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be nice to have the class of compound.&lt;/p&gt;&#xA;" OwnerUserId="1881" LastActivityDate="2013-06-25T17:29:03.707" Title="What is the Molecular Formula for CH3-C≡C-CH2-CH3" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2013-06-26T11:11:53.967" ContentLicense="CC BY-SA 3.0" />
  <row Id="5438" PostTypeId="2" ParentId="5409" CreationDate="2013-06-25T16:58:23.800" Score="3" Body="&lt;p&gt;A few addition comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The production of water would create a problem, as if being used as a  heat source, how is the water going to escape? Steam and condensation can cause many problems, not to mention; there is the issue of steam and the safety risk to end user. All these factors also cause metal pipes, machinery and appliances to (potentially) rust. On top of the points mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hindenberg is an interesting example of the flammability of Hydrogen and the effects of it's products when combusted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an point of interest, I have provided a link for a table of flammable gases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen's &quot;Lower Explosive or Flammable Limit&quot; (LEL/LFL) (%) is 4, in the order of methane and propane 5 and 2.1 respectively, bit it's &quot;Upper Explosive or Flammable Limit&quot; (UEL/UFL) (%) is much higher 75 as compared to 15 and 10.1 for methane and propane, respectively. Making it more highly combustible at concentrated levels, leading to the difficulties in containing such explosions, as mentioned in the other answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still remember clearly a video our Chemistry Lecturer showed us of the Hindenberg and how the sheet of water (produced from the combustion of Hydrogen) saved lives. That was over 30 years ago, so it made an impression and it's a subject I find interesting.&lt;/p&gt;&#xA;" OwnerDisplayName="user27560" LastEditorDisplayName="user27560" LastEditDate="2013-06-25T17:22:58.860" LastActivityDate="2013-06-25T17:22:58.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5439" PostTypeId="2" ParentId="5437" CreationDate="2013-06-25T17:29:03.707" Score="5" Body="&lt;p&gt;Are you sure you didn't mean the name of this molecule? Because strictly speaking, the molecular formula of the molecule you posted would be $\ce{C5H8}$. Pent-2-ene has molecular formula $\ce{C5H10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule you asked about has a triple bond, meaning that it's an &lt;strong&gt;alkyne&lt;/strong&gt; and in this case, pent-2-yne. If you want the wiki page, it's &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Pentyne&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count the number of carbon atoms and the number of hydrogen atoms to be sure.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2013-06-25T17:29:03.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5440" PostTypeId="1" AcceptedAnswerId="5456" CreationDate="2013-06-25T17:46:39.690" Score="1" ViewCount="2316" Body="&lt;p&gt;I have heard that chemical reactions create energy.  I wanted to know how this energy was created (specifically light energy) and how this energy came about.  I wanted to know if the energy produced was made of particles (for example, light energy is made of photons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer&lt;/strong&gt; I already knew about heat, kinetic, potential, and sound energy.  What I was really looking for was light energy, which is caused by a phenomenon known as chemiluminescence.  &lt;/p&gt;&#xA;" OwnerUserId="1046" LastEditorUserId="1046" LastEditDate="2013-07-01T17:44:38.703" LastActivityDate="2013-07-01T17:44:38.703" Title="What is the energy released in chemical reactions?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5441" PostTypeId="2" ParentId="5369" CreationDate="2013-06-25T21:25:30.127" Score="8" Body="&lt;p&gt;I think the answer to your question can be described based primarily on thermodynamic considerations.  Ta has very high fusion (25 kJ/mol), vaporization (760 kJ/mol) and atomization (780 kJ/mol) enthalpies which are a result of strong metal-metal bonding in the solid (see Greenwood and Earnshaw, Chemistry of the Elements).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further understand the protective nature of the native oxide that forms on Ta, it is fruitful to look at an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ellingham_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ellingham diagram&lt;/a&gt;, which describes the stability of a metal oxide as a function of temperature. For more authoritative sources than the Wikipedia page, see a comprehensive Inorganic Chemistry textbook such as those by Shriver and Atkins or Housecroft and Sharpe.  I found &lt;a href=&quot;http://www.engr.sjsu.edu/ellingham/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; particularly interesting for purposes of visualizing Ellingham diagrams and used it to make the following plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0iBNj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies for the color scheme, which was not adjustable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I've plotted the free energy for the oxide formation of tantalum oxide along with oxides of iron and aluminum.  The x-axis is temperature.  The large negative values for each of these oxides indicates that they are all thermodynamically favorable at the temperatures plotted.  Room temperature would be very slightly to the left of the intersection of the x and y-axes.  A complete analysis of the stability of the metal oxides would also require consideration of kinetics, but we can use thermodynamics to postulate that the very large negative values for the free energy of formation of the tantalum oxide are one factor in the stability of the oxide film.  I'll leave it to the OP to play around with thermochemical cycles, which should demonstrate why the oxide films are resistant to many types of acids.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="402" LastEditDate="2017-04-08T22:34:33.397" LastActivityDate="2017-04-08T22:34:33.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5443" PostTypeId="2" ParentId="5406" CreationDate="2013-06-26T00:09:14.473" Score="1" Body="&lt;p&gt;I think the answer is 26 - as suggested from this &lt;a href=&quot;http://www.nmsea.org/Curriculum/Primer/forms_of_energy.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;periodic table of life&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ehQ07.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can estimate that the raw materials in the human body amount to approximately $3.50 (so you probably can't sell off your little brother for more than a buck and a quarter).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-06-26T00:09:14.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5445" PostTypeId="1" CreationDate="2013-06-26T04:52:50.657" Score="2" ViewCount="309" Body="&lt;p&gt;So I was thinking about something to do for a science fair and photosynthesis sounded like a good idea.  I was wondering what the exact process of photosynthesis was on a chemical level.  I know the formula, but how exactly is water and carbon dioxide converted into glucose and oxygen?  And where does sunlight come in, not in the equation but the chemical process.  Finally, any ideas for artificial photosynthesis?&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="1878" LastEditDate="2013-06-26T09:07:23.073" LastActivityDate="2013-06-29T05:43:37.637" Title="Photosynthesis, science fair" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2017-09-19T17:08:08.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="5446" PostTypeId="1" AcceptedAnswerId="5449" CreationDate="2013-06-26T05:07:33.237" Score="1" ViewCount="81" Body="&lt;p&gt;I have been watching some videos on how batteries work and they seem to talk about the cathode being the most important. I also watched another video a long time ago and while my memory is blurred it talked about impurities in the cathode being the only reason the electrical charge doesn't flow constantly. While I found this weird I also wonder how generates seem to get this unlimited supply of electrons also so it's only fair for me to assume that the battery can also.&lt;/p&gt;&#xA;" OwnerUserId="1885" LastActivityDate="2013-07-26T07:43:52.170" Title="In a battery what would happen if you had a perfectly uniform and pure cathode?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5448" PostTypeId="1" CreationDate="2013-06-26T07:04:48.823" Score="9" ViewCount="7130" Body="&lt;p&gt;I have to determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ dioxygen by mass in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have started as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For our convenience we take $\pu{100 g}$ of that iron oxide:&lt;br&gt;&#xA;We have $\pu{69.9 g}$ of iron in it and $\pu{30.1 g}$ dioxygen in it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We convert into amount of substance of each element:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{amount of substance of iron} = 69.9/55.8 = 1.25$&lt;/li&gt;&#xA;&lt;li&gt;$\text{amount of substance of dioxygen} = 30.1/32 = 0.94$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is my method and calculations right? &lt;/p&gt;&#xA;" OwnerUserId="1886" LastEditorUserId="4945" LastEditDate="2018-02-20T10:48:33.907" LastActivityDate="2018-02-20T10:57:05.153" Title="How to determine the empirical formula for an unknown iron oxide?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5449" PostTypeId="2" ParentId="5446" CreationDate="2013-06-26T07:09:36.943" Score="3" Body="&lt;p&gt;I don't think that's accurate. Even the universe doesn't have an &amp;quot;unlimited supply of electrons&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Impurities in the cathode likely give rise to internal resistance inside the battery (among other things), leading to deviation from an ideal cell. I would guess that as your cathode approached purity, your battery would approach the behavior of 'ideal cells', and that discharge rates would approach those of capacitors, and efficiency would approach 100%. To get an ideal cell though, the anode and electrolyte would also have to be 'pure' and some real-world effects would have to be ignored.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Because of internal resistance, the terminal voltage of a cell that is discharging is smaller in magnitude than the open-circuit voltage and the terminal voltage of a cell that is charging exceeds the open-circuit voltage. An ideal cell has negligible internal resistance, so it would maintain a constant terminal voltage of &lt;span class=&quot;math-container&quot;&gt;$ \epsilon $&lt;/span&gt; until exhausted, then dropping to zero. If such a cell maintained 1.5 volts and stored a charge of one coulomb then on complete discharge it would perform 1.5 joule of work. In actual cells, the internal resistance increases under discharge&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Battery_(electricity)#Battery_cell_performance#Principle_of_operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What do you mean by pure, exactly? Complete homogeneity and lack of any other elements? That sounds like a physics problem where you &amp;quot;ignore air resistence&amp;quot; and other real world effects for simplicity. If this is what you're after, search for things like 'Ideal Battery' and 'Ideal Electrochemical Cell'.&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-06-26T07:09:36.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5450" PostTypeId="2" ParentId="5448" CreationDate="2013-06-26T07:40:03.367" Score="6" Body="&lt;p&gt;First assume the mass of an oxide of iron to be $\pu{100 g}$ which implies mass of iron would be $\pu{69.9 g}$ and mass of oxygen would be $\pu{30.1 g}$. Now calculate amount of substance of both the elements by dividing the assumed mass by their atomic number.&lt;br&gt;&#xA;Then calculate relative amount of substance of each element by dividing the obtained amount of substance of each element by the smallest of them. (E.g. suppose you've got $\pu{5 mol}$ of iron and $\pu{2 mol}$ of oxygen, divide 5 by 2 and 2 by 2 to get relative amounts of substance of both.)&lt;br&gt;&#xA;Now just write relative amount of substances in the form of a ratio and bring the obtained ratio to least simplified form. (Remember the ratio should come in whole numbers.)&#xA;After obtaining ratio, write the empirical formula as I did in this example.&#xA;(eg. suppose I got this ratio $2:5$ from last example. So the empirical formula will be $\ce{Fe2O5}$ .)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, you can skip the relative amount of substance step when you've got 2 elements in empirical formula, but you will need to perform all the steps when there are more than 2 elements as the answer comes to be different in some cases.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2018-02-20T10:54:05.723" LastActivityDate="2018-02-20T10:54:05.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5451" PostTypeId="1" CreationDate="2013-06-26T08:47:07.853" Score="-5" ViewCount="3427" Body="&lt;p&gt;$\ce{NH4-}$ is conjugate acid and $\ce{NH4}$ is conjugate base. $\ce{NH4-}$ is conjugate acid because it has one more Hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that is difficult for me to understand is why $\ce{NH4-}$ is anion? Is negative sign refer to whole molecule or just to Hydrogen? How a molecule can be anions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of the same situation is $\ce{HCO3-}$ (conjugate base) and $\ce{H2CO3}$ (Conjugate acid). By adding one Hydrogen to the base the negative sign goes away. Why?&lt;/p&gt;&#xA;" OwnerUserId="1743" LastEditorUserId="4231" LastEditDate="2015-03-04T20:33:08.430" LastActivityDate="2015-03-04T20:33:08.430" Title="Why is NH₄⁻ an anion and NH₃ is not?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-13T21:43:41.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="5452" PostTypeId="2" ParentId="5451" CreationDate="2013-06-26T09:36:27.923" Score="13" Body="&lt;p&gt;I'm quite confused about your post and even its title. So let's start wit title: ammonia  $\ce{NH3}$ is neutral because of electron count. Now the $\ce{NH4-}$ is tricky. You would have to add hydride $\ce{H-}$ to ammonia. As far as I'm aware it's utopia. Hydorgen atoms in ammonia are protic so they would immediately react with  $\ce{H-}$ giving $\ce{H2}$. I never heard that ammonia can react with metal hydrides as  $\ce{KH}$ or lithium triethylborohydride $\ce{Li[HBEt3]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the text. Ammonia is amphoteric (&lt;em&gt;i.e.&lt;/em&gt; it can get proton or loose it; or it can act as a base or acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conjugate acid of ammonia ($\ce{NH3}$) is the ammonium cation $\ce{NH4+}$. Acid, in Brønsted theory,  is a proton donor ($\ce{H+}$) and on loosing proton turns into base. In this case our acid is positively charged, so if it looses proton it turns into neutral molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conjugate base of ammonia is the amide anion $\ce{NH2-}$. If ammonia is an acid then to become base it must loose $\ce{H+}$. The final molecule is negatively charged. If you want to do it experimentally, you need to dissolve alkali metal ($\ce{Na}$) in liquid ammonia. H-radical is replaced by metal and dimerise to hydrogen $\ce{H2}$ (and salt - sodium amide is created $\ce{NaNH2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now your last question: &quot;By adding one Hydrogen to the base the negative sign goes away&quot;. You are not adding Hydrogen (atom) but proton ($\ce{H+}$).&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="327" LastEditDate="2013-06-26T11:46:10.590" LastActivityDate="2013-06-26T11:46:10.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5453" PostTypeId="2" ParentId="5421" CreationDate="2013-06-26T09:45:17.263" Score="6" Body="&lt;p&gt;Actually hydrogen is not at all explosive. The mixture of hydrogen with oxygen (also air) is. As you know hydrogen is stored as liquid (high pressure cylinders) and it's perfectly stable. You can mix it with any amount of argon heat it as high as you want and nothing will happened. OK, if you reach temperature as high as on the Sun then it will explode, but that's an nuclear fusion (&lt;em&gt;i.e.&lt;/em&gt; physics not chemistry). On the other side the mixture of hydrogen with oxygen can very easily by ignited by an electric spark or a cigarette.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-06-26T09:45:17.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5454" PostTypeId="2" ParentId="2820" CreationDate="2013-06-26T10:02:33.180" Score="2" Body="&lt;p&gt;You didn't say anything about size of the system. Also what computer hardware is available for your computations.&#xA;For TM-complexes you need to get better basis than lanl2 and definitely you need much better basis for your ligands. sto3g is a joke. Start with 6-31G(D,P) and LANL2DZ. For good data go for 6-311+G(d) for non-metal and LANL2TZ+f for Ni and CO. Another possibility is def2-svp (it has all 4th raw metals in as well as non-metals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General remark. I hope that you have pre-optimized your ligand before you started with the TM-complex.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-06-26T10:02:33.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5455" PostTypeId="1" CreationDate="2013-06-26T10:31:16.393" Score="8" ViewCount="3146" Body="&lt;p&gt;I can not find any good resources on the Internet except for some shady drug syntheses. I was hoping that someone could provide me with some more educational information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in imine-amine reduction methods, but it was hardly discussed in my class. It would be very nice if someone could explain the exact reaction mechanism of the imine-amine reduction in the presence of an aluminium amalgam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found that the very dangerous compound Mercury(II) chloride was used a lot in this method and I was wondering why for example Mercury sulfide is not used, as I thought it would be a lot safer since it is hardly soluble in water and is therefore less likely to do any physical harm.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="865" LastEditDate="2015-10-16T17:47:27.630" LastActivityDate="2015-10-16T17:47:27.630" Title="Reaction mechanism of imine reduction to amine using an aluminium amalgam" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5456" PostTypeId="2" ParentId="5440" CreationDate="2013-06-26T16:34:29.197" Score="4" Body="&lt;p&gt;Usually it's radiation. Depending on how exotermic reaction was, in which medium was it done, result is light staring with infra-red up to v.short ultra-violet. You can't generate particles like electrons  because  the low of Charge conservation must be obeyed (you transferring the electrons from one substance to the other). &lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2013-06-26T16:34:29.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5457" PostTypeId="2" ParentId="5414" CreationDate="2013-06-26T16:42:09.543" Score="7" Body="&lt;p&gt;Methoxide ($\ce{CH3O-}$) is the conjugate base of methanol. Methanol is very weak acid (e.g. its dissociation constant is very small), so its conjugated base is very strong.&lt;/p&gt;&#xA;" OwnerUserId="1889" LastEditorUserId="13184" LastEditDate="2016-07-14T16:03:23.970" LastActivityDate="2016-07-14T16:03:23.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5458" PostTypeId="2" ParentId="5342" CreationDate="2013-06-26T16:49:35.090" Score="0" Body="&lt;p&gt;If you use strong base for titration the titration curve will be very steep. It makes it easier to observe the equilibrium point.&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2013-06-26T16:49:35.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5459" PostTypeId="2" ParentId="5288" CreationDate="2013-06-26T16:59:26.713" Score="2" Body="&lt;p&gt;No. Per definition you need to have at least local supersaturation. You can archive it by different means (cooling, evaporating of solvent, mixing with other solvent...) but in the end you need supersaturation. See: &lt;a href=&quot;http://xray.chem.ufl.edu/growing%20tips.htm&quot; rel=&quot;nofollow&quot;&gt;http://xray.chem.ufl.edu/growing%20tips.htm&lt;/a&gt; or &lt;a href=&quot;http://web.mit.edu/x-ray/cystallize.html&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/x-ray/cystallize.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can use techniques like sublimation but your question was about solution (btw in case of sublimation you also have supersaturation, only difference is that your solvent is gas).&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2013-06-26T16:59:26.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5460" PostTypeId="2" ParentId="5182" CreationDate="2013-06-26T17:16:30.197" Score="3" Body="&lt;p&gt;$\ce{en}$ is bidenate so it has to take two neighbour corners from octahedron (let say equatorial position). The second $\ce{en}$ can bind with 1st nitrogen also on equator but then the 2nd has a choice: it can bind in equator or on axial position. So in theory it would be possible to see two isomers. The reality has however more to do with energy. $\ce{en}$ is very good ligand and $\ce{Ni++}$ makes very stable quadratic planar complexes. For that reason both $\ce{en}$ goes into equatorial position and the weaker ligands $\ce{Cl-}$ occupy axial positions.&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2013-06-26T17:16:30.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5461" PostTypeId="1" CreationDate="2013-06-26T18:12:21.787" Score="3" ViewCount="5204" Body="&lt;p&gt;I was trying to make Calcium metal by electrolyzing molten $\ce{CaCl_2}$. I used a stainless steel spoon and a thick steel wire as electrodes. Then I attached them to a power supply. I placed some $\ce{CaCl_2}$ in the spoon and melted with a gas torch. I then placed the steel electrode in the molten pool and gradually pulled it out. A black glassy substance formed on the rod. I ran water over it and it fizzed as I would expect from Calcium metal, however, when it was exposed to air for a period of time it start to &lt;code&gt;&quot;melt&quot;&lt;/code&gt; as it absorbed moisture from the air. As I was breaking it off of the electrode, it shattered and I could see that it was the glassy substance all the way through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out what I have made. Could it possibly be a mixture of unreacted $\ce{CaCl_2}$ and a small amount of Calcium?&lt;/p&gt;&#xA;" OwnerUserId="1146" LastEditorUserId="4231" LastEditDate="2015-03-02T18:20:45.587" LastActivityDate="2015-03-02T18:21:22.510" Title="Electrolysis of molten Calcium Chloride: Identifying the product" Tags="&lt;electrochemistry&gt;&lt;synthesis&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5462" PostTypeId="2" ParentId="4932" CreationDate="2013-06-26T18:46:25.530" Score="2" Body="&lt;p&gt;&lt;strong&gt;-ide&lt;/strong&gt; means the lowest oxidation state. For &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt;, if it gains 1 e&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt;, oxidation state is -1 (&lt;em&gt;i.e.&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;); in case of &lt;span class=&quot;math-container&quot;&gt;$\ce{S}$&lt;/span&gt; (16th group) it's 2 e&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt; (&lt;em&gt;i.e.&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{S^{2-}}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;per-&lt;/strong&gt; means higher than usual oxidation state. For &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; it's +7 (&lt;span class=&quot;math-container&quot;&gt;$\ce{ClO4-}$&lt;/span&gt;); for &lt;span class=&quot;math-container&quot;&gt;$\ce{S}$&lt;/span&gt; look at compounds with O-O bond (peroxomonosulfate &lt;span class=&quot;math-container&quot;&gt;$\ce{SO5^{2-}}$&lt;/span&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So per*ide like your perchloride is an oxymoron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Compare&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Chlorite&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1889" LastEditorUserId="83758" LastEditDate="2020-03-21T10:09:45.150" LastActivityDate="2020-03-21T10:09:45.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5463" PostTypeId="1" AcceptedAnswerId="5464" CreationDate="2013-06-26T20:56:24.230" Score="4" ViewCount="1487" Body="&lt;p&gt;Related: &lt;a href=&quot;https://chemistry.stackexchange.com/q/4315/1892&quot;&gt;Covalent Bond/Hydrogen Bond&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the half-life of hydrogen bonds in a glass of water at room temperature, say 18 degrees Celsius? I guess values may be different for distilled water and for tab water, but both may be intersting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempts to get an answer involve searching with google, google scholar, and google books. Complications were that I cannot get access to all articles using my universities database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.wiley.com/college/boyer/0470003790/reviews/pH/ph_non-covalent.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; we have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The half-life of a hydrogen bond is about 10 seconds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="1892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-10-09T19:14:12.740" Title="Halflife of hydrogen bonds" Tags="&lt;water&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5464" PostTypeId="2" ParentId="5463" CreationDate="2013-06-26T23:11:02.457" Score="7" Body="&lt;p&gt;I'm not sure whether the term &quot;half-life&quot; means the same as &quot;lifetime&quot; but I found a very interesting &lt;a href=&quot;http://www.lsbu.ac.uk/water/hbond.html&quot;&gt;website&lt;/a&gt; about hydrogen bonding in water (with a lot of references) where it is stated that the lifetime of a hydrogen bond is in the range of 1 - 20 ps (it gives the following reference for this: &lt;a href=&quot;http://www.pnas.org/content/98/19/10533.full.pdf+html&quot;&gt;F. N. Keutsch and R. J. Saykally, Water clusters: Untangling the mysteries of the liquid, one molecule at a time, &lt;em&gt;Proc. Natl. Acad. Sci. U.S.A.&lt;/em&gt; &lt;strong&gt;98&lt;/strong&gt; (2001) 10533-10540&lt;/a&gt;). That is quite a big difference to the 10 seconds from your reference. Thinking of the very dynamic nature of a hydrogen bond network and of the typical time scale for molecular processes in general I would tend to think, that the breakage and formation of hydrogen bonds on a molecular level should take much less time than 10 seconds. So if you are looking for the lifetime of a hydrogen bond on a molecular level a few picoseconds sound reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;a href=&quot;http://www.chem1.com/acad/webtext/states/water.html&quot;&gt;Here&lt;/a&gt; is another website, which also states a hydrogen bond lifetime of about a few picoseconds.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-06-26T23:20:52.690" LastActivityDate="2013-06-26T23:20:52.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5465" PostTypeId="2" ParentId="5445" CreationDate="2013-06-27T00:01:06.657" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/cctc.201000126/abstract;jsessionid=0FDFD4E6EBF77AA09235958788D2D1CA.d03t02?deniedAccessCustomisedMessage=&amp;amp;userIsAuthenticated=false&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; provides a very good overview of the state of knowledge about the mechanism of water oxidation (which is pretty much the heart of artificial photosynthesis). So if you want to understand the process of oxygen formation from water on a molecular level this is a very good and fairly recent review of the field, which treats the heterogeneous, homogeneous and biological catalytic pathways proposed in the literature. It is solely concerned with oxygen formation and doesn't say anything about the formation of glucose from carbon dioxide - for details on that process you should look into a biology textbook. Furthermore, I want to add that the mechanistic details are still very much a matter of debate.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-06-27T00:01:06.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5466" PostTypeId="1" CreationDate="2013-06-27T11:29:36.677" Score="1" ViewCount="1398" Body="&lt;p&gt;What is the most logical and easy to remember step by step method for drawing Lewis dot structures?&#xA;I'm having trouble drawing a simple one for $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="4231" LastEditDate="2015-02-17T02:20:50.880" LastActivityDate="2015-02-17T02:20:50.880" Title="How to draw Lewis dot structures?" Tags="&lt;inorganic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5467" PostTypeId="2" ParentId="5440" CreationDate="2013-06-27T13:02:17.977" Score="5" Body="&lt;p&gt;Let's start off on common footing by making sure we understand the definition of energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy&lt;/strong&gt; is the capacity to do work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object may have many types of energy, we'll get to those in a second, but the &lt;em&gt;total energy&lt;/em&gt; of an object will be the sum of its &lt;strong&gt;kinetic energy&lt;/strong&gt; and its &lt;strong&gt;potential energy&lt;/strong&gt;.  Kinetic energy is the energy associated with the movement of an object and potential energy is the energy associated with an object's position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a physical example, hold out a book (preferably your heavy, expensive general chemistry textbook) and drop it on the floor (preferably, not on your little brother's toes).  The book falls to the ground and a loud noise is heard.  As the book moved, the &lt;em&gt;potential energy&lt;/em&gt; stored in its original position was converted to &lt;em&gt;kinetic energy&lt;/em&gt; as it fell to the ground.  Upon hitting the ground the individual atoms that make up the book crashed in to the individual atoms that make up the ground, causing them all to move.  This movement of atoms is &lt;strong&gt;thermal energy&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's focus on &lt;em&gt;thermal energy&lt;/em&gt; for a moment, the majority of chemical reactions, including the combustion reactions you refer to, involve the transfer of thermal energy.  Hold a cup of coffee in your hand; what happens to your hand?  Assuming the coffee was hot, and your hand was not, your hand starts to feel warm.  We might say your hand is &lt;em&gt;heating up&lt;/em&gt;.  This thing we refer to as *&lt;em&gt;heat&lt;/em&gt; is the &lt;em&gt;transfer of thermal energy&lt;/em&gt;; the molecules in the cup of coffee were bouncing around, crashing in to the cup, which made the molecules in the cup crash around hitting your hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let's turn to &lt;strong&gt;chemical energy&lt;/strong&gt; which is the &lt;em&gt;energy associated with the position of nuclei and electrons in atoms and molecules.&lt;/em&gt; Since we are talking about (relative) positions of objects, this energy is a form of &lt;em&gt;potential energy&lt;/em&gt;.  Some molecules, such as methane, have nuclei and electrons that are positioned in such a way that they store a large amount of potential energy.  When the positions of these nuclei and electrons are moved, potential energy is converted into kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;in summary&lt;/em&gt;, the energy of chemical reactions is best described in terms of potential and kinetic energy, or in other words the relative position and movement of nuclei and electrons.  Often, the energy we experience in chemical reactions can be adequately described by the transfer of heat, and therefore &lt;strong&gt;thermochemistry&lt;/strong&gt; and &lt;strong&gt;thermodynamics&lt;/strong&gt; take up a large portion of our instruction in Chemistry.  Other types of energy, such as electromagnetic radiation, electrochemical, and nuclear, can all be described in terms of potential and kinetic energy, namely the position and movement of objects. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-06-27T13:02:17.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5468" PostTypeId="2" ParentId="5466" CreationDate="2013-06-27T17:01:06.773" Score="4" Body="&lt;p&gt;&lt;ol&gt;&#xA;&lt;li&gt;Calculate the total number of valance shell electrons ($n_1$) in all constituent atoms or ions. &lt;em&gt;For example, in $\ce{NH_4^+}$, $n_1$ will be 8 (5 of N, 4 from all H's and -1 because the molecule has +1 charge).&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now, calculate the total number of electrons that are required to fill valance shell of every constituent atom or ion ($n_2$). &lt;em&gt;In other words&lt;/em&gt;, calculate the total number of electrons that will be required to fill valance shell of each atom or ion individually. We can use following formula to find $n_2$&lt;br&gt;$n_2 = [\text{# of Hydrogen atoms} \times 2] + [\text{# of all other atoms except Hydrogen} \times 8]$&#xA;&lt;em&gt;In case of $\ce{NH_4^+}$, $n_2$ will be 16 ($[4 \times 2] + [1 \times 8]$).&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now, calculate the total number of electrons that atoms or ions will need to share among each other in order to complete their octets ($n_3$).&#xA;$n_3 = n_2 - n_1$&#xA;&lt;em&gt;For $\ce{NH_4^+}$, $n_3$ will be 8.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now calculate the number of bonds that will be present in our molecule&#xA;$n_4 = \frac{n_3}{2}$&#xA;&lt;em&gt;For $\ce{NH_4^+}$, $n_4$ will be 4.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now let's find the number of unshared electrons, $n_5$&#xA;$n_5 = n_1 - n_3$&lt;br&gt;&lt;em&gt;For $\ce{NH_4^+}$, $n_5$ will be 0.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now find the number of lone pairs, $n_6$, which is nothing but $n_5$ divided by 2.&lt;br&gt;_For $\ce{NH_4^+}$, $n_6$ will be 0.&lt;/li&gt;&#xA;&lt;li&gt;Now find the central atom. &lt;em&gt;Central atom is the one which is either least in count or more electropositive.&lt;/em&gt;&#xA;&lt;strong&gt;Always remember that Hydrogen and Fluorine can never be placed at center.&lt;/strong&gt;&#xA;&lt;em&gt;So in our case, Nitrogen gets to be the central atom.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;After allocating the central atom, place the surrounding atoms such that the molecule is symmetric.&#xA;&lt;em&gt;In our case, geometry of molecule will be tetragonal.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Distribute the bonds between atoms. Remember that the total number of bonds, that you distribute, should be equal to $n_4$.&#xA;&lt;em&gt;In our case, we simply place a bond between Nitrogen and each Hydrogen atom.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now use the lone pairs to complete the octet of atoms. Remember that, by now, every atom will already have 2 electrons in their valance shell because of the bond.&lt;br&gt;&lt;em&gt;In our case, which is $\ce{NH_4^+}$, we don't have any lone pairs. OOPS!&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now comes the &lt;strong&gt;Formal Charge&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Formal Charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. It is not an actual electrical charge but just a mathematical construct.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  $\text{Formal Charge of an atom} = \text{# of valence shell electrons (in Ground state)} - \text{# of unshared/non-bonded electron pairs or lone pairs} - \text{# of bonds}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please note that you must place formal charge on each and every atom when it is a non-zero value for that atom.&lt;/strong&gt;&#xA;&lt;em&gt;This step is little confusing so let's get back to our example. Nitrogen has 5 electrons in its valance shell in ground state. It made 4 bonds in our molecule and has none unshared electron pairs or lone pairs (please note that a single unpaired electron doesn't count here). So formal charge on Nitrogen will be 1 (5 - 0 - 4), whereas formal charge on each Hydrogen atom will be 0 as every Hydrogen atom has 1 electron in its valance shell and made one bond in our molecule.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, you actually obtain the structure in $10^{th}$ step but Formal charge, if a &lt;strong&gt;non-zero&lt;/strong&gt; value, should be represented for each atom in molecule in order to complete its Lewis Dot Structure.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2015-02-16T11:03:49.810" LastActivityDate="2015-02-16T11:03:49.810" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5469" PostTypeId="2" ParentId="5461" CreationDate="2013-06-27T17:59:25.733" Score="2" Body="&lt;p&gt;I would hypothesize that the &quot;black glassy stuff&quot; is nothing more than glassified $\ce{CaCl_2}$. Metal is an excellent heat conductor and the liquid $\ce{CaCl_2}$ likely cooled rapidly to form glass when the cold wire was inserted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things could be done to avoid this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) heat the wire before inserting it into the mixture of molten $\ce{CaCl_2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) heat the $\ce{CaCl_2}$ &lt;a href=&quot;https://chemistry.stackexchange.com/questions/884/how-did-michael-faraday-melt-sodium-chloride&quot;&gt;well above its melting point&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are determined to make Calcium metal it might be worth electrolysing an aqueous solution of $\ce{CaCl_2}$ to evolve the $\ce{Cl2(g)}$ (do this responsibly outside/in fume hood of course as $\ce{Cl2(g)}$ is poisonous.) Calcium Hydroxide should be the result. Evaporating the water will yield solid $\ce{CaOH}$, which should be &lt;a href=&quot;http://yarchive.net/explosives/sodium_make.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;much easier to electrolize using your method&lt;/a&gt; since it has a melting point of 580°C vs. 772°C for $\ce{CaCl_2}$. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-03-02T18:21:22.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5471" PostTypeId="1" CreationDate="2013-06-25T11:32:55.907" Score="14" ViewCount="115930" Body="&lt;p&gt;I &lt;a href=&quot;http://www.ehow.com/how_5982537_revive-lithium-ion-batteries.html&quot; rel=&quot;noreferrer&quot;&gt;have read&lt;/a&gt; that it is possible to revive a dead Li-ion battery by putting it in the freezer for &lt;em&gt;three to seven days&lt;/em&gt;, then letting it get back to room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can this process work, and if so, how does it work?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Somehow the opposite, i.e. heating (cooking) the battery, is more intuitive to me. After all, in the dead battery, there is &quot;something stuck&quot; (crystallized?), and heat could break that. But I don't know if it's a good idea to &lt;em&gt;try that at home&lt;/em&gt;. ;-)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also found that hitting a Li-ion battery, against a table, for example, has some effect. At the moment, I have a battery in an IBM ThinkPad T41 laptop that, although still working, is on it's way to death. When it runs flat and I connect the laptop to mains, then the battery is not detected properly (battery status LED continues blinking orange). So, I take it out, hit it, put it back in, and the problem is solved (LED constantly orange). Also, I have the impression that hitting gives some extra charge. It could be all just a coincidence, of course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am asking in part for a friend whose laptop battery has been dead, I believe, already for some weeks or months. He said that the battery is not that old, but he has let it run flat in cold weather (up in the mountains). Anyhow, whether the process works for this particular battery should not influence my decision on accepting an answer. I am primarily interested in understanding a little bit more about battery chemistry, just to broaden my general knowledge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1607" OwnerDisplayName="feklee" LastEditorUserId="402" LastEditDate="2017-05-08T01:13:02.390" LastActivityDate="2017-05-08T01:13:02.390" Title="Reviving Li-ion battery in freezer?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5472" PostTypeId="1" AcceptedAnswerId="5760" CreationDate="2013-06-28T20:12:26.487" Score="5" ViewCount="382" Body="&lt;p&gt;I read this question (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/5202/flame-temperature-for-cooking-food&quot;&gt;Flame temperature for cooking food&lt;/a&gt;) and came up with the following questions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Could companies color their flames so that products would heat quicker?&lt;/li&gt;&#xA;&lt;li&gt;How would they do this?  &lt;/li&gt;&#xA;&lt;li&gt;Could you use this to make an infrared or ultraviolet flame? If&#xA;you could, how hot would they be?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1896" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-06-14T14:12:44.483" Title="Can you color a flame and what would the effects be?" Tags="&lt;thermodynamics&gt;&lt;color&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5473" PostTypeId="2" ParentId="5445" CreationDate="2013-06-28T20:13:49.160" Score="2" Body="&lt;p&gt;I am assuming you are in grade school.&#xA;The problem is, as was mentioned before, that photosynthesis is very complicated.&#xA;The formula you know of is extremely simplistic and skips over 100 reactions on the way. you can find plenty of information in &lt;a href=&quot;http://en.wikipedia.org/wiki/Photosynthesis&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, or if that is too much you can use the &lt;a href=&quot;http://simple.wikipedia.org/wiki/Photosynthesis&quot; rel=&quot;nofollow&quot;&gt;Simple Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high school we took some kind of water plant or algae and put it in an upside down test tube filled with water, as oxygen was formed it filled the end of the test tube. This is a fairly simple way to monitor photosynthesis and you can use it to study photosynthesis. for example: how does the temperature effect photosynthesis, how does the color of light affect it etc.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-06-28T20:13:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5474" PostTypeId="2" ParentId="5202" CreationDate="2013-06-28T20:46:20.337" Score="1" Body="&lt;p&gt;In fact black body radiation has a small role in the color of the flame:&#xA;putting in 700nm (for blue) in the equation yields over 3500 degrees celsius while the temperature of a gas stove flame is about 1700. The colors are a result of, as was mentioned in the comments, ionization of gas molecules . If i remember correctly from analytical chemistry class, the hottest part of the flame is the middle which has a pale color, this results from a certain fuel(gas)/oxidiser(air) ratio that gives complete combustion.&#xA;also see &lt;a href=&quot;https://en.wikipedia.org/wiki/Flame&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-06-28T20:46:20.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5475" PostTypeId="2" ParentId="5472" CreationDate="2013-06-28T21:34:40.767" Score="3" Body="&lt;p&gt;The flame has different color because it has different temperature. If you add something to change flames color (like source of metal ions) it will not affect temperature of the flame. So in short: they can not.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-06-28T21:34:40.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5476" PostTypeId="2" ParentId="5461" CreationDate="2013-06-29T05:39:04.063" Score="1" Body="&lt;p&gt;Does this sound like what you made: &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_chloride&quot; rel=&quot;nofollow&quot;&gt;Iron (III) Chloride&lt;/a&gt;. Depending on the polarity of the electrode, one constituent of the melt will combine with the material of the electrode, so it sounds like the chlorine combined with the iron in the steel wire.&lt;/p&gt;&#xA;" OwnerUserId="1824" LastActivityDate="2013-06-29T05:39:04.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5477" PostTypeId="1" AcceptedAnswerId="5483" CreationDate="2013-06-29T05:43:07.927" Score="49" ViewCount="355198" Body="&lt;p&gt;I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, will using $\ce{NaCl}$ instead of $\ce{KNO3}$ in making the salt bridge have any effects on voltage/current output of the cell? why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus if it matters, I'm using a Zinc-Copper voltaic cell with a tissue paper soaked in $\ce{KNO3}$ as salt bridge&lt;/p&gt;&#xA;" OwnerUserId="1821" LastEditorUserId="402" LastEditDate="2017-04-02T20:45:44.570" LastActivityDate="2021-02-01T20:05:47.853" Title="Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="5479" PostTypeId="2" ParentId="5472" CreationDate="2013-06-29T08:20:25.507" Score="3" Body="&lt;p&gt;When we talk about the color of flame we should consider its source, too.  If you are burning a transition element then you will see a colored flame. Each element may give different color so the energy caused by this element will also be different. And the heat is also related to this energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the page you have linked, you can see the relationships for these energies. However, if you need further informations you can study related concepts e.g. electronic spectra and quantum mechanics. Since this issue can be explained with more details. &lt;/p&gt;&#xA;" OwnerUserId="792" LastActivityDate="2013-06-29T08:20:25.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5480" PostTypeId="2" ParentId="5466" CreationDate="2013-06-29T08:34:09.923" Score="1" Body="&lt;p&gt;I will give you a simple way, but if you need to study properly, you can follow Ashutosh Gangwar's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C has 4 valance electrons. &#xA;O has 6 valance electrons. &#xA;O has 6 valance electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First distribute these electrons to each element and indicate the covalent bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you know carbon and oxygen need to obey octed rule, so in this case carbon makes 4 bonds with oxygens; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dcfdr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="792" LastActivityDate="2013-06-29T08:34:09.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5481" PostTypeId="1" CreationDate="2013-06-29T09:22:39.623" Score="4" ViewCount="1085" Body="&lt;p&gt;Most people that have ever visited a viticulteur (winemaker) will know that it is common wisdom that vibrations are bad for the wine quality. When I was just a kid the winemaker always told me that we should be silent because the wines needed to sleep, but later a different one explained that this must have been to stop me from running around and causing vibrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even scientific research (e.g. &lt;a href=&quot;http://dx.doi.org/10.1016/j.jfca.2008.07.004&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chung et al. 2008 Journal of Food Composition and Analysis (21)8&lt;/a&gt;) has shown that vibrations will indeed change the wine and, among others, accelerate the aging process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I am wondering is why these vibrations matter?&lt;/strong&gt; Is it a chemical effect, in which the vibrations accelerate certain chemical reactions? Or is it more physical in nature, in which the vibrations increase mixing rates and therefore somehow change the evolution of the wine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps to get your minds working: Vibrating wine in a glass&#xA;&lt;img src=&quot;https://i.stack.imgur.com/01iY8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-06-29T13:33:51.403" Title="Effect of vibration on wine quality" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5482" PostTypeId="2" ParentId="5481" CreationDate="2013-06-29T11:10:22.893" Score="2" Body="&lt;p&gt;One of the effects could be that vibrations increase mixing on phase boundary between wine itself and air above it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other thing is that vibration are generally not good for crystal growing process (instead big and nice monocrystals one gets polycrystalline solid).  Usually you can find nice crystals of tartaric acid in the bottom of bottles and barrels. I do not know chemistry behind it but certainly it will change taste of the wine if the tartaric acid stays longer in wine (slow crystallisation) instead of falling rapidly out of solution when you shake it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if there is still a sediment earlier during fermentation, certainly that will be not good if this mixes with clear wine.&lt;/p&gt;&#xA;" OwnerUserId="1889" LastEditorUserId="1889" LastEditDate="2013-06-29T13:33:51.403" LastActivityDate="2013-06-29T13:33:51.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5483" PostTypeId="2" ParentId="5477" CreationDate="2013-06-29T11:22:14.907" Score="52" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2644&quot;&gt;There's another question&lt;/a&gt; related to salt bridges on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of a salt bridge is &lt;strong&gt;not&lt;/strong&gt; to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLfN9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons flow from the anode to the cathode.  The oxidation reaction that occurs at the anode generates electrons and positively charged ions.  The electrons move through the wire (and your device, which I haven't included in the diagram), leaving the unbalanced positive charge in this vessel.  In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell.  A similar (but reversed) situation is found in the cathodic cell, where $\ce{Cu^{2+}}$ ions are being consumed, and therefore electroneutrality is maintained by the migration of $\ce{K+}$ ions from the salt bridge into this half cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second part of your question, it is important to use a salt with inert ions in your salt bridge. In your case, you probably won't notice a difference between $\ce{NaCl}$ and $\ce{KNO3}$ since the $\ce{Cu^{2+}}$ and $\ce{Zn^{2+}}$ salts of $\ce{Cl-}$ and $\ce{NO3-}$ are soluble. There will be a difference in the &lt;a href=&quot;https://www.youtube.com/watch?v=qQoDML0miVg&quot; rel=&quot;noreferrer&quot;&gt;liquid junction potential&lt;/a&gt;, but that topic is a bit advanced for someone just starting out with voltaic/galvanic cells.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="4945" LastEditDate="2018-08-27T06:28:18.417" LastActivityDate="2018-08-27T06:28:18.417" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5484" PostTypeId="1" AcceptedAnswerId="5485" CreationDate="2013-06-30T02:29:43.897" Score="2" ViewCount="1119" Body="&lt;p&gt;Apart from hydrogen which forms a duet, which element's don't observe the octet rule?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastActivityDate="2013-06-30T16:13:23.640" Title="What elements don't observe the octet rule?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5485" PostTypeId="2" ParentId="5484" CreationDate="2013-06-30T05:58:38.717" Score="2" Body="&lt;p&gt;Octet rule is more advisory than a rule; It's usually obeyed by the main group elements.&#xA;Exception are paramagnetic compounds (obvious reason); first for elements in periodic system (He-configuration), 3-valent boron compounds; depending on counting system boron clusters (f.e. dodecaborane with 6-coordianted boron); transition metal complexes (these usually follow 18e rule, however there are plenty of examples with 14, 16 or even 20 electron complexes).&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="1889" LastEditDate="2013-06-30T16:13:23.640" LastActivityDate="2013-06-30T16:13:23.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5486" PostTypeId="1" AcceptedAnswerId="5487" CreationDate="2013-06-30T08:23:54.253" Score="14" ViewCount="73787" Body="&lt;p&gt;What does it mean to &lt;em&gt;shift a chemical equilibrium?&lt;/em&gt; For example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the equilibrium shifts to the left …&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand that.&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="16683" LastEditDate="2018-09-08T08:30:51.307" LastActivityDate="2018-09-08T08:31:20.137" Title="What does it mean to shift equilibrium?" Tags="&lt;equilibrium&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5487" PostTypeId="2" ParentId="5486" CreationDate="2013-06-30T11:52:21.243" Score="18" Body="&lt;p&gt;Probably you are having problems with Le Chatelier's Principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have an equilibrium established between four substances $\ce{A}$, $\ce{B}$, $\ce{C}$ and $\ce{D}$.&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would happen if you changed the conditions by increasing the&#xA;  concentration of $\ce{A}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of $\ce{A}$ decreases again - by reacting it with $\ce{B}$ and turning it into $\ce{C + D}$. The position of equilibrium moves to the right. Of course, this assumes that there is still some amount of $\ce{B}$ left in the reaction vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For better understanding, refer &lt;a href=&quot;http://www.chemguide.co.uk/physical/equilibria/lechatelier.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-09-08T08:31:20.137" LastActivityDate="2018-09-08T08:31:20.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5489" PostTypeId="1" AcceptedAnswerId="5499" CreationDate="2013-06-30T14:36:23.183" Score="3" ViewCount="880" Body="&lt;p&gt;How does the effect of heat and effect of substituents affect racemization of biphenyl compounds?&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorDisplayName="user1811" LastEditDate="2013-07-02T06:57:23.640" LastActivityDate="2013-07-02T06:57:23.640" Title="Racemization of biphenyl compounds" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5490" PostTypeId="2" ParentId="5486" CreationDate="2013-06-30T15:07:26.293" Score="4" Body="&lt;p&gt;Let’s do it simply. You start from some equilibrium. For example you have simple reaction $\ce{A &amp;lt;=&amp;gt; B}$, and in the equilibrium state there is $60~\%$ of $\ce{A}$ and $40~\%$ of $\ce{B}$. If it said that equilibrium is shifted to the right it means that you have more product $\ce{B}$ (&lt;em&gt;i.e.&lt;/em&gt; more than $40~\%$). If you shifting to the left you gaining more substrate. Now how it's done — look on the previous answers.&lt;/p&gt;&#xA;" OwnerUserId="1889" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2015-10-14T11:37:15.440" LastActivityDate="2015-10-14T11:37:15.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5491" PostTypeId="1" CreationDate="2013-06-30T15:55:07.800" Score="1" ViewCount="284" Body="&lt;p&gt;I have got around to at least read on Wikipedia, and it does not clarify how this works in any sensible manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered how the chemicals make electrons &quot;move&quot;. How is the basic understanding of this?&lt;/p&gt;&#xA;" OwnerDisplayName="user1869" LastActivityDate="2013-08-05T05:19:40.143" Title="How do the stored chemicals in batteries drive electronic current from their terminal to outside devices?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5495" PostTypeId="1" CreationDate="2013-07-01T16:54:30.763" Score="4" ViewCount="8881" Body="&lt;p&gt;I boiled a solution of glycerol and water at about 160 degree Celsius and got a brown liquid much less in volume (boiled off the entire water I think; also not so sure about the accurate temperature). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before mixing glycerol with water it was a colourless liquid, now it is golden brown! Why would that be? Also, I smelled burnt sugar during the last couple of minutes of boiling. So I'd like to know what that gas was. So, do I have glycerin in my hand or any other byproducts?&lt;/p&gt;&#xA;" OwnerUserId="1908" LastEditorUserId="22" LastEditDate="2013-07-02T05:26:01.850" LastActivityDate="2017-06-26T13:21:18.443" Title="What is this brown liquid that I got when I boiled glycerol and water solution?" Tags="&lt;synthesis&gt;&lt;redox&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5496" PostTypeId="1" CreationDate="2013-07-01T17:41:22.697" Score="3" ViewCount="380" Body="&lt;p&gt;I have to bottles of Lamy blue-black fountain pen ink. It turns out that some months ago the composition of the ink was changed by the manufacturer and there is no difference in the package. &#xA;Before, this ink used to contain Iron-Gall. Which makes the ink more permanent (old inks contained most of the times Iron-Gall). As Iron-Gall is an oxidizer, it has been removed from the ink because it can damage the pens if not cleaned regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about chemistry and I wonder if any of you could tell me a way to know if my ink has Iron-Gall and therefore is safe for long lifespan documents.&#xA;I was wondering if litmus paper would do the trick comparing this ink with normal black ink (which does not have Iron-Gall).&lt;/p&gt;&#xA;" OwnerUserId="1909" LastEditorDisplayName="user7951" LastEditDate="2015-10-25T16:18:16.803" LastActivityDate="2015-10-25T16:19:27.707" Title="Testing for Iron-Gall Inks" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5498" PostTypeId="2" ParentId="5418" CreationDate="2013-07-01T22:00:39.890" Score="1" Body="&lt;p&gt;PET bottles are perfectly safe for repeated use; they will not degrade or undergo any chemical change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it IS very important to thoroughly wash (with soap and water), rinse and DRY the bottle so there is no moisture left in the bottle where bacteria might grow.  This is true of any type of bottle, but especially one with a narrow neck where it can be difficult to thoroughly dry the inside of the bottle.&lt;/p&gt;&#xA;" OwnerUserId="1911" LastActivityDate="2013-07-01T22:00:39.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5499" PostTypeId="2" ParentId="5489" CreationDate="2013-07-01T22:59:53.793" Score="5" Body="&lt;p&gt;Preparation: Take the parent compound per se, biphenyl.  The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'.  Now if you label the other carbon atoms in each  phenyl ring seperately in clockwise direction, you'll get &lt;em&gt;C2&lt;/em&gt;, C3, C4, C5, &lt;em&gt;C6&lt;/em&gt; in one cycle; and &lt;em&gt;C2'&lt;/em&gt;, C3', C4', C5', &lt;em&gt;C6'&lt;/em&gt; in the second.  (I haven't a credit to upload a picture yet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other.  Eventually, this may lead to atropisomerism, and the isomers are &lt;a href=&quot;https://en.wikipedia.org/wiki/Atropisomerism&quot; rel=&quot;noreferrer&quot;&gt;atropisomers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, please, sometimes there may be already a preference for one or the other conformation.  If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a &lt;em&gt;s-trans&lt;/em&gt; conformation).  Offering a cation to bind, to complex to, however may lead to a preference towards the &lt;em&gt;s-cis&lt;/em&gt;-like conformation.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditDate="2013-07-01T23:05:42.087" LastActivityDate="2013-07-01T23:05:42.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5500" PostTypeId="1" CreationDate="2013-07-02T02:51:26.973" Score="10" ViewCount="14963" Body="&lt;p&gt;Say we have the following reversible reaction: $$\ce{NaOH{(s) }&amp;lt;=&amp;gt; Na^+{(aq) }$+ $OH^{-}{(aq)}} +10.6 \mathrm{kcal}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add $\ce{OH^-}$ and equilibrium shifts to the left, does that affect the amount of $\ce{NaOH{(s)}}$ present or does it remain constant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused because I remember that pure solids and liquids don't affect equilibrium value, or does that only apply when we're talking about adding pure solids and liquids..not the effect on them by an equilibrium shift?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="41556" LastEditDate="2018-02-24T15:07:48.077" LastActivityDate="2018-03-19T16:15:27.607" Title="How do equilibrium shifts affect solids?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5501" PostTypeId="2" ParentId="5500" CreationDate="2013-07-02T05:11:46.180" Score="14" Body="&lt;p&gt;When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of product relative to reactants and the rate of the reverse reaction will increase relative to the forward reaction until equilibrium is reestablished. This means that the ions will recombine into a crystal lattice and form a precipitate. So, to answer your first question, no, the amount of $\ce{NaOH_{(s)}}$ does not remain constant; more of it will be formed if additional ions are added to a solution already at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why pure solids are not factored into equilibrium expressions is that they are not in fact part of the solution. Any excess precipitate, irrespective of the exact quantity, has no impact on the composition of the solution at equilibrium because it exists outside of the solution by definition. If equilibrium is disturbed due to changing conditions, then more precipitate can form, or some of the existing precipitate may dissolve back into solution. Under any given set of controlled environmental conditions, however, the product of the concentrations of the dissolved species remains constant at equilibrium, which is precisely why the equilibrium will shift if conditions or concentrations change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more rigorous and mathematical explanation, you could read up on dynamic equilibria, equilibrium constants, and thermodynamic activity.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-02T05:11:46.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5503" PostTypeId="1" AcceptedAnswerId="5506" CreationDate="2013-07-02T06:14:38.337" Score="35" ViewCount="27276" Body="&lt;p&gt;I'm not satisfied with the rationale for the intermolecular attraction known as hydrogen bonding. In my book, it states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hydrogen bonding is a special type of intermolecular attraction between the hydrogen atom in a polar bond (particularly H ¬ F, H ¬ O, and H ¬ N) and non-bonding electron pair on a nearby small electronegative ion or atom usually F, O, or N (in another molecule).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that chemists looked at the data and found they needed a 'fudge factor' to fit the higher boiling points of ammonia, water, and hydrogen fluoride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0usus.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't hydrogen bonding apply to other atoms like Sulfur or Chlorine? They seem to be electronegative enough IMO. Is there an explanation that explains &lt;em&gt;why&lt;/em&gt; this is, rather than saying that it's there? What makes Hydrogen so special?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Chlorine is &lt;em&gt;more&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;electronegative&lt;/a&gt; than Nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="1758" OwnerDisplayName="user2116192" LastEditorUserId="9961" LastEditDate="2017-03-28T13:59:14.280" LastActivityDate="2017-03-28T13:59:14.280" Title="What exactly is hydrogen bonding (and does it really need fluorine, oxygen, or nitrogen)?" Tags="&lt;physical-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5504" PostTypeId="2" ParentId="5503" CreationDate="2013-07-02T07:11:47.397" Score="15" Body="&lt;p&gt;N, O, F have atomic numbers $Z=7, 8, 9$, respectively. Chlorine has $Z = 17$ which is much larger. Consequently, the atom is larger as well and it is more diffuse. Equivalently, the lone pairs in chlorine are at the 3-level which is too high. Because the hydrogen bond isn't a real bond but a dipole-dipole attraction and because the force between two dipoles scales like $1/r^4$, a larger atom implies a significantly weaker dipole-dipole attraction which is, in the case of chlorine, beaten by other forces. That's why hydrogen bonds are only observed with light atoms bound to hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="1913" LastActivityDate="2013-07-02T07:11:47.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5506" PostTypeId="2" ParentId="5503" CreationDate="2013-07-02T14:14:16.773" Score="24" Body="&lt;p&gt;There are two main ways to look at hydrogen bonding.  The first is electrostatic, where the electronegativity of the atoms is used to describe the interaction.  Your argument about chlorine being more electronegative than nitrogen is a good one suggesting that the electrostatic argument is only part of the story, and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp004542e&quot;&gt;there is at least one study&lt;/a&gt; that suggests that hydrogen bonding does occur with chlorine in polar molecules such as chloroform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can increase dramatically the number of atoms considered to undergo hydrogen bonding if we take a molecular orbital approach.  My reasoning here is a summary of what can be found in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321811054&quot;&gt;Inorganic Chemistry&lt;/a&gt; by Miessler, Fischer and Tarr.  They, in turn, rely heavily on a &lt;a href=&quot;http://pac.iupac.org/publications/pac/pdf/2011/pdf/8308x1637.pdf&quot;&gt;new definition of hydrogen bonding&lt;/a&gt; recommended by the IUPAC Physical and Biophysical Chemistry Division.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydrogen bond is formed when an $\ce{X-H}$ (where X is more electronegative than H) interacts with a donor atom, $\ce{B}$.  The attraction $\ce{X-H...B}$ can be described as consisting of the following components:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An electrostatic contribution based on the polarity of $\ce{X-H}$&lt;/li&gt;&#xA;&lt;li&gt;A partial covalent character which arises from the donor-acceptor nature of the interaction&lt;/li&gt;&#xA;&lt;li&gt;Dispersion forces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first bullet is typically the only phenomenon discussed in General Chemistry classes, and this is not unreasonable since the second bullet requires the introduction of Lewis Acid/Base concepts which may not have been covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that the new definition includes what I like to call &quot;the proof is in the pudding&quot; where the existence of hydrogen bonding requires experimental evidence, which can be found using a number of methods:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The $\ce{X-H...B}$ bond angle: a bond angle of 180 degrees indicates strong hydrogen bonding and would be accompanied by shord $\ce{H...B}$ bond distances.&lt;/li&gt;&#xA;&lt;li&gt;A red shift in the IR frequency upon formation of $\ce{X-H...B}$&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen bonding results in deshielding of the H atom, which can be observed in NMR spectroscopy.&lt;/li&gt;&#xA;&lt;li&gt;Thermodynamic studies should indicate that $\Delta G$ for the formation of $\ce{X-H...B}$ should be larger than the thermal energy of the system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While electrostatics is the dominant contributor to hydrogen bonding, an analysis of frontier orbitals can also be insightful.  A thorough answer requires a fair amount of MO theory background, but for those interested, looking at the MO diagram of $\ce{FHF-}$ will be helpful.  In brief, the base ($\ce{F-}$ in this case) has a filled $p_z$ orbital that gains access to a delocalized orbital with lower energy, thus validating the formation of a strong hydrogen bond.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-02T14:14:16.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5508" PostTypeId="1" AcceptedAnswerId="5749" CreationDate="2013-07-02T15:07:14.527" Score="5" ViewCount="118" Body="&lt;p&gt;After cementing a wall, water is applied on the surface to keep it moist to maintain its strength for the next few days. Why is this necessary for maintaining the strength of the concrete?&lt;/p&gt;&#xA;" OwnerUserId="1886" LastEditorUserId="16683" LastEditDate="2015-10-23T16:42:01.350" LastActivityDate="2015-10-23T16:42:01.350" Title="Cement strength by moistening" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5509" PostTypeId="1" CreationDate="2013-07-02T22:07:48.437" Score="4" ViewCount="181" Body="&lt;p&gt;Does the electronegativity only qualitatively describe bond polarity, or does it have any quantitative use? Is it possible to describe electronegativity in terms of other periodic properties, like valence electrons, $Z_{eff}$, etc. (thus eliminating the need of a separately defined value)?&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2015-09-13T04:52:35.423" Title="Is the electronegativity value ever used in calculations, or is it just a number assigned to each element?" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5510" PostTypeId="1" AcceptedAnswerId="5524" CreationDate="2013-07-03T05:58:23.703" Score="2" ViewCount="404" Body="&lt;p&gt;How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab) How 'pure' are they? Also: is it possible to create relatively pure substances with household goods?&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2013-07-05T10:59:03.807" Title="How are pure substances (elements, compounds, etc) produced?" Tags="&lt;molecules&gt;&lt;purification&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-10-03T04:28:52.087" ContentLicense="CC BY-SA 3.0" />
  <row Id="5513" PostTypeId="1" AcceptedAnswerId="5780" CreationDate="2013-07-03T15:25:20.303" Score="18" ViewCount="119972" Body="&lt;p&gt;Hydrogen has only one electron, yet it exhibits multiple lines in a spectral series, why is this?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="402" LastEditDate="2017-03-30T21:53:51.403" LastActivityDate="2017-03-30T21:54:06.870" Title="Why are there multiple lines in the hydrogen line spectrum?" Tags="&lt;spectroscopy&gt;&lt;hydrogen&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5514" PostTypeId="2" ParentId="5508" CreationDate="2013-07-03T19:45:18.057" Score="0" Body="&lt;p&gt;V.shortly. Cement is a mixture of $\ce{CaO}$, $\ce{MgO}$ and $\ce{SiO2}$ (sand). The 4th component comes from air - $\ce{CO2}$. You need water for two main reasons. 1st is that you want metal oxides to react with $\ce{SiO2}$. They do it better if they are in form of hydroxide. Also some of the salts that are creted are hydrates (i.e. they have crystaline water). The 2nd reason is that hydroxides of both these metals easier react with $\ce{CO2}$ than oxides. One of the reasons is solubiltiy of $\ce{CO2}$ in water, the other, bigger renwewable reaction surface (reaction between solids are usually slower than between solid and liquid). &lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="1878" LastEditDate="2013-08-03T03:35:15.197" LastActivityDate="2013-08-03T03:35:15.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5515" PostTypeId="1" CreationDate="2013-07-03T21:28:12.523" Score="5" ViewCount="1320" Body="&lt;p&gt;Ca has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.00$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.31$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Mg is smaller and has a higher electronegativity, one would think Mg would have a smaller hydrated radius than Ca, but it doesn't.  Would someone explain to me why this isn't so?  I have read an explanation pertaining to charge density, but I'm not understanding why Ca would have a larger charge density if it is bigger with less electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of understanding this is to understand why the addition of Ca to soil will make it drain better and be looser, less sticky, etc.  And conversely, why highly respected agricultural consultants say soils high in Mg (low in Ca) are &quot;sticky&quot;.  Any help in understanding this would be well appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="16683" LastEditDate="2015-10-23T16:48:08.890" LastActivityDate="2015-10-23T16:48:08.890" Title="Ca and Mg hydrated radius in clay flocculation" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5516" PostTypeId="1" CreationDate="2013-07-03T21:47:11.987" Score="4" ViewCount="149" Body="&lt;p&gt;How could one determine the likelihood of one cation displacing another on a negatively charged exchange site of a clay particle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a soil solution containing equal parts (by number of atoms) &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}$&lt;/span&gt;, which cations will find the most exchange sites and which will ultimately be washed deeper into the soil? (Assuming there are not enough sites for all of them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could two &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; cations displace an adsorbed &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; cation simply because they &quot;bump&quot; into it? Or is there some mechanism where cations are dislodged? Is it easier for two &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; to replace &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; or for &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; to replace two &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1916" LastEditorUserId="41328" LastEditDate="2019-01-19T22:49:41.457" LastActivityDate="2019-01-19T22:49:41.457" Title="Soil-solution cations in competition for exchange sites" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;surface-chemistry&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5520" PostTypeId="1" CreationDate="2013-07-04T03:09:23.313" Score="10" ViewCount="31938" Body="&lt;p&gt;I don't understand why liquids and solids have constant concentrations. Can anyone explain this to be in simple terms?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastActivityDate="2020-06-23T03:02:17.617" Title="Why do liquids and solids have constant concentrations?" Tags="&lt;equilibrium&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5521" PostTypeId="2" ParentId="5520" CreationDate="2013-07-04T06:18:02.463" Score="14" Body="&lt;p&gt;If you think about what concentration, it usually refers to the molar concentration in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{mol/l}$&lt;/span&gt; (at least in the field of chemistry). To calculate this concentration we need to find how many moles of a certain chemical are in 1 liter of the liquid or solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will start with an example and describe the general principle afterwards. Let's say we have pure water at room temperature and atmospheric pressure. This means that the density is &lt;span class=&quot;math-container&quot;&gt;$998\,\mathrm{g/l}$&lt;/span&gt;. We also know that the molar mass of water, consisting of 2 hydrogen and 1 oxygen, is &lt;span class=&quot;math-container&quot;&gt;$18.016\,\mathrm{g/mol}$&lt;/span&gt;. If you look at the units of the molar mass and the density you can already see that we can combine these into something that has units &lt;span class=&quot;math-container&quot;&gt;$\ce{mol/l}$&lt;/span&gt;. If we do that we get that the concentration of water is &lt;span class=&quot;math-container&quot;&gt;$\frac{998\, \mathrm{g/l}}{18.016\,\mathrm{g/mol}}=55.4\,\mathrm{mol/l}$&lt;/span&gt;. So the concentration of water is &lt;span class=&quot;math-container&quot;&gt;$55.4\,\mathrm{mol/l}$&lt;/span&gt;. Notice that I have only used the density and the molar mass of water, which are (at least for equal temperature and pressure) constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you can do this for any solid or liquid (or gas) and find the concentration as &lt;span class=&quot;math-container&quot;&gt;$c=\frac{\rho}{M}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the molar mass and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; the molar concentration. For a given chemical, the molar mass is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now about the ‘fact’ that the molar concentration is constant for liquids and solids: strictly speaking it is not. That is because the density of liquids and solids is temperature and pressure dependent. The reason that the molar concentration is often called constant is twofold: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the density of liquids and solids has a much weaker dependence on the temperature than gases do therefore it can be regarded as approximately constant&lt;/li&gt;&#xA;&lt;li&gt;if you look at solutions, the variation of concentration can be&#xA;orders of magnitude, whereas for e.g. water the concentration&#xA;variation when going from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$100\,\mathrm{^\circ C}$&lt;/span&gt; is only a few percent.&#xA;Therefore, again, the concentration of the liquid is approximately&#xA;constant.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1362" LastEditorDisplayName="user7951" LastEditDate="2019-02-03T11:53:40.240" LastActivityDate="2019-02-03T11:53:40.240" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5522" PostTypeId="1" CreationDate="2013-07-04T08:49:16.623" Score="6" ViewCount="222" Body="&lt;p&gt;I need some practical advise. What would be the easiest approach to compute strengthening or weakening of π-interaction in a bond with multipole-bond character. I'm not really fond of going through coefficients of Kohn–Sham orbitals. I did it on occasion, but it's a lot of work and I just do no have time. Unfortunately the molecules I need to compare have low symmetry ($\ce{C1}$) so I can't make use of symmetry to divide $\sigma$ and $\pi$ contributions. I'm not quite sure but I remember to read that the energy decomposition analysis in ADF could be used to divide in $\sigma$ and $\pi$ components. Let me make it clear, I'm not interested in comparing bond lengths, I need quantitative measure for the strength of $\pi$ interaction: either energy component or relative contribution of $\sigma$ and $\pi$ components orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I welcome any suggestions concerning methodology and/or software that can help.&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="42741" LastEditDate="2017-05-19T04:50:06.697" LastActivityDate="2017-05-19T04:50:06.697" Title="Computing of strengthening of π-interaction" Tags="&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5523" PostTypeId="2" ParentId="5522" CreationDate="2013-07-04T12:38:11.227" Score="5" Body="&lt;p&gt;First off, definitely check out &lt;a href=&quot;http://www.scm.com/Products/Capabilities/ETS-NOCV.html&quot; rel=&quot;noreferrer&quot;&gt;ETS-NOCV&lt;/a&gt;, which I believe is a fairly recent addition to ADF which seems to do exactly what you're interested in. Other NBO-esque software may also suit your needs - I seem to recall NBO5 in Gaussian can generate sigma and pi natural bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding doing things the energy decomposition way:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have not already seen it, this paper goes about doing (in ADF) what you want to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2012/DT/c2dt31845h&quot; rel=&quot;noreferrer&quot;&gt;http://pubs.rsc.org/en/Content/ArticleLanding/2012/DT/c2dt31845h&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(disclosure - I am a member of the Stranger group)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this for a related system a while back but have forgotten the exact details, however fragment bonding analysis is a nifty technique that allows estimation of the energy of single bonds in a molecule as well as decomposition by bond type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadly, you cut your molecule into chunks on either side of the bond you're interested in and generate fragment orbitals from them. (that is, you use the TAPE21 files for these molecular fragments as a basis for the calculation of the full molecule). These fragment orbitals subsume all of the interactions within the fragments so that the total bonding energy only consists of interactions between the fragments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, you can actually remove atomic orbitals from the fragment orbitals. If you prune the orbitals relevant to pi bonding, you can estimate the pi and sigma components by relative energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where you should start: &lt;a href=&quot;http://www.scm.com/Doc/Doc2013/ADF/Examples/page142.html&quot; rel=&quot;noreferrer&quot;&gt;ADF manual - Analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-07-04T12:38:11.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5524" PostTypeId="2" ParentId="5510" CreationDate="2013-07-04T21:34:56.993" Score="2" Body="&lt;p&gt;No lab reagent is 100% pure. But you can get &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/339741&quot; rel=&quot;nofollow&quot;&gt;really close&lt;/a&gt; to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;How 'pure' are they?&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the label or do the analysis. It varies A LOT from one reagent to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Is it possible to create relatively pure substances with household goods?&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depends on the substance, depends how pure you want it. For instance, you can very easily distill water at home, but &lt;a href=&quot;http://www.millipore.com/lab_water/clw4/tutorial&quot; rel=&quot;nofollow&quot;&gt;pure water is hard to get&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need to be more specific. There are entire classes you can take at the uni on the large scale production and purification methods of this and that type of reagents. I see you used the ionic-compounds tag, but that's far from specific. A more answerable question would be &quot;How are compounds X and Y produced and purified?&quot;&lt;/p&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2013-07-04T21:34:56.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5525" PostTypeId="2" ParentId="5509" CreationDate="2013-07-04T22:34:42.600" Score="4" Body="&lt;p&gt;True, with values of electronegativity in hand (and beside the by Pauling, there &lt;em&gt;are&lt;/em&gt; others by Rochow, Mulliken, etc. to name some scales) you describe bond polarity. Beyond this number, infrared vibrational frequencies, chemical shifts in NMR spectroscopy and isomer shifts in Mössbauer spectroscopy are mentioned by the corresponding wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Correlation_of_electronegativity_with_other_properties&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-07-04T22:34:42.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5526" PostTypeId="2" ParentId="5510" CreationDate="2013-07-05T06:07:38.117" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How are large quantities of the substances we use in the lab generally produced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_purification_methods_in_chemistry&quot; rel=&quot;nofollow&quot;&gt;This is answered reasonable well on Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also: is it possible to create relatively pure substances with household goods?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends. Recrystallization can be performed at kitchen, so some substances can be purified at home. However, it is easier to buy pure substance in most cases.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="22" LastEditDate="2013-07-05T10:59:03.807" LastActivityDate="2013-07-05T10:59:03.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5527" PostTypeId="1" CreationDate="2013-07-05T11:19:15.063" Score="1" ViewCount="1176" Body="&lt;p&gt;I diluted sulfuric acid in deionized water in order to get a 10% $\ce{H2SO4}$ solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was going to use this solution as a cathode of a porous sample to measure its electrical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this solution has a quite high resistivity (~200 ohm.cm DC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked several sources that tell me that I should be measuring a lower resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone has an idea of what may be happening?&lt;/p&gt;&#xA;" OwnerUserId="1928" LastEditorUserId="275" LastEditDate="2013-07-06T21:40:15.500" LastActivityDate="2013-07-06T21:40:15.500" Title="Sulfuric Acid Resistance / Conductivity" Tags="&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5529" PostTypeId="1" AcceptedAnswerId="5531" CreationDate="2013-07-05T22:38:35.470" Score="5" ViewCount="889" Body="&lt;p&gt;Several organic compounds have resonance structures, but can a polymer have resonance structures?&lt;/p&gt;&#xA;" OwnerUserId="815" LastActivityDate="2014-09-17T11:32:51.210" Title="Can a polymer have different resonance structures?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5531" PostTypeId="2" ParentId="5529" CreationDate="2013-07-06T05:31:06.293" Score="10" Body="&lt;p&gt;There's no reason why the structural subunits of polymers could not exhibit resonance, provided there are conjugated $\pi$-bonds present. A polymer such as kevlar, for example, certainly benefits from resonance stabilization within each monomeric unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether there exist any polymers in which the resonance is truly distributed over the entire molecule (if that's what you're really asking), however, I'm unsure. From the perspective of valence bond theory, the electron delocalization required for resonance necessitates a planar geometry with unhybridized $p$ orbitals that align and overlap to form $\pi$-bonds perpendicular to the plane of the $\sigma$-bond skeleton of the molecule. In molecular orbital theory, the sort of extensive electron delocalization that resonance implies requires a bonding molecular orbital, distributed over all participating atoms, constructed from a set of symmetrical, in-phase $p$ orbitals, with all planes of symmetry present in the original molecule(s) conserved in the generation of new molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the rigidity of those criteria, I sincerely doubt that polymers of any great length could exist in which the resonance is distributed over the entire polymer chain. However, that could be just a failure of imagination or a lack of knowledge on my part, and I hope somebody else weighs in with a more conclusive answer.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-06T05:31:06.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5532" PostTypeId="1" AcceptedAnswerId="5533" CreationDate="2013-07-06T07:44:13.133" Score="10" ViewCount="239" Body="&lt;p&gt;For fictional use, I'd like to &quot;construct&quot; a clockwork device featuring phosphorus which could reasonably be expected to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) glow consistently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) explode upon some trigger (e.g., influx of air, mixture with another chemical, a spark), with enough explosive force or incendiary strength to kill someone standing very close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constraints:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) this takes place in the early 18th century, so phosphorus has been discovered, but most of the modern types of phosphorus used in weaponry have not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the quantity is no larger than a couple of liters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Progress:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read the relevant chapters in Emsley's &lt;em&gt;13th Element&lt;/em&gt;, and then moved on to the primary sources for the time – mostly records of demonstrations held for the Royal Society and Boyle's &lt;em&gt;Aerial Noctiluca&lt;/em&gt;.&#xA;It's hard to tell exactly what sort of phosphorus any of these early chemists were working with, but it looks like I can achieve goal (A) by just letting in a bit of air now and then (based on Boyle's description / Wikipedia on oxidation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goal (B) is more difficult. Is just oxygen enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slare added sulfuric acid (&quot;oil of vitriol&quot;) and achieved &lt;em&gt;something&lt;/em&gt;, but it doesn't seem to have been an explosion exactly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;being well shiken together did first heat, and then throw up such fiery &lt;em&gt;Balls&lt;/em&gt;, which like so many stars will adhere to the sides of the &lt;em&gt;Glass&lt;/em&gt; and continue to burn for some time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1934" LastEditorUserId="1934" LastEditDate="2016-03-25T16:10:31.920" LastActivityDate="2017-04-08T21:59:25.407" Title="What chemical available in 18th century would explosively react with phosphorus?" Tags="&lt;reaction-mechanism&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5533" PostTypeId="2" ParentId="5532" CreationDate="2013-07-06T08:57:27.093" Score="8" Body="&lt;p&gt;I think if you just add oxygen you'll just end up with a lump of burning phosphorus. I would suggest having a timed explosive charge at the centre of a phosphorus payload to essentially achieve something like a white phosphorus bomb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since white phosphorus is pyrophoric, you only need to disperse it for it to ignite. To do this, you could either use an automated striking mechanism to ignite a fuse to something like black powder, or my Improvised Munitions Handbook suggests a combination sulfuric acid/potassium chlorate or perchlorate (note: not chloride) + sugar device, with the mechanism keeping the two components separate, would be suitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with your design specifications as I see them is having an excuse to have the external surface of phosphorus exposed to the air: white phosphorus will only glow from the air-exposed surface, so if you put it behind glass, for example, it won't glow. A fine wire mesh as your external covering might do the job, but I'm not sure exactly what effect that has on reliability or effectiveness (and may cause the device to spontaneously and accidentally catch fire, especially if not packed properly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near explosion + being covered in burning white phosphorus = much sadness, probable death. See: &lt;a href=&quot;http://en.wikipedia.org/wiki/White_phosphorus#Effects_on_people&quot; rel=&quot;nofollow noreferrer&quot;&gt;White Phosphorus - Effect on People&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Usual disclaimer: I am not a munitions designer.)&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="402" LastEditDate="2017-04-08T21:59:25.407" LastActivityDate="2017-04-08T21:59:25.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5534" PostTypeId="1" CreationDate="2013-07-06T11:17:40.680" Score="1" ViewCount="160" Body="&lt;p&gt;Okay, so I was fiddling about in the chemistry lab, and I added:&lt;br&gt;&#xA;Sodium Nitroprusside (soln) $\ce{Na_2[Fe(CN)_5NO]}$ with Potassium Ferrocyanide (soln) $\ce{K_4[Fe(CN)_6]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na_2[Fe(CN)_5NO]}$ was a yellowish-reddish colour - actually don't remember the &lt;em&gt;exact colour&lt;/em&gt;, while $\ce{K_4[Fe(CN)_6]}$ was pale yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing really happened to the mixture - except the colours mixed giving a weird yellow colour. I assume the only thing that happened was a double decomposition reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_4[Fe(CN)_6] + Na_2[Fe(CN)_5NO] \rightarrow K_2[Fe(CN)_5NO] + Na_4[Fe(CN)_6]}$&lt;br&gt;&#xA;or something like that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I added a bit of dil Hydrochloric Acid. I expected nothing to happen - nothing ever happens with the chemicals they keep in labs. :P However, the colour of the mixture turned a beautiful Emerald Green!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've been trying to figure out what exactly occured to produce the green colour, but I don't know. Could anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P.S.$ My friend's gonna send me a picture of the green we took. I'll edit it to the question when I recieve it, maybe it will help?&lt;/p&gt;&#xA;" OwnerUserId="1858" LastEditorUserId="4231" LastEditDate="2015-03-02T16:14:58.790" LastActivityDate="2015-03-02T16:14:58.790" Title="What reaction took place?" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5535" PostTypeId="2" ParentId="5471" CreationDate="2013-07-06T16:50:23.043" Score="2" Body="&lt;p&gt;There are a number of rumors on the internet about different ways to treat your battery. Some of these involve putting batteries in the refrigerator or freezer for varying lengths of time to improve performance in some way. &lt;a href=&quot;http://www.snopes.com/oldwives/battery.asp&quot; rel=&quot;nofollow&quot;&gt;Snopes&lt;/a&gt; has concluded that there is no basis for this. However, batteries and battery aging mechanisms are very complex and so I think it's difficult to completely rule out the possibility that under some situations there could be some beneficial effect from freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I would recommend that you avoid using your battery in any way that it was not designed for. The reason is that internal damage to the battery (such as from extreme temperatures, moisture, or mechanical stress) could create a safety hazard. Even if cooking, freezing, or bashing does have some positive effect, it is probably not worth putting yourself at risk. The best thing to do is to recycle your old battery and buy a replacement.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-07-06T16:50:23.043" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5536" PostTypeId="2" ParentId="5529" CreationDate="2013-07-06T22:15:27.127" Score="5" Body="&lt;p&gt;I feel it's important when considering this question to remember that resonance forms are a &lt;em&gt;model&lt;/em&gt;, not an observable phenomenon -- they serve to plug the gap between observable electronic effects and Lewis structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that, it should be reasonably easy to see that multiple strengths and regions of conjugation, representable by more than one resonance form, could exist at various effective ratios and levels of robustness (according to chain deformation) in polymeric structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a simple polyacetylene, for example: ${H_2C}\left({CH}\right)_n{CH_2}$. In the ideal case, this would have perfect conjugation along the entire straight chain, representable by two resonance forms. In practice, the electron density is asymmetric where the molecular geometry is also, which can be represented as additional resonance forms with charge separation.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2014-09-17T11:32:51.210" LastActivityDate="2014-09-17T11:32:51.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5539" PostTypeId="1" AcceptedAnswerId="5540" CreationDate="2013-07-07T09:51:28.197" Score="0" ViewCount="487" Body="&lt;p&gt;I use Sodium persulfate to etch a copper board to produce a PCB. I mix 1 dl sodium persulfate with 1 liter hot water. When the water get cold, I heat it up by microwave. After the microwave, I see such crystallized particles at the bottom of the jar. Please note that there were not any solid particles in the acid before the microwave. I wonder what are these particles? is it sodium persulfate? if it is so, why it is crystallized when it is heated up?&lt;/p&gt;&#xA;&lt;p&gt;Solubility of water increases as the temperature of the water increases, right? Then why it causes solid particles in the car?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/soXrOm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Btw, if I put these crystallized particles into the same solvent, they don't dissolve, however, if I put them into a new hot water cup, they dissolve.&lt;/p&gt;&#xA;&lt;p&gt;If it is sodium persulfate and if they are crystallized because of the microwave, can't I reuse them again?&lt;/p&gt;&#xA;&lt;p&gt;If it is copper particles, isn't it supposed to be in &lt;a href=&quot;http://frgmnt.org/w/wp-content/uploads/2013/05/Cu-residue_640.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;such color&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1940" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-11-13T20:19:26.290" Title="sodium persulfate microwave heating cause crystallized particles" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5540" PostTypeId="2" ParentId="5539" CreationDate="2013-07-07T12:57:12.657" Score="3" Body="&lt;p&gt;It looks to me like copper sulphate ($\ce{CuSO4·nH2O}$). If I'm right then you can't use it for etching. For this you need oxidant (persulphate). Try to solve this solid in water. You should get light blue solution. Stick into it a nail (&lt;em&gt;i.e&lt;/em&gt; some metal iron source). If it cover itself with a brownish powder like in wiki-link you gave, then it's a copper(+2) salt.&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="327" LastEditDate="2013-07-07T13:08:39.247" LastActivityDate="2013-07-07T13:08:39.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5541" PostTypeId="1" CreationDate="2013-07-07T14:05:44.513" Score="1" ViewCount="2113" Body="&lt;p&gt;&lt;strong&gt;QUESTION&lt;/strong&gt;&#xA;    Butene, C4H8 is burned in an engine with a fuel-rich air-fuel ratio. Dry  analysis of the exhaust gives the following volume percents:                   $\ce{CO2} = 14.95%$, $\ce{O2} = 0%$, $\ce{CO} = 0%$, $\ce{H2} = 0%$, $\ce{C4H8} = 0.75%$, with the rest being $\ce{N2}$. Higher heating value of this fuel is 46.9 MJ/kg. Write the balanced chemical equation for one mole of this fuel at these conditions. Calculate the air-fuel ratio, equivalence ratio, lower heating value of the fuel, and energy released when one kg of this fuel is burned in the engine with a combustion efficiency of 98%.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt at solution:&lt;/strong&gt;&#xA;I write the stoichiometric equation like so:&#xA;$$ \ce{C4H8 + \frac{1}{\phi}(4+(8/4))(O2 +3.773N_2)\rightarrow n_1 C4H8 +n_2 CO2 +n_3 H2O}$$&#xA;where, $\phi$ is the equivalence ratio(a engineering term used to describe the ratio of $(F/A)_{\text{actual}}$ to $(F/A)_{\text{stoichiometric}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,now I have 3 unknowns, $n_1$, $n_2$ and $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I unfortunately get $n_2=0$ after doing the mole balance of carbons,hydrogen and nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the percentages given by volume are the mass fractions which should later form a separate equation. But, I still get $n_2=0$. Where have I gone wrong?(Should I still consider putting in $\ce{O2}$ and $\ce{CO}$ into the above equation even though their %by volume are zero?)&lt;/p&gt;&#xA;" OwnerUserId="1728" LastEditorUserId="865" LastEditDate="2021-05-22T20:50:48.090" LastActivityDate="2021-05-22T20:50:48.090" Title="Lower Heating value calculation from chemcial combustion equation of Butane" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5542" PostTypeId="2" ParentId="5541" CreationDate="2013-07-07T19:13:57.237" Score="4" Body="&lt;p&gt;I see three flaws in your reasoning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, &lt;a href=&quot;http://en.wikipedia.org/wiki/Butane&quot; rel=&quot;nofollow&quot;&gt;butane&lt;/a&gt; is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of &lt;a href=&quot;http://en.wikipedia.org/wiki/C4H8&quot; rel=&quot;nofollow&quot;&gt;butene&lt;/a&gt;. I will keep using your formula, but you should specify whether you mean &lt;strong&gt;butane&lt;/strong&gt; or &lt;strong&gt;$\ce{C4H8}$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C4H8 + 6O2 -&amp;gt; 4CO2 + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C4H8 + (6-6n_1)O2 -&amp;gt; n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-07-07T19:13:57.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5543" PostTypeId="1" AcceptedAnswerId="5550" CreationDate="2013-07-08T08:55:02.470" Score="5" ViewCount="624" Body="&lt;p&gt;I've recently obtained a couple of electronics units which have suffered corrosion on their copper inductor coils. They were operated in conditions of intense humidity/heat and were placed near the sea. This has resulted in the complete failure in one of these units, where the inductor has come apart due to being eaten away by corrosion, and the other coil shows noticeable amounts of corrosion (patina present).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to operate the latter device in a more forgiving environment (ie: putting a filter in front of it in an air conditioned room) will the copper continue to corrode regardless or will the oxide layer stop spreading further?&lt;/p&gt;&#xA;" OwnerUserId="1943" LastActivityDate="2015-07-18T21:29:40.110" Title="If copper has started to corrode, can it stop?" Tags="&lt;everyday-chemistry&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5544" PostTypeId="1" AcceptedAnswerId="5545" CreationDate="2013-07-07T08:49:23.227" Score="12" ViewCount="18106" Body="&lt;p&gt;I remember that in my primary classed I had an experiment in science to show air contains 20% oxygen. The experiment involved taking a trough of water, lighting a candle in the middle of the trough and inverting a glass jar over the candle so that air supply for the candle is cut of. When candle extinguishes, water level in the glass jar rises. I have seen this happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why should the water raise in the jar, since burning of candle should produce an equivalent amount of carbon dioxide, so the molar amount of gas in jar is constant? Also the heat from the candle should expand the gasses in the jar, water level should go down I think.&lt;/p&gt;&#xA;" OwnerUserId="1946" OwnerDisplayName="Gautam" LastActivityDate="2016-11-24T14:26:46.437" Title="Experiment to show that air contains about 20% oxygen" Tags="&lt;home-experiment&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5545" PostTypeId="2" ParentId="5544" CreationDate="2013-07-07T09:36:43.637" Score="6" Body="&lt;p&gt;The paraffin wax is an alkane ($\ce{C_nH_{2n+2}}$); its combustion reaction is (complete combustion):&#xA;$$\ce{2C_nH_{2n+2} + $(3n+1)$ O2 -&amp;gt; $2(n+1)$ H2O + $2n$ CO2}$$&#xA;or (incomplete combustion):&#xA;$$\ce{2C_nH_{2n+2} + $(2n+1)$ O2 -&amp;gt; $2(n+1)$ H2O + $2n$ CO}$$&#xA;In the two cases the quantity (in moles) of $\ce{O2}$ is greater than the quantity of $\ce{CO2}$ or $\ce{CO}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover the gaseous $\ce{H2O}$ becomes liquid.&lt;/p&gt;&#xA;" OwnerUserId="1899" OwnerDisplayName="metacompactness" LastEditorUserId="16683" LastEditDate="2016-11-24T14:26:46.437" LastActivityDate="2016-11-24T14:26:46.437" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5546" PostTypeId="1" AcceptedAnswerId="5564" CreationDate="2013-07-08T12:17:29.667" Score="3" ViewCount="488" Body="&lt;p&gt;From what I understand, the working agent inside a peltier cooler is Bismuth-Telluride. I've read a toxicity data-sheet regarding Bismuth-Telluride &lt;a href=&quot;http://www.espimetals.com/index.php/msds/50-bismuth-telluride&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to use a peltier cooler inside a sealed cooling system, for example a thermoelectric cooler, will any groceries stored in it be &quot;infused&quot; with Bismuth-Telluride?&lt;/p&gt;&#xA;" OwnerUserId="1945" LastEditorUserId="7448" LastEditDate="2015-07-18T13:46:11.337" LastActivityDate="2015-07-18T13:46:11.337" Title="Peltier Bismuth-Telluride toxicity" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5547" PostTypeId="1" CreationDate="2013-07-08T12:41:23.780" Score="0" ViewCount="101" Body="&lt;p&gt;I've recently had several jerry-cans of oil (awaiting disposal) topple over in the attached garage, and several rooms following influx of water following heavy rain. After the water level receded the floor is left with a thin layer of burnt oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various sites on the net recommend using sawdust, baking soda etc. I'm talking close to 800sq.ft of area split into several levels with a step between different levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have to ask, therefore, must be obvious now. How does one clean large volumes of burnt oil spilt distributed by water over cement, and stone floors? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Nature giveth, and Nature taketh away. Another cloudburst, this time over 100mm of rain in a couple of hours, and muddy waters found their way indoors again. Now the waters have receded, and almost all the oil is replaced with a thin layer of mud. Hopefully washing with water will get rid of the most of it.&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="865" LastEditDate="2015-10-30T17:00:09.263" LastActivityDate="2015-10-30T17:00:09.263" Title="Is there an efficient and effective way to clean oil from floor?" Tags="&lt;everyday-chemistry&gt;&lt;surfactants&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5548" PostTypeId="2" ParentId="5544" CreationDate="2013-07-08T12:46:26.167" Score="7" Body="&lt;p&gt;One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\ce{CO2}$ and water vapor.  If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen.  A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\ce{4Fe(s) + 3O2(g) -&amp;gt; 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This experiment has been reported in the &lt;a href=&quot;http://dx.doi.org/10.1021/ed064p809&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Journal of Chemical Education&lt;/em&gt;&lt;/a&gt; and a brief search of the web found an &lt;a href=&quot;http://chemlabs.uoregon.edu/Classes/Exton/CH228/PercentOxygen.pdf&quot; rel=&quot;noreferrer&quot;&gt;adaptation of this experiment&lt;/a&gt; if you don't have access to the journal.  Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water.  The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution.  Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment.  Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-08T12:46:26.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5549" PostTypeId="2" ParentId="5547" CreationDate="2013-07-08T16:09:52.447" Score="2" Body="&lt;p&gt;Assuming you only want chemistry-related answers, I'd recommend any type of detergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you lift that restriction, however, then I'd recommend &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure_washer&quot; rel=&quot;nofollow&quot;&gt;pressure washers&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1237" LastActivityDate="2013-07-08T16:09:52.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5550" PostTypeId="2" ParentId="5543" CreationDate="2013-07-08T16:14:01.390" Score="2" Body="&lt;p&gt;Unlike rust on iron, patina naturally protects underlying copper, and stops further corrosion directly underneath.&lt;/p&gt;&#xA;" OwnerUserId="1237" LastActivityDate="2013-07-08T16:14:01.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5551" PostTypeId="1" AcceptedAnswerId="5555" CreationDate="2013-07-08T17:41:39.063" Score="5" ViewCount="37889" Body="&lt;p&gt;I have submerged a iron nail in vinegar twice recently. What is puzzling me is the different results each time. The first time I submerged it, it was covered in rust. When I put it in the solution of vinegar, I got a red precipitate dissolved in the solution that I believe was Iron(II) and Iron(III) acetate. However, when I put the newly cleaned nail in ANOTHER bath of the same concentration of vinegar, I did get some bubbles but no red precipitate that was dissolved in the solution. It's only been about 2 days, though. However, there was a black precipitate on the surface of the nail, but I do not believe it is iron acetate because of its color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused as to the results because Iron is higher than hydrogen on the activity series, so it SHOULD be reacting with the vinegar to give me that red precipitate and hydrogen gas. I added some hydrogen peroxide to the solution to oxidize the iron, and the solution is getting a reddish tint to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you think I am not getting that reddish precipitate when I put in the cleaned iron nail in the new vinegar solution? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="418" LastEditDate="2015-01-08T07:58:57.600" LastActivityDate="2015-11-13T20:15:52.273" Title="Different reactions of clean and rusty iron nails in vinegar (acetic acid)" Tags="&lt;home-experiment&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5552" PostTypeId="2" ParentId="5551" CreationDate="2013-07-08T19:00:48.900" Score="1" Body="&lt;p&gt;When you dip a rusty Iron Nail in veniger solution, the Viniger reacts with $\ce{Fe2O3.nH2O}$ (Rusty Iron or oxidized Iron).&#xA;Whereas in case of clean nail, Its directly reacts with iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your product is probably different due to this reason only, if I understood your question right. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : Chemical Equation for pure iron in venigar -&#xA;$\ce{Fe(s) + 2CH3COOH(aq) -&amp;gt; Fe(CH3COO)2(aq) + H2(g)}$&#xA;Chemical Equation for Rusty iron in Venigar -&#xA;$\ce{Fe2O3 + 8CH3COOH(aq) -&amp;gt; 4Fe(CH3COO)2(aq) + O2 + 4H2O}$ (Not sure about this equation.)&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-07-09T06:42:54.293" LastActivityDate="2013-07-09T06:42:54.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5553" PostTypeId="2" ParentId="5108" CreationDate="2013-07-08T19:12:00.930" Score="-1" Body="&lt;p&gt;I agree with @Nicolau Saker Neto that it's complicated. You might want to look at these two papers, if you're really interested in more of the details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R. N. Keller, &lt;em&gt;J. Chem. Ed.&lt;/em&gt; 39(6), 1962. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed039p289&quot; rel=&quot;nofollow&quot;&gt;PDF available here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Robin M. Hochstrasser, &lt;em&gt;J. Chem. Ed.&lt;/em&gt; 42(3), 1965. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed042p154&quot; rel=&quot;nofollow&quot;&gt;PDF available here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The links might be paywalled. Sorry. :-/&lt;/p&gt;&#xA;" OwnerUserId="1949" LastActivityDate="2013-07-08T19:12:00.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5554" PostTypeId="2" ParentId="5544" CreationDate="2013-07-09T00:33:44.210" Score="0" Body="&lt;p&gt;I doubt that the CO2 dissolves quickly enough. My gut feeling is that this experiment is not working as suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is a physical, not chemical, effect. &#xA;When you put the glass above the candle, there is already a column of hot air. When that air cools, the water rises.&#xA;It would be a worthwhile experiment to ignite the candle electrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better experiment would use steel wool, like bobthechemist suggested. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw. it is very easy to electrically ignite steel wool.&lt;/p&gt;&#xA;" OwnerUserId="872" LastActivityDate="2013-07-09T00:33:44.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5555" PostTypeId="2" ParentId="5551" CreationDate="2013-07-09T07:37:36.330" Score="4" Body="&lt;p&gt;In your first experiment, the rust ($\ce{Fe2O3. xH2O}$) probably reacted with vinegar to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(III)_acetate&quot; rel=&quot;nofollow&quot;&gt;Iron (III) acetate&lt;/a&gt; which makes the solution reddish in color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second experiment, the iron on the clean nail surface would have reacted with vinegar to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(II)_acetate&quot; rel=&quot;nofollow&quot;&gt;Iron (II) acetate&lt;/a&gt;. There are two main reasons why the solution might appear to have no change:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Usually, Iron (II) acetate is prepared using &lt;strong&gt;concentrated acetic acid&lt;/strong&gt; and scrap iron (or ferrous oxide/hydroxide). Since &lt;a href=&quot;http://en.wikipedia.org/wiki/Vinegar#Distilled_vinegar&quot; rel=&quot;nofollow&quot;&gt;vinegar&lt;/a&gt; is extremely dilute (around 5-8% acetic acid solution in water), the reaction would be very slow and the conc. of Iron (II) acetate formed will be very low.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Iron (II) acetate happens to be very soluble in water and has a very faint light green color in solution which is not easily detectable in lower concentrations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then again, when you added $\ce{H2O2}$ to the second solution, you mentioned that a reddish color appeared. This shows that the dissolved $\ce{Fe^2+}$ ions are oxidized to $\ce{Fe^3+}$. The $\ce{Fe^3+}$ formed will react with the acetate ions to form red Iron (III) acetate.&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorUserId="10185" LastEditDate="2015-08-19T09:54:11.553" LastActivityDate="2015-08-19T09:54:11.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5556" PostTypeId="2" ParentId="5101" CreationDate="2013-07-09T08:27:52.450" Score="9" Body="&lt;p&gt;I think you’ve made a so called “aluminium soap”. It is a compound of the form $\ce{(RCOO)3Al}$ where $\ce{RCOO-}$ is a fatty acid chain. The $\ce{Al^3+}$ ions from the alum you handled would have displaced the $\ce{Na+}$ ions in soap to form the thick gel like aluminium version of soap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to how $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions form an insoluble scum (which consists of $\ce{(RCOO)2Mg}$ or $\ce{(RCOO)2Ca}$) when soap is used in hard water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume here that you’re talking about potash alum $\ce{KAl(SO4)2.12H2O}$ as @BenNorris has mentioned in his comment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other aluminium containing alum will also have the same effect. But chrome alums (containing chromium instead of aluminium) might have a different effect on soap.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These “aluminium soaps” are used as a thickening agent in oils and greases and also in napalm (as Georg has mentioned in his comment above).&lt;/p&gt;&#xA;" OwnerUserId="1322" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T17:32:51.647" LastActivityDate="2015-04-24T17:32:51.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5558" PostTypeId="1" AcceptedAnswerId="5563" CreationDate="2013-07-09T18:23:01.653" Score="2" ViewCount="849" Body="&lt;p&gt;The process of crystallization of a liquid or any solute occurs when its concentration exceeds the maximum concentration which the solvent can carry or when the temperature drops below the freezing temperature in cases of liquids. Then why is it that these materials require a small &quot;nuclei&quot; or a jagged edge to start forming crystal on, why cant it form crystal anywhere within the bulk of the matter. (eg cloud formation from condensation of water)&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-07-09T23:25:58.433" Title="Crystallization of solutions and liquids" Tags="&lt;aqueous-solution&gt;&lt;crystal-structure&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5563" PostTypeId="2" ParentId="5558" CreationDate="2013-07-09T23:25:58.433" Score="5" Body="&lt;p&gt;Crystallization is rightly called by you &lt;em&gt;a process&lt;/em&gt;.  Simplified, there are two steps to be considered: Nucleation and crystal growth.  For the sake of simplification, let me consider the crystallization of a molecular compound in absence of ionic charges.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nucleation refers to the initial stage, where once isolated molecules of the dissolved substance assemble to a first crystal germ. This is energetically hindered because these particles of especially small size have unfavourably high ratio of particle surface energy per molecule in these just formed particle (or perhaps better: aggregates). This energetic drawback has to be overcompensated. &lt;br/&gt;In addition, molecules loose degrees of freedom accessible in solution, at first: dynamic freedom, and secondly, depending how strictly rules are followed to build the crystal, a number of orientations in space otherwise available and accessible--In a perfect crystal, the structural close environment of one molecule is going to be repeated for all molecules, the very next ones and for their next ones, over ranges of Angstroms to Centimetres.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once such a germ is formed,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Crystal growth more or less refers to simply attachment of additional molecules according to these &quot;(energetically justified) previously determined rules&quot; to yield a bigger entity.  The need of oversaturation of the solute is less prominent than during nucleation, yet still existent to energetically favour this process. In this description, crystal growth is again simplified, too.  It may happen, in the (sub)process of crystal growth, conditions are met that allow crystallisations in realisations even more energetically favourable than at early stages of crystal growth so that a compound preferentially forms a &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymorphism_%28materials_science%29&quot;&gt;polymorph A&lt;/a&gt; initially, then ones in polymorph B.&lt;br/&gt; And often a crystallization does not yield only one (perhaps even single)crystal, rather than multiple ones.  So nucleation may occur in one spot of the solution while simultaneously crystal growth already takes place on germs already formed in other parts of the solution.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Complementary, think this way: it is every time easy to dissolve a perfect crystal (already considering entropy), but easily difficult to bring molecules back into a perfectly ordered crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science and art (like craftsmanship) of crystallization is of high importance in our daily life, spanning areas of electronic devices (semiconductors) till pharmaceutics. Journals like &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryst_Growth_Des&quot;&gt;&lt;em&gt;Crystal Growth &amp;amp; Design&lt;/em&gt;&lt;/a&gt; or &lt;a href=&quot;http://pubs.rsc.org/en/journals/journalissues/ce#!recentarticles&amp;amp;all&quot;&gt;&lt;em&gt;CrystEngComm&lt;/em&gt;&lt;/a&gt; are examples concentrating on this subject.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-07-09T23:25:58.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5564" PostTypeId="2" ParentId="5546" CreationDate="2013-07-09T23:51:54.977" Score="3" Body="&lt;p&gt;Yes there are little cooling boxes, for camping, for example, where one of the walls (partially) is made of a sheet of a food-safe-metal, back-to-back to one side of a Peltier element as a cooling device.  This works fine, for example if attached to the car's battery.  Less efficient (by rate of heat transfer) may be an separation of the Peltier element viz. the inner of the cooling box by glass or food-safe plastic. &lt;br&gt;&#xA;Of course, a &lt;em&gt;complete&lt;/em&gt; immersion of the Peltier device into a cooler or cooling system won't work, as necessarily one of the flat sides of the element is going to become cool, but the other going to be warm.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-07-09T23:51:54.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5565" PostTypeId="1" CreationDate="2013-07-10T16:37:55.297" Score="3" ViewCount="887" Body="&lt;p&gt;I have practical question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I chemically remove $\ce{Sn}$ from $\ce{Sn60Cu40}$ alloy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get something like copper sponge this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like nanoporous gold produced from $\ce{Au50Ag50}$ for heat exchanger, but cheaper.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastEditorUserId="10185" LastEditDate="2015-04-20T09:02:26.530" LastActivityDate="2018-11-20T19:46:34.900" Title="How to remove Sn from Sn60Cu40 alloy?" Tags="&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5566" PostTypeId="2" ParentId="5565" CreationDate="2013-07-10T17:53:44.887" Score="5" Body="&lt;p&gt;You can remove tin from bronze (copper-tin alloys) and leave behind the copper. However, doing so may not leave you with copper in an easy to use form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_series&quot; rel=&quot;nofollow&quot;&gt;less reactive&lt;/a&gt; metal than copper. Tin reacts with all acids, like $\ce{HCl}$, while copper only reacts with oxidizing acids, like $\ce{HNO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Sn(s) +2H+(aq) -&amp;gt; Sn^{2+}(aq) + H2 (g)^ }$$&#xA;$$\ce{Cu(s) + 2H+ (aq) -&amp;gt; No \ Reaction}$$&#xA;$$\ce{Cu(s) + 4H+(aq) + 2NO3- (aq) -&amp;gt; Cu^{2+} (aq) + 2H2O (l) + 2NO2 (g) ^}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soaking your bronze in acid (even vinegar might work) will eventually dissolve the tin away from the copper. Heating, or using a stronger acid like the $\ce{HCl}$ in muriatic acid, will likely increase the rate. However, since the composition of the alloy is uniform throughout, the copper will likely come out of solution as a fine powder.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4231" LastEditDate="2015-03-02T16:15:51.387" LastActivityDate="2015-03-02T16:15:51.387" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5568" PostTypeId="1" AcceptedAnswerId="5571" CreationDate="2013-07-11T07:53:41.707" Score="29" ViewCount="110863" Body="&lt;p&gt;In High School I learned that an exothermic reactions releases energy, while an endothermic reaction needs energy to occur. Now I learned that there is a separate, somewhat similar classification scheme of exergonic and endergonic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between these two classification schemes? Are exothermic reactions always exergonic, and if not, can you give me an example?&lt;/p&gt;&#xA;" OwnerUserId="1964" LastActivityDate="2020-06-16T04:17:55.043" Title="Difference between exothermic and exergonic" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;free-energy&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5569" PostTypeId="1" CreationDate="2013-07-11T09:18:29.810" Score="19" ViewCount="405" Body="&lt;p&gt;Humans typically achieve combustion by starting a chemical reaction of a fuel (often a carbon &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; containing fluid like methane &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt;), which is is brought into an an oxidizer (a gas containing &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;, like air). For example, when you turn on a lighter.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What would happen if the situation were turned around?&lt;/p&gt;&#xA;&lt;p&gt;We have an open space entirely filled with fuel and we let in some oxygen-containing fluid. How would the combustion look?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know the (adiabatic?) temperature maximum of a flame will be achieved at conditions which are slightly on the fat side. But I have no idea how, for example, the production of unburned &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;'s, soot production and so on, would look.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-04-21T17:08:52.353" Title="Open flame combustion in bizarro world" Tags="&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5570" PostTypeId="1" AcceptedAnswerId="5581" CreationDate="2013-07-11T10:54:36.810" Score="2" ViewCount="626" Body="&lt;p&gt;Tight temperature controls are needed to ensure that the filament used in fabrication does not begin to &quot;ooze&quot;. In places where the air pressure rises and drops during the course of the day there may be more or less ooze taking place. Controlling ooze is paramount to obtain better print quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Given Boyle's law, shouldn't the melting point of PLA and other plastics used in the Fused Filament Fabrication 3D printing methods change based on the ambient air pressure of the machine's location? Should air pressure be taken into account when calculating how much heat to apply to the print nozzles? Does humidity affect PLA's melting point?&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastActivityDate="2013-07-13T14:44:08.260" Title="Does variations in air pressure affect the melting point of PLA and ABS while 3D Printing?" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5571" PostTypeId="2" ParentId="5568" CreationDate="2013-07-11T11:46:06.827" Score="40" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The classifications &lt;em&gt;endothermic&lt;/em&gt; and &lt;em&gt;exothermic&lt;/em&gt; refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications &lt;em&gt;endergonic&lt;/em&gt; and &lt;em&gt;exergonic&lt;/em&gt; refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are three cases to distinguish:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}H &amp;lt; 0$, an exothermic reaction that releases heat to the surroundings (temperature increases) &lt;/li&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat  &lt;/li&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}H &amp;gt; 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In 1876, &lt;a href=&quot;https://en.wikipedia.org/wiki/Julius_Thomsen&quot; rel=&quot;noreferrer&quot;&gt;Thomson&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Marcellin_Berthelot&quot; rel=&quot;noreferrer&quot;&gt;Berthelot&lt;/a&gt; described this driving force in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thomsen-Berthelot_principle&quot; rel=&quot;noreferrer&quot;&gt;principle&lt;/a&gt; regarding affinities of reactions. According to them, only exothermic reactions were possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, &lt;em&gt;even&lt;/em&gt; during cold winter? Thanks to works by &lt;a href=&quot;https://en.wikipedia.org/wiki/Hermann_von_Helmholtz&quot; rel=&quot;noreferrer&quot;&gt;von Helmholtz&lt;/a&gt;,  &lt;a href=&quot;https://en.wikipedia.org/wiki/Van%27t_Hoff,_Jacobus_Hendricus&quot; rel=&quot;noreferrer&quot;&gt;van't Hoff&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann&quot; rel=&quot;noreferrer&quot;&gt;Boltzmann&lt;/a&gt; (and others) we may do. &lt;a href=&quot;https://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;noreferrer&quot;&gt;Entropy&lt;/a&gt; $S$, depending on the number of accessible realisations of the reactants (&quot;describing the degree of order&quot;) necessarily is to be taken into account, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two contribute to the maximum work a reaction may produce, described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Gibbs free energy&lt;/a&gt; $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants (&quot;degree or order&quot;) may change ($\Delta_\mathrm{R} S$ may be large).  For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are three cases to distinguish:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}G &amp;lt; 0$, an exergonic reaction, &quot;running voluntarily&quot; from the left to the right side of the reaction equation (react is spontaneous as written)&lt;/li&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or&lt;/li&gt;&#xA;&lt;li&gt;$\Delta{}_{\mathrm{R}}G &amp;gt; 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -&amp;gt; 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H &amp;lt; 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S &amp;gt; 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G &amp;lt; 0$).&lt;/li&gt;&#xA;&lt;li&gt;Example, reaction of dioxygen to ozone, $\ce{3 O2 -&amp;gt; 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G &amp;gt; 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S &amp;lt; 0$) and &lt;em&gt;simultaneously&lt;/em&gt; it is endothermic ($\Delta_{\mathrm{R}}H &amp;gt; 0$), too.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Water-gas_shift_reaction&quot; rel=&quot;noreferrer&quot;&gt;Water gas reaction&lt;/a&gt;, where water vapour is guided over solid carbon $\ce{H2O + C &amp;lt;=&amp;gt; CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S &amp;gt; \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic.&lt;/li&gt;&#xA;&lt;li&gt;Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -&amp;gt; 2 H2O}$.  This is an exothermic reaction ($\Delta_{\mathrm{R}}H &amp;lt; 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S &amp;lt; 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| &amp;lt; |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction.  In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="56320" LastEditDate="2017-12-12T23:46:42.010" LastActivityDate="2017-12-12T23:46:42.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5572" PostTypeId="1" CreationDate="2013-07-11T12:42:40.450" Score="2" ViewCount="590" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Sodium persulfate to etch a copper board to manufacture home made Printed Circuit Board (PCB). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the etching powder's plastic, it is written that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1.5-2dl of etch powder in 1 L of hot water or corresponding volumes &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I add 2 cup powder for 10 cup hot water. In the first time usage it works perfect and it etches the board (double layer 8x16 cm FP4 board) in 20-30 min. However, when I use it for the second time (in another day), the solution works tooo slow. The last time I waited 13 hours for board to be ready. As the board stayed too long on the enchant, the acid ate away many traces on the PCB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I add extra sodium persulfate (around 2 cup again) when I use the solution second time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use microwave (900 W) to warm the enchant again. I have not measured the temperature of the acid but I guess it might be around 40-60 C and I don't have any temperature stabilization so that the solution get colds in 5-10 min again and I repeat the same procedure maybe 6-8 times per process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it takes too long to etch the copper when I use the solution second time although I add the same amount (2 cups) of sodium persulfate to the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do to speed up the etching process apart from mixing the solution?&lt;/p&gt;&#xA;" OwnerUserId="1940" LastEditorUserId="9961" LastEditDate="2015-11-13T19:35:57.470" LastActivityDate="2015-11-13T19:35:57.470" Title="speeding up etching process of a copper board" Tags="&lt;redox&gt;&lt;applied-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5573" PostTypeId="1" CreationDate="2013-07-12T00:07:36.207" Score="25" ViewCount="258" Body="&lt;p&gt;Looking at avidity between a ligand-receptor, you're looking at an enhancement of the $K_d$ compared to a lone ligand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it more appropriate to compare the $K_d$ using the concentration of the &quot;free&quot; ligand or the concentration of the free complex?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ligand is odd at the lower regimes since the ligand efficency is fairly low but using the nanoparticle changes how you have to calculate the effective concentration. I've seen papers that do both and I'm curious which is most appropriate/correct&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="1362" LastEditDate="2014-01-30T21:26:17.130" LastActivityDate="2018-08-25T20:00:53.087" Title="When measuring avidity should you use the ligand concentration or the nanoparticle concentration?" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5574" PostTypeId="1" AcceptedAnswerId="5575" CreationDate="2013-07-12T00:25:49.277" Score="2" ViewCount="2808" Body="&lt;p&gt;In molecular orbitals diagram, sigma ($\sigma$) and pi ($\pi$) symbols are used. What do these symbols really mean?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="7448" LastEditDate="2015-07-17T20:28:35.250" LastActivityDate="2015-07-17T20:28:35.250" Title="Sigma and pi symbols in molecular orbital theory" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5575" PostTypeId="2" ParentId="5574" CreationDate="2013-07-12T00:45:29.023" Score="8" Body="&lt;p&gt;These symbols are for bonding orbitals. A covalent bond is formed when two orbitals in different atoms hybridize. $\sigma$ and $\pi$ refer to two different configurations in which this can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a $\sigma$ bond, two $s$ orbitals or two $p$ orbitals oriented towards each other join into one orbital, with the electron density concentrated between the two atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a $\pi$ bond, two $p$ orbitals oriented perpendicular to the bond axis merge, and, in the resulting orbital, electron density is concentrated above and below the bond axis.&lt;/p&gt;&#xA;" OwnerUserId="1815" LastActivityDate="2013-07-12T00:45:29.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5577" PostTypeId="1" AcceptedAnswerId="5579" CreationDate="2013-07-12T14:49:05.723" Score="9" ViewCount="850" Body="&lt;p&gt;I'm trying to remove water from an imidazolium salt (i.e. dry the salt).  However I want to know if it is safe to remove the water under vacuum.  Is it safe to condense water into a liquid nitrogen trap?  Wouldn't the water freeze in the trap since the freezing temperature of water is 0 degrees Celcius?  Just curious b/c I was informed that it was ok to do it this way.&lt;/p&gt;&#xA;" OwnerUserId="1972" LastActivityDate="2013-07-13T00:54:38.937" Title="Is it safe to condense water into a liquid nitrogen trap?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;experimental-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5578" PostTypeId="2" ParentId="5018" CreationDate="2013-07-12T23:59:37.033" Score="2" Body="&lt;p&gt;Most likely that your teacher meant to say phosphoric acid ($\ce{H3PO4}$) rather than the hydrogen phosphate anion ($\ce{HPO4^{2-}}$) so lets go with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ 3CaCO3 + 2H3PO4 -&amp;gt; Ca3(PO4)2 + 3CO2 + 3H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the carbon dioxide and water would exist in an equilibrium state with carbonic acid, $\ce{H2CO3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ H2CO3 (aq) &amp;lt;=&amp;gt; H2O (l) + CO2(aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="1941" LastEditorUserId="16683" LastEditDate="2017-04-07T12:39:52.687" LastActivityDate="2017-04-07T12:39:52.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5579" PostTypeId="2" ParentId="5577" CreationDate="2013-07-13T00:54:38.937" Score="11" Body="&lt;p&gt;One use of a liquid nitrogen cold trap is to collect organic and water vapors that are removed from a system under vacuum.  In this case, the vapors are collected as ices (solids).  See &lt;a href=&quot;http://operatingexperience.doe-hss.wikispaces.net/file/view/Schlenk+Line+Safety.pdf&quot;&gt;this reference&lt;/a&gt; which has some useful information on Schlenk line safety and use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you want to be cautious is when your collection is finished.  You do not want to keep a liquid nitrogen cold trap exposed to the atmosphere because oxygen will condense and cause an explosion hazard.  Typical procedures would include shutting of the vacuum pump, dismantling of the vacuum apparatus and allowing the trap to warm to room temperature (in a fume hood if organic vapors were collected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional references might be helpful, including &lt;a href=&quot;http://kubiak.ucsd.edu/manual/schlenktech.php&quot;&gt;this one&lt;/a&gt; that not only describes proper Schlenk line technique but also provides some good references; and you can see how liquid nitrogen traps are used for several industrial purposes &lt;a href=&quot;http://vacaero.com/information-resources/the-heat-treat-doctor/1225-cold-traps.html&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.sesco-inc.com/product_listing.aspx?Category=badcdc16-c31f-4484-9fa7-0af5d8ea82c1&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-13T00:54:38.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5580" PostTypeId="2" ParentId="5569" CreationDate="2013-07-13T01:21:28.877" Score="10" Body="&lt;p&gt;I'll leave the science fiction/fantasy to more creative contributors, but I think your conditions can be replicated in our not-so-bizzaro world (although those over at the &lt;a href=&quot;https://www.youtube.com/channel/HCvMAz8uP7SYo&quot; rel=&quot;noreferrer&quot;&gt;Chemistry World Blog&lt;/a&gt; might be able to convince you that our world is a bit bizzare).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you returned from watching some videos, consider the following chemical equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + 2O2 -&amp;gt; 2H2O + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly a combustion reaction with two reagents: a fuel, $\ce{CH4}$, and an oxidant, $\ce{O2}$.  It is important to emphasize that &lt;strong&gt;the reaction requires both reactants in order to go to completion&lt;/strong&gt;. If we look closely at the stoichiometry, we require two molecules of oxygen for every one molecule of methane.  The question posed was&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we have an open space filled with fuel, and a little amount of oxygen is added, what would it look like?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, perhaps it would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tNqVw.png&quot; alt=&quot;pre&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the blue dots represent $\ce{CH4}$ and the red dots represent $\ce{O2}$.  There is an excess amount of methane (fuel) for the amount of oxygen (oxidant) in this box.  In General Chemistry class we refer to this as a limiting reagent problem.  The oxygen is a limiting reagent, and the reaction will only go on so long as there is a sufficient amount of both reagents.  When all the oxygen is used up, we have this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FQljZ.png&quot; alt=&quot;post&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the orange dot represents $\ce{CO2}$ and the green dots represent $\ce{H2O}$.  Nothing else is likely to happen at this point.  So getting back to your question, &lt;em&gt;what would this world look like?&lt;/em&gt;  Well it would look a little wetter and a little hotter.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-13T01:21:28.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5581" PostTypeId="2" ParentId="5570" CreationDate="2013-07-13T13:13:07.467" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Boyle%27s_law&quot; rel=&quot;nofollow&quot;&gt;Boyle's law&lt;/a&gt; refers to properties of an idealized gas, where the product of pressure and volume remains constant.  Taking into account your description, as well from filament-based &lt;a href=&quot;https://en.wikipedia.org/wiki/3D_printing#Extrusion_deposition&quot; rel=&quot;nofollow&quot;&gt;3D printers&lt;/a&gt; seen so far, my answer is &lt;em&gt;no, Boyle's law does not apply here&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If running &lt;a href=&quot;https://en.wikipedia.org/wiki/Fused_deposition_modeling&quot; rel=&quot;nofollow&quot;&gt;by extrusion&lt;/a&gt;, polylactic acid (&lt;a href=&quot;https://en.wikipedia.org/wiki/Polylactic_acid&quot; rel=&quot;nofollow&quot;&gt;PLA&lt;/a&gt;) is heated to overpass melt, however this liquid is not heated to become a gas. Temperature control clearly is of importance, because the viscosity of the polymer depends on it.  And viscosity co-influences the precision of details you may print, actually down to the scale of less as a tenth of a millimetre.  Without experience to run one 3D print in at sea level, and the same again during an intercontinental flight, I assume it is &lt;em&gt;more likely&lt;/em&gt;, machine control takes into account the &lt;em&gt;pressure difference&lt;/em&gt; between ambient air and polymer melt, and not (yet) the (in comparison) lesser variations of air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical structure of PLA shown in Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Polylactic_acid&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; shows a relatively low number of functional groups susceptible to water, relatively low to the inner backbone of the polymer (in the structural scheme, this is written in the square brackets).  The article refers to typical molecular weights of 130 kDa when PLA is produced.  These two let me say, humidity does not play an important role.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="73" LastEditDate="2013-07-13T14:44:08.260" LastActivityDate="2013-07-13T14:44:08.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5583" PostTypeId="1" AcceptedAnswerId="5584" CreationDate="2013-07-14T16:56:53.980" Score="4" ViewCount="4680" Body="&lt;p&gt;The wave function for a particle in a infinite potential well located at $0$ and $L$ is $$\psi_n=\sqrt{\frac2L}\sin\frac{n\pi x}L$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find our wave function for the same situation where the walls are located at $-L$ and $+L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can write the Schrödinger equation, but I am unable to apply the boundary conditions, so I am unable to find the wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find the wave function in case of walls at $-L$ and $+L$?&lt;/p&gt;&#xA;" OwnerUserId="1854" LastEditorUserId="1085" LastEditDate="2013-07-15T08:36:35.340" LastActivityDate="2013-08-16T13:53:52.510" Title="Wave function for particle in a infinite well located at -L and +L" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5584" PostTypeId="2" ParentId="5583" CreationDate="2013-07-14T18:55:02.490" Score="4" Body="&lt;p&gt;I will answer this question by copying a passage out of the very good QM book bei Shankar (see &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0306447908&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) which I can recommend wholeheartedly to anyone trying to learn quantum mechanics. I considered writing up a solution myself but since the passage answers your question exactly (ok, not quite exactly since the calculation is performed for the infinite wells being located at $-\frac{L}{2}$ and $\frac{L}{2}$ but that doesn't change anything important) it would have been a waste of time and effort to do so. If you have any questions about the derivation I'm happy to help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPvxJ.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KLWKt.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/abvWG.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UfjUu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-07-14T19:10:13.250" LastActivityDate="2013-07-14T19:10:13.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5585" PostTypeId="1" AcceptedAnswerId="5586" CreationDate="2013-07-14T23:15:32.157" Score="-5" ViewCount="876" Body="&lt;p&gt;I'm trying to figure out why the answer is C. Can anyone explain to me why that is the answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMcYDm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1975" LastEditorUserId="22" LastEditDate="2013-07-15T12:38:11.340" LastActivityDate="2015-07-17T20:07:42.043" Title="Which compound reacts most slowly?" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2013-07-15T12:39:20.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="5586" PostTypeId="2" ParentId="5585" CreationDate="2013-07-15T00:17:44.503" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitration is a general class of chemical process for the introduction of a nitro  group into an organic chemical compound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For introduction of Nitro group, the compound must have either partial charge or proper charge (ions) or must form a radical to form the bond between Nitro group and compund. Compound III (&lt;code&gt;option C&lt;/code&gt;) is more stable and ease of charge generation or free radical generation will be less due to less difference in electronegavities of atoms.&#xA;     Due to less electronegativity difference, the shared pair of electron contributes equally to the electron density of both atoms. If electronegativity difference is higher, ease of Charge generation is higher, eg. $\ce{HCL}$ (both have Partial charges, $H^{\delta+}$ &amp;amp; $Cl^{\delta-}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the only reason for which the reaction will be slowest for compound C.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="7448" LastEditDate="2015-07-17T20:07:42.043" LastActivityDate="2015-07-17T20:07:42.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5587" PostTypeId="2" ParentId="5448" CreationDate="2013-07-15T04:21:40.373" Score="6" Body="&lt;p&gt;You've been doing everything correctly so far - what we have now is the relative amount of substance in moles of each element in this $\pu{100 g}$ compound. As you correctly calculated, we have $\pu{1.25 mol}$ of $\ce{Fe}$ and $\pu{0.94 mol}$ of $\ce{O2}$ - for the sake of simplicity we'll say it's $\pu{1.88 mol}$ of $\ce{O}$ (I multiplied by 2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just take this into a calculator! $1.88/1.25$ is approximately $1.5$, so the correct answer would be $\ce{Fe2O3}$. The reason we divide large by small is that it's easier to see whole number ratios from the values that are larger than one - $0.66$ being converted to $2/3$ may not be as obvious as $1.5$ being converted to $3/2$, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did we divide? so we've calculated the relative ratio of the amounts of substances in this compound - this means for any compound of any mass, we'll always get the ratio of $3/2$ of $\ce{O/Fe}$, right? Which means the empirical formula is $\ce{Fe2O3}$.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="4945" LastEditDate="2018-02-20T10:57:05.153" LastActivityDate="2018-02-20T10:57:05.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5590" PostTypeId="2" ParentId="5585" CreationDate="2013-07-15T12:37:45.370" Score="5" Body="&lt;p&gt;Electrophilic aromatic substitution, of which nitration is one specific example, is often a kinetically controlled process, i.e., the distribution of products is determined by the energies of the transition states and intermediates of the respective reaction pathways available. Under conditions where thermodynamic equilibrium cannot be established, the predominant product will be the one with the lowest activation energy for the rate-determining step, and more stable transition states and intermediates imply lower energy of activation. This is also the product that forms most rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, even under thermodynamic conditions where equilibrium can be established, it's frequently the case that the major pathway will involve the initial kinetic product in equilibrium with the thermodynamic product. If I remember correctly, this is the case with, e.g., the isomerization that occurs with certain Friedel-Crafts polyalkylations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, in order to answer your question, you need to examine the transition states of the nitration process for each of your reactants. If you consider the mechanism shown on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution#Ortho.2Fpara_directors&quot;&gt;Wikipedia page for electrophilic aromatic substitutions&lt;/a&gt;, you'll see that the cationic intermediate of the reaction is additionally stabilized when electron-donating groups are present. In your specific question, molecules I and II are both very strongly electron-donating by means of $\pi$-conjugation, which is nicely illustrated by the resonance structures one could draw for those molecules. Molecule IV is stabilized by a similar effect, although it's slightly less efficient due to the lone pair on the nitrogen also being conjugated to the electron-withdrawing carbonyl group. Molecule V is very weakly donating, but by different means. In the case of that molecule, it donates electrons by induction, and by hyperconjugation of the vacant $p$ orbital with a $\sigma$ carbon-hydrogen bond. Because the energies of $p$ orbitals and bonding $\sigma$ molecular orbitals are vastly different, and because the orbital overlap is incomplete and sensitive to conformation, the stabilization provided by the methyl substituent is comparatively minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I recently became aware of another, somewhat orthogonal explanation for the reactivity of substituted aromatics, namely the frontier orbital theory approach. Fleming, in his excellent book on molecular orbital theory, explains (by analogy with the benzyl anion) that conjugation of the benzene ring with an electron-donating substituent disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals, and yields a non-bonding HOMO with nodes at alternating atoms. The locations of those nodes results in relatively large coefficients on the atoms in the ortho and para positions, and very low (idealized to zero, but experimentally, of course, non-zero) coefficients on the atoms in the meta positions. I suppose this can serve as a complementary theoretical justification for the observed kinetics and perhaps regioselectivity of these types of reactions.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-15T12:37:45.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5591" PostTypeId="1" CreationDate="2013-07-15T14:39:55.637" Score="2" ViewCount="101" Body="&lt;p&gt;My smartphone gets over heated where I use mobile data for long period. Last day I found a cooler of an old cpu, the thing silver in colour to absorb heat(with eventually gives out heat as it has more surface area on the other side). I used to keep my phone above it and it gets cooled. So my question is whether it will cause any problem for the battery if I tries to cool it using the same? I am asking this because there are a lot of thermodynamics involved in batteries and would it cause any problem to my battery life?&lt;/p&gt;&#xA;" OwnerUserId="1978" LastActivityDate="2013-07-15T15:32:23.380" Title="Would cooling of a battery using cooler affect its life time?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5592" PostTypeId="2" ParentId="5591" CreationDate="2013-07-15T15:32:23.380" Score="2" Body="&lt;p&gt;Let's see what I know. Not really sure whether this is a terrific answer to your question, but it's what I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through knowledge of using laptops, I know that lithium-ion batteries will decrease in performance when overheated. This probably applies to most redox reaction batteries for the reason that the reaction proceeds more quickly when temperature is increased, and as a consequence, kinetic energy of the molecules are increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the temperature of the battery also increases internal resistivity, which is somewhat more of a physics question. The electrons collide with the material that they are flowing through much more frequently in a higher temperature situation as opposed to a lower temperature situation. This doesn't mean, however, that you can dip your phone in LN2 and have it work forever. There's obvious physical limitations to everything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're doing to your phone (a.k.a. cooling it) shouldn't have a problem at all on your phone's performance, mainly because you're doing to it what others do to their laptops - placing it on a cooling pad.&lt;/p&gt;&#xA;" OwnerUserId="767" LastActivityDate="2013-07-15T15:32:23.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5594" PostTypeId="1" AcceptedAnswerId="5595" CreationDate="2013-07-15T18:32:22.333" Score="-3" ViewCount="405" Body="&lt;p&gt;$\ce{1-octene -&amp;gt; octan-1-ol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction occurs in the presence of $\ce{ROOR}$ and $\ce{HBr}$.&#xA;There is no information about temperature.&lt;/p&gt;&#xA;" OwnerUserId="1979" LastEditorUserId="22" LastEditDate="2013-07-16T05:24:58.440" LastActivityDate="2013-07-16T05:46:56.287" Title="Can anyone tell me the mechanism behind this reaction?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-07-16T16:14:18.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="5595" PostTypeId="2" ParentId="5594" CreationDate="2013-07-15T18:54:30.770" Score="1" Body="&lt;p&gt;I have only trivial knowledge of organic chemistry, but here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming ROOR represents a peroxide, we have homolytic cleavage of ROOR to form $\ce RO^{.} and ^{.}OR$. Afterwards, these radicals interact with HBr to produce ROH and a Br radical. The ROH then is protonated on the alcohol group to make a better leaving group, and the $\ce {H2O}$ is cleaved, leaving a radical R group, which is then attacked by a Br radical. Water is then added to the alkene oct-1-ene to form octan-1-ol.&lt;/p&gt;&#xA;" OwnerUserId="767" LastActivityDate="2013-07-15T18:54:30.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5597" PostTypeId="1" AcceptedAnswerId="5598" CreationDate="2013-07-15T23:04:06.903" Score="23" ViewCount="8067" Body="&lt;p&gt;I came across a naming convention which I haven't seen before. I let ChemDraw name the following compound for me and got a name containing lowercase &quot;r&quot; and &quot;s&quot; configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1qdUu.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me what this means and something about the actual convention behind it? I think it must have something to do with rings and maybe with some kind of &quot;pseudo-chirality&quot; due to a certain conformation of the ethyl group which is attached to the ring, but I'm only guessing.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="544" LastEditDate="2018-09-13T15:10:10.230" LastActivityDate="2018-09-13T15:10:10.230" Title="What does lowercase r-s notation mean?" Tags="&lt;nomenclature&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5598" PostTypeId="2" ParentId="5597" CreationDate="2013-07-16T01:08:23.043" Score="13" Body="&lt;p&gt;This notation is used to designate so-called &quot;pseudoasymmetric carbons.&quot; This occurs in instances in which precisely two of the groups bonded to a tetrahedral carbon are structurally indistinguishable in terms of connectivity (i.e., the same atoms, bonded with the same multiplicity, and in the same order moving outward from the carbon whose absolute configuration is being assigned), but contain chiral centers which have opposite configurations. See the &lt;a href=&quot;http://goldbook.iupac.org/P04921.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC Gold Book entry on pseudo-asymmetric carbons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the molecule in your example, it is named incorrectly. There is an inherent symmetry present in 1,4-disubstituted cyclohexanes that prevents them from being chiral (except in cases where the substituents on the ring themselves contain chiral centers, which methyl and ethyl groups do not), and without any chiral centers in the molecule, there obviously can be no pseudoasymmetric carbons either. I would add, however, that &lt;em&gt;cis/trans&lt;/em&gt; isomerism is, of course, possible for 1,4-disubstituted cyclohexanes, but that's entirely different from the property of chirality. The only stereochemical designator that could reasonably be added to the molecule from your example is &quot;&lt;em&gt;cis-&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="7448" LastEditDate="2016-08-28T16:17:22.290" LastActivityDate="2016-08-28T16:17:22.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5599" PostTypeId="1" AcceptedAnswerId="5600" CreationDate="2013-07-16T05:57:16.463" Score="11" ViewCount="9409" Body="&lt;p&gt;Unit cells are divided into two main types&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Primitive&lt;/li&gt;&#xA;&lt;li&gt;Non-primitive&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Primitive includes simple cubic lattice whereas non-primitive includes fcc bcc end centered.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the seven types of crystal systems orthorhomic crystal system posses both primitive as well as non-primitive unit cells i.e. face centred body centred rnd centred and simple cubic but the cubic system posses primitive , body centred and face centred arrangements only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if face centred is possible in cubic system why is not end centred type of unit cell arrangement possible in cubic system if there is no sort of hindrance in vacant space as in orthorhombic which posses all types of unit cell arrangements.&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorUserId="22" LastEditDate="2013-07-16T08:34:48.513" LastActivityDate="2014-06-29T13:06:42.783" Title="Confusion in unit cells of crystal system" Tags="&lt;solid-state-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5600" PostTypeId="2" ParentId="5599" CreationDate="2013-07-16T12:40:13.027" Score="9" Body="&lt;p&gt;First, a note: the classification of crystal systems, or the reduction of possible lattice types into primitive Bravais lattices, has nothing to do with “space available” or “hindrance”. It is purely a mathematical property having to do with the symmetry of the lattices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the &lt;strong&gt;base-centered cubic lattice is not a Bravais lattice&lt;/strong&gt;, because it is &lt;strong&gt;&lt;a href=&quot;http://wings.buffalo.edu/eng/mae/courses/mae381/chapter3/3_03.pdf&quot; rel=&quot;noreferrer&quot;&gt;equivalent to a simple tetragonal lattice&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;      &lt;img src=&quot;https://i.stack.imgur.com/WdHnU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-07-16T12:40:13.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5601" PostTypeId="1" CreationDate="2013-07-16T18:17:36.103" Score="6" ViewCount="246" Body="&lt;p&gt;I'm trying to remove titanium filling from a tantalum crucible for a high-temperature effusion cell.  I've managed to get a majority of the material out with tweezers, but some remains and is difficult to remove.  I understand that sulfuric acid will corrode titanium, but not tantalum, and was considering this as a method for extraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has some advice on a method to (safely) remove $\ce{Ti}$ from the $\ce{Ta}$ crucible without destroying it, I would very much appreciate it.  I am certainly not a chemist and could use some help.&lt;/p&gt;&#xA;" OwnerUserId="1690" LastEditorUserId="4945" LastEditDate="2014-07-30T04:50:43.333" LastActivityDate="2017-04-08T21:58:47.620" Title="Removing titanium from tantalum crucible" Tags="&lt;experimental-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;purification&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5602" PostTypeId="1" CreationDate="2013-07-17T04:53:46.977" Score="7" ViewCount="1526" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How would you compare the acidity of the most acidic proton in each of the following molecules?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;diphenylmethane &lt;/li&gt;&#xA;  &lt;li&gt;toluene&lt;/li&gt;&#xA;  &lt;li&gt;triphenylmethane &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer says 3 is the most acidic, followed by 1, and finally 2 is the least acidic. Can someone please tell me how to determine this? I know the benzene group is electron donating group, but am not quite sure how to correlate this knowledge with determining the relative acidities.&lt;/p&gt;&#xA;" OwnerUserId="1986" LastEditorUserId="5026" LastEditDate="2018-03-02T12:07:16.843" LastActivityDate="2018-03-02T12:07:16.843" Title="Which of mono-, di-, triphenylmethane is the most acidic?" Tags="&lt;acid-base&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5604" PostTypeId="2" ParentId="5602" CreationDate="2013-07-17T06:10:53.980" Score="9" Body="&lt;p&gt;You should consider the stability of carbanion that is left behind, after removing $\ce{H+}$.   One could make arguments for resonance stabilization of each of the carbanions. You could &quot;push&quot; the minus charge around the phenyl rings, and there is a general rule that the more &quot;good&quot; resonance structures that one can draw, the more stable the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the tertiary carbanion in 2) is flanked by three phenyl groups (thus many good resonance structures compared to 1) and 3) ).&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2013-07-18T01:53:42.357" LastActivityDate="2013-07-18T01:53:42.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5605" PostTypeId="1" CreationDate="2013-07-17T12:30:13.000" Score="1" ViewCount="101" Body="&lt;p&gt;I'm currently building a toy chemistry model based around Gillespie's algorithm. Starting from a food set of {H,O,N,C} I let my system evolve, I have been storing my molecules as strings in SMILES format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[O] reacts with [H] --&gt; [O] ([H])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[O] ([H]) reacts with [H] --&gt; [O] ([H]) ([H])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using square brackets I've been able to keep the molecules in my system consistent with their SMILES interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this system of storage is only allowing me to construct linear molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I create, for instance, a carbon ring, so that it is interpreted in SMILES as only a ring of carbon atoms, without hydrogen atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance C1CCCCC1 gives &lt;a href=&quot;http://www.chemicalize.org/structure/#!mol=C1CCCCC1&amp;amp;source=calculate&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalize.org/structure/#!mol=C1CCCCC1&amp;amp;source=calculate&lt;/a&gt;&#xA;How would I generate 6 carbon atoms in a ring without the 12 hydrogen?&lt;/p&gt;&#xA;" OwnerUserId="1990" LastEditorUserId="4945" LastEditDate="2015-09-26T20:26:18.943" LastActivityDate="2015-09-26T20:26:18.943" Title="How to store molecule data using SMILES when building a toy chemistry model" Tags="&lt;cheminformatics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5607" PostTypeId="2" ParentId="5495" CreationDate="2013-07-17T15:41:46.787" Score="9" Body="&lt;p&gt;If your mixture reached 160 degrees C then you must have boiled off all the water. If there was some water left, it would stay at around 100 degrees C and all (most of) the heat would be used to evaporate water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiling point of glycerol is about 290 degrees C and so you didn't boil that off. I disagree with @Skeptical Chymist about you making the caramel (for the sake of god, don't eat it to test it out). Not all that gets brown is caramel and there were no sugars there I presume. If i had to guess I'd say that at 160 C you've managed to completely obliterate any complicated organic structures and what's left is glycerol, charcoaled bits of any organic impurities that you happened to have in the starting material and possibly some &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrolein&quot;&gt;acrolein&lt;/a&gt; which is a common product of dehydration of glycerol (you can recognize it by it's smell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall; it's hard to tell what one gets, when they heat the hell out of something.&lt;/p&gt;&#xA;" OwnerUserId="1915" LastActivityDate="2013-07-17T15:41:46.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5608" PostTypeId="2" ParentId="5386" CreationDate="2013-07-17T15:48:30.593" Score="1" Body="&lt;p&gt;Yes. I don't have much time to go into details (sorry) but temperature changes the pKa of water which in turn affects the concentrations of [$\ce{H3O+}$] and [$\ce{OH-}$] and that affects both reduction and oxidation potential of water. The effect it negligible for most intentions and purposes over the 100 C of liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Pourbaix_diagram&quot; rel=&quot;nofollow&quot;&gt;Pourbaix diagrams&lt;/a&gt; on some of them you will see the dashed lines. Those indicate the reduction and oxidation potentials of water as they change  with the pH.&lt;/p&gt;&#xA;" OwnerUserId="1915" LastEditorUserId="1878" LastEditDate="2013-07-17T17:38:15.713" LastActivityDate="2013-07-17T17:38:15.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5609" PostTypeId="2" ParentId="5386" CreationDate="2013-07-17T18:40:43.070" Score="3" Body="&lt;p&gt;I see two questions here.  The one that is stated (temperature effects on ORP) and a hidden one about water ionizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the temperature effect on ORP.  The thermodynamics of oxidation and reduction reactions is governed by the Nernst equation:&#xA;$$E(ORP) = E_o - \frac{2.3RT}{nF}\times log\frac{[reductant] }{[oxidant]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation, R and F are the gas law and Faraday's constants, respectively, $E_o$ is the standard reduction potential of the system of interest, T is temperature, n is the number of electrons transferred in the reaction and the concentrations of the reductant and oxidant are found in the log term.  The coefficient $\frac{2.3RT}{nF}$ in front of the log term can be simplified to $\frac{0.00198\times T}{n}$ and when we perform dimensional analysis on the term we find that it will provide units in Volts.  If we are looking for the &lt;strong&gt;largest&lt;/strong&gt; effect that temperature would have on ORP then we want to use the &lt;strong&gt;smallest&lt;/strong&gt; value of n, which would be 1 electron, further simplifying the coefficient to approximately $0.002T$.  T has to be in Kelvin so if we take our two extreme values as the freezing (273 K) and boiling (373 K) points of water, we find that the difference in ORP between these two extremes (all else being the same) &lt;em&gt;would be about 200 millivolts&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the hidden question about water ionizers.  There's a lot of hype about these systems with little scientific evidence to back up the claims.  There is at least one website useful for &lt;a href=&quot;http://www.naturalchoice.net/articles/Art12_WaterIonization.htm&quot; rel=&quot;nofollow&quot;&gt;exploring the science behind water ionizers&lt;/a&gt; that is (I think) accessible to the general public.  A number of claims made by the manufacturers of these products, such as ion streams and water clusters simply don't stand up to scientific experimentation, and consumers need to be wary of the health benefit claims.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-17T18:40:43.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5610" PostTypeId="1" AcceptedAnswerId="5704" CreationDate="2013-07-17T20:24:53.560" Score="3" ViewCount="223" Body="&lt;p&gt;I came across the following relation in a book (written as a fact, with no proof), for a mixture of ideal gases both at the same pressure and temperature.  The final volume of the mixture of gases after mixing is:&#xA;$$ V_\text{mix} = \frac {M_1V_1 + M_2V_2}{M_1+M_2} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M_1$, $M_2$ are the molar masses of the individual gases and $V_1$, $V_2$ are the respective volumes.  I could not figure out its proof.  Is this relation valid at all?&#xA;Please do &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; confuse with a similar looking relation,with $ M $ corresponding to molarity, in case of a &lt;strong&gt;&lt;em&gt;solution&lt;/em&gt;&lt;/strong&gt; mixture.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorDisplayName="user7951" LastEditDate="2015-10-02T17:53:40.997" LastActivityDate="2015-10-02T17:53:40.997" Title="Is this relation in case of a mixture of ideal gases valid?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5611" PostTypeId="1" CreationDate="2013-07-17T20:32:40.287" Score="4" ViewCount="279" Body="&lt;p&gt;I have had a question that has been bugging me for a couple of years in my graduate school career, and I was wondering if perhaps someone could help explain it to me. Given an endpoint assay like this one I've shown below (an ELISA assay, but I assume it could work for any endpoint), how does the endpoint signal (the signal at the highest concentration) relate to binding? And how does $EC_{50}$ (the concentration between the maximum and minimum signal) relate to the equilibrium constant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/92m4Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For years, biologists have used the $EC_{50}$ to give a rough approximation for the equilibrium binding constant $K_d$. However, it always bothered me that all curves features on the graph could produce nearly the same $EC_{50}$, but with different levels of maximum absorbance. For instance two of the curves are saturated at around 1.2 and the others are around 2.5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do all of these curves have roughly the same $K_d$, or is it worrisome that we don't get equivalence signal for the maximum concentrations?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="73" LastEditDate="2013-07-17T21:17:05.333" LastActivityDate="2015-07-21T07:25:02.740" Title="What does an endpoint saturated signal mean. How does it relate to equilibrium constant?" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5613" PostTypeId="1" CreationDate="2013-07-18T05:30:39.210" Score="4" ViewCount="3362" Body="&lt;p&gt;I'm not chemist, and I'm looking for easiest way to measure how much oxygen plant in closed container will produce per hour [can measure over time for increased accuracy].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea was to measure percentage of oxygen in air with automotive lambda sensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't expect good precision of measurement like this, and this is a little complicated (lambda sensor needs high temperature, so I need to heat it up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there is some chemical way to perform measurement like this.&lt;/p&gt;&#xA;" OwnerUserId="1960" LastEditorUserId="57118" LastEditDate="2018-01-05T20:46:32.273" LastActivityDate="2018-01-05T20:46:32.273" Title="How to measure amount of oxygen (or carbon dioxide) in air?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5615" PostTypeId="1" AcceptedAnswerId="5636" CreationDate="2013-07-18T08:48:38.477" Score="5" ViewCount="2475" Body="&lt;p&gt;I'm experimenting with synthesis in microreactors and I'm using 2 reactants. I'd like to estimate the Reynolds number of the flow containing two miscible fluids. I have the values of their properties (density, viscosity, flowrate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I just average both property values for both liquids or is there a more intricate relation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance for your solutions and suggestions!&lt;/p&gt;&#xA;" OwnerUserId="1997" LastActivityDate="2013-07-20T03:48:12.843" Title="How to determine a Reynolds number for mixture of 2 liquids" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5616" PostTypeId="1" CreationDate="2013-07-18T12:52:09.700" Score="4" ViewCount="489" Body="&lt;p&gt;I am currently investigating the assembly of gold nanoparticles in a mixture of water and oil. The thing is the following: If we have gold nanoparticles (35nm diameter) dispersed in water and we put some oil on it. After that we shake it, then the gold will be kept between the layer between water and oil and now I was wondering why this happens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is it not possible that the gold nanoparticles stay solely in the water or go over in the oil completely? Or it could be possible that the gold particles agglomerate, why does this not happen? Or maybe it does happen and I do not know this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does somebody have some background information on this topic?&lt;/p&gt;&#xA;" OwnerUserId="1998" LastActivityDate="2013-09-17T18:13:09.887" Title="Gold nanoparticles in a water/oil environment" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;&lt;intermolecular-forces&gt;&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5617" PostTypeId="1" CreationDate="2013-07-18T13:31:15.390" Score="0" ViewCount="1559" Body="&lt;p&gt;Say you are given a set of particular elements, such as Dy, C, I, Er, P. Is it possible to determine whether a compound exists with all of the given elements and no more? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is it possible to find this out if given the number of a particular element, such as; two hydrogen and one oxygen where you could come to the conclusion that $\ce{H2O}$ is possible. &lt;/p&gt;&#xA;" OwnerUserId="1999" LastEditorUserId="10185" LastEditDate="2015-09-28T10:29:15.110" LastActivityDate="2015-09-28T10:29:15.110" Title="Can you determine whether a compound can form given particular elements?" Tags="&lt;molecules&gt;&lt;theoretical-chemistry&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5619" PostTypeId="1" AcceptedAnswerId="5627" CreationDate="2013-07-18T15:37:53.670" Score="1" ViewCount="151" Body="&lt;p&gt;When a battery is subject to overcharging or overdischarging (including attempts to recharge a primary cell), it may vent hydrogen.  It seems that every type of electrochemical battery I've read about or used &lt;em&gt;specifically emits hydrogen&lt;/em&gt; under these circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is hydrogen emitted in all cases, and not some other gas or chemical? What role does hydrogen play in electrochemical batteries that other chemicals can't?&lt;/p&gt;&#xA;" OwnerUserId="888" LastActivityDate="2013-07-19T14:15:06.803" Title="Why do batteries specifically vent *hydrogen* in the event of abuse?" Tags="&lt;electrochemistry&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5620" PostTypeId="2" ParentId="5617" CreationDate="2013-07-18T21:25:13.510" Score="2" Body="&lt;p&gt;Answer on first question is no, on second is yes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider this short list &lt;code&gt;C&lt;/code&gt;, &lt;code&gt;H&lt;/code&gt;, &lt;code&gt;O&lt;/code&gt;, &lt;code&gt;N&lt;/code&gt; this elements can produce unlimited number of compounds. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have short formula like C6H6 and you assuming only covalent bonds in molecule, define valence 4 for Carbon and 1 for Hydrogen you can make mathematical calculation according to graph theory and you will get 217 structures. But most of them doesn't exist due to bonds strain and intersection. If you want to obtain existing molecules you can filter this 217 models using for instance quantum mechanics calculation.     &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1985" LastEditorUserId="1985" LastEditDate="2013-07-18T22:40:31.157" LastActivityDate="2013-07-18T22:40:31.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5621" PostTypeId="1" AcceptedAnswerId="5623" CreationDate="2013-07-19T01:46:29.770" Score="4" ViewCount="12933" Body="&lt;p&gt;I'm trying to figure out the patterns for Ionization Energies. I am familiar with the periodic trend, however things become quite different when we hit the 1st I.E. For example, Na has an I.E(1) of 495.8 kJ while its second I.E. rockets up to 4562 kJ while the atoms towards the right are much lower than this. The trend says that the I.E. increase up and right of the periodic table, which is not the case here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is, in order to get a better estimate, would it be safe to say that the Zeff charge and atom size relate to the I.E? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example: Effective Charges&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg = 2&#xA;Al = 1&#xA;S = 4&#xA;Si = 2&#xA;Na = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the strongest pull towards the charged center would be &lt;strong&gt;Al&lt;/strong&gt; and &lt;strong&gt;Na&lt;/strong&gt; in this case. However, having Al being the smaller atom would require more energy to remove the electron from its valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the case of first Ionization Energy we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg = 3&#xA;Al = 2&#xA;S = 5&#xA;Si = 3&#xA;Na = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;strong&gt;Na&lt;/strong&gt; now has reduced its size due to the fact it jumped from n=3 to n=2 level and &lt;strong&gt;Al&lt;/strong&gt; also reduced in size but is still a bigger atom than &lt;strong&gt;Na&lt;/strong&gt; due to our trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is this approach fairly accurate or should I be looking somewhere else?&lt;/p&gt;&#xA;" OwnerUserId="2002" LastActivityDate="2013-07-19T07:23:35.887" Title="Z* effective charge and Ionization Energy" Tags="&lt;atoms&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5622" PostTypeId="2" ParentId="5613" CreationDate="2013-07-19T05:27:39.207" Score="2" Body="&lt;p&gt;This is untested, but I think it might work. You could light some fuel on fire inside the closed container and see how much of it burns (just put the cup of fuel on a kitchen scale) before it dies out and than back calculate how much oxygen was present. Preferably you should take a fuel that burns easily to make sure that the total combustion reaction is the only reaction that occurs. Also if you want to repeat the experiment every hour or so, you want to be sure that the fuel does not evaporate (substantially) because that might cause the entire container to light up when you try to burn the fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not really sure which 'household' fuel would be a good option, but maybe one of the other people on this site have suggestions?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably use one of those candle lighters and stick the long side through a small hole in your container and than seal the rest of the hole to be able to light a fire inside, without having air flowing in or out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just be careful not to set the plant (or anything other than the fuel) on fire!&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-07-19T05:27:39.207" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5623" PostTypeId="2" ParentId="5621" CreationDate="2013-07-19T07:15:38.773" Score="4" Body="&lt;p&gt;&lt;strong&gt;Using the common idiom: &quot;full subshells are stable&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is a little more compact to use the incorrect explanation and correct it than to explain in terms of correct. Also, when you hear this incorrect explanation, you will understand what is meant.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, size is a factor (thus the &quot;up&quot; part- IE is higher in lower periods), but the observed anomaly in second-ionization energies in $\ce{Na}$ and $\ce{Al}$ is better related to removal of an electron from a full subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na}$ has an electron configuration of $\ce{[Ne] 3s^1}.$ The first ionization gives $\ce{Na}$ makes it $\ce{[Ne]}$, the same as a noble gas- full 2p subshell. Because full subshells are stable, it takes a lot of energy to remove the first electron from it. $\ce{Al}$ is analogous; the second ionization means removing an electron from a full subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is observed in the first ionization energies (in $kJ/mol$):&#xA;$$\ce{Na}:495$$$$\ce{Mg}:737$$$$\ce{Al}:577$$&#xA;Ionization of $\ce{Mg}$ removes an electron from a full $\ce{3s}$ subshell, so it is a little higher than the trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many factors that contribute to exact values, but these are difficult to predict, which is why they taught a simple version in your class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correcting the myth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is common to refer to full subshells as being more stable, it isn't really what is going on. In reality, the next electron added in destabilized. This occurs because of &lt;em&gt;shielding.&lt;/em&gt; Outer electrons feel repulsion from inner electrons, thus outer electrons are easier to remove than inner ones. The saying that &lt;em&gt;&quot;half-filled subshells are stable&quot;&lt;/em&gt; is also used. Again, it is that the next added electron is destabilized, but in this case it is because the energy needed to &lt;em&gt;spin pair&lt;/em&gt; the two electrons.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-07-19T07:15:38.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5624" PostTypeId="2" ParentId="5621" CreationDate="2013-07-19T07:23:35.887" Score="3" Body="&lt;p&gt;$Z_{eff}$ is certainly a major factor in determining ionization energies, however atomic and ionic radius probably shouldn't be viewed as having a direct causative relationship to ionization energy. It's more correct to say that ionization energy and atomic/ionic radius have some of the same underpinnings (namely, $Z_{eff}$ and various electron-electron interactions). Because many of the same underlying forces are at work, there is some correlation (i.e., ionization energy increases in the same direction as atomic radius decreases, certain specific anomalies aside), but don't confuse that correlation with causation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one moves down a group on the periodic table, $Z_{eff}$ obviously remains constant, however the number of core electrons shielding the nucleus increases, and consequently the valence level electrons become increasingly energetic as their distance from the nucleus grows. This fact largely accounts for the increase in atomic radius and the decrease in ionization energy that occurs as one moves down along any given group on the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, as one moves right along a period, the number of core electrons remains constant, while the $Z_{eff}$ increases. It's reasonable to expect, therefore, that valence electrons will experience a stronger electrostatic attraction to the nucleus as one moves right along a period, causing both an increase in ionization energy and a decrease in atomic radius. The ionization energy trend mostly conforms to that expectation, with the notable exceptions of transitioning from group IIA to group IIIA, and from group VA to group VIA, where ionization energy (perhaps unexpectedly) drops. To explain these exceptions, compare the electron configurations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When removing the first group IIIA valence electron, it is being removed from a $p$ orbital, while the first group IIA valence electron would be removed from an $s$ orbital. Electrons in $p$ orbitals are somewhat more energetic due to the nuclear charge being partially shielded by the electrons of the preceding $s$ orbital (in addition to more complex quantum mechanical effects), hence they are easier to remove.&lt;/li&gt;&#xA;&lt;li&gt;The first group VIA electron to be removed is paired with another electron in the same $p$ orbital, while all $p$ orbital electrons of group VA elements are unpaired (in accordance with Hund's rule). Electron pairing causes some mutual electron-electron repulsion, making these electrons more energetic, resulting in a drop in ionization energy for group VI by comparison to group V.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you move further down the periodic table, the contributions of electrons in $d$ and $f$ orbitals become significant, the energy gaps between subsequent principal energy levels get narrower, and the ionization energy trend becomes more strictly linear for main group elements (specifically, the exceptions I described above no longer apply once you reach principal energy level five).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-19T07:23:35.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5625" PostTypeId="1" AcceptedAnswerId="5637" CreationDate="2013-07-19T09:31:32.203" Score="3" ViewCount="2678" Body="&lt;p&gt;I'm not a chemist, but I need some insight into chemical thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book that I'm reading &lt;a href=&quot;http://www.amazon.co.uk/Physical-Chemical-Equilibrium-Engineers/dp/0470927100/ref=sr_1_1?ie=UTF8&amp;amp;qid=1374225908&amp;amp;sr=8-1&amp;amp;keywords=physical%20and%20chemical%20equilibrium%20for%20chemical%20engineers&quot; rel=&quot;nofollow&quot;&gt;(Nevers, 2012)&lt;/a&gt;, equation (4.28) relates the chemical potential of a &lt;strong&gt;pure species&lt;/strong&gt; to it's Gibbs energy per mol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\qquad\mu_a^{(1)} = \bigg( \dfrac{\partial G}{\partial n_a} \bigg)_{T,P,n_b...} = g_a^{(1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the subscript stands for the species $a$, the superscript stands for it's phase, $\mu$ is the chemical potential and $g$ is the Gibbs energy per mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the first equal sign. Though, I don't understand the second equal sign. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="2004" LastEditorUserId="9961" LastEditDate="2015-10-23T17:21:57.447" LastActivityDate="2015-10-23T17:21:57.447" Title="Why are partial molar Gibbs energy and Gibbs energy per mol equal?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5626" PostTypeId="1" AcceptedAnswerId="5628" CreationDate="2013-07-19T11:31:59.230" Score="2" ViewCount="203" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfqyS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{10}~\mathrm{pg}/\mathrm{ml}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand the scale.  How do I convert this to a standard concentration (without the logs)?&lt;/p&gt;&#xA;" OwnerUserId="2005" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T18:02:17.467" LastActivityDate="2015-10-23T18:02:17.467" Title="What does this scale mean?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;&lt;units&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5627" PostTypeId="2" ParentId="5619" CreationDate="2013-07-19T12:44:14.473" Score="4" Body="&lt;p&gt;Lead storage batteries use an electrolyte of sulfuric acid and water. On overcharging you are electrolysing the water with emission of $\ce{H2}$ in one electrode and $\ce{O2}$ in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other batteries use molecules with H and Li. The idea is to use light elements so you can accumulate more charge for the weight.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="1300" LastEditDate="2013-07-19T14:15:06.803" LastActivityDate="2013-07-19T14:15:06.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5628" PostTypeId="2" ParentId="5626" CreationDate="2013-07-19T14:12:06.120" Score="3" Body="&lt;p&gt;To expand upon the comments, the y axis in this graph is simply the log of the concentration in pg/ml.  To convert you use the equation $$C = 10^{y}$$ where y is the y value and C is the concentration.  Note that in older literature, log was oftentimes assumed to be base e whereas nowadays we assume base 10.  The figure (thankfully) clarifies this potential discrepancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final note: log values are &lt;em&gt;unitless&lt;/em&gt; so the y axis has no units.  The units are reported here so that you know what the appropriate units of the converted value should be.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-19T14:12:06.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5629" PostTypeId="1" AcceptedAnswerId="5633" CreationDate="2013-07-19T15:03:19.010" Score="3" ViewCount="6078" Body="&lt;p&gt;I found &lt;a href=&quot;http://www.youtube.com/watch?v=QzjPva_MUAg&quot; rel=&quot;nofollow&quot;&gt;this reaction&lt;/a&gt; of $\ce{NH4Cr2O7 + Hg(SCN)2 -&amp;gt;[\Delta]\ ?}$ on youtube, and I'm curious as to how it works. I tried searching online, but didn't find much on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1959" LastEditorUserId="418" LastEditDate="2014-03-20T14:26:56.043" LastActivityDate="2014-03-20T14:26:56.043" Title="What happens in the reaction of ammonium dichromate and mercury(II) thiocyanate?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5630" PostTypeId="2" ParentId="5616" CreationDate="2013-07-19T15:31:07.237" Score="1" Body="&lt;p&gt;I believe this would depend more on the capping agent (citrate in Turkevich method, TOAB in Brust method etc. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Colloidal_gold&quot; rel=&quot;nofollow&quot;&gt;Colloidal gold&lt;/a&gt;) capping agents are amphiphilic and will therefore concentrate at the oil-water interface.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-07-19T15:31:07.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5631" PostTypeId="1" AcceptedAnswerId="5635" CreationDate="2013-07-19T15:43:35.697" Score="5" ViewCount="758" Body="&lt;p&gt;The documentation for the indigo module can be found here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ggasoftware.com/opensource/indigo/api#inputoutput&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ggasoftware.com/opensource/indigo/api#inputoutput&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for instance if I have a molecule object for some SMILES string, e.g. &lt;code&gt;&quot;[C](=[O])&quot;&lt;/code&gt;, I wish to calculate the valency of each atom, for instance here the desired output would be &lt;code&gt;[atom=C, unbound_electrons=2],[atom=O, valency=0]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I consider the atom &lt;code&gt;&quot;[C]&quot;&lt;/code&gt;&#xA;Can anyone explain why is this code printing &lt;code&gt;[atom=C, unbound_electrons=0]&lt;/code&gt; not &lt;code&gt;[atom=C, unbound_electrons=4]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre class=&quot;lang-python prettyprint-override&quot;&gt;&lt;code&gt;from indigo import *&#xA;indigo = Indigo()&#xA;&#xA;mol=indigo.loadMolecule(&quot;[C]&quot;)&#xA;&#xA;print(mol.grossFormula(),&quot;\n&quot;)&#xA;&#xA;for atom in mol.iterateAtoms():&#xA;        print([atom.symbol(),atom.radicalElectrons()])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;EDIT: I could work it out if I could generate a list of the types of bonds on the atom in conjuction with &lt;code&gt;atom.atomicNumber()&lt;/code&gt;. E.g. if I could say &lt;code&gt;[C]&lt;/code&gt; has a double bond I could take it's atomic number - 2 (second shell) - 2 (double bond)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT#2: This might be useful for visualising what i'm talking about&lt;/p&gt;&#xA;&#xA;&lt;pre class=&quot;lang-python prettyprint-override&quot;&gt;&lt;code&gt;from indigo_renderer import *&#xA;renderer = IndigoRenderer(indigo)&#xA;renderer.renderToFile(mol,&quot;mol.png&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;EDIT#3: I am not a chemist, so might have got some concepts wrong&lt;/p&gt;&#xA;" OwnerUserId="1990" LastEditorUserId="194" LastEditDate="2017-08-07T16:02:21.183" LastActivityDate="2017-08-07T16:02:21.183" Title="Counting valency of atoms, in a molecule with python" Tags="&lt;bond&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5632" PostTypeId="1" AcceptedAnswerId="5638" CreationDate="2013-07-19T17:30:43.117" Score="4" ViewCount="153" Body="&lt;p&gt;I've recently started undertaking a project to construct a small fume hood. So far I'm considering its use for soldering, 3d printing with ABS and polystyrene, and PCB etching using a muriatic acid/hydrogen peroxide solution, however I'd also like to leave my options open. I want to test the fume hood once completed - I've seen some videos that use dry ice for this purpose, and I figure I might also be able to test the hood by smell using bathroom deodorant, but I'm wondering what sort of concentration these would have to reach in order for me to detect them, and whether there are any other tests I might consider.&lt;/p&gt;&#xA;" OwnerUserId="1982" LastEditorUserId="1982" LastEditDate="2013-07-19T19:55:51.403" LastActivityDate="2013-07-20T01:42:33.250" Title="Safe, easily detectable gas or aerosol for testing fume hoods" Tags="&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5633" PostTypeId="2" ParentId="5629" CreationDate="2013-07-19T19:04:52.877" Score="5" Body="&lt;p&gt;This is a combination of two reactions.  The first is the &lt;a href=&quot;http://www.angelo.edu/faculty/kboudrea/demos/volcano/volcano.htm&quot; rel=&quot;noreferrer&quot;&gt;ammonium dichromate vocano&lt;/a&gt; and the second is called &lt;a href=&quot;http://chemistry.about.com/od/fireworksprojects/a/pharaohs-snakes.htm&quot; rel=&quot;noreferrer&quot;&gt;Pharaoh's snakes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The volcano&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reaction occurring here is: $$\ce{(NH4)2Cr2O7(s) -&amp;gt; Cr2O3(s) + N2(g) + 4 H2O(g)}$$  The reference above provides several pictures and a description of the chemistry.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Pharaoh's snakes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Again, the reference above describes what is happening, so only a summary of the chemical reactions is given here:&#xA;$$\ce{2 Hg(SCN)2-&amp;gt;2 HgS + CS2 + C3N4}$$ $$\ce{CS2 + 3O2 -&amp;gt; CO2 + 2SO2}$$ $$\ce{2C3N4-&amp;gt;3(CN)2 + N2}$$ $$\ce{HgS + O2-&amp;gt;Hg + SO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pharaoh's snakes is a rather hazardous demonstration and should not be performed without the proper ventilation, protective gear and waste disposal equipment.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-19T19:04:52.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5634" PostTypeId="1" AcceptedAnswerId="5642" CreationDate="2013-07-19T19:39:12.440" Score="7" ViewCount="6199" Body="&lt;p&gt;How can I figure out the degeneracy of the d orbitals for a site that has a given point group? Specifically I'm interested in $D_{3d}$ and $D_{3h}$, but it would be good to know how to do it in the general case.&lt;/p&gt;&#xA;" OwnerUserId="955" LastEditorUserId="4945" LastEditDate="2014-06-12T08:42:42.563" LastActivityDate="2014-06-12T08:42:42.563" Title="How do I determine the crystal field splitting for an arbitrary point group?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5635" PostTypeId="2" ParentId="5631" CreationDate="2013-07-19T19:46:00.827" Score="5" Body="&lt;p&gt;I was looking into this, and the results I was getting seemed very anomalous.  Consider this iteration from a single carbon atom at -4 through +4, this is the output I receive (python code at the bottom of the post):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[C+4]&#xA;C valence= 4 radicalE= 0 charge= 4 implicitH= 0&#xA;[C+3] |^1:0|&#xA;C valence= 4 radicalE= 1 charge= 3 implicitH= 0&#xA;[C+2] |^3:0|&#xA;C valence= 4 radicalE= 2 charge= 2 implicitH= 0&#xA;[C+]&#xA;C valence= 0 radicalE= 0 charge= 1 implicitH= 0&#xA;[C]&#xA;C valence= 0 radicalE= 0 charge= 0 implicitH= 0&#xA;[C-]&#xA;C valence= 0 radicalE= 0 charge= -1 implicitH= 0&#xA;[C-2] |^3:0|&#xA;C valence= 4 radicalE= 2 charge= -2 implicitH= 0&#xA;[C-3] |^1:0|&#xA;C valence= 4 radicalE= 1 charge= -3 implicitH= 0&#xA;[C-4]&#xA;C valence= 4 radicalE= 0 charge= -4 implicitH= 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I printed them using the canonicalSmiles format.  After I realized something strange was going on (neutral, and +/- 1 charges were 0, but everything else was fine), I dug into SMILES documentation to find out that currently it appears only &lt;strong&gt;up to&lt;/strong&gt; divalent radicals are supported.  Something like a lone carbon atom has technically a tetra-valent radical, so it defaults to nothing, because it cannot be interpreted by SMILES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pulled this from &lt;a href=&quot;http://www.chemaxon.com/marvin/help/formats/cxsmiles-doc.html&quot; rel=&quot;noreferrer&quot; title=&quot;ChemAxon&quot;&gt;ChemAxon&lt;/a&gt; which explains what the notation after the atom means. I'm going to assume that |^1:0| is the notation for a mono-valent system.  The 0 indicates the index of the atom, which is always zero for this example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Radical numbers&lt;/strong&gt;&lt;br&gt;&#xA;  Atom indexes with:&lt;br&gt;&#xA;  - divalent radical center are written after &quot;^2:&quot;&lt;br&gt;&#xA;  - divalent singlet radical center are written after &quot;^3:&quot;&lt;br&gt;&#xA;  - divalent triplet radical center are written after &quot;^4:&quot;&lt;br&gt;&#xA;  - trivalent radical center are written after &quot;^5:&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And from the &lt;a href=&quot;http://www.ggasoftware.com/opensource/indigo/concepts&quot; rel=&quot;noreferrer&quot; title=&quot;Indigo website&quot;&gt;Indigo website&lt;/a&gt; itself it states under the supported extensions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radical numbers: monovalent, divalent singlet, and divalent triplet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I don't think you've done anything wrong, you've just simply hit a limitation of the SMILES system and the interpreter of the indigo software.  Here is the code I used below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from indigo import *&#xA;indigo = Indigo()&#xA;&#xA;from indigo_renderer import *&#xA;&#xA;renderer = IndigoRenderer(indigo)&#xA;mols = []&#xA;mols.append(indigo.loadMolecule(&quot;[C+4]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C+3]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C+2]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C+1]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C-]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C-2]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C-3]&quot;))&#xA;mols.append(indigo.loadMolecule(&quot;[C-4]&quot;))&#xA;&#xA;i = 0&#xA;for mol in mols:&#xA;    print mol.canonicalSmiles()&#xA;    for atom in mol.iterateAtoms():&#xA;      print atom.symbol(),&quot;valence=&quot;,atom.valence(),&quot;radicalE=&quot;,atom.radicalElectrons(),&quot;charge=&quot;,atom.charge(),&quot;implicitH=&quot;,atom.countImplicitHydrogens()&#xA;    filename = 'mol'+str(i)+'.png'&#xA;    renderer.renderToFile(mol,filename)&#xA;    i += 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="446" LastActivityDate="2013-07-19T19:46:00.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5636" PostTypeId="2" ParentId="5615" CreationDate="2013-07-19T21:03:13.803" Score="3" Body="&lt;p&gt;As a very rough estimate, you could do that and just weight them by the fluids' respective concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how ideally the two fluids mix, you may want to look up or calculate the actual properties of the particular mixture you're working with. If you can't find the values in the literature but have access to physical property estimation software such as Aspen, you could calculate these values yourself.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastActivityDate="2013-07-19T21:03:13.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5637" PostTypeId="2" ParentId="5625" CreationDate="2013-07-19T21:22:15.257" Score="4" Body="&lt;p&gt;The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as&#xA;$$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$&#xA;where &lt;em&gt;N&lt;/em&gt; is the number of particles and &lt;em&gt;g&lt;/em&gt; is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then &lt;em&gt;G&lt;/em&gt; will also double. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above expression you can see that the chemical potential is just the Gibb's free energy per particle: &#xA;$$\mu\left(T,P\right) = g\left(T,P\right)$$&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-07-19T21:22:15.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5638" PostTypeId="2" ParentId="5632" CreationDate="2013-07-20T01:42:33.250" Score="4" Body="&lt;p&gt;Smoke generation is preferred over smell, as the visual test is a bit more objective. See &lt;a href=&quot;http://www.wcsu.edu/efs/hs_pdffiles/s113_lfhpt.pdf&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; for more info, although $\ce{TiCl4}$is not available for home use. There are hobbyist alternatives such as &lt;a href=&quot;http://www.amazon.com/Copperfield-41509-Smoke-Test-Matches/&quot; rel=&quot;nofollow&quot;&gt;smoke matches&lt;/a&gt; which will help you test the fume hood performance. (The above link is just an example, I did see better prices at less-known websites.)  An anemometer might be of help as well.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-20T01:42:33.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5639" PostTypeId="2" ParentId="5615" CreationDate="2013-07-20T02:23:04.560" Score="6" Body="&lt;p&gt;The Reynolds Number is the ratio of inertial to viscous forces and is calculated according to this equation: $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density, $\mu$ is the dynamic viscosity, $v$ is the fluid velocity and $L$ is a characteristic length that depends on the flow being modeled. When estimating the Reynolds number of a mixture, the term $v L$ would not change but we need to make the assumption that the term $\frac{\rho}{\mu}$ of the mixture can be derived from the properties of the pure components.  This assumption will be valid if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The components have similar structure&lt;/li&gt;&#xA;&lt;li&gt;The components are non-polar&lt;/li&gt;&#xA;&lt;li&gt;The components are not associated&lt;/li&gt;&#xA;&lt;li&gt;The difference in the kinematic viscosities is small (~15 cP)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some of these conditions could be relaxed if one species is in significant excess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Densities are additive so long as the volume of the mixture equals the sum of the volumes of the pure solutions.  (Search for volume weighted densities for more information.)  Viscosities of mixtures is a much harder thing to estimate.  One method is the &lt;strong&gt;Grunberg-Nissan&lt;/strong&gt; equation: $$Log\eta_m=x_1Log\eta_1+x_2Log\eta_2+x_1x_2d$$ Where $\eta_i$ is the dynamic viscosity of component $i$, $x_i$ is the mole fraction of component $i$ and $d$ is an interaction coefficient.  If you assume $d=0$ then you have the &lt;strong&gt;Arrhenius equation&lt;/strong&gt; for the viscosity of a mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at the end of the day, if you state you are working under the assumptions that &lt;em&gt;volume weighted densities equal mass weighted densities&lt;/em&gt; and use the &lt;em&gt;Arrhenius equation&lt;/em&gt; to determine the viscosity of the mixture then you can average the weighted properties of the pure components to get an estimate of the mixture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just a word of caution&lt;/strong&gt; under these assumptions you should take the &lt;strong&gt;mass&lt;/strong&gt; weighted average of the densities but the &lt;strong&gt;mole&lt;/strong&gt; weighted average of the viscosities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I relied on &lt;a href=&quot;http://link.springer.com/content/pdf/10.1007%2F978-1-4020-5482-2_5.pdf&quot;&gt;this reference&lt;/a&gt; for much of this answer.  It will require a subscription to the service to access the PDF, but some of it can be found on google books.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="1573" LastEditDate="2013-07-20T03:48:12.843" LastActivityDate="2013-07-20T03:48:12.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5641" PostTypeId="1" AcceptedAnswerId="5695" CreationDate="2013-07-20T20:38:06.777" Score="3" ViewCount="1325" Body="&lt;p&gt;A question was asked on another stackexchange site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/308/decay-of-a-corpse-in-space&quot;&gt;This is not nice perspective, but eventually it will happen. An astronaut falls out of spaceship because of damage caused by collision with other object, or because of suit decontamination. The fluids from the body would evaporate, and if any bacteria would survive, than only as spores. &lt;strong&gt;Does it mean the perfect mummification of the body&lt;/strong&gt;? Or there will be some decay, caused by enzymes from damaged cells, for example?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that the water in the body will outgass taking with it much of the rest of the body, are there any cited sources that specify the pressure at which proteins, fats and bones will break down and become vapour?&lt;/p&gt;&#xA;" OwnerDisplayName="user310" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.897" LastActivityDate="2013-07-26T06:35:41.127" Title="At what temperature will proteins and fats boil in a vacuum?" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5642" PostTypeId="2" ParentId="5634" CreationDate="2013-07-21T10:23:45.623" Score="8" Body="&lt;p&gt;If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470853484&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; one) and on &lt;a href=&quot;http://www.webqc.org/symmetry.php&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g. $\ce{d}$ orbitals). In the following picture I highlighted the important parts in the character table of the $\ce{D_{3\mathrm{h}}}$ group.&#xA;So in the $\ce{D_{3\mathrm{h}}}$ group the $\ce{d_{z^2}}$ orbital transforms as the irreducible representation $A_{1}^{'}$ and the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals transform as the irreducible representation $\ce{E^{'}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fxq14.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you have identified the irreducible representations of the $\ce{d}$ orbitals you can tell their degeneracy by identifying which $\ce{d}$ orbitals belong to the same irreduzible representation. If two or three $\ce{d}$ orbitals belong to the same irreducible representation they are degenerate. In the case of the $\ce{D_{3\mathrm{h}}}$ group shown below this means that $\ce{d_{z^2}}$ forms an energy level of its own, while $\ce{d_{x^2-y^2}}$ and $\ce{d_{xy}}$ (both belonging to $\ce{E^{'}}$) are degenerate and $\ce{d_{xz}}$ and $\ce{d_{yz}}$ (both belonging to $\ce{E^{''}}$) are degenerate. So, your $\ce{d}$ orbitals split into three energy levels, two of which are doubly degenerate, when your site has $\ce{D_{3\mathrm{h}}}$ symmetry.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-07-21T10:23:45.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5643" PostTypeId="1" CreationDate="2013-07-21T13:08:01.230" Score="2" ViewCount="7795" Body="&lt;p&gt;What are the oxidation states of the sulfur in the tetrathionate ion &lt;span class=&quot;math-container&quot;&gt;$\ce{S4O6^{2-}}$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l5kDW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l5kDW.png&quot; alt=&quot;Structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1977" LastEditorUserId="4945" LastEditDate="2018-11-13T13:24:39.983" LastActivityDate="2018-11-16T19:34:56.290" Title="What are the oxidation states of sulfur in the tetrathionate ion?" Tags="&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5644" PostTypeId="2" ParentId="5643" CreationDate="2013-07-21T13:51:16.190" Score="7" Body="&lt;p&gt;Consider the structure of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrathionate&quot;&gt;tetrathionate&lt;/a&gt;. The two central sulfurs each have two lone pairs and are assigned half of the electrons from the two bonds they make, since the electrons of bonds between atoms of the same element must be distributed evenly (due to there being, by definition, an electronegativity difference of zero between two atoms of the same element). A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6 - 4 - \frac{1}{2}(4) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminal sulfurs, on the other hand, have no lone pairs, and all the electrons from the sulfur-oxygen bonds are assigned to oxygen, since it's the more electronegative element. The only electrons assigned to those sulfurs are half of those from the sulfur-sulfur single bonds. Hence, their oxidation state is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6 - \frac{1}{2}(2) = +5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, six electrons in neutral sulfur, minus half of the two sulfur-sulfur bonding electrons, giving an oxidation state of +5.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-21T13:51:16.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5645" PostTypeId="2" ParentId="286" CreationDate="2013-07-21T16:48:11.303" Score="3" Body="&lt;p&gt;2D periodic systems can be mapped to toruses, but not spheres. This is easy to see because in a sphere, parallel lines always intersect. In the periodic system parallel lines never intersect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your bonus question: I don't know of anyone who has tried to use a periodic model to study a sphere or torus. But people have kind of gone the other way around, and &lt;a href=&quot;http://jcp.aip.org/resource/1/jcpsa6/v94/i1/p597_s1?isAuthorized=no&quot; rel=&quot;nofollow&quot;&gt;replaced a 3D periodic model with the surface of a 4D sphere&lt;/a&gt;. This allows you to avoid complications associated with long-range Coulomb interactions.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-07-21T16:48:11.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5646" PostTypeId="2" ParentId="681" CreationDate="2013-07-21T17:28:33.690" Score="4" Body="&lt;p&gt;At least six of these seven point groups been encountered. You can see examples on the &lt;a href=&quot;http://symmetry.otterbein.edu/gallery/index.html&quot; rel=&quot;nofollow&quot;&gt;Otterbein symmetry gallery&lt;/a&gt;. The one that's missing from this gallery is $I$, although as Richard pointed out, people have &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chir.20545/abstract&quot; rel=&quot;nofollow&quot;&gt;theoretically predicted&lt;/a&gt; some molecules with this symmetry to be stable. &lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2013-07-21T17:28:33.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5647" PostTypeId="1" AcceptedAnswerId="5676" CreationDate="2013-07-21T17:51:45.667" Score="2" ViewCount="113" Body="&lt;p&gt;Cathodic protection prevents the removal of positive form the metal surface by applying a negative charge to the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is is possible to prevent the reaction of an active metal (i.e. sodium) with water by aggressively forcing current into the metal using a DC source? We allow the &quot;water&quot; to be a dilute solution (~1%) of the metal hydroxide. There is also an anode that completes the circuit but its OK if the anode disintegrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I will let the answer stand but it begs one question: If the power supply forces the sodium metal to be at a strongly positive potential, it may end up &lt;em&gt;coated&lt;/em&gt; in layer(s) of hydroxide (or even oxide?), which may prevent the water from reaching it. But if the metal is liquid (NaK) then it may cause the metal to have &quot;negative surface tension&quot; making it impossible to maintain a high potential as the capacitance runs away to &quot;infinity&quot;. So it's not completely settled.&lt;/p&gt;&#xA;" OwnerUserId="1211" LastEditorUserId="1211" LastEditDate="2019-05-14T17:44:51.317" LastActivityDate="2019-05-14T17:47:25.883" Title="Cathodic protection of active metals" Tags="&lt;electrochemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5648" PostTypeId="2" ParentId="4281" CreationDate="2013-07-21T18:15:56.110" Score="3" Body="&lt;p&gt;You shouldn't need 4. The only dangerous thing is the barium, and permeakra is right that you should keep an eye on the equipment so that it doesn't silently fail on you. You should check perhaps once a month. It's only ~3.5 times the EPA limit, so a failure won't be immediately be a hazard (unless the sequestered barium is released all at once, but that seems unlikely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other 3 problems and the hardness are, at those levels in the water, are simply nuisances. So they will discolor your water, produce a weak sulfurous odor, and eventually clog the pipes, but they will not harm anyone. If you fixed the iron and sulfur, that would probably improve the color, so only 3 systems are needed. And you don't have to worry about the system failing silently since you will see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check these statements before making a decision, this is too important to trust only one source!&lt;/p&gt;&#xA;" OwnerUserId="1211" LastActivityDate="2013-07-21T18:15:56.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5649" PostTypeId="2" ParentId="2825" CreationDate="2013-07-21T18:23:04.613" Score="0" Body="&lt;p&gt;Ethene does not normally undergo nucleophile substitution reaction. Halogenation is an exception as the mechanism through which it goes forms a cyclic intermediate of Halonium ion and hence it substitutes electrophilically to $\pi$-bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in $\ce{CN}$, no such case arises. (probably due to it being less nucleophilic than halogens)&lt;/p&gt;&#xA;" OwnerUserId="2012" LastEditorUserId="1984" LastEditDate="2013-08-20T20:24:19.507" LastActivityDate="2013-08-20T20:24:19.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5650" PostTypeId="1" AcceptedAnswerId="5654" CreationDate="2013-07-21T18:32:40.357" Score="3" ViewCount="186" Body="&lt;p&gt;If one were to dissolve magnesium citrate (about $100~\mathrm{mg}$) in water (about $70~\mathrm{ml}$) and let the solution sit for 48 hours, would the solution be exactly the same as one dissolved and consumed right away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one explain the difference in taste from a 48-hour solution and a 5-minute solution?&lt;/p&gt;&#xA;" OwnerUserId="861" LastEditorUserId="7475" LastEditDate="2015-10-30T17:21:37.327" LastActivityDate="2015-10-30T17:21:37.327" Title="Would dissolving a salt and letting it sit for 48 hours have an effect on the solution?" Tags="&lt;aqueous-solution&gt;&lt;ionic-compounds&gt;&lt;taste&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5651" PostTypeId="1" AcceptedAnswerId="5653" CreationDate="2013-07-21T21:41:53.013" Score="-12" ViewCount="3236" Body="&lt;p&gt;What are 6 atoms commonly found in organic molecules?&lt;/p&gt;&#xA;" OwnerUserId="2013" LastActivityDate="2013-07-21T22:04:32.807" Title="(Organic Chemistry) Common Atoms in Organic Molecules" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-07-24T09:58:56.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="5652" PostTypeId="2" ParentId="5650" CreationDate="2013-07-21T22:01:47.310" Score="1" Body="&lt;p&gt;The difference in taste is due to a difference in molarity of the magnesium citrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its solubility in water is 20 g/100ml (&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_citrate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnesium_citrate&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over a 48-hour period a significant amount of water will evaporate from the container meaning the concentration will increase. Based on the numbers provided it is unlikely that the concentration will get the the point of precipitation; the solution will visually be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporation is dependent on surface area so the larger the surface area of exposed water the more profound the difference in taste will be.&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2013-07-21T22:01:47.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5653" PostTypeId="2" ParentId="5651" CreationDate="2013-07-21T22:04:32.807" Score="2" Body="&lt;p&gt;Carbon, Hydrogen, Nitrogen, Oxygen, Phosphorus, and Sulfur. Best way to remember this is CHNOPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wiki.answers.com/Q/The_six_most_common_elements_in_organic_molecules_are&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/The_six_most_common_elements_in_organic_molecules_are&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2013-07-21T22:04:32.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5654" PostTypeId="2" ParentId="5650" CreationDate="2013-07-21T22:07:01.333" Score="7" Body="&lt;p&gt;The only reactions that I would expect to occur to any significant extent with pure water are acid-base reactions between the citrate and water, where water, being amphiprotic, could act as either an acid or a base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H6O7^2- + H2O &amp;lt;=&amp;gt; C6H7O7- + OH-}$$&#xA;$$\ce{C6H6O7^2- + H2O &amp;lt;=&amp;gt; C6H5O7^3- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citrate, being the conjugate base of a weak acid, is itself a weak base (according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid&quot; rel=&quot;nofollow&quot;&gt;citric acid Wikipedia page&lt;/a&gt;, the $\mathrm pK_\mathrm a$ for the dissociation of the second proton is 4.75, and for the third between 5.41 and 6.40). Citrate is tripotic, and in the magnesium salt citrate possesses one undissociated acidic proton, which makes it an amphiprotic molecule as well. Given that the $\mathrm pK_\mathrm a$ of water is approximately 15.7, and the $\mathrm pK_\mathrm a$ of hydronium ($\ce{H3O+}$) is −1.74, both reactions above favor the reactants over the products by many orders of magnitude, though the second reaction is favored above the first (i.e., the dibasic magnesium citrate salt should still be very slightly acidic, overall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acid-base reactions typically reach equilibrium quite quickly, so there should be very little change in pH and no pH-attributable differences in taste over time. If any change does occur, I would expect it to be the result of a change in concentration as the water gradually evaporates. Another possibility is that, if the water is impure (which tap water certainly is), additional side-reactions might occur. What those are would depend on the exact composition of the water as well as environmental conditions. It's possible for the magnesium to form a precipitate, for example, in combination with other anions dissolved in the water (unlikely given the quantity in your example, but possible). If the magnesium citrate is part of a commercial product containing other ingredients, then it may well be that other agents are responsible for any effects you observe.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorDisplayName="user7951" LastEditDate="2015-10-02T18:22:16.590" LastActivityDate="2015-10-02T18:22:16.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5655" PostTypeId="1" CreationDate="2013-07-22T00:36:35.053" Score="8" ViewCount="16485" Body="&lt;p&gt;Can we determine whether a reaction is endothermic or exothermic? For example if we are given the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2SO2 + O2 -&amp;gt; 2SO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 + I2 -&amp;gt;2HI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2 + 3H2 -&amp;gt; 2NH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we determine weather these reactions are endothermic or exothermic or do we have to just memorize this? Are there any general trends that could make memorizing this information more manageable?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="305" LastEditDate="2013-07-23T05:24:57.100" LastActivityDate="2020-04-30T13:56:32.587" Title="Exothermic or Endothermic reactions" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5656" PostTypeId="2" ParentId="5655" CreationDate="2013-07-22T01:09:43.570" Score="10" Body="&lt;p&gt;In general just by looking at these equations there is no surefire way to know which is endo or exothermic. However you can look at the bond strength or the ∆&lt;em&gt;H&lt;/em&gt; of formation for each molecule. In short you have to know something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5iOzA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy stored in the molecules(specifically in the bonds) is lower before the reaction in an endothermic reaction and higher in an exothermic reaction. This means if you know relatively how strong the bonds or how much energy it takes to form each of the molecules then you can know whether the reaction is endo or exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a few ∆&lt;em&gt;H&lt;/em&gt; of formation values you can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess&amp;#39;s_law&quot; rel=&quot;noreferrer&quot;&gt;Hess's Law&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2014" LastEditorUserId="30" LastEditDate="2013-07-23T06:59:52.313" LastActivityDate="2013-07-23T06:59:52.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5657" PostTypeId="1" CreationDate="2013-07-22T02:22:57.793" Score="3" ViewCount="967" Body="&lt;p&gt;What does it mean for our bond of interest if we have low bond length and high frequency? Trying to make this connection but am not understanding. If anyone could explain it intuitively, that would be much appreciated. Say we have the bond of interest of heptane which is C-C. What is the variable that is changing (bond order, atom mass, etc.)? Also if the IR frequency is 1400, what is the effect of variable? Also what does this say about the physical bonding in the molecular bonds of interest?&lt;/p&gt;&#xA;" OwnerUserId="1396" LastEditorUserId="305" LastEditDate="2013-07-23T07:07:53.817" LastActivityDate="2017-04-27T01:48:44.530" Title="Bond of interest" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5658" PostTypeId="2" ParentId="5569" CreationDate="2013-07-22T10:30:27.557" Score="11" Body="&lt;p&gt;tl;dr: You are likely to see more soot, $\ce{CO}$, $\ce{H2}$, less of $\ce{CO2}$ and same amount of $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different type of combustion that you can talk about. For example when you mix fuel and oxidizer before burning them, it is referred to as premixed system. For example a gas cooking range, gas turbine engines, IC engines, etc. On the other hand if you do mix the fuel and oxidizer you get what's called a diffusion system. For example a candle flame, flame thrower, apartment fires, etc. Typically premixed systems will produce a blue flame, whereas diffusion systems will produce yellow-orangish flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to distinguish between a premixed and diffusion flame system because they will behave differently. A lighter produces a diffusion flame. For this answer we will focus on diffusion flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important characteristic of a diffusion flame is that it burns where there is just the right amount of fuel and oxidizer. That means irrespective of whether there is fuel entering oxidizer or oxidizer entering fuel, the flame will only be present where there is just the right amount of fuel and oxidizer. So for example for a methane-oxygen system at the flame we get, $$\ce{CH4 + 2O2 -&amp;gt; CO2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in both the cases we would expect to see same amount of $\ce{CO2}$ and $\ce{H2O}$ begin produced when same amount of fuel is burnt. That doesn't sound interesting at all. Plus our ideal situation ignored the effect of heat produced and its effect on the surrounding.&#xA;What happens when we take this in account?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Soot&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Well in most methane systems it is very difficult to produce soot, so let us consider something like propane $(\ce{C3H8})$ or higher order hydrocarbon (anything with more carbon atoms), which is more likely to soot. Soot is produced when unburned $\ce{C}$-atoms combine to form larger molecules, which combine to form particles. How soot is produced is still an active research area. But we do know that most soot is produced at relatively low temperatures (compared to the flame temperature) in fuel rich regions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a candle/lighter flame the soot is produced near the wick, but most of it is oxidized as it reaches the flame. Only some close to the center escapes from the tip because there is not enough oxidizer there to burn it all. This is the main reason diffusion flame appear yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine your lighter which starts off with oxidizer in the center, there won't be any soot particle inside the flame. So the flame will appear bluish. But the heat produced from the flame may be enough to break (pyrolyse) the fuel in the surrounding region. This will cause soot to be produced. Also because there is no reason for the soot to approach the flame it will not get oxidized easily and you will be left with a lot more soot.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Unburned fuel&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Next consider other unburned combustion products. When you burn methane in oxygen you produce more than just carbon dioxide and water. The unbalanced reaction can written as:&#xA;$$\ce{CH4 + O2 -&amp;gt; CO2 + H2O + CO + O2 + H2 + C + ...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unburned $\ce{C}$ forms the soot. A side effect is production of $\ce{H2}$. So now not only will you see more soot, there will be excess hydrogen too. Also in case of higher hydrocarbon fuels you might see some pyrolyzed fuel which is yet to form soot or get oxidized fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a premixed flame system when you have lot of fuel lying around you are more likely to produce $\ce{CO}$ rather than $\ce{CO2}$. A simple reason being you first need to produce $\ce{CO}$ before you get $\ce{CO2}$. A side effect of this is that the temperature will be lower. This is because most heat is released when $\ce{CO}$ is oxidized to $\ce{CO2}$.&#xA;So what happens in a diffusion flame system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned earlier in diffusion flame systems, the flame is around where there is just the right amount of fuel and oxidizer. The products formed at the flame diffuse out (into the fuel and oxidizer) and since there is scarcity of oxygen, $\ce{CO2}$ will get reduced to $\ce{CO}$ i.e. you will see increase in $\ce{CO}$ production and decrease in amount of $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: Saturn's moon Titan has an &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Titan&quot;&gt;atmosphere&lt;/a&gt; which primarily consists of nitrogen (98.4%). The next abundant gas is methane (1.4%) followed by hydrogen (0.1-0.2%). So it's not a bizarre world after all.&lt;/p&gt;&#xA;" OwnerUserId="655" LastActivityDate="2013-07-22T10:30:27.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5660" PostTypeId="1" CreationDate="2013-07-22T15:25:54.913" Score="10" ViewCount="3086" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Reaction of toluene at high temperatures gives us (o-m)toluene; whereas at normal conditions of electrophilic attack, it gives us (o-p) directing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this happen at high temp although as methyl group is activating and o-p directing?&lt;/p&gt;&#xA;" OwnerUserId="2018" LastEditorUserId="42741" LastEditDate="2017-04-28T14:56:33.997" LastActivityDate="2017-04-28T14:56:33.997" Title="How does toluene react at higher temperatures and why?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5661" PostTypeId="2" ParentId="708" CreationDate="2013-07-22T21:34:26.757" Score="4" Body="&lt;p&gt;I also unsuccessfully tried to look for such a paper in the past but below are some useful resources I've found. The only definitive answer I can give you is that traditional Ewald summation will be slower than a particle-mesh technique for any relevant system sizes (this is backed up in several of the references below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 4.9 of the &lt;a href=&quot;ftp://ftp.gromacs.org/pub/manual/manual-4.6.3.pdf&quot; rel=&quot;nofollow&quot;&gt;GROMACS user manual&lt;/a&gt; has some very nice discussion on their various implemenations of long-range electrostatics and appropriate references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://lammps.sandia.gov/doc/kspace_style.html&quot; rel=&quot;nofollow&quot;&gt;LAMMPS documentation&lt;/a&gt; also has quite a few references and some bits of helpful discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try digging through the &lt;a href=&quot;http://ambermd.org/doc11/Amber11.pdf&quot; rel=&quot;nofollow&quot;&gt;AMBER manual&lt;/a&gt; which uses a PME implementation but I haven't read any of this myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;http://codeblue.umich.edu/hoomd-blue/doc/classhoomd__script_1_1charge_1_1pppm.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; you can find implementation details for the PPPM method in HOOMD (helpful if you're interested in GPGPU implementations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the most relevant bit of discussion is probably this excerpt in the GROMACS manual:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Particle-Particle Particle-Mesh methods of Hockney &amp;amp; Eastwood can&#xA;  also be applied in GROMACS for the treatment of long range&#xA;  electrostatic interactions [99]. Although the P3M method was the ﬁrst&#xA;  efﬁcient long-range electrostatics method for molecular simulation,&#xA;  the smooth PME (SPME) method has largely replaced P3M as the method of&#xA;  choice in atomistic simulations. One performance disadvantage of the&#xA;  original P3M method was that it required 3 3D-FFT back transforms to&#xA;  obtain the forces on the particles. But this is not required for P3M&#xA;  and the forces can be derived through analytical differentiation of&#xA;  the potential, as done in PME. The resulting method is termed P3M-AD.&#xA;  The only remaining difference between P3M-AD and PME is the&#xA;  optimization of the lattice Green inﬂuence function for error&#xA;  minimization that P3M uses. However, in 2012 it has been shown that&#xA;  the SPME inﬂuence function can be modiﬁed to obtain P3M [100]. This&#xA;  means that the advantage of error minimization in P3M-AD can be used&#xA;  at the same computational cost and with the same code as PME, just by&#xA;  adding a few lines to modify the inﬂuence function. However, at&#xA;  optimal parameter setting the effect of error minimization in P3M-AD&#xA;  is less than 10%. P3M-AD does show large accuracy gains with&#xA;  interlaced (also known as staggered) grids, but that is not supported&#xA;  in GROMACS (yet).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're really looking to dive in, you can look at the source code for these packages (I believe you need a license for AMBER) to see the actual implementations.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastEditorUserId="2008" LastEditDate="2013-07-22T21:45:17.383" LastActivityDate="2013-07-22T21:45:17.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5664" PostTypeId="2" ParentId="5657" CreationDate="2013-07-23T04:08:58.340" Score="7" Body="&lt;p&gt;In IR, the vibrational frequencies of a simple diatomic molecule can be described by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\bar\nu = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $k$ is the force constant, $\mu$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reduced_mass&quot;&gt;reduced mass&lt;/a&gt; of the molecule, and $c$ is the speed of light (which results in appropriate units for reporting the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wavenumber&quot;&gt;wavenumber&lt;/a&gt;, rather than ordinary units of frequency, e.g., Hz.). This equation is derived from treating the molecule as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Harmonic_oscillator&quot;&gt;simple harmonic oscillator&lt;/a&gt;, using Hooke's law, Newton's laws of motion, and the classic technique of reducing a two-body problem to a one-body problem.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation of the force constant is complex, but suffice to say that it can be viewed as being roughly proportional to the strength of the bond. Since the wavenumber is proportional to the square root of the force constant it can be said that, &lt;em&gt;ceteris paribus&lt;/em&gt;, stronger bonds tend to yield higher wavenumbers. Since higher bond multiplicity and shorter bond length correlate positively to bond strength, those factors can also be correlated to higher wavenumbers. (There are also positive correlations between metrics such as, e.g., bond energies and wavenumber, etc.) These factors can be used to rationalize the trend that wavenumbers of triple bonds tend to be greater than those of double bonds, which in turn tend to be greater than those of single bonds. Similarly, the diminished wavenumber of conjugated bonds vs. analogous unconjugated bonds can be explained by the bonds of conjugated systems having lengths intermediate between single and double bonds. &lt;a href=&quot;http://books.google.com/books?id=oZeFG6QDNekC&amp;amp;pg=PA287&amp;amp;lpg=PA287#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;One source&lt;/a&gt; indicates the following approximate proportionality relationship between bond length, $r$, and the force constant, $k$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;k \propto \frac{1}{r^6}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, even minute differences in bond length produce large variance in the force constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, wavenumber is inversely proportional to the square root of the reduced mass. Hence, as reduced mass grows, wavenumbers decrease. This largely explains, for example, why the highest wavenumbers are those of bonds between hydrogen and some other element, since hydrogen is the lightest element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;*I think the math involved is probably well beyond the necessary scope of this answer, so I won't elaborate on it here. I dealt with these topics in introductory physics and differential equations classes. McQuarrie's physical chemistry textbook specifically deals with the harmonic oscillator model vis-a-vis spectroscopy, and I can recommend it for anybody interested in further details.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-23T04:19:09.227" LastActivityDate="2013-07-23T04:19:09.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5665" PostTypeId="1" AcceptedAnswerId="5669" CreationDate="2013-07-23T05:11:47.963" Score="5" ViewCount="1598" Body="&lt;p&gt;The question may be off-topic here. If my concept is flawed, please poke me in the rib (+:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20100721064824AA6hmHM&quot;&gt;Sources&lt;/a&gt; on the internet indicate a minimum 1.2Volts is adequate for electrolysis of water; nominally 6V. Applying a DC source to water provides adequate energy to break the hydrogen bond. Apparently some heat is also released in the process..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, one were to collect 100ml (an arbitrary figure) oxygen by electrolysis applying 6V DC to 1 Litre of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the energy to break the hydrogen bond be provided by applying heat to water? All other things being identical, how many calories would be required to collect the same amount of oxygen from the same volume of water?&lt;/p&gt;&#xA;" OwnerUserId="765" LastActivityDate="2013-07-23T11:50:51.830" Title="What is the heat energy required to perform electrolysis equivalent?" Tags="&lt;heat&gt;&lt;electrolysis&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5666" PostTypeId="2" ParentId="5655" CreationDate="2013-07-23T05:24:28.520" Score="9" Body="&lt;p&gt;As anglinb pointed out, the only completely reliable method of determining whether the reactions are endo- or exothermic is by calculation. That said, certain generalizations may allow reasonable predictions about changes in thermodynamic parameters to be made:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reactions that produce a net increase in $\sigma$-bonds between particular pairs of atoms that were $\pi$-bonded in the reactants tend to be exothermic, since $\sigma$-bonds tend to be more stable than $\pi$-bonds. Conversely, reactions in which the generation of $\pi$-bonds and reduction in number of equivalent $\sigma$-bonds occurs tend to be endothermic.&lt;/li&gt;&#xA;&lt;li&gt;Reactions that involve a net reduction in the total number of molecules of products as compared to reactants, and/or involve the generation of products in more highly ordered states (for example, the generation of solids from liquids or gases), tend to occur with a reduction in entropy. Reactions that involve the generation of heteroatomic compounds from homoatomic substances tend to increase entropy. The spontaneity of a reaction is governed by the equation $\Delta G = \Delta H - T\Delta S$, where $\Delta G &amp;lt; 0$ for a spontaneous process. Therefore, if a reaction occurs spontaneously while the entropy change ($\Delta S$) is negative, the reaction &lt;em&gt;must be exothermic&lt;/em&gt; by default; an endothermic process with negative entropy change cannot be spontaneous.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Remember that these are very rough guidelines, and ultimately only performing calculations with concrete values for the relevant thermodynamic parameters can be totally definitive. The first and third reactions from your question can be predicted to have negative $\Delta S$ by the above guidelines, and therefore must be exothermic (if they're supposed to be spontaneous at all). Crunching the numbers using thermodynamic data does in fact confirm that.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-23T05:47:11.770" LastActivityDate="2013-07-23T05:47:11.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5667" PostTypeId="2" ParentId="5660" CreationDate="2013-07-23T08:15:11.380" Score="11" Body="&lt;p&gt;Assuming you're talking about Friedel-Crafts alkylation of toluene, this can be explained by the reaction being reversible and thermodynamic control taking precedence at sufficiently high temperatures. The directing ability of substituents in electrophilic aromatic substitution reactions is a function of the reaction being under kinetic control at lower temperatures. According to Fleming's book on MO theory, a single electron-donating substituent on benzene conjugated to the ring's $\pi$ system disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals of unsubstituted benzene. Comparing the intermediates of an EDG-substituted benzene undergoing electrophilic attack, the HOMO of the $\pi$ system, $\psi_3$, is higher in energy when attack occurs at the meta- position than when it occurs at the ortho- or para- positions. Using the concept of resonance, this can be illustrated in an abstract sense by considering the resonance structures of the intermediates resulting from attack at the various positions; an additional resonance structure exists for ortho- and para- attack if the substituent is conjugated to the $\pi$ electron system of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With alkyl groups, the electrons are donated by hyperconjugation, which is a weaker effect (and not apparent from resonance structures), but nonetheless has a similar impact on regioselectivity and the $\pi$ molecular orbitals. Additionally, the coefficients on the atoms of the HOMO of the $\pi$ system are lowest at the meta positions when an electron donating group is present (including alkyl groups), indicating electron density is higher at the ortho- and para- positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I should clarify, of course, that resonance is not an actual physical process/phenomenon, merely an abstraction which provides a qualitative picture of the extent of electron delocalization and which often correlates well with the actual distribution of electron density in the molecular orbitals.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, at low temperatures, when the reaction is kinetically controlled, the favored pathways are the ones in which the transition states and intermediates are stabilized, leading to ortho-/para- substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher temperatures, the preferred product is the thermodynamic one, and the advantage of meta- substitution is that there is less crowding and reduced steric hindrance in, e.g., a 1,3,5-trisubstituted benzene by comparison to a 1,2,4-trisubstituted benzene. Additionally, Friedel-Crafts is actually reversible at high temperatures. Even if the kinetic product forms initially, then it can decompose by dealkylation in the presence of Lewis-Acid catalysts and acid at high temperatures, at which point equilibrium can be established with the thermodynamic product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to March, experimental evidence suggests that methyl groups will mainly migrate intramolecularly in the isomerization reaction to give the thermodynamically preferred meta-substituted benzenes, while both intramolecular rearrangements and complete dealkylation with subsequent intermolecular reactions are possible with other types of alkyl groups. &lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-23T10:24:15.397" LastActivityDate="2013-07-23T10:24:15.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5668" PostTypeId="1" CreationDate="2013-07-23T11:21:42.733" Score="2" ViewCount="203" Body="&lt;p&gt;I'm currently working on building a simulator for the early earth to examine the spontaneous occurrence of food generated autocatalytic sets. No one in my group is a chemist, so the bottle neck for the project is implementing some sort of realistic chemistry in our model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are considering using the software outlined in this paper:&#xA;&lt;a href=&quot;http://arxiv.org/abs/1304.1356&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/abs/1304.1356&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will allow us to create a reaction network from an initial set, based on a set of general rules. The paper demonstrates well how the formose reactions are executed, which we intend to include. What other reaction groups would it be important to include? Our initial set of molecules would be comprised of the food molecules for each set of reaction rules included, we don't intend to start from H,O,N,C for example, so fundamental chemical rules should not be too crucial, as they should be generalised into the major reaction groups we will use.&lt;/p&gt;&#xA;" OwnerUserId="1990" LastActivityDate="2013-07-24T20:47:18.990" Title="To model early earth chemical reactions which types of reactions should be included?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5669" PostTypeId="2" ParentId="5665" CreationDate="2013-07-23T11:50:51.830" Score="6" Body="&lt;p&gt;I won't do any rib poking, but there is some flawed thinking here.  Certainly not off topic though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nomenclature:&lt;/strong&gt; First, when you pass a current through water, you are not breaking the hydrogen bonds, which are intermolecular forces, but (a) breaking the bond between hydrogen and oxygen and (b) forming molecular oxygen and molecular hydrogen, $\ce{2H2O -&amp;gt; 2H2 + O2}$, and this reaction can be broken down into two half reactions: $$\ce{4H2O + 4 e- -&amp;gt; 2H2 + 4OH-}$$ $$\ce{2H2O -&amp;gt; O2 + 4H+ + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Potentials:&lt;/strong&gt; The 1.2 V you refer to likely arises from the &lt;em&gt;standard reduction potential&lt;/em&gt; of oxygen, which is the reverse of the second reaction above.  I won't go in to the thermodynamics here, but suffice it to say that a brief look at a &lt;a href=&quot;http://www.jesuitnola.org/upload/clark/refs/red_pot.htm&quot;&gt;standard reduction potential table&lt;/a&gt; indicates that the reduction of water (first equation) has a potential of -0.83 V and the oxidation of water has a potential of -1.23 V for a grand total of ~ -2 V. (Note, these values are pH dependent and standard reduction potential tables report values at a pH of 0.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrochemistry&lt;/strong&gt; When you apply an overpotential to a system (the potential beyond what is needed to get the oxidation/reduction reactions to go forward) then two things will happen (a) the reaction will go faster and (b) heat will be generated.  This heat would be analogous to heat due to friction in a physical system and is considered wasted energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermochemistry&lt;/strong&gt; Finally, we get to the answer to your question, which is &quot;probably not, but the way you asked it, yes&quot;  (hunh?)  Recall the nomenclature issue above.  Adding heat to water &lt;em&gt;will break the &lt;strong&gt;hydrogen bonds&lt;/strong&gt; in water&lt;/em&gt; eventually leading to the water changing from a liquid to a gaseous state.  However, adding heat to water would not result in the oxidation/reduction reactions outlined above because an important ingredient is missing: electrons.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-07-23T11:50:51.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5670" PostTypeId="1" CreationDate="2013-07-23T11:55:42.103" Score="3" ViewCount="124" Body="&lt;p&gt;I am creating an automated chemical component mixing system (DIY project). One of the features required is continuous monitoring of pH value in my main solution. I do not need it to be very precise, precision of 0.2 would be enough. Also, I want it to be as low maintenance as possible (cleaning the probe once a month would be great). pH of the solution will vary from $pH = 4$ to $pH = 7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is, which of this testing ways will require less maintenance of pH probe:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Constant submersion in the solution.&lt;/li&gt;&#xA;&lt;li&gt;Once per hour dipping in the solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2021" LastEditorUserId="418" LastEditDate="2014-03-20T14:15:23.130" LastActivityDate="2014-03-20T14:15:23.130" Title="Less maintenance-dependent continuous pH testing" Tags="&lt;ph&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5671" PostTypeId="2" ParentId="5657" CreationDate="2013-07-23T12:02:23.660" Score="7" Body="&lt;p&gt;To add on to Greg's well-written answer (since this is too long for a comment) we can write Hooke's Law in a slightly more practical form for purposes of predicting IR stretching frequencies:$$\bar\nu (cm^{-1}) = 4.12\sqrt{\frac{k}{\mu}}$$ with $k = 5, 10 \text{, or } 15$ $\times10^5$ dynes/cm for a single, double or triple bond.  For example, a $\ce{C=C}$ bond: k = $10 \times 10^5$, $\mu=\frac{(12)(12)}{12+12}=6$ giving $\bar\nu=1682 cm^{-1}$ where the experimental value is $1650$.  We can leave it to the OP to see that a $\ce{C-H}$ bond has a calculated frequency of $3032 cm^{-1}$ (experimental is $3000$) and a $\ce{C-D}$ bond has a calculated frequency of $2228 cm^{-1}$ (experimental $2206$).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="41556" LastEditDate="2017-04-27T01:48:44.530" LastActivityDate="2017-04-27T01:48:44.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5672" PostTypeId="1" CreationDate="2013-07-23T12:17:11.667" Score="9" ViewCount="44425" Body="&lt;p&gt;I have read that paint contains VOCs(Volatile Organic Compounds), and solvents generally, and I could wait for them to dissipate, but i've read there are ways to remove the smell of paint such as half an onion in a bucket of water, such as slices of lemon on a plate, such as small bowls of distilled white vinegar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these techniques actually remove all the nasty elements from the paint fumes, or just some of the nasty elements from them?  Do they remove all the paint fumes?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't want to remove the smell but leave the VOC or other bad elements there.  If that's the case i'd rather just leave a window open in the room, closing it periodically, seeing if there's finally a time when the paint smell doesn't build up when the window is closed, then i'll know the paint isn't emitting any more fumes.&lt;/p&gt;&#xA;" OwnerUserId="2020" LastEditorUserId="12860" LastEditDate="2015-10-30T17:15:42.680" LastActivityDate="2015-10-30T17:15:42.680" Title="Does an onion completely remove &quot;fresh&quot; paint VOCs and other chemicals emitted?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5673" PostTypeId="1" CreationDate="2013-07-23T14:04:49.560" Score="4" ViewCount="272" Body="&lt;p&gt;I'm looking for a material that I can use to make a mesh netting.  It should be flexible and have reasonable tensile strength when cold but brittle when hot (&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{400^\circ F}$&lt;/span&gt;).  I want to use it to contain charcoal when cold but disintegrate when charcoal is burning and some force is applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="2023" LastEditorUserId="69977" LastEditDate="2020-04-29T12:01:20.177" LastActivityDate="2020-04-29T12:01:20.177" Title="Material that is flexible when cold but becomes brittle when hot" Tags="&lt;materials&gt;&lt;mechanical-properties&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="5674" PostTypeId="2" ParentId="5668" CreationDate="2013-07-23T14:40:55.387" Score="6" Body="&lt;p&gt;For the reactions look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller%E2%80%93Urey_experiment&quot;&gt;Stanley Miller's experiment&lt;/a&gt;. The experiment used water (H2O), methane (CH4), ammonia (NH3), and hydrogen (H2). I think you should start with that.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-07-23T14:40:55.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5675" PostTypeId="2" ParentId="5673" CreationDate="2013-07-23T14:48:42.170" Score="2" Body="&lt;p&gt;Clay or ceramic materials. Not flexible, but worth a try. It has to be completely dry before heating or it can explode on heating projecting pieces everywhere.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-07-23T14:48:42.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5676" PostTypeId="2" ParentId="5647" CreationDate="2013-07-23T17:34:35.600" Score="2" Body="&lt;p&gt;It is unlikely that a potential applied to a sodium electrode in water would prevent the spontaneous oxidation of sodium by water.  Consider the following reduction potentials:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Na+ + 2e- -&amp;gt; 2Na(s)}\ E^0=-2.71\ \mathrm V\tag{1}$$&lt;/span&gt; &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 H2O + 2e- -&amp;gt;H2 + 2 OH-}\ E^0=-0.83\ \mathrm V\tag{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction of water will always occur before the reduction of sodium ions.  If we reverse equation (1) so we can combine the two and get the full redox equation for your system, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 Na(s) + 2 H2O &amp;lt;=&amp;gt; 2Na+(aq) + H2 + 2 OH-}\ E^0= 1.88\ \mathrm V\tag{3} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A positive potential indicates a spontaneous reaction under the given conditions so thermodynamically, it will be near impossible to prevent the reaction from being spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How impossible?  We can do a thought experiment and ask the question &lt;em&gt;at what pH would this reaction be non spontaneous&lt;/em&gt;? In other words, when would the Nernst Equation for this system evaluate to zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E=E^0-\frac{RT}{nF}\ln\frac{\text{Products}}{\text{Reactants}}\tag{4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entering our values, converting Ln to Log and substituting &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = -\log[\ce{H}]$&lt;/span&gt; (and assuming all other concentrations are fixed at 1 M or 1 atm):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E=1.88+0.059(14-\mathrm{pH})\tag{5}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction would not be spontaneous until the pH reached &lt;strong&gt;45 units&lt;/strong&gt;.  Something that can't be done in water.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorDisplayName="user7951" LastEditDate="2019-05-14T17:47:25.883" LastActivityDate="2019-05-14T17:47:25.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5677" PostTypeId="1" AcceptedAnswerId="18414" CreationDate="2013-07-23T17:46:45.920" Score="4" ViewCount="156" Body="&lt;p&gt;This is an outdoors &amp;amp; chemistry question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an old and functional water bladder, shown below. On it, there are markings that &lt;em&gt;coffee, beer, wine, hot drinks&lt;/em&gt; etc. must NOT be stored inside, only water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I realize that wine is acidic, damaging the bladder and becoming poisonous, beer seems to me quite close to water. Furthermore, I indend to test it with weisbeer, which does not contain such a large amount of gases as to explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Would be safe for me to store non-fizzy beer for up to a day every day in a &quot;do-not-store-beer-inside&quot; water bottle and then safely drink it?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T3o3Jm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2026" LastEditorUserId="4325" LastEditDate="2015-07-20T15:12:29.477" LastActivityDate="2015-07-20T15:12:29.477" Title="&quot;Drinking water&quot; plastic - can I store beer in this?" Tags="&lt;safety&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5678" PostTypeId="2" ParentId="5677" CreationDate="2013-07-23T20:09:58.717" Score="1" Body="&lt;p&gt;From the description, the plastic will capture volatile compounds and you will never eliminate the smell of beer and the alcohol could disolve the plastic a bit and give the beer a plastic smell/taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not a good idea.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-07-23T20:09:58.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5679" PostTypeId="1" CreationDate="2013-07-23T21:37:11.720" Score="2" ViewCount="87" Body="&lt;p&gt;Usually maximum time we see then after cooking by gas there is no black spot on the pot. But continually burning can affect the pot surface and makes black powder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the most acceptable reason that black powder(carbon) take places on the pot surface. Is it related to the burning temperature? or fuel reacts with the pot surface?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2028" LastActivityDate="2013-07-24T19:19:39.373" Title="Reaction with carbon" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5681" PostTypeId="1" AcceptedAnswerId="5682" CreationDate="2013-07-24T09:48:34.270" Score="16" ViewCount="41220" Body="&lt;p&gt;How does the addition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Alum&quot;&gt;alum&lt;/a&gt; — potassium aluminium sulfate or $\ce{KAl(SO4)2}$ — help settle the dust particles in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was such a statement in my Year 9 textbook, and so I was wondering how exactly alum does it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I have posted an answer below based on what I've found, but I'd love it if some else could verify and fill in the missing gaps in it.&lt;/p&gt;&#xA;" OwnerUserId="2030" LastEditorUserId="1362" LastEditDate="2014-11-10T06:42:44.617" LastActivityDate="2015-03-26T05:36:24.833" Title="How does alum improve the settling of particles to produce 'clean' water?" Tags="&lt;purification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5682" PostTypeId="2" ParentId="5681" CreationDate="2013-07-24T10:00:55.513" Score="13" Body="&lt;p&gt;OK, so after a bit of researching this is what I've found. &lt;hr&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First of all,why does the dust not settle in the water in the first place?&lt;/strong&gt; Well, dust forms a kind of a suspension.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The particles are all similar and ionized, so therefore they have a similar charge (positive, I think) causing them  repel each other and stay separate. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And because they're so small, the gravitational force acting on them isn't very strong either. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, the dust particles, previously dispersed in the air, have a lot of kinetic energy already. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All these 3 reasons coupled cause the particles to remain suspended for so long.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what does adding alum do?&lt;/strong&gt; Alum has a negative charge and tends to disperse in water very fast and very well (Why so? Because it's a colloid?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes it to join up with all of the offending particles and neutralize them. Now that the particles don't have any repelling charges, they tend to clump together into &lt;a href=&quot;http://en.wikipedia.org/wiki/Flocculation&quot;&gt;'flocs'&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increased size as well as the lack of repelling charges cause the alum particles to settle down at the bottom or rise up and float in the water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do the particles clump together and settle?&lt;/strong&gt; Apparently even this isn't as simple as it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the particles are neutralized, they clump together because of something called the &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot;&gt;London Dispersion Force&lt;/a&gt; which are part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot;&gt;Van der Waals forces&lt;/a&gt;. Basically, weak intermolecular force arising from quantum-induced instantaneous polarization multipoles in molecules causes even nonpolar particles to attract each other due to the correlated movements of the electrons in interacting molecules. More info can be found inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why they then settle, I think it can be sufficiently explained by gravity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oh well, that's all. All for explaining something &lt;a href=&quot;http://en.wikipedia.org/wiki/Alum#In_antiquity&quot;&gt;even the ancients knew&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2030" LastEditorUserId="1389" LastEditDate="2013-07-24T21:40:36.297" LastActivityDate="2013-07-24T21:40:36.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5683" PostTypeId="2" ParentId="5670" CreationDate="2013-07-24T10:53:20.747" Score="3" Body="&lt;p&gt;The most common approach is to measure periodically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH glass electrodes (the most common ones) need to be &quot;wet&quot; in order to be used, so if you are going to remove it from the solution every hour you'll need to keep it submerged in the appropriate solution. Not only that, you need to wash it (preferably with the most pure water you can get) between measurements. Read section 2.2 of &lt;a href=&quot;http://partners.metrohm.com/GetDocument?action=get_dms_document&amp;amp;docid=955207&quot; rel=&quot;nofollow&quot;&gt;this monograph&lt;/a&gt; to understand why it needs to be &quot;wet&quot; and what solution to store it in. There's information on a lot of types of electrodes. &lt;a href=&quot;https://en.wikipedia.org/wiki/PH_meter#The_probe&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; explains how to wash it. Another problem you might face with continuous measuring is not the electrode, but the meter. From that monograph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The condition of the pH measuring assembly is subject to fluctuations over the course of time. Depending on the accuracy demands, a pH calibration should therefore be performed periodically, e.g. once a day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'd have to test and see what accuracy you get with your equipment.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="1342" LastEditDate="2013-07-24T10:58:42.113" LastActivityDate="2013-07-24T10:58:42.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5684" PostTypeId="2" ParentId="5679" CreationDate="2013-07-24T12:29:01.243" Score="1" Body="&lt;p&gt;For combustion to occur you need 3 things at the same place: fuel, oxygen and heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the air intake is dirty or not big enough, you get a flame that has an orange color, and on a cooler surface deposits C before it has a chance to burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction with the pot is only a physical one of heat transfer and maybe C deposition.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="1300" LastEditDate="2013-07-24T19:19:39.373" LastActivityDate="2013-07-24T19:19:39.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5685" PostTypeId="1" CreationDate="2013-07-24T19:41:46.683" Score="2" ViewCount="794" Body="&lt;p&gt;What are the technical/scientific details behind the process of water deionization? How does deionization machines like those manufactured by &lt;a href=&quot;http://www.millipore.com/&quot; rel=&quot;nofollow&quot;&gt;Millipore&lt;/a&gt; work?&lt;/p&gt;&#xA;" OwnerUserId="2033" LastEditorUserId="21964" LastEditDate="2016-08-16T21:27:31.753" LastActivityDate="2016-08-16T21:27:31.753" Title="How does water deionization process work?" Tags="&lt;water&gt;&lt;ions&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5686" PostTypeId="2" ParentId="5668" CreationDate="2013-07-24T20:47:18.990" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Formose_reaction&quot; rel=&quot;noreferrer&quot;&gt;formose reaction&lt;/a&gt; is one of the proposed pathways that carbohydrates could form. Two equivalents of formaldehyde ($\ce{CH2O}$) react to form glycoaldehyde ($\ce{HOCH2CHO}$), which reacts with further equivalents of formaledehyde or gycloaldehyde in &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldol_reaction&quot; rel=&quot;noreferrer&quot;&gt;aldol condensations&lt;/a&gt; to produce the 3-, 4-, 5-, and 6-carbon sugars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH2O + CH2O -&amp;gt; HOCH2CHO}$$&#xA;$$\ce{CH2O + HOCH2CHO -&amp;gt; HOCH2CH(OH)CHO}$$&#xA;$$\ce{HOCH2CHO + HOCH2CHO -&amp;gt; HOCH2CH(OH)CH(OH)CHO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formaldehyde can be produced from hydrogen and carbon monoxide or dioxide, themselves the products of &lt;a href=&quot;http://en.wikipedia.org/wiki/Methane_reformer&quot; rel=&quot;noreferrer&quot;&gt;steam reforming of methane&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Water-gas_shift_reaction&quot; rel=&quot;noreferrer&quot;&gt;water-gas shift reaction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam-reformation:&#xA;$$\ce{CH4 + H2O &amp;lt;=&amp;gt; CO + 3H2}$$&#xA;Water-gas shift:&#xA;$$\ce{CO + H2O &amp;lt;=&amp;gt; CO2 + H2}$$&#xA;Fomaldehyde synthesis:&#xA;$$\ce{H2 + CO &amp;lt;=&amp;gt; H2CO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the sugars, the carbonyl group and the alcohols give you the option to bring all of organic chemistry to bear and convert the sugars into anything you want through substitutions, additions, eliminations, and redox. Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chichibabin_pyridine_synthesis&quot; rel=&quot;noreferrer&quot;&gt;Chichibabin pyridine synthesis&lt;/a&gt;, which can operate with as simple molecules as acetaldehyde ($\ce{CH3CHO}$), formaldehyde, and ammonia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2CO + CH3CHO &amp;lt;=&amp;gt; H2C=CHCHO + H2O}$$&#xA;$$\ce{H2C=CHCHO + CH3CHO + NH3 + 1/2 O2 &amp;lt;=&amp;gt; C5H5N + 3H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-07-24T20:47:18.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5687" PostTypeId="1" CreationDate="2013-07-25T03:01:31.700" Score="1" ViewCount="203" Body="&lt;p&gt;I am attempting to optimize interactions for a series of random and test molecules to a protein using docking software. The software uses QM/MMGBSA to calculate free energy in joules (yes it is not a measurement but it is very corollary). Larger molecules automatically have more free energy because they have greater potential energy. I would like to divide the free energy by the amount of atoms in the molecule or the molecules total Dalton. The idea would be to normalize the energy and provide the amount of calories associated with the bond. I asked a professor at a university about measuring the efficiency of a bond in this manner and he said he was unsure what joule/mass would provide. It does appear that the amount of bond calories is able to properly classify with correct magnitudes of viability 2 positive controls and 1 negative but I need this idea to pass peer review. Is my rationalization of the joule/Dalton reasonable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am measuring the binding energy of azttp (AZT triphosphate), gtp (Guanine triphosphate) and AZTppppA (AZTP+AMP) on HIV Reverse Transcriptase wild type (8ICO). I am used 8 high quality conformers and independent docks with QM/MMGBSA for my negative sample (AZT+AMP), 5 conformers for my major positive sample (AZTTP) and 1 conformer for the minor positive sample (gtp). I have two or three high quality conformers through the same processes for each of my test samples.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HIV RT has a tunnel where nucleotide analogs pass through with a pocket where if a molecule binds effectively it is accepted but if it fails it is phosphorylated and excreted. I am attempting to create a metric which can be used to measure the effectiveness of a binding and allowing me to infer (only) probability of rejection. Since the window/box for the docking is small and does not change for each test subject I would like to consider HIV-RT a constant. &lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorDisplayName="user1464" LastEditDate="2013-07-25T04:46:28.410" LastActivityDate="2013-07-25T04:46:28.410" Title="bond energy calories" Tags="&lt;bond&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="0" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5690" PostTypeId="1" AcceptedAnswerId="5691" CreationDate="2013-07-25T15:00:11.833" Score="15" ViewCount="19084" Body="&lt;p&gt;In nucleophilic aromatic substitution reactions, why do fluorides react faster than bromides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordinarily bromide is a better leaving group than fluoride, e.g. in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; reactions, so why isn't this the case here? The only thing I can think of is that fluorine is more electron-withdrawing (via the inductive effect), which could stabilise the Meisenheimer complex formed as an intermediate.&lt;/p&gt;&#xA;" OwnerUserId="2018" LastEditorUserId="16683" LastEditDate="2018-10-17T22:48:05.320" LastActivityDate="2018-10-17T22:49:40.093" Title="Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;heterocyclic-compounds&gt;&lt;chemoselectivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5691" PostTypeId="2" ParentId="5690" CreationDate="2013-07-25T15:52:03.207" Score="20" Body="&lt;p&gt;The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it S&lt;sub&gt;N&lt;/sub&gt;Ar in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uvM8U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the first step (= addition) is very slow as aromaticity is lost and thus the energy barrier is very high. The second step (= elimination of the leaving group) is quite fast as aromaticity is restored. So, since the elimination step is fast compared to the addition step, the actual quality of the leaving group is not very important, because even if you use a very good leaving group (e.g. iodine), which speeds up the elimination step, the overall reaction rate will not increase as the addition step is the bottleneck of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about fluorine? Fluorine is not a good leaving group, but that doesn't matter as I said before. It is not the leaving group ability of &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; which makes the reaction go faster than with, say, bromine or chlorine, but its very high negative inductive effect (due to its large electronegativity). This negative inductive effect helps to stabilize the negative charge in the Meisenheimer complex and thus lowers the activation barrier of the (slow) addition step.&#xA;Since this step is the bottleneck of the overall reaction, a speedup here, speeds up the whole reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving groups with a negative mesomeric effect (but little negative inductive effect) are not good at stabilizing the Meisenheimer complex, because the negative charge can't be delocalized to them.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-10-17T22:49:40.093" LastActivityDate="2018-10-17T22:49:40.093" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="5692" PostTypeId="2" ParentId="5690" CreationDate="2013-07-25T15:52:13.670" Score="6" Body="&lt;p&gt;Fluorine is the most electronegative element, and the fluoride anion is also much smaller and less polarizable than any of the other halogen anions, making its activity much more dependent on solvent effects. In protic solvents, fluoride tends to be very strongly solvated as a hydrogen bond acceptor and is thereby stabilized, resulting in high leaving-group tendency and very low nucleophilicity. The larger halogens can be better leaving groups in aprotic solvents, however, where their bulk and polarizability results in greater comparative stability, and the stabilizing effect of hydrogen bonds is not present. I guess it's conceivable that the presence of fluorine atoms would better stabilize the Meisenheimer complex than other halogens, though I don't know for sure. Still, since all halogens are electron-withdrawing by induction or negative hyperconjugation, the presence of more strongly electron-withdrawing groups (e.g., nitro) conjugated to the $\pi$ system of the ring would still be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What follows is not a direct answer to your question, but a slight segue into a related subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate-determining step in SnAr mechanisms is typically the initial nucleophilic attack resulting in formation of the Meisenheimer complex, or occasionally the final proton transfer if neutral nucleophiles are used. The loss of the leaving group is normally a fast step, given the high energy of the Meisenheimer complex intermediate. Once the complex is formed, loss of the leaving group should be rapid. Therefore, I suspect aprotic solvents would be preferable in many situations (particularly with neutral nucleophiles), irrespective of the effect on the leaving group, since they don't hamper the activity of the nucleophile in the way that protic solvents can.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-25T16:02:58.997" LastActivityDate="2013-07-25T16:02:58.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5694" PostTypeId="1" CreationDate="2013-07-26T03:50:34.810" Score="3" ViewCount="85" Body="&lt;p&gt;I'm trying to elucidate current peaks for benzenesulfonamide, using an acetic acid-sodium acetate 0.1M buffer at pH 5.5 as electrolyte. When doing cyclic voltammetry, I get a voltammogram similar to the platinum (Electroanalytical Chemistry and lnterfacial Electrochemistry, 43 (1973) 9-36), because I use platinum as a working electrode. So, I can not see any current peak related to benzenesulfonamide, and I have not found any article related to this. I've been thinking about using a glassy carbon electrode instead of platinum.&lt;/p&gt;&#xA;" OwnerUserId="1111" LastActivityDate="2014-02-02T12:17:48.403" Title="Electrochemical characterization of benzenesulfonamide" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5695" PostTypeId="2" ParentId="5641" CreationDate="2013-07-26T06:35:41.127" Score="6" Body="&lt;p&gt;You're talking about the human body being exposed to outer space and asking about what would happen to it's internal structure. Bones, fats and proteins are within the enclosed system that is the human body, they are not subjected to the low pressures of space. The answer to &lt;code&gt;the pressure at which proteins, fats and bones will break down and become vapour&lt;/code&gt; is not related to the answer to &lt;code&gt;what would happen to them in the human body if the body is exposed to the vacuum of space&lt;/code&gt;. Nevertheless, I'll try to provide both answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the second question (short-term) &lt;a href=&quot;http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970603.html&quot;&gt;can be found on NASA's website at &quot;ask an astrophysicist&quot;&lt;/a&gt;. It's actually quite surprising for someone used to sci-fi movies:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You do not explode. Your blood does not boil. You do not freeze. You do not instantly lose consciousness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question you linked talks about &quot;decay&quot;, or the effects of long time exposure. That's hard to predict, but at some point the body would be attracted by a planet and collide and you wouldn't have anything to boil anymore. Now to the chemistry part of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't much (if any) data on the boiling point or vapor pressure of the compounds you're asking - it's hard to reproduce and it's more a curiosity than anything else. So let's talk about what would happen if you have fat, bones and proteins &lt;strong&gt;directly exposed&lt;/strong&gt; to outer space pressures and temperatures, which seems to be your point anyway. Temperatures in space go from -270°C (in outer space, unexposed to sunlight) to +120°C (near the earth, exposed to sunlight). Since you're asking about boiling point let's talk about what will happen at the highest temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the easier part - the bones. It is safe to assume that the inorganic components will be tougher than the organic ones. The inorganic part of bones is mostly a specific type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Apatite&quot;&gt;apatite&lt;/a&gt;, which is a rock. It can be found in moon rocks. It will not boil. So at least the inorganic part of the bone would remain somewhat intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fat is made of &lt;a href=&quot;https://en.wikipedia.org/wiki/Triglyceride&quot;&gt;triglycerides&lt;/a&gt;. It'll probably break down to glycerol and the correspondent acid. &lt;a href=&quot;http://biodiesel.infopop.cc/eve/forums/a/tpc/f/739605551/m/111106862?r=357109862#357109862&quot;&gt;Then it would decompose and boil&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance.&lt;/p&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2013-07-26T06:35:41.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5696" PostTypeId="1" AcceptedAnswerId="5697" CreationDate="2013-07-26T07:28:44.310" Score="-7" ViewCount="15546" Body="&lt;p&gt;A metal oxide contains $60$% metal.The equivalent weight of metal is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) $12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) $60$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) $40$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) $24$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the equivalent weight of metal oxide = equivalent weight of metal + 8.But how can i arrive at the answer with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1855" LastActivityDate="2018-01-29T12:59:05.627" Title="Equivalent weight of metal" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2013-07-26T09:41:35.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="5697" PostTypeId="2" ParentId="5696" CreationDate="2013-07-26T08:06:33.693" Score="1" Body="&lt;p&gt;Equivalent mass of a substance(element or compound) is defined as the number of parts by mass of the substance which combine or displace directly or indirectly $1.008$ parts by mass of hydrogen or $8$ parts by mass of oxygen or $35.5$ parts by mass of chlorine or $108$ parts by mass of silver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalent mass is a pure number.When a known mass if element is changed into oxide directly or indirectly , the mass of oxide is noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MASS OF OXYGEN= Mass of oxide-mass of element&#xA;Thus, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Eq. Mass = \frac { Mass of element * 8} {Mass of oxide} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here in this question you need ti assume the mass of oxide to be $100g$.&#xA;So, the mass of metal is $60$ and mass of oxygen is $40$.&#xA;Substituting these values in the above formula you get $\frac {60*8} {40}$$ = 12$.&#xA;Hence the ans is $12$.&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorUserId="58254" LastEditDate="2018-01-29T12:59:05.627" LastActivityDate="2018-01-29T12:59:05.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5699" PostTypeId="1" AcceptedAnswerId="5700" CreationDate="2013-07-26T11:46:25.667" Score="8" ViewCount="638" Body="&lt;p&gt;How did chemists find how many electrons, neutrons, and protons each element had?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I can understand the Mendeleev's table but without really knowing how you make one...&lt;/p&gt;&#xA;" OwnerUserId="2041" LastEditorUserId="1573" LastEditDate="2013-07-26T12:30:46.587" LastActivityDate="2013-07-26T12:30:46.587" Title="How chemists find out the electronic properties of elements (atomic numbers)" Tags="&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5700" PostTypeId="2" ParentId="5699" CreationDate="2013-07-26T12:07:23.927" Score="8" Body="&lt;p&gt;One could do it in the following way.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The atomic number $Z$ can be determined by examining the x-ray spectrum of an atom.&#xA;The frequency $f$ of the so-called &lt;em&gt;K-alpha line&lt;/em&gt; is related to the atomic number $Z$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Moseley%27s_law&quot;&gt;Moseley's law&lt;/a&gt;&#xA;$$&#xA;    \sqrt f = k_1 \cdot \left(Z - k_2\right) \, .&#xA;$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The atom is electrically neutral, so the number of electrons is obviously calculated from the atomic number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And the number of neutrons is also simply to choose: the only requirement at this stage would be the correct atomic weight for the element in question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastActivityDate="2013-07-26T12:07:23.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5701" PostTypeId="1" AcceptedAnswerId="5705" CreationDate="2013-07-26T16:59:57.630" Score="6" ViewCount="11792" Body="&lt;p&gt;I have just recently learnt the theory of fractional and normal distillations and the basics of maximum and minimum boiling azeotropes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most books say that the components of the mixture forming an azeotrope cannot be separated by distillation because on heating, the residue (in case of negative or maximum boiling azeotropes) and the distillate (in case of positive or minimum boiling azeotropes) is closer in composition to the azeotropic composition and always approaches it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being understood, we at least separate one component from the mixture as the residue (for positive azeotropes) and the distillate(for negative azeotropes) is richer in one component than the other since the remaining part approaches the azeotropic concentration thereby increasing the concentration of one of the components depending on which side of the azeotropic &quot;point&quot; are we on the graph of boiling temperature and composition.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, for any mixture, we can at least theoretically obtain one purified component (provided we have enough quantities of the mixture)??&lt;br&gt;&#xA;Reference &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;nofollow&quot;&gt;azeotrope basic&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="418" LastEditDate="2014-03-20T14:06:49.377" LastActivityDate="2014-03-20T14:31:02.780" Title="Azeotropes and separation by distillation?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;purification&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5702" PostTypeId="1" CreationDate="2013-07-26T17:06:32.777" Score="3" ViewCount="823" Body="&lt;p&gt;I realise that tear gas can be several different things (Pepper spray / CS Gas / CR gas, PAVA, etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good way to detect some or any of these in the air?&#xA;Ideally using an electronic sensor that can be reused?&lt;/p&gt;&#xA;" OwnerUserId="2042" LastEditorUserId="1573" LastEditDate="2013-07-26T19:10:02.183" LastActivityDate="2014-01-30T18:08:09.693" Title="What is the best way to detect the presence of tear gas?" Tags="&lt;analytical-chemistry&gt;&lt;toxicity&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5703" PostTypeId="2" ParentId="5610" CreationDate="2013-07-26T17:42:07.630" Score="3" Body="&lt;p&gt;This relation is not valid if the gases are assumed to be ideal. see &lt;a href=&quot;http://en.wikipedia.org/wiki/Amagat%27s_law&quot; rel=&quot;nofollow&quot;&gt;Amagat's law of partial volumes&lt;/a&gt;&#xA;Moreover, if the gases are non ideal, the relations between the volumes of the combining gases and the final resultant volume is very complex and the given relation is too simple and includes too less variables to even begin describing the actual non-ideal gas mixing.&lt;br&gt;&#xA;It, however unlikely, might represent a certain special case that can only be judged from the context wherein the aforementioned equation was taken from.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-07-26T17:42:07.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5704" PostTypeId="2" ParentId="5610" CreationDate="2013-07-26T20:43:44.327" Score="3" Body="&lt;p&gt;The relation appears to be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ pV = nRT = \frac{m}{M} RT $$&#xA;This can be expressed as:&#xA;$$ VM = m \left(\frac{RT}{p}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for mixture of two ideal gases at same pressure $(p)$ and temperature $(T)$:&#xA;$$ m_\text{mix} = m_1 + m_2 \tag{mass conservation}$$&#xA;$$ m_\text{mix} \left(\frac{RT}{p}\right) = m_1 \left(\frac{RT}{p}\right) + m_2 \left(\frac{RT}{p}\right) $$&#xA;$$ V_\text{mix} M_\text{mix} = V_1 M_1 + V_2 M_2 $$&#xA;$$ V_\text{mix} = \frac{V_1 M_1 + V_2 M_2}{M_\text{mix}} \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not same as the mentioned relation,&#xA;$$ V_\text{mix} = \frac {M_1V_1 + M_2V_2}{M_1+M_2} \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because this require $ M_\text{mix} = M_1 + M_2 $. Which is not correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, imagine mixting two gases such that $ M_1 = M_2 $ and $ V_1 = V_2 $, then equation (2) gives $ V_\text{mix} = V_1 = V_2 $ instead of the correct final volume of $ V_\text{mix} = 2 V_1 = 2 V_2 $ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think the relation should have read:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M_\text{mix} = \frac { M_1 V_1 + M_2 V_2 }{ V_1 + V_2 } $$&#xA;which follows from equation (1) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Amagat%27s_law&quot; rel=&quot;nofollow&quot;&gt;Amagat's law&lt;/a&gt; which states $V_\text{mix} = V_1 + V_2$.&lt;/p&gt;&#xA;" OwnerUserId="655" LastActivityDate="2013-07-26T20:43:44.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5705" PostTypeId="2" ParentId="5701" CreationDate="2013-07-26T21:10:21.017" Score="5" Body="&lt;p&gt;You seem to have the right idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate, take the specific example of the positive azeotrope of (roughly) 96% ethanol and 4% water. If you begin with a mixture containing less than 96% ethanol, distillation will result in a distillate more abundant in ethanol and nearer to the azeotropic composition. That implies the residue will necessarily be proportionally lower in ethanol. If you continuously repeat the distillation procedure using the remaining pot residue, it will approach 100% water. On the other hand, if you begin with a concentration above 96% ethanol, the resulting distillate will actually be lower in ethanol (and therefore nearer to the azeotropic point), while the residue will actually become more concentrated in ethanol. Hence, in theory, with many repeated distillations you can approach purity in the residue when dealing with positive azeotropes, and the ultimate composition of the residue will depend on which side of the curve you're on to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with negative azeotropes, the situation is analogous but the particulars are basically reversed. Take the example of hydrofluoric acid and water, which form an azeotrope at approximately 37% HF. If you initially begin distilling a mixture comprising less than 37% HF, you'll obtain a distillate more dilute in HF, while the residue will contain more HF and be nearer to the azeotrope. If you were to collect the distillate and repeat the distillation, you would again end up with an even more dilute distillate, with the concentration of the distillate approaching 100% pure water. If you begin distilling a mixture containing more than 37% HF, then the residue will end up being more dilute as it approaches the azeotrope, while the distillate will actually be richer in HF. If you repeat this process many times over, you can eventually obtain a distillate approaching 100% pure HF (in theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, you're correct to suggest that one of the components can be isolated with something approaching complete purity from an azeotropic mixture, assuming a sufficient volume to enable an adequate number of simple distillation cycles or theoretical plates in fractional distillation. Of course, the problem is that you really have no choice in determining which of the two components you can isolate, as that depends entirely on the starting composition of the mixture. In practice, you can often separate azeotropeic mixtures by modifying the distillation process in various ways.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-26T21:16:32.397" LastActivityDate="2013-07-26T21:16:32.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5706" PostTypeId="1" AcceptedAnswerId="5707" CreationDate="2013-07-27T02:42:41.887" Score="2" ViewCount="3229" Body="&lt;p&gt;If we take a test-tube of bromine and invert an upside-down test tube of air on top of it, it'll diffuse with the air, until the bromine is thoroughly 'mixed' with the air in both test tubes. But why doesn't the same thing happen with two immiscible liquids? Say oil and water. &#xA;When you put a layer of oil on top of water, the two don't mix, but they should. By diffusion. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="189" LastEditDate="2013-07-27T12:18:30.913" LastActivityDate="2013-07-27T21:59:53.503" Title="Why doesn't diffusion take place between immiscible liquids?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5707" PostTypeId="2" ParentId="5706" CreationDate="2013-07-27T21:59:53.503" Score="4" Body="&lt;p&gt;In gaseous phases, we generally assume that there's very little or no bonding between individual molecules (or atoms in, say, a noble gas). This isn't &lt;em&gt;strictly&lt;/em&gt; the case, but it's a pretty good approximation, and many good approximate techniques rely on it. A gas that behaves entirely in this way is known as an 'ideal gas'. (See also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, liquids have bonding between individual molecules, and their behaviour is strongly affected by molecular properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the examples you gave:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Bromine and air&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, imagine that you have a bunch of ping pong balls bouncing around at a reasonably high speed. (Ping pong balls are often used as examples for ideal gases.) Since the balls are already travelling at high speed (given a sensible temperature), and they only get in each other's way by bouncing off each other, it's relatively easy to get the bromine and the air 'balls' to mix, and there's really nothing apart from gravity to make them separate again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Water and oil&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water molecules form pretty good bonds to other water molecules. (See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;hydrogen bond&lt;/a&gt;.) Bonding in oils is generally weaker, but also occurs via a very different mechanism to the bonds in water. (See: &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;Van der Waals force&lt;/a&gt;.) This means that oil does not form good bonds to water, and any water to diffuse into the bulk oil would have to break the bonds to all the neighbouring water molecules, and any oil to diffuse into the bulk water would have to separate a few water molecules to make a space to do so. Not just that, but since the water and oil molecules can reform their bonds if they manage to connect to other water and oil molecules, if you mix them up, the droplets will tend to recombine into the separate phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison to this example, let's check out two liquids that &lt;em&gt;do&lt;/em&gt; mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Water and ethanol&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the water, we have basically the same situation as before -- water molecules forming good bonds to each other. The ethanol, though, has an -OH group that can form bonds to the water in the same way that the water does (though not as well). This means that ethanol that mixes with water (and vice versa) will tend to stay mixed, and given that the liquids are being mixed around just by random motions, means that you'll get one mixing with the other just as a matter of statistics.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-07-27T21:59:53.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5708" PostTypeId="1" AcceptedAnswerId="5709" CreationDate="2013-07-28T01:50:02.413" Score="14" ViewCount="36550" Body="&lt;p&gt;It's there in every high-school textbook: $\ce{O2}$ is the supporter of combustion. Without $\ce{O2}$ combustion cannot take place. Why? And why only $\ce{O2}$? Why not some other element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, what happens when a combustible gas burns in air? Say $\ce{H2}$? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="7475" LastEditDate="2015-09-19T17:57:25.083" LastActivityDate="2017-06-10T12:52:39.907" Title="Why is O2 the supporter of combustion?" Tags="&lt;combustion&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5709" PostTypeId="2" ParentId="5708" CreationDate="2013-07-28T02:09:14.263" Score="21" Body="&lt;p&gt;Oxygen is not strictly needed, it just happens to be very, very good at what it does!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, combustion requires three things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;Fuel&lt;/strong&gt;: The thing that burns. This is is often a hydrocarbon, or other organic molecule. The simplest possible fuel is pure hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;Energy&lt;/strong&gt;: What gets the reaction started, which is true of most chemical reactions.  Note that this so-called &lt;em&gt;activation energy&lt;/em&gt; is usually much less than the energy ultimately released from combustion.  This is like rolling a boulder some distance in order to roll it down a hill, which releases much potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;strong&gt;Oxidizer&lt;/strong&gt;: The molecule that accepts electrons.  It turns out that combustion requires the fuel to be &lt;em&gt;oxidized&lt;/em&gt;, that is, it donates electrons.  So we need something to accept the electrons, and that's the oxidizer, which is then &lt;em&gt;reduced&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is a great oxidizer because it is so &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot;&gt;electronegative&lt;/a&gt;, which means it really wants to accept electrons.  Only fluorine is more electronegative, and fluorine is a superb oxidizer: blow fluorine gas at nearly any substance and it bursts into flames.  For example, see this video: &lt;a href=&quot;http://www.periodicvideos.com/videos/009.htm&quot;&gt;Fluorine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is in many ways a perfect oxidizer for supporting life: it doesn't set things on fire like fluorine, but otherwise can oxidize very many things.  When you eat, oxygen is oxidizing the food (fuel) you ingested, to generate energy you need to live.  So there is &quot;combustion&quot; going on inside you!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-07-28T02:09:14.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5710" PostTypeId="1" AcceptedAnswerId="5711" CreationDate="2013-07-28T03:18:40.847" Score="8" ViewCount="14388" Body="&lt;p&gt;The reaction of sodium with water is quite infamous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="73" LastEditDate="2013-11-28T05:50:58.493" LastActivityDate="2020-05-07T16:52:57.687" Title="Why does sodium react with water?" Tags="&lt;redox&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5711" PostTypeId="2" ParentId="5710" CreationDate="2013-07-28T04:45:41.557" Score="11" Body="&lt;p&gt;Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorUserId="5026" LastEditDate="2018-05-14T16:04:43.953" LastActivityDate="2018-05-14T16:04:43.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5712" PostTypeId="2" ParentId="4031" CreationDate="2013-07-28T09:05:48.783" Score="2" Body="&lt;p&gt;As Paul J. Gans correctly pointed out in his comment, in general it is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice you can calculate reaction rates only for quite restrictive class of reactions - the &lt;em&gt;unimolecular gas-phase&lt;/em&gt; reactions - using &lt;em&gt;RRKM theory&lt;/em&gt;.&#xA;The whole potential energy surface is actually not needed, as far as I know, you just need to identify reactant, product, and transition state.&#xA;RRKM is based on few assumptions, with the first one being quite obvious: since we are talking about potential energy surfaces, the electronic motion and the nuclear motion is separated by the Born-Oppenheimer approximation.&#xA;The nuclear motion is also assumed to be well separated into rotational, vibrational, and translational one. The internal degrees of freedom are treated classically with quantum mechanical corrections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RRKM is quite useful for predicting gas phase dissociation and isomerization rates  under low pressure (low enough so that reaction can be safely assumed to be unimolecular). For that reason, this theory is ofthen used in mass-spectrometry, since the above mentioned conditions are quite often met here, but it is of no use for reactions in the condensed phase.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2013-07-28T09:05:48.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5713" PostTypeId="1" CreationDate="2013-07-28T10:33:36.027" Score="0" ViewCount="732" Body="&lt;p&gt;I'm going to study the thermal expansion of graphite. For this goal, I should change the temperature continuously. But I do not know how to code it by &lt;a href=&quot;http://lammps.sandia.gov&quot; rel=&quot;nofollow&quot;&gt;Lammps&lt;/a&gt;? &lt;/p&gt;&#xA;" OwnerUserId="2048" LastEditorUserId="73" LastEditDate="2013-07-28T20:24:13.507" LastActivityDate="2017-11-11T16:40:31.873" Title="How would I write an input Lammps command for continuous change in temperature?" Tags="&lt;physical-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2017-11-12T02:53:43.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="5714" PostTypeId="2" ParentId="5713" CreationDate="2013-07-28T11:30:14.227" Score="3" Body="&lt;p&gt;&lt;em&gt;(Disclosure: I'm not a LAMMPS user, but I've used other MD packages.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll probably want to use one of the thermostat settings such as the Nose-Hoover type described &lt;a href=&quot;http://lammps.sandia.gov/doc/fix_nh.html&quot; rel=&quot;nofollow&quot;&gt;here in the LAMMPS documentation&lt;/a&gt;, but there's a &lt;a href=&quot;http://lammps.sandia.gov/doc/Section_howto.html#howto_16&quot; rel=&quot;nofollow&quot;&gt;more general HowTo guide on temperature control and calculation&lt;/a&gt; in the docs as well that looks pretty useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would &lt;em&gt;strongly&lt;/em&gt; recommend reading the sections of the manual concerning any given thermostat type you're going to use fully before running serious calculations with it, though, since there are some things to look out for, such as the tendency for setting the thermostat interval too low to cause large oscillations in effective temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nose-Hoover type (and some of the others, but I only took a quick look) also lets you step up or down the temperature with a given interval. Depending on exactly what type of thermal expansion you're looking at, I'd just like to note: you really need to set the stepping interval appropriately to let the structure equilibrate at each temperature you want to look at, otherwise your results won't be meaningful. (Or, at least, won't have the meaning that you might want.) Relatedly, I'd also suggest that if you weren't planning to already, you run each simulation both ways: one stepping the temperature up, and one bringing the temperature down. If the behaviour matches at the temperature steps in each, you can be relatively confident that you've given the structure enough time at each temperature.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-07-28T11:30:14.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5716" PostTypeId="1" CreationDate="2013-07-28T22:02:06.470" Score="5" ViewCount="17539" Body="&lt;p&gt;How do you determine concentration of a sample using wavelength and absorbency? Say our wavelength for a sample of Red 40 is 508.50 nm and absorbance is 0.283 when we use 6ml water and 4 ml Red 40. What would be our concentration and the slope ε? This is really frustrating me so any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2053" LastEditorUserId="73" LastEditDate="2013-11-27T02:51:38.697" LastActivityDate="2013-11-27T02:51:38.697" Title="Determine concentration Using Beer's law" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5718" PostTypeId="1" CreationDate="2013-07-29T06:37:25.173" Score="5" ViewCount="159" Body="&lt;p&gt;How can you accurately qualitatively and quantitatively determine the HCN in a volatile product from water?&lt;/p&gt;&#xA;" OwnerUserId="2055" LastEditorUserId="7448" LastEditDate="2015-10-02T16:26:09.020" LastActivityDate="2015-10-02T16:26:09.020" Title="How to estimate HCN content in a volatile mixture?" Tags="&lt;organic-chemistry&gt;&lt;solutions&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5719" PostTypeId="2" ParentId="5128" CreationDate="2013-07-29T07:16:06.883" Score="5" Body="&lt;p&gt;$\ce{Na_2CO_3}$ forms several hydrates with decomposition point close to room temperature (30-40 Celsium). Moreover, the solubility of decahydrate decreases with temperature decrease, and solubility of anhydrous salt almost does not depend on temperature and is much higher than solubility of decahydrate.  The relevant pictures are googleable by words 'sodium carbonate solubility curve', see 'images' google service, not general search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I believe you have prepared a solution of the salt at the higher temperature at day, but the salt mostly precipitated in form of decahydrate overnight, because the solution cooled significantly at night. Bonus points for this version if you have significant changes in temperature at your work during the nychthemeron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second version is that you indeed prepared seemingly oversaturated solution because it is not saturated for anhydrous $\ce{Na_2CO_3}$, but overnight it crystallized, because it is saturated for decahydrate. I'm a bit sceptical about such possibility, though, because significantly oversaturated solution crystallize even on slightest imperfections, and the still not-dissolved salt is a great collection of such imperfections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, if the solution stayed overnight &lt;em&gt;open&lt;/em&gt;, it indeed may consumed some amount of atmospheric $\ce{CO_2}$ and partially transform into sodium hydrocarbonate with much lesser solubility.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4231" LastEditDate="2015-03-09T15:29:15.430" LastActivityDate="2015-03-09T15:29:15.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5720" PostTypeId="2" ParentId="5716" CreationDate="2013-07-29T09:50:53.667" Score="5" Body="&lt;p&gt;I don't know anything about Red 40, but you will need the extinction coefficient.  Perhaps you determined this from a plot of concentration &lt;em&gt;vs&lt;/em&gt; absorbance in the lab?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to attempt to answer you question with reference to &lt;a href=&quot;http://en.wikipedia.org/wiki/Nicotinamide_adenine_dinucleotide&quot; rel=&quot;noreferrer&quot;&gt;NADH&lt;/a&gt;, which has an accepted extinction coefficient of 6220 M&lt;sup&gt;-1&lt;/sup&gt;.cm&lt;sup&gt;-1&lt;/sup&gt; at 340 nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a solution of NADH in a test-tube, I take 4 ml of this and dilute with water to 10 ml.  I then take this diluted solution and measure its absorbance at 340 nm in a spectrophotometer using a cuvette with a path-length of 1 cm.  The value I obtain is 0.4.&#xA;(Edit.  In the above measurement, I take care to 'blank' the spectrophotometer against water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the concentration of NADH in the original solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Beer-Lambert law (where c is concentration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.4 = 6220 x c x 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = 0.64 x 10&lt;sup&gt;-6&lt;/sup&gt; M  = 0.64 &lt;em&gt;&amp;micro;&lt;/em&gt; M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the solution on which the measurement was made was&#xA;diluted 4 in 10 or 1 in 2.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the concentration of NADH in the original solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = 2.5 x 0.64 x 10&lt;sup&gt;-6&lt;/sup&gt; M = 160.77 x 10&lt;sup&gt;-6&lt;/sup&gt; M = 160.77 &lt;em&gt;&amp;micro;&lt;/em&gt; M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the concentration of NADH in the original solution is about 161 &lt;em&gt;&amp;micro;&lt;/em&gt; M (micro-Molar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am making a number of assumptions here, the main one being that the spectrophotometer obeys the Beer-Lambert law in the absorbance region of interest.  You should determine that this is so by measuring  the absorbance of series of NADH solutions and make sure that the resulting graph of absorbance vs concentration &lt;strong&gt;is linear and through the origin&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very least, doubling the concentration of NADH should exactly double the absorbance (0.8) and halving it should give exactly half the absorbance (0.2).  In addition, measurement of water (or whatever you are 'blanking against) should give zero - that is the spectrophotometer should be properly 'blanked'.  If not you will need to subtract this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens if we do not dilute our original sample, but just measure the absorbance at 340nm directly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Beer-Lambert&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = 6220 x 160.77 x 10&lt;sup&gt;-6&lt;/sup&gt; = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original solution is predicted to have an absorbance of 1, but IMO even with modern spectrophotometers you are 'pushing your luck' to assume that Beer-Lambert is obeyed at such a high absorbance.  A 'rule-of-thumb' is to keep the absorbance below 0.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final comment.  It is very difficult to accurately determine an extinction coefficient from absorbance versus concentration plots, unless you accurately know the concentration of the original material.  If this figure relies on a 'dry-weight' measurement, you will need to be sure that the compound is pure, for a start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading your question again, I think we need a bit more experimental detail, and if you could supply it, I'll have another go, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NADH is widely used in the biochemical sciences where (among many other things) the relatively high extinction coefficient of 6220 M&lt;sup&gt;-1&lt;/sup&gt;.cm&lt;sup&gt;-1&lt;/sup&gt; is exploited in continuous enzyme assays.  If an enzyme produces or consumes NADH (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Nicotinamide_adenine_dinucleotide_phosphate&quot; rel=&quot;noreferrer&quot;&gt;NADPH&lt;/a&gt;) then it is (usually) very easy to design an assay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would appear that the accepted extintion coefficient (Molar Absorption Coefficient) for &lt;a href=&quot;http://en.wikipedia.org/wiki/Allura_Red_AC&quot; rel=&quot;noreferrer&quot;&gt;Allura Red AC&lt;/a&gt; (Red 40) is about 25 000 M&lt;sup&gt;-1&lt;/sup&gt;.cm&lt;sup&gt;-1&lt;/sup&gt; (or 25 000 mol&lt;sup&gt;-1&lt;/sup&gt; L cm&lt;sup&gt;-1&lt;/sup&gt;) at about 505 nm. The best reference I can get it &lt;a href=&quot;http://books.google.ie/books?id=E5SBRSKR53EC&amp;amp;pg=PA223&amp;amp;lpg=PA223&amp;amp;dq=Allura+Red+AC+Molar+absorption+coefficient&amp;amp;source=bl&amp;amp;ots=bekWjzOMn7&amp;amp;sig=8j98De6NIG34XZKvgieG0sHIVq8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=_nf2UbPrMIiR7Aaq2IHoDg&amp;amp;ved=0CCwQ6AEwADgK#v=onepage&amp;amp;q=Allura%20Red%20AC%20Molar%20absorption%20coefficient&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Azo dyes tend to have very high extinction coeffficients, so this 'looks about right'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diluted solution has an absorbance of 0.283&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the above as the extinction coefficient, and in addition assuming the measurement was made in a cuvette with a path length of 1 cm (l =1), one can calculate from Beer-Lambert that the concentration of Allura Red in the diluted solution is about 11.32 x 10&lt;sup&gt;-6&lt;/sup&gt; M.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dilution factor is 4 in 10, or 1 in 2.5.  Therefore, the concentration of the original solution is about 28.3 x 10&lt;sup&gt;-6&lt;/sup&gt; M. &lt;/p&gt;&#xA;" OwnerUserId="1791" LastEditorUserId="1791" LastEditDate="2013-07-29T15:37:53.560" LastActivityDate="2013-07-29T15:37:53.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5721" PostTypeId="2" ParentId="4517" CreationDate="2013-07-29T14:13:48.040" Score="5" Body="&lt;p&gt;There is no such thing as conservation of orbital energies at different fixed values of $R$, the internuclear distance. Obviously, at $R \rightarrow \infty$ the energies are the same regardless of the linear combinations you use (there is no interaction and the &quot;molecular&quot; energy is exactly the same as the sum of the &quot;atomic&quot; energies); obviously too, the energies raise a lot if you make $R$ very small: you have the fusion barrier that doesn't allow you to convert your initial atomic orbitals of atoms A and B into the atomic orbitals of atom AB. Energy is obviously conserved in a dynamical process where two atoms approach and you have kinetic energy in the collisional coordinate and so on and so forth (you can immediately see that two atoms cannot form a diatomic molecule is something else does not take away energy from the system).&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2013-07-29T14:13:48.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5722" PostTypeId="1" AcceptedAnswerId="5723" CreationDate="2013-07-29T23:49:52.987" Score="3" ViewCount="744" Body="&lt;p&gt;I'm not sure whether the caffeine contained in otc caffeine pills is made synthetically or isolated from coffee and other plants, but if the latter is the case, is it likely that it contains the chemical harmane?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's claimed that isolated caffeine is 99% pure and I'm wondering whether the remaining 1% is likely to contain any appreciable amount of harmane.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="73" LastEditDate="2013-07-30T05:09:18.647" LastActivityDate="2013-07-30T05:09:18.647" Title="Does caffeine isolated from coffee contain harmane?" Tags="&lt;purification&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5723" PostTypeId="2" ParentId="5722" CreationDate="2013-07-30T02:13:29.700" Score="3" Body="&lt;p&gt;Given that caffeine can be readily extracted from a variety of abundant and exceedingly cheap natural sources, I doubt it would be economical to synthesize. Various decaffeinated products are popular, and the caffeine removed during the decaffeination process is normally recovered, purified, and re-sold for use in various products, which makes the synthetic route even less economically attractive. As far as I'm aware, the vast majority of caffeine sold commercially is extracted from plant sources. You can do your own research to determine the exact statistics, but since your concern relates to caffeine extracted from natural sources that might also contain harmala alkaloids, I'll address that case specifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different methods of caffeine extraction (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Decaffeination&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on decaffeination&lt;/a&gt;), but all use solvents and/or adsorbents at some stage in the separation and purification processes. From the perspective of chemistry, then, the question is how similar are the solubilities of caffeine and harmala alkaloids, and what percentage by mass do they constitute of the natural caffeine sources used? I was able to find &lt;a href=&quot;http://www.elsevierdirect.com/brochures/conap2/PDFs/Vol3_Chemistry_of_Coffee_lr.pdf&quot; rel=&quot;nofollow&quot;&gt;one article&lt;/a&gt; that quantified the chemical constituents of coffee beans. If you examine the data table therein, you'll see that the main constituents of those beans are carbohydrates, lipids, amino acids, and chlorogenic acids, not one of which is a category that applies to the harmala alkaloids. Lipids, carbohydrates, and amino acids are generally the main constituents of virtually all plants, so I think it's probably safe to extend those statistics to other natural sources besides coffee beans. Purely as a matter of probability, it's reasonable to assume the impurities are likely to consist mostly of substances from those categories. With respect to solubility, the harmala alkaloids are mostly somewhat heavier nitrogenous bases which are less polar than the caffeine molecule (at least in their neutral, &quot;free base&quot; form). Given that, I'd expect that methods using water extraction, which appear to be popular, would separate caffeine from harmala alkaloids fairly efficiently (I found &lt;a href=&quot;http://www.ajur.uni.edu/v8n2-3/Brobst%20et%20al%20pp%201-4.pdf&quot; rel=&quot;nofollow&quot;&gt;one paper&lt;/a&gt; indicating that harmala alkaloids are water insoluble, while caffeine is anywhere from moderately to highly soluble depending on temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impurity of 1% is already quite small, and that 1% comprises a variety of compounds, both organic and inorganic (the overwhelming bulk of which are likely not harmala alkaloids, given the natural composition of caffeine-containing plant sources I described above). Based on the above information, I'd suspect that only a tiny fraction of those impurities, if any, would be likely to consist of harmala alkaloids at a detectable level.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-30T03:06:48.020" LastActivityDate="2013-07-30T03:06:48.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5724" PostTypeId="1" AcceptedAnswerId="5725" CreationDate="2013-07-30T04:27:12.140" Score="3" ViewCount="382" Body="&lt;p&gt;So say I have a reaction where there are x moles of gas on the reactant side, and y moles of gas on the product side, where x&gt;y. The volume decreases, favoring the products side. The textbook I am using says that the equilibrium concentration of the reactants would decrease, but wouldn't it end up increasing due to the lowered volume?&lt;/p&gt;&#xA;" OwnerUserId="2063" LastEditorUserId="305" LastEditDate="2013-07-30T08:23:38.817" LastActivityDate="2017-04-25T20:26:05.353" Title="Would concentration of a compound in equilibrium increase even if it's not favored by a volume decrease?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5725" PostTypeId="2" ParentId="5724" CreationDate="2013-07-30T06:55:16.570" Score="3" Body="&lt;p&gt;Suppose you have the following generic reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{xA_{(g)} &amp;lt;=&amp;gt; yB_{(g)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the condition, $x &amp;gt; y$. The reaction quotient, written in terms of concentration would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = \frac{(\frac{n_B}{V})^y}{(\frac{n_A}{V})^x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $n_A$ and $n_B$ are moles of the respective species and $V$ is volume. With some algebraic manipulation, this can be rewritten as follows (note here I'm explicitly treating $Q$ as dimensionless, q.v., the relevant &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_constant#Units&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on equilibrium constant units&lt;/a&gt;, which applies to the reaction quotient as well):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = \frac{n_B^y}{n_A^x} \cdot V^{x-y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Written in that form, it should be apparent that, if $x &amp;gt; y$, then $x - y &amp;gt; 0$, and therefore $Q$, the ratio of products to reactants, varies directly with $V$. In other words, as volume grows, $Q$ grows, and as volume decreases, $Q$ decreases. Suppose that, prior to some volume change, the reaction is at equilibrium, i.e., $Q = K$. If volume instantaneously begins to increase, $Q$ becomes instantaneously greater than $K$, and therefore the rate of the reverse reaction becomes greater than the rate of the forward reaction, favoring generation of $A$ until equilibrium is reestablished. Conversely, if volume instantaneously begins to decrease, the opposite happens and production of $B$ is favored. This is precisely what one would expect based on an intuitive understanding of Le Chatelier's principle: if volume decreases, equilibrium shifts to favor production of fewer moles of gas to counteract the increasing pressure; if volume increases, equilibrium shifts to favor production of more moles of gas to counteract the decrease in pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I misread part of your question initially. In short, if the volume decreases, generation of product $B$ becomes favored, so the concentration of reactant $A$ must decrease relative to $B$ (since $x$ moles of $A$ are consumed in the production of $y$ moles of $B$, and the reaction quotient approaches the equilibrium constant over time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to clarify this further, it's true that, in an absolute sense, if volume decreases in an instantaneous fashion, then the concentrations of the reactant and product, considered separately, would both have to increase. However, in &lt;em&gt;relative&lt;/em&gt; terms, the concentration of $B$ must increase &lt;em&gt;relative&lt;/em&gt; to $A$ as the reaction proceeds. The ratio between the two must grow to compensate for the diminished volume, otherwise $Q$ will remain perpetually smaller than $K$, when in fact $Q$ always converges to $K$, for a given set of conditions, over time.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-30T08:15:00.437" LastActivityDate="2013-07-30T08:15:00.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5726" PostTypeId="1" AcceptedAnswerId="9484" CreationDate="2013-07-30T16:27:03.400" Score="8" ViewCount="436" Body="&lt;p&gt;I am running experiments using reagents preserved with 0.1% sodium azide, which are applied to a gold film.  Unfortunately, there are very few options for using reagents without sodium azide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that sodium azide reacts with gold to form heavy metal azides, which are explosive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How long will it take for sodium azide to react with gold?&lt;/li&gt;&#xA;&lt;li&gt;Generally, what is going to happen to the gold film?&lt;/li&gt;&#xA;&lt;li&gt;Any other info you can think of&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="4645" LastEditDate="2016-01-21T21:29:32.980" LastActivityDate="2016-01-21T21:29:32.980" Title="Consequences of using reagent preserved with sodium azide on gold film" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;safety&gt;&lt;metal&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5727" PostTypeId="1" CreationDate="2013-07-30T17:45:16.370" Score="4" ViewCount="176" Body="&lt;p&gt;I am interested to know the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which gems can be prepared at home (without sophisticated instruments)?&lt;/li&gt;&#xA;&lt;li&gt;What will be the quality (appearance) of the product for a certain method?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want these information just for hobby-scale purposes(no academic/commercial production). In addition to common lab equipment, I have $\pu{1000 K}$ of temperature and few atmospheres of pressure at disposal.    &lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="41328" LastEditDate="2017-08-06T05:34:34.080" LastActivityDate="2017-08-06T05:34:34.080" Title="Which gems can be prepared on a small scale basis?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5728" PostTypeId="2" ParentId="5727" CreationDate="2013-07-30T21:22:57.130" Score="3" Body="&lt;p&gt;Most likely the only gems you will be able to make at home are &lt;a href=&quot;http://www.youtube.com/watch?v=u-BEya1xsjA&quot; rel=&quot;nofollow&quot;&gt;plastic gems&lt;/a&gt;.&#xA;Wikipedia sheds some light on the production of some synthetic gems, &lt;a href=&quot;http://en.wikipedia.org/wiki/Corundum#Synthetic_corundum&quot; rel=&quot;nofollow&quot;&gt;Corundum&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_zirconia#Synthesis&quot; rel=&quot;nofollow&quot;&gt;Cubic Zirconia&lt;/a&gt;, both of them not possible at home (or at least my home).&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-07-30T21:22:57.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5729" PostTypeId="1" CreationDate="2013-07-31T03:55:44.927" Score="1" ViewCount="307" Body="&lt;p&gt;I use a silver based disinfectant in my hot tub. It is in the form of silver nitrate beads. There is chlorine in the tub that I presume becomes chloride ion. Why doesn't the silver form a precipitate of silver chloride in the water?&lt;/p&gt;&#xA;" OwnerUserId="658" LastEditorUserId="2095" LastEditDate="2013-08-09T07:46:08.557" LastActivityDate="2013-08-09T07:46:08.557" Title="How can silver work as a disinfectant in a bathtub if precipitates in the presence of chloride?" Tags="&lt;everyday-chemistry&gt;&lt;metal&gt;&lt;toxicity&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5730" PostTypeId="1" AcceptedAnswerId="5740" CreationDate="2013-07-31T05:18:09.627" Score="6" ViewCount="3202" Body="&lt;p&gt;My friend recently bought a ruby gem from a store and asked me how to test whether its an original one or a duplicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any ways to test it using only domestic materials?&lt;/p&gt;&#xA;" OwnerUserId="1315" LastEditorUserId="5017" LastEditDate="2015-07-16T22:09:11.633" LastActivityDate="2016-08-26T18:23:05.557" Title="How can one test if a ruby is real?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5731" PostTypeId="2" ParentId="5729" CreationDate="2013-07-31T06:20:30.997" Score="4" Body="&lt;p&gt;If you're referring to chlorination of public water supplies, keep in mind that this is likely not the main source of chloride ion. Elemental chlorine disproportionates via the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2 + H2O &amp;lt;=&amp;gt; HCl + HOCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most chloride ion probably derives from various natural and industrial sources, as well as miscellaneous human activity. &lt;a href=&quot;http://www.hc-sc.gc.ca/ewh-semt/pubs/water-eau/chloride-chlorure/index-eng.php#fnb12-ref&quot; rel=&quot;nofollow&quot;&gt;One source&lt;/a&gt; (somewhat outdated) I found indicates an average chloride ion concentration in U.S. public water of $11.5$ mg/L, which equates to a molarity of $3.24 \cdot 10^{-4}$. Using the $K_{sp}$ of $AgCl$ you can calculate the concentration of silver ion necessary to produce a precipitate, but given that it has a value of $1.77 \cdot 10^{-10}$ under standard conditions, you can assume that a fairly minimal amount of silver would be required, even with the large volume of a tub and increasing solubility in hot water. It's likely the case that any precipitate is just finely dispersed in solution and not clearly visible.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-07-31T06:20:30.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5732" PostTypeId="1" AcceptedAnswerId="5733" CreationDate="2013-07-31T07:58:24.783" Score="1" ViewCount="2146" Body="&lt;p&gt;I just bought a butter dish which is a millimeter or two too small (the problems are that the corners are rounded which they forgot to mention before I ordered it and asked for the dimensions) to fit a piece of butter as it is sold in the supermarket. Since I don't want to return it my idea was to freeze the butter before I put it in this butter dish but for different reasons I can't make a &quot;field test&quot; just now (will update the question when I get the opportunity to do that). Instead I have to fall back to on theoretical answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much does butter shrink when it's cooled from approximately 4°C (39°F) to -18°C (-0.4°F)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically butter has about 1% salt, 82% fat and the rest is water.&lt;/p&gt;&#xA;" OwnerUserId="2069" LastActivityDate="2013-07-31T11:21:49.163" Title="How much does butter shrink when frozen?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5733" PostTypeId="2" ParentId="5732" CreationDate="2013-07-31T10:42:09.400" Score="2" Body="&lt;p&gt;Given that fat (consisting mainly of assorted triglycerides of varying saturation) is the major constituent by a wide margin, it's probably reasonable to compare the change in density w/r/t temperature of butter to the change measured in various vegetable oils, since they have roughly comparable chemical composition (and because extensive data on them actually exists). I found &lt;a href=&quot;http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1016&amp;amp;context=chemeng_biomaterials&quot; rel=&quot;nofollow&quot;&gt;a study&lt;/a&gt; in which the densities of various vegetable oils were measured over a range of temperatures. On average, the increase in density resulting from cooling from 110.0&amp;deg;C down to 23.9&amp;deg;C was roughly 6-7%. The data points graphing density against temperature arrange very linearly, and the regression lines generated fit them exceedingly well. I see no reason to suppose that the trend wouldn't continue as temperature decreases further. Assuming the comparison is reasonably valid, you could extrapolate that the increase in density you'd observe over the temperature range you specified should be on the order of approximately 2% at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the amount of shrinkage, strictly considering volume, the ratio of densities (assuming the 2% approximation is reasonable), would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{D_2}{D_1} = 1.02$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the density after cooling, $D_2$, is 2% greater than $D_1$. That implies, holding mass constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_1}{1.02} = V_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, given that volume is in units of cubic length, you can get a rough idea of how much the length of each side/edge should shrink by taking the cube root of both sides of the equation (obviously, this is only completely accurate for a perfect cube, but it emphasizes that the shrinkage in each dimension is smaller than the shrinkage in volume):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt[3]{V_2} \approx 0.993\sqrt[3]{V_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless your stick of butter is really gargantuan, a decrease of 0.7% on an edge is not going to account for 1-2mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Additionally, I've read that, in general, the magnitude of relative thermal expansion/contraction that liquids experience is usually greater than that experienced by solids, meaning that liquid density varies more over a given temperature range than solid density. Of course, I'll add the caveat: that's a very broad statement that admits of various exceptions. Nevertheless, this suggests that the estimate of 2% I made above by comparison to data on liquid vegetable oils is, if anything, more likely to be too large than too small.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-31T11:21:49.163" LastActivityDate="2013-07-31T11:21:49.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5734" PostTypeId="1" AcceptedAnswerId="5735" CreationDate="2013-07-31T12:26:40.153" Score="12" ViewCount="18754" Body="&lt;p&gt;I really don't understand why metals form metallic bonds. I mean, it makes no sense. It would make much more sense for them to form covalent bonds with themselves and have a 'pseudo-full' outer shell. How does freeing off electrons make them any more stable?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="81" LastEditDate="2014-01-25T15:27:41.020" LastActivityDate="2018-02-24T11:20:13.370" Title="Why don't metals form covalent bonds in bulk metal?" Tags="&lt;bond&gt;&lt;metal&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5735" PostTypeId="2" ParentId="5734" CreationDate="2013-07-31T13:50:56.267" Score="12" Body="&lt;p&gt;This is due to the low ionization energies of the metals. It's easier for them to release few electrons from the outer shell to obtain a noble gas configuration rather than consuming several ones. However, the difference between an ionic and a polar covalent bond is always fringe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's not true that metals don't form covalent bonds at all. I guess you never heard of &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadruple_bond&quot;&gt;quadruple bond&lt;/a&gt; or δ-bond. There are several examples such as &lt;em&gt;$\ce{K2[Re2Cl8]·2H2O}$&lt;/em&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium%28II%29_acetate&quot;&gt;Chromium(II) acetate hydrate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="1878" LastEditDate="2013-07-31T23:20:24.987" LastActivityDate="2013-07-31T23:20:24.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5737" PostTypeId="1" AcceptedAnswerId="5738" CreationDate="2013-07-31T18:58:49.087" Score="110" ViewCount="66058" Body="&lt;p&gt;A friend of mine was looking over the definition of pH, and wondering if it is possible to have a negative pH level. From the equation below, it certainly seems possible—just have a $1.1$ (or something $\gt 1$) molar solution of $\ce{H+}$ ions:&#xA;$$\text{pH} = -\log([\ce{H+}])$$&#xA;(Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$, then $\mathrm{pH} = -\log(1.1) \approx -0.095 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is &lt;em&gt;theoretically&lt;/em&gt; possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?&lt;/p&gt;&#xA;" OwnerUserId="2073" LastEditorUserId="15410" LastEditDate="2016-05-02T19:16:29.813" LastActivityDate="2019-08-10T08:56:45.993" Title="Is a negative pH level physically possible?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="7" CommentCount="4" FavoriteCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="5738" PostTypeId="2" ParentId="5737" CreationDate="2013-07-31T20:03:01.630" Score="122" Body="&lt;p&gt;One publication for you: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed083p1465&quot;&gt;“Negative pH Does Exist”, K. F. Lim, &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;83&lt;/em&gt;, 1465&lt;/a&gt;. Quoting the abstract in full:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article text provides some counterexamples:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastEditorUserId="16683" LastEditDate="2016-05-02T16:04:38.117" LastActivityDate="2016-05-02T16:04:38.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5739" PostTypeId="2" ParentId="5737" CreationDate="2013-07-31T20:31:51.847" Score="24" Body="&lt;p&gt;It is possible to have $\mathrm{pH}&amp;lt;0$ and you don't need to create any substance. Take a concentrated solution of one of the strong inorganic acids (i.e. one with dissociation constant above 1000 like sulfuric acid) and here you are. &lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="402" LastEditDate="2017-04-08T21:59:05.213" LastActivityDate="2017-04-08T21:59:05.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5740" PostTypeId="2" ParentId="5730" CreationDate="2013-07-31T20:45:37.507" Score="10" Body="&lt;p&gt;To test if it's ruby at home check it hardness. If it's harder than 7 even if it's not ruby it still something precious. :-) You can also try to find refraction index however for this you need some equipment. If you really want 100% prove, ask on a local chemistry faculty if they could &quot;as a teaching example&quot; make single crystal diffraction experiment. Simple run with 30 frames would be enough to find a cell what is enough to prove it's ruby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fitting your ruby on diffractometer would be challenge as a new diffractometers are working on crystals with size of 10--300 micrometers. To find if it natural or synthetic you need to go to specialize jeweler. They look for imperfections (natural stone are usually less perfect than synthetic).&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="7448" LastEditDate="2015-07-15T19:17:28.953" LastActivityDate="2015-07-15T19:17:28.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5741" PostTypeId="2" ParentId="5737" CreationDate="2013-07-31T20:47:43.117" Score="93" Body="&lt;p&gt;It's certainly possible theoretically. Solve for $\ce{pH &amp;lt; 0}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{-log[H+] &amp;lt; 0\\&#xA;log[H+] &amp;gt; 0\\&#xA;[H+] &amp;gt; 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you said, a solution in which the hydrogen ion concentration exceeds one should theoretically have a negative $\ce{pH}$. That said, at those extremes of concentration, the utility and accuracy of the $\ce{pH}$ scale breaks down for various reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even acids conventionally categorized as &quot;strong&quot; do not in fact dissociate 100%. In reality, their dissociation is also essentially an equilibrium process, though this only becomes apparent at surpassingly high concentrations. As the solution becomes more concentrated, any additional acid cannot be as thoroughly solvated, and the chemical equilibrium begins to favor dissociation progressively less and less. Hence, as the solution becomes increasingly saturated, the extent of dissociation begins to plateau and the hydrogen ion concentration approaches some practical upper limit. Furthermore, $\ce{pH}$ measured via molar concentration as a proxy for thermodynamic activity is inherently inaccurate at the extremes of concentration. Other phenomena, such as the formation of distinct chemical species by self-ionization in a concentration-dependent manner further complicate things (e.g., generation of $\ce{H3SO4+}$ in concentrated sulfuric acid, $\ce{H2F+}$ in concentrated hydrofluoric acid, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For highly concentrated solutions of strong acids, alternatives/extensions to $\ce{pH}$ exist that are functional beyond the limits of $\ce{pH}$ (see, for example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hammett_acidity_function&quot;&gt;Hammett acidity function&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether solutions of negative $\ce{pH}$ have actually been experimentally prepared or observed, the answer is yes. Here's a link to &lt;a href=&quot;http://www.pnas.org/content/96/7/3455.full.pdf&quot;&gt;one article&lt;/a&gt; describing the measurement of $\ce{pH}$ in acidic mine waters, which cites a figure of $-3.6$.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-07-31T20:52:48.560" LastActivityDate="2013-07-31T20:52:48.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5742" PostTypeId="1" CreationDate="2013-08-02T04:14:33.433" Score="-2" ViewCount="852" Body="&lt;p&gt;Also, can you explain why, as in which solubility rules apply in this case?&lt;/p&gt;&#xA;" OwnerUserId="2077" LastEditorUserId="189" LastEditDate="2013-08-21T15:43:51.773" LastActivityDate="2013-08-21T15:43:51.773" Title="Is Mn(C₂H₃O₂)₂ soluble in water?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-03-28T03:27:06.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="5743" PostTypeId="2" ParentId="5742" CreationDate="2013-08-02T09:06:40.893" Score="4" Body="&lt;p&gt;&lt;strong&gt;Yes it is&lt;/strong&gt;. But have no idea on its &lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ksp}$&lt;/span&gt;&lt;/em&gt; or &lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{g/L}$&lt;/span&gt;&lt;/em&gt; value! If I'm not wrong, it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese%28II%29_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manganese(II) acetate&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;'Which rule applies' is the following :&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;All nitrates, acetates and perchlorates are soluble in water.&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Possibility of &lt;strong&gt;H&lt;/strong&gt;-bonding adds to acetates' solubility in water.  Have a look at the first few lines of this &lt;a href=&quot;http://adsabs.harvard.edu/abs/2001JChEd..78..111V&quot; rel=&quot;nofollow noreferrer&quot;&gt;abstract&lt;/a&gt;, it's quite explanatory.  However, few acetates like those of &lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt;&lt;/em&gt;, &lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Hg}$&lt;/span&gt;&lt;/em&gt; or  &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Pb}$&lt;/span&gt;&lt;/strong&gt; are slightly soluble.&lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-08-02T14:04:36.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5744" PostTypeId="1" CreationDate="2013-08-02T15:18:28.277" Score="1" ViewCount="74" Body="&lt;p&gt;Which synthetic resins are suitable for fabrication of plumbing components e.g pipe sections?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find an easy way to mold/house/encase a custom designed data logger.&#xA;The water temperature ranges between 10 to 50 degrees C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm planning this for both indoor &amp;amp; out door use. Under direct sunlight.&lt;/p&gt;&#xA;" OwnerUserId="2078" LastEditorUserId="2078" LastEditDate="2013-08-03T12:06:26.860" LastActivityDate="2013-08-03T12:06:26.860" Title="Suitability of liquid plastic resin for plumbing application" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5745" PostTypeId="2" ParentId="5744" CreationDate="2013-08-02T15:54:22.517" Score="2" Body="&lt;p&gt;Epoxy should work OK and you have different options that are easy to get.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-08-02T15:54:22.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5746" PostTypeId="1" CreationDate="2013-08-02T17:21:42.473" Score="3" ViewCount="112" Body="&lt;p&gt;I'm working on a lab report that has some specific questions that need to be included. The lab work itself dealt with GC–MS.&lt;/p&gt;&#xA;&lt;p&gt;What is the effect of the temperature selected during cryofocusing/cryotrapping? What is the effect of making this temperature warmer or colder?&lt;/p&gt;&#xA;&lt;p&gt;My understanding was that the gas based samples would adsorb to the sample loop and then rapidly desorb when the temperature is increased. However, I am not sure what the effect of changing the cold temperature is.&lt;/p&gt;&#xA;" OwnerUserId="2079" LastEditorUserId="41328" LastEditDate="2021-05-20T07:32:35.873" LastActivityDate="2021-05-20T07:32:35.873" Title="What is the effect of temperature during cryotrapping?" Tags="&lt;analytical-chemistry&gt;&lt;temperature&gt;&lt;chromatography&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5747" PostTypeId="2" ParentId="5672" CreationDate="2013-08-02T20:53:32.407" Score="9" Body="&lt;p&gt;I've never heard of onions, lemons or vinegar removing a paint smell, but I would guess that they'd just cover up the unpleasant odor with something nicer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your later comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I wouldn't want to remove the smell but leave the VOC or other bad elements there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;VOC stands for &quot;volatile organic compound.&quot; (See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Volatile_organic_compound&quot;&gt;Wikipedia page on VOCs&lt;/a&gt;) Volatile compounds are ones that evaporate easily. Technically, the smell of that onion (or lemon, or vinegar) is a VOC, just not a particularly harmful one. (Not knowing the particular chemicals that make up a onion's scent, I can't say for sure that it's a volatile &lt;strong&gt;organic&lt;/strong&gt; compound, but it's a pretty good bet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That fresh paint smell comes from VOCs (and potentially from other volatile compounds), so by the time the smell is gone, any VOCs you &lt;em&gt;couldn't&lt;/em&gt; smell are likely gone too.&lt;/p&gt;&#xA;" OwnerUserId="1949" LastActivityDate="2013-08-02T20:53:32.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5748" PostTypeId="1" CreationDate="2013-08-02T21:43:49.867" Score="47" ViewCount="474010" Body="&lt;p&gt;It is well known that when you add salt to ice, the ice not only melts but will actually get colder. From chemistry books, I've learned that salt will lower the freezing point of water. But I’m a little confused as to why it results in a drop in temperature instead of just ending up with water at 0 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is occurring when salt melts the ice to make the temperature lower?&lt;/p&gt;&#xA;" OwnerUserId="2081" LastEditorUserId="402" LastEditDate="2017-01-22T09:55:18.403" LastActivityDate="2019-08-13T08:15:46.927" Title="Why does ice water get colder when salt is added?" Tags="&lt;water&gt;&lt;solutions&gt;&lt;heat&gt;&lt;phase&gt;" AnswerCount="9" CommentCount="3" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5749" PostTypeId="2" ParentId="5508" CreationDate="2013-08-02T22:22:45.777" Score="4" Body="&lt;p&gt;I'm not sure what you mean by &quot;cementing&quot; a wall. However, cement (typically Portland cement) is the paste that binds together the aggregate (gravel and sand) in concrete. The better the bond between the cement and the aggregate, the higher the resultant compressive strength of the concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cement is mixed with water in concrete it binds to the aggregate through a chemical reaction known as &quot;hydration.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydration_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydration_reaction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydration reaction consumes water as a reactant, which is why you must keep concrete wet for several days after it is placed. If the concrete dries out then it has no water to continue the hydration reaction. The end result then is that you're left with cement that is not fully adhered to the aggregate (less compressive strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Careful though - you only need to keep the concrete wet after it has set up and is curing. Adding too much water to the mix (giving it a high water to cement ratio, w/c) actually drastically reduces the compressive strength of the concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curing compounds are especially helpful for properly curing concrete. (speaking from a heavy civil general contractor's point of view) We use this product all all the time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wrmeadows.com/1100-clear-resin-water-based-concrete-curing-compound/&quot; rel=&quot;nofollow&quot;&gt;http://www.wrmeadows.com/1100-clear-resin-water-based-concrete-curing-compound/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2082" LastActivityDate="2013-08-02T22:22:45.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5750" PostTypeId="1" AcceptedAnswerId="5753" CreationDate="2013-08-03T04:44:37.220" Score="3" ViewCount="479" Body="&lt;p&gt;How is $\ce{HSO4-}$ formed? If Hydrogen is 1+ and $\ce{SO4}$ has 2- charge? I don't understand this. Could someone please clarify?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="4231" LastEditDate="2015-03-02T22:44:15.527" LastActivityDate="2015-03-02T22:44:15.527" Title="How is the acid HSO₄⁻ formed?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5752" PostTypeId="1" CreationDate="2013-08-03T07:25:24.980" Score="3" ViewCount="295" Body="&lt;p&gt;Here's is what I understand so far and the following is referenced directly from UNSW CHEM1061 lecture notes by Professor Pall Thordarson:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;order of the reaction SHOULD be equal to the number of molecules of that reactant involved in each collision ( however this is not always the case)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it is only possible to predict the order of a reaction is it occurs in a single step (elementary reactions)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;reaction order is experimentally derived from rate equation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so I understand all of the above but what I fail to fully comprehend is that actual DEFINITION of 'reaction order' what is it in it's entirety, what does it's value represent if it only sometimes correlates with molecularity (no. of molecules which react in an elementary reaction) ? If someone can answer this question could they please provide a qualitative answer that explains the value conceptually? Or is it just a value to relate concentration to reaction rate? that helps us work out more information?&lt;/p&gt;&#xA;" OwnerUserId="1548" LastEditorUserId="1548" LastEditDate="2013-08-04T10:03:22.230" LastActivityDate="2015-07-15T19:11:07.080" Title="What is the reaction order of a reaction?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5753" PostTypeId="2" ParentId="5750" CreationDate="2013-08-03T07:48:36.967" Score="3" Body="&lt;p&gt;$\ce{HSO4-}$ look at the valencies of the hydrogen and sulphate ion, because of the numbers don't match i.e. hydrogen has a +1 charge that means it is deficient by 1 electron while sulphate has a -2 charge which means it has two extra electrons (it can also be viewed that it wants to 'give away' these two electrons) these valence numbers basically show the amount of bonds the molecules can form (since at this level of chemistry bonds can be seen as the sharing of electrons) thus the sulphate can form two while the hydrogen can only form one and in this case since it forms $\ce{HSO4-}$ that means there is one 'unhappy electron' or one that is unbonded since the bond between the sulphate and hydrogen makes hydrogen's charge neutral and takes an electron from the sulphate ion leaving the unbonded electron thus giving the molecule an overall charge of -1. Note that here I am just explaining the bonding that is occurring between the molecules but $\ce{H2SO4}$ DOES form I am just explaining the charges and values involved and also note that bonding and chemical reactions usually involve the VALENCE electrons!.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this molecule ($\ce{H2SO4}$) will disassociate in water or solvent of some sort, and form ions as the person before me pointed out. into $\ce{H+}$ ions and $\ce{HSO4^-}$ thus forms the conjugate base of the acid ($\ce{H+}$ is the basis of most acidic reactions) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're confused as to why $\ce{H2SO4}$ doesn't form, it does but $\ce{HSO4-1}$ is what happens after it disassociates in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know which part of the question or problem is where your confusion lies, hope I answered or helped!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am only a first year university student if my answer is wrong please notify me but it correlates with my notes, textbooks and what I was taught back in high school)&lt;/p&gt;&#xA;" OwnerUserId="1548" LastEditorUserId="1878" LastEditDate="2013-08-03T09:16:30.723" LastActivityDate="2013-08-03T09:16:30.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5754" PostTypeId="2" ParentId="5752" CreationDate="2013-08-03T08:40:53.843" Score="5" Body="&lt;p&gt;Very Shortly, Rate of reaction is directly proportional to product of the active masses of reactants &lt;em&gt;raised to some power which is Order of the Reaction&lt;/em&gt;.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an elementry chemical reaction&lt;br&gt;&#xA;$$\ce{2A(g) + B(s) -&amp;gt; C }$$ &lt;br&gt;&#xA;As reaction is elementry, stoichiometric coefficient and order will be same. The order of the reaction will be 2 (as the other component is solid, therefore will not affect concentrations). The rate law for reaction will be given as &lt;br&gt;&#xA;$${R = K[A]^2}$$ &lt;br&gt;&#xA;where K is reaction constant.&#xA;[A] is concentration of A.&lt;br&gt;&lt;br&gt;&#xA;If the order of the reaction would be 0, then Rate would be equal to reaction constant itself as the rate law would be given as &lt;br&gt;&#xA;$${R = K[A]^0}$$ &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains the maximum use of Order of the reaction and also what order of reaction is. &lt;br&gt;&#xA;Edit: for more wide explaination on this, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_of_reaction&quot; rel=&quot;nofollow&quot;&gt;wikipedia.&lt;/a&gt; (Same explaination is provided but a widely expanded.)&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="7448" LastEditDate="2015-07-15T19:11:07.080" LastActivityDate="2015-07-15T19:11:07.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5755" PostTypeId="1" CreationDate="2013-08-03T12:43:22.827" Score="1" ViewCount="1476" Body="&lt;p&gt;Why is the methoxide ($\ce{CH_{3}O^{-}}$) anion negatively charged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has 13 valence electrons and has 13 free electrons. In spite of that the methoxide anion is negatively charged.&lt;/p&gt;&#xA;" OwnerUserId="2084" LastEditorUserId="4231" LastEditDate="2015-03-02T23:13:29.087" LastActivityDate="2015-03-02T23:13:29.087" Title="Negative charge of methoxide anion" Tags="&lt;organic-chemistry&gt;&lt;electronic-configuration&gt;&lt;covalent-compounds&gt;&lt;formal-charge&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5756" PostTypeId="2" ParentId="5755" CreationDate="2013-08-03T19:37:47.530" Score="4" Body="&lt;p&gt;The negative charge is due to the unpaired electron in the &lt;strong&gt;&lt;em&gt;oxygen&lt;/em&gt;&lt;/strong&gt; atom, remained after all bonds(molecular orbitals) are filled. &lt;strong&gt;&lt;em&gt;Oxygen&lt;/em&gt;&lt;/strong&gt; has 2 unpaired electron in its valence shell and is singly bonded to &lt;strong&gt;&lt;em&gt;carbon&lt;/em&gt;&lt;/strong&gt; in this molecule. Hence being un-bonded, the other electron exhibits the charge.          &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this &lt;a href=&quot;http://flashchem.nelson.com/demo/LS-methoxide_Demo.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;, hover the cursor on the $\ce{O}$-atom, you will get a graphical representation of the same. &lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="4231" LastEditDate="2015-03-02T23:13:26.507" LastActivityDate="2015-03-02T23:13:26.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5757" PostTypeId="1" CreationDate="2013-08-03T20:45:59.857" Score="3" ViewCount="183" Body="&lt;p&gt;If I have a system in which I introduce $\mathrm{100~e.u.F}$ (energy units) derived from chemical energy (i.e. fuel) and generate thermal energy to heat water what are e.u. absorbed by water? I.e. the efficiency of the transformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I would like to know the &quot;average/general&quot; efficiency of energy transformation for a district heating plant. Considered that the only energy input is fuel and the only energy output is heat (no electricity)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the energy loss?&lt;/p&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="7448" LastEditDate="2015-07-15T19:08:02.157" LastActivityDate="2015-07-15T19:08:02.157" Title="energy efficiency, transformation from chemical to thermal energy, for district heating plants" Tags="&lt;energy&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5758" PostTypeId="2" ParentId="5748" CreationDate="2013-08-03T21:19:54.827" Score="2" Body="&lt;p&gt;When you dissolve $\ce{NaCl}$ in water, it will have to take energy from the system to break its structure so it can dissolve in water. This is the reason the water gets colder because the salt uses the energy from the water to solve it.&#xA;Now let's look at why ice melts when salt is added. This is based on a so-called colligative attribute. These attributes are only dependent on the amount of substance. When you add particles to a solvent, its vapor pressure lowers. This will result in a higher boiling point(using salt for cooking) and a lower freezing temperature for the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this gives a starting point for further reading consult books on physical chemistry(for e.g. Atkins).&lt;/p&gt;&#xA;" OwnerUserId="2086" LastEditorUserId="-1" LastEditDate="2018-05-29T17:14:13.553" LastActivityDate="2018-05-29T17:14:13.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5759" PostTypeId="2" ParentId="5529" CreationDate="2013-08-04T02:36:42.287" Score="6" Body="&lt;p&gt;I rather like Greg E.'s answer. I'd like to add that there is at least one polymer that has resonances distributed over the entire chain: polyacetylene. When doped with iodine, its conductivity can reach values comparable to silver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like this answer might not have been what you were looking for, since I&#xA;usually think of resonance forms as being distinct from each other, whereas for polyacetylene, the resonance forms look pretty much the same (just shifted over one bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Polyacetylene&quot;&gt;https://en.wikipedia.org/wiki/Polyacetylene&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-08-04T02:36:42.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5760" PostTypeId="2" ParentId="5472" CreationDate="2013-08-04T02:55:48.937" Score="8" Body="&lt;p&gt;Unfortunately not, I'm afraid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you might have gotten cause and effect mixed up here. Blue flames are hotter than yellow flames not because of their color, but because &lt;em&gt;hotter flames generate blue light and cooler ones generate yellow/red light&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take an example of this kind of problem, I know that very red tomatoes tend to be tastier than very pale tomatoes. Does this mean that we could make tastier tomatoes by injecting food coloring into them? Well, probably not, because it's not the color that causes the flavor. Rather, tomatoes that generate more color tend to also generate more things like sugar and flavors, and so flavorful tomatoes tend to have more color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With flames, it's even more black-and-white because while its possible to have a pale tomato that happens to taste good, the color of the flame is set by the universe: unless you add colors yourself, it is &lt;em&gt;impossible&lt;/em&gt; to have a blue flame that is the same color as a yellow one. The color of the flame is determined by something called &quot;blackbody radiation,&quot; and if you're curious about why flames are colored the way they are, it's definitely something to take a look at.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-08-04T02:55:48.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5761" PostTypeId="1" AcceptedAnswerId="5787" CreationDate="2013-08-04T03:45:13.837" Score="9" ViewCount="598" Body="&lt;p&gt;I am trying to learn simulation myself and I have installed Matlab on my machine for that purpose. In web I have gone through some of the basics of &lt;a href=&quot;http://web.chem.ucsb.edu/~kalju/MonteCarlo_1.html&quot;&gt;Monte Carlo simulation&lt;/a&gt;, but I am unable to find a simple Monte Carlo simulation experiment for chemistry. I am trying to simulate growth of a nano particle during a wet synthesis. I have tried searching research articles, but as I am new to simulations most of them I didn't understand. If possible can you explain a simple Monte Carlo experiment in chemistry, which can be done in Matlab. Is it possible to simulate the growth of nano particles using the Monte Carlo method?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="4945" LastEditDate="2015-05-14T04:59:07.227" LastActivityDate="2015-05-14T04:59:07.227" Title="What are simple applications of Monte Carlo simulations in chemistry, that can be done in Matlab?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5762" PostTypeId="1" AcceptedAnswerId="5763" CreationDate="2013-08-04T05:12:47.223" Score="3" ViewCount="245" Body="&lt;p&gt;For a research project, I'm looking into existing  chemical molecules that have the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is mostly chemically inert, safe for animals and children&lt;/li&gt;&#xA;&lt;li&gt;It exhibits piezoelectricity -  production of high electric oscillating potentials under pressure&lt;/li&gt;&#xA;&lt;li&gt;paramagnetic moment a plus, but not required&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are there known candidates for this?&lt;/p&gt;&#xA;" OwnerUserId="627" LastEditorUserId="305" LastEditDate="2013-08-04T07:00:22.453" LastActivityDate="2013-08-04T07:00:22.453" Title="chemically inert molecule with piezoelectric properties" Tags="&lt;electrochemistry&gt;&lt;materials&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5763" PostTypeId="2" ParentId="5762" CreationDate="2013-08-04T06:54:27.477" Score="8" Body="&lt;p&gt;The ur-piezoelectric crystal is quartz. Being composed of $\ce{SiO2}$, it more or less epitomizes chemical/biological inertness. It has the additional advantage of being comparatively cheap and common. There are a number of other natural minerals which are piezoelectric (e.g., various &lt;a href=&quot;http://en.wikipedia.org/wiki/Perovskite_%28structure%29&quot;&gt;perovskites&lt;/a&gt;). Most of those minerals are, as far as I know, relatively unreactive and non-toxic. Naturally, if toxicity is a concern, avoid the common lead-based compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen mention of materials that exhibit both ferrimagnetic and piezoelectric properties. If ferrimagnetism is acceptable to you instead of paramagentism, then you may wish to consider that as an avenue for research. If you need paramagnetic materials specifically, then it's probably not worth your time (unless your project is compatible with subjecting those materials to their respective Curie temperatures).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-08-04T06:54:27.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5764" PostTypeId="2" ParentId="5748" CreationDate="2013-08-04T07:13:55.450" Score="30" Body="&lt;p&gt;When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the above point can be confusing if you're new to thinking about phase transitions. An ice cube melting will take up energy, while an ice cube freezing will give off energy. I like to think of it in terms of Le Chatelier's principle: if you need to lower the temperature to freeze an ice cube, this means that the water gives off heat as it freezes.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cooling you get, therefore, comes from the fact that some of the bonds in the ice are broken to form water, taking energy with them. The loss of energy from the ice cube is what causes it to cool.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-08-04T07:13:55.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5765" PostTypeId="2" ParentId="5748" CreationDate="2013-08-04T10:16:43.770" Score="0" Body="&lt;p&gt;When you add salt to the ice it melts, I won't go into why since you didn't ask that; all you need to do is that it does if you don't believe me -&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.howstuffworks.com/nature/climate-weather/atmospheric/road-salt.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://science.howstuffworks.com/nature/climate-weather/atmospheric/road-salt.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, whenever a substance undergoes a phase change it's temperature does not rise and usually stays relatively constant, if you looked at a graph of most substances when undergoing different phase changes (i.e. solid to liquid to gas) you will observe regions that are 'flat' or horizontal this is because the energy is no longer causing a rise in temperature but a change in state. Since you have dissolved salt in the ice it will lower the freezing point (note that freezing and melting point of any substance is the same they can be seen as mirrors for one another) this means that water can now exist at lower temperatures and not turn into ice or in other words it will begin to melt at lower temperatures this could attribute as to why the temperature would LOWER as it no longer needs to reach as high a temperature to begin to melt. I haven't explained it very clearly but I hope you understand it consulting a physics and chemistry text book as the person above has suggested is a good idea.&lt;/p&gt;&#xA;" OwnerUserId="1548" LastEditorUserId="402" LastEditDate="2017-01-22T10:13:29.283" LastActivityDate="2017-01-22T10:13:29.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5766" PostTypeId="1" CreationDate="2013-08-04T14:01:35.157" Score="4" ViewCount="83" Body="&lt;p&gt;Why is the cathodic reaction of $I^-/AgI/Ag$ is $$\ce{AgI + e-  -&amp;gt; Ag + I-}$$ but not,&#xA;$$\ce{Ag+ + e- -&amp;gt; Ag }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas, in $Ag/AgCl(s)$, $KCl$(0.2M); it is &#xA;$$\ce{Ag+ + e- -&amp;gt; Ag}$$&lt;/p&gt;&#xA;" OwnerUserId="2090" LastEditorUserId="946" LastEditDate="2013-08-07T04:12:51.293" LastActivityDate="2013-08-07T04:12:51.293" Title="Reaction of an electrode" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5767" PostTypeId="1" AcceptedAnswerId="5793" CreationDate="2013-08-04T18:15:40.090" Score="2" ViewCount="399" Body="&lt;p&gt;I have seen vacuum take-off adapters that are essentially plugs, which could be placed at the top of a three-way above the fractionating column, and also straight and bent vacuum take-offs with an inner stem that seem meant to be placed right before the collection vessel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be any difference in which place is better to vacuum from? Would it be best to maybe use a T-connector on the vacuum line and vacuum the top of the fractionating column as well as the collecting flask?&lt;/p&gt;&#xA;" OwnerUserId="2067" LastActivityDate="2013-08-06T20:58:48.690" Title="Best place to take off vacuum for distillation" Tags="&lt;purification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5768" PostTypeId="2" ParentId="5491" CreationDate="2013-08-05T05:19:40.143" Score="1" Body="&lt;p&gt;A battery, at its heart, is what we call a redox cell: some compounds are stripped of their electrons, and others gain those electrons. The movement of electrons between these two compounds can be harnessed to provide electricity, but that's taking us a step too far. What we want to understand here is why the electrons move. To do this, we'll have to understand the redox cell. As much as I hate to assume things, you're going to have to understand a bit about ions and equilibrium to understand this explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the explanation I'm about to use is heavily taken from the excellent Chemguide site, run by Jim Clark. I'll include a link at the end.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I place a bar of magnesium in a water bath. I'll take a look at the magnesium bar a short while after I place it in the water. If I could visualize atoms and electrons, this is what I'd see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ak8J8.gif&quot; alt=&quot;A bar of magnesium takes a bath&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the atoms of magnesium leave the bar and become ions in the solution, solvated by water (and other things, but that's not crucial to this explanation). Since they become ions, they leave electrons on the metal bar. Because the ions are positively charged, they are attracted to the negative charge on the metal bar, and tend to hang around there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wait a little longer, I'll see more and more ions coming off the bar. As they start to build up in the water around the metal, every once in a while, one of the ions will hit the bar and reclaim its electrons. This is what we'll see then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdD1c.gif&quot; alt=&quot;ASSIMILATION COMPLETE!!!&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wait long enough, eventually, the number of atoms coming off the metal will be equal to the number of ions reabsorbing onto the metal. If I take out the &quot;bites&quot; in the metal bar to simplify the situation, this is what I see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWkvT.gif&quot; alt=&quot;The magnesium bar is much more bar-like now&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how there's quite a buildup of electrons on the magnesium. This is because magnesium is a very active metal, like lithium. What if we used a much less active metal, one that doesn't have a large propensity to lose atoms into solution? An example of such a metal would be copper. Here's a side-by-side of the two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6geu.gif&quot; alt=&quot;Magnesium and copper. A match made in...what&amp;#39;s this sludgy stuff?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the magnesium is a lot more active, it tends to lose ions more, and so it has more electron buildup. If we connect the two with a wire, we give the electrons a way to flow from the high concentration to the low concentration. We can then place something along the wire in the middle (say, a lightbulb) and harvest some energy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4SzLq.jpg&quot; alt=&quot;The world&amp;#39;s worst battery&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, is a hilariously oversimplified scheme. For one, as the electrons go to the copper, more and more magnesium ions break off to replace them. At some point, there are so many positive ions in the solution that they draw the electrons back toward them, through electrostatic attractions (which is a fancy way of saying that positive and negative charges attract each other). To solve this, we'd put a salt bridge between the solutions that lets ions migrate back and forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other problems too: for example, this setup wouldn't work in real life because there are no ions in solution: real cells use salt solutions of the metal e.g. $\ce{MgCl2}$ or $\ce{CuSO4}$. There's also the issue that this battery is liquid and wouldn't do so well if you tipped it over, the fact that the voltage (which is really the difference between the charge on the metal bars) is only about 1.1 volts, yada yada. That's not central to the explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that certain compounds are more likely to gain electrons and some are more likely to lose them is the central reason why things like batteries can exist. If you'd like to read more about it, I pulled this explanation from chemguide.co.uk, run by Jim Clark. I am currently ready to graduate with a B.Sc in chemistry, and his website is still one of the best resources for conceptual learning that I've come across---as a matter of fact, I learned about things on that website (eutetic mixtures and azeotropes) that I still have yet to come across in four years of university-level study. If you'd like to read more into redox reactions, the redox equilibria page is &lt;a href=&quot;http://www.chemguide.co.uk/physical/redoxeqiamenu.html#top&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this link right here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-08-05T05:19:40.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5769" PostTypeId="5" CreationDate="2013-08-05T07:07:38.860" Score="0" Body="&lt;p&gt;A thermodynamic state function describing the total energy content of a system. The sum of the system's internal energy, $U$, and the product of its volume and pressure, $pV$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H = U + pV$&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-05T11:02:30.307" LastActivityDate="2013-08-05T11:02:30.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5770" PostTypeId="4" CreationDate="2013-08-05T07:07:38.860" Score="0" Body="A thermodynamic state function describing the total energy content of a system." OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-05T11:02:53.323" LastActivityDate="2013-08-05T11:02:53.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5771" PostTypeId="5" CreationDate="2013-08-05T07:12:22.363" Score="0" Body="&lt;p&gt;The branch of chemistry concerned with identifying and quantifying the chemical constituents of matter. Possible subtopics include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Spectroscopy&lt;/li&gt;&#xA;&lt;li&gt;Spectrometry&lt;/li&gt;&#xA;&lt;li&gt;Calorimetry&lt;/li&gt;&#xA;&lt;li&gt;Chromatography&lt;/li&gt;&#xA;&lt;li&gt;Microscopy&lt;/li&gt;&#xA;&lt;li&gt;Classical combustion techniques for elemental analysis&lt;/li&gt;&#xA;&lt;li&gt;Numeric analysis&lt;/li&gt;&#xA;&lt;li&gt;Preparative techniques such as digestion and decomposition&lt;/li&gt;&#xA;&lt;li&gt;Titration techniques&lt;/li&gt;&#xA;&lt;li&gt;Chemical speciation, e.g. determination of oxidation states &lt;/li&gt;&#xA;&lt;li&gt;Surface tecniques, e.g. electron microscopy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="305" LastEditorUserId="39539" LastEditDate="2017-05-16T08:36:32.827" LastActivityDate="2017-05-16T08:36:32.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5772" PostTypeId="4" CreationDate="2013-08-05T07:12:22.363" Score="0" Body="The branch of chemistry concerned with identifying and quantifying the chemical constituents of matter." OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-05T11:02:58.247" LastActivityDate="2013-08-05T11:02:58.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5773" PostTypeId="5" CreationDate="2013-08-05T07:23:13.520" Score="0" Body="&lt;p&gt;Physical and chemical laws describing the properties and behavior of gases; relationships between the pressure, volume, and temperatures of gases. Specific examples include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ideal gas law&lt;/strong&gt;: that gives us the ideal gas equation: &lt;em&gt;pV&lt;/em&gt; = &lt;em&gt;nRT&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Henry's law&lt;/strong&gt;: relates the mole fraction of a gas dissolved in solution to its partial pressure&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dalton's law&lt;/strong&gt;: relates the partial pressure of gases to their mole fraction in the container&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Raoult's law&lt;/strong&gt;: relates the vapour pressure of immiscible solvents with their mole fractions&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Charles' law&lt;/strong&gt;: relates volume to temperature at constant pressure&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Gay-Lussac's law&lt;/strong&gt;: refers to two different laws: 1) The ratio between the volumes of the reactant gases and the gaseous products can be expressed in simple whole numbers. 2) The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas's absolute temperature.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Avogadro's law&lt;/strong&gt;: equal volumes of all gases, at the same temperature and pressure, have the same number of molecules&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Amagat's law&lt;/strong&gt;: According to Amagat’s law of partial volume, the total volume of a non-reacting mixture of gases at constant temperature and pressure should be equal to the sum of the individual partial volumes of the constituent gases&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Graham's law&lt;/strong&gt;: relates the rate of diffusion of a gas to its molar mass&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="305" LastEditorDisplayName="user7951" LastEditDate="2018-03-17T08:31:41.867" LastActivityDate="2018-03-17T08:31:41.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5774" PostTypeId="4" CreationDate="2013-08-05T07:23:13.520" Score="0" Body="Physical and chemical laws describing the properties and behavior of gases. This tag currently applies to properties and laws concerning both ideal and real gases." OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-17T07:57:55.403" LastActivityDate="2018-03-17T07:57:55.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5775" PostTypeId="5" CreationDate="2013-08-05T07:29:52.627" Score="0" Body="&lt;p&gt;The forces, either attractive or repulsive, that exist between molecules due to electric charges or varying electron cloud distribution. The major types of inter-molecular forces are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dipole-dipole&lt;/li&gt;&#xA;&lt;li&gt;Ion-dipole&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen bonding&lt;/li&gt;&#xA;&lt;li&gt;Van der Waals forces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-19T14:43:18.247" LastActivityDate="2018-03-19T14:43:18.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5776" PostTypeId="4" CreationDate="2013-08-05T07:29:52.627" Score="0" Body="The forces, either attractive or repulsive, that exist between molecules due to electric charges or varying electron cloud distribution." OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-19T14:43:41.643" LastActivityDate="2018-03-19T14:43:41.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5777" PostTypeId="5" CreationDate="2013-08-05T07:43:18.557" Score="0" Body="&lt;p&gt;An atom (or aggregation of atoms) of a particular element identified by mass number and distinguished by the number of neutrons in its nucleus. Any atom of any given element can be classified as an isotope, so questions should be limited specifically to effects related to variance in mass number. Examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Radioactivity and other nuclear properties&lt;/li&gt;&#xA;&lt;li&gt;Kinetic effects&lt;/li&gt;&#xA;&lt;li&gt;Specific relevance to spectroscopy (e.g., NMR, IR)&lt;/li&gt;&#xA;&lt;li&gt;Isotope separation&lt;/li&gt;&#xA;&lt;li&gt;Isotopic analysis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-05T11:03:45.203" LastActivityDate="2013-08-05T11:03:45.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5778" PostTypeId="4" CreationDate="2013-08-05T07:43:18.557" Score="0" Body="This tag should only be applied where the effects are specifically related to variance in mass number with the same atomic number (existence of isotopes). Examples include: radioactivity and other nuclear properties, kinetic effects, specific relevance to spectroscopy (e.g., NMR, IR), isotope separation, isotopic analysis." OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-14T13:40:16.690" LastActivityDate="2018-03-14T13:40:16.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5779" PostTypeId="1" CreationDate="2013-08-05T11:02:46.057" Score="4" ViewCount="2305" Body="&lt;p&gt;In many cultures, human dead bodies are burnt to ashes, which is then thrown in rivers and some time buried in earth.&#xA;Are human ashes after burning dead bodies biodegradable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Earlier I posted this question &lt;a href=&quot;https://area51.meta.stackexchange.com/questions/5712/human-ashes-biodegradable-or-not?noredirect=1#comment18223_5712&quot;&gt;here&lt;/a&gt;&#xA;But I was suggested to post it here.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="-1" LastEditDate="2017-05-03T08:41:06.673" LastActivityDate="2013-08-05T22:00:13.190" Title="Dead-body Ashes: biodegradable or not?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5780" PostTypeId="2" ParentId="5513" CreationDate="2013-08-05T20:12:24.600" Score="15" Body="&lt;p&gt;As the electrons fall from higher levels to lower levels, they release photons.  Different &quot;falls&quot; create different colors of light.  A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visible wavelengths are caused a by single electron making the different transitions shown below.  There are even more transitions that release invisible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Wavelength   Transition  Color              &#xA; (nm)                                        &#xA;---------------------------------------------&#xA; 383.5384      9 -&amp;gt; 2    Ultra Violet&#xA; 388.9049      8 -&amp;gt; 2    Ultra Violet&#xA; 397.0072      7 -&amp;gt; 2    Ultra Violet&#xA; 410.174       6 -&amp;gt; 2    Violet&#xA; 434.047       5 -&amp;gt; 2    Violet&#xA; 486.133       4 -&amp;gt; 2    Bluegreen (cyan)&#xA; 656.272       3 -&amp;gt; 2    Red&#xA; 656.2852      3 -&amp;gt; 2    Red&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;*Values taken from &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html&quot;&gt;Hyperphysics: Hydrogen Energies and Spectrum&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why are there two different 3-&gt; 2 transitions?  See here: &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html#c1&quot;&gt;Hydrogen Fine Structure &lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1929" LastEditorUserId="10119" LastEditDate="2014-12-04T14:44:47.057" LastActivityDate="2014-12-04T14:44:47.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5781" PostTypeId="2" ParentId="5766" CreationDate="2013-08-05T20:20:54.300" Score="6" Body="&lt;p&gt;The real reaction is $\ce{Ag+ + e- -&amp;gt; Ag}$ for both cases. Equation written in this way, ${E^0}$ will be equal to the reduction potential of Ag(I) in its 1 M solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is written as $\ce{AgI + e- -&amp;gt; Ag + I-}$, then ${E^0}$ will be equal to the reduction potential of Ag(I) in 1 M $\ce{I-}$ solution -or in other terms in ${K_{sp}(AgI)}$ Molar Ag(I). solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, technically, $\ce{Ag+ + e- -&amp;gt; Ag}$ is same with $\ce{AgI + e- -&amp;gt; Ag + I-}$. Latter is more descriptive and practically more relevant, however, because the presence of $\ce{I- /Cl-}$ ions in the medium limits the amount of $\ce{Ag+}$ cations via ${K_{sp}}$ and the notation readily implies this.&lt;/p&gt;&#xA;" OwnerUserId="946" LastActivityDate="2013-08-05T20:20:54.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5782" PostTypeId="2" ParentId="5779" CreationDate="2013-08-05T22:00:13.190" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;http://books.google.com/books?id=Q7Pb2wXV2woC&amp;amp;pg=PA90#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;one source&lt;/a&gt;, the major elemental constituents of cremation remains are calcium and phosphorus, presumably from the hydroxylapatite in bone (which consists primarily of calcium phosphate). The rest is mainly assorted trace metals, which exist in relatively minute concentrations in the human body and don't vaporize at the temperatures typical of a crematorium furnace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtually all organic matter, comprising mainly oxygen, carbon, hydrogen, nitrogen, and sulfur, is oxidized into a variety of simple compounds and is removed as a mixture of gases in the furnace fumes (and perhaps some finely dispersed, light solid particulates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, what remains is largely inorganic and non-calorific in nature. While the mineral and metal content of the ash can be dissolved and subsequently absorbed by various organisms in trace amounts, there's probably little or no organic material left that can be metabolically decomposed by bacteria, fungi, or plants as a direct energy source. Of course, there are chemical processes that can further break down the ash and isolate specific components via redox reactions (dissolution in strong acids to give salts, electrolysis, etc.), but obviously that's distinct from biodegradation. &lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-08-05T22:00:13.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5783" PostTypeId="2" ParentId="4091" CreationDate="2013-08-05T22:19:09.293" Score="4" Body="&lt;p&gt;The major issue with the blood-brain barrier is that aside from small hydrophobic/lipophilic compounds (including $\ce{O_{2}}$ and $\ce{CO_{2}}$), almost everything else is prevented from passing due to cellular tight junctions and will only (potentially) pass through via active transport.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for small lipophilic molecules passing through a phospholipid bilayer (such as on a cell surface in the blood-brain barrier), this post provides information on equations: &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK21626/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ncbi.nlm.nih.gov/books/NBK21626/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dn}{dt}$ is diffusion rate (mol/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is the partition coefficient (the equilibrium constant for its partition between oil and water) - e.g. for Urea, K = 0.0002 while for diethylurea which is more hydrophobic, K = 0.01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C^{m}$ is the concentration in the hydrophobic part of the phosopholipid bilayer (mol/L)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C^{aq}$ is the concentration in aqueous solution (mol/L)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ is permeability coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ is diffusion coefficient of the substance &lt;em&gt;within&lt;/em&gt; the phospholipid membrane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ is the area of the membrane (in $m^{2}$ I think?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ is the membrane thickness (in m I think?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{1}^{aq}$ and $C_{2}^{aq}$ are the aqueous concentrations on either side of the layer (one will be the serum or CSF concentration, the other will be the intracellular concentration)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$K = \frac{C^{m}}{C^{aq}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{dn}{dt}=PA(C_{1}^{aq}-C_{2}^{aq})$ and $P = \frac{KD}{x}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{dn}{dt}=A\frac{KD}{x}(C_{1}^{aq}-C_{2}^{aq})$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fP5Kv.jpg&quot; alt=&quot;A simple model for passive diffusion of small hydrophobic molecules directly across the hydrocarbon core of a pure phospholipid bilayer of thickness x in centimeters and area A in square centimeters from Section 15.1, Molecular Cell Biology. 4th edition. Lodish H, Berk A, Zipursky SL, et al. New York: W. H. Freeman; 2000.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have a similar equation to 2. on the other side of the cells lining the blood-brain barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems from my reading that $P$ tends to be better known than $D$, which is actually usually calculated using equation 2. from $P$, $K$ and $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="876" LastEditDate="2015-05-01T22:00:28.903" LastActivityDate="2015-05-01T22:00:28.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5784" PostTypeId="2" ParentId="5755" CreationDate="2013-08-05T22:38:17.283" Score="3" Body="&lt;p&gt;Simply put: there are more electrons than protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrogen x 3 = 1 x 3 = 3&lt;/li&gt;&#xA;&lt;li&gt;Oxygen x 1 = 8 x 1 = 8&lt;/li&gt;&#xA;&lt;li&gt;Carbon x 1 = 6 x 1 = 6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;3 + 8 + 6 = 17 protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the electron shells and covalent bonds in the Lewis diagram provided by blackSmith, you have the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 single covalent bonds = 4 x 2 = 8&lt;/li&gt;&#xA;&lt;li&gt;3 lone pairs on oxygen = 3 x 2 = 6&lt;/li&gt;&#xA;&lt;li&gt;Unseen $1s^{2}$ pair on both carbon and oxygen = 2 x 2 = 4&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;8 + 6 + 4 = 18 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there is 1 more electron than proton, it has a charge of -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to where that negative charge tends to be, one must create and look at the Lewis diagram where you see that the oxygen atom has an extra valence electron.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastActivityDate="2013-08-05T22:38:17.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5785" PostTypeId="1" CreationDate="2013-08-06T00:10:30.083" Score="0" ViewCount="254" Body="&lt;p&gt;In a Pourbaix diagram (E-pH diagram) where the regions are given , but the assignment of these to the elements we have is not done yet , how can the oxidation numbers helps me out assigning each element in its appropriate region?&lt;/p&gt;&#xA;" OwnerUserId="2096" LastEditorUserId="7448" LastEditDate="2015-07-15T16:23:29.517" LastActivityDate="2015-07-15T16:23:29.517" Title="Oxidation number and Pourbaix diagrams" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5786" PostTypeId="2" ParentId="5785" CreationDate="2013-08-06T00:18:40.783" Score="2" Body="&lt;p&gt;If you do not have a list of the species that belong in the Pourbaix diagram, then it is not possible to accurately predict what species occur where, unless you have extensive experience with the system in question.  If you have a list of species and need to place them in the appropriate regions of a Pourbaix diagram (fill in the blank style), then the higher oxidation state species occur at the top of the diagram and the more basic (e.g. OH containing) species occur towards the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have Latimer diagrams of all the species in the Pourbaix diagram (this will require the Latimer diagrams for both acidic and basic conditions, typically) then you can more accurately predict the structure of a Pourbaix diagram by (a) writing balanced half reactions and (b) adjusting them for the appropriate pH using the Nernst equation.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-08-06T00:18:40.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5787" PostTypeId="2" ParentId="5761" CreationDate="2013-08-06T03:57:16.317" Score="6" Body="&lt;p&gt;I've done a little simulation as a senior year elective, though not in Matlab. Can you be a little more specific about how/what you're trying to model with nanoparticle growth? Perhaps direct us to an abstract or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first Monte Carlo simulation you should attempt is the drunken sailor problem. The basic idea is that you have a drunken sailor who has equal probability of taking a step forwards or backwards, and you let him stagger around on a one dimensional line for some number of steps (1000) and repeat the experiment a large number of times. You then get some summary statistics and plots around your experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you're comfortable with that concept, you can look at increasing complexity into 2 or 3 dimensional random walks, changing the weighting function (not 50-50 for each direction), or some specific algorithm such as Metropolis-Hastings Monte Carlo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps. If you're looking for specific programming help, posting on stack exchange would be a good idea.&lt;/p&gt;&#xA;" OwnerUserId="2097" LastActivityDate="2013-08-06T03:57:16.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5789" PostTypeId="1" AcceptedAnswerId="5790" CreationDate="2013-08-06T13:21:53.197" Score="17" ViewCount="19691" Body="&lt;p&gt;We know that the molecule of H₂O is V-shaped. This is what makes it a dipole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdU57.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is that? I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other, until they are diametrically opposite to the oxygen molecule. But that doesn't happen. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="30" LastEditDate="2013-08-25T10:05:53.973" LastActivityDate="2016-07-18T15:50:18.443" Title="Why is H₂O V shaped?" Tags="&lt;water&gt;&lt;bond&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ClosedDate="2017-09-25T00:18:14.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="5790" PostTypeId="2" ParentId="5789" CreationDate="2013-08-06T17:33:35.540" Score="13" Body="&lt;p&gt;The oxygen atom fills its octet rule by forming two bonds. It shares 1 electron in a covalent bond with each hydrogen and has 4 remaining valence electrons. It is &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation#sp3_hybrids&quot;&gt; sp$^3$ hybridized&lt;/a&gt; and has the 4 non-bonding electrons in two lone pairs. An sp$^3$ hybridized atom has four attachment points spaced approximately 109$^\circ$ apart and has the shape of a pyramid with a triangular base. See &lt;a href=&quot;http://image.tutorvista.com/cms/images/44/bond-angles-in-water.JPG&quot;&gt;this image&lt;/a&gt; for an example. You don't typically see the water molecule drawn in 3 dimensional space with the lone pairs of electrons, all you see is a planar molecule with H-O-H bond angle of ~109$^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="2097" LastEditorUserId="2097" LastEditDate="2013-08-07T01:47:56.793" LastActivityDate="2013-08-07T01:47:56.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5792" PostTypeId="2" ParentId="5755" CreationDate="2013-08-06T20:39:06.110" Score="1" Body="&lt;p&gt;From what you have given, this question can not be answered in terms of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is meaningless to do any electron calculation, because you have to know it has one negative charge before you can do the electron counting. How can you get a correct Lewis structure without knowing that it has one extra electron? The negative charge is part of the given condition, not your result you get from calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I saw you are talking about 13 valence electrons, I guess you want to count valence electrons. If you add valence electron from all atoms, you will get 13 valence electron: 4 (C) + 3 * 1 (H) + 6 (O) = 13. Then because you have one negative charge, you have to add one more electron and then you get 13 + 1 = 14 valence electrons. Now everything agree with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out which atom the charge is on, you have to do the formal charge calculation. Find it &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-06T20:39:06.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5793" PostTypeId="2" ParentId="5767" CreationDate="2013-08-06T20:58:48.690" Score="2" Body="&lt;p&gt;Assuming that you are using a typical distillation set up (boiling container, fractionating column, distillation head, condensation tube, collecting adapter, and collecting vessel), and the the &quot;three-way&quot; you are talking about is the distillation head. It is much better to put the vacuum take off adapter right before your collecting vessel. The reason is this set up creates a vapor flow that goes through your condensation tube, and most of your fractions will be condensed and collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you attach the vacuum take off adapter on top of your distillation head, most of your vapor will take shortcut to your pump through your adapter instead of going to the condensation tube and the collecting vessel (this direction is a dead end). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the top of the distillation head is the place to put your thermometer to measure the temperature of vapor, not for anything else.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-06T20:58:48.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5794" PostTypeId="2" ParentId="5737" CreationDate="2013-08-06T21:03:32.307" Score="34" Body="&lt;p&gt;Any strong acid solution with concentration more than 1 mol/L has the negative pH. Think about any concentrate commonly used strong acid solution such as 3M $\ce{HCl}$, 6M $\ce{HNO3}$. Negative pH is actually very common.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorDisplayName="user15489" LastEditDate="2015-04-15T03:45:44.903" LastActivityDate="2015-04-15T03:45:44.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5795" PostTypeId="2" ParentId="5734" CreationDate="2013-08-06T21:10:46.283" Score="1" Body="&lt;p&gt;First of all! Metal do form covalent bond. It is very common in transition metal like platinum, palladium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not the way you are talking about. Typically, when pure metal atoms bond together, they prefer metallic bond. &lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-06T21:10:46.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5796" PostTypeId="2" ParentId="5685" CreationDate="2013-08-06T21:42:41.877" Score="2" Body="&lt;p&gt;Most cost/energy efficient method to make deionized water is reverse osmosis. The mechanism is to apply external pressure to water and force it to go through a layer or layers of semipermeable membrane.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-06T21:42:41.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5798" PostTypeId="2" ParentId="5789" CreationDate="2013-08-07T00:08:30.833" Score="0" Body="&lt;p&gt;In the bonding of Oxygen with Hydrogen, Oxygen ends up with 2 extra electrons, giving it what is called the &quot;bent&quot; shape. This is because the attachments and the electrons all repel each other but the electrons are not drawn except in a Lewis Dot structure such as the one here: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Water-2D-flat.png&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Water-2D-flat.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2013-08-07T00:08:30.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5799" PostTypeId="1" AcceptedAnswerId="5800" CreationDate="2013-08-07T00:14:58.570" Score="14" ViewCount="2576" Body="&lt;p&gt;Why is carbon dioxide that comes from dry ice opaque and smoky while the carbon dioxide that we exhale transparent?&lt;/p&gt;&#xA;" OwnerUserId="1262" LastEditorUserId="2095" LastEditDate="2013-08-07T04:13:23.720" LastActivityDate="2015-06-10T15:41:10.297" Title="Why is the carbon dioxide we exhale transparent?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5800" PostTypeId="2" ParentId="5799" CreationDate="2013-08-07T00:55:07.067" Score="24" Body="&lt;p&gt;Essentially, because the carbon dioxide sublimates from solid (dry ice) to gas at a very low temperature (roughly −78 °C at 1 atm), it causes water vapour in the air to condense, causing a visible fog. Thus what you are seeing is &lt;em&gt;not&lt;/em&gt; carbon dioxide, but rather water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we exhale and it is reasonably warm, the carbon dioxide expelled is roughly body temperature (36–37.5 °C) and so we do not see any fog forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when we exhale in cold weather is a bit different. Our breath contains a high percentage of water, higher than the percentage of water that can stay in gaseous form at a low temperature. Thus the ‘extra’ water condenses into droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, here is the phase diagram of $\ce{CO2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FMIbL.gif&quot; alt=&quot;Phase Diagram of carbon dioxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the phase diagram of $\ce{H2O}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4jOWX.jpg&quot; alt=&quot;Phase Diagram of H2O&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the saturation fraction of water at various temperatures at 1 atm (sea level):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eu1x1.jpg&quot; alt=&quot;Saturation fraction of H2O by temperature (dew point)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorDisplayName="user7951" LastEditDate="2015-06-10T15:41:10.297" LastActivityDate="2015-06-10T15:41:10.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5801" PostTypeId="2" ParentId="691" CreationDate="2013-08-07T01:19:24.123" Score="2" Body="&lt;p&gt;A combination of osmosis and Gibbs-Donnan effect is the correct explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a semi-permeable membrane such as the one you are using, or a cell membrane, or in dialysis, water can move freely as can chloride and other small ions, however large molecules such as proteins cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplified explanation is this: in osmosis, the tendency is for the osmotic pressure and charge to equalise on either side of the membrane. Proteins usually have a negative charge and exert a large osmotic pressure. Thus, to balance this out, small anions will diffuse to the opposite side to where the proteins are. Water will diffuse until the osmolality is equal on either side of the membrane unless there is a hydrostatic pressure forcing it to the other side. In the absence of hydrostatic pressure, you will end up with a higher concentration chloride solution on one side and a higher concentration protein solution on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qT99j.jpg&quot; alt=&quot;osmosis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the small particles you can see are chloride and the big particles are your proteins, you can see that the chloride concentration in the permeate is higher than the retentate and higher even than the feed.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="2095" LastEditDate="2013-08-07T01:27:42.520" LastActivityDate="2013-08-07T01:27:42.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5802" PostTypeId="1" CreationDate="2013-08-07T02:02:29.880" Score="15" ViewCount="8626" Body="&lt;p&gt;I have read that in VSEPR theory, multiple bonds are considered or treated as single bonds when predicting the geometry of a molecule? I've read in Yahoo! Answers that it is because only sigma bonds are used in determining their shape. What's the reasoning behind this?&lt;/p&gt;&#xA;" OwnerUserId="2102" LastEditorUserId="5026" LastEditDate="2018-06-13T08:24:10.627" LastActivityDate="2018-06-13T08:24:10.627" Title="Why are only sigma bonds used in determining the geometry and bond angle in VSEPR theory? Why aren't pi bonds used?" Tags="&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5803" PostTypeId="2" ParentId="1187" CreationDate="2013-08-07T02:21:53.690" Score="4" Body="&lt;p&gt;Simply, $\ce{CO_{2}}$ is dissolved in those solvents in a similar way to gases dissolved in water. The molecules do not react with the solvent but mix. The higher the pressure of the gas you pump into the liquid, the more that dissolves in it (rather than turning into bubbles and escaping). In fact, amount of gases dissolved in liquids can be measured both as a concentration (mol/L) or as a partial pressure (Pa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henry's law states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;At a constant temperature, the amount of a given gas that dissolves&#xA;  in a given type and volume of liquid is directly proportional to the&#xA;  partial pressure of that gas in equilibrium with that liquid.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p = k_{H}c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is partial pressure of the gas in solution, $c$ is its concentration and $k_{H}$ is a constant based on what the solute is (in this case $\ce{CO_{2}}$), what the solvent is (eg methanol) and the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The harder you push, the tighter you can pack it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yYE3i.gif&quot; alt=&quot;Henry&amp;#39;s Law&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, with the amine-based solvents, the $\ce{CO_{2}}$ actually reacts, which is why a degasification process needs to happen to release it.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="4231" LastEditDate="2015-03-02T23:13:08.260" LastActivityDate="2015-03-02T23:13:08.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5804" PostTypeId="2" ParentId="5802" CreationDate="2013-08-07T02:35:32.043" Score="1" Body="&lt;p&gt;Because any $\pi$ bond is always affiliated to a $\sigma$ bond, which participates the determination of the geometry. It is no way that a $\pi$ bond can be different orientation to its corresponding $\sigma$ bond.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="2093" LastEditDate="2013-08-07T02:51:25.940" LastActivityDate="2013-08-07T02:51:25.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5805" PostTypeId="2" ParentId="5455" CreationDate="2013-08-07T03:46:44.537" Score="6" Body="&lt;p&gt;The mechanism is electron transfer. The imine will take one electron from metal to form a radical anion ($\ce{.CR2-NR'^{-}}$) intermediate; the $\ce{-NR'^{-}}$ will grab one proton from solvent to give you the neutral radical ($\ce{.CR2-NR'H}$); the carbon radical grab one hydrogen atom ($\ce{.H}$) from solvent to form your amine product and a solvent radical ($\ce{.OH}$ if water used as solvent); and the resulting solvent radical will take another electron from aluminum to finish the whole process. The reason to add mercury is to destroy the relatively less reactive aluminum oxide layer. Mercury also serve as the medium for electron transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HgCl2}$ is not that dangerous, since it is not volatile at all like metal mercury or alkylmercury. After reaction, simply add some sulfide or sulfur to transfer all of them to $\ce{HgS}$ which is not soluble and not reactive in non-oxidative condition. This is also the reason that you can not use $\ce{HgS}$ instead, because it is neither soluble nor reactive.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-07T03:46:44.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5806" PostTypeId="2" ParentId="5802" CreationDate="2013-08-07T03:53:24.567" Score="5" Body="&lt;p&gt;VSEPR is used to predict highly idealized geometries corresponding to Platonic solids or combinations thereof (e.g., the trigonal bipyramidal geometry being essentially the result of fusing two faces of a tetrahedron). The basic rationale behind its use, and the reason for its success in many situations, is the recognition that electrons, being like-charged, repel one another electrostatically. Hence, as a first approximation, it's reasonable to assume that the most stable spatial configuration of any molecule would have the atoms spaced as uniformly and as distant from one another as possible. VSEPR achieves this by placing the central atom of a simple molecule (or some molecular sub-structure) at the center of a geometrically perfect polyhedron having as many vertices as non-central atoms and lone pairs, and then placing the non-central atoms and lone pairs at those vertices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that $\sigma$ bonds (and lone pairs) determine the geometry is that they form the basic skeleton of the molecule. Recall that $\sigma$ bonds are formed by head-on overlap of atomic orbitals, meaning that they are oriented along the imaginary axis connecting two atomic nuclei, and hence concentrate electron density in the region directly between the two nuclei. $\pi$ bonds, on the other hand, are essentially orthogonal to the $\sigma$ bond skeleton, and are substantially weaker. Moreover, $\pi$ bonds do not exist in isolation, meaning any $\pi$ bond between two given atoms is always formed secondarily to the $\sigma$ bond between said atoms. As such, $\pi$ bonds do not alter the basic idealized geometry of a molecule as dictated by $\sigma$ bonding, although in practice, because they do actually introduce additional electron density and require closer orbital overlap, bond multiplicity affects bond length and bond angles (as does, for that matter, the size of the atoms involved). The magnitude of those deviations from the idealized VSEPR geometry could, in principle, be quite large, although most often those deviations are slight enough that VSEPR remains useful as a first approximation.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-07T05:43:05.167" LastActivityDate="2013-08-07T05:43:05.167" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5807" PostTypeId="2" ParentId="5802" CreationDate="2013-08-07T04:01:35.040" Score="3" Body="&lt;p&gt;Question actually implies a misunderstanding. VSEPR has nothing to do with number of sigma bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSEPR is a primitive theory, relying only on simple electrostatic effects. In this view, nature of the bond is not relevant but only the number of groups surrounding the central atom is. This is the number of bound atoms, an number of unbound electron pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All VSEPR says is that, groups surrounding the central atom should be separated as much as they can. This yields linear shape for two groups, triangular for three, tetrahedral for four etc...&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="946" LastEditDate="2018-06-12T20:17:26.530" LastActivityDate="2018-06-12T20:17:26.530" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5808" PostTypeId="2" ParentId="5789" CreationDate="2013-08-07T04:05:25.070" Score="3" Body="&lt;p&gt;It is not the + charges on hydrogens, but the negative charges on the O-H bonds, and on the O it self that is dominant in determining the molecule's shape. There are four non-bonding electrons on O -two lone pairs- on top of the two O-H bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should think of four groups around O, not two.&lt;/p&gt;&#xA;" OwnerUserId="946" LastActivityDate="2013-08-07T04:05:25.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5810" PostTypeId="1" AcceptedAnswerId="5812" CreationDate="2013-08-07T12:00:23.480" Score="1" ViewCount="11565" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In the titration of a solid acid, an endpoint is reached after 22.0mL&#xA;  of 0.120 M $\ce{NaOH}$ has been added. Assuming that each acid particle&#xA;  contains two acidic hydrogens and both are neutralized, what is the&#xA;  molar mass of the unknown acid?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to solve this. I've figured out how to calculate the moles of acid were in the sample but don't see how to find the molar mass from that...any ideas?&lt;/p&gt;&#xA;" OwnerUserId="1850" LastEditorUserId="4231" LastEditDate="2015-03-02T21:03:27.403" LastActivityDate="2015-03-02T21:03:27.403" Title="Finding molar mass of an unknown acid" Tags="&lt;physical-chemistry&gt;&lt;titration&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5811" PostTypeId="1" AcceptedAnswerId="5814" CreationDate="2013-08-07T12:20:48.437" Score="49" ViewCount="3889" Body="&lt;p&gt;The following equation is standard in thermodynamics: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta G^\circ=-RT\log(K)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K$ is the equilibrium constant.  In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless.  But $K$ may have dimensions (depending on the particular equilibrium).  Why is this OK?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; working out dimensions explicitly gives: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;[R]&amp;amp;=EN^{-1}\Theta^{-1}\\&#xA;[T]&amp;amp;=\Theta\\&#xA;[\Delta G^\circ]&amp;amp;=EN^{-1}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$.  From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well.  But it isn't always.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say).  Suppose now that we scale the units for number by a factor $a$.  Then we get new values for $K,R,\Delta G^\circ$, given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\hat K&amp;amp;=K/a\\&#xA;\hat R&amp;amp;=aR\\&#xA;\hat{\Delta G^\circ}&amp;amp;=a\Delta G^\circ&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our original equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\Delta G^\circ&amp;amp;=-RT\log(K)\\&#xA;a\Delta G^\circ&amp;amp;=-aRT\log(K)\\&#xA;\hat{\Delta G^\circ}&amp;amp;=-\hat RT\log(a\hat K)\\&#xA;\hat{\Delta G^\circ}&amp;amp;=-\hat RT\log(\hat K)-\hat RT\log(a)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the new quantities do not satisfy the old equation.  Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on.  What is going on here?&lt;/p&gt;&#xA;" OwnerUserId="2106" LastEditorUserId="2106" LastEditDate="2016-09-10T03:40:42.103" LastActivityDate="2019-06-24T00:47:15.567" Title="Unit of the equilibrium constant: contradiction of Bridgman's theorem?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="5812" PostTypeId="2" ParentId="5810" CreationDate="2013-08-07T15:36:42.850" Score="5" Body="&lt;p&gt;Your data is not enough to calculate molecular weight. You may have missed the mass of the solid acid you have weighed.&#xA;Let's assume you have weighed $a$ gram of acid. Then:&#xA;$Molecular Weight = \frac{Weight}{Moles_{acid}} = \frac{Weight}{2 \times Moles_{NaOH}} =  \frac{Weight}{2 \times Molarity_{NaOH} \times Volume_{NaOH}} = \frac{a}{2 \times 0.120 \times \frac{22.0}{1000}}$&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-08-07T15:36:42.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5814" PostTypeId="2" ParentId="5811" CreationDate="2013-08-07T16:46:57.693" Score="48" Body="&lt;p&gt;The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From thermodynamics it is known that the Gibbs free energy of reation is given by&#xA;\begin{equation}&#xA; \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.&#xA;Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.&#xA;Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely&#xA;\begin{equation}&#xA; \mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).&#xA;\end{equation}&#xA;From Maxwell's relations it is known that&#xA;\begin{equation}&#xA; \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to&#xA;\begin{equation}&#xA; \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one gets&#xA;\begin{equation}&#xA; \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.&#xA;Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get&#xA;\begin{equation}&#xA; \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p   \ ,&#xA;\end{equation}&#xA;so that, introducing the mole fraction $x_{i}$,&#xA;\begin{equation}&#xA; \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .&#xA;\end{equation}&#xA;Please, note that there is a dimensionless quantity inside the logarithm.&#xA;Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$&#xA;\begin{equation}&#xA;a_{i} = \frac{\varphi_{i} p_{i}}{p^0}&#xA;\end{equation}&#xA;where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.&#xA;When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises&#xA;\begin{equation}&#xA; \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .&#xA;\end{equation}&#xA;Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets&#xA;\begin{equation}&#xA; \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,&#xA;\end{equation}&#xA;where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.&#xA;Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that&#xA;\begin{equation}&#xA;\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .&#xA;\end{equation}&#xA;So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant&#xA;\begin{equation}&#xA; K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .&#xA;\end{equation}&#xA;which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the &quot;real&quot; dimensionless equilibrium constant via&#xA;\begin{equation}&#xA; K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .&#xA;\end{equation}&#xA;it is more or less proportional to $K$ and thus gives qualitatively the same information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If the solution at hand behaves like an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_solution&quot;&gt;ideal solution&lt;/a&gt; then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$&#xA;\begin{equation}&#xA; K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .&#xA;\end{equation}&#xA;one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I forgot to mention there are also &quot;versions&quot; of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those &quot;versions&quot; can be traced back to the original equilibrium constant.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-04-22T19:35:12.380" LastActivityDate="2014-04-22T19:35:12.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5815" PostTypeId="1" AcceptedAnswerId="5818" CreationDate="2013-08-07T17:39:55.793" Score="9" ViewCount="8859" Body="&lt;p&gt;I am not from chemistry background. Could you tell me in simple words what the activity of hydrogen $a_{\ce{H+}}$ is? It arises in the context that the negative logarithm of the activity of hydrogen gives the pH: $\text{pH} = -\log_{10} a_{\ce{H+}}$.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="189" LastEditDate="2013-08-08T04:25:01.313" LastActivityDate="2017-05-02T23:22:12.990" Title="Activity of hydrogen ion" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5816" PostTypeId="2" ParentId="5815" CreationDate="2013-08-07T17:43:42.650" Score="0" Body="&lt;p&gt;I'm not quite sure what your source means by &quot;activity.&quot; Generally, the formula we use for pH is: $\mathrm{pH} = -\log[\ce{H}^{+}]$ where $[\ce{H}^{+}]$ is the concentration of the hydrogen ion, in moles per liter.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="402" LastEditDate="2017-05-02T23:22:12.990" LastActivityDate="2017-05-02T23:22:12.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5817" PostTypeId="2" ParentId="5815" CreationDate="2013-08-07T19:00:46.360" Score="7" Body="&lt;p&gt;The activity is some sort of dimensionless measure for the concentration. In thermodynamics, a lot of important equations are derived for the so-called ideal gas, which is a model system where the particles are treated as if they had no own size (point particles) and don't interact with one another (apart from elastic collisions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model is a very crude simplification of reality but nevertheless gives very good results, when the temperature is not too high and the pressure is low. So, the equations one can derive from this model are rather simple and yet quite usable for real systems. But of course, the applicability of these equation has limits, since in real systems the particles interact a lot with each another (via electrostatic and quantum mechanical forces) and have a certain size. But, in order to retain the simple form of the equations derived for ideal gases, some physical quantities like the pressure or the concentration are scaled by empirical coefficients to reflect the deviation from the ideal behavior and get the right results for real systems. An example is the activity $a$, which is equal to the dimensionless concentration, the so-called mole fraction $x$, scaled by the activity coefficient $f$&#xA;\begin{equation}&#xA;a = f x \ .&#xA;\end{equation}&#xA;So, for example, instead of the concentration (or more correctly the mole fraction) one uses the activity in some equations derived for ideal gases to describe real systems accurately. &lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="402" LastEditDate="2017-05-02T23:21:33.863" LastActivityDate="2017-05-02T23:21:33.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5818" PostTypeId="2" ParentId="5815" CreationDate="2013-08-07T22:22:08.057" Score="8" Body="&lt;p&gt;Building on @Philipp and @chipbuster's answers-&lt;/p&gt;&#xA;&lt;h2&gt;What is Activity?&lt;/h2&gt;&#xA;&lt;p&gt;When we have dilute solutions or small quantities of a gas or other substance, the substance behaves as if the particles are just points, rather than having a size. In these settings, we often use a simplified set of equations.&lt;/p&gt;&#xA;&lt;p&gt;For example, with the hydrogen ion, we use &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = -\log_{10}[\ce{H+}]$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt; is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).&lt;/p&gt;&#xA;&lt;p&gt;Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).&lt;/p&gt;&#xA;&lt;p&gt;In these circumstances, we talk about ‘activity’, which is a way to correct for this. Basically, it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words, it uses a correction factor for the simplified laws.&lt;/p&gt;&#xA;&lt;h2&gt;Activity, Concentration and Activity Coefficient&lt;/h2&gt;&#xA;&lt;p&gt;The thermodynamic definition of activity is:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a_{i} = \operatorname{e}^{\frac{\mu_i - \mu_i^⦵ }{RT}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Which in &lt;em&gt;low partial pressures&lt;/em&gt; and &lt;em&gt;dilute solutions&lt;/em&gt; is simplified to:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a = \gamma\cdot\frac{c}{c^⦵}$&lt;/span&gt; for a solution&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a = \gamma\cdot\frac{p}{p^⦵}$&lt;/span&gt; for a gas&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the activity&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the activity coefficient (a constant number based on what the substance is, what the solvent is if there is one, the temperature and other factors)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the concentration (for something dissolved); &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the partial pressure (for a gas)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$c^⦵$&lt;/span&gt; is the standard state concentration; &lt;span class=&quot;math-container&quot;&gt;$p^⦵$&lt;/span&gt; is the standard state pressure&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;pH and Activity&lt;/h2&gt;&#xA;&lt;p&gt;The original version of pH as defined by Sørensen is:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = -\log_{10}[\ce{H+}]$&lt;/span&gt; (hereafter referred to as &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}[\ce{H+}]$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;However, the modern definition that takes into account activity is:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = -\log_{10}a_{\ce{H+}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is found experimentally. In the case of the hydrogen ion, it is found using electrochemical methods.&lt;/p&gt;&#xA;&lt;h2&gt;Activity of Hydrogen Ion in Dilute Solutions&lt;/h2&gt;&#xA;&lt;p&gt;Standard thermodynamic state for &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; is considered to be 1 mol/L (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}[\ce{H+}]$&lt;/span&gt; 0); thus:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a = \gamma\cdot\frac{c}{c^⦵}$&lt;/span&gt; (as a dilute solution)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a_{\ce{H+}} = \gamma\cdot\frac{c}{c^⦵} = \gamma\cdot\frac{[\ce{H+}]}{1\ \mathrm{mol/L}} = \gamma\cdot [\ce{H+}]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Also, in dilute solutions, activity can be approximated/calculated using an extension of the Debye-Hückel equation, the Güntelberg equation, which applies to concentrations of &amp;lt; 0.1 mol/L:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\log_{10} \gamma_i=\frac{-Az_{i}^{2}\sqrt{I}}{1+\sqrt{I}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; are constants that depend on solvent and temperature (0.51 and &lt;span class=&quot;math-container&quot;&gt;$0.33\times10^8$&lt;/span&gt; respectively for water at 25 °C)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the charge of the ion (1 for &lt;span class=&quot;math-container&quot;&gt;$H^+$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is concentration of the ion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; in a solution of water at 25 °C this would be:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\log_{10}\gamma=\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Substituting, we get:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a_{\ce{H+}}=\gamma[\ce{H+}]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\log_{10}a_{\ce{H+}}=\log_{10}[\ce{H+}]+\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$-\log_{10}a_{\ce{H+}}=-\log_{10}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;and so &lt;em&gt;for dilute solutions only&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = \mathrm{p}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-05-02T23:21:11.997" CommentCount="1" CommunityOwnedDate="2013-09-08T18:22:37.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="5819" PostTypeId="2" ParentId="4099" CreationDate="2013-08-07T23:51:29.347" Score="2" Body="&lt;h2&gt;Caustic Ingestion Injury&lt;/h2&gt;&#xA;&lt;p&gt;Triton X-100 has a pH of 9.7 at 100% concentration (according to Wikipedia) which can cause alkaline caustic injury, however it is less likely to cause severe full thickness injury, as those injuries tend to happen with ingested products of pH &amp;lt;2 or &amp;gt;12.5 or with other corrosive properties that Triton X-100 does not have. Ingestions at pH 6.0-8.0 are unlikely to cause injury unless there is long term exposure (as in gastro-oesophageal reflux disease or bile reflux); note that saliva itself often has a pH of 7.8!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://emedicine.medscape.com/article/813772-overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alkaline caustic injury has a large spectrum&lt;/a&gt;. The alkali can cause liquefactive necrosis via saponification and protein dissolution. In the best case scenario, there is only superficial injury to mucous membranes, causing pain and exudates. In the worst case scenario there is full thickness involvement, usually involving the oesophagus, where you get a perforation. One of the worst cases I have had any involvement in was a young lady who drank a large quantity of Drano (concentrated NaOH with pH 11.5-13.5) and had oesophageal, gastric and duodenal perforation with extensive chemical damage to her pancreas, liver and fortunately to a much lesser extent, mediastinum.&lt;/p&gt;&#xA;&lt;p&gt;For reference, Zargar's grading system for caustic ingestion (usually based on findings on gastro-oesophagoscopy +/ pharyngolaryngoscopy) is as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Grade 0: Normal examination&lt;/li&gt;&#xA;&lt;li&gt;Grade 1: Oedema and hypermia of the mucosa&lt;/li&gt;&#xA;&lt;li&gt;Grade 2a: Superficial ulceration, erosions, friability, blisters, exudates, hemorrhages, whitish membranes&lt;/li&gt;&#xA;&lt;li&gt;Grade 2b: Grade 2a plus deep discrete or circumferential ulcerations&lt;/li&gt;&#xA;&lt;li&gt;Grade 3a: Small scattered areas of multiple ulceration and areas of necrosis with brown-black or greyish discoloration&lt;/li&gt;&#xA;&lt;li&gt;Grade 3b: Extensive necrosis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Triton X-100 would most likely cause Grade 1-2b injury. Grade 3 injury is unlikely but I suppose it depends on dose as well.&lt;/p&gt;&#xA;&lt;h2&gt;Ingestion at lower concentrations and effects not related to caustic ingestion:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://ec.europa.eu/health/opinions/products-resembling-food/en/l-3/4-products-dangerous-swallow.htm#2p3&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; contains information on the general toxicity of non-ionic surfactants such as Triton X-100.&lt;/p&gt;&#xA;&lt;p&gt;In summary:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They cause mild irritation to mucous membranes&lt;/li&gt;&#xA;&lt;li&gt;This is primarily due to foaming&lt;/li&gt;&#xA;&lt;li&gt;Foaming and irritation can cause vomiting and diarrhoea&lt;/li&gt;&#xA;&lt;li&gt;Foam can also be aspirated causing chemical pneumonitis&lt;/li&gt;&#xA;&lt;li&gt;Alcohol ethoxylates (as Triton X-100 appears to be) are readily absorbed into the bloodstream via mucous membranes and is likely to have toxic effects in this way&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://scholar.google.com.au/scholar?hl=en&amp;amp;q=triton%20x-100%20toxicity&amp;amp;btnG=Submit&amp;amp;as_sdt=1,5&amp;amp;as_sdtp=&quot; rel=&quot;nofollow noreferrer&quot;&gt;rest of the evidence I could find&lt;/a&gt; also states:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They can cause haemolysis- makes sense, as surfactants in the circulation would come into contact with red cell membranes&lt;/li&gt;&#xA;&lt;li&gt;Triton X-100 causes direct cytotoxicity due to cell membrane disruption&lt;/li&gt;&#xA;&lt;li&gt;It also inhibits NAD(P)H, which is a coenzyme involved in cellular respiration&lt;/li&gt;&#xA;&lt;li&gt;There is dose-dependent destruction of the cytoskeleton&lt;/li&gt;&#xA;&lt;li&gt;There is also some associated double-stranded DNA breakage&lt;/li&gt;&#xA;&lt;li&gt;It is possible for cells to recover from this poisoning&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Thus, the likely scenario from swallowing dilute Triton X-100 is this:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First you get foaming, oral &amp;amp; oesophageal pain and abdominal pain&lt;/li&gt;&#xA;&lt;li&gt;Next you get vomiting and diarrhoea, at which point you may inhale some of the foam&lt;/li&gt;&#xA;&lt;li&gt;If you've inhaled any foam, you develop respiratory distress&lt;/li&gt;&#xA;&lt;li&gt;You develop dose-dependent haemolysis which presents as tiredness (and/or shortness of breath, chest pain, palpitations), red or dark brown urine, jaundice&lt;/li&gt;&#xA;&lt;li&gt;In severe cases potential multi-organ failure as a result of direct cytotoxicity, haemolysis, anaemia and/or chemical pneumonitis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;It seems it'd be a huge spectrum all the way from general foaminess and some pain all the way to multi-organ failure and death. I'm not entirely sure what doses would be toxic as it appears no-one has forced animals or humans to drink the stuff!&lt;/p&gt;&#xA;&lt;h2&gt;Regarding whether there are more toxic and corrosive but chemically similar compounds out there&lt;/h2&gt;&#xA;&lt;p&gt;yes.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-08-08T00:36:04.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5820" PostTypeId="2" ParentId="5547" CreationDate="2013-08-08T00:53:23.370" Score="0" Body="&lt;p&gt;If it's a very large amount of oil and area, I'd recommend mopping up the bulk of it using something like this: &lt;a href=&quot;http://www.eco-absorb.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.eco-absorb.com/&lt;/a&gt; or &lt;a href=&quot;http://www.clear-passage.com/absorb.html&quot; rel=&quot;nofollow&quot;&gt;http://www.clear-passage.com/absorb.html&lt;/a&gt; or &lt;a href=&quot;http://www.oildri.net/Gran_QuickSorb.html&quot; rel=&quot;nofollow&quot;&gt;http://www.oildri.net/Gran_QuickSorb.html&lt;/a&gt; (NB I have no idea if these particular products any good, but it came up when I did a google search)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, you can use a pressure washer and/or detergent for the rest.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastActivityDate="2013-08-08T00:53:23.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5821" PostTypeId="1" CreationDate="2013-08-07T07:07:58.577" Score="0" ViewCount="63383" Body="&lt;p&gt;I need to understand the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering an element Sulfur - S which has 16 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we calculate the number of valence electrons of S?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am wrong:2+2+6+2+4=16. So, the valence electron = 4+2=6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that always valence electron is the addition of the last two shells? In that case for Na, sodium, if we take 11 and arrange it as 2+2+6+1, then the valence electron is 1 or 6+1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, considering the fact valence electron is 1, if we go by, up down up down rule, then as it is 1, hence the spin is up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 6: is it up(1) down(2) up(3) down(4) up(5) down(6) as it is both up/down hence it would be +-1/2 spin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Phosphorous 2,8,5, what would be the valence electron? 5 or 13?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 5 then up down up down up, will it have an up spin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that for even numbers like 4,6 the spin will be both up/down i.e. +-1/2?&lt;/p&gt;&#xA;" OwnerDisplayName="Shounak Bhattacharya" LastEditorUserId="402" LastEditDate="2017-01-14T09:08:31.470" LastActivityDate="2019-08-02T20:43:37.127" Title="How to find the number of valence electrons?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5822" PostTypeId="2" ParentId="5821" CreationDate="2013-08-07T12:08:51.237" Score="10" Body="&lt;p&gt;'p' and 'd'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.&lt;br&gt;&#xA;There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.&lt;br&gt;&#xA;There are &lt;strong&gt;five&lt;/strong&gt; types of 'd' orbitals (10 electrons) and &lt;strong&gt;seven&lt;/strong&gt; types of 'f' orbitals (14 electrons). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PuP8x.png&quot; alt=&quot;orbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So let's take the example of sulfur.&lt;br&gt;&#xA;The total number of electrons is 16. We'll distribute them in the following manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons &lt;em&gt;in the last (3rd) orbital&lt;/em&gt;. Therefore the valence number is 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not simply the last 2 numbers while writing the electronic configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Phosphorous, we have:&#xA;$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Lithium (3 electrons), we have:&#xA;$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the spin, just remember that you have to fill up the entire &lt;strong&gt;&lt;em&gt;orbital&lt;/em&gt;&lt;/strong&gt; with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.&lt;/p&gt;&#xA;" OwnerUserId="1858" OwnerDisplayName="mikhailcazi" LastEditorUserId="402" LastEditDate="2017-01-14T09:04:54.653" LastActivityDate="2017-01-14T09:04:54.653" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5823" PostTypeId="2" ParentId="5811" CreationDate="2013-08-08T05:32:37.497" Score="11" Body="&lt;p&gt;(this is a comment to Phillip answer; upon editing it failed to get saved on due time)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically $K_c$ is valid as long as the solutions show colligative properties. These &quot;concentration&quot; constants can (and should) be defined dimensionless, as they properly emerge as a limit for the dimensionless activity in &quot;ideal&quot; systems, and the activity $a_j$ measures how the chemical potential $\mu_j$ of the substance $j$ changes as its concentration (or in general the chemical composition of the system) changes from some reference state: $a_j = [c]_j/c^\ominus$ where $[c]_j$ is the molar concentration and $c^\ominus = 1$ M.&#xA;Therefore to remember that inside the logarithm everything is dimensionless one should not forget to write the reference magnitude $c^\ominus$. These constants can use different units for the concentration too, for instance $m_j/m^\ominus$ where $m_j$ is the molal concentration and the reference magnitude $m^\ominus$ is $1$ mol/kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the limits of validity, $K_c$ fails miserably for electrolytes, where the true thermodynamic constant depends on all the charges in the system (ionic strength of the solution), that is, the activity of the electrolyte is not only proportional to its number (concentration, etc.).&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="1100" LastEditDate="2013-08-08T07:49:05.107" LastActivityDate="2013-08-08T07:49:05.107" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5824" PostTypeId="1" AcceptedAnswerId="5825" CreationDate="2013-08-08T10:22:44.240" Score="6" ViewCount="50476" Body="&lt;p&gt;just a simple question from me today...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at the structure of Norepinephrine for a school project and was curious about the diagrams on the Wikipedia page for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qoQ46.png&quot; alt=&quot;noradrenaline/norepinephrine structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering why hexagons are used in this diagram and what the purpose of the three lines on the inside of the hexagon mean.&lt;/p&gt;&#xA;" OwnerUserId="2111" LastEditorUserId="30" LastEditDate="2013-08-29T09:20:27.317" LastActivityDate="2013-08-29T09:20:27.317" Title="Lines on inside of hexagonal shapes of structure diagrams" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5825" PostTypeId="2" ParentId="5824" CreationDate="2013-08-08T11:01:58.483" Score="12" Body="&lt;p&gt;Noradrenaline/norepinephrine is an aromatic organic compound containing a benzene ring (the hexagon). Essentially, the vertices in the diagram are carbon atoms and the number of lines represents the covalent bond order; 1 line (eg connecting an OH to the benzene ring) means a single covalent bond. Any 'spare' bonds are where a C-H bond is. So it can be rewritten like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XeKra.png&quot; alt=&quot;Redrawn Noradrenaline with all bonds, atoms and lone pairs&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER, when it comes to a benzene ring (the hexagon), you do not simply have alternating single and double bonds. In fact you have what is known as a 'resonance structure', where the electrons/bond is shared around the entire ring. One way of thinking of it is that there are 1.5 bonds between each carbon atom in the ring. Another way of thinking of it is that pi-orbitals from each carbon atom merge to form a ring combined orbital structure, so instead of existing in that figure-8 type area around one carbon nucleus, those 6 electrons can be in a donut shaped area above and below the carbon nuclei:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VeXJR.png&quot; alt=&quot;benzenes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why you often see a hexagon with a circle inside it to represent a benzene ring as it is less misleading than the alternate single and double bonds. In that case, you would redraw the noradrenaline diagram like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ngox7.png&quot; alt=&quot;Noradrenaline redrawn in aromatic form&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage, though, of having the alternate double and single bonds is in drawing diagrams when working out reactions.&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="2095" LastEditDate="2013-08-29T04:49:01.823" LastActivityDate="2013-08-29T04:49:01.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5827" PostTypeId="1" AcceptedAnswerId="5840" CreationDate="2013-08-08T16:25:30.907" Score="7" ViewCount="4563" Body="&lt;p&gt;The oncotic pressure or colloidal osmotic pressure is the osmotic pressure developed due to the presence of colloids in a solution. But since the colloids are not true solution, why should the colloids be termed as solutes soluble in the solvent and consequently capable of producing any osmotic pressure? Should they not behave like insoluble substances who do not affect the osmolarity of the solution? And theses colloidal osmotic pressure plays a very important role in trans-capillary transfer dynamics.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the colloids do affect the osmotic properties, is the expression to measure the colloidal osmotic pressure same as that for true solutions(vant-hoff equation)?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-02-24T07:20:26.730" LastActivityDate="2018-02-24T07:20:26.730" Title="What is the colloidal osmotic pressure?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;osmosis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5829" PostTypeId="1" CreationDate="2013-08-09T04:19:59.750" Score="3" ViewCount="549" Body="&lt;p&gt;A stoichiometric fuel-air mixture flowing in a Bunsen burner forms a well-defined conical flame. The mixture is then made leaner. For the same flow velocity in the tube, how does the cone angle change? &lt;/p&gt;&#xA;" OwnerUserId="2114" LastEditorUserId="22" LastEditDate="2013-08-09T07:50:07.980" LastActivityDate="2013-08-10T06:09:32.557" Title="Variation of cone angle with concentration of fuel-air mixture" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5832" PostTypeId="1" AcceptedAnswerId="9452" CreationDate="2013-08-09T09:37:45.143" Score="2" ViewCount="300" Body="&lt;p&gt;Do ions also form special electron configurations? (just as Copper and Chromium do)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that because copper has 1 electron in its s-orbital (highest) because it wants to have a full d-orbital (d^10)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do ions also act this way?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-03-25T01:54:31.393" Title="Do ions also form special electron configurations? (just as Copper and Chromium do)" Tags="&lt;theoretical-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5834" PostTypeId="1" AcceptedAnswerId="5835" CreationDate="2013-08-09T14:05:12.103" Score="1" ViewCount="337" Body="&lt;p&gt;I am really confused about what the size of particles of dispersed phase in a colloid is: one of my textbook says its between $1 - 100nm$, another says its between $1-1000nm$, and Wikipedia says its between $2-500nm$. Is there any reliable source giving the exact information?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2013-08-09T21:00:44.043" Title="Size of dispersed phase particles in colloid?" Tags="&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5835" PostTypeId="2" ParentId="5834" CreationDate="2013-08-09T14:27:55.660" Score="2" Body="&lt;p&gt;Its between $1-1000nm$ as stated on &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Colloid&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; in a paragraph name &lt;code&gt;IUPAC definition&lt;/code&gt;.&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Colloid: Short synonym for colloidal system.&lt;br&gt;&#xA;  Note: Quotation from refs.[3][4]&lt;br&gt;&#xA;  Colloidal: State of subdivision such that the molecules or polymolecular particles dispersed in a medium have at least one dimension between approximately 1 nm and 1 μm, or that in a system discontinuities are found at distances of that order.[5]&lt;br&gt;&#xA;  Note: Quotation from refs. [3],[4]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Since it is mention under IUPAC definition that size lies between $1nm-1000nm$, therefore this must be the correct size.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-08-09T21:00:44.043" LastActivityDate="2013-08-09T21:00:44.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5836" PostTypeId="1" CreationDate="2013-08-09T15:56:30.103" Score="8" ViewCount="226" Body="&lt;p&gt;I have written some molecular dynamics code that I want to use to model known chemical reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By known I mean that it has been observed in a lab that $\ce{A +B-&amp;gt;C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not interested in finding out if it is feasible that two things react (i.e. is there a favorable energy configuration?), rather I am interested in finding out what the concentration over time looks like given a volume, concentration, temperature and a pair of chemicals that I &lt;strong&gt;know&lt;/strong&gt; will react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I just check if they are very close to each other. This produces reasonable looking concentration curves. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this is a correct way of doing things? I am looking for publications so it has to be defend-able, not just OK.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am also interested in modelling large bio-molecules in a coarse grained simulation. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there any particular potential functions that I should use for large biomolecules, or will LJ do?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2118" LastEditorUserId="10185" LastEditDate="2015-09-27T17:27:37.840" LastActivityDate="2015-09-27T17:27:37.840" Title="How should I modify my molecular dynamics simulation so that known chemical reactions can be modelled?" Tags="&lt;kinetics&gt;&lt;computational-chemistry&gt;&lt;concentration&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5837" PostTypeId="1" AcceptedAnswerId="5839" CreationDate="2013-08-09T16:27:05.603" Score="6" ViewCount="3251" Body="&lt;p&gt;One reaction used to form sodium hypochlorite (A type of bleach) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NaOH(aq) +Cl2(g)-&amp;gt;NaOCl(aq) + NaCl(aq) + H2O(l)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided the sodium hydroxide is dissolved in some substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the sodium hydroxide dissociates to form the $\ce{Na+}$ and $\ce{OH-}$ ions, but I do not know why the $\ce{Cl2}$ breaks down into 2 separate atoms instead of just staying the way it is. My questions are: what are the driving forces behind this reaction and why does the $\ce{Cl2}$ break down? I see how it happens, I just don't know why. &lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="16683" LastEditDate="2016-09-30T09:46:32.693" LastActivityDate="2018-03-10T13:04:47.183" Title="Why does chlorine undergo disproportionation in alkaline media?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5839" PostTypeId="2" ParentId="5837" CreationDate="2013-08-09T20:17:48.597" Score="6" Body="&lt;p&gt;At constant pressure and temperature, the spontaneity of all chemical reactions is controlled by the difference in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;nofollow&quot;&gt;Gibbs free energy&lt;/a&gt; ($\Delta G$), as described by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = \Delta H - T\Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change, and $T$ is the absolute temperature. For a reaction to take place spontaneously, the free energy change must be negative (i.e., $\Delta G &amp;lt; 0$). In the specific reaction you're asking about, it's very probable that the change in entropy is negative, since a gas is being consumed and an equal total number of moles products are being formed as moles of reactants, so that the end result is an increase in order with respect to the reaction mixture. However, the formation of products results in a net release of heat (i.e., $\Delta H$ is negative), which increases the entropy of the surroundings sufficiently, such that, overall, entropy is increased in a global sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, what drives this reactions forward is its exothermic nature (under standard conditions). In this reaction, the formation of stable chemical bonds releases energy as heat. If one were to attempt this reaction at very high temperatures, it would eventually fail, as the increase of the $T\Delta S$ term would at some point dwarf the enthalpy change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other factors at work as well, including &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_equilibrium&quot; rel=&quot;nofollow&quot;&gt;chemical equilibrium&lt;/a&gt;. By &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier's principle&lt;/a&gt;, the process can be made favorable by, e.g., maintaining a continual excess of reactants, or removing products formed from the reaction mixture as it proceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Here's &lt;a href=&quot;http://www.northstarchemical.com/prod_sodiumhypochlorite_prof.htm&quot; rel=&quot;nofollow&quot;&gt;one source&lt;/a&gt; which indicates 348cal/g energy released, to put a specific number on the process. You could also calculate the enthalpy change using tables of standard thermodynamic data.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-09T20:36:57.017" LastActivityDate="2013-08-09T20:36:57.017" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5840" PostTypeId="2" ParentId="5827" CreationDate="2013-08-09T20:30:57.850" Score="5" Body="&lt;p&gt;I think, first I should clarify what causes the osmotic pressure:&#xA;Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species, e.g. the solvent, through it but not others, e.g. the solute.&#xA;So, there is a concentration gradient between the two solutions which would lead to a diffusion from the side with the higher to the side with the lower concentration.&#xA;But since the solute cannot pass through the membrane the diffusive flow cannot even the concentration differences out and thus constantly presses against the membrane exerting a force on it which leads to a pressure.&#xA;The pressure originating from this tendency for osmotic flow to occur is called the osmotic pressure $\Pi$ of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: Is the theory behind diffusion applicable to colloidal solutions?&#xA;To reason about this, I will first address the matter of whether colloids can be considered &quot;real&quot; solutes or not. Colloidal particles are usually some polymer particle coated with a tenside layer. So, it is the molecules of those tenside layers that interact with the solvent molecules and they are chosen in such a way that the colloidal particles neither lump together (coalesce) nor sink to the ground or float to the top - that keeps the colloidal solution stable. So, a colloidal particle is surrounded by solvent molecules which solvate the tenside layer. Of course, the situation is a bit different from a &quot;normal&quot; solution since the colloidal particle is a lot bigger than, say, an alcohol molecule and thus the the ratio between solute particle size and the size of its solvation sphere is much smaller. But, in essence, a colloidal particle and an alcohol molecule are solvated in pretty much the same way. &#xA;Furthermore it can also be assumed that interactions (e.g. electrostatic forces) between the colloidal particles are weak.&#xA;This means that colloidal particles are thus not subject to external forces which influence their movement. They simply move through the solution by Brownian motion. This is important because then their motion can mathematically be described by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stochastic_process&quot; rel=&quot;nofollow&quot;&gt;stochastic process&lt;/a&gt; which is the basis for the thermodynamic description of diffusion.&#xA;And, as I described above, the reason for diffusion to happen is the same as the origin of the osmotic pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your last question: Yes, I think, the description for osmotic pressure caused by colloidal particles should be the same as for &quot;normal&quot; solutes.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-10T02:12:32.597" LastActivityDate="2013-08-10T02:12:32.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5841" PostTypeId="1" AcceptedAnswerId="5845" CreationDate="2013-08-10T01:08:54.653" Score="2" ViewCount="3058" Body="&lt;p&gt;In ionic compounds, where does the extra electron go? Or, in case of a cation, from which shell does it come?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for example, $\ce{OH^{-}}$, where did this extra electron go in the electronic configuration? And, what about $\ce{OH^{+}}$? There exists no such ion. Why?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="4231" LastEditDate="2015-03-02T22:19:07.533" LastActivityDate="2015-03-02T22:19:07.533" Title="Where does the extra electron go in ionic compounds?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5842" PostTypeId="1" AcceptedAnswerId="5843" CreationDate="2013-08-10T01:48:31.583" Score="2" ViewCount="1412" Body="&lt;p&gt;If you look at $\ce{Na}$, it has a very low ionization energy, but the question is, why does it have an ionization energy at all? If it can get to the electronic configuration of $\ce{Ne}$, by losing an electron, which is a very stable configuration, then it should release energy. Consider the analogy to gravitational potential energy, you can consider $\ce{Na}$ to be analogous to standing on a very high ledge. And $\ce{Na^+}$, is the position at the ground. Obviously, getting to the ground is much more stable. And if I jump down from the ledge I convert my potential energy to kinetic energy and ultimately to heat, thus releasing energy. Why does sodium require energy to ionize? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, consider ferric chloride $\ce{FeCl_3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VcUqW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does iron form this bond? I mean, it makes sense from chlorine's point of view to share an electron with iron. All the chlorines have a full outer shell. But what about iron? It had $2$ valence electrons, now it has $5$. It would make much more sense for iron to bond with chlorine to form $\ce{FeCl_6}$, since then it, too would have a full outer shell. Or, $\ce{Fe}$ could give away its $2$ electrons to $2$ chlorines and form ionic bonds. Why does this not happen? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="1888" LastEditDate="2013-08-10T02:25:55.277" LastActivityDate="2013-08-10T02:41:16.390" Title="Why does sodium require energy to ionize?" Tags="&lt;ions&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5843" PostTypeId="2" ParentId="5842" CreationDate="2013-08-10T02:27:14.100" Score="9" Body="&lt;p&gt;There are two very different questions here so let's break things down a bit&lt;/p&gt;&#xA;&lt;h2&gt;Why does sodium have an ionization energy?&lt;/h2&gt;&#xA;&lt;p&gt;Since electrons are negatively charged and nuclei (owing to the presence of protons) are positively charged, electrons will typically be drawn to the nucleus.  The situation gets a bit more complicated by (a) the quantum nature of energy and (b) the presence of other electrons.  Complication (a) is the reason why electrons &lt;strong&gt;do not behave like planets orbiting the sun&lt;/strong&gt; because if they did the electron would eventually be pulled in to the nucleus.  (More details probably requires a different question.)  The second complication is more relevant to your question, because additional electrons will both repel the electron we are trying to eject through ionization and also shield some of the positive charge of the nucleus.  (Think about wearing sunglasses; they don't make the sun less bright, they just make less light reach your eyes.)&lt;/p&gt;&#xA;&lt;p&gt;Your comparison of Na to that of an object on a cliff is not correct; the sunglasses analogy I used above is better (although it's flawed as well).  The reason why sodium, or any other atom/ion for that matter, has an ionization energy is because it requires energy to separate two oppositely charged particles.  The attraction may be weak, as it is in sodium, but it is still non-zero.&lt;/p&gt;&#xA;&lt;h2&gt;Why does &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; bond the way it does?&lt;/h2&gt;&#xA;&lt;p&gt;First, it is a good idea to look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(III)_chloride&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt; for iron(III) chloride, which shows that the molecule actually binds in an octahedral fashion, where two additional chlorides from neighboring molecules bind in a bridging fashion.  This is a type of bonding that is typically not discussed until one's third year of chemistry education.  Valence shell electron counting works really well for identifying &amp;quot;stable&amp;quot; electron configurations of ions and covalent molecules made out of lighter elements (C, N, O, for example).  The presence of d-orbitals, plus the ionic nature of the interaction between Fe(III) and chloride ions, &lt;em&gt;plus&lt;/em&gt; the Lewis acid/base concepts that do a better job at describing and rationalizing bonding in transition metal complexes, makes the concept of valence electrons not the best model to explain bonding in &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Without access to molecular orbital theory, it is perhaps best to explain the bonding in iron(III) chloride from a strictly ionic bond nature.  Fe(III) is a +3 cation and therefore requires three -1 anions to balance the charge.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-08-10T02:41:16.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5844" PostTypeId="1" AcceptedAnswerId="5846" CreationDate="2013-08-10T02:45:22.737" Score="47" ViewCount="41076" Body="&lt;p&gt;It seems that every website on sexual health advises against using oil-based lubricants with condoms. It is claimed that &quot;oil breaks down latex&quot;. One source claimed that a latex condom completely breaks down in only 60 seconds. It made me curious, so I made an experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a piece of rubber latex condom and soaked it into regular canola oil I found in the kitchen. I checked the condom after 1 minute, nothing changed. So I let it soak for about 5 minutes more, and then 5 hours more, still nothing. It was able to hold a large amount of water without leaking or breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering, is it really true that oil degrades latex? What sort of chemical reaction is supposed to happen? What properties of the latex material and the oil influence this reaction?&lt;/p&gt;&#xA;" OwnerUserId="2121" LastEditorUserId="7448" LastEditDate="2015-10-30T16:48:58.173" LastActivityDate="2019-06-24T21:55:59.907" Title="Does oil break down condoms?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5845" PostTypeId="2" ParentId="5841" CreationDate="2013-08-10T03:21:20.467" Score="3" Body="&lt;p&gt;In $\ce{OH-}$, that extra electron is on the Oxygen atom.  Oxygen has a very large electronegativity, so you can be sure that extra electron is with the $\ce{O}$ not the $\ce{H}$.  A lone oxygen has 6 electrons in its valence shell (2s2, 2p4), and wants to get to 8 (2s2, 2p6).  In the $\ce{OH-}$ ion, both atoms are relatively &quot;happy&quot;, the $\ce{H}$ has 2 electrons in its valence shell (1s2) shared with Oxygen in a bond, and the Oxygen has 8 electrons in its valence shell (2s2, 2p6), 6 unpaired electrons and 2 shared with hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, in an anion the extra electron will be going onto the most electronegative atom.  In a cation the positive charge will go to the lest electronegative atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a helpful picture, your extra electron is in red:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pNbkO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;picture reference: &lt;a href=&quot;http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question about $\ce{OH+}$, it would never be stable because that oxygen is going to tear some electrons off of the first thing it sees, because it REALLY wants that full octet.  If you somehow made some $\ce{OH+}$, my guess would be that the $\ce{H+}$ would just dissociate, and your $\ce{O}$ would find another $\ce{OH+}$'s $\ce{O}$ and make $\ce{O2}$ immediately.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="1878" LastEditDate="2013-08-10T10:20:29.257" LastActivityDate="2013-08-10T10:20:29.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5846" PostTypeId="2" ParentId="5844" CreationDate="2013-08-10T04:58:51.790" Score="42" Body="&lt;p&gt;First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a day. Of course, if you're lubricating your condoms with hexanes, you're a) an idiot or b) absolutely insane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/2535978&quot; rel=&quot;noreferrer&quot;&gt;A paper I managed to find suggests that there really isn't too much direct data on condoms&lt;/a&gt;, and it muses that the warnings might have arisen from industry, where nonpolar solvents decidedly &lt;em&gt;do&lt;/em&gt; degrade latex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out, they did a burst experiment with condoms that had been treated with various oils. Glycerol and Vaseline-treated condoms showed a very, very minor decrease in strength, while mineral oil/baby oil-treated ones burst at less than 10% of the volume of an untreated condom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also found that 10 month-old condoms have half the burst volume of 1-month old ones, so you could argue that using 1-month-old condoms that have been slathered in Vaseline is still much safer than using older ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the actual chemistry of the weakening, I honestly don't know. If I were to hazard a guess, I would note that the latex looks like a bunch of ethylenes glued together, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so my guess would be that the solvents get between the chains and force them apart, weakening them. For this to happen, the solvent must be nonpolar, but still small enough to slip between the chains of the polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's probably why vaseline and canola oil don't have much of an effect---they're just too big to fit between the chains. Again though, I don't know for sure, so don't quote me on this last paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L4paH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="60112" LastEditDate="2019-04-19T20:06:55.847" LastActivityDate="2019-04-19T20:06:55.847" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="5847" PostTypeId="2" ParentId="5829" CreationDate="2013-08-10T06:09:32.557" Score="3" Body="&lt;p&gt;The cone angle of the Bunsen flame is determined by the flame speed of the mixture and the average flow velocity. To first approximation one can write: $$ \sin \alpha = \frac{S}{U} $$&#xA;where $\alpha$ is the half cone angle, $S$ is flame speed and $U$ is flow velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mixture is made fuel-lean, depending on how flame speed changes the cone angle will change accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually flame speed for fuel-lean mixture is less than that for stoichiometric mixture. Therefore we can expect the cone angle to &lt;em&gt;decrease&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="655" LastActivityDate="2013-08-10T06:09:32.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5848" PostTypeId="2" ParentId="5220" CreationDate="2013-08-10T10:06:50.333" Score="8" Body="&lt;p&gt;I'm really not sure this is a chemistry question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, evaporative cooling will do the job for you. Here on the Indian sub-continent a clay/earthenware water container works wonders using the technique of evaporative cooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of earthenware, wrap two pieces of thick cloth of the appropriate size around your containers. Keep both cloth wet/moist, and let nature do the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also consider taking a bit of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pot-in-pot_refrigerator&quot;&gt;pot luck&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="765" LastActivityDate="2013-08-10T10:06:50.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5849" PostTypeId="1" CreationDate="2013-08-10T22:10:02.000" Score="3" ViewCount="877" Body="&lt;p&gt;How can I calculate the energy of a solution by knowing all of it's elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have 2 liters of salting water $\mathbf{NaCl_{(aq)}+H_2O_{(l)}}$ contains 1.5 g of salt, how can I determine the internal energy stored in it?&lt;/p&gt;&#xA;" OwnerUserId="1883" LastEditorUserId="1883" LastEditDate="2013-08-11T21:18:21.167" LastActivityDate="2015-09-18T01:38:44.577" Title="How can I calculate the energy of a solution by knowing all of it's elements?" Tags="&lt;energy&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5851" PostTypeId="1" AcceptedAnswerId="6267" CreationDate="2013-08-11T05:42:51.253" Score="17" ViewCount="1001" Body="&lt;p&gt;In his book on molecular orbital theory, &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions&lt;/em&gt;, Ian Fleming notes that Pauling formulated an early alternative model to Huckel theory for explaining the bonding of simple conjugated polyenes. Evidently, this came to be called $\tau$-bonding. Fleming describes it as a modification to, or offshoot from, the hybridization model, in which orbitals similar to sp3 hybridized orbitals are combined. Fleming says that the $\tau$-bond model makes the extent of the conjugation less obvious, but that the model&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] might have some virtues, not present in the Huckel model, especially in trying to explain some aspects of stereochemistry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have access to the relevant primary literature, and the treatment of the subject in Fleming's book amounts to one paragraph with an accompanying diagram. A search online was relatively fruitless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is $\tau$-bonding just the historical precursor to the bent/banana-bond model, or does it have some unique independent significance that makes it directly relevant to modern chemists?&lt;/li&gt;&#xA;&lt;li&gt;What are the specific &quot;virtues&quot; that Fleming might have been referring to, and what, if any, advantages does the $\tau$-bond model have over Huckel theory and/or other approaches grounded in MO theory?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-11T05:54:08.427" LastActivityDate="2014-08-01T10:32:25.503" Title="What utility does the Tau bond model of orbital overlap have?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;bent-bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5852" PostTypeId="1" CreationDate="2013-08-11T06:21:54.920" Score="3" ViewCount="413" Body="&lt;p&gt;I know that transference number depends on Ionic mobility, which depends on temperature. But is there any equation which describes the relation of Transference number and temperature, any ions like H+.&lt;/p&gt;&#xA;" OwnerUserId="2128" LastActivityDate="2016-03-06T05:33:50.417" Title="How will the transference number of H+ ion in a dilute solution vary as temperature increases?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5853" PostTypeId="1" CreationDate="2013-08-11T08:23:07.560" Score="6" ViewCount="25530" Body="&lt;p&gt;How do you calculate how much energy is needed to break a water molecule into its constituent atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note&lt;/em&gt;&lt;br&gt;&#xA;This is not homework. I just want to know what is required to get hydrogen out of water.&lt;/p&gt;&#xA;" OwnerUserId="2129" LastEditorUserId="2129" LastEditDate="2013-08-11T11:39:09.697" LastActivityDate="2016-10-13T12:31:35.173" Title="How much energy is needed to break a water molecule?" Tags="&lt;energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5854" PostTypeId="1" AcceptedAnswerId="5856" CreationDate="2013-08-11T16:38:22.833" Score="1" ViewCount="41" Body="&lt;p&gt;What state of matter are phosphomonoesterases?  I am not a biochemist; while I've found various lists of such enzymes, and plenty of (to me) barely comprehensible articles describing their production, it's not clear to me whether the final product in isolation is a solid, liquid, or gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, is there some reference work in which one can look up the state of matter (at standard temperature and pressure) for a given chemical?&lt;/p&gt;&#xA;" OwnerUserId="2130" LastActivityDate="2013-08-11T18:58:23.503" Title="Are phosphomonoesterases solid?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5855" PostTypeId="1" AcceptedAnswerId="5858" CreationDate="2013-08-11T17:21:38.327" Score="3" ViewCount="1229" Body="&lt;p&gt;My questions relates to the fundamental concept of electrochemistry, more specifically the electrode potentials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; First, why is there a potential difference at the interface of two phases? Considering the simplest case of a metal rod dipped in the solution of its ions, why should there be a potential difference. Out of many ways, one of the ways of understanding this provided by &lt;a href=&quot;http://www.chemguide.co.uk/physical/redoxeqia/introduction.html&quot; rel=&quot;nofollow&quot;&gt;this site here,&lt;/a&gt;is in terms of equilibrium between the metal atoms and ions. But then, why is the presence of the other phase (ions for the metal piece) required to effect this equilibrium? For example, for the equilibria of ammonia and nitrogen/hydrogen, even if we take pure ammonia, the equilibrium is set up according to the surrounding conditions. Next,then can this equilibrium be described in terms of the equation $\Delta G=-RT\ln(K)$. Lastly, if this is an equilibrium, then there must also be an equilibrium between the metal and the hydrogen or the hydroxyl ions in solution as these might also take up the electrons from the metals?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; When we measure the difference in electrode potentials of two different half cells, we are just measuring the potential difference between the two metal electrodes. How can we get an idea of the potential difference of the metal piece and its solution as the absolute value of potential of the two solutions might also differ. That is, if the two solutions are at different electrostatic potentials ( though not measurable), the potential difference between the two electrodes will not be the difference of their electrode potentials(which are the potential difference between the electrode and the electrolyte in each case). So, in order for the potential difference of the electrodes be equal to the difference in electrode potentials of the two half-cells, the solutions should be at the same potential?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-08-12T02:36:01.807" LastActivityDate="2013-08-12T02:36:01.807" Title="Electrode potentials at interfaces?" Tags="&lt;electrochemistry&gt;&lt;thermodynamics&gt;&lt;redox&gt;&lt;theoretical-chemistry&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5856" PostTypeId="2" ParentId="5854" CreationDate="2013-08-11T18:56:39.180" Score="5" Body="&lt;p&gt;Enzymes are very, very large in comparison to the typical molecules you encounter in chemistry (excluding any polymers). A 10 kD protein is considered to be small, and there are proteins of several hundred kD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove all the water from a typical protein you end up with a solid. But for most enzymes this would be an irreversible step, and they would lose their activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually enzymes are kept in aqueous solution, often with some glycerol or other cryoprotectants, and frozen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, they would probably be solid, but likely inactive in that form so I'm not sure if it counts. The relevant form of an enzyme is dissolved in water.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-08-11T18:56:39.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5857" PostTypeId="2" ParentId="5854" CreationDate="2013-08-11T18:58:23.503" Score="2" Body="&lt;p&gt;Phosphomonoesterases are proteins, i.e. long polymer chains of amino acid monomers. When they are active in living cells they are dissolved in an aqueous solution (or found in lipid membranes, which is sorta like a 2D liquid).  You can get purified proteins in solution with a variety of techniques, and if you were to evaporate away all the water you would get dried protein solid.  Most all biopolymers are solids if you remove the water the are naturally found in.  You never have pure protein liquids or protein gasses, if you heated the solids I think you would break down the polymers into the constituent monomers long before you got a proper liquid.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2013-08-11T18:58:23.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5858" PostTypeId="2" ParentId="5855" CreationDate="2013-08-11T20:14:01.150" Score="7" Body="&lt;p&gt;There are a bunch of of sub parts to these questions; let me see if I can tease some of them out.&lt;/p&gt;&#xA;&lt;h2&gt;Why does a potential difference occur at an interface?&lt;/h2&gt;&#xA;&lt;p&gt;The easiest way to answer this question is through thermodynamics.  If we consider the situation of a reactive metal rod (e.g. magnesium) being exposed to water, the reaction &lt;span class=&quot;math-container&quot;&gt;$$\ce{Mg -&amp;gt; Mg^{2+} + 2 e-}$$&lt;/span&gt; is favorable.  In other words, the &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; for this reaction will be less than zero.  I've done nothing more than summarize the &lt;a href=&quot;http://www.chemguide.co.uk/physical/redoxeqia/introduction.html&quot; rel=&quot;noreferrer&quot;&gt;website you reference&lt;/a&gt; in your question, which does a nice job at describing what happens when a piece of metal comes in contact with water, so I don't feel the need to repeat that work.  We then get to your next question:&lt;/p&gt;&#xA;&lt;h2&gt;Why is a second phase required to generate a potential difference?&lt;/h2&gt;&#xA;&lt;p&gt;If we have a universe that consists just of magnesium, then there would be no place for the above reaction to occur.  In this trivialized case, there is no driving force to make magnesium give up its electrons.  That doesn't mean it would not happen, only that any dynamic equilibrium that is established would highly favor the pure metal.  The meat of question one, then, is (may be?):&lt;/p&gt;&#xA;&lt;h2&gt;Why doesn't a potential difference between a metal and gas occur?&lt;/h2&gt;&#xA;&lt;p&gt;Actually it does.  There's an &lt;a href=&quot;http://link.springer.com/content/pdf/10.1023%2FA%3A1019092024757.pdf&quot; rel=&quot;noreferrer&quot;&gt;interesting paper&lt;/a&gt; (which is behind a paywall if you are not at a university that subscribes to this journal) that describes the measurement of potentials across metal/gas interfaces.  The challenge in studying metal/gas potentials is the ability of controlling this potential externally.  With the metal/liquid scenario, an electrochemist (or budding electrochemist) can add another electrode to the solution and apply a potential between the electrodes.  It's much harder to do the same thing with a gas.  (Well, technically it's *&lt;em&gt;not&lt;/em&gt; harder to connect two electrodes to one another and hold them in air, but the redox work functions, resistance and capacitance issues that arise in gas-phase electrochemistry make this a really challenging feat.)&lt;/p&gt;&#xA;&lt;p&gt;The paper I reference above uses chemisorbed species that have active IR or Raman vibration modes to measure the surface potential using spectroscopy and something called &lt;a href=&quot;http://webb.cm.utexas.edu/research/papers/Webb_Proteins_2011.pdf&quot; rel=&quot;noreferrer&quot;&gt;Stark tuning&lt;/a&gt; (not the best reference, but it's also not behind a paywall) which states that the vibrational frequency of an harmonic oscillator is proportional to an applied electric field.  Not many metal/gas interfaces have been explored using this method, but those that have display surface potentials fairly close to 0 vs. SCE or 5 eV vs. vacuum.&lt;/p&gt;&#xA;&lt;h2&gt;What about the rest of the stuff in solution?&lt;/h2&gt;&#xA;&lt;p&gt;Back to the metal/liquid scenario, the last part of your first question asks about other ions in solution.  They most definitely play a role in defining the potential of a metal in solution.  In order to accurately predict electrode potentials, one must know concentrations and possible reactions (including states of matter) for all species present.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What about question 2?&lt;/strong&gt;&#xA;I'm afraid question 2 might be based on the false premise that we can measure the potential of an electrode in the absence of anything else (liquid, gas, ions or molecules).  When electrochemists refer to a 'half cell', they have in mind both the reduced (in this case, metal electrode) and oxidized (metal ions) forms of a redox couple contained within the same system.  (Think about a beaker with a copper wire placed in it and the beaker is filled with a solution containing 0.1 M &lt;span class=&quot;math-container&quot;&gt;$CuCl_2$&lt;/span&gt;.  That half-cell or system has a given potential based upon (primarily) the standard reduction potential of &lt;span class=&quot;math-container&quot;&gt;$Cu^{2+}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$Cu$&lt;/span&gt; and the concentration of &lt;span class=&quot;math-container&quot;&gt;$CuCl_2$&lt;/span&gt;.  If I have misread your second question, please edit and I will update my answer as needed.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-08-11T20:14:01.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5859" PostTypeId="2" ParentId="5853" CreationDate="2013-08-11T21:39:05.463" Score="7" Body="&lt;p&gt;The answer is given by the standard enthalpy of formation, $\Delta H_\mathrm f^\circ$ for water: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{H2 + 1/2O2 &amp;amp;-&amp;gt; H2O(l)}\quad&amp;amp;\Delta H_\mathrm f^\circ&amp;amp;=-285.8\ \mathrm{kJ/mol}\\[6pt] &#xA;\ce{H2 + 1/2O2 &amp;amp;-&amp;gt; H2O(g)}\quad&amp;amp;\Delta H_\mathrm f^\circ&amp;amp;=-241.8\ \mathrm{kJ/mol}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy required for the following reaction, your point of interest:&#xA;$\ce{H2O -&amp;gt; H2 + 1/2O2}$ is simply a positive value for either one of the above $\Delta H_\mathrm f^\circ$ values depending on the state of the reactant. It means that per mole water splitting requires at least $241.8\ \mathrm{kJ}$ of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any method(thermal,electrolysis etc) followed, this much of energy will be always required. When using electrolysis, it may be convenient to express it in $\mathrm{eV}$.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding calculation, I have to mention that $\Delta H_\mathrm f^\circ$s are experimental values. To know how they are calculated, &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_enthalpy_of_formation#Mechanics&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is a nice example of $\ce{NaCl}$.&lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorDisplayName="user7951" LastEditDate="2016-10-13T12:31:35.173" LastActivityDate="2016-10-13T12:31:35.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5860" PostTypeId="1" AcceptedAnswerId="5861" CreationDate="2013-08-12T05:39:03.810" Score="10" ViewCount="2766" Body="&lt;p&gt;I really don't know anything about chemistry, but I was wondering about the following vague question: has any substance been created that does not contain any metallic elements, but is metallic?  In fact, the only nonelemental metals I know of are alloys, so I may as well ask whether we have ever created any metallic &lt;em&gt;compounds&lt;/em&gt;, with or without metallic constituents (without preferred).  Both Wikipedia and the tag wiki here claim that metallic compounds do exist but don't give examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My vague understanding of what makes an element metallic is that it has something to do with the sharing of electron clouds between atoms, and this question is motivated by wondering whether this kind of quantum effect can be induced from nothing in an appropriately constructed molecule.&lt;/p&gt;&#xA;" OwnerUserId="2132" LastEditorUserId="2095" LastEditDate="2013-08-12T10:41:06.150" LastActivityDate="2020-04-23T21:13:36.990" Title="Have we ever created artificial metals?" Tags="&lt;metal&gt;&lt;materials&gt;&lt;quantum-chemistry&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5861" PostTypeId="2" ParentId="5860" CreationDate="2013-08-12T11:19:43.863" Score="17" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metal&quot;&gt;Metals&lt;/a&gt; have several general properties in common (to some degree):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lustre_%28mineralogy%29&quot;&gt;Luster&lt;/a&gt; - metals are shiny&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity&quot;&gt;Electrical conductivity&lt;/a&gt; - metals conduct electricity&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_conductivity&quot;&gt;Thermal conductivity&lt;/a&gt; - metals conduct heat&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ductility&quot;&gt;Ductility&lt;/a&gt; - metals can be drawn into wires&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ductility&quot;&gt;Malleability&lt;/a&gt; - metals can be beaten into shapes &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fusible_alloy&quot;&gt;Fusibility&lt;/a&gt; - metals can be melted and forged&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these properties derive from the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bond&quot;&gt;metallic bond&lt;/a&gt;&quot;, a delocalized sharing of electrons throughout the metal lattice, which is the key necessity for displaying macroscopic metallic properties. All metals do not behave the same way for these properties. For example, gold is more malleable than iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Metalloid&quot;&gt;Metalloids&lt;/a&gt; are those elements with one or more &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropy&quot;&gt;allotropes&lt;/a&gt; that have metallic characteristics. Few if any metalloids display all of the metallic characteristics (otherwise they would be metals!). For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite&quot;&gt;graphite&lt;/a&gt; is lustrous and electrically conductive, but not thermally conductive, malleable, ductile, or fusible. The metalloid with the most metallic behavior is &lt;a href=&quot;http://en.wikipedia.org/wiki/Tellurium&quot;&gt;tellurium&lt;/a&gt;, which is fusible and ductile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for compounds, many are like the metalloids - they have some properties of metals but not all. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_carbide&quot;&gt;silicon carbide&lt;/a&gt; is lustrous, but that's about it. Otherwise it is hard, brittle, and an insulator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyacetylene&quot;&gt;Polyacetylene&lt;/a&gt; and derivative polymers are sometimes call &quot;synthetic metals&quot; because of their electric conductivity and sometimes luster. However, again, they are thermally insulating. They also tend to burn instead of fuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metalloid elements and compounds are not quite metallic because their electron delocalization is in only one (polyacetylene) or two (graphite) dimensions, not all three.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Update!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on Snipergirl's comment, it appears that we are closer than I originally thought to preparing true synthetic metals. &lt;a href=&quot;http://phys.org/news/2013-02-metallic-behavior-material-heralds-electronic.html&quot;&gt;This highlight at phys-org&lt;/a&gt; suggests that supramolecular lattices of redox-coupled conjugated organic ions can exhibit metallic behavior at low temperature. The full article is at &lt;em&gt;&lt;a href=&quot;http://www.nature.com/ncomms/journal/v3/n9/full/ncomms2079.html&quot;&gt;Nature Communications&lt;/a&gt;&lt;/em&gt; and has been made open access. These compounds appear to be close to achieving delocalization in all three dimensions. The behavior of this organic material is not surprising: metals and metalloids become &lt;a href=&quot;http://en.wikipedia.org/wiki/Superconductivity&quot;&gt;even more metallic at cryogenic temperatures&lt;/a&gt;. The induced superconductivity allows for such cool things as magnetic levitation and MRI. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2013-08-13T12:43:14.897" LastActivityDate="2013-08-13T12:43:14.897" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5863" PostTypeId="1" CreationDate="2013-08-12T19:09:55.683" Score="0" ViewCount="5918" Body="&lt;p&gt;One common cure for dirty pennies is to put them in vinegar and salt. The vinegar should react with the copper oxide on the penny in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2CH3COOH + CuO -&amp;gt; (CH3COO)2Cu + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nurdrage, on YouTube, did a video on it, and he did say that salt speeds up the reaction. However, he did not mention the way in which it sped up the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same exact question is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2765/salt-in-vinegar-to-clean-pennies.&quot;&gt;here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answers did conflict and there was no clear correct answer. I am re-asking the question, in hope of getting a clear(er) answer. &lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2013-08-12T23:48:56.677" Title="Why Add Salt to Vinegar to Clean Pennies?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2013-08-12T23:50:55.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="5864" PostTypeId="2" ParentId="2765" CreationDate="2013-08-12T22:24:04.250" Score="-1" Body="&lt;p&gt;The salt is added to vinegar to speed up the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition of NaCl to vinegar shifts the equilibrium of acid dissociation, i.e. &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH &amp;lt;=&amp;gt; CH3COO- + H+}$&lt;/span&gt;, to right therefore concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COO-}$&lt;/span&gt; increases. Due to this, &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; reacts with &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt; to give &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl + CH3COOH &amp;lt;=&amp;gt; CH3COONa + HCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooper Oxide (base) reacts with acids to form water and copper salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt;), which is a weak acid, will react slowly with Copper oxide as compared to &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; which is strong acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when use vinegar-salt solution to clean penny, Copper Oxide parellely reacts with HCl and CH3COOH both and gives Copper salt and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CuO + CH3COOH -&amp;gt; 2H2O + Cu(C2H3O2)2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CuO + 2HCl -&amp;gt; H2O + CuCl2}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="23561" LastEditDate="2019-04-10T16:47:50.457" LastActivityDate="2019-04-10T16:47:50.457" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="5865" PostTypeId="1" AcceptedAnswerId="5866" CreationDate="2013-08-12T23:25:26.237" Score="4" ViewCount="2839" Body="&lt;p&gt;I am studying for a final tomorrow and I was going through some problems on older exams and came across one which I wasn't really sure how to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to write the balance equation when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;managnese(IV)oxide is reduced to manganese(II) oxide by hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a few approaches to this, because at first I did not realize it say reduced &quot;by&quot; hydrogen gas. This part threw me off because I initially added $\ce{H2(g)}$ to the product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have:&#xA;$\ce{MnO2(s) + H2(g)-&amp;gt; MnO(s) + H2O(l)}$ which is not balanced yet. I noticed in this case the conventional type of balancing the equation does not work. Would this be a redox rxn in which I find each half reaction and then balance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am completely stuck at this point, and I don't think my formula is correct. Does anyone have any insight or good examples ? &lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="4231" LastEditDate="2015-03-02T22:19:17.580" LastActivityDate="2015-03-02T22:19:17.580" Title="Balancing the equation when MnO₂ is reduced to MnO by H₂(g)" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5866" PostTypeId="2" ParentId="5865" CreationDate="2013-08-12T23:51:54.860" Score="4" Body="&lt;p&gt;This is balanced: I count the same number of elements on each side, which is the definition of &lt;em&gt;balanced&lt;/em&gt;. What is not explicitly stated is that your hydrogen goes from $0$ to $+1$, and is oxidized, thus reducing your manganese from $+4$ to $+2$. But this all balances!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-08-12T23:51:54.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5868" PostTypeId="1" CreationDate="2013-08-13T02:14:46.643" Score="3" ViewCount="794" Body="&lt;p&gt;My friend once told me that technically , we should not draw Lewis dot structure of molecules containing no central atom(for example for benzene,&#xA;acetic acid).&lt;/p&gt;&#xA;&lt;p&gt;Is there an explanation for this?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="95133" LastEditDate="2020-09-11T19:57:38.093" LastActivityDate="2020-09-11T19:57:38.093" Title="When should we not draw a Lewis dot structure?" Tags="&lt;structural-formula&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5869" PostTypeId="2" ParentId="5868" CreationDate="2013-08-13T02:51:27.670" Score="1" Body="&lt;p&gt;Lewis dot structures are rhetorical devices to understand bonding and molecular structure (using the VSEPR model). Any molecule can be drawn using a Lewis dot structure if it helps you  correctly draw the structure. Often molecules (especially large molecules) are not drawn with Lewis dots to de-clutter.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-08-13T02:51:27.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5870" PostTypeId="2" ParentId="5868" CreationDate="2013-08-13T03:06:56.987" Score="4" Body="&lt;p&gt;Short answer: no, there's nothing preventing you from drawing arbitrarily complex molecules according to the conventions of Lewis dot structures. They are not special, and the rules for drawing them are settled by convention and have evolved since they were first developed in Lewis's original paper. They're just one type of diagrammatic rendering of molecular structure which (probably due to their simplicity and explicitness) has caught on in pedagogy. They don't see much use beyond that. Read on for further details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worthwhile to precisely define our terms here. The IUPAC Gold Book provides the following definition for a central atom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The atom in a coordination entity which binds other atoms or groups of atoms (ligands) to itself, thereby occupying a central position in the coordination entity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By that definition, the term is reserved almost exclusively for coordination complexes with a distinct metal atom at the center and ligands bound to it (with some exceptions for purely molecular Lewis adducts, e.g., amine oxides), where the electrons of the bond(s) originate on the ligands, rather than each atom donating an equal number of electrons in the formation of the bond(s). For non-coordinate covalent bonds, if the bonds were to be split homolytically (i.e., such that an equal number of electrons are distributed to each atom), the result would be separate, neutral atoms. Were coordinate covalent bonds to be split this way, that would not be the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, that definition is highly limited, and it's not the prevailing definition in my experience. The other usage of the term is essentially to refer to any non-terminal atom (i.e., one that has two or more bonds to separate atoms). There's nothing inherent to Lewis structures that prevents them being used to represent molecules containing multiple such central atoms. They're simply one representation of bonding. In practice, they don't see much use, chiefly because they're substantially less economical than condensed structural formulas or the skeletal bond-line structures (introduced in their prototypical form by Kekule), and because pure Lewis dot structures, as they were originally conceived, are exceedingly poor at representing stereochemistry.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-13T03:17:21.703" LastActivityDate="2013-08-13T03:17:21.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5871" PostTypeId="1" CreationDate="2013-08-13T16:33:17.427" Score="0" ViewCount="101" Body="&lt;p&gt;Anode: $\ce{2 H2O_{(l)} -&amp;gt; O2 + 4 H+_{(aq)} + 4e^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{ 2 H2O_{(l)} + 2 e- -&amp;gt; H2_{(g)} + 2 OH-_{(aq)}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone show me the steps for adding the two. For the overall reaction, it is $\ce{2 H2O_{(l)} -&amp;gt; 2 H2_{(g)} + O2_{(g)}}$&lt;/p&gt;&#xA;" OwnerUserId="1587" LastEditorUserId="1389" LastEditDate="2013-08-13T21:33:35.553" LastActivityDate="2016-03-13T20:04:10.993" Title="Balancing Oxidation-Reduction equations" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5872" PostTypeId="1" AcceptedAnswerId="5882" CreationDate="2013-08-13T21:19:07.203" Score="12" ViewCount="30009" Body="&lt;p&gt;Lets say you have the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{$a$A + $b$B -&amp;gt; $c$C + $d$D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of the reaction is given by the change in initial concentration over the change in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\Delta \left[\ce A \right]}{a\Delta t} = -\frac{\Delta \left[\ce B \right]}{b\Delta t} = \frac{\Delta \left[ \ce C \right]}{c\Delta t} = \frac{\Delta \left[\ce D \right]}{d\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is also the rate expression, which is of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{reaction rate} = k \left[\ce  A \right]^{m}\left[\ce  B \right]^{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between these two, and what purpose does each one serve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, what does the $k$ represent?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="16683" LastEditDate="2016-01-07T16:24:12.430" LastActivityDate="2018-04-02T12:43:52.380" Title="Difference between Reaction Rate and Rate Law?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5873" PostTypeId="2" ParentId="5871" CreationDate="2013-08-13T21:33:35.040" Score="2" Body="&lt;p&gt;You cant really combine those because for the top one you're making an acid, and for the bottom one you're making a base.  You've either got to make an acid on one side and consume it on the other, or consume a base on one side and make it on the other.  Here are the two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anode: $\ce{2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{2H+(aq) + 2e- ⟶ H2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply bottom by two and the $\ce{H+}$ and $\ce{e-}$ cancel out with the top leaving the overall reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anode: $\ce{4OH- (aq) ⟶ 2H2O(l) + O2(g) + 4e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{2H2O(l) + 2e- ⟶ H2(g) + 2OH- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply bottom by two and again get the overall reaction.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorUserId="2101" LastEditDate="2013-08-14T00:32:05.847" LastActivityDate="2013-08-14T00:32:05.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5877" PostTypeId="2" ParentId="5868" CreationDate="2013-08-13T23:47:58.297" Score="1" Body="&lt;p&gt;Adding to GregE's answer, one particular class of molecules where a Lewis diagram could be deceptive are &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;resonance structures&lt;/a&gt;.  In such molecules, there are multiple valid Lewis diagrams because the actual bonds are &lt;a href=&quot;https://en.wikipedia.org/wiki/Delocalized_electron&quot; rel=&quot;nofollow noreferrer&quot;&gt;delocalized&lt;/a&gt;, meaning that the electrons are free to roam over several bonds.  One of the most vivid examples of resonance is &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzene&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzene&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some representations of benzene from the wiki (licensed under Wikimedia Commons):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPauD.png&quot; alt=&quot;Benzene representations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the diagram in the middle of the top row.  It shows that there are two Kekule structures, which could be rendered as Lewis diagrams easily.  But the reason you need both is because the double bonds shown between the carbons are notional: there is no reason why either structure is preferred.  And the reality is that a benzene molecule is more like the &quot;average&quot; of the two structures.  The bottom row shows various representations showing the delocalized electrons, and all of these are &quot;better&quot; than the static Lewis or Kekule diagrams.  But the Kekule diagram is a lot easier to write down, and any chemist will understand that resonance is involved.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-08-13T23:47:58.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5878" PostTypeId="1" CreationDate="2013-08-14T00:02:14.057" Score="2" ViewCount="220" Body="&lt;p&gt;A half-cell consisting of palladium rod dipping into a $1\ M$ $\ce{Pd(NO_3)_2}$ solution is connected with a standard hydrogen half-cell. The cell voltage is $0.99\ volt$ and the platinum electrode in the hydrogen half-cell is the anode. Determine $E^ο$ for the reaction $\ce{Pd\longrightarrow Pd^{2+} + 2e}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is: can I find a numerical value of the needed $E^ο$, if I can't use half-cell potential of $\ce{Pt^{2+}\!−\!Pt}$?&lt;/p&gt;&#xA;" OwnerUserId="2144" LastEditorDisplayName="user15489" LastEditDate="2015-04-25T03:14:40.250" LastActivityDate="2015-04-25T03:14:40.250" Title="Finding one of the two unknown potentials with a given voltage of a net reaction" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5879" PostTypeId="2" ParentId="5871" CreationDate="2013-08-14T02:06:02.007" Score="4" Body="&lt;p&gt;It is not entirely inconceivable that a Galvanic cell could be created with an acidic anode and basic cathode.  Having both half-reactions occurring in the same beaker, however, is unlikely.  That has implications when determining the thermodynamics (cell potential) of the reaction but not in balancing the half reactions.  Given:&#xA;$$\begin{align} &#xA;\ce{2 H2O_{(l)} &amp;amp; -&amp;gt; O2 + 4 H+_{(aq)} + 4e^-} \tag{1}\\&#xA;\ce{ 2 H2O_{(l)} + 2 e- &amp;amp; -&amp;gt; H2_{(g)} + 2 OH-_{(aq)}} \tag{2}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply equation 2 by 2 and add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} &#xA;\ce{2 H2O_{(l)} &amp;amp; -&amp;gt; O2 + 4 H+_{(aq)} + 4e^-} \tag{3}\\&#xA;\ce{4 H2O_{(l)} + 4 e- &amp;amp; -&amp;gt; 2 H2_{(g)} + 4 OH-_{(aq)}} \tag{4} \\&#xA;\ce{6 H2O_{(l)} + 4 e- &amp;amp; -&amp;gt; O2 + 2 H2 + 4 H+ + 4 OH- + 4 e^-} \tag{5}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that the  $\ce{4H+ + 4OH-}$ showing up on the same side equation 5 will yield $\ce{4H2O}$.  Combining acid and base (to make $\ce{4 H2O}$ on the product side) and canceling the $\ce{4e-}$ and $\ce{4 H2O}$ that occur on both sides of the chemical equation gives the desired equation.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="24052" LastEditDate="2016-03-13T20:04:10.993" LastActivityDate="2016-03-13T20:04:10.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5882" PostTypeId="2" ParentId="5872" CreationDate="2013-08-14T13:13:25.310" Score="15" Body="&lt;p&gt;Let's use the following gaseous chemical reaction as an example:&#xA;$$\ce{H2 + I2 -&amp;gt; 2HI}$$&#xA;When we study reaction kinetics, we oftentimes can't observe the changes in concentration of each reactant and product.  Therefore the &lt;strong&gt;rate expression&lt;/strong&gt; helps us to relate the rates of appearance/disappearance of the species present:&#xA;$$-\frac{1}{1}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\; &#xA;= \; -\frac{1}{1}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\; &#xA;=\; \frac{1}{2}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{rate expression}\label{rateex}$$&#xA;Note that each rate is modified by the inverse of the stoichiometric coefficient.  I explicitly put $\frac{1}{1}$ in the expression just to emphasize this point.  We can see that &lt;em&gt;multiplying the coefficients of the reaction will not change the rate expression&lt;/em&gt; in the following example: let's multiply the coefficients in the chemical reaction by 2 and then write the corresponding rate expression.&#xA;$$\ce{2H2 + 2I2 -&amp;gt; 4HI}$$&#xA;$$-\frac{1}{2}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\; &#xA;= \; -\frac{1}{2}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\; &#xA;=\; \frac{1}{4}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{new expression}\label{newex}$$&#xA;The $\ref{rateex}$ and $\ref{newex}$ are numerically equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full disclosure here: the term &lt;em&gt;rate expression&lt;/em&gt; is ambiguous and is also used as a synonym for the &lt;em&gt;rate law&lt;/em&gt;. I've also seen the rate expression referred to as the &lt;em&gt;rate relationship&lt;/em&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rate law&lt;/strong&gt; or &lt;em&gt;differential rate law&lt;/em&gt; relates the rate of a reaction to the concentration (or pressure) of the reactants.  The rate of a reaction is proportional to the concentration (or pressure) of the reactants modified by some experimentally determined number called the &lt;strong&gt;reaction order&lt;/strong&gt;.&#xA;$$\begin{align}&#xA;-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}&#xA;&amp;amp;\propto\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\\&#xA;-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}&#xA;&amp;amp;= k\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\tag{rate law}&#xA;\end{align}$$&#xA;Here, $m$ and $n$ are the reaction orders for $\ce{H2}$ and $\ce{I2}$, respectively, and the sum, $m + n$ yields the &lt;strong&gt;overall order&lt;/strong&gt; of the reaction.  The proportionality $a\propto b$ can be replaced with an equality $a = k b$.  In kinetics, this constant of proportionality is referred to as the &lt;strong&gt;rate constant&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In summary: the &lt;strong&gt;rate expression&lt;/strong&gt; tells us how the appearance/disappearance rates of the products and reactants relate to one another; the &lt;strong&gt;rate law&lt;/strong&gt; tells us how the rate is related to the concentrations (or pressures) of reactants and the &lt;strong&gt;rate constant&lt;/strong&gt; is a constant of proportionality that comes out of the rate law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is a summary of information found at the &lt;a href=&quot;https://www.boundless.com/chemistry/chemical-kinetics/&quot; rel=&quot;noreferrer&quot;&gt;boundless website&lt;/a&gt; and from &lt;a href=&quot;http://www.amazon.com/Principles-Chemistry-Molecular-Approach-MasteringChemistry/&quot; rel=&quot;noreferrer&quot;&gt;Tro's&lt;/a&gt; most recent general chemistry textbook.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="4945" LastEditDate="2016-09-30T05:54:22.567" LastActivityDate="2016-09-30T05:54:22.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5883" PostTypeId="1" AcceptedAnswerId="5884" CreationDate="2013-08-14T16:30:03.840" Score="2" ViewCount="8042" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aldehydes#Specialty_methods&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says that esters can be reduced to aldehydes using DIBAl-H. Can anyone please explain how this reaction occurs?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="1573" LastEditDate="2013-08-14T20:22:31.930" LastActivityDate="2020-11-16T12:17:43.103" Title="How does partial reduction of esters to aldehydes occur?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5884" PostTypeId="2" ParentId="5883" CreationDate="2013-08-14T18:53:03.113" Score="7" Body="&lt;p&gt;DIBAL exists as a bridged dimer, and it becomes a reducing agent only after it has formed a Lewis acid–base complex, so it reduces electron-rich carbonyl groups most rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JMFGi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DIBAL will reduce esters even at –70 °C, and at this temperature the tetrahedral intermediate that forms during the reaction may be stable. Only in the aqueous acidic work-up does it collapse to the aldehyde (by expelling EtOH in the example reaction below) when excess DIBAL has been destroyed so that no further reduction to the alcohol is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G2BKx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stable tetrahedral intermediate is more likely in the reduction of lactones, and DIBAL is most reliable in the reduction of lactones to lactols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As requested by the OP here are some more details about the collapse of the tetrahedral intermediate: During the aqueous acidic work-up the alcohol group is turned into a good leaving group via protonation. Thus, it is easily expelled from the molecule, when a water molecule removes the $\ce{Al^{$i$}Bu_2}$ group and the $\ce{C=O}$ double bond is reformed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nx45X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A different route, where the former carbonyl-oxygen is protonated twice - thus being turned into a good leaving group - and afterwards expelled from the molecule to give an oxonium ion is possible too. But such an oxonium ion is rather unstable and highly reactive, so that the reverse reaction is very fast and the oxonium ion is not produced in any appreciable amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mKxJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-15T18:29:06.603" LastActivityDate="2013-08-15T18:29:06.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5885" PostTypeId="1" CreationDate="2013-08-14T19:28:24.747" Score="0" ViewCount="282" Body="&lt;p&gt;If I were to buy/build a large still, that was air tight, and could hold a vacuum, and filled the reservoir with vegetable glycerin (which has a boiling point higher than the majority or desired organic compounds) and I made the atmospheric pressure inside less than normal atmospheric pressure, could I distill chemicals out of organic matter without altering them? Such as distilling willow bark to get an aspirin containing solution, or distill mint leaves without changing the chemical composition of the medicinal compounds in the plant/organic matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be a cool project. It would enable us to explore further into natural medicine. But would this procedure work?&lt;/p&gt;&#xA;" OwnerUserId="2147" LastEditorUserId="865" LastEditDate="2015-10-02T16:48:06.167" LastActivityDate="2015-10-02T16:48:06.167" Title="Vacuum Distilled Medicine" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5886" PostTypeId="2" ParentId="5885" CreationDate="2013-08-14T21:05:20.783" Score="0" Body="&lt;p&gt;Not really. There are too many different types of compounds for only one procedure to work. If it was that easy there wouldn't be so many different ways of preparing extracts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to take one plant/compound to work with at a time.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-08-14T21:05:20.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5887" PostTypeId="1" AcceptedAnswerId="5895" CreationDate="2013-08-15T09:08:06.360" Score="2" ViewCount="106" Body="&lt;p&gt;I have to calculate the enthalpy of the reaction&#xA;$$\ce{Cs(l) + 1/2I2(s) -&amp;gt; CsI(s)}$$&#xA;and the answer is&#xA;$$\Delta H_f = \Delta H_{vap}^{\ce{Cs}} + \frac{1}{2}\Delta H_{subl}^{\ce{I2}} + \frac{1}{2}E_{diss}^{\ce{I2}} + I_1^{\ce{Cs}} + E_A^{\ce{I}} + E_N.$$&#xA;I can see that the molecules are first put to gas state before creating an ionic bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it have to do it ?&#xA;In which case do the molecules have to first go in gas state before reacting ?&#xA;Is it only in case of ionic bond ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E_{diss}$ is the energy of dissociation between the to atoms $\ce{I}$.&lt;/li&gt;&#xA;&lt;li&gt;$I_1$ is the energy of first ionisation.&lt;/li&gt;&#xA;&lt;li&gt;$E_A$ is the energy of electron affinity.&lt;/li&gt;&#xA;&lt;li&gt;$E_N$ is the energy of neutralisation between $\ce{Cs+}$ ans $\ce{I-}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2149" LastEditorUserId="2149" LastEditDate="2013-08-15T19:10:14.853" LastActivityDate="2013-08-15T20:47:11.307" Title="Why do molecules have to first transform to gas before doing a reaction?" Tags="&lt;enthalpy&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5888" PostTypeId="1" CreationDate="2013-08-15T15:11:16.233" Score="2" ViewCount="122" Body="&lt;p&gt;If we change the null half-potential, would it change half-potentials of other half-cell reactions by the same amount, so that voltages of net reactions stay the same?&lt;/p&gt;&#xA;" OwnerUserId="2144" LastEditorUserId="7448" LastEditDate="2015-07-20T19:41:59.767" LastActivityDate="2015-07-20T19:41:59.767" Title="Shift of the null half-reaction potential" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5889" PostTypeId="2" ParentId="1088" CreationDate="2013-08-15T15:33:16.037" Score="1" Body="&lt;p&gt;There is oxidation numbers method. Look it up.&#xA;Shortly, you give any hydrogen atom (in a compound which also has one different element, aside from oxygen) +1 charge, any oxygen atom (in a compound which also has one different element, aside from hydrogen) -- -2 charge. Using these and molecule charge you determine unknown atom's charge. Difference between charges of the same element on the left and on the right should equal difference between charges of another element on the left and on the right, multiplied by -1. Then you balance a reaction so that charge is conserved. &#xA;Of course, before that you need to use the information given to complete your reaction with compounds, that are initially absent. For example, in this case you see, that there should be oxygen on the right. Also it's said that reaction proceeds in the acidic solution, which means protons are used (add them on the left side), which means they will react with oxygen, which means on the right should be not oxygen, but water.&#xA;Walkthrough:&#xA;So we have&#xA;$MnO^−_4+Fe^{2+}⟶Fe^{3+}+Mn^{2+}$.&#xA;Obviously, we need to add $4O^{-2}$ on the right side and $H^+$ on the left side (and, consequently, on the right side too):&#xA;$MnO^−_4+Fe^{2+}+H^+⟶Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$.&#xA;As we establish by the oxidation numbers,&#xA;$Mn$'s charge (in $MnO_4^-$) equals $-1 - -2(4) = 7$. While on the right side it's charge is $+2$. $2 - 7 = -5$.&#xA;Iron also changes it's charge, so we see, that we need to find a coefficient to make a difference between charges of Fe on the right and Fe on the left equal $+5$. That would be $5$:&#xA;$MnO^−_4+5Fe^{2+}+H^+⟶5Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$&#xA;Now only thing, that's left, is to add a coefficient to the hydrogen, so water can be formed:&#xA;$MnO^−_4+5Fe^{2+}+8H^+⟶5Fe^{3+}+Mn^{2+}+4H_2O$.&lt;/p&gt;&#xA;" OwnerUserId="2144" LastEditorUserId="2144" LastEditDate="2013-08-15T17:30:18.363" LastActivityDate="2013-08-15T17:30:18.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5890" PostTypeId="1" CreationDate="2013-08-15T17:10:51.063" Score="2" ViewCount="495" Body="&lt;p&gt;I was wondering if anyone had any insight as to why the reaction of chlorosulfonic acid (colorless) and 1-dodecanol (colorless) produces a deep dark-red-colored solution? The reaction should give hydrogen lauryl sulfate, which according to &lt;a href=&quot;http://www.guidechem.com/dictionary/151-41-7.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;, is a white solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried adding $\ce{NaCl}$ to the chlorosulfonic first, and then adding the 1-dodecanol, which surprisingly gives the sodium dodecyl sulfate soap as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this color from over sulfonating the hydrocarbon chain on the hydrogen lauryl sulfate?  If so, what would this side product be called, and why isn't it seen much in industry if it is created along with sodium dodecyl sulfate?&lt;/p&gt;&#xA;" OwnerUserId="1265" LastEditorUserId="7448" LastEditDate="2015-07-20T19:40:56.127" LastActivityDate="2015-07-20T19:40:56.127" Title="Why does the reaction of chlorosulfonic acid (colorless) and 1-dodecanol (colorless) produces a deep dark-red-colored solution?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;spectroscopy&gt;&lt;surfactants&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5891" PostTypeId="2" ParentId="5860" CreationDate="2013-08-15T17:16:09.493" Score="1" Body="&lt;p&gt;Carbon Nanotubes and graphene can be metallic in the sense that they conduct electricity very well, they are produced mainly by humans (not seen much in nature), and they are created from carbon, which in it's graphitic form is not very conductive at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer doesn't answer all of your questions but it takes a small stab at the first.&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2013-08-15T17:16:09.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5892" PostTypeId="1" AcceptedAnswerId="5893" CreationDate="2013-08-15T18:13:59.493" Score="3" ViewCount="2888" Body="&lt;p&gt;Wiki says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The oxidation state of a free element (uncombined element) is zero&#xA;    for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2&#xA;          the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like $\ce{K2UO4}$? &lt;/p&gt;&#xA;" OwnerUserId="2144" LastEditorUserId="7448" LastEditDate="2015-07-20T19:39:15.930" LastActivityDate="2019-07-14T11:52:52.667" Title="How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen?" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5893" PostTypeId="2" ParentId="5892" CreationDate="2013-08-15T18:22:11.340" Score="3" Body="&lt;p&gt;In this case, the potassium ions are +1 each, and the oxygens are -2 as normal.  That leaves uranium in a highly oxidized +6 state.  The piece of information I think is missing is that Group 1 ions are almost always in the +1 state, because they so readily lose their outer valence electron.  A molecule that is not an ion, like your example, must have an overall charge of zero.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-08-15T18:22:11.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5894" PostTypeId="1" CreationDate="2013-08-15T18:58:35.027" Score="4" ViewCount="87" Body="&lt;p&gt;If MALDI is used as an ionization method, then TOF as an analyzer is suitable since the MALDI creates singly charged ions which have high m/z ratio and only TOF can analyze that ions, and multichannel plate should be used because the TOF is very quick and MCP's duty time is enough to capture all ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a kind of table about configurations similar to the example I gave above? A table with reasons would be great but not certainly necessary.&lt;/p&gt;&#xA;" OwnerUserId="1060" LastEditorUserId="7448" LastEditDate="2015-07-20T19:36:17.983" LastActivityDate="2015-07-20T19:36:17.983" Title="Is there a kind of table about configurations similar to the example I gave below?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5895" PostTypeId="2" ParentId="5887" CreationDate="2013-08-15T20:47:11.307" Score="2" Body="&lt;p&gt;Not all reactions require vaporization. This process looks like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born-Haber_cycle&quot; rel=&quot;nofollow&quot;&gt;Born-Haber cycle&lt;/a&gt; for cesium iodide, an exercise in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess%27_Law&quot; rel=&quot;nofollow&quot;&gt;Hess's Law&lt;/a&gt; usually used to fill in one of the missing values in the cycle that might be hard to measure (usually lattice energy, but not always). The Born-Haber cycle can also be used to drive general chemistry students crazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $E_N$ &quot;energy of neutralization&quot; looks like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_energy&quot; rel=&quot;nofollow&quot;&gt;lattice energy&lt;/a&gt;, which is &lt;strong&gt;defined&lt;/strong&gt; as the enthalpy/energy change accompanying the formation of a solid crystal lattice from the gas phase ions. In your case, the lattice energy of $\ce{CsI}$ is defined as the energy change accompanying the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cs+ (g) + I^- (g) -&amp;gt; CsI(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the lattice energy is defined for gas phase ions, you need the vaporization enthalpies for this process. Not all reactions require this vaporization step.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-08-15T20:47:11.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5896" PostTypeId="2" ParentId="5892" CreationDate="2013-08-15T21:27:43.240" Score="1" Body="&lt;p&gt;Technically (which may not be the answer you want) you compute the oxidation numbers in a molecule by first drawing its Lewis (dot) diagram showing all of the valence electrons.  Then you assign the electron pair in each bond to the more electronegative atom.  Then refigure the &quot;charges&quot; on each atom.  The resulting &quot;charge&quot; will be the oxidation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the years folks have worked out shortcuts to this.  For example, alkali metals easily lose an electron in almost all situations.  Thus we have for example that &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; has an oxidation number of +1.&lt;/p&gt;&#xA;" OwnerUserId="675" LastEditorUserId="72973" LastEditDate="2019-07-14T11:52:52.667" LastActivityDate="2019-07-14T11:52:52.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5897" PostTypeId="1" AcceptedAnswerId="5914" CreationDate="2013-08-16T00:47:12.270" Score="6" ViewCount="440" Body="&lt;p&gt;Does anyone know the analytical solutions to the following rate laws:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\frac{\mathrm{d}\rho_\text{T}}{\mathrm{d}t} = \rho_\text{W} \cdot K_2 - (K_4+K_5) \cdot \rho_\text{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) $\frac{\mathrm{d}\rho_\text{C}}{\mathrm{d}t} = \rho_\text{W} \cdot K_3 + \rho_\text{T} \cdot K_5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, I have determined the following analytical solution for this rate law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3) \qquad \Rightarrow \qquad \rho_\text{W} = \rho_{\text{W}0} \cdot \operatorname{e}^{-(K_1+K_2+K_3) \cdot t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho_{\text{W}0} = $ initial density, $K = $ reaction rate constant, $t =$ time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scheme for the reactions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{wood} \xrightarrow{K_{1}}{} \ce{gas}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{wood} \xrightarrow{K_{2}}{} \ce{tar}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{wood} \xrightarrow{K_{3}}{} \ce{char}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{tar} \xrightarrow{K_{4}}{} \ce{gas}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{tar} \xrightarrow{K_{5}}{} \ce{char}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help with the three rate laws above would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorDisplayName="user7951" LastEditDate="2015-03-26T16:18:01.313" LastActivityDate="2015-03-26T16:18:01.313" Title="Analytical solution to first-order rate laws" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;analytical-chemistry&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5899" PostTypeId="1" AcceptedAnswerId="5927" CreationDate="2013-08-16T11:20:18.813" Score="4" ViewCount="5939" Body="&lt;p&gt;When I have a gadolinium ion ($\ce{Gd^3+}$), how can I calculate its electron configuration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Gd} : \mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to first subtract the 2 electrons in the $\mathrm{s}$-orbital and then the $\mathrm{d}$-/$\mathrm{f}$-orbital?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="41328" LastEditDate="2017-12-10T12:26:32.070" LastActivityDate="2017-12-10T12:30:27.287" Title="Finding electron configuration of lanthanide ions" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5900" PostTypeId="1" AcceptedAnswerId="5919" CreationDate="2013-08-16T11:25:27.777" Score="3" ViewCount="167" Body="&lt;p&gt;Gold and Chromium have weird electron configurations, because electrons are differently divided : eg an electron jumps from a s-orbital to a d-orbital to have a half-filled s orbital and a fully/half filled d-orbital, which is much more stable.&#xA;But does this also occur with the p-orbitals? eg why does Fluorine not transfer an electron from its s orbital to its p orbital to have a  half-filled s orbital and a fully filled p-orbital, that would be more stable?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2013-08-18T18:35:00.293" Title="Special electron configurations in p block on periodic table?" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5901" PostTypeId="1" CreationDate="2013-08-16T12:24:39.313" Score="12" ViewCount="798" Body="&lt;p&gt;Was there any logic or thinking behind the experiments performed by Cannizzaro, Beckmann, Gattermann, Koch etc.? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did they thoughtfully perform experiments or just add any two reactants and luckily discovered the reactions which are known by their name today? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I meant to ask that in Physics all problems are treated mathematically and logically; But in case of Organic Chemistry, there are lot of exceptions to all statements which are made by Cannizzaro etc. It is said that if a single observation is against a scientific law; the law is abandoned and new perfect law takes its place. But it seems that Organic Chemistry is unaware of this and &quot;perfect&quot; statements are not part of it.&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="7448" LastEditDate="2015-07-20T19:23:57.393" LastActivityDate="2015-07-20T19:23:57.393" Title="Regarding logic behind Organic Chemistry" Tags="&lt;organic-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ClosedDate="2017-08-22T19:13:29.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="5902" PostTypeId="2" ParentId="5583" CreationDate="2013-08-16T13:53:52.510" Score="5" Body="&lt;p&gt;It is very simple, there is no need to solve it again since you already know the solution of infinite square well. The idea is very simple, the trap looks the same, but wider now $2L$. So first you should extend it by a factor of two and then move it back to the center, it is given by the rescaling and translation:&#xA;$$L \mapsto 2L \tag{1}$$&#xA;$$x \mapsto x-L \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is:&#xA;$$\psi_{n}=\sqrt{\frac{1}{L}}\sin\left(\frac{n\pi x}{2L}-\frac{n\pi}{2}\right) \tag{3}$$&#xA;$$E_{n}=\frac{n^2\pi^2 \hbar^2}{8mL^2}$$&#xA;As expected. Note that $\sin(\pi/2-x)=\cos(x)$ so that you can split it into odd and even series, but the Eq (3) given above is more succinct.&lt;/p&gt;&#xA;" OwnerUserId="2154" LastActivityDate="2013-08-16T13:53:52.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5903" PostTypeId="1" AcceptedAnswerId="5905" CreationDate="2013-08-16T17:55:37.497" Score="7" ViewCount="299" Body="&lt;p&gt;Uranyl formate has a structure something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{UO2-(OCHO)2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the structural formulae I've seen have two double-bonded oxygens directly attached to the U and each formate group is attached by single bonds to the oxygens. Like &lt;a href=&quot;http://images.chemnet.com/suppliers/chembase/cas3/cas16984-59-1.gif&quot; rel=&quot;nofollow&quot;&gt;this.&lt;/a&gt;&#xA;The empirical formula for the dry powder is written as a monohydrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain how this thing gets solvated in water and what the solvated ions are, exactly? I just realized I was assuming it dissociates into a 'uranyl' ion (whatever that looks like) and two formate ions$\ldots$but then I'm confused about where the formate's oxygen goes.&lt;/p&gt;&#xA;" OwnerUserId="912" LastEditorUserId="7448" LastEditDate="2015-07-20T19:21:46.953" LastActivityDate="2015-07-20T19:21:46.953" Title="Solvation of uranyl formate/acetate" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5904" PostTypeId="1" AcceptedAnswerId="111304" CreationDate="2013-08-16T20:46:14.500" Score="9" ViewCount="6196" Body="&lt;p&gt;If the fluorescence is the re-emitting of a photon with a larger wavelenght due to the transition from a higher energy state to a lower energy state the case of resonance Raman (where there aren't any virtual states involved) seems be equal to the fluorescence. Which differences are there?&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2013-09-30T21:23:28.877" LastActivityDate="2019-03-21T03:16:20.240" Title="Which are the differences between resonance Raman and fluorescence?" Tags="&lt;spectroscopy&gt;&lt;resonance&gt;&lt;raman&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5905" PostTypeId="2" ParentId="5903" CreationDate="2013-08-16T22:25:16.440" Score="7" Body="&lt;p&gt;This is a uranyl ion, one of the most common forms of dissolved uranium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u2yvF.png&quot; alt=&quot;Uranyl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two U=O bonds can form a single delocalised region, there's a strong tendency towards linearity (&lt;em&gt;c.f.&lt;/em&gt; carbon dioxide). It tends to dissolve in water by gaining equatorial water ligands (4 or 5), some of which it may then hydrolyse, depending on the pH:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRokQ.png&quot; alt=&quot;Hydrated uranyl and hydrolysed form&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial oxygen atoms are also valid targets for hydrogen bonding, typically gaining one hydrogen bond each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sgHuo.png&quot; alt=&quot;Axial oxygen hydrogen bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in dissolving uranyl formate in particular suggests that the formate ion is quite well-bound, and it seems that in this case you get a combination of water and formate ligands in the complex around the uranyl. I was expecting the formate to be a bi-dentate ligand, but according to the &lt;a href=&quot;http://dx.doi.org/10.1039/C3DT51711J&quot; rel=&quot;nofollow noreferrer&quot;&gt;results of a combined computational and experimental study&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/u2yvF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, it's uni-dentate (which incidentally allows it to form polymeric chains in solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture from that study showing a computationally obtained structure of a mixed water/formate-ligated uranyl structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QO9Mc.gif&quot; alt=&quot;Combination water and formate uranyl ion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;2013&lt;/strong&gt;, Dalton Trans., Christian Lucks, André Rossberg, Satoru Tsushima, Harald Foerstendorf, Karim Fahmya, Gert Bernharda, &lt;a href=&quot;http://dx.doi.org/10.1039/C3DT51711J&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI:10.1039/C3DT51711J&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="39" LastEditorUserId="39" LastEditDate="2013-08-17T12:11:48.577" LastActivityDate="2013-08-17T12:11:48.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5906" PostTypeId="2" ParentId="5901" CreationDate="2013-08-16T22:49:42.933" Score="19" Body="&lt;p&gt;Sometimes it was a lucky accident, sometimes there was a lot of math and logic going in the designing experiments.  The organic chemistry of that era, though, was a lot more applied and &lt;em&gt;empirical&lt;/em&gt; than the physics of today.  They usually weren't setting out to find the absolute laws-of-nature governing chemistry, they were just trying to make better dyes, medicines, rubber, etc, and stumbled upon new phenomena along the way.&lt;/p&gt;&#xA;&lt;h2&gt;Further Explanation&lt;/h2&gt;&#xA;&lt;p&gt;Almost all these guys worked in Germany in the late 1800s, when chemical industry was the hot new thing (think of biotechnology in today's world).  There was a lot of money to be made in making better dyes, medicines, rubber, etc, so people would hack/engineer the established methods for making that stuff, and see if the resulting products were any better.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;p&gt;Chemical A cures headaches really well, but causes nausea.  Chemical B gives a little relief from headaches, but doesn't have any side effects.  I bet I could make chemical C (which no one had ever made before), which has a structure that's sort of a combination of chemicals A and B, and maybe that will cure headaches with no side effects.&lt;/p&gt;&#xA;&lt;p&gt;==&amp;gt; {made the chemical and did some tests to find out the structure}.&lt;/p&gt;&#xA;&lt;p&gt;What the heck!  This has a completely different structure than what I was expecting, &lt;em&gt;it must have undergone a second reaction to rearrange itself&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;==&amp;gt; {publish paper},{get famous},{people 100 years later question the logic of your experiments} :)&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-11-16T09:50:52.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5907" PostTypeId="2" ParentId="5901" CreationDate="2013-08-17T02:26:05.607" Score="8" Body="&lt;p&gt;The problem with organic chemical reactions is that they are &lt;em&gt;hard&lt;/em&gt;.  Organic molecules have many covalent and polar covalent bonds, which require much energy to break.  Specifying which bond you want to break may require very specific chemical conditions.  It usually requires various catalysts to speed up the reactions.  So to determine exactly what organic reaction you want can be rendered &quot;logical&quot;, but it usually requires a lot of experimentation, which may well be sped up by &quot;luck&quot;.  But once the experimental results have been replicated, you have a &quot;logical&quot; procedure for the future.  It may have not involved mathematics, but it was proven by experiment, which is the ultimate criterion in science, even in physics!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-08-17T02:26:05.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5908" PostTypeId="1" CreationDate="2013-08-17T12:06:00.537" Score="2" ViewCount="8274" Body="&lt;p&gt;Is there any difference between &lt;a href=&quot;http://en.wikipedia.org/wiki/18-Electron_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;18-electron rule&lt;/a&gt; and &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/179961/effective-atomic-number-EAN#ref6518&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective atomic number rule&lt;/a&gt;? Both rules are based on attainment of a noble gas configuraion as a criteria for stability of complexes. Besides, both rules were proposed by Sidgwick. I was wondering if the mere fact that 18-electron rule applies mainly to organometallic complexes is a good enough difference.&lt;/p&gt;&#xA;" OwnerUserId="2158" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-08-18T15:55:44.093" Title="Difference between 18 electron rule and effective atomic number rule" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5910" PostTypeId="1" CreationDate="2013-08-17T18:07:00.800" Score="2" ViewCount="1394" Body="&lt;p&gt;For an ideal chain, Flory Huggins theory gives us a distribution which is like a Gaussian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(l, r) \sim \left(\frac{3}{2 \pi l^2}\right)^{3/2} e^{\frac{-3 r^2}{2 l^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $l$ is the monomer to monomer distance(also known as Kuhn length) and $R$ is the size of the monomer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is great but what are the approximation away from an ideal chain polymer? I want to understand how one can go beyond Flory Huggins Theory.&lt;/p&gt;&#xA;" OwnerUserId="2160" LastActivityDate="2014-04-05T14:39:20.240" Title="Flory Huggins Theory" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5911" PostTypeId="2" ParentId="5194" CreationDate="2013-08-18T02:10:44.337" Score="2" Body="&lt;h3&gt;Valence electrons&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Valence electrons are electrons in the outermost shell of an atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Valency&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Valency is the number of electrons involved in the formation of a chemical bond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Coming to your question&lt;/h3&gt;&#xA;&lt;p&gt;The number of valence electrons of helium is 2 because of its configuration which is &lt;span class=&quot;math-container&quot;&gt;$1s^2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hovewer, the valency is 0 because it is already stable and none of the outermost electrons are involved in the formation of chemical bonds.&lt;/p&gt;&#xA;" OwnerUserId="1912" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-08-18T04:37:25.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5913" PostTypeId="2" ParentId="2881" CreationDate="2013-08-18T06:12:24.627" Score="6" Body="&lt;p&gt;Kinetic - how unwillingly something reacts.&#xA;Thermodynamic - the energies involved, amount of gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, aluminium metal is kinetically much more stable than iron metal in damp air (it doesn't rust), but thermodynamically much less stable (if it did oxidise, it would give out more energy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are thermodynamically unstable in air, but don't worry, you're not going to spontaneously combust, which is another way of saying that you are kinetically stable.&lt;/p&gt;&#xA;" OwnerUserId="2161" LastActivityDate="2013-08-18T06:12:24.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5914" PostTypeId="2" ParentId="5897" CreationDate="2013-08-18T07:14:55.783" Score="9" Body="&lt;p&gt;This is the solution I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\rho _T(t) &amp;amp;= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &amp;amp;= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &amp;amp;=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$.&#xA;Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-08-18T07:14:55.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5915" PostTypeId="1" AcceptedAnswerId="5916" CreationDate="2013-08-18T12:44:48.113" Score="27" ViewCount="27227" Body="&lt;p&gt;If we look at solubility of salts in water, we are told that they disassociate, because the positive $\ce{Na}$ ion is attracted to the partially negative oxygen in water and the negative  $\ce{Cl}$ is attracted to the partially positive hydrogen. But why does this happen? I mean, the couloumb force should be much stronger between the $+1$ and $-1$ charges on the $\ce{Na}$ and $\ce{Cl}$ than the partial charges on the polar water molecule. Why does $\ce{NaCl}$, then, disassociate? It would make much more sense for the two ions to stick together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when you cut $\ce{NaCl}$, shouldn't it stick to its pieces? Look at salt crystals: they don't have any force of attraction between them. But, if you have positive sodium ions and negative chloride ions, then they should stick together. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="16683" LastEditDate="2016-12-29T13:16:00.593" LastActivityDate="2018-07-23T08:06:07.970" Title="Why do salts such as NaCl dissolve?" Tags="&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5916" PostTypeId="2" ParentId="5915" CreationDate="2013-08-18T15:17:15.513" Score="21" Body="&lt;p&gt;As it happens, the enthalpy of solution of $\ce{NaCl}$ in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, i.e., it is an endothermic process. At a constant temperature and pressure, these kinds of thermodynamic processes are dictated by the change in Gibbs free energy, described by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta G = \Delta H - T\Delta S&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Delta G &amp;lt; 0$ is a necessary criterion for a spontaneous process. Given that $\Delta H$ is positive, $\Delta S$ must be positive as well, otherwise the process wouldn't occur spontaneously (which is to say, not without input of work from the surroundings, or coupling to some other strongly favorable reaction, neither of which is the case for the dissolution of salt). In other words, this is a process that is driven by the increase in entropy, which is entirely to be expected when moving from a highly ordered state (i.e., a crystalline solid) to a less ordered liquid solution. Indeed, in ideal solutions, where intermolecular forces of attraction are taken to be equal between all components, enthalpy change is necessarily zero, and entropy is always positive, so that the process of mixing in ideal solutions is always spontaneous. Of course, in real solutions, this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your intuition that the Coulombic forces between ions should be stronger is correct in this particular instance, as indicated by the positive enthalpy (meaning, the breaking of ionic bonds in the crystal lattice, as well as intermolecular bonds between solvent molecules, requires more energy than is released in the forming of ion-dipole bonds). The input of energy required for this process comes in the form of heat, drawn from the solvent. However, there are numerous examples of salts for which the enthalpy of solution is negative under a wide range of conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement &quot;[l]ook at salt crystals: they don't have any force of attraction between them&quot; is inaccurate, though. In fact, there is strong ionic bonding. The very fact that the crystals remain solid except at very high temperatures is sufficient evidence of this, and the strength of the bonding in the crystal lattice can be quantified by calculating the lattice energy. It's a mistake to consider any substance in isolation. When you talk about &quot;cutting&quot; $\ce{NaCl}$, what you're talking about is applying physical force to the crystals, exposing them to the complex mixtures of gasses and mositure in the air, and bringing them in contact with, say, the metal in the knife. If left in relative isolation, the salt remains undisturbed, packed in a crystal lattice. A gas, however, which has weak forces of attraction between individual particles, will expand to fill a container without any work being done to it (at least as far as atmospheric pressure allows). A liquid will experience capillary forces, either climbing the walls of a narrow container, resulting in a concavity at its surface, or pooling away from the walls of the container, resulting in a convexity at its surface (which of the two happens depends mainly on the comparative strengths of the forces of attraction between the molecules of the liquid and those of the substance comprising the container).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-18T17:26:37.953" LastActivityDate="2013-08-18T17:26:37.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5917" PostTypeId="2" ParentId="5897" CreationDate="2013-08-18T15:23:44.200" Score="9" Body="&lt;p&gt;For what it's worth, problems like this can be solved very nicely with Mathematica and I find that it provides access to some really cool chemistry even when the math is a bit overwhelming for some students.  One can obtain the same solution as @ManishEarth with the following Mathematica code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;soln = DSolve[&#xA;  {&#xA;   wood'[t] == -wood[t] (k1 + k2 + k3),&#xA;   tar'[t] == wood[t]*k2 - (k4 + k5)*tar[t],&#xA;   char'[t] == wood[t]*k3 + tar[t]*k5,&#xA;   wood[0] == wood0,&#xA;   tar[0] == 0,&#xA;   char[0] == 0},&#xA;  {wood[t], tar[t], char[t]}, t]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The fun part is how easy it is to visualize the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mnds8.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are ways to dynamically manipulate the plot such that one can explore the effect of the rate constants on the time-dependent concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RT6D.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical rigor of my solution is naturally a bit weak, but oftentimes my emphasis is on teaching the science.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-08-18T15:23:44.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5918" PostTypeId="2" ParentId="5908" CreationDate="2013-08-18T15:55:44.093" Score="5" Body="&lt;p&gt;Both of these rules are equivalent and are related to transition metals attaining a noble gas configuration of electrons in coordination complexes. The difference between them is the number of electrons given to the central metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective atomic number uses all of the metals electrons. For example, in an $Fe^{2+}$ complex, the central iron atom has 24 electrons, and you would want the ligands coordinated to the metal to contribute another 12 electrons to reach 36 electrons, the next noble gas configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 18-electron rule is a simplification of the EAN. There is always a noble gas configuration of electrons for the metals core orbitals (e.g. 1s2s2p3s3p for $Fe^{2+}$), so you can ignore these electrons in the counting scheme. For transition metals, this means that the only important orbitals are the valence nd(n+1)s(n+1)p orbitals, which can hold a total of 18 electrons. For $Fe^{2+}$, there are 6 electrons in these orbitals. Again, the ligands would need to donate 12 electrons to reach the 18 electron limit.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-08-18T15:55:44.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5919" PostTypeId="2" ParentId="5900" CreationDate="2013-08-18T18:35:00.293" Score="3" Body="&lt;p&gt;The simplest answer to this problem is related to relative energy differences between the Nd and (N+1)s orbitals in transition metals and Ns and Np orbitals. The Nd orbitals in transition metals are at a slightly lower energy than the (N+1)s orbitals, so for certain atoms there are some unusual changes in the electron configurations, e.g. Cr and Cu. However, the energy differences between the Ns and Np orbitals are quite large, with Np being higher in energy. In these cases, there is a substantial energetic penalty in moving an electron from the Ns to Np orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron configuration problem for transition metals is complex, but &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed071p469&quot; rel=&quot;nofollow&quot;&gt;this J Chem Ed article&lt;/a&gt; has a nice description of the theory, although it is somewhat technical.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-08-18T18:35:00.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5920" PostTypeId="2" ParentId="5904" CreationDate="2013-08-18T19:45:08.407" Score="7" Body="&lt;p&gt;The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of the molecule can relax to accommodate this new electron configuration. However, this new geometry is not equivalent to the lowest energy structure, so when the molecule returns to the ground state, the energy of the system will be slightly higher than where it started. This means that the fluorescence energy is always lower than the excitation energy (this is true for molecules, not atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Raman scattering, light absorption promotes the molecule to a very short-lived 'virtual' excited state. When this state relaxes, it can return to a different vibrational or rotational sublevel of the ground state than where the molecule started. This leads to slightly different energies than the elastically scattered light, which is at the same energy as the incoming radiation source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance Raman is a special case where the excitation energy is tuned to a particular electronic transition of a molecule. This has the advantage that vibrational transitions associated with the electronic change are substantially enhanced compared to other Raman transitions, which simplifies the Raman spectrum. Because you are exciting a specific electronic transition, though, it is possible to get a substantial fluorescence background as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed081p705&quot; rel=&quot;noreferrer&quot;&gt;J Chem Ed article&lt;/a&gt; has a nice description of Raman versus fluorescence. ChemWiki has a page devoted to the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Raman_Spectroscopy/Resonant_vs._Nonresonant_Raman_Spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;differences between Raman and Resonace Raman&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-08-18T19:45:08.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5921" PostTypeId="1" CreationDate="2013-08-19T01:08:12.310" Score="6" ViewCount="9709" Body="&lt;p&gt;I was taking a chemistry test and I encountered the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many electrons can an orbital of type f hold?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. 6&lt;br&gt;&#xA;  B. 10&lt;br&gt;&#xA;  C. 2&lt;br&gt;&#xA;  D. 14&lt;br&gt;&#xA;  E. 1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there can be [-ℓ, ℓ] orientations and since the orbital type f has ℓ = 3, we should have 7 possible orientations with 2 spins, so ${7 \times 2 = 14}$, so I thought the correct answer was D (14).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I got it wrong and the correct answer is marked as C (2). Is it an error in the test, or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="1878" LastEditDate="2013-08-19T18:04:14.563" LastActivityDate="2013-08-19T18:04:14.563" Title="How many electrons can an orbital of type f hold?" Tags="&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5922" PostTypeId="1" CreationDate="2013-08-19T01:38:50.163" Score="19" ViewCount="49241" Body="&lt;p&gt;We all know that pure water can't conduct electricity. But during electrolysis, if add a small amount of $\ce{HCl}$ acid inside, water could be decomposed to hydrogen gas and oxygen gas, as it is reduced at the cathode and oxidised at the anode. So why can't pure water just conduct electricity by electrolysis? &lt;/p&gt;&#xA;" OwnerUserId="2134" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T22:40:55.087" LastActivityDate="2021-03-19T16:09:31.387" Title="Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode?" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5923" PostTypeId="2" ParentId="5921" CreationDate="2013-08-19T01:44:41.947" Score="10" Body="&lt;p&gt;You're correct that there are seven possible spatial orientations for an f-type orbital, and hence seven possible orbitals in one f-type sub-shell. However, the question specifically asks for the maximum number of electrons in one such orbital, and &lt;em&gt;any&lt;/em&gt; single atomic orbital, regardless of the sub-shell type specified by $l$, can only hold two electrons. This is by virtue of the Pauli exclusion principle. Fourteen would be the maximum number of electrons across an entire f-type sub-shell, but the question only asks about one orbital.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-08-19T01:44:41.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5924" PostTypeId="2" ParentId="5922" CreationDate="2013-08-19T02:26:39.173" Score="10" Body="&lt;p&gt;Pure water has a very high resistivity, on the order of 18 MOhm/cm.  This effectively means that any applied potential is going to be converted into thermal energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side note, oxidation of an $\ce{HCl}$ (or any salt with chloride ions) solution will produce $\ce{Cl_2}$ gas, so electrolysis of these solutions should be performed in a well ventilated area.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T22:41:26.503" LastActivityDate="2015-05-03T22:41:26.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5925" PostTypeId="1" AcceptedAnswerId="5930" CreationDate="2013-08-19T05:45:06.520" Score="17" ViewCount="78198" Body="&lt;p&gt;Recently the &lt;a href=&quot;http://articles.mercola.com/sites/articles/archive/2013/08/18/exclusion-zone-water.aspx&quot;&gt;quack medicine&lt;/a&gt; folks online have been promoting the research of a certain &lt;a href=&quot;http://faculty.washington.edu/ghp/&quot;&gt;Dr. Gerald Pollack&lt;/a&gt; who claims to have discovered a &quot;forth phase of water&quot;, and who has recently published a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0962689548&quot;&gt;book&lt;/a&gt; on the said topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's clear that his research is being deliberately misinterpreted, it's not clear to me whether or not his own claims are valid in the first place. Although some papers on the topic were published in peer reviewed journals, it doesn't seem like anyone else in the scientific community has acknowledged or replicated his results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is &quot;EZ-water&quot; a real breakthrough or just the brainchild of a deluded scientist?&lt;/p&gt;&#xA;" OwnerUserId="1754" LastEditorUserId="21964" LastEditDate="2016-08-16T21:27:27.910" LastActivityDate="2021-05-02T18:45:03.477" Title="EZ-Water - Fraud or breakthrough?" Tags="&lt;water&gt;&lt;energy&gt;&lt;reference-request&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5926" PostTypeId="1" CreationDate="2013-08-19T08:26:46.170" Score="2" ViewCount="3160" Body="&lt;p&gt;Looking at Wikipedia, Gold &amp;amp; Platinum are in adjacent seats.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Platinum                                Gold&#xA;[Xe] 4f14 5d9 6s1                     [Xe] 4f14 5d10 6s1&#xA;Atomic number of 78                   Atomic number of 79  &#xA;Atomic pass of 195.084u               Atomic mass of 196.96657 ± 0.000004 u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How does that single electron make such a huge difference?&lt;/p&gt;&#xA;" OwnerUserId="765" LastActivityDate="2015-04-30T15:23:59.373" Title="What makes Gold so different from Platinum?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ClosedDate="2017-09-05T12:56:31.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="5927" PostTypeId="2" ParentId="5899" CreationDate="2013-08-19T09:26:30.643" Score="5" Body="&lt;p&gt;I agree with &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5899/finding-electron-configuration-of-lanthanide-ions#comment8649_5899&quot;&gt;bobthechemist's comment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are determining the electronic configuration of anions/cations, you will first need to determine electronic configuration of neutral atom. Like in the case above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Gd} : \mathrm{[Xe] (6s)^2 (4f)^7 (5d)^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now arrange this configuration according to ascending value of $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{[Xe] (4f)^7 (5d)^1 (6s)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, according to your need start subtracting electrons from highest value of $n$, like for $\ce{Gd^3+}$, the electronic configuration will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Gd^3+} : \mathrm{[Xe] (4f)^7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_Theory/Trapped_Particles/Atoms/Electron_Configuration#Electronic_Configurations_of_Cations_and_Anions&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; to know more on calculating electronic configuration of cations and anions.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="41328" LastEditDate="2017-12-10T12:30:27.287" LastActivityDate="2017-12-10T12:30:27.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5928" PostTypeId="1" CreationDate="2013-08-19T11:30:48.093" Score="1" ViewCount="31032" Body="&lt;p&gt;Copper have $[Ar] 3d^{10} 4s^1$ electron configuration. And Zinc have $[Ar] 3d^{10} 4s^2$ electron configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the right quantum numbers of last electron of copper? Is it $$a. n=3,l=2,m=+1,s=-1/2$$ or $$b. n=3,l=2,m=+2,s=-1/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use A, the electron configuration of copper is $[Ar] 3d^{9} 4s^2$ and it isn't stable. But if I use B quantum numbers, it is the same with the last electron of zinc.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2013-08-19T12:39:40.210" Title="What is the right quantum numbers of Copper?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5929" PostTypeId="2" ParentId="5928" CreationDate="2013-08-19T12:33:34.870" Score="3" Body="&lt;p&gt;The correct quantum numbers for last electron of Copper ($\ce{Cu}$) would be $B. n=3,l=2,m=+2,s=-1/2$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if I use B quantum numbers, it is the same with the last electron of zinc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, last electrons of Both can have same quantum numbers because Pauli Exclusion law says that no two electrons &lt;strong&gt;in an atom can have all four quantum numbers same&lt;/strong&gt; but it doesn't say that two electrons of different atoms can't have all four quantum number same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer &lt;a href=&quot;http://www.kentchemistry.com/links/AtomicStructure/quantumnumbers.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for more.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-08-19T12:39:40.210" LastActivityDate="2013-08-19T12:39:40.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5930" PostTypeId="2" ParentId="5925" CreationDate="2013-08-19T12:39:17.177" Score="26" Body="&lt;p&gt;EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking &amp;quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ionized_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionized water&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The original &amp;quot;article&amp;quot; linked, contains a number of misleading statements.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water molecules make up 99% of your body. This is true, but misleading. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Body_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;various sources&lt;/a&gt;, the human body is between 57% and 75% water by mass. Since water has a much lower molar mass than most biomolecules, 99% of the molecules in your body may well be water.&lt;/li&gt;&#xA;&lt;li&gt;The journal listed in which the research is published is &lt;em&gt;&lt;a href=&quot;http://www.mdpi.com/journal/water&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water&lt;/a&gt;&lt;/em&gt;, which is not a high impact journal, despite its name. &lt;em&gt;Water&lt;/em&gt; had an &lt;a href=&quot;http://en.wikipedia.org/wiki/Impact_factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;impact factor&lt;/a&gt; of 0.973 in 2012. That means that articles published in 2011 in &lt;em&gt;Water&lt;/em&gt; were cited less than 1 time each by &lt;strong&gt;anyone&lt;/strong&gt; in 2012. Compare with &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cell_%28journal%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cell&lt;/a&gt;&lt;/em&gt;, which had in impact factor of 31.957 in the same period - each article in &lt;em&gt;Cell&lt;/em&gt; is cited an average of nearly 32 times.&lt;/li&gt;&#xA;&lt;li&gt;Water has a &amp;quot;fourth phase&amp;quot; and no single boiling point or melting point. This is true! Any student of chemistry should be able to tell you this. Take a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_%28data_page%29#Phase_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase diagram of water&lt;/a&gt; below. Water apparently has at least 18 phases (15 solid phases, plus liquid, vapor, and supercritical fluid). You can also see that the solid-liquid and liquid-vapor phase boundaries are not single points. They are curves varying in temperature and pressure. Those curves represent the conditions in which the two phases have the same &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure#Relation_to_boiling_point_of_liquids&quot; rel=&quot;nofollow noreferrer&quot;&gt;vapor pressure&lt;/a&gt;. This is not a new discovery or a startling secret. This behavior has been known for many decades (if not centuries).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqyR0.png&quot; alt=&quot;phase diagram of water&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Later on the &amp;quot;article&amp;quot; tells us that EZ-water is not &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; (note that they cannot be bothered to format their chemical formulas) but &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O2}$&lt;/span&gt;, which is alkaline with a negative charge. Hmmm... If it has a different formula and different properties, it must be a different compound. Note that this &amp;quot;formula&amp;quot; is essentially &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O2 = H2O + OH}$&lt;/span&gt;, and hints at what happens inside these water ionizers.&lt;/li&gt;&#xA;&lt;li&gt;EZ-water has a negative charge. It is possible for objects to build up charge, but it is not possible for them to remain permanently charged. Overall, the universe is balanced in charge. Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Lightning&quot; rel=&quot;nofollow noreferrer&quot;&gt;lightning&lt;/a&gt; as a consequence of attempting to maintain separation of charge. This &amp;quot;negative charge&amp;quot; comes from a misunderstanding of pH. The water definitely has a higher pH, which means more &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions than &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions, but something (probably &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;) needs to replace the missing &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I can go on, but you get the point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;A water ionizer &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrolyzes&lt;/a&gt; water, which requires electricity (not light as suggested by Mercola). I had the good fortune to be involved in the testing and dismantling of one of these ionizers recently. The device claimed to take ordinary tap water and convert it to a &amp;quot;super-cleaning solution that contained only negatively charged oxygen particles&amp;quot; or something like that. From experiments:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ionizer increases the pH of the water.&lt;/li&gt;&#xA;&lt;li&gt;The ionizer does not work with deionized or distilled water!&lt;/li&gt;&#xA;&lt;li&gt;The ionizer does not produce measurable oxygen gas (a smoldering splint will not relight in the vicinity)&lt;/li&gt;&#xA;&lt;li&gt;The device contains an electrolysis cell and an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_exchange_resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;ion exchange resin&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Here is what happens:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Water is electrolyzed.&lt;br /&gt;&#xA;Doing so requires &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5922/why-cant-pure-water-conduct-electricity-since-it-can-be-reduced-at-cathode-and&quot;&gt;ions in the water already&lt;/a&gt; to conduct the electricity, which is why the device must use tap water (and not purified water). At the cathode, water is reduced to hydrogen and hydroxide. At the anode, it is oxidized to oxygen and protons.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2H2O +2e- &amp;amp;-&amp;gt; H2 + 2OH-}\\&#xA;\ce{2H2O &amp;amp;-&amp;gt; O2 +4H+ + 4e-}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Protons are exchanged for sodium ions&lt;br /&gt;&#xA;To prevent the recombination of hydroxide and protons to form water and neutralize the solution, there is an ion exchange membrane between the anode and the rest of the solution.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + OH- -&amp;gt; H2O}$$&lt;/span&gt;&#xA;The ion exchange resin collects the protons and releases sodium ions into solution to maintain charge neutrality, which must happen or you would have a lightning gun and not a mystic device to produce super water.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Ultimately, drinking this stuff would probably have no effect on your health. But we are no doctors, so don’t listen to us either. It might help or it might hurt, depending on your condition, it also might not matter.&lt;/p&gt;&#xA;&lt;p&gt;First, your stomach secretes acid, so any extra base would be neutralized immediately. Let's say an ionizer gets the water's pH to 8 (slightly basic, and more basic than your blood). This means that the concentration of protons is &lt;span class=&quot;math-container&quot;&gt;$\ce{[H+]}=10^{-\text{M}pH}=10^{-8} \text{ M}$&lt;/span&gt; (remember that &lt;a href=&quot;http://en.wikipedia.org/wiki/PH&quot; rel=&quot;nofollow noreferrer&quot;&gt;pH is a logarithmic scale&lt;/a&gt;. The concentration of hydroxide is &lt;span class=&quot;math-container&quot;&gt;$10^{-14} / 10^{-8} = 10^{-6} \text{ M}$&lt;/span&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Stomach_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;pH in your stomach&lt;/a&gt; ranges from 1.5 - 3.5, which represent proton concentrations between &lt;span class=&quot;math-container&quot;&gt;$10^{-3.5}= 3.16\times 10^{-4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10^{-1.5}=3.16\times 10^{-2}$&lt;/span&gt;. If we use the higher pH stomach acid to neutralize an equal volume of EZ-water, the final pH would be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;[\ce{H+}]_f &amp;amp;= \frac{3.16\times 10^{-4} - 1\times 10^{-6}}{2}=\frac{3.15\times 10^{-4}}{2}=1.07_5 \times 10^{-4}\\&#xA;\text{pH} &amp;amp;=-\log[\ce{H+}] =-\log(1.07_5 \times 10^{-4})=3.97&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since your stomach acid would likely have a greater volume than the ionized water, the change in pH would be even smaller. This calculation assumes your stomach acid is not a buffer, which it is.&lt;/p&gt;&#xA;&lt;p&gt;Second, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid-base_homeostasis&quot; rel=&quot;nofollow noreferrer&quot;&gt;acid-base homeostasis&lt;/a&gt; of your blood prevents the pH of your body from changing. Yes, your blood pH is 7.4 or so (which is slightly basic), but that does not mean your blood is negatively charged. Outside of range of 7.38 - 7.42 all of your blood proteins (like hemoglobin) start to denature, which will kill you.&lt;br /&gt;&#xA;Well, to be more precise, you will suffocate before that happens as hemoglobin/myoglobin oxygen binding capacity is pH sensitive. And if anything, we can tolerate acidic blood more robustly than basic blood pH (down to 6.8 or 7, but only up to about 7.6) before death. You might want to look up acidosis.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2021-05-02T18:45:03.477" LastActivityDate="2021-05-02T18:45:03.477" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="5931" PostTypeId="2" ParentId="5601" CreationDate="2013-08-19T15:05:27.997" Score="1" Body="&lt;p&gt;Titanium and Tantalum are quite nonreactive, so using 'safe' acids, such as diluted acetic acid won't make any difference. That's why using strong acids such as (sometimes has to be hot ) sulfuric acid will do it. If you are looking for not having to work with sulfuric acid, you could always use hydrogen iodide: that is even stronger than sulfuric acid... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's documented &lt;a href=&quot;https://www.google.nl/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=3&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CC4QFjACahUKEwja4rqz3uLGAhVBGywKHbZLBbc&amp;amp;url=http%3A%2F%2Fwww.chem.ualberta.ca%2F~chem33x%2Fchandouts%2Flecture26_files%2Flecture26.ppt&amp;amp;ei=rUGpVZrYMsG2sAG2l5W4Cw&amp;amp;usg=AFQjCNE8Tu7Zq3WNGk9EWbJC2szx7VVezQ&amp;amp;sig2=6hdekN7n5wU6tc91SB8TVg&amp;amp;bvm=bv.98197061,d.bGg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tantalum appear not to be attacked by many acids at room temperature&#xA;  but does dissolve in  hydrofluoric acid, $\ce{HF}$, or oleum (a solution of&#xA;  sulfur trioxide, $\ce{SO3}$, in sulfuric   acid, $\ce{H2SO4}$, also known as&#xA;  fuming sulfuric acid). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Reaction of tantalum with water&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Tantalum does not react with water under normal conditions. The&#xA;  surface of tantalum   metal is protected by a thin oxide layer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then I read for titanium (&lt;a href=&quot;https://www.webelements.com/titanium/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reaction of titanium with water&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Titanium metal is coated with an oxide layer that usually renders it&#xA;  inactive. However, titanium will react with steam form the&#xA;  dioxide, titanium(IV) oxide, $\ce{TiO2}$,   and hydrogen, $\ce{H2}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{Ti(s) + 2H2O(g) → TiO2(s) + 2H2(g)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So Tantalum doesn't react with steam so you might be able to use steam for that separation as seen as $\ce{TiO2}$ is white and tantalum not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source : &lt;a href=&quot;http://caldersgr9science.wikispaces.com/73+Tantalum&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://caldersgr9science.wikispaces.com/73+Tantalum&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="402" LastEditDate="2017-04-08T21:58:47.620" LastActivityDate="2017-04-08T21:58:47.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5932" PostTypeId="2" ParentId="2429" CreationDate="2013-08-19T15:33:43.627" Score="0" Body="&lt;p&gt;I've read in a book that Europium is used in the red phosphors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See more at this wikipedia page : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphor&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Phosphor&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;on this part of the webpage you will find info about used substances : &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphor#Phosphor_degradation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Phosphor#Phosphor_degradation&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphor#Materials&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Phosphor#Materials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2013-08-19T15:33:43.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5934" PostTypeId="1" AcceptedAnswerId="5945" CreationDate="2013-08-19T17:28:40.170" Score="8" ViewCount="2693" Body="&lt;p&gt;I was linked to a Kickstarter for a cooling ball to be placed in drinks which claims with constant marketing hyperbole how much it relies on &quot;Phase Change&quot; ! which is a neat concept and seems not unreasonable, but in thinking about it I'm failing to imagine (as I'm no chemist, and of ill education) what refrigerant could be inside that meets the named requirements of non-toxic food safe, as well as passing from solid to liquid (or liquid to gas) such that a typical freezer would freeze it (−2 °C to −5 °C typically?), and it would liquefy around a temperature you'd want to keep your drink (say 1–3 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously water meets these requirements, however due to the expansion pressures water commits I have a very hard time imagining a small stainless steel ball containing that expansion through many freeze-thaw cycles without causing eventual deformities which would be continually weakened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is in the title, and to note the container is a sealed solid stainless steel ball of unknown thickness 2&quot; in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this question is off-topic, more than anything I'm just curious so I realize sometimes questions not based on a problem are poor fits for SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do note: I am not interested in actually trying to do something with this, it is simply a curiosity, so don't be concerned if the things that come to mind aren't FDA tested, I'll settle for answers that simply show there are such things that are likely non-toxic regardless of thorough safety testing. Also the phase change from liquid to gas is a possible one too if you know of any safe chemicals here that wouldn't cause uncontrollable internal pressure and would have that phase change at the temperature range specified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just trying to come up with an idea for a compound that could meet their marketing hyperbole and finding I come up short.&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T14:10:24.610" LastActivityDate="2015-04-26T14:10:24.610" Title="What non-toxic non-water substances have a freezing point very close to water's?" Tags="&lt;enthalpy&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5937" PostTypeId="2" ParentId="5934" CreationDate="2013-08-20T03:06:14.863" Score="1" Body="&lt;p&gt;Jimmy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are most likely many chemicals that have a melting point range in that vicinity, but I can't comment on their toxicity. However, I can comment on some of the science here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, water is one of very few molecules that &lt;em&gt;expands&lt;/em&gt; when it freezes. That has to do with the hydrogen bond network formed between water molecules in the solid state. This network is slightly less dense than low-temperature liquid water, so water will expand when it is frozen. (This is also why solid water floats on liquid water.) Most compounds form frozen solids that are more dense than their liquid state, so they will &lt;em&gt;contract&lt;/em&gt; on freezing. This is why it is fine to freeze these metal balls if they are filled with something besides water. Although they could have water inside, but just not be completely full, I suppose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were going to find a water substitute, then you would want to find a chemical that has a high heat capacity and enthalpy of fusion. Heat capacity is the amount of energy required to heat a given substance by a certain temperature. If a substance has a high heat capacity, then it take more energy to change the temperature of that material. (It gets this energy by removing heat from your beverage.) The enthalpy of fusion is the amount of energy it takes to change a substance from a solid to a liquid. Again, you would want something with a high enthalpy of fusion for the same reasons as for heat capacity. Unfortunately, water is really great on both of these fronts, which is why ice cubes are so great for cooling a drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I really answered your question, but I hope that helps a little.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-08-20T03:06:14.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5939" PostTypeId="2" ParentId="5934" CreationDate="2013-08-20T08:37:54.203" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Glycerol&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Glycerol&lt;/a&gt; has a melting point of 20°c (in comparison to water's melting point +20°c) and it is used as ingredient in 'Bounty' sweets (E number  : E422)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suggest that you have a look at the E-number list. There are a lot of allowed products for human consumption (however not all are safe in high doses) you can find the list here : &lt;a href=&quot;http://en.wikipedia.org/wiki/E_number&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/E_number&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An exact answer to the question I don't have :(&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2013-08-20T08:37:54.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5943" PostTypeId="2" ParentId="5789" CreationDate="2013-08-20T17:10:12.090" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, this is not a complete answer to your question, but this could be a factor I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond pair – lone pair repulsion is always greater than bond pair – bond pair repulsion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to say is that &lt;strong&gt;the repulsion between the O---H bond and the lone pair&lt;/strong&gt; is so heavy such that it is really hard for two hydrogen atoms to get away each other as you are expecting to happen.&lt;/p&gt;&#xA;" OwnerUserId="1912" LastEditorUserId="177" LastEditDate="2014-12-10T17:05:59.707" LastActivityDate="2014-12-10T17:05:59.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5945" PostTypeId="2" ParentId="5934" CreationDate="2013-08-20T19:46:54.197" Score="10" Body="&lt;p&gt;I tried to use the curated dataset from Mathematica to answer this question.  You can ignore the following if you have no interest in Mathematica and just jump down to the end for my ideas.&lt;/p&gt;&#xA;&lt;p&gt;The following code gets all of chemicals currently in the curated dataset and grabs the melting points and NFPA health rating.  The second line finds the chemicals with melting points between -7 and 0 degrees and the last line looks for chemicals with a health rating of 0, which is low risk.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;out = {#,  ChemicalData[#,&amp;quot;Name&amp;quot;], ChemicalData[#,&amp;quot;MeltingPoint&amp;quot;],&#xA;   ChemicalData[#,&amp;quot;NFPAHealthRating&amp;quot;])&amp;amp; ChemicalData[];&#xA;cull1 = Cases[out, x_ /; -7 &amp;lt; x[[3]] &amp;lt; 0];&#xA;cull2 = Cases[cull1, x_ /; x[[4]] == 0];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;We go from about 43000 chemicals in the first list to 220 in the second and finally 6 in the last.  I'm not sure I agree with the health ratings of these compounds but here you go (melting points in parentheses):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;2-methyl-2-propanethiol (-0.5)&lt;/p&gt;&#xA;&lt;p&gt;prehnitene (-6)&lt;/p&gt;&#xA;&lt;p&gt;trans-1,2-dichlorocyclohexane (-6)&lt;/p&gt;&#xA;&lt;p&gt;hexadecene (-6)&lt;/p&gt;&#xA;&lt;p&gt;trans-1,2-dibromocyclohexane (-6)&lt;/p&gt;&#xA;&lt;p&gt;olive oil (-6)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The only chemical I would deem &amp;quot;food safe&amp;quot; would be the last one.  Perhaps some type of food-grade oil is the mystery substance.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-08-21T00:51:23.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5946" PostTypeId="1" CreationDate="2013-08-20T19:48:43.103" Score="20" ViewCount="4059" Body="&lt;p&gt;I'm curious, I was trying to look into the affect of freezing a solution with water even when the solution is completely miscible. I came across something that detailed this regarding salt water and basically said when it gets low enough to freeze, the water forms it's tell-tale lattice basically pushing the salts out so the ice itself is pure water and less dense (considerably less dense because the salt solution allowed the density to increase all the way to the lowered freezing point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious, is this a general fact of the behaviour of any solution with water regardless of miscibility? Are there water solutions which break up the lattice while still allowing freezing, or does the affect of blocking the formation of that lattice completely block the solution from freezing altogether no matter what? Are there water solutions which will freeze solid even with the lattice without just separating the non-water compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I recognize amorphous ice can be created with wizardry, but my question is more in regards to the typical behaviour on earth.&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditDate="2013-08-22T14:45:49.823" LastActivityDate="2020-08-22T16:50:27.567" Title="Does freezing a solution with water always cause the water to separate and form the ice lattice?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5947" PostTypeId="1" AcceptedAnswerId="5948" CreationDate="2013-08-21T05:00:30.183" Score="5" ViewCount="5354" Body="&lt;p&gt;I know the Cannizzaro reaction takes place in carboxyl compounds without $\alpha$-H with the regular examples being formaldehyde and benzaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can this reaction take place with say diphenylmethanone or di ter-butyl ketone (2,2,4,4-tetramethylpentan-3-one) or 2,2-dimethylpropanal ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any examples of these compounds showing the cannizzaro reaction.&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="5026" LastEditDate="2018-07-07T11:36:35.817" LastActivityDate="2018-07-07T13:03:31.883" Title="Can substituted acetone undergo Cannizzaro reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5948" PostTypeId="2" ParentId="5947" CreationDate="2013-08-21T05:48:40.720" Score="9" Body="&lt;p&gt;As far as I'm aware, the Cannizzaro reaction can't occur with ketones, only aldehydes. The reaction mechanism involves the transfer of hydride from the former carbonyl carbon (of the tetrahedral intermediate resulting from hydroxide's nucleophilic addition to the carbonyl carbon), during which step the carbonyl reforms, yielding a carboxylic acid. Since a ketone, by definition, has no hydrogens bonded to the carbonyl carbon, it obviously can't undergo such a reaction. I would expect ketones with no $\alpha$-hydrogens to be essentially unreactive with hydroxide, since aldol reactions and keto-enol tautomerism are impossible with no $\alpha$-hydrogens, and most hydrates are unstable.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-08-21T05:58:13.087" LastActivityDate="2013-08-21T05:58:13.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5950" PostTypeId="1" CreationDate="2013-08-21T06:57:15.943" Score="5" ViewCount="12084" Body="&lt;p&gt;If a compound is placed before you and you are asked to determine what it is and what it's structure is, how do you actually do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a baby question for most of you but I am not from chemistry background and so this amazes me!&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="402" LastEditDate="2017-05-02T23:23:33.730" LastActivityDate="2017-05-02T23:28:15.140" Title="How do we actually know the chemical structure of a compound?" Tags="&lt;analytical-chemistry&gt;&lt;identification&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" ClosedDate="2017-06-11T12:33:57.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="5951" PostTypeId="2" ParentId="5950" CreationDate="2013-08-21T07:45:28.327" Score="2" Body="&lt;p&gt;The easiest way is to inquire where the compound originated, observe the compound with generic senses (color, texture, phase-gas/liquid/solid) and then make educated guesses based on that! If there remains any doubt as to the chemical or if that info is not available then a systematic process of ruling out the most likely possibilities is conducted. The single largest category in chemistry would be organic or carbon containing compounds. They may be analyzed using NMR to determine their structure with good accuracy. Any decent sized chemistry department will have an NMR available. For inorganic compounds, some simple acid-base tests could narrow the possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the most coveted and expensive tools, it is easy to tell what elements compose a sample. I once visited a forensic laboratory where they had a scanning electron microscope that could display individual grains of gunshot residue (invisible to the naked eye)for which police swab the hands of suspects. The particular SEM I saw in action had an x-ray source connected to it, which allowed any point including an individual microscopic dot of &lt;a href=&quot;http://www.forensicmag.com/articles/2008/12/innovations-scanning-electron-microscopy-preserve-sample-integrity-and-validity-forensic-analysis#.UhRsx7yUdpQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;suspected gunshot residue&lt;/a&gt; to be analyzed for its elemental composition. The guy there clicked on any part of the SEM image to display what element it was made of... completely automated! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the longest laboratory experiments I did in school was determining the percentage of iron in a mixture of iron and clay. It took 12 hrs and used only wet chemistry from 100 yrs ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many techniques that could be used but it really comes down to which one is most promising or available. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="402" LastEditDate="2017-05-02T23:27:21.130" LastActivityDate="2017-05-02T23:27:21.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5952" PostTypeId="1" AcceptedAnswerId="5955" CreationDate="2013-08-21T08:17:50.850" Score="4" ViewCount="269" Body="&lt;p&gt;I apologize for a perhaps newbie question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to compute the potential energy surface for several small molecular ions, like $\ce{H2+}$ and $\ce{H3+}$, using quantum chemistry software. What are the proper method and basis set to do that and why? I am thinking about ROHF/STO-3G or UHF.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="39" LastEditDate="2013-08-21T21:35:46.070" LastActivityDate="2013-08-22T09:14:14.137" Title="Which computational method/basis set should I use for small molecular ions?" Tags="&lt;ions&gt;&lt;quantum-chemistry&gt;&lt;basis-set&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5953" PostTypeId="1" CreationDate="2013-08-21T09:53:27.443" Score="2" ViewCount="899" Body="&lt;p&gt;I must confess, I am not a chemistry student but have always been a fan of it, so I am not sure if this is a blunder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have often wondered if Thermoplastics can ever be converted to or made to behave like a Thermosetting plastic. In other words, can a Thermoplastics be made more heat resistant? &lt;/p&gt;&#xA;" OwnerUserId="2181" LastActivityDate="2013-09-23T22:26:41.353" Title="Thermoplatics into Thermosetting plastics" Tags="&lt;organic-chemistry&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5954" PostTypeId="2" ParentId="4018" CreationDate="2013-08-21T15:05:45.613" Score="3" Body="&lt;p&gt;I was completely unfamiliar with the notion of HHO before this question, so +1 just for bringing it to our attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have much to add to the previous answers, but for those interested in learning more about the origins of HHO (and if you have access to these journals through your university), head on over to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0360319905003344&quot; rel=&quot;nofollow&quot;&gt;The main article&lt;/a&gt; by Santilli in the &lt;em&gt;International Journal of Hydrogen Energy&lt;/em&gt; volume 31 (2006) pages 113-1128.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0360319906005581&quot; rel=&quot;nofollow&quot;&gt;discussion&lt;/a&gt; by J. M. Cato in the same journal, volume 32 (2007) pages 1309-1312 which points out some of the issues with interpreting the data in the main article. (Note there's some great pedagogical opportunities in this work for General Chemistry courses: the Santilli paper uses non-SI units that can be converted, and the basic thermochemistry of evaporation and oxidation/reduction reactions can be incorporated into Hess' Law type activities.)&lt;/li&gt;&#xA;&lt;li&gt;There are two follow up discussions, in the same journal again, one by &lt;a href=&quot;http://dx.doi.org/10.1016/j.ijhydene.2007.11.009&quot; rel=&quot;nofollow&quot;&gt;Cloonan&lt;/a&gt; volume 21 page 1113 and one by &lt;a href=&quot;http://dx.doi.org/10.1016/j.ijhydene.2007.10.030&quot; rel=&quot;nofollow&quot;&gt;Kadeisvili&lt;/a&gt; which serve as a rebuttal to Cato's arguments.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This whole discussion makes for a great activity on the scientific method, understanding of analytical techniques, proper interpretation of data and the inherent resistance to change within the scientific community.  This reminds me of the cold fusion debacle, which is very well described in Gary Taubes book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0394584562&quot; rel=&quot;nofollow&quot;&gt;Bad Science&lt;/a&gt;.  Good reading for anyone interested in these types of controversial experiments.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-08-21T15:05:45.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5955" PostTypeId="2" ParentId="5952" CreationDate="2013-08-21T16:09:30.653" Score="4" Body="&lt;p&gt;$\ce{H2+}$ or $\ce{H3+}$ have unpaired electrons because they are ions. Due to this, you should be using an unrestricted shell theoretical model which doesn't restrict electrons to paired (closed shell) orbitals. You should start with unrestricted Hartree-Fock (UHF) and can increase the level of theory to include electron correlation with something like 2nd order Møller Plesset (MP2) or 4th order (MP4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron correlation models acknowledge the fact that electron locations are not totally random, repulsion between electrons is a real effect and can give more realistic results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to basis set, STO-3G would be considered a starting point. If time and resources allow, you could build up to something like a 6-31G(d,p) which adds p-orbitals to the hydrogen atoms allowing more optimization of molecular orbitals. Increasing the basis set quality approximates the radial components of the atomic orbitals with more Gaussian curves, thereby increasing the possible accuracy of the calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0963676938&quot; rel=&quot;nofollow&quot;&gt;Exploring Chemistry with Electronic Structure Methods - Foresman and Frisch&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2097" LastEditorUserId="2097" LastEditDate="2013-08-21T20:30:29.343" LastActivityDate="2013-08-21T20:30:29.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5956" PostTypeId="1" CreationDate="2013-08-21T20:28:25.847" Score="6" ViewCount="5848" Body="&lt;p&gt;I have this problem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid bromine is added to a container of sodium iodide crystals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I interpreted it with this molecular formula $\ce{Br_{2(l)} + 2NaI_{(s)} \rightarrow 2NaBr_{(s)} + I_{2(g)}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, since none of the products nor any of the reactants is aqueous, there shouldn't be a net ionic equation. However, my friend has proceeded with the problem as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br_2 + 2Na^+ + 2I^- \rightarrow 2Na^+ + 2Br^- + I_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody explain which one of us is correct and why?&lt;/p&gt;&#xA;" OwnerUserId="2185" LastEditorUserId="9961" LastEditorDisplayName="user4076" LastEditDate="2018-03-03T23:50:11.853" LastActivityDate="2018-03-03T23:50:11.853" Title="What is the net ionic equation for rection of liquid bromide with sodium iodide crystals?" Tags="&lt;redox&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5958" PostTypeId="2" ParentId="5956" CreationDate="2013-08-22T06:58:41.457" Score="1" Body="&lt;p&gt;Technically you are correct. How can there be any ions when the reactants are not in aqueous medium ? But here it is a matter of convention and maybe an illustration of how to write the ionic equations from molecular ones. Likewise, if you have done balancing of redox reactions , you might notice half of the reactions given to balance are completely impossible according to the laws of chemical combination. You'll always be at a loss when you overthink in chemistry, even if your argument is right.  Taking it as just a convention would be a better idea . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped ! &lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2013-08-22T06:58:41.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5959" PostTypeId="2" ParentId="5952" CreationDate="2013-08-22T09:14:14.137" Score="1" Body="&lt;p&gt;if u accurately want to make ur computations a bigger basis set and high level are necessery (CI(SDQ) or CAS), however so bigger basis set so less human readable are results. If the only aim is to see nice orbitals, minimal basis and DFT or even semiemprical methods should be enough. As stated before unrestricted methods are necessary unless u want to play with half-occupations.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2013-08-22T09:14:14.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5960" PostTypeId="2" ParentId="5950" CreationDate="2013-08-22T09:18:48.967" Score="1" Body="&lt;p&gt;The easiest and most precise method: grow single crystals and do x-ray diffraction experiment. as long the question is not which lanthanide I have, some chemical &quot;common sense&quot; + x-ray structure give u full answer. Of course, you still need to do other characterisations as NMR and IR spectroscopy. &lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="402" LastEditDate="2017-05-02T23:27:58.263" LastActivityDate="2017-05-02T23:27:58.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5961" PostTypeId="1" CreationDate="2013-08-22T12:28:44.873" Score="1" ViewCount="291" Body="&lt;p&gt;Let's assume a setup with a static linear molecule with three identical atoms connected by bonds and a single atom, identical to the other three, being shot at the molecule. Let's also assume that everything happens in 1D and can therefore be illustrated by this simple scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(projectile) A -&gt;    A-A-A (molecule)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the direction of the arrow is the direction of motion of the projectile and the dash signs in the molecule are the bonds between atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[EDIT - must change next paragraph as the description does not fit what would happen: stand-by, no time to do it right now!]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a low-energy, non-relativistic, perfectly elastic collision, the projectile will stop dead in its track while the molecule will be set in motion at 1/3rd of the projectile's speed, as it has 3 times the projectile's mass (conservation of Momentum). Balancing also the equations for the Kinetic Energy easily reveals that 2/3rds of projectile's kinetic energy must have been locked away in internal vibrations of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's easy for us to calculate these numbers by considering the projectile and the molecule as just two separate systems and conveniently attaching a total mass to each. But in reality the universe doesn't group atoms into molecules, does it? Each bond in the setup doesn't &quot;know&quot; about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, not &quot;knowing&quot; total masses, how does the universe &quot;calculates&quot; that in this case the molecule must store 2/3rds of the projectile kinetic energy into internal vibrations, no more, no less?&lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorUserId="2189" LastEditDate="2013-08-30T14:21:56.607" LastActivityDate="2013-08-30T14:21:56.607" Title="Storing kinetic energy in bonds" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;kinetics&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5962" PostTypeId="2" ParentId="5922" CreationDate="2013-08-22T12:40:58.417" Score="18" Body="&lt;p&gt;Indeed, water can conduct electricity by electrolysis. But this is hampered by the fact that water has very high resistance. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water&quot;&gt;Electrolysis of water&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrolysis of pure water requires excess energy in the form of&#xA;  overpotential to overcome various activation barriers. Without the&#xA;  excess energy the electrolysis of pure water occurs very slowly or not&#xA;  at all. This is in part due to the limited self-ionization of water.&#xA;  Pure water has an electrical conductivity about one millionth that of&#xA;  seawater.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And from &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot;&gt;Self-ionization of water&lt;/a&gt; we can see how small it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water molecules dissociate into equal amounts of $\ce{H3O+}$ and $\ce{OH-}$, so their concentrations&#xA;  are equal to 1.00×10&lt;sup&gt;−7&lt;/sup&gt; mol∙dm&lt;sup&gt;-3&lt;/sup&gt; at 25 °C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="50" LastEditorUserId="50" LastEditDate="2013-08-24T21:05:36.733" LastActivityDate="2013-08-24T21:05:36.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5963" PostTypeId="1" CreationDate="2013-08-22T13:56:38.213" Score="1" ViewCount="307" Body="&lt;p&gt;We have seen that $\ce{HBr}$ shows an anti-Markovnikov effect when it reacts with alkenes in the presence of peroxides. Why does only $\ce{HBr}$ shows this effect? How about the other hydrogen halides $\ce{HX}~(\ce{X} = \ce{F}, \ce{Cl}, \ce{I})$?&lt;/p&gt;&#xA;" OwnerUserId="2153" LastEditorUserId="16683" LastEditDate="2015-10-09T16:25:35.097" LastActivityDate="2015-10-09T16:25:35.097" Title="Anti-Markovnikov Effect" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5964" PostTypeId="1" AcceptedAnswerId="5978" CreationDate="2013-08-22T14:39:58.060" Score="4" ViewCount="8635" Body="&lt;p&gt;Look at the structure for sulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kh9Ow.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does sulfur form so many covalent bonds. Its valency is only $-2$, it only needs two electrons, yet here it's getting $6$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens with phosphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nK1ga.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus has a valency of $-3$, yet, it forms $5$ covalent bonds. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="73" LastEditDate="2013-08-22T19:29:03.913" LastActivityDate="2020-05-27T14:52:47.447" Title="Why does sulfate have this structure?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5965" PostTypeId="2" ParentId="5964" CreationDate="2013-08-22T21:31:51.850" Score="3" Body="&lt;p&gt;It's a highly debated topic and seems that none has won. I thought it to be fully explainable through MOT. However, while searching I found this &lt;a href=&quot;http://www.madsci.org/posts/archives/2000-01/949098457.Ch.r.html#3&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. It cleared few of my misunderstanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look. You should be able to find the answer yourself. If not leave a reply, I will update the answer accordingly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again thanks to Nick for citing the (3c,4e)-bond model. &lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2013-08-22T21:31:51.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5966" PostTypeId="2" ParentId="5963" CreationDate="2013-08-22T22:07:14.730" Score="4" Body="&lt;p&gt;Main reason is behind is the reaction mechanism involved. &lt;em&gt;$\ce{HBr}$&lt;/em&gt; or other halides usually react with alkenes through an &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/eladd/symhbr.html&quot; rel=&quot;nofollow&quot;&gt;electrophilic addition&lt;/a&gt;. Due to the weak &lt;em&gt;$\ce{O-O}$&lt;/em&gt; bond, peroxides act as a source of free radicals. Thus in the presence of peroxides, a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Radical_Additions--Anti-Markovnikov_Product_Formation&quot; rel=&quot;nofollow&quot;&gt;free radical chain pathway&lt;/a&gt; is followed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The other hydrogen halides are reluctant to this reaction, because &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/freerad/alkenehbr.html&quot; rel=&quot;nofollow&quot;&gt;the chain reaction can't proceed due to energy requirements for the generation of the corresponding halogen free radicals&lt;/a&gt;.&lt;/strong&gt;     &lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2013-08-22T22:07:14.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5967" PostTypeId="2" ParentId="5961" CreationDate="2013-08-22T22:43:47.163" Score="4" Body="&lt;p&gt;The universe doesn't calculate anything. Energy and it's conservation and everything that follows from it is a human-made concept, a physical model. It describes the things that happen in nature quite well (in the right circumstances), but it is only a model. In how far this lends some kind of &quot;reality&quot; to the concept/model I couldn't say. It's more of a philosophical question. Physical laws only describe how nature works, but not why it works the way it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the physical content of the collision you described: I think of it in the following way: The projectile hits the first atom of the molecule, let's call it A1, and transfers its momentum to it. Now, this atom moves in the direction along the bond axis directly towards the next atom, A2. While it does so, it has to work against the spring force of the bond between A1 and A2. Thus, kinetic energy of movement is converted to kinetic energy of vibration. A2 will now be accelerated in the direction of the next atom, A3, and it has to work against the spring force of the bond between A2 and A3 &lt;strong&gt;and(!)&lt;/strong&gt; the restoring force brought about by the bond between A1 and A2. Again kinetic energy is converted to vibrational energy. But now you have a system of coupled oscillators. This process will be repeated for every atom in the molecule until every bond vibrates. So, it is not true that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each bond in the setup doesn't &quot;know&quot; about the existence of the other&#xA;  bond and each atom could be said to be unaware of all but the atoms&#xA;  bonded or colliding with it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are coupled through the spring forces of the bonds and thus each atom &quot;feels&quot; the mass of every other atom in the molecule as well as the bond strength of every bond in the molecule.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-22T23:13:38.287" LastActivityDate="2013-08-22T23:13:38.287" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="5973" PostTypeId="1" AcceptedAnswerId="5975" CreationDate="2013-08-23T18:07:16.743" Score="2" ViewCount="424" Body="&lt;p&gt;The ratio of the heat capacities $\frac{C_\mathrm p}{C_\mathrm v}$ for one mole of a gas is $1.67$. The gas is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a) $\ce{He}$&lt;br&gt;&#xA;  b) $\ce{H2}$&lt;br&gt;&#xA;  c) $\ce{CO2}$&lt;br&gt;&#xA;  d) $\ce{CH4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know how to answer this by analysing the degrees of freedom of the gas molecule. Is there any other way to derive the ratio?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="5026" LastEditDate="2018-02-25T02:52:44.790" LastActivityDate="2018-02-25T02:52:44.790" Title="Identifying the gas given ratio of its specific heats" Tags="&lt;physical-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5974" PostTypeId="1" CreationDate="2013-08-23T19:16:35.433" Score="3" ViewCount="6564" Body="&lt;p&gt;Is chlorine in there that accumulates (at the anode) and emerge out as chlorine gas? I was worried since I added a pinch of salt to the steamer in order to hasten the heating of tap water when I caught a cold (I should've just boiled a pot of water on the stove). I'm worried since chlorine gas is bad for the respiratory system. Also I'm curious to know what is this brown precipitate under the solution - hydroxides or cupric-oxides from the electrode?&lt;/p&gt;&#xA;" OwnerUserId="1908" LastEditorDisplayName="user15489" LastEditDate="2015-06-23T17:39:02.090" LastActivityDate="2019-12-01T11:27:59.730" Title="Is it dangerous to inhale the steam coming from electrolysis of salt-water solution" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;safety&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ClosedDate="2019-12-02T15:22:39.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="5975" PostTypeId="2" ParentId="5973" CreationDate="2013-08-23T20:13:37.250" Score="4" Body="&lt;p&gt;I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  C_{p} - C_{v} = R&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\&#xA;  \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\&#xA;  \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each &lt;strong&gt;excited&lt;/strong&gt; quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$.&#xA;Since $C_{v} = \frac{dU}{dT}$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-25T18:04:00.743" LastActivityDate="2013-08-25T18:04:00.743" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5977" PostTypeId="1" AcceptedAnswerId="5979" CreationDate="2013-08-24T01:58:23.357" Score="8" ViewCount="9682" Body="&lt;p&gt;Optical isomerism in normal organic molecules can be found out by looking for chiral carbon, and the number of isomers can be found out by drawing different spacial orientations of the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I don't know how to check whether a given coordination compound is optically active or not, and how many optical isomers does it have. I don't know how to draw different spacial orientations with all those polydentate ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to learn the general guidelines for finding optical isomerism in co-ordination compounds. I would like to solve questions like how many no. of optical isomers are there for $\ce{[CoCl2(en)2]^+}$&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="5026" LastEditDate="2018-05-08T03:22:46.360" LastActivityDate="2018-05-08T14:29:51.673" Title="Optical isomerism in Co-ordination Compounds" Tags="&lt;stereochemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5978" PostTypeId="2" ParentId="5964" CreationDate="2013-08-24T02:29:24.157" Score="4" Body="&lt;p&gt;I think an important point to mention here is that Lewis dot structures, and the octet rule, are simply &lt;em&gt;models that describe experimental observables.&lt;/em&gt; For example, the concept of valency works well with organic molecules but was challenged by &lt;a href=&quot;http://en.wikipedia.org/wiki/Alfred_Werner&quot; rel=&quot;nofollow&quot;&gt;Alfred Werner&lt;/a&gt; in the development of bonding models that adequately described coordination compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron-dot model of bonding is so meaningful because it can be adapted to explain various experimental observables.  You can draw two perfectly reasonable electron dot structures for $\ce{SO4^{2-}}$: one with all $\ce{S-O}$ single bonds and one with two $\ce{S=O}$ double bonds (and the additional resonance structures).  The question is, which electron-dot structure best represents the &lt;strong&gt;real&lt;/strong&gt; structure?  We can use the concept of &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/09/24/how-to-calculate-formal-charge/&quot; rel=&quot;nofollow&quot;&gt;formal charges &lt;/a&gt; to predict that the structure with double bonds is most likely closer to reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then look at the experimental data (which is reported on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfate&quot; rel=&quot;nofollow&quot;&gt;sulfate wikipedia page&lt;/a&gt;) for S-O single and double bond lengths.  The S-O bond length in $\ce{SO4^{2-}}$ is 149 pm, which is shorter than that observed in sulfuric acid (157 pm) and very close to the gas-phase bond length of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_monoxide&quot; rel=&quot;nofollow&quot;&gt;sulfur monoxide&lt;/a&gt; which is 148 pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at the end of the day, an electron dot structure is a type of &lt;em&gt;model&lt;/em&gt; that describes bonding.  It has some additional features such as &lt;em&gt;expanded octets&lt;/em&gt; and &lt;em&gt;formal charge&lt;/em&gt; that help broaden its applicability to more situations; however, models such as this one don't tell us what an atom &quot;needs&quot;; rather, it provides an explanation for how we observe atoms &quot;behaving&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-08-24T02:29:24.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5979" PostTypeId="2" ParentId="5977" CreationDate="2013-08-24T02:50:19.073" Score="9" Body="&lt;p&gt;Any 3rd-year inorganic chemistry textbook worth its weight in salt will provide some information about isomers in coordination compounds.  An optically active compound contains &lt;strong&gt;no rotation-reflection symmetry element&lt;/strong&gt;.  Typically, this means that if a molecule has an &lt;strong&gt;inversion center&lt;/strong&gt; or a &lt;strong&gt;mirror plane&lt;/strong&gt; then it is not chiral.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes a systematic &lt;em&gt;brute force and ignorance&lt;/em&gt; approach to determining the number of isomers a coordination compound has, and then a bit more visual thinking to determine if any of the isomers are enantiomers.  The following figure is taken from Miessler, Fischer, Tarr 5th edition and is useful for identifying isomers of 6-coordinate complexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eF4UR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a 6-coordinate complex with 6 unique ligands in an octahedral field would be enantiomers. In fact, there are 15 different isomers, &lt;em&gt;each of which is chiral&lt;/em&gt; resulting in a total of 30 isomers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of determining isomers this way is to formulate a structure of the type M where  indicates two ligands that are trans to each other.  Then, you start varying a, b, c, ... until you've come up with as many unique combinations as possible.  For your case, you have 2 instances of a bidentate ligand.  Bidentate ligands are typically designated with capital letters in this notation, so the formula for your molecule would be $\ce{M(AA)_2b_2}$.  Further, since bidentate ligands are too small to coordinate in a &lt;em&gt;trans&lt;/em&gt; conformation, you will decrease the number of possible isomers.  One isomer, for example, would be M, where I use A and A' to designate the different $\ce{(en)}$ ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question, then, is that there are two isomers, and they are enantiomers, since you can't rearrange the A, A' and b in the formula above to come up with another unique configuration. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="5026" LastEditDate="2018-05-08T03:24:13.353" LastActivityDate="2018-05-08T03:24:13.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5980" PostTypeId="1" AcceptedAnswerId="109132" CreationDate="2013-08-24T09:03:00.577" Score="6" ViewCount="696" Body="&lt;p&gt;What is the structure of the molecule $\ce{N5P3}$? $\ce{N5P3}$ is not documented online, so could anyone please comment on the structure of the molecule: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many ligands does each phosphorous atom have?&lt;/li&gt;&#xA;&lt;li&gt;How many ligands does each nitrogen atom have?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="41328" LastEditDate="2019-02-07T03:27:08.320" LastActivityDate="2019-02-07T03:27:08.320" Title="What is the structure of N₅P₃?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;molecules&gt;&lt;solid-state-chemistry&gt;&lt;pnictogen&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5981" PostTypeId="2" ParentId="4588" CreationDate="2013-08-24T09:12:20.513" Score="2" Body="&lt;p&gt;I guess that a long time ago there was a lot of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; in the atmosphere and then the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; could have reacted with that lead oxide to form  &lt;span class=&quot;math-container&quot;&gt;$\ce{PbS}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;On wikipedia it says (under Lead(II)oxide,reactions) :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The red and yellow forms of this material are related by a small change in enthalpy:&lt;/p&gt;&#xA;&lt;p&gt;PbO(red) → PbO(yellow) ΔH = 1.6 kJ/mol&lt;/p&gt;&#xA;&lt;p&gt;PbO is amphoteric, which means that it reacts with both acids and with bases. With acids,it forms salts of &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb2+}$&lt;/span&gt; via the intermediacy of oxoclusters such as &lt;span class=&quot;math-container&quot;&gt;$\ce{[Pb6O(OH)6]4+}$&lt;/span&gt;. With strong base, &lt;span class=&quot;math-container&quot;&gt;$\ce{PbO}$&lt;/span&gt; dissolves to form plumbite(II) salts:[6]&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{PbO + H2O + OH- -&amp;gt; [Pb(OH)3]-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It could have reacted with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; following this reaction : (in water)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Pb2+ + H2S -&amp;gt; PbS + 2 H+}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-09-23T11:27:07.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5982" PostTypeId="2" ParentId="5926" CreationDate="2013-08-24T09:27:08.010" Score="1" Body="&lt;p&gt;That one electron is making sure that Gold has a complete d-orbital, and therefor is much more stable. Elements want to achieve noble gas configurations, and if they can't, then they want to have a half-filled orbitals. In the case of gold the d-orbital is filled completely and the s-orbital is half filled. Whereas Platinum doesn't have that complete d-orbital filled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemical properties (such as reactivity) are influenced by electrons.&lt;/strong&gt; The amount of neutrons will determine if it can be radioactive or not(,...) look at Tritium : that is a hydrogen atom with 2 neutrons in it and it is radioactive. Whilst protium (the most common isotope of Hydrogen with 1 proton and 0 neutrons) is not radioactive. &lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2013-08-24T09:27:08.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5983" PostTypeId="2" ParentId="5980" CreationDate="2013-08-24T11:34:51.223" Score="13" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N5P3}$&lt;/span&gt; is more commonly written as &lt;span class=&quot;math-container&quot;&gt;$\ce{P3N5}$&lt;/span&gt;, and known as &lt;strong&gt;&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C12136913&quot; rel=&quot;noreferrer&quot;&gt;triphosphorus pentanitride&lt;/a&gt;&lt;/strong&gt;. It's a &lt;strong&gt;crystalline solid&lt;/strong&gt; at ambient conditions and &lt;strong&gt;not&lt;/strong&gt; a molecular compound. From the first publication that reported the pure compound and its structure &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cm950385y&quot; rel=&quot;noreferrer&quot;&gt;[1]&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the solid a three-dimensional cross-linked network structure of corner sharing &lt;span class=&quot;math-container&quot;&gt;$\ce{PN4}$&lt;/span&gt; tetrahedra has been identified with 2/5 of the nitrogen atoms bonded to three P atoms and 3/5 of the nitrogen atoms bonded to two P atoms.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uEY2T.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Only the combination of spectroscopical and diffraction methods, especially electron microscopy for structural analysis, enables a detailed structural and crystallographic characterization of &lt;span class=&quot;math-container&quot;&gt;$\ce{P3N5}$&lt;/span&gt; as no conventional single-crystal data for structure determination have been available for this compound.&lt;/p&gt;&#xA;&lt;p&gt;A representative for an unique binary structure type was found for the first time which combines a tetrahedral structure with the very rare molar ratio of 3:5 for the constituent elements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Schnick, W.; Lücke, J.; Krumeich, F. &lt;em&gt;Chem. Mater.&lt;/em&gt; &lt;strong&gt;1996&lt;/strong&gt;, 8 (1), 281–286 DOI: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cm950385y&quot; rel=&quot;noreferrer&quot;&gt;10.1021/cm950385y&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-09-21T07:02:12.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5984" PostTypeId="1" CreationDate="2013-08-24T12:30:47.757" Score="1" ViewCount="6173" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction between &lt;em&gt;p&lt;/em&gt;-methylbenzaldehyde and $\ce{NaOH}$ is an example of:&lt;br&gt;&#xA;  &lt;strong&gt;A&lt;/strong&gt;) Cannizzaro reaction&lt;br&gt;&#xA;  &lt;strong&gt;B&lt;/strong&gt;) Aldol condensation&lt;br&gt;&#xA;  &lt;strong&gt;C&lt;/strong&gt;) Hydrolysis&lt;br&gt;&#xA;  &lt;strong&gt;D&lt;/strong&gt;) Haloform reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I selected &lt;strong&gt;B&lt;/strong&gt;, aldol condensation, as the given compound is an aldehyde and it reacts with $\ce{NaOH}$. But the given answer is &lt;strong&gt;A&lt;/strong&gt;. Why is it so? The $\ce{NaOH}$ is not mentioned to be concentrated.&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="41328" LastEditDate="2017-08-15T22:00:39.423" LastActivityDate="2017-08-15T22:00:39.423" Title="Cannizzaro or aldol condensation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5985" PostTypeId="1" AcceptedAnswerId="5987" CreationDate="2013-08-24T13:15:01.153" Score="6" ViewCount="16930" Body="&lt;p&gt;I know that compounds like alcohols can be further oxidized to aldehydes (or ketones) and to carboxylic acids. But, is carboxylic acid the best oxidized form of a given compound? &#xA;What happens when carboxylic acids further reacts with oxidizing agents like $\ce{KMnO4}$?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="10185" LastEditDate="2015-10-02T16:53:17.987" LastActivityDate="2018-02-02T14:50:03.477" Title="Can carboxylic acids be oxidised further?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5986" PostTypeId="1" AcceptedAnswerId="5995" CreationDate="2013-08-24T13:50:57.263" Score="4" ViewCount="2291" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Entropy change is positive, but enthalpy change is negative.&lt;/li&gt;&#xA;  &lt;li&gt;Entropy change and enthalpy changes are both positive.&lt;/li&gt;&#xA;  &lt;li&gt;Entropy change is positive and enthalpy does not change.&lt;/li&gt;&#xA;  &lt;li&gt;Entropy change is negative and enthalpy change is positive.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The final conc. of the mixture would be less than the initial concentration of the concentration $\ce{CuSO4}$ given.&lt;br&gt;&#xA;Therefore  I think the entropy change is positive. But, what about enthalpy?  What factors should I look into to find answer to these types of questions?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="5026" LastEditDate="2018-02-28T14:59:00.200" LastActivityDate="2018-02-28T14:59:13.743" Title="Entropy and enthalpy change on mixing liquids" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5987" PostTypeId="2" ParentId="5985" CreationDate="2013-08-24T14:00:02.217" Score="11" Body="&lt;p&gt;In carboxylic acid, the carbon atom of the COOH group is already in high oxidation state, namely +3. In order &lt;strong&gt;to reach a higher oxidation state (+4), it requires breaking the C—C bond&lt;/strong&gt;, typically to form molecular CO2. Thus, &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm#crbacd7b&quot; rel=&quot;noreferrer&quot;&gt;oxidation of carboxylic acid in strongly oxidizing conditions is known as a decarboxylation reaction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-08-24T14:00:02.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5988" PostTypeId="1" AcceptedAnswerId="5990" CreationDate="2013-08-24T14:13:03.977" Score="5" ViewCount="12841" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Cyclohexene is reacted with bromine in carbon tetrachloride in the dark. The product of the reaction is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EWxNV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EWxNV.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that reaction with bromine in &lt;span class=&quot;math-container&quot;&gt;$\ce{CCl4}$&lt;/span&gt; will result in vicinal dibromocyclohexane. So the answer is either A or B. But how to decide between A and B? Since its an SN2 reaction it would lead to inversion of the reacting compound. Does this have to do anything with the answer?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-24T11:30:14.013" Title="Bromination of cyclohexene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5989" PostTypeId="1" CreationDate="2013-08-24T15:32:42.353" Score="5" ViewCount="433" Body="&lt;p&gt;The phenomenon of deliquescence wherein a water soluble substance absorbs water from atmosphere and dissolve in it forming a saturated solution, has thermodynamic roots. I have read the fact that the vapour pressure of the saturated solution should be lower than the the vapour pressure of water at that temperature. Why is this so?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since deliquescence is a spontaneous process and therefore thermodynamically favourable, there must be a negative $\Delta G$ for the absorption of water vapour and conversion of solid substance into its saturated solution? How can we derive, understand or quantify various thermodynamic parameters ($\Delta G, \Delta S, \Delta H$ etc) for the process of deliquescence?&lt;br&gt;&#xA;I am comfortable with basic thermodynamic manipulations, binary phase diagrams, multiphase systems and equilibria.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-08-25T09:50:00.010" Title="Thermodynamics behind Deliquescence?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5990" PostTypeId="2" ParentId="5988" CreationDate="2013-08-24T16:06:57.193" Score="9" Body="&lt;p&gt;To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the &lt;em&gt;trans&lt;/em&gt; product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z8ZAw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQNW6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-24T16:33:10.057" LastActivityDate="2013-08-24T16:33:10.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5991" PostTypeId="2" ParentId="5986" CreationDate="2013-08-24T16:13:44.187" Score="1" Body="&lt;p&gt;You could argue that the copper and sulphate ions are (on average) better solvated in a more dilute solution. Furthermore, the ions with the same charge can &quot;go out of each other's way&quot; more easily in a dilute solution, so that the correlations between the ions are weaker. Both of these effects would lead to an energetically more favourable state, so that the enthalpy change should be negative (exothermic reaction).&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-24T16:37:13.073" LastActivityDate="2013-08-24T16:37:13.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5992" PostTypeId="2" ParentId="5984" CreationDate="2013-08-24T16:23:34.090" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Aldol_condensation&quot; rel=&quot;noreferrer&quot;&gt;aldol condensation&lt;/a&gt; needs $\alpha$-hydrogens next to the carbonyl group in order to proceed. Check out this reaction equation from the Wikipedia article:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/bARJO.png&quot; alt=&quot;aldol reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these hydrogen atoms is removed to form the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enolate#Enolates&quot; rel=&quot;noreferrer&quot;&gt;enolate&lt;/a&gt;. See this mechanism (also from the Wikipedia article). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a5pjB.png&quot; alt=&quot;Aldol mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, look at the structure of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tolualdehyde&quot; rel=&quot;noreferrer&quot;&gt;p-methylbenzaldehyde&lt;/a&gt;. This molecule does not have any $\alpha$-enolizable hydrogen atoms. The $\alpha$-position is an $sp^2$-hybridized carbon atom in the aromatic ring. The aldol condensation is thus not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHYzc.png&quot; alt=&quot;4-methylbenzaldehyde&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also rule out the haloform reaction because there are no halogen compounds given. There are also no hydrolyzable functional groups (halides, ethers, esters, anhydrides, amides, nitriles, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only remaining possibility is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cannizaro_reaction&quot; rel=&quot;noreferrer&quot;&gt;Cannizzarro reaction&lt;/a&gt;, which requires a benzaldehyde derivative or other aldehyde lacking $\alpha$-enolizable hydrogen atoms.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-08-24T16:23:34.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5994" PostTypeId="1" AcceptedAnswerId="6000" CreationDate="2013-08-24T17:15:24.483" Score="4" ViewCount="2937" Body="&lt;p&gt;I know that $\ce{OH-}$ is usually a bad leaving group, but why does the following reaction occur via SN1 in presence of anhydrous $\ce{ZnCl2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oTXum.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="487" LastEditorUserId="9961" LastEditDate="2017-04-08T16:40:10.333" LastActivityDate="2017-04-08T16:40:10.333" Title="Halogenation of alcohols in the presence of ZnCl2" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5995" PostTypeId="2" ParentId="5986" CreationDate="2013-08-24T17:55:36.550" Score="4" Body="&lt;p&gt;This is difficult, because we do not know what is your current understanding and at what “level” of approximation you expect the question to be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a undergraduate level, or in the absence of any details (or thermochemical data or access to thermochemical tables), I would go with option &lt;strong&gt;(3)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Enthalpy: we are not given any indication at what is the enthalpy of mixing. In the simplest approximation, the enthalpy of mixing is zero, as in an ideal solution. It would correspond to a regime where the enthalpy of dissolution ($\Delta H_{\text{sol}}$) does not depend on the concentration of the solution. It's probably a good approximation, until you reach high concentrations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Entropy: the entropy of mixture is positive, i.e. the process of mixing the two solutions is spontaneous. It can be explained by &lt;a href=&quot;https://en.wikipedia.org/wiki/Entropy_of_mixing#Ideal_and_regular_solutions&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reasoning&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5026" LastEditDate="2018-02-28T14:59:13.743" LastActivityDate="2018-02-28T14:59:13.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5996" PostTypeId="1" AcceptedAnswerId="5998" CreationDate="2013-08-24T18:04:07.393" Score="2" ViewCount="3948" Body="&lt;p&gt;By a dicarbonyl ester I mean something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oOIXX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R1 and R2 are alkyl substituents and the reacting carboxylic acids are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eRk3I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GnHHo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2014-06-05T11:51:42.143" LastActivityDate="2014-06-05T11:51:42.143" Title="Is it possible to react two distinct carboxylic acids to form a dicarbonyl ester?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5997" PostTypeId="2" ParentId="2589" CreationDate="2013-08-24T18:06:39.507" Score="1" Body="&lt;p&gt;I can't post a comment, so this is not a true answer but I think can help. If I've understand what you mean you want to perform a &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature-programmed_reduction&quot; rel=&quot;nofollow&quot;&gt;Temperature-programmed reduction (TPR)&lt;/a&gt; followed by a temperature-programmed oxidation (TPO). In this case I think this old book can help: &lt;a href=&quot;http://books.google.it/books?id=n03ZL1U0BbQC&amp;amp;dq=TPR+Temperature-programmable+reduction&amp;amp;hl=it&amp;amp;source=gbs_book_similarbooks&quot; rel=&quot;nofollow&quot;&gt;Temperature-programmed Reduction for Solid Materials Characterization&lt;/a&gt;. I don't know much about single crystal but I think that many things from TiO2 studies of nanomaterials can be consider valid for your case, this &lt;a href=&quot;http://etd.fcla.edu/CF/CFE0001432/QIU_SHIPENG_200612_MS.pdf&quot; rel=&quot;nofollow&quot;&gt;online book&lt;/a&gt; can be helpful to understand all the variables.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2013-08-24T18:06:39.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5998" PostTypeId="2" ParentId="5996" CreationDate="2013-08-24T18:58:26.870" Score="7" Body="&lt;p&gt;The standard term for such molecules is &lt;em&gt;acid anhydride&lt;/em&gt;, as they can be viewed as the product of a condensation reaction between two carboxylic acids, with concomitant loss of $\ce{H2O}$. Asymmetric anhydrides (i.e., ones that decompose into two distinct carboxylic acids if hydrolyzed) can certainly be prepared. One possible route is to synthesize an acid chloride from carboxylic acid #1 (thionyl chloride, $\ce{SOCl2}$, being one common reagent used to make the acid chloride) and then react it with the carboxylate salt of carboxylic acid #2. In that reaction, the carboxylate functions as a nucleophile and attacks the carbonyl carbon of the acid chloride, with chloride anion expelled as the leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it's an efficient laboratory method to prepare asymmetric anhydrides by simple dehydration of stoichiometric mixtures of distinct carboxylic acids. I suspect that would probably yield a (roughly) statistical distribution of products, at least in the ideal case. In reality, various factors might skew the distribution one way or another.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-08-24T18:58:26.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5999" PostTypeId="1" AcceptedAnswerId="6001" CreationDate="2013-08-24T19:50:48.557" Score="11" ViewCount="598" Body="&lt;p&gt;I've had my nose in the early literature on GFP (Green Fluorescent Protein), which contains a &lt;a href=&quot;http://en.wikipedia.org/wiki/Green_fluorescent_protein#Structure&quot;&gt;4-(p-hydroxybenzylidene)imidazolidin-5-one&lt;/a&gt; fluorophore, and &lt;a href=&quot;http://www.nature.com/nbt/journal/v14/n10/pdf/nbt1096-1246.pdf&quot;&gt;in one paper&lt;/a&gt; the author refers to the &quot;imidazolidone ring&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that this change of spelling was equivalent, but I can't find anything that looks reliable on Google (there are also pages referring to imidazolinone, which looks like it refers to an aromatic ring with Nitrogens at positions 2 and 4...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I'd know quite where to look on the IUPAC website for the info so trying here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me understand the variations?&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="30" LastEditDate="2013-08-24T20:58:29.823" LastActivityDate="2016-08-29T13:54:49.973" Title="Is there a difference between imidazolidinone and imidazolidone?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6000" PostTypeId="2" ParentId="5994" CreationDate="2013-08-24T19:53:31.853" Score="8" Body="&lt;p&gt;You are right, $\ce{OH-}$ is a bad leaving group. But the $\ce{HCl}$ in the reaction mixture protonates it. Thus, the bad leaving group $\ce{OH-}$ is converted to the good leaving group $\ce{H2O}$. Anhydrous $\ce{ZnCl2}$ might additionally facilitate the reaction by acting as a Lewis acid and it binds the expelled water thus preventing the back reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m4Ny1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-24T20:25:39.850" LastActivityDate="2013-08-24T20:25:39.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6001" PostTypeId="2" ParentId="5999" CreationDate="2013-08-24T21:18:29.693" Score="15" Body="&lt;p&gt;According to the &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C120934&amp;amp;Units=SI&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST ChemBook&lt;/a&gt;, &lt;strong&gt;2-imidazolidinone and 2-imidazolidone are synonymous&lt;/strong&gt;. The official IUPAC name is imidazolidin-2-one. 1,3-ethyleneurea is also another name for this same molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;    &lt;img src=&quot;https://i.stack.imgur.com/EcuPu.gif&quot; alt=&quot;imidazolidin-2-one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But, as you note, tiny changes in spelling can sometimes be much more tricky than this. To take an example from this family: &lt;strong&gt;imidazolidinone and imidazolinone&lt;/strong&gt; (note that the later ends in &lt;em&gt;–inone&lt;/em&gt;, not &lt;em&gt;–idone&lt;/em&gt;) &lt;strong&gt;are ketones formed from imidazolidine and imidazoline&lt;/strong&gt;, respectively. &lt;a href=&quot;https://en.wikipedia.org/wiki/Imidazolidine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Imidazolidine and imidazoline&lt;/a&gt; are partly saturated and fully saturated imidazole derivatives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;   &lt;img src=&quot;https://i.stack.imgur.com/cBROn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imidazolidine is a proper IUPAC name, but the official name of imidazoline is &lt;a href=&quot;https://en.wikipedia.org/wiki/Imidazoline&quot; rel=&quot;nofollow noreferrer&quot;&gt;4,5-dihydro-1⁠&lt;em&gt;H&lt;/em&gt;-imidazole&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorDisplayName="user7951" LastEditDate="2016-08-29T13:54:49.973" LastActivityDate="2016-08-29T13:54:49.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6002" PostTypeId="1" CreationDate="2013-08-25T03:43:39.123" Score="8" ViewCount="1948" Body="&lt;p&gt;Water tends to leach metals near the surface and deposit them deeper through precipitation in two layers. The upper layer is usually made of metal carbonates, oxides, and occasionally pure or native metals, while the second and deeper layer is metal sulfides. The following excerpt is summarized above.&lt;/p&gt;&#xA;&lt;p&gt;Source: &amp;quot;&lt;a href=&quot;http://www.jstor.org/stable/30055814&quot; rel=&quot;nofollow noreferrer&quot;&gt;Secondary Enrichment of Ore Deposits&lt;/a&gt;&amp;quot; by Samuel Franklin Emmons, which was published in the American Institue of Mining Engineers, Transaction 30, (1901), p. 177-201.&lt;/p&gt;&#xA;&lt;p&gt;The paper does not go into detailed chemistry for the most part but mentions that the chemistry varies from one morphology of soil/rock to another. The author postulates a thermodynamic reason along the lines of reduction potential that facilitates this stratification:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The actual changes observed by me in a great body of pyrite (&lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrite&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$FeS_2$&lt;/span&gt;&lt;/a&gt;) carrying galena (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lead%28II%29_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;PbS or Lead(II)Sulfide&lt;/a&gt;) in a limestone (&lt;a href=&quot;http://en.wikipedia.org/wiki/Limestone&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$CaCO_3$&lt;/span&gt;&lt;/a&gt;)country-rock, which had undergone partial decomposition from the periphery inwards, are as follows: the original fresh pyrite or marcasite crystals are first disintegrated and slightly pitted on the surface, then changed to  melanterite or hydrated ferrous sulfate and the galena becomes anglesite. In the outer or more fully oxidized zone, the iron-vitriol has changed in part to yellow, basic sulfate; in part to limonite with a separation of native sulfur.&lt;/p&gt;&#xA;&lt;p&gt;The theoretical changes that are &lt;strong&gt;assumed&lt;/strong&gt; to take place by the action of waters carrying oxygen or oxidizing agents are: first, an alteration of the iron sulfide (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$FeS$&lt;/span&gt;&lt;/a&gt;) to ferrous sulfate (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron(II)Sulfate or &lt;span class=&quot;math-container&quot;&gt;$FeSO_4$&lt;/span&gt;&lt;/a&gt;) with the formation of sulfuretted hydrogen (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Sulfide? &lt;span class=&quot;math-container&quot;&gt;$H_2S$&lt;/span&gt;&lt;/a&gt;) and sulfur which may have oxidized to sulfuric (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$H_2SO_4$&lt;/span&gt;&lt;/a&gt;)or sulfurous acid (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfurous_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$H_2SO_3$&lt;/span&gt;&lt;/a&gt;). By further oxidation the ferrous sulfate will become, in part at least, ferric sulfate (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron(III)Sulfate or &lt;span class=&quot;math-container&quot;&gt;$Fe_2(SO_4)_3$&lt;/span&gt;&lt;/a&gt;), and this in its turn will react upon the remaining ferrous sulfate, or upon the sulfides, and form more ferrous sulfate or sulfates of the other metals which are present. By this cycle of reactions, a supply of both ferric and ferrous sulfates would seem to be provided in the oxidized zone, but the extending downwards of the ferric salts would decrease as the supply of oxygen in the waters became less abundant.&#xA;It may be assumed that the sulfates of the metals thus formed would be transported for greater or lower distances, generally in proportion to their solubility, the iron sulfates being the most soluble; next, those of copper and zinc; silver sulfate is less soluble and also more readily decomposed, while lead sulfate is extremely insoluble.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="114" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-04-08T21:57:16.567" Title="Why do carbonates, oxides, and pure metals precipitate before metal sulfides?" Tags="&lt;geochemistry&gt;&lt;precipitation&gt;&lt;reduction-potential&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6003" PostTypeId="1" AcceptedAnswerId="6004" CreationDate="2013-08-25T05:31:02.043" Score="5" ViewCount="3470" Body="&lt;p&gt;As we all know, metallic ions are surrounded by a sea of electrons. If we continually bend or stretch a metal, say, iron, it will break. Does that mean we break the metallic bond in subatomic level?&lt;/p&gt;&#xA;" OwnerUserId="2205" LastEditorUserId="22" LastEditDate="2013-08-25T09:42:35.473" LastActivityDate="2018-04-05T15:16:45.783" Title="Do you break the metallic bond when you break a metal into two pieces?" Tags="&lt;bond&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6004" PostTypeId="2" ParentId="6003" CreationDate="2013-08-25T09:34:35.653" Score="4" Body="&lt;p&gt;Yes, you are breaking metallic bonds. Similarly, when you crush a crystal of salt, you are breaking ionic bonds. There aren't that many good examples of breaking covalent bonds&lt;sup&gt;1&lt;/sup&gt;, though breaking a polymer like nylon usually will do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;physical&quot; breaking of bonds is not a very exotic thing. Remember that while the energy required to break every bond in a gram of substance is a lot, the energy required to just break a slice of those bonds isn't much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very common question that arises is &quot;what happens with the open valencies left behind when you physically break bonds?&quot;. Usually, gases from the atmosphere adsorb onto the metal surface, forming hydrogen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note: When you do this, there is nothing going on at the subatmoic level, only at the atomic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. This is not due to the strength of covalent bonds -- indeed, these are usually weaker than ionic bonds. This happens because most covalent substances are made up of smaller molecules (as opposed to macroscopic lattices) held together by hydrogen bonding/Van der Waals forces, and it is these bonds that break when you break a covalent substance.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-08-25T09:34:35.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6005" PostTypeId="2" ParentId="5989" CreationDate="2013-08-25T09:50:00.010" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://www.chem.purdue.edu/gchelp/solutions/colligv.html&quot; rel=&quot;nofollow&quot;&gt;vapour pressure lowering&lt;/a&gt; is a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Raoult%27s_law&quot; rel=&quot;nofollow&quot;&gt;Raoult's law&lt;/a&gt; with a non-volatile solute, i.e. a solvent that has zero vapour pressure (or at least, lower vapour pressure than the solvent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second part (&lt;em&gt;“How can we derive, understand or quantify various thermodynamic parameters (ΔG,ΔS,ΔH etc) for the process of deliquescence?”&lt;/em&gt;), I am not sure exactly what you are looking for. This is a complex phenomenon, and there is a lot of research literature of the topic. To give only one example, see&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“&lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.food.080708.100915&quot; rel=&quot;nofollow&quot;&gt;Water-Solids Interactions: Deliquescence&lt;/a&gt;”, Lisa J. Mauer and Lynne S. Taylor, &lt;em&gt;Ann. Rev. of Food Sci. Technol.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;1&lt;/em&gt;, 41-63.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-08-25T09:50:00.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6006" PostTypeId="2" ParentId="6003" CreationDate="2013-08-25T10:01:24.893" Score="7" Body="&lt;p&gt;In addition to ManishEarth’s answer, I'll add that this is only strictly true if you take a pure single crystal of metal and break it. If you take a real-life piece of iron, it is not a pure crystal but contains a large number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_defect&quot; rel=&quot;nofollow noreferrer&quot;&gt;crystalline defects&lt;/a&gt;. The most common (and easiest to think about) in a polycrystalline material is the presence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Grain_boundary&quot; rel=&quot;nofollow noreferrer&quot;&gt;grain boundaries&lt;/a&gt; between the many &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallite&quot; rel=&quot;nofollow noreferrer&quot;&gt;crystallites&lt;/a&gt; that make up the metal particle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:CrystalGrain.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4ppJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  &lt;em&gt;Micrograph of a polycrystalline metal; grain boundaries evidenced by acid etching&lt;/em&gt; (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Grain_boundary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, your metal piece is not a single crystal, and &lt;strong&gt;crystalline defects are key in the understanding of mechanical response to applied stress, including the case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Fracture&quot; rel=&quot;nofollow noreferrer&quot;&gt;fracture&lt;/a&gt;&lt;/strong&gt; in which you are interested. So, by breaking a piece of metal in two, you may not actually break so many metallic bonds :)&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="5026" LastEditDate="2018-04-05T15:16:45.783" LastActivityDate="2018-04-05T15:16:45.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6009" PostTypeId="1" CreationDate="2013-08-25T18:15:32.390" Score="4" ViewCount="1393" Body="&lt;p&gt;Applying an electric potential to impure water splits water into its constituents H/O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the temperature of the electrolyte change during this electrolysis?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="30" LastEditDate="2013-08-25T19:06:36.337" LastActivityDate="2016-07-19T06:37:09.970" Title="Does the temperature of water change during electrolysis?" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6010" PostTypeId="2" ParentId="5922" CreationDate="2013-08-25T19:21:48.403" Score="1" Body="&lt;p&gt;A solution conducts current when positive and negative ions are randomly dissolved in it. Pure water doesn't have enough ions but when you add $\ce{HCl}$ you add a strong electrolyte so, after the addition, solution can conduct electricity very well. However you can perform an electrolysis even without $\ce{HCl}$ but this is a very slow process because even if there are positive and negatives ions (and this could induce you to think that the pure water will conduct current!) the positive ions are near the negative electrode (the cathode) to be reduce on the other hand the negative ions are near the positive electrode (the anode) to be oxidise. So in fact you mustn't imagine a solution of positive and negative ions like in a solution of a true strong electrolyte but two bunch of ions, positive in the anode and negative near the cathode, so the current can't flow very well. This is why pure water can't conduct electricity even during the electrolysis when there are many ions in the solution.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T22:42:21.437" LastActivityDate="2015-05-03T22:42:21.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6016" PostTypeId="2" ParentId="4639" CreationDate="2013-08-26T01:37:09.667" Score="1" Body="&lt;p&gt;From a paper I wrote on the algebraic method for balancing chemical equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the inspection method, shown in most chemistry texts, a more general method of balancing chemical reaction equations is the algebraic method, proposed by the Russian academician, V. A. Kistiakovskij, in 1919. Let’s illustrate the method using the following chemical reaction, which is a complete oxidation (or burning) of pentane fuel in pure oxygen to give carbon dioxide and water as the only products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C5H12 + O2 = CO2 + H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to write down the answer, only using algebraic letters instead of numbers, and then solve the system of equations resulting to obtain the numerical values of the coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aC5H12 + bO2 = cCO2 + dH2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that most of the time, there is one more variable (the algebraic coefficients we set) than equations (different atom balances), and in these cases, we can set a = 1 and solve the equations using this starting value. This works because as all good chemists know, the compounds in a chemical equation react and form in ratio amounts, always (the mole concept).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step in the process is to set up the system of equations to solve, which are just balances of atoms of each element on the left and right sides of the original equation (required according to the law of conservation of mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, the number of atoms of a particular element is equal to the coefficient in front of the compound containing that element multiplied by the subscript on the element in that compound, added to any other instances of the element in other compounds on either side of the equal sign in the original equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example, we have balances on carbon (C), hydrogen (H) and oxygen (O):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: 5a = c&lt;br&gt;&#xA;H: 12a = 2d&lt;br&gt;&#xA;O: 2b = 2c + d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting a = 1, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 = c (so c = 5)&lt;br&gt;&#xA;12 = 2d (so d = 6)&lt;br&gt;&#xA;2b = 2•5 + 6 = 16 (so b = 8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the values obtained from the coefficients back into the original equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C5H12 + 8O2 = 5CO2 + 6H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last step is to do a quick check to make sure our algebra was done correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: 5 = 5 (check)&lt;br&gt;&#xA;H: 12 = 6•2 (check)&lt;br&gt;&#xA;O: 8•2 = 5•2 + 6 (check)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method can be employed for the vast majority of chemical reactions studied in high school and even beginning college chemistry, in both the inorganic and organic worlds. More importantly, it can be automated, and so it is used as the algorithm that does the calculating for on-line and off-line, computer-based chemical equation balancers. But it is simple enough that for most chemical reactions studied in high school, it can be used by students with some knowledge of algebra, given a pencil and paper, and for more difficult reactions, can be completed faster than the inspection method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that fractional coefficients are often obtained using the substitution&#xA;a = 1 shown above. Since the convention is that coefficients of chemical equations are whole numbers, to get the final answer it is often required to multiply the set of coefficients through by the lowest common multiple (LCM or LCD) of all the denominators of the fractional values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example used in several papers in the Journal of Chemical Education from the 1930s considers the famous experiment of putting a copper penny in a flask of nitric acid, which dissolves the penny, producing copper nitrate, while giving off nitric oxide (which quickly oxidizes in air to the brown gas nitrogen dioxide):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cu + HNO3 = NO + Cu(NO3)2 + H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the algebraic method, we write down the answer using algebraic letters as coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aCu + bHNO3 = cNO + dCu(NO3)2 + eH2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then do the atomic balances on each element (I always go from left to right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cu (copper): a = d&lt;br&gt;&#xA;H (hydrogen): b = 2e&lt;br&gt;&#xA;N (nitrogen): b = c + 2d (remember that the subscripts occurring after parentheses are distributed into the elements inside like exponents in algebra)&lt;br&gt;&#xA;O (oxygen): 3b = c + 6d + e&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now set a = 1 and solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = d (so d = 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting b = 2e and d = 1 into the nitrogen and oxygen balances gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5e = c + 6 (oxygen balance)&#xA; – 2e = c + 2 (nitrogen balance)&#xA;   3e = 0 + 4 (subtracting the bottom equation from the top one)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So e = 4/3, and therefore b = 8/3 (from the hydrogen balance) and c = b – 2d (from the nitrogen balance) or c = 8/3 – 2 = 2/3. The coefficients are therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 1&lt;br&gt;&#xA;b = 8/3&lt;br&gt;&#xA;c = 2/3&lt;br&gt;&#xA;d = 1&lt;br&gt;&#xA;e = 4/3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LCM of the denominators of the fractions is 3, so we multiply through by 3, giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 3&lt;br&gt;&#xA;b = 8&lt;br&gt;&#xA;c = 2&lt;br&gt;&#xA;d = 3&lt;br&gt;&#xA;e = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3Cu + 8HNO3 = 2NO + 3Cu(NO3)2 + 4H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our quick check shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cu: 3 =3 (check)&lt;br&gt;&#xA;H: 8 = 4•2 (check)&lt;br&gt;&#xA;N: 8 = 2 + 2•3 (check)&lt;br&gt;&#xA;O: 8•3 = 2 + 6•3 + 4 (check)&lt;/p&gt;&#xA;" OwnerUserId="2208" LastActivityDate="2013-08-26T01:37:09.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6017" PostTypeId="1" CreationDate="2013-08-26T09:09:02.570" Score="8" ViewCount="2091" Body="&lt;p&gt;I need to know, is real water made up of heavy water which is formed with deuterium, and light water which is water by protium?&lt;br&gt;&#xA;We know that hydrogen has three major isotopes, including protium and deuterium.&lt;br&gt;&#xA;We also know that water, $\ce{H2O}$, is made up by hydrogen and oxygen.&lt;br&gt;&#xA;All this might mean that real water itself can naturally contain some heavy water - Deuterium oxide. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="367" LastEditDate="2013-08-27T15:51:17.310" LastActivityDate="2014-10-15T05:39:32.217" Title="Is real water made up of heavy water and light water?" Tags="&lt;water&gt;&lt;hydrogen&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6019" PostTypeId="2" ParentId="6017" CreationDate="2013-08-26T09:54:01.943" Score="16" Body="&lt;p&gt;Water has formula H&lt;sub&gt;2&lt;/sub&gt;O. Oxygen has 3 stable isotopes (99.76% &lt;sup&gt;16&lt;/sup&gt;O, 0.039% &lt;sup&gt;17&lt;/sup&gt;O, 0.201% &lt;sup&gt;18&lt;/sup&gt;O), and hydrogen has two (99.985% &lt;sup&gt;1&lt;/sup&gt;H, 0.015% &lt;sup&gt;2&lt;/sup&gt;H). Thus, there are &lt;strong&gt;9 natural isotopic configurations for water&lt;/strong&gt;: 3 possibilities for oxygen, multiplied by 3 possibilities for 2 hydrogens with 2 possible isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of those 9 possible configurations, &lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/applychem/waterchem.html&quot; rel=&quot;noreferrer&quot;&gt;only 4 have a natural abundance larger than 1 ppm&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;                        &lt;img src=&quot;https://i.stack.imgur.com/C3bQ6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, pure water is a mixture of 9 different isotopic configurations. You will note that &lt;strong&gt;D&lt;sub&gt;2&lt;/sub&gt;O, which is what is usually called &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heavy_water&quot; rel=&quot;noreferrer&quot;&gt;heavy water&lt;/a&gt;&lt;/em&gt;, is not actually among the 4 more common isotopic configurations given by natural abundance&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also of note: water that is not “heavy water” is sometimes called “light water”, but never “soft water”. &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_softening&quot; rel=&quot;noreferrer&quot;&gt;“Soft water”&lt;/a&gt; is a term used for a chemical process of natural (non pure) water that has been depleted from certain metal cations. (It is used in contrast to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_water&quot; rel=&quot;noreferrer&quot;&gt;“hard water”&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-08-26T09:54:01.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6022" PostTypeId="2" ParentId="5946" CreationDate="2013-08-26T18:45:17.353" Score="16" Body="&lt;p&gt;The purification that occurs when an aqueous solution freezes is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_freezing&quot; rel=&quot;noreferrer&quot;&gt;Fractional Freezing&lt;/a&gt; and is based on the idea that impurities of a solvent (and this can be any solvent, not just water) have a much lower solubility in the solid phase of the solvent relative to the liquid phase.   Several techniques for performing fractional freezing for purification purposes can be found &lt;a href=&quot;http://archive.org/stream/jresv32n6p253/jresv32n6p253_A1b_djvu.txt&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.   As the solid is formed, the impurities are expelled into the liquid that hasn't yet solidified.  The purified crystals can then be removed by physical methods (e.g. filtration).  In the case of water, where ice is less dense than the liquid phase, the purified crystals float to the top of the impurity-laden liquid.  This is probably the reason why &lt;a href=&quot;http://www.thenakedscientists.com/forum/index.php?topic=19783.0&quot; rel=&quot;noreferrer&quot;&gt;some people&lt;/a&gt; have observed a layer of impurities deep within a block of ice.  As mentioned above, the process is not limited to just water, and freeze crystallization is use to eliminate the purities of silicon wafers and has been known for many years as a way to make &lt;a href=&quot;http://www.popsci.com/science/article/2013-01/beersci-how-make-strong-beer-stronger&quot; rel=&quot;noreferrer&quot;&gt;hard liquor&lt;/a&gt;.  This latter use of fractional freezing is close to the OPs question, which inquires about miscible substances (in this case, ethanol and water).  Despite their miscibility, ethanol and water can be separated by fractional freezing, just like they can be separated by fractional distillation.  Certain substances cannot be separated completely using distillation, and these substances are called &lt;a href=&quot;http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Azeotrope.html&quot; rel=&quot;noreferrer&quot;&gt;azeotropes&lt;/a&gt;.  Water and ethanol make an azeotropic mixture that is approximately 95% water.  (So the moonshine you were thinking of making using the reference above will be limited to 190 proof.)  What I find interesting is the following figure, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sIAjK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows the freezing point of water/ethanol mixtures as a function of ethanol content.  At a point around 95% there is a significant dip in the freezing point.  At this point, called the eutectic point, the solution behaves as if it is a single component (much like an azeotrope) and will not likely be separable through 'simple' fractional freezing means.  Note I use the term simple because azeotropes can be separated using &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope#Separation_of_constituents&quot; rel=&quot;noreferrer&quot;&gt;special distillation techniques&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So in summary:&lt;/strong&gt; fractional freezing will work with any aqueous solution, regardless of the miscibility of the components, unless the component ratio is close the eutectic point.  We don't need to limit ourselves to aqueous solutions, as many species can be purified in this manner.  The purification mechanism is not so much a factor of the crystal lattice, but the relative solute solubilities in the solid and liquid phases of the solvent.  Finally, many water solutions will freeze solid even with impurities.  This and other types of separations work on equilibrium processes, and multiple iterations are necessary to &quot;completely&quot; remove impurities.  (or more correct, remove the impurities to a level that is acceptable for your application.)&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-08-26T18:45:17.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6024" PostTypeId="2" ParentId="5946" CreationDate="2013-08-26T18:47:34.817" Score="9" Body="&lt;p&gt;The situation you are describing is known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;eutectic system&lt;/a&gt;. Whether a system is at the eutectic point depends upon both the materials composing the solution as well as the relative amounts of each. A solution of &lt;span class=&quot;math-container&quot;&gt;$\bf{A}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\bf{B}$&lt;/span&gt; that are not at eutectic compositions will freeze out one component until a eutectic composition is achieved. While a eutectic system will simultaneously solidify both components, many systems will freeze solid &lt;span class=&quot;math-container&quot;&gt;$\bf{A}$&lt;/span&gt; and solid &lt;span class=&quot;math-container&quot;&gt;$\bf{B}$&lt;/span&gt; (the two being interdeispersed) and not solid &lt;span class=&quot;math-container&quot;&gt;$\bf{AB}$&lt;/span&gt;, though some systems do solidify as mixed (metal alloys, for example).&lt;/p&gt;&#xA;&lt;p&gt;Water and sodium chloride form a eutectic at about 23% &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; by weight. You can see a phase diagram for water and &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; at &lt;a href=&quot;http://people.ucalgary.ca/%7Ekmuldrew/cryo_course/cryo_chap6_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this University of Calgary page&lt;/a&gt;. Note that an eutectic can consist of more than two components, as mentioned by the &lt;a href=&quot;http://people.ucalgary.ca/%7Ekmuldrew/cryo_course/cryo_chap6_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Calgary page&lt;/a&gt;. A typical phase diagram of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl/H2O}$&lt;/span&gt; solution is as depicted below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wOc4N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wOc4N.png&quot; alt=&quot;Phase diagram of NaCl/H2O solution&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; An eutectic composition is around 27% &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; (by &lt;span class=&quot;math-container&quot;&gt;$w/v$&lt;/span&gt;) dissolved in the water (keep in mind that saturated &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; solution is around 30% at room temperature).&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="61578" LastEditDate="2020-08-21T17:02:28.790" LastActivityDate="2020-08-21T17:02:28.790" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="6025" PostTypeId="1" AcceptedAnswerId="6027" CreationDate="2013-08-27T12:31:19.013" Score="4" ViewCount="3445" Body="&lt;p&gt;In AlMg3, magnesium atoms outnumber aluminum atoms by a factor of three. In terms of mass, the ratio of aluminium to magnesium is a mere $27.0 / (3\times24.3) = 0.37$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Shouldn't AlMg3 be better called magnesium alloy?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1607" LastEditorDisplayName="user7951" LastEditDate="2015-03-04T17:28:25.510" LastActivityDate="2015-03-04T17:36:15.893" Title="Why is AlMg3 called aluminum alloy?" Tags="&lt;metal&gt;&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6026" PostTypeId="2" ParentId="6002" CreationDate="2013-08-27T12:46:18.907" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why Do Carbonates, Oxides, and Pure Metals Precipitate Before Metal Sulfides?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't. I cannot get to the particular paper you read, but I can provide both a logical and a quantitative argument that the metal sulfides precipitate first. The thermodynamics of this situation do not involve reduction potentials so much as simple solubilities. Metal sulfides are much less soluble in water than metal carbonates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description, I imagine a system like the following image. There are two strata of sedimentation: an upper layer of metal oxides and a lower layer of metal sulfides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TH5zZ.png&quot; alt=&quot;Sedimentation strata diagram showing two strata: a metal sulfide layer on the bottom and a metal oxide and carbonate layer on the top&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three reasonable explanations why the sulfides are on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sulfides did not precipitate from the water; they were always there. The carbonates and oxides deposited on top of them.&lt;/li&gt;&#xA;&lt;li&gt;The carbonates and oxides were deposited first, and the sulfides were deposited second, settling through the carbonates and oxides to form a deeper layer.&lt;/li&gt;&#xA;&lt;li&gt;The sulfides were deposited first, and the carbonates and oxides were deposited second.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Explanation 1 is very reasonable, but counterproductive. We will ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation 2 is the one supposed by the OP in the question, and perhaps it was also provided in the paper. Let us examine this idea a little. In order for this to happen, the metal sulfides would have to be denser than the metal carbonates and oxides to settle through. All it takes the counterexample of iron to see that this is not always the case. Iron (II) sulfide is denser than iron (II) carbonate but less dense than iron (II) oxide. &#xA; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron (II) oxide&lt;/a&gt;, $\ce{FeO}$, 5.745 g/mL&#xA; - &lt;a href=&quot;http://www.mindat.org/min-3647.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron (II) carbonate&lt;/a&gt;, $\ce{FeCO3}$, 3.961 g/mL&#xA; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron (II) sulfide&lt;/a&gt;, $\ce{FeS}$, 4.84 g/mL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, if the sulfides were settling through, the layers would not be so well stratified - the boundary between them would be indistinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Y0bL.png&quot; alt=&quot;Sedimentation strata diagram showing three strata: a metal sulfide layer on the bottom, a metal oxide and carbonate layer on the top, and a mixed layer with all three in the middle.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal sulfides are much less soluble in water than metal carbonates. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That leaves with Explanation 3, which is the correct one. Metal sulfides are almost universally less soluble that metal carbonates. We'll get to where the oxides come from later. Both metal carbonates and metal sulfides are nearly insoluble in water, so the minute amounts present are hard to quantify in the &quot;grams of solute per liter of solution&quot; kind of way. The saturated solution may only have few parts per billion of the compound. Instead, we talk in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_product&quot; rel=&quot;nofollow noreferrer&quot;&gt;solubility equilibrium&lt;/a&gt;, in particular, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_product#Dissolution_with_dissociation&quot; rel=&quot;nofollow noreferrer&quot;&gt;solubility product&lt;/a&gt; (constant), $k_{sp}$, the equilibrium constant associated with the combination dissolution and dissociation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can determine the concentration from the $k_{sp}$. Consider &lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;barium carbonate&lt;/a&gt; $k_{sp}=5.1\times10^{-9} \text{ M}^3$. The solubility equilibrium is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BaCO3(s) &amp;lt;=&amp;gt;Ba^{2+}(aq) + CO3^{2-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the equilibrium constant is (the concentration of solid barium hydroxide is constant and is thus part of $k_{sp}=k_{obs}$.&#xA;$$k_{sp} = [\ce{Ba^{2+}}][\ce{CO3^{2-}}]=5.1 \times 10^{-9} \text{ M}^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $[\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]$ (mass balance):&#xA;$$[\ce{Ba^{2+}}][\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]^2$$&#xA;$$[\ce{Ba^{2+}}]^2=5.1\times 10^{-9} \text{ M}^3$$&#xA;$$[\ce{Ba^{2+}}]= 7.14\times 10^{-5} \text{ M}=&amp;gt;0.0141 \text{ grams per liter}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's look at carbonates and sulfides. All $k_{sp}$ came from &lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cadmium:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carbonate $\ce{CdCO3}$, $k_{sp}=5.2\times 10^{-12} \text{ M}^2$, $[\ce{Cd^{2+}}]=2.28\times 10^{-6} \text{ M}$&lt;/li&gt;&#xA;&lt;li&gt;Sulfide $\ce{CdS}$, $k_{sp}=8\times 10^{-27} \text{ M}^2$, $[\ce{Cd^{2+}}]=8.94\times 10^{-14} \text{ M}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cadmium sulfide is less soluble than cadmium carbonate by a factor $10^8$ (100 million). The same behavior exists for cobalt, copper, iron, lead, manganese, nickel, silver, and zinc (all metals for which data for both the carbonate and sulfide are listed).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why might this paper suppose that the sulfides are deposited second when the sulfides are much less soluble than the carbonates?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors of the paper probably did not know that (and perhaps few others did either). That paper was published in 1901. &lt;a href=&quot;http://en.wikipedia.org/wiki/Svante_Arrhenius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Svante Arrhenius&lt;/a&gt; presented his thesis on the behavior of salts in water in 1884. It was not well thought of initially. He received a Nobel Prize for the work in 1903 (after the paper was published), but there is little reason to believe the authors (mining engineers) would have been taught controversial and cutting edge physical chemistry in whatever minimal introductory chemistry courses they had taken (if any). I may mention recent Nobel work in chemistry in my intro courses, but I certainly don't &lt;strong&gt;teach&lt;/strong&gt; recent Nobel work in chemistry in into courses. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do the metal oxides come from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Metal oxides form either by decomposition of the carbonates or by oxidation of pure metals (which probably occur from the decomposition of the sulfides):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decomposition of sulfides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{FeS(s) &amp;lt;=&amp;gt; Fe(s) + S(S)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation of metal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{6Fe(s) + 3O2(aq) -&amp;gt; 3Fe2O3(s)}$$&#xA;$$\ce{2Fe(s) + 3H2O(l) -&amp;gt; Fe2O3(s) + 3H2} \text{ (acid is helpful here)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decomposition of the carbonate. Since the carbon dioxide would be carried away by the water, this equilibrium is continually pushed toward the oxide.&#xA;$$\ce{FeCO3(s) &amp;lt;=&amp;gt; FeO(s) + CO2(aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="402" LastEditDate="2017-04-08T21:57:16.567" LastActivityDate="2017-04-08T21:57:16.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6027" PostTypeId="2" ParentId="6025" CreationDate="2013-08-27T13:48:42.407" Score="14" Body="&lt;p&gt;&lt;strong&gt;AlMg3 is not the chemical formula of the alloy&lt;/strong&gt;, it's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_alloy#Wrought_alloys&quot; rel=&quot;noreferrer&quot;&gt;product name for a wrought alloy&lt;/a&gt;. Its &lt;a href=&quot;http://www.stella-welding.com/st/AlMg3.pdf&quot; rel=&quot;noreferrer&quot;&gt;technical sheet&lt;/a&gt; specifies the following composition:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Product name&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;AlMg3&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Class of product&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;Al-Mg alloy for MIG/TIG welding.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Corresponding standards&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;DIN 1732, SG-AlMg3, AWS A5.10, ER 5754&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Nominal composition (weight %)&lt;/strong&gt;&lt;/em&gt;&lt;br /&gt;&#xA;Al: Bal. Si: 0.4 Mg: 3 Mn: 0.5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, &lt;strong&gt;the 3 in the product name refers to 3 % (in weight) of magnesium.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-03-04T17:36:15.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6028" PostTypeId="2" ParentId="6017" CreationDate="2013-08-27T16:16:07.590" Score="1" Body="&lt;p&gt;Theoretically, you are correct; water, being chemically composed of hydrogen and oxygen, can contain any isotope of hydrogen (or oxygen) and so there are many isotopic combinations of water. However, if you look at hydrogen's molar mass (which is calculated based on a weighted average of the atomic masses of the two stable isotopes), you'll see that it's much closer to 1g than to 2. This is because Earth's hydrogen content is dominated by the protium isotope, with an atomic mass of 1 dalton. If there were a 50-50 mix between protium and deuterium we'd expect the molar mass to be 1.5g, but as it is, Earth's hydrogen content is only approximately 0.015% deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we could expect, on average, a slightly lower percentage of water to have at least one deuterium atom, because a very few have two. As shown by F'x's chart, the concentration of naturally-occurring deuterium oxide is measured in the parts per billion (22ppb). That's actually a lot of heavy water, if you were to get all of it in one place; there are about 31 trillion kilograms of naturally occurring deuterium oxide on the planet (31 km&lt;sup&gt;3&lt;/sup&gt;). That's enough to fill Lake Mead, created by the Hoover Dam. However, to get one kilogram, or liter, of $\ce{D2O}$ by separating it from light water, you would need to process 45 million kilograms of water. Put another way, there is a little over one gallon of natural deuterium oxide (5L to be more precise) in a typical Olympic swimming pool.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-03T15:36:40.000" LastActivityDate="2013-09-03T15:36:40.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6030" PostTypeId="1" CreationDate="2013-08-27T18:12:11.397" Score="4" ViewCount="535" Body="&lt;p&gt;Nicotinoyl chloride is only available as its hydrochloride salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't nicotinoyl chloride be synthesized in free form? How can we seperate nicotinoyl chloride from its salt form? Can we try any base for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="2214" LastEditorUserId="22" LastEditDate="2013-08-27T18:57:47.737" LastActivityDate="2013-12-29T12:32:56.707" Title="Why is nicotinoyl chloride available as its hydrochloride salt? How can we separate them?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6031" PostTypeId="1" AcceptedAnswerId="6060" CreationDate="2013-08-28T01:29:32.867" Score="3" ViewCount="8947" Body="&lt;p&gt;I know that for a carbon to be chiral four different groups should be attached to that particular carbon. But, what about in the case of cyclic compounds? What's the method to check chirality if the carbon is not $\mathrm{sp}^3$ hybridized and only three, or two groups are attached to it (through double or triple bonds)?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="30836" LastEditDate="2020-04-06T11:53:43.397" LastActivityDate="2020-04-06T11:53:43.397" Title="Finding chiral carbons in aromatic or cyclic compounds" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;stereochemistry&gt;&lt;chirality&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6032" PostTypeId="1" AcceptedAnswerId="6044" CreationDate="2013-08-28T01:50:16.093" Score="2" ViewCount="5805" Body="&lt;p&gt;Why can't you make potassium chlorate using the chloride anion from the potassium chloride and having oxygen bond with it from hydrogen peroxide?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="41328" LastEditDate="2017-10-13T13:37:05.743" LastActivityDate="2020-11-21T19:58:40.163" Title="Potassium chlorate from potassium chloride and hydrogen peroxide" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6033" PostTypeId="1" CreationDate="2013-08-28T02:45:35.913" Score="1" ViewCount="139" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You are given compounds $x, y$ and $z$. $x$ conducts electricity in aqueous solution as well as in molten state. $y$ conducts electricity in aqueous form only. $z$ conducts electricity in neither the aqueous nor the molten state. Predict the intramolecular bonds in each compound. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compound $x$ is clearly ionic. In water, the ions disassociate and form free charges for flow of electricity. In the molten state, the ionic bonds break, and again free charges can flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about the other two compounds. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="5026" LastEditDate="2018-02-27T02:46:21.767" LastActivityDate="2018-02-27T02:46:21.767" Title="Predicting bonds from properties of compounds" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6034" PostTypeId="2" ParentId="6032" CreationDate="2013-08-28T02:58:37.473" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_chlorate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Potassium chlorate ($\ce{KClO3}$)&lt;/a&gt; has &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;covalent bonds&lt;/a&gt; between the chloride and the oxygen, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;potassium chloride ($\ce{KCl}$)&lt;/a&gt; lacks those covalent bonds to oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide decomposes into water ($\ce{H2O}$) and oxygen gas ($\ce{O2}$) while oxidizing (LEO: Loss of Electrons is Oxidation) its environment. Peroxide is useful primarily for sucking electrons from a solution, which is very different from forming covalent bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wished to dissolve solid potassium in water to make KOH peroxide would likely perform nicely as the peroxide would suck electrons from the solid K and thereby help it form ions ($\ce{K+}$) which would combine with $\ce{OH-}$ from the water to from KOH. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="41328" LastEditDate="2017-10-13T13:40:14.277" LastActivityDate="2017-10-13T13:40:14.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6035" PostTypeId="1" CreationDate="2013-08-28T05:30:12.677" Score="7" ViewCount="2795" Body="&lt;p&gt;Look at the molecular structure of benzene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTvAI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a perfect hexagon. Why aren't there any molecules arranged in a triangular fashion with bonds forming the edges and the molecules the vertices? Or, for that matter, square molecules? And what about other n-gonal molecules? octagons? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2014-11-29T13:49:07.220" Title="Why aren't there any triangular molecules?" Tags="&lt;orbitals&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6036" PostTypeId="2" ParentId="6035" CreationDate="2013-08-28T06:13:04.063" Score="8" Body="&lt;p&gt;There exists triangular molecule, eg. &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopropane&quot; rel=&quot;nofollow&quot;&gt;cyclopropane&lt;/a&gt;, but they are highly reactive due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_strain&quot; rel=&quot;nofollow&quot;&gt;ring strain&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ring strain results from a combination of angle strain, conformational strain or Pitzer strain, and transannular strain or van der Waals strain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;C-C-C bond angle in rings should be approximately 109.5° degrees which in case of small rings like cyclopropane, bond angles &lt;strong&gt;are 60° whereas tetrahedral 109.5° bond angles are expected&lt;/strong&gt;. The intense angle strain leads to nonlinear orbital overlap of its sp3 orbitals. Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this strain they exists in higher energy state and therefore &lt;strong&gt;Because of their high strain, the heat of combustion for these small rings is elevated.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kindly see &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_strain&quot; rel=&quot;nofollow&quot;&gt;Ring Strain&lt;/a&gt;, it will help you to understand about these kinds of molecules.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-09-08T08:47:12.507" LastActivityDate="2014-09-08T08:47:12.507" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6037" PostTypeId="1" CreationDate="2013-08-28T09:38:18.350" Score="9" ViewCount="233" Body="&lt;p&gt;In chemistry we use ultrasonication for cleaning vessels ,synthesis of chemical compounds (sonochemistry) etc.I know that during ultrasonication high temperature and pressure is formed &quot;locally&quot;.But I want to know whether there is any equations which relates between ultrasonication and temperature .&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2015-01-21T07:40:08.850" Title="Is there any relationship between ultrasonication and temperature" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6038" PostTypeId="2" ParentId="6002" CreationDate="2013-08-28T15:32:31.037" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In marine environment, reducing conditions generally occur a short distance below the depositional interface and sulfur is available in solution and organic matter. The result is &lt;strong&gt;reduction of iron, first to mono-sulfides and then to pyrite&lt;/strong&gt; (see Chap. 19). p.403.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Iron minerals are forming today in several environments: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) tidal flats and bog lakes&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) oceanic shelves in tropical climates&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) deep basins with restricted circulation, such as fjords and the Black Sea&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The &lt;strong&gt;formation of pyrite in muds&lt;/strong&gt; has been the subject of several theoretical, experimental, and field studies. Sulfur isotope analysis has helped to throw light on the origin of the sulfur. These studies have been summarized by Berner (1970)  and his conclusions are given below. He deduced that the fulfur in the pyrite of  modern muds comes from two sources (a) organic matter and (b) sulfate dissolved in seawater. Although organic matter forms up to 10% of some modern marine muds, this carbonaceous material contains only 1% sulfur. Yet the muds frequently contain more than 1% pyrite and so it is clear that an additional source of sulfur is required. &lt;strong&gt;The other source is the bacterial reduction of dissolved sulfate&lt;/strong&gt;, which has been shown by sulfur isotope studies to be supplied by diffusion into the sediment from the overlying seawater. Where dissolved $H_2S$ exists, pyrite is generally the thermodynamically stable form of iron; pyrite is not formed directly however. &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The first iron compounds to appear are black, finely disseminated&#xA;  iron mono-sulfides.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;They can tribute much of the black color generally seen just beneath the surface of tidal flats and organic-rich lakes and stagnant basins. These compounds, which are thermodynamically meta-stable, are observed to change to pyrite a few inches below the sediment-water interface. Berner has presented evidence indicating that &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;the formation of pyrite takes place mainly by the reaction of&#xA;  mono-sulfides with elemental sulfur produced from the oxidation of&#xA;  $H_2S$ either inorganically or through the action  of&#xA;  sulfur-oxidizing bacteria.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If the bottom waters contain dissolved oxygen, elemental sulfur  is&#xA;  produced by the reaction of FeS and $H_2S$ with dissolved oxygen&#xA;  stirred into the sediment by storms, currents, and burrowing&#xA;  organisms.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The fact that free $H_2S$ is present in a sediment, or even in the deeper parts of basins with restricted circulation, does not necessarily mean that iron sulfides will be the only iron minerals present.&lt;/strong&gt; In the Black Sea deeper waters have a  slightly higher salinity (22 ppt) than the surface waters (18 ppt). The result of the density stratification has been to prevent mixing between the surface, oxygenated waters, and the waters below a depth of 150m with the result that the waters from 150m to depths in excess of 200 m contains dissolved $H_2S$. Muds in the deeper parts of teh Black Sea contain 3 to 5% total iron, on a carbonate-free basis. Normally 30 to 50% of this iron is present as iron sulfide. Not all the iron is present as pyrite because over 95% of the iron supplied to the Black Sea is in the form of very fine clastic particles (magnetite, hematite, chlorite), which are not converted to sulfides as they  settle through the deeper waters despite the high content of $H_2S$ in these waters. Once buried below the sediment interface, the conversion of the iron compounds to sulfides is limited by the availability of organic matter, sulfur, and perhaps also by slow reaction rates. This situation occurs in spite of the high content of organic matter, which reaches values in excess of 5% organic carbon. (To me this is confusing because &quot;organic matter&quot; first causes pyrite and then doesn't... perhaps the word &quot;normally&quot; was meant to apply to pyrite formation and the latter to this special case of increasing salinity at depth.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The &quot;&lt;strong&gt;bog iron ores&lt;/strong&gt;&quot; of modern swamps and lakes represent one of the few examples of modern iron-rich sediments. They are found in pororly drained, recently glaciated regions or areas of older iron-rich rocks. Two general groups have been recognized: lake ores and bog ores. Lake ores consist of oolitic or pisolitic grains, cemented together into discs up to a foot or more across. They are commonly found in water only a few feet deep, at the border of the lake. Bog ores form thin layers of earthy-to-pisolitic iron oxides at the surface or below several meters of peat. &lt;strong&gt;Both types appear to be formed by the migration of acidic organic-rich groundwater. Precipitation of ferric iron takes place where the migrating soil or swamp-derived waters  enter a more oxidizing, less acidic environment.&lt;/strong&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;p.603-605. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Blatt, Harvey, Gerard V. Middleton, and Raymond C. Murray. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0136427103&quot; rel=&quot;nofollow&quot;&gt;Origin of Sedimentary Rocks&lt;/a&gt;. Englewood Cliffs, N.J: Prentice-Hall, 1972. Print.&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-08-28T15:32:31.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6039" PostTypeId="2" ParentId="4636" CreationDate="2013-08-28T17:20:51.693" Score="1" Body="&lt;p&gt;They stink because the nasal receptors activated by those compounds trigger nerve pathways that activate neurons corresponding to the perception of stench. The question is phrased as if the elements themselves were responsible for stench, when really it is the nasal receptor cells/proteins and neuronal pathways that are to blame. This questions belongs on biology.se unless the question is rephrased to inquire about the mechanism of protein receptor binding to said elements. &lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2013-08-28T17:20:51.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6040" PostTypeId="1" AcceptedAnswerId="6042" CreationDate="2013-08-28T19:20:55.713" Score="6" ViewCount="8556" Body="&lt;p&gt;In a recent Mythbusters episode, where they tested some Breaking Bad chemistry assertions, they debunked the hydrofluoric acid body-dissolving myth referenced elsewhere in Chemistry.SE in the question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/4877/is-it-actually-possible-to-dispose-of-a-body-with-hydrofluoric-acid/4915#4915&quot;&gt;Is it actually possible to dispose of a body with hydrofluoric acid?&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no, not really, and for the same reasons as my answer to the linked question; fluorine's just too electronegative to let the proton go do its acid thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To attempt to replicate the results (in this case dissolving out the bottom of the bathtub they were using), they moved to a much stronger acid, sulfuric, and then &quot;supercharged&quot; it with what they simply called their &quot;special sauce&quot; (as they said, they're not in the business of showing people how to dispose of bodies). They gave it away in the end, however, by saying the special sauce was &quot;basically a lot of hydrogen and oxygen&quot;, and since it wouldn't be water the obvious candidate is hydrogen peroxide, a strong oxidizer (in high concentrations at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a well-known trick of household chemistry; add household peroxide to cream of tartar or lemon juice, for instance, and watch those rust stains come right out. It also seems to be the bracingly direct way to form perchlorates in the lab; add hydrogen peroxide to hydrochloric acid, and behold perchloric acid (or a fuming crater, if you weren't managing the exothermic nature of this reaction properly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;first question, what's the technical term for this trick?&lt;/strong&gt; It can't just be &quot;supercharging&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second and related question, does it work for all acids and oxidizers&lt;/strong&gt;, or do only certain combinations of acids and oxidizers have this effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, for purely academic purposes, &lt;strong&gt;what would the basic reaction have been between sulfuric acid, hydrogen peroxide and organic tissue&lt;/strong&gt; ($\ce{HxCyOz}$)? The sulfate anion is pretty highly oxygenated as it is, so I'd imagine the combination, unlike with a hydrohalic acid forming an oxoacid, would cause the reaction to be less acid-base, and more rapid oxidation and &quot;combustion&quot; of the organic tissues, leaving behind mainly carbon in water and off-gassing a large amount of steam and sulfur dioxide.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2013-08-29T01:58:04.713" Title="What's the technical term for supercharging an acid with an oxidizer? And does it always work?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6041" PostTypeId="1" AcceptedAnswerId="6043" CreationDate="2013-08-28T21:25:38.500" Score="1" ViewCount="386" Body="&lt;p&gt;For a homework assignment on significant digits, I have to find the number of significant digits in the number $0.40 \times 10^3$. I know that in scientific notation, the number of significant digits is equal to the number of digits in the stem. However, I've also learned that for a number to be in scientific notation, it must have a stem between $1$ and $10$. What is the rule for this situation? Is the number of Significant digits $2$ or $3$?&lt;/p&gt;&#xA;" OwnerUserId="2220" LastEditorUserId="69830" LastEditDate="2021-05-08T12:33:37.860" LastActivityDate="2021-05-08T12:33:37.860" Title="Significant Figures in almost-scientific notation" Tags="&lt;terminology&gt;&lt;notation&gt;&lt;significant-figures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6042" PostTypeId="2" ParentId="6040" CreationDate="2013-08-28T21:47:38.797" Score="6" Body="&lt;p&gt;I don't know if it's called anything, but when you use an oxidizer in solution, you usually need a source of hydrogen ions for the oxidized oxygen to be reduced.  In the case of hydrogen peroxide, $\ce{H2O2}$, the oxygen is in the $-1$ state rather than the normal $-2$.  The usual way $\ce{H2O2}$ acts as an oxidizer is to be reduced to water in the presence of 2 hydrogen ions, usually from a strong acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+ + 2e- + H2O2 -&amp;gt; 2H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the reaction consumes 2 electrons on the left, which would come from other reactants that have been oxidized (i.e. lost electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that a strong acid by itself has only a corrosive effect, and a highly dehydrating effect in the case of sulfuric acid.  But if you add an oxidizer, then in the presence of a fuel (like an organic compound) plus sufficient energy (say from the heat of reaction), then you are effectively getting a &lt;em&gt;combustion&lt;/em&gt; reaction as well.  This is the sort of &quot;combustion&quot; that goes on inside your body to use food as fuel, for example.  I supposed this could be called &quot;supercharged&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So to answer your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) No technical term that I know of;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Yes, assuming the acid is strong enough to donate protons;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In most organic molecules, carbon is more electronegative than hydrogen, so it can be oxidized by oxygen.  And so you get a &quot;combustion&quot; reaction where molecular carbon is oxidized to elemental carbon (i.e. burnt).&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-08-29T01:58:04.713" LastActivityDate="2013-08-29T01:58:04.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6043" PostTypeId="2" ParentId="6041" CreationDate="2013-08-28T22:28:34.233" Score="2" Body="&lt;p&gt;Most of the answers to your questions can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; wikipedia article. You are right if you say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've also learned that for a number to be in scientific notation, it&#xA;  must have a stem between 1 and 10.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the convention in scientific notation. The number format of $0.40 \times 10^3$ is called exponential notation (at least it was called that way in a book about the programming language Fortran I'm reading at the moment): Here the number has a stem between 0 and 1 followed by a power of 10. This is quite common in computer science.&#xA;As to the significant digits: &lt;a href=&quot;http://en.wikipedia.org/wiki/Significant_figures#Identifying_significant_digits&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; wikipedia article (especially the paragraph &quot;Scientific notation&quot;) contains the information you need. According to its information I would say that the number $0.40 \times 10^3$ has $2$ significant digits.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-08-28T22:38:22.123" LastActivityDate="2013-08-28T22:38:22.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6044" PostTypeId="2" ParentId="6032" CreationDate="2013-08-29T02:14:10.670" Score="6" Body="&lt;p&gt;&lt;strong&gt;Do not attempt!&lt;/strong&gt; Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redox reactions are spontaneous if the two half reactions would produce a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;noreferrer&quot;&gt;positive potential in a galvanic cell.&lt;/a&gt; Most redox half reactions have measured &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_standard_electrode_potentials&quot; rel=&quot;noreferrer&quot;&gt;standard electrode potentials&lt;/a&gt; against a known standard so that they can be compared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction you want to do is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{3H2O2 + Cl- -&amp;gt; 3H2O + ClO3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two half-reactions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Oxidation:   $\ce{Cl- +3H2O -&amp;gt; ClO3- +6H+ +6e-}$&lt;/li&gt;&#xA;&lt;li&gt;Reduction:   $\ce{H2O2 +2H+ +2e- -&amp;gt; 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Hess's Law&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ 2ClO3- + 12H+ + 10e- -&amp;gt;Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$&#xA;$$\ce{Cl2 +2e- -&amp;gt; 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$&#xA;$$\ce{2ClO3- 12H+ + 12e- -&amp;gt; 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since standard potentials do not scale by stoichiometric factor (it's in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;noreferrer&quot;&gt;Nernst equation&lt;/a&gt; instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the &quot;reduction half-reaction&quot;, we need a negative sign:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Oxidation:   $\ce{Cl- +3H2O -&amp;gt; ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$&lt;/li&gt;&#xA;&lt;li&gt;Reduction:   $\ce{H2O2 +2H+ +2e- -&amp;gt; 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$&lt;/li&gt;&#xA;&lt;li&gt;Overall: $\ce{3H2O2 + Cl- -&amp;gt; 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  &lt;strong&gt;Do not attempt!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reduction:   $\ce{H2O2 +2H+ +2e- -&amp;gt; 2H2O} \ \ \ \ E^ \circ =  +1.78&#xA;   \text{ V}$ &lt;/li&gt;&#xA;&lt;li&gt;Oxidation:   $\ce{2Cl- + 2e- -&amp;gt; Cl2} \ \ \ \ E^ \circ = &#xA;   -1.36 \text{ V}$ &lt;/li&gt;&#xA;&lt;li&gt;Overall: $\ce{ 2Cl- + H2O2 + 2H+ -&amp;gt; Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-08-29T02:14:10.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6045" PostTypeId="1" AcceptedAnswerId="6046" CreationDate="2013-08-29T05:10:12.000" Score="3" ViewCount="2697" Body="&lt;p&gt;What does &lt;strong&gt;5%&lt;/strong&gt; aqueous solution of glucose mean? Note that it's just &lt;strong&gt;5%&lt;/strong&gt; and not &lt;strong&gt;5%w/w&lt;/strong&gt;. Do we assume it to be w/w?&lt;/p&gt;&#xA;&lt;p&gt;The question is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If the osmotic pressure of 5% aqueous soultion of glucose is &lt;span class=&quot;math-container&quot;&gt;$\pi_1$&lt;/span&gt; and that of 5% aqueous solution of urea is &lt;span class=&quot;math-container&quot;&gt;$\pi_2$&lt;/span&gt; then:&lt;/p&gt;&#xA;&lt;p&gt;A) &lt;span class=&quot;math-container&quot;&gt;$\pi_1=\pi_2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;B) &lt;span class=&quot;math-container&quot;&gt;$\pi_1&amp;lt;\pi_2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;C) &lt;span class=&quot;math-container&quot;&gt;$\pi_1&amp;gt;\pi_2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Does 5% mean that the solutions are equi-molar/molal?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-08-29T10:06:43.410" Title="Notation &quot;5% solution&quot;" Tags="&lt;aqueous-solution&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6046" PostTypeId="2" ParentId="6045" CreationDate="2013-08-29T06:52:57.470" Score="5" Body="&lt;p&gt;Generally speaking, percentage aqueous solutions like that are either weight for weight or weight for volume. Either way, in this example, you are comparing how many osmoles there for the same weight of urea ($CO(NH_2)_2$) and glucose ($C_6H_{12}O_6$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mole_fraction#Mole_percentage&quot; rel=&quot;nofollow&quot;&gt;Mole fraction as expressed in a percentage&lt;/a&gt; does exist though I have not seen that notation much myself and is usually designated n/n%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the wording your question I'm guessing that it's a physiology or biochemistry question, in which case they're almost certainly referring to the 5% aqueous solution of glucose as the common IV hydration fluid &lt;a href=&quot;https://en.wikipedia.org/wiki/Intravenous_sugar_solution&quot; rel=&quot;nofollow&quot;&gt;5% Dextrose Solution&lt;/a&gt;, which is w/w.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, it's hard to guess as it's ambiguous!&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="2095" LastEditDate="2013-08-29T10:06:43.410" LastActivityDate="2013-08-29T10:06:43.410" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6047" PostTypeId="1" AcceptedAnswerId="6048" CreationDate="2013-08-29T10:10:29.700" Score="4" ViewCount="1184" Body="&lt;p&gt;I ask this question because if acid anhydrides are always unstable in water then my interest, as a student of pharmacology, would be greatly diminished. Acid anhydrides are, in my books, at least, the result of a dehydratation reaction between two carboxylic acids and hence their structures look like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/P56BN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where R1 and R2 are organic substituents.&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2014-05-21T07:30:16.537" LastActivityDate="2020-07-01T12:52:38.127" Title="Is it possible to create an acid anhydride that's stable in water and if so what conditions must its structure satisfy?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6048" PostTypeId="2" ParentId="6047" CreationDate="2013-08-29T13:10:42.150" Score="4" Body="&lt;p&gt;I don't know of any acid anhydride that has completely stability in water, but I think &lt;strong&gt;the best way to stabilize them would be to choose a cyclic one&lt;/strong&gt;, which would therefore favor the anhydride form because it makes a closed ring structure. Looking in that direction, it seems that &lt;a href=&quot;http://en.wikipedia.org/wiki/Phthalic_anhydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;phthalic anhydride&lt;/a&gt; in particular has a  rather slow reaction with water (to the point that solubility values are available!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;                                                                 &lt;img src=&quot;https://i.stack.imgur.com/ikjyb.png&quot; alt=&quot;phthalic anhydride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if you want a larger ring, &lt;a href=&quot;http://www.orgsyn.org/orgsyn/orgsyn/prepcontent.asp?prep=cv3p0449&quot; rel=&quot;nofollow noreferrer&quot;&gt;homophtalic anhydride&lt;/a&gt; (also &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.12274.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;                                                          &lt;img src=&quot;https://i.stack.imgur.com/u3gRI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't find any data on its stability/solubility in water.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-08-29T13:43:38.183" LastActivityDate="2013-08-29T13:43:38.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6050" PostTypeId="1" AcceptedAnswerId="6058" CreationDate="2013-08-29T13:39:42.080" Score="5" ViewCount="219" Body="&lt;p&gt;Looking for the structure of 2-methylimidazolate, I find conflicting results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this page on &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=87677887&amp;amp;viewopt=PubChem&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium 2-methylimidazolate&lt;/a&gt; gives:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RgaNe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RgaNe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;this other page on &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.24770682.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cadmium 2-methylimidazolate&lt;/a&gt; gives another anion, namely a carboxylate derivative of 2-methylimidazole:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dqYhs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dqYhs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which is it? Where in IUPAC nomenclature rules can I get the definitive answer?&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="41328" LastEditDate="2017-08-12T13:01:28.407" LastActivityDate="2017-08-13T03:51:31.953" Title="What is the structure of 2-methylimidazolate?" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6051" PostTypeId="1" AcceptedAnswerId="6052" CreationDate="2013-08-29T13:52:11.813" Score="1" ViewCount="3311" Body="&lt;p&gt;I guess that sodium loses an electron so has an oxidation state of +1 and that oxygen has an oxidation state of −2. But what oxidation state does the aluminium atom have in sodium aluminate ($\ce{NaAlO2}$)?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="16683" LastEditDate="2020-09-14T09:10:55.933" LastActivityDate="2020-09-14T09:11:22.470" Title="What is the oxidation state of aluminium in NaAlO₂?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2020-09-14T09:10:39.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="6052" PostTypeId="2" ParentId="6051" CreationDate="2013-08-29T13:56:51.367" Score="3" Body="&lt;p&gt;For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-08-29T13:56:51.367" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6053" PostTypeId="1" AcceptedAnswerId="6057" CreationDate="2013-08-29T15:55:37.817" Score="10" ViewCount="200" Body="&lt;p&gt;I've read &lt;a href=&quot;http://www.sciencedaily.com/releases/2013/08/130827091636.htm&quot;&gt;this&lt;/a&gt; today. From the little I know, they discover new elements by bombarding some atoms in particle accelerators (I guess) and these new elements last a few miliseconds. So what's the insight they reveal to us? Why are these discoveries useful?&lt;/p&gt;&#xA;" OwnerUserId="1311" LastEditorUserId="189" LastEditDate="2013-08-29T16:39:31.213" LastActivityDate="2013-08-29T19:15:37.630" Title="Why search new elements?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6055" PostTypeId="1" CreationDate="2013-08-29T16:47:00.340" Score="7" ViewCount="8072" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If pH of the solution is zero, what would be the normality of $250 \, \mathrm{ml}$ $\ce{H2SO4}$ solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My work:&lt;/strong&gt;&#xA;$\mathrm{pH} = 0$, hence the concentration of $\ce{H+}$ ions is $\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\pu{1M}$ hence the normality of the solution must be $2~\mathrm{N}$.  Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="1813" LastEditorUserId="5026" LastEditDate="2018-02-26T05:04:06.170" LastActivityDate="2018-02-26T05:37:17.350" Title="If the pH is zero, what would be the normality of a sulphuric acid solution?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6057" PostTypeId="2" ParentId="6053" CreationDate="2013-08-29T19:15:37.630" Score="15" Body="&lt;p&gt;There are multiple reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;To improve our experimental prowess:&lt;/strong&gt; There's no single catch-all process to create higher elements. While the basic process is to smash stuff together, once can't get much more specific than that. For each new element, the techniques used differ. And, as with any experiment, the complications which arise differ. Solving these lead to a deeper understanding of how to conduct experiments. Note that there have been gaps of years between the synthesis of various transuranium elements. This is not because scientists find it a useless task; it is because designing an experiment to synthesize these elements is &lt;em&gt;hard&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;To test stability:&lt;/strong&gt; We have various nuclear models that can predict the stability of the elements. Creating new atoms helps validate these theories.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Islands of stability:&lt;/strong&gt; Some higher elements are considered &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot;&gt;islands of stability&lt;/a&gt; in that they are more stable than their neighbors. If we can improve our experimental techniques to the extent that these can be generated, we may find some use for them. Of course, these may or may not be useful &amp;mdash; these still are short-lived, just not &lt;em&gt;extremely&lt;/em&gt; short-lived&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Because we can!&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-08-29T19:15:37.630" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6058" PostTypeId="2" ParentId="6050" CreationDate="2013-08-29T20:20:32.213" Score="6" Body="&lt;p&gt;As far as I understand the IUPAC nomenclature, 2-methylimidazolate would not be the correct name for either structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term imidazolate refers to the anion of imidazole, though it is not really the correct suffix according to the proposed revision of the &lt;a href=&quot;http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter7.pdf&quot; rel=&quot;noreferrer&quot;&gt;IUPAC blue book&lt;/a&gt; (P72.1):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Method (1) leads to preferred IUPAC names. Some names and some&#xA;  contracted names are retained as preferred IUPAC names and for use in&#xA;  general nomenclature. The following suffixes are used: &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;‘ide’ (corresponding to removal of a hydron, H+), &lt;/li&gt;&#xA;  &lt;li&gt;‘uide’ (corresponding to the addition of a hydride, H−), &lt;/li&gt;&#xA;  &lt;li&gt;‘elide’ (corresponding to the addition of an electron) &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;The endings ‘ate’ and ‘ite’ are used to indicate removal of a proton&#xA;  from a −OH group of acids and hydroxy compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct term for the imidazole anion would be imidazolide, and not imidazolate as imidazol is neither an acid nor a hydroxy compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both database entries the IUPAC name mentioned is not imidazolate, this seems to be a trivial name for those structures. So my suspicion would be that both have been called imidazolate at some point, but this does not seem to be the correct IUPAC nomenclature for either of them.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-08-29T20:20:32.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6059" PostTypeId="1" CreationDate="2013-08-29T20:45:57.963" Score="1" ViewCount="2321" Body="&lt;p&gt;I'm writing up a practical and $\beta$-mercaptoethanol was used, with no concentration given. I've found I don't need the BME molarity since I'm giving percentage (v/v) concentration, however I'm slightly confused as to why there are different values for its concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/m6250&quot; rel=&quot;nofollow&quot;&gt;Sigma-Aldrich provide BME concentration 14.3 M&lt;/a&gt; (pure liquid), yet the &lt;a href=&quot;http://openwetware.org/wiki/SDS_sample_buffer&quot; rel=&quot;nofollow&quot;&gt;openwetware.org protocol for SDS sample buffers&lt;/a&gt; gives a molarity of 14.7 M. I can't see precisely how this is derived from its density, or why there'd be different molarities floating around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not that it matters for my writing here, but since I don't know the molarity I'm just a bit puzzled now as to what it may have actually been, and would rather understand now than have to work it out further down the line. Any answers greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="4945" LastEditDate="2014-07-30T02:22:54.977" LastActivityDate="2015-05-15T12:30:04.367" Title="$\beta$-mercaptoethanol molarity" Tags="&lt;experimental-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6060" PostTypeId="2" ParentId="6031" CreationDate="2013-08-29T21:20:15.297" Score="6" Body="&lt;p&gt;If an atom is part of a double or a triple bond it is never a stereogenic center because then this atom and the atoms bound to it form something planar, so that you can always find a symmetry plane.&#xA;But an atom that is part of a double bond can be &lt;a href=&quot;https://en.wikipedia.org/wiki/Prochirality&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;prochiral&lt;/strong&gt;&lt;/a&gt;, if it can be converted from to a stereogenic center in a single step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B8Lho.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-08-29T21:20:15.297" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6062" PostTypeId="2" ParentId="5718" CreationDate="2013-08-30T08:04:26.160" Score="1" Body="&lt;p&gt;From a quick search of the literature, it seems that as you correctly thought, gas chromatography mass spectrography is a method that can be used to quantitate small amounts of cyanide in vapour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I am certainly no expert on this field or the precise method that you would use, this paper, which describes quantitation of cyanide and other chemicals in tobacco smoke may be of interest: &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/%28SICI%291520-667X%282000%2912:3%3C142%3a%3aAID-MCS3%3E3.0.CO;2-A/abstract&quot; rel=&quot;nofollow&quot;&gt;A simple GC–MS technique for the analysis of vapor phase mainstream cigarette smoke&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2095" LastActivityDate="2013-08-30T08:04:26.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6064" PostTypeId="1" AcceptedAnswerId="6075" CreationDate="2013-08-30T12:03:59.973" Score="1" ViewCount="1018" Body="&lt;p&gt;Dicarbonyl amides, that is carboxylic acid anhydrides with their single-bonded oxygen between their carbonyl groups replaced with an amide group, I'm curious as to whether they're stable in water or unstable like carboxylic acid anhydrides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I mean by a dicarbonyl amide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yuF7x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R1 and R2 are organic substituents/hydrogen. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2013-08-31T01:52:10.123" Title="Are dicarbonyl amides stable in water?" Tags="&lt;stability&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6065" PostTypeId="2" ParentId="6064" CreationDate="2013-08-30T12:53:53.380" Score="0" Body="&lt;p&gt;The standard compound with your formula is &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C625774&amp;amp;Units=CAL&amp;amp;Mask=8&quot; rel=&quot;nofollow noreferrer&quot;&gt;diacetylamide&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;   &lt;img src=&quot;https://i.stack.imgur.com/fbFlu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect it to be unstable in water, just like anhydrides. I cannot, however, find a definitive reference on that. The NIST page linked above has a value for the $\Delta_r H°$ of the hydrolysis reaction in gas phase, but no $\Delta_r G°$ or aqueous solution data.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-08-30T12:53:53.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6066" PostTypeId="1" CreationDate="2013-08-30T15:12:00.180" Score="1" ViewCount="176" Body="&lt;p&gt;How can we prepare nicotinoyl glycine  from nicotinic acid ?Is that possible by reaction between nicotinic acid and glycine?&lt;/p&gt;&#xA;" OwnerUserId="2214" LastActivityDate="2013-08-31T18:34:40.803" Title="preparation methods of nicotinoyl glycine" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6067" PostTypeId="2" ParentId="6064" CreationDate="2013-08-30T16:47:30.573" Score="2" Body="&lt;p&gt;Turns out that I knew of a dicarbonyl amide that's stable in water prior to asking this question but I just hadn't realised it: &lt;strong&gt;aniracetam&lt;/strong&gt;. How I know it is stable in water is the fact that it is orally bioavailable in humans and since this would require them to be water-stable I consider it a safe assumption that it is water-stable. But I agree with F'x odds are most dicarbonyl amides are unstable in water and this dicarbonyl amide is probably only stable in water because its amide group is contained within a 5-membered cyclic ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the molecular structure of aniracetam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cqi4I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2013-08-30T16:47:30.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6068" PostTypeId="1" CreationDate="2013-08-30T18:14:09.423" Score="37" ViewCount="24929" Body="&lt;p&gt;The wiki tells me that if you heat carbon at atmospheric pressure it eventually turns directly into a gas without being liquid first.  At what pressure can you make liquid carbon?  Has anyone actually ever studied liquid carbon?  Surprisingly a web search did not answer these questions.&lt;/p&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="4945" LastEditDate="2014-05-13T09:56:15.417" LastActivityDate="2021-04-17T09:30:41.183" Title="What is known about liquid carbon?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;sublimation&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6069" PostTypeId="2" ParentId="6068" CreationDate="2013-08-30T19:02:38.207" Score="32" Body="&lt;p&gt;The carbon phase diagram shows that liquid carbon is only achievable at high pressure (~100 atm) and high temperature (~4500 K). This means that liquid carbon probably has very limited applications, and not many will be researching it. Funding is hard to come by when the applications of the research are not apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZ4Dt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lbruno.home.cern.ch/lbruno/documents/Bibliography/LHC_Note_78.pdf&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T11:03:17.770" LastActivityDate="2015-05-02T11:03:17.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6070" PostTypeId="2" ParentId="6068" CreationDate="2013-08-30T21:06:39.543" Score="20" Body="&lt;p&gt;&lt;em&gt;“Surprisingly a web search did not answer these questions”&lt;/em&gt; — Well, a web search for liquid carbon turns a few results:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A 2005 study of liquid carbon by x-ray absorption spectroscopy [&lt;a href=&quot;http://prl.aps.org/abstract/PRL/v94/i5/e057407&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] (second Google search result) (the &lt;a href=&quot;http://prl.aps.org/abstract/PRL/v94/i5/e057407&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper itself&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;low-density liquid carbon contains predominantly twofold-coordinated chain structures (&lt;em&gt;sp&lt;/em&gt; hybridization). As the density increased to that of solid forms, bond hybridization increased and threefold-coordinated (graphite-like) and fourfold-coordinated (diamond-like) bonds become more prevalent (indicating &lt;em&gt;sp&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt; and &lt;em&gt;sp&lt;/em&gt;&lt;sup&gt;3&lt;/sup&gt; hybridization, respectively). These observations are consistent with molecular dynamics calculations that rely on a tight-binding model of interatomic bonding. The fits also suggest that the bond length between carbon atoms in the liquid is significantly shorter than those in the solid, an observation also consistent with simulations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An earlier study (&lt;a href=&quot;http://www.nature.com/news/1999/990617/full/news990617-10.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1999&lt;/a&gt;) suggest two different liquid phases&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So it's still a pretty open research question…&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="17175" LastEditDate="2021-02-11T14:08:07.550" LastActivityDate="2021-02-11T14:08:07.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6071" PostTypeId="1" CreationDate="2013-08-30T21:13:15.013" Score="5" ViewCount="2265" Body="&lt;p&gt;Carbon has no liquid form at atmospheric pressure. Is this unique amongst naturally occurring elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more interesting and vague question is, is this property somehow important or related to the many other amazing properties carbon has on Earth or just a coincidence?&lt;/p&gt;&#xA;" OwnerUserId="1764" LastEditorUserId="84" LastEditDate="2013-08-31T02:27:15.063" LastActivityDate="2013-09-01T07:06:24.340" Title="Which elements have no liquid form at atmospheric pressure?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;sublimation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6072" PostTypeId="1" AcceptedAnswerId="6074" CreationDate="2013-08-30T21:25:52.080" Score="3" ViewCount="679" Body="&lt;p&gt;I am trying to make a copper sulfate electroplating solution. I came across a &lt;a href=&quot;http://www2.bren.ucsb.edu/~dturney/port/papers/Modern%20Electroplating/02.pdf&quot; rel=&quot;nofollow&quot;&gt;document&lt;/a&gt; (Schlesinger and Paunovic 2011) which mentions the amount of copper sulfate that should be added to water to create such a solution. In Table 2.1 on the bottom of page 35, it has two columns: one for conventional solutions and one for high throw solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read another &lt;a href=&quot;http://electronics.macdermid.com/documents/resources/Innovative%20High%20Throw%20Copper%20Electrolytic%20Process.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; (Nikolova and Watkowski) which talks about high throw copper electroplating process. All I could glean from the paper was that high throw solutions enable higher aspect ratios for copper PCBs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reply if anyone knows what high throw solutions are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nikolova, Maria, and Jim Watkowski. “INNOVATIVE HIGH THROW COPPER ELECTROPLATING PROCESS FOR METALLIZATION OF PCB.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schlesinger, M., and M. Paunovic. 2011. Modern Electroplating. The ECS Series of Texts and Monographs. Wiley. &lt;a href=&quot;http://books.google.com/books?id=j3OSKTCuO00C&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=j3OSKTCuO00C&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="736" LastEditorUserId="736" LastEditDate="2013-08-31T17:40:16.940" LastActivityDate="2013-11-13T22:49:02.200" Title="What are high throw solutions?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6073" PostTypeId="2" ParentId="6071" CreationDate="2013-08-30T22:58:14.557" Score="7" Body="&lt;p&gt;Have a look at a generic phase diagram, for a compound with one solid phase, one liquid phase and a gas phase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;            &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_diagram#2D_phase_diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Phase-diag2.svg/530px-Phase-diag2.svg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the solid to go directly into the vapor phase (&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sublimation_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;sublimation&lt;/a&gt;&lt;/strong&gt;) at ambient pressure upon heating, you thus need its &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;triple point&lt;/a&gt;&lt;/strong&gt; to have a higher pressure than atmospheric pressure, i.e. ~1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at &lt;a href=&quot;http://www.knowledgedoor.com/2/elements_handbook/triple_point.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this partial table&lt;/a&gt; of element triple points, &lt;strong&gt;arsenic&lt;/strong&gt; is another element that sublimes at ambient pressure. Iron would seem to qualify, but it has more phases that reported in the table, and the full phase diagram indicates that it does not sublime at ambient pressure.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.447" LastActivityDate="2013-09-01T07:06:24.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6074" PostTypeId="2" ParentId="6072" CreationDate="2013-08-30T23:36:04.063" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.plating.com/platingtechnical/poorthrowingpower.htm&quot;&gt;This page&lt;/a&gt; suggests that throwing power is the ability to coat a surface evenly regardless of current density differences, and thus to produce an even plating thickness across the whole of a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution with high throwing power would thus have the ability to produce a good, close-to-uniform coating on more difficult surface shapes.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-08-30T23:36:04.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6075" PostTypeId="2" ParentId="6064" CreationDate="2013-08-31T01:52:10.123" Score="4" Body="&lt;p&gt;This functional group is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Imide&quot; rel=&quot;nofollow noreferrer&quot;&gt;imide&lt;/a&gt;. Imides are stable to water and require significantly more effort to hydrolyze than anhydrides (from personal experience). The acyl nitrogen group is a better resonance donor and poorer leaving group than acyl oxygen groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the wikipedia page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Being highly polar, imides exhibit good solubility in polar media. The N-H center for imides derived from ammonia is acidic and can participate in hydrogen bonding. Unlike the structurally related acid anhydrides, they resist hydrolysis and some can even be recrystallized from boiling water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Succinimide&quot; rel=&quot;nofollow noreferrer&quot;&gt;succinimide&lt;/a&gt;, a cyclic imide has a solubility in water of 1 g for every 3 mL - if succinimide reacted with water, such a high solubility would not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The related &lt;a href=&quot;http://en.wikipedia.org/wiki/N-bromosuccinimide&quot; rel=&quot;nofollow noreferrer&quot;&gt;N-bromosuccinimide&lt;/a&gt; (show below) is also water stable despite that unfriendly looking $\ce{N-Br}$ bond. It is prepared from succinimide in water and an be &lt;a href=&quot;http://dx.doi.org/10.1021/ja01527a027&quot; rel=&quot;nofollow noreferrer&quot;&gt;recrystallized from near-boiling water&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RCNgv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-08-31T01:52:10.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6076" PostTypeId="1" AcceptedAnswerId="6080" CreationDate="2013-08-31T03:17:24.380" Score="5" ViewCount="11820" Body="&lt;p&gt;The net cell reaction of an electrochemical cell and its standard potential is given below:&#xA;$$\ce{ Mg + 2Ag+ -&amp;gt;Mg^{2+} + 2Ag} \ \ \ \ \ \ \ \ E^\circ=3.17\:\mathrm{V}$$ &#xA;The question is to find the maximum work obtainable from this electrochemical cell if the initial concentrations of $\ce{Mg^{2+}}=0.1\ \mathrm{M}$ and of $\ce{Ag+}=1\ \mathrm{M}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution just uses the Nernst equation to find the potential at this concentrations and uses $\Delta G=-nFE$ to calculate the Gibbs free energy change which is then equated to the maximum work obtainable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just the work obtainable per mole &lt;em&gt;at this concentration only&lt;/em&gt;. As soon as the reaction proceeds, the concentration change and so does the value of $\Delta G$ per mole and therefore the maximum work obtainable is different for different concentrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to find the maximum work obtainable, shouldn't we first calculate the equilibrium constant of this reaction (I calculated it to be $K_c=1.89\times10^{107}$), then the equilibrium concentrations(which is probably $\ce{Mg^{2+}}=0.6\ \mathrm{M}, \ \ce{Ag+}=0.178\times10^{-53}\ \mathrm{M}$) and then use something like:&#xA;$$&#xA;\int_{\mathrm{in}}^{\mathrm{eq}}\Delta G* \mathrm{d\,M}&#xA;$$ &#xA;where $\Delta G$ is per mole and the $dM$ is a small amount in moles that the reaction proceeds at that concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me verify my method and to proceed further.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="275" LastEditDate="2013-08-31T14:39:38.563" LastActivityDate="2019-05-21T12:54:59.963" Title="Express the maximum work from a voltaic cell" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;thermodynamics&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6078" PostTypeId="1" AcceptedAnswerId="6079" CreationDate="2013-08-31T14:31:16.937" Score="3" ViewCount="4062" Body="&lt;p&gt;Beryllium chloride's geometric structure is linear, with a bond angle of 180 degrees. Drawing the Lewis dot structure for this covalent compound, one can see that the compound is electron deficient and requires two lone pairs to achieve stability. As a result it is a strong Lewis acid. Since water has two lone pairs and is a strong Lewis base, mixing beryllium chloride with water results in stability for beryllium chloride and the evolution of energy.&lt;br&gt;&lt;br&gt;&#xA;My question is, why don't two lone pairs from either chlorine atom or even just a single chlorine atom make a coordinate bond with beryllium, as can be seen in the case of carbon monoxide where one of oxygen's lone pairs is donated to carbon to achieve nearest noble gas configuration of neon. If such a compound does not naturally exist, why does it not? Is it a theoretical defect or is there some explanation?&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2017-03-11T03:46:05.667" Title="Beryllium chloride - Highly hydrolysable?" Tags="&lt;covalent-compounds&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6079" PostTypeId="2" ParentId="6078" CreationDate="2013-08-31T15:14:39.067" Score="5" Body="&lt;p&gt;The structures of $\ce{BeX_2}$ where $\ce{X = F}$ or $\ce{Cl}$ can be found in A.F. Wells, &lt;em&gt;Structural Inorganic Chemistry&lt;/em&gt; 5th edition, 1984 pages 412 and 1047.  There you will see that while the predicted structures for $\ce{BeX_2}$ are linear (sometimes rationalized with Lewis dot structures displaying double bonds and ignoring formal charge considerations of the halogens).  However, the actual structure of $\ce{BeCl_2}$ in the vapor-phase is a 3-coordinate dimer that converts to the linear structure at higher temperatures.  This behavior supports your 'gut feeling' that the electron deficiency of $\ce{Be}$ should be compensated for by another $\ce{Cl}$ atom.  The dimer network formed in the solid state allows lone pairs from $\ce{Cl}$ to be donated to the $\ce{Be}$ in an attempt to fill the valence shell of beryllium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of a double bond to describe the nature of the $\ce{Be-Cl}$ bond is supported experimental data which shows that the bond length is shorter than expected for a single bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below is taken from Inorganic Chemistry by Miessler, Fischer and Tarr, 5th edition.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5VWro.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="17175" LastEditDate="2017-03-11T03:46:05.667" LastActivityDate="2017-03-11T03:46:05.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6080" PostTypeId="2" ParentId="6076" CreationDate="2013-08-31T15:15:41.347" Score="5" Body="&lt;p&gt;There is a simpler approach to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, since both initial concentrations are not 1 M, you need to determine &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; for the initial conditions, where &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_quotient&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction quotient&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; for this reaction is &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt;, and assuming &lt;span class=&quot;math-container&quot;&gt;$T=298\ \mathrm{K}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then go and convert this potential into free energy or maximum work or whatever else you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilbrium &lt;span class=&quot;math-container&quot;&gt;$E=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q=K_c$&lt;/span&gt;, so we can calculate the equilibrium constant if we want to compare it to &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; and determine which direction is favored. We don't need to. Since &lt;span class=&quot;math-container&quot;&gt;$E_i&amp;gt;0$&lt;/span&gt;, this reaction procedures toward the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as the reaction proceeds the concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^{2+}}$&lt;/span&gt; will change. From the law of mass action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some time &lt;span class=&quot;math-container&quot;&gt;$\Delta \ce{[Ag+]} = 0.25 \text{ M}$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{Ag+}]=0.75 \text{ M}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{Mg{2+}}]=0.225 \text{ M}$&lt;/span&gt;. The value of &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; has changed and so has the value of &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the reaction proceeds, the value of &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; decreases as the reaction proceeds, so too does the work the cell can do at any moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the &lt;strong&gt;total&lt;/strong&gt; work, then you need to integrate with respect to &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;, not worrying about the actual concentrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="72973" LastEditDate="2019-04-07T14:16:58.053" LastActivityDate="2019-04-07T14:16:58.053" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6081" PostTypeId="2" ParentId="6066" CreationDate="2013-08-31T18:34:40.803" Score="1" Body="&lt;p&gt;As a microscale laboratory method, it should be possible to react nicotinic acid with, e.g., thionyl chloride, yielding nicotinoyl chloride (the acid chloride of nicotinic acid). This can be reacted with glycine to yield the desired product. Unfortunately, glycine would exist predominantly in zwitterionic form at neutral pH, so I would guess that pre-treatment of the glycine with base would probably be required prior to the reaction with nicotinoyl chloride.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-08-31T18:34:40.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6082" PostTypeId="1" CreationDate="2013-08-31T19:11:19.023" Score="1" ViewCount="1040" Body="&lt;p&gt;Water has a very high heat of vaporization (2260 kJ per kg). Why not just gradually sprinkle water on your roof to keep the building cool in opposed to using air conditioners?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wouldn't take much water to keep the building a few degrees cooler for a long time. Air conditioners use a lot of electricity which is bad for the environment. I read some places ban water vaporization cooling systems because of droughts and because tap water is government subsidized but that seems silly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a society it would be cheaper and more environmentally friendly to use water. The water will simply evaporate and return as rain which it will do anyway regardless. The energy air conditioning uses puts more greenhouse gases  in the environment which (allegedly) causes more global warming and puts more chemicals in rain water.&lt;/p&gt;&#xA;" OwnerUserId="2232" LastEditorUserId="2095" LastEditDate="2013-08-31T21:29:17.897" LastActivityDate="2013-09-03T18:12:38.240" Title="Why is water evaporation not more commonly used to cool buildings" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="6" ClosedDate="2013-09-03T22:37:27.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="6083" PostTypeId="1" CreationDate="2013-08-31T22:56:55.370" Score="3" ViewCount="1889" Body="&lt;p&gt;I don't really get the concept of &lt;em&gt;index of hydrogen deficiency&lt;/em&gt; (IHD). Can someone please use an example to explain what $\mathrm{H_{reference}}$ and $\mathrm{H_{molecule}}$ are? I know the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{IHD} = \frac{\mathrm{H_{reference}} - \mathrm{H_{molecule}}}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="16683" LastEditDate="2015-10-09T16:12:47.980" LastActivityDate="2015-10-09T16:12:47.980" Title="index of hydrogen deficiency" Tags="&lt;organic-chemistry&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6084" PostTypeId="2" ParentId="6083" CreationDate="2013-09-01T00:00:56.547" Score="5" Body="&lt;p&gt;You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$IHD = \frac{2C + 2 - H - X + N}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + 2}}$, meaning that given $n$ carbons, a simple acyclic alkane will have $2n + 2$ hydrogens. Every increment in the IHD represents either a degree of unsaturation due to $\pi$-bonds, or the presence of a ring. In your formula, $H_{reference}$ should be twice the number of carbons in your molecule, plus two. $H_{molecule}$, on the other hand, is the actual number of hydrogens in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if your molecule has the formula $\ce{C6H8}$, $H_{reference} = 2(6) + 2 = 14$ and $H_{molecule} = 8$. The IHD is then $\frac{14 - 8}{2} = 3$, suggesting that the total number of $\pi$-bonds plus rings in the molecule is three. Note that this isn't sufficient to determine the actual structure of the molecule, but it's helpful when combined with spectroscopy data (that is, it could have one ring and two $\pi$-bonds, or just three $\pi$-bonds, or some other combination).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-09-01T00:00:56.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6085" PostTypeId="2" ParentId="6082" CreationDate="2013-09-01T00:24:06.633" Score="1" Body="&lt;p&gt;&quot;This question seems rhetorically posed: how can you justify calling your solution better and cheaper without having run the numbers? – Aesin 1 hour ago&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A central air conditioning system for a house takes about 3500 watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://michaelbluejay.com/electricity/howmuch.html&quot; rel=&quot;nofollow&quot;&gt;http://michaelbluejay.com/electricity/howmuch.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 watt = 1 J/sec so that's about 1,260,0000 J/hour&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gallon of water weighs about 3.7854118 Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so at 2260 kJ per kg you can extract 6,780,000 J of energy per gallon of water which is more energy than the air conditioning system even uses and the air conditioning system doesn't extract the amount of heat energy from a building than the energy it uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run the numbers even with a gallon of water the energy efficiency of water usage is much superior than that of an air conditioning system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high heat of vaporization of water is how sweat keeps you cool. Ancient civilizations used to pour water on their tents or tepees or whatever to keep themselves cool. It's cheap and efficient. The superiority of this method when opposed to air conditioning, in dry areas at least, is very obvious. I have yet to see one good reason why it's not used more often.&lt;/p&gt;&#xA;" OwnerUserId="2233" LastActivityDate="2013-09-01T00:24:06.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6086" PostTypeId="1" AcceptedAnswerId="6088" CreationDate="2013-09-01T00:59:36.467" Score="8" ViewCount="4657" Body="&lt;p&gt;I'm curious to know, what exactly does the 3,4 part of 3,4-methylenedioxymethamphetamine represent?&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="418" LastEditDate="2014-01-24T18:48:05.370" LastActivityDate="2017-03-09T22:36:45.213" Title="What does the &quot;3,4&quot; mean in 3,4-methylenedioxymethamphetamine (MDMA)?" Tags="&lt;nomenclature&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6088" PostTypeId="2" ParentId="6086" CreationDate="2013-09-01T01:18:49.833" Score="12" Body="&lt;p&gt;Numbers in organic molecules tell you where so-called &lt;em&gt;functional groups&lt;/em&gt; (important configurations of elements) are attached to a central structure.  The central structure in MDMA is a benzene ring $\ce{(C6H6)}$, with the main functional group attached at a carbon which makes methamphetamine.  This carbon is labeled 1 as the &quot;most important&quot; functional group.  To get to MDMA from methamphetamine, you add a methelynedioxyl group, and this attaches at carbons 3 and 4.  So that's where those numbers come from!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Below is a molecule of MDMA:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a8NWK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hexagon shape near the center is the benzene ring.  The group to the right, attached to the benzene ring at about 2 o'clock is the main functional group that turns benzene into methamphetamine.  That position on the ring is labeled 1.  Now all other functional groups are numbered sequentially in the direction that makes the next functional group nearest to 1.  So that means we go counterclockwise, and the group on the far left with the 2 O's (oxygens) is attached at positions 3 and 4.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2013-09-01T01:40:32.987" LastActivityDate="2013-09-01T01:40:32.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6089" PostTypeId="1" AcceptedAnswerId="6100" CreationDate="2013-09-01T02:28:59.080" Score="9" ViewCount="2609" Body="&lt;p&gt;My textbook begins the Chemical Kinetics section by describing the &quot;Initial Rates Method&quot; of determining the rate of reaction. I understood it as the following (for a first order reaction):&#xA;$$\ce{A -&amp;gt; B+C}$$&#xA;$$\text{rate}_\mathrm{i}=k[\ce{A}]_\mathrm{i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subscript $\text{i}$ shows that it is the initial rate. This I agree with, as it is suggested by experimental data. However, on the next page, the textbook makes a leap and (implicitly) states that this relationship holds at any time:&#xA;$$\text{rate}=-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=k[\ce{A}]$$&#xA;This becomes a differential equation, the justification for which the textbook did not provide. How do we know that the change of concentration with respect to time depends on the concentration itself? I don't see how this can be assumed from the initial rates method.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="5026" LastEditDate="2018-03-20T14:00:55.370" LastActivityDate="2018-03-20T14:03:17.890" Title="How does initial rate of reaction imply rate of reaction at any time?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6091" PostTypeId="1" AcceptedAnswerId="6101" CreationDate="2013-09-01T04:38:49.163" Score="6" ViewCount="25621" Body="&lt;p&gt;As the title suggest can you explain how DRS can be used for finding band gap of a material. Below is an DRS output result: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kti4L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the figure we can see that at about 465 nm the reflectance drops steeply with respect to wavelength and attain a plateau at around 420 nm. Using above data we can assume that the band gap is around 2.76 eV (450 nm).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can you explain this figure in  physical sense (&lt;em&gt;i.e.&lt;/em&gt; with respect to the material)?&lt;/li&gt;&#xA;&lt;li&gt;Why do we take the energy of a ~450 nm photon as the band gap? What is the rationale behind it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="139" LastEditorUserId="84" LastEditDate="2013-09-02T01:30:22.780" LastActivityDate="2013-09-02T01:30:22.780" Title="What is the principle of diffuse reflectance spectroscopy behind band gap measurement?" Tags="&lt;spectroscopy&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6092" PostTypeId="1" AcceptedAnswerId="6097" CreationDate="2013-09-01T07:03:12.077" Score="2" ViewCount="7531" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;One mole of a hydrocarbon is combusted. The products obtained are cooled down to STP and occupy a volume of &lt;span class=&quot;math-container&quot;&gt;$89.6\ \mathrm l$&lt;/span&gt;. Oxygen required for combustion was &lt;span class=&quot;math-container&quot;&gt;$145.6\ \mathrm l$&lt;/span&gt; at STP. Find the molecular formula of the hydrocarbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let the hydrocarbon be &lt;span class=&quot;math-container&quot;&gt;$\ce{C}_x\ce{H}_y$&lt;/span&gt;. The combustion reaction can be written as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{C_xH_y + O2 -&amp;gt; H2O + CO2}&#xA;$$&lt;/span&gt;&#xA;Keep in mind that this is not yet balanced.&lt;/p&gt;&#xA;&lt;p&gt;We can now find out the amount of reactants and products. Since one mole of any gas occupies &lt;span class=&quot;math-container&quot;&gt;$22.4\ \mathrm l$&lt;/span&gt; of space at STP, we can say that we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{89.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 4\ \mathrm{mol}&#xA;$$&lt;/span&gt;&#xA;of products. As for the oxygen reacted:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{145.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 6.5\ \mathrm{mol}&#xA;$$&lt;/span&gt;.&#xA;We can now deduce that the hydrocarbon will react with &lt;span class=&quot;math-container&quot;&gt;$13\ce{O}_2$&lt;/span&gt;. Therefore, the equation becomes.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{C_xH_y + 13O2 -&amp;gt; H2O + CO2}&#xA;$$&lt;/span&gt;&#xA;But, There are &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;s one the left hand side so the right hand side also has &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;s:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{C_xH_y + 13O2 -&amp;gt; H2O + xCO2}&#xA;$$&lt;/span&gt;&#xA;There are &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;'s on the left hand side. Same for the right hand-side:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{C_xH_y + 13O2 -&amp;gt; y/2H2O + xCO2}&#xA;$$&lt;/span&gt;&#xA;Since the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; has to balance as well, we have &lt;span class=&quot;math-container&quot;&gt;$26$&lt;/span&gt; oxygens on the LHS and therefore &lt;span class=&quot;math-container&quot;&gt;$26$&lt;/span&gt; on the RHS. Which means:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;26 = \frac{y}{2} + 2x \implies \frac{y}{2} = 26 - 2x&#xA;$$&lt;/span&gt;&#xA;We have already figured out that &lt;span class=&quot;math-container&quot;&gt;$4\:\mathrm{mol}$&lt;/span&gt; of products are obtained. i.e.:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;4 = \frac{y}{2} + x = 26 - 2x + x = 26 - x \implies x = 26 - 4 = 22&#xA;$$&lt;/span&gt;&#xA;But this would mean that &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is negative. Where did I go wrong?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="95133" LastEditDate="2020-09-20T19:46:48.283" LastActivityDate="2020-09-20T19:46:48.283" Title="Figuring out formula of a hydrocarbon from amount of products obtained and reactants used" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6093" PostTypeId="1" AcceptedAnswerId="6099" CreationDate="2013-09-01T07:58:29.093" Score="4" ViewCount="714" Body="&lt;p&gt;Do all metals in metal oxo complexes (eg $\ce{Fe}$ in Ferrate, $\ce{Mo}$ in $\ce{MoO4}$,...) have positive oxidation states?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="1878" LastEditDate="2013-09-01T18:11:30.107" LastActivityDate="2013-09-02T01:26:52.860" Title="Oxidation state of metal in metal oxo complexes always positive?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6097" PostTypeId="2" ParentId="6092" CreationDate="2013-09-01T11:50:18.990" Score="1" Body="&lt;p&gt;From your interpretation I have to say that giving the stochiometric coefficent of $13$ is wrong because this number is not related to the molecules of the species (therefore the suspect that you multiplied the moles obtained by two is where lies the main error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are talking about a combustion, we can safely assume that every reagent is completely oxidised, so we can write the complete reaction with the &lt;em&gt;generic&lt;/em&gt; coefficients:&#xA;$$&#xA;\ce{C}_{\color{blue}{x}}\ce{H}_{\color{blue}{y}(g)}+\left[\color{blue}{\frac{y}{4}}+\color{blue}{x}\right]\ce{O2_{($g$)}}\to\color{blue}{\frac{y}{2}}\ce{H2O}_{(g)}+\color{blue}{x}\ce{CO2_{($g$)}}&#xA;$$&#xA;Note that the rule of given saturated hydrocarbons implies that $y=2x+2$, but this is left intentionally generalized &lt;em&gt;because the presence of specific double/triple bonds is unknown&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of writing this equation is to specify the relation between the reactants, so now we can assume the molar relation between the oxygen needed to the first hydrocarbon to obtain a simple equation in terms of $x$ and $y$:&#xA;$$&#xA;n_{\ce{O2}}=\left[\frac{y}{4}+x\right]n_{\ce{C}_{x}\ce{H}_{y}}&#xA;$$&#xA;Since, as you calculated, there are needed $6.5\:\mathrm{mol}$ of oxygen to oxidise $1\:\mathrm{mol}$ of the unknown hydrocarbon; we can substitute those values inside the above equation, giving:&#xA;$$&#xA;6.5=\frac{y}{4}+x\tag{1}&#xA;$$&#xA;After the complete combustion, the total amount of moles is indicated as you derived correctly, and there can be expressed in relation to the unknown reagent, and then added together to obtain the second equation as follows:&#xA;$$&#xA;n_{\ce{H2O}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}\quad n_{\ce{CO2}}=x\:n_{\ce{C}_{x}\ce{H}_{y}}&#xA;$$&#xA;Since the total volume is additive, then (because $\ce{C}_{x}\ce{H}_{y}=1:\mathrm{mol}$):&#xA;$$&#xA;4\:\mathrm{mol}=n_{\ce{H2O}}+n_{\ce{CO2}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}+x\:n_{\ce{C}_{x}\ce{H}_{y}}=\frac{y}{2}+x&#xA;$$&#xA;It now produces the second needed equation:&#xA;$$&#xA;4=\frac{y}{2}+x\tag{2}&#xA;$$&#xA;Solving the equations $(1)$ and $(2)$ gives the values of $x$ and $y$ needed to derive the brute formula for the unknown hydrocarbon:&#xA;$$&#xA;y=-10,\quad x=9&#xA;$$&#xA;Since the result is negative for one term, we can assume that the water produced is &lt;em&gt;not&lt;/em&gt; in the vapor phase, hence we eliminate water term inside the equation $(2)$, $y/2\:n_{\ce{C}_{x}\ce{H}_{y}}$, giving the immediate value of $x$ because there is no effort given by the water:&#xA;$$&#xA;x=4,\quad y=10\to \ce{C}_{4}\ce{H}_{10}\quad\mathrm{butane}&#xA;$$&#xA;This result is describing a saturated hydrocarbon, because it follows the rule $\ce{C}_{n}\ce{H}_{2n+2}$.&lt;/p&gt;&#xA;" OwnerUserId="1984" LastEditorUserId="1984" LastEditDate="2013-09-01T18:07:58.573" LastActivityDate="2013-09-01T18:07:58.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6098" PostTypeId="2" ParentId="5123" CreationDate="2013-09-01T11:55:58.003" Score="1" Body="&lt;p&gt;There are a number of commercial kits available to determine the presence of egg in commercial products and they are based on a technique called &lt;a href=&quot;http://www.elisa-antibody.com/ELISA-Introduction/ELISA-types/sandwich-elisa&quot; rel=&quot;nofollow&quot;&gt;sandwich ELISA&lt;/a&gt; (enzyme linked immunosorbent assay).  A recent article in Food Research International by &lt;a href=&quot;http://dx.doi.org/10.1016/j.foodres.2013.01.019&quot; rel=&quot;nofollow&quot;&gt;Gomaa and Boye&lt;/a&gt; discusses the impact of thermal processing on allergen detection (in this case casein, egg, gluten and soy).  Egg recoveries with the kits are low, ranging from 48% down to 0.  It appears that in some cases detecting egg in thermally processed samples is either very difficult to do or the egg products are transformed significantly that they can no longer be called egg (I did not read the paper in enough detail to determine if the authors speak to this point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kits tested in this study were from the following companies: Morinaga, Ridascreen and Ridascreen.  I web search will reveal these products and additional information, but I'll leave that to the viewer as I do not want this question to be an endorsement or critique of any commercial products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative to these kits appears to be &lt;a href=&quot;http://probes.invitrogen.com/resources/education/tutorials/4Intro_Flow/player.html&quot; rel=&quot;nofollow&quot;&gt;flow cytommetry&lt;/a&gt;, which has better recoveries, but does not fit the OPs conditions of &lt;em&gt;economical for a mid-income business without access to a laboratory&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-09-01T11:55:58.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6099" PostTypeId="2" ParentId="6093" CreationDate="2013-09-01T16:31:13.237" Score="4" Body="&lt;p&gt;Yes. Oxygen is much more electronegative than all of the transition metals, so it is given a &lt;em&gt;formal&lt;/em&gt; -2 charge. Keep in mind, though, that these values are not intrinsic properties of these elements/molecules. Their use is intended to help us rationalize some of the chemistry and properties associated with these materials. As an interesting example, consider oxygen difluoride, $\ce{OF_2}$. Fluorine is the most electronegative element (that forms compounds with other elements) and is given a formal charge of -1. To balance the formal charges then, oxygen must have a +2 charge!&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="84" LastEditDate="2013-09-02T01:26:52.860" LastActivityDate="2013-09-02T01:26:52.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6100" PostTypeId="2" ParentId="6089" CreationDate="2013-09-01T17:14:08.673" Score="12" Body="&lt;p&gt;&lt;strong&gt;First things first&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about the rate of a reaction, we mean the rate at which the products appear. This means that it is also equal to the disappearance of the reactants. So, when we quantify the rate in terms of reactants, they must be in opposing magnitude. $$\text{rate }= \frac {\Delta [\text{product}]}{\Delta t}=-\frac{\Delta [\text{reactant}]}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From A to B (and C)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that I am running the reaction $\ce{A -&amp;gt;B +C}$ in two separate vessels, flask 1 and flask 2, each with a different concentration of $\ce{A}$. We'll say that flask 1 is more concentrated. If you agree that the initial rate is given by $\text{rate}_\mathrm{i} = k[\ce{A}]_\mathrm{i}$, then it follows that for each the initial rate is $\text{rate}_{i,fl1} = k[\ce{A}]_{i, fl1}$ and $\text{rate}_{i,fl2} = k[\ce{A}]_{i, fl2},$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S58bx.png&quot; alt=&quot;initial reaction conditions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to add something to flask 2, would it change the reaction rate? As long as we didn't add any $\ce{A},$ the rate would be the same. The rate is dependent on $\ce{A}$ but nothing else. So let's add a little bit of $\ce{B}$ and a little bit oc $\ce{C}$. We'll add the same amount of both. Let's call that amount $x$ and set it equal to the difference in concentrations in $\ce{A}$ in the two flasks. $$x=[\ce{A}]_{i,f1}-[\ce{A}]_{i.f2}$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fnxMf.png&quot; alt=&quot;adjusted initial reaction conditions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's let the reaction in flask 1 proceed until the concentration of $\ce{A}$ is equal to the amount in flask 2. To distinguish the conditions now from the initial conditions, we'll call this time $t=1$ (arbitrary units). By the stoichiometry of the reaction, we know the amount of $\ce{A}$ and $\ce{A}$ will be $\Delta[\ce{A}] = [\ce{A}]_{i,f1}-[\ce{A}]_{i,f2}=x$ There is no perceivable difference between the two flasks, and the reaction rate is thus the same in flask 1 as in flask 2. Because we have chosen arbitrary concentrations, we can abstract it to any concentration. $$\text{rate}_{t=1,f1}=\text{rate}_{i,f2}=k[a\ce{A}]_{i,f2}=k[\ce{A}]$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qXKNf.png&quot; alt=&quot;final reaction conditions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we want the instant rate. We do this by decreasing the time interval to an infinitesimal amount. $$\lim_{\Delta t \to 0}\frac{-\Delta[\ce{A}]}{\Delta t}=-\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t}=k[\ce{A}]$$&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="5026" LastEditDate="2018-03-20T14:02:21.457" LastActivityDate="2018-03-20T14:02:21.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6101" PostTypeId="2" ParentId="6091" CreationDate="2013-09-01T17:22:59.960" Score="7" Body="&lt;p&gt;Reflectance spectroscopy is very closely related to UV/Vis spectroscopy, in that both of these techniques use visible light to excite valence electrons to empty orbitals. The difference in these techniques is that in UV/Vis spectroscopy one measures the relative change of transmittance of light as it passes through a solution, whereas in diffuse reflectance, one measures the relative change in the amount of reflected light off of a surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution that is completely clear and colorless has essentially 100% transmission of all visible wavelengths of light, which means that it does not contain any dissolved components that have (allowed) electronic transitions over that energy range. By the same line of reasoning, a white powder effectively reflects 100% of all visible wavelengths of light that interacts with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the material has electronic energy levels that are separated by an energy in the visible region, then it may absorb some of light energy to move electrons from the filled energy level (valence band) into this empty level (conduction band). This causes an relative decrease in the amount of light at that particular energy, relative to a reference source. In other words, the % transmission/reflectance will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your data suggests that the separation between the filled valence band and the empty conduction band is around 2.76 eV. Because when light of that energy or higher interacts with the material, it is absorbed resulting in an excitation of the electrons from the valence bands into the conduction band. This lowers the relative reflectance of the material (i.e. lower % transmittance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't this that entirely answered all your questions, but I hope it helps a little.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-09-01T17:22:59.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6102" PostTypeId="1" AcceptedAnswerId="6103" CreationDate="2013-09-01T19:47:28.190" Score="3" ViewCount="528" Body="&lt;p&gt;Metal coated plastics are commercially abundant now-a-days. I want to know how they're plated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before posting, I did a quick search and found a &lt;a href=&quot;http://www.thomasnet.com/articles/custom-manufacturing-fabricating/how-to-metalize-plastic&quot; rel=&quot;nofollow&quot;&gt;nice link&lt;/a&gt;. It refers to this process as &lt;em&gt;metallization&lt;/em&gt;, although I'm not convinced by the term. Is it right ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That site already provided the basic information. However, it would be great if someone can mention a few examples(which metal/plastic used) for the various methods deployed.   &lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2015-03-09T13:30:01.283" Title="How metal-plating on plastics is performed?" Tags="&lt;inorganic-chemistry&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6103" PostTypeId="2" ParentId="6102" CreationDate="2013-09-01T22:35:34.813" Score="4" Body="&lt;p&gt;I couldn't comment on the name.&#xA;I am familiar with vacuum deposition, but i have only come across deposition on silicon, not plastic. the substrate ad metal are put in a vacuum chamber, the vacuum means that now the metal can be evaporated at a much lower temperature, and is then deposited onto the substrate.(see link)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did an experiment of electroless plating of plastic with nickel. I can't say what plastic it was, probably PS or PP. usually electro plating requires electrodes and current to flow between them. electroless plating uses a electrochemical reaction in which a metal precipitates. Due to the treatment of the plastic the metal prefers to precipitate onto the plastic (this is a very simplistic information see liks for more information if you're interested)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the method I used in my experiment:&#xA;the plastic is cleaned by diping it in a base bath (sodium carbonate, trisodium phosphate, sodium hydroxide), this is mostly to remove oily debris .&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Etching: ...etches the plastic to create microscopic pores that improve&#xA;adhesion of metal and following steps &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sensibilization: plastic is&#xA;dipped in ti solution, keeps the palladium at its reduced state (next&#xA;step)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Activation: plastic is dipped in a palladium solution, Palladium&#xA;catalyses the formation of nickel particles. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Plating: since there is no current flowing another particle has to be oxidized for the nickel to be reduced. in this case the reaction used is oxidation of  hypophosphite to phosphite&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The result was a very nice metallic finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;more information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electroless_nickel_plating&quot; rel=&quot;nofollow&quot;&gt;Electroless nickel plating on Wikipedia&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.caswellplating.com/electroplating-anodizing/nickel-plating-kits/electroless-nickel-plating-kits.html?p=2&quot; rel=&quot;nofollow&quot;&gt;plating kit&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tau.ac.il/~chemlaba/Files/Electroless/12777_01.pdf&quot; rel=&quot;nofollow&quot;&gt;plenty of information on electroless deposition&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Evaporation_%28deposition%29&quot; rel=&quot;nofollow&quot;&gt;evaporation-deposition&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-09-01T22:35:34.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6105" PostTypeId="2" ParentId="6089" CreationDate="2013-09-02T07:14:38.103" Score="6" Body="&lt;p&gt;A first order reaction is defined as a reaction in which the reaction rate obeys the equation &#xA;$$r(t) \equiv \mathrm{d}[A(t)] / \mathrm{d}t = k[A(t)] \tag{1}$$ &#xA;Your first equation&#xA;$$r(0) \equiv \mathrm{d}[A(0)] / \mathrm{d}t = k[A(0)]   \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is just a particular situation. We cannot know a priori if a reaction that obeys (2) will obey (1). We can only measure how $r(t)$ varies over time. If $r(t)$ follows equation (1) then it is said that the kinetics follow a first-order equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemical reactions of the type $\ce{A -&amp;gt; B + C}$ it is natural to expect a first order reaction at initial times. As long as the reaction doesn't need any other molecule or agent in the medium (aside from perhaps other essentially constant factors) than $\ce{A}$, and doesn't involve collisions between two or more molecules to ignite the reaction, one can expect the rate to be proportional to the number of molecules present at each time, hence to $[\ce{A}(t)]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a long time, when the reactive concentration is depleted and the concentration of the products builds in (assuming no diffusion of the products) the first order law is typically violated, since the back reaction $\ce{B + C -&amp;gt; A}$ (aside from other collisions) can occur. In particular, is is not possible to reach equilibrium conditions under the first order law.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastEditorUserId="5026" LastEditDate="2018-03-20T14:03:17.890" LastActivityDate="2018-03-20T14:03:17.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6106" PostTypeId="1" AcceptedAnswerId="6113" CreationDate="2013-09-02T16:22:37.277" Score="2" ViewCount="4041" Body="&lt;p&gt;Can we predict in which direction a given molecule will bend plane polarized light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a molecule with R form in R-S nomenclature does it mean that the molecule will be dextrorotatory or is it to be found by experiments like finding out rate of a reaction by experiments?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2013-09-02T19:17:55.870" Title="Can we predict Dextrorotatory and levorotatory character of an organic molecule?" Tags="&lt;organic-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6107" PostTypeId="1" CreationDate="2013-09-02T16:48:19.643" Score="4" ViewCount="513" Body="&lt;p&gt;Take any solution, then start freezing it at sufficiently low temperature. Soon a small amount of ice forms. I'd like to know how the composition of dissolved salts and gases in the bit of ice differs from that in the water that's still liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the solution is far from saturation and whatever other extreme conditions there can be. I don't have any particular solution in mind - it's the general principles that I'm interested in.&lt;/p&gt;&#xA;" OwnerUserId="2243" LastEditorUserId="5026" LastEditDate="2018-04-05T15:32:26.567" LastActivityDate="2018-04-05T15:32:26.567" Title="How does the composition of ice differ from that of water in a partly frozen solution?" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6108" PostTypeId="1" AcceptedAnswerId="6111" CreationDate="2013-09-02T16:51:56.560" Score="4" ViewCount="936" Body="&lt;p&gt;What does it mean in asking the number of lone pairs of central atom when it has an odd number of electrons? &#xA;As lone pairs means the no. of non bonding &lt;strong&gt;pairs&lt;/strong&gt; of electrons does it exclude the one odd electron if present in the central atom?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="1878" LastEditDate="2013-09-02T18:54:16.013" LastActivityDate="2020-08-30T10:27:19.893" Title="What is the number of lone pairs of central atom when it has an odd number of electrons?" Tags="&lt;electronic-configuration&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6109" PostTypeId="1" AcceptedAnswerId="6112" CreationDate="2013-09-02T17:19:57.107" Score="17" ViewCount="5863" Body="&lt;p&gt;This question arose in my mind because it's given in a book that &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrene&quot; rel=&quot;noreferrer&quot;&gt;pyrene&lt;/a&gt; is aromatic. How can it be with 16 π electrons? So, it should be actually anti-aromatic or is there any special rule for these types of compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mH2qg.png&quot; alt=&quot;pyrene&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="30" LastEditDate="2013-09-02T19:12:01.560" LastActivityDate="2021-01-10T15:16:25.510" Title="Is there any special rules for checking the aromaticity of polycyclic compounds?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6111" PostTypeId="2" ParentId="6108" CreationDate="2013-09-02T19:03:31.533" Score="5" Body="&lt;p&gt;This seems to be a tricky bit of chemical terminology that may not have a lot of practical relevance, so I wouldn't get too hung up on this, if I were you. Technically, I would say that you are correct in that lone pairs should be pairs of electrons, as this is how &lt;a href=&quot;http://goldbook.iupac.org/L03618.html&quot;&gt;the IUPAC defines them&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-09-02T19:03:31.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6112" PostTypeId="2" ParentId="6109" CreationDate="2013-09-02T19:10:56.507" Score="21" Body="&lt;p&gt;First, &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule&quot; rel=&quot;noreferrer&quot;&gt;Hückel's rule&lt;/a&gt; is only valid for monocyclic systems&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic even though they fail the 4n + 2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Pyrene is a commonly used counter-example of Hückel's rule. A refinement specific to &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycyclic_aromatic_hydrocarbon&quot; rel=&quot;noreferrer&quot;&gt;Polycyclic Aromatic Hydrocarbons&lt;/a&gt; (PAH) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycyclic_aromatic_hydrocarbon#Aromaticity&quot; rel=&quot;noreferrer&quot;&gt;Clar's rule&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The case of pyrene is explained in detail in the literature by a &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/10916460601054206&quot; rel=&quot;noreferrer&quot;&gt;2007 paper in &lt;em&gt;Petroleum Science and Technology&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This representation agrees with Clar’s sextet rule in which the molecule is represented as having two aromatic sextets, one for each of the rings a and d, and two fixed double bonds on rings b and c (Hernández-Trujillo et al., 2005).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P1eMD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-09-02T19:10:56.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6113" PostTypeId="2" ParentId="6106" CreationDate="2013-09-02T19:14:48.253" Score="4" Body="&lt;p&gt;There is no simple set of rules that one can use to determine the absolute configuration from the specific rotation, or the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick web search did dig up &lt;a href=&quot;http://dx.doi.org/10.1016/S0957-4166%2800%2900178-6&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt; that deals with calculating the specific rotation for arbitrary molecules, but it requires some fairly sophisticated algorithms. In principle, you could reproduce their methods, but it would require a substantial time commitment unless you are familiar with the software they describe. (Which is &lt;a href=&quot;http://www.gaussian.com/&quot; rel=&quot;nofollow&quot;&gt;Gaussian&lt;/a&gt;, if you're interested.)&lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="30" LastEditDate="2013-09-02T19:17:55.870" LastActivityDate="2013-09-02T19:17:55.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6114" PostTypeId="2" ParentId="6107" CreationDate="2013-09-02T21:31:52.263" Score="3" Body="&lt;p&gt;It sounds like you're talking about freezing-point depression.  If the solid component of a system is pure (oft assumed so; crystals are good at excluding impurities), the equilibrium is shifted because the chemical potential of the solution drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system with two pure species, the chemical change from say, water ($\star$ denoting pure) to ice is in equilibrium when their chemical potentials (denoted $\mu$) are equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{\ce{H_2O}(\text{s})} = \mu^\star_{\ce{H_2O}(\text{l})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Note: Chemical potentials are always dependent, to varying degrees, on both temperature and pressure, so I'm not going to bother writing $\mu(T, P)$ everywhere&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the following graph that's at $T_0$; the usual freezing point of water where the &quot;pure water&quot; and &quot;ice&quot; curves intersect.  If water exists at a temperature lower than $T_0$, its higher chemical potential translates to a higher free energy, so it's thermodynamically driven towards freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gexqg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gexqg.png&quot; alt=&quot;Ice, pure water, and solution chemical potential graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;image source: &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/solutions/images/mu-vs-t.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;a website&lt;/a&gt; that is currently down; retrieved via the Web Archive&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something non-volatile (e.g. salt) is dissolved in the water, its chemical potential changes to $\mu_i^\star+RT\ln \chi_i$, where $\mu_i^\star$ is the chemical potential of the pure species $i$, $R$ is the gas constant, and $\chi_i$ is the molar fraction of species $i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, the fraction was $1.0$ (it was pure), and because $\ln1 = 0$, that term was ignored.  But now, because some of the solution species is something else, $\chi_i$ drops, thus $\ln\chi_i$ becomes negative, so the term lowers the overall chemical potential.  Because the chemical potential is lower, its free energy is lower, so it is thermodynamically more stable at lower temperatures.  At some $T$, we will find that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{\ce{H_2O}(\text{s})} = \mu_{\text{solution}} = \mu^\star_{\ce{H_2O}(\text{l})} + RT\ln\chi_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there, the two species will be in thermodynamic equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more, check &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/solutions/faq/print-thermo-explanation-of-freezingpoint-depression.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, or Wikipedia articles on chemical potential and freezing point depression (a basic Physical Chemistry text book, or chapter in a Gen. Chem. one might be better; the WP articles are not that great to get started with).&lt;/p&gt;&#xA;" OwnerUserId="11" LastEditorUserId="5026" LastEditDate="2018-04-05T15:30:56.520" LastActivityDate="2018-04-05T15:30:56.520" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6115" PostTypeId="1" AcceptedAnswerId="6128" CreationDate="2013-09-02T23:41:01.433" Score="3" ViewCount="6572" Body="&lt;p&gt;If I have a hydrate, such as copper (II) sulfate pentahydrate, $\ce{CuSO_4 \cdot 5H_2O}$, what is the amount of heat to required to dehydrate it? I assume that it is equal to or greater than the amount of heat required to evaporate the water molecules. Does the required heat vary for every hydrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for dehydrating copper (II) sulfate pentahydrate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuSO_4 \cdot 5H_2O -&amp;gt;[\Delta] CuSO_4 +5H_2O}$$&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="143" LastEditDate="2013-09-12T17:49:15.947" LastActivityDate="2021-04-22T17:55:57.690" Title="How much heat is required to dehydrate a hydrate?" Tags="&lt;water&gt;&lt;crystal-structure&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6116" PostTypeId="1" CreationDate="2013-09-03T02:57:52.117" Score="7" ViewCount="3098" Body="&lt;p&gt;Why is there a diagonal line cutting through some of the nonmetals and metalloids on the periodic table in groups IIIA, IVA, VA, VIA, and VIIA?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3zpgH.jpg&quot; alt=&quot;Periodic table of elements&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any historical background about it?&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="305" LastEditDate="2013-09-03T03:19:28.317" LastActivityDate="2017-08-01T23:45:30.700" Title="Diagonal line going through groups IIIA-VIA on the periodic table" Tags="&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6117" PostTypeId="1" CreationDate="2013-09-03T07:22:55.120" Score="2" ViewCount="1923" Body="&lt;p&gt;I am using a rotary evaporator ('rotovap') in my lab mainly with dichloromethane as a solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually between experiments I clean the receiving and boiling flasks with a brush using soap and water and finally rinse with alcohol.&#xA;For the condenser I try to do the same, but it is next to impossible to brush the water serpentine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to try and distil some coffee and tea for consumption, but I am afraid of any possible contaminants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an efficient and safe method to clean the glassware?&lt;/p&gt;&#xA;" OwnerUserId="1513" LastEditorUserId="84" LastEditDate="2013-09-03T09:53:52.790" LastActivityDate="2014-04-26T11:45:29.883" Title="Rotary Evaporator Cleaning" Tags="&lt;safety&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6118" PostTypeId="1" AcceptedAnswerId="6338" CreationDate="2013-09-03T08:13:31.003" Score="5" ViewCount="661" Body="&lt;p&gt;Is there any reduction in size of the orbital for a half-filled orbital? Is the probability at any point of finding an electron doubled if there are two electrons instead of one? Is there any difference at all? I'm trying to understand the quantum mechanical model. Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2013-09-04T00:54:39.490" LastActivityDate="2013-09-26T11:38:37.387" Title="Is there any difference between a completely filled orbital and an half-filled one?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6119" PostTypeId="1" AcceptedAnswerId="6120" CreationDate="2013-09-03T08:24:38.650" Score="2" ViewCount="419" Body="&lt;p&gt;I can see how a double bond and a ring structure can lead to geometric isomerism but can't see the same for steric factors. Could you help? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2013-09-03T14:54:21.583" Title="How can geometric isomerism arise due to steric factors?" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6120" PostTypeId="2" ParentId="6119" CreationDate="2013-09-03T14:54:21.583" Score="6" Body="&lt;p&gt;Let's take a biphenyl compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcvPp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to steric factors, the rings cannot be planar. Instead, they have an approximate right angle between them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/isq8U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(&lt;a href=&quot;http://www.uscibooks.com/Anslyn/AD06rt.pdf&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for the right angle to &quot;flip&quot; over to the other side, the $\ce{NO2}$ and $\ce{CH3}$ would have to pass. These groups are bulky enough to prevent that from happening&lt;sup&gt;1&lt;/sup&gt; unless sufficient energy is provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ATpwR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, try to look from it along the axis. You'll notice that for this particular molecule, the nitro group farther from your eyes is a clockwise step from the closer nitro group. This happens regardless of which side of the axis (left/right) you look at it from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if we twisted the molecule so that the angle is again a right angle on the other side, there is now an anticlockwise step from the closer nitro to the farther nitro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two configurations of the compound are stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the steric barrier can be overcome with sufficient energy, such compounds usually racemize over time (though when the groups themselves  are as bulky as a phenyl, this doesn't happen at any appreciable timescale, IIRC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. I actually had some trouble using ChemDoodle to draw the molecule without any overlapping bits because of that. Doesn't exactly prove the bulkiness, but one gets the idea.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-09-03T14:54:21.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6121" PostTypeId="1" CreationDate="2013-09-03T15:00:01.170" Score="3" ViewCount="22740" Body="&lt;p&gt;I put 98% white fuming nitric acid and then I pour some sodium bicarbonate expecting $\ce{NaNO3}$ being formed, I'm not really using much water because it will make evaporation takes ages, even so $\ce{NaHCO3}$ should give $\ce{NaNO3}$ when in contact with $\ce{HNO3}$ and it seems so, $\ce{CO2}$ is formed very quickly so I suppose the reaction is happening... and I can see those white crystals (as the solution is over-saturated of $\ce{NaNO3}$) I shake, shake and shake, and pour more Sodium bicarbonate tills $\ce{HNO3}$ gets a pH around 7 (usually a bit less like 5 or 2) when I notice that it doesn't want to react anymore, however, I try to add mol per mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there's some unreacted $\ce{HNO3}$ but it should decompose into $\ce{H2O} + \ce{NO2}$ so I put it to start drying in a hot plate, and then the solution becomes yellow (decomposing) it's suddenly brown... and then my &quot;supposed&quot; $\ce{NaNO3}$ is yellowish/pink... and I wonder that seems like a $\ce{NaNO3} + \ce{NaNO2}$ mix but it makes no much sense, I tried with $\ce{NaOH}$ instead $\ce{NaHCO3}$ and it's the same ending color; it's not working... what did I do wrong? how to do it properly?&lt;/p&gt;&#xA;" OwnerUserId="2247" LastEditorUserId="57118" LastEditDate="2018-01-05T16:05:05.773" LastActivityDate="2018-07-02T09:26:03.220" Title="NaHCO₃ + HNO₃ → NaNO₃ + H₂O + CO₂ ... but something weird happens to me in practice" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6122" PostTypeId="1" AcceptedAnswerId="6127" CreationDate="2013-09-03T16:25:16.930" Score="5" ViewCount="4125" Body="&lt;p&gt;In hydrolysis of polysaccharides, does the hydrolysis speed increase when more $\ce{H+}$ ions are present in the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I were to use $\ce{H2SO4}$ instead of $\ce{HCl}$ (same volumes) then $\ce{H2SO4}$ will be the most effective?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="7448" LastEditDate="2017-04-03T19:20:12.833" LastActivityDate="2017-04-03T19:20:12.833" Title="Does the hydrolysis speed increase when more H+ ions are in the solution?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;ions&gt;&lt;kinetics&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6124" PostTypeId="2" ParentId="6082" CreationDate="2013-09-03T17:51:26.347" Score="0" Body="&lt;p&gt;Thank you buckminst. Your link was useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think much of the reason this method isn't used more often in the U.S. (in dry areas of course) has to do with marketing or perhaps bad laws passed for some convoluted political reasons. If more people used this in opposed to regular air conditioners it would easily substantially lower energy costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wanted to add some more numbers to bring perspective to how inefficient regular air conditioners are compared with evaporative cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia &quot;a 5000 BTU/h air-conditioning unit, with a SEER of 10, would consume 5000/10 = 500 Watts of power on average.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Air_conditioning#Seasonal_energy_efficiency_ratio&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Air_conditioning#Seasonal_energy_efficiency_ratio&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;500 W = 1,800,000 J per hour&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5000 BTU = 527,500 J of energy removed from the environment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the two and you get about 30 percent efficiency, which is about the efficiency of a good regular air conditioner (so their example is realistic). A single gallon of water removes 6,780,000 J of heat energy from its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the central air conditioner uses about 12,600,000 J/hour (sorry I placed my commas wrong in the last post which confused me and drew me to an incorrect conclusion) and (assuming about 30 percent efficiency) will remove only about 3,780,000 J of energy from its environment. A single gallon of water will remove 6,780,000 J of energy from its environment which is enough to almost last two hours in opposed to using the central air conditioner for two hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet even in dry areas I do not see this used. I see regular air conditioners used and I think that needs to change.&lt;/p&gt;&#xA;" OwnerUserId="2249" LastEditorUserId="2249" LastEditDate="2013-09-03T18:12:38.240" LastActivityDate="2013-09-03T18:12:38.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6125" PostTypeId="1" AcceptedAnswerId="6129" CreationDate="2013-09-03T18:13:09.473" Score="2" ViewCount="143" Body="&lt;p&gt;I want to convert cellulose to another form (crystalline =&gt; amorphous). Cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in water.(Is the pressure achievable in the lab?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this doable in the lab and how should I begin on reproducing that situation (which tools would I need) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that a bunsen burner would produce high enough temperatures (but I'm not quite sure about the pressure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to my question, following up the advice of bobthechemist...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystalline substance to amorphous - Other ways?&#xA;You can convert crystalline cellulose to the amorphous, more reactive form by heating it to a temperature of around 320°C under a pressure of 25MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there other, less extreme ways, to achieve that amorphous form of the compound?&#xA;So because I can't achieve these high temperatures, how can I achieve it in another way?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2013-09-06T16:27:20.187" LastActivityDate="2015-05-22T15:52:42.557" Title="Is the current situation possible for recreaction in a lab?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6126" PostTypeId="2" ParentId="6121" CreationDate="2013-09-03T18:17:21.653" Score="5" Body="&lt;p&gt;Let's look at the equation first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaHCO3 + HNO3 \to NaNO3 + H2O + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when added in equal parts, you'd ideally expect the bicarbonate anion to be protonated by the nitric acid, forming weaker carbonic acid (which will dissociate into water and carbon dioxide under temperature or even agitation), and the remaining ions will pair up into sodium nitrate (soda niter) solution in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, nitric acid is tricky, and doesn't like to act &quot;ideally&quot;. First, you already know that nitric acid decomposes into nitrogen dioxide and water; this happens whether you want it to or not. Highly exothermic reactions (most nitric acid reactions, including this one, are so) ramp up this inevitable side reaction with the extra heat. This will reduce your yield, but I'd expect it to go the other way than you apparently did; 1:1 mixtures of acid and bicarbonate will inevitably leave unreacted bicarbonate in the solution (not acid) along with some dissolved nitrogen dioxide (aka nitrite hydrate, $\ce{H2NO3}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, &quot;pure&quot; WFNA is defined as anywhere north of 98% nitric acid, so pretty much any quantity of nitric acid you could get is likely to have impurities. So, I think your first mistake was to keep adding sodium bicarbonate; if you mixed a mass of &quot;pure&quot; (&gt;98%) WFNA (even if you made it from scratch) with equivalent molar weight of sodium bicarb, you'll have a small excess of bicarb due to impurities in the nitric acid (pre-existing and created), but the solution will still read acidic (because of dissolved $\ce{CO2}$ and $\ce{NO2}$) after you've added more bicarb than will dissolve in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, heating the equilibrium solution to obtain the dry solid salts will remove the water from the dissolved nitrogen dioxide, and it will start acting like the nitrite ion it is, reacting with any unreacted bicarbonate to form yellow sodium nitrite, water and more $\ce{CO2}$. This sounds like exactly what happened to you, resulting in a more yellowish solid than the pure white soda niter you were expecting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not careful with the heat while evaporating off the water, parts of the reaction, namely the precipitating solid collecting on the bottom of the flask, can be superheated beyond boiling, and reach the temperature at which both of the niter salts will decompose into sodium oxide and more oxides of nitrogen. This is typically accomplished only at temperatures in excess of those capable with the average benchtop rig, but it becomes more thermodynamically favorable in the presence of excess sodium (which you have in the sodium bicarbonate). I'm not saying this happened - it's more likely that the nitric acid decomposed more than you might have wanted - but it's a theoretical possibility.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-04T14:06:27.237" LastActivityDate="2013-09-04T14:06:27.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6127" PostTypeId="2" ParentId="6122" CreationDate="2013-09-03T18:30:12.193" Score="6" Body="&lt;p&gt;The glycosidic bond of a polysaccharide chain is essentially an acetal moiety, and hydrolysis of acetals proceeds under acid-catalyzed conditions. Hence, the addition of acid, within some ambit of concentration, will accelerate the hydrolysis reaction. The optimum proportion of acid, however, is probably something that needs to be determined experimentally on a case-by-case basis. It isn't necessarily true that the rate of the reaction will continuously increase, without bound, as the concentration of catalyst increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism proceeds through a series of proton transfers and SN1 reactions. As far as I'm aware, the rate-determining step is typically the actual cleavage of the glycosidic bond after an initial protonation of the oxygen of the leaving group. Given that, I would expect increases in acid concentration to have marginal effect beyond some threshold value. That's just a surmise on my part, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find &lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/789/1/JSIR%2067%284%29%20%282008%29%20295-298.pdf&quot;&gt;one study&lt;/a&gt; that examined the kinetics of starch hydrolysis, which found the optimum pH to be 3 (with values of 2-5 being tested).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-09-03T18:30:12.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6128" PostTypeId="2" ParentId="6115" CreationDate="2013-09-03T18:35:14.253" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Copper%28II%29_sulfate#Chemical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, copper(II) sulfate will dehydrate from penta- to trihydrate at 63 °C, then to monohydrate at 109 °C and then finally to the anhydrate at 200 °C. It melts at 150 °C, but won't decompose into ions until 650 °C which is beyond most benchtop rigs (your Pyrex flask would give way at about 500 °C).&lt;/p&gt;&#xA;&lt;p&gt;The required heat of decomposition, as you can see, does vary for each hydrate (especially for the same salt), but doesn't always have to be greater than the boiling point; water has a significant vapor pressure at nearly all ambient Earth temperatures (even at freezing), and so in a dry environment, water will evaporate rapidly at temperatures well below boiling.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2021-04-22T17:55:57.690" LastActivityDate="2021-04-22T17:55:57.690" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6129" PostTypeId="2" ParentId="6125" CreationDate="2013-09-04T00:40:49.433" Score="5" Body="&lt;p&gt;25MPa is just north of 3600psi for those of us that haven't converted to SI units; 320*C ~= 608*F. Your average lecture bottle is rated to about 3000psi, while compressed gas cylinders can go as high as 6000psi. Trouble is, most of them don't stand up well to being heated by an open flame. Temperature is another beast; your Bunsen burner can easily achieve a flame temperature well in excess of 600*F (and a simple variation, the Meker burner, can hit more than twice that), but your Pyrex glassware is only rated to 500*F, and that for short periods. Bunsen burners are often used for glassworking on the lab bench, creating custom tubing rigs by softening straight tubing and then bending or stretching to create bends and bottlenecks. If the glass can be molded over a Bunsen burner, it's not going to be able to sustain the temperatures you're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short of it is that both the temperature and pressure are possible, but neither are achievable with a benchtop apparatus in a fume hood; you'll need an HTHP reactor vessel rated for your experimental conditions. &lt;a href=&quot;http://www.highpressure.com/pdfs/section/pv.pdf&quot; rel=&quot;nofollow&quot;&gt;They exist&lt;/a&gt;, but whether you can get access to one is another question only your superior can answer. If you were planning on visually studying the change in the cellulose, you'll need a sapphire viewport, and these get very expensive very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the right tool for the job, it becomes a rather academic task of calculating how much volume of reactants are required to achieve the desired pressure in the vessel when heated to temperature, then filling the vessel and evenly heating it. Only two of the vessels in the linked column hold any appreciable amount of material at the proper temp and pressure, and one of them will &lt;em&gt;just barely&lt;/em&gt; meet your needs; if you unevenly heat any of these vessels, by the time the far end reaches the proper temperature the spot being heated will have failed.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2015-05-22T15:52:42.557" LastActivityDate="2015-05-22T15:52:42.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6130" PostTypeId="1" CreationDate="2013-09-04T01:00:58.213" Score="6" ViewCount="238" Body="&lt;p&gt;From &lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/covalent.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemguide.co.uk&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the phosphorus is going to form $\ce{PCl5}$ it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the $\mathrm{3s}$ orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the $\mathrm{3d}$ orbitals. You might have expected it to use the $\mathrm{4s}$ orbital because this is the orbital that fills before the $\mathrm{3s}$ when atoms are being built from scratch. Not so! &lt;strong&gt;Apart from when you are building the atoms in the first place, the $\mathrm{3s}$ always counts as the lower energy orbital&lt;/strong&gt;.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could you please explain why this is so?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="41328" LastEditDate="2017-08-08T22:08:20.803" LastActivityDate="2017-08-08T22:08:20.803" Title="Why apart from when building atoms in the first place, the 3d is the lower energy orbital?" Tags="&lt;orbitals&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6131" PostTypeId="2" ParentId="5672" CreationDate="2013-09-04T02:17:54.897" Score="5" Body="&lt;p&gt;VOCs, as BLHaas said, are &quot;volatile organic compounds&quot;; that is, compounds that have a high vapor pressure and so readily evaporate into the surrounding air, and are mainly hydrogen-oxygen-carbon. The ones we smell and identify as being VOCs are most often &quot;aromatics&quot;, which have one or more six-carbon rings (phenyl/benzyl, napthyl, etc). These aren't all bad; the smell of cooking meat, of flowers, herbs, fresh-cut grass, lemons, and yes even onions, are all VOCs. The VOCs in paint are what you smell, and also what are bad for you (at least some of them), so theoretically, getting rid of the smell means you've gotten rid of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The VOCs typically associated with latex and oil paints are mineral spirits, typically an amalgam of 7- to 12-carbon hydrocarbons produced by petroleum refinement. Oil paints require a high percentage of these in order to properly dissolve the pigment and achieve the desired consistency. Latex paints are based on an acrylic resin, which for many decades now has required relatively low VOC content in favor of a simple water base. However, acrylic acid, the basis of the acrylic resins holding the pigments and binders of the paint, can easily degrade into another example of a stinky organic chemical class; aldehydes. Aldehydes are caused by hydrolysis and subsequent hydrogenation of carboxylates, of which acrylics are a member. The simplest one is formaldehyde, and it's not great news; it's carcinogenic and acutely toxic in sufficient concentrations (higher than those in the average freshly painted room). Aldehydes also have a relatively low odor threshold; humans can detect aldehydes in concentrations measured in the low parts per billion. So it doesn't take much to stink up a room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually answer your question, the most common trick I could find was to cut an onion in half and place each half, cut-side-up, in opposite corners of the room. The chemistry behind that is apparently inherent in the active ingredient of onions that irritates your eyes when you cut them. This ingredient is produced by enzymatic alteration of amino acid sulfoxides, and is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Syn-propanethial-S-oxide&quot; rel=&quot;nofollow&quot;&gt;Syn-propanethial-S-oxide&lt;/a&gt;. Notice the &quot;thial&quot; portion of the name. Thials are to aldehydes as thiols are to alcohols; the switch from an oxygen to a sulfur, as with thiols, typically increases a thial's reactivity and lowers its boiling point compared to an aldehyde. So, you have a highly reactive sulfur-based chemical that yearns to evaporate into the air and irritate your mucous membranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing it will apparently do, interestingly enough, is undergo a &lt;a href=&quot;https://en.wikipedia.org/wiki/1,3-Dipolar_cycloaddition&quot; rel=&quot;nofollow&quot;&gt;dipolar cycloaddition reaction&lt;/a&gt; with aldehydes&lt;sup&gt;[&lt;a href=&quot;http://www.reddit.com/r/askscience/comments/1alluh/why_do_onions_neutralize_the_smell_of_paint_fumes/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;]&lt;/sup&gt;, like the ones produced by evaporation of impurities in acrylic paint. This rearranges the thial and the aldehyde into a combined substance with a five-membered, oxygenated ring, which ostensibly does not offend the senses as much as either precursor. Because aldehydes are offensive in such low concentrations, it similarly doesn't take much Syn-propanethial-S-oxide to remove them and reduce the level below the odor threshold.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-04T14:03:53.900" LastActivityDate="2013-09-04T14:03:53.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6132" PostTypeId="2" ParentId="6116" CreationDate="2013-09-04T03:48:26.877" Score="4" Body="&lt;p&gt;Those elements are considered to be good semiconductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.f.kth.se/~f95-pax/refs/PeriodicTable/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this periodic table&lt;/a&gt; that has them designated as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they are no longer emphasized on the periodic table because there are so many other semiconducting materials that lie outside of that particular pattern.  For example, Gallium is often paired with Arsenic (in Ga-As semiconductors), but only As is along the artificial &quot;metal/non-metal&quot; horizon designated by the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that from &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_semiconductor_materials&quot; rel=&quot;nofollow noreferrer&quot;&gt;this list&lt;/a&gt; aluminum phosphide is listed as a semiconductor, but neither element is situated on that line.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2013-09-04T03:48:26.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6133" PostTypeId="2" ParentId="6116" CreationDate="2013-09-04T07:27:41.837" Score="5" Body="&lt;p&gt;&lt;strong&gt;The red line is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dividing_line_between_metals_and_nonmetals&quot; rel=&quot;noreferrer&quot;&gt;dividing line between metals and non-metals&lt;/a&gt;&lt;/strong&gt;. It is rather arbitrary; as Mendeleev himself wrote: &lt;em&gt;“It is...impossible to draw a strict line of demarcation between metals and nonmetals, there being many intermediate substances.”&lt;/em&gt; Most modern periodic tables do not feature it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The elements with a blue background are the &lt;a href=&quot;http://en.wikipedia.org/wiki/Metalloid&quot; rel=&quot;noreferrer&quot;&gt;metalloids&lt;/a&gt;&lt;/strong&gt;. Against, it's a rather arbitrary choice, but it corresponds to elements that have some properties of metals. Wikipedia describes them in the following way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Physically, metalloids usually have a metallic appearance but they are brittle and only fair conductors of electricity; chemically, they mostly behave as (weak) nonmetals. They can, however, form alloys with metals. Ordinarily, most of the other physical and chemical properties of metalloids are intermediate in nature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-09-04T07:27:41.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6134" PostTypeId="1" AcceptedAnswerId="6143" CreationDate="2013-09-04T07:35:19.887" Score="3" ViewCount="1862" Body="&lt;p&gt;When the lone pair of an amino group $\ce{R-NH2}$ is involved in resonance, its basicity decreases. But, between conjugation with carbonyl group $\ce{R}=\ce{R'CO}$ or resonance with benzene $\ce{R} = \ce{Ph}$, which one will reduce the basicity of amino group to a greater extent? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2018-02-24T11:37:02.363" LastActivityDate="2018-02-24T11:38:30.553" Title="Are aromatic amines or amides less basic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6135" PostTypeId="1" AcceptedAnswerId="6136" CreationDate="2013-09-04T07:41:48.867" Score="3" ViewCount="386" Body="&lt;p&gt;Is it possible to heat a material above its solidus without melting occurring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, at a heating rate of approximately 200K/s, will a metal alloy exhibit superheating and if so by what order of magnitude i.e. fractions of a degree versus tens of degrees. What is the mechanism behind this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of information about supercooling but I am not certain whether this applies!&lt;/p&gt;&#xA;" OwnerUserId="2250" LastActivityDate="2013-09-04T08:39:40.777" Title="Superheating metal alloy above solidus" Tags="&lt;physical-chemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6136" PostTypeId="2" ParentId="6135" CreationDate="2013-09-04T08:39:40.777" Score="2" Body="&lt;p&gt;Yes, superheating can happen in some alloys. Whether it occurs at all, and its extent if it occurs, is very specific to each alloy (and heating rate dependent, obviously). It also depends on things like sample geometry (particles, solid, etc.) and size. &lt;a href=&quot;http://www.fen.bilkent.edu.tr/~bengu/chem523/PprRev2/OKK%20SAYIN%20Melting%20and%20superheating%20of%20metals%20and%20alloys.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; reports delays in melting by up to 100 K for some Al-Pb and Cu-Pb alloys.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-09-04T08:39:40.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6138" PostTypeId="1" CreationDate="2013-09-04T13:03:53.400" Score="2" ViewCount="4133" Body="&lt;p&gt;If we add an impurity atom (doping) to the crystalline struture of another material and take the &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_Diffraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;XRD&lt;/a&gt; of the doped material what kind of peak shift can we expect. For the better description of my question i have added an image below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/niDPt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image &lt;em&gt;black bold line&lt;/em&gt; ($2\theta $) is the standard XRD peak of a material; &lt;em&gt;blue&lt;/em&gt; ($2\theta ^{L}$) and &lt;em&gt;green&lt;/em&gt; ($2\theta ^{R}$) dotted lines are the peak shift caused due to the doping of impurity atom. Can you tell me if there is any relationship between dopent size and peak shift (right/left dotted lines).&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="2608" LastEditDate="2013-11-02T13:35:47.600" LastActivityDate="2019-06-01T15:07:36.290" Title="Effect of dopant size on XRD peak shift" Tags="&lt;spectroscopy&gt;&lt;crystal-structure&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6142" PostTypeId="1" CreationDate="2013-09-04T18:24:16.470" Score="5" ViewCount="8006" Body="&lt;p&gt;Sulfur trioxide violates the octet rule. Upon drawing the Lewis dot structure for sulfur trioxide, we see that the central sulfur atom is bonded to three other oxygen atoms by double covalent bonds. Therefore, sulfur is in fact surrouded by 12 electrons and not 8 as per the octet rule. This is explained by the fact that the excited state of sulfur has the following electronic configuration: $1s^22s^22p^63s^13p^34s^03d^2$. As there are 3 oxygen atoms, $sp^2$ hybridised orbitals take part in single $\sigma$ covalent bonds with the unpaired $p_y$ orbitals of the oxygen atoms. In sulfur, there still exist one unpaired $p$ orbital and 2 unpaired $d$ orbitals. These remaining unpaired orbitals take part in $\pi$ covalent bonds with the unpaired $p_z$ orbitals of the oxygen atoms. My question is, why do the electrons in sulfur from the $3s$ and $3p$ orbitals in the ground state promote themselves to the $d$ orbitals rather than one electron going into the intermediate $4s$ orbital and the next electron going into the first $3d$ orbital? After all, according to the Aufbau Principle, $4s$ is less energetic when compared to $3d$.  &lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2018-12-06T13:00:39.113" Title="Sulfur trioxide - vacant d-orbitals" Tags="&lt;electrons&gt;&lt;orbitals&gt;&lt;atoms&gt;&lt;covalent-compounds&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6143" PostTypeId="2" ParentId="6134" CreationDate="2013-09-04T21:45:40.107" Score="7" Body="&lt;h2&gt;Measures for acid and base strength&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, you need to be clear about 'strong' and 'weak' acids.&#xA;In order to measure the strength of an acid relative to water and find out how effective a proton donor it is, you must look at the equilibrium constant for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{HA (aq) + H2O(l) &amp;lt;=&amp;gt; H3O+ (aq) + A- (aq)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of equilibrium is measured by the equilibrium constant for this reaction $K_{\text{eq}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\text{eq}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{H2O}][\ce{HA}]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of water remains essentially constant (at $55.56 \, \text{mol} \, \text{dm}^{–3}$) with dilute solutions&#xA;of acids wherever the equilibrium may be and a new equilibrium constant, $K_{\text{a}}$, is defined and called the &lt;strong&gt;acidity constant&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\text{a}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like $\text{pH}$, this is also expressed in a logarithmic form, $\text{p}K_{\text{a}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \text{p}K_{\text{a}} = -\log (K_{\text{a}})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the minus sign in this definition, the lower the $\text{p}K_{\text{a}}$, the larger the&#xA;equilibrium constant, $K_{\text{a}}$, is and hence the stronger the acid. The $\text{p}K_{\text{a}}$ of the acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists as undissociated $\ce{HA}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An acid's $\text{p}K_{\text{a}}$ depends on the stability of its conjugate base.&#xA;$\ce{HCl}$ is a much stronger acid than acetic acid: the $\text{p}K_{\text{a}}$ of $\ce{HCl}$ is around $–7$ compared to $4.76$ for acetic acid. This tells you that in solution $K_{\text{a}}$ for hydrogen chloride is $10^{7} \, \text{mol} \, \text{dm}^{–3}$ whilst for acetic acid it is only $10^{–4.76} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3}$. But, why are the equilibria so different? Why does hydrogen chloride fully dissociate but acetic acid do so only partially?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{HCl (aq) + H2O (l) &amp;lt;=&amp;gt;&amp;gt; H3O+ (aq) + Cl- (aq)} \quad \qquad \qquad K_{\text{a}} = 10^{7} \, \text{mol} \, \text{dm}^{–3} \\&#xA;  \ce{CH3COOH (aq) + H2O (l) &amp;lt;&amp;lt;=&amp;gt; H3O+ (aq) + CH3COO- (aq)} \qquad K_{\text{a}} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3} \\&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer must have something to do with the conjugate base $\ce{A^{–}}$ of each acid $\ce{HA}$, since this is the only thing that varies from one acid to another. In both the equilibria above, water acts as a base by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other hand, is easily protonated by $\ce{H3O+}$ to give neutral acetic acid, which means that acetate must be a stronger base than chloride ion.&#xA;A nice thing to remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The stronger the acid $\ce{HA}$, the weaker its conjugate base, $\ce{A^{–}}$&lt;/li&gt;&#xA;&lt;li&gt;The stronger the base $\ce{A^{–}}$, the weaker its conjugate acid $\ce{AH}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the $\text{p}K_{\text{a}}$ is a good measure for the strength of an acid: the lower the $\text{p}K_{\text{a}}$ the stronger the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about bases? To say something about the strength of a base it is convenient to take the following viewpoint: to what extent does a protonated base want to keep its proton or in other words: how big is the acid strength of its conjugate acid? For example if you want to know which is the stronger base - say, formate anion or acetylide anion - you look up the $\text{p}K_{\text{a}}$s for their conjugate acids. You find that the $\text{p}K_{\text{a}}$ for formic acid ($\ce{HCO2H}$) is $3.7$, whilst the $\text{p}K_{\text{a}}$ for acetylene is around $25$. This means that acetylene is much more reluctant to part with its proton, that is, acetylide is much more basic than formate. This is all very well for anions - you simply look up the $\text{p}K_{\text{a}}$ value for the neutral conjugate acid, but what if you want to know the basicity of ammonia? If you look up the $\text{p}K_{\text{a}}$ for ammonia you find a value around $33$ but this is the value for deprotonating neutral ammonia to give the amide ion, $\ce{NH2^{–}}$.&#xA;If you want to know the basicity of ammonia, you must look up the $\text{p}K_{\text{a}}$ of its conjugate acid, the ammonium cation, $\ce{NH4+}$, protonated ammonia. Its $\text{p}K_{\text{a}}$ is $9.24$ which means that ammonia is a weaker base than hydroxide - the $\text{p}K_{\text{a}}$ for water (the conjugate acid of hydroxide) is $15.74$.&#xA;In conclusion: the higher the $\text{p}K_{\text{a}}$ of the conjugate acid the stronger the base.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The answer to your question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The $\text{p}K_{\text{a}}$ of the conjugate acid of aniline is $4.6$. So, since this value is much lower than the $9.24$ you had for ammonia, it is much less basic than ammonia, as is expected since the lone pair of aniline is delocalized into the phenyl ring. But for amides the $\text{p}K_{\text{a}}$ values of the conjugate acid are typically smaller than $0$, so an amide is much less basic than aniline. This is to be expected because the carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring which can interact more efficiently with the nitrogen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-02-24T11:38:30.553" LastActivityDate="2018-02-24T11:38:30.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6144" PostTypeId="1" AcceptedAnswerId="7663" CreationDate="2013-09-05T03:28:53.923" Score="2" ViewCount="126" Body="&lt;p&gt;This time I am interested in a specific type of acid anhydride, &lt;strong&gt;sulfonyl-carboxylates&lt;/strong&gt;, i.e. compounds with the following molecular structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/70uji.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{R2}$ can be an organic substituents &lt;em&gt;or&lt;/em&gt; hydrogen atoms. $\ce{R1}$ can be an organic substituent only. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2015-05-01T21:26:04.003" LastActivityDate="2015-05-01T21:50:41.137" Title="Are sulfonyl-carboxylates stable in water?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6145" PostTypeId="1" AcceptedAnswerId="6938" CreationDate="2013-09-05T04:01:51.140" Score="7" ViewCount="327" Body="&lt;p&gt;Here are some &lt;strong&gt;standard electrode potentials&lt;/strong&gt;:- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO_4^{2-} + 4H^+ + 2e^- -&amp;gt; SO_2 + 2H_2O}\\E^o=0.17\mathrm{V}\\  &#xA;\ce{NO_3^- + 4H^+ + 3e^- -&amp;gt; NO + 2H_2O}\\E^o=0.97\mathrm{V}\\&#xA;\ce{NO_3^- + 2H^+ + e^- -&amp;gt; NO_2 + H_2O}\\E^o=0.78\mathrm{V}$$ &#xA;When nitric acid reacts with a metal, either $\ce{NO_2}$ or $\ce{NO}$ is formed? How do you &lt;strong&gt;predict which is thermodynamically favourable&lt;/strong&gt; and which reaction will actually take place? For example, when dilute nitric acid reacts with copper, Nitric oxide is liberated and when it is concentrated, Nitrogen dioxide is evolved? As Nitric acid formation is more favourable, shouldn't it proceed in all cases as increasing the concentration will decrease/increase the potential of both the reaction simultaneousy(&lt;em&gt;Nernst equation&lt;/em&gt;)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of sulfate ion, although the reduction of sulfate is &lt;em&gt;more favourable&lt;/em&gt; than that of hydrogen, it is hydrogen ions which are getting reduced when a metal reacts at dilute concentrations? why is this so? Moreover, at high concentrations, once again sulfur dioxide is evolved? How can we understand such occurrences? Is overpotential involved in these observances? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of permanganate ions, there are so many possible redox pathways, but a different one is preferred in a different case based on the conditions. For example, in acidic medium, permanganate is reduced to $\ce{Mn^{2+}}$ while in neutral/slightly alkaline solutions, it reduces to $\ce{MnO_2}$. Why is this so? &lt;a href=&quot;http://www.wou.edu/las/physci/ch462/redox.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here are the redox potential values for permanganate conversions (latimer diagrams)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the general way to explain all these phenomena, as to what drives the reaction to give one product under certain condition and another in different conditions. Explaining any one example would be enough to set the general method which I can extrapolate to understand the remaining Examples.I am comfortable with basic electrochemical equations and those of thermodynamics.  &lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="402" LastEditDate="2017-01-14T00:28:25.037" LastActivityDate="2017-01-14T00:28:54.440" Title="Prediction of ionic reduction?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6147" PostTypeId="1" CreationDate="2013-09-05T11:08:56.440" Score="0" ViewCount="2021" Body="&lt;p&gt;I believe that Volumetric Analysis and Tetration Theory are not the same subject.&lt;br&gt;&#xA;But I need to know what is a main difference between those subjects and what is a main intersection?&lt;/p&gt;&#xA;" OwnerUserId="796" LastEditorUserId="796" LastEditDate="2013-09-06T12:19:32.790" LastActivityDate="2016-10-05T18:51:07.123" Title="What is a relationship between Volumetric Analysis and Titration Theory?" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6148" PostTypeId="2" ParentId="6147" CreationDate="2013-09-05T12:46:36.700" Score="1" Body="&lt;p&gt;I think you mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Titration&quot; rel=&quot;nofollow&quot;&gt;titration&lt;/a&gt;, not “tetration”. “Volumetric analysis” is another (somewhat dated) name for titration, as explained the first paragraph of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Titration&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-09-05T12:46:36.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6150" PostTypeId="1" AcceptedAnswerId="6151" CreationDate="2013-09-05T17:18:43.237" Score="7" ViewCount="99" Body="&lt;p&gt;I am particularly interested in knowing about glucose breakdown. If we artificially burn glucose, will the energy released be the same if it is broken down in steps(like in a living cell)? &lt;/p&gt;&#xA;" OwnerUserId="2108" LastActivityDate="2017-05-02T23:22:39.260" Title="Does it make a difference in energy if a reaction is done in many steps or in one step?" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6151" PostTypeId="2" ParentId="6150" CreationDate="2013-09-05T17:32:31.963" Score="6" Body="&lt;p&gt;You might want to take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hess's law&lt;/a&gt;. It states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps. But that covers only the thermodynamic aspect. In reality, you also have kinetic barriers (activation energy) which add to the energy needed for a reaction to run (or released during a reaction). That's why catalysts are so useful: They break a reaction down into steps (involving different intermediates) with lower activation barriers than the original reaction so that the reaction as a whole can proceed at a lower temperature.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="402" LastEditDate="2017-05-02T23:22:39.260" LastActivityDate="2017-05-02T23:22:39.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6152" PostTypeId="1" CreationDate="2013-09-05T21:46:39.117" Score="1" ViewCount="63" Body="&lt;p&gt;More questions for my DIY hot air balloon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some very common (i.e. easy to purchase) plastic film materials that have very good flame-retardant properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the &quot;degree of flame retarding&quot; of a plastic film material rated, so that I can compare materials by their manufacturer's rating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any common &quot;spray on&quot; flame retardant treatments that could work? Or is a flame retardant plastic film the only way to go to really stop it from bursting into flames?&lt;/p&gt;&#xA;" OwnerUserId="2258" LastActivityDate="2013-09-06T03:24:47.550" Title="Common flame-retardant plastic film materials?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6153" PostTypeId="2" ParentId="6152" CreationDate="2013-09-06T03:24:47.550" Score="1" Body="&lt;p&gt;What have you found so far? In a situation like this honestly I would go ahead and research what hot air balloons are commonly made with already. Find somebody more experienced in the hobby and ask them if you can, or even an experienced fire fighter you can get in touch with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Hot_air_balloon&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;: &quot;For modern sport balloons the envelope is generally made from nylon fabric and the inlet of the balloon (closest to the burner flame) is made from fire resistant material such as Nomex&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The type of measurement you are asking for seems a bit more complicated than something like measuring mass or volume, it seems to be commonly represented with images, videos of tests, graphs, etc. You may feel more at ease by testing how useful materials are before and after flame testing them yourself, that is what I would do anyways. I hope this helps you.&lt;/p&gt;&#xA;" OwnerUserId="584" LastActivityDate="2013-09-06T03:24:47.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6154" PostTypeId="1" CreationDate="2013-09-06T08:18:11.420" Score="5" ViewCount="29699" Body="&lt;p&gt;I have a practical question: today I have to clean the oven, and it doesn't have self cleaning burn down mode, nor an official scraper I can use...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's full of charred grease and sugar. &lt;em&gt;I managed to remove 98% of the charred grease with cold grease remover&lt;/em&gt;. But it seems that reagent has no effect on the remaining 2%, that's black fully charred carbon or sugar. I also tried vinegar, but it doesn't have any effect on it either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to scrape it because I fear that I would scratch the oven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which reagent can I use to remove those pesky carbon deposits without damaging the oven, and what will be the chemical reaction?&lt;/p&gt;&#xA;" OwnerUserId="2259" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:05:45.747" LastActivityDate="2017-05-10T15:06:29.517" Title="Which reagent can I use to remove carbon from oven?" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6155" PostTypeId="1" AcceptedAnswerId="6160" CreationDate="2013-09-06T10:19:00.413" Score="25" ViewCount="33485" Body="&lt;p&gt;Acetic acid (ethanoic acid, $\mathrm pK_\mathrm a \approx 5$) is more acidic than phenol ($\mathrm pK_\mathrm a \approx 10$), which is reflected in their reactivity with a weak base such as sodium carbonate (adding acetic acid leads to effervescence of carbon dioxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we look at the structures of acetate and phenoxide ions, the delocalization of negative charge should be greater in the phenoxide ion as the charge moves inside the benzene ring. On the other hand, in the acetate ion, the negative charge is delocalized between two oxygen atoms only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible that acetic acid is still a stronger acid, even when its conjugate base has fewer resonance structures?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorUserId="16683" LastEditDate="2018-09-07T17:59:59.867" LastActivityDate="2021-02-23T14:56:49.240" Title="Why is acetic acid more acidic than phenol?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;phenols&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="6156" PostTypeId="2" ParentId="6155" CreationDate="2013-09-06T10:32:42.353" Score="9" Body="&lt;p&gt;The $\text{p}K_{\text{a}}$ of ethanoic acid is $4.8$ while the $\text{p}K_{\text{a}}$ of phenol is $10$. So ethanoic acid is indead more acidic than phenol (for an explanation of $\text{p}K_{\text{a}}$ values and their connection to acid strength see my answer &lt;a href=&quot;https://chemistry.stackexchange.com/a/6143/189&quot;&gt;here&lt;/a&gt;). And this is only to be expected from an intuitive point of view: one might say phenol is only an alcohol (although the phenyl ring has a lot of influence of course, which can be seen at the fact that normal alcohols have $\text{p}K_{\text{a}}$ values around $16$) and one wouldn't expect an alcohol to be more acidic than a carboxylic acid. The reason for this is simple: It not only the &quot;delocalization radius&quot; (over how many atoms the negative charge is delocalize) that matters when determining how well a negative charge is stabilized by delocalization but the effectiveness of the delocalization. The carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring. This LUMO can interact more efficiently with the oxygen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.&#xA;You might say, in a phenolate ion most of the negative charge is still located at the oxygen and only a relatively small fraction of it is delocalized inside the phenyl ring but in an ethanoate ion the negative charge is fully delocalized over the two oxygen atoms. Furthermore the two oxygen atoms in ethanoate are more electronegative than the carbon atoms in a phenyl ring and thus are able to stabilize the electron density more efficiently.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2013-09-06T11:17:49.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6157" PostTypeId="1" AcceptedAnswerId="6158" CreationDate="2013-09-06T14:04:25.407" Score="14" ViewCount="4750" Body="&lt;p&gt;We have products $\ce{A + B}$ combining to form $\ce{C + D}$ through the transition state $\ce{X}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are all reactions at least virtually reversible? Is the difference between a reversible reaction and an irreversible reaction simply that for the former, $\Delta G$ is small, where as the irreversible reaction will have a difference in free energy in products &lt;em&gt;vs&lt;/em&gt;. reactants, that the reverse reaction simply doesn't occur, but theoretically could?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do concentrations affect the free energy? Is it because you sum the free energy for all reactant molecules, as well as for all the product molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is from my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] the transition state is too unstable to exist for long. It collapses to either substrate or product, but which of the two accumulates is determined only by the energy difference between the substrate and the product -- the $\Delta G$ of the reaction. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would appreciate if someone could clarify this for me. Feel free to be very nitpicky, I don't want an understanding that is &quot;&lt;em&gt;approximately&lt;/em&gt;&quot; correct or &quot;&lt;em&gt;functionally&lt;/em&gt;&quot; good.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="41328" LastEditDate="2017-08-09T00:51:23.143" LastActivityDate="2017-08-09T00:51:23.143" Title="Transition state and free energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;&lt;transition-state-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6158" PostTypeId="2" ParentId="6157" CreationDate="2013-09-06T15:54:53.093" Score="15" Body="&lt;p&gt;According to the principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Microscopic_reversibility&quot; rel=&quot;noreferrer&quot;&gt;Microscopic Reversibility&lt;/a&gt; all elementary reactions, i.e. those that proceed in a single step, are reversible.&#xA;And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LzUMc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation.&#xA;The thing that determines which way your reaction goes is how easy it is to make the transition state or to &quot;get the ball to the top&quot;. In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to &quot;climb&quot; the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are &quot;elevated&quot; to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back &quot;on top of the hill&quot; in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates.&#xA;When you have a reversible reaction and $\Delta G$ is very small the &quot;hill to climb to the transition state&quot; is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-09-06T16:54:57.877" LastActivityDate="2013-09-06T16:54:57.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6159" PostTypeId="2" ParentId="6157" CreationDate="2013-09-06T16:01:14.623" Score="2" Body="&lt;p&gt;First question:&#xA;The answer is true from a reaction view. You are right. No matter how the reaction is favored in terms of Gibbs free energy change. The reverse process will happen in small or tiny amount. &#xA;From a thermodynamics view, they are different. Similar to the reversible and irreversible process in physics, reversible reaction has no energy loss but irreversible reaction has energy loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question:&#xA;Gibbs free energy change is a function depending on concentration for solution and partial pressure for gas. So concentration will obviously affect it.&#xA;Do not confuse it with the &lt;strong&gt;standard&lt;/strong&gt; Gibbs free energy change, which assume that you have a standard condition (1 Pa pressure or 1 mol/L concentration). Since the concentration or partial pressure is a constant (precondition), the standard Gibbs free energy change is independent to concentration or partial pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of your question is not related to transition state. Transition state is part of the chemical kinetics, not thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-09-06T16:01:14.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6160" PostTypeId="2" ParentId="6155" CreationDate="2013-09-06T16:19:25.687" Score="17" Body="&lt;p&gt;There are actually several factors that affect the stabilization of negative charge. The first is what kind of atom centers the charge can be delocalized on and the second factor is how many atom centers the charge can be delocalized on. The first factor is more important than the second. Just think it like quality is more important than quantity. Symmetry is another factor but not very important in your case.&lt;/p&gt;&#xA;&lt;p&gt;Let's look at the ethanoic acid vs phenol case. In the acetate ion, the negative charge is delocalized on two oxygen atom centers, while in the phenoxide ion, the charge is delocalized on one oxygen and three carbon atom centers. Because oxygen is much more electronegative than carbon, the delocalization of negative charge over two oxygens is better than delocalization over one oxygen and three carbons. Quality beats quantity here.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CTXlk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CTXlk.jpg&quot; alt=&quot;Resonance forms of acetate and phenoxide ions&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you have 2,4,6-trichlorophenol, because the strong election withdrawing chlorine atoms is helping carbon center to stabilize negative charge, it is a much stronger acid than phenol itself. This is an example of the inductive effect at work, as opposed to resonance.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="72973" LastEditDate="2021-02-23T14:56:49.240" LastActivityDate="2021-02-23T14:56:49.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6161" PostTypeId="2" ParentId="6134" CreationDate="2013-09-06T16:36:47.007" Score="1" Body="&lt;p&gt;Firstly, as the result, amide ($\ce{R}=\ce{R'CO}$) is much less basic than aniline ($\ce{R}=\ce{Ph}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is carbonyl group is able to stabilize amino lone pair not only by a better resonance (negative charge on oxygen) but also by inductive effect as a strong electro-withdrawing group while phenyl group only have resonance. In a lot of cases the quality of resonance structure is more important than the amount of resonance structures.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="2093" LastEditDate="2013-09-06T21:54:01.973" LastActivityDate="2013-09-06T21:54:01.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6162" PostTypeId="2" ParentId="6144" CreationDate="2013-09-06T16:43:48.270" Score="1" Body="&lt;p&gt;No, it hydrolyzes easily to a carboxylic acid ($\ce{R2COOH}$) and a sulfonic acid ($\ce{R1SO3H}$). $\ce{R1}$ should not be a hydrogen atom by the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is common to use sulfonyl choride or sulfonyl anhydride to react with carboxylic acid to make mixed anhydride which then react with amine or alcohol (or alkoxide) to generate amide or ester. This reaction can be one pot but have to be dry otherwise hydrolysis will reduce the yield a lot. Have done it many times.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="876" LastEditDate="2015-05-01T21:50:41.137" LastActivityDate="2015-05-01T21:50:41.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6163" PostTypeId="2" ParentId="6142" CreationDate="2013-09-06T17:12:11.597" Score="3" Body="&lt;p&gt;I feel you are confused on atomic orbital and molecular orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the excited state of a single sulfur atom, electron do fill 4s first. You are right here. However, it is a different story in molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sulfur trioxide the bonding electrons are filling the molecular orbital, not the atomic orbital such as 4s or 3d. It is forbidden that both 3s and 4s participate any hybridization or formation of molecular orbital. So no matter how low is the energy of 4s, because 3s is already participating, 4s can not be part of the bonding.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-09-06T17:12:11.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6164" PostTypeId="2" ParentId="6122" CreationDate="2013-09-06T17:25:16.043" Score="1" Body="&lt;p&gt;There are acid promoted hydrolysis and base promoted hydrolysis depending on the two different mechanisms. More $\ce{H+}$ do increase the reaction rate of the former but will decrease the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give me your specific reaction so I can figure out what kind of mechanism it undergo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the comparison of sulfuric acid vs hydrochloric acid. Sulfuric acid is available in much higher concentration compare to hydrochloric acid (around 12M maximum). It is preferred when you need very concentrate acid. When using at same concentration, sulfuric acid may be little more better than hydrochloric acid in terms of $\ce{H+}$ amount but will not be big difference because the second deprotonation is not significant under highly acidic condition.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2013-09-06T17:25:16.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6169" PostTypeId="1" AcceptedAnswerId="6170" CreationDate="2013-09-07T13:26:33.153" Score="6" ViewCount="808" Body="&lt;p&gt;I have a Wikipedia account and the people there have a tendency for getting upset over chemical structures that I created using MarvinSketch not conforming to the ACS drawing style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to make MarvinSketch use the ACS drawing style?&lt;/p&gt;&#xA;" OwnerUserId="876" OwnerDisplayName="Brenton Horne" LastEditorUserId="22" LastEditDate="2013-09-07T22:20:21.100" LastActivityDate="2013-09-08T22:01:04.717" Title="How do I get MarvinSketch to use the ACS style of drawing?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6170" PostTypeId="2" ParentId="6169" CreationDate="2013-09-07T20:24:56.113" Score="8" Body="&lt;p&gt;You can find some information on JACS style &lt;a href=&quot;https://en.wikipedia.org/wiki/Wikipedia%3aManual_of_Style_%28chemistry%29/Structure_drawing&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are using Marvin Sketch 5.1, first clean your structure, then got to &lt;code&gt;edit&lt;/code&gt; then &lt;code&gt;format&lt;/code&gt;. Click the &lt;code&gt;Apply changes for all the atoms / bonds&lt;/code&gt; and then &lt;code&gt;Load Style&lt;/code&gt;. Pick J. Am. Chem. Soc and &lt;code&gt;OK&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Marvin Sketch 6 open &lt;code&gt;File&lt;/code&gt; then &lt;code&gt;Document Style&lt;/code&gt; and continue as above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see the documentation &lt;a href=&quot;http://www.chemaxon.com/marvin/help/sketch/sketch-basic.html#styles&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should work&lt;/p&gt;&#xA;" OwnerUserId="1055" OwnerDisplayName="Daniel" LastEditorUserId="1055" LastEditDate="2013-09-08T22:01:04.717" LastActivityDate="2013-09-08T22:01:04.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6171" PostTypeId="2" ParentId="2859" CreationDate="2013-09-08T06:50:57.177" Score="3" Body="&lt;p&gt;Flat coke is less acidic than carbonated coke. I did a lab report on this topic, and found that Carbonated coke had a concentration of around 0.131M whereas flat coke had a concentration of 0.064M. I guess we can assume that the carbonic acid that's dissolved in Coke makes up a large proportion of the acidity.&lt;/p&gt;&#xA;" OwnerUserId="2267" LastActivityDate="2013-09-08T06:50:57.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6172" PostTypeId="1" CreationDate="2013-09-08T10:38:57.893" Score="-1" ViewCount="17136" Body="&lt;p&gt;Does the chemical element neon consist of a molecule? For example, oxygen consist of two atoms together. But is that the same with neon?&lt;/p&gt;&#xA;" OwnerUserId="2268" LastActivityDate="2016-05-17T04:46:06.520" Title="Is neon a molecular element?" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6173" PostTypeId="2" ParentId="6172" CreationDate="2013-09-08T10:46:12.697" Score="5" Body="&lt;p&gt;Since neon is a noble gas, it has its full share of valence electrons, which makes it unlikely to bond with other atoms.  Under certain conditions other noble gases, specifically xenon and krypton, can form compounds in extreme temperatures and pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_gas&quot;&gt;Wikipedia&lt;/a&gt; article on noble gases,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[the noble gases] are all monatomic gases under standard conditions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and, according to the article&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neutral compounds in which helium and neon are involved in chemical bonds have not been formed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastEditorUserId="22" LastEditDate="2013-09-08T10:49:31.997" LastActivityDate="2013-09-08T10:49:31.997" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6174" PostTypeId="1" CreationDate="2013-09-08T12:13:03.430" Score="1" ViewCount="196" Body="&lt;p&gt;Is it possible to predict wether an organic compound will photochemically decompose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to predict the products which it will decompose into?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="7448" LastEditDate="2015-10-09T15:58:28.157" LastActivityDate="2015-10-09T15:58:28.157" Title="Predict photochemical decomposition of organic compound?" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6177" PostTypeId="1" AcceptedAnswerId="6178" CreationDate="2013-09-09T19:05:59.607" Score="3" ViewCount="3841" Body="&lt;p&gt;I recently bought in Italy a bottle of &quot;Alcool puro&quot; which is in fact 95% Alcohol and 5 % Water. So alcohol has a freezing point of -114°C and water a freezing point of 0°C. How can I calculate then the freezing point of the &quot;Alcool puro&quot;?&lt;/p&gt;&#xA;" OwnerUserId="2275" LastEditorUserId="2275" LastEditDate="2016-01-17T19:14:19.683" LastActivityDate="2018-01-15T03:03:09.060" Title="How to get the freeze point of mixed liquids?" Tags="&lt;water&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6178" PostTypeId="2" ParentId="6177" CreationDate="2013-09-10T05:30:55.197" Score="5" Body="&lt;p&gt;In general it is extremely hard to get the freezing point of a mixture purely by calculation. The reason is that the specific molecular interactions of the two substances you have mixed cause pretty much every mixture to need a unique formula for the calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are some ways to get a rough estimate but these estimates mainly work for (linear) hydrocarbons because those are (fairly) simple to generalize. (google for &quot;freezing point hydrocarbon mixture&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, because it is ethanol-water your best way to find out the freezing point is to look up the experimental results of someone else. &lt;a href=&quot;http://www.engineeringtoolbox.com/ethanol-water-d_989.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The engineering toolbox&lt;/a&gt; has a dataset on ethanol-water which you could use to get a reasonable estimate for your case (they don't list 95v%, but they do have the whole range from 0-100v% in steps of 10v%).&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="2182" LastEditDate="2018-01-15T03:03:09.060" LastActivityDate="2018-01-15T03:03:09.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6179" PostTypeId="1" AcceptedAnswerId="6184" CreationDate="2013-09-10T08:37:24.407" Score="3" ViewCount="11005" Body="&lt;p&gt;For example, this reaction: $$\ce{2NaOH + H2SO4 -&amp;gt; Na2SO4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose both concentrations are $1\ \mathrm{mol\ l^{-1}}$ (standard condition). There is $0.02\ \mathrm{mol}$ ($20\ \mathrm{cm^3}$) of $\ce{NaOH}$, and $0.01\ \mathrm{mol}$ ($10\ \mathrm{cm^3}$) of $\ce{H2SO4}$. So neither reactant is the limiting reactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I know one way is to find the standard enthalpy of formation of all components and subtract the ones of products from reactants'. My question is if only experiment data is given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For temperature change of $\Delta T$,&lt;br&gt;&#xA;$\Delta H = mc\Delta T = (0.03)\times(4200)\times\Delta T$,&lt;br&gt;&#xA;where $m$ is the mass of the solution. Since the density should be similar to water $1000\ \mathrm{kg\ m^{-3}}$,&lt;br&gt;&#xA;$20\ \mathrm{cm^3} + 10\ \mathrm{cm^3} = 30\ \mathrm{cm^3} = 0.00003\ \mathrm{m^3}$&lt;br&gt;&#xA;which translates to $0.03\ \mathrm{kg}$. $c$ is just the specific heat capacity of water, it was assumed that the specific heat capacity of the mixture is more or less the same as the water's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the standard one $\Delta_\mathrm{r}H^\circ$ requires division by the amount of substance, but which one – $0.02\ \mathrm{mol}$ or $0.01\ \mathrm{mol}$? What if $\ce{NaOH}$ is $0.021\ \mathrm{mol}$? Or $\ce{H2SO4}$ is $0.011\ \mathrm{mol}$?&lt;/p&gt;&#xA;" OwnerUserId="2276" LastEditorDisplayName="user7951" LastEditDate="2015-07-14T09:08:51.533" LastActivityDate="2015-07-14T09:08:51.533" Title="Calculating standard enthalpy of reaction, which reactant to choose for the amount of substance" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6181" PostTypeId="1" AcceptedAnswerId="6189" CreationDate="2013-09-10T15:57:45.963" Score="2" ViewCount="6756" Body="&lt;p&gt;I found a question in a biochemistry book.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How can osmosis be used to measure molecular weight of macromolecules? Is it applied practically?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A general answer without the usage of many technical terms would be most suitable for me.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="83758" LastEditDate="2020-09-11T20:17:18.907" LastActivityDate="2020-09-11T20:17:18.907" Title="Use of osmosis to measure molecular weight of macromolecules" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;osmosis&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6182" PostTypeId="1" CreationDate="2013-09-10T17:47:27.503" Score="12" ViewCount="5895" Body="&lt;p&gt;$$\ce{H2O~(l) + NaCl~(s) -&amp;gt;[\Delta] Na+~(aq) + Cl^{-}~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When table salt is placed in water, it dissolves due to the polarity of water molecules. When solvation takes place, negatively polar sides of water molecules attach to $\ce{Na+}$ ions, and positively polar sides of water molecules attach to $\ce{Cl-}$ ions. The ions are then carried away by water molecules through diffusion. Since the sodium is now isolated from the chlorine, what keeps it from &lt;a href=&quot;http://www.youtube.com/watch?v=HY7mTCMvpEM&quot; rel=&quot;nofollow&quot;&gt;violently exploding with water&lt;/a&gt; in the following chemical reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na + 2H2O -&amp;gt; 2 NaOH + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since chlorine is poisonous, wouldn't that make drinking salted water extremely lethal?&lt;/p&gt;&#xA;" OwnerUserId="143" LastEditorUserId="16622" LastEditDate="2015-08-02T17:28:49.763" LastActivityDate="2015-08-06T16:06:30.717" Title="Why is it not dangerous to dissolve NaCl?" Tags="&lt;water&gt;&lt;aqueous-solution&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6183" PostTypeId="1" AcceptedAnswerId="57536" CreationDate="2013-09-10T18:46:06.287" Score="3" ViewCount="6447" Body="&lt;p&gt;In our lab, we commonly use a mixture of sulfuric acid and nitric acid to clean borosilicate glassware.  I'd like to clean some PTFE pieces in there, too, but not sure if they will dissolve or not.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any first-hand experience with this, and know whether or not a sulfuric acid/nitric acid mixture dissolve PTFE?&lt;/p&gt;&#xA;" OwnerUserId="1040" LastActivityDate="2016-08-17T18:53:22.963" Title="Does a sulfuric acid/nitric acid mixture dissolve PTFE (ie. teflon)?" Tags="&lt;experimental-chemistry&gt;&lt;cleaning&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6184" PostTypeId="2" ParentId="6179" CreationDate="2013-09-10T19:14:25.293" Score="4" Body="&lt;h2&gt;The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term &quot;Standard&quot; refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases.&#xA;So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How to calculate $\Delta_{\mathrm{r}} H$?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereby the extent of reaction $\xi$ is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively.&#xA;Consider the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{\nu_{A} A + \nu_{B} B &amp;lt;=&amp;gt; \nu_{C} C + \nu_{D} D}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= &quot;conversion of one formula unit&quot;).&#xA;So, using the differential form of your calorimetric formula (under the assumptions you made) and &quot;dividing&quot; it by $\mathrm{d} \xi$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mathrm{d} H = m C \mathrm{d}T \\&#xA;  \frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\&#xA;  \Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity.&#xA;Considering finite instead of infinitesimal changes of $n_{i}$ leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the target compound is to be formed by reacting the (pure) elements&#xA;it consists of, whereby each element is expected to be in its most&#xA;stable modification for the given temperature.&lt;/li&gt;&#xA;&lt;li&gt;The reaction equation is to be formulated in such a way that $1 \,&#xA;   \text{mol}$ of the compound in question is formed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-05-17T17:33:20.370" LastActivityDate="2014-05-17T17:33:20.370" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6185" PostTypeId="2" ParentId="6182" CreationDate="2013-09-10T19:14:34.620" Score="20" Body="&lt;p&gt;[A &lt;em&gt;really-good&lt;/em&gt; answer to this question should include quantitative evidence, but here is my hand-waving version...]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvated sodium ions ($\ce{Na+}$) aren't all that reactive.  Surrounded by a lot of polar water molecules, they're actually quite stable.  What you're thinking of is &lt;strong&gt;solid&lt;/strong&gt; sodium (Na) which will &quot;react violently with water&quot; until it's all dissolved - dissolved that is, into stable sodium ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same kind of logic applies to chlorine.  Solvated chloride ions ($\ce{Cl-}$) are stable and not nearly as poisonous as chlorine gas. (Note:  &lt;em&gt;everything&lt;/em&gt; is poisonous at some level, &lt;em&gt;i.e.&lt;/em&gt; &quot;the poison is in the dose&quot;.  In saying chlorine gas is more poison, that really just means it is a more 'potent' poison:  it will kill at lower doses compared to chloride ions.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way to think of it: &#xA;Solid sodium and chlorine gas are at high energy states.  Since they can reach much lower energy states by ionzing in the presence  with water, the WILL break down and ionize in the presence of water  &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="84" LastEditDate="2013-09-11T02:57:05.170" LastActivityDate="2013-09-11T02:57:05.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6187" PostTypeId="2" ParentId="6183" CreationDate="2013-09-10T21:12:05.647" Score="4" Body="&lt;p&gt;I can't claim to have any personal experience with that sulfuric acid/nitric acid cleaning mixture specifically, but, as far as I'm aware, PTFE is generally inert towards acids. As a testament to that, it is apparently the material used for storing &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot; rel=&quot;nofollow&quot;&gt;fluoroantimonic acid&lt;/a&gt;. I found a &lt;a href=&quot;http://www.zeusinc.com/technicalservices/technicalbulletins/chemicalresistanceofpolymers/chemicalresistancechartptfe.aspx&quot; rel=&quot;nofollow&quot;&gt;chemical resistance chart&lt;/a&gt; confirming its stability against strong acids. It is evidently resistant to concentrated sulfuric and nitric acids, as well as aqua regia (a concentrated mixture of nitric acid and hydrochloric acid). I have used the last of those for cleaning PTFE lab equipment before with no problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all that, it would be wise to consult the handling and safety data for your particular labware to be certain.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-09-10T21:12:05.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6189" PostTypeId="2" ParentId="6181" CreationDate="2013-09-11T03:16:39.043" Score="4" Body="&lt;p&gt;It has to do with the fact, that osmotic pressure of any solution is dependent only on the relative number of particles in the solution (in case of dilute solutions), irrespective of their nature. Hence osmotic pressure is called a colligative property.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, you take a certain known weight of the substance you want to calculate the molecular mass of, prepare a solution and measure the osmotic pressure. Then you compare this to the known osmotic pressure of standard solution which tells you the relative number of particles required to  create such a pressure. From this and the knowledge of the weight you used to prepare the solution, you can get the required molecular weight (weight of Avogadro's number of particles). In practice, certain complications, like dissociation and association might complicate the matter a bit.&lt;/p&gt;&#xA;&lt;p&gt;This method is widely used to calculate the molecular weight (especially) of polymers. See &lt;a href=&quot;https://www.qsstudy.com/chemistry/determination-of-osmotic-pressure-by-berkeley-and-hartleys-method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Berkeley and Hartley method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="83758" LastEditDate="2020-09-11T20:15:49.623" LastActivityDate="2020-09-11T20:15:49.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6191" PostTypeId="1" CreationDate="2013-09-11T12:05:55.813" Score="1" ViewCount="179" Body="&lt;p&gt;I'm going to try out the transesterification of propylene carbonate and methanol. I know that the mechanism is as follows : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQvWc.png&quot; alt=&quot;enter Transesterification&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-What are the products yielded after the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-And which catalyst should I use? (reaction is happening at normal air pressure and at room temperature)&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="186" LastEditDate="2015-10-02T16:37:34.220" LastActivityDate="2017-04-08T16:48:59.383" Title="Transesterification of propylene carbonate and methanol" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6192" PostTypeId="2" ParentId="6191" CreationDate="2013-09-11T15:08:03.163" Score="3" Body="&lt;p&gt;I know it is used to obtain  dimethyl carbonate (DMC) I think the others products should vary from which catalyst you choose. Have a look to &lt;em&gt;Transesterification of propylene carbonate with methanol using $\ce{MgAlCO3}$&#xA;hydrotalcite as solid base catalyst&lt;/em&gt;: &lt;em&gt;&lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/10172/1/IJCA%2049A%289%29%201182-1188.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Indian Journal of Chemistry&#xA;Vol. 49A, September 2010, pp. 1182-1188&lt;/a&gt;&lt;/em&gt; to choose the catalyst. I think there are two little typos in Scheme 2 where the mechanism is shown: miss $\ce{^-}$ from $\ce{CH_3O^-}$ (first step) and miss $\ce{C}$ from one part ($\ce{CH_3CO}$) of DMC formula (second step) but is a good article. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="9961" LastEditDate="2017-04-08T16:48:59.383" LastActivityDate="2017-04-08T16:48:59.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6193" PostTypeId="1" AcceptedAnswerId="6195" CreationDate="2013-09-11T19:19:06.017" Score="2" ViewCount="126" Body="&lt;p&gt;I am not a chemist, I develop software. I'm hoping for a general explanation I can understand in order to assist me with a project I'm working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Customers viewing a product online such as peroxide see &lt;strong&gt;97% (dry wt.), wet with 25% water&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? What would you expect the product to be in the form of? A powder, liquid/gel, solid in water? Any specific type of container?&lt;/p&gt;&#xA;" OwnerUserId="2285" LastEditorUserId="4945" LastEditDate="2014-10-24T10:39:21.460" LastActivityDate="2014-10-24T20:23:59.167" Title="What does the commercial chemical composition notation &quot;Percent dry, wet with water&quot;mean?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6194" PostTypeId="1" CreationDate="2013-09-11T19:46:40.730" Score="3" ViewCount="952" Body="&lt;p&gt;The Lugol solution is a mixture of $\ce{(K+, I- )}$ and $\ce{I2}$ such that $\ce{I3-}$ are formed by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I- + I2 -&amp;gt; I3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the brown color comes from $\ce{I2}$, but in this solution one have $\ce{I3-}$ not $\ce{I2}$, why is it brown?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I mix Lugol with $\ce{H2O2}$ without the presence of $\ce{H3O+}$ there is a variation in the conductivity of the mixture but the brown color doesn't disappear. If I add $\ce{H3O+}$ the solution becomes transparent after few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that either $\ce{I2}$ is reduced by $\ce{H2O2}$ (hence the disappearing of the brown color) or a redox reaction between $\ce{H2O2}$ and $\ce{I-}$, but in the later case I don't understant why the brown color disapear instead of appearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you enlighten me please, thanks.&lt;/p&gt;&#xA;" OwnerUserId="2286" LastEditorDisplayName="user7951" LastEditDate="2015-09-04T18:58:49.020" LastActivityDate="2015-09-04T19:58:08.770" Title="Which reaction between Lugol and Hydrogen peroxide?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6195" PostTypeId="2" ParentId="6193" CreationDate="2013-09-11T21:22:28.237" Score="2" Body="&lt;p&gt;I believe there is no general rule.&#xA;In your case (assuming you mean &lt;a href=&quot;https://us.vwr.com/store/catalog/product.jsp?product_id=9898342&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;) it is a solid. materials can have 25% and even more water and still be solids.&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2013-09-11T21:22:28.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6196" PostTypeId="2" ParentId="6194" CreationDate="2013-09-11T21:28:39.490" Score="1" Body="&lt;p&gt;I'll try to figure out the color of every compound separately, but there is a strong interaction between them and the different parameters so this could be quite tricky:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{I+ , IO3-}$ don't absorb in the visible region&lt;/li&gt;&#xA;&lt;li&gt;$\ce{I3-}$  I think should be &lt;a href=&quot;http://rstarchem.weebly.com/experiment-ana2.html&quot; rel=&quot;nofollow&quot;&gt;red&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\ce{I2}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine&quot; rel=&quot;nofollow&quot;&gt;blue&lt;/a&gt; if solid but has different colors in solution depending from the type of solvent and $\ce{H+}$ concentration see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01931a005&quot; rel=&quot;nofollow&quot;&gt;Hildebrand (1909)&lt;/a&gt;. In polar solvent is brown, however I don't exclude this is due to the interaction with $\ce{I3-}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because there is an equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{I_2 + I^- &amp;lt;=&amp;gt; I_3^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is always a little bit of iodine in the solution and this could cause the brown color (however keep in mind that is $I$ is not so soluble in water).&#xA;Adding $H_2O_2$ should oxidize the iodide in this way:&#xA;$$ \ce{2I^- + 2H^+ + H_2O_2 -&amp;gt; I_2 + 2H_2O}$$&#xA;It occurs the reverse reaction too but should be negligible. In this case is probably a cinetic matter,  you can make the solution transparent reducing the colored compounds. If you have a look to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_clock_reaction&quot; rel=&quot;nofollow&quot;&gt;iodine clock reactions&lt;/a&gt; iodine can be reduced by bisulfite: &#xA;$$ \ce{I_2 + HSO_3^- + H2O -&amp;gt; 2I^- + HSO4^- +2H^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I exclude that a reduction could occur in an oxidizing ambient like your. So my hypothesis are two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FIRST&lt;/strong&gt;: You are oxidizing the colored species. Adding $\ce{H2SO4}$ and $\ce{H2O2}$ you could obtain one of the most oxidant agent, Peroxymonosulfuric acid ($\ce{H2SO5}$),  see &lt;a href=&quot;http://en.wikipedia.org/wiki/Piranha_solution&quot; rel=&quot;nofollow&quot;&gt;Piranha solution&lt;/a&gt; (&lt;strong&gt;I think this could be very dangerous please take all the precautions!&lt;/strong&gt;). But now I don't think that the equation that if I wrote before (see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01352a006?journalCode=jacsat&quot; rel=&quot;nofollow&quot;&gt;J. Am. Chem. Soc., 1931, 53 (1), pp 38–44&lt;/a&gt; and&#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00469a014&quot; rel=&quot;nofollow&quot;&gt;J. Am. Chem. Soc., 1978, 100 (1), pp 87–91&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{I_2 + 5H2O2 -&amp;gt; 2H^+ + 2IO3^- + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could occur. I know iodic acid $\ce{HIO3}$can be produced oxidating $\ce{I2}$ but I never found a reaction involving $\ce{H2SO4}$ and hydrogen peroxide. It could be a more complex reaction maybe involving the others ions. I was thinking about potassium hydrogen iodate ($\ce{KH(IO3)2}$). I try to figure out a stechiometric reaction but is quite an hell!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECOND&lt;/strong&gt;: simply you are  modifying the interaction between the solvent and $\ce{I2}$ and $\ce{I3-}$ avoiding the  light absorption to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry this is now quite not an answer, but I hope it helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine color reaction are wonderful (but very complex)!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="10185" LastEditDate="2015-09-04T19:58:08.770" LastActivityDate="2015-09-04T19:58:08.770" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6197" PostTypeId="1" CreationDate="2013-09-12T01:04:44.590" Score="0" ViewCount="820" Body="&lt;p&gt;I need to find the ratio of a substance as it converts from a bicarbonate to a carbonate:&lt;/p&gt;&#xA;&lt;p&gt;ex:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(HCO3)2 -&amp;gt; FeCO3 + CO2 + H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(HCO3)2} / \ce{FeCO3}$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$3/2$&lt;/span&gt; *using molar masses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But what if the substances aren't  the same number of moles?&lt;/p&gt;&#xA;&lt;p&gt;ex:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2Fe(HCO3)2 -&amp;gt; FeCO3 + CO2 + H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Would the ratio be:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2Fe(HCO3)2 / FeCO3} = x$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;or still:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(HCO3)2} / \ce{FeCO3} = 3/2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How do moles factor into this equation and why?&lt;/p&gt;&#xA;" OwnerUserId="2287" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-09-12T11:52:41.593" Title="Law of Definite Proportion: Do the numbers indicating moles count?" Tags="&lt;homework&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6198" PostTypeId="1" AcceptedAnswerId="6200" CreationDate="2013-09-12T01:59:36.363" Score="8" ViewCount="431" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e2sac.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e2sac.png&quot; alt=&quot;Diazotisation of 1,2-diaminobenzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, &lt;span class=&quot;math-container&quot;&gt;$\ce{NaNO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; are the reagents used to generate an aryldiazonium chloride from an aniline. However, in this compound there are two amine groups adjacent to each other, so it seems unlikely that a diazonium chloride would form. Is it possible for the two amine groups to react during the reaction and form some cyclic product?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2019-03-08T00:40:39.083" LastActivityDate="2019-03-10T23:10:07.363" Title="What happens in the attempted diazotisation of 1,2-diaminobenzene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6200" PostTypeId="2" ParentId="6198" CreationDate="2013-09-12T11:37:10.223" Score="10" Body="&lt;p&gt;Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ue6UV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ue6UV.png&quot; alt=&quot;Diazotisation of one amine in 1,2-diaminobenzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in addition to substitution reactions by loss of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt;, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are &lt;a href=&quot;http://en.wikipedia.org/wiki/Diazonium_compound#Diazo_coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;diazo couplings&lt;/a&gt;, where the nucleophile is an activated arene. However, other nucleophiles, especially intramolecular nucleophiles, can attack diazonium cations. The other amine is poised to do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/goPWc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/goPWc.png&quot; alt=&quot;Cyclisation to form benzotriazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2019-03-10T23:10:07.363" LastActivityDate="2019-03-10T23:10:07.363" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="6201" PostTypeId="2" ParentId="6197" CreationDate="2013-09-12T11:52:41.593" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The stoichiometric coefficients are important. They make sure you have a valid relationship between reactants and products that obeys the Law of Conservation of Mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Jerry hints at in his comment, there is another fundamental law that you need to consider: &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_conservation_of_mass&quot; rel=&quot;nofollow&quot;&gt;the Law of Conservation of Mass&lt;/a&gt;. Matter cannot be created or destroyed (outside of nuclear reactions and relativist physics). In chemistry this means you need the same number of atoms of each element on both sides of the equation. You cannot create additional atoms from nothing, nor can extra atoms of something vanish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider your first equation:&#xA;$$\ce{Fe(HCO3)2 -&amp;gt; FeCO3 + CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is balanced - it obeys the law of conservation of mass. Each side of the equation has one $\ce{Fe}$, two $\ce{H}$, two $\ce{C}$, and six $\ce{O}$ atoms. &lt;strong&gt;This is a valid chemical equation representing a real reaction relationship.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second equation is not balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe(HCO3)2 -&amp;gt; FeCO3 + CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are now twice as many atoms of each element on the reactant side of the equation as on the product side. What happens to this extra atoms? They cannot vanish! This is not a valid chemical relationship. Balancing this equation generates a scalar multiple of your first equation, which will yield the same ratio of iron (ii) bicarbonate to iron (ii) carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe(HCO3)2 -&amp;gt; 2FeCO3 + 2CO2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_definite_proportions&quot; rel=&quot;nofollow&quot;&gt;Law of Definite Proportions&lt;/a&gt; tells you that chemical compounds always contain the same elemental compositions by mass. I am not sure how that is relevant to the ratio of product to reactant in your stoichiometry problem. In other words, it tells you that iron (ii) carbonate is always $\ce{FeCO3}$ and never $\ce{Fe3(CO3)4}$ or $\ce{Fe2(CO3)3}$ or $\ce{FeCO4}$ or anything else. Those other formulas may refer to real compounds (the first two are real), but they are not iron (ii) carbonate. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-12T11:52:41.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6203" PostTypeId="1" CreationDate="2013-09-12T16:46:37.533" Score="1" ViewCount="693" Body="&lt;p&gt;I want to remove carbonate ions from dissolved potassium carbonate (potash) by letting an organic compound react with the carbonate ions. I don't have a very good knowledge of organic chemistry, but are there certain types of organic molecules that can react with these carbonate ions to form new compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, are there reactions in which the carbon atom from the carbonate ion can be inserted into the main carbon chain of another organic compound (making, for example, propane from ethane)?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="16683" LastEditDate="2015-10-09T16:18:28.213" LastActivityDate="2015-10-09T16:18:28.213" Title="Organic compounds that react with carbonate ions" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6204" PostTypeId="1" AcceptedAnswerId="6205" CreationDate="2013-09-12T18:07:48.317" Score="3" ViewCount="2242" Body="&lt;p&gt;Say you have $\ce{NaOH}$ and $\ce{CH3COOH}$ and want to make a buffer.&#xA;Theory teaches 6 mols of $\ce{CH3COOH}$ and half, 3 mols, of $\ce{NaOH}$ would be needed.&#xA;Then you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$ and 3 mols of $\ce{CH3COOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what if you mixed 3 mols of $\ce{NaOH}$ with 12 mols of $\ce{CH3COOH}$? According to my calculations, you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$, and 9 mols of $\ce{CH3COOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter example there is a clear imbalance in the salt:acid ratio. However, won't this still act as a buffer, while not as strong as an equally concentrated 1:1 buffer? So it will work, it's just not ideal?&lt;/p&gt;&#xA;" OwnerUserId="1344" LastEditorUserId="1878" LastEditDate="2013-09-12T19:49:22.630" LastActivityDate="2020-06-08T09:54:54.913" Title="Why do buffers need to be composed of equal amounts of the acid and salt?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6205" PostTypeId="2" ParentId="6204" CreationDate="2013-09-12T20:52:58.807" Score="3" Body="&lt;p&gt;They don't &lt;em&gt;need&lt;/em&gt; to be comprised of equal amounts.  However, when you are near the 50:50 point, the buffer is &quot;unbiased&quot; (a contrived term), and it can equally accommodate assaults to its pH going in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, as in your first example, you're sitting at equal concentrations of sodium acetate and acetic acid, dropping in a 1 mole of either NaOH or HCl won't overwhelm the buffer capacity (pH will change, but not as quickly as it would when you exceed capacity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, in your second example, if you add 4 moles of NaOH you'll still be buffered, (acetate will rise to 7 moles, acid fall to 5), but if you add 4 moles of HCl, you'll protonate all the acetate back to acetic acid and the buffering effect will collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while they don't need to be at the 50:50 point, it's generally a good idea to stay in that order of magnitude (which is why recommended buffering ranges are roughly 1 pH unit); 10:90 is iffy (used appropriately, it works fine), but 1:99 is nearly pointless.  The pKa is the theoretical pH when at a 50:50 mix, so different chemicals are used for different pH buffers (&lt;a href=&quot;http://en.wikipedia.org/wiki/Good%27s_buffers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Good's buffers&lt;/a&gt; shown below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4l8eq.jpg&quot; alt=&quot;Good&amp;#39;s Buffer table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11" LastActivityDate="2013-09-12T20:52:58.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6208" PostTypeId="1" CreationDate="2013-09-13T21:05:15.577" Score="8" ViewCount="38065" Body="&lt;p&gt;I was under the impression that if I put a semi full bottle of vinegar in the freezer the water would freeze but the vinegar would not until a lower temp.  Therefore I could remove the bottle after the water freezes and pour off the vinegar. Apparently the water and vinegar doesn't separate nicely before freezing so this didn't work so well.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen boiling suggested but I'm afraid I'll lose some of the vinegar in the process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the simplest way to concentrate the vinegar at home?    &lt;/p&gt;&#xA;" OwnerUserId="2301" LastActivityDate="2018-05-20T16:54:18.387" Title="How can I separate the vinegar from a 5% white vinegar to make a more concentrated solution." Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6209" PostTypeId="2" ParentId="6208" CreationDate="2013-09-14T03:16:12.490" Score="7" Body="&lt;p&gt;If you have access to a still you can distil the vinegar solution, however this can only go so far. You may find &lt;a href=&quot;https://en.wikipedia.org/wiki/Azeotrope#Zeotropic_mixtures&quot;&gt;this section of the wikipedia article on azeotropy&lt;/a&gt; edifying - to summarise, adding ethyl acetate to a vinegar solution generates a water:ethyl acetate azetrope with a depressed boiling point that makes distillation more effective, however this may be beyond the pale for a home experiment and would obviously render the vinegar non-food-safe.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-09-14T03:16:12.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6210" PostTypeId="1" AcceptedAnswerId="6219" CreationDate="2013-09-14T08:19:28.790" Score="5" ViewCount="5821" Body="&lt;p&gt;I'm interested in making a liquid freeze spray, for a fun science project. My only experience with this comes in spraying inverting canned air dusters, which spray a -60F mist/liquid (liquid if you spray while the can is inverted). I'd like to make a spray that I can store and use. And if it can't be stored more than a few hours, then hopefully something easy to make on the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry understanding is high-school level basic, so I do not have a working knowledge of chemistry. On one end of the spectrum, I know there is liquid nitrogen, but it's expensive and complex to store correctly (high-pressure canisters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_cooling_baths&quot;&gt;this list of cooling bath configurations&lt;/a&gt;. Could someone point out which configuration can be stored long term (months), stored cheaply (e.g. a plastic or thin metal thermos; not a heavy high-pressure fire-extinguisher container), and be a liquid when &quot;sprayed&quot; (assuming keeping conditions in our pressure and room temperature)? I'd like a spray temperature of less than -60F, so ice and water won't cut it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of going with dry ice and &gt; 90% concentration ethanol. I have two questions regarding this approach and a random question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a solution of half dry ice and half 99% ethanol, how can you keep it at a low temperature? Won't the solution slowly return to room temperature? My canned air dusters always spray out at the -60F; I'd like something like that if possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can this solution be stored in a cheap plastic or thin metal container? Will there be some pressure buildup from the dry ice that will eventually crack or explode the container? Assuming a 10 oz container with a solution of half dry ice and half &gt; 90% ethanol.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are canned air dusters always able to spray their liquid solution (if inverted while sprayed) at -60F, even if they are stored for months at room temperature? Is it because the contents of the can are under pressure and that this pressure makes the liquid cold?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="750" LastActivityDate="2017-03-04T14:52:29.203" Title="How would one go about making a cheap freeze spray?" Tags="&lt;home-experiment&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6212" PostTypeId="1" AcceptedAnswerId="7302" CreationDate="2013-09-14T08:51:56.593" Score="6" ViewCount="18282" Body="&lt;p&gt;&quot;Magnesium has a very slight reaction with cold water, but burns in steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very clean coil of magnesium dropped into cold water eventually gets covered in small bubbles of hydrogen which float it to the surface. Magnesium hydroxide is formed as a very thin layer on the magnesium and this tends to stop the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium burns in steam with its typical white flame to produce white magnesium oxide and hydrogen.&quot; - From Chemguide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May I know why the reactions of Magnesium in steam and cold water differ?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2014-01-10T08:38:46.617" Title="Why do the reactions of Magnesium in steam and cold water differ?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6216" PostTypeId="1" CreationDate="2013-09-14T09:40:55.827" Score="1" ViewCount="18911" Body="&lt;p&gt;A solution prepared by dissolving $2.8 g$ of lime, $\ce{CaO}$ in enough water to make $1.00 l$ of lime water ($\ce{Ca(OH)_{2(aq)}}$). If solubility of $\ce{Ca(OH)2}$ in water is $1.48 g$. The $p\ce H$ of the solution obtained will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[$\log 2 = 0.3$, Atomic masses are $\ce Ca$ = $40$ , $\ce O$ = $16$, $\ce H$ = $1$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note : $\ce {Ca(OH)2}$ is considered to be a weak base because  of its low solubility but it is 100% dissociated]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) $12.3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) $12.6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) $1.3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D) $13$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;strong&gt;As $2.8g$ of $\ce {CaO}$ dissolves in $1l$ water the molarity of the solution is 0.05M. So,the concentration of OH- ions in the solution is $0.1M$.So $p\ce {OH}$ is $1$ and hence $p\ce H$ is $13$. But, the answer is given to be $12.6$.What am I doing wrong?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="1431" LastEditDate="2013-10-11T17:53:46.413" LastActivityDate="2013-11-10T18:16:11.103" Title="Calculating the pH of a solution of Ca(OH)2" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6218" PostTypeId="1" AcceptedAnswerId="6227" CreationDate="2013-09-14T14:51:49.523" Score="1" ViewCount="5719" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't get the right answer. I did this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\text{total}} = 5.4\;\mathrm{kg} =5400\;\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=420\;\mathrm{cm^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{\ce{Cu}} = 8.96\;\mathrm{g} ~\mathrm{cm^-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V\times{\mathrm{\rho_{\ce{Cu}}}} = 3763.2\;\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m_{\ce{Cu}}}{m_{\text{total}}}=69.9\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me how to approach this problem?&lt;/p&gt;&#xA;" OwnerUserId="2306" LastEditorUserId="5026" LastEditDate="2018-02-25T02:38:12.763" LastActivityDate="2018-02-25T02:43:34.470" Title="Find the number of copper spheres in a box" Tags="&lt;physical-chemistry&gt;&lt;density&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6219" PostTypeId="2" ParentId="6210" CreationDate="2013-09-14T17:52:51.487" Score="9" Body="&lt;p&gt;The reason the 'canned air' dusters and other aerosol-based things like anti-perspirant cans produce a cryogenic effect is from the decompression involved. They are stored at room temperature, but the can is pressurised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two effects that happen when you spray out some aerosol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, by the ideal gas law, $PV=nRT$, when you release the pressure (i.e. lower $P$), the temperature ($T$) also drops. (The increase in volume $V$ mostly compensates for this, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the contents of the can are stored under enough pressure to keep at least one of the parts of the mixture in a liquid state. When you release the pressure by spraying it out, the liquid boils and evaporates, and takes energy in the form of heat to do so. You'll notice that the contents of a canned air duster, if you hold them upside-down when you spray, include more than just air -- there's also a liquid in there which is designed to be mostly inert to electronics and to quickly evaporate just as described above when the pressure inside the can drops, maintaining the pressure needed to produce a gas jet. If you spray for an extended period of time, you'll normally notice a boiling sound coming from inside the can, and the can getting cold, as that liquid evaporates. (As a safety note, these substances are often flammable, so don't use these air dusters near naked flames or ignition sources unless you want fire.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people don't have the equipment needed to produce this sort of thing at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the cooling baths in that list are designed to be left as-is for months at a time -- they're designed to be made up as and when needed, with occasional addition of liquid nitrogen or dry ice to keep them cool. While you &lt;em&gt;could&lt;/em&gt; produce a cold pressurised spray on the spot using dry ice and ethanol, there's a high probability that any sort of robust switched-release mechanism could make the vessel you're using explode under the pressure of sublimating carbon dioxide. Also, unless you relied on a pressure source other than the carbon dioxide itself, your spray probably wouldn't have time to get very cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I don't recommend it.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-09-14T17:52:51.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6220" PostTypeId="1" CreationDate="2013-09-15T00:00:02.187" Score="4" ViewCount="429" Body="&lt;p&gt;I have to do research on a drug called Amifostine: 2-(3-aminopropylamino)ethylsulfanyl&#xA;phosphonic acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jx2gt.png&quot; alt=&quot;Structure of Amifostine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm supposed to say how the organic reaction works inside the body.  I understand that it's hydrolyzed by an enzyme into its active form; 2-((aminopropyl)amino)ethanethiol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nSCZw.png&quot; alt=&quot;2-((aminopropyl)amino)ethanethiol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get what happens to the phosphonic acid functional group. Does the water break apart the &lt;span class=&quot;math-container&quot;&gt;$\ce{S-P}$&lt;/span&gt; bond because it is a weak bond? And what does the acid group become? &lt;span class=&quot;math-container&quot;&gt;$\ce{H3PO4}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2308" LastEditorUserId="83758" LastEditDate="2020-06-06T11:05:44.107" LastActivityDate="2020-06-06T11:05:44.107" Title="What is the reaction mechanism for the hydrolysis of Amifostine?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6221" PostTypeId="1" CreationDate="2013-09-15T06:33:06.513" Score="2" ViewCount="468" Body="&lt;p&gt;I want to extract the $\ce{KI}$ component of the Lugol solution, is there a method to do it? If not, how can I make a $\ce{KI}$ solution? &lt;/p&gt;&#xA;" OwnerUserId="2286" LastEditorDisplayName="user7951" LastEditDate="2015-09-20T16:06:42.407" LastActivityDate="2015-09-20T16:10:10.077" Title="Can we extract KI from Lugol solution?" Tags="&lt;aqueous-solution&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6222" PostTypeId="2" ParentId="6221" CreationDate="2013-09-15T12:23:23.193" Score="2" Body="&lt;p&gt;As you know you have to remove the others species like $\ce{I2}$ and $\ce{I3-}$. $\ce{I3-}$ I'm quite sure it exists only in solution so simply evaporating the water from Lugol 's solution you should eliminate it. To remove completely $\ce{I2}$ you have to do some others steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KI}$ is produced industrially by treating iodine with $\ce{KOH}$.(see Wikipedia for &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_iodide#cite_note-Ullmann-3&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{3I2 + 6KOH -&amp;gt; 5KI + KIO3 + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the simple way is to treat Lugol's solution with $\ce{KOH}$ make an ice bath to precipitate $\ce{KIO3}$, filter the solution to remove it and than evaporate water. &#xA;You can also try to remove iodine by means of active carbon I know this method is used to purify water see &lt;a href=&quot;http://www.activated-carbon-manufacturer.com/press-release/company-news/111-the-properties-of-activated-carbon.html&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt; and &lt;a href=&quot;http://www.lenntech.com/periodic/water/iodine/iodine-and-water.htm#ixzz2exjruJNl&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;.&#xA;You can try evaporating the Lugol's solution and re-crystalize it and than eventually  try different solvents to solubilize one of the compounds and leave the other as precipitate but if think is quite difficult to find a better solvent than  water (as you know iodine is very insoluble in water instead $\ce{KI}$ is very soluble).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make the $\ce{KI}$ from your self here there are two Home experience:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://wiki.answers.com/Q/How_do_you_extract_potassium_iodide_from_salt&quot; rel=&quot;nofollow&quot;&gt;extract potassium iodide from salt&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://pharamceuticals.wonderhowto.com/how-to/make-potassium-iodide-ki-from-elemental-iodine-i-and-potassium-hydroxide-koh-0126105/&quot; rel=&quot;nofollow&quot;&gt;make $\ce{KI}$ from $\ce{I2}$ and $\ce{KOH}$&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user7951" LastEditDate="2015-09-20T16:10:10.077" LastActivityDate="2015-09-20T16:10:10.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6224" PostTypeId="2" ParentId="6208" CreationDate="2013-09-15T18:05:44.513" Score="5" Body="&lt;p&gt;Yes,you are right. What you are talking about is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_freezing&quot; rel=&quot;noreferrer&quot;&gt;Fractional Freezing&lt;/a&gt;. With acetic acid at low concentration you will have ice and a solution of acetic acid more and more concentrated  until you reach the eutectic point. The max concentration you will have is about about 58%. If your freezer go to -18°C you can reach only 38%. Take a look to this &lt;a href=&quot;http://ijset.com/ijset/publication/v2s4/paper5.pdf&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2013-09-15T18:05:44.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6225" PostTypeId="1" AcceptedAnswerId="6226" CreationDate="2013-09-15T19:27:27.740" Score="12" ViewCount="5458" Body="&lt;p&gt;I am currently studying stoichiometry (high school level) and we came across the following equation in school:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2 + 1O2 -&amp;gt; 2H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is clear to me. We learned in school that reactions in chemistry only happen if there are enough atoms/molecules to react (so every molecule/atom can react), therefore the following reaction won't do anything:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{1H2 + 1O2 -&amp;gt; nothing happens}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now my first question: Why does the following not happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{1H2 + 1O2 -&amp;gt; 1H2O + 1O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as a follow up question, why does the following not work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2 + O2 -&amp;gt; H2O + 1/2O2 (Left) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I do wrong? (maybe I simply misunderstood our teacher)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any explanations for this? I'm especially interested in the answer of my example with $1$ mole of $\ce{H2}$ and $1$ mole of $\ce{O2}$. According to our teacher nothing will happen at all. I would assume that at least all $\ce{H2}$ molecules react with half of the oxygen molecules to $\ce{H2O}$ - leaving half a mole of $\ce{O2}$ left over.&#xA;(otherwise this would mean that all molecules &quot;know&quot; each other so they are &quot;aware&quot; that not all molecules/atoms can react) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help me? Or point out what I am doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2310" LastEditorUserId="41556" LastEditDate="2017-04-04T04:30:48.100" LastActivityDate="2017-04-04T04:30:48.100" Title="Can 1 mole of H2 react partially with one mole of O2?" Tags="&lt;water&gt;&lt;hydrogen&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6226" PostTypeId="2" ParentId="6225" CreationDate="2013-09-15T20:34:22.333" Score="22" Body="&lt;p&gt;As you have said, you are studying stoichiometry at &lt;strong&gt;High School Level&lt;/strong&gt;. From this I can guess, that you have probably not studied the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Limiting_reagent&quot; rel=&quot;noreferrer&quot;&gt;concept of Limiting Reagent&lt;/a&gt;&lt;/strong&gt; yet.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;strong&gt;What is Limiting Reagent?&lt;/strong&gt;&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a chemical reaction, the limiting reagent, also known as the &amp;quot;limiting reactant&amp;quot;, is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As per your question, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; is limiting reagent. Hence it will get totally consumed when reaction happens and 1/2 Moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; will be left.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1 mole of H2 and 1 mole of O2 won't make 1 mole of H2O and 1/2 mole O2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;1 mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; and 1 mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; &lt;strong&gt;will make&lt;/strong&gt; 1 mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; and 1/2 mole &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; will be left.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2 + O2 -&amp;gt; H2O + 1/2O2 (Left) }$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And here is an advice for you. If your teacher is saying that this will not happen at all, tell him to be sure that reactant in reaction are single &lt;strong&gt;molecules&lt;/strong&gt;, not 1 &lt;strong&gt;mole&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Read &lt;a href=&quot;http://en.wikipedia.org/wiki/Limiting_reagent&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; to be more clear on concept of limiting reagent.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-12-17T09:38:58.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6227" PostTypeId="2" ParentId="6218" CreationDate="2013-09-16T06:56:50.880" Score="1" Body="&lt;p&gt;Here is your formulation :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic. &lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="5026" LastEditDate="2018-02-25T02:43:34.470" LastActivityDate="2018-02-25T02:43:34.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6229" PostTypeId="1" CreationDate="2013-09-16T11:05:53.617" Score="1" ViewCount="9209" Body="&lt;p&gt;In the reaction where $\ce{Zn}$ reacts with $\ce{HCl}$ to give off Hydrogen gas &amp;amp; Zinc Chloride solution, does the zinc turn a dull black color after a while? If so, why?&lt;/p&gt;&#xA;" OwnerUserId="2314" LastEditorUserId="22" LastEditDate="2013-09-16T13:23:01.777" LastActivityDate="2017-04-25T23:52:50.453" Title="Does Zinc become black &amp; dull after reacting with dil. HCl?" Tags="&lt;acid-base&gt;&lt;metal&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6231" PostTypeId="1" CreationDate="2013-09-16T23:30:49.427" Score="1" ViewCount="782" Body="&lt;p&gt;I am advising a student on her senior math thesis.  She would like to involve chemistry somehow (her other major).  I know no chemistry.  I have read in several places that representation theory features prominently in some areas of chemistry.  When I try to read anything online I am not sure how much chemistry one needs to assume.  So, if my student is in her senior year of a chemistry major, are there any places where she can start to look for a project in group theory and chemistry?&lt;/p&gt;&#xA;" OwnerUserId="2292" LastEditorUserId="73" LastEditDate="2013-09-17T00:58:59.997" LastActivityDate="2013-09-19T00:33:54.903" Title="Group Theory and Chemistry for Undergraduate Thesis" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6233" PostTypeId="1" AcceptedAnswerId="6235" CreationDate="2013-09-17T10:08:30.673" Score="19" ViewCount="2075" Body="&lt;p&gt;There are elements like neptunium and plutonium in the periodic table. Did their discovery have anything to do with Neptune and Pluto? Or are they randomly assigned with such names?&lt;/p&gt;&#xA;" OwnerUserId="2159" LastEditorDisplayName="user7951" LastEditDate="2017-08-17T17:48:13.813" LastActivityDate="2017-08-17T17:48:13.813" Title="Do the names of some elements have anything to do with planets?" Tags="&lt;nomenclature&gt;&lt;elements&gt;&lt;history-of-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6234" PostTypeId="1" CreationDate="2013-09-17T10:50:55.920" Score="3" ViewCount="3131" Body="&lt;p&gt;For a chemistry lab, I'm required to calculate the vapor pressure of some organic solvents... methanol, ethanol, and propanol, in addition to an &quot;unknown&quot; sample of some solvent. In the lab, we were instructed to heat these solvents in a sealed syringe along with 5 mL of air. While they were being heated, I recorded the volume and temperature of the air and solvent in the syringe. Now, from that data, I need to calculate the vapor pressure of these substances at the temperatures at which I recorded the data. How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="2320" LastActivityDate="2014-01-26T10:43:53.513" Title="Calculate vapor pressure from volume" Tags="&lt;gas-laws&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6235" PostTypeId="2" ParentId="6233" CreationDate="2013-09-17T11:00:33.993" Score="21" Body="&lt;p&gt;Neptunium (93) and plutonium (94) were so named because they followed uranium (92) on the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium#Discovery&quot;&gt;Uranium was named after Uranus&lt;/a&gt;. German chemist Martin Heinrich Klaproth discovered uranium in 1789, eight years after William Herschel discovered Uranus. At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="84" LastEditDate="2013-09-17T11:26:57.257" LastActivityDate="2013-09-17T11:26:57.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6236" PostTypeId="2" ParentId="6231" CreationDate="2013-09-17T11:07:08.227" Score="2" Body="&lt;p&gt;The first 4 chapters of &lt;em&gt;&lt;a href=&quot;http://www.mediafire.com/download/cenj3cbyxmc8q34/Chemical_applications_of_group_theory_3ed_-_Cotton.djvu&quot; rel=&quot;nofollow&quot;&gt;Chemical Applications of Group Theory&lt;/a&gt;&lt;/em&gt; by &lt;em&gt;F A Cotton&lt;/em&gt; should be enough for your student. However there are numerous books available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again &lt;em&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Symmetry/Group_Theory%3a_Theory&quot; rel=&quot;nofollow&quot;&gt;chemwiki&lt;/a&gt;&lt;/em&gt; has a good documentation on group theory. &lt;/p&gt;&#xA;" OwnerUserId="2059" LastActivityDate="2013-09-17T11:07:08.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6237" PostTypeId="1" AcceptedAnswerId="6240" CreationDate="2013-09-17T11:45:29.117" Score="3" ViewCount="1060" Body="&lt;p&gt;More than dozen of heavy elements of periodic table do not occur naturally but are prepared artificially at nuclear reactors.. These elements are said to disappear in a flash..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why invest lot of money for the discovery of such elements? What are the advantages of such elements which disappear in a flash? &lt;/p&gt;&#xA;" OwnerUserId="2159" LastActivityDate="2013-09-17T15:36:32.850" Title="What is advantage of creating elements that disappear in a flash?" Tags="&lt;physical-chemistry&gt;&lt;periodic-table&gt;&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6238" PostTypeId="1" AcceptedAnswerId="6239" CreationDate="2013-09-17T13:10:26.193" Score="7" ViewCount="17547" Body="&lt;p&gt;Halides are group 7 elements. They generally have a low melting and boiling point. They are highly reactive. Silver chloride and silver bromide are not soluble in water. So they will be precipitates. However, I do know that they have varying solubilities. So, they will form different amounts of precipitates. I want to know why they have different solubilities and which is the most soluble halide.&lt;/p&gt;&#xA;" OwnerUserId="856" LastActivityDate="2013-09-17T15:19:22.823" Title="Which silver halide is the most soluble?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;silver&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6239" PostTypeId="2" ParentId="6238" CreationDate="2013-09-17T13:58:43.500" Score="10" Body="&lt;p&gt;The solubilities of silver halides decreases down the periodic table:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{AgF}\ : K_{sp} = 205$ &lt;/li&gt;&#xA;&lt;li&gt;$\ce{AgCl}: K_{sp} = 1.8\times10^{-10}$ &lt;/li&gt;&#xA;&lt;li&gt;$\ce{AgBr}: K_{sp} =5.2\times10^{-13}$ &lt;/li&gt;&#xA;&lt;li&gt;$\ce{AgI}\ \ \ : K_{sp} =8.3\times10^{-17}$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rationale for this trend is typically described using the concept of hard/soft acids and bases.  (See &lt;em&gt;J. Chem. Educ.&lt;/em&gt;, &lt;strong&gt;1968&lt;/strong&gt;, 45, page 581 and also page 643 for early discussions on HSAB theory.)  Briefly, ions with similar size and high polarizability will lend themselves to more covalent interactions, and therefore will have decreased solubility in water.  HSAB classifies ions as either hard or soft, with soft being large and polarizable and hard being small and highly charged.  H-H and S-S interactions are strong while H-S or S-H interactions are weaker.  Ag is classified as soft and the halides go from hard (F-) to soft (I-).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="275" LastEditDate="2013-09-17T15:19:22.823" LastActivityDate="2013-09-17T15:19:22.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6240" PostTypeId="2" ParentId="6237" CreationDate="2013-09-17T15:36:32.850" Score="6" Body="&lt;p&gt;Researchers are looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Island_of_stability&quot;&gt;Island of Stability&lt;/a&gt;, a predicted group of heavy nuclei with halflives in the minutes or even days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we have some understanding that certain nuclear configurations and compositions are more stable, we are currently unable to reliably to &lt;em&gt;a priori&lt;/em&gt; predict the stability, the half-life, or even the decay mechanisms of a nuclide. Thus, when a new element or isotope is produced, no one knows how long it will be around. Every time a new nuclide is produced, observations of its decay help refine the models of nuclear stability and decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Island of Stability is believed to be centered around 120 protons and 180 neutrons. Element 118 $\ce{^194Uuo}$ has been produced, and while its most stable isotope has a half-life of only 0.89 ms, it is very close to the island of stability and there may be a heavier, more stable isotope of $\ce{Uuo}$ out there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in many areas of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pure_science&quot;&gt;pure science&lt;/a&gt;, the benefits to society (or even other fields) may not be evident except for those few who are actually doing the work. The outcome of pure science research is knowledge and information. Consider, for example, that research on &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathode_ray#History&quot;&gt;cathode rays&lt;/a&gt; led to the development of the television decades later. &#xA;The scientists studying cathode rays likely could not even envision their work being applied to the projection of moving pictures (especially those transferred over the airwaves as radio signals), as such technology did not exist or was very new at the time.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-17T15:36:32.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6241" PostTypeId="2" ParentId="6233" CreationDate="2013-09-17T17:35:48.660" Score="16" Body="&lt;p&gt;Asked and answered, but this is a good place to include other astronomically named elements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Ce&quot;&gt;cerium&lt;/a&gt;&lt;/strong&gt; named after the asteroid Ceres (or the Roman goddess of agriculture)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=He&quot;&gt;helium&lt;/a&gt;&lt;/strong&gt; from the Greek &lt;em&gt;helios&lt;/em&gt; or &quot;Sun&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Hg&quot;&gt;mercury&lt;/a&gt;&lt;/strong&gt; OK, named after the Roman god, but so was the planet&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Pd&quot;&gt;palladium&lt;/a&gt;&lt;/strong&gt; named after the asteroid Pallas, discovered the year that Pd was first isolated&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Se&quot;&gt;selenium&lt;/a&gt;&lt;/strong&gt; from the Greek &quot;selene&quot; for &quot;Moon&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Te&quot;&gt;tellurium&lt;/a&gt;&lt;/strong&gt; from the Greek &quot;tellus&quot; for &quot;Earth&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/element.php?sym=Kr&quot;&gt;krypton&lt;/a&gt;&lt;/strong&gt; Sorry comic fans, the name predates Superman&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Naming information is from &lt;a href=&quot;http://www.angelo.edu/faculty/kboudrea/periodic/hist_names.htm&quot;&gt;here&lt;/a&gt; and a very nice &lt;a href=&quot;http://elements.vanderkrogt.net/&quot;&gt;element etymology&lt;/a&gt; website.  Now that you know something, you should probably take a &lt;a href=&quot;http://www.sporcle.com/games/jb225/elements-peoplecountryplanet&quot;&gt;mentally stimulating diversion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="2341" LastEditDate="2013-09-20T20:47:42.963" LastActivityDate="2013-09-20T20:47:42.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6242" PostTypeId="2" ParentId="5616" CreationDate="2013-09-17T18:13:09.887" Score="3" Body="&lt;p&gt;It is caused by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wetting&quot; rel=&quot;nofollow noreferrer&quot;&gt;wettability&lt;/a&gt; of the surface of the nanoparticles. For naked particles this will be determined by the properties of gold but, as Daniel mentioned, it is likely that there is a capping agent, in which case the wettability will be governed by the properties of the capping agent. From this point on I will just call this the solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collecting at the interface&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the surface energies between the solid and water (sw), water and oil (wo) and solid and oil (so) you can have 2 situations both with their own extreme case. We can understand which of the three by looking at the famous &lt;a href=&quot;http://www.utwente.nl/tnw/pcf/education/jmbc_course_on_capillarity_driv/Articles/anton_darhuber/surface_tension_etc/Thomas_Young_PhilTrans1805_cap_eq.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Young's equation&lt;/a&gt;$^1$ :&#xA;$$\tag{1} \gamma_{so}-\gamma_{sw}=\gamma_{wo} \cos \theta $$&#xA;where $\theta$ is called the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Contact_Angles&quot; rel=&quot;nofollow noreferrer&quot;&gt;contact angle&lt;/a&gt; and in this case I have defined it through the water phase as shown in the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QtxlX.png&quot; alt=&quot;wetting and non-wetting particles&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the surface energies are such that a high contact angle occurs then the solid particles will be mostly on the oil side of the interface (situation 1). But given the surface energies it is still beneficial (energetically) to have some contact with the water, hence the angle. If the surface energies are such that $\cos \theta &amp;lt; -1 $ it means that there is no longer a contact angle and it is preferable for the solid particles to be completely submerged in the oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situation 2 is exactly the opposite, the solid 'likes' the water resulting in a low contact angle. Here again an extreme case can occur in which the equation requires $\cos \theta &amp;gt; 1$ which means that the particles will fully submerge in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because many liquid-liquid-solid combinations have at least some contact angle (i.e. not 0 and not 180) the particles tend to collect at the interface of the two fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Agglomeration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning for whether agglomeration will occur is in fact similar. Here again you will look at the surface energies and determine whether it is energetically favorable to have reduce the solid-liquid area by making solid-solid contact or whether it is favorable to have the solid in contact with the liquid. Capping agents are typically designed to repel themselves thus making sure that agglomeration doesn't occur (because it is usually not desired)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1: note that the original essay doesn't contain the equation in mathematical form, only in wording&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-09-17T18:13:09.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6243" PostTypeId="1" CreationDate="2013-09-18T04:20:49.857" Score="12" ViewCount="5293" Body="&lt;p&gt;For a reaction of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{aA + bB &amp;lt;=&amp;gt; cC + dD}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the equilibrium constant is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_c=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regardless of the mechanism of the reaction. Why is this the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen derivations that use a reaction in one elementary step to demonstrate this, but this obviously doesn't work in general. The sample reaction used was &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4 &amp;lt;=&amp;gt; 2NO2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="41328" LastEditDate="2019-03-30T20:58:20.993" LastActivityDate="2019-03-30T21:46:54.220" Title="How is it that the equilibrium constant does not depend on the mechanism?" Tags="&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="6244" PostTypeId="1" AcceptedAnswerId="6248" CreationDate="2013-09-18T05:31:04.017" Score="13" ViewCount="86333" Body="&lt;p&gt;Why are sigma bonds stronger than pi bonds? &lt;/p&gt;&#xA;" OwnerUserId="1977" LastEditorUserId="4945" LastEditDate="2014-06-19T02:53:29.203" LastActivityDate="2014-06-19T07:00:27.520" Title="Why are sigma bonds stronger than pi bonds?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;theoretical-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6246" PostTypeId="1" AcceptedAnswerId="6256" CreationDate="2013-09-18T11:18:27.983" Score="5" ViewCount="13470" Body="&lt;p&gt;I read that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Take a cupful of water in a beaker and add a few drops of dilute&#xA;  sulfuric acid. Heat the water. When it starts boiling add copper&#xA;  sulfate powder slowly while stirring continuously.&#xA;  Continue adding copper sulfate powder till no more powder can be&#xA;  dissolved. Filter the solution. Allow it to cool. Do not disturb the&#xA;  solution when it is cooling. Look at the solution after some time. Can&#xA;  you see the crystals of copper sulfate? If not, wait for some more&#xA;  time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So why is sulfuric acid required in this experiment?&lt;/p&gt;&#xA;" OwnerUserId="2326" LastEditorUserId="402" LastEditDate="2016-12-04T19:52:37.467" LastActivityDate="2016-12-04T19:52:37.467" Title="Why is sulfuric acid used in crystallization" Tags="&lt;acid-base&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6248" PostTypeId="2" ParentId="6244" CreationDate="2013-09-18T13:02:03.770" Score="12" Body="&lt;p&gt;The reason behind this is the orientation of the overlapped orbitals. Sigma bonds result from head-on(co-axial) overlapping while pi bonds are outcome of lateral(para-axial) overlapping. Here is a pictorial representation of ethene(sp&lt;sup&gt;2&lt;/sup&gt; hybridized C atoms) :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7mqW.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The greater the extent of overlapping, the higher the probability of finding the valence  electrons in between the nuclei and hence the bond will be stronger &amp;amp; shorter.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MOT, this can be explained using &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_overlap&quot; rel=&quot;noreferrer&quot;&gt;Overlap Integral&lt;/a&gt;. This is how Atkins depicts it :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJicn.png&quot; alt=&quot;enter image description here&quot;&gt;       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;In simple terms, after forming a sigma-bond (a pre-requisite for pi-bonds), the two atoms get locked along the inter-nuclear axis. As a result, the orbitals available for pi-bonding can only partially overlap, thus forming a weaker bond.&lt;/em&gt;&lt;/strong&gt;   &lt;/p&gt;&#xA;" OwnerUserId="2059" LastEditorUserId="402" LastEditDate="2014-06-19T07:00:27.520" LastActivityDate="2014-06-19T07:00:27.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6249" PostTypeId="1" CreationDate="2013-09-18T13:53:31.670" Score="1" ViewCount="66" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cQp6m.png&quot; alt=&quot;enter image description here&quot;&gt;I have been looking for a physical explanation for entropy and enthalpy variations. An example would be the air system of a vehicle, where people say we have a strong enthalpy and entropy variations along the exhaust line, which is not the case in the fresh air line. How could that be possible?&lt;/p&gt;&#xA;" OwnerUserId="2327" LastEditorUserId="2327" LastEditDate="2013-09-23T14:33:19.853" LastActivityDate="2013-09-23T14:33:19.853" Title="Enthalpy: when and how does it vary?" Tags="&lt;enthalpy&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2017-04-16T14:10:40.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="6250" PostTypeId="1" CreationDate="2013-09-18T13:54:16.413" Score="4" ViewCount="779" Body="&lt;p&gt;I am trying to compute the ionization energy of $\ce{H2+}$ molecule ion from the electronic energy spectrum. The question is whether one should use the purely electronic Hamiltonian or the Hamiltonian with the internuclear repulsion term to do that? I illustrated two ways to compute the ionization energy below. Which one is correct? &#xA;&lt;img src=&quot;https://i.stack.imgur.com/QAPwr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, nuclei are not at the equilibrium distance since in my problem I can manage this distance by external forces.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorDisplayName="user7951" LastEditDate="2015-07-08T19:28:21.660" LastActivityDate="2015-07-08T20:45:43.070" Title="The ionization energy of H₂⁺ molecule ion" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;hydrogen&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6251" PostTypeId="1" AcceptedAnswerId="6254" CreationDate="2013-09-18T15:20:09.273" Score="7" ViewCount="43817" Body="&lt;p&gt;According to the law of conservation of mass, during any physical or chemical changes, the total mass of the products is equal to the total mass of the reactants. But in nuclear fusion reaction if considered an example, energy that sun emits in its core is due to collision of hydrogen nuclei and formation of helium nuclei. Here conservation of mass is not obeyed as certain part of mass is converted into energy. So, law of conservation of mass is violated here. Then law of conservation must be redefined as: during any physical or chemical change, the total mass of the products is equal to the total mass of reactants provided mass has not undergone conversion into energy.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2018-11-20T19:09:05.263" LastActivityDate="2018-11-20T19:09:05.263" Title="Nuclear fusion reactions violate law of conservation of mass, but it has not yet been taken into account. Why?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;nuclear&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6252" PostTypeId="2" ParentId="6243" CreationDate="2013-09-18T16:34:54.560" Score="10" Body="&lt;p&gt;You should indeed be able to write every equilibrium reaction in this simple form. I guess that the reason for this might lie in the principle of microscopic reversibility (shameless self-plug: see &lt;a href=&quot;https://chemistry.stackexchange.com/a/6158/189&quot;&gt;this&lt;/a&gt; answer of mine). But for non-elementary reactions, i.e. those whose reaction mechanism consists of more than one step, you get a more complicated equilibrium constant that is the product of the equilibrium constants of all the intermediate reaction steps (if the reaction mechanism has no branches). An example: Consider the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{H2(g) + I2(g) &amp;lt;=&amp;gt;&amp;gt; 2HI(g)} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its mechanism consists of three reaction steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{ I2 &amp;lt;&amp;lt;=&amp;gt; 2I } \qquad \qquad \qquad \qquad K_{1} = \frac{[\ce{I}]^{2}}{[\ce{I2}]}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{I + H2 &amp;lt;&amp;lt;=&amp;gt; H2I} \ \ \quad \qquad \qquad K_{2} = \frac{[\ce{H2I}]}{[\ce{I}] [\ce{H2}]}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H2I + I &amp;lt;=&amp;gt;&amp;gt; 2HI} \quad \qquad \qquad K_{3} = \frac{[\ce{HI}]^{2}}{[\ce{I}] [\ce{H2I}]}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, take a look at the equilibrium constant of the overall reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\mathrm{tot}} = \frac{[\ce{HI}]^{2}}{[\ce{I2}] [\ce{H2}]} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then make the following substitutions: From the third reaction step's equilibrium equation you get $[\ce{HI}]^{2} = K_{3} [\ce{I}] [\ce{H2I}]$. Substituting this into the equation for $K_{\mathrm{tot}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] [\ce{H2I}]}{[\ce{I2}] [\ce{H2}]} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From reaction step 2 you get $[\ce{H2I}] = K_{2} [\ce{I}] [\ce{H2}]$ which then gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} [\ce{I}]^{2}}{[\ce{I2}]} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, from reaction step 1 you get $[\ce{I}]^{2} = K_{1} [\ce{I2}]$. Substituting this into the equation for $K_{\mathrm{tot}}$ you see that the overall equilibrium constant reduces to the product of the equilibrium constants of the reaction steps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{\mathrm{tot}} = \frac{K_{3} [\ce{I}] K_{2} [\ce{I}] [\ce{H2}] }{[\ce{I2}] [\ce{H2}]} = \frac{K_{3} K_{2} K_{1} [\ce{I2}]}{[\ce{I2}]} = K_{1} K_{2} K_{3} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a reaction that doesn't proceed along a chain of reaction steps but branches out the situation will become more complicated but can be treated in an analogous way as above.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2013-09-18T18:28:32.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6253" PostTypeId="2" ParentId="6243" CreationDate="2013-09-18T18:31:35.833" Score="8" Body="&lt;p&gt;Philipp's answer is a nice application of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Hess's Law&lt;/a&gt; to equilibrium. Let's look at why this approach works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of equilibrium (as given by the equilibrium constant $K$) is &lt;a href=&quot;http://en.wikipedia.org/wiki/Van%27t_Hoff_equation&quot; rel=&quot;noreferrer&quot;&gt;related to the standard Gibbs free energy change&lt;/a&gt; that occurs during the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rG^\circ=-RT\ln{K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Gibbs free energy&lt;/a&gt; $G$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/State_function&quot; rel=&quot;noreferrer&quot;&gt;state function&lt;/a&gt; - a property that depends only on the current state of the system, not the pathway by which the system reached that state. The value of $\Delta_rG^\circ$ depends only on the values of $G^\circ$ (or $\Delta_f G^\circ$ if you prefer) of the reactants and the products, which depend only on the identity and amounts of the substances. $\Delta_fG^\circ$ is independent of the way in which any of the substances were prepared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, $K$ is a state function. $K$ describes the compositions of the mixture which qualify as meeting the requirements for equilibrium. The value of $K$ is independent of the amounts of any specific substance or the pathway (mechanism) by which equilibrium is reached (see Philipp's answer). The individual equilibrium concentrations, however, in the system are not state functions. $[\ce{A}]_{eq}$ depends on the initial concentrations of all species in the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rate&lt;/strong&gt; of reaction is dependent on the mechanism. Different mechanisms may have different &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_order&quot; rel=&quot;noreferrer&quot;&gt;orders&lt;/a&gt;, different &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation&quot; rel=&quot;noreferrer&quot;&gt;activation energies&lt;/a&gt;, etc. Reaction rate (and consequently the rate constant $k$) are not necessarily state functions. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-18T18:31:35.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6254" PostTypeId="2" ParentId="6251" CreationDate="2013-09-18T18:54:06.890" Score="14" Body="&lt;p&gt;Nuclear reactions appear to violate both the Laws of Conservation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservation_of_mass&quot;&gt;Mass&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservation_of_energy&quot;&gt;Energy&lt;/a&gt; because mass is converted into energy or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass-energy_equivalence&quot;&gt;mass-energy equivalence&lt;/a&gt; that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conservation_of_mass-energy#Conservation_of_mass_and_energy&quot;&gt;Law of Conservation of Mass-Energy&lt;/a&gt;. Since mass is now a form of energy, we can convert back and forth as long as there is no net loss between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalence of mass and energy comes from Einstein's famous $E=mc^2$, where $c$ is the speed of light (in a vacuum).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-18T18:54:06.890" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6255" PostTypeId="1" CreationDate="2013-09-18T19:21:06.317" Score="5" ViewCount="1312" Body="&lt;p&gt;Imagine a container of fixed volume is taken and hydrogen gas is filled in it, as a result the container floats in air. But if in the same container is filled with so much hydrogen that the hydrogen gas molecules had to compress to fit in the container, will the container still float in air ?&lt;/p&gt;&#xA;" OwnerUserId="2329" LastActivityDate="2016-05-10T15:52:36.390" Title="Will compressed hydrogen float in air?" Tags="&lt;hydrogen&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6256" PostTypeId="2" ParentId="6246" CreationDate="2013-09-18T20:04:43.703" Score="3" Body="&lt;p&gt;The sulfuric acid will dissolve any residual copper metal or any copper oxides that exist as impurities in the copper sulfate, converting those impurities into copper sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4(aq) + Cu(s) -&amp;gt; CuSO4(aq) + H2(g)}$$&#xA;$$\ce{H2SO4(aq) + CuO(s) -&amp;gt; CuSO4(aq) + H2O(l)}$$&#xA;$$\ce{2H2SO4(aq) + Cu2O(s) -&amp;gt; 2CuSO4(aq) + H2O(l) +H2(g)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-18T20:04:43.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6257" PostTypeId="2" ParentId="6255" CreationDate="2013-09-18T20:47:21.520" Score="3" Body="&lt;p&gt;Here is a short answer to your question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compression will result in increament of density of hydrogen in the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen will float in the air until its density becomes equal to air through the compression.&#xA;Hydrogen's density (0.08988 g/L) is  &lt;strong&gt;very much less&lt;/strong&gt; than that of air's (1.225 kg/m3).&#xA;As soon as the density of hydrogen becomes more than that of air, it will settle down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20090908232018AAUC7no&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; to know how does gas density increases due to compression.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-09-18T20:57:44.997" LastActivityDate="2013-09-18T20:57:44.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6258" PostTypeId="2" ParentId="6255" CreationDate="2013-09-18T21:37:39.693" Score="7" Body="&lt;p&gt;Yes. A simple example is an ordinary balloon, which places the gas inside it under pressure (because the tension of the rubber surface adds to the air pressure from outside the balloon), yet the balloon still floats readily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're probably talking about is a theoretical container that can hold highly-compressed hydrogen, like from a supply cylinder, yet has negligible weight. At what point, then, would the hydrogen weigh more than air and the container would settle to the ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer that, we'll look at the ideal gas law: $PV = nRT$, or equivalently, $P = \dfrac{nRT}{V}$. For our purposes, we'll ignore the temperature increase of the compressed gas by saying this container, while having negligible weight, also has an efficient cooling system so the temperature delta is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At STP (273*K, 100kPa), one liter of volume filled with hydrogen gas contains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$100 = \dfrac{n(8.3145)(273)}{1}$$&#xA;$$100 = 2269.8482n$$&#xA;$$n = .0441\text{ mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With hydrogen having a molar weight of 1.00794 g/mol (and $\ce{H2}$ gas having double that), the density in g/L of hydrogen gas at STP is .08890 g/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borrowing from ashu's answer, since his math and mine are within 1% error, the density of air at STP is 1.225g/L (1 m3 = 1000 L). So, in order to have the same mass of gas in the same volume, and thus the same density, you would need 13.78 times the amount of gas than you'd have in the same volume at STP (pretty much 1.225 mol). Because, given a unit volume, double the gas equals double the pressure, the resulting pressure of a container of negligible mass holding enough compressed hydrogen to not float in air is 1377.953kPa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SAE units, that's actually not much in relative terms; only 199.855 psi (call it 200 so it would actually sink in air and not be neutrally buoyant). That's the water pressure at a depth of about 460 feet (140.72m), which human freedivers have matched using dive sleds.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-19T17:06:49.917" LastActivityDate="2013-09-19T17:06:49.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6259" PostTypeId="2" ParentId="6231" CreationDate="2013-09-19T00:33:54.903" Score="2" Body="&lt;p&gt;Several answers and comments have suggested general texts that have good foundation information; however if a student is writing a thesis, these sources will be insufficient since they are geared towards the &lt;em&gt;teaching&lt;/em&gt; of group theory rather than the &lt;em&gt;application&lt;/em&gt; of group theory to generate new knowledge.  Even at an undergraduate level, there should be an attempt to generate new knowledge in a thesis.  To do this, the student would need to get her nose in the chemical literature, and I assume any student in an ACS accredited undergraduate program has experience with literature searches.  My very brief survey of group theory reviews came up with the following potentially interesting areas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;understanding the structure of natural and synthetic perovskites, which have a number of technological applications. Journal of Physics: Condensed Matter (2013), 25(17), 175902&lt;/li&gt;&#xA;&lt;li&gt;Stereochemistry of medicinal compounds:  Helvetica Chimica Acta (2013), 96(1), 4-30.&lt;/li&gt;&#xA;&lt;li&gt;the role of group theory in furthering our understanding of superconductivity:  Comptes Rendus Physique (2011), 12(5-6), 532-541. &lt;/li&gt;&#xA;&lt;li&gt;Group theory and biomolecular conformations: Journal of Physics: Condensed Matter (2010), 22(32), 323103/1-323103/21.&lt;/li&gt;&#xA;&lt;li&gt;A somewhat esoteric area of nuclear point groups:  Acta Physica Polonica B (2013), 44(3), 305-318. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Almost certainly, a senior has encountered the importance of symmetry in Inorganic Chemistry and IR spectroscopy, and these areas may be more accessible to the student.  In the end, however, good questions and topics for a senior thesis are going to require getting ones nose in the literature.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-09-19T00:33:54.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6265" PostTypeId="2" ParentId="6250" CreationDate="2013-09-19T20:37:32.453" Score="4" Body="&lt;p&gt;Finally, I have found the answer. The correct way is the second one, presented in the right-hand side figure. Thus, the ionization energy is the difference between zero level and pure electronic energy. One could compute the ionization energy from the potential curve as well, but, in this case, the ionization energy is a difference between the potential energy of $\ce{H2^+}$ and $\ce{H2^{2+}}$. The potential curve for $\ce{H2^{2+}}$ just the Coulomb repulsive potential and, thus, we subtract the repulsion potential from the potential energy curve that leads us to the pure electronic energy. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/k9rzm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2013-09-19T20:46:39.640" LastActivityDate="2013-09-19T20:46:39.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6266" PostTypeId="1" CreationDate="2013-09-19T20:38:22.920" Score="6" ViewCount="592" Body="&lt;p&gt;Is there proof or a theory that shows that the first amino acid syntheses on Earth were influenced by a change in heat, gravity, radiation, or something similar?&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="865" LastEditDate="2015-08-05T16:00:36.290" LastActivityDate="2015-08-05T16:00:36.290" Title="How did the first amino acids appear on Earth?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6267" PostTypeId="2" ParentId="5851" CreationDate="2013-09-19T22:26:19.340" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;Tau-bonds provide an alternative valid description of electron density in alkenes and alkynes. Even in the modern chemistry&lt;/li&gt;&#xA;&lt;li&gt;A good example is the conformational preferences in propene (lower energy of the &quot;eclipsed&quot; conformation). You can find more information here: &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2011/OB/c1ob05393k#!divAbstract&quot;&gt;http://pubs.rsc.org/en/Content/ArticleLanding/2011/OB/c1ob05393k#!divAbstract&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2337" LastActivityDate="2013-09-19T22:26:19.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6269" PostTypeId="1" CreationDate="2013-09-20T05:42:15.130" Score="19" ViewCount="16066" Body="&lt;p&gt;In short, I do not understand how or why the migratory aptitude is as listed in every textbook I've read. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, they talk about a positive charge being built up that is stabilised by the movement of the most substituted group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot fathom such a build up in my mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me understand this in a mechanistic fashion.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="16683" LastEditDate="2020-11-09T12:23:01.433" LastActivityDate="2020-11-09T12:23:01.433" Title="Migratory aptitude in Baeyer-Villiger reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;organic-oxidation&gt;&lt;rearrangements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6270" PostTypeId="1" AcceptedAnswerId="6289" CreationDate="2013-09-20T08:58:26.223" Score="2" ViewCount="198" Body="&lt;p&gt;Pectin is an example of a food safe polyelectrolyte. It acts as an acid because it is a proton donor and develops negative charges along the polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of any examples of food safe polyelectrolytes that act as bases and develop positive charges along the polymer?&lt;/p&gt;&#xA;" OwnerUserId="799" LastActivityDate="2013-09-22T01:56:46.933" Title="Food safe basic polyelectrolytes?" Tags="&lt;acid-base&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6271" PostTypeId="1" AcceptedAnswerId="6277" CreationDate="2013-09-20T11:01:16.397" Score="22" ViewCount="4821" Body="&lt;p&gt;Why does having molecular chirality result in optical rotation? The dissymetry or chirality of molecules translates to the rotation of plane polarized light, the magnitude and direction depending on the concentration and the nature of the substance. But why does molecular chirality cause the rotation of plane polarized light.&lt;br&gt;&#xA;If this rotation is due to the assymetry around the bond, the different molecules are oriented in all possible directions(randomly) and hence the light will fall on them in all different orientations, and should ultimately be undeflected since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.&lt;br&gt;&#xA;Moreover, why does inverting the configuration invariably reverses the direction of rotation of light? &lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2015-11-11T09:20:16.803" Title="Molecular chirality and optical rotation" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;symmetry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6272" PostTypeId="2" ParentId="6269" CreationDate="2013-09-20T11:25:18.037" Score="17" Body="&lt;p&gt;The migratory aptitude list for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baeyer-Villager oxidation&lt;/a&gt; is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations came first.&lt;/p&gt;&#xA;&lt;p&gt;The migratory aptitude list &lt;span class=&quot;math-container&quot;&gt;$(\ce{H&amp;gt;R3C&amp;gt;R2HC}\approx\ce{Ph(&amp;gt;H)&amp;gt;RCH2&amp;gt;Me})$&lt;/span&gt; is the result of many experiments. Nor is it a linear list as is usually presented. &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; migrates more readily than tertiary groups, which migrate more readily than secondary groups, which migrate more readily than phenyl groups (which migrate more readily than &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The mechanism and theoretical explanations came later. The partially complete mechanism from the Wikipedia page is shown below. The key step (center) is a concerted collapse of the tetrahedral intermediate with a [1,2]-sigmatropic shift of &lt;span class=&quot;math-container&quot;&gt;$\ce{R}^1$&lt;/span&gt; and ejection of the carboxylate leaving group. Note that this mechanism shows none of the proton transfer steps.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZWnO.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since the whole thing is supposedly concerted, I can understand your objection to the build-up of &lt;span class=&quot;math-container&quot;&gt;$\delta^+$&lt;/span&gt; on &lt;span class=&quot;math-container&quot;&gt;$\ce{R}^1$&lt;/span&gt;. However, with the exceptions of the anomalous behavior of &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph}$&lt;/span&gt;, the migratory aptitude list is the same as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation#Structure_and_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbocation stability&lt;/a&gt; list. Thus, researchers looked for a link between the two phenomena and a way to relate one to the other. The link may not be real, but the trends are there. To a student learning organic chemistry, it is much friendlier at first to apply an existing trend in reactivity to a new reaction (with an exception), than to learn a whole new trend that is unrelated but cosmetically similar.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="41686" LastEditDate="2020-11-09T11:50:40.353" LastActivityDate="2020-11-09T11:50:40.353" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="6273" PostTypeId="1" CreationDate="2013-09-20T17:19:07.947" Score="1" ViewCount="315" Body="&lt;p&gt;With regards to complexes, I read that thermodynamic products are favored over kinetic products, which are labile. Why wouldn't the first product formed be the most prominent?&lt;/p&gt;&#xA;" OwnerUserId="2339" LastEditorUserId="189" LastEditDate="2013-09-20T19:24:32.800" LastActivityDate="2013-09-23T15:41:20.577" Title="Thermodynamic vs. Kinetic products" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6276" PostTypeId="1" AcceptedAnswerId="6280" CreationDate="2013-09-20T19:13:34.743" Score="7" ViewCount="11809" Body="&lt;p&gt;I used to be under the (wrong) assumption that the density of an element correlates with it's atomic number $\mathrm{Z}$, I thought that since having more protons meant the atom weighed more; but of course that's wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does make osmium $\mathrm{Os}$, with an atomic number of $\mathrm{Z}_{\mathrm{Os}}=76$ more dense than say uranium $\mathrm{U}$, with an atomic number of $\mathrm{Z}_{\mathrm{U}}=92$?&lt;/p&gt;&#xA;" OwnerUserId="2340" LastEditorUserId="2340" LastEditDate="2013-09-20T19:36:13.363" LastActivityDate="2013-09-26T04:42:35.737" Title="What determines the density of an element?" Tags="&lt;atoms&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6277" PostTypeId="2" ParentId="6271" CreationDate="2013-09-20T19:55:12.273" Score="25" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is false.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take 2-butanol. For this stereoisomer, light is turning clockwise when viewed from the right side (I'm not sure of this, but we can assume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ehi2m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the disks are there to clarify the direction from which we are viewing the entire setup. Here, the &quot;face&quot; of the disks is to the right so we are viewing it from the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the principle of reversibility of light, if light enters from the other side, we get the reverse path:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Se6W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotate this diagram (by $180^\circ$ around an axis perpendicular to your screen):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fn2MX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the disks are being viewed from the other side (the left) now. But we can flip them:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Aqd8X.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aqd8X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a close look at this. This is the &lt;em&gt;same&lt;/em&gt; molecule as the first, just oriented backwards. And looking from the right, it still gives a clockwise twist to the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while rays of light incident on different orientations of a molecule may differ in the exact value of the angle of rotation, rays of light incident on a reversed (&lt;em&gt;not&lt;/em&gt; stereoisomeric, just spatially rotated) molecule have the same direction and value of the angle of rotation and do not cancel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[8]: &lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2015-11-11T09:20:16.803" LastActivityDate="2015-11-11T09:20:16.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6279" PostTypeId="2" ParentId="6276" CreationDate="2013-09-21T15:49:14.490" Score="1" Body="&lt;p&gt;Since the density of matter is the mass per unit volume, I think that, from the chemical point of view, lengths of chemical bonds matter. If we consider same element forming different substances, the substance with shorter bond length should be denser. The term &quot;density of element&quot; seems confusing for me. The density depends on the state of matter (solids, liquids, gas) and conditions including temperature etc.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2013-09-21T15:49:14.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6280" PostTypeId="2" ParentId="6276" CreationDate="2013-09-21T16:45:46.430" Score="9" Body="&lt;p&gt;There are a couple of factors here that you need to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, increasing Z does mean that there are more particles in the nucleus; however, you're also adding electrons to the systems as well. Electrons dictate the size (i.e. volumen) of the atom, so the size of the atom &lt;em&gt;could&lt;/em&gt; increase with increasing Z, which would mean that fewer atoms could fit into the same volume. To use your example, the &lt;a href=&quot;http://www.webelements.com/periodicity/covalent_radius_2008/&quot;&gt;covalent radius&lt;/a&gt;  of Os is 144 pm; the covalent radius of U is 196 pm. So just from this property alone, you would expect more Os than U atoms to pack into the same volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another factor to consider as well: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_system&quot;&gt;crystal system&lt;/a&gt; of the solid. At standard temperature and pressure, atoms for different elements in the solid state pack into different arrangements, which are called crystal systems. Some of these are much more efficient than others. For example, there are three variants of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_crystal_system&quot;&gt;cubic crystal system&lt;/a&gt;: primative, body-centered, and face-centered. From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming one atom per lattice point, in a primitive cubic lattice with&#xA;  cube side length &lt;em&gt;a&lt;/em&gt;, the sphere radius would be &lt;em&gt;a&lt;/em&gt;⁄2 and the atomic&#xA;  packing factor turns out to be about 0.524 (which is quite low).&#xA;  Similarly, in a &lt;em&gt;bcc&lt;/em&gt; lattice, the atomic packing factor is 0.680, and&#xA;  in &lt;em&gt;fcc&lt;/em&gt; it is 0.740. The &lt;em&gt;fcc&lt;/em&gt; value is the highest theoretically&#xA;  possible value for any lattice, although there are other lattices&#xA;  which also achieve the same value, such as hexagonal close packed and&#xA;  one version of tetrahedral &lt;em&gt;bcc&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, depending on which crystal system the atoms in the solid take, you could get more or less atoms per unit volume. Unfortunately, determining which crystal system is favored for different elements is not straightforward, and requires some knowledge of the bonding between the atoms and other factors.&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2013-09-21T16:45:46.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6281" PostTypeId="1" CreationDate="2013-09-21T17:53:09.357" Score="5" ViewCount="4300" Body="&lt;p&gt;How do we represent a molecule in text form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A molecule, as a computer scientist I see it as a connected graph of atoms. A connected graph means that beginning at any atom, there is a path to any other atom in the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting question is, how do we represent these as text?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is $\ce{H2O}$ different than $\ce{OH2}$? That is, does order matter, or what does order mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What does this mean $\ce{NH4NO3}$? Why do we write N two times? Is this the same thing $\ce{NNH4O3}$?&lt;/p&gt;&#xA;" OwnerUserId="2343" LastEditorUserId="73" LastEditDate="2013-11-27T02:49:46.827" LastActivityDate="2013-11-27T02:49:46.827" Title="How do we represent a molecule in text form?" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6283" PostTypeId="2" ParentId="6281" CreationDate="2013-09-21T19:19:50.387" Score="2" Body="&lt;p&gt;Correct me If I am wrong and also I am not sure that this answers your full question or not as it only provides an answer to your specific questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wiki.answers.com/Q/What_are_the_rules_in_writing_chemical_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; are some rules for writing chemical formulas. Kindly read them first.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Is $\ce{H2O}$ different than $\ce{OH2}$? That is, does order matter, or what does order mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://2.bp.blogspot.com/-bCD8sjOMLaM/T7cjQnKd16I/AAAAAAAAAg0/dnqlx0GUxFc/s1600/water.jpg&quot; alt=&quot;stucture of H2O&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literally in these type of molecules, it doesn't matter that you write it as $\ce{H2O}$ or $\ce{OH2}$ because it will have no effect on its structure. These type of molecules follow &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill System&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) What does this mean $\ce{NH4NO3}$? Why do we write N two times? Is this the same thing $\ce{NNH4O3}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, $\ce{NH4+}$ (Ammonium) and $\ce{NO3-}$ (Nitrate) are two different ions which makes $\ce{NH4NO3}$.&#xA;In these type of compounds, cation is written first and then the anion. In the formation of Ammonium Nitrate, $\ce{NH3}$ reacts with $\ce{HNO3}$ and then cation &amp;amp; anion formation takes place and reaction proceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8e/Ammonium-nitrate-2D.png/200px-Ammonium-nitrate-2D.png&quot; alt=&quot;Ammonium Nitrate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically these two N in ammonium nitrate came from two different molecules. By 'from different molecules' I mean that each N is bonded with different atoms, this is why we need to write it twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't think that $\ce{NNH4O3}$ is the same thing. (I am not sure about it so kindly correct me if I am wrong here. and leave a comment with explaination.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: As commented by @F'x, the order of elements in a chemical formula is a complex matter. See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/537/why-is-arsenous-acid-denoted-h3aso3/538#538&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/974/why-is-methanes-molecular-formula-conventionally-ch4-while-water-is-h20-a&quot;&gt;this&lt;/a&gt;, there are some good answers as mentioned by @F'x. I also found them useful.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2013-09-21T20:05:31.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6284" PostTypeId="2" ParentId="6281" CreationDate="2013-09-21T21:57:16.937" Score="4" Body="&lt;p&gt;There are many text representations of molecules. SMILES is commonly used but has issues in representing stereochemistry, to the MDL (now Accelrys) CTAB format, to InChi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly a molecule can be represented as a graph, in which the atom are nodes and the bonds edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the molecular formulae you asked about, for a simple molecule like water, a chemist would recognise H2O and OH2 as the same thing, but things quickly get out of hand when more complex molecules are represented in this fashion, and order becomes important. Any chemist would recognise NH4NO3 as ammonium nitrate, but may fail to immediately see it when represented as N2H4O3. We write the N twice because it makes the context easier to determine - ie that this is a salt, comprised of the ammonium anion and the nitrate cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid that there are a number of contextual complexities when approaching cheminformatics from the viewpoint of a computer scientist - a knowledge of chemistry is just as important :)&lt;/p&gt;&#xA;" OwnerUserId="2251" LastActivityDate="2013-09-21T21:57:16.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6285" PostTypeId="2" ParentId="6281" CreationDate="2013-09-21T22:51:36.670" Score="7" Body="&lt;p&gt;The representation of a molecule in text form actually has a name: the chemical formula (pl. formulae).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula we use for everyday purposes is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Empirical_formula&quot;&gt;empirical formula&lt;/a&gt;. That article should give you more information, but in a nutshell, it is a simple way of identifying a compound by the ratio of its constituent elements. Since ionic compounds and other compounds such as macromolecules can have huge absolute numbers of atoms which are technically bonded but in practice are subdivided into their simplest components, empirical formulae are useful for unambiguously identifying most compounds. Note, however, that we run into trouble when talking about isomers or other compounds with the same elemental ratio but different structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these compounds, we must disambiguate by using a structural formula, which is basically a graph of the molecule. Translating this into text can be done many different ways, using a formal system for specifying branches and stereochemistry. The three most widely used formats for chemical database applications are SMILES, PDB, and CML. You can get more information on all these from &lt;a href=&quot;http://openbabel.org/wiki/Category%3aFormats&quot;&gt;Open Babel&lt;/a&gt; (alternately OpenBabel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a middle ground, using a formula similar to SMILES, variously called a molecular formula or condensed formula. It implies structure by counting every atom and splitting individual elements apart to show functional groups like hydroxyl ($\ce{-OH}$) or amino ($\ce{-NH2}$) bonded to the main structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is what your question is really about. Yes, in a condensed formula, the order of elements and functional groups matters, because you are trying to communicate information about a compound's structure with a minimum of overhead. It wouldn't make sense to say that butyraldehyde is $\ce{C2H4CHOCH3}$ because the formyl group is at the end. Therefore we say butyraldehyde is $\ce{CH3(CH2)2CHO}$ to better represent the compound's structure. This helps distinguish isomers (but not stereoisomers). Even using the Hill system, one must still be careful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this clears things up for you on how chemists use different formulae in different situations, and how a condensed formula works.&lt;/p&gt;&#xA;" OwnerUserId="2345" LastActivityDate="2013-09-21T22:51:36.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6286" PostTypeId="2" ParentId="5950" CreationDate="2013-09-21T23:21:29.683" Score="2" Body="&lt;p&gt;Actually, it's quite simple for most substances. Use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_chromatography&quot; rel=&quot;nofollow&quot;&gt;gas chromatograph&lt;/a&gt;. If the substance can be vaporized at reasonable temperatures, this will give you enough information to pin down the chemical compounds, or at least rule out a wide range of compounds. We can tell because every compound has a unique retention time based on its mass and attraction to the medium coating the column. With enough reference data and good precision, it's easy to figure out firstly how many separate compounds are in the mixture and secondly what their chemical formulae are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, there are other kinds of chromatography, not to mention &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;mass spectrometry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2345" LastActivityDate="2013-09-21T23:21:29.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6287" PostTypeId="1" AcceptedAnswerId="6288" CreationDate="2013-09-22T00:20:07.887" Score="14" ViewCount="967" Body="&lt;p&gt;I'm working on my own 3D molecule viewer for my AP Chem class. So far everything's going together nicely. There's a couple of things I'm not sure about though... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a certain rule of thumb or standard for what colors each element inside a molecule should be? For example, &lt;a href=&quot;http://jmol.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jmol&lt;/a&gt; renders oxygen red and hydrogen white:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7VunS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my app they are currently grey:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXcZC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how can I check if there are single bonds, double bonds, etc.? As you can see here, I haven't programmed that yet into my app. &lt;a href=&quot;https://en.wikipedia.org/wiki/Paracetamol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tylenol&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v7lVZ.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/U6yiW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look up coordinates for each molecule in an &lt;a href=&quot;http://cactus.nci.nih.gov/chemical/structure/water/file?format=xyz&amp;amp;get3d=True&quot; rel=&quot;nofollow noreferrer&quot;&gt;NCI/CADD group server online database&lt;/a&gt;. This database gives me the number of elements, and the &lt;code&gt;X&lt;/code&gt;, &lt;code&gt;Y&lt;/code&gt;, and &lt;code&gt;Z&lt;/code&gt; coordinates for where I should put each element. For example, water:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;3&#xA;H2O&#xA;O   -0.0000   -0.0589   -0.0000&#xA;H   -0.8110    0.4677    0.0000&#xA;H    0.8110    0.4677    0.0000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It doesn't tell me anything about where bonds go. More info about &lt;code&gt;.xyz&lt;/code&gt; file format &lt;a href=&quot;https://en.wikipedia.org/wiki/XYZ_file_format&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2347" LastEditorUserId="41328" LastEditDate="2017-08-09T01:15:29.897" LastActivityDate="2017-08-09T01:15:29.897" Title="How should elements be colored? And how many bonds does an atom have?" Tags="&lt;bond&gt;&lt;computational-chemistry&gt;&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6288" PostTypeId="2" ParentId="6287" CreationDate="2013-09-22T01:07:01.043" Score="16" Body="&lt;p&gt;For the coloring part I would suggest you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/CPK_coloring&quot; rel=&quot;nofollow noreferrer&quot;&gt;CPK&lt;/a&gt; coloring convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the decision whether a bond is a single or a double bond: You could make this decision based on the bond length between two atoms. But you will need to find typical values for single, double and triple bonds for every combination of elements which seems a bit too much trouble. On the other hand, the depiction of double and triple bonds is used mostly for bonds between carbon, oxygen, sulfur, phosphorus and nitrogen (for other elements like metals the bond order is not readily identifiable, so I would leave those as single bonds). So, the amount of data to be collected might not be that large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very naive way to make the decision how many bonds an atom has could be the following (though this might not be very efficient):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;calculate the distances between the atoms&lt;/li&gt;&#xA;&lt;li&gt;for each element set a maximum distance another atom can be away from an atom of the element in question to be considered a bonding partner&lt;/li&gt;&#xA;&lt;li&gt;for each element set a maximum number of bonds $N_{\mathrm{max}}$ an atom of this element can have, e.g. 4 for carbon, 4 for nitrogen, 3 for oxygen, etc. (for metals this might be a bit difficult to decide)&lt;/li&gt;&#xA;&lt;li&gt;for each atom in the system order the list of atoms within bonding distance from lowest to highest distance. From this list choose only the first to $N_{\mathrm{max}}$th atoms as bonding partners. Then look whether some of the bonding partners qualify for double or triple bonds - if so you have to kick out some of the bonding members from the end of your bond candidates so that you don't exceed your maximum bond number.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There will be better ways out there to do this but this one worked for me in the process of creating &quot;static&quot; pictures of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked what size you should use for your atoms. Again, there is no definite answer to this - use whatever looks best for you. Nevertheless, I found that the sizes calculated according to the following formula give rather good looking atom size ratios for the different elements (maybe you have to multiply the resultant values by a constant to give good default sizes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \text{size} = \frac{1}{2 \sqrt{2}} \cdot \sqrt[6 \,]{\text{atomic number}} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, here is a nice source for some typical bond distances. Other good resources would be the &lt;a href=&quot;http://www.amazon.de/Handbook-Chemistry-Physics-93rd-Edition/dp/1439880492/ref=sr_1_2?ie=UTF8&amp;amp;qid=1345240538&amp;amp;sr=8-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;CRC Handbook of Chemistry and Physics&lt;/a&gt; or the book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0123526515&quot; rel=&quot;nofollow noreferrer&quot;&gt;Holleman and Wiberg&lt;/a&gt; if you can lay your hands on them. Those might come in handy if you try to adapt my method and look for values for the maximum bond distances. Another advice concerning the maximum bond distances would be that typical values for the bond lengths of bonds containing (transition) metals are between 200 and 300 pm, so a value of 350 or 400 pm might be a good choice for the maximum bond distance for those species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the maximum bond number each element can have: here, the typical rules for drawing Lewis structures apply. You can find good resources for this on this site, e.g. &lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit&quot;&gt;here&lt;/a&gt;, and in every introductory general chemistry text. But you should be mindful of some special cases. You should not forget the cationic forms of the elements, when thinking about their maximum bond number. For example, carbon usually has 4 bonds, but there are very rare cases where it comes with 5 bonds ($\ce{CH5^{+}}$) although those are practically not important and some not so rare cases where it has 3 bonds (carbenes, carbocations, carbanions), so you might consider a maximum bond number of 5 for carbon, although for nearly all practical purposes 4 might do. Another example is oxygen. It usually has 2 bonds but when it is protonated or in Meerwein's reagent ($\ce{Me3O^{+}BF4^{-}}$) it has 3 bonds. Transition metals are very flexible when it comes to the number of bonds they have - it can be everything from 1 to 6 sometimes even more, so you should choose a high number, e.g. 8, for them. &lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2013-09-24T00:49:31.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6289" PostTypeId="2" ParentId="6270" CreationDate="2013-09-22T01:56:46.933" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pectin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pectin&lt;/a&gt; is a heteropolysaccharide rich in &lt;a href=&quot;http://en.wikipedia.org/wiki/D-galacturonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;galacturonic acid&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GBrVq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carboxylic acid in galacturonic acid gives pectin the ionizable character. Pectin can react with bases to form carboxylate anions along the polysaccharide chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exists a sugar-derivative and polysaccharide with opposite behavior - they react with acids to produce cations along the polysaccharide chain. The modified sugar is &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucosamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;glucosamine&lt;/a&gt;, and the polysaccharides are &lt;a href=&quot;http://en.wikipedia.org/wiki/Chitin&quot; rel=&quot;nofollow noreferrer&quot;&gt;chitin&lt;/a&gt; and chitosan(&lt;a href=&quot;http://en.wikipedia.org/wiki/Chitosan&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Chitosan&lt;/a&gt;). The amine ($\ce{NH2}$) group on glucosamine is basic and will react with acids to become cationic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jVrbg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glucosamine is used as a dietary supplement, and chitin and chitosan are found in various foods, like fungi and shellfish. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-22T01:56:46.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6290" PostTypeId="1" AcceptedAnswerId="6295" CreationDate="2013-09-22T04:00:55.363" Score="17" ViewCount="8087" Body="&lt;p&gt;We've just spilled the can of used meat-grease we collect so that it doesn't go down the drain and it turns out it was rotten. Now the granite counter-top and the wood-laminate floor of the kitchen smell very bad even after repeated cleaning with (1) dish-soap, (2) vinegar and (3) Murphy's Oil Soap as well as an enzymatic product called anti-icky-poo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extensive googling suggests that the smelly chemical is either putrescine or cadaverine. Am I right about that? How can I clean my kitchen such that it does not smell so bad?&lt;/p&gt;&#xA;" OwnerUserId="2348" LastEditorUserId="402" LastEditDate="2019-05-19T20:40:20.603" LastActivityDate="2019-05-19T20:43:00.727" Title="What causes the smell in rotten meat-grease and how can I get rid of it thoroughly?" Tags="&lt;organic-chemistry&gt;&lt;cleaning&gt;&lt;smell&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6292" PostTypeId="1" CreationDate="2013-09-22T09:45:16.513" Score="2" ViewCount="13483" Body="&lt;p&gt;It has been stated that the molar conductance($\Lambda_m$) of strong electrolytes is not affected to a greater extent on dilution and so to find the limiting value of molar concentration($\Lambda_m^0$) we can extrapolate the graph of molar conductance vs the concentration ($\sqrt c$). But as the concentration would near zero shouldn't the molar conductance decrease sharply to the range of say $10^{-8}$ which is the conductance of distilled water as the amount of solute that is the strong electrolyte would be negligible to consider?Also in this case extrapolating the graph should not be possible?&lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="1431" LastEditDate="2013-09-22T13:18:40.657" LastActivityDate="2017-05-20T00:08:08.453" Title="Effect of concentration on molar conductance" Tags="&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6293" PostTypeId="1" CreationDate="2013-09-22T09:47:25.070" Score="6" ViewCount="9075" Body="&lt;p&gt;Why does $\ce{BF4^-}$ have a larger bond length than $\ce{BF3}?$ Usually in molecules the bond angle increases to reduce repulsion but, here the bond length increases. Why is it so? Does the bond length increase with the bond angle in other cases of such kinds of molecules? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:01.130" LastActivityDate="2020-08-14T06:30:57.527" Title="Increase in bond length of BF₄⁻ compared to BF₃" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6294" PostTypeId="2" ParentId="6292" CreationDate="2013-09-22T11:07:09.337" Score="2" Body="&lt;p&gt;On increasing the dilution the conductivity($\kappa$) does decrease as it is proportional to the number of ions in a unit volume. But the molar conductance is the product of $\Lambda_m=\kappa V$ where $V$ is the volume containing one mole of the dissolved electrolyte.It is proportional to the number of ions in $V$ volume of the electrolytic solution which includes all the ions in one mole of the electrolyte plus the additional contribution from water. Hence, on increasing the dilution $\kappa$ goes down but $V$ goes up in such a way that its product goes up. Therefore in increasing the dilution, the conductivity will fall to the level of distilled water but the molar conductance still includes the volume containing one mole electrolyte and hence will have a value higher than that of distilled water and the concentrated solution of the electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[The general term Conductance (as defined by $C=\frac{1}{R}$) is different from the quantity molar conductance. The former, deals with a electrolytic setup with a definite cell constant and fixed condition. The other is equal to the Conductance of $V$ volume of electrolyte between two electrodes placed 1 UNIT APART.]&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-09-22T11:07:09.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6295" PostTypeId="2" ParentId="6290" CreationDate="2013-09-22T11:22:02.667" Score="17" Body="&lt;p&gt;If the stench is caused by &lt;a href=&quot;http://en.wikipedia.org/wiki/Putrescine&quot; rel=&quot;nofollow noreferrer&quot;&gt;putrescine&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Cadaverine&quot; rel=&quot;nofollow noreferrer&quot;&gt;cadaverine&lt;/a&gt;, then you are in luck! Both putrescine and cadaverine are amines (&lt;span class=&quot;math-container&quot;&gt;$\ce{RNH2}$&lt;/span&gt;), which react with acids to form water-soluble ammonium salts (&lt;span class=&quot;math-container&quot;&gt;$\ce{RNH3+}$&lt;/span&gt;). Most soaps, surfactants, and other cleaning products are basic to help remove fats and oils (by hydrolysis). Try white vinegar, which contains acetic acid. If that fails, you can buy muriatic acid (a dilute solution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Muriatic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrochloric acid&lt;/a&gt;), which is used to clean pools. Find it wherever pool supplies are found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caution&lt;/strong&gt;: While both of these solutions are fine for the granite, muriatic acid may not be fine for your floor - test in a small hidden area first. Vinegar should be okay for the floor as long as you don't let it soak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putrescine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/89cNN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/89cNN.png&quot; alt=&quot;Putrescine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cadaverine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P6umS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P6umS.png&quot; alt=&quot;Cadaverine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="402" LastEditDate="2019-05-19T20:41:03.897" LastActivityDate="2019-05-19T20:41:03.897" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6296" PostTypeId="2" ParentId="6290" CreationDate="2013-09-22T13:23:55.113" Score="13" Body="&lt;p&gt;If you are right, and the smell is coming from an amine, then you may need to clean a lot more than you have.  Even if you think you've done a thorough cleaning, amines have very low odour thresholds.  I can't find specific numbers for putrescine and cadaverine; however, another amine generated in decaying animal protein, trimethylamine, has an odour threshold of 0.00021 ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do a back of the envelope calculation here.  Volatile organic amines need to be &lt;a href=&quot;http://digitalcommons.usu.edu/cgi/viewcontent.cgi?article=1594&amp;amp;context=etd&quot; rel=&quot;nofollow noreferrer&quot;&gt;at a concentration of 1 ppm&lt;/a&gt; in order for the vapor pressure to be high enough to generate an odour.  Your odour is &quot;strong&quot; so let's choose a concentration of 10 ppm.  The goal is to use our cleaning product to get the concentration of amines from their current level (10 ppm) to the desired level (0.00021 ppm) so that we don't smell it any longer.  I read on many bottles of commercial cleaning products &quot;kills 99.99% of the germs!&quot; so let's go with that number as the cleaning efficiency.  If I get rid of 99.99% of the amine in one wash, that's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$10-10\cdot0.9999=0.001~\text{ppm remaining}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means we need &lt;strong&gt;two good cleanings&lt;/strong&gt; to get the amines to a level where they cannot be detected by our noses.  &lt;em&gt;Note that this is a simplified problem&lt;/em&gt; and assumes that the fats don't interfere with the cleaning efficiency and that you are able to &lt;em&gt;clean every surface exposed to the grease with 99.99% efficiency&lt;/em&gt;.  These are typically the limiting factors in cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of odours are a serious issue for swine and poultry farmers and one of their approaches to eliminating odours is through proper air flow.  Get some fans going to dilute the smell, as this will make life a bit more manageable and will also help in ultimately eliminating the odour.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="402" LastEditDate="2019-05-19T20:41:30.563" LastActivityDate="2019-05-19T20:41:30.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6297" PostTypeId="1" CreationDate="2013-09-22T19:01:14.277" Score="1" ViewCount="96" Body="&lt;p&gt;As per topic, I was wondering why the Tm of the polymers PGA, PLA and PLGA are in the following order. Does it correlate to their ability to undergo hydrolysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PGA&gt;PLA&gt;PLGA&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="30836" LastEditDate="2020-04-30T08:26:52.227" LastActivityDate="2020-04-30T08:26:52.227" Title="Is there a correlation between the susceptibility of a polymer to undergo hydrolysis and its melting point?" Tags="&lt;polymers&gt;&lt;melting-point&gt;&lt;hydrolysis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6298" PostTypeId="1" AcceptedAnswerId="6305" CreationDate="2013-09-22T23:32:19.647" Score="2" ViewCount="2930" Body="&lt;p&gt;Suppose I have a white paper originally. Then it is placed in a trunk, where soda spills occasionally and gasoline gas/gasoline also spills due to no gas tank cover. Over time, this paper becomes discoloured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I make it white again? What equipment do I use?&lt;/p&gt;&#xA;" OwnerUserId="2355" LastActivityDate="2015-01-29T10:26:41.210" Title="How do I make a paper white again?" Tags="&lt;chromatography&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6299" PostTypeId="1" AcceptedAnswerId="6302" CreationDate="2013-09-23T00:57:09.260" Score="1" ViewCount="4147" Body="&lt;p&gt;Consider the following comproportionation reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4NO3 -&amp;gt; N2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;How would I go about balancing this using the half reaction/ion-electron method?&lt;/p&gt;&#xA;" OwnerUserId="2356" LastEditorUserId="16683" LastEditDate="2021-05-22T16:31:03.797" LastActivityDate="2021-05-22T16:43:29.350" Title="Balancing a redox reaction with only one product" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6300" PostTypeId="1" AcceptedAnswerId="6339" CreationDate="2013-09-23T08:19:55.803" Score="1" ViewCount="1363" Body="&lt;p&gt;If I prepare the silver nitrate solution or any other solution, how can I know the shelf life of this solution? I know that it depends on the stability of the substance, but maybe there is some document which give information and is interesning how realy it was found.&lt;/p&gt;&#xA;" OwnerUserId="802" LastEditorUserId="802" LastEditDate="2013-09-26T06:42:49.630" LastActivityDate="2013-09-27T01:29:26.627" Title="Shelf-life of solutions" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6301" PostTypeId="2" ParentId="4588" CreationDate="2013-09-23T11:31:46.967" Score="5" Body="&lt;p&gt;Lead(IV) and sulfide ions are large and polarizable, making them &lt;em&gt;soft&lt;/em&gt; in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds.  The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(IV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HSAB theory has been discussed on Chem.SE &lt;a href=&quot;https://chemistry.stackexchange.com/q/4875/1573&quot;&gt;in this question&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/a/6239/1573&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2013-09-23T11:31:46.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6302" PostTypeId="2" ParentId="6299" CreationDate="2013-09-23T11:33:00.180" Score="9" Body="&lt;p&gt;You should use the the same product species in both half -reactions. In the example you gave, this is &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4NO3-&amp;gt; N2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{NH4+ &amp;amp;-&amp;gt; N2} \tag{1} \\&#xA;\ce{NO3- &amp;amp;-&amp;gt; N2} \tag{2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Balancing the reduction half-equation &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2NO3- &amp;amp;-&amp;gt; N2} &amp;amp; &amp;amp; \text{(balance N)} \\&#xA;\ce{2NO3- &amp;amp;-&amp;gt; N2 \color{red}{+ 6H2O}} &amp;amp; &amp;amp; \text{(add } \ce{H2O} \text{ to balance O)} \\&#xA;\ce{2NO3- \color{red}{+ 12H+} &amp;amp;-&amp;gt; N2 + 6H2O} &amp;amp; &amp;amp; \text{(add } \ce{H+} \text{ to balance H)} \\&#xA;\ce{2NO3- + 12H+ \color{red}{+ 10e-} &amp;amp;-&amp;gt; N2 + 6H2O} &amp;amp; &amp;amp; \text{(add } \ce{e-} \text{ to balance charge / oxidation state)}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And balancing the oxidation half-reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2NH4+ &amp;amp;-&amp;gt; N2} &amp;amp; &amp;amp; \text{(balance N)} \\&#xA;\ce{2NH4+ &amp;amp;-&amp;gt; N2 \color{red}{+ 8H+}} &amp;amp; &amp;amp; \text{(add } \ce{H+} \text{ to balance H)} \\&#xA;\ce{2NH4+ &amp;amp;-&amp;gt; N2 + 8H+ \color{red}{+ 8e-}} &amp;amp; &amp;amp; \text{(add } \ce{e-} \text{ to balance charge / oxidation state)} \\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In order to combine the two balanced half-equations, we need to multiply the first by &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; and the second by &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{8NO3- + 48H+ + 40e- &amp;amp;-&amp;gt; 4N2 + 24H2O} \\&#xA;\ce{10NH4+ &amp;amp;-&amp;gt; 5N2 + 40H+ + e-} \\ \hline&#xA;\ce{8NO3- + 10NH4+ + 48H+ + 40e- + 10H2O &amp;amp;-&amp;gt; 9N2 + 24H2O + 40H+ + 40e-}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which finally reduces to&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{8NO3- + 10NH4+ + 8H+ -&amp;gt; 9N2 + 14H2O}.$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2021-05-22T16:43:29.350" LastActivityDate="2021-05-22T16:43:29.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6303" PostTypeId="1" AcceptedAnswerId="6306" CreationDate="2013-09-23T14:27:48.627" Score="5" ViewCount="1094" Body="&lt;p&gt;Would Chlorine or Chloramine gas show up on a VOC test?  My apartment was cleaned using chlorox and lysol, and there has been a smell that's been lingering for about 10 days.  The air was tested with a VOC meter but nothing came up.  Would Chlorine and Chloramine have shown up on this test?&lt;/p&gt;&#xA;" OwnerUserId="2358" LastActivityDate="2013-09-23T17:38:22.530" Title="Would Chlorine or Chloramine gas show up on a VOC test?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6304" PostTypeId="2" ParentId="6273" CreationDate="2013-09-23T14:51:22.490" Score="3" Body="&lt;p&gt;The difference between Thermodynamic and Kinetic products is the difference in activation energy and the difference in (Gibbs)-Free-Energy.&#xA;When the activation energy is low, the reaction proceeds fast, while the product might not have as low a Free-Energy as an other product (with higher activation energy). When the reaction conditions allow however, the thermodynamically more stable product (with lower Free-Energy) will be formed. Even though the kinetic product first forms, most complexes are form in equilibrium reactions. Therefore, the kinetic product reverts to starting materials and then, slowly forms the thermodynamic product. This thermo dynamic product, though it is also in equilibrium, will not revert to starting materials as easily as the kinetic product will, due to the lower Free-Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude: First, the kinetically favored product will form. Then, when the reaction conditions allow it, the equilibrium shifts towards the thermodynamic product.&#xA;Note that the equilibrium is essential for the thermodynamically favored product to form.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="1573" LastEditDate="2013-09-23T15:41:20.577" LastActivityDate="2013-09-23T15:41:20.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6305" PostTypeId="2" ParentId="6298" CreationDate="2013-09-23T16:47:08.090" Score="4" Body="&lt;p&gt;One approach would be to use various solvents. You would require solvents in which your various stains were soluble in. Alcohols / water would be suitable to dissolve the soda. Whereas an organic solvent would be suitable for washing away the gasoline. Though gasoline is typically colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach would be to expose the paper to copious UV irradiation, materials are coloured due to absorptions in the visible spectrum which trigger electronic excitation. These absorptions bias the transmission and reflection of otherwise white light and give the materials their colour. UV radiation is of higher energy than visible light radiation, and therefore the chemical excitations involved can prompt reactions which result in the destruction of the materials themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar approach to invoke this chemical degradation of the coloured materials could be achieved through using a strong acid or base. This would be akin to the effect of bleach on household surfaces. Again, exposure to acids and bases can prompt reactions which result in the degradation of coloured materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that all of these methods will have some affect on the paper. Continued washing / drying of the paper with solvents will likely cause the paper to become brittle and affect the absorption of the paper, which may affect your ability to write / print on it. A similar effect may be noticed under UV irradiation, though I expect it wont be to the same extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sufficiently strong acid or base may be able to completely degrade your paper as well as your coloured contaminants.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2013-09-23T16:47:08.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6306" PostTypeId="2" ParentId="6303" CreationDate="2013-09-23T17:29:53.243" Score="3" Body="&lt;p&gt;Chlorine, no. Chloramine, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monochloramine, $\ce{NH2Cl}$, is an inorganic compound, and not very volatile (when dissolved in water it forms more or less of an azeotrope, evaporating slightly faster than the water itself). So, not a VOC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the way it does its job as a disinfectant is to basically shred organic compounds, like cell walls, key enzymes, etc, forming &quot;organic chloramines&quot; where one or more of the remaining hydrogens is replaced with an organic functional group. These now-organic compounds have a higher vapor pressure, so by design, chloramine is used to &lt;em&gt;create&lt;/em&gt; VOCs. The idea, when used in tap water or pool/spa chlorination, is to allow the organic chloramines to boil off and disperse in the air, leaving clean water behind. That &quot;chlorine smell&quot; of a chlorinated pool is actually mostly organic chloramines evaporating out. In a more enclosed space like an interior room, these compounds can't disperse and so the smell builds.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-09-23T17:38:22.530" LastActivityDate="2013-09-23T17:38:22.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6307" PostTypeId="2" ParentId="5953" CreationDate="2013-09-23T22:26:41.353" Score="1" Body="&lt;p&gt;Depending on the plastic, it is possible to convert a thermoplastic to a thermoset. The way to achieve this, is by linking the polymer chains in a thermoplastic to one-another and is called 'cross-linking'. Then, the chains cannot 'slide' along-side each other and will therefor become a thermoset. This means the plastic will not melt and start to degrade when the temperature becomes to high for the molecular bonds to hold.&#xA;Note that cross linking the polymer chains is a chemical reaction which needs certain functional molecules. When the thermoplastic chains have functional groups left to attach a 'cross linker' to, a bridge between two chains can be made. When the thermoplastic has no functional groups left however, it might never be possible to convert the thermoplastic into a thermoset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I would like to subscribe to the idea that there are no stupid questions (or blunders). Just people stupid enough not to ask them. Good job asking this one :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped!&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-09-23T22:26:41.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6308" PostTypeId="2" ParentId="4904" CreationDate="2013-09-23T22:51:13.123" Score="2" Body="&lt;p&gt;If I'm not mistaken a good example of the inductive effect through multiple bonds is trifluoro-acetic-acid, or $\ce{F3CCOOH}$. The proton is so electron deficient that it is easily abstracted by anything that can act as a base. The not 'fluorinated' equivalent, acetic acid, is not nearly as strong an acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the difference in acidity, is the difference in electro negativity between hydrogen atoms and fluorine. Due to the very large electro-negativity of fluorine atoms the electrons in the acid protons are 'pulled' towards it, leading to a clear example of the inductive effect over multiple bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment on one of the comments below the question: Electron displacement does occur in double bonds. A good example is the carbonyl functional group (C=O) in a molecule. The inductive effect results in a slightly more positive charge on the Carbon atom and slightly higher at the O-atom. This is also due to electro-negativity and results in the reactivity of the C=O functional group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped!&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-09-23T22:51:13.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6309" PostTypeId="2" ParentId="6293" CreationDate="2013-09-24T01:30:04.300" Score="12" Body="&lt;p&gt;There seems to be a bit of confusion between bond length and bond angle here.  It is true that VSEPR theory predicts that bonding pairs of electrons (and lone pairs, for that matter) will repel one another as much as possible, thereby increasing bond angles, but this has little to do with bond length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to explain the different bond lengths in $\ce{BF_3}$ and $\ce{BF_4^-}$ is by looking at the molecular orbital diagram of $\ce{BF_3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZjQe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked diagram isn't great since it does not show the filled orbitals: the $1a_2'$ $1e''$ and $4e'$ orbitals are all filled, making the lowest unoccupied orbital (LUMO) the $2a_2''$.  If you haven't wrapped your head around molecular orbital diagrams of this complexity, don't sweat it.  The bottom line is, the $2a_2''$ orbital has antibodning interactions with the $2p_z$ orbitals of boron and the $2p_z$ orbitals of the surrounding fluorines.  Thus, adding electrons to this orbital makes the $\ce{B-F}$ bonds weaker, and therefore longer.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="4231" LastEditDate="2015-03-06T23:21:58.070" LastActivityDate="2015-03-06T23:21:58.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6311" PostTypeId="1" CreationDate="2013-09-24T14:58:11.500" Score="7" ViewCount="20241" Body="&lt;p&gt;Given the carbanion, $\ce{R3C-}$, the carbon is $\mathrm{sp^3}$ hybridized, unless it is participating in resonance. This is clear from its steric number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having trouble in drawing its orbital diagram. I know that if this was a neutral carbon atom, it would look like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{3s}$: _&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{2p}$: ↑ ↑ _&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{2s}$: ↑↓&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{1s}$: ↑↓&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which would hybridize to this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{3s}$: _&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{2sp^3}$: ↑ ↑ ↑ ↑&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{1s}$: ↑↓&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now if I was to examine the carbanionic state, would I simply be adding one electron to the top most diagram, then re-hybridize that? Or would I add two electrons because the carbon is not sharing the electrons with any other atom? That's another question I have with orbital diagrams. Do we break each bond homolytically, and base our diagram off of that? This would give:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{3s}$: _&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{2p}$: ↑ ↑ ↑&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{2s}$: ↑↓&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{1s}$: ↑↓&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not sure how I would rehybridize that. Have I drawn the radical form?&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="5026" LastEditDate="2018-04-03T10:05:22.330" LastActivityDate="2018-04-03T10:05:22.330" Title="Hybridisation of the negatively charged carbon atom in a carbanion" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6312" PostTypeId="1" CreationDate="2013-09-24T16:22:53.120" Score="0" ViewCount="6538" Body="&lt;p&gt;We had analytical chemistry experiments to determine acidic radicals from a mixture of two salts. After the chloroform-chlorine water test for distinction between halides $(\ce{salt\ solution + H2SO4 + CHCl3 + Cl2_{(aq)}})$ when I emptied the test-tube in the sink, all of it suddenly turned yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What reaction could have occurred to cause this?&#xA;The sink mostly contained diluted $\ce{H2SO4}$ and $\ce{HNO3}$ along with small quantities of $\ce{AgNO3, BaCl2, NH4OH}$, lead acetate and ammonium molybdate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these would mostly be in small quantities since I had rinsed the sink with water before emptying the test-tube.&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="418" LastEditDate="2014-03-21T23:37:28.520" LastActivityDate="2018-03-10T23:09:18.090" Title="Yellow precipitate in the presence of ammonium molybdate?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6314" PostTypeId="1" CreationDate="2013-09-24T17:39:45.710" Score="5" ViewCount="1547" Body="&lt;p&gt;While working at CNC engraving, I had at my disposal a small bottle of liquid used to stain brass for creating signs etc - engrave a writing, stain the engraving, then grind/polish the flat topmost layer to remove the colored layer and leave it only in engraved depressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottle didn't reveal much about composition of the liquid, beyond some warnings about toxicity. It was a brass staining liquid, period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coloration it created was dark brown, that seemed black by comparison with the surrounding metal. The reaction took only a few seconds until full pitch of coloration was achieved and the liquid itself was transparent with very mild cyan-blue hue. The coloration was quite persistent; even roughly 10 years afterwards it's still good as new on the pieces I kept and required rather deep brushing to remove from the topmost surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me what substance it was?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(below a sample with dark parts achieved through that liquid)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3buwU.jpg&quot; alt=&quot;sample&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I found a very similar liquid on sale, but the page is equally devoid of composition details. The instructions say: [the liquid] is to be used in room temperature. It allows the finish in tones between black and light brown. The time of reaction is almost immediate. The color and speed can be controlled through dilution in water up to 50%.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SSFuu.jpg&quot; alt=&quot;the liquid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1676" LastEditorUserId="1676" LastEditDate="2014-12-09T09:39:14.827" LastActivityDate="2014-12-09T09:39:14.827" Title="What substance is used for staining brass?" Tags="&lt;everyday-chemistry&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6315" PostTypeId="2" ParentId="6314" CreationDate="2013-09-24T18:47:09.630" Score="2" Body="&lt;p&gt;This site regarding patina forming solutions might help:  &lt;a href=&quot;http://www.sciencecompany.com/-W160.aspx#1&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencecompany.com/-W160.aspx#1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1689" LastActivityDate="2013-09-24T18:47:09.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6318" PostTypeId="1" CreationDate="2013-09-24T23:20:33.257" Score="5" ViewCount="949" Body="&lt;p&gt;I was having a conversation when the topic of sublimation came up. Prior to the discussion, I had always thought that sublimation was caused by a great pressure/temperature difference, but another person stated that whether or not an element sublimates depends on its composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will any element sublimate under the right conditions? Or are there elements which will always pass through a liquid state?&lt;/p&gt;&#xA;" OwnerUserId="2375" LastEditorUserId="84" LastEditDate="2013-09-25T14:31:16.790" LastActivityDate="2014-01-24T02:22:58.913" Title="Is sublimation caused by an environmental difference or an element's composition?" Tags="&lt;physical-chemistry&gt;&lt;temperature&gt;&lt;sublimation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6319" PostTypeId="1" CreationDate="2013-09-24T23:27:51.363" Score="3" ViewCount="1145" Body="&lt;p&gt;I know that a molecule is more stable when a free radical is on a secondary carbon over a primary carbon, but how about when there is a lone pair of elections that can be placed on a secondary or primary? Is there any difference in the stability of the molecule or does that only apply to free radicals?&lt;/p&gt;&#xA;" OwnerUserId="2363" LastEditorUserId="189" LastEditDate="2013-09-24T23:58:01.803" LastActivityDate="2014-05-10T16:55:36.037" Title="Lone pair electrons on secondary carbon vs primary carbon" Tags="&lt;organic-chemistry&gt;&lt;electronic-configuration&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6320" PostTypeId="2" ParentId="6319" CreationDate="2013-09-25T01:21:14.880" Score="0" Body="&lt;p&gt;Yes, there is a difference.  You should look into &quot;carbene&quot; stability: &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbene&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Carbene&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A secondary carbene will be more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other interesting carbenes that demonstrate carbene stability; called &quot;persistent carbenes&quot;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Persistent_carbene&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Persistent_carbene&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1689" LastActivityDate="2013-09-25T01:21:14.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6321" PostTypeId="1" CreationDate="2013-09-25T02:34:24.577" Score="3" ViewCount="1050" Body="&lt;p&gt;My instructor has been drawing 3D molecules that show the directionality of non-bonding electron pairs and p-orbitals. I've been trying to find references online that show this process, but I'm having a hard time finding anything useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain the process or link to a source that does? I've included two pictures of molecules that show what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dJV61.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XSixY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2376" LastEditorUserId="189" LastEditDate="2013-09-25T03:15:50.167" LastActivityDate="2014-09-16T06:10:03.413" Title="References to draw 3D molecules with directionality of non-bonding electron pairs and p-orbitals" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6322" PostTypeId="1" AcceptedAnswerId="6325" CreationDate="2013-09-25T06:26:43.283" Score="9" ViewCount="6233" Body="&lt;p&gt;$$\ce{2Na(l) +2NH3 (g) -&amp;gt;2NaNH2(s) + H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3Cl (alc) +NH3 (alc) -&amp;gt; CH3NH2 (alc) +HCl(alc)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't ammonia donate a proton in both cases?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="5591" LastEditDate="2015-11-13T19:04:04.560" LastActivityDate="2015-11-13T19:04:04.560" Title="Why does ammonia act as an acid in only the first of these two reactions?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6323" PostTypeId="2" ParentId="6318" CreationDate="2013-09-25T11:25:30.573" Score="1" Body="&lt;p&gt;Depending on the field of study sublimation is either always possible, or dependent upon the substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Organic chemistry, sublimation only occurs if the phase diagram (&lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_diagram&quot; rel=&quot;nofollow&quot;&gt;wikipedia, phase diagram&lt;/a&gt;) would allow it. If a phase change from solid towards gas is possible in the phase diagram, sublimation is possible when changing temperature and pressure in the right way. When no transition can be made between the two, sublimation is not possible. It all depends upon thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the field of inorganic chemistry however, sublimation-like techniques are almost alway possible. If I'm not mistaken, with low pressure and high temperature, sputtering (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sputter_deposition&quot; rel=&quot;nofollow&quot;&gt;wikipedia, sputtering&lt;/a&gt;) can be used to deposit single elements in a film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am an organic chemist myself, so the details of the sputtering I don't particularly know, sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this does answer your question.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-09-25T11:25:30.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6325" PostTypeId="2" ParentId="6322" CreationDate="2013-09-25T16:27:27.207" Score="7" Body="&lt;p&gt;Ammonia has a $\mathrm{p}K_{\mathrm{a}}$ value (a description what that means can be found in &lt;a href=&quot;https://chemistry.stackexchange.com/a/6143/189&quot;&gt;this&lt;/a&gt; answer of mine) of 33 and is thus not acidic at all - you'd need a $\mathrm{pH}$ value around 30 to deprotonate $\ce{NH3}$ to any appreciable amount! This means that $\ce{NH3}$ very rarely acts as a proton donor itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens in your first reaction is the following: It is widely known that when you dissolve $\ce{Na}$ in liquid ammonia you get &lt;em&gt;solvated electrons&lt;/em&gt; (in the form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electride&quot; rel=&quot;nofollow noreferrer&quot;&gt;electride&lt;/a&gt; complexes). But electrons have of course very strong reducing power, so after a while they attack $\ce{NH3}$ abstracting one of its $\ce{H}$ atoms under the liberation of hydrogen gas. The important point here is that there is no dissociation of $\ce{NH3}$ here, i.e. no reaction $\ce{NH3 &amp;lt;=&amp;gt; NH2^{-} + H^{+}}$, so $\ce{NH3}$ does not act as a proton donor in the same sense a Bronsted acid would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second reaction is an $\mathrm{S}_{\mathrm{N}}2$ reaction where $\ce{NH3}$ acts as a nucleophile. So, it doesn't act as a proton donor in this reaction neither. What it does is that it replaces (substitutes) the $\ce{Cl}$ in $\ce{CH3Cl}$, thus forming a methylammonium salt, $\ce{CH3NH3^{+}Cl^{-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vXQnA.png&quot; height=&quot;115&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the $\mathrm{p}K_{\mathrm{a}}$ of an alkylammonium cation lies somewhere around 10 (the $\mathrm{p}K_{\mathrm{a}}$ of $\ce{NH4^{+}}$ is 9.24). So, it is much more acidic than $\ce{NH3}$ and also more acidic than the surrounding alcohol (I assume that &quot;(alc)&quot; means that the reaction is done in alcohol), which has a $\mathrm{p}K_{\mathrm{a}}$ between 15 and 17, and thus it (the ammonium salt) acts as a proton donor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ke8nv.png&quot; height=&quot;100&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2013-09-25T20:16:23.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6326" PostTypeId="2" ParentId="6276" CreationDate="2013-09-26T04:42:35.737" Score="1" Body="&lt;p&gt;While neucleons (protons and neutrons) are responsible for the mass of the atom, electrons are responsible for its size. The neucleons are in the tiny nucleus of the atom. The electrons take up a vast area around this necleus. If you keep on increasing the subatomic particles (electrons, protons, etc), defintely the atomic mass increased, but the volume (size) of the atom does not necessarily. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you know electrons are kind of orbitting around the neucleus (this simple model well suffices for this explanation) in orbits: s,p,d,f, etc... Each orbit means a shell which determines how far those electrons in a particular shell. A shell can have several fixed number of electrons, so when you keep on adding subatomic particles the mass necessarily increases, while the volume is necessarily not. What happens if a particular shell is already filled with its maximum?? Then the new electrons go to the next higher level/orbit which means a larger volume. If you think in this line of thought, you will be able to understand the densitiy does not linearly goes with the atomic mass, but there is a logic still.&lt;/p&gt;&#xA;" OwnerUserId="2378" LastActivityDate="2013-09-26T04:42:35.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6327" PostTypeId="1" CreationDate="2013-09-26T05:21:14.687" Score="1" ViewCount="695" Body="&lt;p&gt;I was wondering if someone could help me find references available that will help me solve the following problems. These are &lt;em&gt;not&lt;/em&gt; homework problems, but are suggested problems that were given to me by my instructor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fD2nb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm specifically interested in learning how to solve question A. (b). How can I know whether a reaction will occur between the two products and further, how can I predict what the product will be if a reaction does occur (what atoms bond together, and how can I know which atoms bond together in any organic reaction)?&lt;/p&gt;&#xA;" OwnerUserId="2376" LastActivityDate="2013-12-25T18:34:22.107" Title="References for learning about acid and bases in organic chemistry" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;reaction-mechanism&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6328" PostTypeId="2" ParentId="6327" CreationDate="2013-09-26T06:00:30.480" Score="2" Body="&lt;p&gt;Concerning your first question: I found that the  book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199270295&quot; rel=&quot;nofollow&quot;&gt;&quot;Organic Chemistry&quot; by Clayden, Warren, Wothers and Greeves&lt;/a&gt; and here especially the 8th chapter contains a very good and readable introduction to acidity and basicity. After you have worked through this chapter you should be able to explain why certain molecules are as basic/acidic as they are and you see a lot of typical $\mathrm{p}K_{\mathrm{a}}$ values for a lot of substance classes so that you get some knowledge about what $\mathrm{p}K_{\mathrm{a}}$ values you can expect for other substances. As for your second question: If you want to know why reactions occur in the way they do you'll have to take courses in Organic Chemistry. Apart from that, I wholeheartedly recommend the aforementioned book again. If you work through this book (this will take you a while: it has 1200 pages... but then again, that is 1200 very readable pages :)) you'll have a very solid basis in organic reaction mechanisms and organic chemistry in general. If you want to dig a little deeper &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0121381102&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; book by Bruckner is very thorough but a little dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One particular perspective that I found very helpful in explaining why reactions happen in the way they do was the frontier orbital theory. A very good and readable book about its use in organic chemistry is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470746599&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one by Fleming. The book by Clayden et al. also uses this theory a lot in explaining reaction mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-09-26T15:57:47.653" LastActivityDate="2013-09-26T15:57:47.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6330" PostTypeId="1" AcceptedAnswerId="6331" CreationDate="2013-09-26T08:28:32.200" Score="3" ViewCount="284" Body="&lt;p&gt;I see that Gaussian has feature to set periodic boundary conditions specifying the parameter Tv in the input file. Does it do it via the plane wave basis set? Also, the question is whether it is efficient to use Gaussian to compute the relaxed lattice constant, band structure and DOS for solids or it is better to use programs like ABINIT.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2013-09-26T10:35:47.020" LastActivityDate="2013-09-26T10:35:47.020" Title="Gaussian vs ABINIT for solids" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6331" PostTypeId="2" ParentId="6330" CreationDate="2013-09-26T08:47:43.950" Score="6" Body="&lt;p&gt;Even if it is possible to calculate periodic structures in Gaussian, I would not suggest it. &#xA;Better option is to use codes devoted to calculation of solids, among others the ABINIT you mentioned, or e.g. Quantum Espresso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaussian does not use plane wave basis set, will be slower and in general less powerful than codes devoted to solids.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-09-26T08:47:43.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6333" PostTypeId="1" CreationDate="2013-09-26T09:59:29.677" Score="2" ViewCount="91" Body="&lt;p&gt;I have some questions on the nomenclature-&lt;br&gt;&#xA;(1) In $H_2O$ we have $H$ written first and $O$ is second but when it comes to Hydroxyl group $OH$ the $O$ comes first, Why?&#xA;&lt;br&gt;&#xA;(2)Why is Propanal called Propion-aldehyde and not Propane- aldehyde? &lt;/p&gt;&#xA;" OwnerUserId="2382" LastActivityDate="2013-09-26T10:59:00.677" Title="On nomenclature of certain compounds" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6335" PostTypeId="2" ParentId="6333" CreationDate="2013-09-26T10:24:44.320" Score="2" Body="&lt;p&gt;To answer (2) first, propionaldehyde is the common (Trivial) name and hence is derived and held-onto only due to historical reasons and now obsolete nomenclature reasons. The 3-C aliphatic saturated qroup was historically called Propion- and hence propanoic acid corresponds to propionic acid and propanal corresponds to propionaldehyde, and propanoate(ester) corresponds to propionate in trivial nomenclature. It is a consequence of the history of the original naming of the 3-C aliphatic group at the time of its discovery.&lt;br&gt;&#xA;As for (1), it is a consequence of pure convention; just as water is not called hydrogen hydroxide! &lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-09-26T10:59:00.677" LastActivityDate="2013-09-26T10:59:00.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6336" PostTypeId="2" ParentId="6118" CreationDate="2013-09-26T10:54:50.160" Score="0" Body="&lt;p&gt;Orbital is &quot;one electron wavefunction&quot;, introduced in order to make the treatment of full wavefunction tractable, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater_determinant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Slater_determinant&lt;/a&gt;. You should not take an orbital away from the complete wavefunction and ask, what-if. Because the complete wave function and therefore all the orbitals will adopt to the new conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, by adding electron, you change the overall charge. What the change is depends on the exact nature of the process, so it is different if you start with radical-cation and end up with neutral molecule, or start with neutral radical and end up with anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best option is to just play around with any reasonably simple code for electronic structure calculations, be it semi-empiric for simplicity and speed, e.g. &lt;a href=&quot;http://www.openmopac.net&quot; rel=&quot;nofollow&quot;&gt;OpenMOPAC&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-09-26T10:54:50.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6337" PostTypeId="1" CreationDate="2013-09-26T10:56:24.040" Score="5" ViewCount="1393" Body="&lt;p&gt;The thermodynamic efficiency of any cell (especially Fuel cells) is given as $$\frac{\Delta G}{\Delta H} \times 100$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood this partly, that since $\Delta G$ is the useful work obtained in the ideal case, the efficiency expression must have it in the numerator. But I fail to see the intuition (or maths) behind the $\Delta H$ term in the denominator. I have two questions regarding this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;On applying this equation to ordinary (not fuel cells), for a general reaction $$\ce{A^+ {(aq)} + B {(s)} -&amp;gt; A {(s)} + B^+ {(aq)}}$$ the $\Delta G$ term is given as $-nF E_{\text{cell}}$ where $E_{\text{cell}}$ depends on the concentration of the two ionic species as given by the Nernst equation $\left(E=E^\circ-\frac{RT}{nF}\ln\frac{[\ce B^+]}{[\ce A^+]}\right)$. On the other hand, is the $\Delta H$ term independent of the concentration? and if not, how would it (and consequently efficiency) depend on the ionic concentrations?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For fuel cells, suppose the cell reaction is such that the $\Delta S$ is positive and since $\Delta G=\Delta H -T\Delta S$, therefore $|\Delta G|&amp;gt;|\Delta H|$ since the second term in the equation for $\Delta G$ becomes negative due to positive entropy change. Would this predict an efficiency greater than 100%?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-05-17T11:12:10.140" LastActivityDate="2018-05-17T11:12:10.140" Title="Clarifications on thermodynamic cell efficiency" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6338" PostTypeId="2" ParentId="6118" CreationDate="2013-09-26T11:38:37.387" Score="4" Body="&lt;p&gt;While what ssavec says about not taking orbitals away from a structure is true, it's possible to make some generalisations, particularly for lone atoms where the symmetry constraints make the orbital divisions more clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you generate orbitals for an atom, you get the set of &lt;em&gt;s&lt;/em&gt;, &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;d&lt;/em&gt;, &lt;em&gt;f&lt;/em&gt; (and so on) orbitals separated out, and you can &lt;em&gt;generally&lt;/em&gt; say that for the same principal quantum number (usually denoted &lt;em&gt;n&lt;/em&gt;), a doubly-filled orbital will produce more diffuse orbitals than a singly-filled one. This is an obvious consequence of Coulombic repulsion. But, again, all the other orbitals you generate for the altered wave-function will be affected by the change in occupancy and overall charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be possible to construct a somewhat artificial pair of calculations in which you kept fixed all but one of the occupied orbitals, and then optimised the wavefunction using one electron in one calculation and two in the other, and I would expect you to find that in the calculation with two electrons, the varied orbital would be more diffuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two orbitals would be different in spatial form, you wouldn't expect the value at a given point to be a multiple of the other.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-09-26T11:38:37.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6339" PostTypeId="2" ParentId="6300" CreationDate="2013-09-27T01:29:26.627" Score="4" Body="&lt;p&gt;You won't find a general formula for the shelf life of a solution.  A silver nitrate solution will have a very short shelf life if it is kept in the sunlight, whereas potassium nitrate could last for years.  The shelf life is also going to be very strongly dependent on the quality of the container.  &lt;a href=&quot;http://www.sommelierjournal.com/articles/article.aspx?year=2008&amp;amp;month=6&amp;amp;articlenum=47&quot; rel=&quot;nofollow&quot;&gt;Think about what happens&lt;/a&gt; to the aging of a wine bottle if the cork does not provide a sufficient seal .  Interactions between the solution and the container have an impact on the solution shelf life as well.  To stick with the wine theme, &lt;a href=&quot;http://www.thekitchn.com/what-is-corked-wine-what-does-corked-wine-taste-like-164148&quot; rel=&quot;nofollow&quot;&gt;corked wine&lt;/a&gt; is due to contamination of the wine with 2,4,6-trichloroanisole which is generated by natural fungi in the cork that have been exposed to chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is important to understand the chemistry of the solution with which you are working in order to determine its shelf life.  &lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-09-27T01:29:26.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6340" PostTypeId="1" CreationDate="2013-09-27T03:18:28.530" Score="2" ViewCount="817" Body="&lt;p&gt;What is the proper procedure for working with KHMDS?  I would like to use it to deprotonate a compound and have a bottle of KHMDS in toluene.  Does it need to be handled under inert atmosphere?  And is it soluble in halogenated solvents?&lt;/p&gt;&#xA;" OwnerUserId="1972" LastActivityDate="2013-09-27T12:21:25.260" Title="How to properly work with potassium HMDS" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6341" PostTypeId="1" CreationDate="2013-09-27T04:08:00.213" Score="6" ViewCount="36428" Body="&lt;p&gt;If the wavefunction of the electron in a box of length $L$ is $$\psi{(x)}=\sqrt{\frac{2}{L}}\cdot\sin{\frac{n\pi x}{L}}$$&#xA;What would be the probability of finding the electron between $0.25L$ and $0.75L$?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From postulate 1 of quantum mechanics I know that $$1=\int_{0}^{L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying out a similar reasoning leads me to think that the required probability is the integral $$ \int_{0.25L}^{0.75L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}$$ which gives the answer as $0.5$. But the book gives the answer as $0.82$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the problem with my method and can anyone explain me why the answer is $0.82$?&lt;/p&gt;&#xA;" OwnerUserId="2385" LastEditorDisplayName="user7951" LastEditDate="2018-07-27T18:56:50.727" LastActivityDate="2019-08-13T20:19:18.273" Title="How to calculate the probability of finding an electron in a box between 0.25L and 0.75L?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6342" PostTypeId="2" ParentId="6341" CreationDate="2013-09-27T07:35:34.773" Score="8" Body="&lt;p&gt;You have the reasoning correct, so I suspect you made a mistake in your integral calculation. The probably of finding the particle between $a$ and $b$, for the $n$-th state, is indeed given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_n(a,b) = \int_a^b \left|\psi_n(x)\right|^2\, \mathrm{d}x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if I do the calculation for the ground state ($n = 1$), I find the same results as your textbook:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818 $$&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-09-27T07:35:34.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6343" PostTypeId="1" AcceptedAnswerId="6346" CreationDate="2013-09-27T08:48:08.053" Score="4" ViewCount="539" Body="&lt;p&gt;What complicated the synthesis of upsalite (anhydrous $\ce{MgCO3}$), for it to be synthesized and published in 2013 as opposed to the 20th century, as one would expect for such a simple salt?&lt;/p&gt;&#xA;" OwnerUserId="660" LastEditorUserId="4231" LastEditDate="2015-03-05T15:50:44.520" LastActivityDate="2015-03-05T15:50:44.520" Title="Why was anhydrous magnesium carbonate “upsalite” discovered so late?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6344" PostTypeId="2" ParentId="6290" CreationDate="2013-09-27T08:50:13.547" Score="1" Body="&lt;p&gt;I think that the problem is that you have meat-grease that creates a hydrophobic layer inside tiny holes of your granite and wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend, with all the necessary precautions (eyeglasses and gloves), a solution of hot caustic soda for your granite. This will saponify the fat so you can remove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the amines, you have to use an acid: HCl and vinegar are good choices but I would try citric acid too. Hot solutions tend to be more efficient.    &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="402" LastEditDate="2019-05-19T20:43:00.727" LastActivityDate="2019-05-19T20:43:00.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6345" PostTypeId="2" ParentId="6340" CreationDate="2013-09-27T12:21:25.260" Score="2" Body="&lt;p&gt;If you would look up the properties of KHMDS from a chemical supplier, you will find it will react violently with with water. So do keep it under inert atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for solubility, it is a non-nucleophilic base, so probable no substitution and addition like reactions will occur. I wont suspect it will be a problem there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment on the use though. If you plan to use this, explore the dangers and possibilities of a substance like this a bit further than asking it here. For safety considerations. Look up the MSDS sheets from your supplier and be sure to only use this in a properly equipped lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-09-27T12:21:25.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6346" PostTypeId="2" ParentId="6343" CreationDate="2013-09-27T13:55:55.873" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0068486#s1&quot; rel=&quot;nofollow&quot;&gt;paper describing the synthesis of Upsalite&lt;/a&gt; has a clear introduction explaining the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Magnesium is the eighth most abundant element in the earth’s crust and essential to most living species. It can form several structures of hydrated carbonates such as nesquehonite ($\ce{MgCO3·3H2O}$), and lansfordite ($\ce{MgCO3·5H2O}$), a number of basic carbonates such as hydromagnesite ($\ce{4 MgCO3·Mg(OH)2·4 H2O}$), and dypingite ($\ce{4 MgCO3·Mg(OH)2·5 H2O}$), as well as the anhydrous and rarely encountered magnesite (MgCO3) [5]. In contrast to other alkali earth metal carbonates, chemists have found anhydrous magnesium carbonate difficult to produce, particularly at low temperatures. Above 100°C, magnesite (crystalline $\ce{MgCO3}$) can be obtained from $\ce{Mg(HCO3)2}$ solutions by precipitation. However, at lower temperatures, hydrated magnesium carbonates tend to form, giving rise to what has been referred to as “the magnesite problem” [6].Yet, not only chemists have been intrigued by magnesium carbonates. Although abundant in nature, where crystalline forms exist as traces in most geological structures, pure magnesium carbonate is seldom found on its own in larger deposits, a fact that has puzzled geologist for more than a century [7].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, anhydrous crystalline magnesium carbonate has been synthesized for a long time (and occurs in nature), in the form of a dense phase: &lt;em&gt;magnesite&lt;/em&gt;. The new material reported under the name of &lt;strong&gt;upsalite is simply an porous amorphous &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymorphism_%28materials_science%29&quot; rel=&quot;nofollow&quot;&gt;polymorph&lt;/a&gt; of magnesite&lt;/strong&gt;. New polymorphs of inorganic structures are regularly discovered/synthesized by researchers, and reported in the literature. The most striking example may be the family of &lt;a href=&quot;http://en.wikipedia.org/wiki/Zeolite&quot; rel=&quot;nofollow&quot;&gt;zeolites&lt;/a&gt;, with more than 200 polymorphs of aluminosilicate.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="4231" LastEditDate="2015-03-05T15:50:39.760" LastActivityDate="2015-03-05T15:50:39.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6347" PostTypeId="1" AcceptedAnswerId="6357" CreationDate="2013-09-27T16:03:20.387" Score="1" ViewCount="5375" Body="&lt;p&gt;I have came across the following statement : &quot;the oxidation state $+II$ is stable for the elements of the alkalin earth metals&quot; , the same is true for the elements of the first group , i've looked for a possible explanation and heard that it has a relation with the charge densities of the elements , if that's true what is that relation ? otherwise , what could explain the oxidation states for each of the groups in the periodic table ?&lt;/p&gt;&#xA;" OwnerUserId="1463" LastEditorUserId="1463" LastEditDate="2013-09-28T18:03:19.300" LastActivityDate="2013-09-28T18:03:19.300" Title="Oxidation state of group 2 elements" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6348" PostTypeId="1" AcceptedAnswerId="6349" CreationDate="2013-09-27T17:34:46.627" Score="3" ViewCount="714" Body="&lt;p&gt;I use Sodium persulfate to etch a copper board to manufacture home made Printed Circuit Board (PCB), and I use the same etching tank on the below photo which has no cover on the top. I wonder if it is okay to keep it like that at the room temperature or would it evaporate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6AZn.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="1940" LastActivityDate="2016-02-16T13:09:40.190" Title="does Sodium persulfate evaporate at room temperature" Tags="&lt;acid-base&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6349" PostTypeId="2" ParentId="6348" CreationDate="2013-09-27T18:05:58.577" Score="3" Body="&lt;p&gt;Regardless of evaporation, it is an oxidizer and unhealthy so it is recommended to keep it in a closed container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that being said, you most likely have a solution of sodium persulfate in water.  Water will evaporate so again, yes, keep it closed.&lt;/p&gt;&#xA;" OwnerUserId="1689" LastActivityDate="2013-09-27T18:05:58.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6350" PostTypeId="1" AcceptedAnswerId="6356" CreationDate="2013-09-27T23:03:10.377" Score="7" ViewCount="7862" Body="&lt;p&gt;In &lt;a href=&quot;http://www.imdb.com/title/tt1487706/&quot;&gt;The Big Bang Theory Season 3 Episode 1 &quot;The Electric Can Opener Fluctuation&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leonard gives Penny a &lt;a href=&quot;http://en.wikipedia.org/wiki/Snowflake&quot;&gt;snowflake&lt;/a&gt; that will never melt. He said some type of chemical gas name when he try to explain, but I don't understand what he said exactly so I can't do any research about that. You can watch it on this &lt;a href=&quot;http://www.youtube.com/watch?v=J4zZ-fzZX74&quot;&gt;YouTube video&lt;/a&gt; as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Google for some searches like &lt;em&gt;doesn't melt snowflake&lt;/em&gt; and &lt;em&gt;insoluble snowflake&lt;/em&gt; but results gave me nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is this possible to save a snowflake preserved in glass with a chemical element, gas or something?&lt;/p&gt;&#xA;" OwnerUserId="680" LastEditorUserId="73" LastEditDate="2013-09-28T03:04:01.830" LastActivityDate="2016-01-27T21:22:23.167" Title="Is it possible to save an insoluble snowflake?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6351" PostTypeId="1" AcceptedAnswerId="6360" CreationDate="2013-09-27T23:10:54.670" Score="5" ViewCount="1264" Body="&lt;p&gt;I am learning quantum chemistry and I am trying to understand the statement in a .pdf I'm reading which states&#xA;&quot;The energy transition associated with the spectral transitions observed is:&#xA;$$\Delta E = E_{LUMO} - E_{HOMO} = \frac{(N+1)h^2}{8mL^2}$$&#xA;where $L$ is the length of the &quot;box&quot; (it's talking about particle in a box), $m$ is the mass of the electron, and $N$ is the number of $\pi$-electrons in the system. I know due to the solution of the Schrodinger equation that $$E = \frac{n^2 h^2}{8mL}$$&#xA;So, $$\Delta E = E_{HOMO} - E_{LUMO} = E_{k+1} - E_{k} = \frac{(2k+1)h^2}{8mL^2}$$&#xA;So, I want to figure out why $2k = N$?&lt;/p&gt;&#xA;" OwnerUserId="2386" LastEditorUserId="4231" LastEditDate="2015-03-05T15:50:32.147" LastActivityDate="2015-03-05T15:50:32.147" Title="Number of pi-electrons in an energy state" Tags="&lt;energy&gt;&lt;quantum-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6352" PostTypeId="1" CreationDate="2013-09-28T01:52:48.507" Score="0" ViewCount="5003" Body="&lt;p&gt;Can someone please explain the intuition around the answers to this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j9xjJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm finding it very confusing because it seems like you have to nitpick between two resonance structures to get the desired results. Do we just always look for the &quot;lowest&quot; possible VSEPR in any resonance structure? Why isn't the observable geometry for &quot;X&quot; linear instead of none?&lt;/p&gt;&#xA;" OwnerUserId="2376" LastEditorUserId="275" LastEditDate="2013-09-28T02:20:24.370" LastActivityDate="2013-09-28T14:00:55.413" Title="Predicting electronic geometry, observable geometry, and hybridization for any atom in a molecule" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;hybridization&gt;&lt;resonance&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6353" PostTypeId="1" CreationDate="2013-09-28T01:59:06.193" Score="0" ViewCount="3175" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/koSM2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we deprotonate that specific hydrogen as opposed to any other one? What information does the table on the right give us? It seems unrelated to citric acid - where does the fluorine come from?&lt;/p&gt;&#xA;" OwnerUserId="2376" LastEditorUserId="4945" LastEditDate="2015-02-25T04:55:56.977" LastActivityDate="2015-02-25T04:55:56.977" Title="How will citric acid be deprotonated by pyridine?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6354" PostTypeId="2" ParentId="6353" CreationDate="2013-09-28T02:18:48.157" Score="5" Body="&lt;p&gt;The table on the right looks like a standard group of acids shown to illustrate the effect of inductive withdrawing on carboxylic acid pKa. The first two structures are relevant to the question. The table may have been copied from somewhere else because it had both structures on it. Two of the pKa values in the table are every similar to two of the pKa values given in the question. The fluoroacetic acid is a fine distraction; you are correct in thinking that it is not involved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As for figuring out which of citric acid's three carboxylic acids is deprotonated first...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Citric acid has three carboxylic acid groups &lt;strong&gt;of two different types&lt;/strong&gt;. Can you see that? One of the acids is closer to the OH group than the other two. Which acid or acids look more like the first entry in the table (except jammed onto the citric acid structure)? Which acid or acids look more like the second entry on the table? Once you have made that distinction, you should be able to use the pKa data on the table to determine which of the three carboxylic acid groups in citric acid is the most acidic (goes with the 3.13 pKa).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-28T02:18:48.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6356" PostTypeId="2" ParentId="6350" CreationDate="2013-09-28T10:49:10.660" Score="11" Body="&lt;p&gt;With gas no absolutely. To preserve ice at room temperature using the pressure of a gas you should reach a pressure of one billion Pascals. Too dangerous for Penny safe...&lt;/p&gt;&#xA;&lt;p&gt;In fact there are some imprecisions in the video Leonard cites two different methods to &amp;quot;preserve&amp;quot; snow flakes with an organic polymer. In fact these methods create a snow flakes replicas, they are called &lt;strong&gt;snow crystal fossils&lt;/strong&gt;, because there is no more ice inside them. Both methods are describe &lt;a href=&quot;http://www.its.caltech.edu/%7Eatomic/snowcrystals/preserve/preserve.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, have a look!&lt;/p&gt;&#xA;&lt;h1&gt;Formvar Replicas&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Leonard: &amp;quot;it will last forever I preserve it in a 1% solution of&#xA;polyvinyl acetal resin&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is the Schaefer and Day technique [1] the solution Leonard is talking about is made from 1% of polyvinyl acetal resin in ethylene dichloride.&lt;/p&gt;&#xA;&lt;h1&gt;Superglue Replicas&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Leonard: &amp;quot;in fact is a pretty simple process, you see, Cyanoacrylate&#xA;are monomers which polymerize mmm... &amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Cyanoacrylate are the main components of &lt;em&gt;Super Glue&lt;/em&gt;  this is easy method propose by the chemist Tryggvi Emilsson that you can do at home.&lt;/p&gt;&#xA;&lt;p&gt;You have to choose one of these methods, you can't do both of them as Leonard (it seems) did... Take in account that Leonard is a physicist he doesn't know much about this stuff...;-)&lt;/p&gt;&#xA;&lt;p&gt;[1] V. J. Schaefer and J. A. Day, Peterson Field Guides: Atmosphere (Houghton Mifflin, 1981).&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-09-28T11:24:12.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6357" PostTypeId="2" ParentId="6347" CreationDate="2013-09-28T13:58:41.700" Score="2" Body="&lt;p&gt;The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOOWr.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for the alkali metals, each element has one more electron than a noble-gas configuration.  This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MlcKS.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much harder to remove the 2nd electron, since it is contained within a completely filled shell.  (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thought process can be applied to the alkaline earth metals.  Here are the first three ionization eneriges for each:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/98uOY.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove.  Thus, the metals in this group tend to stay in the +2 oxidation state.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-09-28T13:58:41.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6358" PostTypeId="2" ParentId="6352" CreationDate="2013-09-28T14:00:47.630" Score="1" Body="&lt;p&gt;I've never seen this distinction made between &quot;VSEPR geometry&quot; and &quot;observed geometry&quot;, and it does not make much sense to me. It would seem more correct that what they call &quot;VSEPR geometry&quot; includes the lone pairs, while their &quot;observed geometry&quot; is only the arrangement of atoms, ignoring the lone pairs. Thus, geometry around X is ill-defined (&quot;none&quot;), because it has only one neighbor (linear geometry require two neighbors, like the central C in O=C=O).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-09-28T14:00:47.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6359" PostTypeId="2" ParentId="6352" CreationDate="2013-09-28T14:00:55.413" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we just always look for the &quot;lowest&quot; possible VSEPR in any resonance structure? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Consider the two resonance contributors below for a N,N-dimethylacetamide, which has a functional group in common with your molecule. The rule of thumb I teach is: **if an atom appears to be $sp^2$-hybridized in one resonance contributor and $sp^3$ in another, then that atom is $sp^2$-hybridized always. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DFJen.png&quot; alt=&quot;resonance contributors to amide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure above, I have used the curved arrow notation to represent how the lone pair on the nitrogen atom is delocalized into the $\pi$-system. In order for this delocalization to occur, that lone pair needs to be in a $p$ orbital. In order for that nitrogen atom to have a $p$ orbital, it must be $sp^2$-hybridized (or possibly $sp$-hybridized if applicable). All three of the atoms (N, C, and O) in the delocalized system need a p orbital, and thus must be $sp^2$-hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have hybridization, then you have limited your choices of geometry:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$sp^3 \implies$ tetrahedral geometry group: tetrahedral, trigonal pyramidal, bent, &quot;none&quot;&lt;/li&gt;&#xA;&lt;li&gt;$sp^2 \implies$ trigonal planar group: trigonal planar, bent, &quot;none&quot;&lt;/li&gt;&#xA;&lt;li&gt;$sp^{\ \ \  }  \implies$ linear group: linear, &quot;none&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why isn't the observable geometry for &quot;X&quot; linear instead of none?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are confusing observable geometry at an atom with molecular geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observable geometries &lt;strong&gt;at an atom&lt;/strong&gt; are a relationship describing the relative geometric positions of other atoms bonded to that central atom. The central atom is not counted.A linear arrangement means that the atom in question is bonded to two other atoms or groups and those two atoms or groups are $180^\circ$ apart. If there is only one atom or group bonded to the atom in question, we cannot define the geometric relationship between that atom or group and the other atoms or groups bonded to that atom because there are not other atoms or groups. Consider the following examples from the tetrahedral group. The last case is &quot;none&quot; or &quot;undefined, because you need at least two things to have a relationship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GtE7m.png&quot; alt=&quot;tetrahedral group geometries&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is different than molecular geometry which deals with the shape of the entire molecule. Let's look at HF. The shape of the molecule is &lt;strong&gt;linear&lt;/strong&gt;, because the two points define a line (segment). The geometry around F is &quot;none&quot; or undefined. We cannot describe the arrangement in space of the hydrogen atom relative to other atoms bonded to fluorine because there are none.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-09-28T14:00:55.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6360" PostTypeId="2" ParentId="6351" CreationDate="2013-09-28T14:04:39.273" Score="3" Body="&lt;p&gt;Your calculation is indeed correct. However, because each electron in the $\pi$ system can have two spin states, there are two electrons per energy level of the “particle in a box” Hamiltonian. Hence, $N=2k$.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-09-28T15:49:59.773" LastActivityDate="2013-09-28T15:49:59.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6361" PostTypeId="2" ParentId="4033" CreationDate="2013-09-28T22:48:47.133" Score="3" Body="&lt;p&gt;It might be possible to deposit a metallic film on a sheet of plastic. It is however, a problem with respect to the bonding of that layer. Depending on the plastic used, it might not have a strong bond with the metal, which would then over time develop defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it might be possible, when carefully manufacturing this, that the plastic has 'functional groups' (parts of the molecule with a specific function) that would bind a metal. Note however, that in this case, the functional groups might also react with other chemicals. Almost no functional group will only under go one reaction, especially not when metals are involved. Therefore, this would not be the most durable of options. Furthermore, because of the interface between, most likely, an amorphous solid (&lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_solid&quot; rel=&quot;nofollow&quot;&gt;wikipedia, amorphous solids&lt;/a&gt;) and a metal structure is bound to create some defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A last option I can think of, is the use of conductive polymers. These polymers, because in many ways they mimic metals, also have a similar 'shininess' to them. However, I'm not sure about their really being a mirror. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-09-28T22:48:47.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6366" PostTypeId="1" CreationDate="2013-09-30T01:10:04.683" Score="1" ViewCount="269" Body="&lt;p&gt;Can someone explain chelation and its relation to monodentate and tridentate ligands? I dont understand what they mean by chelate effect and chelating rings. &lt;/p&gt;&#xA;" OwnerUserId="1554" LastActivityDate="2014-04-01T18:58:39.690" Title="Chelation and its relation to transition metal ligands" Tags="&lt;periodic-table&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6367" PostTypeId="1" AcceptedAnswerId="6569" CreationDate="2013-09-30T01:30:43.237" Score="23" ViewCount="40026" Body="&lt;p&gt;In organic chemistry we can classify carbon atoms as primary, secondary, tertiary, or quaternary based on the number of additional carbon atoms bonded to the carbon atom of interest. The definitions in the textbook I teach from (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471756148&quot;&gt;Klein, 1st Ed. Wiley, 2011&lt;/a&gt;), as well as in several others I have, and from from &lt;a href=&quot;http://en.wikipedia.org/wiki/Secondary_carbon&quot;&gt;several locations&lt;/a&gt; &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/06/16/1%C2%B0-2%C2%B0-3%C2%B0-4%C2%B0/&quot;&gt;around the web&lt;/a&gt;, are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Primary $(1^\circ)$ carbon atom - bonded to one other carbon atom, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H2CH3}$&lt;/li&gt;&#xA;&lt;li&gt;Secondary $(2^\circ)$ carbon atom - bonded to two other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H(CH3)2}$&lt;/li&gt;&#xA;&lt;li&gt;Tertiary $(3^\circ)$ carbon atom - bonded to three other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{(CH3)3}$&lt;/li&gt;&#xA;&lt;li&gt;Quaternary $(4^\circ)$ carbon atom - bonded to four other carbon atoms, e.g. ${\bf \color{red}{\ce{C}}}\ce{(CH3)4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this labeling system limited to $sp^3$-hybridized carbon atoms? I do not see anything in the definitions that exclude $sp^2$- or $sp$-hybridized carbon atoms. Nor have I seen or read anything that explicitly excludes these atoms. However, the images in textbooks and around the web all show and discuss only $sp^3$-hybridized carbon atoms in this context. Are the following valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central atom in propene $\ce{CH3 -}{\bf\color{red}{\ce{C}}}\ce{H=CH2}$ is a $2^\circ$ carbon atom because it is bonded to two other carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon atoms in acetylene $\ce{HC#CH}$ are $1^\circ$ carbon atoms because they each are bonded to one other carbon atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ipso carbon in toluene $\ce{C6H5CH3}$ (the carbon atom where the $\ce{CH3}$ is attached) is a $3^\circ$ carbon atom because it is bonded to three other carbon atoms.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2017-05-13T14:36:45.440" LastActivityDate="2017-05-13T14:36:45.440" Title="Is the nomenclature &quot;primary&quot;, &quot;secondary&quot;, &quot;tertiary&quot;, and &quot;quaternary&quot; carbon atoms limited to sp3 carbon atoms?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6369" PostTypeId="1" CreationDate="2013-09-30T14:15:38.377" Score="9" ViewCount="12542" Body="&lt;p&gt;Can it be isolated at room temperature?  Or even at any temperature?&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorUserId="5026" LastEditDate="2018-06-13T09:46:34.120" LastActivityDate="2018-06-13T09:46:59.580" Title="Why is aluminium carbonate unstable?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6370" PostTypeId="1" CreationDate="2013-09-30T16:23:00.197" Score="3" ViewCount="148" Body="&lt;p&gt;A person at wikipedia is getting on my case for creating this image because the N and H symbols are so separated even though this seems to be the default of &lt;a href=&quot;http://www.acdlabs.com/resources/freeware/chemsketch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemSketch&lt;/a&gt; (one of the softwares that's recommended for chem structure drawing in the Wikipedia manual of style) and I'd like to know how to make them happy and get the separation between the N and H shortened without changing the style and letter sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qzTEM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2013-09-30T16:23:00.197" Title="How can I get ChemSketch to get N and H atoms closer while still operating in ACS style and without adjusting atom sizes?" Tags="&lt;software&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6372" PostTypeId="5" CreationDate="2013-09-30T21:31:55.230" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-09-30T21:31:55.230" LastActivityDate="2013-09-30T21:31:55.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6373" PostTypeId="4" CreationDate="2013-09-30T21:31:55.230" Score="0" Body="For questions relating to Raman scattering/effect and its applications as spectroscopy technique." OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2013-09-30T22:18:45.273" LastActivityDate="2013-09-30T22:18:45.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6374" PostTypeId="2" ParentId="6369" CreationDate="2013-10-01T10:13:49.640" Score="12" Body="&lt;p&gt;Aluminium carbonate reacts easily with water/water vapour to form aluminium hydroxide as you can see on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_carbonate#Decomposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al2(CO3)3 + 3 H2O → 2 Al(OH)3 + 3 CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to store aluminium carbonate, you have to make sure it is free from moisture.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why it is unstable:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller and more charged metal cations (i.e. metal cations with high charge density) tend to distort the electron cloud of the carbonate ions (polarisation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polarisation eventually leads to abstraction of oxygen from the carbonate ion, making one of the $\ce{C-O}$ bond in the carbonate weak, which makes it very prone to attack by water and heat (similar to the decomposition of copper (II) carbonate when compared to alkali metal carbonates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction itself doesn't release a lot of energy; hence why it is rather unstable and not reactive in chemistry terms.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="5026" LastEditDate="2018-06-13T09:46:59.580" LastActivityDate="2018-06-13T09:46:59.580" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6375" PostTypeId="1" AcceptedAnswerId="6376" CreationDate="2013-10-01T18:24:47.573" Score="0" ViewCount="143" Body="&lt;p&gt;$\ce{H2C=C=O}$ what is the group of the  molecule? How to characterize it and how  is this dissimilar to ketone? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?&lt;/p&gt;&#xA;" OwnerUserId="2400" LastEditorUserId="418" LastEditDate="2014-03-21T22:47:38.910" LastActivityDate="2014-03-21T22:47:38.910" Title="Naming of a molecule" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2013-10-01T21:51:12.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="6376" PostTypeId="2" ParentId="6375" CreationDate="2013-10-01T21:42:44.397" Score="2" Body="&lt;p&gt;As, your question isn't much clear and it needs to be focused little more.&#xA;From current state of this question, I can provide you the answers for specific questions that &lt;strong&gt;I get to understand from its current state.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H_2 \ \ C=C=O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What is the group of the  molecule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Well this is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Ketene&quot; rel=&quot;nofollow noreferrer&quot;&gt;ketene&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;how to characterize it and how this is dissimilar to ketone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/6/69/Ketenes.png&quot; alt=&quot;Structure of ketene&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ketenes are generally very reactive, and participate in various cycloadditions. They will also undergo [2+2] cycloaddition reactions with electron-rich alkynes to form cyclobutenones, or carbonyl groups to form beta-lactones.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/40/Ketone-group-2D-skeletal.svg&quot; alt=&quot;Structure of ketone&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Ketones engage in many organic reactions. The most important reactions follow from the susceptibility of the carbonyl carbon toward nucleophilic addition and the tendency for the enolates to add to electrophiles.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ketene&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ketone&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If we add &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; with it then will we get &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No! I hope structure of ketene and wikipedia articles make it clear to you why. :)&lt;/p&gt;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Ketene_cycloaddition&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-10-01T21:42:44.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6377" PostTypeId="1" AcceptedAnswerId="6379" CreationDate="2013-10-02T08:17:08.600" Score="4" ViewCount="1526" Body="&lt;p&gt;Every now and then you reach into the bathroom cabinet to find that some pill or ointment you where planning to use has gone past its expiration date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested whether this expiration date has any value. In a short search I found &lt;a href=&quot;http://www.health.harvard.edu/fhg/updates/update1103a.shtml&quot; rel=&quot;nofollow&quot;&gt;this article from Harvard medical school&lt;/a&gt; stating that some test found that 90% of all drugs where still perfectly active 15(!) years after the expiration date. The article then continues to give some vague statements on possible reasons for the expiration date (including financial reasons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am wondering is whether there would be any chemical reasons for drugs to become less effective or even become harmful over time, which could warrant the existence of an expiration date?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a specific example, in case it is difficult to answer this question in general, let's take an ointment meant to treat muscle pains which use &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzydamine&quot; rel=&quot;nofollow&quot;&gt;Benzydaminehydrochloride&lt;/a&gt; as an active component.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-10-02T08:45:55.173" Title="Why do pharmaceutical drugs have a expiration date?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6378" PostTypeId="2" ParentId="6377" CreationDate="2013-10-02T08:43:40.050" Score="3" Body="&lt;p&gt;A good question, I'd say!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the reason drugs 'go bad', it is true, that if the molecules are stable, they could last for a long, long time and in some cases it is perfectly possible that they would not harm anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, molecules used as medication etc. are often highly functional. This means, they contains parts that perform a reaction, bind to proteins in your body or something like that. The reason the drugs work, is this functionality. Is is, however a very good possibility for a functional group to perform a reaction that is not desired. The reason for performing the wrong reaction, is that the functional groups of the molecules are not 100% selective for their use in the body. So if any other reaction goes on with the drug, it loses/changes functionality and could even become harmful. Another term used for this process is 'degrading'. Part of the active medicine in the drug has changed into something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a wide range of drugs available, it is not possible to always know which degradation product will form. Therefore, it is also not possible to ensure the drugs work, and are safe, for years and years. For health and safety reasons then, a 'best before date' or expiration date is added. This date works both ways, warns you that the drugs might be harmful, and protects the manufacturer of the drugs against misuse and lawsuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped!&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2013-10-02T08:43:40.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6379" PostTypeId="2" ParentId="6377" CreationDate="2013-10-02T08:45:55.173" Score="6" Body="&lt;p&gt;It's a good question, and I cannot answer your specific example but can give some generic reasons why pharmaceuticals have an expiration date:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Degradation&lt;/strong&gt; of the compound/formulation over time. This obviously comes to mind, but there is more to it than meets the eye. For example, you gave the example of an ointment and cited the active ingredient: however, the product you buy (and use) is more than just the active molecule. The whole formulation has to be stable, not only the benzydamine. The ointment is likely to be a complex &lt;a href=&quot;http://en.wikipedia.org/wiki/Emulsion&quot; rel=&quot;noreferrer&quot;&gt;emulsion&lt;/a&gt;, designed to target many different physical and chemical properties, including chemical stability, thermal stability, but also viscosity (so it's easy to apply), color, etc. I suppose in that particular case, the emulsion might be less stable than the active ingredient taken alone.&lt;/p&gt;&#xA;&lt;p&gt;Also to note: many pharmaceuticals are sensitive to air, in many cases because they are unstable in the long term in presence of oxygen, or moisture. Such products are sold enclosed in a plastic, metallic or composite packaging that protects them. There, the limited stability in time might simply be due to the limit of stability of the (cheap) packaging itself: e.g., after some time of possible light exposure, plastic will become porous and let moisture in.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Being conservative, because we're talking about health.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Being conservative, because the company selling the product doesn't want to be liable.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Economics: it takes money to make absolutely sure the drugs are safe for a longer period of time. Are customers willing to pay more? Most people don't store drugs for use years later (nor is it recommended), so they won't see the point in paying more for longer expiration.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-10-02T08:45:55.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6380" PostTypeId="2" ParentId="5197" CreationDate="2013-10-02T08:54:59.747" Score="0" Body="&lt;p&gt;Yes, wavefunctions for helium can be calculated numerically with various levels of approximation with &lt;a href=&quot;http://en.wikipedia.org/wiki/Ab_initio_quantum_chemistry_methods&quot; rel=&quot;nofollow&quot;&gt;quantum chemistry methods&lt;/a&gt;. Be aware, though, that wavefunctions depend on a specific representation of the system (the use of molecular orbitals) but are not physical observables of the system (only the electronic density is an observable).&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-02T08:54:59.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6381" PostTypeId="1" AcceptedAnswerId="7317" CreationDate="2013-10-02T08:57:48.900" Score="26" ViewCount="9902" Body="&lt;p&gt;According to IUPAC rules, names of chemical elements should not be capitalized. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_element#Element_names&quot;&gt;Wikipedia’s take&lt;/a&gt; on the issue:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to IUPAC, chemical elements are not proper nouns in English; consequently, the full name of an element is not routinely capitalized in English&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I commonly see people (even in research articles) capitalize the element names, and I wonder why. Is there a historical trend of doing so? Did the rule change at some point? Are element names capitalized in other languages?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2021-02-15T15:05:49.837" Title="Why do people often capitalize element names?" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6382" PostTypeId="1" AcceptedAnswerId="6400" CreationDate="2013-10-02T13:30:44.143" Score="5" ViewCount="4087" Body="&lt;p&gt;I'm not able to understand how the following compounds are aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When should the lone pairs on heteroatoms be taken into consideration when counting the number of π electrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PedLs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PedLs.png&quot; alt=&quot;Heterocycles: isoquinoline, thiophene, imidazole, thiazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2018-01-06T17:52:25.497" LastActivityDate="2020-05-24T13:54:19.467" Title="Do lone pairs contribute towards the pi electron count of heterocycles?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6383" PostTypeId="1" CreationDate="2013-10-02T13:51:30.793" Score="2" ViewCount="169" Body="&lt;p&gt;I want to simulate, at a very coarse grained level, the interaction of water molecules with metal ions, specifically Calcium and Potassium. Is there a standard force field for this. I was told to use to the Martini potential but it doesn't seem to have metal ions...&lt;/p&gt;&#xA;" OwnerUserId="2118" LastEditorUserId="30" LastEditDate="2013-10-03T15:09:12.233" LastActivityDate="2014-02-01T21:19:13.577" Title="Coarse-grained force field for ion–water interactions" Tags="&lt;computational-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6385" PostTypeId="1" AcceptedAnswerId="6398" CreationDate="2013-10-02T14:30:42.000" Score="8" ViewCount="2134" Body="&lt;p&gt;In the extraction of $\ce{Sn}$ from $\ce{SnO2}$ by carbon reduction method, why is  &lt;strong&gt;$\ce{CO}$ formed instead of $\ce{CO2}$&lt;/strong&gt; as per the following reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SnO2 + C -&amp;gt; Sn + CO}$&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2014-05-22T11:56:27.190" LastActivityDate="2015-10-08T00:44:32.647" Title="Extraction of Sn from SnO2" Tags="&lt;redox&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6386" PostTypeId="1" AcceptedAnswerId="6397" CreationDate="2013-10-02T14:47:40.150" Score="6" ViewCount="1792" Body="&lt;p&gt;My 7th grader is embarking on a long-term science project and I'm on tap to &quot;help.&quot; He is showing a strong interest into finding ways to get drinking water to the masses. His idea is that we need to find inexpensive and efficient desalination processes. So, we are exploring current desalination processes to see what makes it &quot;expensive&quot; and &quot;inefficient.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we live in the Rocky Mountains and don't have access to &lt;em&gt;proper&lt;/em&gt; sea water. We started to explore simply adding salt to tap water and researching &lt;em&gt;how much&lt;/em&gt;. Then, we discovered &quot;Artificial Sea Water&quot; or &quot;ASW&quot; is a common creation in science. Because sea water &lt;a href=&quot;https://en.wikipedia.org/wiki/Seawater#Origin&quot; rel=&quot;noreferrer&quot;&gt;contains more than just salt&lt;/a&gt;. It contains chemicals like fluoride, strontium, boron, bromide, inorganic carbon, potassium, calcium, sulfate, magnesium, chlorine and of course, sodium. &lt;strong&gt;I am not a chemist&lt;/strong&gt;, so this is starting to look daunting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here are my questions to counsel my son:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;To experiment with desalination, is adding salt to tap water&#xA;sufficient or do we need to investigate finding official ASW?&lt;/li&gt;&#xA;&lt;li&gt;How do we test the salinity of water? Do I really need to invest&#xA;hundreds of dollars on a salinity meter?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2399" LastActivityDate="2013-10-03T10:55:44.717" Title="How to make artificial sea water" Tags="&lt;water&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6387" PostTypeId="1" CreationDate="2013-10-02T14:52:58.293" Score="5" ViewCount="19846" Body="&lt;p&gt;I have no problem with identifying whether a given molecule has geometrical isomerism if it's not cyclic. But cyclic compounds are confusing for me. For example, how to check whether the following molecule posses geometrical isomerism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QWBVA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QWBVA.png&quot; alt=&quot; 2,2,5,5‐tetramethyl‐4‐[(3Z)‐thiolan‐3‐ylidene]oxane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a molecule to show geometrical isomerism different molecules should be attached to the double bonded carbons. So, I'm sure that geometrical isomerism (if it is present) is along the double bonded carbons. But how to check whether different groups are attached to it when the double bonded carbons are the part of a cycle? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="41328" LastEditDate="2020-05-31T07:40:25.463" LastActivityDate="2020-06-01T14:28:59.133" Title="How to check for geometrical isomerism in cyclic compounds?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6390" PostTypeId="1" CreationDate="2013-10-02T19:01:18.067" Score="9" ViewCount="4493" Body="&lt;p&gt;Is there a way to safely cool a heated test tube quickly without cracking it due to sudden thermal stress in the glass? One thing that I tried was blowing on it, cooling it in slightly lukewarm water followed by cold water but this method kind of involves a &quot;hope for the best factor&quot;. Is there a better technique?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="23561" LastEditDate="2019-03-20T21:33:52.920" LastActivityDate="2019-03-20T21:33:52.920" Title="How to quickly cool a test tube without cracking it?" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;&lt;equipment&gt;&lt;glass&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6391" PostTypeId="1" AcceptedAnswerId="6393" CreationDate="2013-10-03T00:46:36.477" Score="1" ViewCount="10676" Body="&lt;p&gt;I don't know what identical molecules are but these are by thoughts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since from the word identical the molecules should have the same number and kind of atoms attached in the same way. Since in stereo-chemistry the isomerism is due to spacial orientation all stereo-chemical isomers are identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thinking correct? What factors determine whether two molecules are identical or not?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="39153" LastEditDate="2017-10-06T14:44:53.913" LastActivityDate="2017-10-06T14:44:53.913" Title="What are identical molecules?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6393" PostTypeId="2" ParentId="6391" CreationDate="2013-10-03T02:38:43.317" Score="8" Body="&lt;p&gt;Identical compounds have the same chemical and physical properties (and other properties too). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the two types of stereoisomers, diastereomers (non-mirror image stereoisomers) are clearly nonidentical. Consider the isomers of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tartaric_acid&quot;&gt;tartaric acid&lt;/a&gt;. Meso-tartaric acid and (+)-tartaric acid have different melting points (165 $^\circ$C and 171 $^\circ$C, respectively). They also have different acidities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (+) and (-) isomers (enantiomers) are more challenging, but they are distinguishable by one physical property - the way they rotate plane-polarized light. If the two molecules were identical, they would not be isomers - we would not be able to distinguish them. If they are distinguishable, they are not identical.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-03T02:38:43.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6395" PostTypeId="2" ParentId="4027" CreationDate="2013-10-03T05:11:23.470" Score="18" Body="&lt;p&gt;The hypervalent carbon do exist. Look at this article:&#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja043802t&quot; rel=&quot;noreferrer&quot;&gt;Kin-ya Akiba, &lt;em&gt;et. al.&lt;/em&gt;, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;127&lt;/em&gt; (16), 5893–5901. &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FJ8l3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ray measurements confirmed the 10-C-5 structure.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="4945" LastEditDate="2015-04-30T05:38:18.317" LastActivityDate="2015-04-30T05:38:18.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6396" PostTypeId="2" ParentId="6387" CreationDate="2013-10-03T08:17:59.983" Score="2" Body="&lt;p&gt;Two ways to think about it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intuitively: Take one side of the double bond (say, the sulfur-containing one) and imagine rotating it by 180° (so what was up is now down, in your drawing). Is this molecule the same? If it is, then you don't have isomerism. If the procedure gives a new, different molecule, then you have isomerism.&lt;/li&gt;&#xA;&lt;li&gt;More formal: use the Cahn–Ingold–Prelog rules to rank both sides of the two carbon atoms of the double bond. If priorities are different, then you have different isomers (and you can actually assign them a name.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In your particular example, the compound has another isomer, and the one you have drawn has the &lt;a href=&quot;http://en.wikipedia.org/wiki/E-Z_notation&quot; rel=&quot;nofollow&quot;&gt;(&lt;em&gt;E&lt;/em&gt;)&lt;/a&gt; configuration.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-03T08:17:59.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6397" PostTypeId="2" ParentId="6386" CreationDate="2013-10-03T10:55:44.717" Score="5" Body="&lt;p&gt;If you look at the (average) &lt;a href=&quot;http://en.wikipedia.org/wiki/Seawater&quot; rel=&quot;noreferrer&quot;&gt;composition of seawater&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;       &lt;img src=&quot;https://i.stack.imgur.com/UHWkQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will find that NaCl is the main solute of this complex solution. Its concentration is 10 times higher than that of the next constituents (magnesium and sulfate). So, it seems to me like a simple NaCl solution is a good approximation for your experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding testing salinity of water, I don't have a definite answer on that, I do have two suggestions to offer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measure salinity by electric conductance of the solution, e.g. by making a simple conductometer and testing it with solutions of known concentration of salt.&lt;/li&gt;&#xA;&lt;li&gt;Measure salinity by the quantity of salt (solid residue) once all water is evaporated. You can speed the evaporation by heating up.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-03T10:55:44.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6398" PostTypeId="2" ParentId="6385" CreationDate="2013-10-03T11:46:44.730" Score="16" Body="&lt;p&gt;Above a $1000\ \mathrm K$, the $\Delta_\mathrm fG$ of $\ce {CO}$ from $\ce C$ is more negative than that of $\ce{CO2}$ formation from $\ce C$. Therefore, during smelting, when coke ($\ce C$) reacts with $\ce {SnO2}$, the formation of $\ce {CO}$ rather than $\ce {CO2}$ is thermodynamically preferable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below that temperature, the reduction with carbon is not possible since $\Delta_\mathrm fG(\ce{SnO2})\lt\Delta_\mathrm fG(\ce{CO2})\lt\Delta_\mathrm fG(\ce{CO})$. Therefore at all temperatures where this reduction is possible, thermodynamically carbon monoxide formation would be more favourable as compared to carbon dioxide formation. To see for yourself, the variation of $\Delta_\mathrm fG$ of various compounds, have a look at Ellingham Diagrams, a link to a comprehensive set of which is &lt;a href=&quot;http://showard.sdsmt.edu/MET320/Handouts/EllinghamDiagrams/Ellingham_v22_Macro.pdf&quot; rel=&quot;nofollow&quot;&gt;Given Here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_\mathrm fG$ is the Gibbs free energy of formation.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorDisplayName="user7951" LastEditDate="2015-10-08T00:44:32.647" LastActivityDate="2015-10-08T00:44:32.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6399" PostTypeId="1" CreationDate="2013-10-03T13:57:16.247" Score="0" ViewCount="177" Body="&lt;p&gt;Suppose that I have a cast-iron pot which my awful roommate has vigorously scrubbed with soap and left to soak in soapy water over the weekend. The seasoning has been destroyed and replaced with awful-tasting soapy contaminants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the simplest and best way to deal with this is to heat the pot over a campfire until only black carbon remains, then clean and reseason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing exactly what is in my awful roommate's dish soap, and being a very fussy foodie, I want to be conservative here... so how hot would I have to heat the pot, and for how long, to be quite sure that virtually nothing but carbon remains?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Compare &lt;a href=&quot;http://lifehacker.com/5845822/remove-rust-from-cast-iron-cookware-in-your-oven&quot; rel=&quot;nofollow&quot;&gt;using the self-cleaning cycle of an oven&lt;/a&gt; for similar purpose, reported to work well...&lt;/p&gt;&#xA;" OwnerUserId="2412" LastEditorUserId="2412" LastEditDate="2013-10-05T19:34:35.877" LastActivityDate="2013-10-05T19:34:35.877" Title="Temperature &amp; time necessary to carbonize all organics" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6400" PostTypeId="2" ParentId="6382" CreationDate="2013-10-03T14:51:07.640" Score="11" Body="&lt;p&gt;Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in compound &lt;strong&gt;2&lt;/strong&gt; (thiophene), there are two lone pairs on sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bnTqT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnTqT.png&quot; alt=&quot;Lone pairs on sulfur in thiophene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://en.wikipedia.org/wiki/Conjugated_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Conjugated system&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="16683" LastEditDate="2018-01-08T16:15:59.543" LastActivityDate="2018-01-08T16:15:59.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6401" PostTypeId="2" ParentId="6366" CreationDate="2013-10-03T15:52:55.560" Score="1" Body="&lt;p&gt;I'm giving the answer thinking that you are familiar with ligands and Denticity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polydentate ligands form two or more coordinate bond with the central metal atom. For eg:EDTA can form up-to 6 coordinate bond with the central atom. In these situations a single ligand encloses the metal ion like enclosing it in a cage. This is called chelation. Co-ordination compounds with chelated metal ions are more stable then those in which the metal ions are surrounded by many monodentate ligands.If you want more information read the Wikipedia article on this &lt;a href=&quot;http://en.wikipedia.org/wiki/Chelation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chelation&lt;/a&gt;.       &lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2013-10-03T15:52:55.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6402" PostTypeId="1" AcceptedAnswerId="6403" CreationDate="2013-10-03T16:14:51.333" Score="4" ViewCount="3587" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The following reduction is best accomplished with:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Zn/Hg, conc. HCl&lt;br&gt;&#xA;  (b) N&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;4&lt;/sub&gt;, C&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;5&lt;/sub&gt;ONa&lt;br&gt;&#xA;  (c) H&lt;sub&gt;2&lt;/sub&gt;, Pd&lt;br&gt;&#xA;  (d) All of the above&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ejs3a.png&quot; height=&quot;100&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thoughts: By looking at the question it is evident that only the C=O group should be affected. H&lt;sub&gt;2&lt;/sub&gt;/Pd is used for hydrogenation which might affect the C=C double bond as well. Option (a) gives the reagents which are used for the Clemmensen reduction which is used to convert carbonyl groups to alkanes. So, I think (a) is the correct answer. Is that right?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2019-05-09T14:11:36.970" LastActivityDate="2019-05-09T14:13:31.260" Title="Best reagent for reducing a ketone in the presence of an alkene" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6403" PostTypeId="2" ParentId="6402" CreationDate="2013-10-03T16:33:37.507" Score="10" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Clemmensen_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clemmensen reduction&lt;/a&gt; is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choice (b) is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wolff-Kishner_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolff–Kishner&lt;/a&gt; reaction, which also converts ketones into alkenes. It requires strong base (not acid) and works on almost all ketones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choice (c) will react with the alkene also, and so cannot be correct.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2019-05-09T14:13:31.260" LastActivityDate="2019-05-09T14:13:31.260" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6404" PostTypeId="1" CreationDate="2013-10-03T16:38:32.353" Score="16" ViewCount="11018" Body="&lt;p&gt;$\mathrm{S}_{\mathrm{N}}1$ reaction involves only one molecule in the rate determining step. So, the molecule which undergoes $\mathrm{S}_{\mathrm{N}}1$ reaction should be stable when it forms a positive ion. $\ce{(CH3)3C^+}$ is considered stable due to inductive effect from the three carbon atoms and also hyper-conjugation. But, what about bridged compounds like 1-bromobicyclo[2.2.1]heptane:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PpsMV.png&quot; alt=&quot;1-bromobicyclo[2.2.1]heptane&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\ce{Br^-}$ is abstracted the positive ion formed will be stable and therefore this will undergo $\mathrm{S}_{\mathrm{N}}1$ reaction. Am I thinking right? Please help me correct myself. &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2016-12-20T10:01:13.293" LastActivityDate="2016-12-20T10:01:13.293" Title="Will bridged compounds the undergo SN1 reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;carbocation&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6406" PostTypeId="2" ParentId="6399" CreationDate="2013-10-04T00:24:09.630" Score="3" Body="&lt;p&gt;There's a lot of exciting, and not well defined, chemistry that happens with seasoning cast-iron cookware.  Essentially, you are creating a cross-linked polymer film that is hydrophobic, providing a &quot;stick-free&quot; surface.  There are many many websites claiming the &quot;best way&quot; to season (or re-season) cast-iron cookware.  Few of these sites (and I'll let the OP and visitors search for those on their own) are based on science.  (Although I am not going to question tried and true family secrets, passed down from generation to generation.)  &lt;a href=&quot;http://sherylcanter.com/wordpress/2010/01/a-science-based-technique-for-seasoning-cast-iron/&quot; rel=&quot;nofollow&quot;&gt;One site in particular&lt;/a&gt; makes a fairly convincing argument for seasoning that is science-based.  I suggest this approach can help restore your pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect what has happened is that the scrubbing did remove the seasoning and exposed base metal to water and oxidation - your pot rusted.  The &quot;soapy contaminants&quot; is probably iron oxide rather than something from the dishwater, although there are such products as &lt;a href=&quot;http://www.baerlocher.com/products/metal-soaps/&quot; rel=&quot;nofollow&quot;&gt;metal soaps&lt;/a&gt; which are metals chelated to long-chain fatty acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for heating the pot to burn off all the residue - I doubt this is the most effective use of your time.  Scrubbing with acidified water to remove the iron oxide and a long seasoning session with an easily cross-linkable oil should get your pot back to (or better than) original condition.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-10-04T00:24:09.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6407" PostTypeId="1" AcceptedAnswerId="6414" CreationDate="2013-10-04T02:40:34.673" Score="4" ViewCount="21489" Body="&lt;p&gt;There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at the half reaction for copper:&#xA;\begin{align}&#xA;  \ce{Cu (s) &amp;amp;-&amp;gt; Cu^2+ (aq) + 2e-}&amp;amp;&#xA;  (E^\circ &amp;amp;= \pu{+0.34 V})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to nitrate:&#xA;\begin{align}&#xA;  \ce{NO3- (aq) + 4H+ (aq) + 3e- &amp;amp;-&amp;gt; NO (g) + 2H2O (l)}&amp;amp;&#xA;  (E^\circ &amp;amp;= \pu{+0.96 V})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it produces more energy, I believe the second half reaction will occur instead.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I begin by finding all the species/ions I have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Cu^2+}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NO3^-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taking a look at the &lt;em&gt;Standard Reduction Potentials at 25°C&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$):&#xA;\begin{align}&#xA;  \ce{NO3^- + 4H+ + 3e- &amp;amp;-&amp;gt; NO + 2H2O}&amp;amp;&#xA;  (E^\circ = \pu{+0.96 V})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$):&#xA;\begin{align}&#xA;  \ce{2H2O &amp;amp;-&amp;gt; O2 + 4H+ + 4e-}&amp;amp;&#xA;  (E^\circ &amp;amp;= \pu{-1.23 V})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the balanced reaction I have:&#xA;\begin{align}&#xA;  \ce{4NO3- + 4H+ &amp;amp;-&amp;gt; 4NO + 2H2O + 3O2}&amp;amp;&#xA;  (E^\circ &amp;amp;= \pu{-0.27 V})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some observations that can be made are that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitric gas is forming at the anode&lt;/li&gt;&#xA;&lt;li&gt;Oxygen gas is forming at the cathode&lt;/li&gt;&#xA;&lt;li&gt;At least $\pu{0.27 V}$ must be put in&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.&lt;/li&gt;&#xA;&lt;li&gt;What would happen if you placed a necklace at the cathode?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2356" LastEditorUserId="4945" LastEditDate="2017-05-09T09:18:25.670" LastActivityDate="2020-06-24T15:48:55.907" Title="Electrolysis of aqueous copper (II) nitrate" Tags="&lt;redox&gt;&lt;aqueous-solution&gt;&lt;electrolysis&gt;&lt;reduction-potential&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6408" PostTypeId="1" AcceptedAnswerId="6410" CreationDate="2013-10-04T04:32:21.757" Score="0" ViewCount="573" Body="&lt;p&gt;Antlerite is $\ce{Cu3(SO4)(OH)4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get the charge on the copper ion in this substance.  So far I have the mass of $1$ mole of antlerite, its percent oxygen content, and the number of individual hydroxide ions in a $\pu{955 mg}$ bead of the substance, if that information helps at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help to solve this question.&lt;/p&gt;&#xA;" OwnerUserId="2417" LastEditorUserId="5026" LastEditDate="2018-02-17T08:51:39.720" LastActivityDate="2018-02-17T08:51:39.720" Title="What is the charge on the copper ion in antlerite?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6410" PostTypeId="2" ParentId="6408" CreationDate="2013-10-04T06:37:28.570" Score="1" Body="&lt;p&gt;Ok, here is a short answer. Since you already know the chemical formula it's actually quite easy provided you are familiar with the process of determining the oxidation numbers in chemical compound. From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Antlerite&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; (and the way the chemical formula is written indicates that too) you can get the information that the $\ce{SO4}$ part of the formula is sulfate and sulfate ions carry a charge of $-2$, i.e. $\ce{SO4^{2-}}$. Furthermore, in most cases $\ce{H}$ gets an oxidation number of $+1$ and $\ce{O}$ gets an oxidation number of $-2$ so that the 4 $\ce{OH}$ units in the compound each have a charge of $1 - 2 = -1$, so you have $\ce{OH^{-}}$, i.e. hydroxyl groups - this is supported by the way those $\ce{O}$'s and $\ce{H}$'s are grouped, if you see &quot;$\ce{OH}$&quot; in a compound it usually means &quot;hydroxyl group&quot;. So, you have one sulfate group with a charge of $-2$ and four hydroxyl groups with a charge of $-1$ each, giving you a charge of $-6$ in total. But your compound $\ce{Cu3(SO4)(OH)4}$ is neutral, i.e. has a charge of $0$, so the three $\ce{Cu}$ have to counteract the negative charge exerted by the anions, in other words you have to distribute a charge of $+6$ between the three copper atoms. If you assume that the charge is evenly distributed, i.e. that all the $\ce{Cu}$ are the same (this need not necessarily be the case, you can also have mixed oxides like $\ce{Fe3O4}$ where you have $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ within the same mineral), you get to a charge of $+2$ for each copper ion, i.e. $\ce{Cu^{2+}}$. &lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-10-04T06:37:28.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6411" PostTypeId="2" ParentId="6404" CreationDate="2013-10-04T07:07:12.267" Score="12" Body="&lt;p&gt;Very interesting question… I would give two answers, at different levels:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At undergrad level, I think the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbenium_ion&quot;&gt;carbenium ion&lt;/a&gt; formed at the bridge head is less stabilized that “regular” ternary carbenium ions, because it cannot adopt the planar trigonal geometry which would be ideal for it.&lt;/li&gt;&#xA;&lt;li&gt;At higher level, this is much more complicated: the norboryl cation is an example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-classical_ion&quot;&gt;non-classical ion&lt;/a&gt;…&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-04T07:07:12.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6413" PostTypeId="2" ParentId="6383" CreationDate="2013-10-04T19:03:07.557" Score="1" Body="&lt;p&gt;Forcefields for alkali ions in water have been developed for decades now, so the literature is rich with examples… and yet there still are significant developments (see &lt;a href=&quot;http://link.aip.org/link/doi/10.1063/1.3692965&quot; rel=&quot;nofollow&quot;&gt;this very new paper&lt;/a&gt; for only one such example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to have a forcefield for alkali ions that mix well with the TIP3P water model, you better look at what existing studies on such systems have done… &lt;a href=&quot;http://www.engineering.ucsb.edu/~baronp/images/Poster_mStatMech.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is an example, but you can find many others with searching through the literature.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-04T19:03:07.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6414" PostTypeId="2" ParentId="6407" CreationDate="2013-10-04T21:10:50.447" Score="3" Body="&lt;p&gt;There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.&#xA;\begin{align}&#xA;  \ce{Cu (s) &amp;amp;-&amp;gt; Cu^2+ (aq) + 2e-}&amp;amp;&#xA;  E^\circ_\mathrm{ox} &amp;amp;= \pu{-0.34 V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second half-reaction is a reduction (as written)&#xA;\begin{align}&#xA;  \ce{NO3- (aq) + 4H+ (aq) +3e- &amp;amp;-&amp;gt; NO (g) + 2H2O (l)}&amp;amp;&#xA;  E^\circ_\mathrm{red} &amp;amp;= \pu{+0.96 V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -&amp;gt; 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\ &#xA;E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this change your analysis?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2017-05-09T09:23:39.203" LastActivityDate="2017-05-09T09:23:39.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6416" PostTypeId="1" CreationDate="2013-10-05T09:47:43.413" Score="1" ViewCount="81" Body="&lt;p&gt;We have two bottle named &lt;strong&gt;$\ce{H2SO4~(aq)}$&lt;/strong&gt; and solution of &lt;strong&gt;$\ce{NH3}$&lt;/strong&gt; but we don't know their name actually. What I have to do to detect these solution  by only using : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{CuCl2}$         &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;$\ce{NaCO3_{(s)}}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2400" LastEditorUserId="4231" LastEditDate="2015-03-05T20:16:39.430" LastActivityDate="2015-03-05T20:16:39.430" Title="How to detect H₂SO₄ (aq) and NH₃" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6417" PostTypeId="1" CreationDate="2013-10-05T10:42:44.987" Score="9" ViewCount="25278" Body="&lt;p&gt;Mnemonics help to by-heart things which are difficult to do so and which are important. I found mnemonics for studying the periodic table, electromagnetic series etc which were of great help. So, is there a mnemonic to study the spectrochemical series since knowing it will be of great help to find the geometry of a given co-ordination compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrochemical_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;spectrochemical series&lt;/a&gt; is a list of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ligand#Examples_of_common_ligands_.28by_field_strength.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;common ligands&lt;/a&gt; for transition metals in order of increasing ligand-field splitting parameter $\Delta$, which is the energy difference between the sets of $\ce{d}$ orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I-} &amp;lt; \ce{Br-} &amp;lt; \ce{S^2-} &amp;lt; \ce{SCN-} &amp;lt; \ce{Cl-} &amp;lt; \ce{NO3-} &amp;lt; \ce{N3-} &amp;lt; \ce{F-} &amp;lt; \ce{OH-} &amp;lt; \ce{H2O} \approx \ce{C2O4^2-} \text{ (oxalate)} &amp;lt; \ce{NCS-} &amp;lt; \ce{CH3CN} &amp;lt; \ce{py}\text{ (pyridine)} &amp;lt; \ce{NH3} &amp;lt; \ce{en}\text{ (ethylenediamine)} &amp;lt; \ce{bpy}\text{ (2,2'-bipydiridyl)} &amp;lt; \ce{phen}\text{ (1,10-phenanthroline)} &amp;lt; \ce{NO2-} &amp;lt; \ce{PPh3} &amp;lt; \ce{CN-} \approx \ce{CO}$$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a mnemonic device for remembering this series? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="41328" LastEditDate="2017-10-06T17:17:41.253" LastActivityDate="2021-04-06T11:57:51.110" Title="Is there a mnemonic to study the spectrochemical series?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6418" PostTypeId="2" ParentId="6416" CreationDate="2013-10-05T12:35:18.023" Score="5" Body="&lt;p&gt;On mixing 2 parts of $\ce{NH3}$ with 1 part of $\ce{CuCl2}$, following reaction will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuCl2 + 2NH3 -&amp;gt; Cu(NH3)2^{2+} + 2Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you &lt;strong&gt;soluble deep-blue copper-ammonia&lt;/strong&gt; complex: $\ce{Cu(NH3)2^{2+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also, $\ce{NH3}$ does not react with $\ce{Na2SO4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\ce{H2SO4}$ with $\ce{Na2CO3}$ will follow the reaction below : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4_{(aq)} + Na2CO3_{(s)} -&amp;gt; H2O + CO2_{(g)} + Na2SO4_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\ce{Na2SO4}$ &lt;strong&gt;solution is colorless&lt;/strong&gt;, so you can perform any appropriate test to identify the evolved gas i.e. $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple method used to test $\ce{CO2}$ is :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bubble the gas through limewater (calcium hydroxide). If carbon dioxide is present, the limewater will turn cloudy and a milky white precipitate will form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-10-06T07:08:30.357" LastActivityDate="2013-10-06T07:08:30.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6419" PostTypeId="1" AcceptedAnswerId="6519" CreationDate="2013-10-05T15:24:02.863" Score="5" ViewCount="13470" Body="&lt;p&gt;I'm to find the violation of a certain rule or principle in the following electronic configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$n+l$ rule is not violated because: for $\ce{1s^2}$, $n+l=1+0=1$ so it is filled first, for $\ce{2s^2}$, $n+l=2+0=2$ so it is filled second, and so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aufbau Principle is not violated because: the orbitals with minimum energy are filled first with electrons and $\ce{1s &amp;lt; 2s &amp;lt; 2p}$ which is the sequence of increasing energy orbital judged by the $n+l$ rule. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm quoting here a paragraph from my chemistry textbook of the chapter &lt;em&gt;Atomic Structure&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The orbitals given by a particular value of $l$ if $n$ is same, have the same energy and such orbitals are called degenerate. Like $l=1$ gives three $p$ orbitals specified as $\ce{p_x, p_y, p_z}$ and the p orbitals are said to be three fold degenerate. These orbitals are filled according to Hund's Rule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The numerical value of the principle quantum number $n$ for $\ce{2s}$ and $\ce{2p}$ is the same; however, they don't have a particular value of the azimuthal quantum number $l$. For $\ce{2s}$, $l=0$, and for $\ce{2p}$, $l=1$, so $\ce{2s}$ and $\ce{2p}$ are not degenerate. So none of the three principles or rules whether Aufbau's, $n+l$ or Hund's is violated in this electronic configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my friend says that there is a violation of the Hund's Rule in this electronic configuration, and according to him the right configuration is: $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$, which in turn violates the Aufbau Principle, because the orbitals of lower energy level have to filled first completely and $\ce{2s &amp;lt; 2p}$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, which one of the electronic configurations is right, $\ce{1s^2 2s^2 2p^1_x 2p^0_y 2p^0_z}$ or $\ce{1s^2 2s^1 2p^1_x 2p^1_y 2p^0_z}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: The required electronic configuration is for the electrons in their ground state.&lt;/p&gt;&#xA;" OwnerUserId="1667" LastEditorUserId="5026" LastEditDate="2018-07-07T16:20:56.447" LastActivityDate="2020-03-01T21:28:05.703" Title="The violation of a certain rule or principle in the following electronic configuration" Tags="&lt;quantum-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6421" PostTypeId="1" AcceptedAnswerId="6422" CreationDate="2013-10-06T00:49:58.680" Score="5" ViewCount="349" Body="&lt;p&gt;For example the electron configuration for oxygen is 1s&lt;sup&gt;2&lt;/sup&gt; 2s&lt;sup&gt;2&lt;/sup&gt; 2p&lt;sup&gt;4&lt;/sup&gt;. But if an element can never have an element in the p orbital unless it has 2 full s orbitals*, then why can't we write 2p&lt;sup&gt;4&lt;/sup&gt; as the electron configuration for oxygen? Is it not implied that it has 1s&lt;sup&gt;2&lt;/sup&gt; 2s&lt;sup&gt;2&lt;/sup&gt;? I would think so because the noble gas configuration works on a similar principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*My teacher told me this.&lt;/p&gt;&#xA;" OwnerUserId="2427" LastActivityDate="2013-10-07T16:55:40.333" Title="Is the electron configuration redundant?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6422" PostTypeId="2" ParentId="6421" CreationDate="2013-10-06T02:01:00.970" Score="12" Body="&lt;p&gt;The configuration you describe is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ground_state&quot;&gt;ground state&lt;/a&gt; configuration, the configuration of oxygen in which all electrons are in their lowest allowed energy states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substances can also be their &lt;a href=&quot;http://en.wikipedia.org/wiki/Excited_state&quot;&gt;excited states&lt;/a&gt;, where electrons are not in the lowest allowed energy states. Excited states happen when atoms or molecules absorb energy, usually electromagnetic radiation (but not always).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excited state for oxygen may involve one of the $2s$ electrons being promoted to one of the remaining $2p$ spots:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;\ce{O} + E &amp;amp;\ce{-&amp;gt; O}^* \\&#xA;1s^2 2s^2 2p^4 &amp;amp;\ce{-&amp;gt; }1s^2 2s^1 2p^5&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This configuration still describes an oxygen atom, but now the $2s$ subshell is partially empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason you can't always assume that the lower subshells are always filled is that there are some exceptions to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot;&gt;aufbau principle&lt;/a&gt;. Some elements do not follow it, especially the lanthanides and actinides.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium&quot;&gt;Chromium&lt;/a&gt;: $\ce{Cr} - \ce{[Ar]}4s^1 3d^5$ ($s$ is not filled so that $d$ can be half-full)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Copper&quot;&gt;Copper&lt;/a&gt;: $\ce{Cu} - \ce{[Ar]}4s^1 3d^{10}$ ($s$ is not filled so that $d$ can be full)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Palladium&quot;&gt;Palladium&lt;/a&gt;: $\ce{Pd} - \ce{[Kr]}4d^{10}$ ($s$ is empty so that $d$ can be full)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cerium&quot;&gt;Cerium&lt;/a&gt;: $\ce{Ce} - \ce{[Xe]} 6s^2 5d^1 4f^1$ ($f$-block elements are weird)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="283" LastEditDate="2013-10-07T16:55:40.333" LastActivityDate="2013-10-07T16:55:40.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6423" PostTypeId="1" CreationDate="2013-10-06T03:30:26.943" Score="4" ViewCount="8834" Body="&lt;p&gt;I have two volatile compounds (such as ethyl acetate and ethyl propionate). I am going to separate these compounds by simple and fractional distillation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate mole ratios of the two compounds and determine the boiling point temperatures of the first drop based on the liquid and vapor phase diagram?&lt;/p&gt;&#xA;" OwnerUserId="2428" LastEditorUserId="865" LastEditDate="2015-07-31T23:20:42.640" LastActivityDate="2015-07-31T23:20:42.640" Title="How to calculate mole ratio of two volatile compounds in distillation?" Tags="&lt;organic-chemistry&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6425" PostTypeId="2" ParentId="6423" CreationDate="2013-10-06T10:39:59.780" Score="5" Body="&lt;p&gt;If you have the vapor-liquid phase diagram, then it is easy to determine the vapor composition above a boiling liquid mixture (which should approximate the composition of your first drop in a simple distillation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll assume you diagram looks like the one below, which is from the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor%E2%80%93liquid_equilibrium&quot; rel=&quot;noreferrer&quot;&gt;vapor-liquid equilibrium&lt;/a&gt;(and is mostly at the physical chemistry level):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qbzI5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a mixture of components 1 and 2 that is 40 mole percent (mole fraction = 0.4) component one (approximately the blue line labeled liquid on the diagram). Draw a line vertically from that composition to the bubble point (or liquid) line. Then draw a horizontal isotherm at that point (red line on the diagram above). Where that isotherm crosses the dew point (or vapor) line, you can read down to get the composition of the vapor above that boiling liquid (blue line labeled vapor), about 15% Component 1 (and 85% Component 2). This represents a single vaporization-condensation cycle, which is what happens in a simple distillation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a fractional distillation, you can have multiple vaporization-condensation cycles (called &lt;a href=&quot;http://en.wikipedia.org/wiki/Theoretical_plate#Distillation_columns&quot; rel=&quot;noreferrer&quot;&gt;theoretical plates&lt;/a&gt;). If you know how many theoretical plates your distillation column will have, then you repeat the process for the simple distillation several times. For example, if there are two theoretical plates, you have two vaporization-condensation cycles. After the first one, start over with a new liquid composition equal to the vapor composition after the first cycle. Draw a line up to the liquid line, draw the isotherm to the vapor line, and draw a line back down to composition. If you use the example I have been working with on the diagram I have shown, you should get about 5% Component 1. Repeat as necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not know how many theoretical plates you have in your fractionating column, then you can only make an educated guess.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-06T10:39:59.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6426" PostTypeId="1" CreationDate="2013-10-06T12:36:26.663" Score="2" ViewCount="667" Body="&lt;p&gt;In $\ce{BeCl2}$ the number of orbitals on central atom, i.e. on beryllium, are 2. In $\ce{BF3}$, the number of orbitals on central atom , i.e. on boron, are 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly in $\ce{NH3}$ there are 4, on $\ce{PCl5}$, $\ce{SF4}$, and $\ce{ClF3}$ there are 5. But how? &lt;/p&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="73" LastEditDate="2013-10-06T12:39:54.930" LastActivityDate="2013-10-06T20:56:11.320" Title="Finding the number of orbitals on a central atom" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;atoms&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6428" PostTypeId="1" CreationDate="2013-10-06T15:51:57.753" Score="1" ViewCount="584" Body="&lt;p&gt;Organic semiconductors differ from inorganic semiconductors. In organic semiconductors the molecules are held together by weak van der Waals interactions and in inorganic semiconductors by covalent bonds. So the bonds are different. How do you express the main difference between the twos? Is it the electronic structure or rather the electronic configuration, or something else? &lt;/p&gt;&#xA;" OwnerUserId="2431" LastEditorUserId="2431" LastEditDate="2013-10-10T12:42:46.217" LastActivityDate="2013-10-10T12:42:46.217" Title="Difference between inorganic and organic semiconductors: electronic structure or configuration, or?" Tags="&lt;electronic-configuration&gt;&lt;terminology&gt;&lt;bond&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6429" PostTypeId="1" AcceptedAnswerId="6434" CreationDate="2013-10-06T16:12:00.873" Score="4" ViewCount="37492" Body="&lt;p&gt;Which element has more metallic character - aluminium or magnesium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that metallic character decreases along a period (from left to right) and increases down a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium comes after Magnesium in the same period (period 3) in the periodic table. So, I thought Magnesium would have greater metallic character. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I read in a book that Aluminium has greater metallic character. Is this true? If this is true, why?&lt;/p&gt;&#xA;" OwnerUserId="2425" LastEditorUserId="402" LastEditDate="2017-08-10T23:13:27.017" LastActivityDate="2017-08-10T23:13:27.017" Title="Which has more metallic character: aluminium or magnesium?" Tags="&lt;metal&gt;&lt;periodic-trends&gt;&lt;metallurgy&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6431" PostTypeId="1" AcceptedAnswerId="6437" CreationDate="2013-10-06T18:32:55.823" Score="6" ViewCount="22938" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The energy required to remove both electrons from the helium atom in its ground state is &lt;span class=&quot;math-container&quot;&gt;$79.0\;\mathrm{eV}$&lt;/span&gt;. The energy required to ionize helium (i.e. to remove one electron) is - &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A)   &lt;span class=&quot;math-container&quot;&gt;$24.6\;\mathrm{eV}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(B)   &lt;span class=&quot;math-container&quot;&gt;$39.5\;\mathrm{eV}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(C)   &lt;span class=&quot;math-container&quot;&gt;$51.8\;\mathrm{eV}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(D)   &lt;span class=&quot;math-container&quot;&gt;$54.4\;\mathrm{eV}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My thoughts&lt;/strong&gt;: What I actually did was to take half of the given energy value since energy required to remove electrons depends only on the principal quantum number(and also on atomic number). But, I found my answer to be wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand just calculating the energy required to remove one electron from &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt; state gives the correct answer. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="83758" LastEditDate="2020-03-22T19:36:28.227" LastActivityDate="2020-03-22T23:19:38.313" Title="Energy required to remove an electron from He" Tags="&lt;electrons&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6432" PostTypeId="2" ParentId="6426" CreationDate="2013-10-06T20:50:38.890" Score="3" Body="&lt;p&gt;&lt;strong&gt;According to the question, it is totally not clear that whether you want to know filled orbitals or total orbital (filled + vacant).&lt;/strong&gt; Well, but after seeing the examples you gave, it can be concluded that you are talking only about filled orbitals. &lt;strong&gt;As I'm not clear and assuming that you are only talking about filled orbitals, so if you are talking about total, leave a comment.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As each &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow&quot;&gt;orbital&lt;/a&gt; contains 2 electrons, therefore it is easy to find number of filled orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of filled orbitals in central atom will be equal to &lt;strong&gt;the sum of number of bonds and number of lone pairs present on central atom.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in case of $\ce{BeCl2}$, $\ce{Be}$ makes 2 bonds (both with $\ce{Cl}$) and have no lone pair, therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{No. of filled orbitals} = 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let us take another example of $\ce{NH3}$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of $\ce{NH3}$, $\ce{N}$ makes 3 bonds (each with $\ce{H}$) and have 1 lone pair, therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{No. of filled orbitals} = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Take another example of $\ce{ClF3}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the central atom $\ce{Cl}$ makes 3 bonds with $\ce{F}$ by &lt;strong&gt;sharing its one electron in each bond&lt;/strong&gt; and have 2 lone pairs on it. (As $\ce{Cl}$ have 7 electrons, 3 shared in bonding and 4 left = 2 lone pairs.) Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{No. of filled orbitals} = 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope, now everything is clear to you!&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-10-06T20:56:11.320" LastActivityDate="2013-10-06T20:56:11.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6433" PostTypeId="1" CreationDate="2013-10-06T21:12:08.340" Score="3" ViewCount="1482" Body="&lt;p&gt;The hair gel is called &quot;lets Jam Gel&quot;. It says it uses micro emulsion technology.&#xA;What chemicals can you use to break it down or undo the bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get it out of my head. I washed my hair with T sal and cut my hair off but it still grows back patchy and easy to style like it has gel in it. I think it may be in the epidermis or even dermis by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ingredients: 42692 SS11 - Aqua / Water Isocteth-20 Paraffinum Liquidum / Mineral oil Glycerin Oleth-2 Propylene Glycol Parfum / fragrance Sorbic Acid Methylparaben DMDM Hydantoin Ethylparaben Limonene Hydroxycitronellal Geraniol Amyl cinnamal Linalool Simethicone CI 19140 / Yellow 5 CI 14700 / Red 4, CI 172200 / Red 3. F.I.L D45949/1&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2013-10-07T15:25:57.183" LastActivityDate="2013-10-09T22:57:06.150" Title="What can you use to remove hair gel?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2013-10-10T11:09:27.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="6434" PostTypeId="2" ParentId="6429" CreationDate="2013-10-06T21:22:48.083" Score="4" Body="&lt;p&gt;Yes, you are right that $\ce{Al}$ is more metallic than $\ce{Mg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is because, &lt;strong&gt;$\ce{Mg}$ have fully filled $3\text{s}$ subshell&lt;/strong&gt;. Due to this, the first &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionisation energy$^{[1]}$&lt;/a&gt; of $\ce{Al}$ (578 kJ/mol) is less than that of $\ce{Mg}$ (738 kJ/mol).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Filled and half-filled subshells show a small increase in stability in the same way that filled shells show increased stability. So, when trying to remove an electron from one of these filled or half-filled subshells, a slightly higher ionization energy is found.&lt;a href=&quot;http://www.grandinetti.org/ionization-energy-trends&quot; rel=&quot;nofollow noreferrer&quot;&gt;$^{[2]}$&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can also say that higher levels of subshells are further away from the nucleus which is why it is easier to remove an electron from $\ce{Al}$ than from $\ce{Mg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;br&gt;&#xA; 1. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Ionization_energy&lt;/a&gt;&lt;br&gt;&#xA; 2. &lt;a href=&quot;http://www.grandinetti.org/ionization-energy-trends&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.grandinetti.org/ionization-energy-trends&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Updated reference(s)&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2017-01-11T16:10:33.423" LastActivityDate="2017-01-11T16:10:33.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6437" PostTypeId="2" ParentId="6431" CreationDate="2013-10-07T11:32:51.833" Score="10" Body="&lt;p&gt;We can use Bohr's atomic model to calculate the ionization energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{He+}$&lt;/span&gt;.&#xA;The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives &lt;span class=&quot;math-container&quot;&gt;$IE_2=E_{\infty}-E_1$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$E_1$&lt;/span&gt; is the energy in the first bohr-orbit (principal quantum number=&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;). Conventionally taking &lt;span class=&quot;math-container&quot;&gt;$E_{\infty}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; we use the Bohr's equation to find the energy &lt;span class=&quot;math-container&quot;&gt;$E_1$&lt;/span&gt;:-&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$&lt;/span&gt;&#xA; where &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;= atomic number, &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt;= charge on electron, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;= principal quantum number, &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;= Planck's constant, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;= mass of electron, &lt;span class=&quot;math-container&quot;&gt;$\epsilon_0$&lt;/span&gt;= permittivity of free space.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Z=2$&lt;/span&gt; and the appropriate values for other constants we get &lt;span class=&quot;math-container&quot;&gt;$IE_2=-E_n=\pu{54.4eV}$&lt;/span&gt;.&#xA;Now the energy for removing both electron &lt;span class=&quot;math-container&quot;&gt;$IE_{net}=IE_1+IE_2$&lt;/span&gt;. Using &lt;span class=&quot;math-container&quot;&gt;$IE_{net}=\pu{79eV}$&lt;/span&gt;and &lt;span class=&quot;math-container&quot;&gt;$IE_2=\pu{54.4eV}$&lt;/span&gt; we get, &lt;span class=&quot;math-container&quot;&gt;$$IE_1=\pu{24.6eV}$$&lt;/span&gt;&#xA;which corresponds to option (A), which I assume is typed wrong by you in the question.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="83758" LastEditDate="2020-03-22T23:19:38.313" LastActivityDate="2020-03-22T23:19:38.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6438" PostTypeId="1" CreationDate="2013-10-07T11:48:42.253" Score="1" ViewCount="82" Body="&lt;p&gt;Is there an appropriate alternative to DMSO? I mean for embryo or oocyte freezing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a PhD student at present. I am going to study this substance and I'm very interested in this field.&lt;/p&gt;&#xA;" OwnerUserId="2436" LastEditorUserId="84" LastEditDate="2013-10-07T12:14:05.287" LastActivityDate="2013-10-07T12:14:05.287" Title="An alternative to dimethylsulfoxide (DMSO) for cryoprotection?" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6439" PostTypeId="1" AcceptedAnswerId="6440" CreationDate="2013-10-07T16:33:23.027" Score="8" ViewCount="10272" Body="&lt;p&gt;I have learnt that only chiral or unsymmetrical molecules can rotate the plane of linearly polarized light. But, why is it so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can molecules rotate the plane or what does it actually mean by 'rotating'?&#xA;Does it mean that the electric field vectors are tilted or turned left or right depending on the type of isomer?   &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="7475" LastEditDate="2015-11-10T10:27:33.593" LastActivityDate="2015-11-10T10:27:33.593" Title="Why do only chiral molecules rotate the plane of polarized light and how do they rotate it?" Tags="&lt;stereochemistry&gt;&lt;optical-properties&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6440" PostTypeId="2" ParentId="6439" CreationDate="2013-10-07T16:54:38.530" Score="7" Body="&lt;p&gt;See also: &lt;a href=&quot;https://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light&quot;&gt;How do Optically Active Compounds Rotate Plane Polarized Light?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because optical rotation is a chiral phenomenon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a molecule, and draw arrows depicting the polarization of incoming and outgoing light on it. Even if the molecule is achiral, the molecule with the arrows is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chirality can't spring out of nowhere, it can only arise from a chiral process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, see my answer &lt;a href=&quot;https://chemistry.stackexchange.com/q/6271/22&quot;&gt;here&lt;/a&gt;. Since molecules will exist in all rotations in a given fluid, for an achiral molecule the mirror image molecules cancel each other out.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2013-10-07T17:39:07.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6441" PostTypeId="1" AcceptedAnswerId="6442" CreationDate="2013-10-07T18:38:59.163" Score="11" ViewCount="22491" Body="&lt;p&gt;What I understand by a spontaneous reaction is that - A reaction whose G is negative and which occurs without the input of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gasoline reacts with oxygen only if a spark is provided. So how can it be a spontaneous reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do nonspontaneous reactions occur in nature? If so, are they always slow? If they are slow then what is the difference between a slow spontaneous reaction and a nonspontaneous reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are nonspontaneous reactions active energies more than that of spontaneous reactions?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="402" LastEditDate="2017-05-02T23:15:48.977" LastActivityDate="2017-05-02T23:15:48.977" Title="What exactly is a spontaneous reaction?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6442" PostTypeId="2" ParentId="6441" CreationDate="2013-10-07T19:26:23.907" Score="15" Body="&lt;p&gt;Spontaneous reactions simply mean that the products will be lower in free energy than the reactants ($\Delta G$ is negative). If $\Delta G$ is negative, then it is energetically favourable for the reaction to occur - in other words, there will be a release of energy due to the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of $\Delta G$ (how large it is) does not say anything about the reaction speed. For example, the reaction between gasoline and oxygen in the atmosphere at room temperature will be extremely slow, despite a large, negative $\Delta G$. This is because this reaction has a large activation energy, $E_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hold on. When two reactants collide, they interact to form an extremely unstable structure called the transition state. The transition state is high in energy and is NOT energetically favourable, and thus the transition state quickly collapses again and becomes either the reactants or the products. You can see this in the following picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/99/Rxn_coordinate_diagram_5.PNG&quot; alt=&quot;Transition state&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the kinetic energy of the reactants, they might not reach the top of the curve, in which case they will fall back to being reactants. But if they collide fast enough and are positioned appropriately with respect to each other, the top of the curve is reached, and the products will form. The energy required to form the transition state is the activation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With gasoline and oxygen, the activation energy is high. So even though a tremendous amount of energy is released during the reaction, it needs a spark to overcome its activation energy. From here on, the heat generated from the reaction supplies the activation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this reaction was not spontaneous, we might be able to force the reaction along, but the reaction would not continue on its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, not all reactions run to completion. This is because Gibbs free energy is dependent on concentrations of reactants and products, so as the products accumulate, and the reactants are used, $\Delta G$ becomes closer to 0, eventually reaching equilibrium, where $\Delta G = 0$. Visualise this scenario on the above picture. If there is no difference in Gibbs free energy, the reactants will still get to the transition state and turn into products. But the energy required for products to reach the transition state and becomes reactants is equivalent, and thus the rate of the forward reaction is equal to the rate of the reverse reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between reactions that will reach an equilibrium and the ones that run to completion is the magnitude of the $\Delta G$ at standard conditions (denoted $\Delta G°$). This is a measure of how energetically favourable the reaction &quot;inherently&quot; is. If $\Delta G°$ is large and negative, the reaction will run to completion. If it is smaller, the reaction will establish an equilibrium at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's at least generally true for reactions, where all of the species stay in the same phase I think. If you envision a reaction where one of the products is a gas, that escapes the solution of reactants, then the products will be unable to reform into reactants.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="402" LastEditDate="2017-05-02T23:15:02.797" LastActivityDate="2017-05-02T23:15:02.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6443" PostTypeId="1" CreationDate="2013-10-07T19:42:23.287" Score="0" ViewCount="1034" Body="&lt;p&gt;So, I was doing my homework and had to write out the complete equation and net ionic equation for a combination of Lead (II) Iodate and Sodium Hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my equation is given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb(IO3)2_{(aq)} + NaOH_{(aq)} -&amp;gt; ?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my dilemma is that both of these compounds &lt;em&gt;seem&lt;/em&gt; to be insolvable and I'm not really sure that I can either write the complete equation or even the net ionic equation, since I have no idea what to expect. If I do it on &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Pb%28IO3%292%20%2b%20NaOH%20-%3E&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;Wolfram Alpha&lt;/code&gt;&lt;/a&gt; I get the resulting products as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2 + I2 + 2Na + 4O2 + Pb}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I have no idea if this is right/and or what's even going on here?&lt;/p&gt;&#xA;" OwnerUserId="2315" LastEditorUserId="2315" LastEditDate="2013-10-07T19:56:05.250" LastActivityDate="2014-10-29T13:06:53.260" Title="How do I write the complete and net ionic equations for Insoluble compounds?" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6444" PostTypeId="2" ParentId="6433" CreationDate="2013-10-07T20:05:46.630" Score="2" Body="&lt;p&gt;Yikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I'm aware, a thorough application of ordinary soap or washing-up liquid should be able to wash this off using the normal &lt;a href=&quot;http://en.wikipedia.org/wiki/Surfactant&quot; rel=&quot;nofollow&quot;&gt;surfactant&lt;/a&gt; processes (binding to organic components with lipophilic regions of the molecule). If you're experiencing long-lived after-effects (say, more than a week maybe), you might want to see a doctor -- it's not impossible you could have had some sort of allergic reaction. Oh, and if you're going to wash hair with soap or washing-up liquid, I'd suggest using conditioner afterwards, to reduce the long-term damage incurred by stripping the natural fatty acids and oils off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a medical professional or a tricologist, though, so please don't take this as Official Medical Advice.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-10-07T20:05:46.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6445" PostTypeId="1" AcceptedAnswerId="6457" CreationDate="2013-10-07T23:02:22.977" Score="4" ViewCount="308" Body="&lt;p&gt;I plan to do a string of experiments on the kinetics of the decomposition of hydrogen peroxide. I’m going to see the effect of catalyst concentration on rate and the effect of varying hydrogen peroxide concentration on rate, which will then allow me to formulate a rate equation for the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to calculate the activation enthalpy of the reaction specifically using the Arrhenius Equation $k=A\operatorname{e}^\tfrac{-E_\mathrm{a}}{RT}$. How would I do this? As far as I’m aware I should use varying temperatures in the reaction to get different values of the rate, but I’m not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what kind of experiment should I do (varying temperature to see the change in rate?)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I do with my results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the value of $E_\mathrm{a}$ obtained from analysing a specific graph that will be drawn up, and if so what graph would this be?&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorDisplayName="user7951" LastEditDate="2015-06-27T08:02:59.677" LastActivityDate="2015-06-27T08:02:59.677" Title="How can I go about obtaining the activation enthalpy of a reaction using the Arrhenius Equation experimentally?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6446" PostTypeId="2" ParentId="6431" CreationDate="2013-10-08T04:30:59.173" Score="8" Body="&lt;p&gt;Taking half the total value of both ionization energies isn't a bad initial approach. You were just one step shy of finding the answer (without calculations). Each additional ionization energy will be greater than the last because the potential will be greater; the fewer the electrons, the less electron-electron repulsive force. Thus, the first ionization energy is less than half the total. Only answer A satisfies that requirement, so that must be the answer.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-10-08T04:30:59.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6447" PostTypeId="1" CreationDate="2013-10-08T05:38:04.130" Score="2" ViewCount="3037" Body="&lt;p&gt;It seems that most people draw $\ce{ClO4^{-}}$ with three double bonds and the negative charge on the oxygen:&#xA;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.109953.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemspider.com/Chemical-Structure.109953.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it would seem that it is more stable for the negative charge to be on the chlorine as it is more electronegative? (thus giving 4 double bonds to each oxygen)&lt;/p&gt;&#xA;" OwnerUserId="2438" LastEditorUserId="4231" LastEditDate="2015-03-10T14:47:02.327" LastActivityDate="2015-03-10T14:47:04.663" Title="Why does ClO₄⁻ only have 3 double bonds?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6450" PostTypeId="1" AcceptedAnswerId="6454" CreationDate="2013-10-08T14:51:31.757" Score="5" ViewCount="4321" Body="&lt;p&gt;I learned that all strong acids have a 100% degree of ionization and that there are no degrees of strength in acids; however, I also read that fluoroantimonic acid is &lt;strong&gt;20 quintillion&lt;/strong&gt; times stronger than 100% sulfuric acid !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please explain the chemistry behind this?&lt;/p&gt;&#xA;" OwnerUserId="2442" LastEditorUserId="5137" LastEditDate="2014-04-29T17:04:15.170" LastActivityDate="2015-02-20T19:41:49.230" Title="Why is fluoroantimonic acid so strong?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6451" PostTypeId="1" AcceptedAnswerId="6452" CreationDate="2013-10-08T15:23:27.850" Score="2" ViewCount="8340" Body="&lt;p&gt;What product will be obtained get when a sucrose and potassium nitrate mix, which are the ingredients for smoke bomb, are burned in a combustion container? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{KNO3 + C12H22O11 -&amp;gt; ?}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2443" LastEditorUserId="4945" LastEditDate="2018-12-11T13:18:27.897" LastActivityDate="2018-12-11T13:23:05.297" Title="What product will be produced when a sucrose and potassium nitrate mix burns?" Tags="&lt;inorganic-chemistry&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2013-10-11T00:21:21.623" ContentLicense="CC BY-SA 4.0" />
  <row Id="6452" PostTypeId="2" ParentId="6451" CreationDate="2013-10-08T16:34:59.080" Score="1" Body="&lt;p&gt;You will get potassium carbonate (&lt;span class=&quot;math-container&quot;&gt;$\ce{K2CO3}$&lt;/span&gt;), carbon dioxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;), water (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;), and nitrogen gas (&lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt;):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{48KNO3 + 5C12H22O11 -&amp;gt; 24K2CO3 + 36CO2 + 55H2O + 24N2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20110505190952AAGfyEa&quot; rel=&quot;nofollow noreferrer&quot;&gt;this yahoo question&lt;/a&gt; and &lt;a href=&quot;http://www.scienceforums.net/topic/19364-kno3-and-sugar-smoke-bomb/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this science forums thread&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2018-12-11T13:21:24.220" LastActivityDate="2018-12-11T13:21:24.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6454" PostTypeId="2" ParentId="6450" CreationDate="2013-10-08T22:33:09.463" Score="11" Body="&lt;p&gt;For many common purposes we say there's weak acids and strong acids, and define strong acids to be ones that dissociate completely in water. This is because if the acid dissociates completely in water, then every proton dissociated will just protonate water, and turn it into $\ce{H_3O^+}$ (hydronium ion). The hydronium ion is the strongest acid that can exist in water as far as I understand, and hence why strong acids are simply strong acids in some contexts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're describing is a super acid, that is not used in aqueous solution Take HCl for example. This is a classic strong acid, that will protonate a wide range of structures. But if we take something that is a VERY, VERY poor base (something that really does not want a proton), such as an alkane, HCl can't protonate it. The proton will rather stay with chlorine than protonate the alkane. That is one of the things you can do with the acid you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of an acid depends on many different factors, but it all comes down to how strongly the $\ce{H^+}$ is bound to the acid structure. If it is very loosely bound, the structure is high in energy, and the proton would like to find a more stable structure to partake in. In this case protonating an alkane to form a carbocation and molecular hydrogen is low enough in energy, that the acid can protonate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But remember that protonation is an equilibrium. If we achieve a vastly lower energy state by having the acid protonate another structure, then the protonation will be complete. If it is the same energy, half of the acid will have protonated (assuming equal concentrations of acid and base).&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-10-08T22:33:09.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6455" PostTypeId="2" ParentId="5023" CreationDate="2013-10-09T01:42:13.330" Score="4" Body="&lt;p&gt;Calcium chloride produces an exothermic reaction when mixed with water or when placed on top of ice, which means that it literally heats up when it comes in contact with ice. Sodium chloride does not produce this exothermic reaction.&lt;/p&gt;&#xA;" OwnerUserId="2445" LastActivityDate="2013-10-09T01:42:13.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6456" PostTypeId="1" CreationDate="2013-10-09T02:40:27.370" Score="2" ViewCount="265" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kK6ih.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping I could get help finding ways to sort information in the problem. What kind of information does &quot;the conversion of toluene to benzene...&quot; give us that is relevant towards solving the problem for values $R_G$ and $R_L$?&lt;/p&gt;&#xA;" OwnerUserId="2376" LastActivityDate="2014-03-16T11:52:04.633" Title="Balancing chemical systems - searching and sorting preliminary information" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6457" PostTypeId="2" ParentId="6445" CreationDate="2013-10-09T03:06:58.790" Score="6" Body="&lt;p&gt;&lt;strong&gt;Experiments at various temperatures will allow you to calculate $E_a$.&lt;/strong&gt; This is clear from a putting the Arrhenius equation in a different form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln{k}=\ln{A} + \frac{-E_a}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plot (or linear regression) of $\ln{k}$ vs $\frac{1}{T}$ yields $\frac{-E_a}{R}$ as the slope. The same experiment with a catalyst will show the activation energy is decreased.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-10-09T03:06:58.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6458" PostTypeId="1" CreationDate="2013-10-09T07:53:48.437" Score="8" ViewCount="18669" Body="&lt;p&gt;In an experiment in school, we heated copper wire (Cu) and yellow sulfur powder together in a crucible. I was wondering why this reaction cannot produce Copper Sulfate. I know it produces Copper Sulfide but I want to know why don't oxygen from the air, sulfur, and copper combine to form copper sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain this in simple high school terminology?&lt;/p&gt;&#xA;" OwnerUserId="2446" LastEditorUserId="402" LastEditDate="2017-04-01T01:30:54.813" LastActivityDate="2017-07-12T15:40:57.793" Title="Heating copper wire and sulfur powder" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6459" PostTypeId="2" ParentId="6456" CreationDate="2013-10-09T11:19:48.870" Score="2" Body="&lt;p&gt;From your diagram, $R_G$ is the recycled gas - the hydrogen that has not been consumed plus the methane that has been produced. $R_L$ is the recycled liquid - the unreacted toluene. Both go back into the mixer with fresh toluene added at the rate given and fresh hydrogen added to bring the mole ratio of hydrogen to toluene (and hydrogen to methane) to what is given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_L$ is easier. The problem tells you that 80% of the toluene is converted into benzene and 8% of the toluene is converted into biphenyl. What percentage of toluene is left over? The original feed flow of toluene is 3450 lb/hr, but the amount of toluene entering the gross feed to the reactor is higher (includes $R_L$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_{\text{tol}}=F+R_L$$&#xA;$$R_L = 0.12\times G_{\text{tol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $R_G$, you need to know how much hydrogen you have going into the mixer. You are told that the mole ratio of hydrogen to toluene in the gross feed is 4:1. Now that you have $R_L$, you can figure out the total toluene in the gross feed, and then do the stoichiometry calculation to get to total hydrogen in the gross feed. Since you have the balanced equations, you can determine how much hydrogen is consumed. If 80% of the toluene is consumed in the generation of benzene, how much hydrogen is consumed? How much methane is produced? Do the same with the conversion of toluene to biphenyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you are able to obtain the molar flow rates of all molecules out from the reactor, you should be able to calculate the flow rate of the gas exiting from the separator. This gas stream (Let's call it $K$) contains only $\ce{CH4}$ and $\ce{H2}$ of mol fractions $y$ and $1-y$ respectively. The mol fractions can be calculated considering that all $\ce{CH4}$ and $\ce{H2}$ that exists the reactor leaves the separator in $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar composition of the $R_G$ and $P$ streams that leave $K$ is the same. This means that you are now able to use a molar flow rate balance on the mixing point between $M$, $R_G$ and the stream to the mixer (Lets's call this $Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y = M + R_G$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing a balance on methane, you can find $R_G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0.5)*(Y) = (1-y)(R_G) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you know $Y$ and $y$, you will then be able to find $R_G$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2014-03-16T11:52:04.633" LastActivityDate="2014-03-16T11:52:04.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6460" PostTypeId="1" CreationDate="2013-10-09T11:48:48.310" Score="20" ViewCount="10996" Body="&lt;p&gt;Are there positively charged ligands which can bind to a central metal atom to form coordination compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that ligands are &lt;a href=&quot;https://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;noreferrer&quot;&gt;Lewis bases&lt;/a&gt; which donate a pair of electrons, and the central metal atom is usually a Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a positive charge on a species usually means it is electron-deficient. In that case, there would be no suitable electrons for it to bond to the central metal atom. So, even if positively charged ligands do exist, I think there would be very few. In such cases, it might bond with central atom which is electron rich forming the coordinate bond in the opposite way. Is that possible?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2018-12-26T23:10:54.847" LastActivityDate="2019-05-29T15:19:04.707" Title="Do ligands with a positive charge exist?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;coordination-compounds&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6461" PostTypeId="2" ParentId="6460" CreationDate="2013-10-09T13:01:53.383" Score="12" Body="&lt;p&gt;I would answer as a tentative yes.  Your reasoning is correct that a ligand, which behaves as a Lewis base, is unlikely to have a positive charge (and therefore be electron deficient).  The earliest reference to a positively-charged ion behaving as a ligand (that I've found) is from the journal &lt;em&gt;Inorganic Chemistry&lt;/em&gt; (1969, volume 8 page 2331) which describes the synthesis and characterization of Cu(II) and Mn(II) complexes containing a N-methyldabconium cation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tiJET.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the ligand is positively charged, &lt;strong&gt;but&lt;/strong&gt; there is also a lone pair on the non-methylated nitrogen that is most likely forming the coordinative covalent bond with the metal center.  It is this property (a sufficiently large ligand that has a localized positive charge that is remote from a region with Lewis base character) that would result in a positively charged ligand, and we don't need to invoke a switching of the metal roles (where the metal becomes the Lewis base) as you suggest.  That said, one of the exciting aspects of Inorganic Chemistry is the constant effort to &quot;break the rules&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-10-09T13:01:53.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6462" PostTypeId="1" AcceptedAnswerId="6464" CreationDate="2013-10-09T21:13:58.707" Score="4" ViewCount="1292" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Consider the reaction &lt;span class=&quot;math-container&quot;&gt;$$\ce{2A + B -&amp;gt; Products},$$&lt;/span&gt; when the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; alone was doubled, the half-life did not change. When the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; alone was doubled, the rate increased by two times. The units of the rate constant is:&lt;/p&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;b) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{L\ mol^{-1}\ s^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;c) Unitless&lt;/p&gt;&#xA;&lt;p&gt;d) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{mol\ L^{-1}\ s^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have tried it solving through this method:&lt;/p&gt;&#xA;&lt;p&gt;According to me, when the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; alone was doubled, reaction rate also increased by two time, implies that reaction is first order according to &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;. In same way, the reaction should be 0th order according to &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt;. So the net order would be 1. So this gives me answer &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s^{-1}}$&lt;/span&gt; but the answer is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{L\ mol^{-1}\ s^{-1}}$&lt;/span&gt;. Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-10T16:11:11.950" Title="How to find the units of the rate constant?" Tags="&lt;kinetics&gt;&lt;units&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6463" PostTypeId="2" ParentId="6451" CreationDate="2013-10-09T21:16:44.273" Score="1" Body="&lt;p&gt;The reaction equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{9.6 KNO3 + C12H22O11 -&amp;gt; 4.8 K2CO3 + 7.2 CO2 + 11 H2O + 4.8 N2}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$\pu{970 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KNO3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{342 g}$&lt;/span&gt; sucrose, which is &lt;span class=&quot;math-container&quot;&gt;$74\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KNO3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$26\%$&lt;/span&gt; sucrose. This produces &lt;span class=&quot;math-container&quot;&gt;$\pu{662 g}$&lt;/span&gt; of potassium carbonate.&lt;/p&gt;&#xA;" OwnerUserId="2450" LastEditorUserId="4945" LastEditDate="2018-12-11T13:23:05.297" LastActivityDate="2018-12-11T13:23:05.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6464" PostTypeId="2" ParentId="6462" CreationDate="2013-10-09T21:27:30.083" Score="3" Body="&lt;p&gt;You are doing it wrong in situation &lt;strong&gt;when concentration of B alone was doubled but half life did not change.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, the reaction order should be 1 according to B as Half life in first order reaction doesn't depend upon initial concentration of reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t_{1/2} = \ln2 / k $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the final order will be &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; and the answer of the question will be b.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="16683" LastEditDate="2020-02-10T16:11:11.950" LastActivityDate="2020-02-10T16:11:11.950" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6465" PostTypeId="1" CreationDate="2013-10-09T22:18:27.567" Score="2" ViewCount="4034" Body="&lt;p&gt;I've graphed a distillation curve (temperature against volume of distillate) from the data obtained from a simple distillation experiment. The substance can be any of methanol, ethanol, 1-propanol or a  1-propanol and water azeotrope). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find the boiling point from this curve to determine the identity of the compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also mention that there was leftover nonvolatile dye in the round bottom flask.&lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorUserId="360" LastEditDate="2013-10-09T22:31:29.623" LastActivityDate="2021-03-29T13:01:21.117" Title="How can I determine the boiling point of a compound using simple distillation?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6467" PostTypeId="1" AcceptedAnswerId="6476" CreationDate="2013-10-10T00:03:13.947" Score="10" ViewCount="809" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tzWWt.png&quot; alt=&quot;Question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound I is obviously the Z isomer. I thought it was obviously the less stable one too, given that there's a phenyl and t-butyl on the same side. The answer says that since the phenyl is planar, it actually exhibits less steric hindrance than a methyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because the planar phenyl ring can basically rotate its bulkiness away from the t-butyl, where as the methyl, while also free to rotate, always ends up with a hydrogen in the t-butyl's face?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2452" LastEditorUserId="2452" LastEditDate="2013-10-10T02:03:07.883" LastActivityDate="2013-10-11T09:26:25.100" Title="Phenyl vs methyl cis/trans stability" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6468" PostTypeId="1" CreationDate="2013-10-10T01:28:07.203" Score="0" ViewCount="241" Body="&lt;p&gt;In a liquid crystal cell used for LCD the conductive substrates, usually made of &lt;a href=&quot;http://en.wikipedia.org/wiki/Indium_tin_oxide&quot; rel=&quot;nofollow&quot;&gt;ITO&lt;/a&gt;, are in contact with the liquid crystal. The question is: when a voltage is applied to the electrodes to reorient the molecules is there any current flowing through the liquid? The answer should be positive because LCs are conductive materials. But I know that the molecules should reorient themselves under an applied electric field for example between two plates like a capacitor even without being in contact with the LC. So why the electrodes are placed in contact with the LC and why LCDs consume electricity?&lt;/p&gt;&#xA;" OwnerUserId="2448" LastEditorUserId="5078" LastEditDate="2014-08-10T22:40:13.737" LastActivityDate="2014-08-10T22:40:13.737" Title="Liquid crystal current flow" Tags="&lt;electrochemistry&gt;&lt;crystal-structure&gt;&lt;chirality&gt;&lt;liquid-crystals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6469" PostTypeId="1" CreationDate="2013-10-10T03:54:53.627" Score="16" ViewCount="4824" Body="&lt;p&gt;Stoichiometric defects (&lt;a href=&quot;http://en.wikipedia.org/wiki/Schottky_defect&quot; rel=&quot;nofollow&quot;&gt;Schottky&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Frenkel_defect&quot; rel=&quot;nofollow&quot;&gt;Frenkel&lt;/a&gt; defects) in ionic crystals are formed due to irregularities in the ionic arrangement which maintain the stoichiometry and electroneutrality of the solid. It is known that $\ce{AgBr}$ shows both Frenkel and Schottky defects while most other compounds  show only one of these defects; for example, $\ce{AgCl}$ shows only Frenkel defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the sizes of the cation and anion determines which type of defect is shown by the crystal. If the ratio is closer to unity, Schottky defects are more common, and if it deviates significantly from one, Frenkel defects are more common. Intuitively, this is correct since Frenkel defects involve the movement of the cation (or the smaller ion) into interstitial sites, which is indeed easier if the ion is very small as compared to the other ion, making it more mobile. Generally, the activation energy for cationic movement is lower. And if the ionic sizes are comparable, filling the interstitial sites is not feasible, and hence Schottky defects are more likely to be exhibited by the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{Cl-}$ ion is smaller than $\ce{Br-}$ (i.e. the ratio $r_+/r_-$ for $\ce{AgCl}$ is closer to 1 than it is for $\ce{AgBr}$). Yet, $\ce{AgBr}$ displays Schottky defects while $\ce {AgCl}$ does not. Why is that so? (The size and hence the mobility of $\ce{Ag+}$ should be the same in both cases.)&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2016-04-17T21:28:41.153" LastActivityDate="2020-06-30T15:47:16.100" Title="Stoichiometric defects in silver bromide" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6470" PostTypeId="2" ParentId="6465" CreationDate="2013-10-10T05:42:00.227" Score="1" Body="&lt;p&gt;Do you mean that the substance could be pure methanol, ethanol or 1-propanol? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case there is a simple way to distinguish between the four options. A pure substance will have a T-V curve that consists of 2 straight lines, 1 running from $T=T_{start}$ to $T=T_{b}$ at $V=0$ and one running from $V=0$ to $V=V_{total}$ at $T=T_{b}$. Probably you will have some impurities which cause some small deviations from this perfect hook, but that will be the general shape. Determining the boiling temperature is just a matter of checking where the curve goes straight up. Depending on the temperature at which this happens you can identify methanol ($T_{b}\approx65$), ethanol ($T_{b}\approx78$) or 1-propanol ($T_{b}\approx97$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The azeotropic mixture makes it a bit tricky, because the boiling points of 1-propanol and water are so close together, but the curve should show a distinct difference not going vertically up at a certain temperature.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-10-10T05:42:00.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6472" PostTypeId="2" ParentId="5023" CreationDate="2013-10-10T12:00:43.893" Score="2" Body="&lt;p&gt;In addition to the colligative and thermochemical properties, there are practical considerations as to why calcium chloride or sodium chloride would be used to melt ice in a public area. Calcium chloride will make a surface slippery under reasonably cold ($&amp;lt;0~^\circ\mathrm{F}$) conditions presumably due to the hygroscopic nature of the salt.  On the other hand, sodium chloride is fairly corrosive and can damage concrete and vegetation.  I'm not a fan of sending visitors to other Q&amp;amp;A sites, but &lt;a href=&quot;http://chemistry.about.com/cs/howthingswork/a/aa120703a.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this answer&lt;/a&gt; has a nice table of the pros and cons of several salts used in melting ice.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="16683" LastEditDate="2016-11-28T18:23:14.913" LastActivityDate="2016-11-28T18:23:14.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6473" PostTypeId="1" CreationDate="2013-10-10T15:45:16.677" Score="6" ViewCount="5430" Body="&lt;p&gt;Considering water as a given, the third most important ingredient in a copper electroplating solution, as a short survey of related recipe webpages will show you, is hydrochloric acid (HCl). Here's a sample link: &lt;a href=&quot;http://www.thinktink.com/faqs/cupltfaq/cupltf01.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.thinktink.com/faqs/cupltfaq/cupltf01.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess that the role of sulfuric acid (the second most important ingredient; the first being copper sulfate) is to provide you with sufficient anions to allow copper cations to easily enter the solution at the anode. But in almost all recipes that I have seen, a small amount of hydrochloric acid is added. I wanted to know how useful it is and what is the opportunity cost associated with it.&lt;/p&gt;&#xA;" OwnerUserId="736" LastEditorUserId="402" LastEditDate="2017-01-14T08:55:17.903" LastActivityDate="2019-12-03T05:49:22.363" Title="What is the role of hydrochloric acid in a copper electroplating bath?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6475" PostTypeId="1" CreationDate="2013-10-10T21:36:59.907" Score="13" ViewCount="485" Body="&lt;p&gt;In Szabo and Ostlund's Modern Quantum Chemistry, the procedure of single determinant energy minimization is presented. Omitting the whole procedure, I have a question about functional variation during the derivation of the Hartree-Fock equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the single determinant $| \Psi_0 \rangle = | \chi_1 \chi_2 \ldots \chi_a \chi_b \ldots \chi_N \rangle$, the energy $E_0 = \langle \Psi_0 | \mathscr{H} | \Psi_0 \rangle$ is a functional of the spin orbitals $\{ \chi_a \}$. $E_0$ is the expectation value of the single determinant $\vert \Psi_0 \rangle$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E_0[\{ \chi_a \} ] = \sum\limits_{a=1}^N [a\vert h \vert a] + \frac{1}{2} \sum\limits_{a=1}^N \sum\limits_{b=1}^N [aa \vert bb] - [ab \vert ba].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation can be varied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\delta E_0 &amp;amp;= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] + [\chi_a | h | \delta \chi_a] \\&#xA;&amp;amp;+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \delta \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] + [\chi_a \chi_a | \chi_b \delta \chi_b] \\&#xA;&amp;amp;-\frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + [ \chi_a \chi_b | \delta \chi_b \chi_a] + [\chi_a \chi_b | \chi_b \delta \chi_a] \label{1}\tag{1}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Authors suggest to the reader as an exercise to manipulate this equation for $\delta E_0$ to show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\delta E_0 = \sum \limits_{a=1}^N [\delta \chi_a | h |\chi_a] + \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] - [\delta \chi_a \chi_b | \chi_b \chi_a] + \text{c.c.} \label{2}\tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the first sum in $\eqref{1}$ can be easily converted to that in $\eqref{2}$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\delta \chi_a | h | \chi_a]^* = [\chi_a | h | \delta \chi_a]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, for the second sum in $\eqref{1}$ one can show that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\delta \chi_a \chi_a | \chi_b \chi_b]^* = [\chi_a \delta \chi_a | \chi_b \chi_b]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\chi_a \chi_a | \delta \chi_b \chi_b]^* = [\chi_a \chi_a | \chi_b \delta\chi_b].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the same manipulations with the third sum in $\eqref{1}$ and obtained&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\delta E_0 &amp;amp;= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] \\&#xA;&amp;amp;+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] \\&#xA;&amp;amp;- \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + \text{c.c.} \label{3}\tag{3}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, comparing $\eqref{3}$ and $\eqref{2}$ I suppose that one should show that both terms in each sum are equal, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\delta \chi_a \chi_a | \chi_b \chi_b] = [ \chi_a \chi_a | \delta \chi_b \chi_b]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\delta \chi_a \chi_b | \chi_b \chi_a] = [ \chi_a \delta \chi_b | \chi_b \chi_a].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how to do that?&lt;/p&gt;&#xA;" OwnerUserId="2456" LastEditorUserId="194" LastEditDate="2018-02-02T01:00:11.680" LastActivityDate="2018-02-02T01:00:11.680" Title="Functional variation in derivation of the Hartree-Fock equations" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6476" PostTypeId="2" ParentId="6467" CreationDate="2013-10-11T09:26:25.100" Score="11" Body="&lt;p&gt;The molecule contains conjugated system of double bonds, being essentially a substituted styrene. The phenyl cannot therefore freely rotate and its steric requirements are difficult to estimate just by hand-waving arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct answer needs some quantitative argument, either e.g. heats of formation or calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second, the energies come out using MMFF94 (using Avogadro):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Z: 177.6 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;E: 175.6 kJ/mol &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, both the isomers rotated the styrene substantially from planarity. The difference in results is quite small and can easily change when using other method. According to the numbers I'd agree with your answer, Z is less stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not consider this a valid question in test, the answer is not straightforward.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-11T09:26:25.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6477" PostTypeId="1" AcceptedAnswerId="6479" CreationDate="2013-10-11T14:33:19.160" Score="22" ViewCount="13890" Body="&lt;p&gt;How is 1-bromo-3-chloropropa-1,2-diene ($\ce{BrHC=C=CHCl}$) chiral? It doesn't have any carbon that is connected to four different groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XJyg5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJyg5.png&quot; alt=&quot;Chemical structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="30836" LastEditDate="2020-04-05T13:02:09.747" LastActivityDate="2020-04-05T13:02:09.747" Title="Why are allenes chiral?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;&lt;allenes&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6478" PostTypeId="2" ParentId="6216" CreationDate="2013-10-11T17:20:43.460" Score="4" Body="&lt;p&gt;You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-&lt;br&gt;&#xA;The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of &#xA;$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-10-11T17:20:43.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6479" PostTypeId="2" ParentId="6477" CreationDate="2013-10-11T20:40:50.820" Score="21" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;Chirality&lt;/a&gt; is a property of objects in which they lack certain symmetry operations, specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Improper_rotation&quot; rel=&quot;noreferrer&quot;&gt;improper rotations&lt;/a&gt;, including the mirror plane and inversion operations. For example, 3-dimensional chiral objects lack mirror symmetry. According to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The feature that is most often the cause of chirality in molecules is the presence of an asymmetric carbon atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not the only cause of chirality. Any molecule of low enough symmetry to lack a mirror plane is chiral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &lt;a href=&quot;http://en.wikipedia.org/wiki/Stereogenic_center&quot; rel=&quot;noreferrer&quot;&gt;stereogenic centers&lt;/a&gt; need not be carbon atoms (or any atoms), they just need to be a point in the molecule's space which generates chirality for the molecule. The central carbon atom in &lt;a href=&quot;http://en.wikipedia.org/wiki/Allene#Symmetry&quot; rel=&quot;noreferrer&quot;&gt;low symmetry allenes&lt;/a&gt; serves this purpose. This is also an example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_chirality&quot; rel=&quot;noreferrer&quot;&gt;axial chirality&lt;/a&gt;, which is chirality about an axis (the $\ce{C=C=C}$ axis) instead of about a point (which is point chirality — the more familiar kind).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7475" LastEditDate="2015-12-30T21:36:02.980" LastActivityDate="2015-12-30T21:36:02.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6480" PostTypeId="2" ParentId="6477" CreationDate="2013-10-12T02:02:26.313" Score="29" Body="&lt;p&gt;Asked and answered, but I think one thing that's missing is that allenes are not planar like alkenes or alkynes are. &lt;a href=&quot;https://chemistry.stackexchange.com/questions/49670/shape-of-allene-molecule&quot;&gt;You can refer to this question for an explanation.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/14wVX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/14wVX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dihedral angle between the two halogens is 90 degrees (ideally). Here's an animation hopefully providing a better view of the 3D structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbsyJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in the two mirror images being non-superimposable upon each other, which means the molecule satisfies the criterion for chirality. If you rotate the right-hand molecule by $90^\circ$ such that the blue and white atoms at the top coincide, the green and white atoms will not match each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2AWkT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2AWkT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any allene with the formula $\ce{R^1R^2C=C=CR^3R^4}$ will be chiral as long as $\ce{R^1}\neq\ce{R^2}$ and $\ce{R^3}\neq\ce{R^4}$. In this case we have $\ce{H} \neq \ce{Br}$ and $\ce{H} \neq \ce{Cl}$, so the compound in the question is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{R^1} = \ce{R^2}$ or $\ce{R^3} = \ce{R^4}$, then the compound will possess a mirror plane and will therefore be achiral.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2016-09-07T15:38:36.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6481" PostTypeId="1" CreationDate="2013-10-12T06:20:04.770" Score="0" ViewCount="551" Body="&lt;p&gt;I want some information regarding the protocol for determining any type of impurity in pharmaceutical substances or drug product.&#xA;Which type of impurities can be identified by HPLC and MS?&#xA;what is the difference between related substances and impurity?&lt;/p&gt;&#xA;" OwnerUserId="2463" LastActivityDate="2013-12-13T19:22:59.740" Title="which are the possible ways to detect impurity in drugs?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6482" PostTypeId="1" AcceptedAnswerId="6484" CreationDate="2013-10-12T09:14:19.483" Score="2" ViewCount="7979" Body="&lt;p&gt;My attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on symmetry: I think that looking for symmetry phenol has a higher boiling point than 1,2-dihydroxybenzene because the two -OH groups projecting out from the benzene ring of 1,2-dihydroxybenzene will not allow close packing of them which results in a lower boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on hydrogen bonding: In this sense phenol would have a lower boiling point because 1,2-dihydroxybenzene would have more inter-molecular hydrogen bonding than phenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know which factor wins?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="-1" LastEditDate="2014-09-24T18:26:54.393" LastActivityDate="2014-09-24T18:26:54.393" Title="Which among phenol and 1,2-dihydroxybenzene has the higher boiling point?" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6483" PostTypeId="2" ParentId="6482" CreationDate="2013-10-12T10:50:52.590" Score="1" Body="&lt;p&gt;Predicting boiling points based on similar structures can be a little hand-wavy, given the factors that can contribute. I'm not sure that comparing symmetry between phenol and 1,2-dihydroxybenzene (catechol) is strictly a direct comparison. Not only do have have an additional functional group to contend with, but you are also introducing molecular weight changes too. Perhaps comparing catechol to 1,3-dihydroxybenzene (resorcinol) or 1,4-dihydroxybenzene (hydroquinone) might allow for better arguments for symmetry: catechol (bp 245); resorcinol (bp 277); hydroquinone (bp 277).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, having more functional groups present which increase the opportunity for intermolecular forces to do their bit is more likely to influence boiling point than symmetry.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-10-12T10:50:52.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6484" PostTypeId="2" ParentId="6482" CreationDate="2013-10-12T11:48:57.253" Score="6" Body="&lt;p&gt;Your structures (grabbed from Wikimedia Commons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jR4OF.png&quot; alt=&quot;enter image description here&quot;&gt;          &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catechol (1,2-dihydroxybenzene)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZICkh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three main influences on boiling point are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shape - molecules with a more spherical shape have lower boiling points than molecules that are more cylindrical, &lt;strong&gt;all else as equal as possible&lt;/strong&gt;. Spheres have the lowest surface area to volume ratio of any of the solids. This is where symmetry comes in. More symmetric molecules are more spherical. The shape/symmetry argument does not work here since the two compounds are not isomers. The issue of close packing is relevant to the melting points of solids, since in a solid, most particles do not have translational motion - they are locked in a crystal lattice. In a liquid, such random translational motion is common. In the liquid state, molecules of most compounds are so disorderly arranged that it does not matter how well they pack as a solid. In your case, both molecules are rather like flat disks with a little bit of polarness sticking off one side. Their shapes are not so different as far as surface area to volume goes. Boiling point will be dominated by mass and polarity. &lt;/li&gt;&#xA;&lt;li&gt;Molecular mass - bigger molecules require more energy to vaporize because they have more mass, and $KE=\frac{1}{2}mv^2$, &lt;strong&gt;all else as equal as possible.&lt;/strong&gt; Mass difference applies in your case because the two molecules are similar enough in shape and polarity, except one has an extra oxygen in its formula.&lt;/li&gt;&#xA;&lt;li&gt;Polarity - polarity controls the types and number of intermolecular forces, including hydrogen bonding. The more and stronger intermolecular forces between two molecules, the higher the boiling point, &lt;strong&gt;all else as equal as possible.&lt;/strong&gt; Since catechol has two hydroxy groups and phenol has one, you have your strong argument for increased intermolecular forces. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The other easy thing to do is use the internet to look up the boiling points, and then explain the difference using the three factors. Once you know what the boiling points are, it is easy to choose the structural features that support the difference. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol&quot; rel=&quot;noreferrer&quot;&gt;Phenol&lt;/a&gt; - $181.7 \ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Catechol&quot; rel=&quot;noreferrer&quot;&gt;Catechol&lt;/a&gt; - $245.5\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-12T11:48:57.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6485" PostTypeId="2" ParentId="5256" CreationDate="2013-10-12T13:22:44.843" Score="7" Body="&lt;p&gt;It depends a little on your definition of combustion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most combustion we see involves things reacting with the oxygen in air. Most combustion reactions are a self-sustaining exothermic reaction of some substance with oxygen gas. But atmospheric oxygen is relatively dilute (only about 21% of air is oxygen) and this limits what will burn sustainably in air (also, at slightly lower concentrations many commonly flammable things won't stay lit). The requirement for things to be combustible in air is, crudely, a reaction path with oxygen that releases energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples include many types of compound. Hydrogen reacts with oxygen to give water; hydrocarbons react to give carbon dioxide and water. Many metals react to give metal oxides (eg magnesium, spectacularly when lit; iron, only when hot and finely divided). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many things won't burn in air for two reasons: there are activation barriers to a potential oxidation meaning a potential reaction can't be sustained (aluminium has a strong oxide layer on its surface protecting the reactive metal from further oxidation, for example); other compounds are already highly oxidised an no further oxidation can happen (eg silicon dioxide, a major ingredient of sand and glass). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many more things that will combust in pure oxygen as &lt;a href=&quot;http://en.wikipedia.org/wiki/Apollo_1&quot; rel=&quot;nofollow&quot;&gt;Apollo 1 discovered&lt;/a&gt;, tragically. And as many demonstrations show spectacularly (see demo &lt;a href=&quot;http://www.youtube.com/watch?v=WuG5WTId-IY&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). But highly oxidised things like silica can't be oxidised further and won't burn not matter how concentrated the oxygen is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But oxygen isn't the only gas that can support exothermic reactions. Rather more things will burn in fluorine, for example. Even relatively unreactive substances can support combustion if there is an available reaction path that releases enough energy and a reaction can get started. One of the best known examples is that magnesium will burn in carbon dioxide (the reaction producing magnesium oxide releases energy) and even nitrogen (producing magnesium nitride). See the explanation &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and the video &lt;a href=&quot;http://www.youtube.com/watch?v=wqErrNvns4o&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to return to the original question, are there any substances that won't combust? Whatever the definition of combustion is the answer is still yes. Noble gases, for example, though some can be coerced into reacting under extreme circumstances, never combust or support combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we restrict our definition to combust in air, many substances won't combust though many of them might do so under different circumstances.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2016-07-30T21:58:33.017" LastActivityDate="2016-07-30T21:58:33.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6486" PostTypeId="1" AcceptedAnswerId="6488" CreationDate="2013-10-12T16:12:18.887" Score="8" ViewCount="2304" Body="&lt;p&gt;In a common interpretation of the Arrhenius rate equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = A\exp\left(-\frac{E_\mathrm a}{RT}\right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the activation energy $E_\mathrm a$ is understood to represent the &lt;em&gt;difference&lt;/em&gt; in the energy of the reactants and the minimum energy required to form the activated complex or the transition state, as shown in the following diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yfVOB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this must be wrong! As $E_\mathrm a$ is the &lt;em&gt;minimum energy itself&lt;/em&gt; which the molecules must possess to reach the activated complex. That is, it is not the &lt;em&gt;difference&lt;/em&gt; but the &lt;em&gt;minimum energy level itself&lt;/em&gt;. This is the only way in which it will remain constant i.e. independent of temperature (otherwise with increase in $T$, $E_\mathrm a$ will go down) and can be used in the Arrhenius Equation where the exponential term denotes the fraction of molecules having energy higher than or equal to $E_\mathrm a$ according to Maxwell-Boltzmann distribution. Sometimes, threshold energy ($E_\mathrm t$) is defined as the minimum energy required for reaction to proceed and $E_\mathrm a = E_\mathrm t-\mathrm{KE_{average}}$. If this definition is true, then threshold energy and not activation energy should be in the Arrhenius equation. I therefore need to clarify the actual definition of activation energy $E_\mathrm a$.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2018-02-05T23:33:47.773" LastActivityDate="2018-02-05T23:34:30.493" Title="How exactly is activation energy defined?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6488" PostTypeId="2" ParentId="6486" CreationDate="2013-10-12T17:59:01.583" Score="9" Body="&lt;p&gt;You are right that the energy of the &lt;em&gt;initial state&lt;/em&gt; is a function of temperature. But in practice for this equation the energy difference &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm a$&lt;/span&gt; usually is fixed for an arbitrary temperature (in the middle of the diapason of experimental measurements) and assumed temperature-independent.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy#Temperature_independence_and_the_relation_to_the_Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;While this equation suggests that the activation energy is dependent&#xA;on temperature, in regimes in which the Arrhenius equation is valid&#xA;this is cancelled by the temperature dependence of &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;. Thus, &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm a$&lt;/span&gt; can be&#xA;evaluated from the reaction rate coefficient at any temperature&#xA;(within the validity of the Arrhenius equation).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As to the physical meaning of activation energy, &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation#Limitations_of_the_idea_of_Arrhenius_activation_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;citing Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Both the Arrhenius activation energy and the rate constant &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; are&#xA;experimentally determined, and represent macroscopic reaction-specific&#xA;parameters that are not simply related to threshold energies and the&#xA;success of individual collisions at the molecular level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In other words, activation energy &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm a$&lt;/span&gt; is a semi-empirical quantity and is not considered to have a strictly defined physical meaning (at least on molecular level). Note that the Arrhenius equation&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k=A\exp\left(-\frac{E_\mathrm a}{RT}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;is itself considered to be a semi-empirical relationship.&lt;/p&gt;&#xA;&lt;p&gt;In the IUPAC Gold Book (&lt;a href=&quot;https://doi.org/10.1351/goldbook.A00102&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1351/goldbook.A00102&lt;/a&gt;), activation energy is also defined to be an empirically determined quantity, in line with the above:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An empirical parameter characterizing the exponential temperature dependence of the rate coefficient &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm a = RT^2 \frac{\mathrm d (\ln k)}{\mathrm dT}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="957" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-05T23:34:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6489" PostTypeId="1" AcceptedAnswerId="6493" CreationDate="2013-10-12T18:04:55.763" Score="7" ViewCount="103120" Body="&lt;p&gt;How to calculate it? I know that it depends on the coordination compound and the number of electrons present in t2g and eg orbitals and the final answer is expressed in terms of crystal field splitting parameter. But, how to know how many electrons are there in t2g and eg orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think rather than a broad answer and example would help. So, consider the compound $\ce{[Ni(CO)4]}$ and how to calculate CFSE for it?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="275" LastEditDate="2013-10-13T11:27:39.940" LastActivityDate="2017-10-12T16:33:18.030" Title="How to calculate crystal field stabilisation energy?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;crystal-structure&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6490" PostTypeId="1" AcceptedAnswerId="6492" CreationDate="2013-10-12T20:00:35.263" Score="4" ViewCount="182" Body="&lt;p&gt;I've recently read that the Carbon takes a unique position as a reducing agent , and i think it has to do with the very special form of its Ellingham diagram  , but i am still not able to make a connection , So what makes exaclty the carbon a good reducing agent ?&lt;/p&gt;&#xA;" OwnerUserId="1463" LastActivityDate="2015-05-19T16:12:15.240" Title="Carbon Free energy vs Temperature Diagram" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6491" PostTypeId="1" AcceptedAnswerId="6498" CreationDate="2013-10-13T09:22:29.913" Score="7" ViewCount="1962" Body="&lt;p&gt;When I use $\ce{FeCl3}$ solution for etching circuits it gives off an acidic smell.&#xA;I was wondering what is the cause of the smell.&#xA;I don't think it's caused by the $\ce{Cl2}$ from $\ce{2FeCl3-&amp;gt;2FeCl2 + Cl2}$ because chlorine has a different smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product used is the one in &lt;a href=&quot;http://www.vitacom.ro/images/products/PictureVitacom/media/c/CLORURA%20F38_500.jpg&quot;&gt;this picture&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2469" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:57.317" LastActivityDate="2016-12-23T08:45:37.420" Title="why does FeCl₃ have such a specific smell?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6492" PostTypeId="2" ParentId="6490" CreationDate="2013-10-13T11:25:48.080" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://showard.sdsmt.edu/MET320/Handouts/EllinghamDiagrams/Ellingham_v22_Macro.pdf&quot; rel=&quot;nofollow&quot;&gt;Here is the link to a set of Ellingham Diagrams&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high temperatures, smelting and carbon reductions have several practical challenges some of which are energy efficiency, formation of metal carbides, equipment concerns etc. Therefore, to assess the reducing ability of any reducing agent, we must work within practical limits. Let us select a temperature of $1600\ \mathrm{K}$ as our upper bound beyond (which is still quite high) which, practical concerns outweigh any benefit of having a good reductant. At this temperature $\Delta_\text fG(\ce{CO2})$ is approx $-394.39\ \mathrm{kJ/mol}$ and $\Delta_\text fG(\ce{CO})$ is $-504\ \mathrm{kJ/mol}$. These values are high enough and can already reduce several oxides of metals like $\ce{Co, Fe, Cu, Pb, Ni, Bi}$ and $\ce{MnO2, CrO2}$ etc (subject to practical constraints). This gives it a good reducing character, better than $\ce{H2}$ or ammonia.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to notice, is that, although the variation of $\Delta_\text fG$ for carbon dioxide is flat, for carbon monoxide it slopes downward and hence with increasing temperature, it can reduce many more oxides further expanding its capacity to reduce metal oxides. $\ce{V2O5, VO2, SnO2, WO3, Mn3O4, MoO3, GeO2}$ can be reduced by carbon. (All this just implies thermodynamic favourability; it can still be unfavourable kinetically, or due to practical restraints.) Hence, the negative sloping $\Delta_\text fG$ for $\ce{CO}$ makes carbon a very good reducing agent, and it can reduce a huge variety of metal oxides, within a considerable temperature limit.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T16:12:15.240" LastActivityDate="2015-05-19T16:12:15.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6493" PostTypeId="2" ParentId="6489" CreationDate="2013-10-13T11:51:43.283" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 1: Look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel_carbonyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nickel Carbonyl&lt;/a&gt; and find out what geometry it has. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdZA2.png&quot; alt=&quot;nickel carbonyl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 2: Find the appropriate &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_field_theory#Crystal_field_splitting_diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;crystal field splitting diagram&lt;/a&gt; for this geometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXd7i.png&quot; alt=&quot;tetrahedral ligand field splitting diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 3: Figure out how many $d$ electrons there are. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nickel&lt;/a&gt; is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state &lt;strong&gt;atomic&lt;/strong&gt; electron configurations just group state &lt;strong&gt;molecular&lt;/strong&gt; electron configurations. For more information on $d$ electron count, go to &lt;a href=&quot;http://en.wikipedia.org/wiki/D_electron_count#d10&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 4: Fill in the $d$ electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 5: Determine $\Delta E$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_field_theory#Crystal_field_stabilization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)&lt;/li&gt;&#xA;&lt;li&gt;$t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="40029" LastEditorDisplayName="user5764" LastEditDate="2017-03-08T17:21:59.227" LastActivityDate="2017-03-08T17:21:59.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6494" PostTypeId="1" AcceptedAnswerId="6496" CreationDate="2013-10-13T13:47:47.047" Score="3" ViewCount="3344" Body="&lt;p&gt;I know that because of the absence of the hydrogen ions in a reaction ($n_{H+} = 0 $) , the correspondent E-ph line have to be horizental , but what exactly the types of reactions (if there is) that characterizes the vertical lines of an E-ph plot ?&lt;/p&gt;&#xA;" OwnerUserId="1463" LastEditorUserId="275" LastEditDate="2013-10-13T15:20:49.423" LastActivityDate="2013-10-13T15:20:49.423" Title="Vertical lines in a Pourbaix Diagram" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6495" PostTypeId="1" AcceptedAnswerId="6500" CreationDate="2013-10-13T15:11:14.057" Score="1" ViewCount="8851" Body="&lt;p&gt;I'm conducting an experiment using iron(III) chloride in aqueous solution, but am also aware that it's usually a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is iron(III) chloride experimentally converted to a solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can you please explain how the conversion happens and its mechanism.&lt;/p&gt;&#xA;" OwnerUserId="667" LastEditorUserId="10185" LastEditDate="2015-03-25T13:29:22.590" LastActivityDate="2018-06-08T20:24:45.277" Title="How is iron(III) chloride changed from solid to solution?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6496" PostTypeId="2" ParentId="6494" CreationDate="2013-10-13T15:20:39.500" Score="6" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Pourbaix_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pourbaix diagram&lt;/a&gt; is a special kind of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase diagram&lt;/a&gt;. As in a phase diagram, the lines of all kinds represent phase boundaries. Each line represents the conditions in which two phases are in dynamic equilibrium. Without information on what the phases enclosed by lines look like, a phase diagram is pretty meaningless. Below is the phase diagram of water. Imagine how hard it would be to interpret without the labels. We still need to go look up what is meant by Solid IX - is it &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_IX&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice IX&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_9&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice 9&lt;/a&gt;?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rxoFq.png&quot; alt=&quot;phase diagram of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the iron-&lt;a href=&quot;http://en.wikipedia.org/wiki/Steel#Material_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon (steel)&lt;/a&gt; phase diagram. Again, the phases are labeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8e/Steel_pd.svg/500px-Steel_pd.svg.png&quot; alt=&quot;steel phase diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Pourbaix diagram, the axes are pH and applied electric potential $(E_0)$ in volts instead of temperature and pressure (pure substance) or temperature, pressure (sometimes omitted), and composition for mixtures. The lines represent boundaries, but the &quot;phases&quot; are now the predominant iron-containing ion in solution. These diagrams seem to be primarily for aqueous systems, and so there is never an absence of $\ce{H+}$ ions. They may not be needed in the redox equation, but they are in solution. Otherwise, how would we have different pH along the horizontal axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the Pourbaix diagram for aqueous solutions of iron. Notice that not all lines are horizontal or vertical. Each line represents a set of conditions ($E_0$ and $\text{pH}$) where two ionic species are in dynamic equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wf0Rc.png&quot; alt=&quot;Pourbaix diagram of iron&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at $E_0=1.0\ \text{V}$ and $\text{pH}=2$, the predominating species is $\ce{Fe^{3+}(aq)}$. If we increase the $\text{pH}$ to $5$, the predominating iron species becomes $\ce{Fe2O3}\cdot\ce{nH2O}(s)$, rust. The reaction might be (we increase the pH by added $\ce{OH-}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe^{3+}(aq) +6OH- (aq) -&amp;gt; Fe2O3(s) + 3H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this reaction is redox neutral - all elements remain in the same oxidation state. Vertical lines represent redox neutral transitions/reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start at the same place $E_0=1.0\ \text{V}$ and $\text{pH}=2$, and change $E_0$ from $1.0 \ \text{V}$ to $0.4 \ \text{V}$, the predominating species becomes $\ce{Fe^{2+}(aq)}$. When we lower $E_0$, it is like adding electrons to the system (it is really encouraging electrons to flow through the system in a specific way), we get this reaction, which is pH-neutral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{3+}(aq) + e- -&amp;gt; Fe^{2+}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, there is no $\ce{H+(aq)}$ in this reaction. Horizontal lines are pH-neutral transitions/reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagonal or curved line is neither pH or redox neutral. For example, let's go from $E_0=0\ \text{V}$ and $\text{pH}=6 \ (\ce{Fe^{2+}(aq)}$ to $E_0=0\ \text{V}$ and $\text{pH}=10 \ (\ce{Fe2O3\cdot nH2O(s)}$ crossing a diagonal line. The reaction now involves a change in oxidation state and in pH. Since we are traveling horizontally, the electrons lost by $\ce{Fe^{2+}}$ must go somewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe^{2+}(aq) + 4HO- (aq) -&amp;gt; Fe2O3(s) + 2H2(g) + H2O(l)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.523" LastActivityDate="2013-10-13T15:20:39.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6497" PostTypeId="2" ParentId="5174" CreationDate="2013-10-13T16:59:57.473" Score="1" Body="&lt;p&gt;The key to fluorescence is an extended Pi system. In hematoxylin, the two aromatic rings are far apart, while in Eosin, the conjuated Pi system is uninterrupted and continuous. You need at least four conjugated groups to absorb in the visible light region. Aromatic systems absorb strongly in the UV region. In addition, the attachment of halogens (i.e. Br) reduces fluorescence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things that reduce fluorescence besides halogens are $NO_2^-$ (nitro groups) and flexibility of the molecules. Rigidity increases fluorescence.&lt;/p&gt;&#xA;" OwnerUserId="2471" LastEditorUserId="1573" LastEditDate="2013-10-14T01:34:18.983" LastActivityDate="2013-10-14T01:34:18.983" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6498" PostTypeId="2" ParentId="6491" CreationDate="2013-10-13T17:44:01.603" Score="9" Body="&lt;p&gt;The only clue I have been able to find is a one-off sentence in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_chloride#Reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; for $\ce{FeCl3}$, for which no reference is given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Iron(III) chloride undergoes hydrolysis to give an acidic solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so, then the reactions are probably those along the conversion of $\ce{FeCl3}$ to $\ce{Fe2O3}$, producing hydrochloric acid $(\ce{HCl})$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen chloride gas&lt;/a&gt; that escapes from the solution has a pungent acidic acrid smell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible reactions include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{FeCl3 + H2O &amp;lt;=&amp;gt; FeOCl + 2HCl}$$&#xA;$$\ce{2FeCl3 + H2O &amp;lt;=&amp;gt; Cl2FeOFeCl2 + 2HCl}$$&#xA;$$\ce{2FeOCl + 2H2O &amp;lt;=&amp;gt; Fe2O3(s) + 2HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because these reactions are in equilibrium, solutions of $\ce{FeCl3}$ can be stabilized by addition of $\ce{HCl}$. Addition of $\ce{HCl}$ also promotes the formation of the tetrachloroferrate ion $\ce{FeCl4-}$, which is more resistant to hydrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{FeCl3 + HCl -&amp;gt; FeCl4- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ferric_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ferric oxide&lt;/a&gt; (Iron(III) oxide $\ce{Fe2O3}$), is insoluble. Does your solution develop reddish precipitate as it ages? That would be  evidence in favor of hydrolysis as the source of $\ce{HCl}$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="17368" LastEditDate="2016-12-23T08:45:37.420" LastActivityDate="2016-12-23T08:45:37.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6499" PostTypeId="1" CreationDate="2013-10-13T18:02:57.453" Score="4" ViewCount="577" Body="&lt;p&gt;&lt;a href=&quot;http://www.uu.se/en/media/press-release-document/?id=2736&amp;amp;area=3&amp;amp;typ=artikel&amp;amp;na&amp;amp;lang=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It turned out that Upsalite had the highest surface area measured for an alkali earth metal carbonate; 800 square meters per gram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that a sphere may enclose the maximum volume for a given surface area, the radius for a sphere enclosed by $\pu{800 m^2}$ is $\pu{7.9788 m}$. It's corresponding volume is $\pu{2127.656 m^3}$ - which leads to the incredible figure of something in excess of 2000 litres of water ... per gram of dessicant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much water vapour can a gram of Uppsalite absorb at NTP?&lt;/p&gt;&#xA;" OwnerUserId="765" LastEditorUserId="5026" LastEditDate="2018-05-18T08:35:12.353" LastActivityDate="2019-10-29T22:23:02.250" Title="How much water vapour can a gram of Uppsalite absorb at NTP?" Tags="&lt;water&gt;&lt;absorption&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6500" PostTypeId="2" ParentId="6495" CreationDate="2013-10-13T18:15:03.817" Score="1" Body="&lt;p&gt;You can prepare a solution of $\ce{FeCl3}$ by dissolving the appropriate amount in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you want a $0.1 \text{ M}$ solution of $\ce{FeCl3}$. You would do this by dissolving 0.1 &lt;a href=&quot;http://en.wikipedia.org/wiki/Mole_(unit)&quot; rel=&quot;nofollow noreferrer&quot;&gt;mole&lt;/a&gt; of $\ce{FeCl3}$ in 1 liter of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.1000 \text{ mol} \times \dfrac{162.2\text{ g}}{1\text{ mol}}=16.22 \text{ g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to dissolve 16.22 g of $\ce{FeCl3}$ in enough water to make 1 liter of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solution is not stable over time. $\ce{FeCl3}$ hydrolyzes in water. See my answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6491/why-does-fecl3-have-such-a-specific-smell/6498#6498&quot;&gt;this question&lt;/a&gt;. The hydrolysis can be slowed/reversed by adding $\ce{HCl}$. Since solutions of $\ce{FeCl3}$ are already acidic, there is no harm usually in adding more acid. For this 0.1 molar solution, you might choose to dissolve 16.22 g of $\ce{FeCl3}$ in enough 0.1 M HCl to make one liter of solution.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2013-10-13T18:15:03.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6501" PostTypeId="2" ParentId="6499" CreationDate="2013-10-13T20:27:01.823" Score="8" Body="&lt;p&gt;One fallacy in your question is assuming that the particles 1) are spherical and 2) only accessible on the outside. The press release linked says that this material is &lt;a href=&quot;http://en.wikipedia.org/wiki/Nanoporous&quot; rel=&quot;nofollow noreferrer&quot;&gt;nanoporous&lt;/a&gt; - the particles are perforated by pores &quot;all having a diameter smaller than 10 nano meters&quot;. The extra surface area is inside these pores. Spheres have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface-area-to-volume_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;lowest surface-area-to-volume ratio&lt;/a&gt;. Other shapes have higher ratios (in other words, more surface area for less volume). These nanoporous materials maximize the surface area to volume ration - minimizing the volume. If these particles are $\ce{MgCO3}$, then their density is $\pu{2.958 g/cm^3}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_carbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). One gram of this material has a volume of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pu{1 g}\times \pu{1 cm^3 g^-1}\times \left(\frac{\pu{1 m}}{\pu{10^2 cm}}\right)^2=\pu{3.38E-7 m^3}=\pu{3.38E-3 L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface-area-to-volume ratio is then:&#xA;$$\frac{A}{V}=\dfrac{\pu{800 m^2}}{\pu{3.38E-7 m^3}}=\pu{2.37E9 m^-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second fallacy is that the volume of water that can be absorbed is the same as the volume of the particle. The particle is not absorbing water per se. The water is not taken into the interior volume where the $\ce{MgCO3}$ is. That would require displacement of the magnesium carbonate, and where would it go? The water goes into the pore of the material - it adsorbs to the surface inside those pores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about the science at the original research article:  A Template-Free, Ultra-Adsorbing, High Surface Area Carbonate Nanostructure. &lt;em&gt;PLoS ONE&lt;/em&gt; &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt; (7). &lt;a href=&quot;https://doi.org/10.1371/journal.pone.0068486&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1371/journal.pone.0068486&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-05-18T08:32:28.523" LastActivityDate="2018-05-18T08:32:28.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6503" PostTypeId="2" ParentId="6468" CreationDate="2013-10-14T01:01:58.770" Score="1" Body="&lt;p&gt;LCs materials are dielectric. But they are not perfect insulators and have some non-zero conductivity, allowing a leakage current to flow, slowly discharging the capacitor.&#xA;&lt;code&gt;This is caused by mobile ions in the liquid crystal and inside the polyimide layer [2]&lt;/code&gt;.&#xA;Infact a liquid crystal cell can be modeled electrically as this circuit:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;               Rl&#xA;         _____WWW_____&#xA;         |           |&#xA;         |     Cl    |&#xA;     O--------||---------O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;References:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.lintech.org/comp-per/03lcd.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lintech.org/comp-per/03lcd.pdf&lt;/a&gt;‎ p.10-11&lt;br&gt;&#xA;&lt;a href=&quot;http://www.emdmillipore.com/showBrochure?id=200908.037&quot; rel=&quot;nofollow&quot;&gt;http://www.emdmillipore.com/showBrochure?id=200908.037&lt;/a&gt; p.9&lt;/p&gt;&#xA;" OwnerUserId="2448" LastEditorUserId="2448" LastEditDate="2013-10-14T01:13:44.750" LastActivityDate="2013-10-14T01:13:44.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6505" PostTypeId="2" ParentId="6439" CreationDate="2013-10-14T06:37:15.840" Score="3" Body="&lt;p&gt;ManishEarth's &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6271/molecular-chirality-and-optical-rotation&quot;&gt;second link&lt;/a&gt; provides an absolutely excellent explanation for why chiral molecules in solution cannot simply be reoriented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add, however, that technically, molecules of the D2d symmetry can rotate light **only if ** they are in a crystal. These are known as birefringent crystals, and they only show optical activity in the crystal form: light going in the x-axis is rotated one way, light going in the y-axis is rotated the other way, and the z-axis is averaged to zero rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6271/molecular-chirality-and-optical-rotation&quot;&gt;paper I've read on this subject&lt;/a&gt;, but it's nowhere near basic (I don't even pretend to understand half of it).&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2013-10-14T06:37:15.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6507" PostTypeId="1" AcceptedAnswerId="6509" CreationDate="2013-10-14T13:44:32.000" Score="1" ViewCount="203" Body="&lt;p&gt;In any spontaneous process entropy increases.In biological systems disordered elements are converted into highly ordered tissues etc(-ve entropy change) which make up living organisms. But, it requires energy from the sun(ultimately). So,does that mean all processes that consume energy from outside result in negative entropy change? Endothermic reactions consume energy in the form of heat from outside so, is all endothermic reactions causing the entropy to decrease?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2013-10-14T15:47:08.250" Title="Does all processes that consume energy from outside result in negative entropy change?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6508" PostTypeId="1" AcceptedAnswerId="6510" CreationDate="2013-10-14T13:58:37.320" Score="4" ViewCount="225" Body="&lt;p&gt;I have a very basic question concerning solvation. If a solid (solute) was dissolved in a liquid (solvent), is the solute still in the solid state? or does it convert into another state of matter when this happens?&lt;/p&gt;&#xA;" OwnerUserId="1787" LastActivityDate="2013-10-15T23:37:20.130" Title="Solvated solids" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;solvents&gt;&lt;solid-state-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6509" PostTypeId="2" ParentId="6507" CreationDate="2013-10-14T15:30:57.833" Score="3" Body="&lt;p&gt;&lt;strong&gt;Entropy increases in a spontaneous evolution of an isolated system… other than that, there is no general law on the evolution of entropy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does that mean all processes that consume energy from outside result in negative entropy change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. When you heat a solid, it increases its entropy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;endothermic reactions causing the entropy to decrease?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Endothermic reactions which cause entropy to decrease have both ∆H positive and ∆S negative, so they will have a positive ∆G, i.e. their are not thermodynamically favorable.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-10-14T15:47:08.250" LastActivityDate="2013-10-14T15:47:08.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6510" PostTypeId="2" ParentId="6508" CreationDate="2013-10-14T17:37:03.887" Score="4" Body="&lt;p&gt;Good question. Theoretically, the physical phase of the matter does not change; crystalline salt dissolved into water does not become a liquid or gas. The dissolution reaction is theoretically a kinetic change, not a thermodynamic one as the traditional solid-liquid-gas phase changes are (at least at 1atm). However, the solute really isn't a &quot;solid&quot; anymore, either; depending on the nature of the solute and solvent, the two materials may interact chemically, and in fact when dissolution occurs, it's because it is thermodynamically favorable for the solute to be in a dissolved state as compared to a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often useful to describe dissolved solutes as being in this dissolved state, especially if the solvent is water, in which case it is in an &quot;aqueous&quot; state. You see this in chemical equations where the states of the reactants and products are of interest, such as when a solid or gas precipitates from an aqueous reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na(s) + 2H2O(l) \to 2NaOH(aq) + H2(g)\uparrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the sodium hydroxide produced by addition of sodium metal into water is virtually never &quot;pure&quot;, even if the reactants are mixed in ideal ratio. The thermodynamically-favorable product is an aqueous mixture, which is treated as a liquid even though &quot;pure&quot; sodium hydroxide is an off-white solid powder.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-10-15T23:37:20.130" LastActivityDate="2013-10-15T23:37:20.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6511" PostTypeId="1" CreationDate="2013-10-14T18:28:21.873" Score="3" ViewCount="14982" Body="&lt;p&gt;Why is hot concentrated sulfuric acid sometimes used to react with metals instead of concentrated sulfuric acid which is at room temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ascertained that concentrated sulfuric acid has quite strong oxidizing properties.  Still, some metals which are below the hydrogen in electrochemical series do not react with concentrated sulfuric acid which is kept at room temperature.  When the acid is warm/hot, the reaction takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I want to know is how does temperature affects its oxidizing properties?&lt;/p&gt;&#xA;" OwnerUserId="2478" LastEditorUserId="9961" LastEditDate="2015-11-13T19:33:01.637" LastActivityDate="2018-03-22T10:13:34.037" Title="Why is hot concentrated sulfuric acid preferable to cold?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6512" PostTypeId="2" ParentId="6481" CreationDate="2013-10-14T18:59:33.063" Score="2" Body="&lt;p&gt;There are a lot of types of impurity. Basically, much as any plant that's growing where you don't want it is a weed, any material in your sample that isn't what you want to have in it is an impurity.&lt;/p&gt;&#xA;&lt;p&gt;Since you asked a question in simple terms I will attempt to answer it in equally simple terms:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Byproduct contamination&lt;/strong&gt; - The theoretical reaction is usually not the only possible one, especially in organic chemistry such as pharmaceutical work. Areas of the reaction which didn't have just the right mixture of reactants, or wasn't quite at the right temperature, could have produced different products. Additionally, devations from the ideal when adding reactants can result in unreacted quantities of those reactants.&lt;/p&gt;&#xA;&lt;p&gt;Various forms of chromatography, such as HPLC, CC, GC, MS, etc can detect and even separate these various byproducts from the ideal product. How effective any one of these techniques is depends on the type and combination of the products and byproducts. One of the most basic aspects of any drug research chemist's job is to isolate their product from all of the byproducts of their reaction.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Structural isomer contamination&lt;/strong&gt; - Two compounds are structural isomers if they have the same molecular formula (number and type of atoms), but have different molecular structures. Tetramethylbutane, for instance, is based on a 4-carbon alkane, to which two additional methyl groups each are added to the carbons in the &amp;quot;2&amp;quot; and &amp;quot;3&amp;quot; positions of the alkane. The resulting molecule has all the same carbons and hydrogens as the totally linear octane molecule, but its structure causes very different behaviors (for instance, it has a much narrower range of temperatures at which it is a liquid).&lt;/p&gt;&#xA;&lt;p&gt;Your average petrochemist doesn't care too much about the difference between any of the structural isomers of these hydrocarbon alkanes; their primary reaction of interest, combustion, only cares about the required stoichiometric ratio of fuel to air, and the resulting rate and heat of combustion. However, such isomers pose problems in pharmacology for the above reasons; a reaction that was theorized to produce a molecule with one arrangement may in fact produce an entirely different isomer, with similar physical properties as the desired molecule, but much different reaction chemistry including biochemical effects. Chromatography methods based on molecular weight or physical properties like boiling point or solubility may or may not be able to tell you whether you have the right isomer, but other methods might show a difference in the physical properties of the chemical allowing separation.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Chirality and Racemization&lt;/strong&gt; - Even if two isomers have all the same atoms, and each atom is bonded to the &amp;quot;correct&amp;quot; adjoining atoms, the molecules can have different shapes because the atoms are bonded to different relative positions of their adjoining atoms. These &amp;quot;spatial isomers&amp;quot; or &amp;quot;stereoisomers&amp;quot; have many forms, from structures that exhibit the same &amp;quot;mirror-image&amp;quot; symmetry that your left and right hand do (enantiomers), or that have certain functional groups transposed by rotation around a &amp;quot;chiral center&amp;quot; (diasteromers, including cis-trans and most detro-levo isomers). Again, like structural isomers, these stereoisomers can cause problems for pharmacologists because the different isomers can have different reactive properties in the complex biochemical soup that is the human body.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes these chiral isomers can be separated from each other; enantiomers and other spatial isomers are often &amp;quot;optically active&amp;quot;, meaning they rotate polarized light in opposing directions depending on their chiral arrangement. Certain &amp;quot;chiral resolving agents&amp;quot; can also separate these isomers based on a differing affinity for one isomer over another, and these agents are often used as &amp;quot;stationary phases&amp;quot; of column chromatography including HPLC that can separate and thus &amp;quot;purify&amp;quot; the desired isomer from its chiral mirror. However, a few troublesome spatial isomers will &amp;quot;racemize&amp;quot;; that is, they will readily rearrange between the two stereoisomers when given the right environment, and so any attempt to produce a pure sample of one enantiomer is ultimately pointless.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="367" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-10-14T18:59:33.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6513" PostTypeId="1" AcceptedAnswerId="6514" CreationDate="2013-10-14T20:14:03.417" Score="2" ViewCount="34221" Body="&lt;p&gt;Recently, we covered metallic bonding in chemistry, and frankly, I understood little.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Metals bond to each other via metallic bonding  &lt;/li&gt;&#xA;&lt;li&gt;Electricity can flow via free or delocalized electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But, I do not understand why the metal atoms turn into ions and delocalize the electrons, why don't the metal atoms stay as atoms?&lt;/p&gt;&#xA;" OwnerUserId="1896" LastEditorUserId="73" LastEditDate="2013-10-14T20:18:04.070" LastActivityDate="2018-01-21T15:26:11.347" Title="Why are there free electrons in a metal?" Tags="&lt;bond&gt;&lt;metal&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6514" PostTypeId="2" ParentId="6513" CreationDate="2013-10-14T21:21:13.207" Score="6" Body="&lt;p&gt;When electricity flows, the electrons are considered &quot;free&quot; only because there are more electrons than there should be, and because the transition metals, such as iron, copper, lead, zinc, aluminum, gold etc. are willing to transiently accept and give up electrons from the &lt;em&gt;d&lt;/em&gt;-orbitals of their valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition metals are defined in part by their stability in a wide range of &quot;oxidation states&quot;; that is, in several combinations of having too many or too few electrons compared to protons. This is thought to be because of the &lt;em&gt;d&lt;/em&gt; orbital in their valence shells. Compared to the &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;p&lt;/em&gt; orbitals at a particular energy level, electrons in the &lt;em&gt;d&lt;/em&gt; shell are in a relatively high energy state, and by that token they have a relatively &quot;loose&quot; connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material, usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely). This impetus can be caused by many things, from mechanical impact to chemical reactions to electromagnetic radiation (aka light, though not all of it visible); antennas work to capture radio frequencies, because the light at those frequencies induces an electric current in the wire of the antenna. Now, in the absence of a continuous force keeping the electron in this higher energy state, the electron (and the metal atoms) will naturally settle into a state of equilibrium. Electricity is generated when just such a force is acting on the metal, giving energy to the electrons in the &lt;em&gt;d&lt;/em&gt; orbital and forcing them to move in a certain direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This impetus can come from many sources, as discussed, be it the movement of a magnet within a coil of wire, or a chemical redox reaction in a battery creating a relative imbalance of electrons at each of two electrodes. The end result is that the electrons, given additional energy from this voltage source, are ejected from their &quot;parent&quot; atom and are captured by another. The &quot;holes&quot; left behind by these electrons are filled by other electrons coming in behind them from further back in the circuit. Thus, the energy provided by the voltage source is carried along the wire by the transfer of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy typically made is to the flow of water, and it generally holds in many circumstances; the &quot;voltage source&quot; can be thought of as being like a pump or a reservoir, from which water flows through pipes, and the amount of water and the pressure it's placed under (by the pump or by gravity) can be harnessed to do work, before draining back to a lower reservoir. The pipes are similar to wires in many ways; the larger the diameter, and the smoother the inside of the pipe, the more and the faster water can flow through it (equivalent in many ways to the thickness and conductivity of the metal wire), and when under enough pressure (high enough voltage), the pipes will actually expand slightly and hold more water than they would at low pressure (this is a property of wires and other electrical conductors called &quot;capacitance&quot;; the ability to store a charge while under voltage and to discharge it after the voltage is released).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-10-14T21:28:34.707" LastActivityDate="2013-10-14T21:28:34.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6515" PostTypeId="2" ParentId="6511" CreationDate="2013-10-14T22:41:21.093" Score="6" Body="&lt;p&gt;A really short answer is kinetics. Heating up chemicals adds more energy to the system, allowing molecules to colide with more force increasing the probably of a reaction taking place. &lt;/p&gt;&#xA;" OwnerUserId="316" LastActivityDate="2013-10-14T22:41:21.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6516" PostTypeId="1" CreationDate="2013-10-14T23:46:11.080" Score="3" ViewCount="172" Body="&lt;p&gt;I want to change $\ce{CO2}$ coming out of the tailpipe of a car to some other compound that is considerably less harmful to the environment with a filter of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some ways that this is possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ to carbon and oxygen?&lt;/p&gt;&#xA;" OwnerUserId="2481" LastEditorUserId="4231" LastEditDate="2015-03-09T17:40:40.937" LastActivityDate="2015-03-09T17:40:40.937" Title="How to change CO₂ to a less polluting gas?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6517" PostTypeId="1" AcceptedAnswerId="6523" CreationDate="2013-10-15T01:59:54.507" Score="7" ViewCount="11684" Body="&lt;p&gt;Alcohols can undergo nucleophilic substitution with $\ce{PCl5, POCl3, HCl}$.  Halogenoalkanes can undergo nucleophilic substitution with $\ce{NH3}$. The $\ce{C-O}$ (in alcohol) and $\ce{C-Cl}$ (in chloroalkanes) bond energies are comparable. Why don't alcohols undergo nucleophilic substitution with $\ce{NH3}$?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="10185" LastEditDate="2015-10-02T16:41:57.537" LastActivityDate="2016-10-13T23:17:41.603" Title="Why don't alcohols undergo nucleophilic substitution with ammonia?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6519" PostTypeId="2" ParentId="6419" CreationDate="2013-10-15T05:45:44.740" Score="2" Body="&lt;p&gt;If we talk only about electronic configurations and rules violation then you are correct. The 2nd one (excited state) is referred for hybridization that is sp2. An excited state  configuration is meant only when the particular atom gets ready or excited for forming a bond. Like boron does it with 3 fluorine atoms to form $\ce{BF3}$. So, if we talk about electronic config and rule violations we must go for ground states exceptional cases are $\ce{Cu}$ , $\ce{Cr}$ and $\ce{Mo}$.&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="22" LastEditDate="2013-10-15T09:05:07.420" LastActivityDate="2013-10-15T09:05:07.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6520" PostTypeId="2" ParentId="6517" CreationDate="2013-10-15T11:34:28.160" Score="0" Body="&lt;p&gt;In substitution, you must think not only about nucleophile, but also about leaving group. That means, alcohols react only with very good nucleophiles, because $\rm{OH^-}$ is so bad leaving group. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-15T11:34:28.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6521" PostTypeId="1" CreationDate="2013-10-15T13:13:41.620" Score="0" ViewCount="163" Body="&lt;p&gt;I recently freezed a huge bunch of apple slices, and thawed them again a few days later. After thawing, I noticed a few things compared to a batch that I had put in the fridge (no freezing)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The apples turned brown far quicker&lt;/li&gt;&#xA;&lt;li&gt;After pureeing, they tasted sweeter&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that slow freezingg can destroy cell walls in animals due to crystallization. My guess what has happened here is that the same mechanism also broke up long chain sugar polymers. Thus the sugars are chemically more active, oxidize more easily after contact with air, and taste stronger than in the other batch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is my guess plausible? Do polymers actually break under those mechanical stresses? Is this or a similiar effect observed elsewhere?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="275" LastEditDate="2013-10-16T00:37:51.440" LastActivityDate="2013-10-16T00:37:51.440" Title="Can freezing break up sugar polymers?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;&lt;polymers&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6522" PostTypeId="2" ParentId="6517" CreationDate="2013-10-15T13:31:39.553" Score="5" Body="&lt;p&gt;The $\ce{OH-}$ is a very poor leaving group, however, $\ce{H2O}$ is a very good leaving group. Of course, one only needs to acidify to convert one group to the other. For chloride as the nucleophile, this poses no problems; $\ce{HCl}$ is a strong acid and $\ce{Cl-}$ is a weak conjugate base. However, ammonia is a pretty good base, and it converts to the ammonium ion $\ce{NH4+}$ at acidic pH. This greatly reduces its capability as a nucleophile, and the reaction does not proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5XELi.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2013-10-15T13:31:39.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6523" PostTypeId="2" ParentId="6517" CreationDate="2013-10-15T13:32:42.107" Score="7" Body="&lt;p&gt;To build on ssavec's answer: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nucleophilic substitution reactions&lt;/a&gt; require two species: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophile&quot; rel=&quot;nofollow noreferrer&quot;&gt;nucleophile&lt;/a&gt; (a Lewis Base) and a substrate with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophile&quot; rel=&quot;nofollow noreferrer&quot;&gt;leaving group&lt;/a&gt;. Nucleophiles are often generically represented as $\ce{Nu}$ and leavings groups as $\ce{LG}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All leaving groups are not created equal. The leaving group ability of a leaving group is defined as the relative rates of a particular analogous series of substitutions. For example, the relative leaving group ability might be based on the following reaction, with rates determined for various leaving groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{EtOH + PhCH(LG)CH3 -&amp;gt; PhCH(OEt)CH3 + H-LG}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Sn1&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm{S_N1}$ substitution&lt;/a&gt;, so the first (and rate determining) step of the mechanism is loss of the leaving group (and is independent of the nucleophile):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PhCH(LG)CH3 -&amp;gt; PhCH+CH3 + LG-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative rates of this reaction are influenced by the stability of the $\ce{LG-}$ anion (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammond_postulate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammond Postulate&lt;/a&gt;, which proposes that the transition state of an endothermic process resembles the products). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability of the $\ce{LG-}$ anion can be predicted from the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of $\ce{LG-}$, which is controlled by the same ability to stabilize a negative charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate acid of $\ce{Cl-}$ is $\ce{HCl}$, which is a strong acid. Using some Hess's Law trickery, the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCl}$ has been predicted to be around $-8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate acid of $\ce{OH-}$ is $\ce{H2O}$, which has a $\mathrm{p}K_\mathrm{a}$ around $+16$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O}$ is a much weaker acid than $\ce{HCl}$, which means that the negative charge on  $\ce{OH-}$ is less stabilized than $\ce{Cl-}$. Thus, $\ce{OH-}$ is a poorer leaving group than $\ce{Cl-}$ (by a lot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good discussion of leaving groups (and the $\mathrm{p}K_\mathrm{a}$ data used above) &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/04/12/what-makes-a-good-leaving-group/&quot; rel=&quot;nofollow noreferrer&quot;&gt;come from this site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example reaction (ammonia + ethanol), the product of the reaction has a better leaving group ($\ce{NH3}$, conjugate base of $\ce{NH4+}$, which has a $\mathrm{p}K_\mathrm{a}$ of $+9.75$) than the $\ce{OH-}$ leaving group in the reactant, so the reaction will also run in reverse, and the equilibrium will strongly favor the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2OH + NH3 &amp;lt;=&amp;gt; CH3CH2}\color{red}{\ce{NH3+}}\ce{+ OH-}\ \ K_\mathrm{a} \ll 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your confusion about bond energies...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond dissociation energies are defined for &lt;a href=&quot;http://en.wikipedia.org/wiki/Homolysis_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;homolytic bond cleavage&lt;/a&gt;, in which each atom keeps one electron when the bond breaks (see image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oNZG8.png&quot; alt=&quot;homolysis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution reactions involve &lt;a href=&quot;http://en.wikipedia.org/wiki/Heterolysis_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;heterolytic bond cleavage&lt;/a&gt;, in which one atom gets both electrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nxh3W.png&quot; alt=&quot;heterolysis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7475" LastEditDate="2016-10-13T23:17:41.603" LastActivityDate="2016-10-13T23:17:41.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6524" PostTypeId="2" ParentId="6521" CreationDate="2013-10-15T15:09:36.573" Score="4" Body="&lt;p&gt;Freezing breaks the cell structure of an apple, not individual chains of short polysaccharides you can feel the taste. What happens is probably enzymes which the apple contains in cell structures go out of these structures and split long polysaccharides (which have no taste) into short sugars which you feel as sweet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the apple digest itself by its own enzyme systems after breaking the cell structures.&lt;/p&gt;&#xA;" OwnerUserId="957" LastEditorUserId="957" LastEditDate="2013-10-15T15:14:37.360" LastActivityDate="2013-10-15T15:14:37.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6525" PostTypeId="1" AcceptedAnswerId="6971" CreationDate="2013-10-15T16:25:46.810" Score="0" ViewCount="338" Body="&lt;p&gt;I couldn't find this in any textbooks, so I guess I will ask this here. What is the physical state of $\ce{HX}$ when reacting with alcohol (nucleophilic substitution) and alkenes (electrophilic addition) respectively?   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For electrophilic substitution of benzene with bromine, what is the state of bromine?  (e.g. aqueous, liquid, solid)     &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If possible, can you explain why they have the particular state?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="10185" LastEditDate="2015-10-09T16:57:31.143" LastActivityDate="2017-04-08T22:26:43.373" Title="What are the states of HX when reacting with alcohol and alkenes respectively, and Br2 when reacting with benzene?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6527" PostTypeId="1" AcceptedAnswerId="6539" CreationDate="2013-10-16T01:49:21.577" Score="5" ViewCount="24407" Body="&lt;p&gt;According to my teacher, Na in ethanol reduces carbonyl compounds (e.g. aldehydes and ketones) to alcohols. However, I can't see where the hydrogen needed for reduction comes from. How does the reduction take place? (or is my teacher wrong?)&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="16683" LastEditDate="2016-03-12T23:43:29.953" LastActivityDate="2020-06-05T12:36:58.523" Title="How does sodium in ethanol reduce carbonyl compounds?" Tags="&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6529" PostTypeId="1" CreationDate="2013-10-16T06:03:52.060" Score="2" ViewCount="55" Body="&lt;p&gt;I understand that deflagration is slower and smoother than a detonation, but can't a deflagration be used to initiate detonations of other explosive agents?&lt;/p&gt;&#xA;" OwnerUserId="815" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:50:25.090" LastActivityDate="2015-10-30T16:50:25.090" Title="Detonation and deflagration" Tags="&lt;explosives&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6530" PostTypeId="1" CreationDate="2013-10-16T06:39:34.817" Score="13" ViewCount="93073" Body="&lt;p&gt;Halogens bonded to benzene ring has three lone pairs. These three electron pairs can cause resonance in benzene ring. But, halogens are also highly electronegative and thus they have strong -I effect. So, they are deactivating groups. But, why are they ortho para directing?  &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="2328" LastEditDate="2013-11-17T11:23:57.820" LastActivityDate="2018-12-29T09:12:17.090" Title="Why are halogens ortho para directing even though deactivating?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6531" PostTypeId="1" AcceptedAnswerId="6540" CreationDate="2013-10-16T06:51:04.173" Score="1" ViewCount="4029" Body="&lt;p&gt;I have this redox reaction that I can't balance using half reaction method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + HNO3 -&amp;gt; Mg(NO3)2 + N2O + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I get at the end is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Mg + 6HNO3 -&amp;gt; 2Mg(NO3)2 + 3N2O +3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't get the $\ce{N}$ balanced.&lt;/p&gt;&#xA;" OwnerUserId="2492" LastEditorUserId="5026" LastEditDate="2018-02-27T04:42:19.820" LastActivityDate="2018-02-27T04:42:19.820" Title="Attempting to balance a redox reaction using half reactions leaves an unbalanced component" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6532" PostTypeId="1" AcceptedAnswerId="6534" CreationDate="2013-10-16T07:33:43.660" Score="-4" ViewCount="2480" Body="&lt;p&gt;Every one have observed that when we pour cool water into a transparent glass (or simply glass), some droplets accumulate on the outside part. Did the droplets come pass through the glass? If they didn't, why is there accumulation of water droplets outside the glass?&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LePJl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2013-11-16T13:30:46.903" LastActivityDate="2014-09-02T13:55:14.887" Title="Strange observation!" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6534" PostTypeId="2" ParentId="6532" CreationDate="2013-10-16T08:11:26.267" Score="8" Body="&lt;p&gt;This is happening because of &lt;strong&gt;moisture present in air&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you'll pour the cold water in a glass, the outer surface will also show fall in its temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know already know that the &lt;strong&gt;heat flows from one body to another body until the temperature of both becomes equal&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in our case the &lt;strong&gt;outer surface of the glass will absorb heat from surrounding, thereby causing in &lt;a href=&quot;http://en.wikipedia.org/wiki/Condensation&quot; rel=&quot;nofollow&quot;&gt;condensation&lt;/a&gt; of water present in air as moisture.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing, you might notice that when you take out the bottle of cold water from a refrigerator, it have no droplets present on its outer surface because air inside the refrigerator is dry. But after when you place the same bottle out for a few minutes, you will observe the droplets on the outer surface of bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try to perform your experiment again by pouring normal water (at room temperature) in a glass and then placing it in refrigerator and then take it out after 30-40 minutes. Observe that if there are droplets present on the glass &lt;strong&gt;just after you take it out&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://voices.yahoo.com/why-water-appears-outside-cold-drinks-6347900.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-09-02T13:55:14.887" LastActivityDate="2014-09-02T13:55:14.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6536" PostTypeId="1" CreationDate="2013-10-16T09:18:47.073" Score="4" ViewCount="119" Body="&lt;p&gt;Both are m-directors, and Electron Withdrawing groups, but which one is more reactive? At first sight, one might think clearly $\ce{-N(R)_3^+}$ is more reactive, but when you factor in that $\ce R$ is actually an o,p-director, hence activator, then I think $\ce{-N(R)_3^+}$ is indeed stronger than $\ce{ NH_3^+}$. I can't seem to find any verification on the internet for this.&lt;/p&gt;&#xA;" OwnerUserId="2494" LastEditorUserId="4945" LastEditDate="2014-09-10T03:29:50.087" LastActivityDate="2014-09-10T03:29:50.087" Title="Is the reactivity of Benzene with an -N(R)₃⁺ group greater than its reactivity with an -NH₃⁺ group?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6537" PostTypeId="1" CreationDate="2013-10-16T10:14:01.773" Score="2" ViewCount="14751" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{molarity} = \frac{\text{amount of solute}}{\text{volume of solution}} $$&lt;/span&gt; and amount of substance is based on quantity (larger mass means larger amount), so how come it is an intensive property. Shouldn't it be an extensive property?&lt;/p&gt;&#xA;" OwnerUserId="2495" LastEditorUserId="72973" LastEditDate="2019-07-28T22:07:24.003" LastActivityDate="2019-07-28T22:07:24.003" Title="Is molarity an intensive property?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6538" PostTypeId="2" ParentId="6537" CreationDate="2013-10-16T11:47:09.497" Score="6" Body="&lt;p&gt;Concentration is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Intensive_property#Intensive_properties&quot; rel=&quot;noreferrer&quot;&gt;intensive property&lt;/a&gt;. The value of the property does not change with scale. Let me give you an example: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us say you had a homogenous mixture (solution) of sodium carbonate in water prepared from 112 g of sodium carbonate dissolved in 1031 g of water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The concentration (in mass percent, or mass of solute per mass of solution) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac{112\text{ g solute}}{(112+1031)\text{g solution}}=0.09799 =9.799\%\text{ sodium carbonate by mass}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration is the ratio of sodium carbonate to the total mass of the solution, which does not change if you are dealing with the entire 1143 g of the solution or if you dispense some of that solution into another vessel. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you dispense 11.7 g of that solution into a flask for a reaction, what is the concentration of sodium carbonate in that flask? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is still 9.799% by mass. The ratio of the mass of sodium carbonate present to the total mass present has not changed. The &lt;strong&gt;actual&lt;/strong&gt; mass of sodium carbonate has changed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.09799\dfrac{\text{g solute}}{\text{g solution}}\times11.7\text{ g solution}=1.15\text{ g solute}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration is a property dependent only on the &lt;strong&gt;concentration&lt;/strong&gt; of the solution, not the &lt;strong&gt;amount&lt;/strong&gt; of solution you have. The concentration of a solution with defined composition is independent of the size of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, any property that is a ratio of two extensive properties becomes an intensive property, since both extensive properties will scale similarly with increasing or decreasing size of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Concentration&quot; rel=&quot;noreferrer&quot;&gt;Concentration&lt;/a&gt; (including molarity) - ratio of amount of solute (mass, volume, or moles) to amount of solution (mass or volume usually)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Density&quot; rel=&quot;noreferrer&quot;&gt;Density&lt;/a&gt; - ratio of mass of a sample to the volume of the sample&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;noreferrer&quot;&gt;Specific heat&lt;/a&gt; - ratio of heat transferred to a sample to the amount of the sample (mass or moles usually, but volume also)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each of these intensive properties is a ratio of an extensive property we care about (amount of solute, mass of sample, heat transferred) divided by the scale of the system (amount of stuff usually). This is like finding the slope of a graph showing the relationship between two extensive properties. The graph is linear and the value of slope does not change based on how much stuff you have - thus the slope (the ratio) is an intensive property.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-16T11:47:09.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6539" PostTypeId="2" ParentId="6527" CreationDate="2013-10-16T12:02:03.097" Score="10" Body="&lt;p&gt;Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bouveault%E2%80%93Blanc_reduction&quot; rel=&quot;noreferrer&quot;&gt;Bouveault-Blanc reduction&lt;/a&gt;. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 1 - one electron transfer from sodium to carbonyl group:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na. + RCH=O-&amp;gt;Na+ + RCH.-O^-}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 2 - abstraction of a proton from ethanol:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2OH + RCH.-O^- &amp;lt;=&amp;gt; CH3CH2O- + RCH.-OH}$$&#xA;This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 3 - one electron transfer from a second sodium atom to the organic radical:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na. +RCH.-OH -&amp;gt; Na+ + RCH:^{-}{}-OH}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 4 - second proton transfer from ethanol:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2OH + RCH:^{-}{}-OH -&amp;gt; CH3CH2O- + RCH2OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na + RCH=O + 2CH3CH2OH -&amp;gt; RCH2OH + 2CH3CH2ONa}$$&#xA;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="24052" LastEditDate="2016-03-12T23:39:33.833" LastActivityDate="2016-03-12T23:39:33.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6540" PostTypeId="2" ParentId="6531" CreationDate="2013-10-16T12:06:29.583" Score="5" Body="&lt;p&gt;It seems likely that you are starting with the wrong half reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this reaction happens in water, separate the ionic species into their ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg(NO3)2 -&amp;gt; Mg^{2+} + 2NO3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3 -&amp;gt; H+ + NO3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one half reaction should involve magnesium (and have no nitrogen anywhere). The other half reaction should involve nitrogen only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg -&amp;gt; Mg^{2+}}$$&#xA;$$\ce{NO3- -&amp;gt; N2O}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-16T12:06:29.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6544" PostTypeId="2" ParentId="6530" CreationDate="2013-10-16T17:07:52.130" Score="5" Body="&lt;p&gt;+R effect of halogens which increase electron density on o and p position while &#xA;overall electron density has reduced on ring because of -I effect.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2013-10-16T17:07:52.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6547" PostTypeId="2" ParentId="6537" CreationDate="2013-10-17T08:02:14.973" Score="0" Body="&lt;p&gt;Consider the following picture:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1Dw6y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break the ice block shown in the picture into two equal halves.Now I hope you would be able to answer the following questions:&lt;br&gt;&#xA;1.What are the physical properties of ice block which got halved?&lt;br&gt;&#xA;Absolutely mass,volume,etc.(These are all extensive properties.)&lt;br&gt;&#xA;2.What are the physical properties of ice block which remained same?&lt;br&gt;&#xA;Density,etc.(These are all intensive property.)&lt;br&gt;&#xA;If you have the doubt so as to why the density remained same,here is the explanation:&lt;br&gt;&#xA;I hope you know basically even if block got halved,mass per unit volume remains the same in either of the pieces.All the way it mean that density remained the same(mass per unit volume).Thus it is an intensive property.&lt;br&gt;&#xA;&lt;strong&gt;Similarly if you imagine solution instead of ice block,you will find that molarity remains the same even if you divide solution into two equal halves.Thus molarity is a intensive property.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2013-10-17T08:02:14.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6548" PostTypeId="1" AcceptedAnswerId="6549" CreationDate="2013-10-17T08:14:10.207" Score="1" ViewCount="146" Body="&lt;p&gt;Please see the following figure. This is from the high school chemistry book of my son. He wants to know what does that /2 after $\ce{-COOCa}$ mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gbhrr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book is a non English one. And I have changed the relevant caption to English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primary searches over Google have not revealed anything usable. And my major not being Chemistry might have to do something with this.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="7475" LastEditDate="2015-10-09T16:30:10.417" LastActivityDate="2015-10-09T16:30:10.417" Title="What does &quot;/2&quot; in the structure of calcium benzoate mean?" Tags="&lt;organic-chemistry&gt;&lt;ionic-compounds&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6549" PostTypeId="2" ParentId="6548" CreationDate="2013-10-17T09:54:00.503" Score="4" Body="&lt;p&gt;It means half of calcium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organic acid is $\rm{COO}^-$, whereas calcium is $\rm{Ca}^{2+}$. For the structure to be fully correct, there are two benzoic acids per one calcium. Other way round this would be $\rm{(C_6H_5COO)_2Ca}$. Full drawing of this will be way to messy, so it is simpler to put half calcium to one acid.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-17T09:54:00.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6550" PostTypeId="1" CreationDate="2013-10-17T10:02:19.467" Score="2" ViewCount="243" Body="&lt;p&gt;Is there a diagram/flow chart available that shows which atoms fused together to make each of the elements?  I can find examples for some elements, but nothing which shows all the combinations of atoms which make up each of the naturally occurring elements.&lt;/p&gt;&#xA;" OwnerUserId="2505" LastActivityDate="2021-05-01T17:20:35.587" Title="Which atoms form which elements?" Tags="&lt;bond&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6551" PostTypeId="2" ParentId="6550" CreationDate="2013-10-17T12:07:08.280" Score="8" Body="&lt;p&gt;The general consensus is that each element is made up of unique atoms of that element's type. I address nucleosynthesis below, but this is not how elements become elements.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/oxygen&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxygen&lt;/a&gt; is made of oxygen atoms. &lt;a href=&quot;http://en.wikipedia.org/wiki/titanium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titanium&lt;/a&gt; is made of titanium atoms. &lt;a href=&quot;http://en.wikipedia.org/wiki/neon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neon&lt;/a&gt; is made of neon atoms.&lt;/p&gt;&#xA;&lt;p&gt;The first good &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomic theory&lt;/a&gt;, formulated by John Dalton and refined by Amadeo Avogadro and others has a the following principles (some of which are invalidated by modern experiments):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All matter consists of atoms as fundamental building blocks (we now know that subatomic particles are the real fundamental building blocks).&lt;/li&gt;&#xA;&lt;li&gt;Atoms cannot be created or destroyed (we now have the ability to do this by nuclear fusion and fission).&lt;/li&gt;&#xA;&lt;li&gt;Each element has its own characteristic type of atom (still valid).&lt;/li&gt;&#xA;&lt;li&gt;All atoms of the same element element are chemically and physically indistinguishable (the discovery of isotopes invalidated this one).&lt;/li&gt;&#xA;&lt;li&gt;Chemical reactions occur by the rearrangement of atoms (this one is still valid).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;More modern atomic theory takes into account the discoveries of the subatomic particles and the development of quantum mechanics.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atom&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atoms&lt;/a&gt; are now defined by their composition of the three principle subatomic particles: protons, neutrons, and electrons. Any atom can be characterized by three numbers derived from this composition:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atomic Number&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;) - equal to the number of protons.&lt;/li&gt;&#xA;&lt;li&gt;Mass Number (&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;) - equal to the sum of the numbers of protons and neutrons (&lt;span class=&quot;math-container&quot;&gt;$A=Z+N$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Charge (&lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;) - equal to the difference between the number of protons and the number of electrons (&lt;span class=&quot;math-container&quot;&gt;$Q=Z-E$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The atomic number, &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;, defines to which element the atom belongs. For example, all atoms with &lt;span class=&quot;math-container&quot;&gt;$Z=1$&lt;/span&gt; are hydrogen atoms. All atoms with &lt;span class=&quot;math-container&quot;&gt;$Z=6$&lt;/span&gt; are carbon atoms. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;periodic table&lt;/a&gt; is arranged by increasing atomic number. The atomic number controls the number of electrons an atom prefers (see below), and therefor controls the majority of the chemistry of an atom.&lt;/p&gt;&#xA;&lt;p&gt;The number of neutrons (mass number) defines the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotope&quot; rel=&quot;nofollow noreferrer&quot;&gt;isotope&lt;/a&gt;. For example, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_hydrogen#Hydrogen-1_.28protium.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;three isotopes of hydrogen&lt;/a&gt;, with &lt;span class=&quot;math-container&quot;&gt;$A=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A=2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$A=3$&lt;/span&gt;: protium, deuterium, and tritium. These three isotopes have 0, 1, and 2 neutrons respectively, but they each have only one proton, so each of them is a hydrogen atom. The mass differences influence the properties of these isotopes subtly, but not in such a way that is usually noticeable to the average human.&lt;/p&gt;&#xA;&lt;p&gt;The number of electrons (or more precisely the charge) determines whether the atom is neutral or an ion(&lt;a href=&quot;http://en.wikipedia.org/wiki/Monatomic_ion&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Monatomic_ion&lt;/a&gt;). The number of electrons controls the behavior of an atom chemically (for example the number of covalent bonds that can be formed), and the number of electrons of a neutral atom is the same as the atomic number &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;. For example, a carbon atom has 6 electrons (and you can go further into the quantum mechanics of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_configuration&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron configuration&lt;/a&gt; to learn that two of those electrons are less accessible than the other four, which accounts for most of carbon's chemical behavior). Carbon can have a number of electrons not equal to &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;. If carbon were to have 7 electrons, the charge on carbon would be &lt;span class=&quot;math-container&quot;&gt;$Q=Z-E=6-7=-1$&lt;/span&gt;, making carbon an anion. Carbon behaves differently as an anion than it does as a neutral atom.&lt;/p&gt;&#xA;&lt;p&gt;Atoms of some elements can be made by &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;nuclear reactions&lt;/a&gt;. For example, in the sun &lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen is converted into helium&lt;/a&gt; by the following pathway (and others). The notation below follows the generic example &lt;span class=&quot;math-container&quot;&gt;$^A_Z \ce{X}^Q$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is the atomic symbol (and here &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; means a neutron with &lt;span class=&quot;math-container&quot;&gt;$A=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Z=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; means an electron with &lt;span class=&quot;math-container&quot;&gt;$A=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Z=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q=\pm1$&lt;/span&gt;). A proton is represented as &lt;span class=&quot;math-container&quot;&gt;$^1_1 \ce{H}^+$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^1_1H + ^1_1H -&amp;gt; ^2_2He}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^2_2He -&amp;gt; ^2_1H + e+}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^2_1H + ^1_1H -&amp;gt; ^3_2He}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^3_2He + ^3_2He -&amp;gt; ^4_2He + ^1_1H + ^1_1H}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In addition to the nuclear transformations, gamma rays, neutrinos, and heat and being generated and flung around like it was nothing.&lt;/p&gt;&#xA;&lt;p&gt;The number of possible nuclear reactions is humongous. I would go check out a book on nuclear reactions or nuclear chemistry from a library to learn more.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="24052" LastEditDate="2021-05-01T17:20:35.587" LastActivityDate="2021-05-01T17:20:35.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6552" PostTypeId="1" AcceptedAnswerId="6571" CreationDate="2013-10-17T15:06:46.077" Score="4" ViewCount="118" Body="&lt;p&gt;Why, in the photochemical substitution of a ligand in a carbonyl complex, is a water-cooled lamp jacket required to separate the solution and the mercury vapour lamp? It's shown in Figure 22.16 of Atkins' Inorganic Chemistry.&lt;/p&gt;&#xA;" OwnerUserId="2506" LastEditorUserId="4945" LastEditDate="2015-03-23T09:29:19.147" LastActivityDate="2015-03-23T09:29:19.147" Title="Photochemical ligand substitution of metal carbonyls - role of water cooled lamp-jacket?" Tags="&lt;inorganic-chemistry&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6553" PostTypeId="1" CreationDate="2013-10-17T15:36:37.240" Score="4" ViewCount="1404" Body="&lt;p&gt;I've taught in my class that if a molecule posses high degree of symmetry it's stability will be more,in other words symmetry and stability are directly proportional. Is this statement correct? And if this is correct than why Jahn teller distortion stabilizes a molecule when it's symmetry gets low eventually because of the distortion.  &lt;/p&gt;&#xA;" OwnerUserId="2478" LastEditorUserId="2478" LastEditDate="2013-10-17T16:12:15.733" LastActivityDate="2017-07-18T16:20:35.997" Title="Symmetry and stability of a molecule (Jahn-Teller Distortion)" Tags="&lt;coordination-compounds&gt;&lt;symmetry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6555" PostTypeId="1" AcceptedAnswerId="6561" CreationDate="2013-10-17T16:48:51.107" Score="1" ViewCount="909" Body="&lt;p&gt;Is there a database or a website with a list of oil : water partition coefficients of substances ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually want the oil:water partition coefficients of :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cholesterol, starch , methyl citrate , citric acid , iso- citrate , methyl urea , carotene&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="4945" LastEditDate="2014-05-22T09:08:16.703" LastActivityDate="2014-05-22T09:08:16.703" Title="Oil :water partition coefficients" Tags="&lt;solutions&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6556" PostTypeId="1" CreationDate="2013-10-18T00:28:33.223" Score="0" ViewCount="1287" Body="&lt;p&gt;I am interested in Non-Stoichiometry and how it relates to crystal structures; In particular I wish to learn about surfaces of metal oxides. I am aware that point defects in such crystals are a result of an entropy driven reactions. So my question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any good introductory resources on this subject? I have basic undergrad chemistry and physics background. &lt;/p&gt;&#xA;" OwnerUserId="2509" LastEditorUserId="2328" LastEditDate="2013-11-02T10:51:23.023" LastActivityDate="2013-11-02T10:51:23.023" Title="How do point defects in a crystal relate to Entropy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;crystal-structure&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6557" PostTypeId="1" AcceptedAnswerId="6560" CreationDate="2013-10-18T02:53:01.070" Score="1" ViewCount="636" Body="&lt;p&gt;How is a reduction in the number of molecules associated with a decrease in Entropy? For example, stoichiometric equations like $\ce{H2 + O -&amp;gt; H2O}$ are enthalpy driven if I am correct. But it also implies that entropy is decreasing because the system is becoming &quot;more orderly.&quot; Is this intuition correct?&lt;/p&gt;&#xA;" OwnerUserId="2511" LastEditorUserId="1878" LastEditDate="2013-10-18T07:41:40.530" LastActivityDate="2013-10-18T12:15:27.717" Title="How is a reduction in the number of molecules associated with a decrease in Entropy?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6559" PostTypeId="2" ParentId="6557" CreationDate="2013-10-18T07:50:19.883" Score="2" Body="&lt;p&gt;&lt;strong&gt;Gases have much larger molar entropy than liquids and solids&lt;/strong&gt;, owing to the contribution of translational entropy (gas molecules have much less hindered translational movement than molecules in liquids or solids). Thus, &lt;strong&gt;for a reaction that diminishes the total number of gas molecules, you can expect to have $\Delta_r S° &amp;lt; 0$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also true to some extent with other phases, as long as all reactants and products are in the same phase. But, if you have a reaction with some products or reactants in the gas phase, those will determine the sign of the molar reaction entropy.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-18T07:50:19.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6560" PostTypeId="2" ParentId="6557" CreationDate="2013-10-18T12:15:27.717" Score="4" Body="&lt;p&gt;Interpreting &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt; as &quot;disorder&quot;, which comes from thermodynamics applications, is only one way to interpret entropy. Entropy can also be described &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy#Statistical_mechanics&quot; rel=&quot;nofollow&quot;&gt;according to statistical mechanics&lt;/a&gt; as the statistical distribution of possible degenerate microstates: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=k_B \ln\Omega$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k_B$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;nofollow&quot;&gt;Boltzman's constant&lt;/a&gt; and $\Omega$ is the number of microstates. Thus it is a measure of uncertainty. The more equivalent/degenerate microstates there are, the less likely we know exactly which one of them the system is in, and thus, the higher the entropy. This probabilistic interpretation provides a more intuitive and satisfactory understanding of the relationship between number of molecules and entropy of the system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A decrease in the number of molecules leads to a decrease in the total number of degenerate microstates, which leads to a decrease in entropy. There are more degenerate ways to distribute N objects in M spaces than there are degenerate ways to distribute (N-1) objects in M spaces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's look at a very simple example, in which we have three molecules $\ce{2H2 + O2}$ in a container with room for four molecules. How many ways can we fill four spaces with these three molecules? There are four spaces in which we could put the $\ce{O2}$ molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}&#xA;\ce{O2} &amp;amp; ... \\ &#xA;... &amp;amp; ... \\ &#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;... &amp;amp; \ce{O2} \\ &#xA;... &amp;amp; ... \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;... &amp;amp; ... \\ &#xA;\ce{O2} &amp;amp; ... \\ &#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;... &amp;amp; ... \\ &#xA;... &amp;amp; \ce{O2} \\ &#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many ways can we put two $\ce{H2}$ molecules in one of these arrangements? In order to answer this, we need to temporarily distinguish one $\ce{H2}$ from the other $\ce{H2^{\circ}}$ (or to distinguish the one we place first from the one we place second) in order to generate all of the degenerate states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{pmatrix}&#xA;\ce{O2} &amp;amp; \ce{H2} \\ &#xA;\ce{H2^{\circ}} &amp;amp; ... \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;\ce{O2} &amp;amp; \ce{H2} \\ &#xA;... &amp;amp; \ce{H2^{\circ}} \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;\ce{O2} &amp;amp; \ce{H2^{\circ}} \\ &#xA;\ce{H2} &amp;amp; ... \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;\ce{O2} &amp;amp; ... \\ &#xA;\ce{H2} &amp;amp; \ce{H2^{\circ}} \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;\ce{O2} &amp;amp; \ce{H2^{\circ}} \\ &#xA;... &amp;amp; \ce{H2} \\ &#xA;\end{pmatrix}\begin{pmatrix}&#xA;\ce{O2} &amp;amp; ... \\ &#xA;\ce{H2^{\circ}} &amp;amp; \ce{H2} \\ &#xA;\end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are six arrangements of two $\ce{H2}$ molecules for each location of $\ce{O2}$ for a total of $\Omega=24$ arrangements. The entropy $S=k_B\ln(24)=4.39\times10^{-23}) \frac{\text{J}}{\text{K}}$. Note that this happened to be $\Omega=4\times3\times2=12$, or 4 ways of placing the first molecule, 3 ways of placing the second, and two ways of placing the third. We will use this shortcut next time we do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us run the reaction, converting hydrogen and oxygen to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2 + O2 -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let us repeat the exercise. How many ways can we fit two $\ce{H2O}$ molecules in four spots? There are 4 ways to place the first one, and 3 ways to place the second for each of those 4 placements of the first. Thus, $\Omega=4\times3=12$ and the entropy is $S=k_B\ln(12)=3.43\times10^{-23}) \frac{\text{J}}{\text{K}}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that there is a subtle flaw in my explanation. The molecules themselves have an intrinsic entropy hidden in their structures as well. According to &lt;a href=&quot;http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Standard-Molar-Entropies-984.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; the standard molar entropies of $\ce{O2},\ \ce{H2},$ and $\ce{H2O}$ are (in$\frac{\text{J}}{\text{K}\cdot\text{mol}}$) 205.1, 130.7, and 188.8, respectively. Thus, we have an entropy change of reaction:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rS^\circ=\sum S^\circ_{products} - \sum S^\circ_{reactants}$$ $$ \Delta_rS^\circ =\left(2\times188.8 - (205.1+2\times130.7)\right)\frac{\text{J}}{\text{K}\cdot\text{mol}}=-40\frac{\text{J}}{\text{K}\cdot\text{mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing standard entropies, we also get a decrease in entropy. Standard entropies are easier to work with because we are frequently working on the molar scale. If we changed our simplified example to be moles of $\ce{H2}, \ \ce{O2},$ and $\ce{H2O}$, in 4 moles of spaces we would have the original number of microstates be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega=(4N_A)(4N_A-1)(4N_A-2)=64N_A^3-48N_A^2+8N_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you plug in $N_A=6.022\times10^{23}$, all but the most staunch calculators and computer programs are going to give up. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%284%2a6.022e23%29%284%2a6.022e23-1%29%284%2a6.022e23-2%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; can do it because it does so symbolically, and we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega=1.398\times10^{73}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\ln(1.398\times10^{73})=1.400\times10^3\frac{\text{J}}{\text{K}\cdot\text{mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(At the molar scale we replace $K_B$ with $R$, the ideal as constant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another apparent weakness of the statistical approach is that it assumes no phase change. As F'x mentioned, there is an entropy change associated with a phase change. Roughly, the number of available spaces increases or decreases.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-18T12:15:27.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6561" PostTypeId="2" ParentId="6555" CreationDate="2013-10-18T13:34:49.120" Score="1" Body="&lt;p&gt;Two places where you can find predicted values of LogP (the oil:water partitioning coefficient) are chemicalize.org and Wolfram Alpha.  Both provide a predicted value for most of the compounds you are interested in (chemicalize provides a complete set, assuming you mean betacarotene).&lt;/p&gt;&#xA;&lt;h2&gt;Examples&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemicalize.org/structure/#!mol=cholesterol&amp;amp;source=calculate&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; provides a direct link to the output of the cholesterol page from Chemicalize.org and shows the logP value on the left hand side (7.11)&lt;/p&gt;&#xA;&lt;p&gt;To get the logp from Wolfram alpha you must search for logp  and then chose the predicted link that appears when no results are returned.  &lt;a href=&quot;http://www.wolframalpha.com/input/?i=logp+cholesterol&amp;amp;a=*DPClash.ChemicalE.cholesterol-_*Cholesterol.dflt-&amp;amp;a=*DPClash.ChemicalP.logp-_*PredictedLogPHydrophobicity-&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; should get you to the value obtained from Wolfram Alpha (7.02).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-10-18T13:34:49.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6562" PostTypeId="2" ParentId="5143" CreationDate="2013-10-18T19:33:36.757" Score="5" Body="&lt;p&gt;I happen to have a simple FCI code with detailed documentation of implementation. You can find it at &lt;a href=&quot;http://scholar.princeton.edu/boxiao/software/fci-201302&quot; rel=&quot;noreferrer&quot;&gt;http://scholar.princeton.edu/boxiao/software/fci-201302&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2013-10-18T19:33:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6563" PostTypeId="1" CreationDate="2013-10-19T07:54:47.553" Score="3" ViewCount="121" Body="&lt;p&gt;Chemical nomenclature seems to be the way chemistry &quot;talks&quot;, allied with the mathemathics. But where to start?... and what reasons and theories are behind? Some &quot;names&quot; come from &lt;em&gt;latin&lt;/em&gt;... I know there are rules IUPAC etc, but it is so demanding and exaustive to learn all that rules at once, for organic and inorganic chemistry. Maybe there are reasons and rules that help us to name a chemical or compound... Since I am interested in metallurgy, can you give me some advice to start? To associate theory to the formula, ad strutural formula, it would be nice. The implicit &quot;why&quot; in everthing included is always welcome.&lt;/p&gt;&#xA;" OwnerUserId="2516" LastActivityDate="2013-10-19T22:33:39.123" Title="How do I start studying chemical nomenclature?" Tags="&lt;nomenclature&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6564" PostTypeId="1" CreationDate="2013-10-19T12:50:21.767" Score="1" ViewCount="1544" Body="&lt;p&gt;Aren't alpha halo carbonyl compounds extremely fast and reactive via $S_N2$ (the fastest i guess)? If this is so, why isn't the major product in the Favorskii rearrangement an alpha hydroxy carbonyl compound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the alpha carbon(to which the halide is attached) is tertiary, there should be a high chance of an elimination product via $E_2$ (leading to the formation of alpha beta unsaturated carbonyl), right?  &lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="22" LastEditDate="2013-10-19T18:51:56.807" LastActivityDate="2014-07-30T07:10:00.653" Title="Favorskii rearrangement" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6568" PostTypeId="1" CreationDate="2013-10-19T21:55:32.550" Score="4" ViewCount="11590" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;:&#xA;Well, it's summer here in Brazil, and a great way to keep cool is drinking a so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Terer%C3%A9&quot; rel=&quot;nofollow&quot;&gt;Tereré&lt;/a&gt;, and, of course, we do need cold water for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;: I was looking for endothermic spontaneous reactions so I can cool the water in a bottle using an inner aluminum bottle in which the endothermic reaction occurs. I already knew that &lt;a href=&quot;http://en.wikipedia.org/wiki/Urea&quot; rel=&quot;nofollow&quot;&gt;Urea&lt;/a&gt;'s dissolution is endothermic and spontaneous, but I couldn't find how much endothermic it is, i.e., how many Joules it absorbs per Mol of urea dissolved. Another problem related with urea is that it's dissolution is a slow process (I think, not sure), additionally, the urea I can find is the one use as fertilizer, and this one has lots of impurities that I'll need to take into account when calculating how much I'll need to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know is that the urea dissolves in the same amount of water in weight, that the reaction time increases as the concentration increases, but have no idea how much energy it absorbs neither the how much time it takes to react, of course it depends on the concentration, granularity, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Considering all of that, are there any other reagent(s) I can buy or synthesize without being a licensed chemist that can be used to cool the water? Or, do you have the information that is missing about urea's dissolution (Joules/Mol absorbed in dissolution, approximated reaction time) so I can figure out if I can use it for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="2522" LastEditorDisplayName="user467" LastEditDate="2013-10-20T00:36:00.657" LastActivityDate="2016-10-21T06:13:48.573" Title="Homemade reactor for water cooling" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;home-experiment&gt;" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6569" PostTypeId="2" ParentId="6367" CreationDate="2013-10-19T22:19:09.850" Score="11" Body="&lt;p&gt;To best of my knowledge, the classification primary to quaternary is used for discussing reactivity of hydrocarbons, i.e. alkanes. There the number of substituents matters, as it governs carbocation stability, steric hindrance, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all other cases, the electronic structure of the molecule rules the reactivity. As you mentioned alkenes, alkynes or aromates, there it makes no sense to discuss such subtle effect as number of neighbouring carbons, when you have $\pi$-system at hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/A00228.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC Gold Book entry for alkyl groups&lt;/a&gt; supports this distinction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The groups $\ce{RCH2}$, $\ce{R2CH}\ (\ce{R} ≠ \ce{H})$, and $\ce{R3C}\ (\ce{R} ≠ \ce{H})$ are primary, secondary and tertiary alkyl groups, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="24052" LastEditDate="2016-03-13T20:09:08.207" LastActivityDate="2016-03-13T20:09:08.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6570" PostTypeId="2" ParentId="6563" CreationDate="2013-10-19T22:33:39.123" Score="2" Body="&lt;p&gt;Chemical nomenclature is mixture of historical traits, empirical knowledge and little bit of systematics.&#xA;And as you mentioned, it is exhaustive to learn all at once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand the nomenclature, you must understand the systems themselves. Because it's their structure and reactivity which is reflected in the name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes an unfortunate cyclic dependency. To start studying any given chemistry branch, you must know its language. But without knowing the science behind, the language does not make sense. It's like that and you probably cannot do anything about it, just cut this circle and crunch through the nomenclature. It will start to make sense, once you get sufficiently deep. Additionally, the &quot;why&quot; is good indicator. Once you have enough knowledge to understand the answer why it is named so, than you can probably answer yourself.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-19T22:33:39.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6571" PostTypeId="2" ParentId="6552" CreationDate="2013-10-19T22:39:48.247" Score="4" Body="&lt;p&gt;The main point is to stop the heat flow from the lamp, while allowing UV through. You want to do photochemistry, not thermochemistry. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-19T22:39:48.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6572" PostTypeId="2" ParentId="6568" CreationDate="2013-10-20T00:44:57.733" Score="3" Body="&lt;p&gt;Unfortunately, many aqueous endothermic reactions involve some chemicals you may wish to avoid.  Ammonium nitrate is the classic example, but in the US it is becoming more and more unavailable because of its potential use as an explosive.  Another frequently mentioned possibility is reacting hydrated barium hydroxide with ammonium chloride, but $\ce{Ba(OH)2}$ is very caustic and poisonous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of what I've read, your best choices might be mixing water with a chloride salt like ammonium or potassium.  I do not know how cold this will get your container. I suspect it will not be as good as ammonium nitrate, which would probably be ideal given the former use of this compound in medicinal cold packs. &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2013-10-20T00:44:57.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6573" PostTypeId="1" AcceptedAnswerId="6576" CreationDate="2013-10-20T04:14:22.407" Score="5" ViewCount="161" Body="&lt;p&gt;Can whatever is special about quasicrystals be understood at A-Level standard or slightly above it? The material I saw online is either too layman or too advanced for me to follow. What practical uses do they have and why does their structure lead to the properties that allow them to have these practical uses?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-06-10T18:44:17.563" Title="What is so special about quasicrystals?" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6574" PostTypeId="1" CreationDate="2013-10-20T07:55:48.847" Score="3" ViewCount="69" Body="&lt;p&gt;If the Equilibrium constant stands for the extent of reaction at equilibrium, whether more of forward or backward reaction is taking place, then why is it not affected by factors like concentration of reactants and product? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you increase concentration of reactant, equilibrium will shift to the right, which means more of forward reaction is taking place, so shouldn'tthe  equilibrium constant change?&lt;/p&gt;&#xA;" OwnerUserId="2391" LastEditorUserId="22" LastEditDate="2013-10-20T11:16:53.957" LastActivityDate="2013-10-20T11:16:53.957" Title="What Does the Equilibrium Constant represent?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2013-10-20T21:51:06.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="6576" PostTypeId="2" ParentId="6573" CreationDate="2013-10-20T08:10:29.820" Score="4" Body="&lt;p&gt;It's hard to know exactly what “level” you want address, as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasicrystal#Materials_science_of_quasicrystals&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry&lt;/a&gt; seems a decent introduction to me. But in any case, here's what I consider “special” about quasicrystals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From the fundamental point of view:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Though they are regular (they are formed by symmetry operations that allow them to repeat throughout space), they are not periodic (the symmetry operations are not purely translations). Although there's nothing inherently “wrong” or “impossible” about that from theoretical physics and chemistry, it goes against the classical classification of solids: crystalline or amorphous.&lt;/li&gt;&#xA;&lt;li&gt;Though hundreds of quasicrystals have been synthesized, there is only naturally occurring quasicrystal phase (&lt;a href=&quot;http://en.wikipedia.org/wiki/Icosahedrite&quot; rel=&quot;nofollow&quot;&gt;icosahedrite&lt;/a&gt;)!&lt;/li&gt;&#xA;&lt;li&gt;All quasicrystals are metal alloys (most of them being binary or ternary intermetallic alloys): given that there are other types of crystalline solids, why aren't there non-metallic quasicrystals?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding applications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quasicrystals have found, so far, few practical and industrial applications. In these applications, they are used as other alloys would, and do not have any “magic quasicrystal property”.&lt;/li&gt;&#xA;&lt;li&gt;One specific feature they have, though, is that quasicrystals tend to be rather strong alloys and that they age well, due to the &lt;a href=&quot;http://www.aip.org/tip/INPHFA/vol-2/iss-4/p26.pdf&quot; rel=&quot;nofollow&quot;&gt;way they are synthesized&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-20T08:10:29.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6577" PostTypeId="1" CreationDate="2013-10-20T12:13:31.533" Score="1" ViewCount="3425" Body="&lt;p&gt;I have heard that acids taste sour because of hydrogen (I don't really know exactly).  According to the definition of acids:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An acid (from the Latin acidus/acēre meaning sour) is a chemical substance whose aqueous solutions are characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some acids may not contain hydrogen.  Would they also taste sour?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="73" LastEditDate="2013-10-20T12:16:37.537" LastActivityDate="2018-11-05T15:55:57.370" Title="Do acids without hydrogen taste sour?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6579" PostTypeId="2" ParentId="6568" CreationDate="2013-10-20T16:37:33.643" Score="3" Body="&lt;p&gt;There are some &lt;strong&gt;Homemade reactor for water cooling&lt;/strong&gt;  that can do what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mohammed Bah Abba&lt;/strong&gt; had the same problem in Niger so he invented the &lt;strong&gt;Abba refrigerator&lt;/strong&gt;: a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pot-in-pot_refrigerator&quot; rel=&quot;nofollow noreferrer&quot;&gt;pot-in-pot refrigerator&lt;/a&gt; that uses the evaporation of the water from a porous material to refrigerate the things inside the pot.&#xA;The image belowe is not to of the Abba refrigerator but the principle is quite similar and here is well explained.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Oya6h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can make a mod &lt;strong&gt;cuia/guampa&lt;/strong&gt; with a pot-in-pot refrigerator embedded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To speed up your cooling I should use a pressurized gas, this can cold the thing in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the gas expansion is an endothermic process&lt;/li&gt;&#xA;&lt;li&gt;if the flow of gas go above the wet surface you can boost the evaporation process&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE: the Abba refrigerator works well in dry and ventilated ambient, I hope you are not in Amazonia!!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2013-10-21T09:11:27.113" LastActivityDate="2013-10-21T09:11:27.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6580" PostTypeId="1" CreationDate="2013-10-20T17:41:25.077" Score="10" ViewCount="2965" Body="&lt;p&gt;So when combining atomic orbitals to form molecular orbitals, you can either add the wave functions or subtract them. But at the same time, orbitals can exist in opposite phases (say one lobe of the p orbital is '+' and the other '-'). So what happens when you subtract two 1s orbitals of opposite phase, is that the same thing as constructively interference of the wave functions?&lt;/p&gt;&#xA;" OwnerUserId="2527" LastEditorUserId="4945" LastEditDate="2014-05-07T01:48:31.943" LastActivityDate="2014-08-11T10:14:57.237" Title="LCAO (Linear Combination of Atomic Orbitals) and Phases" Tags="&lt;theoretical-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6581" PostTypeId="1" AcceptedAnswerId="6584" CreationDate="2013-10-20T18:47:02.683" Score="4" ViewCount="186" Body="&lt;p&gt;In an experiment we conducted, we heated 1.6 grams of copper dust, and got 1.7 grams of oxidised copper, (it probably reacted with 0.1 grams of $\ce{O2}$)&#xA;and our goal was to find which of the two forms, $\ce{CuO}$ or $\ce{Cu2O}$ we got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lll}&#xA;\hline&#xA; &amp;amp; \ce{Cu} &amp;amp; \ce{O2} \\&#xA;\hline&#xA;m\ \text{(in g)}&amp;amp; 1.61 &amp;amp; 0.1 \\ &#xA;M\ \text{(in g/mol)}&amp;amp; 63.5 &amp;amp; 32 \\ &#xA;n\ \text{(in mol)}&amp;amp; 0.025 &amp;amp; 0.003125 \\ &#xA;\text{ratio}&amp;amp; 8 &amp;amp; 1 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the ratio between copper atoms to oxygen atoms is $4:1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it seems the form is $\ce{Cu4O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where is my mistake?&lt;/p&gt;&#xA;" OwnerUserId="2528" LastEditorDisplayName="user7951" LastEditDate="2016-10-15T18:39:49.507" LastActivityDate="2016-10-15T18:39:49.507" Title="Copper oxidation" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6583" PostTypeId="2" ParentId="5143" CreationDate="2013-10-20T21:27:38.800" Score="3" Body="&lt;p&gt;Useful links include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://vergil.chemistry.gatech.edu/notes/ci/node23.html&quot; rel=&quot;nofollow&quot;&gt;http://vergil.chemistry.gatech.edu/notes/ci/node23.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009261489874640&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0009261489874640&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/98/4/10.1063/1.464087&quot; rel=&quot;nofollow&quot;&gt;http://scitation.aip.org/content/aip/journal/jcp/98/4/10.1063/1.464087&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0010465589900337&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0010465589900337&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009261488874128&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0009261488874128&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/000926149085633N&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/000926149085633N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/000926148485513X&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/000926148485513X&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0301010480800450&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/0301010480800450&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/002689798168303#.UmRKqJRgbVk&quot; rel=&quot;nofollow&quot;&gt;http://www.tandfonline.com/doi/abs/10.1080/002689798168303#.UmRKqJRgbVk&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And there is the more recent Monte Carlo FCI method:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/132/4/10.1063/1.3302277&quot; rel=&quot;nofollow&quot;&gt;http://scitation.aip.org/content/aip/journal/jcp/132/4/10.1063/1.3302277&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;and many more courtesy of Google Scholar&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2013-10-20T21:27:38.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6584" PostTypeId="2" ParentId="6581" CreationDate="2013-10-20T21:43:37.280" Score="1" Body="&lt;p&gt;Your calculation looks good. Maybe the issue is not with the calculation, but with the experiment. Did you wait long enough so that all the copper is oxidized?&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-20T21:43:37.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6586" PostTypeId="2" ParentId="6577" CreationDate="2013-10-20T21:49:15.403" Score="5" Body="&lt;p&gt;Look at the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Taste#Sourness&quot;&gt;sourness&lt;/a&gt; for an explanation of that. But you also really need to really read a basic introduction to acid/base chemistry, because the question does not make much sense… &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid&quot;&gt;Acids&lt;/a&gt; are, by definition, substances that increase the concentration of H+ ions in an aqueous solution.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-20T21:49:15.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6587" PostTypeId="1" CreationDate="2013-10-21T05:11:20.207" Score="2" ViewCount="40469" Body="&lt;p&gt;I am asking this because I know that acids are normally aqueous but according to the solubility rules, all carbonates are insoluble.&lt;/p&gt;&#xA;" OwnerUserId="2531" LastEditorUserId="30" LastEditDate="2013-10-21T09:07:35.587" LastActivityDate="2016-02-22T12:15:28.020" Title="Is H2CO3 (carbonic acid) soluble or insoluble in water?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6589" PostTypeId="2" ParentId="6568" CreationDate="2013-10-21T09:05:45.107" Score="0" Body="&lt;p&gt;Recently has been developed an ammonium nitrate free cold pack. It consists of a mixing of Water, Urea, and Carbamakool™ this is the secret ingredient you should search for. This is a safe alternative to posioning chemical. &#xA;See &lt;a href=&quot;http://www.nortechlabs.com/instakool-faq.html&quot; rel=&quot;nofollow&quot;&gt;this site for more information&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can try to analyze one of this pack to find the information you need.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2013-10-21T09:15:49.867" LastActivityDate="2013-10-21T09:15:49.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6590" PostTypeId="2" ParentId="6581" CreationDate="2013-10-21T09:40:05.937" Score="5" Body="&lt;p&gt;When I was at university I used to oxidize metal oxides. In my case was difficult to fully oxidize a metal with only one step. This is what I should do.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First oxidation&lt;/li&gt;&#xA;&lt;li&gt;weigh it&lt;/li&gt;&#xA;&lt;li&gt;crush with a mortar what you have oxidized&lt;/li&gt;&#xA;&lt;li&gt;weigh it&lt;/li&gt;&#xA;&lt;li&gt;second oxidation&lt;/li&gt;&#xA;&lt;li&gt;weigh it&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the weight 4. and 6. are the same you can do the calculation if not you have to re-oxidize again. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2013-10-21T09:40:05.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6593" PostTypeId="2" ParentId="5673" CreationDate="2013-10-21T17:16:44.030" Score="3" Body="&lt;p&gt;It's indeed not common, as the brittle/ductile transition is usually the other way around. You can, however, find counter-examples (in a limited range of temperature) by searching for &lt;a href=&quot;http://www.google.com/?q=inverse+brittle-to-ductile+transition&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;“inverse brittle-to-ductile transition”&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-21T17:16:44.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6595" PostTypeId="1" AcceptedAnswerId="6596" CreationDate="2013-10-21T21:40:38.350" Score="11" ViewCount="411" Body="&lt;p&gt;I'd just like to ask for a little clarification here due to confusion from interdisciplinary studies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently reading the 1976 paper related to the recent 2013 Nobel Prize for Chemistry, by 2 of the laureates Warshel and Levitt — &lt;a href=&quot;http://dx.doi.org/10.1016/0022-2836%2876%2990311-9&quot;&gt;&lt;em&gt;Theoretical Studies of Enzymic Reactions: Dielectric, Electrostatic and Steric Stabilisation of the Carbonium Ion in the Reaction of Lysosyme&lt;/em&gt;&lt;/a&gt; — and have recently been reading into quantum chemistry, as well as studying biochemistry and have had further lectures in Physics to boot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if the points made here are comparable to points I've heard in mechanics lectures, about a cannonball not being modelled as a quantum object, since every subatomic particle would be too computationally intensive, so a coarse-grain model is used (i.e. Newton's &quot;classical&quot; laws).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't be so quick to draw parallels between mechanics lectures and enzyme biochemistry but it seems to be in the language they are using here (see highlights below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand (in mechanics) only 3 of the 4 'fundamental forces' (strong, weak, and EM) only strictly apply at the quantum/microscopic scale, and scientists are as it stands struggling to work gravity into the setup. Is this worth considering in this paper - if an enzyme is considered all of a sudden due to its size as a quantum object does this entail ignoring gravity QM now in favour of 'Newtonian', &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot;&gt;inertial frame-of-reference&lt;/a&gt; type mechanics, or considering it in some other 'Newtonian' manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The introductory section is as follows (&lt;a href=&quot;http://dx.doi.org/10.1016/0022-2836%2876%2990311-9&quot;&gt;full paper here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;sub&gt;The conformation of the enzyme-substrate complex can be adequately studied using empirical energy functions based on the “classical” contributions of bond stretching, bond angle bending, bond twisting, non-bonded interactions, etc. The mechanism and energetics of the enzymic reaction can only be studied using &lt;strong&gt;a quantum mechanical approach&lt;/strong&gt;. Previous quantum mechanical calculations on enzymic reactions have been limited in several respects. In the first place, they deal with an over-simplified model system consisting of only small fractions of the atoms involved in the real enzyme-substrate interaction. Second, all available quantum mechanical methods treat the reaction as if in a vacuum and are, therefore, unable to include the dielectric, which has a very important effect on the energy and charge distribution of the system. This limitation does not depend on the actual quantum mechanical scheme, being equally valid for the simplest Hiickel treatment and most extensive &lt;em&gt;ab initio&lt;/em&gt; calculation.&lt;/sub&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;sub&gt;Because the &lt;strong&gt;treatment of the whole enzyme-substrate complex quantum mechanically is computationally impossible&lt;/strong&gt;, it is necessary to simplify the problem and use a &lt;strong&gt;hybrid classical/quantum mechanical approach&lt;/strong&gt;. The method described here considers the whole enzyme-substrate complex: the energy and charge distribution of those atoms that are directly involved in the reaction are evaluated quantum mechanically, while the potential energy surfac e of the rest of the system, including the urrounding solvent, is evaluated classically. An important feature of the method is the treatment of the “dielectric” effect due to both the protein atoms and the surrounding water molecules. Our dielectric model is based on a direct calculation of the electrostatic field due to the dipoles induced by polarizing the protein atoms, and the dipoles induced by orienting the surrounding water molecules. With the inclusion of the microscopic dielectric effect we feel that our model for enzymic reactions includes all the important factors that may contribute to the potential surface of the enzymic reaction. Because we use analytical expressions to evaluate the energy and first derivative with respect to the atomic Cartesian co-ordinates, it is possible to allow the system to relax using a convergent energy minimization method. Here we use the method to study the mechanism of the lysozyme reaction, with particular emphasis on factors contributing to the stabilization of the carbonium ion intermediate. We first outline our theoretical scheme, paying attention to the &lt;strong&gt;partition of the potential surface into classical and quantum mechanical parts and to details of the microscopic dielectric model&lt;/strong&gt;. Then we use the method to investigate factors that contribute to the stabilization of the carbonium ion intermediate and deal with the relative importance of each factor. The effect of steric strain is considered for both the ground and the carbonium states and found to be of minor importance. &lt;strong&gt;Electrostatic stabilization&lt;/strong&gt; is treated at some length with special emphasis on the balance between stabilization by induced dipoles (polarizability) and stabilization by charge- charge interactions. The results demonstrate the importance of electrostatic stabilization for this reaction.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="2119" LastEditDate="2013-10-21T21:50:15.737" LastActivityDate="2013-12-04T09:38:29.287" Title="How can enzyme/substrate reactions that adhere (largely) to quantum theory also require 'Newtonian' consideration of gravity?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6596" PostTypeId="2" ParentId="6595" CreationDate="2013-10-21T22:23:32.980" Score="8" Body="&lt;p&gt;Unfortunately there's been something of a nomenclature mismatch here. While classical mechanics refers to Newtonian SUVAT-style things, &quot;classical&quot; molecular dynamics or geometry optimisation refers to, as the paper puts it: &quot;empirical energy functions based on the 'classical' contributions of bond stretching, bond angle bending, bond twisting, non-bonded interactions, etc.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pure &quot;classical force fields&quot;, these functions use parameters produced from analysis of various known geometries, and are collectively optimised using linear algebra optimisation techniques to obtain an approximation of a molecular or ionic system. Simple point-wise electrostatics may also be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper refers to the use of a combination of quantum mechanical techniques (modelling the complex reaction centre) with classical force field techniques (modelling the rest of the protein) and a dielectric model, modelling the interaction between the two partitions and the effect of the bulk solvent on the protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to look into this in more detail, I'd suggest looking for mention of &quot;QM/MM&quot; (Quantum Mechanics/Molecular Mechanics) or &quot;ONIOM&quot; techniques, as such models are now known. The use of simulated dielectrics for solvents is now largely covered under &quot;PCM&quot; (polarisable continuum model) and &quot;COSMO&quot; (Conductor-like Screening Model) techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're curious about &quot;classical&quot; molecular force-fields, I'd suggest looking up one of the packages that perform such calculations, such as &lt;a href=&quot;http://www.stfc.ac.uk/cse/25526.aspx&quot;&gt;DL POLY&lt;/a&gt; or &lt;a href=&quot;http://www.gromacs.org/&quot;&gt;GROMACS&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-10-21T22:23:32.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6597" PostTypeId="1" AcceptedAnswerId="6598" CreationDate="2013-10-21T22:49:50.620" Score="7" ViewCount="233" Body="&lt;p&gt;Having never given it a though before, I recently discovered (in a different context) that the prefix halo- actually means 'salt' or 'sea' and the suffix -gen means 'to form' or 'to generate'. So the Halogens are the elements that 'form salts'. But there are salts that do not have halogens in them, like $\ce{Na2S}$ or $\ce{(NH4)2SO4}$. Was it known at the time that other salts existed, and what other names for this group might have been considered? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, why aren't Group I metals called halogens? Doesn't it take two to tango (I mean, to form salts?)&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorDisplayName="user7951" LastEditDate="2016-10-15T18:26:38.780" LastActivityDate="2016-10-15T18:26:38.780" Title="How did Halogens become known as Halogens? They are not the only elements that form salts!" Tags="&lt;nomenclature&gt;&lt;ionic-compounds&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6598" PostTypeId="2" ParentId="6597" CreationDate="2013-10-22T02:34:50.390" Score="11" Body="&lt;p&gt;This seems like a bit of a rhetorical question, so this isn't a terribly formal or authoritative answer, but anyhow - a lot of chemical nomenclature is like lava flow. It solidified and people just worked around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The halogens are so named because they have a rich chemistry of ionic compounds (fluorine through iodine, anyhow). However, both the halogens and the group I metals can form a wide range of things that aren't salts. The noble gases can form compounds with elements of low birth. The rare earth elements aren't particularly rare. Oxygen ('acid-former') is not a necessary component of acids. Technetium ('artificial' + &lt;em&gt;ium&lt;/em&gt;) is produced in nature in significant quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, a lot of chemistry defies our human efforts to succintly categorise things, so at some point chemists have tried their best to find pragmatic general descriptors that unite groups of elements or molecules on the basis of the properties or constitution. The divide between organic and inorganic chemistry and the resulting exceptions and edge cases in classification (such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Mellitic_anhydride&quot;&gt;mellitic anhydride&lt;/a&gt;) is a good example of this. &lt;em&gt;The map is not the territory.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2013-10-22T02:34:50.390" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6599" PostTypeId="1" AcceptedAnswerId="6600" CreationDate="2013-10-22T04:09:39.167" Score="3" ViewCount="7007" Body="&lt;p&gt;I have some weird, yellow-amberish pebbles that have been filtered by my Zero waterfilter. Any idea what they are? Copper maybe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black stuff is charcoal from the filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quality of the water in my area is fairly good, with a TDS of 40-60. With the filter, I have a TDS of 0, which is excellent. But I'm still curious as to what those things are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TtHhV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3DOWH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x0882.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2540" LastEditorUserId="84" LastEditDate="2013-10-22T04:47:57.053" LastActivityDate="2019-06-12T14:49:12.127" Title="Yellow-amber debris above water filter: What is it?" Tags="&lt;filtering&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6600" PostTypeId="2" ParentId="6599" CreationDate="2013-10-22T06:00:00.450" Score="4" Body="&lt;p&gt;It looks like &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_exchange&quot; rel=&quot;nofollow&quot;&gt;ion exchange resin&lt;/a&gt;, it is used to specifically capture ions from water and replace them by others (either $\ce{H^+}$ or alkali metal ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, it is probably also part of the filter.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-22T06:00:00.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6602" PostTypeId="1" CreationDate="2013-10-22T08:32:54.230" Score="0" ViewCount="15011" Body="&lt;p&gt;I completed a titration experiment in a lab. The aim was to find the concentration of the titrant $\ce{KMnO4}$ solution by titrating it with a sodium oxalate solution of a known concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first made a standard solution of sodium oxalate solution, using $\pu{0.826 g}$ of sodium oxalate powder, then poured $\pu{50ml}$ of distilled water along with $15$-$\pu{20ml}$ of $\pu{3 M} \,\,  \ce{H2SO4}$. Then I pipetted out $\pu{25 ml}$ of the solution and poured it in to a volumetric flask. Then I filled the rest of the volumetric flask with distilled water, to form the standard solution. I then did the Mcbride experiment to find $75\%$ of the volume of $\ce{KMnO4}$ need to titrate the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then did the experiment three times to get an average volume of $\ce{KMnO4}$ to be $\pu{25.48 ml}$. And the volume of sodium oxalate solution is $\pu{25 ml}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should find the concentration of $\ce{KMnO4}$ now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the molar ratio is 2:5. How do I calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to find the moles of sodium oxalate then I multiply it by $\frac{2}{5}$ and then divide that value by the volume of $\ce{KMnO4}$ which is $\pu{25.48 ml}$. But how do I calculate the moles of sodium oxalate? What would the concentration be? I know the volume is $\ce{25 ml}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have done, but I am not sure if it is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles of sodium oxalate = $\frac{\pu{0.8265 g}}{\pu{133.96 g mol-1}} = \pu{0.006169752 mol}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{0.006169752}{\pu{0.025L}}$ = concentration of sodium oxalate is therefore $0.246790086$, $0.0248$ to 4 s.f. &lt;br&gt;&#xA;($\pu{0.025 L}$ is the volume of the sodium oxalate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then multiplied that by $\frac25$.  $0.0248 \cdot \frac25$ gives me the concentration of $\ce{KMnO4}$ which is $0.09872$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think this is wrong because I have not used the volume of $\ce{KMnO4}$ used, which was the whole point of the experiment.&lt;/p&gt;&#xA;" OwnerUserId="2409" LastEditorUserId="19616" LastEditDate="2017-04-21T18:26:31.423" LastActivityDate="2017-04-21T18:46:49.977" Title="How to find the concentration of potassium permanganate in this experiment?" Tags="&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6603" PostTypeId="1" AcceptedAnswerId="6608" CreationDate="2013-10-22T10:00:58.377" Score="1" ViewCount="1095" Body="&lt;p&gt;We have learnt that the energy evel of both R and S optical isomer of a compoud is same, just as well the entropy of each isomer is also same. If we consider the conversion of R form of a compound to it's S isomer we observe that the change in gibbs free energy is &quot;0&quot;. This implies that the two isomers must exist in equilibrium. But we also know that any compound does not change from its R configuration to its S configuration for example inversion of configuration in Sn1 mechanisms lead to only one configuration and not both are obtained after the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is that why don't any compounds R isomer convert into its S isomer on its own so as to maintain equilibrium ?&lt;/p&gt;&#xA;" OwnerUserId="2534" LastActivityDate="2013-10-22T13:58:29.990" Title="Equilibrium between R and S configuration?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6604" PostTypeId="2" ParentId="6587" CreationDate="2013-10-22T10:12:24.400" Score="6" Body="&lt;p&gt;Yes, it is soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the oceans are constantly getting $\ce{CO2}$ from the atmosphere that gets converted into carbonic acid and thus decreasing pH on a global scale. Or the soft drinks we drink contain carbonic acid in them in dissolved form only due to mixing of $\ce{CO2}$ gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, not all carbonates are insoluble as &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6587/is-h2co3-carbonic-acid-soluble-or-insoluble-in-water/6604#comment9672_6587&quot;&gt;Ben Norris&lt;/a&gt; has commented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard and fast 'rules' in chemistry tend to be rare.  Consider them guidelines.&lt;/p&gt;&#xA;" OwnerUserId="2534" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2016-02-22T12:15:28.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6605" PostTypeId="2" ParentId="6473" CreationDate="2013-10-22T10:42:26.980" Score="1" Body="&lt;p&gt;As already mentioned by @GM it indeed is used to take out $\ce{Cu^{2+}}$, but it is also used to take out $\ce{Cu+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this because I once used a mixture of $\ce{HCl}$ and $\ce{H2O2}$ to etch copper boards, in that process both $\ce{Cu^{2+}}$ and $\ce{Cu+}$ were dissolved by the formation of $\ce{CuCl2}$ and $\ce{CuCl}$ with $\ce{Cl-}$ from $\ce{HCl}$.&lt;/p&gt;&#xA;" OwnerUserId="2534" LastEditorUserId="41328" LastEditDate="2017-07-24T15:50:12.767" LastActivityDate="2017-07-24T15:50:12.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6606" PostTypeId="2" ParentId="6229" CreationDate="2013-10-22T11:03:19.660" Score="0" Body="&lt;p&gt;The $\ce{Zn}$ cation polarises the $\ce{Cl}$ anions to a good extent and reduces the energy gap between certain energy levels. Therefore the electrons in $\ce{ZnCl2}$ are able to jump this gap by absorbing certain radiation, when this radiation is absorbed all the radiations except these are reflected and can be seen as complimentary color which in this case happens to be black.&lt;/p&gt;&#xA;" OwnerUserId="2534" LastEditorUserId="73" LastEditDate="2013-10-23T16:05:39.043" LastActivityDate="2013-10-23T16:05:39.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6607" PostTypeId="1" CreationDate="2013-10-22T12:54:42.343" Score="1" ViewCount="123" Body="&lt;p&gt;How fast does $\ce{BaCl2}$ bind water from its surroundings, becoming a hydrate, compared with same process with $\ce{Na2CO3}$?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="30" LastEditDate="2013-10-22T15:06:14.993" LastActivityDate="2013-10-22T15:06:14.993" Title="How effectively bound Na2CO3 and BaCL2 water" Tags="&lt;physical-chemistry&gt;&lt;water&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6608" PostTypeId="2" ParentId="6603" CreationDate="2013-10-22T13:58:29.990" Score="5" Body="&lt;p&gt;Because even though neither the R nor the S isomer is thermodynamically favored (in an achiral environment), the free energy barrier (or &lt;em&gt;activation energy&lt;/em&gt;) between those two states is huge. This is the most common case, when R/S interconversion requires breaking and reforming a chemical bond. Then, even though the thermodynamic equilibrium would be to have a racemic mixture, kinetics prevent the solution from reaching this equilibrium state.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-22T13:58:29.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6609" PostTypeId="1" CreationDate="2013-10-22T18:59:51.710" Score="2" ViewCount="311" Body="&lt;p&gt;We're asked often to calculate the density of a crystal, and there is a known formula for that related to the unit cell volume.  Now, for simple geometric structures such as the bcc structure (and since it is close packed), I know how to relate the different radii of the constituent compounds with the unit cell's (unique in this case) dimension (the length a), but how do I relate it for such other complex structures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general given a certain crystal geometry (cubic , tetragonal , orthorhombic), how does one relate the unit cell dimension with the radii of the constituent atoms?&lt;/p&gt;&#xA;" OwnerUserId="1463" LastEditorUserId="73" LastEditDate="2013-10-22T20:26:17.050" LastActivityDate="2014-05-18T21:25:29.170" Title="Relating a compound's radii and unit cell dimensions" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6610" PostTypeId="1" CreationDate="2013-10-22T21:22:10.860" Score="14" ViewCount="2338" Body="&lt;p&gt;So often I see total energies of transition states corrected for zero-point vibrational energy which always confuses me.  Zero point energy is the lowest energy that a &lt;strong&gt;ground state&lt;/strong&gt; minimum energy configuration can have at 0K in a vacuum.  Transition states are indeed &lt;strong&gt;not&lt;/strong&gt; minima, can never be isolated, have one imaginary mode of vibration, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do people constantly factor in zero-point vibrational energy to the total energy for a transition state?  Obviously this is done to have 'comparable' minima and TS energetics (if you are applying ZPE to the minima) but is this even physical?  If zero-point correcting the TS, should the zero-point energy of the imaginary mode be included in the summation or excluded (or should both quantities be provided)?  &lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-29T14:13:44.173" LastActivityDate="2014-09-29T14:13:44.173" Title="Why are the total energies of transition states so commonly corrected for zero-point vibrational energy?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;transition-state-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6612" PostTypeId="2" ParentId="6580" CreationDate="2013-10-23T02:37:55.063" Score="0" Body="&lt;p&gt;Add or substract the wavefunctions is the first step, then you need to normalize the new wavefunction, so its norm becomes 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you get twice 1s wavefunction before normalization, and result in 1s afterwards. So usually there's no point to combine the same orbitals in LACO.&lt;/p&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2013-10-23T02:37:55.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6615" PostTypeId="1" AcceptedAnswerId="6616" CreationDate="2013-10-23T09:39:03.453" Score="4" ViewCount="531" Body="&lt;p&gt;We always hear about the different arrangements of molecules that lead to the formation of polymorphs, my question here is why would a polymorph form in the first place? The material was given enough time to solidify in  a proper arrangement, why did it choose to solidify in a form that isn't the most stable?&lt;/p&gt;&#xA;" OwnerUserId="1787" LastActivityDate="2013-10-23T09:44:21.040" Title="Why polymorphs form?" Tags="&lt;molecules&gt;&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6616" PostTypeId="2" ParentId="6615" CreationDate="2013-10-23T09:44:21.040" Score="3" Body="&lt;p&gt;This is a very broad and very interesting question, but the basic point is: polymorphs exist either because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;their formation is controlled by kinetics (rather than thermodynamics, which would favor the most stable phase), or&lt;/li&gt;&#xA;&lt;li&gt;the thermodynamic equilibrium between the various polymorphs is influenced by the synthesis conditions (temperature, pressure, presence of templating agents and other species in solutions…), and once the material has crystallized in a given form, it does not spontaneously undergo a transition to the most stable phase (again, because of kinetics)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-23T09:44:21.040" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6617" PostTypeId="2" ParentId="6610" CreationDate="2013-10-23T09:50:08.793" Score="10" Body="&lt;p&gt;At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of the transition state, you have to include the contributions of ground state energy of all the other vibration modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more rigorous way of looking at it is to use statistical physics, and to consider the partition function of the transition state. See &lt;a href=&quot;http://vallance.chem.ox.ac.uk/pdfs/TSTnotes.pdf&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://www.chem.umn.edu/groups/taton/chem4011/Lecture%20Notes/10_23.pdf&quot;&gt;there&lt;/a&gt; for such treatments.&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="30" LastEditDate="2013-10-23T12:48:28.503" LastActivityDate="2013-10-23T12:48:28.503" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6618" PostTypeId="1" AcceptedAnswerId="6623" CreationDate="2013-10-23T14:14:36.017" Score="6" ViewCount="3207" Body="&lt;p&gt;Recently I have watched a &lt;a href=&quot;http://izismile.com/2013/10/22/can_mercury_be_absorbed_with_a_sponge.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; demonstration with a sponge and mercury and the sponge can't absorb mercury like it does with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that?&lt;/p&gt;&#xA;" OwnerUserId="2546" LastEditorUserId="402" LastEditDate="2017-07-20T05:37:25.773" LastActivityDate="2017-07-20T05:37:25.773" Title="Why sponge can't absorb mercury" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6619" PostTypeId="2" ParentId="6618" CreationDate="2013-10-23T15:29:41.167" Score="2" Body="&lt;p&gt;You are asking wrong way. Sponge does not absorb almost anything. Apples, sand, dust, wires,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is: Why can sponge absorb water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the interaction of water and the sponge are similarly or even more favorable than interaction inside water. The nature of this effect is &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot; rel=&quot;nofollow&quot;&gt;capilary action&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-23T15:29:41.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6620" PostTypeId="1" CreationDate="2013-10-23T16:09:55.220" Score="2" ViewCount="366" Body="&lt;p&gt;When formulating a rate a equation two experiments must take place: one in which the reactant concentrations are varied, and one in which the catalyst concentrations are varied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am given a sample of a solid catalyst, such as $MnO_2$, how can I change its concentration? Because I can't dilute a solid and increase the volume, which is how I'd do it for a catalyst in solution.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2014-01-22T10:22:49.347" Title="How can I formulate a rate equation using a solid catalyst?" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6621" PostTypeId="1" CreationDate="2013-10-23T17:55:45.703" Score="4" ViewCount="3142" Body="&lt;p&gt;I'm reading up on molecular modelling and have come across some terms I should probably have learnt a long time ago but always scared me off a bit with the Greek letters. I guess I'm a big boy now so it's probably time to clarify...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook uses the lowercase symbol omega ($\omega$) for torsion (a.k.a. dihedral) angle while &lt;a href=&quot;http://en.wikipedia.org/wiki/Dihedral_angle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives the uppercase, $\Omega$, plus both upper and lower case phi ($\Phi$, $\phi$) and psi ($\Psi$, $\psi$) in its description which from reading I'm starting to understand designate particular geometrical cases from the article, though it looks like it might be a little difficult to remember at first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook also gives in the list of symbols and physical constants [lower case] tau, $\tau$, but I can't see any mention of this on the Wiki page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly I wanted to ask is the use of upper/lower case totally arbitrary, and secondly is tau just a non-specific way of expressing torsion angle compared to the other 3 (which is the only use I can imagine for it?) Or is this use totally confined to molecular modelling?&lt;/p&gt;&#xA;" OwnerUserId="2119" LastActivityDate="2013-10-24T09:11:44.547" Title="Torsion angle symbol" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6622" PostTypeId="2" ParentId="6620" CreationDate="2013-10-23T19:59:14.050" Score="1" Body="&lt;p&gt;You will perform the reaction inside a container or solution of some particular volume. This means that you can change the catalyst concentration inside that volume by adding more or less catalyst in different experiments. So you don't increase the volume of the solid (i.e. change its density), you change the mass of solid inside the reaction volume.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-10-23T19:59:14.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6623" PostTypeId="2" ParentId="6618" CreationDate="2013-10-23T21:00:17.303" Score="9" Body="&lt;p&gt;I think the key feature is the surface tension of mercury-air which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface-tension_values&quot; rel=&quot;noreferrer&quot;&gt;extremely high&lt;/a&gt; (about 7 times water-air which is already high), combined with the chemical nature of the sponge, which is typically &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot; rel=&quot;noreferrer&quot;&gt;cellulose&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since cellulose is a big polysaccharide it has a reasonably favorable interaction with both water (due to the OH groups) and alcohols and alkanes (due to the carbon chain). This causes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot; rel=&quot;noreferrer&quot;&gt;contact angle&lt;/a&gt; of water, alcohol and alkanes with the sponge to be low (at least lower than 90 degrees) thus allowing capillary action to suck the liquid into the sponges pores. The metallic nature of mercury has an unfavorable interaction with the sponge thus resulting in a contact angle above 90 degrees. This will not allow &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot; rel=&quot;noreferrer&quot;&gt;capillary action&lt;/a&gt; because the cosine of angles above 90 degrees is negative, therefore the sponge cannot soak up the mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sponge made out of metal (silver is common I believe) would be able to soak up the mercury because it has a favorable interaction in terms of surface energy.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-10-25T08:13:15.000" LastActivityDate="2013-10-25T08:13:15.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6624" PostTypeId="1" AcceptedAnswerId="6633" CreationDate="2013-10-23T23:23:36.153" Score="1" ViewCount="93" Body="&lt;p&gt;Given a set of partial charges for the atoms in a molecule, which are educated guesses but do not add up to zero, we would like to zero them in an intelligent way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the net charge is negative and we see a +0.1 atom bonded to a -0.9 atom, we should probably shift the -0.9 towards zero to make that section of the molecule more neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also chemical information which can be used, eg, keep oxygen atoms positive and don't let the charge on hydrogen be greater than 1.0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this class of problem? What algorithms can be applied?&lt;/p&gt;&#xA;" OwnerUserId="1207" LastActivityDate="2013-12-23T15:33:21.510" Title="What is an algorithm for making a net neutral molecule from a set of guessed positive charges?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-03-05T10:30:30.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="6625" PostTypeId="1" AcceptedAnswerId="24621" CreationDate="2013-10-23T23:32:47.410" Score="13" ViewCount="11318" Body="&lt;p&gt;A molecule of phenol is much more inclined to under go an electrophilic substitution reaction than a molecule of benzene because the $\ce{-OH}$ group is highly reaction favoring. From what I understand, this is because when the $\ce{-OH}$ group is attached, it contributes a resonance structure involving a double bond between the oxygen and the ring, which stabilizes the intermediate arenium ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from what I also understand, oxygen is also a highly electronegative atom and therefore inductively draws electron density away from the benzene ring. Despite these two competing effects (electron donating via resonance and electron withdrawing via induction), the overall effect of the $\ce{-OH}$ group is electron donating. This also applies to $\ce{-NH2}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in concluding that resonance effects are stronger than inductive effects, if not in general, then at least for electrophilic substitution of substituted aryls? &lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="7448" LastEditDate="2015-02-03T16:47:48.467" LastActivityDate="2015-02-11T04:49:06.460" Title="Competing resonance and inductive effects in a substituted benzene" Tags="&lt;aromatic-compounds&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6626" PostTypeId="1" CreationDate="2013-10-24T01:23:49.220" Score="2" ViewCount="415" Body="&lt;p&gt;I'm trying to find a free automatic organic synthesis Linux command line program which take mol or smile files as input. I have random molecules and I'd like to know how difficult above a baseline they will be to synthesize. An implementation of synthetic accessibility and not a full step wise synthesis program would also be acceptable. Thanks for the help.&lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorUserId="5017" LastEditDate="2015-02-06T16:19:18.687" LastActivityDate="2015-02-06T16:19:18.687" Title="Free automatic organic synthesis Linux command line program" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6627" PostTypeId="1" CreationDate="2013-10-24T02:29:29.723" Score="3" ViewCount="11818" Body="&lt;p&gt;The molecular orbital diagram for $\ce{O2}$ says that the  sigma 2p bonding molecular orbital is lower in energy than the pi 2p bonding molecular orbital. Why is this not the case in the $\ce{B2}$ MO diagram?&lt;/p&gt;&#xA;" OwnerUserId="2550" LastEditorUserId="4356" LastEditDate="2014-10-26T08:47:39.320" LastActivityDate="2018-02-27T19:45:47.667" Title="Why is there a difference between O2 and B2 sigma 2p molecular orbitals in diagrams?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6628" PostTypeId="1" CreationDate="2013-10-24T06:05:53.160" Score="3" ViewCount="261" Body="&lt;p&gt;A massive apartment complex near where I live is being fumigated with Vikane gas (sulphuryl flouride, $\ce{SO2F2}$).  The process will take a long time (weeks) because there are a lot of buildings, and their tents appear to have rips in them, clearly visible: several neighbors have complained of symptoms of exposure to the gas, and I'm worried about it in my home.  Obviously we should try to get them to stop / etc., but that's a matter of law and politics not meant for a chemistry site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime, what I would like to know is if there is any safe way I can use to pull the gas out of the air if it should waft its way into my home.  My understanding is that activated charcoal will not work for this molecule.  I've heard that either strong bases (usually photo-developing chemicals are mentioned) or heavy water vapor reacts with the gas; but I'm not sure what that reaction is, or if it would result in something better or worse than the gas itself.  I've tried Googling but with just a high school chemistry exposure some decades ago, I can't make much headway.  Please note that I do not have access to any chemicals or machines I can't readily buy in a nearby store.  My goal is to turn it into some liquid or solid, or bind it to such, so that I can't breath it in.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastEditorUserId="2551" LastEditDate="2013-10-24T10:29:04.893" LastActivityDate="2014-08-08T12:03:37.593" Title="Are there any tricks I can use to pull SO2F2 out of the air?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6631" PostTypeId="2" ParentId="6621" CreationDate="2013-10-24T09:11:44.547" Score="4" Body="&lt;p&gt;It is a habit to assign greek letters (most often lowercase) to angles. In case of &quot;normal&quot; angles, it's mostly $\alpha, \beta, \gamma$, whereas for dihedrals some higher letters, as you mentioned, $\phi, \psi, \omega$. As noted in comments, whatever you use, clearly state what you mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between dihedral and torsional angle is more conceptual, then formal. In torsion you discuss rotation around a bond, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dihedral_angle&quot; rel=&quot;nofollow&quot;&gt;dihedral&lt;/a&gt; is general angle of two planes, as mentioned in the $\ce{CH_4}$ example. The formal expression for both of them is the same, indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be noted, there are some well known widely used dihedral angles, where you should stick to the common notation. The most prominent example is the protein backbone arrangement, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramachandran_plot&quot; rel=&quot;nofollow&quot;&gt;Ramachandran plot&lt;/a&gt;. In this case you should not invent your own notation.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-24T09:11:44.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6633" PostTypeId="2" ParentId="6624" CreationDate="2013-10-24T11:34:46.933" Score="1" Body="&lt;p&gt;Although you did not mentioned specifically, it is likely that you want to perform some Molecular mechanics force field calculations. The charges are important quantity, so you need more than educated guess to obtain usable results. Nonetheless, the problem of charge equilibration is solved by &quot;Polarizable force fields&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted, that they are often more expensive, less reliable and less accurate than standard FFs.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-24T11:34:46.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6636" PostTypeId="2" ParentId="6626" CreationDate="2013-10-24T20:12:15.553" Score="2" Body="&lt;p&gt;The SAscore is now implemented in RDkit and available for &lt;a href=&quot;http://www.mail-archive.com/rdkit-discuss@lists.sourceforge.net/msg03477.html&quot; rel=&quot;nofollow&quot;&gt;download&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastActivityDate="2013-10-24T20:12:15.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6637" PostTypeId="1" CreationDate="2013-10-25T06:40:35.357" Score="10" ViewCount="2695" Body="&lt;p&gt;I am currently doing doping studies in a system similar to this ($\ce{A_{x}B_{1-x}}$) and I have to find atomic weight percentage of $\ce{B}$ element using atomic absorption spectroscopy(AAS). I have prepared different PPM of solutions according to the weight percentage of $\ce{B}$ element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example to prepare 10 atomic weight percentage of $\ce{B}$ in $5~\mathrm{ppm}$ solution, I have used this equation to calculate molecular weight:&#xA;$$\text{Actual mol weight}=\text{mol.wt}(A)\cdot x + \text{mol.wt}(B) \cdot (1-x).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case $x = 0.9$ and $1-x = 0.1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I have prepared $5~\mathrm{ppm}$ of a solution of the doped material assuming that $10\%$ of $\ce{B}$ is doped completely (but in reality only very low percentage is doped).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the below data from AAS:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)&#xA;    10          5                 1.5&#xA;    20          5                 0.9      &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From the above data, I have to calculate actual atomic weight percentage of $\ce{B}$.&#xA;My colleague has suggested this method in order to find $\text{Actual atomic weight percentage}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{act.At.Wt.%} =\frac{\text{Obs.AAS.Conc(µg/ml)}}{\text{Sol.Conc(ppm)}}\cdot \text{At.Wt.percentage.of.}\ce{B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this I got the results below.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B (At wt %) Conc of sol(ppm)    AAS Conc(µg/ml)   Actual B (At wt %)&#xA;10                5                  1.25            2.5&#xA;20                5                  0.9             3.6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I have doubt in this method as you can see in the table the observed AAS concentration for 10 and 20 At wt % $\ce{B}$ is 1.25 and 0.9 (µg/ml) respectively. You can see that the AAS concentration of 20 At.wt % is lower than 10 A.wt %. But when we calculate actual $\ce{B}$ At.wt.% we can see that the 10 A.wt %  is lower comparing 20 A.wt %. Can you tell me whether the equation suggested by my colleague is correct or not? If not can you suggest any other method in order to find the actual atomic weight percentage?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="402" LastEditDate="2017-01-12T02:40:16.597" LastActivityDate="2021-05-02T19:05:05.590" Title="Calculation of Atomic weight percentage in AAS" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6638" PostTypeId="1" AcceptedAnswerId="6686" CreationDate="2013-10-25T11:45:21.837" Score="14" ViewCount="20778" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Which of the contributing structures of the resonance is more stable?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Y13D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm watching a video lecture by a professor of my college where he puts this question to the class. The class unanimously says &lt;strong&gt;B&lt;/strong&gt;. At first I thought they were wrong, but then the professor agreed with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? The reason he gave was that oxygen's octet would be complete in &lt;strong&gt;B&lt;/strong&gt;, hence it contributes more. But I don't understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Isn't the octet complete in &lt;strong&gt;A&lt;/strong&gt; too? It has 2 covalent bonds - &lt;em&gt;4 electrons (2 from sharing)&lt;/em&gt; and 2 lone pairs. Adding up to 8.&lt;br&gt;&#xA;2) Oxygen being more electronegative than carbon, would pull the electrons towards itself, leading me to think that it would be more comfortable with a lone pair rather than another bond. Also, won't it prefer to lose out that '+' charge which the bonding's burdened onto it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing which made me believe that Prof. may have made a mistake while talking about the more relevant contributor, is that when he added Br&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt; to the compound, he added it to the canonical structure A. And don't you explain reactions with the most relevant contributor, i.e. B?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The presence of an extra bond in the canonical structure &lt;strong&gt;B&lt;/strong&gt;, along with the completion of the carbon valency makes it more stable than &lt;strong&gt;A&lt;/strong&gt;. The reaction with Br&lt;span class=&quot;math-container&quot;&gt;$^-$&lt;/span&gt; is best explained looking at &lt;strong&gt;A&lt;/strong&gt;, though, because the incoming nucleophile ends up bound to carbon, not oxygen.&lt;br&gt;&#xA;Thanks, everyone!&lt;/p&gt;&#xA;" OwnerUserId="1858" LastEditorUserId="72973" LastEditDate="2020-02-06T16:40:23.307" LastActivityDate="2020-02-06T16:40:23.307" Title="Oxygen or carbon with a positive charge? (comparison of canonical structures)" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;resonance&gt;" AnswerCount="5" CommentCount="9" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="6640" PostTypeId="1" AcceptedAnswerId="9751" CreationDate="2013-10-25T20:53:57.240" Score="1" ViewCount="1958" Body="&lt;p&gt;My book gives me the following problem:&#xA;&quot;A mixture of $125.0 g$ $\ce{N2}$ and $32.0 g$ $\ce{H2}$ reacts to $36.5 g$ $\ce{NH3}$. Calculate the efficiency.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My method to do this is - one I learned from my teacher, to put every given molecule's amount of substance (moles) in a table with 3 steps: start (S), gone (G)/formed (F) and rest (R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$125g\:N_2 \leftrightarrow n=4.462$ moles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$32g\:\left(3\right)H_2\leftrightarrow n=16.13$ moles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{matrix} &amp;amp;N_2&amp;amp;+&amp;amp;3H_2&amp;amp;\longrightarrow &amp;amp;2NH_3\\S&amp;amp;4.462&amp;amp;&amp;amp;16.13&amp;amp;&amp;amp;\\G&amp;amp;4.462&amp;amp;&amp;amp;13.386&amp;amp;F&amp;amp;8.924\\R&amp;amp;0&amp;amp;&amp;amp;2.74&amp;amp;&amp;amp;8.924\end{matrix}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\rightarrow$ abundance of $H_2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?&lt;/p&gt;&#xA;" OwnerUserId="2560" LastEditorUserId="1431" LastEditDate="2013-10-26T16:08:35.987" LastActivityDate="2014-04-05T15:34:52.983" Title="Stoichiometry: efficiency" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6641" PostTypeId="1" CreationDate="2013-10-25T20:55:28.860" Score="6" ViewCount="17272" Body="&lt;p&gt;As far as I know, a large equilibrium constant shows that the a reaction like the one below  wants to complete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B &amp;lt;=&amp;gt; A}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If A and B are gases, then K is:&#xA;$$K_1=\frac{[A]}{[B]}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but when I multiply the reaction by 20 , K is&#xA; $$K_{20}=\frac{[A]^{20}}{[B]^{20}}$$ while the same number of moles of B are consumed. If $\small K_1 = 10$ then $\small K_{20} = $$10^{20}$ so the reaction must be approximately complete. Why does this happen while the reaction materials didn't change?&lt;/p&gt;&#xA;" OwnerUserId="2561" LastEditorUserId="39" LastEditDate="2013-11-02T11:16:57.670" LastActivityDate="2013-11-03T04:58:06.770" Title="Why does equilibrium constant change when the reaction is multiplied by a number and nothing in the real has been changed?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6642" PostTypeId="2" ParentId="6641" CreationDate="2013-10-25T23:36:29.307" Score="6" Body="&lt;p&gt;First of all, you switched the numerator and denominator in the definition of $K$. If $\ce{A &amp;lt;=&amp;gt; B}$ , then $K=\frac{[B]}{[A]}$. Products over reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that, if $K$ is large, then mathematically it is necessary for either the numerator to be large and/or the denominator to be small. Therefore, qualitatively, if $K$ is large, then you either have a lot of products, or little left of reagents, both of which imply a reaction which occurs close to completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the problem - we were speaking &lt;em&gt;qualitatively&lt;/em&gt;. Your first sentence (&quot;a big equilibrium constant shows that the reaction wants to get complete&quot;) is only a &lt;em&gt;rule of thumb&lt;/em&gt;. It happens to work well because in reactions where equilibria descriptions are quantitatively reliable/exist at all, the stoichiometric coefficients are relatively small when written with the lowest integers (using the lowest integer coefficients is almost always done). The rule of thumb is borne out of repeated calculations under these implied conditions. They are &lt;em&gt;not&lt;/em&gt; the case for your second equilibrium example, so such a rule should not be expected to still be valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, though $10^{20}$ is an impressive number at first, when you actually go do the calculations to find the relative amounts of reactants and products, you will at some point have to take its &lt;em&gt;twentieth root&lt;/em&gt;, promptly turning it back into a much smaller number. Another way to think about it is that with such large stoichiometric coefficients, even a small change in concentrations is amplified enormously by the large exponents and has a huge impact on the value of the reaction quotient; a small increase in products will go a long way to reaching the equilibrium condition.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-10-25T23:36:29.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6643" PostTypeId="2" ParentId="6318" CreationDate="2013-10-26T00:11:27.673" Score="3" Body="&lt;p&gt;To a first approximation, any pure substance can sublimate. However, complex substances may decompose or change their structure before any appropriate sublimation conditions are reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only (and &lt;em&gt;very&lt;/em&gt; remarkable) exception is &lt;a href=&quot;http://ltl.tkk.fi/research/theory/helium.html&quot; rel=&quot;nofollow&quot;&gt;helium&lt;/a&gt; (in fact, helium is so special that helium-3 and helium-4 have quite different phase diagrams even though they're isotopes!). Note that due to the presence of superfluid phases, there is no direct transition between the solid and gas phases in either isotope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, there is nothing prohibiting a substance to exhibit a solid-gas transition curve crossing a region of mild temperature and pressure conditions. Though unusual, an interesting example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_heptafluoride&quot; rel=&quot;nofollow&quot;&gt;iodine heptafluoride&lt;/a&gt;. It has a very narrow liquid range at ambient pressures (meaning a small decrease in pressure is enough to go below the triple point pressure), and the liquid range happens to fall near very reasonable temperatures.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2013-10-26T00:23:23.040" LastActivityDate="2013-10-26T00:23:23.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6644" PostTypeId="1" CreationDate="2013-10-26T03:48:31.577" Score="1" ViewCount="7048" Body="&lt;p&gt;Why do the more easily oxidized elements displace the less easily oxidized elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question inspired by Activity Series of Metals&lt;/p&gt;&#xA;" OwnerUserId="2531" LastActivityDate="2013-12-31T20:52:14.190" Title="Why are more easily oxidized elements more reactive than less easily oxidized elements?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6646" PostTypeId="1" CreationDate="2013-10-26T06:02:20.637" Score="0" ViewCount="1158" Body="&lt;p&gt;I'm having trouble drawing this molecule because I don't understand how the functional group Phenyl (by -ene suffix) can also be a substituent (by 1-phenyl). Also, what does the 1,3-butadiene mean?&lt;/p&gt;&#xA;" OwnerUserId="2376" LastActivityDate="2013-10-26T13:54:04.437" Title="How do you draw (E)-1-phenyl-1,3-butadiene?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6647" PostTypeId="1" CreationDate="2013-10-26T06:06:48.887" Score="8" ViewCount="11816" Body="&lt;p&gt;The most stable conformation of optically inactive butane-2,3-diol is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(A)&lt;/strong&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZII1f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&amp;nbsp;&#xA;&lt;strong&gt;(B)&lt;/strong&gt;&lt;img src=&quot;https://i.stack.imgur.com/9kIe9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&amp;nbsp;&#xA;&lt;strong&gt;(C)&lt;/strong&gt;&lt;img src=&quot;https://i.stack.imgur.com/csEBD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&amp;nbsp;&#xA;&lt;strong&gt;(D)&lt;/strong&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Rdij.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose C as the correct answer but, in the answer key it is given to be B. C is the staggered form and the most stable one. It can also have hydrogen bonding between -OH and -H groups. B can also have hydrogen bonding but there would be repulsion between the two Me groups. So, which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="4945" LastEditDate="2015-09-15T08:43:30.787" LastActivityDate="2019-07-22T14:04:01.223" Title="Which is most stable conformation of optically inactive butane-2,3-diol?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;conformers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6648" PostTypeId="2" ParentId="6646" CreationDate="2013-10-26T07:47:00.897" Score="2" Body="&lt;p&gt;Firstly, the phenyl group is just a benzene with a hydrogen removed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7XdGn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, $\ce{C6H5 -}$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the functional group Phenyl (by -ene suffix) can also be a substituent (by 1-phenyl).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The -ene suffix is about the double bonds in the system, in this case it applies to the butane backbone. It has nothing to do with the phenyl.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, what does the 1,3-butadiene mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1,3-butadiene is a butane backbone with two double bonds at carbon 1 and 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVLia.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final molecule will look lie this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0KEh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-10-26T07:47:00.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6649" PostTypeId="2" ParentId="6647" CreationDate="2013-10-26T11:59:58.883" Score="15" Body="&lt;p&gt;Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's &lt;a href=&quot;http://en.wikipedia.org/wiki/A_value&quot; rel=&quot;noreferrer&quot;&gt;A value&lt;/a&gt;, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;noreferrer&quot;&gt;hydrogen boning&lt;/a&gt;, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer cannot be A or D since they are representing a chiral diastereromer.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2016-01-02T14:20:15.360" LastActivityDate="2016-01-02T14:20:15.360" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6650" PostTypeId="1" CreationDate="2013-10-26T12:03:56.180" Score="3" ViewCount="943" Body="&lt;p&gt;I am looking for open-source tools which can generate 2d molecular structure from the MOL or SMILES file. Further, it SHOULD number the atoms in the order of occurrence of atoms in the connection table of MOL file or SMILE string. I used MarvinSketch 6.1.2 but the position of numbering is not very good, as it get cluttered with the bonds.&lt;img src=&quot;https://i.stack.imgur.com/XcZEo.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Please refer some tools with which I can draw molecule similar to the one shown in figure but numbering is also clear. In the figure, all numbering except 1-7 are ok.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2013-10-26T14:41:22.430" Title="Tools to draw molecular structure (wireframe) along with atom numbering" Tags="&lt;software&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6651" PostTypeId="2" ParentId="6646" CreationDate="2013-10-26T13:54:04.437" Score="3" Body="&lt;p&gt;ManishEarth put together the pieces to understand the &lt;em&gt;connection&lt;/em&gt; of the atoms to form the molecule.  Yet since the molecular structure requested in the original post is about &lt;a href=&quot;http://www.orgsyn.org/orgsyn/orgsyn/prepContent.asp?prep=cv4p0771&quot; rel=&quot;nofollow noreferrer&quot;&gt;(&lt;em&gt;E&lt;/em&gt;)-1-phenyl-1,3-butadiene&lt;/a&gt;, the answer actually is opposite &lt;em&gt;configuration&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VLuCp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By convention, alkenes posses a reference plane orthogonal to the plane of drawing.  In case the alkene is substituted twice by groups attached at the termini of the alkene -- in the upper example a phenyl and an alkyl chain -- these may be either at the same or the opposite side of this reference plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ease of convenience, two drawings simplifying the substitutents by &quot;R&quot; demonstrate this. In the first example, they are opposing each other, so their relationship is described either by (&lt;em&gt;E&lt;/em&gt;) or (&lt;em&gt;trans&lt;/em&gt;). The second example shows them at the same side of the reference plane, therefore, this alkene has a (&lt;em&gt;Z&lt;/em&gt;)- or (&lt;em&gt;cis&lt;/em&gt;)-configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mfbGY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask about the origin of (&lt;em&gt;E&lt;/em&gt;) and (&lt;em&gt;Z&lt;/em&gt;).  They are derived from German &lt;em&gt;entgegen&lt;/em&gt; and &lt;em&gt;zusammen&lt;/em&gt;; similar like the Latin (&lt;em&gt;trans&lt;/em&gt;) for &lt;em&gt;across&lt;/em&gt;, and (&lt;em&gt;cis&lt;/em&gt;) for &lt;em&gt;nearby&lt;/em&gt;.  Both ways to describe these &lt;em&gt;configurational&lt;/em&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomers&quot; rel=&quot;nofollow noreferrer&quot;&gt;isomers&lt;/a&gt; are found in the literature for one and the same molecule, yet in an article or report, you should stick to either the (&lt;em&gt;E&lt;/em&gt;)/(&lt;em&gt;Z&lt;/em&gt;) or (&lt;em&gt;cis&lt;/em&gt;)/(&lt;em&gt;trans&lt;/em&gt;). (And yes, in italics.)&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-10-26T13:54:04.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6652" PostTypeId="2" ParentId="6650" CreationDate="2013-10-26T14:41:22.430" Score="2" Body="&lt;p&gt;To answer the question, I assumed you would like to stick with Marvin if possible. Likely, the options described and found functional in version 5.2.4 met at &lt;a href=&quot;http://www.webqc.org/moleculareditor2.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; are functional at later versions, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I drew from scratch a similar representation of your molecule.  And indeed, there is overlap by labels and the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5oEs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet the software allows you to &lt;em&gt;print&lt;/em&gt; the drawings.  So, thanks to the installation of HP-1102P printer and &lt;em&gt;printing into a file&lt;/em&gt; I got a postscript file.  This vector format is of course beneficial, you may &lt;em&gt;import&lt;/em&gt; it in &lt;a href=&quot;http://inkscape.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;inkscape&lt;/a&gt;, for example.  Note two potential pitfalls: select &quot;import&quot; (instead of &quot;open&quot;) and &quot;import text as text&quot;.  Twice you have to ungroup the drawing before being able to move the labels around as you prefer.  Or to change their colour (&lt;em&gt;vide infra&lt;/em&gt;), or substitute them.  Eventually, you may either save it in numerous file formats or like in this case, export the resulting bitmap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UAZ26.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-10-26T14:41:22.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6655" PostTypeId="1" AcceptedAnswerId="6663" CreationDate="2013-10-26T16:15:52.483" Score="14" ViewCount="13967" Body="&lt;p&gt;Aluminium oxide is amphoteric. It is easy to see that it is a Brønsted-Lowry base through the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aluminum oxide splits and the oxygen accepts a proton, forming water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the reaction with a base? In my textbook, they say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al_2O_3 + NaOH \rightarrow 2NaAlO_2 + H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the textbook claims that Aluminium oxide is an acid merely because it reacts with a base to form a salt and water, as is characteristic of a neutralization reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not satisfied with this definition. I mean, acids aren't defined as 'things that neutralize bases', we have well-established definitions for them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to figure out for myself how this could be. Clearly, the Brønsted-Lowry theory cannot be applied here since the compound in question has no protons to donate. Therefore, the only alternative is the Lewis concept. I cannot see how that is applicable in this case.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="5026" LastEditDate="2018-03-03T11:21:46.527" LastActivityDate="2018-06-10T02:55:55.627" Title="How can aluminium oxide be called an acid?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6656" PostTypeId="1" AcceptedAnswerId="6658" CreationDate="2013-10-26T20:03:29.090" Score="1" ViewCount="204" Body="&lt;p&gt;Given the similarity in formula between pentane ($\small\ce{CH3(CH2)3CH3}$, b.p. 36.2 °C) and neopentane ($\small\ce{C(CH3)4}$, b.p. 9.5 °C), why are their boiling points so different?&lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="7475" LastEditDate="2015-10-16T17:14:46.660" LastActivityDate="2015-10-16T17:14:46.660" Title="Boiling points of v. similar organic molecules" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6658" PostTypeId="2" ParentId="6656" CreationDate="2013-10-26T23:08:49.203" Score="3" Body="&lt;p&gt;Solids and liquids are hold together by cohesion. Intermolecular interactions contributing to cohesion depend on the geometry of the molecules in interaction,too.  Here, &lt;em&gt;n&lt;/em&gt;-pentane with an elongated molecular structure exhibits a much larger molecular surface as compared to &lt;em&gt;neo&lt;/em&gt;-pentane of a more sphere-like structure*).  As a result, other criteria kept constant (like the molecular mass), the lesser the molecular surface area of molecules of a substance, the lower the boiling temperature of this substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*)True, &lt;em&gt;neo&lt;/em&gt;-pentane's methyl groups point towards the vertices of a tetrahedron. Actually I recommend to build a molecular model of both compounds and place them side-by-side to see it.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-10-26T23:08:49.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6659" PostTypeId="1" AcceptedAnswerId="6661" CreationDate="2013-10-27T00:07:14.980" Score="11" ViewCount="18651" Body="&lt;p&gt;I know that the mole is widely used in chemistry instead of units of mass or volume as a convenient way to express amounts of reactants or of products of chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why people in chemistry still excessively use it for their measurement? To be backward-compatible and consistent with traditional textbooks? Why they don't simply express their quantity measurement per atom, per unit volume, per molecule or etc.?&lt;/p&gt;&#xA;" OwnerUserId="2566" LastActivityDate="2016-11-14T18:32:47.420" Title="Why do people still use the mole (unit) in chemistry?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;mole&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6661" PostTypeId="2" ParentId="6659" CreationDate="2013-10-27T01:55:42.557" Score="25" Body="&lt;p&gt;Simply speaking, because it's an appropriate unit to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine I wanted to measure the length of a rope. What would be an appropriate length to use? Inches? Centimeters? Feet, maybe? It would really be awkward to express it as 0.000189393 miles, or as 304,800,000 nanometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: if you can't see why these units are awkward, take any page discussing things like this (e.g. biomass of certain species) and change all the units so that they're nonsense like this. Then put it away for a week and try to read it later.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's say I've changed my mind and I'd like to measure all the rope created in the world in a year. Now an appropriate unit is almost certainly miles or kilometers, and not inches or centimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider something else: so far, I've been using length to measure ropes. Would it make sense to measure their combined mass instead? Maybe not for small amounts, since I think I would throttle you if you told me to cut off half a pound of rope, but for global-scale things, tons or metric tons may make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, using measurements like the average density of the rope or the combined diameters of all the ropes really wouldn't be much use at all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What we've seen here is that when we're measuring things, there are measures that make sense (for rope, length, maybe weight) and some measures that really don't (color, average diameter, etc.). In those measures, there are units that are convenient (inches, feet), and some that aren't (nanometers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the problem with chemical units, but much more magnified. When I measure the energy released on hydrolysis of a sample, I'm not measuring the energy of one bond, or two bonds, or a thousand bonds, or even a million. I'm measuring a collection of so insanely many molecules that there's really no number in common language for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an appropriate way to measure things in these molecules? If I'm only interested in how much I have and not what's in it, per gram or per volume is often a good way to do it. Since when I'm measuring the temperature change of something, I don't particularly care what it's made of, measures of this (specific heat, for instance) are done in units of something per gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if I &lt;em&gt;do&lt;/em&gt; care what my sample is made of, then I need a better way to measure it. For example, sodium chloride and calcium chloride look similar, but will have very different energies since there's three ions in a $\ce{CaCl2}$ unit and only two in an $\ce{NaCl}$ unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can't have half a molecule, the simplest way to do it is to count the darn things. Unfortunately, measuring things per atom is a really awkward way to do things because of the problems I described above. The energy it takes to melt ice is 0.000000000000000000001029 J/molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we need then, is some count of molecules that's convenient. We could go by 1000 molecules, or a million molecules, but it's a pain in the butt to convert between that and macroscopic units (how much does 1 million calcium atoms weigh?) Now it just so happens that $6.022 \times 10^{23}$ atoms of carbon-12 weigh 12 grams, and a single carbon-12 weighs 12 amu. A single molecule of water weighs roughly 18 amu, $6.022 \times 10^{23}$ molecules of water weigh 18 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a look at what using $6.022 \times 10^{23}$ molecules offers us over other units:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is appropriate. $6.022 \times 10^{23}$ molecules will typically&lt;br&gt;&#xA;fall in the gram to kilogram range of substance, which can easily be &#xA;measured out on a balance. It's also within the range of what you'd&#xA;usually use in a lab.&lt;/li&gt;&#xA;&lt;li&gt;It allows for easy conversion. You can easily tell when you have&#xA;$6.022 \times 10^{23}$ molecules because your atomic mass is equal to&#xA;your macroscopic mass.&lt;/li&gt;&#xA;&lt;li&gt;It makes sense. Intuitively, we'd like to measure things per atom or &#xA;per molecule, but doing so leads to some ugly units. Instead, if we&lt;br&gt;&#xA;measure per $6.022 \times 10^{23}$ molecules, we still are doing a&lt;br&gt;&#xA;counting-based measurement, but the numbers themselves are a lot&lt;br&gt;&#xA;nicer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You probably know this already, but we call $6.022 \times 10^{23}$ molecules a mole. These are the advantages of using the mole as a unit, and not for backwards-compatibility. I am of the opinion that if we shed all our units today and started from scratch, we would start off with grams and liters, but we would probably start using the measure of a mole again within a year. It's just a very powerful, useful way to measure things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This is much longer than I was initially intending and was not written while I was in the most awake state of mind. Please let me know in the comments if you think this answer is unclear, or stupid, or just plain wrong. &lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-10-27T01:55:42.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6663" PostTypeId="2" ParentId="6655" CreationDate="2013-10-27T06:02:13.017" Score="9" Body="&lt;p&gt;The most basic definition of &quot;acid&quot; is that it is a proton donor (or one which accepts a lone pair)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that $\ce{OH-}$ is available to react), we get the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al2O3 + OH- -&amp;gt; 2AlO2- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al2O3 + 6H+ -&amp;gt; 2Al^3+ + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where it is acting like a proton acceptor (base).&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T18:33:37.610" LastActivityDate="2015-05-03T18:33:37.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6664" PostTypeId="2" ParentId="6659" CreationDate="2013-10-27T14:29:58.097" Score="10" Body="&lt;p&gt;Your statement that the &quot;mole is widely used in chemistry &lt;strong&gt;instead&lt;/strong&gt; of units of mass or volume..&quot; is misleading.  There are &lt;a href=&quot;http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf&quot;&gt;seven base units of measurement&lt;/a&gt; in the SI system and your question alludes to two, mass and amount.  The other 5 are length, time, temperature, electrical current, and luminosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider going out for a typical (American) breakfast of bacon and eggs.  There are two diners that you like to visit; one serves small-sized eggs and bacon strips while the other uses jumbo eggs and offers canadian bacon.  At each place, you order &quot;Two scrambled eggs with three slices of bacon.&quot;  But do you get the same amount of food at each diner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rash of bacon weights about 22 grams and a slice of canadian bacon is a bit closer to 50 g, so three slices of each weigh 66 and 150 grams, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small chicken egg weighs about 43 grams while a jumbo egg weighs 71 g.  These weights are with the shell, and you visit good diners that give you scrambled eggs without the shell, but the point should be clear: &lt;em&gt;while the &lt;strong&gt;amount&lt;/strong&gt; of items ordered is the same, the &lt;strong&gt;weight&lt;/strong&gt; of items is different&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colloquially, we use the term &lt;em&gt;amount&lt;/em&gt; when we are interested in answering a question beginning with &quot;How much...&quot;.  The units of our response oftentimes depends on practical considerations about how we measure the substance.  From the deli, I'd order a pound of ham and two balls of mozzarella.  In this case using units of weights or amount could be valid (I'd like 12 slices of ham and 100 grams of mozzarella.)  If I'm waiting in line, however, I would say &quot;there are 3 patrons before me&quot; not &quot;I have to wait for 425 pounds of people before I can be served.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last a comment about the mole.  The International Bureau of Weights and Measures has a &lt;a href=&quot;http://www.bipm.org/en/si/si_brochure/chapter2/2-1/mole.html&quot;&gt;good definition&lt;/a&gt; of the mole that doesn't need to be repeated here.  To carry out the food analogy just one more step, the mole is the same type of word as a dozen.  When you go to the store to prepare for a large breakfast, do you think to yourself, &quot;I'm going to buy twenty four eggs&quot; or &quot;I'm going to by 2 dozen eggs&quot;.  The word dozen, which stands for 12, is a way to simplify or group our numbers.  Along the same lines, it's much easier to say 12 grams of carbon contains 1 mol of carbon atoms instead of $6.022 \times 10^{23}$ atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary&lt;/strong&gt; The mole is used extensively in the sciences because we need a unit that describes an &lt;em&gt;amount&lt;/em&gt; of substance, which is different from the &lt;em&gt;mass&lt;/em&gt; of a substance or how much space a given amount of substance occupies.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-10-27T14:29:58.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6665" PostTypeId="1" CreationDate="2013-10-27T22:24:25.320" Score="3" ViewCount="1902" Body="&lt;p&gt;Is the rate of heat transfer between two bodies brought into thermal contact dependent upon the different in temperature between them, assuming they each have the same mass? &#xA;Imagine two bodies of water connected to each other by an impermeable glass barrier. If the difference between the temperatures is 10 degrees, will the system reach its equilibrium state faster per degree than if the difference is one degree? &lt;/p&gt;&#xA;" OwnerUserId="658" LastActivityDate="2013-10-28T01:31:40.297" Title="Heat transfer dependence on temperature gradient?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6666" PostTypeId="2" ParentId="6665" CreationDate="2013-10-27T22:56:57.440" Score="1" Body="&lt;p&gt;Yes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe according to Stefan-Boltzmann's law the hotter something is, then the more energy it will radiate per surface area, causing more energy to be lost faster at higher temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2067" LastActivityDate="2013-10-27T22:56:57.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6667" PostTypeId="2" ParentId="667" CreationDate="2013-10-27T23:14:34.813" Score="3" Body="&lt;p&gt;BET analysis on &lt;a href=&quot;https://en.wikipedia.org/wiki/Zeolit&quot; rel=&quot;nofollow&quot;&gt;zeolithes&lt;/a&gt; represents a typical example.  Dependent on the size of the particles you have, you select a glass cell of about a centimetre; for a powder a smaller cell, for more coarse pellets you don't want to grind a bigger one.  You need to outgas the samples first, i.e. by applying vacuum and elevated temperatures already adsorbed gases and moisture is removed from the surface of the material to be analyzed as well as the inner walls of the sample cell. &lt;em&gt;This&lt;/em&gt; already may take time to be accomplish -- because you do not want to destroy or alter the material in question by too harsh conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual measurement requires the sample cell being kept at a constant temperature.  An initial determination takes place while the sample cell is immersed in a Dewar of liquid nitrogen.  This low temperature provides better conditions for the adsorption experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first blank is measured with Helium gas, entered via a graduated piston. Helium typically &lt;em&gt;does not&lt;/em&gt; adsorb on surfaces (at this temperature).  This allows to determine the dead volume of the measurement cell.  In the second step, Nitrogen gas enters the measurement cell, again in a controlled manner (graduated piston, pressure control).  Again, this takes time as you want to establish the equilibrium between the adsorbent and the amount of gas adsorbed in function of gas pressure and sample temperature.&#xA;A controlled entry of Helium gas is necessary as a second determination of the dead volume to complete the calibration eventually leading to the read-out for our substance at given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, dependent on what kind of isotherme (BET/Langmuir) is to be determined, Nitrogen gas as a load may be replaced by other gases, like methane.  Beside determination of the surface area of a solid, the adsorption properties may be determined in a series of more elevated temperatures, too.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-10-27T23:14:34.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6669" PostTypeId="2" ParentId="6665" CreationDate="2013-10-28T01:31:40.297" Score="3" Body="&lt;p&gt;Yes, it is very much dependent on temperature gradient. The stefan-boltzman law is only for radiative dissipation of heat and not for general heat transfer by conduction(contact).&#xA;The equation for &lt;em&gt;steady state&lt;/em&gt; heat flow is&#xA;$$\vec h=-k \vec{\nabla T}$$ where $\vec \nabla$ is the gradient operator which is based on the temperature difference between the bodies and $k$ is the thermal conductivity. $\vec h$ is the heat per unit area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is not in steady state, then the heat diffusion equation becomes relavent:- &#xA;$$\frac{dT}{dt}=\frac{k}{c}\nabla^2 T$$ where $c$ is the specific heat of the body.&lt;br&gt;&#xA;Lastly, a to a crude approximation, the rate of cooling in atmosphere at low difference of temperature of atmosphere and the body is given by Newton's law of cooling (partly by radiation):- &#xA;$$\frac{dT}{dt}=-K(T_{body}-T_{surrounding})$$ where $K$ is a constant.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all these equation, it is the temperature gradient between the bodies in contact, which drives the rate of heat transfer (or temperature change). &lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-10-28T01:31:40.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6670" PostTypeId="1" CreationDate="2013-10-28T04:23:17.740" Score="1" ViewCount="65" Body="&lt;p&gt;Is there a compound that turns to gas at a certain temperature but if heated more turns back to liquid and if heated more than turn to gas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it exist naturally? Has a such a substance be found? Can such a substance exist? if yes, then why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why couldn't at some temperature molecules interact in such a harmony that they actually cling together? &lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="4945" LastEditDate="2015-10-23T17:13:26.593" LastActivityDate="2015-10-23T17:13:26.593" Title="Is there a element that shows this pattern of boiling point?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6671" PostTypeId="2" ParentId="6234" CreationDate="2013-10-28T09:00:24.440" Score="2" Body="&lt;p&gt;You should treat everything as ideal gas. Then you calculate the volume change from the air itself, using ideal gas law and pressure being constant (atmospheric, maybe). Calculate also the number of moles of air in the syringe, it will come handy soon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining change in volume is then attributed to the solvent, again using ideal gas law, you put in the given volume, temperature, pressure and obtain number of moles, which turned into vapor. &#xA;Having this, you know number of moles of solvent and number of moles of air, so again assuming everything ideal (Raoult's law), you know the partial pressures (i.e. vapor pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For added fancines, you can take the decrease of the liquid volume into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: While using ideal gas and ideal mixture of ideal gases approximation, all quantities are nicely additive, so don't be afraid.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-28T09:00:24.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6672" PostTypeId="2" ParentId="6670" CreationDate="2013-10-28T09:06:37.633" Score="1" Body="&lt;p&gt;No. If you take a molecular system in the gas state, and you heat it up (at constant pressure), there is no reason for it to go into a denser, more organized state, such as liquid or solid. As you increase temperature, the interaction energy between molecules becomes smaller compared to thermal energy.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-10-28T09:06:37.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6673" PostTypeId="1" CreationDate="2013-10-28T13:22:13.653" Score="2" ViewCount="1137" Body="&lt;p&gt;I need to explain (chemically) why expired tetracyclines should not be used, relating to the changes in tetracycline structure that occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can find is that they cause nephrosis, but not why. &lt;/p&gt;&#xA;" OwnerUserId="2573" LastEditorUserId="41890" LastEditDate="2017-11-27T17:58:01.503" LastActivityDate="2017-11-27T17:58:01.503" Title="What happens to the structure of tetracyclines when they expire?" Tags="&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6674" PostTypeId="1" CreationDate="2013-10-28T14:40:59.170" Score="3" ViewCount="2789" Body="&lt;p&gt;In the following figure &lt;img src=&quot;https://i.stack.imgur.com/NQylA.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can see that the p-orbitals overlap 1s orbital (though relatively very little). How can an electron in p-orbital, be simultaneously in the 1s orbital at any given point in space and time?&lt;/p&gt;&#xA;" OwnerUserId="2575" LastEditorUserId="1431" LastEditDate="2013-10-28T16:03:51.680" LastActivityDate="2015-12-17T16:26:45.330" Title="How do 1s and 2p orbitals overlap?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6675" PostTypeId="1" CreationDate="2013-10-28T17:36:03.080" Score="1" ViewCount="738" Body="&lt;p&gt;I know element in left two columns have 1+,2+... and right ones 1-,2-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in middle is all over the place. And with multiple charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What defines that if this element is gonna have this charge. All i have been told is that elements want to be close to as much possible to noble gasses because they are stable. Which doesn't explain much as to why such behavior is present. What does it mean to be stable. And why elements in middle find it stable just after 3 +/- charge even though they are far from noble gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, there are few questions in here but all very related i think. &lt;/p&gt;&#xA;" OwnerUserId="2491" LastActivityDate="2013-10-28T18:41:13.250" Title="why there is no simple rule to predict the possible ionic charges of element in middle of periodic table" Tags="&lt;ions&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6676" PostTypeId="1" CreationDate="2013-10-28T18:03:30.313" Score="4" ViewCount="115443" Body="&lt;p&gt;Consider the following sets of problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8CNGb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping someone could clarify what each checkbox means exactly. All of these answers are incorrect (except the first one with the check marks signifies the correct answers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, why are the first pair different molecules? What makes something a conformation/constitution/structural isomer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean to say something is an &quot;identical conformation of the same molecule&quot; vs &quot;different conformation of the same molecule?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When can I definitiviely tell if something is an enantiomer or just a diastereomer? It seems like (for an enantiomer) you can't just do the check of &quot;switch all dashes to wedges and all wedges to dashes&quot; because, as in the third pair of newman projections, they are enantiomers of each other but I don't understand why that is.&lt;/p&gt;&#xA;" OwnerUserId="2376" LastEditorUserId="10185" LastEditDate="2015-10-02T16:44:28.790" LastActivityDate="2016-04-13T04:53:45.550" Title="Difference between conformational, constitutional, and structural isomers and the same and different molecules" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6677" PostTypeId="2" ParentId="6675" CreationDate="2013-10-28T18:41:13.250" Score="2" Body="&lt;p&gt;TL;DR - The upshot is that there just happens to be a simple approximation governing small atoms but as the atoms get bigger and more complex, those approximations become less good and there is not another simple rule that works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more depth, the issue is a combination of two factors: distance from the nucleus and full/half-full valence orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first place, valence (outermost) electrons that are further away from the nucleus are more weakly bound than valence electrons that are closer. This means that the energy required to add or remove an electron is less, and so the charge given by a given electron matters less to the stability of the atom. Because of this, the further down the periodic table you go, the less the &quot;normal&quot; charges matter. As an example, Germanium is two rows below Carbon and is naturally found to have charges of +4 and +2 (as well as sometimes having +1, +3 and -4!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second factor is that elements find a special stability when their valence orbitals are either full (preferable) or half-full (less preferable, but still better than neither). So an element like Cr, which has almost no hope of gaining or losing enough electrons to look like a noble gas, &quot;wants&quot; to gain lose the 4 3d electrons, getting it down to a full 4s orbital. The tricky bit is that unlike the p orbitals, not all of the d orbitals have the same energy, so there are some that are more likely to be removed than others. There just isn't a convenient rule of thumb about which is which (nature is governed by physics, not approximations).&lt;/p&gt;&#xA;" OwnerUserId="2542" LastActivityDate="2013-10-28T18:41:13.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6678" PostTypeId="2" ParentId="6674" CreationDate="2013-10-28T18:46:25.537" Score="0" Body="&lt;p&gt;Remember that the orbitals are really just approximations showing the most likely places for an electron with a given set of quantum numbers to be. If you don't apply a cutoff (like less than a 1/1000000 chance), all orbitals would encompass the entire universe. We just make pictures like this by identifying the places that an electron could be within some margin of probability.&lt;/p&gt;&#xA;" OwnerUserId="2542" LastActivityDate="2013-10-28T18:46:25.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6679" PostTypeId="2" ParentId="6638" CreationDate="2013-10-28T18:50:05.757" Score="2" Body="&lt;p&gt;The difference between the two structures is just the amount of space available to the lone pair. If you think about it in terms of the particle in a box approximation, the path length for that lone pair is much longer when in the double bond state, so it will be a lower energy state. &lt;/p&gt;&#xA;" OwnerUserId="2542" LastActivityDate="2013-10-28T18:50:05.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6680" PostTypeId="1" CreationDate="2013-10-28T18:58:46.390" Score="1" ViewCount="695" Body="&lt;p&gt;I have graphite turning into diamond.I need to find the thermal effect of this reaction.How would one write that?I'm not sure how to proceed here.The teacher told us to use oxygen?&lt;/p&gt;&#xA;" OwnerUserId="2577" LastActivityDate="2013-10-28T19:24:04.747" Title="How do I find the thermal effect of this reaction?" Tags="&lt;reaction-mechanism&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6681" PostTypeId="2" ParentId="6680" CreationDate="2013-10-28T19:24:04.747" Score="1" Body="&lt;p&gt;Look at the difference between the enthalpies of formation. Not sure why you'd need oxygen, unless there's something I'm missing.&lt;/p&gt;&#xA;" OwnerUserId="2542" LastActivityDate="2013-10-28T19:24:04.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6682" PostTypeId="2" ParentId="6674" CreationDate="2013-10-28T21:41:20.657" Score="4" Body="&lt;p&gt;The question is interesting, but slightly ill-defined.&#xA;If you take s and p orbitals as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen-like_atom&quot; rel=&quot;nofollow&quot;&gt;solution of Schrodinger equation of hydrogen-like atom&lt;/a&gt;, then they are eigenfunctions of Hamiltonian, therefore are orthogonal. That means - no overlap. Do not be misled by the figure, if you do the math properly, $\langle1s|2p\rangle = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you speak of just one electron, the solution is reasonably simple as stated in &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen-like_atom&quot; rel=&quot;nofollow&quot;&gt;the article&lt;/a&gt; and the electron does not simultaneously reside in both orbitals. Even if you would think of interaction with excited states, it would not help, as single excitations do not interact with ground state. You can of course excite it, but than it is not &quot;simultaneous&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many-electron system, the picture gets more complicated, but this was not a question.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorDisplayName="user23061" LastEditDate="2015-12-17T16:26:45.330" LastActivityDate="2015-12-17T16:26:45.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6683" PostTypeId="1" CreationDate="2013-10-28T22:35:25.730" Score="3" ViewCount="183" Body="&lt;p&gt;What exact nomenclature rule define the naming of glucose 6-phosphate?&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="15410" LastEditDate="2016-05-23T07:21:16.760" LastActivityDate="2016-05-23T11:24:15.350" Title="Why is the nomenclature glucose 6-phosphate, not 6-phosphateglucose?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6684" PostTypeId="2" ParentId="4877" CreationDate="2013-10-28T23:28:24.957" Score="8" Body="&lt;p&gt;I believe dilute boiling hydrochloric acid is quite destructive to all flesh and bones. You don't want a concentrated acid as it has annoying oxidation effects which slow the reaction. You don't want hydrofluoric as its hideous dangerous to handle at all temperature (a small amount on your skin will be absorbed and lead to inevitable death if you don't inject calcium glutamate into you) and the formation of insoluble calcium fluoride (and others) slows everything down.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="402" LastEditDate="2017-01-22T09:58:16.267" LastActivityDate="2017-01-22T09:58:16.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6685" PostTypeId="2" ParentId="6683" CreationDate="2013-10-29T04:49:15.037" Score="2" Body="&lt;p&gt;Reading &lt;a href=&quot;http://www.chem.qmul.ac.uk/iupac/misc/phospho.html#p2&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; closely, technically, the compound should be 6-phosphonoglucose but glucose 6- phosphate is acceptable usage, probably derived from considering it as a salt analogue rather the ester that it is.   &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="2581" LastEditDate="2013-10-29T12:46:48.757" LastActivityDate="2013-10-29T12:46:48.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6686" PostTypeId="2" ParentId="6638" CreationDate="2013-10-29T11:10:10.290" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the contributing structures of the resonance below is more stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the question is asking is &quot;Which contributor is more important in describing the structure and behavior of the hybrid?&quot; Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-29T11:10:10.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6687" PostTypeId="1" CreationDate="2013-10-29T12:56:40.203" Score="1" ViewCount="650" Body="&lt;p&gt;How can &lt;a href=&quot;http://en.wiktionary.org/wiki/epimerization&quot; rel=&quot;nofollow&quot;&gt;epimerization&lt;/a&gt; during peptide synthesis be detected? And how can we change or modify the coupling conditions to stop epimerization occurring?  &lt;/p&gt;&#xA;" OwnerUserId="2583" LastEditorUserId="2328" LastEditDate="2013-10-31T21:08:33.727" LastActivityDate="2016-05-18T05:20:07.600" Title="Epimerization in peptide synthesis" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6688" PostTypeId="2" ParentId="6674" CreationDate="2013-10-29T16:18:42.257" Score="2" Body="&lt;p&gt;It is important to know that the overlap must be performed with the wave functions $\psi_j(\vec{r})$, not the density, $\rho_j(\vec{r}) = \psi_j(\vec{r})^* \psi_j(\vec{r})$. Wave functions have signs; they oscillate, as these oscillations encode the momentum distribution $\rho_j(\vec{p})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you calculate the overlap between $\psi_s(\vec{r})$ and $\psi_p(\vec{r})$, $\int_V \psi_s(\vec{r}) \psi_p(\vec{r}) dV$ ($V$ is the volume) you should remember that the lobes of the &quot;p&quot; wave function have different signs, while the spherical $s$ wave function keeps the same sign over the surface of the sphere. Therefore, the integral has two identical but opposite (different sign) contributions that cancel the overlap.&lt;/p&gt;&#xA;" OwnerUserId="1100" LastActivityDate="2013-10-29T16:18:42.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6689" PostTypeId="1" AcceptedAnswerId="6699" CreationDate="2013-10-29T16:37:44.570" Score="1" ViewCount="7298" Body="&lt;p&gt;I'm no chemist and I probably wouldn't understand any technical answer but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm investigating the environmental impacts of various materials as compared to PVC and this is one that crops up regularly as a more environmental alternative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately there does not seem to be any authoritative information on the web concerning this material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm beginning to think that this is a green-washing propaganda move by the phthalate/heavy metal free PVC industry as a move avoid mentioning PVC in product advertising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway I would welcome any information on what exactly PER is and what it is made from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="2584" LastActivityDate="2016-04-05T09:35:59.097" Title="What is Polymer Environmental Resin (PER)?" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6690" PostTypeId="1" AcceptedAnswerId="6706" CreationDate="2013-10-29T17:23:50.497" Score="13" ViewCount="62707" Body="&lt;p&gt;According to the definition of organic compounds on &lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_compound&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An organic compound is any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carbonates, simple oxides of carbon, and cyanide molecules do have carbon, but they are not regarded as organic.  The same Wikipedia article on organic compounds reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The modern meaning of &quot;organic compound&quot; is any one of them that contains a significant amount of carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current main definition for organic compounds doesn't fit for carbonates, simple oxides of carbon, steel, etc. So, what should be the redefined definition for organic compounds?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="11367" LastEditDate="2016-01-05T15:21:57.733" LastActivityDate="2016-02-29T10:29:14.150" Title="What is the definition of organic compounds?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6692" PostTypeId="1" CreationDate="2013-10-29T17:48:17.320" Score="6" ViewCount="4140" Body="&lt;p&gt;I've looked at the other similarly worded questions but I can't seem to make heads or tails out of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my case. I need to find the heat capacity of the calorimeter (in Joule/degree Celsius), and I've been provided with the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH (aq) + HCl (aq) -&amp;gt; NaCl (aq) + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Delta T = 11.5\ ^\circ \ \text{C}$&lt;/li&gt;&#xA;&lt;li&gt;$\Delta_r H = -58.3 \text{ kJ/mol}$&lt;/li&gt;&#xA;&lt;li&gt;$s_{p,\ce{NaCl}} = 3.91 \text{ J/(g}\cdot \ ^\circ \text{C)}$&lt;/li&gt;&#xA;&lt;li&gt;$d_{1\ \text{M NaCl}}=1.0037\text{ g/mL}$&lt;/li&gt;&#xA;&lt;li&gt;$V= 100 \text{ mL}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far I've calculated the mass of $\ce{NaCl}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = d\cdot V = (1.037 \text{ g/mL})(100 \text{ mL}) = 103.7 \text{ g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I calculated the heat capacity of $\ce{NaCl}$...at least I think that's what I did. I'm not even sure if I was supposed to do this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$q = m\cdot s_{p,\ce{NaCl}}\cdot \Delta T$&lt;/li&gt;&#xA;&lt;li&gt;$q = (103.7 \text{ g})(3.91 \text{J/(g} \cdot\  ^\circ \text{C}))(11.5 \ ^\circ \text{ C})$&lt;/li&gt;&#xA;&lt;li&gt;$q = 4662.8 \text{ J}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now I'm confused as to the next step, and I don't know how to use $\Delta_r H$ in this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be immensely appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Below I've added more things I've tried in my attempt to find the answer.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this equation in a textbook so I gave it a shot:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$q = - \Delta H\cdot n_{\text{limiting reactant}}$&lt;/li&gt;&#xA;&lt;li&gt;$q = (- 58,300 \text{ J/mol})(1.62 \text{ mol HCl})$&lt;/li&gt;&#xA;&lt;li&gt;$q = 94,446 \text{ J}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I don't know how/if this fits anywhere in the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another equation I tried: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$q = C_P\Delta T$&lt;/li&gt;&#xA;&lt;li&gt;$-58,300 \text{ J}= C (11.5 \ ^\circ \text{C})$&lt;/li&gt;&#xA;&lt;li&gt;$C = 58,300 \text{ J} / 11.5 \ ^\circ \text{C}= 5069.5 \text{ J/}\ ^\circ \text{C}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which is at least the correct unit of the heat capacity I'm supposed to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I tried subtracting the heat capacity I found for $\ce{NaCl} \ (4662.8 \text{ J})$ from the supposed heat capacity of the system $(5069.5 \text{ J})$, I get $406.76 \text{ J}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm kind of at a loss...&lt;/p&gt;&#xA;" OwnerUserId="2585" LastEditorUserId="4945" LastEditDate="2019-07-11T16:11:22.363" LastActivityDate="2019-07-11T16:11:22.363" Title="How to calculate the heat capacity of a calorimeter based on a neutralisation reaction?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6693" PostTypeId="1" AcceptedAnswerId="6719" CreationDate="2013-10-29T17:55:01.240" Score="0" ViewCount="2710" Body="&lt;p&gt;How you you calculate/estimate the pH of a solution with 3 components all of different pHs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a paper describing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;0.1M Buffer PCTP at pH 9.5&#xA;1.6M Sodium Dihydrogen Phosphate at pH 3.39&#xA;0.2M Sodium Acetate at pH 9.0255&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measured pH is ~5.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i take a weighted average i.e (0.1 * 9.5 + 1.6*3.39 + 0.2 * 9.0255)/ 1.9 the pH is estimated at 4.311.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the buffer &quot;drags&quot; up the pH but how to estimate this?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="1362" LastEditDate="2013-10-31T12:30:08.650" LastActivityDate="2013-10-31T12:30:08.650" Title="How to estimate pH?" Tags="&lt;homework&gt;&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6694" PostTypeId="1" AcceptedAnswerId="6698" CreationDate="2013-10-29T17:58:32.093" Score="8" ViewCount="12798" Body="&lt;p&gt;We know that intermolecular hydrogen bonding is seen in hydrogen peroxide $(\ce{H2O2})$. But is intramolecular H-bonding, i.e. hydrogen bond between an oxygen atom and a hydrogen atom attached to the adjacent oxygen atom possible? How strong will it be, compared to intermolecular H-bonding and why?&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorDisplayName="user7951" LastEditDate="2015-08-19T23:51:38.957" LastActivityDate="2015-08-20T00:12:07.893" Title="Does hydrogen peroxide exhibit intramolecular hydrogen bonding interactions?" Tags="&lt;inorganic-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6695" PostTypeId="1" AcceptedAnswerId="6696" CreationDate="2013-10-29T00:57:05.297" Score="0" ViewCount="738" Body="&lt;p&gt;Almost every share-house I've lived in has had some nasty looking solids collecting inside.  I always regret looking inside the kettle because a cup of tea never tastes quite as good after you've seen this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ATJJ2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;is&lt;/em&gt; that stuff?  What's the physical process that causes it to collect and why is it usually concentrated around the heating element?  Is it related to calcium in tap water, or is there any trick to prevent it appearing in your kettle?  Is it dangerous to ingest and should be removed, or is it harmless?&lt;/p&gt;&#xA;" OwnerUserId="648" OwnerDisplayName="wim" LastActivityDate="2021-05-26T08:18:13.267" Title="What is the solid crud that appears inside my kettle?" Tags="&lt;water&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6696" PostTypeId="2" ParentId="6695" CreationDate="2013-10-29T08:27:05.867" Score="5" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Limescale&quot;&gt;limescale&lt;/a&gt; and you have it because your tap water is rich in minerals (mainly, calcium, as you note). You could remove these minerals with an ion exchanger, but you wouldn't to it with a tap water since these minerals give it a better taste. Limescale is not dangerous to ingest, but certainly not pleasant either (and I see on the photo that there is a slot for a mesh that should filter scales). Thick layer of limescale is also bad for the heating element – it hinders heat transfer to water and leads to overheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a good idea to remove limescale once in a while. You can dissolve limescale with any weak acid, like vinegar, but vinegar smells, so I suggest buying a descaling agent.&lt;/p&gt;&#xA;" OwnerDisplayName="gigacyan" LastActivityDate="2013-10-29T08:27:05.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6698" PostTypeId="2" ParentId="6694" CreationDate="2013-10-29T20:52:26.283" Score="12" Body="&lt;p&gt;&lt;strong&gt;Preamble:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives.&#xA;I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$.  In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric).  For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital.  I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methods:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure &lt;strong&gt;A&lt;/strong&gt;). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions.  The basis set is denoted henceforth as haTZ.  The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$.  The frozen-core approximation was invoked for all computations.&#xA;A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density.&#xA;This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure &lt;strong&gt;B&lt;/strong&gt;).&#xA;All computations were performed with the G09 (Rev. D01) software package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Results:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure &lt;strong&gt;A&lt;/strong&gt; is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure &lt;strong&gt;B&lt;/strong&gt; is a transition state with one imaginary mode of vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NBO analysis of structure &lt;strong&gt;A&lt;/strong&gt; reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity.  The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure &lt;strong&gt;A&lt;/strong&gt;.  The occupation number for this antibond, 0.00351, indicates a very small overlap.  The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$.  We note that this is merely an estimated measure of the strength of the interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sXsLC.jpg&quot; alt=&quot;H2O2 nsigma* overlap&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure &lt;strong&gt;B&lt;/strong&gt;), the $n\rightarrow\sigma^{*}$ interaction disappears.  The intramolecular hydrogen bonding  may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure &lt;strong&gt;A&lt;/strong&gt;) as the transition state, structure &lt;strong&gt;B&lt;/strong&gt;, does not contain this interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward.  Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex.  Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number.  To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond.   The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds.  These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex.&#xA;The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDENDUM:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to avoid calling this intramolecular interaction a hydrogen bond.  Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$.  If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond.  Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upon reflection of these issues, I conclude that there is, in fact, &lt;em&gt;no intramolecular hydrogen bonding&lt;/em&gt; in H$_2$O$_2$.&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2013-12-05T18:09:39.357" LastActivityDate="2013-12-05T18:09:39.357" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6699" PostTypeId="2" ParentId="6689" CreationDate="2013-10-29T22:24:47.010" Score="6" Body="&lt;p&gt;PER is essentially PVC which has been plasticized and stabilized with acetyl tributyl citrate, instead of the host of phthalate-based plasticizers such as bis-diethylhexyl phthalate (DEHP) whose endocrine-disrupting properties give PVC its well-deserved bad reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citrate salts tend to be readily water soluble, and therefore this material is touted to be more biodegradeable than phthalate-plasticized PVC. I certainly cannot find any information to refute that claim. ATBC will further degrade into other acetates and butyrates, most of which are relatively common and harmless in the environment (though butyl compounds are infamously stinky; butyric acid is a favorite less-lethal offensive-odor weapon of law enforcement and groups like the Sea Shepherds). All PVC will eventually degrade as the plasticizer leaches out over time, unlike other types of plastics that are so ridiculously inert they'll likely be the fossils of our modern era, but again, the phthalates just aren't a great idea to have in the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, neither is the vinyl chloride monomer. What many proponents of PER fail to mention is that the monomer is worse for your health than the plasticizers. It's carcinogenic in relatively small doses, for one thing. It's also toxic to the liver in even smaller doses, measured in the low parts per million of body mass for an acute exposure. Lastly, the monomer is extremely flammable, though this is mainly an industrial hazard; the polymer is actually among the more flame-retardant commercial plastics, due to its chlorine content (in other words it's already highly oxidized, as opposed to something like polyethylene). None of this changes when the plasticizer does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, while PER is a step forward, there are still plenty of inherent hazards in the production, long-term use and disposal of PER products. PER is also a relative specialty product, whose primary use is a foam product marketed for children's toys and yoga mats. It's not a silver bullet to replace PVC wholesale. &lt;/p&gt;&#xA;" OwnerUserId="367" LastActivityDate="2013-10-29T22:24:47.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6701" PostTypeId="2" ParentId="6676" CreationDate="2013-10-30T02:02:08.353" Score="3" Body="&lt;p&gt;I'm going to assume that you don't need an explanation for the checkboxes &lt;code&gt;the same molecule&lt;/code&gt; and &lt;code&gt;difference molecules&lt;/code&gt;, so that leaves the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &lt;strong&gt;constitutional isomers&lt;/strong&gt; and &lt;strong&gt;stereoisomers&lt;/strong&gt; make up two broad categories of isomers (molecules with the same chemical formula).  Constitutional isomers typically have different connectivities and stereoisomers have the same connectivities but differ in spatial arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;enantiomers&lt;/strong&gt; are stereoisomers that are non-superimposeable mirror images of one another.  Your hands are (roughly) enantiomers.  &lt;strong&gt;both are optically active&lt;/strong&gt; is true of enantiomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;diastereomers&lt;/strong&gt; also known as geometric isomers are also stereoisomers that are not mirror images of one another. cis- and trans-dichloroethene are diastereomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do a web image search of isomer flowchart you'll see a lot of informative information on isomers.  Linked below is the public domain version from &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; which is an example (although not the best in my opinion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rU4ZK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-10-30T02:02:08.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6702" PostTypeId="1" CreationDate="2013-10-30T06:50:33.703" Score="2" ViewCount="5615" Body="&lt;p&gt;So I have several descriptions of reactions and I have to give the net ionic equation. Are these just the equations or are they the proper net ionic ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid calcium is added to warm water. &#xA;$$\ce{Ca_{(s)} + 2 H_{2}O_{(l)} -&amp;gt; Ca^{2+}_{(aq)} + 2 OH_{(aq)}^- + H2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powdered magnesium oxide is added to a container of carbon dioxide gas. &#xA;$$\ce{MgO_{(s)} + CO2_{(g)} -&amp;gt; MgCO3_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I do those right, or am I on the wrong track?&lt;/p&gt;&#xA;" OwnerUserId="2589" LastEditorUserId="95133" LastEditDate="2021-05-25T12:43:07.420" LastActivityDate="2021-05-25T12:43:07.420" Title="Reacting calcium with warm water and magnesium oxide with carbon dioxide, what are the products?" Tags="&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6703" PostTypeId="2" ParentId="6702" CreationDate="2013-10-30T07:51:51.210" Score="1" Body="&lt;p&gt;Purpose of ionic equation is to describe ions in (aqueous) solution, without the need to bother about details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation looks reasonable, you could possibly end up with $\ce{Ca(OH)_2}$, which is not very water soluble, but it's IMHO the type of detail you should not bother with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is more difficult, though. At first, there is no solution to speak about, therefore the request for ionic equation itself is invalid. The reaction written is correct, as long as it describes possibility. If it would really happen is question of reaction conditions, therefore totally outside the scope of homework. As such, I'd say, correct.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-30T07:51:51.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6704" PostTypeId="1" AcceptedAnswerId="6705" CreationDate="2013-10-30T08:50:36.653" Score="0" ViewCount="249" Body="&lt;p&gt;Let us consider a non-isothermal process inside a metal container. If we somehow are able to make the temperatures on the inside and outside the system remain same throughout the process, can it be called an adiabatic process?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastActivityDate="2013-10-30T10:55:46.343" Title="Can we have an adiabatic process in a sealed metal container?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6705" PostTypeId="2" ParentId="6704" CreationDate="2013-10-30T10:55:46.343" Score="1" Body="&lt;p&gt;&lt;code&gt;Adiabatic&lt;/code&gt; means that energy is transferred between the system and the surroundings only as work and not as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at your process. You have a non-isothermal process, so $\Delta T_{sys} \ne 0$ for the process. An adiabatic process does not need to be isothermal, so this condition is okay. What makes your process non-adiabatic is the condition that constant temperature is somehow maintained (likely through a constant temperature bath which serves as the surroundings). This requires heat exchange between the system inside the metal container and the surroundings outside the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the constraint that the overall temperature is held constant, then you can have an adiabatic process. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-30T10:55:46.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6706" PostTypeId="2" ParentId="6690" CreationDate="2013-10-30T11:30:19.050" Score="19" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, what should be the redefined definition for organic compounds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We should not define organic compounds. As F'x pointed out in the comments, the distinction between organic and inorganic is largely arbitrary and mostly based on a distortion of historical precedent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, we should classify compounds by the type(s) of bonding interaction present in the substance. By this criterion, we have several classes of compounds:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Molecular (or covalent) compounds - the particles that make up these compounds are molecules held together by covalent bonds. This class of compound includes traditionally inorganic $\ce{CO2}$ and organic $\ce{CH4}$ carbon compounds. Covalent compounds also the majority of &lt;a href=&quot;http://en.wikipedia.org/wiki/Organometallic_chemistry&quot;&gt;organometallic compounds&lt;/a&gt;, so we do not need a separate category for these. Other traditionally &quot;inorganic&quot; compounds, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass&quot;&gt;glass&lt;/a&gt;, also belong in this category.&lt;/li&gt;&#xA;&lt;li&gt;Ionic compounds - the particles that make up these compounds are largely ions held together by Coulombic forces. For the sake of simplicity most people would include in this class of compounds those ionic compounds where one or both ions is a small molecular ion held together by covalent bonds $($e.g. the $\ce{NH4^+}$ ion in $\ce{NH4Cl}$ or the  $\ce{CO3^{2-}}$ ion in $\ce{Na2CO3})$. &lt;/li&gt;&#xA;&lt;li&gt;Mixed covalent/ionic compounds - those in which there are some ionic functional groups, but mostly covalent bonds, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_orange&quot;&gt;methyl orange&lt;/a&gt;. Many folks might list these substances in Category 1. This category would include most &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_complex&quot;&gt;coordination compounds&lt;/a&gt; including such things as &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal-Organic_Framework&quot;&gt;metal-organic frameworks&lt;/a&gt; and probably many composite and/or ceramic materials.&lt;/li&gt;&#xA;&lt;li&gt;Metallic compounds - those substances with extremely delocalized bonding, typical of &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bond&quot;&gt;metals&lt;/a&gt;. Alloys, including steel, belong in this category. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Under this system, what we traditionally study as &lt;code&gt;organic&lt;/code&gt; chemistry becomes redefined as the study of the structure, properties, and reactions of covalent compounds (Classes 1 and 3), especially the nature of the covalent bond and how it can be formed and broken selectively. What you might learn about the structure, properties, and reactions of  &lt;code&gt;organic compounds&lt;/code&gt; can easily be extended to the structure, properties, and reactions of molecular &lt;code&gt;inorganic compounds&lt;/code&gt;. For example, we do not need two different branches of chemistry to study &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzene&quot;&gt;benzene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Borazine&quot;&gt;borazine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnpUb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXVcs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-30T11:30:19.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6707" PostTypeId="2" ParentId="6641" CreationDate="2013-10-30T13:33:52.133" Score="1" Body="&lt;p&gt;You're a bit muddled up.&lt;br&gt;&#xA;Let's say I have the following reaction:&#xA;$$\ce{A &amp;lt;=&amp;gt; B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This signifies that &lt;em&gt;one molecule&lt;/em&gt; of $A$ reacts to give &lt;em&gt;one molecule&lt;/em&gt; of $B$. The molecularity of this reaction is 1.&#xA;The equilibrium constant (say $K_{eq}$) is &lt;em&gt;defined&lt;/em&gt; as the ratio of concentration of products to that of reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. $K_{eq} = \frac{[B]}{[A]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a reaction is written as:&#xA;$$\ce{2A &amp;lt;=&amp;gt; 2B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This signifies that &lt;strong&gt;&lt;em&gt;two molecules of A&lt;/em&gt;&lt;/strong&gt; react with each other to produce &lt;strong&gt;&lt;em&gt;two molecules of B&lt;/em&gt;&lt;/strong&gt;. It is &lt;em&gt;not the same&lt;/em&gt; as our first reaction!&lt;br&gt;&#xA;The equilibrium constant (say $K'_{eq}$) will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K'_{eq} = \frac{[B]\cdot[B]}{[A]\cdot[A]} = \frac{[B]^2}{[A]^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant is used to get an idea of how much the reaction is favoured. If $K = 13$, then the concentrations of the products are &lt;em&gt;thirteen&lt;/em&gt; times more than the concentrations of reactants. Therefore, any value of $K$ greater than one can signify that the reaction wants to proceed forward. &lt;/p&gt;&#xA;" OwnerUserId="1858" LastEditorUserId="1858" LastEditDate="2013-11-03T04:58:06.770" LastActivityDate="2013-11-03T04:58:06.770" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6709" PostTypeId="1" CreationDate="2013-10-30T17:06:33.927" Score="13" ViewCount="3544" Body="&lt;p&gt;I read in several papers that DFT is notoriously bad at describing free oxygen molecules but in none of them an explanation or reference concerning the problem has been provided. I guess this has something to do with its triplet ground state but I haven't come across a good explanation yet. Could you help me out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: If someone wants to know the papers I'm refering to, here are the links and the passages that are most relevant (both groups use gradient-corrected (GGA) density functionals, so maybe this isn't a general problem of DFT but only of GGA- or LDA-functionals):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0301010405002053&quot;&gt;Rossmeisl et al.&lt;/a&gt; have devised a method for calculating the Gibbs Free Energy of the intermediate in the oxygen evolution reaction. The intermediates are treated with DFT but in order to calculate the Free Energy of the final products, including desorbed oxygen molecules, they use tabulated values.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The free energy change of total&#xA;  reaction: $\ce{H2O -&amp;gt; 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are other groups studying water splitting that use the experimental value too, instead of calculating it with DFT.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The paper by &lt;a href=&quot;http://prb.aps.org/abstract/PRB/v62/i10/p6158_1&quot;&gt;Bloechl&lt;/a&gt; includes calculations of the binding energies of a hydrogen molecule, which are in rather good agreement with experiment (4.338 eV (DFT) vs. 4.448 eV (exp.)), and of an oxygen molecule, where the DFT value is off by 0.795 eV.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The binding energy of an oxygen molecule is calculated to be 5.912 eV, including the zero-point vibartion energy of 0.110 eV, which is comparable to the atomization energy of 5.906 eV, obtained with the Becke-Perdew-Wang gradient-corrected density functional. At 5.116 eV, the experimental binding energy is substantially smaller (by 0.795 eV) than the theoretical prediction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2013-10-31T02:03:19.513" LastActivityDate="2021-05-03T23:05:58.937" Title="Why is Density Functional Theory notoriously bad at describing oxygen molecules?" Tags="&lt;density-functional-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6710" PostTypeId="1" AcceptedAnswerId="6711" CreationDate="2013-10-30T18:52:55.467" Score="6" ViewCount="445" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to IUPAC recommendations for naming of inorganic compounds, the more electropositive element is written first with a suitable prefix, and the more electronegative element is written at the last with a suffix -&lt;em&gt;de&lt;/em&gt; and a prefix. For example, $\ce{H2O}$ (Water) can be called &lt;em&gt;dihydrogen monooxide&lt;/em&gt;. [this is an modified statement extracted from the wiki encyclopedia-&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;IUPAC nomenclature of inorganic chemistry&lt;/a&gt;]&lt;br&gt;&#xA;  In the above example, $\ce{H}$ (Pauling scale = 2.20 or Allen scale = 2.300) is less electronegative than $\ce{O}$ (Pauling scale = 3.44 or Allen scale = 3.610). So, $\ce{H}$ is placed first, followed by $\ce{O}$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If suppose we were needed to name the compounds containing  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Ru}$ (Pauling scale = 2.2) and $\ce{Os}$ (Pauling scale = 2.2)  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{Mo}$ (Allen scale = 1.47) and $\ce{W}$ (Allen scale = 1.47)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;, in what order should we write the elements?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use pauling scale as referance, we would face difficulty in naming $\ce{Ru}$ and $\ce{Os}$, if we use allen scale, we would be in trouble with respect to the elements like $\ce{Mo}$ and $\ce{W}$, etc, as we can see that elements have same electronegativity value. I don't know whether the compounds with the above mentioned elements exist or not, but if we suppose, assume that, they exist, in what order should they be named? &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorDisplayName="user7951" LastEditDate="2015-08-14T19:15:04.230" LastActivityDate="2015-08-14T19:15:04.230" Title="In what order should the compounds containing Ru and Os, or Mo and W be named?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6711" PostTypeId="2" ParentId="6710" CreationDate="2013-10-30T21:45:45.317" Score="8" Body="&lt;p&gt;When IUPAC rules refer to electronegativity (and electropositivity), they don't mean &lt;em&gt;actual&lt;/em&gt; electronegativity measured by one or another scale. IUPAC decided, for nomenclature's sake, on a &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry_2005#Sequencing_elements.E2.80.94the_.22electronegativity.22_list&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;formal&lt;/em&gt; ordering of electronegativity&lt;/a&gt;, which goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;    &lt;img src=&quot;https://i.stack.imgur.com/VCSzWl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is Table VI, on page 260 of IUPAC's very own 2005 &lt;a href=&quot;http://media.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Red Book&lt;/a&gt;, aka &lt;em&gt;Nomenclature of inorganic chemistry&lt;/em&gt;. The cases in which it is used are detailed in IR-2.15.3.1 (page 42).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation of the changes from earlier IUPAC recommendations is in IR-1.6.3 of the Red Book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The element sequence of Table VI&lt;/strong&gt;&lt;br&gt;&#xA;  In &lt;em&gt;Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990&lt;/em&gt; (Ref. 11), the position of oxygen in certain element sequences was treated as an exception. Such exceptions have been removed and the element sequence of Table VI is now strictly adhered to. In particular, oxygen is treated as the electropositive component relative to any halogen for constructing compositional names (Section IR-5.2) and corresponding formulae (Section IR-4.4.3) for binary compounds. This results in, for example, the formula O2Cl and the name dioxygen chloride rather than the formula ClO2 and the name chlorine dioxide.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In Ref. 11, the formulae for intermetallic compounds were also subject to an exceptional rule although no guidance was given for naming such compounds, and the term ‘intermetallic compound’ was not defined. The problem is to define the term ‘metal’. Therefore, no attempt is now made to make a separate prescription for either the formulae or the names of intermetallic compounds. It is stressed, however, that the present recommendations allow some flexibility regarding formulae and compositional names of ternary, quaternary, etc. compounds. Several ordering principles are often equally acceptable (see Sections IR-4.4.2 and IR-4.4.3).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The element sequence of Table VI is also adhered to when ordering central atoms in polynuclear compounds for the purpose of constructing additive names (see Section IR-1.6.6).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For more on this topic, you've got to go read (and upvote) &lt;a href=&quot;https://chemistry.stackexchange.com/a/538/30&quot;&gt;this excellent answer by Manishearth&lt;/a&gt;, which is where I first learnt that useful and intriguing nugget of IUPAC idiosyncrasy!&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2013-10-30T21:53:40.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6712" PostTypeId="1" CreationDate="2013-10-30T22:04:19.293" Score="1" ViewCount="366" Body="&lt;p&gt;Silicon has been known to form analogues of organic compounds. For example, silicon with 4 hydrogen atoms forms silane as opposed to methane. This continues up the alkane list, with disilane for ethane, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silanes also have with hydroxyl groups to form &quot;silanol.&quot; Thus, disilanol is the silicon analogue of ethanol. ($\ce{H3Si-SiH2OH}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How toxic is disilanol?&lt;/p&gt;&#xA;" OwnerUserId="2592" LastEditorUserId="22" LastEditDate="2013-10-31T06:58:57.080" LastActivityDate="2013-12-30T12:42:43.023" Title="How toxic is disilanol?" Tags="&lt;inorganic-chemistry&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6713" PostTypeId="2" ParentId="853" CreationDate="2013-10-30T23:40:33.647" Score="5" Body="&lt;p&gt;This is an area of corrosion science. Chloride ions do not aid general corrosion  of stainless but rather the special processes called crevice corrosion and pitting corrosion. General corrosion is of large areas. Crevice corrosion is like　a hole in the metal and local corrosion effects occur. Pitting corrosion is corrosive fluid forms its own hole and similar local corrosion effects occur. The local corrosion effects for chloride crevice and pitting corrosion corrosion are complex and dependant on the fluid but a major one is the increase in acidity from  ferric ions (which normal complex with water to be acid) as the chloride ions displace a water molecule and the new complex is more acidic. This acidity damages the complex chromium iron oxide layers. Another effect is that the chloride penetrates the oxide layer and opens it up to allow corrosive species mainly oxygen to penetrate to base metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickel gives some benefit to corrosion protection of  stainless steels by substituting for iron in the iron-chromium-oxide but that's not main reason its included but rather allows the stainless steel to be austenitic- basically more ductile and easier to weld.&#xA;Molybdenum is main element that is added to improve pitting/corrosion resistance. 304 is the most common basic number stainless steel 16-18% Chromium and 8-10% Nickel  (they add letters to give more information) . 316 is second as it contain  16-18% Chromium, 8-10% Nickel and 2-3% Molybdenum- greatly reduces pitting corrosion- conditions need to be more severe to have pitting and crevice corrosion.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-10-30T23:40:33.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6714" PostTypeId="2" ParentId="5134" CreationDate="2013-10-30T23:47:43.647" Score="3" Body="&lt;p&gt;By definition, rust includes water in its structure. Its not rust if it doesn't contain water.&lt;br&gt;&#xA;Rust is porous- it has holes so corrosion of iron keeps going a lot longer if it rusts rather it forms a straight oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron forms oxides instantly which are more protective than rust- not a lot more. Hot iron in oxygen environments corrodes by forming iron oxide.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-10-30T23:47:43.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6715" PostTypeId="2" ParentId="1215" CreationDate="2013-10-31T00:02:11.747" Score="1" Body="&lt;p&gt;While virtually all metals from a basic oxide or amphoteric (reacts with both bases and acids) oxide like aluminium and both Tin(II) and Tin(IV), some also form acidic oxides like Chromium(VI) oxide, Iron(II) oxide.  With oxides its better to describe they according to their reactivity with acids and bases as you are not trying to convert them to the hydrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not incredible for Polonium to form only an acid oxide.    &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-10-31T00:02:11.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6716" PostTypeId="1" AcceptedAnswerId="6717" CreationDate="2013-10-31T01:24:07.347" Score="2" ViewCount="2664" Body="&lt;p&gt;If $15.0\ \mathrm{mol}$ of nitrogen are reacted with $30.0\ \mathrm{mol}$ of hydrogen, how much ammonia will be produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2 + 3H2 -&amp;gt; 2NH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I’ve tried is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{15.0\ \mathrm{mol}}{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{x}{30\ \mathrm{mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not sure how to proceed from here. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2594" LastEditorDisplayName="user7951" LastEditDate="2016-06-30T16:24:55.083" LastActivityDate="2016-06-30T16:35:23.123" Title="Amount of substance produced in chemical reaction" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6717" PostTypeId="2" ParentId="6716" CreationDate="2013-10-31T04:44:38.903" Score="4" Body="&lt;p&gt;In the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N_2 + 3H_2 \to 2NH_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$&#xA;From this&#xA;$$x=\frac{30}{3}=10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="7475" LastEditDate="2016-06-30T16:35:23.123" LastActivityDate="2016-06-30T16:35:23.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6718" PostTypeId="2" ParentId="6712" CreationDate="2013-10-31T10:56:45.873" Score="3" Body="&lt;p&gt;Difficult to say. After a fair amount of Googling, I was unable to find any reliable sources on the compound you ask about, $\ce{H3Si-SiH2OH}$. From the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Silanol&quot; rel=&quot;nofollow&quot;&gt;silanols&lt;/a&gt;, it appears that the simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Silanol#Parent_silanols&quot; rel=&quot;nofollow&quot;&gt;silanols are not stable&lt;/a&gt;, but no reference is provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It stands to reason that if $\ce{H3SiOH}$ is not stable, there is a good chance that $\ce{H3Si-SiH2OH}$ is also not very stable. Disiloxane $\ce{H3Si-O-SiH3}$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Disiloxane&quot; rel=&quot;nofollow&quot;&gt;much more stable&lt;/a&gt;, and silanols likely convert into siloxanes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good knowledge about the health hazards of a substance requires thorough investigation, which is not possible when the substance is not stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkyl silanols, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethylsilanol&quot; rel=&quot;nofollow&quot;&gt;trimethylsilanol&lt;/a&gt; are stable and used for a variety of applications. Acutley, &lt;a href=&quot;http://www.oakwoodchemical.com/ShowMSDS.aspx?Number=078729&quot; rel=&quot;nofollow&quot;&gt;trimethylsilanol(pdf)&lt;/a&gt; is a volatile liquid that irritates skin, eyes, and mucus membranes (e.g. the lungs). The language on the MSDS is &quot;may be harmful if...&quot; This language suggests that you want to avoid contact with it but that some exposure will not cause serious harm. Note the &lt;a href=&quot;http://www.nafaa.org/ethanol.pdf&quot; rel=&quot;nofollow&quot;&gt;msds for ethanol(pdf)&lt;/a&gt; has much scarier language.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-10-31T10:56:45.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6719" PostTypeId="2" ParentId="6693" CreationDate="2013-10-31T12:15:54.297" Score="1" Body="&lt;p&gt;You can't just take an average of pHs- supposing it was a simple dissolution question rather than one with chemical reactions. pH=-log(H). H=10^(-pH)  First convert all the pH values to concentrations of hydrogen ions and then take an average of those values and then convert to pH- that will at least get you close the correct value.   &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="2581" LastEditDate="2013-10-31T12:24:44.037" LastActivityDate="2013-10-31T12:24:44.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6720" PostTypeId="1" AcceptedAnswerId="6721" CreationDate="2013-10-31T13:15:22.013" Score="26" ViewCount="56517" Body="&lt;p&gt;This isn't particularly a question about chemistry itself, but more-so the prospect of chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of opportunities arise when you overlap chemistry and computer science? I'm a computer science major, but prior to going to university I had (and still have) a deep passion for chemistry. My highest marks were always in chemistry, but I decided to go down the compsci route because likewise it is a passion of mine. An opportunity has presented itself to double major in compsci and chemistry and I'm curious to know whether it's worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of chemistry is fairly barebones (a limitation of highschool education), but as I said, it was my strongest subject and for some reason it just clicks with me. I'd often spent sleepless nights perusing chemistry textbooks and online periodicals because it fascinates me to no end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for greater job prospects (I already work in both compsci as a programmer and chemistry in an electronic cigarette company), I'm more-so looking for a more fulfilling pursuit of knowledge. While irrelevant, my university recently opened a new building dedicated primarily to chemistry, so facilities are recent and faculty is adept and education is modern. I'd very much like to take advantage of the opportunity, but I'm just unsure if a university-level chemistry program will scratch that itch for me.&lt;/p&gt;&#xA;" OwnerUserId="2596" LastEditorUserId="4945" LastEditDate="2015-10-27T07:06:02.827" LastActivityDate="2018-11-12T01:30:48.777" Title="Overlap between computer science and chemistry?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6721" PostTypeId="2" ParentId="6720" CreationDate="2013-10-31T13:55:38.980" Score="17" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Category%3aComputational_chemistry_software&quot;&gt;list of computational chemistry software&lt;/a&gt;. It gives you overview, how the computers are used in chemistry. Many of them fall more into the chemical physics field, some into the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cheminformatics&quot;&gt;chemoinformatics&lt;/a&gt;, to name just few. So definitely there are lot of places to contribute to use of computers in chemistry. If you are able to mix the software engineering and the science itself in a good way, possibilities are immense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the last question - almost whatever field of chemistry you choose, you can contribute to the programs used within it. There is no such thing like completed scientific code. And university level is a necessary prerequisite to do reasonable science. Just concentrate on the models and mathematics behind the scenes and everything will follow. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-10-31T13:55:38.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6722" PostTypeId="1" AcceptedAnswerId="6724" CreationDate="2013-10-31T17:09:43.733" Score="2" ViewCount="1376" Body="&lt;p&gt;So, I am attempting to calculate the equilibrium constant for two separate reactions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A + B&amp;lt;=&amp;gt;C + D}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A + B &amp;lt;=&amp;gt; C + E}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have figured out the final pressures at equilibrium for &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{D}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{E}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I was wondering if the equilibrium constant is calculated as normal using just the final pressures and molar ratios in the equations, or if the final pressures have to be adjusted somehow to compensate for the fact that the second reaction is also consuming the same products and producing one of the same reactants.&lt;/p&gt;&#xA;" OwnerUserId="2597" LastEditorUserId="95133" LastEditDate="2021-05-21T07:32:27.027" LastActivityDate="2021-05-21T07:36:11.757" Title="Calculating the equilibrium constants for parallel reactions" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6723" PostTypeId="1" AcceptedAnswerId="6726" CreationDate="2013-10-31T17:17:23.810" Score="23" ViewCount="11559" Body="&lt;p&gt;Why does water dissociate to&#xA;$\ce{H_{3}O^{+} + OH^{-}}$&#xA;instead of $\ce{H^{+} + OH^{-}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question came to surface when I was learning about acids and bases, and learned this definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{pH=}-\log_{10}[\ce H^{+}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which after looking at wikipedia, looks like a simplified definition, but it got me thinking - how can water have a pH of around seven if it dissociates into $\ce{H_{3}O^{+}}$ instead of $\ce{H^{+}}$? Why does it do this instead of the expected dissociation?&lt;/p&gt;&#xA;" OwnerUserId="842" LastEditorUserId="2328" LastEditDate="2013-10-31T21:08:38.400" LastActivityDate="2015-06-26T15:29:04.213" Title="Why does water dissociate to H3O+ and OH- rather than H+ and OH-?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6724" PostTypeId="2" ParentId="6722" CreationDate="2013-10-31T19:41:50.057" Score="2" Body="&lt;p&gt;You can combine the two chemical reactions into a single global reaction scheme by adding them:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{A + B&amp;amp;&amp;lt;=&amp;gt;C + D}\tag{1}\\&#xA;\ce{A + B&amp;amp;&amp;lt;=&amp;gt;C + E}\tag{2} \\ \hline&#xA;\ce{2A + 2B &amp;amp;&amp;lt;=&amp;gt; 2C + D + E}   \tag{Global}&#xA;\end{align}&lt;/span&gt;&#xA;and the equilibrium constant for this overall reaction scheme is equal to the products of the two equilibrium constants:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;K_\mathrm{eq,global} &amp;amp;= K_\mathrm{eq,1} K_\mathrm{eq,2} \\&#xA;&amp;amp;= \frac{\ce{[C]^2[D][E]} } {\ce{[A]^2[B]^2}}&#xA;\end{align}&lt;/span&gt;&#xA;(where &lt;span class=&quot;math-container&quot;&gt;$\ce{[i]}$&lt;/span&gt; denotes either molar concentration or partial pressure of component i). You do not need to adjust the final total pressure or mole amounts.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="95133" LastEditDate="2021-05-21T07:36:11.757" LastActivityDate="2021-05-21T07:36:11.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6725" PostTypeId="2" ParentId="6720" CreationDate="2013-10-31T19:59:45.433" Score="11" Body="&lt;p&gt;Computational chemistry is a huge field. The 2013 Nobel Prize in chemistry was awarded to researchers who laid the foundations for modern computational chemistry to advance forward, and is essentially a commentary by the Chemistry Nobel Prize committee on the importance of computational chemistry. Computer models are particularly critical for developing an understanding of systems that are hard (or impossible) to access experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This essentially involves creating mathematical models for how molecules and atoms behave, writing programs to solve them, and running them on (usually massive) computers. Molecular dynamics is a common activity, and involves simulating a handful of molecules that follow a handful of rules (how do they collide, how do they spin, how do they attract or repel one another, etc.). The more computing power you have, the more molecules you can simulate, and so the better your extrapolations to larger scales are. Simulations that are run on supercomputers can provide a direct link between the microscopic (molecules) and the macroscopic (fluid properties, like viscosity or conductivity). Writing programs that will run (and run efficiently) on supercomputers involves a lot of moving parts, most of which are pure computer science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a computer science background and you enter a chemistry degree program, or go to work at a chemical company where you're working with chemists, you'll have an edge because of your computational skills, and will have the intuition to think about problems in a computational way - the value of which can't be overstated. As mentioned above, computer models are used often in chemistry, so programming fluency is important. On the other hand, you'll have plenty of catching up to do on the math and physics, which is a different kind of intuition you may not have that others will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;National laboratories and universities do a significant amount of computational chemistry work, less so in industry R&amp;amp;D (caused by, among other reasons, lack of access to enough computing resources and difficulty of obtaining immediately applicable results). Pharmaceutical companies are the 800 pound gorillas of industrial computational chemistry. If you're looking for work in industry, though, consider chemical engineering. Chemical engineering is another huge field with an equally huge number of opportunities (read: low-hanging fruit) for someone with a computational bent.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2013-10-31T19:59:45.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6726" PostTypeId="2" ParentId="6723" CreationDate="2013-10-31T20:13:51.927" Score="14" Body="&lt;p&gt;You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly &lt;strong&gt;self&lt;/strong&gt; but a matter of &quot;couple-ionization&quot;: you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen bonding. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/fO0vE.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;From there the electron from Hydrogen forms a lone pair with the oxygen of the same molecule forming $\ce{OH^-}$. Hydron ($\ce{H^+}$) is in fact a proton, a very reactive specie so is not so simple found it uncoupled and moreover the hydrogen is suppose to be bonded with an hydrogen bonding with the other water molecule so is more accurate use hydronium ($\ce{H_3O^+}$) instead of hydron...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you don't have to worry about it for your pH calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{pH=}-\log_{10}[\ce {H^{+}}]= -\log_{10}[\ce {H_3O^{+}}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, in fact, you have the same ratio between basic species ($\ce{OH^-}$) and acid species( $\ce{H_3O^+}$ or $\ce{H^+}$ as you prefer) in both cases! &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T14:09:47.230" LastActivityDate="2015-06-26T14:09:47.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6727" PostTypeId="2" ParentId="6568" CreationDate="2013-10-31T22:38:28.783" Score="3" Body="&lt;p&gt;Beside construction of a solar power driven DIY cooler for beverages (like &lt;a href=&quot;https://www.youtube.com/watch?v=ToH27OK_CgY&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), G M's post directing to use the heat of evaporation as a vehicle to cool a liquid may be indeed advantageous.  What about using molar sieves in an outer shell of a barrel -- again, at the place of consumption, no electricity is needed. Yet, CoolKeg claims to be scaleable (7 to 20 L) and both barrel and the active ingredient (zeolithe) is able to be recycled. A &lt;a href=&quot;https://www.youtube.com/watch?v=s_HFZff69yQ&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; describes how this may be implanted into a production chain, this &lt;a href=&quot;http://www.brewtique.com.au/kegs-draught-beer/coolkeg/55&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; around beer provides some to read to understand the function. &lt;br/&gt;&#xA;Of course, to regenerate the molar sieves, alternative / more ecofriendly sources of heat may be used, too.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2013-10-31T22:38:28.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6728" PostTypeId="1" CreationDate="2013-11-01T01:19:00.617" Score="1" ViewCount="8496" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many p-orbitals are there in each period?&lt;br&gt;a. 1&lt;br&gt;b. 3&lt;br&gt;c. 5&lt;br&gt;d. 7&lt;br&gt;e. 14&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I take this to mean &quot;how many p-orbitals are there in each period of the periodic table?&quot; Looking at the p-block on the periodic table, it looks like there are 6 elements with p-orbitals per period. This isn't an answer, so I'm a little lost.&lt;/p&gt;&#xA;" OwnerUserId="2600" LastEditorUserId="5026" LastEditDate="2018-02-24T02:27:56.277" LastActivityDate="2018-02-24T02:27:56.277" Title="How many p-orbitals are there in each period?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6729" PostTypeId="1" CreationDate="2013-11-01T02:21:45.897" Score="4" ViewCount="159" Body="&lt;p&gt;This is a question from the College Entrance Test in China.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a closed, constant temperature container, reaction &lt;span class=&quot;math-container&quot;&gt;$\ce{A(s)&amp;lt;=&amp;gt;B +C(g)}$&lt;/span&gt; reached chemical equilibrium. If the volume of the container is reduced, when the equilibrium is reached again, the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; equals the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; before the reduction.&#xA;Is it possible that &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; is in gas state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The official answer is impossible.&lt;/p&gt;&#xA;&lt;p&gt;And the official explanation to this is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; is in gas state, we have &lt;span class=&quot;math-container&quot;&gt;$K=C_\ce{B} \cdot C_\ce{C}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Otherwise, if &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; is in liquid or solid state, we have &lt;span class=&quot;math-container&quot;&gt;$K=C_\ce{C}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; does not change after the reduction, in order to remain the &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; value unchanged, &lt;span class=&quot;math-container&quot;&gt;$C_\ce{B}$&lt;/span&gt; must not be involved in this process. So &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; is impossible to be in gas state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But I don't think it makes sense.&lt;/p&gt;&#xA;&lt;p&gt;Consider a container with a volume of &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{B(g)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{C(g)}$&lt;/span&gt; is inside it.&lt;/p&gt;&#xA;&lt;p&gt;If   &lt;span class=&quot;math-container&quot;&gt;$K_{\text{pre}}=K_{\text{post}}$&lt;/span&gt;&#xA;, we have &lt;span class=&quot;math-container&quot;&gt;$C_{\ce{B},\text{pre}} \cdot C_{\ce{C},\text{pre}}=C_{\ce{B},\text{post}} \cdot C_{\ce{C},\text{post}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;When the container's volume is reduced by &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt;, if &lt;span class=&quot;math-container&quot;&gt;$\dfrac{b_{\text{post}}}{b_{\text{pre}}}=\dfrac{c_{\text{post}}}{c_{\text{pre}}}=\dfrac{1}{2}$&lt;/span&gt;, the equilibrium can be reached, with all the requirements stated in the question.&lt;/p&gt;&#xA;&lt;p&gt;Conclusion: &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; can be in gas state.&lt;/p&gt;&#xA;&lt;p&gt;So, is the official answer wrong or I just didn't make the right conclusion?&lt;/p&gt;&#xA;" OwnerUserId="2601" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-09T09:38:11.670" Title="How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6730" PostTypeId="2" ParentId="5369" CreationDate="2013-11-01T02:42:35.383" Score="0" Body="&lt;p&gt;Other reactive metals like Tantalum with highly protective oxide layers are Zirconium, Hafnium (though most Zirconium contains Hafnium), Titanium (though it needs oxidising ions or platinum alloying to handle acid solutions), Chromium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium and even iron will handle concentrated nitric and sulfuric acids due to protective oxide layers formed. Tantalum is generally the best and most all-around active metal to handle acid and other solutions. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="402" LastEditDate="2017-04-08T22:34:05.707" LastActivityDate="2017-04-08T22:34:05.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6732" PostTypeId="1" CreationDate="2013-11-01T06:49:54.303" Score="4" ViewCount="18722" Body="&lt;p&gt;According to the definition of ionic bond:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBN16.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the above figure showing formation of ionic bond, due to the transfer of electron from sodium atom to fluorine.It demonstrates clearly, how electrostatic interaction takes place.&lt;br&gt;&#xA;In general, electrostatic interaction takes place between two stationary charges.Moreover, electrons would not be static as demonstrated by the above figure.Electrons will be moving at high speed, producing magnetic field around them, due to their motion; and even will also have electric field around them.Thus, we can expect electromagnetic interaction between oppositely charged ions along with electrostatic interaction due to stationary protons.So, does it mean that ionic bond is due to electromagnetic interaction and not just by electrostatic interaction?&lt;br&gt;&#xA;[All the statements made are up to my view on the concept.Any correction advisory is welcome]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;REFERENCES&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Ionic_bond&lt;/a&gt;   (definition and diagram has been extracted from this link)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Electromagnetic_radiation&lt;/a&gt; (provides notion of charge motion effects)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrostatics&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Electrostatics&lt;/a&gt; (gives notion of electrostatic force)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2013-11-03T07:03:23.500" LastActivityDate="2013-11-03T08:45:54.290" Title="Is ionic bond just formed by &quot;electrostatic&quot; interaction between two oppositely charged ions?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6733" PostTypeId="2" ParentId="5084" CreationDate="2013-11-01T08:24:08.260" Score="0" Body="&lt;p&gt;Lithium 6 and 7 have different electrolysis properties such that given an appropriate membrane - Lithium 7 can be enriched. Common is the mercury process.&#xA;Alternatively this process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jolisfukyu.tokai-sc.jaea.go.jp/fukyu/mirai-en/2012/4_10.html&quot; rel=&quot;nofollow&quot;&gt;http://jolisfukyu.tokai-sc.jaea.go.jp/fukyu/mirai-en/2012/4_10.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-11-01T08:24:08.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6734" PostTypeId="2" ParentId="6732" CreationDate="2013-11-01T09:57:03.357" Score="1" Body="&lt;p&gt;First to the magnetic properties - if you would be happy with Schroedinger equation and paired electrons, which is perfectly appropriate for the example show, then just forget about magnetism at all, it is negligible. All the information needed is in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronic_molecular_Hamiltonian&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electronic_molecular_Hamiltonian&lt;/a&gt;, which is valid for all bonding situations.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-01T09:57:03.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6735" PostTypeId="1" AcceptedAnswerId="6737" CreationDate="2013-11-01T10:38:03.457" Score="5" ViewCount="36065" Body="&lt;p&gt;Physical chemistry is easy to study as I have to just understand the concepts (like in physics) and apply some equations for calculations. In organic chemistry, reaction mechanisms helpsin remembering a lot of reactions and also doing conversion problems help a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, in the case of inorganic chemistry it is very difficult for me to remember those equations. My high school syllabus includes a lot of preparation methods of different compounds and reactions showing their chemical properties. But, unlike organic chemistry there is no problems based on it for deep understanding. So, what is actually needed is a good memory power. But, I'm not able to remember all those equations. So, how should I study them? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="73" LastEditDate="2013-11-02T00:22:38.080" LastActivityDate="2015-03-15T14:42:47.587" Title="How to study (by heart) all the equations in inorganic chemistry?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="3" ClosedDate="2015-03-15T17:44:04.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="6736" PostTypeId="2" ParentId="6732" CreationDate="2013-11-01T11:49:55.583" Score="0" Body="&lt;p&gt;Yes, in fact electrostatic attraction is an electromagnetic interaction. So there is no need for thinking about different approach. If you mean only magnetic interaction, so you are searching for the relation with magnetism, I have some doubts about the possibility that they are related. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2013-11-01T11:49:55.583" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6737" PostTypeId="2" ParentId="6735" CreationDate="2013-11-01T13:11:47.470" Score="13" Body="&lt;p&gt;In organic chemistry, everything is based around C, H, O and N. Therefore to understand reactivity (somehow), there is just limited number of options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, inorganic chemistry deals with the rest of periodic table, and each of the elements has its special properties. But there are general trends as well, but are more difficult to decipher and are often expressed by set of typical reactions, you should learn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually they concern the stability of (oxidation) states, how to go from less stable to more stable one to obtain the desired compound and few processes how tho achieve the least stable state from the one found in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to state some simple rules, but there are several most typical anions and cations, and the chemistry of given element is just interactions with them. During the learning it helped me to have &quot;all&quot; the reactions (several hundreds) on a paper and for each one try to understand what happens and if there are others in the set, which are similar, and why. This helps you to spot the similarities. And - it is better to have more reactions then less, because then you have higher probability to spot the trends.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-01T13:11:47.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6738" PostTypeId="1" AcceptedAnswerId="6890" CreationDate="2013-11-01T13:28:14.487" Score="5" ViewCount="378" Body="&lt;p&gt;I know that for a particular, immiscible solute, there is a limit on how much of it can be dissolved in a solvent at a given temperature. However, I was wondering if there is only so much &quot;total solute&quot; that a given solvent can dissolve. As a simple example, suppose you saturate water with salt, then start dissolving sugar into it. Does the fact that it is saturated with salt limit how much sugar it can dissolve? My mental image is one of each water molecule being able to suport only so many solvation structures, regardless of the chemicals participating in the structure.&lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastEditorDisplayName="user2603" LastEditDate="2013-11-01T18:43:21.497" LastActivityDate="2013-11-11T11:27:18.600" Title="Absolute limits on solubility of immiscible solutes?" Tags="&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6740" PostTypeId="1" CreationDate="2013-11-01T18:37:54.083" Score="1" ViewCount="428" Body="&lt;p&gt;I'm an A-level student studying the F324 unit of OCR GCE Chemistry 'A'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Explain how condensation polymers may be photodegradable. Equations needed!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What a condensation polymer is.&lt;/li&gt;&#xA;&lt;li&gt;Photodegradation may be possible by either a) Light-sensitive additives which catalyse a (slow) reaction in the presence of UV OR b) carbonyl bonds (&lt;code&gt;C=O&lt;/code&gt;) breaking from the light energy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What my teacher wants me to do for option 'a' in terms of equations. I'm inclined to leave that out, as it seems to be well outside the scope of the course.&lt;/li&gt;&#xA;&lt;li&gt;Once the carbonyl bond breaks, what happens then? My best guess is that the oxygen atom receives the electron pair from the bond, then $H^{+}$ ions from 'somewhere' arrive to form water and finish the reaction. If I didn't know better, I'd say you needed a reducing agent for things to photodegrade.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, what I'd ideally like is an equation (and explanation, if possible) showing what happens &lt;em&gt;after&lt;/em&gt; the carbonyl bond breaks from UV.&lt;/p&gt;&#xA;" OwnerUserId="2606" LastEditorUserId="73" LastEditDate="2013-11-02T04:28:57.030" LastActivityDate="2013-12-04T17:56:06.967" Title="Equations for UV degradation of condensation polymers" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;polymers&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6741" PostTypeId="1" CreationDate="2013-11-01T18:40:53.027" Score="3" ViewCount="1662" Body="&lt;p&gt;I am working on a project in my chemistry class, and even though we've already finished the lab, I'm still confused as to why and how $\ce{MgCl2}$ melts ice/ lowers the freezing temperature of water. Can anybody tell me why this happens?&lt;/p&gt;&#xA;" OwnerUserId="2607" LastEditorUserId="4945" LastEditDate="2014-07-29T11:48:26.703" LastActivityDate="2014-07-29T11:50:25.720" Title="Why does MgCl2 lower the freezing temperature, and how does it do so?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6742" PostTypeId="2" ParentId="6741" CreationDate="2013-11-01T18:47:01.273" Score="4" Body="&lt;p&gt;The $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$ ions disrupt hydrogen bonding, preventing ice from forming a lattice structure. Therefor, the water must be colder so those weak hydrogen bonds will be acting on water molecules with less kinetic energy (i.e., colder), finally allowing them to create a crystal around the dissolved &quot;impurities&quot; $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a basic colligative property. Its more effective for electrolytes because they dissociate, creating more lattice disruptions. As a general resource, see: Ebbing, Darrell D., General Chemistry, 3rd ed, Houghton Mifflin, 1990.&lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastEditorUserId="4945" LastEditDate="2014-07-29T11:50:25.720" LastActivityDate="2014-07-29T11:50:25.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6743" PostTypeId="2" ParentId="6644" CreationDate="2013-11-01T18:55:05.213" Score="2" Body="&lt;p&gt;Ease of oxidation is how easily it loses electrons. Electrons are the basis of chemical reactions. If chemical A is more easily oxidized than B, then B is a relative oxidizer to A, so in terms of the activity series, the elements that are more easily oxidized can react with more chemicals, since they will be able to act as reducing agents to more chemicals.&lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastActivityDate="2013-11-01T18:55:05.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6744" PostTypeId="2" ParentId="6556" CreationDate="2013-11-01T22:18:39.793" Score="1" Body="&lt;p&gt;I would strongly recommend this resource:&#xA;&lt;a href=&quot;http://www.tf.uni-kiel.de/matwis/amat/def_en/&quot; rel=&quot;nofollow&quot;&gt;http://www.tf.uni-kiel.de/matwis/amat/def_en/&lt;/a&gt; &#xA;It is a good complete on line introduction to defects theory in crystals.&#xA;Some key points listed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of point defects $c_{PD}$ at equilibrium can be expressed by an Arrhenius formula:&#xA;$$c_{PD}=A \cdot e^{-\frac{G_F}{kT}}=A \cdot e^{-\frac{S_F}{kT}}\cdot  e^{-\frac{H_F}{kT}}$$&#xA;where $G_F$ is the free energy of formation, $H_F$ the enthalpy of formation and $S_F$ the entropy of formation. In solids  $H_F$ can be confused with the energy of formation, $S_F$ is associated the variation of the vibration entropy of the crystal due to the introduction of a point defect. &#xA;Formation enthalpies, and thus formation energies can be obtained by: &#xA;$$ H_F(D) = E_{tot}(D) - E_{tot}(X)+\sum_i n_i \mu_i +q (E_F -E_V) $$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E_{tot}(D)$  and $E_{tot}(X)$: total energies of the system with and&#xA;without the defect D. &lt;/li&gt;&#xA;&lt;li&gt;$n_i$: number of atom removed from the system,&#xA;a negative value denote the addition of one atom. &lt;/li&gt;&#xA;&lt;li&gt;$\mu_i$: chemical potential of the atom $i$. This should consider the chemical environment at the equilibrium condition (which reservoir is to take into account).&lt;/li&gt;&#xA;&lt;li&gt;$E_F -E_V$: Fermi energy relative to the valence band maximum. &lt;/li&gt;&#xA;&lt;li&gt;$q (E_F -E_V)$: energy change due to the exchange of electrons and holes with the carrier reservoirs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the equilibrium concentration in the multiple defects case an additional entropy term comes from the standard configuration entropy which can be calculated from the Boltzmann formula&#xA;$$S=k_B \cdot \ln P$$&#xA;where P is the number of configurations for the same macrostate.&lt;/p&gt;&#xA;" OwnerUserId="2608" LastEditorUserId="2608" LastEditDate="2013-11-02T08:43:15.853" LastActivityDate="2013-11-02T08:43:15.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6745" PostTypeId="2" ParentId="6138" CreationDate="2013-11-02T11:31:45.727" Score="4" Body="&lt;p&gt;Doping, not to confuse with alloying, introduces a low amount of impurities or more generally point defects in a crystal. The overall periodicity of the system is thus not changed and no shift should be expected in XRD. However doping introduces some disorder that makes XRD peaks broader.&lt;/p&gt;&#xA;" OwnerUserId="2608" LastActivityDate="2013-11-02T11:31:45.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6748" PostTypeId="2" ParentId="6723" CreationDate="2013-11-02T18:47:00.500" Score="6" Body="&lt;p&gt;Single water molecule dissociates into  $\ce{H^+ + OH^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in @G M's answer, $\ce{H^+}$ is very reactive species and some people do not like it and write it as less provoking $\ce{H_3O^+}$, for discussion see first paragraph of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydronium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short - $\ce{H^+}$ organizes water molecules around itself and it's behavior is complicated and interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point important point stated by @G M is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{pH=}-\log_{10}[\ce H^{+}]= -\log_{10}[\ce{H3O+}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is correct, because the concentration of $\ce{H^+}$  is so low, that using $\ce{H_3O^+}$ does not change the concentration of $\ce{H_2O}$.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="7448" LastEditDate="2015-06-26T15:29:04.213" LastActivityDate="2015-06-26T15:29:04.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6749" PostTypeId="2" ParentId="6709" CreationDate="2013-11-02T18:48:37.043" Score="9" Body="&lt;p&gt;The answer is most certainly the &lt;strong&gt;Self interaction error&lt;/strong&gt; of an electron with itself in the exchange term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would probably be more helpful than what I can write here, as I have little time to make this answer more complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://prb.aps.org/abstract/PRB/v78/i14/e140404&quot; rel=&quot;noreferrer&quot;&gt;A. Droghetti PRB 2008&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also have a look at&#xA;&lt;a href=&quot;https://notendur.hi.is/~hj/papers/paperODDmolecJCP12.pdf&quot; rel=&quot;noreferrer&quot;&gt;Klupfel2012&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="230" LastActivityDate="2013-11-02T18:48:37.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6750" PostTypeId="1" AcceptedAnswerId="6815" CreationDate="2013-11-02T19:02:32.820" Score="3" ViewCount="8953" Body="&lt;p&gt;In my chemistry class, we had a problem akin to the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an object with mass bla and specific heat bla is placed into a bla g calorimeter with an initial temperature of bla and a specific heat of bla, and the temperature of the calorimeter rises to bla at the end of the experiment, &lt;strong&gt;how much heat was lost by the object?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I performed the calculations and got an answer of, let's say, $8 J$. When I submitted this answer, I was told that it was wrong, and the correct answer would have been $-8 J$. Wouldn't this response to the above question mean that the object &lt;em&gt;gained&lt;/em&gt; heat?&lt;/p&gt;&#xA;" OwnerUserId="1974" LastActivityDate="2013-11-05T10:23:08.283" Title="Is it customary to express heat loss as a negative number?" Tags="&lt;energy&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6751" PostTypeId="1" CreationDate="2013-11-02T19:13:16.447" Score="25" ViewCount="175475" Body="&lt;p&gt;Let's take the $\ce{KOH}$ molecule into account. I know it is a base from literature, but how would one go about determining if a molecule is acidic or basic simply based on the structure of the molecule? How about amphoteric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I understand that there are two complementary systems - Lewis and Brønsted-Lowry theories. How do they work and how do they fit together?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="5026" LastEditDate="2018-03-03T11:23:41.293" LastActivityDate="2019-03-13T21:52:40.643" Title="How does one tell if a specific molecule is acidic or basic?" Tags="&lt;acid-base&gt;&lt;molecular-structure&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6753" PostTypeId="1" AcceptedAnswerId="6754" CreationDate="2013-11-02T19:47:46.983" Score="2" ViewCount="6719" Body="&lt;p&gt;I have some questions regarding how to read a MALDI mass spectrum in a paper. I am a biologist by training, so I would like some analytical chemists to help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The researchers ran a western blot, and immunoblotted for a protein of interest. They cut out this band from the western blot, subjected the protein to trypsin digest, and used these peptide fragments for MALDI-TOF analysis. I know for a fact there will be other proteins at the same apparent molecular mass which would have also been included in the analysis, which should be taken as contaminants in the resulting spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They produced a graph with the &lt;em&gt;absorbance intensity x 1,000&lt;/em&gt; on the dependent axis, and &lt;em&gt;mass:charge (m/z) ratio&lt;/em&gt; on the independent axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors point to two peaks, which they say correspond to two peptide fragments from the tryptic digest of their protein of interest. They provide the supposed peptide fragment sequences and m/z ratio data for these two peaks, and I checked to make sure they are the same as reported. The authors expect to see two peaks: protein of interest is digested to two fragments, A + B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Peak 'A' and peak 'B' are of different intensities. Peak 'A' is 20-fold greater than peak 'B'. However, the tryptic digest must yield &lt;em&gt;equimolar amounts&lt;/em&gt; of peptide fragments A and B. Is it possible that these intensity peaks can be 20-fold different? Wouldn't the absorbance intensity correlate to total analyzed peptide mass?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it normal to multiple absorbance intensity by a (seemingly arbitrary) factor, in this case, 1000? Since the absorbance intensity of peak 'B' is &amp;lt; 0.5 (after scaling up), could this represent &lt;em&gt;noise&lt;/em&gt; rather than &lt;em&gt;signal&lt;/em&gt; in the analysis?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user2614" LastActivityDate="2018-06-15T13:15:29.440" Title="How to interpret a MALDI mass spectrum?" Tags="&lt;analytical-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6754" PostTypeId="2" ParentId="6753" CreationDate="2013-11-02T20:58:06.620" Score="3" Body="&lt;p&gt;Please see the wiki articles for the general method:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_spectrum&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Mass_spectrum&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then let's start with question 2: yes, you can and you should. The spectrometer measures in arbitrary units, therefore you can speak only about relative abundances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And question 1: it depends how easily is the given molecule ionized. So if the fragment A is rich in polar aminoacids, whereas fragment B is mostly hydrophobic and difficult to ionize, than you should expect this big difference in signal intensities.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-02T20:58:06.620" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6755" PostTypeId="1" AcceptedAnswerId="6889" CreationDate="2013-11-03T06:17:00.930" Score="1" ViewCount="1771" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the number of unpaired electrons present in the d orbital (whose lobes are present along the axis) for the complex $\ce{[Co(SCN)4]^{2-}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{{SCN}^{-}}$ is a weak ligand I did not pair the electrons, but still I got the wrong answer. The correct answer is 0. But how?&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="-1" LastEditDate="2018-05-18T04:14:24.037" LastActivityDate="2018-05-18T04:14:24.037" Title="What is the number of unpaired electrons in cobalt(II) tetrathiocyanate?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6756" PostTypeId="2" ParentId="6732" CreationDate="2013-11-03T08:45:54.290" Score="3" Body="&lt;p&gt;You are absolutely correct that the electron has an angular momentum, which can be thought of like a magnetic field and which interacts with magnetic fields.&#xA;In &quot;light&quot; elements, however, the magnetic interaction is almost completely negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html&quot; rel=&quot;nofollow&quot;&gt;Hyperphysics page on the Zeeman Effect&lt;/a&gt; shows that in a hydrogen atom, a magnetic field of one tesla (which is fairly strong) splits energy levels by $5.78 \times 10^{-5} eV$, or roughly one-millionth of the ionization energy of a hydrogen atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From just having done calcuations on the anomalous Zeeman effect for homework (you can use the formula in the link I provided and check it yourself), I can tell you that in a potassium atom, the energy levels of a p-electron split by 0.006 Hartree/Tesla, or 1/10,000 of the ionization energy of a hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, while there &lt;em&gt;is&lt;/em&gt; a magnetic field (or what can be treated as one) generated by electron motion, it's fairly safe to ignore it in lighter elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as heavy elements go...well, you undergo what's called j-j coupling, caused by the fact that the electrons are going so fast that their motion becomes relativistic. To describe that motion requires the Dirac equation, which I have not studied in any level of detail. Instinct says that there are other considerations which are much more significant than magnetic interactions, but I have no proof whatsoever to back it up.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-11-03T08:45:54.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6757" PostTypeId="1" AcceptedAnswerId="6760" CreationDate="2013-11-03T09:37:13.493" Score="4" ViewCount="805" Body="&lt;p&gt;I'm aware that tritium undergoes beta decay forming Helium-3. But in the case of HTO, what does it decay into? I don't suppose it decays into HeHO.&lt;/p&gt;&#xA;" OwnerUserId="2620" LastEditorUserId="9961" LastEditDate="2020-06-27T13:30:00.240" LastActivityDate="2020-06-27T13:30:00.240" Title="What does HTO decay into?" Tags="&lt;water&gt;&lt;hydrogen&gt;&lt;isotope&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6758" PostTypeId="2" ParentId="6602" CreationDate="2013-11-03T10:53:18.200" Score="2" Body="&lt;p&gt;I think your experiment has one setup mistake.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're mixing $\pu{0.8265 g}$ of oxalate with $\pu{50 mL}$ of water and $15$-$\pu{20 mL}$ of sulfuric acid. So your $\pu{25 mL}$ oxalate solution does not contain $\pu{0.8265 g}$  of oxalate (more or less $\frac{25 \cdot 0.8265}{ 65-\pu{70 mL}}$). So you can't know the exact oxalate concentration of this solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you to use a $\pu{100 mL}$ volumetric flask to prepare the oxalate solution. You can then calculate its concentration and the amount of oxalate you have in $\pu{25 mL}$.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="19616" LastEditDate="2017-04-21T18:46:49.977" LastActivityDate="2017-04-21T18:46:49.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6759" PostTypeId="1" AcceptedAnswerId="6763" CreationDate="2013-11-03T09:01:57.637" Score="8" ViewCount="23946" Body="&lt;p&gt;In movies cool guys don't need a matchbox − they just light a match by scratching it on whatever hard surface − concrete walls, wooden bar tables, leather belts, etc. Is it just a movie trick, or is there actually a type of matches you can do that with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've tried scratching usual matches on a ton of different surfaces and the heads just smear without even a smoke.&lt;/p&gt;&#xA;" OwnerUserId="2618" OwnerDisplayName="user1306322" LastEditorUserId="7475" LastEditDate="2015-10-30T17:08:23.303" LastActivityDate="2016-02-21T07:02:25.347" Title="Is it possible to light a match by scratching it on hard surfaces, like they do in movies?" Tags="&lt;everyday-chemistry&gt;&lt;energy&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6760" PostTypeId="2" ParentId="6757" CreationDate="2013-11-03T11:05:24.883" Score="9" Body="&lt;p&gt;Very shortly after a radioactive element decays in a molecule, you could consider that for a brief moment, it does indeed form a molecule with the new element in the same place as its originating radioisotope. So yes, after the decay of $HTO$ you could get $HeOH^+$ via the unusual reaction $TOH → HeOH^+ + e^- + \bar\nu_e$. Of course, though, this is a very unstable compound and it would immediately participate in further chemical reactions, very likely releasing the helium atom at some point. Even if the molecule were relatively stable after having one of its atoms transmuted, the molecule might still decompose as the nuclear decay releases a lot of energy, which can tear the comparatively fragile molecule to pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A curious tidbit is that, in fact, the decay of tritiated compounds has been considered as a means to produce legitimate helium compounds, given that directly reacting helium with other substances has been fruitless other than for very trivial species, such as $HeH^+$. The problem is that tritium has a rather long half life of about 11 years, so it would be quite hard to produce reasonable quantities of helium compounds in a reasonable timespan, along with the likelihood of the helium compound being very poorly bound and the emission of vast quantities of nuclear decay energy which could easily decompose the fleeting substance.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-03T11:05:24.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6763" PostTypeId="2" ParentId="6759" CreationDate="2013-11-03T13:44:56.107" Score="16" Body="&lt;p&gt;Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e.  the brownish paint thing down the side of the matchbox or across the front of the matchbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most &quot;regular&quot; matches now are safety matches. The brownish strip has red phosphorous in it. The match head has potassium chlorate $\ce{KClO3}$. Combine those two with friction, and you get fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are still striking matches, with heads of phosphorus sesquisulfide $\ce{P4S3}$, still around. Though largely in countries where consumer protection is not yet well-developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With sufficient energy, you can still ignite a safety match without the striking strip. But you'll need more energy than the friction of scratching it on a wall or bit of stubble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS Don't play with fire. Get a friendly neighbourhood chemist to play with fire on your behalf.)&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2013-11-03T13:50:05.760" LastActivityDate="2013-11-03T13:50:05.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6764" PostTypeId="2" ParentId="6751" CreationDate="2013-11-03T14:21:23.630" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does one tell if a specific molecule is acidic or basic?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An acid (from the Latin acidus/acēre meaning sour) is a chemical substance whose aqueous solutions are characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry, a base is a substance that, in aqueous solution, is slippery to the touch, tastes bitter, changes the colour of indicators (e.g., turns red litmus paper blue), reacts with acids to form salts, and promotes certain chemical reactions (base catalysis).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do have $3$ different definitions from Arrhenius, Brønsted-Lowry, and Lewis. We would be in trouble, in some cases, where one definition accepts a molecule as acid or base and other not. So, we do have a common definition for acids and bases as mentioned above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to amphoteric substance, consider $\ce{H2O}$ which has no taste.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;LINKS&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Acid&lt;/a&gt; (definition of acid is extracted from this link)  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Base_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Base_(chemistry)&lt;/a&gt; (definition of base is extracted from this link)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="5026" LastEditDate="2018-03-03T11:26:24.580" LastActivityDate="2018-03-03T11:26:24.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6766" PostTypeId="2" ParentId="6720" CreationDate="2013-11-03T17:27:44.987" Score="14" Body="&lt;p&gt;I'm late to the party and everyone else here has suggested how computer science experience can contribue to chemistry. I'm going to do a bit of that, but also point out how chemistry experience could potentially contribute to some (very select) CS opportunities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find myself in a similar position to you. I recently declared a double major in Chemistry and Applied Math-Comp.Sci emphasis because I enjoyed both subjects. I was initially planning to double in pure Comp. Sci., but our CS department is full of horrible lecturers and the math is full of amazing lecturers so I kinda got shoehorned into the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational chemistry is one of many fields in which CS and chem will overlap these days. It's not obvious, but there are many, many others, as people who are good with computers and science are rarer than they should be. Many labs, even those that are not computational in nature, rely on some basic modeling. I currently do some work with a nanoparticles group, and the PI was recently complaining that most of his grad students couldn't program anything to save their lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, our field could always use some good software devs. Perhaps I'm a little spoiled, but some our so-called &quot;cutting edge&quot; software suites have to be partially recomplied every time they are launched and suffer from spontaneously having communications modules crash, necessitating reboots of several computers in order to get things up and running again. It's pretty obvious to me that a lot of the stuff was designed by a software developer and not by a practicing scientist--such as the million button presses needed to load data and work with it, and the general recalcitrance of such software to spit out various data types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to work on that problem, we're still struggling with the issue of how to share code and models/data. Everything is moving to computers, and research is no exception. Researchers are still trying to figure out how to share files easily, store them, and publish them (more than one study has gone to press as a letter without any of the code attached to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I don't know what you want. Based off your last few paragraphs, this probably isn't what you were looking for, but since you opened with a question about the overlap between two fields, and not how one could contribute to the other, I'm leaving these here as ideas about how several fields can potentially interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. For an interesting example, Philip Guo's PhD thesis revolved around the question of how the same code could give the same results when run on different systems, in spite of different environments. The question led him to create a tool (I believe Python) which allows simulations to run in a safely sandboxed (and thus identical) environement without loss of performance. &lt;a href=&quot;http://www.pgbovine.net/PhD-memoir.htm&quot;&gt;http://www.pgbovine.net/PhD-memoir.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-11-03T17:27:44.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6767" PostTypeId="2" ParentId="6751" CreationDate="2013-11-03T20:45:34.153" Score="20" Body="&lt;p&gt;It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered.  An important point to consider when dealing with acids and bases is that &lt;em&gt;acid/base strength is inherently tied to the solvent&lt;/em&gt;.  For this answer, I'm going to limit the discussion to acids and bases in an aqueous environment.&lt;/p&gt;&#xA;&lt;p&gt;It is helpful to consider the terms acid and base as a means to &lt;em&gt;classify&lt;/em&gt; substances.  This way, Chemists can explain chemical reactivity and structure-function relationships of substances.  Very early classification systems depended on our senses (acids are sour, bases are slippery to the touch) and more recent classification systems utilize structural characterization tools such as NMR or crystallography.  Many classification systems have been proposed over the years, and only a few of them have found sufficiently widespread use to end up in textbooks used in the standard Chemistry curriculum.  Below are a few systems, taken from Miessler &amp;amp; Tarr's Inorganic Chemistry textbook.  The 2nd through 4th entries are those in common use today.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;NAME (YEAR) acid definition [example]; base definition [example]&lt;/li&gt;&#xA;&lt;li&gt;Liebig (~1776) Acid: an oxide of N, P, S [&lt;span class=&quot;math-container&quot;&gt;$\ce{SO_3}$&lt;/span&gt;]; Base: Reacts with&#xA;acid [&lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Arrhenius (1894) Acid: Forms hydronium ion [&lt;span class=&quot;math-container&quot;&gt;$\ce{HNO_3}$&lt;/span&gt;]; Base: Forms hydroxide ion [&lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Brønsted (1923) Acid: Proton donor [&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;]; Base: Proton acceptor [&lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Lewis (1923) Acid: Electron-pair acceptor [&lt;span class=&quot;math-container&quot;&gt;$\ce{Ag^+}$&lt;/span&gt;]; Base: Electron-pair donor [&lt;span class=&quot;math-container&quot;&gt;$\ce{NH_3}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Ingold-Robinson (1932) Acid: Electrophile [&lt;span class=&quot;math-container&quot;&gt;$\ce{BF_3}$&lt;/span&gt;]; Base: Nucleophile [&lt;span class=&quot;math-container&quot;&gt;$\ce{NH_3}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Lux-Flood (1939) Acid: Oxide ion acceptor [&lt;span class=&quot;math-container&quot;&gt;$\ce{SiO_2}$&lt;/span&gt;]; Base: Oxide ion donor [&lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Usanovich (1939) Acid: Electron acceptor [&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl_2}$&lt;/span&gt;]; Base: Electron donor [&lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Solvent system (1950s) Acid: Solvent cation [&lt;span class=&quot;math-container&quot;&gt;$\ce{BrF_2^+}$&lt;/span&gt;]; Base: Solvent anion [&lt;span class=&quot;math-container&quot;&gt;$\ce{BrF_4^-}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;li&gt;Frontier Orbitals (1960s) Acid: LUMO of acceptor [&lt;span class=&quot;math-container&quot;&gt;$\ce{BrF_3}$&lt;/span&gt;]; Base: HOMO of donor [&lt;span class=&quot;math-container&quot;&gt;$\ce{NH_3}$&lt;/span&gt;]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I find the various acid/base systems very enlightening.  Note how From Arrhenius through Lewis there was a broadening of the acid/base classification system; Arrhenius can't be used to describe non-aqueous acids and bases and Brønsted can't be used with aprotic substances.  Yet, the Lewis definition incorporates the previous two (a Brønsted acid &lt;em&gt;is also&lt;/em&gt; a Lewis acid; an Arrhenius base &lt;em&gt;is also&lt;/em&gt; a Lewis base).  One then may ask, what's up with the Lux-Flood definition then?  This definition is counter to the trend of broadening the classification system, and yet it is useful in describing anhydrous solid-state chemistry and is used to describe geochemical reactions as well as the chemistry of high-temperature melts.  My point being: classification of substances into acids and bases is only meaningful if it helps explain chemical phenomena.&lt;/p&gt;&#xA;&lt;p&gt;Which brings us to the Usanovich definition, which essentially states that &lt;em&gt;every reaction is an acid-base reaction&lt;/em&gt;.  Such a broad definition is not overly helpful, and to those of us with an affinity towards electrochemistry (ahem), is somewhat arrogant :-)&lt;/p&gt;&#xA;&lt;p&gt;I do believe that determining if a substance will behave as an acid or a base requires a bit of chemical intuition (or a Socratic method).  For example, you may have performed an experiment in which &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt; served as a base; you know that, like potassium, sodium is an alkali metal; therefore you presume that &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; would be a base as well.&lt;/p&gt;&#xA;&lt;h2&gt;Back to the question at hand&lt;/h2&gt;&#xA;&lt;p&gt;So how do I suggest one use this information to &lt;strong&gt;predict whether a substance will behave as an acid or a base using its structure alone&lt;/strong&gt;?  Personally, I find the Lewis theory as the most useful classification system in answering this type of question.  If the structure of a compound is set before me and I am to predict its acid/base chemistry, I will ask two questions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there any lone pair electrons that can be donated?&lt;/li&gt;&#xA;&lt;li&gt;Are there any electron-deficient atoms that could serve as electron pair acceptors?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If the answer to question 1 is yes, then the molecule can behave as a base.  If the answer to question 2 is yes, then the molecule is an acid.  If both are true, then I have an amphoteric substance.&lt;/p&gt;&#xA;&lt;p&gt;In proofreading this answer, I realize I said I would restrict myself to aqueous systems, in which case using the Brønsted system may be more helpful.  In this case, the questions become:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a hydrogen that can be donated? (You'll be right more often than you are wrong if you rephrase this question as &amp;quot;Is there a hydrogen that is attached to something other than carbon?&amp;quot;).&lt;/li&gt;&#xA;&lt;li&gt;Is there a lone pair that can accept a proton?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-03T11:26:10.313" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6768" PostTypeId="1" AcceptedAnswerId="6769" CreationDate="2013-11-03T21:06:06.617" Score="0" ViewCount="58" Body="&lt;p&gt;Usually I have the habit of keeping the hot milk for a while, until it becomes cool. I have noticed many a times that, drinking it, after keeping for a while, gives the mixture of hot (relatively less hot than before) and cool milk.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Almost $3$ by $4$ (approx) part of the milk at the top, is relatively found to be hot than the last $1$ by $4$ (approx) part of the milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for upper part to be hot and lower to be cooler?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2013-11-03T21:16:21.760" Title="Hot and cool milk" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6769" PostTypeId="2" ParentId="6768" CreationDate="2013-11-03T21:16:21.760" Score="1" Body="&lt;p&gt;By convection heat transfer. Hot milk goes up from the bottom of the cup, while the milk at top download, in contact with air, gets cooler and descents to the bottom of the cup.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-03T21:16:21.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6772" PostTypeId="1" CreationDate="2013-11-04T00:05:33.107" Score="2" ViewCount="93" Body="&lt;p&gt;I am working on a Master degree and need to know the following data :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The static (low frequency) dielectric constant of isopropanol at 25 &amp;deg;C&lt;/li&gt;&#xA;&lt;li&gt;The Optical (high frequency) dielectric constant of isopropanol at 25 &amp;deg;C&lt;/li&gt;&#xA;&lt;li&gt;Dielectric relaxation time (τ) of isopropanol at 25 &amp;deg;C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Could someone please provide me with these values and a reference or link to a book contains these data?&lt;/p&gt;&#xA;" OwnerUserId="2628" LastEditorUserId="4945" LastEditDate="2015-10-23T17:14:24.373" LastActivityDate="2015-10-23T17:14:24.373" Title="Reference to find dielectric constants and dielectric relaxation time" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6773" PostTypeId="1" AcceptedAnswerId="6776" CreationDate="2013-11-04T04:24:03.260" Score="2" ViewCount="492" Body="&lt;p&gt;Take the electrolysis of Lead(II) bromide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write it as two half-reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Pb^{2+}(l) + 2e^{-} \rightarrow Pb(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Br^{-} \rightarrow Br + e^{-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do they become neutral after they reach the electrodes. Why do they gain electrons at only the electrodes to become neutral? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2013-11-04T07:18:58.870" Title="During electrolysis, why are the products attracted to the cathode?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6774" PostTypeId="1" AcceptedAnswerId="10823" CreationDate="2013-11-04T04:29:26.973" Score="6" ViewCount="26371" Body="&lt;p&gt;In the Hall–Héroult process of extracting aluminium from aluminium oxide, the electrolyte is a liquid mixture of cryolite &lt;span class=&quot;math-container&quot;&gt;$(\ce{NaAlF6})$&lt;/span&gt; and aluminium oxide &lt;span class=&quot;math-container&quot;&gt;$(\ce{Al2O3}).$&lt;/span&gt; The cathodes and the anodes are made of graphite.&lt;/p&gt;&#xA;&lt;p&gt;During the electrolysis process, aluminium is deposited at the cathode and oxygen is liberated at the anode. Some of this oxygen reacts with the carbon in the graphite to form carbon-dioxide, thus slowly burning away the anodes. Thus, the anodes have to replaced periodically.&lt;/p&gt;&#xA;&lt;p&gt;But if this is the case, why do they use graphite anodes at all? Why don't they use anodes of some other, less reactive material? I'm pretty sure there are many out there that would fit the bill.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="41328" LastEditDate="2021-02-14T13:45:07.150" LastActivityDate="2021-02-14T13:48:06.210" Title="Why do they use graphite electrodes in the Hall–Héroult process?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;electrolysis&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6775" PostTypeId="1" CreationDate="2013-11-04T06:24:26.727" Score="11" ViewCount="2779" Body="&lt;p&gt;Why does oxygen form a double bond in $\ce{O2}$ but sulfur, also from group 16, forms single bonds in $\ce{S8}$?&lt;/p&gt;&#xA;" OwnerUserId="2547" LastEditorUserId="7448" LastEditDate="2015-06-19T10:51:53.177" LastActivityDate="2020-10-27T08:58:06.277" Title="Oxygen and sulfur bonding" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6776" PostTypeId="2" ParentId="6773" CreationDate="2013-11-04T07:18:58.870" Score="2" Body="&lt;p&gt;You have to think about what exactly a cathode and anode are. You are falling victim to this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A widespread misconception is that anode polarity is always positive (+). This is often incorrectly inferred from the correct fact that in all electrochemical devices, negatively charged anions move towards the anode and positively charged cations move away from it. In fact anode polarity depends on the device type, and sometimes even in which mode it operates, as per the above electric current direction-based universal definition. Consequently, as can be seen from the following examples, the anode is positive in a device that consumes power, and the anode is negative in a device that provides power.&lt;/em&gt; - Wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's your conditioning from other chemical nomenclature--&quot;cation&quot; and &quot;anion&quot;--which are absolute about their charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So beyond misleading nomenclature, what does cause this to happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this picture for example: &lt;a href=&quot;http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif&quot; rel=&quot;nofollow&quot;&gt;http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If E°cell &gt; 0, then the process is spontaneous (galvanic cell)&#xA;If E°cell &amp;lt; 0, then the process is nonspontaneous (electrolytic cell)&#xA;where E°cell = E°(V) reduction + E°(V) oxidation&#xA;In your case, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pb → Pb2+ + 2e- = 0.13 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2Br- → Br2 + 2e- = -1.06&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E°cell = .13 - 1.06&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E°cell = -.93&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the process is nonspontaneous (makes sense, think about trying to reduce ionized bromine), so you're looking at an electrolytic cell, which is battery driven (it needs work put in to operate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, again, to directly answer your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lead is formed at the anode because work from a battery was put in that pushed the reaction in that direction. Before work was put in, the reaction would have proceeded in the opposite direction (think exothermic vs endothermic, but in this case spontaneous cell potential vs nonspontaneous cell potential.) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But why do they become neutral after they reach the electrodes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They become neutral at the electrodes because that is where the reaction happens, and that is where the electrons are transferred. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do they gain electrons at only the electrodes to become neutral?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, where else could they gain electrons? They are sitting in a solution contain themselves and their respective anions, which they have already reacted with, and will no longer be transferring electrons with.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2013-11-04T07:18:58.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6778" PostTypeId="1" CreationDate="2013-11-04T08:19:06.063" Score="1" ViewCount="7845" Body="&lt;p&gt;During the &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/extraction/copper.html&quot; rel=&quot;nofollow&quot;&gt;electrolytic purification of copper&lt;/a&gt;, the electrolyte is taken as copper(II) sulphate ($\ce{CuSO_4}$). Is there any particular reason for this? Could another salt of copper, say $\ce{CuCl_2}$ work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, I don't see any reason why it shouldn't. The only difference will be, that at the anode chlorine instead of oxygen will be evolved, which doesn't seem to be such a big thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, then, is $\ce{CuSO_4}$ preferred?  &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="73" LastEditDate="2013-12-15T07:08:33.347" LastActivityDate="2014-08-10T18:25:09.827" Title="Why is CuSO4 used as electrolyte while purifying copper?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6780" PostTypeId="1" AcceptedAnswerId="6781" CreationDate="2013-11-04T12:00:30.010" Score="9" ViewCount="32256" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Larger the difference in electronegativities of bonded atoms, the larger the dipole moment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(this statement has been extracted from &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Intermolecular_Forces/Dipole_moments&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemwiki-dipole moment&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;The electronegativity difference between carbon and fluorine is about &lt;span class=&quot;math-container&quot;&gt;$1.43$&lt;/span&gt; (Pauling scale), &lt;span class=&quot;math-container&quot;&gt;$1.649$&lt;/span&gt; (Allen scale). The electronegativity difference between carbon and chlorine is about &lt;span class=&quot;math-container&quot;&gt;$0.61$&lt;/span&gt; (Pauling scale), &lt;span class=&quot;math-container&quot;&gt;$0.325$&lt;/span&gt; (Allen scale). Although, the electronegativity difference of &lt;span class=&quot;math-container&quot;&gt;$\ce{C-F}$&lt;/span&gt; bond is greater than electronegativity difference of &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bond, dipole moment of &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bond (&lt;span class=&quot;math-container&quot;&gt;$\pu{1.860 Debye}$&lt;/span&gt;) in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_3-Cl}$&lt;/span&gt; is greater than dipole moment of &lt;span class=&quot;math-container&quot;&gt;$\ce{C-F}$&lt;/span&gt; bond (&lt;span class=&quot;math-container&quot;&gt;$\pu{1.847 Debye}$&lt;/span&gt;) in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3-F}$&lt;/span&gt;, why?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="69830" LastEditDate="2021-05-22T07:53:28.560" LastActivityDate="2021-05-22T07:53:28.560" Title="Why do C-F bond has less dipole moment than C-Cl bond?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6781" PostTypeId="2" ParentId="6780" CreationDate="2013-11-04T12:34:46.083" Score="12" Body="&lt;p&gt;The bond &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond_dipole_moment&quot; rel=&quot;noreferrer&quot;&gt;dipole moment&lt;/a&gt; ($\mu$) of a chemical bond is equal to the magnitude of the separation of charge ($\delta$) times the displacement of the charge (or the distance over which the charges are spearated) $d$, or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\mu}=\delta \vec{d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, there are two factors controlling dipole moment. The charge separation can be estimated from the electronegativity difference, but not perfectly. Disparities in size and polarizability also effect the separation of charge between two atoms. The displacement is the length of the bond. The $\ce{C-Cl}$ bond length (176 pm) is longer than the $\ce{C-F}$ bond length (134 pm) - data from &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond_length#Bond_lengths_of_carbon_with_other_elements&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C-F}$, the electronegativity difference on the Pauling scale is 1.43, and the bond length is 134 pm: $1.43\times 134 = 191$ (ignoring units at the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C-Cl}$, the electronegativity different on the Pauling scale is 0.61, and the bond length is 174 pm: $0.61\times 174=106$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my quick and dirty method did not work so well. However the disparity between these two bonds by my method $(106/191)=0.55)$ is smaller than by just comparing electronegativities $(0.61/1.43=0.42)$&#xA; and that is because of the other factor mentioned above: size and polarizability. Chlorine is bigger and more polarizable than fluorine (and carbon). The carbon-chlorine bond is polarized toward chlorine not just because of the electronegativity difference. Chlorine's larger size decreases the effectiveness of orbital overlap in bonding, and chlorine's increase polarizability implies that it can support more partial negative charge than suggested by its electronegativity alone.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-04T12:34:46.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6783" PostTypeId="2" ParentId="6580" CreationDate="2013-11-04T15:14:49.500" Score="0" Body="&lt;p&gt;Since you are talking about s-orbitals and p-orbitals only(which would be a $ \sigma$bond),there is a head-on combination of the orbitals.This means that the orbitals would combine along the &quot;axes&quot;(each p orbital has both of its lobes aligned along the x-axis,or y-axis or z-axis,mathematically).The s-orbitals always combine head-on,since they are spherical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now,if two s-orbitals combine,it would be constructive interference if both are of same phase.If they are of opposite phases,there will be destructive interference.What you are talking about when you say &quot;add&quot; or &quot;subtract&quot; means &lt;strong&gt;constructive&lt;/strong&gt; and &lt;strong&gt;destructive&lt;/strong&gt; interference,respectively when the orbitals are of same phase;and &lt;strong&gt;destructive&lt;/strong&gt; and &lt;strong&gt;constructive&lt;/strong&gt; interference,respectively when orbitals are of different phases.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="660" LastEditDate="2013-11-07T20:17:15.463" LastActivityDate="2013-11-07T20:17:15.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6784" PostTypeId="1" CreationDate="2013-11-04T15:29:06.217" Score="1" ViewCount="115" Body="&lt;p&gt;What method did John Dalton use to measure the mass of one mole of gases, for example, chlorine and hydrogen, to get their relative masses?&lt;/p&gt;&#xA;" OwnerUserId="2634" LastEditorUserId="1573" LastEditDate="2013-11-04T16:24:22.250" LastActivityDate="2013-11-07T20:16:43.603" Title="Mass of a gas sample" Tags="&lt;history-of-chemistry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6785" PostTypeId="1" AcceptedAnswerId="44377" CreationDate="2013-11-04T15:47:21.523" Score="20" ViewCount="10228" Body="&lt;p&gt;I know that the VSEPR theory explains $\ce{IF_7}$ (iodine heptafluoride)'s structure as a pentagonal bi-pyramidal one. The valence bond theory can be used to say that it has $\mathrm{sp^3d^3}$ hybridisation (I think). How can the bonding in it be explained using the LCAO (Linear Combination of atomic orbitals) method in molecular orbital theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this doubt because there are only two &quot;axial&quot; d-orbitals ($\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$) among the 5 present, which can have head-on overlap with p-orbitals (non-axial orbitals can only have sideways overlap). In the excited state, the iodine atom has 7 half filled atomic orbitals (an s-orbital, three p-orbitals, and 3 d-orbitals) and according to LCAO each of them has to combine with one p-orbital of each fluorine atom to form molecular orbitals by &amp;#x3c3; and &amp;#x3c0; bonds. But, the formation of only 6 &amp;#x3c3; bonds can happen between an iodine and 6 fluorine atoms, as &amp;#x3c3; bonds involve head-on overlap of orbitals, and there are only 6 orbitals capable of doing that (the s and p orbitals, and the two 'axial' d-orbitals). Therefore, the seventh fluorine atom must be in p&amp;#x3c0;-d&amp;#x3c0; bond with fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't explain &quot;five equivalent bonds and two bonds of equal length (all &amp;#x3c3; bonds)&quot; as predicted by VSEPR theory. How many &amp;#x3c3; bonds does it really have?&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="402" LastEditDate="2017-04-17T00:55:29.353" LastActivityDate="2017-04-17T03:16:34.660" Title="How can the Bonding in IF7 be explained using LCAO method?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6787" PostTypeId="2" ParentId="6740" CreationDate="2013-11-04T17:13:16.997" Score="1" Body="&lt;p&gt;Having spoken to my Chemistry teacher, the idea is that the UV causes the molecules to split; after the fission, the molecules may rejoin or react with 'other radicals' to break the polymer down, hence degrading it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. for a polyester:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_1COOR_2 \xrightarrow{hv} R_1CO\boldsymbol\cdot + R_2O\boldsymbol\cdot$&lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2013-11-04T17:13:16.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6789" PostTypeId="2" ParentId="6784" CreationDate="2013-11-04T18:36:09.357" Score="2" Body="&lt;p&gt;The concept of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mole_%28unit%29#History&quot; rel=&quot;nofollow&quot;&gt;mole&lt;/a&gt;(c. 1805), and &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro%27s_number#History&quot; rel=&quot;nofollow&quot;&gt;Avogadro's number&lt;/a&gt;(1811) occurred after Dalton formulated his &lt;a href=&quot;http://en.wikipedia.org/wiki/Dalton%27s_law&quot; rel=&quot;nofollow&quot;&gt;law&lt;/a&gt; (1801). So he would not have set out to measure a &quot;mole of gas&quot; as the modern formulation sets out. This history of Dalton may be useful to you: &lt;a href=&quot;http://books.google.com/books?id=fIo-AAAAYAAJ&amp;amp;pg=PA26&amp;amp;lpg=PA26&amp;amp;dq=On%20the%20Absorption%20of%20Gases%20by%20Water%20and%20Other%20Liquids,%20John%20Dalton&amp;amp;source=bl&amp;amp;ots=8RTWb6v64Z&amp;amp;sig=VvRZBfCa0txIFgyKoOgKkoAOB9c&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=j-h3UqqqGqnXygHbioCgAw&amp;amp;ved=0CFMQ6AEwCQ#v=onepage&amp;amp;q=On%20the%20Absorption%20of%20Gases%20by%20Water%20and%20Other%20Liquids,%20John%20Dalton&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Google Book Search&lt;/a&gt; Most of his work centered around pressure readings, as his early interests were meterological.&lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastEditorDisplayName="user2603" LastEditDate="2013-11-04T18:41:42.343" LastActivityDate="2013-11-04T18:41:42.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6791" PostTypeId="1" CreationDate="2013-11-04T19:46:31.127" Score="7" ViewCount="37830" Body="&lt;p&gt;The reactions of primary and secondary alcohols with halogen acids ($\ce{HX}$, where $\ce{X}$=halogen) require the presence of a catalyst, $\ce{ZnCl_2}$. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated $\ce{HCl}$ at room temperature. Which mean that, tertiary alcohols are more reactive than primary and secondary alcohols. The reason is said to be, due to the inductive effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how the inductive effect would explain that, tertiary alcohol is more reactive than primary and secondary. &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2017-11-09T12:46:12.130" LastActivityDate="2019-07-05T11:08:55.607" Title="How does the inductive effect, explain the reactivity of tertiary alcohols with halogen acids?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;alcohols&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6792" PostTypeId="1" AcceptedAnswerId="6801" CreationDate="2013-11-04T20:16:07.350" Score="0" ViewCount="72" Body="&lt;p&gt;Is there a function $f(A)$ or $f(Z, A)$ that gets the energy of nuclear binding of one nuclide in atom? Or other way to get it? My chemistry book says only to get it from the chart which is very impractical. &lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2013-11-05T21:46:29.050" Title="A function for getting the nuclide's nuclear binding energy?" Tags="&lt;physical-chemistry&gt;&lt;nuclear&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6793" PostTypeId="2" ParentId="6784" CreationDate="2013-11-04T20:33:53.683" Score="0" Body="&lt;p&gt;To set '&lt;em&gt;relative&lt;/em&gt;' weights, Dalton took the lightest element known, hydrogen, and set an atomic weight of 1. Then he started to weigh other compounds,e.g., water and determined the ratios of the other elements. From this data, Dalton set the relative weight of several elements.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="660" LastEditDate="2013-11-07T20:16:43.603" LastActivityDate="2013-11-07T20:16:43.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6794" PostTypeId="1" CreationDate="2013-11-04T20:54:34.933" Score="0" ViewCount="331" Body="&lt;p&gt;So somebody told me that iron metal (zero charge) cannot exist at all when $\ce{H2O}$ is present because it will all turn into rust.&lt;br&gt;&#xA;How is this possible? Isn't there some sort of equilibrium?&lt;br&gt;&#xA;Does it have to do with the pH/pE? And if iron really can't exist, then what happens if I put an excess of iron into the water?&lt;/p&gt;&#xA;" OwnerUserId="2638" LastEditorUserId="660" LastEditDate="2013-11-07T20:17:00.883" LastActivityDate="2013-11-07T20:17:00.883" Title="Is it true that iron (Fe 0+) cannot exist when there's water around?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2021-04-18T16:59:56.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="6795" PostTypeId="2" ParentId="6775" CreationDate="2013-11-04T20:54:58.487" Score="0" Body="&lt;p&gt;The relative difference in energy between the double bond and single bond is much greater in oxygen than in sulfur, and so is the more stable $\ce{O2}$ diatomic molecule with a double bond, than molecules with multiple single bonds like $\ce{S8}$.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="402" LastEditDate="2017-04-08T22:30:59.873" LastActivityDate="2017-04-08T22:30:59.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6796" PostTypeId="1" CreationDate="2013-11-04T21:04:06.197" Score="3" ViewCount="3329" Body="&lt;p&gt;Alkyl halides are best prepared from alcohols ($\ce{R-OH}$, where $\ce{R}$ is alkyl group), which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acid ($\ce{HX}$, where $\ce{X}$ is halogen), to get alkyl halides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we don't need boiling the reactants. But, for the preparation of alkyl bromide requires constant boiling of alcohol with $\ce{HBr}$ ($48\%$). Assume that alcohol is tertiary, so that, we can avoid using catalyst. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought was, alcohol can develop inductive effect, and thus, can get detached with $\ce{OH}$, so that, the bromine from $\ce{HBr}$ can add to $\ce{R}$ group to form alkyl bromide. So, I thought there is no requirement of boiling of reactants, as like we do for preparation of alkyl chloride. Then, why do we need boiling in case of $\ce{HBr}$ and not in case of $\ce{HCl}$?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="41328" LastEditDate="2017-08-06T07:44:15.253" LastActivityDate="2017-08-06T07:44:15.253" Title="Why is alcohol boiled with HBr for preparing alkyl bromide?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6797" PostTypeId="2" ParentId="6791" CreationDate="2013-11-04T21:19:47.457" Score="10" Body="&lt;p&gt;Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following &lt;a href=&quot;http://en.wikipedia.org/wiki/SN1_reaction&quot; rel=&quot;noreferrer&quot;&gt;S&lt;sub&gt;N&lt;/sub&gt;1&lt;/a&gt; mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ROH &amp;lt;=&amp;gt; HX + ROH_2^+ + X^-} \tag{1}$$&#xA;$$\ce{ROH_2^+ &amp;lt;=&amp;gt; R^+ +H_2O} \tag{2} $$&#xA;$$\ce{R+ +X- \rightarrow  RX} \tag{3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VpCUx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack! &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="11551" LastEditDate="2015-01-04T10:25:31.967" LastActivityDate="2015-01-04T10:25:31.967" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6798" PostTypeId="1" CreationDate="2013-11-04T22:27:22.913" Score="3" ViewCount="1992" Body="&lt;p&gt;My book says, we can't prepare aryl halides ($\ce{RX}$, where $\ce{R}$ is aryl group and $\ce{X}$ is any halogen) by treating phenol with halogen acids ($\ce{HX}$, where $\ce{X}$ is halogen). It is said to be, because of partial double bond character of carbon-oxygen bond in phenols ($\ce{C6H5OH}$), which is difficult to break (I don't know how there is partial double bond character). What I thought was, if we assume there is no partial double bond character between $\ce{C}$ and $\ce{O}$, we can heat the reactants (i.e. phenol and halogen acid), so that $\ce{OH}$ bond eliminates from the phenol (which has the least bond dissociation energy than $\ce{CH}$), and $\ce{X}$ can get dissociated from the $\ce{HX}$, then it can add up to $\ce{R}$ (aryl group) to form aryl halide. So, can we prepare aryl halides by heating with halogen acids?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorDisplayName="user7951" LastEditDate="2015-12-31T20:37:47.127" LastActivityDate="2016-01-01T01:19:53.373" Title="Can we prepare aryl halides by heating with halogen acids?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6799" PostTypeId="1" CreationDate="2013-11-04T22:37:23.107" Score="3" ViewCount="1943" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In which orbital, 4f or 6s, would an electron have a greater likelihood of being near the nucleus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I figured that the 6s would be closest, since it is filled in first when using the Aufbau principle.&lt;/p&gt;&#xA;&lt;p&gt;However, does that necessarily imply that it is closer to the nucleus? Or would 4f be closer, since its principal quantum number is smaller?&lt;/p&gt;&#xA;" OwnerUserId="2220" LastEditorUserId="69830" LastEditDate="2021-05-26T06:28:25.480" LastActivityDate="2021-05-26T06:28:25.480" Title="Are orbitals always filled in from closest to nucleus to farthest away?" Tags="&lt;electrons&gt;&lt;orbitals&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6800" PostTypeId="2" ParentId="917" CreationDate="2013-11-04T22:58:19.557" Score="1" Body="&lt;p&gt;This article should be useful for you...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Material Recovery and Characterization of PCB from Electronic Waste&lt;/strong&gt;&lt;br&gt;&#xA;  Jessica Hanafi, Eric Jobiliong, Agustina Christiani, Dhamma C. Soenarta, Juwan Kurniawan, Januar Irawan, (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1877042812046563&quot; rel=&quot;nofollow&quot;&gt;doi:10.1016/j.sbspro.2012.09.1194&lt;/a&gt;)&lt;br&gt;&#xA;  &lt;em&gt;Abstract&lt;/em&gt;&lt;br&gt;&#xA;  Printed Circuit Boards are commonly exist in computers, laptops and mobile phones. Since it contains of a variety of metals and semiconductors, it is difficult to recover the materials in PCB. This paper aims to characterize the composition of these materials and a method to properly recycle and recover the materials in them. This paper utilized simple mechanical and chemical procedures to recover PCBs. Collected PCBs were disassembled, pulverized and separated by using density and magnetic separation method. By using the chemical recycling method, 98.82% purity of CuSO4 hydrate and Al2(SO4)3 hydrate were recovered from the PCBs. It is found that the milling methods, size of the sample and the density of the separating solution determine the effectiveness of the materials recovered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2639" LastEditorUserId="4945" LastEditDate="2014-11-17T07:35:50.690" LastActivityDate="2014-11-17T07:35:50.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6801" PostTypeId="2" ParentId="6792" CreationDate="2013-11-04T23:19:31.910" Score="1" Body="&lt;p&gt;Nuclear physics is very complicated and full of unsolved problems. For example, we seem to understand rather poorly the effects of &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-body_force&quot; rel=&quot;nofollow&quot;&gt;three-body forces&lt;/a&gt; between nucleons, and describing them quantitatively is probably out of the question. Therefore, I expect that nuclei binding energies for most isotopes are solely determined empirically and have no good theoretical curve to originate from. Maybe there are statistically fitted curves to the data we have, though they may well not have a simple form and likely are only good approximations for relatively small intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As it turns out, &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-empirical_mass_formula&quot; rel=&quot;nofollow&quot;&gt;equations indeed exist&lt;/a&gt;, though as expected they are quite unwieldy and are semi-empirical.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2013-11-05T21:06:17.407" LastActivityDate="2013-11-05T21:06:17.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6802" PostTypeId="1" CreationDate="2013-11-05T02:03:59.187" Score="1" ViewCount="7853" Body="&lt;p&gt;I can't find a simplified explanation on this topic. My textbook just adds to the confusion and all I'm really left with are more questions rather than answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And adding to that question, does polarity have an effect on electron localization/delocalization?&lt;/p&gt;&#xA;" OwnerUserId="2640" LastEditorUserId="73" LastEditDate="2013-11-05T02:53:39.270" LastActivityDate="2014-02-04T18:43:25.447" Title="How does electron localization/delocalization work?" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6805" PostTypeId="1" AcceptedAnswerId="6813" CreationDate="2013-11-05T02:27:05.313" Score="2" ViewCount="46521" Body="&lt;p&gt;The concentrations of $\ce{NaCl}$ and $\ce{KI}$ are at $\pu{0.120 M}$ and $\pu{0.090 M}$, respectively, in the following electrochemical cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu_{(s)}|CuI_{(s)}|I^{–}_{(aq)}||Cl^{–}_{(aq)}| AgCl_{(s)}|Ag_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the cell voltage using the following half-reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{Cu+ + e- &amp;amp;&amp;lt;=&amp;gt; Cu(s) &amp;amp; E^\circ &amp;amp;= \pu{0.518 V}} \\&#xA;\ce{Ag+ + e- &amp;amp;&amp;lt;=&amp;gt; Ag(s) &amp;amp; E^\circ &amp;amp;= \pu{0.7993 V}} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility products $(K_{sp})$ for $\ce{AgCl}$ and $\ce{CuI}$ are $\pu{1.8e–10}$ and $\pu{1.0e-12}$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="2642" LastEditorUserId="41328" LastEditDate="2017-08-05T21:27:09.167" LastActivityDate="2017-08-05T21:27:09.167" Title="How do I calculate the cell voltage using the following half-reactions and solubility products?" Tags="&lt;electrochemistry&gt;&lt;solubility&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6806" PostTypeId="2" ParentId="5734" CreationDate="2013-11-05T02:28:56.703" Score="4" Body="&lt;p&gt;Certain metals under certain conditions form highly covalently bonded structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain allotropes (phases) of metals have no metal qualities such as alpha-tin. Formed at about 13.2 degrees celcius with pure tin- it has no metallic qualities and pure covalent bonds. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-11-05T02:28:56.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6808" PostTypeId="1" CreationDate="2013-11-05T06:23:32.227" Score="3" ViewCount="1815" Body="&lt;p&gt;Say, you conducted two separate reactions; in one, you combined $\mathrm{50~mmol}$ saturated $\ce{KCl}$ solution with $\mathrm{10~mmol}$ 1-bromooctane (assume heating), and in the other you combined $\mathrm{10~mmol}$ saturated $\ce{KI}$ solution with $\mathrm{10~mmol}$ 1-bromoctane. In both cases $\mathrm{0.5~mmol}$ hexadecyltributylphosphonium bromide was used as a catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, due to relative charge densities and the fact that the electrophillic carbon is primary, you would have a low yield of 1-chloroocatane vs starting material in reaction 1 and you would have a high yield of 1-iodooctane vs starting material in reaction 2, both cases obtained via SN2 reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose you would have repeated the same two reactions, only this time you would have reacted the $\ce{KCl}$ and $\ce{KI}$ with 2-bromo, 2-methyloctane, how exactly would the results differ from the results of the previous two reactions, or would they have been the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the reaction would have no &quot;choice&quot; but to react via the SN1 (or perhaps E1) route, as the electrophillic carbon is tertiary and thus very sterically hindered. Also, I know that $\ce{I-}$ ions are good nucleophiles, but that doesn't matter in this case, &lt;em&gt;does it?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this reaction would go slower than the previous, owing to the fact that it requires the $\ce{Br-}$ to dissociate, and therefore there would be lower yields of both haloalkane products. But then again, wouldn't it go faster owing to the fact that it is in a polar protic solvent, and that the formed carbocation would be tertiary?&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="10185" LastEditDate="2015-10-02T16:55:59.003" LastActivityDate="2015-10-02T16:55:59.003" Title="How would substitution reaction results differ between reacting halides with a tertiary bromoalkane vs reacting it with a primary bromoalkane?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6809" PostTypeId="2" ParentId="6799" CreationDate="2013-11-05T07:11:27.347" Score="2" Body="&lt;p&gt;As you have studied for the quiz, you should know the expected answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In full-scope reality, the question has nothing to do with the two explanations you offered, but with the shape of atomic orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f0aC2.gif&quot; alt=&quot;Radial distribution of selected orbitals in $\ce{Sm^{3+}}$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial distribution of selected orbitals in $\ce{Sm^{3+}}$, &lt;a href=&quot;http://pubs.rsc.org/en/content/articlehtml/2004/dt/b407173e&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.rsc.org/en/content/articlehtml/2004/dt/b407173e&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-05T07:11:27.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6810" PostTypeId="2" ParentId="6802" CreationDate="2013-11-05T07:24:17.373" Score="1" Body="&lt;p&gt;There is no simplified explanation, and especially not of &quot;how it works&quot;. The behavior of electrons is governed by quantum mechanics. So electrons do what they do, because it is energetically favorable. It should become quite clear upon obtaining master degree in theoretical chemistry ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a teaser, I'd at least show the pictures:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Localized sigma bond, localized means it is as compact in space, as possible, now connecting two carbon atoms in ethane:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mL6x0.gif&quot; alt=&quot;Localized sigma bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Delocalized pi electrons in benzene, shared by 6 atoms:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/neeRn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-05T07:24:17.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6811" PostTypeId="2" ParentId="6808" CreationDate="2013-11-05T07:43:19.460" Score="1" Body="&lt;p&gt;There are (at least) four major factors to consider in predicting the results of these reactions: (a) the nature of the substrate, (b) the strength of the nucleophile, (c) the quality of the leaving group, and (d) solvent effects. Halides are able to function, to varying degrees, as both good leaving groups and strong nucleophiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In protic solvents, negative charges are stabilized by the $\ce{\delta^{+}}$ of the solvent hydrogens bonded to some electronegative atom (conventionally limited to $\ce{N, O, F}$). Since nucleophiles carry either a full or partial negative charge, protic solvents tend to reduce the reactivity of nucleophiles. Because smaller nucleophiles are more easily and thoroughly solvated than large and polarizable nucleophiles, the effect on them is more drastic. Consequently, in protic solvents, the relative nucleophilicity of the halides is generally given as $\ce{I- &amp;gt; Br- &amp;gt; Cl- &amp;gt; F-}$. In aprotic solvents, the reactivity is precisely the opposite. Hence, the proper selection of solvent is crucial. In your first example, there is no particular reason I can think of to suppose that the yield of 1-chlorooctane would inevitably be low. Indeed, in an aprotic solvent, the usual choice for SN2 reactions, $\ce{Cl-}$ can be expected to be a stronger nucleophile and poorer leaving group than $\ce{Br-}$. I would expect the yield to be good, especially with an excess of $\ce{KCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the matter of the substrate, with a tertiary carbon, SN2 is all but completely impossible. In that situation, elimination reactions are typically favored if some base is present. The halides, however, are extremely weak bases (being the conjugate bases of strong acids), so if a non-basic solvent is selected, substitution by an SN1 mechanism seems most probable. Selection of a relatively non-basic, non-nucleophilic protic solvent would be ideal. (Off the top of my head, DMSO comes to mind, which is only weakly nucleophilic, especially by comparison to the halides.) You're right that the strength of the nucleophile is not the deciding factor in an SN1, since the rate-determining step is the loss of the leaving group and the resultant carbocation formation. In a protic solvent, $\ce{I-}$ should theoretically be more effective as a nucleophile than $\ce{Cl-}$. However, I would still expect the latter reaction to work, particularly if the concentration of $\ce{Cl-}$ is high. I'm fairly confident that the difference in reactivity between, e.g., $\ce{Cl-}$ and $\ce{Br-}$ would be largely negated if the former is available in vastly larger concentration in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: On a second reading, I just noticed that you labeled the halide solutions as aqueous. I don't know if that's deliberate, but I don't think that would make sense. For one thing, the reagents (namely, the haloalkanes) would be largely insoluble in water. Second, because water is an amphoteric species, it could be potentially reactive.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-11-05T07:52:10.143" LastActivityDate="2013-11-05T07:52:10.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6813" PostTypeId="2" ParentId="6805" CreationDate="2013-11-05T08:21:02.370" Score="8" Body="&lt;p&gt;I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $R$ is the familiar gas law constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = 298.15\ \mathrm{K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z$ is the number of moles of electrons transferred in the half-reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $F$ is Faraday’s constant, which is measured in coulombs per mole: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$&#xA;$$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$&#xA;$$1.0 × 10 ^{-12} = [X][0.090 + X]$$&#xA;Do the quadratic formula&#xA;$$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$&#xA;$$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$&#xA;$$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$&#xA;$$1.8 \times 10^{-10} = [X][0.120 + X]$$&#xA;Do the quadratic formula&#xA;$$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$&#xA;$$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=0.00740$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, plugging this all into the Nernst equation, the answer should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=-0.155\ \mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will give the same answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$&#xA;$$E_\text{cell}=-0.155\ \mathrm{V}$$&lt;/p&gt;&#xA;" OwnerUserId="2630" LastEditorDisplayName="user7951" LastEditDate="2015-05-16T15:57:58.500" LastActivityDate="2015-05-16T15:57:58.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6814" PostTypeId="2" ParentId="6794" CreationDate="2013-11-05T10:09:24.220" Score="1" Body="&lt;p&gt;Iron metal can exist in presence of water. As you stated, important point is the $\ce{Fe^0 -&amp;gt; Fe^3+}$ reaction. Water is not an oxidizing agent, therefore is not capable of inducing this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create rust, you need also oxygen, or as you noted, electrochemical potential.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-05T10:09:24.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6815" PostTypeId="2" ParentId="6750" CreationDate="2013-11-05T10:23:08.283" Score="2" Body="&lt;p&gt;&lt;strong&gt;Your assumptions are correct&lt;/strong&gt;, and from the wording of the question, the &quot;correct&quot; answer doesn't make sense, unless like you said, the object gained heat, which is possible; people have negative profits all the time. Sign conventions are not flawed in chemistry, but you just have to be careful when wording a question. To answer the title of your question, yes it is not only customary, it is &lt;em&gt;necessary&lt;/em&gt; to express heat loss as a negative number, but the question still has to be phrased correctly--using the term &quot;lost&quot; serves as a negation in this use of language, the same way the unary operator in -1 serves as negation for a number. If heat loss were not expressed negatively, it wouldn't make sense because the system of an exothermic reaction must lose energy for its surroundings to gain it.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2013-11-05T10:23:08.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6816" PostTypeId="2" ParentId="6729" CreationDate="2013-11-05T10:59:23.723" Score="4" Body="&lt;p&gt;&lt;strong&gt;Distinction:&lt;/strong&gt; you are only considering Kp, which is an equilibrium constant that doesn't take concentration into account. You're going wrong here because the question asks for concentration, which has a unit of $L^{-1}$, not mole fraction, which is just in units of moles. &#xA;$$Kc=[C]c[D]d / [A]a[B]b$$&#xA;$$Kp=(C)c(D)d / (A)a(B)b$$&#xA;Think about their answer to the question. It states that $K=Cc_{pre}$ and $K=Cc_{post}$. The only way for this to make sense is if $Cc$ is in units of concentration because then as the total moles of c go up, the fact that the volume is decreasing doesn't change the value of Kp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If   $K_{pre}=K_{post}$ &#xA;  , we have $C_{\ce{B},pre} \cdot C_{\ce{C},pre}=C_{\ce{B},post} \cdot C_{\ce{C},post}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{post}}&#xA;{b_{pre}}=\dfrac{c_{post}}{c_{pre}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you do not take concentration into account, or the fact that $Cc_{pre} = Cc_{post}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For then you cannot restate your equation in terms of concentration: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2013-11-05T10:59:23.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6817" PostTypeId="1" CreationDate="2013-11-05T12:00:45.500" Score="0" ViewCount="10134" Body="&lt;p&gt;What is the best alcohol for your teeth according to the pH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that if the acidity of the drink is higher (and ph is lower) than it's worse for your teeth (enamel erosion + bacteria friendly acidic environment) than if the alcohol drink is less acidic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does vodka have the highest pH of all the alcohol drinks out there and therefore is the best for your teeth?&lt;/p&gt;&#xA;" OwnerUserId="2649" LastEditorUserId="2649" LastEditDate="2013-11-05T12:09:53.077" LastActivityDate="2014-01-22T16:45:43.760" Title="Teeth friendly alcohol drinks" Tags="&lt;ph&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6818" PostTypeId="2" ParentId="6785" CreationDate="2013-11-05T13:40:31.880" Score="7" Body="&lt;p&gt;The structure of $\ce{IF_7}$ cannot be understood as simple static pentagonal bipyramid. The full explanation is given in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00057a044&quot; rel=&quot;nofollow&quot;&gt;Karl O. Christe, E. C. Curtis, David A. Dixon, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1993,&lt;/strong&gt; &lt;em&gt;115&lt;/em&gt; (4), 1520–1526.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last sentence of abstract states clearly describes the bonding situation: These features can be explained, however, by a bonding scheme involving a planar, delocalized pxy hybrid on the central atom for the formation of five equatorial, semi-ionic, 6-center 10-electron bonds and an sp hybrid for the formation of two mainly covalent axial bonds. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="4945" LastEditDate="2016-01-20T06:05:50.360" LastActivityDate="2016-01-20T06:05:50.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6820" PostTypeId="2" ParentId="6817" CreationDate="2013-11-05T14:32:08.987" Score="1" Body="&lt;p&gt;Is quite reductive think that pH is the onlyfactor that determinate the health of your teeth, the content of carbohydrate have a strong effect too.&#xA;However in fact to my knowledge clear vodaka have one of the highest pH and very few carbohydrates so indeed is one of the most &quot;teeth friendly alcohol drink&quot;.&#xA;These is take from &lt;em&gt;Victor R. Preedy, Ronald Ross Watson Academic Press, 24/dic/2004 Comprehensive Handbook of Alcohol Related Pathology&lt;/em&gt; with the adding of Vodka and Coca Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|}\hline&#xA;Alcolic \space beverage &amp;amp; Carboydrate \space content(\%) &amp;amp; pH \\ \hline&#xA;Beer               &amp;amp; 3.0-5.0   &amp;amp;  4.1-4.5    \\ \hline&#xA;Wine &amp;amp; 0.0-12.9 &amp;amp;  2.8-3.8     \\ \hline&#xA;Sweet \space liquor           &amp;amp; 30.0-31.0 &amp;amp; 3.3-3.9 \\ \hline&#xA;Strong \space alcohol &amp;amp; 0.0-1.2 &amp;amp;   6.5-6.9\\ \hline&#xA;Vodka                &amp;amp; very \space few       &amp;amp; 6.0-7.0        \\ \hline&#xA;Coca Cola &amp;amp; 10.6(?) &amp;amp; 2.8\\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However keep out  from Vodka lemon!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2013-11-16T16:43:42.950" LastActivityDate="2013-11-16T16:43:42.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6821" PostTypeId="1" CreationDate="2013-11-05T20:44:21.197" Score="0" ViewCount="9929" Body="&lt;p&gt;I'm studying organic chemistry at the moment and it's really confusing. What could ever happen between benzene and an alkene? and why is $\ce{H2SO4}$ there? &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="418" LastEditDate="2014-01-05T08:35:34.433" LastActivityDate="2014-02-04T20:36:42.180" Title="Reaction of Benzene with 2-methylpropene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6822" PostTypeId="2" ParentId="6817" CreationDate="2013-11-05T21:13:14.453" Score="1" Body="&lt;p&gt;Sugar content will be the determining factor in tooth decay, my dentist informed me when I asked her this very question that it's not the inherent acidity of the drink / food stuff that does the damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does cause damage is the bacteria in your mouth digesting simple sugars over a long period of time. From &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Dental_caries&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tooth decay disease is caused by specific types of bacteria that&#xA;  produce acid in the presence of fermentable carbohydrates such as&#xA;  sucrose, fructose, and glucose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took this advice as meaning I should take all my drinks neat of course.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2013-11-05T21:13:14.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6823" PostTypeId="2" ParentId="6821" CreationDate="2013-11-05T21:21:38.410" Score="0" Body="&lt;p&gt;First, I think the name of your molecule must be malformed. That said, the reaction of an alkene with strong acid can generate a carbocation. That carbocation is electrophilic, and can react with benzene (as the nucleophile) in an electrophilic aromatic substitution reaction. The final product is the (potentially poly-)alkylated benzene.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-11-05T21:21:38.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6824" PostTypeId="2" ParentId="6792" CreationDate="2013-11-05T21:46:29.050" Score="1" Body="&lt;p&gt;I hope to be not wrong, but believe recalling this to be one of the holy grails in theoretical nuclear physics and a testing benchmark for most comprehensive [field] theories on that subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A given theory's proper prediction of binding energies/ masses is also a pre requirement towards the concept of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Theory_of_Everything&quot; rel=&quot;nofollow&quot;&gt;Theory of Everything&lt;/a&gt; - should it even be feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember as an undergrad having been fascinated for months on end by the obscure &lt;a href=&quot;http://en.wikipedia.org/wiki/Heim_theory&quot; rel=&quot;nofollow&quot;&gt;Heim theory&lt;/a&gt;, trying to approach its mathematics. It took me a while to un-glamour myself and realize that it is virtually impossible for a theory so old and unaccepted to be as good as its advocates state. Statistically speaking, there is always a chance that you can derive at magical results with a lucky mixture of the right constants and formulas. But these new constants would then need to reveal meaning in other natural concepts as well, in order to make sense in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much of the status quo, but I would surmise deriving the binding energies is still a matter of more or less competing theories, each with successes in their own right, but none covering the entire chemical elemental spectrum solely based on fundamental physical constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I recall there are &lt;strong&gt;several semi-empirical formulas&lt;/strong&gt;. Nicolau Saker Neto, already pointed out one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model_%28mathematical_formulation%29#Mass_terms_and_the_Higgs_mechanism&quot; rel=&quot;nofollow&quot;&gt;Standard Model&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="660" LastActivityDate="2013-11-05T21:46:29.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6825" PostTypeId="2" ParentId="6775" CreationDate="2013-11-05T22:28:21.313" Score="10" Body="&lt;p&gt;This is essentially the same question as to why carbon concatenates through up to four bonds with binding energies low enough to be broken again yet stable enough to last given typical terrestrial environmental conditions. Great for entities like us.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, the binding energies of Silicon, Germanium, from the same group IV, for instance, are higher. The binding energies are directly related to the bond length/bond distance, which is great for determining the energies through crystallography, along with proper model fitting.&lt;/p&gt;&#xA;&lt;p&gt;Sulfur is one of few elements amenable to catenation, - forming long molecules with each other.&lt;/p&gt;&#xA;&lt;p&gt;So whichever configuration releases the greatest enthalpy is typically the most stable one, but depends upon the initial energies thus giving rise to allotropes - elemental structures of different molecular configurations. The bound system typically has lower potential energy.&#xA;If a double bond lowers the overall energy of the bound system more it will typically form over a single bonded one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I found this (see also &lt;a href=&quot;http://www.q1.fcen.uba.ar/materias/qi1/Tablas/disocia.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table of dissociation energies&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update 2:&lt;/strong&gt; I'd like to add that from a quantum mechanical perspective, which is a viewpoint much closer to nature, the bonds arise from 3D space where electron densities are highest - which means most energetically favorable.&lt;/p&gt;&#xA;&lt;p&gt;Since for whatever weird reason, nature is really quantized, and I cannot stress that &amp;quot;really&amp;quot; part enough, one gets standing waves for the electron equations around a given nucleus - rather than a continuum - as is the case of a sphere, like say a falling waterdrop to add a few exterior forces into the mix as well. The classical and the quantum object have only in common the &amp;quot;energetically favorable&amp;quot; state.&lt;/p&gt;&#xA;&lt;p&gt;Somehwat simplified then, those standing waves become &amp;quot;orbitals&amp;quot; when doing the calculations for 3D space. The orbitals change in shape with increasing interference with other electrons and higher potential energies due to larger nuclei , meaning higher positive proton counts.&lt;/p&gt;&#xA;&lt;p&gt;The energies involved result in very different packing i.e. different lattices. A necessity for life is breaking bonds or &amp;quot;bond accessibility&amp;quot;. There simply isn't &amp;quot;enough energy&amp;quot; in today's protein-scaffold based catalyists on earth, to use anything else than the constituents of the periodic table with few protons in case of homogenous bonds (i.e. atoms of the same species like say Oxygen O2).&lt;/p&gt;&#xA;&lt;p&gt;It does not have to be this way. We from earth are biased by our own biochemistry we observe. Every human generation breaks down the biases of the previous ones and encounters new ones.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Bond energies&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 S8: 1808 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;1 O8: 1168 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;4 S2: 1408 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;4 O2: 1976 kJ/mol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="660" LastEditorUserId="660" LastEditDate="2020-10-27T08:58:06.277" LastActivityDate="2020-10-27T08:58:06.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6826" PostTypeId="1" AcceptedAnswerId="6828" CreationDate="2013-11-06T07:11:40.030" Score="8" ViewCount="16717" Body="&lt;p&gt;Why is &lt;span class=&quot;math-container&quot;&gt;$\Delta G = -nFE$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what the motivation is behind this definition. Was it derived or just given? The textbook provides no justification for this equation. In fact, much of the book associated with the Gibbs free energy provides no justification and just says, &lt;em&gt;'This is how it is. Now go and solve some problems.'&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="4945" LastEditDate="2020-06-02T10:32:50.653" LastActivityDate="2021-01-23T07:33:11.900" Title="Derivation of relationship between Gibbs free energy and electrochemical cell potential" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;free-energy&gt;&lt;reduction-potential&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6827" PostTypeId="1" CreationDate="2013-11-06T07:35:25.683" Score="6" ViewCount="20053" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is $\Delta G=\Delta G^o+RT\ln Q?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It feels like all online sources were written for introductory Chemistry students! Where do I find a rigorous proof of this identity? Greatly appreciate it!&lt;/p&gt;&#xA;" OwnerUserId="1758" LastEditorUserId="1758" LastEditDate="2013-11-07T03:58:16.900" LastActivityDate="2020-02-18T13:41:05.940" Title="Derivation of relationship between equilibrium constant and Gibbs free energy change" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6828" PostTypeId="2" ParentId="6826" CreationDate="2013-11-06T08:09:52.993" Score="7" Body="&lt;p&gt;There are two ways to understand this equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is to realise that (reversible ideal case) &lt;span class=&quot;math-container&quot;&gt;$\Delta G = W_\text{non-exp}$&lt;/span&gt; (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;, to move one mole electrons across the external circuit will be &lt;span class=&quot;math-container&quot;&gt;$FE$&lt;/span&gt;, which must be equal to the decrease in gibbs free energy of the system. Hence for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; mole electrons transferred at the &lt;em&gt;same&lt;/em&gt; potential, &lt;span class=&quot;math-container&quot;&gt;$W_\text{non-exp} = \Delta G = -nFE$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that &lt;span class=&quot;math-container&quot;&gt;$\Delta G = W_\text{non-exp}$&lt;/span&gt; can be derived as under:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\mathrm{d}S &#xA;  &amp;amp;= \frac{\delta q}{T}  &amp;amp;&amp;amp; \text{(reversible case)}\\&#xA;\mathrm{d}U &#xA;  &amp;amp;= \delta q + W_\text{non-exp} + W_\text{exp}\\&#xA;W_\text{non-exp} &#xA;  &amp;amp;= \mathrm{d}U - W_\text{exp} - \delta q \\&#xA;  &amp;amp;= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S &#xA;  &amp;amp;&amp;amp; \text{(const. $p$ and $T$)} \\&#xA;  &amp;amp;= \mathrm{d}H - T\,\mathrm{d}S \\&#xA;  &amp;amp;= \mathrm{d}G&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to use &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; is the electric potential at the point, &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the charge on the species. Using this and &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$&lt;/span&gt;&#xA;we get the required equation for &lt;span class=&quot;math-container&quot;&gt;$z = 1$&lt;/span&gt; for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a &lt;em&gt;rigorous&lt;/em&gt; derivation for the electrochemical potential but &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrochemical_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has some basics on it.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2020-06-02T16:23:27.570" LastActivityDate="2020-06-02T16:23:27.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6829" PostTypeId="1" CreationDate="2013-11-06T09:08:30.913" Score="5" ViewCount="3707" Body="&lt;p&gt;A venerable question, as old as LCDs themselves. I assume it has something to do with pressure-induced changes in the path length across the display, but that's just a guess. Of particular interest to me as I press my fingernail into my (TFT LCD) phone screen&lt;sup&gt;1&lt;/sup&gt; are the observations of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A symmetrical bilobate distortion in the direction of movement of my fingernail.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ripples of colour distortion that radiate outwards at about a centimeter per second.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A trail of slight colour distortion that with a persistence of a few seconds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any ideas as to the cause of this rich behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]&lt;/sup&gt; Don't worry, it's a nokia n95 and as such will likely outlive the human race if not the solar system.&lt;/p&gt;&#xA;" OwnerUserId="84" LastActivityDate="2019-12-26T07:55:56.820" Title="Why do liquid crystal displays (LCDs) visually distort under pressure?" Tags="&lt;physical-chemistry&gt;&lt;liquid-crystals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6830" PostTypeId="2" ParentId="6829" CreationDate="2013-11-06T10:15:26.560" Score="4" Body="&lt;p&gt;Liquid crystal displays work by using liquid crystals where the direction of alignment of the molecules in a plane rotates as you go from layer to layer. Usually this is done with &lt;a href=&quot;http://en.wikipedia.org/wiki/Twisted_nematic_field_effect&quot; rel=&quot;nofollow&quot;&gt;twisted nematic&lt;/a&gt; crystals, however &lt;a href=&quot;http://en.wikipedia.org/wiki/Cholesteric_liquid_crystal&quot; rel=&quot;nofollow&quot;&gt;cholesteric LCs&lt;/a&gt; can also be used to achieve similar effects (I don't think smectic LCs are useful here as they don't align).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a liquid crystal cell, twisted nematic liquid crystal is filled between two crossed polarizers&lt;sup&gt;1&lt;/sup&gt;, and a backlight is sent through. The important thing about liquid crystals is that the specific rotation is easy to control by varying the potential difference (more potential difference leads to more anisotropic disorder). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you make a grid of these and put color filters on them, we get an RGB LCD display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when you press on them? You change the concentration/length of the liquid crystal, as well as making it bulge out slightly&lt;sup&gt;2.&lt;/sup&gt;. This changes the optical rotation it induces, causing the amount of light that comes out to change (as the amount of light is dependant on how aligned the rotated light is with the second polarizer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not too sure of the exact reasons behind the phenomena you list, but I can guess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the color changes are caused because each type of crystal cell is affected differently by the pressure. I think there's a difference in concentration between the different color cells to better suit human eyes. When I do it to LCD displays accessible to me, I only get a darkening, not a color change, so it seems to not be a universal feature of LCD displays. If the rotation of each color cell is affected differently by pressure, then pressure can easily cause color distortions as each color will not darken by the same amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the ripples are caused by high pressure where you change the optical rotation of the inner cells (the ones closer to your finger) so much that they go through a at least one full 360&lt;sup&gt;o&lt;/sup&gt; rotation, leading to regions of both bright and dark light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The persistence is because liquid crystals are viscous &amp;mdash; they have to be, as they consist of long molecules. They seem to align quickly to potential differences (potential differences align the molecules without moving the liquid), but when we add pressure (where the liquid itself distorts) I think it takes some time to bounce back. The ripples move due to this same persistence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. If you're ever at a 3D movie, have a look at your phone screen with the 3D polarised glasses on. It looks pretty cool!&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2. If this did not occur, then the rotation would stay the same as $\theta\propto lc$ and $lc$ does not change if there is no bulge.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-11-06T10:15:26.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6831" PostTypeId="2" ParentId="6826" CreationDate="2013-11-06T12:18:23.210" Score="6" Body="&lt;p&gt;I'm surprised your textbook did not derive this equation from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_equilibrium#Historical_introduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction isotherm relationship&lt;/a&gt; between &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; and the reaction quotient &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nernst equation&lt;/a&gt;. The derivation is not hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction isotherm equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_\mathrm{r} G =\Delta_\mathrm{r} G^\circ + RT\ln Q$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nernst equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{RT}{nF}\ln Q$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we solve both equations for &lt;span class=&quot;math-container&quot;&gt;$RT\ln Q$&lt;/span&gt;, we get your equation (almost).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  RT\ln Q &#xA;    &amp;amp;= \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ\\&#xA;  RT\ln Q &#xA;    &amp;amp;= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell}\\&#xA;  \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ &#xA;    &amp;amp;= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is my equation not as simple as the one you started with? Your equation is at equilibrium, and I assumed that we might not be at equilibrium. At equilibrium, the following are true, which simplify the relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;Q &amp;amp;= K\\&#xA;\Delta_\mathrm{r} G &amp;amp;= 0\\&#xA;E_\mathrm{cell} &amp;amp;= 0&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm{r} G = 0$&lt;/span&gt; because the reaction has achieved a minimum energy state — the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical potential&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\mu_i = \left(\frac{\partial G}{\partial N_i}\right)_{T,P}$&lt;/span&gt; is also &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; because there is no net change of state at equilibrium. Similarly &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{cell} = 0$&lt;/span&gt; at equilibrium. There is no change of state, and thus the redox reaction has ground to a halt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, the final relationship is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_RG^\circ = nFE^\circ_\mathrm{cell}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2020-06-02T10:35:16.377" LastActivityDate="2020-06-02T10:35:16.377" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6832" PostTypeId="1" CreationDate="2013-11-06T12:50:18.867" Score="2" ViewCount="394" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c1cHj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mom gave puffed rice to eat, I had quite wet hands. I saw that, each puffed rice pieces started hanging to the hand, when I brought my hand in contact with them. I thought about electrostatics initially to be the cause. Actually, the electrostatics usually deals with rough surfaces. Here due to hand being wet, it has got smooth surface. Then, what might be reason? I started thinking, suddenly I thought it to be because of hydrophillic nature of kadale puri (puffed rice). Is it really because of hydrophillic nature of puffed rice?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2013-11-10T07:19:11.707" LastActivityDate="2013-11-10T07:19:11.707" Title="Why do puffed rice stick to wet hands?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;&lt;bond&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6834" PostTypeId="1" AcceptedAnswerId="6836" CreationDate="2013-11-06T17:03:10.790" Score="24" ViewCount="81317" Body="&lt;p&gt;Is it possible to melt diamonds into a liquid? I mean if you heat diamond at open air it will start to burn around 700 degrees Celsius, reacting with oxygen to produce carbon dioxide gas. In the absence of oxygen it will transform into graphite, a more stable form of carbon, long before turning into graphite. My question is, is it possible to melt diamonds into liquid? If so, then, how? If not, then why?&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="7448" LastEditDate="2015-10-16T00:10:38.070" LastActivityDate="2017-05-29T18:24:04.193" Title="Is it possible to melt diamonds into a liquid?" Tags="&lt;bond&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6835" PostTypeId="2" ParentId="6834" CreationDate="2013-11-06T17:46:19.903" Score="8" Body="&lt;p&gt;Carbon, which makes up diamond, can certainly exist as a liquid and phase transition between the diamond phase and the liquid phase seems to be possible based on &lt;a href=&quot;http://phycomp.technion.ac.il/~anastasy/teza/teza/node5.html&quot;&gt;this phase diagram&lt;/a&gt;. Whether diamonds melting can actually be observed depends on the kinetics though: The phase transition may be too slow to be observed on a human time scale.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditDate="2013-11-06T18:00:29.497" LastActivityDate="2013-11-06T18:00:29.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6836" PostTypeId="2" ParentId="6834" CreationDate="2013-11-06T17:53:34.537" Score="27" Body="&lt;p&gt;Liquid carbon does indeed exist, but perhaps surprisingly, relatively little is known about it. It only exists above around $4000\ \mathrm{K}$ and $100\ \mathrm{atm}$, which are not trivial conditions to sustain and probe. There certainly are many theoretical studies into the properties of liquid carbon, though. You can find a phase diagram for carbon &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Carbon-phase-diagramp.svg&quot;&gt;here&lt;/a&gt;, though it is likely just a rough guide. The specific bonding properties of carbon atoms in the liquid phase still seems uncertain to some extent because carbon is very versatile and can create very dissimilar structures (easily seen comparing its two most common phases, graphite and diamond). &lt;a href=&quot;http://www-als.lbl.gov/index.php/contact/276-time-resolved-study-of-bonding-in-liquid-carbon.html&quot;&gt;This source&lt;/a&gt; suggests that bonding in the liquid varies continuously between linear acetylenic chains, graphite-like and diamond-like structures, depending on the exact conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Now I realize that obtaining liquid carbon and melting diamonds are two slightly different things. Liquid carbon is &lt;em&gt;trivial&lt;/em&gt; to obtain compared to melting a diamond. As you mention, heating diamond at relatively low pressures will cause it to convert into graphite before it could melt. The only way to inhibit that would be to exploit the slight difference in density between graphite and diamond; as diamond is denser, the increased pressure will stabilize it slightly, decreasing the tendency for diamond to revert back to graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need a lot of pressure though! Rigorously speaking, according to the phase diagram above, to obtain liquid carbon directly from melting a diamond would require a pressure of at least $10\ \mathrm{GPa}$, which is about $10^5\ \mathrm{atm}$. To put that into perspective, it's about $1/30$ of the pressure at the centre of the Earth. Amazingly, we actually can reach such absurd pressures in the lab by &lt;a href=&quot;http://en.wikipedia.org/wiki/Diamond_anvil_cell&quot;&gt;compressing two diamonds against each other&lt;/a&gt;. However, if the conditions required conditions for melting diamond (and sustaining it in molten form) are reached, it would probably compromise the integrity of the instrument and cause catastrophic (possibly explosive) failure, so I don't think anyone will be purposefully melting diamonds any time soon! Other techniques could be used (&lt;a href=&quot;http://en.wikipedia.org/wiki/Inertial_confinement_fusion#ICF_mechanism_of_action&quot;&gt;laser inertial confinement&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Light_gas_gun&quot;&gt;light-gas guns&lt;/a&gt;, etc), but I believe most would only transiently produce liquid diamond.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T12:36:54.610" LastActivityDate="2015-04-19T12:36:54.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6838" PostTypeId="2" ParentId="6827" CreationDate="2013-11-07T01:54:26.500" Score="4" Body="&lt;p&gt;Using the fundamental equations for the state function (and its natural variables): &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \mathrm{d}G &amp;amp;= -S\mathrm{d}T + V\mathrm{d}P\\&#xA;  V &amp;amp;= \left(\frac{\partial G}{\partial P}\right)_T\\&#xA;  \bar{G}(T,P_2) &amp;amp;= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p&#xA;\end{align}&lt;/span&gt;&#xA;Here &lt;span class=&quot;math-container&quot;&gt;$\bar{x}$&lt;/span&gt; represents molar &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, i.e. &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; per mole&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \bar{V} &amp;amp;= \frac{RT}{P}\\&#xA;  \bar{G}(T,P_2) &amp;amp;= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1}&#xA;\end{align}&lt;/span&gt;&#xA;Defining standard state as &lt;span class=&quot;math-container&quot;&gt;$P = \pu{1 bar}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\bar{G}=\mu$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$&lt;/span&gt;&#xA;consider the general gaseous reaction &lt;span class=&quot;math-container&quot;&gt;$\ce{a A + b B -&amp;gt; c C + d D}$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$&lt;/span&gt;&#xA;for &quot;unit progress&quot; in reaction. Using &lt;span class=&quot;math-container&quot;&gt;$\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\Delta G &amp;amp;= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\&#xA;\Delta G &amp;amp;= \Delta G^\circ + RT\ln Q&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="4945" LastEditDate="2020-02-18T13:41:05.940" LastActivityDate="2020-02-18T13:41:05.940" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6839" PostTypeId="1" CreationDate="2013-11-07T02:37:45.243" Score="3" ViewCount="1521" Body="&lt;p&gt;When roasting vegetables in a oven, after a while there is a nice caramel smell. After that, if they start burning, there is a smokey smell, and if they continue to burn, the smokey smell gets more and more acrid. The smell really sticks in the back of your throat/sinuses, and induces gagging if it's strong enough. What chemicals make up the gasses and particles that cause that acrid smell?&lt;/p&gt;&#xA;" OwnerUserId="2654" LastActivityDate="2013-11-10T18:44:45.670" Title="What chemicals cause the acrid smell of burning food?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6840" PostTypeId="1" AcceptedAnswerId="6841" CreationDate="2013-11-07T08:07:47.357" Score="0" ViewCount="2709" Body="&lt;p&gt;If so, is it because alchohols have hydrogen bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which type of compound evaporates faster, and why?&lt;/p&gt;&#xA;" OwnerUserId="2656" LastEditorUserId="5591" LastEditDate="2015-10-09T17:55:08.437" LastActivityDate="2015-10-09T17:55:08.437" Title="Do alkanes evaporate faster than alcohols?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6841" PostTypeId="2" ParentId="6840" CreationDate="2013-11-07T13:37:12.480" Score="2" Body="&lt;p&gt;Yes, alkanes do evaporate faster than their corresponding alcohols (i.e. an alkane boils at a lower temperature than an alcohol with the same number of carbon atoms). This effect is due to hydrogen bonding between the alcohol hydroxyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alcohol#Physical_and_chemical_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of hydrogen bonding, alcohols tend to have higher boiling points than comparable hydrocarbons and ethers. The boiling point of the alcohol ethanol is 78.29 °C, compared to 69 °C for the hydrocarbon hexane (a common constituent of gasoline), and 34.6 °C for diethyl ether.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1774" LastActivityDate="2013-11-07T13:37:12.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6842" PostTypeId="2" ParentId="6832" CreationDate="2013-11-07T14:34:07.803" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://etheses.nottingham.ac.uk/1620/1/366288.pdf&quot; rel=&quot;noreferrer&quot;&gt;This thesis&lt;/a&gt;, (Texture and Hydration of Expanded Rice, Clive Norton, September 1998), has a vast amount of information on rice, including puffed rice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rice is mostly starch, which when heated is &lt;a href=&quot;http://en.wikipedia.org/wiki/Starch_gelatinization&quot; rel=&quot;noreferrer&quot;&gt;gelatinised&lt;/a&gt;. This causes &quot;the breaking of hydrogen bonds between poly-α-1,4-glucan chains, leading to loss of crystallinity&quot; allowing those hydrogen bonding sites to absorb more water. The loss of crstallinity also leads to air pockets making puffed rice sponge like. The starches become &quot;interspersed with gas pockets&quot; which are filled by water through capillary action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, puffing makes the starch more hydrophillic, exposing hydrogen bonding sites, and the air pockets act as capillaries. The puffing also increases the ratio of surface area to weight. These factors cause the rice to be liftable with a &quot;water chain&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="2067" LastEditDate="2013-11-07T18:42:21.680" LastActivityDate="2013-11-07T18:42:21.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6843" PostTypeId="1" AcceptedAnswerId="6844" CreationDate="2013-11-07T14:36:57.050" Score="4" ViewCount="724" Body="&lt;p&gt;I'm working on a project making a product line more sustainable, environmentally friendly and ready for recycling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm searching for good and suitable biopolymers for a part of the product that is made from PVC and one for PU (Polyurethane). For PES I already have chosen PLA. Is there someone who can help me select the right biopolymers with similar characteristics of PVC and PU?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know, thank you!&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="24093" LastEditDate="2016-08-03T10:41:54.613" LastActivityDate="2016-08-03T10:41:54.613" Title="Biopolymers as alternative for PVC and PU?" Tags="&lt;polymers&gt;&lt;plastics&gt;&lt;green-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6844" PostTypeId="2" ParentId="6843" CreationDate="2013-11-07T15:54:55.233" Score="2" Body="&lt;p&gt;Ethylene vinyl acetate (also known as EVA) is the copolymer of ethylene and vinyl acetate and is a replacement for PVC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biomass &gt; (fermentation) &gt; &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/elim/dhethanol.html&quot; rel=&quot;nofollow&quot;&gt;Bioethanol &gt; Ethene&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethylene + &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetic_acid#Oxidative_fermentation&quot; rel=&quot;nofollow&quot;&gt;acetic acid&lt;/a&gt; + 1/2 O2 → Vinyl acetate + H2O&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tung_oil&quot; rel=&quot;nofollow&quot;&gt;Tung oil&lt;/a&gt; can be used to create a plastic like water proofing film as a replacement for PU&lt;/p&gt;&#xA;" OwnerUserId="2067" LastActivityDate="2013-11-07T15:54:55.233" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6846" PostTypeId="5" CreationDate="2013-11-07T23:41:28.240" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2013-11-07T23:41:28.240" LastActivityDate="2013-11-07T23:41:28.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6847" PostTypeId="4" CreationDate="2013-11-07T23:41:28.240" Score="0" Body="Density functional theory (DFT) is a quantum mechanical model used to estimate the electronic structure of molecules and condensed matter. In broad terms, DFT works by treating all the electrons in the system as a single electron density, and computing physical quantities of interest as functionals of that electron density. (from Physics.SE)" OwnerUserId="323" LastEditorUserId="323" LastEditDate="2013-11-07T23:52:48.360" LastActivityDate="2013-11-07T23:52:48.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6848" PostTypeId="1" CreationDate="2013-11-08T06:18:53.790" Score="-1" ViewCount="112" Body="&lt;p&gt;I was helping out in my beginner's chemistry class today, as I'm a &quot;lab assistant&quot; of sorts, and was reading on a bottle of HCl: 1M. I'm assuming that is the concentration of how much hydrocloric acid is in the bottle, in moles, but how would you determine the concentration of the solution (thanks Ove)? What is the actual percentage of HCl to water(and whatever else is in the bottle)?&lt;/p&gt;&#xA;" OwnerUserId="2661" LastEditorUserId="2661" LastEditDate="2013-11-08T06:47:08.667" LastActivityDate="2013-11-09T10:42:02.717" Title="How to Understand Concentrations Measurement" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6849" PostTypeId="1" AcceptedAnswerId="6855" CreationDate="2013-11-08T07:09:39.660" Score="10" ViewCount="12654" Body="&lt;p&gt;I'm currently studying some organic chemistry at highschool level. I understand that when naming an organic compound, the names for substituents should be done in alphabetical order. For instance &lt;strong&gt;3-bromo-3,4-dimethylhexane&lt;/strong&gt; (B in bromo before M in methyl), or &lt;strong&gt;3-chloro-3-ethyl-4-methylhexane&lt;/strong&gt; (C in chloro before E in ethyl before M in methyl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it doesn't seem to be the same for functional group suffixes. For instance, for one compound I wrote &lt;strong&gt;but-1-amin-3-ene&lt;/strong&gt;, while the correct answer was apparently &lt;strong&gt;but-3-en-1-amine&lt;/strong&gt;. What's more, these are referred to as &lt;em&gt;suffixes&lt;/em&gt;, while what I thought was &lt;strong&gt;prop-3-amin-1-ol&lt;/strong&gt; was in fact &lt;strong&gt;3-aminopropan-1-ol&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there some kind of pattern or rule as to the order of writing functional group &quot;suffixes&quot; in organic compounds? At first I thought perhaps it follows the priority the groups have over each other in numbering the parent chain carbons (i.e. alcohol &gt; amine &gt; alkene, etc), but this doesn't seem to be it either. It's possible that it'll be explained further on in the course, but I want to make sure I haven't missed anything.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="7" LastEditDate="2013-11-08T09:53:29.113" LastActivityDate="2015-04-17T17:56:13.810" Title="Functional group naming order?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6850" PostTypeId="2" ParentId="6832" CreationDate="2013-11-08T07:40:53.557" Score="1" Body="&lt;p&gt;Hydrophilic surface and capillary action is sertainly part of the answer together with the surface tension of water trying to minimize its surface area. But another part is probably the very low weight of puffed rice! That makes the effect of the beforementioned forces more noticeable. If the rice grains would be heavier (denser) you would probably not notice the same stickyness. &lt;/p&gt;&#xA;" OwnerUserId="2663" LastActivityDate="2013-11-08T07:40:53.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6851" PostTypeId="1" AcceptedAnswerId="6852" CreationDate="2013-11-08T09:30:15.300" Score="3" ViewCount="1844" Body="&lt;p&gt;Is there any way to prevent petrol from being flammable without losing it's solvent properties? Preferably some affordable way that is applicable for DIY. &#xA;In my case, petrol is used for bearing cleaning, and there is a risk of inflammation due to accidental spark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no affordable way exists, what octane number is safer (less flammable)?&lt;/p&gt;&#xA;" OwnerUserId="2664" LastEditorUserId="2664" LastEditDate="2013-11-08T09:49:30.293" LastActivityDate="2021-03-18T14:58:13.847" Title="How to prevent petrol from being flammable?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6852" PostTypeId="2" ParentId="6851" CreationDate="2013-11-08T09:50:30.630" Score="4" Body="&lt;p&gt;Nearly every solvent is flammable, some exceptions are halogenated hydrocarbons. But plain hydrocarbons are pretty much all flammable, though some are more dangerous than others like e.g. diethyl ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloroform is non-flammable, but due to the toxicity I would not recommend to use it. There are some other non-flammable solvents for cleaning mentioned in &lt;a href=&quot;http://tdglobal.ksc.nasa.gov/servlet/sm.web.Fetch/TEERM_LOWREY_Nonflammable_Solvents_for_Oxygen_Systems12.6.pdf?rhid=1000&amp;amp;did=882267&amp;amp;type=released&quot; rel=&quot;nofollow&quot;&gt;this presentation from NASA on the search for non-flammable solvents&lt;/a&gt;. But it seems that many candidates are being phased out because they are a hazard to the ozone layer.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2013-11-08T09:50:30.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6853" PostTypeId="2" ParentId="6848" CreationDate="2013-11-08T10:04:15.633" Score="1" Body="&lt;p&gt;The unit M is an abbreviated form of mol/L: 1 M = 1 mol/L, i.e. there is 1 mol of HCl per liter of solution. The M is pronounced &quot;molar&quot;. For example, a solution with concentration 1 mol/L is said to be a &quot;1 molar&quot; solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calculate the fraction of HCl in a 1 M solution, some information is needed. The density of the solution varies with concentration, so I looked up &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1%20M%20HCl&quot; rel=&quot;nofollow&quot;&gt;1 M HCl on Wolfram Alpha&lt;/a&gt; and found $\rho = 1.016~g/cm^3=1.016~kg/L$. Also, the molar mass of HCl is $M_{HCl}=36.46~g/mol$. Thus, 1 L of solution has a mass of 1.016 kg and 1 mol of HCl weighs $$m_{HCl}=1~mol\cdot36.46~g/mol=36.46~g$$ The mass of water in the solution must therefore be $$m_{H_2O}=1.016\cdot10^3-36.46=979.5~g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the mass fraction of HCl in the solution can be calculated:&#xA;$$w_{HCl}=\frac{m_{HCl}}{m_{tot}}=\frac{m_{HCl}}{m_{HCl}+m_{H_2O}}$$&#xA;Similarily, the molar fraction can be determined:&#xA;$$x_{HCl}=\frac{n_{HCl}}{n_{tot}}=\frac{n_{HCl}}{n_{HCl}+n_{H_2O}}=\frac{n_{HCl}}{n_{HCl}+\frac{m_{H_2O}}{M_{H_2O}}}$$&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="1774" LastEditDate="2013-11-09T10:42:02.717" LastActivityDate="2013-11-09T10:42:02.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6854" PostTypeId="1" CreationDate="2013-11-08T10:48:38.033" Score="27" ViewCount="858" Body="&lt;p&gt;Many general chemistry textbooks introduced the concept &quot;hybridization&quot; to construct a symmetry-adapted VB-type wavefunction. In the textbooks, usually the minimal basis is used and without optimizing all resonance forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, in the ab initio valence bond calculation with extended basis set and optimizing many resonance forms, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) do people still use hybridization to construct the wavefuntion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked the question (ii), since NBO analysis tends to give non-equivalent composition for lone pairs in $\ce{H2O}$, one sp, the other p. I am wondering what is the answer in ab initio valence bond method.&lt;/p&gt;&#xA;" OwnerUserId="2629" LastEditorUserId="16683" LastEditDate="2016-04-21T11:16:40.770" LastActivityDate="2020-05-08T21:02:55.303" Title="Is hybridization used in ab initio valence bond calculation?" Tags="&lt;computational-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6855" PostTypeId="2" ParentId="6849" CreationDate="2013-11-08T12:42:46.707" Score="10" Body="&lt;p&gt;Generating the systematic names of organic compounds follows the rules for &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&quot;&gt;IUPAC nomenclature&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this system, a name has several parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prefix (meth-, eth-, prop-, but-, pent-, etc.) - which indicates the number of carbon atoms in the parent chain.&lt;/li&gt;&#xA;&lt;li&gt;Infix (-an-, -en-, -yn-) - which indicates the presence or absence of carbon-carbon multiple bonds in the parent chain.&lt;/li&gt;&#xA;&lt;li&gt;Suffix (-e, -ol, -amine, -oic acid, -al, etc.) - which indicates the highest priority functional group in the parent chain.&lt;/li&gt;&#xA;&lt;li&gt;Substituents (bromo-, methyl-, hydroxy-) - indicate the presence of other things on the parent chain.&lt;/li&gt;&#xA;&lt;li&gt;Locants (the numbers) - indicate where all of these things happen.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You are confusing the roles of infixes and suffixes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Infixes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The base infix is &lt;strong&gt;-an-&lt;/strong&gt; indicating no carbon-carbon multiple bonds. The presence of alkenes or alkynes in the parent chain changes the infix to -en- or -yn-, respectively. No other functional group can change the infix, as there are no carbon-carbon bonds of higher order than 3 or lower order than 1. It is also possible to have a complex -enyn- infix if you have both, but let us not worry about that now. In your example compounds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3-bromo-3,4-dimethylhexane - the infix is -an- because there are no carbon-carbon double or triple bonds. &lt;/li&gt;&#xA;&lt;li&gt;but-3-en-1-amine - the infix is -en- because there is a carbon-carbon double bond.&lt;/li&gt;&#xA;&lt;li&gt;3-aminopropan-1-ol - the infix is -an- again because there are no carbon-carbon multiple bonds. The amine and alcohol functional groups do not consist of carbon-carbon multiple bonds so &lt;strong&gt;they cannot change the infix.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suffixes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The base suffix is &lt;strong&gt;-e&lt;/strong&gt;, signifying a hydrocarbon with some substituents. Many functional groups change the suffix of a molecule's name, but the name can have only one suffix. If two or more suffix-changing functional groups are found on the parent chain, then the one with the &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot;&gt;highest priority&lt;/a&gt; is chosen to be the suffix &lt;strong&gt;and all others become substituents.&lt;/strong&gt; The common functional groups by decreasing priority are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;functional group     suffix         substituent name&#xA;carboxylate anion    -oate         &#xA;carboxylic acid      -oic acid      carboxy&#xA;ester                alkyl -oate    alkoxycarbonyl or alkylcarboxy&#xA;acid halide          -oyl halide    halocarbonyl&#xA;amide                -amide         amido&#xA;nitrile              -enitrile      cyano&#xA;aldehyde             -al            oxo&#xA;ketone               -one           oxo&#xA;alcohol              -ol            hydroxy&#xA;thiol                -ethiol        mercapto&#xA;amine                -amine         amino&#xA;hydrocarbon          -e             varies by group&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The name &lt;strong&gt;3-aminopropan-1-ol&lt;/strong&gt; is correct. Both the alcohol and the amine cannot change the suffix. Only the higher priority group (alcohol) changes the suffix. The other (amine) becomes a substituent. As mentioned before, neither group changes the infix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Substituents&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In addition to the substituents listed above, several functional groups are always substituents (which are listed alphabetically in front of the parent name):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;functional group     substituent name&#xA;alkyl                alkyl&#xA;alkyl halides        halo&#xA;ethers               alkoxy&#xA;nitro                nitro&#xA;sulfides             alkylthio&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are some other common substituents that deal with exception cases, like phenyl, aryl, vinyl, methylene, formyl, allyl, and acetyl. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-08T12:42:46.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6858" PostTypeId="1" CreationDate="2013-11-08T18:41:25.957" Score="9" ViewCount="7143" Body="&lt;p&gt;My book says alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Why do they develop colour when exposed to light? It is important to note that alkyl fluorides, and alkyl chlorides are not said to develop colour.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="186" LastEditDate="2015-10-02T16:24:25.870" LastActivityDate="2017-07-08T16:42:29.553" Title="Why do alkyl bromides and iodides develop colour when exposed to light?" Tags="&lt;organic-chemistry&gt;&lt;periodic-trends&gt;&lt;color&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6859" PostTypeId="1" CreationDate="2013-11-08T18:51:36.587" Score="4" ViewCount="36192" Body="&lt;ul&gt;&#xA;&lt;li&gt;My book says, methyl chloride, methyl bromide, ethyl chloride and some cholorofluromethanes are gases at room temperature, whereas higher members are liquids or solids. Why?  &lt;/li&gt;&#xA;&lt;li&gt;With this question, I got another question for which the former would be subset. How to determine whether a compound is gas or liquid or solid (assume room temperature)? Thinking about this question, I thought molecular force between the molecules is going to be the key for answer of this query door. But how to determine, whether a compound has stronger intermolecular force between the molecules or not?  &lt;/li&gt;&#xA;&lt;li&gt;We have another way to check if a compound is gas or solid or liquid, i.e. entropy (randomness), which is given by heat gained or lost divided by temperature. In general, compound which has greater randomness will be gas. We can expect most of the compounds to be gases, which have standard molar entropy greater than $200\ \mathrm{J K^{-1} mol^{-1}}$ at $298\ \mathrm{K}$.  &lt;/li&gt;&#xA;&lt;li&gt;Entropy concept brings many questions under light, for example, $\ce{H2}$ gas has standard molar entropy of $130.7\ \mathrm{JK^{-1}mol^{-1}}$, which is lesser than $\ce{Br2}$ liquid, which has standard molar entropy of $152.3\ \mathrm{JK^{-1}mol^{-1}}$. We would had expected $\ce{H2}$ gas to have greater randomness than $\ce{Br2}$ liquid, because gases should have greater randomness than liquids. Anyway it becomes difficult, to determine a compound as solid or liquid or gas using entropy.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;LINKS&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt&quot; rel=&quot;nofollow&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt&lt;/a&gt; (This link helps to determine a compound as gas, but leads to still more questions like: Why most non metals are gases? Why all the gases are covalent bonded that contain two or more non metals? Why gases are defined to have low molecular weights, in spite of knowing that molecular interactions matter? Can’t ionic bonded compounds be gases?)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I will try to find list of solid and liquid compounds, but now I have just the above link. If any one have links related to this particular topic, they are welcome.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastEditorDisplayName="user7951" LastEditDate="2015-06-02T10:41:56.250" LastActivityDate="2015-10-11T07:11:56.337" Title="How to determine whether a compound is gas or liquid or solid?" Tags="&lt;physical-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;gas-laws&gt;&lt;intermolecular-forces&gt;&lt;liquid-crystals&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-08-19T04:24:22.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="6860" PostTypeId="1" AcceptedAnswerId="6869" CreationDate="2013-11-08T19:37:38.597" Score="2" ViewCount="25880" Body="&lt;p&gt;For the reaction between alkyl halide and a nucleophile, following the S&lt;sub&gt;N&lt;/sub&gt;2 mechanism, the reactivity of alkyl halides is in the order: primary halide &gt; secondary halide &gt; tertiary halide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the reactivity of alcohols with halogen acids to form alkyl halides, the reactivity of alcohols is in the order: tertiary alcohol &gt; secondary alcohol &gt; primary alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a difference in order of reactivity in both the cases?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="5026" LastEditDate="2018-03-16T15:28:19.537" LastActivityDate="2018-03-16T15:30:04.543" Title="Why is the reactivity of primary alkyl halides with nucleophiles (SN2 mechanism) greater than secondary and tertiary alkyl halides?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;halides&gt;&lt;reactivity&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6861" PostTypeId="1" CreationDate="2013-11-08T21:24:20.310" Score="0" ViewCount="317" Body="&lt;p&gt;Before a sodium atom and a chlorine atom combine together to form the ionic compound $NaCl$, they are just neutral atoms. However, when an electron is transferred from the outer energy level of the sodium atom to the chlorine atom, both of these gain a charge on them; the sodium atom or ion has a charge of $1+$ and the chlorine ion has a charge of $1-$. Both of these ions attract each other and an ionic bond is formed between them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But what happens when an atom of an element say $Ca$ combines with a polyatomic ion say $OH^-$. Is this calcium atom an ion, is it &lt;strong&gt;&lt;em&gt;already&lt;/em&gt;&lt;/strong&gt; a charged atom when it interacts with the hydroxide ion or does it gain the charge by any transfer of electrons to the $OH^-$ ion and becomes $Ca^{2+}$ to bond with the $OH^-$ ion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the polyatomic ions say $OH^-$ is already an ion and not an uncharged compound of elements that gains charge as in the case of $NaCl$, and the bond between an ion and a polyatomic ion is due to the attractive forces between them if already they have opposite charges.  &lt;/p&gt;&#xA;" OwnerUserId="1667" LastEditorUserId="1667" LastEditDate="2013-11-08T23:06:27.437" LastActivityDate="2013-11-09T14:48:18.687" Title="Ions and Polyatomic Ions" Tags="&lt;bond&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2017-08-02T19:03:18.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="6862" PostTypeId="2" ParentId="6859" CreationDate="2013-11-09T01:53:16.803" Score="3" Body="&lt;p&gt;As the molecular weight increases (more halogens or carbon atoms), so does the electron density and consequently the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;van-der-waals&lt;/a&gt; forces(&lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;dispersion forces&lt;/a&gt;) between the molecules rises. These forces increase intermolecular attraction and therefore increase the boiling point. T^his is valid in case of most other organic compounds' homologous series, not just alkyl halides.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-11-09T01:53:16.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6863" PostTypeId="2" ParentId="6858" CreationDate="2013-11-09T01:57:07.850" Score="8" Body="&lt;p&gt;This is because of spontaneous breakdown in case of iodides and bromides which have a lower bond-dissociation enthalpy as compared to fluorides and chlorides. This low enthalpy means that even at normal temperature or on exposure to sunlight, the halides (iodides and bromides) decompose to a certain extent forming free iodine and bromine, both of which are coloured and hence impart colour to the original halides. &lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-11-09T01:57:07.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6864" PostTypeId="1" AcceptedAnswerId="6867" CreationDate="2013-11-09T04:40:51.290" Score="3" ViewCount="12251" Body="&lt;p&gt;So I have an equilibrium phase diagram of steel and I am asked to 'Calculate the proportion of pearlite in the microstructure of 0.4 wt% C steel just below the eutectoid temperature (727 °C).'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been using the forumlas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Wa=\frac{Cb-Co}{Cb-Ca}$%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Wb=100-Wa$%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Ca=weight$% of (Carbon in this example) at the first intersection point on the eutectoid line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Cb=weight$% of (Carbon in this example) at the second intersection point on the eutectoid line or the bottom of the 'V'. I don't know how to explain any better sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So looking at my diagram, I see that $Ca=0.022wt$% Carbon and $Cb=0.76wt$% Carbon, and $Co$ is just 0.4%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $Wa=\frac{0.76-0.4}{0.76-0.022}$%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=0.487805%$%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I am confused. What exactly is Wa? Is that how much ferrite/austenite/cementite that is in the pearlite then? Or how much pearlite is at that point compared to ferrite? And in any case, what does that mean Wb is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be much appreciated. Thanks &lt;/p&gt;&#xA;" OwnerUserId="2172" LastActivityDate="2015-04-01T05:59:37.197" Title="Calculating percentages of microstructures in equilibrium phase diagrams (engineering)" Tags="&lt;metal&gt;&lt;transition-metals&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6867" PostTypeId="2" ParentId="6864" CreationDate="2013-11-09T14:19:45.807" Score="3" Body="&lt;p&gt;Assuming you mean &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/8/8e/Steel_pd.svg&quot; rel=&quot;nofollow&quot;&gt;this phase diagram&lt;/a&gt;, it looks like your calculations are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my understanding of the diagram, at $727\ ^\circ \text{C}$, you have 100% pearlite at $C_b = 0.76\ \%C$ and 0% pearlite at $C_a=0.022\ \%C$. $W_a$ then appears to be how far $C_o$ is between $C_a$ and $C_b$, so $W_a$ should be the amount of pearlite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of $C_a$, $C_b$, and $C_o$ are $\% C\  by\  mass$. That means when you calculate $W_a$ initially, you are calculating the &lt;strong&gt;fraction&lt;/strong&gt; of pearlite and not the &lt;strong&gt;percent&lt;/strong&gt; of pearlite. All of the $\% C\  by\  mass$ units cancel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_a=\frac{0.76\ \%C-0.4\ \%C}{0.76\ \%C - 0.022\ \%C}=\frac{{\%C}(0.76-0.022)}{{\%C}(0.76-0.022)}=0.487805\implies48.7805\%$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-09T14:19:45.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6868" PostTypeId="2" ParentId="6861" CreationDate="2013-11-09T14:48:18.687" Score="4" Body="&lt;p&gt;I risk not directly answering the question as I have not perfectly understood it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your first example, you consider the reaction between two neutral atoms, $\ce{Na}$ and $\ce{Cl}$. An electron is transferred generating $\ce{NaCl}$ where the species are now charged ($\ce{Na^+ \  Cl^-}$). Then you follow up with a hypothetical reaction between a neutral atom ($\ce{Ca}$) and an already charged species ($\ce{OH^-}$). The two reactions are fundamentally different, but not because only the latter has polyatomic ions; the difference is in the very nature of the reactions you constructed (two neutral species reacting in the former, compared to one neutral species and an ion reacting in the latter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most direct way to obtain $\ce{NaCl}$ is the reaction between sodium metal (composed of neutral atoms, $\ce{Na^0}$) and chlorine gas (composed of neutral diatomic molecules, $\ce{Cl2}$). One then has: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na^0_{(s)} + Cl2_{(g)} → 2NaCl_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain $\ce{Ca(OH)2}$, a similarly simple reaction would be that between calcium metal (composed of neutral atoms, $\ce{Ca^0}$) and water (composed of neutral molecules, $\ce{H_2O}$), where the water serves as a source of hydroxyl ions according to the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ca^0_{(s)} + 2H2O_{(l)} → Ca(OH)2_{(s)} + H2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine you want to force a reaction between a neutral atom and an ion. There are many cases of this, but we can stick to something similar to what you wanted. In order to expose a substance directly to ions, a conceptually easy way is to add it to a molten salt. For now, let us consider pumping sodium vapour into molten caesium hydroxide ($\ce{CsOH}$). You seem to expect some interaction between the neutral $\ce{Na^0}$ and charged $\ce{OH^-}$, but in fact it is $\ce{Cs^+}$ that reacts with the sodium, according to the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na^0_{(g)} + CsOH_{(l)} → NaOH_{(l)} + Cs^0_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have simple exchange reaction. The hydroxide ion is actually only a spectator, and can be omitted without loss of understanding in an ionic equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na^0_{(g)} + Cs^+_{(l)} → Na^+_{(l)} + Cs^0_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the reaction between sodium and molten caesium chloride instead, where chloride ions would be present, we have again a simple exchange reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na^0_{(g)} + CsCl_{(l)} → NaCl_{(l)} + Cs^0_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, it can be described by the same ionic equation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Disclaimer: I actually am not sure the reactions between sodium and caesium compounds works as above. I don't have any understanding of the relative electrochemical potentials in molten salts, though I believe I've read that the reaction with molten chloride is a legitimate way to obtain metallic caesium. But that is mostly beside the point, either way. I avoided exemplifying with calcium and calcium hydroxide because I think the reaction wouldn't work, and the hydroxide is thermally unstable)&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-09T14:48:18.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6869" PostTypeId="2" ParentId="6860" CreationDate="2013-11-09T14:49:50.500" Score="7" Body="&lt;p&gt;The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an S&lt;sub&gt;N&lt;/sub&gt;2 mechanism. A $\ce{C-Nu}$ bond forms and a $\ce{C-X}$ bond breaks at the same time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2Br + NaOH -&amp;gt;[\text{acetone}] Na}\bigg[\ce{CH3CH2(Br)(OH)}\bigg]^{^\ddagger_{-}}\ce{-&amp;gt;CH3CH2OH +NaBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All else equal, the rates of these reactions are controlled by sterics on the alkyl halide, which control how well the nucleophile can react the $\ce{C-X}\ \sigma^*$ orbital. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Methyl}&amp;gt;1^\circ&amp;gt;2^\circ&amp;gt;&amp;gt;3^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions of alcohols with hydrohalic acids follow a different mechanism (mostly). These reactions happen in a protic medium ($\ce{HX}$ is protic) with a poor leaving group ($\ce{OH}$), at least initially. These are conditions that favor the step-wise S&lt;sub&gt;N&lt;/sub&gt;1 mechanism, where $\ce{C-X}$ bond breaking occurs before $\ce{C-Nu}$ bond forming. A carbocation is formed as a key intermediate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)3COH + HBr -&amp;gt; [(CH3)3COH2+]Br- -&amp;gt; [(CH3)3C+]Br- + H2O -&amp;gt; (CH3)3CBr + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammond_postulate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammond postulate&lt;/a&gt;, the stability of this carbocation influences the rate of reaction. More stable carbocation intermediates suggest lower energy transition states for the formation of those carbocations, which implies lower activation energies and faster rates. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3^\circ&amp;gt;2^\circ&amp;gt;&amp;gt;1^\circ,\text{Methyl}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-16T15:30:04.543" LastActivityDate="2018-03-16T15:30:04.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6870" PostTypeId="1" AcceptedAnswerId="6888" CreationDate="2013-11-09T16:10:01.643" Score="3" ViewCount="1775" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbhJN.png&quot; alt=&quot;Mass spectrum of the unknown compound &amp;quot;MS88&amp;quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm given the above spectrum. &lt;span class=&quot;math-container&quot;&gt;$M^+ = 102$&lt;/span&gt;, probably &lt;span class=&quot;math-container&quot;&gt;$\ce{C_6H_14O}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{C_5H_10O_2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I'm thinking of &lt;a href=&quot;https://i.imgur.com/2miTTv0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;ethyl propionate&lt;/a&gt;. (that picture also shows the fragments I expect). I can account for the following peaks:&#xA;m/z = 73: Loss of ethyl from acyl side (expected lowish intensity)&#xA;m/z = 57: Loss of alkoxy to form &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CO+}$&lt;/span&gt; (moderate intensity)&#xA;m/z = 45: Alkoxy ion (weak)&#xA;m/z = 29: ethyl carbocation, moderate intensity.&lt;/p&gt;&#xA;&lt;p&gt;I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.&lt;/p&gt;&#xA;&lt;p&gt;I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...&lt;/p&gt;&#xA;&lt;p&gt;Can someone help me out, I've come to a dead end for hours now.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="24052" LastEditDate="2021-05-16T14:44:45.783" LastActivityDate="2021-05-16T14:44:45.783" Title="Interpretation of the following mass spectrum" Tags="&lt;analytical-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6871" PostTypeId="1" CreationDate="2013-11-09T16:27:13.207" Score="3" ViewCount="982" Body="&lt;p&gt;I was observing from many days, how hairs seems to be sticking to water, when made to fall on it. I made an attempt to know what really was going on. I kept a jug of stationary water under the light, and kept my dry hairs combing, to make hairs fell on it. I observed once again that hairs appeared to be sticking to water.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I started thinking about the cause for this observation. At first, what went to my mind was about hairs being rubbed with comb. Usually hairs get positively charged (I don't know exactly. I thought it because plastic becomes negatively charged when rubbed with most of  other suitable surface) when rubbed with plastic comb. We also know that water has more negative charge because of oxygen atom (I came to know about this from the diagram of wiki encyclopedia-&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;electronegativity&lt;/a&gt;). So, I thought positively charged hairs might be attracted by negative charge nature of water molecules, which might be making hairs to stick with water.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is it really because of positively charged hairs and negatively charged water molecules, which is making hairs to be sticking to water?  &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2013-11-10T07:16:09.523" LastActivityDate="2014-02-27T21:31:07.327" Title="Why do hairs (rubbed with plastic comb) fallen on water seems to be sticking?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;&lt;formal-charge&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6872" PostTypeId="1" CreationDate="2013-11-09T18:04:52.577" Score="-2" ViewCount="128973" Body="&lt;p&gt;It's easy to do atomic electron configuration, but ionic is confusing.&lt;/p&gt;&#xA;" OwnerUserId="3667" LastEditorUserId="-1" LastEditDate="2016-09-26T10:53:52.377" LastActivityDate="2016-09-26T10:53:52.377" Title="How can I figure out the electron configuration of Br-?" Tags="&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-09-01T23:50:01.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="6873" PostTypeId="2" ParentId="6872" CreationDate="2013-11-09T19:26:26.873" Score="7" Body="&lt;p&gt;The electronic configuration for $\ce{Br-}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it have one more electron than bromine, which ends its electronic configuration with $\mathrm{4p^5}$.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="186" LastEditDate="2015-09-29T13:36:29.387" LastActivityDate="2015-09-29T13:36:29.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6874" PostTypeId="2" ParentId="6870" CreationDate="2013-11-09T19:34:36.390" Score="-1" Body="&lt;p&gt;For m/z 74 it could be methyl ester:  in the loss of a &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=CH2}$&lt;/span&gt; by McLafferty rearrangement.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=C(OH)OCH3+}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="95133" LastEditDate="2021-05-16T12:37:40.730" LastActivityDate="2021-05-16T12:37:40.730" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6876" PostTypeId="1" CreationDate="2013-11-10T05:33:46.227" Score="4" ViewCount="7588" Body="&lt;p&gt;From my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Halogens ($\ce{Cl_2}$ and $\ce{Br_2}$ ) react with alkanes in the presence of ultraviolet light to form haloalkanes.This reaction is free radical substitution and gives a mixture of mono, di, or polysubstituted haloalkanes which are difficult to separate into pure components. Moreover, the yield of any one compound is low because of the formation of other compounds. For example, in case of butane, a mixture of 1-chlorobutane (28%) and 2-chlorobutane (72%) is obtained, where the secondary alkyl halide is obtained as the major product rather than the primary one. &#xA;       $$\ce{CH_3CH_2CH_2CH_3-&amp;gt;[\ce{Cl_2,~UV~light}]CH_3CH_2CH_2CH_2Cl + CH3CH_2CHClCH_3}$$&lt;br&gt;&#xA;    In general, the ease of substitution of various hydrogen follows the sequence:&lt;br&gt;&#xA;      $${\rm tertiary &amp;gt; secondary &amp;gt;primary}$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After reading through all the above statements in my book, I now have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are tertiary alkyl halides obtained as major products instead of secondary or primary ones? It seems that a hydrogen from a tertiary carbon can be more easily replaced by a halogen than that of a secondary or primary carbon. If it is so, why is hydrogen from a tertiary carbon more easily replaced than a hydrogen from a secondary or primary carbon?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are fluorine and iodine not considered in case of free radical halogenation (the first two lines of the above passage extracted from the book mentions only chlorine and bromine)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is a tertiary alkyl halide not produced in the above example? Even if we think it might be produced in a lesser amount, there should have been more of it than the primary or secondary alkyl halides. It seems the reason for tertiary alkyl halide not being in the equation is because it might not have formed or it might have formed in a lesser amount than primary or secondary alkyl halides. If it is either any case, it would violate the order mentioned in the book.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="865" LastEditDate="2016-03-13T20:19:41.787" LastActivityDate="2016-03-13T20:19:41.787" Title="Why are tertiary alkyl halides obtained in higher yield than secondary or primary during free radical halogenation of alkanes?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6877" PostTypeId="1" CreationDate="2013-11-10T09:25:08.197" Score="1" ViewCount="446" Body="&lt;p&gt;I have a compound potassium oxide($\ce{K2O}$) and I am trying to convert it into potassium nitride ($\ce{K3N}$).&#xA;Here are a few possible reaction methods : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Split potassium oxide into the constituent elements potassium and oxygen  (Not very good  as a lot of energy is required)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{K2O + N2 -&amp;gt; K3N + O2}$ (however this won't happen(?) because nitrogen has a lower electronegativity value than oxygen does)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how can I convert the oxide to the nitride, releasing oxygen in the process, by not having to use a lot of energy-requiring processes?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="2094" LastEditDate="2013-11-10T09:59:35.003" LastActivityDate="2013-11-10T09:59:35.003" Title="Most economical method to convert potassium oxide to potassium nitride." Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6878" PostTypeId="1" CreationDate="2013-11-10T16:35:29.993" Score="3" ViewCount="36573" Body="&lt;p&gt;I'm an IB Chemistry student and I had to do a back titration lab where 1g of a powdered antacid tablet were mixed with 50mL 1M HCl, and the mixture was titrated using 0.2 M NaOH; the NaOH was to neutralize the leftover HCl that hadn't reacted with the antacid. The active ingredient in the antacid is CaCO3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took me about 40mL of NaOH to titrate the leftover HCl to endpoint; 40 mL at 0.2 mol/L translate to 0.008 moles of NaOH. In the equation for reaction between NaOH and HCl the ratio of the two reactants is 1:1; therefore there were 0.008 mol of unreacted HCl in the mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were 50mL of HCl (1 mol/L) in the mixture originally; 50mL * 1 mol/1000mL = 0.05 mol HCl total. Subtracting from this the molar quantity of unreacted (leftover) HCl yields: 0.05 - 0.008 = 0.042 mol HCl that did react with the CaCO3 in the antacid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --&gt; CaCl2 + CO2 +H2O&#xA;Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a problem - i can't be having 210% composition by mass (the sample was 1g; I calculated 2.1 g CaCO3 present). So somewhere along the line i made a mistake, but I've checked my math so I don't know if it was a calculation error or if my method is dead wrong? Can anyone help me with this?&lt;/p&gt;&#xA;" OwnerUserId="3671" LastActivityDate="2014-02-08T20:26:11.610" Title="Finding the percent of calcium carbonate in commercial antacid by using back titration" Tags="&lt;stoichiometry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6879" PostTypeId="2" ParentId="1030" CreationDate="2013-11-10T16:45:15.047" Score="-1" Body="&lt;p&gt;Sulfur has four $sp^3$ orbitals (four $\sigma$-bonds) with oxygen and two $\pi$-bonds ($d$ orbitals).&#xA;In the case of $\ce{SO4^{2-}}$ ion, also sulfur also has four $sp^3$ orbitals and four $\pi$-bonds with $d$ orbitals. The electron from hydrogen is moves to sulfur's d orbitals&lt;img src=&quot;https://i.stack.imgur.com/IQIIP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3670" LastEditorUserId="15410" LastEditDate="2016-11-20T01:15:11.150" LastActivityDate="2016-11-20T01:15:11.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6880" PostTypeId="2" ParentId="6878" CreationDate="2013-11-10T17:40:41.713" Score="2" Body="&lt;p&gt;In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, in order to find where is the error you can check:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="2578" LastEditDate="2013-11-10T18:25:25.637" LastActivityDate="2013-11-10T18:25:25.637" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6881" PostTypeId="2" ParentId="6839" CreationDate="2013-11-10T18:44:45.670" Score="1" Body="&lt;p&gt;Take a look at this link about Maillard reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.landfood.ubc.ca/courses/fnh/301/brown/brown_prin.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.landfood.ubc.ca/courses/fnh/301/brown/brown_prin.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what occurs if continue burning: pyrolisis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.healthy-food-site.com/burnt-food.html&quot; rel=&quot;nofollow&quot;&gt;http://www.healthy-food-site.com/burnt-food.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-10T18:44:45.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6882" PostTypeId="1" CreationDate="2013-11-10T21:14:23.457" Score="3" ViewCount="343" Body="&lt;p&gt;In terms of producing gold nano flakes, Guo et al. [&lt;a href=&quot;https://doi.org/10.1016/j.colsurfa.2005.11.075.&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] added aniline to the solution of chloroauric acid in ethylene glycol at 95 °C. However, I couldn't figure out the step by step chemical reaction in it. Can anyone please explain me the reaction procedure or guide me to the relevant source?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Guo, Z.; Zhang, Y.; DuanMu, Y.; Xu, L.; Xie, S.; Gu, N. Facile Synthesis of Micrometer-Sized Gold Nanoplates through an Aniline-Assisted Route in Ethylene Glycol Solution. &lt;em&gt;Colloids and Surfaces A: Physicochemical and Engineering Aspects&lt;/em&gt; &lt;strong&gt;2006&lt;/strong&gt;, 278 (1), 33–38. &lt;a href=&quot;https://doi.org/10.1016/j.colsurfa.2005.11.075&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1016/j.colsurfa.2005.11.075&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3672" LastEditorUserId="41328" LastEditDate="2019-05-13T15:57:15.657" LastActivityDate="2019-05-13T16:04:04.380" Title="Producing gold nano flakes" Tags="&lt;inorganic-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6884" PostTypeId="1" AcceptedAnswerId="6891" CreationDate="2013-11-11T01:26:26.493" Score="3" ViewCount="405" Body="&lt;p&gt;I searched on internet about rocket fuel and the most common fuel is $\ce{KNO3}$ + sugar (rocket candy), but I think that I could use another fuel methods, like using can with gas or using $\ce{HCl + Al}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Which one ($\ce{KNO3}$ + sugar, can with gas or $\ce{HCl + Al}$) gets more performance? Can I increase the performance by increasing the pressure and the durability?&lt;/p&gt;&#xA;" OwnerUserId="3674" LastEditorDisplayName="user7951" LastEditDate="2015-06-19T09:03:57.383" LastActivityDate="2018-09-26T17:08:45.690" Title="Which is the best rocket fuel?" Tags="&lt;fuel&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6888" PostTypeId="2" ParentId="6870" CreationDate="2013-11-11T10:48:56.907" Score="4" Body="&lt;p&gt;I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The spectre is of:&lt;/strong&gt; 2-methyl-butanoic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;M-17 and M-45:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha cleavage of carbonyl =&gt; loss of hydroxyl radical or carboxyl radical&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;m/z = 74:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loss of ethene by McLafferty rearrangement&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;m/z = 45:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{COOH^+}$ fragment&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;M-15 and M-29:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loss of methyl radical or ethyl radical&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;m/z = 29:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethyl carbocation&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-11-11T10:48:56.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6889" PostTypeId="2" ParentId="6755" CreationDate="2013-11-11T11:17:01.213" Score="2" Body="&lt;p&gt;According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastActivityDate="2013-11-11T11:17:01.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6890" PostTypeId="2" ParentId="6738" CreationDate="2013-11-11T11:27:18.600" Score="0" Body="&lt;p&gt;If you speak about solubility, you are out of domain of simple approximations and the only correct answer is the experiment. For the example given, see &lt;a href=&quot;http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the difficulty to quantify the effect of temperature, which is very important here.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-11T11:27:18.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6891" PostTypeId="2" ParentId="6884" CreationDate="2013-11-11T11:46:49.527" Score="4" Body="&lt;p&gt;I used &lt;code&gt;Zn&lt;/code&gt; + &lt;code&gt;S&lt;/code&gt; as a teenager. It is a lot safer than Al, and you can mix it cold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stick to solid propelants because it's simpler, safer and smaller. You don't want an explosion but something very quick burning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mix the two fine powders well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can mix them in a thick plastic bag or wide mouth bottle and shake until you see an even color in all the mix with no lumps in sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can make a paste with alcohol 100% but I never tried this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then tap it in with wood or plastic. You don't want a spark! Nozzle reduction is 1/3 and not smaller. The mix is 3:1. Light with a filament, it is safer than a spark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In those days all I had was a book but you can find a lot of info out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We switched to this after a (not too serious) accident with rocket candy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safety first allways, the rocket can go anywhere!&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="1300" LastEditDate="2013-11-13T11:50:05.907" LastActivityDate="2013-11-13T11:50:05.907" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6892" PostTypeId="2" ParentId="6871" CreationDate="2013-11-11T12:03:59.157" Score="0" Body="&lt;p&gt;Maybe it's &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow&quot;&gt;surface tension&lt;/a&gt;. See if hairs without electrical charge react the same way.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-11-11T12:03:59.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6893" PostTypeId="1" AcceptedAnswerId="6898" CreationDate="2013-11-11T14:02:07.163" Score="4" ViewCount="4964" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are ionic reactions extremely fast,  as compared to other gaseous&#xA;  reactions?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ionic reactions (precipitations, neutralizations etc.) are extremely fast (somewhere about $10^{12}$ times faster) than gas-phase reactions and other ordinary reactions in solutions (inversion, substitution etc). Why is this so?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My thoughts&lt;/strong&gt;&lt;br&gt;&#xA;One possible reason might be the relative freedom of steric orientation in case of simple ionic reactions as compared to organic reactions and complicated gas reactions, but this isn't the case with simple gas reactions or formation and dissociation of ionic complexes, which do require specific steric orientations. The activation energy could be very low for ionic reactions, if this is so, then what is the reason behind it?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2016-05-15T23:31:44.973" Title="Speed of ionic reactions?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6894" PostTypeId="2" ParentId="6893" CreationDate="2013-11-11T14:25:48.057" Score="4" Body="&lt;p&gt;In ionic reactions there is no electronic transfer between ions and the number of collisions between ions with opposite charge in high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When reaction occurs between neutral compounds, there is electronic transfer between atoms or molecules, and also there is bonding reorientation, son the speed in such reactions are lower than in ionic reactions.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-11T14:25:48.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6898" PostTypeId="2" ParentId="6893" CreationDate="2013-11-11T19:11:08.080" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_rate&quot;&gt;Reaction rate&lt;/a&gt; is inversely proportional to activation energy for a given process. The examples given have very low barriers and often the diffusion in solution is the rate limiting process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of reaction of anion with cation, than there should be no barrier at all, because the interaction between oppositely charged species is strongly attractive.  &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-11T19:11:08.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6900" PostTypeId="1" AcceptedAnswerId="6903" CreationDate="2013-11-12T03:47:44.497" Score="3" ViewCount="184" Body="&lt;p&gt;I was taught that when adding/subtracting measurements, add the absolute uncertainties, and when multiplying/dividing measurements, add the relative uncertainties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem in a lab where I need to find the difference between two measurements that are very close together, and are each very precise; however, when I propagate the absolute uncertainties, I get a huge relative uncertainty for the final result (upwards of 50%). Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0.050 ± 5%) - (0.044 ± 3%) = ?&#xA;= (0.050 - 0.044) ± ((0.050 * 5/100) + (0.044 * 3/100))&#xA;= 0.006 ± 0.00382&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ± 0.00382 corresponds to 63.7% relative uncertainty; am I doing something wrong? Both of the original measurements were pretty accurate (&amp;lt;12% RU). In this case can I just add their relative uncertainties instead of converting to absolute, which is very large in comparison with the difference between the measurements?&lt;/p&gt;&#xA;" OwnerUserId="3671" LastEditorUserId="83758" LastEditDate="2020-04-03T11:47:07.593" LastActivityDate="2020-04-03T11:47:07.593" Title="Propagation of uncertainty" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6901" PostTypeId="1" AcceptedAnswerId="6904" CreationDate="2013-11-12T03:50:26.227" Score="23" ViewCount="133821" Body="&lt;p&gt;Since fluorine has its valence electrons in the n=2 energy level, and since chlorine has its valence electrons in the n=3 energy level, one would initially expect that an electron rushing towards fluorine would release more energy, as it would land in the n=2 energy level, whereas in chlorine, the electron would land only in the n=3 energy level, and would then not release as much energy.  Thus, one would expect fluorine to have a greater electron affinity than chlorine.  However, why is it that chlorine has a higher electron affinity (349 kJ/mol)  than fluorine (328.165 kJ/mol)?&lt;/p&gt;&#xA;" OwnerUserId="3682" LastActivityDate="2015-05-18T07:38:11.467" Title="Why does chlorine have a higher electron affinity than fluorine?" Tags="&lt;periodic-trends&gt;&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6902" PostTypeId="1" AcceptedAnswerId="6905" CreationDate="2013-11-12T04:00:11.643" Score="2" ViewCount="5776" Body="&lt;p&gt;Please describe why copper only reacts with oxygen and not nitrogen in the air. Isn't nitrogen a negative ion, and copper positive, so wouldn't that mean that copper and nitrogen would want to form bonds?&lt;/p&gt;&#xA;" OwnerUserId="3684" LastEditorUserId="10185" LastEditDate="2015-06-15T18:25:44.927" LastActivityDate="2015-06-15T18:25:44.927" Title="Why does copper not react with nitrogen?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6903" PostTypeId="2" ParentId="6900" CreationDate="2013-11-12T04:23:00.123" Score="6" Body="&lt;p&gt;There is nothing wrong with the ±0.004 uncertainty you obtained. The relative uncertainty of the difference between close results is supposed to be far larger than that of the original measurements. One way of seeing why is stretching your scenario a bit: suppose you got 0.048 rather than 0.044 as the result of the second measurement. The difference, 0.002, would be small enough to be meaningless, as the second result would encroach into the uncertainty range of the first measurement. Your example is not so extreme; still, the large relative uncertainty is justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Aside: an analogous effect can affect computer calculations done with limited precision. Suggestively enough, it is referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Catastrophic_cancellation&quot;&gt;catastrophic cancellation&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="3683" LastActivityDate="2013-11-12T04:23:00.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6904" PostTypeId="2" ParentId="6901" CreationDate="2013-11-12T05:11:21.967" Score="15" Body="&lt;p&gt;Fluorine, though higher than chlorine in the periodic table, &lt;strong&gt;has  a very small atomic size&lt;/strong&gt;. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20090424075637AA1Dz7r&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-09-02T13:57:18.157" LastActivityDate="2014-09-02T13:57:18.157" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6905" PostTypeId="2" ParentId="6902" CreationDate="2013-11-12T05:18:16.087" Score="7" Body="&lt;p&gt;Sketching with broad strokes: the first key point is that oxygen and nitrogen are major components of the atmosphere in the form of $\ce{O2}$ and $\ce{N2}$ molecules, and not as ions. $\ce{N2}$ is a very stable compound because the two nitrogen atoms in its molecules are united by a particularly strong triple bond. $\ce{O2}$ is overall more reactive than $\ce{N2}$, largely as a reflex of the bond between the atoms not being as strong. Specifically concerning oxidation reactions (of which those you suggest are typical examples), oxygen atoms having more affinity for electrons than nitrogen atoms contributes towards making $\ce{O2}$ being a better oxidiser than $\ce{N2}$.&lt;/p&gt;&#xA;" OwnerUserId="3683" LastEditorDisplayName="user7951" LastEditDate="2015-06-15T17:40:57.160" LastActivityDate="2015-06-15T17:40:57.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6906" PostTypeId="1" CreationDate="2013-11-12T06:18:34.093" Score="5" ViewCount="465" Body="&lt;p&gt;In quantum mechanics, while writing the wave functions we can take a product of the spatial and the spin parts.But what does it means? What is the meaning of the spatial wave function and the spin wave function?&lt;/p&gt;&#xA;" OwnerUserId="3685" LastEditorUserId="1431" LastEditDate="2013-11-12T06:36:00.443" LastActivityDate="2014-02-27T20:19:22.910" Title="How can we write the wave function in quantum mechanics?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6907" PostTypeId="1" AcceptedAnswerId="6908" CreationDate="2013-11-12T11:04:46.033" Score="4" ViewCount="1037" Body="&lt;p&gt;I frequently encounter two ways of writing enthalpies of some process, e.g. for the enthalpy of formation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Delta_{r} G$ (or $\Delta_{R} G$)&lt;/li&gt;&#xA;&lt;li&gt;$\Delta G_{r}$ (or $\Delta G_{R}$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On the IUPAC Gold Book webpage they used the first notation (although the use of upper- and lower-case descriptors varies, see &lt;a href=&quot;http://goldbook.iupac.org/I03313.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://goldbook.iupac.org/GT07388.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). Is this the accepted way of doing it (and the other one is simply a relic) or might this be a matter of preference. And how about the upper- or lower-case descriptor: Is there any convention for that?&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2013-11-12T12:55:22.013" Title="What's the convention for the symbols of enthalpies of formation, reaction, etc.?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6908" PostTypeId="2" ParentId="6907" CreationDate="2013-11-12T12:55:22.013" Score="6" Body="&lt;p&gt;I think you have already answered your question… IUPAC's usage should be a pretty good indicator of the proper notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, however, an even better reference than the Gold Book: &lt;strong&gt;IUPAC's &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantities,_Units_and_Symbols_in_Physical_Chemistry&quot;&gt;Green Book&lt;/a&gt;, or &lt;em&gt;Quantities, Units and Symbols in Physical Chemistry&lt;/em&gt;&lt;/strong&gt; (&lt;a href=&quot;http://www.iupac.org/fileadmin/user_upload/publications/e-resources/ONLINE-IUPAC-GB3-2ndPrinting-Online-Sep2012.pdf&quot;&gt;full text here&lt;/a&gt;), on section 2.11.1 Other symbols and conventions in chemical thermodynamics (page 59 and following):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(iii) Examples of the use of the symbol ∆&lt;br&gt;&#xA;  The symbol ∆ denotes a change in an extensive thermodynamic quantity for a process. The addition of a subscript to the ∆ denotes a change in the property.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It says later:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[∆&lt;sub&gt;vap&lt;/sub&gt;&lt;em&gt;H&lt;/em&gt;] can also be written ∆&lt;em&gt;H&lt;/em&gt;&lt;sub&gt;vap&lt;/sub&gt;, but &lt;strong&gt;this usage is not recommended&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason is that ∆&lt;em&gt;H&lt;/em&gt;&lt;sub&gt;vap&lt;/sub&gt; could be understand as a change in &lt;em&gt;H&lt;/em&gt;&lt;sub&gt;vap&lt;/sub&gt;, which it is not. It a change of &lt;em&gt;H&lt;/em&gt; during a process, so the subscript should be on the ∆.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-11-12T12:55:22.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6911" PostTypeId="1" AcceptedAnswerId="6912" CreationDate="2013-11-12T18:06:10.790" Score="6" ViewCount="1477" Body="&lt;p&gt;I am a mathematics student doing a course in thermodynamics and am working on the basics now. The question I am asked is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose two states differ in energy by one electronvolt. What can be said about the ratio of the population at &lt;span class=&quot;math-container&quot;&gt;$T = \pu{300 K}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now it is clear that I will use the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{N_i}{N_j} = \exp\left(-\frac{E_i-E_j}{kT}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k = \pu{1.381 \times 10^−23 J K-1}$&lt;/span&gt;. Now if let &lt;span class=&quot;math-container&quot;&gt;$E_i$&lt;/span&gt; be the higher energy state, then am I right in saying that &lt;span class=&quot;math-container&quot;&gt;$E_i-E_j = \pu{1 eV}$&lt;/span&gt;, or should I convert this to some other unit?&lt;/p&gt;&#xA;" OwnerUserId="3688" LastEditorUserId="16683" LastEditDate="2019-05-09T01:28:54.753" LastActivityDate="2019-05-09T01:28:54.753" Title="What units do I use for energy in the Boltzmann distribution?" Tags="&lt;energy&gt;&lt;units&gt;&lt;statistical-mechanics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6912" PostTypeId="2" ParentId="6911" CreationDate="2013-11-12T18:48:45.407" Score="6" Body="&lt;p&gt;Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that &lt;span class=&quot;math-container&quot;&gt;$kT$&lt;/span&gt; has to have the same units as &lt;span class=&quot;math-container&quot;&gt;$E_i-E_j$&lt;/span&gt;, so the terms in the exponent cancel. If you're measuring temperature in &lt;span class=&quot;math-container&quot;&gt;$\pu{K}$&lt;/span&gt; and Boltzmann's constant in &lt;span class=&quot;math-container&quot;&gt;$\pu{J/K}$&lt;/span&gt;, then the units of energy you have to use are joules, &lt;span class=&quot;math-container&quot;&gt;$\pu{J}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="47" LastEditorUserId="16683" LastEditDate="2019-05-09T01:27:03.773" LastActivityDate="2019-05-09T01:27:03.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6913" PostTypeId="1" CreationDate="2013-11-13T00:10:58.860" Score="3" ViewCount="858" Body="&lt;p&gt;For solids where the compound contains defects, the stoichiometric equation deviates from the &quot;pure&quot; one. How are these compounds determined? I am looking in my chemistry book but I do not see anything on it. Are there some resources on this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing some thinking, I considered the &lt;a href=&quot;https://en.wikipedia.org/wiki/Auger_electron_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Auger Electron Spectroscopy&lt;/a&gt; technique used to study the composition of surfaces in materials science. These , I believe, are typically used to study the amount of defects in materials. I do not know how this would be applied to the situation I described above though (as this is only applicable to surfaces). I will continue searching. &lt;/p&gt;&#xA;" OwnerUserId="2511" LastEditorUserId="4945" LastEditDate="2014-05-12T13:43:14.583" LastActivityDate="2014-10-04T00:15:07.870" Title="How are non-Stoichiometric compounds determined?" Tags="&lt;spectroscopy&gt;&lt;solid-state-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6916" PostTypeId="1" CreationDate="2013-11-13T10:01:55.900" Score="0" ViewCount="26004" Body="&lt;p&gt;Calculate the molality, molarity and mole fraction of a $\ce{CuSO4}$&#xA;in a 17% m/m aqueous solution (density of the solution is 1.367 g/mole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did .17 * 1.367 to find grams of $\ce{CuSO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and .83 * 1.367 to find grams of $\ce{H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then used to them calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;molarity = 0.00146 M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;molality = 1.287 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mole fraction = 0.023&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if I'm doing it right?&lt;/p&gt;&#xA;" OwnerUserId="3692" LastEditorUserId="1878" LastEditDate="2013-11-13T10:56:16.030" LastActivityDate="2014-01-13T09:28:52.260" Title="How to calculate molality, molarity and mole fraction given density (g/mole)?" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6917" PostTypeId="1" CreationDate="2013-11-13T10:55:31.887" Score="0" ViewCount="106" Body="&lt;p&gt;If you facilitate a weak acid base reaction. i.e. $\ce{NaOH + H2SO4}$. All of it may not react and the solutions will not neutralize. How can you determine how much actually reacted?&lt;/p&gt;&#xA;" OwnerUserId="3693" LastEditorUserId="418" LastEditDate="2014-03-21T12:16:07.510" LastActivityDate="2014-03-24T18:17:04.490" Title="How can you tell how much actually reacted in an acid base reaction" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6919" PostTypeId="1" CreationDate="2013-11-13T14:49:02.070" Score="3" ViewCount="1125" Body="&lt;p&gt;Is there an online database of acid and basic dissociation constants, all in one place, free to use?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="4356" LastEditDate="2015-05-15T21:25:08.647" LastActivityDate="2015-05-16T18:01:13.450" Title="Is there an online database of acid and basic dissociation constants?" Tags="&lt;acid-base&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6920" PostTypeId="2" ParentId="6072" CreationDate="2013-11-13T17:10:21.463" Score="0" Body="&lt;p&gt;I googled and learned from a &lt;a href=&quot;http://www.ami.ac.uk/courses/topics/0223_plate/index.html&quot; rel=&quot;nofollow&quot;&gt;webpage&lt;/a&gt; that &quot;The ability of a solution to plate evenly over a large area (‘covering power’), and to plate into holes (‘throwing power’) depends on the current density, the concentration of acid and copper ions, the agitation of the solution, the anode-to-cathode separation, and the bath additives.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3695" LastEditorUserId="275" LastEditDate="2013-11-13T22:49:02.200" LastActivityDate="2013-11-13T22:49:02.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6921" PostTypeId="1" CreationDate="2013-11-13T18:10:58.137" Score="4" ViewCount="8188" Body="&lt;p&gt;So recently I have undertaken some research with some organometallic chemistry involved, specifically regarding that of ruthenium-aryl compounds. Trying to understand the chemistry of the central metal ion, I noticed something weird a couple of weeks ago, from Wikipedia: Ruthenium only has one electron in that orbital ($4d^75s^1$), rather than the two one would expect a transition metal to have, going with a naive Aufbau across the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware, I think, in the 4th period period transition metals, of a tendency to promote their 4s electrons to the 3d orbitals, when possible, so that they have either &quot;half-filled&quot; (5 electrons) or completely filled (10 electron) shells.  These are &quot;magic numbers&quot; for transition metals, apparently; just another one of those heuristic things they teach you in inorganic chemistry that I wish I understood more about. But alas, ruthenium has an asymmetric 7 electrons in its 4d orbital, blasted thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know anything? I tried using Google Scholar and a few databases, but when I type in &quot;orbital&quot; pretty much all I see is molecular orbital calculations specific to particular molecules. I'd really love to see papers pointing at some faint ideas I may be dimly aware about...&lt;/p&gt;&#xA;" OwnerUserId="711" LastEditorUserId="711" LastEditDate="2013-11-13T18:25:04.640" LastActivityDate="2013-11-14T08:19:19.563" Title="Why does Ruthenium only have one electron in its 5s orbital in the neutral oxidation state?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6923" PostTypeId="1" AcceptedAnswerId="6939" CreationDate="2013-11-13T20:59:57.597" Score="3" ViewCount="25198" Body="&lt;p&gt;My teacher in college says that bonds between metals and nonmetals are ionic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \ce{Metal - Metal} $ $\Rightarrow$ Metalic bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Non metal - Non metal}$ $\Rightarrow$ Covalent bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to write about $\ce{CuCl2}$, and found in Wikipedia that if you subtract electronegativity you get what bond it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi (\ce{Cl}) = 3.16, \quad \chi (\ce{Cu}) = 1.90$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi = 3.16 - 1.90 = 1.26$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\chi &amp;lt; 0.4 \Rightarrow$ covalent non-polar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.4 &amp;lt; \chi &amp;lt; 1.7 \Rightarrow$ covalent polar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\chi &amp;gt; 1.7 \Rightarrow$ ionic&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So it should be really a covalent bond. Which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="3696" LastEditorUserId="41328" LastEditDate="2017-08-05T21:35:38.663" LastActivityDate="2017-11-24T10:25:33.460" Title="Bonds between metals and non-metals" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6924" PostTypeId="1" AcceptedAnswerId="6927" CreationDate="2013-11-13T21:17:44.467" Score="20" ViewCount="1793" Body="&lt;p&gt;As you may know, the reaction quotient $Q_c$ is defined by the equation&#xA;$$&#xA;Q_c = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}&#xA;$$&#xA;for the chemical reaction&#xA;$$&#xA;\alpha A + \beta B \rightarrow \gamma C + \delta D.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something I've been struggling to understand at an intuitive level. Why is $Q_c$ the ratio of the product of the reaction's product concentrations and the product of the reaction's reactant concentrations (sorry for the confusing terminology; I can't think of a more precise way to phrase that), rather than the ratio of the sums of concentrations, like this:&#xA;$$&#xA;Q_c' = \frac{\gamma[C]+\delta[D]}{\alpha[A]+\beta[B]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This second, incorrect equation makes more sense to me because concentrations of solutes add up. Is the ratio defined as such so that it is directly proportional to the concentrations of each product and inversely proportional to the concentrations of each reactant? Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?&lt;/p&gt;&#xA;" OwnerUserId="3697" LastEditorUserId="3697" LastEditDate="2013-11-17T06:48:47.963" LastActivityDate="2017-05-14T14:14:47.797" Title="Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6927" PostTypeId="2" ParentId="6924" CreationDate="2013-11-14T02:54:58.983" Score="22" Body="&lt;p&gt;The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me &lt;em&gt;try&lt;/em&gt; to explain how.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a &lt;em&gt;abstract&lt;/em&gt; manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an &lt;em&gt;unfulfilled&lt;/em&gt; equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself).&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2017-05-14T14:14:47.797" LastActivityDate="2017-05-14T14:14:47.797" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6928" PostTypeId="2" ParentId="6921" CreationDate="2013-11-14T08:19:19.563" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://books.google.com/books?id=fJWpg4ZJ2esC&amp;amp;pg=PA219&amp;amp;lpg=PA219&amp;amp;dq=4d75s1%204d65s2&amp;amp;source=bl&amp;amp;ots=W7aOpR-Hl4&amp;amp;sig=AewlseP1QY0Vy-g07DYAcAcOErQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=VtqDUqTRHIiQiAfTs4GQAg&amp;amp;ved=0CHsQ6AEwCQ#v=onepage&amp;amp;q=4d75s1%204d65s2&amp;amp;f=false&quot;&gt;Oxtoby et al, Principles of Modern Chemistry, 7th edi p219&lt;/a&gt; and Levine, Quantum Chemistry, 5th edi, p314, the orbital energies of Sc and the elements beyond, 3d&amp;lt;4s. The electrostatic repulsion makes energy of 3d$^3$ higher than 3d$^1$4s$^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just my pure guess, Ru is bigger than Fe, electrostatic repulsion is smaller, so 5s$^1$. However, the 6s$^2$ electrons in Os are extra stable due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_quantum_chemistry&quot;&gt;relativistic effect&lt;/a&gt;, so again 6s$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantitative calculation will be highly appreciate for justifying this hand-waving argument&lt;/p&gt;&#xA;" OwnerUserId="2629" LastActivityDate="2013-11-14T08:19:19.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6929" PostTypeId="1" AcceptedAnswerId="17041" CreationDate="2013-11-14T09:22:31.130" Score="7" ViewCount="1592" Body="&lt;p&gt;To explain the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/xrayc.html&quot; rel=&quot;nofollow&quot;&gt;characteristic X-ray&lt;/a&gt; emission peaks for various elemental targets, a formula was developed which was similar in construct to the Rydberg equation for H-atom as derived by the Bohr-model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde\nu=R\left(\frac{1}{n_\mathrm{f}^2}-\frac{1}{n_1^2}\right)\left(Z-\sigma\right)^2$$ &#xA;where $\tilde\nu$ is the wavenumber, $\sigma=1$ for the $K_\alpha$ line and $7.4$ for the $L_\alpha$ line. My question is:&lt;br&gt;&#xA;Although Bohr model is wrong in light of the modern quantum mechanical concept of atomic structure, why does the energy for transition (and X-ray emission) of complicated atoms with complicated screening effect be represented by a simple adjustment of the $Z$ factor in the Rydberg equation with the value of $R$ used in case of H atom? Why isn’t there a need of complicated corrections (instead of simple subtraction of an experimentally obtained $\sigma$) to predict the X-ray wavelength?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T18:54:18.983" LastActivityDate="2015-07-02T18:54:18.983" Title="Understanding Moseley’s law from the Rydberg-type equation?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6932" PostTypeId="1" CreationDate="2013-11-14T16:28:41.327" Score="3" ViewCount="1470" Body="&lt;p&gt;I have a solution with &lt;strong&gt;only&lt;/strong&gt; $\ce{CO_3^{2-}}$ ions and water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen with the carbonate ions if I were to evaporate the water of that solution (in a vacuum)?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T13:22:41.940" LastActivityDate="2015-06-26T13:23:34.707" Title="Evaporating the water of a solution of only carbonate ions and water gives?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6933" PostTypeId="1" AcceptedAnswerId="6945" CreationDate="2013-11-14T17:39:44.693" Score="2" ViewCount="212" Body="&lt;p&gt;Currently i am doing some research on optical properties of  a oxide material and for that purpose i have done some photoluminescence study on it. Because this is an ongoing research i cant divulge much information here. I have used &quot;$Ex$&quot; excitation wavelength for getting emission spectra for my oxide sample. I have prepared two types of oxide samples one is heat treated and the other not heat treated. After taking emission spectra data for both the samples I got peaks similar to this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Samples               Peak wavelength (nm)&#xA;&#xA;Non Heat treated          a, b&#xA;&#xA;Heat treated              b,c,d,e&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;*a,b,c,d &amp;amp; e are wavelength where emission peaks detected&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both samples (heat treated and non heat treated) has same chemical composition which i confirmed using powder XRD. As you can see from the data the heat treated sample has more emission peaks comparing to non heated samples. I can understand from this data that heat treatment to the sample causes increased defects in the sample. what i thought was that after heat treating a crystalline material; the defect in that material should decrease and not the other way round? Is my rationale wrong. Why heat treated sample has more defects?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2013-11-15T09:13:53.427" Title="Changes in defects in a material due to heat treatment" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;spectroscopy&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6934" PostTypeId="2" ParentId="6882" CreationDate="2013-11-14T18:18:05.430" Score="5" Body="&lt;p&gt;I believe that ethylene glycol in this reaction is used as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diethylene_glycol&quot; rel=&quot;nofollow noreferrer&quot;&gt;solvent&lt;/a&gt; and &lt;a href=&quot;http://www.scientific.net/SSP.121-123.251&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aniline&lt;/a&gt; is used as a reducing agent. Ethylene glycol solution heated with chloroauric acid produces &lt;span class=&quot;math-container&quot;&gt;$Au^{3+}$&lt;/span&gt; ions (first 20 min) which is then reduced to &lt;span class=&quot;math-container&quot;&gt;$Au^{0}$&lt;/span&gt;  (3hr reaction) in the presence of aniline. Because there is no stirring during those 3 hrs (so less agitation of the reaction mixture)there is less formation of gold nanoparticle and more gold nano flakes are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be this &lt;a href=&quot;http://iopscience.iop.org/0957-4484/23/1/015602/article&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper will&lt;/a&gt; help you, it has a similar mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bScs8.gif&quot; alt=&quot;enter image description&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="1234" LastEditDate="2019-05-13T16:04:04.380" LastActivityDate="2019-05-13T16:04:04.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6936" PostTypeId="2" ParentId="6932" CreationDate="2013-11-14T18:32:51.270" Score="2" Body="&lt;p&gt;As a theoretical experiment, you must think on different aquieus equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O &amp;lt;=&amp;gt; H+ + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO3^{2-} &amp;lt;=&amp;gt; HCO3- &amp;lt;=&amp;gt; H2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2CO3}$ will be formed with water's $\ce{H+}$, and in vacuum, possibly will decompose to water and $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, may be a mixture of $\ce{OH-}$ and $\ce{CO3^{2-}}$ will remain.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="4945" LastEditDate="2014-04-25T03:17:15.463" LastActivityDate="2014-04-25T03:17:15.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6937" PostTypeId="2" ParentId="6932" CreationDate="2013-11-14T20:18:10.117" Score="4" Body="&lt;p&gt;Theoretically you should have an explosion. Because every $\ce{CO_3^{2-}}$ ions exerts a force against the other. You should be able to calculate it with coulomb equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|F|=K_e\frac{q_{\ce{CO_3^{2-}}}^2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you have to consider the force exerts from every particle to each others, summing every force field assuming a certain          distance: a very hard calculation. &#xA;By the way your assumption are quite hard to reach in great scale as say Raur Ferro. But a practical application of electrostatic forces that could be enlightening is Electrospray ionization (ESI), a technique used in mass spectrometry to produce ions. &#xA; In this technique the solvent from an areosols that hold the analyte is forced to evaporate (with various expedients) from a charged droplet until the drop becomes unstable due the charges. At a certain point the droplet undergoes to what is called Coulomb fission, so the original droplet explodes creating many smaller droplets. I know this is not the same thing but this can show you how electrostatic forces can act. I think however that this question is more related to physic than chemistry.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T13:23:34.707" LastActivityDate="2015-06-26T13:23:34.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6938" PostTypeId="2" ParentId="6145" CreationDate="2013-11-14T23:12:12.920" Score="6" Body="&lt;p&gt;Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the &lt;strong&gt;standard&lt;/strong&gt; free energies $\Delta G_r^o$ alone, as most reactions are &lt;em&gt;not&lt;/em&gt; performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that, &lt;em&gt;in standard conditions&lt;/em&gt;, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-01-14T00:28:54.440" LastActivityDate="2017-01-14T00:28:54.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6939" PostTypeId="2" ParentId="6923" CreationDate="2013-11-15T00:39:59.120" Score="15" Body="&lt;p&gt;You should be careful with simple associations such as &quot;metal + non-metal = ionic bond&quot;. These tend to throw out the idea of understanding the chemistry involved in favour of rote memorization. Note for example that mixing caesium metal with gold will produce a &lt;em&gt;salt&lt;/em&gt; instead of an alloy, &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_auride&quot; rel=&quot;nofollow noreferrer&quot;&gt;caesium auride&lt;/a&gt; ($\ce{Cs^+ Au^{-}}$). Mixing barium metal and platinum can also produce salts, though their structures are somewhat more complex. One can also argue that there is significant ionic character in solid xenon difluoride, even though both atoms are non-metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of using electronegativity to determine covalent/ionic character is also meant as a helpful guide, not as a strict rule with black-and-white limits. Firstly, all bonds have both ionic and covalent character; both concepts are an oversimplification, and in reality it is more correct to say that a bond has a certain contribution from each type of bonding. This means there is a smooth transition from compounds with mostly ionic character and those with mostly covalent character. Also, the inequalities you mention rely on Pauling electronegativities. Electronegativity is surprisingly still a hotly debated topic, as we continue to search more general, more fundamental and more precise ways of defining it. Pauling electronegativities are based on empirical thermodynamic data regarding bond energies after applying a certain equation that was &quot;picked&quot;, not derived from scratch. The values are particularly poorly defined for transition elements, such as the $\ce{Cu}$ in your problem. You get some not-so-easy to explain situations, like $\ce{HF}$ as a gas that is a borderline ionic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in light of these comments, the answer to your question is that bonding in $\ce{CuCl_2}$ (I'm pretty sure that's what you actually meant to write) has intermediate characteristics between a purely ionic and a polar covalent bond, with similar contributions (though pinpointing which is highest sounds like an exercise in futility). A good way to study it more in depth is to analyze &lt;a href=&quot;http://en.wikipedia.org/wiki/Fajans%27_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fajans' rules&lt;/a&gt;. After a little self-calibration, you can get a good feel for the degree of ionicity and covalency of a compound. Some further but less certain evidence (lots of caveats!) for the intermediate character of $\ce{CuCl_2}$ can be found by looking at the substances' melting and boiling points ($\pu{498°C}$ and $\pu{993°C}$ [decomposition], respectively, according to Wikipedia). They are both quite high compared to substances with polar covalent bonds (dimethylformamide boils at around $\pu{150°C}$), but rather low compared to substances with very ionic bonds ($\ce{NaCl}$ boils over $\pu{1400°C}$).&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="7475" LastEditDate="2017-11-24T10:25:33.460" LastActivityDate="2017-11-24T10:25:33.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6941" PostTypeId="1" CreationDate="2013-11-15T05:55:44.940" Score="3" ViewCount="1073" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My book says:&lt;br&gt;&#xA;  Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.&#xA;  $$\ce{R-X + NaI -&amp;gt;[\ce{acetone,~reflux}] R-I + NaX}\\&#xA;(\ce{X}=\ce{Cl,Br};\ \ce{R}=\text{alkyl group})$$&#xA;    &lt;em&gt;NOTE&lt;/em&gt;: Refluxing is a process of heating a liquid in a flask provided with a condenser. The vapours of the liquid which rise up get condensed back into the flask.&lt;br&gt;&#xA;    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;MY QUERY&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?&lt;br&gt;&#xA;As we can see, first line reads that Finkelstein reaction is particularly useful for preparing iodoalkanes. The lines followed by it signify that $\ce{NaCl}$ and $\ce{NaBr}$ are less soluble in acetone. If we want to prepare alkyl chloride or alkyl bromide, we will be needed to dissolve $\ce{NaCl}$ or $\ce{NaBr}$ respectively in acetone, which is not so possible. Is it true that, we can't prepare alkyl chloride or alkyl bromide by Finkelstein reaction because of the above reasons?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="7475" LastEditDate="2016-12-01T22:36:28.310" LastActivityDate="2016-12-01T22:36:28.310" Title="Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6943" PostTypeId="1" AcceptedAnswerId="6952" CreationDate="2013-11-15T07:24:03.650" Score="9" ViewCount="2344" Body="&lt;p&gt;For this case, let's say there are two identical chambers connected by a small tube, constant pressure and temperature. There's a closed valve that separates the two chambers. Within the chambers there are two ideal gases (doesn't matter which, I think; they're ideal gases). I release the valve and the contents of the two chambers mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that heat of mixing for ideal gases is 0 conceptually, because the molecules don't have interaction forces between them so introducing the gases to each other doesn't do much with respect to this. &lt;strong&gt;But what is the mathematical basis for heat of mixing being 0 for an ideal gas?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;quick edit: above might not be clear. I mean that dH=TdS+VdP. In isothermal, isobaric conditions, dP could be zero, but dS isn't necessarily, that'd imply adiabatic conditions. So how is dH=0 for mixing an ideal gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the above line of thought apply for mixing? If so, do we take the terms before or after the volume increases? (in this case, final volume = 2*initial volume) I feel this is a frame of reference issue, one that comes up a lot for me in thermo.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-10-22T12:06:34.393" Title="Mathematical basis of why enthalpy of mixing is 0 for ideal gas" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6944" PostTypeId="1" CreationDate="2013-11-15T08:27:36.973" Score="26" ViewCount="230194" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The branched chain compounds have lower boiling points than the corresponding straight chain isomers. For example,  &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$\ce{CH_3CH_2CH_2CH_2CH_3}$ - &#xA;  No branching-Pentane (n-pentane) ($\mathrm{b.p.}=309~\mathrm{K}$)  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{CH_3CH(CH_3)CH_2CH_3}$ - &#xA;  One branching-2-Methylbutane (Iso-pentane) ($\mathrm{b.p.}=301~\mathrm{K}$)  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{C(CH_3)_4}$ - Two branches-2,2-Dimethylpropane (Neo-pentane) ($\mathrm{b.p.}=282.5\mathrm{K~}$)  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface area. Therefore, the intermolecular attractive forces which depend upon the surface area, also become small in magnitude on account of branching. Consequently, &lt;em&gt;the boiling points of the branched chain alkanes are less than the straight chain isomers.&lt;/em&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above extract from my book, mentions clearly that branching makes the molecule more compact and thereby decreases the surface area. Even for isomeric alkyl halides, the boiling points decrease with branching. The reason is said to be because of the decrease in surface area, same as explained before. How does decrease in the surface area make the intermolecular forces small in magnitude?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="402" LastEditDate="2017-04-02T20:21:20.963" LastActivityDate="2019-11-22T17:43:27.500" Title="Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6945" PostTypeId="2" ParentId="6933" CreationDate="2013-11-15T09:13:53.427" Score="3" Body="&lt;p&gt;It seems, that your oxide can have different crystal polymorphs and probably also other non-crystalline phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You synthesize your material as mixture of those, where the composition depends on the preparation process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon heating and slow cooling you probably recrystallize the species to more thermodynamically stable polymorph (new peaks c,d,e) on the expense of less stable one (a).&#xA;It can also depend on the speed of cooling, so you could obtain some metastable phases, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then, treat your material always as an unkonwn mixture of uknown number of polymorphs and let the spectra tell you in which way the composition shifted.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-11-15T09:13:53.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6946" PostTypeId="1" CreationDate="2013-11-15T10:42:20.227" Score="5" ViewCount="6637" Body="&lt;p&gt;Crystal Field Theory explains colors of Coordination compounds as follows : A d-orbital splits into multiple orbitals, the process being called crystal field splitting. When white light falls on the compound, an electron makes a transition into a higher state thus absorbing a particular wavelength of light. The rest of the light is reflected. This reflected light is responsible for the color of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Suppose, all electrons have now transit into a higher state. If more light falls on the compound, there should be no more transitions as all the electrons are already in the highest state. This means the compounds should lose color after some time in the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How is the light absorbed again and again? If we suppose that the electron falls back to lower state, that would emit the same wavelength it absorbed. Even in this case, how do we explain color?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S.&lt;/em&gt; Also, if they can absorb only a particular wavelength(given quantisation), that should not bring about a color change. There must be a range of wavelengths(however narrow) being absorbed. What is the reason for that?&lt;/p&gt;&#xA;" OwnerUserId="3708" LastActivityDate="2014-02-27T19:25:18.003" Title="Crystal Field Explanation of Colors of Coordination Compounds" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6947" PostTypeId="1" CreationDate="2013-11-15T13:36:10.710" Score="7" ViewCount="26944" Body="&lt;p&gt;&lt;strong&gt;My book states that:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.&lt;br&gt;&#xA;  $$\ce{R-X~+~NaI-&amp;gt;[\ce{acetone,~reflux}]R-I~+NaX}$$&#xA;  $(\ce{X=Cl,~Br;~R=alkyl ~group})$&lt;br&gt;&#xA;    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are $\ce{NaCl}$ and $\ce{NaBr}$ not soluble, whereas $\ce{NaI}$ is soluble in acetone?&#xA;All we can observe is that, sodium iodide has greater molecular weight than sodium chloride or sodium bromide. But, how do this account for solubility of sodium iodide?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="5026" LastEditDate="2018-02-21T14:28:21.127" LastActivityDate="2018-02-21T14:28:21.127" Title="Why are NaCl and NaBr not soluble, whereas NaI is soluble in acetone (in the Finkelstein reaction)?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6948" PostTypeId="1" CreationDate="2013-11-15T13:47:16.140" Score="9" ViewCount="7243" Body="&lt;p&gt;&lt;strong&gt;My book states that:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.&lt;br&gt;&#xA;  $$\ce{R-X~+~NaI-&amp;gt;[\ce{acetone,~reflux}]R-I~+NaX}$$&#xA;  $(\ce{X=Cl,~Br;~R=alkyl ~group})$&lt;br&gt;&#xA;  Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chartelier's principle. The reaction gives the best results with primary halides.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we prepare alkyl fluoride (not aryl fluoride) by Finkelstein reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book mentions about another reaction called Swarts reaction, through which alkyl fluoride is prepared. In Swarts reaction, alkyl fluorides are prepared by heating alkyl chlorides or bromides with salts such as mercurous fluoride ($\ce{Hg2F2}$), silver fluoride ($\ce{AgF}$), cobalt fluoride ($\ce{CoF3}$) or antimony trifluoride ($\ce{SbF3}$). Why can't we prepare alkyl fluoride by treating sodium fluoride with alkyl halide (Finkelstein reaction)?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="5026" LastEditDate="2018-02-21T14:31:20.237" LastActivityDate="2019-10-19T02:38:10.187" Title="Can we prepare alkyl Fluoride by Finkelstein reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6950" PostTypeId="1" CreationDate="2013-11-15T00:39:28.567" Score="6" ViewCount="79" Body="&lt;p&gt;I see tons of papers referencing this structure but none show what it looks like.&lt;/p&gt;&#xA;" OwnerUserId="4449" OwnerDisplayName="user17338" LastEditorUserId="30" LastEditDate="2013-11-15T16:50:32.217" LastActivityDate="2014-03-26T13:31:43.270" Title="What does the Sm5Ge4 orthorhombic structure look like?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6951" PostTypeId="2" ParentId="6950" CreationDate="2013-11-15T03:28:57.897" Score="5" Body="&lt;p&gt;You can look at the $\text{Sm}_5 \text{Ge}_4$ crystal structure at Materials Project &lt;a href=&quot;https://materialsproject.org/materials/mp-1506/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EmQ9l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user23660" LastActivityDate="2013-11-15T03:28:57.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6952" PostTypeId="2" ParentId="6943" CreationDate="2013-11-15T15:42:13.973" Score="11" Body="&lt;p&gt;A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}}H = \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S$ later): for isothermal and isobaric conditions it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.&#xA;So, now we need an expression for the chemical potential of an ideal gas.&#xA;It can be shown that this is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{i}}{p^{0}}\Bigr) \hspace{18em} (2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $p^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).&#xA;For ideal gases we can apply Raoult's law stating that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  p_{i} = x_{i} p_{\mathrm{tot}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_{\mathrm{tot}}$ is the total pressure of the system.&#xA;Using this with equation (2) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{x_{i} p_{\mathrm{tot}}}{p^{0}}\Bigr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \hspace{6em} (3)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  p_{\mathrm{tot}} = \sum_{i} p_{i}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.&#xA;At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.&#xA;Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the ambient/total pressure $p_{\mathrm{tot}}$ in the system, since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).&#xA;According to equation (1) the Gibbs free energy of the $i$th chamber is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{i} &amp;amp;= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\&#xA;        &amp;amp;= n_{i} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Gibbs free energy of the total system at the initial state is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{\mathrm{ini}} &amp;amp;= \sum_{i} G_{i} \\&#xA;                 &amp;amp;= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\&#xA;                 &amp;amp;= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\&#xA;                 &amp;amp;= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\&#xA;                 &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.&#xA;So now, the complete system of connected chambers must be treated via equation (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{\mathrm{final}} &amp;amp;= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Delta_{\mathrm{mix}}G &amp;amp;= G_{\mathrm{final}} - G_{\mathrm{ini}} \\&#xA;                       &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  \\&#xA;                       &amp;amp;= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} \hspace{22em} (4)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating equation (4) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} = - n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you get the desired result by using the famous equation $G = H - TS$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Delta_{\mathrm{mix}}H &amp;amp;= \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S \\&#xA;                        &amp;amp;= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} - T n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i} \\&#xA;                        &amp;amp;= 0&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-22T12:06:34.393" LastActivityDate="2014-10-22T12:06:34.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6954" PostTypeId="1" AcceptedAnswerId="7004" CreationDate="2013-11-15T18:03:47.797" Score="5" ViewCount="3359" Body="&lt;p&gt;According to this article called&#xA;&lt;a href=&quot;http://www.samharris.org/blog/item/the-fireplace-delusion&quot; rel=&quot;nofollow&quot;&gt;The Fireplace Delusion&lt;/a&gt;&#xA;a wood-fire will create benzene. This surprised me and I can't find an article online that explains the chemical reaction behind that. &lt;/p&gt;&#xA;" OwnerUserId="3709" LastEditorUserId="2599" LastEditDate="2013-11-20T00:11:52.743" LastActivityDate="2020-01-28T06:52:30.697" Title="Why does a wood fire create benzene?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6955" PostTypeId="1" AcceptedAnswerId="6962" CreationDate="2013-11-15T20:57:11.323" Score="4" ViewCount="13192" Body="&lt;p&gt;A solution made of two or more compounds have adhesive and cohesive forces. Cohesive forces are the attractive forces between the same molecules. Adhesive forces are the attractive forces between different molecules of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real solution shows positive deviation from Raoult's law when the cohesive forces are stronger then adhesive forces. It shows negative deviation when the adhesive forces are stronger than cohesive forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do stronger cohesive forces in real solution make positive deviation and stronger adhesive forces make negative deviation from an ideal solution?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="1343" LastEditDate="2013-11-16T20:05:04.413" LastActivityDate="2020-05-24T14:20:40.200" Title="Deviation from Raoult's law because of adhesive and cohesive forces" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6956" PostTypeId="1" CreationDate="2013-11-15T21:02:50.080" Score="4" ViewCount="152" Body="&lt;p&gt;For example can I introduce salts that have a chaotropic or kosmotropic affect on ethanol? Or is this phenomenon restricted to water? If this phenomenon does occur in other polar solvents, does it follow the Hofmeister series as well? Are there any papers out there on this phenomenon in polar non-aqueous solvents?&lt;/p&gt;&#xA;" OwnerUserId="3710" LastActivityDate="2014-06-18T18:11:24.593" Title="Do chaotropic and kosmotropic affects occur in polar non-aqueous media?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6957" PostTypeId="1" AcceptedAnswerId="44133" CreationDate="2013-11-16T05:39:43.520" Score="9" ViewCount="307" Body="&lt;p&gt;When you drop a bottle of soda, the dissolved $\ce{CO_2}$ starts nucleating. Opening it before it's 'settled' will trigger more nucleation and usually makes a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the shock trigger nucleation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, is it the same mechanism in play when you shake a bottle? And what about the physical state changes as it 'settles'? i.e., why does the $\ce{CO_2}$ rapidly nucleate if you open the bottle right after you drop it, but it doesn't if you wait a bit?&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorDisplayName="user15489" LastEditDate="2015-06-23T10:58:29.303" LastActivityDate="2016-01-23T21:29:08.137" Title="How does shock trigger nucleation of gases dissolved in a liquid?" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6962" PostTypeId="2" ParentId="6955" CreationDate="2013-11-16T18:08:09.223" Score="14" Body="&lt;h1&gt;Foreword&lt;/h1&gt;&#xA;&lt;p&gt;This topic is treated in most books and sites so I've tried to explain it to you in a more unusual and friendly way to grasp the concept but maybe the canonicals explanations are more suited for exams and serious conversation.&lt;/p&gt;&#xA;&lt;h1&gt;First step idealize and simplify the Raoult law&lt;/h1&gt;&#xA;&lt;p&gt;You can see Raoult law as the most basic attempt to predict the partial vapour pressure of a multi-components solution. If &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is your first component with higher vapour pressure an &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is the second component with lower vapour pressure, it seems quite obvious that if you mix an equal amount of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; you have a vapour pressure equal to the mean of the two.&lt;/p&gt;&#xA;&lt;h1&gt;Back to reality&lt;/h1&gt;&#xA;&lt;p&gt;In fact, the reality is more complex and you have to take into account the force between the molecules of the two components. If you have &lt;strong&gt;strong adhesive&lt;/strong&gt; forces the molecules of the first component near the molecules of the second keep themself more stuck together. So this creates a sort of &amp;quot;net&amp;quot; that doesn't allow the molecules to &amp;quot;escape&amp;quot; from the solution and so the vapour pressure (that in fact quantify the ratio of molecules that &amp;quot;escape&amp;quot; from the bulk) is lower and finally  you have a &lt;strong&gt;negative deviation&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But what if the interaction between the molecules of the same component is greater than the interaction between the two components? As you state, this is the case of &lt;strong&gt;strong cohesive&lt;/strong&gt; forces. In this case you have to imagine that the components act as if they were separate.  Molecules of the component A lower the positive interaction between the molecules of the component B interposing them self and avoiding the formation of bonds between molecules of component B and finally increasing the vapour pressure, a &lt;strong&gt;positive deviation&lt;/strong&gt;!&lt;/p&gt;&#xA;&lt;h1&gt;Deep into fantasy&lt;/h1&gt;&#xA;&lt;p&gt;So if you think the molecules as humans exiting from an elevator, humans A are more claustrophobic and so tend to exit more rapidly than humans B when the doors open. If they interact positively with humans B they will keep calm when exiting. If humans B don't find funny humans A they preserve their panic attitude and at the same time separate B humans from each other and rushing out from the elevator pull humans B out more rapidly too!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-05-24T14:20:40.200" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6965" PostTypeId="1" CreationDate="2013-11-17T03:24:36.197" Score="3" ViewCount="550" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt; required to prepare &lt;span class=&quot;math-container&quot;&gt;$\pu{150 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.75 N}$&lt;/span&gt; solution based on the reaction&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{BrO3- + 6 H+ + 6e- -&amp;gt; Br- + 3 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{150 ml}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.75 N}$&lt;/span&gt; means &lt;span class=&quot;math-container&quot;&gt;$0.1125$&lt;/span&gt; equivalents of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt; are present in the solution. Since, according to the given equation, n-factor (of equivalence) of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$0.1125$&lt;/span&gt; equivalents means &lt;span class=&quot;math-container&quot;&gt;$0.01875$&lt;/span&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt;. The molecular mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$151$&lt;/span&gt;. Thus, I get the mass as &lt;span class=&quot;math-container&quot;&gt;$\pu{2.83 g}$&lt;/span&gt; approximately. Am I correct? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the book says the answer should be &lt;span class=&quot;math-container&quot;&gt;$\pu{1.42 g}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="83758" LastEditDate="2020-05-28T09:06:13.173" LastActivityDate="2020-05-28T09:06:13.173" Title="How do I calculate the normality of NaBrO3?" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6966" PostTypeId="1" CreationDate="2013-11-17T04:48:06.060" Score="2" ViewCount="1709" Body="&lt;p&gt;To understand the commonly quoted magnetic values of coordination complexes (central ion) we use &#xA;$$m_l=\sqrt{n(n+2)} \text{BM where BM}=\frac{e\hbar}{2m_e}\text{JT}^{-1}$$ &#xA;$n$=number of unpaired electrons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How did we derive this equation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the orbital angular momentum of electron is given by $\sqrt{l(l+1)}\hbar$, spin angular momentum is given by $\sqrt{s(s+1)}\hbar$ and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gyromagnetic_ratio&quot; rel=&quot;nofollow&quot;&gt;gyromagnetic ratio&lt;/a&gt; for electron is $$\vec \mu_B=\frac{-e}{2m_e}\vec L$$. There is also the &quot;total&quot; angular momentum given by $n_0\hbar$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I fail to see how these magnetic moments and angular momentums are&#xA; related, and how they can be added or related to get the corresponding&#xA; values for an entire atom or a complex?&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also &lt;em&gt;heard&lt;/em&gt; that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Land%C3%A9_g-factor&quot; rel=&quot;nofollow&quot;&gt;Lande g-factor&lt;/a&gt; is relevant here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key:-&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_0$=principal quantum number, $l$=azimuthal quantum number,$s$=spin (half for an electron), $\hbar=\frac{h}{2\pi}$, $S=\Sigma s+1$(spin multiplicity), $\vec L$=angular momentum, $\vec \mu_B$=Magnetic moment.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="22" LastEditDate="2014-02-27T21:04:51.877" LastActivityDate="2014-02-27T21:04:51.877" Title="Magnetic moment of coordination complexes?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;coordination-compounds&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6968" PostTypeId="2" ParentId="6965" CreationDate="2013-11-17T15:55:19.797" Score="2" Body="&lt;p&gt;To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-17T15:55:19.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6969" PostTypeId="1" AcceptedAnswerId="6970" CreationDate="2013-11-17T19:27:23.050" Score="7" ViewCount="153" Body="&lt;p&gt;I'm referring to the Shuttle Battery recently announced (read more on it &lt;a href=&quot;http://www.connexxsys.com/en/en_shuttle.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) that suggests that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The energy density of the device is up to 7,700 Wh/L. It is more than 5 times &#xA;  higher than state-of-the-art (SOA) Li-ion (550Wh/L), and comparable to fossil &#xA;  fuel (10,000 Wh/L).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes on to say that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is rechargeable because it is based on a simple reversible chemical reaction, &lt;code&gt;3Fe + 4H2O = 4H2 + Fe3O4 and 2H2 + O2 = 2H2O&lt;/code&gt;. The proof of concept cell demonstrated zero capacity loss after 200 cycles of discharge and charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the question is: Is this even remotely possible? Or is there a red flag here somewhere that I can't seem to pick up?&lt;/p&gt;&#xA;" OwnerUserId="3722" LastEditorUserId="73" LastEditDate="2013-11-18T03:29:01.840" LastActivityDate="2013-11-21T23:00:42.573" Title="Can a battery powered by iron and air really provide a feasible power source for automobiles?" Tags="&lt;redox&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6970" PostTypeId="2" ParentId="6969" CreationDate="2013-11-18T03:54:20.753" Score="5" Body="&lt;p&gt;This is certainly possible.  In fact, there are a large number of metal-air batteries being developed all over the world.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, however, strange that such high energy density is achieved, because iron ion itself would not offer very high energy density.  My guess is that likely they did it by having an extremely high matter density.  So this battery may be small in size for its capacity, but at the same time it may also be much heavier than a conventional battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That could be why they brand it as a shuttle power supply(well, energy density isn't really relevant for a bus anyways).  A very heavy battery on a hand held device is never good.   Cost is the more important factor there, but it is hard to tell at the moment.  0 loss over 200 cycles seems like a fairy tale, but I will give them the benefit of the doubt at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This battery will also likely to have a considerable loss of charge when not in use.   Again, this might not be a huge problem for buses, where you keep driving non-stop during days then kept it in the charging station overnight.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, don't expect this to be on your iPhone 10 or your laptop.  Even if it works as presented, it might not be that ideal for portable electronic devices, where people worry much more about weight and stand-by time than if it occasionally explodes, sadly.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-11-18T03:54:20.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6971" PostTypeId="2" ParentId="6525" CreationDate="2013-11-18T04:55:44.283" Score="2" Body="&lt;p&gt;The first reaction ($\ce{HX}$+Ethanol) happens in solution.  High concentration of HX mixed with ethanol will produce the product. However, for $\ce{X}$ other than $\ce{Cl}$, more often one put concentrated phosphoric or sulfuric acid with alcohol and a salt of X (for example, $\ce{KBr}$ or $\ce{KI}$).  Concentrated phosphoric/sulfuric acid are easy to obtain and store in high concentration, as opposed to $\ce{HX}$. It is, therefore, a liquid phase reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ethene + $\ce{HX}$, it can be done in gas-phase, in an aqueous environment, or you can also add $\ce{HX}$ gas to an ethene solution in a solvent such as $\ce{CCl4}$. So, it can be either gas phase or liquid phase.  It is almost always more preferable to do a reaction in solution in a laboratory set up. (So much that we call the experimental part of the chemistry wet chemistry.)  This constraint is, however, lifted in the industrial world, with the availability of expansive specialized reactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for benzene + $\ce{Br2}$, both reactants are liquid and $\ce{Br2}$ is also soluble in benzene, so the reaction is in also the liquid phase.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="402" LastEditDate="2017-04-08T22:26:43.373" LastActivityDate="2017-04-08T22:26:43.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6972" PostTypeId="1" CreationDate="2013-11-18T05:36:36.187" Score="1" ViewCount="1021" Body="&lt;p&gt;I have been considering building a fluid dynamics demonstration.  Basically, it would be a thin plexiglass box, with some turbulence inducing elements attached to the back face, and a pump to move the fluid around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody recommend a pair of fluids that will:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;not mix/dissolve permanently&lt;/li&gt;&#xA;&lt;li&gt;can be dyed different colors, while respecting (1)&lt;/li&gt;&#xA;&lt;li&gt;won't react with plexiglas (or, alternatively, polycarbonate)&lt;/li&gt;&#xA;&lt;li&gt;isn't toxic or particularly flammable&lt;/li&gt;&#xA;&lt;li&gt;can be pumped with a &quot;very&quot; low power pump at temperatures suitable for humans (say, 60-100F)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The point of the demonstration is to create a &quot;sustainable&quot; chaotic system.  I might even point an HD camera at it, and extract random numbers from the video stream (maybe).&lt;/p&gt;&#xA;" OwnerUserId="3724" LastActivityDate="2013-12-18T08:35:04.303" Title="Imiscible fluids for demonstration" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6973" PostTypeId="2" ParentId="6972" CreationDate="2013-11-18T06:52:34.487" Score="1" Body="&lt;p&gt;You could use water and some low viscosity oil like olive oil or sunflower oil. Water can be dyed easily with food-coloring that you can buy at the local pharmacy or sometimes grocery store. Both substances are eatable and fairly inert (on the timescale of days at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a very small part of the oil will dissolve in the water and after saturation there is no problem any more. After thorough stirring the demixing process might be a bit slow since both oils have a density around 920 $\ce{kg/m^3}$ which is not that far from water, but you will have to try whether you can for example run the experiment by day and let the whole thing demix at night. If it doesn't properly demix you could consider going for a lower density oil. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-11-18T06:52:34.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6976" PostTypeId="1" CreationDate="2013-11-18T12:35:29.753" Score="13" ViewCount="65893" Body="&lt;p&gt;Both aldehydes and ketones have a carbonyl group, but since the carbonyl group of ketone is between alkyl groups, wouldn't it be more difficult for it to form hydrogen bonds than aldehydes? Why then, do aldehydes have a higher boiling points than ketones?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="7475" LastEditDate="2017-11-16T08:55:35.443" LastActivityDate="2017-11-16T08:55:35.443" Title="Why are the boiling points of aldehydes less than those of ketones?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;boiling-point&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6977" PostTypeId="1" AcceptedAnswerId="7001" CreationDate="2013-11-18T13:01:49.353" Score="18" ViewCount="40027" Body="&lt;p&gt;I have been going through reduction of aldehydes using $\ce{LiAlH4}$ and $\ce{NaBH4}$. If there is a double bond conjugated with the carbonyl group, $\ce{LiAlH4}$ doesn't reduce it, leading to an allylic alcohol. However, using $\ce{NaBH4}$, some of the fully reduced alcohol will also be formed. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="16683" LastEditDate="2017-12-10T22:59:19.767" LastActivityDate="2020-08-21T12:45:33.463" Title="Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not?" Tags="&lt;organic-chemistry&gt;&lt;regioselectivity&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6979" PostTypeId="1" AcceptedAnswerId="12235" CreationDate="2013-11-18T14:32:40.990" Score="6" ViewCount="310" Body="&lt;p&gt;It is well known that the usage of the Gaussian basis set, in contrast to Slater basis set, leads relative simple semi-analytical expressions for the two electron repulsion integral $(ab|cd)$. Could you propose any other basis sets allowing simple computations of two-electron integrals? This is quite general questions meaning that these functions should not fit atomic orbitals well (the problem is formulated for wider class of potentials rather than for the Coulomb one only).&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="4945" LastEditDate="2014-06-01T13:29:40.273" LastActivityDate="2014-06-02T05:38:27.333" Title="An alternative basis set for analytical computation of two electron integrals" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;basis-set&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6980" PostTypeId="1" AcceptedAnswerId="7103" CreationDate="2013-11-18T14:56:34.317" Score="1" ViewCount="1270" Body="&lt;p&gt;I have a mixture with 1 part of gas A, and 7 parts gas B. This ratio is in terms of mass.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Gas   Molar Mass (kg/mol)&#xA;A     0.17034&#xA;B     0.034&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find the specific gas constant of the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;R_{\text{sp}} = \frac{R}{M} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it might be&#xA;$$&#xA;M = 0.17034\frac{1}{8} + 0.034\frac{7}{8}&#xA;$$&#xA;but I can't work out if that makes sense given the units, or whether I should be using atomic mass or something.&lt;/p&gt;&#xA;" OwnerUserId="3728" LastActivityDate="2013-11-27T15:28:40.923" Title="Specific gas constant for two gases" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6982" PostTypeId="1" CreationDate="2013-11-18T16:04:35.247" Score="3" ViewCount="1870" Body="&lt;p&gt;I use filtered water to make my coffee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I mix a few drops of said coffee with our tap water, it turns a grayish color; it's much darker than the coffee would turn the water by itself (I compared by doing a side by side comparison and adding an equal amount of coffee to filtered water, resulting in a much lighter brown (not gray) color that is closer to what I'd expect from coffee). It's a very pronounced effect (i.e. easily visible when you rinse out a coffee mug that has just a few drops of coffee at the bottom). Because of the distinct color change and the fact that it's significantly darker than it should be from simple mixing, it's obviously a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What common tap water contaminants would react with coffee in this manner? Iron perhaps?&lt;/p&gt;&#xA;" OwnerUserId="3731" LastEditorUserId="275" LastEditDate="2013-11-18T16:45:41.710" LastActivityDate="2016-08-29T22:44:48.803" Title="Coffee and tap water reaction - results in gray tint" Tags="&lt;water&gt;&lt;home-experiment&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6983" PostTypeId="1" CreationDate="2013-11-18T17:31:13.723" Score="1" ViewCount="856" Body="&lt;p&gt;Can you split carbonate ions into their constituent elements &lt;strong&gt;directly&lt;/strong&gt; in the carbonate form? (so not first producing $\ce{CO2}$ and then decomposing the carbon dioxide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pQShX.png&quot; width=&quot;130&quot; height=&quot;100&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that when running electricity through an aqueous solution containing carbonate ions and other element ions, the carbonate ions will act as individual groups and will not split (only after chemical reactions (for example the thermolysis of metal carbonates will produce carbon dioxide)) So this was just out of interest - how can the carbonate ion itself be split into its constituent elements if it can't be split with electricity (or at high voltage, could it then be split?)?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="4231" LastEditDate="2015-03-05T20:13:31.120" LastActivityDate="2015-03-05T20:13:31.120" Title="Split carbonate ions?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6985" PostTypeId="1" CreationDate="2013-11-18T21:04:05.473" Score="4" ViewCount="2425" Body="&lt;p&gt;To extract essential oils from plant matter (amongst other purposes), steam distillation is often used. I've seen a lot of different glassware setups for this. I've sometimes seen them broken into two categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Internal steam distillation&lt;/li&gt;&#xA;&lt;li&gt;External steam distillation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's my (admittedly shaky) understanding of each:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Internal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With internal steam distillation, you have one boiling flask containing plant matter and liquid water. You simply boil the mixture and send the gas to to the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;External&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With external steam distillation, you have two flasks. One contains water only. You heat this one and route the steam into the second flask. The second flask contains a mixture of water and plant matter. The steam from the first flask combines with the liquid mixture in the second flask. Optionally, the second flask also has a heating mantle. The output of the second flask then goes to the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Differences?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the effect of using one or the other setup? External is substantially more complex, so I can't imagine it would be used if it didn't offer some advantages in some circumstances. When would you use external, and why?&lt;/p&gt;&#xA;" OwnerUserId="3733" LastEditorUserId="7475" LastEditDate="2015-10-09T15:47:04.433" LastActivityDate="2015-10-09T15:47:04.433" Title="Internal vs external steam distillation" Tags="&lt;experimental-chemistry&gt;&lt;distillation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6986" PostTypeId="2" ParentId="6116" CreationDate="2013-11-18T21:36:08.813" Score="4" Body="&lt;p&gt;Boron (B), Silicon (Si), Arsenic (As), Tellurium (Te), Germanium (Ge), and Antimony (Sb) are considered to be the semi-metals or more professionally known as metalloids. Metalloids are half-metallic, half-nonmetallic types of elements. The ones on the right side of the steps that I mentioned are more non-metallic than the ones on the left. The stairs also represent the seperation between metals and non-metals so you can see why the ones I name are adjacent to the highlighted line. They don't have a specific placement that's why the put them adjacently to the line and on one side or the other depending on which is more metallic or nonmetallic. Hope this helped you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I learned this today in Chemistry and I pay attention in class. If I'm wrong somebody can correct me but &lt;strong&gt;I'm 100% sure I'm right&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3734" LastEditorUserId="-1" LastEditDate="2017-08-01T23:45:30.700" LastActivityDate="2017-08-01T23:45:30.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6987" PostTypeId="1" AcceptedAnswerId="6988" CreationDate="2013-11-18T22:50:34.877" Score="1" ViewCount="589" Body="&lt;p&gt;My question is related to the edge between Subatomic Physics and Chemistry, but I decided that here is the best place to discuss it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we all know, Theoretical Physics is a well developed field of knowledge, able to describe the very tiny forms of interactions. However, after some researches for some kind of &lt;strong&gt;formula&lt;/strong&gt; to perfectly describe the interactions on the atomic level (not subatomic), more specifically, a way to calculate the strength of chemical bounds, I keep comming across frustrating highschool explanations, such as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ionic &gt; covalent &gt; H-bond &gt; dipole-dipole &gt; van der Waals &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most precise data that I was able to find comes at page number 20 and 21 of &lt;a href=&quot;http://www.smallscalechemistry.colostate.edu/PowerfulPictures/ChemicalBonding.pdf&quot; rel=&quot;nofollow&quot;&gt;this pdf&lt;/a&gt;, that apparently shows some experimental results (I'm guessing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is my question: is there really a unified formula? What if I want to simulate chemical interactions (proton-by-proton, electron-by-electron) on a computer, is it that complex to do, even assuming that I want a simple simulation of about 20 atoms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: When I reffer to calculating chemical bonds strength, I also include predicting many other chemical properties, as &lt;strong&gt;electron affinity&lt;/strong&gt; or &lt;strong&gt;reactivity of the elements&lt;/strong&gt;, which are all related. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I can accept as answer any pdf that clarifies my question, because I wasn't able to find. Maybe I'm looking it with the wrong key-words. &lt;/p&gt;&#xA;" OwnerUserId="3736" LastEditorUserId="3736" LastEditDate="2013-11-18T23:01:56.170" LastActivityDate="2014-05-01T18:23:44.820" Title="Unified theory of chemical bonds" Tags="&lt;bond&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6988" PostTypeId="2" ParentId="6987" CreationDate="2013-11-19T05:07:05.123" Score="4" Body="&lt;p&gt;The short answer: no. We do not know how to solve the Schrodinger equation analytically, even for systems as simple as the He atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The much longer answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Schrodinger equation is the basis for almost all theoretical chemistry: chemistry that attempts to predict properties of compounds without an actual experiment. The problem is that the Schrodinger equation is unsolvable for any compound with more than two electrons, and so we must make some approximations to solve it. This leads to methods such as Hartree-Fock and Density Functional Theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these approximations are usually pretty good, they are still only approximations, and sometimes stop working correctly. There is an added problem that the Schrodinger equation is only an approximation, and the proper theory to use to describe light-matter interaction is Quantum Electrodynamics, which is incredibly difficult to solve for a single photon reflecting off of a mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short: there may or may not be a &quot;unified theory&quot; for describing exactly how all bonds behave in every situation, but there are currently methods that are good enough to predict a large portion of the behavior that we observe.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2013-11-19T05:07:05.123" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6990" PostTypeId="2" ParentId="6292" CreationDate="2013-11-19T07:50:53.247" Score="0" Body="&lt;p&gt;Since molar conductivity is directly proportional to the number of ions ,reduction in concentration reduces the number of ions per unit volume hence reduction in conductivity.&lt;/p&gt;&#xA;" OwnerUserId="3741" LastActivityDate="2013-11-19T07:50:53.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6991" PostTypeId="2" ParentId="6976" CreationDate="2013-11-19T09:04:50.613" Score="0" Body="&lt;p&gt;For ketones and aldehydes of similar molecular mass, ketones have higher boiling point due to the fact that its carbonyl group is more polarized than in aldehydes. So, interactions between molecules of ketones is stronger than between molecules of aldehydes, and that gives a higher boiling point.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-19T09:04:50.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6992" PostTypeId="2" ParentId="6985" CreationDate="2013-11-19T10:36:07.307" Score="3" Body="&lt;p&gt;The main purpose is to distill the component (essential oil) with quite high boiling point. &#xA;You have to think of distillation of immiscible liquids, &lt;a href=&quot;http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html&lt;/a&gt;&#xA;where you can take advantage of the equillibria in liquid and gas phase separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short - two very beneficial things happen: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;boiling point of the mixture is lower than boiling point of water (and substantially lower than BP of the oil)&lt;/li&gt;&#xA;&lt;li&gt;the composition of vapor is independent of composition of liquid, i.e. you can distill out even the smallest fraction of oil (c.f. &lt;a href=&quot;http://en.wikipedia.org/wiki/Raoult%27s_law&quot; rel=&quot;nofollow&quot;&gt;Raoult's law&lt;/a&gt; for miscible liquids)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the partial pressure of the oil is small, let's say 5% that of the water, therefore you need to evaporate and condense 20x more of water than of oil. To achieve this, you can either put all the needed water in the flask and do the &quot;internal&quot; distillation, or add just a small amount of hot water and use steam from external source in &quot;external&quot; setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is simpler to just boil pure water (or even have a direct source of steam) and use it for distillation, than to take care of the whole process in one flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, I would see the &quot;external&quot; setup simpler and more robust, even though it needs little bit more glassware.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditDate="2013-11-22T09:20:16.467" LastActivityDate="2013-11-22T09:20:16.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6993" PostTypeId="2" ParentId="6292" CreationDate="2013-11-19T12:13:52.713" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Solution_conductivity#Theory&quot; rel=&quot;noreferrer&quot;&gt;molar conductivity&lt;/a&gt; of a strong electrolyte $(\Lambda_m)$ is given as the ratio of measured conductivity $(\kappa)$ to the molar concentration $(c)$:&#xA;$$\Lambda_m=\frac{\kappa}{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you note, the value of $\Lambda_m$ is not independent of concentration. The reason for this is that $\kappa$ does not scale linearly with concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Lambda_m = \Lambda_m^\circ -K\sqrt{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation, $\Lambda_m^\circ$ is the limiting or intrinsic conductivity of the electrolyte and $K$ is an empirical constant. The interpretation of $\Lambda_m^\circ$ is the molar conductivity at infinite dilution as $c\rightarrow0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{c\rightarrow 0}{\Lambda_m}=\Lambda_m^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinite dilution scenario is different from the scenario when there were no ions in solution to begin with, just as mathematically $\lim_{x \rightarrow 0}{f(x)}$ does not always equal $f(0)$ (often because $f(0)$ is undefined).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limiting case of $c\rightarrow 0$, the ion concentration never reaches zero, but it does keep getting close. You can never dilute a solution containing a solute to zero concentration (i.e. pure solvent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you graph $\Lambda_m$ as a function of $\sqrt{c}$, you should get a straight line with a non-zero intercept $(\Lambda_m^\circ)$. This value is an empirical parameter used to model data. It can also tell you something about the intrinsic conductivity of ions (see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Solution_conductivity#Strong_electrolytes&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;). However, as you say, at exactly $c=0$, he conductivity should be negligible. At any other value of $c$, however the conductivity should follow from the equation given. Thus, the conductivity is not necessary linear in the infinitesimal range around $c=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Lambda_m=\begin{cases}&#xA;10^{-8}\approx0, &amp;amp;c=0\\&#xA;\Lambda_m^\circ - K\sqrt{c}, &amp;amp;c&amp;gt;0 \end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-19T12:13:52.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6994" PostTypeId="1" AcceptedAnswerId="6998" CreationDate="2013-11-19T13:34:44.040" Score="3" ViewCount="121" Body="&lt;p&gt;Are there any two distinct types of atom or molecule that have identical emission or absorption spectra?&lt;/p&gt;&#xA;" OwnerUserId="3742" LastActivityDate="2013-11-19T18:17:12.617" Title="Are there any molecules with the same spectrum?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6995" PostTypeId="1" AcceptedAnswerId="7010" CreationDate="2013-11-19T16:23:40.507" Score="3" ViewCount="145" Body="&lt;p&gt;How did we derive the relation between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isoelectric_point&quot; rel=&quot;nofollow&quot;&gt;isoelectric pH&lt;/a&gt; and the dissociation constants for the amino and carboxyl groups on zwitterionic amino acids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pI=\frac {pK_{a1}+pK_{a2}}{2}$$&#xA;where $$\ce{NH3^+-R2(COO^{-})&amp;lt;=&amp;gt;NH3^+-R2(COOH)-K_{a1}}$$&#xA;$$\ce{NH3^+-R2(COO^{-})&amp;lt;=&amp;gt;NH2-R2(COO^{-})-K_{a2}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complicated cases (charged side groups) and their isoelectric pH, &lt;a href=&quot;http://www.mhhe.com/physsci/chemistry/carey5e/Ch27/ch27-1-4.html&quot; rel=&quot;nofollow&quot;&gt;here is one site&lt;/a&gt; which just posts the equations without any derivation. Besides this, there is no reference I could find for this derivation.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-11-20T11:34:00.493" LastActivityDate="2013-11-20T14:04:08.047" Title="Isoelectric pH of colloids?" Tags="&lt;biochemistry&gt;&lt;equilibrium&gt;&lt;ions&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6996" PostTypeId="1" CreationDate="2013-11-19T17:08:43.093" Score="2" ViewCount="265" Body="&lt;p&gt;I have a question about electrolysis. How do we know what will be on anode and cathode if we have for example NaCl diluted in water. How do we know if it's gonna be Na+ or H+??&lt;/p&gt;&#xA;" OwnerUserId="3717" LastActivityDate="2013-11-20T16:06:47.303" Title="electrolysis - selection of what is going to be on anode and cathode" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6997" PostTypeId="1" AcceptedAnswerId="7000" CreationDate="2013-11-19T17:22:38.437" Score="5" ViewCount="43149" Body="&lt;p&gt;Is a pH value outside 0 - 14 possible?&#xA;I asked my teacher who said: yes, it is, but very difficult to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on the internet, I found multiple answers, one saying it is but because of a fault in the pH glass meter we can't be sure, the other said that it is not possible because of the intrinsic value of water to ionize.&#xA;As you can see, I'm getting mixed answers from different sources and I'm really confused now. Could somebody here give a concise answer whether or not it is possible? A deep argument will be valuable.&lt;/p&gt;&#xA;" OwnerUserId="3743" LastEditorUserId="402" LastEditDate="2017-03-17T16:44:11.080" LastActivityDate="2018-04-17T12:48:51.543" Title="pH range outside conventional 0-14" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ClosedDate="2013-11-19T21:05:38.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="6998" PostTypeId="2" ParentId="6994" CreationDate="2013-11-19T18:17:12.617" Score="4" Body="&lt;p&gt;In addition to the R/S chiral molecules mentioned by @user26143. There are also molecules with a distinct atomic composition that have experimentally indistinguishable spectra, even though theoretically there will be minor differences. One example, according to &lt;a href=&quot;http://link.aps.org/doi/10.1103/PhysRevLett.102.253001&quot; rel=&quot;nofollow&quot;&gt;Rabitz et al. 2009 PRL&lt;/a&gt; (freely available for viewing) are &lt;a href=&quot;http://en.wikipedia.org/wiki/Flavin_mononucleotide&quot; rel=&quot;nofollow&quot;&gt;flavin mononucleotide&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Riboflavin&quot; rel=&quot;nofollow&quot;&gt;riboflavine&lt;/a&gt; and in fact many other flavins. Due to there very similar structure you need unconventional techniques to be able to distinguish them.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-11-19T18:17:12.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6999" PostTypeId="2" ParentId="6997" CreationDate="2013-11-19T18:29:23.180" Score="0" Body="&lt;p&gt;From its definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH = -log [H+]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If [H+] is greater than 1, the pH will be negative. For example, HCl at 40% minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pH greater than 14 are necessary highly concentrated basic solutions.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-19T18:29:23.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7000" PostTypeId="2" ParentId="6997" CreationDate="2013-11-19T19:56:11.627" Score="15" Body="&lt;p&gt;Let me add a bit more to the answers already given. As has been said, $\mathrm{pH}$ is nothing but a measure of the activity of protons ($\ce{H+}$) in a solvent - $\displaystyle\mathrm{pH} = -\log_{10} \ce{\ a_{H_{\text{solvated}}^{+}}}$. In dilute solutions, a solute's activity is approximately equal to its concentration, and so you can get away with saying $\mathrm{pH} = -\log_{10} \ce{[H_{solvated}^{+}]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;normal&quot; aqueous $\mathrm{pH}$ scale that goes from 0-14 delimits the region where both $\ce{[H^{+}]}$ and $\ce{[OH^{-}]}$ are lower than $\pu{1 mol L^{-1}}$, which is about the upper limit where concentrations and activities of solutes in solution are approximately the same and they can be used interchangeably without introducing too much error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The reason this is true is because a solute in a very dilute solution behaves as if it surrounded by an infinite amount of solvent; there are so few solute species that they interact very little with each other on average. This view breaks down at high concentrations because now the solute species aren't far enough apart on average, and so the intermolecular interactions aren't quite the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only is it rarer to talk about pH for very highly concentrated solutions, because activity has a much more subtle definition than concentration, but also because you have to produce quite concentrated solutions to get unusual values of pH. For example, &lt;strong&gt;very approximately&lt;/strong&gt;, to get a litre of an aqueous solution of $\mathrm{pH} = -1$ using hydrogen perchlorate ($\ce{HClO4}$, a very strong monoprotic acid) you would need to dissolve around $\pu{1 kg}$ of $\ce{HClO4}$ in a few hundred $\pu{mL}$ of water. A litre of $\mathrm{pH} = 15$ aqueous solution using potassium hydroxide ($\ce{KOH}$) would also require several hundred grams of the hydroxide in about as much water. It's so much acid/base in a small amount of solvent that many acids/bases can't even dissolve well enough to reach the required concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another way to come across unusual values of pH. We almost always talk about acids in bases in water, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid-base_theory#Solvent_system_definition&quot; rel=&quot;noreferrer&quot;&gt;they also exist in other media&lt;/a&gt;. Notice that in the definition of pH, no direct reference is made to water. It just happens that, for water, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O (l) &amp;lt;=&amp;gt; H^+ (aq) + OH^{-} (aq)} \\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{w}=a_{\ce{H^+(aq)}} \times a_{\ce{OH^- (aq)} } \simeq \ce{[H^{+}]} \times \ce{[OH^{-}]}&#xA; = 1.01\times 10^{-14}$$&#xA;$$\mathrm{pH}+\mathrm{pOH}=-\log\ k_\mathrm{w} \simeq 14 \ (25^\circ C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for liquid ammonia, one has: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH3{(l)} &amp;lt;=&amp;gt; H^+{(am)} + NH2^{-}{(am)}}&#xA;\\&#xA;K^{\pu{-50^\circ C}}_{\text{autodissociation}}=k^{\pu{-50^\circ C}}_\mathrm{am}=a_{\ce{H^+(am)}} \times a_{\ce{NH2^- (am)} } \simeq \ce{[H^{+}]} \times \ce{[NH2^{-}]} = 10^{-33}$$&#xA;$$\mathrm{pH}+\mathrm{p}\ce{NH2}=-\log\ k_{\mathrm{am}} \simeq 33 \ (-\pu{50^\circ C})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\ce{(am)}$ stands for a substance solvated by ammonia). Hence, in liquid ammonia at $-50^\circ C$, the pH scale can easily go all the way from 0 to 33 (of course, it can go a little lower and higher still, but again now activities become important), and that neutral pH is actually 16.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For pure liquid hydrogen sulfate (some difficulties arise as $\ce{H2SO4}$ is a diprotic acid and tends to decompose itself when pure, but putting that aside and looking only at the first dissociation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4{(l)} &amp;lt;=&amp;gt; H^+{(hs)} + HSO4^{-}{(hs)}}\\ &#xA;K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{hs}=a_{\ce{H^+(hs)}} \times a_{\ce{HSO4^- (hs)} } \simeq \ce{[H^{+}]} \times \ce{[HSO4-]} \simeq 10^{-3}$$&#xA;$$\mathrm{pH}+\mathrm{p}\ce{HSO4}=-\log\ k_{\mathrm{hs}} \simeq 3 \ (\pu{25^\circ C})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\ce{(hs)}$ stands for a substance solvated by hydrogen sulfate). Thus, the pH in liquid hydrogen sulfate has a much smaller range than in water and ammonia, not going much further than interval from 0 to 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also that not only does the range of $\mathrm{pH}$s change in each solvent, but there is no direct relationship between the values of pH between different solvents; formic acid in ammonia would behave as a strong acid and as such a $\pu{1 M}$ solution in ammonia would have a pH close to 0, but formic acid is actually a base in liquid hydrogen sulfate, and as such a $\pu{1 M}$ solution would have a pH above 1.5. An approximate comparison between the pH ranges and their relative positions in different solvents can be found in the figure below, from &lt;a href=&quot;https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201000252&quot; rel=&quot;noreferrer&quot;&gt;A Unified pH Scale for All Phases&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QJxfr.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2018-04-17T12:48:51.543" LastActivityDate="2018-04-17T12:48:51.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7001" PostTypeId="2" ParentId="6977" CreationDate="2013-11-19T20:30:58.710" Score="21" Body="&lt;p&gt;Neither &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH_4}$&lt;/span&gt; nor &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4}$&lt;/span&gt; are able to reduce an isolated &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;HSAB Principle&lt;/a&gt; the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.&lt;/p&gt;&#xA;&lt;p&gt;Now, &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4}$&lt;/span&gt; is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4}$&lt;/span&gt; in a second step.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; in a second step.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="41328" LastEditDate="2020-08-21T12:45:33.463" LastActivityDate="2020-08-21T12:45:33.463" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="7002" PostTypeId="2" ParentId="6997" CreationDate="2013-11-19T21:09:11.760" Score="2" Body="&lt;p&gt;In short, it is possible to have pH outside of this range.  However, in such cases, pH fails to be a useful or even meaningful measure. In concentrated solutions, one should use concentration directly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical point of view, pH helps to bring extremely small H+ concentrations to a more familiar range of numbers, but as the concentrations become larger this practice loses its value and becomes counterproductive instead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a theoretical point of view, pH as an acidity measure is based on the assumption that water does not have a varying concentration.  However in concentrated solutions, because of the lowering concentration of H2O itself you may find H+ concentration decrease as more acid is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further complicating this, in water H+ is a simplified notation for H3O+ as H+ by itself actually ever exist in any solution, in concentrated solutions, it becomes ambiguous what H+ may refer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the notation &quot;98%(w/w) sulfuric acid&quot; would be much more practically useful than &quot;a sulfuric acid with a pH of -2.25&quot;.  Not to mention there is no such thing as a &quot;H+&quot; concentration in that situation because it is not clear which one of the many species (H3O+, H3SO4+, HSO3+,...) would be referred to as H+, and anyway you choose that H+ concentration will most likely turn out to be even lower than a 50% sulfuric acid solution.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="402" LastEditDate="2017-04-08T22:22:23.020" LastActivityDate="2017-04-08T22:22:23.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7003" PostTypeId="1" AcceptedAnswerId="7009" CreationDate="2013-11-19T21:56:44.467" Score="4" ViewCount="13527" Body="&lt;p&gt;I'm going to be aliquoting out ethanol-stabilized chloroform into smaller containers. It's presently in a large bottle sealed with a rubber(y) sheet so I can only draw it out with a needle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How do I seal them?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I'm aliquoting to minimize exposure of the full bottle to air, can I get rid of the rubber sheet?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any restrictions on the sort of cap I can use on the aliquot bottle?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2341" LastEditorUserId="7448" LastEditDate="2015-04-24T15:44:34.483" LastActivityDate="2015-08-31T08:04:03.333" Title="How to store chloroform?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7004" PostTypeId="2" ParentId="6954" CreationDate="2013-11-19T22:21:12.220" Score="14" Body="&lt;p&gt;This is actually just one part of a much more general phenomena, which is the release of large, polycyclic aromatic compounds due to pyrolysis or combustion of organic matter. It is not specific to wood. So, to understand why these compounds form in a wood fire, or any other kind of fire, we must first understand what wood is, and then we must understand how wood burns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wood is biomass - that is, matter composed primarily of carbons, hydrogens, and oxygens. We can generalize here, and talk about any biomass - wood, dead or dry grass, leaves, animal carcasses, fibers, sugar, fat, and even things like coal (which is fossilized biomass with special properties). All of these things contain cellulose or similar molecules, which consists of long chains of aromatic rings (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot;&gt;wikipedia article on cellulose&lt;/a&gt; is a good place to start, and any textbook on biochemistry will give much more detail). In the context of combustion, this is called &quot;volatile organic matter.&quot; Volatile just means that if you heat it up, it burns, gives off heat, and turns into gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Living things also have non-volatile and non-organic matter. As an example, people require some trace amounts of metals for biochemical functions, so humans contain some percentage of mineral matter. In the context of combustion, this material is often called &quot;ash.&quot; If you have a fuel like wood, this ash is what's left behind after the wood burns, and composes much of what's released into the air in the form of smoke. Some fuels can have large percentages of ash (as an example, when organic matter fossilizes and forms coal, a large amount of rock or dirt interspersed with the organic material can lead to coals with very high ash content).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you heat up organic matter, what you're doing is dumping a large amount of thermal energy into the molecules composing that matter, and after a certain point (depends on the fuel, but we're talking about 400 or 500 C) the large chains that compose organic matter begin to break apart. If this happens in the presence of oxygen, the material will also begin to oxidize, and the process is called combustion. If it happens in the absence of oxygen, the material will gasify. As the temperature gets hotter and hotter, a soup of big organic macromolecules is formed. (When I say big, I mean, BIG - hundreds of carbon, hydrogen, and oxygen atoms, and thousands of side chains and functional groups). It is in this macromolecular soup that you can find benzene and its many, many, many cousins. (These molecules are too complex to have names, and are so large that they can't be characterized exactly - instead, chemists can only statistically describe the molecules, using techniques like nuclear magnetic resonance (NMR), mass spectrometers (MBMS), or infrared spectrometers (FTIR).) This process of solid matter turning into volatile gases is called pyrolysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big polycyclic molecules form when there is not enough oxygen (if there were enough oxygen, all the carbon and hydrogen would form carbon dioxide and water). They are huge and hydrophobic, and agglomerate to form soot and tar. When you see sticky brown films on cigarettes or smokers' hands or in pipes, this is caused by these macromolecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you looked very closely at a wood fire, and were able to see the actual flame-wood interface, what you would see is jets of hot, flammable gases shooting out of the pores of the wood. It is these gases that ignite - not the wood itself. (Of course, as the wood heats up, and the organic matter is burned, the mineral matter is left behind, and that becomes extremely hot but does not vaporize, so you still have radiant, red-hot material.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can still find benzene, even in a hot fire, because flames are not perfectly mixed, and you can get pockets of cold gas interspersed in a cloud of hot gas. If it is cold enough, benzene (which can be very stable) will not be decomposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the next time you watch a campfire, or smoke a cigarette, or set your dinner on fire in the oven, you should be able to better understand what you're seeing - or know where to look.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2013-11-19T22:21:12.220" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7005" PostTypeId="2" ParentId="5388" CreationDate="2013-11-20T00:03:20.623" Score="3" Body="&lt;p&gt;Hydroxyl radicals are strong proton acceptors, due to an oxygen with additional electrons, and will therefore &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom_abstraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;abstract hydrogens&lt;/a&gt; from anything that has hydrogens, e.g., silane or methane. This attack can be enhanced, depending on the electronic properties of the atom sitting at the center (silicon or carbon). The more stable the atom, the more stable the 3-hydrogen radical will be, the more thermodynamically favorable the hydrogen abstraction reaction will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of silane combustion, this leads to the formation of $\ce{^.SiH3}$ radicals, which react with molecular oxygen ($\ce{O2}$) in a chain-branching step. Depending on whether the temperature is low or high, this step can occur along various different pathways. It generally occurs following the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^.SiH3 + O2 -&amp;gt; Si(O)OH + H + H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the intermediates change depending on temperature and the energetics of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information in a couple of papers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Babushok, Tsang, Burgess Jr., Zachariah, &quot;Numerical study of low- and high-temperature silane combustion.&quot; Twenty-Seventh International Symposium on Combustion (1998) pp. 2431-2439.&lt;/li&gt;&#xA;&lt;li&gt;Kondo et al., &quot;Ab Initio Energetic Calculations of Elementary Reactions Relevant to Low-Temperature Silane Oxidation by Gaussian-2 Theory.&quot; Journal of Physical Chemistry A (1997) pp. 6015–6022&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="41328" LastEditDate="2017-08-15T22:06:57.603" LastActivityDate="2017-08-15T22:06:57.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7006" PostTypeId="2" ParentId="4667" CreationDate="2013-11-20T01:13:59.017" Score="1" Body="&lt;p&gt;Transition elements have partially filled d orbitals. We also know that when electrons jump from one orbital to another light is emitted due to which the compounds of transition elements seem to be colored compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This can also happen in some organic compounds though in this case, it is p orbitals and not d orbitals.&lt;/p&gt;&#xA;" OwnerUserId="3744" LastEditorUserId="402" LastEditDate="2017-04-02T20:38:29.527" LastActivityDate="2017-04-02T20:38:29.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7007" PostTypeId="1" AcceptedAnswerId="7012" CreationDate="2013-11-20T06:17:17.267" Score="3" ViewCount="1043" Body="&lt;p&gt;If there is more overlap of the same sign wave function, this apparently leads to lower energy. I don't understand why. My professor said that there would be fewer nodes, and fewer nodes means lower energy. However, why would overlap in general cause the energy to decrease?&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="3747" LastEditDate="2019-02-28T16:49:00.813" LastActivityDate="2019-02-28T16:49:00.813" Title="How does molecular orbital overlap affect energy?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7009" PostTypeId="2" ParentId="7003" CreationDate="2013-11-20T10:23:01.567" Score="6" Body="&lt;p&gt;A septum sealed chloroform bottle is likely to have been packed under nitrogen to increase shelf life. Unless you are doing this under a dry nitrogen atmosphere, I'm not sure that decanting into smaller containers is likely to improve shelf life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloroform needs to be stored in a cool, dry, well-ventilated place, out of direct light. Your choice of storage container is likely to be influenced by how long you intend to store it for, but an amber glass bottle with a ptfe-lined cap is probably best suited for longer storage. Rigid polyethylene and polypropylene do not have the best chemical resistance for long-term storage, but if they are air-tight containers will be probably be ok only for short term storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure to read and understand the MSDS for chloroform fully before doing this. You will find storage recommendations and incompatibilities listed. &lt;strong&gt;DO NOT&lt;/strong&gt; use containers or lids made from or containing aluminium, magnesium and other metals.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-11-20T10:23:01.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7010" PostTypeId="2" ParentId="6995" CreationDate="2013-11-20T14:04:08.047" Score="1" Body="&lt;p&gt;Consider the equilibria in dilute aqueous solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2A+ \rightleftharpoons HA + H+} \ \ \ \ \ \ \ K_{a_1}=\frac{[HA][H^+]}{[H_2A^+]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA \rightleftharpoons A^- + H+} \ \ \ \ \ \ \ K_{a_2}=\frac{[A^-][H^+]}{[HA]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the isoelectric point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[H_2A^+]=[A^-]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the expressions for $K_{a_1}$ and $K_{a_2}$, we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{{[HA]}[H^+]}{K_{a_1}}=\frac{K_{a_2}[HA]}{[H^+]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[H^+]^2 = K_{a_1}K_{a_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pH_{isoelectric}=\frac{pK_{a_1}+pK_{a_2}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_dissociation_constant#Isoelectric_point&quot; rel=&quot;nofollow&quot;&gt;this section&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-20T14:04:08.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7011" PostTypeId="2" ParentId="6996" CreationDate="2013-11-20T16:06:47.303" Score="3" Body="&lt;p&gt;It depends on its oxidation/reduction potentials. For example, in the electrolysis of a NaCl solution (in water), the reduction potential of water to H2 is -0.83 V while the reduction potential for Na+ to Na is -2.71 V. As the potential for water is greater, water will be reduced instead Na+.&#xA;Another example, a solution of CuSO4 in water, the potential reduction for Cu2+ to Cu is +0.34 V, greater than the -0.83 V for water, so the Cu2+ will be reduced.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-20T16:06:47.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7012" PostTypeId="2" ParentId="7007" CreationDate="2013-11-20T17:10:33.127" Score="4" Body="&lt;p&gt;Although the nature of orbital and wave functions can be puzzling, by the end of the day chemical bonds are hold together by the familiar electrostatic forces.  Although the electrons don't have a well-defined position, the forces felt by the nuclei is dictated by the spacial probability distribution of the electron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To form a covalent bond, electron density has to accumulate between the two atoms.   The energy is lowered because the electron is close to both nuclei.   In a chemical bond such electron-nuclear attraction cancels out the repulsion between nuclei and holds the atoms together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbital overlap changes electron distributions.   Orbital with different signs cancel each other in and lower the relative probability electron appear between the bonding atoms.  In the most extreme case of nodal plane, electron density is completely annihilated.  Because the total number of electron is unchanged during interaction, the lowering of electron density between nuclei is accompanied by an increase of density to the far side.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when you have orbital with canceling phases interacting, electron density will move away from the bonding area, resulting in higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a strengthening interaction increase the electron density between nuclei, thus lowering the energy and creating a pulling force between nuclei.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-11-20T17:10:33.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7015" PostTypeId="1" CreationDate="2013-11-21T01:55:56.937" Score="4" ViewCount="5741" Body="&lt;p&gt;I'm having difficulty understand the relative stabilities of the compounds&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2,4-dimethyl-3-methylidenepentane &amp;lt; (3⁠&lt;em&gt;Z&lt;/em&gt;)-2,5-dimethylhex-3-ene &amp;lt; (3⁠&lt;em&gt;E&lt;/em&gt;)-2,5-dimethylhex-3-ene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which, for simplicity’s sake, can be described as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1,1-diisopropylethene &amp;lt; (&lt;em&gt;Z&lt;/em&gt;)-1,2-diisopropylethene &amp;lt; (&lt;em&gt;E&lt;/em&gt;)-1,2-diisopropylethene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The compounds are listed in order of increasing stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what does the $\ce{C=C-C}$ bond angle have do to with stability? &lt;/p&gt;&#xA;" OwnerUserId="360" LastEditorDisplayName="user7951" LastEditDate="2016-11-25T14:17:01.677" LastActivityDate="2016-11-25T14:17:01.677" Title="Stability of branched alkenes" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7016" PostTypeId="1" CreationDate="2013-11-21T04:58:52.850" Score="2" ViewCount="729" Body="&lt;p&gt;I've been researching sodium polyacrylate, which normally comes as those &quot;aqua gems&quot; for plants, or fake snow. Specifically, I've been hoping to find information on molding, casting, or otherwise giving shape to the material - nothing terribly complex, just things like simple stars for example (while hoping no one mistakes them for Jell-o jigglers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I've only been able to determine that sodium polyacrylate is &lt;em&gt;not&lt;/em&gt; a thermoplastic, so the obvious solution won't work. I heard from one source that those toys that grow in water may be made of the same stuff combined with another polymer, but couldn't get any further leads on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, what solutions are available for giving sodium polyacrylate a defined shape, while keeping most of its absorption/growth properties?&lt;/p&gt;&#xA;" OwnerUserId="3754" LastActivityDate="2016-03-16T22:12:46.240" Title="Molding/shaping sodium polyacrylate?" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7017" PostTypeId="2" ParentId="7015" CreationDate="2013-11-21T05:03:57.560" Score="6" Body="&lt;p&gt;The &lt;em&gt;Z&lt;/em&gt; isomer is destabilized by comparison to the &lt;em&gt;E&lt;/em&gt; isomer due to steric strain. This is a consequence of the isopropyl groups being on the same side of the double bond in the &lt;em&gt;Z&lt;/em&gt; isomer, resulting in electron-electron repulsion between the groups. This also, at least partially, accounts for the increase in the bond angle that should be observed in the &lt;em&gt;Z&lt;/em&gt; isomer: increasing that angle also increases the distance between the isopropyl groups, minimizing the mutual repulsion between them (admittedly, this is at the expense of the other angles, but the bulkiness of the isopropyls by comparison to hydrogens necessitates that distortion from idealized geometry). A similar effect occurs with the geminal isopropyl groups in the 1,1- isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the degree of substitution of an alkene has a dramatic effect on stability. The presence of alkyl groups on the carbons of an alkene is generally stabilizing (as can be empirically confirmed by examining thermodynamic data, i.e., heats of hydrogenation and heats of combustion). Evidently, calculations grounded in molecular orbital theory indicate that the mixing of $\sigma$ type molecular orbitals of alkanes with the $\pi$ orbitals of less-substituted alkenes is typically energy-lowering overall, and that the $\pi^{*}$ antibonding orbitals are sufficiently close in energy to the $\sigma$ orbitals of alkanes that hyperconjugative stabilization occurs (albeit relatively small in magnitude, since the energies are not very close). In the 1,1- isomer, only one carbon of the alkene can be stabilized by hyperconjugation, while both carbons should be able to benefit from it in the 1,2- isomers. (Speculatively, I would also think that the extent of orbital overlap should be poorer in the 1,1- isomer, since the geminal isopropyl groups would likely be staggered with respect to one another, which might prevent proper alignment of the alkyl group orbitals with those of the alkene $\pi$ system.)&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2013-11-21T05:31:19.687" LastActivityDate="2013-11-21T05:31:19.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7018" PostTypeId="1" CreationDate="2013-11-21T05:53:01.650" Score="1" ViewCount="1228" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An egg turning hard when boiled.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A spoon full of batter cooking on a hot griddle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I said that they both cannot be reversed, because I think if you throw both the egg and batter into the freezer, it will freeze, but it will not be in its original form before you cooked both of them. Then again, I guess I could try this for real on my own but I don't know if that would be a good idea...&lt;/p&gt;&#xA;" OwnerUserId="3756" LastActivityDate="2014-03-04T16:56:22.993" Title="Can these changes be reversed by changing temperature?" Tags="&lt;phase&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7019" PostTypeId="2" ParentId="7018" CreationDate="2013-11-21T06:42:28.777" Score="-1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When you boil an egg, you denature its proteins, changing their structure. It can't be reversed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once the butter is melted by heat, you can solidify back by freezing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="1431" LastEditDate="2013-11-21T12:50:37.467" LastActivityDate="2013-11-21T12:50:37.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7020" PostTypeId="1" AcceptedAnswerId="7821" CreationDate="2013-11-21T10:33:54.133" Score="0" ViewCount="41" Body="&lt;p&gt;What are the crystallisation conditions for Equine Myoglobin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried repeating some from the PDB but no success.&lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2014-01-14T11:42:42.783" Title="What conditions to use to crystallise Equine Myoglobin?" Tags="&lt;crystal-structure&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7022" PostTypeId="1" CreationDate="2013-11-21T14:40:21.867" Score="1" ViewCount="2060" Body="&lt;p&gt;On the periodic table, period 2 and 3, 4 and 5, 6 and 7 and so on have similar blocks, identical length and groups. What are such pairs of analogous periods called?&lt;/p&gt;&#xA;" OwnerUserId="3760" LastActivityDate="2014-03-01T07:12:44.170" Title="Pair of analogous periods on the periodic table" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7023" PostTypeId="1" AcceptedAnswerId="7025" CreationDate="2013-11-21T14:46:29.167" Score="0" ViewCount="38" Body="&lt;p&gt;Hi im searching for atmospheric data on the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Air quality, Dust Particle Concentration, Pollution, Average Temperature etc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;for various cities such as Frankfurt, London, Hong Kong, Bangkok...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any websites that can give me this data (or similar atmospheric measurements) for each city?&lt;/p&gt;&#xA;" OwnerUserId="3761" LastEditorUserId="865" LastEditDate="2015-10-16T17:42:14.940" LastActivityDate="2015-10-16T17:42:14.940" Title="Where to find Atmospheric data for various cities" Tags="&lt;organic-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-10-16T22:02:35.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="7025" PostTypeId="2" ParentId="7023" CreationDate="2013-11-21T16:12:17.117" Score="2" Body="&lt;p&gt;Most cities used to have own pages about pollution data. But there are other sources. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London: &lt;a href=&quot;http://www.londonair.org.uk/LondonAir/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.londonair.org.uk/LondonAir/Default.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bangkok: &lt;a href=&quot;http://www.pcd.go.th/airquality/bangkok/&quot; rel=&quot;nofollow&quot;&gt;http://www.pcd.go.th/airquality/bangkok/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a single page, try this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aqicn.org/city/all/&quot; rel=&quot;nofollow&quot;&gt;http://aqicn.org/city/all/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-21T16:12:17.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7026" PostTypeId="1" CreationDate="2013-11-21T17:22:23.423" Score="5" ViewCount="524" Body="&lt;p&gt;How do I keep hydrogen or oxygen in a test tube when pulling them out of water? I have a Styrofoam cup with two metal thumbtacks pushed though the bottom, and about 100 ml of distilled water mixed with 6 grams of sodium hydroxide in the cup. The cup is sitting on a 9 volt battery which separates the hydrogen and oxygen in the cup through the thumbtacks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put a test tube over the thumbtack that is producing oxygen and one over the thumbtack producing hydrogen. I'm trying to capture the hydrogen and oxygen in the test tubes and light them on fire. My problem is that when I pull out the test tube, the little water left in the tube pulls out all the gas when it falls out. How do I keep that from happening? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=HQ9Fhd7P_HA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a video of what I'm trying to do&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3762" LastEditorUserId="39539" LastEditDate="2017-03-29T06:00:29.430" LastActivityDate="2017-07-27T08:50:17.383" Title="Electrolysis Lab" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7027" PostTypeId="1" CreationDate="2013-11-21T22:54:51.423" Score="1" ViewCount="1176" Body="&lt;p&gt;I was wondering if anyone had any experience with PFBS or PFOS (perflurooctane sulfonic acid)?&#xA;I am thinking about trying to purchase some of these fluorosurfactants in bulk in order to (hopefully) use as a solvent.  However, I have read that PFOS is now banned due to the bad effects on people's health and the environment.  However, they claim that PFBS is a little better (half like of ~1 month in humans compared to ~5.5 years).&lt;br&gt;&#xA;So, I was wondering how terrible it would be to be dealing with the pure liquid on a daily basis, with the use of some gloves and a hood. The hood isn't great in terms of suction, but it's at least something. I should also add that the hood is of course open to the air, so there's moisture in the air. &#xA;I do deal with chlorosulfonic acid in the hood pretty regularly, and it isn't a huge problem, although I'm not aware of HCl getting trapped in your system for extended periods of time either.&#xA;I also hope to be contacting these liquids with chlorosulfonic acid, where I hope that no reactions will occur, but I am not so great at chemistry so I don't know if that is a fair assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also noticed that they state in the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfluorooctanesulfonic_acid&quot; rel=&quot;nofollow&quot;&gt;PFOS&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfluorobutanesulfonic_acid&quot; rel=&quot;nofollow&quot;&gt;PFBS&lt;/a&gt; that these substances are used in some consumer items.  So, if they are used in consumer items without degrading, how dilute do they have to be to not have strong acute or mild long term health effects?&#xA;I figure they would have to be pretty bad in order for 3M to ban PFOS, but I don't know if that was mainly in order to stop the mass use of them in consumer items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=US&amp;amp;language=en&amp;amp;productNumber=562629&amp;amp;brand=ALDRICH&amp;amp;PageToGoToURL=http%3A%2F%2Fwww.sigmaaldrich.com%2Fcatalog%2Fproduct%2Faldrich%2F562629%3Flang%3Den&quot; rel=&quot;nofollow&quot;&gt;The MSDS for PFBS states that the LD50 for mice is 430 mg/kg, and it shouldn't be contacted with any moisture in air at all&lt;/a&gt;  While the &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/571024?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;Chlorosulfonic acid MSDS states some similar, but more tame, statements about moisture, and it's LD50 in mice is 50mg/kg.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I guess my main question is, has anyone used this substance in it's pure form in a normal hood type of environment?&lt;/p&gt;&#xA;" OwnerUserId="1265" LastActivityDate="2013-11-21T22:54:51.423" Title="Perfluorobutanesulfonic acid (PFBS) Safety and Toxicity" Tags="&lt;acid-base&gt;&lt;safety&gt;&lt;toxicity&gt;&lt;surfactants&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7028" PostTypeId="2" ParentId="6969" CreationDate="2013-11-21T23:00:42.573" Score="4" Body="&lt;p&gt;Current Li-ion cells have $\ce{LiCoO4}$ or $\ce{LiFePO4}$ cathodes where the $\ce{Co}$ or $\ce{Fe}$ are only oxidised or reduced by 1 electron per atom, so if they have managed to make a cell with iron giving up $\frac{8}{3}$ electrons per atom they will get a higher capacity. I find it hard to believe they have got this reaction to happen reversibly and at high current density (i.e. high power/weight ratio) at room temperature, though - if you put iron in water at room temperature it doesn't react very quickly, and hydrogen-oxygen fuel cells also can't get high power densities at moderate temperatures.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2013-11-21T23:00:42.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7029" PostTypeId="1" AcceptedAnswerId="7031" CreationDate="2013-11-22T05:42:57.070" Score="3" ViewCount="3846" Body="&lt;p&gt;It seems that all wave functions studied in physical chemistry are orthogonal (e.g. particle in a box, hydrogen atomic orbitals). Does this come about because we purposefully make them orthogonal, or are they derived that way naturally? Can there be useful wave functions that are not orthogonal?&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="25" LastEditDate="2013-11-22T11:46:06.187" LastActivityDate="2019-09-15T13:45:01.730" Title="Why are wave functions orthogonal?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7030" PostTypeId="2" ParentId="4903" CreationDate="2013-11-22T05:56:58.440" Score="2" Body="&lt;p&gt;My textbook states that the energy of the p orbitals &lt;em&gt;will&lt;/em&gt; split in a magnetic field. (Physical Chemistry, McQuarrie, p.210)&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2013-11-22T05:56:58.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7031" PostTypeId="2" ParentId="7029" CreationDate="2013-11-22T06:33:43.177" Score="4" Body="&lt;p&gt;In general, orthogonal wavefunctions are much easier to treat. In some cases they appear naturally, but usually, the orthogonality is imposed as a constrain while constructing the wavefunction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you construct electronic wavefunction in the atomic orbital basis, you try to construct the orthogonal basis. This guarantees that the AOs are linearly independent. (Implication, not equivalence). Would you fail to fulfill this, the solution might still be possible, but much more difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you manage to solve the eigenvalue - eigenvector problem, the solutions are by definition orthogonal to each other. This is the case for the examples you provided.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditDate="2013-11-22T06:55:10.867" LastActivityDate="2013-11-22T06:55:10.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7032" PostTypeId="1" AcceptedAnswerId="7773" CreationDate="2013-11-22T07:26:21.430" Score="10" ViewCount="354" Body="&lt;p&gt;Anaerobic digestion and methanisation of organic feedstocks is a complex series of processes that result in biogas. One way to look at this is a very top-down, stochiometric approach as in the Buswell equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C_aH_bO_c + \left($a - \frac{b}{4} - \frac{c}{2}$\right)H2O -&amp;gt; \left($\frac{a}{2} + \frac{b}{8} - \frac{c}{4}$\right)CH4 + \left($\frac{a}{2} - \frac{b}{8} + \frac{c}{4}$\right)CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a &lt;a href=&quot;http://www.renewable-energy-concepts.com/biomass-bioenergy/biogas-basics/buswell-and-boyle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buswell-Boyle equation&lt;/a&gt; that takes nitrogen and sulfur into account, but right now I just want to understand one aspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the process, at some point, water is dissociated. When does this happen (hydrolysis, acidification, ...), what role does the water play and what are sample reactions that illustrate this?&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="41328" LastEditDate="2020-04-06T11:02:05.927" LastActivityDate="2020-04-06T11:02:05.927" Title="When and how in anaerobic digestion is water dissociated?" Tags="&lt;water&gt;&lt;biochemistry&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7033" PostTypeId="1" AcceptedAnswerId="7035" CreationDate="2013-11-22T08:15:14.677" Score="16" ViewCount="24557" Body="&lt;p&gt;Charcoal (also &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot;&gt;activated charcoal&lt;/a&gt;) is known to adsorb a huge variety of substances including a variety of paints, dyes and many different kinds of ions. Moreover, the amount of &lt;a href=&quot;http://en.wikipedia.org/wiki/Adsorption&quot;&gt;adsorption&lt;/a&gt; at normal pressure and temperature is high. I know physisorption requires high capacity of Van-der-waal bonding with the adsorbate which is favoured by surface atoms with large, easily-dispersed electron clouds and chemisorption requires actual chemical bond formation which requires highly reactive surface atoms. Charged species get adsorbed to form a &lt;a href=&quot;http://en.wikipedia.org/wiki/Double_layer_%28interfacial%29&quot;&gt;double layer&lt;/a&gt; when the surface is charged, either due to ionization or due to polarity.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But charcoal possesses none of the structural specialities required to&#xA;  be a great adsorbent. Neither it has charged centres (unlike colloidal&#xA;  proteins, or metal sols) nor it has a very high reactivity (Charcoal&#xA;  is almost the final product of combustion of wood, bone, sugar etc and&#xA;  is supposed to be immune to further oxidation even at considerable&#xA;  temperature). Why is it then an excellent adsorbent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2013-12-05T16:39:05.100" LastActivityDate="2018-06-26T02:36:23.627" Title="Why is Charcoal such an excellent adsorbent?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;bond&gt;&lt;surface-chemistry&gt;&lt;adsorption&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7034" PostTypeId="1" AcceptedAnswerId="7036" CreationDate="2013-11-22T08:22:56.690" Score="3" ViewCount="10725" Body="&lt;p&gt;I was reviewing the thermodynamics of micelle formation from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamics_of_micellization&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; and &lt;a href=&quot;http://books.google.co.in/books?id=ep7U5G4O3mQC&amp;amp;pg=PA17&amp;amp;lpg=PA17&amp;amp;dq=thermodynamics%20of%20micellization%20entropy%20change%20positive&amp;amp;source=bl&amp;amp;ots=3ePpuUjLFk&amp;amp;sig=xgQF2mopsrsO1LfYcJ-THmnANgI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=vxKPUu-PM8PorAe_7IGADA&amp;amp;ved=0CF0Q6AEwBQ#v=onepage&amp;amp;q=thermodynamics%20of%20micellization%20entropy%20change%20positive&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this book&lt;/a&gt;. I understood most of the arguments but micellization is supposed to be an entropy driven process with a positive entropy change in the process of micellization. But isn't this a bit counter-intuitive?  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the unassociated form, the surfactant molecules are free to roam&#xA;  about in the solution much like normal solutes. But in the form of a&#xA;  micelle, they get associated, grouped into an associated whole, which&#xA;  reduces the freedom of independent movement and should therefore&#xA;  decrease the entropy (randomness) of the system? The entropy change&#xA;  should be negative during micellization. Why is it positive?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understood the mathematical arguments leading to the develpment of the approximate formula $\Delta G=RT\ln(\text{CMC})$.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-03-23T11:58:28.737" LastActivityDate="2018-03-23T11:58:28.737" Title="The entropy change during micellization" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;&lt;surface-chemistry&gt;&lt;surfactants&gt;&lt;colloids&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7035" PostTypeId="2" ParentId="7033" CreationDate="2013-11-22T08:51:00.457" Score="6" Body="&lt;p&gt;I see three reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Activated carbon is commonly used to adsorb organics, that should bind well on coal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For many purposes, activated carbon is treated with potassium or iodine to provide ions for charged centers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Charring organic stuff is a cheap and easy way to create large surfaces for adsorption.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="102" LastEditorUserId="45441" LastEditDate="2018-06-26T02:36:23.627" LastActivityDate="2018-06-26T02:36:23.627" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7036" PostTypeId="2" ParentId="7034" CreationDate="2013-11-22T15:24:00.180" Score="4" Body="&lt;p&gt;The micellization releases water molecules which initially solvate the hydrophilic head groups. The release of these waters greatly increases the entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Dong et al., Chem. Rev. 2010, 110, 4978-5022.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="5026" LastEditDate="2018-03-23T11:53:21.280" LastActivityDate="2018-03-23T11:53:21.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7038" PostTypeId="2" ParentId="6475" CreationDate="2013-11-22T15:51:16.393" Score="2" Body="&lt;p&gt;Is it not just because of the fact that $[aa|bb]$ is in chemists notation, meaning that $[aa|bb] = [a(1)a(1)|b(2)b(2)]$ and so you can just switch them around?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-11-22T15:51:16.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7039" PostTypeId="1" AcceptedAnswerId="7041" CreationDate="2013-11-22T20:22:37.247" Score="1" ViewCount="188" Body="&lt;p&gt;Assume that we have a space with just two $\ce{H}$ atoms and their distance to each other is $d$. Let's say they don't have initial velocity. What is the force with which they will act on each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question might have some mistakes in itself, but the main thing I'm trying to learn is how an atom of element $X$ interact with an atom of element $Y$.&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="73" LastEditDate="2013-11-23T21:03:41.353" LastActivityDate="2013-11-23T21:03:41.353" Title="How will these hydrogen atoms interact?" Tags="&lt;atoms&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7041" PostTypeId="2" ParentId="7039" CreationDate="2013-11-23T00:52:57.623" Score="4" Body="&lt;p&gt;If you just have two H atoms, nothing else then the only forces acting are gravity and &lt;a href=&quot;http://ph.qmul.ac.uk/sites/default/files/CM_PHY108_Lecture2_Interatomic%20Forces1b.pdf&quot; rel=&quot;nofollow&quot;&gt;interatomic forces&lt;/a&gt;, in particular, the van-der-waals force (note slide 5 which says the van der waals potential is applicable to monatomic gases) . The van der walls force varies and the inverse sixth power, so it will be zero until the hydrogen atoms are quite close, so weak ol' gravity will be doing most of the work until then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you should set up your potential around the center of mass of the system with U including the gravitational and van-der-waals potentials. For all but the closest distances, be prepared for very slow dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know how to use potentials, then take the negative of the derivative of the potential for the van der waals and then add in the force of gravity, both acting in the same direction.&lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastActivityDate="2013-11-23T00:52:57.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7042" PostTypeId="2" ParentId="4904" CreationDate="2013-11-23T04:30:23.110" Score="2" Body="&lt;p&gt;The argument is false.  Inductive effects happen through any covalent bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that as far as systems with multiple bonds are concerned, &lt;em&gt;conjugation, hyper conjugation, or resonance effects&lt;/em&gt; are often more significant than inductive effects.  These effect may reach a much larger distance and have richer chemistry, and frankly are more interesting. As a result, text books focus mainly on these effects in double bonded systems. Presence of a myriad of strong effects also makes it harder to identify the role of induction in system with multiple bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this by no means eliminates the presence of inductive effect and in fact induction often do have significant contribution to reactivity in such systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best example might be fluorobenzene (Ph-F).  Electrophilic substitutions of fluorobenzene is significantly slower than those of benzene, and almost predominantly the para- product is obtained.  This is caused by the induction of F atom.  This can be attributed solely to inductive effects because the conjugation effect of F atom is electron donating and would cause an increase in reactivity instead, and would cause near statistical distribution between para and ortho substitution products(which is the case in chlorobenzene and bromobenzene).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact inductive effect also occur through C=O.  You could not find an example only because indicator of electron charge on the oxygen atom of C=O is harder to come by.  The examples do exist.  For example, pKa of cation acid $\textrm{R2C=O-H}^+{\longleftrightarrow}\textrm{R2C=O + H}^+$ is strongly influenced by substitutions on the R groups.   &lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-11-23T04:30:23.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7043" PostTypeId="2" ParentId="7039" CreationDate="2013-11-23T04:41:24.340" Score="5" Body="&lt;p&gt;How one atom interact with another atom is an difficult question and requires quite some calculations, even for a system with only two hydrogen atoms.  Basically, this is done by solving the Schrödinger equation for the system.  The answer will also depend on the electronics state of the system, especially the spin state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is in a triplet electronic state, the two atom will experience a repulsive force that approach infinity as you pull the two atoms closer, and that force approaches zero as the distance approaches infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is in a singlet state, the force is zero at the equilibrium bond length of $H_2$ atom(~0.74Å).  As the distance gets smaller the atoms experience an increasing repulsive force which approaches infinity at 0.  As the two atoms are pulled apart from the equilibrium there is a decreasing attractive force that approaches 0 at infinity.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-11-23T04:41:24.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7046" PostTypeId="1" CreationDate="2013-11-23T07:38:02.460" Score="0" ViewCount="164" Body="&lt;p&gt;I wanted to know about the substitution reaction of alkanes particularly interested in the reaction of methane with fluorine and chlorine.&lt;/p&gt;&#xA;" OwnerUserId="3769" LastEditorUserId="418" LastEditDate="2014-03-23T11:25:14.657" LastActivityDate="2014-03-23T11:25:14.657" Title="Safety concerns for radical halogenations of methane in sunlight?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7048" PostTypeId="2" ParentId="4867" CreationDate="2013-11-23T09:02:23.513" Score="1" Body="&lt;p&gt;The equation above is a decent approximation of one group of reactions. Another one given here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{C_{c}H_{h}O_{o} -&amp;gt; z CO + x H2 + w H2O + v C_{n}H_{m}O_{p} + u C_{(s)}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only the pyrolysis part of the process. There's a host of reduction reactions, I only give the ones that produces methane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CO + 3 H2 &amp;lt;-&amp;gt; CH4 + H2O} \\&#xA;\ce{C_{(s)} + 2 H2 &amp;lt;-&amp;gt; CH4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, methane is formed when there's hydrogen in the gas phase. Besides the pyrolysis there's two reactions that generate free hydrogen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CO + H2O -&amp;gt; CO2 + H2} \\&#xA;\ce{C_{(s)} + H2O &amp;lt;-&amp;gt; CH4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So indeed, you need water vapor to generate the hydrogen for your methane. If I recall correctly, methane is only a minor component in pyrolysis gas most of the time. By caloric value, the biggest component is either carbon monoxide or tar (long, complex hydrocarbons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least you have a host of oxisation reactions (both in the gas phase and on the solid stuff) that power the pyrolysis - I won't go into them here.&lt;/p&gt;&#xA;" OwnerUserId="102" LastEditorUserId="194" LastEditDate="2018-01-27T19:00:02.460" LastActivityDate="2018-01-27T19:00:02.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7049" PostTypeId="1" CreationDate="2013-11-23T10:08:29.237" Score="1" ViewCount="102" Body="&lt;p&gt;For example, why do some solutions not cause any injuries at pH 2 while HCl does?&lt;/p&gt;&#xA;" OwnerUserId="3770" LastEditorUserId="73" LastEditDate="2013-11-23T21:07:29.230" LastActivityDate="2013-11-23T21:07:29.230" Title="What are the properties of various solutions at pH &lt;2 that may cause injuries?" Tags="&lt;acid-base&gt;&lt;safety&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7050" PostTypeId="2" ParentId="7049" CreationDate="2013-11-23T10:50:29.070" Score="5" Body="&lt;p&gt;Injury is not solely related to how strongly acidic a solution is, as there are several mechanisms by which organic matter is damaged from exposure to acids, not just protonation. It is also highly dependent on the type and length of exposure. Aqueous $\ce{HF}$ is likely considerably corrosive after skin contact even at pH 2, mostly because of $\ce{F^-}$ which is a rather reactive ion with a strong tendency to abstract $\ce{Ca^{2+}}$ (in fact, the fact that  $\ce{HF}$ is a weak acid makes it even &lt;em&gt;more&lt;/em&gt; dangerous because the undissociated $\ce{HF}$ is much less polar and the molecules can cross the skin barrier more easily, allowing direct access to muscle and bone which can be severely damaged). Aqueous $\ce{HCN}$ also forms weak acids, but again $\ce{CN^-}$ ions are extremely toxic as they block cell respiration. The risk is compounded by the fact that $\ce{HCN}$ is a very volatile substance and can be released as a gas from concentrated or acidic solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to $\ce{HCl}$, you need to be a bit more specific. What kind of injury? $\ce{HCl}$ itself is a relatively safe acid in dilute solution, even though it's a strong acid, as it does not have strongly oxidising/reducing/dehydrating/toxic properties and is only volatile at rather high concentrations. Remember that $\ce{HCl}$ is even used as an integral part of our digestive system, being &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrochloric_acid#Presence_in_living_organisms&quot; rel=&quot;noreferrer&quot;&gt;the species that allows stomach acids to reach pH 1-2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-23T10:50:29.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7051" PostTypeId="1" AcceptedAnswerId="7088" CreationDate="2013-11-23T11:43:53.920" Score="7" ViewCount="1750" Body="&lt;p&gt;My teacher gave me a shortcut to calculate the resultant pH if 2 or 3 solutions are mixed, which are relevant only if the difference between the pH of individual solutions is 1.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If there are two solutions, the pH of the mixture &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; (the mean pH − 0.24).&lt;/li&gt;&#xA;&lt;li&gt;If there are three solutions, the pH of the mixture &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; (the mean pH + 0.56).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;What is the logic behind this approximation?&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="865" LastEditDate="2021-05-25T14:37:59.903" LastActivityDate="2021-05-25T14:37:59.903" Title="Shortcut for calculating pH of mixtures" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7052" PostTypeId="1" CreationDate="2013-11-23T11:45:01.020" Score="1" ViewCount="102" Body="&lt;p&gt;What technology has the radioisotope Cobalt 60 replaced in medicine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.,PET preferred over CT scans&lt;/p&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="73" LastEditDate="2013-11-23T21:12:28.223" LastActivityDate="2014-07-22T05:04:22.363" Title="New technology that has replaced the old radioisotope" Tags="&lt;nuclear&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7053" PostTypeId="1" AcceptedAnswerId="7056" CreationDate="2013-11-23T12:07:46.203" Score="1" ViewCount="8602" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8pTvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8pTvd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher explained it as restricted rotation between the double bonds. But I think there is more to it. After all, why is it considered to be a type of isomerism itself? Whether an alkene is &lt;em&gt;cis&lt;/em&gt; or &lt;em&gt;trans&lt;/em&gt; only depends on how you make the structure on paper. So what is the basic idea behind it?&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="16683" LastEditDate="2015-10-16T16:42:39.007" LastActivityDate="2016-07-19T00:52:51.750" Title="What is the actual reason for geometrical isomerism?" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7055" PostTypeId="1" AcceptedAnswerId="7075" CreationDate="2013-11-23T14:25:10.150" Score="0" ViewCount="5187" Body="&lt;p&gt;I'm wondering what the abbreviation &quot;Anal. Calcd. stands for &lt;em&gt;exactly&lt;/em&gt;. It is put in front of calculated (expected) values for elemental analyses. &quot;Analytically calculated&quot; makes no sense for me, but maybe it's more of a language problem for me in this particular case. According to a quick Google research, some are listing it in their lists of abbreviations as &quot;Analytical Calculated&quot;. Using the adjective instead of an adverb doesn't make it any more correct to me.&lt;/p&gt;&#xA;" OwnerUserId="661" LastActivityDate="2013-11-24T23:49:12.403" Title="What does &quot;Anal. Calcd.&quot; stand for written out?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7056" PostTypeId="2" ParentId="7053" CreationDate="2013-11-23T15:00:04.480" Score="9" Body="&lt;p&gt;Geometric isomerism occurs when two structures with the same connectivity are not interconvertible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cis-trans_isomerism&quot; rel=&quot;noreferrer&quot;&gt;Cis-Trans isomerism&lt;/a&gt; is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\pi$-bond located perpendicular to the plane of the molecule. In the image below (created by &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Doppelbindung2.png&quot; rel=&quot;noreferrer&quot;&gt;David Pilz for Wikipedia&lt;/a&gt;), you can see the $\pi$-bond in a perpendicular plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DwsAL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\pi$-bond is made from $p$-orbitals, so it cannot have the same symmetry as the $\sigma$-bond. You can see the formation of the $\pi$-bond from $p$-orbitals in this picture by &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Pi-Bond.svg&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia user JoJan&lt;/a&gt;. In order for the $\pi$-bond to exist the $p$-orbitals must be aligned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLq8D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the molecule to rotate from $cis$ to $trans$, the bond axis would need to rotate so that the $p$-orbitals are no longer aligned and can no longer form the $\pi$-bond.  &lt;a href=&quot;http://www.roanestate.edu/faculty/chemistry/chemistryslideshows/hybridIII/hybridIII.pdf&quot; rel=&quot;noreferrer&quot;&gt;This slide show from James Condon at Roane State(PDF)&lt;/a&gt; explains the same concept. Rotation around the double bond requires enough energy input to break the $\pi$-bond, at least $250 \ \text{kJ/mol}$, which is a lot for room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clearest demonstration that geometric isomers are clearly different molecules is to look at some examples with drastically different properties. It is important to remember that the experimental observation that these compounds with very similar structures have very different properties predates the explanation that the double bond has limited rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two isomers of &lt;a href=&quot;http://en.wikipedia.org/wiki/1,2-Dichloroethene&quot; rel=&quot;noreferrer&quot;&gt;1,2-dichloroethene&lt;/a&gt; have different physical properties due both the difference in polarity and the difference in shape:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$cis$-1,2-dichloroethane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g1NfK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dipole moment = $1.9 \ \text{D}$&lt;/li&gt;&#xA;&lt;li&gt;Boiling point = $60.2 \ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;Melting point = $-81.47 \ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;Density = $1.28 \ \text{g/cm}^3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$trans$-1,2-dichloroethane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mimiY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dipole moment = $0.0 \ \text{D}$&lt;/li&gt;&#xA;&lt;li&gt;Boiling point = $48.5 \ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;Melting point = $-49.44 \ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;Density = $1.26 \ \text{g/cm}^3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, my favorites are &lt;a href=&quot;http://en.wikipedia.org/wiki/Maleic_acid&quot; rel=&quot;noreferrer&quot;&gt;maleic acid&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fumaric_acid&quot; rel=&quot;noreferrer&quot;&gt;fumaric acid&lt;/a&gt;. Notice the extreme difference in water solubility as well as the difference in acidity. These molecules behave very different &lt;em&gt;chemically&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maleic Acid $cis$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eccrl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Melting point = $135 \ ^\circ\text{C}$ (decomposes)&lt;/li&gt;&#xA;&lt;li&gt;Density = $1.59 \ \text{g/cm}^3$&lt;/li&gt;&#xA;&lt;li&gt;Solubility in water $788 \ \text{g/L}$&lt;/li&gt;&#xA;&lt;li&gt;Acidity $\text{p}K_{a_1} = 1.9$, $\text{p}K_{a_2} = 6.07$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fumaric Acid $trans$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJiLV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Melting point = $287 \ ^\circ\text{C}$ (does not decompose)&lt;/li&gt;&#xA;&lt;li&gt;Density = $1.635 \ \text{g/cm}^3$&lt;/li&gt;&#xA;&lt;li&gt;Solubility in water $6.3 \ \text{g/L}$&lt;/li&gt;&#xA;&lt;li&gt;Acidity $\text{p}K_{a_1} = 3.03$, $\text{p}K_{a_2} = 4.44$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-23T15:00:04.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7057" PostTypeId="1" AcceptedAnswerId="8855" CreationDate="2013-11-23T15:42:13.147" Score="2" ViewCount="194" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mixture weighing $4.08\ \mathrm g$ of $\ce{BaO}$ ($M(\ce{Ba})$ = $138\ \mathrm{g\ mol^{-1}}$) and an unknown carbonate $\ce{XCO3}$ was heated strongly. The residue weighed $3.64\ \mathrm g$. This was dissolved in $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$. The excess acid required $16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ solution for complete neutralization. What is the molar mass of $\ce{X}?$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$16\ \mathrm{ml}$ of $2.5\ \mathrm M$ $\ce{NaOH}$ means $0.04\ \mathrm{mol}$ of $\ce{NaOH}$. Thus, the excess acid was $0.04\ \mathrm{mol}$ of $\ce{HCl}$. $100\ \mathrm{ml}$ of $1\ \mathrm M$ $\ce{HCl}$ means $0.1\ \mathrm{mol}$ of $\ce{HCl}$. Thus, $0.06\ \mathrm{mol}$ of $\ce{HCl}$ reacted with the residue. Now, how can I find the ratio in which the residue reacted with $\ce{HCl}?$ What is the composition of the residue?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="69830" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T07:42:46.793" LastActivityDate="2021-05-22T07:42:46.793" Title="How do I find the ratio in which the residue of a reaction combines with HCl?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7061" PostTypeId="1" CreationDate="2013-11-23T23:30:43.800" Score="1" ViewCount="1022" Body="&lt;p&gt;Ok, so let's say I have raw silver which is polluted with a bit of gold, copper and lead.&#xA;So through electrorefining I want to make pure silver out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have it all in an acidic solution. What happens when I turn on the Voltage?&#xA;How would I prevent the other elements from getting electrolyzed?&#xA;Because when I have the electrorefining of copper it gets clean even when polluted with silver, so why doesn't the copper get electrolyzed as well? &lt;/p&gt;&#xA;" OwnerUserId="3775" LastActivityDate="2014-01-16T12:07:58.113" Title="Silver electrorefining" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7063" PostTypeId="2" ParentId="7052" CreationDate="2013-11-24T00:50:27.990" Score="2" Body="&lt;p&gt;After searching on Google (now there's an idea)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking about what technology was used in medicine before Co-60 therapy was developed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Co-60 therapy was developed in the early 1950's and was seen as an alternative to other forms of X-ray radiotherapy, or radiation therapy. It is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/External_beam_radiation_therapy&quot; rel=&quot;nofollow&quot;&gt;external beam radiation therapy&lt;/a&gt;. Radiation sources at the time (mostly derived from radium) required very high energy sources, and proved largely ineffective against deep-set cancers. Co-60 therapy was able to penetrate deeper at lower energies, and therefore was more effective against a range of previously untreatable cancer types. The first Co-60 machine was from the The University of Saskatchewan, and they have an excellent &lt;a href=&quot;http://www.usask.ca/cobalt60/&quot; rel=&quot;nofollow&quot;&gt;site devoted to the history of Co-60 treatment&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asking about what technology is now being used in place of Co-60 therapy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventional X-ray sources were improved with the development the linear accelerator which could produce X-ray energies exceeding 8MV. Nowadays, commercial units can produce photons up to 25MV. &lt;a href=&quot;https://www.ecri.org/PatientSafety/Pages/Linear_Accelerators_Radiotherapy_Units_Cobalt.aspx&quot; rel=&quot;nofollow&quot;&gt;Here is a useful link&lt;/a&gt; to compare linear accelerator and Co-60 radiation sources for medical applications. There are still many developing areas of use, with some experts claiming there is &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2805891/&quot; rel=&quot;nofollow&quot;&gt;still very much a place for Co-60 therapy&lt;/a&gt;. Co-60 therapy can be &lt;a href=&quot;http://www.abta.org/care-treatment/treatments/stereotactic-radiosurgery/cobalt-60-therapy.html&quot; rel=&quot;nofollow&quot;&gt;applied with high precision with low side-effects&lt;/a&gt; and is just one of many approaches available, often used in tandem with other medical treatments.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-11-24T00:50:27.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7065" PostTypeId="1" AcceptedAnswerId="7069" CreationDate="2013-11-23T21:15:28.730" Score="3" ViewCount="2214" Body="&lt;p&gt;So a teacher at school told me that strong acids do not create an equilibrium in general, while weak acids do (thats the difference, strong acids namely ionize completely). However, I lately stumbled upon a passage reporting that when experimenting with extreme pH, so extreme concentrations of acids, strange things start to occur and strong acids are not able to ionize completely, resulting in an equilibrium. I would explain this by the extreme concentration -&gt; less water molecules in the vicinity of acid to react -&gt; less chance to ionize.&#xA;But then I thought, are strong acids in fact not also creating an equilibrium, but because they react so strongly, they immediately release their proton. The conjugate bases are extremely weak, but they should be able to pick up a proton right? They simply become and acid, and release it immediately again. Please clarify this entire situation for me, thanks already.&lt;/p&gt;&#xA;" OwnerUserId="3743" OwnerDisplayName="user209347" LastEditorUserId="22" LastEditDate="2013-11-24T11:37:02.097" LastActivityDate="2017-08-06T07:36:45.123" Title="Strong acid equilibrium" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7066" PostTypeId="1" AcceptedAnswerId="7077" CreationDate="2013-11-24T09:01:18.153" Score="4" ViewCount="445" Body="&lt;p&gt;When we want the bad smell in toilet to go away we use a toilet spray to clean the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does it really cleans it, or just the elements in that spray have better effect on our smell sense. And we think that the bad smell went away while it is not, and the air contains a mix of bed and good smells?&lt;/p&gt;&#xA;" OwnerUserId="3776" LastEditorUserId="19616" LastEditDate="2015-10-30T17:26:32.533" LastActivityDate="2015-10-30T17:26:32.533" Title="Toilet spray effect" Tags="&lt;reaction-mechanism&gt;&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7068" PostTypeId="2" ParentId="7065" CreationDate="2013-11-24T12:02:56.740" Score="4" Body="&lt;p&gt;Whether a substance is a strong or weak acid/base depends on the solvent it's in. For example, $\ce{HNO_3}$ is a strong acid in water. This is partly because $\ce{HNO_3}$ has a relatively high tendency to donate a proton (the conjugate base $\ce{NO^{-}_3}$ is quite stable and therefore has a low tendency to accept a proton back), and partly because water has a certain basic character, accepting protons to create $\ce{H3O+}$. However, the same $\ce{HNO_3}$ in pure liquid $\ce{H3CCOOH}$ (also known as glacial acetic acid) is actually a weak acid. This change in character is largely due to the fact that $\ce{H3CCOOH}$ is a much, much weaker base than water, forming the $\ce{H3CCOOH^+_2}$ ion with significantly more reluctance. Even though $\ce{HNO_3}$ can easily lose its proton, $\ce{H3CCOOH}$ is even moreso unwilling to accept it, and thus $\ce{HNO_3}$ becomes a weak acid. An amusing situation arises in pure liquid $\ce{H_2SO_4}$. Adding $\ce{HNO_3}$ there, we find that $\ce{HNO_3}$ actually &lt;em&gt;behaves as a weak base&lt;/em&gt;. $\ce{H_2SO_4}$ itself also has a high tendency to donate protons, and of the two possible reaction products from protonation, $\ce{H_2NO^+_3}$ is somewhat more stable than $\ce{H_3SO^+_4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, acidity is dependent on the solvent, not only on the solute. A strong acid such as $\ce{HNO_3}$ in aqueous solution at high dilution is completely dissociated, as we expect of a strong acid. However, if more and more water is removed, then the solution starts to behave differently. Even though $\ce{HNO_3}$ has a strong tendency to protonate water, you can have a solution that contains more moles of $\ce{HNO_3}$ than water. Then it becomes obvious that not all $\ce{HNO_3}$ molecules are deprotonated. Practically all of the water left in the solution will be in the form of $\ce{H_3O^+}$, but there will be left over undissociated $\ce{HNO_3}$, so most of the excess stays undissociated, except for a small amount that undergoes self-protonation, generating a bit of $\ce{H_2NO^+_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, in extremely concentrated solutions, even strong acids/bases will not show 100% dissociation, and saying that &quot;strong acids completely ionize&quot; is just a rule of thumb which shouldn't be taken too far.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-24T12:02:56.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7069" PostTypeId="2" ParentId="7065" CreationDate="2013-11-24T15:28:37.460" Score="4" Body="&lt;p&gt;To be more strict, a strong acid is defined as one that completely ionize in aquatic solutions. And as always (in chemistry), completely means almost completely.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a strict sense the strong acid do &quot;have an equilibrium&quot; because any reaction you can write down and balance will have an equilibrium constant, calculable through Nernst equation. However, the equilibrium is inconsequential unless in extreme conditions. Any acid with $\mathrm{p}K_\mathrm{a} &amp;lt; 0$, or equivalently $K_\mathrm{a}&amp;gt;1$ is a strong acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry is a practical science and there is a tendency to use practical instead of strict terms.  So when your teacher say something like &quot;fully&quot;, &quot;completely&quot;, &quot;do not have&quot;, etc. in a chemistry classroom, it implies approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember your teacher's discussions will be all focused on highly diluted solutions.  Even concentrations otherwise the situation becomes extremely difficult to treat and unfit for a textbook example.  Therefore in the context of your textbook, $\mathrm{p}K_\mathrm{a} &amp;lt; 0$ guarantees essentially complete dissociation.   &lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="41328" LastEditDate="2017-08-06T07:36:45.123" LastActivityDate="2017-08-06T07:36:45.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7070" PostTypeId="1" CreationDate="2013-11-24T17:47:54.563" Score="3" ViewCount="745" Body="&lt;p&gt;The figure is a portion of the titanium copper phase diagram for which only single-phase regions are labeled. Specify all temperature composition points at which eutectics, eutectoids, peritectics and congruent phase transformations occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; i've found the eutectic and congruent point. (aprox 960 and 980 celsius) but how do I find the eutectoid and peritectic point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://bildr.no/image/a2JVWGZB.jpeg&quot; alt=&quot;lol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for blurry image!&lt;/p&gt;&#xA;" OwnerUserId="3778" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.540" LastActivityDate="2014-12-08T19:22:16.040" Title="Phase diagram analysis" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7071" PostTypeId="1" CreationDate="2013-11-24T18:39:54.213" Score="1" ViewCount="5643" Body="&lt;p&gt;I conducted a lab in which different concentrations of $\ce{HCl}$ were used and $\ce{Na2CO3}$ was added. The pressure change was measured for a reaction with the different concentrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am going to find the reaction rate for the reactions. I am thinking that using the pressure data, I can find the reaction rate by doing the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider: &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;${PV = nRT}$ ($R$ and $T$ are constant so they can be taken out of the equation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${PV = n}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = {\frac nV}$ (Knowing that $\text{c}={\frac nV}$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${P} = \text{c}$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Seeing as pressure is directly related to concentration, then couldn't I just use the pressure data to find the rate and then relate that to the concentration change over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative would be to use ${PV=nRT}$ for each second of the reaction and find the concentration change that way. However, I want to consider the possibility of doing it this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem I forecast would be the units: kPa/s vs. mols/L &lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="39539" LastEditDate="2017-03-30T19:36:37.053" LastActivityDate="2017-03-30T19:36:37.053" Title="Finding reaction rate from pressure change" Tags="&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2017-09-04T19:21:52.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="7073" PostTypeId="1" CreationDate="2013-11-24T19:31:14.830" Score="1" ViewCount="84" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NKuIN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's from the start of part 3 / chapter 19 in Lewin's Genes X (see &lt;a href=&quot;http://goo.gl/ThZu7t&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page in Google Books&lt;/a&gt;)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks a bit like a Ramachandran plot(?) but the axes only go to 100 / 70ish, but I don't even know what molecule it is since it doesn't say and there's no image attribution page at the front/back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sleuths out there?&lt;/p&gt;&#xA;" OwnerUserId="2119" LastActivityDate="2013-11-24T19:31:14.830" Title="Can anyone identify this uncaptioned diagram?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7075" PostTypeId="2" ParentId="7055" CreationDate="2013-11-24T23:49:12.403" Score="4" Body="&lt;p&gt;They are two distinct terms, that just happen to frequently be found next to each other. Their definitions can be found in the &lt;a href=&quot;http://pubs.acs.org/paragonplus/submission/joceah/joceah_authguide.pdf&quot; rel=&quot;nofollow&quot;&gt;Journal of Organic Chemistry Guidelines for Authors&lt;/a&gt; (see page 29).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anal. - abbreviation for combustion elemental analysis.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calcd - abbreviation for calculated.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not actually a single phrase, as you've suggested here. Combustion elemental analysis is what people often mean when referring to 'getting an analytical done'. Writing Anal. signifies the type of experimental data that is about to be listed, just as you might specify 1H NMR, IR or MS. A common alternative to combustion elemental analysis is high resolution MS, which is often written in experimental sections as HRMS (ESI) &lt;em&gt;m/z&lt;/em&gt; calcd &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The format for presenting elemental analysis is journal specific, and therefore varies. The format you're asking about is consistent with the &lt;a href=&quot;http://pubs.acs.org/&quot; rel=&quot;nofollow&quot;&gt;ACS publication recommendations&lt;/a&gt;. &lt;a href=&quot;http://www.rsc.org/images/Author_guidelines_tcm18-186308.pdf&quot; rel=&quot;nofollow&quot;&gt;RSC guidelines&lt;/a&gt; accept the the data in a slightly different format.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-11-24T23:49:12.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7077" PostTypeId="2" ParentId="7066" CreationDate="2013-11-25T01:50:22.527" Score="6" Body="&lt;p&gt;Some toilet sprays can be classed as deodorizers (or air sanitizers in the US?), while others call themselves air fresheners. So it may depend on which brand you are using, but  generally the deodorizer spray is working a little harder than the air freshener. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odor control products can work through a number of different methods:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chemical transformation. Typically oxidization. This is why bleach is a good product for eliminating smell. &lt;/li&gt;&#xA;&lt;li&gt;Adsorption or Absorption. Trapping airborne smelly molecules is a great way to keep them from your nostrils. Activated charcoal in-soles for your shoes are a great example of this. Zeolites are also used. An air purifier would be what you'd need to use in a bathroom to eliminate smells using this method. Absorption into other media can then also be used in conjunction with other methods (chemical)&lt;/li&gt;&#xA;&lt;li&gt;Sanitizer or &quot;biocide&quot; agents. These kill odor-causing bacteria, fungi and other microorganisms. The main culprit for 'toilet smell' comes from the results of bacterial activity in your gut. Whilst they do reduce the quantity of airborne bacteria, they require a constant applcation, and so can't be classed as an antiseptic.&lt;/li&gt;&#xA;&lt;li&gt;Masking. This what air fresheners do. Add a stronger, nicer smelling fragrance. Humans only have a few million olfactory receptors. If you keep them busy with volatile pleasant fragrances, malodorous molecules can't get a look in.&lt;/li&gt;&#xA;&lt;li&gt;Sealing. Painting over or applying a sealant to a urine-soaked concrete floor or smoke-laden timber panel is often the only way to mask some smells.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most air spray deodorizers (sanitizers) use a combination of the above methods. They almost all use some form of masking (except those that claim to be fragrance-free), as well as acting as a biocide and in some cases promoting chemical transformation as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are growing concerns about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Air_freshener#Toxicity&quot;&gt;safety of using air sprays&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just as a wonderful aside; one of the main causes of toilet odor is a molecule called skatole (3-methylindole), which also happens to be used in perfumery and also as a flavoring in ice cream! &lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-11-25T01:50:22.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7079" PostTypeId="1" CreationDate="2013-11-25T03:30:49.220" Score="0" ViewCount="2665" Body="&lt;p&gt;How could you test that there's oxalic acid in potatoes.For example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what chemical reacts to oxalic acid and then what will happen if it&#xA;  reacts?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3783" LastEditorUserId="1431" LastEditDate="2013-11-25T04:50:34.623" LastActivityDate="2015-01-17T16:44:57.127" Title="How to test oxalic acid in potatoes" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7080" PostTypeId="1" AcceptedAnswerId="7083" CreationDate="2013-11-25T03:46:08.347" Score="2" ViewCount="475" Body="&lt;p&gt;What triggers sodium to convert from elemental Na to Na+? I know that it wants to have a full valence shell and all that, but how does it just eject the electron out?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2013-11-25T04:44:44.910" Title="What prompts sodium to give up an electron?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7081" PostTypeId="2" ParentId="5339" CreationDate="2013-11-25T03:53:14.220" Score="1" Body="&lt;p&gt;The amount of adsorbed protein is proportional to the area of the 100 nm islands. Figure 4E proves it. The authors misinterpreted their own figure.&lt;/p&gt;&#xA;" OwnerDisplayName="user3784" LastActivityDate="2013-11-25T03:53:14.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7083" PostTypeId="2" ParentId="7080" CreationDate="2013-11-25T04:44:44.910" Score="2" Body="&lt;p&gt;To asses the possibility of a chemical change, we need to look into the spontaneity as given by the thermodynamic Gibbs' Free energy change at constant T and P. For your particular case, the change occurs when an electron accepting ion is present to form the crystalline compound after the electron exchange, in solution where the hydration releases energy or in certain solutions such as ammonia, where solvation of free electrons also take part.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case (1). The changes involved are $$\ce{Na_{(s)}-&amp;gt;Na_{(g)}^+ +e^{-}}$$, $$\ce{M_{(g)} +e^{-}-&amp;gt;M^{-}}$$ $$\ce{Na_{(g)}^+ + M_{(g)}^{-}-&amp;gt;NaM_{(s)}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M$ is an electronegative atom.The ionization energy ($\Delta H_{I}$) and the sublimation energy ($\Delta H_S$) required in the first transformation need to be compensated for by the electron gain enthalpy($\Delta H_{eg}$) released in the second transformation and the lattice enthalpy($\Delta H_{F}\text{lattice}$) released in the final transformation. If it is compensated, the change will be spontaneous. Moreover, the enthalpy released (negative) must also compensate for the lost entropy since gaseous $M$ is converted to an ordered crystal. $\Delta G=\Delta H-T\Delta S$ tells you the compensation required for $\Delta G$ to be negative.&lt;br&gt;&#xA;($|\Delta H_{eg}+\Delta H_F|-|\Delta H_I+\Delta H_S|&amp;gt;|T\Delta S|$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case (2) Elemental sodium in water might lose electrons to the hydrogen evolving hydrogen gas and forming dissolved sodium hydroxide. &#xA;$$\ce{Na_{(s)}-&amp;gt;Na_{(aq)}^+ + e^{-}}$$&#xA;$$\ce{H2O +e^{-}-&amp;gt;\frac 12H2 +OH-}$$&#xA;here, the relesed hydration energy compensates the required sublimation energy in the first step and the negative electrode potential in the second step. There is net positive entropy change which favours the reaction, making $\Delta G$ negative.  ($|\Delta H_{hyd}(\ce{OH^-,Na^+})|+|T\Delta S|=|\Delta G|$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case (3) very similar to case (2) but here the solvation is both of free electrons and that of metal ions by ammonia (liquid). The equations are similar except that now instead of the hydration enthalpy, solvation enthalpy will be released. the entropy change is also positive here, supporting the change.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-11-25T04:44:44.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7084" PostTypeId="1" AcceptedAnswerId="7087" CreationDate="2013-11-25T04:55:45.623" Score="4" ViewCount="4895" Body="&lt;p&gt;What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?  I noticed when I look at a periodic table of elements, the elements of the F-block between group 3 and 4 doesn’t have a group number.  Are they just refer as Lan­thanide,and Actinide or Inner Transition Elements,  But why doesn't they have a group number, it least it appear that way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to a wide version of Periodic table of Elements so you can see what I am talking about.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Template:Periodic_table_%28wide_large%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Template:Periodic_table_%28wide_large%29&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3786" LastActivityDate="2016-05-13T02:29:05.123" Title="What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7085" PostTypeId="1" AcceptedAnswerId="7086" CreationDate="2013-11-25T05:21:44.873" Score="4" ViewCount="58795" Body="&lt;p&gt;What explains why electronegativity increases as you move across a period? Does it have something to do with the shielding effect of added electrons?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="73" LastEditDate="2013-11-26T09:21:35.400" LastActivityDate="2019-05-18T08:16:16.147" Title="Why does electronegativity increase across a period?" Tags="&lt;periodic-trends&gt;" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7086" PostTypeId="2" ParentId="7085" CreationDate="2013-11-25T06:15:38.540" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The electronegativity is the tendency of an atom or a functional group to attract electrons (or electron density) towards itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because the attraction of bonding electrons by an atom increases with nuclear charge (Atomic Number) and decrease of atomic radius. Both these factors operate as we move to the right in period.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it have something to do with the shielding effect of added electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The electron cloud in the inner orbits act as a shield and reduces the nuclear attraction to the outer orbits. Because of the shielding effect, the tendency of nuclear attraction reduces and thus electronegativity reduces.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2013-11-25T06:15:38.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7087" PostTypeId="2" ParentId="7084" CreationDate="2013-11-25T11:28:36.003" Score="7" Body="&lt;p&gt;I suppose IUPAC hasn't bothered with group numbers for the lanthanides/actinides quite yet because there would only be two elements for each group, and because there are few vertical similarities among the f-block elements. The entire row of lanthanides behaves very similarly due to the &lt;a href=&quot;http://depa.fquim.unam.mx/amyd/archivero/LecturasobreNodosRadiales_12854.pdf&quot;&gt;4f orbitals being anomalously compact&lt;/a&gt; and not being available for bonding or ionization. The first half of the actinides is quite different to the lanthanides, as the 5f orbitals there can actually participate in chemical interactions, though the second half becomes more similar as the 5f orbitals are once again strongly trapped to the atom, this time by the very high effective nuclear charge.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-25T11:28:36.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7088" PostTypeId="2" ParentId="7051" CreationDate="2013-11-25T12:24:47.850" Score="9" Body="&lt;p&gt;&lt;strong&gt;Case 1: Mixing equal volumes of two solutions with a pH difference of 1.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Initially we have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{|c|c|c|}&#xA;\hline &#xA;&amp;amp;[\ce{H+}]&amp;amp; \mathrm{pH} &amp;amp;n_\ce{H+}\\ \hline&#xA;a&amp;amp; x &amp;amp; \log x &amp;amp; xV\\&#xA;b&amp;amp;\frac{x}{10} &amp;amp; -\log x + 1 &amp;amp; \frac{xV}{10}\\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;After complete mixing, we obtain:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\&#xA;[\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\&#xA;\mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Before mixing:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{|c|c|c|}&#xA;\hline &#xA;&amp;amp;[\ce{H+}]&amp;amp; \mathrm{pH} &amp;amp;n_\ce{H+}\\ \hline&#xA;a&amp;amp; x &amp;amp; \log x &amp;amp; xV\\&#xA;b&amp;amp;\frac{x}{10} &amp;amp; -\log x + 1 &amp;amp; \frac{xV}{10}\\&#xA;c&amp;amp;\frac{x}{100} &amp;amp; -\log x + 2 &amp;amp; \frac{xV}{100}\\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;After complete mixing, we obtain:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\&#xA;[\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\&#xA;\mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(It seems you got the sign wrong on your last equation.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Caveats:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of 0 and above &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of 14.&lt;/p&gt;&#xA;&lt;p&gt;You also get a slight deviation if you mix solutions that have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; too close to neutral, because the auto-dissociation of water affects the equilibrium values of &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For example, if we mix two equal amounts of liquid, one being pure water(&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 7$&lt;/span&gt;) and the other being aqueous &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; solution with &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCl}] = \pu{10^{-6}M}$&lt;/span&gt;. Considering the auto-dissociation constant for water as &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{w} = 10^{-14}$&lt;/span&gt;, the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; for the last solution is actually approximately &lt;span class=&quot;math-container&quot;&gt;$5.9957$&lt;/span&gt; because water auto-dissociation already affects it slightly.&lt;/p&gt;&#xA;&lt;p&gt;Neglecting that, we mix the liquids creating a solution with &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCl}] = \pu{5\times 10^{-7} M}$&lt;/span&gt;. Directly using &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$&lt;/span&gt; yields a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$6.3010$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Using the trick, we get the pH for &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$6.2596$&lt;/span&gt;. However, the real value for &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt;, taking into account the shifting of the water auto-dissociation equilibrium, is &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$&lt;/span&gt;, such that the real &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$6.28461$&lt;/span&gt;. A small deviation surely, but it's present.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="16683" LastEditDate="2021-05-25T11:17:58.500" LastActivityDate="2021-05-25T11:17:58.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7089" PostTypeId="1" CreationDate="2013-11-25T17:59:20.940" Score="-2" ViewCount="3176" Body="&lt;p&gt;What pressure difference must be generated across the length of a $15\ \mathrm{cm}$ vertical drinking straw in order to drink a water-like liquid of density $1.0\ \mathrm{g\ cm^{−3}}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really unsure how to proceed on this one. I was thinking of using the ideal gas law for the air inside the straw initially, and then to increase the pressure difference so the volume of gas in the straw decreases. But this doesn't get me far at all. If any could help me with this one I would be very grateful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This was probably clear but I am very new to chemistry, I am a mathematics student.&lt;/p&gt;&#xA;" OwnerUserId="3688" LastEditorDisplayName="user7951" LastEditDate="2017-07-16T13:43:40.843" LastActivityDate="2017-07-16T13:43:40.843" Title="What pressure difference is needed across the length of a 15 cm vertical drinking straw in order to drink a liquid of density 1.0 g/cm³?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7091" PostTypeId="1" CreationDate="2013-11-25T23:12:06.810" Score="5" ViewCount="742" Body="&lt;p&gt;There is supposedly a joke shown in this picture relating to periodic trends.&#xA;Does anyone know what it is? And could they also explain where the elements on the eye exam came from?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/z0QYZ.png&quot; alt=&quot;The joke&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3790" LastEditorUserId="46306" LastEditDate="2018-02-24T11:59:51.517" LastActivityDate="2018-02-24T11:59:51.517" Title="What is the joke that is shown in this picture?" Tags="&lt;periodic-trends&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7093" PostTypeId="2" ParentId="7091" CreationDate="2013-11-26T10:59:09.480" Score="6" Body="&lt;p&gt;As commented by @long, &lt;strong&gt;the elements shown are not completely randomised&lt;/strong&gt;. The elements in all the rows, except in the 3rd row ($\ce{Na, K, Ca}$), are arranged in same order as they are given in periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After studying them on periodic table for 15 minutes, it seemed to have no meaning. There are no periodic trends. $\ce{H, Li, Na, K, Rb}$ belongs to IA group, $\ce{Be, Ca, Sr}$ belongs to IIA group. And they are arranged in a random way, having no meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The picture is clearly a joke showing that when scientist have eye check-up, they have to read elements instead of alphabets.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-09-03T03:32:40.170" LastActivityDate="2014-09-03T03:32:40.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7094" PostTypeId="1" AcceptedAnswerId="7097" CreationDate="2013-11-26T16:09:02.457" Score="8" ViewCount="51005" Body="&lt;p&gt;I have been given $\mathrm pK_\mathrm a$ values of an amino group, a carboxyl group and a side chain of cysteine. How can I find the ionic charge on it at different $\mathrm{pH}$ values?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="11367" LastEditDate="2017-04-24T17:12:44.223" LastActivityDate="2018-06-05T08:04:04.213" Title="Calculating charge on amino acid from pKa" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;amino-acids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7095" PostTypeId="2" ParentId="7085" CreationDate="2013-11-26T16:09:54.230" Score="5" Body="&lt;p&gt;ashu's answer is good, but it does not get to the real reason that electronegativity increases across periods. &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot;&gt;Electronegativity&lt;/a&gt; is dependent on other properties that are more intuitively dependent on position within a period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute (&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Mulliken_electronegativity&quot;&gt;Mulliken&lt;/a&gt;) electronegativity $\chi$ (which correlates with the more familiar, but more abstract, Pauling electronegativity) of an atom is the arithmetic mean of its first &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot;&gt;ionization energy&lt;/a&gt; and its first &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity&quot;&gt;electron affinity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_A = (E_{i,A}+E_{ea,A})/2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These values behave in a predictable manner related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot;&gt;effective nuclear charge&lt;/a&gt; idea ashu mentions. Consider the data from the second period (all values in kJ/mol):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Ei      Ea     X&#xA;Li  520     60     290&#xA;Be  900     0      450&#xA;B   800     27     414&#xA;C   1087    122    605&#xA;N   1402    ~0     701&#xA;O   1314    141    728&#xA;F   1681    328    1005&#xA;Ne  2081    ~0     1041&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that $E_i$ increases and $E_{ea}$ also increases (though not as regularly) from lithium to neon. The electronegativities also increase.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-11-26T16:09:54.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7096" PostTypeId="1" CreationDate="2013-11-26T17:49:01.130" Score="2" ViewCount="133" Body="&lt;p&gt;I want to do some spectrometry experiments with glow in the dark paints. The red paints I've read product details for all seem to require a uv light to charge. Are there any red glow in the dark paints that can be charged with visible light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also wondering why most red phosphorescent pigments don't have the ability to glow as long as green pigments.&lt;/p&gt;&#xA;" OwnerUserId="3742" LastEditorUserId="10119" LastEditDate="2014-12-11T17:27:15.477" LastActivityDate="2014-12-11T17:27:15.477" Title="Red glow in the dark pigments" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7097" PostTypeId="2" ParentId="7094" CreationDate="2013-11-26T17:53:17.483" Score="16" Body="&lt;p&gt;For this, you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt;. Using the degree of dissociation, $\alpha$, this can be written as&#xA;$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\alpha}{1-  \alpha}$$&#xA;Rewriting to solve for $\alpha$:&#xA;$$\alpha = \frac{1}{10^{\mathrm{p}K_\mathrm{a} - \mathrm{pH}} + 1}$$&#xA;As stated above, $\alpha$ is the degree of dissociation, meaning the degree at which $\ce{H+}$ is dissociated from the ionizable group. At $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, $\alpha = 0.5$, meaning that $50\%$ of the ionizable groups in question are deprotonated. If $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 1$, about $90\%$ of all groups are deprotonated and if $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 2$, about $99\%$ of all groups are deprotonated. (If you want to, take a look at the shape of the $\alpha$ distribution for $\ce{-SH}$ in cysteine at &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1/%281%2b10%5E%2810.5-x%29%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://books.google.ch/books?id=qDWZiFcbS0EC&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow noreferrer&quot;&gt;a textbook&lt;/a&gt; I found the following $\mathrm{p}K_\mathrm{a}$ values for cysteine:&#xA;\begin{align}&#xA; \mathrm{p}K_\mathrm{a}(\ce{-COOH})   &amp;amp;= 1.9\\&#xA; \mathrm{p}K_\mathrm{a}(\ce{-NH_3+}) &amp;amp;= 8.35\\&#xA; \mathrm{p}K_\mathrm{a}(\ce{-SH})     &amp;amp;= 10.5&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these values, $\alpha$ can be calculated for each ionizable group at the desired pH and this will give you the net charge of the amino acid. Upon deprotonation, the following changes in charge occur for the ionizable groups:&#xA;\begin{array}{lclcr}&#xA; \ce{-COOH} &amp;amp;:&amp;amp; 0 &amp;amp;\rightarrow &amp;amp;-\\&#xA; \ce{-NH_3+}&amp;amp;:&amp;amp; + &amp;amp;\rightarrow &amp;amp;0\\&#xA; \ce{-SH}   &amp;amp;:&amp;amp; 0 &amp;amp;\rightarrow &amp;amp;-\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's calculate the charge of cysteine at pH 10. Using the Henderson-Hasselbalch in a spreadsheet yields the following $\alpha$ values:&#xA;\begin{array}{lcr}&#xA; \ce{-COOH}  &amp;amp;:&amp;amp; 0.9999\\&#xA; \ce{-NH_3+} &amp;amp;:&amp;amp; 0.9781\\&#xA; \ce{-SH}    &amp;amp;:&amp;amp; 0.2403\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the following total charge for cysteine at pH 10:&#xA;$$(-1)\cdot0.9999+(+1)\cdot(1-0.9781)+(-1)\cdot0.2403=-1.218$$&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="4945" LastEditDate="2018-06-05T08:04:04.213" LastActivityDate="2018-06-05T08:04:04.213" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="7098" PostTypeId="1" CreationDate="2013-11-26T19:18:41.917" Score="4" ViewCount="1828" Body="&lt;p&gt;Is there any substance that's a 0-0-0 on the NFPA diamond? Basically a perfectly safe chemical.&lt;/p&gt;&#xA;" OwnerUserId="3795" LastEditorUserId="73" LastEditDate="2013-11-26T21:53:12.833" LastActivityDate="2014-08-24T22:01:40.653" Title="Is there any substance that's a 0-0-0 on the NFPA diamond?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7099" PostTypeId="2" ParentId="7098" CreationDate="2013-11-26T21:31:02.580" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9927321&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Water&lt;/strong&gt;&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uH6Ixm.png&quot; alt=&quot;Water MSDS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can find plenty of others if you perform &lt;a href=&quot;http://en.wikipedia.org/w/index.php?search=%22NFPA-H+%3D+0%22+%22NFPA-F+%3D+0%22+%22NFPA-R+%3D+0%22&amp;amp;title=Special%3ASearch&quot; rel=&quot;nofollow noreferrer&quot;&gt;this search on Wikipedia&lt;/a&gt; (but then, go check the MSDS for the individual compounds before doing anything serious with them).&lt;/p&gt;&#xA;" OwnerUserId="30" LastEditorUserId="-1" LastEditDate="2014-08-24T22:01:40.653" LastActivityDate="2014-08-24T22:01:40.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7103" PostTypeId="2" ParentId="6980" CreationDate="2013-11-27T15:28:40.923" Score="1" Body="&lt;p&gt;The molar mass $(M)$ of a mixture with known mass fractions $(w_{i})$ is calculated from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{M} = \sum_{i}\frac{w_{i}}{M_{i}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case&#xA;$$&#xA;\frac{1}{M} = \frac{1/8}{0.17034}+\frac{7/8}{0.034} = 26.469\ldots\\&#xA;M = \frac{1}{26.469\ldots} = 0.037779872\ \mathrm{kg}/\mathrm{mol}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3728" LastActivityDate="2013-11-27T15:28:40.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7106" PostTypeId="1" AcceptedAnswerId="10934" CreationDate="2013-11-28T06:27:09.093" Score="1" ViewCount="908" Body="&lt;p&gt;I just got a final test and one of the question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Ni^2+}$ + 6$\ce{NH_3}$ $\ce {&amp;lt;=&amp;gt;}$ $\ce{[Ni(NH3)6]^2+}$&#xA;&lt;br&gt;If we add some aqueous $\ce{KOH}$ after the reaction reach equilibrium, where will the reaction shifts?&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-05-19T19:32:31.593" Title="If I add some KOH, where will the reaction shifts?" Tags="&lt;equilibrium&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7108" PostTypeId="1" AcceptedAnswerId="7127" CreationDate="2013-11-28T10:00:40.083" Score="9" ViewCount="69138" Body="&lt;p&gt;I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you have little energy available, would you rather use an&#xA;  isothermal or an adiabatic process to compress a gas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My analysis is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=-\int_{V_0}^{V}PdV$$&#xA;For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above:&#xA;$$W=-\int_{V_0}^{V}\frac{nRT}{V}dV=-nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V_0}{V}\right)$$&#xA;For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=-\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV=-\frac{P_0V_0^\gamma}{-\gamma+1}\left(V^{-\gamma+1}-V_0^{-\gamma+1}\right)=\frac{PV-P_0V_0}{\gamma-1}$$ &#xA;       $$W=-\frac{P_0V_0-PV}{\gamma-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, finally, to my question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed?&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorUserId="29628" LastEditDate="2018-05-15T17:46:37.860" LastActivityDate="2018-05-16T03:45:39.023" Title="Isothermal vs. adiabatic compression of gas in terms of required energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="7109" PostTypeId="2" ParentId="5718" CreationDate="2013-11-28T11:27:28.227" Score="0" Body="&lt;p&gt;A specific $CN^-$ electrode will detect and measure from about 50 ppb of cyanide in water.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="1300" LastEditDate="2013-11-28T13:09:49.170" LastActivityDate="2013-11-28T13:09:49.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7111" PostTypeId="1" AcceptedAnswerId="7113" CreationDate="2013-11-28T13:21:44.923" Score="2" ViewCount="1347" Body="&lt;p&gt;I've been trying to find out about this for some time now and when I searched online, I couldn't get any proper reasons.&lt;/p&gt;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What is the reason for the characteristic colours of elements?&lt;/strong&gt; For eg: Why does Sodium have a characteristic colour of silvery-white?&lt;/p&gt;&#xA;&lt;p&gt;Is there any solid scientific reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="2425" LastEditorDisplayName="user99496" LastEditDate="2020-10-14T15:39:48.700" LastActivityDate="2020-10-14T15:39:48.700" Title="What is the reason behind the characteristic colours of elements?" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7112" PostTypeId="1" AcceptedAnswerId="7116" CreationDate="2013-11-28T13:31:42.437" Score="6" ViewCount="6486" Body="&lt;p&gt;In a crossed aldol condensation (condensation between aldehyde and a ketone), the primary products are such, that the carbonyl group of the ketone remains intact, and the $\alpha$-hydrogen of ketone participates, i.e. gets removed by the base to form the resonance stabilised nucleophilic &lt;a href=&quot;http://www.chemtube3d.com/model/conjugate_addition/enolates/acetaldehyde/enolization.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;enolate&lt;/a&gt; which later attacks the aldehyde. These are formed together with the self-condensation products, but the reverse products (aldehyde-derived enolate + ketone) are rarely formed.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a look at &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the mechanism&lt;/a&gt;, and also how a base catalyses condensation by favouring enolate formation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the ketone involved in enolate formation and not aldehyde in&#xA;  crossed/mixed aldol-condensation reactions?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EmoDW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2017-10-12T09:42:05.587" LastActivityDate="2017-10-12T09:42:05.587" Title="Selectivity in aldol condensation between pivaldehyde and acetone" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aldol-reaction&gt;&lt;chemoselectivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7113" PostTypeId="2" ParentId="7111" CreationDate="2013-11-28T13:56:26.987" Score="4" Body="&lt;p&gt;Colour is a property of the light reflected or transmitted by any substance. Depending on the substance's characteristics (energy levels of constituent levels, bonding, intermolecular forces), there are certain available energy levels (molecular orbitals, bands etc) for the electrons. Whenever any wide spectrum energy is available to a compound (considering only the visible spectrum), then depending on the various energy gaps between the available to the electrons, different wavelengths will be selectively absorbed and re-emitted, but now in any random direction, therefore the reflected and the transmitted light will be partially deficient n these wavelengths. The absorption and re-emission is very characteristic of all substances, and hence, the wavelengths absorbed and the resultant spectrum (transmitted or reflected) will be characteristic. Depending on this spectrum, various colours can be seen for different compounds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the energy gap between the electron energy level is very high and therefore the light in the visible spectrum cannot excite the electron. No wavelength will be absorbed and it will act as a transparent glass. If say, the gap corresponds to a energy gap of $E=h(4000A)$, then the blue light will be absorbed and the remaining light will be white-blue=yellow. This is the general way in which colors are produced. The energy gaps can be anything, the band gaps in metals, molecular orbital energy difference in covalent compounds, orbital energy difference in isolated elements, or simply the energy difference between the various vibrational modes in a gas/solid.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-11-28T13:56:26.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7114" PostTypeId="2" ParentId="7089" CreationDate="2013-11-28T14:11:46.750" Score="1" Body="&lt;p&gt;The formula is $p =dgh$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d$ is density = $1000\ \mathrm{kg/m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ is gravity = $9.80665\ \mathrm{m/s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ is height = $0.15\ \mathrm m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = 1.0\ \mathrm{g\ cm^{-3}} \times 9.80665\ \mathrm{m\ s^{-2}} \times 15\ \mathrm{cm} = 1471\ \mathrm{N/m^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the pressure you must overcome (about 0.21335 psi) to drink your water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7089/what-pressure-difference-is-needed-across-the-length-of-a-15-cm-vertical-drink/7114#comment10393_7089&quot;&gt;note by @Nicolau Saker Neto&lt;/a&gt; for references.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorDisplayName="user7951" LastEditDate="2017-07-16T13:37:41.717" LastActivityDate="2017-07-16T13:37:41.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7115" PostTypeId="1" CreationDate="2013-11-28T16:43:31.173" Score="21" ViewCount="5671" Body="&lt;p&gt;The chemical structure of a diamond is defined as an endless lattice in which each carbon atom is covalently bonded to four other carbon atoms situated at the four ends of a tetrahedron.  But of course they aren't endless; diamonds have edges and faces.  So what happens when the lattice reaches the edge and there aren't any other carbon atoms to bond to?  What is the &quot;surface layer&quot; of a diamond composed of?&lt;/p&gt;&#xA;" OwnerUserId="3807" LastEditorUserId="1431" LastEditDate="2013-11-30T05:06:50.003" LastActivityDate="2018-03-01T18:56:58.623" Title="What is the edge of a diamond like?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;crystal-structure&gt;&lt;surface-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7116" PostTypeId="2" ParentId="7112" CreationDate="2013-11-28T19:58:36.503" Score="9" Body="&lt;p&gt;Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, I think it's also equally (if not more) valid to consider the reactivity from the perspective of the frontier orbitals. Alkyl substituents tend to slightly lower the energy of the HOMO of the carbonyl (by comparison to hydrogens), but concomitantly they raise the energy of the $\pi^{*}$ LUMO of the carbonyl, into which the electrons of the enolate would flow, resulting in lower electrophilicity (since the energy of the tetrahedral intermediate and transition state resulting from attack by the enolate are going to be correspondingly higher).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-11-28T19:58:36.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7117" PostTypeId="1" CreationDate="2013-11-28T21:18:56.387" Score="17" ViewCount="34783" Body="&lt;p&gt;What are the precise conditions under which benzene undergoes ozonolysis? All anyone can tell me is that it undergoes ozonolysis under &quot;vigorous&quot; conditions. But, what vigorous conditions?&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="5026" LastEditDate="2018-03-06T07:26:26.610" LastActivityDate="2018-04-25T16:51:24.963" Title="Ozonolysis conditions for benzene" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7118" PostTypeId="1" AcceptedAnswerId="7133" CreationDate="2013-11-28T22:44:48.820" Score="4" ViewCount="13352" Body="&lt;p&gt;Sorry if this is not the place for asking this question, but I'm going to do a real life project and don't know where else to ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to prepare and sell snow cones next summer. For making snow cones, I was told by ice cone makers that ice should come from carbonated water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in my country is quite difficult to find a proper ice factory that could produce ice with clean water, I'm planning to make my own ice for snow cones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking to buy bottles of carbonated water, pour them in many plastic bags and freeze those bags in my refrigirator. In that way I could get carbonated ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you think that pouring carbonated water in plastic bags for producing ice would be a good idea? Or could I get adverse effects like water loosing sodium ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3810" LastEditorUserId="4356" LastEditDate="2015-03-04T21:19:00.460" LastActivityDate="2017-12-01T18:40:52.667" Title="Making ice from carbonated water" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7119" PostTypeId="1" AcceptedAnswerId="7123" CreationDate="2013-11-29T00:02:32.857" Score="1" ViewCount="2282" Body="&lt;p&gt;An aqueous solution of sodium hydrogen carbonate corrodes aluminum foil quite readily. How do you explain this?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2013-11-29T12:35:30.880" Title="Corrosion of Aluminum by Baking soda" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7120" PostTypeId="2" ParentId="6511" CreationDate="2013-11-29T05:39:43.403" Score="2" Body="&lt;p&gt;Firstly these metal are not non-reactive with concentrated sulfuric acid- rather they form protective layers whether sulfate or oxide which reduce the speed of the reaction (kinetics).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of heating the sulfuric acid depends on the metal (e.g. steel is good for up to 25C 60-95% sulfuric acid) but there the trans-passive region where metals such as iron no longer have a protective layer. This is due to instability of the oxide/sulfate layer so it dissolves or falls off. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-11-29T05:39:43.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7122" PostTypeId="1" CreationDate="2013-11-29T07:14:38.537" Score="2" ViewCount="108" Body="&lt;p&gt;I'm trying to make a solution of $\ce{TiO2}$ that will coat stuff in as dense a monolayer as possible.  To do so, I'm using nano sized $\ce{TiO}$ particles and trying to get them to get as close to each other as possible.  From what I understand, the way to get these little guys to get close to each other is to get the solution to its isoelectric point thereby eliminating any repulsion caused by charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: what pH should I set my water and $\ce{TiO2}$ powder at to attain my isoelectric point?  I ask because finding such a number has proven to be elusive.  I've found pH numbers from 3.8 to 9 to attain isolectricity for such a solution.  Does anyone have any experience with the stuff such that a true pH to attain isoelectricity revealed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the meter to which I have access for pH also measures mV.  Is there a way to relate mV to the isoelectric point?  Is it as obvious as it seems i.e. keeping adding an acid until my little meter says that mV equals zero?  Or is there something more going on (there always is)?  Are this mV value and the isoelectric point even related?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="874" LastEditorUserId="4231" LastEditDate="2015-03-06T23:23:17.120" LastActivityDate="2015-03-06T23:23:17.120" Title="Isoelectric point pH, mV, for a metal oxide nanopowder" Tags="&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7123" PostTypeId="2" ParentId="7119" CreationDate="2013-11-29T12:35:30.880" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_aluminate&quot; rel=&quot;nofollow&quot;&gt;Sodium aluminate&lt;/a&gt; is produced. The reaction is slower than the usual $NaOH + Al$ and liberates $CO_2$.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-11-29T12:35:30.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7124" PostTypeId="1" AcceptedAnswerId="56985" CreationDate="2013-11-29T13:26:31.060" Score="13" ViewCount="15131" Body="&lt;p&gt;The symbol from the periodic table for Potassium is &quot;K&quot;. This fits my native language (dutch), where Potassium is called &quot;Kalium&quot;. The same question applies to Sodium (Na), which again in Dutch is called &quot;Natrium&quot;. A quick look at Wikipedia showed that Potassium is mainly used as a word in Roman based languages (French, Spanish) and to my surprise in English. What is the history behind this?&lt;/p&gt;&#xA;" OwnerUserId="3813" LastEditorUserId="22" LastEditDate="2013-12-01T01:08:14.300" LastActivityDate="2016-08-05T20:48:53.903" Title="Why is Potassium in French and English not called Kalium?" Tags="&lt;nomenclature&gt;&lt;history-of-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7125" PostTypeId="2" ParentId="7124" CreationDate="2013-11-29T13:34:58.963" Score="5" Body="&lt;p&gt;It derive from the word &quot;potash&quot; (pot ashes - ashes of vegetable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium hydroxide, which was obtained from the ashes of marine plants, are confused over time with potassium hydroxide, from the ashes of land plants and both were called &quot;alqili&quot;, an Arabic word meaning &quot;ashes of plants.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, both substances were differed between &quot;vegetable alkali&quot; for land plants and &quot;mineral alkali&quot; for marine plants. Later, they received the English names &quot;potash&quot; and &quot;soda&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a symbol for potassium is take the first letter of the Latin word &quot;kalium&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastActivityDate="2013-11-29T13:34:58.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7126" PostTypeId="1" CreationDate="2013-11-29T13:56:49.813" Score="1" ViewCount="783" Body="&lt;p&gt;My teacher taught about Rutherford's gold foil experiment today. Sir said that, Rutherford used gold foil of thickness $10^{-7}$ m and concluded atom to be of size $10^{-10}$ m. Sir also said that Rutherford assumed foil to be consisting of only single layer of gold atoms.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought was, if Rutherford used gold foil of $10^{-7}$ thickness, which is greater than the size of atom [$10^{-10}$] which he had concluded. So, gold foil should be made of more than single layer of atoms. But this would be against his assumption. If we assume gold foil to be consisting of more than one layer of gold atoms. Then we will be in need to know the structure of gold, i.e to know whether it has atoms in ABAB.. or AAAA.. or ABCABC... type of arrangement. My book says that gold has ABCABC... or ccp arrangement, it mean that there is no vertical allignment of atoms, thus there will be no vertical allignment of nuclei. It makes significant changes to Rutherford's predictions with respect to deviations. Won't it?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2013-11-29T18:54:27.103" LastActivityDate="2015-04-16T17:49:50.900" Title="Wasn't Rutherford's assumption of gold foil to be made of single layer wrong?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7127" PostTypeId="2" ParentId="7108" CreationDate="2013-11-29T16:19:19.483" Score="6" Body="&lt;p&gt;To solve this, try to use what I call the &quot;graphical apparatus&quot;. For an isothermal process:&#xA;$$&#xA;\begin{align}&#xA;PV&amp;amp;=\text{constant}\\&#xA;P\mathrm{d}V&amp;amp;=-V\mathrm{d}P\\&#xA;\frac{\mathrm{d}P}{\mathrm{d}V}&amp;amp;=-\frac{P}{V}\\&#xA;\end{align}$$&#xA;for adiabatic process:&#xA;$$&#xA;\begin{align}&#xA;PV^\gamma&amp;amp;=\text{constant}\\&#xA;\frac{\mathrm{d}P}{\mathrm{d}V}&amp;amp;=-\gamma\frac PV&#xA;\end{align}$$&#xA;Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (&lt;em&gt;the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure&lt;/em&gt;), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume. &lt;img src=&quot;https://i.stack.imgur.com/ttkqM.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma&amp;gt;1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-05-16T03:45:39.023" LastActivityDate="2018-05-16T03:45:39.023" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7128" PostTypeId="1" AcceptedAnswerId="7131" CreationDate="2013-11-29T16:56:16.770" Score="4" ViewCount="118" Body="&lt;p&gt;I'm supposed to balance a redox equation. I arrive at&#xA;$$&#xA;\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;official&quot; solution is&#xA;$$&#xA;4 \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow 4\ce{NH3} + 2\ce{S} + 2\ce{H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no information about if this is occuring in basic or acidic medium. But IMHO in the first equation the charges and all atoms are balanced, so are these equations equivalent?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-11-29T20:43:22.977" Title="Are these two reactions equivalent?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7129" PostTypeId="2" ParentId="7126" CreationDate="2013-11-29T17:09:43.133" Score="8" Body="&lt;p&gt;You were confused about the result of the experiment. It did not measure the size of the atom, but the size of the nuclei. Nucleus is the area of concentrated positive charge in the core of the atom, and is only a very small fraction the size of the atom itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the result, $10^{-15}\ \mathrm{m}$, was the size of the nuclei, while the size of gold atoms has the order of magnitude of ångströms ($1\ \mathrm{\mathring{A}}=10^{-10}\ \mathrm{m}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurement comes from the analysis deflection angle of alpha particles when hitting the nuclei. The one-layer-of-atom assumption leaded to Rutherford to use a mathematical model for one single collision to analyze the angular distribution. Because the nucleus is so small, even if there were a small number of layers, most alpha particles will still only have one collision and the result will be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the alpha particle was passing through some hundreds of atoms and many particles actually were deflected multiple times. The calculation would be much more complicated. Even if he knew the number of layers of atoms and tried to do it the right way, he might not be able to arrive at the correct number. Of course people back then would not be able to know that, but nevertheless, the existence of nucleus is still evident and Rutherford did almost get the order of magnitude right.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorDisplayName="user7951" LastEditDate="2015-04-16T17:49:50.900" LastActivityDate="2015-04-16T17:49:50.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7131" PostTypeId="2" ParentId="7128" CreationDate="2013-11-29T20:43:22.977" Score="6" Body="&lt;p&gt;Let's start with the question: How do we get from your answer to the official answer?  We can transform your answer into the official answer by first adding 3 ammonium ions to each side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp;amp; \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\&#xA;+ 3 \ce{NH3}+3\ce{H^+} &amp;amp; \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\&#xA;\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp;amp; \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-}&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xA;\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp;amp; \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\&#xA;-\ce{H2O} &amp;amp; \rightarrow - \ce{H2O}\\&#xA;4\ce{NH4+}+2\ce{S^2-}+\ce{O2} &amp;amp; \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O}&#xA;\end{aligned}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Producing what is the 'official' result.  Now we can ask: are the two answers equivalent?  Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation.  I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-11-29T20:43:22.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7132" PostTypeId="1" CreationDate="2013-11-29T20:57:16.540" Score="5" ViewCount="893" Body="&lt;p&gt;Motorola has a &lt;a href=&quot;http://endthelie.com/2013/05/31/google-owned-motorola-reveals-stomach-acid-powered-tablet-that-turns-your-body-into-a-password/#axzz2m4Pmynsx&quot; rel=&quot;nofollow&quot;&gt;device that is powered by stomach acid&lt;/a&gt;. Given that a battery consumes the electrolyte as it generates a charge, how much power could be generated using a device such as this? Let's assume the electrodes are large enough that the electrolyte is the limiting factor in this reaction. Would the body just generate more acid to compensate? Would there be any physiological side-effects from using such a battery?&lt;/p&gt;&#xA;" OwnerUserId="3815" LastEditorUserId="3815" LastEditDate="2014-01-20T17:00:36.253" LastActivityDate="2014-01-20T17:00:36.253" Title="How much electricity could be generated by a person's stomach acid?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7133" PostTypeId="2" ParentId="7118" CreationDate="2013-11-29T22:32:55.650" Score="4" Body="&lt;p&gt;There should be no issue with putting carbonated water in plastic bags and freezing them (considering they're clean and sterile, of course). I imagine the benefit of freezing carbonated water instead of regular clean water has to do with the fact that there is much more dissolved gas in carbonated water. As the water freezes, the gas is expelled from the ice (only the water molecules will solidify), creating many pockets inside the solid ice which would weaken its structural integrity. This would likely make it much easier to shave or break into small pieces for a snow cone. I've never frozen carbonated water myself though. You should be a bit careful because as the carbonated water freezes, the carbon dioxide it releases will create pressure in a closed container, so you may end up with torn plastic bags, ruptured bottles, or a damaged freezer in the worst case.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-29T22:32:55.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7135" PostTypeId="2" ParentId="7115" CreationDate="2013-11-30T04:56:22.960" Score="14" Body="&lt;p&gt;This is a nice question. Although I do not speak with authority on diamonds, most crystal surfaces have several imperfections which mitigate the hanging valencies that would otherwise appear for the atoms (constituents in general) at the surface.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Impurities&lt;/strong&gt;. This is the primary method which helps satisfy the unfulfilled valencies at the surface. Many metallic, non-metallic or compound impurities are usually incorporated throughout the crystal, but are usually high at the surface. By bonding with these excess impurities, the surface carbon atoms can fulfill their valencies. An extremely pure crystal surfaces (lots of &quot;hanging&quot; valencies) will quickly &lt;a href=&quot;http://en.wikipedia.org/wiki/Adsorption&quot; rel=&quot;noreferrer&quot;&gt;adsorb&lt;/a&gt; most materials that it comes in contact with. Chemisorption or physisorption depends on the nature of the crystal, metals prefer physisorption while non-metals have at least unimolecular chemisorption usually.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another way to mitigate surface valencies is &lt;strong&gt;distortion&lt;/strong&gt;. The surface layer might not follow the exact space-filling &lt;a href=&quot;http://en.wikipedia.org/wiki/Bravais_lattice&quot; rel=&quot;noreferrer&quot;&gt;Bravais lattice&lt;/a&gt; structure and may form short range distortions in structure to try to make the unfulfilled valencies satisfied. There might be kinks, depressions, steps etc which at least help in completing the valencies of a few atoms (at the boundary of distortion, say, a &quot;step&quot; structure).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If none of these operate, there is an inherent &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_energy&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;surface energy&lt;/strong&gt;&lt;/a&gt; or surface tension between any two phases which quantifies the desire of the surface atoms to fulfill their hanging valencies to resemble the atoms in bulk which are stably attracted. In case of metallic and ionic crystals, this is comparatively low, but may be high in a covalent crystal like diamond.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In all cases, the surface layer of constituents (actually first few layers) are different from the atoms in bulk in energy, stability, forces acting (unbalanced on the surface and balanced in the bulk) and their structure/orientation.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="194" LastEditDate="2018-03-01T18:56:58.623" LastActivityDate="2018-03-01T18:56:58.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7137" PostTypeId="1" CreationDate="2013-11-30T01:50:51.967" Score="0" ViewCount="68" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At $25 ^\circ C$, $K=0.090$ for the reaction &#xA;  $$\ce{H2O (g) + Cl2O (g) -&amp;gt; 2HOCl (g)}$$ &#xA;  calculate the concentrations of all species at equilibrium for the case: $\pu{1.0 g} \ce{H2O}$ and $\pu{2.0 g} \ce{Cl2O}$ are mixed in a $\pu{1.0 L}$ flask.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know where I go wrong in my calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $\pu{0.0555 M}$ $\ce{H2O}$ and $\pu{0.023 M}$ $\ce{Cl2O}$. So at equilibrium, &#xA;\begin{align}&#xA;  \ce{[H2O]} &amp;amp;= 0.0555 - x,\\&#xA;  \ce{[Cl2O]} &amp;amp;= 0.023 - x\\&#xA;  \ce{[HOCl]} &amp;amp;= 2x\\&#xA;\therefore K &amp;amp;= \frac{[\ce{HOCl}]^2}{[\ce{H2O}][\ce{Cl2O}]}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some moving about I get to &#xA;$$-3.91x^2 - 7.1\times10^{-3}x + 1.15\times10^{-4} = 0$$ &#xA;Where I end up getting $x = -8.74\times10^{-3}$ or $x=8.4\times10^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where it all goes wrong, doing some quick calculations, I see that it does not match what the book says the answer is. So I'm doing something wrong, but I don't know what it is or where it's at.&lt;/p&gt;&#xA;" OwnerUserId="3822" OwnerDisplayName="user100620" LastEditorUserId="4945" LastEditDate="2018-02-21T10:13:14.013" LastActivityDate="2018-02-21T10:27:54.450" Title="Concentrations at equililibrium" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7138" PostTypeId="1" CreationDate="2013-11-30T09:06:09.307" Score="3" ViewCount="72072" Body="&lt;p&gt;Wikipedia has two different pages for &lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy_of_combustion&quot; rel=&quot;nofollow&quot;&gt;enthalpy of combustion&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_of_combustion&quot; rel=&quot;nofollow&quot;&gt;heat of combustion&lt;/a&gt;. Their definition is the same and they use the same symbol $\Delta H_c^\circ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that enthalpies of combustion are always negative as these reactions are exothermic. The heat of combustion tables list positive values for all substances.&#xA;Why do these values differ in sign?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2018-10-11T10:33:51.193" Title="What's the difference between enthalpy of combustion and heat of combustion?" Tags="&lt;thermodynamics&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7139" PostTypeId="2" ParentId="7137" CreationDate="2013-11-30T13:45:22.850" Score="3" Body="&lt;p&gt;I suggest you to use this schematic: &#xA;\begin{array}{l*{5}{c}}\hline&#xA;  \text{Equation} &amp;amp; &#xA;    \ce{H2O} &amp;amp; \ce{Cl2O} &amp;amp; \ce{&amp;lt;=&amp;gt; 2 HClO}\\[2ex]\hline&#xA;  \text{Initial concentration} /\pu{M} &amp;amp; &#xA;    \pu{0.055 mol//L} &amp;amp; \pu{0.023 mol//L} &amp;amp; 0  \\[2ex] &#xA;  \text{Equilibrium variation} /\pu{M} &amp;amp;&#xA;    - x &amp;amp; -x &amp;amp; +2x \\[2ex]&#xA;  \text{Equilibrium concentration} /\pu{M} &amp;amp; &#xA;    \pu{0.055 M} - x &amp;amp; \pu{0.023 M} - x &amp;amp; 2x \\[2ex] \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you have to substitute into the equilibrium equation you mentioned:&#xA;\begin{align}&#xA;  0.090 &amp;amp;= \frac{(2x)^{2}}{(0.055 - x)(0.023 - x)}\\&#xA;  0.090 &amp;amp;= \frac{4x^{2}}{0.001265 - 0.055x - 0.023x + x^2}\\&#xA;  4x^2  &amp;amp;= 0.090 (0.001265 - 0.055x - 0.023x + x^2)\\&#xA;  4x^2  &amp;amp;= 0.090 (0.001265 - 0.078x + x^2)\\&#xA;  4x^2  &amp;amp;= 0.000114 - 0.007x + 0.09x^2\\&#xA;  0     &amp;amp;=3.91x^2 + 0.007x - 0.000114\\&#xA;\end{align}&#xA;So the positive result is:&#xA;$$ x_2 = \frac{-0.007 + \sqrt{4.9\cdot10^{-5} + 0.00178}}{7.8}\approx 0.0046$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An then equilibrium concentration are for $\ce{[H2O]} = 0.055 - 0.0046 = 0.050$, $\ce{[Cl2O]} = 0.018$ and $\ce{[HClO]} = 2x = 0.092$. This are my results.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="4945" LastEditDate="2018-02-21T10:27:54.450" LastActivityDate="2018-02-21T10:27:54.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7140" PostTypeId="1" AcceptedAnswerId="7141" CreationDate="2013-11-30T15:52:17.077" Score="9" ViewCount="365" Body="&lt;p&gt;I'll preface this by saying I haven't had a chemistry class in 10 years and molecular structures are not my thing. But I'm submitting a revision to a journal paper where I need to include figures of molecules and I need to make sure I get it right so I don't get nit-picked apart for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard/guide/list of rules for how to create chemical molecule figures like &lt;a href=&quot;http://en.wikipedia.org/wiki/HMX&quot; rel=&quot;nofollow noreferrer&quot;&gt;the one below&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JgWzs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JgWzs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a comprehensive list of rules but as an example of things I would need to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Should I include the carbon atoms in the ring drawing?  &lt;/li&gt;&#xA;&lt;li&gt;At what angle should the entire molecule be drawn (I've seen rigid-body rotated versions of the same molecule before)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Rather than answering those specific questions, I'd like to know if anybody (IUPAC maybe?) has a standard for drawing these figures.  &lt;/p&gt;&#xA;" OwnerUserId="785" LastEditorUserId="41328" LastEditDate="2017-09-24T10:04:34.423" LastActivityDate="2017-09-24T10:04:34.423" Title="Standards for drawing chemical molecules" Tags="&lt;molecules&gt;&lt;structural-formula&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7141" PostTypeId="2" ParentId="7140" CreationDate="2013-11-30T17:05:29.260" Score="10" Body="&lt;p&gt;As it happens, there are a set of quite involved documents from the &quot;IUPAC Chemical Nomenclature and Structure Representation Division&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iupac.org/publications/pac/keywords/IUPAC_Chemical_Nomenclature_and_Structure_Representation_Division/&quot;&gt;There's a list here&lt;/a&gt;, but to avoid later link-rot, they are currently:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jonathan Brecher Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) &lt;em&gt;2006&lt;/em&gt;, Vol. 78, Issue 10, pp. 1897-1970&lt;/li&gt;&#xA;&lt;li&gt;Richard M. Hartshorn, Evamarie Hey-Hawkins, René Kalio and G. Jeffery Leigh: Representation of configuration in coordination polyhedra and the extension of current methodology to coordination numbers greater than six (IUPAC Technical Report)&#xA;&lt;em&gt;2007&lt;/em&gt;, Vol. 79, Issue 10, pp. 1779-1799&lt;/li&gt;&#xA;&lt;li&gt;W. Mormann and K.-H. Hellwich: Structure-based nomenclature for cyclic organic macromolecules (IUPAC Recommendations 2008) &lt;em&gt;2008&lt;/em&gt;, Vol. 80, Issue 2, pp. 201-232&lt;/li&gt;&#xA;&lt;li&gt;Jonathan Brecher: Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) &lt;em&gt;2008&lt;/em&gt;, Vol. 80, Issue 2, pp. 277-410&lt;/li&gt;&#xA;&lt;li&gt;Andrey Yerin, Edward S. Wilks, Gerard P. Moss and Akira Harada: Nomenclature for rotaxanes and pseudorotaxanes (IUPAC Recommendations 2008) &lt;em&gt;2008&lt;/em&gt;, Vol. 80, Issue 9, pp. 2041-2068&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-11-30T17:05:29.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7142" PostTypeId="1" AcceptedAnswerId="7150" CreationDate="2013-11-30T17:18:58.883" Score="17" ViewCount="1106" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot;&gt;Effective nuclear charge&lt;/a&gt; is a very important concept in chemistry, and is the basis for the qualitative explanation of many observed chemical and physical properties, including several periodic trends. Hence it seems to me to be of significant importance that these values are accurately determined (not only for neutral atoms, but for monoatomic ions too). &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%27s_rules&quot;&gt;Slater's rules&lt;/a&gt; for the calculation of effective nuclear charges, from the 1930s, are in practically any university-level general or inorganic chemistry book, though they are heavily outdated and inaccurate, being mostly of interest due to their simplicity and their historical value. Many books go on to mention updated values for atomic screening constants, calculated by Clementi and Raimondi in &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/38/11/10.1063/1.1733573&quot;&gt;1963&lt;/a&gt; and &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/47/4/10.1063/1.1712084&quot;&gt;1967&lt;/a&gt;. This updated table of effective nuclear charges is significantly more interesting, but not used as often probably because all the values for each orbital in each atom must be looked up individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's been 50 years since Clementi and Raimondi's first publication on the topic, and I have seen no clear reference to a more recent improved calculation. Surely it is not because the values they found are not without flaws. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Screening constants have only been calculated up to $Z = 86$, leaving out a portion of the table, in particular the actinides which have interesting electronic properties&lt;/li&gt;&#xA;&lt;li&gt;No relativistic terms whatsoever were considered in the calculation of the screening constants, which is deleterious especially for atoms beyond about $Z \simeq 70$&lt;/li&gt;&#xA;&lt;li&gt;The calculations are not completely reliable as they are, having not converged well especially some lanthanides, creating odd kinks in screening capacities for electrons in the $4f$ subshell &lt;/li&gt;&#xA;&lt;li&gt;The calculations converged to inaccurate electron distributions in several atoms, such as $[\ce{Kr}]4d^85s^2$ for $\ce{Pd}$ instead of the actual $[\ce{Kr}]4d^{10}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By now we should be able to do their calculations again with much more accuracy and much more speed, both due to the development of physical and mathematical theory behind the calculations, and due to the stupendous increase in affordable computational capacity. So why has there been no improvement to effective nuclear charge calculations? Has no one bothered? Or have there been several attempts, none of which have gained wide acceptance since then (sort of like what happens with electronegativity)?&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="4945" LastEditDate="2015-05-25T06:12:39.877" LastActivityDate="2015-05-25T06:12:39.877" Title="Have there been no advances in the determination of effective nuclear charges since Clementi and Raimondi in the 60s?" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;periodic-trends&gt;&lt;relativistic&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7143" PostTypeId="2" ParentId="7142" CreationDate="2013-11-30T19:34:25.150" Score="6" Body="&lt;p&gt;As you say, effectively nuclear charges are an interesting tool to provide simple handwaving-type explanations of chemical and physical properties. The reality is more complex, and effective nuclear charges are not a useful tool for heavy elements, where more advanced quantum chemical methods are used. In short: the values of effective nuclear charges we have are good enough for what we want to do with them.&lt;/p&gt;&#xA;" OwnerUserId="30" LastActivityDate="2013-11-30T19:34:25.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7145" PostTypeId="1" AcceptedAnswerId="7170" CreationDate="2013-11-30T20:35:32.440" Score="0" ViewCount="158" Body="&lt;p&gt;I have access to a GC/MS, FID, and a spectrophotometer. My first thought was to derivatize with trimethylsilylating agent, then run it by GC. Unfortunately the water is problematic. Heating it is likely to destroy the ascorbic. I don't have access to a manifold freeze-drier. I do have access to a rotovap, but I don't think that works well for removing water. I do have access to a freezer, but no way to vacuum the sample down to 0.006 atmospheres. Titration is out of the question because I don't have a clue what else is in the sample.&lt;/p&gt;&#xA;" OwnerUserId="2490" LastActivityDate="2013-12-02T13:00:59.373" Title="Quantifying Ascorbic Acid in Soft Drink" Tags="&lt;water&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7146" PostTypeId="1" CreationDate="2013-11-30T20:44:09.540" Score="0" ViewCount="863" Body="&lt;p&gt;I have the reaction $\ce{CO_{(g)} + Cl2_{(g)} &amp;lt;=&amp;gt; COCl2_{(g)}}$ where $K=4.5\times10^9$ at $100\ \mathrm{^\circ C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question was to find $K_p$ which I know is $K_p=K(RT)^{\Delta n}$ where $R=0.08206$, $T=273+100=373$ and $n=1-2=-1$. Put in the numbers and get $K_p\approx1.47\times10^8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the second question and this is where I get stuck:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Equal moles of $\ce{CO}$ and $\ce{Cl2}$ are reacted at $100\ \mathrm{^\circ C}$. If the total pressure at equilibrium is 5.0 atm calculate the equilibrium partial pressures of the gases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried several solutions, where I either end up with a wrong answer or hit a dead end. I just don't understand how to solve this.&lt;/p&gt;&#xA;" OwnerUserId="3822" LastEditorUserId="5026" LastEditDate="2018-02-22T02:28:56.873" LastActivityDate="2018-02-22T02:28:56.873" Title="Calculate partial pressure of gases at equilibrium" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7147" PostTypeId="1" AcceptedAnswerId="7148" CreationDate="2013-11-30T21:10:21.843" Score="3" ViewCount="1846" Body="&lt;p&gt;I have yet to see an example of a molecule that's both polar and hydrophobic or that's both non-polar an hydrophilic. If no such molecule exists, why use the terms &quot;hydrophilic&quot; and &quot;hydrophobic&quot; at all if we can just refer to them as being polar or non-polar? If there are such molecules, what are some examples and why do they exhibit non-polarity and hydrophilicty, or polarity and hydrophobicity?&lt;/p&gt;&#xA;" OwnerUserId="2549" LastActivityDate="2013-11-30T21:51:26.603" Title="Why distinguish between polarity and hydrophobicity?" Tags="&lt;water&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7148" PostTypeId="2" ParentId="7147" CreationDate="2013-11-30T21:51:26.603" Score="4" Body="&lt;p&gt;Hydrophilicity/hydrophobicity/lipophilicity/lipophobicity and polarity/non-polarity are terms that overlap to a great extent in simple substances, but which are not strictly equal. The former terms can have a much more involved explanation depending on several factors, even if polarity is usually the largest by far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philicity and phobicity are determined by &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot; rel=&quot;nofollow&quot;&gt;contact angles&lt;/a&gt; between substances. It is possible to vary how a same substance gets wetted by exploiting &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aMicrostruct_superhydrophobic.png&quot; rel=&quot;nofollow&quot;&gt;surface patterning&lt;/a&gt;, which fiddles with the surface tension between the substances in contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting example I can think of is that some large &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorocarbon&quot; rel=&quot;nofollow&quot;&gt;perfluorocarbons&lt;/a&gt; are non-polar, but they are &lt;em&gt;also&lt;/em&gt; lipophobic. It is possible to mix water, oil and a perfluorinated solvent, shake, and then watch as three clearly distinct phases separate.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-11-30T21:51:26.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7149" PostTypeId="1" CreationDate="2013-12-01T00:24:00.043" Score="2" ViewCount="370" Body="&lt;p&gt;I'm studying the XRF method cause I'm about to use a WDXRF spectrometer at college so I'm getting all the information about the physics and chemistry behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just curious about one thing, as you know when a beam of x-ray hits the electron of an inner shell if the energy is just about lower than the electron binding energy, that electron is expelled and the hole is filled with an electron from an outer shell and a photon (fluorescence) is emitted with the same energy as the difference between the shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the electron that has been expelled from the atom, where does it go and what happens to it?&lt;/p&gt;&#xA;" OwnerUserId="3823" LastActivityDate="2015-12-21T02:48:22.980" Title="X-ray fluorescence: where do expelled electrons go?" Tags="&lt;energy&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7150" PostTypeId="2" ParentId="7142" CreationDate="2013-12-01T05:17:04.250" Score="13" Body="&lt;p&gt;I would say the most important reason why no development for effective nuclear charges is that the concept of effective nuclear charges has fallen to disuse in the recent years.  I can think of two important reasons: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Effective nuclear charges is intended to be used to explain electronic structure of &lt;strong&gt;isolated atoms&lt;/strong&gt;.  To this day some atoms can still pose a challenge, but those are the atoms that experience heavy relativistic effects and would need other special treatments than shielding effect along.  However, for the most part, chemists are not interested in atoms anymore.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Its theoretical framework is too simple.  Chemist need their results to be extremely accurate.  A 0.1 eV change can easily be a game changer in an organic reaction.  That, unfortunately, cannot be achieved by a model that has only one adjustable constant. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, the idea still lives on.  For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pseudopotential&quot;&gt;pseudo-potential&lt;/a&gt; or &lt;a href=&quot;http://www.nwchem-sw.org/index.php/ECP&quot;&gt;Effective Core Potentials&lt;/a&gt; can be considered as more general and modern version of effective nuclear charge concept.  Instead of simply changing the charge but keeping the Coulomb potential unchanged, ECP provides the entire effective potential of core electrons. Valence electrons and sometimes even sub-valence electrons has to be excluded from such treatment to be able to produce results with sufficient accuracy. These pseudo-potentials are actively developed and widely used today, can provide chemically relevant accuracy in complex molecular systems and may qualify as &quot;improvement to effective nuclear charge&quot;, although they are evolved into different species already.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-01T05:17:04.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7151" PostTypeId="2" ParentId="7146" CreationDate="2013-12-01T10:24:41.203" Score="4" Body="&lt;p&gt;The equations you need are&#xA;$$K_p=\frac{p_{\ce{COCl2}}}{p_{\ce{CO}} p_{\ce{Cl2}}}=1.47\cdot10^8$$&#xA;and&#xA;$$p_\text{tot}=p_{\ce{COCl2}}+p_{\ce{CO}}+p_{\ce{Cl2}}=5.0~\text{atm}$$&#xA;Additionally, $n_{\ce{CO}}=n_{\ce{Cl2}}\implies p_{\ce{CO}}=p_{\ce{Cl2}}$, making the system solvable.&lt;/p&gt;&#xA;" OwnerUserId="1774" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T21:34:05.527" LastActivityDate="2015-05-01T21:34:05.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7153" PostTypeId="2" ParentId="7138" CreationDate="2013-12-01T16:03:08.137" Score="3" Body="&lt;p&gt;Looking at Wikipedia for the definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;standard enthalpy of combustion&lt;/strong&gt; is the &lt;strong&gt;enthalpy change&lt;/strong&gt; when &lt;strong&gt;one mole of a reactant&lt;/strong&gt; completely burns in excess oxygen &lt;strong&gt;under standard thermodynamic conditions&lt;/strong&gt; (although experimental values are usually obtained under different conditions and subsequently adjusted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of combustion $(\Delta H_c^\circ)$ is the &lt;strong&gt;energy released as heat&lt;/strong&gt; when &lt;strong&gt;a compound&lt;/strong&gt; undergoes complete combustion with oxygen &lt;strong&gt;under standard conditions.&lt;/strong&gt; The chemical reaction is typically a hydrocarbon reacting with oxygen to form carbon dioxide, water and heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On first inspection, the two seem the same, however, there is a very slight different use in terminology and wording.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) What are standard conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referring back to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard conditions for temperature and pressure are standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm, 1 bar)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you know, the standard enthalpy of combustion can be referred to as as $\Delta H ^{\circ} _{\mathrm{total}}$ whose units are in kJ. The molar enthalpy of combustion is how much energy is released per kJ/mol. How is the standard enthalpy of combustion found?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the individual enthalpies of formation, and Hess's law, we can calculate the standard enthalpy of combustion, or $\Delta H ^{\circ} _{\mathrm{total}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_{reaction}^\ominus = \sum \Delta H_{\mathrm f \,(products)}^{\ominus} - \sum \Delta H_{\mathrm f \,(reactants)}^{\ominus}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard enthalpies of formation, or the negative values are located here &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)&lt;/a&gt; and as well, generally in your textbook in an appendix along with the standard entropy of formation, and Gibbs free energy of formation, as all three are state functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Why do the signs differ in the values from your textbook compared to the table in Wikipedia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the heat of combustion, it is generally expressed in units of higher heating value, lower heating value, and gross heating value. As the article states in Wikipedia, the heating value (or energy value or calorific value) of a substance, usually a fuel or food (see food energy), is the amount of heat released during the combustion of a specified amount of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the sign difference that I see is due to different terminology, they are specifically referring to units of HHV, LHV, and GHV and expressing those in the table below. Although, heat of combustion may also be calculated as the difference between the heat of formation $\big( \Delta H_f^\circ \big)$ of the products and reactants just as the standard enthalpy of combustion, it seems to me that the heat of combustion is specialized in a way to be applied to fuels, and as such, different from the standard enthalpy of combustion that we are accustomed to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps someone could clarify a little more on this?&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2013-12-01T16:13:45.487" LastActivityDate="2013-12-01T16:13:45.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7154" PostTypeId="1" AcceptedAnswerId="7158" CreationDate="2013-12-01T17:37:54.043" Score="0" ViewCount="100" Body="&lt;p&gt;In describing proteins, I've written &quot;the proteins detached from the [Ni-IDA] matrix and &lt;strong&gt;were eluted&lt;/strong&gt; from the column&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use &quot;were&quot; or can something simply &quot;elute&quot; ?&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="1939" LastEditDate="2013-12-01T19:43:42.627" LastActivityDate="2013-12-01T19:59:20.193" Title="The proteins &quot;eluted&quot; or &quot;were eluted&quot;?" Tags="&lt;purification&gt;&lt;terminology&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7155" PostTypeId="1" AcceptedAnswerId="7157" CreationDate="2013-12-01T17:50:42.803" Score="6" ViewCount="980" Body="&lt;p&gt;A neat party trick (typically for outdoors) is to build a pyramid of tealights, let them burn for a while and then pour a bit of water over it and see a large jet flame shoot out (see e.g. &lt;a href=&quot;http://www.youtube.com/watch?v=rrbeswlpj_A#t=46&quot;&gt;this youtube video&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was always told that this is caused by the water spreading the liquid wax out allow more of it to burn at the same time, but I am not completely satisfied with that answer because, in my opinion, that cannot explain the large jet flame. Can someone explain why this jet flame occurs? Is it the wax vapour that catches fire, which can form at 370 $^o\ce{C}$, and is this the reason you need to build a close packed pyramid? But what is the role of the water then?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2013-12-01T18:00:38.710" LastActivityDate="2013-12-01T18:40:40.740" Title="Why does water cause a jet flame when poured over a bunch of tealights?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7156" PostTypeId="2" ParentId="7155" CreationDate="2013-12-01T18:11:36.977" Score="1" Body="&lt;p&gt;I think it's a mixture of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually get the same sort of flame just using a pyramid of tealights, but it takes a while longer. I think the role of the water is get the process started, throwing around some wax vapour that can then ignite and self-sustain.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-12-01T18:11:36.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7157" PostTypeId="2" ParentId="7155" CreationDate="2013-12-01T18:40:40.740" Score="13" Body="&lt;p&gt;The wax doesn't need to become gaseous to cause the column of flame (if that's what you were thinking). Water, when in contact with the liquid wax at a couple of hundred degrees above its boiling point, will flash-boil in a very chaotic manner as if it were &quot;exploding&quot; (&quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aHeat_transfer_leading_to_Leidenfrost_effect_for_water_at_1_atm.png&quot;&gt;transition boiling&lt;/a&gt;&quot; in this graph). The water closest to the wax very quickly vaporises and tries to escape, but is temporarily blocked by the layer of liquid water above it, so the system ends up being vigorously churned to make way for the vapour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the turmoil will make tiny &lt;em&gt;liquid&lt;/em&gt; wax droplets airborne, as if wax were being shot out of an atomizer, creating a wax aerosol. Suddenly, the same amount of wax has its surface area increased tremendously and is exposed to much more oxygen. Even though the wax in the aerosol quickly loses heat and may become solid again in a fraction of a second, if even a small region of the wax aerosol reaches ignition temperature, the flame quickly propagates as a huge amount of heat is dumped at once in the vicinity and triggers the ignition of all the airborne wax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trick is effectively the same as what happens if you try to put out a grease fire with water. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-12-01T18:40:40.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7158" PostTypeId="2" ParentId="7154" CreationDate="2013-12-01T19:59:20.193" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;eˈlute&lt;/strong&gt;, &lt;em&gt;v.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;trans.&lt;/em&gt; To wash out, cleanse. Now spec.  [after German eluieren, in same sense] to wash (adsorbed matter) away from the substance that has adsorbed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It's transitive, so you elute something, rather than something eluting. Your original &amp;quot;were eluted&amp;quot; is correct.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-12-01T19:59:20.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7159" PostTypeId="1" AcceptedAnswerId="7182" CreationDate="2013-12-01T22:27:35.740" Score="3" ViewCount="4893" Body="&lt;p&gt;I'm studying radioisotopes at the moment and balancing nuclear reactions isn't making sense in that more matter is coming out of the equation in negative β⁻ decay equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{_6^{14}C -&amp;gt; _7^{14}N + e- + \bar{\nu_e}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the original element has decayed into a new element with an unchanged mass number but an atomic number that has increased by one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In positive β⁺ decay equations, it makes sense: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{_{12}^{23}Mg -&amp;gt; _{11}^{23}Na + e+ + \nu_e}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you create something with the same mass but with another proton?&lt;/p&gt;&#xA;" OwnerUserId="3828" LastEditorUserId="41328" LastEditDate="2020-04-15T06:44:09.337" LastActivityDate="2020-04-15T06:45:39.473" Title="In nuclear chemistry, how does a neutron split to form a proton and an electron?" Tags="&lt;energy&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;nuclear&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7160" PostTypeId="2" ParentId="7115" CreationDate="2013-12-01T23:04:08.713" Score="9" Body="&lt;p&gt;At the surface of a diamond, the carbon atoms are generally bound to various other atoms (or groups of atoms) with lower &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;valence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, pretty much anything that could act as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Substituent&quot; rel=&quot;noreferrer&quot;&gt;substituent group&lt;/a&gt; in organic chemistry could also be bonded to the surface of diamond, but in practice, you're most likely to see simple groups with low reactivity, mostly either plain hydrogen or various oxygen-containing groups (I suspect mainly hydroxyls and/or ketones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Material_properties_of_diamond&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; (which itself cites Harlow 1998, &lt;em&gt;The nature of diamonds&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although the diamond lattice itself is very stable and inert, these surface groups can participate in chemical reactions, and indeed, it's possible to alter the composition of the surface layer by reacting it with appropriate chemicals.  For an example, see &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1998/FT/A707905B#!divAbstract&quot; rel=&quot;noreferrer&quot;&gt;this random paper I found on Google&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="283" LastActivityDate="2013-12-01T23:04:08.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7161" PostTypeId="2" ParentId="7159" CreationDate="2013-12-01T23:45:42.770" Score="1" Body="&lt;p&gt;What you are missing is that the atomic mass has not changed — mass is conserved — the top number on the isotope notation 14 for both carbon and nitrogen in first case (in fact, it goes down very slightly, but they rounded it up). The increasing atomic number (bottom number) just means a neutron became a proton. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="41328" LastEditDate="2020-04-15T06:45:39.473" LastActivityDate="2020-04-15T06:45:39.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7162" PostTypeId="1" AcceptedAnswerId="7222" CreationDate="2013-12-02T01:12:16.970" Score="-1" ViewCount="4112" Body="&lt;p&gt;Why should colorimetric titration of weak acid with a weak base (or vice versa) not be done in laboratory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I draw a pH graph for this?&lt;/p&gt;&#xA;" OwnerUserId="1977" LastEditorUserId="4945" LastEditDate="2018-10-08T09:37:31.817" LastActivityDate="2018-10-08T09:37:31.817" Title="Why should colorimetric titration of a weak acid with a weak base not be done in laboratory?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-02T04:21:32.700" ContentLicense="CC BY-SA 4.0" />
  <row Id="7163" PostTypeId="1" CreationDate="2013-12-02T02:09:32.483" Score="-2" ViewCount="1671" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{15.8 g}$&lt;/span&gt; strip of zinc metal was placed in &lt;span class=&quot;math-container&quot;&gt;$\pu{100 ml}$&lt;/span&gt; of silver nitrate. When the reaction was complete, the strip of zinc had a mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{13.1 g}$&lt;/span&gt;. What was the concentration of the silver nitrate solution? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am confused on writing the equation for this problem, because doesn't single displacement occur? So, how can zinc metal still be there?&lt;/p&gt;&#xA;" OwnerUserId="2594" LastEditorUserId="4945" LastEditDate="2018-10-08T09:48:13.017" LastActivityDate="2018-10-08T10:01:09.943" Title="How can zinc metal still be present after the reaction?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-03-23T15:49:22.963" ContentLicense="CC BY-SA 4.0" />
  <row Id="7164" PostTypeId="2" ParentId="7163" CreationDate="2013-12-02T03:08:46.930" Score="3" Body="&lt;p&gt;As we can clearly see, the silver nitrate is the limiting reactant in this problem as there are still 13.1 grams of zinc left that have not reacted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that &lt;span class=&quot;math-container&quot;&gt;$\pu{15.8 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; metal - &lt;span class=&quot;math-container&quot;&gt;$\pu{13.1 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; metal = &lt;span class=&quot;math-container&quot;&gt;$\pu{2.7 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; metal has reacted with all of the &lt;span class=&quot;math-container&quot;&gt;$\ce{100 mL}$&lt;/span&gt; silver nitrate solution with &lt;span class=&quot;math-container&quot;&gt;$\pu{13.1 g}$&lt;/span&gt; excess zinc metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are however given &lt;span class=&quot;math-container&quot;&gt;$\pu{100 mL}$&lt;/span&gt; of silver nitrate, and we are interested to find out how much product is formed. Looking online, the density of silver nitrate is &lt;span class=&quot;math-container&quot;&gt;$\pu{5.35 g/cm3}$&lt;/span&gt; and keeping in mind that &lt;span class=&quot;math-container&quot;&gt;$\pu{1 cm3} = \pu{1 mL}$&lt;/span&gt;. Note that since the silver nitrate is in solution, the density of the silver nitrate in solution is unknown, but for references sake, we will refer to the density of the solid silver nitrate which is the 5.35 grams per cubic centimeter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho = m/V,$$&lt;/span&gt;&#xA;we can find the mass by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\pu{5.35 g/cm^3} = \frac{m}{\pu{100 mL} \times \pu{1 cm^3/mL}},$$&lt;/span&gt;&#xA;therefore our mass of silver nitrate (theoretically) is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m = \pu{535 g}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In looking at the reactivity series, zinc is higher than silver, so single displacement does occur just as you predicted and therefore zinc &quot;should not&quot; appear as it then performs a redox reaction with the nitrate in silver nitrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for the reactivity series of the metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we write the chemical equation for this reaction we can see that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{AgNO3(aq) + Zn(s) -&amp;gt; ZnNO3(aq) + Ag(s)},$$&lt;/span&gt;&#xA;and the balanced equation is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2AgNO3(aq) + Zn(s) -&amp;gt; Zn(NO3)2(aq) + Ag(s)}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's quantitatively calculate what would happen if we reacted each of our reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that zinc is the limiting reactant and we have silver nitrate in excess:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\pu{15.8 g}\ \ce{Zn} \times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{65.38 g}\ \ce{Zn}} &#xA;\times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{1 mol}\ \ce{Ag}} &#xA;\times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{107.9 g}\ \ce{Ag}} &#xA;= \pu{0.002240 g}\ \ce{Ag}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that silver nitrate is the limiting reactant and we have zinc in excess:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\pu{535 g}\ \ce{AgNO3} &#xA;\times \frac{\pu{1 mol}\ \ce{AgNO3}}{\pu{169.9 g}\ \ce{AgNO3}} &#xA;\times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{2 mol}\ \ce{AgNO3}} &#xA;\times \frac{\pu{107.9 g}\ \ce{Ag}}{\pu{1 mol}\ \ce{Ag}} &#xA;= \pu{169.9 g}\ \ce{Ag}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, we expect to see aqueous zinc nitrate to form, and solid silver to precipitate. In looking at my example, all of the zinc is consumed in the reaction so we expect to not see any zinc metal left in the first reaction since that is the limiting reactant, because there is not enough zinc metal to react with the silver nitrate. In the second reaction, we see that silver nitrate is in excess because we have a much greater yield for silver, however, this reaction is &quot;limited&quot; due to not having enough zinc metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in &lt;strong&gt;your&lt;/strong&gt; case, 13.1 grams of zinc metal are still remaining. Therefore, zinc must be the reactant &lt;strong&gt;in excess&lt;/strong&gt; and thus the mass of silver nitrate is &lt;strong&gt;all consumed&lt;/strong&gt; in the reaction. Therefore although you expected the zinc metal to bind to the nitrate in the redox reaction, it does! However, we have too much zinc metal and not enough silver nitrate, therefore we still see some zinc metal left along with our usual products of zinc nitrate and silver. I just did a basic stoichiometric limiting reactant problem so you can quantitatively see how this type of reaction works. In reality, I do not know the density of the silver nitrate solution, and thus the mass of the silver nitrate, which is much lower in the context of your problem.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2018-10-08T10:01:09.943" LastActivityDate="2018-10-08T10:01:09.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7170" PostTypeId="2" ParentId="7145" CreationDate="2013-12-02T13:00:59.373" Score="4" Body="&lt;p&gt;Titration is in the question because ascorbic acid is likely the principal, if not only, &lt;strong&gt;reducing agent&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using iodine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, [this site] describes a titration using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Starch_indicator&quot; rel=&quot;nofollow&quot;&gt;iodine-starch&lt;/a&gt; redox indicator (the presence of iodine and starch together creates a blue/black color).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The active oxidant is $\ce{I3-}$, formed from $\ce{I2}$ and $\ce{I-}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I2 + I- &amp;lt;=&amp;gt; I3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redox reaction then is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H8O6 + I3- -&amp;gt;C6H6O6 + 3I- + 2H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a small amount of a saturated solution of soluble starch to your soft drink sample and then titrate against a known solution that is equimolar in $\ce{NaI}$ or \$\ce{KI}$ and $\ce{I2}$. When the dark color persists in your soft drink sample, you have your end point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.outreach.canterbury.ac.nz/chemistry/documents/vitaminc_iodine.pdf&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; provides a full procedure (PDF).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using $\ce{KMnO4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potassium permanganate is its own indicator, since $\ce{KMnO4}$ solutions are purple and the reduced version is colorless (technically an insoluble brown solid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your soft drink is dark-colored, you may want to dilute it so that you can see the color change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redox equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2MnO4- +6H+ 5C6H8O6-&amp;gt; 2Mn^{2+} + 8H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spectrophotometrically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ascorbic acid absorbs UV light (but then so does other things in your soft drink, like the preservative sodium benzoate). However, you could use ascorbic acid's reducing power to turn on an absorbance in the visible. Here is &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jccs.200500072/abstract&quot; rel=&quot;nofollow&quot;&gt;one example&lt;/a&gt;. This paper may be behind a paywall, but the abstract contains the essential idea. In this paper, the authors describe a system using copper(II) and phenanthroline(phen). Ascorbic acid reduces copper(II) to (copper(I), which creates a coordination complex with phenanthroline that has $\lambda_{\text{max}}=460 \ \text{nm}$. The authors claim a detection limit of 40 $\mu\text{g/L}$, which is pretty good (40 parts per million). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Cu^{2+} + C6H8O6 -&amp;gt; 2Cu+ + C6H6O6 +2H+}$$&#xA;$$\ce{Cu+ + 3phen -&amp;gt;Cu(phen)3+}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another colorimetric assay&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Folin-Ciocalteu_reagent&quot; rel=&quot;nofollow&quot;&gt;Folin-Ciocalteu&lt;/a&gt; assay is another nice colorimetric assay. Reducing agents react with the phosphomolybdate and phosphotungstate compounds in the reagent mixture and produce a blue color. This method needs to be calibrated against some ascorbic acid standards because the redox chemistry is complex and not fully understood. The absorbance max of the reduced FC reagent is in the 670ish range, if I remember correctly.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-12-02T13:00:59.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7173" PostTypeId="1" CreationDate="2013-12-02T16:40:08.237" Score="3" ViewCount="3509" Body="&lt;p&gt;Let $\ce{R-OH}$ be a group with a $pK_a$ of 16. Are the following two statements correct?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In principle, the group can protonate another acid with $pK_a &amp;gt; 16$, and so (put differently), any &quot;acid&quot; with $pK_a &amp;gt; 16$ can be used to &quot;deprotonate&quot; the group&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I had a table with bases were only $pK_b$ values were given, in order to select these bases from the table which are able to deprotonate the group, I would need to choose the bases with $pK_b &amp;lt; -2$ because the $pK_a$ values of their conjugate acids (i.e. $pK_{a\text{H}}$) will be 16 or higher.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="4945" LastEditDate="2014-10-07T01:53:57.513" LastActivityDate="2014-10-07T01:53:57.513" Title="De-/protonation of R-OH group by acids and bases" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7174" PostTypeId="1" AcceptedAnswerId="7191" CreationDate="2013-12-02T17:50:06.077" Score="18" ViewCount="81884" Body="&lt;p&gt;So I was checking the structure of nitric acid in Wikipedia, however I couldn't quite fathom why it looked like that because it seemed to contradict the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N6XgC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we draw it like the one on the right, we get rid of formal charges and the structure is said to be more &quot;stable&quot;. Why does this concept not work in this case?&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorDisplayName="user7951" LastEditDate="2015-05-22T17:37:50.657" LastActivityDate="2017-12-05T17:52:33.153" Title="The Lewis structure of HNO3" Tags="&lt;lewis-structure&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7175" PostTypeId="2" ParentId="7174" CreationDate="2013-12-02T18:09:55.450" Score="1" Body="&lt;p&gt;Even though you may have a filled octet, if the central atom has a positive formal charge, we generally form double bonds until the formal charge is reduced to 0 as close as possible as that will be the more stable configuration. However, there are some exceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take a look at the chlorate ion. For chlorate, you would expect to leave four single bonded oxygens to the chlorine, however, we are left with a formal charge of +3 on the chlorine and -1 on each of the oxygens. Therefore, we form double bonds until the formal charge is removed and are left with only a formal charge of -1 on the single bonded oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: looking at your question the structure on the left is preferred even though the one on the right has a lower formal charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps Wikipedia has the wrong structure for the Nitric acid page? It looks like all the oxygen atoms are single bonded in the figure to the top right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nitric_acid&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Nitric_acid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for images, Wikipedia does have the right resonance structures with the double bonded oxygen, however, there is only one double bond contrary to what we expect which is two double bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Nitric-acid-resonance-A.png&quot; rel=&quot;nofollow&quot;&gt;https://commons.wikimedia.org/wiki/File:Nitric-acid-resonance-A.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="1878" LastEditDate="2013-12-02T20:14:28.613" LastActivityDate="2013-12-02T20:14:28.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7176" PostTypeId="1" AcceptedAnswerId="7178" CreationDate="2013-12-02T18:57:03.727" Score="13" ViewCount="4043" Body="&lt;p&gt;Where can one find examples of raw data for an 1H-NMR spectrum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may sound as an odd request, so I'll give some context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found lots of images of spectra such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzPTx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where you can see an image of the peaks. What I am interested is a spreadsheet&#xA;of the $(x,y)$-values used to draw the spectre. The data is needed for a student mathematics/chemistry project (high school). The mathematics section will be on calculating the area under a curve using numerical methods. To get a connection to the chemistry part, I'd like the student to use realistic data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am aware that it is the ratio of the areas of the peaks that are interesting, so the actual scaling of the y-axis is unimportant.)&lt;/p&gt;&#xA;" OwnerUserId="3841" LastEditorUserId="81" LastEditDate="2015-12-20T12:29:48.793" LastActivityDate="2019-12-11T16:41:42.707" Title="Sources for raw 1H-NMR data?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7177" PostTypeId="2" ParentId="7174" CreationDate="2013-12-02T20:11:36.347" Score="6" Body="&lt;p&gt;Well! here is a short quick reply..!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was pretty sure that the lewis stucture of $\ce{HNO3}$ would be the one with 0 formal charge and than I found &lt;a href=&quot;http://wiki.answers.com/Q/Calculate_the_net_formal_charge_of_this_species_HNO3&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;. Though this link only shows how to calculate formal charge on $\ce{HNO3}$ but it gives a little hint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly a thought came to my mind that if you will focus on Resonance, there will be double bond character on both bonds between N &amp;amp; O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/6/69/Nitric-acid-3D-balls-B.png&quot; alt=&quot;HNO3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-12-02T20:33:08.573" LastActivityDate="2013-12-02T20:33:08.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7178" PostTypeId="2" ParentId="7176" CreationDate="2013-12-02T23:27:21.247" Score="13" Body="&lt;p&gt;I might suggest that you write a short program to simulate the spectra.  The problem with using real data is the amount of points (upwards of 32k) that make publishing CSV files of raw spectra unreasonable.  Below is a quick attempt to give you what you are looking for.  I apologize for those who do not use Mathematica; however I tried an Excel version and it became unwieldy very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have three functions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Clear[feature];&#xA;feature[shift_, n_, coupling_, intensity_] := Module[{},&#xA;  Plus @@ &#xA;   Table[Binomial[n, k]*intensity/(n + 1)*&#xA;     Cos[(x)*(shift + coupling (k - n/2))], {k, 0, n}]]&#xA;Clear[fid]&#xA;makefid[feature_, broadening_] := Module[{},&#xA;  Exp[-broadening x] Plus @@ {feature}]&#xA;Clear[transform]&#xA;transform[fid_] := Module[{resolution = 4096, $y, $x},&#xA;  $y = Re[Fourier[Table[fid, {x, 0, resolution, 0.1}]]];&#xA;  $y = Take[$y, Round[Length[$y]/2]];&#xA;  $x = Range[Length[$y]]/(resolution/(2 Pi));&#xA;  Transpose@{$x, $y}&#xA;  ]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The function &lt;code&gt;feature&lt;/code&gt; creates a peak with a chemical shift as the first argument, the splitting as the second, the coupling constant (use 0.1 for starters) and the intensity (its integral).  &lt;code&gt;makefid&lt;/code&gt; creates a simulated FID based on the feature that it is passed and &lt;code&gt;transform&lt;/code&gt; makes the spectrum.  It can be used like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = feature[1.2, 2, 0.1, 3];&#xA;b = feature[2, 0, 0.1, 3];&#xA;c = feature[4.1, 3, 0.1, 2];&#xA;fid = makefid[Plus @@ {a, b, c}, 0.01];&#xA;ListLinePlot[transform[fid], PlotRange -&amp;gt; {{0, 5}, All}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The features &lt;code&gt;{a,b,c}&lt;/code&gt; are simulated ethyl acetate yielding the following spectrum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dgmEM.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data you want is in &lt;code&gt;transform[fid]&lt;/code&gt; which can be exported as a CSV file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Export[&quot;myfile.csv&quot;,transform[fid]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which will make it accessible to other software packages and you can perform your math lesson plan.  To make sure this is working properly in Mathematica, we can make a function out of our data and numerically integrate the three peaks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;data = Interpolation[transform[fid]];&#xA;NIntegrate[data[x], {x, 0.9, 1.5}]&#xA;NIntegrate[data[x], {x, 1.5, 2.2}]&#xA;NIntegrate[data[x], {x, 3.7, 4.5}]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This produces areas of 0.32, 0.25, and 0.33, respectively, which is what we expected.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-12-02T23:27:21.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7179" PostTypeId="2" ParentId="7176" CreationDate="2013-12-03T03:23:23.053" Score="11" Body="&lt;p&gt;Raw NMR data is usually stored in proprietary format, which is vendor specific, and even though most NMR data processing software can handle multiple formats, it is not a straight forward process to get data into an x-y format for use with other software (Excel, Matlab, Origin etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;However, to answer your question:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few sites where you can access raw 1H NMR data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly the most useful is the &lt;a href=&quot;http://harned.chem.umn.edu/&quot;&gt;Spectral Database for Instructors&lt;/a&gt;, which has a small database of 1H, 13C and 2D spectra available for download. The data are in Bruker format (fid, acqus for 1D and ser acqus, acqu2s for 2D). This site has some useful information about how to access and process this data, for that is another question altogether!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can trawl your way through &lt;a href=&quot;http://www.chemspider.com/&quot;&gt;Chemspider&lt;/a&gt;; you may get lucky with an odd spectrum or two, but I don't think this is worth the effort&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few old sources of NMR data in jcamp format that still exist. Jcamp was/is the universal data format proposed for many data sources (widely used for IR, UV for example). All major NMR software has the potential to read and write in jcamp format, although you can run in to problems using different jcamp version. Examples of sites with jcamp data include: &lt;a href=&quot;http://www.acornnmr.com/sample_data_files.htm&quot;&gt;AcornNMR&lt;/a&gt;, &lt;a href=&quot;http://pslc.uwsp.edu/&quot;&gt;Polymer Science Learning Center&lt;/a&gt;, &lt;a href=&quot;http://wwwchem.uwimona.edu.jm/spectra/testdata/&quot;&gt;JCamp test data&lt;/a&gt; at the University of West Indies (with links to what jcamp is all about) are just some. You can probably google nmr jcamp to find more. I can't vouch that these data are in the public domain, and I have omitted a couple of databases that I know are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want real examples of real molecules run by real researchers (and not just for teaching purposes), you probably can't go past Open Science researchers like the &lt;a href=&quot;http://opensourcemalaria.org/&quot;&gt;Open Source Malaria project&lt;/a&gt;. You can follow the &lt;a href=&quot;http://malaria.ourexperiment.org/osm_procedures/group/Summary%20Page&quot;&gt;links to the compounds&lt;/a&gt; and save the raw NMR data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a lot will depend on the types of molecules you would like to get data for, and how many different structures you need. It may be easier to approach a friendly NMR Facility Manager at your nearby College or University and ask very nicely for some data for some simple molecules. I'd be happy to provide further information about that if you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your data, you will need to be able to convert it into a format that you can make sense of. Again the best approach here will depend on how many data sets you need and how long you are likely to use the data for. You can download demo versions of NMR processing software like &lt;a href=&quot;http://mestrelab.com/&quot;&gt;Mestrenova&lt;/a&gt; or &lt;a href=&quot;http://home.cc.umanitoba.ca/~wolowiec/spinworks/&quot;&gt;Spinworks&lt;/a&gt; which should allow to read in raw data formats and export in a more desirable (ASCII X,y) format.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-12-03T03:23:23.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7180" PostTypeId="1" CreationDate="2013-12-03T04:54:45.797" Score="2" ViewCount="71900" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I worked the problem I got $\pu{29.9 g/mol}$ is that right?&lt;/p&gt;&#xA;" OwnerUserId="3843" LastEditorUserId="4945" LastEditDate="2017-11-27T11:50:34.067" LastActivityDate="2017-11-27T12:14:43.623" Title="How do I find molar mass using gas laws?" Tags="&lt;gas-laws&gt;" AnswerCount="4" CommentCount="0" ClosedDate="2015-06-15T02:36:45.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="7181" PostTypeId="2" ParentId="7180" CreationDate="2013-12-03T06:09:13.487" Score="2" Body="&lt;p&gt;From gas equation:&#xA;$$ P V = n  R  T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $n = m / M$ &#xA;and  density $d = m / V$,&#xA;$$P = d  R  T / M,$$&#xA;so&#xA;$$M = 0.977 \cdot 0.082 \cdot 373 / (710/760) = \pu{32.1 g/mol}.$$&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="4945" LastEditDate="2017-11-27T12:14:43.623" LastActivityDate="2017-11-27T12:14:43.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7182" PostTypeId="2" ParentId="7159" CreationDate="2013-12-03T06:27:08.320" Score="2" Body="&lt;p&gt;One neutron has changed into a proton, that's what has happened. Our convention in chemistry is to identify nuclear species by the proton number (or the &quot;atomic number&quot;) and that is why a new nuclide with an &lt;em&gt;increased&lt;/em&gt; atomic number is formed. Both the &lt;span class=&quot;math-container&quot;&gt;$\beta^+$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta^-$&lt;/span&gt; decay follow identical paths: in &lt;span class=&quot;math-container&quot;&gt;$\beta^-$&lt;/span&gt; decay, a neutron changes into a proton (thus giving &lt;em&gt;no change&lt;/em&gt; in mass number and &lt;em&gt;increase&lt;/em&gt; in atomic number), whereas in &lt;span class=&quot;math-container&quot;&gt;$\beta^+$&lt;/span&gt; decay, a proton decays into a neutron giving &lt;em&gt;no change&lt;/em&gt; in the mass number and a &lt;em&gt;decrease&lt;/em&gt; in the atomic number.&lt;/p&gt;&#xA;" OwnerUserId="3844" LastEditorUserId="41328" LastEditDate="2020-04-15T06:33:40.800" LastActivityDate="2020-04-15T06:33:40.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7183" PostTypeId="1" CreationDate="2013-12-03T07:04:35.197" Score="0" ViewCount="5360" Body="&lt;p&gt;Consider the following reaction : $\ce{H2 + Cl2 -&amp;gt; 2HCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) If you have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine at STP, how many liters of HCl would you produce?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) What would be the limiting reagent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) How many moles of excess reagent would remain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Molarity = moles/Liters&#xA;to check for $\ce{H2}$: M = 2/15.00 = .13334&#xA;to check for $\ce{Cl2}$: M = (35.45 x 2)/10.00 = 7.09&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I do not think that is correct. can somebody show me the correct way to do it.&lt;/p&gt;&#xA;" OwnerUserId="3843" LastEditorUserId="418" LastEditDate="2014-01-31T22:46:49.253" LastActivityDate="2015-05-19T06:29:24.083" Title="Gas Law problems with stoichiometry" Tags="&lt;homework&gt;&lt;gas-laws&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7184" PostTypeId="1" CreationDate="2013-12-03T07:10:23.320" Score="0" ViewCount="184" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A sample of solid magnesium reacts with excess $\ce{HCl}$ solution to form hydrogen gas. The sample of gas measures $1.628\ \mathrm{L}$ at $22.0\ \mathrm{^\circ C}$ and $0.930\ \mathrm{atm}$ of pressure.&lt;br&gt;&#xA;  a) how many moles of gas were collected?&lt;br&gt;&#xA;  b) how many grams of magnesium did you start out with?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First I stated by writing the chemical equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg + HCl -&amp;gt; H2 + MgCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then convert the temperature to kelvin and pressure to atm and volume to L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = 1.628\ \mathrm{L}$&lt;br&gt;&#xA;$T = 295.15\ \mathrm{K}$&lt;br&gt;&#xA;$P = 0.930\ \mathrm{atm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PV = nRT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.930\ \mathrm{atm} \cdot 1.628\ \mathrm{l} = n \cdot 0.08206\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}} \cdot 295.15\ \mathrm{K}$&lt;br&gt;&#xA;which will give that $n = 16.0\ \mathrm{mol}$. Did I do this correctly?&lt;/p&gt;&#xA;" OwnerUserId="3843" LastEditorUserId="5026" LastEditDate="2018-03-06T14:12:16.390" LastActivityDate="2018-03-06T14:12:16.390" Title="Gas law and moles" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7186" PostTypeId="1" CreationDate="2013-12-03T07:25:50.570" Score="0" ViewCount="50" Body="&lt;p&gt;How many grams of oxygen are present at 25.0 C in a 5.00L tank of oxygen enriched air? The total pressure is 30.0 atm and the only other gas is nitrogen with a pressure of 17.0 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After converting the temperature to Kalvin, I do not really know what to do. &#xA;How can I find the grams of oxygen with another molecule in the container? &lt;/p&gt;&#xA;" OwnerUserId="3843" LastActivityDate="2014-01-03T07:52:24.380" Title="Gas law with molecules at the same time" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7187" PostTypeId="1" AcceptedAnswerId="8512" CreationDate="2013-12-03T13:36:59.887" Score="2" ViewCount="2547" Body="&lt;p&gt;In the course of my undergraduate studies it has become clearer and clearer that it is quite helpful to know the ground electronic state of transition metals by heart, if only to speed up the process of analysing a reaction or to evaluate a compound in respect to its number of valence electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Note that when I say ground electronic state, I don't mean the state in vacuum, but when the metal is bonding to ligands. As a consequence all the electrons are found in the $d$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Method:&lt;/strong&gt;&#xA;I have now put the $d$-Block of the periodic table as a wallpaper on my screen background so I get constantly exposed to the arrangement of the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your way to memorize what metals have which $d^x$ ground state?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2013-12-03T22:41:24.957" LastActivityDate="2014-02-16T13:20:21.027" Title="Memorize Transition Metal Electronic Ground State" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7188" PostTypeId="1" CreationDate="2013-12-03T15:02:54.067" Score="1" ViewCount="3662" Body="&lt;p&gt;Today my teacher taught about amorphous and crystalline substances. Crystalline substances are said to have different values for physical properties (like speed of light, electrical resistance, etc) in different directions. Crystalline substances are said to have orderly arranged constituents, however this orderly arrangement is not said to be same in different directions. Thus, they are said to show different values for physical properties in different directions.&lt;br&gt;&#xA;  Amorphous are said to be isotropic in nature, i.e they have same value for all the physical properties in different directions. I was stuck here, amorphous substances have disordered arrangement of constituents. So, they should strictly have different values in different directions. Is it we speak about average values of physical properties in case of amorphous substances? or is it that, they strictly have same values for the physical properties in all directions?&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2013-12-05T10:23:06.973" Title="Why do amorphous substances have different values of physical properties in different directions?" Tags="&lt;physical-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7189" PostTypeId="1" AcceptedAnswerId="7192" CreationDate="2013-12-03T15:12:21.140" Score="1" ViewCount="139" Body="&lt;p&gt;I am looking for a full set of data on every element on the periodic table. (and if possible those only speculated about)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say full I mean everything. Electron configurations, Valence electrons, melting point, Atomic mass, Isotopes and what they decay into and all the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a database anywhere?&lt;/p&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2013-12-03T18:00:02.197" Title="A full verion of the periodic table (For computer game)" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7190" PostTypeId="1" AcceptedAnswerId="7193" CreationDate="2013-12-03T17:09:51.110" Score="3" ViewCount="3849" Body="&lt;p&gt;I am reading a book which is saying about the Syn-Anti nomenclature of oximes. It is showing this molecule to be Syn-methyl-ethyl ketoxime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4rWxJ.png&quot; alt=&quot;Syn-methyl-ethyl ketoxime&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand what is a ketoxime and how is there nomenclature done. Can somebody explain please ?&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorDisplayName="user7951" LastEditDate="2015-05-18T11:02:28.980" LastActivityDate="2017-04-13T13:42:02.480" Title="What are ketoximes?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7191" PostTypeId="2" ParentId="7174" CreationDate="2013-12-03T17:55:34.890" Score="11" Body="&lt;p&gt;Nitrogen is in the second row with no $d$ orbital in the valence shell.  It obeys the octet rule and cannot have more than 8 electron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are exceptions to the octet rule.  Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule.   However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot;&gt;valence bond theory&lt;/a&gt;, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts.  However, their contributions are not even.  some of them are more favorable than others.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds.  Based on octet rule alone, there are 3 possible resonance structures that are favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JYVhS.png&quot; alt=&quot;Resonance structures of HNO3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.  As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also common to write a mixture of resonance structures as the &lt;em&gt;&lt;strong&gt;hybrid form&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dis5C.png&quot; alt=&quot;Hybrid form of HNO3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;strong&gt;this representation is not a single Lewis structure&lt;/strong&gt;, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms.  The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-03T17:55:34.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7192" PostTypeId="2" ParentId="7189" CreationDate="2013-12-03T18:00:02.197" Score="4" Body="&lt;p&gt;You can find all these information on &lt;a href=&quot;http://www.wolframalpha.com&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt;.  You can use the website directly to obtain these information, and you can call their &lt;a href=&quot;http://products.wolframalpha.com/api/&quot; rel=&quot;nofollow&quot;&gt;API&lt;/a&gt; to directly fetch the data online from your own program.  Of course, you can also use Mathematica to write your program and query the database with &lt;a href=&quot;http://reference.wolfram.com/mathematica/ref/WolframAlpha.html&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; function.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-03T18:00:02.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7193" PostTypeId="2" ParentId="7190" CreationDate="2013-12-03T18:17:30.900" Score="7" Body="&lt;p&gt;An oxime is a chemical compound belonging to the imines, with the general formula $\ce{(R^1)(R^2)C=N-O H}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/byRLZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the nomenclature of methyl-ethyl ketoxime done? &lt;br&gt;&lt;br&gt;&#xA;The answer is :&lt;br&gt;&#xA;Look at your photo.&lt;br&gt; &#xA;1. On the left side you can see &quot;$\ce{CH3-CH2 -}$&quot; (ethyl).&lt;br&gt;&#xA;2. On the right side you can see &quot;$\ce{CH3 -}$&quot; (methyl). &lt;br&gt;&#xA;3. On the middle you can see &quot;$\ce{C=N-O-H}$&quot; (ketoxime).&lt;br&gt;&#xA;Now assemble (1 &amp;amp; 2 &amp;amp; 3 ) ..then you will have the name &quot;methyl-ethyl ketoxime&quot;.&lt;br&gt;&lt;br&gt;&#xA;For more you can  see  : &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxime&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Oxime&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="4945" LastEditDate="2015-05-18T07:27:14.340" LastActivityDate="2015-05-18T07:27:14.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7194" PostTypeId="2" ParentId="7184" CreationDate="2013-12-03T20:09:47.337" Score="2" Body="&lt;p&gt;Although it doesn’t affect the result, your equation is incorrect. Magnesium tends to form $\ce{Mg^2+}$ ions and chlorine tends to form $\ce{Cl-}$ ions. That’s why magnesium chloride compounds are found in 1:2 proportions. After balancing the coefficients your equation should look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg + 2HCl -&amp;gt; MgCl2 + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using the formula $pV = nRT$ we get $n\approx0.063\ \mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n(\ce{Mg})=n(\ce{H2})$ we have $0.063\ \mathrm{mol}$ of magnesium, which is approximately $1.531\ \mathrm{g}$.&lt;/p&gt;&#xA;" OwnerUserId="3840" LastEditorDisplayName="user7951" LastEditDate="2015-05-10T20:26:33.677" LastActivityDate="2015-05-10T20:26:33.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7198" PostTypeId="1" CreationDate="2013-12-04T03:32:23.543" Score="2" ViewCount="1485" Body="&lt;p&gt;I am a bit confused about what chemical shift value signifies in proton NMR spectroscopy. If the proton of the standard substance TMS absorbs X Joules of energy to go into resonance, than a substance with a chemical shift value of 2.5ppm means it absorbs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;energy absorbed= X- (2.5X/10^6)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please tell me if this is what chemical shift value is mathematically. In my book, it says the substance absorbs &quot;2.5 millionth less amount of energy than TMS to go into resonance&quot;... but I don't really understand this.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2016-12-03T10:33:54.023" Title="Chemical shift value and proton NMR spectroscopy" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7200" PostTypeId="1" AcceptedAnswerId="7207" CreationDate="2013-12-04T05:30:26.473" Score="1" ViewCount="889" Body="&lt;p&gt;I'm having trouble understanding how exactly coordination complexes are formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds such as $\ce{[Cu(H2O)6]++}$ are the ligands filled into the empty 3 of 4p orbitals+ 1 of 5s orbital + 2 of 4d orbitals or how? &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2013-12-04T18:07:36.220" Title="Formation of coordination complexes?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7201" PostTypeId="2" ParentId="7186" CreationDate="2013-12-04T07:19:22.760" Score="3" Body="&lt;p&gt;We have 2 gases in the container, oxygen ($O_2$) and nitrogen ($N_2$). Now we are given that $N_2$ has a pressure of $17$ atm out of $30$ atm, the total volume of the container is $5.00$ L and that the temperature is $25^o$C or $298$ K. So somehow, we need to find out the volume that oxygen occupies and from there we will be able to find out the grams of oxygen present in the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's find out the mole fraction $N_2$.&lt;br&gt;&#xA;You can find mole fraction by the following formula,&lt;br&gt;&#xA; $$Mole\, Fraction = \frac{Partial\,Pressure\,of\,Gas}{Total\,Pressure}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;br&gt;&#xA;Mole fraction of $N_2$= $\frac{17}{30}$=$0.6$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore mole fraction of $O_2$=$1-0.6$=$0.4$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to find out the total no. of moles by the formula,&lt;br&gt;&#xA;$$PV={nRT}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the values given for the whole system,&#xA;$$30\times5=n\times0.0821\times298$$&#xA;$$\Rightarrow\,n=6.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So moles of oxygen in container = $0.4\times6.1$ = $2.5$&lt;br&gt;&#xA;Consequently, grams of $O_2$ present in container = $2.5\times32$ = $78.4$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that is how you get the grams of oxygen. Hope you understood everything. If not, feel free to ask by commenting. Good Luck!  &lt;/p&gt;&#xA;" OwnerUserId="3826" LastActivityDate="2013-12-04T07:19:22.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7202" PostTypeId="2" ParentId="6595" CreationDate="2013-12-04T09:38:29.287" Score="1" Body="&lt;p&gt;What they mean by &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the treatment of the whole enzyme-substrate complex quantum mechanically is computationally impossible, [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is basically a consequence of the results from the inner workings of the quantum chemical formulation of the theory. If one goes through the equations (it takes a while), one arrives at an expression that tells you that the required number of calculations scales with $N^4$, where $N$ is the number of &lt;em&gt;basis functions&lt;/em&gt;. Imagine the wavefunction being a (hyperdimensional) vector in a coordinate system. Say you had a 3-dimensional vector but only two basis vectors to express it, then your representation of the vector would be very inaccurate, so you add another basis vector to your set of basis vectors and then you can accurately describe the vector by $\vec{v} = a\vec{e}_x + b\vec{e}_y + c\vec{e}_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In enzymes, the huge amount of atoms would lead to gigantic numbers of required calculations which make it largely impossible to calculate completely quantum mechanically. So it's an impracticality, not a theoretical problem.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-12-04T09:38:29.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7203" PostTypeId="2" ParentId="7198" CreationDate="2013-12-04T11:55:02.177" Score="4" Body="&lt;p&gt;Firstly, the ppm scale is a dimensionless scale that describes the energy level of a signal relative to a reference (usually TMS). Significantly, it is magnetic field independent, and so allows comparison of signals recorded on different systems (not just a 300MHz vs a 600MHz spectrometer, but between all different 600MHz spectrometers, because no two will have been set up at identical Bo fields).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical shift scale is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta =(v-v_r )*10^6/v_r$$ , where $v_r$ is the spectrometer reference frequency (of TMS, say)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a 600MHz NMR spectrum, each ppm represents 600Hz. So, if the resonance frequency of a signal at 0ppm is 600,000,000Hz, the signal at 1ppm represents an energy of 600,000,600Hz. At 2.5ppm, the frequency is 600,001,500Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical shift scale is often confusing in its labelling, but the thing to remember is that as you go to higher chemical shift values, you go to higher frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher ppm == downfield == Deshielded == higher frequency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower ppm == upfield == Shielded == lower frequency&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2013-12-04T12:04:30.983" LastActivityDate="2013-12-04T12:04:30.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7205" PostTypeId="2" ParentId="6404" CreationDate="2013-12-04T16:37:46.757" Score="6" Body="&lt;p&gt;Well, t-butyl cation isn't really stabilized by inductive effects only. Interaction of empty $p$ orbital of the central carbon with the $\ce{C-H}$ sigma bonds are really important in that manner. And they are not present in norbornyl cation due to strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: no, it is very unlikely for that compound to undergo an $S_N1$ reaction.&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T07:29:31.423" LastActivityDate="2015-06-26T07:29:31.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7207" PostTypeId="2" ParentId="7200" CreationDate="2013-12-04T18:07:36.220" Score="1" Body="&lt;p&gt;If you want to analyse a compound, it is best to start with the $d$-electron count of the metal. For $\ce{Cu^{2+}}$ we get $d^9$, since $\ce{Cu^0}$ would be $d^{11}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider the following (simplified) molecular orbital diagram for a $d^1$ compound:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RgbQQ.png&quot; alt=&quot;Molecular orbital diagram for a $d^1$ complex.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can clearly see, the ligand electron pairs mix with some of the $d$, the $s$ and $p$ orbitals of the metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, instead of $d^1$ we have $d^9$, so we just start filling in 8 more electrons into the MOs. This gives us a full $t_\text{2g}$ MO and 3 electrons in the $e_\text{g}$ MO. ($t_\text{2g}$ and $e_\text{g}$ are ways of describing the symmetry of the orbitals. This comes from applied group theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see, the $5s$ orbital never comes into play here, only the $3d$, $4s$ and $4p$ orbitals of the metal take part in bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Ligand_field_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ligand Field Theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular Orbital Theory&lt;/a&gt; to further your knowledge in inorganic chemistry. Not everything is as simple as the &quot;fill in the boxes&quot; method.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-04T18:07:36.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7210" PostTypeId="1" CreationDate="2013-12-04T20:41:44.663" Score="-1" ViewCount="104" Body="&lt;p&gt;I have a precipitation reaction of this form: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ca^{2+} + CO_{3}^{2-} → CaCO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many mole of $\ce{Ca^{2+}}$ ion is contained in a 200ml solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how to calculate this, thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="3854" LastEditorUserId="1878" LastEditDate="2013-12-04T22:27:19.980" LastActivityDate="2013-12-16T02:35:19.670" Title="ions proportion" Tags="&lt;reaction-mechanism&gt;&lt;ions&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2013-12-05T00:21:05.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="7211" PostTypeId="2" ParentId="7210" CreationDate="2013-12-04T22:08:22.347" Score="2" Body="&lt;p&gt;Carbonates are usually insoluble in water except for when combined with group 1 elements and and ammonium ions, however, as you may already know, almost everything is slightly soluble in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally in order to see a precipitate form, a solution of sodium carbonate and aqueous calcium chloride are mixed. The calcium ions and carbonate ions are then in contact and can form the calcium carbonate precipitate. The other product is sodium chloride which is known to be soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please keep in mind that $ M = \frac{moles \; solute}{liter \; solution}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is expressed as mol per liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of calcium carbonate is very dependent upon the pH of its environment. Assuming that this reaction occurs in water, the pH will be very close to 7. Tap water generally contains many ions, chloride for instance that will interfere with our calculation in the moles of calcium ions that we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we need the solubility constant product denoted as $K_{sp}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Solubilty/Solubility_Product_Constant,_Ksp&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Solubilty/Solubility_Product_Constant,_Ksp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from the UC Davis ChemWiki &quot;the Solubility Product Constant, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. A more a substance dissolves, the higher the Ksp value it has.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for the solubility product of calcium carbonate is $$[Ca^{2+}][CO_3^{2-}] = 8.7 * 10^{-9} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of each ion is 1:1 as shown in our equation, so there is no need to square the concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If S mol of $CaCO_3$ dissolved in 1L, the equilibrium concentration of $Ca^{2+}$ will be $[Ca^{2+}] = S$. The concentration of $CO_3^{2-}$ ions will be $[CO_3^{2-}] = S$ again, because each mol of $CaCO_3$ each produces one mol of $Ca^{2+}$ ions and one mol of $CO_3^{2-}$ ions. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[Ca^{2+}][CO_3^{2-}] = S * S = S^2 = K_sp = 8.7 * 10^{-9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $S^2$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S^2 = 8.7 * 10^{-9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the square root on each side of this equation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt{S^2} = \sqrt{8.7 * 10^{-9}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = 9.327*10^{-5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium concentrations are therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ [Ca^{2+}] = S = 9.327*10^{-5} M $$&#xA;$$ [CO_3^{2-}] = S = 9.327*10^{-5} M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have our concentration of $Ca^{2+}$ ions, we can then calculate the number of moles of ions that we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \frac{moles \; solute}{liter \; solution}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9.327*10^{-5} M = \frac{moles \; solute}{1 L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ mol = 9.327*10^{-5} M * L^{-1} * \frac{1L}{1000mL} * 200 = 1.8654*10^{-5} \; mol \; of \; Ca^{2+} ions $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For completeness sake, lets take a look of the number of moles in grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the molar mass of $CaCO_3$ is $100.0869 g/mol$, using our our definition of molarity from above, we can calculate the gram solubility which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ gram \; solubility = (9.327*10^{-5} \; mol \; L^{-1})(100.0869 \; g/mol) = 0.009335 \; g/L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting this to our 200mL solution,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0.009335 \frac{g}{L} * \frac{1L}{1000mL} * 200mL = 0.001867 g$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Last note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $$CaCO_3(s) \rightarrow Ca(aq) + CO_3(aq)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some of the carbonate hydrolyses to produce HCO3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$CO_3(aq) + H_2O(l) \rightarrow HCO_3(l) + OH(aq)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this instance, the hydrolysis of carbonate decreases the concentration of carbonate, which pulls the solubility equilibrium to the right making calcium carbonate more soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.colby.edu/chemistry/CH142/CH142B/SolubilityCalciumCarbonate.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.colby.edu/chemistry/CH142/CH142B/SolubilityCalciumCarbonate.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please take a look at the link provided, as the actual number of moles will vary due to the pH of the environment. They nicely derive the formula at eq. 11 which you can then find the number of moles of $Ca^{2+}$ ions.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2013-12-16T02:35:19.670" LastActivityDate="2013-12-16T02:35:19.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7212" PostTypeId="2" ParentId="6404" CreationDate="2013-12-04T22:51:03.467" Score="4" Body="&lt;p&gt;The bridgehead carbon in the pictured compound is $sp^3$ hybridized. Upon solvolysis of the bromide ion, it should rehybridize to $sp^2$. However, this does not occur because the bridgehead carbon is too strained. Therefore, a bridged compound will not undergo an $S_N1$ mechanism st the bridgehead carbon.&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T07:29:57.677" LastActivityDate="2015-06-26T07:29:57.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7213" PostTypeId="1" CreationDate="2013-12-04T23:42:31.653" Score="-1" ViewCount="195" Body="&lt;p&gt;Does the hydrogen ozonide exist? I mean if this compound could be found in Nature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$HO_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the theoretical or phenomenological of such a chemical compound...&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="73" LastEditDate="2013-12-05T00:24:30.907" LastActivityDate="2014-01-05T21:20:56.820" Title="Hydrogen ozonide existence" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7214" PostTypeId="2" ParentId="7213" CreationDate="2013-12-05T00:31:42.113" Score="3" Body="&lt;p&gt;Yes.  It exists in the atmosphere, is proposed to be intermediate in various atmospheric reactions and has been experimentally detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencemag.org/content/285/5424/81.abstract&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Science&lt;/em&gt; Vol*&lt;em&gt;285&lt;/em&gt;* No.5424(1999): p81&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  Full text is available after free registration.   &lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="3723" LastEditDate="2014-01-05T21:20:56.820" LastActivityDate="2014-01-05T21:20:56.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7215" PostTypeId="1" AcceptedAnswerId="7435" CreationDate="2013-12-05T05:15:55.067" Score="1" ViewCount="263" Body="&lt;p&gt;Can anyone direct me to a good book with lots of molarity exercises, i really want to practice this and would like to find as many types of exercises involving molarity. I of course did google this and there are some, but usually they are three or four exercises per resource.  Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="275" LastEditDate="2013-12-19T13:08:18.303" LastActivityDate="2013-12-19T13:08:18.303" Title="Molarity exercises" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2013-12-19T13:08:09.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="7216" PostTypeId="1" AcceptedAnswerId="7220" CreationDate="2013-12-05T09:29:59.150" Score="18" ViewCount="59187" Body="&lt;p&gt;My question is: Is hydrogen-rich water true? I am aware of the alkaline water. I have read several articles in the Internet regarding this kind of water but I cannot seem to be convinced. Wikipedia isn't even talking about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am puzzled with how can a water be enriched with hydrogen. It's definitely $\ce{H2O}$ and everybody knows it. It's not water anymore if it's not $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.hydrogenwater-stick.com/&quot;&gt;link&lt;/a&gt; is one of the examples of websites that I read about hydrogen-rich water.&lt;/p&gt;&#xA;" OwnerUserId="3859" LastEditorUserId="4945" LastEditDate="2014-05-22T09:20:43.663" LastActivityDate="2018-09-13T01:01:57.720" Title="What is hydrogen-rich water?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7217" PostTypeId="2" ParentId="7188" CreationDate="2013-12-05T09:52:24.060" Score="0" Body="&lt;p&gt;There is probably some misunderstanding here... As an macroscopic example, take a book. And then split it in half. There is only one way you can do it with small effort. That's anisotropy. Then run the book through paper shredder and you obtain isotropic material (&lt;a href=&quot;http://www.compatibletoner.org/wp-content/uploads/2009/09/shredded-papers.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.compatibletoner.org/wp-content/uploads/2009/09/shredded-papers.jpg&lt;/a&gt;). You can split it easily in half in all directions and also other properties are independent of the direction (compared to the book).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-12-05T09:52:24.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7218" PostTypeId="2" ParentId="7188" CreationDate="2013-12-05T10:23:06.973" Score="1" Body="&lt;p&gt;Yes, you’re exactly right that we are talking about an average. If we could somehow look at a small enough portion of an amorphous solid, we would see differences between the properties in different directions. Indeed, a small enough portion of, say, amorphous ice is a single water molecule, which is certainly not isotropic: we can talk about the direction perpendicular to the molecular plane or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when we talk about real, physical materials, as usual in chemistry, we are talking about Avogadro’s number (plus or minus a few orders of magnitude) of molecules. So taking the spatial average of an amorphous structure makes perfect sense, and the law of large numbers means that the tiny differences between properties in different directions will be utterly undetectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an aside, it’s worth noting that the perfect crystalline structure we imagine is &lt;em&gt;also&lt;/em&gt; an average over every unit cell in the material – it’s just that because each unit cell is oriented in more or less the same way, different directions are not “mixed” as part of this average, so crystals can have anisotropic properties.)&lt;/p&gt;&#xA;" OwnerUserId="390" LastActivityDate="2013-12-05T10:23:06.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7219" PostTypeId="2" ParentId="7216" CreationDate="2013-12-05T11:33:47.607" Score="12" Body="&lt;p&gt;I was completely unaware of the concept of hydrogen-rich water, and although there is a lot of misinformation and quackery around it, a quick search on Google afforded &lt;a href=&quot;http://www.medicalgasresearch.com/content/1/1/11&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article&lt;/a&gt; (whose quality I cannot vouch for). It would seem that hydrogen-rich water is nothing but regular water with dissolved hydrogen gas. You could simply pump $\ce{H_{2(g)}}$ into some water, but there is a smarter way to do it, by producing the hydrogen &lt;em&gt;in situ&lt;/em&gt; via the addition of reactive metals such as magnesium, according to the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg_{(s)} + 2H2O_{(l)} → Mg(OH)2_{(aq)} + H2_{(g)}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Producing the hydrogen directly inside the water also seems to temporarily supersaturate the solution and increase the amount of hydrogen dissolved, because the equilibrium solubility of hydrogen gas in water is tiny, &lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/solubilities-gases&quot; rel=&quot;nofollow noreferrer&quot;&gt;only around 16 mg per kg of water at 20°C&lt;/a&gt;. The exact mechanism by which hydrogen-rich water has medicinal properties I do not know, and that would probably be best answered on &lt;a href=&quot;https://biology.stackexchange.com/&quot;&gt;Biology.SE&lt;/a&gt; if you're curious. It is conceivable that the dissolved hydrogen gas could act as an antioxidant, though relying on the distribution of a gas across the body seems a bit strange (there are gaseous substances known to be important for our metabolism, such as $\ce{NO}$). I wouldn't rule out the possibility of it being nonsense before reading into it a lot more, though.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2018-09-13T01:01:57.720" LastActivityDate="2018-09-13T01:01:57.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7220" PostTypeId="2" ParentId="7216" CreationDate="2013-12-05T11:54:39.877" Score="19" Body="&lt;p&gt;Hydrogen water is a &lt;em&gt;solution&lt;/em&gt; of molecular hydrogen $\ce{H2}$ in water, which shouldn't be confused with acidic or basic solutions.  The first scientific reference to hydrogen water appears to be in the journal, &lt;em&gt;Nature Medicine&lt;/em&gt; which can be accessed &lt;a href=&quot;http://dx.doi.org/10.1038/nm1577&quot;&gt;here&lt;/a&gt; if you have a personal or institutional subscription to the journal. The claim is that molecular hydrogen can serve as a mild reducing agent and therefore has antioxidant properties; namely, it can reduce hydroxyl radicals:&#xA;$$\ce{H2 +2 OH\cdot -&amp;gt; 2H2O}$$&#xA;The article has generated a considerable amount of research, with well over 300 journal articles citing this work originally published in 2007.  There seems to be some merit to the claim although in my brief scan of the abstracts, there does not appear to be a large body of evidence suggesting that commercial products are beneficial.  One such article, which is publicly accessible is &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2831093/&quot;&gt;here&lt;/a&gt;.  As mentioned in another answer, the commercial product approach to generating hydrogen water appears to be the oxidation of magnesium in water, which generates $\ce{Mg^2+}$ ions.  As with any other nutritional supplement, I would be very wary of entering this type of treatment without consulting a professional.  Magnesium is an important nutrient, however at high doses it does &lt;a href=&quot;http://www.livestrong.com/article/372709-signs-symptoms-of-magnesium-toxicity/&quot;&gt;show signs of toxicity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-12-05T11:54:39.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7221" PostTypeId="1" AcceptedAnswerId="7223" CreationDate="2013-12-05T13:06:40.377" Score="0" ViewCount="132" Body="&lt;p&gt;In the electrolysis of $\ce{Na2SO4}$, is it valid to write the anode half-reaction (oxidation) as&#xA;$$&#xA;\ce{OH-} \rightarrow \ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}&#xA;$$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book says it's&#xA;$$&#xA;\ce{H2O} \rightarrow 2\ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than the additional $\ce{H+}$, there's no difference between the two.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2013-12-05T13:17:47.217" Title="Electrolysis Anode equation" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7222" PostTypeId="2" ParentId="7162" CreationDate="2013-12-05T13:09:51.543" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When a weak acid reacts with a weak base, the equivalence point solution will be basic if the base is stronger and acidic if the acid is stronger. If both are of equal strength, then the equivalence pH will be neutral. However, weak acids are not often titrated against weak bases because the colour change shown with the indicator is often quick, and therefore very difficult for the observer to see the change of colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid%E2%80%93base_titration&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing the titration curve is hard, because you generate a buffer system which buffers against acids and bases at the same time. Around the equivalence point the buffer system works equally well against acids and bases (assuming their $\text{p}K_\text{a}$ values are the same), resulting in a slow rise of the pH. You don't get the usual &quot;jump&quot; in the curve at the equivalence point, which makes it hard to detect the actual equivalence point (your indicator should fulfill $\text{p}K_\text{a}^\text{Ind} = \text{pH}_\text{equiv.}$, which is highly unlikely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reasons given above I disagree with the last part of the Wikipedia quote, but having never done the experiment myself I cannot be sure.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-05T13:09:51.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7223" PostTypeId="2" ParentId="7221" CreationDate="2013-12-05T13:17:47.217" Score="2" Body="&lt;p&gt;This depends on the medium in which you do the electrolysis. In the case of a basic medium your equation would be more correct, but in the case of an acidic medium your answer would be wrong, since the concentration of $\ce{OH-}$ ions would be far too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally: In &lt;strong&gt;acidic&lt;/strong&gt; medium you balance your redox equations using $\ce{H+}$ and $\ce{H2O}$, in &lt;strong&gt;basic&lt;/strong&gt; medium you balance with $\ce{OH-}$ and $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Since you generate $\ce{H+}$ ions it is possible to argue that starting with a solution at $\text{pH}=7$ gives you an acidic medium, resulting in the equation in the book.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-05T13:17:47.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7225" PostTypeId="1" AcceptedAnswerId="17244" CreationDate="2013-12-05T17:05:57.120" Score="3" ViewCount="3759" Body="&lt;p&gt;The question is simple:  What human-consumption safe liquid has the best caffeine solubility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 200mg/10mL of water.  That's a lot of water for one dose!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some ramblings on a fourm stating it's &quot;low&quot; solubility in propylene glycol and glycerine (glycol) but no sources or exact numbers, so that doesn't mean much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In boiling water, it seems much higher, 66g/100mL, which is more of the range I am hoping.  So what happens if I boil a liter of water, dump 100g into it, and then let it cool? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping for something in the range of 200mg in 1mL-3mL, but as low as possible, and relatively easy to acquire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All help very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3862" LastActivityDate="2015-02-08T20:47:05.833" Title="Best consumption-safe liquid for caffeine solubility?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7226" PostTypeId="1" CreationDate="2013-12-05T17:11:37.753" Score="0" ViewCount="1125" Body="&lt;p&gt;Why is it that sp3 hybridized carbonium ions do not exist? Or why are they very difficult to form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, nucleophilic substitution in bicyclo compounds at the bridge head carbon is not very easy. I don't understand the explanation given in the book. Thanks for any help.  &lt;/p&gt;&#xA;" OwnerUserId="3863" LastActivityDate="2014-02-03T22:19:32.883" Title="Why are carbocations with sp3 hybridisation highly unstable?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-02-08T10:00:52.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="7227" PostTypeId="1" AcceptedAnswerId="14657" CreationDate="2013-12-05T19:16:36.077" Score="4" ViewCount="1514" Body="&lt;p&gt;I want to quantify a biotin-protein conjugate after chemical biotinylation. After biotinylation, the solution contains both biotinylated protein and free biotin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to use a HABA-avidin biotin quantitation kit like &lt;a href=&quot;http://www.piercenet.com/product/pierce-biotin-quantitation-kit&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; from Pierce. Since the biotinylated protein and the free biotin will both displace HABA and decrease the absorbance, I need to remove the free biotin so that I only quantify the biotinylated protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I separate the free biotin from the solution?&lt;/p&gt;&#xA;" OwnerUserId="60" LastActivityDate="2014-07-27T05:43:32.893" Title="How do I remove free biotin after biotin-protein conjugation?" Tags="&lt;biochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7228" PostTypeId="1" CreationDate="2013-12-05T19:42:43.600" Score="2" ViewCount="64" Body="&lt;p&gt;I am looking for references (review articles, reports) about non-stoichometric compounds. Any idea of where should I look?&lt;/p&gt;&#xA;" OwnerUserId="3857" LastActivityDate="2013-12-24T23:41:20.147" Title="Non-stoichometric compounds" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7229" PostTypeId="2" ParentId="7226" CreationDate="2013-12-05T21:35:02.170" Score="2" Body="&lt;p&gt;When you hybridize towards $sp^3$, you use one $p$ orbital more compared to the $sp^2$ hybridization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional $p$ orbital raises the energy of the 4 $sp^3$ orbitals higher with respect to the energy level of $sp^2$-hybridized orbitals ($p$ orbitals are higher in energy than $s$ orbitals), which in the energy-minimizing world of chemistry is not favoured by the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is minimized by hybridizing in a $sp^2$ fashion (in which all the $\sigma$-bonding electrons go) and adopting a planar geometry.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-05T21:35:02.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7233" PostTypeId="1" AcceptedAnswerId="7597" CreationDate="2013-12-06T00:49:34.220" Score="5" ViewCount="11884" Body="&lt;p&gt;If silicon dioxide tends to form a crystal lattice with four $\ce{O}$'s around a central $\ce{Si}$, why isn't the molecular formula $\ce{SiO4}$ then? I'm confused why it's unique in that its molecular formula doesn't match up with its geometry.&lt;/p&gt;&#xA;" OwnerUserId="3866" LastEditorUserId="5017" LastEditDate="2015-03-11T16:20:30.313" LastActivityDate="2016-03-21T18:40:24.247" Title="What's up with SiO₂ being tetrahedral?" Tags="&lt;crystal-structure&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7234" PostTypeId="2" ParentId="7233" CreationDate="2013-12-06T01:09:28.157" Score="2" Body="&lt;p&gt;As a short answer, chemical formulae are only really meant to show the proportions between different atoms in a compound, and in general they do not carry any information regarding the way the atoms are bonded. In quartz, there are simply two oxygen atoms for each silicon atom, hence $\ce{SiO_2}$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-12-06T01:09:28.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7235" PostTypeId="1" CreationDate="2013-12-06T01:13:37.410" Score="3" ViewCount="3991" Body="&lt;p&gt;Question is rather self-explanatory. Came up during a lecture without a concrete answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but &lt;strong&gt;why do fluorescent molecules necessarily overlap in their excitation and emission spectra?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastActivityDate="2016-07-12T21:13:27.173" Title="Why does the excitation and emission spectrum of a fluorescent molecule have overlap?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7236" PostTypeId="1" AcceptedAnswerId="9025" CreationDate="2013-12-06T02:53:52.670" Score="4" ViewCount="2780" Body="&lt;p&gt;Recently, our water from the artesian well started to smell (again). It only smells bad for only few seconds than the smell disappear.  We have sulfur-iron removal system, water softener, UV lamp. Only the water from the system I listed above smell, because we also have a Reverse Osmosis installed in the kitchen and the water filtered from it does NOT smell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We did a hydrocarbon test that was negative. We also did a normal &quot;drinkable water&quot; analysis and it passed with nothing ususual. My question is what are the possibilities of that smell? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. I can confirm the smell is not sulfur (different from rotten egg)&lt;br&gt;&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/4281/do-i-really-need-4-different-water-treatment-systems-with-this-water-analysis&quot;&gt;See this post&lt;/a&gt; for our water results before installing the system if it may be useful&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2020-04-19T02:55:59.000" Title="What can cause artesian well water to smell?" Tags="&lt;water&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7238" PostTypeId="2" ParentId="5925" CreationDate="2013-12-06T05:59:22.450" Score="6" Body="&lt;p&gt;According to the website of Dr Pollack himself, the health effect of EZ water is a &lt;strong&gt;proposed&lt;/strong&gt; research project currently &lt;strong&gt;applying for funding&lt;/strong&gt;.  He has made no claim of any proven health effect, and he even claimed explicitly that he does not know if EZ water has health benefits.  He also explicitly stated that the mechanism of formation of EZ water is still under investigation.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His book is not a scientific work.  He presents his proposed yet untested ideas and theories regarding phenomena that may be attributed to EZ water; it however does not claim anything is proven at all.  It is a good thing to have ample imagination; however at least at this moment he is somewhat behind the schedule on proving these ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, he is currently carrying out research on many of these subjects.  According to what I saw on his website, his research seem to be sound, but at this moment &lt;strong&gt;no conclusive results are produced yet&lt;/strong&gt; about any of his big hypotheses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it is fairly common to describe a proposed research to have the potential to have a revolutionary impact on the life of every man on earth.  Very few write a book like him though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not call him a deluded scientist because he did not really claim anything nonconventional being discovered.  However he does put himself in a position where immense commercial profits are generated solely through exploiting and distorting results of his research.  &lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-06T05:59:22.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7239" PostTypeId="2" ParentId="7233" CreationDate="2013-12-06T06:11:29.060" Score="7" Body="&lt;p&gt;Every $\boxed{^{14}\textrm{Si}}$ atom is surrounded by &lt;strong&gt;4&lt;/strong&gt; $\boxed{^8\textrm{O}}$ atoms.   On the other hand, every $\boxed{^8\textrm{O}}$ atom is surrounded by &lt;strong&gt;2&lt;/strong&gt; $\boxed{^{14}\textrm{Si}}$ atoms.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the number of $\textrm{Si-O}$ bonds is $n$, then the number of $\boxed{^{14}\textrm{Si}}$ atoms is $n/4$, while the number of $\boxed{^8\textrm{O}}$ atoms is $n/2$.  Therefore you arrive at the formula $\textrm{Si}\textrm{O}_2$.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="5017" LastEditDate="2015-03-11T16:21:12.277" LastActivityDate="2015-03-11T16:21:12.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7242" PostTypeId="2" ParentId="5477" CreationDate="2013-12-06T09:04:50.647" Score="15" Body="&lt;p&gt;Without the salt bridge, the solution in the anode compartment would become positively charged and the solution in the cathode compartment would become negatively charged, because of the charge imbalance, the electrode reaction would quickly come to a halt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to maintain the flow of electrons from the oxidation half-cell to a reduction half cell, this completes the circuit.&lt;/p&gt;&#xA;" OwnerUserId="3873" LastEditorUserId="402" LastEditDate="2017-04-02T20:12:53.350" LastActivityDate="2017-04-02T20:12:53.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7244" PostTypeId="1" CreationDate="2013-12-06T12:52:48.243" Score="4" ViewCount="6937" Body="&lt;p&gt;I have been reading up (on &lt;a href=&quot;http://www.brighthubengineering.com/thermodynamics/4616-what-are-reversible-and-irreversible-processes/&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;) about the difference between reversible and irreversible thermodynamic process and it defines them saying that reversible processes are ideal processes while irreversible processes are the natural processes. Are these definitions correct?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not how are they defined?&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorUserId="22" LastEditDate="2013-12-06T12:56:56.040" LastActivityDate="2015-07-05T01:15:08.540" Title="Difference between reversible and irreversible thermodynamic process" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7245" PostTypeId="1" CreationDate="2013-12-06T15:16:31.787" Score="-1" ViewCount="1814" Body="&lt;p&gt;I'm trying to do compound optimization for acetic acid on a EI-MS.&#xA;Later on it would be also run via HPLC to the MS. &#xA;But I do only get fragments of 53/54 and 32.&#xA;Also I need a high concentration to be able to do a compound optimization for the 32 fragment (...around 1mM solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up other mass spectra of acetic acid online.... and they got nice fragments around 43/45 and 29&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...what am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: for clarification of the question&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We work with ESI, also tried it with APCI&lt;/li&gt;&#xA;&lt;li&gt;Acetic acid as a liquid, with mobile phase of water and formic acid&lt;/li&gt;&#xA;&lt;li&gt;yes, it is acetic acid&lt;/li&gt;&#xA;&lt;li&gt;in the meantime we did a blank (thanks for the suggestion!) and we also got the 32 fragment... but non at 60/61 (for positiv mode)&#xA;so it should actually work, if we first make Q1 for 61 and then look for fragments&lt;/li&gt;&#xA;&lt;li&gt;we re-run a fresh dilution with UV-water ... and got some nicer fragments... but unstable turbo spray signal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3877" LastEditorUserId="3877" LastEditDate="2013-12-11T14:40:37.363" LastActivityDate="2014-07-09T18:10:04.103" Title="Mass spectrometry: trouble with acetic acid fragmentation" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7246" PostTypeId="2" ParentId="6447" CreationDate="2013-12-06T18:15:51.287" Score="3" Body="&lt;p&gt;I think the best argument comes from resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing perchlorate with three $\ce{Cl-O}$ double bonds and the negative charge on the oxygen allows for three resonance structures to be drawn (four total). This resonance stabilization  dominates the Lewis structure over simply drawing four double bonds to chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, energetic benefit of charge spread out over four oxygens &gt; charge localized on one chlorine.&lt;/p&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="4231" LastEditDate="2015-03-10T14:47:04.663" LastActivityDate="2015-03-10T14:47:04.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7247" PostTypeId="2" ParentId="7244" CreationDate="2013-12-06T18:16:27.510" Score="-1" Body="&lt;p&gt;These are not definitions of reversible and irreversible processes, but it is a description of one of the differences between the processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real world situations, however, for some processes, the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle. (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Reversible_process_%28thermodynamics%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A (...) definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings. In thermodynamic terms, a process &quot;taking place&quot; would refer to its transition from its initial state to its final state. &lt;a href=&quot;https://en.wikipedia.org/wiki/Reversible_process_%28thermodynamics%29&quot; rel=&quot;nofollow&quot;&gt;again, wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="87" LastActivityDate="2013-12-06T18:16:27.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7248" PostTypeId="1" CreationDate="2013-12-06T19:36:42.320" Score="3" ViewCount="1538" Body="&lt;p&gt;I have the following question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The potential energy of two atoms, a distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; apart, is: &lt;span class=&quot;math-container&quot;&gt;$$U = -Ar^{-2} + Br^{-10}$$&lt;/span&gt; Given that the atoms form a stable molecule at a separation of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3 nm}$&lt;/span&gt; with an energy of &lt;span class=&quot;math-container&quot;&gt;$\pu{-4 eV}$&lt;/span&gt;, calculate &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;. Also find the force required to break the molecule, and the critical separation at which the molecule breaks. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Molecule is stable at the distance when the first derivative is zero. But how to find the critical separation at which the molecule breaks?&lt;/p&gt;&#xA;" OwnerUserId="3878" LastEditorUserId="83758" LastEditDate="2020-06-03T12:38:23.873" LastActivityDate="2020-06-03T16:20:42.133" Title="Critical separation at which molecule breaks" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7249" PostTypeId="2" ParentId="7248" CreationDate="2013-12-06T22:24:13.083" Score="1" Body="&lt;p&gt;... As far as I know, in the absence of an external influence, the molecule never actually breaks. I mean, most people wouldn't call two atoms a metre apart a molecule as such, but there's no negative curvature stationary point in a potential like this at which you could say &quot;if $r &amp;gt; x$, this molecule is broken and will not reform&quot;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-12-06T22:24:13.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7250" PostTypeId="2" ParentId="409" CreationDate="2013-12-06T23:46:24.323" Score="1" Body="&lt;p&gt;What about an ammonia-based fluid mix? I saw one guy on youtube that said he made it with Windex (the liquid was a clear greenish color) and another guy said he made it with Windolene(he's in the UK) but it went cloudy after a while. Meanwhile doing some other research I found someone who says their product (they sell ferrofluid desk toys) should be kept at room temp and will go cloudy if kept too cold. My guess is the guy in the UK had it accidentally get too cold.. because it's the UK, lol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are my ideas, hope they help. :)&lt;/p&gt;&#xA;" OwnerUserId="3881" LastActivityDate="2013-12-06T23:46:24.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7254" PostTypeId="1" CreationDate="2013-12-07T12:54:27.847" Score="-1" ViewCount="2603" Body="&lt;p&gt;According to our book, when the solution is dilute, the mole fraction of solute can be given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X2 = molality (m) x mass of solvent (M1) / 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X2 = m/(1000/M1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason i can think of is that as the solution is dilute, X2 should almost be equal to zero. As the solution is dilute so, molality is very low and it is also divided by 1000 which will give us a value near to zero (for X2). But i know its a poor explanation. So, how does this equation work? &lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="1343" LastEditDate="2013-12-07T13:43:20.927" LastActivityDate="2013-12-07T17:04:52.537" Title="Mole fraction of solute of a dilute solution" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7255" PostTypeId="2" ParentId="7254" CreationDate="2013-12-07T14:08:10.103" Score="2" Body="&lt;p&gt;The formula is correct, being M1 the &lt;strong&gt;molar&lt;/strong&gt; mass of the solvent. I'll call it below $ M_{solvent} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the definition of molar fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ X = \frac{mol_{solute}}{mol_{solvent}+mol_{solute}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For diluted solutions, we can simply write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ X = \frac{mol_{solute}}{mol_{solvent}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To introduce molality into the defintion of molar fraction, one has to consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ mol_{solvent} = \frac{g_{solvent}}{g_{solvent}/mol_{solvent}}=\frac{g_{solvent}}{M_{solvent}} = \frac{kg_{solvent}}{M_{solvent}}1000$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing the latter into the definition of molar fraction, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ X = \frac{mol_{solute}M_{solvent}}{kg_{solvent}1000} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that $mol_{solute}/kg_{solvent}$ is the molality $m$, you get your formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X= \frac{mM_{solvent}}{1000} $&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2013-12-07T17:04:52.537" LastActivityDate="2013-12-07T17:04:52.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7257" PostTypeId="1" AcceptedAnswerId="7258" CreationDate="2013-12-07T21:18:19.177" Score="2" ViewCount="91635" Body="&lt;p&gt;If one considers boiling points (in °C) of primary alcohols, one finds the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;methanol: 65&lt;/li&gt;&#xA;&lt;li&gt;ethanol: 79&lt;/li&gt;&#xA;&lt;li&gt;1-propanol: 97&lt;/li&gt;&#xA;&lt;li&gt;1-butanol: 117&lt;/li&gt;&#xA;&lt;li&gt;1-pentanol: 138&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This trend is due to Van der Waals forces increasing with molecular weight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if one focuses on melting points, I would expect the same trend, but these are the experimental values (in °C):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;methanol: −98&lt;/li&gt;&#xA;&lt;li&gt;ethanol: −117&lt;/li&gt;&#xA;&lt;li&gt;1-propanol: &lt;strong&gt;−127&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;1-butanol: −90&lt;/li&gt;&#xA;&lt;li&gt;1-pentanol: −79&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;the lowest value being the melting point of 1-propanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general idea stating that with all else being equal, more massive molecules have higher boiling and melting points, does not fully apply for this set of alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question: why melting and boiling points of primary alcohols do not follow the same trend?&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T12:49:21.357" LastActivityDate="2015-04-03T12:49:21.357" Title="Melting points and boiling points of primary alcohols do not follow the same trend" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7258" PostTypeId="2" ParentId="7257" CreationDate="2013-12-07T22:37:46.373" Score="3" Body="&lt;p&gt;Melting points are a bit trickier to compare than boiling points, especially when you're looking at the lightest examples of a group of molecules. &lt;a href=&quot;http://en.wikipedia.org/wiki/File:AlkaneBoilingMeltingPoint.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;This graph&lt;/a&gt; for the unbranched alkanes exemplifies nicely how the trends differ in smoothness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase the weight of molecules, the observed trend depends on &lt;em&gt;how&lt;/em&gt; you increase the weight. For example, if you start with methanol and increase the weight by adding more $\ce{-HCOH -}$ units (methanol, ethylene glycol, glycerol, and so on) you would find a very clear increase in both melting and boiling points. In some sense this is a more legitimate trend to analyse, because the relative amounts of different types of intermolecular interactions stays approximately constant (in particular, all molecules have one hydroxyl per carbon atom, and hence they all have about the same amount of hydrogen bonding per atom in the molecule). In your sequence of alcohols, the strong hydrogen bonding allowed by the hydroxyl group becomes &quot;diluted&quot; as the molecule gains an ever larger alkyl chain, which can only sustain much weaker van der Waals interactions. In the limit of alcohols with very large alkyl chains, the alcohols' melting and boiling points do not differ tremendously from their parent saturated hydrocarbons. Compare the melting and boiling points of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cetyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-hexadecanol&lt;/a&gt; (49°C and 344°C) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cetane&quot; rel=&quot;nofollow noreferrer&quot;&gt;hexadecane&lt;/a&gt; (18°C and 287°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the temperature at which a substance freezes is dependent not only on the strength of intermolecular interactions, but how the solid packs. Even substances with strong interactions may only freeze at very low temperatures if they pack very poorly when solidified. Extreme examples to which this effect contributes are &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_liquid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic liquids&lt;/a&gt;. They tend to have a very low vapour pressure at reasonable temperatures (as you would expect of many salts), but some melt even below 0°C. The freezing of such bulky entities would entail lots of ordering and therefore a very large decrease in entropy, so it is disfavoured (or equivalently, their melting is highly entropically favoured, so the free energy of the liquid phase becomes lower than the free energy of the solid phase even at relatively low temperatures). The exact geometry of the molecules in the solid is also important, as some favourable interactions may be suppressed in the solid state due to steric hindrance which may be less present in the more freely-moving molecules of a liquid. Furthermore, molecules with a high amount of symmetry tend to have high melting points, because freezing is entropically favoured due to molecules more readily &lt;em&gt;fitting&lt;/em&gt; in the solid lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another effect to consider is how large a surface area each molecule has available for intermolecular interactions. This is often brought up to explain differences between melting and boiling points of branched and unbranched organic compounds; when comparing structural isomers, branched compounds have a lower available area for interaction than unbranched compounds, so the latter tend to have higher &lt;s&gt;melting and&lt;/s&gt; boiling points (branching may sometimes create a combination of opposing packing and surface area effects, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does your sequence of alcohols behave the way it does? The fact that the trend in boiling point is monotonic while the melting point is not suggests that even though there is a relative shift in importance of types of intermolecular interaction as the alkyl chain increase (from hydrogen bonding to van der Waals interactions), it is probably not the source of the melting point drop. Hence it likely has to do with geometric factors in the solid. The alcohol's alkyl chain probably disrupts the hydrogen bonding network of the solid as it grows from methyl to propyl in a very severe fashion, by making the solid pack less well or by partially hindering the number or strength of hydrogen bonds in the geometry of the solid. Perhaps looking at the crystalline structures of the solid alcohols may provide further insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I wrote a wrong answer initially. I've updated it, but now I realize I'm still not quite sure it's correct. I originally meant to write just a comment, and as I kept writing more I forgot that my very first thought was actually incorrect, so the whole text wasn't built as well as it should be. Anyone is free to pick apart or repurpose what I wrote!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: User Uncle Al posted &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8777/why-does-sif4-have-a-higher-melting-point-than-sf4/8784#8784&quot;&gt;an interesting list of compounds&lt;/a&gt; and their melting/boiling points, showing the importance of molecular symmetry and solid packing.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-03-08T18:27:22.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7259" PostTypeId="1" CreationDate="2013-12-08T01:38:12.793" Score="1" ViewCount="73" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the exothermic reaction of nitrogen gas with iodine solid to produce nitrogen triodide gas. Write the balanced equation including heat and select the overall direction of shift in the reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have balanced the equation: $\ce{N2 (g) + 3 I2 (g) &amp;lt;=&amp;gt; 2 NI3}$&#xA;Is this equation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused on how to find the overall direction of shift. Should I use the Le Chatelier's principle? If so, then I think the overall shift is a &quot;shift to the left&quot;/reverse reaction. Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="3889" LastEditorUserId="5026" LastEditDate="2018-02-18T06:11:04.193" LastActivityDate="2018-02-18T06:11:04.193" Title="Determine the position of an equilibrium process" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7260" PostTypeId="1" CreationDate="2013-12-08T03:50:41.850" Score="5" ViewCount="136" Body="&lt;p&gt;My intuition says no, but I am having trouble coming up with concrete examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that minimizing Gibbs free energy predicts a state of equilibrium, while minimizing kinetic or internal energy does not correspond to equilibrium. However, I am not so sure about potential energy--- when PE is at a minimum, is the system also at equilibrium? &lt;/p&gt;&#xA;" OwnerUserId="3690" LastActivityDate="2014-05-27T20:58:36.840" Title="Does minimizing ANY type of energy ALWAYS predict a state of equilibrium?" Tags="&lt;equilibrium&gt;&lt;energy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7261" PostTypeId="1" AcceptedAnswerId="7263" CreationDate="2013-12-08T06:58:27.900" Score="1" ViewCount="12477" Body="&lt;p&gt;What is the meaning of &lt;code&gt;pre-concentration&lt;/code&gt; both in general and specific to this context of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) &quot;In addition to consistent sampling protocols,.. pre-concentration, and analysis of breath samples require standard methodology, calibration standards...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) &quot;Offline breath analysis involves some form of pre-concentration of analytes followed by a separation step using high-resolution GC-MS based detection&quot;&lt;/p&gt;&#xA;" OwnerUserId="3890" LastActivityDate="2017-07-01T08:18:19.667" Title="What is pre-concentration (context: Breath analysis)?" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7262" PostTypeId="1" CreationDate="2013-12-08T10:28:30.343" Score="3" ViewCount="179" Body="&lt;p&gt;I'm curious to know what's the mechanism of:&#xA;$$\ce{HF + D^+ \rightarrow DF + H^+}$$&lt;/p&gt;&#xA;" OwnerUserId="3875" LastEditorUserId="1858" LastEditDate="2013-12-08T12:49:28.657" LastActivityDate="2016-05-17T20:16:13.190" Title="What's the mechanism for this isotopic substitution reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7263" PostTypeId="2" ParentId="7261" CreationDate="2013-12-08T11:23:38.280" Score="2" Body="&lt;p&gt;Pre-concentration just refers to the process of concentrating a sample before analysis, so that trace components won't be overlooked. If you're looking for something that's only present at a few parts-per-million, you could easily miss it if you used the raw sample, so ideally you want a way to separate off either large, common sources of noise/error, or to just extract the exact components you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick look suggests, for example, that a key first component to remove in breath analysis is moisture, as some volatile components will dissolve readily in the water and make it harder to detect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this paper on a few pre-concentration and detection techniques might be informative: &lt;a href=&quot;http://dx.doi.org/10.1016/j.jchromb.2009.05.010&quot; rel=&quot;nofollow&quot;&gt;Mochalski P, Wzorek B, Sliwka I, Amann A; Chromatogr B Analyt Technol Biomed Life Sci. 2009 Jul 1;877(20-21):1856-66&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2013-12-08T11:23:38.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7264" PostTypeId="1" CreationDate="2013-12-08T11:46:36.417" Score="0" ViewCount="979" Body="&lt;p&gt;I would like to ask how many valence electrons took part in creating chemical bond in $\ce{OH-}$? Is the minus related to $\ce{O}$ or to $\ce{H}$? Are there any spare valence electrons which didnt take part in creating chemical bond in $\ce{OH-}$?&lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:51.110" LastActivityDate="2015-03-06T23:22:51.110" Title="Chemical bond-covalent-valence electrons OH-" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7265" PostTypeId="1" AcceptedAnswerId="7271" CreationDate="2013-12-08T12:14:06.783" Score="2" ViewCount="11910" Body="&lt;p&gt;I would like to ask about formula of phosphorus oxide(V). Should it be &lt;span class=&quot;math-container&quot;&gt;$\ce{P4O10}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt;? Which is right when I have to write the molecular formula? &lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="41328" LastEditDate="2019-05-07T10:16:32.993" LastActivityDate="2019-05-07T10:16:32.993" Title="Is the proper formula for phosphorus oxide(V) P4O10 or P2O5?" Tags="&lt;molecules&gt;&lt;pnictogen&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7266" PostTypeId="2" ParentId="7264" CreationDate="2013-12-08T13:42:31.400" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BtYqt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photo the orange electrons are non bond electrons.&lt;br&gt;&#xA;And the black electrons are bond electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 electrons took part in creating chemical bond in $\ce{OH}$.&#xA;The first one is from oxygen.&lt;br&gt;&#xA;The second is from hydrogen.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of bond is called &quot;Covalent bond&quot;.&#xA;For more about Covalent bond : &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Covalent_bond&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minus is  related to  the black and white electrons that are free (in Oxygen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white electron is come from other element.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:47.383" LastActivityDate="2015-03-06T23:22:47.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7268" PostTypeId="2" ParentId="7262" CreationDate="2013-12-08T14:52:38.107" Score="2" Body="&lt;p&gt;To answer this question you should understand the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot; rel=&quot;nofollow&quot;&gt;kinetic isotope effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you deliver no real data about reaction rate constants I assume that the dissociation of $\ce{HF}$ is faster (has a larger $k_\text{dis}$) than the dissociation of $\ce{DF}$. I postulate the following mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HF + D+ &amp;lt;=&amp;gt;[k_\text{dis}^\ce{HF}] H+ + F- + D+ &amp;lt;=&amp;gt;[][k_{\text{dis}}^\ce{DF}] DF + H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially it is a pretty simple reaction mechanism, but it tends to forming $\ce{DF}$ over $\ce{HF}$ because the deuterium dissociates more slowly.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-08T14:52:38.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7269" PostTypeId="1" CreationDate="2013-12-08T18:45:30.543" Score="0" ViewCount="132" Body="&lt;p&gt;This may be a very broad question.&#xA;I always asked myself how scientists manage to find out the exact molecular structures of for instance water or carbohydrates. How do they know? How do they know the angle between the bonds in water? Is there a classical example demonstrating and substantiating current knowledge on molecular level of structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The if we go to atoms, how do people know what charge an ion has? How can be know we have atoms with 7 protons for instance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably there are zillion ways required to find out, but could you give me some general examples?&lt;/p&gt;&#xA;" OwnerUserId="3743" LastActivityDate="2013-12-08T19:47:50.573" Title="Atomic properties" Tags="&lt;molecules&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2020-01-03T02:59:00.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="7271" PostTypeId="2" ParentId="7265" CreationDate="2013-12-08T19:18:32.597" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The simplest types of chemical formulas are called empirical formulas, which use only letters and numbers indicating atomic proportional ratios (the numerical proportions of atoms of one type to those of other types). &lt;strong&gt;Molecular formulas&lt;/strong&gt; indicate the &lt;strong&gt;simple numbers&lt;/strong&gt; of each type of atom in a molecule of a molecular substance, and are thus sometimes the same as empirical formulas (for molecules that only have one atom of a particular type), and at other times require larger numbers than do empirical formulas. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_formula&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For phosphorus pentoxide the molecule &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorus_pentoxide&quot; rel=&quot;noreferrer&quot;&gt;seems to be&lt;/a&gt; an adamantane-like compound with 4 phosphorus atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct molecular formula of the compound would be $\ce{P4O10}$.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-08T19:18:32.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7272" PostTypeId="2" ParentId="1254" CreationDate="2013-12-08T19:38:57.427" Score="3" Body="&lt;p&gt;Ben's answer covers it well. There is another aspect which affects the products of fire; that is the temperature of the fire, which varies according to the phase of the fire, among other things. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;I found the following list in some educational materials for fire fighters:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Temperature               Reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3920F                       Production of water vapor, carbon dioxide, formic and    (2000C)                       acetic acids&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3920-5360F                  Less water vapor - some carbon monoxide - still primarily an&#xA;(2000-2800C)               endothermic reaction (absorbing heat)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5360-9320F                  Exothermic reaction (giving off heat) with flammable vapors&#xA;(2800-5000C)               and particulates; some secondary reaction from charcoal formed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 9320F                  Residue primarily charcoal with notable catalytic action&#xA;(5000C)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fire is fascinating. As already noted in answers above, fire is a mixture of matter and energy. Fire consists of fuel, oxygen (or another oxidizer) and heat interacting in a chain reaction yielding more heat and oxidized byproducts. &lt;/p&gt;&#xA;" OwnerUserId="3894" LastActivityDate="2013-12-08T19:38:57.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7273" PostTypeId="2" ParentId="7269" CreationDate="2013-12-08T19:47:50.573" Score="4" Body="&lt;p&gt;First it is very important to know that molecules don't freeze at one geometry and stay there forever.  Even at absolute zero temperature, a molecule can go through any geometry imaginable, and what you call molecular structure is one of the most likely structures, which would be more accurately termed &lt;strong&gt;equilibrium structure&lt;/strong&gt;.  Even more sophisticated definitions are also present, depending on wether you measure the vibrational average or thermal average or the actually most statically likely geometry, but I believe this is beyond the scope of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With finite temperature this is more of a problem, in many cases the notion of a single structure may be completely inappropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of ways to determine molecular structure.  However, they all have limitations.  You may be surprised to learn that for many molecules, the structures are not yet determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various experiments can provide information regarding equilibrium structure.   For small molecules, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotational_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;microwave spectroscopy&lt;/a&gt; can be used to get geometry information.   Other spectroscopy methods that can achieve high enough precision to resolve rotational states can also be used to obtain a direct measurement of molecular structure.  These are especially useful if the molecule in question is difficult to prepare or unstable, such as reaction intermediates, ions and free radicals. Examples include &lt;a href=&quot;http://bromine.cchem.berkeley.edu/sevi.htm&quot; rel=&quot;nofollow&quot;&gt;Slow Photoelectron Velocity-Map Imaging&lt;/a&gt;, cryogenic IR spectroscopy and cryogenic ion spectroscopy.  Note that these method only work for isolated molecules in gas phase, and all these are limited to very small molecules.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger molecules, if you can obtain a crystal of it, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;nofollow&quot;&gt;X-ray crystallography&lt;/a&gt; to determine the coordinates of atoms inside.  However, it is important to realize that crystal structure of a molecule can be qualitatively different from the structure in gas phase or solution; also, it is not always possible to obtain a crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you absorb an molecule on a surface, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_force_microscopy&quot; rel=&quot;nofollow&quot;&gt;atomic force microscopy&lt;/a&gt; to directly measure the molecule. However, similar to crystallization process, absorption and AFM itself may disturb the structure, and you cannot just put anything you want on a surface.  AFM is also limited to the atoms on the surface and cannot see atoms wrapped inside, so it is also limited to rather small molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solution, the situation is usually much more complicated and there is not even a clear definition of equilibrium geometry anymore.  For example, even a relatively small number of water molecules can form thousands of different water clusters, each having different equilibrium structure; in bulk water that number is astronomical.  There will be no measurement of a quantity if you cannot even clearly define it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_chemistry&quot; rel=&quot;nofollow&quot;&gt;quantum chemistry&lt;/a&gt; methods to calculate the geometry of a molecule.  This can be done fairly accurately for a moderately large molecule, provided that the electronic structure of that molecule is not too complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently one of the most reliable method of predicting molecule structure is &lt;a href=&quot;https://en.wikipedia.org/wiki/Coupled_cluster&quot; rel=&quot;nofollow&quot;&gt;coupled cluster method&lt;/a&gt;. For small molecules (let us say up to 100 atoms) CCSDT/CBS is usually describe as &quot;gold standard&quot; and in many cases can provide similar precision with experimental measurements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large molecules (for example, up to a few hundred atoms) one usually use &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;density functional theory&lt;/a&gt;, which less reliable but in many cases can still be surprisingly accurate.  However, if not used correctly it can easily provide results that are not even remotely correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even larger molecules such as protein, &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_mechanics&quot; rel=&quot;nofollow&quot;&gt;molecular mechanics&lt;/a&gt; methods or &lt;a href=&quot;https://en.wikipedia.org/wiki/QM/MM&quot; rel=&quot;nofollow&quot;&gt;a mixture of quantum and classical mechanics&lt;/a&gt; can be used, which by the way won &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/2013/&quot; rel=&quot;nofollow&quot;&gt;the 2013 Nobel Prize in Chemistry&lt;/a&gt;.   This is, however, so dependent on the details of implementation and expertise of the users that any results must be viewed with utmost discretion.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-08T19:47:50.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7279" PostTypeId="1" CreationDate="2013-12-09T10:17:35.110" Score="4" ViewCount="245" Body="&lt;p&gt;There are many types of plastics that can be used for food packaging, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Polyethylene terephthalate (PET) &lt;/li&gt;&#xA;&lt;li&gt;Polyethylene (PE) &lt;/li&gt;&#xA;&lt;li&gt;Polyvinyl chloride (PVC) &lt;/li&gt;&#xA;&lt;li&gt;Polypropylene (PP)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are the differences between them?&#xA;Which one is most suitable for packaging pastries, for example, biscuits?&lt;/p&gt;&#xA;" OwnerUserId="3899" LastActivityDate="2014-03-01T07:41:09.597" Title="Plastics for food packaging" Tags="&lt;plastics&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7281" PostTypeId="1" AcceptedAnswerId="7297" CreationDate="2013-12-09T15:42:48.363" Score="2" ViewCount="884" Body="&lt;p&gt;Take acetylene, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NI1Qe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is defined that the two outer bonds are at 180 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible to theoretically define all $\ce{C-H}$ bonds in acetylene as being 180 degrees?&lt;/p&gt;&#xA;" OwnerUserId="154" LastEditorUserId="4231" LastEditDate="2015-03-06T23:22:58.907" LastActivityDate="2015-03-06T23:22:58.907" Title="How can it be certain that some bonds are 180 degrees?" Tags="&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7282" PostTypeId="1" CreationDate="2013-12-09T20:29:08.217" Score="3" ViewCount="1271" Body="&lt;p&gt;I'm currently seeking your expertise while creating a line of environmentally friendly yoga mats, and I apologize in advance if any chemistry knowledge of mine has been absent since college and I lack the technical capability of this forum. I typically have a scientific approach to problem solving, and I'm hoping I can apply the same methodology to finalizing the material composition of my product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of blindly accepting the industry suggestions of natural rubber, NBR, TPE, and PER, I wanted to start with the physical characteristics required and work backwards to determine the ideal (perhaps innovative) material that would work best. My requirements are: 1) Eco-Friendly, 2) high fidelity printing compatibility via heat transfer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Environmentally friendly: PVC is not an option for the reasons I'm sure you would know. One would assume natural rubber would be the obvious choice, however if it degrades and has to be replaced every 2 years or so, I'd think the long term production effects might actually be more harmful then something more sustainable. From reading an article on PER it seems like it is essentially phthalate free PVC, and although processed different to be phthalate free the vinyl chloride monomer still exists and could be viewed as toxic, although to a lesser degree then PVC. Nitrile Rubber (NBR) is another option but it sounds like although it might be more heat resistant then natural rubber, it degrades similar to natural rubber. Various types of TPEs are also an option, but I've found minimum answers towards their actual eco-friendliness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Heat resistant printing - Sublimation printing has typical been used on polyester, but from what I've researched should be compatible with any polymer. Unfortunately it requires a heat transfer of about 400 deg F, which I believe would scorch/melt natural rubber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know its a stretch, but I would love to get your input on the above materials or any other polymer that might be an option. I don't want to blindly commit to an &quot;eco-friendly&quot; material without fully investigating their chemical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for your help, and I'm definitely open to all of your respected critiques and suggestions.&#xA;Regards,&#xA;Shane&lt;/p&gt;&#xA;" OwnerUserId="3902" LastActivityDate="2014-03-11T17:04:35.830" Title="A chemistry approach to environmentally friendly yoga mats… NBR, NR, TPE, PER" Tags="&lt;thermodynamics&gt;&lt;polymers&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-03-23T15:19:37.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="7283" PostTypeId="1" AcceptedAnswerId="7284" CreationDate="2013-12-09T23:40:55.703" Score="3" ViewCount="655" Body="&lt;p&gt;In a similar vein and for pretty much the same reason as I post &lt;a href=&quot;https://physics.stackexchange.com/questions/76650/explaining-uv-radiation-to-a-6-year-old&quot;&gt;this question on the Physics.SE site&lt;/a&gt;, what would be a an effective means to promote understanding about redox reactions to a 6 year old?  In particular, what (if any) simulations are available?&lt;/p&gt;&#xA;" OwnerDisplayName="user3901" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2013-12-13T19:00:52.123" Title="Explaining redox reactions to a 6 year old?" Tags="&lt;redox&gt;&lt;teaching-lab&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7284" PostTypeId="2" ParentId="7283" CreationDate="2013-12-10T00:44:20.147" Score="4" Body="&lt;p&gt;I might suggest you take a look at &lt;em&gt;What's Chemistry All About&lt;/em&gt; by Frith, Gillespie and Larkum.  (ISBN: 1409547078)  It is a very approachable book that in my mind pays close attention to the science jargon introduced to its audience.  Redox reactions are described in the section &quot;Let's make things happen&quot; under the types of reactions.  A bit of history about where the name oxidation comes from and the safety symbol that you will see on an oxidizing agent are both included in the brief discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example reactions given are combustion and firecrackers, which I think are appropriate and accessible to a child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for experiments to do, I'll suggest another book, the &lt;em&gt;Illustrated Guide to Home Chemistry Experiments&lt;/em&gt; by Thompson. (ISBN: 0596514921) There is a chapter on electrochemistry with two experiments that might be of interest to a young learner: the electrolysis of water to make oxygen and hydrogen gas, and the oxidation of iron.  Of course, there's also the &quot;make your battery&quot; type experiments which are always fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for simulations, I would suggest heading on over to the &lt;a href=&quot;http://phet.ucolorado.edu&quot; rel=&quot;nofollow&quot;&gt;PHet&lt;/a&gt; website and have a stroll through all the activities there.  I don't recall off hand what they have for electrochemistry, but there's plenty to keep a budding scientist interested.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-12-10T00:44:20.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7285" PostTypeId="2" ParentId="7235" CreationDate="2013-12-10T02:27:10.343" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mnC9H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Energy-Level Diagram&lt;/a&gt;&#xA;&lt;br&gt;There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).&#xA;&lt;br&gt;&#xA;&lt;br&gt;I've attached a HE-Ne laser schematic from &lt;a href=&quot;http://books.google.com/books/about/Laser_electronics.html?id=WvFRAAAAMAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Verdeyen's Laser Electronics book&lt;/a&gt;. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2. &#xA;&lt;br&gt;&#xA;&lt;br&gt;However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum. &lt;/p&gt;&#xA;" OwnerUserId="3903" LastActivityDate="2013-12-10T02:27:10.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7289" PostTypeId="1" AcceptedAnswerId="7367" CreationDate="2013-12-10T05:06:53.037" Score="8" ViewCount="186" Body="&lt;p&gt;I admit that my knowledge of collision theory may be lacking, but, as I understand it, when particles collide, a reaction will not occur without overcoming the activation energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, as the temperature of the environment in which the collision takes place is decreased, I believe it is logical that the kinetic energy of these particles will also decrease. Hence, to me, it would make sense if none of these particles had the energy required to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: Why do reactions still occur when the environment's temperature around a collision decreases? That is, shouldn't there be a point (such as the temperature in a freezer, perhaps?) in which the activation energy cannot be overcome at all?&lt;/p&gt;&#xA;" OwnerUserId="581" LastEditorUserId="16683" LastEditDate="2016-09-30T10:11:12.117" LastActivityDate="2016-09-30T10:11:12.117" Title="When temperature is decreased, why do reactions occur at all?" Tags="&lt;energy&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7290" PostTypeId="2" ParentId="7245" CreationDate="2013-12-10T08:28:20.533" Score="2" Body="&lt;p&gt;I ran this question past our mass spec. guy, and he offers the following suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have you run a blank to ensure you are not just measuring some other random compound; Or contaminants from the instrument; Or gas system.&lt;/li&gt;&#xA;&lt;li&gt;Be aware when measuring ions less than 45m/z that you are not just measuring ions from air leaking in. For example: 18/19 is H2O; 28 is N2; 32 is O2; 40 is Ar; 44 is CO2. There are probably other ions to watch as well.&lt;/li&gt;&#xA;&lt;li&gt;Is the ionization energy set to 70eV and it is measured in positive ion mode as that is what most online libraries use&lt;/li&gt;&#xA;&lt;li&gt;Is too much sample being injected in the instrument as you do measure different fragment patterns when a lot of sample is injected as the molecular ion may be a charge bound cluster (similar to what is seen in ESI).&lt;/li&gt;&#xA;&lt;li&gt;Is the acetic acid reacting with anything in the sample introduction path which is giving another compound? If it is a water injection into a GCMS, not only are you destroying your column but the solvent will expand in the injection port and goes back into your gas system as water has a very high expansion when going from liquid to gas.&lt;/li&gt;&#xA;&lt;li&gt;Are you sure, really, really sure you are using acetic acid and not something else?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-12-10T08:28:20.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7293" PostTypeId="1" AcceptedAnswerId="24060" CreationDate="2013-12-10T13:03:19.557" Score="9" ViewCount="4705" Body="&lt;p&gt;How is the spontaneity of an exothermic reaction changed when temperature is decreased? &#xA;Assuming the entropy is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Gibbs free energy equation, $\Delta G=\Delta H-T\Delta S$, $\Delta G$ should become less negative as temperature is decreased meaning the reaction is less spontaneous. However, using Le Chatelier's principle, as temperature decreases the reaction should shift towards the right towards the products, becoming more spontaneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3908" LastEditorUserId="1431" LastEditDate="2013-12-11T14:08:58.097" LastActivityDate="2015-10-07T13:19:30.907" Title="Spontaneity of an exothermic reaction when temperature is decreased" Tags="&lt;equilibrium&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7294" PostTypeId="1" CreationDate="2013-12-10T13:25:41.747" Score="1" ViewCount="878" Body="&lt;p&gt;Silver compounds such as silver fluoride are light sensitive (as are all other silver halides) so they will decompose when exposed to light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck with a problem : I want to expose silver(I)fluoride to light to then produce elemental silver and fluorine gas - I'm worried that because of the high reactivity of fluorine silver(I)fluoride will be formed again...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how could I successfully split silver(I)fluoride by light into elemental silver and fluorine gas without the fluorine reacting again with the silver?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="1499" LastEditDate="2014-03-08T01:19:19.753" LastActivityDate="2014-03-08T01:19:19.753" Title="Light sensitive AgF decomposition reaction" Tags="&lt;inorganic-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7295" PostTypeId="1" AcceptedAnswerId="7296" CreationDate="2013-12-10T16:26:02.833" Score="6" ViewCount="446" Body="&lt;p&gt;If we replace two oxygen in $\ce{-COOH}$ what is the name of the compound we get?&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="4945" LastEditDate="2014-09-11T07:12:05.050" LastActivityDate="2014-09-11T07:12:05.050" Title="Which functional group is -CSSH?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7296" PostTypeId="2" ParentId="7295" CreationDate="2013-12-10T18:30:43.140" Score="5" Body="&lt;p&gt;Is a dithioic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a benzene with two -CSSH hubstituents in 1, 2 positions is a benzene-1,2-dicarbodithioic acid.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="2578" LastEditDate="2013-12-10T18:59:46.223" LastActivityDate="2013-12-10T18:59:46.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7297" PostTypeId="2" ParentId="7281" CreationDate="2013-12-10T19:26:46.230" Score="0" Body="&lt;p&gt;If we look at the molecule we see it is symmetrical. The forces acting upon the 2 hydrogen atoms are identical. As you know, when hydrogen forms a covalent bond with carbon, it's electron is drawn more strongly by the carbon than by the hydrogen, which leaves the hydrogen with a net positive charge (carbon is more electronegative than hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the 3 covalent bonds between the carbons are kind of bowed outward in the middle. The 3 bonds between the carbons constitute the most negative part of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally, the positive and negative charged moieties are drawn toward each other as much as possible. But, given that the force of attraction is equal on all sides of the hydrogen, it will be aligned equidistant on all sides to the negative triple bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I am painting a good word picture here, but the forces exerted on the hydrogens place them exactly on a straight line, i.e. their bonds are 180 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are other forces in play, namely the forces of neighboring molecules. However, they will also align themselves as much as possible so as to minimize the repulsion of having hydrogen atoms close to one another and to maximize proximity of hydrogen atoms to the triple bonds of neighboring molecules. This arrangement will be preserved as much as possible as the molecules move about. So, the net forces upon the hydrogen-carbon bonds will always force those bonds to remain 180 degrees.     &lt;/p&gt;&#xA;" OwnerUserId="3894" LastActivityDate="2013-12-10T19:26:46.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7299" PostTypeId="2" ParentId="7283" CreationDate="2013-12-11T06:19:24.120" Score="1" Body="&lt;p&gt;I was taught using the OIL RIG acronym, where oxidation is loss, and reduction is gain. I would also utilise a picture of some sort showing how a oxidant is reduced, and a reducer is oxidised. I would also mention that most oxidising agents have oxygen in them, whereas reducing agents can have hydrogen. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="3912" LastActivityDate="2013-12-11T06:19:24.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7301" PostTypeId="2" ParentId="7132" CreationDate="2013-12-11T06:44:43.483" Score="2" Body="&lt;p&gt;Well a battery works on the principle of dissolving one of the electrodes with the electrolyte. This means that you would have a solution of the metal and electrolyte in your stomach. If consumed in large quantities or in quick succession, then you most probably would notice some physiological effects, however, I would assume they'd be very minor given the mass of the tablet in question. Other than under those circumstances, I would expect the physiological effect to be non-existant. Given this has been approved by the FDA, you can be pretty confident that it's safe. &lt;/p&gt;&#xA;" OwnerUserId="3912" LastActivityDate="2013-12-11T06:44:43.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7302" PostTypeId="2" ParentId="6212" CreationDate="2013-12-11T07:56:52.467" Score="2" Body="&lt;p&gt;Magnesium is a very active metal- highest activity of all engineering metals. Your question gives you part of the answer. In cold water it forms a decent protective layer. Given greater activation energy- simple as hot water, that layers is simply damaged the oxidation of the magnesium proceeds unimpeded at a greater rate. Once its in steam the oxidation is extremely rapid.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-12-11T07:56:52.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7303" PostTypeId="2" ParentId="2598" CreationDate="2013-12-11T08:10:46.990" Score="0" Body="&lt;p&gt;Magnesium's oxide layer is not protective- its not tenacious like aluminium oxide on aluminium. Its simply that virtually all metals form an oxide layer almost instantaneous with cold dry air- the activation potential is small because the surface of metals has high energy due to deformed metal bonds to form the surface. After the initial oxide is formed, the reaction always slows down as a greater activation energy is needed for it to continue- both iron and magnesium are still corroding but just its not obvious.  Also don't confuse iron oxide (which would initially FeO) with rust Fe2O3.xH20- with the normal thickness of FeO on iron its just about clear.          &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-12-11T08:10:46.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7304" PostTypeId="1" AcceptedAnswerId="8025" CreationDate="2013-12-11T10:09:11.040" Score="7" ViewCount="1221" Body="&lt;p&gt;In class, I learned that when $\ce{NaCl}$ is dissolved in water, the partially negative oxygen part of the water molecules surround the $\ce{Na+}$ ion and energy is given off (hydration enthalpy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, when a transition metal ion like $\ce{Fe^2+}$ is dissolved in water, water molecules surround the ion. However, in this case, we say that a complex ion is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please tell if there is really a difference in the two cases and if yes, could someone specify them.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4231" LastEditDate="2015-03-09T15:29:09.877" LastActivityDate="2015-03-09T15:29:09.877" Title="Difference between sodium ion and a transition metal ion dissolving in water?" Tags="&lt;ions&gt;&lt;energy&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7305" PostTypeId="2" ParentId="2883" CreationDate="2013-12-11T10:59:22.220" Score="3" Body="&lt;h1&gt;Charge density&lt;/h1&gt;&#xA;&lt;p&gt;Charge density is the density of charge around the ion. Silberberg (Chemistry: The Molecular Nature of Matter and Change. 4th ed. 2006) define it as:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The ratio of the charge of an ion to its volume.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The factors playing into the strength of an ion's charge density are therefore the &lt;em&gt;charge of the ion&lt;/em&gt; (e.g. 2+ for Mg, 1+ for Na) and the &lt;em&gt;effective volume which that charge acts over&lt;/em&gt; - which is where the ionic radius comes in. (Note this is different to atomic radius.)&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Charge density = charge/volume&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;As the magnesium ion and sodium ion both have the same electronic configuration i.e. that of neon (because Mg has lost 2 electrons and Na has lost 1 and they now both have 10 electrons in total) the charge of the electrons around the ions is the same. But - magnesium has one more proton in its nucleus. This means magnesium has a smaller radius (think more condensed due to the greater pull of the nucleus - see &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective nuclear charge&lt;/a&gt;). &lt;span class=&quot;math-container&quot;&gt;$Na^+$&lt;/span&gt; has an ionic radius of 102 pm and &lt;span class=&quot;math-container&quot;&gt;$Mg^{2+}$&lt;/span&gt; has an ionic radius of 72 pm. (Also from Silberberg)&lt;/p&gt;&#xA;&lt;p&gt;Since the volume is less and the charge is greater for magnesium, the charge density will be greater for magnesium than for sodium.&lt;/p&gt;&#xA;&lt;h1&gt;Periodic trend&lt;/h1&gt;&#xA;&lt;p&gt;The periodic trend is that &lt;strong&gt;charge density will decrease going &lt;em&gt;down&lt;/em&gt; the periodic table&lt;/strong&gt; as the charge stays the same but the size increases. Going across (as for Mg and Na) the &lt;strong&gt;charge density &lt;em&gt;increases&lt;/em&gt; going to the right&lt;/strong&gt; as the charge increases and the size decreases.&lt;/p&gt;&#xA;&lt;h1&gt;...&lt;/h1&gt;&#xA;&lt;p&gt;What do you mean by &amp;quot;What does charge density show?&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2013-12-11T10:59:22.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7306" PostTypeId="1" CreationDate="2013-12-11T12:23:50.877" Score="10" ViewCount="218" Body="&lt;p&gt;For these two metals, the standard reduction potentials are:&#xA;$$&#xA;\ce{2e-} + \ce{Cd^{2+}} \rightarrow \ce{Cd}:\mbox{ } E^0 = -0.403 V\\&#xA;\ce{2e-} + \ce{Ni^{2+}} \rightarrow \ce{Ni}:\mbox{ } E^0 = -0.25 V&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the simple ionization energies are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{Ni} \rightarrow  \ce{Ni^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2490 \ce{kJ. mol^{-1}}$$ $$ \ce{Cd} \rightarrow  \ce{Cd^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2500\ce{kJ. mol^{-1}}$$&#xA;The difference in $E^\theta$ is greater than the difference in ionization energy of both elements.&lt;br&gt;&#xA;How can this be explained?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="5017" LastEditDate="2014-09-22T17:42:08.373" LastActivityDate="2014-09-22T18:57:53.503" Title="How can normal potentials be explained?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7307" PostTypeId="2" ParentId="7282" CreationDate="2013-12-11T13:57:09.867" Score="1" Body="&lt;p&gt;You will need a natural woven fiber. Any film will not help the skin in contact with it.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2013-12-11T13:57:09.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7308" PostTypeId="1" AcceptedAnswerId="7311" CreationDate="2013-12-11T15:35:44.077" Score="17" ViewCount="243131" Body="&lt;p&gt;What's the difference between $\alpha$-D-glucose and $\beta$-D-glucose? Are they isomers? Or do they differ in their elemental composition?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="7475" LastEditDate="2016-10-30T23:07:43.897" LastActivityDate="2017-04-02T20:15:17.633" Title="What's the difference between alpha-glucose and beta-glucose?" Tags="&lt;stereochemistry&gt;&lt;isomers&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7309" PostTypeId="1" CreationDate="2013-12-11T16:19:41.853" Score="12" ViewCount="7993" Body="&lt;p&gt;My book says that benzoic acid does not show resonance as the carboxylate anion and the benzene ring are not in the same plane due to steric effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there aren't any large groups in the ortho positions to cause steric inhibition of resonance. Then, is the info given in the book wrong? If not, where does the steric effect come from?&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorUserId="5026" LastEditDate="2018-04-25T08:28:12.987" LastActivityDate="2018-04-25T08:28:43.330" Title="Does unsubstituted benzoic acid not show resonance due to steric effects?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7310" PostTypeId="1" CreationDate="2013-12-11T16:26:04.933" Score="3" ViewCount="592" Body="&lt;p&gt;I would like to understand the temperature at which a substance will vaporize when dissolved in a liquid. I have researched this online for hours, but haven't found a conclusive answer. Is it the boiling point of the dissolved substance? I'm attempting to find the temperature at which caffeine vaporizes when dissolved in water or other vegetable glycerin. Many thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="3904" LastActivityDate="2013-12-13T14:01:58.543" Title="How do I determine the temperature at which a dissolved substance evaporates?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7311" PostTypeId="2" ParentId="7308" CreationDate="2013-12-11T16:34:26.173" Score="29" Body="&lt;p&gt;$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\alpha$-D-glucose &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MYOEj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\beta$-D-glucose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JUzVl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is &lt;strong&gt;down&lt;/strong&gt;, and, in the $\beta$ form, the $\ce{OH}$ group on the far right is &lt;strong&gt;up&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, they are a class of stereoisomer called an &lt;a href=&quot;http://en.wikipedia.org/wiki/Anomer&quot; rel=&quot;noreferrer&quot;&gt;anomer&lt;/a&gt;. Anomers are capable of interconverting in solution. All cyclic structures of monosaccharides exhibit anomeric $\alpha$ (down) and $\beta$ (up) versions. These differences occur at the anomeric acetal carbon (the only carbon with two $\ce{C-O}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two forms exist because all monosaccharides also have an open-chain form with &lt;strong&gt;one fewer stereocenter&lt;/strong&gt;. When the chain closes to the cyclic structure, the aldehyde or ketone carbon becomes a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29#Stereogenic_centers&quot; rel=&quot;noreferrer&quot;&gt;stereocenter&lt;/a&gt;, and it can do so in either configuration. One configuration is preferred ($\beta$), but both exist. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Open chain form of glucose:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/71elQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of acid or base (although water can fulfill this role if need be), the two anomers interconvert through the open form until dynamic equilibrium is established. The mechanism below starts with $\alpha$ in the upper left and finishes with $\beta$ in the lower right. The open-chain form is in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ah5Ov.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="402" LastEditDate="2017-04-02T20:15:17.633" LastActivityDate="2017-04-02T20:15:17.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7312" PostTypeId="2" ParentId="7309" CreationDate="2013-12-11T18:44:11.647" Score="9" Body="&lt;p&gt;I have never heard of this effect occurring with an ordinary, unsubstituted benzoic acid. There is, however, a phenomenon known as the &lt;em&gt;ortho effect&lt;/em&gt; that occurs when bulky groups occupy the ortho positions on the ring. In that case, the steric hindrance between the bulky groups and the carboxylic acid moiety forces the carboxylic acid group to rotate out of the plane of the aromatic ring, thereby breaking the conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the consequences of this, which may not be readily apparent, is that this often produces a substantial increase in the acidity of the carboxylic acid, even when the substituents are not electron-withdrawing (viz. 2,6-dimethylbenzoic acid, which is disubstituted at the two ortho positions, having a $\mathrm{p}K_\mathrm{a}$ roughly one unit lower than ordinary benzoic acid, while similar para- and meta-substituted benzoic acids have higher $\mathrm{p}K_\mathrm{a}$ values &amp;mdash; that is, are less acidic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I've seen references to this effect in multiple sources, I've never actually found a thorough explanation for it. I think it could be argued that the benzene ring is electron-donating within the $\pi$-system (i.e., by resonance), which would only serve to destabilize the carboxylate conjugate base already bearing a negative charge. By rotating out of the plane, the electron-donating effect is removed. Rather than consider the stability of the resulting carboxylate, I suppose one could also consider the stability of the carboxylic acid directly. Here, I'd expect that the reactivity of the carboxylic acid flanked by bulky substituents should be greater, as the loss of $\pi$-conjugation with the carboxylic acid moiety is inherently destabilizing. My reasoning may be flawed, though, so if anybody has some corrections or a better explanation to offer, that would be entirely welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this effect occurs with ordinary benzoic acid or not, I can't be certain. At least one good reason I can think of to doubt that it does is the observed regioselectivity of electrophilic aromatic substitution reactions on benzene rings conjugated to carbonyl groups. They are known to be deactivating (i.e., they withdraw electron density) via their conjugation with the $\pi$ system of the aromatic ring. If they were not coplanar with the ring, the conjugation would be impossible, and the deactivating effect would not be observed.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-04-25T08:28:43.330" LastActivityDate="2018-04-25T08:28:43.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7313" PostTypeId="1" AcceptedAnswerId="7329" CreationDate="2013-12-11T20:38:50.197" Score="3" ViewCount="560" Body="&lt;p&gt;I produce high density balsamic vinegar, when it's winter and the temperatures are very low, high density products tends to crystallize. think this is a problem of over saturation of the glucose. The main sugars are glucose and fructose the ratio is 1:0.95.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to make products that are lower than 70 brix, or 1,35 density and 6% acidity, composed by $\ce{CH3COOH}$ and tartaric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I exactly calculate the solubility threshold of this mixture? How can I measure it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prevent the crystallization to happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&#xA;&lt;em&gt;I can't sleep at night because of this problem !!!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3915" LastEditorUserId="4945" LastEditDate="2015-02-27T06:29:56.327" LastActivityDate="2015-02-27T06:29:56.327" Title="How can I avoid crystallisation of sugars in vinegar at low temperatures?" Tags="&lt;solubility&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7314" PostTypeId="1" AcceptedAnswerId="7319" CreationDate="2013-12-11T21:27:31.317" Score="2" ViewCount="5441" Body="&lt;p&gt;Would you add in $\ce{NaOH}$, deprotonating the acid, and then wash the mixture in water, thereby removing the $\ce{Na^+}$ and $\ce{CH3O^-}$ ions in an aqueous layer? But then wouldn't the ethanol be dissolved in the water as well, as it is polar? In general, how do you separate two polar substances from each other using extraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't use extraction, how else would you separate the two, aside from distillation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All answers are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="4231" LastEditDate="2015-03-06T23:23:08.580" LastActivityDate="2020-08-30T18:06:33.720" Title="How does one separate ethanol from acetic acid using aqueous extraction?" Tags="&lt;extraction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7315" PostTypeId="1" AcceptedAnswerId="7320" CreationDate="2013-12-11T21:37:33.307" Score="2" ViewCount="2346" Body="&lt;p&gt;I know outer electrons include (n-1)d...+ all n'ss.. But i don't understand, never been told what unpaired electron is supposed to be? is it the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so in the following electron configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4s^2\: 3d^{10}\: 4p^3$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OuterElectrons: $15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ValenceElectrons: $5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from what I remember someone told me unpaired electrons are $15$ too for this...which is the same as  outer electrons so I guess it's the same thing. But I am not sure so please help. &lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="3826" LastEditDate="2013-12-12T12:24:27.833" LastActivityDate="2013-12-12T12:24:27.833" Title="what is a difference between outer electrons and unpaired electrons" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7317" PostTypeId="2" ParentId="6381" CreationDate="2013-12-11T22:05:16.457" Score="18" Body="&lt;p&gt;This is a &lt;a href=&quot;https://books.google.com/ngrams/graph?content=(Zinc+-+_START_+Zinc)%2Czinc%2C(Iodine+-+_START_+Iodine)%2Ciodine%2Cnitrogen%2C(Nitrogen+-+_START_+Nitrogen)%2Ccarbon%2C(Carbon+-+_START_+Carbon)&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t1%3B%2C%28Zinc%20-%20%20_START_%20Zinc%29%3B%2Cc0%3B.t1%3B%2Czinc%3B%2Cc0%3B.t1%3B%2C%28Iodine%20-%20%20_START_%20Iodine%29%3B%2Cc0%3B.t1%3B%2Ciodine%3B%2Cc0%3B.t1%3B%2Cnitrogen%3B%2Cc0%3B.t1%3B%2C%28Nitrogen%20-%20%20_START_%20Nitrogen%29%3B%2Cc0%3B.t1%3B%2Ccarbon%3B%2Cc0%3B.t1%3B%2C%28Carbon%20-%20%20_START_%20Carbon%29%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google ngram&lt;/a&gt; of the usage of the words: iodine, carbon, nitrogen, zinc and Iodine, Carbon, Nitrogen, Zinc (not at the beginning of the phrase).&lt;/p&gt;&#xA;&lt;p&gt;There is no historical trend, at least not an usage of an archaic form (we can see that the maximum  usage of the capitalized element name is from 1900 to 1960 but normalized to the overall usage is steady).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6Uim8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can make some hypothesis about why someone could choose to write capitalized element names:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we must write element abbreviation capitalized (Zn, Cu, C, N) I  can think I have to write the element capitalized even if is not abbreviated.&lt;/li&gt;&#xA;&lt;li&gt;Someone (me too) may think that if I write Zinc it is more unambiguous and means I'm talking about the concept of element &lt;a href=&quot;http://goldbook.iupac.org/C01022.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&amp;quot;all atoms with the same number of protons in the atomic nucleus&amp;quot;&lt;/strong&gt;&lt;/a&gt; and not about the metal zinc.  If you write report with some elemental techniques (XRF, AES) sometimes you feel the urge to specify to the reader (that maybe is not a chemist): &amp;quot;when I write that I've found zinc I mean that in the sample could be present &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; but also &lt;span class=&quot;math-container&quot;&gt;$\ce{ZnO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^{2+}}$&lt;/span&gt;&amp;quot;.&lt;/li&gt;&#xA;&lt;li&gt;The fact the atoms are identical and the definition of a proper noun is: &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_noun&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;&amp;quot;a noun that in its primary application refers to a unique entity&amp;quot;&lt;/em&gt;&lt;/a&gt; could induce to think that elements may fit the definition, and so gain the privilege to be capitalized.&lt;/li&gt;&#xA;&lt;li&gt;In alchemy elements such water and fire were capitalized.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wrORr.png&quot; alt=&quot;enter image description here&quot; /&gt;&#xA;However up to my knowledge the form not capitalized is preferred even in the others languages, in Italian I'm sure.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="9961" LastEditDate="2021-02-15T15:05:49.837" LastActivityDate="2021-02-15T15:05:49.837" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7318" PostTypeId="1" CreationDate="2013-12-11T22:47:48.487" Score="1" ViewCount="129" Body="&lt;p&gt;In &lt;a href=&quot;https://www.youtube.com/watch?v=yBrp8uvNAhI&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;, Sal Khan says that an electron reaches a higher energy state if the electron possesses more energy. So I assume that, say in the case of carbon, the electrons not in the S1 orbital have more energy. Where does this energy come from?&lt;/p&gt;&#xA;" OwnerUserId="3916" LastActivityDate="2014-01-11T06:43:35.297" Title="Where do the electrons not in the S1 orbital get this &quot;extra energy&quot; from?" Tags="&lt;energy&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7319" PostTypeId="2" ParentId="7314" CreationDate="2013-12-11T23:42:31.323" Score="3" Body="&lt;p&gt;Well, distillation is almost certainly the best method. However, since you've explicitly ruled that out, there are only a few alternatives I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ethanol can be separated from water with reasonable efficiency (at least on the microscale) by a process of &quot;salting out.&quot; The ethanol/water mixture is saturated with $\ce{NaCl}$. The solvation of the salt ions is a highly thermodynamically favorable process, and as that occurs the molecules of water become increasingly less able and available to form hydrogen bonds with ethanol. I've never tried this with acetic acid specifically, but I can't think of any reason why it wouldn't work.&lt;/li&gt;&#xA;&lt;li&gt;I would imagine with the right choice of eluent, separation by column chromatography using silica or alumina should be feasible.&lt;/li&gt;&#xA;&lt;li&gt;Fractional freezing may be possible in theory, although I have no idea as to its efficacy for this specific case.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm also assuming here that you're looking strictly for methods of physical separation, not chemical (so, strictly as a general example: adding something like calcium oxide to consume water in the generation of insoluble calcium hydroxide would be out of the question).&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-11T23:42:31.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7320" PostTypeId="2" ParentId="7315" CreationDate="2013-12-12T04:31:18.760" Score="1" Body="&lt;p&gt;Outer electrons and unpaired electrons are not the same. Outer electrons are, as you said, $(n-1)d$ and all $n$'s. But unpaired electrons are different. To understand what they are, you must know what is electron spin (which, I hope you do). If you don't, read on.&lt;br&gt;&#xA;Every electron in an atom has a quantum spin state (either clockwise or anticlockwise, nothing else) denoted by $+\frac{1}{2}$ and $-\frac{1}{2}$. These spin quantum numbers have no classical analogue, i.e., you can't relate them to any phenomenon in real life situations that you can visually perceive.&lt;br&gt;&#xA;So, you see, every electron is paired with another electron with opposite spin. Together, the two of them fill one orbital, as for example, $2p_x$, $3d_{xy}$ etc. These orbitals are found out by solving the Schrodinger Equation.&lt;br&gt;&#xA;Now, whenever an electron in an orbital does not have another electron with opposite spin, it is called an unpaired electron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic configuration that is given is $4s^2\: 3d^{10}\: 4p^3$. As, you can see, the $4p$ is filled with $3$ electrons whereas its maximum capacity is $6$. $4p$ has $3$ orbitals and so, each orbital is filled with one electron each, as given (according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hund%27s_rule_of_maximum_multiplicity&quot; rel=&quot;nofollow&quot;&gt;Hund's Rule&lt;/a&gt;). So, these three are unpaired electrons. All other shells are fully filled and so, there are no more unpaired electrons.&lt;br&gt;&#xA;So, number of unpaired electrons is 3 not 15.&lt;/p&gt;&#xA;" OwnerUserId="3826" LastActivityDate="2013-12-12T04:31:18.760" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7321" PostTypeId="2" ParentId="5601" CreationDate="2013-12-12T04:51:32.090" Score="2" Body="&lt;p&gt;Straight hydrochloric acid- Say 20% is virtually  inert to Tantalum- but significantly corrosive to titanium. If you heat it to boiling point of hydrochloric- it will eat through the titanium.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.parrinst.com/wp-content/uploads/downloads/2011/07/Parr_Titanium-Corrosion-Info.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.parrinst.com/wp-content/uploads/downloads/2011/07/Parr_Titanium-Corrosion-Info.pdf&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.tantaline.com/Tantalum-Corrosion-71.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.tantaline.com/Tantalum-Corrosion-71.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-12-12T04:51:32.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7322" PostTypeId="2" ParentId="7318" CreationDate="2013-12-12T06:40:05.087" Score="1" Body="&lt;p&gt;In simple terms: The further apart the negatively charged electrons are from the positively charged nucleus, the higher is the potential energy of said electrons. When increasing the quantum number $n$, you automatically get further away from the nucleus. That is why the $2s$ electrons have a higher energy than $1s$ electrons. (That, and adding that the $1s$ electrons shield some of the nuclear charge.)&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-12T06:40:05.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7323" PostTypeId="1" CreationDate="2013-12-12T07:01:14.583" Score="10" ViewCount="1564" Body="&lt;p&gt;When do covalent bonds of a molecule $M$ extend or compress while no bonds in $M$ are broken? I can conceive of some possibilities but I don't know how common they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Temperature changes&lt;/li&gt;&#xA;&lt;li&gt;Inter-molecular forces around $M$ which may stretch or compress $M$ in various directions&lt;/li&gt;&#xA;&lt;li&gt;The transition from $M$ into a different conformation isomer (perhaps some stretching and twisting is required to overcome the energy barrier for a large change in conformation?)&lt;/li&gt;&#xA;&lt;li&gt;Perhaps bonds can store energy like springs?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How likely are the above to happen, and are there other situations which induce bond deformation? Could I also have some references or topics to search on related topics? As far as I know, covalent bonds are relatively rigid and don't deform much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: one answer.&lt;/strong&gt; Upon some &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00174a012&quot; rel=&quot;noreferrer&quot;&gt;literature review&lt;/a&gt;, it seems that &lt;a href=&quot;http://www.cliffsnotes.com/sciences/chemistry/organic-chemistry-i/structure-of-organic-molecules/free-rotation-around-single-bonds&quot; rel=&quot;noreferrer&quot;&gt;rotation about $\sigma$-bonds&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;noreferrer&quot;&gt;molecular inversion&lt;/a&gt; involve temporary bond deformations because the intermediate states (&lt;a href=&quot;http://en.wikipedia.org/wiki/Eclipsed_conformation&quot; rel=&quot;noreferrer&quot;&gt;eclipsed conformations&lt;/a&gt; for the former, and the planar/linear form for the latter) involve a change in potential energy. After the transformation, the bond lengths revert to the original.&lt;/p&gt;&#xA;" OwnerUserId="3917" LastEditorUserId="3917" LastEditDate="2013-12-15T09:29:25.470" LastActivityDate="2013-12-15T09:29:25.470" Title="Covalent bond deformation" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;stereochemistry&gt;&lt;conformers&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7326" PostTypeId="1" CreationDate="2013-12-12T08:55:21.460" Score="2" ViewCount="5500" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bnWO5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused about how this chair conformation was constructed and why that particular Cl leaves. Why is one of the Cl's drawn axial while the others are drawn equatorial (it seems like the others could have been drawn axial as well)? And, to reiterate, why is it that Cl that's favored to leave over the other two?&lt;/p&gt;&#xA;" OwnerUserId="3919" LastEditorUserId="186" LastEditDate="2015-10-02T16:23:43.953" LastActivityDate="2015-10-02T16:23:43.953" Title="Why is this chair conformation the most stable and which leaving group is best?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;conformers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7327" PostTypeId="2" ParentId="7326" CreationDate="2013-12-12T09:12:36.333" Score="1" Body="&lt;p&gt;In cyclohexane, the equatorial position is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclohexane_conformation#Chair_conformation&quot; rel=&quot;nofollow&quot;&gt;energetically favored&lt;/a&gt; over the axial position. So the lowest energy conformer is the one where the most substituents are in equatorial position. (Or rather: Where you minimize the energy according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/A_value&quot; rel=&quot;nofollow&quot;&gt;A Value&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial Cl is favored as leaving group because of the elimination reaction mechanism. Normally, we see $\beta_{1,2}$ elimination in an anti-fashion. Which is what is proposed by your image.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-12T09:12:36.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7328" PostTypeId="2" ParentId="7326" CreationDate="2013-12-12T09:18:47.670" Score="1" Body="&lt;h2&gt;The chair conformation is the most stable due to the following reasons:&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The carbons in cyclohexane have sp3 hybridization and hence they tend to have an angle of 109.5 degrees. The chair conformation allows it to have the closest possible angle which would be 110.9 degrees.Due to this the strain reduces and results in a lowered energy. The two chair conformations have the lowest total energy, and are therefore the most stable.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since cyclohexane's planar form results into bond angle being 120 degrees&#xA;(which means a higher deviation from 109.5). Hence the energy of the planar form is higher and it always has the tendency to convert to chair form.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="3920" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-12-12T09:18:47.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7329" PostTypeId="2" ParentId="7313" CreationDate="2013-12-12T09:40:20.893" Score="2" Body="&lt;p&gt;It is difficult to correctly measure the solubility, especially if it goes to multicomponent solutions, as your vinegar. It is possible to find tables with some components of your mixture, but you won't get the full picture, and it looks as the moment, when the hands-on experience is most important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I found some facts for you, namely:&#xA;the solubility of a-D-glucose in all the solvents increases with&#xA;increasing temperature and decreases with increasing volume fraction&#xA;of acetic acid, from &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0021961413001833&quot; rel=&quot;nofollow&quot;&gt;Solubilities of {α-d-glucose in water + (acetic acid or propionic acid)} mixtures at atmospheric pressure and different temperatures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point is that the supersaturated solution can be stabilized by various means, and I think it already happens, as your vinegar contains lot of &quot;impurities&quot; from the grapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second point, how to prevent it, the simplest solution will be to keep the temperature high enough, so that the crystallization does not occur. If this is not possible, the supersaturated solution will crystallize, sooner or later. If you than ask, how to slow down the process, until spring comes, filtration could help to remove the crystallization nuclei.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-12-12T09:40:20.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7330" PostTypeId="2" ParentId="7310" CreationDate="2013-12-12T13:23:48.817" Score="4" Body="&lt;p&gt;First, it is a good idea to clarify the meaning of the word &lt;strong&gt;vaporize&lt;/strong&gt; from &lt;a href=&quot;http://www.thefreedictionary.com/vaporize&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;(Physics / General Physics) to change or cause to change into vapour or into the gaseous state&lt;/li&gt;&#xA;  &lt;li&gt;to evaporate or disappear or cause to evaporate or disappear, esp suddenly&lt;/li&gt;&#xA;  &lt;li&gt;(Physics / General Physics) to destroy or be destroyed by being turned into a gas as a result of extreme heat (for example, generated&#xA;  by a nuclear explosion)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemists should be careful when using this word because it can be used to describe evaporation, sublimation, and boiling.  Note the third definition, which suggests that the vaporization of matter typically involves a large amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the mixture you refer to in your question is in a liquid state, I am going to focus on liquid boiling.  (Note, we can say that boiling is the rapid vaporization of a liquid.)  The &lt;strong&gt;boiling point&lt;/strong&gt; of a liquid is the &lt;em&gt;point at which the vapor pressure of the liquid is equal to the surrounding pressure&lt;/em&gt;.  The boiling point of a liquid is affected by intermolecular forces such as hydrogen bonding, dispersion and dipole-dipole interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a solute is dissolved in a pure liquid (your example of caffeine dissolved in water), these intermolecular forces are altered.  The end result is an elevation of the boiling point.  This change can be predicted quite accurately using concepts typically covered in the first year of a college-level general chemistry class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we know now that a mixture of caffeine and water will have a boiling point higher than pure water; however what happens as the solution continues to boil?  I ignored an important point in the previous paragraph; &lt;em&gt;boiling point elevation due to addition of a solute will only occur if the solute is nonvolatile.&lt;/em&gt; Caffeine is such a solute.  Therefore, as you boi your mixture, the water will be turned into a gas, but the caffeine will be left behind.  The volume of solvent is decreasing; therefore the solution concentration is increasing.  Once the caffeine solubility limit is reached, it will start to crystallize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are left with a mound of caffeine you can continue to heat it.  As you heat it above about $230\,^{\circ}\mathrm{C}$ it will start to sublime, or in more general terms, &lt;em&gt;vaporize&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested, the experimental procedure for the extraction and purification of caffeine from tea leaves is described at an undergraduate level &lt;a href=&quot;http://chem-courses.ucsd.edu/CoursePages/Uglabs/143A_Weizman/expt_3N.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="1573" LastEditDate="2013-12-13T14:01:58.543" LastActivityDate="2013-12-13T14:01:58.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7335" PostTypeId="1" AcceptedAnswerId="7337" CreationDate="2013-12-12T18:57:26.487" Score="0" ViewCount="895" Body="&lt;p&gt;I really want to make liquid nitrogen(or oxygen) without using dry ice, a specially made freezer, or other chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a mixture of some sponges, rotten grass, sugar and water, which makes pressure in the bottle(produces $\ce{CO2}$). Which temperatures can I expect if I freeze it and then quickly opens the bottle after 24 hours of freezing. Is it possible to make liquid nitrogen out of it?&lt;/p&gt;&#xA;" OwnerUserId="3923" LastEditorUserId="1389" LastEditDate="2013-12-13T01:17:36.353" LastActivityDate="2013-12-13T01:44:46.750" Title="Is it possible to make liquid nitrogen with a bottle of CO2 gas and a freezer?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7336" PostTypeId="1" CreationDate="2013-12-12T23:42:29.090" Score="0" ViewCount="455" Body="&lt;p&gt;If you have an aromatic ring that requires substitution of some molecule, how do you know where this molecule is supposed to substitute on the ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ihAw6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (i) and (ii) how do we select the ring and position that will receive the molecule (HSO3 in (i) and Cl in (ii))? In (iii), how do we know where to attach the I?&lt;/p&gt;&#xA;" OwnerUserId="3919" LastEditorUserId="3919" LastEditDate="2013-12-12T23:49:11.640" LastActivityDate="2014-04-15T19:24:53.110" Title="Where do things connect in EAS reactions?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7337" PostTypeId="2" ParentId="7335" CreationDate="2013-12-13T01:44:46.750" Score="5" Body="&lt;p&gt;Short answer, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method you are suggesting here is to use Joule-Thompson expansion to cool the air as it escapes and hope that it liquefies. This is a valid way to make liquid nitrogen and is used in both lab-scale and industrial air liquefaction. Unfortunately, you will be many orders of magnitude off from where you need to be. Plans I've seen for liquefying air on a lab scale require starting pressures of around 3000 psi. That air is sent through a long copper tube where it is rapidly expanded through a nozzle into an insulated vessel to atmospheric pressure (14.7 psi). Even then it does not liquefy. The cooled air is vented through another tube concentric with the first tube, where it chills the pressurized gas entering the vessel. My recollection is that if you start with a full-sized cylinder of nitrogen, after about 1/2 the tank is gone, you will start to condense liquid nitrogen and when the tank is empty, you'll be left with about a half-liter of liquid nitrogen. Those numbers are probably off, but the order of magnitude is correct. (Note: if you try this based on these &quot;instructions&quot;, you will probably get yourself killed given the pressures involved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen liquefies at 77 K, or -196 C. Dry ice is only −78.5 C, and my freezer at home sits at about -5 C. I think a PET Coke bottle gets pressurized to 80 psi and a soda keg is rated to 120 psi. As you can see, the temperatures, pressures, and volumes you're likely to be able to achieve at home are off by one or more orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry.&lt;/p&gt;&#xA;" OwnerUserId="1390" LastActivityDate="2013-12-13T01:44:46.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7338" PostTypeId="1" CreationDate="2013-12-13T03:49:39.697" Score="5" ViewCount="22892" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FejjQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My task is to arrange the bonds from longest to shortest. Although I have the answer (C,D,B,A), I'm having a very hard time understanding how to derive this answer. Any clues would be helpful, especially those that help me in other related problems.&lt;/p&gt;&#xA;" OwnerUserId="3919" LastEditorUserId="5026" LastEditDate="2018-03-03T10:41:46.773" LastActivityDate="2018-03-03T10:41:46.773" Title="How can I generally determine which bonds are longest/shortest in a molecule?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7339" PostTypeId="1" AcceptedAnswerId="7349" CreationDate="2013-12-13T06:32:55.950" Score="2" ViewCount="312" Body="&lt;p&gt;I understand the principal of the Cis-Trans naming system (at least I think so), until I saw this example in my learning material. My teacher doesn't know either so I thought I would post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the molecule:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/T03Y8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ethyl groups both on the bottom side of the double bond, so it could be cis right?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6Ltsj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently not, the book says this is the answer:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/m6wn8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain the reasoning behind this? This particular example is unusual because there are three ethyl groups attached to the double bond so I cannot determine which one should be the parent chain. The book doesn't explain.&lt;/p&gt;&#xA;" OwnerUserId="3926" LastActivityDate="2016-05-17T18:04:17.453" Title="Nomenclature for stereoisomer example possible wrong?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7341" PostTypeId="2" ParentId="7339" CreationDate="2013-12-13T07:31:52.513" Score="1" Body="&lt;p&gt;Cis/trans nomenclature for aklenes is usable only for the simplest cases, so you can use it safely when there are two hydrogens on the double bond. If you want to be fully correct, use the E/Z, it's the only bulletproof way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cis/trans is also widely used, if you have e.g. chlorine on the double bond, which makes the E/Z nomenclature counter-intuitive (but still correct).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-12-13T07:31:52.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7342" PostTypeId="2" ParentId="7339" CreationDate="2013-12-13T09:00:56.713" Score="4" Body="&lt;p&gt;Unless I'm very much mistaken, this molecule should not be labelled with &lt;em&gt;any&lt;/em&gt; stereochemical descriptors, whether they be relative (cis/trans) or absolute (E/Z). The double bond has three identical ethyl substituents, therefore there is no stereoisomerism possible. By way of explanation, imagine if the ethyl group and the hydrogen on carbon #4 were interchanged: the resulting molecule would be completely identical to the one pictured. This is always the case when an alkene bears two identical substituents on the same carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To correctly use cis/trans or E/Z nomenclature, it is necessary that there not be two identical groups on the same carbon of the alkene; if there are, the labels are meaningless, since interchanging the two groups results in an identical molecule.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-13T09:00:56.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7343" PostTypeId="1" CreationDate="2013-12-13T09:34:24.240" Score="5" ViewCount="1211" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3pyEO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3pyEO.png&quot; alt=&quot;Structure of azete&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m confused, because it seems like Hückel’s rule is satisfied (two electrons from the nitrogen lone pair and four from the double bonds) and that the molecule should be aromatic. Where am I going wrong here?&lt;/p&gt;&#xA;" OwnerUserId="3919" LastEditorUserId="16683" LastEditDate="2018-01-28T00:15:59.827" LastActivityDate="2018-01-28T01:06:35.340" Title="Why isn’t azete aromatic?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7345" PostTypeId="2" ParentId="7343" CreationDate="2013-12-13T10:54:52.873" Score="10" Body="&lt;p&gt;The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3dQvw.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3dQvw.png&quot; alt=&quot;Orbitals in azete&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the total number of π-electrons is only four: two from each double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please read this for a better understanding : &lt;a href=&quot;http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="16683" LastEditDate="2018-01-28T00:14:58.307" LastActivityDate="2018-01-28T00:14:58.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7347" PostTypeId="1" AcceptedAnswerId="7383" CreationDate="2013-12-13T12:45:58.080" Score="8" ViewCount="1182" Body="&lt;p&gt;In science, the negative logarithm of the acid dissociation constant is denoted $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{A}$ depending on the source (lowercase &quot;a&quot; or uppercase &quot;A&quot;). Since it is related to the acid dissociation constant, defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{A} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would write $\mathrm{p}K_\mathrm{A}$ (the LaTeX package &lt;a href=&quot;https://www.ctan.org/pkg/chemmacros&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;chemmacros&lt;/code&gt;&lt;/a&gt; renders it so, but doesn't give any source it in &lt;a href=&quot;http://mirrors.mi.ras.ru/CTAN/macros/latex/contrib/chemmacros/chemmacros_en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;its documentation&lt;/a&gt;). It seems to me that $\mathrm{p}K_\mathrm{a}$ is the &quot;modern&quot; version of &lt;code&gt;pKa&lt;/code&gt;, which was written like this in the last century because of the lack of subscript character on typesetting machines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any source or styling guide saying explicitly which one should be used &lt;em&gt;today&lt;/em&gt; (like the IUPAC Green Book), and why.&lt;/p&gt;&#xA;" OwnerUserId="3932" LastEditorUserId="16683" LastEditDate="2018-09-11T08:55:57.390" LastActivityDate="2018-09-11T08:55:57.390" Title="Should the 'a' in the acid dissociation constant (Ka or pKa) be capitalised?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7348" PostTypeId="1" AcceptedAnswerId="7358" CreationDate="2013-12-13T12:48:22.217" Score="3" ViewCount="3721" Body="&lt;p&gt;I have pretty much zero experience in chemistry whatsoever but I wonder about how much crystal will I get from fully dissolved element when I reduce water temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variables&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;at $\mathrm{40^\circ C,\ 171}\pu{g}$ of $\ce{Pb(NO3)2}$ is fully dissolved in water&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;at $\mathrm{18^\circ C,\ }\ce{Pb(NO3)2}$ - How much is dissolved now? Everything that was not dissolved turned to crystals?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that it probably has something to do with this graph right here but I can't turn my mind to right direction. Can someone explain this problem and how can I find solution?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dOBXk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3931" LastEditorUserId="23561" LastEditDate="2018-08-12T22:09:20.073" LastActivityDate="2018-08-12T22:09:20.073" Title="Reducing temperature on dissolved Pb(NO3)2" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="7349" PostTypeId="2" ParentId="7339" CreationDate="2013-12-13T12:53:39.470" Score="3" Body="&lt;p&gt;In my previous (deleted) answer, I overlooked the fact that this molecule has indeed three identical substituents and not two. Therefore, I would consider Greg E's answer as correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, C3 has 2 identical substituents on the same sp2 carbon, so cis/trans is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Example from organic chemistry 6th edition Paula Yurkanis Bruice:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hq3X4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your book is wrong!&lt;/p&gt;&#xA;" OwnerUserId="3929" LastEditorUserId="3929" LastEditDate="2013-12-13T12:59:24.573" LastActivityDate="2013-12-13T12:59:24.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7350" PostTypeId="2" ParentId="7336" CreationDate="2013-12-13T13:22:10.863" Score="1" Body="&lt;p&gt;In (i), it is marked correctly, in (ii) Cl will go only to the &quot;right&quot; benzene, not to the nitro- substituted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (iii), I do not see any reasonable reaction. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;electrophilic aromatic substitution - there is no Lewis acid.&lt;/li&gt;&#xA;&lt;li&gt;nucleophilic aromatic substitution - not enough activation by electron-withdrawing substituents.&lt;/li&gt;&#xA;&lt;li&gt;nucleophilic substitution of Br of $\ce{-CH_2Br}$ - not theromdynamically feasible. Maybe under Finklestein conditions. &lt;a href=&quot;http://en.wikipedia.org/wiki/Finkelstein_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Finkelstein_reaction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2013-12-13T13:22:10.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7353" PostTypeId="1" AcceptedAnswerId="7357" CreationDate="2013-12-13T18:12:43.997" Score="5" ViewCount="4579" Body="&lt;p&gt;I learned that the order of $+M$ effect (mesomeric effect) for halogens is as follows: $\ce{F} &amp;gt; \ce{Cl} &amp;gt; \ce{Br} &amp;gt; \ce I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why this is so?&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="7475" LastEditDate="2017-11-15T13:51:16.160" LastActivityDate="2017-11-15T13:51:16.160" Title="Order of positive mesomeric effect of halides" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7354" PostTypeId="2" ParentId="7283" CreationDate="2013-12-13T19:00:52.123" Score="0" Body="&lt;p&gt;Let's speak the Redox with an exempla : Corrosion off a piece of Iron.And Oxygene around you and me. As child of 6 years old had probably seen this, it will directly rely to everyday 's live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say this to him : You cannot see the &quot;messy movements&quot; in this Iron because your eyes are too big.......and because you can see the moon and stars.......but not all stars in the universe because we are too small. &#xA;The idea off beeing somewhere in the middle off &quot;The Scale&quot; is an important step to start imaging the all around and inside off you. Back the Redox.&#xA;The iron, as all the rest is made of 3 sweeties, called Neutron...not an easy name but this is exiting for a child the knonw some specials words. As 3 new words is fine at C yers old,continue with Electron, an other one to love and Proton. Those 3 sweeties are bricks to create all you can see, fill and  smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Redox is when sweeties are given and received.I can not receive if nothing is given. Both are related and if one does not happens, the other one too.&#xA;A child should ask how it is possible to have only 3 sweeties too have all around with différences. More or less you have always those &#xA;3 sweeties together exept when speaking e.g. nuclear explosion. Just catch his meaning with powerfull stuffs. Back on redox,now to have Iron you need those 3 sweeties with 25 other teams of those 3 members: E-, P+ and N°. This make 26 teams to make Iron, the very kind Iron can give some Electrons too Oxygene we need to be alive. And the volume off Iron increase as the oxygene who received some electrons must stay around the iron or leave. As they are now togheter, they will stay togheter as long as you do not found and other team off sweeties able to give back electrons too Iron or to stoll the electron owned by oxygene. Use off this ugly stuff called chlorine can make it true.&#xA;Next : Explain the exeption off nuclear phenomena ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards.&lt;/p&gt;&#xA;" OwnerUserId="3933" LastActivityDate="2013-12-13T19:00:52.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7355" PostTypeId="2" ParentId="6916" CreationDate="2013-12-14T08:52:22.160" Score="1" Body="&lt;p&gt;First let me correct you, m/m means mass/mass % I think you assumed it to be mole/mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have 100g of solution in which Cu is 17g and solvent is 83 gm. 17 gm copper means 0.268 moles of Cu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;code&gt;density = Total weight/Total moles = 100/(0.268 + x) = 1.367&lt;/code&gt; where x is moles of solvent, from here you will get &lt;code&gt;x = 72.8849&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, mole fraction of Cu is &lt;code&gt;0.268/(0.268+72.8849) = 0.00366&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Molality = Moles of solute/Weight of Solvent(in Kg) = 0.268/0.083 = 3.229 m&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, density in terms of g/l is not given I don't think it's possible to calculate molarity.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2013-12-14T08:52:22.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7356" PostTypeId="1" CreationDate="2013-12-14T11:31:54.630" Score="2" ViewCount="104" Body="&lt;p&gt;How can I theoretically predict the time needed to photolyze n moles of $\ce{AgF}$? (with UV) (this is not a homework question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photolysis of $\ce{AgF}$ requires UV light - but I don't know which wavelength the UV must be at for the optimal photolysis. Then if the wavelength is known I can calculate the energy of one UV photon of a specific wavelength. After that I can calculate the required amounts of photons required for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of this method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First I need to find out which wavelength the UV must be at for the optimal photolysis of $\ce{AgF}$ (How?)&lt;/li&gt;&#xA;&lt;li&gt;Then I need to calculate the amount of energy required for the photolysis of AgF : on wolframalpha I entered this equation: $\ce{2AgF -&amp;gt; 2Ag + F2}$ and it gave me a $\Delta{}H=409.2\:\mathrm{kJ/mol}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And somehow I must be able to then figure out which lamp I could use and how long I should use it.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="4945" LastEditDate="2014-06-12T08:22:32.010" LastActivityDate="2014-06-12T08:22:32.010" Title="Calculating photolysis time of AgF" Tags="&lt;thermodynamics&gt;&lt;photochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7357" PostTypeId="2" ParentId="7353" CreationDate="2013-12-14T11:57:41.837" Score="2" Body="&lt;p&gt;Found the answer, it is because of size difference that smaller halogens are able to make more effective p orbital overlap with carbon.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2013-12-14T14:08:44.973" LastActivityDate="2013-12-14T14:08:44.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7358" PostTypeId="2" ParentId="7348" CreationDate="2013-12-14T13:59:44.853" Score="4" Body="&lt;p&gt;Assuming that the solution is &lt;strong&gt;saturated&lt;/strong&gt;, meaning that if you add more salt to the solution it will not dissolve, and neglecting eventual evaporation of water, we have a mass of water of about 228g for 171g of salt (assuming solubility of 75g in 100g at 40°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have &#xA;$$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining 45.6g of the compound will have precipitated. Whether they form nice crystals or not depends on various other things, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallization&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the solution is not saturated (meaning your salt can still be fully dissolved, but there may be room for more in the solution) the calculation is impossible if you don't know the mass of the water at the beginning.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-14T13:59:44.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7362" PostTypeId="1" AcceptedAnswerId="7363" CreationDate="2013-12-14T20:36:25.140" Score="5" ViewCount="522" Body="&lt;p&gt;Why the dihydroxyacetone phosphate is called that way when it has only one OH ?&lt;/p&gt;&#xA;" OwnerUserId="3940" OwnerDisplayName="Pao" LastActivityDate="2013-12-15T14:45:25.460" Title="Nomenclature of Dihydroxyacetone Phosphate" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7363" PostTypeId="2" ParentId="7362" CreationDate="2013-12-15T03:17:02.427" Score="2" Body="&lt;p&gt;Because it is the phosphoester of dihydroxyacetone. One of the -OH groups was used in the esterification reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://guweb2.gonzaga.edu/faculty/cronk/biochem/D-index.cfm?definition=DHAP&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3939" LastEditorUserId="1431" LastEditDate="2013-12-15T14:45:25.460" LastActivityDate="2013-12-15T14:45:25.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7364" PostTypeId="2" ParentId="7323" CreationDate="2013-12-15T04:04:40.970" Score="5" Body="&lt;p&gt;There are two distinct phenomena, which are somewhat related. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dynamical effect of &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_vibration&quot;&gt;molecular vibration&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bonds do vibrate like springs, and as you said they can store energy by changing its vibrational state. Like a string, the more energy stored in the vibrational motion, the larger the magnitude of such vibration.  Different from a spring, such vibrations are associated with discrete quantum numbers, and even at absolute zero temperature, such vibration still possess a &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero-point_energy&quot;&gt;zero-point energy&lt;/a&gt; and can never stop.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Non-dynamical effect that change the statistical average of bond length&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Molecular vibration often (but not always) occurs at an extremely short time frame, and the vibration itself is often (again not always) of a very small magnitude, so when we refer to bond length we often mean the average bond length over time.&lt;br&gt;&#xA;    Normally a molecule stays near the bond length where the energy is the lowest.  There can be a few reasons for a the average bond length to change while keeping the connectivity of the molecule intact, with some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each conformational isomers will have different bond lengths.  Usually this is a small number.  Examples of large bond length change between conformational isomers with the same bond connectivity are those with &lt;a href=&quot;https://en.wikipedia.org/wiki/Jahn%E2%80%93Teller_effect&quot;&gt;Jahn-Teller effect&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By changing the vibrational state of the molecule, the average bond length can be different.  Usually, because of &lt;a href=&quot;https://en.wikipedia.org/wiki/Anharmonicity&quot;&gt;anharmonicity&lt;/a&gt;, the higher the vibrational state, the higher the average bond length.  Temperature change, pressure change, collision or photon excitation (especially infrared light) are the most common causes of a change in vibrational state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Through inter-molecular interactions or external forces, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Potential_energy_surface&quot;&gt;potential energy surface&lt;/a&gt; of the molecule can be changed so that the energy minimum happens at a different distance.  In this case, the bond length will change, but the energy stored in the bond will not change.  This can be achieved through external electronic field, or intermolecular forces such as electrostatic force from an ion, hydrogen bonds, charge transfer, electron donator/acceptor interaction, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-15T04:04:40.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7365" PostTypeId="2" ParentId="5708" CreationDate="2013-12-15T10:39:31.623" Score="2" Body="&lt;p&gt;The main benefit of oxygen is that its the most commonly available oxidizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main component of air that support combustion is oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However titanium burns in nitrogen and so does lithium, and also some other active metals such as finely divided aluminium. These form nitrides. Similarly most of these active metals also burn in hydrogen forming hydrides such as lithium.    &lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="30998" LastEditDate="2017-06-10T12:52:39.907" LastActivityDate="2017-06-10T12:52:39.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7366" PostTypeId="1" AcceptedAnswerId="7368" CreationDate="2013-12-15T11:18:26.087" Score="7" ViewCount="7556" Body="&lt;p&gt;$\ce{Cl-}$ is a monodentate ligand. It has four lone pairs then why it can't donate two lone pairs to central atom and act as a bidentate ligand?&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="4231" LastEditDate="2015-03-08T17:04:32.560" LastActivityDate="2018-03-25T19:42:17.047" Title="Why Cl⁻ can't act as bidentate ligand?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7367" PostTypeId="2" ParentId="7289" CreationDate="2013-12-15T12:07:02.613" Score="4" Body="&lt;p&gt;The activation energy varies a lot for various reactions. As long as T &gt; 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy varies with each reaction.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2013-12-15T12:07:02.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7368" PostTypeId="2" ParentId="7366" CreationDate="2013-12-15T14:57:39.043" Score="17" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Denticity#Classes_of_denticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;bidentate ligand&lt;/a&gt; is one that forms bonds to the metal center through two different atoms, not through two lone pairs. While chloride does have more than one pair of valence electrons, it only has one atom. Thus it cannot be bidentate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chloride shares a second pair of electrons with the metal center, there would still only be one bond site between the ligand and the metal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[M+]-Cl &amp;lt;-&amp;gt; [M]=Cl+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bidentate ligand needs two lone pairs on two separate atoms. Consider ethylenediamine (en), one of the classic bidentate ligands: $\ce{H2NCH2CH2NH2}$. Each nitrogen atom has a lone pair that can be used to form a bond to the metal center, creating two bonds from two different sites on the ligand. See the structure of the tris(ethylenediamine)cobalt(III) ion from tris(ethylenediamine)cobalt(III) chloride $(\ce{Co(en)3Cl3})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lafMo.png&quot; alt=&quot;tris(ethylenediamine)cobalt(III) &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&#xA;As Nicolau points out in a comment, chloride can &lt;strong&gt;bridge&lt;/strong&gt; two metal centers. This is not the denticity phenomenon, which is one ligand with two bonding sites to the same metal. This is one ligand with bonds to two metal centers. See the structure of niobium pentachloride (really $\ce{Nb2Cl10}$ or better $\ce{(NbCl5)2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EREQq.png&quot; alt=&quot;Niobium pentachloride&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7786" LastEditDate="2018-03-25T19:42:17.047" LastActivityDate="2018-03-25T19:42:17.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7369" PostTypeId="1" CreationDate="2013-12-15T15:57:50.740" Score="13" ViewCount="949" Body="&lt;p&gt;While refreshing myself on denticity to answer another question, I started to read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Hapticity&quot; rel=&quot;noreferrer&quot;&gt;hapticity&lt;/a&gt;, which is the coordination of a ligand through multiple continuous atoms. A famous case is ferrocene (bis[$\eta^5$-cyclopentadienyl]iron, $\ce{Cp2Fe}$), where all five carbon atoms in each cyclopentadienyl ligand (Cp) are coordinated to the iron center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene can also be a polyhaptic ligand, forming $\eta^4$ or $\eta^6$ complexes depending on the oxidation state of the metal. In particular, bis(benzene)ruthenium complexes can undergo redox shuttling, where one benzene ligand switches between $\eta^4$ and $\eta^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pPnWU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\eta^4$ complex on the right, the top benzene ligand is now bent out of planarity, implying a break of aromaticity. This compound looks like a good dienophile for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diels_Alder&quot; rel=&quot;noreferrer&quot;&gt;Diels-Alder reaction&lt;/a&gt;. The &lt;a href=&quot;http://dx.doi.org/10.1002/anie.197105561&quot; rel=&quot;noreferrer&quot;&gt;article linked&lt;/a&gt; in the Wikipedia page on this switching phenomenon is published in Angewandte Chemie, International Edition, which is behind a paywall for me. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does coordination of benzene to a metal in the $\eta^4$ mode activate it toward cycloadditions? Because of paywalls, I would appreciate summaries of articles in addition to links to them. My attempts to find articles using Google Scholar led me to &lt;a href=&quot;https://www.google.com/search?q=metal%20mediated%20cycloaddition%20benzene&amp;amp;ie=utf-8&amp;amp;oe=utf-8&amp;amp;aq=t&amp;amp;rls=org.mozilla:en-US:official&amp;amp;client=firefox-a#q=metal%20mediated%20cycloaddition%20of%20benzene&amp;amp;rls=org.mozilla:en-US:official&quot; rel=&quot;noreferrer&quot;&gt;articles about the construction of benzenes by metal mediated cycloadditions&lt;/a&gt;, and not to reactions of benzene itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2017-07-09T02:53:03.313" LastActivityDate="2017-07-09T02:53:03.313" Title="Do eta-4 benzene ligands participate in Diels Alder or other cycloadditions?" Tags="&lt;organic-chemistry&gt;&lt;organometallic-compounds&gt;&lt;pericyclic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7370" PostTypeId="2" ParentId="7366" CreationDate="2013-12-15T17:31:32.027" Score="-1" Body="&lt;p&gt;Chlorine ion is monodentate.For bidentate ligand it should have atleast two bonding sites.In chlorine it has only one donatable lone pair.So it can't act as bidentate ligand.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2013-12-15T17:31:32.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7371" PostTypeId="1" AcceptedAnswerId="7373" CreationDate="2013-12-15T17:34:42.193" Score="1" ViewCount="326" Body="&lt;p&gt;Why we don't get sharp point at equivalence points in case of acid base titration other than the case of strong acid and strong base titration?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2013-12-15T20:24:57.987" Title="Acid base titration curve" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7373" PostTypeId="2" ParentId="7371" CreationDate="2013-12-15T20:24:57.987" Score="0" Body="&lt;p&gt;In the case of strong acid-strong base titration at equivalence point with the addition of very small amount of titrant (like a drop) the pH change is very large. That means the range over which indicator changes its color is likely to be passed in a drop of titrant, resulting in a sharp end point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas in the other cases the change of pH at equivalence point is less dramatic and there will be a smaller change in pH at equivalence point. So, the indicator will change its color slowly resulting in a wider end point.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2013-12-15T20:24:57.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7374" PostTypeId="1" CreationDate="2013-12-15T21:21:29.907" Score="2" ViewCount="833" Body="&lt;p&gt;I was wondering what the crystal structures of FCC, BCC and HCP actually look like, and where atoms are placed in their configuration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I'm looking at elemental nickel, iron and cobalt in their lowest temperature lattice type. (Below their curie temperature to be exact.)&lt;/p&gt;&#xA;" OwnerUserId="3946" LastActivityDate="2013-12-15T21:21:29.907" Title="Bond Lengths in FCC, BCC and HCP Structures" Tags="&lt;crystal-structure&gt;&lt;transition-metals&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7375" PostTypeId="1" CreationDate="2013-12-15T23:47:05.017" Score="6" ViewCount="1291" Body="&lt;p&gt;Many years ago, I used to work as a gold exploration geologist.  One of the main locations was a salt lake in Western Australia detailed in the document &lt;a href=&quot;http://crcleme.org.au/Pubs/Monographs/regolith98/12-carey&amp;amp;dusci.pdf&quot;&gt;WMC Resources Exploration Successes inN Lake Terrains - Applications of Element Dispersion, Kambalda W.A&lt;/a&gt; (Carey and Dusci, 1993).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the article, and in my research at the time, it is stated that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;gold disperses laterally along iron redox fronts that tend to occur higher in the regolith profile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the area is arid (up to 50C or 122F in summer), but was not always like this, fossil records suggest far milder conditions occurred in the past.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how does the ambient temperature affect the rate of redox reaction in a salt lake, such as in the example provided?&lt;/p&gt;&#xA;" OwnerDisplayName="user3901" LastActivityDate="2014-09-28T06:48:42.837" Title="How does the ambient temperature affect the rate of redox reactions in a salt lake?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7376" PostTypeId="1" CreationDate="2013-12-15T23:59:07.853" Score="4" ViewCount="6131" Body="&lt;p&gt;I did an experiment where I dissolved 30 grams of zinc sulfate in water and boiled it with strips of solid zinc metal and strips of copper metal and a layer of zinc was deposited on the copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if you place solid copper in zinc sulfate aqueous without the solid zinc nothing would happen because zinc is more reactive than copper.  Why does the addition of zinc sulfate cause the zinc to coat the copper?  What is happening? &lt;/p&gt;&#xA;" OwnerUserId="3947" LastEditorUserId="7475" LastEditDate="2017-10-30T03:37:42.547" LastActivityDate="2017-10-30T05:25:39.830" Title="Why is zinc deposited on copper when boiling an aqueous zinc sulfate solution?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;transition-metals&gt;&lt;ionic-compounds&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7378" PostTypeId="2" ParentId="4877" CreationDate="2013-12-16T01:12:59.917" Score="8" Body="&lt;p&gt;A man here in Australia dissolved his wife's body in Hydrochloric acid in a wheelie bin.&#xA;He put her body in the bin and poured 20 liters of Hydrochloric Acid in it.&#xA;Then 2 days later went and bought another 40 liters and topped it up until everything was liquid.&#xA;He then poured it down a storm drain and rinsed it all out with water.&#xA;The only thing they found was her prosthetic teeth.&#xA;Basically, once the Acid stops dissolving and becomes weak, you have to keep adding more.&#xA;But it will eventually dissolve a whole human body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cairns man Klaus Andres, who admitted to dissolving wife's body in acid, guilty of murder, jailed for life&#xA;&lt;a href=&quot;http://www.abc.net.au/news/2013-12-12/cairns-man-guilty-of-murdering-wife/5152488&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.abc.net.au/news/2013-12-12/cairns-man-guilty-of-murdering-wife/5152488&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3948" LastEditorUserId="402" LastEditDate="2017-01-22T09:59:46.853" LastActivityDate="2017-01-22T09:59:46.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7380" PostTypeId="2" ParentId="7376" CreationDate="2013-12-16T01:46:20.393" Score="1" Body="&lt;p&gt;In looking at the reaction, it seems our reactants are zinc sulfate, zinc metal, and copper metal. Qualitatively, this reaction yielded what looks like a zinc layer deposited on to the copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One familiar reaction is that of zinc and copper sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(s) + CuSO4(aq) -&amp;gt; ZnSO4(aq) + Cu(s) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc metal is above copper in the activity series and therefore, zinc will replace copper in solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zinc metal is being oxidized as it loses electrons, and the copper(II) ions are reduced to copper metal as they acquire electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is interesting is that you are correct, since zinc metal is above copper in terms of the reactivity series, if you just placed copper metal in zinc sulfate solution, you would not expect the reaction to go in reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Wikipedia's definition, a galvanic cell, or voltaic cell, named after Luigi Galvani, or Alessandro Volta respectively, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals connected by a salt bridge, or individual half-cells separated by a porous membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zinc metal in the zinc sulfate solution along with the copper metal, zinc metal loses its electrons more easily, and its electrons are received by copper. We can write these reactions by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\ce{Zn(s) &amp;amp;-&amp;gt; Zn^{2+}(aq) + 2e^-}\\&#xA;\ce{Cu^{2+}(aq) + 2e^- &amp;amp;-&amp;gt; Cu(s)}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can see that the zinc metal that we placed in our solution is oxidized and our copper metal is reduced. This reaction is able to proceed as the energy released from the oxidation of zinc is more than that of the reduction of the copper metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As zinc loses its electrons, it forms ions, and I assume that this is the deposited zinc metal that you see on the copper metal that gains electrons.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="7475" LastEditDate="2017-10-30T03:46:10.533" LastActivityDate="2017-10-30T03:46:10.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7383" PostTypeId="2" ParentId="7347" CreationDate="2013-12-16T06:19:15.390" Score="8" Body="&lt;p&gt;Good go-to references for this kind of problem are the IUPAC books (Gold, Green, etc.). I haven't found $\mathrm{p}K_\mathrm{a}$ directly but &lt;a href=&quot;http://goldbook.iupac.org/A00080.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; you can find their convention for the acidity constant. They use a lower-case letter, i.e. $K_\mathrm{a}$. But I don't think using an upper-case letter would be wrong; I'd say it's a matter of taste.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorDisplayName="user7951" LastEditDate="2017-09-22T08:09:39.533" LastActivityDate="2017-09-22T08:09:39.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7384" PostTypeId="2" ParentId="7376" CreationDate="2013-12-16T09:29:06.223" Score="3" Body="&lt;p&gt;Zinc, being in contact with copper, is easily oxidized, because copper drives electrons out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metallic copper becomes electron-rich and is therefore capable to yield electrons to $\ce{Zn^2+}$ (stemming from $\ce{ZnSO4}$), which reduces to $\ce{Zn^0}$. The latter is finally observed to adsorb at the copper strips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, copper metal works as an auxiliary electrode whose function is to transport electrons, without being $-$ by no means $-$ transformed during the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the overall reaction uniquely involves zinc and its ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to what is observed when copper, in contact with zinc, is dipped into an acidic solutions. Hydrogen bubbles are seen even at the copper surface, without the latter being oxidized.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="7475" LastEditDate="2017-10-30T03:38:30.523" LastActivityDate="2017-10-30T03:38:30.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7385" PostTypeId="1" CreationDate="2013-12-16T09:30:42.050" Score="2" ViewCount="9062" Body="&lt;p&gt;First of all what is exactly Gibbs free energy in regards to an electrochemical process? How can we visualise it? How does Gibbs free energy change in an electrochemical reaction? How is it related to the cell potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't potential the amount of work done per unit charge to bring it from infinite to that point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the ionisation in cell can be thought of as occurring in multiple steps, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;\ce{Zn(s)} &amp;amp; \ce{-&amp;gt;Zn(g)} &amp;amp;&amp;amp; (1) \\ &#xA;\ce{Zn(g)} &amp;amp; \ce{-&amp;gt;Zn+(g) + e-} &amp;amp;&amp;amp;(2) \end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the work done to remove the charge should be the change in Gibbs free energy of second reaction and that energy should be equal to potential energy, but we take the combined change in Gibbs free energy as change in electric potential energy of the electron. Why is that so?&lt;/p&gt;&#xA;" OwnerUserId="3950" LastEditorUserId="275" LastEditDate="2013-12-16T17:30:30.960" LastActivityDate="2014-02-12T07:56:23.253" Title="how is the relation between Gibbs free energy and cell potential?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;free-energy&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7386" PostTypeId="2" ParentId="7376" CreationDate="2013-12-16T09:35:27.677" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I am not sure an electrochemical reaction is occurring; it could be a simple dissolution and deposition. It happens all the time but it is so slow at room temperature that it is not noticeable. Measure the weight of the zinc  and actual copper before and after the experiment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it is an electrochemical reaction, then the answer is in 3 parts&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Standard electrochemical half cell potential (usually expressed as reduction such as the first reaction but I'll reverse it to give the oxidation reaction that may occur)  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\ce{Zn^2+(aq) + 2e- &amp;amp;-&amp;gt; Zn (s)}&amp;amp;&amp;amp; E^\circ=\pu{-0.7628V}\\&#xA;\ce{Cu(s)&amp;amp;-&amp;gt;Cu^2+ (aq) + 2e-} &amp;amp;&amp;amp; E^\circ=\pu{-0.3402 V}\\&#xA;\ce{Zn^2+ (aq) + Cu(s) &amp;amp;-&amp;gt; Cu^2+ (aq) + Zn(s)} &amp;amp;&amp;amp; E^\circ=\pu{-1.103V}\end{align}$$ so this would not occur spontaneously; electrical energy is needed for it to occur.   &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;This occurs at standard condition which are usually $\pu{1mol/L}, \pu{25 ^\circ C}, \pu{1 atm}$ (other ions also have an effect) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To understand it properly you will need to calculate the actual concentration in mol/L (&lt;a href=&quot;http://drfus.com/chapter-20/20-6-cell-potentials-under-nonstandard-conditions&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://drfus.com/chapter-20/20-6-cell-potentials-under-nonstandard-conditions&lt;/a&gt;)&#xA;or you can just measure the actual potential difference between the pieces of  metal to find the potential difference. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="41328" LastEditDate="2017-10-30T05:25:39.830" LastActivityDate="2017-10-30T05:25:39.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7388" PostTypeId="2" ParentId="7117" CreationDate="2013-12-16T12:28:15.747" Score="-7" Body="&lt;p&gt;Ozonolysis of benzene can be done at very high temperature or by its reaction with  Zinc and water.&lt;img src=&quot;https://i.stack.imgur.com/KirDr.jpg&quot; alt=&quot;image of Ozonolysis of benzene&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="3943" LastEditDate="2013-12-16T14:17:24.160" LastActivityDate="2013-12-16T14:17:24.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7390" PostTypeId="1" AcceptedAnswerId="7399" CreationDate="2013-12-16T14:30:13.967" Score="1" ViewCount="26395" Body="&lt;p&gt;How to convert glucose to fructose? Moreover, can there be direct conversion(not necessarily in a single step), or we have to use indirect means, such as from sucrose, or other higher carbohydrates, and then separating fructose from the products?&lt;/p&gt;&#xA;" OwnerUserId="3863" LastEditorUserId="13184" LastEditDate="2015-02-19T11:23:35.917" LastActivityDate="2019-01-25T15:39:19.073" Title="How to convert glucose to fructose" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7391" PostTypeId="1" CreationDate="2013-12-16T15:11:17.063" Score="10" ViewCount="8297" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Apvfg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;An ionic bond is the bonding between a non-metal and a metal, that occurs when charged atoms (ions) attract after one loses one or more of its electrons,and gives it to the other molecule, for example sodium and chlorine. This makes the bond stronger and harder to break.&lt;br&gt;&#xA;  In other words, an ionic bond is the electrostatic force of attraction between two oppositely charged ions. The positive ion is called cation, and the negative ion is the anion. It is like the north and south poles of a magnet.  &lt;/li&gt;&#xA;  &lt;li&gt;A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. The stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding. For many molecules, the sharing of electrons allows each atom to attain the equivalent of a full outer shell, corresponding to a stable electronic configuration.  &lt;/li&gt;&#xA;  &lt;li&gt;A dipolar bond, also known as a dative covalent bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.  &lt;/li&gt;&#xA;  &lt;li&gt;NOTE: &lt;em&gt;It is important to recognize that pure ionic bonding - in which one atom &quot;steals&quot; an electron from another - cannot exist: all ionic compounds have some degree of covalent bonding, or electron sharing. Thus, the term &quot;ionic bond&quot; is given to a bond in which the ionic character is greater than the covalent character - that is, a bond in which a large electronegativity difference exists between the two atoms, causing the bond to be more polar (ionic) than other forms of covalent bonding where electrons are shared more equally. Bonds with partially ionic and partially covalent character are called polar covalent bonds. Nevertheless, ionic bonding is considered to be a form of noncovalent bonding.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{equation}\tag{1}F=\frac{1}{4\pi\epsilon}\cdot\frac{q_1q_2}{r^2}\end{equation}  $$&#xA; According to Coulomb's law equation $(1)$, the force between two charges depend on the medium in which they are placed. The value of $\epsilon_r$ (relative permitivity) of air is found to be approx $1$, whereas relative permittivity of water is found to approx $80$ at room temperature.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought Coulombs law could answer why ionic compounds dissociate into ions whereas other compounds of different bonds won't. I thought, ionic compounds are formed by the electrostatic interaction of two oppositely charged ions (I assumed these to be charges), they will be stable in air, as relative permittivity value is not much effective. When, the same is dissolved in water, I thought they will dissociate as relative permittivity value of water effects more. I found the explanation, so as to why ionic compounds dissociate into ions when dissolved in water. In case of coordinate bond, I thought an atom losses a pair of electron and thus can be called as cation, whereas the other becomes anion. So, even here we can expect electrostatic interaction, and as discussed above must dissociate into ions when dissolved in water. But &lt;em&gt;coordination complex of coordinate compound won't dissociate?&lt;/em&gt; I think, I have misunderstood somewhere or else there must be some chemistry theory which could explain, &lt;em&gt;why coordinate complex won't dissociate?&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in case of covalent bond we can see the bond to be formed by stable electrostatic attraction and repulsion between electrons and protons, so I thought, even covalent bonds should break, when dissolved in water, as electrostatic interaction is going to be affected. I thought, there must be some other forces which must be keeping them bonded. If so what are they?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know whether I am right or wrong, if any was the case please explain.   &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="7475" LastEditDate="2016-09-27T10:14:19.200" LastActivityDate="2016-09-27T10:17:48.853" Title="Why do ionic compounds dissociate whereas coordinate complexes won't?" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7392" PostTypeId="1" CreationDate="2013-12-16T15:16:38.370" Score="0" ViewCount="220" Body="&lt;p&gt;I'm interested in chemical reaction mechanism with more exotic particles. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropes_of_oxygen#Atomic_oxygen&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; seems to imply that the normal atomic oxygen is $O(^3P)$, is that right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also came across $O(^1D),O(^1S),O(^5P)$ and some longer strings like $O(3S3s)$. What are the atomic oxygen species, or at least these which we can expect to see in the parts per millions range or a little below, in oxydation processes specifically? How to know which I left out?&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="275" LastEditDate="2013-12-16T17:35:49.130" LastActivityDate="2013-12-17T02:11:57.947" Title="What are the known atomic oxygen species?" Tags="&lt;atoms&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7395" PostTypeId="1" AcceptedAnswerId="7396" CreationDate="2013-12-16T19:24:03.970" Score="6" ViewCount="129" Body="&lt;p&gt;I am writing an essay on the origin of life and I was reading an article on hydrothermal vents. &lt;a href=&quot;http://www.sciencemag.org/content/281/5377/670.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Peptides by Activation of Amino Acids with CO on (Ni,Fe)S Surfaces: Implications for the Origin of Life)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that (Ni,Fe)S has played an important role in protein synthesis, but I wanted to know more about the substance its self. It is a compound, a group of chemicals? Or is it simply nickel, iron, and sulfur individually? I'm not very clued up when it comes to chemistry and I haven't come across that sort of notation before.&lt;/p&gt;&#xA;" OwnerUserId="3957" LastEditorUserId="402" LastEditDate="2017-04-08T22:21:38.560" LastActivityDate="2017-04-08T22:21:57.750" Title="What is 'coprecipitated (Ni,Fe)S'?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7396" PostTypeId="2" ParentId="7395" CreationDate="2013-12-16T21:49:33.317" Score="8" Body="&lt;p&gt;When two materials precipitate at the same time, you'll often get an extremely intimate mix of the two. I'm assuming that the notation you've got here means a precipitate consisting of a mixture of both nickel (II) ions, iron (II) ions, and sulfide ions, or a mix of nickel and iron sulfides (NiS &amp;amp; FeS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron and nickel (II) have similar ionic radii and charge densities, so it should be possible to create crystal domains containing both metal ions interchangeably, in any ratio.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="402" LastEditDate="2017-04-08T22:21:50.257" LastActivityDate="2017-04-08T22:21:50.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7397" PostTypeId="1" CreationDate="2013-12-17T01:18:11.970" Score="3" ViewCount="848" Body="&lt;p&gt;I have a Chemistry lab in which I must design and carry out an experiment to determine the compound in each of eight numbered vials. Each vial contains one of these ionic compounds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;BaCl&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;CaCO&lt;sub&gt;3&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ca(OH)&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;KI&lt;/li&gt;&#xA;&lt;li&gt;NaCl&lt;/li&gt;&#xA;&lt;li&gt;NaHCO&lt;sub&gt;3&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;Na&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;Pb(NO&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Materials:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Distilled water&lt;/li&gt;&#xA;&lt;li&gt;Several microwell reaction plates&lt;/li&gt;&#xA;&lt;li&gt;Tooth picks (to stir)&lt;/li&gt;&#xA;&lt;li&gt;8x 1mL Beral style pipets&lt;/li&gt;&#xA;&lt;li&gt;The contents of the lab drawer (beakers, test tubes, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition to the materials above, &lt;b&gt;one&lt;/b&gt; of the following four additional reagents may be used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;6M H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;li&gt;6M HCl&lt;/li&gt;&#xA;&lt;li&gt;6M NaOH&lt;/li&gt;&#xA;&lt;li&gt;phenolphthalein indicator solution&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know how to identify a couple of them. For instance, CaCO&lt;sub&gt;3&lt;/sub&gt; can be identified by its insolubility in water, but as for the rest, I'm not completely sure. I know some can be identified by how and if they react with other compounds, but I don't know exactly how to go about that. &lt;/p&gt;&#xA;" OwnerUserId="3960" LastEditorUserId="3960" LastEditDate="2013-12-17T04:21:57.313" LastActivityDate="2014-01-31T22:42:47.997" Title="How to identify ionic compounds" Tags="&lt;inorganic-chemistry&gt;&lt;ionic-compounds&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7398" PostTypeId="2" ParentId="7392" CreationDate="2013-12-17T02:11:57.947" Score="3" Body="&lt;p&gt;In this content &lt;strong&gt;atom&lt;/strong&gt; means &lt;strong&gt;isolated atom&lt;/strong&gt;, which have defined quantum states.  This is to be distinguished with &lt;strong&gt;atom in a molecule&lt;/strong&gt;, which is an entirely conceptual construction that have not strict definition or physical meaning.  Atoms in a molecule have no defined quantum states.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, only an isolated oxygen atom not in a molecule can have designations like $O(^3P)$, etc.   For molecules, term symbols belong to the entire molecule, not individual atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In oxidation reactions atomic oxygen is rarely involved.  It can only be formed when very large amount of excess energy is available, such as in combustion or ignition processes, or in a radiation rich environment such as outer space.  There, the majority of atomic oxygen should be the most stable species $O(^3P)$.  However, in these high energy scenarios, all the other low lying states of oxygen will also be present in smaller quantities.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-17T02:11:57.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7399" PostTypeId="2" ParentId="7390" CreationDate="2013-12-17T02:34:36.517" Score="2" Body="&lt;p&gt;After searching on Google about &lt;a href=&quot;http://www.google.com/search?q=glucose+isomerase+to+fructose&quot; rel=&quot;nofollow&quot;&gt;Isomerization of Glucose to Fructose&lt;/a&gt;, you can find so many sources that explain the process. I think that &lt;a href=&quot;http://www.rpi.edu/dept/chem-eng/Biotech-Environ/IMMOB/poppezz/hfcs.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one is easy to understand among them. It explains that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The isomerization of glucose to fructose is part of the glycolysis cycle that converts glucose to pyruvate. The way this is done is to isomerize the aldehyde (hemiacetal) glucose to the ketone (as a hemiacetal) fructose,and make another phosphate ester. The isomerization takes advantage of the ease of breakage of a C-H bond which involves a carbon next to a carbonyl carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives a clear Idea on the mechanism of Glucose Isomerage to fructose. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose-6-phosphate_isomerase&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2013-12-17T02:34:36.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7400" PostTypeId="1" AcceptedAnswerId="7410" CreationDate="2013-12-17T08:12:45.777" Score="1" ViewCount="373" Body="&lt;p&gt;Also, are there any exceptions that you know of?&lt;/p&gt;&#xA;" OwnerUserId="3730" LastActivityDate="2013-12-17T16:07:51.570" Title="Do superhydrophobic/hydrophobic materials generally have a lot of friction or only a little friction" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;materials&gt;&lt;intermolecular-forces&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7401" PostTypeId="1" CreationDate="2013-12-17T10:17:11.620" Score="3" ViewCount="330" Body="&lt;p&gt;Is Coupling Reactions, Heck Reaction for example, different from Coupled Reactions like the production of Glucose 6-phosphate in living organism?&lt;/p&gt;&#xA;" OwnerUserId="3906" LastActivityDate="2021-05-14T10:09:00.363" Title="Coupled and Coupling Reactions" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7402" PostTypeId="2" ParentId="7391" CreationDate="2013-12-17T11:56:56.187" Score="10" Body="&lt;p&gt;First of all, who says coordination complexes don't dissociate? The interesting chemistry happens right there, and without dissociation no catalytic coordination compounds would be possible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the way I see it, when you move your compound, be it ionic, covalent or coordination, from air to a condensed phase, it will automatically interact with said phase. I think that you assume this phase is water, but it might as well be dichloromethane or ammonia or diethyl ether or anything else that is liquid (basically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes a bond happen (or not) is the overall energy of the system. See, in chemistry (as in physics) the energy minimum is the goal of the system, since it is the most stable. Using this line of reasoning we can derive that bonds are formed because the overall energy (of the bonding electrons) is lower than if they were not bonded. Note that it doesn't matter where these electrons come from (they might be shared by both bonding partners or made available by only one of them). A good and easy way to visualize this is if you scratch the surface of &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot;&gt;Molecular Orbital Theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do ionic compounds dissolve? Mostly due to the solvent-compound interactions. See more &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I also think that your approach using Coulomb theory doesn't work that well, since the change of the dielectric constant &lt;strong&gt;around&lt;/strong&gt; the compounds has nothing to do with what is &lt;strong&gt;between&lt;/strong&gt; the bonding partners, thus affecting attractive forces between the two partners.)&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-17T11:56:56.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7403" PostTypeId="2" ParentId="7397" CreationDate="2013-12-17T12:11:42.047" Score="4" Body="&lt;p&gt;First, you can use phenolphthalein indicator solution, $\ce{Ca(OH)2}$ and $\ce{NaHCO3}$ will turn red. Then use 6M $\ce{HCl}$ and the $\ce{NaHCO3}$ bubbles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaOH}$ to identify $\ce{Pb(NO3)2}$: it generates precipitations. Use $\ce{Pb(NO3)2}$ to identify $\ce{KI}$: generate golden precipitations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4}$ with $\ce{BaCl2}$ gives white precipitations, $\ce{Na2SO4}$ with $\ce{BaCl2}$ likewise. So the rest is $\ce{NaCl}$. &lt;/p&gt;&#xA;" OwnerUserId="3963" LastEditorUserId="73" LastEditDate="2013-12-17T21:41:49.347" LastActivityDate="2013-12-17T21:41:49.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7404" PostTypeId="2" ParentId="7385" CreationDate="2013-12-17T12:25:00.483" Score="2" Body="&lt;p&gt;The Gibbs energy is just a way to measure the thermodynamically favoured direction of a reaction. It doesn't tell you anything about the kinetics, so it might well be that the activation energy is too high for a reaction to happen, although it is thermodynamically favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs free energy is related to the cell potential $\Delta E$ via&#xA;$$ \Delta G = -nF\Delta E\; ,$$&#xA;where $n$ is the number of Electrons transferred per mole product and $F$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_constant&quot; rel=&quot;nofollow&quot;&gt;Faraday constant&lt;/a&gt;. It follows that the larger the cell potential is, the more negative the Gibbs free energy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cell potential can be calculated using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow&quot;&gt;Nernst equation&lt;/a&gt;:&#xA;$$ \Delta E = \Delta E^0 - \frac{RT}{nF}\ln\left( \frac{[\text{Red}]}{[\text{Ox}]}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\text{Red}]$ and $[\text{Ox}]$ refer to the concentrations of the reduced and oxidized species respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this information allows you to determine in which direction a given redox reaction will develop.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2014-02-12T07:56:23.253" LastActivityDate="2014-02-12T07:56:23.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7406" PostTypeId="1" AcceptedAnswerId="9414" CreationDate="2013-12-17T13:48:03.487" Score="9" ViewCount="11785" Body="&lt;p&gt;I do know that the lanthanides start with the element lanthanum, but why? Lanthanum doesn't even have an $f$ orbital, so why isn't it considered a transition metal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same way with actinium, which doesn't even have any electrons in its $5f$ orbital, why isn't it also a transition metal?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorDisplayName="user3901" LastEditDate="2013-12-27T09:00:06.980" LastActivityDate="2019-04-25T06:14:11.727" Title="Why is lanthanum a lanthanide and actinium an actinide?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7409" PostTypeId="1" AcceptedAnswerId="7414" CreationDate="2013-12-17T15:56:04.060" Score="4" ViewCount="2198" Body="&lt;p&gt;Why did Thomson assume that electrons are embedded in a positively charged sphere? Why did he not assume that they are positively charged particles embedded in a negatively charged sphere?&lt;/p&gt;&#xA;" OwnerUserId="3967" LastEditorUserId="73" LastEditDate="2014-08-23T03:14:27.177" LastActivityDate="2014-08-23T03:14:27.177" Title="In Thomson's atomic model, why did he not assume that there are positively charged particles embedded in a negatively charged sphere?" Tags="&lt;electrons&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7410" PostTypeId="2" ParentId="7400" CreationDate="2013-12-17T16:07:51.570" Score="0" Body="&lt;p&gt;Since you are talking about hydrophobicity I am assuming you are interested in liquid-solid friction. If you are interested in solid-solid friction I can say only one thing: hydrophobicity and solid-solid friction are in general unrelated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the liquid-solid case. I am going to give you an answer that gives some insight, but no concrete rules, because, as far as I know, there aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Superhydrophobe&quot; rel=&quot;nofollow&quot;&gt;Superhydrophobic surfaces&lt;/a&gt; typically have some surface structure that allows the liquid to 'float' partly on air and partly on top of the surface structure . Friction is related to the amount of contact area where there is a no-slip boundary so that will decrease therefore friction will be lower. A cool application of this effect can be found in this &lt;a href=&quot;http://www.pnas.org/content/early/2013/05/01/1304403110&quot; rel=&quot;nofollow&quot;&gt;2013 PNAS article by Karatay et al.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you have a droplet moving, instead of a full liquid stream, then there is a counter-acting effect which is called contact angle hysteresis, which allows a droplet to sustain a difference in contact angle at the front and back of the droplet. This causes a force counter-acting motion of the order of the surface tension of the liquid $\gamma$ and the difference between the contact angles $\Delta \theta$: $F=\gamma l \Delta \theta$, where $l$ is a lengthscale of the order of the droplet radius. This force basically acts as a static friction on the droplet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $\Delta \theta$ is a complex function of the amount of surface roughness, being 0 for a atomically smooth surface but also approaching 0 for roughness with a small typical lengthscale. Therefore it is hard to make a general statement whether superhydrophobic surfaces will have a lot or a little friction. It all depends on the details of the surface roughness.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2013-12-17T16:07:51.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7411" PostTypeId="1" AcceptedAnswerId="7413" CreationDate="2013-12-17T17:26:19.783" Score="11" ViewCount="5322" Body="&lt;p&gt;I have understood the &lt;a href=&quot;http://www.name-reaction.com/cannizzaro-reaction&quot; rel=&quot;noreferrer&quot;&gt;basic mechanism&lt;/a&gt; of cannizzaro reaction. But I am unable to understand the products formed in case of a crossed Cannizzaro reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose Benzaldehyde and Formaldehyde are heated in presence of $\ce{NaOH}$ to carry out Cannizzaro transformation. I have read that it is Benzyl Alcohol and Formic Acid formed in the greatest proportion and not Benzoic acid and Methyl alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so? How can we understand this in terms of the stability&#xA; of Cannnizzaro intermediates?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="7448" LastEditDate="2015-10-02T16:44:16.510" LastActivityDate="2018-09-08T05:58:21.077" Title="Crossed Cannizzaro products" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7412" PostTypeId="1" AcceptedAnswerId="7419" CreationDate="2013-12-17T17:32:39.033" Score="9" ViewCount="7848" Body="&lt;p&gt;One of the questions in my textbook was to arrange the following in order of their reactivity for nucleophilic addition reaction.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{(p-NO2Ph)_2CO}$  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{(CH3)_2CO}$  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{(p-NO2Ph)COH}$  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{CH3COH}$  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know there are two factors in play here. One is the steric factor (4&gt;3&gt;2&gt;1) and annother is the electropositive nature of the carbonyl carbon, which is the site of nucleophilic attack (1&gt;3&gt;4&gt;2). I am unable to decide which of these factors shall overpower the other and what shall be the final order?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="46306" LastEditDate="2018-02-20T10:58:17.157" LastActivityDate="2018-02-20T10:58:17.157" Title="Order of reactivity of carbonyl compounds to Nucleophilic addition reaction" Tags="&lt;organic-chemistry&gt;&lt;c-c-addition&gt;&lt;c-x-addition&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7413" PostTypeId="2" ParentId="7411" CreationDate="2013-12-17T17:53:43.597" Score="8" Body="&lt;p&gt;The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;&amp;amp;\ \ \ \ \  \ce{O} &amp;amp;&amp;amp; \ \ \ \ \ \ce{O-} \\&#xA;&amp;amp;\ \ \ \ \  \parallel &amp;amp;&amp;amp; \ \ \ \ \ \mid \\&#xA;&amp;amp;\ce{R-C-H +OH- -&amp;gt;}  &amp;amp;&amp;amp; \ce{R-C-H} \\&#xA;&amp;amp; \ &amp;amp;&amp;amp; \ \ \ \ \ \mid \\&#xA;&amp;amp; \ &amp;amp;&amp;amp; \ \ \ \ \ \ce{OH} &#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde&lt;/li&gt;&#xA;&lt;li&gt;Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-12-17T17:53:43.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7414" PostTypeId="2" ParentId="7409" CreationDate="2013-12-17T18:26:10.860" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.chemheritage.org/discover/online-resources/chemistry-in-history/themes/atomic-and-nuclear-structure/thomson.aspx&quot;&gt;J. J. Thomson&lt;/a&gt; is recognized with blowing apart a long-held belief that atoms were indivisible through his discovery of electrons while studying the deflection of discharges from cathode ray tubes.  He postulated that the particles formed in this way were small with a very high charge to mass ratio.  At this point he had two pieces of information: (1) an atom has an overall neutral charge and (2) part of the atom consisted of these small negatively charged entities which we now call electrons.  He therefore postulated that the rest of the atom consists of positive charge to balance the negative charge of electrons.  &lt;a href=&quot;http://www.chemheritage.org/discover/media/distillations/054-holiday-greetings-2008.aspx&quot;&gt;It is believed&lt;/a&gt; that the plum pudding analogy stems from his proficiency as an instructor and a desire to relate the atomic model to something his students could understand.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-12-17T18:26:10.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7415" PostTypeId="1" AcceptedAnswerId="7418" CreationDate="2013-12-17T18:48:46.030" Score="4" ViewCount="403" Body="&lt;p&gt;In order to create a simple program which calculates titration curves (also to provide some closure in the comments &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7162/why-should-colorimetric-titration-of-weak-acid-and-weak-base-should-not-be-done&quot;&gt;here&lt;/a&gt;) I am trying to come up with all the equations so I can calculate the equilibrium proton concentration for any given concentrations of any two given weak acids/bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a collection of what I've come up with so far and what I assume I still need. If someone could help me back on track I'd kindly appreciate it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We're looking at the following reactions:&#xA;$$\begin{align} &#xA;\ce{HA &amp;amp;&amp;lt;=&amp;gt;[$K$_{\text{a, 1}}][$K$_{\text{b, 1}}] H+ + A- &amp;amp;&amp;amp; (1) }\\ &#xA;\ce{H+ + B- &amp;amp;&amp;lt;=&amp;gt;[$K$_{\text{b, 2}}][$K$_{\text{a, 2}}] HB &amp;amp;&amp;amp; (2)} &#xA;\end{align}$$&#xA;A is the weak acid and B the weak base with the according equilibrium constants (the indices 1 and 2 refer to the acid and base respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the total masses I've got the following:&#xA;$$\begin{align}&#xA;\ce{[A]_{tot} &amp;amp;= [HA] + [A- ] &amp;amp;&amp;amp; (3) }\\&#xA;\ce{[B]_{tot} &amp;amp;= [HB] + [B- ] &amp;amp;&amp;amp; (4) }&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constants are linked via&#xA;$$ K_\text{a, 1} \times K_\text{b, 2} = \frac{\ce{[A- ][HB]}}{\ce{[HA][B- ]}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore:&#xA;$$&#xA;\ce{[H+ ]_{eq}} = \frac{\ce{[A]_{tot}} K_\text{a, 1}}{\ce{[A- ]}} - K_\text{a, 1}&#xA; = \frac{\ce{[HB]}}{K_\text{b, 2} (\ce{[B]_{tot} - [HB]})}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the equilibrium proton concentration I've derived&#xA;$$ \ce{[H+ ]_{eq}} = K_\text{a, 1} \ce{[HA]} + K_\text{a, 2} \ce{[HB]} + K_\text{w} \ce{[H2O]}$$&#xA;where I neglect the influence of the water (the additional term $[\dots] + K_\text{w} \ce{[H2O]}$) since it is so small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm struggling with the following:&#xA;$$ \ce{[H+ ]_{tot}} = ? $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I believe that with the last equation I will be able to solve for $\ce{[H+ ]_{eq}}$. But maybe I'm wrong there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any pointers.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2016-08-09T20:42:53.807" Title="How does one calculate the pH of an aqueous solution of a weak acid and a weak base?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7416" PostTypeId="2" ParentId="7415" CreationDate="2013-12-17T19:34:00.157" Score="0" Body="&lt;p&gt;[H+]tot would have to equal [H+]eq. All of your sources of H+ are involved in the equilibria. If you were working with a weak acid and it's conjugate base you could just use the Henderson-Hasselbach equation, eliminating the need for a Kb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, this is a titration of a weak acid by a weak base (or vice versa). Therefore H+ would be neutralized immediately, as you add base. So the pH would start low (~pKa, I believe), then slowly increase to where all the acid would be used, and continue to slowly increase until ~pKb. But wow, that endpoint would be hard to find any other way than graphically!&lt;/p&gt;&#xA;" OwnerUserId="3968" LastActivityDate="2013-12-17T19:34:00.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7417" PostTypeId="2" ParentId="7415" CreationDate="2013-12-17T23:41:20.583" Score="3" Body="&lt;p&gt;This is the way I would tackle the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} &#xA;\ce{HA &amp;amp;&amp;lt;=&amp;gt;[K_\text{a}] H+ + A- &amp;amp;&amp;amp; (1) }\\ &#xA;\ce{B +H2O&amp;amp;&amp;lt;=&amp;gt;[K_\text{b}] BH+ + OH- &amp;amp;&amp;amp; (2)} &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass balance writes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{[A]_{tot} &amp;amp;= [HA] + [A- ] &amp;amp;&amp;amp; (3) }\\&#xA;\ce{[B]_{tot} &amp;amp;= [B] + [BH+ ] &amp;amp;&amp;amp; (4) }&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge balance writes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{[A- ] + [OH- ]&amp;amp;=[H+] + [BH+ ] &amp;amp;&amp;amp; (5) }&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, by introducing eq. (3) into the definition of $K_\text{a}$, one gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{[A- ]} = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And using the charge balance (eq. 5), one writes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{[BH+ ] +[H+ ]}-\frac{K_\text{w}}{\ce{[H+ ] }}  = \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same can be done for the base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{[BH+ ]} = \frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the last two equations together, one obtains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{K_\text{w}}{\ce{[H+ ]}}  + \frac{\ce{[A]_{tot}} K_\text{a}}{\ce{[H+ ]}+K_\text{a}}=\frac{\ce{[H+ ]}\ce{[B]_{tot}} K_\text{b}}{K_\text{b}\ce{[H+ ] }+K_\text{w}}+\ce{[H+ ]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the desired equation with $\ce{[H+ ]}$ as the unique unknown variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2013-12-17T23:41:20.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7418" PostTypeId="2" ParentId="7415" CreationDate="2013-12-17T23:55:30.673" Score="5" Body="&lt;h2&gt;Equations and variables&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To calculate the proton concentration in a solution as given in the question the following five equations are used for the following five unknown variables: $\ce{[H+ ]}$, $\ce{[HA]}$, $\ce{[A- ]}$, $\ce{[HB]}$ and $\ce{[B- ]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ion balance:&lt;/strong&gt;&#xA;$$ \ce{[H+ ]} = \ce{[A- ] + [B- ]} + \frac{K_\text{w}}{\ce{[H+ ]}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mass balances:&lt;/strong&gt;&#xA;$$\begin{align}&#xA;\ce{[A]_{tot}} &amp;amp;= \ce{[HA] + [A- ]} \\&#xA;\ce{[B]_{tot}} &amp;amp;= \ce{[HB] + [B- ]} &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Law of mass action:&lt;/strong&gt;&#xA;$$\begin{align}&#xA;K_\text{a, 1} &amp;amp;= \frac{\ce{[H+ ][A- ]}}{\ce{[HA]}} \\&#xA;K_\text{a, 2} &amp;amp;= \frac{\ce{[H+ ][B- ]}}{\ce{[HB]}}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Combining the above equations we get:&#xA;$$&#xA;\ce{[H+ ]} = \frac{K_\text{a, 1} \ce{[A]_{tot}} }{\ce{[H+ ]}+K_\text{a, 1}} + \frac{K_\text{a, 2} \ce{[B]_{tot}} }{\ce{[H+ ]}+K_\text{a, 2}} + \frac{K_\text{w}}{\ce{[H+ ]} }&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some reforming we get a fourth-order polynomial for the proton concentration:&#xA;$$\begin{align}&#xA;0 = &amp;amp; \ce{[H+ ]^4} + \ce{[H+ ]^3}(K_\text{a, 1} + K_\text{a, 2}) \\&#xA;  &amp;amp;- \ce{[H+ ]^2}(\ce{[A]_{tot}} K_\text{a, 1} + \ce{[B]_{tot}} K_\text{a, 2} - K_\text{a, 1} K_\text{a, 2} + K_\text{w}) \\&#xA; &amp;amp; -\ce{[H+ ]}\left[ (\ce{[A]_{tot}} + \ce{[B]_{tot}}) K_\text{a, 1} K_\text{a, 2} + K_\text{w}(K_\text{a, 1} K_\text{a, 2}) \right] \\&#xA; &amp;amp; - K_\text{w} K_\text{a, 1} K_\text{a, 2}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2013-12-18T00:01:51.850" LastActivityDate="2013-12-18T00:01:51.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7419" PostTypeId="2" ParentId="7412" CreationDate="2013-12-18T04:27:53.320" Score="7" Body="&lt;p&gt;My instinct is that there are probably too many variables here for a definitive and accurate answer to be given strictly from first principles. When comparing molecules in which there is only one key difference, especially when that difference leads to steric and electronic effects which are essentially additive, it's fairly easy to draw accurate conclusions. This question, however, is a bit thornier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing acetone, $\ce{(CH3)2CO}$, to acetaldehyde, $\ce{CH3COH}$, is quite straightforward: acetaldehyde should be the more electrophilic of the two. Firstly, the greater steric bulk of acetone's methyl group slightly hinders nucleophilic attack. Secondly, the additional methyl group is electron-donating by hyperconjugation, comparatively reducing acetone's electrophilicity. Here, both steric effects and electronic effects are synergistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other comparisons are not quite as straightforward, as they require (in the absence of empirical data) qualitative judgment calls to be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider &lt;em&gt;p&lt;/em&gt;-nitrobenzaldehyde. Sterically, reactivity should be reduced due to the obvious bulk of the aromatic ring. Electronically, the picture seems to me to be more complex. From the perspective of valence bond theory, the $sp^2$ carbons may be electron-withdrawing by induction, particularly in concert with the electronegative nitro group. On the other hand, the carbonyl group is also highly electronegative and should remove electron density from the ring, primarily via the $\pi$ system, and I would expect that effect to be greater. Finally, from the MO theory perspective, considering the frontier orbitals, conjugation of the carbonyl to the aromatic ring should result in lowering the energy of the $\pi^*$ LUMO, which ought to enhance reactivity. That, however, may be mitigated by any overall increase in electron density around the carbonyl group. By comparison to acetaldehyde, the steric effects are clearly unfavorable, while the electronic effects are slightly murky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the 4,4'-dinitrobenzophenone, the analysis of electronic effects is roughly the same as for &lt;em&gt;p&lt;/em&gt;-nitrobenzaldehyde, but here sterics is undoubtedly going to be the overriding factor, and should result in massively diminished reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I would predict the reactivity order to be acetaldehyde &gt;&gt; &lt;em&gt;p&lt;/em&gt;-nitrobenzaldehyde &gt; acetone &gt;&gt; 4,4'-dinitrobenzophenone. I don't claim this is in any way authoritative, so I hope others weigh in as well. If I can find some relevant literature containing, e.g., relative rates, I'll update this answer with a link.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-18T04:27:53.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7420" PostTypeId="1" AcceptedAnswerId="7421" CreationDate="2013-12-18T04:51:10.717" Score="0" ViewCount="439" Body="&lt;p&gt;This is my first question in this site. I am referring to this &lt;a href=&quot;https://bicycles.stackexchange.com/questions/16001/is-there-a-way-to-fill-your-tires-with-nitrogen&quot;&gt;question&lt;/a&gt;.The linked question what is the advantage of filling nitrogen in bike tubes over air and I got some answers such as:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;The biggest advantage to filling your tires with Nitrogen is consistency.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Air pressure fluctuates based on temperature, whereas Nitrogen fluctuates to a much much lower degree&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Nitrogen changes pressure with temperature slightly less than water does&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But I want to know the reason. Please help me understand this scientifically.&lt;/p&gt;&#xA;" OwnerUserId="3971" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-01-03T17:34:45.180" Title="Does nitrogen pressure in tyres fluctuate to much a lower degree than air with changes in temperature?" Tags="&lt;inorganic-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7421" PostTypeId="2" ParentId="7420" CreationDate="2013-12-18T05:23:40.063" Score="5" Body="&lt;p&gt;There can be no scientific explanation, because the argument itself is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most part &lt;em&gt;air&lt;/em&gt; is &lt;em&gt;nitrogen&lt;/em&gt;.  To be specific, about 80% of dry air is made up from nitrogen.  In fact both nitrogen and oxygen, the second major components of air that takes up almost all the other 20%, behave extremely like ideal gas.   Pressures of &lt;strong&gt;dry air&lt;/strong&gt; and &lt;strong&gt;pure nitrogen gas&lt;/strong&gt; at different temperatures, from far below freezing point to above boiling point, have indeed been measured to incredibly high accuracy.  &lt;strong&gt;The difference between these two gases is below 0.1%.&lt;/strong&gt;   In fact, near room temperature, many types of gas are close to ideal gas and have very similar temperature profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure why people are comparing nitrogen with water in that question thread you mentioned.  Water is a minor component of air and its concentration varies so dramatically across places.  However, even in a summer day, at the maximum humidity of 100%, the amount of water in the air is only 10% of that of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, &lt;strong&gt;pressure of nitrogen gas do not fluctuate less with change of temperature compared to normal air&lt;/strong&gt;.   &lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="3723" LastEditDate="2013-12-18T05:44:13.207" LastActivityDate="2013-12-18T05:44:13.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7422" PostTypeId="1" CreationDate="2013-12-18T06:47:39.177" Score="6" ViewCount="4084" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$ in water at $25\ \mathrm{^\circ C}$. What is the standard Gibbs energy of dissolution of the salt?&#xA;  The reaction is&#xA;  $$\ce{Hg2I2(s) -&amp;gt; Hg2^2+(aq) + 2I- (aq)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I think I need in order to solve the question:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G=-RT\ln(K)$$ where $K$ will be the solubility constant. When the question says “solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$” I don’t exactly understand what this means.&lt;br&gt;&#xA;The units make it seem like a concentration to me.    &lt;/p&gt;&#xA;" OwnerUserId="3972" LastEditorUserId="46306" LastEditDate="2018-03-02T03:23:12.007" LastActivityDate="2018-03-02T03:23:12.007" Title="Calculating Standard Gibbs Energy of Dissolution" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7423" PostTypeId="2" ParentId="6972" CreationDate="2013-12-18T08:35:04.303" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/toc/jceda8/87/12&quot; rel=&quot;nofollow&quot;&gt;Have a look at this paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;put food coloring to 35% isopropyl alcohol in water.&lt;/li&gt;&#xA;&lt;li&gt;add ammonium sulfate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What happens ? Once ammonium sulfate added and the solution mixed, the ammonium sulfate dissolves, pushing the isopropyl alcohol out of the aqueous solution (and the food coloring as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separate layer thus forms, that is clearly visible thanks to the food coloring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weak point of the whole story is that isopropyl alcohol is flammable...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PS: I would have preferred to reply to @Michiel's answer and @Nicolau Saker Neto's comment: but I do not have enough reputation to do that...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2013-12-18T08:35:04.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7424" PostTypeId="1" CreationDate="2013-12-18T13:01:01.857" Score="6" ViewCount="608" Body="&lt;p&gt;In Clayden et al., &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Chemistry&lt;/a&gt;, p. 332, the authors compare the reaction energy profile for ester formation from a) acyl chloride and b) anhydride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TOkjY.png&quot; alt=&quot;Clayden et al., Organic Chemistry, OUP 2001&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They then write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The energy levels of the starting materials, the transition state, and the intermediate are all lower in the anhydride reaction than in the chloride reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are they really comparing absolute energies between different species?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="5026" LastEditDate="2018-04-26T02:33:50.857" LastActivityDate="2018-04-26T02:33:50.857" Title="Is this really a comparison of the absolute energies in the reaction energy profile for ester formation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7425" PostTypeId="2" ParentId="7422" CreationDate="2013-12-18T14:37:38.180" Score="4" Body="&lt;p&gt;You are correct that you need $\Delta G^\circ =-RT\ln{K}$ and that the equilibrium constant in this case is $K_{sp}$ the solubility product constant, which is defined for a process like this one to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A_{m} B_{n} (s) &amp;lt;=&amp;gt; mA^{n+}(aq) + nB^{m-}(aq)}$$&#xA;$$K_{sp} = \ce{[A^{n+}]^{m} [B^{m-}]^{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case $K_{sp}=[\ce{Hg2^{2+}}][\ce{I-}]^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are given the solubility (maximum obtainable concentration) of $\ce{Hg2I2}$ in water at $25 \ ^\circ \text{C}$. How can you use that information to find $[\ce{Hg2^{2+}}]$ and $[\ce{I-}]$ when the solution is saturated?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-12-18T14:37:38.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7426" PostTypeId="1" AcceptedAnswerId="22319" CreationDate="2013-12-18T15:37:18.887" Score="10" ViewCount="19375" Body="&lt;p&gt;According to both caffeine's &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=2519#x27&quot; rel=&quot;nofollow noreferrer&quot;&gt;PubChem page&lt;/a&gt; and &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.2424.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemSpider page&lt;/a&gt; its boiling point lies at &lt;span class=&quot;math-container&quot;&gt;$\pu{173 °C}$&lt;/span&gt; and its melting point at &lt;span class=&quot;math-container&quot;&gt;$\pu{\sim 235 °C}.$&lt;/span&gt; How can it melt at that temperature if it already sublimated at &lt;span class=&quot;math-container&quot;&gt;$\pu{173 °C}?$&lt;/span&gt; Or is that the temperature of the triple point? If so, how come none of those pages gives any pressure specifications for that temperature?&lt;/p&gt;&#xA;&lt;p&gt;I heated up some anhydrous caffeine at home, and at my home atmospheric pressure it appears to melt (not sublimate) and starts creating vapors immediately after (but not before) melting. This is based just on what I can see with the naked eye.&lt;/p&gt;&#xA;" OwnerUserId="3974" LastEditorUserId="41328" LastEditDate="2021-03-11T07:41:34.523" LastActivityDate="2021-03-11T07:41:34.523" Title="Caffeine boiling and melting points" Tags="&lt;phase&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7427" PostTypeId="1" CreationDate="2013-12-18T16:54:04.247" Score="3" ViewCount="1754" Body="&lt;p&gt;Which bond is stronger, ionic or covalent? I have a lot of confusion about this. &lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="73" LastEditDate="2013-12-18T18:26:53.307" LastActivityDate="2014-07-23T21:32:57.133" Title="Ionic or covalent bond" Tags="&lt;bond&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7428" PostTypeId="1" CreationDate="2013-12-18T16:42:48.963" Score="1" ViewCount="771" Body="&lt;p&gt;Does a grain of salt have exactly equal amount of sodium and chlorine? I know that the formula unit of salt is 1:1, but for example if there is 10000000 atoms of sodium is it possible that that there is 9999999 atoms chlorine? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4012" OwnerDisplayName="Kevin" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:51:09.230" LastActivityDate="2015-10-23T16:51:09.230" Title="Does a grain of salt have exactly equal amount of sodium and chlorine?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;elemental-analysis&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7429" PostTypeId="2" ParentId="7428" CreationDate="2013-12-18T18:00:44.277" Score="6" Body="&lt;p&gt;Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Point defects &lt;/li&gt;&#xA;&lt;li&gt;Line defects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One type of point defect is a vacancy defect in which some of the lattice sites in the crystal are vacant. This means that some atoms of Na or Cl or both are missing from their lattice sites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is, yes. That scenario is very much possible.&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="2321" LastEditDate="2013-12-20T06:11:24.473" LastActivityDate="2013-12-20T06:11:24.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7430" PostTypeId="1" AcceptedAnswerId="7434" CreationDate="2013-12-18T18:20:51.580" Score="4" ViewCount="6498" Body="&lt;p&gt;Why has no one ever been able to synthesise cyclopentyne as a stable molecule? I've been thinking about this for quite some time and I can't seem to figure it out. Something to do with angular strain in the molecule?&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="7475" LastEditDate="2015-10-09T15:54:22.730" LastActivityDate="2016-12-22T16:39:39.690" Title="Why is cyclopentyne unstable?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7431" PostTypeId="1" CreationDate="2013-12-18T18:55:43.037" Score="2" ViewCount="1045" Body="&lt;p&gt;I performed an electrolysis of an aqueous sodium carbonate solution with a platinum electrode and a chromium-vanadium-steel-alloy electrode. The solution turned yellow at the chromium-vanadium-alloy electrode (I suspect that it is sodium metavanadate (NaVO3)) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I also performed an electrolysis in an aqueous sodium carbonate solution with a platinum electrode and an inox electrode (the inox electrode is composed of iron, chromium,...) . A green compound (insoluble) was formed and when I later added acetic acid, a brown compound (I think Iron(III) acetate ) was formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in both cases there were no chromium compounds formed (at least of what I know) So why didn't the chromium react (why didn't the chromium get oxidized)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information:&#xA;Voltage was around 13 Volts, the inox electrode was stainless steel (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Stainless_steel&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Stainless_steel&lt;/a&gt;) and has according to wikipedia a minimum of 10.5% chromium content by mass. I read online that a regular chromium-vanadium-steel-alloy has a vanadium  content  of approximately  0.181%  and a chromium  content  approximately  1.00%. (however, I'm not sure about these last two figures) What is true is that there wasn't a lot of yellow sodium metavanadate after an electrolysis of around 20 minutes. ) The electrolysis solution contained sodium carbonate(Na2CO3) and water(H2O) the electrolysis was performed at normal conditions (1atm - 25°C) &lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2013-12-19T14:11:11.610" LastActivityDate="2018-09-01T11:33:18.393" Title="Chromium from electrode not oxidizing in electrolysis" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7433" PostTypeId="2" ParentId="6489" CreationDate="2013-12-18T19:46:52.183" Score="2" Body="&lt;p&gt;We have to see the $d$-orbital of the metal atom how many electrons there are, but in &lt;em&gt;excited state&lt;/em&gt; not in ground state; then see the spins whether it is low spin or high spin because $d$-orbital is divided into $t_{2g}$ and $e_g$ with respect to their energies difference then apply the formula. The formula is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\mathrm{no.\ of\ electrons\ in\ t_{2g}\cdot(-0.4)}+\mathrm{no.\ of\ electrons\ in\ e_g\cdot(0.6)}$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Example: I have $\mathrm{[Co(F)_6]^{3-}}$ now see cobalt has $3d^7$, $4s^2$ system in ground state but in excited state it loses three electrons in the formation of ions and two elctrons from $4s$ and one from $3d$ orbital so thus Cobalt gets $3d^6$ configuration. Now it is of high spin so $4$ electrons go to $t_{2g}$ orbital and $2$ electrons go to $e_g$ orbital. By applying formula,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$4(-0.4)+2(0.6)=-1.6+1.2=-0.4$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3977" LastEditorUserId="670" LastEditDate="2015-03-29T20:04:14.613" LastActivityDate="2015-03-29T20:04:14.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7434" PostTypeId="2" ParentId="7430" CreationDate="2013-12-18T20:22:05.327" Score="11" Body="&lt;p&gt;Stable alkynes are normally linear, that is, they have bond bond angles of 180°. That is necessitated by the fact that alkynes are formed by two π-bonds and one σ-bond, which must be mutually perpendicular. If cyclopentyne were to exist, it would be forced to adopt bond angles of roughly 108°, equal to those of the interior angles of a regular pentagon. This deviation from ideal geometry causes extreme instability by virtue of poor orbital overlap. The rigidity of the alkyne moiety would also exacerbate the inherent conformational inflexibility of small rings, wherein the inability of bonds to freely rotate results in eclipsing between atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only cyclic alkyne I'm aware of is &lt;a href=&quot;http://en.wikipedia.org/wiki/Aryne&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzyne&lt;/a&gt;. However, it's far from stable, existing only as a fleeting intermediate in specific nucleophilic aromatic substitution and pericyclic cycloaddition reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: After a quick search, I found the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycloalkyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;cycloalkyne Wikipedia page&lt;/a&gt;, which claims that cyclooctyne is the smallest cycloalkyne stable enough to isolate and observe. I don't know whether this information is completely current, however.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorDisplayName="user7951" LastEditDate="2016-12-22T16:39:39.690" LastActivityDate="2016-12-22T16:39:39.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7435" PostTypeId="2" ParentId="7215" CreationDate="2013-12-19T04:33:27.693" Score="1" Body="&lt;p&gt;U can try rc mukherjee,an indian author for preliminary level.and schaum publishers for higher level.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2013-12-19T04:33:27.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7436" PostTypeId="2" ParentId="7428" CreationDate="2013-12-19T07:12:22.723" Score="0" Body="&lt;p&gt;&lt;strike&gt;Yes. The amounts of Sodium and Chlorine are the same. The reason being you need to have the charge balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The defects mentioned in the above answer do exist, but always in pairs. That is, for every missing Sodium atom, you have a Chlorine atom missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;p&gt;What can be said is that in 58.44 g (Molar Mass) of NaCl, you won't necessarily have Avagadro number of Sodium or Chlorine atoms. But, whatever their number, there will be equal amounts of Sodium and Chlorine.&lt;/p&gt;&lt;/strike&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditDate="2013-12-20T05:50:45.470" LastActivityDate="2013-12-20T05:50:45.470" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7437" PostTypeId="1" CreationDate="2013-12-19T08:56:25.707" Score="4" ViewCount="446" Body="&lt;p&gt;Imagine some S&lt;sub&gt;N&lt;/sub&gt;2 reaction involving an aliphatic sp&lt;sup&gt;3&lt;/sup&gt; carbon with a halide leaving group (e.g. iodide or bromide). Given the rise of femtosecond spectroscopy (e.g. using a mode-locked Ti:sapphire laser) and even attosecond spectroscopy, I'm curious if there are any studies that have measured the properties and approximate lifetime of the S&lt;sub&gt;N&lt;/sub&gt;2 transition state complex? I guess that this is in the $\approx$ picosecond regime alongside the timescales for a bond stretching dynamics (e.g. between oxygen and hydrogen on a water molecule), however, is there a more precise value that is now known?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am be curious about results from quantum mechanical simulations (using e.g. Jaguar from the Schrodinger computational chemistry suite) for S&lt;sub&gt;N&lt;/sub&gt;2 reactions?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll specify that I'm referring to a &quot;concerted&quot; S&lt;sub&gt;N&lt;/sub&gt;2 reaction, not, for example, the kind of reaction described in: [Chandrasekhar, J., Jorgensen, W. L. Energy profile for a non-concerted SN2 reaction in solution. J. Am. Chem. Soc. 107, pp. 2974 - 2975 (1985).]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, provided that non-concerted S&lt;sub&gt;N&lt;/sub&gt;2 reactions exist (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00296a024?journalCode=jacsat&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;) perhaps my question is underspecified. Perhaps, I should ask instead if there are any interesting long-lived S&lt;sub&gt;N&lt;/sub&gt;2 interaction complexes, or tables for transition state lifetimes from ultrafast spectroscopy studies?&lt;/p&gt;&#xA;" OwnerUserId="3984" LastEditorUserId="5026" LastEditDate="2018-03-17T07:43:41.133" LastActivityDate="2018-03-17T07:44:30.163" Title="What is the approximate lifetime of an SN2 transition state complex?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7438" PostTypeId="1" CreationDate="2013-12-19T10:19:42.433" Score="6" ViewCount="340" Body="&lt;p&gt;Textile fabrics are made out of cellulose fibres, that can be derived from cotton or bamboo. As far as I know all grasses contain cellulose, so why are only a few grasses used? What makes them so special? Why could you not use for example gamba grass/andropogon gayanus for that, which contains 30%  cellulose and 21% hemicellulose (&lt;a href=&quot;http://www.lrrd.org/lrrd20/2/oded20027.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.lrrd.org/lrrd20/2/oded20027.htm&lt;/a&gt;)? Any insight on that would be very helpful&lt;/p&gt;&#xA;" OwnerUserId="3985" LastEditorUserId="9961" LastEditDate="2016-02-04T00:01:02.003" LastActivityDate="2016-02-04T00:01:02.003" Title="What chemical properties make some grasses usable for producing textile fabrics?" Tags="&lt;polymers&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7439" PostTypeId="1" AcceptedAnswerId="7491" CreationDate="2013-12-19T10:25:37.260" Score="34" ViewCount="98742" Body="&lt;p&gt;Is there a scientific basis for the coffee making equipment which Gale Boetticher describes in Breaking Bad?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXXSf.jpg&quot; alt=&quot;Picture of the &amp;quot;Ultimate&amp;quot; coffee maker&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He talks about maintaining the right conditions for bringing out the coffee flavor without degrading it.&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="402" LastEditDate="2016-12-29T00:41:15.573" LastActivityDate="2018-03-03T03:03:43.853" Title="Chemistry behind Gale's coffee maker in Breaking Bad" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="7440" PostTypeId="1" CreationDate="2013-12-19T12:43:54.813" Score="3" ViewCount="1152" Body="&lt;p&gt;In this diagram (Evans et al., J. Chem. Edu.) the current is non-zero, but the potential is still below the normal potential. How can the compound become oxidized at the working electrode, when the applied potential is still below the normal potential of the compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nh3mg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="4356" LastEditDate="2015-07-10T18:59:52.200" LastActivityDate="2015-07-10T18:59:52.200" Title="Why is current non-zero when applied potential is still below E$^0$ in cyclic voltammetry?" Tags="&lt;electrochemistry&gt;&lt;cyclic-voltammetry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7441" PostTypeId="1" AcceptedAnswerId="7445" CreationDate="2013-12-19T14:39:38.860" Score="-1" ViewCount="5647" Body="&lt;p&gt;What are the products formed when phenol reacts with iodomethane in presence of dilute alkali? Also, a reaction with a mechanism would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Phenol reacts with NaOH to give a salt, sodium phenoxide. And yes, I was thinking about addition of iodine to the ring, but I don't know how. &lt;/p&gt;&#xA;" OwnerUserId="3863" LastEditorUserId="9961" LastEditDate="2017-04-08T16:53:03.340" LastActivityDate="2017-04-08T16:53:03.340" Title="How does phenol react with iodomethane in presence of dilute alkali?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7442" PostTypeId="1" CreationDate="2013-12-19T15:16:27.390" Score="4" ViewCount="3878" Body="&lt;p&gt;The $\ce{H-H}$ bond enthalpy is $436 \mathrm{kJ/mol}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 (g) -&amp;gt; 2H (g)} \qquad \Delta H = +436 \mathrm{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this only apply at the standard temperature of $298 \text{K}$, i.e. would the bond enthalpy vary if I decrease or increase the temperature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is it practically just so that, for a temperature $T$, as long as you pump the bond enthalpy’s worth of kilojoules into the substance, it will be atomized?&lt;/p&gt;&#xA;" OwnerUserId="3989" LastEditorUserId="41556" LastEditDate="2017-04-01T21:35:50.930" LastActivityDate="2017-04-16T01:26:10.343" Title="Does the bond dissociation enthalpy vary with temperature?" Tags="&lt;bond&gt;&lt;enthalpy&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7443" PostTypeId="1" CreationDate="2013-12-19T20:12:03.437" Score="2" ViewCount="192" Body="&lt;p&gt;Say for example, you have isopropyl alcohol. Is there any reason why if you react it with bromine that it won't form a radical, as one of the bromines removes one of the protons adjacent to the hydroxyl group on the alkane, and then propagate into 2-bromo,2-propanol?&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="-1" LastEditDate="2013-12-20T16:02:11.533" LastActivityDate="2014-07-29T16:00:05.680" Title="Is there any reason why halogenation of substituted alkanes wouldn't occur through free radical substitution?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-10-02T16:55:19.790" ContentLicense="CC BY-SA 3.0" />
  <row Id="7444" PostTypeId="1" CreationDate="2013-12-19T20:23:27.983" Score="3" ViewCount="503" Body="&lt;p&gt;Is there such a thing as a dye that can stain both oil and water?  In other words, does anyone know of an amphiphilic dye?  Maybe a surfactant that is also a pretty shade of blue?  Or maybe turn one color in oil, but another in water, but stain both well?&lt;br&gt;&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="874" LastEditorUserId="418" LastEditDate="2014-03-12T08:58:25.563" LastActivityDate="2021-02-27T11:12:18.200" Title="Dyes that can stain oil and water" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;surfactants&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7445" PostTypeId="2" ParentId="7441" CreationDate="2013-12-19T22:41:30.083" Score="1" Body="&lt;p&gt;A good approach to this type of problem is to break it down into smaller problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know we have phenol $\ce{PhOH}$, methyl iodide $\ce{CH3I}$, and base, let's say $\ce{NaOH(aq)}$. Instead of trying to figure out what these three things do all together, can we determine what might happen with just two of them?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Phenol and methyl iodide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since this is the main thrust of the question, we probably cannot decide what happens here. If we knew this one, then we would know the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Phenol and NaoH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol&quot; rel=&quot;nofollow&quot;&gt;Phenol&lt;/a&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol#Acidity&quot; rel=&quot;nofollow&quot;&gt;acidic&lt;/a&gt;, more acidic than aliphatic alcohols. NaOH is a base and probably reacts with phenol. What would the product of this acid base reaction be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PhOH + NaOH -&amp;gt; ??}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NaOH and methyl iodide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reaction between these two is probably unproductive, since it would happen without the phenol present.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you can figure out what happens when phenol reacts with NaOH, then can you figure out what one of the products of that reaction might do when it encounters $\ce{CH3I}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-12-19T22:41:30.083" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7447" PostTypeId="2" ParentId="7428" CreationDate="2013-12-19T23:49:07.730" Score="-1" Body="&lt;p&gt;If it doesn't it isn't &quot;salt&quot; or at least it isn't NaCl, which requires one of each per molecule.&lt;/p&gt;&#xA;" OwnerUserId="3995" LastActivityDate="2013-12-19T23:49:07.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7448" PostTypeId="1" CreationDate="2013-12-20T02:07:46.553" Score="0" ViewCount="2594" Body="&lt;p&gt;What are some white compounds that look like powder (like flour) and immediately black after being heated? It turns immediately black and creates a odorous gas. &lt;/p&gt;&#xA;" OwnerUserId="3996" LastEditorUserId="4945" LastEditDate="2014-10-28T10:49:08.083" LastActivityDate="2014-10-28T10:49:08.083" Title="What white powdery compound turn black immediately after heating?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7449" PostTypeId="1" AcceptedAnswerId="7450" CreationDate="2013-12-20T06:28:32.177" Score="25" ViewCount="75589" Body="&lt;p&gt;When water temperature reaches $100\ ^\circ \mathrm{C}$, the molecules get so excited that the hydrogen atoms lose the bonds to the oxygen atom and therefore the water starts to become gas. I get that, but at room temperature ($23\ ^\circ \mathrm{C}$), is there no excitation in the atoms or is there? &lt;/p&gt;&#xA;" OwnerUserId="3998" LastEditorDisplayName="user7951" LastEditDate="2015-12-14T15:00:47.137" LastActivityDate="2021-03-08T05:49:30.820" Title="Why does water evaporate at room temperature?" Tags="&lt;water&gt;&lt;phase&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7450" PostTypeId="2" ParentId="7449" CreationDate="2013-12-20T06:51:31.537" Score="35" Body="&lt;p&gt;First, I think I should make it clear that when water boils, the bonds in the water molecule linking the hydrogen and oxygen atom are &lt;em&gt;not&lt;/em&gt; broken. During boiling, the intermolecular bonds in water are the ones that get broken, that is the bonds that link the water molecules together.&lt;/p&gt;&#xA;&lt;p&gt;At room temperature, there is evaporation (I wouldn't call it excitation). This is because there are a few molecules of water which can manage to muster enough energy to escape from the large body of molecules and escape into air.&lt;/p&gt;&#xA;&lt;p&gt;This can be explained through a graph depicting the distribution of speed among water molecules worked out by Maxwell and Boltzmann.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRZZc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can probably see, there are a lot of water molecules with lower kinetic energy than with higher kinetic energy. Those that have the higher kinetic energy are the ones that are able to break through the water surface to become vapour.&lt;/p&gt;&#xA;&lt;p&gt;Even at low temperatures, there are &lt;em&gt;some&lt;/em&gt; water molecules are have enough energy to escape and that's why evaporation in water can occur at any temperature (yes, even if the water is in ice).&lt;/p&gt;&#xA;&lt;p&gt;When the temperature increases, there are more molecules with higher kinetic energy and thus, more water can evaporate.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2021-03-08T05:49:30.820" LastActivityDate="2021-03-08T05:49:30.820" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="7451" PostTypeId="2" ParentId="7449" CreationDate="2013-12-20T07:35:03.433" Score="9" Body="&lt;p&gt;To add to Jerry's answer, the amount of evaporation of water also depends on pressure.&lt;p&gt;Infact, one way of defining boiling point is the temperature at which the vapour pressure equals atmospheric pressure. So, you can actually &lt;a href=&quot;http://www.youtube.com/watch?v=Ks4VuXTTKmo&quot;&gt;boil water at room temperature&lt;/a&gt;.&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastActivityDate="2013-12-20T07:35:03.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7452" PostTypeId="1" AcceptedAnswerId="7453" CreationDate="2013-12-20T07:44:50.363" Score="4" ViewCount="6583" Body="&lt;p&gt;The following is a problem from 'Fundamentals of Organic Chemistry' by McMurry-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw two constitutional isomers of &lt;em&gt;cis&lt;/em&gt;-1,2-dibromocyclopentane. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Constitutional isomers must have different connections. But the question specifies everything including the spatial arrangement. I don't get how there can be &lt;em&gt;constitutional&lt;/em&gt; isomers for the above-mentioned compound. &lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="83758" LastEditDate="2020-02-13T18:15:12.427" LastActivityDate="2020-02-13T18:15:12.427" Title="Constitutional isomers of cis-1,2-dibromocyclopentane" Tags="&lt;organic-chemistry&gt;&lt;molecular-structure&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7453" PostTypeId="2" ParentId="7452" CreationDate="2013-12-20T09:23:46.860" Score="4" Body="&lt;p&gt;If I understood you correctly, your objection is to the relative &lt;em&gt;cis&lt;/em&gt;- designation in the name of the compound. As far as I can tell, the use of any stereochemical designation within the context of that question is either deliberately misleading, incorrect, or simply pointless. Per the definition of constitutional isomerism, the only criteria that the two molecules in your answer must satisfy is that they have &lt;strong&gt;(a)&lt;/strong&gt; the same molecular formula, and &lt;strong&gt;(b)&lt;/strong&gt; different structural connectivity. Stereochemistry, whether relative or absolute, is totally irrelevant. Note the wording in the IUPAC Gold Book entry for &lt;a href=&quot;http://goldbook.iupac.org/C01282.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;constitution&lt;/em&gt;&lt;/a&gt; (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The description of the identity and connectivity (and corresponding bond multiplicities) of the atoms in a molecular entity (&lt;strong&gt;omitting any distinction arising from their spatial arrangement&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, any molecule with the formula $\ce{C5H8Br2}$ that differs in connectivity from a cyclopentane with two vicinal bromine atoms should be a valid answer. Two such isomers are given by simply permuting the positions of the bromines (i.e., 1,1-dibromocyclopentane and 1,3-dibromocyclopentane). Of course, there are numerous other possible answers.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-20T09:23:46.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7454" PostTypeId="2" ParentId="7431" CreationDate="2013-12-20T09:39:09.900" Score="2" Body="&lt;p&gt;Some of the ways you aren't at standard condition.&#xA;Metal solution so the activity of the metals are at their concentration in the alloys &#xA;The solutions have far below 1mol/L of their respective metal ions &#xA;The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre-&#xA;whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were at standard condition with low voltage then the first experiment makes perfect sense-&lt;br&gt;&#xA;From wikipedia full list of electropotentials&#xA;V2+ + 2 e−&#xA;    V(s)     −1.13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cr3+ + 3 e−&#xA;    Cr(s)    −0.74&#xA;Vanadium is preferred. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2013-12-20T09:39:09.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7455" PostTypeId="1" AcceptedAnswerId="7457" CreationDate="2013-12-20T09:59:45.230" Score="1" ViewCount="207" Body="&lt;p&gt;&lt;strike&gt;What sort of a &lt;b&gt;microstructure&lt;/b&gt; would you get on bubbling Aluminium vapors through a large tank of Helium?&lt;/strike&gt;&lt;p&gt;Would it be a glassy microstructure and/or would you get nano-powders? &lt;strike&gt;and/or a new crystal structure formation because of the excess helium which would be present in the aluminium matrix? Ignore metal coolant reactions, if any.&lt;/strike&gt;&lt;/p&gt;&lt;p&gt; Aluminum vapour would consist of Aluminium atoms, unlike liquid aluminum which consists of clusters. So, on rapid cooling from gas to solid, would you still get an FCC crystal structure for aluminium? (There would be no time for diffusion)&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditDate="2013-12-20T15:28:04.350" LastActivityDate="2013-12-20T16:36:00.970" Title="Rapidly cooling Aluminum vapour to solid state?" Tags="&lt;crystal-structure&gt;&lt;materials&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7456" PostTypeId="2" ParentId="7448" CreationDate="2013-12-20T15:38:18.243" Score="2" Body="&lt;p&gt;The answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2428/question-about-fire-and-colors&quot;&gt;this question&lt;/a&gt; partially answers yours.&lt;p&gt;Let me speculate. The compound could be some crystallized organic compound, making it appear white. On heating, it degrades and becomes solid carbon, which is black, giving of some carbon dioxide and water vapour. It wouldn't have any sulfurous groups, as it would give off sulfur dioxide.&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2013-12-20T16:06:41.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7457" PostTypeId="2" ParentId="7455" CreationDate="2013-12-20T16:36:00.970" Score="1" Body="&lt;p&gt;There are a number of actual production processes that involve coating surfaces with metals via what are effectively vapour-phase metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sputter_deposition&quot; rel=&quot;nofollow&quot;&gt;Sputtering&lt;/a&gt; seems to be preferred to simple evaporation because it gives better results and allows for better control over the properties of the deposited metal coating (or non-metal, a very wide variety of compounds can be deposited this way including some created by vapour phase reactions). Sputtering creates aluminium (the same applies to many other elements or compounds) vapour by bombarding solid aluminium with inert atoms and allowing the atoms to hit the coated surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There don't appear to be any unusual crystal structures in the resulting coatings, though the grain size and other factors can be controlled by altering the conditions.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-20T16:36:00.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7459" PostTypeId="1" AcceptedAnswerId="7793" CreationDate="2013-12-21T02:53:44.863" Score="-2" ViewCount="2030" Body="&lt;p&gt;What is the major product for the reaction of (1&lt;i&gt;R&lt;/i&gt;,2&lt;i&gt;R&lt;/i&gt;)-2-methylcyclohexanol&lt;sup&gt;[1,2]&lt;/sup&gt;  with hydrogen bromide? What will happen to the stereocenters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DHXuu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DHXuu.png&quot; alt=&quot;(1R,2R)-2-methylcyclohexanol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only know the &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; will somehow substitute the alcohol, but I'm not sure about the position or stereo chemistry of the final product.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CSID:22437, &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.22437.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemspider.com/Chemical-Structure.22437.html&lt;/a&gt; (accessed 14:03, Oct 12, 2018)&lt;/li&gt;&#xA;&lt;li&gt;National Center for Biotechnology Information. PubChem Compound Database; CID=24004, &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/24004&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pubchem.ncbi.nlm.nih.gov/compound/24004&lt;/a&gt; (accessed Oct. 12, 2018).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="4945" LastEditDate="2018-10-12T13:13:27.877" LastActivityDate="2018-10-12T13:13:27.877" Title="What is the major product for the reaction of (1R,2R)-2-methylcyclohexanol with hydrogen bromide?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-01-14T10:28:23.903" ContentLicense="CC BY-SA 4.0" />
  <row Id="7460" PostTypeId="1" CreationDate="2013-12-21T06:13:37.737" Score="109" ViewCount="4897" Body="&lt;p&gt;There have been various explanations posited for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_effect&quot; rel=&quot;noreferrer&quot;&gt;α-effect&lt;/a&gt;. The α-effect refers to a phenomenon wherein nucleophiles with lone pairs on atoms adjacent (i.e., in the α- position) to the atom bearing the reacting lone pair sometimes exhibit dramatically higher reactivity than similar nucleophiles without α-electrons. This effect is especially adduced when no associated increase in Brønsted basicity occurs. For example, hydroperoxide ($\ce{HOO-}$) experimental reaction rate constants are orders of magnitude greater&lt;sup&gt;[1]&lt;/sup&gt; those of hydroxide ($\ce{HO-}$) with various electrophilic substrates, despite the former exhibiting lower Brønsted basicity. There is also a thermodynamic α-effect, in which equilibrium constants are enhanced&lt;sup&gt;[2]&lt;/sup&gt;. It is currently on the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_chemistry&quot; rel=&quot;noreferrer&quot;&gt;list of unsolved problems in chemistry&lt;/a&gt; on Wikipedia, but, due to a lack of references to that effect, I'm not entirely convinced it really should be listed there. Here's the summary of my research on the topic thus far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I read Ren, Y. &amp;amp; Yamataka, H.&lt;sup&gt;[3]&lt;/sup&gt;, &lt;em&gt;&quot;The alpha-effect in gas-phase S&lt;sub&gt;N&lt;/sub&gt;2 reactions revisited&quot;&lt;/em&gt;. In it, they claim that explanations based on ground-state destabilization (presumably due to repulsion between the electrons of the nucleophilic atom and the α-electrons) are not correct. Their reasoning is that this would result in a difference in the $\Delta G$ between reactants and products, leading to thermodynamic equilibrium effects. They argue that a correct explanation should be one exclusively involving stabilization of the transition state (i.e., minimization of $\Delta G^{\ddagger}$), and go on to offer some explanation for how this may occur (along with experimental data). Intuitively, their conclusion seems reasonable to me, and it also (at least to my naive comprehension) seems eminently testable. I don't know whether equilibrium effects consistent with ground-state destabilization have actually been observed or not; however, if they haven't, shouldn't that put the nail in the coffin of that theory? Or is it simply that the authors are searching for a purely kinetic α-effect, so that a distinction between a thermodynamic one needs to be made?&lt;/li&gt;&#xA;&lt;li&gt;Fleming devotes a section to the effect in his book, &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions&lt;/em&gt;. He notes that the presence of the α-lone pair should raise the energy of the HOMO of the nucleophile, but also points that experimental results don't correlate sufficiently well with the HOMO energies of various α-nucleophiles. In particular, certain soft electrophiles (per HSAB theory), such as alkyl halides, apparently show an anomalous low preference for α-nucleophiles. In the context of SET mechanisms, Fleming says that the higher energy of the HOMO and the availability of α-electrons (which can stabilize a radical intermediate) ought to have a highly favorable effect on the rate of reaction, and notes that experimental results have borne this out. My interpretation of this is that, while the picture is perhaps murky for anionic mechanisms, transition-state stabilization clearly seems to be operative in SET mechanisms.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've also read the original 1962 paper by Pearson and Edwards&lt;sup&gt;[4]&lt;/sup&gt;, which also largely argued for transition-state stabilization as the primary explanatory mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, from my reading thus far, it seems that transition-state stabilization has been most consistently invoked and has the largest wealth of evidence and the most plausible arguments supporting it. What I'd like to ask is, &lt;strong&gt;(a)&lt;/strong&gt; are there flaws in my reasoning or understanding of the material, and &lt;strong&gt;(b)&lt;/strong&gt; is this truly a fundamentally unsolved problem, or is there actually some emerging consensus among experts?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes and References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fleming provides a small table with relative rates ($k_\mathrm{rel} = k_{\ce{HOO-}}/k_{\ce{HO-}}$) in his book. For example, he gives $k_\mathrm{rel} \approx 10^5$ for reaction with $\ce{PhCN}$ and $k_\mathrm{rel} \approx 50$ for $\ce{PhCH2Br}$, while $k_\mathrm{rel} \approx 10^{-4}$ for reaction with $\ce{H3O+}$. The rate of reaction correlates in the expected way with Brønsted basicity only in the case of proton transfer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, citing Fleming, he gives the example of the reaction of &lt;em&gt;N&lt;/em&gt;-acetylimidazole with hydroxylamines, in which both rate and equilibrium constants are positively affected. Qualitatively, he explains this by noting that the α-electrons raise the energy of the lone pair conjugated to the π-system, making overlap of said lone pair with the π* LUMO more effective. Additionally, he claims &lt;em&gt;both&lt;/em&gt; ground-state stabilization and transition-state destabilization as being factors in the reduced electrophilicity of oximes and hydrazones relative to (most) other standard imines.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ren, Y.; Yamataka, H. The α-Effect in Gas-Phase S&lt;sub&gt;N&lt;/sub&gt;2 Reactions Revisited. &lt;em&gt;Org. Lett.&lt;/em&gt; &lt;strong&gt;2006,&lt;/strong&gt; &lt;em&gt;8&lt;/em&gt; (1), 119–121. &lt;a href=&quot;https://doi.org/10.1021/ol0526930&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/ol0526930&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Edwards, J. O.; Pearson, R. G. The Factors Determining Nucleophilic Reactivities. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1962,&lt;/strong&gt; &lt;em&gt;84&lt;/em&gt; (1), 16–24. &lt;a href=&quot;https://doi.org/10.1021/ja00860a005&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/ja00860a005&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-03T11:25:29.307" LastActivityDate="2019-05-31T10:34:22.660" Title="Is there a general consensus on the causes of the alpha-effect?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="7461" PostTypeId="2" ParentId="6976" CreationDate="2013-12-21T10:04:39.680" Score="3" Body="&lt;p&gt;Amongst aldehydes and ketones, ketones have higher boiling point. This is due to the presence of two electron donating alkyl groups around the $\ce{C=O}$ group which makes them more polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: the boiling point of $\ce{CH3-CHO}$ is 322 K and dipole moment is 2.52 D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling point of $\ce{CH3-CO-CH3}$ is 329 K and dipole moment is 2.88D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dipole moment of $\ce{CH3-CO-CH3}$ is greater than of $\ce{CH3-CHO}$. This because there are two electron donating $\ce{CH3}$ groups around $\ce{C=O}$ bond whereas there is only one $\ce{CH3}$ group around $\ce{C=O}$ in $\ce{CH3CHO}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As dipole moment is greater so it is more polar and hence has higher boiling point.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorDisplayName="user10259" LastEditDate="2015-05-16T08:01:24.270" LastActivityDate="2015-05-16T08:01:24.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7462" PostTypeId="1" CreationDate="2013-12-21T13:10:12.177" Score="3" ViewCount="789" Body="&lt;p&gt;What is the most rare element in periodic table? I have a lot of confusion about this.&lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="305" LastEditDate="2013-12-21T13:53:25.960" LastActivityDate="2013-12-21T15:08:50.127" Title="What is the most rare element in periodic table" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7463" PostTypeId="2" ParentId="7462" CreationDate="2013-12-21T14:47:29.660" Score="10" Body="&lt;p&gt;Well, the first question to answer would be: where?  The rarest element in the universe? the sun? in meteorites? on earth? in the ocean? in humans?  We can tap in to the curated datasets provided by Mathematica to get the answer:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wSWX.png&quot; alt=&quot;Mathematica graphics&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Elements up to atomic number 98 have been found in trace quantities; however, the dataset does not contain any information about the trans uranium elements which were &lt;a href=&quot;http://en.wikipedia.org/wiki/Synthetic_element&quot; rel=&quot;noreferrer&quot;&gt;once thought to be synthetic&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Here's a plot of abundance as a function of atomic number:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N4Hrv.png&quot; alt=&quot;Mathematica graphics&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can see that the elements beyond about Z = 82 (lead) have a marked decrease in abundance, and this is consistent with all of the elements with atomic number above 82 being radioactive.&lt;/p&gt;&#xA;&lt;h1&gt;Where do we come from?&lt;/h1&gt;&#xA;&lt;p&gt;We can compare the elemental abundance of these individual &amp;quot;spaces&amp;quot; to speculate on where humans come from.  Below are plots showing the correlation of the elemental abundances.  The plot labels give the best fit equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUPeq.png&quot; alt=&quot;Mathematica graphics&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The best slope is observed when comparing our abundance to that of the Earth's crust; however it does appear as if the scatter is smallest for the Us vs. the Ocean plot.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-12-21T15:08:50.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7464" PostTypeId="1" AcceptedAnswerId="7487" CreationDate="2013-12-21T15:14:54.433" Score="0" ViewCount="897" Body="&lt;p&gt;I'm driving from Sydney to Melbourne (approximately 12 hours trip) and I'm wondering if it's safe to have this Galileo's Thermometer in my car during the trip if it's going to be a scorcher of a day. It could easily reach 35 degrees Celsius or over during the trip. I'm a little worried that it might explode. Any help much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4005" LastActivityDate="2016-07-16T23:29:32.297" Title="Will a Galileo's Thermometer containing liquid paraffin explode in my car if the temperature is above 35 degrees Celsius?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7465" PostTypeId="1" CreationDate="2013-12-21T17:56:50.853" Score="1" ViewCount="750" Body="&lt;p&gt;According to CFT, the degenerate d-orbitals split into $eg$ and $t_2 g$ groups. Are these the outermost nd orbitals which are vacant(4-d in case of 4th period) or the penultimate filled (n-1)d orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are the (n-1)d orbitals, how can they be degenerate if they are not completely filled? If they are (vacant) nd orbitals, how do they split into $eg$ and $t_2 g$ groups if they have no electrons to be repelled by the incoming ligands (CFT is based purely on electrostatic forces)?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastActivityDate="2014-02-20T13:37:07.277" Title="Which d-orbitals split according to CFT?" Tags="&lt;coordination-compounds&gt;&lt;molecular-orbital-theory&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7466" PostTypeId="1" AcceptedAnswerId="7468" CreationDate="2013-12-21T22:26:08.183" Score="4" ViewCount="115" Body="&lt;p&gt;I was trying to remove the electroplating from a aluminum rod using inverse electrolysis and everything went wrong. I'm definitely not a chemistry expert, so I'd like to know what happened here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used hydrochloric acid (30%) as the electrolyte, the aluminum rod as the cathode and a common copper wire as the anode. I ran it through a 12V 1A power supply, in a small glass jar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I turned the power on a lot of bubbles started to appear at the surface of the rod, as expected. After aproximately 30 seconds the bubbling was too strong, so I decided to turn off the power supply, however, the bubbling didn't stop. As I know that electrolysis of hydrochloric acid releases chlorine, I decided to put the jar to outside, with both electrodes still inside it. After aproximately five minutes, to my surprise, the jar was bubbling extremely strong, spilling over and the electrolyte had turned black, and quite frankly it was kind of scary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I removed the electrodes and the reaction ceased, but I still have no clue of what happened.&lt;/p&gt;&#xA;" OwnerUserId="4006" LastActivityDate="2013-12-22T02:02:45.323" Title="What is going on in this reaction?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7467" PostTypeId="1" AcceptedAnswerId="7473" CreationDate="2013-12-22T01:32:51.147" Score="18" ViewCount="42441" Body="&lt;p&gt;In other &lt;a href=&quot;https://outdoors.stackexchange.com/questions/4929/how-safe-is-drinking-distilled-water/4930#4930&quot;&gt;site&lt;/a&gt;, someone claimed that pure water is very corrosive.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Water distilled for some laboratory uses (like the Z-pinch reactor) is distilled to 0.00000009% purity, and is caustic enough to burn through most things.&lt;/p&gt;&#xA;&lt;p&gt;Water, purified enough, cannot be distributed through stainless steel pipes. Instead, they use glass pipes to transport it within the plants where it's made.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For sure, caustic is not the correct word, since water is neutral.&lt;/p&gt;&#xA;&lt;p&gt;But, is it that corrosive? Or is it even less than water with salt?&lt;/p&gt;&#xA;" OwnerUserId="1680" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-01-21T18:18:14.927" Title="Is pure water very corrosive?" Tags="&lt;water&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7468" PostTypeId="2" ParentId="7466" CreationDate="2013-12-22T02:02:45.323" Score="3" Body="&lt;p&gt;As I recall, $\ce{Cu}$ is inert to $\ce{HCl}$, however $\ce{Al}$ will react with it to produce $\ce{H2}$ and $\ce{AlCl3}$, resulting in a turbid, dark gray solution that bubbles vigorously as the $\ce{H2}$ gas is evolved. The reaction may not have been instantaneous, especially if the $\ce{Al}$ had a passivation layer of oxides on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once initiated, that reaction would've continued even after removal of the external current, and is probably the source of the black precipitate you describe.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-22T02:02:45.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7470" PostTypeId="1" CreationDate="2013-12-22T03:53:02.117" Score="2" ViewCount="2376" Body="&lt;p&gt;On what basis can the SRP value of different elements in the periodic table be explained?&#xA;For example why does Cu have a greater tendency to accept an electron and Zn have a tendency to lose electrons in a Daniel cell?&lt;/p&gt;&#xA;" OwnerUserId="4008" LastActivityDate="2014-02-12T03:26:46.900" Title="Electrochemical cell-SRP value" Tags="&lt;electrons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7471" PostTypeId="2" ParentId="7470" CreationDate="2013-12-22T05:36:14.940" Score="3" Body="&lt;p&gt;There are many factors and the ideal way to understand this would be with the quantum mechanical description of atoms, which gets too complicated for answering such questions. But in general, following are few simple-to-understand face that helps in understanding the SRP values- &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Electronegativity and Electron gain enthalpy:&lt;/strong&gt; Higher the electronegativity, greater is the tendency to accept electrons, be reduced and hence greater (more positive) is the Reduction potential. For example, reduction potential of fluorine is greater than chlorine and that is greater than bromine. As we move down the halogen group, electronegativity decreases. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Ionization potential:&lt;/strong&gt; This is a consideration for metals, which usually loose electrons and hence electronegativity is not the right quantity to consider. The lesser the IP, easier is the oxidation, and more negative (lower) is the SRP. In gas phase, the SRP of cesium is the most negative (in the group) and it becomes less negative as we move upwards.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Hydration:&lt;/strong&gt; This is an important fact to consider since most electrochemical reactions occur in aqueous phase and hence the commonly cited SRP values corresponds to the aqueous phase. Greater the hydration of the ion formed after oxidation/reduction, greater is the tendency to lose/gain electrons and hence lower/higher is the SRP values. Lithium ion has very high hydration enthalpy and hence lithium has a strong tendency to lose an electron to form an ion which can be excellently solvated. This lends it one of the most negative SRP in the periodic table. The same holds true for fluorine, which is excellently solvated in the ionic form by water and hence has a high SRP.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Size, EC etc:&lt;/strong&gt; Several factors like ionic size, it's electronic configuration, the inner orbital structure and the nuclear force of attraction determines how the incoming electron will be received and consequently the SRP values. Higher atomic size, higher inner electron repulsion, almost-stable configuration(just 1 electron more) and lower nuclear force, all favour lower SRP values.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these consideration, in varying proportions can be applied to the transition elements, but they act differently and to different extent in different compounds. Hence, predicting the SRP order without experimental data is very difficult and not trustworthy.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="22" LastEditDate="2014-02-12T03:26:46.900" LastActivityDate="2014-02-12T03:26:46.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7472" PostTypeId="1" CreationDate="2013-12-22T06:11:29.997" Score="0" ViewCount="297" Body="&lt;p&gt;What is chemical free energy?  How can we practically explain its existence?  How can it be used to do some useful work like in an electrochemical cell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't understand clearly about it when I referred a few books&lt;/p&gt;&#xA;" OwnerUserId="4008" LastEditorUserId="19616" LastEditDate="2015-10-30T18:28:20.780" LastActivityDate="2015-10-30T18:28:20.780" Title="What is chemical free energy?" Tags="&lt;physical-chemistry&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7473" PostTypeId="2" ParentId="7467" CreationDate="2013-12-22T07:42:36.990" Score="25" Body="&lt;p&gt;That quote had false statements all over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type I ultra pure water (Milli-Q water, or others) are fairly common among all labs.  They are not corrosive to stainless steel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @chipbuster said, purified water is never distributed through stainless steel partly because concern of contamination.  Metal reacts not only with oxygen, but also many pollutants in the air.  Overtime on metal surface will build up a layer that has all kind of compounds in them.  When water flows through, ions get picked up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the most major reason would be that glass is better in all respect.  The reason glasswares are commonly used in the lab have little to do with pure water.   Glasswares are relatively stable across wide range of pH and very hard to be oxidized or reduced.   Plus, glasswares are cheap, extremely easy to manufacture, and transparent, so you can see through.  You will really only need metal if extreme pressure or temperature is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, it is also unwise to use glassware to distribute ultrapure water--you will get ion contaminations from glass as well.  Purified water are actually almost always stored in plastic containers and distributed through plastic tubes.  &lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="3723" LastEditDate="2013-12-24T22:07:59.533" LastActivityDate="2013-12-24T22:07:59.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7474" PostTypeId="2" ParentId="7465" CreationDate="2013-12-22T12:54:31.690" Score="1" Body="&lt;p&gt;The $d$ orbitals that split are the $(n-1)$ $d$-orbitals because they are the relevant ones that do chemistry (because they have the electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the orbitals are &lt;a href=&quot;http://en.wikipedia.org/wiki/Degenerate_energy_levels&quot; rel=&quot;nofollow&quot;&gt;degenerate&lt;/a&gt;, it doesn't matter whether they're filled or not, the energy level is the same for all the degenerate states. So it also doesn't matter if they're filled or not, the energy states will split up according to CFT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound only &quot;feels&quot; this raising or lowering of energy depending on how many electrons are filled in into the different orbitals after splitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maybe clarify some confusion: The existence of an orbital doesn't mean that the electron has to be in there, it is only a mathematical construct.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-22T12:54:31.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7475" PostTypeId="2" ParentId="7472" CreationDate="2013-12-22T13:23:04.750" Score="2" Body="&lt;p&gt;&lt;strong&gt;Chemical Free Energy or Gibbs Free Energy is best understood as potential energy for reactions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a lot of Gibbs Free Energy, then I have a lot of potential energy.&#xA;Kind of like if I were on top of a hill, it's very easy for me to roll down that hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often what's talked about in books is dG or the change in Gibbs Free Energy.&lt;br&gt; If we think of it is simply being on top of a hill (High Gibbs Free Energy/Potential Energy) G(initial)&#xA;&lt;br&gt; going to the bottom of the hill (Low Gibbs Free Energy/Potential Energy) G(final)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the change in Gibbs Free Energy will be negative for something spontaneous (like the rolling down a hill example). &#xA;&lt;br&gt;Change is always calculated as: change = final - initial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs Free Energy is used in the same was as potential energy. But instead of translating to kinetic energy, it translates to Enthalpy (heat) and/or Electrochemical Energy (being able to move electrons). Both Enthalpy and Electrochemical Energy can then be converted to mechanical work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to think about it is not so much that you can &lt;strong&gt;&lt;em&gt;use&lt;/em&gt;&lt;/strong&gt; Gibbs Free Energy, but rather it's just the innate property of whatever situation you're working with. Molecules have a tendency to be reduced or oxidized under different conditions, the Gibbs Free Energy just tells you which way is spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dG = dH - TdS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets kind of complicated, since dG is affected by many factors, but it gives you an idea, that the Change in Gibbs Free Energy can become positive or negative under different conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that &lt;strong&gt;Gibbs Free Energy&lt;/strong&gt; isn't a real thing, but rather a concept that helps explain why things happen.&#xA;&lt;br&gt; Why do gases expand?&#xA;&lt;br&gt; I don't know, but the universe has a tendency to do this. So we say that it is spontaneous, just like how we would say things roll down a hill because of gravity.&#xA;&lt;br&gt; What's gravity?&#xA;&lt;br&gt; It's our way of explaining &lt;em&gt;why&lt;/em&gt; things roll down hills. :p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps a little bit.&lt;/p&gt;&#xA;" OwnerUserId="4013" LastEditorUserId="4013" LastEditDate="2013-12-22T13:31:47.643" LastActivityDate="2013-12-22T13:31:47.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7476" PostTypeId="2" ParentId="7440" CreationDate="2013-12-22T14:01:24.050" Score="1" Body="&lt;p&gt;I'm having a hard time understanding fully what the question is asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I think the answer to the question is that there is an energy associated with the full oxidation of whatever chemical you are working with. So basically, the more you oxidize the working electrode the easier it becomes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situations I'm familiar with, you form a layer on the working electrode. This layer has Van der Waals forces associated with it, so the more molecules that participate, the stronger the interaction. This makes it easier/less negative to form a skin/oxide on the working electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also makes the reduction harder. If the skin/oxide you have on the working electrode has Van der Waals forces associated with it, it's not going to be as easy as just applying the normal potential to remove/reduce the oxide; you have to apply a little bit more negative potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I completely misunderstood the question, just let me know; I'm pretty familiar with CVs.&lt;/p&gt;&#xA;" OwnerUserId="4013" LastEditorUserId="4013" LastEditDate="2013-12-25T03:05:57.827" LastActivityDate="2013-12-25T03:05:57.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7477" PostTypeId="1" AcceptedAnswerId="17206" CreationDate="2013-12-22T15:14:11.797" Score="11" ViewCount="17943" Body="&lt;p&gt;Several of my friends, when they do the dishes manually, do not rinse the dishes after cleaning but merely dip them in the dishwater and then dry them (using a towel or a dish rack). I myself on the other hand always rinse the dishes with fresh (hot) water from the tap before drying them, mostly because the idea of having a dose of soap with my next meal freaks me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there some guideline as to how much of the soap you can ingest until something goes wrong, be it the taste of the next meal or some poisonous reaction? The soap used for doing the dishes manually doesn't have any warning labels on it... I've found &lt;a href=&quot;http://www.nlm.nih.gov/medlineplus/ency/article/002729.htm&quot;&gt;this&lt;/a&gt; describing poisonous effects of automatic dishwater soap, and &lt;a href=&quot;http://wiki.answers.com/Q/Is_Dawn_Dish_Soap_harmful_if_swallowed?#slide=2&quot;&gt;this&lt;/a&gt; pertaining to a dish soap that is used for manual dishwashing.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="5017" LastEditDate="2014-10-08T03:00:12.930" LastActivityDate="2014-10-08T22:15:49.090" Title="At what amount is digesting soap unhealthy?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7478" PostTypeId="1" CreationDate="2013-12-22T15:25:24.080" Score="5" ViewCount="376" Body="&lt;p&gt;How do you determine the migratory &quot;aptitude&quot; of various groups during carbocation rearrangements? Is there a experimentally determined order?&lt;br&gt;&#xA; For example, what will be the product in case of the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Pinacol_rearrangement&quot; rel=&quot;nofollow&quot;&gt;pinacol-pinacolone type rearrangement&lt;/a&gt;:&#xA;$$\ce{(cyclo-pentyl)CH3C(OH)-C(OH)(CH3)(cyclo-pentyl) -&amp;gt;products}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the methyl group be transferred or the cyclopentyl moiety? The same question arises if the cyclopentyl is replaced with phenyl group? In general, which shifts, methyl, hydrogen, phenyl or any other group, will take place? Or equivalently, &lt;strong&gt;what determines which group has a greater tendency to undergo shifts during carbocation rearrangements?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="7475" LastEditDate="2015-10-02T16:39:49.203" LastActivityDate="2015-10-02T16:39:49.203" Title="Determination of products in reactions involving carbocation rearrangement?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7479" PostTypeId="1" AcceptedAnswerId="7482" CreationDate="2013-12-22T15:27:52.420" Score="2" ViewCount="5001" Body="&lt;p&gt;I am finding a good way to liquefy or extract water from superabsorbent polymer. From what I remember it was salt that can liquefy it. Since salt when dissolve will have sodium cation and there for naturalize the polymer chain making it contract.  Is that true?. Thank very much.  &lt;/p&gt;&#xA;" OwnerUserId="4012" LastActivityDate="2017-07-14T17:37:43.473" Title="How to liquefy or extract water from superabsorbent polymer" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;polymers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7480" PostTypeId="2" ParentId="7440" CreationDate="2013-12-22T15:40:46.703" Score="2" Body="&lt;p&gt;I think you may be coming at this question from the wrong angle. Since I'm not entirely knowledgeable about electrochemistry, I'll give a somewhat qualitative answer, the general idea of which I hope to be understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentally, what you are doing is taking a system containing a species, say $\ce{A}$, and its oxidized form $\ce{A+}$ (of course this same argument applies for the reduced form $\ce{A-}$ if you sweep in the other direction of voltage). In the absence of an applied voltage, the system is at equilibrium with a certain concentration of $\ce{A}$ and $\ce{A+}$. Let us consider that $\ce{[A]}$ $\gg $ $\ce{[A+]}$. This implies that the equilibrium constant for $\ce{A &amp;lt;=&amp;gt; A+ + e-}$ is much smaller than 1, and therefore that the oxidation reaction has a high positive standard reaction free energy change $\Delta _r G^{o} \gg 0$ (loosely speaking, the reaction is &quot;not spontaneous&quot; in &lt;em&gt;standard&lt;/em&gt; conditions, though this sort of thinking is misleading, as I explain later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you start to sweep the voltage, you begin adding &lt;em&gt;electrical free energy&lt;/em&gt; into the system, such that the reaction free energy (in &lt;em&gt;non-standard conditions&lt;/em&gt;) now becomes a sum of the initial chemical free energy change and the electrical free energy $\Delta _r G = \Delta _r G^{o} + G_{electrical} = \Delta _r G^{o} + nFE$ (I think that you can argue that by altering the voltage, you are altering the free energy of the electron in the equilibrium). If you sweep in the right direction, then the electrical free energy term can contribute to decrease the reaction free energy. At some defined voltage, the two factors cancel out, giving $\Delta _r G = 0\ \ \rightarrow \ \ \ \Delta _r G^{o} = -nFE^*$, where to $E^*$ we give the special name &quot;oxidation potential&quot;, $E^{ox}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to notice is that all this time, the potential sweep is altering the reaction free energy incrementally. Since the free energy is shifting, the associated equilibrium constant is also changing. At voltages much than lower $E^{ox}$,  $\Delta _r G = \Delta _r G^{o} + nFE \gg 0$, and so even though the reaction free energy is lowering, it is still quite positive and the associated equilibrium constant is still much smaller than 1. Thus the reaction is only slowly shifting towards products, and therefore there are few electrons being passed into the circuit, and the current is close to zero. However, when the voltage applied starts to get close to the oxidation potential, even if $\Delta _r G &amp;lt; 0$, the equilibrium constant will be smaller than but somewhat close to 1, and there is some conversion of $\ce{A}$ into $\ce{A+}$, creating a current below the formal oxidation potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole problem comes from thinking that as long as $\Delta _r G &amp;lt; 0$, a reaction &quot;happens&quot;, while for $\Delta _r G &amp;gt; 0$, the reaction &quot;doesn't happen&quot;. In truth, there is a gradual change of equilibrium concentrations as a reaction goes from one regime to the other. Even if $\Delta _r G &amp;gt; 0$, if it sufficiently close to zero then the equilibrium constant is only a bit smaller than 1, and you can expect appreciable amounts of products. Notice that in the voltammogram, the current is practically zero until about 0.1 V away from the oxidation potential, and it only really starts to pick up around a mere 0.05 V away. Even though these values mean $\Delta _r G &amp;gt; 0$, they are sufficiently close to the oxidation potential that the equilibrium constants aren't so small, and so you can actually measure a significant amount of species being oxidized.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2013-12-22T15:40:46.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7481" PostTypeId="1" AcceptedAnswerId="7483" CreationDate="2013-12-22T17:09:07.417" Score="7" ViewCount="28399" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wqyuS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wqyuS.png&quot; alt=&quot;5-ethyl-3,3-dimethylheptane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering why the above image is named 5-ethyl-3,3-dimethylheptane rather than 3-ethyl-5,5-dimethylheptane. I thought an ethyl substituent has a higher priority (I know this is true when arranging the final name, but what about numbering the parent chain)? &lt;/p&gt;&#xA;" OwnerUserId="4015" LastEditorUserId="4945" LastEditDate="2017-04-27T12:32:41.603" LastActivityDate="2019-06-13T06:00:18.333" Title="What is the priority for numbering alkyl side chains?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7482" PostTypeId="2" ParentId="7479" CreationDate="2013-12-22T20:17:46.853" Score="2" Body="&lt;p&gt;This may be true but I suggest another type of explanation.  If a super absorbent polymer absorbs its maximum in distilled water it is called a fully swollen hydrogel.  There are two ways of decreasing the amount of water stored without using heat.  The first one is to increase the salt load of additional water and place the hydrogel within this salted water.  The salt in the water will decrease the total water absorbed by the hydrogel and the hyrogel will shrink, giving up some of its water.  A more complete and full way to make the hydrogel to give up its water is the addition of an acid.  Lowering the pH of the hydrogel to below 3.5 will convert all the sodium carboxylates to the free acid form which does not absorb water.  These hydro gels are pH sensitive and if you plot pH vs absorbed fluid you would find a bell curve that starts around 4.5 or 5.0, tops out around 7.5 and decreases as you increase the pH until around 10.5 to 11.0.&lt;/p&gt;&#xA;" OwnerUserId="4017" LastEditorUserId="-1" LastEditDate="2017-07-14T17:37:43.473" LastActivityDate="2017-07-14T17:37:43.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7483" PostTypeId="2" ParentId="7481" CreationDate="2013-12-22T21:44:37.697" Score="10" Body="&lt;p&gt;The process for naming &lt;em&gt;this&lt;/em&gt; molecule correctly follows the steps listed here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Identify the parent hydrocarbon, which will be the one with the longest chain. It is correctly identified as the 7-carbon chain, and as such is a derivative of heptane.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Identify the side-chain functional groups. We have 2 methyl groups and an ethyl group.  When listing them, they will be &lt;strong&gt;written alphabetically&lt;/strong&gt;.  So, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-ethyl-Y,Y-dimethylheptane and not Y,Y-dimethyl-X-ethylheptane&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Numbering (which is what you are asking about) is determined by numbering the carbon atoms consecutively from the end of the chain nearer the functional group of the highest priority. In terms of molecule &lt;strong&gt;numbering&lt;/strong&gt;, alkane groups have the &lt;strong&gt;same functional group priority&lt;/strong&gt;. We would only need to consider relative priorities of the ethyl and methyl groups if there was &lt;em&gt;only one&lt;/em&gt; methyl substituent, in which case you would have &lt;em&gt;3-ethyl-5-methylheptane&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, the process is to count in from the ends of the primary chain. Here, the first substituents occur at an equal distance from each end of the chain, both the ethyl and the methyl occur at position 3. We now need to invoke one of the IUPAC sub-rules for &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/&quot; rel=&quot;noreferrer&quot;&gt;nomenclature&lt;/a&gt;. &lt;strong&gt;&lt;em&gt;If the first substituents occur at an equal distance from each end of the chain, we need to number from the end nearer a second substituent.&lt;/em&gt;&lt;/strong&gt; The second substituent is the extra methyl group, and so this is the end that gains numbering priority.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are many nice examples of nomenclature on the internet. One very good example, which highlights your question is &lt;a href=&quot;http://padakshep.org/otp/subjects/chemistry/organic-chemistry/alkanes-nomenclature/&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt;. Have a look at Rule 2 and sub-rule 2&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2013-12-22T21:44:37.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7484" PostTypeId="1" CreationDate="2013-12-23T03:59:26.220" Score="7" ViewCount="2000" Body="&lt;p&gt;I know the &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/aldol-condensation.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;mechanism of aldol condensation&lt;/a&gt; and general method to determine products in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7112/crossed-aldol-condensation-products&quot;&gt;crossed aldol reactions&lt;/a&gt;, or reactions between dissimilar molecules. But the following statement in my class notes baffles me: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In presence of a base (alkaline medium) the $\alpha$ hydrogen of the lower aldehyde is more acidic and hence is removed in the first step, while in presence of an acid, the $\alpha$ hydrogen of the higher aldehyde is more acidic and hence migrates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to this, on reaction of propanal with ethanal, basic medium yields 3-hydroxy pentanal while acidic medium yields 3-hydroxy 2-methyl butanal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this statement true (if it is)? And can aldol condensation proceed in an acidic medium, since the presence of the base is necessary for the removal of the slightly acidic $\alpha$-hydrogen to start the reaction?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-03-16T12:23:22.030" LastActivityDate="2018-03-16T12:23:22.030" Title="Does aldol condensation give different products in different conditions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;erratum&gt;&lt;aldol-reaction&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7485" PostTypeId="2" ParentId="6620" CreationDate="2013-12-23T10:06:20.763" Score="1" Body="&lt;p&gt;The rate equation is actually defined for the activity of each reactant and product. In most cases, the activity can be approximated to the concentration.&lt;p&gt;In the case of solid catalysts, you'll have to use activity. I'm not sure how the actual value of activity is calculated, but the solid catalyst's activity is increased when you increase its surface area. That can be one way to change its effect on the rate equation.&lt;/p&gt;&lt;p&gt;&lt;a href=&quot;http://home.ku.edu.tr/~okeskin/chbi502/chbi502-chapter_3.pdf&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; has some mention of the activity based rate equation and some examples.&lt;/p&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastActivityDate="2013-12-23T10:06:20.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7486" PostTypeId="1" AcceptedAnswerId="8134" CreationDate="2013-12-23T11:58:24.447" Score="17" ViewCount="21298" Body="&lt;p&gt;I'm clear with the concepts of crystal field theory. But I can't figure out the exact reason why the hybridisation of manganese in potassium permanganate ($\ce{KMnO4}$) is $\mathrm{d^3s}$. Can anyone explain please?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="16683" LastEditDate="2016-08-03T21:23:08.573" LastActivityDate="2019-02-01T06:31:09.823" Title="Hybridisation of Mn in potassium permanganate" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7487" PostTypeId="2" ParentId="7464" CreationDate="2013-12-23T13:59:49.133" Score="2" Body="&lt;p&gt;Do not confuse the &lt;a href=&quot;http://en.wikipedia.org/wiki/Flash_point&quot; rel=&quot;nofollow&quot;&gt;flash point&lt;/a&gt; with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Autoignition_temperature&quot; rel=&quot;nofollow&quot;&gt;autoignition temperature&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what atmosphere you have inside the thermometer, but either way I think it's hardly the case that you reach the autoignition temperature of paraffin (gasoline undergoes autoignition at 246-280 °C).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-23T13:59:49.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7488" PostTypeId="2" ParentId="4877" CreationDate="2013-12-23T15:30:34.127" Score="31" Body="&lt;p&gt;I think the use of Hydrofluoric Acid was script-driven rather than fact driven: it sounds scary rather than being a good choice. Also, it allows for the possibility of the darkly comic bathtub scene where the acid dissolves a ceramic bath because Jessie ignores Walter's instructions (which establishes Walter's expertise and Jessie's lack of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no good reason why Pinkman and White pharmaceuticals needed to have hydrofluoric acid, therefore using large quantities of it is somewhat implausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, it probably wouldn't work as well as several alternatives. Hydrofluoric acid is very nasty stuff, but it isn't a strong acid. Even when dilute it will etch glass and ceramics, but it won't dissolve or burn flesh. I once saw a demonstration where a lecturer showed this by spilling some dilute hydrofluoric acid on his hand and then onto a glass surface. The surface was frosted, his hand unharmed (he was very careful to wash the acid off quickly and take appropriate precautions and I don't recommend trying this at home!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its danger to people is its toxicity, not its ability to burn: it insinuates itself into the body and destroys connective tissue and bone slowly by interfering with anything containing calcium. Its danger is worse because it doesn't cause &lt;em&gt;immediate&lt;/em&gt; damage and you may receive a dangerous dose without noticing. So it is scary but not corrosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other alternatives are better. Concentrated alkalis such as Sodium Hydroxide are readily available and are very good at dissolving flesh (which is why they are commonly used as drain cleaners). But alkalis don't do a good job on bone. Concentrated sulfuric acid is even better as it does a good job on flesh and will, eventually, dissolve the bone as well. Murderers have used both methods to try to dispose of evidence. For example, John George Haigh who used sulfuric acid and left little other than gallstones (&lt;a href=&quot;http://en.wikipedia.org/wiki/John_George_Haigh&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/John_George_Haigh&lt;/a&gt;). Using alkali is often done but tends to leave bone fragments even with sophisticated processes that pressure cook the solution (see &lt;a href=&quot;http://www.slate.com/articles/news_and_politics/explainer/2009/12/soluble_dilemma.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.slate.com/articles/news_and_politics/explainer/2009/12/soluble_dilemma.html&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the answer is that HF solutions are not a good choice for body disposal as it probably doesn't work well compared to known alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the above is &lt;em&gt;theory&lt;/em&gt; but good scientists do experiments. So Periodic Videos decided to test this very idea using chicken legs as a model. They compared what happens when raw chicken legs are suspended in strong solutions of HCl, H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; and HF. The HF was the least impressive for flesh-dissolving characteristics, though it did seem to cause other, more subtle damage, to the components of the flesh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the actual results &lt;a href=&quot;https://www.youtube.com/watch?v=oipksRhISfM&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="402" LastEditDate="2017-04-08T22:20:52.893" LastActivityDate="2017-04-08T22:20:52.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7489" PostTypeId="1" AcceptedAnswerId="7490" CreationDate="2013-12-23T16:09:22.517" Score="64" ViewCount="19935" Body="&lt;p&gt;In the fantastic TV series Breaking Bad, Walter White, the dying chemistry teacher, takes to making crystal meth (d-methamphetamine) by several routes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, he uses the common small-scale route starting with pseudo-ephedrine which uses a stereospecific starring material to make a stereospecific end product (d-methamphetamine is the active compound). But doing this on a large scale is limited because it is hard to get large quantities of the pseudo-ephedrine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walter uses his knowledge of chemistry to move to an alternative route starting with phenyl-acetone(or P2P as it often called):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fVJi6.png&quot; alt=&quot;wikipedia reaction method for P2P routes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways of doing the reduction to produce the final product (White first seems to use a Thorium dioxide dehydrocarboxylation in a tube furnace but later seems to favour the aluminium amalgam reduction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all the P2P methods produce racemic products. White clearly knows this as the issue of getting the right enantiomer is raised more than once in the show. His product is considered to be exceptionally pure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does he get an enantiomerically pure product? Or perhaps, since the show might not want to tell us the answer, what &lt;em&gt;possible&lt;/em&gt; methods could he have used?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="7448" LastEditDate="2015-08-30T09:48:14.887" LastActivityDate="2021-03-03T12:33:05.957" Title="How does Walter White make pure crystal meth using a non-stereospecific reaction?" Tags="&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7490" PostTypeId="2" ParentId="7489" CreationDate="2013-12-23T20:32:36.030" Score="53" Body="&lt;p&gt;Intriguing question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as &quot;asymmetric synthesis&quot; reactions, because they favor production of one enantiomer over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pseudoephedrine method for methamphetamine employs one of the more common methods of asymmetric synthesis, called &quot;chiral pool resolution&quot;. As you state, starting with an enantiomerically-pure sample of a chiral reagent (pseudoephedrine) as the starting point allows you to preserve the chirality of the finished product, provided the chiral point is not part of any &quot;leaving group&quot; during the reaction. However, again as you show, phenylacetone is achiral, and so the P2P process cannot take advantage of this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other methods of asymmetric synthesis, however none of them seem applicable to the chemistry shown or described on TV either; none of the reagents or catalysts mentioned would work as chiral catalysts, nor are they bio- or organocatalysts. Metal complexes with chiral ligands can be used to selectively catalyze production of one enantiomer, however the aluminum-mercury amalgam is again achiral. I don't remember any mention of using organocatalysis or biocatalysis, but these are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining route, then, is chiral resolution; let the reaction produce the 50-50 split, then separate the two enantiomers by some means of reactionary and/or physical chemistry. This seems to be the way it works in the real world. The advantage is that most of the methods are pretty cheap and easy; the disadvantage is that your maximum possible yield is 50% (unless you can then run a racemization reaction on the undesireable half to &quot;reshuffle&quot; the chirality of that half; then your yield increases by 50% of the last increase each time you run this step on the undesirable product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of methamphetamine, this resolution is among the easiest, because methamphetamine forms a &quot;racemic conglomerate&quot; when crystallized. This means, for the non-chemists, that each enantiomer molecule prefers to crystallize with others of the same chiral species, so as the solution cools and the solvent is evaporated off, the D-methamphetamine will form one set of homogeneous crystals and the L-methamphetamine will form another set. This means that all White has to do is slow the evaporation of solvent and subsequent cooling of the pan, letting the largest possible crystals form. Then, the only remaining trick is identifying which crystals have which enantiomer (and as these crystals are translucent and &quot;optically active&quot;, observing the polarization pattern of light shone through the crystals will identify which are which).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2013-12-23T20:42:46.130" LastActivityDate="2013-12-23T20:42:46.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7491" PostTypeId="2" ParentId="7439" CreationDate="2013-12-23T22:01:16.013" Score="27" Body="&lt;p&gt;Yes, there is a scientific basis, but I think you can't do anything with the apparatus shown in the figure. &lt;a href=&quot;http://weakinteractions.wordpress.com/2010/05/01/the-science-of-breaking-bad-sunset/&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; you can find much information about it. I will try to summarize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that extracting coffee at a lower temperature makes it tastes better. Gale thinks that the amount of quinic acid is the key variable that makes good coffee. In fact, some studies (McCamey, D. A.; Thorpe, T. M.; and McCarthy, J. P. Coffee Bitterness. In &quot;Developments in Food Science.&quot; Vol 25. 169-182. 1990.) report that bitterness of coffee is due to quinic acid but is not the only substance involved (see &lt;a href=&quot;http://www.coffeeresearch.org/science/bittermain.htm&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt;). I don't have any reference about the dependence between temperature and method of extraction and quantitative of quinic acid extracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the apparatus should extract the coffee in a more effective way. The vacuum pump seems to suggest the use of the vacuum to make the water boil at a lower temperature. But the connections in my opinion make no sense, so I will try to describe it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3zLS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All start with an autoclave that is connected with a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007IBO2I0&quot; rel=&quot;noreferrer&quot;&gt;Gast vacuum pump&lt;/a&gt;. This is connected with an &lt;a href=&quot;http://en.wikipedia.org/wiki/Condenser_(laboratory)#Allihn_condenser&quot; rel=&quot;noreferrer&quot;&gt;Allihn condenser&lt;/a&gt; (not the right thing to do because in there you should connect the water used to refrigerate), and then to a heated Erlenmeyer flask that is connected with a strange steel cylinder (I don't know what it is, maybe a filter), then to a Florence flask and then to another steel cylinder (...). The inspiration comes from a Florence Siphon but this is much simpler and is generally something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ktu3F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://makezine.com/projects/the-florence-siphon/&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; you can read and &lt;a href=&quot;http://www.youtube.com/watch?v=rHjVKnsOpBw&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; you can watch how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is not so nice and intriguing as the previous apparatus and is not a geeky thing that makes us think we are dealing with a great chemist, so the authors thought to make the apparatus more appealing but completely useless!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user5264" LastEditDate="2016-12-28T21:17:21.610" LastActivityDate="2016-12-28T21:17:21.610" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7493" PostTypeId="1" CreationDate="2013-12-24T12:26:12.620" Score="0" ViewCount="4550" Body="&lt;p&gt;How to compute the electron configuration of an (electrically neutral) atom ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when given the atom number how does one compute the electron configuration ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the sum of electrons must equal the atomic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know the first shells fill up as $2n^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how to compute quantum numbers when given the atom number.&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="2328" LastEditDate="2013-12-24T19:17:01.073" LastActivityDate="2013-12-24T19:17:01.073" Title="How to compute the electron configuration of an atom?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7494" PostTypeId="1" AcceptedAnswerId="7495" CreationDate="2013-12-24T15:00:10.200" Score="6" ViewCount="612" Body="&lt;p&gt;I'm studying civil engineer in my first year. We have a course called &quot;General Chemistry&quot;. One of the chapters was about chemical balance of a reaction (called equilibrium in English, I think). We got exercises on this were we had to calculate the amount of chemicals got converted to others, using acidity constants Ka.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the equations we had to solve were easy and looked like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_a = \frac{(x)^2}{(0.1 - x)(0.2 - x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These exercises were easy. But then came the point were we combined two reactions at the same time and had to calculate the balance for both at the same time. The equations we get were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BMcGv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RusAE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tutor used to make simplifications by getting rid of the &quot;$-x$&quot; and &quot;$-y$&quot; in the denominator, assuming x and y would be so small it would almost make no difference. I don't like making such simplifications: I want the correct answer. So, what I did was &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%7B1.77e-4%20=%20%28x%29%28x%2by%29/%280.1-x%29,%203.3e-4=%28y%29%28x%2by%29/%280.1-y%29%7D&quot; rel=&quot;nofollow noreferrer&quot;&gt;putting the equations in WolframAlpha&lt;/a&gt; and let it compute it, since I don't know how to solve such a system of equations. The result shocked me: the solutions for x and y were complex (so, with a real and imaginary part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WolframAlpha says this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jp96D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the tutor found (due to simplification):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EOhL6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tutor's solution is in the complex plane not really far away from the correct solution to the equation (WolframAlpha's solution #3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: How do you explain these roots are in fact complex? I don't think a complex amount of chemicals can react right?&lt;/p&gt;&#xA;" OwnerUserId="4025" LastEditorUserId="4025" LastEditDate="2013-12-24T15:44:46.953" LastActivityDate="2019-04-03T01:33:09.447" Title="Equilibrium: Equations with complex roots" Tags="&lt;equilibrium&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7495" PostTypeId="2" ParentId="7494" CreationDate="2013-12-24T16:29:52.323" Score="9" Body="&lt;p&gt;Complex roots to mass/concentration balance equations should be discarded as they have no physical interpretation. If there are only complex roots, then odds are there's some error in the manipulation of variables that lead to the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case seems different though. Notice the very small magnitude of the imaginary part of the complex roots; the number is &quot;almost real&quot;. This happens at times simply because of Mathematica's numerical root search algorithms, which aren't always guaranteed to reach the exact value. I'm actually a bit surprised that it got caught up in such a simple calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As proof that the imaginary part of the root is nothing but an issue with the calculation, I simplified the equations until I found a better representation for Wolfram Alpha to parse, hoping the complex root would disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let $A=1.77\times 10^{-4}$ and $B=3.3\times 10^{-4}$ for concision. Isolating $x$ in the second equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{B(0.1-y)-y^2}{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing $x$ for the above expression in the first equation then yields, after some tinkering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(A+B)y^3+(B^2-0.1A-0.1B-AB)y^2+(0.1AB-0.2B^2)y+0.01B^2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aha, a simple polynomial. Alpha should be able to deal with that just fine. Replacing the values of $A$ and $B$ produces &lt;a href=&quot;http://www.wolframalpha.com/input/?i=0.000507%2ay%5E3-0.00005064951%2ay%5E2-0.000000015939%2ay%2b0.000000001089=0&quot; rel=&quot;noreferrer&quot;&gt;the following equation&lt;/a&gt;, which Alpha solves with no imaginary part. The physically meaningful solution out of the three roots is $y\simeq 0.00458505$. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%283.3E-4%2a%280.1-0.00458505%29-0.00458505%5E2%29/0.00458505&quot; rel=&quot;noreferrer&quot;&gt;Replacing that value in the equation for $x$&lt;/a&gt; completes the problem, with $x\simeq 0.00228225$, very close to the approximate solutions.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2013-12-24T16:38:44.443" LastActivityDate="2013-12-24T16:38:44.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7496" PostTypeId="2" ParentId="7493" CreationDate="2013-12-24T17:02:46.447" Score="2" Body="&lt;p&gt;To calculate electronic configurarion you need to know several thing first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;nofollow&quot;&gt;Electron Shell&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemistry and atomic physics, an electron shell, also called a principal energy level may be thought of as an orbit followed by electrons around an atom's nucleus. The closest shell to the nucleus is called the &quot;1 shell&quot; (also called &quot;K shell&quot;), followed by the &quot;2 shell&quot; (or &quot;L shell&quot;), then the &quot;3 shell&quot; (or &quot;M shell&quot;), and so on farther and farther from the nucleus. &lt;strong&gt;The shells correspond with the principal quantum numbers (1, 2, 3, 4 ...) or are labeled alphabetically with letters used in the X-ray notation (K, L, M, …).&lt;/strong&gt;&#xA;  Each shell can contain only a fixed number of electrons: The 1st shell can hold up to two electrons, the 2nd shell can hold up to eight (2 + 6) electrons, the 3rd shell can hold up to 18 (2 + 6 + 10), and the 4th shell can hold up to 32 (2 + 6 + 10 + 14) and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Subshells&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called &quot;1s&quot;; the second (L) shell has two subshells, called &quot;2s&quot; and &quot;2p&quot;; the third shell has &quot;3s&quot;, &quot;3p&quot;, and &quot;3d&quot;; the fourth shell has &quot;4s&quot;, &quot;4p&quot;, &quot;4d&quot; and &quot;4f&quot;; the fifth shell has &quot;5s&quot;, &quot;5p&quot;, &quot;5d&quot;, and &quot;5f&quot; and can theoretically hold more but the &quot;5f&quot; subshell, although occupied in actinides, is not filled in any element occurring naturally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See 'Number of electrons in each shell' section under &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow&quot;&gt;Atomic Orbitals&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. Each orbital in an atom is characterized by a unique set of values of the three quantum numbers n, ℓ, and m, which correspond to the electron's energy, angular momentum, and an angular momentum vector component, respectively. Any orbital can be occupied by a maximum of two electrons, each with its own spin quantum number. &lt;strong&gt;The simple names s orbital, p orbital, d orbital and f orbital refer to orbitals with angular momentum quantum number ℓ = 0, 1, 2 and 3 respectively. These names, together with the value of n, are used to describe the electron configurations.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Understand electron configuration notation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron configurations are written so as to clearly display the number of electrons in the atom as well as the number of electrons in each orbital. Each orbital is written in sequence, with the number of atoms in each orbital written in superscript to the right of the orbital name. The final electron configuration is a single string of orbital names and superscripts. &#xA;For example, here is a simple electron configuration: $1s^2 2s^2 2p^6$. This configuration shows that there are two electrons in the 1s orbital set, two electrons in the 2s orbital set, and six electrons in the 2p orbital set. 2 + 2 + 6 = 10 electrons total. This electron configuration is for an uncharged neon atom (neon's atomic number is 10.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Memorize the order of the orbitals.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that orbital sets are numbered by electron shell, but ordered in terms of energy. For instance, a filled $4s^2$ is lower energy (or less potentially volatile) than a partially-filled or filled $3d^{10}$, so the 4s shell is listed first. Once you know the order of orbitals, you can simply fill them according to the number of electrons in the atom. *&lt;em&gt;The order for filling orbitals is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s. *&lt;/em&gt;&#xA;An electron configuration for an atom with every orbital completely filled would be written: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6 8s^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above list, if all the shells were filled, would be the electron configuration for Uuo (ununoctium), 118, the highest-numbered atom on the periodic table - so this electron configuration contains every currently known electron shell for a neutrally charged atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can fill in the orbitals according to the number of electrons in your atom. For instance, if we want to write an electron configuration for an uncharged calcium atom, we'll begin by finding its atomic number on the periodic table. Its atomic number is 20, so we'll write a configuration for an atom with 20 electrons according to the order above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill up orbitals according to the order above until you reach twenty total electrons. The 1s orbital gets two electrons, the 2s gets two, the 2p gets six, the 3s gets two, the 3p gets 6, and the 4s gets 2 (2 + 2 + 6 +2 +6 + 2 = 20.) Thus, the electron configuration for calcium is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Energy level changes as you go up. For example, when you are about to go up to the 4th energy level, it becomes 4s first, then 3d. After the fourth energy level, you'll move onto the 5th where it follows the order once again. This only happens after the 3rd energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 'How to find quantumn numbers?', see &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20081207115901AArKUxw&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2013-12-24T19:07:17.717" LastActivityDate="2013-12-24T19:07:17.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7497" PostTypeId="2" ParentId="7489" CreationDate="2013-12-24T18:59:18.430" Score="22" Body="&lt;p&gt;Meth doesn't have to be optically pure to be &quot;pure&quot;.  A mixture of d,l-methamphetamine is still pure, but I get where you're going with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has a few options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Chiral resolution - he could make the racemic meth and they resolve it by selectively crystallizing out the desired enantiomer.  Chiral acids like tartaric acid can be used to do this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;He could perform a stereoselective reduction of the intermediate imine.  You couldn't do this with Al/Hg, but there are other methods to do it (stereoselective hydrogenation using a chiral catalyst).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, the thorium oxide furnace was not used to reduce the imine intermediate, it was used to make the P2P.  Phenyl acetic acid + acetic anhydride over thorium oxide will give you P2P.&lt;/p&gt;&#xA;" OwnerUserId="4029" LastActivityDate="2013-12-24T18:59:18.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7499" PostTypeId="2" ParentId="7228" CreationDate="2013-12-24T23:41:20.147" Score="3" Body="&lt;p&gt;There are several resources available for this, for example, searching in &lt;a href=&quot;http://scholar.google.com&quot; rel=&quot;nofollow&quot;&gt;Google Scholar&lt;/a&gt; yields over 37,000 articles (this is without any filters for age or specifics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organisations such as the American Chemical Society have a searchable &lt;a href=&quot;http://www.acs.org/content/acs/en/publications.html&quot; rel=&quot;nofollow&quot;&gt;publications link&lt;/a&gt;; related, CAS has a &lt;a href=&quot;http://www.cas.org/content/references&quot; rel=&quot;nofollow&quot;&gt;searchable reference list&lt;/a&gt; and &lt;a href=&quot;https://pc.acschemworx.acs.org/platinum/LoginPage.jsp?home=chemworx&amp;amp;code=1000#home&quot; rel=&quot;nofollow&quot;&gt;ChemWorx&lt;/a&gt; which is a way to manage your research.  A similar set up is available with the &lt;a href=&quot;http://www.rsc.org/publishing/currentawareness/&quot; rel=&quot;nofollow&quot;&gt;Royal Society of Chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual universities also have searches/tutorials - for example, The University of California, Davis have a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Chemical_Compounds/Case_Study%3a_Non-Stoichometric_Bonding_in_Materials&quot; rel=&quot;nofollow&quot;&gt;'ChemWiki'&lt;/a&gt;.  Additionally, university libraries also are often accessible to the public and have search functions for journal articles, books.  Some universities, such as the University of Connecticut have &lt;a href=&quot;http://rdl.lib.uconn.edu/subjects/1878&quot; rel=&quot;nofollow&quot;&gt;Research Database Locator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments, Wikipedia, particularly if a page is written well, have several links to journal articles and books etc.&lt;/p&gt;&#xA;" OwnerDisplayName="user3901" LastActivityDate="2013-12-24T23:41:20.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7500" PostTypeId="1" AcceptedAnswerId="7504" CreationDate="2013-12-25T00:16:54.427" Score="3" ViewCount="3550" Body="&lt;p&gt;I understand we can calculate the maximum work done by the electromotive force (driving electrons through a cell) by multiplying the potential of the cell with the charge of one mole of electrons (Faraday's constant) and with &quot;the number of moles of electrons involved in the reaction&quot;. I thought that in the case of a hydrogen cell: the reaction $\ce{H2 - 2e- -&amp;gt; 2H+}$ involves two electrons but I can see on wikipedia that in the Nernst equation for the standard hydrogen electrode $z$ (the number of moles involved) is considered to be 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this $z$, or expression &quot;electrons involved&quot; actually mean? &lt;/p&gt;&#xA;" OwnerUserId="3940" LastEditorUserId="73" LastEditDate="2013-12-25T10:32:51.900" LastActivityDate="2013-12-25T10:32:51.900" Title="&quot;Number of moles of electrons involved in the reaction&quot; for hydrogen cell" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7501" PostTypeId="2" ParentId="4193" CreationDate="2013-12-25T01:40:26.060" Score="1" Body="&lt;p&gt;This is a partial answer to your question.  I have used chromaticity coordinates for various scientific endeavours (air quality mainly).  I found that it is a very powerful means of determining colour differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit of searching and there is somewhat of a precedent for using chromaticity for blood plasma.  An example is the article &lt;a href=&quot;http://file.scirp.org/Html/9-2100279_19401.htm&quot; rel=&quot;nofollow&quot;&gt;Features of Color Characteristics of Blood Plasma of Patients with Liver Cirrhosis in the Colorimetric System XYZ as Compared with Healthy People&lt;/a&gt; (Dolomatov et al. 2012) - even though it may be a different context to your study, but does show that it can and has been done before.  Specifically, their method was&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;by colorimetric method, based on absorption electromagnetic radiation in visible area.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several more examples of how it can be used.&lt;/p&gt;&#xA;" OwnerDisplayName="user3901" LastActivityDate="2013-12-25T01:40:26.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7502" PostTypeId="1" CreationDate="2013-12-25T03:26:42.083" Score="13" ViewCount="9730" Body="&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/AKrtiDD.png&quot; alt=&quot;ester vs. thioester lewis structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem set solution cites: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Although sulfur is less electronegative than oxygen, overlap is a&#xA;  more important consideration.  Sulfur is a member of the third period.&#xA;  As a consequence, a sulfur atom is larger than an oxygen  atom and the&#xA;  C–S bond is longer than the C–O bond, contributing to poorer overlap.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I don't quite understand why a longer bond length would contribute to poor orbital overlap, and how greater overlap contributes to more stability.&lt;/p&gt;&#xA;" OwnerUserId="4032" LastEditorUserId="10185" LastEditDate="2015-09-12T19:38:34.913" LastActivityDate="2019-03-02T20:00:50.420" Title="Resonance stability: ester vs. thioester" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;stability&gt;&lt;resonance&gt;&lt;esters&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7504" PostTypeId="2" ParentId="7500" CreationDate="2013-12-25T10:30:18.510" Score="1" Body="&lt;p&gt;Nernst equation is used for finding the cell potential when the various reactants, i.e. anodic and cathodic electrolytes are not in their standard states ($298\text{K}, 1\text{mol l}^{-1}$). Thus, in such cases, you first write down any balanced form of equation, which for our hydrogen electrode, I choose to write down $\ce{3H2 -&amp;gt; 6H+ +6e-}$ and these electrons are used in a standard $\ce{H2}$ cell on the other electrode. Hence the Nernst equation would give the cell potential for this &lt;a href=&quot;http://en.wikipedia.org/wiki/Concentration_cell&quot; rel=&quot;nofollow&quot;&gt;concentration cell&lt;/a&gt; as ($E^\circ=0$ since the the standard potentials of both sides is the same since the materials are same, and the non-standard hydrogen cell is the anode)&#xA;$$E_{cell}=-\frac{RT}{6F}\ln\frac{[H^+]^6}{1}$$&#xA;By using the properties of logarithm, you can show this is equivalent to $-\frac{RT}{1F}\ln[H^+]^1$, and this is the equation if you took only 1 electron transfer in the reaction.&lt;br&gt;&#xA;It does not matter what equation you write, since in the final analysis, the number of electrons taken are compensated by the power of the concentration term in the logarithm.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2013-12-25T10:30:18.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7507" PostTypeId="2" ParentId="5197" CreationDate="2013-12-26T06:41:29.923" Score="2" Body="&lt;p&gt;If you want a quick and dirty expressions in terms of hydrogen atom wave functions, you can try this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For ground state you can get a quick approximation by applying effective nuclear charge.  Then you get a hydrogen-like wave function with a modified nuclear charge between 1 and 2;  solve that and get orbital.  It will be doubly occupied, spin singlet.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The lowest excited states of He are all Rydberg states.  The roughest approximation would be to consider it He+ plus a Rydberg electron, and you can even use hydrogen-like atom formula to get the Rydberg orbital.  The total wave function would be antisymmetrized product of these two orbitals, coupled to be either singlet or triplet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not sure you want transition density matrix or transition dipole matrix.  Either way, its just quite straightforward integration from there.  I think there might be ways to do this analytically, but frankly I do not remember and you can probably try that on Mathematica.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2013-12-26T06:41:29.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7508" PostTypeId="1" CreationDate="2013-12-26T06:50:14.683" Score="4" ViewCount="5144" Body="&lt;p&gt;My book says that electrophilic bromination of toluene (methylbenzene) is faster than that of ethylbenzene. How is that possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doubting it because ethyl is a better $+\ce{I}$ group than methyl, so ethyl would increase the electron density in the benzene ring more than methyl. As a result, bromine, which is an electrophile, should react with ethylbenzene faster than methylbenzene.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is the bromination of methylbenzene faster than ethylbenzene and please give the reaction mechanism, if possible?&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorUserId="7475" LastEditDate="2015-10-02T16:34:38.440" LastActivityDate="2018-03-21T00:40:35.393" Title="Rate of electrophilic bromination of toluene and ethylbenzene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7509" PostTypeId="2" ParentId="6553" CreationDate="2013-12-26T07:10:45.140" Score="2" Body="&lt;p&gt;Well generally you can say that symmetry contributed to the stability. But in most cases you will find exception. The Jahn–Teller effect is most often encountered in octahedral complexes of the transition metals. In octahedral complexes, the Jahn–Teller effect is most pronounced when an odd number of electrons occupy the eg orbitals. This situation arises in complexes with the configurations d9, low-spin d7 or high-spin d4 complexes, all of which have doubly degenerate ground states. In such compounds the eg orbitals involved in the degeneracy point directly at the ligands, so distortion can result in a large energetic stabilisation and is not related to symmetry.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="9961" LastEditDate="2017-07-18T15:30:17.357" LastActivityDate="2017-07-18T15:30:17.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7510" PostTypeId="2" ParentId="7508" CreationDate="2013-12-26T08:43:23.380" Score="3" Body="&lt;p&gt;First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider &lt;strong&gt;sterics&lt;/strong&gt;, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(a)&lt;/strong&gt; electrostatic repulsion toward the electrophile, particularly in the case of &lt;em&gt;ortho&lt;/em&gt;- attack, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b)&lt;/strong&gt; inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the standpoint of &lt;strong&gt;electronic effects&lt;/strong&gt;, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;strong&gt;quantitative approach&lt;/strong&gt;, you may want to take a look at the substituent constants derived for various alkyl groups from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammett equation&lt;/a&gt;. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one &lt;a href=&quot;http://www.bluffton.edu/~bergerd/classes/cem311/handouts/sigmas.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;table of said substituent constants&lt;/a&gt;. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as to the &lt;strong&gt;reaction mechanism&lt;/strong&gt;, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_halogenation&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrophilic halogenation&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrophilic aromatic substitution&lt;/a&gt; more generally.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="51553" LastEditDate="2018-03-21T00:40:35.393" LastActivityDate="2018-03-21T00:40:35.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7511" PostTypeId="2" ParentId="7502" CreationDate="2013-12-26T10:40:11.617" Score="6" Body="&lt;p&gt;Try to understand it in a layman's way, greater the extent of overlap between the atoms, the closer their nucleus will be. The stability of the molecule will thus increase as stability of a molecule is inversely proportional to the bond length of the atoms in the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is applicable only upto some extent like your situation but if the nucleus of two atoms get too close a stronger force which is nuclear force starts acting on them which repels them, it is a very short length force.&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="13339" LastEditDate="2015-05-18T09:44:00.187" LastActivityDate="2015-05-18T09:44:00.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7512" PostTypeId="1" AcceptedAnswerId="7513" CreationDate="2013-12-26T13:13:45.497" Score="3" ViewCount="976" Body="&lt;p&gt;Is it an ion or a neutral atom that is created through beta decay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: $_6^{14}\mathrm C \to ~ _7^{14}\mathrm N + \mathrm e^- + \mathrm{v{_e}}^{-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't $_7^{14}N$ an ion since neutron gives us only a proton and an electron, not a proton and 2 electrons? If so, why isn't it stated in the example above (it's from Wikipedia)?&lt;/p&gt;&#xA;" OwnerUserId="4042" LastEditorUserId="402" LastEditDate="2017-04-09T17:26:44.927" LastActivityDate="2017-04-09T17:26:44.927" Title="Notation of beta decay" Tags="&lt;ions&gt;&lt;nuclear&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7513" PostTypeId="2" ParentId="7512" CreationDate="2013-12-26T13:22:34.647" Score="5" Body="&lt;p&gt;An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about &lt;em&gt;nuclei&lt;/em&gt;, not the entire atom. The reaction is to be read as &quot;A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the neutrino has no charge.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2013-12-26T13:22:34.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7514" PostTypeId="1" AcceptedAnswerId="7527" CreationDate="2013-12-26T14:15:56.787" Score="2" ViewCount="14032" Body="&lt;p&gt;Having mixed powder of $\ce{KCl}$, $\ce{NaCl}$, and $\ce{MgCl2}$ (and, perhaps, some crap), how can the components be separated?&lt;/p&gt;&#xA;" OwnerUserId="4043" LastEditorUserId="4231" LastEditDate="2015-03-08T17:04:48.123" LastActivityDate="2015-09-08T02:59:28.587" Title="How to separate KCl from NaCl and MgCl₂?" Tags="&lt;inorganic-chemistry&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7515" PostTypeId="1" AcceptedAnswerId="7706" CreationDate="2013-12-26T17:04:40.353" Score="2" ViewCount="1340" Body="&lt;p&gt;Can bromide ion reduce ferric ion ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a query.&#xA;In my opinion it can since bromide ion can undergo oxidation.&#xA;Is my answer correct?&lt;/p&gt;&#xA;" OwnerUserId="4044" LastEditorUserId="3943" LastEditDate="2013-12-26T18:27:38.813" LastActivityDate="2014-01-08T10:39:16.827" Title="Can bromide ion reduce ferric ion?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7516" PostTypeId="2" ParentId="7515" CreationDate="2013-12-26T18:21:58.203" Score="0" Body="&lt;p&gt;According to you the reaction will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2Br- + 2Fe(III)  ---&gt; Br2 + 2Fe(II) by balancing.&#xA;Spontaneity can be checked out by using standard EMF of the cell formed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence standard emf E(cell)=E(Reduction) for {Fe(III)/Fe(II)}  +  E(oxidation) for {Br-/Br2}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using standard values we get  E(Reduction) for {Fe(III)/Fe(II)}=+0.77 and  E(oxidation) for {Br-/Br2}=-1.07&#xA;hence the E(cell) we get is =.77-1.07=-.3(negative)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have (Delta)G=-n*F*E(cell) where (Delta)G is change in Gibb's Free Energy change&#xA;n is no. of electrons(here n=2),F is Faraday's constant and E is Emf of cell(Standard)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (Delta)G is negative then reaction is spontaneous and if (Delta)G is positive then non spontaneous.&#xA;Clearly here E(cell) is negative hence (Delta)G is positive hence reaction is non-spontaneous and will not occur[NOTE THAT HERE CONCENTRATIONS OF IONS AND PRESSURE OF GAS IS TAKEN UNITY IN ITS SI UNITS.WE CAN MAKE IT SPONTANEOUS BY VARYING THE CONCENTRATION OF ION OR PARTIAL PRESSURE OF GAS]&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="3943" LastEditDate="2013-12-26T18:34:30.723" LastActivityDate="2013-12-26T18:34:30.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7517" PostTypeId="2" ParentId="5922" CreationDate="2013-12-26T18:48:49.567" Score="1" Body="&lt;p&gt;IN pure water there is no chance for electricity conduction as it has no free ions.It contains only $\ce{H2O}$ molecules. For electricity conductance, there should be ions present. In the case you are talking about some drops of $\ce{H2SO4}$ is added for the conduction of electricity.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="15429" LastEditDate="2015-05-04T03:06:01.120" LastActivityDate="2015-05-04T03:06:01.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7518" PostTypeId="1" AcceptedAnswerId="7525" CreationDate="2013-12-27T00:26:26.160" Score="2" ViewCount="1111" Body="&lt;p&gt;How come when a non polar molecule increases in size, the attractive forces between the molecules of a substance also increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that when a molecule increases in size, the atomic radius increases, which creates a bigger space between particles; thus, the force of attraction would be weaker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, why does the force of attraction increase when a non polar molecule increases in size? &lt;/p&gt;&#xA;" OwnerUserId="2594" LastEditorDisplayName="user3901" LastEditDate="2013-12-27T09:00:23.310" LastActivityDate="2013-12-27T13:46:02.197" Title="Non polar molecular and attractive forces" Tags="&lt;molecules&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7519" PostTypeId="2" ParentId="7512" CreationDate="2013-12-27T02:11:24.580" Score="4" Body="&lt;p&gt;Further to @ManishEarth's answer, note that the distinction is important in doing mass-energy calculations.  In the above example, you can find the energy released by subtracting the &lt;strong&gt;&lt;em&gt;tabulated&lt;/em&gt;&lt;/strong&gt; mass of a $N^{14}$ atom from the &lt;strong&gt;&lt;em&gt;tabulated&lt;/em&gt;&lt;/strong&gt; mass of the $C^{14}$ atom, and ignore the electron mass.  Why?  Because the actual Nitrogen atom produced only has 6 orbital electrons, while tabulated values include the mass of 7 orbital electrons.  So, in using the tabulated mass, you already include an extra electron...&lt;/p&gt;&#xA;" OwnerUserId="3668" LastActivityDate="2013-12-27T02:11:24.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7520" PostTypeId="1" CreationDate="2013-12-27T05:23:52.517" Score="27" ViewCount="12350" Body="&lt;p&gt;I was recently asked the question &quot;Why are noble gases stable? with the expectation of providing an answer beyond the general explanation of &quot;they have full valence layers&quot; and I couldn't think of one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to hear a full description of the this stability, I have a solid background in quantum mechanics, so feel free to speak of wave functions or the like if necessary.&lt;/p&gt;&#xA;" OwnerUserId="4045" LastEditorDisplayName="user3901" LastEditDate="2013-12-27T09:00:13.247" LastActivityDate="2020-02-25T01:40:54.930" Title="Why are noble gases stable" Tags="&lt;quantum-chemistry&gt;&lt;stability&gt;&lt;noble-gases&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7521" PostTypeId="1" CreationDate="2013-12-27T06:54:54.270" Score="2" ViewCount="105" Body="&lt;p&gt;Sulfur belongs to the same group as oxygen. Then why oxygen exists as an $\ce{O2}$ molecule while sulfur exists as $\ce{S8}$ molecule? What's the chemistry behind this?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="418" LastEditDate="2016-12-04T17:14:49.927" LastActivityDate="2016-12-04T17:14:49.927" Title="Why does sulfur exist as octasulfur?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2013-12-27T19:16:15.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="7522" PostTypeId="1" CreationDate="2013-12-27T09:48:22.737" Score="1" ViewCount="1377" Body="&lt;p&gt;Why do we only consider the $\mathrm{p}K_\text{a}$ value of a drug regardless of whether it is an acid or base? Why do we not use the $\mathrm{p}K_\text{b}$ value?&lt;/p&gt;&#xA;" OwnerUserId="4047" LastEditorUserId="41890" LastEditDate="2017-11-27T17:59:17.097" LastActivityDate="2017-11-27T18:05:47.207" Title="What is the importance of using a pKa value instead of a pKb value when describing drug chemistry?" Tags="&lt;ph&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7523" PostTypeId="2" ParentId="7522" CreationDate="2013-12-27T11:07:41.343" Score="1" Body="&lt;p&gt;For easier comparability: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing two pKa values is much easier than comparing a pKa with a pKb value.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-27T11:07:41.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7525" PostTypeId="2" ParentId="7518" CreationDate="2013-12-27T12:54:07.103" Score="1" Body="&lt;p&gt;There is more than one factor that matters in the forces between molecules. The tradeoff between these different factors is complex and can't be reduced to a single simple equation or factor. So some of the factors become more important because of molecular or atomic size counterbalancing the factor of distance between the entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collective force that cause molecules and atoms to attract are often referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;van der Waal's force&lt;/a&gt; (though some sources use a more narrow definition that doesn't include all of the components below). The key non-bonding forces involved are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;forces between permanent dipoles&lt;/li&gt;&#xA;&lt;li&gt;forces between permanent dipoles and induced dipoles&lt;/li&gt;&#xA;&lt;li&gt;forces between instantaneously induced dipoles caused by quantum fluctuations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In non-polar molecules it is the third of these that dominates. Crudely stated the force arises because of uneven distribution of the electron cloud in a molecule or atom that happens as a result of quantum fluctuations in the cloud. An uneven distribution of electron density is the equivalent of a dipole moment. These dipole moments create induced dipoles in neighbouring atoms and can interact with other quantum fluctuations in the neighbouring atoms. These create small electrostatic forces between the atoms or molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why larger atoms or molecules seem to have larger forces is two fold. Larger molecules, crudely, have larger electron clouds and more ways to interact (shape matters too and large flat molecules &quot;stick&quot; together more than irregular lumpy ones: compare the melting points of benzene and toluene). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In larger atoms the explanation is simpler. Larger electron clouds are less tightly held by there nuclei than those in smaller atoms. Crudely, larger, floppier clouds can have more scope for the quantum fluctuations than tighter-held smaller clouds (also they have a larger area and that maters to the total force). So the size effect increases the strength of the force rather than reducing it as might be expected from simpler arguments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm aware that this is a slightly superficial explanation, but a better one would probably involve enough quantum mechanics to make the average reader's head explode. I think i've captured the key intuitions that matter.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2013-12-27T13:46:02.197" LastActivityDate="2013-12-27T13:46:02.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7527" PostTypeId="2" ParentId="7514" CreationDate="2013-12-27T13:38:54.053" Score="9" Body="&lt;p&gt;As I mentioned in the comments, a quantitative separation of the three compounds will probably not be possible outside of a proper laboratory, however a coarse separation is possible. A quick search on the internet yields &lt;a href=&quot;http://www.sciencemadness.org/talk/viewthread.php?tid=3196&quot;&gt;this link&lt;/a&gt; for the separation of $\ce{NaCl}$ and $\ce{KCl}$. You didn't give us the exact composition of the table salt substitute, so I assume $\ce{MgCl2}$ is present in much smaller amounts than the first two compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest strategy is to use the difference between the solubilities of the salts in water or ethanol. Using water is much more convenient, so we'll stick to that. I'll outline the method to obtain the highest purity $\ce{KCl}$. First, add a very large amount of table salt substitute to a container with boiling water (you'll want to add some 200-250 g per 100 g of water) and let the mixture dissolve as much as possible (though it probably won't even come close to dissolving completely). What you have then is a boiling water solution saturated with both $\ce{NaCl}$ and $\ce{KCl}$. Now, while the water is still very hot, quickly filter off the excess salt (might be doable with a fine sieve, or a funnel with some cotton would work better) and reserve the liquid in another container. Let the liquid cool down (using an ice bath is optional but increases yield and purity), and it will start to precipitate the salts. Now you perform another filtration, but this time you keep the solid retained on the filter, not the liquid. This yields purified $\ce{KCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strategy works because it turns out that &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table&quot;&gt;$\ce{NaCl}$ and $\ce{KCl}$ behave rather differently in how their solubility in water is affected by temperature&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl}$ soubility in water:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;35.6 g/100 g water (0°C)&lt;/li&gt;&#xA;&lt;li&gt;35.9 g/100 g water (20°C)&lt;/li&gt;&#xA;&lt;li&gt;39.0 g/100 g water (100°C)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{KCl}$ soubility in water:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;28.0 g/100 g water (0°C)&lt;/li&gt;&#xA;&lt;li&gt;34.2 g/100 g water (20°C)&lt;/li&gt;&#xA;&lt;li&gt;56.3 g/100 g water (100°C)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Evidently, as the filtered liquid cools down, $\ce{NaCl}$ remains almost as soluble in cooler water as in hot water, while $\ce{KCl}$ is only about half as soluble. This means that a large amount of $\ce{KCl}$ will precipitate as the filtered liquid cools, while only a bit of $\ce{NaCl}$ will follow. Hence, ideally, from 100 g of water saturated with both salts at 100°C (and containing nothing else), you will be able to extract 25.0 g of a mixture of salts containing 87.6% $\ce{KCl}$ if you perform the second filtration at 20°C. If you filter at 0°C, then the yield increases to 31.7 g of a mixture containing 89.3% $\ce{KCl}$. If for some reason you don't want to do the second filtration, you can just let the water evaporate completely after the first filtration, producing 95.3 g of a mixture containing 59.1% $\ce{KCl}$ from 100 g of saturated boiling water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice your yield and the purity of the $\ce{KCl}$ will likely be somewhat lower due to losses in manipulation and the presence of other impurities. The procedure may be complicated by the very large initial amount of table salt mixture that you have to dissolve in water. You can reduce the amount of the mixture, but at some point the purification yield will start to go down.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2013-12-27T13:45:33.557" LastActivityDate="2013-12-27T13:45:33.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7528" PostTypeId="2" ParentId="7522" CreationDate="2013-12-27T15:27:12.227" Score="1" Body="&lt;p&gt;The reason we only usually consider the $\mathrm p K_\mathrm a$ is that in a constant medium (in the case of drugs, water) there is a simple relationship between $\mathrm p K_\mathrm b$ and $\mathrm p K_\mathrm a$ so we don't need two scales for acidity and basicity and can define a continuous scale from strong acid to strong base using just one number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm p K_\mathrm b = \mathrm p K_\mathrm w - \mathrm p K_\mathrm a$$ where $\mathrm p K_\mathrm w = -\log_{10}([\ce{H+}][\ce{OH-}])$, roughly equal to 14 under normal conditions.  &lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="16683" LastEditDate="2017-11-27T18:05:47.207" LastActivityDate="2017-11-27T18:05:47.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7529" PostTypeId="1" AcceptedAnswerId="7530" CreationDate="2013-12-27T16:06:16.233" Score="5" ViewCount="429" Body="&lt;p&gt;When I name an organic compound, I follow a preferential order like so: &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot;&gt;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&lt;/a&gt; But I don't understand why this order exists. And intuitively (to me at least), a triple bond has a higher preference than a double bond... then why is the order like in the link, i.e., double bond preferred over triple bond? Is there any chemistry behind this?&lt;/p&gt;&#xA;" OwnerUserId="4050" LastActivityDate="2013-12-27T17:08:14.237" Title="Nomenclature of organic compounds" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7530" PostTypeId="2" ParentId="7529" CreationDate="2013-12-27T16:59:15.113" Score="4" Body="&lt;p&gt;The naming of organic compounds are governed by &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/&quot; rel=&quot;noreferrer&quot;&gt;IUPAC guidelines&lt;/a&gt;, specifically the way the names are formed are by some general principles, which are given on the IUPAC website at their page &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_317.htm&quot; rel=&quot;noreferrer&quot;&gt;Guide to Name Construction: General Principles&lt;/a&gt;, with &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_334.htm&quot; rel=&quot;noreferrer&quot;&gt;naming examples here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Another rather nice summary is found on the  webpage &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature&lt;/a&gt; (Reusch, 1999), states that a organic compound name will have the following features, based on several common chemical characteristics of organic compounds:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;•  A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.&lt;/p&gt;&#xA;&lt;p&gt;•  A suffix or other element(s) designating functional groups that may be present in the compound.&lt;/p&gt;&#xA;&lt;p&gt;•  Names of substituent groups, other than hydrogen, that complete the molecular structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(Specific IUPAC guidelines for different hydrocarbons are included in the link).&lt;/p&gt;&#xA;&lt;p&gt;Another summary is found on the handout &lt;a href=&quot;http://www.angelo.edu/faculty/kboudrea/organic/IUPAC_Handout.pdf&quot; rel=&quot;noreferrer&quot;&gt;Short Summary of IUPAC Nomenclature of Organic Compounds&lt;/a&gt; (Dr. Jan Simek, California Polytechnic State University at San Luis Obispo), which state that the general nomenclature principle is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;IUPAC nomenclature is based on naming a molecule’s longest chain of carbons connected by&#xA;single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or&#xA;atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a&#xA;specific set of priorities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user3901" LastEditorUserId="-1" LastEditorDisplayName="user3901" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-12-27T17:08:14.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7532" PostTypeId="1" CreationDate="2013-12-28T00:19:15.410" Score="1" ViewCount="545" Body="&lt;p&gt;This data was taken for my science fair project, and I would like to know the relationship between spectrophotometer absorbance and cell density (taken manually) of algae after a toxicity test. The different levels of toxicity are IC 0 (no chemical), IC 25, IC 50, IC 100 (the most chemical). The IC value is basically the estimated about of the chemical needed to destroy a certain amount of the population. I took the manual density and absorbancy of four samples of different levels of toxicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IC value/Cell densities(cells/ml)/absorbancy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IC 100/28000/0.408&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IC 50/42600/0.302&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IC 25/70000/0.01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IC 0/138000/0.79&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based off of these points, I would like to make a calibration curve. But when I do this, the highest r^2 is quadratic. Is this an acceptable calibration curve for finding other densities based off of known absorbances? Or would it have to be linear with a very low r^2?&lt;/p&gt;&#xA;" OwnerUserId="4055" LastActivityDate="2013-12-28T13:24:39.417" Title="Does my data show quadratic regression in a calibration curve?" Tags="&lt;spectroscopy&gt;&lt;toxicity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7534" PostTypeId="2" ParentId="7532" CreationDate="2013-12-28T03:36:35.280" Score="0" Body="&lt;p&gt;Your absorbency values seem anomalous compared with the cell density. One would assume that as cell density increased, so would absorbency. Your absorbency decreases first and then leaps up to a maximum. I can't say from your data whether there is any relationship between cell density and absorbency. On the other hand, there seems to be an inverse-square relationship between IC value and cell density. Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2013-12-28T03:36:35.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7535" PostTypeId="1" CreationDate="2013-12-28T03:40:55.787" Score="10" ViewCount="481" Body="&lt;p&gt;It doesn't seem as though the hybridization model adds anything useful to the discussion of molecular geometry as predicted by the VSEPR model. It's just another way of labeling linear, trigonal planar, and tetrahedral geometry (as $sp$, $sp^2$, and $sp^3$, respectively). Considering that orbital hybridization may not be the most accurate view of molecular bonding, what is the utility of having it around?&lt;/p&gt;&#xA;" OwnerUserId="1758" LastActivityDate="2013-12-28T12:39:44.597" Title="Does the hybridization model gives us any further insight on Molecular Geometry?" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;hybridization&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7537" PostTypeId="1" AcceptedAnswerId="7547" CreationDate="2013-12-28T09:40:55.217" Score="10" ViewCount="1226" Body="&lt;p&gt;Following is the question that was asked in my exam. a,b,c,d are the various options, one of which is the answer. I am familiar with the &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/reformatsky-reaction.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reformatsky reaction&lt;/a&gt;, Aldol condensation, &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/knoevenagel-condensation.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knoevenagel's reaction&lt;/a&gt; and general base catalysed carbonyl reaction. 1 and 2 are my attempts at deriving the mechanism and hence getting at the products. But none of those succeeded to bring about the required products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help in how to proceed through in this case and derive the products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KfhEK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4059" LastEditorUserId="5026" LastEditDate="2018-02-25T07:44:53.673" LastActivityDate="2018-02-25T07:44:53.673" Title="Prediction of products in reaction of dicarbonyl with alpha-haloketone" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7539" PostTypeId="2" ParentId="7535" CreationDate="2013-12-28T12:39:44.597" Score="8" Body="&lt;p&gt;I find orbital hybridization of very limited use.  Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom.  Presently, we have no evidence that supports the claim that orbital hybridization happens.  The reason why I am not in favor of using orbital hybridization to describe chemical phenomena is because it is &lt;strong&gt;not a predictive tool&lt;/strong&gt;.  We only know the hybridization of an atom &lt;strong&gt;after we know the structure&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSEPR, however, &lt;strong&gt;is&lt;/strong&gt; a predictive tool, although it has no significant quantum mechanical basis.  The VSEPR model is based upon the idea that pairs of electrons (whether bonding or not) will repel one another.  Geometrical factors are considered to determine how far away pairs of electrons can be from one another (e.g 180 degrees apart for two pairs, and 120 degrees apart for 3 pairs).  The model accurately predicts the H-C-H angle in methane and also predicts that the H-N-H angle in ammonia is closer to 109 degrees than 120 (as is the case in a molecule such as $BF_3$.  Because VSEPR assumes ideal conditions, the model starts to fall apart when considering complex molecules.  For example, while the VSEPR model correctly predicts that $SF_4$ is seesaw shaped, it can only predict that the structure won't be precisely seesaw (with F-S-F angles of 180, 120 and 90 degrees).  It only allows us to suggest that the lone pair on the sulfur will be a &quot;space hog&quot; in a manner of speaking and repel the bonding electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to the last part of your question &lt;em&gt;why do we still use it&lt;/em&gt;?  (This is somewhat speculative and opinion-based.) We have to blame the organic chemists here.  orbital hybridization terminology ($sp^2, sp^3$, ...) provides a lexicon for describing the bonding environments of carbon atoms that is compact and visually constructive for organic molecules.  If I state that there is an $sp^2$ carbon in a molecule, you can begin to visualize what that part of the molecule looks like, that it is a region of planarity, that it cannot be a chiral center, that it likely has a double bond and therefore may undergo certain types of chemical reactions, etc. So while orbital hybridization doesn't provide any insight into molecular geometry, it provides a basis for terminology that chemists can use to transmit meaningful information.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2013-12-28T12:39:44.597" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7540" PostTypeId="2" ParentId="7532" CreationDate="2013-12-28T13:24:39.417" Score="2" Body="&lt;p&gt;As you have correctly assumed, the relationship between absorbance and absorbant concentration is linear (law of &lt;a href=&quot;http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambert-Beer&lt;/a&gt;):&#xA;$$ A = \varepsilon \, l \times c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should give you a straight line through (0,0) and your data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a graphical analysis of data, let's not look at the numbers themselves but on what kind of a graph it gives us. So after 2 minutes with gnuplot I get something like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/C40Jf.jpg&quot; alt=&quot;Plot of the data provided by the OP.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are two things that I note here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The absorbance of the $\mathrm{IC_{25}}$ value is very very low (nearly 0) which indicates some wrong measurement. Maybe the cells weren't in the pathway of the light beam, only the medium?&lt;/li&gt;&#xA;&lt;li&gt;You have &lt;strong&gt;four&lt;/strong&gt; data points.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My advice is the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Double check your measurements (the absorbance and the cell density). It would be even better if you did several measurements for the different inhibitor concentrations and then calculate the mean (and standard deviation if you so desire).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Generate more data.&lt;/strong&gt; It is very wonky to try and justify a linear regression through these 4 data points (although mathematically allowed). However, if all intermittent points are equally chaotically distributed along a straight line the linear relationship still holds (with a fairly high $R^2$). What this does to the interpretation of the data is a completely different story.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2013-12-28T13:24:39.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7541" PostTypeId="1" AcceptedAnswerId="7910" CreationDate="2013-12-28T15:40:49.500" Score="16" ViewCount="57636" Body="&lt;p&gt;This may be a silly question but why is water wet? &#xA;All the previous questions on water do not explain the reason why is water wet. They assume that its a natural property. However we need to understand why is it wet in the first place. This could apply to other liquids such as milk, juices but the essential element is water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly another basic property which we take for granted is heat caused by fire. What causes fire to be hot as an inherent property? For the purpose of discussion consider simple fire by burning wood or paper.&lt;/p&gt;&#xA;" OwnerUserId="4060" LastActivityDate="2018-01-22T15:10:53.767" Title="Why is water wet and fire hot?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7542" PostTypeId="1" AcceptedAnswerId="7544" CreationDate="2013-12-28T17:42:54.570" Score="7" ViewCount="8211" Body="&lt;p&gt;I have read about the Brønsted-Lowry and Arrhenius concepts to define a substance as an acid or a base. Our teacher told us that there are substances which behave both as an acid and a base, using water and ammonia as examples. He also asked us to find other substances with the same behavior. Does anyone know any more substances which are amphoteric in nature?&lt;/p&gt;&#xA;" OwnerUserId="4052" LastEditorUserId="5026" LastEditDate="2018-03-03T10:48:38.320" LastActivityDate="2019-01-04T15:45:19.610" Title="Are there any amphoteric substances other than water and ammonia?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7544" PostTypeId="2" ParentId="7542" CreationDate="2013-12-28T18:48:57.120" Score="11" Body="&lt;p&gt;First, it may be worth elucidating the nomenclature a bit and distinguishing between &lt;em&gt;amphiprotic&lt;/em&gt; and &lt;em&gt;amphoteric&lt;/em&gt;  substances. The former are those that can react as either acids or bases in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Br%C3%B8nsted-Lowry_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brønsted-Lowry&lt;/a&gt; sense, while the latter are those that react in either manner per the broader &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis&lt;/a&gt; definition. As you mention, water and ammonia are amphiprotic (and thus also amphoteric by definition), which is evident even from the fact that both undergo auto-ionization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + OH-}\\&#xA;\ce{NH3 + NH3 &amp;lt;=&amp;gt; NH4+ + NH2-}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, a large number of protic substances exhibit this type of behavior to an appreciable degree, including concentrated solutions of various acids (e.g., sulfuric, hydrofluoric, etc.). Another canonical example of an extremely common amphiprotic molecule prevalent in nature is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bicarbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;bicarbonate&lt;/a&gt;, which you find in countless consumer products, as well as within the human body as a physiologic buffering agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the right conditions, a very wide array of molecules of diverse types can be amphoteric. The two necessary and sufficient criteria for amphoterism are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The molecule possesses lone pairs, so that it can act as a base.&lt;/li&gt;&#xA;&lt;li&gt;The molecule can accept electron pairs, so that it can act an acid.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Those are not rare properties, hence amphoterism isn't really an exotic phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="69629" LastEditDate="2019-01-04T15:45:19.610" LastActivityDate="2019-01-04T15:45:19.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7546" PostTypeId="1" CreationDate="2013-12-28T19:17:59.787" Score="7" ViewCount="149" Body="&lt;p&gt;There are some interesting laboratory reactions that use liquid ammonia as a solvent (for example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Birch_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birch Reduction&lt;/a&gt; where an alkali metal is dissolved in liquid ammonia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have also been some high volume industrial processes that use liquid ammonia as a solvent (the weedkiller paraquat has two, quite different, production routes using liquid ammonia producing thousands of tonnes a year). One uses dissolving sodium (like the Birch reduction):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cUNyr.png&quot; alt=&quot;wikipedia description of one paraquat synthesis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there any other significant chemical manufacturing processes that use it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB I would include the production of pharmaceuticals or their ingredients as well as any bulk chemicals.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="4945" LastEditDate="2014-05-21T08:02:07.943" LastActivityDate="2014-05-21T08:02:07.943" Title="Are there any significant industrial processes that use liquid ammonia as a solvent?" Tags="&lt;synthesis&gt;&lt;solvents&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7547" PostTypeId="2" ParentId="7537" CreationDate="2013-12-28T21:36:35.870" Score="13" Body="&lt;p&gt;As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3.&#xA;The enolate of &lt;strong&gt;1&lt;/strong&gt; (sorry for the circle in the structure, i'm to lazy to fix it) and &lt;strong&gt;2&lt;/strong&gt; undergo an aldol reaction yielding &lt;strong&gt;3&lt;/strong&gt;. With a bit of charge shifting, we get again a nicely stabilized enolate (&lt;strong&gt;4&lt;/strong&gt;), which dehalogenates (you might call that an O-alkylation as well) to &lt;strong&gt;5&lt;/strong&gt;. Subsequent elimination under aqueous conditions results in the formation of &lt;strong&gt;6&lt;/strong&gt;. I'd say this is, more or less, an example of the &lt;em&gt;Feist-Bénary&lt;/em&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVefH.png&quot; alt=&quot;Feist-Bénary reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, I created the drawings with &lt;a href=&quot;http://bkchem.zirael.org/&quot; rel=&quot;noreferrer&quot;&gt;BKChem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A conceivable depiction for the &quot;transformation&quot; of &lt;strong&gt;3&lt;/strong&gt; to &lt;strong&gt;4&lt;/strong&gt; is given below, but please note that this is just a model to please our eyes and minds ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0oH6Z.png&quot; alt=&quot;intermediates&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2013-12-29T09:39:55.933" LastActivityDate="2013-12-29T09:39:55.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7548" PostTypeId="1" AcceptedAnswerId="7549" CreationDate="2013-12-28T21:41:01.180" Score="7" ViewCount="1566" Body="&lt;p&gt;EDTA (Ethylenediaminetetraacetic acid) has a structure like this&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LSKnz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why, as the name 'Ethylene' suggests, doesn't it should have a double bond between carbons?&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="81" LastEditDate="2013-12-29T13:26:59.397" LastActivityDate="2013-12-29T13:26:59.397" Title="Why doesn't EDTA have a double bond?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;nomenclature&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7549" PostTypeId="2" ParentId="7548" CreationDate="2013-12-28T22:38:35.707" Score="15" Body="&lt;p&gt;A lot of non-systematic chemical names don't describe the composition of the compound, they describe what the compound is synthesised &lt;em&gt;from&lt;/em&gt;. That is the case here since EDTA is derived from ethylene by adding two units of the diamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't that uncommon in chemical names. Polyethylene, for example, is (simplifying only slightly) a long chain of carbons connected by single bonds. But it is made by polymerising ethylene. Hence has a name that doesn't describe what its structure &lt;em&gt;is&lt;/em&gt; but where it &lt;em&gt;came from.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-28T22:38:35.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7550" PostTypeId="2" ParentId="7537" CreationDate="2013-12-28T22:47:53.363" Score="9" Body="&lt;p&gt;Klaus Warzecha has already provided the correct mechanism, but I'd like to offer a supplementary discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach here would be to attempt to eliminate some of the options based simply on first principles. Choices &lt;strong&gt;(b)&lt;/strong&gt; and &lt;strong&gt;(c)&lt;/strong&gt; seem improbable &lt;em&gt;prima facie&lt;/em&gt; for reasons of thermodynamics: the carbonyl moiety is not conjugated to the aromatic furan ring, so these should be intrinsically less stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure &lt;strong&gt;(b)&lt;/strong&gt; has an additional problem, which is that the position of the ether linkage is &lt;em&gt;para-&lt;/em&gt; to the remaining carbonyl. A similar problem occurs in structure &lt;strong&gt;(d)&lt;/strong&gt;, except in that case the ether linkage is &lt;em&gt;ortho-&lt;/em&gt; to the carbonyl moiety. Finding plausible mechanisms that would lead to either of those regiochemical outcomes is difficult. The greater acidity of the 1,3-cyclohexanedione makes it the likelier enolate nucleophile, whereas the 2-chloroacetaldehyde is the likelier electrophile. This means that the oxygens of the original 1,3-cyclohexanedione should remain in place, not undergoing elimination, so they should maintain a &lt;em&gt;meta-&lt;/em&gt; relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the conclusion of that analysis, only &lt;strong&gt;(a)&lt;/strong&gt; survives as the correct choice. The mechanism is fairly rote, and has already been given. The only subtle question, to my mind, is which site on the enolate reacts with which site on the electrophile (i.e., does the carbon of the enolate add to the the carbonyl, or does it substitute the chlorine?); either mechanism can lead to the same product. The enolate carbon is the softer nucleophilic site and should prefer the softer electrophile. Fleming, in his book &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions&lt;/em&gt;, provides a table of the absolute hardness, $\eta$, of various Lewis acids. Acetaldehyde is given to be considerably softer than chloromethane. Assuming it's valid to compare those two molecules as a proxy for this reaction, it seems probable that the enolate will attack the carbonyl from the carbon, while the chlorine will be displaced in an SN2 attack by the oxygen. Were the leaving group one of the larger, softer halogens (i.e., $\ce{Br-}$ or $\ce{I-}$), the order might well be the opposite. Aside from the superior leaving group quality of those larger halogens, another mechanistic advantage of the other order is that it results in the hydroxide leaving group of the final intermediate being shifted over by one carbon, so that the E1cB process leading to the final product results from abstraction of a more acidic allylic $\beta$-hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2013-12-28T22:47:53.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7551" PostTypeId="2" ParentId="5109" CreationDate="2013-12-29T12:24:17.253" Score="4" Body="&lt;p&gt;There is nothing in the chemistry of what they describe that is different in any way to a conventional Lead-Acid battery. What they really seem to claim is improvement in materials in the battery not to change the electrochemistry but to improve other aspects of battery operation. &lt;a href=&quot;http://megglobal.com/about.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;For example&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;MEG's E-SOURCE™ power cell technology is based on a lead-fiberglass composite material that is one-third the weight and 50 times the strength of metal alloys used in conventional batteries. To date, more than $100 million has been invested in research, development, manufacturing, and market testing of this technology in the United States and several international markets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This suggests they have built a better lead-acid battery rather than something more radical. This is not ridiculous. They claim the following advantages:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Extraordinary manufacturing and user cost savings&lt;/li&gt;&#xA;&lt;li&gt;Significant weight reduction compared to conventional lead-acid    batteries&lt;/li&gt;&#xA;&lt;li&gt;Exceptionally long service life&lt;/li&gt;&#xA;&lt;li&gt;Dramatically fast recharge speeds at all operating temperatures&lt;/li&gt;&#xA;&lt;li&gt;Safe, simple, and low-cost maintenance&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It is not impossible to imagine clever overall design and improved microstructural components and electrode materials could deliver these sorts of improvements and it is certainly far easier than inventing a whole new battery electrochemistry. For example, charging speed is often dominated by the physical material properties of the electrodes and not the underlying electrode chemistry, so clever design can mitigate the physical limits on charging speed such as by increasing the effective surface area.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2013-12-29T12:24:17.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7552" PostTypeId="2" ParentId="6030" CreationDate="2013-12-29T12:32:56.707" Score="4" Body="&lt;p&gt;Unfortunately, i don't have access to the original reference from 1928 &lt;a href=&quot;http://dx.doi.org/10.1002/cber.19280610925&quot; rel=&quot;nofollow&quot;&gt;Über die Einwirkung von Thionylchlorid auf die Pyridin-monocarbonsäuren&lt;/a&gt; by Meyer and Graf, but it's cited in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01113a530&quot; rel=&quot;nofollow&quot;&gt;The Preparation of Nicotinyl Chloride&lt;/a&gt; from 1953.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this later note, where dried and grounded potassium nicotinate was reacted with oxalyl chloride in anhydrous benzene and the resulting nicotinyl chloride eventually destilled off under reduced pressure (75-90 °C, 10-12 mm), the authors state: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On exposure to the air, the oily acyl chloride soon solidified to nicotinic acid hydrochloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With other words, the combination of moisture and the basicity of the pyridine moiety results in the decomposition of the product. This is apparently prevented by deactivation of the pyridine upon conversion to its hydrochloride.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-12-29T12:32:56.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7553" PostTypeId="2" ParentId="5885" CreationDate="2013-12-29T12:34:07.170" Score="1" Body="&lt;p&gt;In general the answer would be no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are many distillable compounds you could extract this way (not sure how many medicinal ones, though) there are many more that would decompose or simply not be easily extracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why there are many other ways chemists commonly extract things from mixture or from plants. For example, if you wanted to extract caffeine from coffee or from coffee beans, you could use solvent-based extraction (methylene chloride works well, for example, and has been used in industrial processes to create decaffeinated coffee). There is no single easy solution, though, and often much trial and error will be required to find the right solvent to differentially extract what you want.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-29T12:34:07.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7554" PostTypeId="1" AcceptedAnswerId="7558" CreationDate="2013-12-29T12:37:44.313" Score="2" ViewCount="591" Body="&lt;p&gt;Note: This question has come out of curiosity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: We all have read that atoms are mostly empty, they are generally compared to a football stadium in which a pea is kept in the center (nucleus) and flies buzzing in stands (electrons). Then why is it that we can interact with things?&#xA;It would be of great help if anyone can provide me with any article because I can't find any.&lt;/p&gt;&#xA;" OwnerUserId="4052" LastEditorUserId="9961" LastEditDate="2016-11-05T10:46:20.127" LastActivityDate="2016-11-05T10:46:20.127" Title="If atoms are mostly empty space then how are we able to interact with things?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7555" PostTypeId="1" AcceptedAnswerId="7557" CreationDate="2013-12-29T13:09:36.283" Score="0" ViewCount="395" Body="&lt;p&gt;Can anyone please clear the difference between ozonide and ozonoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already googled it but came up with nothing&lt;/p&gt;&#xA;" OwnerUserId="4065" LastActivityDate="2013-12-30T12:32:02.823" Title="Ozonide v/s Ozonoid" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7556" PostTypeId="2" ParentId="5107" CreationDate="2013-12-29T15:25:42.367" Score="6" Body="&lt;p&gt;This looks like a typical &lt;em&gt;Kharasch&lt;/em&gt; (chain) reaction. The reaction is initiated by the formation of peroxide radicals (by homolytic cleavage of the weak O–O bond). These peroxide radicals then go on to abstract Br• from the starting material:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ScjyR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ScjyR.png&quot; alt=&quot;Initiation of Kharasch reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the alternative generation of Br• is unfeasible, as radicals tend to attack at less hindered sites: hence the phenyl radical attacks the bromine which only has one attached atom, and not the carbon which has four flanking halogens. The weaker C–Br bond is also preferentially broken over the other three C–Cl bonds, so there is no chance of generating a CCl&lt;sub&gt;2&lt;/sub&gt;Br• radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trichloromethyl radical then adds to C-1 of the alkene, resulting in the more stable secondary radical. This can now abstract a bromine atom from unreacted bromotrichloromethane to give the product and a new Cl&lt;sub&gt;3&lt;/sub&gt;C• radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iRR5a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iRR5a.png&quot; alt=&quot;Propagation of Kharasch reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, with additions of X–Y across a C=C double bond (as opposed to additions of H–X), the designations Markovnikov and anti-Markovnikov lose their meaning. A Markovnikov addition of H–X would be one where the hydrogen adds to the less substituted carbon. However, here it is not clear whether to define &quot;Markovnikov&quot; as the regioselectivity where X adds to the less substituted carbon or where Y adds to the less substituted carbon.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2018-10-19T00:02:19.217" LastActivityDate="2018-10-19T00:02:19.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7557" PostTypeId="2" ParentId="7555" CreationDate="2013-12-29T16:16:51.967" Score="1" Body="&lt;p&gt;I've seen the term &lt;em&gt;ozonoide&lt;/em&gt; now and then, both in German and English texts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was used for the the primary, as well as for the secondary ozonide. Interestingly, throughout these texts, you'll find this form only once or twice while the same authors mostly use the term &lt;em&gt;ozonide&lt;/em&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm roughly familiar with the Criegee mechanism, I'm not an expert on ozonolysis. Nevertheless, I'd dare to say there's no difference between these terms (at least, there seemingly was none for the authors) and I'd recommend to stay with the established term &lt;em&gt;ozonide&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="2321" LastEditDate="2013-12-30T12:32:02.823" LastActivityDate="2013-12-30T12:32:02.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7558" PostTypeId="2" ParentId="7554" CreationDate="2013-12-29T17:34:25.297" Score="6" Body="&lt;p&gt;When we think about the structure of atoms and molecules we need to use a range of simplifying analogies as the objects we are talking about are too small to see and behave according to quantum rules we are unfamiliar with in the macroscopic world. None of the analogies we use are perfect and may mislead unless they are brought back to reality by recognising their limitations an checking their implications again the mathematics we know that does describe their behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the problem here. On one analogy atoms are built from nuclei containing protons and neutrons surrounded by electrons orbiting a long way away. This analogy is useful for explaining what happens when, for example, alpha particles are fired at a gold film a few layers of atoms thick. This is the famous &lt;a href=&quot;http://en.wikipedia.org/wiki/Rutherford_scattering&quot; rel=&quot;nofollow&quot;&gt;Rutherford scattering experiment&lt;/a&gt; first done in 1909. This experiment led to the &quot;solar system&quot; analogy for what atoms look like. But really it only explains what they look like from the point of view of fast alpha particles. The mass of atoms is concentrated in the small nucleus and it is the nucleus that alpha particles bounce off, occasionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is only one, imperfect, way to describe what atoms and molecules look like. It doesn't describe how individual atoms join to form molecules, for example. For that we have to look at how electrons behave. Here a different range of analogies or mathematical descriptions are needed. The solar-system view of electron behaviour is deeply inadequate to describe what electrons actually do. And it is a very poor description of where electrons &lt;em&gt;are&lt;/em&gt;. It is better to think of electrons as occupying 'clouds' of probability than being at a single point in space. Those clouds have a range of shapes from simple spheres to multi-lobed, multi-layer things in atoms with many electrons. Many of these have some probability of the electron being anywhere inside the cloud even, in some cases, right into the nucleus. In this analogy atoms are not empty at all but are filled with some probabilistic amounts of electron. No more than two electrons can occupy any orbital which imposes some strong constraints on what can happen when the clouds of two atoms or molecules meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all the interactions we experience in normal macroscopic life involve those clouds of electrons interacting with each other. There are a range of different ways they can do this, but at the heart of things they all consist of electrostatic interactions at the atomic or molecular level where electron clouds push (or pull) each other. The low density of electron clouds doesn't matter since the quantum mechanical rules determine what can happen when the clouds meet. Normal interactions in the real world are just the macroscopic impact of these interactions.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-29T17:34:25.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7559" PostTypeId="1" AcceptedAnswerId="7565" CreationDate="2013-12-29T17:35:11.667" Score="1" ViewCount="153" Body="&lt;p&gt;Our city sprays brine on the roads in the winter to prevent people from sliding around. It works but it makes it impossible to keep vehicles clean. One thing I've noticed is that the brine sticks to the city buses outlining the frame underneath the body panels, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQg2Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing some research, these buses are made out of stainless steel to prevent rusting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a chemical reason for this happening?  &lt;/p&gt;&#xA;" OwnerUserId="4068" LastEditorUserId="865" LastEditDate="2015-10-30T16:54:54.287" LastActivityDate="2015-10-30T16:54:54.287" Title="Why does road salt reveal this bus's frame?" Tags="&lt;everyday-chemistry&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7560" PostTypeId="1" AcceptedAnswerId="9412" CreationDate="2013-12-29T19:14:18.210" Score="7" ViewCount="10560" Body="&lt;p&gt;I am trying to figure out the shape of the iodate ion or $\ce{IO3^-}$. No matter how many times I draw it, I keep getting iodine in the centre, and 3 oxygen atoms surrounding it, 2 with single bonds, and 1 with a double bond. This would make it a trigonal planar shape. It would mean that it is not a &quot;bent&quot; shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to double check on the internet, and I am now confused. On Wikipedia it says it is in fact polar and a pyramidal shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does iodine have a pair of lone electrons in its shell? Why would it not combine with one of the oxygen atoms to make a double bond?&lt;/p&gt;&#xA;" OwnerUserId="4069" LastEditorUserId="5026" LastEditorDisplayName="user4076" LastEditDate="2018-05-14T01:28:07.337" LastActivityDate="2018-05-14T01:28:07.337" Title="Shape and polarity of the iodate ion" Tags="&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="7561" PostTypeId="2" ParentId="7546" CreationDate="2013-12-29T19:34:18.490" Score="1" Body="&lt;p&gt;I might be terribly wrong, but I'm  not aware of any other industrial process involving ammonia as a reactant (fertilizers, caprolactam, amino acids) where liquid ammonia is used as a solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, liquefication at least plays a role over here (Germany) for the transport in &lt;a href=&quot;http://www.abload.de/image.php?img=dsc_9733s8udq.jpg&quot; rel=&quot;nofollow&quot;&gt;rail tank cars&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-12-29T19:34:18.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7563" PostTypeId="2" ParentId="4901" CreationDate="2013-12-29T23:47:37.157" Score="1" Body="&lt;p&gt;I wouldn't worry at all about the oxidative dimerization of cysteine to cystine when you consume freshly prepared solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lab, you'd use oxidants, such as $\ce{FeCl3}$ (in the presence of air) or hydrogen peroxide to achieve this reaction; both of which you're not likely to add to your amino acid diet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the article which you mentioned is concerned: The stabilization methods in question will only be relevant to maintain the composition of prefabricated and sterile-packed solutions over a long period of time.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-12-29T23:47:37.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7564" PostTypeId="1" CreationDate="2013-12-30T02:00:22.850" Score="0" ViewCount="218" Body="&lt;p&gt;Is it realistic to have a natural oxygen source in a cave? I am writing a story, about people who live underground in a land that is freezing outside. Heat is taken care of by geothermally heated groundwater. How do I make it realistic for these people to breath in the caves?&lt;/p&gt;&#xA;" OwnerUserId="4072" LastEditorUserId="4945" LastEditDate="2015-10-23T18:02:37.507" LastActivityDate="2015-10-23T18:02:37.507" Title="Is it realistic to have a natural oxygen source in a cave?" Tags="&lt;physical-chemistry&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7565" PostTypeId="2" ParentId="7559" CreationDate="2013-12-30T02:07:13.637" Score="3" Body="&lt;p&gt;No. the best two explanations don't need any chemistry at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look closely at the panelling on a bus like this you will find that there is physical  evidence on the outer skin about where the structural frame is. The skin immediately over the frame tends to stick out a little. These physical features are probably enough to ensure that more dirt sticks to those parts of the skin. You can see a similar effect above warm radiators when small bumps on the wall above tend to pick up a disproportionate amount of the dust in the moving air caused by the convection currents from the warm radiator, becoming darker as a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible contributor would be the fact that the vibrations in the panel skin of the bus will also be higher at the centre of each panel farthest away from the frame. This might contribute something as well.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-30T02:07:13.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7566" PostTypeId="1" AcceptedAnswerId="9446" CreationDate="2013-12-30T03:02:34.777" Score="4" ViewCount="153" Body="&lt;p&gt;Suppose that I have 10 kg of plastic/polymer that is non solvable in water. I then divide the plastic/polymer into two equal parts.  Let us name the parts A and B in order to identify them.  I then proceed to make beads from each part. I make $n_A$ beads of radius $r_A$ from part A. I also make $n_B$ beads of radius $r_B$ from part B. Here $r_A&amp;gt;r_B$ and consequently $n_B&amp;gt;n_A$. Suppose that I put the beads from part A into 10 liters of water and I do the same for the beads from part B.  Which one is more viscous, the &quot;solution&quot; from part A or from part B? Note: here $n_A,n_B$ stand for some arbitrary number.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="9961" LastEditDate="2015-10-23T16:46:14.173" LastActivityDate="2015-10-23T16:46:14.173" Title="Viscosity of two different solutions" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7567" PostTypeId="1" AcceptedAnswerId="7568" CreationDate="2013-12-30T05:59:12.977" Score="5" ViewCount="394" Body="&lt;p&gt;I want to make some homemade ink suitable for amateurs to practice calligraphy. I have access to many household chemicals, including acetic acid, sodium hydrogen carbonate, sugar, sodium chloride, and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give me an easy recipe for a colored (any color will be fine) ink that I can make at home? (preferably basic, but slightly acidic is fine too I guess)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Caramelized sugar --- nope, too expensive and smelly.&lt;/li&gt;&#xA;&lt;li&gt;Something with rust? (iron)&lt;/li&gt;&#xA;&lt;li&gt;Something like india ink. (pigments are acceptable too)&lt;/li&gt;&#xA;&lt;li&gt;Cornstarch and iodine.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="1362" LastEditDate="2014-11-07T16:40:08.030" LastActivityDate="2019-09-29T21:37:55.737" Title="How can I make homemade ink?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7568" PostTypeId="2" ParentId="7567" CreationDate="2013-12-30T08:47:06.403" Score="7" Body="&lt;p&gt;I don't have a full recipe, but if you're fine with black ink, try to boil some crushed or grounded organic material containing large amounts of gallic acid (or other hydrolyzable tannins) with rusty iron and some vinegar (acetic acid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suitable organic materials that come to my mind are oak bark or green tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should end up with an ink that darkens over time. Note that this ink will probably clog fountain pens but should be fine for calligraphic pens.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2013-12-30T08:47:06.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7570" PostTypeId="2" ParentId="7564" CreationDate="2013-12-30T17:11:37.613" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://wasg.iinet.net.au/Co2paper.html&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;, the CO2 buildup in the cave might also be a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if there was a river that ran on the surface, then oxygen and nitrogen would dissolve into the water. The stream could then pass into the cave and release some of the dissolved oxygen into the oxygen deficient environment in the cave. It would also remove the CO2. Of course, if the surface of the planet is frozen, then the river would have to avoid freezing somehow. But perhaps there is a hot spring on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that the rate of oxygen dissolving out of the water would be very slow and it could be increased by increasing the surface area of the water. In the human lungs, there are many, many tiny branches (Alveoli) that allow the lungs to have a huge surface area so that the slow diffusion of oxygen from the air into our blood can go fast enough for us to survive. Perhaps the cave-people can accomplish something similar if they let their oxygen-rich stream pass over a tall, broad waterfall that would cause the river to break up into tiny droplets, thereby increasing the surface area. This has the added benefit of looking super cool in case your story gets made into a movie :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other options: &#xA;1) they could generate electricity and use that to make hydrogen and oxygen from water. &#xA;2) &lt;a href=&quot;http://www.mpg.de/621120/pressRelease201003241?filter_order=L&quot; rel=&quot;nofollow&quot;&gt;http://www.mpg.de/621120/pressRelease201003241?filter_order=L&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4037" LastActivityDate="2013-12-30T17:11:37.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7571" PostTypeId="1" AcceptedAnswerId="7610" CreationDate="2013-12-30T17:24:12.413" Score="18" ViewCount="2926" Body="&lt;p&gt;We know that pure water does not conduct electricity, but salt water is a decent conductor. This is commonly explained by saying that “the ions carry the current through the solution”, which is an incomplete explanation, because it does not  indicate what happens when all of the ions have migrated to the electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complete explanations of conduction through a salty solution (&lt;a href=&quot;https://physics.stackexchange.com/questions/52267/why-does-a-salt-solution-conduct-electrical-current&quot;&gt;like this one&lt;/a&gt; or &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#aq&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;) rationalize the conduction of electricity in terms of a reduction reaction that takes place at the anode and an oxidation reaction that takes place at the cathode. In the case of salt water, chlorine gas ($\ce{Cl2}$) is formed at the anode and hydrogen gas ($\ce{H2}$) is formed at the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation seems reasonable, but it implies that the conduction of electricity through a solution is fundamentally different than the conduction of electricity through a wire. A copper wire is (usually) unchanged, even after a large amount of electricity passes through it. In contrast, when electricity passes through salt water, two chemical reactions occur (one at each electrode), which fundamentally change the composition material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that it is not possible for an aqueous solution to conduct electricity forever. Since we are driving a chemical reaction, we are either consuming our salt (forming $\ce{Cl2}$ gas in the case of an NaCl solution or plating it onto the electrodes in other cases), or we are consuming the water by forming $\ce{H2}$ or $\ce{O2}$ gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is surprising to me. So, I’m asking if my thinking is correct: &lt;strong&gt;is it possible for an aqueous salt-containing solution to conduct electricity forever&lt;/strong&gt;, or it will it always eventually consume the reactants and stop conducting electricity as I have surmised?&lt;/p&gt;&#xA;" OwnerUserId="4037" LastEditorUserId="4037" LastEditDate="2018-05-06T15:46:30.230" LastActivityDate="2018-05-06T15:46:30.230" Title="Can an aqueous solution conduct electricity forever?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;water&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7572" PostTypeId="1" AcceptedAnswerId="7573" CreationDate="2013-12-30T18:18:19.460" Score="1" ViewCount="219" Body="&lt;p&gt;Do you know some books or articles that elaborate on the chemistry behind fertilizers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken organic chemistry so far, and  inorganic chemistry as well. I am also familiar with the Haber-Bosch process. &lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="30836" LastEditDate="2019-06-22T13:46:40.387" LastActivityDate="2019-06-22T13:46:40.387" Title="The science behind fertilizers" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7573" PostTypeId="2" ParentId="7572" CreationDate="2013-12-30T22:30:24.743" Score="4" Body="&lt;p&gt;Asides from a general search on &lt;a href=&quot;http://scholar.google.com&quot; rel=&quot;nofollow&quot;&gt;Google Scholar&lt;/a&gt;, where, unfortunately, many resources are behind paywalls, here are some book references that may be of use for your broad-ranging request:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methods in Agricultural Chemical Analysis: A Practical Handbook&lt;/strong&gt; (Google eBook) (Faithfull, 2002) - some of this is available online through Google Books&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fertilizer Manual&lt;/strong&gt; (UNIDO, 1998) - this, according to its contents has some information about the chemistry of frtilizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemistry For Agriculture And Ecology&lt;/strong&gt; (Mido and Satake, 1994), A lot of this book goes back to basics, but chapter 18 is specifically about the chemistry of fertilizers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Agricultural Chemicals and the Environment&lt;/strong&gt; (Hester and Harrison, 1996) - once again, has some information about the chemical processes of fertilizers, with a focus on environmental concerns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope these are useful.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2013-12-30T22:30:24.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7574" PostTypeId="1" AcceptedAnswerId="9158" CreationDate="2013-12-31T01:17:12.083" Score="6" ViewCount="1910" Body="&lt;p&gt;The oxidation state of the iron doesn't matter too much; I am just trying to make an attractive red-brown ink. Here are some instructions I prescribed to a friend.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's an approximately decent iron acetate recipe. I don't know if it'll work, though. Maybe you'll want to heat the acetic acid. The optional first step is to oxidize the iron by hydrogen peroxide or by heating in air. Pour enough heated acid to corrode the iron plenty. Wait. Filter what is now the aqueous solution of iron acetate and acetic acid through some filter paper to remove the undissolved iron. Let the (I hope so) red-brown solution dry: this is because you want to evaporate the acetic acid. When it is dry, you may dissolve what I hope to be a dark red-brown powder in as little distilled water as possible and some alcohol if you so desire. Adjust dilution to suit your hue/shading preference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would this synthesize iron (I or preferably II) acetate?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2014-03-12T21:05:01.860" Title="Can I synthesize iron acetate like this?" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;&lt;home-experiment&gt;&lt;aqueous-solution&gt;&lt;color&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7575" PostTypeId="2" ParentId="6944" CreationDate="2013-12-31T01:32:54.287" Score="16" Body="&lt;p&gt;The simple explanation is that weak intermolecular forces (the forces that make something condense to a liquid when things are cold enough) depend on the surface area (as well as many other things). But in the case of relatively similar non-polar isomers (where weak intermolecular forces are the dominant forces), the larger surface area will lead to the large force and hence the highest boilling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,2 dimethyl propane is a compact, almost spherical molecule. Pentane is long and &quot;floppier&quot; so will experience more forces between molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying, but slightly simplified, explanation of this is that intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. These lead to short-lived dipole moments that can also induce dipoles in neighboring molecules to which the original dipole is attracted. The larger the surface area, the more opportunity for such dipoles to exist and therefore a stronger force.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2017-05-28T22:03:18.117" LastActivityDate="2017-05-28T22:03:18.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7576" PostTypeId="1" AcceptedAnswerId="7723" CreationDate="2013-12-31T04:37:11.040" Score="16" ViewCount="27843" Body="&lt;p&gt;I noticed sometime ago that bubbles formed in puddles/streets whenever it rained. I am wondering if someone could tell me what could cause such bubbles. I examined the bubbles again today more closely and they didn't look &quot;soapy&quot;. I know that is not a really valid answer to rule out that leftover detergent/soap could be the cause of such bubble formation. However, the bubbles are short-lived. Furthermore, there aren't &quot;frothy&quot;. In other words, there are no clumps of small bubbles. &lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T18:04:31.907" LastActivityDate="2016-07-19T15:54:35.777" Title="Why do bubbles form when it rains?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7577" PostTypeId="2" ParentId="7486" CreationDate="2013-12-31T05:51:32.940" Score="-1" Body="&lt;p&gt;yes because there is d3s hybridisation in this complex&#xA;and in d3s hybridisation only dxy, dyz and dzx atomic orbitals can only participates in this hybridisation so only three AO's out of five are utilised in this&#xA;and according to CFT energy difference between t2g and next s orbital is less so hybridisation can be easily occurs&lt;/p&gt;&#xA;" OwnerUserId="4078" LastActivityDate="2013-12-31T05:51:32.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7578" PostTypeId="1" CreationDate="2013-12-31T07:41:13.157" Score="3" ViewCount="1830" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/7574/can-i-synthesize-iron-acetate-like-this&quot;&gt;I recently asked about the synthesis of iron (III) acetate.&lt;/a&gt; I want a rather concentrated solution to use as a calligraphy ink (which will be dissolved in alcohol/water). Will it be okay to heat the solution of iron(III) acetate, hydrogen peroxide, acetic acid, and water to evaporate all of the other reagents? That is, will heat cause the decomposition of the iron acetate?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="41328" LastEditDate="2019-01-21T16:06:56.053" LastActivityDate="2019-01-21T16:06:56.053" Title="Decomposition of iron(III) acetate?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7579" PostTypeId="1" AcceptedAnswerId="7582" CreationDate="2013-12-31T08:15:02.867" Score="5" ViewCount="842" Body="&lt;p&gt;I have learnt the ideal gas equation $PV=nRT$. However, I have some doubts regarding the units of the quantities in this equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we take the gas constant $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$, what should be the units of $P$ and $V$? &lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T14:41:20.700" LastActivityDate="2016-11-06T14:41:20.700" Title="What are the units of P and V when R is expressed in terms of J/(mol K)?" Tags="&lt;gas-laws&gt;&lt;units&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7580" PostTypeId="1" CreationDate="2013-12-31T08:46:06.850" Score="4" ViewCount="70821" Body="&lt;p&gt;In the van der Waal's formula for real gases $\left( p+\frac{a}{V^2}\right)(V-b)=nRT$, what are the dimensions of $a$ and $b$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learnt that $a$ is related to the average force of attraction between the molecules and $b$ is related to the total volume of the molecules, but I fail to understand their dimensions.&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="73" LastEditorDisplayName="user4076" LastEditDate="2014-08-23T03:18:04.310" LastActivityDate="2014-08-23T03:18:04.310" Title="What are the dimensions of a and b in van der Waal's real gas formula?" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7581" PostTypeId="2" ParentId="7579" CreationDate="2013-12-31T08:49:38.643" Score="6" Body="&lt;p&gt;Would you agree that pressure is expressed as a force acting on a surface? If so, let's juggle with units and start with the &lt;em&gt;pascal&lt;/em&gt;. &#xA;$1\ \mathrm{Pa} = 1\ \mathrm{Nm^{-2}}$&#xA;will certainly be familiar to you ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you further agree that we can express the &lt;em&gt;joule&lt;/em&gt; as $1\ \mathrm J = 1\ \mathrm{Nm}$?&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T14:29:56.993" LastActivityDate="2016-11-06T14:29:56.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7582" PostTypeId="2" ParentId="7579" CreationDate="2013-12-31T08:53:13.220" Score="5" Body="&lt;p&gt;The units for pressure $P$ are pascals ($\mathrm{Pa}$) and volume $V$ is metres cubed ($\mathrm m^3$). The proof can be found by the derivation provided from the University of Waterloo's page &lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/c120/idealgas.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Ideal Gas Law&lt;/a&gt;, reproduced below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the formula to make $R$ the subject, hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = \frac{PV}{nT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rewrite the Ideal Gas Law in terms of its units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{J\ mol^{-1}\ K^{-1}} = \frac{\mathrm{Pa\ m^3}}{\mathrm{mol\ K}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{Pa\ m^3\ mol^{-1}\ K^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $\mathrm{Pa} = \mathrm{N\ m^{-2}}$, rewrite that into the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{N\ m^{-2}\ m^3\ mol^{-1}\ K^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the term $\mathrm{N\ m^{-2}\ m^3}$ can be simplified to $\mathrm{N\ m} = \mathrm J$  as work (unit = $\mathrm J$) = force x distance ($\mathrm{N\ m}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the right hand side becomes $\mathrm{J\ mol^{-1}\ K^{-1}}$ as required&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T14:26:11.107" LastActivityDate="2016-11-06T14:26:11.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7583" PostTypeId="2" ParentId="6447" CreationDate="2013-12-31T13:07:34.350" Score="3" Body="&lt;p&gt;You have to be careful to distinguish between the formal structures we draw to count electrons and the actual structure of the molecule. In reality the 4 bonds are the same in the perchlorate ion. But when we are counting electrons we find it bothersome to put fractional charges on atoms because that makes adding up harder. And we want to have single or double bonds for the same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we draw complete bonds which have two electrons each and balance out the total number of electrons with charge. Moreover the charge should be on the more electronegative atom, hence it goes on oxygen. However, this formalism gives an incomplete picture of the true molecular orbitals that make up the bonding (crudely, nature isn't as fussy as us about doing calculations with fractions and is happy distributing that negative charge over all the oxygens). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more sophisticated version of our whole-number accounting formalism recognises this via an extension of the simple formalism to include &quot;resonance&quot;. Here we pretend that a number of structures &quot;resonate&quot; to give the true structure. In this case we can draw the single bond and negative charge on each of the four oxygens and pretend that each of those structures contributes equally to the true molecule structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of resonance is often used to bridge the simple accounting model where bonds have two electrons, with real world structures where the electrons and bonds are spread across several atoms.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2013-12-31T13:07:34.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7584" PostTypeId="2" ParentId="911" CreationDate="2013-12-31T13:17:35.987" Score="4" Body="&lt;p&gt;This is a part answer - an annotated short list of references regarding burst nucleation and mechanisms of nanoparticle growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/0707.2380&quot; rel=&quot;nofollow noreferrer&quot;&gt;Model of nanocrystal formation in solution by burst nucleation and diffusional growth&lt;/a&gt; (Robb and Privman, 2007), in this article the authors&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;present a model with the assumptions of instantaneous thermalization below the critical nucleus size and irreversible diffusive growth above the critical size&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A main feature of this article is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for the first time formulates LaMer's explanation of burst nucleation in a manner allowing quantitative calculations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is followed up by one of the authors the next year with the article &lt;a href=&quot;http://arxiv.org/abs/0806.4644&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diffusional Nucleation of Nanocrystals and Their Self-Assembly into Uniform Colloids&lt;/a&gt; (Privman, 2008), where he &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;review theoretical explanation of mechanisms of control of uniformity in growth of nanosize particles and colloids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An explanatory diagram of LaMer's explanation (referred to in the 1st article) is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pHjP4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is from the article &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2013/cs/c2cs35289c#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Controlled synthesis of colloidal silver nanoparticles in organic solutions: empirical rules for nucleation engineering&lt;/a&gt; (Sun, 2012) - unfortunately, the article is behind a paywall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a sample chapter explaining the process in full entitled &lt;a href=&quot;http://www.wiley-vch.de/books/sample/3527316647_c01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Growth of Nanocrystals in Solution&lt;/a&gt; (Viswanatha and Sarma), provides a greater detail about the processes involved with nanoparticle growth.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2013-12-31T13:17:35.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7585" PostTypeId="2" ParentId="7580" CreationDate="2013-12-31T14:52:37.200" Score="6" Body="&lt;p&gt;Since the van der Waals equation is derived from the ideal gas equation, the empirical constants $a$ and $b$ need to have dimensions so that the units work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal gas constant $R$ has dimensions of $\dfrac{\text{pressure}\cdot\text{volume}}{\text{moles}\cdot\text{temperature}}$ since $R=\frac{pV}{nT}$ for an ideal gas. This is one of the definitions of $R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the van der Waals equation, now $R=\dfrac{(p-\frac{a}{V^2})(V-b)}{nT}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have $(p-\frac{a}{V^2})$, the term $\frac{a}{V^2}$ needs to have units of pressure for this subtraction to not screw up the units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore \dfrac{a}{V^2}\equiv\text{pressure}\implies a\equiv \text{pressure}\cdot(\text{volume})^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we guarantee that $(p-\frac{a}{V^2})$ has units of pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(p-\dfrac{a}{V^2}\equiv (\text{pressure}-\dfrac{\text{pressure}\cdot\text{volume}^2}{\text{volume}^2} \equiv \left(\text{pressure}-\text{pressure}\cdot \dfrac{\text{volume}^2}{\text{volume}^2} \right)\equiv (\text{pressure}-\text{pressure}) \equiv \text{pressure}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant $b$ is easier. $b$ needs to have dimensions of volume so that $V-b$ still has dimensions of volume.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2013-12-31T14:52:37.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7586" PostTypeId="2" ParentId="7580" CreationDate="2013-12-31T14:59:10.693" Score="8" Body="&lt;p&gt;Part of your confusion is in the formula you have written.  According to the Tanner's General Chemistry webpage &lt;a href=&quot;http://www.tannerm.com/vander.htm&quot;&gt;The van der Waals Equation&lt;/a&gt;, if you are considering 1 mole of a substance, then the formula is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(P+a/V^2)(V-b) = RT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n$ moles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(P+an^2/V^2)(V-nb) = nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Science HQ website &lt;a href=&quot;http://www.sciencehq.com/chemistry/van-der-waals-equation.html&quot;&gt;Van der waal’s equation&lt;/a&gt;, the constants $a$ and $b$ represent pressure and volume corrections respectively.  The units are based on the formula for $n$ moles of substance (the 2nd equation above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking each term on the left hand side individually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(P+an^2/V^2)$ is the van der Waals equation equivalent for $P$ in the ideal gas equation, thus this therm must have the same unit, so using $Pa$ as the unit for pressure would require that the units of $a$ to be $Pa L^2 mol^{-2}$  (the website uses the unit $atm$, but the same idea still applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar manner, the term $(V-nb)$ is the van der Waals equivalent of $V$ in the ideal gas equation.  So to keep the unit as for volume in the second term, the unit for $b$ must be $Lmol^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to always be careful with, as there are a number of units used for both pressure and volume, it is imperative to keep these constant throughout your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, volume can be expressed as $m^3$ or $L$ and several others (hence the $L^2$ term above would be the same as $(m^3)^2$ or simply $m^6$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final document is a master list of $a$ and $b$ constants for several chemicals (once again be careful with the units - unit conversion are provided in the document), &lt;a href=&quot;http://courses.chem.indiana.edu/c360/documents/vanderwaalsconstants.pdf&quot;&gt;van der Waals constants for gases.&lt;/a&gt;. Also further explanations of the above can be found on this &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/deviation5.html&quot;&gt;Purdue University page entitled Deviations from Ideal Gas Law Behavior&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2013-12-31T15:09:59.163" LastActivityDate="2013-12-31T15:09:59.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7587" PostTypeId="1" CreationDate="2013-12-31T23:17:58.360" Score="12" ViewCount="4529" Body="&lt;p&gt;Can someone explain clearly to me why some things are explosive?&lt;/p&gt;&#xA;" OwnerUserId="4079" LastEditorUserId="402" LastEditDate="2016-11-20T20:13:54.310" LastActivityDate="2018-03-05T12:54:36.310" Title="What makes substances explosive" Tags="&lt;physical-chemistry&gt;&lt;explosives&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7588" PostTypeId="1" CreationDate="2014-01-01T10:25:44.643" Score="1" ViewCount="1168" Body="&lt;p&gt;I have a solution of only sulfate ions (no other positive ions belonging to the sulfate ions, because the positive ions are attached to the intermediate in a catalytic reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to convert the sulfate ions to sulfur trioxide by using $\ce{V^5+}$ ions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what if $\ce{V^5+}$ ions could be present and could gain 2 electrons from one sulfate ion to produce $\ce{V^3+}$ and $\ce{SO3}$ (however I'm not even sure if $\ce{SO3}$ would be formed)? Here the two negatively charged oxygen atoms will each lose 1 electron to the $\ce{V^5+}$ ion, so the oxidation state of the central sulfur atom is still $+6$, but the oxidation states of the 2 oxygen atoms are $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be possible? And if so, would it be $\ce{SO3}$ that is formed, or other sulfur oxides?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nVMHD.png&quot; alt=&quot;The mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="41328" LastEditDate="2017-08-05T18:32:35.057" LastActivityDate="2017-08-05T18:32:35.057" Title="Converting sulfate ion to sulfur trioxide?" Tags="&lt;redox&gt;&lt;solutions&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7589" PostTypeId="1" AcceptedAnswerId="7590" CreationDate="2014-01-01T11:06:46.867" Score="3" ViewCount="15559" Body="&lt;p&gt;Inspired in part by the question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1227/boiling-point-of-vodka&quot;&gt;boiling point of vodka&lt;/a&gt; - given that the exact ingredients of Coca-Cola are kept secret, based on what is known, what would a viable prediction of the boiling point of Coca-Cola be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be significantly different from related products (e.g. Diet Coke, Coke Zero, etc)?&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2018-07-04T11:27:44.717" Title="Boiling point of Coca-Cola" Tags="&lt;everyday-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7590" PostTypeId="2" ParentId="7589" CreationDate="2014-01-01T14:11:04.643" Score="13" Body="&lt;p&gt;Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling-point_elevation&quot;&gt;standard calculation&lt;/a&gt; using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-01T14:11:04.643" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7591" PostTypeId="1" AcceptedAnswerId="7594" CreationDate="2014-01-01T17:43:26.837" Score="22" ViewCount="9628" Body="&lt;p&gt;When comparing the acidities of carboxylic acids, we primarily see the electropositivity of the carboxylic acid carbons, i.e. we see how effectively the negative charge on the carboxylate ion is dispersed upon ionization. Electron withdrawing groups increase the acidity by stabilising the anion &lt;em&gt;via&lt;/em&gt; electron withdrawal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following two comparisons:-   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Formic acid ($\ce{HCOOH}$) and benzoic acid($\ce{C6H5COOH}$)  &lt;/li&gt;&#xA;  &lt;li&gt;Benzoic acid and 3-chloropropanoic acid ($\ce{ClCH2CH2COOH}$)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c}&#xA;\hline&#xA;\text{Species} &amp;amp; \mathrm{p}K_\mathrm{a} \\&#xA;\hline&#xA;\ce{HCOOH} &amp;amp; 3.75\mathrm{-}3.77 \\&#xA;\ce{PhCOOH} &amp;amp; 4.19 \\&#xA;\ce{ClCH2CH2COOH} &amp;amp; 3.98 \\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zirchrom.com/organic.htm&quot; rel=&quot;noreferrer&quot;&gt;(source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Why is 3-chloropropanoic acid less acidic than formic acid. The negative inductive effect of chlorine helps in dispersing the negative charge on the carboxylate ion. It is not conjugated and too far to exert positive mesomeric effect. Therefore, the negative inductive effect should increase the acidity by improving the stability of the conjugate base.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Why is benzoic acid weaker than formic acid? Unless I am mistaken, in several other places, the phenylic moiety is treated as slightly electron withdrawing, as in case of alcohol acidity, acidity of hydrogens in base catalysed reactions etc. I am unable to understand the resonance here or any other factor, that might cause this anomaly.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2016-09-09T08:38:46.607" LastActivityDate="2016-09-09T11:30:56.277" Title="Comparing acidities of substituted and aromatic carboxylic acids" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7592" PostTypeId="2" ParentId="2588" CreationDate="2014-01-01T17:45:14.597" Score="1" Body="&lt;p&gt;The answer by Ben Norris is nice but it seems to have an additional confusion.  The difference between reduction potentials of copper and zinc can be calculated based on the empirical ionization data that Ben describes, but this doesn't mean that the difference has nothing to do the electronegativity of the two different elements.  A nice discussion of this issue is provided at the following &lt;a href=&quot;http://blogs.scientificamerican.com/degrees-of-freedom/2011/10/13/the-periodic-table-and-batteries/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4082" LastEditorUserId="7448" LastEditDate="2015-07-06T20:10:05.650" LastActivityDate="2015-07-06T20:10:05.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7593" PostTypeId="2" ParentId="7591" CreationDate="2014-01-01T18:15:46.040" Score="11" Body="&lt;p&gt;Actually, the chlorine atom in 3-chloropropanonic acid does have a significant &lt;em&gt;-I&lt;/em&gt; effect; the $\mathrm{p}K_\mathrm{a}$ of propanoic acid is $4.86$. Your expectations were just a bit too high. If you want a stronger effect, you need to get a bit closer:&#xA;the $\mathrm{p}K_\mathrm{a}$ of 2-chloropropanoic acid is $2.83$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-09-09T11:30:27.917" LastActivityDate="2016-09-09T11:30:27.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7594" PostTypeId="2" ParentId="7591" CreationDate="2014-01-01T18:52:31.173" Score="17" Body="&lt;p&gt;In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Due to steric hindrance, it may be the case that the negative charge is less accessible to solvent molecules, and thus more weakly stabilized. I suspect the difference in enthalpy here is actually negligible, but I'm not able to find a reference at this time.&lt;/li&gt;&#xA;&lt;li&gt;Perhaps more importantly, there is an entropic effect, in that larger molecules tend to require a larger number of solvent molecules to solvate them. These solvent molecules form highly ordered, cage-like structures, which is entropically unfavorable. I believe similar underlying mechanisms are at work in, e.g., the hydrophobic effect, micellization, protein folding, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As Klaus Warzecha points out, if the chlorine were not as far-removed from the carboxylate moiety (since induction rapidly diminishes with growing distance), the strength of the inductive effect would trump the solvent effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of benzoic acid, there are probably similar solvent effects occurring. Additionally, the benzene ring is actually electron-donating within the $\pi$-electron system of the molecule, which can only be destabilizing to the carboxylate conjugate base. Unlike in, e.g., phenol or aniline, the excess negative charge cannot actually be delocalized into the ring. Experimentally, the electron-donating effect of the ring (equivalently, the electron-withdrawing effect of the carboxylate) is confirmed by the fact that benzoic acid is severely deactivated in electrophilic aromatic substitution reactions by comparison to ordinary benzene. The benzene ring's electron-withdrawing effect within the $\sigma$-framework (rationalized, according to valence bond theory, by the greater s-character of $\mathrm{sp^2}$ carbons) of the molecule is probably quite weak. There may also be Coulombic destabilization between the strong $\delta^+$ of the carboxylate carbon atom and the $\mathrm{sp^2}$ carbons of the aromatic ring, again due to greater s-character, although that's speculative.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="16683" LastEditDate="2016-09-09T11:30:56.277" LastActivityDate="2016-09-09T11:30:56.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7595" PostTypeId="1" CreationDate="2014-01-01T20:03:07.503" Score="7" ViewCount="12602" Body="&lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(s) | Cr^3+ || Cd^2+ | Cd(s)}$&lt;/span&gt; galvanic cell has redox half reactions:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\ce{Cd^2+ + 2e- &amp;amp;-&amp;gt; Cd(s)} &amp;amp;\quad E^\circ &amp;amp;= \pu{-0.4 V}\tag{R1}\\&#xA;\ce{Cr^3+ + 3e- &amp;amp;-&amp;gt; Cr(s)} &amp;amp;\quad E^\circ &amp;amp;= \pu{-0.7 V}\tag{R2}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Balanced reaction equation then is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{3 Cd^2+ + 2 Cr(s) -&amp;gt; 2 Cr^3+ + 3 Cd(s)}\tag{R3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{cell} = E_\mathrm{red} + E_\mathrm{ox} = \pu{-0.4 V} + \pu{0.7 V} = \pu{0.3 V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Why is &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{cell}^\circ$&lt;/span&gt; independent of molar coefficients &lt;span class=&quot;math-container&quot;&gt;$(3,2)?$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4083" LastEditorUserId="41328" LastEditDate="2021-01-29T03:27:47.863" LastActivityDate="2021-01-29T12:54:22.950" Title="Why can we disregard stoichiometric coefficients when computing reduction potentials?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;reduction-potential&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="7596" PostTypeId="2" ParentId="7428" CreationDate="2014-01-01T20:11:04.417" Score="2" Body="&lt;p&gt;It depends what you mean by &quot;exactly&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single gram of salt contains about 10&lt;sup&gt;22&lt;/sup&gt; atoms of sodium and chlorine. The important question is how much of an imbalance would be physically &lt;em&gt;noticeable&lt;/em&gt;. If you can't detect the difference it isn't significant. A single atom discrepancy would be impossible to detect with the best equipment available to science. So it doesn't really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly most real world salt will contain a great deal of other stuff. Standard table salt might be 99.9% pure (meaning it could contain 10&lt;sup&gt;19&lt;/sup&gt; atoms of say, K or Br). But, for most purposes this is very pure and nobody will notice the extraneous elements. Sea salt is sometimes only 98% NaCl and we barely notice that difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some context consider one of the purest objects ever made: the silicon sphere being developed as a new, improved, standard for weight. We have extremely good methods for making pure silicon (because of its importance in the semiconductor industry). As &lt;a href=&quot;http://www.nature.com/news/2007/070528/full/news070528-2.html&quot; rel=&quot;nofollow&quot;&gt;this nature article&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The researchers spent six months eliminating contaminating elements by repeatedly melting the silicon in an apparatus that does not touch the material. The resulting crystal is thought to contain &lt;strong&gt;one foreign atom to every 10 million atoms of silicon.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This state of the art level of purity would allow a gramme of salt to have about 10&lt;sup&gt;15&lt;/sup&gt; atoms of impurities. Doing better than that is going to be tough to measure: a larger mismatch in atom count is going to be totally insignificant in any practical situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might notice a significant effect if there were a significant imbalance of positive and negative ions, but it wouldn't last long as any crystal would attract opposite charged ions from the environment to make it neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall salt contains the same number of Cl and Na ions within any practical limits of detection or effect.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2014-01-05T19:51:16.973" LastActivityDate="2014-01-05T19:51:16.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7597" PostTypeId="2" ParentId="7233" CreationDate="2014-01-01T21:00:32.453" Score="6" Body="&lt;p&gt;The problem you are having with the formula is based on a misunderstanding about how far you can take molecular and structural formulae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compounds that form distinct molecules it is often worth writing the molecular formula in a way that helps you understand the structure of the molecule. But this is a convenience not a generalisation that can apply to all possible compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't apply to $\ce{SiO2}$ because there is no such thing as an silica &lt;em&gt;molecule.&lt;/em&gt; Silica, like many other minerals, is a 3D network of bonds with no discrete molecular components. In silica each Silicon is bonded to 4 Oxygens, but each oxygen is shared with two silicons. This gives the $\ce{SiO2}$ formula. This also explains why mineralogy is harder than chemistry.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="4231" LastEditDate="2015-03-08T17:05:13.350" LastActivityDate="2015-03-08T17:05:13.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7598" PostTypeId="1" CreationDate="2014-01-01T21:20:23.137" Score="4" ViewCount="194" Body="&lt;p&gt;I was reading &lt;em&gt;&lt;a href=&quot;https://medium.com/matter/6cfeae2f4b53&quot; rel=&quot;nofollow&quot;&gt;How radioactive poison became the assassin’s weapon of choice&lt;/a&gt;&lt;/em&gt; and it opens describing how&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The men eventually left, and Andrade cleared the table. As he poured the remaining tea away, he noticed that the consistency of the liquid that tipped into the sink was strange. Gooey. He couldn’t have known it as he puzzled over its weird yellow tinge, but the man who’d been sipping the tea was a 43-year-old Russian dissident called Alexander Litvinenko, and the tea itself, draining away into the London sewers, was lethally radioactive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later in the article it mentions the total amount he ingested was about a microgram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, skipping past the part on how you wouldn't notice you are drinking gooey tea, could micrograms of polonium change the viscosity of a whole tea kettle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't say how much he drunk, or the size of the teapot, so lets take some guesses. Lets assume he drank a very small amount of tea, say 10 mL to give us a decent upper bound on the concentration of the solution. So that would be 0.1 micrograms/mL or about 0.5 micromolar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a solution in this concentration range change the viscosity of water enough for a casual observer to notice?&lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="-1" LastEditDate="2014-04-22T16:29:49.063" LastActivityDate="2014-04-22T16:29:49.063" Title="Could micrograms of polonium really change the viscosity of a whole teapot?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7599" PostTypeId="1" CreationDate="2014-01-02T02:55:54.197" Score="1" ViewCount="81" Body="&lt;p&gt;Does EDTA take $\ce{Mg^{2+}}$ and $\ce{Mn^{2+}}$ out of solution, so that it can be removed by a filter? Or does it just take it out of any chemical reactions going on in the solution?&lt;/p&gt;&#xA;" OwnerUserId="658" LastEditorDisplayName="user4076" LastEditDate="2014-01-02T03:50:20.270" LastActivityDate="2014-02-01T14:19:13.727" Title="Does chelation take ions out of solution" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7600" PostTypeId="2" ParentId="4135" CreationDate="2014-01-02T04:02:57.010" Score="8" Body="&lt;p&gt;Ethyl nitrite is rated 4-4-4 but it does not have anything on the white square.&#xA;Pentaborane has a good 4-4-4-W. However it's not an oxidizer.&lt;/p&gt;&#xA;" OwnerUserId="4086" LastActivityDate="2014-01-02T04:02:57.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7601" PostTypeId="1" CreationDate="2014-01-02T05:38:11.453" Score="3" ViewCount="150" Body="&lt;p&gt;If I connect a zinc half cell (Left hand side) and a copper half cell (right hand side), I get an E cell value of 1.1V taken from the voltmeter. If I replace the voltmeter with a light bulb, I will get light and heat energy of the bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that the above set up is equivalent to adding Zinc to copper sulphate solution. However, in this case, there is no wire for the electron to flow...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MY QUESTION:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Then, where does the energy go when I add Zinc to copper sulphate solution (that in the first scenario I converted to electrical energy)? Is it released as heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.As the electrons flow from the Zinc half cell to the copper half cell through the light bulb, will the voltage between the electrodes decrease as the charges on the electrodes are supposed to decrease. Then what would the formula for total energy released be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if any of my concepts are wrong. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="2595" LastEditDate="2014-01-02T09:16:02.307" LastActivityDate="2014-01-03T14:17:12.543" Title="Intuition for electrode potential" Tags="&lt;electrochemistry&gt;&lt;electrons&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7602" PostTypeId="1" AcceptedAnswerId="7612" CreationDate="2014-01-02T08:42:23.900" Score="3" ViewCount="33092" Body="&lt;p&gt;According to Werner's theory, metal ions have two types of valency-primary and secondary, where the primary valency is said to be satisfied by negative ions only, and  the secondary valency can be satisfied by positive ligand, negative or neutral molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question here is:&#xA;While formation of complex compounds, which of these valencies is satisfied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm not able to interpret from this is whether there is a 'possibility' for both the valencies being satisfied or is it necessary that both are satisfied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if the later is true how is it possible? Can it be explained taking some example as reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm not able to interpret it and understand its details.&lt;/p&gt;&#xA;" OwnerUserId="2003" LastEditorUserId="4945" LastEditDate="2020-01-24T14:27:47.153" LastActivityDate="2020-01-24T14:30:34.087" Title="Werner's theory and complex salts" Tags="&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7603" PostTypeId="1" AcceptedAnswerId="7604" CreationDate="2014-01-02T10:54:15.830" Score="4" ViewCount="518" Body="&lt;p&gt;The Belousov-Zhabotinsky reaction is a very fascinating one. The Chemistry behind the reaction concerns thermodynamics and equilibrium. I want to perform the experiment, but I'm not able to find detailed instructions about the reaction. Like, proportions of the chemicals, explanation of the reaction, duration of the reaction, sequence of mixing the reagents, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.youtube.com/watch?v=IBa4kgXI4Cg&quot; rel=&quot;nofollow&quot;&gt;Belousov-Zhabotinsky reaction&lt;/a&gt; is a video I found on YouTube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that does not give any details at all. Can anyone give me a detailed reaction procedure?&lt;/p&gt;&#xA;" OwnerUserId="2321" LastActivityDate="2014-01-02T11:14:21.890" Title="Belousov-Zhabotinsky reaction" Tags="&lt;equilibrium&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7604" PostTypeId="2" ParentId="7603" CreationDate="2014-01-02T11:14:21.890" Score="5" Body="&lt;p&gt;That is indeed a fascinating reaction!  The following is a part answer, providing links to some resources that may be of use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following resources may be of assistance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://linus.chem.ku.edu/genchemlab/1892008/Click%20Here%20for%20BZ%20Reaction%20Info.htm&quot;&gt;Belousov-Zhabotinsky&lt;/a&gt; page from the Kansas Academy for Mathematics and Science Education provides a bit about the background, chemical processes and most relevant to your request, an experimental procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mary Newton of the Georgia Regent University has provided a reasonably detailed procedure on the webpage &lt;a href=&quot;http://www.aug.edu/nonlinear/bz.htm&quot;&gt;Procedure for B-Z reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly different, is the information provided in the &lt;a href=&quot;http://home2.fvcc.edu/~dhicketh/DiffEqns/spring05projects/ErikLeigh/BZ.htm&quot;&gt;Belousov-Zhabotinsky Reaction&lt;/a&gt; (Leigh) page from the Flathead Valley Community College takes a different approach, with a focus on the mathematical derivations relevant to the experiment. &lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-02T11:14:21.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7605" PostTypeId="1" CreationDate="2014-01-02T12:14:20.490" Score="10" ViewCount="11413" Body="&lt;p&gt;1,3 dithianes are used in umpolung reactions and can be deprotected by mercury containing compounds - but could anyone help explain what the exact mechanism is? Many texts seem to skim over this by explaining it is similar to acetal hydrolysis, but I'd really like a more detailed explanation as to the role of the compounds. A reaction I have seen involves $\ce{HgO}$, $\ce{HgCl2}$, $\ce{CH3CN}$ and $\ce{H2O}$. If it is similar to acetal hydrolysis, the $\ce{H2O}$ is probably to complete the final step of attacking the $\ce{S=H}$ bond, but what about the initial step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming one sulfur atom forms a bond to the $\ce{Hg}$, but which Hg compound would it be in this case, the $\ce{HgO}$ or the $\ce{HgCl2}$? What is the role of the $\ce{CH3CN}$ in this case? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="17368" LastEditDate="2016-12-05T05:51:04.837" LastActivityDate="2017-06-22T12:47:13.523" Title="Mercury assisted deprotection of dithiane" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7606" PostTypeId="2" ParentId="7599" CreationDate="2014-01-02T13:24:11.173" Score="1" Body="&lt;p&gt;No- still too small for normal filters. Those ions are in water complexes anyway. It doesn't stop (prevent) chemical reactions rather alter the thermodynamics. EDTA and other organic complexes are the common way to allow animals to absorb ions that would otherwise tend to form oxides when swallowed so its biologically (chemically) available.&lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-01-02T13:24:11.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7607" PostTypeId="2" ParentId="7605" CreationDate="2014-01-02T15:32:12.677" Score="-1" Body="&lt;p&gt;Ok, I'm speculating a bit here: The mixture of acetonitrile and water is probably just used to keep all components in solution. HgO serves as an oxidant; alone it might be able to convert the 1,3-dithiane to the corresponding 1-oxide, which doesn't necessarily results in its cleavage. $\ce{ Hg^{2+}}$ on the other hand has a high affinity to (di)thiols.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="15235" LastEditDate="2017-06-22T12:47:13.523" LastActivityDate="2017-06-22T12:47:13.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7608" PostTypeId="2" ParentId="7180" CreationDate="2014-01-02T15:43:46.343" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ideal gas law states $$pV = nRT\tag{1}\label{id-gas}$$ where $p$ is defined as the pressure, any units, $V$ is the volume in liters, $n$ is the amount of substance of gas, $R$ is the universal gas constant, and $T$ is temperature in kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information given to us is the density of $\pu{0.977 g/L}$, a pressure of $\pu{710.0 torr}$, and temperature of $\pu{100 ^\circ C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that amount of substance, $$n = \frac{M}{m},$$ where $M$ is molecular mass in $\pu{g/mol}$, and $m$ is the mass of gas in gram, we can see the unit gram cancels out and we are left with the unit mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we are given a density, $$d = \frac{m}{V} = \pu{0.977 g/L}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging into $\eqref{id-gas}$ the molecular mass and mass for $n$ we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ pV = \frac{M}{m}RT .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isolating $m$ in density, $m = d\cdot V$, and isolating $m$ in $n = m/M\Leftrightarrow m = n\cdot M$, we can combine those two equations for $m$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n\cdot M = d\cdot V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isolate $n$, and plug back into our ideal gas equation $\eqref{id-gas}$:&#xA;\begin{align}&#xA; n &amp;amp;= \frac{dV}{M}&amp;amp;&#xA; \implies&amp;amp;&amp;amp;&#xA; pV &amp;amp;= \frac{dV}{M}RT&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume cancels out, and we isolate for $M$:&#xA;$$ M = d \cdot \frac{RT}{p}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in our known values, knowing that $\pu{1 torr} = \pu{1 mm Hg}$ and $\pu{760 mm Hg} = \pu{1 atm}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = \pu{0.977 g//L} \cdot \frac{&#xA;  \pu{0.082057 L * atm // mol * K}\cdot\pu{373.15K}}{&#xA;  \pu{710 torr}\cdot \pu{mmHg//torr}\cdot\pu{{1atm}//760mmHg}}&#xA;     = \pu{32.0 g//mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And our answer has three significant figures.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2017-11-27T12:11:28.097" LastActivityDate="2017-11-27T12:11:28.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7610" PostTypeId="2" ParentId="7571" CreationDate="2014-01-02T16:32:48.917" Score="5" Body="&lt;p&gt;Since nobody has posted a counter example, I will go ahead and say that my claim is true for direct current (which is what I originally had in mind). The current drives a chemical reaction and, at some point, the reactants will run out and the solution will no longer have enough ions to effectively conduct electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully someone will post a counter-example and make things more interesting :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of alternating current, it seems possible to plate material onto one electrode and remove it from the other. When the current switches direction, this process would reverse and the electrodes would remain the same mass. In theory, this could go forever, but in reality I wonder if the electrodes would deteriorate and break due to the many cycles of deposition and erosion.&lt;/p&gt;&#xA;" OwnerUserId="4037" LastEditorUserId="4037" LastEditDate="2015-08-19T16:44:48.537" LastActivityDate="2015-08-19T16:44:48.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7611" PostTypeId="1" CreationDate="2014-01-02T17:17:53.170" Score="2" ViewCount="22111" Body="&lt;p&gt;A buffer consists of a weak acid and its salt or weak base and its salt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the ratio of weak acid and its salt in a buffer (or the ration of weak base and its salt) is equal to 1, we say that the buffer capacity is maximum. How does this tell us that buffer capacity is maximum?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="402" LastEditDate="2016-12-17T02:44:46.610" LastActivityDate="2016-12-17T02:50:59.513" Title="Why is buffer capacity at a maximum when the ratio of its components is 1?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2017-11-09T01:17:46.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="7612" PostTypeId="2" ParentId="7602" CreationDate="2014-01-02T17:55:01.883" Score="6" Body="&lt;p&gt;A certain number of ions, atoms or molecules closely associate around a central atom leading to the formation of distinct entity called &lt;em&gt;coordination complex&lt;/em&gt;.&#xA;The groups atoms or molecules linked to the central atom are said to be coordinated with the latter and their number gives the &lt;em&gt;coordination number&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;non-ionisable&lt;/em&gt; units lie in the primary zone or sphere, whereas the  &lt;em&gt;ionisable&lt;/em&gt;  units lie in the secondary sphere and appear as ions in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hexammine platinic chloride &lt;span class=&quot;math-container&quot;&gt;$\ce{[Pt(NH3)6]Cl4}$&lt;/span&gt;  and in potassium ferrocyanide &lt;span class=&quot;math-container&quot;&gt;$\ce{K4[Fe(CN)6]}$&lt;/span&gt; four chlorine ions of the former and four potassium ions of the latter ionize in the aqueous solution, whereas six ammonia molecules of the former and six cyanogen groups of the latter belong to the primary sphere or non-ionizable sphere and thus fail to respond to their characteristic tests or fail to ionize. Or you could also say that the chlorine ions in the former and the potassium ions in the latter occupy the primary valencies and hence are ionizable and the ammonia and cyanogen group occupy the secondary valencies  and are non-ionizable, while the &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt; atom in the latter is present in the &lt;em&gt;primary Sphere&lt;/em&gt; and thus fails to ionize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per your question during the formation of a compound say &lt;span class=&quot;math-container&quot;&gt;$\ce{[Pt(NH3)5Cl]Cl3}$&lt;/span&gt;, five molecules of ammonia and the chlorine ion (linked by secondary valencies) are present in the &lt;strong&gt;primary Sphere&lt;/strong&gt; and the rest of the three chlorine atoms lie in the &lt;strong&gt;secondary sphere&lt;/strong&gt;. Thus the coordination number of the compound would be  &lt;strong&gt;six&lt;/strong&gt; as five ammonia molecules and one chlorine atom are attached to the central metal atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, every element tends to satisfy both its primary and secondary valencies. A negative ion tends to satisfy both these valencies i.e. primary as well as the secondary valencies.&#xA;The presence of a negative ion in the coordination sphere reduces the amount of charge on the complex ion by the amount of charge present on it while the negative ions in the primary sphere or the coordination sphere are not ionized.&lt;/p&gt;&#xA;" OwnerDisplayName="user1811" LastEditorUserId="4945" LastEditDate="2020-01-24T14:30:34.087" LastActivityDate="2020-01-24T14:30:34.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7613" PostTypeId="2" ParentId="7180" CreationDate="2014-01-02T18:11:01.720" Score="3" Body="&lt;p&gt;No! The answer $\pu{29.9 gm/mol}$ is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to solve these type of problems is Ideal Gas Equation. You need to do a little modification as provided in answer by Rauru Ferro, i.e.&#xA;$$P V = n R T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ n = m / M $, we can also write above as&#xA;$$P V = \left( \frac{m}{M} \right) R T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since density $d = m/V$,&#xA;$$P = \left( \frac{d}{M} \right) R T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can easily apply this modified equation in your problem.&#xA;&lt;strong&gt;Before applying this, all you need to do is to convert the pressure from torr to atm:&lt;/strong&gt;&#xA;$$\pu{1 atm} = \pu{760 torr}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also you will need to see the units of gas constant and other quantities as these type of problems seems much easier, but not-focusing-on-units can lead you to the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for your problem will be $\pu{31.9 g/mol}$ approximately.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2017-11-27T11:55:36.070" LastActivityDate="2017-11-27T11:55:36.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7614" PostTypeId="2" ParentId="7180" CreationDate="2014-01-02T21:44:31.693" Score="2" Body="&lt;p&gt;Density is $\pu{0.977 g/l}$ at specified conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the density if you increase the pressure from $710$ to $760$?  This tells you to multiply by $\frac{760}{710}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the density if you cool it from $373$ to $273$?  This tells you to multiply by $\frac{373}{273}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\pu{1 l}$  has this mass, what is the mass of $\pu{22.414 l}$?&lt;/p&gt;&#xA;" OwnerUserId="3668" LastEditorUserId="4945" LastEditDate="2017-11-27T11:51:54.230" LastActivityDate="2017-11-27T11:51:54.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7615" PostTypeId="2" ParentId="7598" CreationDate="2014-01-02T21:50:30.803" Score="6" Body="&lt;p&gt;I think it is highly unlikely polonium makes water notably more viscous. The concentration is just too low, and radiation doesn't make water more viscous, controlrods are cooled using water and they produce a lot more radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what can happen in the case of tea is that the radiation induces the formation of free radicals that in turn initiate polymerization or other reactions that form larger molecules which would make the solution more viscous, this still sounds highly improbable though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are multiple studies on radiation initiated polymerisation:&#xA;&lt;a href=&quot;http://large.stanford.edu/courses/2012/ph241/weil1/&quot;&gt;http://large.stanford.edu/courses/2012/ph241/weil1/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4090" LastActivityDate="2014-01-02T21:50:30.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7616" PostTypeId="1" CreationDate="2014-01-03T01:01:58.717" Score="2" ViewCount="199" Body="&lt;p&gt;As I understand it, the branch of chemistry dealing with identifying substances present in a sample is called &quot;analytical chemistry&quot;. This branch looks vast and complicated, with &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_chemical_analysis_methods&quot; rel=&quot;nofollow&quot;&gt;lots and lots of individual techniques&lt;/a&gt; applicable in specific circumstances. I'm hardly even a beginner in chemistry, so it's very difficult to get a good grasp of what can be done with current technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;em&gt;in general&lt;/em&gt;, how well can one identify the substances in a completely unknown sample? Does one need to have at least a vague idea of what's in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differently put, let's say I give you a container with a mixture of ten &lt;em&gt;randomly chosen&lt;/em&gt; chemicals from the entire CAS Registry, and suppose they didn't react with each other. How difficult would it be to identify all ten of them?&lt;/p&gt;&#xA;" OwnerUserId="1205" LastActivityDate="2014-01-03T13:08:30.937" Title="What are the limits of analytical chemistry?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7617" PostTypeId="1" CreationDate="2014-01-03T01:18:13.277" Score="4" ViewCount="489" Body="&lt;p&gt;As an artist, some of the work I do involves the use of seashells, sea rocks/stones, etc. &#xA;I picked up a black rock from shore at one of the beaches here in Florida and it looked somewhat clumpy with other rocks and debris encrusted on it. As I do with everything I collect, I thoroughly washed all the rocks with fresh water and then placed them in a bucket filled with water and a good dose of bleach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After drying the rocks, I noticed that the black one had small, reddish, powdery pimples all over it. It looked odd, so I broke the rock in half with a hammer,but there was nothing red inside. It actually looked like carbon, but the rock isn't brittle or powdery. I stuck the two halves back in a bucket, filled it half way with water, and them put more bleach in it. The water turned dark, almost muddy, and a reddish, almost gelatinous precipitate appeared and it was quite a bit of it. Any idea what the heck is in that rock that can cause that? I didn't notice any odor other than the bleach, albeit the rock did smell a  bit sulfury. Any idea what I have soaking in the bucket?&lt;/p&gt;&#xA;" OwnerUserId="4092" LastEditorUserId="8083" LastEditDate="2016-02-02T11:22:24.730" LastActivityDate="2016-02-02T11:39:35.330" Title="What's the reddish powder that oozes out of a black rock?" Tags="&lt;everyday-chemistry&gt;&lt;precipitation&gt;&lt;geochemistry&gt;&lt;rocks&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7618" PostTypeId="1" AcceptedAnswerId="7622" CreationDate="2014-01-03T01:38:36.593" Score="1" ViewCount="12838" Body="&lt;p&gt;I know that the first shell can hold up to 2, the second and third can hold up to 8, and the fourth 18. But what about after that?&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="23561" LastEditDate="2019-11-09T19:12:42.710" LastActivityDate="2019-11-09T19:12:42.710" Title="How many electrons can each shell in an atom hold?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7619" PostTypeId="1" AcceptedAnswerId="7621" CreationDate="2014-01-03T01:43:49.053" Score="2" ViewCount="1203" Body="&lt;p&gt;Besides alpha, beta, and gamma, are there any other types of radioactive decay? If so, what are they? Is there any type of radioactive decay that is more powerful than gamma?&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="10185" LastEditDate="2016-07-16T16:17:04.510" LastActivityDate="2016-07-16T16:17:04.510" Title="Types of radioactive decay" Tags="&lt;atoms&gt;&lt;nuclear&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7620" PostTypeId="2" ParentId="7617" CreationDate="2014-01-03T01:46:02.147" Score="2" Body="&lt;p&gt;Is your rock magnetic?  It might not be, but what you are reporting sounds like iron oxide (i.e. rust), and so your rock could be an iron ore like taconite, or even magnetite.  Such an ore could look like carbon.  Even if not magnetic, it could still be an iron ore.  &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-01-03T01:46:02.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7621" PostTypeId="2" ParentId="7619" CreationDate="2014-01-03T02:28:32.273" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radioactive_decay#Decay_modes_in_table_form&quot; rel=&quot;nofollow&quot;&gt;Here's a list&lt;/a&gt; of types of radioactive decay. The most notable types of decay that are not among the classic three involve the direct emission of a free proton or a neutron, the emission of atomic clusters other than helium nuclei (alpha particles), the absorption of one of the innermost shell electrons into the nucleus, or spontaneous fission of the unstable nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the question about gamma ray energies, it is first important to notice that &quot;gamma ray&quot; is a generic term that refers to any electromagnetic energy (photons) coming from a nucleus or subatomic particle decays, and has no direct reference to the energy of the photon (&lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_ray#Naming_conventions_and_overlap_in_terminology&quot; rel=&quot;nofollow&quot;&gt;though in the past it did&lt;/a&gt;). For example, a nuclear excited state of thorium-229, $^{229m}\ce{Th}$, decays through the emission of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_thorium#Thorium-229m&quot; rel=&quot;nofollow&quot;&gt;gamma rays with an energy of 7.6 eV&lt;/a&gt;, which corresponds to photons in the ultraviolet region of the electromagnetic spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the answer depends on exactly what you mean by &quot;power&quot;. Assuming you're comparing kinetic energies, Wikipedia says &lt;a href=&quot;http://en.wikipedia.org/wiki/Gamma_ray&quot; rel=&quot;nofollow&quot;&gt;gamma rays from nuclear decay rarely have energy above 10 MeV&lt;/a&gt;, and another source says &lt;a href=&quot;http://www.nucleonica.net/wiki/index.php?title=Beta_decay&quot; rel=&quot;nofollow&quot;&gt;electrons from beta decay usually have energies up to 4 MeV&lt;/a&gt;. The energies are all of the same order of magnitude, which makes sense because all the particles are coming from the same source (an unstable nucleus). Theoretically gamma rays can have a slightly higher kinetic energy than other decay particles because photons have zero rest mass, therefore all the decay energy is converted into kinetic energy, whilst for electrons 511 keV is &quot;wasted&quot; on creating the rest mass &lt;strike&gt;(and even higher &quot;wastes&quot; for heavier particles such as protons/neutrons/alpha particles etc)&lt;/strike&gt;. I think one can also consider that some energy is used up in &quot;climbing out&quot; of the electrical potential and residual strong nuclear potential wells of the nucleus, neither of which applies to gamma ray photons (though positively charged species would in fact gain energy, being electrically accelerated away from the nucleus by repulsion, but this contribution is probably smaller than the nuclear force one). However, different types of nucleus have different allowed decay paths, and depending on complicated details the most energetic beta particles or proton/neutron emissions might have a slightly higher energy than the most energetic nuclear decay gamma ray. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note, it is also possible to create even more energetic photons than any that could be produced via nuclear decay by putting electrons in a sufficiently powerful particle accelerator. Any charged particle under acceleration/deceleration loses some energy in the form of electromagnetic radiation. For ultrarelativistic charged particles, the photons emitted can be of extremely high energy. These are a type of x-ray termed &quot;hard x-rays&quot; (more energetic than so called &quot;soft x-rays&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Correction regarding energy lost in rest masses of particles from decay. Energy is used up in the creation of electrons, but protons/neutrons/clusters already exist inside the nucleus, and so are not created. These particles could never be produced directly as a single one would have upwards of 930 MeV, far above typical nuclear decay energies.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-01-03T10:28:48.860" LastActivityDate="2014-01-03T10:28:48.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7622" PostTypeId="2" ParentId="7618" CreationDate="2014-01-03T03:43:59.347" Score="7" Body="&lt;p&gt;$$\text{Electron Capacity for a shell} = 2n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, n represents the Principal Quantum Number, a.k.a. energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level because beyond the fourth energy level, you would get into theoretical subshells.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://education.jlab.org/qa/electron_number.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2015-03-10T20:04:47.817" LastActivityDate="2015-03-10T20:04:47.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7623" PostTypeId="1" AcceptedAnswerId="7626" CreationDate="2014-01-03T05:30:29.277" Score="3" ViewCount="10987" Body="&lt;p&gt;Apparently it has $sp^3$ hybridization, but I don't understand why. Ammonia ($\ce{NH3}$) seems to me to not require $sp$ hybridization because all of its bond lengths are already equal. It has 3 hydrogens bonded to the $p$ orbitals. Why can't the lone electron pair in nitrogen's $2s$ shell stay put where it is?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="4231" LastEditDate="2015-03-08T17:04:59.447" LastActivityDate="2015-03-08T17:05:02.757" Title="Why does ammonia have sp hybridization?" Tags="&lt;molecules&gt;&lt;electronic-configuration&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7626" PostTypeId="2" ParentId="7623" CreationDate="2014-01-03T07:45:30.660" Score="5" Body="&lt;p&gt;Well, I can think of these reasons that can justify the hybridization in ammonia molecule.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bond angle&lt;/strong&gt;: If the molecule did not have hybrid orbits, and instead had unhybridized p-orbitals taking part in the bond formation, then the bond angle between the orbitals would be 90 degrees. And as for the real situation the bond angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pair-lone pair repulsion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Energy&lt;/strong&gt;: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost equal, thereby increasing the stability of the molecule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Geometry&lt;/strong&gt;: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different from that of the orbitals containing bond pair of electrons. Also if hybridization would not be taking place, explaining the geometry of the molecule, that is it's trigonal pyramidal shape, would not be possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2003" LastActivityDate="2014-01-03T07:45:30.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7627" PostTypeId="2" ParentId="7623" CreationDate="2014-01-03T07:52:32.907" Score="3" Body="&lt;p&gt;This question has been answered before. Look here: &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20130803221208AAYOCBr&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20130803221208AAYOCBr&lt;/a&gt;&#xA;If you look at the MO diagram for $\ce{NH_3}$ you'll find that the orbital which the lone pair resides in is higher in energy (in the $2s$ antibonding orbital &lt;a href=&quot;http://www.d.umn.edu/~pkiprof/ChemWebV2/Bonding/MO-ammonia/index.html&quot; rel=&quot;nofollow&quot;&gt;http://www.d.umn.edu/~pkiprof/ChemWebV2/Bonding/MO-ammonia/index.html&lt;/a&gt;). Also, if I recall well, hybridization is not entirely correct. In hybridization, you typically assume that all of the hybridized molecular orbitals are degenerate. This is not always the case due to the different types overlap of the orbitals needed to create the molecular orbitals. Some hybridized molecular orbitals (even the ones of the same &quot;type&quot; such as $sp^3$ or $sp^2d$, etc) are higher in energy even though all of them are bonding molecular orbitals. You'll learn more of this in the second semester of inorganic chemistry.  I think , ammonia is not really $sp^3$ due to the poor overlap (methane has a better overlap between the $s$ orbitals and $p$ oribitals). Take it with a grain of salt. It has been a while since I took anything related to MO's.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="4231" LastEditDate="2015-03-08T17:05:02.757" LastActivityDate="2015-03-08T17:05:02.757" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7628" PostTypeId="2" ParentId="7616" CreationDate="2014-01-03T08:05:01.823" Score="2" Body="&lt;p&gt;I would prefer questions that are more specific, rather than broad. Nevertheless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends. I think you would first need to separate the compounds from each other.You could try    chromatography (I believe that some exotic chromatographs can even separate enantiomers). You could then try to use NMR to determine its connectivity. You could also use atomic emission spectroscopy along with mass spectroscopy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of the sample you have is important as well. Many of the analytical tools destroy the sample. In addition, not knowing how many compounds are present in a mixture makes the process more difficult since you wouldn't be able to tell whether the separation process was effective. &#xA;Regardless, it would be awfully expensive to do so.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastActivityDate="2014-01-03T08:05:01.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7629" PostTypeId="1" CreationDate="2014-01-03T10:41:42.387" Score="3" ViewCount="514" Body="&lt;p&gt;During a discussion of NMR, my lecturer told me to think about which amino acids contain protons in an aromatic ring, and whether any are likely to be affected by changes in pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 aromatic amino acids are His, Phe, Tyr and Trp. I've marked Histidine as such in my notes, but can't remember why I thought this at the time. Contrary to this, Wikipedia states explicitly that “&lt;em&gt;the imidazole ring of histidine is aromatic at all pH values&lt;/em&gt;”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also marked Histidine as &quot;+ve&quot; (as in, positive) but I'm stumped as to what I meant by it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any information on the others, but am guessing perhaps the fact that these have phenyl (rather than imidazole) rings means it's possible they respond differently to changing pH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice or hints would be appreciated, as I can't find any further information online and nothing in the structures jumps out at me to suggest an answer.&lt;/p&gt;&#xA;" OwnerUserId="2119" LastActivityDate="2014-12-11T19:29:17.220" Title="Is aromaticity of the 4 aromatic amino acids affected by pH?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7630" PostTypeId="2" ParentId="7629" CreationDate="2014-01-03T11:04:52.460" Score="5" Body="&lt;p&gt;There are two nitrogen atoms in the imidazole ring in histidine, &lt;a href=&quot;http://en.wikipedia.org/wiki/Imidazole#Amphoterism&quot;&gt;but the $sp^2$ nitrogen is significantly more basic than the $sp^3$ nitrogen&lt;/a&gt;, as the electron pair in the latter is delocalized over the ring. Protonating the $sp^3$ nitrogen would destroy the aromaticity, but that would require very strongly acidic conditions, possibly not attainable in aqueous solution. Protonating the $sp^2$ nitrogen does not affect aromaticity because the electron pair is approximately perpendicular to the plane of the ring, meaning it cannot overlap with other orbitals in the ring and therefore does not participate in mesomerism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tryptophan&quot;&gt;Tryptophan&lt;/a&gt; also has a $sp^3$ nitrogen atom in an aromatic ring, but like in the imidazole ring &lt;a href=&quot;http://en.wikipedia.org/wiki/Indole#Basicity&quot;&gt;it is not very basic&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-01-03T11:12:16.517" LastActivityDate="2014-01-03T11:12:16.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7631" PostTypeId="2" ParentId="7616" CreationDate="2014-01-03T13:08:30.937" Score="5" Body="&lt;p&gt;The trouble with your question is the use of the phrase &lt;em&gt;in general&lt;/em&gt;. Most analytical chemistry is directed towards specific goals and isn't about trying to find what is in a sample &lt;em&gt;in general&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the range of techniques applied are highly dependent on the problem you want to solve. If you want to detect, for example, pesticide residues in foodstuffs you probably already know the list of things you want to detect and the range of techniques appropriate to identifying them. If you seek the active ingredient of a herbal remedy, a different range of techniques apply. Sometimes you want to verify that a pure substance you have made is what you think it is, this involves a different range of techniques again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common thread in much analytical chemistry is separation: if you seek to identify something in a mixture you want to separate the mixture into its different components. So a variety of types of chromatography may be used in the bulk or on a micro scale. Identifying very small amounts of contaminants in food (e.g. pesticides) probably involves some form of extraction into a solvent followed by separation using analytical chromatography (gas or high pressure liquid chromatography (HPLC)). Where the range of compounds is not known in advance this can be supplemented by mass spectroscopy (the chromatography separates the components, the mass spectrograph helps identify the fragments of the molecules in each component).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know what you are looking for in advance, it isn't easy to elect the right separation technique and a great deal of experimentation is likely to be required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is that some things are much easier to detect than others. Chromatographic and detection tools like the electron capture detector can identify organohalides to below parts per trillion levels. Atomic emission spectroscopy can identify some metals easily and at low levels. Other techniques can help identify structure; NMR is (sometimes) good at identifying structures, but usually needs pure compounds in relatively large quantities. X-ray techniques can be useful for minerals. Some compounds have very distinct features in their IR spectra or their UV absorption. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is there are no &lt;em&gt;general&lt;/em&gt; techniques. There are a wide range of tools to apply depending on the circumstances and the problem you are trying to solve. Almost everything starts with separation. Then, for simple problems, you may be able to apply a single known technique to find out what you want. For harder problems, a rage of tools may be required but the tools will be fairly specific to the problem.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-03T13:08:30.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7632" PostTypeId="2" ParentId="7601" CreationDate="2014-01-03T14:08:16.717" Score="1" Body="&lt;p&gt;Ad 1) I am not sure what do you mean by &quot;equivalent&quot;. But you add Zinc in a copper sulphate solution, the &quot;energy&quot; will be consumed through &lt;a href=&quot;http://en.wikipedia.org/wiki/Cementation_%28metallurgy%29&quot; rel=&quot;nofollow&quot;&gt;cementation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to a recation Cu$^2$$^+$(aq) + Zn(s) → Cu(s) + Zn$^2$$^+$(aq)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ad 2) Please see &lt;a href=&quot;http://chemistry.about.com/od/workedchemistryproblems/a/Equilibrium-Constant-Of-An-Electrochemical-Cell-Reaction-Example-Problem.htm&quot; rel=&quot;nofollow&quot;&gt;Equilibrium Constant of an Electrochemical Cell Reaction Example Problem&lt;/a&gt; where you can find ansewer to your second question. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-01-03T14:17:12.543" LastActivityDate="2014-01-03T14:17:12.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7634" PostTypeId="2" ParentId="7571" CreationDate="2014-01-03T14:55:42.437" Score="4" Body="&lt;p&gt;It can be possible, but it depends on the experimental conditions/set up.&#xA;I think the following setup could work “forever”:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two electrodes made of $\ce{Pt}$ in solution of $\ce{CuSO4}$ and $\ce{H2SO4}$ provided with AC source, the bath must be sealed – to avoid evaporation, applying reasonable voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The copper is plating onto one electrode and depleting it to the other electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction: $\ce{Cu^2+ + 2e- -&amp;gt; Cu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation: $\ce{Cu $-$ 2e- -&amp;gt; Cu^2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrodes made form $\ce{Pt}$ will avoid change shape of the electrodes during the endless process. The additional $\ce{H2SO4}$ makes the electrolyte more conductive.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorDisplayName="user7951" LastEditDate="2015-08-15T15:12:01.263" LastActivityDate="2015-08-15T15:12:01.263" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7635" PostTypeId="1" AcceptedAnswerId="8319" CreationDate="2014-01-03T16:08:41.983" Score="2" ViewCount="299" Body="&lt;p&gt;Soaps are amphiphilic molecules that, practically, removes body oils from the surface of the skin. The body, on the other hand, always replenishes sebum through the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sebaceous_gland&quot; rel=&quot;nofollow&quot;&gt;sebaceous glands&lt;/a&gt; after a certain amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Le Chatelier's principle to this complex chemical equilibrium, where the equilibrium shifts towards the right hand side, where oil is being extracted from the equation, essentially so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(body · oil)  ⇌  body + oil       # RHS preferred&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Because the process of sebum regeneration must be endothermic, this means the body is losing energy in this overall process:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;body + (lipids + heat)  ⇌  body + oil  ⇌  (body · oil)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are my assumptions correct?&lt;/li&gt;&#xA;&lt;li&gt;Does removing oil from the body more readily, by means of a stronger detergent/soap, reduce body mass?&lt;/li&gt;&#xA;&lt;li&gt;Does the temperature of the water bath used in the experiment affect the rate of sebum dissociation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1237" LastEditorUserId="23561" LastEditDate="2018-11-19T19:05:17.313" LastActivityDate="2018-11-19T19:05:17.313" Title="Does using stonger soaps help me lose weight?" Tags="&lt;equilibrium&gt;&lt;biochemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7637" PostTypeId="1" CreationDate="2014-01-03T17:14:56.907" Score="1" ViewCount="933" Body="&lt;p&gt;Paraquat is 4,4'dipyridinium chloride. There are two common ways to synthesise it starting with pyridine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one reaction that can be done on a large scale in water or liquid ammonia, methyl-pyridinium chloride is reacted with a base and catalytic amounts of sodium cyanide in the solvent to give a dimeric intermediate that oxidises easily into paraquat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the chemical mechanism of the dimerisation reaction?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="22" LastEditDate="2014-05-08T07:36:34.217" LastActivityDate="2014-05-08T07:36:34.217" Title="The weedkiller paraquat can be produced from methyl-pyridinium chloride using a cyanide catalyst: what is the mechanism?" Tags="&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7638" PostTypeId="1" CreationDate="2014-01-03T18:18:38.913" Score="0" ViewCount="2033" Body="&lt;p&gt;$$\ce{C_{(s)} + H_2O_{(g)} &amp;lt;=&amp;gt; CO_{(g)} + H_{2(g)}}$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = +131\ \mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction is in a state of equilibrium under standard conditions, what can be done to increase the value of $K_\text{eq}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is would you have to increase or decrease the temperature to increase $K_\text{eq}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the basic principle that you have to decrease the temperature to increase $K_\text{eq}$, but in my notes that was for $\ce{A +B&amp;lt;=&amp;gt;C}$ and the reaction above is not in that form, Instead it has two products. Would this still apply? &lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T09:28:33.987" LastActivityDate="2015-04-26T11:08:38.720" Title="how to increase Keq(equilibrium) with temperatrue" Tags="&lt;equilibrium&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7639" PostTypeId="2" ParentId="7637" CreationDate="2014-01-03T19:05:58.893" Score="1" Body="&lt;p&gt;I remember PET (photoinduced electron transfer) works by the Mariano group in which N-dimethylallyl pyridinium perchlorate underwent intramolecular radical cyclizations upon uv irradiation and in fact, I've seen N-methyl chinolinium salts undergo dimerization under these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the N-methylpyridinium cation is irradiated at $\lambda$ = 300 nm. The resulting $S_1$ state oxidizes cyanide to a cyanogen radical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The N-methylpyridinium cation is thereby reduced to a &lt;strong&gt;pyridyl radical&lt;/strong&gt;.&#xA;Radical dimerization furnishes a 4-4'-dihydrobiaryl (see drawing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3qwAu.png&quot; alt=&quot;intermediate dimer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromatization to paraquat requires the loss of two hydrogen atoms and two electrons; the latter are being used to &lt;strong&gt;regenerate cyanide&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism is a bit speculative but consistent and reflects the catalytic use of cyanide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;em&gt;Note: I would have liked to add a sketch for the mechanism but BKChem died on me several time while doing so.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-03T21:58:07.580" LastActivityDate="2014-02-03T21:58:07.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7640" PostTypeId="1" AcceptedAnswerId="7644" CreationDate="2014-01-03T19:32:56.070" Score="1" ViewCount="1173" Body="&lt;p&gt;Consider the following equilibrium: $\ce{A2\ _{(g)} + 2B_{(g)} &amp;lt;=&amp;gt; 2AB_{(g)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chemist puts 8.0 moles of $\ce{A2}$ and 10.0 moles of $\ce{B}$ into an evacuated 1.0 L container. The system reaches equilibrium at a certain temp according to the process outlined above. When the system is at equilibrium the chemist determines $\ce{[AB]}$ is 8.0 M. What is the $K_c$ for this process at this temperature (278 K). I set up a nice chart and everything and I cant figure this out. My answer choices are 10, 3.0, 1.0, 4.0, and .50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I thought that The Kc was [AB]^2/([A2][B^2]) but I put in the values and didn't get any of the answers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then i set up an ice chart &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to show on here but I then ended up with [8+2x]^2/([8-x][10-2x]^2) but that didn't really get me anywhere&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="4101" LastEditDate="2014-01-03T22:55:29.850" LastActivityDate="2014-01-04T08:48:41.470" Title="Equilibrium constant calculation from concentrations" Tags="&lt;equilibrium&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7641" PostTypeId="1" CreationDate="2014-01-04T03:36:57.693" Score="9" ViewCount="185" Body="&lt;p&gt;I am reading this article &lt;a href=&quot;http://pubs.rsc.org/en/content/articlehtml/2011/cp/c1cp21527b&quot; rel=&quot;noreferrer&quot;&gt;Phys. Chem. Chem. Phys. , 2011, 13, 14902-14905&lt;/a&gt; and seen that the authors have drawn a image of defect level using PL data&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PmYgG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain me how this diagram can be drawn from PL data. And if possible can you explain this using some other simplest PL data.&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2014-01-04T03:36:57.693" Title="How to get defect-energy level diagram from PL data" Tags="&lt;spectroscopy&gt;&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7642" PostTypeId="1" AcceptedAnswerId="9614" CreationDate="2014-01-04T04:07:31.083" Score="1" ViewCount="280" Body="&lt;p&gt;Imagine an arrangement of atomic orbitals in an alternate universe, in which the s subshell contains 2 orbitals instead of 1, and the p subshell contains 4 orbitals rather than 3. In which groups would you expect the anomalously low electron affinities to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are groups 3 and 7, due to the fact that they are &quot;one away&quot; from filling up their subshell. The &lt;em&gt;groups&lt;/em&gt; would look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s: _ _ _ _  ... p: _ _ _ _ _ _ _ _&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer says 4 and 8. Is there something I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: in terms of electron configuration blanks, it would be&#xA;s: _ _ ... p: _ _ _ _&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="3747" LastEditDate="2014-01-05T05:19:27.843" LastActivityDate="2015-05-18T08:04:32.590" Title="Electron Affinity trends" Tags="&lt;inorganic-chemistry&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7644" PostTypeId="2" ParentId="7640" CreationDate="2014-01-04T08:48:41.470" Score="1" Body="&lt;p&gt;Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else do we know?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In equilibrium, $[AB]$ = 8.0M&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;For every mole of $B$, one mole of $AB$ is produced.&lt;/li&gt;&#xA;&lt;li&gt;For every mole of $A_2$, &lt;strong&gt;two&lt;/strong&gt; moles of $AB$ are produced.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With just one cup of coffee so far, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-04T08:48:41.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7645" PostTypeId="1" CreationDate="2014-01-04T08:58:23.893" Score="0" ViewCount="1626" Body="&lt;p&gt;What is the nature of the bonds between P and O in oxoacids of phosphorous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a double bond or coordinate bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says that it is &quot;double bond&quot; but many times I encountered &quot;coordinate bond&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4103" LastEditorUserId="-1" LastEditDate="2014-02-11T21:25:51.707" LastActivityDate="2016-07-23T17:05:28.737" Title="Nature of bond in oxyacids of phosphorus" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7646" PostTypeId="2" ParentId="7645" CreationDate="2014-01-04T13:38:49.507" Score="0" Body="&lt;p&gt;Numerous oxoacids of phosphorus their respective ions or salts are known.Some points which could be enumerated from all the structures are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All phosphorous atoms in the oxoacids and the oxoanions are 4-coordinated and consists of at least one P=O bond. &lt;/li&gt;&#xA;&lt;li&gt;All the phosphorous atoms in the oxoacids contain at least one P-OH bond which is ionizable.&lt;/li&gt;&#xA;&lt;li&gt;Some oxoacids have one direct P-H bond which is non-ionizable&lt;/li&gt;&#xA;&lt;li&gt;Catenation takes place via P-O-P links or direct P-P bonds. The former either exists in linear chains or cyclic species.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user1811" LastActivityDate="2014-01-04T13:38:49.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7647" PostTypeId="1" CreationDate="2014-01-04T13:54:09.293" Score="0" ViewCount="5588" Body="&lt;p&gt;$$\ce{Cu^{2+} + Zn&amp;lt;=&amp;gt;Zn^{2+} + Cu}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_c=1.9*10^{37}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{cell}=1.1V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that they both suggest that the forward reaction is going to completion but can someone please tell me a bit more about what both these values are suggesting and how these two values concord with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, many books quote such reactions in only one direction and gives a value for $E_{cell}$. However, isn't $E_{cell}$ just another way of illustrating the condition of an equilibrium (i.e. in which direction the reaction would occur if left alone)... so how can a one way reation be shown&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="73" LastEditDate="2014-01-05T00:22:25.860" LastActivityDate="2014-01-05T15:02:12.823" Title="Similarity between Ecell and equilibrium constant Kc" Tags="&lt;electrochemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7648" PostTypeId="1" CreationDate="2014-01-04T15:51:19.833" Score="21" ViewCount="14792" Body="&lt;p&gt;In my textbook, a footnote says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In case of weak acids, on dilution the total number of $\ce{H^{+}}$ ions in solution increases because dissociation of the weak acid increases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This didn't make sense to me however. Consider the general dissociation reaction of a weak acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA (aq)  + H_2O \rightleftharpoons H_3O^{+} (aq) + A^{-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Chatelier's principle tells us that if we add more reactants ($\ce{H_2O}$), the reaction will proceed more to the right, and so the dissociation of the acid will increase, but the part about the total number of $\ce{H^{+}}$ ions in water being higher was still unclear to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delving into the math a bit, we can see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_a = \frac{[\ce{H_3O^{+}}][\ce{A^{-}}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the concentrations of the two products are equal, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H_3O^{+}}] = \sqrt{[\ce{HA}]\cdot K_a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us assume that we have a $1 \ M$ solution of the acid in question dissolved in 1 litre of water, plugging this into the equation, we have a $\sqrt{K_a}$ molar solution of $H^{+}$ ions. Since only 1 litre of solvent is there, we have $\sqrt{K_a}$ moles of ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, say we dump 1 litre of water into this solution. The concentration of our acid will be halved. Plugging this into the equation, we get the concentration of $\ce{H_3O^{+}}$ ions as approx. $0.7 \sqrt{K_a} \ M$. We have two litres of solvent, so we have $1.4 \sqrt{K_a}$ moles of ions, which is considerably higher. So, the numbers all work out, but what's the intuition behind it?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="16290" LastEditDate="2015-06-11T14:20:18.810" LastActivityDate="2018-12-14T23:13:34.360" Title="Does the number of H+ ions in solution go up on dilution of a weak acid?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7649" PostTypeId="1" AcceptedAnswerId="7650" CreationDate="2014-01-04T16:51:44.820" Score="1" ViewCount="79" Body="&lt;p&gt;$\ce{Heat\ +\ PCl5(g)\ \rightleftharpoons\ PCl3(g)\ +\ Cl2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the process above, which of the following can be done to increase the number of moles of $\ce{Cl2}$ in a sealed container at equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reducing temp&lt;/li&gt;&#xA;&lt;li&gt;Increasing the pressure at a constant temp&lt;/li&gt;&#xA;&lt;li&gt;Adding more moles of $\ce{PCl5}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure it's not 1. but I'm stuck between 2. and 3. Are they both true or only one of them?&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="4231" LastEditDate="2015-02-27T17:20:09.057" LastActivityDate="2015-02-27T17:20:14.313" Title="How to increase equilibrium of Cl₂" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7650" PostTypeId="2" ParentId="7649" CreationDate="2014-01-04T17:02:32.640" Score="1" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 2. Try to apply Le Chatelier's principle. Increasing pressure will make the system try to decrease it and there is one side of the equilibrium with an unequal number of moles.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2014-01-04T17:02:32.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7651" PostTypeId="2" ParentId="7033" CreationDate="2014-01-04T17:22:01.223" Score="11" Body="&lt;p&gt;Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface area to mass ratio.&lt;/p&gt;&#xA;" OwnerUserId="4105" LastActivityDate="2014-01-04T17:22:01.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7652" PostTypeId="1" CreationDate="2014-01-04T17:43:42.317" Score="2" ViewCount="1256" Body="&lt;p&gt;$\ce{2CH4(g) + O2(g) \rightleftharpoons 2CO(g) + 4H2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Equal numbers of moles of $\ce{CH4}$ and $\ce{O2}$ are drawn into a&#xA;  sealed container and the system is left until it reaches equilibrium.&#xA;  Which of the following relationships pertaining to this particular&#xA;  system are true at equilibrium?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{[CH4]&amp;lt;[H2]}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{[CH4]&amp;gt;[O2]}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{[H2]&amp;gt;[CO]}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the equation for $\ce{K_{eq}}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\dfrac{[H2]^4[CO]^2}{[CH4]^2[O2]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that 3. is wrong because if $\ce{CH4}$ and $\ce{O2}$ are equal, then $\ce{H2}$ and $\ce{CO}$ have to be the same. Now I'm confused about 1 and 2. I'm not sure how to distinguish between the two.&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="1409" LastEditDate="2014-01-04T17:51:38.747" LastActivityDate="2014-02-03T18:19:32.100" Title="Relationship in equilibrium constants" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7653" PostTypeId="2" ParentId="7652" CreationDate="2014-01-04T18:00:13.353" Score="2" Body="&lt;p&gt;Let's say that the amount of $\ce{CH4}$ and $\ce{O2}$ at the start is $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the reaction is complete, let's say that $y$ moles of $\ce{O2}$ reacted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of moles of $\ce{CH4}$ left is $x-2y$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of moles of $\ce{O2}$ left is $x-y$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of moles of $\ce{CO}$ formed is $2y$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of moles of $\ce{H2}$ formed is $4y$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should make one of the choices clearly false, one clearly true and one might not be so obvious, unless you assume that the forward reaction path is less favourable ($\ce{CO}$ is not very stable)&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2014-01-04T18:00:13.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7655" PostTypeId="1" AcceptedAnswerId="7656" CreationDate="2014-01-04T22:21:16.453" Score="12" ViewCount="63234" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following must be true for adiabatic processes?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C_V = C_p$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H = 0$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta U = 0$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta S = 0$&lt;/span&gt; &lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$q = 0$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Source: Chemistry GRE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;span class=&quot;math-container&quot;&gt;$q = 0$&lt;/span&gt;. From what I can find, an adiabatic process is when there is no transfer of heat, but then why is the enthalpy change not zero?&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorDisplayName="user7951" LastEditDate="2019-10-16T20:14:06.997" LastActivityDate="2019-10-17T07:48:57.277" Title="Why is the enthalpy change not zero in an adiabatic process?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="7656" PostTypeId="2" ParentId="7655" CreationDate="2014-01-05T04:09:01.950" Score="16" Body="&lt;p&gt;The Enthalpy $H$ is defined as $H=U+PV$. Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H=\Delta U + P\Delta V +V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an adiabatic process, $q=0$. therefore from the first law of thermodynamics, &#xA;$$\Delta U = q +w =q-P\Delta V$$&#xA;$$\Delta U=w=-P\Delta V$$&#xA;Substituting this in the first equation you get, &#xA;$$\Delta H=V\Delta P$$&#xA;If $\Delta P$ is zero during the process (Isobaric) then the enthalpy change will indeed be zero, but it will be non-zero if the process is not isobaric.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="275" LastEditDate="2014-05-28T10:12:38.817" LastActivityDate="2014-05-28T10:12:38.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7657" PostTypeId="2" ParentId="6821" CreationDate="2014-01-05T08:48:34.380" Score="0" Body="&lt;p&gt;I took the liberty and corrected the title of your question, assuming that the &quot;ethylene with two methyls&quot; is 2-methylpropene (isobutene). Over here, this classic reaction is known as &lt;em&gt;Friedel-Crafts&lt;/em&gt; alkylation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-05T19:13:40.913" LastActivityDate="2014-01-05T19:13:40.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7660" PostTypeId="1" AcceptedAnswerId="7680" CreationDate="2014-01-05T10:09:41.973" Score="3" ViewCount="3810" Body="&lt;p&gt;In an exercise I have the following structures &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Moclobemide&lt;/strong&gt;: &lt;img src=&quot;https://i.stack.imgur.com/0eHku.png&quot; alt=&quot;moclobemide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Paracetamol&lt;/strong&gt;: &lt;img src=&quot;https://i.stack.imgur.com/vpjxX.png&quot; alt=&quot;paracetamol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hexobarbital&lt;/strong&gt;: &lt;img src=&quot;https://i.stack.imgur.com/veyI9.png&quot; alt=&quot;hexobarbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for example by heart that &lt;strong&gt;paracetamol&lt;/strong&gt; is an &lt;strong&gt;acid&lt;/strong&gt;. What is the best approach in determination of acidity of a compound based on it's structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know for example if these compounds are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Basic&lt;/li&gt;&#xA;&lt;li&gt;Acidic&lt;/li&gt;&#xA;&lt;li&gt;Amphoteric&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know for example that many metals (Zn, Pb, etc) can be amphoteric (act as $H^+$ donors or acceptors). But I don't know how to determine what a complex structure is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please offer a detailed explanation, with links if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards,&lt;/p&gt;&#xA;" OwnerUserId="2248" LastEditorUserId="41890" LastEditDate="2017-11-27T18:03:18.340" LastActivityDate="2017-11-27T18:03:18.340" Title="Determine compound acidity based on structure" Tags="&lt;organic-chemistry&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7661" PostTypeId="2" ParentId="7649" CreationDate="2014-01-05T10:22:25.290" Score="0" Body="&lt;p&gt;yest the answer 3. is correct. It will be even more effective to adding more $\ce{PCl5}$ and increasing temperature too.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4231" LastEditDate="2015-02-27T17:20:14.313" LastActivityDate="2015-02-27T17:20:14.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7662" PostTypeId="1" AcceptedAnswerId="7664" CreationDate="2014-01-05T11:31:32.383" Score="4" ViewCount="1029" Body="&lt;p&gt;The question asks to write the theoretical transformation steps to convert ethanol to crotonic acid in considerable yield. One constraint is that &lt;strong&gt;you cannot use any external organic compounds but inorganic catalysts and reagents are allowed and aldol condensation is prohibited.&lt;/strong&gt; I do not know the answer, but I came up with the following steps:- &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Treatment with $\ce{HBr}$ to form $\ce{CH3CH2Br}$ via $S_N2$.&lt;/li&gt;&#xA;&lt;li&gt;Treatment with $\ce{Mg}$ in ether, to form corresponding Grignard reagent.&lt;/li&gt;&#xA;&lt;li&gt;Oxidation of ethanol by Pyridinium chlorochromate to ethanal, and then treating it with the formed Grignard reaction to form $\ce{CH3CHOHCH2CH3}$. &lt;/li&gt;&#xA;&lt;li&gt;Acid catalysed dehydration to form $\ce{CH3CH=CHCH3}$.&lt;/li&gt;&#xA;&lt;li&gt;Using NBS (N-Bromo Succinimide) as an allylic halogenating agent, followed by treatment with aq. $\ce{KOH}$ to give $\ce{CH3CH=CHCH2OH}$.&lt;/li&gt;&#xA;&lt;li&gt;Oxidising this with PCC followed by Tollen's reagent to selectively oxidise alcohol to acid, to finally yield crotonic acid.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am unsure if any of these steps are incorrect, i.e if any of these steps will not give appropriate yields of the desired product owing to side reactions. Another problem is that this transformation is too long. &lt;strong&gt;Is there any shorter way to carry out this conversion in proper yields?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="10185" LastEditDate="2015-10-02T16:46:53.677" LastActivityDate="2015-10-02T16:46:53.677" Title="What route to take in converting ethanol to crotonic acid?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7663" PostTypeId="2" ParentId="6144" CreationDate="2014-01-05T11:57:38.357" Score="3" Body="&lt;p&gt;In addition to the procedures described by Ian Fang, these mixed anhydrides (with $\ce{R_1, R_2 \ne H}$) have been used to prepare symmetric anhydrides of carboxylic acid in solid state reactions. If that's interesting to anyone, I can look up the reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the effect of water is concerned, I do remember a procedure where the mixed anhydrides in question were actually prepared in a biphasic system of &lt;strong&gt;water&lt;/strong&gt; and dichloromethane in the presence of a phase-transfer catalyst. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="876" LastEditDate="2015-05-01T21:24:53.407" LastActivityDate="2015-05-01T21:24:53.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7664" PostTypeId="2" ParentId="7662" CreationDate="2014-01-05T12:10:53.940" Score="3" Body="&lt;p&gt;Allylic bromination and subsequent hydrolysis might be replaced by an allylic oxidation of the butene using selenium dioxide - if that's inorganic enough for you ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm wondering whether there's a shorter way. What about the Grignard reaction you described, followed by oxidation (PCC on silica or manganese dioxide) to 2-butenone. Subsequent bromination with bromine in hydrobromic acid should yield an $\alpha,\alpha'$-dibromoketone which, upon treatment with base (Favorskii rearrangement) and, eventually acidic workup should yield crotonic acid.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-05T14:09:42.363" LastActivityDate="2014-01-05T14:09:42.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7665" PostTypeId="2" ParentId="7647" CreationDate="2014-01-05T15:02:12.823" Score="5" Body="&lt;p&gt;$E^\circ_{cell}$ and $K_c$ are related. Both indicate which direction of the reaction is spontaneous. Both tell you something about the position of equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $E^\circ_{cell}$ is a directly measurable value. If you construct an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrochemical_cell&quot; rel=&quot;noreferrer&quot;&gt;electrochemical cell&lt;/a&gt; as using a zinc electrode immersed in a 1.0 M solution of say zinc chloride and a copper electrode immersed in a 1.0 M solution of copper (II) chloride with a salt bride of sodium chloride connecting the solutions, a voltmeter connected between the two electrodes will read the value of $E^\circ_{cell}$, at least initially. The voltage will change as the reaction progresses, but the initial voltageis $E^\circ_{cell}$ which is defined for standard conditions of 1.0 M concentration, 1 bar of pressure, and 25 $^\circ$C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn|ZnCl2}(1.0 \text{ M})\parallel\ce{CuCl2}(1.0 \text{ M})|\ce{Cu} \ \ \ \ E^\circ_{cell}=1.1\text{ V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between $E^\circ_{cell}$ and $K_c$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;noreferrer&quot;&gt;Nernst Equation&lt;/a&gt;. Note the use of both $E^\circ_{cell}$, the &lt;strong&gt;standard&lt;/strong&gt; cell potential and $E_{cell}$, the &lt;strong&gt;observed&lt;/strong&gt; cell potential under other conditions. The remainder of the alphabet soup is defined below.&#xA;$$E_{cell}=E^\circ_{cell}-\frac{RT}{zF}\ln{Q}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E^\circ_{cell}$ - Standard cell potential&lt;/li&gt;&#xA;&lt;li&gt;$E_{cell}$ - nonstandard cell potential at other conditions&lt;/li&gt;&#xA;&lt;li&gt;$R$ - the &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_gas_constant&quot; rel=&quot;noreferrer&quot;&gt;ideal gas constant&lt;/a&gt; (which appears here because it is equal to &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;noreferrer&quot;&gt;Avogadro's number&lt;/a&gt; times &lt;a href=&quot;http://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;noreferrer&quot;&gt;Boltzmann's constant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$T$ - the temperature in kelvins&lt;/li&gt;&#xA;&lt;li&gt;$z$ - the number of &lt;strong&gt;moles&lt;/strong&gt; of electrons being transferred in this redox process. If we wanted to use the number of individual electrons, then we would need to replace $R$ with $k_B$&lt;/li&gt;&#xA;&lt;li&gt;$F$ - &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_constant&quot; rel=&quot;noreferrer&quot;&gt;Faraday's constant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;$Q$ - the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_quotient&quot; rel=&quot;noreferrer&quot;&gt;reaction quotient&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We need to use $Q$ in the Nernst equation, because this equation can relate the cell potential to any range of conditions, not just equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At equilibrium $Q = K_c$ and $E_{cell} = 0 \text{ V}$ (there is no net change in the reaction composition, so there is no potential difference across the half-cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the Nernst Equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=E^\circ_{cell}-\frac{RT}{zF}\ln{K_c}$$&#xA;$$E^\circ_{cell} = \frac{RT}{zF}\ln{K_c}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To determine the values of $E^\circ_{cell}$ and $K_c$ for the reverse reaction...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For $E^\circ_{cell}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_{cell} = -E^\circ_{rev}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $K_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{rev} = \frac{1}{K_c}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-05T15:02:12.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7666" PostTypeId="1" CreationDate="2014-01-05T16:47:28.440" Score="0" ViewCount="81" Body="&lt;p&gt;I learned that deprotonation is the process in which a proton is lost from ligand $\ce{H2O}$ molecules surrounding a central metal ion (I am referring to hexaaqua ions). Also, the process continues until the charge of the complex ion is neutralized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&#xA;Why does it only continue until the charge is neutralized, why don't all the six ligand water molecules lose a proton?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="3846" LastEditDate="2014-01-05T17:39:43.747" LastActivityDate="2014-01-05T17:39:43.747" Title="Does deprotonation of a hexaaquaion continue until neutral charge" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7667" PostTypeId="2" ParentId="7666" CreationDate="2014-01-05T17:26:50.047" Score="3" Body="&lt;p&gt;Let's look at an example: Hexaaqua aluminium, or $\ce{[Al(OH2)6]^{3+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is acidic and will undergo deprotonation of a ligand water molecule spontaneously:&#xA;$$ \ce{[Al(OH2)6]^{3+} &amp;lt;=&amp;gt; H+ + [Al(OH)(OH2)5]^{2+}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even get it to deprotonate twice:&#xA;$$ \ce{[Al(OH)(OH2)5]^{2+} &amp;lt;=&amp;gt; H+ + [Al(OH)2(OH2)4]^{+}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as indicated, both reactions are &lt;strong&gt;equilibrium&lt;/strong&gt; reactions. So if the concentration of protons in solution is high enough, the hydroxo ligands will be reprotonated and become neutral aqua ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add some hydroxide to the solution, you &lt;a href=&quot;http://alevelchem.com/aqa_a_level_chemistry/unit3.5/s355/03.htm&quot; rel=&quot;nofollow&quot;&gt;can precipitate&lt;/a&gt; aluminium hydroxide, since it is not soluble in water. While this might be a driving force for the deprotonation, it doesn't seem to be the case here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have shown that the process does not necessarily continue until the charge of the complex ion is neutralized.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, concerning your question why this doesn't happen until all aqua ligands have lost a proton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, a part of the solution is the equilibrium idea brought forward in the upper part. But there's another way to think about it: If you have a negatively charged ion, you also need a positively charged counterion. The only counterion that exists in the solution of water and the aluminium compounds is the freshly abstracted proton, which happily will bind back to the hydroxoligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also $\pi$-donor effects that are stronger in the hydroxo than the aqua ligand. So it might be favourable to have no more than three hydroxo ligands in an octahedral complex so no two $\pi$-donors are across each other.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-05T17:26:50.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7668" PostTypeId="1" CreationDate="2014-01-05T20:51:36.393" Score="1" ViewCount="353" Body="&lt;p&gt;Say I have a soft, possibly porous object. I want to coat it with a material,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Resistant to boiling water and some scraping&lt;/li&gt;&#xA;&lt;li&gt;Not hazardous, at least after the coating&lt;/li&gt;&#xA;&lt;li&gt;Largely transparent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The idea is to take, say, a fruit with the insides removed and to make a cup or a vase out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I could buy or make at home for this?&lt;/p&gt;&#xA;" OwnerUserId="2243" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T18:17:14.690" LastActivityDate="2015-10-30T18:17:14.690" Title="Hard coating for soft surface" Tags="&lt;materials&gt;" AnswerCount="4" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="7669" PostTypeId="1" AcceptedAnswerId="7672" CreationDate="2014-01-05T21:07:14.100" Score="5" ViewCount="66018" Body="&lt;p&gt;$\ce{CO(g) + 2H2(g) +O2(g) &amp;lt;=&amp;gt; 2CH3OH(g)}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H=-128\ \mathrm{kJ/mol}$&lt;br&gt;&#xA;$\Delta S=-409.2\ \mathrm{J/(mol \cdot K)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine if the reaction if thermodynamically favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wanted someone to see if my logic is right. Since there are 4 moles on the reactants side and and 2 on the products side the reaction would not be thermodynamically favored because the reaction decreases the amount of entities, which decreases the entropy.&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T15:43:15.967" LastActivityDate="2018-05-06T18:51:02.033" Title="Is the reaction thermodynamically favored or not?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7671" PostTypeId="1" CreationDate="2014-01-05T21:14:27.027" Score="0" ViewCount="1312" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A student is provided with 500 mL of 600 ppm solutions of fructose.  What volume of this solution in millilitres contains 0.15 fructose?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean that I just have to use 600 ppm as the concentration and the 0.15 as the weight? Like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;{600 \space ppm}&#xA;&amp;amp;=\frac {0.15 g}{x}\cdot 10^{6} \\&#xA;x&amp;amp;=\frac {0.15 }{600}\cdot 10^{6}  \\ &#xA;x&amp;amp;=250 \space mL\\&#xA;&amp;amp;&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when can 500 mL be used?&lt;/p&gt;&#xA;" OwnerUserId="2188" LastEditorUserId="73" LastEditDate="2014-01-05T21:18:52.140" LastActivityDate="2014-02-06T14:20:18.960" Title="A student is provided with 500 mL of 600 ppm solutions of fructose. What volume of this solution in millilitres contains 0.15 fructose?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-03-20T23:37:10.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="7672" PostTypeId="2" ParentId="7669" CreationDate="2014-01-05T21:40:23.000" Score="4" Body="&lt;p&gt;Your reasoning is correct but there is more to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at it from the point of view of Gibbs Free Energy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = \Delta H - T \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given reaction if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G &amp;gt;0$, the reaction is nonspontaneous in the forward direction, &lt;em&gt;not thermodynamically favourable&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G &amp;lt; 0 $, the reaction is spontaneous in the forward direction, &lt;em&gt;thermodynamically favourable&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we substitute the values for $\Delta H$ and $\Delta S$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this,  the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).&lt;/p&gt;&#xA;" OwnerUserId="4115" LastEditorUserId="-1" LastEditDate="2018-05-06T18:51:02.033" LastActivityDate="2018-05-06T18:51:02.033" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="7673" PostTypeId="1" CreationDate="2014-01-06T02:58:51.987" Score="0" ViewCount="609" Body="&lt;p&gt;I have a ventless propane heater newly installed in my home, which has been an increasingly source of paranoia for me (constant condensation on the windows, worries about carbon monoxide poisoning).  It's been a interesting few weeks of trying to find the right balance of windows to crack; needless to say I'm looking for a new heater. &#xA;Today, I noticed that NONE of my windows had condensation on them, even though the heater had been running for hours and the relative humidity outside was 100% (~30F) with fog and sleet.  I would expect it to feel like a swamp in here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the chemistry question- I know that the incomplete combustion of propane in a low oxygen environment would produce carbon monoxide, but is it typical for the air to become more dry as well?  Does the lack of oxygen also produce less water, and maybe more H2 or something like that?  What would cause the unusual drop (or perceived drop?) in humidity in my house?&lt;/p&gt;&#xA;" OwnerUserId="4116" LastEditorUserId="73" LastEditDate="2014-01-06T08:41:35.767" LastActivityDate="2014-01-06T08:41:35.767" Title="What would cause a sudden drop in humidity in my home?" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7674" PostTypeId="1" AcceptedAnswerId="7676" CreationDate="2014-01-06T03:54:02.043" Score="4" ViewCount="256425" Body="&lt;p&gt;I want an high-school level answer. What I mean with &quot;valence electrons&quot; is the outermost electrons in that atoms' electronic arrangement? For example, 3 in an atom with an electronic arrangement of 2, 8, 3.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without actually writing the electron configuration, or orbitals, and knowing electron number, is there any way to find this number easily?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there an easy way to find if certain atom fits &quot;atomic/orbital symmetry&quot; given the numbers of electrons? (I don't know the english words for this and couldn't find them. So I translated it literally, but what I mean is, if orbital is half-full or full i.e  &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s^1}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p^3}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p^6}$&lt;/span&gt; and so on... hope you understand what I am trying to say)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this because sometimes questions give me electron numbers in a range of 40+ and just writing it out doesn't seem like a good way.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="402" LastEditDate="2019-05-17T06:45:49.297" LastActivityDate="2019-05-17T06:45:49.297" Title="Is there an easy way to find number of valence electrons&quot;" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7676" PostTypeId="2" ParentId="7674" CreationDate="2014-01-06T06:27:11.727" Score="5" Body="&lt;p&gt;You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrFqx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Group 1 atoms have 1 valence electron.&lt;/li&gt;&#xA;&lt;li&gt;The Group 2 atoms have 2 valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;The Group 3 atoms have 3 valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;The Group 4 atoms have 4 valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;The Group 5 atoms have 5 valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;The Group 6 atoms have 6 valence electrons.&lt;/li&gt;&#xA;&lt;li&gt;The Group 7 atoms have 7 valence electrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The atom with electric configuration 2, 8, 3 is Aluminium and you can find it in Group 3 of the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periodic table is usually given in exams so...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first two groups of the periodic table are the groups concerned with the filling of the $s$ orbital. So, all Group 1 atoms will have an outermost $s$ orbital filled with a single electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, Group 3 - Group 8 (Group 8 is also referred to as Group 0) are where the $p$ orbital is being filled up. All the Group 5 atoms thus all have an outermost $p$ orbital filled with 3 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Periods in the periodic table can be used as an indication of the number of shells the atom has. Sulfur for instance is in Period 3 and thus has 3 shells.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastEditorUserId="1409" LastEditDate="2015-10-14T09:39:12.377" LastActivityDate="2015-10-14T09:39:12.377" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7677" PostTypeId="2" ParentId="7673" CreationDate="2014-01-06T08:01:40.620" Score="1" Body="&lt;p&gt;the humidity is caused mainly from sources of water in your home such as plants, pets and of course humans. As you breath out water is coming out of you. And the condensation of water on windows is caused by different temperature outside/inside especially at the interface glass/air.&#xA;There is no way that heater can reduce humidity in sealed room (in means of decomposition of water as you suggested). Imho if you have CO alarm you have nothing to be afraid of.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-01-06T08:01:40.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7678" PostTypeId="2" ParentId="7668" CreationDate="2014-01-06T08:12:03.407" Score="0" Body="&lt;p&gt;you can use epoxy resin. It is transparent, can be boiled, easy to use and obtainable.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-01-06T08:33:52.623" LastActivityDate="2014-01-06T08:33:52.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7679" PostTypeId="1" CreationDate="2014-01-06T09:45:40.690" Score="1" ViewCount="274" Body="&lt;p&gt;I want to learn chemistry. I did not have the benefit of studying it when I was in highschool, and I was always intimidated by it (I feel it's more rote memorization rather than understanding - I'm more of a Physics and Math kinda guy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, what are the concepts that I need to learn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How, and what resources are you going to use to homeschool Chemistry to a teenager?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that studying it would put the student on par with a normal highschool student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Nice Stuff:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Topics for a Normal Highschool Chemistry Curriculum&#xA;and Maybe AP and Honors Program Curriculums?&#xA;Comprehensive Textbooks&#xA;Exams and Self Testing Resources&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4119" LastEditorUserId="4119" LastEditDate="2014-01-06T12:24:17.460" LastActivityDate="2014-01-06T12:24:17.460" Title="Chemistry Self Education" Tags="&lt;materials&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2014-01-06T12:14:06.987" ContentLicense="CC BY-SA 3.0" />
  <row Id="7680" PostTypeId="2" ParentId="7660" CreationDate="2014-01-06T10:42:53.617" Score="2" Body="&lt;p&gt;Panagiotis, in order to solve the exercise, one needs to identify the relevant functional groups and explore their environment a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed that we're talking about &quot;standard conditions&quot;, i.e., what is likely to happen if the dissolve/suspend the compounds in water, the situation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the structure of &lt;strong&gt;Moclobemide&lt;/strong&gt;, we spot an amide, and - in the morpholin moiety - an ether and a tertiary amine. Amides do have a lone pair on the nitrogen atom, but due to the conjugation with the carbonyl group, the electron density on the N atom is strongly reduced, as compared to an amine. Ethers are rather lousy bases, in fact, the $pK_a$ of protonated tetrahydrofuran is as low as -2.05.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining tertiary amine thus is the functional group that influences the pH: tertiary amines show a $pK_b$ around 3.3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;Paracetamol&lt;/strong&gt; we identify an amide (that we've already excluded as irrelevant) and a phenol group. The latter, once deprotonated, will stabilize the negative charge by resonance with the $\pi$-system (G M has already mentioned this). Therefore, phenols are far more acidic than alkanols. Consequently, the compound definitely acts as an acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that you've already noticed two imide units in the case of &lt;strong&gt;Hexobarbital&lt;/strong&gt; and in fact, it is possible to abstract the proton from the nitrogen atom. However, you'd need a base such as a carbonate under anhydrous conditions to deprotonate here. As a result, &lt;strong&gt;Hexobarbital&lt;/strong&gt; will hardly do anything at all in water.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-06T10:42:53.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7681" PostTypeId="2" ParentId="7679" CreationDate="2014-01-06T10:53:13.173" Score="2" Body="&lt;p&gt;You might want to have a look at free online courses, e.g. at &lt;a href=&quot;https://www.khanacademy.org/science/organic-chemistry&quot; rel=&quot;nofollow&quot;&gt;Khan Academy&lt;/a&gt;. To my opinion, this can however not replace the real experience of endless, often frustrating hours in the lab ;) - chemistry is an experimental science.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-06T10:53:13.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7682" PostTypeId="2" ParentId="7679" CreationDate="2014-01-06T10:56:11.517" Score="2" Body="&lt;p&gt;There are several concepts you need to learn, and the direction these will take will depend on your area(s) of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To that end, the following are some resources that could be of help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.khanacademy.org/science/chemistry&quot; rel=&quot;nofollow&quot;&gt;Khan Academy tutorials&lt;/a&gt;, these start with the fundamental concepts of the atom, periodic table, orbitals and electrons, chemical reactions, stoichiometry, states of matter, gas laws etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemtutor.com/&quot; rel=&quot;nofollow&quot;&gt;ChemTutor&lt;/a&gt; has some basic concepts to start of with, including measurement and pretty much all that Khan Academy has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something, as someone in education, I need to emphasise, and that is to test yourself - there are some online self-testing sites, such as &lt;a href=&quot;http://www.chem.utoronto.ca/undergrad/webexams.php&quot; rel=&quot;nofollow&quot;&gt;this one from the University of Toronto&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of all, don't fear chemistry, it is a lot of fun and requires substantial problem solving and experimentation - you can buy some of these at toy stores and many experiments can be done at home - such as these from &lt;a href=&quot;http://www.hometrainingtools.com/chemistry-kits/c/11/&quot; rel=&quot;nofollow&quot;&gt;Home Science Tools&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-06T10:56:11.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7683" PostTypeId="1" AcceptedAnswerId="7685" CreationDate="2014-01-06T12:41:04.833" Score="72" ViewCount="49966" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQoPu.png&quot; width=&quot;72&quot; alt=&quot;2-methylaniline&quot;&gt;  &lt;img src=&quot;https://i.stack.imgur.com/j6qW6.png&quot; width=&quot;83&quot; alt=&quot;2-methylbenzoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When comparing &lt;em&gt;o,m,p&lt;/em&gt;-toluidine basicities, the &lt;em&gt;ortho&lt;/em&gt; effect is believed to explain why &lt;em&gt;o&lt;/em&gt;-toluidine is weaker. But when comparing &lt;em&gt;o,m,p&lt;/em&gt;-toluic acid basicities, the &lt;em&gt;ortho&lt;/em&gt; effect is stated as a reason why &lt;em&gt;o&lt;/em&gt;-toluic acid is stronger acid. I was told that the &lt;em&gt;ortho&lt;/em&gt; effect is a phenomenon in which an &lt;em&gt;ortho-&lt;/em&gt; group causes steric hindrance, forcing the $\ce{-COOH}$, $\ce{-NH2}$ or some other bulky group to move out of the plane, inhibiting resonance. Then, if the &lt;em&gt;ortho&lt;/em&gt; effect inhibits resonance, why is &lt;em&gt;o&lt;/em&gt;-toluic acid the strongest and &lt;em&gt;o&lt;/em&gt;-toluidine the weakest?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I going wrong in my understanding of the &lt;em&gt;ortho&lt;/em&gt; effect?&lt;/p&gt;&#xA;" OwnerUserId="4114" LastEditorUserId="402" LastEditDate="2018-05-27T03:23:22.420" LastActivityDate="2020-05-01T04:26:49.387" Title="Ortho-effect in substituted aromatic acids and bases" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;amines&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="32" ContentLicense="CC BY-SA 4.0" />
  <row Id="7684" PostTypeId="1" CreationDate="2014-01-06T13:01:59.267" Score="4" ViewCount="3455" Body="&lt;p&gt;In Allenes, the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C=C}$&lt;/span&gt; structure makes sure that the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;-orbital of central carbon overlaps with 2 other carbons and the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;-orbital of central carbon spreads out over the 3 carbons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Alkynes, the &lt;span class=&quot;math-container&quot;&gt;$\ce{C#C-C}$&lt;/span&gt; structure makes sure that the energy released during formation of alkynes.&lt;br&gt;&#xA;I also read that more the number of non-delocalized &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-electrons on an atoms, the lesser its stability, though I do not know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, it comes to comparing Allenes and their isomeric Alkynes,it has always been stated that  &quot;Allenes are more stable than Alkynes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason I found was that &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-electrons in Allenes are less strongly bound than those in Alkynes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why are &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-electrons less strongly bound?  If the reason in incorrect, what is the actual reason? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is it that as the number of non-delocalized &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-electrons increases on a carbon, the smaller its stability?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4114" LastEditorUserId="30836" LastEditDate="2020-04-05T19:07:15.750" LastActivityDate="2020-04-05T19:07:15.750" Title="Stability of Allenes and Alkynes," Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;allenes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7685" PostTypeId="2" ParentId="7683" CreationDate="2014-01-06T14:07:55.680" Score="63" Body="&lt;p&gt;I'd like to throw a tentative explanation for the &lt;em&gt;ortho&lt;/em&gt; effect into the ring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWtTN.png&quot; alt=&quot;ortho effect&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively,  can be assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, the electron density on the N atom is (slightly) reduced and thus the basicity of &lt;em&gt;o&lt;/em&gt;-toluidine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter case, a similar interaction provides additional stabilisation of the carboxylate. As a result, &lt;em&gt;o&lt;/em&gt;-toluic acid is more acidic than the isomers.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-06T14:07:55.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7687" PostTypeId="2" ParentId="6843" CreationDate="2014-01-06T19:03:43.803" Score="1" Body="&lt;p&gt;The common properties of PVC vary wildly with the concentrations of plasticizers and filling reinforcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common properties of PU vary wildly with the choice of isocyanate and cross-link density. PU comes in foams, fibers and hard plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modifications, copolymers and alloy blends are common with both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When selecting a plastic, it is useful to gather requirements rather than emulate another plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some physical properties to look into are: Glass transition, melting/vicat softening point, degradation onset temperature, thermoset/thermoplastic, solvent resistance, transparency, UV-resistance, food/medical-safety, tensile strength and Young's modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to manufacturing some properties to look into are: pot life, working time, tack time, demold time, cure time, cost at scale, shipping and sourcing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to your question you're also considering if they are biobased(ASTM D6866), their recyclability and their compostability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only bioplastic that is highly recycled is HDPE. The only other plastic that is highly recycled is PET. There are probably more recycling operations for other plastics, but they are much smaller in scope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compostibility is dominated by starch and cellulose plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a start, you should visit &lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=f4237c328ae74ef6a64737034747e193&quot; rel=&quot;nofollow&quot;&gt;Matweb&lt;/a&gt; for the basic properties of various plastics.&lt;/p&gt;&#xA;" OwnerUserId="2490" LastActivityDate="2014-01-06T19:03:43.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7688" PostTypeId="1" CreationDate="2014-01-07T00:08:32.920" Score="5" ViewCount="108" Body="&lt;p&gt;I have two reactions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + C \rightleftharpoons AC}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B + C \rightleftharpoons BC}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to estimate the relative steady state proportions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{BC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{AC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;without numerically solving any differential equations, given $[\ce{A}]_{init} = [\ce{B}]_{init}$ and thermodynamic properties: $\Delta G, \Delta H, $ and $\Delta S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping that there is a way to generalize this for more than two reactions in a linear way. It is for a DNA microarray optimization problem in which computational speed is important. Any thoughts? &lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="30836" LastEditorDisplayName="user15489" LastEditDate="2017-10-14T10:43:58.197" LastActivityDate="2017-10-14T10:43:58.197" Title="How to estimate reaction products of competitive binding?" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7689" PostTypeId="1" CreationDate="2014-01-07T02:12:59.343" Score="0" ViewCount="938" Body="&lt;p&gt;Will mixing these two together cause a reaction of any sort? If so, is it safe and is there an order in which one should be added to the other?&lt;/p&gt;&#xA;" OwnerUserId="4123" LastEditorUserId="73" LastEditDate="2014-01-07T05:29:52.340" LastActivityDate="2020-11-17T15:11:00.600" Title="Is it safe to mix potassium alum powder with magnesium oil (made from water and magnesium flakes)?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7690" PostTypeId="1" AcceptedAnswerId="7692" CreationDate="2014-01-07T04:59:41.490" Score="3" ViewCount="2235" Body="&lt;p&gt;The SN1 is a two-step mechanism which begins with heterolysis of the C-X bond. What triggers this? Are there external causes that prompt the leaving group to simply leave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is more conceptual than anything, but I can't seem to find answers anywhere.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2016-09-28T07:26:22.227" Title="What causes the SN1 reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7691" PostTypeId="2" ParentId="7689" CreationDate="2014-01-07T07:55:06.047" Score="2" Body="&lt;p&gt;If you are not sure and you have to do the reaction, try it with small sample of $\ce{Mg}$ such as one flake. $\ce{Mg}$ flakes should react in water and &lt;em&gt;will&lt;/em&gt; react with potassium alum.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="73" LastEditDate="2014-02-16T12:59:48.250" LastActivityDate="2014-02-16T12:59:48.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7692" PostTypeId="2" ParentId="7690" CreationDate="2014-01-07T08:11:15.467" Score="3" Body="&lt;p&gt;Recall that SN1 heavily favors protic solvents (polar/H+). In such cases, the nucleophile to-be-leaving-group (to be Nu-) is going to be attracted to the positively charged molecules of the protic solvent (...H+) and away from the molecule it is attached to. The power of the attraction comes from the dipole between C-Nu, where Nu has the negative bit of the dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of this step, when the nucleophile attached group leaves, you often see &quot;Nu-&quot; hanging around, but really it's with the solvent, e.g. H3O+Br- , and it's fairly common to omit the solvent here and in other similar ionic situations, if I'm not mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples that invoke hydrogen bonding make this more apparent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F-C...(tertiary) + H2O &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dipole here looks like (δ-)F-C...(δ+) because F is more electronegative&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The F is &lt;em&gt;strongly&lt;/em&gt; attracted to the closer H: (δ+)H-(δ-)O-&lt;strong&gt;H(δ+)(δ-)F&lt;/strong&gt;-C...(δ+) (pretend the HOH is bent :) &lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-01-07T08:11:15.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7695" PostTypeId="1" CreationDate="2014-01-07T12:25:53.520" Score="6" ViewCount="511" Body="&lt;p&gt;Why is the ionization enthalpy of francium greater than that of cesium, even though it has a larger size? I found no Google result regarding this.&lt;/p&gt;&#xA;" OwnerUserId="4103" LastEditorUserId="5026" LastEditDate="2018-02-22T10:05:56.803" LastActivityDate="2018-02-22T10:05:56.803" Title="Why is the ionization enthalpy of francium greater than that of cesium?" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;&lt;atoms&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7696" PostTypeId="1" CreationDate="2014-01-07T12:50:38.407" Score="8" ViewCount="16873" Body="&lt;p&gt;I am finding it hard to distinguish between the two. Can someone explain the two terms and the difference between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried learning from Wikipedia, but it said for both the above terms that it is the total energy contained by a thermodynamic system.&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="30894" LastEditDate="2018-02-14T19:38:30.667" LastActivityDate="2020-03-18T15:48:07.983" Title="How do I distinguish between internal energy and enthalpy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7697" PostTypeId="2" ParentId="7671" CreationDate="2014-01-07T13:04:22.187" Score="1" Body="&lt;p&gt;Assuming it's $0.15gm$ of fructose,$$ppm=\frac{Weight\;of\;solute_{(in\;gm)}}{Weight\;of\;solution_{(in\;gm)}}10^6$$&#xA;Since density of water is $1gm/ml$, $500ml$ of water would weigh $500gm$&#xA;$$\therefore600=\frac{W_b}{500}10^6$$&#xA;$$\therefore W_b=0.3gm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $500ml$ of water contains $0.3gm$ of fructose. You can find out what volume will contain $0.15gm$ of it.&lt;/p&gt;&#xA;" OwnerUserId="3718" LastActivityDate="2014-01-07T13:04:22.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7699" PostTypeId="1" CreationDate="2014-01-07T13:52:23.850" Score="0" ViewCount="1634" Body="&lt;p&gt;Consider the reaction $\ce{Sb2S3 + F2 -&amp;gt; SbF5 + SF6}$. If the milliequivalents of antimony, sulfur and fluorine are x, y and z respectively, then prove that $x+y=z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I know: Number of equivalents of each reactant in the reaction is equal, and that number of equivalents is (mass of species) /(equivalent weight). The problem I'm facing is that $\ce{Sb}$ and $\ce{S}$ are in the same compound, so I think $x$ should be equal to $y$ (this is clearly incorrect). Also, the mass of reactants used is not mentioned, so I don't know how to proceed with the question.&lt;/p&gt;&#xA;" OwnerUserId="4127" LastEditorUserId="402" LastEditDate="2017-04-25T00:43:30.937" LastActivityDate="2017-04-25T00:43:30.937" Title="Relation between number of equivalents of reactants" Tags="&lt;inorganic-chemistry&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7700" PostTypeId="2" ParentId="7696" CreationDate="2014-01-07T14:33:08.543" Score="13" Body="&lt;p&gt;Internal energy, $U$, &lt;strong&gt;is&lt;/strong&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Internal_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;total energy contained in a thermodynamic system&lt;/a&gt;. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy also (like radiation), but these two are common. Thus, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU = q + w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transfer can be determined easily by &lt;a href=&quot;http://en.wikipedia.org/wiki/Calorimetry&quot; rel=&quot;nofollow noreferrer&quot;&gt;calorimetry&lt;/a&gt;, where $q=cdT$, where $c$ is the heat capacity of the system and $T$ is the temperature of the system. Alternatively, it may be easier to measure the temperature change of surroundings with known heat capacity, since $q_{sys}=-q_{surr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is a little more challenging to determine, since work is defined as the change in volume at constant pressure. $w=-PdV$ (the minus sign is there to show that when the volume of the system increases, the system does work on the surroundings). However, unless you have a closed system, volume changes are difficult to measure. Additionally, pressure may not be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=q+w=cdT-PdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal energy changes can be determined easily at constant volume $dV=0$ so $w=0$ so $dU=cdT$. Internal energy changes can also be determined in closed systems at constant pressure (so $PdV$ is evaluatable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in open systems (like a beaker), $dV$ is hard to determine, making $w$ hard to determine. A new thermodynamic state function was devised to remedy this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy is a &lt;strong&gt;measure&lt;/strong&gt; of the total energy of a system. More precisely, it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamic potential&lt;/a&gt;. Enthalpy is defined as $H=U+PV$. $PV$ is a state function, even though $P$, $V$, $PdV$, and $VdP$ are not state functions. Thus the change in enthalpy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH=dU+d(PV)=cdT-PdV+PdV+VdP=cdT+VdP$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy changes are easy to determine at constant pressure (most open systems are constant pressure systems, which is why enthalpy is so useful to most chemists), since $dP=0$ and $dH=cdT$. Enthalpy changes can also be determined in closed systems of constant volume where the pressure change can be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, internal energy and enthalpy are used to estimate the thermodynamic potential of the system. There are other such estimates, like the Gibbs free energy $G$. Which one you choose is determined by the conditions and how easy it is to determine pressure and volume changes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At constant volume: $dU=cdT$ and $dH=cdT+VdP$ &lt;/li&gt;&#xA;&lt;li&gt;At constant pressure: $dU = cdT-PdV$ and $dH=cdT$ &lt;/li&gt;&#xA;&lt;li&gt;At constant volume and pressure: $dU=dH=cdT$ &lt;/li&gt;&#xA;&lt;li&gt;At variable volume and pressure, $dU$ and $dH$ are both hard to&#xA;determine (but $dH$ may be easier since you may be able to numerically estimate $d(PV)$ from pressure-volume data).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="30894" LastEditDate="2018-02-14T20:24:43.493" LastActivityDate="2018-02-14T20:24:43.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7701" PostTypeId="2" ParentId="7699" CreationDate="2014-01-07T14:56:31.893" Score="2" Body="&lt;p&gt;Step 1: Balance the equation. Your equation as written $\ce{ Sb2S3 + F2 -&amp;gt; SbF5 + SF6}$ is not balanced. You do not have the same numbers of sulfur, antimony, or fluorine atoms on each side of the equation. You cannot say anything about equivalents or stoichiometry with an unbalanced equation. Hint: Start with $\ce{Sb}$ and $\ce{S}$ and balance $\ce{F}$ last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Determine the relationship between $x$ and $y$. Since $x$ is antimony and $y$ is sulfur and they are in the same compound initially, they have a relationship (but they are not equal). In $\ce{Sb2S3}$, there are 2 sulfur atoms for every 3 antimony atoms. What does that tell you that the ratio of $\frac{x}{y}$ must be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Use what you now know about $x$ and $y$ and the correct stoichiometric equivalents (which should give you $z$) to show that $x+y=z$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-07T14:56:31.893" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7702" PostTypeId="1" AcceptedAnswerId="7703" CreationDate="2014-01-07T17:44:00.370" Score="4" ViewCount="1485" Body="&lt;p&gt;I know that once you have sugars, you can ferment, distill, and dehydrate to create 95+% ethanol. I've also read about cellulosic ethanol, with the new step there being the breakdown of the cellulose into simpler sugars.  Is there a relatively cheap way to do that yet, or it still only in expensive research labs? &lt;/p&gt;&#xA;" OwnerUserId="4121" LastEditorUserId="30836" LastEditDate="2019-04-14T15:19:22.033" LastActivityDate="2019-04-14T15:19:22.033" Title="Is there a way to turn cellulose into sugar at home?" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;&lt;carbohydrates&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7703" PostTypeId="2" ParentId="7702" CreationDate="2014-01-07T18:03:43.670" Score="3" Body="&lt;p&gt;A method used to break down cellulose into sugars is hydrolysis : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use enzymes to hydrolyze the cellulose to produce glucose, but enzymes cost a lot and as you are looking for a cheap way, using enzymes is not an option. A second method is to use an acid (e.g. sulfuric acid) to produce the glucose (just google &quot;cellulose acid hydrolysis&quot;  and you'll find some useful links)&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-01-07T18:03:43.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7704" PostTypeId="1" AcceptedAnswerId="7709" CreationDate="2014-01-07T18:27:00.497" Score="16" ViewCount="881" Body="&lt;p&gt;Why does plutonium have more oxidation states than samarium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron configuration of Pu: $\ce{[Rn] 5f^6 7s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron configuration of Sm: $\ce{[Xe] 4f^6 6s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..$\ce{f^6}$..$\ce{s^2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements such as scandium (Sc),Yttrium (Y) and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..$\ce{d^1}$..$\ce{s^2}$ , but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have to do with the ionization energy, or the energy level or ....?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user4076" LastEditDate="2014-02-10T21:47:54.177" LastActivityDate="2014-02-10T21:47:54.177" Title="Plutonium having more oxidation states than samarium?" Tags="&lt;electronic-configuration&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7705" PostTypeId="1" AcceptedAnswerId="7715" CreationDate="2014-01-07T18:52:04.093" Score="0" ViewCount="3111" Body="&lt;p&gt;Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr]  4d^{10} 5s^0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$. &#xA;How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$? &lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user4076" LastEditDate="2014-01-07T21:44:26.520" LastActivityDate="2014-01-08T13:57:24.140" Title="Explaining the oxidation states of palladium" Tags="&lt;electronic-configuration&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7706" PostTypeId="2" ParentId="7515" CreationDate="2014-01-07T18:55:15.347" Score="3" Body="&lt;p&gt;You can predict this with the help of electrochemical series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5u6Sh.png&quot; alt=&quot;Electrochemical Series&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On going up the series reducing power increases, for you case it can be seen that Fe is a better reducing agent than bromide ion as Fe is above Br2 so bromide can't reduce ferric ion.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2014-01-08T10:39:16.827" LastActivityDate="2014-01-08T10:39:16.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7707" PostTypeId="1" AcceptedAnswerId="7708" CreationDate="2014-01-07T19:16:26.593" Score="1" ViewCount="827" Body="&lt;p&gt;Which of the following is true at equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T \Delta{S} = \Delta{H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T \Delta{s} = \Delta {G}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's the first one but I just wanted to be sure. &lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="73" LastEditDate="2014-01-08T04:53:22.830" LastActivityDate="2014-01-08T11:02:21.790" Title="Which of these thermodynamic relationships is true for a system at equilibrium?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7708" PostTypeId="2" ParentId="7707" CreationDate="2014-01-07T19:18:23.677" Score="1" Body="&lt;p&gt;At equilibrium, $\Delta {G}$ is zero, so according to the equation $\Delta {G} = \Delta {H} - T \Delta {S}$ the first option is correct.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2014-01-08T11:02:21.790" LastActivityDate="2014-01-08T11:02:21.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7709" PostTypeId="2" ParentId="7704" CreationDate="2014-01-07T19:23:44.927" Score="17" Body="&lt;p&gt;This is yet another interesting effect of &lt;a href=&quot;http://depa.fquim.unam.mx/amyd/archivero/LecturasobreNodosRadiales_12854.pdf&quot;&gt;the anomalous compactness of orbitals in the first appearance of each type of subshell&lt;/a&gt; ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with &lt;em&gt;no radial nodes&lt;/em&gt;. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers, and have significant coordination chemistry using $f$ orbitals. Curiously, as you go form curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-01-07T19:23:44.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7710" PostTypeId="1" CreationDate="2014-01-07T22:32:35.283" Score="16" ViewCount="1140" Body="&lt;p&gt;A range of very different compounds are used in medicine as anaesthetics. They don't seem to have much in common chemically but they all seem to keep people asleep while medics are doing nasty things to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inhaled anaesthetics include a variety of fluorocarbon gases such as halothane and sevoflurane plus barbiturates like thiopental, oddballs like propofol and even xenon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctWi7.png&quot; alt=&quot;anaesthetic structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do these heterogeneous compounds have in common and how do they work?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="41890" LastEditDate="2017-11-17T19:09:37.840" LastActivityDate="2017-11-21T18:44:35.083" Title="What is the mechanism of action of anaesthetics?" Tags="&lt;biochemistry&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;&lt;pharmacology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7711" PostTypeId="2" ParentId="7710" CreationDate="2014-01-07T23:33:34.640" Score="10" Body="&lt;p&gt;This answer is a general answer, based on some research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanisms behind anaesthetics is not entirely understood, according to the article &lt;a href=&quot;http://www.nature.com/npp/journal/v28/n1s/full/1300142a.html&quot;&gt;Mechanisms of Anesthesia: Towards Integrating Network, Cellular, and Molecular Level Modeling&lt;/a&gt; (Arhem et al. 2003), primarily due to, as you alluded to in your question, the fact that anaesthetics come from a very diverse group of chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the fragmentary data presented in the article, the conclusions the authors made were that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;general anesthetics modulate the activity of ion channels, the main targets being GABAA and NMDA channels and possibly voltage-gated and background channels, thereby directly or indirectly hyperpolarizing neurons in thalamocortical loops, and thereby disrupting coherent oscillatory activity in the cortex. In our view, it does not seem unreasonable that the ultimate target is NMDA channels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a suggestion that the unconsciousness that results is in a way like sleep, an adaptive response to the chemical input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting perspectives are provided in the article &lt;a href=&quot;http://anesth.medicine.arizona.edu/system/files/pdfs/Why%20anesthetic%20mechanism%20research%20has%20failed.pdf&quot;&gt;Why anesthetic mechanism research &#xA;has failed, and what to do about it&lt;/a&gt; (Hammeroff), where the author suggests, research is needed in investigating anaesthetic mechanisms by&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Look closely at the quantum nature of anesthetic binding by van der Waals &#xA;  London forces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In terms of local anaesthetics, the slides included in the presentation &lt;a href=&quot;http://www.drexelmed.edu/documents/pharmacology_physiology/lectures/TallentLocalAnestheticsSlides.pdf&quot;&gt;Local Anesthetics: Overview&lt;/a&gt; state that the mechanism here is due to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ionized form of weak base &#xA;  blocks $\ce{Na^+}$&#xA;  channels&#xA;  by binding to an internal &#xA;  sequence involved in channel inactivation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-07T23:33:34.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7712" PostTypeId="2" ParentId="687" CreationDate="2014-01-08T08:31:05.043" Score="5" Body="&lt;p&gt;As Ben Norris already pointed out, these processes are typical a bit complex. It is however conceivable that the direct oxidation of the thioether &lt;strong&gt;1&lt;/strong&gt; is among the first steps involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/upnZW.png&quot; alt=&quot;oxidation of thioether&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the oxidative cleavage of PMB (= 4-methoxybenzyl) ethers, often applied as protecting groups, is assumed to take place in the methoxy-substituted arene, here the electron transfer mostly likely proceeds from the sulfur atom, leading to the sulfur-centred radical cation &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of a nucleophile, this species is likely to break down en route to the radical &lt;strong&gt;3&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOYf9.png&quot; alt=&quot;cleavage of radical cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether a disulfide, formed upon dimerization of &lt;strong&gt;3&lt;/strong&gt;, will take part in the further course of the reaction is beyond these musings.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-08T08:31:05.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7713" PostTypeId="1" AcceptedAnswerId="10631" CreationDate="2014-01-08T09:15:22.733" Score="4" ViewCount="1631" Body="&lt;p&gt;Like in title: Why $\ce{Fe(CO)5}$ exists, but $\ce{Fe(NH3)5}$ doesn't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jan Dvorak pointed out, it may exist. If so: why is it so unstable?&lt;/p&gt;&#xA;" OwnerUserId="1133" LastEditorUserId="1133" LastEditDate="2014-01-08T10:24:30.533" LastActivityDate="2014-05-11T17:25:18.593" Title="Why Fe(CO)5 exists, but Fe(NH3)5 doesn't?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7715" PostTypeId="2" ParentId="7705" CreationDate="2014-01-08T13:57:24.140" Score="3" Body="&lt;p&gt;Ions of the transition metals do not often occur &quot;naked&quot; the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sodium nitrate, $\ce{NaNO3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $\ce{NaNO3}$ dissolves in water, the two ions dissociate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaNO3 -&amp;gt; Na+(aq) + NO3- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ions are &lt;strong&gt;solvated&lt;/strong&gt; by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider &quot;bonds&quot; (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be &quot;just&quot; sodium ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6d2b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Palladium nitrate, $\ce{Pd(NO3)2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Transition metal ions are different. They tend to form complex ions by &lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_ion&quot; rel=&quot;nofollow noreferrer&quot;&gt;coordinating&lt;/a&gt; solvent molecules. These &lt;a href=&quot;http://en.wikipedia.org/wiki/Dative_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;dative interactions&lt;/a&gt; are far stronger than the ion-dipole interactions in solvated sodium ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ligand_field_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron configuration of that complex ion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Pd(NO3)2 + nH2O -&amp;gt; Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-08T13:57:24.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7716" PostTypeId="1" AcceptedAnswerId="7731" CreationDate="2014-01-08T14:20:42.040" Score="2" ViewCount="1459" Body="&lt;p&gt;I'm studying thermodynamics and I don't understand, why, when we are asked which effect has a rise in temperature on the equilibrium constant of a reaction, we can't use the formula: $\Delta G= \Delta H-T \Delta S$, and say that if the reation is entropic, the change in free energy will more negative, therefore the reaction will be more spontaneous and the equilibrium shifts towards the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my exercices books, it is said, that, according to le Chatelier's principle,the equilibrium will try to &quot;fight&quot; the new constraint and if it is exothermic,the equilibrium will shift towards the reactants, and if it is endothermic towards the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle seems intuitive to me for the compression of a gas: when compressed, the molecules are now more narrow and they combine faster than they separate. But I don't see why it can be used in the raise of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance &lt;/p&gt;&#xA;" OwnerUserId="3940" LastEditorUserId="2087" LastEditDate="2014-01-09T08:28:12.257" LastActivityDate="2014-01-09T11:10:56.903" Title="Thermodynamics effect of change of temperature on equilibrium" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7717" PostTypeId="1" CreationDate="2014-01-08T14:25:50.090" Score="3" ViewCount="1664" Body="&lt;p&gt;I just wanted to make sure how we can know whether it is $K_c$ (equilibrium constant of concentrations) or $K_p$ (equilibrium constant of pressures) which &quot;comes out&quot; of the equation, $\Delta G^o= -RT \ln(K)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now, I have always taken $K_c$ when the reaction was in a solution or $K_p$ if it was a gas. But as there is a difference between their values, I'd like to know exactly which one to take.&lt;/p&gt;&#xA;" OwnerUserId="3940" LastEditorUserId="22" LastEditDate="2014-01-09T19:13:05.133" LastActivityDate="2018-04-16T07:15:29.787" Title="Thermodynamics and equilibrium constant" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ClosedDate="2018-04-16T15:54:35.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="7718" PostTypeId="1" AcceptedAnswerId="7725" CreationDate="2014-01-08T14:58:51.730" Score="1" ViewCount="10178" Body="&lt;p&gt;Why does iodine exhibit metallic properties at room temperature although being a non-metal? Are there any similar elements like this?&lt;/p&gt;&#xA;" OwnerUserId="3840" LastActivityDate="2016-10-05T16:58:29.667" Title="Metallic properties of iodine" Tags="&lt;elements&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-10-05T20:36:41.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="7719" PostTypeId="1" CreationDate="2014-01-08T18:41:59.597" Score="4" ViewCount="1069" Body="&lt;p&gt;What is the major product formed when 1-chloro-3-iodopropane reacts with diphenyl sulfide in the presence of one equivalent of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgBF4}$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pWdT4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pWdT4.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that sulfur is nucleophilic and that it can displace a halide in a S&lt;sub&gt;N&lt;/sub&gt;2 reaction, but the resulting molecule will have a positive charge. Is that okay? What possible use does the product have?&lt;/p&gt;&#xA;" OwnerUserId="4132" LastEditorUserId="16683" LastEditorDisplayName="user4076" LastEditDate="2018-12-24T01:25:58.447" LastActivityDate="2019-01-08T16:47:29.653" Title="Silver-promoted reaction of alkyl halide with diphenyl sulfide" Tags="&lt;organic-chemistry&gt;&lt;nucleophilic-substitution&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7720" PostTypeId="1" CreationDate="2014-01-08T19:21:24.503" Score="8" ViewCount="42554" Body="&lt;p&gt;How do you determine the chemical formula of a compound given its crystal structure?&lt;/p&gt;&#xA;" OwnerUserId="1463" LastEditorUserId="73" LastEditDate="2014-01-10T00:20:29.013" LastActivityDate="2017-07-30T16:07:40.753" Title="How do you determine a chemical formula given the molecule's crystal structure?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7722" PostTypeId="2" ParentId="7719" CreationDate="2014-01-08T19:49:36.183" Score="10" Body="&lt;p&gt;In the presence of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgBF4}$&lt;/span&gt;, diphenylsulfide is alkylated to 3-chloropropyl diphenylsulfonium tetrafluoroborate. The sulfonium salt is stable and can be isolated as a crystalline product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xle40.png&quot; alt=&quot;alkylation of disulfide&quot;&gt;&#xA;&lt;hr&gt;&#xA;This sufonium salt can be deprotonated (e.g. with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaH}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{THF}$&lt;/span&gt;) to the corresponding sulfonium ylide.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/u7Ev5.png&quot; alt=&quot;cyclization&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter, intramolecular displacement of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; yields cyclopropyl diphenylsulfonium tetrafluoroborate, which again is a stable solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cyclopropyl derivative, once converted to its sulfonium ylide &lt;strong&gt;S&lt;/strong&gt; may serve as a reagent in the &lt;em&gt;Corey–Chaykovsky&lt;/em&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZJ2y.png&quot; alt=&quot;Corey–Chaykovsky epoxidation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2018-12-24T02:48:33.173" LastActivityDate="2018-12-24T02:48:33.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7723" PostTypeId="2" ParentId="7576" CreationDate="2014-01-08T20:23:59.637" Score="12" Body="&lt;p&gt;Bubbles consist of a gas trapped by a liquid, where the liquid has a surface tension high enough to encapsulate the gas. This gives bubbles three ingredients: the liquid, the gas, and surface tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first ingredient, the liquid, is obviously &lt;strong&gt;the rainwater&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second ingredient, the gas, is (also obviously) &lt;strong&gt;air&lt;/strong&gt;. But what isn't obvious is how the gas becomes entrained in the liquid. Asphalt is porous, meaning the microstructure of asphalt consists of lots and lots of tiny random tubes through the solid material. Thanks to the capillary effect, water is sucked into those tiny tubes, and gas is forced out. This gas being forced out of the porous asphalt material and through the liquid on the surface creates bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface tension is the third ingredient, and controls the bubbles' shape, size, and strength. Surface tension can be enhanced by surfactants (big molecules which, when added to a liquid, agglomerate and increase the surface tension). The surface of a road accumulates lots of surfactants - &lt;strong&gt;mostly, oil&lt;/strong&gt;. Oil consists of many long-chain hydrocarbons, which are perfect surfactants (most detergents are actually long-chain hydrocarbons that are polarized at one end). Surfactants tend to agglomerate (that's why, if you pour a little oil on top of water, it will tend to clump together as it spreads out over the surface of the water). Having oil act as a surfactant increases the surface tension of the rain water on the surface of the road, and makes any bubbles formed in the water stronger. That explains the large bubbles you saw.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditDate="2015-06-15T06:49:05.193" LastActivityDate="2015-06-15T06:49:05.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7724" PostTypeId="1" CreationDate="2014-01-08T21:40:51.920" Score="11" ViewCount="1369" Body="&lt;p&gt;There are many compounds that taste sweet. Many of the artificial ones taste much sweeter than natural sugars (which is why they contribute little to energy intake when use as sweeteners). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A range of sweeteners is shown in the figure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTfX0.png&quot; alt=&quot;structures of common sweetners&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structurally there are common features for some. We can see why sucralose (a non-metabolisable sweetener based on sucrose) might taste sweet. Rebaudioside a (a major ingredient of the sugar substitute Steria) shares some of those features. But the very sweet artificial sweeteners seem to have less in common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we know what makes molecules generate the sensation of sweetness? Which features matter and why are some sweeter than others?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="22" LastEditDate="2014-01-09T19:22:04.743" LastActivityDate="2014-01-10T08:54:10.580" Title="What molecular features create the sensation of sweetness?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7725" PostTypeId="2" ParentId="7718" CreationDate="2014-01-08T22:57:47.160" Score="3" Body="&lt;p&gt;Iodine is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metalloid&quot; rel=&quot;nofollow&quot;&gt;metalloid&lt;/a&gt;. Metalloid elements have one or more allotropes with properties intermediate between those of a metal and a nonmetal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals are lustrous, ductile, and conductive of heat and electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonmetals are matte, brittle, and insulators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metalloids may be &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiconductor&quot; rel=&quot;nofollow&quot;&gt;semiconductors&lt;/a&gt;, like silicon, germanium, arsenic, and carbon (graphite). They may be lustrous, but brittle, like iodine and tellurium.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-08T22:57:47.160" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7726" PostTypeId="1" AcceptedAnswerId="7732" CreationDate="2014-01-09T05:03:59.573" Score="1" ViewCount="4186" Body="&lt;p&gt;A current of 2.25 A is applied to  $\ce{NiCl2}$ solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Write the balanced half reaction that takes place at the anode&#xA;B. Write the balanced half reaction that takes place at the cathode&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone check if my answers seem logical? I'm not quite sure if I did the problem correctly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anode: $\ce{O2 + 4H+ + 4e- -&amp;gt; 2H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{2Ni^{2+} + 4e- -&amp;gt; 4Ni}$&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="22" LastEditDate="2014-01-09T19:13:47.573" LastActivityDate="2014-01-09T19:13:47.573" Title="Balanced reactions at the anode and cathode" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2014-01-09T19:13:56.183" ContentLicense="CC BY-SA 3.0" />
  <row Id="7727" PostTypeId="1" CreationDate="2014-01-09T08:22:29.520" Score="0" ViewCount="954" Body="&lt;p&gt;$$C_x H_y O_z$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.What are the units of x,y,z.I've read that they're termed as 'stoichiometric coefficients'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.What is the unit of mass proportion of carbon(c) in the fuel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c=\frac{M_{carbon}}{M_{fuel}} .x$$&lt;/p&gt;&#xA;" OwnerUserId="1728" LastActivityDate="2014-01-09T09:58:20.513" Title="Units in stoichiometry" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7728" PostTypeId="2" ParentId="7726" CreationDate="2014-01-09T09:37:04.847" Score="3" Body="&lt;p&gt;The reaction at cathode seems OK, but at the anode another reaction will occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode, 4 e$^-$ are consumed. The electrons have to be &quot;generated&quot; in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark:&#xA;The value of current applied is irrelevant in this kind of question. More relevant will be the applied potential. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-01-09T09:37:04.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7729" PostTypeId="2" ParentId="7727" CreationDate="2014-01-09T09:58:20.513" Score="1" Body="&lt;p&gt;The stoichiometric coefficients is something else than you suggest.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.boundless.com/chemistry/definition/stoichiometric-coefficient/&quot; rel=&quot;nofollow noreferrer&quot;&gt;stoichiometric coefficients&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;The number of molecules of a given component that participate in the reaction as written.&lt;/p&gt;&#xA;&lt;p&gt;The indexes &lt;span class=&quot;math-container&quot;&gt;$x, y, z$&lt;/span&gt; represent quantities of atoms in a compound. So they do not have any units.&lt;/p&gt;&#xA;&lt;p&gt;@Q2 Do you know what are &lt;span class=&quot;math-container&quot;&gt;$c, M$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in the equation? If you do, you can answer your question. If you do not, than you should find it out.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-09T09:58:20.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7730" PostTypeId="1" CreationDate="2014-01-09T10:21:44.920" Score="12" ViewCount="25370" Body="&lt;p&gt;Why does ZnO's color become yellow when it is hot? I have read that it is because of physical properties, but I need a more elaborate explanation.&lt;/p&gt;&#xA;" OwnerUserId="4012" LastEditorUserId="73" LastEditDate="2014-01-10T03:31:29.813" LastActivityDate="2017-04-29T16:51:50.837" Title="Why does ZnO turn yellow when it is hot?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;color&gt;" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7731" PostTypeId="2" ParentId="7716" CreationDate="2014-01-09T11:10:56.903" Score="2" Body="&lt;p&gt;First of all, spontaneity of a reaction depends on the kinetics, not thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I always imagine LeChatelier's principle to work like this: When you raise the temperature for an endothermic reaction, it has more temperature to work with so the equilibrium shifts to the right. And if you lower the temperature for an exothermic reaction, that means there's more room for the energy created by the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to use $\Delta G = \Delta H - T\Delta S$ to check how the equilibrium behaves, you will only see the effect the temperature has on the free energy. It relates to the equilibrium via $$ \Delta G = -RT\ln K \Longleftrightarrow K = \exp\left(-\frac{\Delta G}{RT}\right)$$&#xA;This means that if you raise the temperature the expression in the exponent gets closer to 0, which in turn means that the overall $K$ becomes larger. So the equilibrium shifts towards the products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have neglected the influence of the entropy above. The formula should be, when completely written out:&#xA;$$ K = \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with this you can always see whereto the equilibrium will shift.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-09T11:10:56.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7732" PostTypeId="2" ParentId="7726" CreationDate="2014-01-09T11:16:05.887" Score="3" Body="&lt;p&gt;At the &lt;strong&gt;A&lt;/strong&gt;-node you &lt;strong&gt;O&lt;/strong&gt;-xidize, at the &lt;strong&gt;C&lt;/strong&gt;-athode you &lt;strong&gt;R&lt;/strong&gt;-educe. (Vowel to vowel, consonant to consonant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this, you will have to generate electrons at the anode, which doesn't happen with your reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What reaction will happen depends on the normal potential of each reaction, but I assume it will be something along the lines of&#xA;$$\ce{2Cl- -&amp;gt; Cl2 ^ + 2e-} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I just noticed that the cathode reaction can be simplified to&#xA;$$ \ce{Ni^{2+} + 2e- -&amp;gt; Ni} $$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-01-09T12:47:09.697" LastActivityDate="2014-01-09T12:47:09.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7733" PostTypeId="2" ParentId="7730" CreationDate="2014-01-09T12:40:14.853" Score="2" Body="&lt;p&gt;Most probably because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Band_gap&quot; rel=&quot;nofollow&quot;&gt;band gap&lt;/a&gt; closes, thus allowing electrons from the valence band to be elevated into the conduction band (by absorbing blue light, which makes the reflected color yellow).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-09T12:40:14.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7735" PostTypeId="2" ParentId="7730" CreationDate="2014-01-09T14:16:25.207" Score="15" Body="&lt;p&gt;The &lt;strong&gt;thermochromism&lt;/strong&gt; of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T15:22:09.620" LastActivityDate="2015-04-02T15:22:09.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7736" PostTypeId="2" ParentId="7717" CreationDate="2014-01-09T18:26:41.410" Score="4" Body="&lt;p&gt;What you are doing is right. The value to use depends on the definition of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard state.&lt;/a&gt; The values of $K_c$ and $K_p$ are indeed different, but what you are not accounting for is the fact that $\Delta G^o$ also changes with the equilibrium constant used.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naught over $G$, representing &quot;standard state&quot; establishes some rules. For homogeneous gaseous reaction, the standard $\Delta G$ corresponds to the standard state of all gases, which is defined in terms of pressure $P=\pu{1atm}$, and hence $K_p$ needs to be used with the quoted value of $\Delta G^o$.&#xA;For reactions involving solutes and pure liquids, the standard states are defined differently, in terms of their concentrations, and hence $K_c$ is to be used. Since the definition of standard state affects the value of $\Delta G$, we must choose that form of the equilibrium constant which uses the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow noreferrer&quot;&gt;activity&lt;/a&gt; of the reacting species in terms of the quantity used to define the standard state of that species. This maintains uniformity.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-04-16T07:15:29.787" LastActivityDate="2018-04-16T07:15:29.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7737" PostTypeId="2" ParentId="7587" CreationDate="2014-01-09T22:34:04.153" Score="21" Body="&lt;p&gt;I could say explosives are things that blow up, but this would not a &lt;em&gt;helpful&lt;/em&gt; answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also not that helpful to describe explosives as compounds with highly exothermic decomposition. This characteristic alone does not define an explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What matters is &lt;em&gt;rate&lt;/em&gt; of the reaction in the bulk compound. This rate defines two key types of explosive. If the reaction, once started, proceeds through the bulk compound faster than the speed of sound we call it &lt;strong&gt;detonation.&lt;/strong&gt; If the reaction is slower than the speed of sound we describe it as &lt;strong&gt;deflagration.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distinction matters because of the different uses explosives can be put to. Propellants, such as the explosives used to drive bullets, tend to deflagrate (as you want a relatively slow, controlled reaction to drive the shell out of the barrel without blowing up the barrel). But if you want something to really go bang and create a mess (as the military do) or if you want something to break up rock (as quarriers and miners do) then you want something that will detonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structures of some well-known explosives are shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rvneF.png&quot; alt=&quot;structures of common explosives&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most well known real world explosive compounds are made from these compounds plus some added materials to make them easier to handle. C-4, a well known military explosive is mostly RDX with plasticisers to make it easy to mould like plasticine. The favourite terrorist plastic explosive is Semtex, similar to C-4 but made from a mixture of PETN and RDX. TNT is often used in high-explosive military shells (partially because it can be melted by steam and poured into shell casings). The first modern explosive, dynamite, is based on nitroglycerine mixed with diatomaceous earth (which makes it much easier to handle than the dangerously shock-sensitive pure compound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical reactions typically involve the production of CO&lt;sub&gt;2&lt;/sub&gt;, H&lt;sub&gt;2&lt;/sub&gt;O and, sometimes, N&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another characteristic that matters is how easy the reaction is to set off. Some compounds are very shock sensitive like mercury fulminate or nitroglycerin. This is only useful in detonators (in small quantities) or for chemists who have a death wish. Many bulk explosives require a detonator to trigger an explosion, making them much safer to handle and transport. C-4 in small quantities will actually burn slowly (and was sometimes used by soldiers in the Vietnam war to warm rations).&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-09T22:34:04.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7738" PostTypeId="2" ParentId="7033" CreationDate="2014-01-09T23:06:37.323" Score="7" Body="&lt;p&gt;Charcoal isn't a particularly good adsorbent even though it is chemically very similar to activated carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activated carbon is usually made by more specialised processes that guarantee the final product will have a very large surface area (often &gt;1,000m&lt;sup&gt;2&lt;/sup&gt;/g). Manufacture usually involves pyrolysation with hot gasses, but many forms are also further activated by partially oxidising the surface (using gas containing oxygen) or bay adding chemicals such as phosphoric acid or potassium hydroxide before pyrolysis. These activation reactions create some functional groups on the surface that can enhance specific types of adsorption on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the primary feature that matters is the high surface area, which normal methods for making charcoal do not achieve.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-09T23:06:37.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7739" PostTypeId="2" ParentId="7587" CreationDate="2014-01-10T01:31:41.947" Score="3" Body="&lt;p&gt;A material that has low reaction rate in bulk are also possible to be explosive when increase its surface area which increase its reaction rate such as dust explosion.&lt;/p&gt;&#xA;" OwnerUserId="4012" LastActivityDate="2014-01-10T01:31:41.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7740" PostTypeId="2" ParentId="7587" CreationDate="2014-01-10T01:56:39.977" Score="7" Body="&lt;p&gt;One thing that the answers so far have not mentioned is that an explosion is a &lt;em&gt;chemical reaction&lt;/em&gt;.  So the question is: why does the explosion reaction occur?  And like so many chemical reactions, the answer is that the post-explosion state is lower energy than the pre-explosion state. That would seem to be obvious, given all the energy released in an explosion!  But what's more interesting is that an explosive compound can remain for extended periods in a &lt;em&gt;metastable&lt;/em&gt; condition (literally, in an energy state higher than its lowest possible state).  This metastability is precisely why explosive compounds can be useful: they don't just spontaneously blow up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this begs the question: why are explosive compounds metastable?  This gets to questions of activation energy or introduction of reactants.  In the first case, it may be required that more energy is required to break the metastability.  This is what is needed for explosive mixtures of hydrogen and oxygen to detonate, for example.  In the second case, the introduction of a reactant may be just what is needed for explosive decomposition.  For example, the explosive $\ce{XeO_3}$ is triggered by mere contact with organic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary: explosions are particularly vigorous examples of chemical reactions moving to a lower energy state, but the &quot;why&quot; and &quot;how&quot; can be complex.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-01-10T01:56:39.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7741" PostTypeId="2" ParentId="7684" CreationDate="2014-01-10T02:17:25.503" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The shape of the two bonds (see the first page of this &lt;a href=&quot;http://cronodon.com/files/Hydrocarbons_-_Alkenes_and_Arenes.pdf&quot; rel=&quot;nofollow&quot;&gt;.PDF&lt;/a&gt;) shows why pi bond electrons are less tightly bound. Sigma bonds are formed pretty much along the line connecting the two atomic nuclei, whereas the pi bonds are formed above and below that same line. Naturally, then, the distribution of the latter less concentrated (having more space to cover) than that of the former. Loosely stated, the shape of the pi bond distribution is more roundabout whereas the sigma bond directly touches the two nuclei which makes the electrons in it less strongly bound.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When it can, delocalization of electrons provides greater stability for the molecule because the electron density is distributed well over the delocalized regions, rather than, say, confining it in a small high-energy region surrounded by low-energy regions. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Think about &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;resonant&lt;/a&gt; structures. Benzene (C6H6) is a good example. So the inverse of that is: less delocalization, less stability. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-01-10T02:17:25.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7742" PostTypeId="1" AcceptedAnswerId="8394" CreationDate="2014-01-10T03:55:57.580" Score="1" ViewCount="6819" Body="&lt;p&gt;I have an old oil cooler of  an old Volkswagen engine (the type where the oil itself is cooled, and uses no water) stained with decades of usage. If I pass water through it, it comes out clear, but if I pass a clear oil through it, it comes out black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best method to clean it that uses cheap and easy to find substances? I want to use it to cool clean mineral oil, I want my oil to remain clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My father suggested to pass mineral oil through it until what comes out is clean, then install it in the system, but that would be horribly expensive and messy, and I would need to discard all that dirty oil properly afterwards. I myself thought of using gasoline, filling halfways, shakink and discarding, until clean enough, but although that would be less expensive, it would be as messy as the first option, and probably dangerous. My sister suggested using water and kitchen wash-up liquid. That would be much less messier, because it is water soluble, but I don know the proportion of water and kitchen detergent tI must use for it to be effective. My last idea was to use some grese remover like Ajax, that would not be so.much messy, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what do you suggest?&lt;/p&gt;&#xA;" OwnerUserId="4139" LastActivityDate="2014-02-09T19:51:14.173" Title="Cheap and not messy solvent for motor oil" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;&lt;cleaning&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7743" PostTypeId="2" ParentId="7724" CreationDate="2014-01-10T05:00:51.167" Score="6" Body="&lt;p&gt;For humans, the &quot;sweet taste buds&quot; are the T1R2+T1R3 &amp;amp; T1R3+T1R3 G-protein coupled receptors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemicals that are sweet are the ones that can bind to these receptors. We know some structures that tend to do this, namely &lt;strong&gt;ketones and aldehydes (carbonyl groups)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are chemicals that are sweet that we do not know the mechanism for, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Curculin&quot; rel=&quot;noreferrer&quot;&gt;Curculin&lt;/a&gt;, which is up to ~2000 times sweeter than sucrose by weight.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="3704" LastEditDate="2014-01-10T08:54:10.580" LastActivityDate="2014-01-10T08:54:10.580" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7744" PostTypeId="1" CreationDate="2014-01-10T05:16:16.987" Score="2" ViewCount="78" Body="&lt;p&gt;Say I have a straightforward &lt;a href=&quot;http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics&quot; rel=&quot;nofollow&quot;&gt;Michaelis-Menten reaction&lt;/a&gt; where an enzyme $E$ at some fixed concentration $[E]$ forms an interaction complex $ES$ with a substrate $S$ (with initial concentration $[S]_0$), and at a rate $k_{cat}$ can irreversibly convert the substrate $S$ to a product $P$.  The rate of interaction complex formation $ES$ is $k_f$ and its dissociation rate is $k_r$, such that overall, we have the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E + S \rightleftharpoons^{k_f}_{k_r} ES \to_{k_{cat}} E + P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explicit expression for the time $\tau_{v}$ to convert some percentage $v$ of the substrate $S$ to product $P$?  In lieu of an explicit expression for $\tau_{v}$, what scaling for this time should we expect for $\tau_{v}$ as a function $[S]$ (holding $[E]$ fixed)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, for $\tau_{v} \approx 1$ I suppose the steady-state approximation will begin to fail?&lt;/p&gt;&#xA;" OwnerUserId="4140" LastEditorUserId="1362" LastEditDate="2014-11-02T14:32:56.650" LastActivityDate="2014-11-02T14:32:56.650" Title="What is the scaling for % completion time in a Michaelis–Menten scenario as a function of substrate concentration?" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7745" PostTypeId="1" CreationDate="2014-01-10T11:05:24.177" Score="2" ViewCount="63" Body="&lt;p&gt;At least to me, the contents of chapters 5 &amp;amp; 6 &lt;a href=&quot;http://books.google.ie/books?id=xUtLAgAAQBAJ&amp;amp;lpg=PR1&amp;amp;pg=PR11#v=onepage&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;given here&lt;/a&gt; are representative of discussions on chemical bonding given in nearly all intro general &amp;amp; organic chem books, the statements in those chapters are mainly memorization (to me anyway). I basically don't find it rigorous enough, leads to more rather than less confusion and doesn't give any in depth discussion (similar to the way the quantum mechanics sections of those books are rote assertion ignoring all of the math). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in the middle of a physics-based quantum mechanics course, so I'm developing the conceptual &amp;amp; mathematical tools to read a serious explanation of lewis bonding, polar-covalent bonding etc... but &lt;a href=&quot;http://books.google.ie/books?id=J9ui6KwC4mMC&amp;amp;lpg=PP1&amp;amp;pg=PA309#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;my book&lt;/a&gt; doesn't mention much in detail on this topic either, and apparently not even &lt;a href=&quot;http://books.google.ie/books?id=D48aGQTkfLgC&amp;amp;lpg=PA377&amp;amp;pg=PA377#v=onepage&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Pauling's QM book&lt;/a&gt; mentions much on the topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really hoping somebody would have a quick look at the contents of chapters 5 &amp;amp; 6 of the first link &amp;amp; offer a recommendation of a rigorous reference or four on the contents of those chapters - I really can't face the headache I'll get reading chapters like those &lt;em&gt;again&lt;/em&gt; due to all the confusion they lead to by dumbing it down, I need something more rigorous if such a thing exists, thanks a lot :)&lt;/p&gt;&#xA;" OwnerUserId="2379" LastActivityDate="2014-01-10T13:11:12.697" Title="Quantum Valence Theory Reference" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7746" PostTypeId="2" ParentId="7742" CreationDate="2014-01-10T11:31:30.370" Score="0" Body="&lt;p&gt;You can use kerosene instead of gasoline in a well ventilated area, but your best bet will probably be grease remover.&#xA;Trying different things will let you know what is effective, and each pass will clean the cooler a bit. &#xA;Just don't leave it a long time with water, as that might rust it.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2014-01-10T11:31:30.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7747" PostTypeId="2" ParentId="7744" CreationDate="2014-01-10T12:01:25.987" Score="2" Body="&lt;p&gt;In short, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the relation between τ and v, we need the actual time evolution of the concentration of the product by solving the differential equations (DEs) and then some way to get steady-state concentration. The DEs are non-linear. So we have to find approximate solutions. &lt;a href=&quot;http://resolver.scholarsportal.info/resolve/03014622/v16i0002/173_asocnre.xml&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; develops a solution. It's not nice..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will probably come back to add more. For the time being, I highly recommend those interested to browse through the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what do you mean by τ≈1 leading to failure of steady-state approximation please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The tittle of the paper is &quot;Analytical solution of coupled nonlinear rate equations : I. Michaelis-menten kinetics&quot; by Phillipson&lt;/p&gt;&#xA;" OwnerUserId="4142" LastEditorUserId="4142" LastEditDate="2014-01-11T17:59:42.613" LastActivityDate="2014-01-11T17:59:42.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7748" PostTypeId="2" ParentId="5849" CreationDate="2014-01-10T12:19:12.170" Score="1" Body="&lt;p&gt;Since you ask specifically for internal energy, would it be that you already have some ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is internal energy? It is the energy due to the interaction among the species plus the energy due to the interaction between the species and any external field. That is, the energy it takes to &quot;assemble&quot; the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for gaseous species, it would be mainly the bonding energies of the molecules. Those are usually inferred from experiments. And you can look up tables in textbooks and IUPAC website.&lt;br&gt;&#xA;For ionic crystals, you would reason this way:  first ionize the species and them find the Columb interaction among them in crystal. Ionization energies are measured. Columb interaction is probably calculated and is a nightmare to calculate.&lt;br&gt;&#xA;The further problem is from NaCl(s) to (aq) where the solvation involves quite a bit of interactions between the ions and water. For the case of Na+ Cl-, water molecules do not form regular cages around them, even though we CAN study the ideal case of regular cages. I think there has been some minimal models which can possibly guide a simulation study for the solvation energy. Though I don't think there has been any simulation study or any theoretical study that is sufficient to predict solvation energy.  I can be wrong. On the other hand, direct measurement should at least be able to measure the heat change during solvation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in digression, I don't think you can find the complete energy too well.&lt;/p&gt;&#xA;" OwnerUserId="4142" LastEditorUserId="4142" LastEditDate="2014-01-11T17:53:24.720" LastActivityDate="2014-01-11T17:53:24.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7749" PostTypeId="2" ParentId="847" CreationDate="2014-01-10T12:33:45.587" Score="2" Body="&lt;p&gt;I think the easiest way is to sputter coat the bars, if you can access one. One that can rotate your bars such that the rotation is vertical to the direction of sputtering would be the best. If not, switch sides manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depends on the composition of your bars, you might want to coat something else before coating zinc. A wild guess: nickel.  I suggest so because I don't recall zinc being sticky. So the coating might come off easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'd try sputter slow but for a long time. I think low sputtering rate tends to produce more densely packed and uniform coating. I am not sure. But dense and uniform coating provides good foundation of the zinc coat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electric coating, I think low voltage and uniform coating is still nice. I don't think you want formation of gas at the bar electrode, as that disturbs the structure of the coating. Not sure what's influencing the nucleation sites on the bars. I would guess that the current is not uniform on the bars, say there more on the hill tops. Still consider coating something else first. I would try whatever is available. I would try to get the solutions as pure as possible though. Organic compounds can act as surfactants which tend to aggregate the particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whichever your method is, search for some journal articles characterizing similar processes. That might help.&lt;/p&gt;&#xA;" OwnerUserId="4142" LastEditorUserId="4142" LastEditDate="2014-01-10T12:53:00.043" LastActivityDate="2014-01-10T12:53:00.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7750" PostTypeId="2" ParentId="7061" CreationDate="2014-01-10T12:39:10.687" Score="1" Body="&lt;p&gt;It depeds on the acid. If it is HCl then you will get &lt;a href=&quot;http://info.goldavenue.com/info_site/in_glos/in_glos_wolhill.html&quot; rel=&quot;nofollow&quot;&gt;Wohlwill Process&lt;/a&gt; only the silver will be not disolved. So baisiclly what you want. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-01-16T12:07:58.113" LastActivityDate="2014-01-16T12:07:58.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7751" PostTypeId="2" ParentId="7745" CreationDate="2014-01-10T13:11:12.697" Score="2" Body="&lt;p&gt;Your &lt;em&gt;Essentials of Physical Chemistry&lt;/em&gt; seems to be heavy on the &quot;essentials&quot; and light on the &quot;physical chemistry&quot;. A textbook used in a &lt;strong&gt;physical chemistry course&lt;/strong&gt; (and not a &quot;physical chemistry for &lt;em&gt;_&lt;/em&gt;_&quot; course) is what you need. The second portion of the traditional physical chemistry sequence is devoted entirely to the implications of quantum mechanics in chemical bonding. Pauling's book on QM will not cover much modern bonding theory since that book predates most modern bonding theory. The applications of QM to chemical bonding in all but the simplest molecules require computer programs to slog through the maths, since multielectron wave functions suffer from the three(+) body problem and cannot be solved exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in &lt;a href=&quot;http://books.google.ie/books?id=lk2PzH9LmS8C&amp;amp;printsec=frontcover&amp;amp;dq=physical+chemistry&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=SO_PUsSmI_GgsQSUoYHoCw&amp;amp;ved=0CDkQ6AEwAA#v=onepage&amp;amp;q=physical%20chemistry&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Physical Chemistry&lt;/em&gt; by Peter Atkins and Julio De Paula (W. H. Freeman &amp;amp; Company)&lt;/a&gt;, Chapters 8 through 20 deal with chemical bonding and phenomena first with quantum mechanics and then with statistical mechanics. &lt;a href=&quot;https://www.google.ie/search?tbm=bks&amp;amp;hl=en&amp;amp;q=physical+chemistry&amp;amp;oq=physical+chem&quot; rel=&quot;nofollow&quot;&gt;There are other physical chemistry texts&lt;/a&gt; out there as well. These texts will also cover the computational methods seeking to find 1) exact solutions to approximate multielectron wavefunctions (treating multielectron wavefunctions as linear combinations of single electron wavefunctions) and 2) approximate solutions to exact multielectron wavefunctions.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-10T13:11:12.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7752" PostTypeId="1" CreationDate="2014-01-10T13:49:40.933" Score="3" ViewCount="93" Body="&lt;p&gt;I am an immunology master student without prior knowledge of biochemistry.&lt;/p&gt;&#xA;&lt;p&gt;Could someone explain to me the following 6 reactions?&lt;/p&gt;&#xA;&lt;p&gt;What I want to understand is why exactly the reactions happen at &lt;strong&gt;this&lt;/strong&gt; site ( in terms of donor/acceptor ).  What is the donor, what is the acceptor?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m4X67.png&quot; alt=&quot;protein modifications1&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qlmfH.png&quot; alt=&quot;protein modifications2&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4143" LastEditorUserId="69830" LastEditDate="2021-05-19T08:13:29.800" LastActivityDate="2021-05-19T08:13:29.800" Title="Explaining a biochemical reaction (protein labelling, e.g. with fluorescent probes)" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;proteins&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7753" PostTypeId="1" AcceptedAnswerId="7754" CreationDate="2014-01-10T14:19:27.903" Score="1" ViewCount="25160" Body="&lt;p&gt;What are the $\Delta E$'s of the transitions of an electron from $n=5$ to $n=1$ and from $n=5$ to $n=2$ in a Bohr hydrogen atom? The wavelength of the first electron transition is $\lambda_1=409~\mathrm{nm}$ and the other electron transition $\lambda_2=1091~\mathrm{nm}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I solve this using those two given wavelengths?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="4144" LastEditorUserId="73" LastEditDate="2014-01-10T16:22:07.530" LastActivityDate="2014-01-10T16:22:07.530" Title="How do I calculate the change in energy of an electron transition?" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;orbitals&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7754" PostTypeId="2" ParentId="7753" CreationDate="2014-01-10T14:53:43.817" Score="1" Body="&lt;p&gt;When an electron transitions from a higher energy state to a lower energy state, the energy has to go somewhere. Usually it is emitted as light of a given wavelength which corresponds directly to the change in energy of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of light at a given frequency $\nu$ is &#xA;$$ E = h\nu\; , $$&#xA;where $h$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Planck_constant&quot; rel=&quot;nofollow&quot;&gt;Planck's constant&lt;/a&gt;.&#xA;The frequency of light is directly related to the wavelength $\lambda$ and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light&quot; rel=&quot;nofollow&quot;&gt;speed of light&lt;/a&gt; $c$ via&#xA;$$ c = \nu\lambda\; . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these two equations we get&#xA;$$ E = \frac{hc}{\lambda}\; , $$&#xA;with which you should be able to calculate the $\Delta E$ values for the transitions.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-10T14:53:43.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7755" PostTypeId="1" AcceptedAnswerId="7789" CreationDate="2014-01-10T15:22:21.883" Score="3" ViewCount="344" Body="&lt;p&gt;I am planning an RPG adventure where the characters might end up getting a large cubic lump of &quot;gold&quot;, which in reality, as you might have guessed, is in fact pyrite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wanted to plan for the contingency that the group wouldn't notice that it's not the real deal, so I thought I could maybe let one of the characters cut himself on the crystal. A quick google search yields images such as this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FHR18.jpg&quot; alt=&quot;Large pyrite crystals&quot;&gt;&#xA;(&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:2780M-pyrite1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks pretty sharp and possibly blood-drawing, but is there a way to know for sure? Any recommendations as to how to handle sharp minerals? Any first-hand experiences?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-12T05:56:27.120" Title="Is it possible to cut yourself on pyrite?" Tags="&lt;safety&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7756" PostTypeId="2" ParentId="7752" CreationDate="2014-01-10T16:39:38.653" Score="3" Body="&lt;p&gt;All reactions you gave are typically used in protein labelling, e.g. with fluorescent probes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to allow further measurements on the labeled protein, the following conditions have to be met:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds formed have to be stable like the peptide bonds in your protein. You don't want the probe flying around somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proteins tend to be a bit sensitive to degradation; boiling in acetic acid is obviously not an option to run the labeling. As a consequence, the functional group to be attached needs to be &lt;em&gt;activated&lt;/em&gt;, e.g. with good a &lt;em&gt;leaving group&lt;/em&gt;, to allow the labelling of the protein under mild conditions. The succimidyl ester in the first reaction is a good example.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-10T16:39:38.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7757" PostTypeId="1" AcceptedAnswerId="7760" CreationDate="2014-01-10T18:03:13.437" Score="0" ViewCount="317" Body="&lt;p&gt;What is the pKa of 2-methyl-2,4-pentanediol? Is there a reference that contains a list of pKa values for various molecules that I can consult or is there a way to determine/estimate the pKa of this molecule if such a reference does not exist?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="1362" LastEditDate="2014-01-17T06:35:03.160" LastActivityDate="2014-01-17T06:35:03.330" Title="What is the pKa of 2-methyl-2,4-pentanediol?" Tags="&lt;organic-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7758" PostTypeId="1" AcceptedAnswerId="7763" CreationDate="2014-01-10T18:03:37.010" Score="1" ViewCount="1015" Body="&lt;p&gt;In all the chemistry books and eBooks it says that &quot;like dissolves like&quot;... but when it comes to some molecules like ether, cellulose, vitamin A and $\ce{CO_2}$, it confuses me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take ether, it is true that it will not form hydrogen bonds with water when it dissolves, but they have polar bonds between $\ce{O}$ and $\ce{C}$ and it's not in a linear form like $\ce{CO_2}$ where the dipoles cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vitamin A also has $\ce{-OH}$ group but it still is hydrophobic. In some book, I read that this is because it has more hydrophobic molecules or that they have more $\ce{C-H}$ bonds than $\ce{-OH}$. If it is so, octadecanol and nonadecanol also have way more $\ce{C-H}$ bonds than $\ce{-OH}$ groups, so  why is it still soluble? $\ce{CO_2}$ is dipolar &amp;mdash; the dipoles cancel out, so why is it still soluble? Cellulose also has a bunch of $\ce{-OH}$ but still it is soluble. I am very confused by this.&lt;/p&gt;&#xA;" OwnerUserId="4147" LastEditorUserId="22" LastEditDate="2014-01-12T14:47:50.653" LastActivityDate="2014-01-12T14:47:50.653" Title="Solutions, solvents and solutes" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7759" PostTypeId="1" AcceptedAnswerId="9095" CreationDate="2014-01-10T18:39:09.783" Score="11" ViewCount="10298" Body="&lt;p&gt;Manganese dioxide catalyzes the decomposition of hydrogen peroxide to water and oxygen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the intermediates in this catalyzed reaction?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="5026" LastEditDate="2018-04-09T02:58:33.137" LastActivityDate="2019-02-26T09:04:41.973" Title="Reaction intermediates of MnO2 catalyzed H2O2 decomposition reaction" Tags="&lt;inorganic-chemistry&gt;&lt;catalysis&gt;&lt;decomposition&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7760" PostTypeId="2" ParentId="7757" CreationDate="2014-01-10T19:03:41.160" Score="2" Body="&lt;p&gt;Do you mean 2-methyl-2,4-pentanediol? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $pK_a$ values of 2-propanol (isopropanol) and 2-methylpropan-2-ol (tert. butanol) are 16.5 and 17, respectively and I'd expect your diol to be in this range range, unless deprotonation is featured by intramolecular hydrogen bonding. In that case, the $pK_a$ might be somewhat smaller, around 15.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-10T19:03:41.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7761" PostTypeId="1" CreationDate="2014-01-10T19:36:22.807" Score="4" ViewCount="76" Body="&lt;p&gt;The easiest carcinogenic thing for me to grasp is radiation, as it directly messes with DNA. Then it seems there are other compounds that simply mimic hormones, but these shouldn't necessarily cause cancer, right? Then there are things that can accelerate mitosis or mess with other processes, which can inadvertently mess with DNA. So my question is, in general, how many different flavors of carcinogens are there? I know there are a lot, but can someone list a good few? Maybe some examples? I'm most interested in how plastics and things like smoke cause cancer. I think a few examples would be best in aiding my understanding.&#xA;Thank you!&#xA;Edit: I also just realized this is the Chemistry Stack Exchange, I think I am more interested in the chemistry side of this question, but it might need to be migrated.&lt;/p&gt;&#xA;" OwnerDisplayName="user1638" LastActivityDate="2014-01-10T20:01:41.567" Title="Carcinogens, how do they work?" Tags="&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7762" PostTypeId="2" ParentId="7761" CreationDate="2014-01-10T20:01:41.567" Score="3" Body="&lt;p&gt;There are at least two different radiationless pathways to wreak havoc on the DNA:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;intercalaction: compounds like benzo[a]yrene or ethidium molecules slide between DNA strands and cause transcription errors&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;alkylation: compounds like N-Nitroso-N-methylurea may transfer alkyl groups to nucleobases&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But the list is by no means complete.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-10T20:01:41.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7763" PostTypeId="2" ParentId="7758" CreationDate="2014-01-10T20:54:20.243" Score="4" Body="&lt;hr /&gt;&#xA;&lt;h1&gt;When you make assumption for guessing the solubility of the molecule in some solvent you should consider the whole molecule, his structure and his size.&lt;/h1&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;We can start with the ether, in the below figure (from wikipedia the one at right ) is shown a diethylether molecule.&lt;/p&gt;&#xA;&lt;p&gt;                           &lt;img src=&quot;https://i.stack.imgur.com/B9NLr.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see that is quite a non-polar molecule (charge 0,green, in the most part of the surface) so this explains why has a low solubility in a polar solvent like water. Same thing for Vitamin A:&lt;/p&gt;&#xA;&lt;p&gt;                           &lt;img src=&quot;https://i.stack.imgur.com/fUp7n.png&quot; alt=&quot;vitamin a electrostaic potential with Avogadro&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The most part of the molecule is not polar. Only one &lt;span class=&quot;math-container&quot;&gt;$-OH$&lt;/span&gt; can't make the difference. And so is for &lt;a href=&quot;http://en.wikipedia.org/wiki/Stearyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-Octadecanol&lt;/a&gt; that has a solubility in water of &lt;span class=&quot;math-container&quot;&gt;$1.1  \times 10^{-3} mg/L$&lt;/span&gt; and &lt;a href=&quot;http://www.thegoodscentscompany.com/data/rw1155861.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nonadecyl alcohol (1-nonadecanol)&lt;/a&gt; seems something like &lt;span class=&quot;math-container&quot;&gt;$4.8  \times 10^{-3} mg/L$&lt;/span&gt; both up to my references have very low solubility in water (where did you find that were not?). For the &lt;span class=&quot;math-container&quot;&gt;$CO2$&lt;/span&gt; is true what you have stated it has not &lt;strong&gt;a very strong&lt;/strong&gt; polarity (but in fact it has a little bit!) because is a linear molecule.&lt;/p&gt;&#xA;&lt;p&gt;                                       &lt;img src=&quot;https://i.stack.imgur.com/oJwrn.png&quot; alt=&quot;carbon dioxide electrostaic potential with Avogadro&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This last example let me introduce to an important factor that influence the solubility: &lt;strong&gt;molecule size&lt;/strong&gt;.&#xA;Little molecule are more easily solvatated by solvent molecule, this bring us to the case of cellulose:&lt;/p&gt;&#xA;&lt;p&gt;                  &lt;img src=&quot;https://i.stack.imgur.com/JNeqV.png&quot; alt=&quot;cellulose with Avogadro&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Cellulose is very &lt;strong&gt;hydrophilic&lt;/strong&gt; and so should be soluble in water, but this doesn't happen because cellulose is a polymer (in figure is show his monomer) that can have a molecular weight very hight. Polymers may swell but can't be solubilize if their chains are bonded with chemicals bonds until the bonds are break.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-10T20:54:20.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7764" PostTypeId="1" CreationDate="2014-01-10T21:35:55.383" Score="2" ViewCount="257" Body="&lt;p&gt;&lt;a href=&quot;http://www.chemicalelements.com/groups/noblegases.html&quot; rel=&quot;nofollow&quot;&gt;Noble gases&lt;/a&gt; were considered to be inert until compounds that include them, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Xenon_trioxide&quot; rel=&quot;nofollow&quot;&gt;xenon trioxide&lt;/a&gt; (as an example) were found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what &lt;strong&gt;natural&lt;/strong&gt; conditions allow the formation of noble gas compounds?&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="81" LastEditDate="2014-01-14T00:42:31.173" LastActivityDate="2014-01-14T00:42:31.173" Title="Are there natural conditions that could enable the formation of noble gas compounds?" Tags="&lt;reaction-mechanism&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7765" PostTypeId="1" AcceptedAnswerId="7771" CreationDate="2014-01-11T00:48:22.900" Score="11" ViewCount="9863" Body="&lt;p&gt;To explain - a friend and I have been working on a program that would allow you to draw organic molecules, indicate the conditions, and predict the products.  This would be primarily aimed at basic, first year undergrad organic chemistry (SN1/2, E1/2, applications, potentially (retro)synthesis, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've realized that it is probably worthwhile to figure out if there are similar existing programs out there.  From what we've read there doesn't seem to be something that meets what we're planning - everything we've found seems to be aimed at professional/higher level chemists, not assisting a beginner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it comes down to a couple main points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are you aware of similar software?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you think that this sort of software would be used (either by yourself or students or whoever).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If/when we have a working version, would users here be interested in using/testing it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I apologize if this doesn't meet the guidelines/standards for questions here, I was kind of iffy about asking it.&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorUserId="30836" LastEditDate="2021-03-03T19:38:34.043" LastActivityDate="2021-03-03T19:38:34.043" Title="Program that simulates basic reactions in organic chemistry" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7766" PostTypeId="2" ParentId="7755" CreationDate="2014-01-11T01:31:41.020" Score="2" Body="&lt;p&gt;Anything that can have an edge will cut you, assuming it is thin enough. This is why some science fiction novels will describe 'super sharp' swords as having a width of a single atom/molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address pyrite, specifically, we'd have to know the shape of your planned fool's gold.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it a lump of it, a bunch of coins or some other 'non-sharp' form?  Those probably wouldn't cut anyone. &lt;/li&gt;&#xA;&lt;li&gt;Is it a 'golden' sword?  If it was thin (sharp) enough then it probably could.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To address your question about the handling of sharp materials - wear gloves, and don't touch the sharp side.  If you can avoid touching it entirely (with tongs, or by lifting whatever it is resting in/on) even better.  As a general rule of thumb contact with the pointy bits is more likely to make you lose that thumb than no contact.&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorUserId="4148" LastEditDate="2014-01-11T02:01:59.723" LastActivityDate="2014-01-11T02:01:59.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7767" PostTypeId="1" CreationDate="2014-01-11T02:59:02.553" Score="4" ViewCount="32135" Body="&lt;p&gt;If I add $aq. \ce{KOH}$ to both of the compounds substitution takes place in the case of benzyl chloride and then if I oxidize with $\ce{KMnO4}$ I will get benzene carboxylic acid which will give effervescence on reaction with $\ce{NaHCO3}$. But, does $\ce{KMnO4}$ have any effect on chloro benzene? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="418" LastEditDate="2014-01-11T06:54:49.347" LastActivityDate="2016-01-21T06:57:07.953" Title="How to distinguish between benzyl chloride and chloro benzene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7768" PostTypeId="1" AcceptedAnswerId="7775" CreationDate="2014-01-11T03:10:16.790" Score="4" ViewCount="2814" Body="&lt;p&gt;To make the question clear I would like to take an example. $\ce{K_3[Fe(CN)6]}$ and $\ce{K3[FeF6]}$ shows different colors in solutions. It is given in my textbook that it is because of different number of unpaired electrons. &#xA;&lt;strong&gt;But, how can the no. of unpaired electrons affect the frequency of light emitted?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colors of co-ordination compounds are due to d-d transition of electrons. $\ce{CN^-}$ is a strong field ligand while $\ce{F^-}$ is a weak field ligand. Thus, there will be difference in the crystal field splitting energy in both of the given compounds. This, results in a different energy gap between the eg and t2g energy levels in them. Thus, when electrons undergo transition different frequencies corresponding to different energy gap is emitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping this as the answer. But, in my textbook the reason is attributed to only the difference in no. of unpaired electrons. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorDisplayName="user4076" LastEditDate="2014-01-11T08:48:29.273" LastActivityDate="2014-01-11T12:50:22.230" Title="How can different no. of unpaired electrons lead to different colours of co-ordination compounds?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7769" PostTypeId="1" CreationDate="2014-01-11T06:09:54.267" Score="4" ViewCount="11733" Body="&lt;p&gt;I've looked at a number of different colloidal/ionic &quot;silver solution generators.&quot;  After my research I'm under a few impressions and if they're wrong please let me know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also am wondering what the product is that one generator sells to add to the ionic silver to create colloidal silver and is it safe to ingest?  I've already seen enough information to believe that ionic silver is safe. I'm wondering if the additive that turns ionic silver into colloidal silver is safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impressions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you set up a silver generator, either homemade or store bought, you send a current through two pieces of silver in distilled water.  The particles released into the water are ONLY ionic silver.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will colloidal silver form over time?  Will ionic silver precipitate or remain suspended? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could easily build a machine to do this rather than buy one from a dealer that seems to withhold info to either prevent comparisons or to prevent laymen from building their own.  I'm looking for any and all comments related to this topic.  Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reply to GM's answer: I hesitate to put up pictures or links to the commercial generators since I'm not sure if that would violate the TOU and I have no interest in advertising for anyone.  I imagine if the commercial generators, which work on same principal as the home made ones, didn't generate at least ionic silver they'd have some explaining (lawsuits) to do to someone and they've been on the market for years.&lt;/p&gt;&#xA;" OwnerUserId="2301" LastEditorUserId="402" LastEditDate="2015-02-14T22:45:35.657" LastActivityDate="2018-05-04T20:55:29.487" Title="What is the process of turning ionic silver solution into colloidal silver solution?" Tags="&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7770" PostTypeId="2" ParentId="7767" CreationDate="2014-01-11T07:08:03.767" Score="6" Body="&lt;p&gt;Your observations are correct and on the short term, i.e. in the time scale of typical reactions in the lab, you'll hardly see an effect of the permanganate on chlorobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there seemingly have been attempts to decontaminate soil, e.g. on production sites, using permanganate for the complete oxidation of pollutants to carbon dioxide. But these processes are supposed to take much longer ( &gt; 1 year) and usually don't play a role in considerations on what will happen under typical lab conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But referring to the title, I'd rather suggest to simply record $^1H$ NMR spectra of the compounds. Even on an ancient 90 MHz spectrometer, you'll be able to spot the signals of the benzylic protons at $\delta$ = 4.5 ppm.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-11T07:08:03.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7771" PostTypeId="2" ParentId="7765" CreationDate="2014-01-11T08:24:35.627" Score="10" Body="&lt;p&gt;As you mentioned, there've been quite some attempts on CAOS (Computer-Assisted Organic Synthesis), such as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www2.chemie.uni-erlangen.de/software/wodca/index.html&quot;&gt;WODCA&lt;/a&gt; from the Gasteiger group &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://osiris.corg.umontreal.ca/chiron.shtml&quot;&gt;CHIRON&lt;/a&gt; from the Hanessian group &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cheminf.cmbi.ru.nl/cheminf/lhasa/&quot;&gt;LHASA&lt;/a&gt; from the Corey group&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I might be awfully wrong, but I think they were/are all closed source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you might want to have a look at &lt;a href=&quot;http://ivan.tubert.org/caos/&quot;&gt;OSET&lt;/a&gt;, an open source Java project on which &lt;a href=&quot;http://ivan.tubert.org/&quot;&gt;Ivan Tubert-Brohman&lt;/a&gt; worked some ten years ago.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-11T08:24:35.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7772" PostTypeId="2" ParentId="7401" CreationDate="2014-01-11T09:38:27.887" Score="1" Body="&lt;p&gt;A simplistic answer is as follows (though, please do refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupling_reaction&quot; rel=&quot;nofollow&quot;&gt;the&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Endergonic_reaction&quot; rel=&quot;nofollow&quot;&gt;links&lt;/a&gt; given by Greg in a comment to the question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the University of Wisconsin page &lt;a href=&quot;http://www.chem.wisc.edu/deptfiles/genchem/netorial/modules/thermodynamics/altdefs/coupling.htm&quot; rel=&quot;nofollow&quot;&gt;Coupling Reactions&lt;/a&gt;, &lt;strong&gt;coupling reactions&lt;/strong&gt; are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a reactant-favored reaction is linked to a product-favored reaction so that both reactions yield products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A main point in the related &lt;a href=&quot;http://www.chem.wisc.edu/deptfiles/genchem/netorial/modules/thermodynamics/free/free05.htm&quot; rel=&quot;nofollow&quot;&gt;University of Wisconsin website&lt;/a&gt; is that if the reaction is reactant-favoured ($\Delta{G} &amp;gt; 0$), then it is not going to happen, so it is coupled with a product-favoured reaction ($\Delta{G} &amp;lt; 0$), leading to &lt;strong&gt;coupled reactions&lt;/strong&gt; are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When chemical reactions are coupled, the free energy released by the exergonic reaction is used to drive the endergonic one toward the formation of products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the overall reation is product-favoured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The website uses an analogy of a coupled train).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great explanation is in &lt;a href=&quot;http://www.youtube.com/watch?v=-ZndltxrNqY&quot; rel=&quot;nofollow&quot;&gt;this YouTube clip&lt;/a&gt; from Dr. John Pollard, of the University of Arizona.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="22" LastEditDate="2014-01-11T10:12:36.257" LastActivityDate="2014-01-11T10:12:36.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7773" PostTypeId="2" ParentId="7032" CreationDate="2014-01-11T10:50:55.770" Score="6" Body="&lt;p&gt;This is not a trivial question to answer, as many diagrams are often simplified and do not state the full chemical reactions that occur, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4H3vg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://water.me.vccs.edu/courses/ENV149/lesson4b.htm&quot; rel=&quot;noreferrer&quot;&gt;Aerobic and Anaerobic Digestion and &#xA;Types of Decomposition&lt;/a&gt; education webpage from the Mountain Empire Community College.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to this image (below) from the &lt;a href=&quot;http://www.ecocorp.com/Other_Pages/anaerobicdigestion.htm&quot; rel=&quot;noreferrer&quot;&gt;Ecocorp page - Metabolic Chain of the Anaerobic Process for the Utilization of Organic Wastes&lt;/a&gt;, water is a reactant in the &lt;strong&gt;acetogenesis&lt;/strong&gt;&#xA; stage and is removed as a product in the &lt;strong&gt;methanogenesis&lt;/strong&gt; stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sample reactions are included in the image)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/laseW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Just in case the image becomes a dead link), the chemical processes &lt;em&gt;involving water&lt;/em&gt; from the diagram are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;acetogenesis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2COOH + 2H2O -&amp;gt; CH3COOH + CO2 + 3H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;methanogensis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4H2 + CO2 -&amp;gt; CH4 + 2H2O}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-11T10:50:55.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7774" PostTypeId="1" AcceptedAnswerId="7777" CreationDate="2014-01-11T12:39:37.220" Score="9" ViewCount="5594" Body="&lt;p&gt;In my textbook the molecularity is defined as(along with the standard definition) the order of the rate determining step(RDS). It is also given that the order can be a fraction but molecularity is always a whole number. But, then if the order of RDS is a fraction molecularity will be a fraction which is a against the definition. So, where is the problem? Is it that the order of RDS can only be a whole number? &lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2017-10-14T10:44:01.080" Title="How can molecularity be the order of the rate determining step?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7775" PostTypeId="2" ParentId="7768" CreationDate="2014-01-11T12:50:22.230" Score="3" Body="&lt;p&gt;The crystal field splitting does definitely contribute.But it is not the only thing to contribute to the colour of the coordinate complex solution. The &lt;em&gt;colour&lt;/em&gt; is equivalent to the &lt;em&gt;spectra&lt;/em&gt; (emission or the absorption) of the solution, and hence by enumerating the factors contributing to the emission spectra of a solution, we can compare the colours of two different solutions.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It usually is an unpaired electron which shifts between the various available orbitals and paired electrons rarely do so, since the energy required to unpair acts as a barrier for its' transitions. Moreover, there might also be spin inversions (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperfine_structure&quot; rel=&quot;nofollow&quot;&gt;Hyperfine structure&lt;/a&gt;) which might give additional lines in the spectra. Presence of unpaired electrons helps in producing additional energy levels in the electric and magnetic fields of the ligands (zeeman and stark effect), providing additional lines in the spectra. All these establish the uniqueness of the spectral distribution which the solution absorbs, and hence the colour.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of unpaired electrons also determines the resultant magnetic field of other species in the solution, which alters the energy levels of the solvating molecules.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-01-11T12:50:22.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7776" PostTypeId="1" AcceptedAnswerId="8376" CreationDate="2014-01-11T13:58:34.950" Score="11" ViewCount="255" Body="&lt;p&gt;&lt;a href=&quot;http://www.baenebooks.com/chapters/0743498747/0743498747___6.htm&quot;&gt;A Pail of Air&lt;/a&gt; is a classic science fiction short story by Fritz Leiber from 1951 describing how a family survives on an Earth that has left its orbit around the sun, using desparate low-tech methods: a shelter isolated by many layers of blankets and a coal fire to thaw frozen oxygen gathered in the eponymous pails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious how realistic this scenario is from a basic chemistry and biology perspective. I think not for a number of reasons, but don't know enough to quantify these factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would a coal fire produce enough energy to thaw the oxygen the fire itself uses?&lt;/li&gt;&#xA;&lt;li&gt;Would the resulting atmosphere (at sufficient pressure for the oxygen to support human life) contain toxic levels of carbon dioxide?&lt;/li&gt;&#xA;&lt;li&gt;...or carbon monoxide?&lt;/li&gt;&#xA;&lt;li&gt;How does oxygen at lower pressures affect flammability? Wouldn't it all go up in a blaze at the drop of a hat?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4149" LastEditorUserId="4149" LastEditDate="2014-01-12T01:03:16.443" LastActivityDate="2014-02-09T13:44:43.857" Title="Would it be possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen?" Tags="&lt;redox&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7777" PostTypeId="2" ParentId="7774" CreationDate="2014-01-11T14:34:06.083" Score="11" Body="&lt;p&gt;What your book, and most other elementary books present, is at best a generalization, not applicable to all cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecularity, as per the collision theory, is defined only for &lt;em&gt;elementary reactions&lt;/em&gt;. One of the things that define an elementary reaction is the fact that the entire transformation from reactants to products occurs simultaneously, and follows from a properly oriented, appropriately energetic &lt;em&gt;collision&lt;/em&gt; of the reactant species. If the reaction is not a direct consequence of a single collision, it is not an elementary reaction, and hence molecularity is meaningless.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what usually happens is that out of several elementary steps constituting a reaction, one is very slow as compared to others. So whatever the rate of that &lt;strong&gt;elementary&lt;/strong&gt; reaction, becomes practically the rate of the entire reaction. And since it is an elementary reaction, its order is its molecularity, and to an approximation, the molecularity of the slowest step then becomes the rate of the equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, in most cases, the slowest step is not slow enough for it to entirely determine the rate. It is the slowest step, and it is an elementary reaction and therefore its molecularity is its order. But now, its order is not necessarily the order of the entire reaction, since that was an approximation valid only in cases where the slowest step was &lt;em&gt;so slow&lt;/em&gt; that it entirely determines the rate. Then, the order of the entire reaction, due to the influence of the other steps (elementary in itself), can be fractional, while the order of the slowest step, is still its integer molecularity. ( here the term &lt;em&gt;slowest step&lt;/em&gt; is preferred over the misnomer &lt;em&gt;rate determining step&lt;/em&gt;, as it does not entirely determine the rate).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind, that many a times what people refer to as the &quot;rate determining step&quot; or the &quot;slowest step&quot; is not an elementary reaction. then its order need not be an integer, even if it is slow enough for the order of the net reaction to be equal to the order of that step.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="30836" LastEditDate="2017-10-14T10:44:01.080" LastActivityDate="2017-10-14T10:44:01.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7778" PostTypeId="2" ParentId="7769" CreationDate="2014-01-11T14:55:08.460" Score="3" Body="&lt;p&gt;If you provide a scheme of this mysterious  &amp;quot;colloidal silver generator&amp;quot; would be very interesting, from what I see from googling I see only things like this.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjL5S.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Up to my knowledge this &amp;quot;colloidal silver generator&amp;quot; has no sense. From instructions notes it seems there are two silver bar in a solution of pure water, as you say, this in fact should create a very slow (because there are no electrolytes) electrolysis of water into &lt;span class=&quot;math-container&quot;&gt;$O_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H_2$&lt;/span&gt; but I don't see any reasons for the producing of colloidal silver.&lt;/p&gt;&#xA;&lt;p&gt;In chemistry laboratory to produce colloidal silver usually you start from a solution of ionic silver &lt;span class=&quot;math-container&quot;&gt;$Ag^{+}$&lt;/span&gt; and reduce it to metallic silver &lt;span class=&quot;math-container&quot;&gt;$Ag$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ag^{+} + e^{-} -&amp;gt; Ag} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Caution the following is a reaction not a recipe do not ingest the ultimate product of this reaction!!!&lt;/h1&gt;&#xA;&lt;p&gt;Once time I've made a true colloidal silver with Lee-Meisel method (&lt;em&gt;J.Phys. CHem 1982,86,3391-3395&lt;/em&gt;) if you look at the original paper there are many methods however the following (&lt;strong&gt;method c&lt;/strong&gt;) is used most.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(c) &lt;span class=&quot;math-container&quot;&gt;$AgNO_3$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$90 mg$&lt;/span&gt;) was dissolved in &lt;span class=&quot;math-container&quot;&gt;$500 mL$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$H_2O$&lt;/span&gt; and brought to boiling. A solution of &lt;span class=&quot;math-container&quot;&gt;$1 %$&lt;/span&gt; sodium citrate (&lt;span class=&quot;math-container&quot;&gt;$10 mL$&lt;/span&gt;) was added. The solution was kept on boiling for cs. 1h.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The solution should be greenish yellow. If you examine the reaction the &lt;span class=&quot;math-container&quot;&gt;$AgNO_3$&lt;/span&gt; is the source of &lt;span class=&quot;math-container&quot;&gt;$Ag^{+}$&lt;/span&gt;, sodium citrate is the reducing agent that reduce &lt;span class=&quot;math-container&quot;&gt;$Ag^{+}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$Ag$&lt;/span&gt; and stabilize it. I think colloidal silver generator would be for anyone who makes colloidal solution a dream I assure you is a very complex and tedious process.&lt;/p&gt;&#xA;&lt;p&gt;Moreover both &lt;span class=&quot;math-container&quot;&gt;$Ag^{+}$&lt;/span&gt; and colloidal silver are very difficult to store, the first may  undergo photo-reduction to the metal the second is unstable as many colloidal solution because colloidal solution have grater energy due their larger surface (More energy many times means more instability). Without any stabilizer (such as t-BuONa or citric acid) I don't think the colloid will last long.&lt;/p&gt;&#xA;&lt;p&gt;However I don't see why you should drink something like this. In fact &lt;span class=&quot;math-container&quot;&gt;$Ag^{+}$&lt;/span&gt; has antiseptic properties such as  Mercurochrome,HydrogenPeroxide or Lysol(R) but this don't mean drinking Lysol, Hydrogenperoxide or mercurochorme every morning will make you feel better. I strongly suggest you to do not drink such thing ever.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-12T10:13:03.547" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7780" PostTypeId="2" ParentId="6404" CreationDate="2014-01-11T16:20:27.437" Score="9" Body="&lt;p&gt;As already pointed out by previous contributors, 1-bromobicyclo[2.2.1]heptane will not undergo substitution via $S_{N}1$. A fully solvated and relaxed bridgehead cation is unlike to be formed. Whether a short-lived ion pair in a common solvent cage is possible... who knows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But fortunately, the title compound isn't as dead as it seems: I'm sure I've seen it converted to the corresponding Grignard reagent and then quenched with $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-11T16:20:27.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7781" PostTypeId="1" CreationDate="2014-01-11T16:20:29.887" Score="3" ViewCount="617" Body="&lt;p&gt;I'm trying to get my hands on some raw interferogram data / raw FID data from an FT-IR / FT-NIR / FT-NMR so I can run some tests using FFT with it (it needs to be real data). Here's a picture below of the data I'm looking for &lt;strong&gt;highlighted in yellow&lt;/strong&gt;.  I'm looking for Alkali metals (Sodium, Potassium, but also Magnesium, Gold, Silver, Platinum and Palladium).  Any ideas where I can get these or if someone can send them to me as a text/data file?  I've searched NIST but everything is already converted&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jHAal.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="554" LastEditorUserId="554" LastEditDate="2014-01-12T00:52:03.237" LastActivityDate="2014-02-17T22:32:17.230" Title="Looking for raw interferogram data / raw FID data from FT-IR / FT-NIR / FT-NMR" Tags="&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7782" PostTypeId="1" CreationDate="2014-01-11T17:16:38.237" Score="2" ViewCount="1295" Body="&lt;p&gt;Let's say I have several pieces of trash and each is made with a different type of material like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plastic bottle : High Density Polyethylene (HDPE) &lt;/li&gt;&#xA;&lt;li&gt;Cereal box : Paperboard &amp;amp; Epichlorohydrin&lt;/li&gt;&#xA;&lt;li&gt;Shrink Wrap : Polyolefin&lt;/li&gt;&#xA;&lt;li&gt;Kitchen utensil : Melamine-resin&lt;/li&gt;&#xA;&lt;li&gt;Textiles : Urea-formaldehyde&lt;/li&gt;&#xA;&lt;li&gt;Crown Cork Bottle Cap : Metal and Plastisol lining&lt;/li&gt;&#xA;&lt;li&gt;Rock&lt;/li&gt;&#xA;&lt;li&gt;Plasterboard &lt;/li&gt;&#xA;&lt;li&gt;Coffee cup : Ceramic and Glaze (silica + metal oxides)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My objective is to build a system complete with sensors that automatically determines the material (chemical makeup) of each object. Are there material sensors currently on the market that can identify all of the materials that make up these objects?&lt;/p&gt;&#xA;" OwnerUserId="4151" LastEditorUserId="4151" LastEditDate="2014-01-12T12:04:40.637" LastActivityDate="2014-01-12T13:48:13.793" Title="What type of sensor technologies exist to determine the chemical composition of a polymer?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;polymers&gt;&lt;materials&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7783" PostTypeId="2" ParentId="7782" CreationDate="2014-01-11T18:44:07.010" Score="1" Body="&lt;p&gt;Well, it seems the question has emerged to &lt;em&gt;How to design a polymer recycling plant?&lt;/em&gt; ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocks and pieces of plasterboard can be separated from the rest of material by streams of compressed air along or across perforated sheets of metal in a conveyor belt: they'll keep sitting there - like a rock ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastics should be milled down to flakes then. The mixture can be submitted to a swim-and-sink step: PE whould swim on water, PS and PVC will sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it is technically useful to repeat this step using NaCl solutions of different concentration to achieve further separation is beyond my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IR spectroscopy will provide a lot of information on the type of polymer measured. Aren't InGaAs line cameras/sensors used in this context? Pattern matching with stored spectra might be used to control streams of air to blow identified material off a conveyor belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry that I can't provide more precise information, but I'm not an engineer. However, I wouldn't be astonished if full-fledged setups are commerically available nowadays.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-12T13:48:13.793" LastActivityDate="2014-01-12T13:48:13.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7784" PostTypeId="1" AcceptedAnswerId="7792" CreationDate="2014-01-11T20:31:24.033" Score="3" ViewCount="1094" Body="&lt;p&gt;I am wondering if any of you know about glue/adhesives that do not harden over time after they are applied or glue/adhesives that take a really long time to harden after application. I need the glue to temporarily hold polymer sheets before they are sewed together.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="19616" LastEditDate="2015-10-30T17:40:12.150" LastActivityDate="2015-10-30T17:40:12.150" Title="Non-hardening adhesive" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7785" PostTypeId="2" ParentId="7782" CreationDate="2014-01-11T20:41:50.637" Score="4" Body="&lt;p&gt;You can identify many polymers without any clever sensor technologies at all but by following a number of simple physical tests. Sensor technology might help for some of the very fine distinctions, but if you can make a broad categorisation first with simple physical tests then that is what you should do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, most plastic opponents are actually labelled with symbols to aid in recycling so you can often just look to identify the chemical nature of the polymer. For reference the symbols are shown below (V identifies chlorine-containing polymers like PVC):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EOJON.jpg&quot; alt=&quot;polymer recycling symbols&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps required to identify polymers, if you don't have a label, involve some simple physical tests. First identify whether the polymer is elastic or hard. Then identify whether it is a thermoplastic or a thermoset (by using a hot object like a soldering iron to see whether it remelts). Then a variety of tests can be applied: does it float in water or not? Does it burn and if so how does it behave and what does the flame look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One flowchart for this is available &lt;a href=&quot;http://www.loyresearchgroup.com/uploads/1/0/2/2/10227236/new_id_chart_.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as a pdf. Another simpler one is &lt;a href=&quot;http://www.mtec.or.th/images/users/78/FAQ_identificationPlastic/Polymer%20Identification.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; also as a pdf but sufficient to distinguish the major classes of polymer. There are even interactive versions of such flowcharts such as &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/artefact/flash/polymeridchart.swf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; (flash).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included below is another version originally produced by the major, but now defunct, British chemical company ICI:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqrZO.jpg&quot; alt=&quot;ICI polymer flowchart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;￼￼&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-11T20:41:50.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7786" PostTypeId="1" CreationDate="2014-01-11T21:40:42.963" Score="7" ViewCount="884" Body="&lt;p&gt;There are a variety of different types of glues, some derived from animal or natural materials and some designed by chemists. Some glues need to be activated before they harden and stick (many epoxies, for example) but others seem to stick very quickly without any obvious activation (like &lt;a href=&quot;http://www.loctiteproducts.com/super-glue.shtml&quot; rel=&quot;nofollow&quot;&gt;Loctite Superglue&lt;/a&gt; or &lt;a href=&quot;http://www.supergluecorp.com/super-glue&quot; rel=&quot;nofollow&quot;&gt;The Original Super Glue&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical reactions make these glues work?&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="4945" LastEditDate="2014-06-12T08:23:24.773" LastActivityDate="2014-06-12T08:23:24.773" Title="What is the chemical basis for superglues?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7787" PostTypeId="2" ParentId="7786" CreationDate="2014-01-11T22:07:46.053" Score="13" Body="&lt;p&gt;This is a very general answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Chemistry World podcast transcript, &lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/06/cyanoacrylate-superglue-adhesive-glue-podcast&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyanoacrylate&lt;/a&gt; and the article &lt;a href=&quot;http://voh.chem.ucla.edu/vohtar/spring00/30H/pdf/Brubaker.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Chemistry of Superglue&lt;/a&gt; (Brubaker, 2000), the main ingredient &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyanoacrylate&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyanoacrylate&lt;/a&gt;, which is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a type of plastic known as an acrylic resin. It exists in the tube as single molecules in liquid form. But when they come into contact with water, the molecules react with hydroxide ions to form long polymer chains that set into a hard solid. As there is a thin layer of moisture covering most surfaces, cyanoacrylate can bond them together very effectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This also explains why superglue tends to harden very quickly when in contact with skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram of cyanoacrylate polymerisation in the presence of water is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wmbhl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the &lt;a href=&quot;http://chem.chem.rochester.edu/~chem421/anionic.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Rochester&lt;/a&gt;, who also state that asides from water, that the presence of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Weak bases (such as those on the proteins in skin) work even better&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for polymerising the cyanoacrylate.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-11T22:07:46.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7789" PostTypeId="2" ParentId="7755" CreationDate="2014-01-12T05:56:27.120" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohs_scale_of_mineral_hardness&quot;&gt;Wikipedia page for Mohs scale&lt;/a&gt; shows that pyrite has a hardness of 6-7 (according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyrite&quot;&gt;the page on pyrite&lt;/a&gt;, it's 6-6.5) while plate glass has a hardness of 5.5 and a pocketknife has 5.1. I've cut myself on both knives and glass before, so I'd wager that pyrite should be able to slice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch is that I've never cut myself on a cube of glass or a cube of steel before--a 90 degree edge just doesn't generate enough pressure for that. So either your pyrite must have very nice serrations on the edge or it must have an edge sharper than 90 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that possible? I don't know. I suppose at this point you could invoke the leeway granted to you by fiction.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2014-01-12T05:56:27.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7790" PostTypeId="1" CreationDate="2014-01-12T08:02:36.650" Score="0" ViewCount="72" Body="&lt;p&gt;I rent an office and want an alternative to a whiteboard and continual dry-erase marker purchases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can shut out the light and use the surface under UV lamps only, that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commonly listed organic glowing elements are blood, urine and semen. None of which, I think would be untraceable by eyes (or nose) during the day!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface I would like to use contains typical white interior paint, applied last year. &lt;/p&gt;&#xA;" OwnerUserId="4153" LastActivityDate="2014-01-12T10:43:47.097" Title="What eco-friendly substance can I use to write and erase on a wall for UV glowing without detection in daylight?" Tags="&lt;organic-chemistry&gt;&lt;materials&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7791" PostTypeId="2" ParentId="7790" CreationDate="2014-01-12T10:43:47.097" Score="4" Body="&lt;p&gt;I think that in a wall you can't write and erase so easily. You can write and then re-paint the surface in this case you can use Zinc White (make sure that what you use is $ZnO$). It was a common white pigment before $TiO_2$ discovery and has a strong fluorescence under UV light so if your wall is white (best $TiO_2$ painted, $CaCO_3$ may have some fluorescence) you will not notice $ZnO$ area in visible light but only under UV light.  &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-01-12T10:43:47.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7792" PostTypeId="2" ParentId="7784" CreationDate="2014-01-12T15:07:51.723" Score="3" Body="&lt;p&gt;I think you could use some types of &lt;a href=&quot;http://en.wikipedia.org/wiki/Butyl_rubber&quot; rel=&quot;nofollow noreferrer&quot;&gt;Butylrubber&lt;/a&gt;. If I'm not wrong from Isobutylene is derived the Post-it adhesive. However you can find Butyl tape easily in every big hardware store (especially nautical) or online.&lt;/p&gt;&#xA;&lt;h1&gt;EDIT&lt;/h1&gt;&#xA;&lt;p&gt;At this &lt;a href=&quot;http://www.exxonmobilchemical.com/Chem-English/yourindustry/butyl-rubber-applications-adhesives.aspx?ln=productsservices&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; you can find some more information this is an excerpt:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tapes and flooring adhesives  Exxon butyl rubber can be used in&#xA;adhesive formulations for transparent tape, hot-melt pressure&#xA;sensitive adhesives, mastic for pipe wrap tape and vinyl floor tile&#xA;adhesives. It can also be used for semiconducting electrical splicing&#xA;tape. A rubber splicing tape must be self-fusing; i.e., when wrapped&#xA;upon itself, it must fuse to the underlying layer, forming a permanent&#xA;bond. The self-fusing is a result of compounding and eliminates the&#xA;need for application of a separate adhesive coat. These compounds have&#xA;very high tensile strengths and exhibit some degree of recovery or&#xA;elasticity. Electrical splicing tapes based on Exxon butyl rubber have&#xA;moisture resistance, outstanding aging resistance and good electrical&#xA;properties. Polyisobutylene elastomers have long been successfully&#xA;used in pressure sensitive adhesives both as primary elastomers and as&#xA;tackifiers and modifiers. Exxon chlorobutyl rubber can be used alone&#xA;or in blends with other polymers to achieve special properties in&#xA;pressure sensitive adhesives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-01-12T19:23:07.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7793" PostTypeId="2" ParentId="7459" CreationDate="2014-01-12T17:15:59.850" Score="4" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; is a fairly strong acid; protonation of (1&lt;i&gt;R&lt;/i&gt;,2&lt;i&gt;R&lt;/i&gt;)-2-methylcyclohexanol (&lt;strong&gt;1&lt;/strong&gt;) gives &lt;strong&gt;1a&lt;/strong&gt;. Water is a pretty good leaving group and that's what it does, resulting in the secondary cation &lt;strong&gt;1b&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLApY.png&quot; alt=&quot;2-methylcyclohexanol-protonation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad news: You have just lost one stereocentre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might remember that tertiary cations are more stable than secondary ones.&#xA;A 1,2 hydride shift converts &lt;strong&gt;1b&lt;/strong&gt; into the more stable &lt;strong&gt;1c&lt;/strong&gt;.&lt;br&gt;&#xA;More bad news: The rearrangement has flattened the tertiary centre.&lt;br&gt;&#xA;The nucleophile &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; can add from both sides of cation, yielding &lt;strong&gt;2&lt;/strong&gt; as the major product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KnR8h.png&quot; alt=&quot;hydrogen shift&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the reaction conditions, minor products will arise from the addition of &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; to &lt;strong&gt;1b&lt;/strong&gt;, or from elimination reactions. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2018-10-12T13:11:44.963" LastActivityDate="2018-10-12T13:11:44.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7794" PostTypeId="1" CreationDate="2014-01-13T00:29:11.387" Score="3" ViewCount="105" Body="&lt;p&gt;The majority of current flat screen TVs and computer monitors are based on Liquid Crystal displays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid crystals are compounds that orient under an applied electric field and alter the polarisation of transmitted light. This can be used to build what are effectively valves that selectively transmit light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are the typical chemical structures of the chemicals that have these properties and how do the structures relates to how they work?&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-13T00:29:11.387" Title="What are the key chemical structures of liquid crystals used in display technology?" Tags="&lt;organic-chemistry&gt;&lt;liquid-crystals&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7795" PostTypeId="1" CreationDate="2014-01-13T01:52:56.070" Score="2" ViewCount="4653" Body="&lt;p&gt;My question is, what makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances? &lt;/p&gt;&#xA;" OwnerUserId="4155" LastEditorUserId="81" LastEditDate="2014-01-13T23:41:17.747" LastActivityDate="2014-01-15T12:24:07.693" Title="What makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances?" Tags="&lt;adsorption&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7796" PostTypeId="1" AcceptedAnswerId="8780" CreationDate="2014-01-13T02:40:15.263" Score="2" ViewCount="14398" Body="&lt;p&gt;Why is $M_1V_1 = M_2V_2$ only used for diluting solutions? Why can it not be used to figure out how many moles to add to make a solution more dilute?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="73" LastEditDate="2014-01-13T04:28:40.603" LastActivityDate="2014-04-07T05:44:49.030" Title="Why is M1*V1 = M2*V2 only used for diluting solutions?" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7797" PostTypeId="1" CreationDate="2014-01-13T04:22:11.707" Score="1" ViewCount="3067" Body="&lt;p&gt;I know that you can't always trust Wikipedia, so I just want to check something regarding their &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot; rel=&quot;nofollow&quot;&gt;kinetic isotope effect page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example implies that $\sideset{^{12}}\ C$ and $\sideset{^{13}}\ C$ can be used to determine whether a reaction is $S_N1$ or $S_N2$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...In the nucleophilic substitution reaction of methyl bromide with&#xA;  cyanide, the kinetic isotope effect of the methyl carbon, in this case&#xA;  defined as $k_{12}/k_{13}$, was found to be 1.082 ± 0.008 ... The observed&#xA;  methyl carbon kinetic isotope effect is indicative of an $S_N2$&#xA;  mechanism.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No explanation appears to be offered for this inference. I know that the reaction will not proceed via $S_N1$ because of the unstable carbocation but (how) does the KIE datum predict an $S_N2$ mechanism?&lt;/p&gt;&#xA;" OwnerUserId="4158" LastEditorDisplayName="user4076" LastEditDate="2014-01-13T05:24:28.817" LastActivityDate="2014-01-17T00:05:41.837" Title="SN1 or SN2, the kinetic isotope effect" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7798" PostTypeId="1" AcceptedAnswerId="7817" CreationDate="2014-01-13T08:57:01.760" Score="10" ViewCount="1543" Body="&lt;p&gt;I know that CCSD(T) with the aug-cc-pVTZ basis set for geometry optimization of omeprazole (5-methoxy-2-[(4-methoxy-3,5-dimethylpyridin-2-yl)methanesulfinyl]-1⁠&lt;em&gt;H&lt;/em&gt;-benzimidazole, the active substance in Losec) is not suitable using normal computers. The substance is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TCIu2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TCIu2.gif&quot; alt=&quot;omeprazole&quot;&gt;&lt;/a&gt;&#xA;(omeprazole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me what exactly makes these methods too expensive for this substance? &lt;/p&gt;&#xA;" OwnerUserId="4162" LastEditorDisplayName="user7951" LastEditDate="2016-12-11T19:55:41.490" LastActivityDate="2016-12-11T19:55:41.490" Title="Why is CCSD(T)/aug-cc-pVTZ not a suitable method for the geometry optimisation of omeprazole?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7799" PostTypeId="2" ParentId="6948" CreationDate="2014-01-13T09:36:16.173" Score="7" Body="&lt;p&gt;As far as the Swarts reaction is concerned: There are indeed reports using $\ce{SbF3}$ alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, however, the active component in the Swarts reaction for the fluorination through halogene exchange is not an $\ce{Sb(III)}$ fluoride, but a mixed $\ce{Sb(V)}$ halide, such as $\ce{SbF3Cl2}$. This may be prepared by reaction of $\ce{SbF3}$ with chlorine (either previously or &lt;em&gt;in situ&lt;/em&gt;), or with $\ce{SbCl5}$ (typically &lt;em&gt;in situ&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'd give the reaction of an alkyl iodide with sodium fluoride in dry DMF a try. Alternatively, I'd start from a tosylate.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-20T18:17:06.653" LastActivityDate="2014-03-20T18:17:06.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7800" PostTypeId="1" CreationDate="2014-01-13T09:47:29.380" Score="16" ViewCount="9542" Body="&lt;p&gt;Why can't fluorine be the central atom in inter-halogen compounds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\ce{F-F}$ bond is weaker than a $\ce{F-X}$ bond and thus, fluorine should be happy to form inter-halogen compounds. But, why doesn't it act as the central atom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my textbook, the reason given is because of the high electronegativity of fluorine atoms. But, how does that affect fluorine being the central atom or not? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If fluorine is the central atom, it can draw electrons from other less electronegative halogens and be happy. Does this lead to instability of the compound?&lt;/p&gt;&#xA;" OwnerUserId="2196" LastEditorUserId="9961" LastEditorDisplayName="user4076" LastEditDate="2017-07-04T14:06:25.700" LastActivityDate="2017-07-04T14:53:53.687" Title="Can fluorine act as the central atom in interhalogen compounds?" Tags="&lt;inorganic-chemistry&gt;&lt;molecules&gt;&lt;halides&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7801" PostTypeId="1" CreationDate="2014-01-13T10:05:14.563" Score="1" ViewCount="3029" Body="&lt;p&gt;In my textbook it is written that gold sol is used for intramuscular injection. But, actually what is its use? Why would people use such a precious element for injection. Can other colloids be used like silver sol? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:I posted this question to get a deeper understanding of what is given in my text about the use of gold sol in intramuscular injection. Googling it didn't help.  &lt;/p&gt;&#xA;" OwnerUserId="2196" LastActivityDate="2014-01-13T18:19:24.817" Title="Why is gold sol used for intra-muscular injection?" Tags="&lt;surface-chemistry&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7802" PostTypeId="2" ParentId="7117" CreationDate="2014-01-13T10:09:42.897" Score="19" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What are the precise conditions under which Benzene undergoes ozonolysis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In &lt;em&gt;Über das Ozobenzol&lt;/em&gt;, an early German publication on the reaction of benzene with ozone, &lt;em&gt;Ber. Dtsch. Chem. Ges.&lt;/em&gt;, &lt;strong&gt;1904&lt;/strong&gt;, &lt;em&gt;337&lt;/em&gt;, 3431-3433 (&lt;a href=&quot;http://dx.doi.org/10.1002/cber.190403703157&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;)&#xA;, the authors write&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Leitet man in solches Benzol 1–2 Stunden lang bei einer Temperatur von 5–10° einen 5-procentigen Ozonstrom, so scheidet sich eine gelatinöse Masse aus, welche stark opalisirt. […]&lt;/p&gt;&#xA;&lt;p&gt;Lässt man das unangegriffene Benzol verdunsten, so bildet es eine weisse, amorphe Masse von furchtbar explosiven Eigenschaften. Selbst beim Uebergiessen mit warmem Wasser detonirte diese Substanz heftig.&lt;/p&gt;&#xA;&lt;p&gt;Uebergiesst man aber das gelatinöse Product mit eiskaltem Wasser,&#xA;so entsteht eine krystallinische Modification, die wohl möglich noch&#xA;heftiger als die amorphe bei der geringsten Berührung, ähnlich wie&#xA;Jodstickstoff, explodirt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To sum this up:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The authors prepared the primary triozonide of benzene by passing a stream of ozone though it at a temperature of 5–10 °C.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once the remaining benzene evaporated at room temperature, a white, amorphous mass with &lt;em&gt;frightfully explosive properties&lt;/em&gt; was obtained.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When warm water was poured over it, this mass detonated.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When cold water was poured over the amorphous, jelly-like mass, a crystalline modification was obtained, which detonated on the slightest touch, similar to to nitrogen triiodide.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In the 1950s, a group of Dutch researchers published a series of articles about their kinetic experiments on the ozonization reactions.&lt;/p&gt;&#xA;&lt;p&gt;The second article in the series, &lt;em&gt;Recl. Trav. Chim. Pays-Bas&lt;/em&gt;, &lt;strong&gt;1951&lt;/strong&gt;, &lt;em&gt;70&lt;/em&gt;, 1005–1019, (&lt;a href=&quot;http://dx.doi.org/10.1002/recl.19510701109&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;) notes that the ozonisation of benzene was performed in chloroform at a temperature of 242 K (–31 °C).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In the late 1960s, ozonization of benzene again gained some interest as a possible source for glyoxal in industrial synthesis.&lt;/p&gt;&#xA;&lt;p&gt;In &lt;em&gt;Glyoxal from Ozonolysis of Benzene&lt;/em&gt;, published in &lt;em&gt;Ind. Eng. Chem. Prod. Res. Dev.&lt;/em&gt;, &lt;strong&gt;1969&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 89–92&#xA;&lt;a href=&quot;http://dx.doi.org/10.1021/i360029a017&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;), the authors describe the ozonolysis of benzene in a mixture of acetic acid and nitromethane at temperatures between –20 °C and 15 °C.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;All anyone can tell me is that it undergoes ozonolysis under &amp;quot;vigorous&amp;quot; conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Apparently, your sources never bothered to check the literature! While the ozonide itself may detonate on touch, &lt;strong&gt;claims on the &amp;quot;vigorous&amp;quot; conditions necessary for its synthesis seem to be fully unsubstantiated by the literature!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-11-16T05:43:10.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7803" PostTypeId="2" ParentId="7801" CreationDate="2014-01-13T10:38:22.653" Score="5" Body="&lt;p&gt;Will this help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2988217/&quot; rel=&quot;nofollow&quot;&gt;Toxicity and cellular uptake of gold nanoparticles: what we have learned so far?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold has a long tradition as a disease-modifying antirheumatic drug in the treatment of rheumatoid arthritis and it seems to relieve pain and decreases swelling of the joints.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-13T18:19:24.817" LastActivityDate="2014-01-13T18:19:24.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7804" PostTypeId="1" AcceptedAnswerId="7814" CreationDate="2014-01-13T12:32:02.873" Score="7" ViewCount="233" Body="&lt;p&gt;I know that catalysts don't change chemical &lt;em&gt;equilibrium&lt;/em&gt; because they accelerate both the reactions in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand why CFC catalysts accelerate ozone destruction but not ozone formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you help me?&lt;/p&gt;&#xA;" OwnerUserId="4163" LastEditorUserId="81" LastEditDate="2014-01-13T23:41:42.747" LastActivityDate="2014-08-20T11:31:24.697" Title="Ozone catalysis" Tags="&lt;catalysis&gt;&lt;atmospheric-chemistry&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7805" PostTypeId="1" AcceptedAnswerId="8796" CreationDate="2014-01-13T15:57:10.583" Score="3" ViewCount="4620" Body="&lt;p&gt;On Wikipedia, it says that polyethylene terephthalate (PET) decomposes at temperatures above 350°C. &#xA;What are these decomposition products? I read further that acetaldehyde is produced, but is that the only decomposition product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what compounds are produced when PET is burned? ($\ce{+O2-&amp;gt;?}$)&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user4076" LastEditDate="2014-02-28T07:53:09.023" LastActivityDate="2014-02-28T07:53:09.023" Title="Decomposition products of PET" Tags="&lt;organic-chemistry&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7806" PostTypeId="2" ParentId="2706" CreationDate="2014-01-13T16:21:27.467" Score="1" Body="&lt;p&gt;If you're willing to start with an ester, the following might serve as a side note to the excellent answer given by Ben Norris:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/C2CC32142D&quot; rel=&quot;nofollow&quot;&gt;Matthias Beller&lt;/a&gt; and coworkers reported on the conversion of esters to ethers using silanes $\ce{R3SiH}$ in the presence of $\ce{Fe3(CO)12}$ as catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reductions of esters to ethers using $\ce{LiAlH4}$ or $\ce{NaBH4}$ can be carried out in the presence of $\ce{BF3\cdot Et2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ester can be cyclic, i.e. a lactone, trichlorsilane in the presence of a radical initiator or under uv light will yield cyclic ethers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're willing to add a carbon atom to your ester or lactone and convert it to an enol ether, the &lt;em&gt;Tebbe olefination&lt;/em&gt; is your friend. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-13T16:21:27.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7807" PostTypeId="2" ParentId="7478" CreationDate="2014-01-13T18:12:58.663" Score="4" Body="&lt;p&gt;All else equal, the smaller group migrates preferientially. The energy required to more the group is related to the mass of the group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=\int_0^\vec{x}\vec{F}\cdot d\vec{x}$$&#xA;$$\frac{d\vec{x}}{dt}=\vec{v}   \ \ \ \ \ \therefore \ \ \ \ \ d\vec{x}=\vec{v}dt$$&#xA;$$W=\int_0^t\left(\vec{F}\cdot \vec{v}\right)dt&#xA;$$$$\vec{F}=m\vec{a}$$&#xA;$$W=m\int_o^t \left(\vec{a}\cdot\vec{v}\right)dt=m\int\left(\vec{v}\cdot\vec{a}dt\right)$$&#xA;$$\vec{a}=\frac{d\vec{v}}{dt}$$&#xA;$$\vec{a}dt=d\vec{v}$$&#xA;$$W=m\int_0^\vec{v}\vec{v}\cdot d\vec{v}=\frac{1}{2}m\vec{v}^2\vert_0^\vec{v}=\frac{1}{2}m\vec{v}^2$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-13T18:12:58.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7808" PostTypeId="2" ParentId="7764" CreationDate="2014-01-13T18:26:28.683" Score="3" Body="&lt;p&gt;As noticed in the comments, there are probably no natural conditions that would allow formation of noble gas compounds; although it should be noted that definition of 'natural' conditions should be specified. Lightning bolts surely carry enough energy to force such a reaction, but it is impossible for them to meet right mixture of gases, due to scarcity of both noble gases and super-strong oxidants and abundance of other substrates; as noted in comments (again), substrate able to create a compound with noble gas will create it with everything else first, and noble gas last.&lt;/p&gt;&#xA;" OwnerUserId="4105" LastActivityDate="2014-01-13T18:26:28.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7809" PostTypeId="1" CreationDate="2014-01-13T19:22:42.043" Score="2" ViewCount="375" Body="&lt;p&gt;$\ce{ROH + NaBr + H2SO4 -&amp;gt; RBr + NaHSO4 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is exactly the reaction mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it $\ce{NaBr + H2SO4 -&amp;gt; NaHSO4 + HBr}$ and then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HBr}$ reacts with $\ce{ROH}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes the why in the reaction of $\ce{NaBr}$ and $\ce{H2SO4}$ does $\ce{H2SO4}$ oxidise $\ce{NaBr}$?&lt;/p&gt;&#xA;" OwnerUserId="4133" LastEditorUserId="418" LastEditDate="2014-01-13T20:14:26.590" LastActivityDate="2014-01-13T20:22:28.977" Title="explain the reaction mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-01-14T09:56:34.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="7810" PostTypeId="1" CreationDate="2014-01-13T20:03:04.110" Score="1" ViewCount="218" Body="&lt;p&gt;For an ideal gas, how can someone obtain this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{V}dP = RdT- P \bar{dV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which it can de deducted, that for a constant temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dP}{\bar{dV}} = -RT $$&lt;/p&gt;&#xA;" OwnerUserId="4165" LastActivityDate="2014-01-14T13:03:57.647" Title="Ideal gas derivations" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7811" PostTypeId="2" ParentId="7809" CreationDate="2014-01-13T20:22:28.977" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;HBr reacts with ROH?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly. It's just a nucleophilic substitution.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-13T20:22:28.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7812" PostTypeId="1" CreationDate="2014-01-13T20:26:01.537" Score="5" ViewCount="1414" Body="&lt;p&gt;Does anyone know how to draw (predict) a hypothetical COSY NMR spectra for AMX spin system (the spin of A M X nuclei is 1/2)?&lt;/p&gt;&#xA;" OwnerUserId="4166" LastEditorUserId="4945" LastEditDate="2015-02-26T07:43:53.207" LastActivityDate="2015-02-26T07:43:53.207" Title="How to predict a COSY NMR spectra for an AMX spin system?" Tags="&lt;analytical-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7813" PostTypeId="2" ParentId="7812" CreationDate="2014-01-13T21:05:05.390" Score="6" Body="&lt;p&gt;An AMX spin system represents three nuclei coupled to each other, with their signals separated by a large chemical shifts compared to the coupling between them (such that $\Delta \delta &amp;gt;&amp;gt; 3J$). This system is said to be first-order, and the A, M and X signals are each a doublet of doublets, and the couplings can be extracted by inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple COSY spectrum of this will show the three core signals along the diagonal, with cross peaks appearing for all signals ie each signal will have two cross peaks. Below is a (very basic) schematic for an AMX COSY.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XgAjT.jpg&quot; alt=&quot;COSY of AMX spin system&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The appearance of the cross peaks will depend on the type of COSY experiment you are running, such as a COSY-45 or COSY-dqf etc, but this should give you the general picture for what you are asking.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-01-13T21:05:05.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7814" PostTypeId="2" ParentId="7804" CreationDate="2014-01-13T22:59:47.690" Score="10" Body="&lt;p&gt;You make the assumption that the ozone concentration in the upper atmosphere is in equilibrium. It isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O3}$ is a much less stable molecule than $\ce{O2}$ (the heat of formation from $\ce{3/2 O2}$ is $143~\mathrm{kJ/mol}$) and the concentration at equilibrium would be &lt;em&gt;very&lt;/em&gt; low. A significant concentration exists in the upper atmosphere because  $\ce{O3}$ is continuously formed from the reaction of $\ce{O2}$ and UV light. Also, though it is a highly reactive molecule, there are still kinetic barriers to its conversion back to $\ce{O2}$. There is no simple chemical route back to normal oxygen and the UV absorption of the $\ce{O2}$ molecule that creates ozone is different to the UV absorption of the ozone molecule, so the driving force of ozone creation doesn't destroy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CFCs release $\ce{Cl}$ atoms in the upper atmosphere via a photochemical reaction. The $\ce{Cl}$ atoms react with ozone in a variety of reactions the simplest of which are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl + O3 -&amp;gt; ClO + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ClO + O3 -&amp;gt; Cl + 2 O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, strictly speaking, the $\ce{Cl}$ from CFCs is not directly catalysing the equilibrium between $\ce{O3}$ and $\ce{O2}$ but providing a new reaction pathway via several intermediates that, in a roundabout way, allows the &quot;equilibrium&quot; to be reestablished. Another way to think about this is that the presence of $\ce{Cl}$ facilitates a lower kinetic barrier route to ozone destruction. A lower kinetic barrier to the reaction allows the relative concentrations to get closer to the natural equilibrium. So even if there were a reverse pathway (which there isn't in these complex reactions) the result would be a much lower concentration of ozone.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="4945" LastEditDate="2014-08-20T11:31:24.697" LastActivityDate="2014-08-20T11:31:24.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7815" PostTypeId="1" CreationDate="2014-01-13T23:47:16.917" Score="5" ViewCount="766" Body="&lt;p&gt;I know that the best source of lye is hardwood. A brief search tells me that Bamboo is considered a hardwood of the monocotyledon variety. I've done a lot of searching to find this answer, but I can't seem to find if I can use the same process with bamboo ash as I would with, say, Apple ash, to extract lye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is two-fold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is it possible to extract lye from bamboo ash?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if so, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Would this lye be a quality lye, capable of being used to create bar soap?&lt;/p&gt;&#xA;" OwnerUserId="4167" LastEditorUserId="4167" LastEditDate="2014-01-14T00:04:11.577" LastActivityDate="2016-09-01T13:33:57.493" Title="Can lye be extracted from bamboo ash?" Tags="&lt;home-experiment&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7817" PostTypeId="2" ParentId="7798" CreationDate="2014-01-14T07:34:09.847" Score="13" Body="&lt;p&gt;To transform the question into something simpler, let's assume CISD instead CCSD(T), which is cheaper and share the common expensive step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule has 182 electrons and 1545 basis functions in the basis you specified, that means approx. 1400 virtual orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the number of double excitation $ij \rightarrow ab$ is approximately $180 \cdot 180 \cdot 1400 \cdot 1400 = 32400 \cdot 1960000 = 63504\cdot10^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, in each of this excitation you have to calculate integral of type $(ab|cd)$, and there are $1400^4$ of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a big problem is out of reach for traditional CCSD, but tractable with more approximate methods.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="194" LastEditDate="2016-12-11T19:42:06.383" LastActivityDate="2016-12-11T19:42:06.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7818" PostTypeId="1" CreationDate="2014-01-13T06:08:29.000" Score="4" ViewCount="380" Body="&lt;p&gt;I have read many books on standard electrode potential and they define the term in various ways but I just can’t seem to get an intuitive idea of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^2+ + 2e- &amp;lt;=&amp;gt; Cu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard electrode potential, $E= 0.34\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to copper half cell, $0.34\ \mathrm{J}$ of energy is released?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn^2+ + 2e- &amp;lt;=&amp;gt; Zn}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard electrode potential, $E= -0.76\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to zinc half cell, $0.76\ \mathrm{J}$ is absorbed?&lt;/p&gt;&#xA;" OwnerUserId="2595" OwnerDisplayName="Eliza" LastEditorUserId="10185" LastEditDate="2015-06-11T17:36:33.977" LastActivityDate="2015-06-11T17:36:33.977" Title="Intuitive understanding of standard electrode potential" Tags="&lt;electrochemistry&gt;&lt;energy&gt;&lt;reduction-potential&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7819" PostTypeId="2" ParentId="7818" CreationDate="2014-01-14T07:36:57.010" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_electrode_potential&quot; rel=&quot;nofollow&quot;&gt;absolute electrode potential&lt;/a&gt;, i.e. the actual volts between electrode and electrolyte, is incredibly difficult to measure accurately.    Instead we simplify: declare the hydrogen electrode potential to be zero, when actually it's roughly +4.4V @25C deg.    Then measure other electrode potentials using a standard hydrogen electrode to contact the electrolyte.&lt;/p&gt;&#xA;" OwnerUserId="33761" OwnerDisplayName="wbeaty" LastActivityDate="2014-01-14T07:36:57.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7820" PostTypeId="2" ParentId="7369" CreationDate="2014-01-14T10:02:03.843" Score="4" Body="&lt;p&gt;I'm definitely not an expert in organometallic chemistry, and I don't know which aspect ($\eta^{\mathbf{4}}$, any other hapticity, or benzene as a dienophile or diene) is more important for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here's a &lt;a href=&quot;http://dx.doi.org/10.1021/ja011689q&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;JACS&lt;/em&gt; paper&lt;/a&gt;, which describes the the Diels-Alder reaction of N-methyl maleimide with an $\eta^2$-rhenium complex of benzene as the &lt;strong&gt;diene&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-14T10:02:03.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7821" PostTypeId="2" ParentId="7020" CreationDate="2014-01-14T11:42:42.783" Score="0" Body="&lt;p&gt;Does the 1958 Nature paper by &lt;a href=&quot;http://dx.doi.org/10.1038/181662a0&quot; rel=&quot;nofollow&quot;&gt;Kendrew&lt;/a&gt; include any useful details? (Sorry, I'm behind a paywall right now).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-14T11:42:42.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7822" PostTypeId="2" ParentId="7810" CreationDate="2014-01-14T13:03:57.647" Score="3" Body="&lt;p&gt;This looks like the differential of the ideal gas equation. I'm going to use &lt;span class=&quot;math-container&quot;&gt;$V_m$&lt;/span&gt; for the molar volume.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$PV_m=RT$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we convert each side to the differential. We keep &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_m$&lt;/span&gt; together since they are both variable. We can move &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; out of the differential because it is a constant.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d\left(PV_m\right)=RdT$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can evaluate &lt;span class=&quot;math-container&quot;&gt;$d\left(PV_m\right)$&lt;/span&gt; using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Product_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;product rule&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d\left(PV_m\right) = PdV_m + V_mdP$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$PdV_m + V_mdP = RdT$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which is equivalent to your first equation.&lt;/p&gt;&#xA;&lt;p&gt;**However, at constant temperature &lt;span class=&quot;math-container&quot;&gt;$dT=0$&lt;/span&gt;, so at constant temperature, the equation simplifies to:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$PdV_m + V_mdP=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V_mdP = - PdV_m$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dP}{dV_m}=-\frac{P}{V_m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which is not equal to your second equation. The closest I can get is the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dP}{dV_m}=-\frac{P}{V_m}\cdot\frac{V_m}{V_m}=-\frac{PV_m}{V^2_m}=-\frac{RT}{V^2_m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which is what we would get by evaulating:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\partial}{\partial V_m}\right)_T \left(P=\frac{RT}{V_m}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-14T13:03:57.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7823" PostTypeId="1" AcceptedAnswerId="7826" CreationDate="2014-01-14T14:22:33.910" Score="3" ViewCount="1366" Body="&lt;p&gt;I'm trying to read &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/e3768?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow noreferrer&quot;&gt;this chemical formula&lt;/a&gt; but I'm a bit confused. I'm hoping that someone could help break this down for me to explain what the various parts mean. Someone purchased &lt;em&gt;epigallocatechin&lt;/em&gt; in a purchase request and I found that it's synonymous with the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(−)-&lt;em&gt;cis&lt;/em&gt;-2-(3,4,5-Trihydroxyphenyl)-3,4-dihydro-1(2&lt;i&gt;H&lt;/i&gt;)-benzopyran-3,5,7-triol&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the minus up front refers to the way that the polarity changes as light passes through the molecule. What does the 3,4,5 refer to and the 2?&lt;/p&gt;&#xA;" OwnerUserId="4171" LastEditorUserId="544" LastEditDate="2018-11-18T23:50:39.700" LastActivityDate="2018-11-18T23:55:56.970" Title="How to read a given chemical formula?" Tags="&lt;nomenclature&gt;&lt;identification&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7824" PostTypeId="2" ParentId="7823" CreationDate="2014-01-14T15:03:28.997" Score="0" Body="&lt;p&gt;3,4,5 are the locants of phenolic OH on the trihydroxyphenyl substituent and 2- is its position on the 1-benzopyran. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-01-14T15:03:28.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7825" PostTypeId="2" ParentId="7823" CreationDate="2014-01-14T16:10:43.727" Score="5" Body="&lt;p&gt;In addition to what ssavec already outlined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cis&lt;/em&gt; refers to the relative stereochemistry of the substituents at C2 and C3 of the benzopyran, which is the core structure (and thus name-giving) of the title compound. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3,4-dihydro-1(2H)-benzopyran&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This tells us a lot!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Benzopyrans come in two isomers, which are distinguished by the position of the $\ce{O}$ atom &lt;strong&gt;in&lt;/strong&gt; the pyrane ring. Apparently this is a 1-benzopyran, the $\ce{O}$ is close to the benzene ring.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Benzopyrans have a $\ce{C=C}$ double in the pyrane ring as well as a saturated centre, the one with the two $\ce{H}$ atoms. The position of the double bond is determined by the saturated centre. Exactly this is denoted with (&lt;em&gt;2H&lt;/em&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a matter of fact, our benzopyrane doesn't have a double bond ;) 3,4-dihydro denotes were it was, i.e. where a formal hydrogenation has taken place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-14T16:10:43.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7826" PostTypeId="2" ParentId="7823" CreationDate="2014-01-14T17:43:50.727" Score="10" Body="&lt;p&gt;Klaus's answer has all the right explanations but I always find these things hard to follow without pictures, so here are some to clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the pictures below we don't explicitly draw carbons or the hydrogens attached to them (we just mentally add enough hydrogens to make the number of bonds to each carbon equal 4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First what is benzopyran?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HJumW.png&quot; alt=&quot;benzopyran numbering&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two numbers on the first structure tell us where the oxygen appears in the pyran ring and where the saturated carbon (with an extra hydrogen) appears. Both could be in different places. The second picture showns the numbering convention for each atom in the ring so we can keep track of where anything else is attached. The third picture shows what happens if we notionally add extra hydrogens on carbons 3 and 4 to give a fully saturated ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add an OH (hydroxy) in certain positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KFndl.png&quot; alt=&quot;parts of epigallocatechin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture on the right shows how we number the parts of trihydroxy phenyl when it is attached to something else (it will be stuck on the 2-carbon on the benzopyran in the final compound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we have the epigallocatechin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1HDlg.png&quot; alt=&quot;epigallocatechin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;cis&quot; indicates that the two hydrogens (mentally fill them in in positions 2 and 3) are on the same side of the benzopyran (the opposite side to the two wedge-like bonds on the OH and phenyl groups). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that all makes some sort of sense. Note that none of the above bears any relation to how the molecules are made: this is all notional deconstruction for the purpose of explaining how the molecule is named.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-14T17:43:50.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7827" PostTypeId="1" AcceptedAnswerId="7829" CreationDate="2014-01-14T22:12:43.980" Score="3" ViewCount="1485" Body="&lt;p&gt;It seems that the terms mass spectrometry, mass spectroscopy, and mass spectrography are mostly used like synonyms. Is this correct? Or is there a difference?&lt;/p&gt;&#xA;" OwnerUserId="4173" LastActivityDate="2014-01-15T07:03:57.137" Title="What's the difference between mass spectrometry, mass spectroscopy, and mass spectrography?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7829" PostTypeId="2" ParentId="7827" CreationDate="2014-01-15T07:03:57.137" Score="2" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/368325/mass-spectrometry#ref619843&quot; rel=&quot;nofollow&quot;&gt;Encyclopaedia Britannica&lt;/a&gt;, mass spectrographs use(d) photographic plates or films for the detection of ions, as in the apparatus thought out by &lt;a href=&quot;http://www-outreach.phy.cam.ac.uk/camphys/massspectrograph/massspectrograph_index.htm&quot; rel=&quot;nofollow&quot;&gt;F. W. Aston&lt;/a&gt; in 1919.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-15T07:03:57.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7830" PostTypeId="1" AcceptedAnswerId="23911" CreationDate="2014-01-15T11:35:06.400" Score="2" ViewCount="158" Body="&lt;p&gt;Based on some physics background, I want to know whether have scientists synthesized these two materials: $\ce{Na_{$x$}IrO2}$ and $\ce{NaOsO2}$ in the lab, if so, what properties do they have? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be very appreciative if someone could recommend me some relevant articles. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would you be willing to share your interest in these two materials? For example, is there a specific application in mind? If so, we might be able to suggest alternate materials should these two not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to seek a material that has triangular lattice structure, strong spin-orbital coupling (e.g. 5d element) and large (or moderate) coulomb interaction between electrons. For example, $\ce{ABO2}$ type, $\ce{NaOsO2}$ compounds, where $\ce{A}$ and $\ce{B}$ are alkali and transition-metal ions, respectively. So far, I know $\ce{IrTe2}$ may be a one. &#xA;Do you mean that both $\ce{NaxIrO2}$ and $\ce{NaOsO2}$ do not exist?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any information on iridium and osmium oxides in these low oxidation states?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, $\ce{Na2IrO3}$ and $\ce{Li2IrO3}$. But these two belong to the honeycomb lattice structure, and what about the triangular lattice structure?&lt;/p&gt;&#xA;" OwnerUserId="4176" LastEditorUserId="4945" LastEditorDisplayName="user4076" LastEditDate="2015-01-19T02:34:54.120" LastActivityDate="2019-05-24T12:04:18.990" Title="Do the compounds NaₓIrO₂ and NaOsO₂ exist?" Tags="&lt;physical-chemistry&gt;&lt;synthesis&gt;&lt;materials&gt;&lt;transition-metals&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7831" PostTypeId="2" ParentId="7795" CreationDate="2014-01-15T12:24:07.693" Score="4" Body="&lt;p&gt;Charcoal (although &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;activated carbon&lt;/a&gt;, a specially processed microporous charcoal product is better) is a material that is mostly carbon. The black allotrope of carbon is &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite&quot; rel=&quot;nofollow noreferrer&quot;&gt;graphite&lt;/a&gt;, which is made of overlapping planar sheets of &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;nofollow noreferrer&quot;&gt;graphene&lt;/a&gt;. Graphite is very nonpolar (since it has only carbon atoms). Thus, nonpolar substances will adsorb. Most organic substances impurities are nonpolar substances (though not all, for example, sugars). Additionally, the graphene sheets are aromatic. The ring current generates a dipolar magnetic field and a quadrupolar electric field that attract other aromatic substances through &lt;a href=&quot;http://en.wikipedia.org/wiki/Pi_stacking&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\pi$-stacking&lt;/a&gt;. Thus graphite is best at removing aromatic substances, which include all organic dyes. &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_filtering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbon filtering&lt;/a&gt; can also remove contaminants like chlorine, although this is likely a reaction leading to chlorination of the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silica $(\ce{SiO2})$ and alumina $(\ce{Al2O3})$ are polar materials. They are polar for three reasons. 1) They are composed of polar $\ce{Si-O}$ and $\ce{Al-O}$ covalent bonds. 2) The geometries of these minerals especially around the oxygen atoms are not perfectly in-line with VSEPR predictions, so polarity is not canceled by symmetry (see the silica unit image below). &#xA;&lt;img src=&quot;https://i.stack.imgur.com/N5XDk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Both materials are hygroscopic and absorb/adsorb water from the atmosphere. The $\ce{Si-O-Si}$ and $\ce{Al-O-Al}$ covalent network on the surface is partially hydrolyzed so that the surfaces of these materials actually present $(\ce{-OH})$ groups, increasing the polarity of the surface. Polar substances are attracted to silica and alumina by &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole-dipole_interaction#Dipole-dipole_interactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;dipole-dipole interactions&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen bonding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{-Si-O-Si - + H2O -&amp;gt; -Si-OH + HO-Si -}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-15T12:24:07.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7832" PostTypeId="1" CreationDate="2014-01-15T12:38:35.767" Score="3" ViewCount="2039" Body="&lt;p&gt;I want to convert terephthalic acid (TPA) to an alcohol (however I think the correct term would be &quot;a phenol&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMz88.png&quot; alt=&quot;Terephthalic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that lithium tetrahydridoaluminate ($\ce{LiAlH4}$) can produce an alcohol from a carboxylic acid. And, according to Wikipedia, dicarboxylic acids (such as TPA) react around the same. So does $\ce{LiAlH4}$ reduce TPA to a phenol (or an other product)?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="5026" LastEditDate="2018-07-14T09:13:56.960" LastActivityDate="2018-07-14T09:13:56.960" Title="Reduction of terephthalic acid to an alcohol (or phenol?)" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7833" PostTypeId="1" AcceptedAnswerId="7841" CreationDate="2014-01-15T12:48:57.533" Score="3" ViewCount="15442" Body="&lt;p&gt;I want to know what is an equivalent and how is it different from equivalent mass. How to calculate and why is it used in chemistry for knowing concentration?&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="9961" LastEditDate="2018-12-06T22:49:27.613" LastActivityDate="2021-05-21T20:31:06.183" Title="What is an equivalent?" Tags="&lt;solutions&gt;&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7835" PostTypeId="2" ParentId="7832" CreationDate="2014-01-15T14:47:33.640" Score="8" Body="&lt;p&gt;I want to make sure that you understand the different between two commonly confused molecules (their names don't help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phenol&lt;/a&gt; is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QntoD.png&quot; alt=&quot;phenol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Benzyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Benzyl alcohol&lt;/a&gt; is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/fb/Alkohol_benzylowy.svg/150px-Alkohol_benzylowy.svg.png&quot; alt=&quot;benzyl alcohol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversion of benzoic acid to phenol requires loss of a carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ph{\bf C}O2H -&amp;gt; PhOH + {\bf CO}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_aluminium_hydride#Use_in_organic_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lithium aluminum hydride&lt;/a&gt; converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, &lt;strong&gt;stop here.&lt;/strong&gt; If you want the phenol, read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{-&amp;gt;}}}\ce {PhCH2OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a &lt;a href=&quot;http://en.wikipedia.org/wiki/Swern_oxidation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Swern oxidation&lt;/a&gt;. Then benzaldehyde can be converted to phenyl formate using a &lt;a href=&quot;http://www.organic-chemistry.org/abstracts/lit2/958.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baeyer-Villiger Oxidation&lt;/a&gt;. The formate could be hydrolyzed in aqueous base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PhCH2OH \overset{Swern}{-&amp;gt;} PhCHO}$$&#xA;$$\ce{PhCHO \overset{Baeyer-Villiger}{-&amp;gt;} PhOCHO}$$&#xA;$$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \  H2O}}{\underset{2. \ \ce{H3O+}}{\ce{-&amp;gt;}}} \ce{PhOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it would be easier to buy the phenol. In your case, you want &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroquinone&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydroquinone.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/Hydrochinon2.svg/200px-Hydrochinon2.svg.png&quot; alt=&quot;hydroquinone&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.570" LastActivityDate="2014-01-15T16:49:55.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7837" PostTypeId="2" ParentId="6627" CreationDate="2014-01-15T16:10:44.247" Score="4" Body="&lt;p&gt;This topic is really fun. You see in case of boron molecule $\sigma_{2s}$ and $\sigma_{2p_z}$ (being on the intermolecular axis) will have the same &lt;strong&gt;symmetry&lt;/strong&gt; and hence the orbitals will overlap and result in the formation of 2 bonded $sp_z$ orbitals. So the $\sigma_{2p_z}$ orbital will obtain somewhat $s$ character and it will become unstable or in other words will be of higher energy. The $\sigma_{2s}$ MO will obtain somewhat $p$ character. So in a energy level MO diagram $\sigma_{2p}$ bonding orbital is placed above the $\pi_{ 2p}$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in case of $\ce{O2}$, due to the higher nuclear charge of oxygen, the &lt;strong&gt;energy difference&lt;/strong&gt; of $2s$ and $2p$ is large so there is no such kind of intermixing between the bonded MOs. Whereas in boron, the energy difference of $2s$ and $2p$ is &lt;em&gt;less, facilitating intermixing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find this topic in any higher level chem books. Hope you got my point; thank you.&lt;/p&gt;&#xA;" OwnerUserId="4179" LastEditorUserId="-1" LastEditDate="2018-02-27T19:45:47.667" LastActivityDate="2018-02-27T19:45:47.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7838" PostTypeId="1" AcceptedAnswerId="7972" CreationDate="2014-01-15T16:37:16.467" Score="2" ViewCount="12112" Body="&lt;p&gt;If the pKa of Glycerol is 14.15. How do you calculate the pH for it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the Henderson-Hasselbalch derivative that works for weak acids and bases is not applicable here.&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="418" LastEditDate="2014-03-18T10:23:32.503" LastActivityDate="2014-03-18T10:23:32.503" Title="What is the pH of 1M Glycerol?" Tags="&lt;ph&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7839" PostTypeId="1" CreationDate="2014-01-15T16:51:41.467" Score="4" ViewCount="656" Body="&lt;p&gt;The following two questions prompted me to ask this question. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the order of the following compounds according to reactivity in Williamson Ether Synthesis.&lt;br&gt;&#xA;  a. $\ce{CH3CH2CH2Br}$&lt;br&gt;&#xA;  b. $\ce{CH2=CHCH2Cl}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of these represent the ether($\ce{CH2=CHCH2OC3H7}$) on forming peroxide in air&lt;br&gt;&#xA;  a. $\ce{CH2=CHCH2OCH(OOH)CH2CH3}$&lt;br&gt;&#xA;  b. $\ce{CH2=CH-CH(OOH)OC3H7}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In both the case, I think the question is the relative ease of nucleophilic attack on carbon with $\alpha$-unsaturation. In the former, the bromine is a better leaving group. But, the presence of $\alpha$-unsaturation in the chloro compound, changes the order to b&gt;a. In the latter, the attack of oxygen depends on the most stable electrophilic centre.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:-  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is $\alpha$-unsaturation favourable for nucleophilic attack?&lt;/strong&gt; Relevant factors would be the increased electrophilicity of the carbon attached to the $sp^2$ hybridised ethylinic carbon and the repulsion of the negative nucleophile by the pi electrons. &lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="7448" LastEditDate="2015-10-09T15:37:08.867" LastActivityDate="2015-10-09T15:37:08.867" Title="SN2 nucleophilic attack with alpha unsaturation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7840" PostTypeId="1" AcceptedAnswerId="7844" CreationDate="2014-01-15T17:11:07.307" Score="-2" ViewCount="411" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A cylinder, 0.5 m long and of cross section &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, closed at each end is fitted with a smooth thermally insulating piston, so as to contain gas in each end A and B. Initially, the piston is in equilibrium at a distance of 0.2 m from one end of the cylinder, and the gas in each compartment is at temperature 27 °C. The temperature of the smaller volume of gas A is now increased to 177 °C, while that of gas in the larger volume B is maintained at 27 °C. The piston moves through a distance of &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; until a position is obtained.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) Write an equation relating the initial and final pressure, volume and temperature of the gas in A.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) Write an equation relating the initial and final pressure, volume and temperature of the gas in B.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) Hence find the new position of equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;P.S. This is not my assignment or homework so you could feel free to explain.&lt;/p&gt;&#xA;" OwnerUserId="4180" LastEditorDisplayName="user7951" LastEditDate="2019-05-19T08:55:42.447" LastActivityDate="2019-05-19T08:55:42.447" Title="Thermodynamics Problems!" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-05-19T08:55:51.260" ContentLicense="CC BY-SA 4.0" />
  <row Id="7841" PostTypeId="2" ParentId="7833" CreationDate="2014-01-15T17:50:35.990" Score="5" Body="&lt;p&gt;&lt;em&gt;In reaction stoichiometry, it is the amount of one substance that reacts with one mole of another substance.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Equivalent mass is molar mass divided by the &lt;em&gt;n factor&lt;/em&gt; which isn't always dependent on the other reactant. As a rule of thumb &amp;quot;One equivalence of one reactant reacts completely with one equivalence of another.&amp;quot;&#xA;These concepts of equivalence are extremely important in acid-base titration to determine the neutralisation point.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$xM_1V_1 = yM_2V_2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;here &lt;span class=&quot;math-container&quot;&gt;$M_i$&lt;/span&gt; is molarity.&lt;/p&gt;&#xA;&lt;p&gt;For a relation in terms of equivalence use normality.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Normality is a measure of concentration that is equal to the gram equivalent weight per liter of solution. Gram equivalent weight is a measure of the reactive capacity of a molecule.&lt;/p&gt;&#xA;&lt;p&gt;As an example a solution containing &lt;span class=&quot;math-container&quot;&gt;$\pu{1M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; is actually &lt;span class=&quot;math-container&quot;&gt;$\pu{2N}$&lt;/span&gt; as each molecule dissociates to give 2 &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions.&lt;/p&gt;&#xA;&lt;p&gt;I hope that clears it.&lt;/p&gt;&#xA;" OwnerUserId="4035" LastEditorUserId="39388" LastEditDate="2021-04-15T22:56:01.583" LastActivityDate="2021-04-15T22:56:01.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7843" PostTypeId="1" CreationDate="2014-01-15T19:06:17.640" Score="3" ViewCount="402" Body="&lt;p&gt;I've been toying with different potential recipes for DIY photoresist. My current favored design is a chemically amplified resist based on PVA (poly(vinyl alcohol)). For those who are unaware, a CAR works like this: a polymer is functionalized with a protecting group, and is mixed with a photoacid generator (PAG), which as the name suggests, generates acid upon exposure to light. The acid then cleaves the protecting group (in this case, t-Boc), rendering the polymer soluble in the developer. Basically, my question is how to functionalize PVA with the t-Boc group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References are scant, but I did find one, which is unfortunately extremely vague. It says to react fully hydrolyzed PVA with di-tert-butyl dicarbonate (boc anhydride) and sodium hydride in N-methylpyrrolidone. However, &lt;a href=&quot;http://proceedings.spiedigitallibrary.org/proceeding.aspx?articleid=941239&quot; rel=&quot;nofollow&quot;&gt;the paper itself&lt;/a&gt; says that this reaction is not very predictable, and the authors are (were?) looking for better options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being satisfied, I decided to lookup the reaction used for functionalizing poly(hydroxystyrene), which is the polymer used in the original CAR, APEX. Unfortunately, I wasn't really able to find anything on this either, save for even more vague references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, I have looked up information on how amino acids are protected with t-Boc, but the problem here is that I'm not sure if that would even work on PVA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any idea how I would I go about doing this? Even more preferable would be a method not involving dangerous/explosive and for that matter, expensive substances such as Sodium hydride, or dangerous solvents (although NMP is not really that bad).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find &lt;a href=&quot;http://www.arkat-usa.org/get-file/28638/&quot; rel=&quot;nofollow&quot;&gt;one option&lt;/a&gt;  that simply involves reacting amino acids directly with molten boc anhydride. Is it likely this procedure would work with PVA?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2014-01-15T20:09:11.677" Title="How to synthesize t-Boc protected derivative of PVA" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7844" PostTypeId="2" ParentId="7840" CreationDate="2014-01-15T19:25:47.827" Score="6" Body="&lt;p&gt;The volume $V$ of a cylinder is given as $V=\pi r^2 h$ where $h$ is the height and $r$ the radius. Since the cross-section of the cylinder is $S$, the radius is given as $r=S/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that the piston is at equilibrium in the initial position, and the equilibrium in this case is $p_\text{A} = p_\text{B}$ for the pressures of the corresponding gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know the ideal gas law, which, when adding the information above (and assuming the gases A and B behave ideally) is: $$pV = nRT \Longleftrightarrow p = \frac{nRT}{V} = \frac{nRT}{\pi (S/2)^2 x} = \frac{4nRT}{\pi S^2 x}$$&#xA;For the distance $x$ of the piston to one wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gas A we have the initial and final conditions:&#xA;$$ p_\text{i}^{(\text{A})} = \frac{4n^{(\text{A})} RT_\text{i}}{\pi S^2 x_\text{i}} \; \text{and} \;\; p_\text{f}^{(\text{A})} = \frac{4n^{(\text{A})} RT_\text{f}}{\pi S^2 x_\text{f}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for B (with a total height of the cylinder of $h$):&#xA;$$ p_\text{i}^{(\text{B})} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{i})} \; \text{and} \;\; p_\text{f}^{(\text{B})} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{f})} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the initial and final pressures were the same (equilibrium condition):&#xA;$$ \frac{4n^{(\text{A})} RT_\text{i}}{\pi S^2 x_\text{i}} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{i})} \Longleftrightarrow \frac{n^{\text{(A)}}}{x_\text{i}} = \frac{n^{\text{(B)}}}{h - x_\text{i}}$$&#xA;$$ \frac{4n^{(\text{A})} RT_\text{f}}{\pi S^2 x_\text{f}} = \frac{4n^{(\text{B})} RT_\text{i}}{\pi S^2 (h-x_\text{f})} \Longleftrightarrow \frac{n^{\text{(A)}}T_\text{f}}{x_\text{f}} = \frac{n^{\text{(B)}}T_\text{i}}{h-x_\text{f}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now solve both equations for $\frac{n^\text{(A)}}{n^\text{(B)}}$ and equate them:&#xA;$$ \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x_\text{i}}{h-x_\text{i}} \; \text{and} \;\; \frac{n^\text{(A)}}{n^\text{(B)}} = \frac{x_\text{f}T_\text{i}}{T_\text{f}(h-x_\text{f})} $$&#xA;$$ \frac{x_\text{i}}{h-x_\text{i}} = \frac{x_\text{f}T_\text{i}}{T_\text{f}(h-x_\text{f})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $x_\text{f}$ gives us the final equation:&#xA;$$ x_\text{f} = \frac{\frac{hx_\text{i}T_\text{f}}{T_\text{i}(h-x_\text{i})}}{1 + \frac{x_\text{i}T_\text{f}}{T_\text{i}(h-x_\text{i})}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the values provided by you I get $x_\text{f} = 0.25~\mathrm{m}$, which is exactly in the middle.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-15T19:25:47.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7845" PostTypeId="1" AcceptedAnswerId="7859" CreationDate="2014-01-15T20:00:07.880" Score="3" ViewCount="1850" Body="&lt;p&gt;I'm trying to find an alternative to simple ring stands to hold a simple electrochemical cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that it is difficult to adjust the height to the right position to within even a couple inches. If you try to screw the stand in it can drop or lean, and I need to be able to adjust it to within at least an inch without it falling down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use a micrometer to adjust an electrode coming in from the top, and I don't have a lot of play to push down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also not easy to adjust the stand from under a few hood, due to the nature of the screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be nice if the clamp could be adjusted along the y-axis to move it closer or farther from the stand for different situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NOQrj.jpg&quot; alt=&quot;Current Ring Stand Situation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4013" LastEditorUserId="73" LastEditDate="2014-05-28T00:10:15.603" LastActivityDate="2016-03-24T10:31:30.520" Title="Alternatives to Ring Stands" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7846" PostTypeId="2" ParentId="7843" CreationDate="2014-01-15T20:09:11.677" Score="2" Body="&lt;p&gt;Interesting, and i like the part with the photoacid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't read the full details of the cited article (paywall), but since the authors aren't very pleased with the functionalization of the PVA, how about a change in the order of events:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ITQCW.png&quot; alt=&quot;vinyl tBOC&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;prepare or buy the monomer (vinyl tBOC)&lt;/li&gt;&#xA;&lt;li&gt;polymerize it according to &lt;a href=&quot;http://dx.doi.org/10.1007/BF00955876&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; (or a better source)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-15T20:09:11.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7847" PostTypeId="1" CreationDate="2014-01-15T20:18:40.877" Score="1" ViewCount="231" Body="&lt;p&gt;The detailed version of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we burn a strip of magnesium in air, there will be two main products, magnesium oxide and magnesium &#xA;nitride.  If we then treat that mixture with water, the nitride ion is converted to ammonia.  After boiling away &#xA;the water and ammonia, we would be left with only magnesium oxide.  Knowing that air is approximately &#xA;78% nitrogen and 20% oxygen, we might expect that the mixture that forms in the burning process would be &#xA;a similar distribution of magnesium nitride and magnesium oxide.  In fact, the ratio is closer to 75% of the &#xA;nitride to 25% of the oxide.  Assuming that the reactions are 100% efficient (i.e., no loss of magnesium), &#xA;how much magnesium oxide was formed initially, if the final product contains 3.22 grams of magnesium &#xA;oxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking about this for a while and working it out, it seems like the other information provided was irrelevant to finding the amount of MgO. I believe since the reactions are 100% efficient, that the final amount of 3.22 grams of MgO would be exactly the same for the initial amount of MgO, which was left over at the part when ammonia and water was evaporated. However, I am not sure; I would like to know, do the given percentages of nitrogen and oxygen given affect the unknown amount of MgO?&lt;/p&gt;&#xA;" OwnerUserId="4182" LastEditorUserId="4182" LastEditDate="2014-01-15T22:17:28.390" LastActivityDate="2014-02-15T04:00:34.187" Title="How to find how much there is of a compound given the following information?" Tags="&lt;reaction-mechanism&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7848" PostTypeId="2" ParentId="7845" CreationDate="2014-01-15T20:29:49.937" Score="1" Body="&lt;p&gt;Does your university have a workshop with some precision mechanics, who can handle a milling machine and a lathe to process metals and teflon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talk to the guys and explain them what you need. If they know their job (and I'm certain they do) they'll figure out all technical details and build it.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-15T20:29:49.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7849" PostTypeId="2" ParentId="7847" CreationDate="2014-01-15T20:45:08.340" Score="1" Body="&lt;p&gt;Maybe I'm just too tired right now, but didn't you mention that the absolute amount of magnesium oxide doesn't change during aqueous workup? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't thus the amount of 3.22 g $\ce{MgO}$ exactly what was formed initially?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't it be more reasonable to let the students calculate how much magnesium metal was burned initially, taking the $\ce{N2}$/$\ce{O2}$ ratio in air and the stoichiometry of the magnesium nitride into account? ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-15T20:45:08.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7850" PostTypeId="1" CreationDate="2014-01-15T20:45:44.140" Score="1" ViewCount="355" Body="&lt;p&gt;For a solution of chromimum phosphate when putting in to solution aluminum there is removal of the aluminum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to prevent a galvanic cell, stop the etching of the aluminum, by putting in a sacrificial anode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation I have access to metals like tantalum, titanium, tungsten and copper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of these would stop the oxidation of the aluminum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have slightly familiar with electronegativity and I understand that aluminum is 1.6.  I just do not know if you want some thing with a higher or lower electronegativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4184" LastEditorUserId="4945" LastEditDate="2015-10-23T16:49:48.930" LastActivityDate="2015-10-23T16:49:48.930" Title="What can I use as a sacrificial electrode to stop the oxidation of aluminium?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7851" PostTypeId="1" CreationDate="2014-01-15T21:20:11.927" Score="6" ViewCount="134104" Body="&lt;p&gt;Am I correct in assuming that the following is true?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathrm pK_\mathrm a &amp;lt; 3$ is for a strong acid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$3 &amp;lt; \mathrm pK_\mathrm a &amp;lt; 7$ is for a weak acid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$7 &amp;lt; \mathrm pK_\mathrm a &amp;lt; 11$ is for a weak base&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\mathrm pK_\mathrm a &amp;gt; 11$ is for a strong base&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2587" LastEditorDisplayName="user7951" LastEditDate="2015-10-02T23:17:21.183" LastActivityDate="2017-08-23T16:12:45.527" Title="What is the pKa Range for weak acids and bases?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7852" PostTypeId="2" ParentId="7838" CreationDate="2014-01-15T21:32:08.690" Score="0" Body="&lt;p&gt;Correct, Henderson-Hasselbalch is for when you have a buffered solution of both an acid and its conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you want to imagine dissolving one mole of Glycerol in one liter of water. When we say that the pKa of glycerol is 14.15 we mean that the Ka for the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_3H_7O_2OH \Rightarrow C_3H_7O_2O^- + H^+$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is $10^{-14.15}$. This is a very small number, which indicates that while glycerol dissolved in water does produce some $H^+$, it likely won't decrease the pH very much. So here you would expect a pH slightly less than 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate this value, I'd construct what many call an &quot;ICE chart&quot;&#xA;Example 2 in &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/Le_Chatelier%27s_Principle/Ice_Tables&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt; seems similar to the problem you're working on. &lt;/p&gt;&#xA;" OwnerUserId="4185" LastActivityDate="2014-01-15T21:32:08.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7853" PostTypeId="2" ParentId="7832" CreationDate="2014-01-15T23:07:20.880" Score="2" Body="&lt;p&gt;As nice and thoroughly described by Ben Norris, 1,4-(bishydroxymemethyl)benzene is available via reduction of the starting material with $\ce{LiAlH4}$. In order to minimize the amount of alanate used, one might consider to first convert the terephthalic acid to a corresponding ester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're heading for the phenol, which doesn't make a lot of sense since hydroquinone is cheap as dirt, the following &quot;strategy&quot; might work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;brominate terephthalic acid to yield 2,5-dibromo terephthalic acid&lt;/li&gt;&#xA;&lt;li&gt;heat the dibromo acid in an aqueous solution of sodium hydroxide, cupric bromide and 1,2-diaminoethane. You'd end up with 2,5-dihydroxy terephthalic acid, probably through a copper-catalyzed nucleophilic substitution&lt;/li&gt;&#xA;&lt;li&gt;decarboxylate the 2,5-dihydroxy terephthalic acid (or its disodium salt) by heating with calcium oxide.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You'll get hydroquinone, and unless your arene was labeled, you won't notice that you haven't replaced the $\ce{COOH}$ with $\ce{OH}$ but actually changed the positions of the substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all of this is nothing but a mind game ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-15T23:31:25.280" LastActivityDate="2014-01-15T23:31:25.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7854" PostTypeId="1" CreationDate="2014-01-15T23:08:09.383" Score="5" ViewCount="12982" Body="&lt;p&gt;I have rather basic knowledge of chemistry (at the level of a first year college class). But maybe somebody can still clarify with some simple words: What makes platinum a good catalyst for the electrodes of fuel cells? And why can't other materials be used? &lt;/p&gt;&#xA;" OwnerUserId="4186" LastEditorUserId="81" LastEditDate="2016-04-17T13:46:57.150" LastActivityDate="2016-04-17T13:46:57.150" Title="Why is platinum used as a catalyst in fuel cell electrodes (and elsewhere)?" Tags="&lt;hydrogen&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7855" PostTypeId="2" ParentId="7854" CreationDate="2014-01-15T23:29:29.490" Score="7" Body="&lt;p&gt;Imagine that a fuel cell is like &lt;em&gt;electrolysis - the other way around&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the electrolysis of water, you'd apply a certain potential (a voltage) to the electrodes and observe hydrogen and oxygen formed at the electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, you need a particular potential to run the reaction. In reality - and this depends on phenomena on the surface of the electrode material - you often need more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With platinum, however, this &lt;strong&gt;overpotential&lt;/strong&gt; for hydrogen is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you have no losses, neither in electrolysis, nor in the fuel cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be oversimplified but I hope it's not completely wrong ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-15T23:29:29.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7856" PostTypeId="2" ParentId="7845" CreationDate="2014-01-15T23:54:55.480" Score="2" Body="&lt;p&gt;On the more improvised/DIY side, you may be able to get away with something like a gooseneck clamp of the sort commonly used for microphones; something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.walimex.biz/bilder/produkte/gross/walimex-pro-Studio-Klemme-MX-mit-Schwanenhals.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;https://www.walimex.biz/walimex-pro-Studio-Clamp-MX-with-Gooseneck&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously they're not really designed to be exposed to hazardous chemicals, though, and you'll want to make sure that it can handle the weight you need to put on it without bending further.&lt;/p&gt;&#xA;" OwnerUserId="39" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.590" LastActivityDate="2014-01-15T23:54:55.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7859" PostTypeId="2" ParentId="7845" CreationDate="2014-01-16T09:04:07.703" Score="2" Body="&lt;p&gt;What about puting the whole stand on the lab-lift, hopefully there will be also some precision version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wgh3S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-01-16T09:04:07.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7860" PostTypeId="2" ParentId="7850" CreationDate="2014-01-16T09:05:49.643" Score="2" Body="&lt;p&gt;I think that you are interested in the corrosion protection of aluminum. You have basically two choices depending on your experiment setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use an impressed current protection or a sacrificial anode method. The sacrificial anode method uses a more reactive metal (less noble) in contact with the metal requiring protecting. The less noble metal reacts first within corrosive environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impressed current protection uses electric current to charge the metal requiring protecting with excess electrons. It changes the electrode potential in the negative direction until the metal is not reacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use the sacrificial anode method look in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; where you are looking for metal that is positioned higher than your aluminium (is more electronegative - so non of the mentioned metals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More useful information can be found here &lt;a href=&quot;http://nzic.org.nz/ChemProcesses/metals/8J.pdf&quot; rel=&quot;nofollow&quot;&gt;CORROSION PROTECTION OF METALS &#xA;&lt;/a&gt; or here &lt;a href=&quot;http://metals.about.com/od/metallurgy/a/Corrosion-Prevention.htm&quot; rel=&quot;nofollow&quot;&gt;metals.about&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-01-16T12:11:39.350" LastActivityDate="2014-01-16T12:11:39.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7861" PostTypeId="2" ParentId="7851" CreationDate="2014-01-16T10:06:51.137" Score="6" Body="&lt;p&gt;Not quite, what you are describing is the &lt;a href=&quot;http://en.wikipedia.org/wiki/PH&quot; rel=&quot;nofollow noreferrer&quot;&gt;pH scale&lt;/a&gt;, although the two terms are related, according to the Columbia University document &lt;a href=&quot;http://www.columbia.edu/%7Ecrg2133/Files/CambridgeIA/Chemistry/AcidityBasicitykPa.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acidity, Basicity and &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt; is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It turns that that the &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt; of an acid is the &lt;span class=&quot;math-container&quot;&gt;$pH$&lt;/span&gt; at which it is exactly half dissociated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The full derivation is on the document (and a bit long to post here with explanations).&lt;/p&gt;&#xA;&lt;p&gt;But another explanation is provided in the document &lt;a href=&quot;http://kchem.weebly.com/uploads/8/6/0/4/8604167/6_ph_and_pka.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$pH$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;&lt;/a&gt;, where they state&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt; tells you if a given molecule is going to either&#xA;give a proton to water at a certain &lt;span class=&quot;math-container&quot;&gt;$pH$&lt;/span&gt;, or remove a&#xA;proton&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, to answer your question, in terms of &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;, strong and weak acids and bases can be defined by the following table of examples:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qzj49.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://pharmacy.nova.edu/Pharmacodynamics/Modules/AcidsSaltsBases/Main_P3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strength of Acids and Bases&lt;/a&gt;, a key point is that there is no clear boundary that defines strong from weak acids or bases.&lt;/p&gt;&#xA;&lt;p&gt;They conclude with the following very generalised rule for &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For acids: the stronger the acid, the smaller the &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For bases: the stronger the base, the larger the &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="-1" LastEditorDisplayName="user4076" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-16T10:19:56.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7862" PostTypeId="1" AcceptedAnswerId="7901" CreationDate="2014-01-16T12:55:27.727" Score="1" ViewCount="1364" Body="&lt;ul&gt;&#xA;&lt;li&gt;Is there any mnemonic that helps remembering the names of the different geometries (Shapes) that we obtain from applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow&quot;&gt;AXE method&lt;/a&gt; ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1463" LastActivityDate="2014-01-19T17:23:01.557" Title="is there any mnemonic for the AXE method?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7863" PostTypeId="1" CreationDate="2014-01-16T17:03:43.993" Score="2" ViewCount="2955" Body="&lt;p&gt;When we reduce Vanadium (V) using iron(II), why are we adding 85% Phosphoric acid?&#xA;After the above reduction, if we are going to titrate the mixture with Potassium permanganate, why are we adding dipotassiumpersulfate to the mixture before the titratioin?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's the procedure:&lt;br&gt;&#xA;  To 5 ml of vanadium (V) solution, add 2 ml of 6 M sulfuric acid, 1 ml of phosphoric acid and 0.20 g of hydrated ferrous sulfate to reduce vanadium (V). Then swirl the solution and allow it to stand for 3 minutes. Then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. Stir the contents vigorously and titrate with 0.02 standard potassium permanganate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4191" LastEditorUserId="275" LastEditDate="2014-01-17T15:28:15.110" LastActivityDate="2014-01-17T15:28:15.110" Title="reaction between Iron(ii) and Vanadium(v)" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7864" PostTypeId="2" ParentId="7851" CreationDate="2014-01-16T18:22:59.333" Score="12" Body="&lt;p&gt;$\mathrm pK_\mathrm a$ is negative log of the acid dissociation constant ($K_\mathrm a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm pK_\mathrm a = -\log K_\mathrm a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acid dissociation constant is the equilibrium constant of the dissociation of ions of an acid in an aqueous solution. Consider a weak acid $\ce{HA}$. Weak acids do not dissociate completely in aqueous solution. The equilibrium for the dissociation of such acids can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{HA + H2O &amp;lt;=&amp;gt; H3O+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant for this reaction(dissociation) will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm a = \frac{\ce{[H3O+][A^{-}]}}{\ce{[HA]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen that the greater is the extent of dissociation, the greater will be the value of $K_\mathrm a$. Or, the stronger is the acid, the greater will be its $K_\mathrm a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, $\mathrm pK_\mathrm a$ is negative log of $K_\mathrm a$, it's values will be greater for weaker acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no sharp boundary between &lt;em&gt;weak&lt;/em&gt; and &lt;em&gt;strong&lt;/em&gt; acid. Wikipedia defines strong acids as &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;acids which ionize completely in aqueous solution&lt;/em&gt;&lt;/a&gt;. So, an acid, say $\ce{HA}$ , is said to be strong if one mole of this acid dissociates in aqueous solution to give one mole of $\ce{H+}$ and one mole of $\ce{A-}$. Now $K_\mathrm a$ of such an acid will be 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, $\mathrm{pH}$ is used to measure the acid strength, which is negative log of $\ce{H+}$ ion concentration. $$\mathrm{pH} = -\log\ce{[H+]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of $\ce{H+}$ ions in water is $10^{-7}$ (It has been found out experimentally). Hence, its $\mathrm{pH}$ is $-\log(10^{-7}) = 7$.&#xA;For acids stronger than $\ce{H2O}$, the concentration of $\ce{H+}$ ions, $\ce{[H+]} &amp;gt; 10^{-7}$. So, $\mathrm{pH} &amp;lt; 7$ for acids and $\mathrm{pH} &amp;gt; 7$ for bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are referring to may actually be $\mathrm{pH}$ and in that case the ranges you've given &lt;s&gt;are&lt;/s&gt; may be correct. But once again there is no clear difference between &lt;em&gt;weak&lt;/em&gt; and &lt;em&gt;strong&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple relation between $\mathrm{pH}$ and $\mathrm pK_\mathrm a$,&#xA;$$\mathrm{pH} = \frac{1}{2}[\mathrm pK_\mathrm a - \log c]$$&#xA;where $c$ is concentration of the acid. You can derive this relation using &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_dilution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ostwald's dilution law&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we have $K_\mathrm b$, $\mathrm pK_\mathrm b$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/PH#pOH&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm{pOH}$&lt;/a&gt; for bases which are basic analogues of $K_\mathrm a$, $\mathrm pK_\mathrm a$ and $\mathrm{pH}$. These variables show the strength of a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $K_\mathrm a$, the stronger will be the acid and the weaker will be the base.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $\mathrm pK_\mathrm a$, the weaker will be the acid and the stronger will be the base.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $\mathrm{pH}$, the weaker will be the acid and the stronger will be the base. For acids, $\mathrm{pH} &amp;lt; 7$ and for bases, $\mathrm{pH} &amp;gt; 7$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $K_\mathrm b$, the stronger will be the base and the weaker will be the acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $\mathrm pK_\mathrm b$, the weaker will be the base and the stronger will be the acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The greater is the value of $\mathrm{pOH}$, the weaker will be the base and the stronger will be the acid. For bases, $\mathrm{pOH} &amp;lt; 7$ and for acids, $\mathrm{pOH} &amp;gt; 7$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2335" LastEditorUserId="16683" LastEditDate="2017-06-05T14:08:08.027" LastActivityDate="2017-06-05T14:08:08.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7865" PostTypeId="2" ParentId="7863" CreationDate="2014-01-16T21:40:30.543" Score="3" Body="&lt;p&gt;I think I realize (a part of) the answer, concerning the conditions of a redox titration of vanadium(V), for which a metavanadate, such as $\ce{NH4VO3}$ is typically used as starting material. Under acidic conditions, i.e. in the presence of 85% $\ce{H3PO4}$, the colourless $\ce{VO2+}$ is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{VO3- + 2 H+ -&amp;gt; VO2+ + H2O} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again under acidic conditions, the latter species is reduced by Fe(II) to the blue $\ce{VO^{2+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{VO2+ + Fe^{2+} + 2 H+ -&amp;gt; VO^{2+} + Fe^{3+} + H2O} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is is conceivable that the peroxodisulfate is used to oxidize the excess of Fe(II), which otherwise would be oxidized by the permanganate?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-16T21:40:30.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7866" PostTypeId="1" CreationDate="2014-01-16T22:58:15.217" Score="5" ViewCount="779" Body="&lt;p&gt;In many electrochemistry scholarly articles/studies they talk about certain specific reactions occurring at certain voltage potentials, while in other articles they talk about current density as the determiner for which reactions happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? From what I know about electrical engineering you can either set the voltage, and the current will adjust accordingly, or you can set the current and the voltage will adjust to meet the current. You cannot set both of them at the same time. That would defy the laws of physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what do many of these articles mean when they talk about, for instance: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reported reactions of this type often occur at potentials smaller than -3 V&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this relate to the current density that is set by a power supply for instance? &lt;/p&gt;&#xA;" OwnerUserId="4192" LastEditorUserId="418" LastEditDate="2016-11-03T10:20:37.010" LastActivityDate="2016-11-03T10:20:37.010" Title="Voltage vs current density - why are both parameters important to control an electrochemical reaction?" Tags="&lt;organic-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7867" PostTypeId="2" ParentId="7797" CreationDate="2014-01-17T00:05:41.837" Score="3" Body="&lt;p&gt;The rate constant depends on the energy difference between the ground state of the reactants and the transition state. The ratio of the rate constants due to the kinetic isotope effect therefore depends on the difference in vibrational zero point energy between the initial state and the transition state. Vibrational ZPE depends on bond stiffness and reduced mass. The change in reduced mass from $\ce{^{12}C}$ to $\ce{^{13}C}$ causes the kinetic isotope effect, and the bond stiffness determines the size of the KIE. Bond stiffness is to a first approximation proportional to bond order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both $S_N1$ and $S_N2$, the initial state has a single bond between the methyl $\ce{C}$ and $\ce{Br}$  (as well as some $\ce{C-H}$ bonds which stay the same throughout the reaction).&#xA;For $S_N2$, the transition state has partial bonds between the methyl $\ce{C}$ and both $\ce{Br}$ and the $\ce{C}$ of $\ce{CN}$.&#xA;In the $S_N1$ transition state, the methyl $\ce{C}$ has only a partial bond to $\ce{Br}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the change in bond order is greater between initial state and $S_N1$ transition state than between ground state and $S_N2$ transition state, $S_N1$ will have a greater kinetic isotope effect. Measuring the KIE of an unknown reaction and comparing it to the KIEs of reactions with known mechanisms will therefore distinguish between $S_N1$ and $S_N2$.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-01-17T00:05:41.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7868" PostTypeId="1" CreationDate="2014-01-17T00:34:08.683" Score="5" ViewCount="280" Body="&lt;p&gt;It goes without saying that every reduction agent has different reducing properties and strengths in terms of what sort of bond it can reduce/hydrogenate. But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction (in the presence of the necessary hydrogen or $\ce{H}$-donor of course)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it about the properties of oxygen that makes it so ready to disassociate from the molecule when it gains an electron, allowing hydrogen to take its place? I'm sure there is something simple that I'm missing, but any explanation is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4193" LastEditorDisplayName="user4076" LastEditDate="2014-01-17T08:03:48.857" LastActivityDate="2016-03-15T06:03:09.970" Title="Why does hydrogen so easily take the place of oxygen in organic molecules during reduction reactions?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7869" PostTypeId="1" CreationDate="2014-01-17T03:16:26.720" Score="1" ViewCount="4427" Body="&lt;p&gt;Even though it has more $\ce{OH}$ groups and still it is very weak in terms of dissolving?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please clarify to me in terms of polar and non polar solubility, how to determine whether a compound is soluble or not in water.&lt;/p&gt;&#xA;" OwnerUserId="4147" LastEditorUserId="4231" LastEditorDisplayName="user4076" LastEditDate="2015-03-09T15:29:28.393" LastActivityDate="2015-03-09T15:29:28.393" Title="why are Ca(OH)₂ and Al(OH)₃ less soluble than Na(OH) in water" Tags="&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7870" PostTypeId="2" ParentId="7866" CreationDate="2014-01-17T03:18:51.747" Score="2" Body="&lt;p&gt;Reaction rate in electrochemistry is limited by diffusion of reactants from solution towards electrode. Once all reactants in near proximity for given type of reactions are exhausted, and if voltage is still applied, other reactions (if voltage is strong enough) may occur. For example, consider positive electrode in $\ce{KCl}$ solution (leaving aside paired negative electrode). It is true, that $\ce{Cl^-}$ anion is oxidized easier then water. However, once all chloride near the electrode is oxidised, the next possible substrate to be oxidised is $\ce{OH^-}$ that is always present is water via reaction $\ce{2H_2O &amp;lt;=&amp;gt; H_3O^+ + OH^-}$. Thus, depending on the current density on the electrode you will have different major products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: resistance of the electrochemical cell is a complex function of temperature, voltage, current and solution composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current density in electrochemistry may be adjusted not only by adjusting voltage applied, but also by adjusting surface area of the electrodes used. Also, it is critical to maintain conditions for effective transportation of reactants towards electrodes.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="13339" LastEditDate="2015-05-21T12:26:29.433" LastActivityDate="2015-05-21T12:26:29.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7871" PostTypeId="2" ParentId="7869" CreationDate="2014-01-17T03:34:24.020" Score="6" Body="&lt;p&gt;$\ce{OH}$ is small anion and bonds effectively with small cations. Thus, in general, the bigger (and less specific charge) of the cation, the higher is solubility of the corresponding base in polar solvents, meaning that $\ce{Ca(OH)_2}$ and $\ce{Al(OH)_3}$ having much smaller and more charged cations are almost covalent in nature and thus, do not dissociate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, solubility of salt in water is determined by energy balance between breaking of the ionic bonds and hydratation of anions formed. Thus, the most soluble salts are usually formed by small and big ions, that cannot bond effectively, but at least one of them can be hydrated effectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, note: solubility for molecular compounds, like sugars and some non-dissociating compounds that looks like salts (like $\ce{Cu(CH_3COO)_2*H_2O}$) is an entirely different matter and determined by relative strength of intermolecular interactions (read: the molecule have to be able to form hydrogen bonds)&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user4076" LastEditDate="2014-01-17T06:33:25.107" LastActivityDate="2014-01-17T06:33:25.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7872" PostTypeId="1" CreationDate="2014-01-17T11:43:03.690" Score="3" ViewCount="941" Body="&lt;p&gt;The six electronic configurations below could all be found in the matrix elements of the CI matrix at CISD level. How do I construct the 6 x 6 CI matrix composed of these six configurations and which of the elements will be zero?&#xA;How different would it be if I used MP4(SDTQ) method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYNtV.png&quot; alt=&quot;electronic configurations&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4162" LastEditorUserId="15751" LastEditDate="2018-12-28T23:36:24.963" LastActivityDate="2018-12-28T23:36:24.963" Title="Configuration Interaction matrix composition from electronic configurations" Tags="&lt;computational-chemistry&gt;&lt;electronic-configuration&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7873" PostTypeId="2" ParentId="2429" CreationDate="2014-01-17T12:28:36.703" Score="4" Body="&lt;p&gt;You should ask yourself: &quot;For which application?&quot;. In fact there are different types of &quot;phosphors&quot; (most of them have no &quot;Phosphorus&quot; inside) for different applications and different compound were used during the evolution of CRT monitor. Some kind of phosphors may have advantages in terms of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Efficiency of the luminescence induced.&lt;/li&gt;&#xA;&lt;li&gt;Cheaper price.&lt;/li&gt;&#xA;&lt;li&gt;Longer life.&lt;/li&gt;&#xA;&lt;li&gt;Better color.&lt;/li&gt;&#xA;&lt;li&gt;Shorter decay time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note however that in many application longer decay times are prefer because allow to the user to see transient signal one of this application is &lt;strong&gt;radar&lt;/strong&gt; in this case is used the &lt;strong&gt;GB&lt;/strong&gt; phosphors (it usually used the &lt;strong&gt;WTDS&lt;/strong&gt; symbol to refer to a specific compound. In &lt;a href=&quot;http://books.google.it/books?id=I9O1K20-uo4C&amp;amp;pg=PA607&amp;amp;lpg=PA607&amp;amp;dq=WTDS++phosphor&amp;amp;source=bl&amp;amp;ots=8IIe6n9DLi&amp;amp;sig=4A9Le-rml6u_ble8COK7hN5XOFc&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=8xXZUqSLJOLcygOJ6YGIDA&amp;amp;ved=0CFUQ6AEwBA#v=onepage&amp;amp;q=WTDS%20%20phosphor&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Phosphor Handbook&#xA; by Shigeo Shionoya,William M. Yen,Hajime Yamamoto&lt;/a&gt; p. 607 you can find the relative compound in this case is  $(Zn,Cd)S:Cu + (Ca,Mg)SiO_3:Ti)$) that has a time decay between 100 ms and 1 s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are  interested in RGB monitor and you want a list of the phosphors manly used. &#xA;I've copied the the following table from &lt;a href=&quot;http://books.google.it/books?id=b61jkO-ZKkwC&amp;amp;pg=PA232&amp;amp;dq=CRT+phosphor+composition&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=TQ_ZUpa8EobDygOHwoDgBg&amp;amp;ved=0CFYQ6AEwBQ#v=onepage&amp;amp;q=CRT%20phosphor%20composition&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Practical Applications of Phosphors by William M. Yen,Shigeo Shionoya,Hajime Yamamoto&lt;/a&gt; p. 232 showing how phosphors compound for every RGB channel changed during time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c c c  c}\hline&#xA;&amp;amp;Blue  &amp;amp; Green  &amp;amp; Red \\ \hline&#xA;1               &amp;amp; ZnS:Ag^{2+}   &amp;amp;  Zn_2SiO_4:Mn^{2+}   &amp;amp; Zn_3(PO_4)_2:Mn^{2+}\\ &#xA;2 &amp;amp; ZnS:Ag^{2+} &amp;amp;  (Zn,Cd)S:Ag    &amp;amp;  (Zn,Cd)S:Ag  \\ &#xA;3&amp;amp; ZnS:Ag^{2+} &amp;amp; (Zn,Cd)S:Ag &amp;amp; YVO_4:Eu^{3+}\\ &#xA;4 &amp;amp; ZnS:Ag^{2+} &amp;amp;   (Zn,Cd)S:Cu,Al &amp;amp; Y_2O_2S:Eu^{3+}or\space Y_2O_3:Eu^{3+} \\ &#xA;Current &amp;amp; ZnS:Ag^{2+}       &amp;amp; ZnS:Au,Cu,Al  \space or \space ZnS:CU,Al    &amp;amp;Y_2O_2S:Eu^{3+}    \\ \hline&#xA; \hline&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-08-02T12:32:43.310" LastActivityDate="2014-08-02T12:32:43.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7875" PostTypeId="1" CreationDate="2014-01-17T16:45:36.460" Score="2" ViewCount="1672" Body="&lt;p&gt;I am looking into the subject of water repelling and have been led to the hydrophobic subject area. I have two questions regarding this subject.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a hydrophobic gas ? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a hydrophobic liquid is heated to a gas does it loose its water repellent qualities. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help guys &lt;/p&gt;&#xA;" OwnerUserId="4200" LastActivityDate="2015-05-03T14:48:52.547" Title="Is there a hydrophobic gas?" Tags="&lt;water&gt;&lt;gas-laws&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7876" PostTypeId="1" AcceptedAnswerId="7879" CreationDate="2014-01-17T17:13:25.757" Score="4" ViewCount="177" Body="&lt;p&gt;I live in Charleston, WV.  On 1/9/2014 we were told not to use our tap water for anything other than flushing the toilet after a storage facility leaked over 7,400 gallons of 4-methylcyclohexane methanol into the Elk river, less than two miles up-river from the water company's main in-take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CDC said that concentrations under 1ppm were safe.  After a week of bathing in my sink and drinking bottled water, my area was determined to be safe, however my water still has a strong odor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much information about this chemical's health effects.  While the chemical itself is low-toxicity, I wonder if it might react with items in the water supply to create other compounds which may be more toxic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, would this chemical react with chlorine in the water, or cast iron, PVC, and copper in the pipes?  Does anyone have any experience with this compound?&lt;/p&gt;&#xA;" OwnerUserId="4199" LastEditorUserId="14" LastEditDate="2014-01-17T18:51:18.567" LastActivityDate="2014-01-17T21:42:35.320" Title="Chemical Spill in Charleston, West Virginia" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7877" PostTypeId="1" CreationDate="2014-01-17T17:20:09.293" Score="2" ViewCount="252" Body="&lt;p&gt;I'm doing a science project on agarose gel electrophoresis and would like to know if it is safe to store Tris/Borate/EDTA buffer solution inside a PETE container since I will be making about 10L of the buffer solution and do not have access to a spigoted carboy or any such container.&lt;/p&gt;&#xA;" OwnerUserId="4201" LastActivityDate="2014-01-17T17:20:09.293" Title="Can TBE electrophoresis buffer be stored in a PETE container?" Tags="&lt;biochemistry&gt;&lt;plastics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7879" PostTypeId="2" ParentId="7876" CreationDate="2014-01-17T21:42:35.320" Score="5" Body="&lt;p&gt;The name MCHM or &quot;4-methycycloheaxe methanol&quot; (as newspapers reported it) is not entirely a systematic name and is a little ambiguous. The actual compound is better called  4-methylcyclohexylmethanol and the structure is shown below:&lt;img src=&quot;https://i.stack.imgur.com/QL0OV.png&quot; alt=&quot;MCHM structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific details about its toxicity are not easy to find, though it has been patented as an air freshener ingredient. It has also been around for a long time and is used in the coal industry as a frothing agent in significant quantities so if it had some particularly nasty property we would probably know by now. Some &lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=US&amp;amp;language=en&amp;amp;productNumber=125598&amp;amp;brand=ALDRICH&quot; rel=&quot;nofollow noreferrer&quot;&gt;safety information is available&lt;/a&gt; (pdf) on the related 1,4-cyclohexanedimethanol and that isn't particularly toxic. So there doesn't seem to be any strong reason to worry about it, though it will taint the flavour of water. There are no obvious paths for it to known nasties in low concentrations in water, so the biggest worry is probably the taint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all this, the company should not have let it get into the water supply.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-17T21:42:35.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7883" PostTypeId="1" CreationDate="2014-01-18T06:54:55.207" Score="1" ViewCount="6488" Body="&lt;p&gt;How to know how many electrons are there in $T_{2g}$ and $E_g$ orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think rather than a broad answer, an example would help. So, consider the compound $\ce{Ni(CO)4}$ and how to calculate CFSE for it?&lt;/p&gt;&#xA;" OwnerUserId="4207" LastEditorUserId="4945" LastEditDate="2015-10-23T16:50:29.477" LastActivityDate="2020-11-11T15:52:09.870" Title="How do I determine crystal field splitting (for example, for [Ni(CO)4])?" Tags="&lt;physical-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7884" PostTypeId="2" ParentId="7868" CreationDate="2014-01-18T10:26:16.810" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok, if at all, it's an oxygen atom ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see that general trend towards complete deoxygenation, neither in catalytic hydrogenations, nor in hydride transfer reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It usually takes&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;rather drastical conditions &lt;/li&gt;&#xA;&lt;li&gt;quite some special reagents&lt;/li&gt;&#xA;&lt;li&gt;a particular type of derivatisation &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;or a combination of any of these to convert oxygen-containing compounds to alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R1-CO-R2 -&amp;gt; R1-CH2-R2}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Ketones and aldehydes may be directly reduced to alkanes in the &lt;em&gt;Clemmensen&lt;/em&gt; reduction. The &quot;mildest&quot; variants use anhydrous $\ce{HCl}$ and amalgamated $\ce{Zn}$ in organic solvents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The clasical &lt;em&gt;Wolff-Kishner&lt;/em&gt; reduction, using hydrazine and sodium in diglyme, involves the &lt;em&gt;in situ&lt;/em&gt; functionalization of the carbonyl and isn't a prototype for a mild reaction either. Admittedly, milder variants exist, using N-&lt;em&gt;tert&lt;/em&gt;-butyldimethylsilylhydrazone and catalytical amounts of a Lewis acid, e.g. $\ce{Sc(OTf)3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tosylhydrazones may be converted to alkanes using $\ce{NaCNBH3}$, $\ce{Na(CH3COO)3BH}$, or boranes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Catalytic hydrogenation over Raney nickel is possible after converting the carbonyl to a dithiane; so it's actually a desulfurization.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{R1-CH2-OH -&amp;gt; R1-CH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any transformation that does not involve functionalization.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Alkyl tosylates can be deoxygenated using $\ce{LiEt3BH}$.&lt;/li&gt;&#xA;&lt;li&gt;Acylation of alkohols with thiocarbonyl chlorides furnish thiocarbonates, which may be deoxygenated under using either tin hydrides ($\ce{R3SnH}$, &lt;em&gt;Barton&lt;/em&gt; deoxygenation) or $\ce{B(CH3)3}$ in water/benzene.&lt;/li&gt;&#xA;&lt;li&gt;Alkanes may also be obtained from alkohols under &lt;em&gt;Mitsunobu&lt;/em&gt; conditions ($\ce{Ph3P}$, $\ce{DEAD}$, &lt;em&gt;o&lt;/em&gt;-nitrobenzenesulfonylhydrazine).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{R1-CH2-COOH -&amp;gt; R1-CH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Direct&lt;/strong&gt; transformations of carboxylic acids to alkanes typically involve &lt;em&gt;decarboxylation&lt;/em&gt;, i.e. the loss of $\ce{CO2}$. This may be achieved either by the &lt;em&gt;Barton&lt;/em&gt; decarboxylation, or via photoinduced electron transfer reactions with electronically excited (electron) acceptors. In the course of the latter reactions, a carboxylate is oxidized to an acyloxy radical, which then undergoes decarboxylation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it up: &lt;strong&gt;It's not that easy ;)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-18T10:26:16.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7885" PostTypeId="1" CreationDate="2014-01-18T10:33:50.353" Score="1" ViewCount="126" Body="&lt;p&gt;When I work through the following problem it is unclear to me why 0.00120 mmHg is converted to 0.00332 mmHg. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the temperature of 0.00332 g of mercury in the gas phase with pressure of 0.00120 mmHg and a volume of 435 L?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I converted the grams of mercury to moles and then used the ideal gas law constant of 62.36 to solve for T. Is there any reason that the pressure of 0.00120 mmHg should be converted to 0.00332 mmHg?&lt;/p&gt;&#xA;" OwnerUserId="4208" LastEditorUserId="5026" LastEditDate="2018-02-28T14:07:08.100" LastActivityDate="2018-02-28T14:07:08.100" Title="Why is the pressure value altered in this gas law numerical?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7886" PostTypeId="2" ParentId="7885" CreationDate="2014-01-18T10:58:13.303" Score="2" Body="&lt;p&gt;Your tutor is a master of confusion :-D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the exercise be easier if the pressure was given in stones per square acre? ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write down the equation for the ideal gas law again. Pay attention to the units! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't let the mercury fool you ;-)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-18T10:58:13.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7887" PostTypeId="2" ParentId="7875" CreationDate="2014-01-18T11:52:38.733" Score="4" Body="&lt;p&gt;Sure many gases are hydrophobic! A hydrophobic molecule is in fact mainly a non polar molecule. So every non polar gas without $H$ bond site is a hydrophobic molecule. The alkanes for example are all hydrophobic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Rzu3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact I will not say that if a hydrophobic liquid is heated to a gas  it loses its water repellent qualities, because in fact it is still a non polar molecule even if he has greater kinetic energy however in gas state molecules have a really low interaction between each other so it doesn't really have a strong influence in what happen if the polarity is different. For this reason I don't know if is really useful to speak of hydrophobic gas (even if someone use it see e.g. &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2008/CP/B801220B#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;The interface between water and a hydrophobic gas&lt;/a&gt;) because you can't measure contact angle or use other tests usually perform to characterize hydrophobic materials.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="1362" LastEditDate="2014-01-19T15:50:55.030" LastActivityDate="2014-01-19T15:50:55.030" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7888" PostTypeId="2" ParentId="7875" CreationDate="2014-01-18T12:46:19.493" Score="8" Body="&lt;p&gt;The trouble with the question is that hydrophobicity is not a meaningful concept in a gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept is mostly useful when characterising how surfaces behave. Typical hydrophobic compounds don't interact strongly with water (e.g. non polar hydrocarbon waxes) and tend to disrupt water's hydrogen bond network. So water doesn't &quot;wet&quot; the surface and tends to form droplets that seem to be trying to minimise the contact area with the surface. The contact angle that a droplet makes with the surface is used as a measure of static hydrophobicity (if the droplet spreads out because the surface likes to interact with water, the angle is low, if the surface reels water the angle is high).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is less clear that describing liquid phases as hydrophobic is meaningful, though it is not uncommon in casual usage. Compounds that would be hydrophobic as solids will probably exist as liquids that don't mix with water or can form micelles where they cluster together in very small droplets in the water phase. The bulk effect that matters is that hydrophobic compounds don't mix with water and form separate phases. The hydrophobicity of the molecules describes why the separation happens, but the effect is immiscibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this matters in the gas phase. The point of gases (compared to solids or liquids) is that the bulk properties are not determined by molecular interactions. The molecules of a gas don't interact much and are, to a good approximation, interchangeable whatever the individual molecules are made of. Gas equations describing gas behaviour can effectively ignore the molecular interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that if you take a compound that would be hydrophobic as a solid and immiscible with water as a liquid, it will mix quite happily with water vapour. So a hydrophobic compound can be a gas, but a gas can't be hydrophobic.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-18T12:46:19.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7890" PostTypeId="1" AcceptedAnswerId="7909" CreationDate="2014-01-18T16:08:28.283" Score="4" ViewCount="1397" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In the following questions use a value of 3 for &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$6 \times 10^{23}$&lt;/span&gt; for Avogadro’s number and &lt;span class=&quot;math-container&quot;&gt;$660$&lt;/span&gt; for the molecular weight of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 bp}$&lt;/span&gt; of DNA. The volume of a sphere of radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$4/3\,πr^3$&lt;/span&gt;. A bacterium has a single copy of a &lt;span class=&quot;math-container&quot;&gt;$\pu{4 \times 10^6 bp}$&lt;/span&gt; circular genomic DNA.&lt;/p&gt;&#xA;&lt;p&gt;If the diameter of this spherical cell is 1 micrometer, what would be the molar concentration of DNA in this cell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The volume comes to be &lt;span class=&quot;math-container&quot;&gt;$\pu{2 \times 10^{-5} L}$&lt;/span&gt;.&#xA;Amount of substance is &lt;span class=&quot;math-container&quot;&gt;$\pu{6.7 \times 10^-18 mol}$&lt;/span&gt;.&#xA;Dividing this by the volume, my answer comes &lt;span class=&quot;math-container&quot;&gt;$\pu{3.3 \times 10^{-13} M}$&lt;/span&gt;, but the answer given in my book is &lt;span class=&quot;math-container&quot;&gt;$\pu{3.3 \times 10^{-9} M}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="69830" LastEditDate="2020-06-18T12:33:02.577" LastActivityDate="2021-03-01T08:44:51.053" Title="Calculating the molarity of DNA in a cell" Tags="&lt;concentration&gt;&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7893" PostTypeId="1" CreationDate="2014-01-18T21:11:26.170" Score="2" ViewCount="820" Body="&lt;p&gt;When we cook food, we add several spices and condiments which enhance color, taste and smell. Are these spices and condiments just adsorbed on the food particles? Or are they absorbed? For example using pepper while cooking a piece of chicken, would it just stay on the top of the leg (adsorbed)? Or get absorbed?&lt;/p&gt;&#xA;" OwnerUserId="2534" LastEditorUserId="7475" LastEditDate="2015-10-30T17:23:27.277" LastActivityDate="2015-10-30T17:23:27.277" Title="Spices are absorbed or adsorbed in food?" Tags="&lt;food-chemistry&gt;&lt;adsorption&gt;&lt;absorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7894" PostTypeId="2" ParentId="7893" CreationDate="2014-01-18T22:53:19.340" Score="2" Body="&lt;p&gt;The answer varies with the type of food and the type of condiment. Both are very complicated mixtures, so a same combination of food and condiment can display varying levels of adsorption/absorption depending on the region/compound being analyzed, but in general, I expect that while all of chemical/physical adsorption and chemical/physical absorption happen to some extent (cooking is quite a messy collection of physical and chemical interactions!), either physical absorption is the most prevalent, or &lt;em&gt;none&lt;/em&gt; of them happens to the largest part of the spices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, adsorption is a surface phenomenon, which fundamentally limits how much matter can be considered adsorbed onto an object. Adsorbed films are usually very thin (probably not much more than a few nanometers thick, often a single molecule thick). Regular macroscopic objects have very little surface area relative to their volume, so it takes a comparatively tiny amount of substance to completely cover the object in a thin adsorbed layer, compared to say, activated charcoal. Thus we can likely exclude adsorption as the main factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to consider that most food is cooked with some amount of solvent (water or oil/fat), and in fact solvents come out of the food while it's being cooked. This means that a lot of the spices can very well end up as part of a liquid (albeit often quite concentrated) solution, which remains stuck on the surface simply due to its high viscosity and significant friction with the uneven surface of the food. I bet this is what goes on in most cases. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-01-18T22:53:19.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7897" PostTypeId="2" ParentId="7872" CreationDate="2014-01-19T05:41:31.040" Score="7" Body="&lt;p&gt;Given those six configurations, the CI matrix with be a six-by-six matrix with rows and columns defined by the respective configuration. The matrix elements themselves will be the Hamiltonian expressed in those configurations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say column one corresponds to $\psi_0$, or $\left|0\right&amp;gt;$ in bra-ket notation. Similarly, row one corresponds to $\psi_0$, or $\left&amp;lt;0\right|$ in bra-ket notation. The CI matrix element where these coincide, namely $\left&amp;lt;0|H|0\right&amp;gt;$, where H is your Hamiltonian, is simply the ground state energy (the Hartree-Fock energy, if the configurations are the result of HF orbitals). The other matrix elements, like $\left&amp;lt;1|H|0\right&amp;gt;$ or $\left&amp;lt;0|H|2\right&amp;gt;$ have rules for determining them. The best reference is the book Modern Quantum Chemistry, by Szabo and Ostlund. Dr. David Sherrill has a good reference as well, which you can find here: &lt;a href=&quot;http://vergil.chemistry.gatech.edu/notes/ci/ci.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://vergil.chemistry.gatech.edu/notes/ci/ci.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, to answer the second part of your question. There are a few rules for determining CI matrix elements. One of the first ones is that singly excited configurations do not mix with the ground state. This is Brillouin's theorem. I don't see any singly excited configurations in the six you gave, so we won't have any of those. The other rule is that configurations that differ by more than two orbital occupations are zero. Since you truncated to doubles, you also won't have this case. The other thing you need to check is spin. If spin is not conserved, you'll have a zero matrix element. I think you'll have this case with the first configuration and the last one, which have different multiplicities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your last question, I'm not real sure what you are asking. MP4(SDQ) is a perturbative method, which is different from CI, which is a variational method. In other words you do a CI by diagonalization, whereas MP4 is not. So I think CISDQ would be a better question...anyway, the point is that you are increasing your configuration space from singles and doubles (SD) to singles, doubles and quadruples (SDQ). The point I made in the preceding paragraph still holds, and expressing your Hamiltonian between ground state and quadruply excited configurations will be zero because their occupation differs by more than two. Same for singly excited configurations and quadruples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Just had the thought, you should also search for &quot;Slater's rules&quot; when trying to construct CI matrices. These are the rules I mentioned earlier.&lt;/p&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="194" LastEditDate="2016-12-14T15:39:30.600" LastActivityDate="2016-12-14T15:39:30.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7898" PostTypeId="2" ParentId="7798" CreationDate="2014-01-19T05:59:20.223" Score="12" Body="&lt;p&gt;I'd also like to add to ssavec's response. In addition to the insane amount of flops that you have to carry out to run the computation, you also run into problems of storage for the computation. In a CCSD calculation you have to store single and doubly excited amplitudes. ssavec pointed out that there are ${6.3\cdot10^{10}}$ double excitations...assuming 8 bytes per double precision floating point, that would mean storing the doubles would require 450 GB of storage! (Not to mention the I/O between iterations...)&lt;/p&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="17657" LastEditDate="2016-11-18T04:46:34.373" LastActivityDate="2016-11-18T04:46:34.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7899" PostTypeId="1" AcceptedAnswerId="7904" CreationDate="2014-01-19T07:48:17.397" Score="1" ViewCount="378" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two gases in a container: krypton and carbon dioxide. If the mass of the gases is 35 grams, and total pressure of the container is 0.708 atm, and the pressure of krypton is 0.250 atm. What is the mass of the krypton?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've found the mole fraction of krypton is 0.3531. First, I think if I multiply mole fraction of krypton and mass of the gases, I'd get the mass of krypton. But my teacher said if the answer is false. So what is the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="5026" LastEditDate="2018-03-08T09:37:48.223" LastActivityDate="2018-03-08T09:37:48.223" Title="Calculating mass of krypton from mole fraction" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7900" PostTypeId="2" ParentId="7560" CreationDate="2014-01-19T08:14:43.090" Score="-2" Body="&lt;p&gt;In iodate ions there are three double bonds between iodine and oxygen and there is presence of one lone pair.iodine is central atom.the shape of molecule can be seen as pyramidal with corners of base triange be oxygen atoms and iodine at top.And there is no meaning for polarity for ions as they already have a charge and hence polar.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2014-01-19T08:14:43.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7901" PostTypeId="2" ParentId="7862" CreationDate="2014-01-19T10:11:43.817" Score="3" Body="&lt;p&gt;I think you shouldn't abuse with mnemonic. In this case you should start with this rule of &lt;strong&gt;VSPR&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;each atom in a molecule will be positioned so that there is minimal&#xA;repulsion between the valence electrons of that atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now you should begin to train your three dimensional thinking!&#xA;This is the method I've adopted to learn AXE, of course everyone has his own but maybe it could help to you to find yours:&lt;/p&gt;&#xA;&lt;h1&gt;Learn to draw the &lt;span class=&quot;math-container&quot;&gt;$AX_nE_0$&lt;/span&gt; series:&lt;/h1&gt;&#xA;&lt;p&gt;This for me was the first step: learn to draw the basic geometry 0 lone pair. This is quite intuitive knowing the previous law. Begin with the &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; then add the &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;s. Note that the linear module could help you to remember it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jq3sY.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Add 1,2,3 lone pairs series&lt;/h1&gt;&#xA;&lt;p&gt;This is the next step. I've brutally modified the wikipedia table to show how you can easily do this step from the &lt;em&gt;Basic geometry 0 lone pair&lt;/em&gt;. Simply begin to substitute &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;: one &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; for &lt;em&gt;1 lone pair&lt;/em&gt;, two for &lt;em&gt;2 lone pair&lt;/em&gt;...&lt;/p&gt;&#xA;&lt;h1&gt;Nomenclature&lt;/h1&gt;&#xA;&lt;p&gt;The last step is to add the nomenclature. First of all you should take in account that the lone pair (&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;) don't appear as a part of the final shape from which the nomenclature is derived. If not considering the &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; group you obtain a &lt;em&gt;planar&lt;/em&gt; molecule you have to add the word &lt;strong&gt;planar&lt;/strong&gt; after the name, if you obtain a pyramidal shape you should use &lt;strong&gt;pyramidal&lt;/strong&gt; after the name of the planar pyramid's base.&#xA;So imagine that &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; groups are invisible and try use this method if you  can't find it only with pyramids try to figure out which of the following figures is more close to the actual shape of the molecule.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pJxYQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The first two are platonic solids  (Octhaedron, and Tetrahedron) then there is a &lt;strong&gt;T&lt;/strong&gt; for a &lt;em&gt;T-shaped&lt;/em&gt;, and a bent figure. In the original wikipedia table respective molecule configuration is already rotated to the right position, I think you can do it on your own, with your 3D thinking.&#xA;For the last, &lt;strong&gt;seesaw&lt;/strong&gt; shaped, there is a powerful &amp;quot;mem&amp;quot;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZguYy.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is an old photo of two boys in a &lt;span class=&quot;math-container&quot;&gt;$SF_4$&lt;/span&gt; molecule.Good ol' times!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-01-19T17:23:01.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7902" PostTypeId="1" CreationDate="2014-01-19T10:36:22.390" Score="5" ViewCount="761" Body="&lt;p&gt;This is a fundamental question, though I didn't really get the meaning/idea what does it mean by aromaticity or degree of aromaticity. Does it mean the ability of a molecule to be aromatic?&lt;/p&gt;&#xA;" OwnerUserId="2498" LastActivityDate="2019-04-08T19:55:58.110" Title="What does it mean by degree of aromaticity of a molecule?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7904" PostTypeId="2" ParentId="7899" CreationDate="2014-01-19T13:10:58.833" Score="4" Body="&lt;p&gt;Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can get a second equation in both variables, you can solved the system for each variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were able to determine the mole fractions $\chi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$&#xA;$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$&#xA;$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From mole fractions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$&#xA;$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-19T13:10:58.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7905" PostTypeId="1" CreationDate="2014-01-19T14:02:12.413" Score="1" ViewCount="1353" Body="&lt;p&gt;Why is the ionic product for water ($K_\text{w}$) temperature dependent? The concentration of $\ce{H+}$ and $\ce{OH-}$ ions in pure water is $[\ce{H+}] = [\ce{OH-}] = 1.0\times 10^{-7}\ \mathrm{mol/l}$ at $25\ \mathrm{^\circ C}$, and therefore the ionic product of water at $25\ \mathrm{^\circ C}$ is $1.0\times 10^{-14}$. Why it is so? What is the effect of temperature on this value?&lt;/p&gt;&#xA;" OwnerUserId="4220" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T11:12:59.253" LastActivityDate="2020-02-23T03:52:05.593" Title="Ionic Product of Water" Tags="&lt;equilibrium&gt;&lt;water&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7906" PostTypeId="1" CreationDate="2014-01-19T14:37:01.663" Score="0" ViewCount="62" Body="&lt;p&gt;What is fire? Is it a material or energy or something else? Please explain it to me. My friend asked me this. I am stuck with this problem. &lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="81" LastEditDate="2014-01-19T20:05:47.957" LastActivityDate="2014-01-19T20:05:47.957" Title="What is the nature and character of fire?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2014-01-21T18:56:23.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="7907" PostTypeId="2" ParentId="1251" CreationDate="2014-01-19T14:39:20.730" Score="5" Body="&lt;p&gt;I think you made a counting mistake. It is 6-methylcyclohex-2-en-1-ol&lt;/p&gt;&#xA;" OwnerUserId="4221" LastActivityDate="2014-01-19T14:39:20.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7908" PostTypeId="2" ParentId="7905" CreationDate="2014-01-19T17:46:34.243" Score="4" Body="&lt;p&gt;I think &lt;a href=&quot;http://www.chem.ufl.edu/~itl/2045/lectures/lec_v.html&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; should really help you out in understanding how the equilibrium constant depends on temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, considering the case of water dissociation:&#xA;$$\ce{H2O &amp;lt;=&amp;gt; H+ + HO-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the free energy of the reaction ($\Delta_\text{r}G$), using the Gibbs energy for the reactant and the products (in aqueous solution):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G(\ce{H2O}) = -237.13\ \mathrm{kJ/mol}$$&#xA;$$\Delta G(\ce{H+}) = 0\ \mathrm{kJ/mol}$$&#xA;$$\Delta G(\ce{HO-}) = -157.24\ \mathrm{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_\text{r}G = \Delta G(\ce{H+}) + \Delta G(\ce{HO-}) - \Delta G(\ce{H2O}) = -157.24\ \mathrm{kJ/mol} + 237.13\ \mathrm{kJ/mol} = 79.89\ \mathrm{kJ/mol}$$&#xA;The values of the Gibbs energies were taken from the 8th edition of Atkins' Physical Chemistry.&lt;br&gt;&#xA;Now, applying the formula below (which can be obtained from $\Delta_\text{r}G = -RT\ln K_\text{eq}$) for $T = 298\ \mathrm{K}$ (or $25\ \mathrm{^\circ C}$):&#xA;$$K_\text{eq} = \operatorname{e}^{-\frac{\Delta_\text{r}G}{RT}}$$&#xA;we get something like this : &#xA;$$K_\text{eq} = 2.71^{-\frac{79890\ \mathrm{J\ mol^{-1}}}{8.3145\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm{K}}} = 9.93 \times 10^{-15}$$&#xA;This value can be approximated to ${10^{-14}}$.&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T11:10:27.013" LastActivityDate="2015-05-03T11:10:27.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7909" PostTypeId="2" ParentId="7890" CreationDate="2014-01-19T18:06:25.587" Score="1" Body="&lt;p&gt;If the stated bacterium's cell has a diameter of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 \mu m}$&lt;/span&gt;, the volume can be derived in terms of liters remembering the linear relation between cubic meters:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  V_\text{cell}&#xA;    =\frac{4}{3}\pi \left(\pu{0.5\times10^-6 m}\right)^3&#xA;    =\frac{4}{3}\pi \left(\pu{0.5\times10^-5 dm}\right)^3&#xA;    =\pu{5\times 10^-16 L}&#xA;$$&lt;/span&gt;&#xA;Inside this volume, the organism contains a certain number of base pairs; so the total amount of substance has to be known.&lt;/p&gt;&#xA;&lt;p&gt;From Avogadro's Number, a mole is defined to be that portion (number of particles) of every substance in a defined physical phase; since this is an aqueous solution, the volume, as well as the molecular weight is meaningless: if one mole is defined by a certain number of particles, a different number of particles defines a different amount of substance:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  n_{\text{tot bp}}&#xA;  =\frac{N_{\text{bp}}}{N_{\text{A}}}&#xA;  =\frac{\pu{4\times 10^6 molecules}}{\pu{6\times 10^23 molecules mol-1}}&#xA;  =\pu{6.7\times10^-18 mol}&#xA;$$&lt;/span&gt;&#xA;Then, by the definition of molarity, dividing the amount of substance contained inside the cell by its volume gives a decent number, for a cell:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  M_\text{DNA}&#xA;  =\frac{n_\text{tot bp}}{V_\text{cell}} &#xA;  =\frac{\pu{6.7 \times 10^-18 mol}}{\pu{5 \times 10^-16 L}}&#xA;  =\pu{1.34 \times 10^-2 M}&#xA;$$&lt;/span&gt;&#xA;I think that the wrong result is due to the volume, because to me it seems rather surprising that one sphere of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 \mu m}$&lt;/span&gt; in diameter has:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \pu{2\times 10^-5 L}&#xA;  =\pu{2\times 10^-2 mL}&#xA;  =\pu{20 \mu L}&#xA;  =\pu{20 mm3} &#xA;  \neq \pu{5 \times 10^-8 mm3}&#xA;$$&lt;/span&gt;&#xA;of occupied volume. I suspect that something went wrong with the conversions, because I don't see (for now) any errors in my derivation.&lt;/p&gt;&#xA;" OwnerUserId="1984" LastEditorUserId="4945" LastEditDate="2020-06-18T11:41:28.257" LastActivityDate="2020-06-18T11:41:28.257" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7910" PostTypeId="2" ParentId="7541" CreationDate="2014-01-19T19:08:18.283" Score="23" Body="&lt;h1&gt;Why is water wet?&lt;/h1&gt;&#xA;&lt;p&gt;Chemically speaking saying that water is wet has not much sense.We could, however, say that &lt;em&gt;&amp;quot;Wetting process&amp;quot;&lt;/em&gt; is caused by the presence of a thin liquid particles layer over a material. In our specific case, this depends on the chemistry of water and the chemistry of skin. So the main causes of the fact that water is wet, is its wetting properties determined by a force balance between adhesive and cohesive forces over the skin. Mercury, although is a liquid, will not &amp;quot;wet&amp;quot; you because it has a strong cohesive force that doesn't permit the molecules to get in touch with the sensory cells.  When you touch water or a water wetted material part of the water molecules are transfer to your hands (a hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be caused however in some case of parasthesias even if there are no liquid particles. In fact, there is no &amp;quot;wet receptor&amp;quot; on your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in your skin.&#xA;Try to play with plasticine after a while handling it your hands will become hydrophobic you will no more fell the wet sensation when you wash them even if water is still &amp;quot;wet&amp;quot;!&lt;/p&gt;&#xA;&lt;h1&gt;What causes fire to be hot as an inherent property?&lt;/h1&gt;&#xA;&lt;p&gt;Fire is an exothermic process. So energy stored in a chemical bond is released as kinetic energy and electromagnetic wave. The fire has a high temperature because the molecules have great kinetic energies when you get in touch with fire (of course wormed air could be a medium as well for short distance) this kinetic energy is transfer to your skin and so to your &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermoreceptor&quot; rel=&quot;noreferrer&quot;&gt;thermoreceptors&lt;/a&gt; . And you feel what human call hot. In fact, you can feel the sensation of hot even without touching the fire because it even irradiates energy as electromagnetic waves that can be absorbed by your skin and so be feel for his receptors.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-11-10T10:17:05.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7911" PostTypeId="2" ParentId="7132" CreationDate="2014-01-19T19:39:17.563" Score="1" Body="&lt;p&gt;The electricity generate depends mainly from the electrodes. This is the same misconception of lemons battery. In fact the energy of every battery comes from the metals and not from the acid use as electrolyte, as you can't generate energy from lemon juice but from the different potential between copper and zinc coin so you can't generate energy from a stomach acid but with it. However $HCl$ is not used normally for battery so I think you could not produce a big amount of energy even if you  ingest very big electrodes.  &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-01-19T19:39:17.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7912" PostTypeId="1" AcceptedAnswerId="8637" CreationDate="2014-01-19T22:56:49.210" Score="5" ViewCount="2674" Body="&lt;p&gt;I usually make tea by adding two spoons of tea powder to water at room temp. and start to heat it. My friend said it will be better to add the powder once the water starts boiling. She said that it will dissolve tea more easily because of the sudden temperature difference experienced by tea powder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make a difference? Why?&lt;/p&gt;&#xA;" OwnerUserId="4223" LastEditorUserId="1362" LastEditDate="2014-02-21T18:14:44.067" LastActivityDate="2014-02-21T18:14:44.067" Title="Is tea powder better dissolved when put in hot water?" Tags="&lt;water&gt;&lt;solubility&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7913" PostTypeId="1" AcceptedAnswerId="7923" CreationDate="2014-01-20T00:31:23.807" Score="3" ViewCount="3385" Body="&lt;p&gt;From my understanding hydrogen peroxide only reacts with bacteria in a way that it should burn them more or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is true, can I then safely use hydrogen peroxide to clean my bathroom sink which are made of porcelain?&lt;/p&gt;&#xA;" OwnerUserId="4224" LastActivityDate="2014-01-20T19:47:38.027" Title="Is hydrogen peroxide safe for cleaning porcelain?" Tags="&lt;reaction-mechanism&gt;&lt;safety&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7914" PostTypeId="1" CreationDate="2014-01-20T02:12:33.363" Score="0" ViewCount="153" Body="&lt;p&gt;These days, we as a community what to polish and make our world a greener environment. I found a way to accomplish this task, however I don't know if it is applicable. I am wondering if the use of diatonic carbon would do the trick to produce a process that could regenerate the fuel in our vehicles and eliminate the releasing of carbon dioxide into the air. The purpose of this reaction would be to remove the Carbon Dioxide waste that is created by gasoline. Gasoline waste contains one less carbon than was originally in the reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(UNDER CONSTRUCTION)&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="4225" LastEditorUserId="4225" LastEditDate="2014-01-23T02:01:07.443" LastActivityDate="2014-01-31T10:08:42.043" Title="Is it possible to use diatomic carbon to utilize a new type of fuel that takes away the waste product of carbon dioxide being released into the air?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-03-20T10:16:46.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="7915" PostTypeId="1" CreationDate="2014-01-20T03:13:14.307" Score="0" ViewCount="9142" Body="&lt;p&gt;I was doing a chemistry project to see if the pH of a carbonated solution effects the time taken for it to settle after being disrupted, and I am trying to map out the reaction when the components are combined.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to figure out a few, but I got stuck on carbonic acid and lemon juice.  For some reason, I was getting ethane as a by product of the reaction, and I'm sure that's not right.  Could someone explain what I'm doing wrong?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the formula is $\ce{H2CO3 + C6H8O7}$&lt;/p&gt;&#xA;" OwnerUserId="4226" LastEditorDisplayName="user4076" LastEditDate="2014-01-20T15:29:16.777" LastActivityDate="2018-06-11T03:21:43.673" Title="Lemon juice and carbonic acid reaction" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7916" PostTypeId="1" AcceptedAnswerId="7917" CreationDate="2014-01-20T07:19:53.793" Score="216" ViewCount="62613" Body="&lt;p&gt;If I can smell an object, it means that molecules of it are getting separated from it, so they can reach my nose. As far as I know, metals don't sublimate, especially not in room temperature. However, copper has a very strong and characteristic smell. Does it mean that copper will degrade pretty quickly, or are just we so sensitive to it that a few molecules are sufficient? I assume it has to do something with oxidization, but it doesn't oxidize that much naturally as other metals, for example, iron.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="7448" LastEditDate="2015-08-03T13:46:04.200" LastActivityDate="2018-04-17T09:42:31.157" Title="Why can we smell copper?" Tags="&lt;everyday-chemistry&gt;&lt;metal&gt;&lt;transition-metals&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="60" ContentLicense="CC BY-SA 3.0" />
  <row Id="7917" PostTypeId="2" ParentId="7916" CreationDate="2014-01-20T11:20:12.960" Score="190" Body="&lt;p&gt;This is a nice question, as it confronts a very replicable and common experience with a well established yet seemingly contradictory fact. As you expected, the smell of metal has nothing to do with the metal actually getting into your nose, as most metals have far too low of a vapor pressure at ordinary temperatures to allow direct detection. The characteristic smell of metal, in fact, is caused by organic substances!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been the focus on the specific case of the &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.200602100/pdf&quot; rel=&quot;noreferrer&quot;&gt;smell of iron&lt;/a&gt; (free-access article!). There are at least two ways in which iron produces a metallic smell. Firstly, acidic substances are capable of corroding iron and steel, releasing phosphorus and carbon atoms present in the metal or alloy. These can react to form volatile organophosphorus compounds such as methylphosphine ($\ce{H3CPH2}$ which have a garlic/metallic odor at small concentrations. From the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The “garlic” metallic odor (see Supporting Information) of the gas product from the acidic dissolution of cast iron is dominated by these organophosphines. We measured an extremely low odor threshold for two key odorants, methylphosphine and dimethylphosphine (6 and 3 ng P/m³, respectively, garlic-metallic odor), which belong therefore to the most potent odorants known. Phosphine ($\ce{PH3}$) is not important for this odor because we found it has a much higher odor detection threshold (&gt;10⁶ ng/m³). A “calcium carbide” (or “burned lime”/“cement”) attribute of the general “garlic” odor is probably caused by unsaturated hydrocarbons (alkynes, alkadienes) that are linked to a high carbon content of iron (Table 1, see Supporting Information).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, it turns out that $\ce{Fe^{2+}}$ ions (but not $\ce{Fe^{3+}}$) are capable of oxidizing substances present in oils produced by the skin, namely lipid peroxides. A small amount of $\ce{Fe^{2+}}$ ions are produced when iron comes into contact with acids in sweat. These then decompose the oils releasing a mixture of ketones and aldehydes with carbon chains between 6 and 10 atoms long. In particular, most of the smell of metal comes from the unsaturated ketone &lt;a href=&quot;http://en.wikipedia.org/wiki/Oct-1-en-3-one&quot; rel=&quot;noreferrer&quot;&gt;1-octen-3-one&lt;/a&gt;, which has a fungal/metallic odour even in concentrations as low as $1\ \mu g\ m^{-3}$ . In short:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sweaty skin corrodes iron metal to form reactive $\ce{Fe^{2+}}$ ions that are oxidized within seconds to $\ce{Fe^{3+}}$ ions while simultaneously reducing and decomposing existing skin lipid peroxides to odorous carbonyl hydrocarbons that are perceived as a metallic odor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.wiley-vch.de/contents/jc_2002/2006/z602100_s.pdf&quot; rel=&quot;noreferrer&quot;&gt;the supporting information for the article&lt;/a&gt; (also free-access), the authors describe experiments performed with other metals, including copper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparison of iron metal with other metals (copper, brass, zinc, etc.): When solid copper metal or brass (copper-zinc alloy) was contacted with the skin instead of iron, a similar metallic odor and GC-peak pattern of carbonyl hydrocarbons was produced and up to one μmole/dm² of monovalent cuprous ion [$\ce{Cu+}$] was detected as a corrosion product (Supporting Figs. S3 to S6). Zinc, a metal that forms $\ce{Zn^{2+}}$ but no stable $\ce{Zn+}$, was hesitant to form metallic odor, except on very strong rubbing of the metal versus skin (that could produce metastable monovalent $\ce{Zn+}$). The use of common color-tests to demonstrate directly on human palm skin the presence of low-valence ions (ferrous and cuprous) from the corrosion of iron, copper and brass alloys is shown in Supporting Figure S6. Alumina powder rubbed on skin did not produce significant odorants. These results provide additional evidence that it is not metal evaporation, but skin lipid peroxide reduction and decomposition by low valence metal ions that produces the odorants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last paragraphs of the article summarize the findings:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In conclusion: 1) The typical “musty” metallic odor of iron metal touching skin (epidermis) is caused by volatile carbonyl compounds (aldehydes, ketones) produced through&#xA;  the reaction of skin peroxides with ferrous ions ($\ce{Fe^{2+}}$) that are formed in the sweat-mediated corrosion of iron. $\ce{Fe^{2+}}$ ion containing metal surfaces, rust, drinking water, blood etc., but also copper and brass, give rise to a similar odor on contact with the skin. The human ability to detect this odor is probably a result of the evolutionarily developed but largely dormant ability to smell blood (“blood scent”).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) The “garlic-carbide” metallic odor of phosphorus- and carbon-rich cast iron and steel under attack by acid, is dominated by volatile organophosphines. Corroding cast iron&#xA;  is an environmental source of C–P compounds that may lead to confusion in the verification and monitoring of the Chemical Weapons Convention (see also ref. [15])&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an aside, this may be why sometimes people recommend getting strong smells off your hands by rubbing them against a metal object. While it probably doesn't work for some metals and for some smelly compounds, it's possible that the metal catalyzes the decomposition of the malodorous substances into less strongly smelling ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read a little more &lt;a href=&quot;http://www.nature.com/news/2006/061025/full/news061023-7.html&quot; rel=&quot;noreferrer&quot;&gt;in this press article&lt;/a&gt; on the study. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="51553" LastEditDate="2018-04-17T09:42:31.157" LastActivityDate="2018-04-17T09:42:31.157" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7918" PostTypeId="1" AcceptedAnswerId="7925" CreationDate="2014-01-20T14:27:53.423" Score="1" ViewCount="3509" Body="&lt;p&gt;I am trying to determine how long a system that constantly converts water into H and O and then back will last, as in roughly how many cycles. Now this system can either be made of say a reversible fuel cell or a non-reversible fuel cell combined with a dedicated electrolysis machine (if there is a difference in lifetime). As far as I can tell, platinum is used (almost) everywhere as catalyst and as far I can determine it seems to corrode somewhat throughout use, especially in terms of splitting the oxygen in the fuel cell. I am no professional though and I can only find limited information so I would appreciate it if someone could please tell me roughly how long a system like this will last and how.&lt;/p&gt;&#xA;" OwnerUserId="4229" LastActivityDate="2019-06-11T03:57:12.370" Title="Lifetime of a (reversible) hydrogen fuel cell system" Tags="&lt;hydrogen&gt;&lt;electrolysis&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7920" PostTypeId="1" AcceptedAnswerId="7952" CreationDate="2014-01-20T17:29:41.697" Score="28" ViewCount="2736" Body="&lt;p&gt;Please help me with any/all of the cases below. In the following cases, the named method and basis set are not suitable for the chemical systems. Why aren't they? Could you suggest a suitable method/basis set for each?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;1. MP2/6-31G(d) for the calculation of reaction energy of the reaction below (H2C: should be calculated in the triplet state):&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/e7D9s.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;The carbene H2C: has two singlet configurations that contribute to the ground state energy, so we need a multireference method to recover the single excitations. MP2 is not good for this open-shell systems since they have large spin contamination and thus large error. The basis set also needs to include diffuse functions for electron correlation methods.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2. CISD/cc-pVTZ for the calculation of the reaction energy of:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kjCyF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kjCyF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;We have bond elongation in the reactant to the point of breakage. CISD is not size-consistent. It recovers less electron correlation energy for the larger reactant than for the products since the reactant is larger, so CISD cannot be used. The basis set however is fine. I don't know what method can work best for this system but would go arbitrarily with CCSD(T).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;3. BLYP/6-31G(d) for geometry optimization and calculation of binding energy of the neon dimer, $\ce{Ne\bond{...}Ne}$.&lt;/h3&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;The most important interactions in the Ne dimer are London forces, or van der Waals (vdW) interactions. BLYP is a hybrid DFT method that includes exact exchange term in exchange-correlation functional. However, the method is local (in LDA, the external potential is the uniform electron gas, and in a GGA, its gradient is included), and vdW interaction requires fully non-local functionals. The basis set is also too small and should include diffuse functions. I don't know what method to recommend.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;4. MP2/6-311++G(2df,p) for the singlet-triplet energy splitting of para-benzyne. Note that for both singlet and triplet states unrestricted MP2 is needed.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6wTqt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6wTqt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;The para-related electrons have low correlation, meaning they have small singlet-triplet separations of the frontier orbitals. MPn energy terms are related to the inverse of this separation and thus for a very small s-t separation, this term is too large for MP2 to handle. The basis set is probably fine. I think coupled cluster methods or DFT would work here.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;5. CCSD/3-21G for geometry optimization of the DNA base adenine.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ikcyJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ikcyJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;The basis set is too small for this molecule and a correlated method (no diffuse or polarization functions are used). Furthermore, CCSD is very costly and not good for this big molecule. Maybe handling the conjugated pi bonds is important. Suggestions?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;6. CISD/6-311G(d,p) to compare the CC bond lengths of oligoenes ($2&amp;lt;n&amp;lt;20$).&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u14g1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u14g1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;CISD is not size-extensive, so as $n$ increases, the calculations become prohibitive.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;7. MP3/6-311G(d,p) for the electron affinity of F2 i.e. the process:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MtMTb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MtMTb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;MP3 is not good when we have an open-shell species or a system with electron congestion. Convergence will be very slow and not guaranteed at all. We assume in MPn that the zeroth-order wavefunction is a reasonable approximation, and the electron correlation contribution can be added as a small perturbation. But if HF describes this molecule very poorly, then MPn won't improve upon the zeroth-order wavefunction dramatically.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;8. LSDA/6-31G(d) for the relative energies of the various conformations of n-butane obtained through rotation about the central C-C bond.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K2RKh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K2RKh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;LSDA is equal to LDA for closed shell systems. Thus the external potential is the uniformly distributed positive charge and can be used only if the &quot;constant&quot; electron density is different for each point in space. However, here the electron density due to bond twisting is not spatially uniform. I think a GGA or B3LYP should be used. Furthermore, I think the basis set is too small and should include extra polarization and diffuse functions (I don't know why, just saying).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;What method/basis set would you suggest for the following:&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;9. The activation energy of the Diels-Alder reaction between cyclopentadiene and acrylonitrile (reaction does not need static electron correlation treatment).&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/La176.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/La176.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;Pure LDAs or GGAs overestimate bond length, so I suggest the hybrid B3LYP with a small basis set.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;10. The interaction energy between two Ar atoms in the $\ce{Ar\bond{...}Ar}$ van der Waals complex.&lt;/h3&gt;&#xA;&#xA;MY ATTEMPTED ANSWER&#xA;&#xA;&lt;p&gt;Use CCSD with a medium basis set, since vdW is repulsion of electron clouds at large distances as they interpenetrate and CCSD has traces of higher excitations. A small basis set can cause basis set superposition error (BSSE). Inclusion of higher order virtual orbitals as in CCSD improves the dispersion description.&lt;/p&gt;&#xA;&#xA;  &#xA;" OwnerUserId="4162" LastEditorUserId="194" LastEditDate="2018-03-17T16:45:55.730" LastActivityDate="2018-03-17T16:52:34.077" Title="Suggest methods and basis sets for a variety of systems" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="7921" PostTypeId="2" ParentId="7478" CreationDate="2014-01-20T17:44:44.323" Score="0" Body="&lt;p&gt;The group best able to sustain a positive charge migrates (i.e. H &gt; Me3C &gt; Me2HC &gt; Et &gt; Me)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-20T17:44:44.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7922" PostTypeId="2" ParentId="7918" CreationDate="2014-01-20T19:34:45.157" Score="2" Body="&lt;p&gt;Poisoning of the catalyst, e.g. by $\ce{CO}$, is only one aspect that reduces the lifetime of PEM fuel cells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nafion polymer membrane can suffer due to mechanical stress as a result of rapid temperature changes or dehydration.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-20T19:34:45.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7923" PostTypeId="2" ParentId="7913" CreationDate="2014-01-20T19:47:38.027" Score="3" Body="&lt;p&gt;While hydrogen peroxide is not normally used for cleaning surfaces, there is nothing in normal glazed porcelain that it would damage. Porcelain is about as unreactive to most things as glass, which is unreactive enough to be used for the bottles that store hydrogen peroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you should probably be careful with it as it can be dangerous if the solution is a strong one. Normally available hydrogen peroxide should not be as strong as the solutions used in the lab, though, which is often a 30% solution in water and should only be handled if you know what you are doing (see the &lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=GB&amp;amp;language=en&amp;amp;productNumber=216763&amp;amp;brand=SIAL&amp;amp;PageToGoToURL=http://www.sigmaaldrich.com/catalog/search?interface=All&amp;amp;term=hydrogen%2bperoxide&amp;amp;lang=en&amp;amp;region=GB&amp;amp;focus=product&amp;amp;N=0%2b220003048%2b219853121%2b219853286&amp;amp;mode=match%2520partialmax&quot; rel=&quot;nofollow&quot;&gt;safety data sheet&lt;/a&gt;). Consumer hydrogen peroxide is usually a 3-6% solution and isn't any more dangerous that hypochlorite bleach (and is a safer way to bleach hair). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard household (hypochlorite) bleach is usually better for cleaning, though.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-20T19:47:38.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7924" PostTypeId="2" ParentId="7915" CreationDate="2014-01-20T20:24:21.703" Score="1" Body="&lt;p&gt;These two weak acids have similar disassociation constants in water.  A low Ka indicates a relatively small number of acid molecules will form an anion stabilized by water.  Two weak acids with similar disassociation constants will coexist in a solution without interacting.  If two weak acids have very different Ka then there can be a reaction.  In this case nothing would happen.   For more generalized information search for mixture of weak acids.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbonic Acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2CO3 + H2O &amp;lt;=&amp;gt; HCO3- + H3O+}$$&#xA;$$K_a = 4.3 \times 10^{-7} = \frac{[\ce{H3O+}][\ce{HCO3^-}]}{[\ce{H2CO3}]}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Citric Acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C6H5O7 + H2O &amp;lt;=&amp;gt; H2C6H5O7- + H3O+}$$&#xA;$$K_a = 1.8 \times 10^{-6}  = \frac{[\ce{H3O+}][\ce{H2C6H5O7^-}]}{[\ce{H3C6H5O7}]}$$&lt;/p&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="275" LastEditDate="2014-01-21T03:11:15.880" LastActivityDate="2014-01-21T03:11:15.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7925" PostTypeId="2" ParentId="7918" CreationDate="2014-01-20T21:12:10.470" Score="4" Body="&lt;p&gt;Fuel cell manufacturers quote a range of different lifetimes which depend on many factors. Backup cells tend to have lower lifetimes as the start-stop process degrades them compared to continuous operation. On the other hand, use as engines in vehicles requires a reasonably long life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impure fuel will damage the platinum (CO for example will poison the catalytic surface). But platinum is recyclable so either poisoning or other factors won't necessarily be a fundamental barrier to use (some companies take used cells back at the end of life and reuse the platinum). Membranes and electrodes also degrade mechanically depending on duty cycle and any mechanical stabilisation in the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nedstack.com/faq/about-pem-fuel-cells&quot; rel=&quot;nofollow&quot;&gt;Nedstack&lt;/a&gt; quotes lifetimes ranging from 2,000 hours for backup cells (which presumably means years of actual operation as they only get used when other power is cut which isn't a routine happening) and 8,000-16,000 hours for continuous operation (which is two years). &lt;a href=&quot;http://www.hydrogen.energy.gov/pdfs/epact_743_fuel_cell_school_bus.pdf&quot; rel=&quot;nofollow&quot;&gt;Tests&lt;/a&gt; of fuel cells in vehicles have reported lifetimes of 7,500 hours (which presumably relates to several years of use in practice), though they seem to need more maintenance than diesel engines.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-20T21:12:10.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7926" PostTypeId="1" CreationDate="2014-01-20T22:05:00.730" Score="0" ViewCount="224" Body="&lt;p&gt;Recycling Al cans to produce electricity,i.e. to reverse the process of electrolysis and produce electricity&lt;/p&gt;&#xA;" OwnerUserId="4235" LastActivityDate="2017-01-30T05:05:26.733" Title="I want to make electricity of aluminiun cans i.e. reverse the process of refinning aluminium. Any ideas?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7927" PostTypeId="2" ParentId="7926" CreationDate="2014-01-21T01:03:52.113" Score="3" Body="&lt;p&gt;You could &quot;burn&quot; the aluminum in the presence of oxygen to generate aluminum oxide.  For every two moles of aluminum consumed, a bit less than 1700 kJ of energy would be released.  You could use the energy to drive a turbine and generate electricity.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-01-21T02:23:18.917" LastActivityDate="2014-01-21T02:23:18.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7930" PostTypeId="1" AcceptedAnswerId="7937" CreationDate="2014-01-21T06:36:23.757" Score="1" ViewCount="1110" Body="&lt;p&gt;My question is, if &quot;gas&quot; belongs to &quot;State of Matter&quot; then what does &quot;Metal&quot; belong to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas can be substituted with any state of matter(in the analogy); metal can be substituted with metalloid or nonmetal(in the analogy). I've researched this and I can't seem to find an answer to this. So, this could read as &quot;if 'solid' belongs to 'State of Matter', then what does 'nonmetal' belong to'&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not trying to assert that metal is a state of matter. &#xA;As most of y'all (hopefully all) know, the elements on the periodic table can be a metal, metalloid, or nonmetal. Would &quot;Metal&quot; belong to &quot;Metal Status&quot;? I'm asking because I'm writing a program to represent the PeriodicTable and I need an appropriate name for my variable. I may just opt for &quot;Metal Status&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4239" LastEditorDisplayName="user7951" LastEditDate="2017-01-14T12:05:13.413" LastActivityDate="2017-01-14T12:05:13.413" Title="If &quot;Gas&quot; belongs to &quot;State of Matter&quot; then what does &quot;Metal&quot; belong to?" Tags="&lt;metal&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7931" PostTypeId="1" CreationDate="2014-01-21T08:23:48.300" Score="0" ViewCount="520" Body="&lt;p&gt;I have the molecule 1,4-benzenedimethanol: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T5ArL.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I produce two methanol molecules and one benzene molecule? (Is there a catalytic reaction method available for this?)&#xA;I'm guessing that hydrogen ions could be used to split off the methanol molecules on both sides?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="7448" LastEditDate="2015-10-16T17:12:42.943" LastActivityDate="2015-10-16T17:12:42.943" Title="Converting 1,4-benzenedimethanol to benzene and methanol?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7932" PostTypeId="2" ParentId="7930" CreationDate="2014-01-21T09:46:33.837" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;h1&gt;What is &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal&quot; rel=&quot;nofollow&quot;&gt;Metal&lt;/a&gt;?&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;A metal (from Greek &quot;μέταλλον&quot; – métallon, &quot;mine, quarry, metal&quot;) is a solid material (an element, compound, or alloy) that is typically hard, opaque, shiny, and features good electrical and thermal conductivity.  &lt;/p&gt;&#xA;  &#xA;  &lt;h1&gt;What is &lt;a href=&quot;http://en.wikipedia.org/wiki/State_of_matter&quot; rel=&quot;nofollow&quot;&gt;State of matter&lt;/a&gt;?&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;In physics, a state of matter is one of the distinct forms that different phases of matter take on. Four states of matter are observable in everyday life: solid, liquid, gas, and plasma.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the above definitions, it is clear that gas is a state of matter and metal is also the state of matter with certain specific properties. The distinction is that, all solids are not metals but all metals are solids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An element or compound or alloy is said to be metal if it is  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solid  &lt;/li&gt;&#xA;&lt;li&gt;hard  &lt;/li&gt;&#xA;&lt;li&gt;opaque  &lt;/li&gt;&#xA;&lt;li&gt;shiny  &lt;/li&gt;&#xA;&lt;li&gt;good electrical and thermal conductor.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Gold has all the above properties at standard conditions for temperature and pressure, it can be regarded as metal only at that temperature and pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the elements, compounds, or alloys are regarded as metal, at standard conditions for temperature and pressure.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, mercury is regarded as metal only below the temperature of $-38.83$ degree celsius and is not regarded as metal at standard conitions for temperature and pressure. Thus, &lt;strong&gt;we say mercury as the only metal which exist as liquid at standard conditions for temperature and pressure. Remember it is wrong to say mercury as the only liquid metal.&lt;/strong&gt;   &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2014-01-21T13:23:33.390" LastActivityDate="2014-01-21T13:23:33.390" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7933" PostTypeId="1" AcceptedAnswerId="7935" CreationDate="2014-01-21T09:54:03.623" Score="4" ViewCount="421" Body="&lt;p&gt;I had this question from the day (9 years old, now 16) that I learned about states of matter. I have asked many of my teachers, some of them told me it's a gas some that it's a plasma. Can anyone answer my question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I've learned that the plasma state is obtained when all the electrons from the atom are removed. Obviously it's present in the sun due to high temperatures.&lt;/p&gt;&#xA;" OwnerUserId="4194" LastEditorUserId="9961" LastEditDate="2015-10-23T17:23:44.607" LastActivityDate="2015-10-23T17:23:44.607" Title="To which state of matter does the flame belong to?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7934" PostTypeId="2" ParentId="7931" CreationDate="2014-01-21T10:06:18.590" Score="1" Body="&lt;p&gt;I will not ask why you would want to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that the benzylic alkohol can be deprotonated. But that won't help you for the intended reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the possible mechanisms, I suggest to do it on paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the starting material and split the bonds. Draw the fragments that result from splitting the $\ce{Ph\bond{-}CH2OH}$ bond. Do this first for the two heterolytical pathways possible in theory. Inspect the fragments and ask yourself, how stable these might be in general, and in the presence of other molecules of the starting material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they likely? ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare with the following experiment on paper: Which reactions are possible with your starting material in the presence of strong acids? (Assuming that their corresponding bases are lousy nucleophiles.) Yes, it has nothing to do with the reaction you'd like to perform. But it works ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's inspect the homolytical case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cut the &quot;lecture&quot; short, it has indeed been observed that &lt;a href=&quot;http://cpb.pharm.or.jp/cpb/200009/C09_1299.pdf&quot; rel=&quot;nofollow&quot;&gt;benzylic alcohol furnishes benzene upon sonication (PDF)&lt;/a&gt;. The yields however are only in the ppm range and I'm not aware of a comparable study on the title compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the question in your last comment, have a look at the prices for methanol and benzene at the chemical supplier of your choice:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methanol (z.S.)         2,5 l   14.00 €&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene  (reinst)       1,0 l   32.90 €&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,4-Benzenedimethanol   50 g  294.30 €&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's just say that the production of methanol and benzene the way you suggested is not a promising concept. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-02T16:14:11.007" LastActivityDate="2014-02-02T16:14:11.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7935" PostTypeId="2" ParentId="7933" CreationDate="2014-01-21T10:26:32.257" Score="6" Body="&lt;p&gt;A candle flame is the product of the oxidation of wax to $\ce{CO2}$ and other small organic compounds. Since these small compounds are energetically lower, energy is released during the reaction, in the form of light and heat. So really, we start out with &lt;strong&gt;gas&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, flames can also &lt;a href=&quot;http://www.usc.edu/CSSF/History/2006/Projects/J0714.pdf&quot; rel=&quot;noreferrer&quot;&gt;conduct electricity&lt;/a&gt;, and the conductivity varies with the location in the flame. The linked science fair project summary states that the lowest resistance (highest conductivity) was found at the edge of the bright yellow part of the flame the conductivity was the largest, meaning that the concentration of ions was the highest. The lowest resistance value found was $80~\Omega$, which is still very large in comparison to the resistance of plasma, which is generally considered to be $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, summarizing my answer: &lt;strong&gt;A flame is a gas with lightly ionized portions, but does not qualify as a real plasma.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-21T10:26:32.257" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7936" PostTypeId="1" CreationDate="2014-01-21T10:30:29.257" Score="3" ViewCount="89" Body="&lt;p&gt;Would trans hydride ligands on a metal complex be beneficial for a catalyst?&#xA;Such as alkene metathesis or hydrodefluorination&lt;/p&gt;&#xA;" OwnerUserId="4240" LastActivityDate="2014-03-07T17:03:44.140" Title="Would trans hydride ligands on a metal complex be beneficial for a catalyst?" Tags="&lt;inorganic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7937" PostTypeId="2" ParentId="7930" CreationDate="2014-01-21T10:52:40.997" Score="7" Body="&lt;p&gt;&quot;Metal&quot; is a classification used by chemists and usually means something along the lines of &quot;this electrically conductive, &lt;strong&gt;usually solid&lt;/strong&gt; element that is deformable&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way around: Chemists use &quot;gas&quot; to refer to a state of matter, not some defined gas (for example the atmosphere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these two together: &lt;strong&gt;&quot;Metal&quot; belongs to the periodic table of elements&lt;/strong&gt;, while &quot;solid&quot;, &quot;liquid&quot;, &quot;gas&quot; and &quot;plasma&quot; are merely words to describe what state the element or compound is in. It is possible to have liquid, solid or even gaseous metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If you want a name for a variable that is either in the state Metal, Half-Metal or Metalloid and Non-Metal, I'm not sure such a name even exists. Feel free to make one up that suits your needs, comment your code sufficiently and it will not be a problem. &quot;Metallicness&quot; or something along those lines would be my choice.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-01-23T06:15:17.377" LastActivityDate="2014-01-23T06:15:17.377" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7938" PostTypeId="1" CreationDate="2014-01-21T10:55:31.993" Score="1" ViewCount="3220" Body="&lt;p&gt;I've heard that electrons are released from the clutches of nuclei in the plasma state that is present in the sun. My question is: &lt;strong&gt;Where do these electrons go?&lt;/strong&gt;  Do they go to the product nuclei of the nuclear fusion reaction in sun (ie. He)? Or do they remain on the surface of sun?&lt;/p&gt;&#xA;" OwnerUserId="4194" LastEditorUserId="3846" LastEditDate="2014-01-21T11:52:26.710" LastActivityDate="2020-06-09T21:42:39.300" Title="Where do the electrons that are released in the sun go?" Tags="&lt;nuclear&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7939" PostTypeId="2" ParentId="7938" CreationDate="2014-01-21T11:24:12.917" Score="7" Body="&lt;p&gt;Some of the electrons will react with positrons formed during the fusion processes; they annihilate each other and give off high-energy radiation (also called gamma rays):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NrR9F.jpg&quot; alt=&quot;p-p Solar Fusion Chain&quot;&gt;&#xA;&lt;a href=&quot;http://www.cpepweb.org/fusion_chart_view.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this process  happens at the core, and there is no way that those electrons get out to the surface. So my best guess is that the electrons which don't get annihilated just stay there in the core plasma. I do not think that they combine with the $\ce{^{4}_2 He^{2+}}$ particles formed by this chain, because the temperature is still too high. So you start out with 6 protons and 6 electrons (for electrical neutrality reasons), two electrons get annihilated in the process and you end up with a doubly positively charged helium particle and two protons and four electrons. So the electrical neutrality is not violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not only the core of the sun but also the outer layers and the corona, from where we observe the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_wind&quot; rel=&quot;noreferrer&quot;&gt;solar wind&lt;/a&gt;, which is a stream of particles, mainly electrons and protons. So there you have some electrons that escape the sun. But these are not the electrons from the core, they come from the surface of the sun.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-21T11:24:12.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7940" PostTypeId="1" CreationDate="2014-01-21T12:01:16.887" Score="6" ViewCount="46593" Body="&lt;p&gt;How is &lt;em&gt;solubility&lt;/em&gt; different from &lt;em&gt;dissolution&lt;/em&gt;? And how are they both different from &lt;em&gt;solvation&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="3846" LastEditDate="2014-01-21T13:17:54.420" LastActivityDate="2019-07-02T02:51:23.283" Title="Difference between solubility and dissolution?" Tags="&lt;solutions&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7941" PostTypeId="2" ParentId="7940" CreationDate="2014-01-21T12:39:03.970" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Solvation&lt;/strong&gt; [...] is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules.&#xA;  [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Dissolution&lt;/strong&gt; is a kinetic process, and is quantified by its rate. &lt;strong&gt;Solubility&lt;/strong&gt; quantifies the dynamic equilibrium state achieved when the rate of dissolution equals the rate of precipitation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quote from &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-21T12:39:03.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7942" PostTypeId="1" CreationDate="2014-01-21T13:31:06.147" Score="3" ViewCount="483" Body="&lt;p&gt;I'm sort of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7679/chemistry-self-education&quot;&gt;self studying Chemistry&lt;/a&gt;. I'm now getting stuck wrapping my head around a concept (and I realize it was the same concept I had difficulty grasping years ago) and that is &lt;strong&gt;Electron Configurations&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that perhaps due to the nature of chemistry it is non-intuitive. However, I want an explanation that is more than &quot;This is a bunch of numbers and letters that follow a pattern. Live with it&quot; - which seems to just be the explanations I'm getting when I went to school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, what are Electron Configurations, and how would you explain them intuitively?&lt;/p&gt;&#xA;" OwnerUserId="4119" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2016-04-29T18:27:55.650" Title="What is an intuitive explanation for Electron Configurations" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7943" PostTypeId="2" ParentId="7942" CreationDate="2014-01-21T13:56:28.193" Score="3" Body="&lt;p&gt;Electron configurations are a bunch of numbers and letters, and they also follow a pattern. But there is also a logic behind the whole thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, chemists use electron configurations to see in which &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow&quot;&gt;atomic orbitals&lt;/a&gt; the electrons are found. From this several things can be derived, most of all the reactivity of said electrons (which is where the chemistry happens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intuitive answer is this: Electron configurations are a way of telling in which atomic orbitals the electrons are put in, according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow&quot;&gt;Aufbau principle&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;Pauli exclusion principle&lt;/a&gt;. Decoding it is pretty easy, as it always follows this pattern:&#xA;$n$$(s|p|d|f)^x$ where $n$ is the main quantum number (or shell number), $(s|p|d|f)$ denotes the atomic orbitals with the quantum number and $x$ is the number of electrons found in the orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital &quot;groups&quot; ($p$, $d$, $f$) contain several &quot;orbitals&quot; (side note: an orbital is strictly seen a one-electron function) where two electrons with opposite spin fit in. $p$ has 3 ($p_x,\; p_y,\; p_z$), $d$ has 5 ($d_{xy},\; d_{xz},\; d_{yz},\; d_{x^2-y^2},\; d_{z^2}$) and $f$ even more. This means that the maximal number of electrons in the $p$ group is $6$, and in the $d$ group $10$. So when you fill in the electrons according to the aufbau principle, you have to go to the next orbital group when you reach this number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line:&lt;/strong&gt; Every electron in an atom is defined by the four quantum numbers $n$ (main or principal quantum number), $l$ (azimuthal), $m_l$ (magnetic) and $m_s$ (spin). Using the electron configuration we can derive the first three of these four numbers.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-21T13:56:28.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7944" PostTypeId="1" AcceptedAnswerId="7945" CreationDate="2014-01-21T16:35:56.693" Score="2" ViewCount="667" Body="&lt;p&gt;Coming from a DFT background, I'm used to the concept that the DFT eigenvalues do not correspond to excitation energies (i.e., the band gap, ionization potentials, etc.). To correct for these it's common to go to many-body approaches like the GW approximation to the self-energy, which is said to explicitly deal with quasiparticle addition/removal energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious how excited states are calculated within Hartree Fock or CI. Can one do a ground state calculation in HF or CI and then get excitation energies from the difference in these ground-state eigenvalues? Or is this similarly not possible and one needs to explicitly consider excited states in order to get the excitation spectrum (band gap, band eigenvalues for removal/addition of electron,etc.).&lt;/p&gt;&#xA;" OwnerUserId="4243" LastActivityDate="2014-01-21T16:56:02.517" Title="Excited states and electron addition/removal energies within Hartree-Fock, CI, etc" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7945" PostTypeId="2" ParentId="7944" CreationDate="2014-01-21T16:56:02.517" Score="3" Body="&lt;p&gt;There are actually quite a few methods post-HF that you can do to calculate excitation energies. The simplest one you will want to look at is Configuration Interaction Singles (CIS). To do this, you express your Hamiltonian in the basis of singly-excited HF determinants and diagonalize it. It doesn't involve correlation and is roughly the same cost as a HF calculation. The matrix elements are given in Szabo and Ostlund,  Modern Quantum Chemistry pg 236.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are already familiar that for DFT, the most common way to get excitation energies is through a time-dependent DFT calculation (TDDFT). You can do a time-dependent Hartree-Fock (TDHF) as well, though you may see this in the literature as the Random Phase Approximation (RPA). They are essentially the same thing. TDHF and TDDFT are pretty much identical save for the fact that TDDFT is based off Kohn-Sham density, and TDHF is based off of Hartree-Fock density. The equations are almost the same, and it isn't that much more expensive to compute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to include electron correlation in wavefunction-based methods to a greater degree, look for EOM-CCSD like methods. These are based off a HF, followed by a coupled-cluster ground state calculation, followed by an extension to get excited state energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My graduate research uses and develops a lot of these methods. In fact, I wrote about these methods not too long ago --- you can read more about it here: &lt;a href=&quot;http://bit.ly/1anA9dZ&quot; rel=&quot;nofollow&quot;&gt;http://bit.ly/1anA9dZ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2014-01-21T16:56:02.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7952" PostTypeId="2" ParentId="7920" CreationDate="2014-01-21T19:31:08.440" Score="31" Body="&lt;p&gt;This is a great question and I love it.  Thank you for taking the time to not only post the question but each attempted response.  I will try to address the parts of this question that I can at the moment.  I will list, in order, your attempted answer along with my response.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Carbene H2C: has two singlet configurations that contribute to the ground state energy we need a multireference method to recover the single excitations. MP2 is not good for this open-shell systems since they have large spin contamination and thus large error. The basis set also needs to include diffuse functions in the electron correlation methods.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, a multireference method will need to be used and a complete active space will need to be defined.  Therefore, MP2 will NOT work.  The 6-31(d) basis set is atrocious and should never be used for production level work given the growth of computing power in the last decade.  This double-$\zeta$ quality basis set is much too small and at minimum a triple-$\zeta$ basis set should be used.  Diffuse functions are critical to properly describe the spatial electronic region of a weakly bound electron far away from the nucleus, but in this case, I would only include diffuse functions on non-hydrogen atoms given that the unpaired electrons are on carbon and not hydrogen (as per your picture). Lastly, polarization functions should be included on every atom, and not just the non-hydrogen atoms.  Therefore I would use a basis set such as 6-311+G(d,p), 6-311+G(2df,2pd), or even heavy-aug-cc-pVTZ.  Quadruple-$\zeta$ basis sets should be used if you can afford it but may not be feasible in this case.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;We have bond elongation in the Reactant to the point of breackage. CISD is not size-consistent. It recovers less electron correlation energy for the larger reactant as for the products since reactant is larger and thus cannot be used. The basis set however is fine. I don't know what method can work best for this system but would go arbitrarily with CCSD(T).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;CISD does introduce size-consistency problems but there are 'workarounds' you can use (&lt;a href=&quot;http://iopscience.iop.org/0004-637X/489/1/113/fulltext/&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;). Obviously we want to avoid using more and more approximations so it would be prudent to avoid CISD in this case.  The basis set could be dramatically improved by adding diffuse functions to the non-hydrogen atoms (e.g. heavy-aug-cc-pVTZ).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Most important interactions in Ne dimer are London forces of van der Waals. BLYP is a hybrid DFT method that includes exact exchange term in exchange-correlation functional. However, the method is local (in LDA, external potential is the uniform electron gas and in GGA its gradients included) but vdW interaction requires fully non-local functionals. The basis set is as well too small and should included diffuse functions. I don't know what method to recommend.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The neon dimer is a very interesting system.  How does two atoms with full valency (8 electrons) bind?  This is where electron correlation comes into play.  It is due to the correlated motions of the electrons that allow these two atoms to form a bond, albeit a very, very weak bond.  This is known as non-covalent interactions or dispersion interactions.  For these types of systems, very expensive correlated methods are absolutely required to get it right.  DFT methods fail at properly describing dispersion forces though a lot of advancements have been made to build in a dispersion correction to the canonical functionals.  These corrections are usually denoted as -DX, where X is the generation of that correction (currently we are on -D3, or the 3rd generation of these types of corrections).  BLYP has no dispersion corrections built in and therefore will likely tell you that the neon dimer is UNBOUND (i.e. the binding energy is positive at all interatomic separations).  Unsurprising.  Therefore you will have to use a correlated method such as MP2.  But even MP2 can do a poor job of describing dispersion interactions (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.23522/abstract&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt;). In this example, MP2 overbinds the PCCP dimer by as much as 5 kcal mol$^{-1}$ with respect to CCSD(T). CCSD(T) is what we generally consider to be the correct answer.  Given that CCSD(T) binds the PCCP dimer by only 1-2 kcal mol$^{-1}$, you can see the problem here. The error of the MP2 method is about 5 times as large as the actual binding energy, and that is a major problem.  The basis set is also too small and you MUST include diffuse functions on both Neon atoms.  Without these diffuse functions, even correlated methods will have a hard time determining if this thing actually binds at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;The basis set is too small for this molecule and a correlated method (no diffuse or polarization functions are used). Furthermore, CCSD is very costly and not good for this big molecule. Maybe also conjugated pi bonds is of essence. Suggestion?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question deliberately states 'optimization'.  Therefore, nearly any commonly used density functional could be used (e.g. B3LYP).  If you want a slightly better result, go for some newer functionals such as the Minnesota functionals from Truhlar's group (e.g. M06-2X, M08-2X, etc.) or try some dispersion-corrected functionals.  If the question asked what you would use to determine the energetics, THEN move on to something like MP2 or CCSD(T).  I wouldn't trust the DFT energetics unless they have been extensively calibrated in previous work.  (Protip: never rely on DFT methods which haven't be calibrated for your type of system... in fact, just don't trust DFT at all :P).  The basis set is much too small.  A DFT optimization will go quick enough that a triple-$\zeta$ basis set should be used.  It goes without saying that you should include polarization functions on ALL atoms and diffuse functions on the non-hydrogen atoms (e.g. heavy-aug-cc-pVTZ, 6-311+G(2df,2pd), etc.).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;6&quot;&gt;&#xA;  &lt;li&gt;CISD is not size-extensive so as n increases the calculations become prohibitive.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be honest, I have a hard time coming up with an answer to this as the &quot;question&quot; is not very clear as to what the goal is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;7&quot;&gt;&#xA;  &lt;li&gt;MP3 is not good when we have an open-shell species or a system with electron congestion. Convergence will be very slow and not guaranteed at all. We assume in MPn that the zeroeth order wavefunction is reasonable approximation and the add the small perturbation contribution. But if HF describes this molecule very poorly then MPn won't improve it dramatically.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, MP3 is not guaranteed to converge at all which is pretty much the only reason (in my opinion) that you should need.  By the time you figure out the convergence/divergence/oscillation of the MPn series, you would have been better off sticking to something that is more expensive but guaranteed to converge.  Stick to CC methods.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;8&quot;&gt;&#xA;  &lt;li&gt;LSDA is equal to LDA for closed shell systems. Thus eternal potential is the uniformly distributed positive charge and can be used only if the &quot;constant&quot; electron density is different for each point in space. But here electron density due to bond twisting is not spatially uniform. I think GGA or B3LYP should be used. Furthermore I think the basis set is too small and should include polarization and diffusion (I don't know why, just saying).&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the fact that we've come a long way since LSDA, any use of LSDA for any type of production level work should just downright be avoided.  Here I would use GGAs or meta-GGAs.  They are much better.  Also, a larger basis set should be used. (polarization functions for all atoms, diffuse on non-hydrogen atoms, blah blah blah).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;9&quot;&gt;&#xA;  &lt;li&gt;pure LDA or GGA overestimate bond length so I suggest hybrid B3LYP with small basis set&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question deliberately asks how you would go about determining the activation ENERGY of this reaction. Therefore, do NOT rely on any DFT method when it comes to energetics UNLESS extensive calibration has been done.  Here's what I would do for this if I were going to turn this into a research paper.  First and foremost, optimize these guys with MP2/heavy-aug-cc-pVTZ.  They are small enough that this could be done.  Now, if I wanted to use DFT, I would pick a slew of DFT methods of various flavors including hybrids, dispersion-corrected, etc. and optimize each geometry as well.  Now, for each MP2 optimized geometry, I would do some high level single point energy computations using explicitly correlated methods such as MP2-F12 AND CCSD(T)-F12.  (The explicitly correlated methods accelerate the convergence of the correlation to the CBS limit over canonical implementations.  Essentially, MP2-F12/haTZ energetics can be on par with MP2/ha5Z energetics.  Same goes for CCSD(T).)  Given the size of these systems, I would (at the very least) bust out Dunning's correlation-consistent series of basis sets and push them as far as I could go.  From eyeballing the system, I think I could get away with doing heavy-aug-cc-pVDZ and heavy-aug-cc-pVTZ right off the bat.  I'm pretty sure I could do heavy-aug-cc-pVQZ as well.  Once I've done this, I have a darn good idea how close to the CBS limit my energetics are.  Now I can compare the DFT binding energies to the 'right answer' and boom, I've got DFT results I can rely on.  Of course, if the MP2 and CCSD(T) energetics show significant deviation, then we would have to revisit our optimization procedures but I don't think it would be an issue in this case.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;10&quot;&gt;&#xA;  &lt;li&gt;CCSD with medium basis set. Since vdW is repulsion of electron clouds at large distances as they interpenetrate and CCSD has traces of higher excitations. A small basis set can cause basis set superposition error (BSSE). Inclusion of higher order virtual orbitals as in CCSD improves dispersion description.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would not recommend CCSD.  You might as well do an MP2 calculation to be honest but for this type of system with a large dispersion component, you will want to do CCSD(T) energetics.  System is small enough to use a quadruple-$\zeta$ quality basis set.  Add diffuse and polarization functions.  Also, the frozen-core approximation should be tested.  You may want to correlate ALL the electrons to determine core-core and core-valence electron effects (use something like aug-cc-pCVQZ).&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="194" LastEditDate="2018-03-17T16:52:34.077" LastActivityDate="2018-03-17T16:52:34.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7953" PostTypeId="1" CreationDate="2014-01-21T23:00:25.967" Score="2" ViewCount="601" Body="&lt;p&gt;I am trying to verify the results of EPA method 26A (&lt;a href=&quot;http://www.epa.gov/ttnemc01/promgate/m-26a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.epa.gov/ttnemc01/promgate/m-26a.pdf&lt;/a&gt;). The Method claims to be able to separate and quantify a mixed halogen/halide gas stream. Although I believe the chemistry holds true for Cl₂/HCl and Br₂/HBr, I'm not so sure about F₂/HF. Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂. I am dubious that the F₂ results are real and am trying to develop a chemistry argument to support that position, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the Method takes a mixed gas stream (F₂/HF) and passes it through a 0.1 N H₂SO₄ solution, then a 0.1 NaOH solution; sodium thiosulfate is later added to the alkaline potion. The theory is that the HF is captured in the acid solution, while the F₂ passes through unreacted. In the alkaline solution, the F₂ reacts to form HF and the hypohalous acid (HOF, I assume). The thiosulfate converts the HOF to another HF. When the acid and alkaline solutions are analyzed for fluoride ion, the quantity is assumed to be the HF and F₂ in the original sample, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that F₂ is so reactive, I would think that any F₂ in the gas stream would react with the water in the acid solution and that theoretically no F₂ would make it to the alkaline solution. Is this a reasonable assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any F₂ did get to the alkaline solution, I am not sure that the conditions are favorable for the creation of HOF; F₂ reacting with water to produce HF seems more probable. Unless the acid solution is ignored by the F₂, I don't think that a true separation is taking place and any fluoride detected in the alkaline solution would represent blow through or some other error. Does this seem like a reasonable conclusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any input!&lt;/p&gt;&#xA;" OwnerUserId="4247" LastActivityDate="2014-07-28T08:13:18.263" Title="Can F2/HF be separated using acid/base solutions in series?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7955" PostTypeId="1" AcceptedAnswerId="8287" CreationDate="2014-01-22T03:15:27.817" Score="2" ViewCount="263" Body="&lt;p&gt;There are a number of things I don't understand about ionic equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, when you do net ionic equations, is it correct that you're only focusing on precipitates? The way I understood it, you put &quot;no reaction&quot; if there are no solid or gaseous products. To clarify, if all of the products of the reaction are aqueous, we look at it as if no reaction had taken place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I went and looked in my book, and that made it seem like all products that weren't covalent were spectator ions, i.e., covalent molecules were the only ones you note.  Now I'm even more confused.&lt;/p&gt;&#xA;" OwnerUserId="4249" LastEditorUserId="4945" LastEditDate="2018-09-27T12:19:32.463" LastActivityDate="2018-09-27T12:19:32.463" Title="Which ions are accounted for in total and net ionic equations?" Tags="&lt;ions&gt;&lt;stoichiometry&gt;&lt;terminology&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7959" PostTypeId="1" AcceptedAnswerId="7961" CreationDate="2014-01-22T09:30:26.310" Score="3" ViewCount="31142" Body="&lt;p&gt;Below is a copy of the method I am using to determine electron configurations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3mbWm.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the full and short electron configuration for tellurium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My final answer was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Full Configuration:&lt;/b&gt; 1s&lt;sup&gt;2&lt;/sup&gt; 2s&lt;sup&gt;2&lt;/sup&gt; 2p&lt;sup&gt;6&lt;/sup&gt; 3s&lt;sup&gt;2&lt;/sup&gt; 3p&lt;sup&gt;6&lt;/sup&gt; 4s&lt;sup&gt;2&lt;/sup&gt; 3d&lt;sup&gt;10&lt;/sup&gt; 4p&lt;sup&gt;6&lt;/sup&gt; 5s&lt;sup&gt;2&lt;/sup&gt; 4d&lt;sup&gt;10&lt;/sup&gt; 5p&lt;sup&gt;4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Short:&lt;/b&gt; $[\ce{Kr}]~\ce{5s^2 4d^10 5p^4}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, checking with &lt;a href=&quot;https://www.google.com/search?q=electron+configuration+of+tellurium&quot; rel=&quot;nofollow noreferrer&quot;&gt;google&lt;/a&gt;, it seems the answer is actually: $[\ce{Kr}]~\ce{4d^10 5s^2 5p^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the silly &quot;follow the yellow brick road&quot; thing faulty? Or is it that order doesn't matter in this situation?&lt;/p&gt;&#xA;" OwnerUserId="2356" LastEditorUserId="5026" LastEditDate="2018-02-25T03:24:26.177" LastActivityDate="2018-02-25T03:24:26.177" Title="Electronic configuration of tellurium" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7960" PostTypeId="2" ParentId="7839" CreationDate="2014-01-22T09:39:04.287" Score="2" Body="&lt;p&gt;As Ben Norris said, the aspects of nucleophilic and electrophilic attacks are not relevant in the second case.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about the compound given in question 1(b), the first step of nucleophilic attack is opposed by the conditions you mentioned, i.e., of the electron withdrawal effect of the double bond and replusion of nucleophile by $\pi$ electron cloud.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But talking in terms of yield of the product, it is necessary to consider two things:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Chlorine is a good leaving group.  &lt;/li&gt;&#xA;&lt;li&gt;The stability of the carbocation intermediate formed by $SN_1$ mechanism (steric factors don't favour $SN_2$). The carbocation formed will be very stable due to conjugation with the double bond. Hence, the yield of the product is not much low.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While solving the above question, we should look at the stability of carbocation intermediate formed,which is a more dominating effect. Though Bromine is a better leaving group, and there is decrease in tendency for nucleophilic attack due to $\alpha $-unsaturation; both factors which support higher nucleophilic attack for the first compound, stabilisation of carbocation by resonance (conjugation) is &lt;em&gt;highly&lt;/em&gt; stabilising, and this shadows the opposing factors.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorDisplayName="user10259" LastEditDate="2015-05-20T01:44:38.443" LastActivityDate="2015-05-20T01:44:38.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7961" PostTypeId="2" ParentId="7959" CreationDate="2014-01-22T09:51:25.133" Score="6" Body="&lt;p&gt;I haven't done that in decades (or so), but I'd bet that you did right! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result from google only lists the final outcome, not the order in which the orbitals are filled. Except for some examples $\ce{La}$, $\ce{Ac}$, etc., the &lt;em&gt;Aufbau principle&lt;/em&gt; is reliable. So: Follow the yellow brick road :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-22T09:51:25.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7962" PostTypeId="2" ParentId="7912" CreationDate="2014-01-22T12:20:30.697" Score="0" Body="&lt;p&gt;Your friend is right. As the temprature increases, the solubility of tea increases. Hence your tea will dissolve more easily when the water is heated before adding the tea powder, rather than your method.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="4589" LastEditDate="2014-02-21T15:57:14.873" LastActivityDate="2014-02-21T15:57:14.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7963" PostTypeId="2" ParentId="7838" CreationDate="2014-01-22T12:24:17.010" Score="0" Body="&lt;p&gt;I know a formula relating $K_a$,dissociation constant A and concentration C for weak acids and bases as $K_a$=$A^2$C.This may help you out.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2014-01-22T12:24:17.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7967" PostTypeId="1" CreationDate="2014-01-22T20:21:34.863" Score="3" ViewCount="2151" Body="&lt;p&gt;How does titanium occur in nature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it available as a native metal or only in the form of ores?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which precesses are typically used to convert the natural resources to the solid metal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on where to find further information are welcome. Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="4255" LastEditorUserId="4255" LastEditDate="2014-01-29T19:26:01.470" LastActivityDate="2018-05-18T16:45:12.180" Title="How is titanium metal obtained from natural resources?" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7970" PostTypeId="2" ParentId="5128" CreationDate="2014-01-22T22:13:28.543" Score="1" Body="&lt;p&gt;The reason it crystallized was probably because the temperature of the solvent decreased, lowering the solubility level. Also solubility of many solutes are not linearly proportional. Lots of these are solubility curves, so adding 66g to 300ml would not be the same as 33g to 100 ml. When displayed in a graph, the amount of a solute dissolved in a given amount of solvent at the give amount of temperatures, is shown as a quadratic equation. It appears as half a parabola. &lt;/p&gt;&#xA;" OwnerUserId="4257" LastEditorUserId="22" LastEditDate="2014-01-23T05:31:47.207" LastActivityDate="2014-01-23T05:31:47.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7971" PostTypeId="1" AcceptedAnswerId="54592" CreationDate="2014-01-22T22:14:08.493" Score="13" ViewCount="2778" Body="&lt;p&gt;Lysergic acid diethylamide (LSD) is a well known hallucinogen and (usually) illegal psychoactive drug with the structure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O2TLP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drug is active is doses measured in &lt;em&gt;micrograms&lt;/em&gt; in contrast to many other active compounds where the dose is milligrams or even grams. The effective dose of Mescaline, a related psychoactive, is something like 200mg (more than 1,000 times more than for LSD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is LSD so active in small doses? Do we know which features of the structure that contribute to its activity?&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2016-07-07T14:36:43.970" Title="What chemical properties make LSD so psychoactive?" Tags="&lt;organic-chemistry&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7972" PostTypeId="2" ParentId="7838" CreationDate="2014-01-22T22:40:16.837" Score="9" Body="&lt;p&gt;The answer is approximately 6.88. Therefore a $1\  M$ solution of glycerol in water will be ever so slightly acidic (considering that the hydroxyls are much weaker bases than acids, i.e. that the equilibrium constant for the reaction $\ce{C3H7O2OH + H2O &amp;lt;=&amp;gt; C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O &amp;lt;=&amp;gt; C3H7O2OH2^+ + OH^-}$, which may or may not be true. Inclusion of the second equilibrium will drive the true pH even closer to 7)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very weak or very dilute acids/bases, solving a dissociation problem is a bit more tricky because the self-dissociation equilibrium of water must be taken into account. It is often a valid approximation to forget about it, as water has only a small tendency to self-dissociate and so its effect on equilibrium concentrations of $\ce{H^+_{(aq)}}$ and $\ce{OH^{-}_{(aq)}}$ is often negligible, but this is not your case; glycerol in water seems to be about as weak an acid as water itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to solve equilibrium problems is to set up a large equation including all factors by combining smaller equations. We need to make an equation that takes into account all the species in the medium ($\ce{H^+_{(aq)}}$, $\ce{OH^{-}_{(aq)}}$, $\ce{C3H7O2O^{-}_{(aq)}}$ and $\ce{C3H7O2OH_{(aq)}}$ {for convenience I shall label the latter two $\ce{A^-}$ and $\ce{HA}$}) and the related equilibrium constants ($K_a$ and $k_w$). The smaller equations are found by balancing charges (one equation) and the amount of matter of each substance (one or more equations depending on the problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge balance: $[H^+]=[OH^-]+[A^-]$ (I)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter balance: $[HA]+[A^-]=C_{acid}=1\ M$ (II)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for the equilibrium constants are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-14.15}$ (III)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_w=[H^+][OH^-]=1\times 10^{-14}$ (IV)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem, we need to play with the equalities until we get a single equation with a single variable. It's convenient to obtain an equation in $[H^+]$ because we are able to find the pH directly by applying $-log$ to the answer. Notice that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[OH^-]=\frac{k_w}{[H^+]}$ (V)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[HA]=\frac{[H^+][A^-]}{K_a}$ (VI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting (VI) in (II) yields, after some algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[A^-]=\frac{C_{acid}K_a}{[H^+]+K_a}$ (VII)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now insert (V), (VI) and (VII) in (I) so that you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H^+]=\frac{k_w}{[H^+]}+\frac{C_{acid}K_a}{[H^+]+K_a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little bit of persistence will get you to the following polynomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H^+]^3+K_a[H^+]^2-(C_{acid}K_a+k_w)[H^+]-K_ak_w=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing the values of $K_a$, $k_w$ and $C_{acid}$, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E3%2b10%5E%28-14.15%29%2ax%5E2-%2810%5E%28-14.15%29%2b10%5E%28-14%29%29%2ax-10%5E%28-28.15%29=0&quot; rel=&quot;noreferrer&quot;&gt;the only positive root for the equation is $[H^+]=1.30688\times 10^{-7}$&lt;/a&gt;, which after applying the antilogarithm results in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=-log%281.30688x10%5E%28-7%29%29&amp;amp;a=%2aFunClash.log-_%2aLog10-&quot; rel=&quot;noreferrer&quot;&gt;$pH = 6.88$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-01-24T12:36:37.127" LastActivityDate="2014-01-24T12:36:37.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7973" PostTypeId="1" CreationDate="2014-01-22T22:43:50.820" Score="3" ViewCount="1293" Body="&lt;p&gt;Do natural or human engineered enzymes exist that can speed up the break down metals (element, compounds &amp;amp; alloys)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If no, then what biological or chemical construct can break down metals in the same way that enzymes break down food in a person's stomach?&lt;/p&gt;&#xA;" OwnerUserId="4151" LastEditorUserId="418" LastEditDate="2016-11-07T08:08:08.283" LastActivityDate="2016-11-07T08:08:08.283" Title="Are there natural or artificial enzymes that break down metals?" Tags="&lt;metal&gt;&lt;elements&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7974" PostTypeId="1" CreationDate="2014-01-22T22:53:40.957" Score="2" ViewCount="228" Body="&lt;p&gt;An electrode can be plated by having a solution full of electrolyte, using the the metal that will receive the plating as the cathode, and the metal doing the plating as the anode. Current is put through the cell and the anode dissolves oxidized metal ions into the solution, which gravitate over to the cathode, and then &quot;plate&quot; onto the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what prevents this process from happening when doing an electrolytic organic reduction or oxidation? Imagine you plate an electrode with a specific metal to be used as an anode in a different reaction, what prevents that anode from immediately corroding and having the metal ions fall off into the solution when current is applied? Does the presence of an organic species to be reduced/oxidized take priority over the anode corrosion, thus preventing it?&lt;/p&gt;&#xA;" OwnerUserId="4259" LastActivityDate="2014-01-29T02:36:42.183" Title="What prevents the anode from corroding in electrolytic organic reactions?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7975" PostTypeId="1" AcceptedAnswerId="37193" CreationDate="2014-01-22T23:33:35.740" Score="15" ViewCount="1144" Body="&lt;p&gt;Is there an algorithm for finding the number of cyclic isomers a given compound might have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a general algorithm for all isomers: &lt;a href=&quot;http://www-jmg.ch.cam.ac.uk/data/isomercount/isojava.html&quot;&gt;link&lt;/a&gt;, but I'm not sure how to apply it for cyclic isomers only.&lt;/p&gt;&#xA;" OwnerUserId="4260" LastEditorUserId="19616" LastEditDate="2015-10-02T16:04:43.363" LastActivityDate="2015-10-19T14:42:08.700" Title="Finding the number of cyclic isomers algorithmically" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7976" PostTypeId="2" ParentId="7971" CreationDate="2014-01-23T00:24:36.613" Score="3" Body="&lt;p&gt;Very interesting question. I always thought that most of the hallucinogens would somehow act as agonists for serotonine receptors in the frontal cortex and thus trigger visual &lt;em&gt;enhancements&lt;/em&gt;, but i've never been so much into bio- and neurochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I just found &lt;a href=&quot;http://dx.doi.org/10.1371/journal.pone.0009019&quot; rel=&quot;nofollow&quot;&gt;Psychedelics and the Human Receptorome&lt;/a&gt; at &lt;strong&gt;PLOS ONE&lt;/strong&gt;, which might be an interesting survey, but I'm too tired to read and comment on it.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-23T00:24:36.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7977" PostTypeId="1" AcceptedAnswerId="7979" CreationDate="2014-01-23T03:22:48.283" Score="0" ViewCount="2877" Body="&lt;p&gt;Consider two alcohols:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm{CH_3 - CH_2 - OH }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm{CH_3 - CHOH - H }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second alcohol, the hydroxyl radical instead of sticking out from the molecule, is bonded below the terminal carbon, like this&#xA;$$\rm{H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|$$&#xA;$$\rm{R- C- H}$$&#xA;$$|$$&#xA;$$\rm{OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this doesn't affect the nomenclature (as far as I know), both will be called $\rm{1-ethanol}$. Clearly, these are isomers. But IUPAC nomenclature is supposed to be completely unambiguous. Or did I go wrong somewhere and there actually &lt;em&gt;is&lt;/em&gt; a difference in their naming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Is there any way to produce structural formulas more efficiently in latex on chemistry stackexchange? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2014-01-23T03:44:46.397" Title="Does the position of the functional group on the terminal carbon change the nomenclature?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7978" PostTypeId="2" ParentId="7977" CreationDate="2014-01-23T03:33:31.523" Score="1" Body="&lt;p&gt;They are not isomers, they are the same molecule, ethanol (the &quot;1&quot; is superfluous, there is no 2-ethanol), CH3CH2OH&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-23T03:33:31.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7979" PostTypeId="2" ParentId="7977" CreationDate="2014-01-23T03:44:46.397" Score="1" Body="&lt;p&gt;Actually, both of these molecules are exactly the same. The way you write it on paper is just an approximate representation and does not represent the actual structure of the molecule.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By writing the hydroxyl group below the carbon or jutting out from it with a single bond makes absolutely no difference to the actual structure of ethanol.( Ethanol has no isomeric alcohol.) Therefore, it does not matter how you represent it on paper, it will be the same molecule.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in certain chiral compounds, by using some predetermined rules of representation, writing different groups on different sides does change the compound. But this is true only in cases where stereoisomeric forms of the compound exist. Then, writing the group in a different way changes the compound to one of its stereoisomer. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Z&lt;/em&gt; form&lt;/a&gt; is written with higher priority groups on one side and the opposite is followed while representing the &lt;em&gt;E&lt;/em&gt; form. &lt;a href=&quot;http://en.wikipedia.org/wiki/Fischer_projection&quot; rel=&quot;nofollow&quot;&gt;Fischer projection&lt;/a&gt;, &lt;a href=&quot;http://www.chemeddl.org/resources/stereochem/threed7.htm&quot; rel=&quot;nofollow&quot;&gt;Sawhorse projection&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Newman_projection&quot; rel=&quot;nofollow&quot;&gt;Newman projections&lt;/a&gt; to represent molecules are a few examples where the representation of molecules on paper is sensitive to the way in which different groups are attached to the compound, &lt;strong&gt;But not here&lt;/strong&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-01-23T03:44:46.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7980" PostTypeId="1" AcceptedAnswerId="7981" CreationDate="2014-01-23T04:23:35.520" Score="6" ViewCount="2021" Body="&lt;p&gt;I am asking this because I have a review question in my physics class which gives me a multiple-choice question.  I have heard and seen elsewhere on the internet (such as wikipedia) that chemistry is the study of reactions and the properties of matter.  That is usually a satisfactory answer, but for whatever reason this question makes me choose between reactions, properties, and matter.  Now I find that the general definition I accepted contains basically all of those, as far as I can tell.  While I don't necessarily need the answer, my curiosity has been aroused and I want to know why the question makes me choose between those answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what is the very basic idea behind chemistry?  Would it be classified as the study of reactions, properties, or matter (and you can only select one)?&lt;/p&gt;&#xA;" OwnerUserId="1046" LastActivityDate="2016-02-16T19:41:36.503" Title="What is chemistry considered the study of?" Tags="&lt;history-of-chemistry&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="2" ClosedDate="2016-02-17T00:43:13.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="7981" PostTypeId="2" ParentId="7980" CreationDate="2014-01-23T06:01:27.843" Score="3" Body="&lt;p&gt;Chemistry is the science of matter, especially its chemical reactions, but also its composition, structure, and properties. Chemistry is sometimes called &quot;the central science&quot; because it bridges physics with other natural sciences such as geology and biology.&lt;/p&gt;&#xA;" OwnerUserId="4262" LastEditorUserId="5577" LastEditDate="2015-11-18T21:34:57.907" LastActivityDate="2015-11-18T21:34:57.907" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7982" PostTypeId="1" CreationDate="2014-01-23T06:05:49.553" Score="5" ViewCount="1787" Body="&lt;p&gt;What are the products when tert-butyl ethyl ether is cleaved with concentrated $\ce{HI}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is iodoethane and tert-butanol, but the correct answer is ethanol and 2-iodo-2-methylpropane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the ethyl side goes by an SN2 mechanism because it is a primary R group, and the t-butyl group would be SN1 because it is tertiary. However, why would the SN2 reaction not happen first?&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="4231" LastEditDate="2015-03-09T17:42:55.833" LastActivityDate="2018-01-01T12:17:39.273" Title="In cleaving ethers with both SN1 and SN2 mechanisms possible, which occurs first?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7985" PostTypeId="1" AcceptedAnswerId="8030" CreationDate="2014-01-23T06:25:41.043" Score="9" ViewCount="74555" Body="&lt;p&gt;Was looking at Henry's law and Raoult's law&#xA;constants and there seemto be lots of&#xA;equations involved.&#xA;Henry's law involves partial pressure and the latter involves the vapor pressure.&#xA;Wondering what the difference is?&lt;/p&gt;&#xA;" OwnerUserId="4080" LastActivityDate="2017-11-19T22:37:11.843" Title="What is the differences between partial pressure and vapour pressure?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7990" PostTypeId="1" AcceptedAnswerId="7997" CreationDate="2014-01-23T08:47:47.293" Score="2" ViewCount="24427" Body="&lt;p&gt;I want to know how the following sigma bond strength values are calculated.&#xA;Like S-S = 1.0, S-P=1.71 And P-P=3.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How these 1.0, 1.71, 3.0 came?&lt;/p&gt;&#xA;" OwnerUserId="4263" LastActivityDate="2014-01-23T17:34:46.290" Title="Sigma bond strengths S-S, S-P, P-P" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2014-01-23T19:10:01.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="7991" PostTypeId="1" CreationDate="2014-01-23T09:28:38.390" Score="3" ViewCount="452" Body="&lt;p&gt;From a chemistry point of view, if dental amalgam is a compound of mercury and other metals, and compounds are elements which are chemically bonded, how would it be possible for mercury to &quot;leak out&quot; of fillings if no further chemical reaction would be taking place?&lt;/p&gt;&#xA;" OwnerUserId="4264" LastEditorUserId="81" LastEditDate="2014-01-23T20:32:18.547" LastActivityDate="2016-10-21T14:39:12.840" Title="Why do some dentists claim mercury &quot;leaks out&quot; of amalgam fillings?" Tags="&lt;bond&gt;&lt;safety&gt;&lt;alloy&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7993" PostTypeId="2" ParentId="7980" CreationDate="2014-01-23T13:27:22.837" Score="0" Body="&lt;p&gt;Chemistry is the study of chemical systems (i.e. matter).&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-01-23T13:27:22.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7994" PostTypeId="2" ParentId="7991" CreationDate="2014-01-23T15:05:41.837" Score="5" Body="&lt;p&gt;'Leak' is an imprecise term here. Amalgam itself could (and probably does) wear away with attrition, and the 'powder' would be ingested. Certainly, fillings have worked loose and been swallowed. The effect of stomach acid on amalgam is probably not negligible.&lt;/p&gt;&#xA;" OwnerUserId="4266" LastActivityDate="2014-01-23T15:05:41.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7995" PostTypeId="2" ParentId="7982" CreationDate="2014-01-23T15:33:56.873" Score="6" Body="&lt;p&gt;One &quot;side&quot; of a molecule doesn't react by one mechanism and the other &quot;side&quot; by another mechanism. Generally speaking, a molecule reacts by a single mechanism.  Now, the mechanism can change depending upon the other reactants that are present as well as the reaction conditions.  In the case of &lt;em&gt;t&lt;/em&gt;-butyl ethyl ether, if the reaction were carried out with $\ce{KI}$ under non-acidic conditions an $\mathrm{S_N2}$ reaction would occur and your answer (iodoethane and tert-butanol) would be correct.  On the other hand, if the reaction were carried out under acidic conditions then an $\mathrm{S_N1}$ reaction would occur.  As you pointed out a tertiary carbocation is more stable than a primary carbocation so the expected products in this case would be ethanol and 2-iodo-2-methylpropane.  Apparently, the reaction conditions in your reaction must have been such that the reaction occurred by the $\mathrm{S_N1}$ mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7475" LastEditDate="2018-01-01T12:14:25.980" LastActivityDate="2018-01-01T12:14:25.980" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7996" PostTypeId="2" ParentId="7991" CreationDate="2014-01-23T17:22:14.940" Score="3" Body="&lt;p&gt;Mercury vapors are released. It has long since been proven that people with mercury in their teeth constantly have small portions of mercury vapor evaporating from the tooth. There's many videos of this taking place. It's not a mystery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=hqIdGwAMxxs&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=hqIdGwAMxxs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4268" LastActivityDate="2014-01-23T17:22:14.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7997" PostTypeId="2" ParentId="7990" CreationDate="2014-01-23T17:34:46.290" Score="5" Body="&lt;p&gt;Just to clarify, these values are relative bond strengths, not absolute bond strength values, which would have units of energy, i.e. Joules, kilocalories, or electron volts. Another detail missed in this question is the fact that these values refer to the bond strength of these types of orbitals directly overlapping along the same axis, as shown in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYFD4.gif&quot; alt=&quot;See here for a picture reference&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that these relative bond strengths corresponds the amount of overlap occurring in the orbitals. This gives you a qualitative description, but to actually calculate the relative overlaps we can use quantum mechanics. Remember that these orbitals are really just the probability distributions electrons and can be solved analytically by squaring the wave function. We have a mathematical description of both S and P orbitals. By integrating over the volume in which they overlap, we can calculate the relative overlap between S-S, S-P, P-P, and any other orbital overlaps we might be interested in. &lt;/p&gt;&#xA;" OwnerUserId="4267" LastActivityDate="2014-01-23T17:34:46.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7998" PostTypeId="1" AcceptedAnswerId="8000" CreationDate="2014-01-23T18:06:49.177" Score="0" ViewCount="2755" Body="&lt;p&gt;I'm looking at purchase orders and I see purchases for chemicals like the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glutaraldehyde, EM Grade, 25%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the &quot;EM Grade&quot; mean in this text? It says that it's recommended for histological or immunological techniques, but how?&lt;/p&gt;&#xA;" OwnerUserId="4171" LastEditorUserId="22" LastEditDate="2014-01-23T19:55:36.053" LastActivityDate="2014-01-23T19:55:36.053" Title="What does &quot;EM grade&quot; mean" Tags="&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7999" PostTypeId="2" ParentId="7998" CreationDate="2014-01-23T19:40:10.400" Score="3" Body="&lt;p&gt;EM grade is a label for chemicals used in electron microscopy.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-23T19:40:10.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8000" PostTypeId="2" ParentId="7998" CreationDate="2014-01-23T19:48:30.557" Score="3" Body="&lt;p&gt;It is highly purified, typically distilled into glass ampoules and sealed under an inert atmosphere.  Suitable for use in *&lt;em&gt;E&lt;/em&gt;*lectron *&lt;em&gt;M&lt;/em&gt;*icroscopy.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-23T19:48:30.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8001" PostTypeId="2" ParentId="7991" CreationDate="2014-01-23T19:57:44.853" Score="12" Body="&lt;p&gt;You are labouring under a misconception about the exact meaning of the word &quot;compound&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally when we describe, for example, ethanol as a compound of hydrogen, oxygen and carbon we don't mean that the component elements can &quot;leak&quot; out of it without a chemical reaction taking place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury amalgams are casually described as &quot;compounds&quot; but not in the same sense as ethanol. &lt;a href=&quot;http://en.wikipedia.org/wiki/Amalgam_%28chemistry%29&quot;&gt;Amalgams&lt;/a&gt; are mercury alloys with varying compositions but they don't, usually, consist of strongly bonded molecules like ethanol with strong chemical bonds preventing the molecules falling apart. They are closer to &lt;em&gt;mixtures&lt;/em&gt; than what we normally think of as compounds. Having said that they are not just mixtures as they can have very different bulk properties than their components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with mercury amalgams used in fillings is that they are mostly, but not completely, inert. Some mercury &lt;em&gt;will&lt;/em&gt; evaporate from them and some chemical and physical processes can increase the amount that evaporates. But good estimates suggest only micrograms/day which is less than typical workplace exposure limits so probably not a source of legitimate worry. The dentists who still use mercury fillings are likely exposed to more than this and this is a better reason to limit their use. The subject is controversial with a lot of disagreement among studies. This is well summarised by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dental_amalgam_controversy&quot;&gt;Wikipedia article&lt;/a&gt; on the controversy.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2014-12-24T16:35:06.233" LastActivityDate="2014-12-24T16:35:06.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8004" PostTypeId="2" ParentId="7980" CreationDate="2014-01-24T01:48:15.583" Score="4" Body="&lt;p&gt;While the answers so far are not wrong, I'd like to add a refinement. Chemistry is the study of matter at the level of &lt;em&gt;electron interaction&lt;/em&gt;. The basic question in chemistry is about &lt;em&gt;chemical bonding&lt;/em&gt;, and as far as we know this is a function of electron interaction. In contrast, radioactivity is also about matter, but is not at all about electron interaction and is not generally considered chemistry (it's considered a part of physics). Also, the chemistry of isotopes (i.e. differing neutron count) is not generally different across different isotopes. This is good, because we need solvents with are deuterium based ($\ce{D=^2H}$) as opposed to hydrogen based for NMR spectroscopy. If we go lower than electrons, we are into quantum mechanics, which is again physics. So electron interaction is a good level at which to characterize chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I realize the original question was for a one-word answer. Frankly, that's not the greatest question, but it's generating answers that make it more interesting, and it's why I answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give a counterexample to my own answer to show the difficulty: &lt;em&gt;gases&lt;/em&gt;. We learn a lot about gases in chemistry which has nothing to do with chemical bonding; e.g. all the gas equations. I'd argue that most of this is &lt;em&gt;not&lt;/em&gt; really chemistry, it's physics. Understanding how temperature, pressure, and volume interact dynamically is a very physical sort of study, not all that different from studying light, or magnetism, or force.  So why do we study gas physics in chemistry? I suspect it's because we &lt;em&gt;use gas physics a lot in chemistry&lt;/em&gt;! You need to understand the gas laws to do any real experimentation involving actual gas chemistry; i.e. how would you measure gas as a reactant or product without knowing the gas laws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to be clear, if you want one word, I agree with &quot;matter&quot;. But I think if you want to get at what chemistry is about, it's not so simple.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2014-01-24T13:14:21.120" LastActivityDate="2014-01-24T13:14:21.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8008" PostTypeId="2" ParentId="7973" CreationDate="2014-01-24T08:12:43.303" Score="6" Body="&lt;h2&gt;Silly answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is no enzyme that grinds down an Airstream, but an alligator might do the job!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Serious (but disappointing) answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In nature, most metals seem to exist as oxides, sulfides, carbonates, halides, etc. and not as bare metals. Consequently, there was/is probably little evolutionary benefit in tackling the bare metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking down something to into very small particles, as you described it, is a mechanical process that leaves the oxidation state of the metal unchanged. It is not a chemical transformation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Nothing cool with enzymes an metals?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There's a lot! Two topics come to my mind:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Chemosynthesis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There's a bunch of bacteria that can oxidize $\ce{Fe(II)}$ to $\ce{Fe(III)}$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at different pH levels&lt;/li&gt;&#xA;&lt;li&gt;in the presence and absence of oxygen&lt;/li&gt;&#xA;&lt;li&gt;in the presence and absence of light&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Acidithiobacillus ferrooxidans&lt;/em&gt; and &lt;em&gt;Leptospirillum ferrooxidans&lt;/em&gt; - look at their family names ;-) - operate at $pH\ &amp;lt; 4$, use the oxygen from the air and munch away &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrite&quot;&gt;pyrite&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Metal hyperaccumulation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Some plants and bacteria harvest enormous amounts of metal ions even from highly contaminates soils and store them. The metals are held by by the action of proteins. If you're interested, here is a freely available article on &lt;a href=&quot;http://www.plospathogens.org/article/info%3Adoi%2F10.1371%2Fjournal.ppat.1001093&quot;&gt;PLOS One Pathogens&lt;/a&gt; with lots of further references.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-24T08:12:43.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8017" PostTypeId="2" ParentId="6390" CreationDate="2014-01-24T11:22:53.293" Score="11" Body="&lt;p&gt;Rather than working on your technique, check whether other test tubes are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, a glass-like material with a &lt;strong&gt;very low thermal expansion coefficent&lt;/strong&gt; would be a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.udel.edu/chem/GlassShop/PhysicalProperties.htm&quot;&gt;glassblowing shop at the University of Delaware&lt;/a&gt; provides a nice overview of the possible materials and their properties, and there are definitely additional online resources, e.g. by manufacturers like Schott, Corning, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a summary (coefficients are given in $cm/cm/°C$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;soda lime glass $89 \times 10^{-7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pyrex, Duran $32.5 \times 10^{-7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vycor  $7.5 \times 10^{-7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fused quarz $5.5 \times 10^{-7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that quarz is both pretty expensive and demands special care! Alkaline solutions and quarz are a bad conbination!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-24T11:22:53.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8019" PostTypeId="1" AcceptedAnswerId="8021" CreationDate="2014-01-24T12:53:14.160" Score="2" ViewCount="5302" Body="&lt;p&gt;What is the general thermal conductivity of animal skin leather?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand there are many different animal's skins (cow, sheep, buffalo, deer, pig), but in general, does natural leather insulate well (compared to, for scale, an esky for food insulation), or not?&lt;/p&gt;&#xA;" OwnerDisplayName="user1469" LastEditorDisplayName="user1469" LastEditDate="2014-01-24T23:01:50.400" LastActivityDate="2014-01-24T23:01:50.400" Title="How well (or poorly) does natural leather insulate?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;conductivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8020" PostTypeId="2" ParentId="8019" CreationDate="2014-01-24T13:25:49.750" Score="1" Body="&lt;p&gt;I never worked in that field and thus had to google it myself ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the general observation that leather is indeed a good heat insulator, there seems to be an article in the &lt;a href=&quot;http://www.eei.upc.es/3web/catA3/journal/repo/2012/V63N3/eng/02.pdf&quot; rel=&quot;nofollow&quot;&gt;Journal of the AQEIC&lt;/a&gt; (Asociación Química Española de la Industria del Cuero) addressing the question quantitatively.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-24T14:53:21.777" LastActivityDate="2014-01-24T14:53:21.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8021" PostTypeId="2" ParentId="8019" CreationDate="2014-01-24T13:37:37.517" Score="6" Body="&lt;p&gt;Interestingley enough, there are &lt;a href=&quot;http://www.nature.com/jid/journal/v69/n3/abs/5616683a.html&quot;&gt;papers&lt;/a&gt; entirely devoted to the subject of measuring the properties of living skin. The paper linked gives a value of $7 \times 10^{-4}~\mathrm{\frac{cal}{cm\, s\, °C}}$ for human skin, or in everyday units used nowadays: $0.29 ~\mathrm{\frac{W}{m\, K}}$. This is only for the dermis, the part of skin that then &lt;a href=&quot;http://www.leatherresource.com/whatisleather.html&quot;&gt;gets transformed into leather&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;a href=&quot;http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html&quot;&gt;this&lt;/a&gt; website has a comprehensive list of many different thermal conductivity values, and dry leather clocks in at $0.14 ~\mathrm{\frac{W}{m\, K}}$, which is about 2 times less the conductivity than that of living skin. What a difference the tanning process can make!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the similarity  observed in the review paper between man, beef and porcine tissue, I think it is fair to say that the value for leather will not change greatly across species. &lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-01-24T13:48:37.637" LastActivityDate="2014-01-24T13:48:37.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8025" PostTypeId="2" ParentId="7304" CreationDate="2014-01-24T16:00:51.603" Score="5" Body="&lt;p&gt;There is a slight difference between solvated ions and aqua complexes: The nature of the bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When solvating a sodium ion, the electron pairs of the oxygen in water never really interact with the sodium in a way that creates a bond of some sort: It's all just electrostatic interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when a transition metal ion gets surrounded with these water molecules, the lone pair electrons of the oxygen form a bond with the ion. This is because they interact with the $d$ orbitals of the transition metal (which usually also contains some electrons), thus changing the energetic state of the ion itself. First and second group elements don't have that energy-stabilizing effect because they don't have electrons in the orbitals that get split up by the formation of the complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unimaginable that there then exist further electrostatic interactions between the transition metal aqua complex and the surrounding water molecules, thus leading to the solvation of said complex.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-24T16:00:51.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8026" PostTypeId="2" ParentId="7638" CreationDate="2014-01-24T16:52:08.910" Score="2" Body="&lt;p&gt;You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation:&#xA;$$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the equation a little bit, we get that:&#xA;$$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can integrate each side of this equation: &#xA; $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$&#xA;Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to :&#xA;$$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$&#xA;Solving the integral we will get this :&#xA;$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\Delta H^\circ &amp;gt; 0$ we have two cases: when $T_2&amp;gt;T_1$ and $T_2&amp;lt;T_1$.&lt;br&gt;&#xA;If $T_2&amp;gt;T_1$ then $\frac{1}{T_2}&amp;lt;\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} &amp;lt; 0$ . Knowing that $\Delta H^\circ&amp;gt;0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive:&#xA; $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)&amp;gt;0$$&#xA;And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)&amp;gt;0 \Leftrightarrow \frac{K_2}{K_1} &amp;gt; 1 \Leftrightarrow K_2 &amp;gt; K_1$$&#xA;i.e. The equilibrium constant increases as the temperature is increased.&lt;br&gt;&#xA;Using the same principle for $T_2 &amp;lt; T_1$, we get that $K_2 &amp;lt; K_1$&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T09:41:46.640" LastActivityDate="2015-04-26T09:41:46.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8030" PostTypeId="2" ParentId="7985" CreationDate="2014-01-24T21:31:09.363" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does this mean? For example, if we have a mixture of gases $A$, $B$ and $C$ in an isolated room, then, according to &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Dalton%27s_law/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dalton's law&lt;/a&gt;, the pressure exerted by the gases will be the sum of their &lt;strong&gt;partial pressures&lt;/strong&gt; :&#xA;$$P = p_A + p_B + p_C$$&#xA;where $p_A$, $p_B$ and $p_C$ are the partial pressures of each gas. Also, if we have a moles of $A$, b moles of $B$ and c moles of $C$, we can express the partial pressure of each gas as below:&#xA;$$p_A = \frac{a}{a+b+c} P$$&#xA;$$p_B = \frac{b}{a+b+c} P$$&#xA;$$p_B = \frac{c}{a+b+c} P$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the definition states, if we took an equilibrium as $\ce{A(l) &amp;lt;=&amp;gt; A(g)}$, where $A(l)$ and $A(g)$ are the liquid and the vapor state of the compound $A$ . The &lt;strong&gt;vapor pressure&lt;/strong&gt; of $A$ is the pressure exerted by the gaseous state ($A(g)$) over the liquid state ($A(l)$). &#xA;Now, if we have a mixture of $A$, $B$ and $C$, we will have the equilibrium between the liquid and gaseous state : $\ce{(A, B, C)(l) &amp;lt;=&amp;gt; (A, B, C)(g)}$ and the vapor pressure $P_v$ (I noted $P_v$ so there won't be any confusion between it and the total pressure $P$ I mentioned before). &#xA;According to &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Raoult%27s_law/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raoult's law&lt;/a&gt;, the vapor pressure $P_v$ is the sum of the products between the vapor pressures of each compound taken separately and the molar fraction of the compound in the mixture (i.e. in the solution) :&#xA;$$P_v = x_A p_{v^A} + x_B p_{v^B} + x_C p_{v^C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may look just like Dalton's law, but there are &lt;strong&gt;different&lt;/strong&gt; pressures involved. &#xA;The &lt;strong&gt;partial pressure&lt;/strong&gt; of a gas is the pressure exerted by a gas in the volume occupied by a mixture of &lt;strong&gt;gases&lt;/strong&gt;, while the &lt;strong&gt;vapor pressure&lt;/strong&gt; of a gas is the pressure exerted by a gas &lt;strong&gt;over its condensed phase&lt;/strong&gt;.&#xA;Although :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The vapor pressure still refers to the pressure of a gas over its liquid (or solid) phase in a phase transition (condensed phase$\ce{&amp;lt;=&amp;gt;}$non-condensed phase), while the notion of partial pressure can be extended to a mixture of gases &lt;em&gt;only&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the definitions are linked to Wikipedia, and they can really cause confusion throughout chemistry learners..&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="2470" LastEditDate="2017-11-19T22:37:11.843" LastActivityDate="2017-11-19T22:37:11.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8031" PostTypeId="1" CreationDate="2014-01-24T23:34:35.157" Score="3" ViewCount="1061" Body="&lt;p&gt;If you titrate some hydrogen bromide solution (0.1 M) to some sodium hydroxide in order to determine the exact concentration of hydrogen bromide (0.1 M), and you use anthocyanins as a pH indicator. This anthocyanin will be red if you have a low pH value and purple if it is slightly acidic until it is slightly basic. It will be blue if the solution is basic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have to titrate until the color of the solution is blue or purple?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium from red to purple: $$\ce{H2X+ &amp;lt;=&amp;gt; HX}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = 1.047*10^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium from purple to blue: $$\ce{HX &amp;lt;=&amp;gt;X^{−}}$$&lt;br&gt;&#xA;$K_a = 3.16 * 10^{-8}$&lt;/p&gt;&#xA;" OwnerUserId="4286" LastEditorUserId="4297" LastEditDate="2014-01-26T04:05:25.500" LastActivityDate="2014-03-22T11:51:00.253" Title="Anthocyanins as pH-indicators: stop the titration at red, purple or blue?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8032" PostTypeId="2" ParentId="7980" CreationDate="2014-01-25T00:17:53.190" Score="1" Body="&lt;p&gt;This is a little review of the online definition I've found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.oxforddictionaries.com/definition/english/chemistry?q=chemistry&quot; rel=&quot;nofollow&quot;&gt;Oxford Dictionaries&lt;/a&gt; definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The branch of science concerned with the substances of which matter is&#xA;  composed, the investigation of their properties and reactions, and the&#xA;  use of such reactions to form new substances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dictionary.cambridge.org/dictionary/british/chemistry_1?q=chemistry&quot; rel=&quot;nofollow&quot;&gt;Cembridge Dictionaries&lt;/a&gt; definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(the part of science which studies) the basic characteristics of&#xA;  substances and the different ways in which they react or combine with&#xA;  other substances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I like the &lt;a href=&quot;http://www.treccani.it/vocabolario/chimica/&quot; rel=&quot;nofollow&quot;&gt;Treccani&lt;/a&gt;, that here I've badly translate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The science concerned with the properties, composition, identification, preparation and the different ways in which the natural and&#xA;  artificial substances react.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;More mystic the &lt;a href=&quot;http://en.wiktionary.org/wiki/chemistry&quot; rel=&quot;nofollow&quot;&gt;Wikitionary&lt;/a&gt; definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The branch of natural science that deals with the composition and&#xA;  constitution of substances and the changes that they undergo as a&#xA;  consequence of alterations in the constitution of their molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point I find fundamental are that Chemistry concerned with &lt;strong&gt;matter&lt;/strong&gt; and &lt;strong&gt;compounds&lt;/strong&gt; and historically studied the methods to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;investigate the forces between them and their properties&lt;/li&gt;&#xA;&lt;li&gt;separate and isolate them&lt;/li&gt;&#xA;&lt;li&gt;analyze, identify and quantify them (and this is why I like Treccani definition because include analytical chemistry)&lt;/li&gt;&#xA;&lt;li&gt;understand and predict the ways in which they combine and react&lt;/li&gt;&#xA;&lt;li&gt;synthesize and produce new materials&lt;/li&gt;&#xA;&lt;li&gt;find the  way to produce the new material industrially &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-01-25T00:17:53.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8033" PostTypeId="1" AcceptedAnswerId="8045" CreationDate="2014-01-25T01:28:41.823" Score="4" ViewCount="1639" Body="&lt;p&gt;Quite often (from my own experience anyway), one of the safety rules in a school (high school or university) laboratory is that the students must wear leather shoes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to demonstrate some of the safety rules, so my question is, &lt;strong&gt;what is an effective and safe experiment to demonstrate the safety of wearing leather shoes while experimenting in the chemical lab?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;This is in comparison to say sneakers or canvas shoes&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="418" LastEditDate="2015-01-08T08:17:21.530" LastActivityDate="2015-01-08T08:17:21.530" Title="How to demonstrate the safety of wearing leather shoes in the chemistry lab?" Tags="&lt;safety&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8039" PostTypeId="2" ParentId="4517" CreationDate="2014-01-25T07:57:02.773" Score="31" Body="&lt;h2&gt;Mathematical Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:&#xA;$$\begin{align}&#xA;E_+ &amp;amp;= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\&#xA;E_- &amp;amp;= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the &lt;em&gt;resonance integral&lt;/em&gt;, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' &amp;gt; k'$. You'll note that $j'-k' &amp;gt; 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Intuitive Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the &lt;strong&gt;energies of the molecular orbitals get shifted upwards&lt;/strong&gt;, which is what creates the asymmetric image that we have for these energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Stick some electrons between them.&lt;/li&gt;&#xA;&lt;li&gt;Don't stick some electrons between them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Further Information&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I highly recommend the following book, from which most of the information above stems: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;P. Atkins and R. Friedman: &lt;em&gt;Molecular Quantum Mechanics&lt;/em&gt;, $5^\text{th}$ ed. Oxford University Press, 2011.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-25T07:57:02.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8041" PostTypeId="2" ParentId="8031" CreationDate="2014-01-25T11:13:24.173" Score="1" Body="&lt;p&gt;First of all, I don't understand why you would titrate the acid of unknown concentration to your base of known concentration, I've learned it the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, following my proposal, we have our solution of $\ce{HBr}$ waiting to be titrated with $\ce{NaOH}$. Since sodium hydroxide is a fairly strong base, we can assume total neutralization:&#xA;$$ \ce{HBr + NaOH -&amp;gt; NaBr + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Operating under this assumption, equilibrium is reached at $\text{pH}=7$, so we have to know when that happens. How do we find out where the indicator changes colour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we calculate the $\mathrm{pK_a}$ values of the indicator:&#xA;$$ \mathrm{pK_a} = -\log \mathrm{K_a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equilibrium (from purple to blue) switches at $\text{pH} = 7.5$, which is about where we need it to. So you need to titrate until you get the blue colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; It is not always easy to see these colour transitions, I myself have overtitrated quite a few solutions because the colour shift was too subtle (and my eyes too untrained).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-01-25T11:13:24.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8042" PostTypeId="2" ParentId="8031" CreationDate="2014-01-25T11:28:34.000" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you titrate some hydrogen bromide solution to some sodium hydroxide [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's probably a bad idea to start with the anthocyane in strongly alkaline solution. Like other pyrylium compounds, flavylium cations do not tolerate (strongly) akaline conditions very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The nucleophilic opening of the pyrylium ring to a chalcone is irreversible under the conditions of the titration.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-25T11:28:34.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8043" PostTypeId="1" CreationDate="2014-01-25T13:29:55.420" Score="1" ViewCount="609" Body="&lt;p&gt;Ethers can be cleaved under acidic conditions by treatment with hydrogen halides (HX, where X = Cl/Br/I). Some of these reactions proceed by S&lt;sub&gt;N&lt;/sub&gt;1 mechanisms and some by S&lt;sub&gt;N&lt;/sub&gt;2 mechanisms. How can I tell which mechanism is being followed?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="16683" LastEditDate="2018-09-12T16:13:06.120" LastActivityDate="2018-09-12T16:18:13.880" Title="Which mechanism do ethers follow when cleaved by hydrogen halides?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nucleophilic-substitution&gt;&lt;ethers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8044" PostTypeId="2" ParentId="8043" CreationDate="2014-01-25T14:11:26.277" Score="6" Body="&lt;p&gt;The first step of this reaction is always a proton transfer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ucequ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ucequ.png&quot; alt=&quot;Protonation of ether by HX&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the substitution follows either the S&lt;sub&gt;N&lt;/sub&gt;2 or S&lt;sub&gt;N&lt;/sub&gt;1 pathway depending on whether R is methyl, primary, secondary, or tertiary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methyl and primary ethers&lt;/strong&gt; undergo acidic cleavage by the S&lt;sub&gt;N&lt;/sub&gt;2 mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9GXZh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9GXZh.png&quot; alt=&quot;SN2 process for methyl or primary ethers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tertiary ethers&lt;/strong&gt; undergo cleavage by the S&lt;sub&gt;N&lt;/sub&gt;1 mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DwZez.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DwZez.png&quot; alt=&quot;SN1 process for tertiary ethers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Secondary ethers&lt;/strong&gt; probably undergo cleavage by both mechanisms competing with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that unless R' is an aryl group, the alcohol R'OH that forms will also be converted into a halide by a similar mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more at &lt;a href=&quot;https://www.google.com/search?q=ether+nucleophilic+substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;any of the sites in this Google search&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2018-09-12T16:18:13.880" LastActivityDate="2018-09-12T16:18:13.880" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8045" PostTypeId="2" ParentId="8033" CreationDate="2014-01-25T14:21:57.697" Score="7" Body="&lt;p&gt;The simplest thing to do is to sacrifice two pairs of shoes, one with leather uppers and the other with canvas/cloth/whatever. Put something inside the shoes to represent &quot;feet&quot;. Do something nasty to each pair, like pour concentrated acid over them (using secondary containment to avoid spills and splashes!). Then, extract your indicator and compare (assuming the different in damage to the shoes was not enough). You can also try lighting them on fire (though canvas shoes probably have fire-retardant compounds added), or seeing how long it takes water to soak through (leather will win here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And, whatever you do, record it to video so that next time you only need to play the video and not buy more shoes.&lt;/strong&gt; If this is for your students, I would make sure you appear in the video to increase &quot;authenticity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-25T14:21:57.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8046" PostTypeId="1" CreationDate="2014-01-25T14:41:28.570" Score="7" ViewCount="7063" Body="&lt;p&gt;I noticed that distilled water works well to clean away fingerprints on touchscreens or oil stains on a ceramic field after frying an egg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But water and oil/fat are not supposed to mix, so why does that even work?&lt;/p&gt;&#xA;" OwnerUserId="4292" LastEditorUserId="34379" LastEditDate="2016-09-03T00:14:33.190" LastActivityDate="2016-09-03T00:14:33.190" Title="Distilled water to clean greasy stains: Why does that work?" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8047" PostTypeId="2" ParentId="7967" CreationDate="2014-01-25T15:09:21.233" Score="5" Body="&lt;p&gt;Titanium is found as an ore, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Titanium#Occurrence&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Titanium Occurrences Wikipedia page&lt;/a&gt;, specifically, the 2 most economical ores are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ores of titanium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mindat.org/min-3486.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Rutile&lt;/strong&gt;&lt;/a&gt; $\ce{TiO2}$ and &lt;a href=&quot;http://www.mindat.org/min-2013.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Ilmenite&lt;/strong&gt;&lt;/a&gt; $\ce{Fe^{2+}TiO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extraction of titanium metal from ore&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the chemguide page &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/extraction/titanium.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titanium&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Titanium can't be extracted by reducing the ore using carbon as a cheap reducing agent. The problem is that titanium forms a carbide, TiC, if it is heated with carbon, so you don't get the pure metal that you need. The presence of the carbide makes the metal very brittle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather, the ore is chlorinated, then reduced as shown in the schematic from the webpage &lt;a href=&quot;http://www.titaniumexposed.com/titanium-industries.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titanium Exposed&lt;/a&gt; (the chemical processes are included):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aBc6w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemguide states that either $Mg$ or $Na$ can be used in the reduction stage, which adds to the cost of production as they need to be extracted themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asides from the links used in this answer, another good resource is &lt;a href=&quot;http://wwwchem.uwimona.edu.jm/courses/titanium.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry of Titanium&lt;/a&gt; (Professor John Lancashire - University of the West Indies)&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-01-25T15:09:21.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8048" PostTypeId="1" CreationDate="2014-01-25T15:13:16.960" Score="12" ViewCount="61554" Body="&lt;p&gt;I learned in class that the equivalence point in an acid-base titration is reached when the solution contains an equal amount of substance of $\ce{OH-}$ and $\ce{H+}$ ions. However, in a weak acid and strong alkali titration, the pH at the equivalence point (when I added the right amount of acid and alkali to make it a neutral solution) is 9. Also, in the titration of a strong acid and weak alkali, the pH of the solution when I have added the correct amounts is 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I add 1 mole of ethanoic acid to 1 mole of sodium hydroxide, I should get a neutral solution and the pH should be 7. However, why is the pH 9? This does not make sense to me as a pH 9 suggest that the concentration of $\ce{OH-}$ is higher than the concentration of $\ce{H+}$. But how is that, I reacted the correct amounts?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2018-01-29T15:32:06.773" LastActivityDate="2018-01-30T14:18:40.743" Title="Shouldn't the pH at the equivalence point always be 7?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8049" PostTypeId="2" ParentId="8046" CreationDate="2014-01-25T15:23:50.957" Score="8" Body="&lt;p&gt;There may be two different factors at play here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it isn't necessarily the water that is doing most of the cleaning. I presume the water is used to wet a cloth and the cloth then wipes the surface. So the cloth may then physically remove surface contaminants. The role of the water may be just to improve the cloth's surface contact hence improving its ability to physically remove the contaminants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, distilled water might do a better job than tap water as it has fewer dissolved solids which would leave other stains on the surface when the water evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the only way to be sure is to do a &lt;em&gt;comparative&lt;/em&gt; experiment using different cleaning fluids. You might try tap water, distilled water and distilled water with a surfactant (e.g. a liquid soap). You might find that the &lt;em&gt;best&lt;/em&gt; result uses the surfactant but that pure distilled water does a less good, but &lt;em&gt;good enough&lt;/em&gt; job. Science isn't about single isolated experiments: we usually need to compare alternatives to know what is best.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-25T15:23:50.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8050" PostTypeId="2" ParentId="8046" CreationDate="2014-01-25T15:24:00.140" Score="2" Body="&lt;p&gt;In fact distilled water can be acid when expose to atmosphere because $CO_2$ dissolve into forming  carbonic acid:&#xA;$$\ce{H_2O(l) + CO2(g) &amp;lt;=&amp;gt; H2CO3}$$&#xA; Although the $pH$ usually in not very low (normally it is around 5) this can help to removing the  greasy stains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think also the diffusion of the salts present in greasy stain in water can improve the water cleaning capabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally one important thing is that distilled water don't leave residue in the surface. Normal water or other detergents usually contain salt or other substance that can leave halos in the surface. So the final result could appear better even if is not the best substance for solubilize fats.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-01-25T15:24:00.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8051" PostTypeId="2" ParentId="8033" CreationDate="2014-01-25T16:37:31.677" Score="5" Body="&lt;p&gt;Use directly shoes could be not so practical and reproducible. I will try to find first to figure out which experiment could be more convincing.  So why don't take many little piece of leather, cotton and synthetic fiber and make some general test (maybe useful even for clothes). These are what is passing through my mind.&lt;/p&gt;&#xA;&lt;h2&gt;Wettability and permeability&lt;/h2&gt;&#xA;&lt;p&gt;One of the  most important thing is that substance drop over the shoes don't get in touch with the skin. So you could measure the contact angle of polar and non polar liquids over different &amp;quot;shoes textile&amp;quot; and find which has the lower one.&#xA;For permeability there are also some simple instrument you can buy.&lt;/p&gt;&#xA;&lt;h2&gt;Thermal conduction and inflammability&lt;/h2&gt;&#xA;&lt;p&gt;Even if this is more difficult to perform you could try to make your own system with a little hot plate and some thermocouples to see which materials do not conduct heat or is less flammable. One simple test could be the &lt;strong&gt;Clothes iron test&lt;/strong&gt; (this in fact doesn't exist) you can palace a piece of different shoes textile and place it beneath the clothes iron measure the time at which it burns.&lt;/p&gt;&#xA;&lt;h2&gt;Acid and bases resistance&lt;/h2&gt;&#xA;&lt;p&gt;Why don't you put a little piece of textile over a piece of paper with an acid-base indicator and then put a known amount of acid or base over it if you do it in a glass table you can monitor when the acid reach the other side watching when the paper change color.&lt;/p&gt;&#xA;&lt;h2&gt;Penetrability&lt;/h2&gt;&#xA;&lt;p&gt;Pieces of broken glass could cut the textile so you could perform some test for determinate which textile is more hard to penetrate.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-01-25T16:37:31.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8052" PostTypeId="2" ParentId="8048" CreationDate="2014-01-25T16:46:52.523" Score="9" Body="&lt;p&gt;Two kinds of reactions have to be considered:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Neutralisation of the weak acid with the strong base&lt;/li&gt;&#xA;&lt;li&gt;Secondary reactions from the products&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The neutralisation is fairly straightforward:&#xA;$$\ce{HA + NaOH -&amp;gt; NaA + H2O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated &lt;strong&gt;base&lt;/strong&gt; of the acid can also react (as a base):&#xA;$$\ce{A- + H2O &amp;lt;=&amp;gt; OH- + HA}\; ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="402" LastEditDate="2017-03-30T21:44:57.247" LastActivityDate="2017-03-30T21:44:57.247" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8053" PostTypeId="5" CreationDate="2014-01-25T16:50:13.897" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-01-25T16:50:13.897" LastActivityDate="2014-01-25T16:50:13.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8054" PostTypeId="4" CreationDate="2014-01-25T16:50:13.897" Score="0" Body="Questions related to the methodologies, equipment requirements and safety considerations of any experimental procedure." OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2014-01-25T17:09:30.857" LastActivityDate="2014-01-25T17:09:30.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8055" PostTypeId="5" CreationDate="2014-01-25T16:52:45.753" Score="0" Body="&lt;p&gt;Reactivity of noble gases rises with the size of their atoms along with polarisability. Xenon has several known compounds, krypton significantly less and argon only a few. Neon and helium have no &quot;true&quot; compounds, but their protonated forms, excimers, and van der Waals molecules are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name &quot;noble gases&quot; is IUPAC approved and it currently consists of the elements helium $\ce{He}$, neon $\ce{Ne}$, argon $\ce{Ar}$, krypton $\ce{Kr}$, xenon $\ce{Xe}$ and radon $\ce{Rn}$. (compare &lt;a href=&quot;https://web.archive.org/web/20160203110828/http://www.iupac.org/nc/home/publications/iupac-books/books-db/book-details.html?tx_wfqbe_pi1[bookid]=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Red Book &lt;strong&gt;2005&lt;/strong&gt;, IR-3.5&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\scriptstyle\Tiny&#xA;\begin{array}{l|cccccccccccccccccc}&#xA;\text{Gr.} &amp;amp;1 &amp;amp;2 &amp;amp;3 &amp;amp;4 &amp;amp;5 &amp;amp;6 &amp;amp;7 &amp;amp;8 &amp;amp;9 &amp;amp;10 &amp;amp;11 &amp;amp;12 &amp;amp;13 &amp;amp;14 &amp;amp;15 &amp;amp;16 &amp;amp;17 &amp;amp;18\\&#xA;\text{Per.}\\\hline&#xA;1&amp;amp; \ce{^1H}	&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\color{\red}{\ce{^2He}}\\&#xA;2&amp;amp; \ce{^3Li}&amp;amp; \ce{^4Be}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^5B}&amp;amp; \ce{^6C}&amp;amp; \ce{^7N}&amp;amp; \ce{^8O}&amp;amp; \ce{^9F}&amp;amp; \color{\red}{\ce{^{10}Ne}}\\&#xA;3&amp;amp; \ce{^{11}Na}&amp;amp; \ce{^{12}Mg}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^{13}Al}&amp;amp; \ce{^{14}Si}&amp;amp; \ce{^{15}P}&amp;amp; \ce{^{16}S}&amp;amp; \ce{^{17}Cl}&amp;amp; \color{\red}{\ce{^{18}Ar}}\\&#xA;4&amp;amp; \ce{^{19}K}&amp;amp; \ce{^{20}Ca}&amp;amp; \ce{^{21}Sc}&amp;amp; \ce{^{22}Ti}&amp;amp; \ce{^{23}V}&amp;amp; \ce{^{24}Cr}&amp;amp; \ce{^{25}Mn}&amp;amp; \ce{^{26}Fe}&amp;amp; \ce{^{27}Co}&amp;amp; \ce{^{28}Ni}&amp;amp; \ce{^{29}Cu}&amp;amp; \ce{^{30}Zn}&amp;amp; \ce{^{31}Ga}&amp;amp; \ce{^{32}Ge}&amp;amp; \ce{^{33}As}&amp;amp; \ce{^{34}Se}&amp;amp; \ce{^{35}Br}&amp;amp; \color{\red}{\ce{^{36}Kr}}\\&#xA;5&amp;amp; \ce{^{37}Rb}&amp;amp; \ce{^{38}Sr}&amp;amp; \ce{^{39}Y}&amp;amp; \ce{^{40}Zr}&amp;amp; \ce{^{41}Nb}&amp;amp; \ce{^{42}Mo}&amp;amp; \ce{^{43}Tc}&amp;amp; \ce{^{44}Ru}&amp;amp; \ce{^{45}Rh}&amp;amp; \ce{^{46}Pd}&amp;amp; \ce{^{47}Ag}&amp;amp; \ce{^{48}Cd}&amp;amp; \ce{^{49}In}&amp;amp; \ce{^{50}Sn}&amp;amp; \ce{^{51}Sb}&amp;amp; \ce{^{52}Te}&amp;amp; \ce{^{53}I}&amp;amp; \color{\red}{\ce{^{54}Xe}}\\&#xA;6&amp;amp;\ce{^{55}Cs}&amp;amp; \ce{^{56}Ba}&amp;amp; \text{*}&amp;amp; \ce{^{72}Hf}&amp;amp; \ce{^{73}Ta}&amp;amp; \ce{^{74}W}&amp;amp; \ce{^{75}Re}&amp;amp; \ce{^{76}Os}&amp;amp; \ce{^{77}Ir}&amp;amp; \ce{^{78}Pt}&amp;amp; \ce{^{79}Au}&amp;amp; \ce{^{80}Hg}&amp;amp; \ce{^{81}Tl}&amp;amp; \ce{^{82}Pb}&amp;amp; \ce{^{83}Bi}&amp;amp; \ce{^{84}Po}&amp;amp; \ce{^{85}At}&amp;amp; \color{\red}{\ce{^{86}Rn}}\\&#xA;7&amp;amp; \ce{^{87}Fr}&amp;amp; \ce{^{88}Ra}&amp;amp; \text{**}&amp;amp; \ce{^{104}Rf}&amp;amp; \ce{^{105}Db}&amp;amp; \ce{^{106}Sg}&amp;amp; \ce{^{107}Bh}&amp;amp; \ce{^{108}Hs}&amp;amp; \ce{^{109}Mt}&amp;amp; \ce{^{110}Ds}&amp;amp; \ce{^{111}Rg}&amp;amp; \ce{^{112}Cn}&amp;amp; \ce{^{113}Nh}&amp;amp; \ce{^{114}Fl}&amp;amp; \ce{^{115}Mc}&amp;amp; \ce{^{116}Lv}&amp;amp; \ce{^{117}Ts}&amp;amp; \ce{^{118}Og}\\\hline&#xA;\end{array}\\&#xA;\scriptstyle\Tiny&#xA;\begin{array}{lc} &#xA;\text{* Lanthanoid}&amp;amp; \ce{^{57}La}&amp;amp; \ce{^{58}Ce}&amp;amp; \ce{^{59}Pr}&amp;amp; \ce{^{60}Nd}&amp;amp; \ce{^{61}Pm}&amp;amp; \ce{^{62}Sm}&amp;amp; \ce{^{63}Eu}&amp;amp; \ce{^{64}Gd}&amp;amp; \ce{^{65}Tb}&amp;amp; \ce{^{66}Dy}&amp;amp; \ce{^{67}Ho}&amp;amp; \ce{^{68}Er}&amp;amp; \ce{^{69}Tm}&amp;amp; \ce{^{70}Yb}&amp;amp; \ce{^{71}Lu}\\&#xA;\text{** Actinoid}&amp;amp; \ce{^{89}Ac}&amp;amp; \ce{^{90}Th}&amp;amp; \ce{^{91}Pa}&amp;amp; \ce{^{92}U}&amp;amp; \ce{^{93}Np}&amp;amp; \ce{^{94}Pu}&amp;amp; \ce{^{95}Am}&amp;amp; \ce{^{96}Cm}&amp;amp; \ce{^{97}Bk}&amp;amp; \ce{^{98}Cf}&amp;amp; \ce{^{99}Es}&amp;amp; \ce{^{100}Fm}&amp;amp; \ce{^{101}Md}&amp;amp; \ce{^{102}No}&amp;amp; \ce{^{103}Lr}	\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.webelements.com/periodicity/group_number/group_18.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Webelements.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="5026" LastEditDate="2018-03-17T03:02:57.033" LastActivityDate="2018-03-17T03:02:57.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8056" PostTypeId="4" CreationDate="2014-01-25T16:52:45.753" Score="0" Body="Questions about elements of group 18 (group 0 in old notation) of the periodic table and their compounds, created despite the low reactivity of these elements." OwnerDisplayName="user4076" LastEditorUserId="5026" LastEditorDisplayName="user4076" LastEditDate="2018-03-17T10:44:25.057" LastActivityDate="2018-03-17T10:44:25.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8057" PostTypeId="1" AcceptedAnswerId="8074" CreationDate="2014-01-25T17:49:35.647" Score="41" ViewCount="46316" Body="&lt;p&gt;Most books refer to a steep rise in pH when a titration reaches the equivalence point. However, I do not understand why … I mean I am adding the same drops of acid to the alkali but just as I near the correct volume (i.e. the volume required to neutralize the alkali), the pH just suddenly increases quickly.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="7475" LastEditDate="2015-10-18T16:06:35.650" LastActivityDate="2020-10-25T07:08:55.023" Title="The reason behind the steep rise in pH in the acid base titration curve" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="30" ContentLicense="CC BY-SA 3.0" />
  <row Id="8059" PostTypeId="2" ParentId="6516" CreationDate="2014-01-26T00:50:29.843" Score="1" Body="&lt;p&gt;It makes no difference where you grab the $\ce{CO2}$.  Don't burden the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plant mammoth sterile forests with gene-gineered high lignin wood for building.  Or, deep bury the trunks in the Sahara Desert of the Empty Quarter.  RuBisCO is the slowest, least efficient enzyme on Earth, the first carbon fixation step in photosynthesis.  Gene-gineer a faster, more efficient (e.g, not poisoned by oxygen) RuBisCO for the forests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cea-jaea-collaboration.net/literature/cea/ICCAP%20Nice%202007%20NaCO2finalpaper.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cea-jaea-collaboration.net/literature/cea/ICCAP%20Nice%202007%20NaCO2finalpaper.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reactor for your tailpipe. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4231" LastEditDate="2015-03-09T17:40:37.980" LastActivityDate="2015-03-09T17:40:37.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8061" PostTypeId="5" CreationDate="2014-01-26T03:58:38.340" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-01-26T03:58:38.340" LastActivityDate="2014-01-26T03:58:38.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8062" PostTypeId="4" CreationDate="2014-01-26T03:58:38.340" Score="0" Body="Questions regarding the development and exploration of chemical ideas, innovations, and discoveries." OwnerDisplayName="user4076" LastEditorUserId="402" LastEditorDisplayName="user4076" LastEditDate="2017-05-04T00:31:45.133" LastActivityDate="2017-05-04T00:31:45.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8063" PostTypeId="1" CreationDate="2014-01-26T04:02:11.263" Score="1" ViewCount="1813" Body="&lt;p&gt;Take this nicotine molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xn83v.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this mass spectrum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9C6yi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to explain all the peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might explain the peak at 42:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qCux.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can this peek als be explained by something else in the molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the decomposition of the ring with the double bonds:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8Mls4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might explain the peak at 84: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NfPZO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm clueless on the peak at 133&lt;/p&gt;&#xA;" OwnerUserId="4298" LastEditorUserId="4298" LastEditDate="2014-01-26T09:16:48.197" LastActivityDate="2018-04-17T06:38:10.247" Title="How does mass spectroscopy of nicotine work?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8064" PostTypeId="2" ParentId="8063" CreationDate="2014-01-26T04:31:38.690" Score="1" Body="&lt;p&gt;The peak at 133 corresponds to a loss of 29 mass units (nicotine is 162).  N-CH3 has a mass of 29&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-26T04:31:38.690" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8066" PostTypeId="2" ParentId="8063" CreationDate="2014-01-26T08:27:23.187" Score="0" Body="&lt;p&gt;I think that your assignment is reasonable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As ron has already pointed out, cleavage of an endocyclic $\ce{C-N}$ bond (= opening of the pyrolidine ring) in nicotine is likely. The resulting open chain fragment then loses $\ce{NCH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak at 84 represents a N-methylpyrolidine fragment and indicates that the bond between the two heterocycles is cleaved. This also makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it it a good idea to work from higher to lower masses and not to spend to much brain on the very small fragments.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-26T08:27:23.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8067" PostTypeId="1" CreationDate="2014-01-26T08:29:47.347" Score="4" ViewCount="2407" Body="&lt;p&gt;I don't know what exactly are the compounds that causes urine to be smelly, but does it (or do they) have higher or lower boiling points than water? If I have both liquid and exposed them to open air, will the water dry up first or will the smell disappear first?&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="418" LastEditDate="2014-01-26T10:11:52.887" LastActivityDate="2014-01-26T10:11:52.887" Title="Smelly matters in urine - do they have lower or higher boiling point than water?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8068" PostTypeId="1" AcceptedAnswerId="8071" CreationDate="2014-01-26T09:30:11.290" Score="27" ViewCount="13748" Body="&lt;p&gt;Assuming a perfect diamond with no impurities. Would this diamond considered to be a single large molecule?&#xA;Browsing the interweb I found several opinions about this but did not find a clear Yes or No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what if gathered so far I'm pretty sure a diamond is a monocrystalline solid. Since all its parts are connected with covalent bonds I would assume that it should be considered a molecule even though it has no fixed number of atoms in its molecule formula.&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="30836" LastEditDate="2019-04-19T12:18:12.087" LastActivityDate="2019-04-19T12:18:12.087" Title="Is a diamond a single molecule?" Tags="&lt;crystal-structure&gt;&lt;molecules&gt;&lt;carbon-allotropes&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8069" PostTypeId="2" ParentId="8067" CreationDate="2014-01-26T09:44:47.207" Score="4" Body="&lt;p&gt;There are several reasons that a smell vanishes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One obvious reason is that a smelly component has a very low boiling point and simply evaporates. Another reason might be a chemical reaction of the &quot;stinker&quot; when exposed to air, which converts it to something untraceable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, other reactions might result in quite the opposite but I rather doubt that you want to perform a test run with open bowls of urine over the next week at home ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an older article from 1971 on the &lt;a href=&quot;http://www.clinchem.org/content/17/7/592.full.pdf&quot; rel=&quot;nofollow&quot;&gt;gaschromatographic analysis of volatile organic compounds in human urine after extraction with diethylether&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a first view, however, the ketones that they found usually have a rather nice smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get a hand on the components in the vapour phase above the urine, others have performed a headspace GC-MS analysis, see this &lt;a href=&quot;http://www.gerstel.com/pdf/p-gc-an-1995-01-ar.pdf%E2%80%8E&quot; rel=&quot;nofollow&quot;&gt;application note&lt;/a&gt;&#xA;and this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0021967399000175&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the ketones usually have a pleasant smell, but they also identified some of the highly volatile &lt;em&gt;bad boys&lt;/em&gt;, such as trimethylamine, methanthiol, and dimethylsulfide. These have boiling points of ~5, ~6 and ~40 °C, respectively.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-26T09:55:04.667" LastActivityDate="2014-01-26T09:55:04.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8071" PostTypeId="2" ParentId="8068" CreationDate="2014-01-26T11:17:09.210" Score="27" Body="&lt;p&gt;Diamond is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Network_solid&quot; rel=&quot;noreferrer&quot;&gt;covalent network solid&lt;/a&gt;, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc. - network solids form one of the two main classes of &lt;a href=&quot;http://en.wikipedia.org/wiki/Macromolecule&quot; rel=&quot;noreferrer&quot;&gt;macromolecules&lt;/a&gt;, the other being &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymer&quot; rel=&quot;noreferrer&quot;&gt;polymers&lt;/a&gt;. You'll note that the Wikipedia article on macromolecules seems to imply that 'macromolecule' and 'polymer' are synonymous. They are not, at least not to those chemists working in the field. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2017-03-13T05:18:22.343" LastActivityDate="2017-03-13T05:18:22.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8072" PostTypeId="1" AcceptedAnswerId="8078" CreationDate="2014-01-26T11:40:56.607" Score="15" ViewCount="4954" Body="&lt;p&gt;In my high school chemistry book, it is written that when ligands approach the central metal ion (transition metal ion) to form dative bonds, the $3d$ orbitals split into two: two which are in higher energy level and the other 3 in lower energy orbital...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that doesn't make intuitive sense to me is that why 3 orbitals move to a lower energy level... if the ligands repel the electrons in the $3d$ orbitals, how can some of the orbitals possibly lose energy and be at a lower enrgy state than they were currently at before...&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zSRgy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2014-01-26T17:12:01.973" Title="Splitting of $d$ orbitals when ligands approach central metal ion" Tags="&lt;coordination-compounds&gt;&lt;orbitals&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8073" PostTypeId="1" CreationDate="2014-01-26T12:52:47.697" Score="54" ViewCount="10714" Body="&lt;p&gt;I was just thinking what can be the last atomic number that can exist within the range of permissible radioactivity limit and considering all other factors in quantum physics and chemical factors.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="1499" LastEditDate="2014-03-04T01:32:19.277" LastActivityDate="2021-02-01T00:58:43.250" Title="The last element's atomic number" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;nuclear&gt;&lt;elements&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="8074" PostTypeId="2" ParentId="8057" CreationDate="2014-01-26T13:02:12.563" Score="44" Body="&lt;p&gt;I've decided to tackle this question in a somewhat different manner. Instead of giving the chemical intuition behind it, I wanted to check for myself if the mathematics actually work out. As far as I understand, this isn't done often, so that's why I wanted to try it, even though it may not make the clearest answer. It turns out to be a bit complicated, and I haven't done much math in a while, so I'm kinda rusty. Hopefully, everything is correct. I would love to have someone check my results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach here is to explicitly find the equation of a general titration curve and figure out from that why the pH varies quickly near the equivalence point. For simplicity, I shall consider the titration to be between a monoprotic acid and base. Explicitly, we have the following equilibria in solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HA &amp;lt;=&amp;gt; H^+ + A^-} \ \ \ → \ \ \ K_\text{a} = \ce{\frac{[H^+][A^-]}{[HA]}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{BOH &amp;lt;=&amp;gt; B^+ + OH^-} \ \ \ → \ \ \ K_\text{b} = \ce{\frac{[OH^-][B^+]}{[BOH]}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O &amp;lt;=&amp;gt; H^+ + OH^-} \ \ \ → \ \ \ K_\text{w} = \ce{[H^+][OH^-]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us imagine adding two solutions, one of the acid &lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt; with volume &lt;span class=&quot;math-container&quot;&gt;$V_\text{A}$&lt;/span&gt; and concentration &lt;span class=&quot;math-container&quot;&gt;$C_\text{A}$&lt;/span&gt;, and another of the base &lt;span class=&quot;math-container&quot;&gt;$\ce{BOH}$&lt;/span&gt; with volume &lt;span class=&quot;math-container&quot;&gt;$V_\text{B}$&lt;/span&gt; and concentration &lt;span class=&quot;math-container&quot;&gt;$C_\text{B}$&lt;/span&gt;. Notice that after mixing the solutions, the number of moles of species containing &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{A^-}$&lt;/span&gt;) is simply &lt;span class=&quot;math-container&quot;&gt;$n_\text{A} = C_\text{A} V_\text{A}$&lt;/span&gt;, while the number of moles of species containing &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{BOH}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{B^+}$&lt;/span&gt;) is &lt;span class=&quot;math-container&quot;&gt;$n_\text{B} = C_\text{B} V_\text{B}$&lt;/span&gt;. Notice that at the equivalence point, &lt;span class=&quot;math-container&quot;&gt;$n_\text{A} = n_\text{B}$&lt;/span&gt; and therefore &lt;span class=&quot;math-container&quot;&gt;$C_\text{A} V_\text{A} = C_\text{B} V_\text{B}$&lt;/span&gt;; this will be important later. We will assume that volumes are additive (total volume &lt;span class=&quot;math-container&quot;&gt;$V_\text{T} = V_\text{A} + V_\text{B}$&lt;/span&gt;), which is close to true for relatively dilute solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In search of an equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem of finding the final equilibrium after adding the solutions, we write out the charge balance and matter balance equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge balance: &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+] + [B^+] = [A^-] + [OH^-]}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter balance for &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \ce{[HA] + [A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matter balance for &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \ce{[BOH] + [B^+]} = \frac{C_\text{B} V_\text{B}} {V_\text{A} + V_\text{B}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A titration curve is given by the pH on the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axis and the volume of added acid/base on the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis. So what we need is to find an equation where the only variables are &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_\text{A}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$V_\text{B}$&lt;/span&gt;. By manipulating the dissociation constant equations and the mass balance equations, we can find the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[HA]} = \frac{\ce{[H^+][A^-]}}{K_\text{a}}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{[BOH]} = \frac{\ce{[B^+]}K_\text{w}}{K_\text{b}\ce{[H^+]}}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{[A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{a}}{K_\text{a} + \ce{[H^+]}}\right)$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{[B^+]} = \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{b}\ce{[H^+]}}{K_\text{b}\ce{[H^+]} + K_\text{w}}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing those identities in the charge balance equation, after a decent bit of algebra, yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]^4} + \left(K_\text{a} + \frac{K_\text{w}}{K_\text{b}} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^3} + \left(\frac{K_\text{a}}{K_\text{b}}K_\text{w} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} K_\text{a} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}K_\text{a} - K_\text{w}\right) \ce{[H^+]^2} - \left(K_\text{a} K_\text{w} + \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\frac{K_\text{a}}{K_\text{b}} K_\text{w} + \frac{K^2_\text{w}}{K_\text{b}}\right) \ce{[H^+]} - \frac{K_\text{a}}{K_\text{b}} K^2_\text{w} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this equation sure looks intimidating, but it is very interesting. For one, this single equation will exactly solve &lt;em&gt;any&lt;/em&gt; equilibrium problem involving the mixture of any monoprotic acid and any monoprotic base, in any concentration (as long as they're not much higher than about &lt;span class=&quot;math-container&quot;&gt;$1~\mathrm{\small M}$&lt;/span&gt;) and any volume. Though it doesn't seem to be possible to separate the variables &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_\text{A}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$V_\text{B}$&lt;/span&gt;, the graph of this equation represents any titration curve (as long as it obeys the previous considerations). Though in its full form it is quite daunting, we can obtain some simpler versions. For example, consider that the mixture is of a weak acid and a strong base. This means that &lt;span class=&quot;math-container&quot;&gt;$K_\text{b} \gg 1$&lt;/span&gt;, and so every term containing &lt;span class=&quot;math-container&quot;&gt;$K_\text{b}$&lt;/span&gt; in the denominator is approximately zero and gets cancelled out. The equation then becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Weak acid and strong base:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]^3} + \left(K_\text{a} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\ce{A} + V_\ce{B}} K_\ce{a} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}K_\ce{a} - K_\ce{w}\right) \ce{[H^+]} - K_\ce{a} K_\ce{w} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a strong acid and weak base (&lt;span class=&quot;math-container&quot;&gt;$K_\text{a} \gg 1$&lt;/span&gt;), you can divide both sides of the equation by &lt;span class=&quot;math-container&quot;&gt;$K_\text{a}$&lt;/span&gt;, and now all terms with &lt;span class=&quot;math-container&quot;&gt;$K_\text{a}$&lt;/span&gt; in the denominator get cancelled out, leaving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strong acid and weak base:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]^3} + \left(\frac{K_\ce{w}}{K_\ce{b}}+\frac{C_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}\right) \ce{[H^+]^2} - \left(K_\ce{w} + \frac{C_\text{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}} \frac{K_\ce{w}}{K_\ce{b}}\right) \ce{[H^+]} - \frac{K^2_\ce{w}}{K_\ce{b}} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest case happens when adding a strong acid to a strong base (&lt;span class=&quot;math-container&quot;&gt;$K_\ce{a} \gg 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_\ce{b} \gg 1$&lt;/span&gt;), in which case all terms containing either in the denominator get cancelled out. The result is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Strong acid and strong base:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]} - K_\ce{w} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be enlightening to draw some example graphs for each equation, but Wolfram Alpha only seems to be able to handle the last one, as the others require more than the standard computation time to display. Still, considering the titration of &lt;span class=&quot;math-container&quot;&gt;$1~\text{L}$&lt;/span&gt; of a &lt;span class=&quot;math-container&quot;&gt;$1~\ce{\small M}$&lt;/span&gt; solution of a strong acid with a &lt;span class=&quot;math-container&quot;&gt;$1~\ce{\small M}$&lt;/span&gt; solution of a strong base, you get &lt;a href=&quot;http://www.wolframalpha.com/input/?i=plot%20%281%2bx%29*%2810%5E%28-y%29%29%5E2%2b%28x-1%29*10%5E%28-y%29-%281%2bx%29*10%5E%28-14%29=0%20from%20x=0%20to%202%20and%20y=0%20to%2014&quot; rel=&quot;noreferrer&quot;&gt;this graph&lt;/a&gt;. The &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis is the volume of base added, in litres, while the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axis is the pH. Notice that the graph is exactly as what you'll find in a textbook!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now what?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the equations figured out, let's study how they work. We want to know why the pH changes quickly near the equivalence point, so a good idea is to analyze the derivative of the equation and figure out where they have a very positive or very negative value, indicating a region where &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+]}$&lt;/span&gt; changes quickly with a slight addition of an acid/base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we want to study the titration of an acid with a base. What we need then is the derivative &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$&lt;/span&gt;. We will obtain this by implicit differentiation of both sides of the equations by &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \frac{\ce{d}}{\ce{d}V_\ce{B}}$&lt;/span&gt;. Starting with the easiest case, the mixture of a strong acid and strong base, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\ce{d[H^+]}}{\ce{d} V_\ce{B}}=  \frac{K_\ce{w} - C_\ce{B} \ce{[H^+] - [H^+]^2}}{2(V_\ce{A} + V_\ce{B}\left) \ce{[H^+]} + (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}\right)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again a complicated looking fraction, but with very interesting properties. The numerator is not too important, it's the denominator where the magic happens. Notice that we have a sum of two terms (&lt;span class=&quot;math-container&quot;&gt;$2(V_\ce{A} + V_\ce{B})\ce{[H^+]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A})$&lt;/span&gt;). The lower this sum is, the higher &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \frac{\mathrm{d}\ce{[H^+]}}{\mathrm{d} V_\ce{B}}$&lt;/span&gt; is and the quicker the pH will change with a small addition of the base. Notice also that, if the solutions aren't very dilute, then the second term quickly dominates the denominator because while adding base, the value of &lt;span class=&quot;math-container&quot;&gt;$[H^+]$&lt;/span&gt; will become quite small compared to &lt;span class=&quot;math-container&quot;&gt;$C_\ce{A}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_\ce{B}$&lt;/span&gt;. Now we have a very interesting situation; a fraction where the major component of the denominator has a subtraction. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=y=1/%281E-3%2b%28x-5%29%29&quot; rel=&quot;noreferrer&quot;&gt;Here's an example of how this sort of function behaves&lt;/a&gt;. When the subtraction ends up giving a result close to zero, the function explodes. This means that the speed at which &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+]}$&lt;/span&gt; changes becomes &lt;em&gt;very&lt;/em&gt; sensitive to small variations of &lt;span class=&quot;math-container&quot;&gt;$V_\ce{B}$&lt;/span&gt; near the critical region. And where does this critical region happen? Well, close to the region where &lt;span class=&quot;math-container&quot;&gt;$C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$&lt;/span&gt; is zero. If you remember the start of the answer, &lt;strong&gt;this is the equivalence point!&lt;/strong&gt;. So there, this proves mathematically that the speed at which the pH changes is maximum at the equivalence point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was only the simplest case though. Let's try something a little harder. Taking the titration equation for a weak acid with strong base, and implicitly differentiating both sides by &lt;span class=&quot;math-container&quot;&gt;$\displaystyle \frac{\ce{d}}{\ce{d} V_\ce{B}}$&lt;/span&gt; again, we get the significantly more fearsome:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}} = \frac{ -\frac{V_\ce{A}}{(V_\ce{A} + V_\ce{B})^2} \ce{[H^+]} (C_\ce{B}\ce{[H^+]} - C_\ce{B} K_\ce{a} + C_\ce{A} K_\ce{a})}{3\ce{[H^+]^2 + 2[H^+]}\left(K_\ce{a} + \frac{C_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}}\right) + \frac{K_\ce{a}}{V_\ce{A} + V_\ce{B}} (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}) -K_\ce{w}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, the term that dominates the behaviour of the complicated denominator is the part containing &lt;span class=&quot;math-container&quot;&gt;$C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$&lt;/span&gt;, and once again the derivative explodes at the equivalence point.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="31775" LastEditDate="2018-10-06T16:17:21.200" LastActivityDate="2018-10-06T16:17:21.200" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="8075" PostTypeId="2" ParentId="8057" CreationDate="2014-01-26T13:06:50.530" Score="19" Body="&lt;p&gt;The pH titration curve is shaped as it is because pH is a logarithmic scale. pH is the negative base ten logarithm of hydrogen ion concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH}=-\log_{10}{\ce{[H+]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, each change of 1 pH unit is a 10-fold change in concentration, and linear of value that 10-fold change gets smaller as the pH increases. For example, going from pH 1 to pH 2 means changing the concentration of $\ce{H+}$ from $10^{-1}$ to $10^{-2}$ or $0.100... -0.010... =0.099...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from pH 2 to pH 3 means means changing the concentration of $\ce{H+}$ from $10^{-2}$ to $10^{-3}$ or $0.0100... -0.0010... =0.0099...$ If you are added fixed amounts of base (which is what you do during a titration), then that fixed amount has a larger effect on the concentration of $\ce{H+}$ as the pH goes up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's walk through a simple titration:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have 10 mL of 0.100 M $\ce{HCl}$ that you want to titrate with 0.100 M $\ce{NaOH}$. We will consider the change to the solution after each mL of base is added. We have to remember to keep track of &lt;strong&gt;moles&lt;/strong&gt; of acid and base, since the volume is changing (we are adding base) and we need to remember that autoionization of water. The concentrations of $\ce{H+}$ and $\ce{OH-}$ never go to zero because of the autoionization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O &amp;lt;=&amp;gt; H+ + OH-} \ \ \ \ K_w=[\ce{H+}][\ce{OH-}]=10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V_b$ be the volume of base added. At $V_b=0$, we have our original acid solution, which contains $\left(\dfrac{10\  \text{mL}}{1000 \ \frac{\text{mL}}{\text{L}}}\right)\left(0.100\ \dfrac{\text{M}}{\text{L}}\right)=0.001 \ \text{moles of HCl}$. Each time we increase $V_b$ by 1 mL, we add $\left(\dfrac{1\  \text{mL}}{1000 \ \frac{\text{mL}}{\text{L}}}\right)\left(0.100\ \dfrac{\text{M}}{\text{L}}\right)=0.0001 \ \text{moles of NaOH}.$ We decrease the moles of $\ce{H+}$present by this amount and recalculate concnetration for the new volume. Then we recalculation pH. We do this until...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we reach $V_b =10 \ \text{mL}$, we will have added exactly as much base as there was acid originally. The concentrations of $\ce{H+}$ and $\ce{OH-}$ are both $10^{-7}\ \text{M}$, controlled by the autoionization. This is the equivalence point. As we add more base, we increase the concentration of $\ce{OH-}$. We can use the $K_w$ to calculate the concentration of $\ce{H+}$ and then the pH.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; V_b     V_T     moles H+     moles OH-     [H+]        [OH-]        pH&#xA; 0 mL    10 mL   1.00E-3      1.00E-15      1.00E-1     1.00E-13     1.00&#xA; 1 mL    11 mL   9.00E-4      1.34E-15      8.18E-2     1.22E-13     1.09&#xA; 2 mL    12 mL   8.00E-4      1.80E-15      6.67E-2     1.50E-13     1.18&#xA; 3 mL    13 mL   7.00E-4      2.41E-15      5.38E-2     1.86E-13     1.27&#xA; 4 mL    14 mL   6.00E-4      3.27E-15      4.29E-2     2.33E-13     1.37&#xA; 5 mL    15 mL   5.00E-4      4.50E-15      3.33E-2     3.00E-13     1.48&#xA; 6 mL    16 mL   4.00E-4      6.40E-15      2.50E-2     4.00E-13     1.60&#xA; 7 mL    17 mL   3.00E-4      9.63E-15      1.76E-2     5.67E-13     1.75&#xA; 8 mL    18 mL   2.00E-4      1.62E-15      1.11E-2     9.00E-13     1.95&#xA; 9 mL    19 mL   1.00E-4      3.61E-15      5.26E-3     1.90E-12     2.28&#xA;10 mL    20 mL   2.00E-9      2.00E-9       1.00E-7     1.00E-7      7.00&#xA;11 mL    21 mL   4.41E-14     1.00E-4       2.10E-12    4.76E-3     11.68&#xA;12 mL    22 mL   2.42E-14     2.00E-4       1.10E-12    9.09E-3     11.96&#xA;13 mL    23 mL   1.76E-14     3.00E-4       7.67E-13    1.30E-2     12.12&#xA;14 mL    24 mL   1.44E-14     4.00E-4       6.00E-13    1.67E-2     12.22&#xA;15 mL    25 mL   1.25E-14     5.00E-4       5.00E-13    2.00E-2     12.30&#xA;16 mL    26 mL   1.13E-14     6.00E-4       4.33E-13    2.31E-2     12.36&#xA;17 mL    27 mL   1.04E-14     7.00E-4       3.86E-13    2.59E-2     12.41&#xA;18 mL    28 mL   9.80E-15     8.00E-4       3.50E-13    2.86E-2     12.45&#xA;19 mL    29 mL   9.34E-15     9.00E-4       3.22E-13    3.10E-2     12.49&#xA;20 mL    30 mL   9.00E-15     1.00E-3       3.00E-13    3.33E-2     12.52&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Notice that between $V_b =9$ and $V_b = 11$, the change in $\ce{[H+]}$ and $\ce{[OH-]}$ are very large compared to their changes before and after the equivalence point. This is because near the equivalence point, the concentrations are very low and controlled by the autoionization and not by the amount of acid or base present. Even if we expand the range around the equivalence point and determine the pH between $V_b =9$ and $V_b = 11$ after 0.1 mL, we still see similar behavior.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-26T13:06:50.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8077" PostTypeId="2" ParentId="8073" CreationDate="2014-01-26T13:41:33.603" Score="52" Body="&lt;p&gt;Nobody really knows. Using the naive Bohr model of the atom, we run into trouble around $Z=137$ as &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_periodic_table#Feynmanium_and_elements_above_the_atomic_number_137&quot;&gt;the innermost electrons would have to be moving above the speed of light&lt;/a&gt;. This result is because the Bohr model doesn't take into account relativity. Solving the Dirac equation, which comes from relativistic quantum mechanics, and taking into account that the nucleus is not a point particle, then there seems to be no real issue with arbitrarily high atomic numbers, although unusual effects start happening above $Z \approx 173$. These results may be overturned by an even deeper analysis with current quantum electrodynamics theory, or a new theory altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as we can tell, however, we will never get anywhere close to such atomic numbers. Very heavy elements are extremely unstable with respect to radioactive decay into lighter elements. Our current method of producing superheavy elements is based on accelerating a certain isotope of an relatively light element and hitting a target made of an isotope of a much heavier element. This process is extremely inefficient, and it takes many months to produce significant amounts of material. For the heaviest elements, it takes years to detect even a handful of atoms. The very short lifetime of the heaviest targets and the very low collision efficiency between projectile and target mean that it will be extremely difficult to go much further than the current 118 elements. It is possible that we may find somewhat more stable superheavy isotopes in the islands of stability around $Z=114$ and $Z=126$, but the predicted most stable isotopes (which even then are not expect to last more than a few minutes) have such a huge amount of neutrons in their nuclei that we have no idea how to produce them; we may be condemned to merely skirt the shores of the islands of stability, while never climbing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Note that the best calculation presented above is based on quantum electrodynamics alone, i.e. only electromagnetic forces are taken into account. Obviously, to predict how nuclei will behave (and therefore how many protons you can stuff into a nucleus before it's impossible to go any further), one needs detailed knowledge of the strong and weak nuclear forces. Unfortunately, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_unsolved_problems_in_physics#Nuclear_physics&quot;&gt;the mathematical description of nuclear forces is still an incredibly tough problem in physics today&lt;/a&gt;, so no one can hope to provide a rigorous answer from that angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be &lt;em&gt;some&lt;/em&gt; limit, as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_force&quot;&gt;residual nuclear forces&lt;/a&gt; are very short-ranged. At some point there will be so many protons and neutrons in the nucleus (and the resulting nucleus will have become so large) that the diametrically opposite parts of the nucleus won't be able to &quot;detect&quot; each other, as they are too far away. Each additional proton or neutron produces a weaker stabilization via the strong nuclear force. Meanwhile, the electrical repulsion between protons has infinite range, so every additional proton will contribute repulsively just the same. This is why heavier elements need higher and higher neutron-to-proton ratios to remain stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, at some atomic number, possibly not much higher than our current record of $Z=118$, the electrical repulsion of the protons will always win against the strong nuclear attractions of protons and neutrons, no matter the configuration of the nucleus. Hence, all sufficiently heavy atomic nuclei will suffer spontaneously fission almost immediately after coming into existence, or all the valid reaction pathways to reach an element will require events which are so fantastically unlikely that if even all the nucleons in the entire observable Universe were being collided with each other since the Big Bang in an attempt to synthesize the heaviest element possible, we would statistically expect some sufficiently heavy atom to not have been produced even once.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-03-04T01:47:57.467" LastActivityDate="2014-03-04T01:47:57.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8078" PostTypeId="2" ParentId="8072" CreationDate="2014-01-26T14:32:50.650" Score="13" Body="&lt;p&gt;The picture you pose is only a part. The full picture is posted below.&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/8/83/Crystal_Field_Splitting_4.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical &lt;strong&gt;Barycentre&lt;/strong&gt;. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;br&gt;&#xA;By conservation of energy, we mean that the energy is conserved &lt;em&gt;during splitting&lt;/em&gt;. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.600" LastActivityDate="2014-01-26T17:12:01.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8079" PostTypeId="1" CreationDate="2014-01-26T15:13:29.127" Score="6" ViewCount="7645" Body="&lt;p&gt;&lt;a href=&quot;http://thermoliciousfood.wordpress.com/2014/01/25/relaxing-handbody-wash/#respond&quot; rel=&quot;nofollow noreferrer&quot;&gt;This recipe&lt;/a&gt; for home-made handwash (liquid soap) consists of bar soap, water, Epsom salts, and coconut oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author specifically warns against using 'artificial' Epsom salts, preferring the naturally occurring type, because the 'artificial' type is &lt;em&gt;supposedly&lt;/em&gt; contaminated by heavy metals incl. arsenic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note – Be sure to get natural salts as some are man made and a full of&#xA;  unwanted chemicals. A lot of Epsom salts are made from synthetic&#xA;  materials, they are manufactured in factories from neutralized&#xA;  Sulfuric Acid, which is very high in heavy metals and arsenic. The&#xA;  REAL Epsom salt is naturally occurring mineral deposit in spring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_sulfate#Manufacturing&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to Wikipedia&lt;/a&gt;, Epsom salts are made by reacting sulfuric acid with magnesium oxide or magnesium carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it difficult to believe that 'artificial' Epsom salts would have appreciable amounts of heavy metal contamination, as neither sulfuric acid nor magnesium oxide are known to be frequently contaminated with heavy metals. A quick search on Google Scholar gives no results on the topic of heavy metal contamination in Epsom salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I am aware that heavy metals might enter the process is if the sulfuric acid was produced from smelter stack gases, which do have appreciable heavy metal content. This is &lt;a href=&quot;http://www.sulphuric-acid.com/sulphuric-acid-on-the-web/acid%20plants/Incitec%20Pivot%20-%20Mount%20Isa.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;done at Mount Isa Mines&lt;/a&gt;, where $\ce{SO2}$ from the copper smelter stack gas is scrubbed out to make sulfuric acid, but the entire output of this plant goes straight to fertilizer production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is the above claim about heavy metal contamination of Epsom salts credible, or just an urban myth?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4305" LastEditorUserId="402" LastEditDate="2017-04-09T18:23:29.620" LastActivityDate="2017-06-08T09:44:37.040" Title="Epsom salts - heavy metal contamination myth?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;ionic-compounds&gt;" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8080" PostTypeId="1" CreationDate="2014-01-26T15:30:00.553" Score="0" ViewCount="182" Body="&lt;p&gt;The transition metals have valence and penultimate shell incompletely filled . so does this mean that the transition metals first fill up their valence shell and then penultimate shell? &lt;/p&gt;&#xA;" OwnerUserId="4306" LastActivityDate="2014-01-26T17:16:51.693" Title="Nature of transition metals" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-01-28T04:53:57.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="8081" PostTypeId="2" ParentId="8080" CreationDate="2014-01-26T17:16:51.693" Score="1" Body="&lt;p&gt;The short answer to your question is, usually, but not always.  The energy differences between these levels are very small, so sometimes other factors come into play that can cause exceptions.  Here are a few exceptions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Predicted configuration   Actual configuration &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Cr    [Ar] 3d4 4s2  ...........  [Ar] 3d5 4s1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cu    [Ar] 3d9 4s2  ...........  [Ar] 3d10 4s1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ag    [Kr] 4d9 5s2  ...........  [Kr] 4d10 5s1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Au    [Xe] 4f14 5d9 6s2  .. [Xe] 4f14 5d10 6s1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a special stability associated with a full shell, in fact that's probably why you suspected that &quot;transition metals first fill up their valence shell and then penultimate shell.&quot;  There also appears to be a stability advantage associated with a half-filled shell and this can explain many of the anomalies.  The reason that a half-filled shell may have some extra stability is that there are no paired electrons (Hund's rule) in such a configuration.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-26T17:16:51.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8082" PostTypeId="2" ParentId="5734" CreationDate="2014-01-26T18:02:31.450" Score="10" Body="&lt;ol&gt;&#xA;&lt;li&gt;Bulk d-metals and especially intermetallic compounds often &lt;em&gt;do&lt;/em&gt; have significant covalent bonding. &lt;/li&gt;&#xA;&lt;li&gt;Strictly speaking, metal bonding is a kind of covalent bonding in a sense. It is a common knowledge, that 3- or more atoms may be bound by one pair of electrons, like $\ce{H3+}$ ion. In bulk metals, similar bonding bond is working for valence electron.&lt;/li&gt;&#xA;&lt;li&gt;Even more strictly speaking, to fully understand the matter you have to consider the matter from position of theory of molecular orbitals.&lt;br&gt;&#xA;Let's assume we have a nanocrystal of 100 3d row element atoms. From 400 orbitals of said atoms 400 molecular orbitals are formed, with strict distribution of their energies depending on shape and size of the crystal and crystal cell. Then, the only difference between metals and non-metals is that metals have semi-full group of orbitals of same energy (so called &lt;em&gt;conductivity zone&lt;/em&gt;, while non-metals have only full and vacant orbitals.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5026" LastEditDate="2018-02-24T11:20:13.370" LastActivityDate="2018-02-24T11:20:13.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8083" PostTypeId="2" ParentId="8079" CreationDate="2014-01-26T18:13:00.440" Score="4" Body="&lt;p&gt;Here in the US, they sell different grades of Epsom salt, from USP down to plant food.  I'd presume that depending upon what the product will ultimately be used for, the manufacturer may use MgO with different purity levels - the lower the purity, the cheaper the product.  Even the &quot;bath salts&quot; grade contain trace amounts of arsenic and selenium (see the first link below).  Among the &quot;medicinal&quot; offerings, some are labeled &quot;for external use&quot; and others are labeled &quot;for external and internal use.&quot;  I suspect that anything with the &quot;for external and internal use&quot; label would be more than safe enough to use in handwash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links you might find helpful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://k.b5z.net/i/u/2182313/f/Tech%20Data%20Sheets/Tech_Data_Sheet-bath_salt-epsom_USP-3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.drugs.com/otc/104235/pepsom-flying-p-epsom-salt.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2017-04-09T18:23:40.933" LastActivityDate="2017-04-09T18:23:40.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8084" PostTypeId="1" AcceptedAnswerId="8085" CreationDate="2014-01-26T20:31:30.813" Score="2" ViewCount="6526" Body="&lt;p&gt;I'm reading on ionic crystal structures and am confused by this from my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the metallic elements give up electrons when ionized, cations are ordinarily smaller than anions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that having a positive charge will pull the electrons closer, but shouldn't the atom from the larger period (row on periodic table), have more electron shells and be the larger atom?&lt;/p&gt;&#xA;" OwnerUserId="4309" LastActivityDate="2015-10-04T10:28:35.630" Title="Why are cations ordinarily smaller than anions?" Tags="&lt;crystal-structure&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8085" PostTypeId="2" ParentId="8084" CreationDate="2014-01-27T00:01:07.900" Score="4" Body="&lt;p&gt;It isn't true that &lt;strong&gt;any&lt;/strong&gt; cation will be smaller than &lt;strong&gt;any&lt;/strong&gt; anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You intuition that atoms with larger electron counts from lower rows of the periodic table will be bigger is correct. Size for atoms is determined by the highest occupied electron shell or orbital as the interactions with other things that define size are, crudely, caused by interactions between the electrons in their outer orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a Rb&lt;sup&gt;+&lt;/sup&gt; cation radius is about 166pM but an F&lt;sup&gt;-&lt;/sup&gt; anion is smaller at about only 119pm. There is a good table of such values in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; on ionic radii which also explains some of the trends and some of the difficulties in actually defining size for ions.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-27T00:01:07.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8086" PostTypeId="1" AcceptedAnswerId="8193" CreationDate="2014-01-27T00:31:49.720" Score="4" ViewCount="432" Body="&lt;p&gt;I think that I understand most of the parts of the symbol shown for ground state oxygen $\ce{{}^{16}O2}$, which is ${}^{3}\Sigma^{-}_{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was hoping that someone could tell me what exactly the negative superscript sign means? And how do you identify between a positive and negative state? I think it has something to do with a mirror plane.&lt;/p&gt;&#xA;" OwnerUserId="4310" LastEditorUserId="4945" LastEditDate="2014-11-17T06:08:28.520" LastActivityDate="2014-11-17T06:08:28.520" Title="What does the negative superscript in the term symbol for ground state Oxygen stand for?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8087" PostTypeId="1" AcceptedAnswerId="8781" CreationDate="2014-01-27T00:37:00.200" Score="2" ViewCount="7310" Body="&lt;p&gt;I just read that acetic acid is an exception to the general idea that the conjugate base to a weak acid is a strong base.  The example says that the conjugate base of a weak acid will only be strong if the acid is a weaker acid than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm curious about though, is if acetate is a weak base due to its resonance structure.  Does the resonance cause acetate to be more stable than hydrogen acetate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true in the more general case where some substituent can bond to the atoms involved in the resonance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2014-02-27T20:06:40.287" Title="Is acetate a weak base due to its resonance structure?" Tags="&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8088" PostTypeId="1" CreationDate="2014-01-27T00:54:45.380" Score="1" ViewCount="1435" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the solubility of $\ce{N2}$ in water exposed to air at $\pu{25^\circ C}$, if the atmospheric pressure is $\pu{1.2 atm}$, assume that the mole fraction of nitrogen is $0.78$ in air and the Henry's Law constant for nitrogen in water at this temperature is $\pu{6.1 \times 10^{-4} M/atm}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice.&lt;/p&gt;&#xA;" OwnerUserId="4312" LastEditorUserId="5026" LastEditorDisplayName="user4076" LastEditDate="2018-02-27T02:36:54.860" LastActivityDate="2018-02-27T02:38:02.613" Title="Solubility, pressure and Henry's law" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8089" PostTypeId="2" ParentId="8088" CreationDate="2014-01-27T02:09:46.583" Score="1" Body="&lt;p&gt;What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-27T02:09:46.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8090" PostTypeId="2" ParentId="8031" CreationDate="2014-01-27T03:58:47.427" Score="0" Body="&lt;p&gt;@tschoppi: 0.1 M HBr (a strong acid) has a pH of 1. The first equilibrium changes its color from red to purple at the pH of 2.9. Why do you keep titrating if the color has already changed?&lt;/p&gt;&#xA;" OwnerUserId="4298" LastActivityDate="2014-01-27T03:58:47.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8093" PostTypeId="2" ParentId="8084" CreationDate="2014-01-27T05:30:01.027" Score="2" Body="&lt;p&gt;Generally we say that cation is smaller than anion in case of &lt;strong&gt;iso-electronic species&lt;/strong&gt; (see the fig). If you consider ions which are not iso-electronic, the statement may not be true always. As you say, if you consider cations consisting of many shells (say 6) and if you compare them with anions consisting of less shells (say 2), in that case cations will be larger than anion.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hlGTO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-01-27T05:30:01.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8095" PostTypeId="5" CreationDate="2014-01-27T05:56:16.833" Score="0" Body="&lt;p&gt;Food chemistry is the study of the chemistry and biochemistry of food.  Broadly speaking this field is concerned with analytical, biochemical, microbiological, chemical, physical, nutritional, sensory, and toxicological aspects of foods, food ingredients, and food processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The journal &lt;em&gt;Food Chemistry&lt;/em&gt; includes the following topics: chemical analysis of food, chemical additives and toxins; chemistry relating to certain aspects of food; structural changes in molecules during the processing and storage of foods; direct effects on foods of the use of agrochemicals; chemical quality in food engineering and technology, and measurement of micronutrients, macronutrients, additives and contaminants in foodstuffs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discipline of food chemistry borrows from organic chemistry, biochemistry, nutritional biochemistry, physical chemistry, toxicology, analytical chemistry, and chemical engineering.&lt;/p&gt;&#xA;" OwnerUserId="21964" LastEditorUserId="21964" LastEditDate="2016-08-26T14:28:17.350" LastActivityDate="2016-08-26T14:28:17.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8096" PostTypeId="4" CreationDate="2014-01-27T05:56:16.833" Score="0" Body="For questions related to the chemistry of food." OwnerDisplayName="user4076" LastEditorUserId="21964" LastEditorDisplayName="user4076" LastEditDate="2016-08-26T14:33:23.537" LastActivityDate="2016-08-26T14:33:23.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8099" PostTypeId="1" AcceptedAnswerId="8100" CreationDate="2014-01-27T13:35:06.937" Score="15" ViewCount="3245" Body="&lt;p&gt;My book writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When white light shines through a solution of a complex ion of a transition metal, photons of a particular frequency are absorbed and their energy promotes an electron from lower energy level to upper energy level. We perceive the complementary color to the light absorbed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I have a question about this. Once the electron is promoted to a higher energy level, won't it fall back down and give off a photon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely the electron cannot remain at the higher energy level forever, as that would mean that no further electronic transitions could taking place, and the solution would turn colourless.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="16683" LastEditDate="2016-12-14T17:18:28.433" LastActivityDate="2020-03-05T23:13:27.747" Title="What happens to electrons in metal complexes after excitation by visible light?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;color&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8100" PostTypeId="2" ParentId="8099" CreationDate="2014-01-27T14:29:47.003" Score="20" Body="&lt;p&gt;Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{S_0 -&amp;gt;[$h\nu_\mathrm{ex}$] S_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, the excited state deactivates through &lt;strong&gt;internal conversion&lt;/strong&gt; in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{S_1 -&amp;gt; S_0 + $h\nu_\mathrm{f}$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This deactivation via radiation is known as &lt;strong&gt;fluorescence&lt;/strong&gt;. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} &amp;gt; \lambda_\mathrm{ex}$). The effect is known as &lt;strong&gt;Stokes shift&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BP9cb.png&quot; alt=&quot;Ru(bpy)3 2+&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2-&lt;em&gt;a&lt;/em&gt;:2',3'-c]phenazine (dppz), and benzo[&lt;em&gt;i&lt;/em&gt;]dipyrido-[3,2-&lt;em&gt;a&lt;/em&gt;:2',3'-c]phenazine (dppn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KCF57.png&quot; alt=&quot;ligands&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other cases, the excited singlet state undergoes a change in the spin multiplicity (&lt;strong&gt;Intersystem crossing&lt;/strong&gt;) to a triplet state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{S_1 -&amp;gt;[ISC] T_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as &lt;strong&gt;phosphorescence&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{T_1 -&amp;gt; S_0 + $h\nu_\mathrm{p}$}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Keeping all these processes in mind, one question arises - particular from younger students or pupils:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The answer is in the enormous numbers! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, we have a huge number of incidents &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The complex absorbs a photon of a certain energy, which means that one colour is &quot;swallowed&quot;. The complex forms a short-lived &quot;excited state&quot;&lt;/li&gt;&#xA;&lt;li&gt;the solution reflects the remaining light&lt;/li&gt;&#xA;&lt;li&gt;the excited state complex deactivates again&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-12-14T17:21:38.653" LastActivityDate="2016-12-14T17:21:38.653" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8101" PostTypeId="1" AcceptedAnswerId="8102" CreationDate="2014-01-27T14:51:51.400" Score="1" ViewCount="912" Body="&lt;p&gt;Are ligand exchange reactions one-way  reactions or reversible? I know this is a very silly question but its not said outright in any place...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in my high school chemistry book, these two ligand exchange reactions are shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Cu(H2O)6]^{2+} (aq) + 4Cl- (aq) &amp;lt;=&amp;gt; [CuCl_4]^{2-} (aq) + 6H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Cu(H2O)6]^{2+} (aq) + 4NH3 (aq)-&amp;gt; [Cu(NH3)4(H2O)2]^{2+} (aq) + 4H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is reversible but the second is one-way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, my teacher stated that ligand exchange reactions are irreversible... this confused me further&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please tell me what is right...&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="10185" LastEditDate="2015-05-31T16:40:41.903" LastActivityDate="2015-05-31T16:40:41.903" Title="Ligand exchange reactions... are they one way or reversible?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8102" PostTypeId="2" ParentId="8101" CreationDate="2014-01-27T15:15:45.820" Score="3" Body="&lt;p&gt;Let's take the binding of $\ce{O2}$ and $\ce{CO}$ to hemoglobin as example. $\ce{CO}$ does bind much better than $\ce{O2}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can consider the reaction of oxygen-loaded hemoglobin with $\ce{CO}$ a ligand exchange reaction. If that would be completely irreversible, the logical consequences in the case of an intoxication would be: [...] six feet under ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, intoxicated patients are treated with oxygen at higher pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, that sounds like &lt;strong&gt;Le Chatelier&lt;/strong&gt; in the ICU and it only makes sense in the case of equilibria.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-27T15:15:45.820" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8107" PostTypeId="1" CreationDate="2014-01-27T16:43:22.587" Score="5" ViewCount="893" Body="&lt;p&gt;I am not a pro, but I have some experience with gaussian. For example, I can optimize a structure, calculate the frequency and rotational energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lab I made a nucleophilic substitution reaction where the Nucleophile $\ce{X}$ reacted with $\ce{R-\text{alkyl}-CH2-I}$ to form $\ce{R-\text{alkyl}-CH2-X}$ (iodide was eliminated). However, my molecule has on another position a benzylic bond &quot;$\ce{R-Ph-CH2-O-R}$&quot; as well and I noticed that a side reaction can occur at that position (depends on the nucleophile $\ce{X}$) where the nucleophile $\ce{X}$ reacted with &quot;$\ce{R-Ph-CH2-O-R}$&quot; to form &quot;$\ce{R-Ph-X}$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are two reactions: (1) The normal reaction with the alkyl $\ce{R-\text{alkyl}-CH2-I}$ to form $\ce{R-\text{alkyl}-CH2-X}$ and the side reaction (2) where the benzylic position in $\ce{R-Ph-CH2-O-R}$ is cleaved by the nucleophile $\ce{X}$ to form $\ce{R-Ph-X}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now trying to understand why the nucleophile $\ce{X}$ reacted with a benzylic $\ce{CH2}$ group at all even when its just a 15% conversion rather than the $\ce{R-\text{alkyl}-CH2-\text{Halogen}}$ group which is the favoured reaction. I think this could be illustrated by MOs, what do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already optimized the structure of my molecule and generated the formatted checkpoint file to display the molecular orbitals with GaussView. When I actually look at the HOMO and LUMO I don't get any useful information. At the benzylic position there is not really a high coefficient (no big MO) and at the $\ce{CH2}$ atom of &quot;$\ce{R-CH2-I}$&quot; there is even a smaller coefficient.&#xA;I had expected to see at least a big MO at the carbon atom of the alkyl $\ce{R-CH2-I}$ - where the nucleophilic reaction is favoured - but thats not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, how can I analyze the reaction behaviour with DFT calculations, especially with Gaussian? I think calculating transition states is to much. It would be nice to just get some useful information about the molecules reactive sides (electrophilic sides where nucleophiles can react). How can I do that? Which calculation should I make?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way I used this B3LYP basis set:&#xA;b3lyp/6-31g(d,p) nosymm geom=connectivity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4322" LastEditorUserId="40029" LastEditDate="2017-06-14T13:54:02.330" LastActivityDate="2017-06-14T13:54:02.330" Title="DFT optimization of a molecule in gaussian and find electrophilic sides" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2014-02-28T18:54:23.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="8108" PostTypeId="1" CreationDate="2014-01-27T16:47:05.470" Score="2" ViewCount="310" Body="&lt;p&gt;I'm reading about ionic crystal structures with planes of close-packed anions and the interstitial sites for cations. There are two types of interstitial sites: tetrahedral and octahedral. That makes perfect sense. What confuses me is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;for each of these anion spheres, one octahedral and two tetrahedral positions exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where does this come from? It would seem that for every new anion sphere added, there is a single new tetrahedral position and a single new octahedral position.&lt;/p&gt;&#xA;" OwnerUserId="4309" LastActivityDate="2014-01-27T16:47:05.470" Title="Why is there one octahedral and two tetrahedral positions for every anion in a close packed structure?" Tags="&lt;crystal-structure&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8109" PostTypeId="2" ParentId="8088" CreationDate="2014-01-27T17:49:11.750" Score="-1" Body="&lt;p&gt;Henry's law is given by the formula $p=K_\mathrm h\chi$. Here $p$ is partial pressure of the gas $K_\mathrm h$ is Henry's constant for that gas and $\chi$ is mole fraction.Using this you can find it.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="5026" LastEditDate="2018-02-27T02:38:02.613" LastActivityDate="2018-02-27T02:38:02.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8111" PostTypeId="1" AcceptedAnswerId="8112" CreationDate="2014-01-27T18:10:02.417" Score="2" ViewCount="3213" Body="&lt;p&gt;There is an R an organic group connected to either Oxygen or Nitrogen. R is such a group which has almost zero electronic effects (Resonance effect and inductive effect). I want to ask if N or O are to donate a lone pair then which one will have higher donating capacity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my opinion  it should be Oxygen as it is more electronegative than Nitrogen but I'm still not confident and want some proof considering other factors also and neither do I have an answer.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="7448" LastEditDate="2015-12-26T21:43:58.487" LastActivityDate="2015-12-26T21:43:58.487" Title="Comparision of donation power" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8112" PostTypeId="2" ParentId="8111" CreationDate="2014-01-27T23:44:49.243" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...According to my opinion it should be Oxygen as it is more electronegative than Nitrogen...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's correct that oxygen is more electronegative than nitrogen. However, this makes an oxygen group a poorer electron donating group than the corresponding nitrogen group. Electronegativity means that the atom is less willing to give up the electrons. Nitrogen is more willing to give up (donate) its electrons because it is less electronegative.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-01-27T23:44:49.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8113" PostTypeId="1" CreationDate="2014-01-28T00:41:34.570" Score="3" ViewCount="110" Body="&lt;p&gt;Has anybody synthesized di- or higher nitrated cyclopentadienes possibly with methyl or other groups in the 1 position (analogs of TNT)?&lt;/p&gt;&#xA;" OwnerUserId="4326" LastEditorUserId="7448" LastEditDate="2015-10-09T16:28:49.873" LastActivityDate="2015-10-09T16:28:49.873" Title="Are highly nitrated cyclopentadiene compounds possible?" Tags="&lt;synthesis&gt;&lt;explosives&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8114" PostTypeId="1" CreationDate="2014-01-28T03:16:28.953" Score="-3" ViewCount="1766" Body="&lt;p&gt;How can I compare the leaving group ability of $\ce{Br^+}$ and $\ce{Cl^+}$ ions for the electrophilic substitution reactions for aromatic compounds?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorDisplayName="user15489" LastEditDate="2015-05-08T08:45:02.863" LastActivityDate="2015-05-08T08:45:02.863" Title="Which is better leaving group" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ClosedDate="2015-05-11T04:16:19.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="8116" PostTypeId="1" AcceptedAnswerId="8121" CreationDate="2014-01-28T05:16:30.193" Score="4" ViewCount="10247" Body="&lt;p&gt;Can two precipitates form in a precipitation reaction that starts with only two compounds in water?&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="5026" LastEditDate="2018-04-23T04:29:25.687" LastActivityDate="2018-04-23T04:29:25.687" Title="Can two precipitates form in a precipitation reaction that starts with only two compounds in water?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2014-01-30T12:46:10.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="8117" PostTypeId="2" ParentId="8114" CreationDate="2014-01-28T07:00:35.763" Score="5" Body="&lt;p&gt;I presume you're referring to $\ce{Br-}$ and $\ce{Cl-}$ in aromatic nucleophilic substitution reactions. In that case, $\ce{Br-}$ is typically the superior leaving group to $\ce{Cl-}$, due to the former's larger size lending it greater polarizability (i.e., the ability to disperse the negative charge over a greater area). In general, when comparing elements in the same group, the dominant factor in determining anion stability is the larger size as one moves down the group. This largely accounts for the order of acidity of the halogen-containing binary acids (i.e., $\ce{HI} &amp;gt; \ce{HBr} &amp;gt; \ce{HCl}$), the greater acidity of thiols over alcohols, etc. Solvent effects are relevant, since smaller anions are more easily and thoroughly solvated in protic solvents. However, as far as I'm aware, this effect is rarely decisive, and $\ce{Br-}$ usually remains the better leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the rate-determining step in nucleophilic aromatic substitution reactions is not the loss of the leaving group, but rather the formation of the negatively charged &lt;a href=&quot;http://en.wikipedia.org/wiki/Meisenheimer_complex&quot;&gt;Meisenheimer complex&lt;/a&gt; resulting from the initial nucleophilic attack. This intermediate is extremely high in energy due to the temporary loss of aromaticity it entails. Thus, the reaction is largely controlled by kinetics, and therefore dependent upon the stability of the intermediates and transition states. $\ce{Cl-}$ is more electronegative than $\ce{Br-}$, and therefore better stabilizes the negative charge in the intermediate complex by induction ($-I$ effect). Consequently, we can expect the chlorobenzene to be more reactive in practice under most conditions, with the disparity in leaving group quality being less important. A cursory search yielded &lt;a href=&quot;http://chemconnections.org/organic/chem227/Nu-Subs/Nu-arom-subs-ques-09.pdf&quot;&gt;one reference&lt;/a&gt; confirming this order, but unfortunately I don't have access at the moment to more authoritative primary sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, while $\ce{Br-}$ is the superior leaving group, its loss is not the rate-determining step. Hence, chlorobenzenes will in fact be more reactive due to chlorine's superior ability to inductively stabilize the intermediate anionic complex.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-01-28T07:00:35.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8118" PostTypeId="2" ParentId="8113" CreationDate="2014-01-28T08:01:03.040" Score="4" Body="&lt;p&gt;Mono-nitrated cyclopentadienes and their benzo-analogues (indene, fluorene) have been prepared in the context of studies on &lt;em&gt;nitro-aci&lt;/em&gt;-tautomerism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as higher substitution is concerned, there seems to be at least a crystal structure of &lt;a href=&quot;http://dx.doi.org/10.1107/S0567740872004637&quot; rel=&quot;nofollow&quot;&gt;potassium dinitrotrichlorocyclopentadienide&lt;/a&gt; reported in &lt;em&gt;Acta Cryst., B&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the abstract, the authors (Y. Otaka, F. Marumo, and Y. Saito) mention the preparation of this compound by &lt;strong&gt;Haga&lt;/strong&gt; in 1969, (sorry, reference is behind paywall) via &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;nitration of hexachlorocyclopentadiene with alkali nitrite in an aqueous or ethanolic solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NOTE: Hexachlorocyclopentadiene once was a favoured starting material in the synthesis of insecticides via Diels-Alder reaction. However, it is a &lt;strong&gt;potent teratogen&lt;/strong&gt; and the &quot;mother&quot; of least five members of the &lt;strong&gt;dirty dozen&lt;/strong&gt; (persistent organic pollutants) nowadays banned for good reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to read the MSDS and then decide that you'd rather prefer to work with something else.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-28T08:32:35.553" LastActivityDate="2014-01-28T08:32:35.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8119" PostTypeId="2" ParentId="5737" CreationDate="2014-01-28T08:20:49.640" Score="8" Body="&lt;p&gt;The pH scale is taken on our reference as 0 to 14 for concentration values from $1~\mathrm{M}$ to $\mathrm{10}^{-14} \, \mathrm{M}$. This range is such that our normal calculations in lab can be worked out easily. It is to be noted that this scale is at room temperature. If you increase the temperature then the limits changes. For example pH of pure water at $100 \, \mathrm{^\circ C}$ is $6.14$ and not $7$. Hence we can see that the scale has shifted with temperature.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="7475" LastEditDate="2015-10-20T12:07:09.547" LastActivityDate="2015-10-20T12:07:09.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8120" PostTypeId="2" ParentId="8116" CreationDate="2014-01-28T08:32:14.533" Score="0" Body="&lt;p&gt;Yes is the answer to your question provided that reactant two molecules are salts.Two salts can react with eact other and form two new stable salts depending on their $\mathrm{K}_{sp}$ values.&lt;/p&gt;&#xA;" OwnerUserId="3943" LastActivityDate="2014-01-28T08:32:14.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8121" PostTypeId="2" ParentId="8116" CreationDate="2014-01-28T08:49:55.350" Score="6" Body="&lt;p&gt;The number of examples is considerably great.&#xA;Consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ba(OH)2(aq) + MgSO4(aq)  -&amp;gt; BaSO4(s) + Mg(OH)2(s)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above reactants are reasonably soluble in water.&#xA;From Wikipedia, $\ce{Ba(OH)2} $ has a solubility of 1.67 g / 100 mL at 20 °C and $\ce{MgSO4} $ a solubility of 25.5 g / 100 mL at 20 °C&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2014-01-28T08:49:55.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8124" PostTypeId="1" CreationDate="2014-01-28T12:57:41.383" Score="6" ViewCount="447" Body="&lt;p&gt;Without appealing to reactions that require photons / light, how do chemists achieve a very high concentration of continually produced, though quite short lived, singlet oxygen in aqueous solution (i.e. $\ce{{}^1O_2}$ where dioxygen is typically in the triplet $\ce{{}^3O_2}$ state)?  I suppose I'll also ask the same question if we have the freedom to select an organic solvent of choice for the buffer?  By &quot;high concentration&quot; I mean at concentration from $\approx 10~\mathrm{\mu{}M} - 100(+)~\mathrm{\mu{}M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess would be that should be a way to do this starting with a solution containing some amount of $\ce{H_2O_2}$, but I'm not quite sure what a reasonable process might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Singlet_oxygen&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; the wikipedia link for singlet oxygen ($\ce{{}^3O_2}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction should happen at or around room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I'm specifying that I'd like to do this in the dark, and to have the singlet oxygen continually produced at room temperature (given its short lifetime, which is $\leq 100~\mathrm{\mu{}s}$ under any conditions I've heard of), is because I'm curious if I could easily set up some buffer that lets me dose a sample with singlet oxygen on a lab bench with a pipette.  Given the short lifetime of singlet oxygen, having to use a continuous radiation source presents difficulties that, while surmountable, lead me to ask this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided the above, I'd like to avoid reactions that vent gases like $\ce{Cl_2}$ and require fume hoods in the sense of &quot;... well, if you value your life ...&quot; vs. &quot;... OSHA says so ...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4332" LastEditorUserId="4945" LastEditDate="2014-07-30T04:46:38.573" LastActivityDate="2017-01-15T21:16:58.400" Title="How do I create a concentrated solution of singlet oxygen (in the dark)?" Tags="&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;concentration&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8125" PostTypeId="2" ParentId="8124" CreationDate="2014-01-28T13:08:04.300" Score="3" Body="&lt;p&gt;Potassium ferricyanide (and other ion complexes, like hemoglobin) will decompose hydrogen peroxide into water and (presumably) singlet oxygen. This process is necessary for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Luminol#Chemiluminescence&quot; rel=&quot;nofollow&quot;&gt;luminol reaction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O2 \overset{Fe}{-&amp;gt;} 2H2O + ^1O2}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-01-28T13:08:04.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8126" PostTypeId="1" AcceptedAnswerId="8782" CreationDate="2014-01-28T13:11:33.190" Score="5" ViewCount="3321" Body="&lt;p&gt;I know the equations, but I have some difficulty getting the difference between the two concepts of &lt;strong&gt;binding energy&lt;/strong&gt; and &lt;strong&gt;adsorption energy&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4333" LastEditorUserId="3846" LastEditDate="2014-02-27T20:19:57.747" LastActivityDate="2015-11-20T13:02:22.850" Title="What is the (conceptual) difference between binding energy and adsorption energy?" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8127" PostTypeId="2" ParentId="8124" CreationDate="2014-01-28T13:17:24.500" Score="2" Body="&lt;p&gt;I'd always go by the old photochemical Schenck method, but since you insist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal decomposition of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Phosphonite ozonides or &lt;/li&gt;&#xA;&lt;li&gt;9,10-diphenylanthracene endoperoxide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;yield $\ce{^1O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, $\ce{^1O2}$, may also be generated at low oxygen pressure (i a vacuum line) by microwave radiation. (&lt;a href=&quot;http://dx.doi.org/10.1016/1010-6030%2888%2980031-5&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as aqueous solutions are concerned, one should keep in mind that the lifetime of singlet oxygen ($^1\Delta$) is around 2 µs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For detection, one typically relies on the luminescence at $\lambda$ = 1270 nm.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-28T21:10:26.440" LastActivityDate="2014-01-28T21:10:26.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8131" PostTypeId="1" CreationDate="2014-01-28T15:16:46.933" Score="2" ViewCount="126" Body="&lt;p&gt;Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum? Asking as a pharmacologist not a patient.&lt;/p&gt;&#xA;" OwnerUserId="4336" LastEditorUserId="4336" LastEditDate="2014-01-28T15:22:45.287" LastActivityDate="2014-01-28T18:50:18.103" Title="Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum" Tags="&lt;pharmacology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8133" PostTypeId="1" CreationDate="2014-01-28T15:55:27.033" Score="4" ViewCount="6100" Body="&lt;p&gt;Two vessels of the same volume, in the&#xA;same location and open to the atmosphere. One is of a small diameter, the&#xA;other is of a larger diameter.&#xA; The question is will the vessel with the&#xA;larger surface area evaporate the liquid&#xA;faster than the smaller one?&#xA;&lt;strong&gt;If not!&lt;/strong&gt; I need answer to other question ,if i am not wrong.&#xA;&lt;strong&gt;Why&lt;/strong&gt; does vapour pressure decreases with decrease in surface area?&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="1431" LastEditDate="2014-02-02T02:34:16.083" LastActivityDate="2017-04-14T10:51:17.700" Title="Does rate of Evaporation change with surface area?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8134" PostTypeId="2" ParentId="7486" CreationDate="2014-01-28T16:05:50.207" Score="17" Body="&lt;p&gt;This is formally a manganese(VII) compound and hence there are no 3d electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral &quot;hybridisation&quot; can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the &quot;hybridisation&quot; is best considered as $\mathrm{d^3s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put &quot;hybridisation&quot; in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure! &lt;/p&gt;&#xA;" OwnerUserId="4337" LastEditorUserId="16683" LastEditDate="2016-08-03T21:22:10.743" LastActivityDate="2016-08-03T21:22:10.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8135" PostTypeId="2" ParentId="7094" CreationDate="2014-01-28T16:55:12.710" Score="3" Body="&lt;p&gt;I just wanted to show you how you could set up the equation if you don't know that :-). If you find an expression for the ionic charge of one side group as a function of pH, you're well on the way. You just do that for all the side groups, and add them together, and say that overall you want that to equal 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can solve for pH in a program that does such a thing. You can also graph this, and find the isoelectric point (pI) by visual inspection. I've done both below, mostly for fun, since oveoyas has provided a sufficient answer :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/au1R2.png&quot; alt=&quot;Plot of charge as function of pH&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="402" LastEditDate="2017-05-02T23:25:48.063" LastActivityDate="2017-05-02T23:25:48.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8136" PostTypeId="2" ParentId="8131" CreationDate="2014-01-28T17:14:23.390" Score="2" Body="&lt;p&gt;My professional knowledge of vitamin A derivatives is limited to preparative aspects and some photochemical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems that the German &lt;a href=&quot;http://www.bfarm.de/&quot; rel=&quot;nofollow&quot;&gt;Bundesinstitut für Arzneimittel und Medizinprodukte (BfArM)&lt;/a&gt; provides some information in german and english.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't know whether a link to the search would survive the session, so the link above is just to their website and you would need to search yourself in case you're interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;Note: I had a look at the &lt;em&gt;Fachinformationen&lt;/em&gt; (RTF in German) among the documents &quot;linked&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this publication, it is very explicitly pointed out that Isotretinoin is a &lt;strong&gt;TERATOGEN&lt;/strong&gt; and efficient contraception therefore is absolutely vital (in the truest sense of the word).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is suggested that patients use &lt;strong&gt;two independent&lt;/strong&gt; methods of contraception, of which one should be a &quot;barrier method&quot;, e.g. the use of condoms.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-28T18:50:18.103" LastActivityDate="2014-01-28T18:50:18.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8137" PostTypeId="1" AcceptedAnswerId="8138" CreationDate="2014-01-28T17:54:19.980" Score="4" ViewCount="2172" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What are the steps in calculating the photosynthetic rate (&lt;span class=&quot;math-container&quot;&gt;$\pu{\mu mole}~ \ce{O2} \pu{m2/ min}$&lt;/span&gt;) for a leaf that has an area of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0025 m2}$&lt;/span&gt; and a &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; evolution&#xA;rate of &lt;span class=&quot;math-container&quot;&gt;$0.096~\ce{O2}/\pu{min}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To do so , the first step is to convert it to &lt;span class=&quot;math-container&quot;&gt;$\mu L$&lt;/span&gt; i.e. &lt;span class=&quot;math-container&quot;&gt;$\pu{960 \mu L} ~\ce{O2}/\pu{min}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The second step should be to convert it to &lt;span class=&quot;math-container&quot;&gt;$\pu{\mu moles}$&lt;/span&gt;. So, how do we do that ?&lt;/p&gt;&#xA;&lt;p&gt;EDIT : In my textbook, the following calculation has been done but I am not able to understand it...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{960 \mu L}~\ce{O2}~\pu{/L} / [(273+23)/273) \times 22.423]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$22.423$&lt;/span&gt; is the volume of a mole of gas and &lt;span class=&quot;math-container&quot;&gt;$\pu{273K}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{0^\circ C}$&lt;/span&gt; but I am not able to understand how that fits here.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="69830" LastEditDate="2021-05-05T20:02:35.397" LastActivityDate="2021-05-05T20:02:35.397" Title="Photosynthetic rate calculation" Tags="&lt;biochemistry&gt;&lt;concentration&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8138" PostTypeId="2" ParentId="8137" CreationDate="2014-01-28T18:31:19.427" Score="5" Body="&lt;p&gt;Using the general case equation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{P_1V_1 }{T_1} = \frac{P_2 V_2} {T_2}$$&lt;/span&gt;&#xA;and given that pressure is constant at &lt;span class=&quot;math-container&quot;&gt;$\pu{1  atm}$&lt;/span&gt; in your experiment, the equation simplifies to&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{V_1 }{T_1} = \frac{ V_2} {T_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We know &lt;span class=&quot;math-container&quot;&gt;$V_1=\pu{960 \mu L}$&lt;/span&gt; and it appears that the experiment is being run at &lt;span class=&quot;math-container&quot;&gt;$\pu{(273+23)K}$&lt;/span&gt;, so we need to calculate what the volume (&lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt;) would be at standard T (&lt;span class=&quot;math-container&quot;&gt;$\pu{273 K}$&lt;/span&gt;).  This is given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_2 = \frac{V_1 T_2}{T_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;if we next divide &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$\pu{22.423E6 \mu~L}$&lt;/span&gt;, we now know how many moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; we have.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="69830" LastEditDate="2021-05-05T19:52:39.997" LastActivityDate="2021-05-05T19:52:39.997" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8139" PostTypeId="2" ParentId="8133" CreationDate="2014-01-28T19:41:31.283" Score="3" Body="&lt;p&gt;It depends what you mean by &quot;rate&quot;. If the question is in which vessel does the liquid disappear fastest, then the answer is yes. But some people might assume that rate means rate &lt;em&gt;per unit area&lt;/em&gt; in which case the answer would be no as that just depends on the vapour pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty easy to observe the effect with any volatile liquid.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-28T19:41:31.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8140" PostTypeId="1" AcceptedAnswerId="8144" CreationDate="2014-01-28T19:44:53.330" Score="11" ViewCount="689" Body="&lt;p&gt;What is a good and free software for this endeavor? Any recommended articles on the subject?&#xA;I want to design dft functionals, trying out different parametrizations and optimizing mixing constants for exchange and correlation contributions for specific data sets. Even add PT2 perturbation energy into the mix if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of Gaussian's user-defined functional options, but would like to have a bit more control in the parameters and contributions, instead of just 6 parameters.&lt;/p&gt;&#xA;" OwnerUserId="4340" LastActivityDate="2014-10-22T15:19:08.413" Title="Defining and testing custom made DFT functionals?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8141" PostTypeId="2" ParentId="785" CreationDate="2014-01-28T20:03:27.477" Score="4" Body="&lt;p&gt;Actually, it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, though, you won't notice it as it will be heavily diluted by the nitrogen and won't ever reach a concentration large enough to worry about. It's different if you pull air through a trap as the oxygen content of the flowing gas will be condensed preferentially leaving a liquid that is concentrated in oxygen. This is really dangerous if there is any organic material in the trap as well which is why those of who have worked with such systems are so wary of pale blue liquids in our traps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if you like dangerous experiments, you can do some that illustrate the presence of oxygen. If you cut of the top of an aluminium can and half fill it with liquid nitrogen, then wait until most has evaporated, the remaining liquid will be mostly oxygen. The safe way to show this is to use a magnet on a stick to pull some out (oxygen is fairly strongly paramagnetic). Or, much less safely, you could drop a lit match in (this is very dangerous as the can will probably burn violently). There is even a &lt;a href=&quot;http://www.youtube.com/watch?v=KlAl5RHAAns&quot; rel=&quot;nofollow&quot;&gt;youtube demonstration&lt;/a&gt; of this (I think it is number 5 of 7 &quot;super cool&quot; demonstrations with liquid nitrogen). I think they used to have a version with burning but may have taken it off for safety reasons.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-01-28T20:03:27.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8142" PostTypeId="2" ParentId="8140" CreationDate="2014-01-28T20:13:45.910" Score="3" Body="&lt;p&gt;Since you mentioned &quot;free&quot;, you maybe want to have a look at &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;nofollow&quot;&gt;GAMESS&lt;/a&gt; from the Gordon group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have really no idea whether it's fit for the task, but last time I checked it was free for academic use.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's of course &lt;a href=&quot;http://www.cosmologic.de/&quot; rel=&quot;nofollow&quot;&gt;Turbomole&lt;/a&gt;, but according to an &lt;a href=&quot;http://www.turbo-forum.com/index.php?topic=120.0&quot; rel=&quot;nofollow&quot;&gt;old thread in a forum&lt;/a&gt; from 2007, they weren't fond of custom functionals either at that time. The statement of one of the developers might be worth reading anyway.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-28T20:32:47.583" LastActivityDate="2014-01-28T20:32:47.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8143" PostTypeId="1" AcceptedAnswerId="8147" CreationDate="2014-01-28T20:18:51.310" Score="4" ViewCount="296" Body="&lt;p&gt;From my perspective these resonance structures allow these specific polyatomic ions to act as donor-acceptor molecules. Many donor-acceptor molecules also tend to be conjugated systems because they have chains of alternating conjugated &amp;#x3c0; orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does this imply that these ions (due to their resonance structures) can act as though they had a conjugated system or am I making too big of an assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curious on others perspectives/if I'm totally of my rocker.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="7475" LastEditDate="2016-05-28T21:25:11.453" LastActivityDate="2016-05-28T21:25:11.453" Title="Can polyatomic ions (CO₃, PO₄, SO₄, NO₃) be considered conjugated systems?" Tags="&lt;inorganic-chemistry&gt;&lt;resonance&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8144" PostTypeId="2" ParentId="8140" CreationDate="2014-01-28T20:43:26.460" Score="14" Body="&lt;p&gt;Depending on what you're after, you might want to look for programs using &lt;a href=&quot;http://www.tddft.org/programs/octopus/wiki/index.php/Libxc&quot;&gt;libxc&lt;/a&gt;, so you can completely define your own functionals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the libxc section of the Octopus wiki, these include:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.abinit.org/&quot;&gt;Abinit&lt;/a&gt; – plane-wave code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.tddft.org/programs/APE&quot;&gt;APE&lt;/a&gt; – an atomic code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://quantumwise.com&quot;&gt;Atomistix ToolKit&lt;/a&gt; – numerical orbitals code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.wfu.edu/~natalie/papers/pwpaw/man.html&quot;&gt;AtomPAW&lt;/a&gt; – projector augmented wave functions generator&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://inac.cea.fr/L_Sim/BigDFT/&quot;&gt;BigDFT&lt;/a&gt; – wavelet code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.cp2k.org/&quot;&gt;CP2K&lt;/a&gt; -  A program to perform atomistic and molecular simulations of solid state, liquid, molecular, and biological systems.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.dp-code.org/&quot;&gt;DP&lt;/a&gt; – Dielectric Properties, a linear response TDDFT code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://elk.sourceforge.net/&quot;&gt;Elk&lt;/a&gt; – FP-LAPW code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://erkale.googlecode.com&quot;&gt;ERKALE&lt;/a&gt; – a DFT/HF molecular electronic structure code based on Gaussian orbitals&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://exciting-code.org/&quot;&gt;exciting&lt;/a&gt; – FP-LAPW code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;https://wiki.fysik.dtu.dk/gpaw&quot;&gt;GPAW&lt;/a&gt; – grid-based projector-augmented wave method&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://sourceforge.net/p/jdftx/wiki/Home/&quot;&gt;JDFTx&lt;/a&gt; – plane-wave code designed for Joint Density Functional Theory&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://code.google.com/p/molgw/&quot;&gt;MOLGW&lt;/a&gt; - a small, but accurate MBPT code for molecules&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.tddft.org/programs/octopus/wiki/index.php/Main_Page&quot;&gt;octopus&lt;/a&gt; – real-space (TD)DFT code&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.yambo-code.org/&quot;&gt;Yambo&lt;/a&gt; – solid state and molecular physics many-body calculations code&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On a more personal note, and this is just my opinion, if you're just arbitrarily mixing functionals, you should really make sure you have a scientific justification for doing this, otherwise you're just generating numbers without basis.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-01-28T20:43:26.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8145" PostTypeId="2" ParentId="7595" CreationDate="2014-01-28T21:19:17.507" Score="1" Body="&lt;p&gt;The standard electrode potentials are measured in their standard states when the concentration of the electrolyte solutions are fixed as $1M$ and temperature is $298K$. However, in actual practice electrochemical cells don't have always fixed concentrations of the electrolyte solutions. The electrode potentials depend on the concentration of the electrolyte solutions. Nernst gave a relationship between electrode potentials and the concentration of the electrolyte solutions known as Nernst equation. &lt;em&gt;Even the electrode potential which you have posted (assuming they are not standard potentials) is obtained by Nernst equation, which shows the dependence of electrode potential on concentration (in other words molar coefficients)&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-01-28T21:19:17.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8146" PostTypeId="2" ParentId="8124" CreationDate="2014-01-28T22:13:25.330" Score="2" Body="&lt;p&gt;Unraveling a stoichiometric organic adduct &lt;I&gt;in situ&lt;/I&gt; is the way to go.  However,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O2 + NaOCl → O2(a1Δg) + NaCl + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is one thing to mix streams of supermarket $\ce{H2O2}$ and laundry bleach to get a dimer nice red glow in a darkened room.  It is quite another thing to mix higher concentrations of each.  Concentrated $\ce{H2O2}$ is exceptionally dangerous on skin!  It diffuses in until it hits living tissue with peroxidase.  It then pops into subcutaneous gas bubbles attended with remarkable pain.  Concentrated hypochlorite is evil, including conproportionation with chloride to chlorine gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full safety equipment: face shield, gloves, apron - and remember to shield your neck.  In a running fume hood with the door down.  Get literature references before you even think about doing it.  A hefty dual barrel syringe pump feeding a static mixer might barely pull it off. Keep the reacting volume small as you rocket through the juices.  You might take singlet oxygen somewhere outside the reaction zone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 mph = 4.47 mm/100 microsecond half-life&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="485" LastEditDate="2014-07-30T05:42:11.957" LastActivityDate="2014-07-30T05:42:11.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8147" PostTypeId="2" ParentId="8143" CreationDate="2014-01-28T23:19:25.020" Score="2" Body="&lt;p&gt;&quot;Conjugated&quot; implies a 1,3-shift to move an electron or hole.  It is a remnant of LCAO modeling that is obviously wrong but fantastically useful short of the Woodward-Hoffmann rules.  I'm not sure inorganikers would like the name as such. MO modeling is accurate but unwieldly.  &quot;Delocalized&quot; or &quot;resonance hybrid&quot; is good, certainly for inorganic systems that may have (virtual) d-orbital participation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider nitrate.  Inorganikers would say $\ce{N^{5+}}$, $\ce{^{-}O-N(=O)2}$, with five bonds to the nitrogen.  The negative charge 1,3-shifts around all three oxygens.  Organikers would see it as $\ce{N^{3+}}$, $\ce{[^{-}O{}-]_2N^{+}=O}%edit$, with four bonds to the nitrogen.  1,3 shifts, etc.  Is nitrate ever a bidentate ligand?  YES!  But it is just another resonance structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inorg. Chem. 35(24) 6964 (1996) &#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic960587b&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ic960587b&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="24052" LastEditDate="2016-05-27T22:18:15.270" LastActivityDate="2016-05-27T22:18:15.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8148" PostTypeId="2" ParentId="785" CreationDate="2014-01-28T23:34:56.607" Score="2" Body="&lt;p&gt;Oxygen can and does condense into liquid nitrogen dewars. Good lab practice dumps vacuum line dewars Friday afternoon, refills and restarts Monday morning. Remember – the universe hates you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dumping old liquid nitrogen in the parking lot can be fun. LOX plus asphalt paving (or wooden floors) is a contact explosive for days. Charcoal briquettes plus LOX was a blasting combo in coal mines. NEVER pour LOX on burning charcoal briquettes. The fireball burns so hot you can get UV burns, assuming you have any skin remaining. Steel wool plus LOX is a thermal lance, minimum 2730 °C, sometimes claimed to be 4500 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take care that anything you wear when working with cryogens can be removed very fast – especially porous insulating gloves.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T09:06:35.503" LastActivityDate="2015-05-01T09:06:35.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8149" PostTypeId="2" ParentId="8073" CreationDate="2014-01-28T23:55:21.090" Score="6" Body="&lt;p&gt;Naively, the nuclear electric field at Z ~ 137 or greater, reciprocal of the Fine Structure constant, would &quot;spark the vacuum.&quot;  Vacuum would be torn into electron-positron pairs.  The electrons go in to convert protons to neutrons plus neutrinos.  As stated above, non-classical treatment suggests we'll never get near a cold nucleus that sparks the vacuum.  RHIC and LHC rip the vacuum hot by colliding deeply relativistic gold or lead nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big problem in making new heavy elements is getting enough neutrons in there while colliding the heaviest nucleus with the lightest to do the job.  Fusion charge repulsion is the product of the two charges.  It must be minimized.  Ca-48 is stable for a footnote, and it's run its course for fusion with available transuranics.  The isotopes produced are way too neutron deficient to stick around. One might do something clever with a big custom-configured H-bomb - lots of compression and neutron density there - but sample retrieval is problematic.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-28T23:55:21.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8150" PostTypeId="1" AcceptedAnswerId="8151" CreationDate="2014-01-28T23:56:22.277" Score="1" ViewCount="891" Body="&lt;p&gt;How can you make an unfavorable reaction go backwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding if we wanted a reaction at equilibrium to go backwards (i.e. form reactants from products), we would add more product into the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how the reaction being an unfavorable one changes the above statement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After making the reaction go backwards, we have changed the reaction rate going backwards correct?&lt;/p&gt;&#xA;" OwnerUserId="4343" LastActivityDate="2014-01-29T00:20:26.083" Title="Unfavorable reactions" Tags="&lt;equilibrium&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8151" PostTypeId="2" ParentId="8150" CreationDate="2014-01-29T00:20:26.083" Score="1" Body="&lt;p&gt;If the process we're discussing is an equilibrium, and if the forward reaction is &quot;unfavorable&quot; (I take that to mean that the products are of higher energy then the reactants), then there are still ways to shift the equilibrium in whichever direction you want.  For example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &amp;lt;-&gt; B + C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using LeChatelier's principle, if we increase the pressure in our vessel, we will push the equilibrium back towards A, the opposite will occur if we decrease the pressure.  Temperature, catalysts, reagent concentrations, etc. can all affect the position of an equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-29T00:20:26.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8152" PostTypeId="2" ParentId="5885" CreationDate="2014-01-29T00:24:27.490" Score="1" Body="&lt;p&gt;Look up the industrial process.  Organic essences are usually steam distilled or pot-to-pot durped, or liquid CO2 or liquidn propane/butane extracted.  We're not making hash oil, are we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fao.org/inpho_archive/content/documents/vlibrary/ad420e/img/AD420e26.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.fao.org/inpho_archive/content/documents/vlibrary/ad420e/img/AD420e26.gif&lt;/a&gt; &#xA;&lt;a href=&quot;http://www.seanmichaelragan.com/img/radial_alembic_diagram.png&quot; rel=&quot;nofollow&quot;&gt;http://www.seanmichaelragan.com/img/radial_alembic_diagram.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no aspirin in willow bark, there is salicylic acid.  Since you can buy everything at high purity for less than small scale process costs, unless you own a planted acre, why bother?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two differences between &quot;natural&quot; and &quot;synthetic.&quot; 1) If the molecule is chiral, the synthetic product will be racemic. R-(–)-carvone is spearmint, its mirror image S-(+)-carvone is caraway.  Big difference. 2) &quot;Natural&quot; contains radioactive carbon-14: 540 Bq/kg carbon.  Anything made from petroleum or coal is long since decayed cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignorance is not a form of knowing things, nor is faith.  The only empiricially valid solutions are engineering solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.erowid.org/archive/rhodium/chemistry/equipment/pictures/nomograph.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.erowid.org/archive/rhodium/chemistry/equipment/pictures/nomograph.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T00:24:27.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8155" PostTypeId="1" AcceptedAnswerId="8156" CreationDate="2014-01-29T01:34:01.290" Score="2" ViewCount="2842" Body="&lt;p&gt;Correct me If I am wrong, given a reaction at equilibrium, if we say add more concentration of products to the reaction, the rate going backwards would increase to try to get back to equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the rate forward stays the same or slows down?&lt;/p&gt;&#xA;" OwnerUserId="4343" LastActivityDate="2014-01-29T17:41:29.037" Title="Equilibrium rate constants forward and backwards" Tags="&lt;equilibrium&gt;&lt;reaction-control&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8156" PostTypeId="2" ParentId="8155" CreationDate="2014-01-29T02:00:32.443" Score="3" Body="&lt;p&gt;In simple cases, the rate is equal to the rate constant &quot;k&quot; times the concentration.  The forward rate constant and the backward rate constant don't change, but the rate of change (rate constant times concentration) will change as the concentration of a reactant or product changes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-29T02:00:32.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8157" PostTypeId="1" AcceptedAnswerId="8160" CreationDate="2014-01-29T02:08:11.040" Score="3" ViewCount="885" Body="&lt;p&gt;I live in the tropics, and have a water tank that sits in the sun, forming algae. Every year, I need to scrub the thing out. The plumber I’ve hired for the job wants to flush my water pipes with HCl to kill any residual algae. The pipes are mostly PVC, with braided metal hose and metal fittings here and there. Should I let him? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="1027" LastEditorUserId="5591" LastEditDate="2015-11-13T19:23:36.520" LastActivityDate="2019-10-03T14:49:02.420" Title="Does HCl kill algae?" Tags="&lt;acid-base&gt;&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8158" PostTypeId="2" ParentId="8079" CreationDate="2014-01-29T02:13:14.110" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;home-made handwash (liquid soap) consists of bar soap, water, Epsom&#xA;  salts, and coconut oil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bar soap - OK.  Might be sodium or potassium fatty acid carboxlates, or syndet.&#xA;Epson salts - not OK.  Bivalent cations turn literal soap into insoluble slime.  Synets don't much like it either.&#xA;Coconut oil - not OK.  Why are you adding grease to soap?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supermarket, soap aisle, find something nice.  I once washed my hands with Dove bar soap.  It took a day to get them clean again.  Maybe that's what you want.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:13:14.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8159" PostTypeId="2" ParentId="8068" CreationDate="2014-01-29T02:22:01.197" Score="3" Body="&lt;p&gt;Single crystal diamond is literally a single molecule.  Every carbon is covalently bonded to the whole.  A modest level of impurities makes no difference.  Consider epoxy.  A well compounded, well measured, well mixed sample will crosslink into a single molecule with every atom bonded to the whole body.  Ionic crystals are collections of interacting formula units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond, graphite, and buckeyballs are not organic compounds, for they are not compounds. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:22:01.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8160" PostTypeId="2" ParentId="8157" CreationDate="2014-01-29T02:26:23.493" Score="4" Body="&lt;p&gt;Are they metal pipes?  If you want to kill algae as such, flush with a diluted quaternary ammononium disinfectant, let sit a day, then flush everything very well indeed, and again.  If you have rubber or polymer fittings, don't do that, either, for the stuff diffuses in.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:26:23.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8161" PostTypeId="2" ParentId="7974" CreationDate="2014-01-29T02:36:42.183" Score="2" Body="&lt;p&gt;One would control the composition of the electrodes (mercury, platinum, Ebonex, boron-doped diamond coating) and the potential of the cell so that the organic reaction went forward while the electrodes remained inert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal oxidation is facilitated by Lewis base ligands that drive reaction by stabilizing the ionic products.  Don't have them in there.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:36:42.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8162" PostTypeId="2" ParentId="7938" CreationDate="2014-01-29T02:42:31.210" Score="2" Body="&lt;p&gt;The sun is electrically neutral.  It is a roughly 70% hydrogen, 30% helium, and various other stuff as plasma - partially and wholly ionized nuclei in bath of electrons charge neutralizing things overall.  That does not change, for charge is conserved.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:42:31.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8163" PostTypeId="2" ParentId="7933" CreationDate="2014-01-29T02:49:06.393" Score="2" Body="&lt;p&gt;A gaseous flame is hot gas seeded with free radicals and ionized species, a dilute plasma.  Note that solid flames exist.  Mix finely divided nickel and aluminum, tantalum and graphite, etc., compress into a compacted body, and heat an edge very hot until things start.  A white hot interface propagates through the body as the components react.  I would hesitate to call that a condensed phase plasma.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T02:49:06.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8165" PostTypeId="2" ParentId="7875" CreationDate="2014-01-29T03:04:59.890" Score="1" Body="&lt;p&gt;Consider some zero dipole moment, &quot;obviously&quot; hydrophobic gases:  $\ce{CH4}$, $\ce{Xe}$, $\ce{SF6}$.  All three readily form water clathrate hydrates,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clathrate_hydrate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Clathrate_hydrate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T14:48:52.547" LastActivityDate="2015-05-03T14:48:52.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8166" PostTypeId="2" ParentId="8157" CreationDate="2014-01-29T03:07:03.217" Score="2" Body="&lt;p&gt;Clean the tank with bleach and flush it several times.  Once clean, put a piece of copper pipe in the tank and that should help it stay clean.  Don't use HCl, too strong for the job.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-01-29T03:07:03.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8168" PostTypeId="1" CreationDate="2014-01-29T04:58:22.750" Score="2" ViewCount="5678" Body="&lt;p&gt;I performed an experiment about the corrosion of iron. I used acid and base in the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the acid speeds up the corrosion of iron. But I don't understand why the base also corrodes iron. I can't find out the answer even after I've googled it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experiment I mix an ulcer drug into a glass of water, then I put in an iron nail. After I let it sit for a few days, the iron nail becomes rusty. I don't know of any reaction between a base and iron that gives rust.&lt;/p&gt;&#xA;" OwnerUserId="4346" LastEditorUserId="69830" LastEditDate="2021-05-11T05:16:15.320" LastActivityDate="2021-05-11T06:28:21.343" Title="How does iron rust in basic conditions?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8171" PostTypeId="2" ParentId="8155" CreationDate="2014-01-29T14:50:31.630" Score="1" Body="&lt;p&gt;Yes, if we add products, then the reverse reaction will try to make more reactants. If we take away products, then the forward reaction will try to make more products. Not on topic: If you increase pressure, the side with less moles will be favoured. And more things. This is all Le Chatelier's Principle.&lt;/p&gt;&#xA;" OwnerUserId="4352" LastEditorUserId="1431" LastEditDate="2014-01-29T17:41:29.037" LastActivityDate="2014-01-29T17:41:29.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8172" PostTypeId="1" CreationDate="2014-01-29T15:04:41.663" Score="0" ViewCount="268" Body="&lt;p&gt;If I have an intermetallic and want to know the specific arrangement of atoms in the lattice, how can I determine that? What I mean is the Cartesian (xyz) position, or the (abc) positions in the lattice. In my case, I am working with Ni3Sn4, which has a space group C2/m. The space group diagram can be found &lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/medium/012ay1.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4353" LastEditorUserId="1431" LastEditDate="2014-01-29T17:40:54.783" LastActivityDate="2014-03-01T00:46:36.287" Title="How does a space group determine crystal structure?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8173" PostTypeId="2" ParentId="8168" CreationDate="2014-01-29T20:21:17.640" Score="5" Body="&lt;p&gt;Iron can oxidize to &amp;quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow noreferrer&quot;&gt;rust&lt;/a&gt;&amp;quot; (which is actually a mixture of hydrated iron(iii) oxide &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2O3*nH2O}$&lt;/span&gt; and iron(iii) oxide hydroxide &lt;span class=&quot;math-container&quot;&gt;$\ce{FeO(OH)}/\ce{Fe(OH)3}$&lt;/span&gt;). The iron hydroxide species look like they could come from basic solutions.&lt;/p&gt;&#xA;&lt;p&gt;Even if consider only the formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2O3}$&lt;/span&gt;, this can occur by several pathways.&lt;/p&gt;&#xA;&lt;p&gt;Using Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard electrode potential data&lt;/a&gt; page as a source of known redox half reactions, the process may be something like this:&lt;/p&gt;&#xA;&lt;h3&gt;In acidic medium&lt;/h3&gt;&#xA;&lt;p&gt;The net reaction in acid is between the iron and the protons in the acid to produce &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{3+}}$&lt;/span&gt;, which reacts with water in a redox neutral (but generates more acid) process to form the insoluble oxides/hydroxides. The initial redox reaction is spontaneous. The water is the net source of oxygen.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Fe(s) + 6H+(aq) - 2Fe^{3+}(aq) + 3H2(g)} \ \ E^\circ=+0.04\ \text{V}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe^{3+}(aq) +3H2O -&amp;gt; Fe(OH)3(s) + 3H+(aq)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe(OH)3(s) -&amp;gt; FeO(OH)(s) + H2O}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2FeO(OH) -&amp;gt; Fe2O3 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;In basic medium&lt;/h3&gt;&#xA;&lt;p&gt;The net reaction in base is between oxygen (in the air, which is not in contact with your submerged nail) and the iron in the nail. The iron initially reacts with the hydroxide ions, which are regenerated by the reduction of oxygen at the surface of the water.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\ce{4Fe(s) + 8OH- &amp;amp;-&amp;gt; 4Fe(OH)2 + 8e-} \ \ \ &amp;amp;&amp;amp;E^\circ=\pu{+0.89 V}\\&#xA;\ce{4Fe(OH)2 + 4OH- &amp;amp;-&amp;gt; 2Fe2O3 +4e-}\ \ \ &amp;amp;&amp;amp;E^\circ=\pu{+0.86 V}\\&#xA;\ce{3O2 + 6H2O + 12e- &amp;amp; -&amp;gt; 12OH-}\ \ \ &amp;amp;&amp;amp;E^\circ=\pu{+0.40 V}\\ \hline&#xA;\ce{4Fe(s) +3O2(g)&amp;amp; -&amp;gt; 2Fe2O3(s)}\ \ \ &amp;amp;&amp;amp;E^\circ=\pu{+1.95 V}\\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="95133" LastEditDate="2021-05-11T06:28:21.343" LastActivityDate="2021-05-11T06:28:21.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8174" PostTypeId="1" CreationDate="2014-01-29T21:50:47.810" Score="-1" ViewCount="730" Body="&lt;p&gt;This question has been intriguing me since 10th or 11th grade. The teacher just told us about it but didn't get into the details of the why. Recently I asked a biochemist but couldn't get an answer. So I bring it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3OH + O2 -&amp;gt; CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the complete combustion equation of methanol. But the problem is it can't be balanced. What is the practical implication, and significance of this fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side question:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mathematical reasoning behind the inability to balance the equation?&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="2094" LastEditDate="2014-01-29T23:17:17.677" LastActivityDate="2014-01-29T23:44:24.240" Title="What is the implication of not being able to balance the complete combustion reaction equation of methanol?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8175" PostTypeId="2" ParentId="8174" CreationDate="2014-01-29T21:58:22.400" Score="3" Body="&lt;p&gt;Here is the balanced eqn...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 CH3OH + 3 O2 → 2 CO2 + 4 H2O}$$ &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="418" LastEditDate="2014-01-29T22:00:24.633" LastActivityDate="2014-01-29T22:00:24.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8177" PostTypeId="1" AcceptedAnswerId="8178" CreationDate="2014-01-29T23:10:00.417" Score="1" ViewCount="7229" Body="&lt;p&gt;The question sounds vague and non-specific because my background in chemistry is limited to undergrad college courses.  However, I am interested in learning the correlations between the atom-by-atom structure of molecules (simple, complex...) and the resultant physical properties thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, it seems most the material online directly referencing the phrases &quot;molecular structure&quot; and &quot;physical properties&quot; is in the form of highly specialized research publications.  Surely, there is a &lt;i&gt;Basics to the Relationships between Molecular Structure and Physical Properties 101&lt;/i&gt; out there, somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, my questions sum up to the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Where may I go to learn these things?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What material may I read?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Who might I ask?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where do I begin?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="4360" LastActivityDate="2020-03-11T10:04:12.293" Title="How does molecular structure affect physical properties?" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8178" PostTypeId="2" ParentId="8177" CreationDate="2014-01-29T23:30:19.330" Score="3" Body="&lt;p&gt;Predicting (approximate) physical properties starting with molecules is complicated, below.  Nobody has ever produced a way to accurately predict how molecules will pack in a crystal structure.  Big Pharma is desperate to predict crystal polymorphs and their properties. Slightly vast amounts of grant funding pursued predicted super-explosive octanitrocubane.  When it was finally synthesized it crystallized in a less dense crystal structure than expected, deeply disappointing the DoD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemspider.com/ACDLabs.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.chemspider.com/ACDLabs.aspx&lt;/a&gt; &#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed064p575&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ed064p575&lt;/a&gt; &#xA;&lt;a href=&quot;http://molsoft.com/mprop/&quot; rel=&quot;nofollow&quot;&gt;http://molsoft.com/mprop/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a way to predict crystal structures given complex formula units, you're a big winner.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T23:30:19.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8179" PostTypeId="2" ParentId="8177" CreationDate="2014-01-29T23:35:45.677" Score="0" Body="&lt;p&gt;Uncle Al is definitively right, especially on polymorphism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe you just want to start on a more basic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check the web for some introductory information on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;boiling points of linear hydrocarbons of different length&lt;/li&gt;&#xA;&lt;li&gt;boiling points of alcohols as compared to their parent hydrocarbons&lt;/li&gt;&#xA;&lt;li&gt;melting points of &lt;em&gt;saturated&lt;/em&gt; vs &lt;em&gt;unsaturated&lt;/em&gt; fatty acids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to get a feeling how structure and different interactions between molecules influence the physical properties.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-29T23:35:45.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8180" PostTypeId="1" CreationDate="2014-01-29T23:42:43.787" Score="2" ViewCount="133" Body="&lt;p&gt;The reduction of water at a cathode is represented by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?&lt;/p&gt;&#xA;" OwnerUserId="4361" LastEditorUserId="3846" LastEditDate="2014-01-30T09:41:05.127" LastActivityDate="2015-03-05T14:05:22.227" Title="What happens to the hydroxide anion during the electroreduction of water?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8181" PostTypeId="2" ParentId="8174" CreationDate="2014-01-29T23:44:24.240" Score="0" Body="&lt;p&gt;Mass is conserved, charge is conserved, ignore spectator ions until the end.  If it is a redox reaction, begin by listing each active atom's reduction or oxidation, e.g., &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mn(+7) + 5e to Mn(+2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cr(+3) - 3e to Cr(+6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and find the least common denominator (here, 15).  That gives you the primary stoichiometry, here 5Cr(+3) (going to chromate) with 3Mn(+7) (coming from permanganate).  Add the other atoms, slop in H2O, OH-, or H+, add back the spectator ions. Confirm the same number of each kind of atom on both sides.  Try it out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CuSCN + KIO3 + HCl to CuSO4 + KCl + HCN + ICl + H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 to K2Cr2O7 + MnSO4 +CO2 + KNO3 + K2SO4+ H2O&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-29T23:44:24.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8182" PostTypeId="2" ParentId="8172" CreationDate="2014-01-30T00:05:36.790" Score="1" Body="&lt;p&gt;To build lattice you need the space group, the three unit cell dimensions, the three unit cell angles, each atom's fractional coordinates (e.g., a *.cif file overall),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acta Cryst. B38, 598 (1982)  doi:10.1107/S056774088200346X&#xA;Ni3Sn4, C2/m, a=12.214, b=4.060, c=5.219 A, beta=105.0 degrees (others 90 degrees by default), V=250 A^3 (verify), Z=2 (formula units/unit cell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.gla.ac.uk/~louis/software/ortep3/&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.gla.ac.uk/~louis/software/ortep3/&lt;/a&gt; or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.shapesoftware.com/00_Website_Homepage/&quot; rel=&quot;nofollow&quot;&gt;http://www.shapesoftware.com/00_Website_Homepage/&lt;/a&gt; or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iucr.org/resources/other-directories/software?result_42405_result_page=A&quot; rel=&quot;nofollow&quot;&gt;http://www.iucr.org/resources/other-directories/software?result_42405_result_page=A&lt;/a&gt;&lt;br&gt;&#xA;the big kahuna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3D visualization, if you are not practiced,&#xA;&lt;a href=&quot;http://www.berezin.com/3d/3dprism.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.berezin.com/3d/3dprism.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-01-30T00:05:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8193" PostTypeId="2" ParentId="8086" CreationDate="2014-01-30T17:50:05.127" Score="2" Body="&lt;p&gt;You're right that it has to do with a mirror plane. The negative sign means that the overall electronic wavefunction changes sign when reflected in a mirror plane that includes the internuclear axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ground-state oxygen, reflecting in a mirror plane will change the sign of one unpaired $\pi$ electron (the one that sticks out of the mirror plane) but not the other, resulting in an overall change in sign. (This is slightly simplified, a more rigorous explanation would take into account the complex-valued orbitals rather than their real-valued linear combinations.)&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-01-30T17:50:05.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8194" PostTypeId="2" ParentId="8180" CreationDate="2014-01-30T17:56:28.420" Score="1" Body="&lt;p&gt;You only show one a half-reaction, reduction at the cathode.  Write the other half-reaction, oxidation at the anode to give oxygen.  A real world water electrolysis cell requires a salt electrolyte for conductivity.  $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$.  $\ce{Na2CO3}$ can be interesting.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4231" LastEditDate="2015-03-05T14:05:22.227" LastActivityDate="2015-03-05T14:05:22.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8195" PostTypeId="2" ParentId="5702" CreationDate="2014-01-30T18:08:09.693" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.owlstonenanotech.com/air-monitor&quot; rel=&quot;nofollow&quot;&gt;Owlstone&lt;/a&gt; make an electronic multi-agent detector - I don't know how much it costs but I would guess somewhere in the range \$10K - \$100K. There might be cheaper single-agent electronic sensors or chemical tests.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-01-30T18:08:09.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8197" PostTypeId="1" CreationDate="2014-01-30T22:26:24.920" Score="1" ViewCount="130" Body="&lt;p&gt;I'm specifically interested in ions that would adsorb to gold surfaces, however I would be incredibly appreciative if someone could direct me to a textbook reference or something more expansive, that i could use to reference in the future.&#xA;&lt;br&gt;&lt;br&gt;I've been doing some reading and found the following regarding Silver:&#xA;&lt;br&gt;&lt;em&gt;Silver Adsorbing&lt;/em&gt;&#xA;&lt;br&gt;Chloride, Bromide, Iodide, Urea, thiocyanate&#xA;&lt;br&gt;&lt;em&gt;Silver Nonadsorbing&lt;/em&gt;&#xA;&lt;br&gt;Fluoride, Perchlorate, Sulfate&#xA;&lt;br&gt;&lt;br&gt;I know Gold leeches with cyanide and I think bisulfide.&#xA;&lt;br&gt;&lt;br&gt; I actually found a good &lt;a href=&quot;http://www.fisica.unige.it/~rocca/Didattica/Surface%20Science%20and%20Nanostructuring/6%20adsorption%20at%20surfaces.pdf&quot; rel=&quot;nofollow&quot;&gt;set of lecture slides&lt;/a&gt; that discusses chemisorption an Oxygen atom across a series of Transition metals. Also I'm also looking for something about physiorption not chemi...&#xA;&lt;br&gt; The slides reference a book called &quot;Surface Science:Foundations of Catalysis and Nanoscience&quot; by Kurt Kolasinski&quot;...I guess I'll start there, but would love to hear about other references anyone else may have.&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="3903" LastEditDate="2014-01-30T22:59:30.277" LastActivityDate="2014-01-30T22:59:30.277" Title="Any references for adsorption of ions to metal surfaces?" Tags="&lt;adsorption&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8200" PostTypeId="1" CreationDate="2014-01-31T02:26:07.873" Score="0" ViewCount="20089" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Calculate the heat produced when a strip of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; metal with a mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0801 g}$&lt;/span&gt; is reacted with &lt;span class=&quot;math-container&quot;&gt;$\pu{50.0 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; to raise the temperature of the water by &lt;span class=&quot;math-container&quot;&gt;$\pu{7.6 °C}$&lt;/span&gt; (change in temperature is &lt;span class=&quot;math-container&quot;&gt;$\pu{7.6 °C})$&lt;/span&gt; Calculate the heat produced when a mole of this metal is used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have this so far:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q = mc\,\Delta T$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m =\pu{0.0801 g}$&lt;/span&gt; and the change in temperature is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta T = T_2 - T_1 = \pu{7.6 °C} - \pu{4.184 K} = \pu{3.416 K}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is that right? I have no idea what the specific heat &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is. Is it the specific heat of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}?$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What do I do with the &lt;span class=&quot;math-container&quot;&gt;$\pu{50.0 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}?$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4377" LastEditorUserId="41328" LastEditDate="2021-05-22T16:46:18.617" LastActivityDate="2021-05-22T16:46:18.617" Title="Heat produced when magnesium is reacted with acid" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8202" PostTypeId="1" AcceptedAnswerId="8398" CreationDate="2014-01-31T04:25:16.907" Score="7" ViewCount="121703" Body="&lt;p&gt;If a plastic is marked PP as in Polypropylene (i.e. &lt;a href=&quot;http://en.wikipedia.org/wiki/Resin_identification_code&quot; rel=&quot;nofollow noreferrer&quot;&gt;resin code&lt;/a&gt; 5), is it automatically safe for steaming and microwaving, such that it wouldn't leach chemicals into the food(besides not melting)?&lt;/p&gt;&#xA;" OwnerUserId="2020" LastEditorUserId="76440" LastEditDate="2019-04-29T14:11:41.963" LastActivityDate="2019-04-29T14:11:41.963" Title="Are all PP(Polypropylene), i.e. resin code 5 plastics, safe for microwave and steaming?" Tags="&lt;safety&gt;&lt;heat&gt;&lt;plastics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8204" PostTypeId="2" ParentId="8200" CreationDate="2014-01-31T08:22:59.127" Score="3" Body="&lt;p&gt;It can help to imagine the experimental setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a thermally insulated reaction container, let's call it &lt;em&gt;calorimeter&lt;/em&gt;, that is filled with 50 ml 1 M $\ce{HCl}$ ( = a solution of hydrogen chloride gas in &lt;strong&gt;water&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to measure a temperature change ($\Delta{} T$), that comes from an exothermic reaction. &lt;strong&gt;Where&lt;/strong&gt; would we best measure this temperature change? Our &lt;strong&gt;reaction medium&lt;/strong&gt; apparently is a good choice. Therefore, there's most likely a thermometer (insulated from the outside) in good thermal contact with the &lt;strong&gt;water&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we let the magnesium strip fall into the &lt;strong&gt;reaction medium&lt;/strong&gt; (using some clever mechanic) while our &lt;em&gt;calorimeter&lt;/em&gt; is closed), let it react and measure the temperature change until the temperature reaches some final (maximum) value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you wrote, the temperature change ($ \Delta{} T$) is +7 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see that an initial temperature was given, but that's not important. We have the $ \Delta{} T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need a &lt;em&gt;specific factor&lt;/em&gt; that correlates the heat liberated in the reaction with the temperature change in a &lt;em&gt;specific medium&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the &lt;em&gt;specific heat capacity&lt;/em&gt; of &lt;strong&gt;water&lt;/strong&gt; comes into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you go.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-31T08:22:59.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8206" PostTypeId="2" ParentId="7914" CreationDate="2014-01-31T09:17:56.683" Score="1" Body="&lt;p&gt;No doubt, diatomic carbon does exist - in blue flames and in comets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic uv absorption pattern (&lt;em&gt;Swan&lt;/em&gt; bands) has already been described almost 100 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low temperature (77 K) it reacts with hydrocarbons by H-abstraction to form $\ce{HC#CH}$, and the addition to $\ce{C=C}$ double bonds has been reported too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't figure out any reasonable way for a reaction with carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-31T10:08:42.043" LastActivityDate="2014-01-31T10:08:42.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8207" PostTypeId="2" ParentId="2669" CreationDate="2014-01-31T09:53:15.907" Score="4" Body="&lt;p&gt;As described by others before, sodium hypochlorite ($\ce{NaOCl}$) is a strong disinfectant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technology for the production is established:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;electrolysis of brine ($\ce{NaCl}$) to yield chlorine ($\ce{Cl2}$) &lt;/li&gt;&#xA;&lt;li&gt;disproportionation of $\ce{Cl2}$ under alkaline conditions:  $\ce{Cl2 +2 OH- -&amp;gt; OCl- + Cl- + H2O}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr/&gt;&#xA;&#xA;&lt;p&gt;If you want halogen-free disinfection of water without bleach, there are two options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ozone treatment&lt;/li&gt;&#xA;&lt;li&gt;uv irradiation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both technologies are widely used too and have a solid scientific background as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can even go &quot;low-tech&quot; to save lifes: A successful initiative of the Swiss Federal Institute of Aquatic Sciences and Technology, &lt;a href=&quot;http://www.sodis.ch/index_EN&quot; rel=&quot;nofollow&quot;&gt;SODIS (Solar Water Disinfection)&lt;/a&gt;, uses sunlight and PET bottles to kill microbes and produce drinkable water in developing regions of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;And then, you can rely on &lt;em&gt;snake oil&lt;/em&gt; technology.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-01-31T12:43:41.730" LastActivityDate="2014-01-31T12:43:41.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8208" PostTypeId="1" CreationDate="2014-01-31T15:25:21.443" Score="1" ViewCount="242" Body="&lt;p&gt;The assignment question is asking us to explain why you can't get more than 92% gold in a gold alloy, and I can't seem to find out why it's specifically 92% in either the textbook or online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4383" LastActivityDate="2014-10-11T18:02:11.240" Title="Why is the maximum % for a gold alloy 92%?" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8209" PostTypeId="1" CreationDate="2014-01-31T15:39:04.280" Score="0" ViewCount="9928" Body="&lt;p&gt;What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide, $\ce{V2O5}$?&lt;/p&gt;&#xA;" OwnerUserId="4384" LastEditorUserId="4945" LastEditDate="2018-08-21T11:54:33.660" LastActivityDate="2018-08-21T12:06:40.080" Title="What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-02-02T02:36:35.673" ContentLicense="CC BY-SA 4.0" />
  <row Id="8211" PostTypeId="2" ParentId="8209" CreationDate="2014-01-31T16:49:21.743" Score="6" Body="&lt;p&gt;A talk given by Ken Waugh named &lt;em&gt;The Detailed Kinetics of the Partial Oxidation Reactions: Benzene/Butane to Maleic Anhydride, Ethylene to Ethylene Oxide,&lt;/em&gt;&lt;sup&gt;[1]&lt;/sup&gt; proposes the following mechanism on page 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/njquL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/njquL.png&quot; alt=&quot;DA Dowden mechanism of benzene oxidation to MA using molecularly adsorbed oxygen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: Acessible as pdf via &lt;a href=&quot;http://www.rsc.org/images/Ken_Waugh_tcm18-236914.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;rsc.org&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2018-08-21T12:06:40.080" LastActivityDate="2018-08-21T12:06:40.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8213" PostTypeId="1" CreationDate="2014-01-31T19:30:28.910" Score="7" ViewCount="8471" Body="&lt;p&gt;Why do molecules of a body exhibit vibrational motion? Is it due to interaction between various molecules of the body, or interatomic interactions between atoms in the same molecule? If it is because of any of them, how can these interaction cause the molecules to vibrate?&lt;/p&gt;&#xA;" OwnerUserId="4387" LastEditorUserId="16683" LastEditDate="2017-02-09T18:32:14.443" LastActivityDate="2020-03-28T04:16:21.583" Title="Reason for vibration of molecules" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;intermolecular-forces&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8214" PostTypeId="2" ParentId="8208" CreationDate="2014-01-31T21:04:35.950" Score="7" Body="&lt;p&gt;It's not that you can't, it's that you usually don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of Gold goes into jewellery with some in coinage (more in the past, we make gold coins now but they don't circulate much). But pure gold is too soft to be useful in most applications (coins have to last, rings have to not scratch too much and they aren't much use if they bend too easily). It is also expensive. So the most used alloys have to make a compromise between usability and cost. There are also some that try to achieve a particular aesthetic effect like a different colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This imposes limits of practicality. At the cheap end alloys that look like gold might be mostly other metals. But at the high end the issue is to get the correct aesthetic effect while retaining the appeal and value of gold without having something  too soft or weak to be practical. Many coloured golds are only about 75% gold (18 karat) because that gives a wide range of colours. But top end &quot;pure&quot; gold objects can be made at 22 karat (92% isn). We could make richer alloys, but they would be too soft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, mostly we don't. We can make many alloys with more gold, including many that would be useless as jewellery like the AuBe alloy mentioned in another answer (it is far too brittle), but you will rarely see them as they are not useful for anything.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2014-10-11T18:02:11.240" LastActivityDate="2014-10-11T18:02:11.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8215" PostTypeId="2" ParentId="8213" CreationDate="2014-01-31T22:01:23.387" Score="5" Body="&lt;p&gt;Here is a picture and an equation that may add to the comment LordStryker made.  Molecules vibrate because they have energy. First the equation, as $T \to 0$ (absolute zero), the first term will vanish, but the molecule will still have an energy given by $h\nu/2$.  So even at absolute zero a molecule will have some vibrational energy, this is called the &quot;zero point energy&quot;.  The parabola (representing the hydrogen well, for example, that two hydrogen atoms will &quot;fall into&quot; when they bond and become a stable hydrogen molecule) gives a pictorial representation of why this must be - energy levels are quantized.  Even the lowest level is above the bottom of the parabola, so even in the lowest energy level at absolute zero, the atom or molecule will still have a non-zero energy which will cause it to vibrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\varepsilon = \frac{h\nu}{\mathrm e^{h\nu/kT} - 1} + \frac{h\nu}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hrH2Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hrH2Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-02-09T18:31:17.817" LastActivityDate="2017-02-09T18:31:17.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8216" PostTypeId="2" ParentId="7397" CreationDate="2014-01-31T22:42:47.997" Score="1" Body="&lt;p&gt;Is there any chance that you can use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Teclu_burner&quot; rel=&quot;nofollow&quot;&gt;Teclu&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bunsen_burner&quot; rel=&quot;nofollow&quot;&gt;Bunsen&lt;/a&gt; burner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, you will need a batch of magnesia ($\ce{MgO}$) sticks too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the cations can be identified by a &lt;strong&gt;flame test&lt;/strong&gt;, since they furnish a typical colour in the &quot;blue&quot; flame of the burners.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Na: intense yellow&lt;/li&gt;&#xA;&lt;li&gt;K: lilac/violet&lt;/li&gt;&#xA;&lt;li&gt;Ca: brick red&lt;/li&gt;&#xA;&lt;li&gt;Ba: pale green&lt;/li&gt;&#xA;&lt;li&gt;Pb: blue/white (I had to look that up, never did it myself)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-01-31T22:42:47.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8218" PostTypeId="2" ParentId="5694" CreationDate="2014-02-01T11:16:14.573" Score="2" Body="&lt;p&gt;I'd change the solvent to acetonitrile (don't forget the quarternary ammonium salt) and give a glassy carbon electrode a try. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to observe the (irreversible) reduction of the sulfonamide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And once you have optimized the conditions, please perform a final run with ferrocene as an internal reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;Edit&lt;/strong&gt;:&#xA;Let me explain the last suggestion. The literature is full of electrochemical data of organic compounds measured in non-aqueous solvents, mostly acetonitrile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was once common (bad) practise to report the potentials vs &lt;strong&gt;SCE&lt;/strong&gt; (saturated calomel electrode), although the latter, an aqueous(!) reference electrode, was atually &lt;strong&gt;never&lt;/strong&gt; used in the measurement. Often rather exotic conditions plus a shaky conversion of data turned good experimental work into a pile of data of limited value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an internal standard is the obvious solution to obtain solvent-independent electrochemical data for publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the use of ferrocene as such a standard isn't my idea! It was originally proposed by &lt;a href=&quot;http://dx.doi.org/10.1351/pac198456040461&quot; rel=&quot;nofollow&quot;&gt;GRITZER and KUTA in &lt;em&gt;Pure Appl. Chem.&lt;/em&gt;&lt;/a&gt; and is a IUPAC recommendation nowadays.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-02T12:17:48.403" LastActivityDate="2014-02-02T12:17:48.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8220" PostTypeId="1" AcceptedAnswerId="8225" CreationDate="2014-02-01T13:45:25.743" Score="6" ViewCount="1080" Body="&lt;p&gt;I heard that there is a standard about cleaned glass, namely&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemically clean glass supports a uniform film of water, with no hanging droplets visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried to clean my dishes at home using detergent. After several efforts, the ceramic-made dish looks like a uniform film of water just after finishing the cleaning. If I hold the dish for a while, the water starts to concentrate at regions, forming a droplet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, in this video&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=m7liHlmv5Kc&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=m7liHlmv5Kc&lt;/a&gt;&#xA;at 2 min, there is water droplet in the flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Does the clean standard have any time range? Does it apply to arbitary time scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If the answer to question (i) is yes, how should I do to make the dishes so clean that there is absolutely no water concentrating at certain regions? I do not have a dishwasher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Video demonstration is highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2629" LastEditorUserId="19616" LastEditorDisplayName="user4076" LastEditDate="2015-10-30T18:27:22.053" LastActivityDate="2015-10-30T18:27:22.053" Title="How to achieve clean dishes and glasses?" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8221" PostTypeId="2" ParentId="7183" CreationDate="2014-02-01T13:52:10.010" Score="0" Body="&lt;p&gt;You do not need the molar masses to do this question. The ideal gas law includes a relationship between volume $V$ and moles $n$ assuming pressure and temperature are constant (they are at STP).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=nRT$$&#xA;$$n=\dfrac{PV}{RT}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-01T13:52:10.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8222" PostTypeId="2" ParentId="8220" CreationDate="2014-02-01T15:08:13.533" Score="3" Body="&lt;p&gt;I am not that familiar with cleaning procedures, but the 'rule' to support a uniform film of water without droplets seems to be an easy proxy for checking whether there are still contaminated spots on the surface. In that case, a phenomenon called &lt;a href=&quot;http://www.utwente.nl/tnw/pcf/education/jmbc_course_on_capillarity_driv/Presentations/snoeijer_hysteresis.pdf&quot; rel=&quot;nofollow&quot;&gt;contact line pinning&lt;/a&gt; will occur, which results in droplets. The uniform film will only be possible if there are no contaminated spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that there are droplets however doesn't necessarily indicate that the glass is not clean, because it can also be a result of just too much water being present. The maximum thickness of the uniform water film is governed by a balance between gravity and surface tension as $\sqrt{\frac{\gamma}{\rho g}}$ and for water is approximately 2.7 mm (see e.g. &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node47.html&quot; rel=&quot;nofollow&quot;&gt;capillary length&lt;/a&gt; or the great book by &lt;a href=&quot;http://books.google.nl/books/about/Capillarity_and_Wetting_Phenomena.html?id=MxLQk8vms-kC&amp;amp;redir_esc=y&quot; rel=&quot;nofollow&quot;&gt;de Gennes on capillarity and wetting phenomena&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the water film becomes thicker then the 2.7 mm it will start to form droplets as well. So the 'cleaning rule' is only useful if you have removed sufficient water from the surface. In the video I'm quite sure that they have not removed sufficient water yet, so the rule cannot be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you are holding the glassware probably such that water starts to drain from 1 side to the other (due to gravity), at the moment that the layer at the bottom exceed the capillary length you will start to see droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional point to consider: drying a water-wet piece of glassware in an oven only yields a clean surface when you have used deionized water (as they do in the video). If you use normal tap water then you will have some calcification due to the salts in the water that end up on your glass surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to try and answer your questions based on these considerations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;i)&lt;/strong&gt; Yes, there is a time range if you have gravitational drainage of the film. For anything that is not completely horizontal you will see this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ii)&lt;/strong&gt; Water concentrating is not the issue as long as it is deionized. If you really want spotlessly clean glasses: use deionized water after the whole cleaning procedure and just dry the glass in a warm place. If you don't have deionized water: you can make reasonably well deionized water by boiling tap water and catching the steam in a clean cup. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-02-01T15:08:13.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8223" PostTypeId="2" ParentId="8220" CreationDate="2014-02-01T16:26:58.230" Score="3" Body="&lt;p&gt;&quot;Clean&quot; depends on what you consider to be dirt.  detergent leaves a spotty adsorbed layer of detergent.  &quot;Clean&quot; glass by removing its surface. Begin by having a lab coat, chemical gloves, and a full face shield.  If it gets on you it will dig holes.  Instant blindness in an eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5-gallon plastic (not vinyl!) bucket with a cover, 4 gallons of biological denatured ethanol, pound jar of of potassium hydroxide pastilles.  Shove crushed ice into the jar with mixing to form a viscous slurry (exotherm!).  Pour with mixing into the alcohol (probably have to do this twice to get all the KOH dissolved).  This is a KOH bath.  (Standard denatured alcohol will react to go red and stinky, but still work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soak Pyrex in a KOH bath overnight to remove everything except hydrocarbon grease and metal oxides.  Soak soda lime glass for an hour or three, for it slowly dissolves.  Rinse with DI water, soak in 0.1 M HCl for an hour [#SiO(-) K(+) to #SiOH], rinse well with DI water, store totally submerged in DI water.  Water will perfectly sheet.  If the surface is allowed to dry, start again.  Never put your hand in there.  Wear long heavy gloves and positive eye protection.  Immediately, thoroughly wash off skin contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NEVER allow glass flats to stack flat under water.  They will bond (a micron of water layer is the hydrogen bonding adhesive).  Try sliding them apart without applying orthogonal pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromate/conc. sulfuric acid) cleaning solution cover the glass with a layer of chromium(VI) species.  Things like NoChromix work, but you still have the conc. sulfuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note added in proof: NEVER put graduated glassware in a base bath, for they lose calibration.  Frosted syringe barrels are iffy. Aluminum will fizz hydrogen, corrode, and disappear.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-02-02T00:52:11.467" LastActivityDate="2014-02-02T00:52:11.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8224" PostTypeId="2" ParentId="8202" CreationDate="2014-02-01T16:36:05.913" Score="3" Body="&lt;p&gt;Consumer polyolefin plastics contain trace mold releases that adversely affect biological studies.  For human food use, one presumes Generally Recognized as Safe as long as the contents stay water wet, below 100 C.  Burning on residues (cheese) allows the temperature to skyrocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyoelfin containers are penetrated by hot lipid and especially hydrocarbons (red lycopene in tomato products).  Do NOT use dishwasher &quot;bleaches&quot; to remove the internal stain.  The active ingredient is dispersed dibenzoyl peroxide (Bondo cure).  Sure, it diffuses in and free-radical breaks up the molecule's conjugation, rendering then colorless.  It also free-radical scissions all plastics in the dishwasher.  Things go brittle and fail.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-01T16:36:05.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8225" PostTypeId="2" ParentId="8220" CreationDate="2014-02-01T17:55:14.383" Score="9" Body="&lt;p&gt;There are several laboratory techniques to rigorously clean glass. Here are a few of the more common ones, to add to the other answers. There are probably more procedures, and it's also common to perform more than one treatment to maximize efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically none of these have any use at home, because they are too dangerous, too expensive, and just outright unnecessary. A soapy scrub will do just fine cleaning anything in 99% of cases, even if water still beads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Ultrasonication in pure solvents or detergent solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, the main impurities on the surface of glass are organic substances which are present at all times in air outside of strictly clean environments, and which strongly adhere to the glass when they come into contact. By dipping them into a solution with some affinity for the organic substances, it is possible to strip them off. Common options are regular dishwashing liquid (usually sulphonic acids with large alkyl chains), solutions of quaternary ammonium salts, aqueous ethanolanime, or pure deionized water, ethanol, isopropanol, acetone, among many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much more effective, however, to couple the aforementioned compounds with ultrasonication rather than simply dipping or scrubbing. In an ultrasonicator, the liquid is vibrated at frequencies higher than $10\ kHz$. Though macroscopically the liquid doesn't seem to change much, on the microscopic level there is chaos. The high frequency vibrations locally produce very fast solvent jets and cavitation bubbles, which effectively sandblast anything closeby. Glass manages to stay mostly unaffected due to its stiffness and strong bonding, but anything not very firmly attached to its surface will be flung away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - Prolonged exposure to high temperatures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common glass is thermally stable and can be heated strongly before it softens, so it can take temperatures at which many other substance decompose, evaporate or burn (if exposed to oxygen). Different types of glass have different maximum temperatures depending on their composition. Also, the thermal expansion while heating and contraction while cooling can be problematic for delicate glassware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - Soak in base bath&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong bases react with many substances and help dissolve them off the surface of the glass. However, while very inert, glass itself will also slowly react with concentrated basic solutions at room temperature. This also helps clean because if very persistent stains manage to survive the basic conditions, the glass surrounding them will begin dissolving slightly, and after a while the stain ends up detaching from the substrate as the glass behind it will have dissolved away. The regular options are concentrated $\ce{NaOH}$ or $\ce{KOH}$ in water or ethanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 - Soak in highly oxidizing solutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass is composted mostly of silicon already in its most oxidized form ($\ce{SiO2}$). Hence exposure to very oxidizing conditions will not damage the surface of the glass, while anything else on it is likely to dissolve, be it organic or inorganic matter. Classic options for dipping the glass are acidic piranha solutions (a mixture of pure $\ce{H2SO4}$ and concentrated $\ce{H2O2}$), basic piranha solutions (concentrated aqueous $\ce{NH4OH}$ and concentrated $\ce{H2O2}$), sulphocromic solution ($\ce{H2SO4}$ + $\ce{H2Cr2O7}$, in disuse due to the toxicity of $\ce{Cr^{6+}}$) and sulphonitric solution ($\ce{H2SO4}$ + $\ce{HNO3}$). These are all pretty heavy-duty mixtures. For example, &lt;a href=&quot;http://www.youtube.com/watch?v=S5BurdByqRc&quot;&gt;here's why piranha solutions earn their name&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 - Treatment with ozone/oxygen plasma&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a step up from even the piranha solutions. If all you want to do is oxidize organic matter on the surface of the glass, then why not cut out the middleman and directly bathe the glass in reactive oxygen? The glass sample is first put into a closed chamber containing oxygen (air works, but pure oxygen is better). If ozone is being used, then the chamber is lit from the inside by a mercury lamp, which produces lots of ultraviolet light. The light is capable of dissociating oxygen molecules and producing ozone, which then is directly exposed to whatever is on the surface of the glass. In the case of oxygen plasma, the chamber is pumped down to a low pressure of pure oxygen gas before a coil produces strongly varying electric and magnetic fields inside the chamber, generating all sorts of oxygen species such as singlet oxygen (an electronically excited form of standard, triplet oxygen), free oxygen atoms, and several ions and radicals. &lt;a href=&quot;http://www.youtube.com/watch?v=atVSxvbiPg0&quot;&gt;Here's a nice video&lt;/a&gt; on how this process works in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all the work you put into cleaning a glass surface, it will immediately start adsorbing impurities if outside a vacuum, and it may take as little as a few minutes before it is too dirty to use in very sensitive and specialized applications. I have found that water already beads significantly again less than 10 minutes after the final treatment.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-02-01T17:55:14.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8226" PostTypeId="1" AcceptedAnswerId="8232" CreationDate="2014-02-01T21:39:55.593" Score="5" ViewCount="396" Body="&lt;p&gt;My book mentioned that the energy barrier for cyclohexane to reach the half-chair conformation is $50.6\ \mathrm{kJ\ mol^{-1}}$. It says that from this value, it was calculated that cyclohexane undergoes $10^5$ flips per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did they calculate this? We can assume that all the conformational states possible will be of the same energy as the chair conformation (so they all encounter the same energy barrier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm more interested in the principles behind the calculation – &lt;em&gt;why&lt;/em&gt; the equation works – than just seeing the calculation done.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorDisplayName="user7951" LastEditDate="2016-08-31T19:10:28.430" LastActivityDate="2016-08-31T19:11:48.113" Title="How to calculate the frequency of cyclohexane conformational interconversion?" Tags="&lt;kinetics&gt;&lt;conformers&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8231" PostTypeId="2" ParentId="8133" CreationDate="2014-02-02T02:33:43.190" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does vapour pressure decreases with decrease in surface area.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}-&amp;gt;H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2014-03-06T08:13:19.927" LastActivityDate="2014-03-06T08:13:19.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8232" PostTypeId="2" ParentId="8226" CreationDate="2014-02-02T03:19:56.313" Score="2" Body="&lt;p&gt;Look up application of the rate of reaction equation given barrier height and energy available at room temp, $kT$ (Boltzmann's constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = A \mathrm e^{ - \frac{E_\mathrm a}{RT} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg&quot; rel=&quot;nofollow&quot;&gt;http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twist-boat leads to mathematically &lt;I&gt;perfect&lt;/I&gt; geometric chirality,&lt;BR&gt; &#xA;&lt;I&gt;Symmetry: Culture and Science&lt;/I&gt; 19(4) 307-316 (2008).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2016-08-31T19:11:48.113" LastActivityDate="2016-08-31T19:11:48.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8234" PostTypeId="1" CreationDate="2014-02-02T07:45:24.230" Score="3" ViewCount="11721" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/99c0P.png&quot; alt=&quot;In the image you can see two resonating structures they are nitric oxide structures.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first resonating structure you can see 5 unpaired electrons and 4 shared electrons on nitrogen, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving?&#xA;In the second resonating structure you can see that oxygen has 5 unpaired electrons on nitrogen and 4 shared electrons, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving? &#xA;Can anyone explain completely about the above questions? If this question is already been asked please comment me the link of that question. Thank you in advance.&#xA;Please explain with boxes if you can.&lt;/p&gt;&#xA;" OwnerUserId="4194" LastEditorUserId="4231" LastEditDate="2015-03-10T13:42:55.877" LastActivityDate="2020-01-31T16:00:11.190" Title="Where does the 9th electron go in a N=O bond?" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8235" PostTypeId="1" AcceptedAnswerId="8236" CreationDate="2014-02-02T09:33:36.653" Score="0" ViewCount="65" Body="&lt;p&gt;Is there a fixed ratio that can be used to calculate the number of neutrons in the nucleus of the most stable isotope of an element by knowing the atomic number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side question: if Z is the symbol for atomic number, are there symbols for the number of neutrons and/or atomic mass?&lt;/p&gt;&#xA;" OwnerUserId="4400" LastEditorUserId="402" LastEditDate="2017-04-02T06:40:37.507" LastActivityDate="2017-04-02T06:40:37.507" Title="Z:#Neutrons Ratio" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;isotope&gt;&lt;neutrons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8236" PostTypeId="2" ParentId="8235" CreationDate="2014-02-02T10:46:10.500" Score="2" Body="&lt;p&gt;The symbols $Z$, $N$, and $A$ are typically used to describe a nucleus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z$ : atomic number (= number of protons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N$ : neutron number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ : mass number ($A = Z + N$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no fixed $Z:N$ ratio throughout the whole periodic table for stable isotopes, but a &lt;em&gt;belt of stability&lt;/em&gt; in the &lt;a href=&quot;http://www.nndc.bnl.gov/chart/&quot; rel=&quot;nofollow&quot;&gt;Chart of Nuclides&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-02T10:46:10.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8237" PostTypeId="2" ParentId="8234" CreationDate="2014-02-02T11:42:06.230" Score="3" Body="&lt;p&gt;Your Lewis structures contain too many electrons. A molecule of $\ce{NO}$ has 11 valence electrons (5 from nitrogen, 6 from oxygen). Your proposed Lewis structures have 13 electrons, so both of them wound up having one atom with 9 electrons to compensate. &lt;s&gt; With simple valence bond theory, the most accurate structure has a bond order of 2.5 resulting from the combination of two canonical structures, one with a double bond and another with a triple bond, and where both structures have the odd electron on the nitrogen atom. &lt;/s&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Ben Norris mentions in his answer, there actually is no satisfactory Lewis structure for $\ce{NO}$, which actually has a bond order of 2.5. The closest representation is a doubly-bonded molecule, and it is not possible to write a canonical structure with triple bond without going over the octet in one of the atoms. This happens because naive valence bond theory lacks the concept of antibonding bonds.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-02-02T14:01:13.223" LastActivityDate="2014-02-02T14:01:13.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8238" PostTypeId="2" ParentId="8234" CreationDate="2014-02-02T12:30:58.773" Score="12" Body="&lt;p&gt;Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a different bonding picture of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; to answer your question about which orbital the unpaired electron is in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Lewis Structure of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mistake you made in drawing those Lewis Structures is that you used &lt;strong&gt;all&lt;/strong&gt; of the electrons in the nitrogen and oxygen atoms and not just the &lt;strong&gt;valence&lt;/strong&gt; or outer-shell electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Nitrogen atom has 7 total electrons in the following configuration: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these seven electrons, only the &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; in the second shell (energy level) are available for bonding. These are the valence electrons. The &lt;span class=&quot;math-container&quot;&gt;$1s$&lt;/span&gt; electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: &lt;span class=&quot;math-container&quot;&gt;$$2s^2 2p_x^2 2p_y^1 2p_z^1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding in &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{aligned}&#xA;&amp;amp;.. \ \ \  .. &amp;amp;&amp;amp;\  &amp;amp;&amp;amp;&amp;amp;\ ..\ \ \ ..\ \\&#xA;\ce{.} &amp;amp;\ce{N=O:} &amp;amp;&amp;amp;\ce{&amp;lt;-&amp;gt;} &amp;amp;&amp;amp;&amp;amp;\ce{:N=O.}&#xA;\end{aligned}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Resonance&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures inter-converting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where is the unpaired electron?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer this question, we need molecular orbital theory. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular orbital diagram&lt;/a&gt; describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;, bonding or anti-bonding. Since this answer is running long already, I won't go into how to construct such a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the MO diagram for nitric oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGh8c.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unpaired electron (highlighted in blue)is in a &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; anti-bonding orbital that is formed from the destructive overlap of out of phase &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbitals, one from nitrogen and one from oxygen.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="52853" LastEditDate="2020-01-31T16:00:11.190" LastActivityDate="2020-01-31T16:00:11.190" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8239" PostTypeId="1" CreationDate="2014-02-02T19:55:01.783" Score="1" ViewCount="3739" Body="&lt;p&gt;Would molecules like $\ce{C2, H2}$, etc., be considered 1 mole of diatomic [element] or 2 moles of [element]?&lt;/p&gt;&#xA;" OwnerUserId="4402" LastEditorUserId="4231" LastEditDate="2015-03-10T17:24:58.297" LastActivityDate="2015-03-10T17:24:58.297" Title="Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon?" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8240" PostTypeId="2" ParentId="8239" CreationDate="2014-02-02T20:25:15.640" Score="6" Body="&lt;p&gt;You can have 1 mole of $\ce{C_2}$ which contains 2 mole of $\ce{C}$ just as you can have for example 1 mole of $\ce{C_2H_5OH}$ which contains 2 mole of $\ce{C}$, 6 mole of $\ce{H}$ and 1 mole of $\ce{O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that you always have to mention whether you are indicating moles of molecule or moles of atoms. That is why a statement like: &quot;I have 1 mole of oxygen&quot; is ambiguous, because it could be referring to oxygen gas (i.e. $\ce{O_2}$) or to atomic oxygen (i.e. $\ce{O}$).&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-02-03T06:23:09.900" LastActivityDate="2014-02-03T06:23:09.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8241" PostTypeId="2" ParentId="8239" CreationDate="2014-02-02T22:21:13.253" Score="1" Body="&lt;p&gt;$\ce{C_2}$ is the blue-green Swan band emission of comets' ion tails and the blue in gas flames. The big question is its electronic structure (presumably not $\ce{•C#C•}$).  Its paper synthesis in cryogenic gas matrix (UV cleavage) or in vacuum (flash pyrolysis or UV) seems to be easy, from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ROO-C(=O)-C#C-C(=O)-OOR, the peresters of acetylenedicarboxylic acid with R = $\ce{-CF_3}$, $\ce{-C(CH_3)_3}$, or $\ce{-Si(CH_3)_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strings of alpha-omega diradical polycarbyne, $\ce{(-C#C-)_n}$ or $\ce{(=C=C=)_n}$ (your choice in LCAO, same in MO), in carbon arcs are suggested to pericyclicly close to fullerenes. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-02T22:21:13.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8242" PostTypeId="1" CreationDate="2014-02-02T22:50:02.210" Score="3" ViewCount="332" Body="&lt;p&gt;If the longest carbon chain is symmetric numbering-wise (meaning the numbering schemes are identical both ways) and has all alkyl substituents, should one prioritize the largest alkyl group?&lt;/p&gt;&#xA;" OwnerUserId="4032" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T18:23:27.807" LastActivityDate="2016-12-29T11:38:58.670" Title="Nearest branch numbering" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8245" PostTypeId="2" ParentId="6353" CreationDate="2014-02-03T04:33:48.170" Score="0" Body="&lt;p&gt;It will deprotonate the acid group in the middle as is more acid the base can easily take it´s proton away, rathe than having problems with the protons in the other acid groups.&lt;/p&gt;&#xA;" OwnerUserId="4405" LastActivityDate="2014-02-03T04:33:48.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8246" PostTypeId="1" CreationDate="2014-02-03T05:01:04.710" Score="0" ViewCount="180" Body="&lt;p&gt;I was comparing boiling points of $\ce{CHCl_3}$ and $\ce{CH_3Cl}$. According to me B.P. of $\ce{CHCl_3}$ should be higher due to it's higher molecular mass than $\ce{CH_3Cl}$, but the answer is opposite ,ie. B.P. of $\ce{CH_3Cl}$ $\gt$ B.P. of $\ce{CHCl_3}$.Why?&lt;/p&gt;&#xA;" OwnerUserId="3943" LastEditorUserId="72973" LastEditorDisplayName="user15489" LastEditDate="2020-03-14T13:46:29.717" LastActivityDate="2020-03-14T13:46:29.717" Title="Comparison of boiling points" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8247" PostTypeId="2" ParentId="8246" CreationDate="2014-02-03T05:46:01.457" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;but the answer is opposite ,ie. B.P. of $\ce{CH3Cl}$ &gt; B.P. of $\ce{CHCl3}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Really? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3Cl}$ has a boiling point of &lt;strong&gt;−24&lt;/strong&gt; °C. It is a &lt;strong&gt;gas&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CHCl3}$ has a boiling point around 61 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Edit: &lt;strong&gt;Your argumentation is valid, there was probably just some confusion in the notes.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-06-26T07:49:10.070" LastActivityDate="2015-06-26T07:49:10.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8249" PostTypeId="1" CreationDate="2014-02-03T08:32:33.467" Score="1" ViewCount="13864" Body="&lt;p&gt;The heat of combustion of 2-propanol at 298.15 K, determined in a bomb calorimeter, is -33.41 kJ/g. For the combustion of one mole of 2-propanol, determine (a) $\Delta U$ and (b) $\Delta H$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I managed to determine $\Delta U$ by first finding the amount grams in one of 2-propanol and then using that number with 33.41 kJ/g...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not able to determine Delta H as I'm not sure what to do... Please help and guide me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4407" LastEditorUserId="1362" LastEditDate="2014-11-04T08:07:28.763" LastActivityDate="2014-11-04T08:07:28.763" Title="How to determine Delta H from the heat of combustion?" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8253" PostTypeId="2" ParentId="8249" CreationDate="2014-02-03T16:05:02.457" Score="2" Body="&lt;p&gt;Are you supposed to determine $\Delta_rH$ for the reaction (which is $\Delta_cH$ for the combustion of 2-propanol) or are you supposed to determine $\Delta_fH$ for the formation of 2-propanol? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the first case, it might be hard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The change in internal energy $U$ is equal to the sum of the heat transferred $q$ and the work done $w$. If we assume that only pressure volume work is done, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta U=q+w = q-p\Delta V $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative sign in front of $p\Delta V$ is to ensure that a decrease in volume is work being done on the system (system gains energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a thermodynamic potential, internal energy is only easy to use if the pressure is constant (and $\Delta V$ is measurable), which is not true in a bomb calorimeter, or the volume is constant (so $\Delta V =0$), which is true in a bomb calorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant pressure: $\Delta U=q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy is defined such that &#xA;$$\Delta H = \Delta U + \Delta(pV) = q - p\Delta V + p\Delta V + V\Delta p $$&#xA;$$\Delta H = q + V\Delta p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, enthalpy change is only easy to use if the volume is constant (true in a bomb calorimeter) &lt;strong&gt;and&lt;/strong&gt; the pressure change is knowable (you were not given that data) or if the pressure does not change (so that $\Delta p=0$).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second case is simpler.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you were asked to determine $\Delta_f H$ for the formation of 2-propanol, you probably are encouraged to assume that for the combustion reaction $\Delta_rH\approx \Delta_cU$ and go from there. Then, you will need a balanced equation (2-propanol + oxygen forms what?) and some standard enthalpies of formation for the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rH \approx \sum_{products}{\Delta_fH}-\sum_{reactants}{\Delta_fH}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-03T16:05:02.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8254" PostTypeId="1" CreationDate="2014-02-03T16:45:21.387" Score="2" ViewCount="1504" Body="&lt;p&gt;When I took up ionic equilibria and titrations after a long break, I found it hard to solve the questions regarding pH calculations of polyprotic acds. Consider these two questions as examples:-  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Find the approximate pH of a solution with $0.4M\text{ }\ce{HCO3-}$ and $0.2M\text{ }\ce{CO3^2-}$$K_{a1}=4\times 10^{-7}$,&#xA;  $K_{a2}=4\times 10^{-11}$&lt;/li&gt;&#xA;  &lt;li&gt;Find the approximate pH of the resulting solution formed after mixing $20ml$ of $0.1M\text{ }\ce{H3PO4}$ and $20ml$ of $0.1M\text{ }\ce{Na3PO4}$. Express in terms of $pK_{a1},pK_{a2},pK_{a3}$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have mentioned 2 questions here because I want to know the method to solve &lt;em&gt;such type&lt;/em&gt; of questions and not these two questions here in particular. The problem arises when there is a polyprotic acid with hydrolysable salts. For the second case, the ongoing equilibria are the dissociation of the three acids and the hydrolysis of their respective salts. For the first one, there are just two dissociation equilibria(and the dissociation of water). Even by writing down the equilibrium expressions of these reactions, I am unable to reach the final expression with pH or pOH. And, all the solutions have to be approximate (using approximations like the negligible contribution of hydrogen from the dissociation of water, low value of the successive ionization constants) and not at all accurate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate if someone explained how to do such numericals, taking one of the above as an example (I already have the answers for the aforementioned questions. I just want to use them to illustrate the general principle behind these questions).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if my question looks too homework-like and off-topic.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2014-02-04T05:23:23.700" LastActivityDate="2014-02-04T05:23:23.700" Title="Calculating approximate pH of polyprotic acids" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ions&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8255" PostTypeId="2" ParentId="8254" CreationDate="2014-02-03T17:18:51.267" Score="4" Body="&lt;p&gt;For the first scenario, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; is probably good enough. In this case, we assume the major species to be $\ce{HCO3-}$ and $\ce{CO3^{2-}}$ and we will use $K_2$ in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH}=\text{p}K_2 + \log_{10}{\frac{[\ce{CO3^{2-}}]}{[\ce{HCO3-}]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was working on a more general approach using systematic treatment of equilibrium, and it was driving me crazy. I will try to come back to it for the second part of you question.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2014-02-03T18:09:25.037" LastActivityDate="2014-02-03T18:09:25.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8256" PostTypeId="1" CreationDate="2014-02-03T17:59:00.773" Score="6" ViewCount="15285" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the following reaction&lt;br&gt;&#xA;           $$\ce{CH3\bond{-}CH\bond{=}CH2 -&amp;gt;T[$\ce{Br2}$/$\ce{NaCl}$]?}$$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would be the product(s) for this reaction? I am confused because there are two nucleophiles here, $\ce{Br-}$ and $\ce{Cl-}$. So, there are two alternatives for the second step. Let me explain what I am saying.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the $\pi$-electrons will attack an electrophile which in this case is a $\ce{Br+}$ which is obtained form polarisation of the $\ce{Br\bond{-}Br}$ and it is added to the less substituted carbon, which complies with Markovnikov's rule, forming the carbocation below:&#xA; $$\ce{CH3\bond{-}CH+\bond{-}CH2\bond{-}Br}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the carbocation will be attacked by a nucleophile. There are 2 nucleophiles in this problem, as I have stated earlier.&lt;br&gt;&#xA;So, my question is which one of them will attack the carbocation and subsequently what will be the product? Will it be only 1,2-dibromopropane or only 1-bromo-2-chloropropane or a mixture of both? Please give the reasons.&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorUserId="5026" LastEditDate="2018-03-03T15:59:07.740" LastActivityDate="2018-03-03T15:59:07.740" Title="Reaction of alkenes with bromine in presence of sodium chloride" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8258" PostTypeId="2" ParentId="8256" CreationDate="2014-02-03T19:02:29.443" Score="8" Body="&lt;p&gt;The first step is $\ce{Br-Br}$ approaching the pi-cloud, being polarized, loosing a $\ce{Br^{-}}$ to form a 3-cyclic bromonium cation.  Given solubility and reactively inert solvent, $\ce{Br^{-}}$ (large poor nucleophile) competes with $\ce{Cl^{-}}$ (smaller, better nucleophile) for ring-opening capture.  Now factor in relative concentrations.  Now factor in thermodynamics (reversibility, $\ce{Cl^{-}}$ wins for the stronger $\ce{C-Cl}$ bond) versus kinetics (smaller $\ce{Cl^{-}}$ vs. initial tight ion pair rapidly collapsing to product).  Temperature probably makes a product ratio difference, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your linear ion intermediate is controlling, the $\ce{Cl}$ comes in at the 2-position by default.  If my cyclic ion intermediate is controlling, the $\ce{Cl}$ comes in at the 1-position by steric hindrance. And if you do it in water, you get halohydrin byproducts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep the reaction way from light or you will get radical abstraction and allylic substitution.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4231" LastEditDate="2015-03-11T19:20:20.810" LastActivityDate="2015-03-11T19:20:20.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8259" PostTypeId="1" CreationDate="2014-02-03T19:03:55.230" Score="5" ViewCount="2925" Body="&lt;p&gt;Recently I tried to hydrolyze some cellulose: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added some acid to it ($\ce{H2SO4}$) then noticed nothing. Even after a few days there was no visible change and after doing some tests for glucose, those were negative too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the end - expecting nothing would happen - I added some sodium hydroxide ($\ce{NaOH}$) to the solution to neutralize the acid and then the most unexpected thing happened, the solution began to warm up and there was an immediate color change (turning from white (from the cellulose) to yellow/brownish) I suspect that color came from the dehydration of the glucose  that was produced at that moment due to instantaneous hydrolysis of cellulose. I finally diluted the solution with more water and the reaction stopped&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So somehow there was no interaction between the sulfuric acid and the cellulose, but when sodium hydroxide was added, the solution did more than just neutralize the acid!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any idea why this is?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="30836" LastEditorDisplayName="user4076" LastEditDate="2019-04-14T09:45:23.557" LastActivityDate="2019-04-14T09:45:23.557" Title="Unexpected reaction during acid hydrolysis of cellulose" Tags="&lt;organic-chemistry&gt;&lt;carbohydrates&gt;&lt;hydrolysis&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8260" PostTypeId="2" ParentId="8246" CreationDate="2014-02-03T19:20:03.187" Score="4" Body="&lt;p&gt;Molecular mass has NO pertinence to BP, water and ammonia.  Look up the BP of perfluorocyclohexane, hexafluorobenzene, and the hydrocarbons.  If I want to volatize iron, ferrocene.  If I want to volatize iron more, $\ce{Fe(hfac)_3}$ where the ligand is the bidentate enolate anion of 1,1,1,5,5,5-hexafluoro-2,4-pentanedione.  Big MW.  What are the MW and BP of $\ce{UF_6}$?  $\ce{-Si(CH3)_3}$ also confers volatility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling points are more about polarizability, intermolecular van der Waals and such. Bigger molecules are overall electronically floppy.  Benzene, toluene, xylenes – nice BP steps, but not because of MW. Look up $\ce{SF_6}$ bp.  $\ce{I(CF3)_7}$ boils near 0 °C.  Calculate that MW, and its vapor’s mass/liter at STP compared to dry air at 29 g/liter. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-06-26T07:47:04.903" LastActivityDate="2015-06-26T07:47:04.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8261" PostTypeId="2" ParentId="8259" CreationDate="2014-02-03T19:30:31.393" Score="2" Body="&lt;p&gt;Aside from the exotherm... Repeat the experiment with warming.  Glucose can air-oxidize, removing the aldehyde with which glucose tests react,  In fact, you can get the peracid from aldehyde free radical air oxidation, ruining reductive glucose assays toward any ring-opened glucose present.  Glucose slow hydrolytic release in a vessel exposed to air can be tested by doing it in seltzer as the reaction medium, then close the jar or cap cap after fizzing subsides, thus purging solution and containment with carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-03T19:30:31.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8262" PostTypeId="2" ParentId="7645" CreationDate="2014-02-03T21:07:43.070" Score="3" Body="&lt;p&gt;In general, I'm not sure double bond and coordinate covalent bond are mutually exclusive labels and depending on the specific chemical you're looking at, they might both apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Reasoning:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Coordinate bonds are bonds where both electrons were donated by one atom. Usually, the concept of a double bond is used in the context of valence bond theory and coordinate bond is used in the context of crystal field theory. There is evidence for pi bonding in coordinate complexes (i.e. back bonding) leading to what could be called a double bond in valence bond theory.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you look at bond length data you can usually determine if there is a double bond. (Or if you have some of the acid and access to an IR, you can do this yourself.) The CRC should contain reference bond lengths.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I 'm not an expert in phosphorous chemistry, but these could be helpful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.colby.edu/chemistry/webmo/mointro.html&quot; rel=&quot;nofollow&quot;&gt;Structural calculations for many compounds, including a few acids&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://simons.hec.utah.edu/papers/317.pdf&quot; rel=&quot;nofollow&quot;&gt;Study on structures of phosphates, not acids&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4412" LastEditorUserId="4412" LastEditDate="2014-02-05T18:33:07.943" LastActivityDate="2014-02-05T18:33:07.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8263" PostTypeId="2" ParentId="4613" CreationDate="2014-02-03T22:22:05.527" Score="2" Body="&lt;p&gt;In my experience, calculated parameters like topological indices are not viewed as significant if considered alone. It's a strategy for finding trends or correlations and, subsequently, candidates from  large volumes of chemical structures. Many studies will publish trends they find in a target chemical groups which are not meant to apply to every chemical group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of how all these seemly useless data is useful - see the Merck Kaggle Challenge. &#xA;&lt;a href=&quot;https://www.kaggle.com/c/MerckActivity&quot; rel=&quot;nofollow&quot;&gt;https://www.kaggle.com/c/MerckActivity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blog.kaggle.com/2012/11/01/deep-learning-how-i-did-it-merck-1st-place-interview/&quot; rel=&quot;nofollow&quot;&gt;http://blog.kaggle.com/2012/11/01/deep-learning-how-i-did-it-merck-1st-place-interview/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical significance is harder to comment on. I work with between 8 and 20 million data points and the idea of how to define statistical significance is evolving in cheminformatics as it evolves in big data analytics. It gets much more complicated because standard methods, such as the t-test, fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;&#xA;I just realized there might be some fundamental confusion. Mol files include atomic information as do most graph based chemical representations. They're stored as properties of the nodes. An algorithm can choose to use or ignore this information. Not all topological parameters ignore atomic information. &lt;/p&gt;&#xA;" OwnerUserId="4412" LastEditorUserId="4412" LastEditDate="2014-02-03T22:27:05.673" LastActivityDate="2014-02-03T22:27:05.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8264" PostTypeId="2" ParentId="4613" CreationDate="2014-02-03T22:50:30.757" Score="2" Body="&lt;p&gt;Carvone, $\ce{C_10H_14_O}$:  &lt;I&gt;R&lt;/I&gt;-(–)-Carvone is spearmint, &lt;I&gt;S&lt;/I&gt;-(+)-carvone is caraway.  Are the stereoisomers' graphs different?  Quantitative indices generate a paradox that well-managed research fairs poorly in basic discovery (absent insubordination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In counterbalance, a single failed reaction is a setback but a million failed reactions are a combinatorial library.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-03T22:50:30.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8265" PostTypeId="1" CreationDate="2014-02-04T01:51:55.683" Score="9" ViewCount="4796" Body="&lt;p&gt;I held gallium under a heat lamp. It didn't melt. I didn't even find a thin film of liquid gallium. Can anyone explain why? Does gallium rust?&lt;/p&gt;&#xA;" OwnerUserId="4416" LastEditorUserId="1782" LastEditDate="2017-04-06T21:02:08.360" LastActivityDate="2021-01-03T15:57:13.503" Title="My sample of gallium does not melt" Tags="&lt;metal&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8268" PostTypeId="2" ParentId="8265" CreationDate="2014-02-04T11:07:45.853" Score="-1" Body="&lt;p&gt;Elemental gallium does not occur in free form in nature, but as the gallium(III) compounds. So it mustn't be gallium but something else. &lt;/p&gt;&#xA;" OwnerUserId="4420" LastEditorUserId="11367" LastEditDate="2017-04-04T20:23:12.690" LastActivityDate="2017-04-04T20:23:12.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8270" PostTypeId="2" ParentId="6144" CreationDate="2014-02-04T18:29:58.040" Score="3" Body="&lt;p&gt;The mixed carboxylic anhydride, unless sterically hindered (the big sulfonyl helps) and electron rich, will be labile toward nucleophilic attack hydrolysis.  Acetic anhydride usefully persists in cold (acidified) water.  It goes exponentially as it exothermically hydrolyzes.  Keep it cold.  As above, use phase transfer for selective reactivity re the nylon rope trick.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-04T18:29:58.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8271" PostTypeId="2" ParentId="6802" CreationDate="2014-02-04T18:43:25.447" Score="2" Body="&lt;p&gt;Draw the LCAO orbitals.  Delocalization is the summation of discrete electron occupations of orbitals, such as a 1,3 shift (a pi-bond flip).  Remember that LCAO is a terrifically wrong black box. Its only virtue is that it simply, beautifully works (until the Woodward-Hoffmann rules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given propene, abstract a methyl proton.  The LCAO anion is half at both ends by a 1,3-shift of the pi bond.  MO shows degenerate orbital occupation still with (nearly) no additional density at the central carbon.  Abstract an allylic proton from 1-butene.  The allylic ends are no longer 50:50 occupied by the anion for the different electronic properties of two attached protons versus a methyl group and a proton.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-04T18:43:25.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8273" PostTypeId="2" ParentId="6821" CreationDate="2014-02-04T20:36:42.180" Score="3" Body="&lt;p&gt;Sulfuric acid protonates the alkene to give the highest substituted carbocation, &lt;I&gt;tert&lt;/I&gt;-butyl.  That electrophilically adds to a benzene carbon atom, to give the obvious next cation.  A proton is eliminated to obtain &lt;I&gt;tert&lt;/I&gt;-butylbenzene.  Akyl groups activate benzene rings toward &lt;I&gt;ortho, para&lt;/I&gt; attack (mild electron donation via  hyperconjugation), thus 1,6-di-&lt;I&gt;tert&lt;/I&gt;-butylbenzene, certainly if the addend is in excess.  Especially stable carbocations can dissociate and wander around the benzene ring under Friedel-Crafts conditions.  If you push it, you get 1,3,5-tri-&lt;I&gt;tert&lt;/I&gt;-butylbenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steric hindrance strongly discourages further reaction.  You can go to the literature to find how many adjacent &lt;I&gt;tert&lt;/I&gt;-butyls can be jammed onto a benzene ring - and how it is done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With due respect, you are in deep trouble for passing the course if the Friedel-Crafts reaction throws you.   Either look at things with the LCAO model, look in a few more textbooks, or talk with folks who have grasped it.  You must understand it.  You cannot survive substituent effects if you cannot flip bonds.  This is easy stuff compared to what will follow.  Chess can be well-played by computer.  Organic synthesis is done in wetware, though computers help help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemtube3d.com/model/electrophilic_aromatic_substitution/friedel-crafts/alkylation/benzene%2Btbu/mechanism.png&quot; rel=&quot;nofollow&quot;&gt;http://www.chemtube3d.com/model/electrophilic_aromatic_substitution/friedel-crafts/alkylation/benzene%2Btbu/mechanism.png&lt;/a&gt; &lt;BR&gt;&#xA;Close enough&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-04T20:36:42.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8274" PostTypeId="1" CreationDate="2014-02-04T22:24:29.623" Score="0" ViewCount="266" Body="&lt;p&gt;I want to know if there is any difference at all between the alkali metals as far as conducting electricity goes.&lt;/p&gt;&#xA;" OwnerUserId="4426" LastActivityDate="2014-02-04T23:15:50.327" Title="Are some alkali metals better conductors than others?" Tags="&lt;electricity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8275" PostTypeId="2" ParentId="8274" CreationDate="2014-02-04T23:10:52.777" Score="1" Body="&lt;p&gt;&lt;PRE&gt;&#xA;Li &#xA;Conductivity 1.08×10^7 S/m&#xA;Resistivity  9.26×10^(-8) mΩ&#xA;Na&#xA;Conductivity 2.18×10^7 S/m&#xA;Resistivity  4.59×10^(-8) mΩ&#xA;K&#xA;Conductivity 1.43×10^7 S/m&#xA;Resistivity  7.0×10^(-8) mΩ&#xA;Rb&#xA;Conductivity 8.0×10^6 S/m&#xA;Resistivity  1.25×10^(-7) mΩ&#xA;Cs&#xA;Conductivity 5.3×10^6 S/m&#xA;Resistivity  1.9×10^(-7) mΩ&#xA;Fr&#xA;Conductivity 1.4×10^6 S/m&#xA;Resistivity  7.1×10^(-7) mΩ&#xA;&lt;/PRE&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-04T23:10:52.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8276" PostTypeId="2" ParentId="8274" CreationDate="2014-02-04T23:15:50.327" Score="1" Body="&lt;p&gt;Here is the &lt;em&gt;total resistivity&lt;/em&gt; &amp;rho; (in $10^{-6}\ \Omega{}\,\textrm{m}$ at 273.16 K) from &lt;a href=&quot;http://dx.doi.org/10.1063/1.555598&quot; rel=&quot;nofollow&quot;&gt;Electrical resistivity of alkaline earth elements&lt;/a&gt;, &#xA;T. C. Chi, &lt;em&gt;J. Phys. Chem. Ref. Data&lt;/em&gt;, &lt;strong&gt;1979&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 439-497:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li  8.53&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Na  4.33&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K   6.49&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rb 11.54&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cs 18.75&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fr 28.00&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-04T23:15:50.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8281" PostTypeId="1" AcceptedAnswerId="8284" CreationDate="2014-02-05T08:04:12.877" Score="9" ViewCount="44995" Body="&lt;p&gt;A covalent bond involves overlapping of orbitals while an Ionic bond involves charge separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are bonds formed by the overlapping of orbitals weaker than charge separation; why is an ionic bond stronger than a covalent bond?&lt;/p&gt;&#xA;" OwnerUserId="4432" LastEditorUserId="9961" LastEditDate="2018-08-11T19:54:24.523" LastActivityDate="2020-03-02T19:02:08.750" Title="Are ionic bonds stronger than covalent bonds?" Tags="&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8282" PostTypeId="1" CreationDate="2014-02-05T10:30:22.100" Score="2" ViewCount="186" Body="&lt;p&gt;How does cooling/refrigeration help in maintaining the taste and shelf life of food products, especially CSD (Carbonated Soft Drinks)? For example, chilled drinks have a nicer taste than a normal drink and may have longer shelf life. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="4433" LastEditorUserId="275" LastEditDate="2014-02-05T13:00:39.100" LastActivityDate="2014-10-02T23:26:50.310" Title="cooling/refrigeration essentiality for carbonated softdrinks" Tags="&lt;thermodynamics&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2014-10-03T07:33:26.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="8283" PostTypeId="1" AcceptedAnswerId="8285" CreationDate="2014-02-05T11:03:07.650" Score="2" ViewCount="602" Body="&lt;p&gt;The position of equilibrium, I read, is dependent only on the relative Gibbs energies of the products and the reactants. Why is this so? Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium. Why, therefore, is the position of equilibrium between products and reactants only dependent on the relative gibbs energies of the products and reactants.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditDate="2014-02-05T11:17:17.457" LastActivityDate="2019-04-26T13:15:55.700" Title="Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8284" PostTypeId="2" ParentId="8281" CreationDate="2014-02-05T11:08:28.180" Score="8" Body="&lt;p&gt;In a fully covalent bond, you only have the mixing of the orbitals, as you said. But it's not only that, you also have coulomb interactions between the nuclei, that slightly raise the energy of the molecular orbitals, which results in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4517/how-can-antibonding-orbitals-be-more-antibonding-than-bonding-orbitals-are-bondi/8039#8039&quot;&gt;asymmetric energies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a fully ionic bond you also have a coulomb interaction, but this time with the different sign, because you have a positive and a negative charge. These want to be as close together as possible, resulting in a stronger bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking: &lt;strong&gt;Coulomb interactions have a greater impact than molecular orbitals on the energy, and thus the strength, of a bond.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's almost like saying that electromagnetic forces are stronger than gravitational forces.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-02-05T11:08:28.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8285" PostTypeId="2" ParentId="8283" CreationDate="2014-02-05T11:27:49.613" Score="0" Body="&lt;p&gt;I quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_of_mixing&quot; rel=&quot;nofollow&quot;&gt;Wikipedia:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In thermodynamics the entropy of mixing is the increase in the total entropy when several &lt;strong&gt;initially separate&lt;/strong&gt; systems of different composition, each in a thermodynamic state of internal equilibrium, are mixed &lt;strong&gt;without chemical reaction&lt;/strong&gt; by the thermodynamic operation of removal of impermeable partition(s) between them, followed by a time for establishment of a new thermodynamic state of internal equilibrium in the new unpartitioned closed system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the scenario which you describe is an already mixed solution of reactants. So you have already maximized the entropy, because for every molecule that changes according to the reaction, you get the new molecule at the same place. The overall entropy with respect to the mixing doesn't change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the position of the equilibrium is only then dependent on the relative gibbs free energies when the reaction is not kinetically hindered (when the activation energy barrier is low enough to be overcome).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-05T11:27:49.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8287" PostTypeId="2" ParentId="7955" CreationDate="2014-02-05T16:03:27.110" Score="4" Body="&lt;p&gt;After reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little &lt;strong&gt;guide to ionic equations&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Total Ionic Equations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 Na3PO4 + 3 CaCl2 -&amp;gt; 6 NaCl + Ca3(PO4)2 v},$$&lt;/span&gt;&#xA;one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -&amp;gt; 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Net Ionic Equations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Net ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 PO4^3- + 3 Ca^2+ -&amp;gt; Ca3(PO4)2 v}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ions that were eliminated are termed &lt;strong&gt;spectator ions&lt;/strong&gt;, because they do not participate in any notable chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If something is still unclear, maybe have a go at &lt;a href=&quot;https://www.occc.edu/kmbailey/Chem1115Tutorials/Net_Ionic_Eqns.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kristy M. Bailey's&#xA;website (Oklahoma City Community College)&lt;/a&gt;, which explains it pretty neatly.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="4945" LastEditDate="2018-09-27T12:18:16.173" LastActivityDate="2018-09-27T12:18:16.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8288" PostTypeId="2" ParentId="8282" CreationDate="2014-02-05T16:17:57.557" Score="0" Body="&lt;p&gt;It has a longer shelf life because the bacteria, (which is the reason it will go bad), cannot reproduce as fast in cold temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as taste goes with carbonated soft-drinks, I think it is mostly just about preference.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2014-02-06T02:03:45.673" LastActivityDate="2014-02-06T02:03:45.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8289" PostTypeId="1" CreationDate="2014-02-05T16:58:25.930" Score="2" ViewCount="6174" Body="&lt;p&gt;The metal carbonyls (and similar organometallic compounds) involve a combination of sigma bond, a pi bond and backbonding. The bond strengths under consideration are the metal-carbon bond and the carbon oxygen bond. Here is the original question:- &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Predict the order of $\ce{C-O}$ bond strength in the following:-&lt;br&gt;&#xA;  &lt;strong&gt;I)&lt;/strong&gt; $[\ce{Mn(CO)_6}]^+$&lt;br&gt;&#xA;  &lt;strong&gt;II)&lt;/strong&gt; $[\ce{Cr(CO)_6}]$&lt;br&gt;&#xA;  &lt;strong&gt;III)&lt;/strong&gt; $[\ce{V(CO)_6}]^-$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of these are isoelectronic, and hence whatever the difference of bond strengths, must arise from the difference in the resultant nuclear force of attraction. Since $\ce{Mn+}$ has the highest charge density, I would assume that the $\ce{M-C}$ bond strength of the manganese compound is the strongest (vanadium being the weakest) and therefore, the $\ce{C-O}$ bond strength must follow the opposite order, as the strengthening of the metal carbonyl bond should weaken the carbon oxygen bond.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer given is &lt;code&gt;III&amp;lt;II&amp;lt;I&lt;/code&gt;, exactly the opposite of my prediction. &lt;strong&gt;Why is it so?&lt;/strong&gt; A relevant factor which I didn't consider is the back bonding, but I am unsure of how to integrate that into my prediction. Also, &lt;strong&gt;how would we make a prediction if the given compounds were not isoelectronic?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2018-11-26T15:39:39.377" Title="Predicting bond-strength of metal carbonyls" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;coordination-compounds&gt;&lt;quantum-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8290" PostTypeId="2" ParentId="7668" CreationDate="2014-02-05T17:49:09.327" Score="2" Body="&lt;p&gt;You could vacuform heated thin PET film over it.  That would not stop biology.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-05T17:49:09.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8292" PostTypeId="1" CreationDate="2014-02-05T19:55:30.820" Score="0" ViewCount="1470" Body="&lt;p&gt;I'm doing a study a study on voltaic cells.&#xA;I've been working with them for a while and one of the questions that I've had is how does the salt bridge affect the cell. I mean, I know it allows ions to move and that it'a a basic component of the pill. But, does its' concentration affect the voltage of the cell itself?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="4439" LastActivityDate="2018-12-31T06:19:50.317" Title="Salinic bridges in voltaic cells" Tags="&lt;electrochemistry&gt;&lt;electricity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8293" PostTypeId="1" AcceptedAnswerId="8300" CreationDate="2014-02-05T20:02:47.907" Score="3" ViewCount="1558" Body="&lt;p&gt;Assume that somebody wants to measure the sugar content of a liqueur. It seems that there are two types of Brix meters for it, ones which use the solution density and ones which use refraction of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a meter intended for measuring the content of a sugar-in-water solution (and assume that we are talking pure sucrose here, not other sugars), but are using it for measuring a solution which has, beside sugar and water, also 15 to 35 percent alcohol and also unknown organic (colored or transparent) substances dissolved: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;which type will give less error, refractional or density-based? &lt;/li&gt;&#xA;&lt;li&gt;how large will the error be? For example, if the true value is 20 g sugar per 100 ml solution, how far are the values delivered by the meter likely to stray from those 20 grams (I am looking for a back of an envelope estimation, not for an exact interval)? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="251" LastEditorUserId="402" LastEditDate="2017-03-30T17:57:29.150" LastActivityDate="2018-01-25T19:02:07.020" Title="Are refractive or density based Brix meters better for use with alcoholic drinks?" Tags="&lt;organic-chemistry&gt;&lt;aqueous-solution&gt;&lt;analytical-chemistry&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8295" PostTypeId="2" ParentId="8292" CreationDate="2014-02-05T22:09:27.887" Score="1" Body="&lt;p&gt;A cell has an internal ohmic-equivalent resistance.  The salt bridge is a resistive path for the flow of ions.  Ions, plus their hydration shells and possible coordination shells, have low drift velocities through a medium.  The salt bridge can constrain the ion current of the cell and its material equilibrium rate.  A poor salt bridge by length, diameter, concentration, and permeability will increasingly adversely affect a cell's current and voltage as discharge rate increases.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-05T22:09:27.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8296" PostTypeId="2" ParentId="6796" CreationDate="2014-02-05T23:30:31.637" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;Assuming i) a primary alcohol $\ce{R-OH}$ to be converted into the corresponding primary alkyl halogenide $\ce{R-X}$ ii) by $\ce{HX}$ acid &lt;em&gt;only&lt;/em&gt;, I would expect a decrease of reactivity to be observed in this line: $\ce{HI} &amp;gt; \ce{HBr} &amp;gt; \ce{HCl} &amp;gt; \ce{HF}$, with $\ce{HF}$ of lowest reactivity.  This is based on the decrease of the nucleophilicity of the anions of these acids.&lt;/li&gt;&#xA;&lt;li&gt;Of course, halogenations with these four acids are reactions of equilibria, and these may be driven forward by increasing the concentration of the reactants, for example.  And I do recall two-step protocols where the alcohol is activated by sulfuric acid before adding hydrobromic acid.  But in both variants, rather $\ce{H3O+}$ then a neutral $\ce{OH}$ would leave the molecule.&#xA;(However secondary and tertiary alcohols that are of higher reactivity than primary ones under these acidic conditions, are more prone to isomerization, too.  Simultaneously, as hinted by ssavec, instead of a mechanism of type $\mathrm{S_{N}1}$, the of  $\mathrm{S_{N}2}$ may become more favorable.)&lt;/li&gt;&#xA;&lt;li&gt;For the preparation of alkyl chlorides, the addition of (dry) $\ce{ZnCl2}$ in presence of $\ce{HCl}$ is sometimes used.  More easy to handle are routes relying on $\ce{PCl3}$ or $\ce{SOCl2}$, for example.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="402" LastEditDate="2017-04-25T00:45:09.430" LastActivityDate="2017-04-25T00:45:09.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8297" PostTypeId="2" ParentId="8293" CreationDate="2014-02-06T02:03:05.723" Score="1" Body="&lt;p&gt;Consider density vs. a Pousse-café.  Refractive index might do less galloping overall - but whoa! for things like Benedictine and absinthe loaded with refractive organic essences.  You need calibration tables either way.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T02:03:05.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8298" PostTypeId="2" ParentId="8289" CreationDate="2014-02-06T02:09:57.477" Score="1" Body="&lt;p&gt;As the π-antibonding orbital of CO is filled by electrons from the metal, they weaken the C-O bond compared with free CO.  The metal-carbon bond is strengthened in kind. That's what the books says.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T02:09:57.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8299" PostTypeId="2" ParentId="7776" CreationDate="2014-02-06T02:26:05.083" Score="1" Body="&lt;p&gt;No, it would not be possible to use a coal fire and frozen oxygen to create a atmosphere able to sustain life for any prolonged amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid oxygen, though not explosive or flammable by itself, would mix with the soot given off by the coal fire and create a highly explosive mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a coal fire gives off a toxic amounts of carbon dioxide, and potentially other gases. Not to mention the soot, which would quickly suffocate anyone breathing it in. &lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorDisplayName="user467" LastEditDate="2014-02-06T18:52:29.793" LastActivityDate="2014-02-06T18:52:29.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8300" PostTypeId="2" ParentId="8293" CreationDate="2014-02-06T02:32:53.107" Score="4" Body="&lt;p&gt;Once I got going, this turned out to be a lengthy answer. So here's the summary: the alcohol will introduce a significant error in both methods. The density measurement will probably be closer to the actual sugar content. Th density measurement will underestimate the sugar, and the refractive index will overestimate the sugar.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Both the refractive index and the density-based methods work for measuring sugar content because adding sugar to water raises the refractive index and the density in amounts proportional to the concentration of the added sugar. As you and others noted in that thread, ethanol interferes with that measurement because it also affects the density and refractive index of the resulting solution. So how large is that effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate this effect, I am going to restrict myself to the 15-35% alcohol (by volume) range you suggested. The charts I found for the &lt;a href=&quot;http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry&quot; rel=&quot;nofollow noreferrer&quot;&gt;density&lt;/a&gt; and &lt;a href=&quot;http://www.refractometer.pl/refraction-datasheet-ethanol&quot; rel=&quot;nofollow noreferrer&quot;&gt;refractive index&lt;/a&gt; of water/ethanol mixtures gives the ratio by weight; your range converts to 12-28% by weight. Pure water has a density of 0.99823 g/mL, and a refractive index of 1.3330. 12% ethanol has a density of 0.97910 g/mL, and a refractive index of 1.3410. 28% ethanol has a density of 0.95710 and a refractive index of 1.3524. Note that adding ethanol &lt;em&gt;increases&lt;/em&gt; the refractive index, but &lt;em&gt;decreases&lt;/em&gt; the density. I'll come back to that at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; gives a table of the index of refraction and density of sucrose solutions in water, as a function of the sucrose weight fraction. The 20 g sucrose per 100 mL solution is approximately 20% by weight, so that will have a refractive index of 1.3639, and a density of 1.0810 g/mL. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density and refractive index changes for both sucrose and ethanol are pretty linear over the range you're interested in, if you take them independently. Assuming that the sucrose effects and the ethanol effects are independent (NB: this may not be a good assumption), the refractive index measurement will be higher by 0.01-0.02, which corresponds to an overestimate of 7-14% sucrose. The density measurement will be &lt;em&gt;lower&lt;/em&gt; by an amount 0.02-0.04 g/mL, which corresponds to an underestimate of 5-10% sucrose. So it looks like the hydrometer measurement will be a bit better, but the two methods will be off in different directions. It's a noticeable effect: depending on the method and the alcohol content, you might be off by somewhere between a quarter and half of your intended sugar content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one neat thing you could do to take advantage of the fact that the alcohol moves the density in the opposite direction to the sugar. If you had a bottle of liquor with unknown alcohol content and unknown sugar content, you could take both the density and the refractive index measurement. Again, assuming the four effects are linear, and the sugar effects are independent of the alcohol, you could write two equations for the refractive index $RI$ and density $\rho$ as functions of the alcohol and sugar: $RI = Ax + By, \rho = Cx + Dy$. $x$ and $y$ are your alcohol and sugar content in whatever units you prefer, and $A, B, C,$ and $D$ are the empirical parameters describing each of the four effects. If you take a measurement for $RI$ and for $\rho$, you can solve for $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: to answer one question suggested by the comments of the [cooking.se] question you linked, density-based brix meters (I'm used to seeing them called hydrometers) are typically used for home brewing beer and wine because the only source of alcohol is the fermenting sugar. You take a density measurement before fermentation begins to get a baseline. Then, at any point after fermentation begins, the difference between the current density and the baseline density is directly proportional to the alcohol content. I'm sure the same reasoning applies to the refractive index method, but I've never used that for beer-brewing (I've only used it for my real research).  See Cutaia, A.J., Reid, A-J. and R.A. Speers,  2009. Examination of the relationships between Original, Real and Apparent Extracts, and Alcohol in Pilot Plant and Commercially Produced Beers. J. Inst. Brew. 115:318-327.&lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="-1" LastEditDate="2017-03-30T17:08:42.293" LastActivityDate="2017-03-30T17:08:42.293" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8301" PostTypeId="2" ParentId="7668" CreationDate="2014-02-06T08:41:22.660" Score="1" Body="&lt;p&gt;In my opinion this is a (at least) two steps process:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make the fruit's peel more durable.&lt;/li&gt;&#xA;&lt;li&gt;Coating the inner part of the object.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h1&gt;Make the fruit's peel more durable.&lt;/h1&gt;&#xA;&lt;p&gt;The conservation of fruit peels (seems strange but is true) is a well know problem in the field of restoration of contemporary art.&#xA;&lt;a href=&quot;http://www.getty.edu/conservation/publications_resources/newsletters/13_2/news1_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zoe Leonard &amp;quot;Strange Fruit (for David)&amp;quot;&lt;/a&gt; is the classical case of study about this topic.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R37bq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Zoe Leonard after the death of a friends of hers begun to sews the peels and do with different material to recreate the original shape of the fruit. This work of course was very degradable so a German conservator Christian Scheidemann developed  a method to treat the fruit peel to make them more durable. He used the Paraloid B72, a well know ethyl-methacrylate copolymer used in conservation of cultural heritage, under vacuum. I'm quoting from the link above:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Early on, her dealer, Paula Cooper, suggested the possibility of preservative intervention for the sculpture. Leonard was amenable and worked for two years with German conservator Christian Scheidemann to devise a way to arrest the decay of the fruit surfaces. After much testing, Scheidemann developed a solution that consisted of shock-freezing the pieces and then penetrating them with Paraloid B72 under vacuum. This solution was complicated by the need to protect the wires, threads, and other elements from the Paraloid B72; in other words, the piece presented the intricacies typical of any mixed-media work. But Scheidemann succeeded in this as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A generic more detailed research about food conservation can be found in this &lt;a href=&quot;http://ceroart.revues.org/2207&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Coating the inner part of the object&lt;/h1&gt;&#xA;&lt;p&gt;As other user stated generally plastic glasses and other for drinking are made of polystyrene, PET or polycarbonate. The third one is more durable and use for stuff used more times. However coat a cup or a vase with plastic on you own is quite dangerous (there are chances of poisoning if done in the wrong way) and expensive and difficult.&#xA;If you don't use vacuum you should use blow molding over a mold of the interior of the fruit (so you have to make a first mold and then another &amp;quot;negative&amp;quot; mold) and the attach the treated peel in a second moment with some adhesive, very expensive process. I could go to a Murano glassmaker with your mold! In this case you can add a plastic or glass edge to your cup I don't think is very nice to drink from a plastic peel.&lt;/p&gt;&#xA;&lt;p&gt;There is an alternative cheapest way, using &lt;strong&gt;Food Grade Silicone Mold Rubber&lt;/strong&gt; this is probably what I would used you can find in internet a lot of types of silicone rubber that can fit your needs.&lt;/p&gt;&#xA;&lt;p&gt;However I don't think you can later wash it in the dishwasher!&lt;/p&gt;&#xA;&lt;h1&gt;Edit&lt;/h1&gt;&#xA;&lt;p&gt;As stated Jaroslav Kotowski you can use even a more expensive epoxy resin however you should pay attention to use a food grade epoxy resin. Non food grade epoxy resin could be very toxic.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-02-06T08:57:52.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8302" PostTypeId="2" ParentId="7689" CreationDate="2014-02-06T10:54:58.317" Score="2" Body="&lt;p&gt;Potassium alum is $\ce{KAl(SO4)2*12H2O}$, the compound that you use to stop bleeding after a bad shave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say that it won't react, but don't put your health in the hands of strangers on the net. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-06T10:54:58.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8303" PostTypeId="2" ParentId="7668" CreationDate="2014-02-06T11:25:05.317" Score="2" Body="&lt;p&gt;This probably doesn't count as a real answer since the underlying technology isn't coating, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is possible to turn deceased people and their organs into durable objects should (in principle) work for a pumpkin or a melon too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of &lt;strong&gt;plastination&lt;/strong&gt;, as used by Gunther von Hagens for his &lt;a href=&quot;http://www.bodyworlds.com/en&quot; rel=&quot;nofollow&quot;&gt;Body Worlds&lt;/a&gt; exhibitions involves the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;dissection and embalming&lt;/em&gt;: cut the fruit open and remove the &quot;flesh&quot;, soak with aqeous formaldehyde to stop decay&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;removal of soluble fats&lt;/em&gt;: place object in an acetone bath&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;impregnation&lt;/em&gt;: place object in a vacuum chamber (exsiccator) filled with a reactive polymer and apply vacuum&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;curing&lt;/em&gt;: depending on the polymer, apply visible or uv light to harden the object&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Admittedly, this is more of a mind game, but it should work :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-06T11:25:05.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8304" PostTypeId="1" CreationDate="2014-02-06T11:41:00.637" Score="-1" ViewCount="16394" Body="&lt;p&gt;Is there any chemical substance, whose solubility increases with decrease in temperature ?&lt;/p&gt;&#xA;" OwnerUserId="4442" LastEditorUserId="1431" LastEditDate="2014-02-06T17:24:56.840" LastActivityDate="2017-11-24T17:27:19.433" Title="Substance whose solubility increases with decrease in temprature" Tags="&lt;solubility&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8305" PostTypeId="2" ParentId="8304" CreationDate="2014-02-06T12:33:18.010" Score="3" Body="&lt;p&gt;The &lt;strong&gt;solubility of a solid&lt;/strong&gt; depends on the temperature in the following way: When the net dissolving heat is negative (i.e. when heat is given off by the dissolving process) the solubility is higher at lower temperatures because the energy is more readily absorbed by the dissolving medium.&#xA;This works analogously the other way around (and is usually more the case, as you seem aware of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;gases&lt;/strong&gt; the relationship is usually the other way around (increase in temperature results in a decrease of solubility).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reason for this gas solubility relationship with temperature is very similar to the reason that vapor pressure increases with temperature. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further explanations (and the source of the quote above) can be found &lt;strong&gt;&lt;a href=&quot;http://www.elmhurst.edu/~chm/vchembook/174temppres.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-06T12:33:18.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8306" PostTypeId="2" ParentId="8259" CreationDate="2014-02-06T13:46:06.587" Score="6" Body="&lt;p&gt;We miss some data to make a good hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The concentration of the $\ce{H_2SO_4}$ solution.&lt;/li&gt;&#xA;&lt;li&gt;The temperatures.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of temperature such that.  However some &lt;a href=&quot;http://www.hindawi.com/isrn/chemical.engineering/2012/428974/&quot; rel=&quot;nofollow&quot;&gt;studies&lt;/a&gt; show that the reaction can occur at lower temperature with concentrated acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible solution is the formation of alkali cellulose.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user15489" LastEditDate="2015-06-08T21:19:09.077" LastActivityDate="2015-06-08T21:19:09.077" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8307" PostTypeId="2" ParentId="8304" CreationDate="2014-02-06T14:01:01.193" Score="5" Body="&lt;p&gt;Gases are usually more soluble at lower temperature, in fact there are some solid substances the show this behavior too.&#xA; The classical example is $Cerium \space III sulphate$ $Ce_2(SO_4)_3$. This behavior is due to the fact that $Ce_2(SO_4)_3$ has an exothermic &lt;strong&gt;Heat of Solution&lt;/strong&gt;. So for the principle of Le Chatelier increasing the temperature cause a decrease of the solubility of the salt that tend to compensate the increment of temperature absorbing energy through the formation of a precipitate, that in this case is an endothermic process.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;These substance are not so common, however if you deal with sulphate, selenate or some strange metals you should take a look in literature checking if they have this behavior. As suggested Nicolau Saker Neto  I've check the wikipedia list, I've made this plot from there, in the Y axis is shown how much the substance is soluble in water as grams dissolved per 100 grams of water:&#xA; &lt;img src=&quot;https://i.stack.imgur.com/E333c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-02-06T15:51:25.583" LastActivityDate="2014-02-06T15:51:25.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8308" PostTypeId="1" CreationDate="2014-02-06T14:07:21.757" Score="4" ViewCount="8921" Body="&lt;p&gt;I understand that all of these are made up of carbon atoms but I don't understand what differentiates them. Since they are carbon they all must have covalent bonds, right? &lt;/p&gt;&#xA;" OwnerUserId="4443" LastEditorUserId="11367" LastEditDate="2015-09-14T12:59:34.530" LastActivityDate="2017-01-17T10:17:49.950" Title="What's the difference between graphite, nanotubes, and fullerenes?" Tags="&lt;nanotechnology&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8309" PostTypeId="2" ParentId="2411" CreationDate="2014-02-06T15:17:26.030" Score="3" Body="&lt;p&gt;$\ce{(en)}$ serves as a ligand that binds to the metal ($\ce{Co}$) in the complex &lt;strong&gt;&lt;em&gt;dinitro bis(ethylenediammine) cobalt (II) chloride&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4444" LastEditorUserId="7448" LastEditDate="2015-07-03T11:05:19.707" LastActivityDate="2015-07-03T11:05:19.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8310" PostTypeId="1" CreationDate="2014-02-06T15:46:00.573" Score="4" ViewCount="7451" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Series &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; below can be viewed in two different ways to annulate an ethano-bridge onto four-, five-, and six-membered rings. Using the &lt;span class=&quot;math-container&quot;&gt;$\Delta_\mathrm{f}H^\circ$&lt;/span&gt; values given (in kcal/mol), how can one calculate the strain energy for each compound?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GypQq.png&quot; alt=&quot;2 different ways to annulate an ethano-bridge onto four-, five-, and six-membered rings&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Could anyone also explain whether the trend seen in each series is consistent with expectations based on additivity of ring strain?&lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-05-31T08:42:47.953" Title="How to calculate ring strain energy?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8311" PostTypeId="2" ParentId="2411" CreationDate="2014-02-06T15:50:57.450" Score="0" Body="&lt;p&gt;&lt;I&gt;en&lt;/I&gt; is $\ce{H2NCH2CH2NH2}$, a bidentate ligand coordinating through its nitrogen lone pairs, one pair/nitrogen.  Compare with oxalate.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T15:50:57.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8313" PostTypeId="2" ParentId="7689" CreationDate="2014-02-06T15:59:31.740" Score="1" Body="&lt;p&gt;Finely divided magnesium in water corrodes, liberating hydrogen.  &quot;magnesium oil?&quot;  That is magnesium chloride in water.  It is New Age hype.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T15:59:31.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8315" PostTypeId="2" ParentId="8310" CreationDate="2014-02-06T16:23:42.787" Score="8" Body="&lt;p&gt;Typically, ring strain energies (RSE) are calculated as the difference of the heat of formation of &lt;em&gt;cyclic&lt;/em&gt; starting materials and &lt;em&gt;less cyclic&lt;/em&gt; products in &lt;strong&gt;homodesmic&lt;/strong&gt; reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cyclopropane, the RSE would be calculated using the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH2)3 + 3 CH3-CH3 -&amp;gt; 3 CH3-CH2-CH3}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-06T16:23:42.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8316" PostTypeId="1" AcceptedAnswerId="8335" CreationDate="2014-02-06T17:30:31.697" Score="1" ViewCount="702" Body="&lt;p&gt;I have a solution containing many different organic compounds (originally from leek and potato soup) and I would like to extract the starch from it. However, starch is insoluble in water. How could I go about extracting the starch?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="16683" LastEditDate="2015-10-09T16:28:06.913" LastActivityDate="2015-10-09T16:28:06.913" Title="Extraction of starch from soup" Tags="&lt;organic-chemistry&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8317" PostTypeId="2" ParentId="8316" CreationDate="2014-02-06T19:28:30.503" Score="0" Body="&lt;p&gt;Doesn't dissolve in water?  Filter it.  Celite is your friend.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T19:28:30.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8318" PostTypeId="1" CreationDate="2014-02-06T19:59:24.563" Score="2" ViewCount="465" Body="&lt;p&gt;Also, why can some elements such as $\ce{Fe}$ have different ionic forms such as $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$? What determines each form?&lt;/p&gt;&#xA;" OwnerUserId="4449" LastEditorUserId="4231" LastEditDate="2015-03-10T14:39:16.933" LastActivityDate="2015-03-10T14:39:16.933" Title="Why don't B, C, Si, have ionic forms?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8319" PostTypeId="2" ParentId="7635" CreationDate="2014-02-06T22:34:12.440" Score="4" Body="&lt;h1&gt;I'm afraid no: you are applying a chemical principle to a complex biological response.&lt;/h1&gt;&#xA;&lt;p&gt;The Le Chatelier principle state:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If a &lt;strong&gt;chemical system at equilibrium&lt;/strong&gt; experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We start with the definition of chemical system at equilibrium :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the temperature dependent condition in which chemical species are present at concentrations which have no further tendency to change with time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So what you need are chemical species: molecules, ions. Although &lt;strong&gt;fatty acids&lt;/strong&gt; could be a product &lt;strong&gt;&amp;quot;body&amp;quot;&lt;/strong&gt; is not a compound or a molecule. So what you have is not a chemical system but a biological one. You could go a little more close if you think something like this:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fatty acid &amp;lt;=&amp;gt; Chemical  \space Precursor_{Y} + Chemical \space Percursor_{X}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However even this is a wrong oversimplification of many complex biochemistry reactions and equilibria.&lt;/p&gt;&#xA;&lt;p&gt;Moreover you can't interact with these equilibria, perturbing them and so applying LC principle because sebum, which is a complex mixture of lipids, is stored inside &lt;em&gt;sebaceous glands&lt;/em&gt; and is formed by the &lt;strong&gt;complete disintegration of the cell&lt;/strong&gt; it self. So to from new sebum you have to from new &lt;em&gt;sebaceous glands&lt;/em&gt;.&#xA;Soap simply remove sebum not seabaceous glands, so using stronger soap by the way don't affect sebum production. At this point is clear that is no more chemistry but physiology. In this case an equivalent of Le Chatelier's principle is the &lt;strong&gt;homeostasis&lt;/strong&gt;. However from what I know the diet is one of the main factors that induce &lt;em&gt;sebaceous glands&lt;/em&gt; to produce sebum so eat better seems to be the best way to lose weight and sebum!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;References&lt;/h1&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2835893/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pub Med Sebaceous gland lipids&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-02-06T22:34:12.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8321" PostTypeId="2" ParentId="8304" CreationDate="2014-02-06T23:19:26.037" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://adsabs.harvard.edu/abs/1992JCrGr.116..427G&quot; rel=&quot;nofollow&quot;&gt;http://adsabs.harvard.edu/abs/1992JCrGr.116..427G&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://patents.justia.com/patent/4481069&quot; rel=&quot;nofollow&quot;&gt;http://patents.justia.com/patent/4481069&lt;/a&gt; &lt;BR&gt;&#xA;Hydrothermal synthesis of berilinite is also retrograde (negative temperature coefficient) solubility.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-06T23:19:26.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8322" PostTypeId="2" ParentId="6796" CreationDate="2014-02-06T23:58:10.267" Score="0" Body="&lt;p&gt;$\ce{HBr}$ is a stronger acid than $\ce{HCl}$ (pKa -9 vs. -7), meaning that $\ce{Br-}$ feels more comfortable in dissociated form than $\ce{Cl-}$. In other words, $\ce{Br-}$ is a better leaving group. I think this has a direct relationship to reactivity via $S_N1$.&lt;/p&gt;&#xA;" OwnerUserId="4452" LastEditorUserId="418" LastEditDate="2014-02-07T06:45:39.373" LastActivityDate="2014-02-07T06:45:39.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8323" PostTypeId="2" ParentId="8281" CreationDate="2014-02-07T00:08:45.433" Score="3" Body="&lt;p&gt;Two atoms will always (with a few corner-case exceptions) form the strongest bond between them that they can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they form a covalent bond then that is because the covalent bond is stronger than the alternative ionic bond (at least after taking into account the ionisation energies). This is either because the covalent bond is strong (good orbital overlap) or the ionisation energies are so large that they would outweigh the ionic lattice enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they form an ionic bond then that is because the ionic bond is stronger than the alternative covalent bond. This is either because the covalent bond is weak (poor orbital overlap / mismatched orbital energies) or the ionisation energies are relatively small compared to the lattice enthalpy.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-02-07T00:08:45.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8324" PostTypeId="2" ParentId="2606" CreationDate="2014-02-07T00:25:20.570" Score="-1" Body="&lt;p&gt;The reaction between alkali metals and water produces hydrogen, which will rise up and mix with the oxygen in the air above. If the reaction generates enough heat, it can ignite this mixture - hydrogen and oxygen will detonate when mixed in the correct ratio. This explosion will either directly spray out the metal (which by then will probably be molten) or force it down into the water where it will produce even more hydrogen gas and steam, propelling it back up and outwards.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-02-07T00:25:20.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8325" PostTypeId="1" AcceptedAnswerId="8329" CreationDate="2014-02-07T01:27:40.787" Score="1" ViewCount="1658" Body="&lt;p&gt;I'm an AP Chemistry student, and we're doing the Decomposition of $\ce{H2O2}$ lab, in which we decompose $\ce{H2O2}$ with $\ce{KI}$ as a catalyst. I've gotten this data so far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+&#xA;| Trial | Volume H2O2 | Initial [H2O2] | Volume KI | Initial [KI] | Temperature | Rate (kPa/s) | Initial rate (M/s) | [H2O2] after mixing | [KI] after mixing |&#xA;+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+&#xA;|     1 |           4 | 0.882          |         1 | 0.5          | 21.2        | 0.4121       | 0.000168           | 0.706               | 0.10              |&#xA;|     2 |           4 | 0.882          |         1 | 0.25         | 21.1        | 0.2011       | 0.000082           | 0.706               | 0.05              |&#xA;|     3 |           4 | 0.441          |         1 | 0.5          | 21.3        | 0.1844       | 0.000075           | 0.353               | 0.10              |&#xA;|     4 |           4 | 0.882          |         1 | 0.5          | 31.2        | 0.8211       | 0.000334           | 0.706               | 0.10ds            |&#xA;+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think that this is a first order reaction with regards to both, since when you halve the amount of each reactant, the rate of the reaction also halves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I am running into trouble is calculating the rate constant (k) of the reaction. I don't really know where to start on this. I thought of dividing the initial rate by the $\ce{H2O2}$, but that wouldn't account for the amount of $\ce{KI}$. How would I go about calculating the rate constant of this reaction? I don't need an answer here, but pointing me in the right direction would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4453" LastEditorUserId="865" LastEditDate="2021-05-22T21:18:30.120" LastActivityDate="2021-05-22T21:18:30.120" Title="How do I calculate the order of a reaction given slope?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8326" PostTypeId="2" ParentId="8318" CreationDate="2014-02-07T01:36:55.753" Score="2" Body="&lt;p&gt;Boron: borates, borides.  $\ce{MgB2}$ is the highest temperature BCS superconductor by a large margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon: carbonates, carbides, hexacoordinate carbon in complexes, organolithiums, Grignard reagents, carbenium ions ($\ce{Ph3C+}$).  $\ce{Al4C3}$ hydrolyzes to methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon: silicates, aluminosilicates, silicides.  Cook up silica gel in ethylene glycol plus potassium hydroxide (&lt;em&gt;NOT&lt;/em&gt; in glass) to get pentacoordinate siliconate, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://patentimages.storage.googleapis.com/WO1994020507A1/imgf000014_0001.png&quot; rel=&quot;nofollow&quot;&gt;http://patentimages.storage.googleapis.com/WO1994020507A1/imgf000014_0001.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked the second half in the Physics Stack.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4231" LastEditDate="2015-03-10T14:39:14.467" LastActivityDate="2015-03-10T14:39:14.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8327" PostTypeId="1" CreationDate="2014-02-07T02:13:50.697" Score="8" ViewCount="378" Body="&lt;p&gt;I'm just wondering, if one were to run a $\ce{^{11}B}$ NMR on a boron containing sample in a borosilicate tube, how could we separate the spectra of the tube and the sample? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from using a different tube - Perhaps through a pulse sequence like BIRD for $\ce{^1H/^13C}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or decoupling - which I've only just started to learn about and don't really understand?&lt;/p&gt;&#xA;" OwnerUserId="4454" LastEditorUserId="16683" LastEditDate="2019-05-09T00:31:46.417" LastActivityDate="2019-05-09T00:31:46.417" Title="¹¹B NMR in a borosilicate tube" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8328" PostTypeId="2" ParentId="8327" CreationDate="2014-02-07T03:15:14.303" Score="2" Body="&lt;p&gt;You'd run the nmr in a quartz tube, or a tube made from some other non-boron containing material.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-02-07T03:15:14.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8329" PostTypeId="2" ParentId="8325" CreationDate="2014-02-07T06:07:04.870" Score="0" Body="&lt;p&gt;As mentioned, you don't want a direct answer so I am just going to ask you some question to think about, that hopefully help in getting you to the answer. Please reply in the comments whenever something is unclear to you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're expecting first order in both reactants, so if you would write down the reaction rate equation, what does that look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the equation you should have one unknown only: $k$ (or two that make up $k$: $k_0$ and one other, which?, if you are interested in the temperature effect as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now calculate $k$ for each reaction separately. Are they the same? Do you expect them to be? If they're not, should you average them?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-02-07T06:07:04.870" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8330" PostTypeId="1" AcceptedAnswerId="8334" CreationDate="2014-02-07T08:29:31.987" Score="2" ViewCount="95" Body="&lt;p&gt;Today I was making a pastille of Niobium Pentafluoride powder. The procedure includes a 6 ton squeezer, &lt;a href=&quot;http://www.kcprofessional.com/products/wipers/specialty/kimtech-science/05511-kimtech-science-precision-wipes-tissue-wipers&quot; rel=&quot;nofollow&quot;&gt;precision wipes&lt;/a&gt;, silica gel, filter papers and acetone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first suspicious part was that on product details, $NbF_5$ was suppossed to be $5\mu m$ powder. Instead it looked like having hair(forgive my bad terminology but I am not a chemist, but a physicist and  I don't know how to describe it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The procedure is the following.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Acetone and wipe papers$\rightarrow$clean squeezers parts(made of stainless steel)&lt;/li&gt;&#xA;&lt;li&gt;Powder gets into the squeezer(using a funnel made of the filter paper) and gets pressed&lt;/li&gt;&#xA;&lt;li&gt;The powder becomes a pastille.&lt;/li&gt;&#xA;&lt;li&gt;It is stored in a plastic bottle, with Silica gel.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The filter paper is also used to fill the bottle with silica gel(as a funnel) as well as a way to store the pastille while filling the bottle with silica gel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made several pastilles with the exact same recipe(i.e. $B,\;ZnF_2,\;LiF$ etc) and everything went fine. This powder however became very sticky. I have also noticed that while it's stored in the silica gel box, there are some drops inside the bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have some questions... First of all why did it became sticky and secondly why the powder wasn't actually a powder?Is it typical for $NbF_5$?&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="4457" LastEditDate="2014-02-07T08:44:31.047" LastActivityDate="2014-02-07T11:41:17.830" Title="Niobium (V) Fluoride becoming sticky" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8333" PostTypeId="2" ParentId="8327" CreationDate="2014-02-07T11:37:38.850" Score="7" Body="&lt;p&gt;The borosilicate tube gives a very broad signal, the signals from the boron compounds you're measuring should be much sharper. If your samples are concentrated enough and have reasonably sharp lines, that should be sufficient to distinguish them from the broad background signal from the NMR tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could of course also measure an identical NMR tube with the same amount of solvent, but without your boron sample. Comparing this with your real sample should make it easy to identify which signals are background from the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course you could use an NMR tube made from a different material.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-02-07T11:37:38.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8334" PostTypeId="2" ParentId="8330" CreationDate="2014-02-07T11:41:17.830" Score="3" Body="&lt;p&gt;From the properties block on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nbf5&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for $\ce{NbF4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Niobium (V) fluoride is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hygroscopic&quot; rel=&quot;nofollow&quot;&gt;hygroscopic&lt;/a&gt; - it absorbs water from the atmosphere. This hygroscopicity is your main problem. When the powder absorbs water, it probably is becoming sticky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The added water is probably also providing the extra mobility the ions need to crystallize from the powder. Niobium (V) fluoride forms &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoclinic&quot; rel=&quot;nofollow&quot;&gt;monoclinic crystals&lt;/a&gt;, which on a very small scale could look like hair or needles.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-07T11:41:17.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8335" PostTypeId="2" ParentId="8316" CreationDate="2014-02-07T14:07:07.113" Score="1" Body="&lt;p&gt;You can't simply filter it and take the feed because all the vegetables fiber remain in the filter with your starch. If you want to make a starch with a decent level of purity you should start directly with the potatoes: grating the potatoes, putting them in hot water (preferably salted: this make more effective the separation from proteins) for a bit of time. Then take the filtered solution (not the residual) make the water evaporate now what you get is the starch with the potato protein, amino acids and minerals (what is called fruit water). There is this &lt;a href=&quot;http://www.youtube.com/watch?v=aMmXm3p48BE&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; if you want to see.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you want to have a more pure starch also you can try to precipitate the proteins warming the solution and adding an acid. If you are at home acetic acid could be enough.&#xA;Eventually you can cool down the solution this make starch precipitate so you can remove the water with a filter and take the precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to isolate starch there are more sophisticated methods see &lt;a href=&quot;http://books.google.it/books?id=Anbz_whRM2YC&amp;amp;pg=PA604&amp;amp;dq=starch+extraction&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=gOD0Uq4m78bsBva-gYgP&amp;amp;ved=0CDAQ6AEwAA#v=onepage&amp;amp;q=starch%20extraction&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://books.google.it/books?id=A-AiYsGTd38C&amp;amp;pg=PA39&amp;amp;dq=starch+extraction&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=6-L0UuWWCcnnywOGpICgDg&amp;amp;ved=0CDcQ6AEwAQ#v=onepage&amp;amp;q=starch%20extraction&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for references.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-02-07T16:17:01.853" LastActivityDate="2014-02-07T16:17:01.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8336" PostTypeId="1" CreationDate="2014-02-06T19:44:26.307" Score="1" ViewCount="101" Body="&lt;p&gt;Like $\ce{Fe}$, it can be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$, same with $\ce{Ni}$ and $\ce{Co}$, what determines whether the ionic compound will be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$?&lt;/p&gt;&#xA;" OwnerUserId="4449" OwnerDisplayName="user17338" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:09:31.417" LastActivityDate="2015-10-23T17:09:31.417" Title="Why can some ions have different valence?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8337" PostTypeId="2" ParentId="8336" CreationDate="2014-02-07T00:22:40.900" Score="2" Body="&lt;p&gt;An empty, half-filled or fully-filled electron shell is especially stable.  $\ce{Fe}$ metal is [Ar] 3d6 4s2.  $\ce{Fe^{2+}}$ emptiess the 4s-shell.  $\ce{Fe^{3+}}$ give that plus a half-filled 3d-shell.  Vanadium II, III, IV, V.  Now try chloride, hypochlorite, chlorite, chlorate, and perchlorate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most stable oxidation state depends on the medium and added ligands, if any.  Cu(I) is a Lewis soft acid. It is stabilized by soft bases like pyridine, $\ce{MeCN}$, olefins, $\ce{I^{-}}$.  $\ce{Cu^{2+}}$ is a hard Lewis acid.  It is stabilized by hard bases like ammonia, water, $\ce{Cl^{-}}$.  Dissolve $\ce{CuCl2}$ in pyridine or $\ce{MeCN}$ and you have a powerful oxidizing agent. Dissolve a $\ce{Cu^{1+}}$ salt in water and it disproportionates into $\ce{Cu}$ metal and $\ce{Cu^{2+}}$.  Put $\ce{Cu}$ metal and $\ce{Cu^{2+}}$ into $\ce{MeCN}$ and you get $\ce{Cu^{1+}}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" OwnerDisplayName="Uncle Al" LastEditorUserId="4231" LastEditDate="2015-03-10T17:22:04.480" LastActivityDate="2015-03-10T17:22:04.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8338" PostTypeId="2" ParentId="4499" CreationDate="2014-02-07T16:06:50.087" Score="2" Body="&lt;p&gt;Yes, congo red bind the cellulose because is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Substantive_dye&quot; rel=&quot;nofollow&quot;&gt;substantive (or direct) dye&lt;/a&gt; these dyes don't need a mordant so it binds suddenly to the cellulose.&#xA; There are others dyes for example safflower, cochineal with this property. In biology Congo Red is frequently substitute with &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/papers/21417582&quot; rel=&quot;nofollow&quot;&gt;Sirus Red F3B&lt;/a&gt; and Tioflavone S. At this &lt;a href=&quot;http://stainsfile.info/StainsFile/stain/amyloid/directdye.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; you can find some alternatives to Congo Red use for demonstrating amyloid, however I don't know if you can apply these dyes to your case, because I don't find any reference about their use as pH indicator. Although is reported that many substantive dyes are used as pH indicator I can't find any list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take in account that acid pigment are not substantive to cellulosic fiber and most of the pH indicator are acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that maybe you can try to use a mordant to obtain the same result with a pH indicator normally used, however I'm afraid that using a mordant you risk to change the behavior of your pH indicator.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-02-07T16:06:50.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8340" PostTypeId="1" AcceptedAnswerId="8341" CreationDate="2014-02-07T18:43:22.667" Score="12" ViewCount="6470" Body="&lt;p&gt;1) In text books, the absorption event is usually described by a figure shown below&lt;sup&gt;1&lt;/sup&gt;: light excites electrons to higher states in atom or molecule. The difference between the ground state and the excited one matches the energy of the absorbed light. But electron states in atoms and molecules should be quantized, so why do we observe absorbance spectra that are smooth and continuous but not spiked and quantized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iTKir.png&quot; alt=&quot;http://www.photobiology.info/Visser-Rolinski.html&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Where does the energy go when the excited electron returns to its normal state? In fluorescence, the electrons return to lower states and excite light at higher wavelength than the absorbed one. But if I had a molecule that did not exhibit fluorescence, just plain absorption, does the molecule emit light at the same wavelength it absorbed or does the energy turn to heat? I mean, the electron cannot be excited forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is the physical phenomenon behind absorption coefficient? I have been thinking that it has something to do with the lifetime of the excited state &amp;ndash; shorter lifetimes increase the probability of a photon to be absorbed, resulting in larger coefficients. Am I completely wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Image taken from &lt;a href=&quot;http://www.photobiology.info/Visser-Rolinski.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.photobiology.info/Visser-Rolinski.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4452" LastEditorUserId="418" LastEditDate="2014-02-07T18:53:02.310" LastActivityDate="2016-11-02T23:29:08.623" Title="Why is absorbance spectrum continuous and not quantized?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8341" PostTypeId="2" ParentId="8340" CreationDate="2014-02-07T18:57:53.820" Score="10" Body="&lt;p&gt;The image shows the absorption and emission spektra of &lt;strong&gt;molecules&lt;/strong&gt;. For isolated atoms, you would indeed observe sharp lines. Think in the &lt;em&gt;Fraunhofer lines&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UV/VIS absorption of molecules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the case of molecules, absorption occurs from the vibrationally relaxed electronic ground state ($S_0$) to various vibration levels of the electronically excited $S_1$ state. This results in a broad absorption band instead of a sharp line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manifold of $S_1$ with different vibrational levels does not live forever. Within the shortest time, relaxation to the &lt;em&gt;vibrational ground state&lt;/em&gt; of $S_1$ occurs. The electron distribution of the excited state is typically significantly different from the $S_0$ state. This results in a completely changed reactivity. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Radiationless deactivation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you don't observe emission from an $S_1$ or $T_1$ state, the excited state is deactivated by &lt;strong&gt;internal conversion&lt;/strong&gt;, i.e. by vibrational exchange with the bulk (solvent).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-07T19:37:01.330" LastActivityDate="2014-02-07T19:37:01.330" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8343" PostTypeId="2" ParentId="7776" CreationDate="2014-02-07T21:14:02.277" Score="4" Body="&lt;p&gt;Yes, it is possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll answer your questions last to first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, low pressure oxygen will not make flammability worse, fires require something to bond with oxygen (often carbon, causing carbon dioxide). Low pressure will make things less flammable as there is less oxygen to bond with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, using an open coal fire in a semi-enclosed space will gain toxic levels of carbon dioxide, carbon monoxide and sulfur dioxide (among a few others) quickly. However with a minor modification from a open fire to a fire in a old fashioned chimney pipe stove (venting outside the hut) would eliminate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, Definitely Yes, a coal fire will produce more than enough energy to thaw the oxygen it uses. To quote figures 1 pound of coal will produce roughly 12,500 btu's (British Thermal Units) Assuming the temperature of the frozen oxygen we are thawing is -455°F (the &quot;temperature of deep space&quot;) and about 3 degrees below the freezing temperature of helium. To thaw 1 gallon of the frozen oxygen will require roughly 900 btu's (if your stove has a 80% efficiency) and produce 860 gallons of pure oxygen at 80°F and atmospheric pressure. Coal will consume 140% of it's weight in pure oxygen, so for 1 pound of coal 135 gallons of your oxygen will be consumed. Leaving you with 11,500 btu's extra heat and 725 gallons of oxygen. The average Human consumes 25 gallons of oxygen per day making this plenty to survive on.&lt;/p&gt;&#xA;" OwnerUserId="4463" LastActivityDate="2014-02-07T21:14:02.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8344" PostTypeId="1" CreationDate="2014-02-08T03:57:49.547" Score="3" ViewCount="2529" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Bromination of an alkene by &lt;em&gt;N&lt;/em&gt;-bromosuccinimide (NBS) in the presence of light or peroxide is a radical reaction and produces an allylic bromide. For the following bromination of 3-methylcyclopentene, which of the following allylic bromides would be products of the reaction?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D5vZt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D5vZt.png&quot; alt=&quot;Allylic bromination of 3-methylcyclopentene&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In some other reactions such as elimination, there is a tendency to form the most substituted double bond, since it is the most stable. If that were the case, then the products should be &lt;strong&gt;A&lt;/strong&gt;, &lt;strong&gt;E&lt;/strong&gt;, &lt;strong&gt;F&lt;/strong&gt;, and &lt;strong&gt;G&lt;/strong&gt;, since these are all trisubstituted alkenes (the others are disubstituted). Is that a valid consideration here, or is there a different overriding rule?&lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-05-29T19:39:31.077" Title="What are the possible products in allylic bromination of 3-methylcyclopentene?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;&lt;radicals&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8345" PostTypeId="1" AcceptedAnswerId="8347" CreationDate="2014-02-08T06:40:56.603" Score="7" ViewCount="314" Body="&lt;p&gt;The atomic radii are estimated using a variety of methods: most of these involve dividing their bond length by 2. But that is a very crude way of measuring atomic radii. I mean, atoms overlap each other when bonded, surely the actual atomic radius is much more than the one obtained by this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't they simply use the Schrodinger's wave equation to calculate atomic radii. If it can predict atomic orbitals to such accuracy, surely it can predict something as simple as atomic radius?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="418" LastEditDate="2014-02-08T10:30:00.653" LastActivityDate="2014-11-11T04:22:21.743" Title="Why aren't atomic radii calculated using Schrödinger's equation?" Tags="&lt;atoms&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8346" PostTypeId="2" ParentId="8344" CreationDate="2014-02-08T07:11:37.773" Score="2" Body="&lt;p&gt;Your comment shows that you are on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have chosen &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt; because they are formed when the initial allylic radical &lt;strong&gt;instantaneously&lt;/strong&gt; reacts with $\ce{Br2}$ in the propagation step of the reaction, isn't it? (&lt;em&gt;Hint: Just say YES!&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now give the &lt;strong&gt;allylic radicals&lt;/strong&gt; corresponding to &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt; a bit of time, just enough to &lt;em&gt;flip the dot&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these other radicals look like? There you have the other two products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an extra point: Which isomer of methylcyclopentene could lead to &lt;strong&gt;G&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-08T07:11:37.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8347" PostTypeId="2" ParentId="8345" CreationDate="2014-02-08T08:19:57.050" Score="5" Body="&lt;pre&gt;&lt;code&gt;Erwin with his psi can do&#xA;Calculations quite a few.&#xA;But one thing has not been seen:&#xA;Just what does psi really mean?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Erich_H%C3%BCckel&quot;&gt;Erich Hückel&lt;/a&gt;, english translation by &lt;a href=&quot;http://en.wikipedia.org/wiki/Felix_Bloch&quot;&gt;Felix Bloch&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not easy to perform the calculation that you called &lt;em&gt;simple&lt;/em&gt;. The Hamilton operator $\hat{H}$ for a multi-body problem of $K$ cores and $N$ electrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\hat{H} = - \frac{1}{2}\sum_a^K\frac{1}{M_a}\Delta_a - \frac{1}{2}\sum_i^N\frac{1}{M_i}\Delta_i - \sum_i^N\sum_a^K \frac{Z_a}{r_{ai}} + \sum_{i &amp;lt; j}^N\frac{1}{r_{ij}}+ \sum_{a &amp;lt; b}^K\frac{Z_aZ_b}{R_{ab}}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is still a beast for $K=1$ with all the core-electron and electron-electron interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while the question is legitimate, the calculation simply cannot be done without a lot of approximations, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot;&gt;Hartree-Fock&lt;/a&gt; method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, for the determination of the atomic radii of solid metals, there's x-ray diffraction, which I wouldn't call crude.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-08T15:39:56.903" LastActivityDate="2014-02-08T15:39:56.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8348" PostTypeId="1" CreationDate="2014-02-08T09:10:15.320" Score="1" ViewCount="967" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;No nodes should be introduced in the rotational interconversion between the &lt;em&gt;s&lt;/em&gt;-&lt;em&gt;trans&lt;/em&gt; and &lt;em&gt;s&lt;/em&gt;-&lt;em&gt;cis&lt;/em&gt; forms of 1,3-butadiene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the explanation for that, if one of the $\pi$-systems should rotate $180^\circ$? It seems to be that two positive and negative lobes turn upside down. &lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="16683" LastEditDate="2015-10-09T16:28:16.687" LastActivityDate="2015-10-09T16:28:16.687" Title="Conversion of s-trans butadiene to s-cis butadiene" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8349" PostTypeId="2" ParentId="8348" CreationDate="2014-02-08T15:01:18.360" Score="1" Body="&lt;p&gt;During the interconversion of s-cis and s-trans 1,3-butadiene, the molecule passes through a geometry where the two ethylenic arms are rotated 90 degrees one from the other.  In this conformation the two ethylenic arms are orthogonal to each other and there is no overlap between them.  In an MO sense they can be treated as two separate ethylenes.  Rotation beyond this conformation allows overlap to reoccur and the phase of the HOMO will be correctly reestablished with only one node occurring between C2 and C3.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-02-08T15:01:18.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8350" PostTypeId="1" AcceptedAnswerId="8353" CreationDate="2014-02-08T16:07:15.197" Score="1" ViewCount="4522" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a reaction: $$\ce {ClF3 &amp;lt;=&amp;gt; ClF + F2}$$ with $\ce {K_\mathrm{c} = 2.48 x 10^{-3}}$ Initially, $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, and $\ce {[F2] = 0.02 M} $. $\ce {F2}$ is added until the concentration of $\ce {F2} $ is $\ce {0.1 M} $.&lt;br&gt;&#xA;  What is the concentration of these substances after equilibrium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt;&#xA;$\ce {ClF3 &amp;lt;=&amp;gt; ClF + F2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial concentration: $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, $\ce {[F2] = 0.1 M} $&lt;br&gt;&#xA;Concentration that react: $\ce {[ClF3] = +x M} $, $\ce {[ClF] = -x M} $, $\ce {[F2] = -x M}$&lt;br&gt;&#xA;Concentration after equilibrium: $\ce {[ClF3] = 0.25+x M} $, $\ce {[ClF] = 0.031-x M} $, $\ce {[F2] = 0.1-x M} $&lt;br&gt;&#xA;So, $\ce {Kc = [ClF].[F2].[ClF3]^{-1}} $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I applied the concentration after equilibrium in $\ce {Kc} $, so I got:&lt;br&gt;&#xA;$\ce {2.48.10^{-3} = [0.1-x].[0.031-x].[0.25+x]^{-1}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculate the equation above, I got $\ce {x = 0.1111 M}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my question is: if $\ce {x = 0.1111 M} $, so $\ce {[ClF] = -0.0801 M} $ and $\ce {[F2] = -0.0111 M} $. The concentration is negative. Is it possible? Are my steps incorrect or the answer is correct?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="5026" LastEditDate="2018-02-20T16:15:01.820" LastActivityDate="2018-02-20T16:15:01.820" Title="Equilibrium calculation gives negative molarity" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8351" PostTypeId="1" AcceptedAnswerId="8388" CreationDate="2014-02-08T16:39:29.990" Score="7" ViewCount="10209" Body="&lt;p&gt;I wondered if it was possible to predict the product(s) of a reaction (or even the equilibrium) based on the reactants and the temperature/pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an answer to a similar question &lt;a href=&quot;https://chemistry.stackexchange.com/a/2809/4447&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No, there is no such thing as chemical reaction are not that easily&#xA;  predictable. You can get result right in some cases, but it is&#xA;  impossible to count all factors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What (significant) factors also influence a reaction? Is it the arrangement or state of the chemicals? Is it too related to other physical properties to ignore them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please name the most important ones.&lt;/p&gt;&#xA;" OwnerUserId="4447" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-04-27T09:22:27.077" Title="Software for predicting chemical reactions" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8353" PostTypeId="2" ParentId="8350" CreationDate="2014-02-08T17:23:09.217" Score="4" Body="&lt;p&gt;The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-02-08T17:23:09.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8355" PostTypeId="1" AcceptedAnswerId="8467" CreationDate="2014-02-08T18:33:31.940" Score="11" ViewCount="7272" Body="&lt;p&gt;Why can a solvent dissolve only a particular amount of solute? &#xA;If we add more solute to the solution, the number of solute particles in contact with water increases. So rate of dissolution should increase, and subsequently, the solubility product should also increase. But why do we have a constant value of solubility product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s. An explanation at the particle level would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="487" LastEditorUserId="4945" LastEditDate="2014-05-13T12:14:58.133" LastActivityDate="2014-11-29T12:06:10.477" Title="Why do solutions become saturated?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8357" PostTypeId="1" AcceptedAnswerId="8426" CreationDate="2014-02-08T21:34:18.827" Score="30" ViewCount="33094" Body="&lt;p&gt;The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes the 3rd shell to start filling up, rather than the fourth?&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2018-06-13T09:21:58.233" Title="Why does the 3rd electron shell start filling up with scandium?" Tags="&lt;electronic-configuration&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="8358" PostTypeId="1" CreationDate="2014-02-08T23:17:42.000" Score="0" ViewCount="1894" Body="&lt;p&gt;What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It keeps mentioning that my regio-selectivity is incorrect. And that I should consider the charge distributions of the diene and dienophile to figure it out correctly, but I honestly have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q2O8G.png&quot; alt=&quot;screenshot of What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="4945" LastEditDate="2019-04-15T14:43:59.093" LastActivityDate="2019-04-16T11:04:58.630" Title="What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8359" PostTypeId="2" ParentId="8357" CreationDate="2014-02-08T23:33:36.560" Score="7" Body="&lt;p&gt;The electron configuration for scandium is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{4s}$ and $\ce{3d}$ orbitals are relatively close in energy, so there are a few anomalies in the order of filling.  A fair question might be, why isn't the scandium electron configuration $\ce{[Ar] 3d^3 4s^0}$.  There is a special stability associated with completely filled orbitals, so the $\ce{3d^1 4s^2}$ configuration is of lower energy than the alternative $\ce{3d^3 4s^0}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-06-13T09:21:58.233" LastActivityDate="2018-06-13T09:21:58.233" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="8360" PostTypeId="2" ParentId="8358" CreationDate="2014-02-09T02:06:18.280" Score="2" Body="&lt;p&gt;These resonance structures and the polarization shown within explain the regioselectivity (1,4 isomer instead of 1,3) in the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uRHG1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2019-04-15T14:45:42.980" LastActivityDate="2019-04-15T14:45:42.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8362" PostTypeId="1" AcceptedAnswerId="8367" CreationDate="2014-02-09T02:52:16.583" Score="3" ViewCount="3191" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSzCD.png&quot; alt=&quot;Finding reactant unsymmetrical diene that yields all four 1,2 and 1,4-addition products and the correct reaction pathways that lead to the 3 products&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I completed 3/4 of the answers, by choosing the correct reaction pathways leading to each of the 3 different products. But I don't know how to draw the reactant diene, I really thought what I initially had was correct, should I make a trans reactant or is there something else missing? &lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="83758" LastEditDate="2020-05-13T15:45:43.983" LastActivityDate="2020-05-13T15:45:43.983" Title="Mono-addition of HCl to unsymmetrical dienes?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8364" PostTypeId="1" CreationDate="2014-02-09T05:30:13.777" Score="3" ViewCount="23482" Body="&lt;p&gt;I am dealing with the dissociation of water into its components hydrogen and oxygen and I am trying to determine the standard Gibbs free energies of each component. My professor said if the substance is an ideal gas in the standard state then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g_{i}^{0}=g_{i}^{\text{pure IG}}(T, P_\text{ref})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this means I can determine it for hydrogen and oxygen because they can be treated as ideal gases, but how would I consider the standard Gibbs free energy for water?&lt;/p&gt;&#xA;" OwnerUserId="616" LastEditorUserId="83758" LastEditDate="2020-05-27T09:10:01.560" LastActivityDate="2020-05-27T09:10:01.560" Title="Determining standard Gibbs free energy for water?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;water&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8365" PostTypeId="1" CreationDate="2014-02-09T06:17:01.193" Score="5" ViewCount="1023" Body="&lt;p&gt;Transition metals can form stable ions with different oxidation states. But I am confused why doesn't only the most stable state exist. Let me clarify my question more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ti^{2+} -&amp;gt; Ti^{3+} + e^-} $$&#xA;With $E^0 = +0.5~\mathrm{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just shows that $\ce{Ti^{3+}}$ is more stable. So my question is that why doesn't $\ce{Ti^{3+}}$ only exist in its most stable form?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="3846" LastEditDate="2014-02-09T15:04:59.253" LastActivityDate="2021-05-17T03:10:50.527" Title="Variable oxidation state of transition metal ions" Tags="&lt;electrochemistry&gt;&lt;ions&gt;&lt;stability&gt;&lt;transition-metals&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8366" PostTypeId="1" CreationDate="2014-02-09T06:20:44.553" Score="5" ViewCount="1800" Body="&lt;p&gt;In most high school chemistry books, it is said that the delocalised electrons in benzene makes it particularly stable. However, how does it contribute to stability? Is there any simple way of understanding this?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="5026" LastEditDate="2018-04-19T01:27:25.650" LastActivityDate="2018-04-19T01:27:25.650" Title="Delocalised electrons and stability of benzene" Tags="&lt;aromatic-compounds&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8367" PostTypeId="2" ParentId="8362" CreationDate="2014-02-09T06:24:24.317" Score="3" Body="&lt;p&gt;Do we agree that the first step ist the addition of a proton to the diene? This is bit different from the addition of HCl to an alkene, where you would rely on the &lt;strong&gt;Markovnikov&lt;/strong&gt; rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, stabilization of the cation through resonance comes into play.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't know how to draw it [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Draw stepwise!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HGzXN.png&quot; alt=&quot;positions to which a proton can add&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, a proton could add to one of the positions &lt;strong&gt;a&lt;/strong&gt; to &lt;strong&gt;d&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do the resulting cations look like and which of them are stabilized by resonance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6K33k.png&quot; alt=&quot;addition of proton to diene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it now a bit more clear?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-09T06:24:24.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8370" PostTypeId="1" AcceptedAnswerId="8797" CreationDate="2014-02-09T09:14:30.907" Score="22" ViewCount="19745" Body="&lt;p&gt;When a pass a construction site where someone is welding two pieces of metal together there is a very distinct smell that I associate with 'burning' metal, although I am not quite sure it is burning in the traditional sense. Also, when someone ignites a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sparkler&quot;&gt;sparkler&lt;/a&gt; this same (or at least similar) smell emerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is causing the smell? Is it the smell of metaloxide particles that are dispersed from the weld site? Or is it something in the air that reacts and causes the smell?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorDisplayName="user7951" LastEditDate="2016-11-29T00:03:53.610" LastActivityDate="2020-09-01T08:19:12.997" Title="What is the smell of 'burning' metal?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;metal&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8374" PostTypeId="1" CreationDate="2014-02-09T12:21:32.387" Score="4" ViewCount="2715" Body="&lt;p&gt;My textbook says that $\ce{[CrF6]^{3-}}$ is an example of $sp^3d^2$ hybridization. But I don't really see how. Chromium has an electronic configuration of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3d: \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give three more electrons and it might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3d: \boxed{\uparrow \downarrow}\boxed{\uparrow \downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow \downarrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the ground state of the electrons. I see no way for electrons to be promoted for the formation of six orbitals. And the electrons have to be promoted all the way up to the $4d$ orbital, which would require a significant amount of energy. What is the exact mechanism of bonding in this ion? Amazingly, I couldn't find anything on the Web about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; My textbook also says that $\ce{[Co(NH3)6]^{3+}}$ is an example of $d^2sp^3$ quite similar to $\ce{[CrF6]^{3-}}$. But when I try to understand the exact mechanism, I encounter a similar problem as a I did above. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="4945" LastEditDate="2015-03-19T06:45:26.987" LastActivityDate="2015-03-19T06:45:26.987" Title="How does CrF₆³⁻ hybridize?" Tags="&lt;bond&gt;&lt;coordination-compounds&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8376" PostTypeId="2" ParentId="7776" CreationDate="2014-02-09T13:44:43.857" Score="4" Body="&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Okay, so after doing some serious number crunching and searching for literature values all over the web, I've come to the following conclusion: &lt;strong&gt;Yes, it is possible to survive in these conditions!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Plausibility of Environment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, let us examine the plausibility of the environment which is described in the short story. Outside the safety of the living space, cordoned off by 30-or-so blankets, lined with aluminium foil, there is a vacuum, and all of the gases in the atmosphere have frozen out and precipitated onto Earth's surface. The exception is helium, which still is liquid. &#xA;If we check the phase diagram of oxygen (note that in the figure below we would have to look at the pressure of 0.001), we see that indeed it is possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a66yt.gif&quot; alt=&quot;Phase diagram of oxygen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.jaea.go.jp/jaeri/english/press/980624/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Maintaining Pressure&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The minimal pressure that humans need to survive (while breathing pure oxygen) is set &lt;a href=&quot;http://cosmoquest.org/forum/showthread.php?121894-Air-Pressure-and-Human-Survival&amp;amp;s=b98b681cdad99d0ddd1b7c1d9cb67297&amp;amp;p=1941432#post1941432&quot; rel=&quot;nofollow noreferrer&quot;&gt;at about 19 kPa&lt;/a&gt;, let's round that up to 20 kPa (0.2 atm). Let's furthermore also assume the space where this small family is living as about $3\times 3\times 2~\mathrm{m^3} = 18~\mathrm{m^3}$. Adding in the factor of a comfortable enough temperature (I pen that at $15~\mathrm{°C} = 288~\mathrm{K}$) we can use the ideal gas law: $$pV = nRT \Longleftrightarrow n=\frac{pV}{RT} = 150~\mathrm{mol\, \ce{O2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need $150~\mathrm{mol}$ of gas &lt;strong&gt;constantly&lt;/strong&gt; in order to maintain pressurization at the desired temperature. Whether 30 heavy rugs really can do that I'll leave to an experimental physicist.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Oxygen Depletion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The fire isn't the only source of oxygen depletion, also the humans inhabiting the small space will use oxygen. Using &lt;a href=&quot;http://health.howstuffworks.com/human-body/systems/respiratory/question98.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; data, an average adult human being will consume $22.5~\mathrm{mol}$ $\ce{O2}$ per day. So we'll round this to $70~\mathrm{mol\, d^{-1}}$ for the whole family (2 adults, 2 kids). This is already corrected for oxygen exhalation, so the system generates an equal amount of $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Total Oxygen Balance and Required Heat&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The only source of oxygen is the pail above the fire, in which the solid oxygen gets heated up to the boiling point (which, at 20 kPa is 77.1 K, see figure below; data source: &lt;a href=&quot;http://webbook.nist.gov/cgi/fluid.cgi?P=0.02&amp;amp;TLow=54.361&amp;amp;THigh=288&amp;amp;TInc=5&amp;amp;Digits=5&amp;amp;ID=C7782447&amp;amp;Action=Load&amp;amp;Type=IsoBar&amp;amp;TUnit=K&amp;amp;PUnit=MPa&amp;amp;DUnit=kg/m3&amp;amp;HUnit=kJ/mol&amp;amp;WUnit=m/s&amp;amp;VisUnit=uPa*s&amp;amp;STUnit=N/m&amp;amp;RefState=DEF&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST&lt;/a&gt;) and then further to ambient temperature. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/taw8T.png&quot; alt=&quot;Density vs T of Oxygen at 0.2 kPa&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here some of the constants that I used for the following calculation: $c_\text{p} = 54~\mathrm{J\, mol^{-1}\, K^{-1}}$ for the liquid and $c_\text{p} = 29~\mathrm{J\, mol^{-1}\, K^{-1}}$ for the vapor phase with $\Delta H_\text{vap}(77.1~\mathrm{K},\; 20~\mathrm{kPa}) = 7.19~\mathrm{kJ\, mol^{-1}}$. The melting heat (at standard pressure, but let's just take this value from &lt;a href=&quot;http://en.wikipedia.org/wiki/Latent_heat#Table_of_latent_heats&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) is $0.445~\mathrm{kJ\, mol^{-1}}$, and the heat capacity of the solid I estimated at $25~\mathrm{J\, mol^{-1}\, K^{-1}}$. Initial temperature is $T_\text{i}=5~\mathrm{K}$, melting temperature (again for standard pressure, but we'll take it) $T_\text{m}=54~\mathrm{K}$, boiling point is given above and final temperature is $288~\mathrm{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total heat needed for heating up $150~\mathrm{mol\, \ce{O2}}$ is $q_\text{tot, 150}=2.43~\mathrm{MJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula given in the paper &lt;a href=&quot;http://fire.nist.gov/bfrlpubs/fire82/art018.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculations of the Heat Release Rate by Oxygen Consumption for Various Applications&lt;/a&gt; I end up with $\dot{V} = 0.177~\mathrm{m^3\, d^{-1}}$, for a oxygen depletion coefficient of $0.8$ in a pure atmosphere of oxygen ($x_\mathrm{\ce{O2}}^0 = 1$, standard pressure is assumed in the formula). This is about 1% of the total volume in the cove, and is also only needed for the initial pressurization. However, since the inhabitants also produce $\ce{CO2}$ at non-negligible levels, and the concentration of said gas should &lt;a href=&quot;http://www.doomedglobe.com/Pages/co2_toxicity.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;stay under 2%&lt;/a&gt; for normal breathing and to avoid the feeling of dizziness, we need to up the oxygen generation rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to keep this whole system going with $x(\ce{CO2}) &amp;lt; 1.5~\%$, we need to generate 4700 mol $\ce{O2}$ every day (and this is without even looking at $\ce{CO2}$ generation by the fire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is inevitable that the exhaust gases from the fire should be led out via the chimney (that is also described in the story) without great mixing with the air in the room. But all in all the heat generation is possible (the oxygen needed to heat the 4700 mol to the temperature is only 46 mol), thus making it a survivable climate.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;By the above estimation-based calculations and considering the limitations, I conclude that it &lt;strong&gt;would indeed be conceivable&lt;/strong&gt; to survive in these conditions.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-09T13:44:43.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8378" PostTypeId="1" CreationDate="2014-02-09T13:55:33.773" Score="1" ViewCount="151" Body="&lt;p&gt;I have a solution containing glucose and some sulfuric acid. I want to distill out the sulfuric acid to then reuse it for other purposes. The problem is the presence of glucose in the solution. I'm afraid that, when increasing the temperature, the sulfuric acid will dehydrate glucose to carbon and other compounds - something that I don't want to happen. Would distillation cause such a reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in case it would - are there any other methods to extract the sulfuric acid?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="16683" LastEditDate="2015-10-09T16:32:30.210" LastActivityDate="2015-10-09T16:32:30.210" Title="Extract acid from a solution" Tags="&lt;organic-chemistry&gt;&lt;extraction&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8379" PostTypeId="1" AcceptedAnswerId="8409" CreationDate="2014-02-09T14:11:38.377" Score="20" ViewCount="19830" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;p&lt;/em&gt;-cresol (4-methylphenol), undergoes the Reimer–Tiemann reaction upon treatment with chloroform ($\ce{CHCl3}$) in alkaline medium. Which of the following is the major and minor product?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. 2-hydroxy-5-methylbenzene-1,3-dicarbaldehyde&lt;br&gt;&#xA;  B. 2-hydroxy-5-methylbenzaldehyde&lt;br&gt;&#xA;  C. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-one&lt;br&gt;&#xA;  D. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-ol   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XzuE2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XzuE2.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer given is that &lt;strong&gt;B&lt;/strong&gt; is the major product, and &lt;strong&gt;C&lt;/strong&gt; the minor product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions about this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is the mono-aldehyde &lt;strong&gt;B&lt;/strong&gt; the major product, and not the dialdehyde &lt;strong&gt;A&lt;/strong&gt;?&lt;br&gt;&#xA;I think both electron-donating groups should greatly increase the reactivity of benzene, allowing the reaction to take place twice. On the other hand, steric factors would oppose the second formylation. I do not know which one predominates.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can one derive the formation of the minor dearomatised product &lt;strong&gt;C&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="30820" LastEditDate="2018-02-17T13:05:40.443" LastActivityDate="2020-08-29T09:38:03.340" Title="Products of Reimer–Tiemann reaction of 4-methylphenol" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8380" PostTypeId="2" ParentId="8374" CreationDate="2014-02-09T14:23:40.177" Score="3" Body="&lt;p&gt;In $\ce{[CrF6]^{3-}}$ the central atom is $\ce{Cr(III)}$ or $\ce{Cr^{3+}}$. Chromium is usually $d^6$, so now (take away 3 electrons) $d^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we start filling in the electron pairs from the ligands, we get the following:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uVP8H.png&quot; alt=&quot;VB model for [CrF6]-&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons from the metal are marked in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons from the ligands (blue) are filled into two $d$, one $s$ and three $p$ orbitals: $d^2sp^3$ hybridization as per solution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try to do the same with $\ce{[Co(NH3)6]^{3+}}$!&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-09T14:23:40.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8381" PostTypeId="2" ParentId="8379" CreationDate="2014-02-09T14:57:25.367" Score="8" Body="&lt;p&gt;The reaction is run under basic conditions.  Early in the reaction, the phenolic proton is removed to generate the phenoxide anion.  If you draw resonance structures, you'll see that negative charge is delocalized to the ortho and para positions.  Dichlorocarbene, an electrophilic species is also generated from the chloroform and the base in this reaction.  The dichlorocarbene will preferentially attack the electron rich ortho and para positions in the phenoxide anion.  The para intermediate is often isolated as a cyclohexadienone (your product C), whereas the ortho intermediate usually reacts further due to the adjacent phenoxide oxygen playing a role, going on to the aldehyde.  Here is an old, but good reference, showing the scope and mechanism of the reaction.  It is generally not a clean or high-yielding reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://alexandria.tue.nl/repository/freearticles/587973.pdf&quot;&gt;http://alexandria.tue.nl/repository/freearticles/587973.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-02-09T14:57:25.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8383" PostTypeId="1" CreationDate="2014-02-09T15:12:46.367" Score="1" ViewCount="1041" Body="&lt;p&gt;Why are lithium salts heavily hydrated? I am not able to understand this because I feel salts of larger cations like Caesium should be more hydrated because of the larger size, they can able to hold more water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the flaw in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="1499" LastEditDate="2014-02-09T15:38:30.087" LastActivityDate="2014-02-09T16:07:13.827" Title="Hydration of lithium salts" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8384" PostTypeId="2" ParentId="8366" CreationDate="2014-02-09T15:13:24.667" Score="5" Body="&lt;p&gt;In physics (or you can google it) you will learn about the particle in a box experiment.  At the atomic level, if you confine a particle to a box, it will have some energy.  It turns out, if you make the box longer, the energy of the particle will be lowered, if you make the box smaller, the particle will have a higher energy.  The same concept applies to benzene.  Basically you start with 3 separate acetylene molecules in which the pi electrons are confined to a 2 carbon segment.  Once you trimerize the acetylene and form benzene, the pi electrons can now roam over 6 carbon atoms, a much longer &quot;box&quot; if you will.  So the pi electrons in benzene will be stabilized (lower energy, longer box) compared to the pi electrons in the starting acetylene (higher energy, shorter box).  Molecular Orbital theory tells us the same thing in a slightly different manner.  In each acetylene (let's just focus on the 2 pi electrons in the acetylene that will be the pi electrons in benzene) we have 2 pi electrons in the HOMO and none in the LUMO.  As we bring these 3 acetylenes closer together to form benzene, the orbitals interact.  The 3 initially degenerate HOMOs in the 3 acetylene molecules split (they are no longer degenerate) as they move closer to becoming benzene and a net energetic stabilization results.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-02-09T15:13:24.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8385" PostTypeId="2" ParentId="8383" CreationDate="2014-02-09T15:35:17.527" Score="3" Body="&lt;p&gt;The larger $\ce{Cs+}$ ion may interact with more water molecules at once than $\ce{Li+}$, &lt;em&gt;but each interaction is weaker&lt;/em&gt;. The prevailing interaction between water molecules and ions are ion-permanent dipole interactions (or just ion-dipole), which are proportional in strength to the amount of charge on the ion and the partial charge on the dipole, and inversely proportional to some power of the distance between the centres of the charges (via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shell_theorem&quot; rel=&quot;nofollow&quot;&gt;shell theorem&lt;/a&gt;). Hence, the closer the water molecules get to the ion, the stronger the interaction. Since $\ce{Cs+}$ is very large, the charge centres don't come so close, and the interaction is relatively weak. $\ce{Li+}$ is much smaller, so there is strong solvation. For sufficiently small, highly charged ions, the first solvation shell comes so close to the ion that it allows covalent bonding to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this succinctly described as $\ce{Cs+}$ having a lower charge density than $\ce{Li+}$, which is about the same thing.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-02-09T16:07:13.827" LastActivityDate="2014-02-09T16:07:13.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8386" PostTypeId="1" CreationDate="2014-02-09T16:17:29.280" Score="3" ViewCount="67" Body="&lt;p&gt;We know that&#xA;$c^2=\frac{\partial p}{\partial ρ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adiabatic compressibility is defined as: $\beta_S=-\frac{1}{V}\frac{\partial V}{\partial p}$ such that the subscript &quot;S&quot; stands for &quot;adiabatic&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I show that $c^2=\frac{1}{\rho \beta_S}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried replacing $V$ by $\frac{m}{\rho}$ but I get for $\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}$&lt;/p&gt;&#xA;" OwnerUserId="4473" LastEditorUserId="69830" LastEditorDisplayName="user2117" LastEditDate="2021-05-19T18:05:24.477" LastActivityDate="2021-05-19T18:05:24.477" Title="Relation between speed of sound c and compressibility" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8387" PostTypeId="1" CreationDate="2014-02-09T16:19:47.633" Score="12" ViewCount="8566" Body="&lt;p&gt;Why are superoxide ions stable only in presence of large cations such as $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorDisplayName="user4076" LastEditDate="2014-02-28T20:55:48.173" LastActivityDate="2019-11-13T18:58:55.280" Title="Stability of superoxides of alkali metals" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8388" PostTypeId="2" ParentId="8351" CreationDate="2014-02-09T17:19:52.143" Score="11" Body="&lt;p&gt;For predicting chemical reaction you should use classical mechanics to take in account the simple interactions between the molecules  and quantum mechanical for all the aspects concerning the molecule itself and complex interactions.&#xA;Knowing the principal parameter such as temperature and pressure that could describe the system with the classical mechanics (kinetic energy etc. etc.) is not enough for predicting the reaction with &lt;em&gt;ab initio&lt;/em&gt; methods. &#xA;Adding the quantum mechanical contribute is mandatory and quite hard to do. A amateur chemist do not have neither the skills nor the computation power require to make such calculations. For the moment you can't do that with any software but the last Nobel was won by three theoretical chemists who are dealing with this topic. Have a look &lt;a href=&quot;http://www.scientificamerican.com/article/2013-chemistry-nobel-for-molecule-computer-models/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case where the reactions are well known you can use a chemical reactions database like &lt;a href=&quot;http://kinetics.nist.gov/kinetics/KineticsSearchForm.jsp&quot; rel=&quot;noreferrer&quot;&gt;the NIST one&lt;/a&gt;.&#xA;This is a simple query:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ERUYx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xx7cu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-02-09T18:35:44.267" LastActivityDate="2014-02-09T18:35:44.267" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8389" PostTypeId="1" AcceptedAnswerId="15330" CreationDate="2014-02-09T17:43:04.790" Score="3" ViewCount="11349" Body="&lt;p&gt;My mom while making Potato chips put some Baking Soda in Oil with it. Just curious to know, it was first time Baking Soda in Oil at that high temperature. Is it safe to eat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer ASAP before my chips get cold.&lt;/p&gt;&#xA;" OwnerUserId="4476" LastEditorUserId="4476" LastEditDate="2014-02-09T18:54:41.383" LastActivityDate="2014-08-16T14:23:26.207" Title="Baking Soda + Oil at high temps" Tags="&lt;heat&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8390" PostTypeId="1" AcceptedAnswerId="8469" CreationDate="2014-02-09T19:10:38.353" Score="0" ViewCount="1338" Body="&lt;p&gt;If $125$ Calories of heat is applied to $60.0\text{g}$ piece of copper at $21.0^\circ\text{C}$, what will the final temperature be? The specific head of copper is $0.0920 \text{ cal/gC}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have to answer this question is a formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SH = \text{heat/(mass}\times \Delta T)$&lt;/p&gt;&#xA;" OwnerUserId="4477" LastEditorUserId="1431" LastEditDate="2014-02-10T05:47:47.683" LastActivityDate="2020-11-29T18:54:30.607" Title="Finding the final temperature" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8391" PostTypeId="2" ParentId="8389" CreationDate="2014-02-09T19:13:03.000" Score="0" Body="&lt;p&gt;The flavor of fine fried potatoes is from browning, the Maillard reaction.  Carbonyls (sugars) condense with amino groups (alanine residees in protein) to Schiff bases, then the fun starts.  The optimum pH for condensation is bicarbonate buffer, slightly basic.  One might then do the experiment of soaking fries in dilute bicarbonate solution, drain, and fry.  Does that taste better than untreated controls?  Push it a little harder by dissolving a jot of honey (or corn syrup) in the bicarb buffer to add reducing carbonyls from glucose and fructose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fine bagel is formed from dough, boiled in honey water (reducing carbonyls), brushed with egg wash (protein), then baked.  It's all about the crust!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Maillard_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Maillard_reaction&lt;/a&gt; &lt;BR&gt;&#xA;Either way, don't overcook toward burned - trace acrylamide formation.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-09T19:13:03.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8392" PostTypeId="2" ParentId="7752" CreationDate="2014-02-09T19:18:19.977" Score="2" Body="&lt;p&gt;1) Good leaving group for displacement. &lt;BR&gt;&#xA;2) Cf: polyurethanes and truck bed liner (polyureas). &lt;BR&gt;&#xA;3) Nucleophile conjugate addition. &lt;BR&gt;&#xA;4) Good leaving group for displacement. &lt;BR&gt;&#xA;5) Good leaving group for displacement.  &lt;BR&gt;&#xA;6) Schiff base formation.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-09T19:18:19.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8394" PostTypeId="2" ParentId="7742" CreationDate="2014-02-09T19:51:14.173" Score="2" Body="&lt;p&gt;In terms of inexpensive, safe, and a little polar to move oxidation products, I'd try soaking in and flushing with hot soybean oil (replace elastomer hoses and seals).  Then flush with kerosene (odorless white spirits, hardware store  paint section), then blow out with compressed air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carburetor cleaner was mixed cresols - dangerousy on skin (heavy gloves), but great against deposits.  If it is really nasty in there, I'd go for gun bore cleaner soaking, specifically Ed's Red without the lanolin and substituting MEK (methyl ethyl ketone) for the acetone to lower flammability. At the end, blow it out, flush with kerosene, blow it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contain all wastes.  Beware of skin contact (gloves) .  Do not breath the fumes (ventilation, and maybe a charcoal latex paint mask, then discard).  Eye protection (goggles).  VAPORS ARE FLAMMABLE AND HEAVIER THAN AIR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=2OfaCCJjBEU&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=2OfaCCJjBEU&lt;/a&gt; &lt;BR&gt;&#xA;Ed's Red Gun Bore Cleaner  (wear gloves!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 part Dexron II, IIe or III ATF, GM Spec. D-20265 or later. &lt;BR&gt;&#xA;1 part Kerosene - deodorized, K1 (can be Kroil or WD-40) &lt;BR&gt;&#xA;1 part Aliphatic Mineral Spirits, Fed. Spec. TT-T-2981F, CAS #64741-49-9, or may substitute &quot;Stoddard Solvent&quot;, CAS #8052-41-3, or equivalent, (aka &quot;Varsol&quot;) &lt;BR&gt;&#xA;1 part Acetone, CAS #67-64-1. &lt;BR&gt;&#xA;1 lb anhydrous lanolin or  Lanolin, Modified, Topical Lubricant. &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Ed's Red is rapid death to plastics, polymers, varnishes, paints, and coatings.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-09T19:51:14.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8395" PostTypeId="1" CreationDate="2014-02-09T19:54:05.817" Score="1" ViewCount="334" Body="&lt;p&gt;The definition of voltage I'm familiar with is the amount of electric potential energy per unit charge at a location. My prof defined the voltage of a battery to be the amount of potential energy difference an unit charge would have at the positive terminal vs. the negative terminal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from what I remember from high school, the &quot;terminals&quot; aren't points, their rods of metal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/txcaA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the voltage is defined at a point, not cylinders or rods what does the voltage of a battery mean? Also, how does the definition connect to the &quot;pushing&quot; power of a battery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrode_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrode Potential&lt;/a&gt; but it was a bit over my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an intuitive explanation, not one using rigorous chemistry concepts. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4430" LastActivityDate="2014-04-26T12:49:32.580" Title="What's the voltage of a battery?" Tags="&lt;electrochemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8396" PostTypeId="2" ParentId="8355" CreationDate="2014-02-09T19:57:55.190" Score="0" Body="&lt;p&gt;Why does anything dissolve at all?  The binding energy (lattice energy) of the solid must be compensated by solvation energy and entropy of the solution.  Like dissolves like (then cheats, like clathrates and micelles).  As the solution concentrates, incremental driving energies diminish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't think about adding water to thiotimoline.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-02-11T19:21:52.267" LastActivityDate="2014-02-11T19:21:52.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8397" PostTypeId="1" AcceptedAnswerId="8416" CreationDate="2014-02-09T21:50:21.537" Score="6" ViewCount="245" Body="&lt;p&gt;When choosing a cup of tea, I can choose a big cup or a small cup. I'm wondering if I get more caffeine with the bigger cup or the same amount of caffeine. (in moles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Teabag is placed in cup first - and hot water poured directly onto the bag&lt;/li&gt;&#xA;&lt;li&gt;Same amount of stirring across both cups (assume three spoon swirls)&lt;/li&gt;&#xA;&lt;li&gt;The main varying parameter is the volume of solvent changing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question behind the question - trying to express it another way&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a kind of 'fixed rate of dissolution' that limits the amount of caffeine that comes out over a period of time - ie for two cups where one is 250ml and one is 300ml - and each have a tea dissolving time of 30s - wouldn't they have the same number of moles of caffeine at the end because of an upper bound on the rate of caffeine dissolving?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4478" LastEditorUserId="4478" LastEditDate="2014-02-10T02:16:01.727" LastActivityDate="2017-04-20T16:41:48.490" Title="For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;equilibrium&gt;&lt;molecules&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8398" PostTypeId="2" ParentId="8202" CreationDate="2014-02-09T21:54:06.500" Score="8" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because the polymer is of one or the other kind does not let you say what industrial process actually synthesized it. Subsequently it is impossible to know what trace chemicals are left in the polymer packaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If if isn't labelled as microwave-safe, it isn't!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They test the package for it's microwave compatibility, not one type of polymer each! It &lt;a href=&quot;http://www.scn.org/~bk269/plastics.html&quot; rel=&quot;noreferrer&quot;&gt;seems&lt;/a&gt; that resin code 5 polymers are often used for microwave-proof packaging, but again: You don't know how it was made, you have no idea what's in there, don't microwave it (together with food or something else that you then ingest).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-09T21:54:06.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8399" PostTypeId="2" ParentId="8386" CreationDate="2014-02-09T22:01:51.293" Score="4" Body="&lt;p&gt;With your result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this and the definition of $c^2$ you get to the desired result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} = &#xA;\underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ .&#xA;\end{equation} &lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-02-11T12:56:38.270" LastActivityDate="2014-02-11T12:56:38.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8400" PostTypeId="1" CreationDate="2014-02-09T22:12:44.910" Score="4" ViewCount="3816" Body="&lt;p&gt;I am curious as to why &lt;strong&gt;ester&lt;/strong&gt; is used often as a solvent in organic chemistry? How is it different that just using water, which is also polar? I'm sorry if this is a basic question, I'm just a student.&lt;/p&gt;&#xA;" OwnerUserId="4479" LastEditorUserId="4945" LastEditDate="2014-07-17T10:38:15.110" LastActivityDate="2014-07-17T10:38:15.110" Title="Ester as a solvent" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8401" PostTypeId="2" ParentId="8400" CreationDate="2014-02-09T22:57:30.760" Score="7" Body="&lt;ul&gt;&#xA;&lt;li&gt;Ethyl acetate, perhaps one of the best examples that would fit in your category &quot;often used solvents in organic chemistry&quot; is prepared at &lt;em&gt;low cost&lt;/em&gt; from ethanol and acetic acid (and these two are cheap, too).&lt;/li&gt;&#xA;&lt;li&gt;It may dissolve a wide range of organic material. Its low solubility in water presents as an advantage when you want to extract something.&lt;/li&gt;&#xA;&lt;li&gt;It has a boiling point of 77 Celsius that is elevated enough to work at moderate temperatures; yet if you want to remove it as solvent you do not spend terrible much energy to distill it off.&lt;/li&gt;&#xA;&lt;li&gt;Set side-by-side with other solvents, it comes with a relative &lt;em&gt;low toxicity&lt;/em&gt; (at the level we know today).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course there is not a &quot;one [solvent] fits for all [purposes/applications]&quot;, there are chemical conditions where other solvents are more appropriate to be deployed than esters. The presence of mineralic acids, or the of  organometallic reagents for example would call for a substitution.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2014-02-09T22:57:30.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8402" PostTypeId="1" CreationDate="2014-02-09T23:17:27.053" Score="0" ViewCount="1247" Body="&lt;p&gt;The question I'm trying to answer is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Calculate the mass of butane gas that would be needed to heat &lt;span class=&quot;math-container&quot;&gt;$724\ \mathrm{cm^3}$&lt;/span&gt; of water from an initial temperature of &lt;span class=&quot;math-container&quot;&gt;$7.44\ \mathrm{^\circ C}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$50.7\ \mathrm{^\circ C}$&lt;/span&gt;. The thermochemical equation for the combustion of butane is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2C4H10(g) + 13O2 -&amp;gt; 8CO2(g) + 10H2O(l)}\quad\Delta H=-5748\ \mathrm{kJ/mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that the equation for enthalpy is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q=m\times c\times\Delta T$$&lt;/span&gt;&#xA;and I calculated this for water:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q=724\times4.18\times43.26$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q=130918.6032$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I don't know where to go from here in order to figure out the mass of butane needed.  Please tell me if I have done something wrong or if you can help me to solve the rest of the problem.&lt;/p&gt;&#xA;" OwnerUserId="4480" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-06-17T17:13:22.440" Title="How do I calculate the enthalpy change when a gas is being used to heat water?" Tags="&lt;physical-chemistry&gt;&lt;homework&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8403" PostTypeId="2" ParentId="8397" CreationDate="2014-02-09T23:36:32.753" Score="3" Body="&lt;p&gt;How do you prepare you your cup of tea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dunking the leaves of tea (solid phase) exposes them the liquid phase (water).  The concentration in the liquid phase of caffeine (and other chemicals/ingredients carried by your leaves of tea) will increase. Because hot water is more efficient than cold one for this extraction, temperature plays a role, too.  Stirring the liquid phase for a homogenous dissipation of extracted matter across the whole solvent volume at your disposition may be to be considered as a parameter, too.  Yet, the more volume of solvent of extraction at your disposition, the more total solvent capacity to get more caffeine out of your leaves of tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another thought may arise, too; if you do not spend &lt;em&gt;all&lt;/em&gt; the volume &lt;em&gt;at once&lt;/em&gt;. As textbooks for physical chemistry, regarding multiple extractions, teach, repeated extraction with a small fraction of solvent of extraction would be then more efficient than deploying the whole solvent of extraction in a single extraction.  (However, by time the composition of the ingredients being extracted by hot water, varies, too.)&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2014-02-09T23:36:32.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8404" PostTypeId="2" ParentId="8378" CreationDate="2014-02-09T23:50:38.217" Score="4" Body="&lt;p&gt;As above, your labor all by itself (even if you are grad student) is worth more than the sulfuric acid.  As a pedagogic exercise, run the solution down a strong anion exchange resin washed OH-.  Only glucose comes out.  Recover the sulfuric acid by displacing it with phosphoric acid.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-09T23:50:38.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8405" PostTypeId="1" CreationDate="2014-02-10T02:47:10.983" Score="2" ViewCount="2873" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The SN1 solvolysis of one of the following alkenyl bromides is in MeOH approximately 100000 times faster than the other one. Which is the more reactive one? Draw the mechanism including the transition state, determine the rate determining step and draw the reaction product of each compound.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a21Mx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now what is apparently important here is the rotation of the substituent in each relative to the aromatic ring. In the second compound (with two methyl groups), the rotational profile shows destabilized molecule except when 90 degrees. That means the rotation in the second molecule helps with the leaving of Br and this (rate determining) step is much faster than when there are no methyl groups (first compound).&lt;/p&gt;&#xA;&lt;p&gt;Is that a complete explanation or I am missing something important? Also is there anything special about the mechanism of the two reactions and the products in each case?&lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="69830" LastEditDate="2021-05-18T06:46:37.087" LastActivityDate="2021-05-18T06:46:37.087" Title="Which of the these SN1 solvolysis of two alkenyl bromides reactions is faster?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8406" PostTypeId="1" CreationDate="2014-02-10T05:28:02.763" Score="1" ViewCount="1426" Body="&lt;p&gt;I am slightly lost on how to proceed with this equation. I know potassium dichromate is a strong oxidizing agent but I am unsure of how to deal with it along with the titanium complex. I was wondering if I could get tips on how to handle these types of equations.&lt;/p&gt;&#xA;" OwnerUserId="4481" LastEditorUserId="13339" LastEditDate="2015-04-01T16:05:17.060" LastActivityDate="2015-04-01T16:05:17.060" Title="What are the products of [Ti(H2O)6]Cl3 + H2SO4 + K2Cr2O7 ---&gt;" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8407" PostTypeId="2" ParentId="8406" CreationDate="2014-02-10T08:51:04.793" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know potassium dichromate is a strong oxidizing agent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have already identified one partner and its role in a redox reaction. Good start!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which component will be oxidized to what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we assume that the reaction is carried out in water? If so, let's have a look at the ions in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll have $\ce{H3O+}$, $\ce{K+}$, $\ce{Cr2O7^{2-}}$, and &quot;that complex&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at $\ce{[Ti(H2O)6]Cl3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, it's a complex of titanium with 6 &lt;strong&gt;neutral&lt;/strong&gt; molecules of water as the ligands and 3 $\ce{Cl-}$ as the counter ions. So, what is the oxidation state of titanium in the complex when it is neutral with 3 $\ce{Cl-}$ as the counter ions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look again: $\ce{[Ti(H2O)6]^{3+}}$. OK, it is obviously $\ce{Ti(III)}$ and it seems that we have found the victim of our stong oxidant in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We remember that most titanium compounds in nature are $\ce{Ti(IV)}$ oxides and conclude that we will have to balance our redox reaction to account for a $\ce{Ti(III) -&amp;gt; Ti(IV)}$ oxidation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;strong&gt;could&lt;/strong&gt; now continue to conclude that the reduction of $\ce{Cr(VI)}$ to $\ce{Cr(III)}$ is accompanied by $\ce{[Ti(H2O)6]^{3+} -&amp;gt; [Ti(H2O)6]^{4+}}$ but then we remember that $\ce{Ti(IV)}$ - tiny and higly charged - is a &lt;strong&gt;hard Lewis acid&lt;/strong&gt; and will &lt;strong&gt;not exist&lt;/strong&gt; as a naked cation, barely covered with just six neutral ligands in a hexaquo complex!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the likely $\ce{Ti(IV)}$ species is the titanyl cation, $\ce{TiO}^{2+}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time to collect the puzzle pieces and write the redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-10T08:51:04.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8408" PostTypeId="2" ParentId="8405" CreationDate="2014-02-10T09:23:25.823" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] SN1 solvolysis [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's recall what we know about these. We have a two-step reaction, consisting of &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;heterolysis of the starting material&lt;/li&gt;&#xA;&lt;li&gt;association of a nucleophile $\ce{Nu}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We remember that the heterolysis is not assisted by $\ce{Nu}$, there is no backside attack of $\ce{Nu}$, pushing out the leaving group. We also remember that the association step (addition of $\ce{Nu}$ to the carbocation intermediate) is usually pretty fast. Consequently, heterolysis, i.e. the formation of a carbocation is the rate-determining step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank's to good old &lt;em&gt;Arrhenius&lt;/em&gt;, we know that the rate is determined by the activation energy of this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The activation energy is lower when the resulting cation is more stable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cation is more stable when the delocalization of the charge is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distribution of charge on the &lt;strong&gt;vinyl cation&lt;/strong&gt; is better when it's actually a &#xA;&lt;strong&gt;styryl cation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWrXU.png&quot; alt=&quot;styryl cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the geometrical requirement for this interaction and what might prevent it?&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-10T09:30:56.723" LastActivityDate="2014-02-10T09:30:56.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8409" PostTypeId="2" ParentId="8379" CreationDate="2014-02-10T10:32:42.973" Score="15" Body="&lt;p&gt;The &lt;strong&gt;Reimer-Tiemann&lt;/strong&gt; reaction is known as the &lt;em&gt;ortho&lt;/em&gt;-&lt;strong&gt;formylation&lt;/strong&gt; of phenols under strongly alkaline conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an addition of a carbene to a double bond, like the &lt;em&gt;Simmons-Smith&lt;/em&gt; reaction, the pyrrole to pyridine ring enlargement or the &lt;em&gt;Buchner&lt;/em&gt; expansion, which is catalyzed by transition metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As ron already pointed out, pH is one crucial requirement, having a phenol (and not just any other arene) is the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F99Bk.png&quot; alt=&quot;mesomerie&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we're actually facing here is the addition of a carbanion to a strongly &lt;strong&gt;electron-deficient&lt;/strong&gt; carbene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As again already described by ron, there is significant charge density at the 4-position of the phenolate, and the initial addition step might indeed happen here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XdNdp.png&quot; alt=&quot;4-addition&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting intermediate can't rearomatize other than in a back reaction and that's what it will mainly do! Only a minor part will grab a proton and yield &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was close, but no cigar! Let's have a look at the addition at the 2-position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XmALs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XmALs.png&quot; alt=&quot;Addition of carbene at C-2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through the interplay of dienone, phenolate and phenol we get rid of one chloride and all further steps to the final benzaldehyde are likewise assisted by this tautomerism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still haven't answered the question why no 1,3-dialdehyde is formed. Let's have a look at the major product &lt;strong&gt;B&lt;/strong&gt; under alkaline conditions and its resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2lCv.png&quot; alt=&quot;resonance ahdehyde&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of them looks like the enolate of a 1,3-dicarbonyl conpound, with additional charge distribution over the $\pi$ system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This species has a significantly lower localized electron density than the anion of the starting material. So, the party stops here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; In a comment, it was asked whether the &lt;em&gt;geminal&lt;/em&gt; dichloride will convert to the corresponding aldehyde under alkaline conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a transformation could involve a nucleophilic substitution, leading to a &lt;em&gt;geminal&lt;/em&gt; halohydrine which then eliminates HCl to furnish the aldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of benzylic dihalides, this is possible even without the intramolecular assistance of a phenolate (cf. this example) using secondary amines as a base and in the case of diphenyldichloromethane, the transformation to benzophenone is achieved by heating in the presence of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;strong&gt;C&lt;/strong&gt;, the simple deprotonation of the dihalide might be a dead-end alternative pathway to substitution.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2017-01-15T07:01:41.653" LastActivityDate="2017-01-15T07:01:41.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8410" PostTypeId="2" ParentId="7018" CreationDate="2014-02-10T11:14:57.620" Score="3" Body="&lt;hr /&gt;&#xA;&lt;p&gt;Both processes involved the denaturation of proteins and complex reactions so is impossible reverse them &lt;strong&gt;only&lt;/strong&gt; with a change of temperature.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Boiling an Egg&lt;/h1&gt;&#xA;&lt;p&gt;Although in most case denaturation is not reversible there are some cases where it is.&#xA;However in the case of Egg white denaturation is irreversible (&lt;a href=&quot;http://books.google.it/books?id=WSf1FFOTIYEC&amp;amp;pg=PA323&amp;amp;dq=egg+denaturation+reversible&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=vav4UuK4GIee4wT83oGABg&amp;amp;ved=0CDAQ6AEwAA#v=onepage&amp;amp;q=egg%20denaturation%20reversible&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;see here&lt;/a&gt;) and so the whole process is by the way irreversible.&lt;/p&gt;&#xA;&lt;h1&gt;Cooking batter&lt;/h1&gt;&#xA;&lt;p&gt;The same for the batter even if is unclear which flours are used, and the composition of the batter  you can be sure that after cooking batter on a hot griddle surely some irreversible reactions occur: denaturation, &lt;a href=&quot;http://books.google.it/books?id=WSf1FFOTIYEC&amp;amp;pg=PA323&amp;amp;dq=egg+denaturation+reversible&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=vav4UuK4GIee4wT83oGABg&amp;amp;ved=0CDAQ6AEwAA#v=onepage&amp;amp;q=egg%20denaturation%20reversible&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maillard reaction&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Amadori_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amadori rearrangment&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-03-04T16:56:22.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8411" PostTypeId="1" CreationDate="2014-02-10T11:53:16.717" Score="2" ViewCount="34941" Body="&lt;p&gt;What are hybridisation states of each carbon atom in the following compounds?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=C=O}$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH=CH2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{(CH3)2CO}$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=CHCN}$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C6H6}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do I find the hybridisation of the orbitals of the carbon atom in such compounds?&lt;/p&gt;&#xA;" OwnerUserId="4484" LastEditorUserId="4945" LastEditDate="2018-10-15T09:25:54.997" LastActivityDate="2018-10-15T09:30:59.960" Title="How do I find the hybridisation of the orbitals of carbon in a molecule?" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2018-10-15T09:26:34.987" ContentLicense="CC BY-SA 4.0" />
  <row Id="8412" PostTypeId="1" AcceptedAnswerId="8420" CreationDate="2014-02-10T13:10:36.603" Score="11" ViewCount="14790" Body="&lt;p&gt;What is hyperconjugation and how do you know if there is hyperconjugation in a compound? Which orbitals are involved in hyperconjugation?&lt;/p&gt;&#xA;" OwnerUserId="4432" LastEditorUserId="10185" LastEditDate="2015-10-02T16:45:10.023" LastActivityDate="2021-03-06T23:19:44.950" Title="What is hyperconjugation?" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;hyperconjugation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8413" PostTypeId="2" ParentId="8402" CreationDate="2014-02-10T13:13:03.900" Score="2" Body="&lt;p&gt;You're on the right track.&#xA;&lt;hr&gt;&#xA;\begin{eqnarray}&#xA;Q &amp;amp; = &amp;amp; m\cdot c\cdot \Delta T\\&#xA;\mathbf{unit\ analysis} \qquad \textrm{kJ} &amp;amp; = &amp;amp; \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K}&#xA;\end{eqnarray}&#xA;&lt;hr&gt;&#xA;For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the mass of $724\ \textrm{cm}^3$ water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the kJ you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. &lt;strong&gt;Note the $n$!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can calculate how many moles of butane you will need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, the calculation of the mass is not difficult either.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2018-06-17T17:13:22.440" LastActivityDate="2018-06-17T17:13:22.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8414" PostTypeId="1" CreationDate="2014-02-10T14:19:47.257" Score="1" ViewCount="2874" Body="&lt;p&gt;Thermal decomposition of calcium carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, we heat $\ce{CaCO3}$ to temperature of approximately 825°C it decomposes into calcium oxide and liberates carbon dioxide gas:*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3} \overset{825\ ^\circ \text{C}}{\ce{ -&amp;gt;}}\ce{CaO + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to heat the calcium carbonate at temperatures below 825 degrees Celsius.&lt;/p&gt;&#xA;" OwnerUserId="4487" LastEditorUserId="275" LastEditDate="2014-02-10T15:15:05.330" LastActivityDate="2019-08-08T19:09:15.133" Title="Thermal decomposition of calcium carbonate" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8415" PostTypeId="1" AcceptedAnswerId="8419" CreationDate="2014-02-10T14:22:54.497" Score="2" ViewCount="5860" Body="&lt;p&gt;I think it should be acid, as I am aware that a compound by the name of Boric acid exists. But how can the oxide be acidic when it is electropositive? Also, other members of its group are well known metals. Why isn't the oxide amphoteric, when its group member Aluminium has an amphoteric oxide? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher told us that the oxide of boron is basic in nature? So what exactly is its nature?&lt;/p&gt;&#xA;" OwnerUserId="4486" LastActivityDate="2017-08-16T19:21:18.530" Title="What is the nature of the oxide of boron?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8416" PostTypeId="2" ParentId="8397" CreationDate="2014-02-10T14:29:58.693" Score="6" Body="&lt;p&gt;From what I remember from &lt;a href=&quot;http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion&quot; rel=&quot;noreferrer&quot;&gt;Fick's laws&lt;/a&gt;, I would say that the factor that limits the rate of dissolution is the concentration of the already-dissolved compound. At first, you don't have any caffeine dissolved so the rate of dissolution is fastest. As time goes by, more and more caffeine becomes solvated and that hinders the dissolution of remaining caffeine from the tea leaves. At any point of time until saturation, the 300 ml cup will have faster dissolution rate because it has more &quot;free space&quot; to solvate the caffeine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: However, after thinking this for a while, I have to add that in practice the difference might be too small to be detected. The solubility (~ saturation point) of caffeine in water is &lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB0202769.htm&quot; rel=&quot;noreferrer&quot;&gt;20 g/l at 20°C&lt;/a&gt; while the concentration of caffeine in tea is &lt;a href=&quot;http://en.wikipedia.org/wiki/Caffeine&quot; rel=&quot;noreferrer&quot;&gt;0.1 - 0.4 g/l&lt;/a&gt; &amp;ndash; huge difference. It might be that from caffeine's point of view, the &quot;free space&quot; is more or less the same in both cups. &lt;/p&gt;&#xA;" OwnerUserId="4452" LastEditorUserId="4452" LastEditDate="2014-02-11T15:23:33.010" LastActivityDate="2014-02-11T15:23:33.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8417" PostTypeId="2" ParentId="374" CreationDate="2014-02-10T15:09:22.267" Score="1" Body="&lt;p&gt;The $\ce{Fe(NO)2}$ fragment has strongly delocalized electrons, it is considered as covalently bound. The semiempirical calculations show that the electron density is polarized in a manner that allows to ascribe the delta+ charge to $\ce{NO}$ groups. Therefore formally Fe is considered +1. Remember this formal assignment is somewhat simplified...&#xA;If interested, read e.g. &#xA;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlehtml/2011/dt/c0dt01244k&quot; rel=&quot;nofollow&quot;&gt;http://pubs.rsc.org/en/content/articlehtml/2011/dt/c0dt01244k&lt;/a&gt;&#xA;and references.&lt;/p&gt;&#xA;" OwnerUserId="4488" LastEditorDisplayName="user4076" LastEditDate="2014-02-10T21:48:06.473" LastActivityDate="2014-02-10T21:48:06.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8418" PostTypeId="2" ParentId="8411" CreationDate="2014-02-10T15:12:32.833" Score="4" Body="&lt;p&gt;My method for finding the hybridization of the orbitals of an atom in a molecule:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Look at the molecular structure (geometry), apply the following: &#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tetrahedral: sp&lt;sup&gt;3&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Trigonal planar: sp&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Linear: sp&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="4945" LastEditDate="2018-10-15T09:21:58.023" LastActivityDate="2018-10-15T09:21:58.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8419" PostTypeId="2" ParentId="8415" CreationDate="2014-02-10T15:28:01.233" Score="6" Body="&lt;p&gt;The most common oxide of boron is &lt;a href=&quot;http://en.wikipedia.org/wiki/Boron_trioxide&quot; rel=&quot;noreferrer&quot;&gt;boron trioxide&lt;/a&gt;, $\ce{B2O3}$, which is the anhydride of &lt;a href=&quot;http://en.wikipedia.org/wiki/Boric_acid&quot; rel=&quot;noreferrer&quot;&gt;boric acid&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3BO3 -&amp;gt; HBO2 + H2O}$$&#xA;$$\ce{2HBO2 -&amp;gt; B2O3 + H2O}$$&#xA;$$\text{net} \ \ce{2H3BO3 -&amp;gt; B2O3 + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the anhydride of an acid, boron trioxide is acidic. It reacts with water to regenerate the acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B2O3 + 3H2O -&amp;gt; 2H3BO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boric acid's acidity does not derive from the dissociation of a proton (or abstraction of a proton by water), as does many acids, for example sulfuric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 &amp;lt;=&amp;gt; H+(aq) +HSO4- (aq)}$$&#xA;$$\ce{H2SO4 + H2P &amp;lt;=&amp;gt; H+(aq) + HSO4- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boric acid (written now as $\ce{B(OH)3}$, which is suggestive of its structure) instead reacts with water by forming a dative complex which is the active acid. A water molecule associates with the empty $p$ orbital on boron. This difference in behavior is enabled also by the electropositive character of boron. The acidic oxides of some metals behave similarly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + B(OH)3 &amp;lt;=&amp;gt;H2OB(OH)3 &amp;lt;=&amp;gt;H+(aq) + B(OH)4^- (aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-10T15:28:01.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8420" PostTypeId="2" ParentId="8412" CreationDate="2014-02-10T16:39:12.043" Score="8" Body="&lt;p&gt;In the ethyl cation, one can write the following resonance structures:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KRgPz.jpg&quot; alt=&quot;Resonance by hyperconjugation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The structure on the right can be actually written 3 times since there are 3 hydrogens available. Note also that the &amp;quot;ethylenic-like&amp;quot; structure on the right involves a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3-p}$&lt;/span&gt; bond (the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized and the cationic carbon has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital), so it is not a true pi bond (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p-p}$&lt;/span&gt;) like that found in ethylene.  The resonance structure on the right is meant to show the &amp;quot;hyperconjugation&amp;quot; effect of the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3-s}$&lt;/span&gt;) with the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital located on the adjacent cationic carbon.  The drawing is a shorthand way of saying that when the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is rotated into the same plane as the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital on the cationic carbon, there can be a stabilizing overlap - and it is called &amp;quot;hyperconjugation.&amp;quot; In the 2-propyl cation, there are 6 of these hyperconjugative resonance structures (there are 6 &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds on the 2 methyl groups adjacent to the cationic carbon and in the &lt;em&gt;t&lt;/em&gt;-butyl cation there are 9 such resonance structures. This can be used to explain why tertiary carbonium ions are more stable than secondary carbonium ions, which in turn are more stable than primary carbonium ions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2021-03-06T23:19:44.950" LastActivityDate="2021-03-06T23:19:44.950" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8421" PostTypeId="2" ParentId="8414" CreationDate="2014-02-10T17:33:41.017" Score="2" Body="&lt;p&gt;Pump a hard vacuum and calcium carbonate will calcine at a lower temperature.  825°C is about one atmosphere of carbon dioxide at equilibrium.  Thermodynamics proposes, kinetics disposes - finely ground goes faster than bulk.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-10T17:33:41.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8423" PostTypeId="1" CreationDate="2014-02-10T18:34:54.730" Score="3" ViewCount="115" Body="&lt;p&gt;equation 1) $\ce{2 Cl^{–} -&amp;gt; Cl_2 + 2e^{-}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;equation 2) $\ce{2 H2O + 2 e^{–} -&amp;gt; H2 + 2OH^{–}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations are the key reactions in the chlor-alkali process. I want to model a reactor for this process based on a CSTR (continuous stirred tank reactor). However I really need reaction rate coefficients (k values) and design equations. Any help would be much appreciated! &lt;/p&gt;&#xA;" OwnerUserId="4460" LastEditorUserId="418" LastEditDate="2014-02-10T21:47:09.807" LastActivityDate="2014-02-10T21:47:09.807" Title="Does anyone know where I can find the reaction rate coefficient and design equation for the electrolysis of brine and sodium hydroxide solution?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8424" PostTypeId="1" CreationDate="2014-02-10T19:53:12.863" Score="2" ViewCount="2605" Body="&lt;p&gt;I'm trying to draw the state correlation diagram for the reaction of &lt;em&gt;cyclopent-2-enyl anion&lt;/em&gt; to &lt;em&gt;pentadienyl anion&lt;/em&gt; but I don't know which symmetry element is preserved in this reaction. What could it be? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="4231" LastEditDate="2014-05-24T23:16:01.127" LastActivityDate="2014-07-24T03:39:55.113" Title="Symmetry element in the coversion of cyclopent-2-enyl anion to pentadienyl anion?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8425" PostTypeId="2" ParentId="8423" CreationDate="2014-02-10T21:08:01.810" Score="4" Body="&lt;p&gt;You should look in some kind of reaction rate database, e.g. &lt;a href=&quot;http://kinetics.nist.gov/kinetics/welcome.jsp&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST&lt;/a&gt;. An example from the database: &lt;img src=&quot;https://i.stack.imgur.com/sc6MF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-02-10T21:08:01.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8426" PostTypeId="2" ParentId="8357" CreationDate="2014-02-10T22:44:13.413" Score="24" Body="&lt;p&gt;The relative energies of the electronic subshells have been calculated for atoms in the vicinity of &lt;span class=&quot;math-container&quot;&gt;$Z=20$&lt;/span&gt; (&lt;a href=&quot;https://doi.org/10.1021/ed071p469&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Chem. Educ.&lt;/em&gt;, &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;71&lt;/em&gt; (6), 469&lt;/a&gt;), and the result is surprising:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pLScJ.png&quot; alt=&quot;Transition Metals and the Aufbau Principle, fig. 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Looking at this graph, by all means the electronic configuration of scandium should in fact be &lt;span class=&quot;math-container&quot;&gt;$\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\color{blue}{\ce{3d^3}}$&lt;/span&gt; in order to minimize orbital energies! The graph does not show the energy of the &lt;span class=&quot;math-container&quot;&gt;$\ce{4p}$&lt;/span&gt; subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). &lt;strong&gt;It turns out&lt;/strong&gt; (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with &lt;span class=&quot;math-container&quot;&gt;$Z=21$&lt;/span&gt;, the configuration &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 4s^2 3d^1}$&lt;/span&gt; is lower in energy than other potential candidates, such as &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 3d^3}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 4s^1 3d^2}$&lt;/span&gt;. Theoretically, the odd configurations &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 4s^2 4p^1}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar] 4s^1 3d^1 4p^1}$&lt;/span&gt; would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy &lt;span class=&quot;math-container&quot;&gt;$\ce{4p}$&lt;/span&gt; subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.&lt;/p&gt;&#xA;&lt;p&gt;In addition to some material I linked in comments, &lt;a href=&quot;http://ericscerri.blogspot.com.br/2012/06/trouble-with-using-aufbau-to-find.html&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the &lt;span class=&quot;math-container&quot;&gt;$\ce{Sc^3+}$&lt;/span&gt; ion (emphasis mine):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This amounts to saying that all three of the final electrons enter &lt;span class=&quot;math-container&quot;&gt;$\ce{3d}$&lt;/span&gt; but two of them are repelled into an energetically less favourable orbital, the &lt;span class=&quot;math-container&quot;&gt;$\ce{4s}$&lt;/span&gt;, because the overall result is more advantageous for the atom as a whole.  &lt;strong&gt;But this is not something that can be predicted.&lt;/strong&gt;  Why is it 2 electrons, rather than one or even none?  In cases like chromium and copper just one electron is pushed into the &lt;span class=&quot;math-container&quot;&gt;$\ce{4s}$&lt;/span&gt; orbital.  In an analogous case from the second transition series, the palladium atom, the competition occurs between the &lt;span class=&quot;math-container&quot;&gt;$\ce{5s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{4d}$&lt;/span&gt; orbitals. In this case none of the electrons are pushed up into the &lt;span class=&quot;math-container&quot;&gt;$\ce{5s}$&lt;/span&gt; orbital and the resulting configuration has an outer shell of &lt;span class=&quot;math-container&quot;&gt;$\ce{[Kr] 4d^10}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;None of this can be predicted in simple terms from a rule of thumb&lt;/strong&gt; and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital.  To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly &lt;em&gt;wrong&lt;/em&gt; for practically every atom with respect to the placement of the orbital energy levels, but incredibly it &lt;em&gt;happens&lt;/em&gt; to predict the configuration of the valence shell for most atoms.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-06-13T09:21:08.190" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="8427" PostTypeId="1" AcceptedAnswerId="8437" CreationDate="2014-02-11T05:40:47.080" Score="2" ViewCount="1607" Body="&lt;p&gt;I was working on my homework which was to write a description of what would occur if zinc and sulfur reacted and what steps would a scientist have to make in order to make them react. When I researched I bit, &lt;a href=&quot;http://s-owl.cengage.com/ebooks/vining_owlbook_prototype/ebook/ch5/Sect5-4-a.html&quot; rel=&quot;nofollow&quot;&gt;I saw that it would be a exothermic reaction&lt;/a&gt;, and I wanted to show this by &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Heat_of_Reaction#How_to_Calculate_.CE.94H_Experimentally&quot; rel=&quot;nofollow&quot;&gt;using enthalpy&lt;/a&gt;. I haven't been taught this however, and as a result, I'm confused. I know that enthalpy equals the heat transfer and there is some sort relationship with bond energy, but I don't know how to apply this to the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc Powder + Sulfur Powder = Zinc Sulfide + Energy &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn_{(s)} + S_{(s)} -&amp;gt; ZnS_{(s)}}$ + $205.98 \text{ kJ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $205&amp;gt;0$ this is an exothermic reaction. But where did this $205.98\text{ kJ}$ come from?&lt;/p&gt;&#xA;" OwnerUserId="4491" LastEditorUserId="69830" LastEditDate="2021-05-07T20:23:57.490" LastActivityDate="2021-05-07T20:23:57.490" Title="Enthalpy of Zinc and Sulfur Reaction" Tags="&lt;reaction-mechanism&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8428" PostTypeId="1" AcceptedAnswerId="8429" CreationDate="2014-02-11T10:57:04.987" Score="3" ViewCount="680" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NcjhC.png&quot; alt=&quot;Reaction Scheme&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am having trouble understanding the reaction of anisole with dihydrofuran-2,5-dione (succinic anhydride) in presence of &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt; and acid works or how an electrophile is formed. I feel like the product should be something like &lt;span class=&quot;math-container&quot;&gt;$\ce{C11H12O4}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4493" LastEditorUserId="69830" LastEditDate="2021-05-18T19:33:27.720" LastActivityDate="2021-05-18T19:33:27.720" Title="Electrophilic aromatic substitution reaction of dihydrofuran-2,5-dione with anisole" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8429" PostTypeId="2" ParentId="8428" CreationDate="2014-02-11T11:32:05.640" Score="4" Body="&lt;p&gt;Nice! You've brought everything you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't identify your electrophile immediately, do it the other way around: Assign function to all components. The one that is left must be your electrophile - or its precursor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you remember the role of $\ce{AlCl3}$ in electrophilic aromatic substitutions? Where could it possible attack here?&#xA;&lt;hr&gt;&#xA;Here is another hint: What is the trivial name for dihydrofuran-2,5-dione?&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-11T14:51:15.777" LastActivityDate="2014-02-11T14:51:15.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8431" PostTypeId="1" AcceptedAnswerId="8990" CreationDate="2014-02-11T12:49:30.043" Score="5" ViewCount="339" Body="&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt; :&#xA;I am currently testing how a certain type of foam adsorbs HAP, PCB and OCP, and to this end I am depositing some on the foam, then extracting it. To know how much was extracted, I need to compare with a standard solution of HAP, PCB and OCP. The problem is that when I look at the chromatograms, the extracted solution's peaks' areas are about 10 times the standard solution's peaks' areas. I don't know where in the experiment I am doing something wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Details&lt;/strong&gt; :&#xA;I have a solution of HAP, PCB and OCP at 10 mg/L. I put 100 µL of this solution on the foam, so that the deposited mass is 1 µg.&#xA;The foam is ASE-extracted with acetonitrile, then the acetonitrile is rotary-evaporated. What is left in the round-bottom flask is then diluted in 1 mL acetonitrile. So at maximum I have 1 µg pollutants in 1 mL acetonitrile.&#xA;To this 1 mL I then add distilled water until I have a 20 mL solution. The concentration of this solution should at maximum be 50 µg/L, right ? (If all pollutants are extracted from the foam and none lost to evaporation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard solution is made by also taking 100 µL from the 10 mg/L solution (so 1 µg pollutants), and adding distilled water until I have a 20 mL solution. So this solution's concentration should also be 50 µg/L, right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I put the two solutions through GC-MSMS, the peaks' areas of the extracted solution are about ten times greater than those of the standard solution. (Not always ten, but it is usually the case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I made a mistake somewhere ? Are my calculations/manipulations wrong somewhere ?&#xA;The internal standard used is naphtalene d8, and in each 20 mL solution I added 10 µL of a 10 mg/L naphtalene d8 solution. (So its concentration is 5 µg/L : is it a problem that the standard is not at the same concentration as the pollutants ?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment was repeated several times, with different foam extracts and different standard solutions, and the results were the same. It is not a problem of contamination, as everything has been extensively washed with distilled water, detergent and acetone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it is relevant : when a solution is done, I sample 4 mL that I extract with a SPME fiber, and then I inject into the GC-MSMS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance, I am still testing and calculating, so any help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To tschoppi&lt;/strong&gt; (since I don't have enough reputation to comment) : I haven't tried with another kind of foam, but it's a good idea (to see at which step the problem lies). I'll see what we have in the lab. I am also pretty sure the foam I used isn't contaminated, because it was given to us straight from the package, and I washed it three times with different solvents and analysed the wash-solvent (which was clean)(but I may check again to be sure).&lt;/p&gt;&#xA;" OwnerUserId="4496" LastEditorUserId="4498" LastEditDate="2014-02-11T17:55:28.383" LastActivityDate="2014-03-07T11:51:36.580" Title="Standard solution and extracted solution : same concentration?" Tags="&lt;analytical-chemistry&gt;&lt;solutions&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8432" PostTypeId="1" CreationDate="2014-02-11T13:18:47.553" Score="2" ViewCount="34982" Body="&lt;p&gt;I see no difference between the two. Both say that the degree of adsorption reaches a limiting value at high pressures and at very low pressures degree of adsorption is directly proportional to the pressure.&lt;/p&gt;&#xA;&lt;p&gt;Where does the difference lie?&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="41328" LastEditDate="2021-01-21T18:11:08.317" LastActivityDate="2021-01-21T18:11:08.317" Title="How is Langmuir adsorption isotherm different from Freundlich's?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8433" PostTypeId="2" ParentId="8340" CreationDate="2014-02-11T15:19:17.960" Score="1" Body="&lt;p&gt;Thank you all for participating. So, the smoothness arises from mainly two factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Molecules have a wide range of energy levels due to bond stretching / vibration etc. that partially absorb the photon energy. Atomic atoms do not have such energy levels so their AAS spectra are spiky.&lt;/li&gt;&#xA;&lt;li&gt;The measurement technology is not precise enough to visualize individual absorptions. To my understanding, the bandwidths of filters / monochromators are too wide to cover only the individual energy levels.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I am still uncomfortable with the scheme I pictured below. It is easy to imagine that the transitions from various $S_1$ levels to the lower $S_1$ levels&#xA;are released to bond vibration etc. because the energy is very small. But if we look at the transition from $S_1$ to $S_0$, the energy is as large as that of the absorbed photon. So, why isn't the energy released as a photon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L4GG7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2014-02-11T15:19:17.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8434" PostTypeId="2" ParentId="8432" CreationDate="2014-02-11T16:26:09.460" Score="2" Body="&lt;p&gt;There is a fundamental difference between both models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Langmuir&lt;/em&gt;'s model was a theoretical construct, while the &lt;em&gt;Freundlich&lt;/em&gt; isotherm is empirical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;Langmuir&lt;/em&gt; model, it is assumed that at maximum coverage, there is only a &lt;strong&gt;monomolecular&lt;/strong&gt; layer on the surface. This means that there is &lt;strong&gt;no stacking&lt;/strong&gt; of adsorbed molecules. The &lt;em&gt;Freundlich&lt;/em&gt; isotherm does not have this restriction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-11T16:26:09.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8437" PostTypeId="2" ParentId="8427" CreationDate="2014-02-11T19:10:38.687" Score="3" Body="&lt;p&gt;The free energies of formation of the elements in their standard states is zero, by definition.  You can look up the (variously reported) standard enthalpy of formation of the product, −204.6 kJ/mol (exothermic!).  It is roughly the binding energy of the crystal lattice less the ionization energies of the inputs.  So, by the numbers, do you obtain spalerite or wurtzite?  ZnS has crystal polymorphs and polytopes with differently populated unit cells&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn_2S_2}$ Wurtzite-2H &lt;BR&gt;&#xA;$\ce{Zn_4S_4}$ Wurtzite-4H and Sphalerite &lt;BR&gt;&#xA;$\ce{Zn_6S_6}$ Wurtzite-6H &lt;BR&gt;&#xA;$\ce{Zn_15S_15}$ Wurtzite-15R &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-11T19:10:38.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8438" PostTypeId="1" AcceptedAnswerId="8439" CreationDate="2014-02-11T20:34:54.030" Score="2" ViewCount="4721" Body="&lt;p&gt;Would the reaction between zinc and sulfur be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn_{(s)} + S_{(s)} -&amp;gt; ZnS_{(s)}}$$&#xA;or&#xA;$$\ce{8 Zn_{(s)} + S_8\ _{(s)} -&amp;gt; 8ZnS_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\ce{S}$ and $\ce{S8}$ are allotropes of sulfur, but which one should I use for a reaction with sulfur at room temperature?&lt;/p&gt;&#xA;" OwnerUserId="4491" LastEditorUserId="402" LastEditDate="2017-04-25T00:47:28.537" LastActivityDate="2017-04-25T00:47:28.537" Title="Reaction between zinc and sulfur" Tags="&lt;reaction-mechanism&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8439" PostTypeId="2" ParentId="8438" CreationDate="2014-02-11T20:59:52.183" Score="5" Body="&lt;p&gt;Either reaction is valid for representing the reaction between zinc and sulfur. The important thing is that both equations stipulate the same stoichiometry (one atom of zinc for one atom of sulfur). The mathematics of chemical equations work just the same whether you add structural information about the reactants or not; the reaction equation is just shorthand for a system of mathematical equations involving the amounts and types of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In truth, neither equation is a correct description of the reaction at the microscopic scale (though I again restate that's not really the point of a reaction equation). Free zinc atoms do not interact with free sulfur atoms or $\ce{S8}$ rings. Rather, after a little bit of initial heating to get the reaction going *, liquid sulfur (a very complex mixture of sulfur chains ranging from $\ce{S6}$ to $\ce{S_{n}}$, &lt;a href=&quot;http://www.gps.caltech.edu/~vijay/Papers/Chemistry/Meyer-76.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;where n is as large as 100,000&lt;/a&gt;) is in contact with a surface of zinc (or possibly the molten metal), and electrons are exchanged by some mechanism which probably hasn't even been elucidated, as electron transfer reactions are very quick and hard to monitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* You mention considering the reaction at room temperature. Even though thermodynamically, the reaction between zinc and sulfur is spontaneous at room temperature, it is kinetically very slow because both reactants are solid. Even if you mix them very very well, the reaction is diffusion-limited in the solid mixture; the atoms can't &quot;reach&quot; each other very well. Heating a little bit creates a small amount of liquid sulfur, which can then mix better with the zinc.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-25T00:45:52.187" LastActivityDate="2017-04-25T00:45:52.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8440" PostTypeId="2" ParentId="8438" CreationDate="2014-02-11T21:05:38.057" Score="2" Body="&lt;p&gt;Crystalline sulfur, at room temperature, exists only as $\ce{S8}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, that's the only option you have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking down $\ce{S8}$ chains is doable but over 160°C, followed by polymerization.&#xA;At 200°C, polymer chains start to break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your question, $\ce{S8}$ is the most stable form for sulfur at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed072p56.1?prevSearch=%255BTitle%253A%2bPhases%2bof%2bsulfur%255D&amp;amp;searchHistoryKey=&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2014-02-11T21:05:38.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8443" PostTypeId="1" CreationDate="2014-02-12T04:32:14.850" Score="5" ViewCount="198" Body="&lt;p&gt;Today I finally mustered the courage to go and talk to the director of the labs to politely ask him if I could use the labs out of class hours. Long story short, our lab sessions &quot;are bad&quot; because we are given a protocol, and like babies we just need to follow it. No brainer, and I am against that. So I got permission to use the lab for experiments that don't pose a mayor risk, as well as a slight restriction of the reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of which reagents we have in the lab, but there aren't that many there. So, now to the point. I was hoping that you could tell me which experiments I should try. I came up with the idea when we worked with ferric ammonia ferrocyanide $[Fe(CN)_6][Fe(NH_4)_2]$ and I had to balance the chemical equation of the reaction that we used. However, I had no idea of what the byproducts where, because I had no way to test for their presence! My idea is to bring an hypothesis and then, starting pretty much from scratch (except reference tables or such) prove or disprove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you share a nice example of something I could test in the lab? If not, where can I find stuff that I can work with, without getting the answer right out of where I got the info. Did I get to express my idea correctly? More or less?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s.: I kindly asked a professor to supervise my work, so I would get help regarding methodology.&lt;/p&gt;&#xA;" OwnerUserId="1282" LastActivityDate="2014-03-14T18:06:52.067" Title="Getting to use the lab!" Tags="&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2017-06-12T21:16:12.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="8444" PostTypeId="2" ParentId="8432" CreationDate="2014-02-12T05:37:52.650" Score="0" Body="&lt;p&gt;Freundlich gives you the amount absorbed per unit mass of absorbate, and has this absorbtion fraction be the nth root of the pressure, with n being a material-specific constnat. Langmiur gives the surface coverage, so it only allows the surface of a material to be used for sorption. Also, the functional relationship is not a root, but an inverse of the pressure multiplied by the pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualitatively, in the limits, both seem similar, but they differ in their details and will give different answers. &lt;/p&gt;&#xA;" OwnerDisplayName="user2603" LastActivityDate="2014-02-12T05:37:52.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8445" PostTypeId="1" CreationDate="2014-02-12T06:48:29.350" Score="-1" ViewCount="1120" Body="&lt;p&gt;Does periodic acid cleave the following diols?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;trans&lt;/em&gt;-cyclopropane-1,2-diol&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;trans&lt;/em&gt;-cyclobutane-1,2-diol&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;trans&lt;/em&gt;-cyclopentane-1,2-diol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4506" LastEditorUserId="9961" LastEditDate="2017-04-08T17:14:20.460" LastActivityDate="2017-04-08T17:14:20.460" Title="Effect of periodic acid on trans diols of 3, 4 and 5 membered rings" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-04-08T18:17:34.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="8447" PostTypeId="2" ParentId="8445" CreationDate="2014-02-12T06:57:13.493" Score="1" Body="&lt;p&gt;I seriously doubt that any of the &lt;strong&gt;trans&lt;/strong&gt; diols is cleaved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is assumed to proceed through a cyclic intermediate, which is unlikely to be formed in any of the cases.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-12T06:57:13.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8448" PostTypeId="1" AcceptedAnswerId="8449" CreationDate="2014-02-12T07:50:42.413" Score="13" ViewCount="13074" Body="&lt;p&gt;In acetylene (ethyne, HC≡CH), the carbons are sp-hybridised, which make the attached protons more acidic than those in alkenes or alkanes. (See also: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32341/relative-acidities-of-alkanes-alkenes-and-alkynes&quot;&gt;Relative acidities of alkanes, alkenes, and alkynes&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't acetylene react with strong bases like NaOH or KOH?&lt;/p&gt;&#xA;" OwnerUserId="3826" LastEditorUserId="16683" LastEditDate="2018-09-06T07:35:04.380" LastActivityDate="2018-09-06T07:38:25.367" Title="Why does acetylene, having an acidic proton, not react with strong bases like NaOH and KOH?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8449" PostTypeId="2" ParentId="8448" CreationDate="2014-02-12T07:59:56.980" Score="16" Body="&lt;p&gt;The $\mathrm{p}K_\mathrm{a}$ of acetylene is $25$. It is very weakly acidic, and hence, $\ce{KOH}$ is not sufficiently alkaline to react with it. Even methanol, with a $\mathrm{p}K_\mathrm{a}$ of $15.5$ does not react much with hydroxide ion. On other hand, $\ce{NaH}$ or $\ce{NaNH2}$ have more basic anions than hydroxide, and weak acids such as acetylene or alcohols can react with them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicting the &lt;em&gt;relative&lt;/em&gt; basicities can be done qualitatively by invoking the Brønsted conjugate acid-base theory. The rule to remember is, a base gives its corresponding conjugate acid in solution, and stronger the conjugate acid, weaker its conjugate base (and vice versa). This can be understood from the following &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA&amp;lt;=&amp;gt;H^+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\ce{A-}$ is the conjugate base of the acid $\ce{HA}$. Now, if $\ce{A-}$ is highly unstable, it will have a stronger tendency to react with a proton and get converted to its conjugate acid which is relatively more stable. Since the equilibrium lies more to the left in that case, we say that the conjugate base is strong (and consequently the conjugate acid is weak). Now consider the following anions for the sake of example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HC#C-} &amp;gt; \ce{NH2-} &amp;gt; \ce{OH-} &amp;gt; \ce{F-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are listed in decreasing order of stability, and hence also their strength as bases. This is because as the electronegativity increases from carbon to fluorine, the negative charge becomes increasingly stable on the atom. More electronegative elements are increasingly stable with the extra electron due to higher effective nuclear force of attraction. Correspondingly, by the dictum &quot;stronger base is weaker conjugate acid&quot;, the order of the acidity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HC#CH} &amp;lt; \ce{NH3} &amp;lt; \ce{H2O} &amp;lt; \ce{HF}$$&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2018-09-06T07:38:25.367" LastActivityDate="2018-09-06T07:38:25.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8453" PostTypeId="1" CreationDate="2014-02-12T10:47:10.917" Score="2" ViewCount="102" Body="&lt;p&gt;We have a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; with a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;. We need to figure out the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{S^2-}$&lt;/span&gt; ions at equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2S &amp;lt;=&amp;gt; HS- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HS- &amp;lt;=&amp;gt; S^2- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Given &lt;span class=&quot;math-container&quot;&gt;$K_{\mathrm a_1} = 9.1 \times 10^{-8}, K_{\mathrm a_2} = 1.2\times 10^{-13}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I have no idea how to do this problem. I can't relate the &lt;span class=&quot;math-container&quot;&gt;$\ce{S^2-}$&lt;/span&gt; concentration to the first equation. I first tried determining the &lt;span class=&quot;math-container&quot;&gt;$\rm{HS^-}$&lt;/span&gt; concentration from the first equation, then putting that into the second one to obtain the required concentration, but then I realized that as the second reaction approaches equilibrium, the concentrations of the &lt;span class=&quot;math-container&quot;&gt;$\ce{HS-}$&lt;/span&gt; ions will change, which will then throw the first reaction out of whack.&lt;/p&gt;&#xA;&lt;p&gt;I tried taking the required concentration as &lt;span class=&quot;math-container&quot;&gt;$\pu{y M}$&lt;/span&gt; and then tried to establish a relationship, but the only thing I could get was that the concentration of the hydrogen ions will go up by &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, which is of no use at all.&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="95133" LastEditDate="2021-05-16T14:00:02.067" LastActivityDate="2021-05-16T14:06:56.093" Title="Figuring out the concentration of weak polyprotic acids in presence of strong acid" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8454" PostTypeId="2" ParentId="8443" CreationDate="2014-02-12T11:05:46.297" Score="1" Body="&lt;p&gt;You can try this &lt;a href=&quot;http://www.vscht.cz/ach/pub/LACH1-manual-en.pdf&quot; rel=&quot;nofollow&quot;&gt;manual&lt;/a&gt;. The manual is used at ICT Prague as a guide for students during inorganic laboratory exercise. The description of experiments is described there with The description of lab. equipment needed for the experiments.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-02-12T11:05:46.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8455" PostTypeId="1" AcceptedAnswerId="8458" CreationDate="2014-02-12T19:00:54.983" Score="3" ViewCount="299" Body="&lt;p&gt;I've read in some book that one of the old chemists was able to transform cheap metals into gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that gold is an element (Au) so we can't transform a metal into Au. But as I read further it says that the chemist doesn't actually transform metals into gold but into something very similar to gold (similar physical and chemical characteristics )&#xA;such that you can't differentiate between it and between gold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this chemically possible? &lt;/p&gt;&#xA;" OwnerUserId="1836" LastActivityDate="2014-02-13T22:59:29.900" Title="Can we goldify metals?" Tags="&lt;transition-metals&gt;" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8456" PostTypeId="2" ParentId="8355" CreationDate="2014-02-12T19:09:07.900" Score="0" Body="&lt;p&gt;I agree that his answer is misleading ^^.&#xA;Water for instance, only has so much space to hold anything inside of it; ie absorbing or becoming saturated with something, say salt. &#xA;So a few things affect Saturation such as density and tempurature(which changes density of course). If the water is hot, it is able to take in more salt.  Because the energy from the heat is expanding the space between the water; more space to take in material(salt). I'm sure you have experienced this with hot water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Saturation; The solution becomes saturated when the Solvant no longer has enough binding energy (lattice energy) for water to keep taking in Salt for instance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add enough salt to water, the water will become saturated with salt&gt; The salt will no longer dissolve and disappear into the water but will just float to the bottom of the solution.&lt;/p&gt;&#xA;" OwnerUserId="4510" LastActivityDate="2014-02-12T19:09:07.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8457" PostTypeId="1" CreationDate="2014-02-12T19:14:56.110" Score="3" ViewCount="15595" Body="&lt;p&gt;this might be very childish question but i am comfused because they say that organic molecules are those that contain H and C if that is so how can you call carbontetrachloride organic, even it hasn't any H atoms&lt;/p&gt;&#xA;" OwnerUserId="4147" LastActivityDate="2015-07-07T17:20:39.743" Title="why CO2 is not organic" Tags="&lt;organic-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ClosedDate="2015-07-07T17:57:50.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="8458" PostTypeId="2" ParentId="8455" CreationDate="2014-02-12T19:57:40.623" Score="9" Body="&lt;p&gt;In the days before chemistry, alchemists searched for the &lt;em&gt;Philosopher's stone&lt;/em&gt; which would enable them to transform base metals into gold. They spent a lot of effort searching for it but we know now that their search was futile as no &lt;em&gt;chemical&lt;/em&gt; process can cause elemental transformation. Only nuclear processes can do that but they cost far, far more than mining the element ever will so this doesn't really count as a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But plenty of people faked a result or were conned by their lack of understanding. Fool's gold, for example &lt;em&gt;looks&lt;/em&gt; a lot like gold, but doesn't share many other properties but, presumably, the name arises because some people are gullible. There are also many metal alloys that look like gold but are not pure (including many alloys that contain some, but not much, gold). It is also possible to use a variety of chemical or electrochemical processes to coat the surface of many metals with a thin film of gold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But careful analysis can reveal most of these methods of fakery (including the alloying of gold with other metals which was very tempting when coins were made of gold). One of the most famous stories of Greek science is based on one of the most reliable methods. Archimedes was faced with the task of distinguishing debased coinage or other gold objects and, in the legend, realised the answer while having a bath, leapt out the bath and ran down the street naked yelling &quot;eureka&quot;. What he realised was that an object immersed in water displaces a volume of water equal to its volume and this can be measured by weighing the displaced fluid. Combined with the weight of the object this can be used to estimate the &lt;em&gt;density&lt;/em&gt; of the object. Since the only other metal with a density close to gold is tungsten (gold is 19.28 g/cm&lt;sup&gt;3&lt;/sup&gt; and tungsten is 19.25 g/cm&lt;sup&gt;3&lt;/sup&gt;) knowing the density is a good way to test the purity of the metal. Common metals used in alloys, like silver (density 10.5 g/cm&lt;sup&gt;3&lt;/sup&gt;) will &lt;em&gt;substantially&lt;/em&gt; lower the density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in short, you can make a lot of things that look like gold, but it is very hard to make something that shares enough properties to convince a skeptical observer who will test more than the appearance.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-02-12T19:57:40.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8459" PostTypeId="2" ParentId="8457" CreationDate="2014-02-12T20:32:47.797" Score="8" Body="&lt;p&gt;Carbon dioxide &#xA;&quot;is&quot; neither organic nor inorganic, it's just carbon dioxide. &#xA;This distinction is nothing but a convention to separate things &#xA;between the textbooks and professors. &#xA;Carbon dioxide forms carbonic acid in water and the usual salts. &#xA;This was/is reason to think of it as inorganic. &#xA;On the other hand, urea, the diamide of carbonic acid was thought&#xA;to be organic when Wöhler synthesized urea for the first time &#xA;in lab, today it is categorized as inorganic.&lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2014-02-12T20:32:47.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8460" PostTypeId="2" ParentId="8457" CreationDate="2014-02-12T21:26:14.777" Score="0" Body="&lt;p&gt;An organic compound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) must be a (stoichiometric) compound.  Graphite, diamond, buckyballs, polycarbyne, etc. are not compounds. &lt;BR&gt;&#xA;2) must contain reduced carbon, at least one C-C or C-H bond.  Carbon tet is inorganic, chloroform is organic.  Urea, guanidine, cyanuric acid, isocyanic acid, melamine are inorganic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon monoxide, carbon dioxide, phosgene, carbonic acid are inorganic.  Ethylene carbonate, vinylene carbonate, oxalic acid, formic acid, dichloroacetylene are organic.  Aluminum carbide, $\ce{Al4C3}$, is inorganic (hydrolyzes to methane).  Calcium carbide, $\ce{CaC2}$ is organic (hydrolyzes to acetylene). &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-12T21:26:14.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8461" PostTypeId="1" CreationDate="2014-02-12T21:48:12.463" Score="1" ViewCount="138" Body="&lt;p&gt;What is a soap-like material that holds a solid shape, but dissolves completely in water over 10-15 minutes. Think disappearing Rubber Ducky.&lt;/p&gt;&#xA;" OwnerUserId="4512" LastActivityDate="2014-02-13T00:40:33.050" Title="How to make something solid that dissolves faster than bar soap in water" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8462" PostTypeId="2" ParentId="8461" CreationDate="2014-02-13T00:40:33.050" Score="2" Body="&lt;p&gt;Potassium soap is soft.  Vary the sodium-potassium ratio to get what you want.  Leave some glycerin in.  Make the soap from polyunsaturated fats.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-13T00:40:33.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8464" PostTypeId="1" CreationDate="2014-02-13T03:31:10.773" Score="1" ViewCount="7965" Body="&lt;p&gt;Just doing a little bit of research of both surfactants but I can't seem to find too much online about them. Hopefully you guys can help out!&lt;/p&gt;&#xA;" OwnerUserId="4514" LastActivityDate="2014-02-13T12:51:13.487" Title="What are the main differences of CTAB and DTAB surfactants? And what are practical applications of them?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;surfactants&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8465" PostTypeId="2" ParentId="8464" CreationDate="2014-02-13T04:17:32.857" Score="4" Body="&lt;p&gt;CTAB, aka &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/fluka/52369?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;Hexadecyltrimethylammonium bromide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DTAB, aka &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sigma/d8638?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;Dodecyltrimethylammonium bromide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sort of a vague question, don't know where to start. To begin, CTAB has a hydrocarbon chain length of 16 (as its name suggests) whereas DTAB has a -CH2- chain length of 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise they have the same hydrophilic/polar &quot;head&quot;; all this is obvious, you can tell just by the two compounds' names. They also have the aforementioned hydrophobic &quot;tails&quot;, which refers to the hydrocarbon chains because they are non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CTAB and DTAB can be used to create &lt;a href=&quot;http://en.wikipedia.org/wiki/Micelle&quot; rel=&quot;nofollow&quot;&gt;micelles&lt;/a&gt;. It's like how a biological cell is separated from its environment with a phospholipid bilayer where the hydrophobic and hydrophilic components of the phospholipids align such that you get an enclosed structure that separates a non-polar &quot;organic&quot; region in a polar environment (e.g. water). The wikipedia link has good diagrams.&#xA;The reverse is called a reverse micelle, where the heads create a polar region separated from a non-polar environment (e.g. when the solvent is hexane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one example of practical usage, I did a lab just today using hexane as a solvent where I used CTAB to create an environment to limit the formation of CdS quantum dots. This worked by limiting the volumes where conditions aren't non-polar. In this case, reverse micelles were formed; tails pointing out, head surrounding to form a more polar center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another: &lt;a href=&quot;http://www.coloss.org/beebook/I/molecular/3/2/1&quot; rel=&quot;nofollow&quot;&gt;In DNA extraction CTAB is used to pick up insoluble particles (e.g. polysaccharides) and exit the working solution.&lt;/a&gt;  You can relate surfactants to soap, which picks up insoluble (i.e. oily) substances and can easily be removed by water due to its hydrophilic outer layer, while containing the oily substances in its hydrophobic center.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="275" LastEditDate="2014-02-13T12:51:13.487" LastActivityDate="2014-02-13T12:51:13.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8466" PostTypeId="2" ParentId="8464" CreationDate="2014-02-13T10:06:53.937" Score="2" Body="&lt;p&gt;As &lt;strong&gt;rch&lt;/strong&gt; has already described, CTAB and DTAB are cationic surfactants that form microheterogenous environments in solution. In aqueous solution, these &lt;strong&gt;micelles&lt;/strong&gt; consist of an unpolar core and a polar boundary that interacts with the bulk (water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference between CTAB and DTAB is the surfactant concentration, at which these aggregates start to form. The &lt;em&gt;CMC&lt;/em&gt; (critical micellar concentration) , determined by &lt;a href=&quot;http://dx.doi.org/10.1039/B305324P&quot; rel=&quot;nofollow&quot;&gt;ESR spectroscopy&lt;/a&gt; and by &lt;a href=&quot;http://dx.doi.org/10.1016/S0021-9797%2802%2900082-6&quot; rel=&quot;nofollow&quot;&gt;fluorescence spectroscopy&lt;/a&gt; are in the range of 1 mM for CTAB and 14 mM for DTAB, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these are just the lower limits for the formation of mostly spherical aggregates. At higher surfactant concentrations, larger rod-shaped micelles are typically observed. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-13T10:06:53.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8467" PostTypeId="2" ParentId="8355" CreationDate="2014-02-13T12:36:35.937" Score="4" Body="&lt;p&gt;Take a slightly more general problem of miscibility, all that in thermodynamic equilibrium. In your question you mix up kinetics and equilibrium, and additionally in a wrong way. Equilibrium is the most important part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put two liquids together and shake them well, than in general, they do not mix. So you will observe a boundary between, say, diethyl ether (upper layer) and water (bottom layer).&#xA;If you separate them and analyze their composition, you will find, that the ether fraction contains little bit of water (1.5g / 100 mL) and that the water fraction contains little bit of ether (6g / 100mL). Knowing this, if you now take 100 mL of water and mix it with 5g of diethyl ether, you will observe, that no boundary is formed, and all ether has dissolved in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, if you shake ethanol with water, they mix in any ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's observation, now the rationale. The particles of pure liquid interact with each other, in attractive way (otherwise you will have no liquid, but gas). Now you mix two pure liquids together (A + B) and look, if some of the particles of A can easily sneak into the B. If their behavior is reasonably similar, they can do it and interact with particles of B also in attractive way. The attraction should be comparable or greater, than within pure A. That is the enthalpy part.&#xA;But as we are dealing with free energy, there is also the entropy, which you have to take into account. On one hand, entropy favors mixing, as you add lot of accessible states. On the other hand, if the attraction of B to A is very strong, it starts to organize the molecules of A around itself, so they are not as free to move as they were in pure A, and limits the miscibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the solubility of solids, it is technically similar, the A is solid and the interaction which hold sit together is the lattice energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, if you ask about solubility, miscibility and such, you should refrain from ideal approximations, as they do not predict such phenomena. Their calculation from pure substance properties is therefore terribly difficult, and physical chemistry tables are the only reasonable answer.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-02-13T12:36:35.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8468" PostTypeId="2" ParentId="4305" CreationDate="2014-02-13T13:57:37.593" Score="23" Body="&lt;h2&gt;First part&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It won't decide the issue but the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198503466&quot;&gt;Organic Chemistry&lt;/a&gt; text by Clayden, Greeves, Warren and Wothers also mentions that the matter might not be as clear-cut as the majority of your textbooks make it seem. This might strengthen the position of the textbook you're using a bit. But again, there are no references given. Here is the relevant passage (especially the last two paragraphs):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tfnl6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/op2pX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second Part&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have found the following passage on the formation of halohydrins from epoxides in the book by &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470462590&quot;&gt;Smith and March (7th Edition)&lt;/a&gt;, chapter 10-50, page 507:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t3m8Y.png&quot; height=&quot;65&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unsymmetrical epoxides are usually opened to give mixtures of regioisomers. In a typical reaction, the halogen is delivered to the less sterically hindered carbon of the epoxide. In the absence of this&#xA;structural feature, and in the absence of a directing group, relatively equal mixtures of&#xA;regioisomeric halohydrins are expected. The phenyl is such a group, and in 1-phenyl-2-&#xA;alkyl epoxides reaction with $\ce{POCl3}/\ce{DMAP}$ ($\ce{DMAP}$ = 4-dimethylaminopyridine) leads to&#xA;the chlorohydrin with the chlorine on the carbon bearing the phenyl.${}^{1231}$ When done in an&#xA;ionic liquid with $\ce{Me3SiCl}$, styrene epoxide gives 2-chloro-2-phenylethanol.${}^{1232}$ The&#xA;reaction of thionyl chloride and poly(vinylpyrrolidinone) converts epoxides to the corresponding&#xA;2-chloro-1-carbinol.${}^{1233}$ Bromine with a phenylhydrazine catalyst, however,&#xA;converts epoxides to the 1-bromo-2-carbinol.${}^{1234}$ An alkenyl group also leads to a&#xA;halohydrin with the halogen on the carbon bearing the $\ce{C=C}$ unit.${}^{1235}$ Epoxy carboxylic&#xA;acids are another example. When $\ce{NaI}$ reacts at pH 4, the major regioisomer is the 2-iodo-3-&#xA;hydroxy compound, but when $\ce{InCl3}$ is added, the major product is the 3-iodo-2-hydroxy&#xA;carboxylic acid.${}^{1236}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1231}$  Sartillo-Piscil, F.; Quinero, L.; Villegas, C.; Santacruz-Juarez, E.; de Parrodi, C.A. &lt;em&gt;Tetrahedron Lett.&lt;/em&gt; &lt;strong&gt;2002&lt;/strong&gt;,&#xA;&lt;em&gt;43&lt;/em&gt;, 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1232}$  Xu, L.-W.; Li, L.; Xia, C.-G.; Zhao, P.-Q. &lt;em&gt;Tetrahedron Lett.&lt;/em&gt; &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt;, 2435.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1233}$  Tamami, B.; Ghazi, I.; Mahdavi, H. &lt;em&gt;Synth. Commun.&lt;/em&gt; &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;32&lt;/em&gt;, 3725.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1234}$  Sharghi, H.; Eskandari, M.M. &lt;em&gt;Synthesis&lt;/em&gt; &lt;strong&gt;2002&lt;/strong&gt;, 1519.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1235}$  Ha, J.D.; Kim, S.Y.; Lee, S.J.; Kang, S.K.; Ahn, J.H.; Kim, S.S.; Choi, J.-K. &lt;em&gt;Tetrahedron Lett.&lt;/em&gt; &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt;, 5969.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{1236}$  Fringuelli, F.; Pizzo, F.; Vaccaro, L. &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;2001&lt;/strong&gt;, &lt;em&gt;66&lt;/em&gt;, 4719. Also see Concellón, J.M.; Bardales, E.;&#xA;Concellón, C.; García-Granda, S.; Díaz, M.R. &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;69&lt;/em&gt;, 6923.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-02-13T14:57:51.723" LastActivityDate="2014-02-13T14:57:51.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8469" PostTypeId="2" ParentId="8390" CreationDate="2014-02-13T15:25:32.483" Score="1" Body="&lt;p&gt;The energy amount used for heating up a material can be calculated according to the following equation:&#xA;$$ E = c_\text{s}\, m\, \Delta T$$&#xA;where $E$ is the energy, $c_\text{s}$ the specific heat capacity, $m$ the mass of the object and $\Delta T$ the temperature difference that is to be achieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the above equation for $\Delta T$ we get:&#xA;$$ \Delta T = \frac{E}{c_\text{s}\, m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the values provided by you, I get $\Delta T = 22.6\;\mathrm{K}$ (or °C, since we're talking about the &lt;em&gt;difference&lt;/em&gt; and the scale is the same, it doesn't matter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding this increase in temperature to the initial temperature gives the final answer:&#xA;$$ \vartheta_\text{f} = \vartheta_\text{i} + \Delta \vartheta = 21.0\; \mathrm{°C} + 22.6\; \mathrm{°C} = 43.6\; \mathrm{°C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Number of significant digits:&lt;/strong&gt; Since the data given consistently had three significant digits, I've rounded to the third digit.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Symbol for temperature:&lt;/strong&gt; I used $\vartheta$ for temperatures in °C and $T$ for temperatures in K. I strongly suggest you follow this example as to never confuse the two.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-13T15:25:32.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8471" PostTypeId="2" ParentId="8453" CreationDate="2014-02-13T18:02:01.647" Score="2" Body="&lt;p&gt;The first step in solving this kind of problems is to write the reactions that take place in your solution, and the equations of the acidity constants :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2S + H2O &amp;lt;=&amp;gt; HS- + H3O+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_{\mathrm a_1}=\frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HS- + H2O &amp;lt;=&amp;gt; S^2- + H3O+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_{\mathrm a_2}=\frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; is a strong acid (&lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a &amp;lt; 0$&lt;/span&gt;), the dissociation equilibrium for it will be totally shifted to right:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$$&lt;/span&gt;&#xA;And the dissociation of water, occuring in any solution:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + HO-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_\mathrm w = [\ce{H3O+}][\ce{HO-}]$$&lt;/span&gt;&#xA;*The concentration of water is not taken into account because it remains approximately constant as it is much bigger than the concentration of the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; dissolved. It is also right to write the equations as you did above, considering the dissociation of the acid, but it's better to use the hydronium ion, &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;, as we know &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; is quite unstable on its own. (In some books &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt; is used to ease writing)&lt;/p&gt;&#xA;&lt;p&gt;As you saw, you cannot calculate the concentrations at equilibrium (written in square brackets, i.e. &lt;span class=&quot;math-container&quot;&gt;$[\ce{HS-}]$&lt;/span&gt;) considering only the first reaction (unless you consider the shift of the second reaction doesn't affect significantly the first, as Nicolau said, and so we use the &lt;em&gt;analytical concentration&lt;/em&gt; of a compound (in this case &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;) which means the specific concentration of the solute when the solution is prepared (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; for both acids, if these are the concentrations in the mixture of acids solution and not the initial concentration of each acid solution). It is usually denoted &lt;span class=&quot;math-container&quot;&gt;$c_{\ce{H2S}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c_{\ce{HCl}}$&lt;/span&gt; respectively.&lt;/p&gt;&#xA;&lt;p&gt;Looking at the above reactions, we can write the formula for the analytical concentrations :&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_{\ce{H2S}} = [\ce{H2S}] + [\ce{HS-}] + [\ce{S^2-}]$$&lt;/span&gt;&#xA;and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_{\ce{HCl}} = [\ce{Cl-}] $$&lt;/span&gt;&#xA;As you might notice no &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; concentration appeared in any of these formulas, because it depends on both &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; concentrations. &lt;/p&gt;&#xA;&lt;p&gt;However, this is not enough to solve our problem, and so the &lt;em&gt;electroneutrality principle&lt;/em&gt; is used, stating that the concentration of the negative ions must be equal to the concentration of the positive ones, and in our case may be written as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{HS-}] + 2[\ce{S^2-}] + [\ce{Cl-}] + [\ce{HO-}] = [\ce{H3O+}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;*The concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{S^2-}$&lt;/span&gt; has the coefficient &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; because of the two negative charges, unlike the other ions, which have only one negative or positive charge and don't need any.&lt;/p&gt;&#xA;&lt;p&gt;Using the equations of the analytical concentrations in the electroneutrality formula, we get this :&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_{\ce{H2S}} - [\ce{H2S}] + [\ce{S^2-}] + c_{\ce{HCl}} + [\ce{HO-}] = [\ce{H3O+}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And, using also the acidity constants and the water constant :&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_{\ce{H2S}}-\frac{[\ce{H3O+}][\ce{HS-}]}{K_{\mathrm a_1}}+\frac{K_{\mathrm a_2}[\ce{HS-}]}{[\ce{H3O+}]} + c_{\ce{HCl}} + \frac{K_\mathrm w}{[\ce{H3O+}]} = [\ce{H3O+}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The equation has 2 unknown concentrations (&lt;span class=&quot;math-container&quot;&gt;$[\ce{H3O+}]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{HS-}]$&lt;/span&gt;) and you may approximate the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$c_{\ce{HCl}}$&lt;/span&gt; since the acidity constants for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; are very small:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{H3O+}]=c_{\ce{HCl}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{HS-}$&lt;/span&gt; can be calculated:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{HS-}]=\frac{c_{\ce{H2S}} + \frac{K_\mathrm w}{c_{\ce{HCl}}}}{\frac{c_{\ce{HCl}}}{K_{\mathrm a_1}} + \frac{K_{\mathrm a_2}}{c_{\ce{HCl}}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, since &lt;span class=&quot;math-container&quot;&gt;$\frac{K_\mathrm w}{c_{\ce{HCl}}} &amp;lt;&amp;lt; c_{\ce{H2S}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{K_{\mathrm a_2}}{c_{\ce{HCl}}} &amp;lt;&amp;lt; \frac{c_{\ce{HCl}}}{K_{\mathrm a_1}}$&lt;/span&gt;, the above equation can be approximated to:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{HS-}] = \frac{c_{\ce{H2S}}K_{\mathrm a_1}}{c_{\ce{HCl}}}$$&lt;/span&gt;&#xA;The approximations lead us to a simple formula, which could be denoted directly from the first equilibrium, but without understanding exactly why. From this the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{S^{2-}}$&lt;/span&gt; can be calculated easily, just like you did first. Even though some would just ignore the second equilibrium for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt;, that is &lt;strong&gt;NOT&lt;/strong&gt; the way to solve analytical chemistry problems. First you consider all of the equilibriums and processes occuring in a solution and then (if possible) approximate, and not the reverse.&lt;/p&gt;&#xA;" OwnerUserId="2470" LastEditorUserId="95133" LastEditDate="2021-05-16T14:06:56.093" LastActivityDate="2021-05-16T14:06:56.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8476" PostTypeId="1" AcceptedAnswerId="8477" CreationDate="2014-02-13T21:35:21.190" Score="1" ViewCount="4156" Body="&lt;p&gt;This is a question I have for Chemistry homework, but I don't understand how to solve it without a given mass. I narrowed the choices down (see below &lt;strong&gt;&lt;em&gt;_&lt;/em&gt;&lt;/strong&gt;) I know a limiting reactant well limits the reaction (so if there are 2 reactants) and the react left is the excess reactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought Propane would be the limiting reactant because the oxygen is in excess (though that was incorrect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I though Oxygen would be the limiting reactant (naturally) since its a combustion reaction (though that is incorrect). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine whether each reaction depends on a limiting reactant. Explain why or why not and identify the limiting reactant. (Select all that apply.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Propane (C3H8) burns in excess oxygen to produce carbon dioxide and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, one reactant is naturally limiting.&#xA;Yes, there are two reactants.&#xA;No, all reactants are in excess.&#xA;No, there is only one reactant.&#xA;propane&#xA;oxygen&#xA;carbon dioxide&#xA;water&#xA;Not enough information to identify which is the limiting reactant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yes, one reactant is naturally limiting.&#xA;Yes, there are two reactants.&#xA;No, all reactants are in excess.&#xA;propane&#xA;oxygen&#xA;Not enough information to identify which is the limiting reactant.&lt;/p&gt;&#xA;" OwnerUserId="4491" LastActivityDate="2014-02-13T22:50:12.673" Title="Limiting Reactants without a given mass" Tags="&lt;reaction-mechanism&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8477" PostTypeId="2" ParentId="8476" CreationDate="2014-02-13T22:33:05.327" Score="1" Body="&lt;p&gt;It seems that you would need information about how much of each reactant you have when conducting this. edit: &quot;Excess&quot; is the key word here; so it hints that Oxygen is NOT the limiting reagent. Considering how combustion reactions are structured, the answer seems to be the other reactant here.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="3704" LastEditDate="2014-02-13T22:50:12.673" LastActivityDate="2014-02-13T22:50:12.673" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8478" PostTypeId="2" ParentId="8455" CreationDate="2014-02-13T22:59:29.900" Score="2" Body="&lt;p&gt;No alchemist can perform transmutation of lead into gold with alembic! But now day we can perform &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_transmutation&quot; rel=&quot;nofollow&quot;&gt;nuclear transmutation&lt;/a&gt;  and gold can be now synthesized  by irradiation of mercury or platinum in a nuclear reactor or with an accelerator. However this is not cost effective!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there are many alloys that looks like gold and an alchemist could make easily most of them. Brass is one metal that can be similar to gold  varying the copper zinc ratio you can make Aich's alloy and Prince's metal two types of brass with an aspect very similar to gold. In ancient rome &lt;em&gt;aurichalcum&lt;/em&gt; was a metal (brass type alloy probably) with an aspect similar to gold. The problem is that usually you don't see if something is made of gold you feel it! If  you try to cheat someone who deal with gold with these metals he suddenly tell you that they are to light to be gold! Gold is in fact heavier than lead! This is why density esteem works great and simply coating the surface with gold doesn't work. &lt;a href=&quot;http://news.bbc.co.uk/2/hi/africa/7294665.stm&quot; rel=&quot;nofollow&quot;&gt;Ethiopian  fake gold&lt;/a&gt; fail for that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to make a good gold sophistication you have to find an isobar of gold, an element with the same weight and density of gold. Tungsten is a perfect candidate, and sometimes you can read about &lt;a href=&quot;http://usnews.nbcnews.com/_news/2012/09/19/13963728-counterfeit-gold-bars-discovered-in-new-york-city?lite&quot; rel=&quot;nofollow&quot;&gt;founding of gold plated tungsten bars&lt;/a&gt; or other theory &lt;a href=&quot;http://www.wallstreetdaily.com/2013/07/16/fort-knox-gold/&quot; rel=&quot;nofollow&quot;&gt;about FortKnox&lt;/a&gt;. &lt;a href=&quot;http://www.popsci.com/diy/article/2008-03/how-make-convincing-fake-gold-bars&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; you can read a good article about gold sophistication. In this case a layer of gold over the tungsten is needed to make it looks like gold and to pass the exams with superficial elemental analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a good looking gold surface I prefer &lt;a href=&quot;http://books.google.it/books?id=Q1Pc0aY-vg4C&amp;amp;pg=PA54&amp;amp;dq=iron%28III%29+mica+pigments&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=5Uv9Uqq4M4zAtAakvYDwDg&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q=iron%28III%29%20mica%20pigments&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;iron(III) oxide mica pigments&lt;/a&gt; very nice color without cheating nobody! &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-02-13T22:59:29.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8479" PostTypeId="1" AcceptedAnswerId="8506" CreationDate="2014-02-14T02:28:35.080" Score="7" ViewCount="1555" Body="&lt;p&gt;Coffee served hot can be &lt;a href=&quot;http://drinks.seriouseats.com/2012/12/should-you-put-salt-in-your-coffee-reduce-bitterness.html&quot; rel=&quot;noreferrer&quot;&gt;quite bitter&lt;/a&gt; (presumably high pH). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several people I have known will add a single grain of &lt;a href=&quot;http://blog.khymos.org/2010/03/21/a-pinch-of-salt-for-your-coffee-sir/&quot; rel=&quot;noreferrer&quot;&gt;salt to the coffee&lt;/a&gt; - which &lt;a href=&quot;http://chemistry.about.com/b/2014/01/02/adding-salt-to-reduce-bitterness.htm&quot; rel=&quot;noreferrer&quot;&gt;reduces the perceived bitterness&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree?&lt;/strong&gt; Can you explain the pH change that is occurring during this reaction?&lt;/p&gt;&#xA;" OwnerUserId="4478" LastEditorUserId="4478" LastEditDate="2014-02-14T23:33:08.923" LastActivityDate="2017-10-19T15:53:07.350" Title="What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;molecules&gt;&lt;home-experiment&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8480" PostTypeId="1" CreationDate="2014-02-14T02:59:28.477" Score="2" ViewCount="144" Body="&lt;p&gt;I'm doing a project with lead (II) sulfide quantum dots because they have optical transitions in the near infrared. I understand the simple &quot;particle in a box&quot; approach where confinement strengthens exponentially when the box shrinks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is how this confinement results in discrete energy levels? Can someone tell me how this shrinking of molecular size and confinement of the electron hole pair (exciton Bohr radius) results in discrete energy levels? Is it the electron shells being changed?   &lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="73" LastEditDate="2014-11-30T12:37:25.207" LastActivityDate="2014-11-30T12:37:25.207" Title="Why does confinement of quantum dots result in discrete energy levels?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8481" PostTypeId="1" CreationDate="2014-02-14T03:03:29.090" Score="2" ViewCount="682" Body="&lt;p&gt;I'm writing a computer program to calculate the order of a chemical reaction with respect to each reactant. When I do this by hand, it's easy to isolate which reactant concentration stays relatively the same while the others change. However, when you don't have an isolation method available, is there a non-guess and check way to determine rate orders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with the sample data of a 1st order reaction with respect to A and B, and second order overall&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+--------+-----+-----+------+&#xA;| Trial# | [A] | [B] | Rate |&#xA;+--------+-----+-----+------+&#xA;|      1 |   1 |   1 |   10 | &#xA;|      2 |   2 |   3 |   60 | &#xA;|      3 |   3 |   2 |   60 |&#xA;+--------+-----+-----+------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There isn't a trial with two constant values, so I'm not sure how you would programmatically determine the orders with respect to each of the reactants without manually trying every possibility until one fit (I realize that this wouldn't take too much computing time to do, but it won't scale well for multiple trials, and I also would prefer a non-brute force solution if possible). If there is no way to determine the rate with respect to each reactant, is there a way to determine the overall rate? This would greatly reduce the number of possibilities of orders when faced with larger rates&lt;/p&gt;&#xA;" OwnerUserId="4455" LastActivityDate="2014-05-05T14:49:10.947" Title="Determining order of a reaction without isolation" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8483" PostTypeId="2" ParentId="6032" CreationDate="2014-02-14T06:08:17.610" Score="-3" Body="&lt;p&gt;Oxygen atoms want nothing more than to share, borrow, or steal a couple extra electrons.  Chlorine has the same desire for an extra electron.  Therefore, they do not react in solution except in special circumstances I don't know about yet. &lt;/p&gt;&#xA;" OwnerUserId="4526" LastActivityDate="2014-02-14T06:08:17.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8484" PostTypeId="1" CreationDate="2014-02-14T06:45:26.133" Score="-2" ViewCount="326" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A hydrogen atom is in a cubic box with side lengths equal to $\require{mediawiki-texvc}\pu{100 \AA{}}$. For what value of $n$ (principal quantum number) will the expectation value of the radius be equal to one-half the box size?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4524" LastEditorUserId="194" LastEditDate="2018-01-29T04:31:06.930" LastActivityDate="2018-01-29T04:31:06.930" Title="Radial expectation value for particle in a cubic box" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-02-14T20:31:19.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="8485" PostTypeId="2" ParentId="8484" CreationDate="2014-02-14T12:34:20.600" Score="3" Body="&lt;p&gt;\[&#xA;\left&amp;lt; r_{nl} \right&gt; = n^2\cdot a_0\ \left\{ 1+ \frac{1}{2} \left[1-\frac{l(l+1)}{n^2}\right] \right\}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should be close to what is applicable to solve to problem in a simple way. If the value is a bit off, blame it on &lt;em&gt;Niels Bohr&lt;/em&gt; or the cabinetmaker who built the box.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-14T12:34:20.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8486" PostTypeId="1" CreationDate="2014-02-14T14:06:00.843" Score="-2" ViewCount="1041" Body="&lt;p&gt;Preparation of alkanes by aldehydes and ketones, its chemical reaction and mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4528" LastActivityDate="2014-02-14T16:55:06.907" Title="How alkanes can be prepared by aldehydes and ketones?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-02-14T20:30:46.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="8487" PostTypeId="2" ParentId="8395" CreationDate="2014-02-14T14:08:22.603" Score="3" Body="&lt;p&gt;Voltage is indeed something that is measured at a particular point, but that doesn't mean its a property of only one point and in fact every point will have some voltage. In particular you can have a series of points that map out an &quot;equipotential&quot; where the potential is equal all along these points. A terminal of a battery is an equipotential so no matter where an electron &quot;is&quot; it will have the same potential energy U = e V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively this is just like potential energy due to gravity. The battery is like a mountain with a flat top and level plane at the base. Everywhere along the top the altitude is equal and a ball will have the same potential energy. Similarly everywhere along the level base of the mountain the ball will have the same potential energy. &lt;/p&gt;&#xA;" OwnerUserId="4527" LastActivityDate="2014-02-14T14:08:22.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8488" PostTypeId="2" ParentId="8457" CreationDate="2014-02-14T14:20:23.900" Score="-1" Body="&lt;p&gt;The C-H bond has a lower bond energy than&#xA;the carbon-oxygen bond in carbon dioxide,&#xA;making carbon dioxide more stable/less&#xA;reactive than the typical organic compound.&#xA;So, when you're determining whether a carbon&#xA;compound is organic or not, look to see&#xA;whether it contains hydrogen in addition to&#xA;carbon and whether the carbon is bonded to&#xA;the hydrogen. &lt;/p&gt;&#xA;" OwnerUserId="4528" LastActivityDate="2014-02-14T14:20:23.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8490" PostTypeId="1" CreationDate="2014-02-14T15:57:04.260" Score="13" ViewCount="156586" Body="&lt;p&gt;I had a recent question on a test that asked what the products would be if sodium hydrogen carbonate were dissolved in water. I had a few candidate answers&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\displaystyle\ce{NaHCO3 -&amp;gt; Na+ + HCO3-}$, but that one doesn't involve water at all&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle\ce{NaHCO3 + H2O -&amp;gt; Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?&lt;/li&gt;&#xA;&lt;li&gt;$\displaystyle\ce{NaHCO3 + H2O -&amp;gt; Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Part two of the question asked whether the solution would be acidic, basic, or neutral. What is the proper reaction for this?&lt;/p&gt;&#xA;" OwnerUserId="4455" LastEditorUserId="4945" LastEditDate="2017-07-24T05:24:02.193" LastActivityDate="2018-07-22T10:27:53.287" Title="What are the products of the dissociation of sodium bicarbonate in water? What is the relative pH of the solution?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8491" PostTypeId="2" ParentId="8490" CreationDate="2014-02-14T16:39:19.880" Score="-1" Body="&lt;p&gt;First, ionize the salt as you have shown. NaOH is a very strong base so Na+ must be a very weak acid. It is solvated. Being a singly-charged, rather large cation, it stops there. If it were $\ce{Al^{3+}}$, coordinated water would be acidic by charge withdrawal from that brutally small trication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bicarbonate is the salt of the first ionization of weak carbonic acid. It will then be a modestly weak base by hydrolysis, as you have shown (reversible). Bicarbonate solutions are weakly basic. If you add acid (even weak acetic acid) they fizz. Boiling the solution will drive out carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbonate is the salt of the second ionization of weak carbonic acid. It is a strong base.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="23561" LastEditDate="2018-07-22T10:27:53.287" LastActivityDate="2018-07-22T10:27:53.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8492" PostTypeId="2" ParentId="8486" CreationDate="2014-02-14T16:55:06.907" Score="3" Body="&lt;p&gt;You want to deoxygenate the carbonyls under conditions where you do not get condensation reactions.  Do you want to pull the oxygen or dimerize to the olefins then hydrogenate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolff-Kishner or Clemmensen reductions, thioacetal then Raney nickel and hydrogen.  McMurry reaction for dimerization.  Given polymethylhydrosiloxane,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Synlett 276 (2009) &lt;BR&gt;&#xA;Org. Lett. 13, 584 (2011)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-14T16:55:06.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8494" PostTypeId="1" AcceptedAnswerId="8495" CreationDate="2014-02-14T21:24:29.560" Score="6" ViewCount="19734" Body="&lt;p&gt;At school we've learned about calculating the voltage needed to perform an electrolysis. The current is how many electrons pass per time. The formula is given &lt;span class=&quot;math-container&quot;&gt;$$I = \frac UR$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that given an electrolysis with &lt;span class=&quot;math-container&quot;&gt;$U=\pu{4.0 V}$&lt;/span&gt; we get &lt;span class=&quot;math-container&quot;&gt;$$I = \frac{\pu{4.0 V}}{R}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am correct, what would decide how fast the electrolysis will go, is the resistance. Is the resistance fixed, or can I do anything to lower it? How would increasing the voltage affect the electrolysis? Would it just make it go faster (because &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is increased)?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorDisplayName="user7951" LastEditDate="2018-09-21T20:27:01.797" LastActivityDate="2018-09-26T15:02:05.333" Title="How does voltage and resistance affect an electrolysis?" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8495" PostTypeId="2" ParentId="8494" CreationDate="2014-02-14T21:52:51.780" Score="9" Body="&lt;p&gt;Voltage determines the kind of chemistry at each electrode.  Gold is purified by plating the anode to the cathode at minimum voltage for the half-reaction, so impurities either do not dissolve and do not plate out.  Copper electrorefining electrode mud contains valuable trace elements in concentrated from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current determines the amount of chemistry, ~96,500 coulombs/mole of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cell internal resistance converts input energy, power multipleid by time, $\ce{I^2Rt}$, into useless heat.  You then want small electrode spacing and good electrolyte stirring.  You then need enough voltage to do the chemistry, plus more to compensate for resistance losses, but not so much voltage that undesirable chemistries are enabled.  Does cell resistance increase (ohmic) or decrease with temperature?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-14T21:52:51.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8496" PostTypeId="1" AcceptedAnswerId="8503" CreationDate="2014-02-14T23:59:42.957" Score="2" ViewCount="116" Body="&lt;p&gt;To make $\ce{NaOH}$ one could electrolyze a solution of $\ce{NaCl}$ in water ($\ce{H2O}$) It would go like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} + 2Na+_{(aq)} -&amp;gt; H2_{(g)} + Cl2_{(g)} + Na+_{(aq)} + 2OH-_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculating a bit, I found that even just a few minutes of electrolyzing this at $10\ \mathrm{A}$ could be deadly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody confirm this?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="23561" LastEditDate="2018-08-12T21:57:23.270" LastActivityDate="2018-08-12T21:57:23.270" Title="Is there a danger in electrolyzing NaCl?" Tags="&lt;home-experiment&gt;&lt;safety&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8497" PostTypeId="1" AcceptedAnswerId="8500" CreationDate="2014-02-15T00:27:36.370" Score="5" ViewCount="12304" Body="&lt;p&gt;I've read in several places that when electrolyzing water into $\ce{O2}$ and $\ce{H2}$, that adding a salt or an acid will speed up the process. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorDisplayName="user4076" LastEditDate="2014-02-15T06:57:11.803" LastActivityDate="2015-11-28T06:13:40.587" Title="Why does adding an acid speed up electrolysis of water?" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8498" PostTypeId="1" AcceptedAnswerId="8515" CreationDate="2014-02-15T00:39:42.200" Score="14" ViewCount="30575" Body="&lt;p&gt;I want to electrolyze water to get oxygen and hydrogen. I want to mix them in a regular balloon and ignite it. How much hydrogen would you need in a 2:1 ratio with air for it to be dangerous? Is this generally a safe experiment?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="1677" LastEditDate="2014-09-07T14:19:01.060" LastActivityDate="2019-03-09T14:07:39.640" Title="How explosive is hydrogen gas?" Tags="&lt;home-experiment&gt;&lt;safety&gt;&lt;hydrogen&gt;&lt;explosives&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8500" PostTypeId="2" ParentId="8497" CreationDate="2014-02-15T02:38:15.563" Score="3" Body="&lt;p&gt;Primarily due to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pure water conducts an electric current very poorly and, for this reason, is difficult to electrolyze.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Source: &lt;a href=&quot;http://www1.lsbu.ac.uk/water/electrolysis.html#intro&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water Electrolysis&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that salt and acids are able to act as an electrolyte catalyst as most salt and acids dissolve (hence dissociate) in water, specifically, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a water-soluble electrolyte is added, the conductivity of the water rises considerably. The electrolyte disassociates into cations and anions; the anions rush towards the anode and neutralize the buildup of positively charged $\ce{H+}$ there; similarly, the cations rush towards the cathode and neutralize the buildup of negatively charged $\ce{OH-}$ there. This allows the continued flow of electricity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A diagram of the electrolysis of water with dissolved salt/acid ions is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oF4ID.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.healthandmed.com/t-How_Does_an_ionizer_work.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information can be found in the article &lt;a href=&quot;http://www.cres.gr/kape/publications/papers/dimosieyseis/ydrogen/A%20REVIEW%20ON%20WATER%20ELECTROLYSIS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Review of Water Electrolysis&lt;/a&gt; (Zoulias et al).&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2014-02-15T10:55:39.833" LastActivityDate="2014-02-15T10:55:39.833" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8502" PostTypeId="2" ParentId="7847" CreationDate="2014-02-15T04:00:34.187" Score="0" Body="&lt;p&gt;The magnesium nitride hydrolyzes to Mg(OH)2 then dehydrates to MgO.  If formation is 3:1, then 4x = 3.22 g MgO.  Solve for x.  All the fun is in the footnotes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-15T04:00:34.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8503" PostTypeId="2" ParentId="8496" CreationDate="2014-02-15T04:13:33.693" Score="3" Body="&lt;p&gt;If the cell is mixed, you would make bleach, sodium hypochlorite and some free chlorine (and probably chew up your anode). Since the solution is not very basic, hypochlorite would probably disproportionate to chlorite, and that possibly to chlorate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One mole of electrons is one faraday, about 96 500 coulombs. One ampere is one coulomb per second. Five minutes passes $(5\ \mathrm{min})(60\ \mathrm{s/min})(10\ \mathrm{C/s}) = 3000\ \mathrm{C}$ or $0.031\ \mathrm{F}$ in a reaction that requires two moles input for each mole of chlorine. $0.01554\ \mathrm{mol}$ chlorine or $350\ \mathrm{ml}$ at STP, tops. It wouldn’t be good for you.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T16:05:20.317" LastActivityDate="2015-06-17T16:05:20.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8506" PostTypeId="2" ParentId="8479" CreationDate="2014-02-15T10:42:21.757" Score="3" Body="&lt;p&gt;Is simply a dissolution no reaction occur and as said Niculau the pH change, if occur, is absolutely negligible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Salt in general affect taste lowering the bitterness of the food enhancing the positive sensory attributes of foods. In fact the reason of this behavior is the sodium ion $Na^+$ (even litium ion works) and his interaction with the receptors in taste buds. So sodium acetate therethically works as well, Gery Beuchamp has studied many of this salt, quoting from &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK50958/#ch3.r19&quot; rel=&quot;nofollow&quot;&gt;NCBI&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;various sodium-containing ingredients have been known to reduce the&#xA;  bitterness of certain compounds found in foods, including quinine&#xA;  hydrochloride, caffeine, magnesium sulfate, and potassium chloride&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/8788095&quot; rel=&quot;nofollow&quot;&gt;This is an article&lt;/a&gt; about the suppression of bitterness by sodium. From here I'm afraid is more a question of physiology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read this the first time here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, Herve. Molecular gastronomy: exploring the science of flavor. Columbia University Press, 2006 p. 94&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-02-15T10:42:21.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8508" PostTypeId="2" ParentId="8490" CreationDate="2014-02-15T12:57:31.147" Score="13" Body="&lt;p&gt;It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaHCO3 -&amp;gt; Na+ + HCO3-}$, but that one doesn't involve water at all&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at this reaction:&#xA;$$\ce{NaCl &amp;lt;=&amp;gt;Na+ +Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this reaction doesn't &quot;involve&quot; water in the schematics but is right, we assume that is a dissociation of a salt in water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaHCO3 + H2O -&amp;gt; Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but in fact this is an equilibrium:&#xA;$$\ce{H2CO3 &amp;lt;=&amp;gt; H2O + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you have to take into account another important equilibrium many time ignored:&#xA;$$\ce{CO2 (aq) &amp;lt;=&amp;gt; CO2 (g)}$$&#xA;So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ &quot;escapes&quot; until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaHCO3 + H2O -&amp;gt; Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Which reaction is not right?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as:&#xA;$$\begin{multline}&#xA;\ce{NaHCO3 + H2O &amp;lt;=&amp;gt;[A] Na+ + HCO3- + H2O &amp;lt;=&amp;gt;[B] Na+ + H2CO3 + OH-} \\&#xA;\ce{ &amp;lt;=&amp;gt;[C] Na+ + OH- + H2O + (CO2(aq)&amp;lt;=&amp;gt;[C_2]CO2 ^ (g))}\end{multline}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question should be more specific asking for &lt;strong&gt;the ultimate&lt;/strong&gt; products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="4945" LastEditDate="2017-07-24T05:34:18.960" LastActivityDate="2017-07-24T05:34:18.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8509" PostTypeId="1" CreationDate="2014-02-15T15:09:20.057" Score="24" ViewCount="4219" Body="&lt;p&gt;In Electrophilic Aromatic Substitution ($\text{S}_{\text{E}}\text{Ar}$) reactions groups that can donate an electron pair into the aromatic ring, e.g. $\ce{-OH}$ or $\ce{-NH2}$, are &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-directing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/66MCG.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{-NH2}$-groups one has to be cautious because if the $\text{pH}$ value of the solution is too low the $\ce{-NH2}$-group will be present mostly in its protonated form, $\ce{-NH3^+}$, which lacks a free electron pair and has a &lt;em&gt;meta&lt;/em&gt;-directing effect because of its strong $-I$ effect. This brings up the question: What is the situation like with protonated $\ce{-OH}$ groups, $\ce{-OH2^{+}}$? In contrast to $\ce{-NH3^+}$ this group still has a free electron pair which could, in principle, interact with the aromatic ring ($\Rightarrow$ &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-directing) but it should also have a very powerful $-I$ effect ($\Rightarrow$ &lt;em&gt;meta&lt;/em&gt;-directing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bCOdy.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say, if the electron pair can be effeciently delocalised into the aromatic ring then its &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-directing power should outweigh the meta-directing power of the inductive effect. But I wonder if there even is any appreciable degree of delocalisation present because the mesomeric structures don't look very favourable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V0Zgb.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about it from a frontier orbital perspective I would say: The orbital of the $\ce{-OH2^{+}}$ group's free electron pair will be much lower in energy than that of the $\ce{-OH}$ group's free electron pair. Thus, the energy difference between free electron pair's orbital and the aromatic ring's $\pi$-system might be too large to facilitate a good interaction. Therefore, I'd expect a protonated $\ce{-OH}$ group to be &lt;em&gt;meta&lt;/em&gt;-directing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the actual situation when phenols or phenol ethers undergo ($\text{S}_{\text{E}}\text{Ar}$) reactions in strongly acidic solutions? Does one get the &lt;em&gt;ortho/para&lt;/em&gt; or the &lt;em&gt;meta&lt;/em&gt; product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up the $\mathrm{pK}_{\mathrm{a}}$ value for protonated phenol: It is approximately $-7$. This makes me wonder if there are any reasonable (real world) situations where the solution is acidic enough to protonate phenol (or phenol ethers) completely (even $\ce{HI}$ with its $\mathrm{pK}_{\mathrm{a}}$ of $-10$ shouldn't be enough for that, right). I'm not sure if even sulfonation or nitration reach such low $\text{pH}$ values.&#xA;But @Uncle Al's answer made me aware that I forgot about Lewis acids and I'd say that Lewis acids like $\ce{AlCl3}$ that coordinate strongly to oxygen might lead to a state that will be very similar to a $\ce{-OH2^{+}}$ group (please, correct me if I'm wrong and the coordination of Lewis acids is not strong enough to be compared to the $\ce{O-H}$-bond). If this is the case, then what about Friedel-Crafts reactions which make heavy use of Lewis acids: Would those Lewis acids interfere with the &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-directing properties of phenols? Does anyone have experience with the directional properties of a $\ce{-OH}$ group in the presence of a strong Lewis acid, maybe even in the context of Friedel-Crafts reactions?&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-01-05T21:38:29.803" LastActivityDate="2018-12-10T20:49:03.997" Title="Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;phenols&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8510" PostTypeId="1" CreationDate="2014-02-15T15:57:12.397" Score="0" ViewCount="187" Body="&lt;p&gt;I'm told the $\ce{KMnO4}$ is a strong electrolyte. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I figure out how the reaction looks like?&lt;/p&gt;&#xA;" OwnerUserId="4537" LastEditorUserId="189" LastEditDate="2014-02-15T16:41:52.847" LastActivityDate="2014-03-04T06:22:13.753" Title="How to determine the reaction of a strong electrolyte?" Tags="&lt;acid-base&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2019-03-09T05:41:12.287" ContentLicense="CC BY-SA 3.0" />
  <row Id="8511" PostTypeId="2" ParentId="8509" CreationDate="2014-02-15T16:01:12.913" Score="2" Body="&lt;p&gt;A four-coordinate $\ce{sp^2}$ oxygen with two positive charges is a very high energy state.  Aniline is a much weaker base than an aliphatic primary amine for leaking electron density into the ring.  Phenol in turn will be a much weaker base than a primary alcohol.  Deactivating the phenol will require a very Brønsted acidic medium indeed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis acids would be more effective.  Anhydrous $\ce{AlCl3}$ is a powerful oxophile. Adding a mole would turn off all the phenols.  One might have quite a time getting the phenol free of the aluminum during workup.  Note the large number of LAH workup recipes to avoid intractable aluminum oxohydroxide gels during workup.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="51553" LastEditDate="2018-03-21T04:21:28.283" LastActivityDate="2018-03-21T04:21:28.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8512" PostTypeId="2" ParentId="7187" CreationDate="2014-02-15T16:31:42.993" Score="2" Body="&lt;p&gt;Wikipedia has a &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Transition_metals&quot; rel=&quot;nofollow noreferrer&quot;&gt;comprehensive list on chemistry mnemonics&lt;/a&gt;, with which I'm now memorizing the position of the transition metals.&lt;/p&gt;&#xA;&lt;p&gt;For example, for the first two periods of the elements (excluding neon):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;H He Li Be B C N O F&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;H&lt;/b&gt;a. &lt;b&gt;He&lt;/b&gt;althy &lt;b&gt;Li&lt;/b&gt;ttle &lt;b&gt;Be&lt;/b&gt;ggar &lt;b&gt;B&lt;/b&gt;oys &lt;b&gt;C&lt;/b&gt;atching &lt;b&gt;N&lt;/b&gt;ewts &lt;b&gt;O&lt;/b&gt;r &lt;b&gt;F&lt;/b&gt;ish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Or my favourite for the first period of transition metals:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sc Ti V Cr Mn Fe Co Ni Cu Zn&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;b&gt;S&lt;/b&gt;aint &lt;b&gt;T&lt;/b&gt;homas &lt;b&gt;V&lt;/b&gt;ery &lt;b&gt;C&lt;/b&gt;a&lt;b&gt;r&lt;/b&gt;efully &lt;b&gt;M&lt;/b&gt;ade &lt;b&gt;F&lt;/b&gt;ive &lt;b&gt;Co&lt;/b&gt;mpletely &lt;b&gt;N&lt;/b&gt;ew &lt;b&gt;C&lt;/b&gt;opies of &lt;b&gt;Z&lt;/b&gt;arathustha.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-02-16T13:20:21.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8513" PostTypeId="1" CreationDate="2014-02-15T16:41:47.597" Score="2" ViewCount="76" Body="&lt;p&gt;What sort of ideal conditions are required - theoretically - to conduct polymerization reactions to make all the macromolecules end up with exactly the same molecular mass, just like the products in normal chemical reactions? How do living organisms manage to synthesize macromolecules, for instance proteins, different than humans do in the lab? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2014-02-15T23:56:18.163" Title="Polymers with no molecular mass distribution?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8515" PostTypeId="2" ParentId="8498" CreationDate="2014-02-15T21:08:45.543" Score="18" Body="&lt;p&gt;First, let me say that I've enjoyed many times exploding soap bubbles of about &lt;strong&gt;one milliliter&lt;/strong&gt; filled with hydrolysis gas. That is 1 cubic centimeter.&#xA;That will give you a sound that rings in your ears in a decent sized living room. You may wish to use ear protection for the experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 ml will have an effect in a lecture hall that not only wakes up everyone, but also may make people complain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now while the explosive limits of hydrogen in air range from about 18 -- 60 % the flammable limits are from 4 -- 75 %, &lt;a href=&quot;http://people.clarkson.edu/~wwilcox/Design/flamlim2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;in oxygen the limit of flammability goes all the way from 4% to 95%&lt;/a&gt; read: &lt;strong&gt;for practical purposes, hydrogen in oxygen is &lt;em&gt;always&lt;/em&gt; at least a flammable mixture&lt;/strong&gt;.&lt;br&gt;&#xA;For comparison, gasoline in air is flammable roughly between 1.5 - 7%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you start with a stochiometric mixture which is &lt;em&gt;ideal&lt;/em&gt; for explosion (after all, that's why you do it, right?), and it is in oxygen, not in air. So even if you &quot;only&quot; have a flammable mixture reaching an ignition source (e.g. electric switch), chances are quite high that as the ignition proceeds through the mixture, it can reach a zone where the gas/air mixture is explosive. Also, because hydrogen is so much lighter than air, it tends to accumulate under the ceiling, so even if there is enough air in the room that an ideal mixture would not be explosive, chances are that there is an explosive layer of gas. (Same with gasoline or, practially more relevant: solvent from glue for parquet floor tiles)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safety relevant points are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the limits of explosion and flammability are, even in air, &lt;em&gt;extremely&lt;/em&gt; wide. &lt;/li&gt;&#xA;&lt;li&gt;limits of flammability are important even if there happens to be no explosive mixture: hydrogen is a small and light molecule that diffuses fast. If the air you breathe happens to have a hydrogen concentration within the flammability limits, any ignition will burn deep into your throat on its way to the lungs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorDisplayName="user7951" LastEditDate="2019-03-09T14:07:39.640" LastActivityDate="2019-03-09T14:07:39.640" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="8516" PostTypeId="1" AcceptedAnswerId="8521" CreationDate="2014-02-15T21:09:23.533" Score="2" ViewCount="269" Body="&lt;p&gt;I am trying to find out what BPA is used for in the industrial production of plastics. On sources such as Wikipedia I only found the information that it is used, but not why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages in using BPA in the production of plastics? Why is it needed?&lt;/p&gt;&#xA;" OwnerUserId="4540" LastActivityDate="2014-02-15T23:37:58.080" Title="What is Bisphenol A used for?" Tags="&lt;plastics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8517" PostTypeId="2" ParentId="8497" CreationDate="2014-02-15T21:28:32.807" Score="1" Body="&lt;p&gt;I'd recommend using a strong base, e.g. NaOH. The reason is that if there's chloride in the solution (e.g NaCl as the easiest accessible salt or HCl as cheap strong acid), you'll produce chlorine gas instead of oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is most suitable for &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloralkali_process&quot; rel=&quot;nofollow&quot;&gt;industrial purposes&lt;/a&gt;, but not for home experiments ;-) &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-02-15T21:28:32.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8518" PostTypeId="1" CreationDate="2014-02-15T22:39:31.723" Score="1" ViewCount="3694" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UN7B2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a basic question on a closed manometer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that in this case the pressure of gas is lower than that of the liquid, which is why the liquid is higher on the left side. Now, what I don't understand is why do we say that the gas is at a negative pressure?  Wouldn't that mean there's no gas at all? Also, how do we know that it's negative? Usually, I only take the difference between the highest point and the lowest.&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorDisplayName="user7951" LastEditDate="2016-11-04T19:42:39.697" LastActivityDate="2016-11-04T19:42:39.697" Title="What does negative pressure mean in the context of this manometer?" Tags="&lt;pressure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8519" PostTypeId="2" ParentId="8516" CreationDate="2014-02-15T23:11:41.900" Score="1" Body="&lt;p&gt;Bisphenol A is the monomer used for synthesizing polycarbonates, sometimes via the following reaction:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KS1FP.png&quot; alt=&quot;Polycarbonate Reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Polycarbonatsynthese.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Bisphenol_A#Use&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;, you can see that it has &lt;em&gt;many different&lt;/em&gt; (read: non-polymer related) application possibilities, and is cheap to produce (making it the most preferred option when selecting the monomers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synthesis of BPA involves acetone and phenol, both very very cheap substances.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-15T23:11:41.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8521" PostTypeId="2" ParentId="8516" CreationDate="2014-02-15T23:37:58.080" Score="2" Body="&lt;p&gt;Bisphenol-A is very inexpensive for its considerable formula weight - from benzene, propylene, air, acid; acetone (or other ketone or aldehyde) and acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Polycarbonate, as above.  If you don't like HCl, dimethyl carbonate will do. &lt;BR&gt;&#xA;2) Epoxy resin, by bis-etherifying with glycidyl chloride.  That is a huge market, including paints and coatings. &lt;BR&gt;&#xA;3) Bis-etherify with propargyl chloride.  That thermally B-stages (removes the shrinkage upon cure) then thermosets to an astounding thermally resistant super-crosslinked resin that adheres to any polar surface. &lt;BR&gt;&#xA;4) Bis-esterify with methacrylic acid to get into heavily cross-linked free radical or UV cures. &lt;BR&gt;&#xA;5) And if you really want something resistant, instead of acetone use fluorenone or a 2,7-disubstituted fluorenone.  Or be fiendishly clever and use 4,4'-dihydroxybenzophenone instead of acetone.  Now there is a tetrahedral tecton for synthesis!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bisphenol-A is a claimed  endrocrine disruptor.  Eastman Kodak has immense production capacity for acetic anhydride (cellulose acetate photographic film).  It offshot into ketene dimer chemistry then mixed &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-2,2,4,4-tetramethyl-1,3-cyclobutanediol as a BPA replacement.  If you do not mind paying more and getting less, stuff as well as performance, the problem is solved.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-15T23:37:58.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8522" PostTypeId="2" ParentId="8518" CreationDate="2014-02-15T23:43:32.780" Score="0" Body="&lt;p&gt;If the manometer were open at both ends, the two fluid levels would be identical.  If the bulbed gas were at ambient pressure, ditto. That the fluid is higher at the bulb end says the pressure in the bulb was lower than ambient pressure, and lower still for the weight of the differential fluid head, height times density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the diameter of the tube, short of capillary tubing, irrelevant?  You can then tell us the &quot;negative&quot; pressure in the bulb.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-15T23:43:32.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8523" PostTypeId="2" ParentId="8510" CreationDate="2014-02-15T23:46:20.870" Score="1" Body="&lt;p&gt;$\ce{HMnO4}$ is s a strong acid (compare oxidation state with perchloric acid).  Its anion is then a weak base.  The salt is completely dissociated in water solution. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="1362" LastEditDate="2014-03-04T06:22:13.753" LastActivityDate="2014-03-04T06:22:13.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8524" PostTypeId="2" ParentId="8513" CreationDate="2014-02-15T23:56:18.163" Score="2" Body="&lt;p&gt;Living polymerizations can afford nearly zero polymer polydispersivity.  You can go anion (Kraton thermoplastic elastomers - and tell us about the one silicon atom in each polymer molecule), free radical (TEMPO or PROXYL capping of the free radical end, thermally activated), redox (copper halide), olefin metathesis polymerization (ROMP and ADMET), and other clevernesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proteins are made from a DNA template via RNA, one AA residue at a time. There is no dispersivity.  Post-synthesis modification, e.g., glycosylation in erythropoietin, can be polydisperse. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-15T23:56:18.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8525" PostTypeId="1" CreationDate="2014-02-16T00:50:24.523" Score="1" ViewCount="1803" Body="&lt;p&gt;I feel my method for solving it is correct and I've checked my calculations yet I can't get the right answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 °C, $K_c$ equals 1.0E-5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N2_{(g)} + O2_{(g)} &amp;lt;=&amp;gt; 2 NO_{(g)}}$     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $2.75 \times 10^{-2}$ mol $N_2$ and $2.75 \times 10^{-2}$ mol $O_2$ are sealed in a $0.886 L$ flask at $1500 °C$, what is the concentration of $\ce{NO_{(g)}}$ when equilibrium is established?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes: $K_c$ = equilibrium constant&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a) 1.38E-7 M&#xA;b) 4.35E-5 M&#xA;c) 3.85E-5 M&#xA;d) 8.70E-5 M&#xA;e) 2.30E+4 M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I set up the problem with $(2x)^2$ as a numerator and $(.03104 - x)^2$ as the denominator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took square root of both sides and did algebra and got $4.9*10^{-5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record the answer is $8.7 \times 10^{-5}$&lt;/p&gt;&#xA;" OwnerUserId="4542" LastEditorUserId="73" LastEditDate="2014-02-16T08:58:11.693" LastActivityDate="2014-02-16T08:58:11.693" Title="How do I determine the equilibrium concentration given Kc and the concentrations of component gases?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8526" PostTypeId="1" AcceptedAnswerId="8554" CreationDate="2014-02-16T02:32:17.093" Score="2" ViewCount="1275" Body="&lt;p&gt;The crystals that I am synthesizing use two different stabilizers CTAB and SDS, but they say one is used as a surfactant and since they are both stabilizers what is the need for both?&lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="186" LastEditDate="2015-10-02T16:30:50.087" LastActivityDate="2015-10-02T16:30:50.087" Title="What is the difference between the roles of a surfactant and a stabilizer in the process of nanocrystal growth?" Tags="&lt;organic-chemistry&gt;&lt;crystal-structure&gt;&lt;stability&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8528" PostTypeId="2" ParentId="8525" CreationDate="2014-02-16T08:58:06.057" Score="1" Body="&lt;p&gt;For a reaction in the form $\ce{ \mathit{a}\,A + \mathit{b}\,B &amp;lt;=&amp;gt; \mathit{c}\,C}$, the quilibrium constant $K_c$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to check yours.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-16T08:58:06.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8529" PostTypeId="1" AcceptedAnswerId="8531" CreationDate="2014-02-16T12:31:43.910" Score="1" ViewCount="294" Body="&lt;p&gt;I just electrolyzed water with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaHSO4 -&amp;gt; HSO4- &amp;lt;=&amp;gt; H+ + SO4^2-}$&lt;/span&gt; with copper electrodes.&lt;/p&gt;&#xA;&lt;p&gt;At the anode this reaction happened:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu -&amp;gt; Cu^2+ + 2e-} \hspace{26 mm} E_0 = -0.34V$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+ + 6H2O -&amp;gt; [Cu(H2O)6]^2+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can anyone confirm this last reaction? There was a blue-greenish substance falling to the bottom.&lt;/p&gt;&#xA;&lt;p&gt;At the cathode this reaction happened:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2H+ + 2e- -&amp;gt; H2} \hspace{27 mm} E_0 = 0V$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2H2O + 2e- -&amp;gt; H2 + 2OH-} \hspace{5 mm} E_0 = -0.83V$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{OH- + H+ -&amp;gt; H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, additionally, the cathode (&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;) turned black. What reaction could have happened there? I electrolyzed at &lt;span class=&quot;math-container&quot;&gt;$12V$&lt;/span&gt; using a car battery charger.&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-02-16T15:59:24.867" Title="What is made at the copper cathode when electrolyzing water?" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8530" PostTypeId="2" ParentId="8518" CreationDate="2014-02-16T12:40:58.760" Score="1" Body="&lt;p&gt;The manometer shown doesn't measure absolute pressure, just &lt;em&gt;relative&lt;/em&gt; pressure. In this car the pressure of the gas in the closed tube versus the pressure of the gas in the bulb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have drawn it, the pressure of the fas in the tube is higher. If you measure pressure in the bulb relative to the pressure in the tube the answer is negative. If you measured absolute pressure (which this manometer can't do because of the lack of any external reference in the closed system) both would be positive but one would be higher than the other.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-02-16T12:40:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8531" PostTypeId="2" ParentId="8529" CreationDate="2014-02-16T15:59:24.867" Score="1" Body="&lt;p&gt;Look at your redox potentials for the half reactions.  12V is tremendous overkill.  At the cathode (reducing) you will make hydrogen to which copper is inert.  Could you reduce sulfate to sulfide or sulfur and have that react with copper?  CuS is black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode (oxidizing) you will oxidize water to oxygen - but one atom at a time.  It will be very reactive to the copper surface.  CuO is black.  You will also oxidize copper to Cu(II).  It coordinates four waters square planar.  The apical waters are distant and weakly bound given the Jahn–Teller effect,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.adichemistry.com/inorganic/cochem/jahnteller/jahn-teller-distortion.html&quot; rel=&quot;nofollow&quot;&gt;http://www.adichemistry.com/inorganic/cochem/jahnteller/jahn-teller-distortion.html&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility is to add a series resistor to lower the working voltage.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-16T15:59:24.867" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8532" PostTypeId="2" ParentId="330" CreationDate="2014-02-16T21:16:57.327" Score="-2" Body="&lt;p&gt;I would say that it is a chemical reaction due to the fact that it is irreversible. Aside from that, the carbon dioxide in the drink reacts to the fat and the bubble (which make ice cream light and fluffy) and wha-la. There is still not solid research as to how it works, but this is my guess due to the fact that most soft drinks have no fat in them, and ice cream has a fair amount. There is research however that states that the carbon dioxide is able to grow in the small air bubbles in the ice cream and this causes the froth. Personally, I think it's both&lt;/p&gt;&#xA;" OwnerUserId="4547" LastActivityDate="2014-02-16T21:16:57.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8534" PostTypeId="1" CreationDate="2014-02-17T02:44:09.900" Score="3" ViewCount="6816" Body="&lt;p&gt;In a problem they gave me some reagents and they want me to figure out the end product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reagents given are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-butene reacting with sodium tert-butoxide and &lt;span class=&quot;math-container&quot;&gt;$\ce{CHBr3}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what I was thinking was that the oxygen would pull the hydrogen from &lt;span class=&quot;math-container&quot;&gt;$\ce{CHBr3}$&lt;/span&gt; making a positive charge on the carbon. (I'm not quite sure if the positive charge would be strong since the net dipole moment on &lt;span class=&quot;math-container&quot;&gt;$\ce{CHBr3}$&lt;/span&gt; would be zero). However, from there, I immediately thought of a carbene reaction, yet, it has 3 bonds with &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, am I on the right track, and what are the next steps from here? Sorry if I didn't give you much to work with.&lt;/p&gt;&#xA;" OwnerUserId="4551" LastEditorUserId="83758" LastEditDate="2020-06-01T13:43:50.697" LastActivityDate="2020-06-01T15:21:58.997" Title="Alkene reactions with sodium tert-butoxide and CHBr₃" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8535" PostTypeId="2" ParentId="8534" CreationDate="2014-02-17T03:05:28.733" Score="7" Body="&lt;p&gt;Strong bases react with chloroform $(\ce{CHCl3})$ and bromoform $(\ce{CHBr3})$ to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Dichlorocarbene&quot; rel=&quot;noreferrer&quot;&gt;dihalocarbenes&lt;/a&gt; by alpha elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net reaction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{RO- + CHCl3 -&amp;gt; ROH + Cl- + CCl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base removes the hydrogen atom from chloroform or bromoform to form an anion. Remember, when you deprotonate something, the electrons stay behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{RO- + Cl3CH &amp;lt;&amp;lt;=&amp;gt; ROH + Cl3C:^{-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trihalomethyl anion is unstable and fragments to lose a halide ion and generate the carbene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl3C:^{-} &amp;lt;=&amp;gt;&amp;gt; Cl- + Cl2C:}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism using arrow-pushing formalism can be found at &lt;a href=&quot;http://www.chemtube3d.com/Carbenes-Formation%28AlphaElimination%29.html&quot; rel=&quot;noreferrer&quot;&gt;this website&lt;/a&gt; from the University of Liverpool.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-17T03:05:28.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8536" PostTypeId="1" AcceptedAnswerId="8546" CreationDate="2014-02-17T03:42:00.520" Score="2" ViewCount="470" Body="&lt;p&gt;Hello I am looking for this information and I did not find anything, the closest thing I have found it is Fluorocarbon, but I think it is wrong, does anybody know?.&#xA;Help really appreciated I am still looking for this &lt;/p&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="418" LastEditDate="2014-02-17T11:27:33.903" LastActivityDate="2017-06-14T22:36:22.870" Title="name of a pharmaceutical drug used in dentistry that has Ionic bonding? and how this bounding is made?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8537" PostTypeId="1" AcceptedAnswerId="8540" CreationDate="2014-02-17T03:44:39.620" Score="17" ViewCount="17149" Body="&lt;p&gt;Is the name &quot;Dihydrogen monoxide&quot; actually what chemists would use to refer to $\ce{H2O}$ (assuming there was no common name, &quot;water&quot;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is all over the internet. I'm a little skeptical though because the similar chemical $\ce{H2S}$ is called &quot;hydrogen sulfide&quot;, not &quot;dihydrogen monosulfide&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4553" LastEditorUserId="16683" LastEditDate="2017-03-13T10:53:06.660" LastActivityDate="2021-04-20T22:07:27.830" Title="Do chemists refer to water as &quot;dihydrogen monoxide&quot;?" Tags="&lt;water&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8538" PostTypeId="1" CreationDate="2014-02-17T04:28:14.600" Score="0" ViewCount="219" Body="&lt;p&gt;Let's say we have a mixture of different ideal gases. I found that the chemical potential of a single component of the mixture is&#xA;$$\mu_{i}=g_{i}^{0}+Ln(\frac{f_{i}}{P_{ref}})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_{i}$ is the fugacity of the component. I also found somewhere else that the chemical potential of pure substances is the molar gibbs property. Would it be safe to then say..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{i}(T,P)=g_{i}(T,P)$$&lt;/p&gt;&#xA;" OwnerUserId="616" LastActivityDate="2014-02-17T21:58:02.797" Title="Chemical Potential of Idea gas?" Tags="&lt;homework&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8539" PostTypeId="1" AcceptedAnswerId="8541" CreationDate="2014-02-17T04:33:33.053" Score="3" ViewCount="6091" Body="&lt;p&gt;We've started learning about NMR in my ochem class and my instructor told us that in Hydrogen NMR an alcohol can have a fairly wide range of possible values - usually from about 2 to 5.  He said that because of this the actual ppm value is not very diagnostic - but the fact that alcohols tend to have a thicker peak is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He told us not to really worry about why/how there is such a range or why the reading tends to be thicker, but now I'm curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I speculate that there can be such a wide range of possible values because Oxygen is fairly electronegative, and depending on its environment it will exhibit different levels of shielding/deshielding.  I have no idea about why it tends to be a thicker peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I guess I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is my speculation as to the range of possible values correct?&lt;/li&gt;&#xA;&lt;li&gt;Why is the peak generally thicker with alcohols than with other functional groups?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4148" LastEditorDisplayName="user7951" LastEditDate="2015-08-21T00:56:29.227" LastActivityDate="2015-08-21T00:56:29.227" Title="Alcohol irregularities in Hydrogen NMR" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8540" PostTypeId="2" ParentId="8537" CreationDate="2014-02-17T04:48:40.767" Score="23" Body="&lt;p&gt;No, it's not. The &amp;quot;dihydrogen monoxide&amp;quot; name is used as part of a hoax. In the scientific community, there are chemical names for water, and which one is used in the literature generally depends on how it interacts with something else (hydroxic acid and hydrogen hydroxide were two I heard most often in acid-base reactions).&lt;/p&gt;&#xA;&lt;p&gt;IUPAC, the standards committee that sets standard names for chemical structures, suggests &amp;quot;oxidane&amp;quot; as a starting point for the construction of other names for chemicals that are derived from water. However, they do not suggest &amp;quot;oxidane&amp;quot; itself be used to refer to plain water.&lt;/p&gt;&#xA;&lt;p&gt;Most chemists would use &amp;quot;water&amp;quot;, even when writing scientific papers.&lt;/p&gt;&#xA;" OwnerUserId="4554" LastEditorUserId="16683" LastEditDate="2021-04-20T19:23:51.843" LastActivityDate="2021-04-20T19:23:51.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8541" PostTypeId="2" ParentId="8539" CreationDate="2014-02-17T05:50:57.973" Score="6" Body="&lt;p&gt;Alcohols have labile protons which are exchangeable through acid-base equilibria with other similar protons, such as other alcohols, amines, amides, thiols and especially water. This reaction is fast enough, &lt;em&gt;on the NMR timescale&lt;/em&gt;, that it can be observed as a reversible exchange process. The most common scenario is that there will be some amount of water in the sample, and so the alcohol and water 1H atoms exchange. The reaction is typically so fast that you &lt;em&gt;don't see the water and alcohol as separate peaks, but as a population-weighted average.&lt;/em&gt; Therefore the actual position of the peak depends on a number of variables; temperature, solvent, starting chemical shifts of the exchanging species, populations of the exchanging species, rate of exchange, solvent and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say 'thicker', I assume you mean a broader linewidth. Again, this is a consequence of the exchange process, and that you are seeing an average peak for the -OH and (say) H2O. Sometimes this broadening is so large that you can only observe it as a broad hump in the baseline of the spectrum. Again, the linewidth will be a function of the starting chemical shifts of the two components, the rates of exchange, and their relative populations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is actually entirely possible to see the alcohol peak with a narrow linewidth, similar to the CH2 and CH3 linewidths. To do this, you need to be able to eliminate the exchange process, either by excluding water from the sample( run ethanol in very dry dmso for instance), or cooling down the sample sufficiently to slow down the exchange (probably not feasible for most common solvents with ethanol and water though). In this case, you will observe your alcohol as a narrow triplet (approx 1.7Hz coupling by memory), with the CH2 appearing as a doublet of quartets. A great example of observing alcohol protons is to run sucrose dissolved in pyridine-d5. All OH protons, and their couplings are observable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is really important to understand that the integrals of these peaks are still valid. In the absence of any exchange (where your alcohol peaks are sharp), the integration of an  ethanol 1H spectrum should still give you 3:2:1. Where there is an exchange process occurring with (say) water, the integrals will be 3:2:X, where X=(sum of -OH plus water). This method of analysis can be used to verify alcohol content of spirits and wine, for instance.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-02-18T21:32:35.340" LastActivityDate="2014-02-18T21:32:35.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8542" PostTypeId="1" CreationDate="2014-02-17T07:13:16.237" Score="3" ViewCount="15374" Body="&lt;p&gt;Aldehydes are easily oxidized to carboxylic acid but ketones are difficult to oxidise. Most books say that the hydrogen directly bonded to the $\ce{C=O}$ in the aldehyde is what aids the oxidation process. However, can someone please tell me why this is so.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorDisplayName="user4076" LastEditDate="2014-03-01T02:30:05.277" LastActivityDate="2019-01-16T13:24:24.127" Title="aldehydes are easily oxidised but ketones are not" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8544" PostTypeId="2" ParentId="8537" CreationDate="2014-02-17T08:25:49.640" Score="21" Body="&lt;p&gt;Like most of the other professionals answering here, I've given a couple of talks on international conferences and published some articles in peer-reviewed journals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never used the terms &lt;em&gt;dihydrogen monoxide&lt;/em&gt; or &lt;em&gt;oxidane&lt;/em&gt; and do not intend to do so in a serious, scientific context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, one talks and writes about &lt;em&gt;water&lt;/em&gt;, &lt;em&gt;aqueous solutions&lt;/em&gt; and even uses a (traditional) term like &lt;em&gt;brine&lt;/em&gt; when referring to an aqueous, saturated solution of sodium chloride.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-17T08:25:49.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8545" PostTypeId="1" AcceptedAnswerId="8551" CreationDate="2014-02-17T09:18:22.290" Score="2" ViewCount="2558" Body="&lt;p&gt;Does kosher salt has a different structure than regular table salt or it is an issue of crystals size?&lt;/p&gt;&#xA;&lt;p&gt;Why some recipes want us to use one instead of another?&lt;/p&gt;&#xA;" OwnerUserId="4548" LastEditorUserId="87609" LastEditDate="2020-06-29T17:31:08.980" LastActivityDate="2020-06-30T05:59:22.987" Title="Table Salt vs Kosher Salt difference" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;&lt;salt&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8546" PostTypeId="2" ParentId="8536" CreationDate="2014-02-17T09:43:26.597" Score="3" Body="&lt;p&gt;The fluorocarbons that you mentioned do have polarized $\ce{C-F}$ bonds, but no ionic bonds. If you think &lt;em&gt;ionic&lt;/em&gt;, think salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a pharmacologist, but what about the local anestethics used at the dentist? Lidocaine (&lt;strong&gt;1&lt;/strong&gt;), mepivacaine (&lt;strong&gt;2&lt;/strong&gt;), and articaine (&lt;strong&gt;3&lt;/strong&gt;) are frequently used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iFKdr.png&quot; alt=&quot;local anesthetics ant the dentist&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these compounds have tertiary and secondary amino groups. (I've marked the relevant nitrogen atoms blue). When these compounds are treated with hydrogen chloride, the amines get protonated and the neutral compounds are converted to &lt;strong&gt;ionic&lt;/strong&gt; hydrochlorides, $\ce{R3N -&amp;gt;[\textrm{HCl}] R3NH+ Cl-}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-17T10:12:32.943" LastActivityDate="2014-02-17T10:12:32.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8547" PostTypeId="2" ParentId="8537" CreationDate="2014-02-17T11:44:06.900" Score="16" Body="&lt;p&gt;I agree with the other answers. No serious chemist uses any word other than &amp;quot;water&amp;quot; in whatever language the chemist uses.&lt;/p&gt;&#xA;&lt;p&gt;However, the name does appear to be following the &lt;a href=&quot;http://pac.iupac.org/publications/pac/pdf/1971/pdf/2801x0001.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;established rules&lt;/a&gt; for the &lt;a href=&quot;http://www.chem.purdue.edu/gchelp/nomenclature/covalent.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;systematic naming&lt;/a&gt; of &lt;a href=&quot;http://www.kentchemistry.com/links/naming/NameCov.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;binary main group&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Binary_compound#Binary_covalent_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;covalent compounds&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Take for example &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O5}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We list the elements in order of increasing electronegativity: nitrogen oxygen&lt;/li&gt;&#xA;&lt;li&gt;We convert the second element's names in &amp;quot;-ide&amp;quot;: nitrogen oxide.&lt;/li&gt;&#xA;&lt;li&gt;We use prefixes to indicate the number of each element. Note the the &amp;quot;-a-&amp;quot; in &amp;quot;penta-&amp;quot; goes away to make pentoxide easier to pronounce: &lt;a href=&quot;http://en.wikipedia.org/wiki/N2O5&quot; rel=&quot;nofollow noreferrer&quot;&gt;dinitrogen pentoxide&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;We need this sort of system to give us unambiguous names for binary compounds, especially when, for example, there are multiple oxides of nitrogen: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrous_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2O}$&lt;/span&gt;&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitric_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dinitrogen_tetroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2O3}$&lt;/span&gt;&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt;&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dinitrogen_trioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt;&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dinitrogen_pentoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2O5}$&lt;/span&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Note that we rarely use the &amp;quot;mono-&amp;quot; prefix: &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; is nitrogen dioxide.&lt;/p&gt;&#xA;&lt;p&gt;And we never use prefixes with the binary hydrides. The acidic ones are all named like ionic compounds, as are the metal hydrides, and the non-acidic ones all have common names that are used so frequently it is silly to use more complex names:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaH}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium hydride&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Borane&quot; rel=&quot;nofollow noreferrer&quot;&gt;borane&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Methane&quot; rel=&quot;nofollow noreferrer&quot;&gt;methane&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonia&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Properties_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;water&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_fluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen fluoride&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{PH3}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphine&quot; rel=&quot;nofollow noreferrer&quot;&gt;phosphine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen sulfide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HN3}$&lt;/span&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrazoic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen azide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Water would thus be &lt;strong&gt;hydrogen oxide&lt;/strong&gt; if anything.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Even though there are other possible oxides of hydrogen (or hydrides of oxygen), they have different names based on the anions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; - &lt;a href=&quot;https://en.wikipedia.org/wiki/Water&quot; rel=&quot;nofollow noreferrer&quot;&gt;water (hydrogen oxide)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HO2}$&lt;/span&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_superoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen superoxide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_peroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen peroxide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="106377" LastEditDate="2021-04-20T22:07:27.830" LastActivityDate="2021-04-20T22:07:27.830" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8548" PostTypeId="1" CreationDate="2014-02-17T12:06:06.520" Score="3" ViewCount="348" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Q) Iodine dissolves in a variety of solvents, forming solutions that are&#xA;purple, brown or any intermediate shade. This property is due to:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hydrolysis of iodine&lt;/li&gt;&#xA;&lt;li&gt;Formation of crystalline adducts&lt;/li&gt;&#xA;&lt;li&gt;Charge transfer between &lt;span class=&quot;math-container&quot;&gt;$\ce{I-I-}$&lt;/span&gt; and the solvent&lt;/li&gt;&#xA;&lt;li&gt;Charge transfer between &lt;span class=&quot;math-container&quot;&gt;$\ce{I-I+}$&lt;/span&gt; and the solvent&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am completely aware of which concept does option 3 and 4 allude to. I know about the formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{I3-}$&lt;/span&gt; when iodine is dissolved in solution of iodides. This fact, according to me, is what option 2 refers to.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;But the answer given is 3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can someone provide an explanation as to what concept is involved here and how can we predict the answer? I read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Charge-transfer_complex&quot; rel=&quot;nofollow noreferrer&quot;&gt;charge transfer complexes&lt;/a&gt;, but the Wikipedia page was too complicated for me to understand, and also the application of that to the question under consideration was not given.&lt;/p&gt;&#xA;" OwnerUserId="4059" LastEditorUserId="83758" LastEditDate="2020-09-08T19:17:39.643" LastActivityDate="2020-09-08T19:17:39.643" Title="Reasons behind the different colors of a solution containing Iodine" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;molecular-orbital-theory&gt;&lt;ions&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8549" PostTypeId="1" CreationDate="2014-02-17T12:17:53.800" Score="5" ViewCount="1846" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Statement 1&lt;/strong&gt;. The $\ce{HF2-}$ ions exist in solid state and also in&#xA;  liquid state but not in aqueous state.&lt;br&gt;&#xA;  &lt;strong&gt;Statement 2&lt;/strong&gt;. The magnitude of hydrogen bonding between $\ce{HF}$ is weaker than that between $\ce{HF}$ and $\ce{H2O}$  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Both are correct and 2 explains 1  &lt;/li&gt;&#xA;  &lt;li&gt;Both are correct but 2 does not explain 1  &lt;/li&gt;&#xA;  &lt;li&gt;1 is correct and 2 false  &lt;/li&gt;&#xA;  &lt;li&gt;2 is correct and 1 false   &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer given is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the second statement to be true. $$\Delta H (\ce{H-F...O})=40{-}45\text{ kJ/mol},\Delta H(\ce{H-F...F})=20{-}23\text{ kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am completely unsure about statement 1 and the relation of the two statements. I think the existence of the afforementioned ion in aqueous solution must not be absent, as I have read somewhere that this ion acts as the conjugate base of the molecule $\ce{H2F2}$ which dissociates into hydrofluoric acid.  &lt;/p&gt;&#xA;" OwnerUserId="4059" LastEditorUserId="5026" LastEditDate="2018-05-14T01:42:22.067" LastActivityDate="2018-05-14T01:43:24.917" Title="In which state does the bifluoride anion exist?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8550" PostTypeId="1" CreationDate="2014-02-17T12:50:52.207" Score="3" ViewCount="618" Body="&lt;p&gt;This is going to seem like a rather daft question, but what is the definitive half-live for protactinium-234 (&lt;sup&gt;234&lt;/sup&gt;Pa)?  I'm seeing lots of different answers in various academic papers etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in Bonotto &amp;amp; Andrews (2000) the half-life of protactinium-234 (&lt;sup&gt;234&lt;/sup&gt;Pa) is given as 1.18 minutes, but in Edwards et al. (2003) they report the &lt;sup&gt;234&lt;/sup&gt;Pa half-life as 6.67 hours.  So which one is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonotto, D.M. &amp;amp; Andrews, J.N. The transfer of uranium isotopes &lt;sup&gt;234&lt;/sup&gt;U and &lt;sup&gt;238&lt;/sup&gt;U to the waters interacting with carbonates from Mendip Hills area (England). &lt;em&gt;Applied Radiation and Isotopes&lt;/em&gt;. 52, 965-983. 2000.  DOI: &lt;a href=&quot;http://dx.doi.org/10.1016/S0969-8043(99)00151-7&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/S0969-8043(99)00151-7&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edwards, R.L., Gallup, C.D. &amp;amp; C Uranium-series Dating of Marine and Lacustrine Carbonates.  &lt;em&gt;Reviews in Mineralogy &amp;amp; Geochemistry&lt;/em&gt;. 52, 363-406 . 2003. URL: &lt;a href=&quot;http://www.minsocam.org/msa/rim/Rim52.html&quot; rel=&quot;nofollow&quot;&gt;http://www.minsocam.org/msa/rim/Rim52.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4556" LastEditorUserId="3846" LastEditDate="2014-02-17T14:43:27.327" LastActivityDate="2014-02-17T14:43:27.327" Title="Protactinium-234 half-life" Tags="&lt;isotope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8551" PostTypeId="2" ParentId="8545" CreationDate="2014-02-17T12:52:51.373" Score="3" Body="&lt;p&gt;There is no chemical difference between table salt and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kosher_salt&quot; rel=&quot;nofollow noreferrer&quot;&gt;kosher salt&lt;/a&gt; - both are sodium chloride. Kosher salt just has bigger crystals. However, some table salt brands are &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodized_Salt&quot; rel=&quot;nofollow noreferrer&quot;&gt;iodized&lt;/a&gt;, meaning that iodide salts have been added to help us maintain a healthy dietary intake of the essential nutrient iodine. Kosher salt is never iodized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sea_salt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sea salt&lt;/a&gt; is a different matter. Since sea salt is obtained from sea water, it may have other minerals in it, which perhaps alters the taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kosher salt is used in some recipes as a matter of taste - the bigger salt granules take longer do dissolve, and so there may still be little pockets of salty crunchiness. Kosher salt is also used when this property is necessary, for example in the curing of some meats. For more on the differences in culinary use between the various types of salt, you should consult &lt;a href=&quot;https://cooking.stackexchange.com/&quot;&gt;Seasoned Advice&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:30.103" LastActivityDate="2014-02-17T12:52:51.373" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8552" PostTypeId="2" ParentId="8550" CreationDate="2014-02-17T13:12:03.150" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So which one is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both! Proactinium-234 exhibits &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_isomer&quot; rel=&quot;nofollow&quot;&gt;nuclear isomerism&lt;/a&gt;&lt;/strong&gt;. The metastable form $\ce{^{234m}Pa}$ has a half-life time in the range of 1 min, whereas $\ce{^{234}Pa}$ shows a half-life time around 6 hrs.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="275" LastEditDate="2014-02-17T13:20:27.120" LastActivityDate="2014-02-17T13:20:27.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8553" PostTypeId="2" ParentId="8549" CreationDate="2014-02-17T15:33:46.893" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bifluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page for for bifluoride anion&lt;/a&gt;. The question's answer appears to be true, with explanation and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluorohydrogenate group $\ce{-HF-}$ in anions such as bifluoride can assimilate a proton by recombination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HF2- + H+ -&amp;gt; H2F2 -&amp;gt; 2 HF}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this capture of a proton ($\ce{H+}$), bifluoride has basic character. Its conjugate acid is the reactive intermediate, &amp;mu;-fluoro-fluorodihydrogen ($\ce{H2F2}$), which subsequently dissociates to become hydrogen fluoride. In solution, most bifluoride ions are dissociated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HF2- + 3H2O}$ is in a favored equilibrium with $\ce{2H3OF + HO-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes for an interesting oxygen coordination sphere re the other lone pair of electrons.  One presumes it is hydronium fluoride in which the fluoride is bonded to a proton with a delocalized charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've worked with aqueous ammonium hydrogen bifluoride as a silicate etchant. The solution has tremendous surface tension.  A drop of Fluorad surfactant was necessary to have things work.  $\ce{HF}$ will reach out and eat you.  $\ce{NH4HF2}$ is fairly benign.  No skin contact!  Don't work in Pyrex.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-05-14T01:43:24.917" LastActivityDate="2018-05-14T01:43:24.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8554" PostTypeId="2" ParentId="8526" CreationDate="2014-02-17T15:47:31.093" Score="2" Body="&lt;p&gt;CTAB is a cationic surfactant, SDS is an anion surfactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These surfactants act to selectively stabilize ionic faces on the crystal as it grows, causing those faces to grow more rapidly. This leads to a change in geometry or crystal shape. CTAB causes one morphology being cationic, while SDS will cause another, its head ion having the opposite charge. Further, a mixture of the two will cause an intermediary morphology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, CTAB will cause an aqueous solution of thioacetamide and Lead(II) acetate to form PbS nanocrystals in a octahedron shape, while SDS will cause them to form a dendritic fern like shape, while a CTAB/SDS mixture (5:1) will cause a stellated octahedron or star shape to form.&lt;/p&gt;&#xA;" OwnerUserId="2067" LastEditorUserId="2067" LastEditDate="2014-02-17T15:53:03.980" LastActivityDate="2014-02-17T15:53:03.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8557" PostTypeId="2" ParentId="8548" CreationDate="2014-02-17T17:24:48.897" Score="3" Body="&lt;p&gt;Iodine in an inert solvent is purple.  Electron donation alters that.  Linear starch (amylose)-iodine is dark brown.  Branched starch (amylopectin)-iodine is red.  The Wikipedia page shows iodine in chlorinated solvent (very weak electron donation) then addition of strong electron donor triphenylphosphine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an electron donor is reactive, then the interacting iodine molecule looks like iodonium (elecron accepter).  Initial interation is (4), post-mixing is (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a tiny iodine crystal into a melting point capillary, evacuate under vacuum, flame closed.  You will see a very pale purple vapor at equilibrium.  Crack the seal and the vapor is suddenly deeper purple.  Why?  Take a sealed cap and an evacuated sealed cap, and insert the vapor volumes into the beam of a green laser pointer. interesting. Take care that the optical geometry does not reflect back into your eyes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-17T17:24:48.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8558" PostTypeId="1" CreationDate="2014-02-17T18:50:42.617" Score="1" ViewCount="117" Body="&lt;p&gt;Suppose we have a T-S diagram of some vapor thermodynamic cycle , say the rankine cycle , and we are interested for example in the composition of the liquid and the vapor in the middle of the condensation process ;  knowing only the entropy of both the liquid and vapor at that point (or the enthalpy) , how exactly do we get the composition at that point ?&lt;/p&gt;&#xA;" OwnerUserId="4560" LastActivityDate="2014-02-17T18:50:42.617" Title="About the T-S diagram" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8559" PostTypeId="1" AcceptedAnswerId="8561" CreationDate="2014-02-17T19:25:31.053" Score="7" ViewCount="1265" Body="&lt;p&gt;Vibrational frequency computations provide the magnitudes of vibrational frequencies for each vibrational mode of a molecule ($3N-6$ degrees of freedom for non-linear molecules and $3N-5$ for linear molecules).  Sometimes these magnitudes can be non-real (i.e. imaginary).  This is common when characterizing a transition state which contain one, and exactly one, imaginary mode of vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some popular software packages will print these magnitudes as a negative number.  This syntax has unfortunately propagated itself into common, everyday practice even if technically incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the best way to denote these imaginary magnitudes, especially in the context of a peer-reviewed journal?  Should we stick to using the minus sign convention (e.g. -15) with a note explicitly defining the meaning of the $-$ sign,  or should they simply be listed with an $i$  (e.g. 15$i$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've just been told to NEVER use the minus sign convention as this will incur the wrath of the reviewers.  Please feel free to verify.)&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-02-17T19:31:57.097" LastActivityDate="2014-02-17T19:58:02.560" Title="How to properly denote imaginary magnitudes of vibrational frequencies" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8560" PostTypeId="2" ParentId="8559" CreationDate="2014-02-17T19:49:25.323" Score="4" Body="&lt;p&gt;I'm completely out of my depth here, but I would, since the minus sign convention seems to be frowned upon, look for other symbols:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As mentioned, with an &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; denoting the imaginarity: &lt;span class=&quot;math-container&quot;&gt;$15i$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is maybe not so beautiful if we're actually talking about the &lt;em&gt;absolute value&lt;/em&gt; of the imaginary magnitude, because in mathematics the &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; clearly denotes the imaginary part of an imaginary number.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If not yet taken, I would give my vote for the asterisk (*): &lt;span class=&quot;math-container&quot;&gt;$15^*$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since we're talking about transition state magnitudes, maybe the double dagger (&lt;span class=&quot;math-container&quot;&gt;$\ddagger$&lt;/span&gt;): &lt;span class=&quot;math-container&quot;&gt;$15^\ddagger$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Of course this notation would have to be defined somewhere towards the beginning of your article, stating that this symbol denotes imaginary (transition state) magnitudes.&lt;/p&gt;&#xA;&lt;p&gt;Using this method you could even have the option of distinguishing between normal state and transition state imaginary magnitudes: Asterisk for the former, double dagger for the latter.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-02-17T19:49:25.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8561" PostTypeId="2" ParentId="8559" CreationDate="2014-02-17T19:58:02.560" Score="5" Body="&lt;p&gt;The suffix &lt;strong&gt;i&lt;/strong&gt; to the &lt;strong&gt;positive&lt;/strong&gt; wavenumber to denote an imaginary frequency is seemingly accepted by reviewers, see &lt;a href=&quot;http://dx.doi.org/10.1016/j.jphotochem.2013.12.011&quot; rel=&quot;noreferrer&quot;&gt;Photochemical transformations of 5-methyltetrazole. Matrix isolation FTIR and DFT studies&lt;/a&gt; as one example I could find in a very quick search.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-17T19:58:02.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8564" PostTypeId="2" ParentId="8538" CreationDate="2014-02-17T21:58:02.797" Score="1" Body="&lt;p&gt;Yes, by definition chemical potential is related to Gibbs potential in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_{j\neq i}} = g_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those two notation are synonyms. You will find very detailed and precise explanation of this in: &lt;strong&gt;&lt;em&gt;Atkins&lt;/strong&gt;, Physical Chemistry&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-02-17T21:58:02.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8565" PostTypeId="2" ParentId="7781" CreationDate="2014-02-17T22:32:17.230" Score="1" Body="&lt;p&gt;In &lt;em&gt;The Journal of Chemical Education&lt;/em&gt;, published by the American Chemical Society, a rather recent &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed300787v&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; presented a related project, hosted by the University of Minnesota.  Intentionally it is about raw FIDs-as-obtained, out of the NMR spectrometer, aiming to teach correct data processing (including spectra with two orthogonal frequency scales). The article is sponsored by the authors (i.e. freely accessible), the database requires an e-mail to register. &lt;br&gt; Reference: J. Chem. Educ., 2013, 90 (7), pp 941–943; DOI: 10.1021/ed300787v&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2014-02-17T22:32:17.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8566" PostTypeId="1" AcceptedAnswerId="9339" CreationDate="2014-02-18T00:16:15.233" Score="4" ViewCount="2051" Body="&lt;p&gt;I propose to carry out an electrolytic etch on a stainless steel plate like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/hw2hbkL.png&quot; alt=&quot;Electrolysis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of electrodes must be the stainless steel plate which is being etched and the surface layer must be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware such an electrolytic etches can produce harmful by-products such as Chlorine (NaCl electrolyte), Chromium compounds and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, What electrolytes and electrodes can I use to prevent the production of these substances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to create safer Chromium compounds?&#xA;If I can't, what other non-electrolytic options do I have?&lt;/p&gt;&#xA;" OwnerUserId="4563" LastEditorUserId="4563" LastEditDate="2014-02-18T00:52:25.007" LastActivityDate="2017-07-21T14:54:40.750" Title="How can I safely carry out an electrolytic etch on stainless steel without producing relatively harmful by-products?" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;safety&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8567" PostTypeId="2" ParentId="8566" CreationDate="2014-02-18T00:45:10.953" Score="0" Body="&lt;p&gt;Sand or shot blast it instead.  Rub it with crushed aluminum foil and saltwater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is harmful.  Voltage determines products.  Use the minimum voltage (slow way to travel).  Don't use a chloride electrolyte.  (TEST FIRST!) Use hydroxide, carbonate, (nitrate - Homeland Severity and OSHA will beat on you), sulfate, electrolyte (note solubility of metallic byproducts after reaction with electrolyte).  The larger the charge on the ion, the slower the kinetics.  Vent the chlorine plus inert air through lye solution to make bleach, or through a solution of reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cr(III) is OK, Cr(VI) is a carcinogen. Reduce spent electrolyte with sulfite, Vitamin C, zinc granules, FeCl2, shredded scrap iron, steel wool, etc - and have a way to test for Cr(VI) residual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you vent or sewer anything, talk with your Safety Officer.  The EPA and OSHA are carnivorous.  it's always a good idea to be a good upstream neighbor.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-18T00:45:10.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8569" PostTypeId="1" AcceptedAnswerId="8570" CreationDate="2014-02-18T04:24:17.540" Score="2" ViewCount="5276" Body="&lt;p&gt;In my organic chemistry class there are several reactions that I’ve been told are generally conducted at $\pu{−78^\circ C}$, which is apparently also the temperature of dry ice in acetone.  This is all well and good, but I don’t really get what that actually means. Do you put the dry ice in the acetone and then toss in your reactants?  Do you add the dry ice and acetone and it makes the surrounding air colder?  Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m just not sure how this is actually accomplished, and if you do add the reactants to the dry ice/acetone mix then how do you avoid reacting with either of those things?&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorUserId="23561" LastEditDate="2018-09-16T14:14:09.840" LastActivityDate="2019-06-08T19:10:04.197" Title="Dry ice and acetone" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8570" PostTypeId="2" ParentId="8569" CreationDate="2014-02-18T05:51:42.733" Score="6" Body="&lt;p&gt;You put broken up Dry Ice in in an evaporating dish in a bigger evaporating dish lined with glass wool, the whole atop a magnetic stirrer base, then carefully add acetone. Fizz.  Add and clamp your flask with stir bar under dynamic inert gas into the slush.  Go for it, solvent(s) and reactants.  Ace glassware has problems with star cracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't try this in a styrofoam bucket.  Note that acetone vapor is wildly flammable.  Cheesecloth atop NOT touching the slush for more insulation and to keep condensing moisture out.  Never plan to fill a flask more than half-full, because the universe hates you.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-18T05:51:42.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8572" PostTypeId="1" AcceptedAnswerId="8574" CreationDate="2014-02-18T09:37:18.490" Score="4" ViewCount="692" Body="&lt;p&gt;Reading about the light emitting reaction of luciferin + ATP, catalysed by luciferase I am confused about the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On &lt;a href=&quot;http://en.wikipedia.org/wiki/Luciferase&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt; the partial formula is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;luciferin + ATP → luciferyl adenylate + PPi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to me like a one way reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on searching on Google &quot;luciferase ATP equilibrium&quot; I found websites like &lt;a href=&quot;http://www.chegg.com/homework-help/questions-and-answers/firefly-luciferase-enzyme-allowsfireflies-illuminate-abdomens-lightgeneration-atp-requirin-q566648&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; and &lt;a href=&quot;http://books.google.nl/books?id=WtHCXlVMH1EC&amp;amp;lpg=PA249&amp;amp;ots=RS2wP2-jeH&amp;amp;dq=E.LH2-AMP&amp;amp;hl=nl&amp;amp;pg=PA249#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; which seem to indicate that the reaction is reversible, as there is an equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for this reaction to take place in reverse? And what about the link on books.google.nl showing a reversible reaction with light involved? What about that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for clarifying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2014-02-25&lt;/strong&gt;:&#xA;This article: &quot;Kinetics of inhibition of firefly luciferase by dehydroluciferyl-coenzyme A, dehydroluciferin and L-luciferin&quot; by Luis Pinto da Silva etc (&lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2011/pp/c0pp00379d#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;) shows the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3c9As.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;strong&gt;Luc&lt;/strong&gt; is firefly luciferase, &lt;strong&gt;LH2&lt;/strong&gt; is luciferin, &lt;strong&gt;L-CoA&lt;/strong&gt; is dehydroluciferyl-coenzyme A, &lt;strong&gt;L&lt;/strong&gt; is dehydroluciferin, &lt;strong&gt;ATP&lt;/strong&gt; is Adenosine triphosphate, &lt;strong&gt;AMP&lt;/strong&gt; is  adenosine-5`-monophosphate, &lt;strong&gt;PPi&lt;/strong&gt; is inorganic pyrophosphate, &lt;strong&gt;O2&lt;/strong&gt; is oxygen, &lt;strong&gt;CO2&lt;/strong&gt; is carbondioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the last equation, it seems it is possible to form ATP as a product, as it is on the right side of the arrow. Remarkable, I would say. still I do not see any luciferin (&lt;strong&gt;LH2&lt;/strong&gt;) on the right side of any equation, so the answer to question of this topic remains &lt;code&gt;no&lt;/code&gt; thus far.&lt;/p&gt;&#xA;" OwnerUserId="4566" LastEditorUserId="4566" LastEditDate="2014-02-25T09:57:56.460" LastActivityDate="2014-02-25T09:57:56.460" Title="Can luciferin + ATP → luciferyl adenylate + PPi be reversed?" Tags="&lt;equilibrium&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8573" PostTypeId="1" CreationDate="2014-02-18T12:05:11.510" Score="0" ViewCount="2483" Body="&lt;p&gt;I've been trying to make Sodium dichromate $\ce{Na2Cr2O7}$ by following these steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dissolve an inox spoon in $\ce{H2SO4}$&lt;/li&gt;&#xA;&lt;li&gt;Filter the solution&lt;/li&gt;&#xA;&lt;li&gt;Dissolve the obtained sludge in $\ce{NaClO}$&lt;/li&gt;&#xA;&lt;li&gt;Heat the solution until slurries appear&lt;/li&gt;&#xA;&lt;li&gt;Filter the solution&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The solution I obtained was yellowish (light yellow), I tried to add some acid to see the orange color of the (hypothetical) dichromate but nothing happened!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added some ammonia before acidifying the solution, nothing as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st: Is this method the right one to obtain chromate/dichromate?&#xA;2nd: If yes, why didn't the orange color appear?&lt;/p&gt;&#xA;" OwnerUserId="2286" LastEditorUserId="2286" LastEditDate="2014-02-18T12:18:32.390" LastActivityDate="2014-02-18T18:56:50.127" Title="How to make Sodium dichromate" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8574" PostTypeId="2" ParentId="8572" CreationDate="2014-02-18T12:22:38.360" Score="2" Body="&lt;p&gt;In theory every chemical reaction is reversible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction that you mention does not emit any light yet, that comes later with the following reaction:&#xA;$$ \text{Luciferyl adenylate} + \ce{O2 -&amp;gt;} \text{Oxyluciferin} + \text{AMP} +h\nu $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction you mentioned occurs under loss of pyrophosphate ($\ce{PP_{i}}$) which is where the energy was stored beforehand, I guess it's somewhat hard to get it back. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It's almost like hoping that&#xA;$$ \ce{2CO2 + 3H2O -&amp;gt; C2H5OH + 3O2} $$&#xA;occurs under room temperature. Which it doesn't, because $\ce{CO2}$ is such a thermodynamical black hole. I think the same applies to your question.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-18T12:22:38.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8575" PostTypeId="2" ParentId="8572" CreationDate="2014-02-18T12:55:13.843" Score="2" Body="&lt;p&gt;Reactions that are metabolically irreversible are usually so, because mechanisms are in place to keep the mass action ratio (Q) far away from the equilibrium constant (K). Recall that equilibrium occurs when Q = K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change in Gibb's free energy for standard conditions ($\Delta G°$) determines if the reaction is &quot;inherently&quot; thermodynamically favourable or not. But in reality the concentrations of reactants will be far away from their assumed 1 M values. Remember that the actual change in Gibb's free energy is related to $\Delta G°$ by the following expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rG=\Delta_rG° + RT \ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular reaction, hydrolysis of ATP to release $\text{PP}_i$ makes the reaction exergonic / spontaneous ($\Delta G&amp;lt;0$). But in a hypothetical scenario with no substrates, and high concentrations of products (which would make $Q$ large, and thus $\Delta_r G$ positive), the reverse reaction would be spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the reaction is practically irreversible, it's because things are being done to prevent such a situation from happening. For example, pyrophosphatase hydrolyses $\text{PP}_i$ to $2\text{P}_i$ at a high rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many metabolic pathways, irreversible reactions are also catalysed by enzymes, that are inhibited by either their product, or a product downstream in the pathway, thereby inactivating the enzyme long before equilibrium is reached.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-18T12:55:13.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8577" PostTypeId="2" ParentId="8569" CreationDate="2014-02-18T15:24:04.837" Score="8" Body="&lt;p&gt;The mixture of dry ice and acetone is used &lt;em&gt;outside&lt;/em&gt; your reaction vessel to keep the vessel cool just as a water bath is sometimes used to provide uniform heat to a reaction vessel. The only role is to provide a uniformly cool environment for the reaction vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetone isn't the only choice here and isn't essential (anything still liquid at the temperature of dry ice will do). Acetone is cheap and (relatively) harmless which is why it is commonly used. &lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-02-18T15:24:04.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8578" PostTypeId="2" ParentId="8573" CreationDate="2014-02-18T17:07:43.797" Score="1" Body="&lt;p&gt;Note that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chromate&quot; rel=&quot;nofollow&quot;&gt;chromate ion&lt;/a&gt; $\ce{CrO4^{2-}}$ has chromium in the +6 oxidation state like dichromate $\ce{Cr2O7^{2-}}$. Chromate is yellow, dichromate is orange. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chromate&quot; rel=&quot;nofollow&quot;&gt;sodium chromate&lt;/a&gt;: sodium chromate is formed by reaction of dichromate with hydroxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na2Cr2O7 + 2NaOH -&amp;gt; 2Na2CrO4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be making sodium dichromate, but the strongly basic conditions are converting it to sodium chromate.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="418" LastEditDate="2014-02-18T18:56:50.127" LastActivityDate="2014-02-18T18:56:50.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8579" PostTypeId="1" CreationDate="2014-02-18T19:52:45.673" Score="1" ViewCount="4825" Body="&lt;p&gt;If I melt $\ce{NaCl}$, I get a molten salt. This means a liquid formed only by ions ($\ce{Cl^-}$ and  $\ce{ Na^+}$). That could mean that the crystal $\ce{NaCl}$ experimented a chemical change:&#xA;$$\ce{ NaCl-&amp;gt;Cl- + Na+}$$ Like a chemical reaction... The ions have now more space to move and interact with much more ions than in the solid phase. I am trying to get an explanaition why the melting temeprature and the melting heat of molten salts show some erratic behaviour, I mean they do not follow a simple pattern such as that found for n-alkanes or n-alkanols.&lt;/p&gt;&#xA;" OwnerUserId="4570" LastEditorUserId="4945" LastEditorDisplayName="user4076" LastEditDate="2015-10-23T17:27:48.340" LastActivityDate="2015-10-23T17:27:48.340" Title="Is melting of NaCl a physical and chemical process?" Tags="&lt;physical-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2020-07-25T19:41:22.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="8581" PostTypeId="2" ParentId="8579" CreationDate="2014-02-18T22:02:03.757" Score="3" Body="&lt;p&gt;In fact the lattice structure of $\ce{NaCl}$ is composed by $\ce{Na^+}$ and $\ce{Cl^-}$ even when is in the solid state we don't write it with the charge only for convention. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds are different from organic compound, they have complex lattice structure and are in fact compose by two species. Breaking a ionic bond required more energy than an organic compound where you deal only with intramolecular forces.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user4076" LastEditDate="2014-02-19T07:00:00.833" LastActivityDate="2014-02-19T07:00:00.833" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8585" PostTypeId="1" CreationDate="2014-02-19T03:37:40.283" Score="4" ViewCount="4593" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Secondary amines form hydrogen bonds, but having nitrogen in the middle of the chain rather than at the end makes the permanent dipole on the molecule slightly less. The lower boiling point is due to the lower dipole-dipole attractions in the dimethylamine compared with ethylamine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone please tell me why nitrogen, being in the middle of the molecule (as in secondary amines) causes the permanent dipole to be slightly less.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorDisplayName="user7951" LastEditDate="2015-08-19T12:24:41.603" LastActivityDate="2015-10-13T13:45:04.303" Title="Intermolecular forces in secondary amines compared to primary amines" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;amines&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8586" PostTypeId="1" CreationDate="2014-02-19T04:50:09.187" Score="4" ViewCount="310" Body="&lt;p&gt;Is there a stable molecule with total number of electrons equal to 13? Molecule can be either organic or inorganic &lt;/p&gt;&#xA;" OwnerUserId="4575" LastActivityDate="2014-02-20T11:37:24.137" Title="Molecule with 13 total electrons" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8587" PostTypeId="2" ParentId="8586" CreationDate="2014-02-19T07:42:09.370" Score="7" Body="&lt;p&gt;There are quite some molecules that are &lt;em&gt;persistent&lt;/em&gt; (= live long enough in solution to allow for a spectroscopical characterization, but cannot be isolated) or even &lt;em&gt;stable&lt;/em&gt; (= can be isolated, bottled and/or used as a starting material) that have an &lt;strong&gt;odd number &lt;em&gt;n&lt;/em&gt; of electrons&lt;/strong&gt;, however $n \ne 13$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/06fhl.png&quot; alt=&quot;stable or persistent radicals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among these &lt;strong&gt;radicals&lt;/strong&gt; are the Gomberg radical (&lt;strong&gt;1&lt;/strong&gt;), the Koelsch radical (&lt;strong&gt;2&lt;/strong&gt;), and the TEMPO radical (&lt;strong&gt;3&lt;/strong&gt;). &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-19T07:42:09.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8590" PostTypeId="1" CreationDate="2014-02-19T12:50:28.277" Score="16" ViewCount="81626" Body="&lt;p&gt;Textbooks and the online reference differ about this and there are more than two answers.&lt;/p&gt;&#xA;" OwnerUserId="4528" LastEditorUserId="9961" LastEditDate="2017-10-21T15:47:40.667" LastActivityDate="2018-05-25T17:36:19.467" Title="What is the melting point of diamond?" Tags="&lt;phase&gt;&lt;elements&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8591" PostTypeId="2" ParentId="8590" CreationDate="2014-02-19T14:50:59.473" Score="22" Body="&lt;p&gt;Diamond (carbon) does not melt at 1 atm. It sublimes to vapor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using carbon's theoretical phase diagram below (from Wikimedia), &quot;liquid diamond&quot; could be achieved at about 10 GPa (99 thousand atmospheres) and 5000 K (4700 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SZ95j.png&quot; alt=&quot;phase diagram of carbon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, heating up diamonds at 1 atm turns them into graphite first. Only further heating would cause sublimation, as expected by observing the phase diagram. Here's a video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=7L7BV3IBfFA&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.youtube.com/watch?v=7L7BV3IBfFA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1342" LastEditorUserId="14813" LastEditDate="2018-05-25T17:36:19.467" LastActivityDate="2018-05-25T17:36:19.467" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="8592" PostTypeId="2" ParentId="8586" CreationDate="2014-02-19T15:47:14.113" Score="3" Body="&lt;p&gt;A cyanide radical has 13 electrons. This stable radical was one of the first detected molecules in the interstellar medium, 1941 by A. McKellar, doi:10.1086/125159&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stable: persists in vacuum; no imaginary IR frequencies (negative eigenvalues). Reactive: diddles with something else, including another one of itself. Given the name, do your chair parade. The unpaired spin mostly resides on the nitrogen. Chemistry is passion, not compassion&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="1362" LastEditDate="2014-02-20T11:37:24.137" LastActivityDate="2014-02-20T11:37:24.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8593" PostTypeId="2" ParentId="4655" CreationDate="2014-02-19T17:55:42.767" Score="0" Body="&lt;p&gt;&lt;strong&gt;Conductivity:&lt;/strong&gt; The degree to which a specified material conducts electricity is called conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Molar conductivity:&lt;/strong&gt; Conductivity of a solution on dilution. Conductance and conductivity both decrease but molar conductivity increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductivity is directly proportional to the concentration.&lt;/p&gt;&#xA;" OwnerUserId="4578" LastEditorUserId="3846" LastEditDate="2014-02-19T21:33:39.987" LastActivityDate="2014-02-19T21:33:39.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8594" PostTypeId="1" CreationDate="2014-02-19T18:12:27.707" Score="3" ViewCount="234" Body="&lt;p&gt;I need to add a known amount of ethanol vapour to a gas mixture (He/Ar). To do that I let the mixture bubble in liquid ethanol kept at a known temperature (14 C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is made of a large steel cylinder (d=20 cm; l=138cm) lying in horizontal position and half filled with liquid ethanol. the gas is bubbled trough the liquid on one side of the cylinder and extracted on the other side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas-flow trough the system is usually pretty low (~100ml/min) but I would like to increase it to ~10l/min and I need to know if this leaves enough time for the ethanol vapour to approach its equilibrium vapour pressure. &lt;/p&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2014-03-22T10:50:47.023" Title="Evaporation velocity of ethanol: adding ethanol vapour to a gas mixture" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8595" PostTypeId="2" ParentId="8594" CreationDate="2014-02-19T18:42:22.413" Score="1" Body="&lt;p&gt;Please note that I have never done this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are you sure that you want to rely on approximation/assumptions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that it needs additional changes to the experimental setup, but is measuring the ethanol concentration out of question? Gas sensors for ethanol are commercially available and a setup, in which they steer mass flow controllers is conceivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, you might be interested to see that ethanol sensors can be build in the lab: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.3390/s130404378&quot; rel=&quot;nofollow&quot;&gt;A Novel Flexible Room Temperature Ethanol Gas Sensor Based on SnO2 Doped Poly-Diallyldimethylammonium Chloride&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-19T18:42:22.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8596" PostTypeId="2" ParentId="6976" CreationDate="2014-02-19T21:40:27.507" Score="14" Body="&lt;p&gt;&lt;strong&gt;It isn't that good a generalization: always look at the data first.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table of most of the aldehydes and ketones with 6 or fewer carbons (the labels are used in the chart later):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vKgwS.jpg&quot; alt=&quot;boiling points aldehydes and ketones&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now plot this on a chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8s91O.jpg&quot; alt=&quot;boiling point chart&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branches is the number of branches in the carbon chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that while for 3 and 4 carbons the ketones do have higher boiling points, it is not clear this is true for 5 carbon compounds and certainly not true for 6-carbon ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd argue the pattern of boiling points is complicated and there is no simple pattern that needs to be explained.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-02-19T21:40:27.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8598" PostTypeId="1" CreationDate="2014-02-20T04:13:52.097" Score="34" ViewCount="1098230" Body="&lt;p&gt;In my textbook, it says that the maximum number of electrons that can fit in any given shell is given by 2n². This would mean 2 electrons could fit in the first shell, 8 could fit in the second shell, 18 in the third shell, and 32 in the fourth shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was previously taught that the maximum number of electrons in the first orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the sixth orbital.  I am fairly sure that orbitals and shells are the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these two methods is correct and should be used to find the number of electrons in an orbital? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in high school so please try to simplify your answer and use fairly basic terms.&lt;/p&gt;&#xA;" OwnerUserId="4583" LastEditorUserId="402" LastEditDate="2017-01-22T09:54:27.180" LastActivityDate="2017-01-22T10:08:52.210" Title="What are the maximum number of electrons in each shell?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="8599" PostTypeId="1" AcceptedAnswerId="8785" CreationDate="2014-02-20T05:34:34.920" Score="10" ViewCount="1006" Body="&lt;p&gt;I've found a Dover edition copy of Pauling's General Chemistry in the bookstore near me. I'd like to know if it's still readable today, I mean, is it still up-to-date? I have no background in chemistry, but the book was cheap and it seems to be a nice read.&lt;/p&gt;&#xA;" OwnerUserId="1311" LastActivityDate="2014-02-27T20:39:22.540" Title="Is Linus Pauling's General Chemistry still readable and up-to-date today?" Tags="&lt;reference-request&gt;&lt;books&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8600" PostTypeId="1" AcceptedAnswerId="8601" CreationDate="2014-02-20T06:50:28.667" Score="3" ViewCount="652" Body="&lt;p&gt;This question was triggered by another question on this site, about molecules with a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8586&quot;&gt;total number of 13 electrons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't think of any (except the boring answer of $\ce{Al}$), and the answers to the question are all about &lt;strong&gt;radicals&lt;/strong&gt;, instead of 'normal' molecules. So that got me thinking: are all organic molecules with an odd total electron count radicals or radical ions? If so, is there any fundamental reason for this?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-02-20T09:29:10.850" Title="Are all organic molecules with odd-numbered electron count radicals?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8601" PostTypeId="2" ParentId="8600" CreationDate="2014-02-20T09:29:10.850" Score="4" Body="&lt;p&gt;By &lt;a href=&quot;http://goldbook.iupac.org/R05066.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC definition of radical&lt;/a&gt;, the organic molecules with odd number of electrons must have at least one of them unpaired and are therefore radicals. The fundamental reason is the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind, that other spin states must be taken into account, as simplest example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triplet_oxygen&quot; rel=&quot;nofollow&quot;&gt;oxygen&lt;/a&gt;. It exists as a triplet, so two unpaired electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation gets different, when you do have a metal in your compound, but it is out of the scope of the question.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-02-20T09:29:10.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8602" PostTypeId="2" ParentId="8598" CreationDate="2014-02-20T12:06:00.647" Score="25" Body="&lt;p&gt;The pattern of maximum possible electrons = $2n^2$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that Brian's answer is good and takes a different approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you learned about &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number#Traditional_nomenclatures&quot; rel=&quot;noreferrer&quot;&gt;quantum numbers&lt;/a&gt; yet? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If not...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each shell (or energy level) has some number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell#Subshells&quot; rel=&quot;noreferrer&quot;&gt;subshells&lt;/a&gt;, which describe the types of atomic orbitals available to electrons in that subshell. For example, the $s$ subshell of any energy level consists of spherical orbitals. The $p$ subshell has dumbbell-shaped orbitals. The orbital shapes start to get weird after that. Each subshell contains a specified number of orbitals, and each orbital can hold two electrons. The types of subshells available to a shell and the number of orbitals in each subshell are mathematically defined by quantum numbers. Quantum numbers are parameters in the wave equation that describes each electron. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;noreferrer&quot;&gt;Pauli Exclusion Principle&lt;/a&gt; states that no two electrons in the same atom can have the exact same set of quantum numbers. A more thorough explanation using quantum numbers can be found below. However, the outcome is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The subshells are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The $s$ subshell has one orbital for a total of 2 electrons&lt;/li&gt;&#xA;&lt;li&gt;The $p$ subshell has three orbitals for a total of 6 electrons&lt;/li&gt;&#xA;&lt;li&gt;The $d$ subshell has five orbitals for a total of 10 electrons&lt;/li&gt;&#xA;&lt;li&gt;The $f$ subshell has seven orbitals for a total of 14 electrons&lt;/li&gt;&#xA;&lt;li&gt;The $g$ subshell has nine orbitals for a total of 18 electrons&lt;/li&gt;&#xA;&lt;li&gt;The $h$ subshell has eleven orbitals for a total of 22 electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each energy level (shell) has more subshells available to it:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first shell only has the $s$ subshell $\implies$ 2 electrons&lt;/li&gt;&#xA;&lt;li&gt;The second shell has the $s$ and $p$ subshells $\implies$ 2 + 6 = 8 electrons&lt;/li&gt;&#xA;&lt;li&gt;The third shell has the $s$, $p$, and $d$ subshells $\implies$ 2 + 6 + 10 = 18 electrons&lt;/li&gt;&#xA;&lt;li&gt;The fourth shell has the $s$, $p$, $d$, and $f$ subshells $\implies$ 2 + 6 + 10 + 14 = 32 electrons&lt;/li&gt;&#xA;&lt;li&gt;The fifth shell has the $s$, $p$, $d$, $f$, and $g$ subshells $\implies$ 2 + 6 + 10 + 14 + 18 = 50 electrons&lt;/li&gt;&#xA;&lt;li&gt;The sixth shell has the $s$, $p$, $d$, $f$, $g$, and $h$ subshells $\implies$ 2 + 6 + 10 + 14 + 18 + 22 = 72 electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The pattern is thus: $2, 8, 18, 32, 50, 72, ...$ or $2n^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, no known atoms have electrons in the $g$ or $h$ subshells, but the quantum mechanical model predicts their existence. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using quantum numbers to explain why the shells have the subshells they do and why the subshells have the number of orbitals they do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electrons in atoms are defined by 4 quantum numbers. The Pauli Exclusion Principle means that no two electrons can share the same quantum numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n$, the principle quantum number defines the shell. The values of $n$ are integers: $n=1,2,3,...$&lt;/li&gt;&#xA;&lt;li&gt;$\ell$, the orbital angular momentum quantum number defines the subshell. This quantum number defines the shape of the orbitals (probability densities) that the electrons reside in. The values of $\ell$ are integers dependent on the value of $n$: $\ell = 0,1,2,...,n-1$&lt;/li&gt;&#xA;&lt;li&gt;$m_{\ell}$, the magnetic quantum number defines the orientation of the orbital in space. This quantum number also determines the number of orbitals per subshell. The values of $m_\ell$ are integers and depend on the value of $\ell$: $m_\ell = -\ell,...,-1,0,1,...,+\ell$&lt;/li&gt;&#xA;&lt;li&gt;$m_s$, the spin angular momentum quantum number defines the spin state of each electron. Since there are only two allowed values of spin, thus there can only be two electrons per orbital. The values of $m_s$ are $m_s=\pm \frac{1}{2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the first shell&lt;/strong&gt;, $n=1$, so only one value of $\ell$ is allowed: $\ell=0$, which is the $s$ subshell. For $\ell=0$ only $m_\ell=0$ is allowed. Thus the $s$ subshell has only 1 orbital. The first shell has 1 subshell, which has 1 orbital with 2 electrons total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the second shell&lt;/strong&gt;, $n=2$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell, and $\ell=1$, which is the $p$ subshell. For $\ell=1$, $m_\ell$ has three possible values: $m_\ell=-1,0,+1$. Thus the $p$ subshell has three orbitals. The second shell has 2 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, and the $p$ subshell, which has 3 orbitals with 6 electrons, for a  total of 4 orbitals and 8 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the third shell&lt;/strong&gt;, $n=3$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell,  $\ell=1$, which is the $p$ subshell, and $\ell=2$, which is the $d$ subshell. For $\ell=2$, $m_\ell$ has five possible values: $m_\ell=-2,-1,0,+1,+2$. Thus the $d$ subshell has five orbitals. The third shell has 3 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, the $p$ subshell, which has 3 orbitals with 6 electrons, and the $d$ subshell, which has 5 orbitals with 10 electrons, for a  total of 9 orbitals and 18 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the fourth shell&lt;/strong&gt;, $n=4$, so the allowed values of $\ell$ are: $\ell=0$, which is the $s$ subshell,  $\ell=1$, which is the $p$ subshell, $\ell=2$, which is the $d$ subshell, and $\ell=3$, which is the $f$ subshell. For $\ell=3$, $m_\ell$ has seven possible values: $m_\ell=-3,-2,-1,0,+1,+2,-3$. Thus the $f$ subshell has seven orbitals. The fourth shell has 4 subshells: the $s$ subshell, which has 1 orbital with 2 electrons, the $p$ subshell, which has 3 orbitals with 6 electrons,  the $d$ subshell, which has 5 orbitals with 10 electrons, and the $f$ subshell, which has 7 orbitals with 14 electrons, for a  total of 16 orbitals and 32 electrons.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="402" LastEditDate="2017-01-22T10:08:52.210" LastActivityDate="2017-01-22T10:08:52.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8603" PostTypeId="2" ParentId="8598" CreationDate="2014-02-20T12:06:18.190" Score="49" Body="&lt;p&gt;Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number &lt;em&gt;n&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first shell (n=1), we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 1s orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the second shell (n=2), we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 2s orbital&lt;/li&gt;&#xA;&lt;li&gt;The 2p orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the third shell (n=3), we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 3s orbital&lt;/li&gt;&#xA;&lt;li&gt;The 3p orbitals&lt;/li&gt;&#xA;&lt;li&gt;The 3d orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the fourth shell (n=4), we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 4s orbital&lt;/li&gt;&#xA;&lt;li&gt;The 4p orbitals&lt;/li&gt;&#xA;&lt;li&gt;The 4d orbitals&lt;/li&gt;&#xA;&lt;li&gt;The 4f orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher &lt;em&gt;n&lt;/em&gt;. The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now for the different kinds of orbitals&lt;/strong&gt;&#xA;Each kind of orbital has a different &quot;shape&quot;, as you can see on the picture below. You can also see that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The s-kind has only one orbital&lt;/li&gt;&#xA;&lt;li&gt;The p-kind has three orbitals&lt;/li&gt;&#xA;&lt;li&gt;The d-kind has five orbitals&lt;/li&gt;&#xA;&lt;li&gt;The f-kind has seven orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XkAjM.jpg&quot; alt=&quot;Orbital shapes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each orbital can hold &lt;em&gt;two&lt;/em&gt; electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold &lt;em&gt;two&lt;/em&gt; electrons because they each have only one orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2p, 3p, 4p, etc., can each hold &lt;em&gt;six&lt;/em&gt; electrons because they each have &lt;em&gt;three&lt;/em&gt; orbitals, that can hold &lt;em&gt;two&lt;/em&gt; electrons each (3*2=6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3d, 4d etc., can each hold &lt;em&gt;ten&lt;/em&gt; electrons, because they each have &lt;em&gt;five&lt;/em&gt; orbitals, and each orbital can hold &lt;em&gt;two&lt;/em&gt; electrons (5*2=10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thus, to find the number of electrons possible per shell&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we look at the &lt;strong&gt;n=1 shell (the first shell)&lt;/strong&gt;. It has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 1s orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;n=2 (second)&lt;/strong&gt; shell has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 2s orbital&lt;/li&gt;&#xA;&lt;li&gt;The 2p orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;n=3 (third)&lt;/strong&gt; shell has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 3s orbital&lt;/li&gt;&#xA;&lt;li&gt;The 3p orbitals&lt;/li&gt;&#xA;&lt;li&gt;The 3d orbitals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, the formula $2n^2$ holds! What is the difference between your two methods?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an important distinction between &lt;em&gt;&quot;the number of electrons possible in a shell&quot;&lt;/em&gt; and &lt;em&gt;&quot;the number of valence electrons possible for a period of elements&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1s &amp;lt; 2s &amp;lt; 2p &amp;lt; 3s &amp;lt; 3p &amp;lt; 4s &amp;lt; 3d &amp;lt; 4p &amp;lt; 5s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to visualize this is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvjvh.gif&quot; alt=&quot;Filling order of orbitals&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="402" LastEditDate="2017-01-22T10:08:24.243" LastActivityDate="2017-01-22T10:08:24.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8604" PostTypeId="1" CreationDate="2014-02-20T13:16:05.783" Score="3" ViewCount="3848" Body="&lt;p&gt;A frozen urine specimen is required for an organic acids lab test, but the urine refuses to freeze after 4 days in a regular freezer turned as low as it will go. What would explain this, and how cold might it need to be to freeze?&lt;/p&gt;&#xA;" OwnerUserId="4588" LastEditorUserId="19616" LastEditDate="2015-10-30T17:26:16.330" LastActivityDate="2015-10-30T17:26:16.330" Title="What would explain urine that resists freezing?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8605" PostTypeId="1" AcceptedAnswerId="14779" CreationDate="2014-02-20T13:32:53.233" Score="1" ViewCount="7484" Body="&lt;p&gt;From Practical skills in Biology by Allan Jones et al.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At high concentrations of certain ionic solutes, $\gamma$ (activity&#xA;  coefficient) may increase to become greater than unity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My questions : If activity coefficient $\gamma$ $=$ $  {a}/{[C]} $ where $a$ is activity which signifies &lt;em&gt;effective&lt;/em&gt; concentration and $[C]$ is concentration, how can $a$  be greater than $[C]$ ?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastActivityDate="2016-06-17T21:35:05.983" Title="Activity coefficient greater than 1" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8606" PostTypeId="2" ParentId="8605" CreationDate="2014-02-20T14:19:15.863" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution.&lt;/p&gt;&#xA;&lt;p&gt;(From &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the main point is that that activity is just a measure of the deviation from the ideal solutions, because the laws you are using in physical chemistry  assume that solution is ideal. But usually it is not true. Laws can work for real life with a certain precision, but you should always remember about their &amp;quot;ideality&amp;quot; and the non-ideality of our world.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-02-21T08:30:40.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8607" PostTypeId="2" ParentId="8604" CreationDate="2014-02-20T14:33:28.623" Score="3" Body="&lt;p&gt;Urine is an aqueous solution of urea, uric acid, sodium salts and whatnot. These solutions usually exhibit &lt;strong&gt;freezing-point depression&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Brian has correctly pointed out in a comment, the effect isn't that strong and a quick and dirty look in the &lt;a href=&quot;http://dx.doi.org/10.1159/000285189&quot; rel=&quot;nofollow&quot;&gt;literature&lt;/a&gt; shows that a freezing point around -1.3 to -2.2 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to try a cooling bath with dry ice in isopropanol or acetone or other drastic methods to achieve shock-freezing.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-20T15:06:25.873" LastActivityDate="2014-02-20T15:06:25.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8608" PostTypeId="2" ParentId="6985" CreationDate="2014-02-20T16:12:47.080" Score="0" Body="&lt;p&gt;Suppose you have 100 tonnes of harvested spearmint from a farm.  It is degrading in air and biologically as you look at it.  Batch internal steam distillation will be a procedural nightmare.  You will burn the biological where it touches the heated inside of the flask.  With external steam distillation, you continuously feed in biological and live steam, capture the mixed distillate (carvone), then discard the exhausted outflow.  You now have excellent process control for the temperature and volume of steam admitted, and efficient continuous throughput.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were peppermint, the menthol product is a solid at room temp.  That is a whole different ballgame re clogging.  A production line requires trained skilled people.  If it requires intelligent people, you only have a pilot plant.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-20T16:12:47.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8609" PostTypeId="1" CreationDate="2014-02-20T16:18:01.403" Score="1" ViewCount="96" Body="&lt;p&gt;I learned that if concentration of reactants or products is changed, the position of equilibrium changes but value of $K_c$ remains the same.&#xA;This just does not make intuitive sense to me... I mean whatever concentration of reactants I start with does not matter, $K_c$ remains same....?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can someone please elaborate the term &lt;em&gt;position  of equilibrium&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2014-02-20T18:13:58.933" Title="Chemical equilibria and changes in concentration" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8610" PostTypeId="2" ParentId="8609" CreationDate="2014-02-20T17:23:02.250" Score="1" Body="&lt;p&gt;Assuming the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B &amp;lt;=&amp;gt; C + D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{c}=1.23\times10^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We say that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{c}=\frac{\ce{[C]*[D]}}{\ce{[A]*[B]}}=1.23\times10^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the reaction will be at equilibrium for any concentrations (actually, it should be activities, which are in essence &lt;em&gt;effective&lt;/em&gt; concentrations) of A, B, C and D that satisfy the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand your confusion. What they mean is... If you have a reaction at equilibrium, and remove half the reactants, then the reaction will proceed towards reactants, until equilibrium is once again reached. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Their way of saying it:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You remove reactants, and the position of equilibrium is shifted towards the left, because the reaction will now proceed towards the left hand side to restore equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's confusing, because equilibrium occurs at the same conditions as before, so you haven't really shifted the equilibrium position in the way it sounds like.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2014-02-20T18:13:58.933" LastActivityDate="2014-02-20T18:13:58.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8611" PostTypeId="1" AcceptedAnswerId="8612" CreationDate="2014-02-20T17:23:51.093" Score="0" ViewCount="1035" Body="&lt;p&gt;@tschoppi &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried taking a stab at this for me and got this much:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-(1'-oxoethyl)-2-(something-something ester)-pyrolidine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D2SF9.png&quot; alt=&quot;http://i.imgur.com/UiIrOa2.png&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="4945" LastEditDate="2014-10-10T15:58:02.987" LastActivityDate="2014-10-10T15:58:02.987" Title="What is the name of this molecule? (C₈O₃H₁₃N)" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8612" PostTypeId="2" ParentId="8611" CreationDate="2014-02-20T17:37:24.043" Score="4" Body="&lt;p&gt;If I managed to see the structure correctly it should be &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.479338.html&quot; rel=&quot;nofollow&quot;&gt;Methyl 1-acetylprolinate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-20T17:37:24.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8613" PostTypeId="1" AcceptedAnswerId="8617" CreationDate="2014-02-20T19:57:23.437" Score="7" ViewCount="457" Body="&lt;p&gt;I know that usually more electronegative anions would displace less electronegative anions but is there a way to get different anions to &quot;share&quot; a cation?&lt;/p&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="418" LastEditDate="2014-02-21T00:16:18.233" LastActivityDate="2014-02-21T00:16:32.173" Title="Is it possible to create a salt with multiple anions with only one cation(e.g. CaBrCl)?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8614" PostTypeId="1" CreationDate="2014-02-20T20:17:24.187" Score="3" ViewCount="6166" Body="&lt;p&gt;I know 3 types of unit cells, the simple cubic, the body centered and the face centered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is a &lt;em&gt;hexagonal&lt;/em&gt; unit cell, and why does it have a hexagon base while others are all cut into a cubic? And how can I calculate the number of atoms in every primitive hexagonal unit cells?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="3846" LastEditDate="2014-02-21T08:26:37.947" LastActivityDate="2015-10-02T04:08:33.417" Title="Hexagonal unit cells" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8615" PostTypeId="2" ParentId="6229" CreationDate="2014-02-20T23:52:32.753" Score="1" Body="&lt;p&gt;Look at it under a microscope.  If the surface is microporous, light goes into the reflecting cavities, bounces around, and is absorbed before it can escape.  Wilkinson Sword sold bolted stacks of double-edged razor blades as low power laser beam dumps,  Each sharp edge was mirror bright with a slight taper toward the edge.  Light bounced in and could not escape.  The sharp sides of the stack were astoundingly black.  375 blades x .004&quot; = 1.5&quot; square target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you have impure zinc and crud is accumulating at the dissolving surface.  Alloy 8111 aluminum foil contains about 2% or so of strengthening dispersed intermetallics.  Dissolving aluminum foil in acid or base leaves black residues. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-20T23:52:32.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8616" PostTypeId="2" ParentId="8613" CreationDate="2014-02-20T23:59:42.677" Score="1" Body="&lt;p&gt;Sure, why not?  In the bad old days geology claiming isomorphous substitution had a hissy fit with chemistry claiming exact stoichiometry.  Rochelle salt is potassium sodium tartrate. Potassium aluminum sulfate (alum) is often doped with colored transition metal ions substituting for some of the aluminum (chrome alum).  Ruby, where Cr(III) substitutes for some (Al(III).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If size is about the same for identical charge, the alien ion goes into the crystal lattice.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-20T23:59:42.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8617" PostTypeId="2" ParentId="8613" CreationDate="2014-02-21T00:16:32.173" Score="6" Body="&lt;p&gt;A cursory search showed some interesting cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oldest reported examples seemingly are mercury halides &lt;a href=&quot;http://dx.doi.org/10.1007/BF00845579&quot;&gt;$\ce{HgXY}$&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{HgClI}$ was prepared by iodination of solid $\ce{HgCl2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HgBrCl}$ was obtained from the chlorination of an ethanolic suspension of $\ce{HgBr2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HgBrI}$ was prepared by mixing ethanolic solutions of $\ce{HgBr2}$ and $\ce{HgI2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The crystal structure of &lt;a href=&quot;http://dx.doi.org/10.1107/S010827019401022X&quot;&gt;$\ce{Cs6Cl3[Fe(CN)6].H2O}$&lt;/a&gt; has been reported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Passing $\ce{H2S}$ through solutions of oxidosulfido molybdates and tungstates let to &lt;a href=&quot;http://dx.doi.org/10.5560/ZNB.2013-3108&quot;&gt;various mixed salts&lt;/a&gt; with halide, hydrogen sulfide and thiosulfate as additional anions beside the trisulfidometalates $\ce{[MoOS3]^{2−}}$ and $\ce{[WOS3]^{2−}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the concept isn't limited to inorganic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystal structures of mixed salts of &lt;a href=&quot;http://dx.doi.org/10.5560/ZNB.2013-3108&quot;&gt;amino acids with different anions&lt;/a&gt; were reported. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the field of organic conductor research, salts of BEDT-TTF (bis(ethylenedithio)tetrathiafulvalene) with mixed anions of $\ce{[CuBr2]-}$ and $\ce{[CuCl2]-}$ are &lt;a href=&quot;http://dx.doi.org/10.3390/cryst2020284&quot;&gt;known&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-21T00:16:32.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8620" PostTypeId="1" CreationDate="2014-02-21T01:21:42.930" Score="5" ViewCount="1453" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A student determined the $\ce{Ca^{2+}}$ ion content in a sample of nonfat dry milk. For complete titration of the $\ce{Ca^{2+}}$ ion present, a sample weighting $1.483 ~\mathrm{g}$ required $41.33 ~\mathrm{mL}$ of $1.183 \times 10^{-2} ~\mathrm{M}$ EDTA solution. Calculate the mass of the $\ce{Ca^{2+}}$ ions in the titrated sample.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$41.33 ~\mathrm{mL} = 0.04133 \mathrm{L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1.183 \times 10^{-2}~\mathrm{mol\, L^{-1}})\times (0.04133 \mathrm{L}) = 4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}) \times 40.078 ~\mathrm{g\, (mol\, \ce{Ca^{2+}})^{-1}} = 0.0196 ~\mathrm{g}\, \ce{Ca^{2+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $25.06$ oz box of the powdered milk analyzed above costs $5.46$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the cost of $\pu{1.00 g}$ of $\ce{Ca^{2+}}$ ion as provided by this brand of dry milk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I do that?&lt;/p&gt;&#xA;" OwnerUserId="4595" LastEditorUserId="51553" LastEditDate="2018-03-23T05:37:41.090" LastActivityDate="2018-03-23T05:37:41.090" Title="What is the cost of 1.00 g of calcium ions as provided by this brand of dry milk?" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8621" PostTypeId="1" AcceptedAnswerId="8623" CreationDate="2014-02-21T01:37:03.810" Score="4" ViewCount="880" Body="&lt;p&gt;I tried preparing a supersaturated solution of caffeine by mixing &lt;a href=&quot;http://curativo.eu/Nahrungsergaenzung/Koffein-Pulver-100-rein-100g.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;anhydrous caffeine powder&lt;/a&gt; with water in a thin Turkish tea glass and heating it up over a candle (desperate times...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This worked as expected and a lot more caffeine dissolved into the clear solution after I slowly swung it back and forth over the flame. The water didn't boil, no visible vapors formed on the surface, and I was holding the glass with my fingers, so I would say that the temperature was well below 100°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the solution cooled down, instead of crystals precipitating, I found the glass filled with a fibrous cotton-like white mass. I pulled it out (it mostly stuck together) and dried it up on sheets of paper. Now I have white flakes which are flexible instead of brittle (still feel very much like cotton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would seem an awful lot like some polymer forming - except the substance tastes a lot like crystalline caffeine, and melts and evaporates &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7426/caffeine-boiling-and-melting-points&quot;&gt;much like caffeine&lt;/a&gt; - with the vapors having the same distinctive acrid (but not burnt) smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the answer to the polymer question is &lt;strong&gt;no&lt;/strong&gt;, but in case I'm right I'd be very curious as to what's actually happening.&lt;/p&gt;&#xA;" OwnerUserId="3974" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-02-21T02:39:09.510" Title="Does supersaturated caffeine polymerize?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;polymers&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8622" PostTypeId="2" ParentId="8614" CreationDate="2014-02-21T02:27:50.303" Score="4" Body="&lt;p&gt;The cubic primitive, cubic body-centered and the cubic face-centered are just the three possible Bravais lattices of the cubic &lt;strong&gt;crystal system&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total, we know seven &lt;strong&gt;crystal systems&lt;/strong&gt;, namely (in order of increasing symmetry) &lt;em&gt;triclinic&lt;/em&gt;, &lt;em&gt;monoclinic&lt;/em&gt;, &lt;em&gt;orthorhombic&lt;/em&gt;, &lt;em&gt;tetragonal&lt;/em&gt;, &lt;em&gt;trigonal&lt;/em&gt;, &lt;em&gt;hexagonal&lt;/em&gt; and &lt;em&gt;cubic&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among these, the &lt;strong&gt;triclinic&lt;/strong&gt; crystal system has the &lt;strong&gt;lowest symmetry&lt;/strong&gt; ($a \ne b \ne c;\ \alpha, \beta, \gamma \ne 90°$), the &lt;strong&gt;cubic&lt;/strong&gt; crystal system has the &lt;strong&gt;highest symmetry&lt;/strong&gt; ($a = b = c;\ \alpha = \beta = \gamma = 90°$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A careful look at illustrations of the hexagonal closest packing ($7+3+7$ atoms in $ABA$ layers) in a hexagonal prism and a bit of thinking gives the number of atoms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$3$ atoms in the $B$ layer are not shared by any other prism. They fully count &#xA;as $\mathbf{3}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$2$ atoms in the centre of the top and bottom $A$ layer are each shared by any other prism (above and below). Each of them counts $\frac{1}{2}$. So there's another $2 \times\frac{1}{2}= \mathbf{1}$ atom.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Top and bottom $A$ layers each have $6$ atoms on corner positions. Each atoms is shared by three neighbors in the the same layer and with another hexagonal prism on top or below. So they count as $2\times6\times\frac{1}{3}\times\frac{1}{2}=\mathbf{2}$ atoms. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The total count thus is $\mathbf{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These diagrams of the hexagonal closest packing, do &lt;strong&gt;not&lt;/strong&gt; represent the the &lt;strong&gt;unit cell&lt;/strong&gt; as the most simple unit from which the crystal is build by &lt;em&gt;symmetry operations&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cutting the hexagonal prism (along the axis through the atoms centering the top and bottom faces) into three pieces gives identical rhombohedra. These can be transferred into each other by rotations along the axis described above.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="21456" LastEditDate="2015-10-02T04:08:33.417" LastActivityDate="2015-10-02T04:08:33.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8623" PostTypeId="2" ParentId="8621" CreationDate="2014-02-21T02:39:09.510" Score="4" Body="&lt;p&gt;Google Images &quot;caffeine crystals&quot;  They are beautiful fibrous masses.  The solvent can influence crystal shape by making various faces grow faster or slower.  Benzil slowly crystallized from hot optically resolved pinene is slender needles.  Crystallized from room temp supersaturated mixed xylenes with a needle seed crystal, it is single crystal blocks. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-21T02:39:09.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8625" PostTypeId="1" AcceptedAnswerId="8628" CreationDate="2014-02-21T03:17:02.123" Score="4" ViewCount="7972" Body="&lt;p&gt;The question asks to carry out the following conversion in not more than 2 steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzoic acid to m-nitrobenzyl alcohol &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first used nitration (conc. $\ce{HNO3}$ /conc.$\ce{H2SO4}$) and then I'm got stuck, as reducing the $\ce{-COOH}$ group would also reduce $\ce{-NO2}$ group (using $\ce{LiAlH4}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to reduce $\ce{-COOH}$ group using $\ce{LiAlH4}$ to $\ce{-CH2OH}$ and then proceed to nitration but got confused whether $\ce{-CH2OH}$ is o/p or m-directing. I think it should be o/p directing.&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4945" LastEditDate="2015-01-08T04:24:40.573" LastActivityDate="2016-09-13T16:11:55.510" Title="How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8628" PostTypeId="2" ParentId="8625" CreationDate="2014-02-21T06:37:28.140" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first. &lt;/li&gt;&#xA;&lt;li&gt;Reduce &lt;em&gt;m&lt;/em&gt;-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of  $\ce{LiAlH4}$ needed, don't forget that you are reacting an &lt;strong&gt;acid&lt;/strong&gt; with a &lt;strong&gt;hydride&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Edit 1&lt;/strong&gt;:&#xA;Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;:&#xA;Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't &lt;em&gt;m&lt;/em&gt;-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) &lt;em&gt;Friedel-Crafts&lt;/em&gt; alkylations or etherification. I'd dare to say that the result is a mess. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-21T08:10:39.840" LastActivityDate="2014-02-21T08:10:39.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8630" PostTypeId="1" CreationDate="2014-02-21T08:55:23.687" Score="2" ViewCount="43" Body="&lt;p&gt;Consider this reaction of alpha decay: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{^{222}_{88}Ra -&amp;gt; ^{218}_{86}Rn + ^4_2 He} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, whether or not the electrons also get distributed in the reaction like the nucleons?&lt;/p&gt;&#xA;" OwnerUserId="4104" LastEditorUserId="3846" LastEditDate="2014-02-21T09:05:10.200" LastActivityDate="2014-02-21T10:14:42.660" Title="Are the electrons also distributed for the alpha decay?" Tags="&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8631" PostTypeId="2" ParentId="8620" CreationDate="2014-02-21T08:57:10.180" Score="3" Body="&lt;p&gt;First you calculate the value $v$ of the powder per gram:&#xA;$$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you calculate the mass of powder ($m$) that contains one gram of calcium ions:&#xA;$$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you multiply the two results for the price $p$:&#xA;$$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-21T08:57:10.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8632" PostTypeId="1" AcceptedAnswerId="8635" CreationDate="2014-02-21T10:08:06.070" Score="3" ViewCount="1070" Body="&lt;p&gt;I am considering the cathode ray tube of the set up, when the cathode rays were discovered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Voltage = very high&lt;/li&gt;&#xA;&lt;li&gt;Gas = any&lt;/li&gt;&#xA;&lt;li&gt;Atmospheric Pressure = changing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At different pressures, there are different observations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1&amp;nbsp;atm: the tube does not glow.&lt;/li&gt;&#xA;&lt;li&gt;0.1&amp;nbsp;atm: the whole tube lightens up&lt;/li&gt;&#xA;&lt;li&gt;0.01&amp;nbsp;atm: the tube glows at the anode end (Glowing is shown as there is a detector coating of ZnS.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: I don't have this set-up, this is just taken by me from theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does, when I am just reducing the pressure from 1&amp;nbsp;atm to 0.01&amp;nbsp;atm, the tube at first not glow, then it glows completely, and only at low pressure it glows just at one end?&lt;/p&gt;&#xA;" OwnerUserId="4602" LastEditorUserId="4945" LastEditDate="2018-09-27T09:33:41.067" LastActivityDate="2018-09-27T09:33:41.067" Title="Why does the cathode ray tube only start glowing at low pressures?" Tags="&lt;physical-chemistry&gt;&lt;pressure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8633" PostTypeId="2" ParentId="8630" CreationDate="2014-02-21T10:14:42.660" Score="1" Body="&lt;p&gt;The equation for the $\alpha$-decay omits the charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $\alpha$-particle actually is $\ce{^4_2He^{2+}}$, the resulting nucleus is initially left with a surplus of two electrons. These are emitted too, probably on the same timescale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer is yes: the $\alpha$-decay of a neutral nucleus yields another neutral nucleus, a double-positively charged $\alpha$-particle and two electrons.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-21T10:14:42.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8634" PostTypeId="1" AcceptedAnswerId="8654" CreationDate="2014-02-21T15:06:07.013" Score="5" ViewCount="1400" Body="&lt;p&gt;Does anyone know what the intermediate calculations are to obtain equation 1.4.14? Taken from Fundamentals and Applications (Bard) page 31&lt;img src=&quot;https://i.stack.imgur.com/qtZzl.jpg&quot; alt=&quot;Fundamentals and Applications (Bard) pg. 31&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;To be more precise, how to obtain (1.4.14) from &lt;span class=&quot;math-container&quot;&gt;$\frac{nFAm_O}{nFAm_R}=\frac{i_l-i}{i}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-02-22T10:37:49.650" Title="How do I derive the Nernst equation in the case in which R is absent?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8635" PostTypeId="2" ParentId="8632" CreationDate="2014-02-21T15:09:33.987" Score="3" Body="&lt;p&gt;At 1 atm. it is difficult create stable plasma, because breakdown voltage is very high.&#xA;Your electrodes are located at least 10-15 cm I think, and it is simply you do not have enough electrical potential between them to create initial plasma channel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then pressure goes down, voltage is fixed. When you reach breakdown voltage you get your glow (the Townsend discharge happens just as glow discharge becomes visible)-normal glow discharge. What happens? You ionized your gas, you have free electrons and ions, which are floating in different directions: ions are floating to negatively charged cathode and electrons flow to positively charged anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because charge carriers are moving in space. They are rubbing against each other, part of the energy is dissipated and goes into heat and other is emitted and you see your glow. Ions are moving relatively slow, electrons, on contrary,  move very fast comparing to ions, because they are light weighted. So this fast free electrons, accelerated by a sufficiently strong electric field, give rise to electrical conduction through a gas by avalanche multiplication caused by the ionization of molecules by ion impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you continue to reduce your pressure more and more, less and less gas molecules can be ionized by free electron flow, less and les collisions are happening (less energy is dissipated), so the intensity of glow is going down, and finally it stops, because the breakdown voltage increased and overcome your applied voltage. (see Paschen Curve, lower is a pressure higher is the breakdown voltage.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/llRQa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-02-24T11:21:42.683" LastActivityDate="2014-02-24T11:21:42.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8636" PostTypeId="2" ParentId="7683" CreationDate="2014-02-21T17:03:29.607" Score="33" Body="&lt;p&gt;It can be explained another way: &lt;em&gt;o&lt;/em&gt;-toluidene is less basic than aniline cause of a different reason. See, what happens is: when aniline acting as a base becomes &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3+}$&lt;/span&gt; (on top of a benzene ring), it is usually stabilised by solvation. But if there is a substituent on the &lt;em&gt;ortho&lt;/em&gt;-position, it inhibits solvation. Thus the tendency to act like a base is reduced.&lt;/p&gt;&#xA;" OwnerUserId="4604" LastEditorUserId="61578" LastEditDate="2020-05-01T04:26:49.387" LastActivityDate="2020-05-01T04:26:49.387" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8637" PostTypeId="2" ParentId="7912" CreationDate="2014-02-21T18:10:26.147" Score="2" Body="&lt;p&gt;&lt;strong&gt;I would say that your method is actually quicker (if that's what you mean by 'better').&lt;/strong&gt; My reasoning is as follows: the solubility of the powder in water increases with increasing temperature, as does the dissolution rate but the dissolution rate remains finite all the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If you add the powder and start heating at $t=0$ the concentration of dissolved powder will start to increase from $c=0$ to $c&amp;gt;0$ immediately. Once the water reaches the boiling point at $t=t_b$ already a certain amount (maybe even all) of the powder has dissolved i.e. some concentration of dissolved powder is present $c=c_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you boil the water first and then add the powder you start with $c=0$ at $t=t_b$, which is a lower concentration than in the powder-first case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Because there is already more powder dissolved at the boiling point ($c=c_1$ vs. $c=0$ at $t=t_b$), the moment at which all powder is dissolved comes sooner for the powder-first case (situation 1).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I approach this question purely from the physical point of view, I have no clue whether the two methods might result in differently tasting tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;P.S. I ignored any possible changes in the boiling point due to the already dissolved powder. For low concentrations, as in the case of tea, this is no problem. If the concentrations become much higher then the change in boiling point might become dominant and the story changes&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-02-21T18:10:26.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8638" PostTypeId="1" AcceptedAnswerId="15044" CreationDate="2014-02-21T19:21:50.560" Score="13" ViewCount="2860" Body="&lt;p&gt;In the organic chemistry textbook by Clayden et al. they mention that the reaction of (&lt;em&gt;E&lt;/em&gt;)-methyl 3-(furan-2-yl)acrylate in acidic methanol yields dimethyl 4-oxoheptanedioate without providing the reaction mechanism and leaving it to the reader to figure it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ohr75.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried for over an hour now but didn't get to this product. I assumed that since the reactant already has the right number of carbon atoms I won't have some kind of dimerization to consider, only reactions involving one molecule of (E)-methyl 3-(furan-2-yl)acrylate. I think, the first step should be a protonation at the furan 3- or 5-position with a subsequent attack of methanol on the carbonyl group that was formed followed by an acetal cleavage. The problem is that I don't know how to get the second ester group and how to get rid of all 3 $\ce{C=C}$ double bonds? Unfortunately, I don't have access to ChemDraw on this computer so I can't provide some example mechanisms I came up with.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2017-02-28T02:00:03.807" LastActivityDate="2020-04-27T14:32:05.927" Title="What's the mechanism of the reaction of (E)-methyl 3-(furan-2-yl)acrylate in acidic methanol?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8640" PostTypeId="2" ParentId="8638" CreationDate="2014-02-21T21:16:08.213" Score="3" Body="&lt;p&gt;Starting furan is also a bis-enol ether hiding a terminal aldehyde and an internal ketone.  Seven skeletal carbon atoms including that conjugated olefin.  Product has seven skeletal carbon atoms and three carbonyls in the right places.  The double bond is reduced and what would be the terminal aldehyde is oxidized to the acid, appearing as the methyl ester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protonate the ester carbonyl.  1,3 shift of the double bond affords a carbenium ion next to the furan ring and =C(OH)(OR), the enol of the ester.  That is downhill in energy for the alpha-cation conjugating to the electron-rich aromatic ring.  1,3 shift of the ring double bond leaves the ring oxygen an enol ether (exocyclic double bond)  and moves the charge into the ring.  1,3 shift moves the charge alpha to the ring oxygen (again downhill in energy - resonance contributors), where it is captured by methanol to make an acetal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Move the endocyclic unsaturation into the acetal. I see some ugly here, so I hope somebody does better.  Work it up in aqueous acid.  Protonate the endocyclic double bond to put the charge alpha to the captured MeO.  1,2-hydrogen shift to put the charge where it is needed, captured by water.  Now hydrolyze everything to the parent carbonyls. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-21T21:16:08.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8641" PostTypeId="2" ParentId="8585" CreationDate="2014-02-21T21:40:36.717" Score="1" Body="&lt;p&gt;I would have gone for a less connected hydrogen bonding network as the primary engine.  Diethyl amine boils at 55.6 °C, 1.11 debyes.  n-Butylamine boils at 77.1 °C, 1.37 debyes (Organic Solvents, 4th Ed., Riddick, et al., 1986). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/organicprops/amines/background.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/organicprops/amines/background.html&lt;/a&gt; &lt;BR&gt;&#xA;   Both appear to be important.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-08-19T12:25:12.040" LastActivityDate="2015-08-19T12:25:12.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8642" PostTypeId="1" AcceptedAnswerId="8662" CreationDate="2014-02-21T22:48:51.360" Score="12" ViewCount="25432" Body="&lt;p&gt;I have read that combining the DC current with a high frequency AC current, the electrolysis of water speeds up. Is this true? In that case, how is less energy wasted as heat? Or does it simply catalyze the process?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="23" LastEditDate="2015-04-07T09:42:43.897" LastActivityDate="2021-03-10T11:57:15.050" Title="How does High Frequency water electrolysis work?" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8643" PostTypeId="2" ParentId="704" CreationDate="2014-02-21T22:49:53.127" Score="0" Body="&lt;p&gt;You could fix/salvage the experiment by first extracting the acetaminophen with DCM, and then perform your base extraction.. although it deviates a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example lab: &lt;a href=&quot;http://www.odinity.com/separation-of-the-components-of-an-analgesic-tablet/&quot; rel=&quot;nofollow&quot;&gt;http://www.odinity.com/separation-of-the-components-of-an-analgesic-tablet/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4607" LastActivityDate="2014-02-21T22:49:53.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8644" PostTypeId="2" ParentId="8625" CreationDate="2014-02-21T23:34:15.077" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo00956a011&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jo00956a011&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required is a reduction that acts upon the carboxyl as such.  Consider borane-THF, or its equivalent sodium borohydride and BF3·Et2O.  Next best is borane-dimethyl sulfide, given the odor.  Aliphatic nitro groups are reduced.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-21T23:34:15.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8645" PostTypeId="1" CreationDate="2014-02-22T00:49:24.713" Score="4" ViewCount="53" Body="&lt;p&gt;I am programming a simulation, and I need a good name for a data structure that stores pairs $(c, R)$, where $c$ is the stoichiometric coefficient of a reactant $R$ in some chemical reaction. Is there a name for these pairs $(c, R)$ in chemistry?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="73" LastEditDate="2014-02-22T00:51:09.923" LastActivityDate="2014-03-07T13:58:42.233" Title="Is there a name for the pair (c, R), where c is the stoichiometric coefficient of a reactant R in some chemical reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8646" PostTypeId="1" AcceptedAnswerId="8647" CreationDate="2014-02-22T03:07:05.807" Score="2" ViewCount="223" Body="&lt;p&gt;What happens when Ammonia  is added to an (aq) solution of copper II chloride?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuCl2 + NH3 -&amp;gt;\; ?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that ligands bond with metals to form ion but I'm not sure about how they work in a reaction. &lt;/p&gt;&#xA;" OwnerUserId="4608" LastEditorUserId="4608" LastEditDate="2014-02-22T03:19:32.240" LastActivityDate="2014-03-25T15:13:19.990" Title="How do ligands and coordination complexes work in this case with copper (II) chloride?" Tags="&lt;reaction-mechanism&gt;&lt;bond&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8647" PostTypeId="2" ParentId="8646" CreationDate="2014-02-22T03:26:16.057" Score="3" Body="&lt;p&gt;When solid $\ce{CuCl2}$ is dissolved in water, &lt;em&gt;aquo&lt;/em&gt; complexes are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuCl2_{(s)} + H2O -&amp;gt; [Cu(H2O)6]^2+ + 2 Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition of ammonia results in a ligand exchange reaction to give &lt;em&gt;tetrammine&lt;/em&gt; complexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Cu(H2O)6]^2+ + 4 NH3 -&amp;gt; [Cu(NH3)4(H2O)2]^2+ + 2 H2O}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-22T03:26:16.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8648" PostTypeId="1" CreationDate="2014-02-22T03:46:11.440" Score="1" ViewCount="1256" Body="&lt;p&gt;Lipids usually consist of 3 fatty acids and 1 glycerol. I looked up what they look like&lt;img src=&quot;https://i.stack.imgur.com/YS0H3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is it not a polar molecule?  The $\ce{-OH}$ side has negative charge, and the hydrocarbon side chain has a positive charge.&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="73" LastEditDate="2014-02-22T04:52:47.037" LastActivityDate="2014-02-22T06:20:46.293" Title="What is the structure of a lipid and why is this example non-polar?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8649" PostTypeId="2" ParentId="8648" CreationDate="2014-02-22T05:53:30.460" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Lipids usually consist of 3 fatty acids and 1 glycerol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's wrong. You're giving the definition of &lt;em&gt;fats&lt;/em&gt;, which are just a subclass of lipids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure in your question is the skeleton of &lt;strong&gt;cholesterol&lt;/strong&gt; - without the stereochemistry.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-22T06:20:46.293" LastActivityDate="2014-02-22T06:20:46.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8650" PostTypeId="1" AcceptedAnswerId="8652" CreationDate="2014-02-22T06:36:01.427" Score="5" ViewCount="2467" Body="&lt;p&gt;I learned that when benzene undergoes hydrogenation, $\pu{208KJ}$ of energy is given off and thus $\pu{152KJ}$ less energy than what would be given off if Kekule's structure was correct. My confusion is that my book says that this $\pu{152KJ}$ of energy is required to break the delocalized electron cloud - this does not make sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the energy discrepancy of value the energy required to break the delocalized electron cloud in the benzene?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="5026" LastEditDate="2018-03-17T04:45:07.783" LastActivityDate="2018-11-18T05:24:10.067" Title="Energy required to break the &quot;delocalised electron cloud&quot; of benzene" Tags="&lt;aromatic-compounds&gt;&lt;energy&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8651" PostTypeId="2" ParentId="8642" CreationDate="2014-02-22T06:36:25.217" Score="7" Body="&lt;p&gt;From the site that you linked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A battery would be used as a source of electrical energy which would separate the hydrogen/oxygen into gases. The gasses would then fuel an internal combustion engine, which would power a generator to continously recharge the battery as well as deliver useable mechanical energy. If this sort of motor can be made to work, the energy crisis on this planet will be over forever.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sounds like somebody proposes a &lt;strong&gt;perpetual motion engine of the first kind&lt;/strong&gt;. I'm &#xA;very much impressed! The rest on that page is techno-babble of the same quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; This doesn't mean that there's might not be serious research on electrolysis using something else than constant DC, e.g. PWM (pulse width modulation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cursory search (there might be other sources) gave:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K. Mazloomi, N. Sulaiman, S. A. Ahmad, N. A. Yunus,&#xA;Analysis of the Frequency Response of a Water Electrolysis cell,&#xA;&lt;em&gt;Int. J. Electrochem. Sci.&lt;/em&gt;, &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 3731-3739, &lt;a href=&quot;http://www.electrochemsci.org/papers/vol8/80303731.pdf&quot;&gt;PDF&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-22T09:45:13.213" LastActivityDate="2014-02-22T09:45:13.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8652" PostTypeId="2" ParentId="8650" CreationDate="2014-02-22T07:00:33.520" Score="3" Body="&lt;p&gt;Let's sum it up :)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When cyclohexene is hydrogenated, $\mathrm{-120\, kJ\cdot mol^{-1}}$ is released.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When 1,3-cyclohexadiene is hydrogenated, $\mathrm{-232\, kJ\cdot mol^{-1}}$ is released. That's pretty close to 240, so it looks that the energies for the hydrogenation of the double bonds &lt;em&gt;sum up&lt;/em&gt;. So far, we're cool.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Based on Kekule's model of alternating single and double bonds in benzene, we would now assume that the energy for hydrogenation is $\mathrm{ 3 \times -120 = -360 \, kJ\cdot mol^{-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as your textbook says it's only $\mathrm{-208\, kJ\cdot mol^{-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must conclude that Kekule was &lt;strong&gt;almost&lt;/strong&gt; right, but the bonding situation in bezene is special. The energy of the molecule is much lower than that of a hypothetical cyclohexatriene. And when it's lower in energy, less is released upon hydrogenation. The difference of $\mathrm{152\, kJ\cdot mol^{-1}}$ is attributed to resonance stabilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt; If that's still a bit unclear with the signs of the energies, we can do some reverse experiments in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take cyclohexane and rip adjacent hydrogen atoms off to generate unsaturated molecules.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We need $\mathrm{120\, kJ\cdot mol^{-1}}$ to make cyclohexane.&lt;/li&gt;&#xA;&lt;li&gt;We need $\mathrm{232\, kJ\cdot mol^{-1}}$ to make 1,3-cyclohexadiene.&lt;/li&gt;&#xA;&lt;li&gt;If benzene would be cyclohexatriene, we would need $\mathrm{360 \, kJ\cdot mol^{-1}}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But we need less energy to make benzene, because it is more stable than a hypothetical cyclohexatriene. Actually, it is even more stable than 1,3-cyclohexadiene!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Benzene's actual structure is stable by 152 KJ w.r.t. Kekule's structure. That 152KJ is used for breaking down the delocalised electron cloud of benzene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Guess what? The teacher is just repeating with other words what we have already worked on our own&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you want to convert benzene to the hypothetical Kekule structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a process that would &quot;consume&quot; energy, $\mathrm{152\, kJ\cdot mol^{-1}}$ to be precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? It's because benzene has the delocalized $\pi$-system that aromatic molecules use to have. If you want to overcome that and push the stable molecule uphill and turn in into the less stable (= not having the delocalisation) cyclohexatriene, you have to do some work.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-22T08:47:23.627" LastActivityDate="2014-02-22T08:47:23.627" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8653" PostTypeId="1" CreationDate="2014-02-22T08:00:49.817" Score="5" ViewCount="2329" Body="&lt;p&gt;In the chapter acid-base, I learned about indicators and how they are actually weak acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equation given in our book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HIn + H_2O &amp;lt;=&amp;gt; H_3O^+ + In^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{in}= \left(\frac{[\ce{H_3O^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\ce{HIn}$ and $\ce{In^-}$ are of different colors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Further in my book, the concentration of $\ce{H_3O^+}$ at which $\ce{[HIn]=[In^{-}]}$ is calculated. Apparently, this is the concentration of $\ce{H_3O^+}$ at which the indicator changes color and thus this can be used to identify end point of a titration. However, how is this the point where the indicator &quot;changes&quot; color, I mean isn't the indicator changing color throughout the titration as the concentration of $\ce{H_3O^+}$ is changing continuously as the alkali is added to the acid for example&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do different indicators have different $pK_{in}$ i.e. &quot;change color&quot; at different concentration of $\ce{H_3O^+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="3846" LastEditDate="2014-02-25T10:45:47.360" LastActivityDate="2014-02-25T10:45:47.360" Title="The significance of the dissociation constant of an indicator" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8654" PostTypeId="2" ParentId="8634" CreationDate="2014-02-22T10:37:49.650" Score="2" Body="&lt;p&gt;Thank you for the comment Philipp. With this clarification, the derivation is simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=E_0+\frac{RT}{nF}ln\frac{C_O(x=0)}{C_R(x=0)}$ (1.4.12) where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_O(x=0)=\frac{i_l-i}{nFAm_O}$ (1.4.11) and $C_R(x=0)=\frac{i}{nFAm_R}$ (1.4.13).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Than, combining equations (1.4.11) and (1.4.13) to (1.4.12), we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-02-22T10:37:49.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8655" PostTypeId="1" AcceptedAnswerId="8670" CreationDate="2014-02-22T11:24:10.640" Score="29" ViewCount="119646" Body="&lt;p&gt;Nodes are the points in space around a nucleus where the probability of finding an electron is zero. However, I heard that there are two kinds of nodes, &lt;strong&gt;radial nodes&lt;/strong&gt; and &lt;strong&gt;angular nodes&lt;/strong&gt;. What are they and what information do they provide of an atom?&lt;/p&gt;&#xA;" OwnerUserId="4432" LastEditorUserId="16683" LastEditDate="2020-10-24T00:59:09.887" LastActivityDate="2020-10-24T00:59:09.887" Title="What are angular and radial nodes?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="8656" PostTypeId="2" ParentId="8653" CreationDate="2014-02-22T12:54:41.113" Score="6" Body="&lt;p&gt;&lt;strong&gt;1. It all depends on the $\text{p}K_{\text{in}}$ of the indicator.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the Henderson-Hasselbalch equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that when $\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and unprotonated state, because when $[\ce{A-}]=[\ce{HA}]$, the ratio between them will be 1, and $\log 1=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you increase $\text{pH}$ by 1, to balance the equation, the logarithm must also be 1 higher, and this happens when the thing you take the logarithm of becomes &lt;em&gt;10&lt;/em&gt; times larger. This means that for a small change in pH, we have a large change in the ratio between protonated and deprotonated forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, with indicators, you say that the colour change happens when $\text{pH} = \text{p}K_{\text{a}} \pm 1$, because outside of that interval, you will quickly have almost purely $\ce{A-}$ or $\ce{HA}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you follow this far, let's try an example with an indicator with a high $\text{p}K_{\text{in}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You put an indicator with $\text{p}K_{\text{in}}=10$ into pure water, which has a $\text{p}K_{\text{a}}=7$. From these values, we can see that water is a stronger acid than the indicator. This means that $[\ce{H3O+}]$ just in pure water will be high enough, that the indicator is largely protonated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7=10+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$&#xA;$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=-3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means $[\ce{HIn}] = 1000*[\ce{In-}]$ - try plugging in 1/1000, 1/100, 1/1, 100/1 etc. into the logarithm on your calculator to see why this is so!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we see that when the indicator has a $\text{p}K_{\text{in}}$ significantly higher than the $\text{p}K_{\text{a}}$ of the solution, it will mostly be protonated. &lt;em&gt;Why?&lt;/em&gt; We will see this in part 2 of the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another example. Let's see what happens if we have a &lt;em&gt;low&lt;/em&gt; $\text{p}K_{\text{in}}$ indicator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You put an indicator with $\text{p}K_{\text{in}}=3$ into pure water ( $\text{p}K_{\text{a}}=7$). This time the indicator is a stronger acid. This means that $[\ce{H3O+}]$ is not high enough to keep the indicator largely protonated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7=3+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$&#xA;$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means $10000*[\ce{HIn}] = [\ce{In-}]$ - Statistically, every time you find a protonated indicator molecule, you will have found 10000 deprotonated indicator molecules!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Titration scenario: acetic acid titrated with NaOH, using an indicator with $\text{p}K_{\text{a}}=10$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acetic acid has a $\text{p}K_{\text{a}}=4.7$. The acetic acid is a stronger acid than water, so it will have increased the concentration of $[\ce{H3O+}]$. To raise the $\text{pH}$, we need to lower this concentration. As we add $\ce{OH-}$ to the solution, the following reaction occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3O+ + OH- &amp;lt;=&amp;gt;&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are removing $\ce{H3O+}$ from the solution. But the acetic acid was only partially deprotonated, and what happens if the remove $\ce{H3O+}$ from the following reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOH + H2O &amp;lt;=&amp;gt; CH3COO- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, by the principle of Le Chatelier, acetic acid and water reacts further, to replenish the $\ce{H3O+}$. This goes on until the acetic acid is deprotonated to a large degree, and then the $\ce{OH-}$ very quickly raises the $\text{pH}$ of the water, since there is no acid to react with anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the $\text{pH}$ gets into the area of the $\text{p}K_{\text{in}}$ of the indicator, the colour change starts to happen. It happens swiftly, because we are not using very much of the indicator. This means that it doesn't take a lot of $\ce{OH-}$ to do to the indicator what we did to the acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Indicators have different equilibrium constants because they have different acid/base energetics&lt;/strong&gt;&#xA;Indicators have different values for $pK_{in}$ for exactly the same reasons as acids have different $\text{p}K_\text{a}$. It depends on the stability of the bond to the hydrogen that can be lost as a proton. If the bond is very unstable (as in HCl), the proton will lose its share of the electrons in the bond, and become free very easily, and the compound will have a low $\text{p}K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a compound such as acetic acid, protons will become free, but protons from $\ce{H3O+}$ can also transfer back to the acetic acid. The more $\ce{H3O+}$ that exists in the environment, the larger the chance, that a deprotonated acetic acid molecule will collide with it and receive the proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the indicator with a high $\text{p}K_{\text{in}}$, if there is a lot of $\ce{H3O+}$, then the indicator molecules will be re-protonated as fast as they can lose their protons. Thus, they overall stay protonated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constants $K_{\text{in}}$, $K_\text{a}$, etc. basically tell you how likely the molecule is to shed its proton.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2014-02-22T13:53:55.893" LastActivityDate="2014-02-22T13:53:55.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8657" PostTypeId="1" CreationDate="2014-02-22T14:06:03.250" Score="0" ViewCount="157" Body="&lt;p&gt;I know the $pH$, pressure, temperature and concentration of the solution. Is there a method I can use to work out the amount of $\ce{H+}$ ions?  &lt;/p&gt;&#xA;" OwnerUserId="4460" LastEditorUserId="10185" LastEditorDisplayName="user4076" LastEditDate="2015-06-14T14:54:57.727" LastActivityDate="2015-06-14T14:54:57.727" Title="How do I work out the amount of H+ ions in a sodium hydroxide and water solution?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8658" PostTypeId="2" ParentId="8657" CreationDate="2014-02-22T14:32:42.713" Score="1" Body="&lt;p&gt;The pH can be directly related to the concentration of $\ce{H3O+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{pH}=-\log{[\ce{H3O+}]}$$&#xA;$$\Downarrow$$&#xA;$$[\ce{H3O+}]=10^{-\text{pH}}$$&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-22T14:32:42.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8659" PostTypeId="1" AcceptedAnswerId="8673" CreationDate="2014-02-22T15:15:49.990" Score="5" ViewCount="2548" Body="&lt;p&gt;$\ce{Ag^{+}_{(aq)}}$ can be written as $\ce{[Ag(H2O)2]+}$ but how would you know that it bonds with 2 water molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't get when ligands randomly form 6, 4, 2 bonds. &lt;/p&gt;&#xA;" OwnerUserId="4608" LastEditorUserId="4608" LastEditDate="2014-02-22T17:07:06.997" LastActivityDate="2014-03-06T08:15:14.433" Title="How is an aqueous Ag+ the same as [Ag(H2O)2]+?" Tags="&lt;aqueous-solution&gt;&lt;coordination-compounds&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8660" PostTypeId="1" AcceptedAnswerId="8661" CreationDate="2014-02-22T15:59:02.943" Score="7" ViewCount="2568" Body="&lt;p&gt;Why does the Hydrogen proton &quot;prefer&quot; to join the $\ce{H2O}$ molecule instead of remaining bonded to the chlorine atom?&lt;/p&gt;&#xA;" OwnerUserId="4610" LastEditorDisplayName="user4076" LastEditDate="2014-02-22T18:57:34.530" LastActivityDate="2014-02-22T18:57:34.530" Title="Why does HCl ionize in water?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8661" PostTypeId="2" ParentId="8660" CreationDate="2014-02-22T16:20:29.097" Score="8" Body="&lt;p&gt;Chlorine is a very large atom compared to hydrogen, which means the overlap between the hydrogen $1s$-orbital and the chlorine $sp^3$ orbital is poor, which translates into a low bond energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, chlorine is very electronegative, which helps pull the H-Cl bonding pair onto the chlorine atom. The chloride anion solvation in water also plays a big part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's simply because it's energetically favourable. Maybe someone knows in greater detail than I do.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-22T16:20:29.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8662" PostTypeId="2" ParentId="8642" CreationDate="2014-02-22T21:38:40.840" Score="20" Body="&lt;p&gt;First of all, have a look at the wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water&quot; rel=&quot;noreferrer&quot;&gt;electrolysis of water&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I also like this review: &lt;a href=&quot;http://www.cres.gr/kape/publications/papers/dimosieyseis/ydrogen/A%20REVIEW%20ON%20WATER%20ELECTROLYSIS.pdf&quot; rel=&quot;noreferrer&quot;&gt;Zoulias &lt;em&gt;et al.&lt;/em&gt;: A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71&lt;/a&gt;&lt;br&gt;&#xA;Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus, speeding up &lt;em&gt;usually&lt;/em&gt; means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pulsed/modulated DC: looking through a few papers I liked this one: &lt;a href=&quot;http://www.tuks.nl/pdf/Reference_Material/Electrolysis_Water_Arc_and_Dielectric_Breakdown/Shimizu%20e.a.%20-%20A%20novel%20method%20of%20hydrogen%20generation%20by%20water%20electrolysis%20using%20an%20altra-short-pulse%20power%20supply.pdf&quot; rel=&quot;noreferrer&quot;&gt;Shimizu &lt;em&gt;et al.&lt;/em&gt;: A novel method of hydrogen generation by water electrolysis using&#xA;an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090-&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.&#xA;Look at these diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oNVaq.png&quot; alt=&quot;H2 output over input power&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XwiCj.png&quot; alt=&quot;efficiency over input power&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%.   However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I found a bunch of &lt;strong&gt;nonsense claims&lt;/strong&gt; in the internet, about the &lt;strong&gt;resonance&lt;/strong&gt; frequency &lt;strong&gt;of water helping to split bonds&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first thing to realize here is that there is no &lt;em&gt;one&lt;/em&gt; resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ($\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the &lt;strong&gt;UV&lt;/strong&gt;, and here we meet bands that lead to &lt;strong&gt;photodissociation&lt;/strong&gt;, e.g. at &lt;strong&gt;166nm&lt;/strong&gt; (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But &lt;strong&gt;resonance frequencies in the kHz range are electrical LC-circuit resonances&lt;/strong&gt; depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To give the &quot;method&quot; you ask about some real world numbers, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %. &lt;/li&gt;&#xA;&lt;li&gt;The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here. &lt;/li&gt;&#xA;&lt;li&gt;we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %.&lt;/li&gt;&#xA;&lt;li&gt;A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia. &lt;/li&gt;&#xA;&lt;li&gt;Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taking these numbers together, I conclude that after going once through the cycle of the proposed &quot;perpetuum mobile&quot;, 8 - 24 % of the energy are retained in a &quot;useful state&quot; while 76 - 92 % became heat. With fuel cell, we may be able to &quot;boost&quot; the energy efficiency to 43%. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Useful general knowledge (in addition to the law of energy conservation)&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument &lt;em&gt;why&lt;/em&gt; it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents to not have this technical check. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;generators (mechanic -&gt; electric conversion) are sources of &lt;em&gt;current&lt;/em&gt;, while batteries (galvanic cells) are sources of &lt;em&gt;voltage&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-02-22T21:38:40.840" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8663" PostTypeId="1" AcceptedAnswerId="8668" CreationDate="2014-02-22T22:11:23.167" Score="3" ViewCount="2151" Body="&lt;p&gt;Most solubility references (e.g. &lt;a href=&quot;http://www5.csudh.edu/oliver/chemdata/solrules.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;) say that there are &quot;few&quot; insoluble salts of alkali metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there &lt;em&gt;any&lt;/em&gt; exceptions?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="15751" LastEditDate="2018-07-14T16:54:01.627" LastActivityDate="2020-06-30T16:54:19.287" Title="Insoluble Group I/ammonium salts" Tags="&lt;solubility&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8664" PostTypeId="2" ParentId="8663" CreationDate="2014-02-22T23:29:25.547" Score="-1" Body="&lt;p&gt;There are some cations whose Group I/ammonium salts are not soluble in water. For example, (thanks Nicolau) the diuranate of ammonium is insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, (as I just found) the hydrides (H-) of alkali metals are not soluble, it seems, in any room-temperature liquid solvent. They hydrolyze violently (&lt;a href=&quot;http://goo.gl/Dl8Bd4&quot; rel=&quot;nofollow&quot;&gt;http://goo.gl/Dl8Bd4&lt;/a&gt;), and the hydride ion has never been observed in solution.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2014-02-22T23:29:25.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8665" PostTypeId="5" CreationDate="2014-02-22T23:51:18.670" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-02-22T23:51:18.670" LastActivityDate="2014-02-22T23:51:18.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8666" PostTypeId="4" CreationDate="2014-02-22T23:51:18.670" Score="0" Body="The mole is a standard amount of a substance A mole of a substance is a number of units (particles, atoms, molecules, ions, peas, hard candies, people, universes) equal to the number of atoms of carbon-12 in one gram of carbon-12, or an Avogadro's number of stuff. The mole allows us to relate real-word measurements to atomic scale phenomena." OwnerDisplayName="user4076" LastEditorUserId="275" LastEditDate="2014-02-23T00:38:08.893" LastActivityDate="2014-02-23T00:38:08.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8667" PostTypeId="1" CreationDate="2014-02-22T23:58:43.790" Score="5" ViewCount="437" Body="&lt;p&gt;I have been told that the multipole moments of rank 0,1,2,3,...n have respectively 1,3,5,7,...2n+1 components which transform like the spherical harmonics of rank 0,1,2,3,...n. How do I use the character tables to determine the lowest rank non-vanishing multipole moment for the following molecules: $\ce{H2O, BF3, CF4, CO2, SF6}$?&lt;/p&gt;&#xA;" OwnerUserId="2228" LastEditorUserId="4945" LastEditorDisplayName="user4076" LastEditDate="2014-06-12T08:42:29.390" LastActivityDate="2014-06-12T08:42:29.390" Title="Determining the lowest rank non-vanishing multipole moment" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ClosedDate="2018-03-17T00:39:10.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="8668" PostTypeId="2" ParentId="8663" CreationDate="2014-02-23T01:01:12.137" Score="4" Body="&lt;p&gt;Lithium carbonate has poor water solubility.  Cesium triphenylcyanoborate is insoluble in water.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-23T01:01:12.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8670" PostTypeId="2" ParentId="8655" CreationDate="2014-02-23T02:59:49.050" Score="27" Body="&lt;p&gt;&lt;strong&gt;1. How to get the number and type of nodes for an orbital&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, nodes are points of zero electron density. From the principal quantum number $n$ and the azimuthal quantum number $\ell$, you can derive the number of nodes, and how many of them are &lt;em&gt;radial&lt;/em&gt; and &lt;em&gt;angular&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{number of nodes}=n-1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{angular nodes}=\ell$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{radial nodes}=(\text{number of nodes})-(\text{angular nodes})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each type of orbital ($s, p, d$ etc) has its own unique, fixed number of angular nodes, and then as $n$ increases, you add radial nodes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Examples: First shell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first shell, $n=1$, which means the number of nodes will be 0.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Examples: Second shell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the second shell, $n=2$, which yields 1 node.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the $2s$ orbital, $\ell = 0$, which means the node will be radial&lt;/li&gt;&#xA;&lt;li&gt;For the $2p$ orbital, $\ell = 1$, which means the node will be angular&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Examples: Third shell&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third shell, $n=3$, yielding $3-1=2$ nodes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The $3s$ orbital still has $\ell = 0$ meaning no angular nodes, and thus the two nodes must be radial&lt;/li&gt;&#xA;&lt;li&gt;The $3p$ orbital still has one angular node, meaning there will be one radial node as well&lt;/li&gt;&#xA;&lt;li&gt;The $3d$ orbital has two angular nodes, and therefore no radial nodes!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. The difference between &lt;em&gt;radial&lt;/em&gt; and &lt;em&gt;angular&lt;/em&gt; nodes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial nodes are nodes inside the orbital lobes as far as I can understand. Its easiest to understand by looking at the $s$-orbitals, which can only have radial nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RkhwT.jpg&quot; alt=&quot;Radial nodes of the s-orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see what an angular node is, then, let's examine the $2p$-orbital - an orbital that has one node, and that node is angular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jMW9A.gif&quot; alt=&quot;Angular nodes of the 2p orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that angular nodes are not internal countours of 0 electron probability, but rather is a plane that goes through the orbital. For the $2p_\text{z}$-orbital, the angular node is the plane spanned by the x- and y-axis. For the $2p_\text{y}$-orbital, the angular node is the plane spanned by the z- and x-axis.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-23T02:59:49.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8672" PostTypeId="1" AcceptedAnswerId="8676" CreationDate="2014-02-23T05:02:44.060" Score="6" ViewCount="3774" Body="&lt;p&gt;When I poured hot tea into a cup, the cup shattered after about three seconds. The cup was made of glass. It was pretty thin. How did the glass and the tea react in terms of chemicals. This is a bit odd because wasn't heat applied to sand in order to make glass, shouldn't the glass break up into the reactants that made it?&lt;/p&gt;&#xA;" OwnerUserId="4614" LastEditorUserId="23561" LastEditDate="2019-03-20T21:30:49.573" LastActivityDate="2019-03-20T21:30:49.573" Title="Why does glass shatter when heat is applied to it?" Tags="&lt;heat&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8673" PostTypeId="2" ParentId="8659" CreationDate="2014-02-23T05:30:00.500" Score="7" Body="&lt;p&gt;There is no way to &lt;em&gt;absolutely&lt;/em&gt; predict the co-ordination number in any complex entity, without knowing say for example, magnetic or geometric properties of the complex. All we can do is give a probable configuration based on intelligent guesses. The co-ordination number depends on the characetristics of the ligands and the central ion.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver ion, $\ce{Ag+}$ usually forms complexes with Co-ordination Number two. This is partly because the silver ion is small sized, making it difficult to accommodate higher co-ordination numbers around itself owing to increased ligand-ligand repulsion, and partly due to the low charge on silver ion which makes it difficult to hold more than two ligands together. Therefore, in most of its complexes, silver ion co-ordinates with two ligands to form a linear entity. But to confirm this, you need to study solvated silver ions spectroscopically, magnetically or geometrically (diffraction), without which the above arguments can only give you, at best, a good guess.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples include $\ce{[Ag(NH3)_2]+}$, and &lt;a href=&quot;http://wwwchem.uwimona.edu.jm/courses/IC10Kcn.html&quot; rel=&quot;nofollow&quot;&gt;the polymeric&lt;/a&gt; $\ce{[Ag(en)]ClO4}$, both of which have co-ordination number of the silver ion 2.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2014-03-06T08:15:14.433" LastActivityDate="2014-03-06T08:15:14.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8674" PostTypeId="1" CreationDate="2014-02-23T08:15:36.183" Score="1" ViewCount="96" Body="&lt;p&gt;Is there a specific way to know how two or more arbitrary molecules will react? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example that prompted this question: $\ce{(HOC6H4)2CMe2 + 2 NaOH → (NaOC6H4)2CMe2 + 2 H2O}$ (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Polycarbonate#Production&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). How can you know how those molecules will react? Is it just a bunch of substitutions, with certain atoms in the reacting molecules reacting better than others? What factors would determine which atom in the molecule would react with the other molecule?&lt;/p&gt;&#xA;" OwnerUserId="4618" LastEditorUserId="1431" LastEditDate="2014-02-23T09:53:25.340" LastActivityDate="2014-02-23T09:53:25.340" Title="What aspects of a molecule's chemical makeup determine which portions of it will react?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8675" PostTypeId="2" ParentId="8674" CreationDate="2014-02-23T08:39:18.377" Score="2" Body="&lt;p&gt;Let me answer with a question (or two):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you identify in an electric circuit which parts are just the wiring and which are &lt;em&gt;functional parts&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How's that for a mechanical engine? Can you tell cogwheels and cranks from &quot;just the basic structure&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about &lt;strong&gt;pattern recognition&lt;/strong&gt; in the first to identify the &lt;strong&gt;functional groups&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step is to know how they react under certain conditions. As a rule of thumb, a group with a high electron density will probably react with something more positively charged (think &lt;em&gt;plus&lt;/em&gt; and &lt;em&gt;minus&lt;/em&gt; attract each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it takes quite a while and learning to predict exactly what will go on exactly in the reaction. But that not different from any other scientific/technical discipline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I tried to keep this as general and simple as possible ;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;Let me work you comment in:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Generally molecules don't contain ions though&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Says who? :D Molecules may contain ionic units, think in zwitterions of amino acids, think in all the mesoionic sydnones, münchnones and whatnot. Well, no. Forget about the special cases ;) But if it has more than one atom and is electrically neutral in total it is a molecule. At least in the less rigid sense of the definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes, the electronegativity is a good point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive me that I'm not going into detail why the example with fluorine and sodium is a not the best one, but in principle, you're right: Such a setup would allow to form a new $\ce{C-C}$ bond between a carbon that is negatively polarized (the one bearing the sodium) and another one that is positively polarized (the one bearing the halogene).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-23T09:50:41.867" LastActivityDate="2014-02-23T09:50:41.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8676" PostTypeId="2" ParentId="8672" CreationDate="2014-02-23T09:05:05.817" Score="8" Body="&lt;p&gt;This is not a chemical process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects typically expand when heated. Think in railway profiles in the summer, think in loosening a stuck nut by heating it with a blowtorch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the material is under stress and a bad heat conductor, unevenly heating will result in cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is a pity for the cup, the breaking of glassware to thousands of rounded (!) pieces upon uneven or punctual (mechanical) impact may be exactly what you want under different circumstances. Think in tempered (toughed) glass used in windows of railroad passenger cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, there's the &lt;em&gt;Bologna Bottle&lt;/em&gt;. In the making, these bottles are slowly cooled on the outside - but very rapidly on the inside, which maximizes the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, you might use the bottle as a hammer to drive a nail into something, but beware if you drop something &lt;strong&gt;in&lt;/strong&gt; the bottle. Here's a &lt;a href=&quot;http://www.youtube.com/watch?v=AMEmR-grjNA&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt; to demonstrate the effect. (&lt;em&gt;Be warned that there's a lot of screaming kids in it.&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-23T11:19:14.303" LastActivityDate="2014-02-23T11:19:14.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8678" PostTypeId="1" CreationDate="2014-02-23T11:20:05.960" Score="2" ViewCount="141" Body="&lt;p&gt;I've read that when a certain amount of UV light is shone on a metal surface, electrons are ejected (the photoelectric effect). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these electrons from the metal atoms themselves? and in case they are - what happens to the metal atoms seen as they have lost an electron? (Are they chemically different?)&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-02-23T20:07:37.143" Title="Photoelectric effect and electron loss" Tags="&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8680" PostTypeId="1" AcceptedAnswerId="8683" CreationDate="2014-02-23T17:23:52.443" Score="4" ViewCount="622" Body="&lt;p&gt;Different websites say different things on the matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; For example: &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/chemistrystudentfaqs/f/What-Is-Pitchblende.htm&quot; rel=&quot;nofollow&quot;&gt; This Website  &lt;/a&gt; Holds that pitchblende is also known by the name Uraninite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thefreedictionary.com/pitchblende&quot; rel=&quot;nofollow&quot;&gt;This Website  &lt;/a&gt; Holds that pitchblende is a type of Uraninite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, which website is correct? I have always known the first site to be a reputable source, so I assume that this is a contentious topic. &lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="73" LastEditDate="2014-02-23T20:07:51.353" LastActivityDate="2014-02-23T20:07:51.353" Title="Is pitchblende the same as Uraninite?" Tags="&lt;nomenclature&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8681" PostTypeId="2" ParentId="5737" CreationDate="2014-02-23T17:32:19.620" Score="31" Body="&lt;p&gt;It is very much possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s say you put 3 moles of $\ce{HCl}$ into 1 mole of water.&#xA;$\ce{HCl}$, being a strong acid dissociates completely into $\ce{H+}$ and $\ce{Cl-}$ ions as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl -&amp;gt; H+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so after complete dissociation, $[\ce{H+}]=3~\mathrm{mol/L}$   (ignoring the very tiny contribution from water itself)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition,&#xA;$$\mathrm{pH} = -\log[\ce{H+}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, $\mathrm{pH}= -\log 3= -0.48$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is very much possible to have solutions of  strong acids whose $\ce{[H+]}$ is 1 molar or more, and thus whose pH is negative.&lt;/p&gt;&#xA;" OwnerUserId="487" LastEditorUserId="7475" LastEditDate="2015-12-09T10:40:32.143" LastActivityDate="2015-12-09T10:40:32.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8683" PostTypeId="2" ParentId="8680" CreationDate="2014-02-23T18:19:51.943" Score="7" Body="&lt;p&gt;If you search online for Uraninite then you find various websites saying that uraninite is called pitchblende.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On wikipedia you find : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Uraninite is a radioactive, uranium-rich mineral and ore with a&#xA;  chemical composition that is largely UO2, but also contains UO3 and&#xA;  oxides of lead, thorium, and rare earth elements. It is most commonly&#xA;  known as pitchblende&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;even the word pitchblende will automatically redirect you to the uraninite wikipedia page .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another trustworthy website is mindat.org : uraninite information can be found on this link:&lt;a href=&quot;http://www.mindat.org/min-4102.html&quot;&gt;http://www.mindat.org/min-4102.html&lt;/a&gt; (It seems that there are various names associated to the mineral):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Re-named in 1845 by Wilhelm Karl von Haidinger for its composition&#xA;  containing uranium. Originally recognized before the element uranium&#xA;  was known and called &quot;schwarz beck-erz&quot; by Franz Ernst Brückmann in&#xA;  1727. Subsequently described with several names including &quot;pseudogalena&quot; and &quot;pitch-like zinc-blende&quot; by Johan Gottschalk&#xA;  Wallerius (Vallerius) in 1747. Named pechblende by Axel Cronstedt in&#xA;  1758. Named &quot;uranerz&quot; by Martin Klaproth after he discovered the element uranium in this mineral. Many spelling variations of the above&#xA;  names known.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-02-23T18:19:51.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8684" PostTypeId="1" CreationDate="2014-02-23T18:45:58.837" Score="2" ViewCount="331" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SkPge.jpg&quot; alt=&quot;red-ox reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the picture, the product is a neutral material, but in the illustration of the molecular structure, the compound is shown in its ions, why we can't show it in its ions even in the equation or if we can't, why in this molecular structure it is shown as ionic although it is neutral?&lt;/p&gt;&#xA;" OwnerUserId="4147" LastEditorDisplayName="user4076" LastEditDate="2014-02-23T22:45:01.423" LastActivityDate="2014-02-27T06:45:34.790" Title="what are reduction oxidation reactions?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8685" PostTypeId="2" ParentId="8684" CreationDate="2014-02-23T19:29:47.317" Score="2" Body="&lt;p&gt;Your questions involve a lot of chemical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An oxydo-reduction is reaction where electrons are exchanged between species including or not molecular rearrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You totally can write down any salt in the following way, emphasizing ion charges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{NaCl}_{(s)} \equiv \mathrm{Na^+Cl^-}_{(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you specify the phase, because when writing down an ionic form the chemist might assume that it is solvated into water. To underline it you may write the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Na^+Cl^-}_{(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further more solid generally have numerous possible distinct phases, therefore you may need to specify which crystalline structure the salt have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following equation is explicit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2\mathrm{Na}^{(0)}_{(s)} + \mathrm{Cl_{2(g)}} = 2\mathrm{Na^+Cl^-}_{(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And underlines that an electron is withdrawn from metallic sodium in order to produce chlorine. Anyway, this equation is global balance this it hides many details of the complete and complex chemical process, especially the formation of the crystalline structure. And it does not provide any information on conditions of reaction.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-02-24T14:36:12.860" LastActivityDate="2014-02-24T14:36:12.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8686" PostTypeId="2" ParentId="8678" CreationDate="2014-02-23T20:07:37.143" Score="2" Body="&lt;p&gt;It is NOT the amount of light (number of photons), it is the energy of each photon (its frequency).  When the energy of a photon exceeds the vacuum work function of the surface, an electron is unbound.  There obtains a charged surface and a free electron.  Unless the surface is part of a circuit (e.g., grounded or connected to the cathode), the next electron out must overcome the vacuum work function plus the electrical potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron is emitted from the conduction band of the bulk metal surface.  The positive charge is delocalized across the surface if it is of constant radius of curvature.   Surface electric field divergence is inversely proportional to radius of curvature.  Unless the surface is very smooth you can get cold cathode emission or electrical breakdown of the encapsulating medium.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-23T20:07:37.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8687" PostTypeId="1" AcceptedAnswerId="51197" CreationDate="2014-02-23T22:02:21.923" Score="13" ViewCount="10083" Body="&lt;p&gt;I have searched all over the web and found a lot of diverse explanations, but none of them are concluding exactly whether $\ce{FeS2}$ (solid - pyrite) is a covalent or an ionic compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From electronegativity, it should be covalent as the $\Delta\chi=0.7$ which is less than $1.5$ and thus said to make covalent bonds and therefore be a covalent compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the definition of ionic bonds, which are bonds between a metal and a non-metal element (whereas covalent bonds are bonds between non-metal elements), it should be an ionic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone know which of those is 'true', or better, if there is another, more detailed explanation?&lt;/p&gt;&#xA;" OwnerUserId="4619" LastEditorUserId="7475" LastEditDate="2016-05-26T10:22:22.510" LastActivityDate="2021-03-27T00:50:06.767" Title="Is pyrite (FeS₂) an ionic or a covalent compound?" Tags="&lt;bond&gt;&lt;ionic-compounds&gt;&lt;transition-metals&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8688" PostTypeId="1" CreationDate="2014-02-24T02:47:05.000" Score="0" ViewCount="1387" Body="&lt;p&gt;The half-life of the first order reaction, A → products, is 33.8 s. What was the original concentration, [A]O in mol/L, if after 2.5 minutes, the concentration of A is 0.0796 mol/L? Round your answer to three significant figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the formular is 1/(A)t = kt + 1/(A)0  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried 1/33.8s = (0.0796)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so it = 1/0.0796.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is not correct, but I think I'm using the right formula.  I looked back at my notes and nothing is really making any sense, but this formula.&lt;/p&gt;&#xA;" OwnerUserId="4620" LastEditorUserId="4945" LastEditDate="2014-10-24T13:26:52.973" LastActivityDate="2014-10-24T13:26:52.973" Title="How to calculate the start concentration for a first order reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8689" PostTypeId="1" CreationDate="2014-02-24T03:38:48.880" Score="2" ViewCount="328" Body="&lt;p&gt;For a very large-scale project, I am trying to create a greenhouse gas demonstration device, which rapidly produces greenhouse gases from atmospheric air using only electricity, or any chemical reactions or chains of chemical reactions that can attain their input energy from electricity. At the intake would be a very powerful fan which rapidly forces air through the system, to speed up the process. How would it be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking nitrogen oxides would be best, considering the atmosphere is approximately 78% nitrogen and 21% oxygen. But what would be the fastest way to produce them, assuming price is not really an issue?&lt;/p&gt;&#xA;" OwnerUserId="4621" LastEditorUserId="4621" LastEditDate="2014-03-02T14:53:35.427" LastActivityDate="2015-07-28T16:37:30.753" Title="How could I create a machine that as rapidly as possible creates greenhouse gases from air?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8690" PostTypeId="2" ParentId="8689" CreationDate="2014-02-24T06:35:09.270" Score="1" Body="&lt;p&gt;I would think that carbon dioxide and methane would be your options due to their abundance. However, they are very difficult to harvest because of their non-reactivity. There are research groups working with metallo-organic frames/cages that capture carbon dioxide. However, making these takes time and significant funds. In addition, they are not that great when capturing carbon dioxide from atmospheric air. You have to take into account that despite of their abundance, the amount of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere is still small. I think the frame works will first be installed on factories funnels which emit large quantities of carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorDisplayName="user4076" LastEditDate="2014-02-24T11:37:11.650" LastActivityDate="2014-02-24T11:37:11.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8691" PostTypeId="1" AcceptedAnswerId="8696" CreationDate="2014-02-24T08:04:59.057" Score="10" ViewCount="1964" Body="&lt;p&gt;Every definition I see of Van der Waals interactions claim that they are short-range, nonspecific interactions between two chemical species. What does &quot;nonspecific&quot; mean in this context?&lt;/p&gt;&#xA;" OwnerUserId="4622" LastEditorUserId="4622" LastEditDate="2020-01-15T21:50:27.860" LastActivityDate="2020-01-15T21:50:27.860" Title="Van der Waals nonspecific interaction definition" Tags="&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8692" PostTypeId="2" ParentId="8688" CreationDate="2014-02-24T08:15:56.343" Score="3" Body="&lt;p&gt;For a first order reaction $\ce{A -&amp;gt;[k]P}$, the rate constant is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k[A] = - \frac{d[A]}{dt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at any time $t$, the concentation $[A]_t$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[A]_t = [A]_0 \cdot e^{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the half-life time $t_{1/2}$, $[A]_t = \frac{1}{2}[A]_0$, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you agree that we can calculate $k$ using $k = \frac{ln 2}{t_{1/2}}\ $?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing $k$ and $[A]_t$, calculation of $[A]_0$ doesn't seem impossible now ;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="9468" LastEditDate="2014-10-24T09:43:44.603" LastActivityDate="2014-10-24T09:43:44.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8693" PostTypeId="2" ParentId="8689" CreationDate="2014-02-24T08:54:04.290" Score="7" Body="&lt;p&gt;Current concentrations of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere are around 390 ppmv and 1.75 ppmv, resp. These concentrations and the wish to obtain large quantities of both gases rapidly seem mutually exclusive, unless you use (at least one) additional carbon source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the options for interconversion into account, $\ce{CH4}$ in pressurized bottles would be an obvious choice ;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-24T08:54:04.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8694" PostTypeId="2" ParentId="8687" CreationDate="2014-02-24T11:37:01.490" Score="8" Body="&lt;p&gt;Your definition of ionic bonding is incorrect as non-metallic polyatomic cations (e.g ammonium) form ionic compounds. Ionic bonding is the electrostatic attraction of cations and anions forming a regular giant lattice. Transition metal compounds do not have simple close packing ionic lattices as they prefer &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_complex&quot; rel=&quot;noreferrer&quot;&gt;coordination compounds&lt;/a&gt; due to the energies of d-orbitals.&#xA;Here is the pyrite structure:&lt;img src=&quot;https://i.stack.imgur.com/SH3HM.png&quot; alt=&quot;pyrite structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the iron has octahedral geometry and sulfur has tetrahedral. These are the most common geometries in coordination chemistry.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2014-02-24T11:37:01.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8695" PostTypeId="2" ParentId="8691" CreationDate="2014-02-24T11:39:36.207" Score="5" Body="&lt;p&gt;There are 3 forces that could be named as van der Waals' force (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot;&gt;wikipedia article&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is common between them? The common between them is that these interactions are electrostatic interactions between dipoles of different nature (permanent dipoles, induced dipoles). And electrostatic interaction is not specific: there is no space specificity, or chemical specificity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electrostatic interaction between dipoles according to Coulomb law decays as $$\sim 1/r^6$$, that is why van der Waals' forces (&lt;a href=&quot;http://en.wikipedia.org/wiki/Keesom_force#Keesom_force&quot;&gt;Keesom force&lt;/a&gt;,&lt;a href=&quot;http://en.wikipedia.org/wiki/Debye_force#Debye_.28induced_dipole.29_force&quot;&gt;Debye force&lt;/a&gt;,&lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot;&gt;London dispersion&lt;/a&gt;) force are short-range.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-02-24T12:08:25.407" LastActivityDate="2014-02-24T12:08:25.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8696" PostTypeId="2" ParentId="8691" CreationDate="2014-02-24T11:42:46.957" Score="13" Body="&lt;p&gt;The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_interactions&quot;&gt;Wikipedia article on the topic&lt;/a&gt;) is much more *&lt;em&gt;specific&lt;/em&gt; about what is and is not a van der Waals force. This definition originates in the International Union of Pure &amp;amp; Applied Chemistry &lt;em&gt;Compendium of Chemical Terminology&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/V06597.html&quot;&gt;van der Waals force&lt;/a&gt; &lt;code&gt;is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a van der Waals force is a catch-all term to include all other very weak interactions between atoms and molecules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Keesom_force#Keesom_.28permanent_dipole.29_force&quot;&gt;Dipole-dipole interactions&lt;/a&gt; between permanent dipoles &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Keesom_force#Debye_.28induced_dipole.29_force&quot;&gt;Dipole-induced dipole interactions&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Keesom_force#London_dispersion_force&quot;&gt;Induced dipole - induced dipole interactions&lt;/a&gt; (London dispersion)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pi_stacking&quot;&gt;$\pi-\pi$ stacking&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Halogen_bond&quot;&gt;Halogen bonding&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Repulsive forces derived from atoms trying to be inside another atom's van der Waals radius (&lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_strain&quot;&gt;Steric strain&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Eclipsed_conformation&quot;&gt;torsional strain&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-24T11:42:46.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8697" PostTypeId="1" AcceptedAnswerId="8698" CreationDate="2014-02-24T12:01:46.380" Score="6" ViewCount="748" Body="&lt;p&gt;I'm curious as to the role of molecular oxygen in Suzuki homocoupling reactions, as discussed &lt;a href=&quot;https://pubs.acs.org/doi/abs/10.1021/ja0569959&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a standard Suzuki cross-coupling reaction, molecular oxygen is not required. However, this paper claims that the homocoupling reaction will not proceed in an Argon environment. Is there a (relatively) straight forward justification for this that I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reaction mechanism from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://pubs.acs.org/appl/literatum/publisher/achs/journals/content/jacsat/2006/jacsat.2006.128.issue-21/ja0569959/production/images/medium/ja0569959n00001.gif&quot; alt=&quot;mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4623" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.613" LastActivityDate="2014-02-24T12:56:16.947" Title="Molecular Oxygen in Suzuki Homocoupling" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8698" PostTypeId="2" ParentId="8697" CreationDate="2014-02-24T12:41:57.520" Score="5" Body="&lt;p&gt;I don't remember a direct palladation of arylboronic acids or their derivatives in any of the couplings. Typically, $\ce{Pd(0)}$ inserts into the $\ce{C-Hal}$ bond of arylhalides and is thus oxidized to $\ce{Pd(II)}$. Ouite often, $\ce{Hal}$ is then displaced with $\ce{OR}$ by the action of a base. Think in $\ce{KOt-Bu}$ often used in the &lt;em&gt;Kumada&lt;/em&gt; coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the homo coupling that you described here, there is - for obvious reasons - no $\ce{Ar-Hal}$ for the &lt;em&gt;oxiative addition&lt;/em&gt; step. Without $\ce{O2}$ for the oxidative activation of $\ce{Pd(Ph3P)2}$, the mixture will thus remain inactive.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-24T12:41:57.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8700" PostTypeId="2" ParentId="8689" CreationDate="2014-02-24T16:31:14.337" Score="6" Body="&lt;p&gt;The most powerful Greenhouse Effect gas of all is water.  Run a vaporizer to 100% humidity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UmiBZ.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;BR&gt;&#xA; &lt;BR&gt;&#xA;&lt;a href=&quot;http://disc.sci.gsfc.nasa.gov/featured-items/images/fig2_air_infrared_spectrum&quot;&gt;http://disc.sci.gsfc.nasa.gov/featured-items/images/fig2_air_infrared_spectrum&lt;/a&gt; &lt;BR&gt;&#xA;to 50 microns &lt;BR&gt;&#xA;&lt;a href=&quot;http://earthobservatory.nasa.gov/Features/Tyndall/Images/absorption.gif&quot;&gt;http://earthobservatory.nasa.gov/Features/Tyndall/Images/absorption.gif&lt;/a&gt; &lt;BR&gt;&#xA;to 15 microns, near IR &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greenhouse effect incremental champ is a tank of $\ce{SF6}$ (insulation and arc quench for high voltage electrical switches and transformers).  Note that industrial grade is often contaminated with $\ce{SF4}$ that is extremely toxic.  However, your job is already being done for you,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.afeas.org/greenhouse_gases.html&quot;&gt;http://www.afeas.org/greenhouse_gases.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Greenhouse_gas&quot;&gt;http://en.wikipedia.org/wiki/Greenhouse_gas&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.eia-international.org/explosion-of-super-greenhouse-gases-expected-over-next-decade&quot;&gt;http://www.eia-international.org/explosion-of-super-greenhouse-gases-expected-over-next-decade&lt;/a&gt; &lt;BR&gt;&#xA;HFC-23, by-product of HCFC-22 manufacture. is 14,800 times $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-24T16:31:14.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8701" PostTypeId="1" AcceptedAnswerId="8704" CreationDate="2014-02-24T20:25:51.937" Score="6" ViewCount="363" Body="&lt;p&gt;I am not sure if this is a chemistry question; however, I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my desk I have two plastic, zip-lock bags.  One contains cinnamon tea, the other contains nuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After about a week of these bags being next to each other, the flavor of the cinnamon somehow leached through both bags and now the nuts taste like cinnamon!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a chemical explanation for this or isn't this a question of chemistry?&lt;/p&gt;&#xA;" OwnerUserId="4625" LastEditorUserId="418" LastEditDate="2014-02-24T20:59:17.880" LastActivityDate="2014-02-24T21:04:21.687" Title="Flavor transference through the air and plastic bags" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8702" PostTypeId="2" ParentId="8701" CreationDate="2014-02-24T20:28:51.080" Score="0" Body="&lt;p&gt;Try taking the nuts out and putting them in a bowl or cup for a little. See if you just smelling the Cinnamon from the bag is making you &quot;taste&quot; it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just an idea.&lt;/p&gt;&#xA;" OwnerUserId="4510" LastActivityDate="2014-02-24T20:28:51.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8703" PostTypeId="2" ParentId="8701" CreationDate="2014-02-24T21:02:32.600" Score="2" Body="&lt;p&gt;Cinnamaldehyde dissolves in and diffuses through most polymers within sensory threshold. Use glass containers with lined metal caps.  Isn't ground up tree bark amazing?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-24T21:02:32.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8704" PostTypeId="2" ParentId="8701" CreationDate="2014-02-24T21:04:21.687" Score="6" Body="&lt;p&gt;Cinnamon tea is most likely flavoured with cinnamalaldehyde (&lt;strong&gt;1&lt;/strong&gt;) and eugenol (&lt;strong&gt;2&lt;/strong&gt;), the main components in cinnamon oil. The latter compound also gives cloves the characteristic aroma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZDUL6.png&quot; alt=&quot;cinnamon aroma&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, ziplock bags are made from polyethylene. They are not meant to be aroma-tight, like vacuum-sealed coffee packages.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-24T21:04:21.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8705" PostTypeId="1" AcceptedAnswerId="8706" CreationDate="2014-02-24T22:16:00.200" Score="7" ViewCount="534" Body="&lt;p&gt;&lt;strong&gt;The Equation of Interest&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy equation derived from the &quot;Particle in a 1-D box&quot; problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=\dfrac{n^2h^2}{8mL^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the principle quantum number, $h$ is Planck's constant, $m$ is the mass of the particle (e.g. electron), and $L$ is the length of region of space (i.e. box).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Goal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to illustrate the harmonious nature of this energy equation derived from the quantum mechanical nature of the system and classical mechanics.  In theory we should be able to define a mass and region of space for a clear comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Example: Electron vs. Car&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compute the lowest possible energy of an electron using the equation above.  We define $m=9.11\times10^{-31} \textrm{kg}$ and the region of space as a reasonable value for the diameter of an atom, $L=0.2 \textrm{nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into our equation along with Planck's constant ($h=6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1}$) we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\textrm{electron}} = \dfrac{(6.626\times10^{-34} \textrm{ m}^2 \textrm{kg s}^{-1})^2}{8 *9.11\times10^{-31} \textrm{ kg} * (2\times 10^{-10} \textrm{m})^2} = 1.51 \times 10^{-18} \textrm{ J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us imagine a car with $m=2000 \textrm{ kg}$ and a region of space as $L=1 \textrm{ m}$.  The resulting energy is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\textrm{car}}=2.74\times10^{-71} \textrm{ J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in both cases, the energy is so small that we might as well say its zero (from a classical standpoint anyway).  Of course $E_{\textrm{car}} &amp;lt;&amp;lt; E_{\textrm{elecron}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;em&gt;So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can, however, convert these energies (which are purely kinetic energies) into something which is clearly distinguishable such as velocity by using the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \dfrac{1}{2}mv^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = \sqrt{\dfrac{2*E}{m}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass of the object and $v$ is the velocity.  If we plug in our previously determined energies we get the following results...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\textrm{electron}} = 1.8\times 10^6 \textrm{ m s}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{\textrm{car}} = 1.7\times 10^{-37} \textrm{ m s}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we see that the velocity of the car is essentially zero (or technically speaking, so slow that it wouldn't ever be observed as moving in billions of lifetimes).  This jives with classical theory.  However, we see that the velocity of the electron is staggering, even though the energy of the electron is 'essentially zero'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Conclusions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to properly relate the energy equation derived from the quantum mechanical system of a particle in a 1-D box, one &lt;em&gt;cannot&lt;/em&gt; simply compare the raw energies of the particle (i.e. electron) to that of an ordinary object such as a car because both energies result in a number that is essentially zero which defeats the point.  As such, one must relate the energy to something appropriate such as the velocity of the object to properly illustrate how the energy equation still relates to everyday life.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are my conclusions correct or can the raw energetics simply be used to make a proper comparison between quantum/classical theory without resorting to recasing the results into a physical property such as velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;#not-homework&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="865" LastEditDate="2015-10-30T17:42:19.523" LastActivityDate="2015-10-30T17:42:19.523" Title="How to properly relate quantum mechanical energy to the classical world" Tags="&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8706" PostTypeId="2" ParentId="8705" CreationDate="2014-02-24T23:20:18.727" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;*So we cannot stop here as quantum theory and classical mechanics are virtually indistinguishable (with respect to raw energies) at this point correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is wrong. Just because you can't immediately, instinctively comprehend the difference between the $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$ and therefore think they are both &quot;indistinguishably small&quot; doesn't mean it's true. The difference is &lt;strong&gt;massive&lt;/strong&gt;. Our brains just haven't evolved to deal with numbers whose logarithms are very far from 0, because for almost all of our existence as a species, it never had to. Perhaps &lt;a href=&quot;http://mrob.com/pub/math/largenum.html&quot;&gt;this site&lt;/a&gt; may give some very qualitative though fun context on how brains perceive the magnitude of a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to cope with this limitation of ours, we have developed some strategies. One can be understood as a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chunking_%28psychology%29&quot;&gt;chunking&lt;/a&gt;, where we simply modify numbers in a way so they become closer to something we understand. You probably won't be able to immediately give an example of an object that weighs 200,000 ounces or $3\times 10^{-27}$ solar masses, though you may find it quite a bit easier if I say 6 tonnes instead. This is because even if we inherently &lt;em&gt;know&lt;/em&gt; the concept of an ounce or a solar mass, we can't really comprehend the magnitudes of 200,000 or $3\times 10^{-27}$ (the latter of course being much worse), and in order to get our heads around it more easily, we instead invent a new unit to measure weights in which we can express the same information using numbers closer to 1. The whole idea of scientific notation for numbers and SI prefixes for units can be understood as a consequence of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is the one you use; instead of reframing the units and numbers, &lt;em&gt;reframe the situation&lt;/em&gt;. You may feel there is no difference between $10^{-18}\  \mathrm{J}$ and $10^{-71}\  \mathrm{J}$, but the 53 orders of magnitude in difference &lt;strong&gt;is about the same as&lt;/strong&gt; &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%28atoms%20in%20observable%20universe%29/%28atoms%20in%20human%20body%29&quot;&gt;the number of atoms in a human body and the number of atoms in the entire observable Universe&lt;/a&gt;! Notice how I simply took the ratio into a slightly more familiar context (you know the Earth is just a small part of the observable Universe, and a human being is puny in comparison to the Earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all of your calculations are correct, but if &lt;em&gt;Homo sapiens&lt;/em&gt; were smarter, they would be entirely unnecessary!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-02-24T23:29:07.140" LastActivityDate="2014-02-24T23:29:07.140" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8708" PostTypeId="2" ParentId="8705" CreationDate="2014-02-25T00:41:59.003" Score="3" Body="&lt;p&gt;You are treating the car as an elemental particle without inner structure, where the only degree of freedom is the overall translation.  Within this model, it is true that QM seems to produce very little effect for cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, cars that have only overall movement but no other structure is not at all useful.  You would not call something a &quot;car&quot; if it does not have internal combustion.  You need materials with different properties to construct different mechanical and electronic components of the car to make it useful, safe and comfortable.   None of these are possible without quantum mechanical effects.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though combustion and ignition seem quite violent and energetic reactions, on a molecular level it is quite rare to achieve energies E-18J.  In fact, using Bolzmann distribution, one find that such energy level actually correspond to extremely high temperature of almost 60000K, that is 10 times the temperature of the center of the sun.   Of course, we rarely actually deal with electron in boxes, and in chemical reactions one is dealing with the CHANGE of energy, not the energies themselves, so normally one come across energy figures magnitudes smaller than that, but nevertheless quantum mechanical effect is enormous on a chemical level.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, without quantum mechanical effect, there will be no stable atoms, no stable molecules, no reaction and no chemistry.   It will just be a bunch of chaotic movement of particles that no one can understand, and there will be no computers to help either.  In fact, there would be no life as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be baffled at the thought of this, because rarely does such small number seem to matter in life.  In life, when we think about energy, we think about events that only happen a small number of times, like monthly electricity bills or a high MPG car, or say the electricity supply of a billion household.  But the number of molecules is much greater, usually ~10^23, and each carrying a piece of quantum mechanical effect.  Put together, a small energy on a microscopic level can be magnified to reach quite amazing results.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2014-02-25T00:41:59.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8709" PostTypeId="1" AcceptedAnswerId="8815" CreationDate="2014-02-25T01:11:54.697" Score="7" ViewCount="666" Body="&lt;p&gt;I am running an immunoassay using biotinylated antibodies and have noticed a decreasing trend from duplicate assay runs (completed within minutes of each other) from the same vial.&lt;/p&gt;&#xA;&lt;p&gt;In between each run:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I switch to a new low protein binding pipette tip.&lt;/li&gt;&#xA;&lt;li&gt;I vortex the biotinylated antibody in solution in a low protein binding vial.&lt;/li&gt;&#xA;&lt;li&gt;Then I prewet the pipette tip and deliver the reagent.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Note that degradation seems unlikely.&lt;/p&gt;&#xA;&lt;p&gt;Any ideas on what may cause this?&#xA;Could it be vortexing or shearing during pipette tip mixing?&lt;/p&gt;&#xA;" OwnerUserId="4645" LastEditorUserId="41328" LastEditDate="2020-08-25T04:38:48.493" LastActivityDate="2020-08-25T04:38:48.493" Title="Can I damage a biotinylated antibody in solution by vortexing or mixing too vigorously?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8710" PostTypeId="1" AcceptedAnswerId="8711" CreationDate="2014-02-25T01:12:05.913" Score="5" ViewCount="2856" Body="&lt;p&gt;In the redox reaction, where &lt;strong&gt;$\ce{HCl}$ is the excess reactant&lt;/strong&gt; do these reactions produce following half reactions?&#xA;\begin{aligned}&#xA;(1)&amp;amp;&amp;amp;\ce{SnCl2 &amp;amp;-&amp;gt; Sn^{4+} + 2e-}\\&#xA;(2)&amp;amp;&amp;amp;\ce{8H+ + KMnO4 + 5e- &amp;amp;-&amp;gt; Mn^{2+} + 4H2O}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the $\ce{HCl}$ come into play here, or do you consider it at the end when you have a final equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a redox reaction where $\ce{KMnO4}$ was added to $\ce{HCl}$ and $\ce{SnCl2}$ (hydrochloric acid was simply used as an excess reagent)&lt;/p&gt;&#xA;" OwnerUserId="4628" LastEditorUserId="4945" LastEditDate="2014-06-16T18:13:36.997" LastActivityDate="2014-06-16T18:13:36.997" Title="What are the products for the redox reaction HCl + SnCl₂ + KMnO₄?" Tags="&lt;redox&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8711" PostTypeId="2" ParentId="8710" CreationDate="2014-02-25T01:53:59.777" Score="2" Body="&lt;p&gt;$$\ce{KMnO4 + SnCl2 + HCl -&amp;gt; KCl + MnCl2 + SnCl4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation states are for reactants:&#xA;\begin{aligned}\ce{&#xA;K  &amp;amp;= +1\\&#xA;Mn &amp;amp;= +7\\&#xA;O  &amp;amp;= -2 \\&#xA;Sn &amp;amp;= +2\\&#xA;Cl &amp;amp;= -1\\&#xA;H  &amp;amp;= +1\\&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation states for products:&#xA;\begin{aligned}\ce{&#xA;K  &amp;amp;= +1\\&#xA;Cl &amp;amp;= -1\\&#xA;Mn &amp;amp;= +2\\&#xA;Sn &amp;amp;= +4\\&#xA;H  &amp;amp;= +1\\&#xA;O  &amp;amp;= -2\\&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a redox reaction carried out in acidic solution. $\ce{SnCl2}$ and $\ce{KMnO4}$ are titrated with hydrochloric acid. What is being oxidized and what is being reduced? Potassium is being reduced, magnesium and tin are oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write out our half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;HCl + SnCl2 &amp;amp;-&amp;gt; SnCl\\&#xA;HCl + KMnO4 &amp;amp;-&amp;gt; MnCl2 + KCl + H2O&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the oxidation states of our reactants, and products, $\ce{H}$ has +1, $\ce{Cl}$ -1, and for the products $\ce{Sn}$ +2, $\ce{Cl}$ -1 for the first reaction. The first reaction gains two electrons and thus is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 2nd reaction, $\ce{H}$ again has +1 and $\ce{Cl}$ -1. Potassium has +1 and the permanganate ion has -2. Specifically, $\ce{Mn}$ has a theoretical oxidation state of +7 and oxygen -2. In the products side, magnesium has an oxidation state of +2 and $\ce{Cl}$ -1, $\ce{H}$ is +1, $\ce{O}$ is 2-, and $\ce{K}$ becomes +1. Chlorine as an element has an oxidation state of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For redox reactions, we have to make sure our reactants and products are first balanced for our elements other than H and O in acidic solution. It is clear that the chlorines are not balanced, therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;2HCl + SnCl2 &amp;amp;-&amp;gt; SnCl4 \\&#xA;3HCl + KMnO4 &amp;amp;-&amp;gt; MnCl2 + KCl + H2O&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We balance oxygen atoms with adding molecules of water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;2HCl + SnCl2 &amp;amp;-&amp;gt; SnCl4\\&#xA;3HCl + KMnO4 &amp;amp;-&amp;gt; MnCl2 + KCl + 4H2O &#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add H+ to balance H:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;2HCl + SnCl2 &amp;amp;-&amp;gt; SnCl4 + 2H+\\&#xA;3HCl + KMnO4 + 5H+ &amp;amp;-&amp;gt; MnCl2 + KCl + 4H2O &#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now add electrons to balance charge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;2HCl + SnCl2 + 2e- &amp;amp;-&amp;gt; SnCl4 + 2H+\\&#xA;3HCl + KMnO4 + 5H+ &amp;amp;-&amp;gt; MnCl2 + KCl + 4H2O + 5e- &#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We multiply our half-reactions by integers so that the electrons are the same number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;10HCl + 5SnCl2 + 10e- &amp;amp;-&amp;gt; 5SnCl4 + 10H+\\&#xA;6 HCl + 2KMnO4 + 10H+ &amp;amp;-&amp;gt; 2MnCl2 + 2KCl + 8H2O + 10e-&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add 'em up and cancel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;16HCl + 5SnCl2 + 2KMnO4 &amp;amp;-&amp;gt; 5SnCl4 + 2MnCl2 + 2KCl + 8H2O&#xA;}\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2014-06-16T02:43:42.937" LastActivityDate="2014-06-16T02:43:42.937" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="8713" PostTypeId="2" ParentId="8705" CreationDate="2014-02-25T03:20:22.377" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;without resorting to recasing the results into a physical property such as velocity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The energy is a perfectly physical property. In fact, for many quantum mechanical systems, the energy is the easiest physical property to measure. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;can the raw energetics simply be used to make a proper comparison between quantum/classical theory &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the scales of the energies are sufficient to make a comparison between quantum and classical systems. As you and Nicolau Saker Neto suggest in teh comments to &lt;a href=&quot;https://chemistry.stackexchange.com/a/8706/323&quot;&gt;his answer&lt;/a&gt;, convert the energy of the electron into eV: $1.5\times 10^{-18}\mathrm{J} \times \frac{1\mathrm{eV}}{1.6\times 10^{-19}\mathrm{J}} \approx 10 \mathrm{eV}$. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_atom#Solution_of_Schr.C3.B6dinger_equation:_Overview_of_results&quot; rel=&quot;nofollow noreferrer&quot;&gt;ground state of the Hydrogen atom&lt;/a&gt; is $-13.6 \mathrm{eV}$, comparable to the electron in the $2.0\mathrm{nm}$ box. So the energy you calculated for the electron is similar to that of most of chemistry, which is dominated by quantum mechanics. So that's a reasonably-sized quantum mechanical energy. But it's not noticeable compared to the joules of macroscopically-sized objects moving at macroscopic speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $1.7\times 10^{-37}\mathrm{J} \approx 1.0 \times 10^{-18}\mathrm{eV}$, on the other hand, is negligible compared to both macroscopic physics and to ordinary chemistry. &lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-02-25T03:20:22.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8714" PostTypeId="2" ParentId="8710" CreationDate="2014-02-25T08:39:10.843" Score="6" Body="&lt;p&gt;When using $\ce{MnO4-}$ as an oxidant, pH is &lt;strong&gt;crucial&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In acidic medium: $\ce{Mn(VII) -&amp;gt; Mn(II)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{8 H3O+ + MnO4^{-} + 5 e- -&amp;gt; Mn^{2+} + 12 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under neutral conditions: $\ce{Mn(VII) -&amp;gt; Mn(IV)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O + MnO4^{−} + 3 e− -&amp;gt; MnO2 + 4 OH^{−}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In strongly alkaline medium: $\ce{Mn(VII) -&amp;gt; Mn(VI)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4^{−} + e− -&amp;gt; MnO4^{2−}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes &#xA;$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-25T08:39:10.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8715" PostTypeId="1" CreationDate="2014-02-25T10:01:52.297" Score="7" ViewCount="7821" Body="&lt;p&gt;I have dilutions of different samples, which are made with 0.1 M $\ce{HNO3}$. Only I left them on the lab table for a few days and one turned yellow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is why would one turn yellow and not the other ones? They all have the same concentration of sample. The matrix in all the samples are the same. The difference is the concentration of $\ce{Cl-}$ in the samples. The one with the lowest concentration of $\ce{Cl-}$ turned yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas why?&lt;/p&gt;&#xA;" OwnerUserId="4633" LastEditorUserId="418" LastEditDate="2014-02-25T10:35:06.487" LastActivityDate="2016-02-17T13:28:15.683" Title="Why does nitric acid not turn yellow everytime?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8716" PostTypeId="2" ParentId="8715" CreationDate="2014-02-25T10:33:31.947" Score="17" Body="&lt;p&gt;Pure nitric acid is colourless, the brown colour is caused by dissolved nitrogen dioxide, $\ce{NO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen dioxide is formed by the slow decomposition of nitric acid, according to $$\ce{4 HNO3 -&amp;gt; 4 NO2 + 2 H2O + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen dioxide is a good oxidant and it does so - supposed that it finds a suitable partner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be chloride. So, the more chloride in solution the more redox chemistry furnishing colourless products.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-25T14:19:34.843" LastActivityDate="2014-02-25T14:19:34.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8717" PostTypeId="1" CreationDate="2014-02-25T15:16:06.160" Score="39" ViewCount="107357" Body="&lt;p&gt;I was reading about the p-block elements and found that the inert pair effect is mentioned everywhere in this topic. However, the book does not explain it very well. So, what is the inert pair effect? Please give a full explanation (and an example would be great!).&lt;/p&gt;&#xA;" OwnerUserId="4104" LastEditorUserId="194" LastEditDate="2018-03-01T18:48:05.553" LastActivityDate="2018-03-01T18:48:05.553" Title="What is the inert pair effect?" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;periodic-trends&gt;&lt;relativistic&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8719" PostTypeId="1" AcceptedAnswerId="8721" CreationDate="2014-02-25T18:21:24.310" Score="6" ViewCount="523" Body="&lt;p&gt;Recently I noticed that some of our glasses and cutlery had developed a blue-ish deposit as you can see for the two sets of spoons below. For each set the spoon on the left has the blue-ish deposit whereas the spoon on the right is the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Xfn4.png&quot; alt=&quot;Blue-ish deposit on spoons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I traced the cause of this blue-ish deposit to a specific effect. It only occurs when the spoon or glass has been used for a medicine called Questran-A, which is a powder that needs to be dissolved in water and then ingested. And more specifically, it only happens when these glasses/spoons are subsequently washed in our dishwasher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to believe that some reaction occurs between the (left-overs) of the medicine and the dishwasher tablets. My question is: &lt;strong&gt;does anyone know what reaction between the chemicals in the medicine and those in the dishwasher tablets could cause this blue-ish deposit?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this will be impossible to say without the ingredients of both, so here is a list of the contents of the dishwasher tablets and the medicine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cmgtcontent.ahold.com.kpnis.nl/cmgtcontent/media//000682400/000/000682452_001_ah_detergenten_nov2010.pdf&quot; rel=&quot;noreferrer&quot;&gt;Dishwasher tablets (Albert Heijn)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pentasodium triphosphate &lt;/li&gt;&#xA;&lt;li&gt;sodium carbonate and sodium carbonate peroxide&lt;/li&gt;&#xA;&lt;li&gt;taed &lt;/li&gt;&#xA;&lt;li&gt;sodium silicate &lt;/li&gt;&#xA;&lt;li&gt;PEG-90 &lt;/li&gt;&#xA;&lt;li&gt;alcohols, C16-18, ethoxylated mind 30 EO &lt;/li&gt;&#xA;&lt;li&gt;acrylic copolymer &lt;/li&gt;&#xA;&lt;li&gt;PEG-4 &lt;/li&gt;&#xA;&lt;li&gt;maleic acid/acrylic acid copolymer,sodium salt &lt;/li&gt;&#xA;&lt;li&gt;alcohols, C12-18 ethoxylated propoxylated &lt;/li&gt;&#xA;&lt;li&gt;zinc sulfate&lt;/li&gt;&#xA;&lt;li&gt;beta-alanine, N-(2-carboxyethyl)-, N-coco alkyl derivs, disodium salt&lt;/li&gt;&#xA;&lt;li&gt;mn-complex &lt;/li&gt;&#xA;&lt;li&gt;benzotriazole &lt;/li&gt;&#xA;&lt;li&gt;subtilisin &lt;/li&gt;&#xA;&lt;li&gt;tetrasodium etidronate &lt;/li&gt;&#xA;&lt;li&gt;alpha-amylase &lt;/li&gt;&#xA;&lt;li&gt;benzisothiazolinone&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://db.cbg-meb.nl/Bijsluiters/h14276.pdf&quot; rel=&quot;noreferrer&quot;&gt;Questran-A&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 mg colestyramine&lt;/li&gt;&#xA;&lt;li&gt;30 mg Aspartame(E 951) equal to 17 mg Phenylalanine&lt;/li&gt;&#xA;&lt;li&gt;Citric acid(E 330)&lt;/li&gt;&#xA;&lt;li&gt;kelcoloïd&lt;/li&gt;&#xA;&lt;li&gt;Silicon dioxide(E551)&lt;/li&gt;&#xA;&lt;li&gt;Orange flavoring&lt;/li&gt;&#xA;&lt;li&gt;Xantham gum (E 415).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-02-26T07:22:19.890" Title="What chemical reaction can cause this blue-ish deposit?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8721" PostTypeId="2" ParentId="8719" CreationDate="2014-02-25T19:02:23.967" Score="5" Body="&lt;p&gt;Let's speculate a bit ;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Colestyramine is a twodimensional polymer network, build from styrene and and 1,4-divinylbenzene. &lt;/li&gt;&#xA;&lt;li&gt;The polymer has a lot of benzyltrimethylammonium groups. That means it can act as an anion exchanger.&lt;/li&gt;&#xA;&lt;li&gt;It is used to &lt;strong&gt;strongly&lt;/strong&gt; bind bile acids&lt;/li&gt;&#xA;&lt;li&gt;Neither the polymer itself, nor the adduct with bile acid are supposed to degrade in the bowel. The loaded polymer is supposed to be excreted.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The dishwasher tablets, on the other hand, contain the sodium salt of a maleic acid/acrylic acid copolymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means we have two polymers, one with a lot of $\ce{Ph-CH2N^+(CH3)3}$ groups and another one with lots of $\ce{-COO-}$ groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like a &lt;em&gt;mass wedding&lt;/em&gt; of two chain gangs :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is conceivable that this deposits a thin film on the spoons and the glassware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to explain the bluish colour, we have two options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The polymer aggregate absorbs in the 580 nm range - we observe the complementary colour. Due to lack of conjugation, I don't see how this could happen here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The colour results from some dispersion effect related to opalescence, as suggested in Michiel's comment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-26T07:22:19.890" LastActivityDate="2014-02-26T07:22:19.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8722" PostTypeId="1" CreationDate="2014-02-25T19:30:31.280" Score="11" ViewCount="1813" Body="&lt;p&gt;Usually the first step in describing the electronic wave function of a molecule or atom is to describe it with a single Slater determinant. I get that this is an independent particle approximation, and that in the context of Hartree-Fock, each electron moves in the average field of the other electrons. That's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when we correlate the electrons, this is done by generating excitations out of the Slater determinant, and then taking weighted combinations of them. For example, configuration interaction mixes the ground state Slater determinant with singly, doubly, triply, etc excited determinants. A fancier way of doing this is coupled cluster, where cluster operators generate excitations out of the Slater determinant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hermann Kummel says that we can think about electron correlation this way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The first thing one may imagine happening is that two particles mutually interact, thereby lifting themselves out of the Fermi sea, so that after the interaction both are in unoccupied orbitals&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Kümmel, Hermann. &quot;Origins of the coupled cluster method.&quot; Theoretica chimica acta 80.2-3 (1991): 81-89.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why can we (or should we) think about electron correlation that way?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2016-09-27T16:35:40.633" Title="Why are excited Slater determinants used to describe electron correlation?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8723" PostTypeId="1" AcceptedAnswerId="8729" CreationDate="2014-02-25T20:34:59.903" Score="3" ViewCount="19444" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$2.0 \cdot 10^{-4}$&lt;/span&gt; moles of dye in &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; of solution is consumed in 188 seconds, what is the average rate of consumption of dye in &lt;span class=&quot;math-container&quot;&gt;$\pu{mol L-1 s-1}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am not sure if I have done this correctly. I use the rate of reaction formula which is &lt;span class=&quot;math-container&quot;&gt;$\text{Rate} = -\frac{\mathrm{d[A]}}{\mathrm dt}$&lt;/span&gt; There is a negative sign because the dye is being consumed and it is disappearing.&lt;/p&gt;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;[dye] = &lt;span class=&quot;math-container&quot;&gt;$2.0 \cdot 10^{-4}$&lt;/span&gt; moles in &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;volume of solution = &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$\pu{0.05 L}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t = \pu{188 s}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Since we have &lt;span class=&quot;math-container&quot;&gt;$2.0 \cdot 10^{-4}$&lt;/span&gt; moles of dye per &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; of solution, that means in &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt; of solution, we get &lt;span class=&quot;math-container&quot;&gt;$2*(2.0 \cdot 10^{-4})$&lt;/span&gt; moles of dye = &lt;span class=&quot;math-container&quot;&gt;$4.0 \cdot 10^{-4}$&lt;/span&gt; moles&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\text{Rate of consumption/disappearance} &amp;amp;= \frac{-4.0 \cdot 10^{-4} \, \pu{mol}}{\pu{188 s}} \\&#xA; &amp;amp;= -2.1 \cdot 10^{-6} \pu{mol L-1 s-1}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is my answer correct? Please point out my mistakes if I did it incorrectly.&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="83758" LastEditDate="2020-06-18T17:57:44.500" LastActivityDate="2020-06-18T17:57:44.500" Title="Finding the average rate of consumption" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;erratum&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8724" PostTypeId="1" AcceptedAnswerId="8726" CreationDate="2014-02-25T20:37:45.617" Score="2" ViewCount="80" Body="&lt;p&gt;I am just about to start my senior project synthesis of PbS quantum dots and we ordered the wrong lead acetate. The method calls for lead (II) acetate and we got lead (IV) acetate. If we use the lead (IV) acetate will our synthesis be contaminated and not work or can we still use that chemical with two extra acetates attached? &lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="7448" LastEditDate="2017-04-08T16:10:49.540" LastActivityDate="2017-04-08T16:10:49.540" Title="In synthesizing PbS nanocrystals will there be any problems if I use lead IV acetate instead of lead II acetate?" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8725" PostTypeId="2" ParentId="8723" CreationDate="2014-02-25T20:49:14.833" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Since we have 2.0*10^-4 moles of dye per 50 mL of solution, that means in 1 L of solution, we get 2*(2.0*10^-4 moles) of dye = 4.0*10^-4 moles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless I'm too tired, the concentration is &#xA;$ 4.0 \times 10^{-3}\ \mathrm{mol} \cdot \mathrm{l}^{-1}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-25T20:49:14.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8726" PostTypeId="2" ParentId="8724" CreationDate="2014-02-25T21:08:55.557" Score="3" Body="&lt;p&gt;You might run into problems. $\ce{Pb(OAc)4}$ is a pretty strong oxidant! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be your sulfide source? Are you sure that it isn't oxidized these conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least N,N'-diarylthioureas are efficiently desulfurized by lead tetraacetate and furnish ureas (and sulfur) via a carbodiimide intermediate, see &lt;em&gt;Bull.Chem. Soc. Jpn.&lt;/em&gt;, &lt;strong&gt;1984&lt;/strong&gt;, &lt;em&gt;57&lt;/em&gt;, 315-316.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&#xA;If you don't mind a personal advice: I wouldn't want to add complexity to the protocol just to maneuver around problems (and create others) I wouldn't have if I were using the &lt;strong&gt;right&lt;/strong&gt; starting materials in the first. (The German term for that is &lt;em&gt;verschlimmbessern&lt;/em&gt;, which translates to something like &lt;em&gt;disimprove&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-25T22:31:14.973" LastActivityDate="2014-02-25T22:31:14.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8728" PostTypeId="2" ParentId="8722" CreationDate="2014-02-25T21:48:58.337" Score="7" Body="&lt;p&gt;It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra &lt;em&gt;chemistry&lt;/em&gt; that we want to add is electron correlation. The extra &lt;em&gt;maths&lt;/em&gt; we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)&lt;/p&gt;&#xA;" OwnerUserId="390" LastEditorUserId="390" LastEditDate="2014-03-02T18:15:39.030" LastActivityDate="2014-03-02T18:15:39.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8729" PostTypeId="2" ParentId="8723" CreationDate="2014-02-25T22:24:29.963" Score="4" Body="&lt;p&gt;Since we have $2.0 \times 10^{-4}\text{ mol}$ per $50\text{ mL}$ of solution, to get the amount of substance per liter, you multiply by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1000\text{ mL}}{50\text{ mL}}=20$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get $4.0 \times 10^{-3}\text{ mol L}^{-1}$ as Klaus says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're being asked to find a rate of dye consumption. This should be a positive number, if the dye is being consumed - which it is. If dye was being consumed at a negative rate, more dye would appear over time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine we begin with $c = 1.0\text{ mol L}^{-1}$. We end with $c = 0.5\text{ mol L}^{-1}$. Thus we might say that $\Delta c = -0.5\text{ mol L}^{-1}$. We see that the change is negative, signifying that dye has been consumed. If this occured over the course of 180 seconds, we might write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{average rate of consumption} = -\frac{\Delta c}{\Delta t} = -\frac{-0.5\text{ mol L}^{-1}}{180\text{ s}} = 2.77\times 10^{-3}\text{ mol L}^{-1}\text{ s}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even though the change in dye concentration is negative, the rate of dye &lt;em&gt;consumption&lt;/em&gt; should be a positive number.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-02-25T22:24:29.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8731" PostTypeId="2" ParentId="8689" CreationDate="2014-02-26T01:25:28.450" Score="0" Body="&lt;p&gt;Not sure what you are specifically looking for, but I would look into the Solvay Process (At least part of it). An old well known way to produce CO2 from an inorganic source (Which you have tagged). Its fairly simple, the Solvay process is a production process for creating Sodium Bicarbonate, One of the steps is to create CO2 from limestone. This is done from heating up limestone to about 1000 degrees C, which when I did the Solvay process in chem lab we used a bunsen burner but could be done with a good electric burner/furnace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your measuring CO2 in PPM in the experiment then this would work.&lt;/p&gt;&#xA;" OwnerUserId="698" LastActivityDate="2014-02-26T01:25:28.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8732" PostTypeId="1" AcceptedAnswerId="8736" CreationDate="2014-02-26T01:47:45.097" Score="5" ViewCount="9484" Body="&lt;p&gt;According to Graham's Law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of the gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider this situation: We have s balloon with 1 mole of neon gas and another with 2 moles of neon gas. The gases are in thermal equilibrium and occupy the same volume. In this scenario, would the rate of effusion be the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of the neon in both balloons is the same, so according to Graham's Law the rates would be the same. But intuitively, thinking about 2 moles versus 1, the pressure is doubled if volume and temperature are kept constant. And if the pressure is doubled, more collisions per second occur, so I would think that despite Graham's Law, the rate of effusion is also affected by the number of moles.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="5026" LastEditDate="2018-04-22T08:13:02.550" LastActivityDate="2018-04-22T08:14:06.547" Title="In Graham's law, is the rate of effusion only dependent on molar mass?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8733" PostTypeId="1" CreationDate="2014-02-26T01:49:46.547" Score="3" ViewCount="1339" Body="&lt;p&gt;According to Wikipedia, the shielding effect only happen in atoms which have more than 1 electron shells. The core electrons repel the electron in the valence shell. However, I have heard some lectures mention the shielding effect on helium atom when they explained why the first ionization energy of $\ce{He}$ is not exactly 2 times more than $\ce{H}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? According to the definition of shielding effect, there shouldn't be one in $\ce{He}$ atom.&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="73" LastEditDate="2014-02-26T02:06:38.980" LastActivityDate="2020-07-09T07:35:27.637" Title="Why does helium seem to exhibit an ionization energy affected by the shielding of electrons?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8734" PostTypeId="1" CreationDate="2014-02-26T03:03:51.260" Score="3" ViewCount="199" Body="&lt;p&gt;I'm trying to have a conversation/debate about the SmartKleen laundry detergent alternative, but lack the knowledge to understand if their &lt;a href=&quot;http://www.smartklean.com/html/science.html&quot; rel=&quot;nofollow&quot;&gt;explanation&lt;/a&gt; is science or pseudoscience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far Infrared emission at cold water temp, smaller water molecule clusters, magnets that increase levels of dissolved oxygen, etc., can someone offer a solid critique?&lt;/p&gt;&#xA;" OwnerUserId="4644" LastEditorUserId="73" LastEditDate="2014-02-26T03:45:27.477" LastActivityDate="2014-04-27T23:09:04.260" Title="Is the scientific explanation behind Smartkleen legit?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8736" PostTypeId="2" ParentId="8732" CreationDate="2014-02-26T05:13:42.830" Score="7" Body="&lt;p&gt;Molar mass is just one of the factors affecting effusion. Graham's law states that &lt;em&gt;everything else remaining same&lt;/em&gt;, the rate of effusion of a gas is inversely proportional to $\sqrt M$. In other words, Graham's law describes the relation of the rate of effusion with only one of the factors involved in determining the overall rate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the general rate of effusion of a gas at pressure $P$, and the hole area $A$ at a temperature $T$, using kinetic molecular theory, it turns out to be: (&lt;a href=&quot;https://physics.stackexchange.com/questions/79843/rate-of-effusion-in-kinetic-molecular-theory&quot;&gt;derivation on Physics.SE&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=\frac{PA}{\sqrt{2\pi MRT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in your question, the balloon with two moles will have a greater pressure (more moles in the same volume) and therefore it will have a greater rate of effusion for the same hole area and temperature and molecular mass.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="5026" LastEditDate="2018-04-22T08:14:06.547" LastActivityDate="2018-04-22T08:14:06.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8737" PostTypeId="1" AcceptedAnswerId="8789" CreationDate="2014-02-26T06:47:16.390" Score="3" ViewCount="14764" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose we found that the reaction of Red#3 with bleach is 1st order in hypochlorite. The observed rate constant ($k_\mathrm{obs}$) for one of the trials was measured to be $\pu{0.00400 s^{-1}}$. If the concentration of hypochlorite was fixed at $\pu{0.134 M}$, calculate the specific rate constant ($k$) for that trial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am very confused in which equation I should use to solve this problem. Am I to use this equation? &#xA;$$\mathrm{Rate} = k[A]^m[B]^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, for trial #1:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Rate} = \pu{0.00400 s^{-1}} \times \pu{0.134 M} = \pu{5.36e-4 mol L^{-1} s^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then what should I do next? This does not sound right. Please point out my mistakes.&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="5026" LastEditDate="2018-03-09T08:59:15.030" LastActivityDate="2018-03-09T08:59:15.030" Title="Calculation of the specific rate constant (k)" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8739" PostTypeId="1" AcceptedAnswerId="8742" CreationDate="2014-02-26T13:54:21.113" Score="2" ViewCount="790" Body="&lt;p&gt;For methylbenzene (with the methyl group at position 1), is it easier for chlorine to substitute the hydrogen at position 2, 3 or 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking somewhere along the lines of different charge densities around the benzene ring, and that the charge density closest to the already existing methyl group has been diminished due to the methyl group itself. However, I am unable to explain it properly enough.&lt;/p&gt;&#xA;" OwnerUserId="4647" LastEditorUserId="418" LastEditDate="2014-02-26T16:21:02.130" LastActivityDate="2014-02-26T16:40:56.047" Title="Formation of 3-chloromethylbenzene VS 3-chloromethylbenzene VS 4-chloromethylbenzene" Tags="&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8740" PostTypeId="1" CreationDate="2014-02-26T16:21:59.213" Score="1" ViewCount="2772" Body="&lt;p&gt;HEPES, the buffer, has the chemical formula $\ce{C8H18N2O4S}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_strength&quot; rel=&quot;nofollow&quot;&gt;Ionic Strength formula from Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated that $0.1M$ of HEPES would have a strength of $1.7M$, would it be possible to confirm whether this is correct or not? (I have little chemistry knowledge).&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorDisplayName="user4076" LastEditDate="2014-02-26T21:37:10.457" LastActivityDate="2014-02-28T13:19:35.643" Title="Ionic Strength of 0.1M of HEPES?" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8741" PostTypeId="2" ParentId="8739" CreationDate="2014-02-26T16:31:57.757" Score="2" Body="&lt;p&gt;It very much depends on the reagents and reaction type you are using to halogenate the aromatic ring, however, the most common method taught during organic chemistry is electrophilic hologenation, which in this case the methyl will be para directing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1689" LastActivityDate="2014-02-26T16:31:57.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8742" PostTypeId="2" ParentId="8739" CreationDate="2014-02-26T16:34:48.990" Score="2" Body="&lt;p&gt;We are talking about the electrophilic aromatic substitution of methylbenzene (&lt;strong&gt;A&lt;/strong&gt;) with chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8r6Es.png&quot; alt=&quot;EAS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible products are 2-methylchlorobenzene (&lt;strong&gt;2&lt;/strong&gt;), 3-methylchlorobenzene (&lt;strong&gt;3&lt;/strong&gt;), and 4-methylchlorobenzene (&lt;strong&gt;4&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloromethylbenzene (&lt;strong&gt;B&lt;/strong&gt;) is something different, also known as benzylchloride. &lt;em&gt;Chloromethyl&lt;/em&gt; is the name for the  $\ce{-CH2Cl}$ substituent :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please have a look at &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; for an animated explanation of the mechanism and the selectivities.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-26T16:40:56.047" LastActivityDate="2014-02-26T16:40:56.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8743" PostTypeId="2" ParentId="8734" CreationDate="2014-02-26T16:48:51.163" Score="0" Body="&lt;p&gt;Do a wash without adding anything.  Given lightly soiled fabric, there is enough detergent residue to get things &quot;sparkling chean.&quot;  Absence of perfume will give a nice linen scent.  Water soluble dirt does not need detergent to disappear.  HOWEVER, oily stuff will stay put and accumulate. Things likes starch and protein generally adhere and accumulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A laundry ball is nonsense no matter what its claimed mechanism.  A nasty accumulation of crud requires weeks or months to accumulate.  The laundry ball &quot;wears out.&quot;  &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-26T16:48:51.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8744" PostTypeId="1" AcceptedAnswerId="8745" CreationDate="2014-02-26T17:51:28.950" Score="2" ViewCount="1561" Body="&lt;p&gt;For example, if I were to draw the structure of 1,2,4-trimethylcyclohexane, then would one or both of these be wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L4ndc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are both right? I'm wondering if it matters where I start to count from 1 in these cases, if there is some convention to it or if it doesn't matter much.&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2014-02-26T18:27:39.333" Title="Does it matter which end of a chain one sets to be 1 when drawing the structure of organic compounds?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8745" PostTypeId="2" ParentId="8744" CreationDate="2014-02-26T17:57:27.327" Score="3" Body="&lt;p&gt;Both of those are the same molecule (flip the one on the left over its horizontal axis, and they are identical).  It does matter where you start counting from, because the numbers chosen for the functional groups should be as low as possible:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Has the lowest-numbered locant (or locants) for the suffix functional group.&lt;/strong&gt; Locants are the numbers on the carbons to which the substituent is directly attached.&lt;/li&gt;&#xA;&lt;li&gt;Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).&lt;/li&gt;&#xA;&lt;li&gt;Has the lowest-numbered locants for double bonds.&lt;/li&gt;&#xA;&lt;li&gt;Has the lowest-numbered locants for prefixes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;sup&gt; From &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC nomenclature&lt;/a&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, given the choice between 1,3,4-trimethylcyclohexane and 1,2,4-trimethylcyclohexane, you would choose the latter (for both molecules).&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-02-26T18:27:39.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8747" PostTypeId="2" ParentId="8734" CreationDate="2014-02-26T21:55:56.510" Score="1" Body="&lt;p&gt;I think the idea of &quot;smaller water clusters&quot; penetrating the cloth more efficiently is actually quite smart as the same idea is utilised in steam cleaners. But the rest of the article is a bit too psychotic. How can they say that colours are preserved while the Ball constantly generates H$_2$O$_2$ (in every second)? And why do they recommend the Stain Remover for stains if the ball was such a miracle...&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2014-02-26T21:55:56.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8748" PostTypeId="1" AcceptedAnswerId="8750" CreationDate="2014-02-26T22:45:17.793" Score="1" ViewCount="1837" Body="&lt;p&gt;How can tin nitrate be synthesized? Mixing nitric acid and tin makes tin oxide (I did this in chemistry class) so how can tin nitrate be created?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="1948" LastEditDate="2014-02-26T23:43:16.627" LastActivityDate="2020-02-22T01:04:04.027" Title="Make tin nitrate?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2020-02-20T16:38:03.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="8750" PostTypeId="2" ParentId="8748" CreationDate="2014-02-26T23:52:23.353" Score="3" Body="&lt;p&gt;A quick look into some search engine results for tin nitrate yielded &lt;a href=&quot;http://www.nhehs.org.uk/chemistry/_Lead%28II%29_nitrate_Tin.html&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt;, which rather interestingly even comes with a video of the reaction. That probably isn't among the main routes to generate tin nitrate, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the direct reaction of tin and nitric acid &lt;a href=&quot;http://www.sciencemadness.org/talk/viewthread.php?tid=8347&quot; rel=&quot;nofollow&quot;&gt;actually can yield tin nitrate&lt;/a&gt;, but only in cold dilute nitric acid, and even then only as a metastable solution. &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1961/jr/jr9610001996#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; goes deeper into the synthesis of tin nitrate.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-02-26T23:52:23.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8751" PostTypeId="1" AcceptedAnswerId="8752" CreationDate="2014-02-27T00:05:01.593" Score="5" ViewCount="21688" Body="&lt;p&gt;In my organic chemistry lab (next week, the experiment has not been conducted yet) we are being asked to react anisole with bromine to get 2-bromoanisole (as well as 2,4-dibromoanisole and all the others).  In our lab instructions we are told to add anisole, bromine and glacial acetic acid to our mixture.  No mention of other compounds (such as $\ce{FeBr3}$) was made.  I was under the impression that bromine is not reactive enough to brominate an aromatic without $\ce{FeBr3}$.  What purpose does the acid serve, and how does it make this reaction work without the iron compound?&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T10:51:54.657" LastActivityDate="2016-09-04T08:38:21.223" Title="Why use glacial acetic acid in bromination of anisole?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8752" PostTypeId="2" ParentId="8751" CreationDate="2014-02-27T00:21:14.157" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] bromine is not reactive enough to brominate an aromatic without FeBr3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's true for benzene, but the methoxy-group in anisole isn't just &lt;em&gt;o/p&lt;/em&gt;-directing. &#xA;&lt;strong&gt;It is activating as well&lt;/strong&gt;. Actually, it's that much activating that anisole will react with bromine even without the presence of a Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why glacial acetic acid? It's a polar solvent and thus nice to stabilize the &lt;em&gt;Wheland&lt;/em&gt; intermediates. Moreover, it's easily removed in aqueous workup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&#xA;Does the acetic acid react with the (mostly undissociated) acetic acid? At least a polarization of bromine ($\ce{Br^{\delta+}-Br^{\delta-}...HOAc}$) is conceivable and I remember that $\ce{BrOAc}$ has been postulated as the electrophilic species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&#xA;I remembered some brominations on dimethoxy-substituted hydroxymethylnaphthalines I did back in the days. The reaction was performed at 60 °C in $\ce{CHCl3}$ in the presence of about 3 equivalents of anhydrous sodium acetate. Excess of bromine was reduced by addition of aqueous sodium sulfite. The yield was in the 85% range and I don't find any notes on dibrominated products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And with excess Bromine you would (theoretically) react all of the anisole?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pretty much, with 4-bromoanisole as the main product.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="7781" LastEditDate="2015-10-05T19:48:50.883" LastActivityDate="2015-10-05T19:48:50.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8754" PostTypeId="1" AcceptedAnswerId="14004" CreationDate="2014-02-27T04:38:33.370" Score="6" ViewCount="1963" Body="&lt;p&gt;I am looking to calculate the minimum heat needed to create vapor from a specific mixture. I have a mixture of 60 % &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H8O3}$&lt;/span&gt; (glycerine) and 40 % &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H8O2} $&lt;/span&gt; (propylene glycol, propane-1,2-diol). Total volume of the mixture is 3 ml. Here are some facts that I know&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; C3H8O2&#xA; Density = 1.04 g/cm³&#xA; Boiling point = 370.8 °F (188.2 °C)&#xA; Molar mass = 76.09 g/mol&#xA; Specific Heat = 2.5 kJ/(kg·K) or 0.60 Btu/(lb·°F)&#xA;&#xA; C3H8O3&#xA; Density = 1.26 g/cm³&#xA; Boiling point = 554 °F (290 °C)&#xA; Molar mass = 92.09382 g/mol&#xA; Viscosity = 1.412 Pa·s&#xA; Specific Heat = 2.43 kJ/(kg·K) or 0.576 Btu/(lb·°F)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At what heat will the mixture begin to create vapor? I am looking for an answer, although I am not looking for someone to come in and just leave an answer. I would rather understand how you came about the answer, i.e. equations, additional info, so that I can adjust the percentages and volume on my own. The percentages of the mixture will vary. The volume of the mixture will vary depending on the amount of heat needed to create vapor, i.e. decrease volume if not enough heat is present. &lt;/p&gt;&#xA;" OwnerUserId="4653" LastEditorDisplayName="user7951" LastEditDate="2019-03-07T09:07:08.437" LastActivityDate="2019-03-07T09:07:08.437" Title="How to determine the vapor pressure of a glycerine and propylene glycol mixture?" Tags="&lt;home-experiment&gt;&lt;heat&gt;&lt;vapor-pressure&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8755" PostTypeId="1" AcceptedAnswerId="8759" CreationDate="2014-02-27T05:02:43.997" Score="5" ViewCount="11157" Body="&lt;p&gt;As we know that $\ce{KOH}$ absorbs $\ce{CO2}$, then why can't it be used to remove the excessive $\ce{CO_2}$ present in the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="4061" LastEditorUserId="418" LastEditDate="2014-02-27T06:34:28.023" LastActivityDate="2014-02-27T14:53:39.987" Title="Why can potassium hydroxide not be used to remove atmospheric carbon dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8756" PostTypeId="1" AcceptedAnswerId="8790" CreationDate="2014-02-27T05:04:37.607" Score="4" ViewCount="1823" Body="&lt;p&gt;Here is a series of related questions that I want to ask.&lt;/p&gt;&#xA;&lt;h2&gt;Background&lt;/h2&gt;&#xA;&lt;p&gt;The activity of acid phosphatase is measured by an enzymatic reaction that converts &lt;em&gt;para&lt;/em&gt;-nitrophenyl phosphate (pNPP) to &lt;em&gt;para&lt;/em&gt;-nitrophenol (pNP), liberating phosphate. The product, pNP, absorbs light whose wavelength is &lt;span class=&quot;math-container&quot;&gt;$\pu{400 nm}$&lt;/span&gt; with an absorption coefficient (&lt;span class=&quot;math-container&quot;&gt;$\pu{400 nm}$&lt;/span&gt;) of &lt;span class=&quot;math-container&quot;&gt;$\pu{19000 M-1 cm-1}$&lt;/span&gt; at extremely alkaline pH. Reaction mixture for an acid phosphatase is slightly acidic. Thus, it must be alkalinized for quantification of pNP.&lt;/p&gt;&#xA;&lt;p&gt;Two enzyme concentrations are to be examined - They are &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{0.1X}$&lt;/span&gt; (The &lt;span class=&quot;math-container&quot;&gt;$\ce{0.1X}$&lt;/span&gt; enzyme is made by mixing &lt;span class=&quot;math-container&quot;&gt;$\pu{1 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; enzyme in &lt;span class=&quot;math-container&quot;&gt;$\pu{9 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;.) Reaction times for each enzyme is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; minutes.&lt;/p&gt;&#xA;&lt;h2&gt;Procedure&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Protocol for measurement of acid phosphatase activity:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mix &lt;span class=&quot;math-container&quot;&gt;$\pu{0.12 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; acetate buffer (pH &lt;span class=&quot;math-container&quot;&gt;$5.6$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.24 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{5 mM}$&lt;/span&gt; pNPP in a test tube. Start the reaction by adding &lt;span class=&quot;math-container&quot;&gt;$\pu{0.24 ml}$&lt;/span&gt; of an enzyme solution.&lt;/li&gt;&#xA;&lt;li&gt;After the reaction times of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\pu{20 min}$&lt;/span&gt;, respectively, stop the reaction by adding &lt;span class=&quot;math-container&quot;&gt;$\pu{0.6 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; stops the reaction and converts the pNP produced into a yellow-colored (A400-absorbing) form.&lt;/li&gt;&#xA;&lt;li&gt;After all reactions are stopped, measure A400 of the samples.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Assay of potato acid phosphatase:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{ll}&#xA;\pu{0.5 M} \text{Na acetate buffer (pH 5.6)} &amp;amp; \pu{0.12 ml}\\&#xA;\pu{5 mM} \ce{pNPP} &amp;amp; \pu{0.24 ml}\\&#xA;\text{Enzyme} &amp;amp; \pu{0.24 ml}\\&#xA;\pu{0.m M} \ce{NaOH} &amp;amp; \pu{0.5 ml}\\\hline&#xA;\text{Sum} &amp;amp; \pu{1.2 ml}&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are some things I am supposed to calculate after I obtain the results.&lt;/p&gt;&#xA;&lt;p&gt;First I am going to plot a graph of absorbance versus time and find the slope for each of the lines (&lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{0.1X}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I have been asked to find the absorbance change/min/1ml of &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; enzyme.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;My answer&lt;/em&gt;: Divide the slope obtained by &lt;span class=&quot;math-container&quot;&gt;$0.24$&lt;/span&gt; (the amount of enzyme).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I have to convert the absorbance change to concentration change when&#xA;&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{1 cm}$&lt;/span&gt; and e400 of pNP is &lt;span class=&quot;math-container&quot;&gt;$\pu{19000 M-1 cm-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;My answer&lt;/em&gt;: This can be found out by &lt;span class=&quot;math-container&quot;&gt;$A = eCL$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The next question is to convert the concentration change to a change in the amount of substance of pNP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have no clue how to do this.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Finally, I have to calculate total activity (in moles per minute) in &lt;span class=&quot;math-container&quot;&gt;$\pu{4 ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; enzyme solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;My answer&lt;/em&gt;: Multiply the answer I obtain in the above question by &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-16T13:47:50.067" Title="Spectrophotometric assay by pNPP" Tags="&lt;biochemistry&gt;&lt;spectrophotometry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8757" PostTypeId="2" ParentId="8684" CreationDate="2014-02-27T06:45:34.790" Score="2" Body="&lt;p&gt;Simply, Oxidation and reduction is the reaction in which one of the species donates eletron and another species accepts the electron .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the reaction involved in formation of Na Cl(s) , the reaction mechanism is as&#xA;Na will release an electron from its outermost orbit to attain octet state forming Na+ cation.Likewise, the Cl atom will accept the electron forming Cl- anion to attain octet state. As a result, a ionic bond will be formed between them which is indeed a very strong chemical bond and finally after multiple bonding between them a crystalline structure will be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly even it is neutral it is presented in ionic form in its molecular structure mainly to specify that there is ionic linkage between these two atoms &lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-02-27T06:45:34.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8758" PostTypeId="2" ParentId="8755" CreationDate="2014-02-27T06:56:21.100" Score="0" Body="&lt;p&gt;&quot;Potassium hydroxide is usually sold as translucent pellets, which will become tacky in air because KOH is hygroscopic.&quot; from Wikipedia. This clears that if KOH is used to absorb the excess carbon dioxide from atmosphere ,there is greater chance of absorption of water molecules from the atmosphere which in turn can change the atmospheric contents resulting in some imbalance atmospheric conditions if KOH is used in excess.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-02-27T06:56:21.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8759" PostTypeId="2" ParentId="8755" CreationDate="2014-02-27T07:21:26.993" Score="10" Body="&lt;p&gt;$\ce{KOH}$ is produced by electrolysis of aqueous $\ce{KCl}$ solutions. Taking into account that the electrical energy still mostly comes from power plants that burn fossile fuels, it doesn't look like the most clever idea to release $\ce{CO2}$, just in order to capture it from the atmosphere again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it doesn't make sense to tackle the &lt;strong&gt;global&lt;/strong&gt; $\ce{CO2}$ problem with $\ce{KOH}$ or other hydroxides, they do nevertheless bind $\ce{CO2}$ just as you said. Consequently, these hydroxides were and are still used for the &lt;strong&gt;individual&lt;/strong&gt; protection against $\ce{CO2}$ (self)poisoning in emergency cases and/or in confined spaces. Rebreathers used by fire brigades, divers or miners are typical examples. Initially, $\ce{KOH}$ was indeed used, but nowadays, these $\ce{CO2}$ filters mostly contain mixtures of sodium hydroxide and calcium hydroxide.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-27T14:53:39.987" LastActivityDate="2014-02-27T14:53:39.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8760" PostTypeId="1" AcceptedAnswerId="8761" CreationDate="2014-02-27T08:27:17.643" Score="2" ViewCount="2104" Body="&lt;p&gt;When I looked up the definition of ionization energy, it says:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it is the energy required to remove one electron from the shell to&#xA;  infinity in an atom in gaseous state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is it measured in gaseous phase and not in any other phase?&lt;/p&gt;&#xA;" OwnerUserId="4516" LastEditorUserId="1431" LastEditDate="2014-02-27T10:04:28.823" LastActivityDate="2014-02-27T10:04:28.823" Title="Why is ionization energy measured in gaseous phase?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8761" PostTypeId="2" ParentId="8760" CreationDate="2014-02-27T08:33:06.040" Score="4" Body="&lt;p&gt;You want to determine the energy for a &lt;strong&gt;single molecule&lt;/strong&gt; and avoid stabilization effects through intermolecular interactions.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-27T08:33:06.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8762" PostTypeId="1" AcceptedAnswerId="8763" CreationDate="2014-02-27T08:35:45.227" Score="12" ViewCount="43794" Body="&lt;p&gt;It seems related to the atomic size but hydrogen has a smaller atomic size than fluorine. Why is fluorine the most electronegative atom?&lt;/p&gt;&#xA;" OwnerUserId="4516" LastEditorUserId="1431" LastEditDate="2014-02-27T10:11:00.517" LastActivityDate="2018-09-01T02:51:22.347" Title="Why is fluorine the most electronegative atom?" Tags="&lt;inorganic-chemistry&gt;&lt;electrons&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8763" PostTypeId="2" ParentId="8762" CreationDate="2014-02-27T11:54:01.593" Score="20" Body="&lt;p&gt;Fluorine is the most electronegative element because the definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;electronegativity&lt;/a&gt; makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \&amp;amp; \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \&amp;amp; \ AB$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from &lt;a href=&quot;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Mulliken_electronegativity&quot; rel=&quot;noreferrer&quot;&gt;Mulliken scale&lt;/a&gt; relates electronegativity to two other properties: &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity&quot; rel=&quot;noreferrer&quot;&gt;electron affinity&lt;/a&gt; $E_{ea}$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;noreferrer&quot;&gt;ionization energy&lt;/a&gt; $E_i$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other scales, including the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Allred.E2.80.93Rochow_electronegativity&quot; rel=&quot;noreferrer&quot;&gt;Allred-Rochow scale&lt;/a&gt;, based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;noreferrer&quot;&gt;effective nuclear charge&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_radius&quot; rel=&quot;noreferrer&quot;&gt;covalent radius&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Sanderson_electronegativity_equalization&quot; rel=&quot;noreferrer&quot;&gt;Sanderson scale&lt;/a&gt; based on atomic size, and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Allen_electronegativity&quot; rel=&quot;noreferrer&quot;&gt;Allen scale&lt;/a&gt; based on the average energy of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_electron&quot; rel=&quot;noreferrer&quot;&gt;valence electrons&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="41556" LastEditDate="2017-05-18T13:55:41.110" LastActivityDate="2017-05-18T13:55:41.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8765" PostTypeId="2" ParentId="8755" CreationDate="2014-02-27T14:22:31.937" Score="1" Body="&lt;p&gt;KOH can affect us when inhaled and by passing throught the skin.It is a highly corrosive chemical and contact can severely irritate and burn the skin and eyes leading to eye damage.Inhaling it can irritate the lungs. Highly exposures may cause a build-up of fluid in the lungs  (pulmonary oedema) a medical emergency.In addition to these adverse effects, it can also cause headache, dizziness, nausea and vomiting and also may cause a skin allergy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by observing these we can surely say that KOH can never be an option to absorb carbondioxide of the atmosphere.If we use it then we will just be gambling our life to death.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-02-27T14:22:31.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8767" PostTypeId="1" AcceptedAnswerId="8768" CreationDate="2014-02-27T15:55:51.880" Score="1" ViewCount="2930" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many moles of the substance are in 2.50 g of lead?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I did:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic mass of Pb: 207.2 amu (or g?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying unitary method, 1 mol of Pb -&gt; 207.2 g; so, x -&gt; 2.5 g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = \frac{2.5}{207.2}= \pu{0.0121 mol}~\text{of}~\ce{Pb}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be like this?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 mol of Pb -&gt; $(\pu{207.2 amu} \times1.66\times(10^{-24}))~\pu{g}$ so x -&gt; $\pu{2.5 g}$&lt;/p&gt;&#xA;" OwnerUserId="4661" LastEditorUserId="5026" LastEditDate="2018-02-25T03:02:59.980" LastActivityDate="2018-02-25T03:03:55.700" Title="How many moles of the substance are in 2.5 g of lead?" Tags="&lt;physical-chemistry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8768" PostTypeId="2" ParentId="8767" CreationDate="2014-02-27T16:58:15.713" Score="2" Body="&lt;p&gt;Do not mix amu with mass of mol of substance! In your last formula you've forgot about Avagadro number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Pb atom weighs $\pu{207.2amu}$ or $\pu{207.2amu} \cdot \pu{1.66\times10^{-24}g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one $\text{mol}$ of Pb weighs $=\pu{207.2amu}\cdot \pu{1.66\times10^{-24}g}\cdot6.02\cdot10^{23}=\pu{207.2g}$&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="5026" LastEditDate="2018-02-25T03:03:55.700" LastActivityDate="2018-02-25T03:03:55.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8769" PostTypeId="2" ParentId="8767" CreationDate="2014-02-27T17:10:14.510" Score="1" Body="&lt;p&gt;A gram-mole is the numeric value of atomic weight of the formula unit times grams.  Given an amount of said stuff in grams, divide by the numeric value of atomic weight of the formula unit to output the number of gram-moles of stuff.  One gram-mole contains Avogadro's number of formula units.  Embroidering the answer atom-by-atom undoes the installed shortcut to arrive at the same answer the hard way.  Avogadro's number is subject to small changes depending on what defines a kilogram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mole is the archetypal 5 pound bag.  A mole of sulfur atoms is about 32 grams.  Rhombic sulfur is eight-membered rings.  A mole of those is (8)(32 grams).  The crystallographic unit cell contains 16 rings.  A mole of unit cells is (16)(8)(32 grams).  If you are reacting zinc and sulfur, all you care about is moles of atoms.  If you want to know how many unit cells you have for x-ray diffraction of a rhombic sulfur microcrystal, you want to take those atoms 128 at a time.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-27T17:10:14.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8770" PostTypeId="2" ParentId="8754" CreationDate="2014-02-27T17:26:21.990" Score="1" Body="&lt;p&gt;Vapor is efficiently evolved when the sum of the vapor pressures is one atmosphere (boiling) given each mole fraction in homogeneous solution, rimmed by non-idealities.  You then must input the latent enthalpies of vaporization to move mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the juice for a vapor inhaler.  The sum of the enthalpies of heating and vaporization will define the battery you need.  However, do you want vapor or micronized (respirable) fog?  Pedagogy aside, it is more efficient to use an ultrasonic nebulizer in which the mist exits at ambient temperature.  Phase changes are expensive to fuel.  Unless all of the fluid parcel is vaporized, you will get one theoretical plate of fractionation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you run hot in air, you might pyrolyze glycerin into glycidol and acrolein, and propylene glycol into propylene oxide and allyl alcohol (then oxidize to the aldehyde).  Nasty for chronic inhalation.  You might thermally degrade inhalable substances (yeah, like &quot;flavoring&quot;) with high boiling points.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-02-27T17:26:21.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8772" PostTypeId="1" AcceptedAnswerId="8773" CreationDate="2014-02-27T18:14:43.963" Score="5" ViewCount="1258" Body="&lt;p&gt;I'm reading a high school chemistry course online, and I'm now reading about geometric isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They provide this example:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/v78Qi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in terms of naming it, they say this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Begin by identifying the parent chain - it is the continuous chain that contains the carbon-carbon double bond. The two groups that belong to the parent chain are on opposite sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then supply this illustration: &lt;img src=&quot;https://i.stack.imgur.com/Lr2BI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with that explains that it's &lt;strong&gt;&lt;em&gt;trans&lt;/em&gt;-3-ethyl-3-hexene&lt;/strong&gt; for that reason. I don't quite get their explanation though, becuase to me it seems like this could also be cis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't it be this one: &lt;img src=&quot;https://i.stack.imgur.com/KwduM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, that's also a valid way of expanding it, because it has the same amount of carbons in the chain and the numbers are still as low as they get. The only difference is that this one would be &lt;strong&gt;&lt;em&gt;cis&lt;/em&gt;-3-ethyl-3-hexene&lt;/strong&gt; instead of &lt;strong&gt;&lt;em&gt;trans&lt;/em&gt;-3-ethyl-3-hexene&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why exactly is it that only one of them is right?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastActivityDate="2015-03-18T20:36:09.833" Title="Why is this organic compound trans, and not cis?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-01-11T19:19:13.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="8773" PostTypeId="2" ParentId="8772" CreationDate="2014-02-27T18:37:02.270" Score="7" Body="&lt;p&gt;&lt;strong&gt;You are right&lt;/strong&gt;, there is no difference between the two molecules that result if you rotate 180 degrees about the $\ce{C=C}$ double bond - they are the same exact molecule.  So for this molecule, the terms &quot;cis&quot; and &quot;trans&quot; would not be used, they have no meaning here.  The name of the molecule is simply 3-ethyl-3-hexene.  If the on-line course says otherwise, &lt;strong&gt;it is wrong&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-18T20:36:09.833" LastActivityDate="2015-03-18T20:36:09.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8774" PostTypeId="2" ParentId="8717" CreationDate="2014-02-27T18:55:06.010" Score="36" Body="&lt;p&gt;The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for the phenomenological part. But what's the reason for this preference? The 1s electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's known from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius, this results in a contraction of the 1s orbital. Now, this contraction of the 1s orbital leads to a decreased degree of shielding for the outer s electrons (the reason for this is a decreased &quot;core repulsion&quot; whose origin is explained in &lt;a href=&quot;https://chemistry.stackexchange.com/a/376/189&quot;&gt;this answer&lt;/a&gt; of mine, see the part titled &quot;Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?&quot;) which in turn leads to a cascade of contractions of those outer s orbitals. The result of this relativistic contraction of the s orbitals is that the valence s electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter p-block elements (like $\ce{Al}$) usually &quot;give away&quot; their s  and p electrons when they form chemical compounds, heavier p-block elements (like $\ce{Tl}$) tend to &quot;give away&quot; their p electrons but keep their s electrons. That's the reason why for example $\ce{Al(III)}$ is preferred over $\ce{Al(I)}$ but $\ce{Tl(I)}$ is preferred over $\ce{Tl(III)}$.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2017-01-07T14:24:10.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8776" PostTypeId="2" ParentId="6946" CreationDate="2014-02-27T19:25:18.003" Score="5" Body="&lt;p&gt;Your assumption that the electron emits a photon when transitioning back to the ground state is not entirely correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The excited electron can also transition back by transferring the excitation energy to heat energy. The compound then warms up ever so slightly and the electron is buzzing away happily in the lower $d$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your &lt;em&gt;P.S.&lt;/em&gt;-ed question, I'll quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Absorption_band#Types_of_absorption_band&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The main factors that cause broadening of the spectral line into an absorption band of a molecular solid are the distributions of vibrational and rotational energies of the molecules in the sample (and also those of their excited states). [...] In gas phase spectroscopy, the fine structure afforded by these factors can be discerned, but in solution-state spectroscopy, the differences in molecular micro environments further broaden the structure to give smooth bands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-27T19:25:18.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8777" PostTypeId="1" AcceptedAnswerId="8784" CreationDate="2014-02-27T19:33:35.673" Score="14" ViewCount="5172" Body="&lt;p&gt;So looking at the Wikipedia pages of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_tetrafluoride&quot; rel=&quot;noreferrer&quot; title=&quot;Sulfur Tetrafluoride&quot;&gt;sulfur tetrafluoride&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon_tetrafluoride&quot; rel=&quot;noreferrer&quot;&gt;silicon tetrafluoride&lt;/a&gt;, the melting points are −121 °C and −90 °C respectively, and so $\ce{SiF4}$ has the higher melting point. However, their boiling points are −38 °C and −86 °C, respectively, giving $\ce{SF4}$ the higher boiling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can justify that $\ce{SF4}$ has the higher boiling point because it is more polar since its Lewis structure has a lone pair on the sulfur atom, and so it experiences greater dipole-dipole forces. However, why does $\ce{SF4}$ have a lower melting point?&lt;/p&gt;&#xA;" OwnerUserId="4663" LastEditorUserId="69490" LastEditDate="2019-07-15T01:14:32.477" LastActivityDate="2019-07-15T01:34:37.393" Title="Why does silicon tetrafluoride have a higher melting point than sulfur tetrafluoride?" Tags="&lt;intermolecular-forces&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8778" PostTypeId="2" ParentId="6941" CreationDate="2014-02-27T19:41:26.637" Score="3" Body="&lt;p&gt;The thing is that you have to start with some compound $\ce{R-X}$ where $\ce{X}$ is a good leaving group. The quality of the leaving group is often thermodynamically defined, and in the Finkelstein reaction that you described, the energy motor for the whole thing is the insolubility of the salt that is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see what favourable starting material you could use for this in order to prepare other haloalkanes, especially since there are other ways for doing that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it would in principle be possible to do&#xA;$$\ce{R-X + NaY -&amp;gt; R-Y + NaX v}$$&#xA;as long as the solubility of $\ce{NaY}$ is larger than the solubility of $\ce{NaX}$.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-27T19:41:26.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8779" PostTypeId="2" ParentId="8733" CreationDate="2014-02-27T19:57:12.933" Score="5" Body="&lt;p&gt;The premise is problematic as there is no 'shielding' effects in an isolated Helium atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Phillipp was correct in explaining why the ionization energy of Helium is not exactly twice that of Hydrogen.  This is because the energies of each electron are higher due to their correlated nature (i.e. they repel each other into higher, non-ground-state-like energies).  Therefore, it takes LESS energy to rip one of those electrons out of the system.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-02-27T19:57:12.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8780" PostTypeId="2" ParentId="7796" CreationDate="2014-02-27T20:05:45.447" Score="4" Body="&lt;p&gt;As Satwik has stated in the comments, the equation relies on the conservation of moles principle.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\dfrac{mol_1}{V_1} \times {V_1} = \dfrac{mol_2}{V_2} \times {V_2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the above, given a constant volume of a solution, adding more moles of solute will only make the concentration increase, not decrease.  Adding in more solvent is the only answer in that case.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="4160" LastEditDate="2014-04-07T05:44:49.030" LastActivityDate="2014-04-07T05:44:49.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8781" PostTypeId="2" ParentId="8087" CreationDate="2014-02-27T20:06:40.287" Score="5" Body="&lt;p&gt;Consider this generic acid-base reaction in water:&#xA;$$\ce{HA &amp;lt;=&amp;gt;[K_\text{a}] A- + H+ }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (thermodynamic) dissociation constant $K_\text{a}$ depends mainly on two things: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The stability of the product $\ce{A-}$&lt;/li&gt;&#xA;&lt;li&gt;The instability of the starting material $\ce{HA}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The higher one of those factors is, the higher the dissociation constant. So in effect, what a chemist states is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;The carboxylate anion is the conjugated base of the carboxylic acid, whose strength is determined by the stability of said carboxylate ion and the instability of the acid. It &lt;/i&gt;may&lt;i&gt; be reasoned that carboxylic acids would be much less strong if it was not for the additional stability via conjugation.&lt;/i&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-27T20:06:40.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8782" PostTypeId="2" ParentId="8126" CreationDate="2014-02-27T20:18:08.787" Score="7" Body="&lt;p&gt;For solids, the binding energy is what makes the crystal (if you will) hold together. It is what makes the individual constituents of the compound &lt;em&gt;bind&lt;/em&gt; together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adsorption however is an entirely different process: Here, some molecules interact with the surface of another compound. Since (strictly speaking) this is no chemistry but mere physics (there is no formation or breaking of &lt;em&gt;bonds&lt;/em&gt;) there are other interactions at play, mainly &lt;em&gt;van der Waals&lt;/em&gt;-type interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these are weaker (it's easier to wipe off the water from your windshield than rip out a silicon atom) and aren't based on the same principles as the binding energy, they get a different name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that cleared things up for you.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-02-27T20:18:08.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8783" PostTypeId="2" ParentId="6906" CreationDate="2014-02-27T20:19:22.910" Score="5" Body="&lt;p&gt;The wavefunction contains all the information about the system of interest.  This is a basic premise of quantum mechanics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the Born-Oppenheimer approximation, we 'index' all the values required to describe where an electron is.  This includes the spatial coordinates, ($\textbf{r}$), and the spin coordinate, $\omega$.  Electrons are characterized by their spin ($\uparrow$ vs. $\downarrow$).  Another way to think about it is this.  The quantum numbers are used to describe everything we need to know about 'where' an electron is ($n$, $l$, $m_l$, $m_s$).  The spatial coordinates (e.g. Cartesian coordinates) take care of the first 3 quantum numbers.  We need the fourth coordinate to characterize $m_s$.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-02-27T20:19:22.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8784" PostTypeId="2" ParentId="8777" CreationDate="2014-02-27T20:36:22.597" Score="19" Body="&lt;p&gt;High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{lrr}&#xA;\hline&#xA;\text{Compound} &amp;amp; \text{m.p.}/\pu{°C} &amp;amp; \text{b.p.}/\pu{°C} \\&#xA;\hline&#xA;\text{pentane} &amp;amp; −130 &amp;amp; 36.1 \\  &#xA;\text{isopentane} &amp;amp; −160 &amp;amp; 27.2 \\ &#xA;\text{neopentane} &amp;amp; −18 &amp;amp; 9.5 \\&#xA;\hline&#xA;\text{octane} &amp;amp; −57 &amp;amp; 125 \\  &#xA;\text{2-methylheptane} &amp;amp; −110 &amp;amp; 117 \\  &#xA;\text{2,2,4-trimethylpentane} &amp;amp; −107 &amp;amp; 99 \\ &#xA;\text{2,2,3,3-tetramethylbutane} &amp;amp; 101 &amp;amp; 107 \\  &#xA;\text{cubane} &amp;amp; 131 &amp;amp; 133 \\&#xA;  &amp;amp; &amp;amp; \text{(sub. at r.t.)*} \\&#xA;\hline&#xA;\text{decane} &amp;amp; −30  &amp;amp; 174 \\  &#xA;\text{isodecane} &amp;amp; −75 &amp;amp; 167 \\  &#xA;\text{adamantane} &amp;amp; 270 &amp;amp; \text{sub.} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* &lt;strong&gt;sub.&lt;/strong&gt; — sublimes; &lt;strong&gt;r.t.&lt;/strong&gt; — room temperature.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="41328" LastEditDate="2019-07-15T01:34:37.393" LastActivityDate="2019-07-15T01:34:37.393" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8785" PostTypeId="2" ParentId="8599" CreationDate="2014-02-27T20:39:22.540" Score="10" Body="&lt;p&gt;I currently have the Dover copy of Pauling's book on my shelf.  It indeed is a nice read but it is in no way 'up-to-date' as it was written many years ago.  That being said, a lot of conceptual information presented in the book still apply just as much today as they did 6 decades ago (I would stray away from using the book as a source for benchmark measurements and such).  It is a book that should find its way onto any chemist's bookshelf, and it makes for a fascinating read into many areas of general chemistry.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-02-27T20:39:22.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8786" PostTypeId="2" ParentId="765" CreationDate="2014-02-27T20:39:32.587" Score="7" Body="&lt;p&gt;I think the most specific you can get on this phenomenon is &lt;em&gt;nucleation&lt;/em&gt; , specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleation#Heterogeneous_nucleation&quot; rel=&quot;noreferrer&quot;&gt;heterogenous nucleation&lt;/a&gt; / surface nucleation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rough surface formed creates scope for nucleation to occur, and  this creates the crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a specific name for this technique as a lab tenchique .&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2014-02-27T20:39:32.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8788" PostTypeId="2" ParentId="6871" CreationDate="2014-02-27T21:31:07.327" Score="4" Body="&lt;p&gt;It doesn't have to do as much with the combing as it does with surface tension. Water on the hair clumps them together, because smaller clumps have more surface area (net surface area) and thus more surface energy. Breaking these large clumps, therefore, requires energy input, which gives us the impression that they &quot;stick&quot; together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for sticking to the water individually, that's surface tension again. The surface is able to withstand a bit of weight before collapsing and letting the object drop through. Hairs are oily, and thus hydrophobic. The repulsion potential energy that would be formed / needed when it sinks outweighs the energy required to bend the surface (and get enough tension force to balance gravity). This is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Gerridae&quot; rel=&quot;nofollow&quot;&gt;water skeeters&lt;/a&gt; work.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2014-02-27T21:31:07.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8789" PostTypeId="2" ParentId="8737" CreationDate="2014-02-27T22:18:04.330" Score="7" Body="&lt;p&gt;We assume 1&lt;sup&gt;st&lt;/sup&gt; order kinetics in hypochlorite and 1&lt;sup&gt;st&lt;/sup&gt; order kinetics in the color. Putting this in the equation we get the following:&#xA;$$\frac{\text{d}[\text{Red}]}{\text{d}t} = -k\, [\text{Red}]^1 \, [\text{bleach}]^1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you keep the concentration of bleach the same throughout the trial, you can write&#xA;$$k_\text{obs} = k\times [\text{bleach}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can continue onwards as follows:&#xA;$$ k_\text{obs} = k \times [\text{bleach}] \Leftrightarrow k = \frac{k_\text{obs}}{[\text{bleach}]} = 0.0299~\mathrm{s^{-1}\; M^{-1}}$$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-03-07T10:11:44.543" LastActivityDate="2014-03-07T10:11:44.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8790" PostTypeId="2" ParentId="8756" CreationDate="2014-02-27T22:47:44.977" Score="3" Body="&lt;p&gt;For the first subquestion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using unit analysis, the answer is very quickly found: The unit of the slope &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;, assuming dimensionless values for the absorbance, is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. Since you need the change in absorption per &lt;span class=&quot;math-container&quot;&gt;$\pu{mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; solution, you divide the slope of the &lt;span class=&quot;math-container&quot;&gt;$\ce{1X}$&lt;/span&gt; reaction by &lt;span class=&quot;math-container&quot;&gt;$\pu{0.24 mL}$&lt;/span&gt; for the correct value.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the subquestion with which you seem to have the most trouble: simply apply &lt;span class=&quot;math-container&quot;&gt;$c=n/V$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find the amount of substance changed via&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="4945" LastEditDate="2020-02-16T13:47:50.067" LastActivityDate="2020-02-16T13:47:50.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8791" PostTypeId="1" AcceptedAnswerId="8794" CreationDate="2014-02-28T00:23:58.143" Score="23" ViewCount="3787" Body="&lt;p&gt;Reading this article on Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Xenon#Medical&quot;&gt;Xenon Medical applications&lt;/a&gt; I see that Xenon can be used as an anesthetic, neuroprotectant and doping agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a noble gas, and thus, chemically stable, how can it have those properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Xenon does not react with any other substance, how does this happen, from a chemical point of view?&lt;/p&gt;&#xA;" OwnerUserId="4651" LastEditorUserId="4651" LastEditDate="2014-02-28T18:46:59.813" LastActivityDate="2020-03-21T13:05:04.593" Title="Xenon and the human body" Tags="&lt;chemical-biology&gt;&lt;noble-gases&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8793" PostTypeId="2" ParentId="8364" CreationDate="2014-02-28T01:54:14.757" Score="4" Body="&lt;p&gt;Let's look at the reaction of the formation of water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2(g) + O2(g) -&amp;gt; 2H2O (l) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibb's Free Energy relates the spontaneity of various reactions by looking at the change in enthalpy, temperature, and entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can relate those by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at our reaction, it's clear that this is an exothermic reaction. Bonds are being formed as hydrogen gas and oxygen gas form water. Moles of gas are decreased and a liquid is formed so entropy is decreasing as well. We know that the formation of water is a spontaneous reaction, so $\Delta G^\circ$ must be negative. Therefore, this reaction must be run at temperatures where the difference of the enthalpy and the product of the entropy and temperature must be &amp;lt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs free energy is also a state function, so looking at $\Delta G^\circ_f$, we can find $\Delta G^\circ_{rxn}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\circ_{rxn} = \Sigma \Delta G^\circ_{products} - \Sigma \Delta G^\circ_{reactants} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both hydrogen gas and oxygen gas are in their standard states and have a Gibbs Free energy of 0. If you look at tabulated standard thermodynamic data at $25^\circ\ \text{C}$, $\ce{H2O (l)}$ has a Gibbs free energy of$ -237.13 \text{kJ}\cdot\text{mol}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\circ_{rxn} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} - 0 \text{ kJ}\cdot\text{mol}^{-1} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} $$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="275" LastEditDate="2014-02-28T11:06:19.687" LastActivityDate="2014-02-28T11:06:19.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8794" PostTypeId="2" ParentId="8791" CreationDate="2014-02-28T02:19:31.933" Score="20" Body="&lt;p&gt;Chemist Neil Bartlett in 1962 discovered that although Xenon was a noble gas, it is able to form compounds with other substances even though it is chemically &amp;quot;inert.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Neil Bartlett, at the time in 1961, produced an unidentified red solid and discovered that the red solid was a reaction between gaseous fluoride, platinum hexafluoride (&lt;span class=&quot;math-container&quot;&gt;$\ce{PtF6}$&lt;/span&gt;) and oxygen that was oxidized. Finally, determining the chemical structure, they found that the red solid was &lt;span class=&quot;math-container&quot;&gt;$\ce{O^{2+}PtF6}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Oxygen is a powerful oxidizing agent, being consumed in many combustion reactions. However, Bartlett believed that the platinum hexafluoride (&lt;span class=&quot;math-container&quot;&gt;$\ce{PtF6}$&lt;/span&gt;) was a stronger oxidizing agent than oxygen, and oxygen is instead reduced. Bartlett then believed if the oxidation of oxygen was possible, he could apply this xenon.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Because my co-workers at that time (March 23, 1962) were still not sufficiently experienced to help me with the glassblowing and the preparation and purification of &lt;span class=&quot;math-container&quot;&gt;$\ce{PtF6}$&lt;/span&gt; [platinum hexafluoride] necessary for the experiment, I was not ready to carry it out until about 7 p.m. on that Friday. When I broke the seal between the red PtF6 gas and the colorless xenon gas, there was an immediate interaction, causing an orange-yellow solid to precipitate. At once, I tried to find someone with whom to share the exciting finding, but it appeared that everyone had left for dinner!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The discovered new compound was xenon hexafluoroplatinate (&lt;span class=&quot;math-container&quot;&gt;$\ce{XePtF6}$&lt;/span&gt;). Bartlett's important contribution to chemistry is a series of noble gas compounds with specific and useful properties.&lt;/p&gt;&#xA;&lt;p&gt;In regards to your question on how xenon can be so diverse, with over &amp;quot;100 noble gas compounds known today...&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Noble gas compounds have already made an impact on our daily lives. &lt;span class=&quot;math-container&quot;&gt;$\ce{XeF2}$&lt;/span&gt; has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents. The reactivity of radon means that it can be chemically scrubbed from the air in uranium mines and other mines. Excimer lasers use compounds of argon, krypton or xenon to produce precise beams of ultraviolet light (when electrically stimulated) that are used to perform eye surgery for vision repair.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Material derived from a highly interesting article from the American Chemical Society.&#xA;&lt;a href=&quot;http://www.acs.org/content/acs/en/education/whatischemistry/landmarks/bartlettnoblegases.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.acs.org/content/acs/en/education/whatischemistry/landmarks/bartlettnoblegases.html&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Looking at the literature:&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Neuroprotection&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;For the use of Xeon during mechanical ventilation, an article examining the neuroprotective effects of xenon,&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Neuroprotection against traumatic brain injury by xenon, but not argon, is mediated by inhibition at the N-methyl-D-aspartate receptor glycine site.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Anti-Tumor Agent&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Noble gas compounds have already made an impact on our daily lives. &lt;span class=&quot;math-container&quot;&gt;$\ce{XeF2}$&lt;/span&gt; has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents.&lt;/p&gt;&#xA;&lt;p&gt;The American Cancer states that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fluorouracil belongs to the class of chemotherapy drugs known as anti-metabolites. It interferes with cells making DNA and RNA, which stops the growth of cancer cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qMuZ3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Anesthesia&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I'm still looking for literature to add here. Some of these cover very complex physiological processes, and this is not my area of expertise. I will try to find more pertinent articles from peer-reviewed journals with high enough impact and relevance. It will be up to the poster of this question to explore in-depth as much as they need by starting at some of the literature I provide.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-03-21T13:05:04.593" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8795" PostTypeId="2" ParentId="8791" CreationDate="2014-02-28T06:30:40.123" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If Xeon does not react with any other chemical, how does this happen, from a chemical point of view?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If molecules of phospholipids do not react with each other, how do they form biomembranes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If Xenon doesn't react, how can it form &lt;a href=&quot;http://www.ens-lyon.fr/CHIMIE/recherche/Teams/Chimie_Organique_et_Materiaux_Nanostructures/supramolecules-materiaux-et-stereochimie/thematiques-de-recherche-1/host-guest-systems-bio-sensoring-catalysis/cryptophanes-chemistry-and-complexation-properties/xenon-cryptophane-complexes&quot; rel=&quot;nofollow noreferrer&quot;&gt;host-guest complexes with cryptophanes&lt;/a&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It's not necessary to form covalent or ionic bonds to hold the pieces together.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak dispersion forces, recently described by Ben Norris &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8691/van-der-waals-nonspecific-interaction-definition&quot;&gt;here&lt;/a&gt; on this site are apparently enough in the examples above. These forces are also responsible for the binding of xenon to proteins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information, you might want to have a look at the work of &lt;em&gt;Seth M. Rubin&lt;/em&gt;, who did a lot on $\ce{^{129}Xe}$ NMR spectroscopy in proteines.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-02-28T06:30:40.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8796" PostTypeId="2" ParentId="7805" CreationDate="2014-02-28T07:50:12.370" Score="5" Body="&lt;p&gt;This is a partial answer to your query.  According to the paper &lt;a href=&quot;http://www.intechopen.com/download/get/type/pdfs/id/39405&quot;&gt;Degradation and Recyclability &#xA;of Poly (Ethylene Terephthalate) &#xA;&lt;/a&gt; (Venkatachalam, 2012), there are a few degradation products, from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Apart from aldehyde, $\ce{CO}$, $\ce{CO2}$, ethylene, benzene, biphenyl are also identified as &#xA;  degradation products &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The article also shows diagrms of the chemical processes of PTE degradation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an important aspect to note is experiments/observations have&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;shown that rate of thermal degradation in the closed system is &#xA;  about three times higher than that in open system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In regards to burning of PET, the products can vary according to the page &lt;a href=&quot;http://www.pypc.com/epk-pet.htm&quot;&gt;EPB 433- Health and Environmental Effects of Burning Waste &#xA;Plastics &#xA;&lt;/a&gt; (Saskatchewan Ministry of Environment) stating that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the composition of byproducts of plastic combustion as to their type and concentration depends &#xA;  on the combustion temperature and the flame residence time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://www.pypc.com/epk-pet.htm&quot;&gt;Peiyu Plastics page&lt;/a&gt; (a reference from Wikipedia), if complete combustion occurs, then the products would be $\ce{CO2}$ and $\ce{H2O}$, incomplete combustion could produce soot and carbon monoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.plasticsmadesimple.com/DataSheets/PET_MSDS.pdf&quot;&gt;Here is the MSDS for PET&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://peer.ccsd.cnrs.fr/docs/00/51/31/99/PDF/PEER_stage2_10.1080%252F00268970802077876.pdf&quot;&gt;A study of thermal decomposition and combustion products &#xA;of disposable polyethylene terephthalate (PET) plastic using &#xA;the High Resolution Fourier Transform Infrared &#xA;Spectroscopy, Selected Ion flow Tube Mass Spectrometry &#xA;and Gas Chromatography Mass Spectrometry &lt;/a&gt; (Sonova et al. 2008).&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-02-28T07:50:12.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8797" PostTypeId="2" ParentId="8370" CreationDate="2014-02-28T08:10:05.877" Score="20" Body="&lt;p&gt;According to the Canadian Centre for Occupational Health and Safety page &lt;a href=&quot;http://www.ccohs.ca/oshanswers/safety_haz/welding/fumes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Welding - Fumes And Gases&lt;/a&gt;, the welding fumes are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a complex mixture of metallic oxides, silicates and fluorides. Fumes are formed when a metal is heated above its boiling point and its vapours condense into very fine, particles (solid particulates).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This composition varies dependent on the composition of metals being welded and fluxes.  The website lists several main types and their associated health risks.  Often associated with metallic oxides are some pungent/noticeable gases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gases produced from the thermal breakdown of coatings, from welding and cutting processes, stated in the article linked as being examples of &quot;welding gases&quot; (examples from the linked website include ozone) &lt;/li&gt;&#xA;&lt;li&gt;And depending on the coating that may be on the metal, organic vapours (including aldehydes, diisocyanates, phosgene and phosphine)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="36638" LastEditorDisplayName="user4076" LastEditDate="2017-11-22T13:37:11.630" LastActivityDate="2017-11-22T13:37:11.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8798" PostTypeId="2" ParentId="5100" CreationDate="2014-02-28T09:16:36.503" Score="3" Body="&lt;p&gt;The notation used does not necessarily represent a triple bond - for example, using one of the examples that you have provided:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\equiv\text{Si}-\text{OH}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which is for the functional group &lt;strong&gt;silanol&lt;/strong&gt;, that is defined by the $\ce{Si-O-H}$ connectivity, it's chemical structure is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zf0NJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Silanol.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this example, the hydrogens represented by the  ≡ symbol can be replaced by other chemicals forming other members of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Silanol&quot; rel=&quot;nofollow noreferrer&quot;&gt;silanol functional group&lt;/a&gt;, for example, Trimethylsilanol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYkz0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-02-28T09:16:36.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8799" PostTypeId="2" ParentId="7595" CreationDate="2014-02-28T11:33:36.027" Score="3" Body="&lt;p&gt;You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions, some of which are mentioned in GODPARTICLE's answer. Standard thermodynamic conditions are usually&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A temperature of 298 K&lt;/li&gt;&#xA;&lt;li&gt;A pressure of gaseous components of 1 atm (or 1 bar)&lt;/li&gt;&#xA;&lt;li&gt;A concentration of 1 M&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reference reduction potentials you quote are all &lt;strong&gt;standard potentials&lt;/strong&gt;. These are the contributions to the cell potential when the half-cells meet the criteria above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, &lt;strong&gt;and stoichiometry&lt;/strong&gt;. Behold the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow&quot;&gt;Nernst equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_gas_constant&quot; rel=&quot;nofollow&quot;&gt;ideal gas constant&lt;/a&gt; (this is thermo after all)&lt;/li&gt;&#xA;&lt;li&gt;$n$ is the number of electrons transferred - here is the stoichiometry!&lt;/li&gt;&#xA;&lt;li&gt;$F$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_constant&quot; rel=&quot;nofollow&quot;&gt;Faraday constant&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;$Q$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_quotient&quot; rel=&quot;nofollow&quot;&gt;reaction quotient&lt;/a&gt; - which accounts for changes in pressure and concentration&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-02-28T11:33:36.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8800" PostTypeId="1" CreationDate="2014-02-28T11:59:13.850" Score="3" ViewCount="65" Body="&lt;p&gt;I was re-sorting books the other day, and thought about how many years it has been since I bought my Merck Index. I'm sure there are new entries in the more recent additions, but I do not have such a need to stay that up-to-date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first thought, it seemed to me that a book of this nature only adds entries, and does not remove any entries. But is this the case? Are compounds de-listed because they are bad science? Other reasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iit would be great to know other details about the editorial process, as well, if relevant. &lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2014-02-28T13:01:56.930" Title="Are entries ever removed from the Merck Index?" Tags="&lt;history-of-chemistry&gt;&lt;books&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8801" PostTypeId="2" ParentId="8800" CreationDate="2014-02-28T13:01:56.930" Score="2" Body="&lt;p&gt;Without several print editions side-by-side, I restrain my answer to the electronic version. This public &lt;a href=&quot;http://library.dialog.com/bluesheets/html/bl0304.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; or &lt;a href=&quot;http://library.dialog.com/bluesheets/pdf/bl0304.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; regarding the online version based on the 14th edition may be read like it may be the case.  To quote:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; &quot;The 14th Edition also contains &quot;retired&quot; records from earlier editions of The Merck Index which still contain valuable information but are no longer being updated with new information.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, beside RSC you may contact them (Merck in Rahway, NJ), too.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2014-02-28T13:01:56.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8802" PostTypeId="2" ParentId="8740" CreationDate="2014-02-28T13:03:39.877" Score="1" Body="&lt;p&gt;The Ionic Strength is calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\dfrac{1}{2}\cdot\sum{c_iz_i^2},$$ where where $c_i$ is the molar concentration of ion i ($M,\,mol/L$), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is your HEPES molecule ($C_8H_{18}N_2O_4S$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uw9ly.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only one proton at $SO_3H$ group can dissociate, as far as I see. So in your solution you will have two different types of ions with charge equal to 1 (assuming that molecule will dissociates 100%, and than:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=0.5\cdot(0.1M\cdot1^2+0.1M\cdot(-1)^2)=0.1M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your calculations are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also be aware that this molecule is Zwitterion! &#xA;READ THIS &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2248451/pdf/nihms38719.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Do zwitterions contribute to the ionic strength of a solution?&lt;/a&gt;&#xA;The resume of the articles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hence, electrostatic effects due to zwitterions are important, even if the zwitterions do&#xA;not contribute directly to ionic strength.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations are primitive. They are also far from the truth, but they show how the principle works for you.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-02-28T13:19:35.643" LastActivityDate="2014-02-28T13:19:35.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8803" PostTypeId="1" CreationDate="2014-02-28T14:01:54.923" Score="1" ViewCount="2148" Body="&lt;p&gt;I am having great difficulty getting spectra to overlay (the option doesn't even appear) after setting to new window mode and selecting each spectrum. &lt;/p&gt;&#xA;" OwnerUserId="4666" LastEditorUserId="73" LastEditDate="2014-02-28T21:18:42.993" LastActivityDate="2014-02-28T21:19:22.017" Title="How do I overlay spectra in an ACD/NMR Processor, version 12.01?" Tags="&lt;software&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8804" PostTypeId="1" CreationDate="2014-02-28T14:41:36.740" Score="2" ViewCount="4148" Body="&lt;p&gt;I know that catalysts don't undergo chemical changes during a reaction, but do they undergo physical changes? If they do, are there any examples?&lt;/p&gt;&#xA;" OwnerUserId="4667" LastEditorUserId="194" LastEditDate="2017-04-01T01:16:07.803" LastActivityDate="2017-04-01T01:16:07.803" Title="Do catalysts undergo physical changes during a reaction?" Tags="&lt;physical-chemistry&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8805" PostTypeId="2" ParentId="8804" CreationDate="2014-02-28T14:56:32.067" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that catalysts don't undergo chemical changes during a reaction [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't? How then, other by forming a complex with the starting material, is the activation energy of that reaction lowered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do however agree that - in the ideal case - this step is reversible and leaves the catalyst unchanged after the catalytic cycle ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-28T14:56:32.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8806" PostTypeId="1" AcceptedAnswerId="8810" CreationDate="2014-02-28T16:37:02.303" Score="5" ViewCount="51091" Body="&lt;p&gt;Surprisingly, there are very few resources online (outside of obscure lab pages) that explain exactly what standardization is (not even Wikipedia!). The context is the phrase &lt;em&gt;standardize the titrant&lt;/em&gt; for a titration. Does this mean you titrate it with a known solution?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="4945" LastEditDate="2014-07-29T10:48:57.860" LastActivityDate="2016-12-17T17:50:08.597" Title="What's Standardization, and why do it?" Tags="&lt;experimental-chemistry&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8807" PostTypeId="2" ParentId="8804" CreationDate="2014-02-28T16:38:09.473" Score="3" Body="&lt;p&gt;I'll just give an example of a catalysis to show you that catalysts do really temporarily react with the starting product and eventually will be regenerated after the catalytic reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the catalytic decomposition of formic acid to carbon monoxide and water by using sulfuric acid as a catalyst. The sulfuric acid donates one proton ($\ce{H+}$ ion) which is then bonded to the central carbon atom in the formic acid molecule. Then next another hydrogen atom from the sulfuric acid reacts with the hydroxyl group present on the formic acid molecule to yield water. In the end the sulphate ion snatches the last two present hydrogen atoms from the formic acid molecule and yields carbon monoxide. The sulfuric acid is regenerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCOOH + H2SO4 -&amp;gt; H2COOH + HSO4- -&amp;gt; H2CO + SO4^{2-} + H2O -&amp;gt; CO + H2SO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So catalysts &lt;strong&gt;do&lt;/strong&gt; undergo chemical changes in the catalytic reaction but are &lt;strong&gt;regenerated&lt;/strong&gt; after the catalytic reaction is finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you are using a catalyst, you can normally reuse it after your catalytic reaction because it has the same chemical (and physical) properties as before the catalytic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also : when you use manganese dioxide as a catalyst to decompose hydrogen peroxide, you will see that the black solid compound is the same after the catalytic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also so called 'sacrificial catalysts' (which aren't regenerated and do physically/chemically change, but &lt;strong&gt;aren't&lt;/strong&gt; real catalysts) -&gt; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalytic_cycle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Catalytic_cycle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="189" LastEditDate="2014-02-28T23:17:11.513" LastActivityDate="2014-02-28T23:17:11.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8808" PostTypeId="2" ParentId="8806" CreationDate="2014-02-28T16:47:28.283" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Standardization is doing a titration to work out the exact&#xA;  concentration of &lt;em&gt;the solution you want to use&lt;/em&gt; to determine the&#xA;  concentration of an unknown solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where &lt;em&gt;the solution you want to use&lt;/em&gt; is the titrant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See : &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20090903183139AAXfQGm&quot; rel=&quot;nofollow&quot;&gt;http://answers.yahoo.com/question/index?qid=20090903183139AAXfQGm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-02-28T16:47:28.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8810" PostTypeId="2" ParentId="8806" CreationDate="2014-02-28T18:51:38.997" Score="8" Body="&lt;p&gt;Imagine that you need to titrate the sample &lt;strong&gt;S&lt;/strong&gt; with a solution of the reagent &lt;strong&gt;R&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, &lt;strong&gt;R&lt;/strong&gt; isn't available as a stock solution and maybe not even in solid form. All you have is it's precursor &lt;strong&gt;P&lt;/strong&gt; and a recipe for the reaction $\ce{P + X + Y -&amp;gt; R}$. You never know how good that works out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, titration of &lt;strong&gt;S&lt;/strong&gt; with &lt;strong&gt;R&lt;/strong&gt; doesn't make any sense - unless you can determine the concentration of &lt;strong&gt;R&lt;/strong&gt;! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, a stock solution with an exact concentration of the helper &lt;strong&gt;H&lt;/strong&gt;, maybe in a sealed vial from a commerical source comes into play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The titration of &lt;strong&gt;R&lt;/strong&gt; with &lt;strong&gt;H&lt;/strong&gt; to give the exact concentration of the solution of &lt;strong&gt;R&lt;/strong&gt; is the &lt;em&gt;standardization of the titrant&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-02-28T21:29:36.910" LastActivityDate="2014-02-28T21:29:36.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8812" PostTypeId="1" AcceptedAnswerId="8818" CreationDate="2014-02-28T20:33:05.853" Score="3" ViewCount="856" Body="&lt;p&gt;I tried writing an analogy to explain the difference between enantiomers and diastereomers. However, I'm not entirely sure if it's actually a correct representation of the differences. It might also be more confusing than it has to be:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To use an everyday object as an illustration of the difference, imagine two forks with four tines each. If fork 1 has its first and third tine longer than the other ones, and fork 2 has its second and fourth tine longer than the other ones, then the two forks would not match completely if you tried to stack them on top of each other. However, if you held fork 2 up to a mirror, then you would see it now has both its elongated tines in the same place as fork 1. These forks would be enantiomers of each other because they are the same thing, but mirror images of each other&lt;/p&gt;&#xA;&lt;p&gt;If you changed fork 2 to have its first and second tine longer than the other ones, instead of the second and the fourth, then the two forks would instead be diastereomers. This is because now they both cannot match completely when stacked on top of each other, and holding either one up to a mirror will not make them look any more similar. This is exactly how diastereomers work too, because it only partially mirrors its counterpart. As you can see, fork 2 now has its first elongated tine actually match the first elongated tine of fork 1, while also retaining the effect that holding it up to a mirror will have its second elongated tine match that of fork 1. That is, in the mirror fork 2 will have its third tine elongated, just like fork 1 does when not viewed through a mirror. In other words, fork 2 has its first elongated tine match fork 1’s, but its second elongated tine is the mirror image of fork 1’s. This makes them diastereomers because they are the same thing, but neither structurally identical nor complete mirror images of each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, is there anything off about my analogy, or does it correctly and comprehensibly explain the difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="4464" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-03-01T01:01:43.293" Title="Is this analogy explaining the difference between enantiomers and diastereomers correct?" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8813" PostTypeId="2" ParentId="8387" CreationDate="2014-02-28T20:53:23.503" Score="8" Body="&lt;p&gt;According to the Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkali_metal_oxide&quot; rel=&quot;nofollow&quot;&gt;Alkali metal oxides&lt;/a&gt;, the superoxide of $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ form when these metals are burnt in air, and according to the Bodner Research Group page &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch9/active.php&quot; rel=&quot;nofollow&quot;&gt;Active Metal Reactions&lt;/a&gt;, they&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;react so rapidly with oxygen&lt;/strong&gt; they form superoxides, in which the alkali metal reacts with $\ce{O2}$ in a 1:1 mole ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given, according to the Chem-Guide blog article &lt;a href=&quot;http://chem-guide.blogspot.com.au/2010/04/alkali-metals.html&quot; rel=&quot;nofollow&quot;&gt;Alkali metals&lt;/a&gt; that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which, according to the blog, results in the  $\ce{Li+}$ ion (which is small, possessing a larger positive field around it) only being able to naturally combine with a small anion $\ce{O^{2-}}$ forming $\ce{Li2O}$, whereas the substantially larger $\ce{K+}$, $\ce{Rb+}$ and $\ce{Cs+}$ have weaker positive fields and can stabilise the larger superoxide anion $\ce{O2^{-}}$ forming superoxides.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="-1" LastEditDate="2015-05-05T17:23:56.760" LastActivityDate="2015-05-05T17:23:56.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8814" PostTypeId="2" ParentId="8803" CreationDate="2014-02-28T21:19:22.017" Score="1" Body="&lt;p&gt;If anyone else is running into the same issue, after selecting which spectra you would like to overlay go to series -&gt; collect spectra and then select &quot;manual offset&quot; if you want to adjust the height.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-02-28T21:19:22.017" CommentCount="0" CommunityOwnedDate="2014-02-28T21:19:22.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="8815" PostTypeId="2" ParentId="8709" CreationDate="2014-02-28T22:57:28.963" Score="2" Body="&lt;p&gt;I never worked in the field but applied mechanical stress (French Press, sonication) while extracting chlorophyll a from cyanobacteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, a cursory search furnished some sources like &lt;a href=&quot;http://www.ebioscience.com/resources/faq/flow-cytometry-faq.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, where vigorous vortexing of antibody samples is advised against. Instead, &lt;strong&gt;gentle&lt;/strong&gt; or &lt;strong&gt;pulse&lt;/strong&gt; vortexing or simple spinning is suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But please take this with a grain of salt since I'm not really familiar with the standard protocols here.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-02-28T22:57:28.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8816" PostTypeId="2" ParentId="8370" CreationDate="2014-03-01T00:00:12.770" Score="5" Body="&lt;p&gt;The smell is often noticed at construction sites, when reinforcing bars are welded or cut. Constructional steel contains about 0.1% carbon and phosphorus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus normally isn't welcome since it renders steels more brittle. On the other hand, it increases the corrosion stability. This is is a desired property for rebars embedded in concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it conceivable that noticable (alkyl)phosphanes are formed when this steel is welded or cut with a disc grinder?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-01T00:00:12.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8817" PostTypeId="1" CreationDate="2014-03-01T00:41:11.113" Score="7" ViewCount="1235" Body="&lt;p&gt;I am wondering what chemicals exactly contribute to the smell of a septic tank system. I read about $\ce{H2SO4}$, $\ce{NH3}$, but I am not sure. Is methane also a factor or is that odorless?&lt;/p&gt;&#xA;" OwnerUserId="4672" LastEditorUserId="402" LastEditorDisplayName="user4076" LastEditDate="2017-04-25T01:19:13.403" LastActivityDate="2017-04-25T01:19:13.403" Title="What chemicals cause the smell of a septic tank?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8818" PostTypeId="2" ParentId="8812" CreationDate="2014-03-01T01:01:43.293" Score="4" Body="&lt;p&gt;Enantiomers are non-superposable mirror images (all the signs of one spatial coordinate are reversed).  Diastereomers are  formed by inverting some but not all stereocenters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&quot; rel=&quot;nofollow&quot;&gt;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&lt;/a&gt; &lt;BR&gt; &#xA; pp. 11-13.  Excellent overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have some fun with your prof!  An acyclic, aliphatic, unstrained, tetrahedral carbon atom is $\ce{sp^3}$-hybridized.  It is an uncharged, valence-saturated singlet.  It bears four freely rotating, rigorously identical substituents, e.g., $\ce{CR4}$, with identical bond lengths to that central carbon.  Said carbon is a chiral center.  It has no ${S_n}$ improper axes of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Show how this can be true. &lt;BR&gt;&#xA;2) Use any system of nomenclature to name left-handed and right-handed stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acyclic, unstrained, etc., trigonal planar $\ce{sp^2}$-hybridized carbons ${R2&amp;gt;C=C&amp;lt;R2}$ are each identical stereocenters.  Each carbon and the olefin overall have no ${S_n}$ improper axes of symmetry.  The olefin is resolved chiral.  (1) and (2) as above. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-01T01:01:43.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8819" PostTypeId="2" ParentId="8817" CreationDate="2014-03-01T01:40:20.880" Score="6" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Septic_system#Environmental_issues&quot; rel=&quot;nofollow noreferrer&quot;&gt;Septic tanks&lt;/a&gt;, as septic tanks are generally anaerobic environments, pungent smells from the formation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen sulfide&lt;/a&gt; $\ce{H2S}$, which is sometimes referred to as 'rotten egg gas'. As you stated, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonia&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonia&lt;/a&gt; $\ce{NH3}$ can be produced - also with a distinct smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, according to the Wikipedia article, fermentation could occur, producing the odourless &lt;a href=&quot;http://en.wikipedia.org/wiki/Methane&quot; rel=&quot;nofollow noreferrer&quot;&gt;methane&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon dioxide&lt;/a&gt; as well as &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphates&quot; rel=&quot;nofollow noreferrer&quot;&gt;phosphates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, $\ce{H2SO4}$is the chemical formula for &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;sulfuric acid&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="402" LastEditDate="2017-04-25T00:46:20.250" LastActivityDate="2017-04-25T00:46:20.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8820" PostTypeId="1" CreationDate="2014-03-01T02:37:17.130" Score="3" ViewCount="595" Body="&lt;p&gt;I am attempting to generate the probability for the specific speed of a molecule using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution&quot; rel=&quot;nofollow&quot;&gt;Maxwell-Boltzmann distribution&lt;/a&gt;, but I cannot decide on which temperature I should use in the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a very, very cold liquid (let's say 70 K) at room temperature (we'll call that 296.15 K), but would I use the temperature of the liquid in the equation, or should I use the temperature of the air around the liquid?&lt;/p&gt;&#xA;" OwnerUserId="581" LastEditorUserId="16683" LastEditDate="2016-09-30T17:49:29.147" LastActivityDate="2016-09-30T17:49:29.147" Title="Temperature in the Maxwell-Boltzmann Distribution" Tags="&lt;temperature&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8821" PostTypeId="2" ParentId="7279" CreationDate="2014-03-01T06:17:11.057" Score="2" Body="&lt;p&gt;This question is a little bit broad, but thankfully, there are many resources available.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to part of your question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which one is most suitable for packaging pastries, for example, biscuits?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Natural Resources Defense Council webpage &lt;a href=&quot;http://www.nrdc.org/living/shoppingwise/food-storage-containers.asp&quot; rel=&quot;nofollow&quot;&gt;Food Storage Containers&lt;/a&gt;, they state that the safest plastics are PET, HDPE, LDPE, and PP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparing the various types of plastic, the &lt;a href=&quot;http://www.containerandpackaging.com/info/plasticology.asp&quot; rel=&quot;nofollow&quot;&gt;Plasticology 101&lt;/a&gt; website compares the types of plastic according to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Temperature tolerances&lt;/li&gt;&#xA;&lt;li&gt;Clarity&lt;/li&gt;&#xA;&lt;li&gt;Chemical, scratch and impact resistance&lt;/li&gt;&#xA;&lt;li&gt;Rigidity&lt;/li&gt;&#xA;&lt;li&gt;Food contact acceptability&lt;/li&gt;&#xA;&lt;li&gt;Recyclable material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They suggest that PET, HDPE, PVC, LDPE, PP and PS are suitable for food, but temperature tolerances would be a major factor in what food they store.  Another chemical comparison that needs to be considered for the storage of food is information about the $\ce{O2}$, $\ce{N2}$, $\ce{CO2}$ gas and moisture transmission rates are included in Evergreen's document &lt;a href=&quot;http://www.evergreen-packaging.com/en/uploadfile/201111/20111122180814336.pdf&quot; rel=&quot;nofollow&quot;&gt;The barrier performance of common plastic film&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically for biscuits, and any other baked goods, according to sevral websites, it depends on what the ingredients are - things to consider are the dairy content, whether the plastic will react with some of the ingredients etc.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2014-03-01T07:41:09.597" LastActivityDate="2014-03-01T07:41:09.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8822" PostTypeId="2" ParentId="7022" CreationDate="2014-03-01T07:12:44.170" Score="3" Body="&lt;p&gt;In a bit of a web search, there does not seem to be any particular names for families of periods in the Periodic Table.  However, according to &lt;a href=&quot;http://www.webelements.com/nexus/chemistry_page/periods-periodic-table&quot; rel=&quot;nofollow noreferrer&quot;&gt;WebElements&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The f-block elements are assigned to Periods 6 (lanthanoids) and 7 (actinoids) since that is where they are located in the full, or extended, version of the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, these, according to the image below, only apply to the Lanthanides and Actinides, not all of periods 6 and 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P5tMr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-03-01T07:12:44.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8824" PostTypeId="1" AcceptedAnswerId="8825" CreationDate="2014-03-01T13:36:10.750" Score="5" ViewCount="3279" Body="&lt;p&gt;Is the following sequence of reactions correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH2CH2 -&amp;gt;[][\ce{NH3},\,\ce{Na}] NaHC=CHNa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCH=CHNa + 2CH3CH2Br-&amp;gt; CH3CH2CH=CHCH2CH3}$$&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="5026" LastEditDate="2018-02-23T02:30:21.653" LastActivityDate="2018-02-23T02:30:21.653" Title="Reaction of ethene with ammonia and sodium" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8825" PostTypeId="2" ParentId="8824" CreationDate="2014-03-01T14:13:40.927" Score="3" Body="&lt;p&gt;No, because for ethene, $\mathrm{p}K_\text{a} = 44$. You won't be able to abstract even one proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe, you were thinking in ethyne ($\ce{HC#CH})$? In that case, you are right! The bis-alkylation would work with sodium amide as a base! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is the simultaneous abstraction of both protons the most plausible reaction then? &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T13:04:04.283" LastActivityDate="2015-04-03T13:04:04.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8826" PostTypeId="1" AcceptedAnswerId="8834" CreationDate="2014-03-01T15:08:31.837" Score="0" ViewCount="8389" Body="&lt;p&gt;I mean like how can I convert ethane into propane, ethene into propene and like this further on&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="865" LastEditDate="2015-10-02T16:45:26.957" LastActivityDate="2015-10-02T16:45:26.957" Title="how to add a methyl group to a straight carbon chain" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" ClosedDate="2015-10-02T16:59:09.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="8827" PostTypeId="1" AcceptedAnswerId="8864" CreationDate="2014-03-01T15:16:08.617" Score="5" ViewCount="16234" Body="&lt;p&gt;What can I do as a non-professional chemist that is legal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a license is not required, then what are the advantages to having one, such as ordering compounds, getting a job, etc.?&lt;/p&gt;&#xA;" OwnerUserId="4251" LastEditorUserId="5456" LastEditDate="2014-10-09T01:29:51.070" LastActivityDate="2014-10-09T07:59:00.273" Title="Do I need a license or certificate to practice chemistry or what are the advantages if did have one?" Tags="&lt;organic-chemistry&gt;" AnswerCount="4" CommentCount="1" ClosedDate="2015-10-09T16:46:10.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="8828" PostTypeId="2" ParentId="8827" CreationDate="2014-03-01T15:26:46.627" Score="3" Body="&lt;p&gt;Over here, that is Germany, you will hardly find a job in chemical industry or academia without an academic degree (B.Sc., M.Sc, Ph.D) or a formal technical education (typically three years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most suppliers nowadys only deliver chemicals to universities, research facilities or other chemical companies.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-01T15:26:46.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8829" PostTypeId="2" ParentId="8827" CreationDate="2014-03-01T15:54:38.543" Score="0" Body="&lt;p&gt;Chemists are not licensed in the US.  A non-degreed &quot;chemist&quot; is at best a technician.  Homeland Severity bars purchase of chemicals (anything requiring a Material Safety Data Sheet) except for academia, corporations, and tis own pseudopodia.  OSHA (Occupational and Safety Health Administration) can declare a &quot;hazard.&quot;  That can be a $10,000/incident-day fine, plus confiscation and closure.  All this is done above the law.  No prior police or court intervention is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the real terror begins - the War on Drugs.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-01T15:54:38.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8830" PostTypeId="1" CreationDate="2014-03-01T16:10:54.627" Score="4" ViewCount="8186" Body="&lt;p&gt;Please explain how to derive the magnetism  of polyatomic molecules? If we derive it using an addition of an oxygen atom to molecular oxygen ($\ce{O2}$) it will come to be paramagnetic like $\ce{O3}$?&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="4325" LastEditorDisplayName="user4076" LastEditDate="2015-03-11T19:29:27.973" LastActivityDate="2015-03-11T19:29:27.973" Title="Why is ozone O₃ diamagnetic?" Tags="&lt;ozone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8831" PostTypeId="2" ParentId="8826" CreationDate="2014-03-01T16:12:36.420" Score="2" Body="&lt;p&gt;Why would you want an expensive process to convert cheap molecules into other cheap molecules?  Brominate ethane, make the organo-lithium, displace methyl iodide. Epoxidize ethylene, react with methyl lithium, eliminate to the olefin. Silly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing something cheap, easy, and fast on a huge scale with unwanted methane from oil wells to form a high energy density liquid for pipeline conveyance will make you rich. Methanol is marginal - and still can't be done.  Perhaps methane plus air into titanium-silicalite (plus hydrogen peroxide, or cobalt- or vanadium-doped plus air) then that mixed product through ZSM-5 into aromatic gasoline.  That could be exciting in various ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.apcata.2008.04.020&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/j.apcata.2008.04.020&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://chemelab.ucsd.edu/methanol/memos/final.html&quot; rel=&quot;nofollow&quot;&gt;http://chemelab.ucsd.edu/methanol/memos/final.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-01T16:12:36.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8832" PostTypeId="2" ParentId="8826" CreationDate="2014-03-01T16:19:04.340" Score="0" Body="&lt;p&gt;Firstly, you cant go ethane to propene, it should be propane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemically, you can do it by following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;halogenate with $\ce{Cl}$ in UV.&lt;/li&gt;&#xA;&lt;li&gt;then use $\ce{KCN}$ to make nitrile - it will replace $\ce{Cl}$.&lt;/li&gt;&#xA;&lt;li&gt;hydrolyse it using $\ce{KOH}$.&lt;/li&gt;&#xA;&lt;li&gt;then it will be a carboxylic acid. &lt;/li&gt;&#xA;&lt;li&gt;reduce it using $\ce{LiAlH4}$ and then, you have got propane.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can go for ethene the same way, but there will be complications.&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorDisplayName="user4076" LastEditDate="2014-03-01T22:33:51.920" LastActivityDate="2014-03-01T22:33:51.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8833" PostTypeId="1" CreationDate="2014-03-01T16:29:53.127" Score="0" ViewCount="5257" Body="&lt;p&gt;Please explain what reaction is this and what are the products of reagent $\ce{(NH4)2CO3}$&#xA;at 100-115 C with ketone.&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorDisplayName="user4076" LastEditDate="2014-03-01T22:27:58.833" LastActivityDate="2014-03-01T22:27:58.833" Title="reaction of diketone with ammonium carbonate at 100-115 degree celsius" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2014-03-03T16:11:19.863" ContentLicense="CC BY-SA 3.0" />
  <row Id="8834" PostTypeId="2" ParentId="8826" CreationDate="2014-03-01T16:36:40.537" Score="1" Body="&lt;p&gt;Direct ethene to propene (ETP) conversion seems &lt;a href=&quot;http://dx.doi.org/10.3390/molecules16097844&quot; rel=&quot;nofollow&quot;&gt;possible&lt;/a&gt; at 400 °C on mesoporous silica (M41) loaded with nickel ions.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-01T16:36:40.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8835" PostTypeId="2" ParentId="8833" CreationDate="2014-03-01T16:48:16.820" Score="13" Body="&lt;p&gt;1,4-diketones are likely to give pyrroles in a &lt;em&gt;Paal-Knorr&lt;/em&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B8Xml.png&quot; alt=&quot;paal-knorr&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-01T20:16:00.087" LastActivityDate="2014-03-01T20:16:00.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8837" PostTypeId="2" ParentId="8632" CreationDate="2014-03-01T17:02:12.087" Score="2" Body="&lt;p&gt;Cathode rays travel in a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was confirmed with a Maltese cross covered with phosphor that illuminates with radiation. As voltage is applied between the cathode and the anode, the cathode rays emitted struck the metal Maltese cross used, creating a very sharp image of the same cross at the end of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crooke's tubes need to be evacuated, as gas particles within would scatter the electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dm5g6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image 1 shows the Crookes tube apparatus. The gas pressure inside is reduced and eventually disappears at very low pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In image 2, even though evacuated, the rays that come from the cathode still persist, proving that the rays are emitted from the cathode and different from light as they travel in straight lines.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-01T17:09:25.803" LastActivityDate="2014-03-01T17:09:25.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8838" PostTypeId="1" CreationDate="2014-03-01T17:27:02.517" Score="2" ViewCount="1538" Body="&lt;p&gt;I am a second year mathematics student. I am mostly interested in statistics. Right now my curriculum doesn't offer me much choice (all second year courses are mandatory) but next semester I will be able to select the courses I want. Among the ones that are offered is a course called &lt;em&gt;Biochemistry&lt;/em&gt;. It is actually a second year class for those how study chemistry and chemical engineering but I guess there must be applications of statistics in this field since it is also offered to third year math students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I am a math student: I have no knowledge or whatsoever in chemistry or biology. Granted that I had some in high school but back then I just learned everything by heart in order to pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With no knowledge of chemistry, is it realistic to undertake a biochemistry class?&lt;/li&gt;&#xA;&lt;li&gt;The course description lists the following prerequisites: &quot;Basic knowledge in general and organic chemistry&quot;. How important are these subjects in biochemistry? What are some good references (books) for self-learning these subjects?&lt;/li&gt;&#xA;&lt;li&gt;Is Stryer's Biochemistry a good book for self-learning?&lt;/li&gt;&#xA;&lt;li&gt;Are biochemistry and biostatistics related at all?&lt;/li&gt;&#xA;&lt;li&gt;Are there many applications of statistics in biochemistry?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have already searched for some information on the internet but I'd like to have some specific answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps here is the &lt;a href=&quot;http://isa.epfl.ch/imoniteur_ISAP/!itffichecours.htm?ww_i_matiere=4144900&amp;amp;ww_x_anneeAcad=2013-2014&amp;amp;ww_i_section=944590&amp;amp;ww_i_niveau=6683117&amp;amp;ww_c_langue=en&quot; rel=&quot;nofollow&quot;&gt;information page&lt;/a&gt; of this class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I hope this kind of question is accepted here, if not please point me to a better place. Thanks in advance!&lt;/p&gt;&#xA;" OwnerDisplayName="user4680" LastEditorDisplayName="user4680" LastEditDate="2014-03-01T17:46:05.393" LastActivityDate="2014-05-29T23:56:51.690" Title="Biochemistry for a mathematics student?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2017-11-09T04:01:45.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="8839" PostTypeId="2" ParentId="8830" CreationDate="2014-03-01T19:44:19.947" Score="2" Body="&lt;p&gt;Ozone's ground state is a singlet.  All electrons therein are paired,&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SnGC9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5Zp0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.springer.com/article/10.1134%2FS0018143906040060#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/article/10.1134%2FS0018143906040060#page-1&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://link.springer.com/article/10.1134%2F1.1870062#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/article/10.1134%2F1.1870062#page-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4231" LastEditDate="2015-03-11T19:19:45.817" LastActivityDate="2015-03-11T19:19:45.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8840" PostTypeId="2" ParentId="8817" CreationDate="2014-03-01T19:47:07.717" Score="5" Body="&lt;p&gt;Some of the &quot;fragrances&quot; are formed in the degradation of different amino acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amaterasu has already described that hydrogen sulfide is an important component in the septic tank perfume. It is formed in the degradation of the sulfur-containing amino acids methionine (&lt;strong&gt;1&lt;/strong&gt;) and cysteine (&lt;strong&gt;2&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmvrQ.png&quot; alt=&quot;sulfur-containing amino acids&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;http://dx.doi.org/10.1016/j.watres.2011.11.043.&quot; rel=&quot;nofollow noreferrer&quot;&gt;anaerobic degradation of methionine&lt;/a&gt;, the initially formed methylmercaptan ($\ce{CH3SH}$, &lt;strong&gt;3&lt;/strong&gt;) is methylated to dimethylsulfide ($\ce{CH3-S-CH3}$, &lt;strong&gt;4&lt;/strong&gt;). Further microbial transformations of &lt;strong&gt;3&lt;/strong&gt; and &lt;strong&gt;4&lt;/strong&gt; eventually yield $\ce{H2S}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/67ok9.png&quot; alt=&quot;tryptophane and skatole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degradation of the amino acid tryptophan (&lt;strong&gt;5&lt;/strong&gt;) furnishes 3-methylindole (&lt;em&gt;skatole&lt;/em&gt;, &lt;strong&gt;6&lt;/strong&gt;), the aroma of feces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for the completion of the sensoric enjoyment: Other amino acids, such as ornithine (&lt;strong&gt;7&lt;/strong&gt;) and lysine (&lt;strong&gt;8&lt;/strong&gt;) form 1,4-diaminobutane (&lt;strong&gt;9&lt;/strong&gt;) and 1,5-diaminobutane (&lt;strong&gt;10&lt;/strong&gt;), respectively upon decarboxylation (= loss of $\ce{CO2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3xyQe.png&quot; alt=&quot;decarboxylation of ornithine and lysine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trivial names of these diamines are &lt;em&gt;putrescine&lt;/em&gt; and &lt;em&gt;cadaverine&lt;/em&gt;, which suggests another odor than that of roses and lemons ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm however not sure if the last two compounds are actually formed in septic tanks.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-06T10:48:53.040" LastActivityDate="2014-03-06T10:48:53.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8841" PostTypeId="1" AcceptedAnswerId="8842" CreationDate="2014-03-01T22:38:47.947" Score="1" ViewCount="342" Body="&lt;p&gt;I'm trying to calculate the $\Delta\mathrm{G}$ of the following reaction. Are my calculations correct? I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the $\Delta\mathrm{G}$ is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CH3OH + H2O --&amp;gt; HCOOH + 4H        Delta H = 99.49 kJ/mol&#xA;&#xA;% Standard enthalpy of formation of every compound&#xA;&#xA;% -238.7 kJ/mol CH3OH&#xA;% -285.830 kJ/mol H2O&#xA;% -425 kJ/mol HCOOH&#xA;%    0 kJ/mol H2&#xA;% -393.51 kJ/mol CO2&#xA;&#xA;% Standard entropy of formation of every compound&#xA;&#xA;% 126.8 J/mol*K CH3OH&#xA;%  69.91 J/mol*K H2O&#xA;% 129 J/mol*K HCOOH&#xA;% 130.6 J/mol*K H2&#xA;% 213.6 J/mol*K CO2&#xA;&#xA;delta_H_1 = -425 - (-238.7 - 285.830);&#xA;delta_S_1 = 129 + 130.6 - (126.8 + 69.91);&#xA;T = 273.15 + 20;&#xA;&#xA;delta_G_1 = delta_H_1 - T * delta_S_1;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Correct version&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;delta_H_1 = -425 - (-238.7 - 285.830);&#xA;delta_S_1 = 129 + 4*130.6 - (126.8 + 69.91);&#xA;T = 273.15 + 25;&#xA;&#xA;delta_G_1 = delta_H_1 - T * delta_S_1;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1829" LastEditorUserId="1829" LastEditDate="2014-03-01T23:50:59.547" LastActivityDate="2014-03-01T23:50:59.547" Title="Clarifications on the calculation of the $\Delta\mathrm{G}$ of a reaction" Tags="&lt;enthalpy&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8842" PostTypeId="2" ParentId="8841" CreationDate="2014-03-01T23:04:47.733" Score="3" Body="&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibbs Free energy describes how much &quot;free energy&quot; is available, or how much work is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^\circ$ can either be negative or positive even if $\Delta H^\circ$ is $+$. Therefore, &#xA;we have to look at all three terms, $\Delta H^\circ$, $T$, and $\Delta S^\circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs Free Energy, enthalpy, and entropy are state functions. Although I won't derive it here, we can find $\Delta G$, or $\Delta H$, or $\Delta S$, by using Hess's Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information provided above gives us the $\Delta H_{rxn}$ which is $99.49 \text{kJ/mol}$ and thus endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We calculate entropy from the a list of experimental tabulated values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S^\circ_{rxn} = (129.0 \text{J}\cdot\text{mol}^{-1}\text{K} + 4*130.6 \text{J}\cdot\text{mol}^{-1}\text{K}) - (126.8 \text{J}\cdot \text{mol}^{-1}\text{K} + 69.91 \text{J}\cdot\text{mol}^{-1}\text{K})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S^\circ_{rxn} = 454.69 \text{J}\cdot\text{mol}^{-1}\text{K} = 0.45469 \text{kJ}\cdot\text{mol}^{-1}\text{K} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While solving for entropy, you forgot to multiply the entropy of hydrogen gas with the number of moles. Also, you should always include the states of your reactants and products, as thermodynamic data for compounds differs for different states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that is incorrect is standard temperature for thermodynamic reactions is at $25^\circ C$ not $20^\circ C$ and thus your temperature should be $298.15 K$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the ΔG is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is where a lot of people make mistakes. It is true, that in general, endothermic reactions with a positive enthalpy usually have a positive $\Delta G$ and are thus not spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we look at our equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to consider our other terms and not only enthalpy, namely that of temperature and entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction is spontaneous if $ \Delta G &amp;lt; 0 $ such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \Delta H &amp;lt; 0$ and (-) and $ \Delta S &amp;gt; 0$ (+), the reaction is spontaneous at any temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \Delta H &amp;gt; 0$ and (+) and $ \Delta S &amp;lt; 0$ (-), the reaction is NOT spontaneous at any temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \Delta H &amp;lt; 0$ and (-) and $ \Delta S &amp;lt; 0$ (-), the reaction is spontaneous at low enough temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ \Delta H &amp;gt; 0$ and (+) and $ \Delta S &amp;gt; 0$ (+), the reaction is spontaneous at high enough temperatures. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-01T23:42:57.287" LastActivityDate="2014-03-01T23:42:57.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8843" PostTypeId="1" AcceptedAnswerId="8846" CreationDate="2014-03-01T23:39:17.433" Score="3" ViewCount="1810" Body="&lt;p&gt;How is Gold's atomic structure and bonding related to the separation techniques used to distinguish it from other substances? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just confused on what I'm actually searching for, I have read so many blogs on Gold extraction, refinement, occurences and more topics, and either I'm skipping the important parts or the answer's just not there. What can I search to find what I'm looking for? The separation technique I chose is flotation, can someone lead me in the correct path and/or give me a few hints on what to start of with? Thanks in advance, much appreciated! &lt;/p&gt;&#xA;" OwnerUserId="4685" LastEditorUserId="4685" LastEditDate="2014-03-06T06:06:16.153" LastActivityDate="2014-09-24T14:52:33.290" Title="Gold's atomic structure and bonding in relation to the separation techniques used" Tags="&lt;bond&gt;&lt;biochemistry&gt;&lt;metal&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8844" PostTypeId="1" AcceptedAnswerId="8845" CreationDate="2014-03-01T23:54:18.713" Score="1" ViewCount="296" Body="&lt;p&gt;According to Wikipedia 4-methyl-1-pentanol is $\ce{C_6H_{14}O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is what I got, which is slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methyl = Think of methyl group, which is $\ce{CH_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pentanol = &lt;em&gt;Penta&lt;/em&gt; means we have 5 carbons, &lt;em&gt;-ane&lt;/em&gt; ending means we have $2n+2$ hydrogens, i.e. 12 hydrogens and &lt;em&gt;-ol&lt;/em&gt; ending means we have a hydroxyl group $\ce{-OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the molecular formula that I arrived at is: $\ce{CH_3C_5H_{12}O}$ = $\ce{C_6H_{15}O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I end up with an extra hydrogen compared to Wikipedia answer?&lt;/p&gt;&#xA;" OwnerUserId="4686" LastEditorUserId="7475" LastEditorDisplayName="user4076" LastEditDate="2015-10-09T16:43:46.537" LastActivityDate="2019-04-08T13:22:42.510" Title="Molecular formula for 4-methyl-1-pentanol?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8845" PostTypeId="2" ParentId="8844" CreationDate="2014-03-02T00:09:11.593" Score="4" Body="&lt;p&gt;Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$.  In any case, in your &quot;2n+2&quot; step, you need to reduce that number of hydrogens by 2.  One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached.  So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user4076" LastEditDate="2014-03-02T00:43:56.427" LastActivityDate="2014-03-02T00:43:56.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8846" PostTypeId="2" ParentId="8843" CreationDate="2014-03-02T00:26:40.223" Score="4" Body="&lt;p&gt;There are differences in gold extraction for different mines - depending on the chemical nature of the raw ore, the impurities present, the availability of resources etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of gold extraction comes from the Super Pit mine in Kalgoorlie, Western Australia; their gold extraction process is outlined in their website &lt;a href=&quot;http://www.superpit.com.au/Production/MineralProcessing/tabid/150/Default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mineral Processing&lt;/a&gt;, which is outlined below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Crushing and grinding:&lt;/strong&gt; this is a physical process where the mined raw ore is crushed, ground and water added to make a slurry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Flotation:&lt;/strong&gt; chemicals are added to the slurry to remove everything other than gold and pyrite.  Water is also removed at this stage.  The slurry is agitated forming air bubbles, which then rise through the slurry, carrying mineral grains.  From &lt;a href=&quot;http://www.cpchem.com/bl/specchem/en-us/Pages/IntroductiontoMineralProcessing.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to Mineral Processing&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ability of a mineral to float depends upon its surface properties. Chemical modification of these properties enables the mineral particles to attach to an air bubble in the flotation cell. The air bubble and mineral particle rise through the pulp to the surface of the froth or foam that is present on the flotation cell. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Roasting:&lt;/strong&gt; water is re-added and the concentrated slurry is roasted to remove $\ce{SO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CIL:&lt;/strong&gt; or carbon in leach, chemicals are added to dissolve the gold and have close to pure gold precipitate on the carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elution:&lt;/strong&gt; A mixture of caustic soda and cyanide is used to liberate the gold from the carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific request for resources about flotation, the example process below is for a sulphide copper ore containing iron bearing pyrite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uhXO7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.jmeech.mining.ubc.ca/MINE290/proces/Fig_cu2.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested additional resources about flotation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.mtu.edu/chem_eng/faculty/kawatra/Flotation_Fundamentals.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Froth Flotation – Fundamental Principles&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great resource is the book &lt;a href=&quot;http://www.sciencedirect.com/science/book/9780444530820&quot; rel=&quot;nofollow noreferrer&quot;&gt;Handbook of Flotation Reagents: Chemistry, Theory and Practice&lt;/a&gt; (chapters are behind a paywall).&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2014-03-02T00:38:29.310" LastActivityDate="2014-03-02T00:38:29.310" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8847" PostTypeId="1" CreationDate="2014-03-02T00:44:26.397" Score="3" ViewCount="1152" Body="&lt;p&gt;This is the boiling-point elevation equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta\mathrm{T}_{b}=ib\mathrm{K}_{eb}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do the limit of the above, for $b\rightarrow\infty$ we find that the boiling-point elevation is infinite. How this could be explained under a physical view?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{b~\rightarrow~\infty}ib\mathrm{K}_{eb}=\infty,~\text{if}~i\wedge b &amp;gt; 0.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we use a solute molality that is so big that the solution would be something like the pure solute, the final boiling-point of the solution, in the case of water as solvent for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{T}_{b}=ib\mathrm{K}_{eb} + 100\mathrm{°C}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wouldn't be like the boiling-point of the pure solute? I mean, something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{b~\rightarrow~\infty}\mathrm{T}_{solution}=\mathrm{T}_{solute}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, please forget what I've said above about $b\rightarrow\infty$. And take a look at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(b) = 0.86b + 64.7.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9qlxB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see is, the boiling-point of a solution of $\mathrm{CH_3OH}$ with $\mathrm{H_2O}$ as solute would be $322.7\mathrm{°C}$ for $b = 300$. This it's weird to me  'cause that temperature would be about three times higher then the boiling-point of the pure water!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{T}_{b}=ib\mathrm{K}_{eb} + \mathrm{T_{solvent}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;holds true if &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;b = \frac{x~\mathrm{g}_{solute}}{1000~\mathrm{g}_{solvent}} \le 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;right? If yes why (in the above case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;0.86\frac{1000~\mathrm{g}}{18.01528~\mathrm{g}/\mathrm{mol}}\frac{1~\mathrm{mol}^2}{1~\mathrm{kg}} + 64.7~\mathrm{°C} = 112.4~\mathrm{°C}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1829" LastEditorUserId="1829" LastEditDate="2014-03-02T14:45:03.320" LastActivityDate="2014-03-02T14:45:03.320" Title="Boiling-point elevation paradox" Tags="&lt;physical-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8848" PostTypeId="2" ParentId="8847" CreationDate="2014-03-02T01:33:16.560" Score="4" Body="&lt;p&gt;The flaw in your reasoning is that the molality $b = n(\text{solute})/m(\text{solvent})$ cannot be infinite even if you &lt;code&gt;use a solute molality that is so big that the solution would be something like the pure solute&lt;/code&gt;. Take water as an example:&#xA;How concentrated is water? Very concentrated you may say - but the concentration is limited. We know that one mole of pure water has a mass of 18 g and occupies 18 cm³. So, in one dm³, there are 1000/18 = 55.56 mol which gives you a water concentration of 55.56 mol/dm³. You cannot get more concentrated water than this (unless you did something drastic like taking it into a black hole!). For the molality it is quite similar: If you dissolve something in water the solute concentration cannot exceed the concentration of water, so $n(\text{solute})$ (and therefore $b$) has an upper bound.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-03-02T02:04:34.700" LastActivityDate="2014-03-02T02:04:34.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8849" PostTypeId="2" ParentId="8847" CreationDate="2014-03-02T01:42:32.580" Score="5" Body="&lt;p&gt;In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the &lt;em&gt;liquid&lt;/em&gt; solvent molecules at the liquid-gas-interface will enter the gas phase. The &lt;em&gt;gaseous&lt;/em&gt; solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of solvent molecules going from gas-&gt;liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid-&gt;gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -&gt; gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower...  So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -&gt; liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why boiling point goes infinite:&lt;/strong&gt; At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="1160" LastEditDate="2014-03-02T02:35:06.973" LastActivityDate="2014-03-02T02:35:06.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8850" PostTypeId="1" CreationDate="2014-03-02T06:02:49.470" Score="3" ViewCount="441" Body="&lt;p&gt;Most combustion happens involves the fuels burning in an oxidant. But there seems to be no reason that it cannot happen in the opposite way. Are there any experiments on oxygen burning in hydrogen? Or some oxidant in solid state (like $\ce{KMnO4,KClO3,}$...) burning in a reductive gas?&lt;/p&gt;&#xA;" OwnerUserId="4675" LastEditorDisplayName="user7951" LastEditDate="2016-12-22T13:59:44.837" LastActivityDate="2016-12-22T13:59:44.837" Title="Can oxygen burn in hydrogen?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;combustion&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8851" PostTypeId="2" ParentId="8850" CreationDate="2014-03-02T06:31:23.687" Score="4" Body="&lt;p&gt;Could you be more clear on what 'burning &lt;strong&gt;in&lt;/strong&gt;' something means? The chemical reactions require both reactants available, so they need to mix - in what sense is Hydrogen mixing and reacting with Oxygen any different that Oxygen mixing and reacting with Hydrogen?&lt;/p&gt;&#xA;&lt;p&gt;In some sense, we do tend to control these reactions via the fuel, not the oxidant (oxygen free environments are difficult to reproduce on Earth's surface). However there may be some examples of controlling the mixture via restricting the oxidant in a fuel-rich environment... even though I can't even get through the sentence without that sounding wasteful.&lt;/p&gt;&#xA;&lt;h3&gt;Nitrous Oxide&lt;/h3&gt;&#xA;&lt;p&gt;Would you consider the example of Nitrous Oxide injection in an internal combustion engine an example of this? Pre-nitrous, there is compressed atmospheric oxygen and hydrocarbons at a set ratio, and to further increase the burn a nitrous system will inject either NO alone (dry system) or NO and more hydrocarbon fuel (wet system) to add more oxidants that would be available through the 21% Oxygen in the atmosphere.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Chemical: &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrous_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Nitrous_oxide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Internal Combustion application: &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrous&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Nitrous&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Candles&lt;/h3&gt;&#xA;&lt;p&gt;How about a candle? There's an abundant fuel source (wax), and we frequently control the shape and characteristics of the burn with shaped glass, not to mention put it out via starving it of oxidant.&lt;/p&gt;&#xA;&lt;p&gt;In either case, if you were to, say take a lump of solid oxygen (nevermind the practical difficulties of dealing with temperatures below 54 Kelvin) and maintained a burn with Hydrogen, what you'd be likely to find would be combustion only happening at the edges, where the oxygen was evaporating and mixing with the reduction agent, similar to lighting a can of gasoline on fire (only the vapors are flammable).&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-03-02T06:31:23.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8853" PostTypeId="1" AcceptedAnswerId="8856" CreationDate="2014-03-02T11:44:10.897" Score="3" ViewCount="521" Body="&lt;p&gt;I have a solution of potassium dichromate ($\ce{K2Cr2O7}$) and I want to convert it to a less toxic solution. What would be the best way to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using the dissolved potassium dichromate as an oxidizer in the oxidation of alcohols to organic acids, to eventually produce the chromium(+III) ion (what is less toxic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there any other, easier methods?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2019-08-07T12:48:51.040" Title="Conversion of potassium dichromate solution to a less toxic solution" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8855" PostTypeId="2" ParentId="7057" CreationDate="2014-03-02T13:18:15.770" Score="4" Body="&lt;p&gt;In the comments I mentioned one way to solve the problem, but it turns out that was the roundabout way! There is a simpler solution, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the neutralization reaction, it is possible to calculate that the mixture reacted with $0.06\ \mathrm{mol}$ of $\ce{H+}$. The carbonate and the oxide react in a proportion of 1:2 with the hydrogen ions, so we know that $n_{\ce{BaO}}+n_{\ce{XO}}=0.03\ \mathrm{mol}$. The mass balance from the thermal decomposition then suggests that $m_{\ce{XCO3}}-m_{\ce{XO}}=0.44\ \mathrm g$. These two equations are already enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the for the molar masses of the unknown oxide and carbonate, we have $M_{\ce{XCO3}}-M_{\ce{XO}}=44\ \mathrm{g/mol}$. If one converts the masses in the second equation into expressions in $M$ and $n$, the result is $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}M_{\ce{XO}}=0.44\ \mathrm g$, which can be turned into $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$. It is easy to see that the thermal decomposition does not alter the number of moles of compound containing $\ce{X}$, so that $n_{\ce{XCO3}}=n_{\ce{XO}}$. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-M_{\ce{XCO3}}+44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{\ce{XCO3}}\times 44\ \mathrm{g/mol}=0.44\ \mathrm g\quad\Rightarrow\quad n_{\ce{XCO3}}=0.01\ \mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the problem unravels easily. We get that  $n_{\ce{BaO}}=0.02\ \mathrm{mol}$, which implies $m_{\ce{BaO}}=3.08\ \mathrm g$ and $m_{\ce{XCO3}}=1.00\ \mathrm g$. Hence, $M_{\ce{XCO3}}=100\ \mathrm{g/mol}$, and from there $M_{\ce{X}}=40\ \mathrm{g/mol}$. A quick look on a periodic table yields calcium as the most likely candidate, and indeed its carbonate has the formula $\ce{CaCO3}$ and a molar mass of $100\ \mathrm{g/mol}$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2016-11-10T16:01:13.587" LastActivityDate="2016-11-10T16:01:13.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8856" PostTypeId="2" ParentId="8853" CreationDate="2014-03-02T14:52:01.700" Score="5" Body="&lt;p&gt;That looks like one way to reduce the dichromate ion which is known to be toxic. MSDS sheets indicate that potassium dichromate has category 1B carcinogenicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way I believe is to just start from an aldehyde and convert it into a carboxylic acid: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3RCHO + 2Cr_2O_7^- + BH^+ -&amp;gt; 3RCOOH + 2Cr^{3+} + 4H_2O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, one can put dichromate in acidic solution with iron:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cr2O7^{2-} + 6Fe^{2+} + 14 H^+ -&amp;gt; 2 Cr^{3+} + 6 Fe^{3+} + 7H_2O }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dichromate oxidizes $ \ce{Fe^{2+}}$ ions and produces $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ ions.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-04T02:34:43.200" LastActivityDate="2014-03-04T02:34:43.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8857" PostTypeId="1" AcceptedAnswerId="8859" CreationDate="2014-03-02T20:09:34.323" Score="2" ViewCount="30202" Body="&lt;p&gt;I think I've managed to write down the correct reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 + KI + H2SO4 -&amp;gt; K2SO4 + I2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to balance this using the ion-electron method in an acidic environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally this is no problem, but I'm a bit confused on how to work with the reduction of $\ce{O}$ from $\ce{-I}$ to $\ce{-II}$ and I'm not sure how to write down the total ionic equation. I've tried the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H^{+1} + O_2^{-2} + K^{+1} + I^{-1} + 2H^{+1} + SO_4^{-2} -&amp;gt; 2K^{+1}  +SO_4^{-2} + 2I^{} + 2H^{+1} + O^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure if that is correct, and if it is correct, I'm not sure what to do from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this reaction and after diluting (with water?), I also need to know what will happen if I add a few drops of a starch-solution (I will actually have to perform this experiment so if it's merely something visible and not necessarily related to the reaction, then this part is less important.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final question is &quot;what property of $\ce{H2O2}$ is demonstrated here?&quot; and I suppose the answer to that is it's function as an oxidizer?&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="5026" LastEditDate="2018-02-23T02:32:54.593" LastActivityDate="2018-02-23T02:32:54.593" Title="Redox reaction with hydrogen peroxide with potassium iodide" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8859" PostTypeId="2" ParentId="8857" CreationDate="2014-03-02T20:56:05.573" Score="2" Body="&lt;p&gt;With all redox reactions, we should first look at all oxidation states of our reactants and products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xA;\text{Element} &amp;amp; \text{Oxidation state in reactants} &amp;amp; \text{Oxidation state in products} \\&#xA;\hline&#xA;\ce{H} &amp;amp; +1 &amp;amp; +1 \\&#xA;\ce{O} &amp;amp; -1\text{ in }\ce{H2O2}; -2\text{ in }\ce{H2SO4} &amp;amp; -2 \\&#xA;\ce{K} &amp;amp; +1 &amp;amp; +1 \\&#xA;\ce{S} &amp;amp; +6 &amp;amp; +6 \\&#xA;\ce{I} &amp;amp; -1 &amp;amp; 0&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have to look at what is being oxidized and reduced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H_2O_2 + KI + H_2SO_4 -&amp;gt; K_2SO_4 + I_2 + H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen has the same oxidation state of 1+.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium has the same oxidation state of 1+.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine is oxidized from -1 to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is reduced from -1 to -2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For redox reactions, we have to write out our half-reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 -&amp;gt; H_O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{KI + H2SO4 -&amp;gt; K2SO4 + I2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our two half-reactions, we first have to make sure everything is balanced other than hydrogen and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our second half-reaction, we need two moles of potassium iodide in order to balance out the potassium and iodines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI + H2SO_4 -&amp;gt; K2SO_4 + I2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to balance oxygens, we add molecules of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI + H2SO4 -&amp;gt; K2SO4 + I2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are in acidic solution, we add protons to balance hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 +2H+ -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI + H2SO4 -&amp;gt; K2SO4 + I2 + 2H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We add electrons to balance charge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 +2H+ -&amp;gt; 2H2O + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI + H2SO4 + 2e- -&amp;gt; K2SO4 + I2 + 2H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add 'em(the half-reactions) both up and cancel out necessary terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI + H2SO4 + H2O2 -&amp;gt; K2SO4 + I2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differs from your original equation since I think you forgot to balance it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide here is reduced to water in acidic solution.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="16683" LastEditDate="2016-04-27T15:31:31.367" LastActivityDate="2016-04-27T15:31:31.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8860" PostTypeId="1" CreationDate="2014-03-02T21:02:15.960" Score="0" ViewCount="2871" Body="&lt;p&gt;If I mix a $\pu{2 M}$ solution with a different $\pu{1 M}$ solution, do the individual solutions reduce in molarity to $\pu{1.33 M}$ and $\ce{0.33 M}$ respectively?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorUserId="41328" LastEditDate="2017-08-15T21:48:44.747" LastActivityDate="2017-08-15T21:54:09.257" Title="Concentrations of mixed solutions" Tags="&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8861" PostTypeId="2" ParentId="8850" CreationDate="2014-03-02T21:51:42.587" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any experiments on oxygen burning in hydrogen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Assuming that burning &lt;em&gt;in&lt;/em&gt; &quot;something&quot; describes that &quot;something&quot; is more/at higher concentration than the stuff that is said to burn)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure oxygen burns &lt;em&gt;in&lt;/em&gt; hydrogen: the upper flammable limit of hydrogen in oxygen is 94 % hydrogen and 6 % oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which also tells you one case where such experiments are conducted (measuring flammability for safety sheets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For other fuels, the upper flammable limit is &lt;em&gt;much&lt;/em&gt; lower, actually fuel concentration &amp;lt; oxygen concentration, so one would rather say the fuel burns in oxygen.)&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-03-02T21:51:42.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8862" PostTypeId="1" CreationDate="2014-03-02T23:51:34.083" Score="6" ViewCount="6750" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equations&#xA;  $$\ce{NO2- + H2O -&amp;gt; ?}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought it would be $\ce{NO2- + H2O -&amp;gt; HNO2 + OH-}$, but my textbook says that the correct products are $\ce{HNO3}$ and $\ce{OH-}$. What am I doing wrong here? The answer that they give doesn't even seem to balance.&lt;/p&gt;&#xA;" OwnerUserId="4455" LastEditorUserId="194" LastEditDate="2018-03-03T15:23:56.377" LastActivityDate="2018-03-03T15:23:56.377" Title="Why isn't nitrous acid a product of nitrite and water?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8863" PostTypeId="2" ParentId="8860" CreationDate="2014-03-02T23:53:04.047" Score="1" Body="&lt;p&gt;You need to know how much of each solution you add and the final total volume. The molarity of each one would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{C_1 \cdot V_1}{V_1 + V_2} \quad \text{and} \quad \frac{C_2 \cdot V_2}{V_1 + V_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_1 = \pu{2 M}$, $C_2 = \pu{1 M}$, $V_1$, $V_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.&lt;/p&gt;&#xA;" OwnerUserId="4696" LastEditorUserId="41328" LastEditDate="2017-08-15T21:54:09.257" LastActivityDate="2017-08-15T21:54:09.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8864" PostTypeId="2" ParentId="8827" CreationDate="2014-03-03T00:00:01.597" Score="9" Body="&lt;p&gt;Note that being a chemist doesn't require any kind of state license or the like in Germany, neither. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But buying dangerous* substances and running a chemical lab requires that you** have certain licenses and implement safety standards. &#xA;Some of these exams a chemist has to pass during studies (e.g. on toxicology and chemical substances legislation) so having a Master/Diplom in chemistry automatically means that you have those licenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having this license (Chemikaliensachkunde) and being reliable (legal meaning) and of age are legal requirements &lt;em&gt;you&lt;/em&gt; must meet in order to be allowed to buy dangerous substances. In practice however, the suppliers put another very practical additional need: they usually do mail orders only B2B and require you to have VAT/USt number.&lt;br&gt;&#xA;So if I as a small-scale freelancer who fulfills all the legal requirements want to buy from say, Sigma, I don't get anything because I don't have the tax number (for my size of side-business it would be too much hassle to go through this, so I use the small business exemtion law and for the supplier it would be too much hassle to make sure the buyer fulfills the legal requirements if it isn't at least a certain business size). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a private person, you can buy chemicals:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at the pharmacy&lt;/li&gt;&#xA;&lt;li&gt;at specialized stores, e.g. &#xA;&lt;ul&gt;&#xA;&lt;li&gt;photography chemicals at a good photography stores &lt;/li&gt;&#xA;&lt;li&gt;carbide at a speleology outfitter's &lt;/li&gt;&#xA;&lt;li&gt;gases (possibly including dry ice) at a gas merchant's, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;you basically cannot buy dangerous substances by mail/online order (too much burocratic hassle: the seller would need to make sure of your identity and Chemikaliensachkunde etc.) &lt;/li&gt;&#xA;&lt;li&gt;I've never tried to buy privately anything that requires the Chemikaliensachkunde e.g. at a pharmacy so I cannot tell how difficult that is in practice. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;* As a rule of thumb, there are restrictions for substances (or mixtures) that are extremely flammable, oxidizing, toxic or very toxic, (suspected of being) carcinogenic, mutagenic or teratogenic (plus a bunch of listed things)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** or some employee of yours&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, you can of course work as a chemist without (much or even any) need to handle dangerous substances. For example, I'm analytical chemist / spectroscopist / chemometrician. About 90 % of my work is in front of a computer analysing measurement data, developing improved ways of doing so and writing reports/papers. At least half of the rest is in front of a computer programming instruments. I've handled exactly one dangerous substance according to chemicals legislation at two days (actually a sample of few μg I got for analysis) during the last year. The everyday stuff I handle (methanol, ethanol, isopropanol, disinfectant solution and cyclo-hexane and paracetamol=acetaminophen as calibration standards) are not legally restricted dangerous chemicals.&lt;br&gt;&#xA;I do handle biohazard material far more often - but that is not what the Chemikaliensachkunde is good for. So for my work, the more relevant safety issues are biological/human material safety and laser safety... It would be quite easy not to have any chemical lab exposure (is seldom enough as it is... - but hey, what did I become chemist for?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I have some colleagues doing the same type of work who are not chemists but e.g. physicists and thus do not have the Chemikaliensachkunde. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For getting a job it is obviously of advantage if you can show some kind of education/training that makes you fit for the job. For jobs in chemical industry/academia that is obviously a chemistry degree or training as chemical lab technician, but also related professions such as physics/biology/pharmacy/toxicology depending on what the job in question is. &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2014-03-03T00:00:01.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8865" PostTypeId="2" ParentId="8862" CreationDate="2014-03-03T00:15:26.667" Score="1" Body="&lt;p&gt;It should form nitrous acid... If it was $NO_3^{-}$, nitrate ion, then it would presumably form nitric acid $HNO_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrous acid reacts with water to form nitrite and hydronium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO_2 + H_2O -&amp;gt; NO_2^- + H_3O^+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is an amphoteric species. Assuming we just have water the nitrite ion, the nitrite ion accepts the hydrogen forming nitrous acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NO_2^- + H_2O -&amp;gt; HNO_2 + OH^-} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is interesting as well since nitrous acid will also dissociate again into nitrite ions and water as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO_2 + H_2O -&amp;gt; NO_2^- + H_3O^+} $$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4325" LastEditDate="2015-09-30T16:37:16.330" LastActivityDate="2015-09-30T16:37:16.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8866" PostTypeId="1" AcceptedAnswerId="10225" CreationDate="2014-03-03T00:51:46.383" Score="5" ViewCount="700" Body="&lt;p&gt;Multiple-chamber artillery can launch a projectile with energies much higher than single-chamber guns.  I'm interested in setting a fundamental upper bound on the energy that a chamber (of a known volume) can impart to a projectile.  This would, for example, set &lt;a href=&quot;https://space.stackexchange.com/questions/3667/could-any-existing-gun-reach-the-karman-line&quot;&gt;a limit on the needed volume for a sub-orbital gun&lt;/a&gt; (given a projectile mass).  Since the pressure drop-off (like with single-chamber) isn't a constraint, the kinetic energy would be theoretically limited by pressure times volume.  But you can't possibly achieve infinite pressure with any chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that limit, some sources &lt;a href=&quot;https://www.kickstarter.com/projects/1682852725/the-starfire-space-cannon&quot; rel=&quot;nofollow noreferrer&quot;&gt;claim that pressures of 60,000 psi can be achieved&lt;/a&gt;.  Coming from a physics perspective, I understand that a reaction should have a defined energy associated with an individual molecular transition.  We've had &lt;a href=&quot;https://physics.stackexchange.com/questions/92013/initial-pressure-of-explosion&quot;&gt;related questions on physics stack exchange&lt;/a&gt;, but my thinking is that chemical equilibrium may be be a necessary tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore reaction kinematics, I believe that the maximum pressure that a reaction can achieve is limited by two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the density of the reactants before reaction and&lt;/li&gt;&#xA;&lt;li&gt;the energy of the reaction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given this, we could simply model the reaction with ideal gas assumptions, resulting in simple algebra.  But I think this is wrong.  Higher pressure would push the reaction to incomplete combustion.  From a chemistry perspective, is there any &lt;em&gt;theoretical&lt;/em&gt; maximum pressure that we could restrict all combustion reactions to?&lt;/p&gt;&#xA;" OwnerUserId="284" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.897" LastActivityDate="2014-04-25T02:35:41.993" Title="Maximum pressure that can be produced by a combustion reaction" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8867" PostTypeId="1" CreationDate="2014-03-03T02:57:42.767" Score="1" ViewCount="3095" Body="&lt;p&gt;Why does gold react with cyanide and sodium hydroxide and what is the chemical reaction's impact on gold's atomic structure and bonding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think is since in the 'flotation' process, when the water, chemicals and other substances are added, gold isn't affected and neither is the pyrite, but when the cyanide and caustic acid is mixed with the slurry, it separated both substances, then the oxygen or 'air' is added into the mix, so then the gold rises to the top and frothing over the slurry. I'm not sure if i'm correct, but I don't understand how it's related to the atomic structure and bonding of gold? This process is from (&lt;a href=&quot;http://www.superpit.com.au/Production/MineralProcessing/tabid/150/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;Super Pit - Kalgoorlie exploration mine&lt;/a&gt;)&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Can someone please explain how the separation of gold from the mixed compounds is related to its atomic structure and bonding? Is it to do with the 'Elution' process? Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="2180" LastActivityDate="2014-03-04T11:54:43.220" Title="Why does gold react with cyanide and sodium hydroxide" Tags="&lt;reaction-mechanism&gt;&lt;metal&gt;&lt;mixtures&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8868" PostTypeId="1" AcceptedAnswerId="8869" CreationDate="2014-03-03T04:44:46.417" Score="7" ViewCount="6628" Body="&lt;p&gt;In my book, to explain the structure of benzene, it referred to the idea of resonance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The idea of &lt;em&gt;resonance&lt;/em&gt; is that the actual molecule is a definite structure that is a hybrid of two or more theoretical structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I partially understood this sentence, specifically how a molecule could have multiple structures, but am confused about what is meant by hybrid. Could someone please elaborate?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="16683" LastEditDate="2018-10-14T23:18:45.757" LastActivityDate="2019-08-08T04:26:14.980" Title="What is the &quot;resonance hybrid&quot; of benzene?" Tags="&lt;aromatic-compounds&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8869" PostTypeId="2" ParentId="8868" CreationDate="2014-03-03T05:38:29.897" Score="8" Body="&lt;p&gt;It means that what we usually consider to be the 'structure' of benzene is not the actual structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there are three π bonds in the structure of benzene, there is actually &lt;strong&gt;resonance&lt;/strong&gt;, or electron delocalization. This means that the electrons are not localised in π bonds between two specific carbons, but distributed throughout the ring. This is one of the most important characteristics of an aromatic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vSUXy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSUXy.png&quot; alt=&quot;Resonance forms and resonance hybrid of benzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;localised&quot; structures &lt;strong&gt;1a&lt;/strong&gt; and &lt;strong&gt;1b&lt;/strong&gt; above are resonance forms of benzene: neither &lt;strong&gt;1a&lt;/strong&gt; nor &lt;strong&gt;1b&lt;/strong&gt; are a true representation of benzene on their own; they are hypothetical, or &quot;theoretical&quot;, structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;resonance hybrid&lt;/em&gt; is a weighted average of the resonance forms &lt;strong&gt;1a&lt;/strong&gt; and &lt;strong&gt;1b&lt;/strong&gt;. In this case, both forms contribute equally to the hybrid. So, each C–C bond has a bond order of (1+2)/2 = 1.5, which we can represent using dotted lines in &lt;strong&gt;2&lt;/strong&gt; (each dotted line represents a fraction of a bond, so one solid and one dotted line means a bond order between 1 and 2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More symbolically, we can also use the depiction with an inscribed circle &lt;strong&gt;3&lt;/strong&gt;. This means that the six π-electrons are shared between all six of the carbon atoms.&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorUserId="16683" LastEditDate="2018-10-14T23:31:22.360" LastActivityDate="2018-10-14T23:31:22.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8870" PostTypeId="1" CreationDate="2014-03-03T06:05:39.840" Score="8" ViewCount="23756" Body="&lt;p&gt;Is there a general way to predict products of a chemical reaction, given only the atomic masses of masses of all elements involved? for example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CuFeS2 + O2-&amp;gt;}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{K2CrO4 + H2SO4-&amp;gt;}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{FeOCr2O3 + K2CO3 + O2-&amp;gt;}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{K2Cr2O7 + H2SO4-&amp;gt;}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If there's no method to do so, are we supposed to memorize &lt;strong&gt;all chemical reactions&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: In case you are writing the mechanism of chemical reactions, please bear in mind that products are not known beforehand.&lt;/p&gt;&#xA;" OwnerUserId="4698" LastEditorDisplayName="user822" LastEditDate="2014-04-28T08:40:58.363" LastActivityDate="2017-02-21T16:23:41.497" Title="Is there a general way to predict products of a chemical reaction" Tags="&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="3" ClosedDate="2017-02-21T16:23:48.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="8871" PostTypeId="1" AcceptedAnswerId="8873" CreationDate="2014-03-03T07:59:14.453" Score="2" ViewCount="3618" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;All substitution reactions of benzene must be carried out in dry conditions with a catalyst that produces a powerful electrophile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was a statement from my book. My question is, why must it be carried out in &lt;strong&gt;&lt;em&gt;dry conditions&lt;/em&gt;&lt;/strong&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will occur if moisture is present?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="23561" LastEditDate="2018-09-06T14:55:58.950" LastActivityDate="2018-09-06T14:55:58.950" Title="Reasons for dry conditions in substituition reactions with benzene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8872" PostTypeId="2" ParentId="8870" CreationDate="2014-03-03T10:49:36.520" Score="5" Body="&lt;p&gt;Often times it is possible due to previous data. Trends appear and you can make deductions from such trends. Reactions may not happen at all. There may be deviations from the trends as well. The deviations may be for unknown reasons or there may be unproven theories that are included in textbooks. Inorganic chemistry reactions are not as clear cut as organic chemistry reactions. You don't have to memorize all the reactions. Know the trends, conduct the experiment, and if your prediction fails, try to think hard about the conditions in which the experiment took place. Sometimes the explanation may lay in the container that you used for the reaction, or it may be stoichiometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in all sciences, you need experiments.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastActivityDate="2014-03-03T10:49:36.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8873" PostTypeId="2" ParentId="8871" CreationDate="2014-03-03T10:56:13.417" Score="0" Body="&lt;p&gt;Benzene and other aromatic hydrocarbons are immisible with water. So, probably there is no problem with benzene if we there is moisture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the substitution reactions which benzene undergo, requires Lewis acid catalyst, like ferric halide or aluminium halide. Most commonly used catalysts are Anhyd. $\ce{AlCl3}$ and $\ce{FeBr3}$, which become inactivated if they react with water. So, this requires dry conditions.    &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="23561" LastEditDate="2018-08-13T03:13:30.070" LastActivityDate="2018-08-13T03:13:30.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8874" PostTypeId="1" CreationDate="2014-03-03T12:44:16.637" Score="2" ViewCount="4620" Body="&lt;p&gt;Boyle's Law states that pressure is inversely proportional to volume.&#xA;However, let's take an example of a balloon: as we fill air in it, its pressure increases, but its volume also increases. Can anyone explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ydh3h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ydh3h.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/8673/Red_Balloon_Effusion.JPG?size=bestfit&amp;amp;width=350&amp;amp;height=209&amp;amp;revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;ucdavis.edu&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4702" LastEditorUserId="41328" LastEditDate="2019-07-18T14:08:35.057" LastActivityDate="2019-07-18T17:42:30.963" Title="Boyle's Law of pressure and volume" Tags="&lt;gas-laws&gt;&lt;solid-state-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8876" PostTypeId="2" ParentId="8874" CreationDate="2014-03-03T14:12:15.297" Score="4" Body="&lt;p&gt;Balloon is not an ideal system to study pressure-volume relationship. Because, on expansion, the elastic skin also expands and there needs to be an additional pressure build-up on the inner side to counter that force, very similar to the extra pressure in a spherical bubble. (where the surface tension acts exactly similar to the elastic balloon skin)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in your arguments, you are forgetting an important point. Blowing a balloon also involves increasing the amount of air in it. Boyle's law holds for only a fixed volume of gas. If you change the number of moles, Boyle's law no longer stays valid.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2014-03-06T08:16:48.173" LastActivityDate="2014-03-06T08:16:48.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8877" PostTypeId="1" CreationDate="2014-03-03T14:19:10.817" Score="3" ViewCount="2416" Body="&lt;p&gt;I was recently studying oxidation and reduction reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found red phosphorus and iodine reduce many different organic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of interest I would like to know the mechanism for these reactions. A concerted path is preferred.&lt;/p&gt;&#xA;" OwnerUserId="4704" LastEditorUserId="4945" LastEditDate="2014-05-22T12:03:47.477" LastActivityDate="2014-05-22T12:03:47.477" Title="What are the mechanisms for the reduction of organic compounds employing red phosphorus and iodine?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8878" PostTypeId="1" CreationDate="2014-03-03T14:26:10.217" Score="2" ViewCount="1299" Body="&lt;p&gt;Could anybody describe which product is formed in the following reaction (o-Toluidine oxidation)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 + o-Toluidine_{uncolored} - Peroxidase -&amp;gt; H2O + 1/2 O2 + o-Toluidine\; Oxidate_{colored}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: Hydrogen peroxide is one of the products of glucose oxidation to glucolactone by glucose oxydase.&lt;/p&gt;&#xA;" OwnerUserId="4705" LastEditorUserId="3846" LastEditDate="2014-03-04T12:58:32.993" LastActivityDate="2014-03-04T13:04:12.667" Title="Products of the o-Toluidine oxidation with glucose oxidase and peroxidase" Tags="&lt;water&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8879" PostTypeId="2" ParentId="8874" CreationDate="2014-03-03T15:16:20.753" Score="0" Body="&lt;p&gt;Let's start with the general gas equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_1V_1 = nRT_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to get to Boyle's law, we assume that both &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; (temperature) and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; (the number of moles of the substance involved in our experiment) are both constant.  Now we can rewrite the above equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_1V_1 = \mathrm{constant}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if we change &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; in our system their product must still produce the same constant, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_1V_1 = P_2V_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can't blow additional air into our balloon (adding more moles of material and violating one of our initial assumptions) and still expect Boyle's law to hold.  In a fixed system like the one we just described where Boyle's law holds, if we reduce the volume of our balloon &lt;span class=&quot;math-container&quot;&gt;$(V_2)$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$1/2$&lt;/span&gt; then the pressure &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt; must increase by a factor of &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; in order to maintain the value of the constant.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2019-07-18T15:08:38.493" LastActivityDate="2019-07-18T15:08:38.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8880" PostTypeId="1" CreationDate="2014-03-03T17:23:34.903" Score="3" ViewCount="1724" Body="&lt;p&gt;By compressing $\ce{CO2 (g)}$ the gas gains a lot of heat. When the hot compressed carbon dioxide is left to cool, it attains the temperature of the surrounding material. When it is released into the pressure outside (about 100kPa), it will get so cold (actual temperature depending on how much the carbon dioxide was compressed), that it will freeze (below 78.5 degrees celcius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much pressure is needed to make dry ice at 293K (20 degrees celcius)?&lt;/p&gt;&#xA;" OwnerUserId="23" LastEditorUserId="23" LastEditDate="2014-03-03T20:44:40.047" LastActivityDate="2020-03-03T20:02:07.983" Title="How much pressure is needed to make dry ice?" Tags="&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8881" PostTypeId="1" CreationDate="2014-03-03T17:23:37.043" Score="5" ViewCount="6717" Body="&lt;p&gt;We know that an activation complex is the temporary unstable compound formed when activation energy is reached. But what dictates whether the products are indeed going to be products or if they become reactants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the activation complex form when orientation is both correct and incorrect, and only forms products when the orientation is correct? Or is the decomposition of the complex purely a chance matter?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="41328" LastEditDate="2019-05-26T11:04:47.003" LastActivityDate="2019-05-26T13:18:49.693" Title="Activated Complex Products/Reactants" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8882" PostTypeId="2" ParentId="8877" CreationDate="2014-03-03T17:25:06.923" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I found Red P and Iodine reduce many different organic compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, it is typically used for the deoxygenation in &lt;strong&gt;benzylic&lt;/strong&gt; positions, using benzylic alcohols or benzoins as starting materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aliphatic alcohols yield iodoalkanes, propargylic alcohols often undergo elimination under these conditions whereas allylic alcohols are typically converted to complex mixtures without synthetic value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deoxygenation is often carried out with catalytic amounts of iodine but an excess of red phosphorous. In principle, catalytic amounts of $\ce{NaI}$ plus one equivalent of $\ce{H3PO3}$ in water would work as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In princpiple, elimination from water from the protonated starting material would yield a stabilized benzylic cation. This might get reduced by a iodide to a benzylic radical and, in a second step to a very shortlived benzylic anion which is instantaneously protonated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both reduction steps yield iodine radicals from iodide and in summary regenerate iodine.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-03T17:25:06.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8883" PostTypeId="2" ParentId="8880" CreationDate="2014-03-03T17:46:17.803" Score="3" Body="&lt;p&gt;Isn't that a bit different? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 293 K, you need about 56 bar to liquefy carbon dioxide. If you then release the pressure, a part of the liquid evaporates again while cooling down the rest until it solidifies at around -80 °C.  &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-03T17:46:17.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8884" PostTypeId="1" CreationDate="2014-03-03T18:01:56.630" Score="2" ViewCount="118" Body="&lt;p&gt;I have $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ sensors and need to calibrate them. Can anybody help me to produce the $\ce{CO2}$, $\ce{CO}$, $\ce{O2}$ in laboratory for calibrating the sensors (the gas should be produced in known in $\mathrm{ppm}$) &lt;/p&gt;&#xA;" OwnerUserId="4707" LastEditorUserId="7475" LastEditorDisplayName="user4076" LastEditDate="2015-10-16T17:42:28.110" LastActivityDate="2015-10-16T17:42:28.110" Title="gas chamber for calibrating gas sensors" Tags="&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8885" PostTypeId="2" ParentId="8871" CreationDate="2014-03-03T18:33:48.753" Score="0" Body="&lt;p&gt;An electrophile goes after electron density, as in the two lone pairs in $\ce{H2O}$ plus poalrization.  You then require conditions wherein water is inert (hypohalite halogenations, maybe samarium triflate), micelles exclude water, or the substituted benzene is so activated (dimethylaniline) that it doesn't care.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-03T18:33:48.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8887" PostTypeId="2" ParentId="8867" CreationDate="2014-03-03T18:50:19.520" Score="2" Body="&lt;p&gt;Look at half-cell redox potentials.  Gold should be chemically inert in water.  Now, add a coordinating ligand and a mild oxidizing agent.  Gold complexes are deep thermodynamic holes. Saltwater and air will eat gold, giving chloroaurate complexes.  Cis-platin was accidentally discovered when &quot;inert&quot; platinum anodes in cell culture met saline and metabolic ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold often looks like a halogen (relativistic diddling of s- and d-orbital energies given probability near the nucleus).  Touching gold with aluminum can over time gives &quot;purple plague&quot; (aluminum-gold intermetallics at microchip circuit connections).  Gold and beryllium give brittle crumbly AuBe.  Gold is only inert under select circumstances.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-03T18:50:19.520" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8889" PostTypeId="2" ParentId="8871" CreationDate="2014-03-03T19:19:01.710" Score="1" Body="&lt;p&gt;Electrophiles&#xA;are reactive cations, they will react with water to form &#xA;eg an alcohol (or some weak acid etc)  and protons. &#xA;No reaction with benzene. &lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2014-03-03T19:19:01.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8891" PostTypeId="2" ParentId="8878" CreationDate="2014-03-03T20:02:10.863" Score="1" Body="&lt;p&gt;The product most likely is 2-methylnitrobenzene (&lt;strong&gt;3&lt;/strong&gt;), which means that the amino group ($\ce{-NH2}$) in o-toluidine (&lt;strong&gt;1&lt;/strong&gt;) is oxidized to a nitro group ($\ce{-NO2}$). The oxidation proceeds via 2-methylnitrosobenzene (&lt;strong&gt;2&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/duKwx.png&quot; alt=&quot;toluidine oxidation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of 2-methylnitrobenzene can be detected photometrically at $\lambda$ = 400 nm, where the aniline does not absorb.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-04T13:04:12.667" LastActivityDate="2014-03-04T13:04:12.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8892" PostTypeId="2" ParentId="8884" CreationDate="2014-03-03T20:39:55.727" Score="3" Body="&lt;p&gt;To be accurate you will need standard gas cylinder and pneumatic connections (pressure regulator, needle valves or better mass flow controllers, pipes). You will need a measuring chamber with an overflow. As a rule of thumb, if the chamber geometry is not exotic (lets say a cube with no dramatic dead volume), you need to flush it, at least  10 times, with each concentration you want to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a 2 points calibration for O2 by measuring atmosphere where O2 concentration can be assumed constant (span) and then flush with pure nitrogen (zero). For CO2 and CO you will need standard cylinders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway a 2 points calibration is a little bit risky because you will not be able to assess linearity and then you will have no idea in which extent your sensor is linear (dynamic range). Properly calibrating sensor requires multiple points calibration and therefore a dilution device and standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caution:&lt;/strong&gt; When dealing with gas you must take care of exhaust and prevents risk such as intoxication (especially CO and in a fewer extent CO2) or asphyxiation (due to a lack of oxygen) and explosion (CO is explosive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should find those materials, resources and skills in a University (Chemistry or Physics Department) or an Air Quality Laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; If your sensors are electrochemical ones, you need also to assess their correlations/deviations with temperature and humidity.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-03-03T20:48:32.980" LastActivityDate="2014-03-03T20:48:32.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8893" PostTypeId="2" ParentId="8868" CreationDate="2014-03-03T22:25:34.337" Score="1" Body="&lt;p&gt;The core feature of the benzene molecule that is 'caused' by resonance is the bond length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know from your previous studied that double bonds are shorted than single bonds.  In the 'resonating' Lewis structures for benzene, you would thus observe two kinds of bonds, three single bonds and three double bonds.  However spectroscopic studies show only one kind of carbon-carbon bond in benzene.  Hence 'resonance' implies some 'averaging' of structure between the two Lewis structures.  You can't really draw a Lewis structure for the true representation of benzene, which is why the circle motif was developed.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-03-03T22:25:34.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8894" PostTypeId="2" ParentId="7484" CreationDate="2014-03-03T22:44:23.597" Score="4" Body="&lt;p&gt;I believe your conception of the need for deprotonation under acidic conditions is misconceived.  Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key concept here is the enol tautomerization.  The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene.  Thus there is likely a higher concentration of this nucleophile in the reaction mixture.  The mechanism of tautomerization is difficult to elucidate, but presumably starts with protonation of the carbonyl in acidic conditions.  Then, likely another carbonyl deprotonates the the recently protonated carbonyl at its alpha position, forming the enol.  This transformation has produced a nucleophile (enol -- 1-propene-1-ol ) and also a strong electrophile (a protonated carbonyl -- protonated ethanal).  The reaction occurs, leading to a protonated aldol product (proton on the carbonyl) which then is deprotonated by the next carbonyl and the process continues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the base catalyzed case, ethanal is more acidic because it does not have the extra methyl group donating electron density disfavoring formation of a negative charge.  Hence ethanal becomes the nucleophile and propanal becomes the electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, these are only major products.  I don't expect the energy differences are large enough to give great selectivity.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2014-08-17T01:06:49.340" LastActivityDate="2014-08-17T01:06:49.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8895" PostTypeId="2" ParentId="6536" CreationDate="2014-03-03T23:04:14.003" Score="2" Body="&lt;p&gt;$\ce{Ph-NR_3^+}$  is likely weaker than $\ce{Ph-NH_3^+}$ because the mechanism of deactivation is based on the amount of positive charge localized on the Nitrogen:  Alkylgroups will stabilized the positive charge through sigma donation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the difference will be small.  Additionally, there are deprotonation side-reactions to consider for anilium ($\ce{Ph-NH_3^+}$) ions, which may compromise any discussion of deactivation.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-03-03T23:04:14.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8897" PostTypeId="2" ParentId="6735" CreationDate="2014-03-04T00:57:53.537" Score="0" Body="&lt;p&gt;I would suggest you to try and understand the equations and then learn them by heart. &#xA;You can try and write them down when you are done reading them.Once written,you can actually go through it once in 2 days.This will further help you memorise the equations properly!&#xA;Best of Luck! &lt;/p&gt;&#xA;" OwnerUserId="4711" LastActivityDate="2014-03-04T00:57:53.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8898" PostTypeId="1" AcceptedAnswerId="8903" CreationDate="2014-03-04T01:17:39.217" Score="5" ViewCount="16409" Body="&lt;p&gt;Does benzene have both isomers and resonance structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it has resonance structures because you can manipulate the position of the double bonds but what about isomers?&lt;/p&gt;&#xA;" OwnerUserId="4608" LastActivityDate="2014-03-04T02:47:24.837" Title="Does benzene have isomers and resonance structures?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8899" PostTypeId="2" ParentId="8881" CreationDate="2014-03-04T01:22:17.203" Score="4" Body="&lt;p&gt;Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated in my general chem textbook &lt;em&gt;Principles of Modern Chemistry&lt;/em&gt; by Oxtoby et. al, p. 685:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The activated complex is assumed to exist as if it were in equilibrium&#xA;  with the reactants and the theory focuses on calculating the rate at which the&#xA;  activated complex passes through the transition state to form products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eyring formula makes it possible to calculate reaction rates using transition state theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{B}T/h$&lt;/span&gt; measures the rate at which the activated complex dissociates to form into it's products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we can relate the Eyring formula to thermodynamics and substitute &lt;span class=&quot;math-container&quot;&gt;$\exp (-\Delta G^†/RT)$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$K^†$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation that Arrhenius derived is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k_r = A\exp (-E_a/RT)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is substituted with &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{B}T/h$&lt;/span&gt;. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="41328" LastEditDate="2019-05-26T11:03:00.120" LastActivityDate="2019-05-26T11:03:00.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8900" PostTypeId="1" CreationDate="2014-03-04T01:57:56.420" Score="3" ViewCount="36596" Body="&lt;p&gt;When do I use each of the above? If I use one or the other they give me difference results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I'm trying to put the Volts into an equation to find standard free energy change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta  G^\circ= -nFE$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use the cathode minus anode I get a different answer to the oxide + reduct.&lt;/p&gt;&#xA;" OwnerUserId="4712" LastEditorUserId="51553" LastEditDate="2018-09-04T17:51:42.963" LastActivityDate="2018-09-04T18:10:48.347" Title="Half cells: difference between E(cell) = E(cathode}-E(anode) &amp; E(cell) = E(oxidation) + E(reduction)" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8901" PostTypeId="2" ParentId="8900" CreationDate="2014-03-04T02:10:13.613" Score="5" Body="&lt;p&gt;Ah, I see what you did wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E^\circ_\text{cell} = E^\circ_\text{reduction} - E^\circ_\text{oxidation} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can write it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E^\circ_\text{cell} = - E^\circ_\text{oxidation} + E^\circ_\text{reduction} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cathode is actually the reduction step, and the anode is the oxidation step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful mnemonic I learned from my chemistry professor is a &lt;strong&gt;red cat&lt;/strong&gt; and &lt;strong&gt;an ox.&lt;/strong&gt; Interestingly, cathodes are red and are positive. So I have found many uses for this along the way.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorDisplayName="user7951" LastEditDate="2018-09-04T18:10:48.347" LastActivityDate="2018-09-04T18:10:48.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8902" PostTypeId="2" ParentId="8898" CreationDate="2014-03-04T02:33:55.060" Score="3" Body="&lt;p&gt;As you note, benzene has resonance structures, but it does not have isomers.  All of the bond lengths in benzene are identical.  That is why 1,2-dideuteriobenzene or  1,2-dimethylbenzene (o-xylene) both exist as single compounds.  Here's an interesting way to look at it.  Look at the energy diagram below on the left.  If benzene had 2 isomers, that would be them as the two stable wells at the bottom of the energy diagram, the &quot;resonance delocalized&quot; version of benzene is shown as the transition state involved in the interconversion of these two &quot;isomers&quot;.  Now look at the energy diagram below on the right.  This is the real case.  It turns out that the &quot;transition state&quot; is more stable than either of the 2 &quot;isomers&quot;!  It could be said that there is a negative activation energy for the formation of delocalized benzene from either of the 2 localized &quot;isomers&quot;.  The negative activation energy would correspond to the resonance stabilization of benzene compared to either of the localized &quot;isomers&quot;.  Hence, benzene exists as a fully delocalized structure with identical bond lengths at the bottom of the potential well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wk9lq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-03-04T02:33:55.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8903" PostTypeId="2" ParentId="8898" CreationDate="2014-03-04T02:34:05.510" Score="7" Body="&lt;p&gt;Well, it depends on what kinds of isomers you have in mind. It certainly has constitutional isomers: For example, the sum formula $\ce{C6H6}$ does describe benzene as well as hex-1,5-en-3-in $\ce{H2C=CH-C#C-CH=CH2}$.&#xA;There are also a lot ring structures that are isomers of benzene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HqKky.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there aren't any configurational isomers of benzene since a six-membered ring doesn't allow for &lt;em&gt;trans&lt;/em&gt; double bonds (this would be sterically impossible).&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-03-04T02:47:24.837" LastActivityDate="2014-03-04T02:47:24.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8905" PostTypeId="1" AcceptedAnswerId="8906" CreationDate="2014-03-04T10:14:45.567" Score="4" ViewCount="2331" Body="&lt;p&gt;I have to do a lab about enzyme catalysis with peroxidase in my chemistry class. We are to use the blank control solution in the experiment. My question is, why can we not use water instead? I did some research about this and someone says something like &amp;quot;It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy.&amp;quot; Is this the exact reason why we do not use water in this experiment?&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="69830" LastEditDate="2021-05-04T08:56:22.643" LastActivityDate="2021-05-04T08:56:22.643" Title="Why can we not use water instead of blank control solution in enzyme catalysis with Peroxidase experiment?" Tags="&lt;experimental-chemistry&gt;&lt;solutions&gt;&lt;spectroscopy&gt;&lt;enzymes&gt;&lt;uv-vis-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8906" PostTypeId="2" ParentId="8905" CreationDate="2014-03-04T10:44:42.820" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is because the water blank absorbs so much light that the usable scale is compressed, thus, causing a poor accuracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For water, the absorption coefficients $\alpha\ [\mathrm{m^{-1}}]$ at $\lambda$ = 680, 580, 535, 480 and 380 nm are 0.47, 0.09, 0.045, 0.013 and 0.011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: R. M. Pope, E. S. Fry, &lt;em&gt;Applied Optics&lt;/em&gt;, &lt;strong&gt;1997&lt;/strong&gt;, &lt;em&gt;36&lt;/em&gt;, 8710-8723)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;strong&gt;not&lt;/strong&gt; a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you never know about impurities in your buffers or other components (apart from your probe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you usually do not measure against &quot;just the solvent&quot;, but against the solution of &quot;everything except the probe&quot; to record the complete background.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-04T10:55:43.657" LastActivityDate="2014-03-04T10:55:43.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8907" PostTypeId="2" ParentId="8867" CreationDate="2014-03-04T11:54:43.220" Score="5" Body="&lt;p&gt;Flotation and treatment of gold with &lt;strong&gt;$\ce{NaCN}$ in the presence of air&lt;/strong&gt; are two distictive steps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flotation yields fine particles of gold (and pyrite) at the surface of a flotation tank, attached to foamy bubbles while the rest of the minerals sink to the bottom of the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flotation does not rely on differences in specific density, but on hydrophobicity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cyanide treatment, the following process - similar to what &lt;em&gt;Uncle Al&lt;/em&gt; described for the seawater case in his answer - occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid gold is oxidized, the cations are bound by cyanide anions and form a &lt;strong&gt;water-soluble complex&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Au + 8NaCN + O2 + H2O -&amp;gt; 4Na[Au(CN)2] + 4NaOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to reobtain solid gold, the $\ce{Au(I)}$ in the cyano complex $\ce{[Au(CN)2]^-}$ has to be reduced to $\ce{Au(0)}$ again. This can be done either by adding zinc dust, according to  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na[Au(CN)2] + Zn -&amp;gt; Na2[Zn(CN)4] + 2Au}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or electrochemically.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-04T11:54:43.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8908" PostTypeId="1" CreationDate="2014-03-04T11:59:07.223" Score="4" ViewCount="1448" Body="&lt;p&gt;I am trying to make a solution of &quot;trace elements&quot;--primarily transition metals-- for use in biological media; bacteria require elements like Cobalt, Molybdenum, iron, etc. in trace amounts. I am not a chemist, but my understanding is that these elements tend to form hydroxides in physiological pH ranges (slightly above 7). Therefore the authors of the protocol I am using employ EDTA as a complexing agent in 1:1 stoichiometric proportion with the metals in solution. The problem is that EDTA solubility (and complexing capability) and transition metal stability occur at opposite sides of the pH scale, and finding a balance to make both types of reagents happy is apparently really tricky; I've tried this 4 different times with no success. I am able to get all materials into solution, but any disturbance of my flask, (e.g. further stirring or shaking) inevitably causes precipitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my recipe and protocol. I'll describe my observations after that. My question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any thought what the precipitate is? How can I avoid it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have scant chemistry experience, so any insight you can provide would be useful.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Recipe&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;190 mM $\ce{EDTA}$ $\ce{Na4.2H2O}$ (equimolar to subsequent di- and trivalent cations)&lt;/li&gt;&#xA;&lt;li&gt;29 mM $\ce{FeCl3.6H2O}$&lt;/li&gt;&#xA;&lt;li&gt;80 mM $\ce{CaCO3}$&lt;/li&gt;&#xA;&lt;li&gt;6 mM $\ce{MnCl2.4H2O}$&lt;/li&gt;&#xA;&lt;li&gt;0.5 mM $\ce{CuSO4.5H2O}$&lt;/li&gt;&#xA;&lt;li&gt;0.5 mM $\ce{CoCl2.6H2O}$&lt;/li&gt;&#xA;&lt;li&gt;5 mM $\ce{ZnO}$&lt;/li&gt;&#xA;&lt;li&gt;2 mM $\ce{H3BO3}$&lt;/li&gt;&#xA;&lt;li&gt;66 mM $\ce{MgCl2.6H2O}$&lt;/li&gt;&#xA;&lt;li&gt;5 mM $\ce{NaMoO4.2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Protocol (making 1 L)&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dissolve EDTA in $\ce{750 mL}$ of water. Bring pH down to about 5.8 with $\ce{HCl}$.&lt;/li&gt;&#xA;&lt;li&gt;Add $\ce{FeCl3}$ and watch pH; don't let it drop below 4.8.&lt;/li&gt;&#xA;&lt;li&gt;Add $\ce{CaCO3}$. Allow to dissolve (i.e. $\ce{CO2}$ to evaporate) by maintaining pH at 4.8 with $\ce{HCl}$ and occasionally shaking vigorously (not sure why the recipe doesn't just call for something more soluble like $\ce{CaCl2}$.)&lt;/li&gt;&#xA;&lt;li&gt;Add remaining elements in order given in the recipe, watching the pH. It should drop to about 4.2 without any addition of acid or base.&lt;/li&gt;&#xA;&lt;li&gt;At this point solution appears clear. Bring volume up to $\ce{1L}$.&lt;/li&gt;&#xA;&lt;li&gt;Autoclave solution (heat under pressure at $123^{\circ}\ce{C}$ for 15 minutes) ---this is required for biological sterilization, and seems to improve solubility for a time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Observations&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Solution remains transparent throughout preparation, stirring constantly throughout at room temperature. Color is orangish-green, due to iron and copper. Once all ingredients have been added, additional stirring before autoclaving leads to gradual (over 15 minutes) accumulation of a white powdery precipitate. I thought this was EDTA coming out, so I added $\ce{NaOH}$, but this only caused the iron to precipitate, forming a brown precipitate. Letting the solution sit also seems to lead to white precipitation, although less rapidly. When I autoclaved the solution, it remained transparent (color had turned to deep rust-orange) even after sitting overnight, but as soon as I disturbed it by transferring from an Ehrlenmeyer flask to a glass bottle it precipitated again. The pH of the precipitated solution seems stable at 4.2, what it was after all ingredients had been added. I have also tried this preparation adding EDTA after the boric acid and keeping pH below 6 with $\ce{HCl}$, but this caused even more rapid precipitation. I have also tried EDTA (fully protonated) after the boric acid and adding NaOH to deprotonate, but the results were even worse.&lt;/p&gt;&#xA;" OwnerUserId="4716" LastActivityDate="2014-03-04T19:26:38.663" Title="Problems with precipitation of transition metal solution" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8909" PostTypeId="2" ParentId="8908" CreationDate="2014-03-04T12:11:15.203" Score="1" Body="&lt;p&gt;Is it conceivable that the white precipitate is $\ce{Zn(OH)2}$? Usually it precipitates more jelly-like, but anyway...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will however dissolve both on lowering the pH, forming colourless $\ce{[Zn(H2O)6]^{2+}}$ and on increasing the pH, yielding the likewise colourless $\ce{[Zn(OH)4]^{2-}}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-04T12:11:15.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8910" PostTypeId="1" AcceptedAnswerId="8913" CreationDate="2014-03-04T13:07:01.193" Score="22" ViewCount="28575" Body="&lt;p&gt;Quoting from &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter__4%3a_Structure_Determination_I/Section_4.3%3a_Ultraviolet_and_visible_spectroscopy&quot;&gt;this site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As conjugated pi systems become larger, the  energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer.   The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also quite easy to see that this is true glancing at various dye absorption tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is this true?  I can understand why the optical cross-section would increase, implying a higher quantum efficiency at some maximum absorbance wavelength (is this generally true?), but why would the energy gap between the ground state and the excited state correspondingly shrink?&lt;/p&gt;&#xA;" OwnerUserId="4717" LastEditorUserId="189" LastEditDate="2014-03-05T10:15:16.230" LastActivityDate="2020-12-20T16:33:42.087" Title="Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="8911" PostTypeId="2" ParentId="8910" CreationDate="2014-03-04T13:52:06.850" Score="8" Body="&lt;p&gt;Let's have an extreme example of 9 conjugated double bonds. That makes your $\pi$ HOMO orbital. The LUMO, $\pi^\star$, has one node, the middle double bond being antibonding, so it will consist of two times 4 conjugated $\pi$ bonds. To say, the difference in energy between HOMO and LUMO is quite small, therefore the excitation energy is low.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-03-04T13:52:06.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8912" PostTypeId="2" ParentId="8910" CreationDate="2014-03-04T13:52:27.477" Score="11" Body="&lt;p&gt;The effect is already &quot;visible&quot; on the comparison of ethene ($\ce{H2C=CH2}$) and 1,3-butadiene ($\ce{H2C=CH-CH=CH2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's build butadiene by combination of the molecular orbitals (MOs) of of ethene.&#xA;Combination of the $\pi$ orbitals gives one new occupied MO without a nodal plane (this is lower in energy than the $\pi$ orbital of ethene) and another occupied one with one nodal plane. This is our &lt;strong&gt;HOMO&lt;/strong&gt; (highest occupied molecular orbital) of butadiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The HOMO of butadiene has a higher energy than that of ethene&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's have look at the antibonding $\pi^*$ orbitals of ethene and their combination. Again, we obtain two new MOs, one higher in energy than the $\pi^*$ of ethene and another lying lower. This is our new &lt;strong&gt;LUMO&lt;/strong&gt; (lowest unoccupied molecular orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The LUMO of butadiene has a lower energy than that of ethene&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, the &lt;strong&gt;HOMO-LUMO gap is lower&lt;/strong&gt;, and the absorption shifted to higher wavelength.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-04T13:52:27.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8913" PostTypeId="2" ParentId="8910" CreationDate="2014-03-04T17:20:35.697" Score="32" Body="&lt;p&gt;Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a conjugated carbon chain or ring system you can think of the $\ce{C}$ atoms as $\text{sp}^{2}$-hybridized. So, each carbon has 3 $\text{sp}^{2}$ orbitals which it uses to form $\sigma$ bonds and 1 $\text{p}$ orbital which is used to form $\pi$ bonds.&#xA;It is the $\text{p}$ orbitals that are responsible for the conjugation and their combinations according to the LCAO model are the interesting part since the HOMO and LUMO of the system will be among the molecular orbitals formed from the conjugated $\text{p}$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a start take ethene, the simplest $\pi$-system, being comprised of only 2 carbon atoms.&#xA;When you combine two atomic orbitals you get two molecular orbitals.&#xA;These result from combining the $\text{p}$ orbitals either in-phase or out-of-phase.&#xA;The in-phase combination is lower in energy than the original $\text{p}$ orbitals and the out-of-phase combination is higher in energy than the original $\text{p}$ orbitals.&#xA;The in-phase combination accounts for the bonding molecular orbital ($\pi$), whilst the out-of-phase combination accounts for the antibonding molecular orbital ($\pi^{*}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9cscv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens when you lengthen the conjugated system by combining two ethene fragments?&#xA;You get to butadiene.&#xA;Butadiene has two $\pi$ bonds and so four electrons in the $\pi$ system.&#xA;Which molecular orbitals are these electrons in?&#xA;Since each molecular orbital can hold two electrons, only the two molecular orbitals&#xA;lowest in energy are filled.&#xA;Let's have a closer look at these orbitals.&#xA;In $\Psi_1$, the lowest-energy bonding orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one continuous orbital.&#xA;There is bonding between all the atoms.&#xA;The other two electrons are in $\Psi_2$.&#xA;This orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an antibonding interaction between carbons 2 and 3.&#xA;Overall, in both the occupied $\pi$ orbitals there are electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between carbons 2 and 3 in $\Psi_2$ partially cancels out the bonding interaction in $\Psi_1$.&#xA;This explains why all the bonds in butadiene are not the same and why the middle bond is more like a single bond while the end bonds are double bonds.&#xA;If we look closely at the coefficients on each atom in orbitals $\Psi_1$ and $\Psi_2$, it can be seen that the bonding interaction between the central carbon atoms in $\Psi_1$ is greater than the antibonding one in $\Psi_2$.&#xA;Thus butadiene does have some double bond character between carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ti3cx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can construct the molecular orbitals of butadiene by combining the molecular orbitals of the two ethene fragments in-phase and out-of-phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DtFt5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method of construction also shows why the HOMO-LUMO gap of butadiene is smaller than that of ethene.&#xA;The molecular orbital $\Psi_2$, which is the HOMO of butadiene, is the out-of-phase combination of two ethene $\pi$ orbitals, which are the HOMO of ethene.&#xA;Thus, the HOMO of butadiene is higher in energy than the HOMO of ethene.&#xA;Furthermore, the molecular orbital $\Psi_3$, which is the LUMO of butadiene, is the in-phase combination of two ethene $\pi^{*}$ orbitals, which are the LUMO of ethene.&#xA;Thus, the LUMO of butadiene is lower in energy than the LUMO of ethene.&#xA;It follows that the HOMO-LUMO energy gap is smaller in butadiene than in ethene and thus butadiene absorbs light with longer wavelenghts than ethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezK9p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you continue to lengthen the $\pi$ system by adding more ethene fragments you will see that the HOMO and LUMO are getting closer and closer together the longer the $\pi$ system becomes.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2016-08-03T09:18:18.833" LastActivityDate="2016-08-03T09:18:18.833" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="8914" PostTypeId="1" AcceptedAnswerId="8915" CreationDate="2014-03-04T17:25:07.370" Score="4" ViewCount="92" Body="&lt;p&gt;We receive structures from a server within our company. Those structures are transferred as &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_table_file#Molfile&quot; rel=&quot;nofollow noreferrer&quot;&gt;MOL files&lt;/a&gt;. Some of these structures contain R and/or S groups. Some of the creators of these structures create a really large empty space between the actual structure and the groups. We have currently ~30.000 structures in our database. Around 2.000 of these structures are flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached how such a structure might look at the bottom, as the image is quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to do now is to remove that empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is currently receiving the MOL files wrapped within an XML file that does have additional information to store along with the structure. That process is written in Java. Up to now we do not have any library within the project to process and manipulate structures, as we have just parsed the MOL as String from the XML file and written it into a CLOB field in the database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that libraries like &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemistry_Development_Kit&quot; rel=&quot;nofollow noreferrer&quot;&gt;the CDK&lt;/a&gt; exist, but right now I do not know if it can help me with that task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have pasted &lt;a href=&quot;http://tny.cz/498268e9&quot; rel=&quot;nofollow noreferrer&quot;&gt;the actual mol file here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we open it with a viewer it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uGgS0.png&quot; alt=&quot;a sample structure with large empty space&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4719" LastEditorUserId="4719" LastEditDate="2014-03-09T10:17:17.293" LastActivityDate="2014-03-09T10:17:17.293" Title="Remove empty space from structures with R and S groups" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8915" PostTypeId="2" ParentId="8914" CreationDate="2014-03-04T17:38:04.140" Score="4" Body="&lt;p&gt;Are you sure that the R and S are in the mol file and not part of the xml envelope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stereochemical &lt;em&gt;R&lt;/em&gt; and &lt;em&gt;S&lt;/em&gt; descriptors would not make any sense for a molecule like anthracene!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it conceivable that the letters come from &lt;strong&gt;Risk and Safety&lt;/strong&gt; statements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if you could upload the original mol file somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The file that you kindly uploaded is supposed to be a V2000 mol file, but it is &lt;strong&gt;not standards-compliant&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The full specifications can be obtained from &lt;a href=&quot;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accelrys&lt;/a&gt; upon registration.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first flaw is already in the header! The header is supposed to have a length of three lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following &lt;strong&gt;CTAB&lt;/strong&gt; should start with the counts line (in line 4 - note that we're not using zero-based indexing here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this flawed format, &lt;a href=&quot;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;OpenBabel&lt;/a&gt;, a standard tool for the conversion of chemical data formats, casts an error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the file, including the stereochemical descriptors, seemed to be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I had inserted the missing blank line in the header, a PNG could be generated using&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;obabel -imol 498268e9.txt -opng -O stuff.png&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ptG3O.png&quot; alt=&quot;converted&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OpenBabel comes with bindings for Python, Java, C++ and whatnot. You might want to talk to your IT about some refactoring ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-05T09:39:37.743" LastActivityDate="2014-03-05T09:39:37.743" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8916" PostTypeId="1" AcceptedAnswerId="8921" CreationDate="2014-03-04T18:02:55.393" Score="2" ViewCount="2943" Body="&lt;p&gt;In our chemistry practical we dropped 0.07g of Magnesium into various concentrations of an acid (sulfuric or hydrochloric - only one was used, not sure which), and measured the rate of reaction by the rate of the H2 produced (used a gas syringe). We then collected data at a suitable interval of the culmulative amount of H2 produced.&#xA;We drew the graph (with multiple concentrations - 0.25, 05, 0.75, 1, 2). Then we measured the steepest point (fastest rate) and took the gradient. &#xA;Plotting rate against concentrations (0.25, 05, 0.75, 1, 2), we came to the conclusion that rate varies exponentially with concentration. &#xA;But the rate law of a typical reaction is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = k * [A] * [B]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;, which would mean doubling [B] would double the rate law, not have an exponential effect on the reaction. Why did we come to the conclusion that concentration has an exponential effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case the rate law, I would think, would be elementary, but we are not dealing with concentrations of both Mg and HCl, just HCl, as the Mg was solid - therefore no concentration of Mg in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: In a reaction between two liquids, is the rate of reaction linearly dependent on the concentration?&#xA;AND: What is the rate law of our reaction? - Why did we come to the conclusion we did?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help!! :)&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2014-03-05T02:45:55.857" Title="Rate of reaction increases exponentially with concentration" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8917" PostTypeId="2" ParentId="8908" CreationDate="2014-03-04T19:26:38.663" Score="2" Body="&lt;p&gt;Here's an idea.  Add the cations as their acetates or nitrates.  Control pH with acetate buffer. Fe(III) is a bad player.  &quot;Ferric acetate,&quot; $\ce{[Fe3O(OAc)6(H2O)3]OAc}$, is water-soluble. You might sparge the solution with argon before sterilizing (sealed vessel) to prevent oxidation, and/or add ascorbate (or erythorbate, often biologically inactive) as anti-oxidant - if that is the problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lactate chelates, and may give greater solubilities.  Fe(II) lactate is E585 via the European Food Safety Authority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that my singular contribution to biochemistry was telling a lab its phosphatase assay bombed because...they used phosphate buffer.  I can't say they were grateful for acquiring the knowledge.)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-04T19:26:38.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8918" PostTypeId="2" ParentId="88" CreationDate="2014-03-04T20:31:55.420" Score="1" Body="&lt;p&gt;Methyl ester was chosen by the aspartame manufacturer for two specific reasons:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The methyl ester molecule binds the Phenylalanine and Aspartic Acid molecules together forming the Aspartame molecule.&lt;/li&gt;&#xA;&lt;li&gt;The unnatural methanol in the methyl ester provides the artificial sweetness of aspartame.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As an artificial sweetener, aspartame’s job is complete when it has tricked our brain into thinking we have consumed glucose when we have not, its toxic waste (methanol) is then swallowed for our body to get rid of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aspartame per se can do us no harm; it does not enter the bloodstream (EFSA 2013)&#xA;At a temperature of 86 degrees F in the GI tract, the methyl ester in aspartame converts back to pure methanol releasing its bond with the two amino acids. Aspartame’s three components are then free to enter the bloodstream to metabolise separately (EFSA 2013)&#xA;Methanol is by far the most toxic of aspartames components – the other two are amino acids.&#xA;To assess the severe metabolic toxicity of methanol as a food in humans is first necessary to establish its ADI. This can then be used to compare this toxicity to the current aspartame ADI. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been mentioned in your comments, the blinding dose of methanol in humans is 10 ml (one tablespoon) and the fatal dose is 30 ml (3 tablespoons) a simple calculation can express these in (mg/kg) as follows :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The blinding dose of methanol in a 70 kg adult is (114 mg/kg) and the fatal dose is (343 mg/kg). To calculate the ADI we need an NOAEL, which must be between (0.0 mg/kg to 114 mg/kg). If we use for example 10 % of the blinding dose = (11.4 mg/kg) to establish the ADI, this must be divided by a safety margin of 100; therefore the ADI of methanol in humans is (0.114 mg/kg). The ADI of aspartame (40 mg/kg) contains 10 % w/w of methanol or (4 mg/kg).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conclusion: &lt;strong&gt;The current ADI of aspartame contains 35 times more methanol than is safe for humans.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4721" LastEditorDisplayName="user7951" LastEditDate="2015-01-26T16:28:36.997" LastActivityDate="2015-01-26T16:28:36.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8919" PostTypeId="1" AcceptedAnswerId="8920" CreationDate="2014-03-04T21:23:32.957" Score="2" ViewCount="233" Body="&lt;p&gt;Take a look at the emission spectrum for an arbitrary fluorescent dye: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?  Consider that there is going to be a spread of eigenstates around a particular excited state (&lt;a href=&quot;http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg&lt;/a&gt;), corresponding to particular molecular vibration or rotational modes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thought is that, if we excite a dye at a wavelength slightly red-shifted with respect to $\lambda_{max}$, or slightly blue-shifted with respect to $\lambda_{max}$, we'd start out at a different eigenstate clustered around the same excited state (say $S_1$).  I imagine there's a probability of decaying from each of these eigenstates in $S_1$, which gives us the spread for the emission spectrum, so if we bias the initial eigenstate, we'd achieve a different emission spectrum (presumably &quot;statistically&quot; red-shifted with respect to the excitation frequency).  Or is the fluorophore taking a rapid random walk through all accessible eigenstates belonging to a particular excited state?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose we could also ask a similar question with respect to the photoelectric effect in terms of the energy provided to the excited electron (conditioned on having a threshold frequency to span a band gap).&lt;/p&gt;&#xA;" OwnerUserId="4717" LastEditorUserId="4717" LastEditDate="2014-03-04T21:29:09.003" LastActivityDate="2016-07-20T17:55:23.097" Title="Does the emission spectrum for a dye depend on the frequency of light used to excite the dye?" Tags="&lt;electronic-configuration&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8920" PostTypeId="2" ParentId="8919" CreationDate="2014-03-04T23:10:56.657" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (&lt;em&gt;Kasha's&lt;/em&gt; rule).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-04T23:10:56.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8921" PostTypeId="2" ParentId="8916" CreationDate="2014-03-05T02:45:55.857" Score="1" Body="&lt;p&gt;Rate equation is experimentally determined. So, you can't directly write $r=k[A][B]$ (where r is rate, k is rate constant). Probably you might have performed first order reaction. For which $r=k[A]=\frac{2.303}{t}log\frac{[R]}{[R_0]}=\frac{2.303}{t}log{e^{kt}}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation suggests that rate is proportional to exponential function.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-03-05T02:45:55.857" CommentCount="4" CommunityOwnedDate="2014-03-05T02:45:55.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="8922" PostTypeId="1" AcceptedAnswerId="8926" CreationDate="2014-03-05T03:14:40.420" Score="7" ViewCount="7439" Body="&lt;p&gt;Just out of curiosity, can dust conduct electricity? The source of dust varies, is there any form of dust that can conduct electricity?&lt;/p&gt;&#xA;" OwnerUserId="4516" LastEditorUserId="19616" LastEditDate="2015-10-30T17:15:36.703" LastActivityDate="2017-05-11T07:53:30.453" Title="Can dust conduct electricity?" Tags="&lt;everyday-chemistry&gt;&lt;conductivity&gt;&lt;electricity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8923" PostTypeId="2" ParentId="8922" CreationDate="2014-03-05T03:24:50.697" Score="2" Body="&lt;p&gt;Yes, dust can carry static charge. That's why people are cautious with dust on the inside of computers as they can short-circuit the electronics or inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, dust particles in those huge grain silos; if given an ignition source - usually a spark - can produce such a large explosion that it blows off the rough of the silo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it wouldn't conduct full current flow per se; the suspended particles are able to carry electric charge and differences between the charges of the particles allows dust to have conductivity. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-03-05T03:24:50.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8924" PostTypeId="2" ParentId="6244" CreationDate="2014-03-05T05:12:47.707" Score="-1" Body="&lt;p&gt;Unsaturated compounds (due to the existence of a double bond) are more reactive than saturated compounds, because the electron density at the internuclear axis is zero in a pi bond (in contrast to the sigma bond).&lt;/p&gt;&#xA;" OwnerUserId="4724" LastEditorUserId="3846" LastEditDate="2014-03-05T09:06:54.370" LastActivityDate="2014-03-05T09:06:54.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8925" PostTypeId="1" AcceptedAnswerId="8928" CreationDate="2014-03-05T07:04:00.783" Score="11" ViewCount="4525" Body="&lt;p&gt;I asked a question previously about &quot;why&quot; it is the case the expanding the size of pi-conjugated systems decreases the required energy to excite an electron from a HOMO to a LUMO band: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8910/why-does-the-energy-gap-for-transitions-shrink-with-the-size-of-the-pi-co&quot;&gt;Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?&lt;/a&gt;.  The user &quot;Philipp&quot; (and others) very patiently and kindly explained the reason for this by pointing out the expansion in the number of possible bonding and anti-bonding configurations for a polymethine-like pi-conjugated system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's something I'm missing that has to do with what is probably my lack of knowledge about chemistry.  If we look at this image: &lt;a href=&quot;https://i.stack.imgur.com/Ti3cx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://i.stack.imgur.com/Ti3cx.png&lt;/a&gt; (provided by &quot;Philipp&quot;) why is that that we have an energy gap between $\Psi_2$ and $\Psi_3$ smaller than the energy gap for a simpler, smaller pi-conjugated system where the HOMO corresponds to a single pair of p-orbitals in a bonding conformation and the LUMO corresponds to the same single pair of p-orbitals in an anti-bonding conformation (i.e. this system: &lt;a href=&quot;https://i.stack.imgur.com/9cscv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://i.stack.imgur.com/9cscv.png&lt;/a&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess that this would be because the bonding interactions that exist in the larger pi-conjugated system mitigate the penalty for having the one anti-bonding interaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update --- Could we justify the above guess using something like an overlap integral (if we know the shape of the two bonded p-orbitals forming a pi-bond)?  If so, this would be quite intuitive.&lt;/p&gt;&#xA;" OwnerUserId="4717" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-03-05T14:55:42.667" Title="The energy gap between a pi-conjugated system with (2 bonding and 1 anti-bonding orbital) and (1 bonding and 2 anti-bonding) orbitals" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8926" PostTypeId="2" ParentId="8922" CreationDate="2014-03-05T09:25:23.997" Score="10" Body="&lt;p&gt;The answer to this question hinges on two key questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What kind of dust are we dealing with?&lt;/li&gt;&#xA;&lt;li&gt;How is the dust arranged (i.e. what is your experimental setup)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Which Dust?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Firstly, I'm going to assume you meant domestic dust (since there are more kinds of dust than you'd imagine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Dust&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dust in homes, offices, and other human environments contains small amounts of plant pollen, human and animal hairs, textile fibers, paper fibers, minerals from outdoor soil, human skin cells, burnt meteorite particles, and many other materials which may be found in the local environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Arrangement&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The arrangement of the dust particles is key. Are we considering an aerosol, or are we carefully laying a line of dust on the table and connecting it with two electrodes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aerosol&lt;/strong&gt; - We can safely assume that the conductance of air has a far higher impact on the measurement than the conductance of the particles in the aerosol (the dust). So here, dust does not conduct electricity. &lt;em&gt;Note: I ignore the &quot;conducting&quot; via static electricity processes here, but they may play a role. I do not consider that &quot;classical&quot; conductance.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Line of Dust&lt;/strong&gt; - Organic material is usually insulating, as is our skin (if it wasn't for the moisture that is needed for, you know, living), paper, textiles, hair and so on. So the main constituents of domestic dust does not conduct electricity. So also here, we will not observe conductance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one exception to the above, where conductance is a possibility: &lt;strong&gt;Metal dust.&lt;/strong&gt; This is kind of cheating, but consider the waste of some metalworking process, in which metal dust (or powder) is generated. This dust, being composed of metals and metal oxides, would likely conduct electricity (when laying it out in a line, still not as an aerosol).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-05T09:25:23.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8928" PostTypeId="2" ParentId="8925" CreationDate="2014-03-05T12:51:21.937" Score="13" Body="&lt;p&gt;If you want to really understand why energy levels of conjugated $\pi$ systems are the way they are I suggest you take a good look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;noreferrer&quot;&gt;Hückel theory&lt;/a&gt; as @ssavec suggested - this method is only an approximation but because of its simplicity it is really good for helping you to understand the underlying concepts.&#xA;Nevertheless, doing as you suggested and looking at the overall overlap between the $\text{p}$ orbitals in a given molecular orbital does indeed seem to work although I don't have a strict justification for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Hückel theory it is known that the coefficients of the $\text{p}$ orbitals in a given molecular orbital can be calculated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  c_{j,r} = \sqrt{\frac{2}{N + 1}} \sin\Bigl(\frac{r j \pi}{N + 1}\Bigr)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving the coefficient $c_{jr}$ for atom $j$ in molecular orbital $r$ of a conjugated system of $N$ atoms (so that $j$ and $r = 1, 2, 3, \dots , n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply this to butadiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fyT2N.png&quot; height=&quot;100&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\pi$ molecular orbitals of butadiene are $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, and $\Psi_{4}$.&#xA;They are built from a linear combination of 4 atomic $\text{p}$ orbitals ($\phi_{1}$, $\phi_{2}$, $\phi_{3}$, and $\phi_{4}$), so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Psi_{r} = c_{1,r} \phi_{1} + c_{2,r} \phi_{2} + c_{3,r} \phi_{3} + c_{4,r} \phi_{4}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you calculate the atomic orbital coefficients for each molecular orbital $r$ then you get to the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kqbhS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the numbers above the different $\text{p}$ orbitals are the coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same for ethene you get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OoB6Q.png&quot; height=&quot;250&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to compare the overall overlap in the HOMOs of ethene and butadiene.&#xA;Since it is already assumed in the Hückel theory only the overlap between neighboring $\text{p}$ orbitals is considered.&#xA;Furthermore, it is useful to set the overlap integral of two &quot;normal&quot; $\text{p}$ orbitals (both with coefficient $c = 1$) to 1 (its exact value doesn't matter because only the relative amounts of overlap in different MOs are important for this comparison), so that you can calculate the overall overlap in a molecular orbital simply by multiplying the coefficients of neighboring $\text{p}$ orbitals and summing over these products.&#xA;So, for the overlap in the ethene and butadiene HOMOs you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  S_{\text{HOMO, ethene}} &amp;amp;= 0.707 \cdot 0.707 \\&#xA;  &amp;amp;= 0.500 \\&#xA;  \\&#xA;  S_{\text{HOMO, butadiene}} &amp;amp;= 0.600 \cdot 0.371 + 0.371 \cdot (-0.371) + (-0.371) \cdot (-0.600) \\&#xA;  &amp;amp;= 0.307&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is less overlap and thus less bonding in the HOMO of butadiene than in the HOMO of ethene which would imply that the HOMO of butadiene is higher in energy than that of ethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about the LUMOs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  S_{\text{LUMO, ethene}} &amp;amp;= 0.707 \cdot (-0.707) \\&#xA;  &amp;amp;= -0.500 \\&#xA;  \\&#xA;  S_{\text{LUMO, butadiene}} &amp;amp;= 0.600 \cdot (-0.371) + (-0.371) \cdot (-0.371) + (-0.371) \cdot 0.600 \\&#xA;  &amp;amp;= -0.307&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is more overlap and thus more bonding (or better to say less antibonding) in the LUMO of butadiene than in the LUMO of ethene which would imply that the LUMO of butadiene is lower in energy than that of ethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the HOMO of butadiene is higher in energy than that of ethene and the LUMO of butadiene is lower in energy than that of ethene.&#xA;Thus, the HOMO-LUMO gap in butadiene is smaller than that in ethene.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-03-05T12:58:47.603" LastActivityDate="2014-03-05T12:58:47.603" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8929" PostTypeId="1" AcceptedAnswerId="8932" CreationDate="2014-03-05T13:26:08.413" Score="10" ViewCount="1810" Body="&lt;p&gt;I've read that radioactive compounds are dangerous because they emit electromagnetic waves (gamma-radiation etc.) that are dangerous for living organisms because they can ionize atoms in molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one water molecule is ionized by using radiation, would you then get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O -&amp;gt; H^+ + OH^-}$&#xA;or would you get free radicals : $\ce{H2O -&amp;gt; H^{.} + OH^{.}}$?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="9961" LastEditDate="2016-06-17T22:38:25.567" LastActivityDate="2016-06-17T22:43:03.560" Title="What is a product of water ionization?" Tags="&lt;water&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8930" PostTypeId="2" ParentId="8929" CreationDate="2014-03-05T14:21:07.537" Score="3" Body="&lt;p&gt;You can get the free radicals upon direct UV irradiation of water at wavelength shorter than $\lambda$ = 190 nm.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-05T14:21:07.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8931" PostTypeId="2" ParentId="8925" CreationDate="2014-03-05T14:55:42.667" Score="8" Body="&lt;p&gt;Here's an alternate approach to answering this question using the &quot;Particle in a Box&quot; approach.  The $N^{th}$ energy level for a particle in a one-dimensional box is given by $$E_n=\frac{n^2h^2}{8mL^2}$$ where h is Planck's constant, m the mass of the particle and L the length of the box.  The energy of the N+1 energy level is equal to $$E_{n+1}=\frac{(n+1)^2h^2}{8mL^2}$$.  Subtracting these two terms gives us the energy separation between these two energy levels as $$\Delta E=\frac{(2n+1)h^2}{8mL^2}=K\frac{2n+1}{L^2}$$.  Pictured below is the beginning of our polyene series, ethylene, butadiene, hexatriene...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dChza.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let's assume all bond lengths are equal.  Then the &quot;length&quot; of our box increases by 2L each time we move from ethylene to butadiene to hexatriene, etc.  If we now compare the difference of the $\Delta E's$ (the $\Delta\Delta E's$) for the $N^{th}$ and $(N+1)^{st}$ energy levels between ethylene and butadiene, we find that the difference in energy level separation is given by $$\Delta\Delta E=K(2n+1)(\frac{1}{L^2}-\frac{1}{9L^2})= K'(\frac{8}{9L^2})$$&#xA;If we perform the same comparison for butadiene and hexatriene we see that $$\Delta\Delta E=K(2n+1)(\frac{1}{9L^2}-\frac{1}{25L^2})= K'(\frac{16}{225L^2})$$ and for hexatriene compared to octatetraene it is $$\Delta\Delta E=K(2n+1)(\frac{1}{25L^2}-\frac{1}{49L^2})= K'(\frac{24}{1225L^2})$$  In other words as we increase the length of our box (increase the length of our polyene), the energetic separation between any corresponding pair of energy levels (HOMO-LUMO being one case) will shrink.  Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-03-05T14:55:42.667" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="8932" PostTypeId="2" ParentId="8929" CreationDate="2014-03-05T17:04:01.600" Score="18" Body="&lt;p&gt;Water, when hit by radioactive particles, will most likely suffer ionization (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiolysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;radiolysis&lt;/a&gt;). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived species which will tend to react with most other substances nearby, even if neutral. The electron, after being slowed down considerably, becomes hydrated and can participate in reactions itself. &lt;a href=&quot;http://jcsp.org.pk/ArticleUpload/2547-11501-1-RV.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; has several examples of reactions triggered by radiolysis, and also separates them by their relevant timescales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some further examples of reactions involving radiolysis of water via beta particles and gamma rays are (from the very well-cited article &quot;&lt;a href=&quot;http://www3.nd.edu/~ndrlrcdc/Compilations/rxn.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Critical review of rate constants for reactions of hydrated electrons, hydrogen atoms and hydroxyl radicals (⋅OH/⋅O−) in aqueous solution&lt;/a&gt;&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlpLm.png&quot; alt=&quot;So many possibilities!&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, not only are the decay particles damaging, but so are the entities produced by their interaction with matter in the vicinity. If the nuclear decay particle doesn't directly ionize organic matter, reactive species such as $\ce{OH^{.}}$ or $\ce{H^{.}}$ could then react with biomolecules, degrading cells and tissues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radiation_damage#Effects_on_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia link&lt;/a&gt; has some details on the radiolysis of water. Interestingly, it seems that dissolved hydrogen in water can efficiently suppress the formation of unusually reactive species, presumably as the side reactions eventually cause the formation of hydroxyl anions (if electrons are present in excess) or hydronium cations (if positive charges are present in excess). &lt;a href=&quot;https://i.stack.imgur.com/LluEa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This image&lt;/a&gt; (&lt;a href=&quot;https://www.mun.ca/biology/scarr/Radiolysis_of_Water.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) shows a collection of possible reactions in sequence. If you search for articles on water radiolysis, there are many other results, such as &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/j100725a027&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.jstor.org/discover/10.2307/3580071?uid=3737664&amp;amp;uid=2129&amp;amp;uid=2&amp;amp;uid=70&amp;amp;uid=4&amp;amp;sid=21103665367673&quot; rel=&quot;nofollow noreferrer&quot;&gt;two&lt;/a&gt;. It's interesting to note that the effect of radiation on water still doesn't seem to be entirely understood, in part due to our incomplete knowledge of the behaviour of hydrated electrons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here's an extra bit of background on radiation and its biological effects (at least for the most common types. &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7619/types-of-radioactive-decay/7621#7621&quot;&gt;There are many ways nuclei can decay&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending how far the radiation source is from you, gamma radiation is either the least or the most dangerous type of nuclear decay. If the radiation is coming from well outside your body, then the danger scale goes &lt;em&gt;alpha &amp;lt; beta &amp;lt; gamma&lt;/em&gt;. If you happen to have a radioactive nuclide in your pocket, the scale may well become &lt;em&gt;alpha &amp;lt; gamma &amp;lt; beta&lt;/em&gt;. However, if the radiation is coming from &lt;em&gt;inside&lt;/em&gt; your body (&lt;a href=&quot;http://en.wikipedia.org/wiki/Goi%C3%A2nia_accident#Ivo_and_his_daughter&quot; rel=&quot;nofollow noreferrer&quot;&gt;avoid ingesting glowing substances&lt;/a&gt;), then the danger scale goes &lt;em&gt;gamma &amp;lt; beta &amp;lt; alpha&lt;/em&gt;. Why does radiation behave like that? There are two main effects in play: &lt;em&gt;penetration capacity&lt;/em&gt; and &lt;em&gt;ionization capacity&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In collision theory, different types of projectile and target can be characterized by something called an collision cross-section (or absorption cross-section, in nuclear physics), which basically gives you a probability of how likely it is for the projectile to hit a certain target. If the combination of projectile and target have a high collision cross-section, they they are very likely to interact, and so almost any projectile will be stopped by even a thin target. If the collision cross-section is low, then the projectile has a low chance of hitting the target, and therefore will tend to pass through it entirely. To compensate, you need to make a very thick target so there are more chances for the projectile and target to interact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different types of radioactive decay have different penetrating capacities. All of matter as we know it is composed of atoms, which contain well-defined oppositely charged regions (the nuclei and the electron shells), and so we can expect them to interact strongly with other charged particles. Alpha and beta(-minus) decays emit particles with a charge of $+2e$ and $-e$, respectively ($+e$ for beta-plus decay). We then expect these particles to interact strongly with matter, being stopped easily. Indeed, most alpha particles are efficiently stopped even by air. Beta particles have half the charge of an alpha particle and are going quite a bit faster (having approximately 1/7000th the mass), so they manage to go further before interacting. A few centimetres of of metal shielding or a concrete wall are usually enough to stop them. However, gamma rays have no charge, so they interact comparatively weakly with matter. These highly energetic photons will go though dozens of centimetres of solid matter. To maximize the odds that they will hit something and be stopped, you want to use a shield made of a very dense substance (which would pack a lot of possible targets in a small space). This is why lead is the substance of choice for radioactive shielding; its high density combined with low cost allows the fabrication of relatively compact shields which are capable of stopping even gamma radiation, when around 10 cm thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the mechanism of how particles from radioactive decay are stopped. For uncharged particles, the absorption cross-section is low because interactions only happen when the uncharged particle comes very close to a target particle. For gamma rays (very high energy photons), when there is an interaction, it is usually due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Compton_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compton scattering&lt;/a&gt;, where an energetic photon interacts with an electron, giving it a large amount of kinetic energy, while the photon is re-emitted with a smaller amount of energy. The target electron (not necessarily a valence electron) often gains enough energy to be completely removed from the atom, causing ionization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For collisions between charged particles (both same and opposite charges), interactions happen much more easily, as charges interact strongly even at comparatively much larger distances. Beta-minus particles (very fast electrons) can more efficiently repel electrons and shove away from atoms, creating cations. Alpha particles (bare helium-4 nuclei) are interesting because their positive charge means they can tear electrons from any atom or ion. In fact, it's defensible that alpha particles are the strongest Lewis acid one can commonly find; they are effectively free protons with twice the charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that penetration capacity and ionization capacity are inversely correlated; if a radioactive decay particle is strongly ionizing, it interacts well with matter and is stopped in short order. If it is weakly ionizing, one of the main interaction mechanisms is suppressed, and the decay particle can go further before it is stopped. In order to not damage you, you want the nuclear decay products to not dump their energy in your body, i.e. you want them to pass right through you. Here, it is clear that alpha particles are the most dangerous, followed by beta and gamma radiation. However, alpha particles interact so strongly with matter that they literally need to be released closer than your skin to dump energy into living tissue, so you need to ingest, breathe or inject alpha emitters. When this happens, &lt;a href=&quot;http://en.wikipedia.org/wiki/Polonium#Biology_and_toxicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;they are &lt;strong&gt;extremely&lt;/strong&gt; deadly&lt;/a&gt;. It is a very rare occurrence, however, since radiation usually comes from far outside. In that case, gamma radiation has a larger chance of reaching you, so it is said to be the most dangerous.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-03-05T23:49:54.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8933" PostTypeId="1" CreationDate="2014-03-05T17:30:29.327" Score="2" ViewCount="14508" Body="&lt;p&gt;I need help with this one homework question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am combining $100\,\text{mL}$ of $0.5~\text{M}~\ce{HCl}$ and $100~\text{mL}$ of $0.5~\text{M}~\ce{NaOH}$ to obtain $200~\text{mL}$ of $0.25~\text{M}~\ce{NaCl}$. The change in temperature of this reaction was $\Delta \vartheta  =3.38~\text{°C}$. I have to calculate the heat of reaction for 1 mole of the reactant. I also need to calculate the $c_\text{total}$ of this reaction. How do I go about doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation is $$\ce{HCl + NaOH -&amp;gt; H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the heats of formation of $\ce{HCl}$, $\ce{NaOH}$, $\ce{H2O}$, and $\ce{NaCl}$ are $-166.19$, $-469.07$, $-285.85$, and $-406.77$, respectively, all in $\text{kJ/mol}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went about doing this calculating the heat $$q=\Delta H=[-(-166.19-469.07)+(-285.85-406.77)]~\text{kJ/mol} = -57.36~\text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then found the mass of the liquids used for the reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}m &amp;amp;= 0.1~\text{L}\times 0.500 ~\text{M}\times 36.4611~\mathrm{g\,mol^{-1}} \\&#xA;&amp;amp;\;\;+0.1~\text{L} \times 0.500~\text{M} \times 39.99715 ~\mathrm{g\,mol^{-1}}\\&#xA;&amp;amp;=3.822~ \mathrm{g} \end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then solved for $c$ using $q=mc\Delta T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.05\times 57360~\mathrm{J} = 3.822\ \mathrm{g}\times c_\text{total}\times 3.38~\mathrm{K} \Rightarrow c_\text{total}=222.01 ~\mathrm{J \, g^{-1}\, K^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I multiplied the $q$ by $.05$ because that's how many moles of reactant I have (since we have the molarity and volume of the reactant). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure I am doing something wrong, but I am not sure what.&lt;/p&gt;&#xA;" OwnerUserId="4663" LastEditorUserId="69830" LastEditDate="2021-05-11T05:13:25.570" LastActivityDate="2021-05-11T05:13:25.570" Title="How do you calculate the total heat capacity of a reaction" Tags="&lt;acid-base&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8934" PostTypeId="1" CreationDate="2014-03-05T18:06:20.187" Score="6" ViewCount="675" Body="&lt;p&gt;What is the chemistry behind why reactions of order &gt;1 exist? And what role does the number of activated complexes formed have in determining reaction rate?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="4642" LastEditDate="2014-03-06T18:39:25.657" LastActivityDate="2020-01-20T12:54:35.847" Title="Reaction Order Greater than 1" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8935" PostTypeId="1" CreationDate="2014-03-05T19:23:59.387" Score="0" ViewCount="825" Body="&lt;p&gt;The question simply says “calculate the temperature of hot air in kelvin” on a data calculation section for this lab. We recorded:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;volume change on cooling = 170 mL&lt;/li&gt;&#xA;&lt;li&gt;original volume of hot air = 20 mL&lt;/li&gt;&#xA;&lt;li&gt;volume of the cooled air = 150 mL&lt;/li&gt;&#xA;&lt;li&gt;temperature of heated water = 100 °C&lt;/li&gt;&#xA;&lt;li&gt;temperature of cool water = 20.4 °C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and that is all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lab was focused on Charles’ law, namely to predict how much air should contract when cooled from 100 °C to room temperature. We took a flask and inserted a rubber stopper with glass tubing through the center. We then heated the flask in a water bath until the water boiled, then keeping a finger over the top of the glass tubing we inverted the flask in a “cool” water bath and held it there for 10 minutes. Then we equalized the pressure within the flask with the atmospheric pressure by raising the flask until the water level inside equaled the outside.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorDisplayName="user7951" LastEditDate="2015-06-14T19:02:59.417" LastActivityDate="2015-06-14T19:02:59.417" Title="How do I calculate the temperature of hot air in kelvin?" Tags="&lt;homework&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8936" PostTypeId="1" AcceptedAnswerId="8939" CreationDate="2014-03-05T22:22:05.287" Score="1" ViewCount="1092" Body="&lt;p&gt;What are some pharmaceutics that use vanillin? Is vanillin just an aroma and taste agent to pharmaceutics?&lt;/p&gt;&#xA;" OwnerUserId="4729" LastEditorUserId="418" LastEditDate="2014-03-05T23:23:10.860" LastActivityDate="2014-03-05T23:28:09.397" Title="What are pharmaceutical uses for Vanillin?" Tags="&lt;organic-chemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8937" PostTypeId="2" ParentId="7815" CreationDate="2014-03-05T22:30:47.243" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that the best source of lye is hardwood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is that really so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;THE SODIUM, POTASSIUM AND PHOSPHORUS CONTENTS OF TREE SPECIES GROWN IN CLOSE STANDS&lt;/em&gt;, published in &lt;em&gt;New Phytologist&lt;/em&gt;, &lt;strong&gt;1958&lt;/strong&gt;, &lt;em&gt;57(3)&lt;/em&gt;, 273–284 (&lt;a href=&quot;http://dx.doi.org/10.1111/j.1469-8137.1958.tb05316.x&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;), J. D. Ovington and H. A. I. Madgwick state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] In nearly all cases, conifers contain greater weights of sodium, potassium&#xA;  and phosphorus than hardwoods of the same age grown under similar conditions. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-05T22:30:47.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8938" PostTypeId="2" ParentId="8936" CreationDate="2014-03-05T22:45:23.153" Score="0" Body="&lt;p&gt;To my knowledge, vanillin nowadays does not have any pharmaceutical uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to Elisabeth Vaupel in &lt;em&gt;Pharmazeutische Zeitung online&lt;/em&gt;, &lt;strong&gt;2002&lt;/strong&gt;, 38 (&lt;a href=&quot;http://www.pharmazeutische-zeitung.de/index.php?id=24471&quot; rel=&quot;nofollow&quot;&gt;SRC&lt;/a&gt;), &lt;a href=&quot;http://en.wikipedia.org/wiki/Michael_Bernhard_Valentini&quot; rel=&quot;nofollow&quot;&gt;Michael Bernhard Valentini&lt;/a&gt; wrote 1714 in his book &lt;em&gt;Museum Museorum, oder vollständige Schau-Bühne aller Materialien und Specereyen, nebst deren natürlichen Beschreibung&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sehr erwärmende und zerteilende, anbei aber auch stärkende Kraft, womit sie den Magen stärken, die Winde zerteilen und dem Gehirn, der Mutter und anderen nervösen Gliedern sehr gut tun. Sie treiben den Harn, befördern die monatliche Reinigung, natürliche Geburt und Schwierungen: Treiben auch die Nachgeburt und tote Kinder fort und kommen also dem weiblichen Geschlecht in ihren meisten Krankheiten wohl zu pass. Ingleichen werden sie gegen die erstarrend-machenden giftigen Bisse and andere dergleichen giftige Sachen gebraucht [...] Am meisten aber werden die Vanillen zu Verfertigung der Chocolaten gebraucht, welche sie anmutiger und kräftiger machen. Die Tabaksbrüder brauchen sie auch, den Tabak wohlriechend zu machen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a starting material in synthesis, vanillin can be converted to &lt;a href=&quot;http://en.wikipedia.org/wiki/MMDA_%28drug%29&quot; rel=&quot;nofollow&quot;&gt;MMDA&lt;/a&gt; in 5 steps.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-05T23:28:09.397" LastActivityDate="2014-03-05T23:28:09.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8939" PostTypeId="2" ParentId="8936" CreationDate="2014-03-05T22:52:34.903" Score="0" Body="&lt;p&gt;Vanillin is a pharmaceutical intermediate and flavoring, but not a pharmaceutical itself.  One can easily convert vanillin into dihydrocapsaicin (15 million Scovilles versus 16 million for capsaicin):  $\ce{NaBH3CN/NH3/NH4Cl}$, then isodecanoic amide of that benzylamine.  Surprise the downwind neighborhood.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-05T22:52:34.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8940" PostTypeId="1" CreationDate="2014-03-06T00:53:00.607" Score="0" ViewCount="2064" Body="&lt;p&gt;Is it due to the aldol reaction that happens? But the product of that reaction is still soluble in acetone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently if you have a solid mixture of 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone with NaOH and Na2CO3 impurites, and you dissolve them in acetone, after which you centrifuge, the solid impurites will settle out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="2645" LastEditDate="2014-04-27T04:00:42.617" LastActivityDate="2014-04-27T04:00:42.617" Title="Why is it possible to separate NaOH from certain compounds using acetone?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-03-11T17:36:50.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="8942" PostTypeId="2" ParentId="6587" CreationDate="2014-03-06T04:24:49.113" Score="3" Body="&lt;p&gt;Yes, it is soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fizzy part of your soda does not form a layer at the bottom of the bottle. (Personally, I find the empirical solutions best for chemistry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbonate rule only applies to alkali metal carbonates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a general rule, anything with the word &quot;acid&quot; is probably soluble in water.  Seeing as acids are sometimes defined as yielding $\ce{H+}$ ions when dissolved.&lt;/p&gt;&#xA;" OwnerUserId="4730" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T08:38:29.820" LastActivityDate="2015-06-26T08:38:29.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8943" PostTypeId="2" ParentId="8935" CreationDate="2014-03-06T04:33:42.557" Score="2" Body="&lt;p&gt;You already know the temperature of the hot air in Celsius: &lt;code&gt;100 degrees&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equation &lt;code&gt;Kelvin = Celsius + (273.15 degrees)&lt;/code&gt;, you would get &lt;code&gt;100 degrees + 273.15 degrees&lt;/code&gt;, or &lt;code&gt;373.15 degrees&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that this is a preliminary step that is required in order to find the temperature to plug into Charles' law (which must be in Kelvin), in which case you can use the formula I mentioned above for your other calculations as well.&lt;/p&gt;&#xA;" OwnerUserId="4730" LastActivityDate="2014-03-06T04:33:42.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8944" PostTypeId="1" AcceptedAnswerId="8946" CreationDate="2014-03-06T06:39:25.047" Score="11" ViewCount="14168" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A dilute solution of which acid is most likely to produce a  reduction&#xA;product other than &lt;span class=&quot;math-container&quot;&gt;$\ce{H_{2}}$&lt;/span&gt; when it reacts with a  metal?&lt;/p&gt;&#xA;&lt;p&gt;(A) &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(B) &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(C) &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(D) &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\tiny{\text{Question from 2012 local chemistry olympiad}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The given answer is (C), &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3} $&lt;/span&gt; but I don't understand why.&lt;/p&gt;&#xA;&lt;p&gt;First off, what does it mean by reduction product -- is it the product that is formed when the acid is reduced in the redox reaction between the metal and the acid?&lt;/p&gt;&#xA;&lt;p&gt;Second, why is the answer &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3} $&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-05-07T02:33:26.873" Title="Why does HNO3 not give off H2 when reacting with Cu?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8945" PostTypeId="2" ParentId="8934" CreationDate="2014-03-06T07:35:18.760" Score="1" Body="&lt;p&gt;The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn’t clear enough. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take &lt;span class=&quot;math-container&quot;&gt;$$\ce{2NO2 \to 2NO + O2}$$&lt;/span&gt; You can (extremely basically) think of this reaction as two &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; particles colliding, (forming a complex), then coming out as &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; molecules, &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; being &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; being &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;. But you needed two &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation.&#xA;I’m not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I’m wrong)&#xA;Three molecules colliding at once (&lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; rd order reaction): way less likely, and that’s why it’s not studied so much, it’s much harder getting &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; molecules to collide (especially in the gas phase without) in the right orientation with enough energy. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By general rate law do you mean finding the constant &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out.&#xA;I think there is more than one way to derive the general rate law’s form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it’s produced (so there’s no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is composed of multiple constants.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It’s true because they told you it was. This gives you details that you wouldn’t have necessarily been able to deduce just by having the reaction equation there. Importantly, it’s there to tell you straight up that the equation you have can’t be broken down to further reactions... Again you can think of it like this: &lt;span class=&quot;math-container&quot;&gt;$\ce{a A}$&lt;/span&gt; molecules have to collide with &lt;span class=&quot;math-container&quot;&gt;$\ce{b B}$&lt;/span&gt; molecules to produce a single activated complex. That’s one step. If they didn’t specify that, maybe there could have been intermediate steps that could change the rate law.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe it'll be easier to see if you consider extremes. Say Reaction &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; can be divided into &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt; reactions, &lt;span class=&quot;math-container&quot;&gt;$r1, r2, r3, r4$&lt;/span&gt;. Now, say &lt;span class=&quot;math-container&quot;&gt;$r1, r2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$r4$&lt;/span&gt; happen in microseconds, while &lt;span class=&quot;math-container&quot;&gt;$r3$&lt;/span&gt; takes &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; seconds. Already you're at the point where you can basically ignore the times of &lt;span class=&quot;math-container&quot;&gt;$r1, r2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$r4,$&lt;/span&gt; because they're so small compared to the time &lt;span class=&quot;math-container&quot;&gt;$r3$&lt;/span&gt; took (comparing &lt;span class=&quot;math-container&quot;&gt;$1\mathrm E-6$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1\mathrm E+2$&lt;/span&gt; here). So as you can see in this example, &lt;span class=&quot;math-container&quot;&gt;$r3$&lt;/span&gt; was literally the rate determining step.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="52853" LastEditDate="2020-01-20T12:54:35.847" LastActivityDate="2020-01-20T12:54:35.847" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8946" PostTypeId="2" ParentId="8944" CreationDate="2014-03-06T08:03:26.133" Score="12" Body="&lt;p&gt;Whenever any metal is added to an acid, the metal is oxidised to one of its positive ions, at the cost of either the acid which is reduced or water which is reduced to hydrogen. The product formed by reduction is what is referred to as the &lt;em&gt;reduction product&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction potential of $\ce{NO3-}$ is $0.8$ which makes the reduction of nitrate instead of hydrogen ion $\ce{H+}$ ($E^\circ=0.0$) more favourable. Thus it is the nitrate in nitric acid which is reduced instead of $\ce{H+}$ or $\ce{H2O}$. But at very low concentration of nitrate ion and low temperature, as explained by Nernst equation, the reduction potential decreases significantly and combined with overpotential effects, it is $\ce{H2}$ which is formed and not the reduction product of nitrate which is $\ce{NO2}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of halic acids, the halogens cannot be further reduced. In the case of the sulfate anion, although it has a reduction potential of $0.17$, owing to overpotential effects, sulfate is not reduced, instead, water is reduced yielding $\ce{H2}$. At high concentration, the reduction potential rises sufficiently to overcome the overpotential effects, allowing the reduction of sulfate to form $\ce{SO2}$ at higher concentrations of sulfuric acid. Hence under normal conditions, only nitric acid gives a product other than hydrogen gas.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reduction potential data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO4^{2-} + 4H+ + 2e^- -&amp;gt; SO2 + 2H2O}(E^\circ=0.17)$$&#xA;$$\ce{NO3- + 2H+ + e- -&amp;gt; NO2 + H2O} (E^\circ=0.8)$$ &#xA;$$\text{Nernst equation } E=E^\circ- \frac{RT}{nF}\ln Q$$&#xA;$Q$= reaction quotient, $E$=redox potential, $E^\circ$=Standard redox potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;br&gt;&#xA;The oxidation of metal at the cost of reduction of another substance occurs whenever it is thermodynamically favourable. Usually, (except silver, copper) metals have negative reduction potentials, which makes their oxidation at the cost of hydrogen thermodynamically favourable as the $\Delta G&amp;lt;0$. This leads to a spontaneous redox reaction. This is not true if there is a kinetic or physical barrier to reaction. For example, Iron is rendered passive by nitric acid due to the formation of an oxide layer at the surface which prevents any further contact and hence stops the reaction.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="402" LastEditDate="2017-05-07T02:33:26.873" LastActivityDate="2017-05-07T02:33:26.873" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8947" PostTypeId="2" ParentId="8940" CreationDate="2014-03-06T09:44:01.670" Score="1" Body="&lt;p&gt;Ask yourself what kind of reaction happens?&#xA;What glassware have you used to remove impurities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By controlling pH and polarities you are able to separate compounds because of different affinities between them and solvents. You change molecular solvatation of your analyte and impurities. Then by controlling boiling point, using a rotavap, you may be able to separate two phases that are miscible. Or by making impurities insoluble you can remove by mechanical operation such as centrifugation. Such post-treatments are very common in chemistry and they scale up very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is no chemical reaction (except acid/base reaction if any) there are physical reactions (solvatation, phase extraction, phase change: evaporation, condensation, crystallisation and sedimentation).&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-03-06T09:44:01.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8948" PostTypeId="1" AcceptedAnswerId="8950" CreationDate="2014-03-06T09:59:47.517" Score="12" ViewCount="15184" Body="&lt;p&gt;I mean, how can we reverse the neutralization reaction of $\ce{HCl}$ and $\ce{NaOH}$ to get back $\ce{HCl}$ and $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to extract $\ce{HCl}$ and $\ce{NaOH}$ from a mixture of common salt and distilled water in equal proportions at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, how can I make this reaction happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + NaCl -&amp;gt; HCl + NaOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What temperature or pressure conditions are required?&lt;br&gt;&#xA;Also, would the extracted components be pure? (I know common salt contains iodine and even distilled water is not pure, that's why I am asking this.)&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="16683" LastEditDate="2017-04-03T15:24:45.983" LastActivityDate="2017-04-03T15:24:45.983" Title="How to separate HCl and NaOH from salt and water solution?" Tags="&lt;acid-base&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8949" PostTypeId="1" AcceptedAnswerId="14549" CreationDate="2014-03-06T10:28:05.163" Score="10" ViewCount="508" Body="&lt;p&gt;In the addition of $\ce{NaHSO3}$ to aldehydes and methyl ketones (higher ketones do not respond well to this reaction), crystalline addition products are formed. An $\ce{-SO3H}$ and an $\ce{-OH}$ group are attached to the same carbon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R2C=O + NaHSO3 -&amp;gt; R2C(OH)(SO3^{-}Na^+) &amp;lt;=&amp;gt; R2C(O^{-}Na^+)(SO3H)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line I read online, bothering me is:- &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The proton transfer equilibrium lies to the right for most ketones&#xA;  and to the left for most aldehydes.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the reason for this (assuming that it is correct in the first place)?&lt;/strong&gt; Since the sulphonic acid group is more acidic than the hydroxyl group, should not the proton tranfer equilibrium lie towards the left for both the cases?&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-07-23T16:49:32.187" Title="Proton transfer equilibrium in bisulphite adducts" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8950" PostTypeId="2" ParentId="8948" CreationDate="2014-03-06T11:10:51.170" Score="21" Body="&lt;p&gt;Unless &lt;em&gt;time travel&lt;/em&gt; is an option, you could &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chloralkali_process&quot;&gt;Electrolyze&lt;/a&gt; the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation  at $\lambda$ &amp;lt; 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the &lt;strong&gt;chain reaction&lt;/strong&gt; to yield $\ce{HCl}$ gas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 -&amp;gt;[h\nu] 2 Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl* +\  H2 -&amp;gt; HCl + H*}$$&#xA;$$\ce{H* +\  Cl2 -&amp;gt; HCl + Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you know exactly about the risk of both processes, &lt;strong&gt;don't do it!&lt;/strong&gt;&#xA;This &lt;a href=&quot;https://www.youtube.com/watch?v=hrFKoq00kBk&quot;&gt;video&lt;/a&gt; gives a nice impression of how &lt;strong&gt;violently&lt;/strong&gt; hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can &lt;strong&gt;not&lt;/strong&gt; simply revert the neutralisation reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The neutralisation is irreversible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the &lt;strong&gt;entropy&lt;/strong&gt; changes. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-06T14:56:41.067" LastActivityDate="2014-03-06T14:56:41.067" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8952" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2014-03-06T14:40:58.397" Score="2" ViewCount="29635" Body="&lt;p&gt;Will &lt;span class=&quot;math-container&quot;&gt;$\ce{NaH2PO4}$&lt;/span&gt; react as as an acid or a base if you make it react with water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got stuck here. I can figure out the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NaH2PO4 + H2O -&amp;gt; Na^+ + H2PO4^- + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I look &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4^{-}}$&lt;/span&gt; up, it says that it can both be an acid and a base. &#xA;Any ideas? &lt;/p&gt;&#xA;" OwnerUserId="1642" LastEditorUserId="4945" LastEditDate="2019-01-16T13:39:35.747" LastActivityDate="2019-01-16T13:43:55.853" Title="Will monosodium dihydrogen phosphate react as base or acid in water?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8953" PostTypeId="1" AcceptedAnswerId="8954" CreationDate="2014-03-06T15:25:58.753" Score="1" ViewCount="210" Body="&lt;p&gt;I'm doing relatively simple Chemistry but I've been given a formula for the velocity of a reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v= k \cdot c_\mathrm A^a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k= A \cdot \mathrm e^{E_\mathrm a/RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, from that formula, surely decreasing the temperature $T$ would increase the velocity of the reaction? As that would increase the value of $k$ and therefore drive up the reaction speed? Also, same thinking, increasing the activation energy $E_\mathrm a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that's wrong, because common sense. But the formula is lying to my face. &lt;/p&gt;&#xA;" OwnerUserId="4738" LastEditorDisplayName="user7951" LastEditDate="2016-01-30T13:01:48.880" LastActivityDate="2016-01-30T13:01:48.880" Title="How to increase the velocity of a reaction?" Tags="&lt;kinetics&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8954" PostTypeId="2" ParentId="8953" CreationDate="2014-03-06T15:33:28.027" Score="4" Body="&lt;p&gt;It should be $-E_a/RT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor, increasing the temperature, lowering the activation energy (catalysis), or increasing the concentration of reactant(s) would increase the reaction velocity.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-03-06T15:33:28.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8955" PostTypeId="2" ParentId="8952" CreationDate="2014-03-06T16:30:40.633" Score="4" Body="&lt;p&gt;Monosodium dihydrogen phosphate, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4^-}$&lt;/span&gt;, is an amphoteric species and it will act as both an acid and a base. &#xA;These are the related chemical equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \ce{H2PO4- + H2O &amp;amp;&amp;lt;=&amp;gt; HPO4^2- + H3O+}  &amp;amp; \mathrm{p}K_\mathrm{a} &amp;amp;= 7.21\\&#xA;  \ce{H2PO4- + H2O &amp;amp;&amp;lt;=&amp;gt; H3PO4 + OH-}     &amp;amp; \mathrm{p}K_\mathrm{b} &amp;amp;= 11.88&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; solution of such a salt will have a pH of about &lt;span class=&quot;math-container&quot;&gt;$4.7$&lt;/span&gt;.&#xA;In other words, the acidic behaviour dominates over the alkaline one. This is due to the value of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}=7.21$&lt;/span&gt; (related to the acidic behaviour) compared to the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{b}=11.88$&lt;/span&gt; (related to the alkaline behaviour).&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="4945" LastEditDate="2019-01-16T13:43:55.853" LastActivityDate="2019-01-16T13:43:55.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8956" PostTypeId="1" AcceptedAnswerId="8957" CreationDate="2014-03-06T17:42:04.180" Score="1" ViewCount="127" Body="&lt;p&gt;First off, please forgive me, I am a chemistry novice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overall Process&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fuel for the system is distilled water&lt;/li&gt;&#xA;&lt;li&gt;Split water into Hydrogen and Oxygen (electrolysis or similar process)&lt;/li&gt;&#xA;&lt;li&gt;Exhaust Oxygen&lt;/li&gt;&#xA;&lt;li&gt;Burn Hydrogen to power a generator/engine&lt;/li&gt;&#xA;&lt;li&gt;Generator/engine charges battery&lt;/li&gt;&#xA;&lt;li&gt;Battery powers Step 2 as well as whatever else we want to power&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Energy gained needs to be greater than the sum of lost energy in all steps, as well as the energy required to perform electrolysis&lt;/li&gt;&#xA;&lt;li&gt;We could compress the exhaust oxygen and use that as an energy source?&lt;/li&gt;&#xA;&lt;li&gt;Is there a danger of running out of hydrogen? (Forgive my naivety, but I've heard about a helium shortage...)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4739" LastActivityDate="2014-03-06T18:03:36.357" Title="Could this be a feasible fuel system?" Tags="&lt;energy&gt;&lt;combustion&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8957" PostTypeId="2" ParentId="8956" CreationDate="2014-03-06T18:03:36.357" Score="4" Body="&lt;p&gt;&quot;&lt;em&gt;Energy gained needs to be greater than the sum of lost energy in all steps&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lhup.edu/~dsimanek/museum/unwork.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.lhup.edu/~dsimanek/museum/unwork.htm&lt;/a&gt; &lt;BR&gt;&#xA; Perpetual motion machines &lt;BR&gt;&#xA;&lt;a href=&quot;http://news.sciencemag.org/chemistry/2014/01/rhubarb-battery-could-store-energy-future&quot; rel=&quot;nofollow&quot;&gt;http://news.sciencemag.org/chemistry/2014/01/rhubarb-battery-could-store-energy-future&lt;/a&gt; &lt;BR&gt;&#xA; Flow battery, which works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You cannot win (First Law of thermodynamics from Noether's theorems and the isotropy of time) &lt;BR&gt;&#xA;2) You can only break even on a very cold day (Second Law of thermodynamics from Carnot cycle and the law of large numbers). &lt;BR&gt;&#xA;3) It never gets that cold (Third Law of thermodynamics.   If there were an entropy difference at absolute zero, T = 0 K could be reached in a finite number of steps. However, at T = 0 K there is no entropy difference, so an infinite number of steps are needed).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-06T18:03:36.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8960" PostTypeId="1" CreationDate="2014-03-06T18:45:58.480" Score="2" ViewCount="135" Body="&lt;p&gt;My textbook says that for a chemical reaction with two reactants, $\ce{A}$ and $\ce{B}$ with equation $\ce{aA + bB -&amp;gt; Products}$, where $a$ and $b$ are coefficients, the general rate law is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Rate} = k[\ce{A}]^m[\ce{B}]^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ and $n$ are the reaction orders for $\ce{A}$ and $\ce{B}$ respectively. Then it says that $m=a$ and $n=b$ if  the reaction between $\ce{A}$ and $\ce{B}$ occurs in a single step and with a single activated complex. Why is this true?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="189" LastEditDate="2014-03-06T19:19:38.267" LastActivityDate="2014-03-07T05:23:41.180" Title="Specific Case of the Rate law" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8961" PostTypeId="1" AcceptedAnswerId="8962" CreationDate="2014-03-06T18:51:46.287" Score="1" ViewCount="174" Body="&lt;p&gt;Why exactly does the slowest reaction limit the instantaneous rate of a reaction even if the faster processes come after? Theoretically, if the products of the slow reaction are required for the fast reaction, the fast reaction limits the rate because even though it is fast, it will not finish in exact synchronization with the end of the slow reaction. There will be a lag between when the net reaction is done and when the slow reaction is done.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="189" LastEditDate="2014-03-06T19:19:12.077" LastActivityDate="2014-03-06T19:19:12.077" Title="Reaction Rate Limiting Reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8962" PostTypeId="2" ParentId="8961" CreationDate="2014-03-06T19:15:09.970" Score="0" Body="&lt;p&gt;It is the measured overall reaction rate that counts and not which reaction finishes the sequence.&#xA;If you measure the rate of a reaction that consists of two elemental reactions with rate constants $k_1$ and $k_2$ where $k_1 \gg k_2$ you will get an overall rate constant $k_{\mathrm{tot}} \approx k_2$.&#xA;This is so, because if the faster reaction takes place after the slower reaction, the faster reaction uses up its reactants (which are the products of the slower reaction) faster than the slower reaction produces its products, so that the faster reaction will consume every new product molecule produced by the slower reaction nearly instantly compared to the time it takes the slower reaction to produce it.&#xA;The time it takes the faster reaction to consume the newly formed molecule is insignificant.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-03-06T19:15:09.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8964" PostTypeId="1" AcceptedAnswerId="8999" CreationDate="2014-03-06T20:03:28.433" Score="9" ViewCount="15435" Body="&lt;p&gt;In the reaction between hydrochloric acid and Magnesium the overall reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + 2HCl -&amp;gt; MgCl2 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proposed mechanism of reaction is that $\ce{HCl}$ dissociates in $\ce{H2O}$ like so:&#xA;$$\ce{H2O + HCl &amp;lt;=&amp;gt; H3O+ + Cl- }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then the $\ce{H3O+}$ oxidises the magnesium like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H3O+ + Mg -&amp;gt; Mg^{2+} + H2O + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, in an experiment to determine the rate of the reaction, measuring the $\ce{H2}$ produced would measure this mechanism, because it is the step in which hydrogen gas is evolved. Presumably these Magnesium ions are then in solution. I would therefore deduce that the step:&#xA;$$\ce{2H3O+ + Mg -&amp;gt; Mg^{2+} + H2O + H2}$$&#xA;Is the rate-determining step, because it is the step in which magnesium ions are formed and go into solution.&#xA;My rate law, deduced from these mechanisms, would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{rate  = k * [H3O+] }$$&#xA;Because it is the concentration of the $\ce{H3O+}$ which determines the formation of the Mg ions. From there, I would assume, the $\ce{Mg^{2+}}$ and the $\ce{Cl-}$ would remain in solution, as the reaction is carried out in the acid, obviously consisting of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, can I say that the rate law for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + 2HCl -&amp;gt; MgCl2 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is merely:&#xA;$$\ce{rate  = k * [H3O+] }$$&#xA;But $\ce{[H3O+]}$ is:&#xA;$$\ce{[H3O+] = (K*[H2O][HCl])/[Cl^{-}] }$$&#xA;Where K is the equilibrium constant of the acid equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can we write our rate law like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{rate = k*(K[H2O][HCl])/[Cl^{-}] }$$&#xA;Where K is the equilibrium constant and k is the rate constant and might look like this (following the Arrhenius equation):&#xA;$$\ce{k = A e^{-E_a/(R T)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make our rate law like this:&#xA;$$\ce{ rate = A e^{-E_a/(R T)} *((K*[H2O][HCl])/[Cl^{-}]) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is my rate law correct for this reaction? If not, where have I gone wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally: This was one of our practicals. We ended up concluding that rate goes up quadratically or exponentially with concentration. This rate law would seem to indicate a linear increase, not exponential. I assume in my rate law there is no contribution from the solid Mg. Obviously in a powder, more surface area is exposed, and a high rate is experienced. Is there a way to factor this into my rate law? &#xA;Thanks for the help! :)&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditDate="2014-10-31T16:42:05.873" LastActivityDate="2014-10-31T16:42:05.873" Title="Rate and mechanism of the reaction of Mg and HCl" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8965" PostTypeId="2" ParentId="8960" CreationDate="2014-03-06T20:39:01.160" Score="2" Body="&lt;p&gt;According to collision theory, for a chemical reaction to occur, the reacting molecules must collide with each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bimolecular reaction (as given by you), collision occurs in the single step and thus becomes the rate determining step, as there are no other steps involved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider reactions like bromide ions reacting with bromate ions in the presence of acid, it involves 12 reacting particles. If all the 12 reacting particles collide simultaneously you can write rate law expression considering it to be elementary, but those events are extremely rare. In reality such reactions are found to be complex i.e reactions which take place through a sequence of two or more consecutive steps. Here, there are many steps involved, and one should consider slowest step to write rate law expression. But, why? That will become another SE question. In case of reactions taking place in single step, as there are no other steps left to compare with, whether it is slower or not, so you need to consider  the theoritical equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing to be noted is that elementary reactions are associated with single transition state and not single activated complex. For more info read this page: &lt;a href=&quot;http://en.wikipedia.org/wiki/Activated_complex&quot; rel=&quot;nofollow&quot;&gt;Activated complex&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-03-06T20:39:01.160" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8967" PostTypeId="1" AcceptedAnswerId="8969" CreationDate="2014-03-07T01:53:19.147" Score="8" ViewCount="71293" Body="&lt;p&gt;Why is the bond between H and F in HF considered polar covalent whereas HCl, HI, and so on are all ionic? The electronegativity difference between them suggests that it too should be ionic, yet all textbooks say that HF is covalent. Is there a good reason why?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2017-11-30T14:06:36.473" Title="Why is HF considered a covalently bonded compound?" Tags="&lt;bond&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8969" PostTypeId="2" ParentId="8967" CreationDate="2014-03-07T02:04:07.347" Score="19" Body="&lt;p&gt;I don't think I've ever seen a reputable source claiming the bonds in HCl, HBr and HI are mainly ionic. The fact that they are all gasses at room temperature clearly suggests otherwise. HF has the most ionic character of all the hydrogen halides, but it too has a rather low boiling point (just below room temperature), which is uncharacteristic of ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is best to think of all hydrogen halides as covalent polar molecules, with the polarity increasing in the order HI &amp;lt; HBr &amp;lt; HCl &amp;lt; HF, as suggested by the electronegativity differences.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-03-07T02:04:07.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8971" PostTypeId="2" ParentId="6203" CreationDate="2014-03-07T02:48:59.020" Score="2" Body="&lt;p&gt;Cook it up with ethylenediamine or diethylenetriamine- but don't bet on it.  What do you plan to have as the counterion for $\ce{K^+}$?  Check your carbon oxidation states for the second half of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, wait!  Add citric acid.  Yeah, that's the ticket.  Plop, plop, fizz, fizz.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-07T02:48:59.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8973" PostTypeId="2" ParentId="8960" CreationDate="2014-03-07T05:23:41.180" Score="2" Body="&lt;p&gt;First some definitions:-&lt;br&gt;&#xA;&lt;strong&gt;Elementary reaction&lt;/strong&gt;= The reaction that takes place in a single step i.e one transition/activated complex according to the transition theory or one single collision event according to the Collision theory of kinetics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Molecularity&lt;/strong&gt;= Number of independent species involve in the collision event in case of an elementary reaction. These all collide together with sufficient energy (threshold) and proper orientation to convert themselves into products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us assume the collision model for gaseous reaction where there are two species $A$ and $B$ which react together in a reaction according to :- &#xA;$$\ce{aA + bB -&amp;gt;}\text{ products}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction will occur whenever $a$ molecules of $A$ collide with $b$ molecules of $B$ in proper orientation. Suppose out of all collisions of $a$ molecules of $A$ with $b$ molecules of $B$, a fraction $P$ occur in proper orientation. The probability of collision of these specified number of reactants is proportional to the $a^{th}$ power of $[A]$ and $b^{th}$ power of $[B]$. This is because the concentration of the reactant gives crudely the probability of finding that reactant in a unit volume. For collision probability, you multiply the probability of finding one reactant to the probability of finding another reactant. Since there are $a$ molecules of $A$, therefore the probability of finding $A$ will be multiplied $a$ times and similarly for $B$. Thus the resultant rate based on collision probabilities will be :-&#xA;$$k=PZ[A]^a[B]^b$$&#xA;Where Z is an appropriate constant (since the probablities are only proportional to the concentrations and not equal to them, there is some appropriate constant for each proportionality which finally multiplies to give $Z$). &lt;strong&gt;This is true only for elementary reactions adhering to the Collision theory-Mechanism of reaction.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-03-07T05:23:41.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8974" PostTypeId="2" ParentId="6059" CreationDate="2014-03-07T05:52:31.927" Score="5" Body="&lt;p&gt;From the Aldrich data:&#xA; $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converted to a more convenient unit:&#xA; $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the molar mass (also from Aldrich) of:&#xA; $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We learn that one liter has:&#xA; $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't speak to why your other source has a different number.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="13184" LastEditDate="2015-05-15T12:30:04.367" LastActivityDate="2015-05-15T12:30:04.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8976" PostTypeId="1" AcceptedAnswerId="8980" CreationDate="2014-03-07T07:14:43.143" Score="3" ViewCount="4453" Body="&lt;p&gt;This is simple question, I think. At college (I'm an Intro Chem. student), the professor explained that oxygen is always found in molecules of 2 oxygen atoms combined together - $\ce{O2}$. On the other hand, molecule of the water is $\ce{H2O}$. So I don't get, how is it that O is single in this molecule, if it's always found coupled? &lt;/p&gt;&#xA;" OwnerUserId="4743" LastEditorUserId="3846" LastEditDate="2014-03-07T08:50:20.227" LastActivityDate="2014-03-07T09:03:58.373" Title="Oxygen in water: How is it single if it's always double?" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8977" PostTypeId="1" AcceptedAnswerId="8997" CreationDate="2014-03-07T08:06:49.710" Score="12" ViewCount="12889" Body="&lt;p&gt;Why is the dipole moment of $\ce{CH2Cl2}$ ($1.60 ~\mathrm{D}$) greater than that of $\ce{CHCl3}$ ($1.08~\mathrm{D}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my knowledge of vectors, I feel it should be the other way around.&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="402" LastEditDate="2017-04-01T21:22:19.717" LastActivityDate="2017-04-01T21:22:19.717" Title="Difference of dipole moments of dichloromethane and trichloromethane?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8978" PostTypeId="1" AcceptedAnswerId="9000" CreationDate="2014-03-07T08:17:46.420" Score="6" ViewCount="2172" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the following reaction, &lt;span class=&quot;math-container&quot;&gt;$K_c=4.0$&lt;/span&gt;:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{C2H5OH + CH3COOH &amp;lt;=&amp;gt; CH3COOC2H5 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A reaction is allowed to occur in a mixture of &lt;span class=&quot;math-container&quot;&gt;$\pu{17.2 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5OH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{23.8 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{48.6 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOC2H5}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\pu{71.2 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) In what direction will a net change occur?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) How many grams of each substance will be present at equilibrium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what I did for part (a):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{C2H5OH}][\ce{CH3COOH}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, get the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5OH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOC2H5}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; using the given amount of each substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align} &#xA;[\ce{C2H5OH}] &#xA;  &amp;amp;= \pu{17.2 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} &amp;amp;&#xA;  &amp;amp;= \pu{0.373 mol} \\&#xA;[\ce{CH3COOH}] &#xA;  &amp;amp;= \pu{23.8 g} \times \frac{\pu{1mol}}{\pu{60.1 g}} &amp;amp;&#xA;  &amp;amp;= \pu{0.396mol} \\&#xA;[\ce{CH3COOC2H5}] &#xA;  &amp;amp;= \pu{48.6 g} \times \frac{\pu{1 mol}}{\pu{88.1 g}} &amp;amp; &#xA;  &amp;amp;= \pu{0.552mol} \\&#xA;[\ce{H2O}] &#xA;  &amp;amp;= \pu{71.2 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} &amp;amp;&#xA;  &amp;amp;= \pu{3.96 mol} \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the overall volume is &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt;, then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_c = \frac{(\pu{0.552 M})(\pu{3.96 M})}{(\pu{0.373 M})(\pu{0.396 M})} &#xA;      = 14.8 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$4.0$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Q_c$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$14.84$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Q_c$&lt;/span&gt; is larger than &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; and the direction of the reaction is to the left, towards the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to solve for (a) but I have no idea how to get the amount of each substance at equilibrium for (b). The answer for (b) should be &lt;span class=&quot;math-container&quot;&gt;$\pu{68 g}$&lt;/span&gt;. I tried doing the ICE chart, but I did not get &lt;span class=&quot;math-container&quot;&gt;$\ce{68 g}$&lt;/span&gt;. Do I use the ICE chart for this part of the question or which method should I be using? &lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="4945" LastEditDate="2019-05-21T11:27:57.967" LastActivityDate="2019-05-21T11:38:02.243" Title="How many grams of each substance will be present at equilibrium in the reaction of ethanol and ethanoic acid?" Tags="&lt;equilibrium&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="8979" PostTypeId="2" ParentId="8976" CreationDate="2014-03-07T09:01:00.097" Score="6" Body="&lt;p&gt;I think you are confusing something here: &lt;strong&gt;Molecular&lt;/strong&gt; oxygen is most commonly found as the gas we know, $\ce{O2}$ and ozone, $\ce{O3}$. However, this doesn't limit it to only that. Oxygen can appear in a wide variety of (bio-) chemicals and there, it only needs two bonds to other atoms (or a double bond to one atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good examples for this are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The aforementioned water $\ce{H2O}$ where the oxygen is bonded to two hydrogen atoms. &lt;/li&gt;&#xA;&lt;li&gt;One of my most loved molecules: Ethanol. Here, the oxygen is bonded to a hydrogen and a carbon atom.&lt;/li&gt;&#xA;&lt;li&gt;In an ester, which usually are quite fragrant compounds, there actually are two oxygen atoms. But they are not connected to each other. Instead, one has a double bond and the other two bonds, each of them to carbon respectively.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope this has made it clear that oxygen doesn't have to bond to itself (but it can, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxide&quot;&gt;peroxides&lt;/a&gt;), and that oxygen doesn't have to come in pair of twos for the molecule to not fall apart.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T09:01:00.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8980" PostTypeId="2" ParentId="8976" CreationDate="2014-03-07T09:03:58.373" Score="4" Body="&lt;p&gt;Free single oxygen atoms are unstable and will bind to another free single oxygen atom that is present in the air because this is more stable.(So you get $\ce{O2}$) When you look at electron configurations then you see that both oxygen atoms share their electrons with each other and so have the same electron configuration as Neon, which is very stable: $\ce{1s^2 2s^2 2p^6}$&#xA;This is also the same for other non metals such as nitrogen gas : $\ce{N2}$ and also chlorine gas ($\ce{Cl2}$) and fluorine gas ($\ce{F2}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the oxygen atom in a water molecule is bonded to not another oxygen molecule, but to two hydrogen atoms.($\ce{H2O}$) The oxygen atom has a &lt;strong&gt;partial&lt;/strong&gt; negative charge of -2 and both hydrogen atoms have a positive &lt;strong&gt;partial&lt;/strong&gt; charge of +1. This has as a result that all atoms once again have stable electron configurations : oxygen once again has a stable electron configuration, equal to Neon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2}$ is the pure element, water is a molecule. But they both achieve stable electron configurations. Just with different atoms.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-03-07T09:03:58.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8981" PostTypeId="1" CreationDate="2014-03-07T09:34:42.013" Score="4" ViewCount="1828" Body="&lt;p&gt;I am a master student of nuclear physics. Due to some measurements I'd like to make, I have to use Magnesium as a beam target. Searching through the net to see it's properties I found that it can be easily oxidized. In addition I've read that it's highly flammable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have in hand two forms of Magnesium:&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/13112?lang=en&amp;amp;region=GR&quot; rel=&quot;nofollow noreferrer&quot;&gt;powder&lt;/a&gt; and &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/13103?lang=en&amp;amp;region=GR&quot; rel=&quot;nofollow noreferrer&quot;&gt;ribbon&lt;/a&gt;. In order to make the targets I have two alternatives.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use the powder to make highly pressured pellets. I thought of flushing a bottle with Argon or Nitrogen, making the pellet as fast as possible using &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8330/niobium-v-fluoride-becoming-sticky&quot;&gt;this procedure&lt;/a&gt;(&lt;strong&gt;which involves acetone&lt;/strong&gt;) in the atmosphere and then after the pellet is formed put it in the Argon or Nitrogen bottle.&lt;/li&gt;&#xA;&lt;li&gt;Use the ribbon as is. The only problem is that it not wide enough, so I will have to make a spiral of this ribbon around something(I don't yet know what the &quot;something&quot; will be, so any idea is welcome!). The ribbon I have has a width of 3mm but I want a width of 6mm. So I thought of spinning the ribbon around an item, to form a spiral.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The goal is to make the target as &lt;strong&gt;safely&lt;/strong&gt; as possible and &lt;strong&gt;avoid Magnesium oxidizing&lt;/strong&gt;, because I don't want to have Oxygen in my target. So my question has three parts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to safely treat magnesium powder/ribbon so as to avoid any possible accident and preventing it from oxidizing?&lt;/li&gt;&#xA;&lt;li&gt;Which of the above would be the most effective way to make the target?&lt;/li&gt;&#xA;&lt;li&gt;How to store the target until the experiment will be held?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that after the target is made, it will go to an evaporator, in order to evaporate some gold. Any advice, help or direction is more than welcome!&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-03-07T16:58:48.977" Title="How to treat Magnesium powder and ribbon" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8982" PostTypeId="2" ParentId="857" CreationDate="2014-03-07T09:56:43.913" Score="4" Body="&lt;p&gt;In addition to the safety precautions mentioned in the question, the following can increase the safety for the experimentator:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Keep the fume hood as closed as possible with the head outside and the arms inside. Usually this keeps fumes from entering the body, since the fume hood airflow goes the other direction.&lt;/li&gt;&#xA;&lt;li&gt;Work quickly. While this not only minimizes side reactions of the reactant with the air and other things, it also ensures less exposure time, minimizing vaporization.&lt;/li&gt;&#xA;&lt;li&gt;As Aesin mentioned in the comment, if you have concerns about the infrastructure, you could always perform the experiment in a glovebox.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the end, you have to feel confident that the safety measures are effective. If you don't, you will have fear of the reaction, and inevitably be less concentrated and more prone to errors. In combination with &quot;work quickly&quot; this may well become a disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maintain a healthy portion of respect, keep quenching agents close by and execute the reagent transfer with a steady hand.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T09:56:43.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8983" PostTypeId="2" ParentId="8981" CreationDate="2014-03-07T10:08:14.737" Score="3" Body="&lt;p&gt;Magnesium powder and ribbon are relatively stable in air at normal conditions, probably because of thin protective layer of oxide. However, ribbon can burn in air with blinding light and powder gives self-burning (i.e. not needing external oxygen sources like air) mixtures with many common substances, some of them explosive. Moreover, burning magnesium may continue to burn in $\ce{CO_2}$ atmosphere, under quartz sand and especially under water (producing hydrogen, that can form explosive mixture). The only common material that can safely extinguish burning magnesium is iron powder. Surprising enough, being stable in cold water for short time, magnesium readily react with methanol (and probably ethanol), chlorinated hydrocarbons and hot water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing is that magnesium ribbon is not easy to lit up, and powder is almost impossible to lit up when on surface of some massive object (common gas lighter flame lit magnesium ribbon after 15-60 seconds of heating)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium must not be heated on air or under nitrogen. Period. Technical grade argon probably contain significant amount of nitrogen, so it should not be used either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium theoretically can be melted under eutectic mixture of $\ce{NaCl}$-$\ce{KCl}$-$\ce{CaCl_2}$, but by doing so without experienced supervision you ask troubles. Still, theoretically speaking, if you find an iron (not silicate, porcelain or corundum) crucible you probably can melt magnesium in it under mentioned mixture of &lt;em&gt;dry&lt;/em&gt; salts, let it crystallize and then carve it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, magnesium alloy sheets are available on the market, so maybe you can find something and cut needed form from commercially available sheets. &lt;strong&gt;Try this first before doing anything requiring heating magnesium&lt;/strong&gt;. I have doubts, that you will be able to press good tablets from magnesium powder, so if commercially available sheets and maybe some magnesium compounds (oxide, sulfide and others) won't not work for you, you probably will end with melting magnesium. I hope you won't need it, it is dangerous. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user4076" LastEditDate="2014-03-07T10:14:01.247" LastActivityDate="2014-03-07T10:14:01.247" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8984" PostTypeId="1" AcceptedAnswerId="8986" CreationDate="2014-03-07T10:28:35.070" Score="8" ViewCount="5103" Body="&lt;p&gt;We know the alpha particle is the nucleus of &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium&quot; rel=&quot;nofollow&quot;&gt;helium&lt;/a&gt;. It contains four subatomic particles - two protons and two neutrons. The protons and neutrons are further made of particles called up and down quarks. So why call an alpha &lt;em&gt;particle&lt;/em&gt; a particle. And similarly, why do we say electrons, protons, neutrons etc. are &lt;em&gt;particles&lt;/em&gt;, even though, they are made of particles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand the case of electrons, protons, and neutrons as when they were discovered and named, quarks were not discovered. But the helium nucleus is a different case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reckon talking about an English word's meaning changes in the frame of reference. We do not call a quark an object, although, an object could be any real or virtual thing. So in chemistry, I don't think calling an alpha &lt;em&gt;particle&lt;/em&gt; a particle is a good practice. I do understand there are dust particles and all but we mention them only while talking about the macroscopic world. Not so in the case of chemistry. I hope you get my point.&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="4734" LastEditDate="2014-03-08T10:59:55.583" LastActivityDate="2014-03-08T10:59:55.583" Title="Why is the alpha particle called a particle when it is made of four particles?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8985" PostTypeId="1" AcceptedAnswerId="8993" CreationDate="2014-03-07T10:46:17.040" Score="17" ViewCount="51504" Body="&lt;p&gt;I was going through alkyl and aryl halides and noted the following statement in my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;p&lt;/em&gt;-dichlorobenzene has higher melting point and solubility than those of &lt;em&gt;o&lt;/em&gt;-dichlorobenzene and &lt;em&gt;m&lt;/em&gt;-dichlorobenzene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought over it but could not come to a conclusion. Why is the above assertion true?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="5026" LastEditDate="2018-02-19T10:47:28.487" LastActivityDate="2020-08-03T19:14:36.443" Title="Why is the melting point of p-dichlorobenzene higher than those of o-dichlorobenzene and m-dichlorobenzene?" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;melting-point&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8986" PostTypeId="2" ParentId="8984" CreationDate="2014-03-07T11:19:26.467" Score="8" Body="&lt;p&gt;Actually, the alpha particle contains two protons and two neutrons. It is what is emitted from the nucleus under alpha decay, giving off what was first classified as alpha radiation by Rutherford at the end of the 19&lt;sup&gt;th&lt;/sup&gt; century (very late 1800's).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1932 the neutron was discovered. Only then did it become clear that the nuclei of atoms contain more than just some positively charged particles (which we know as protons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is this historic reason for the naming of the alpha particle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&quot;Particle&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The word particle does not only pertain to elementary particles (and in [quantum] chemistry, we stop at the level of nuclei as particles, everything that goes deeper is nuclear physics). There are smoke particles, dust particles, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the wave-particle duality. In principle, since we can observe this &quot;alpha radiation&quot; from radioactive processes, there must be a particle that belongs to the wave. With electromagnetic radiation, this is the photon, and in alpha radiation it is the alpha particle.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T11:19:26.467" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8987" PostTypeId="2" ParentId="8984" CreationDate="2014-03-07T11:24:25.573" Score="6" Body="&lt;p&gt;Note that calling something a particle does not imply calling it a &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_particle&quot;&gt;&lt;em&gt;fundamental particle&lt;/em&gt;&lt;/a&gt;. Anything can be a particle (a particle of dust, a particle of sand, a particle of cheese, etc). However, only non-subdivisible particles (particles with no further inner structure) are called fundamental (electrons, quarks, neutrinos, photons, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used to not know what radiation actually was, so we classified them broadly into three groups according to the way their trajectories bent in an electric or magnetic field, and the three main types found were given an effectively placeholder name (alpha, beta, gamma). The names stuck, even though we now know their real identities. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-03-07T11:24:25.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8989" PostTypeId="2" ParentId="8977" CreationDate="2014-03-07T11:39:33.760" Score="8" Body="&lt;p&gt;Since you know about vectors, then the piece you are missing is probably the proper three-dimensional structure of the molecules. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trichloromethane (chloroform)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ID84V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ID84V.png&quot; alt=&quot;Structure of chloroform&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the third chlorine atom in trichloromethane is pointing away from the other two. More precisely, the three chlorine atoms are at the bottom three points of a tetrahedron. Thus the $x$ and $y$ components of the individual $\ce{C-Cl}$ bond dipoles cancel, leaving only the $z$ component. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dichloromethane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kLvc6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kLvc6.png&quot; alt=&quot;Structure of DCM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For dichloromethane, the $y$ components fully cancel, but the $x$ components do not.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2017-04-01T21:15:35.017" LastActivityDate="2017-04-01T21:15:35.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8990" PostTypeId="2" ParentId="8431" CreationDate="2014-03-07T11:51:36.580" Score="1" Body="&lt;p&gt;I found the answer to my question, and I am going to share it just in case.&#xA;Apparently, when extracting the pollutants from the foam, I am also extracting some foam-specific materials that influence the pollutants' behaviour during the SPME part of the extraction.&#xA;I tested that by extracting the blank foam, then adding a standard amount of pollutants to the extraction solvent. This gave the same result as extracting a spiked foam, meaning higher peak areas than a standard solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short : matrix effect from the foam. I don't yet know why or what, but it's a matrix effect.&lt;/p&gt;&#xA;" OwnerUserId="4496" LastActivityDate="2014-03-07T11:51:36.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8991" PostTypeId="2" ParentId="8985" CreationDate="2014-03-07T12:13:26.443" Score="0" Body="&lt;p&gt;The p- isomer is more linear than the o- and m- isomer, hence it has a larger surface area for inter-molecular attraction. &lt;/p&gt;&#xA;" OwnerUserId="4516" LastActivityDate="2014-03-07T12:13:26.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8992" PostTypeId="2" ParentId="8985" CreationDate="2014-03-07T12:46:21.767" Score="2" Body="&lt;p&gt;I think this reasoning will suffice (not sure though):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point than o-and m-isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2014-03-07T12:46:21.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8993" PostTypeId="2" ParentId="8985" CreationDate="2014-03-07T12:54:53.017" Score="21" Body="&lt;p&gt;Generally, the melting point of &lt;em&gt;para&lt;/em&gt;-isomer is quite higher than that of &lt;em&gt;ortho&lt;/em&gt;- or &lt;em&gt;meta&lt;/em&gt;-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and its melts at higher temperature. This question is most likely to be answered easily.&lt;/p&gt;&#xA;&lt;p&gt;If you consider solubility. These are insoluble in water but soluble in organic solvents. Generally, &lt;em&gt;para&lt;/em&gt;-isomers are highly soluble in organic solvents than &lt;em&gt;ortho&lt;/em&gt;-isomers. This is interesting question.&lt;/p&gt;&#xA;&lt;p&gt;Every compound (more specifically solids) does not dissolve in a given compound. In general, a liquid dissolves in a liquid if the intermolecular interactions are similar. This is in accordance with the rule &amp;quot;&lt;em&gt;like dissolves like&lt;/em&gt;,&amp;quot; ionic or polar compounds dissolve more readily in polar solvents. Non-polar compounds (covalent or organic) are soluble in non-polar solvents.&lt;/p&gt;&#xA;&lt;p&gt;If you consider &lt;em&gt;p&lt;/em&gt;-dichlorobenzene, it has zero dipole moment and thus is more non-polar than &lt;em&gt;o&lt;/em&gt;-dichlorobenzene which has dipole moment of about &lt;span class=&quot;math-container&quot;&gt;$\pu{2.54 D}$&lt;/span&gt;. So, non-polar &lt;em&gt;p&lt;/em&gt;-dichlorobenzene dissolves more readily than &lt;em&gt;o&lt;/em&gt;-dichlorobenzene in non-polar (organic) solvents.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="61578" LastEditDate="2020-08-03T19:14:36.443" LastActivityDate="2020-08-03T19:14:36.443" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8994" PostTypeId="1" CreationDate="2014-03-07T13:27:21.503" Score="15" ViewCount="19321" Body="&lt;p&gt;I have seen this reaction many times:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\text {Phenol}+\text{Zn}\ce {-&amp;gt;}\text{Benzene}+\text{ZnO} $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what is the mechanism?! Tried a lot but couldn't cook up one...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help...&lt;/p&gt;&#xA;" OwnerUserId="4160" LastActivityDate="2019-12-20T16:01:08.517" Title="Reaction of phenol with Zinc dust" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="7" ClosedDate="2019-12-03T19:17:15.120" ContentLicense="CC BY-SA 3.0" />
  <row Id="8996" PostTypeId="1" AcceptedAnswerId="9011" CreationDate="2014-03-07T13:42:40.233" Score="2" ViewCount="3597" Body="&lt;p&gt;I know that the $\ce{S8}$ molecule contains 16 pairs of lone electrons. My teacher always says that we must try to reduce the no of lone pairs, so, had there been double bonds between 8 atoms of sulfur then the no of lone pairs would be reduced. Then why aren't there any double bonds?&lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="402" LastEditDate="2017-05-07T02:35:18.843" LastActivityDate="2017-05-07T02:35:18.843" Title="Why aren't there 8 double bonds in octasulfur (S8)?" Tags="&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8997" PostTypeId="2" ParentId="8977" CreationDate="2014-03-07T13:55:24.437" Score="11" Body="&lt;p&gt;Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$&#xA;where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$&#xA;which will be $1A$, lesser than in the case of dichloromethane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualise how can you get those angles and cosines, see &lt;a href=&quot;http://www.ctralie.com/Teaching/Tetrahedron/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="1431" LastEditDate="2017-04-01T20:56:46.680" LastActivityDate="2017-04-01T20:56:46.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8998" PostTypeId="2" ParentId="8645" CreationDate="2014-03-07T13:58:42.233" Score="1" Body="&lt;p&gt;In chemistry, often quantities get denoted by an index $i$, which then stands for some compound ($i=\text{A, B, C, ...}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The stoichiometric coefficient $\nu_i$&lt;/li&gt;&#xA;&lt;li&gt;The concentration $c_i$&lt;/li&gt;&#xA;&lt;li&gt;The amount of substance $n_i$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The notion of grouping together the stoichiometric coefficient with the substance it uses is a novel one for me, and will most likely not have a name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you use a name such as &quot;coefpair&quot; or the like and document your code accordingly.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T13:58:42.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8999" PostTypeId="2" ParentId="8964" CreationDate="2014-03-07T14:10:31.320" Score="5" Body="&lt;p&gt;Your rate law should be&#xA;$$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$&#xA;which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T14:10:31.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9000" PostTypeId="2" ParentId="8978" CreationDate="2014-03-07T14:22:22.543" Score="3" Body="&lt;p&gt;Knowing that &lt;span class=&quot;math-container&quot;&gt;$K_c=4.0$&lt;/span&gt;, equilibrium is achieved when &lt;span class=&quot;math-container&quot;&gt;$Q_c = K_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's set up an ICE table and write down our equilibrium expression like you did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C2H5OH + CH3COOH &amp;lt;=&amp;gt; CH3COOC2H5 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline&#xA;\text{Initial conc.} &amp;amp; \pu{0.373 mol} &amp;amp; \pu{0.396 mol} &amp;amp; \pu{0.552 mol} &amp;amp; \pu{3.96 mol}\\&#xA;\hline&#xA;\text{Change conc.} &amp;amp; -x &amp;amp; -x &amp;amp; +x &amp;amp; +x\\&#xA;\hline&#xA;\text{End conc.} &amp;amp; \pu{0.373 mol} - x &amp;amp; \pu{0.396 mol} - x &amp;amp; \pu{0.552 mol} + x &amp;amp; \pu{3.96 mol} + x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that our reaction is at equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_c = \ce{\frac{[CH3COOC2H5][H2O]}{[C2H5OH][CH3COOH]}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We plug in our values for our change in equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_c = 4.0 = {\frac{[0.552 + x][3.96 + x]}{[0.373 - x][0.396 - x]}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, which is the change in concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;4 &amp;amp;= \frac{x^2+4.512 x+2.18592}{x^2-0.769 x+0.147708}\\&#xA;0 &amp;amp;= 3x^2-7.588 x-1.59509 &#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the quadratic equation, or thankfully polysolv on modern calculators, we get two values for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;x_1 &amp;amp;= 2.72449 \\&#xA;x_2 &amp;amp;= -0.195154&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative zero is chosen, since the reaction will evolve to the left (being &lt;span class=&quot;math-container&quot;&gt;$Q_c&amp;gt;K_c$&lt;/span&gt;), thus increasing the concentration of the reactants. This will only be accounted for, if one takes the negative zero. If one chooses the other zero, one will go to the wrong opposite direction (towards a diminution of reactants concentration). In the particular case of the present exercise, one would even result in a negative concentration. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{[CH3COOH]} &#xA;  &amp;amp;= \pu{0.373 mol} - (\pu{-0.195154 mol}) &amp;amp;&#xA;  &amp;amp;= 0.568154\\&#xA;\ce{[C2H5OH]} &#xA;  &amp;amp;= \pu{0.396 mol} - (\pu{-0.195154 mol}) &amp;amp;&#xA;  &amp;amp;= 0.591154\\&#xA;\ce{[CH3COOC2H5]} &#xA;  &amp;amp;= \pu{0.552 mol} + (\pu{-0.195154 mol}) &amp;amp;&#xA;  &amp;amp;= 0.356846\\&#xA;\ce{[H2O]} &#xA;  &amp;amp;= \pu{3.96 mol} + (\pu{-0.195154 mol}) &amp;amp;&#xA;  &amp;amp;= 3.764846&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\pu{0.568154 mol} \ \ce{CH3COOH} \times \frac{\pu{60.1 g}}{\pu{1 mol} \ \ce{CH3OOH}} &amp;amp;= \pu{34.1 g}\\&#xA;\pu{0.591154 mol} \ \ce{C2H5OH} \times \frac{\pu{46.1 g}}{\pu{1 mol} \ \ce{C2H5OH}} &amp;amp;= \pu{27.3 g}\\&#xA;\pu{0.356846 mol} \ \ce{CH3COOC2H5} \times \frac{\pu{88.1 g}}{\pu{1 mol} \ \ce{CH3COOC2H5}} &amp;amp;= \pu{31.4 g}\\&#xA;\pu{3.764846 mol} \ \ce{H2O} \times \frac{\pu{18.0 g}}{\pu{1 mol} \ \ce{H2O}} &amp;amp;= \pu{67.8 g}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial product sum is &lt;span class=&quot;math-container&quot;&gt;$\pu{160.8 g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium product sum is &lt;span class=&quot;math-container&quot;&gt;$\pu{160.6 g}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\pu{0.2 g}$&lt;/span&gt; is likely due to truncation error in calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is really cruel... whoever assigned it... :/&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2019-05-21T11:38:02.243" LastActivityDate="2019-05-21T11:38:02.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9001" PostTypeId="1" AcceptedAnswerId="15720" CreationDate="2014-03-07T14:24:46.347" Score="25" ViewCount="47316" Body="&lt;p&gt;What is the difference in the functioning of alkaline and acidified $\ce{KMnO4}$ as a reagent in organic chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., in the oxidation of a primary alcohol to carboxylic acid, we use acidified $\ce{KMnO4}$ and not alkaline $\ce{KMnO4}$. Why?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="7475" LastEditDate="2015-10-08T13:56:26.693" LastActivityDate="2021-01-10T05:32:29.940" Title="Difference in the functioning of alkaline and acidified KMnO4" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9002" PostTypeId="2" ParentId="7520" CreationDate="2014-03-07T14:58:08.813" Score="4" Body="&lt;p&gt;They are stable because the energy is lower. (Ah, the universal answer to every chemistry problem!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me elaborate: If noble gases would occur as diatomic elements, the energy should be lower in comparison to their monoatomic form. But when you combine the atomic orbitals of the bonding partners to the molecular orbitals (MO-LCAO), you then fill in the electrons into &lt;em&gt;all&lt;/em&gt; MOs, the bonding as well as the antibonding MOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4517/how-can-antibonding-orbitals-be-more-antibonding-than-bonding-orbitals-are-bondi&quot;&gt;antibonding orbitals are more antibonding than bonding orbitals bonding&lt;/a&gt;, the overall energy of the compound is raised. So this is a state which the system prefers to avoid, giving you monoatomic elements.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-03-07T14:58:08.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9003" PostTypeId="1" AcceptedAnswerId="9005" CreationDate="2014-03-07T15:52:15.937" Score="9" ViewCount="2571" Body="&lt;p&gt;If I have a pen with refillable ink, can I just pour in concentrated Hydrochloric acid or concentrated sulfuric acid in the ink chamber to make it an acidic pen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be better to use the acid of &lt;em&gt;suitable&lt;/em&gt; molarity instead of being concentrated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if I write with it on a paper, will it burn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good idea to play with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it dangerous in any way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it illegal to make such things for experimenting and fun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other safe method to achieve the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the odds that the poured acid will burn the ink chamber itself and start leaking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be used in any constructive way? Like cutting many sheets of paper at once. Or using it in industries suitably to make smooth cuts in materials that can be burned with the used acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it be fun to you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to everyone.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add a few details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the acid is not concentrated but of suitable molarity to just cut the paper. I mean, to pierce through it and burning its edges to just make a nice cut. Like a glowing incense stick makes when made to touch a paper.&#xA;Also, the nip of the pen and the ink chamber are strongly inert; enough to survive the concentrated acid's attack and prevent leaking. Like, the chamber can be made of glass (in which acids are usually kept) and the nip can be made of galvanized/electroplated iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if there is still more, we can just make this into a thought experiment.&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="402" LastEditDate="2017-01-14T08:48:47.737" LastActivityDate="2017-12-07T12:37:19.420" Title="How can I make an acidic pen to burn paper on writing on it?" Tags="&lt;everyday-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9004" PostTypeId="2" ParentId="9003" CreationDate="2014-03-07T16:31:19.237" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulase&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cellulase&lt;/a&gt; &lt;BR&gt;&#xA;Load the pen with a cellulase solution. &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/N-Methylmorpholine_N-oxide&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/N-Methylmorpholine_N-oxide&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://pubs.acs.org/isbn/9780841200067&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/isbn/9780841200067&lt;/a&gt; &lt;BR&gt;&#xA;Load the pen with a cellulose solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper is bulk cut with blades or lasers.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-07T16:31:19.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9005" PostTypeId="2" ParentId="9003" CreationDate="2014-03-07T16:34:38.120" Score="7" Body="&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; pour anything into your pen. Is it a good idea? Most likely &lt;strong&gt;not&lt;/strong&gt;. Is it dangerous? &lt;em&gt;You bet!&lt;/em&gt; Is it a good idea to play with? &lt;strong&gt;NO!&lt;/strong&gt; Whether it is legal or not to create such an object, you'll have to check with legislation in your country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want paper to ignite on touch with this pen, you would need some truly strong acid (and probably undiluted). I'm certain that if this substance has the capability to burn paper, it would also corrode away the feed and nib of the pen. In the worst case, it will even eat through the ink container and leaking everywhere. In any way, this is no viable option to cut paper in a controlled fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a safer method? Well, there would be laser-assisted cutting of paper, but the good ol' knife or saw would still be cheaper.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T16:34:38.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9006" PostTypeId="2" ParentId="7520" CreationDate="2014-03-07T16:42:19.783" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/noblegasfacts/a/Noble-Gas-Compounds.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemistry.about.com/od/noblegasfacts/a/Noble-Gas-Compounds.htm&lt;/a&gt; &lt;BR&gt;&#xA;Noble gases are reactive. Following are some examples, &lt;BR&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5pzQ1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-07T16:42:19.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9007" PostTypeId="1" CreationDate="2014-03-07T16:58:23.003" Score="8" ViewCount="1860" Body="&lt;p&gt;Currently I investigate a reaction where a starting material A reacts with an excess of thiophenol B towards products C and D.  As shown by recurrent TLC analysis of the reaction, the formation of D is slower than the one of C, while A and B are still detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the reactions are consecutive, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A + B -&amp;gt;[k_1] C -&amp;gt;[k_2] D}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or they are simultaneous, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{D &amp;lt;-[k_2] A + B -&amp;gt;[k_1] C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A, B, C, and D are distinguishable by 1H NMR spectroscopy; NMR data of the purified compounds A, B, C, and D were  recorded successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on a recurrent 1H NMR spectroscopy, I would like to use the integrations of the signals to determine the rate constants; perhaps there is evidence for the first case (similar to Bodenstein principle).  The advancement of the thioetherifications alter the multiplicities of the signals observed, and simultaneously indicator signals vanish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:  Prior to get work on my data, where may I find a tutorial of reactions with known starting material(s)/product(s), determined rate constant(s) and NMR data-used-to-determine-these-rate-constants?&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2014-11-17T16:44:08.117" Title="rate determination by 1H NMR spectroscopy" Tags="&lt;organic-chemistry&gt;&lt;kinetics&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9008" PostTypeId="2" ParentId="8981" CreationDate="2014-03-07T16:58:48.977" Score="0" Body="&lt;p&gt;You can buy magnesium fabrications any way you like.  It does surface air oxidize, but only divided metal is hazardous - powder, shavings, chips - unless you grind or saw on it.  Magnesium burns under air, nitrogen, carbon dioxide, and water; and sand.  Near anything containing oxygen or halogen will feed it.  An argon flood should kill it.  Also look at Metal-X extinguisher ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If surface oxidation is a problem, ion mill it clean in vacuum, then deposit a micron of Parylene. Parylene C may be reactive toward naked magnesium.  Parylene as such should work... but ask the coating vendors or test it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Parylene&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Parylene&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.paryleneengineering.com/why_use_parylene.html&quot; rel=&quot;nofollow&quot;&gt;http://www.paryleneengineering.com/why_use_parylene.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://scscoatings.com/what_is_parylene/parylene_properties.aspx&quot; rel=&quot;nofollow&quot;&gt;http://scscoatings.com/what_is_parylene/parylene_properties.aspx&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;https://www.google.com/patents/US5156919&quot; rel=&quot;nofollow&quot;&gt;https://www.google.com/patents/US5156919&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-07T16:58:48.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9009" PostTypeId="2" ParentId="8984" CreationDate="2014-03-07T17:03:27.620" Score="3" Body="&lt;p&gt;The alpha particle is more &quot;particle-like&quot; than many other nuclei because it is so tightly bound together, thereby making it much more difficult to break into its constituent parts. The stability of an alpha particle is a result of the number of neutrons and protons in the nucleus. Nuclei that have complete nuclear shells have a larger binding energy per nucleon, giving rise to so-called &lt;em&gt;magic numbers&lt;/em&gt; of nucleons (2, 8, 20, 28, 50, 82, 126) that result in particularly tightly bound nuclei. In an alpha particle, the two protons form one complete nuclear shell and the two neutrons form a second complete nuclear shell. This makes the alpha particle &lt;em&gt;doubly magic&lt;/em&gt;. The stability and simplicity of the alpha particle makes it energetically favorable to be produced from the fusion of lighter particles and to be emitted in the radioactive decay of heavy nuclei.&lt;/p&gt;&#xA;" OwnerUserId="4752" LastActivityDate="2014-03-07T17:03:27.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9010" PostTypeId="2" ParentId="7936" CreationDate="2014-03-07T17:03:44.140" Score="1" Body="&lt;p&gt;In alkene metathesis you need large ligands that direct the stereoselectivity of the reaction, and a hydride ligand is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for hydrodefluorination I don't see an advantage of a &lt;em&gt;trans&lt;/em&gt; hydride ligand on the catalyst. &lt;em&gt;cis&lt;/em&gt; hydride ligands could be beneficial, as they then take the place of the fluorine atom on the substrate.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-07T17:03:44.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9011" PostTypeId="2" ParentId="8996" CreationDate="2014-03-07T17:32:47.047" Score="3" Body="&lt;p&gt;An $\ce{S}$ atom in a molecule must have 2 bonds to be neutral and in the $\ce{S8}$ ring it needs both of these bonds to continue the chain/ring. If you introduce some double bonded S atoms into the ring each of these will have 3 bonds and will thus be positively charged. So, while reducing the number of free electron pairs you increase the charge of the compound which would be wrong because $\ce{S8}$ isn't charged. You can only use free electron pairs to form double bonds if you have some electron deficient atoms in the neighborhood which can &quot;receive&quot; them.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-03-07T17:32:47.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9012" PostTypeId="1" AcceptedAnswerId="9014" CreationDate="2014-03-07T18:40:45.647" Score="2" ViewCount="220" Body="&lt;p&gt;Tert-butyl alcohol seems unusual among alcohols in that its melting point is high (25°C) while its boiling point is also still low (82°C).  I am looking for more materials with phase-unstable liquid regions like this so I'm curious what makes the liquid phase so unstable relative to the solid and gas phases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the methyl groups align and act like alkanes to stabilize the solid but the molecule is still small enough to have a low boiling point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my larger point, any advice on characteristics of materials that exhibit low liquid phase stability?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2014-03-07T20:04:21.330" Title="Phase stability of alcohols" Tags="&lt;organic-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9013" PostTypeId="2" ParentId="1274" CreationDate="2014-03-07T19:18:44.083" Score="1" Body="&lt;p&gt;Two-photon excitation using femtosecond lasers that emit in the infrared range is a technique used in living cell imaging via fluorescence microscopy. Advantages of the technique, such as the minimized light scattering (~ 1/$\lambda^4$) using long wavelength for excitation and the fluorescent dyes used are discussed &lt;a href=&quot;http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0051980&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large volumetric displays, a &lt;a href=&quot;http://phys.org/news11251.html&quot; rel=&quot;nofollow&quot;&gt;different technique&lt;/a&gt; was/is under development in a joint project of the Japanese National Institute of Advanced Industrial Science and Technology (AIST), the Keio University and an industrial partner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, fluorescence does not play a role and special gases are not needed either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images can be displayed in air and are formed - pixel by pixel - by plasma dots generated in the focal point of pulsed infrared lasers with a pulse width of 1 ns and a repitition rate of 100 Hz.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-07T19:18:44.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9014" PostTypeId="2" ParentId="9012" CreationDate="2014-03-07T20:04:21.330" Score="1" Body="&lt;p&gt;Highly symmetric molecules tightly fit into crystal lattices, give high melting points for their composition.  Their small surface areas minimize London dispersion forces and van der Walls forces, giving high volatility.  They tend to sublime.  Compare mp and bp data for cyclohexane and benzene versus their perfluorinated derivatives. Less exotic: n-octane, iso-octane, 2,2,4-trimethylpentane, 2,2,3,3-tetramethylbutane, and cubane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neopentane&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Neopentane&lt;/a&gt; &lt;BR&gt;&#xA;n-Pentane and isopentane versus neopentane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromium, molybdenum, and tungsetn metal versus their respective hexacarbonyls.  $\ce{-SiMe3}$ and $\ce{-CF3}$ plus symmetry confer remarkable volatility.  $\ce{I(CF3)7}$ melts and boils around 0 C.  That is molecular weight 609.95 - and it has no static molecular structure (Bartell mechanism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.patentgenius.com/patent/3992424.html&quot; rel=&quot;nofollow&quot;&gt;http://www.patentgenius.com/patent/3992424.html&lt;/a&gt; &lt;BR&gt;&#xA;US Patent 3992424&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-07T20:04:21.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9015" PostTypeId="2" ParentId="9003" CreationDate="2014-03-07T21:45:02.050" Score="6" Body="&lt;p&gt;In principle, a solution of white phosphorus in carbon disulfide ($\ce{CS2}$) can be used as an &quot;ink&quot; to set paper on fire. This &lt;a href=&quot;https://www.youtube.com/watch?v=jw8JUlbgOhA&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt; demonstrates the effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a good idea? &lt;strong&gt;NO!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the disgusting smell of carbonyl sulfide ($\ce{COS}$) often found as an impurity is just a nuisance, the solvent itself is &lt;strong&gt;neutotoxic&lt;/strong&gt;. But it gets worse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;White phosphorus may cause awful deep third degree burns on your body!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unless you want to qualify for the &lt;a href=&quot;http://www.darwinawards.com/&quot; rel=&quot;noreferrer&quot;&gt;Darwin Award&lt;/a&gt;, don't mess around with this!  &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-07T21:45:02.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9016" PostTypeId="1" AcceptedAnswerId="9017" CreationDate="2014-03-07T22:04:11.133" Score="9" ViewCount="15670" Body="&lt;p&gt;In my class my teacher showed us how to find the average atomic mass of an element with a method, but he didn't really state a formula one could use. I came  up with a formula of my own, and from what I observed it works:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{i_1x + i_2y}{100} = A,&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the atomic mass, &#xA;&lt;span class=&quot;math-container&quot;&gt;$i_1$&lt;/span&gt; is the first isotope's atomic weight, &#xA;&lt;span class=&quot;math-container&quot;&gt;$i_2$&lt;/span&gt; is the second isotope's atomic weight, &#xA;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; are the percentages of the isotopes, respectively,&#xA;and they add up to 100, i.e. &lt;span class=&quot;math-container&quot;&gt;$x+y = 100$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore: &lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{i_1x + i_2(100-x)}{100} = A&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just wondering, is there an official formula, or is this method it? &lt;/p&gt;&#xA;" OwnerUserId="4685" LastEditorUserId="4945" LastEditDate="2018-10-12T13:42:20.087" LastActivityDate="2018-10-12T14:17:02.930" Title="What is the formula for finding the average atomic mass of an element based on the abundance of its isotopes?" Tags="&lt;periodic-table&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9017" PostTypeId="2" ParentId="9016" CreationDate="2014-03-07T22:12:24.593" Score="8" Body="&lt;p&gt;The average relative atomic mass of an element comprised of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; isotopes with relative atomic masses &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt; and relative fractional abundances &lt;span class=&quot;math-container&quot;&gt;$p_i$&lt;/span&gt; is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;A = p_1 A_1 + p_2 A_2 + \dots + p_n A_n = \sum\limits_{i=1}^n p_i A_i&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example carbon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lrrr}&#xA;\text{Isotope} &amp;amp; &#xA;\text{Isotopic Mass $A$} &amp;amp; &#xA;\text{Abundance $p$} &amp;amp; &#xA;A\times p\\\hline&#xA;\ce{^{12}C}:   &amp;amp; \pu{12.000000 u} &amp;amp; 0.98892 &amp;amp;= \pu{11.867 u}\\&#xA;\ce{^{13}C}:   &amp;amp; \pu{13.003354 u} &amp;amp; 0.01108 &amp;amp;= \pu{00.144 u}&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have &lt;span class=&quot;math-container&quot;&gt;$p_1A_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_2A_2$&lt;/span&gt;, we add those together to find &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, therefore &#xA; the chemical relative atomic mass of carbon is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A = \pu{00.144 u} + \pu{11.867 u} = \pu{12.0011 u}.$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2018-10-12T14:17:02.930" LastActivityDate="2018-10-12T14:17:02.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9018" PostTypeId="1" CreationDate="2014-03-07T22:27:30.427" Score="3" ViewCount="98" Body="&lt;p&gt;We know that for homogeneous mixtures the equilibrium constant is determined using both concentration of reactants and of products. For heterogeneous equilibrium involving a gas and a liquid or a solid, we only use the concentration of the gas because the concentration of liquids and gases are constants, and can thus be omitted from the equation. But what of the heterogeneous equilibrium between a solid and a liquid? Both are incompressible, so what do you do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2014-03-08T00:34:20.590" Title="Heterogeneous Equilibrium Without Gas" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9019" PostTypeId="2" ParentId="9016" CreationDate="2014-03-07T23:25:40.963" Score="5" Body="&lt;p&gt;Once you know how it works, you never want to do that by hand again. For example for bromine you can use this python code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;isotopes = [79, 81]&#xA;abundances = [50.69, 49.31]&#xA;&#xA;weighted = map(lambda x, y: x * y, isotopes, abundances)&#xA;# multiply isotope mass and abundance&#xA;&#xA;weighted = sum(weighted)/100&#xA;# sum them up and divide by 100&#xA;&#xA;print weighted&#xA;&amp;gt;&amp;gt;&amp;gt;&amp;gt; 79.9862&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2018-10-12T13:54:53.233" LastActivityDate="2018-10-12T13:54:53.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9021" PostTypeId="2" ParentId="9018" CreationDate="2014-03-07T23:54:17.183" Score="1" Body="&lt;p&gt;You're right, they don't really have concentrations or pressure to work with, which would lead to 0/0 craziness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can go back a step and look at how the equilibrium constant is defined. When you derive the equilibrium constant you don't use pressure or concentrations. You actually use the substances' &lt;em&gt;activities&lt;/em&gt; which we can relate to change in Standard Gibbs Free Energy for the reaction (which is easy to find with tabulated data):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/b/7/d/b7d8d093750d500cbe0a12dfa630af86.png&quot; alt=&quot;gstar vs k&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a full derivation refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_constant&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's article on the equilibrium constant&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="3704" LastEditDate="2014-03-08T00:34:20.590" LastActivityDate="2014-03-08T00:34:20.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9022" PostTypeId="2" ParentId="4642" CreationDate="2014-03-08T01:05:35.700" Score="4" Body="&lt;p&gt;Chemistry World UK have made &lt;a href=&quot;http://www.youtube.com/watch?v=_oEFwyoWKXo&quot; rel=&quot;nofollow&quot;&gt;this YouTube Clip&lt;/a&gt; that has a professor from UCLA explaining what is needed and how to go about it - even though the video is about 2.5 minutes, but it is a good demonstration.  The associated article is &lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2012/March/graphene-dvd-player-burn-supercapacitor.asp&quot; rel=&quot;nofollow&quot;&gt;DVD player burns graphene to disc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, according to this JensLabs page &lt;a href=&quot;http://jenslabs.com/2013/03/14/diy-graphene/&quot; rel=&quot;nofollow&quot;&gt;DIY Graphene&lt;/a&gt;, it states that the following readily available materials are needed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Graphite oxide&lt;/li&gt;&#xA;  &lt;li&gt;Light scribe capable DVD burner &lt;/li&gt;&#xA;  &lt;li&gt;Light scribe DVD&lt;/li&gt;&#xA;  &lt;li&gt;Substrate&lt;/li&gt;&#xA;  &lt;li&gt;Sonicator - the article suggests that a ultrasonic cleaner would suffice&lt;/li&gt;&#xA;  &lt;li&gt;Pipette&lt;/li&gt;&#xA;  &lt;li&gt;Ion-porous separator&lt;/li&gt;&#xA;  &lt;li&gt;Electrolyte&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-03-08T01:05:35.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9023" PostTypeId="2" ParentId="7294" CreationDate="2014-03-08T01:15:32.550" Score="1" Body="&lt;p&gt;As discussed in the comments, the asker only wanted to create some fluorine &lt;em&gt;in situ&lt;/em&gt; for use in another flask, with no desire to store it. Chemistry with fluorine is difficult due to its dangerously high reactivity. Do not attempt anything without the proper safety procedures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually still have doubts whether it is possible to obtain fluorine gas reliably from the exposure of silver fluoride to sufficiently high energy photons, and if there is any simple way to produce photons with the required energy. It is well known that the other silver halides easily suffer photolysis, especially for the heavier halides, but reactions which produce fluorine gas tend to be extremely endergonic, and therefore don't happen except in rather severe situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it is somehow possible to produce fluorine gas directly by photolysis of silver fluoride, then it is possible to transport the gas to another container where it can react, simply by connecting the flasks with a pipe. Few materials are capable of withstanding exposure to fluorine, however, so the containers must all be made of very well dried gas. Unless the asker has access to a very high power light source, I suspect that the rate at which silver halides decompose is quite slow, which may preclude their efficient usage as halogen generators. This is especially true for silver fluoride, which likely decomposes the slowest, and the fluorine formed in very low concentrations could easily be lost by reaction with impurities or residual water before even reaching the target. I do not know whether there are any catalysts which can be added to increase the photolysis speed of silver fluoride. Presumably some semicondutor with a high bandgap, such as titanium dioxide, could absorb the light and cause photoreduction of the silver/photooxidation of fluoride, but the catalyst would be quickly consumed by the fluorine formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; fluorine gas can be formed by silver halide photolysis fast enough, then by simply linking it to another flask with the desired reactants should cause the fluorine to diffuse and react, followed by a continuous equilibrium shift in the production of more fluorine in the flask containing the silver fluoride, and its consumption in the flask with the other reagents.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-03-08T01:15:32.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9024" PostTypeId="2" ParentId="5000" CreationDate="2014-03-08T01:55:26.817" Score="3" Body="&lt;p&gt;The blue flame from butane (or many other hydrocarbons) is due to the complete combustion of the gas.  From the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Flame#Flame_color&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flame color&lt;/a&gt; it mentions,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Specifically, making the link to the common tool in the laboratory, a Bunsen Burner, which typically has two 'types of flame (as seen below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/21Yuz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://wsc11sci.wikispaces.com/Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt; - which states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Bunsen Burners we use at school use a mixture of alkane gases like propane and butane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The yellow flame is from an incomplete combustion, producing $\ce{CO}$ and the colour is due to (according to the Wikipedia article):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;incandescence of very fine soot particles that are produced in the flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With more oxygen, a complete combustion of butane is possible (without the soot):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C4H10 +13O2 -&amp;gt; 8CO2 + 10H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which provides&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;enough energy to excite and ionize gas molecules in the flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the blue flame colour is due to the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user4076" LastActivityDate="2014-03-08T01:55:26.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9025" PostTypeId="2" ParentId="7236" CreationDate="2014-03-08T02:14:44.003" Score="4" Body="&lt;p&gt;As mentioned in the comments, water is often treated by ozonation in order to sterilize it and ensure it can be drank with no ill effects. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone&quot; rel=&quot;nofollow&quot;&gt;Ozone&lt;/a&gt; is a very reactive gaseous allotrope of oxygen, produced &lt;em&gt;in situ&lt;/em&gt; by exposure of atmospheric oxygen to UV-B light (in fact this happens naturally in the stratosphere, generating a thin &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone_layer&quot; rel=&quot;nofollow&quot;&gt;ozone blanket&lt;/a&gt; covering the world). Ozone can be produced artificially by using a mercury lamp, which emits large amounts of UV light. The water is then brought into contact with the ozone in order to be purified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water is collected after being ozonated, typically a small amount of ozone is still dissolved in the water. If the water is allowed to rest for a few minutes after being collected, the residual dissolved ozone diffuses out. Ozone has a very strong, characteristically sharp odour, which can be detected even at very low concentrations. This is likely why the asker's water had an odd smell. Ozone is toxic, though, so exposure should be avoided.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-03-08T02:14:44.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9026" PostTypeId="2" ParentId="5000" CreationDate="2014-03-08T02:33:07.223" Score="3" Body="&lt;p&gt;The blue color is the Swan lines' emission of $\ce{C2}$.  Rather than acetylene diradical it is more likely to be the olefin singlet dicarbene, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LQybj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-08T02:33:07.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9027" PostTypeId="1" AcceptedAnswerId="9029" CreationDate="2014-03-08T02:54:33.050" Score="-15" ViewCount="2892" Body="&lt;p&gt;How is nickel mined and extracted from its ores?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the appropriate word and chemical equations for this process?&lt;/p&gt;&#xA;" OwnerUserId="4765" LastEditorUserId="16683" LastEditDate="2017-08-06T08:16:06.883" LastActivityDate="2017-08-06T08:16:06.883" Title="Extraction of nickel from its ores" Tags="&lt;metal&gt;&lt;transition-metals&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-03-08T16:17:03.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="9029" PostTypeId="2" ParentId="9027" CreationDate="2014-03-08T03:29:12.853" Score="9" Body="&lt;p&gt;Nickel can be extracted directly from its ore by reduction by hydrogen or carbon monoxide at elevated temperature at $600\ \mathrm{^\circ C}$ to $650\ \mathrm{^\circ C}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NiO + H2 -&amp;gt; Ni + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NiO + CO -&amp;gt; Ni + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it can be extracted by treatment with dilute sulfuric acid following by precipitation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NiO + 2H+ -&amp;gt; Ni^2+ + H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="4516" LastEditorDisplayName="user7951" LastEditDate="2015-07-28T19:50:52.073" LastActivityDate="2015-07-28T19:50:52.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9030" PostTypeId="1" AcceptedAnswerId="9033" CreationDate="2014-03-08T03:43:33.047" Score="1" ViewCount="296" Body="&lt;p&gt;Consider the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH3 + H2O -&amp;gt; NH4+ + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\ce{NH3}$ is a weak base, and thus $\ce{H2O}$ becomes a weak acid. Thus, shouldn't its conjugate be a strong acid? However, $\ce{NH4+}$ is a weak acid, instead of the strong conjugate acid it should be. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2014-03-08T20:31:53.490" Title="Conjugate acids and bases" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9031" PostTypeId="2" ParentId="8984" CreationDate="2014-03-08T04:18:39.777" Score="2" Body="&lt;p&gt;Particle denotes a logical partition of components of a system in a manner which retains an accurate model (on some particular scale) of a physical system while greatly simplifying analysis of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example of the dust particle is analogous to the alpha. To someone studying dust interactions, they might create a useful abstraction by ignoring the microstructure of the dust particle and just regard the particle as a single unit. In the case of analysis of radiation transport of alpha particles in a shielding application, the internal structure of the particle is useful and important, but only to the extent that it affects the external microscopic properties of the particle, such as its reaction cross sections, charge, and kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for almost all intents and purposes nothing is lost, and much is gained, by treating some system as a particle, then it is useful to do so.&lt;/p&gt;&#xA;" OwnerUserId="4767" LastActivityDate="2014-03-08T04:18:39.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9032" PostTypeId="2" ParentId="7767" CreationDate="2014-03-08T05:08:52.363" Score="8" Body="&lt;p&gt;When benzyl chloride is treated with $\ce{NaOH}$ and $\ce{AgNO3}$ , a white precipitate is formed. &#xA;Chlorobenzene does not form white precipitate reacting with $\ce{NaOH}$ and $\ce{AgNO3}$. &lt;/p&gt;&#xA;" OwnerUserId="4769" LastEditorUserId="7960" LastEditorDisplayName="user4076" LastEditDate="2015-07-05T03:38:06.980" LastActivityDate="2015-07-05T03:38:06.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9033" PostTypeId="2" ParentId="9030" CreationDate="2014-03-08T07:25:14.290" Score="3" Body="&lt;p&gt;Consider the $pK_b$ of the base $\ce{NH3}$ and the $pK_a$ of the acid $\ce{NH4+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pK_b$ = $4.8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pK_a$ = $9.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH3}$ is a stronger base than what $\ce{NH4+}$ is as an acid, being $4.8$ smaller than $9.2$: let's compare $pH$ for equal concentrations, say $0.1 M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{NH3}$, you get $11.1$, while for $\ce{NH4+}$, you get $5.1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11.1$ is about 100 times more &quot;distant&quot; from neutral $pH$ than $5.1$, if we move from the $pH$ scale to the &lt;strong&gt;concentration&lt;/strong&gt; scale. That's what also stems from the comparison of the $pK$ values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify the problem, take the $pK_a$ for the acid/base couple, e.g. $\ce{NH4+/NH3}$, that is equal to $9.2$. If that value is larger than $7$, then, within that couple, the base will be stronger than its conjugated acid. The other way round, if $pK_a$ is lower than $7$.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-08T20:31:53.490" LastActivityDate="2014-03-08T20:31:53.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9034" PostTypeId="1" AcceptedAnswerId="9039" CreationDate="2014-03-08T12:06:46.097" Score="13" ViewCount="6159" Body="&lt;p&gt;I read in my textbook that if we multiply a chemical reaction by some factor(let's say $b$) its new equilibrium constant becomes $K^b$.But I don't understand why this happens..&#xA;What is the difference between the reaction $\ce{A} + \ce{B} = \ce{C}$ and $b \ce{A} + b \ce{B} = b \ce{C}$ ? &#xA;Aren't the two reactions same? So why does the equilibrium constant change?&lt;/p&gt;&#xA;" OwnerUserId="4772" LastEditorUserId="189" LastEditDate="2014-09-01T19:37:01.553" LastActivityDate="2014-09-01T19:37:01.553" Title="Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9035" PostTypeId="1" CreationDate="2014-03-08T13:39:44.860" Score="3" ViewCount="4614" Body="&lt;p&gt;To be explicit, $\text{Cr}^{3+}$ and $\text{Sc}$ have the same number of electrons, the only difference is the nuclear charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an isolated anomaly, it seems: the book I'm using claim that with all fourth period transition metal cations, the $4 \text{s}$ electrons are removed first. This indicates that the $4 \text{s}$ electrons are at a higher energy than $3\text{d}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question&lt;/strong&gt;: why is this the case, given that the $4\text{s}$ electrons are 'added' before the $3\text{d}$ electrons as one moves rightwards across the fourth period, indicating that they are of a  lower energy? I cannot reconcile this apparent contradiction.&lt;/p&gt;&#xA;" OwnerUserId="597" LastActivityDate="2014-03-09T10:12:53.470" Title="Why does $\text{Cr}^{3+}$ not have the same electron configuration as $\text{Sc}$?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9036" PostTypeId="2" ParentId="9034" CreationDate="2014-03-08T16:18:51.033" Score="7" Body="&lt;p&gt;Let's take the simplest reaction equation $$\ce {A &amp;lt;=&amp;gt; B}$$&#xA;In this reaction, your stoichiometric coefficients are simply $$1A:1B$$&#xA;Coefficients are the factor you will need to raise your ratio to when you calculate your equilibrium constant. For this reaction, the equilibrium equation would be:&#xA;$$K^1_{eq} = \frac{[B]^1}{[A]^1}$$&#xA;Now, to make things easier to see, lets say your concentrations are:&#xA;$$A=10 : B=5$$&#xA;This would mean that &#xA;$$K^1_{eq}=\frac{10}{5} = 2$$&#xA;Now, let's multiply our equation by a coefficient say, 2:&#xA;$$\ce{2A&amp;lt;=&amp;gt;2B}$$&#xA;Now when you go to calculate your equilibrium constant, your exponents for a will be 2 instead of 1:&#xA;$$K^2_{eq} = \frac{[B]^2}{[A]^2} $$&#xA;Replace A and B with your concentration since they haven't changed, only their stoichiometric coefficients have:&#xA;$$K^2_{eq}=\frac{10^2}{5^2} = \frac{100}{25}=4$$&#xA;Compare this new K value, call it K2 to your first K value, call it K1:&#xA;$$K_2=4 , K_1=2$$&#xA;Remembering that K2 is the square of K1, prove to yourself the reaction is the same by saying:&#xA;$$K_2=K_1\iff K_1^2=K_2 \iff K_2^\frac{1}{2} = K_1$$ which just shows that &#xA;$$4^\frac{1}{2} = 2 $$&#xA;$$2=2$$&#xA;Your rate hasn't necessarily increased, the kinetics of the reaction are just the same, it's just that an equilibrium constant serves as a way to analyze how many moles of your product you are gaining compared to how many you are losing. If you add more moles of your reactant, your product amount has to increase commensurately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prove it to yourself, do the same thing but &#xA;$$K^3_{eq} = \frac{[B]^3}{[A]^3}$$ You should still get a $$K=2$$ if you use $$B=10 , A=5$$&lt;/p&gt;&#xA;" OwnerUserId="4768" LastActivityDate="2014-03-08T16:18:51.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9037" PostTypeId="1" CreationDate="2014-03-08T16:32:59.473" Score="4" ViewCount="550" Body="&lt;p&gt;This is a question I solved in class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 35 degrees, $\ce{N_2O_4}$ is 15% dissociated in a flask at a pressure of 1.2 atm. Calculate &#xA;  $K_p$  Equation:&#xA;  $$\ce{N_2O_4(g)&amp;lt;=&amp;gt;2NO_2 (g)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was how the answer went...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial pressure of $\ce{N_2O_4} = 0.85\times 1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial pressure of $\ce{NO_2} = 0.1\times 1.2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then $K_p$ was calculated using the standard formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I cannot agree to with my teacher is that how is the partial pressure of $NO_2$ equal to 0.15x1.2. If I begin with 1 mole of $\ce{N_2O_4}$, if 15% dissociated at 1.2 atm, I would end up with 0.3 moles of $\ce{NO_2}$... the the partial pressure would be ${0.3\over 1.15}$x$1.2$... this does not go with the general answer stated above&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="275" LastEditDate="2014-12-02T18:38:18.333" LastActivityDate="2014-12-02T18:38:18.333" Title="calculating Kp with partial pressures" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9038" PostTypeId="1" CreationDate="2014-03-08T16:44:37.917" Score="2" ViewCount="3229" Body="&lt;p&gt;It is easy to compare two ionic compounds when one of the ions is same. However, how do we compare two compounds if one of the ions is the same element but just has different charge and the other ion is completely different?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="1434" LastEditDate="2014-03-08T17:31:17.937" LastActivityDate="2016-10-29T14:52:37.133" Title="How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules?" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9039" PostTypeId="2" ParentId="9034" CreationDate="2014-03-08T17:07:20.963" Score="12" Body="&lt;p&gt;Have a look at &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/189&quot;&gt;this answer&lt;/a&gt; of mine.&#xA;There you can find the derivation of the formula that defines the equilibrium constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of a general reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C &amp;lt;=&amp;gt; \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it is also shown how this definition relates to the maybe more familiar definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C &amp;lt;=&amp;gt; \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are not the same but are related to each other via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; K_{\alpha} = K_{1}^{\alpha}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;comes from the definition of $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Extent_of_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;extend of reaction&lt;/a&gt; $\xi = 1 \, \text{mol}$).&#xA;So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc.&#xA;But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-03-08T22:33:11.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9040" PostTypeId="2" ParentId="8985" CreationDate="2014-03-08T18:10:35.773" Score="5" Body="&lt;p&gt;I had some comments to make, but they're too large to actually fit into the comments, so I'm making another answer. I'll take the opportunity to give a bit of background, which may be going a little too far for you right now, but which may come in useful in the future. I'm sorry all my answers tend to be overly lengthy and dry!&lt;/p&gt;&#xA;&lt;p&gt;There are many factors that combine to create a substance's observed physical properties. When talking about solubility, melting points, and boiling points, it's essential to analyse the intermolecular forces between molecules.&lt;/p&gt;&#xA;&lt;p&gt;First let's start with solubility. After defining the concept of polarity for a substance, (based on the strength of the electric dipole present in a molecule) one quickly discovers that, in general, &amp;quot;like dissolves like.&amp;quot; This is a very approximate rule of thumb that relies on a qualitative description of intermolecular forces. For a substance &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; to dissolve in another substance &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, then it is necessary that intermolecular &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; forces and &lt;span class=&quot;math-container&quot;&gt;$B\dotsb B$&lt;/span&gt; forces be broken in order to create &lt;span class=&quot;math-container&quot;&gt;$A\dotsb B$&lt;/span&gt; forces, and this process must be favourable. We say &amp;quot;like dissolves like&amp;quot; because if &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; are similar, then the &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; interactions formed  are likely about as strong as &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\dotsb B$&lt;/span&gt;, and so the molecules of each substance will tend to mingle well. If &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; are very different, then it's less likely that &lt;span class=&quot;math-container&quot;&gt;$A\dotsb B$&lt;/span&gt; interactions will predominate over &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\dotsb B$&lt;/span&gt;, and hence the substances stay apart.&lt;/p&gt;&#xA;&lt;p&gt;Intramolecular interactions are all about electrostatic interactions at long range. From Coloumb's law of electrostatics, it's clear that there are stronger attractions and repulsions the higher the charges involved. Even in neutral molecules, there can often be parts of the molecules with a slightly concentration of negative charge and another part with a slightly concentration of positive charge. We call these oppositely-charged areas in a same object &lt;em&gt;poles&lt;/em&gt;, and when there are two oppositely charged areas connected, they form a &lt;em&gt;dipole&lt;/em&gt;. The dipole is characterized by something called the dipole moment, which is a vector (a quantity that has direction as well as magnitude). Depending on how spatially concentrated the excess negative and positive charges are, the dipole is said to vary in strength. Weak dipoles will have little charge separation, while strong dipoles have lots of charge separation. The strength of the dipole is what dictates a substance's &lt;em&gt;polarity&lt;/em&gt;, which is a quantification of just &lt;em&gt;how polar&lt;/em&gt; it is.&lt;/p&gt;&#xA;&lt;p&gt;As I mentioned, most intermolecular forces are electrostatic in nature. Smaller charges and dipoles tend to cause weaker intermolecular interactions, and larger charges and dipoles tend to cause stronger intermolecular interactions. If you mix substances &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, both with weak dipoles (low polarity) or both with strong dipoles (high polarity), then &lt;span class=&quot;math-container&quot;&gt;$A\dotsb B$&lt;/span&gt; interactions are similar to &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B\dotsb B$&lt;/span&gt;, so they tend to dissolve. If, however, you mix &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; has low polarity and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; has high polarity, then the electrostatic forces between &lt;span class=&quot;math-container&quot;&gt;$B\dotsb B$&lt;/span&gt; will be stronger than &lt;span class=&quot;math-container&quot;&gt;$B\dotsb A$&lt;/span&gt;, and there is no drive for &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; to mix with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;All of this talk is to justify why polarity is something you'd expect to be important in determining whether two substances mix. In your specific case, you're studying mixing water (a high polarity substance) with dichlorobenzenes. It is possible to approximately calculate the dipoles present in molecules by comparing the placement of atoms in the molecule and the atoms' electronegativities (electronegativity quantifies how badly an element wants electrons). For example, carbon and hydrogen atoms have similar electronegativities, so when they bond, they form approximately non-polar bonds. When chlorine and carbon bond, the former has a significantly higher electronegativity, so it will tug on the bond's electrons a bit, such that the chlorine acquires a small negative charge, and the carbon acquires a small positive charge, and we say the bond thus formed is polar.&lt;/p&gt;&#xA;&lt;p&gt;In dichlorobenzenes (DCBs), this means there are dipole moments in the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bonds. In &lt;em&gt;ortho&lt;/em&gt;-DCB, the dipoles moments of each bond are pointing in a similar direction (two equal vectors with a &lt;span class=&quot;math-container&quot;&gt;$60^\circ$&lt;/span&gt; angle between them), so their effect adds up and the substance has some polarity. In &lt;em&gt;meta&lt;/em&gt;-DCB, the dipoles are kind of facing away from each other (two equal vectors with a &lt;span class=&quot;math-container&quot;&gt;$120^\circ$&lt;/span&gt; angle between them), so their effects cancel about a bit and the molecule is only left with a slight polarity. For &lt;em&gt;para&lt;/em&gt;-DCB, the vectors are exactly opposite and equal (two equal vectors with a &lt;span class=&quot;math-container&quot;&gt;$180^\circ$&lt;/span&gt; angle between them), so they cancel out entirely; this means &lt;em&gt;para&lt;/em&gt;-DCB is a &lt;em&gt;non-polar molecule&lt;/em&gt;, even though it has some polar bonds.&lt;/p&gt;&#xA;&lt;p&gt;Since water is a highly polar substance, if we assume that the dichlorobenzenes with higher polarity will be more soluble, then we expect that the solubilities increase in the order &lt;em&gt;p&lt;/em&gt;-DCB &amp;lt; &lt;em&gt;m&lt;/em&gt;-DCB &amp;lt; &lt;em&gt;o&lt;/em&gt;-DCB. &lt;a href=&quot;http://www.atsdr.cdc.gov/toxprofiles/tp10-c4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Experimental data&lt;/a&gt; show that their solubilities in water are approximately &lt;span class=&quot;math-container&quot;&gt;$\pu{80 mg/L}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{125 mg/L}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\pu{156 mg/L}$&lt;/span&gt; for &lt;em&gt;p&lt;/em&gt;-DCB, &lt;em&gt;m&lt;/em&gt;-DCB, and &lt;em&gt;o&lt;/em&gt;-DCB, respectively. Just what we expected! Note that even &lt;em&gt;o&lt;/em&gt;-DCB, the most polar of the three, has a very low solubility in water. This is because the polar regions in the DCB molecules are quite small relative to the non-polar regions.&lt;/p&gt;&#xA;&lt;p&gt;So we got through solubility. Now with the same background in intermolecular forces, explaining the boiling point behaviour is relatively simple. When a substance is heated close to its boiling point, you are giving the molecules almost enough energy to break the intermolecular &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; forces. The stronger the &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; forces, the more energy is needed to overcome them. All other things being equal, &lt;span class=&quot;math-container&quot;&gt;$A\dotsb A$&lt;/span&gt; interactions are stronger for high-polarity molecules than low-polarity molecules, as there are bigger charges and bigger electrostatic interaction in high-polarity molecules.&lt;/p&gt;&#xA;&lt;p&gt;Again, by looking at polarity alone, we expect the boiling point trend to be &lt;em&gt;p&lt;/em&gt;-DCB &amp;lt; &lt;em&gt;m&lt;/em&gt;-DCB &amp;lt; &lt;em&gt;o&lt;/em&gt;-DCB. Experimental data shows that the boiling points are &lt;span class=&quot;math-container&quot;&gt;$\pu{174 ^\circ C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{173 ^\circ C}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\pu{180 ^\circ C}$&lt;/span&gt; for &lt;em&gt;p&lt;/em&gt;-DCB, &lt;em&gt;m&lt;/em&gt;-DCB, and &lt;em&gt;o&lt;/em&gt;-DCB, respectively. The expected and experimental trends are not exactly the same. This is likely because the polarity difference between &lt;em&gt;p&lt;/em&gt;-DCB and &lt;em&gt;m&lt;/em&gt;-DCB is small enough that other factors need to be taken into account.&lt;/p&gt;&#xA;&lt;p&gt;Explaining melting points is quite a bit harder than boiling points. Several things other than polarity can strongly influence melting temperatures. &lt;a href=&quot;https://chemistry.stackexchange.com/q/7257&quot;&gt;I have a previous post on this matter with some general information&lt;/a&gt;. In the case of the dichlorobenzenes, it seems that molecular symmetry and solid packing is the dominant effect. Chem.SE user Uncle Al has provided &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8777/why-does-sif4-have-a-higher-melting-point-than-sf4/8784#8784&quot;&gt;a list of compounds showing how strong this factor can be&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="61578" LastEditDate="2020-08-03T19:00:04.970" LastActivityDate="2020-08-03T19:00:04.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9041" PostTypeId="2" ParentId="9035" CreationDate="2014-03-08T19:08:12.943" Score="2" Body="&lt;p&gt;The electron configuration of chromium is : $\ce{[Ar] 4s^1 3d^5 }$ (chromium is one of those examples of &lt;em&gt;special&lt;/em&gt; electron configurations : the electron configuration is &lt;strong&gt;not&lt;/strong&gt; $\ce{[Ar] 4s^2 3d^4 }$ as many people might suspect, an &lt;strong&gt;electron is moved to the 3d orbital because this configuration is more stable&lt;/strong&gt;  --&gt; that is the reason why chromium has this special electron configuration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the $\ce{Cr^3+}$ ion loses an electron from $\ce{4s}$ and loses 2 electrons from $\ce{3d}$ to give it an electron configuration of $\ce{[Ar] 3d^3 }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scandium has an electron configuration of $\ce{[Ar] 4s^2 3d^1  }$ and isn't one of those exceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the reasons why the chromium(+III) ion doesn't have the same electron configuration as scandium, despite the fact that they have equal amounts of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why the 4s orbital is emptied first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In all the chemistry of the transition elements, the 4s orbital&#xA;  behaves as the outermost, highest energy orbital. The reversed order&#xA;  of the 3d and 4s orbitals only seems to apply to building the atom up&#xA;  in the first place. (Using the Aufbau principle) In all other respects, the 4s electrons are always&#xA;  the electrons you need to think about first.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(this link is interesting for you in case you wanted a more detailed explanation about the 4s / 3d problem :&lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; : Chromium, just like all other elements, tries to obtain more stable configurations by losing certain amounts of electrons. Now, in the case of $\ce{Cr^3+}$ half/fully filled subshells can't be achieved ( you get $\ce{[Ar] 3d^3}$ ) but that doesn't necessarily mean that it is not stable. But chromium does have other ions in which it does have more stable electron configurations. Take a look at the $\ce{Cr^6+}$ ion : here the ion has an electron configuration that is equal to the one of argon, which does have completely filled valence subshells. ($\ce{[Ne] 3s^2 3p^6}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ions can't always achieve these more stable half/fully filled subshells, but that doesn't necessarily mean that they are not stable.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2014-03-09T10:12:53.470" LastActivityDate="2014-03-09T10:12:53.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9042" PostTypeId="1" AcceptedAnswerId="9048" CreationDate="2014-03-08T19:15:04.310" Score="3" ViewCount="391" Body="&lt;p&gt;I know soap molecules form micelles. Do they also form liposomes in water? I need the reasons too.&lt;/p&gt;&#xA;" OwnerUserId="4751" LastActivityDate="2016-12-04T16:02:09.830" Title="Do soap molecules form liposomes?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9044" PostTypeId="1" AcceptedAnswerId="9047" CreationDate="2014-03-09T00:39:39.563" Score="9" ViewCount="371" Body="&lt;p&gt;I'm interested in learning organic chemistry, and am following a text (Organic Chemistry by Clayden) and an online course. I also have access to ChemBio3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel as though the way chemistry is taught conventionally is more difficult than it needs to be. Some online courses utilize industry software when teaching things (such as using MatLab in engineering), which makes visualizing and understanding a lot of the theories and concepts taught in books much more intuitive, since you can replicate the theory directly into a program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a similar course exist somewhere for chemistry? I'm not talking about the little java applets that are available everywhere that mimic certain reactions or visualize concepts, I mean a course that utilizes a powerful software suite such as ChemBio in order to teach organic chemistry? Something that will teach both organic chemistry at a proper undergrad level and a good piece of industry software at the same time?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="5026" LastEditDate="2018-03-03T11:36:31.553" LastActivityDate="2018-03-03T11:36:31.553" Title="Learning chemistry with software?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9047" PostTypeId="2" ParentId="9044" CreationDate="2014-03-09T03:17:22.077" Score="4" Body="&lt;p&gt;Do your chair parade.  See what the Kahn Academy has to offer.  Have more than one texbook.  If organic is not the best thing in the world pursued clothed, consider other options.  Employment at the bench will be under management that has other goals.  Grant funding considers discovery to be unacceptable risk.  Remember what they called the first designed degenerate Cope rearrangement - BULLVALENE.  It worked.  So did its two-step synthesis using no reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LCAO for hybridization, geometry, resonance, and 1,3-shifts.  Nomenclature and chirality.  Woodward-Hoffmann comes later.  Carbocations, carbanions, radicals.  Carbonyl condensations.  Hard and soft acids and bases, Lewis and Brønsted acids and bases.  Reductants and oxidants; reagents in general.  Name reactions (that you already know as pedagogy).  Practice looking into stereograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NMR, IR, UV-Vis (not so much now), mass spec with database and software.  You should have some glassblowing under your belt - Schlenkware;  T-, butt, and ring seals; making an ampoule.  Own your own didymium glasses and Kennametal cutter.  Burn a pencil for its &quot;lead.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Synthesis.  Work backwards via the synthon approach.  We used to read the new Aldrich catalog, and Fieser and Fieser, but both have has gotten out of hand.  Spend a weekend/month skimming the main journals - JACS, Angew. Chemie, JOC, Tetrahedron - and what interests you.  Never fill a reaction flask more than half full, because the universe hates you.  COTTON lab coats, because they don't melt onto your flesh.  Don't do anything stupid twice.  When in doubt - palladium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy HyperChem Lite, oh yes indeed.  Remember to add hydrogens, assign atom types, then minimize well below the recommended cutoff for 10 times the recommended iterations, at least.  Note how the software is boggled when minimizing high symmetry cage ketones &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it isn't fun, you might be a theorist.  There is no shame in that - but you don't get to make crystals.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-03-03T11:28:35.503" LastActivityDate="2018-03-03T11:28:35.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9048" PostTypeId="2" ParentId="9042" CreationDate="2014-03-09T03:28:04.267" Score="1" Body="&lt;p&gt;Liposomes are a lipid bilayer separating an aqueous internal compartment from the aqueous phase. Micelles are closed lipid monolayers with a fatty acid core and polar surface, or a polar core with fatty acids on the surface (inverted micelle, Aerosol-OT).  A soap would then form a micelle but not a lipsoome for its ionic water-soluble head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-ionic surfactant, e.g., diblock short PEO-PPO, could stretch the definitions.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-09T03:28:04.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9049" PostTypeId="1" AcceptedAnswerId="9058" CreationDate="2014-03-09T04:38:10.583" Score="3" ViewCount="3034" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ions $\ce{B}$ and $\ce{C}$ react to form complex $\ce{BC}$. If $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{B}$ is combined with $\pu{15.0 mL}$ of $\pu{1.00 M}$ $\ce{C}$, and our initial concentration of BC is $0$, yet the end in equilibrium for complex $\ce{BC}$ is $\pu{0.00450 mol}$ determine the equilibrium constant for this reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, I converted the total volume $\pu{15.0 mL + 15.0 mL = 30 mL}$ to L and I got $\pu{0.03L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I used the given $\pu{0.00450 mol}$ and divided it by $\pu{0.03 L}$ and I got $\pu{0.15 M}$ for BC &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, I did the ICE chart&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QBpOS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, I used the quadratic formula to solve for $x$ and calculate the concentration for each substance, as well as getting the $K_\mathrm{c}$ at equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWTA7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got $1.53$ for $K_\mathrm{c}$ but I am not sure if the whole process is correctly done. Please point out my mistakes&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="5026" LastEditDate="2018-03-03T10:29:07.120" LastActivityDate="2018-03-03T10:29:07.120" Title="Determine the equilibrium constant for this reaction" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9050" PostTypeId="1" CreationDate="2014-03-09T07:25:40.193" Score="5" ViewCount="99" Body="&lt;p&gt;Apologies for the extremely basic question, but I would like to look up the amount by which the volume (or equivalently, density) of various substances changes when they freeze. However, I don't know the correct term for this, and Google isn't helping much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I'm asking about the vertical line segment on the plot below (from &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Density_of_ice_and_water_%28en%29.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). If the $y$ axis was internal energy then the height of this line segment would be called &quot;latent heat of fusion&quot;, and I could easily look it up for any given substance. I would like to know the equivalent term for the discontinuous volume (or density) change, in whichever units are most standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vQuX7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that unlike latent heat, this will depend quite a bit on the pressure. However, I would hope that it would be tabulated for atmospheric pressure at least.&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2014-03-09T09:17:10.147" Title="What is the term for the amount of volume change at a phase boundary?" Tags="&lt;terminology&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9051" PostTypeId="1" CreationDate="2014-03-09T08:10:19.353" Score="6" ViewCount="7109" Body="&lt;p&gt;Why is francium not included in the reactivity series? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is potassium considered more reactive than francium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that reactivity increases down the group, but why does it not apply here? &lt;/p&gt;&#xA;" OwnerUserId="4784" LastEditorUserId="4945" LastEditDate="2020-01-28T18:05:00.973" LastActivityDate="2020-01-28T18:13:35.023" Title="Why do the trends in reactivity not apply for francium?" Tags="&lt;reactivity&gt;&lt;periodic-trends&gt;&lt;relativistic&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9052" PostTypeId="2" ParentId="9050" CreationDate="2014-03-09T09:17:10.147" Score="1" Body="&lt;p&gt;Phase transition can be distinguished by analysing their chemical potential, such classification has been introduced by &lt;strong&gt;Ehrenfest&lt;/strong&gt;. For a given transition, lets call it $\phi$ transition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{X_{\alpha} \rightleftharpoons X_{\beta}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assess transition molar specific volume in the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta_{\phi}V = v_{\beta} - v_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial p}\right)_{T} - \left(\frac{\partial \mu_{\alpha}}{\partial p}\right)_{T}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, by modelling your potential in term of pressure and temperature, you can assess molar volume change of a phase transition. This is the key to solve your question. In real life application, we use diagrams and tables to assess it. But building a model is always valuable for comprehension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar way, molar entropy is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta_{\phi}S = s_{\beta} -s_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial T}\right)_{p} - \left(\frac{\partial \mu_{\alpha}}{\partial T}\right)_{p} = \frac{\Delta_{\phi}H}{T_\phi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\Delta_{\phi}V$ and $\Delta_{\phi}H$ are not null for first order (in term of first derivative) transition, slope difference is not null and therefore potential slopes are different both side of phase transition when Gibbs parameters vary ($p$ and $T$). Thus there are discontinuities in specific volume, enthalpy and entropy when a first order transition occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ehrenfest did a classification of phase transition based on this criterion which led to: first order, second order (smooth, no slope discontinuity) and $\lambda$ (sharper) transitions regarding how such chemical potential discontinuity happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discontinuity also impact heat capacity $c_p$. In case of water, whose phase transitions can be assumed as a first order, when a phase transition occurs, discontinuity leads to a virtually infinite heat capacity, explaining why fusion or vaporisation happens at constant temperature (when pressure is held constant). This is why we need an extra term called latent heat (temperature independent) which contrast with sensible heat (temperature dependent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find detailed explanation (models, diagram and tables) of those phenomenons in Atkins, Physical Chemistry, Chapters 2-7.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-03-09T09:17:10.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9053" PostTypeId="2" ParentId="9037" CreationDate="2014-03-09T10:30:44.100" Score="2" Body="&lt;p&gt;I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0}  -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will obtain by substitution, molar ratio values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15},  \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-03-12T22:17:34.350" LastActivityDate="2014-03-12T22:17:34.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9054" PostTypeId="2" ParentId="9051" CreationDate="2014-03-09T11:32:12.847" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Francium&quot;&gt;Francium&lt;/a&gt; is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 minutes) would immediately vaporize any viewable quantity of the element. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This extreme radioactivity makes it hard to determine the chemical properties of francium. Some are known. For example, francium's ionization energy is 392.8 kJ/mol (cesium is 375.7 kJ/mol). Since francium has a higher ionization potential than cesium, which it does because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_quantum_chemistry&quot;&gt;relativistic effects&lt;/a&gt;, francium should be less active than cesium. However, since observable quantities are difficult (nearly impossible) to come by. It is hard to say for sure.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-03-09T11:32:12.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9055" PostTypeId="1" CreationDate="2014-03-09T13:09:12.200" Score="3" ViewCount="343" Body="&lt;p&gt;When a thin lead sheet is hammered, what happens to internal energy, does it increase, decrease or remains the same or first increases and then decreases?&lt;/p&gt;&#xA;" OwnerUserId="4786" LastActivityDate="2016-08-05T00:17:11.150" Title="When a thin lead sheet is hammered, what happens to internal energy" Tags="&lt;energy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9056" PostTypeId="1" AcceptedAnswerId="9072" CreationDate="2014-03-09T15:03:29.053" Score="1" ViewCount="173" Body="&lt;p&gt;where can I find thermodynamics properties for octane at saturated Temp.$T_s$?&#xA;I need this properties or where can i find all ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$liquid-density=\rho_l=?$$&#xA;$$vapor-density=\rho_v=?$$ &#xA;$$latent-heat-of-evaporation=h_fg=?$$&#xA;$$liquid-kinematic-viscosity=\mu_l=? $$&#xA;$$liquid-conductivity=k_l=?$$&lt;/p&gt;&#xA;" OwnerUserId="4788" LastEditorUserId="4788" LastEditDate="2014-03-09T15:22:55.337" LastActivityDate="2014-03-10T02:30:17.413" Title="Standard reference for thermodynamic properties for octane" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9057" PostTypeId="1" CreationDate="2014-03-09T15:11:13.347" Score="-1" ViewCount="520" Body="&lt;p&gt;Which one is more stable, alkyl carbocations or aryl carbocations and why?&lt;/p&gt;&#xA;" OwnerUserId="2084" LastEditorUserId="418" LastEditDate="2014-03-09T15:23:45.440" LastActivityDate="2014-03-09T22:51:15.060" Title="Organic chemistry Reaction mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-03-03T12:49:33.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="9058" PostTypeId="2" ParentId="9049" CreationDate="2014-03-09T15:26:54.093" Score="2" Body="&lt;p&gt;You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B + C &amp;lt;=&amp;gt; BC}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array} {|c|c|c|c|}&#xA;\hline&#xA;\text{Initial conc.} &amp;amp; \text{0.015 mol} &amp;amp; \text{0.015 mol} &amp;amp; \text{0 mol}\\&#xA;\hline&#xA;\text{Change conc.} &amp;amp; -x &amp;amp; -x &amp;amp; +x\\&#xA;\hline&#xA;\text{End conc.} &amp;amp; \text{0.015 mol} - x &amp;amp; \text{0.015 mol} - x &amp;amp;  0 + x \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that our reaction is at equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_c = 1.22 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks @Nicolau for pointing out the error in the question and reviewing my work.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-09T16:28:15.033" LastActivityDate="2014-03-09T16:28:15.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9059" PostTypeId="1" AcceptedAnswerId="9062" CreationDate="2014-03-09T15:30:08.087" Score="2" ViewCount="318" Body="&lt;p&gt;I am learning Chemistry from a book Heterocyclic compounds and I have bumped onto a following image with given tasks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;img src=&quot;https://i.stack.imgur.com/IOQo7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the balanced reaction?&lt;br&gt;&#xA;What is the RMM of thiazole?&lt;br&gt;&#xA;What is the fuel-air mass ratio?&lt;br&gt;&#xA;What is the oxygen depletion?&lt;br&gt;&#xA;What is the yield of CO2?&lt;br&gt;&#xA;What is the yield of H2O?&lt;br&gt;&#xA;What is the yield of N2?&lt;br&gt;&#xA;For ideal combustion, what is the yield of carbon-monoxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from this I can see that aformentioned product is Thiazole( checked on wiki ) and that products of combustion are carbon dioxide, water, hydrogen sulphide (H2S) and nitrogen. Now how can I solve this matter having in mind following data that air is consisted of 21% O2 and 79% of N2.&lt;/p&gt;&#xA;" OwnerUserId="4785" LastEditorUserId="16683" LastEditDate="2017-03-28T13:14:05.380" LastActivityDate="2017-03-28T13:14:05.380" Title="Combustion of thiazole" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;combustion&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9061" PostTypeId="2" ParentId="9057" CreationDate="2014-03-09T16:01:53.943" Score="4" Body="&lt;p&gt;If you want it nice and stable, you want charge delocalization, like in allylic and benzylic cations. Note that more aryl residues mean more delocalization, additional &lt;em&gt;+M&lt;/em&gt; substituents provide further stabilization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJBmG.png&quot; alt=&quot;triarylmethane dyes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the cations of &lt;em&gt;malachite green&lt;/em&gt; (&lt;strong&gt;1&lt;/strong&gt;) and &lt;em&gt;crystal violet&lt;/em&gt; (&lt;strong&gt;2&lt;/strong&gt;) as an example and &lt;strong&gt;draw some resonance structures&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare these with an alkyl cation or an aryl cation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-09T16:01:53.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9062" PostTypeId="2" ParentId="9059" CreationDate="2014-03-09T16:02:04.177" Score="3" Body="&lt;p&gt;First I have to ask about the premise of your question- why are you teaching yourself chemistry from a book on heterocyclic compounds?  The rest of your question suggests you've skipped the basics and jumped to fairly advanced stuff.  There are many good introductory chemistry that will walk you through problems similar to this one that will be a far easier starting point than heterocyclic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming your general question is &quot;how do I do calculations on oxidation reactions?&quot;, you fill find whole chapters and even books on the topic.  I would highly recommend starting with something like &quot;What is the balanced reaction of combusting propane (C3H8) in pure oxygen?&quot;  You will need this kind of basis to be able to work up problems like the one you ask about with many possible products and spectator species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will get you started: &lt;a href=&quot;http://en.wikipedia.org/wiki/Combustion#Chemical_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Combustion#Chemical_equations&lt;/a&gt; with answer to the first question.  In general though, pick up a used introduction to chemistry book up off Amazon for $10 and start with those walk throughs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps answer your real question and please let me know if I misunderstood and you had a more specific question about this particular problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: For the yield questions, you can use the the balanced equation and molecular weights to come up with a yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll have something like:&#xA;$$yield=stoichiometricRatio*\frac{MW_{product}}{MW_{thiazole}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{g_{product}}{g_{thiozole}}=\frac{mol_{product}}{mol_{thiozole}}*\frac{g_{product}}{mol_{product}}*\frac{mol_{thiazole}}{g_{thiazole}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they give you a mass (or mass flow rate) of thiazole, you could further convert this to a mass (or mass flow rate) of products.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorUserId="4754" LastEditDate="2014-03-11T01:15:51.043" LastActivityDate="2014-03-11T01:15:51.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9064" PostTypeId="1" AcceptedAnswerId="9065" CreationDate="2014-03-09T18:04:05.173" Score="2" ViewCount="8042" Body="&lt;p&gt;I performed an experiment where I did this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Add one spatula of malachite to a test tube, add dilute hydrochloric acid to the malachite until the test tube is one third full. Note observations and identify gas produced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The green solid turned to a blue solution and there was an effervescence of hydrogen. However, I now realise that copper doesn't react with acid and malachite is a copper carbonate hydroxide mineral... &#xA;I researched online and found that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It dissolves in sulfuric acid to give Copper sulfate, Carbon dioxide&#xA;  and Water.&#xA;  &lt;a href=&quot;http://wiki.answers.com/Q/What_happens_when_malachite_is_dissolved_in_sulphuric_acid?classic=true&quot; rel=&quot;nofollow&quot;&gt;http://wiki.answers.com/Q/What_happens_when_malachite_is_dissolved_in_sulphuric_acid?classic=true&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So how did I get that malachite reacted with acid and did produce hydrogen with a squeaky pop test?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="3686" LastEditorUserId="3686" LastEditDate="2014-03-09T18:31:20.247" LastActivityDate="2017-05-07T02:36:32.373" Title="How does malachite react with acid?" Tags="&lt;hydrogen&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9065" PostTypeId="2" ParentId="9064" CreationDate="2014-03-09T18:43:00.193" Score="3" Body="&lt;p&gt;No, pure malachite $\ce{Cu2CO3(OH)2}$ usually doesn't produce hydrogen because it is a copper carbonate. Copper is already in a higher oxidation state (+2) so it can't reduce hydronium to hydrogen. The right reaction is the following:&#xA;$$\ce{Cu2CO3(OH)2 +2H2SO4 -&amp;gt; 2CuSO4 + CO2 ^ + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there are traces of metallic copper some hydrogen can be produced but not with sulfuric acid because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu + 2H2SO4 -&amp;gt; CuSO4 + 2H2O + SO2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nor with $HCl$ (copper doesn't react with $HCl$)  but only with nitric acid or aqua regia you can produce hydrogen from the metallic form. So, in fact, the gas you see is carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="402" LastEditDate="2017-05-07T02:36:32.373" LastActivityDate="2017-05-07T02:36:32.373" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9067" PostTypeId="1" AcceptedAnswerId="9068" CreationDate="2014-03-09T20:03:39.507" Score="11" ViewCount="132257" Body="&lt;p&gt;I sort of know how carbonated beverages are carbonated: a lot of $\ce{CO2}$ gets pushed into the liquid, and the container is sealed. There are at least two things I don't know. First, how much carbon dioxide is actually dissolved in the liquid? Second, what is the resulitng partial pressure of $\ce{CO2}$ in the headspace and the total pressure in the headspace? I'm interested in cans, plastic bottles, and glass bottles. I know from experience that there is some variation among manufacturers even for the same beverage, so I will be happy with general numbers or a good estimate.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2014-11-05T18:36:01.393" Title="What is the carbon dioxide content of a soda can or bottle?" Tags="&lt;solubility&gt;&lt;gas-laws&gt;&lt;mixtures&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9068" PostTypeId="2" ParentId="9067" CreationDate="2014-03-09T21:30:50.720" Score="9" Body="&lt;p&gt;General estimates have placed a can of Coca-Cola to have 2.2 grams of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2} $&lt;/span&gt; in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; in a can is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So here we can see we have about 0.05 mol/0.355 L or about 0.14 mol of carbon dioxide per liter of soda. Of course this value varies by manufacturer, type of drink, container, etc.&lt;/p&gt;&#xA;&lt;p&gt;Looking at Wikipedia, inside Coca-Cola is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Carbonated water, Sugar (sucrose or high-fructose corn syrup depending on country of origin), Caffeine, Phosphoric acid, Caramel color (E150d), Natural flavorings&lt;/p&gt;&#xA;&lt;p&gt;A can of Coke (12 fl ounces/355 ml) has 39 grams of carbohydrates (all from sugar, approximately 10 teaspoons), 50 mg of sodium, 0 grams fat, 0 grams potassium, and 140 calories.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, we can calculate the pressure of &lt;span class=&quot;math-container&quot;&gt;$CO_2$&lt;/span&gt; gas using the Ideal Gas equation if we store our coke at, say, 20 Celsius:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = \frac{nRT}{V} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ P = \frac{\text{0.05 mol} * \text{0.08206} \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K} } * \text{293.15 K}}{\text{0.355 L}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ P = 3.39 \ \text{atm} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;According to this website, &lt;a href=&quot;http://hypertextbook.com/facts/2000/SeemaMeraj.shtml&quot; rel=&quot;noreferrer&quot;&gt;http://hypertextbook.com/facts/2000/SeemaMeraj.shtml&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;On average, the 12 ounce soda cans sold in the US tend to have a pressure of roughly 120 kPa when canned at 4 °C, and 250 kPa when stored at 20 °C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{3.39 atm} * \frac{\text{760 torr}}{\text{1 atm}} * \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 342.66 \ \text{kPa}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Water vapor exerts it's own partial pressure. Looking at standard tabulated values for water vapor pressure, water exerts a pressure of 17.5 torr at 20 Celsius.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{17.5 torr}* \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 2.3275 \ \text{kPa} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Knowing that our total pressure is the sum of all our pressures:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{total} = \text{342.66 kPa} + \text{2.3275 kPa} = 344.99 \ \text{kPa}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, we are roughly about 100 kPa off from the data provided by the website. This is just an approximation. A more accurate way would be to calculate the moles of each product inside the soda, and knowing the total pressure or partial pressure of one of the parts, we can calculate the pressures more accurately. However, that information is proprietary. It's their secret recipe!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Headspace:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://extension.psu.edu/food/preservation/news/2012/why-allow-headspace&quot; rel=&quot;noreferrer&quot;&gt;http://extension.psu.edu/food/preservation/news/2012/why-allow-headspace&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;We have to determine the volume of the headspace - again, I am not sure of exact data - which is between 1/2 inch to 1 and 1/2 inches depending on the container and what it holds. I will assume that the headpsace occupies 6% of the total volume of the can which is 21.3 mL.&lt;/p&gt;&#xA;&lt;p&gt;At manufacturing and at storage, the can is at different temperatures. Taking the above data, we'll say it is manufactured at &lt;span class=&quot;math-container&quot;&gt;$4^\circ C$&lt;/span&gt; and say, stored at &lt;span class=&quot;math-container&quot;&gt;$20^\circ C$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, when carbon dioxide is solubilized in water, it forms carbonic acid. I will neglect that as the ionization constant is small.&lt;/p&gt;&#xA;&lt;p&gt;Assuming our 2.2 grams of carbon dioxide is the &lt;em&gt;maximum&lt;/em&gt; amount of carbon dioxide that can be placed inside, some of the carbon dioxide is soluble in water while the rest exerts pressure inside the headspace to force the carbon dioxide inside the liquid. The pressure is necessary inside this closed volume as, once you open the cap, the carbon dioxide tries to achieve equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \ce{CO_2 (solution) &amp;lt;=&amp;gt; CO2 (g)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In general, the solubility of gases &lt;em&gt;increases&lt;/em&gt; at lower temperatures and &lt;em&gt;decreases&lt;/em&gt; at higher temperatures. A notable exception is the noble gases. In regards to pressure, Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.&lt;/p&gt;&#xA;&lt;p&gt;The solubility of carbon dioxide in water shifts according to Le Chatlier's principle.&lt;/p&gt;&#xA;&lt;p&gt;For the solution to this problem, we would need to know several things about the manufacturing of soda's, or rather, more quantitative data. At manufacturing, the pressure is high and temperature low, so as much carbon dioxide as possible is solubilized in the water.&lt;/p&gt;&#xA;&lt;p&gt;Once shipped out and stored, the pressure inside the headspace increases as the pressure is decreased from manufacturing, the carbon dioxide then leaves solution and enters the headspace. Increasing temperature also decreases solubility.&lt;/p&gt;&#xA;&lt;p&gt;If you can provide me with more information, I will be more than happy to help. :)&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-09T23:14:18.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9069" PostTypeId="2" ParentId="9067" CreationDate="2014-03-09T21:52:13.317" Score="3" Body="&lt;p&gt;It depends from drink and time, according to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/20457301&quot; rel=&quot;nofollow&quot;&gt;S. Teerasong, Analytica Chimica Acta 668 (2010) 47–53&lt;/a&gt; for a normal Cola is about $3.1 \frac{g_{CO_2}}{L}$ (take a look even to &lt;a href=&quot;http://www.chemicalbulletin.ro/admin/articole/69129art_5.pdf&quot; rel=&quot;nofollow&quot;&gt;Glevitzky, Chem. Bull. &quot;POLITEHNICA&quot; Univ. (Timişoara) Volume 50 (64), 1-2,2005&lt;/a&gt; but his esteem is very high).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the pressure in the headspace this is very difficult to esteem theoretically because you have to take in account not only the temperature but the whole system. &lt;del&gt;:you can shake the bottle and increase the pressure dramatically&lt;/del&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-03-11T08:43:49.230" LastActivityDate="2014-03-11T08:43:49.230" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9071" PostTypeId="2" ParentId="9057" CreationDate="2014-03-09T22:51:15.060" Score="1" Body="&lt;p&gt;Steric hindrance might kinetically stabilize a tris-alkyl carbenium ion, such as (&lt;em&gt;tert&lt;/em&gt;-Bu)$\ce{_3C^+}$.  Delocalized aromatic systems leave it in the dust - &lt;strong&gt;tropylium&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-09T22:51:15.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9072" PostTypeId="2" ParentId="9056" CreationDate="2014-03-10T02:30:17.413" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;http://webbook.nist.gov&quot; rel=&quot;nofollow&quot;&gt;NIST Chemistry Webbook&lt;/a&gt; maintained by the National Institute of Standards and Technology is a freely searchable database of chemical and physical data on a vast array of substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything you want for octane can be found there.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-03-10T02:30:17.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9073" PostTypeId="1" CreationDate="2014-03-10T03:06:02.793" Score="-2" ViewCount="317" Body="&lt;p&gt;Recently people have been reporting that if you blow 'HHO' gas (Oxyhydrogen) into a used catalytic converter it gives off excessive heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look &lt;a href=&quot;https://www.google.com/search?q=hho+catalytic+converter&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for any number of examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://www.e-catworld.com/2014/01/looking-at-the-h-cat-experiment-with-hho-gas-and-a-catalytic-converter/&quot; rel=&quot;nofollow&quot;&gt;suggestion&lt;/a&gt; that it is cold fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could this reaction actually be?&lt;/p&gt;&#xA;" OwnerUserId="4793" LastActivityDate="2014-03-10T17:36:49.900" Title="Oxyhydrogen applied to a catalytic converter" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9074" PostTypeId="1" CreationDate="2014-03-10T03:56:38.020" Score="23" ViewCount="66390" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In phenol, pulling the $\mathrm{p}_z$ electrons from the oxygen atom into the ring causes the hydrogen atom to be more partially positive than it is in aliphatic alcohols. This means it is much more easily lost from phenol than it is from aliphatic alcohols, so phenol has a stronger acidic property than ethanol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain the link between the $\mathrm{p}_z$ electrons of oxygen overlapping with the cloud of delocalised electrons of the phenol ring and the increase in positivity of the bonded hydrogen … which in turns allows phenol to lose a proton more easily?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="10185" LastEditDate="2016-05-25T09:31:55.017" LastActivityDate="2016-05-25T09:31:55.017" Title="Reason for the stronger acidic property of phenol than alcohol" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;orbitals&gt;&lt;phenols&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9075" PostTypeId="2" ParentId="9073" CreationDate="2014-03-10T05:52:51.457" Score="2" Body="&lt;p&gt;I am not aware of any current claim on &quot;cold fusion&quot; that could hold up against a scientific re-examination. Previous reports by Pons and Fleischmann now seem to be (mis)led by wishful thinking and dirt effects,  while those by Taleyarkhan were debunked as fraud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the inventions of Andrea Rossi are concerned, I'd be very careful with his claims. So far, he is apparently more &quot;famous&quot; for leaving behind stockpiled waste and huge tax debts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more recent findings of Rossi and Focardi and the related so-called &lt;a href=&quot;http://arxiv.org/abs/1305.3913v3&quot; rel=&quot;nofollow&quot;&gt;&quot;independent study&quot;&lt;/a&gt; were heavily criticized by &lt;a href=&quot;http://arxiv.org/abs/1306.6364&quot; rel=&quot;nofollow&quot;&gt;other researchers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, all this nicely fits in the category of &lt;strong&gt;pathological science&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-10T05:52:51.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9077" PostTypeId="1" CreationDate="2014-03-10T07:39:34.837" Score="5" ViewCount="46481" Body="&lt;p&gt;When we heat potassium permanganate, the equation of the decomposition of it is: &#xA;$$\ce{2KMnO4 -&amp;gt;[\Delta] K2MnO4 + MnO2 + O2}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So  what will happen to potassium manganate ($\ce{K2MnO4}$) when we continue heating?&lt;/p&gt;&#xA;" OwnerUserId="4795" LastEditorUserId="418" LastEditDate="2014-03-10T09:22:51.217" LastActivityDate="2018-03-15T14:02:18.390" Title="Decomposition of potassium permanganate under prolonged heating" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9079" PostTypeId="2" ParentId="9074" CreationDate="2014-03-10T07:55:08.597" Score="18" Body="&lt;p&gt;The effect is indeed amazing, if you compare the $\mathrm{p}K_\mathrm{a}$ of &lt;em&gt;tert&lt;/em&gt;-butanol (17.0) with that of phenol (9.95).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deprotonation is facilitated when the reaction goes downhill (energywise). In order to obtain stabilization of the anion, the negative charge needs to be distributed over a larger number of centres. This distribution (only) is possible if the participating orbitals overlap in space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which orbitals may play a role here? On the one hand, there is the $\pi$ system of the benzene, formed from the $\mathrm{p}_z$ orbitals on the carbon atoms. On the other hand, there's  the $\mathrm{p}_z$ on the oxygen with the same spatial orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can we get comparable molecules, more acidic than phenol itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is yes, if we allow further substituents that (a) pull the negative off the oxygen by inductive effects or (b) even allow for a further distribution of the negative charge by additional resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QOTSC.png&quot; alt=&quot;resonance of picric acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impressive example for the latter is 2,4,6-trinitrophenol, also known as &lt;em&gt;picric &lt;strong&gt;acid&lt;/em&gt;&lt;/strong&gt; with a $\mathrm{p}K_\mathrm{a}$ of &lt;strong&gt;0.29&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-01-06T09:53:01.753" LastActivityDate="2016-01-06T09:53:01.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9080" PostTypeId="2" ParentId="9074" CreationDate="2014-03-10T07:55:38.393" Score="20" Body="&lt;p&gt;In fact I find a more simple reasoning with resonance structure. When phenol loses the $\ce{H+}$ the phenolate ion is stabilized due to the resonance effect, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bmron.png&quot; alt=&quot;Phenoxide resonance forms&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the dissociated form is lower and so phenol has more chance to be in the solution dissociated with the phenolate ion. Aliphatic alcohols are not stabilized by resonance so they are not very prone to dissociation.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="16683" LastEditDate="2016-01-06T09:53:54.503" LastActivityDate="2016-01-06T09:53:54.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9081" PostTypeId="1" CreationDate="2014-03-10T08:14:26.950" Score="0" ViewCount="119" Body="&lt;p&gt;I am well versed with all the named reactions and all of organic chemistry.  I am quite weak in stereochemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What canonical text can I use to study this topic and connect with other parts of the chemistry? &lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="73" LastEditDate="2014-03-10T12:39:02.577" LastActivityDate="2014-03-10T17:08:20.517" Title="What is the canonical reference for studying stereochemistry?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2019-03-12T02:45:51.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="9082" PostTypeId="2" ParentId="9077" CreationDate="2014-03-10T08:46:33.093" Score="4" Body="&lt;p&gt;Potassium manganate is produced by reacting manganese dioxide with potassium hydroxide and oxygen gas : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would suspect that the reverse process will occur when further heating. (Of course when there is no water available you might get other products, maybe potassium oxide instead of potassium hydroxide) &lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-03-10T08:46:33.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9083" PostTypeId="1" CreationDate="2014-03-10T09:13:48.457" Score="0" ViewCount="880" Body="&lt;p&gt;I came across a few equation and observed that when concentration of the $\ce{HNO3}$ and $\ce{H2SO4}$ with benzene ($\ce{C6H6}$) or its derivative varied the product changed. I did not perform any experiment and all of this I ask here is from the stuff I read in various books.please tell me about this nitrating solution and give me all the cases if possible.&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="418" LastEditorDisplayName="user4076" LastEditDate="2016-10-19T15:15:47.310" LastActivityDate="2016-10-19T15:15:47.310" Title="Does the outcome of aromatic nitrations depend on the concentrations of nitric acid and sulfuric acid?" Tags="&lt;organic-chemistry&gt;&lt;solutions&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9084" PostTypeId="2" ParentId="9081" CreationDate="2014-03-10T09:28:20.980" Score="4" Body="&lt;p&gt;The classical book on stereochemistry is &lt;a href=&quot;http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471016705.html&quot; rel=&quot;nofollow&quot;&gt;Eliel: Stereochemistry of Organic Compounds&lt;/a&gt;, first edition appeared 1962.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find overwhelming amount of information there. I would recommend it, regardless your age and level. It is so well written, that it is worth at least of fast reading over most important parts. &#xA;You should find it in any reasonable library, no need to get the latest edition, even the older ones should cover what you need.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-03-10T09:28:20.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9085" PostTypeId="1" CreationDate="2014-03-10T13:34:07.320" Score="2" ViewCount="8680" Body="&lt;p&gt;I know that Lewis' concept of acid base deals in terms of electrons and Brønsted in terms of $\ce{H+}$ and base which can provide electrons or $\ce{OH-}$. If I am wrong (partially or fully) please explain Lewis and Brønsted-Lowry concept along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me it is a Lewis base because of the lone pair on oxygen of the alcohol and a Brønsted acid.&lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="5026" LastEditDate="2018-03-03T11:06:57.610" LastActivityDate="2018-03-04T02:54:57.120" Title="Is alcohol Lewis base or Brønsted base?" Tags="&lt;acid-base&gt;&lt;alcohols&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9086" PostTypeId="2" ParentId="374" CreationDate="2014-03-10T13:39:25.003" Score="-3" Body="&lt;p&gt;The other answers are all wrong. Draw Molecular orbital diagram of $\ce{NO}$ and you will see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That electron in antibonding orbital, it is so easy to lose, hence it easily goes to $\ce{NO^+}$ and $\ce{Fe^{(II)}}$ to $\ce{Fe^{(I)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember if you read something extraordinary which claims something out of daily routine (especially in chem) then it must be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link : &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrate_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nitrate_test&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia also says Iron is in $+1$&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4945" LastEditDate="2015-01-05T03:07:58.963" LastActivityDate="2015-01-05T03:07:58.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9087" PostTypeId="2" ParentId="9083" CreationDate="2014-03-10T14:19:56.490" Score="2" Body="&lt;p&gt;Nitration occurs as electrophilic aromatic substitution and activity of substrate varies depending on exact structure of ring system and substitutes in the ring. In general electron acceptors (nitro-, acyl-, nitrile- and sulfogroups) deactivates the reaction, while electron donors (hydroxy-, aryl-, alkenyl- and to lesser extent alkylgroups) activates the reaction. As nitration occurs step by step, it will advance to dinitroarene only if mononitroarene is not deactivated too much by introduced nitrogroup, and same for other nitroarenes. However, different nitration agents have different nitration activity with 100% nitric acid in oleum being one of the strongest, usual nitration mixture (70% nitric acid + 98 % sulfuric acid) being strong, 70% nitric acid being moderate, diluted nitric acid being gentle and sodium nitrite in acidic solution being one of the gentlest. The picture is further complicated because strong nitration agents are also strong oxidizers, capable of breaking apart delicate molecules like phenols. Given that, nitration is often performed in several steps, with first 'gentle' nitration to introduce the first nitrogroup without breaking the molecule and then further nitration with stronger agents (see, for example, synthesis of styphnic acid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exhaustive review of the topic is beyond what a simple post can cover.&lt;/strong&gt; For further insight into the reaction I recommend to consult a decent university-grade organic chemistry textbook, like 'March's Advanced Organic Chemistry'.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="418" LastEditDate="2014-03-10T16:43:22.980" LastActivityDate="2014-03-10T16:43:22.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9089" PostTypeId="2" ParentId="9085" CreationDate="2014-03-10T16:15:05.510" Score="3" Body="&lt;p&gt;Alcohols overall are weak Brønsted acids, (water and ethanol pKa 15.7 and 16).  They are middling Lewis (hard) bases (HCl gas ionization in anhydrous ethanol solution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cactus.dixie.edu/smblack/chem2310/summary_pages/pKa_chart.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Relevant page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Al(O-&lt;em&gt;tert&lt;/em&gt;-Bu)${_3}$ and AlMe${_3}$ as simultaneous Brønsted bases and Lewis acids.  Cf: hard and soft acids and bases, solvo-cations and solvo-anions, and even as a redox reaction. It works solid state as intercalation, J. Chem. Ed. 81(6) 819 (2004).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-03-04T02:53:36.487" LastActivityDate="2018-03-04T02:53:36.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9090" PostTypeId="2" ParentId="9081" CreationDate="2014-03-10T17:08:20.517" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&quot; rel=&quot;nofollow&quot;&gt;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&lt;/a&gt;&lt;br&gt;&#xA;   Start here.  Now, the fun stuff off-syllabus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to create chiral centers - including planar chiral centers - whose handedness cannot be named, even in principle, e.g., $\ce{CR4}$ and $\ce{R2&amp;gt;C=C&amp;lt;R2}$. The pictured acyclic carbons are chiral centers.  Show their non-superposable enantiomers that illustrate the anomaly.  (Answer to be posted end of the week - though you already know it, but may not know you know it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optical rotation is not a measure of chirality, or even an indication of chirality. Solution optical rotations ignore atomic mass distribution[1].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-42${d}$, rotates 522°/mm along [100] at 497.4 nm[2].  ${P3_121}$ quartz and ${P3_221}$ berlinite, with opposite sense three-fold screw axes ($3_1$ right-handed, $3_2$ left-handed), are both levorotatory[3].  CIP notation flips with composition at constant geometry (L-cysteine versus most other L-amino acids).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.mazepath.com/uncleal/norbors.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.mazepath.com/uncleal/norbors.gif&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.mazepath.com/uncleal/norone.png&quot; rel=&quot;nofollow&quot;&gt;http://www.mazepath.com/uncleal/norone.png&lt;/a&gt; &lt;BR&gt;&#xA;[2] Appl. Cryst. 33, 126 (2000) &lt;BR&gt;&#xA;[3] J. Appl. Crystallogr. 19, 108 (1986) &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Math. Phys. 40, 4587 (1999) &lt;BR&gt;&#xA;&lt;a href=&quot;http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM&quot; rel=&quot;nofollow&quot;&gt;http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM&lt;/a&gt; &lt;BR&gt;&#xA;Quantitative geometric chirality can be calculated in any number of dimensions for any collection of a countable number of anonymous points with overall finite principle moments of inertia.  Mathematically &lt;em&gt;perfectly&lt;/em&gt; chiral arrays are rotationally symmetric but lacking all ${S_n}$ improper rotation axes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-10T17:08:20.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9092" PostTypeId="2" ParentId="1371" CreationDate="2014-03-10T17:22:17.440" Score="1" Body="&lt;p&gt;Carbon dioxide phase diagram,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9td5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is extremely important.  $Trillion cumulative carbon sequestration demands deep &quot;burial&quot; of captured power plant carbon dioxide.  All proposed reservoirs are above 31 C.  Above its critical temperature, CO2 is an astoundingly effective penetrant (right through thick solid concrete for a one-day accelerated cure).  You can pump it down there, but it will come back up.  (Also note, PV = energy, 101.325 J/liter-atm.  Capturing, compressing, and pumping emitted carbon dioxide must consume &lt;em&gt;at least&lt;/em&gt; 30% of a fossil fuel power plant's output.)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-10T17:22:17.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9093" PostTypeId="2" ParentId="9073" CreationDate="2014-03-10T17:36:49.900" Score="1" Body="&lt;p&gt;If you expose a stoichiometric mixture ((&quot;Brown's gas,&quot; or any mixture between LEL and UEL) of hydrogen and oxygen to a catalyst, $BOOM!!!$.  A static electricity spark or even a cosmic ray can do it. The Hindenberg was not a pre-mixed flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of mass-energy (First Law of Thermodynamics) is enforced by Noether's theorems and the homogeneity of time.  If you want to diddle that conservation, you must break the symmetry or falsify mathematics.  Good luck on locally doing either for any fuel cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physwiki.ucdavis.edu/Classical_Mechanics/Conservation_of_Mass_and_Energy/Noether&quot; rel=&quot;nofollow&quot;&gt;http://physwiki.ucdavis.edu/Classical_Mechanics/Conservation_of_Mass_and_Energy/Noether&lt;/a&gt;'s_Theorem_for_Energy &lt;BR&gt;&#xA;(copy and paste the entire link) &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.sjsu.edu/faculty/watkins/noetherth.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.sjsu.edu/faculty/watkins/noetherth.htm&lt;/a&gt; &lt;BR&gt;&#xA;  The mathematics &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.lhup.edu/~dsimanek/museum/unwork.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.lhup.edu/~dsimanek/museum/unwork.htm&lt;/a&gt; &lt;BR&gt;&#xA; Perpetual motion machines&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-10T17:36:49.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9095" PostTypeId="2" ParentId="7759" CreationDate="2014-03-10T20:55:21.127" Score="12" Body="&lt;h1&gt;Reaction conditions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As Watts &lt;em&gt;et al.&lt;/em&gt; have shown, the decomposition products of this Fenton-like reaction strongly depend on the $\mathrm{pH}$ of the solution.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If performed in acidic conditions, the reaction generates mostly hydroxy radicals, but no reductants (which would be the hydroperoxide and superoxide anions). If, conversely, the reaction is held in neutral conditions, Watts &lt;em&gt;et al.&lt;/em&gt; have shown that the reaction produces significantly more of the aforementioned reductants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do &lt;em&gt;et al.&lt;/em&gt; have conducted further research into the reaction mechanism at $\mathrm{pH}=7$ and concluded that the observed reaction order may be approximated as a pseudo 1&lt;sup&gt;st&lt;/sup&gt; order reaction, where the ratio $\ce{[H2O2]}/\ce{[#MnO2]}$ is vital to the description of the decomposition rate.&lt;sup&gt;2&lt;/sup&gt; They have further shown that reactive intermediates, such as superoxide and hydroperoxide anions are generated by the reaction. After slightly modifying the pH towards alkaline conditions, the production rates for the reactive anions increased drastically.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Reaction mechanism&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.chemosphere.2008.11.075&quot; rel=&quot;nofollow noreferrer&quot;&gt;Do &lt;em&gt;et al.&lt;/em&gt;&lt;/a&gt; have meticulously assembled a table with a proposed reaction pathway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R4p55.png&quot; alt=&quot;Proposed reaction mechanism for decomposition of hydrogen peroxide on manganese oxide.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, I quote one of their closing statements:&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The existence of hydroperoxide/superoxide anion implies that the suggested reaction mechanism could be explained by hydrogen peroxide being decomposed, not only directly on the surface of manganese oxide, but also through a propagation reaction involving intermediates such as hydroperoxide/superoxide anion in solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Watts, R., Sarasa, J., Loge, F., and Teel, A. &lt;strong&gt;Oxidative and Reductive Pathways in Manganese-Catalyzed Fenton’s Reactions.&lt;/strong&gt; &lt;em&gt;J. Environ. Eng.&lt;/em&gt;, 131(1), 2005, 158–164, &lt;a href=&quot;http://dx.doi.org/10.1061/(ASCE)0733-9372(2005)131:1(158)&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI link&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Si-Hyun Do, Bill Batchelor, Hong-Kyun Lee, Sung-Ho Kong. &lt;strong&gt;Hydrogen peroxide decomposition on manganese oxide (pyrolusite): Kinetics, intermediates, and mechanism&lt;/strong&gt;. &lt;em&gt;Chemosphere&lt;/em&gt;, Volume 75, Issue 1, March 2009, Pages 8-12, &lt;a href=&quot;http://dx.doi.org/10.1016/j.chemosphere.2008.11.075&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI link&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="5026" LastEditDate="2018-04-09T03:01:14.090" LastActivityDate="2018-04-09T03:01:14.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9096" PostTypeId="1" AcceptedAnswerId="9102" CreationDate="2014-03-10T20:57:04.857" Score="0" ViewCount="156" Body="&lt;p&gt;This seems like a stupid question, but how would I find the product of a single displacement reaction like the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Pb + Pb(NO3)2 -&amp;gt; \ ?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, a single displacement reaction follows the path of $\ce{A + BX -&amp;gt; AX + B}$. However, what will the equation look like when A and B are the same element?&lt;/p&gt;&#xA;" OwnerDisplayName="user4802" LastEditorUserId="3846" LastEditDate="2014-03-10T21:07:50.753" LastActivityDate="2014-03-10T22:53:15.617" Title="Single Displacement Reactions: A + AX =?" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9102" PostTypeId="2" ParentId="9096" CreationDate="2014-03-10T22:53:15.617" Score="3" Body="&lt;p&gt;Nothing. All you would have is just a mixture of white powder and lead metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, lead (II) nitrate is synthesized by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PbO (s) + 2 HNO3 (aq) -&amp;gt; Pb(NO3)2 (aq) + H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead (II) nitrate is soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PbCO3(s) + 2HNO3(aq) -&amp;gt; Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) A and B must either be different metals OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{2AgNO3(aq) + Zn(s) -&amp;gt; 2Ag(s) + Zn(NO3)2(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) A and B must be a halogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag (s) + Cu(NO3)2 (aq) -&amp;gt; } \text{No reaction}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{I2 + 2KBr -&amp;gt;} \text{No reaction} $$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-03-10T22:53:15.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9104" PostTypeId="2" ParentId="9067" CreationDate="2014-03-10T23:25:52.053" Score="3" Body="&lt;p&gt;Through an &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed100255g?prevSearch=%255BTitle%253A%2bbeverage%255D&amp;amp;searchHistoryKey=&quot; rel=&quot;nofollow&quot;&gt;interesting approach&lt;/a&gt;, authors estimate $\ce{CO2}$ pressure inside carbonated beverages, by measuring the freezing point ($fp$) depression caused by $\ce{CO2}$. In other words, when $\ce{CO2}$ is dissolved in water, a &#xA;solution is formed and the freezing point is lowered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molality of $m_{\ce{CO2}}$ in water can be obtained, according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_{\ce{CO2}}=\Delta{T_{fp}}/k_{fp}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $k_{fp}$ for water $=-1.86\,^{o}C \, kg \, mol^{-1}$. For a particular brand of sparkling water, a value of $\Delta{T_{fp}}$ = $-0.42^oC$ was measured, thus yielding a value for $m_{\ce{CO2}}$ equal to $0.23$ $mol \, kg^{-1}$. This first result allows to estimate the mass of ${\ce{CO2}}$ dissolved in one liter of that kind of beverage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said mass of ${\ce{CO2}}$ is approximately equal to $10\,g$ per liter (or $3.6\,g$ per can content volume).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Henry's law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_{\ce{CO2}}=m_{\ce{CO2}}/k_{H}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $k_{H}$ being Henry's law constant ($~0.077\,mol\,kg^{-1}\,atm^{-1}$ at $0^{o}C$), one finally gets the pressure $P_{\ce{CO2}}$ exerted by the gas over the liquid, at $0^{o}C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_{\ce{CO2}} = 3.0\,\text{atm}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that corresponds to a mass of about $0.1\,g$ of ${\ce{CO2}}$ inside the headspace volume of a can (about $15\,ml$), mass calculated via $PV=nRT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: for most carbonated beverages, $\Delta{T_{fp}}$ should fall at around $-0.2^oC$ (as reported by &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed071p903.2?prevSearch=%255BContrib%253A%2bbrooker%255D&amp;amp;searchHistoryKey=&quot; rel=&quot;nofollow&quot;&gt;Brooker&lt;/a&gt;): that would give a $P_{\ce{CO2}} = 1.5\,\text{atm}$ at $0^{o}C$, that rises at about $3\,\text{atm}$ at $25^{o}C$.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-11T08:16:08.127" LastActivityDate="2014-03-11T08:16:08.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9105" PostTypeId="1" CreationDate="2014-03-11T00:03:24.020" Score="1" ViewCount="50" Body="&lt;p&gt;Here is my second question, I know 1000 grams of water is 1 liter. But does the density affect the result? What if the substance is mercury?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add seed crystal in to a supersaturated solution, the solute comes out. Does this make it saturated? Is what came out the extra solute that make it supersaturated? Or is it all the solute? This is my first question.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="189" LastEditDate="2014-03-11T00:27:24.467" LastActivityDate="2014-03-11T00:33:06.533" Title="Two chemistry inquiries" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-08T23:11:38.957" ContentLicense="CC BY-SA 3.0" />
  <row Id="9106" PostTypeId="2" ParentId="9105" CreationDate="2014-03-11T00:25:47.913" Score="1" Body="&lt;p&gt;$1000 \, \text{g}$ of water occupy a volume of (approximately) $1 \, \text{L}$ because the density of water is (approximately) $1000 \, \text{g/L}$.&#xA;If you had $1000 \, \text{g}$ of mercury, whose density is $13534 \, \text{g/L}$, it would occupy a volume of $0.0739 \, \text{L}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually you can only dissolve so much of a solute into a solvent until a certain threshold is reached.&#xA;When this threshold is reached you have a saturated solution.&#xA;After that point, it is thermodynamically more favourable (under the given conditions, e.g. temperature, pressure, etc.) for any extra solute not to get dissolved in the solvent but clump together and precipitate.&#xA;But if you use some special techniques you can prepare a supersaturated solution that contains more solute than it should, thermodynamically speaking.&#xA;A supersaturated solution is a meta-stable state, meaning that it is only a local energy minimum for the system, not the global minimum.&#xA;But it would like to get to the global minimum if it gets the chance.&#xA;The only thing preventing that, is a small energy barrier that has to be surmounted.&#xA;So, if you disrupt the supersaturated solution in any way, like shaking it or adding a crystal seed, the solution can surmount the energy barrier and attain its global minimum state, i.e. the saturated state.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-03-11T00:33:06.533" LastActivityDate="2014-03-11T00:33:06.533" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="9107" PostTypeId="1" CreationDate="2014-03-11T01:54:36.830" Score="2" ViewCount="114" Body="&lt;p&gt;This is referring to the Cu-Sn phase diagram in which composition varies along a horizontal line at a certain temperature. I understand that as you move along that line different intermetallic or liquid phases form, but from a activity point of view does the activity go up because of different crystallographic arrangements that are present at different compositions?&lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="4505" LastEditDate="2014-03-11T06:49:49.080" LastActivityDate="2014-08-28T01:32:29.240" Title="How does the activity of Sn chemical potential vary as a function of composition at 900 C?" Tags="&lt;structural-formula&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9108" PostTypeId="1" AcceptedAnswerId="9110" CreationDate="2014-03-11T02:37:39.687" Score="9" ViewCount="12264" Body="&lt;p&gt;According to my textbook (and intuitively) certain changes when the aforementioned 3 variables are altered occur in accordance with Le Chatelier's Principle. However, what I don't understand is what makes temperature change alter the equilibrium constant while the other two do not. All three cause changes in the equilibrium position (the state of concentrations at which equilibrium is achieved with lowest possible stress in the system), so why don't they all also change $K_{eq}$?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="189" LastEditDate="2014-03-11T04:18:36.383" LastActivityDate="2014-12-18T16:41:26.973" Title="Why does the equilibrium constant depend on the temperature but not on pressure and concentration?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9110" PostTypeId="2" ParentId="9108" CreationDate="2014-03-11T03:32:07.460" Score="15" Body="&lt;p&gt;Please have a look at &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/189&quot;&gt;this answer&lt;/a&gt; of mine. It contains the derivation of the defining formula of the equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula shows the temperature dependence explicitely, so it is clear that $K$ must depend on $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the temperature this formula also relates $K$ to the standard Gibbs free energy of reaction $\Delta G^{0}$.&#xA;This quantity is &lt;strong&gt;defined&lt;/strong&gt; for specific standard conditions involving standard pressure and equilibrium concentrations.&#xA;Since there aren't any other quantities contained in the formula (apart from the constant $R$) it follows that $K$ is &lt;strong&gt;defined&lt;/strong&gt; for this specific pressure and these specific concentrations too and so it has no pressure or concentration dependence. &lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-03-11T03:52:36.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9111" PostTypeId="1" AcceptedAnswerId="9116" CreationDate="2014-03-11T06:25:08.823" Score="1" ViewCount="384" Body="&lt;p&gt;$n$ liters of $k$ M $\ce{NaCl}$ are poured into $m$ liters of $l$ M $\ce{NaCl}$. Determine $\ce{[NaCl]}$, $\ce{[Na^+]}$, and $\ce{[Cl^{-}]}$.&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorDisplayName="user4076" LastEditDate="2014-03-16T05:03:48.973" LastActivityDate="2014-03-16T05:03:48.973" Title="Calculating concentration of an ionic compound" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9112" PostTypeId="1" CreationDate="2014-03-11T06:35:33.447" Score="1" ViewCount="40" Body="&lt;p&gt;Battery leaks was an issue in the 80s and 90s, and since quality and innovation is constantly improving, I thought battery leak might be slowly going away. But I found that even nowadays, name brand alkaline batteries still leak after a while, even if originally concealed in a tight package of the metal shell. Is it true that the chemical in an alkaline battery inherently expands in volume over time, and make the alkaline battery leak issue unavoidable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also why does the alkaline expand in volume over time?&lt;/p&gt;&#xA;" OwnerUserId="4676" LastActivityDate="2014-03-11T07:52:28.883" Title="why do battery leaks still happen" Tags="&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2014-03-22T03:27:51.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="9113" PostTypeId="2" ParentId="9111" CreationDate="2014-03-11T07:00:06.103" Score="0" Body="&lt;p&gt;Well in my opinion you do need the dissociatiion constant for NaCl&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                         `NaCl          -&amp;gt;Na    +        Cl` &#xA;&#xA;initial conc.             k*n+m*l          0              0 &#xA;change at equilibrium     -a(k*n+m*l)    +a(k*n+m*l)     +a(k*n+m*l) &#xA;final conc.               k*n+m*l(1-a)   k*n+m*l(1+a)    k*n+m*l(1+a)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where a is the degree of dissociation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kb=  2*(k*n+m*l)(1+a)/k*n+m*l(1-a)=   2*(1+a)/(1-a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where kb is the dissociation constant for nacl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so when you have the &quot;kb&quot; the you can find &quot;a&quot; and then put it in the formula under Na or Cl at final conc. to get their concentration&lt;/p&gt;&#xA;" OwnerUserId="4676" LastActivityDate="2014-03-11T07:00:06.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9114" PostTypeId="2" ParentId="985" CreationDate="2014-03-11T07:52:28.883" Score="10" Body="&lt;p&gt;The reason for leaks in alkaline batteries is that as batteries discharge — either through usage or gradual self-discharge — the chemistry of the cells changes and some hydrogen gas is generated. This out-gassing increases pressure in the battery. Eventually, the excess pressure either ruptures the insulating seals at the end of the battery, or the outer metal canister, or both. In addition, as the battery ages, its steel outer canister may gradually corrode (in alkaline batteries there is some potassium hydroxide, a very strong base) or rust, which can further contribute to containment failure.&#xA;(see &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_battery#Leaks&quot;&gt;http://en.wikipedia.org/wiki/Alkaline_battery#Leaks&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this online : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat is the biggest culprit. Storing your flashlights under the seat of your car, or in direct sunlight will cause the batteries to leak. The reason they leak is because they are filled with a chemical paste of which the liquid portion is composed of a base called Potassium Hydroxide. When the batteries get hot, the paste inside expands, and forces chemicals out of the seals in the bottom and top of the battery. &#xA;[...]&#xA;Regardless of the brand you buy,[...] all batteries will leak eventually, so if you don't use the flashlight for extended periods of 3 months or more, remove the batteries, or change them out every couple months.&#xA;(from &lt;a href=&quot;http://answers.yahoo.com/question/index?qid=20071228083136AA0lDrH&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-03-11T07:52:28.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9116" PostTypeId="2" ParentId="9111" CreationDate="2014-03-11T09:35:33.207" Score="1" Body="&lt;p&gt;Keep it as simple as possible :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both initial solutions are not saturated and all the $\ce{NaCl}$ is dissolved, this will not change upon mixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If in doubt with the result of a calculation, have a look at the units!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentrations are given as the quotient of moles and volume, $c = \frac{n}{V}$, the unit apparently is $\mathrm{mol \cdot l^{-1}}$. The absolute amount of $\ce{NaCl}$ in the first solution is therefore given as $n = c\cdot V$. (Note that this is not the n in your question ;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new $\ce{NaCl}$ concentration is the sum of the moles, divided by the new total volume:&#xA;$c = \frac{(k\cdot n)+(l\cdot m)}{m+n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless given further information, it is safe to assume that $\ce{NaCl}$ still is fully dissociated after mixing.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-11T09:35:33.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9117" PostTypeId="1" CreationDate="2014-03-11T13:32:41.817" Score="3" ViewCount="3994" Body="&lt;p&gt;&quot;The Lanthanide Contraction refers to the fact that the 5s and 5p orbitals penetrate the 4f sub-shell so the 4f orbital is not shielded from the increasing nuclear change.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, 5s and 5p are filled before 4f, so if the atom wants to give an electron out, which energy shell does the electron come out of? How 5s and 5p penetrate 4f? Isn't 4f outer than 5s and 5p?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2014-03-11T17:55:57.690" Title="lanthanide contraction" Tags="&lt;electronic-configuration&gt;&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9118" PostTypeId="1" CreationDate="2014-03-11T14:51:24.810" Score="6" ViewCount="3009" Body="&lt;p&gt;We know that if a system's equilibrium is exposed to a stress, the system shifts to relieve that stress. According to my teacher, certain kinds of pressures are stresses and others are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, if we just say that the total pressure of a container with a reaction occuring inside is increased, the reaction shifts toward the side with a small sum of the mole ratios (i.e. $\rm 2A +3B \longleftrightarrow C + D$, if increasing pressure reaction shifts to the right to decrease pressure). I understood this formerly, but another case erases this understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher also said that if we add a noble gas to the container in which this reaction is occurring, although the pressure increases because of a new gas, the equilibrium does not shift to counteract this pressure increase. She says it has to do with the partial pressures of the reactants and products of only the particles &lt;em&gt;In the reaction,&lt;/em&gt; but not in the total balloon. But this makes no sense to me because by increasing the total pressure of the system as described before, how does this cause an equilibrial stress and this case doesn't? How does increasing total pressure even cause an equilibrium shift? By increasing total pressure we have a proportional increase in the collision rates of all particles involved, do we not? If this is true, then why wouldn't the reaction rates both be doubled, and hence cause the equilibrium to not shift?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm missing some fundamental principle, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="4589" LastEditDate="2014-03-11T15:47:15.307" LastActivityDate="2014-03-14T00:20:10.937" Title="Pressure and LeChatelier's Principle" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9119" PostTypeId="2" ParentId="9117" CreationDate="2014-03-11T15:29:16.573" Score="2" Body="&lt;p&gt;According to Aufbau principle, for multi-electron atoms, in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For multi-electron atoms, the following sequence is observed for orbitals in the increasing order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the numbers 1, 2, 3, 4.. indicates shell number. It may be remembered that, though filling of electrons takes place according energy of orbitals, when electrons are removed from the neutral atom, electrons with highest value of n are removed first. &lt;em&gt;So, electrons from 5p or 5s is removed first than from 4f.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand lantanide contraction, lets consider an example. If you consider Pd (of 5th period) and Pt (of 6th period), you would expect Pt to have relatively large radii than Pd. In practice, they have roughly same radii. This is because, Pd has 4d electrons and Pt has 5d and 4f electrons. &lt;em&gt;Due to diffused shaped of 4f electrons, they can't shield 5d electrons of Pt.&lt;/em&gt; Thus, contracting the size of Pt roughly same as Pd.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Considering your last question, no 4f subshell lies closer to the nuclei than  5s and 5p.&lt;/em&gt; For more info, I would suggest you to read this page: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Transition_Metals/The_Lanthanides/Lanthanide_Contraction&quot; rel=&quot;nofollow&quot;&gt;Lanthanide Contraction&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2014-03-11T17:55:57.690" LastActivityDate="2014-03-11T17:55:57.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9120" PostTypeId="2" ParentId="9118" CreationDate="2014-03-11T16:03:23.367" Score="4" Body="&lt;p&gt;Le Chatelier's principle is not about &lt;em&gt;pressure&lt;/em&gt;, it is about &lt;em&gt;concentrations&lt;/em&gt; and &lt;em&gt;temperature&lt;/em&gt; (Please also see the good article on Wikipedia for &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle#Effect_of_change_in_concentration&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier's principle&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle is also called &lt;strong&gt;&quot;The Equilibrium Law&quot;&lt;/strong&gt;, that can be used to predict the effect of a change in conditions on a &lt;em&gt;chemical equilibrium&lt;/em&gt;. &lt;em&gt;Chemical equilibrium&lt;/em&gt; means that reaction is reversible: it can go as forward, so backwards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the &quot;right&quot; - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;&#xA;I'll make a silly comparison. Let's take this reaction:&#xA;$$\rm Boy+Girl\longleftrightarrow Pair$$&#xA;If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...&#xA;&lt;/sup&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Our system is in equilibrium, so we can calculate &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_constant&quot; rel=&quot;nofollow&quot;&gt;Equilibrium constant&lt;/a&gt;: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$&#xA;$$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$&#xA; where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\upsilon'(A)=2\,mol-2x\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(B)=3\,mol-3x\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(C)=1\,mol+x\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(D)=1\,mol+x\,mol$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$X'(A)=(2-2x)/(7-3x)$&lt;/li&gt;&#xA;&lt;li&gt;$X'(B)=(3-3x)/(7-3x)$&lt;/li&gt;&#xA;&lt;li&gt;$X'(C)=(1+x)/(7-3x)$&lt;/li&gt;&#xA;&lt;li&gt;$X'(D)=(1+x)/(7-3x)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving the eqution gives $x=0.325\,mol$, so now:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\upsilon'(A)=1.35\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(B)=2.03\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(C)=1.33\,mol$&lt;/li&gt;&#xA;&lt;li&gt;$\upsilon'(D)=1.33\,mol$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P'(A)=0.45\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P'(B)=0.67\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P'(C)=0.44\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P'(D)=0.44\,atm$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok, but when we add inert gas, what happens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction:&#xA;$$\rm A_g+B_g\longleftrightarrow C_g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure. &#xA;$$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial pressures are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we will add 1 mol of inert gas:&#xA;$$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$&lt;/li&gt;&#xA;&lt;li&gt;$P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;&#xA;Let's take this reaction:&#xA;$$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$&#xA;You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.&#xA;&lt;/sup&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will not do practically anything, because concentrations of reagents will not increase under the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as &quot;product&quot;. If you remove this so called &quot;product&quot; (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we increase temperature, it is like we added more of this so called &quot;product&quot; (heat) in the jar, so reaction is moved to the left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H&amp;gt;0$, you need to think about heat not as &quot;product&quot;, but as &quot;reagent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For better understanding try to &quot;dive&quot; into &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;Equilibrium thermodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-03-12T10:01:36.683" LastActivityDate="2014-03-12T10:01:36.683" CommentCount="7" CommunityOwnedDate="2014-03-12T10:01:36.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="9123" PostTypeId="1" AcceptedAnswerId="9129" CreationDate="2014-03-12T04:07:38.477" Score="1" ViewCount="6410" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which two substances react spontaneously?&lt;/p&gt;&#xA;&lt;p&gt;(A) &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(B) &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(C) &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(D) &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\tiny{\text{Taken from 2012 Chem Olympiad}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The given answer is B, yet I am unclear as to why. I believe it has something to do with reduction potentials/redox or the gibbs free energy. Can someone please explain?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-12T07:02:13.190" Title="Which two substances react spontaneously?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9129" PostTypeId="2" ParentId="9123" CreationDate="2014-03-12T07:02:13.190" Score="2" Body="&lt;p&gt;From the faint outline of electrochemical series I remember $Cu$ is above $Ag$. The substance which is above is able to displace the substance below from its aqueous solution(obviously lower substance is in ionic form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $(B)$ is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were made to learn this outline in lower class :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K Na Ca Mg Al Zn Fe Pb H Cu Hg Ag Au Pt&lt;/p&gt;&#xA;" OwnerUserId="4597" LastActivityDate="2014-03-12T07:02:13.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9130" PostTypeId="1" CreationDate="2014-03-12T07:19:53.330" Score="6" ViewCount="912" Body="&lt;p&gt;To obtain phenylamine after it is formed from the reduction of nitrobenzene with tin, we carried out steam distillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I learned in class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;During the process,steam is injected into the flask (containing the reaction mixture) to create artificial vapour pressure on the surface of the liquid. When vapour pressure is equal to atmospheric pressure, the liquid starts to boil at 95 degrees&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This didn't make much sense to me. So I searched it up on the net and I  got another explanation (&lt;a href=&quot;http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html#top&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the hot steam passes through the mixture it condenses, releasing heat. This will be enough to boil the mixture of water and phenylamine at 98°C provided the volume of the mixture isn't too great&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, it is saying that the heat produced from the condensation of the steam will allow the mixture to boil. I am really confused, I mean I carried out steam distillation but how is it different from normal distillation and how is phenylamine collected at a lower temperature through this process?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="1431" LastEditDate="2014-03-12T10:06:24.843" LastActivityDate="2016-08-27T19:28:08.860" Title="Steam distillation- how it actually works?" Tags="&lt;aromatic-compounds&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9132" PostTypeId="2" ParentId="7444" CreationDate="2014-03-12T09:04:26.130" Score="3" Body="&lt;p&gt;The diazo dye (&lt;strong&gt;1&lt;/strong&gt;), known as &lt;strong&gt;Solvent Yellow 124&lt;/strong&gt; or &lt;strong&gt;Somalia Yellow&lt;/strong&gt; is soluble in aliphatic hydrocarbons. In the European Community, it is used to mark low-taxed diesel fuel (for heating purposes). In order to prevent its use in cars, samples are acidified. This results in acetal cleavage and protonation of the diazo-fragment, yielding the red, water-soluble iminoquinone species (&lt;strong&gt;2&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ovt6A.png&quot; alt=&quot;solvent-yellow-124 hydrolysis and protonation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is probably not exactly what you wanted, what about dyes that exist in a neutral and a zwitterionic form, depending on the polarity of the solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IOetK.png&quot; alt=&quot;brooker&amp;#39; merocyanine: neutral and zwitterionic form&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Brooker's merocyanine&lt;/strong&gt; is one of the most prominent examples and show strong &lt;strong&gt;solvatochromism&lt;/strong&gt;, i.e. the colour changes with the polarity of the solvent&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-12T09:04:26.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9136" PostTypeId="1" CreationDate="2014-03-12T09:41:13.373" Score="1" ViewCount="90" Body="&lt;p&gt;Is there a way (other than collecting water and heating it on the stove ) to get hot/warm water from the tap without having to alter existing plumbing or electrical wiring ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps use a chemical reaction to provide the heat to warm the water ?&lt;/p&gt;&#xA;" OwnerUserId="4817" LastEditorUserId="4945" LastEditDate="2015-10-23T17:03:13.043" LastActivityDate="2015-10-23T17:03:13.043" Title="Is there a way to get hot water without altering plumbing or electrical wiring too much?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9137" PostTypeId="1" CreationDate="2014-03-12T10:11:26.477" Score="0" ViewCount="221" Body="&lt;p&gt;Something that is not reactive to lesser wavelengths of light, specifically IR.&lt;/p&gt;&#xA;" OwnerUserId="4818" LastActivityDate="2014-03-13T09:36:36.147" Title="What are some examples of a clear, liquid, UV photoactive polymer?" Tags="&lt;physical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9138" PostTypeId="1" AcceptedAnswerId="9162" CreationDate="2014-03-12T14:37:55.003" Score="1" ViewCount="190" Body="&lt;p&gt;:) Without being a chemist myself, I need some basic understanding of the ion activities in my buffer solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am wondering whether the ion activity of the K$^+$ and Cl$^-$ ions is different for the buffers with different pH-values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I google for that, all I get is ion activity for the hydrogen ion, which -I think- does not interest me at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ion activity actually IS different, could you point me to some table to look up the correction values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: tl&amp;amp;dr:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buffer solutions contain:&lt;br&gt;&#xA;Buffer A: 250mM KCl, 10mM MOPS&lt;br&gt;&#xA;Buffer B: 020mM KCl, 10mM MOPS  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH is set to 5,6,7 resp., using unknown amounts of Tris.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the ion activity of K$^+$ and Cl$^-$ affected by the pH?&lt;/p&gt;&#xA;" OwnerUserId="4820" LastEditorUserId="4820" LastEditDate="2014-03-12T18:44:51.593" LastActivityDate="2014-03-12T21:50:04.553" Title="Is ion activity in a salt solution pH-dependent?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9139" PostTypeId="1" AcceptedAnswerId="9141" CreationDate="2014-03-12T14:48:15.017" Score="2" ViewCount="1900" Body="&lt;p&gt;If the volume of each sample of gas was reduced to one-tenth of its original size and the temperature remained the same, which gas would have the lower pressure. I'm getting confused because my options are &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H4}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H4}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2}$&lt;/span&gt;. Our teacher told us to pick the polar compound but the problem is all of them are non polar. Would the answer be &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H4}$&lt;/span&gt;? I have a feeling I might be drawing the lewis structure wrong because I'm positive the other four are non polar.&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorDisplayName="user7951" LastEditDate="2020-04-24T18:19:02.107" LastActivityDate="2020-04-24T18:19:02.107" Title="How to determine which gas would have the lowest pressure?" Tags="&lt;molecules&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9141" PostTypeId="2" ParentId="9139" CreationDate="2014-03-12T15:27:47.150" Score="2" Body="&lt;p&gt;Not all of them are polar. The gases  $\rm N_2$, $\rm O_2$,  $\rm Cl_2$ are not polar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\rm HC\equiv C-CH_3$ is polar, but $\rm H_2C=C=CH_2$ should not be and $\rm C_6H_4$ is also polar, if it is right chemical formula you wrote. But I'm not sure which is more polar, but I will say that it is $\rm C_6H_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qfZU6.png&quot; alt=&quot;enter image description here&quot;&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more polar compound there will more interaction between molecules and that is why pressure will be smaller, simply because they &quot;want&quot; to be liquid form - not in gas.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="4589" LastEditDate="2014-03-12T19:20:31.810" LastActivityDate="2014-03-12T19:20:31.810" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9142" PostTypeId="2" ParentId="9136" CreationDate="2014-03-12T16:01:00.467" Score="2" Body="&lt;p&gt;Crystallization of a supersaturated solution of sodium acetate will release heat - &lt;strong&gt;once&lt;/strong&gt;. But then it is just a solid. It works nicely for pocket warmers, but since you have to heat the flexible container in boiling water prior to re-use, this and other comparable processes are of little use for a water supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, you won't get around plumbing to tap an energy source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your options would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solarthermal&lt;/li&gt;&#xA;&lt;li&gt;geothermal&lt;/li&gt;&#xA;&lt;li&gt;heat released from the fermentation of organic waste&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-12T16:01:00.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9143" PostTypeId="1" AcceptedAnswerId="9156" CreationDate="2014-03-12T16:12:04.300" Score="10" ViewCount="1998" Body="&lt;p&gt;I'm a complete non-chemist, so I don't know how to calculate this,  or even what to google for (my searches yielded esoteric research papers). &#xA;I'm doing some hobby experiments, and I'd  like to know what approximate timeframe to expect for relatively pure (distilled or double distilled) water left in open air at around 24 °C/75F to reach whatever the pH equilibrium point is (I read around 5.5 somewhere). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for a giant equation please,  just a quick approximate number.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4822" LastEditorUserId="418" LastEditDate="2014-03-12T16:37:58.563" LastActivityDate="2014-03-13T21:29:24.943" Title="Approximate time for pure water pH to equilibrate in open air?" Tags="&lt;thermodynamics&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9145" PostTypeId="1" CreationDate="2014-03-12T16:31:34.897" Score="3" ViewCount="22823" Body="&lt;p&gt;What is the ground state electron configuration of chromium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it $\ce{[Ar]}4s^23d^4$ or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it $\ce{[Ar]}4s^13d^5$&lt;/p&gt;&#xA;" OwnerUserId="4823" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T03:58:58.650" LastActivityDate="2015-06-30T16:08:24.887" Title="Ground state electron configuration of chromium" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;electronic-configuration&gt;&lt;electrons&gt;&lt;elements&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-06-30T19:52:03.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="9146" PostTypeId="1" CreationDate="2014-03-12T18:12:53.697" Score="3" ViewCount="35939" Body="&lt;p&gt;How is the wavelength chosen to measure the concentration of an ion determined? and What does the wavelength correspond to in an atom/ion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When light is shined on a solution what happens to the ions in solution? Include how and why light is absorbed or emitted from a substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a high or low concentration absorb more light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the beer-lambert law relate to concentration and information from the spectrophotometer? &lt;/p&gt;&#xA;" OwnerUserId="4825" LastEditorUserId="3879" LastEditDate="2014-03-12T22:07:34.910" LastActivityDate="2014-03-12T22:07:34.910" Title="How to chose wavelength in spectrophotometry? How to use Beer-Lambert Law?" Tags="&lt;ions&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9148" PostTypeId="2" ParentId="9139" CreationDate="2014-03-12T19:24:37.510" Score="-1" Body="&lt;p&gt;A homodiatomic gas cannot be polar, by symmetry arguments. At best it can be polarizable by dispersion forces overall.  The lowest pressure would be the most strongly attracting molecules if any (e.g., water and hydrogen bonding), the most inert molecules, the smallest molecules (van der Walls equation coefficients), and then &quot;anomalous&quot; behaviors.  Look up the coefficients (&lt;em&gt;CRC Handbook&lt;/em&gt;)to get the real world answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would guess nitrogen for its being very tightly bound and all electrons bonded.  Look it up, as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;/_\PV = energy, 101.325 J/liter-atm. Adiabatically compressing a gas tenfold will heat it.  A diesel cylinder compresses 15 - 22 to one, minimum 550 C at the top of the stroke.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T19:24:37.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9151" PostTypeId="2" ParentId="9146" CreationDate="2014-03-12T20:01:11.187" Score="4" Body="&lt;p&gt;Working wavelength is chosen by analysing the spectrogram $A(\lambda)$. Such diagram is obtained by measuring absorbance $A$ in function of wavelength $\lambda$. Practically, you test a moderate concentration of your analyte and sweep the complete wavelength bandwidth of your analyser. Then you select wavelength where a maximum of absorbance occurs. If multiple maximum are present, you may prefer the highest one (best sensitivity) or a smaller one with less interferences (best selectivity). Both can happens simultaneously. The main criterion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \underset{\lambda}{\max}\left\{A(\lambda)\right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For information: the wavelength is linked to photon energy $E$ by the following relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = h \nu = h \frac{c}{\lambda} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\nu$ is the frequency of the EM wave, $c$ the speed of light and $h$ the Plank constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absorbance is a property of matter, when an atom/ion/molecule absorbs a photon (visible or nearly visible IR/UV), it can promote electron to a highest energy level (UV, highest energy) or increase vibrational energy (IR, lowest energy) of the compound and the photon is destroyed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spectrophotometer assess the transmittance $T$ (ratio of transmitted $\phi$ and incident flux $\phi_0$ expressed as powers) of light beam through a sample cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \frac{\phi}{\phi_0} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And transmittance (dimensionless) is linked to absorbance (dimensionless) in the following way (functional scaling to achieve linearisation between signal and physical quantity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = -\log_{10}(T)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beer-Lambert Law is a limit law conditioned by several hypothesis and it is only valid for low concentrations. When Beer-Lambert applies you can linearly link concentration $x$ with absorbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(x) = \epsilon(\lambda)lx $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\epsilon$ is a proportional coefficient (with suitable units) called &lt;em&gt;extinction coefficient&lt;/em&gt; which describes how matter absorbs light at wavelength $\lambda$. The quantity $l$ stands for the optical path length through the sample cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerous substances validate Beer-Lambert Law but some does not. Therefore you must assess linearity before quantifying unknown concentrations. This is usually done when performing calibration curve. Then you know your dynamic range and sensitivity of your analytical method and you can regress concentration from absorbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you to read more about those concepts in a reference book such as &lt;strong&gt;&lt;em&gt;Skoog &amp;amp; West&lt;/strong&gt;, Fundamentals of Analytical Chemistry&lt;/em&gt; or &lt;strong&gt;&lt;em&gt;Vogels&lt;/strong&gt;, Textbook of Quantitative Chemical Analysis&lt;/em&gt;. There is also a free electronic book available on internet called &lt;a href=&quot;http://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Welcome.html&quot; rel=&quot;nofollow&quot;&gt;Analytical Chemistry 2.0&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-03-12T21:43:50.453" LastActivityDate="2014-03-12T21:43:50.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9152" PostTypeId="2" ParentId="9137" CreationDate="2014-03-12T20:02:18.840" Score="0" Body="&lt;p&gt;You mean &quot;monomer&quot; or &quot;oligomer&quot; not &quot;polymer.&quot;  You are making the polymer by polymerization or crosslinking.  UV-curing optical adhesives.  You choose nice (mix of) free radical-cure monomers, then dissolve a photoinitiator like camphorquinone, a diarylketone, a benzoin ether, or a benzil ketal.  The molecule fragments to free radicals after absorbing a UV photon. That sets off the polymerization.  Exclude oxygen (triplet diradical quench).  Cage recombination of the intiator is your enemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A microwave-pumped electrodeless mercury bulb the size of your finger can output a kilowatt of UV.  Beware of pinholes in your irradiation chamber, including entry and exit if continuous.  And now, a second of silence for the Trommsdorff effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a polymer as such to do be photoactive, then something like Prog. Polym. Sci. 28 729 (2003).  Pull it up in a university library to get the graphics.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T20:02:18.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9153" PostTypeId="2" ParentId="307" CreationDate="2014-03-12T20:19:02.757" Score="2" Body="&lt;p&gt;&quot;High-temperature phase transformation and low friction behaviour in highly disordered turbostratic graphite,&quot; doi:10.1088/0022-3727/46/39/395305&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Phase Transformation Mechanism of Graphite-Turbostratic Graphite in the Course of Mechanical Grinding,&quot; CHEM. RES. CHINESE U. 19(2), 216 (2003)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.onxlti.com/product-divisions/contract-manufacturing-products/on-x-pyrolytic-carbon/&quot; rel=&quot;nofollow&quot;&gt;http://www.onxlti.com/product-divisions/contract-manufacturing-products/on-x-pyrolytic-carbon/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T20:19:02.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9154" PostTypeId="2" ParentId="9130" CreationDate="2014-03-12T20:28:08.867" Score="2" Body="&lt;p&gt;You boil the water and the product co-distills.  There is an anomalous selective interaction between steam and many hydrophobes.  If you heat the pot you deplete the water contained and may pyrolyze contents on the hot wall.  Blowing in steam limits temp to near 100 C, moves a lot of energy into the flask and out (540 cal/g enthalpy of vaporization), and maintains the water content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could filter off the solids, acidify the aqueous raffinate,  extract non-basics, then alkalinize and extract the aniline.  Since aniline is air sensitive to oxidation and extremely toxic by skin contact, you can appreciate advantages of steam distillation.   &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T20:28:08.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9155" PostTypeId="2" ParentId="741" CreationDate="2014-03-12T20:35:51.843" Score="1" Body="&lt;p&gt;A thermal plasma has its electrons and ions at the same temperature.  A raw plasma has hot electrons and cold ions from the vast difference in masses when microwave excited.  One way to thermalize is to run the plasma through a Helmholtz pair of coils with a strong uniform magnetic field parallel to plasma flow.  The low mass electrons tightly spiral around the parallel lines of force (that, of course, are not there), colliding with the nuclei to thermal equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(That can sound a lot better than it works.)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T20:35:51.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9156" PostTypeId="2" ParentId="9143" CreationDate="2014-03-12T20:39:28.937" Score="8" Body="&lt;p&gt;Assume you have poured your distilled water in a 1 liter container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We write the flux of $n_{\ce{CO2}}$ between air towards your container, as &#xA;$$&#xA;J_{\ce{CO2}}=\frac{dn_{\ce{CO2}}}{dt}\times\frac{1}{A}=\frac{d[\ce{H2CO3}]^{*}}{dt}\times\frac{V}{A}&#xA;$$&#xA;where $n_{\ce{CO2}}$ is the number of ${\ce{CO2}}$ moles in water, $A$ the water surface and $V$ the water volume.&#xA;Assuming $\ce{CO2}$ atmospheric pressure equal to $10^{-3.5}\,atm$, we obtain the concentration $[\ce{CO2}]_{air}$ approximately equal to $10^{-5}\,M$ (the latter value obtained via $PV=nRT$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call $[\ce{H2CO3}]^{*}$ the sum of $[{\ce{CO2}}]_{water}$ and $[\ce{H2CO3}]_{water}$ concentrations in water: at the beginning, we set it equal to $0$. Note that $[\ce{H2CO3}]^{*}$ could be reasonably written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\ce{H2CO3}]^{*}\approx[{\ce{CO2}}]_{water}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;indicating that the majority of carbon dioxide in water is not converted into carbonic acid, being the kinetics for such a conversion very slow, as pointed out by @Nicolau Saker Neto, .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By expressing $J_{\ce{CO2}}$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;J_{\ce{CO2}}=\frac{D_{{\ce{CO2}}}}{Z}\times ([\ce{CO2}]_{air}-[\ce{H2CO3}]^{*})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $D_{\ce{CO2}}$ the diffusion coefficient of ${\ce{CO2}}$ equal to $7.2\times10^{-4}\,dm^{2}h^{-1}$and $Z$ the thickness of superficial layer through which the exchange occurs. For the latter, we take $40\times10^{-5}\,dm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting previous equations together (with $A/V=1\,dm^{-1}$), the following is obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d[\ce{H2CO3}]^{*}}{dt}=1.8\times10^{-5}-1.8\times[\ce{H2CO3}]^{*}&#xA;$$&#xA;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\ce{H2CO3}]^{*}=1\times10^{-5}(1-e^{-1.8t})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph below shows that about $2$ hours later, a plateau is established, for $[\ce{H2CO3}]^{*}= 1\times10^{-5} M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By considering the acid-base equilibrium ($pK_{a}=6.3$):&#xA;$$&#xA;\text{H}_{2}\text{C}\text{O}_{3}^{*} \ce{&amp;lt;=&amp;gt; HCO3- + H+}&#xA;$$&#xA;we obtain a $\text{pH}=5.7$, at the plateau.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/94ZxK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;strong&gt;UPDATE 1&lt;/strong&gt;: the value for $Z$ has of course a big impact on time. A larger value will increase the time needed to reach the plateau. I have used $40\times10^{-5}\,dm$, considered a &quot;typical value&quot; by the authors of &lt;a href=&quot;http://www.dunod.com/sciences-techniques/sciences-fondamentales/chimie-et-genie-chimique/master-et-doctorat-capes-agreg/chimie-des-milieux-aq&quot; rel=&quot;nofollow noreferrer&quot;&gt;this book&lt;/a&gt; (sorry in French, page 181, 3rd edition). Any more accurately derived value for $Z$ is welcome !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE 2&lt;/strong&gt;: I used the notation $\text{H}_{2}\text{C}\text{O}_{3}^{*}$ to represent the two species $\ce{CO2}$ and $\ce{H2CO3}$, in water.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-13T21:29:24.943" LastActivityDate="2014-03-13T21:29:24.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9157" PostTypeId="2" ParentId="2560" CreationDate="2014-03-12T20:57:51.883" Score="1" Body="&lt;p&gt;Palin, D. E.; Powell, H. M. &quot;Hydrogen Bond Linking of Quinol Molecules,&quot; Nature  156, 334 (1946), and DOI: 10.1039/DF9694800181&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White aqueous starch/iod$ide$ tests for oxidizing agents by turning brown, liberated iodine and ionic oligomers intercalated within amylose starch (linear).  Amylopetcin (branched) gives red.  The mix is purple-brown.  Purple-brown aqueous starch/iod$ine$ tests for reducing agents by the reverse.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T20:57:51.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9158" PostTypeId="2" ParentId="7574" CreationDate="2014-03-12T21:05:01.860" Score="5" Body="&lt;p&gt;Red iron oxide, Fe2O3, from a paint or art supply store, or as polishing rouge (no additives).  dissolve in aqueous acetic acid, filter off any solids, and go on from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolving iron in acetic acid will give you Fe(II)acetate.  Air or peroxide oxidation may (probably will) give you Fenton chemistry and muck up your organics.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T21:05:01.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9159" PostTypeId="2" ParentId="7902" CreationDate="2014-03-12T21:21:23.463" Score="1" Body="&lt;p&gt;Aromaticity might be viewed as resonance stabilzation energy, NMR shift, or altered reactivity.  Here is a way of measuring its change among related systems,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chem. Commun.  48, 8144 (2012) &lt;BR&gt;&#xA;J. Am. Chem. Soc. 131, 189 (2009) &lt;BR&gt;&#xA;J. Am. Chem. Soc. 127, 16251 (2005) &lt;BR&gt;&#xA;J. Org. Chem. 67, 8798 (2002) &lt;BR&gt;&#xA;J. Am. Chem. Soc., 124, 13495 (2002) &lt;BR&gt;&#xA;Chem. Rev. 101, 1301 (2001) &lt;BR&gt;&#xA;Tetrahedron 57, 3689 (2001)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nptel.ac.in/courses/IIT-MADRAS/Engineering_Chemistry_III/pdf/1_Aromaticity.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nptel.ac.in/courses/IIT-MADRAS/Engineering_Chemistry_III/pdf/1_Aromaticity.pdf&lt;/a&gt; &lt;BR&gt;&#xA;A more practical approach derives from MO theory and Hund's rule of maximum multiplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=U6FX5RfLjd0&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=U6FX5RfLjd0&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.youtube.com/watch?v=H0so_NJ2aVw&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=H0so_NJ2aVw&lt;/a&gt; &lt;BR&gt;&#xA;Frost circle&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-03-15T14:53:47.070" LastActivityDate="2014-03-15T14:53:47.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9160" PostTypeId="2" ParentId="6220" CreationDate="2014-03-12T21:31:15.630" Score="3" Body="&lt;p&gt;You are expelling a thiol rather than an alcohol via alkaline phosphatase-catalyzed hydrolysis.  One can also view it as a free electron pair on &quot;X&quot; attacking the phosphorus to pop out its =O to an anion, then the charge poking back in to expel thiolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WAwPD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T21:31:15.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9161" PostTypeId="2" ParentId="6033" CreationDate="2014-03-12T21:38:53.770" Score="1" Body="&lt;p&gt;A is an ionic salt like NaCl.  Z is covalent and does not solvate to ions when dissolved in water, e.g., sugar, poly(ethylene glycol).  Y is polar covalent and requires solvation to ionize, e.g., HCl. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T21:38:53.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9162" PostTypeId="2" ParentId="9138" CreationDate="2014-03-12T21:50:04.553" Score="1" Body="&lt;p&gt;Considering the effect of $\text{pH}$ on ion activity, one should take into account that $\ce{K+}$ and $\ce{Cl-}$ are respectively conjugated acid of a strong base and conjugated base of a strong acid. They are therefore both weak, with no influence on the $\text{pH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, I would basically assume that also the other way round is true and that $\text{pH}$ has no effect on their activity. &lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2014-03-12T21:50:04.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9163" PostTypeId="2" ParentId="5037" CreationDate="2014-03-12T21:57:13.353" Score="1" Body="&lt;p&gt;PEG &quot;silane&quot; is $\ce{HOCH2CH2O-PEG-CH2CH2CH2-Si(OCH3)3}$.  (Pt-catalyzed $\ce{HSi(OCH3)3}$ addition to the terminal allyl monoether.)  A metal oxide surface can be hydrated, (bulk)M-OH.  The TMS end hydrolyzes and dehydrates to tetrahedrally covalently bonded $\ce{HOCH2CH20-PEG-0Si(OM#bulk3)}$.  The alcohol polyethoxy end hydrated waves in the water, and the protein ignores the PEGylated surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An especially fine biocompatibilizer rendering nearly any organic surface except silicone rubber invisible to protein deposition and denaturation is poly(N-vinyl pyrrolidone).  It can be linearly radiation grafted into organic surfaces (visualize with Congo Red).  Plexiglass gives a 19 degree average contact angle in saline and implants with neither cellular or humoral reaction.  NVP plasma deposition gives you sludge.  There is no obvious (inexpensive) chemical attachment of its low polymer or oligomer to a surface.  Pity.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-12T21:57:13.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9164" PostTypeId="1" AcceptedAnswerId="9165" CreationDate="2014-03-12T22:41:24.403" Score="1" ViewCount="96" Body="&lt;p&gt;I'm looking for data to create a simple graph of an exothermic or endothermic reaction but can't find any.&lt;/p&gt;&#xA;&lt;p&gt;Could someone help me with some enough data to create a simple graph displaying energy as function of time.&lt;/p&gt;&#xA;" OwnerUserId="4827" LastEditorUserId="69830" LastEditDate="2021-05-12T07:12:33.017" LastActivityDate="2021-05-12T07:12:33.017" Title="How to trace an energy versus time diagram?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;reaction-coordinate&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9165" PostTypeId="2" ParentId="9164" CreationDate="2014-03-12T23:29:47.193" Score="3" Body="&lt;p&gt;Assuming you want to plot heat exchanged by reaction versus time, you need at least to know what reaction you want to study. Then you will have to assess two different quantities: &lt;em&gt;enthalpy&lt;/em&gt; if the process is isobaric (thermodynamic) and reaction &lt;em&gt;rate&lt;/em&gt; (kinetic). This kind of data can be found on &lt;a href=&quot;http://webbook.nist.gov/chemistry/&quot; rel=&quot;nofollow&quot;&gt;NIST Database&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not find reaction enthalpy $\Delta_{R} H$, you can assess it by using enthalpies of formation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta_{R} H = \sum\limits_i \nu_i h_{\mathrm{f},i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $i$ indexes reactants and products, $\nu_i$ are stoichiometric coefficients (note: they are negative for reactant) and $h_{\mathrm{f},i}$ are enthalpies of formation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need to find out which is the kinetic $v$ of your reaction and solve the related differential equation. If you do not find any information you might try the activated complex model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v = \frac{1}{V}\frac{\mathrm{d}\xi}{\mathrm{d}t} = k(T)\prod_j x_j^{\alpha_j} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\xi$ is the molar coordinate of your reaction (De Donder variable), $k(T)$ the kinetic constant, $x_j$ are reactant concentrations and $\alpha_j$ are partial orders of reaction (they can be taken equals to $|\nu_j|$). Solving this differential equation - for a given set of initial conditions - will provide you a relation, maybe explicit, between $\xi$ and $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merging both step results will lead to enthalpy-time expression which can be plotted. Of course those calculation might be perform numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were looking for an hand-made and pedagogic Energy Plot like &lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aActivation_energy.svg&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, you have not searched so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method exposed above does not create this kind of plot which shows internal energy of sum of compounds versus coordinate of a molecular transformation. Gap between the initial and final state is the energy relaxed/absorbed by the reaction (often pure heat). Such plot are arbitrarily plotted because the activated state is often unknown and definitely non measurable. Furthermore everything between initial and final state is also not-well-known because we just have access on macroscopic data so we ignore microscopic details. Last hint: activation energy (maximum) is always higher than the higher state, you need extra energy to initiate the transformation.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-03-12T23:35:47.373" LastActivityDate="2014-03-12T23:35:47.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9166" PostTypeId="1" AcceptedAnswerId="9167" CreationDate="2014-03-13T00:20:36.907" Score="3" ViewCount="463" Body="&lt;p&gt;I am very confused, I think using reduction would be a good start, but LiAl would go straight to an OH. &lt;img src=&quot;https://i.stack.imgur.com/ahn1M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="5026" LastEditDate="2018-02-24T02:40:55.137" LastActivityDate="2018-02-24T02:40:55.137" Title="Which of these reduction reagents would me most appropriate to yield this product?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9167" PostTypeId="2" ParentId="9166" CreationDate="2014-03-13T00:33:05.100" Score="1" Body="&lt;p&gt;1) As you say*, the ketone must be protected.  Protect it. Any propagating naughtiness is irrelevant to the next step. &lt;BR&gt;&#xA;2) Reduce the ester. Beat on it. &lt;BR&gt;&#xA;3) Deprotect the ketone. &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Note that you begin with an alpha-ketoester.  In the old Morrison and Boyd that was a French joke.  LAH would have the same problem.  How the mighty have fallen.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-13T00:33:05.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9168" PostTypeId="1" CreationDate="2014-03-13T01:12:47.797" Score="1" ViewCount="1646" Body="&lt;p&gt;I have the first part correct; however, I can't seem to understand what the 2nd part's answer is. I realize the water leaves, but then what? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pBfNV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBfNV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="5026" LastEditDate="2018-02-25T02:29:27.870" LastActivityDate="2018-02-25T02:29:27.870" Title="Reaction of a vicinal diol with acetone" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9169" PostTypeId="1" AcceptedAnswerId="9178" CreationDate="2014-03-13T01:44:42.633" Score="3" ViewCount="167" Body="&lt;p&gt;For example, given a reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl -&amp;gt; Na+ + Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its $Ksp$ value, I know that you can set up&#xA;$$Ksp\; =\; \left[ \ce{Na^{+}} \right]\left[ \ce{Cl^{-}} \right]$$ and solve from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do you not put $\ce{[NaCl]}$ in the denominator, as isn't that part of the equilibrium expression also?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="1362" LastEditorDisplayName="user4076" LastEditDate="2014-11-02T15:05:44.760" LastActivityDate="2014-11-02T15:05:44.760" Title="What does the equilibrium equation look like when calculating the solubility from Ksp?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9170" PostTypeId="1" AcceptedAnswerId="9187" CreationDate="2014-03-13T01:49:53.383" Score="3" ViewCount="248" Body="&lt;p&gt;For my experiment, I am trying to determine the fluorescence quantum yield of C60-fullerene in toluene (at room temperature).  In order to do that, I need to have a standard sample with a valid and known fluorescence quantum yield. I have been looking for a standard sample with a known quantum yield in toluene, however, I am not sure of what to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be better if I used a different solvent in determining the quantum yield and if so what solvent and standard should I use.&lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2014-03-13T06:43:38.930" Title="Determining Fluorescence Quantum Yield of C60 in Toluene" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9171" PostTypeId="1" AcceptedAnswerId="9181" CreationDate="2014-03-13T02:11:20.603" Score="5" ViewCount="1747" Body="&lt;p&gt;How do I solve problems where I'm asked for &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f$&lt;/span&gt; and given &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f$&lt;/span&gt; values for equations with the involved elements? For example: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f$&lt;/span&gt; of the following reaction:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\ce{C + O2 -&amp;gt; CO2}$&lt;/span&gt;, given:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrO + CO2 -&amp;gt; SrCO3}$&lt;/span&gt; ; &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f = \pu{-234 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2SrO -&amp;gt; 2Sr + O2}$&lt;/span&gt; ; &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f =  \pu{1184 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2SrCO3 -&amp;gt; 2Sr + 2C + 3O2}$&lt;/span&gt; ; &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f = \pu{2440 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that I'm somehow supposed to manipulate the equations to make them &quot;add up&quot; (adding straight down) to the target equation, but I'm not sure how, exactly, to achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there a different and/or better method for doing these? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made some progress on my own, using &lt;a href=&quot;http://chemistry.about.com/od/workedchemistryproblems/a/Hesss-Law-Example-Problem.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this About.com page&lt;/a&gt;. What I'm wondering now, is if elements are locked into compounds, can I split them up, as long as I keep them on the same side? In the example above, I've manipulated it to where I have &lt;span class=&quot;math-container&quot;&gt;$\ce{2CO2 + O2 -&amp;gt; 2C + 3O2}$&lt;/span&gt;; am I allowed to split the &lt;span class=&quot;math-container&quot;&gt;$\ce{2CO2}$&lt;/span&gt; into &lt;span class=&quot;math-container&quot;&gt;$\ce{2C + 2O2}$&lt;/span&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made it as far as getting to &lt;span class=&quot;math-container&quot;&gt;$\ce{2C + 2O2 -&amp;gt; 2CO2}$&lt;/span&gt;. If you take these as molar ratios, it's the same as &lt;span class=&quot;math-container&quot;&gt;$\ce{C + O2 -&amp;gt; CO2}$&lt;/span&gt;, which is the target equation. What do I have to do to the enthalpies to go from the first to the second?&lt;/p&gt;&#xA;" OwnerUserId="4249" LastEditorUserId="83758" LastEditDate="2020-02-27T15:21:46.657" LastActivityDate="2020-02-27T15:21:46.657" Title="Enthalpy of Formation problems using Hess's Law" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9172" PostTypeId="1" AcceptedAnswerId="9175" CreationDate="2014-03-13T02:17:32.860" Score="2" ViewCount="5603" Body="&lt;p&gt;The $\mathrm{p}K_\text{a}$ is the $-\log_{10}$ of the $K_\text{a}$ value for the disassociation of an acid. So how do you get a $\mathrm{p}K_\text{a}$ value for a base? The base produces no hydrogen ions so can’t you only have a $\mathrm{p}K_\text{b}$?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T19:56:42.780" LastActivityDate="2015-06-13T09:51:11.720" Title="How do you have pKa for a base?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9174" PostTypeId="1" CreationDate="2014-03-13T03:17:39.137" Score="4" ViewCount="1420" Body="&lt;p&gt;What is average rate? What is instantaneous rate? How do you find these? My teacher described finding the tangent to a graph. I am not sure what is meant by this.&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorDisplayName="user7951" LastEditDate="2015-09-30T21:37:27.933" LastActivityDate="2015-09-30T21:37:27.933" Title="How do you find the average rate and instantaneous rate given a kinetics plot?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9175" PostTypeId="2" ParentId="9172" CreationDate="2014-03-13T03:20:49.027" Score="4" Body="&lt;p&gt;For bases, you can derive the $\mathrm{p}K_\text{a}$ from the $\mathrm{p}K_\text{b}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s examine ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia has a $K_\text{b}$ of $1.8 \times 10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the negative logarithm, we can derive $\mathrm{p}K_\text{b}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\text{b} = -\log( 1.8 \times 10^{-5}) = 4.74$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w} = 14$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find $\mathrm{p}K_\text{a}$ by manipulating that expression, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\text{a} = 14 - \mathrm{p}K_\text{b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\text{a} = 14 - 4.74 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\text{a} = 9.26 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in very basic solutions, there will be minute quantities of hydronium ions. By taking the anti-logarithm of the $\mathrm{p}K_\text{a}$, you can derive the $K_\text{a}$.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T19:56:38.297" LastActivityDate="2015-05-20T19:56:38.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9178" PostTypeId="2" ParentId="9169" CreationDate="2014-03-13T03:40:03.717" Score="3" Body="&lt;p&gt;The reason why $K_{sp}$ does not include $[\ce{NaCl}]$ is because it is solid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCl(s) -&amp;gt; Na+ (aq) + Cl- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant $K$ would include $[\ce{NaCl}]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\dfrac{[\ce{Na+}][\ce{Cl-}]}{[\ce{NaCl}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the concentration of sodium chloride in solid sodium chloride is constant, thus we can write an equation for the observed rate constant that is based only on the variable concentrations. $K_{sp}$ is this observed rated constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{sp}=K_{obs}=K[\ce{NaCl}]=[\ce{Na+}][\ce{Cl-}]$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-03-13T03:40:03.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9180" PostTypeId="1" AcceptedAnswerId="9184" CreationDate="2014-03-13T04:02:19.177" Score="2" ViewCount="1335" Body="&lt;p&gt;I don't understand question 6, although I took notes and the professor mentioned an explanation in class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help using easier terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2ACP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="418" LastEditDate="2014-03-13T08:59:54.087" LastActivityDate="2017-03-28T15:02:31.073" Title="Diels Alder reaction with maleic anhydride. Reactivity of different dienes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9181" PostTypeId="2" ParentId="9171" CreationDate="2014-03-13T04:05:00.447" Score="6" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Find the &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ_f$&lt;/span&gt; of the following reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C + O2 -&amp;gt; CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We have our list of reactions with known enthalpies:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrO + CO2 -&amp;gt; SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2SrO -&amp;gt; 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2SrCO3 -&amp;gt; 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:&lt;/p&gt;&#xA;&lt;p&gt;Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrCO3 -&amp;gt; SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We need to get rid of &lt;span class=&quot;math-container&quot;&gt;$\ce{SrO}$&lt;/span&gt;, thus, we use our second equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2SrO -&amp;gt; 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrO -&amp;gt; Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove &lt;span class=&quot;math-container&quot;&gt;$\ce{SrCO3}$&lt;/span&gt; and also divide again by 2, since we only have one strontium.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Sr + C + 3/2O2 -&amp;gt; SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrCO3 -&amp;gt; SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrO -&amp;gt; Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Sr + C + 3/2O2 -&amp;gt; SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C + O2 -&amp;gt; CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-26T19:26:40.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9182" PostTypeId="2" ParentId="9174" CreationDate="2014-03-13T04:07:49.707" Score="2" Body="&lt;p&gt;This is more of a question of calculus than chemistry, but in a first order reaction &#xA;$\ce{A -&amp;gt; B}$, the average rate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Delta A}{\Delta t} = \frac{A_2 - A_1}{t_2 - t_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sXmHq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; Graciously borrowed from &lt;a href=&quot;http://science.widener.edu/~svanbram/chem146/ch15/lm_c15_graphs.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the average rate is the slope of a secant line between the two points, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac {\Delta A}{\Delta t} = \frac{1 - 0.5}{400 - 200}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instantaneous rate is the derivative of the plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dA(t)}{dt}$$ or $$\lim_{t\rightarrow 0} \frac{\Delta A}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above first order rate law, the equation is in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(t) = e^{-kt + ln(A_0)} $$ so the derivative is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{dA(t)}{dt} = -ke^{-kt + ln(A_0)} $$&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-03-13T04:07:49.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9183" PostTypeId="1" AcceptedAnswerId="9185" CreationDate="2014-03-13T04:37:43.517" Score="8" ViewCount="5829" Body="&lt;p&gt;This is what I understand so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of a buffer is to minimize any swings in pH. In order to do this you need to create a solution with both acidic and basic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create one, you start with an acid:&#xA;$\ce{HA + H2O -&amp;gt; A- + H3O+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[My first question is why for buffers the acid needs to be weak.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buffer is then created by adding in a salt containing the anion of the base which allows for both the acid and its conjugate base to exist in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, why doesn't the conjugate base react with the acid? Second, why does this help minimize pH changes. I see that since that you have a weak acid the conjugate base will be strong and can react with any acid added. However, if you add base what is going to react with it? Isn't the acid too weak to do so? Third, what happens to the $\ce{H3O+}$ that exists in the solution?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2015-08-05T11:55:04.027" Title="Can someone please explain buffers to me?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9184" PostTypeId="2" ParentId="9180" CreationDate="2014-03-13T05:00:26.267" Score="2" Body="&lt;p&gt;a) The compound can rotate into a trans configuration that is not suitable for Diels Alder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The compound is locked into a cis configuration that is favorable. Furthermore, a carbon group is slightly electron donating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) The compound has an electron withdrawing group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) &lt;s&gt;What compound is that?&lt;/s&gt; It's compounds labelled b) and c)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment: Normal demand Diels Alder is enhanced by an electron withdrawing group in the dienophile and an electron donating group in the diene. Look at reaction mechanism tab (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure as to why that is. I have forgotten.&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="7448" LastEditDate="2017-03-28T15:02:31.073" LastActivityDate="2017-03-28T15:02:31.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9185" PostTypeId="2" ParentId="9183" CreationDate="2014-03-13T06:00:32.580" Score="11" Body="&lt;p&gt;Writing out the equilibrium expression for a weak acid in solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H3O+}] = K_\mathrm a\frac{[\ce{HA}]}{[\ce{A-}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buffers need to keep the concentration of the weak acid to its conjugate base similar in order for it be effective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s say you add a small amount of base. Some of the $\ce{HA}$ molecules are converted into $\ce{A-}$ ions. The ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add a small amount of acid, we generate more $\ce{HA}$. Again, the ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do buffers for the acid needs to be weak?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the expression, the concentration of hydronium ions is directly related to the ratio of the conjugate acid and its base. If you take a strong acid for instance, hydrochloric, it will almost completely deprotonate in water. Thus the ratio of $\frac{[\ce{HA}]}{[\ce{A-}]}$ is close to 0 and is negligible, and we use the $K_\mathrm a$ to directly find the concentration of hydronium ions. As strong acids have very large $K_\mathrm a$ values, we can directly take the concentration and assume that will be the concentration of our hydronium ions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why doesn’t the conjugate base react with the acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on the acid. In buffer solutions, competition arises between acid and the conjugate base. For instance, if the $K_\mathrm a$ is large relative to $10^{-7}$, the acid ionization will proceed and you will have an equilibrium with the conjugate base reacting with its acid resulting in no buffer action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however, the tendency of the acid in donating hydrogen ions, say, to water, is outcompeted by the tendency of the base to accept those hydrogen ions from water, this net reaction produces $\ce{OH-}$ ions, the $K_\mathrm a$ of this acid-base pair is less than $10^{-7}$, and we use $K_\mathrm b$ to find equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivation of the Henderson–Hasselbalch equation:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s use acetic acid and water as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{AcOH + H2O &amp;lt;=&amp;gt; AcO- + H3O+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_\mathrm a = \frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the log of both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \log(K_\mathrm a) = \log\left(\frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the law of logs where $\log(xy) = \log(x) + \log(y) $:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA; \log\left(K_\mathrm a\right) &amp;amp;= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \log([\ce{H3O+}])\\&#xA; -\log([\ce{H3O+}]) &amp;amp;= \log\left([\ce{AcO-}][\ce{AcOH}]\right) - \log(K_\mathrm a) \\&#xA; \mathrm{pH} &amp;amp;= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \mathrm{p}K_\mathrm a \\&#xA; \mathrm{pH} &amp;amp;= \mathrm{p}K_\mathrm a + \log\left(\frac{[\ce{B}]}{[\ce{A}]}\right) \\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With buffer systems, you can calculate the $\mathrm{pH}$ of any buffer solution. The Henderson–Hasselbalch equation has several caveats however.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You need have both the acid and its conjugate base in solution.&lt;/li&gt;&#xA;&lt;li&gt;The extent of ionization must be small enough so that the&#xA;concentrations of hydronium and hydroxide ions are small relative&#xA;$[\ce{HA}]_0$ and $[\ce{A-}]_0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="16622" LastEditDate="2015-08-05T11:55:04.027" LastActivityDate="2015-08-05T11:55:04.027" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9187" PostTypeId="2" ParentId="9170" CreationDate="2014-03-13T06:43:38.930" Score="2" Body="&lt;p&gt;Toluene instead of benzene can be a poor choice when H-atom abstraction by the excited state is possible. Here, you're safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to get a copy of &lt;a href=&quot;http://dx.doi.org/10.1021/ja00067a062&quot; rel=&quot;nofollow&quot;&gt;Fluorescence spectra and quantum yields of buckminsterfullerene (C60) in room-temperature solutions. No excitation wavelength dependence&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors apparently used 9-cyanoanthracene in toluene as a standard to determine $\Phi_f$ of C60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to have a look for review articles by &lt;em&gt;Dirk Guldi&lt;/em&gt; from the time around 2000; I remember him giving a number of talks on C60 around that time.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-13T06:43:38.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9188" PostTypeId="2" ParentId="9168" CreationDate="2014-03-13T08:48:37.600" Score="3" Body="&lt;p&gt;There was this guy who recently &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9166/challenging-reaction-question&quot;&gt;protected a ketone&lt;/a&gt; in the presence of an ester. You might have heard about him ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-03-13T08:48:37.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9189" PostTypeId="2" ParentId="9137" CreationDate="2014-03-13T09:36:36.147" Score="1" Body="&lt;p&gt;Methacrylates make nice starting materials for the lightinduced polymerization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In methyl methacrylate (MMA), camphorquinone (a typical initiator) has an absorption maximum around $\lambda$ = 473 nm.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-13T09:36:36.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9193" PostTypeId="1" AcceptedAnswerId="9195" CreationDate="2014-03-13T22:18:04.617" Score="3" ViewCount="806" Body="&lt;p&gt;I made some pH indicator by boiling and straining cabbage juice and in an effort to keep it from spoiling for a while without having to store it at fridge temperatures I added some wine making stabiliser which consists primarily of Sodium Metabisulphite. I had previously tested some of it to check if adding it would affect the pH of the indicator since I wanted a pH neutral preservative. Well strangely after leaving it for a while I came back to find that the indicator had turned a weird shade of purple I had never seen the indicator take before. At the concentration it is at the original colour was a very dark purple - almost black. Now I know that the metabisulphite ion decomposes to produce SO2 gas which is acidic under aqueous conditions but I would expect that to change the indicator to its normal purple for a not very concentrated acid. But the colour it currently is is very pale and in fact when I pipette out some of the indicator it is almost totally clear. So I guess really my question is does Potassium Metabisulphite have some kind of bleaching effect I'm not aware of? &lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2014-03-14T01:25:30.733" Title="Red Cabbage Indicator colour change with Sodium Metabisulphite" Tags="&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9194" PostTypeId="2" ParentId="9118" CreationDate="2014-03-14T00:20:10.937" Score="0" Body="&lt;p&gt;I finally figured it out using simple logic for myself as follows. Thank you for the answer though!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an equilibrium as described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{eq} = \dfrac{[A]^n}{[B]^m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we double the pressure of the system at large, i.e., halving the volume, we double the concentrations of products and reactants. If $n=m$, Le Chatelier's Principle is proven correct because no change occurs, as the mole sum is the same for products and reactants. If $n\neq m$, Le Chatelier's Principle is also correct; doubling and squaring, for instance, will not yield the same resulting as doubling and cubing the concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we have a short, comprehensive, and intuitive understanding of the reasoning behind Le Chatelier's Principle with regard to concentration changes caused by pressure. &lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2014-03-14T00:20:10.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9195" PostTypeId="2" ParentId="9193" CreationDate="2014-03-14T01:25:30.733" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://mylespower.co.uk/2012/04/06/homemade-ph-indicator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mylespower.co.uk/2012/04/06/homemade-ph-indicator/&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CRGFM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add the reducing agent (model compound).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/swEt6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That lessens conjugation and bleaches the anthocyanine.  &quot;&lt;em&gt;But the colour it currently is is very pale&lt;/em&gt;&quot; If the new compounds' residual optical absorbances are in the green, the solution will be iodine purple by transmission. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T01:25:30.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9197" PostTypeId="1" CreationDate="2014-03-14T06:23:15.793" Score="4" ViewCount="14101" Body="&lt;p&gt;Is there any organic compound that has another type of bond other than the covalent bond?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="10185" LastEditDate="2015-10-02T16:08:39.710" LastActivityDate="2015-10-02T16:08:39.710" Title="Can an organic compound have an ionic bond?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9198" PostTypeId="1" CreationDate="2014-03-14T06:33:30.507" Score="1" ViewCount="200" Body="&lt;p&gt;For example, oxepane is a 7-atom ring. Can there be 8-atom ring or 9-atom ring … or million-atom ring to infinity?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorDisplayName="user7951" LastEditDate="2015-06-12T10:53:17.697" LastActivityDate="2016-09-26T19:07:57.670" Title="Is there a (carbon) heterocyclic compounds named from number three to infinity?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9199" PostTypeId="2" ParentId="9198" CreationDate="2014-03-14T06:44:03.683" Score="4" Body="&lt;p&gt;To my knowledge, the &lt;strong&gt;Hantzsch-Widman&lt;/strong&gt; nomenclature defines stems for ring sizes up to 10 (&lt;em&gt;ecine&lt;/em&gt; for unsaturated, &lt;em&gt;ecane&lt;/em&gt; for saturated heterocycles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Oxepane&lt;/em&gt; is &lt;strong&gt;not&lt;/strong&gt; the &lt;strong&gt;general stem&lt;/strong&gt; for a seven-membered heterocyclic ring, but the suffix for one containing an oxygen atom.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-14T07:41:41.533" LastActivityDate="2014-03-14T07:41:41.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9200" PostTypeId="2" ParentId="9197" CreationDate="2014-03-14T06:48:33.420" Score="3" Body="&lt;p&gt;Of course, organic can make an ionic bond. For example: Acetylene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetylene is relatively acidic, even it's less acidic than water or alcohols. If acetylene reacts with strong base, it will form acetylide. For example: reaction between acetylene and sodium amide ($\ce {Na+ (NH2)-}$) will form sodium acetylide ($\ce {Na+ (H-C#C:)-}$). And this sodium acetylide is ionic and a organic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another compound is a organometallic compound like n-Butyllithium ($\ce{(CH3-CH2-CH2-CH2:)- Li+}$), Grignard reagents ($\ce{(R:)- (MgX)+}$), etc.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="735" LastEditDate="2014-04-02T09:23:31.543" LastActivityDate="2014-04-02T09:23:31.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9201" PostTypeId="2" ParentId="9197" CreationDate="2014-03-14T07:17:53.313" Score="4" Body="&lt;p&gt;There's a large number of salts with organic anions (e.g. &lt;em&gt;potassium hydrogen tartrate&lt;/em&gt;) or organic cations (e.g. &lt;em&gt;tetrabutylammonium hexafluorophosphate&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are even salts, in which both the anion and the cation are organic compounds with covalent bonds. This is often used in pharmaceuticals. Here the active component appears in protonated form and is served as a salt with the conjugated bases of carboxyclic acids (e.g. &lt;em&gt;citrates&lt;/em&gt;, &lt;em&gt;succinates&lt;/em&gt;, etc.)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-14T07:17:53.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9202" PostTypeId="1" CreationDate="2014-03-14T07:37:10.227" Score="1" ViewCount="64" Body="&lt;p&gt;As i understand, when brushing hair, some electron transfer from hair to the comb. Thus, make the comb have different charged from hair and they attract each other.&#xA;It's look likely what i have learn from school about ionic bond. One electron from Na is transfer to F and so we have NaF . Isn't it ??? &lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="275" LastEditDate="2014-03-14T10:19:29.493" LastActivityDate="2014-03-14T10:19:29.493" Title="Comb and hair, is that an example for ionic boding?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9203" PostTypeId="2" ParentId="9202" CreationDate="2014-03-14T07:54:50.860" Score="2" Body="&lt;p&gt;No, not at all. It is an example for &lt;em&gt;triboelectric charging&lt;/em&gt;. A mechanical example for this type of electrostatic generators would be a &lt;strong&gt;Van de Graaff generator&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all this has nothing to do with ionic bonds.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-14T07:54:50.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9204" PostTypeId="1" AcceptedAnswerId="9205" CreationDate="2014-03-14T13:17:16.807" Score="1" ViewCount="4164" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A sample of certain mass of gas was collected and it was found that the number of atoms and the number of molecules in the collected sample are equal. You are required to state the conclusion drawn from this observation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure of the answer. I think it is related to the quantity of a mole being constant.&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="5026" LastEditDate="2018-04-27T15:17:36.823" LastActivityDate="2018-04-27T15:17:36.823" Title="How can the number of molecules and atoms be the same in a gas?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;mole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9205" PostTypeId="2" ParentId="9204" CreationDate="2014-03-14T13:36:14.207" Score="6" Body="&lt;p&gt;The gas was one (or a mixture) of the noble gases, as every other gas is composed of more atoms than molecules.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Helium: $1~\text{mol}~\ce{He}~\text{gas} = 1~\text{mol}~\ce{He}~\text{atoms}$&lt;/li&gt;&#xA;&lt;li&gt;Nitrogen: $1~\text{mol}~\ce{N2}~\text{gas} = 2~\text{mol}~\ce{N}~\text{atoms}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-03-16T15:15:00.483" LastActivityDate="2014-03-16T15:15:00.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9206" PostTypeId="1" AcceptedAnswerId="9207" CreationDate="2014-03-14T16:25:40.707" Score="2" ViewCount="102" Body="&lt;p&gt;I have a pot with some (bottled) water and added (sea) salt. It has a pencil (graphite) and tin foil (aluminum). The pencil is connected to the positive outlet of a solar panel and the tinfoil is connected to the negative outlet of the solar panel (effectively 'charging' the electrolytic cell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure I don't come to harm, I would like to know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) What chemical(s) gas(es) are produced at the pencil?&#xA;B) What chemical(s) gas(es) are produced at the tin foil?&#xA;C) What will the chemical composition of the water generally be (assuming I've been charging it a while)? (Acidic, toxic, etc?).&lt;/p&gt;&#xA;" OwnerUserId="4838" LastActivityDate="2014-03-14T17:07:33.610" Title="Electrolytic cell chemical composition during/after charging?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9207" PostTypeId="2" ParentId="9206" CreationDate="2014-03-14T17:07:33.610" Score="1" Body="&lt;p&gt;Voltage determines the kind of reactions, amperage determines the amounts.  The positive graphite electrode is the oxidizing anode.  It will evolve oxygen and chlorine and be eroded to carbon dioxide, carbon monoxide, and chloro-organics.  The negative aluminum electrode is the reducing cathode.  It will reduce Na+ to elemental sodium that immediately reacts with water to make lye and hydrogen.  The lye will erode the aluminum into more hydrogen and aluminum hydroxide gel, plus some black slime for about 1% alloy-reinforcing precipitated intermetallics in the aluminum foil (usually 8111 alloy).  Lye plus chlorine gives hypochlorite bleach that will erode both electrodes, and do a little disproportionation into chloride plus chlorite, then that perhaps into chloride and chlorate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium carbonate (washing soda) is a more benign electrolyte, but it is an alkali hazard to eyes and skin.  A glass is a more inert container.  Do not short the electrodes, do it outdoors and stay upwind.  Avoid electrolyte or product contact with your eyes and skin.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T17:07:33.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9208" PostTypeId="2" ParentId="9204" CreationDate="2014-03-14T17:13:29.527" Score="2" Body="&lt;p&gt;The conclusion is that the gas is monoatomic - helium, neon, argon, krypton, xenon; or something clever done in a physics lab with low temperature, high vacuum, and lasers. It is a lot of fun to do a little chair parade, learn new things, and present a second answer that is beyond your teacher's understanding.  Then, smile.  Knowledge wants to propagate.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T17:13:29.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9209" PostTypeId="1" AcceptedAnswerId="9220" CreationDate="2014-03-14T18:02:30.707" Score="3" ViewCount="18849" Body="&lt;p&gt;In school, we did an experiment using fat and $\ce{KOH}$ to create 'soft' soap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, we added $\ce{NaOH}$ to change it into 'hard' soap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read on multiple places on the internet that the difference between using Potassium and Sodium is what determines if it becomes a 'hard' soap or 'soft' soap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in the notion that the $\ce{K+}$, resp. $\ce{Na+}$ ions connect to the $\ce{O-}$ ends of the Fatty Acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to skip the first step and create 'hard' soap by using $\ce{NaOH}$ right away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does 'Soft' soap dissolve so much easier than 'Hard' soap? Or in other words: Why is it much easier for the $\ce{K+}$ to fall off than for the $\ce{Na+}$ ?&lt;/p&gt;&#xA;" OwnerUserId="4839" LastEditorUserId="4945" LastEditDate="2014-07-17T10:37:50.357" LastActivityDate="2014-07-17T10:37:50.357" Title="'Hard' soap vs. 'Soft' soap: Why do they work this way?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9210" PostTypeId="2" ParentId="8443" CreationDate="2014-03-14T18:06:52.067" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books/about/Crystal_Growth_in_Gels.html?id=pQ-JuNXEL6wC&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books/about/Crystal_Growth_in_Gels.html?id=pQ-JuNXEL6wC&lt;/a&gt; &lt;BR&gt;&#xA;  The classic text.  Set them up, come back in a few weeks.  (Potentially any gel - polyacrylamide, agarose, agar, gelatin...)  Meanwhile,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Prussian Blue prep works across the transition metals, with different colors.  Alas, it is not crystalline (but maybe in gel...).  Vanadium oxidation states are a spectrum of colors.  Vanadyl, V=O, is especially fine (but not as a food supplement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cobalt can be preparatively more fun than iron.  Co(II) is substitution labile - put anything on it you want, in real time.  Co(III) is substitution inert.  Anything you put on Co(II) then oxidize to Co(III) stays forever (absent a trace of reducing agent to reopen the door).   There is vast and colorful Co coordination chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/bookseries/10.1002/SERIES2146&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/bookseries/10.1002/SERIES2146&lt;/a&gt; &lt;BR&gt;&#xA;  Everything in there works to spec.  Fine some interesting things to do (and scale down to grams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cobalt sepulcrate and especially sarcophogate (Alan Sargeson)! &lt;BR&gt;  Tris(ethylenediamine)cobalt(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 6 186 (1960)) or made entirely as one hand.  Net charge is 3+.  Tris(oxalato)cobaltate(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 8 204 (1966)) or made entirely as one hand.  Net charge is 3-.  Tris(ethylenediamine)cobalt(III) tris(oxalato)cobaltate(III) as a salt with same or opposite hands could be interesting.  The crystal will certainly be tightly bound.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T18:06:52.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9211" PostTypeId="2" ParentId="9197" CreationDate="2014-03-14T18:18:28.990" Score="1" Body="&lt;p&gt;React maleic anhydride with a mole of $\ce{H_2N-CH_2CH_2-N(CH_3)_3^+}$ to afford the zwitterionic amide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protein - with every kind of bonding interaction from the variety of amino acid side groups.  Hydrogen bonding, ionic bonding, covalent bonding, pi-stacking, zinc-fingers, charge transfer, etc.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="10185" LastEditDate="2015-10-02T16:08:18.273" LastActivityDate="2015-10-02T16:08:18.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9212" PostTypeId="1" CreationDate="2014-03-14T18:28:34.570" Score="9" ViewCount="2629" Body="&lt;p&gt;Can organic compounds like hydrocarbons have types of bonds other than covalent bonds? Can they also possess ionic bonds?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="33991" LastEditDate="2016-12-09T19:00:22.377" LastActivityDate="2016-12-09T19:00:22.377" Title="Can a organic compounds such as hydrocarbons contain an ionic bond?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;&lt;hydrocarbons&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9213" PostTypeId="1" AcceptedAnswerId="9218" CreationDate="2014-03-14T18:35:57.710" Score="11" ViewCount="21929" Body="&lt;p&gt;I don't understand the concept of N factor at all. Why is it important?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it is something dependent upon the oxidation number, for example N factor of Na would be 1 since it has oxidation number of +1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For acid and bases we look at the replaceable $\ce{H+}$ or $\ce{OH-}$ present in them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this, I didn't really understand how I can use N factor for calculation or how to find the N factor for whole chemical reactions. Please correct me where I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="4511" LastEditorUserId="7448" LastEditDate="2017-05-17T18:57:33.237" LastActivityDate="2021-05-13T13:11:26.257" Title="How do we calculate N factor and why is it useful in calulations?" Tags="&lt;inorganic-chemistry&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9214" PostTypeId="1" CreationDate="2014-03-14T19:03:25.117" Score="2" ViewCount="51" Body="&lt;p&gt;With the following setup: plastic container, water (impure), pencil (anode), tin foil (aluminum, cathode) and solar panels supplying electrical charge (roughly 36v at about 400-800ma), with the intention of producing lye via electrochemistry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Would it be possible to produce the right kind of lye to a level sufficient to produce handsoap with this current setup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How would you test the lye quality to be sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not possible to make the right quality lye, is there a way to modify the setup so it's able to produce suitable quality lye for soapmaking?&lt;/p&gt;&#xA;" OwnerUserId="4838" LastActivityDate="2014-03-14T19:03:25.117" Title="Electrochemistry lye quality" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9215" PostTypeId="1" CreationDate="2014-03-14T20:25:52.513" Score="5" ViewCount="138" Body="&lt;p&gt;Does somebody know how many lumens we get from lighting 1 gram of flash powder in pure oxygen (or normal) environment? I'm thinking about $\ce{KClO_4} $ (60%) + $\ce{Al}$ (40%)  flash powder and quite fast flaming due to lighting it up with electricity.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorDisplayName="user4076" LastEditDate="2014-03-14T23:00:22.220" LastActivityDate="2014-03-14T23:00:22.220" Title="How many lumens from 1 gram of flash powder" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9216" PostTypeId="1" AcceptedAnswerId="9219" CreationDate="2014-03-14T20:33:20.483" Score="1" ViewCount="206" Body="&lt;p&gt;Do you have an idea how can be flaming of flash powder (probably $\ce{KClO_4} $ (60%) + $\ce{Al}$ (40%) ) can be speeded up? Of course we consider the pure oxygen-in-air conditions. So how? With some kind of catalyst?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorDisplayName="user4076" LastEditDate="2014-03-14T22:58:48.413" LastActivityDate="2014-03-14T22:58:48.413" Title="How to speed up flash powder flaming reaction" Tags="&lt;inorganic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9217" PostTypeId="2" ParentId="9212" CreationDate="2014-03-14T20:46:27.290" Score="6" Body="&lt;p&gt;Yes, of course. Look up for Grignard reagents, it is the most obvious. But there is other type of bounds. Cyan-hydride ligand is another bound type. Generally speaking this kind of not-exactly-covalent-and-close-to-ionic bound occurs with metals.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-03-14T20:46:27.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9218" PostTypeId="2" ParentId="9213" CreationDate="2014-03-14T20:56:44.590" Score="10" Body="&lt;p&gt;There are some different applications.&lt;/p&gt;&#xA;&lt;p&gt;For one, it tells you how many times an acid can lose a proton. For instance, you can have a diprotic (&amp;quot;with &lt;span class=&quot;math-container&quot;&gt;$\ce{2 H+}$&lt;/span&gt;&amp;quot;) molecule, with &lt;span class=&quot;math-container&quot;&gt;$n=2$&lt;/span&gt;. For instance, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;, deprotonates to &lt;span class=&quot;math-container&quot;&gt;$\ce{HSO4-}$&lt;/span&gt;, which can then go to &lt;span class=&quot;math-container&quot;&gt;$\ce{SO4^{-2}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Why is that important? Take an acidic reaction with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;. If the extent of reaction is strongly forward, you would treat &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; as if it had &lt;em&gt;twice&lt;/em&gt; its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.&lt;/p&gt;&#xA;&lt;p&gt;It's also known as equivalent concentration, which might make it sound less foreign to you.&lt;/p&gt;&#xA;&lt;p&gt;In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.&lt;/p&gt;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;https://en.wikipedia.org/wiki/Equivalent_concentration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="98422" LastEditDate="2021-05-13T13:11:26.257" LastActivityDate="2021-05-13T13:11:26.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9219" PostTypeId="2" ParentId="9216" CreationDate="2014-03-14T21:22:42.350" Score="1" Body="&lt;p&gt;First, we postulate safety is not an issue. If you try any of this you will probably die.  If you need a short duration event-triggered flash, explode a wire.  A Pockels cell plus a polarizer is a nanoseconds shutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ball mill potassium perchlorate all by itself.  Mix well with micronized aluminum.  Compact.  (Impact, friction, or a static discharge may kill you.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum has a high boiling point, 2519 C.  Magnesium is 1091 C and burns under nitrogen.  As above, with micronized Mg, adjusting stoichiometry for the new equivalent weight.  Same warning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a more potent oxophile, like titanium, adjusting stoichiometry for the new equivalent weight.  Micronized titanium can flash in air, nitrogen, or carbon dioxide all by itself.  Same warning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DO NOT TRY CHLORATE, as opposed to perchlorate.  There is almost no activation energy barrier to reaction.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T21:22:42.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9220" PostTypeId="2" ParentId="9209" CreationDate="2014-03-14T21:29:42.560" Score="4" Body="&lt;p&gt;Basically, yes, by common terms, soft soap has &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and hard soap has &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;They are both alkali metals, but &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; is below &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; on the periodic table so maybe we should consider periodicity trends. As you go down a table the atoms become more reactive, so &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; is more likely to react with polar parts of water than &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;. Here are solubilities for both (though there are other values that would tell this story, perhaps better than solubility of the product):&lt;/p&gt;&#xA;&lt;p&gt;[Solubility for &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt;] in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; at 25C: 121 g / 100 mL&lt;/p&gt;&#xA;&lt;p&gt;[Solubility for &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;] in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; at 25C: 100 g / 100 mL&lt;/p&gt;&#xA;&lt;p&gt;Regarding your question of going straight into hard soap, based on &lt;a href=&quot;http://www.thesurvivalistblog.net/how-to-make-soap/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; it seems doable; he just used &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; (lye) and oils with no mention of &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt; (potash?) to speak of.&lt;/p&gt;&#xA;&lt;p&gt;And yes, the positive ions connect to the &lt;span class=&quot;math-container&quot;&gt;$\ce{O-}$&lt;/span&gt; of the fatty acid as seen in this diagram.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SRESz.gif&quot; alt=&quot;img&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(img src: &lt;a href=&quot;http://jan.ucc.nau.edu/%7Edoetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jan.ucc.nau.edu/~doetqp-p/courses/env440/env440_2/lectures/lec19/lec19.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-15T14:41:10.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9221" PostTypeId="2" ParentId="9212" CreationDate="2014-03-14T22:01:45.217" Score="13" Body="&lt;p&gt;Azulene has a 1.05 debye dipole moment,&#xA;&lt;img src=&quot;https://i.stack.imgur.com/n8UsH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydrocarbon as such bonded as an isolated cation-anion pair with no heteroatoms could happen in principle given sufficient steric hindrance.   Penta-&lt;em&gt;tert&lt;/em&gt;-butylcyclopentadienide or pentakis(2,4,6-tri-tert-butylphenyl)cyclopentadienide with like fully substituted cyclpropenium or tropylium might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar substitution on fulvalenes' four inner carbons would dihedral angle rotate the rings by steric hindrance.  This prevents the double bond from forming, giving sort of a sigma covalent plus pi-dative bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gmBGr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-14T22:01:45.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9222" PostTypeId="1" AcceptedAnswerId="9223" CreationDate="2014-03-15T04:39:07.407" Score="12" ViewCount="82046" Body="&lt;p&gt;I'm learning about how to recognise whether a bond is ionic or covalent, based on the difference in electronegativity between the two bonding partners, $\Delta \chi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have now is a formula:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $\Delta \chi = 0$, then the bond is nonpolar&lt;/li&gt;&#xA;&lt;li&gt;If $0 &amp;lt; \Delta\chi \leq 1.7$, then the bond is polar covalent&lt;/li&gt;&#xA;&lt;li&gt;If $\Delta \chi &amp;gt; 1.7$, then the bond is ionic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I don't know how scientists determined that formula, the history of it and which experiment indicates that formula.&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="16683" LastEditorDisplayName="user4076" LastEditDate="2017-08-12T01:08:41.080" LastActivityDate="2017-08-12T01:08:41.080" Title="Why are bonds ionic when the electronegativity difference between bonded atoms is greater than 1.7?" Tags="&lt;bond&gt;&lt;electronegativity&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9223" PostTypeId="2" ParentId="9222" CreationDate="2014-03-15T05:22:53.360" Score="20" Body="&lt;p&gt;The first thing to consider is the difference between covalent and ionic bonding, from the UCDavis ChemWiki site &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Ionic_and_Covalent_Bonds&quot; rel=&quot;noreferrer&quot;&gt;Ionic and Covalent Bonds&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms that have the same electronegativity share electrons in covalent bonds since donating or receiving electrons is unfavorable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The electron donor has a low electronegativity and the electron acceptor has a higher elelctronegativity - so there is a difference in electronegativity $\Delta{EN}$, effectively creating a positive and negative end, an example is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QmjsA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://www.spring8.or.jp/en/news_publications/press_release/2011/111020/&quot; rel=&quot;noreferrer&quot;&gt;Japan Synchrotron Radiation Research Institute (JASRI)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the differences in electronegativities of various compounds are graphed against % ionic character, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ikx1x.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Image source: &lt;a href=&quot;http://www.chem.ufl.edu/~itl/4411/lectures/lec_13.html&quot; rel=&quot;noreferrer&quot;&gt;University of Florida Chemical Bonding page&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values of $\Delta{EN}$ greater than 1.7 correspond to an ionic character of greater than 50%, from the University of Florida website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What determines how the electrons are shared is the relative electronegativity (electron greed) of the bonding atoms. The degree of polarity or degree of ionic bonding of any given bond can vary continuosly zero to nearly 100%. We normally say that bonds between atoms with electronegativity difference ($\Delta{EN}$) greater than 1.7 are ionic, although this really means only more than about half ionic in character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another resource is from the University of Washington &lt;a href=&quot;http://courses.washington.edu/bhrchem/c152/Lec23.pdf&quot; rel=&quot;noreferrer&quot;&gt;Lecture 23: Ionic to Covalent Bonds&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorDisplayName="user4076" LastEditDate="2014-03-15T05:40:39.840" LastActivityDate="2014-03-15T05:40:39.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9228" PostTypeId="2" ParentId="9204" CreationDate="2014-03-16T03:35:23.957" Score="1" Body="&lt;p&gt;If in a given volume of gas the number of atoms and molecules is same ,then the only conclusion that can be drawn is that the mixture was of the noble gases.E.g.-&#xA;&lt;strong&gt;&lt;em&gt;He,Ne,Ar,Kr,Xe&lt;/em&gt;&lt;/strong&gt; etc.&lt;/p&gt;&#xA;" OwnerUserId="4831" LastActivityDate="2014-03-16T03:35:23.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9229" PostTypeId="1" AcceptedAnswerId="9253" CreationDate="2014-03-16T05:40:19.760" Score="3" ViewCount="13207" Body="&lt;p&gt;I am not a Chemist or did not learn chemistry as a student. I have faced this phenomena from a natural visualization. So this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Experiment:&lt;/strong&gt; What will be happen if we add a solution of ammonia and water over bad quality iron? When it shall be dried I shall pour again. Several day It shall be done regularly in open nature. Some reaction will take place. Rust will be created on iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common people will not see anything significant except rust. But &lt;strong&gt;As a chemist&lt;/strong&gt; how do you understand after three to four days after completion of the experiment that I pour ammonia solution on it? Assume that this three or four days it was in open nature. Something will be produced after reaction except rust. How much of them is water soluble?&lt;/p&gt;&#xA;" OwnerUserId="4848" LastEditorUserId="4848" LastEditDate="2014-03-16T06:04:22.657" LastActivityDate="2014-03-17T09:58:56.060" Title="Reaction between ammonia and iron" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9230" PostTypeId="1" CreationDate="2014-03-16T07:21:24.183" Score="8" ViewCount="15988" Body="&lt;p&gt;I am reading this &lt;a href=&quot;http://en.wikipedia.org/wiki/Crisscross_method&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; that I don't understand. What I don't understand is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;suppose we have two elements $X$ and $Y$ having oxidation numbers $x$ and $y$ respectively. Can we prove that the compound formed will be $X_yY_x$.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried to understand this method as: I think this method comes from the fact of &lt;em&gt;charge neutrality&lt;/em&gt;. Suppose the compound to be formed has formula $X_{y'}Y_{x'}$. The charge on $y'$ number of $X$ elements is $xy'$. The charge on $x'$ number of $Y$ elements is $x'y$. Now by charge neutrality we should have : $$|x|y'=x'|y|$$&lt;br&gt;&#xA; this implies $\dfrac{|x|}{|y|}=\dfrac{x'}{y'}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But I could not further understand why the we divide $x$ and $y$ with their greatest common factor to find the required solution.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have also found some contradiction examples to this Crisscross rule. Example is : &lt;strong&gt;Solid-state-semiconductor $GaAs$&lt;/strong&gt;, in this Gallium has valancy $3$ and $As$ has valancy $5$.&lt;br&gt;&#xA; The main points on which I want answers are:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;What is the logic behind this method, how to prove this method?&lt;/li&gt;&#xA;  &lt;li&gt;Who, how and when developed this method? &lt;/li&gt;&#xA;  &lt;li&gt;Is this method only applicable to ionic compounds?  &lt;/li&gt;&#xA;  &lt;li&gt;Why compounds like $GaAs$ do not follow this rule, how charge neutrality is preserved in these compounds. I think these don't follow this rule because these are not ionic compounds, if yes then how the chemical formula of compounds like solid state $GaAs$ is known(or decided) and how it is justified, how one atom of $Ga$ balances one atom of $As$ ,their valencies are different, how they are bonded(covalently I guess) together?  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="4609" LastEditDate="2014-03-24T12:13:32.280" LastActivityDate="2014-03-24T15:19:27.700" Title="The Crisscross method for finding the chemical formula" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9231" PostTypeId="1" CreationDate="2014-03-16T08:29:57.370" Score="2" ViewCount="170" Body="&lt;p&gt;In a typical household, the following surfactants are found, among others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Soap / handwash&lt;/li&gt;&#xA;&lt;li&gt;Shower gel&lt;/li&gt;&#xA;&lt;li&gt;Shampoo&lt;/li&gt;&#xA;&lt;li&gt;Dishwashing agent&lt;/li&gt;&#xA;&lt;li&gt;Washing powder / liquid&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From my understanding, the role of all these is to lower the surface tension between water and fatty acids, enabling one to &quot;wash&quot; things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they all the same then, and is the only difference between them marketing? Can I wash my hair with dishwashing agent, or pour shower gel into my washing machine?&lt;/p&gt;&#xA;" OwnerUserId="4850" LastActivityDate="2014-03-16T09:25:01.237" Title="What are the differences between the various household surfactants?" Tags="&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9232" PostTypeId="2" ParentId="9231" CreationDate="2014-03-16T09:25:01.237" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I wash my hair with dishwashing agent [...]!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it's an agent for dishwashing machines: &lt;strong&gt;Please DON'T!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cleaning in dishwashing machines is performed in &lt;strong&gt;alkaline&lt;/strong&gt; medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tabs usually contain (among others):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;alkaline compounds, such as sodium carbonate and sodium hydroxide&lt;/li&gt;&#xA;&lt;li&gt;bleaching agents, such as sodium perborate and sodium percarbonate&lt;/li&gt;&#xA;&lt;li&gt;pentasodium triphosphate (to keep calcium and magnesium in solution)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The solution of these tabs is &lt;strong&gt;not&lt;/strong&gt; meant to get in contact with skin and eyes.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-16T09:25:01.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9233" PostTypeId="1" CreationDate="2014-03-16T09:47:20.947" Score="3" ViewCount="67" Body="&lt;p&gt;I am reading the contents of the oatstraw:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;.. high in vitamins A, C, B complex, LE, G..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am unable to find on the Internet what LE refers to.  &lt;/p&gt;&#xA;" OwnerUserId="4851" LastEditorUserId="7448" LastEditDate="2015-10-09T16:20:49.353" LastActivityDate="2019-01-16T13:21:32.247" Title="What is LE (plant contents)?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9235" PostTypeId="1" AcceptedAnswerId="9240" CreationDate="2014-03-16T13:40:21.843" Score="0" ViewCount="626" Body="&lt;p&gt;The concentration of an aqueous solution of $\ce{Mg(SO4)2}$ is  $0.001\, \textrm{mol} \cdot \textrm{dm}^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out the concentration of solution and the concentation of $\ce{Mg^{2+}}$ ion in &lt;em&gt;ppm&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to do this? How to find moles in ppm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could clearly describe, it would be great. &lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="418" LastEditDate="2014-03-16T13:55:04.953" LastActivityDate="2014-03-16T15:10:33.633" Title="Conversion of units for concentration: mol/l to ppm" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9236" PostTypeId="1" AcceptedAnswerId="9267" CreationDate="2014-03-16T14:00:16.533" Score="3" ViewCount="216" Body="&lt;p&gt;Hiho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question concerning this image (Chemistry, Catherine E. Housecroft et al., Pearson Education, 2010).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FJ4kn.jpg&quot; alt=&quot;diagram of in-phase and out-of-phase energies as a function of distance in H2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the text, it states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Each of the new molecular wavefunctions, $\Psi$(in-phase) and $\Psi$(out-of-phase), has an energy associated with it wich is dependent of the distance apart of the two hydrogen nuclei. As Figure 4.13 [provided image] shows, the energy of $\Psi$(in-phase) is always lower than $\Psi$(out-of-phase).&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The in-phase-function has a stabilized minimum, but the out-of-phase-function seems stable only at an infinite distance. For my intuition, the two graphs should be symmetric, since the two wavefunctions are alike and approach each other equally. Why do they behave like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are those functions not symmetric?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A following question would be, when I combine those two functions, I'll get a minimum at the x-value I marked with &quot;minimum&quot;. Would this be the distance at which a nucleus in a stable diatomic molecule would be found like in this figure?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XaesW.gif&quot; alt=&quot;1s orbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;===&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help and have a nice day :)&lt;/p&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2014-03-17T16:46:39.253" Title="Symmetry of an in-phase and an out-of-phase wavefunction &quot;approaching&quot; each other in an MO" Tags="&lt;physical-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9237" PostTypeId="1" CreationDate="2014-03-16T14:32:21.237" Score="0" ViewCount="72" Body="&lt;p&gt;Molecular systems - are in a simple count systems made up of molecules. But almost every system, surrounding us are molecular than. Considering the molecular systems in terms of chemistry compels us, firstly, impose dimensional constraints on the molecular system (which?), molecular system could not be macroscopic (How big?), otherwise you'll have to talk about it as a solid body; secondly, energies for such systems are limited (&amp;lt;1 eV); in third place we will be interested in properties of such system as whole, but not as sum of its parts properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the main question is raised: &lt;code&gt;What is the range of sizes for molecular systems?&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="4589" LastActivityDate="2014-04-15T16:44:50.300" Title="What is the range of sizes for molecular systems?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9238" PostTypeId="2" ParentId="9230" CreationDate="2014-03-16T15:00:34.753" Score="4" Body="&lt;p&gt;&lt;strong&gt;Does this also work with compounds other than only ionic compounds?&lt;/strong&gt; Yes - take a look at this : in water hydrogen has a ($\ce{+I}$) partially positive oxidation state and oxygen has a ($\ce{-II}$) partially negative oxidation state. So if you work out the rule then you get : $\ce{H2O1}$ (and you can leave the 1 away so you get $\ce{H2O}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples include : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$ :  ox.state (H) = $\ce{+I}$ and ox.state (Cl) = $\ce{-I}$   ... so... : H &lt;strong&gt;1&lt;/strong&gt; Cl &lt;strong&gt;1&lt;/strong&gt; -&gt;remove ('1')-&gt; HCl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rule even works on certain, more complicated molecules : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4}$ : ox.state(H) = $\ce{+I}$ and ox.state(SO4) = $\ce{-II}$ ..... so... : H &lt;strong&gt;2&lt;/strong&gt; SO4 &lt;strong&gt;1&lt;/strong&gt; -&gt;remove('1')-&gt; H2SO4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KMnO4}$ : ox.state(K) = $\ce{+I}$ and ox.state(MnO4) = $\ce{-I}$ .....so.... : K &lt;strong&gt;1&lt;/strong&gt; MnO4 &lt;strong&gt;1&lt;/strong&gt; -&gt; remove ('1') -&gt; KMnO4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(of course all the charges mentioned above are partial(+/-) charges, because they are covalently bonded molecules)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why the greatest common factor is taken... : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at calcium oxide. Here you have one calcium ($\ce{+II}$) ion and one oxygen ($\ce{-II}$) ion. When the rule is applied you get $\ce{Ca2O2}$. Now, you can divide both indexes by 2 to get $\ce{CaO}$. But why? Look at $\ce{CaO}$ : Here the one calcium atom has lost 2 electrons to achieve a stable noble gas configuration. And oxygen has gained 2 electrons to achieve a stable noble gas configuration. Both atoms are 'happy' and there is no need for another calcium atom or oxygen atom. So $\ce{CaO}$ is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the bonding in&lt;/strong&gt; $\ce{GaAs}$, I found this online: (&lt;a href=&quot;http://books.google.be/books?id=riupCD3UkyIC&amp;amp;pg=PA138&amp;amp;lpg=PA138&amp;amp;dq=Gallium%20arsenide%20%28GaAs%29%20could%20be%20formed%20as%20an%20insulator%20by%20transferring%20three%20electrons%20from%20gallium%20to%20arsenic;%20however,&amp;amp;source=bl&amp;amp;ots=NPj0gdB245&amp;amp;sig=hgyH9yMdxyHQm43bennNMt32Lac&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=IZEoU6bkKtTe7AaI2oDwDQ&amp;amp;redir_esc=y#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6KVGL.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the crisscross method &lt;strong&gt;wouldn't work with semiconducting materials&lt;/strong&gt;. However it does (as mentioned above) work for certain &lt;em&gt;normal&lt;/em&gt; molecules such as H2O, HCl,... these semiconducting materials just are exceptions to the rule...&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2014-03-19T12:19:29.450" LastActivityDate="2014-03-19T12:19:29.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9239" PostTypeId="2" ParentId="9237" CreationDate="2014-03-16T15:05:35.970" Score="1" Body="&lt;p&gt;Well apportioned, mixed, and cured epoxy glue is a single covalent molecule.  Single crystal white diamond is a single covalently-bonded molecule.  On the small end, a dihydrogen molecule, is 0.7414 Å (&lt;em&gt;ortho&lt;/em&gt; or &lt;em&gt;para&lt;/em&gt;, your choice to trim the decimal).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest human chromosome is ~220 million base pairs, or about 1.5x10^11 g/mol.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-16T15:05:35.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9240" PostTypeId="2" ParentId="9235" CreationDate="2014-03-16T15:10:33.633" Score="1" Body="&lt;p&gt;Convert mol-dm^2 to grams/volume.  A decimeter is 10 cm.  A dm^3 is then a liter.  Water has density ~1 g/cm^3.  Multiply the g(Mg)/g(water) by a million.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-16T15:10:33.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9242" PostTypeId="1" AcceptedAnswerId="9244" CreationDate="2014-03-16T19:29:35.377" Score="25" ViewCount="2030" Body="&lt;p&gt;I'm a mathematician who's currently teaching a course on differential equations.  Though I don't know much about chemistry, I like to include examples from chemistry in my course, and I prefer for the details to be accurate.  Here is a typical exam problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a container of gaseous nitrogen dioxide is heated above 150 °C, the gas begins to decompose into oxygen and nitric oxide:&#xA;  $$\ce{2NO2 -&amp;gt; O2 + 2NO}$$&#xA;  The rate of this reaction is determined by the equation&#xA;  $$\frac{\mathrm d\left[\ce{NO2}\right]}{\mathrm dt} = -k\left[\ce{NO}_2\right]^2$$&#xA;  where $k$ is a constant.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Find the general solution to the above equation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) A large container holds 50.0 moles of $\ce{NO2}$ at a constant temperature of 600 °C.  After one hour, only 34.3 moles remain.  How much $\ce{NO2}$ will there be after another hour?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the science in this problem reasonably accurate?  Is there anything you would change?  (I looked up a suitable value of the rate constant $k$ to make sure that the time in part b was reasonable.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are some other examples of reactions that are governed by simple rate laws?  Ideally, I'd like to have several examples each of reactions governed by the equations&#xA;$$&#xA;\frac{\mathrm dy}{\mathrm dt} = -ky,\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^2,\qquad\text{and}\qquad \frac{\mathrm dy}{\mathrm dt}=-ky^3.&#xA;$$&#xA;(Is $y^3$ really possible?  Are non-integer powers of $y$ possible?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4853" LastEditorDisplayName="user7951" LastEditDate="2015-11-11T00:53:43.577" LastActivityDate="2015-11-11T00:53:43.577" Title="What are some good examples of rate equations for a math class?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9243" PostTypeId="2" ParentId="9242" CreationDate="2014-03-16T19:49:08.653" Score="5" Body="&lt;p&gt;I have no example in mind but fractional and high partial order (exponent) are possible. When this kind of relationship appears in chemistry they often result from complex reaction schema where more than one steps are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway have look to &lt;a href=&quot;http://kinetics.nist.gov/&quot; rel=&quot;noreferrer&quot;&gt;Kinetic NIST Database&lt;/a&gt;, you will find many reaction kinetic constants and models.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-03-16T19:49:08.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9244" PostTypeId="2" ParentId="9242" CreationDate="2014-03-16T21:14:37.990" Score="28" Body="&lt;p&gt;How thoughtful of you to include chemistry in your differential equations course! We appreciate your effort, especially going the extra mile to make it realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you've seen the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation&quot;&gt;rate equation page&lt;/a&gt; on Wikipedia as it contains a good deal of the mathematics in several examples of reactions. You should also be interested in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Order_of_reaction&quot;&gt;reaction order page&lt;/a&gt; (you can find examples here of reactions with fraction or negative order with respect to some reagents).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me do a quick recap on basic chemical kinetics theory. Consider the general reaction equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{a\ A + b\ B + c\ C + d\ D + ... \longrightarrow w\ W + x\ X + y\ Y + z\ Z + ...}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, where uppercase letters indicate different molecules in a gas or solution, and lowercase letters indicate the stoichiometric coefficients (negative for the reactants, positive for the products). Naively, we can assume the reaction happens because, while the molecules are jostling about, at some point all of the reactants ($a$ molecules of $\ce{A}$, $b$ molecules of $\ce{B}$, $c$ molecules of $\ce{C}$, etc.) will bump into each other all at the same time and undergo reaction. If you think of collisions as independent events with a probability of happening proportional to the concentration of each species, then it isn't too hard to understand that the frequency with which reactions happens is proportional to $[A]^a[B]^b[C]^c[D]^d...$, and so the rate equation for the reaction would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=k[A]^a[B]^b[C]^c[D]^d...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, where &lt;em&gt;r&lt;/em&gt; is the reaction rate and &lt;em&gt;k&lt;/em&gt; is a proportionality constant. The consumption or production rate for any species $\Gamma$ with stoichiometric coefficient $\gamma$ is trivially related to the reaction rate by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d\Gamma}{dt}=\gamma r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, experimentally things aren't so simple (fortunately for nature, unfortunately for our minds). A reaction that requires the simultaneous collision of many molecules would be highly unlikely, meaning that it would happen very, very slowly. Even so, experimentally we observe many reactions that involve a large amount of molecules. For example, the combustion of cyclohexane in air is formally given by the equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12 + 9 O2 → 6CO2 + 6 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction really required ten molecules to bump into eachother at the same time with sufficient energy and in the right geometry, then this reaction probably could not happen in air, as oxygen would be too rarefied to compensate the extremely low proportionality constant &lt;em&gt;k&lt;/em&gt; for a single-step reaction. In actuality, a gaseous mix of cyclohexane and air can react so fast as to cause an explosion, converting all the reactants into products in a miniscule fraction of a second. Clearly something is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out the assumption that reactions occur in one step is incorrect in general. Usually, there is more than one step involved, and indeed there generally is more than one path from reactants to products. The speed at which a reaction occurs is really described by considering all paths at once, with all their steps, and adding their contributions (like a start and end point in a complex tree diagram). In general this is pretty difficult and unnecessarily complex. It is often a reasonable approximation to model the reaction by selecting only the &lt;em&gt;fastest route&lt;/em&gt; from reactants to products, and the reaction rate is bottlenecked by the &lt;em&gt;slowest step&lt;/em&gt; of the fastest route, which becomes responsible for determining the rate equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the reaction rate depends on selecting the fastest route, you shouldn't expect any of the steps in it to contain some very huge bottleneck, such as requiring six molecules to react simultaneously; there would likely be a slightly different route, perhaps with more steps, but in which each step doesn't involve so many molecules interacting at once, and hence being a faster route. Because of this, it turns out that most of the fastest reaction pathways involve slowest steps that very rarely depend on more than two molecules at once. Most reactions which can be modeled simply like this therefore have rate equations of the type $r=k[A]$, $r=k[A]^2$, $r=k[A][B]$ or even $r=k$. Few reactions rates are of the third order, of the type  $r=k[A]^3$,  $r=k[A]^2[B]$,  $r=k[A][B][C]$ or similar. I don't know of any reaction that goes as $r=k[A]^3$ specifically. Fourth order reaction rates are so rare that they are the &lt;a href=&quot;http://www.nature.com/nature/journal/v162/n4124/abs/162777a0.html&quot;&gt;focus of research&lt;/a&gt; &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01304a016&quot;&gt;when found&lt;/a&gt;. I don't expect anyone to know a fifth order reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, kinetic theory is hard enough that reaction kinetics end up being determined simply by parameter fitting from experiments. Even if the underlying reaction mechanism is not completely understood, if a fit of the type $r=k[A]^{3/2}[B]^{-1}$ happens to be good, then so be it. Reactions which have been studied in depth are significantly more complex, such as the &lt;a href=&quot;http://www.umich.edu/~elements/07chap/html/07lr.htm&quot;&gt;misleading simplicity of the reaction between hydrogen and bromine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I don't know why I didn't search for a fifth order reaction rate! Turns out some exist, such as &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1993/C3/c39930000470#!divAbstract&quot;&gt;this one&lt;/a&gt;. The reaction rate is of the type $r=k[A][B]^4$, so you could have an example where $\frac{d[B]}{dt}=k[B]^4$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-03-16T21:32:26.393" LastActivityDate="2014-03-16T21:32:26.393" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9245" PostTypeId="1" CreationDate="2014-03-17T01:32:08.750" Score="1" ViewCount="829" Body="&lt;p&gt;Williamson ether synthesis is conducted using vanillin, $\ce{KOH}$, $\ce{EtOH}$, water and benzyl chloride.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is water needed for the synthesis?&lt;/li&gt;&#xA;&lt;li&gt;Why must the solution be refluxed for an hour to gain the desired product?&lt;/li&gt;&#xA;&lt;li&gt;Why is water added after reflux to create a cloudy solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4856" LastEditorUserId="5026" LastEditDate="2018-02-24T03:41:07.703" LastActivityDate="2018-02-24T03:41:50.447" Title="Williamson ether synthesis" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9246" PostTypeId="1" AcceptedAnswerId="9251" CreationDate="2014-03-17T04:59:08.433" Score="4" ViewCount="907" Body="&lt;p&gt;Question 13 from the &lt;a href=&quot;http://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2011-local-olympiad-exam.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACS chemistry olympiad&lt;/a&gt; exam:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The phase transition from gas to solid is called &#xA;  \begin{align}&#xA;&amp;amp;\text{(A) condensation.} &amp;amp; \text{(B) evaporation.} \\ &#xA;&amp;amp;\text{(C) polymerization.} &amp;amp; \text{(D) sublimation.}&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure the answer is deposition, which is not listed, but yet the answer is (&lt;strong&gt;A&lt;/strong&gt;) &lt;strong&gt;condensation&lt;/strong&gt; [scroll to the end of the question paper to see the answer key].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did the chemistry olympiad make a mistake, or is there something obvious that I am missing?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="41328" LastEditDate="2017-08-06T07:27:25.337" LastActivityDate="2017-08-06T07:27:25.337" Title="Phase transition from gas to solid" Tags="&lt;physical-chemistry&gt;&lt;nomenclature&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9247" PostTypeId="1" AcceptedAnswerId="9250" CreationDate="2014-03-17T05:26:15.117" Score="6" ViewCount="69682" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Changes in which factors affect both the rate and the rate &#xA;  constant of a first order reaction? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I. temperature &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;II. concentration &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A) I only (B) II only &#xA;  (C) Both I and II (D) Neither I nor II &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would think that the answer would be (C), both temperature and concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentration would obviously affect it (from the rate expression), and the temperature would also affect it because the Arrhenius equation for the rate constant is &lt;span class=&quot;math-container&quot;&gt;$$&#xA;k = A e^{-E_a/(R T)}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and thus temperature would have an influence on the rate constant, and since the rate constant is part of the rate equation, it would also affect the rate of the reaction itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the given answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;A&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this is.&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="83758" LastEditDate="2020-01-27T22:05:10.460" LastActivityDate="2020-01-27T22:05:10.460" Title="Changes in which factors affect both the rate and the rate constant of a first order reaction?" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="9250" PostTypeId="2" ParentId="9247" CreationDate="2014-03-17T06:22:54.127" Score="10" Body="&lt;p&gt;If you see the rate equation for the Arrhenius equation:&#xA;$$&#xA;k = A e^{-E_a/(R T)}&#xA;$$&#xA;you can see that the rate constant increases for an increase in temperature (and as activation energy increases, the rate constant decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since the rate constant increases, this implies that the rate of reaction increases with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding concentration: you are correct in that concentration affects the rate of reaction. However, a rate constant does not change according to concentration. You can observe this in the Arrhenius equation, where A is independent of the concentration of the substrate. The fact that the rate increases does not imply the rate constant changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly more clearly: it is implied that increasing the rate constant will increase the rate of reaction. However, increasing the rate does not increase the rate constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase in temperature increases the rate constant and hence the rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase in concentration increases the rate but not the rate constant.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastActivityDate="2014-03-17T06:22:54.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9251" PostTypeId="2" ParentId="9246" CreationDate="2014-03-17T06:36:10.300" Score="4" Body="&lt;p&gt;I guess the diagram of phase changes you have in mind looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1H1J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Indeed this says &lt;em&gt;deposition&lt;/em&gt; is the right name for a gas-to-solid phase transition and I believe that this is the only correct term.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can argue why it is obvious here that condensation is the right answer (at least the closest) is that this is the only answer that gives a transition from fluid to solid. B and D are the other way around, and C is not really a phase change (although it could result in one).&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-03-17T17:32:21.197" LastActivityDate="2014-03-17T17:32:21.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9253" PostTypeId="2" ParentId="9229" CreationDate="2014-03-17T09:21:00.677" Score="4" Body="&lt;p&gt;In fact metallic Iron doesn't react directly with ammonia. However ammonia can influence the corrosion process because tend to increase the pH. From the Wikipedia Pourbaix diagram (below) you can see which are the products of a corrosion at an higher pH.Of course the pH of you solution depends on the ammonia concentration of your solution (a 1.0 M ammonia aqueous solution has a pH of about 11). &#xA;&lt;img src=&quot;https://i.stack.imgur.com/sPAqr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrosion is in fact inhibited at higher pH, and the products of the corrosion are not specific of the ammonia reaction, so (I think) is hard do identify the presence of ammonia from the corrosion process.&#xA;Furthermore ammonia is in fact a gas and many of the compounds it forms tend to became ammonia (e.g. ammonium carbonate) and then flow away. However I think that there are many chances that some ammonium salts are produce by the reaction with other species, (e.g. acids) present in the surface of the metal or in the environment.&#xA;In this case you can look for ammonium ions: $\ce{NH_4^{+}}$ you can identify them with a Ion chromatography or with &lt;a href=&quot;http://en.wikipedia.org/wiki/Nessler%27s_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nessler's reagent&lt;/a&gt; (&lt;strong&gt;keep in mind that Nessler's reagent contain mercurium $\ce{Hg}$ so is toxic and not very environmental friendly&lt;/strong&gt;) that detects even ammonia trace. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user4076" LastEditDate="2014-03-17T09:58:56.060" LastActivityDate="2014-03-17T09:58:56.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9255" PostTypeId="2" ParentId="5163" CreationDate="2014-03-17T09:53:39.413" Score="9" Body="&lt;p&gt;An SDF file is just a bunch of MOL entries in a single file; &lt;code&gt;$$$$&lt;/code&gt; serves as the separator for the individual entries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Splitting into individual files AND simultaneous conversion to graphic file formats (PNG, SVG) is possible with &lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;noreferrer&quot;&gt;Open Babel&lt;/a&gt;, either directly on the command line or through scripts/programmes using the bindings available for various programming languages, such as &lt;a href=&quot;http://journal.chemistrycentral.com/content/2/1/5&quot; rel=&quot;noreferrer&quot;&gt;Pybel&lt;/a&gt; for Python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, the direct generation of PostScript files - probably the nicest option when using $\LaTeX$ - is not possible, so you might want to used SVG as an alternate vector-based format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel -isdf lotsofstructures.sdf -m -osvg -xC -xu -xd -O basename.svg&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will do the job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;-isdf&lt;/code&gt; specifies the input file format, &lt;code&gt;-m&lt;/code&gt; does the splitting, &lt;code&gt;-osvg&lt;/code&gt; specifies the output file format, and &lt;code&gt;-O basename.svg&lt;/code&gt; specifies the prefix and extension for the individual &lt;strong&gt;numbered&lt;/strong&gt; output files that are generated. &lt;code&gt;-xC&lt;/code&gt; prevents that terminal carbon atoms are printed explicitly &lt;em&gt;with&lt;/em&gt; attached hydrogen atoms. Surely a matter of taste, but I wouldn't want to have lots of $\ce{-CH3}$ in the drawings. &lt;code&gt;-xu&lt;/code&gt; prevents the use of colour for heteroatoms and generates a black and white drawing. &lt;code&gt;-xd&lt;/code&gt; suppressed the name of the compound to be written and displayed in the SVG file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should work for SDF files with 2D and 3D coordinates, but in the latter case I'd suggest to check it carefully; I've seen 3D -&gt; 2D transformations on some molecules that went wrong.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-17T09:53:39.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9256" PostTypeId="2" ParentId="9245" CreationDate="2014-03-17T10:03:35.673" Score="2" Body="&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Consider the phases for potassium hydroxide and benzyl chloride at room temperature.&lt;/li&gt;&#xA;&lt;li&gt;Why might you not want to use ethanol as the solvent to form the benzyl alkoxide of vanillin in?&lt;/li&gt;&#xA;&lt;li&gt;(As an aside, think about why you might not want to use water over, say, acetonitrile as a solvent for this reaction.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What governs the rate of the Williamson ether synthesis?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why would you want to heat a chemical system you want to have a reaction in?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What solvents would the product ether dissolve in?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which by-product(s) would dissolve in water but not ethanol?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How would you remove the product ether from the solvent mixture?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm trying not to give away the answers directly but these bullet points should help you work out each point.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastEditorUserId="5026" LastEditDate="2018-02-24T03:41:50.447" LastActivityDate="2018-02-24T03:41:50.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9258" PostTypeId="2" ParentId="903" CreationDate="2014-03-17T11:13:19.737" Score="3" Body="&lt;p&gt;For thermally grown amorphous $\ce{SiO2}$ films, a bandgap of 9.3 eV was measured by &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.19.3107&quot; rel=&quot;nofollow&quot;&gt;Weinberg&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measurements on $\alpha$-quartz by &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.18.2888&quot; rel=&quot;nofollow&quot;&gt;Calabrese&lt;/a&gt; give $E_g(\Gamma)$ = 6.3 eV, compared to 9.2 eV calculated by &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.15.4020&quot; rel=&quot;nofollow&quot;&gt;Chelikowsky and Schlüter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations by &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.69.085102&quot; rel=&quot;nofollow&quot;&gt;Ramos&lt;/a&gt; on $\beta$-cristobalite with the space groups $Fd3m$ (face-centred cubic, fcc), $I\bar{4}2c$ (body-centred tetragonal, bt), and $P2_13$ (simple cubic, sc) gave direct bandgaps of $E_g(\Gamma)$ = 5.48, 5.79 and 5.73 eV, respectively.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-17T14:02:17.773" LastActivityDate="2014-03-17T14:02:17.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9259" PostTypeId="1" CreationDate="2014-03-17T11:40:47.867" Score="1" ViewCount="255" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the products of the thermal decomposition of potassium cyanate ($\ce{KOCN}$) with carbon ($\ce{C}$) at elevated temps ($\ce{300^\circ C - 1300^\circ C}$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess some of the nitrogen is lost and forms something. I'm trying to find out the full equation, but a google search brings up nothing concerning reduction of potassium cyanate with carbon. &lt;/p&gt;&#xA;" OwnerUserId="4864" LastEditorUserId="5026" LastEditDate="2018-02-21T14:55:51.507" LastActivityDate="2018-02-21T14:55:51.507" Title="What are the products of the thermal decomposition of potassium cyanate with carbon at elevated temperature?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9260" PostTypeId="2" ParentId="9259" CreationDate="2014-03-17T12:09:09.973" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] some of the nitrogen is lost and forms something [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carbon is a good and cheap reductant, often used in the processing of oxidic ores - except when metal carbides are formed. Carbon dioxide a reasonable product in these reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2KOCN + C -&amp;gt;[\Delta] 2 KCN + CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's agree that potassium cyanate is reduced to potassium cyanide, while carbon dioxide is released.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-17T12:09:09.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9261" PostTypeId="2" ParentId="4158" CreationDate="2014-03-17T13:12:55.800" Score="2" Body="&lt;p&gt;A drop of sulfuric acid on the sugar will cause its dehydration. The reaction is highly exothermic and yields a highly porous carbon &quot;foam&quot;, blown up by the steam simultaneously released. [You'll find lots of videos of this experiment, more or less reasonably conducted (with/without gloves and proper eye protection), mostly using larger amounts of sulfuric acids.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of the dehydration reaction will furnish the decomposition of the chlorate, releasing oxygen which, in turn, will start to oxidize the porous carbon mass to carbon dioxide. This reaction again is exothermic ($-394\,\mathrm{kJ\cdot mol^{-1}}$). Sugar might start to caramelize and release more water (as steam), more chlorate decomposes to yield oxygen, etc.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-17T13:12:55.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9262" PostTypeId="2" ParentId="7883" CreationDate="2014-03-17T13:13:00.527" Score="2" Body="&lt;p&gt;Firstly, what geometry does the complex have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the geometry, find the spin-state of the $d$-electrons, as well as the number of $d$-electrons applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the crystal field diagram relevant to your geometry. Fill in the $d$-electrons according to the Aufbau principle and to Pauli's exclusion principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your example, tetracarbonylnickel has a tetrahedral geometry and is a $d$10 compound. Carbon monoxide is a strong field ligand causing a high spin complex (note that since $d$ = 10, all of the $t_{2}$ and $e$ orbitals are full and hence spin-state is irrelevant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4a/Tetrahedral.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the Aufbau principle, you start building up from the bottom, placing two electrons in the bottom $t_2$ orbitals and three in the top three $e$ orbitals, giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e$: ↑ ↑ ↑&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_2$: ↑ ↑&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining five electrons go in each half-filled orbital to give 2 filled $t_2$ orbitals and 3 filled $e$ orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e$: ↑↓ ↑↓ ↑↓&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_2$: ↑↓ ↑↓&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the CSFE, the $t_2$ orbitals each stabilise by $\frac{3}{5}$ and the $e$ destabilize by $\frac{2}{5}$. Since $2 * \frac{3}{5} - 3 * \frac{2}{5}$ = 0, the CFSE = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For octahedral complexes, the $e_g$ and $t_{2g}$ orbital populations can be used analogously to find the stabilisation energy.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastEditorUserId="4860" LastEditDate="2014-03-17T13:27:08.117" LastActivityDate="2014-03-17T13:27:08.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9263" PostTypeId="1" AcceptedAnswerId="9265" CreationDate="2014-03-17T15:24:12.207" Score="-1" ViewCount="606" Body="&lt;p&gt;My chemistry teacher in high school didn't know the formula for carbonic acid, which struck me as something someone with a degree in chemistry (as they allegedly do) should know instantly, or at least after some time thinking. However, to give them the benefit of the doubt - is carbonic acid a relativly common chemical, or alternatively is there a well known way to derive its formula from the name (like how you can work out the formula for a molecule called something somethingelseide for instance)? Should someone with a degree in chemistry know it off the top of their head?&lt;/p&gt;&#xA;" OwnerUserId="4866" LastEditorUserId="4866" LastEditDate="2014-03-17T19:36:15.997" LastActivityDate="2014-03-19T12:20:54.553" Title="How well known is the formula of carbonic acid?" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2018-12-28T14:25:23.370" ContentLicense="CC BY-SA 3.0" />
  <row Id="9264" PostTypeId="2" ParentId="2916" CreationDate="2014-03-17T16:06:08.487" Score="10" Body="&lt;p&gt;The following example(s) might help to illustrate the stereoconversion described by &lt;em&gt;Richard Terrett&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m0Kdu.png&quot; alt=&quot;stereoconversion of alcohol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of alcohols, the &quot;classical&quot; method is the &lt;em&gt;Mitsunobu&lt;/em&gt; reaction, using triphenylphosphine ($\ce{Ph3P})$ and diethyl azodicarboxylate (DEAD, $\ce{EtO2C-N=N-CO2Et})$ as reagents, and water as a nucleophile to quench the intermediate with inversion of configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the course of the reaction, $\ce{Ph3P}$ adds to DEAD in the presence of a proton donor to form an intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVegc.png&quot; alt=&quot;Mitsunobu intermediate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter adds to the alcohol, forming a stable $\ce{P-O}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ma8eg.png&quot; alt=&quot;alcohol Ph3P adduct&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of a nucleophile proceeds according to $S_N2$ with inversion of configuration, the formation of triphenylphosphine oxide renders the reaction irreversible.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NWAvS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-17T16:06:08.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9265" PostTypeId="2" ParentId="9263" CreationDate="2014-03-17T16:20:51.487" Score="4" Body="&lt;p&gt;Carbonic acid has the formula $\ce{H2CO3}$ - you can't always immediately derive the formula from the (trivial) name - you have to learn it. And - yes - carbonic acid is a relatively common chemical : you can find it in all types of fizzy drinks (fanta, cola,sprite,....)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can derive the formula from the other name (systematic) of carbonic acid, dihydrogen carbonate =  $\ce{H2CO3}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2014-03-19T12:20:54.553" LastActivityDate="2014-03-19T12:20:54.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9266" PostTypeId="2" ParentId="9263" CreationDate="2014-03-17T16:23:44.953" Score="1" Body="&lt;p&gt;If there is open water and air, there is carbonic acid; carbonated beveradges including beer and champagne; and you are filled with it from aerobic respiration.  pH vs. pKa determines ionization.  (The leading lowercase &quot;p&quot; means &quot;negative logarithm of.&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naming simple acids:  The root is the element- bor(on), car(bon), nitr(ogen), sulfur, chlor(ine) phosphor(us).  Suffix shows oxidation state, ous/ite (low) and ic/ate high.  Prefix intensifies suffix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrochloric, Cl(-1) (&quot;hydro&quot; in context means &quot;no oxygen&quot;) chlorides &lt;BR&gt;&#xA;Hypochlorous, Cl(+1) hypochlorites &lt;BR&gt;&#xA;chlorous, Cl(+3) chlorites &lt;BR&gt;&#xA;chloric, Cl(+3) chlorates &lt;BR&gt;&#xA;perchloric, Cl(+5) perchlorates &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple names become not simple when complex structures form from concatination (peroxysulfuric acid,polyphosphoric acid), element substitution (thiosulfuric acid), substituents (methanesulfonic acid, chlorosulfonic acid), and general diddling around (adenosine triphosphate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Oxyanion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Oxyanion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-17T16:23:44.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9267" PostTypeId="2" ParentId="9236" CreationDate="2014-03-17T16:46:39.253" Score="1" Body="&lt;p&gt;Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one).&#xA;$$ \Psi^+ = c_1 \psi_1 + c_2 \psi_2 $$&#xA;$$ \Psi^- = c_1 \psi_1 - c_2 \psi_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation:&#xA;$$ H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. &lt;strong&gt;Where is the symmetry?&lt;/strong&gt; Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But: The wavefunction in itself is not all that interesting to us, because there is no &quot;real&quot; physical information that can be &lt;em&gt;directly&lt;/em&gt; extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it:&#xA;$$ \rho = |\Psi|^2 = \Psi \Psi^* $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-17T16:46:39.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9268" PostTypeId="2" ParentId="3949" CreationDate="2014-03-17T17:47:32.310" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;Chop the body into smaller pieces, using a sturdy butcher axe.&lt;/li&gt;&#xA;&lt;li&gt;Use a pressurized stainless steel autoclave (preferable with stirrer) or comparable professional kitchen equipment, suspend the pieces in concentrated $\ce{NaOH}$ and apply heat!&lt;/li&gt;&#xA;&lt;li&gt;Cool down, release pressure and check. Saponification of body fats releases fatty acids that might form protective films and thus prevent further rapid decomposition of the &quot;biological material&quot;. If necessary, stir and repeat until the meat is converted to a sludge.&lt;/li&gt;&#xA;&lt;li&gt;Dispose the sludge, wash the bones, crush them to smaller pieces (= larger surface) and dissolve them in conc. $\ce{H2SO4}$.&lt;/li&gt;&#xA;&lt;li&gt;Bring something to read - just in case you have to spend some 30+ years in a locked room - a book on &lt;a href=&quot;http://en.wikipedia.org/wiki/Adolph_Luetgert&quot;&gt;Adolph Luetgert&lt;/a&gt; would be an obvious choice :D&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-17T17:47:32.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9269" PostTypeId="2" ParentId="3949" CreationDate="2014-03-17T18:18:03.173" Score="3" Body="&lt;p&gt;Hot concentrated aqueous lye will dissolve tissue, aided by the process making soap.  Bone is about 50 wt-% hydroxylapatite, calcium phosphate, $\ce{Ca10(PO4)6(OH)2}$.  That is a problem.  Calcine, then make &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bone_china&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bone_china&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/MythBusters_(2013_season)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/MythBusters_(2013_season)&lt;/a&gt; &lt;BR&gt;&#xA;Mythbusters (Episode 208) demonstrated that acid as such is not effective for dissolving bulk flesh. &quot;Piranha solution&quot; (concentrated hydrogen peroxide in concentrated sulfuric acid) eats organics, oh yes indeed, and can explode while doing it; while making it, too.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-03-17T22:13:48.940" LastActivityDate="2014-03-17T22:13:48.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9270" PostTypeId="1" CreationDate="2014-03-17T19:00:39.877" Score="6" ViewCount="1646" Body="&lt;p&gt;I just learned about rate and order... I just want to confirm if these three ideas are correct:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the first elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be EQUAL to the order the concentration of the reactants in the rate equation is raised to?&#xA;e.g. &#xA;$$\begin{align}&#xA;\ce{2A + B &amp;amp;-&amp;gt; C} &amp;amp;(\text{slow})\\ &#xA;\ce{C + B &amp;amp;-&amp;gt; D} &amp;amp;(\text{fast})&#xA;\end{align}$$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Will the order with respect to $[\ce{A}]$ be 2 and $[\ce{B}]$ be 1 if this proposed mechanism is correct? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the balanced equation is the ONLY elementary step in the reaction mechanism, the stoichiometric coefficients of the reactants will be the same as the order the concentration of the reactants are raised in the rate equation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the second elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be different to the order the concentration of the reactants in the rate equation is raised to?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2014-06-12T09:11:30.153" LastActivityDate="2014-07-28T03:08:53.250" Title="Rate and order confusion" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9272" PostTypeId="1" AcceptedAnswerId="9276" CreationDate="2014-03-17T23:16:41.343" Score="5" ViewCount="78316" Body="&lt;p&gt;How does the pH at the equivalence point, as well as the pH range over which the colour of phenolphthalein changes, make it a suitable indicator for titrations between strong acids and strong bases?&lt;/p&gt;&#xA;" OwnerUserId="2295" LastEditorUserId="16683" LastEditDate="2017-08-14T07:14:17.000" LastActivityDate="2017-08-14T07:16:31.320" Title="Why is phenolphthalein an appropriate indicator for titration of a strong acid with a strong base?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9273" PostTypeId="1" AcceptedAnswerId="9285" CreationDate="2014-03-17T23:33:03.073" Score="0" ViewCount="5229" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At the triple point of water how do the entropies of solid, &#xA;  liquid, and gas compare? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that either they will be equal or it will be that gas &gt; liquid &gt; solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if entropy is influenced by the fact that it will be at the triple point. Can someone please elaborate?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2014-03-18T10:46:25.763" Title="Entropy of solid, liquid, and gas at triple point of water" Tags="&lt;entropy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9274" PostTypeId="1" CreationDate="2014-03-17T23:59:05.763" Score="1" ViewCount="506" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which change will decrease the rate of the reaction  between&#xA;  $\ce{I2(s)}$ and $\ce{H2(g)}$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A) Increasing the partial pressure of $\ce{H2(g)}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(B) Adding the $\ce{I2(s)}$ as one piece rather than as several small ones &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(C) Heating the reaction mixture &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(D) Adding a catalyst for the reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can eliminate D as a catalyst would speed up the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about the others though. I could look up the enthalpy for this reaction and eliminate C, but this is from the Chemistry Olympiad and I can't look up enthalpies during the test. Is there any other way to determine that this reaction is endothermic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you eliminate C, you are left with A and B. How would you go about finding the right answer?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="10185" LastEditDate="2015-09-29T16:45:44.573" LastActivityDate="2015-09-29T16:45:44.573" Title="Decreasing rate of reaction between I2(s) and H2(g)?" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9275" PostTypeId="2" ParentId="5477" CreationDate="2014-03-18T00:09:00.087" Score="10" Body="&lt;p&gt;The purpose of the salt bridge is to move &lt;em&gt;ions&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use enough electrolyte solution on both sides, though, it doesn't matter;  in that case, the salt bridge can be neglected.  &lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="402" LastEditDate="2017-04-02T20:13:02.900" LastActivityDate="2017-04-02T20:13:02.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9276" PostTypeId="2" ParentId="9272" CreationDate="2014-03-18T00:19:05.653" Score="10" Body="&lt;p&gt;If you look at the titration curve, which plots the volume of base added vs pH (&lt;a href=&quot;http://www.mikeblaber.org/oldwine/chm1046/notes/AddAqEq/Titrate/Image1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3iGrp.gif&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see that the equivalence point occurs at pH = 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenolphthalein is fuchsia in pH's roughly between 8.2 and 12, and is colorless below pH 8.2. When the number of moles of added base is equal to the number of moles of added acid (or vice versa; example valid for strong monoprotic acids/bases assuming 100% dissolution), the pH is equal to 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say, if the pH needed is 7, and phenonphthalein changes at pH's around 8.2, how can you use this indicator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, again looking at the curve, from pH = 11 to about pH = 4, pH changes very rapidly with from an infinitesimally small change in the volume of base added. Since one drop of added titrant will cause this large change, even though the change in color of phenolphthalein does not occur right on the equivalence point, it is within approximately one drop. This kind of uncertainty is &quot;acceptable uncertainty&quot; in using titration to volumetrically determine concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by &quot;acceptable uncertainty&quot;, you should realise that each of your measurements has some kind of uncertainty to them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When you weighed out the primary standard to titrate against, was the balance perfect?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Was your solution made up precisely to the graduation in the volumetric flask?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Did you pipette the exact volume of the aliquot or were you off by a drop or two?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are you able to state the volume added from the burette to arbitrary precision or is there some uncertainty beyond the two decimal places given by the graduated lines?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the scheme of things, ~1-2 drops will not be a significant factor in getting an accurate result, but you should most definitely acknowledge that there is uncertainty in your answer.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastEditorUserId="16683" LastEditDate="2017-08-14T07:16:31.320" LastActivityDate="2017-08-14T07:16:31.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9278" PostTypeId="1" CreationDate="2014-03-18T01:21:10.523" Score="2" ViewCount="2471" Body="&lt;p&gt;non-pure alkanes have a taste, but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are all pure hydrocarbons, tasteless, colourless and odorless?&#xA;Are there hydrocarbons that do have a taste, colour, odor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what makes these pure hydrocarbons different, even though they are only made of carbon and hydrogen?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="-1" LastEditDate="2017-05-09T07:20:39.450" LastActivityDate="2017-05-09T07:20:39.450" Title="Are all (pure) alkanes, alkenes and alkynes - tasteless, odourless and colourless?" Tags="&lt;organic-chemistry&gt;&lt;hydrogen&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2014-03-18T16:08:39.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="9279" PostTypeId="1" AcceptedAnswerId="9286" CreationDate="2014-03-18T01:37:28.697" Score="3" ViewCount="220" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The boiling points of $\ce{CH3COCH3}$, $\ce{CH3COC2H5}$, and  $\ce{CH3COC3H7}$ are $56˚C$,&#xA;  $80 ˚C$, and $102 ˚C$, respectively.  What factor(s) best account(s) for this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe london dispersion forces is one of them. Is dipole-dipole interactions another one? It seems to me that there &lt;em&gt;are&lt;/em&gt; dipole-dipole interactions since none of the structures of symmetrical, so can they account for this increase in boiling point?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2014-03-18T09:40:20.007" Title="Which factors best account for the increase in boiling point?" Tags="&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9280" PostTypeId="1" AcceptedAnswerId="9281" CreationDate="2014-03-18T01:48:49.790" Score="2" ViewCount="1591" Body="&lt;p&gt;Why do only unsaturated hydrocarbons undergo addition reactions? For example, ethene or ethyne will undergo an addition reaction with chlorine, whereas ethane will not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, why does benzene not undergo addition reactions, even though it is not saturated?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="7448" LastEditDate="2017-04-08T16:16:25.427" LastActivityDate="2017-04-08T16:16:51.513" Title="Why do only unsaturated hydrocarbons undergo addition reactions?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9281" PostTypeId="2" ParentId="9280" CreationDate="2014-03-18T03:44:18.197" Score="3" Body="&lt;p&gt;To what would you add given a saturated hydrocarbon? You must substitute, abstract, or displace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene is aromatic &amp;mdash; different pattern of reactivity.  It photochlorinates.  It UV reacts with maleic anhydride, then adds another mole thermally.  Palladium catalysis will swap appropriate substituents.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="7448" LastEditDate="2017-04-08T16:16:51.513" LastActivityDate="2017-04-08T16:16:51.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9282" PostTypeId="2" ParentId="9278" CreationDate="2014-03-18T03:55:14.527" Score="3" Body="&lt;p&gt;Adamantane smells of camphor.  n-Nonane is in the scent of some roses, see &lt;a href=&quot;http://dx.doi.org/10.1016/S0021-9673%2800%2900863-3&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="418" LastEditDate="2014-03-18T05:17:04.960" LastActivityDate="2014-03-18T05:17:04.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9283" PostTypeId="2" ParentId="9278" CreationDate="2014-03-18T06:31:04.640" Score="4" Body="&lt;p&gt;Why would a &lt;strong&gt;mixture&lt;/strong&gt; of two odourless alkanes have an odour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already pointed out by &lt;em&gt;Greg E&lt;/em&gt; and &lt;em&gt;Uncle Al&lt;/em&gt;, numerous hydrocarbons without any (heteroatom-based) functional groups do show characteristic smells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naphthalene is always remembered for the typical smell of grandma's mothballs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More pleasant smells are found in the group of &lt;strong&gt;monoterpenes&lt;/strong&gt;, i.e. in those molecules with the sum formula $\ce{C10H16}$ that are derived from the combination of two isoprene units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4aKg.png&quot; alt=&quot;monoterpenes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These volatile compounds were/are used as fragrances for perfumes. Typical examples are myrcene (&lt;strong&gt;1&lt;/strong&gt;), $\alpha$-terpinene (&lt;strong&gt;2&lt;/strong&gt;), limonene (&lt;strong&gt;3&lt;/strong&gt;), $\alpha$-pinene (&lt;strong&gt;4&lt;/strong&gt;), $\beta$-pinene (&lt;strong&gt;5&lt;/strong&gt;), $\alpha$-thujene (&lt;strong&gt;6&lt;/strong&gt;), sabinene (&lt;strong&gt;7&lt;/strong&gt;), and camphene (&lt;strong&gt;8&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as the colour of hydrocarbons is concerned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vZcm1.png&quot; alt=&quot;structure of lycopene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sufficiently conjugated system is far from being colourless, take lycopene (&lt;strong&gt;9&lt;/strong&gt;), the &lt;strong&gt;deep red&lt;/strong&gt; coloured dye in tomatoes as an example.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-18T10:08:04.477" LastActivityDate="2014-03-18T10:08:04.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9284" PostTypeId="2" ParentId="2825" CreationDate="2014-03-18T09:27:46.497" Score="4" Body="&lt;h1&gt;The &amp;quot;classical&amp;quot; way&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qG1gk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Direct addition of &lt;span class=&quot;math-container&quot;&gt;$\ce{H-X}$&lt;/span&gt; to alkenes proceeds via the addition of a proton to the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; double bond in the initial step, i.e via a carbocation. This works nicely for strong mineral acids, but not for the weak acid &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; with the rather weak nucleophile &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;Synchronous addition?&lt;/h1&gt;&#xA;&lt;p&gt;Q: Can't &lt;span class=&quot;math-container&quot;&gt;$\ce{H-CN}$&lt;/span&gt; add synchronously add to &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;A: &lt;strong&gt;NO!&lt;/strong&gt; Such a reaction would involve the interaction of the HOMO (highest occupied molecular orbital) of one reaction partner with the LUMO (lowest unoccupied molecular orbital) of the other. The LUMO of an alkene has a nodal plane orthogonal to the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt; bond, while there is - obviously - no nodal plane but a &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond between hydrogen and carbon in &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The synchronous addition of &lt;span class=&quot;math-container&quot;&gt;$\ce{H-CN}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; is &lt;em&gt;symmetry-forbidden&lt;/em&gt;!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Catalysis?&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Yes!&lt;/strong&gt; Since the first catalytic &lt;strong&gt;hydrocyanation&lt;/strong&gt; using &lt;span class=&quot;math-container&quot;&gt;$\ce{Co2(CO)8}$&lt;/span&gt; (see Philipp's excellent comment above), other catalysts have been developed, namely &lt;span class=&quot;math-container&quot;&gt;$\ce{Ni(0)}$&lt;/span&gt; based species using a &lt;a href=&quot;http://dx.doi.org/10.1021/ja074922e&quot; rel=&quot;nofollow noreferrer&quot;&gt;trypticene-derived chelating phosphine ligand&lt;/a&gt; (see image).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mI4M4.png&quot; alt=&quot;trypticene-based ligand&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-18T09:35:58.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9285" PostTypeId="2" ParentId="9273" CreationDate="2014-03-18T09:27:47.570" Score="1" Body="&lt;p&gt;Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such an equilibrium, you simply write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta S= \frac{\Delta H}{T}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that rises from the fact that $\Delta G=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}&#xA;$$&#xA;$$&#xA;\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}&#xA;$$&#xA;$$&#xA;\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one realises that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the entropy of the gas phase is higher than the entropy of the liquid phase.&lt;/li&gt;&#xA;&lt;li&gt;the entropy of the liquid phase is higher than the entropy of the solid phase&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-18T10:46:25.763" LastActivityDate="2014-03-18T10:46:25.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9286" PostTypeId="2" ParentId="9279" CreationDate="2014-03-18T09:40:20.007" Score="1" Body="&lt;p&gt;You're definitely right on London. Between the three molecules, the only thing that's changing is the carbon chain on the right hand side of the Oxygen. That plus the corresponding increase in boiling temperatures together form a good indicator that London forces are at play here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dipole-dipole (or Hydrogen bonding for that matter) are non-significant in this case, because you're not changing the electronegativity of anything here, only adding more of the same repeating atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: This was on the &lt;a href=&quot;http://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2011-local-olympiad-exam.pdf&quot; rel=&quot;nofollow&quot;&gt;ACS 2011 local exam&lt;/a&gt;, question 55.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.masterorganicchemistry.com/2010/10/01/how-intermolecular-forces-affect-boiling-points/&quot; rel=&quot;nofollow&quot;&gt;Further information here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-03-18T09:40:20.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9287" PostTypeId="1" CreationDate="2014-03-18T11:56:53.560" Score="3" ViewCount="835" Body="&lt;p&gt;I made I high pressurized $\ce{MoS2}$ pellet, using a 6 ton pressurizer. I used very thin powder to create the pellet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using this pellet on beam, with some nuclear techniques. In my spectra I can see oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How easy it is for $\ce{MoS2}$ to become oxidized?&lt;/p&gt;&#xA;" OwnerUserId="4457" LastEditorUserId="418" LastEditDate="2014-03-18T12:47:42.720" LastActivityDate="2014-03-18T21:02:46.963" Title="How easy it is for MoS2 to become oxidized" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9288" PostTypeId="2" ParentId="9287" CreationDate="2014-03-18T12:45:37.633" Score="3" Body="&lt;p&gt;It seems that the problem has been examined even 50 years ago, probably in order to determine if/how oxidation would alter the the properties of $\ce{MoS2}$ as a lubricant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://archive.org/details/nasa_techdoc_19930082560&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an old NASA document from 1949, and in &lt;a href=&quot;http://dx.doi.org/10.1021/j150531a020&quot; rel=&quot;nofollow&quot;&gt;Surface Oxidation of Molybdenum Disulfide&lt;/a&gt; from 1955, the authors state in the abstract, that the oxidation is limited to the outmost surface layer, which then passivates and protects the rest of the batch. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-18T12:45:37.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9289" PostTypeId="2" ParentId="6311" CreationDate="2014-03-18T15:51:04.983" Score="1" Body="&lt;p&gt;$\ce{H3C^-}$ is isoelectronic with$\ce{NH3}$, as $\ce{CH4}$ and $\ce{NH4^+}$ correspond. Each of the four $\ce{sp^3}$ hybrid orbitals contain a pair of electrons, in all cases shown.  Nuclear charge versus electron count and the source of the four electrons added to the naked $\ce{sp^3}$ hybrid orbitals are accessories.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-18T15:51:04.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9290" PostTypeId="1" AcceptedAnswerId="9301" CreationDate="2014-03-18T16:36:34.560" Score="3" ViewCount="4139" Body="&lt;p&gt;Why do we typically take $n^2$ scans, even in the basic NMR experiment? I've heard various explanations, including that it helps the fast Fourier transform, and would like to know the real reason.&lt;/p&gt;&#xA;" OwnerUserId="121" LastActivityDate="2014-03-18T22:15:58.763" Title="Why are NMR typically taken for a power of 2 scans?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9292" PostTypeId="2" ParentId="9290" CreationDate="2014-03-18T16:48:13.230" Score="2" Body="&lt;p&gt;The signal to noise ratio (S/N) is proportional to the square root of the number of scans.  Hence the more scans, the better the S/N ration&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-03-18T16:48:13.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9293" PostTypeId="1" AcceptedAnswerId="9295" CreationDate="2014-03-18T17:14:14.370" Score="1" ViewCount="1821" Body="&lt;p&gt;E-liquids are blends of nicotine, propylene glycol, vegetable glycerin, and aroma.&#xA;I need to develop a method for determining the concentration of nicotine in a vial.&#xA;(to verify if the information listed on the bottle are correct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any idea how to proceed?&lt;/p&gt;&#xA;" OwnerUserId="4873" LastActivityDate="2017-08-01T14:16:14.077" Title="Nicotine concentration in e-liquids" Tags="&lt;titration&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9294" PostTypeId="2" ParentId="9290" CreationDate="2014-03-18T17:33:04.993" Score="2" Body="&lt;p&gt;The processing of the FID uses the Cooley-Tukey FFT algorithm, which can only process $2^n$ data points.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-18T17:33:04.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9295" PostTypeId="2" ParentId="9293" CreationDate="2014-03-18T18:24:10.610" Score="2" Body="&lt;p&gt;Nicotine concentration is a serious concern, as the material is a serious poison (dose matters).  Liquid Chromatography/Tandem Mass Spectrometry is a little pricey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://oxfordchemserve.com/e-cigarette-refill-nicotine-quantitative-test-kit/&quot; rel=&quot;nofollow&quot;&gt;http://oxfordchemserve.com/e-cigarette-refill-nicotine-quantitative-test-kit/&lt;/a&gt; &lt;BR&gt;&#xA;  Specific test kit &quot;designed with the non-chemist in mind&quot; &lt;BR&gt;&#xA;&lt;a href=&quot;http://oxfordchemserve.com/wp-content/uploads/products_img/Oxford_Nicotine_Test.pdf&quot; rel=&quot;nofollow&quot;&gt;http://oxfordchemserve.com/wp-content/uploads/products_img/Oxford_Nicotine_Test.pdf&lt;/a&gt; &lt;BR&gt;&#xA;   Explanation - it is a pH titration of nicotine base. &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.eliquidtest.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.eliquidtest.com/&lt;/a&gt; &lt;BR&gt;&#xA;   Another titration kit.  Google for the product spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saliva or urine nicotine test kits detect nicotine persistent metabolite cotinine and are therefore irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.medpagetoday.com/Pulmonology/Smoking/44238&quot; rel=&quot;nofollow&quot;&gt;http://www.medpagetoday.com/Pulmonology/Smoking/44238&lt;/a&gt; &lt;BR&gt;&#xA;Warning about E-liquid refills.  &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.veppocig.com/product_images/uploaded_images/eliquid-nicotine-strength-guide-2.png&quot; rel=&quot;nofollow&quot;&gt;http://www.veppocig.com/product_images/uploaded_images/eliquid-nicotine-strength-guide-2.png&lt;/a&gt; &lt;BR&gt;&#xA;Claimed dosage guide.  &lt;strong&gt;Caveat emptor.&lt;/strong&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.learn.eversmoke.com/nicotine-strength.html&quot; rel=&quot;nofollow&quot;&gt;http://www.learn.eversmoke.com/nicotine-strength.html&lt;/a&gt; &lt;BR&gt;&#xA;A second dosage opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glycerin can pyrolyze to acrolein.  This is not a good thing to inhale.  NEVER inhale butter flavor.  Butanedione destroys lungs: &quot;popcorn lung,&quot; bronchiolitis obliterans, obliterative bronchiolitis, constrictive bronchiolitis are life-threatenting, cumulative, and irreversible.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-18T18:24:10.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9297" PostTypeId="2" ParentId="8933" CreationDate="2014-03-18T19:25:01.997" Score="5" Body="&lt;p&gt;Two things are wrong in your calculation. Well, it's really just one thing, but the other one bugs me more.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The total mass of the solution.&lt;/li&gt;&#xA;&lt;li&gt;Units.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;1. Mass of the Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To calculate the mass of the total solution, I think it is fair to make the following assumption:&#xA;$$ \rho_\ce{HCl} \approx \rho_\ce{NaOH} \approx \rho_\ce{NaCl} \approx \rho_\mathrm{H_2O} \approx 1000~\mathrm{kg\, m^{-3}}$$&#xA;$$ V_\text{tot} = V_1 + V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that we assume the densities of all the solutions to be the same, and that the total volume is the same as the sum of the two initial volumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above assumptions, we can calculate the total mass of the reaction mixture:&#xA;$$ m_\text{tot} = \rho V_\text{tot} = \rho (V_1 + V_2) = 1~\mathrm{g\, cm^{-3}} \times (200 ~\mathrm{cm^{3}}) = 200~\mathrm{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have calculated the reaction enthalpy to be $\Delta H = -57.36~\mathrm{kJ\,mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to calculate the total heat that was released during this reaction, which leads me to&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2. Units&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Using unit analysis (we have $\mathrm{kJ\,mol^{-1}}$ and need $\mathrm{J}$) we know that we need to multiply the reaction enthalpy with some amount of substance (unit: $\text{mol}$). Now that we know what we have to plug in, we need to know how much of it. Since the reaction enthalpy that you calculated is per $\text{mol}$ of reactant, and we have exactly $0.25~\mathrm{M} \times 0.2~\mathrm{L} = 0.05~\mathrm{mol}$ of reactants, we calculate the total heat as follows:&#xA;$$ q_\text{tot} = n_\text{tot} \times -\Delta H_\text{reaction} = 0.05~\mathrm{mol} \times -57.36~\mathrm{kJ\,mol^{-1}} = 2.868~\mathrm{kJ}$$&#xA;Why the $-\Delta H$? Well, the heat that goes &lt;em&gt;out from the reaction&lt;/em&gt; goes &lt;em&gt;into the mixture&lt;/em&gt;. That is why the sign has to change here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also need the temperature difference, which was measured to be $\Delta T = 3.38~\mathrm{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final result (the heat capacity) is calculated via &#xA;$$ q_\text{tot} = m_\text{tot}c_\text{tot}\Delta T \Rightarrow c_\text{tot} = \frac{q_\text{tot}}{m_\text{tot} \Delta T} = \frac{2.868~\mathrm{kJ}}{200~\mathrm{g} \times 3.38~\mathrm{K}} = 4.243~\mathrm{J\, g^{-1}\, K^{-1}&#xA;}$$&#xA;This is a little bit more than the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities&quot; rel=&quot;noreferrer&quot;&gt;specific heat capacity&lt;/a&gt; of water, so we're good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my walk-through has demonstrated all crucial points of where you went wrong, and that it clarified things for you.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-18T19:25:01.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9299" PostTypeId="2" ParentId="9287" CreationDate="2014-03-18T21:02:46.963" Score="3" Body="&lt;p&gt;Just as for almost every compound of d-elements, $MoS_2$ tends to adsorb molecules from gases on its surface. This is especially true for fine powders, that have high surface-per-gram. MoS2 is especially troublesome, because in strongly reductive atmosphere it is reduced to metallic molybdenum. So, ideally, to get a pure MoS2 you have to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;threat powder with $H_2S$ at about 300-400 Celsium before use&lt;/li&gt;&#xA;&lt;li&gt;press powder into pellet under argon&lt;/li&gt;&#xA;&lt;li&gt;not allow it to touch air for any significant amount of time (more that several seconds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general a very brief exposure to air may significantly oxidize $MoS_2$ catalysts, so you can safely assume that if your $MoS_2$ touched air, it contains oxygen, especially if it is in fine powder.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-03-18T21:02:46.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9300" PostTypeId="1" AcceptedAnswerId="9310" CreationDate="2014-03-18T21:50:30.347" Score="2" ViewCount="665" Body="&lt;p&gt;Today, in class, I learned about the two reactions. &#xA;I know connecting bond releases energy, and breaking it requires and aborb/ consumes energy. &#xA;But on my reference table I, 2C+H2 yielding C2H2 realeasing 227.4 kJ of energy. My point is why it releases energy? Also, where the activation energy comes from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paradox exists in that lower energy level means more stability, but nature also likes to form bonds thus to increase stability. However,  sometimes forming bonds increases the energy level.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2016-06-29T07:43:40.257" Title="Endothermic and exothermic reactions" Tags="&lt;energy&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9301" PostTypeId="2" ParentId="9290" CreationDate="2014-03-18T22:15:58.763" Score="7" Body="&lt;p&gt;If you mean why do we &lt;strong&gt;&lt;em&gt;process&lt;/em&gt;&lt;/strong&gt; raw NMR data using $n^2$ data points, then @Klaus's answer is correct. Although, it is actually usually possible to record an arbitrary number of time domain data points; it is just the processing algorithm is limited to how many data points it can handle. You will either lose the extra time domain data (if using less frequency domain data points) or add zero filling (if using more frequency domain points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean the number of scans as the number of transients added together to give the final raw FID, then this comes about from the requirement for phase cycling a pulse sequence in order to remove unwanted NMR signals that may arise from &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;imperfections in hardware phase (giving rise to quad images)&lt;/li&gt;&#xA;&lt;li&gt;coherent noise&lt;/li&gt;&#xA;&lt;li&gt;generation of undesired magnetization arising in multi-pulse experiments&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, though, the number of scans is not limited to $n^2$. Historically, the number of scans for a 1D spectrum would typically have been confined to a multiple of 8, as a phase cycle requirement. As you start to need to run greater numbers of scans due to sensitivity issues, then you would need to consider the accumulation of noise as well as signal, as @ron has alluded to (doubling the number of scans only give a $\sqrt{2}$ improvement in S/N; you need to run 4 times as many scans to double S/N). It is through this combined relationship that the number of scans, then, adheres to a $n^2$ increase. If 8 scans (required by phase cycling) didn't give sufficient S/N, then we need to double or quadruple the number of scans to see a real improvement in S/N. And so of for ns=16 or 32 or 64.&#xA;For more complicated experiments, phase cycling may have required multiples of 16 or more. I remember running a dqf-COSY experiment which had a phase cycle of 24, so the number of scans had to be a multiple of 24. Yes, these experiments took many, many hours to run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, (there's always a however) in &lt;em&gt;modern&lt;/em&gt; spectrometers (anything newer than about 20 years old), the phase cycling restrictions are largely removed through improvements in hardware design and implementation of gradient pulses. So now we see for most basic 1D experiments, such as a 1H, there is a minimum number of scans of 1. We run many high quality, high resolution, quantitative analyses of samples using long acquisition, single scan experiments.&#xA;For multi-pulse experiments, there is still a requirement to phase cycle, but even for many 2D and 3D experiments, this is limited to multiples of 1, 2 or 4. Very few experiments these days require large numbers of scans to account for phase cycling.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-03-18T22:15:58.763" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9302" PostTypeId="1" AcceptedAnswerId="9315" CreationDate="2014-03-19T00:32:07.840" Score="5" ViewCount="3387" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For which pair of species is the difference in radii the &#xA;  greatest? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A) &lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(B) &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{F^-}$&lt;/span&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(C) &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{O^2-}$&lt;/span&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(D) &lt;span class=&quot;math-container&quot;&gt;$\ce{O^2-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{F^-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure it is not (D) as &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; are quite close to each other on the periodic table and should have atomic radii pretty much the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that gaining an electron increases the atomic radius whereas losing one decreases atomic radius. Also, as you move to the right of the periodic table atomic radius decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the answer could be (A) because &lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; is larger than &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt;, and the other choices result in &lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; becoming smaller and &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; becoming larger which decreases the difference in radii, but this is not the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer given was C. How can this be explained?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="16683" LastEditDate="2020-01-28T00:44:28.320" LastActivityDate="2020-01-28T00:45:09.590" Title="For which pair of species is the difference in radii the greatest?" Tags="&lt;ions&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9303" PostTypeId="1" AcceptedAnswerId="9309" CreationDate="2014-03-19T00:47:50.953" Score="5" ViewCount="1465" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which element has the greatest electrical conductivity?&lt;br&gt;&#xA;  &lt;em&gt;(A)&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{As}$&lt;/span&gt;&#xA;  &lt;em&gt;(B)&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ge}$&lt;/span&gt; &#xA;  &lt;em&gt;(C)&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; &#xA;  &lt;em&gt;(D)&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Sn}$&lt;/span&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am unsure of how to approach this problem. Is there some sort of trend for electrical conductivity or some other method to determine which of these answers the question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure it is not &lt;span class=&quot;math-container&quot;&gt;$\ce{Ge}$&lt;/span&gt; as that is used for semi-conductors, and am pretty sure it is not &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; because that is a non-metal but I don't know how to narrow down between the others.&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="4945" LastEditDate="2019-09-16T11:40:46.190" LastActivityDate="2019-09-16T11:40:46.190" Title="Are there any trends for electrical conductivity?" Tags="&lt;periodic-trends&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9304" PostTypeId="1" AcceptedAnswerId="9305" CreationDate="2014-03-19T01:07:34.210" Score="8" ViewCount="9944" Body="&lt;p&gt;Why is the below structure not considered a valid structure for the nitrate ion? Is it because adding a double bond reduces the formal charges present on the nitrogen? Does the above form exist anywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y0cDk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y0cDkm.png&quot; alt=&quot;possible structure of the nitrate ion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="4945" LastEditDate="2016-02-24T04:23:18.133" LastActivityDate="2017-10-24T08:38:39.337" Title="Is this a valid structure for the nitrate ion?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9305" PostTypeId="2" ParentId="9304" CreationDate="2014-03-19T01:21:47.620" Score="3" Body="&lt;p&gt;Nitrate actually exists as a superposition of these three resonance forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9vH4n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9vH4n.png&quot; alt=&quot;resonance forms of the nitrate ion&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(Source: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/8/81/Nitrate-ion-resonance-2D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, a lone pair from an oxygen with a negative formal change forms a double bond to nitrogen. However, as nitrogen is a period II element, and hence must obey the octet rule, it can not have 5 bonds to other atoms. The already-present double bond then turns into a single bond and the pair of electrons moves onto the oxygen. This &quot;movement&quot; can happen with all three of the oxygen atoms and all three of the bonds. This is called &quot;resonance&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to resonance, the energy of the system is minimised making it more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Lewis structure is not far off. The &quot;true&quot; structure of nitrate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lpx8t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpx8t.png&quot; width=&quot;200&quot; alt=&quot;hybrid resonance structure of the nitrate ion&quot;&gt;&lt;/a&gt; &#xA;(Source: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a5/Nitrate-ion-resonance-hybrid-2D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximately, each oxygen atom contributes one third of a bond with the nitrogen, lowering the formal charge on each atom from what you have written, which does not exist in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true structure is sort of an average of the properties of each possibility for where the electrons could be located, including the three resonance forms above, your structure, and any other configuration. However, these other forms are of minuscule probability to be formed (perhaps read up on wavefunctions and electron distributions if you're interested), and so the main contributors to the &quot;real&quot; structure are the three resonance forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that none of these structures ever exist in isolation for more than the most infinitesimal of time periods.&lt;/p&gt;&#xA;" OwnerUserId="4860" LastEditorUserId="4945" LastEditDate="2016-02-24T04:20:35.633" LastActivityDate="2016-02-24T04:20:35.633" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9306" PostTypeId="1" CreationDate="2014-03-19T04:09:54.480" Score="4" ViewCount="900" Body="&lt;p&gt;To determine the silica ($\ce{SiO2}$) content in Portland cement by gravimetric analysis it is digested in $\ce{HCl}$ and the silica residue is eventually measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chief constituents are $\ce{Ca2SiO4}$ (or $\ce{2CaO.SiO2}$), $\ce{4CaO.Al2O3.Fe2O3}$, $\ce{Ca3(AlO3)2}$ (or $\ce{3CaO.Al2O3}$), and $\ce{CaSiO3}$ (or $\ce{CaO.SiO2}$). Gypsum is also present. What species are present in solution after the $\ce{HCl}$ has reacted with the cement? I assume $\ce{CaCl2}$ and $\ce{H2O}$ would be present, but would $\ce{AlCl3}$ also form? How would I approach this problem? &lt;/p&gt;&#xA;" OwnerUserId="4875" LastEditorUserId="73" LastEditDate="2014-03-20T09:12:36.150" LastActivityDate="2014-06-29T11:39:22.347" Title="How do I perform a gravimetric analysis of Portland cement?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9307" PostTypeId="1" AcceptedAnswerId="9308" CreationDate="2014-03-19T04:35:20.963" Score="12" ViewCount="2958" Body="&lt;p&gt;I've read that in Kolbe's reaction of phenol, i.e. reaction of phenol $(\ce {C6H5OH})$ with $\ce{NaOH}$ forming sodium phenoxide $(\ce {C6H5O^{-}Na^{+}})$, followed by reaction with $\ce{CO_2}$ and acidification; salicylic acid (2-hydroxybenzoic acid ) is formed, but when $\ce{KOH}$ is used instead of $\ce{NaOH}$, 4-hydroxybenzoic acid is formed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why two different products are formed in the same reaction using very similar bases. Moreover, there is electrophilic aromatic substitution on the phenoxide ion in both reactions and  $\ce{O^{-}}$ being an ortho-para directing group must form both products in both the reactions.&lt;/p&gt;&#xA;" OwnerUserId="1018" LastEditorUserId="10185" LastEditDate="2016-05-26T13:35:33.007" LastActivityDate="2020-07-27T06:03:53.627" Title="Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9308" PostTypeId="2" ParentId="9307" CreationDate="2014-03-19T05:28:30.447" Score="11" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I don't understand why two different products are formed in the same reaction using very similar bases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the bases make a a difference, there must be a difference between them.&lt;/li&gt;&#xA;&lt;li&gt;If the anion is the same, the cation must make the difference.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The difference between &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; lies in the ionic radii of 0.95 and 1.33 &lt;span class=&quot;math-container&quot;&gt;$\mathring{A}$&lt;/span&gt;, respectively.&lt;/p&gt;&#xA;&lt;p&gt;It is believed that &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; serves as an anchor and guides &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; into the &lt;em&gt;ortho&lt;/em&gt; position during the formation of the &lt;em&gt;Wheland intermediate&lt;/em&gt;. In the latter, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; is chelated between the oxygen atoms of carboxylate and the phenolate. Apparently, this is very effective.&lt;/p&gt;&#xA;&lt;p&gt;This does not mean that the phenolate doesn't attack &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; via C-4 at all. However, the formation of a &lt;em&gt;Wheland intermediate&lt;/em&gt; (= the rate-determining step) is normally (and definitely in the case of the &lt;em&gt;para&lt;/em&gt; attack) reversible.&lt;/p&gt;&#xA;&lt;p&gt;It is open to question if &lt;strong&gt;full reversibility&lt;/strong&gt; can be claimed for the &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; directed &lt;em&gt;ortho&lt;/em&gt; attack under modest reaction conditions.&lt;/p&gt;&#xA;&lt;p&gt;It is however likely that the back reaction (= cleavage &lt;em&gt;Wheland intermediate&lt;/em&gt;) will stronger come into play at elevated temperatures. Consequently, some 4-hydroxybenzoic acid will be formed then.&lt;/p&gt;&#xA;&lt;p&gt;With the bigger &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;, the anchoring and chelating effect is seemingly not possible, &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; doesn't fit and the &lt;em&gt;para&lt;/em&gt; direction effect of the phenolate prevails.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-07-25T18:12:05.900" LastActivityDate="2020-07-25T18:12:05.900" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="9309" PostTypeId="2" ParentId="9303" CreationDate="2014-03-19T06:00:25.740" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am unsure of how to approach this problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you are not! You have already reduced the number of candidates by good reasoning!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, one &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;could&lt;/em&gt; have a look at conductivity tables or remember in which period the metal blocks start for every main group or&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;remember typical electricity-related uses of $\ce{As}$ and $\ce{Sn}$&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Did you ever form a conducting connection between electronic components (or just some wires) by &lt;em&gt;soldering&lt;/em&gt;? $\Leftarrow$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Do you recall that $\ce{GaAs}$ is used as a semiconductor material in thin film solar cells?)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-19T06:00:25.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9310" PostTypeId="2" ParentId="9300" CreationDate="2014-03-19T07:09:54.503" Score="0" Body="&lt;p&gt;Your idea is in principle correct. The reaction you mentioned works well, you break one H-H bond and create 1 C-C bond (triple bond, in fact) and two C-H bonds. The energy is released. And in most crude approximation it also says where the activation energy barrier comes from - you have to break a H-H bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the paradox - that is the principle of chemistry, to make bonds we wish for. If it means increasing the energy level, thus being less favored by nature, you have to pay for it. By sacrificing other bonds. Those are the highly reactive chemicals humbly stated above reaction arrow, which quite often do all the magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the most revealing example in this I consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_acyl_substitution&quot; rel=&quot;nofollow&quot;&gt;Nucleophilic Acyl Substitution reaction&lt;/a&gt;, where you have a series of derivatives of increased stability. You can always and simply make the more stable out of the less stable, having wide choice of reactions. And there are just a few to climb up, using very nasty chemicals.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-03-19T07:09:54.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9311" PostTypeId="2" ParentId="3965" CreationDate="2014-03-19T09:12:04.523" Score="2" Body="&lt;p&gt;I suggest to forget about the enzymes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pepare for a probably lengthy synthesis - you will need patience and a clever protection/deprotection scheme.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is definitely not applicable to all starting materials but might give some directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ECyKj.png&quot; alt=&quot;quercetin numbering and acetal formation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flavonoid contains a catechol moiety, just like &lt;em&gt;quercetin&lt;/em&gt; (&lt;strong&gt;1&lt;/strong&gt;), the vicinal phenolic $\ce{OH}$ groups can be protected as an acetal by heating the starting material with $\alpha,\alpha$-dichlordiphenylmethane in diphenylether (&lt;a href=&quot;http://dx.doi.org/10.3762/bjoc.5.60&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;). The protecting groups is easily removed by hydrogenation over $\ce{Pd/C}$ at 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the PMB (4-methoxybenzyl) group is another protecting group for (phenolic) $\ce{OH}$ that gets off by hydrogenation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as selectivities for the protection of the other $\ce{OH}$ groups in &lt;strong&gt;1&lt;/strong&gt; are concerned, have a look at the 1962 paper by &lt;a href=&quot;http://dx.doi.org/10.1021/jo01051a041&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonard Jurd&lt;/a&gt;, who described that &lt;strong&gt;the methylation of quercetine pentaacetate in dry acetone predominantly furnishes 7-&lt;em&gt;O&lt;/em&gt;-methylquercetin&lt;/strong&gt;. There are quite some more recent articles citing this!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-19T09:12:04.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9313" PostTypeId="1" AcceptedAnswerId="9345" CreationDate="2014-03-19T12:47:02.613" Score="5" ViewCount="13982" Body="&lt;p&gt;For a lab assignment, I combined &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; sodium sulfate solution with &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; calcium chloride solution. The ionic equation I came up with is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + 2Cl^- (aq) -&amp;gt;}\\&#xA;    \ce{CaSO4(s) + 2Na^+ (aq) + 2Cl^- (aq)}&#xA;\end{multline}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was then asked, in a lab question, the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the reaction called for &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; chloride ion with &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; calcium ion, would you still use &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of each solution for a complete reaction? Explain your reasoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To my understanding, I initially had &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt; chloride ion and &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; calcium ion. With only &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; chloride ion instead, the ionic equation would change to this, or so I'm led to believe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + Cl^- (aq) -&amp;gt;}\\&#xA; \ce{CaSO4(s) + 2Na^+ (aq) + Cl^- (aq)}&#xA;\end{multline}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a (balanced) molecular equation, to my understanding, this would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;\ce{Na2SO4 (aq) + Ca (aq) + Cl (aq) -&amp;gt;}\\&#xA;\ce{CaSO4(s) + NaCl (aq) + Na (aq)}&#xA;\end{multline}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a complete reaction? I have seen conflicting definitions of a complete reaction online:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wiki.answers.com/Q/What_is_complete_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;answers.com&lt;/a&gt; says that a complete reaction means all of the reactants are reacted into products.&lt;/li&gt;&#xA;&lt;li&gt;However, &lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=46704.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;someone on chemicalforums.com&lt;/a&gt; says that a complete reaction means that all of the limiting reactant is used up. &lt;a href=&quot;http://upper.usm.k12.wi.us/academics/faculty/gfriedlander/qcp/QCS/Qcs13/Qcs13.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The University School of Milwaukee&lt;/a&gt; seems to back up what the user on chemicalforums.com said (&quot;all of at least one of the available reactants is used up and converted into products&quot;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My main question is thus: &lt;strong&gt;what is a complete reaction?&lt;/strong&gt; No matter my final equation for this lab question, I still need to know what a complete reaction is so that I can identify them in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition says that this is a complete reaction, since all of the calcium ions and the chloride ions are converted to products. Is this correct? The leftover sodium ion makes me suspicious, so I don't know how to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I later spoke to my lab instructor and he clarified that the question is asking for me to come up with two arbitrary aqueous solutions, one which contains &lt;span class=&quot;math-container&quot;&gt;$\pu{1M}$&lt;/span&gt; of chloride ion and another which contains &lt;span class=&quot;math-container&quot;&gt;$\pu{1M}$&lt;/span&gt; of calcium ion, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl(aq)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(NO3)2 (aq)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, I would need twice the volume of potassium chloride to get the number of chloride ions I need to balance the equation and prevent explosions, i.e. no extra sodium ions leftover.&lt;/p&gt;&#xA;" OwnerUserId="4882" LastEditorUserId="4945" LastEditDate="2018-11-12T09:31:42.663" LastActivityDate="2018-11-12T09:41:13.067" Title="What is a complete reaction?" Tags="&lt;aqueous-solution&gt;&lt;concentration&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9315" PostTypeId="2" ParentId="9302" CreationDate="2014-03-19T13:52:45.173" Score="4" Body="&lt;p&gt;Your argument that &lt;span class=&quot;math-container&quot;&gt;$r_\ce{Li} &amp;gt; r_\ce{F}$&lt;/span&gt; is valid only for the neutral atoms.Consider the lithium ion. Its electronic configuration is exactly similar to helium but the nuclear charge is greater by one unit, and hence is even smaller than the helium atom. On gaining an electron, fluorine now has the electronic configuration of neon with lesser nuclear charge and hence is larger than a neon atom. Similar, oxide ion, with even lesser nuclear charge is the largest. The decrease in radius of lithium on losing one electron (reduction by one entire shell), more than compensates for the difference in the radii of neutral lithium and fluorine. Finally, the order appears to be:-  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Li+ &amp;lt; He &amp;lt; F &amp;lt; Li &amp;lt; F- &amp;lt; O^{2-} }$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2020-01-28T00:44:50.753" LastActivityDate="2020-01-28T00:44:50.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9316" PostTypeId="2" ParentId="9230" CreationDate="2014-03-19T14:07:45.610" Score="3" Body="&lt;p&gt;I think the contradiction you mention in your question is only apparent and is due to a misunderstanding around the word &lt;strong&gt;valency&lt;/strong&gt;, which is by itself a rather ambiguous term (&lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_%28chemistry%29#Valence_versus_oxidation_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;valency vs oxidation state&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, valency yields the number of bonds an element tends to have. For instance, oxygen tends to form $2$ chemical bonds: its valency is then $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valency is different from &lt;strong&gt;valence electrons&lt;/strong&gt;: oxygen has $6$ valence electrons: a valence electron is an electron that can participate in the formation of a chemical bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider arsenic, its valency is $3$ and not $5$, as you write in your question. $5$ is the number of valence electrons. Of course, this value of $3$ is based on the octet rule, that's just a rule (that often works) not a law (that should always work).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering $\ce{GaAs}$, $4$ bonds are actually formed between $\ce{As}$ and $\ce{Ga}$, as discussed in the text given by LievenB in his answer. One should more precisely say that each $\ce{As}$ atom is bonded to $4$ $\ce{Ga}$ atoms and each $\ce{Ga}$ atom is bonded to $4$ $\ce{As}$ atoms, in a diamond-like blende structure.  This is discussed in the book &lt;a href=&quot;http://books.google.ch/books?id=sDCNjgshAmoC&amp;amp;pg=PA65&amp;amp;lpg=PA65&amp;amp;dq=covalent%20bonds%20GaAs&amp;amp;source=bl&amp;amp;ots=p2M1Ohklcu&amp;amp;sig=pmYV1LpfsSwL9wwgJFJ_OBEw5bE&amp;amp;hl=fr&amp;amp;sa=X&amp;amp;ei=028oU4z2I9Kv7Aao5IGgCQ&amp;amp;ved=0CEwQ6AEwBA#v=onepage&amp;amp;q=covalent%20bonds%20GaAs&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fundamentals of Solid State Engineering&lt;/a&gt; of M. Razieghi, here below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Da1GH.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text also says that $\ce{As}$ attracts more strongly electrons. It adds also that for the fourth bond, $\ce{As}$ contributes two electrons, while $\ce{Ga}$ contributes none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, knowing that each $\ce{As}$ is bonded to $4$ $\ce{Ga}$ atoms. Let's consider a given $\ce{As}$ atom.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;That $\ce{As}$ atom and one $\ce{Ga}$ will share $2$ electrons to form a covalent bond: these $2$ electrons are closer to $\ce{As}$: oxidation number will become $-1$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The same $\ce{As}$ atom and a second $\ce{Ga}$ will share $2$ electrons to form a second covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again the same $\ce{As}$ atom and a third $\ce{Ga}$ will share $2$ electrons to form a third covalent bond: these $2$ electrons are again closer to $\ce{As}$: oxidation number will become $-3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The last bond, the fourth, will built up with the $2$ remaining valence electrons of our $\ce{As}$. Like the others, those $2$ electrons will stay closer to the $\ce{As}$: as a consequence, no change in  the oxidation will further occur, since those $2$ electrons are both coming from $\ce{As}$ and not from $\ce{Ga}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Such a reasoning would suggest that $\ce{As}$ has an oxidation number of $-3$ and $\ce{Ga}$ of $+3$ (again &lt;a href=&quot;http://en.wikipedia.org/wiki/Gallium_arsenide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, coming back to the so-called CrissCross method: if you apply the definition you quote in your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;suppose we have two elements $X$ and $Y$ having oxidation numbers x and y respectively. Can we prove that the compound formed will be $X_{y}Y_{x}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and you use oxydation number $+3$ for $\ce{Ga}$ and $-3$ for $\ce{As}$, the logic formula $\ce{GaAs}$ is finally obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other &lt;a href=&quot;http://en.wikipedia.org/wiki/Arsenide&quot; rel=&quot;nofollow noreferrer&quot;&gt;arsenides&lt;/a&gt;, this method will not be easy to apply, since the oxidation number may not be clear for the elements involved, especially if one has no idea about the structure by which atoms are organized.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-20T14:53:19.663" LastActivityDate="2014-03-20T14:53:19.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9317" PostTypeId="1" CreationDate="2014-03-19T14:31:10.870" Score="20" ViewCount="9393" Body="&lt;p&gt;In &lt;a href=&quot;https://www.amazon.co.uk/Electronic-Devices-Circuit-Theory-Boylestad/dp/0132622262&quot; rel=&quot;noreferrer&quot;&gt;Boylestad and Nashelsky's &lt;em&gt;Electronic Devices and Circuit Theory&lt;/em&gt;&lt;/a&gt;, the structure of GaAs is shown as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1UFhZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1UFhZ.png&quot; alt=&quot;GaAs structure, 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arsenic is shown making five covalent bonds to neighbouring gallium atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I recently also saw an alternative depiction in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387921672&quot; rel=&quot;noreferrer&quot;&gt;Razeghi's &lt;em&gt;Fundamentals of Solid State Engineering&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QxV4T.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QxV4T.png&quot; alt=&quot;GaAs structure, 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, arsenic is shown making four covalent bonds to neighbouring gallium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which source is correct? Please give appropriate references or experimental proof.&lt;/p&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="16683" LastEditDate="2017-08-30T13:52:37.893" LastActivityDate="2017-08-30T14:12:52.040" Title="How many Ga atoms are connected to one As atom in solid gallium arsenide?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;erratum&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9318" PostTypeId="2" ParentId="9306" CreationDate="2014-03-19T14:36:56.113" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=VeXQAAAAMAAJ&amp;amp;pg=PA2&amp;amp;lpg=PA2&amp;amp;dq=Gravimetric+Analysis+of+Portland+Cement+%22aluminum&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=VeXQAAAAMAAJ&amp;amp;pg=PA2&amp;amp;lpg=PA2&amp;amp;dq=Gravimetric+Analysis+of+Portland+Cement+%22aluminum&lt;/a&gt; &lt;BR&gt;&#xA;Copy whole URL. Page 2 for clickable index, then page 18. &lt;BR&gt;&#xA;DOI: 10.1039/AN9709500051&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digest it in &quot;strong&quot; HCl (tough lab for employment, worse if you service the fume hoods).  Everything except silica comes out as soluble chlorides (hydrated complex ions).  That says something about Portland cement in acid environments and, unless modified, in seawater.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-19T14:36:56.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9319" PostTypeId="2" ParentId="9317" CreationDate="2014-03-19T15:44:34.167" Score="13" Body="&lt;p&gt;The two-dimensional representations you have given are inadequate and should not be taken too seriously; crystal structures are nearly always three-dimensional. Gallium arsenide has the cubic zincblende structure, shown below. Here, blue atoms can are Ga and orange atoms As (it can be the other way round, too, as these atoms are related by symmetry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNrEv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNrEv.png&quot; alt=&quot;Zincblende structure&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;https://www.tf.uni-kiel.de/matwis/amat/semitech_en/kap_2/illustr/i2_1_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Image source)&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a side note, this structure is similar to that of diamond: in diamond, both types of atoms are carbon.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both gallium and arsenic have coordination number 4; the coordination geometry is tetrahedral about both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, 5-fold symmetry is very rare in crystals (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Quasicrystal&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasicrystals&lt;/a&gt; for more information).  &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="16683" LastEditDate="2017-08-30T14:00:18.967" LastActivityDate="2017-08-30T14:00:18.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9321" PostTypeId="1" CreationDate="2014-03-19T18:16:42.650" Score="4" ViewCount="23942" Body="&lt;p&gt;In case of small molecules like water things are more or less clear, but when I see a large complex molecule like &lt;a href=&quot;http://en.wikipedia.org/wiki/Thyroid_hormone&quot; rel=&quot;noreferrer&quot;&gt;Thyroid Hormone&lt;/a&gt; - how should I find out that this molecule is non-polar? &lt;/p&gt;&#xA;" OwnerUserId="4887" LastActivityDate="2018-03-24T07:01:42.193" Title="How to determine polarity in large molecules?" Tags="&lt;molecules&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9322" PostTypeId="2" ParentId="9230" CreationDate="2014-03-19T18:38:31.993" Score="8" Body="&lt;h2&gt;What is the logic behind this method, how to prove this method?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The logic of the method is in fact trivial. Suppose you have to write a compound knowing his oxidation numbers :&#xA;$$\ce{xA^{a} + yB^{b} -&amp;gt; A_{x}B_{y}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a$ and $b$ are &lt;strong&gt;oxidation numbers&lt;/strong&gt;  that have opposite sings,  $A$, $B$ chemical species and $x$, $y$ stoichiometric coefficient $ \in \mathbb{N}$. &#xA;We can start from this assumption: &lt;strong&gt;if a compound is neutral the algebraic sum of the oxidation numbers of his atoms must be zero&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;( point .4 of current IUPAC definition of the oxidation state)&lt;/a&gt; so we have to find when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ax+by=0 \space [1] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we assumed that $a$ and $b$ have opposite sings so we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ax - by=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a single equation with two variables so you can't solve it easily because it can have several solutions. However when $\textbf{x=b}$ and $\textbf{y=a}$ the equation is always solved, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ ab - ab=0$$&#xA;And so you can always write a compound formula that satisfy equation &lt;a href=&quot;https://i.stack.imgur.com/sBtRx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; using this formula:&#xA;$$\ce{xA^{a} + yB^{b} -&amp;gt; A_{b}B_{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works for $\ce{Al^{+3} , O^{-2}}$:&#xA;$$\ce{xAl^{+3} + yO^{-2} -&amp;gt; Al_{2}O_{3}}$$&#xA; but what's about $\ce{Ti^{+4} , O^{-2}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turns out $\ce{Ti2O4}$ and not $\ce{TiO2}$ we need another step, we need to divide the subscripts by their greatest common factor &lt;strong&gt;gcf(a,b)&lt;/strong&gt; because we have found another solution of the equation &lt;a href=&quot;https://i.stack.imgur.com/sBtRx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; with higher coefficient. In this case the gcf(2,4) is 2 and dividing the stoichiometric coefficient by two we get $\ce{TiO2}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Who, how and when developed this method?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've found it the first time in &lt;a href=&quot;http://books.google.it/books?id=NKYNAQAAIAAJ&amp;amp;q=%22crisscross+method%22+chemistry&amp;amp;dq=%22crisscross+method%22+chemistry&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=4uEpU_uNGuK6ygPguYG4Bg&amp;amp;ved=0CGkQ6AEwBzge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry and Human Affairs&lt;/a&gt; by William Evans Price, George Howard Bruce, World Book Company, 1949. I don't know who is the inventor.&#xA;I think you can easily deduce the crisscross method simply looking at many compounds and their oxidation number! Is not so hard to imagine! &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Is this method only applicable to ionic compounds?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;No, it works for every compounds that can be describe with oxidation numbers. Oxidation numbers are the real charge in ionic compounds and a formal (or imaginary) charge in compounds with covalent bonds, look at $TiO_2$ example.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why compounds like $GaAs$ do not follow this rule?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is not right if we assume that we are dealing with $Ga(+3)$ and $As(-3)$ the equation is this:&#xA;$$\ce{xGa^{+3} + yAs^{-3} -&amp;gt; Ga_{3}As_{3}}$$&#xA;if we divide the subscripts by the &lt;strong&gt;gcf(3,3)&lt;/strong&gt;  that is 3, we get $GaAs$. So this method will be always valid until you can describe the compound effectively with oxidation numbers, unfortunately not all the oxidation numbers are so easy to determine.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact I don't find the &lt;strong&gt;criss cross method&lt;/strong&gt; useful I think is better to calculate the least common multiple of $a$ and $b$ and then divide it by $a$ for $x$ and $b$ for $y$.&#xA;I think this photo, from &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed200513y&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. Chem. Educ., 2012, 89 (11), pp 1436–1438&lt;/a&gt;, summarize the best way to do it:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sBtRx.gif&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Edit:&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;The general solution of |a|x=|b|y  s.t.  x,y∈ℕ is x=n|a| and y=n|b| where n∈ℕ, that is we can use x=2|a|,3|a|...n|a| and similarly y=2|b|,3|b|...n|b| then why use x=|a| and y=|b| only?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;(rearranged after the advices in the comment of mannaia)&#xA;In fact you can write $Na_{2}Cl_{2}$ without breaking any charge neutrality or valence rule. NaCl is a crystal with a lattice and you  can't consider $NaCl$ like a stand alone unit, what matter is the charge balance, the proportion between positive and negative ions. Maybe if you are dealing with a crystal made of 23456 atoms of $Na^{+}$ and 23456 atoms of $Cl^{-}$ you could  write $Na_{23456}Cl_{23456}$ this is in fact a solution of the equation and is not theoretically wrong however if you are talking about a generic crystal of NaCl you should write the ratio between $Na^{+}$ and $Cl^{-}$ ions it has no sense to choose randomly one of the solutions because has stated mannaia: &quot;there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed&quot;. Mathematically speaking we write the ratio to Lowest Terms. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why always take the &quot;gcf&quot;?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In fact we don't take the &quot;gcf&quot; we divide for the &quot;gcf&quot; because this is the best way to get the ratio in the lowest terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore as explained in the $TiO_{2}$ example.&#xA;$$&#xA;A=\{(b,a),(2b,2a),(3b,3a),\ldots\}.&#xA;$$&#xA;Is a set of solutions of the equation but not all the solutions. If you have a=6 b=-4 you have the solution (2,3) too, here is the plot of the solutions in $\mathbb{N}^{2}$.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cNu5I.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compounds with covalent bond is different, because you are dealing with different stand alone units called molecule. For example if you have $C^{-4}$ and $H^{+1}$ the solutions not divide by gcf of the equation have no sense. For example $C_{2}H_{8}$ doesn't exist due the valence of the atoms involved.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-03-24T15:19:27.700" LastActivityDate="2014-03-24T15:19:27.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9323" PostTypeId="2" ParentId="9321" CreationDate="2014-03-19T21:10:46.247" Score="8" Body="&lt;p&gt;Once you move up to &quot;larger&quot; molecules like this, the definition of a &quot;polar&quot; molecule becomes a bit fuzzy. Rather, you will find that molecules contain polar and non-polar groups which all contribute to the overall characteristics of the molecule. It really depends on the context that you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are curious about more macroscale properties, such as boiling point or solubility, a variety of &lt;a href=&quot;http://en.wikipedia.org/wiki/Group_contribution_method&quot;&gt;group contribution methods&lt;/a&gt; have been developed to provide ballpark estimates based purely on the contributions of individual components of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While group contribution methods will primarily give you an idea of bulk properties, you may also be interested in lower level behavior such as micellization or specific protein binding which will depend on the location and arrangements of all of the polar and non-polar groups on a molecular level.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastEditorUserId="2008" LastEditDate="2014-03-20T13:56:12.827" LastActivityDate="2014-03-20T13:56:12.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9325" PostTypeId="1" AcceptedAnswerId="9326" CreationDate="2014-03-19T22:14:14.947" Score="0" ViewCount="418" Body="&lt;p&gt;Here's a challenge problem I created! None of my students could get it; can you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25$ grams of table salt (NaCl) is dissolved in $150$ grams of water. The temperature of the solution is raised to its boiling point until sodium chloride begins to precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much water has boiled off when precipitation begins? Also, what would be the temperature of the boiling solution when precipitation begins?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f9/SolubilityVsTemperature.png&quot; alt=&quot;Solubility Table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4428" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.633" LastActivityDate="2015-06-29T09:04:03.770" Title="How much water has boiled off when precipitation begins?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-02-24T00:02:30.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="9326" PostTypeId="2" ParentId="9325" CreationDate="2014-03-19T23:53:33.337" Score="0" Body="&lt;p&gt;One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. &quot;25.28&quot; g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the problem with ammonium cyanate.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-06-29T09:04:03.770" LastActivityDate="2015-06-29T09:04:03.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9327" PostTypeId="1" CreationDate="2014-03-20T03:30:56.643" Score="4" ViewCount="5814" Body="&lt;p&gt;In the lemon battery experiment, as I understand, both $\ce{Zn}$ and $\ce{Cu}$ meet acid in the lemon, cause $\ce{Zn}$ and $\ce{Cu}$ release electrons to form $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do the electrons follow from $\ce{Zn}$ to $\ce{Cu}$ but not from $\ce{Cu}$ to $\ce{Zn}$? &lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="73" LastEditDate="2014-03-20T09:10:35.180" LastActivityDate="2019-05-09T02:11:34.777" Title="Why do electrons follow from Zn to Cu but not Cu to Zn in the lemon battery experiment?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9328" PostTypeId="1" AcceptedAnswerId="9337" CreationDate="2014-03-20T04:58:30.420" Score="5" ViewCount="2821" Body="&lt;p&gt;I'm looking to purchase a UV lamp for initating some radical polymerizations. Is a hand lamp like the type for examining TLC plates (5-10W, like &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/z169609?lang=en&amp;amp;region=US&quot;&gt;this&lt;/a&gt;) enough power to do this, or do I have to buy something a lot more expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, I'm looking to polymerize a fairly small sample (1mm thick, about 100 mm$^2$ of area, encased in 1mm-thick lab slides) and the chemistry will probably be MMA (initiated with AIBN). I'd like to know for my particular case, but it would be great to know if there's some rules of thumb involved (does it depend on the chemistry? does it depend on the amount? Will different power levels give different final results?)&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2016-11-05T17:48:02.133" Title="How much UV lamp power is needed to initiate radical polymerization?" Tags="&lt;polymers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9329" PostTypeId="1" AcceptedAnswerId="9336" CreationDate="2014-03-20T05:11:31.570" Score="7" ViewCount="13584" Body="&lt;p&gt;Redox: $\ce{Fe + 2HCl  -&amp;gt;  FeCl2 + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation:  $\ce{Fe -&amp;gt; Fe^{2+} + 2e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: $\ce{2H+ + 2e- -&amp;gt; H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{HCl}$ and $\ce{H2}$ are formed by covalent bonds. The $\ce{FeCl2}$ forms an ionic bond.&#xA;The iron actually loses 2 electrons to the two chlorines, so the oxidation equation makes sense to me. But the hydrogen is always in a &lt;strong&gt;covalent&lt;/strong&gt; bond with the chlorine or another hydrogen atom, it is always &lt;strong&gt;sharing&lt;/strong&gt; electrons, never losing or gaining, so the reduction equation doesn't make sense to me. When the $\ce{H}$ is bonded with the $\ce{Cl}$, it &quot;has&quot; two electrons because it is sharing a pair with the $\ce{Cl}$. After the reaction, when the $\ce{H}$ is bonded with another $\ce{H}$, it still &quot;has&quot; two electrons because it is sharing a pair with another $\ce{H}$, so no electron was gained as the reduction equation indicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a little a thinking I came up with an answer, which raised even more questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the covalent bond between the $\ce{H}$ and the $\ce{Cl}$ is a polar bond and it behaves more like an ionic compound because of that, when $\ce{HCl}$ is dissolved in water, the $\ce{H}$ &lt;strong&gt;will&lt;/strong&gt; lose an electron to the $\ce{Cl}$, making $\ce{H+}$ and $\ce{Cl-}$ ions, so there is &lt;strong&gt;no&lt;/strong&gt; covalent bond when $\ce{HCl}$ is in water. The $\ce{H+}$ ions can gain one electron and then make a covalent bond with each other, making $\ce{H2}$ molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the $\ce{Cl-}$ ions in the solution? they have already formed an octet in their outer shell by getting an electron off the hydrogen when they got separated, so they are satisfied. Then why $\ce{Fe}$ atoms oxidize? who is the oxidizing agent? The $\ce{H2}$ molecules are already comfortable with their covalent bond, and the $\ce{Cl-}$ ions are also satisfied with their octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very confused.&lt;/p&gt;&#xA;" OwnerUserId="4890" LastEditorUserId="73" LastEditDate="2014-03-20T09:09:00.897" LastActivityDate="2014-03-20T14:36:49.767" Title="What is the oxidizing agent for Fe in the following redox reaction?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9330" PostTypeId="2" ParentId="9328" CreationDate="2014-03-20T05:57:22.780" Score="4" Body="&lt;p&gt;Sounds interesting! These hand lamps typically have two tubes with emission maxima around $\lambda$ = 254nm and 360 nm, respectively. The latter should be just right to excite AIBN and isn't cut off by Pyrex or comparable glas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More power is obtained from a high pressure mercury lamp, such as the classical &lt;a href=&quot;http://www.msscientific.de/hpk125wlamp.htm&quot; rel=&quot;nofollow&quot;&gt;HPK 125W&lt;/a&gt;, but these lamps are normally used in cylindrical irradiation vessels with &lt;strong&gt;immersion cooling&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a complete system with top- or side irradiation, replacable tubes, metal housing and ventilation have a look at the portfolio of companies like &lt;a href=&quot;http://www.luzchem.com/products/photoreactors.php&quot; rel=&quot;nofollow&quot;&gt;LuzChem&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-20T05:57:22.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9331" PostTypeId="2" ParentId="9327" CreationDate="2014-03-20T11:26:41.687" Score="3" Body="&lt;p&gt;The short answer from the website &lt;a href=&quot;http://www.funsci.com/fun3_en/electro/electro.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Experiments in Electrochemistry&lt;/a&gt; (Carboni, 1998) is that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Copper ($\ce{Cu}$) atoms attract electrons more than do the Zinc ($\ce{Zn}$) atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the Naked Scientists website &lt;a href=&quot;http://www.thenakedscientists.com/HTML/content/kitchenscience/exp/charging-ipods-with-lemons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lemon-Powered iPod - Fruit batteries&lt;/a&gt;, they state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Zinc metal can dissolve to form zinc ions, releasing energy; it also loses electrons. If the zinc is connected to the copper by an electric circuit, these electrons can flow around the circuit and reduce copper ions in the lemon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason why this occurs, is well described in the UC Davis Chemwiki &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Standard_Reduction_Potential#The_Activity_Series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard Reduction Potential: Activity Series&lt;/a&gt;, that as copper has a standard reduction potential of 0.34V, hence will be reduced (as described above), as zinc's standard reduction potential of -0.76V is considerably lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some copper ions dissolve into the lemon's citric acid (Naked Scientists).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is shown in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XMKAY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this continues until the electrons (according to the Carboni, 1998 article linked above):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;concentrate on the copper, the electrons repel each other. When the force of repulsion between electrons and the force of attraction of electrons to the copper become equalized, the flow of electrons stops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some more explanation is available in the question and answer at &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2588/electronegativity-of-zinc-vs-copper-in-galvanic-cell&quot;&gt;Electronegativity of zinc vs copper in galvanic cell&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user4076" LastEditorUserId="-1" LastEditorDisplayName="user4076" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-03-20T11:48:28.537" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9332" PostTypeId="2" ParentId="9327" CreationDate="2014-03-20T11:57:44.597" Score="5" Body="&lt;p&gt;In fact &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; doesn't release electrons, only &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; does. For the usual explanation see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lemon_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;, I will try to explain it in other words to make it much easier.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;One important thing that you should know is that energy comes from metal not from the acid.&lt;/p&gt;&#xA;&lt;p&gt;Some species have the tendency to donate electrons other species the tendency to acquire them. Chemically speaking some species have more internal energy when their are in the reduced form (when they have the maximum number of electrons) others species reach their minimum internal energy when they are in the oxidized form (when they have the least number of electron). In this world every system tend to go to a state of lower energy spontaneously so if two species are not in their lowest energy form if they can exchange their electrons freely they acquire (are reduced) or give (are oxidized) electrons until they reach the lowest energy.  There are also species that do not have any particular tendency or are only little more prone to acquire electrons or to donate them. In all these cases the exchange can occur only if the electrons can move easily from one specie to another, for this reason a good electrolyte is many times needed.&lt;/p&gt;&#xA;&lt;h2&gt;An easy comparison&lt;/h2&gt;&#xA;&lt;p&gt;We could imagine chemical species as men and electrons as money.&#xA;Imagine that there are different kind of men at the antipodes there are avaricious men that needs more money and charitable men that feel the urge to give money. I think that you have no doubt which of them do what when one meet the other. The problem is that many times the transition occurs only if there is the possibility that the flow of money goes from who want to give them to who want them so you need a third agents to create the right environment and conditions for the transition. Furthermore the loss of money must be counterbalanced by something similar.&lt;/p&gt;&#xA;&lt;p&gt;Then there are others type of men that simply don't want to give or take money we can call them &lt;strong&gt;noble&lt;/strong&gt; men and then there are other men that can take money without actually feeling the need, they simply take them, I will call them &lt;strong&gt;hydron&lt;/strong&gt; men.&lt;/p&gt;&#xA;&lt;h1&gt;Your case&lt;/h1&gt;&#xA;&lt;p&gt;Lemon batteries are in fact one of the more complex and debated battery you could examine. When you plug the zinc and the copper in your lemon we can say that we have three main active species: copper, zinc and acid (citric acid). Zinc is one of that metals we compared to a charitable man it tends to donate electrons:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn &amp;lt;=&amp;gt; Zn^{2+} + 2e^{-} }$$&lt;/span&gt;&#xA;Copper however is a noble metal, physics explain this because it has the d-bands of the electronic structure filled, so it doesn't accept zinc's electrons. Instead hydrogen from acid, can accept the electrons in this way:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H+ + 2e^{-} -&amp;gt; H2 ^}$$&lt;/span&gt;&#xA;And this is in fact what actual experimental results show, see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed078p516&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2001,&lt;/strong&gt; &lt;em&gt;78&lt;/em&gt; (4), 516&lt;/a&gt; for reference.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-11-09T04:47:05.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9333" PostTypeId="2" ParentId="5837" CreationDate="2014-03-20T13:55:00.337" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, why does it react with the OH− ion after being broken down?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does it? Chlorine is not stable in water, it undergoes disproportionation :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2 + 3 H2O &amp;lt;=&amp;gt; OCl- + Cl- + 2 H3O+}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performing the reaction in alkaline media simply shifts the equilibrium to the right (Le Chatelier principle) and prevents a possible decomposition of hypochlorous acid ($\ce{HOCl}, pK_a = 7.49$) according to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 HOCl &amp;lt;=&amp;gt; Cl2O + H2O}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-20T13:55:00.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9334" PostTypeId="1" CreationDate="2014-03-20T13:55:53.047" Score="2" ViewCount="7973" Body="&lt;p&gt;I am getting confused between the type of mesomeric effect (+M or -M) shown by compounds containing $\ce{-COX}$ where $\ce{X}$ is an electronegative element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples: $\ce{-CONHCH3}$, $\ce{-COOCH3}$, $\ce{-COOH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, all these compounds must show both a +M as well as a -M effect. Since the group (e.g. $\ce{-COX}$) can accept as well as donate $\pi$ electrons (as in $\ce{CH_2=CH-COOH}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the book says these compounds show only -M effect. Please explain me the reason behind it.  &lt;img src=&quot;https://i.stack.imgur.com/2DLpE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example, isn't the group showing +M effect, since it's donating the electrons?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="418" LastEditDate="2014-03-21T05:57:10.813" LastActivityDate="2014-03-21T05:57:10.813" Title="Understanding Mesomeric Effect" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9335" PostTypeId="2" ParentId="5701" CreationDate="2014-03-20T14:31:02.780" Score="2" Body="&lt;p&gt;Alter the azeotrope by changing the ambient pressure.  Add a third component to remove one component a(e.g., ethanol, water, benzene), leaving the other.  Use another mechanism for separation (membrane permeation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI: 10.1021/ba-1973-0116.fw001 &lt;BR&gt;&#xA;   &lt;I&gt;Azeotropic Data—III&lt;/I&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-20T14:31:02.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9336" PostTypeId="2" ParentId="9329" CreationDate="2014-03-20T14:36:49.767" Score="3" Body="&lt;p&gt;To answer your initial question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The oxidizing agent is the reagent that gets reduced.&lt;/strong&gt; Conversely: &lt;strong&gt;The reducing agent is the reagent that gets oxidized.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, addressing your trouble with the concept:&#xA;You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so:&#xA;$$\ce{HCl -&amp;gt;[][\ce{H2O}] H+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution:&#xA;$$\ce{2H+ + 2e- -&amp;gt; H2 ^}$$&#xA;This is basically our electron sink, because that's where they go &quot;to die&quot;, i.e. to vanish from our solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer we already know: Iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.&lt;sup&gt;1)&lt;/sup&gt;&#xA;$$\ce{Fe^0 -&amp;gt; Fe+ + e- -&amp;gt; Fe^{2+} + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better!&#xA;$$\ce{Fe^{2+} + 2Cl- &amp;lt;=&amp;gt; FeCl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Footnotes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -&amp;gt; Fe^{2+} + 2e-}$ and $\ce{Cl- -&amp;gt; Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-03-20T14:36:49.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9337" PostTypeId="2" ParentId="9328" CreationDate="2014-03-20T15:26:33.567" Score="3" Body="&lt;p&gt;A photopolymerization requires a photoinitiator to capture the energy.  Where does AIBN absorb versus your source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LFvO8.jpg&quot; alt=&quot;AIBN optical absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, that looks good.  AIBN has a strong cage effect.  2,2′-Azobis(2-methylpropionitrile) extrudes nitrogen and the radicals combine without initiating polymerization.  The dimer, tetramethylsuccinonitrile or 2,3-dicyano-2,3-dimethylbutane, is quite toxic.  The superior initiatior is 1,1'-azobis(cyclohexanecarbonitrile), VAZO catalyst 88; &lt;em&gt;J. Org. Chem.&lt;/em&gt; 52 2958 (1987) (e.g., Sigma-Aldrich Chemical Company).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Camphorquinone (yellow powder) can replace AIBN, shifting absorption into the visible, 380 - 510 nm.  That is a tungsten-halogen bulb or a middling blue LED, GaN or InGaN maxing at 450 - 460 nm.  The graph's yellow line is camphorquinone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fz32Z.jpg&quot; alt=&quot;Camphorquinione optical absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular oxygen quenches free radical polymerizations.  If things stay gooey, add a little more initiator to the starting mix to burn out the oxygen.  Now, the fun part!  If you treat the clean glass slides with Rain-X, (spray, wait a few minutes, wipe, lightly buff; maybe lock in at 110 C for 15 minutes) you silanize the surface and polymer will not stick.  It's a miracle for treating vacuum polymerization ampoules.  If you want a cleaner surface and can play chemist, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&lt;/a&gt; &lt;BR&gt;&#xA;Gelest  Aquaphobe CM, diluted by you to 8% in dichloromethane or toluene as a 5-minute soak in a closed glass container.  Soak, rinse with solvent, lightly buff, oven.  The active material gives off HCl, including exposure to humidity.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-20T15:26:33.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9338" PostTypeId="1" CreationDate="2014-03-20T16:00:56.693" Score="12" ViewCount="776" Body="&lt;p&gt;I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not know why (I'm a student studying physics)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I currently have a preliminary understanding of this:&lt;br/&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qxWju.png&quot; alt=&quot;$/pi$ orbit&quot;&gt;&lt;br/&gt;&#xA;In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+&amp;gt;-|-&amp;gt;)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my understanding correct?&lt;/p&gt;&#xA;" OwnerUserId="4892" LastEditorUserId="7448" LastEditDate="2015-04-25T18:55:56.720" LastActivityDate="2019-02-25T10:48:48.630" Title="Why is the orbital angular momentum of a pi electron along the axis of two atoms' molecule one?" Tags="&lt;electronic-configuration&gt;&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9339" PostTypeId="2" ParentId="8566" CreationDate="2014-03-20T16:48:38.703" Score="0" Body="&lt;p&gt;$Cl^-/Cl_2$/ electrode has +1.36 potential, that is significantly higher that $Fe/Fe^{2+}$, so you can use $NaCl$ safely, as iron (and chromium) will oxidize before chloride-ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt the electrolyte will have significant amount of $Cr(VI)$ as $Cr(III)/Cr(VI)$ potential is higher than for iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really care, threat waste electrolyte with Zn granules and then excess of $Na_2CO_3$ solution, collect the solid products and handle them as metal-containing garbage.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-03-20T16:48:38.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9340" PostTypeId="1" CreationDate="2014-03-20T17:00:56.627" Score="3" ViewCount="702" Body="&lt;p&gt;I am aware of the definition and equation of IE and EA. I was wondering if there is a direct relation between this two quantities with the energy of the neutral specie.&lt;/p&gt;&#xA;" OwnerUserId="4333" LastActivityDate="2014-03-21T10:45:12.310" Title="Is there a relation between the ionization energy and the electron affinity?" Tags="&lt;reaction-mechanism&gt;&lt;ions&gt;&lt;energy&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9342" PostTypeId="2" ParentId="9334" CreationDate="2014-03-20T17:28:15.737" Score="2" Body="&lt;p&gt;Following up on Permeakra's comment, the relative importance of resonance structures is typically determined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Maximize atoms that fulfill the octet rule.&#xA;2) Electronegative atoms have negative formal charges and avoid positive charges.&#xA;3) Minimize charge separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carboxylic acid derivatives (esters, amides, etc.) have resonance structures that break #2, but since all atoms fulfill the octet rule, they are important contributors to the overall electronic picture of the compounds. Hopefully the image below is legible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qrNi.png&quot; alt=&quot;Resonance structures of carboxylic acid derivatives&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, resonance (mesomeric) structures that contain an electronegative atom that both breaks the octet rule AND bears a formal positive charge are not reasonable resonance structures and do not contribute to the overall electronic picture of the compound. The resonance structure of the carboxylic acid drawn in the original question falls into this category as the 'top' oxygen both breaks the octet rule and contains a positive charge. &lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-03-20T17:28:15.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9345" PostTypeId="2" ParentId="9313" CreationDate="2014-03-20T19:12:08.303" Score="2" Body="&lt;p&gt;I suspect that you would have both these molecular equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2KCl(aq) + Na2SO4(aq) &amp;amp;-&amp;gt; K2SO4(aq) + 2NaCl(aq)}\\&#xA;\ce{Ca(NO3)2(aq) + Na2SO4(aq) &amp;amp;-&amp;gt; 2NaNO3(aq) + CaSO4(aq)}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question with &quot;What is a complete reaction?&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A complete reaction is one where all of at least one of the available reactants is used up and converted into products. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for example &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt;: In this equation you will need &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; in order for all the reactants to react. However, you only have &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; (so &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; chloride ions). This means that, as you mentioned, you will need the double concentration for the &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; solution to achieve a complete reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case the only way for &lt;strong&gt;all &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; molecules to react&lt;/strong&gt; is to add the double amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; solution. Then this is a &lt;strong&gt;complete reaction&lt;/strong&gt; : at least one reactant is used up in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But say we have a situation in which we have a &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; solution, but a &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; solution, then all the &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; would react, but still \pu{1 M} of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; will not have reacted.If in your lab assignment it is the meaning that you react all &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt;, then in this case you don't have a complete reaction. Only when a two times larger volume of the &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; is added (to 2M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt;), then you will once again achieve a complete reaction. Furthermore you would just get a solution of solvated ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(NO3)2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt;: In this equation you will need &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(NO3)2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; in order for all the reactants to react. You have &lt;span class=&quot;math-container&quot;&gt;$\pu{10 ml}$&lt;/span&gt; of both compounds, each &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt; in concentration: so in this case there is no need to use a larger volume of aqueous &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(NO3)2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="4945" LastEditDate="2018-11-12T09:40:58.983" LastActivityDate="2018-11-12T09:40:58.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9347" PostTypeId="1" CreationDate="2014-03-20T20:27:02.083" Score="1" ViewCount="283" Body="&lt;p&gt;In Germanium, if two electrons go from 1s go to 4p is it in its excited state?&lt;/p&gt;&#xA;" OwnerUserId="4628" LastActivityDate="2014-07-18T23:23:20.757" Title="If two electrons move up to a higher energy level in an atom, is it considered excited?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9348" PostTypeId="2" ParentId="9347" CreationDate="2014-03-20T21:40:02.440" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/PhysRefData/Handbook/Tables/germaniumtable5.htm&quot; rel=&quot;nofollow&quot;&gt;http://physics.nist.gov/PhysRefData/Handbook/Tables/germaniumtable5.htm&lt;/a&gt; &#xA;Energy levels of a neutral germanium atom &#xA;&lt;a href=&quot;https://www.webelements.com/germanium/orbital_properties.html&quot; rel=&quot;nofollow&quot;&gt;https://www.webelements.com/germanium/orbital_properties.html&lt;/a&gt; &lt;BR&gt;&#xA;Middle, electron binding energies for germanium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will certainly be in one of its excited states, yes, probably followed by an Auger cascade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;PRE&gt;&#xA;Label   Orbital   eV &#xA;K       1s       11103 &#xA;L I     2s       1414.6 &#xA;L II    2p1/2    1248.1 &#xA;L III   2p3/2    1217 &#xA;M I     3s       180.1 &#xA;M II    3p1/2    124.9 &#xA;M III   3p3/2    120.8 &#xA;M IV    3d3/2    29.8 &#xA;M V     3d5/2    29.2&lt;br&gt;&#xA;&lt;/PRE&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-20T21:40:02.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9350" PostTypeId="2" ParentId="6266" CreationDate="2014-03-20T22:16:07.693" Score="-1" Body="&lt;p&gt;The universe cooled after the Big Bang. There was certainly an epoch when the universe’s overall temperature decreased from 200 °C to −40 °C throughout a &lt;em&gt;very&lt;/em&gt; large volume. There is your test tube (more like a Florence flask) for abiogenesis and then the panspermia reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you suspect the preceding is clever, how and when (what ambient vacuum temperature) did elements heavier than boron arise?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-06-14T11:27:02.993" LastActivityDate="2015-06-14T11:27:02.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9352" PostTypeId="2" ParentId="9340" CreationDate="2014-03-21T02:10:05.887" Score="8" Body="&lt;p&gt;Okay, here's what I did (and I highly encourage everybody to do this along with this tutorial-style answer, just so you get the feel for prepping and looking at data from the world wild web).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Data Acquisition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, I went and copied the data from the tables at Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt; for the ionization energies and &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; for the electron affinities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I had to clean up this data somewhat. For the electron affinity data I first had to get rid of all the unnecessary data columns. This was quickly done with &lt;code&gt;awk&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   awk '{print $1, $2, $4}' raw-affinity.dat &amp;gt; affinity.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next, I filtered out all the lines containing &quot;-&quot; and &quot;?&quot; using inverse search with &lt;code&gt;grep&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    grep -v &quot;?&quot; affinity.dat | grep -v &quot;-&quot; &amp;gt; affinity2.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And finally I got rid of all these pesky statistical uncertainties with either one or two numbers between parentheses with &lt;code&gt;sed&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    sed -e 's/(.)//g' -e 's/(..)//g' affinity2.dat &amp;gt; affinity3.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the data file looks somewhat like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 H  0.754195&#xA;3 Li 0.618049&#xA;5 B  0.279723&#xA;6 C  1.262118&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now for the ionization data file. Since we only want to look at the first ionization energy we can scrap all the other data. Again, we'll use &lt;code&gt;awk&lt;/code&gt; for this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    awk '{print $1, $2}' raw-ionization.dat &amp;gt; ionization1.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Next we'll filter out all the data we don't want in there (again with &lt;code&gt;grep&lt;/code&gt; and inverse searches):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    cat ionization1.dat | grep -v &quot;use&quot; | grep -v &quot;WEL&quot; | grep -v &quot;talk&quot; &amp;gt; ionization2.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In this file we still have some nasty linebreaks... So I loaded the file in &lt;code&gt;vim&lt;/code&gt; and substituted the line break + &quot;CRC&quot; with nothing and saved it as &lt;code&gt;ionization3.dat&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    :%s/\nCRC//g&#xA;    :saveas ionization3.dat&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I also corrected the value for At manually in &lt;code&gt;vim&lt;/code&gt; while I was at it, because it wasn't entered like all the other data in the Wikipedia table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ionization data file now looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 H  13.59844&#xA;2 He 24.58741&#xA;3 Li 5.39172&#xA;4 Be 9.3227&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All right, so we have the data nicely prepared and we're ready for the next step, which is&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Data Analysis Shenanigans&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I did everything in &lt;code&gt;python&lt;/code&gt; using the numpy and matplotlib packages. The code used can be found &lt;a href=&quot;http://pastebin.com/nMJRfKKY&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't really notice some large trend of something you might expect. The plot gets pretty messy somehow. The darker the scatter points, the heavier the element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5eaw.png&quot; alt=&quot;Ionization Energy vs Electron Affinity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone else notices a pattern that I'm missing right now.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-03-21T10:45:12.310" LastActivityDate="2014-03-21T10:45:12.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9354" PostTypeId="2" ParentId="6948" CreationDate="2014-03-21T05:04:44.817" Score="8" Body="&lt;p&gt;&lt;strong&gt;In the future I recommend consulting March's Advanced Organic Chemistry before asking such questions. It is a priceless reference work for organic synthesis methods. The following text is result of brief consultation with the book.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium shift in Finkelstein reaction is governed by relative nucleofility of halogenide-ions. In general, in polar solvents iodine is the strongest nucleophile, because it is poorly solvated, while in nonpolar solvents chlorine is the strongest solvent as it is smaller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluoride in non-polar solvents should be the strongest nucleophile of all halogenide-ions, but it so strongly solvates in protonic solvents that it almost do not migrate in organic phase even if phase-transfer catalists are present. Strong basicity of fluoride does not help either as beta-eliminations becomes significant side-reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0040-4039(02)00376-3&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; article claims to develop practically meaningful procedure of clean fluoro-de-halogenations.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-10-16T19:10:52.080" LastActivityDate="2015-10-16T19:10:52.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9355" PostTypeId="1" CreationDate="2014-03-21T05:07:09.433" Score="6" ViewCount="12167" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the wavelength of the radiation released when an electron in a hydrogen atom moves from $n = 5$ to $n = 2$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I find this wavelength? &lt;/p&gt;&#xA;" OwnerUserId="4896" LastEditorUserId="16683" LastEditDate="2017-08-29T14:48:00.773" LastActivityDate="2017-08-29T15:37:22.520" Title="Calculate the wavelength of the radiation released when an electron moves from n= 5 to n=2" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9357" PostTypeId="2" ParentId="9355" CreationDate="2014-03-21T05:52:53.873" Score="4" Body="&lt;p&gt;The &lt;em&gt;Rydberg&lt;/em&gt; formula (for the hydrogen atom) is usually given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $R$ is the &lt;em&gt;Rydberg&lt;/em&gt; constant; &lt;strong&gt;take care of the unit ($\mathbf{m^{-1}}$  or $\mathbf{cm^{-1}}$) here&lt;/strong&gt;! &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Balmer series&lt;/a&gt;, this value is indeed correct.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2017-08-29T14:56:45.557" LastActivityDate="2017-08-29T14:56:45.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9360" PostTypeId="2" ParentId="9293" CreationDate="2014-03-21T08:35:00.460" Score="0" Body="&lt;p&gt;If you need a quantitative method witch can determine in the same time the quantity of all component, gas chromatography is good way. Only mass spectrometry detection can resolve aroma problems on your curves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most complicate aspect is chromatographic conditions : column, temperature...&#xA;I don't know the best column for these compounds (propylene glycol, glycerin and nicotine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone can help you to determine how to start your method on GC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;;)&lt;/p&gt;&#xA;" OwnerUserId="4899" LastActivityDate="2014-03-21T08:35:00.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9362" PostTypeId="2" ParentId="7046" CreationDate="2014-03-21T11:17:15.667" Score="4" Body="&lt;p&gt;Let's have a look at some fundamental data and the chemistry involved.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Chlorine is an inflammable gas with a boiling point of -35 °C and a nice yellowish-green colour. It is heavier than air.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is highly irritant and toxic. It is a strong oxidant. With water, it partly decomposes to hydrogen chloride and hypochlorous acid. Your mucous membranes are wet. Inhalation of chlorine results in severe corrosive injuries. Inhale a tad longer to face dyspnoea, coughing of blood and asphyxia. Did I mention the pulmonary oedema and fatality?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Methane is a flammable gas with a boiling point around -160 °C and a flash point around -180 °C. The lower explosion limit of methane in air is around 5%.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Open flames and sparks are not your friend here.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the mixture of methane and chlorine, &lt;strong&gt;red light&lt;/strong&gt; districts are safe.&#xA;The bond dissociation energy of chlorine is $243\, \mathrm{kJ \cdot mol^{-1}}$, you will need blue light ($\lambda$ &amp;lt; 490 nm) to initiate the self-propagating reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 -&amp;gt;[h\nu] 2 Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the chain reaction starts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl* +\  CH4 -&amp;gt; HCl + CH3*}$$&#xA;$$\ce{CH3* +\  Cl2 -&amp;gt; CH3Cl + Cl*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that goes on and on; termination steps are radical recombinations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Cl*  -&amp;gt; Cl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 CH3*  -&amp;gt;  C2H6}$$&#xA;$$\ce{ CH3* +Cl* -&amp;gt;  CH3Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently hydrogen chloride gas is is produced in the reaction, together with methyl chloride. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Methyl chloride is a flammable gas (boiling point around -24 °C, flash point around -20 °C, lower explosion limit in air around 8%).&lt;/li&gt;&#xA;&lt;li&gt;GHS hazard statements file it with H-351: suspected of causing cancer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you perform the reaction with a large excess of chlorine, $\ce{Cl*}$ radicals will attack methyl chloride, generate dichloromethane via the chloromethyl radical $\ce{*CH2Cl}$, etc. Tetrachloromethane ($\ce{CCl4}$) would be the final product then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;That was the healthy and boring part!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When fluorine talks to chlorine, it goes like in the musical &quot;&lt;em&gt;Annie, Get Your Gun&lt;/em&gt;&quot;: &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Anything you can do I can do better!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Particularly when it comes to running havoc on your tissue!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the photolysis of fluorine in the presence of methane is concerned: &lt;strong&gt;People have done that!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical conditions were:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja00517a027&quot; rel=&quot;nofollow&quot;&gt;Deposit methane in a fluorine matrix at 15 K&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja00538a007&quot; rel=&quot;nofollow&quot;&gt;Co-deposit methane/argon and fluorine/argon samples at 12 K&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/j100038a024&quot; rel=&quot;nofollow&quot;&gt;Trap fluorine in a methane matrix at 15 K&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes, these guys had no sense for adventure - &lt;strong&gt;but they weren't blown away either!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-21T12:42:38.127" LastActivityDate="2014-03-21T12:42:38.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9365" PostTypeId="2" ParentId="6917" CreationDate="2014-03-21T14:58:48.713" Score="1" Body="&lt;p&gt;As noted by Ben in the comment, $\ce{NaOH}$ and $\ce{H_2SO_4}$ are strong base and acid, respectively, so they will react till one is completely used up. Those react in 1:2 molar ratio, so it would be easiest to determine if you have more moles of one or another, and it will simply react, using one mole of  $\ce{H_2SO_2}$ per two moles of $\ce{NaOH}$, until there is only one left. If there is one left, you will have a non-neutral solution in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Obvious, obvious mistake... Thanks for pointing it!&lt;/p&gt;&#xA;" OwnerUserId="4105" LastEditorUserId="4105" LastEditDate="2014-03-21T15:10:40.830" LastActivityDate="2014-03-21T15:10:40.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9367" PostTypeId="2" ParentId="5092" CreationDate="2014-03-21T15:29:03.217" Score="1" Body="&lt;p&gt;Removing solvent from goo is very difficult.  I had an oil that was a solid melting above 100 C.  I pumped it under good vacuum as a thin film while doing all the usual tricks with ice then Dry Ice.  It finally crystallized - and bubbled while doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Siliconize the glass flask's inside.  Cheap and easy is Rain-X for car windshields.  Spray some inside the clean dry (air dry &lt;em&gt;not&lt;/em&gt; oven dry) flask.  Swirl for a couple of minutes, drain, dry.  Couple of quick rinses with methylene chloride, set coating by baking at 110 C for 15 minutes.  The coating is stripped by alkali and nucleophiles like NaOH or KOH in water, a base bath, aqueous ammonia, aqueous fluoride.  Renew as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is better, make it a formal procedure using Gelest Aquaphobe® CM or Aquaphobe® CF.  Rain-X is $\ce{(RO)3Si}-$terminated polydimethylsiloxane oligomer in alcohol.  It contains some silicone oil.  The Gelest products are $\ce{Cl3Si}-$terminated oligomers.  You do the 8% dilution in dry methylene chloride or toluene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to spend money, there are medical siliconizers for treating blood-handling glassware.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-21T15:29:03.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9371" PostTypeId="2" ParentId="6312" CreationDate="2014-03-21T23:37:35.090" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] diluted $\ce{H2SO4}$ and $\ce{HNO3}$ [...] $\ce{NH4OH}$ [...] and ammonium molybdate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The yellow precipitate indicates that &lt;strong&gt;phosphate&lt;/strong&gt; was present too. Under these conditions a yellow ammonium molybdatophosphate is formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2PO4^- + 22 H+ + 12 MoO4^2- +3 NH4+ -&amp;gt; \underset{yellow}{(NH4)3[P(Mo3O10)4]} +12H2O}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-21T23:37:35.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9372" PostTypeId="2" ParentId="6312" CreationDate="2014-03-22T02:23:37.853" Score="1" Body="&lt;p&gt;$\ce{PbMoO4}$ has solubility 0.00001161 g/ 100 g and is yellow. I believe, that with diluting acidity of the solution fell down, leading to significant deprotonation of $\ce{H2MoO4}$ and subsequent sedimentation of $\ce{PbMoO4}$.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T18:43:40.237" LastActivityDate="2015-04-02T18:43:40.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9379" PostTypeId="2" ParentId="7931" CreationDate="2014-03-22T18:33:10.770" Score="2" Body="&lt;p&gt;Oxidize to the dialdehyde (Jones reagent), decarbonylate with rhodium (do it at the bench before building the pilot plant), reduce the CO to methanol.  Why would you want to do that?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-22T18:33:10.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9380" PostTypeId="1" CreationDate="2014-03-22T19:18:11.120" Score="4" ViewCount="116" Body="&lt;p&gt;I want to produce alginate nanoparticles from NaAlg. I choosed nebulizing NaAlg solution into $\ce{CaCl2}$ solution but 2% (m/v) NaAlg solution is quite viscous. What should I do to make it inviscid? &lt;/p&gt;&#xA;" OwnerUserId="4915" LastEditorUserId="418" LastEditDate="2014-03-23T00:15:38.003" LastActivityDate="2014-10-19T14:34:46.773" Title="Alginate nanoparticles" Tags="&lt;polymers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9381" PostTypeId="2" ParentId="7868" CreationDate="2014-03-22T20:34:45.143" Score="4" Body="&lt;p&gt;In the special case of reduction to an alcohol in acid (or other moiety that becomes a leaving group under reaction conditions), the resulting (stabilized) carbocation can be further reduced to the hydrocarbon (e.g., catalytic reforming using noble metal catalyst on acid zeolite under hydrogen).  Benzylic positions often reduce to (unwanted) hydrocarbon.  You can push this by using Lewis acid $\ce{AlH3}$ rather than $\ce{LiAlH4}$,&#xA;$$\ce{3 LiAlH4 + AlCl3 → 4 AlH3 + 3 LiCl}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another entry into deoxygenation reduction is to use a potent oxophile like titanium, e.g., McMurray reaction, that is astounding effective for coupling hindered carbonyls into olefins by plucking out the oxygens.  Its cousin the Tebbe methylenation does $\ce{R2C=O}$ to $\ce{R2C=CH2}$, convert a diaryl benzil's carbonyls into methylenes. Follow with olefin metathesis and ethylene extrusion (ADMET polymerization) ton give fully functionalized polyacetlyenes as living polymers, allowing block co-polymer builds, before quench.  What fun!  Here is an example with a &lt;em&gt;p&lt;/em&gt;-phenylene spacer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OYtEl.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4945" LastEditDate="2016-03-15T06:03:09.970" LastActivityDate="2016-03-15T06:03:09.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9383" PostTypeId="2" ParentId="9380" CreationDate="2014-03-22T21:01:34.897" Score="2" Body="&lt;p&gt;Dilute the feed solution tenfold, etc.  Inelegant.  Try solution engineering,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/kosmos.html&quot; rel=&quot;nofollow&quot;&gt;http://www1.lsbu.ac.uk/water/kosmos.html&lt;/a&gt; &lt;BR&gt;&#xA;Chaotropes and kosmotropes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Theta_solvent&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Theta_solvent&lt;/a&gt; &lt;BR&gt;&#xA;Add a little &quot;poor&quot; solvent like ethanol or acetone to ball the polymer, disrupting hydrogen bond crosslinking and reducing chain entanglement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a novel solution not obvious to one skilled in the art.  That qualifies it for patent protection.  A large fraction of the hard answers you need already exist.  They were done by other people doing other things.  Accumulate bone piles of knowledge.  The process for creating ceramic dislocation-pinned nickel superalloys works for Plexiglass, which &quot;dissolves&quot; 30 wt-% copper phthalocyanine as a dye masterbatch.  So it can't be done, so what?  Do it the &quot;other&quot; way.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-22T21:01:34.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9384" PostTypeId="2" ParentId="8057" CreationDate="2014-03-22T22:13:53.877" Score="23" Body="&lt;hr /&gt;&#xA;&lt;h1&gt;A PRACTICAL DEMONSTRATION WITH PYTHON&lt;/h1&gt;&#xA;&lt;p&gt;Nicolau answer is an excellent analysis of the function that describes the titration but I would like to follow the Ben Norris answer (that I think to tackle better the problem at its origin)  using Python hoping that this could be useful to become more confident and explore titration on your own.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Acidify pure water and calculate its pH&lt;/h1&gt;&#xA;&lt;p&gt;I think acidify water should be the first experiment you should do to understand pH. Imagine you have 1 L of water and imagine to add drop by drop HCl 0.1M. Assuming the volume of a drop is 1 mL you are adding 0.0001 mol of &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt; to the solution. With every drop, furthermore, you are adding 1 mL of water to the solution, so the concentration of &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt; after the first drop should be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[ H^{+} ]= \frac{mol~H^{+}_{init} + (mol~H^{+}_{drop} \times n)}{V_{init} + (V_{drop} \times n)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &amp;quot;init&amp;quot; indicate the initial amount of &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;olume of water. and n is the number of drops you added.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;span class=&quot;math-container&quot;&gt;$H^{+}_{init}$&lt;/span&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Before you add the first drop of acid there are already &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O^{+}}$&lt;/span&gt; because water dissociate in &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+ ~, OH- }$&lt;/span&gt; and the product of their concentration (&lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] [OH- ]}$&lt;/span&gt; at 25 °C) is &lt;span class=&quot;math-container&quot;&gt;$10^{-14}$&lt;/span&gt;. In pure water there is the same amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[OH- ]}$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] = [OH- ]}$&lt;/span&gt; and we can write &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]^{2} = 10^{-14}}$&lt;/span&gt; that is &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] = 10^{-7}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The following python code simulates the addition of the drops and plots the concentration:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;dropVolume=0.001#L&#xA;molH=0.0001#mol (you add this mol of H+ for every drop)&#xA;molOH=0.12#molt&#xA;intVolume=1#L&#xA;intHmol=10**(-7)#mol&#xA;n = np.arange(0, 30, 0.1)#here we create the drops.&#xA;#Plot concentration of H3O+&#xA;fig = plt.figure()&#xA;ax = fig.add_subplot(111)&#xA;cH=(intHmol+molH*n)/(intVolume+dropVolume*n)&#xA;plt.ylabel(r'&lt;span class=&quot;math-container&quot;&gt;$ [H_{3}O^{+}]$&lt;/span&gt; ') &#xA;plt.xlabel(r'&lt;span class=&quot;math-container&quot;&gt;$mL$&lt;/span&gt; of acid added')&#xA;plt.plot(n,cH)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The result is this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/R0t87.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can notice that the result is quite obvious: adding acid the &lt;span class=&quot;math-container&quot;&gt;$[H_3O^{+}]$&lt;/span&gt; increase proportionally. Note also that &lt;span class=&quot;math-container&quot;&gt;$H^{+}_{init}$&lt;/span&gt; is negligible.&lt;/p&gt;&#xA;&lt;h1&gt;-log&lt;span class=&quot;math-container&quot;&gt;$[H_3O^{+}]$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;Now you can calculate the pH.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{pH = -log[H3O+ ]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#Plot -log [H3O+]&#xA;fig1 = plt.figure()&#xA;ax1 = fig1.add_subplot(111)&#xA;pH=np.log10((intVolume+dropVolume*n)/(intHmol+molH*n))&#xA;pOH=np.log10((intVolume+dropVolume*n)/((10**(-14))/(intHmol+molH*n)))  &#xA;plt.plot(n,pOH,'--',color='red',label='pOH   &lt;span class=&quot;math-container&quot;&gt;$-log[OH^{-}]$&lt;/span&gt;')&#xA;plt.plot(n,pH,label=&amp;quot;pH   &lt;span class=&quot;math-container&quot;&gt;$-log[H_{3}O^{+}]$&lt;/span&gt;&amp;quot;)&#xA;plt.ylabel(r'pH or pOH') &#xA;plt.xlabel(r'&lt;span class=&quot;math-container&quot;&gt;$mL$&lt;/span&gt; of acid added')&#xA;plt.xlim(-1,30)&#xA;plt.grid(color='red')&#xA;plt.legend(loc=10)&#xA;plt.tick_params(length=4)&#xA;plt.xticks(np.arange(30))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7Zku.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is more interesting. You can see that the pH suddenly decreases from 7 to 4 after the first mL of acid added. This is, in fact, something similar what Ben Norris noted in his answer and is only related to the logarithmic operation. After this introduction, we can deal more easily with a real titration. If you want you can try to increase the pH of pure water adding &lt;span class=&quot;math-container&quot;&gt;$OH^{-}$&lt;/span&gt; and then calculating the hydronium concentration &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]}$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] [OH- ] = 10^{-14}}$&lt;/span&gt; in this way &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]  = \frac{10^{-14}}{[OH- ]}}$&lt;/span&gt; and plot the pH. In the example, I've done the inverse with pOH.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Titration of HCl with NaOH&lt;/h1&gt;&#xA;&lt;p&gt;This is a simulation of a  titration of a solution of HCl with NaOH the system at the beginning has a lower pH that will increase adding drop by drop a solution of  NaOH. In this case, there are already present &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt; from HCl and the following neutralization reaction will occur until all the HCl is neutralized:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + Cl- + Na+ + OH- -&amp;gt; H2O + NaCl}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So we should divide the process into three steps:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ &amp;gt; OH-}$$&lt;/span&gt; until this preposition is true the number of moles of &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt; will lower down by the same amount of &lt;span class=&quot;math-container&quot;&gt;$OH^{-}$&lt;/span&gt; added due to the neutralization reaction: &lt;span class=&quot;math-container&quot;&gt;$$\ce{[ H_{3}O^{+} ]= \frac{mol~H^{+}_{HCl} - (mol~OH^{-}_{drop} \times n)}{V_{init} + (V_{drop} \times n)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ = OH-}$$&lt;/span&gt; at the &lt;strong&gt;equivalence point&lt;/strong&gt;  &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] = [OH- ]}$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] = 10^{-7}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ &amp;lt; OH-}$$&lt;/span&gt; After the equivalence point we are actually &amp;quot;basifying&amp;quot; brine (solution of water and salts), so we can get &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]}$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+]  = \frac{10^{-14}}{[OH- ]}}$&lt;/span&gt; we will use absolute value to avoid negative concentration: &lt;span class=&quot;math-container&quot;&gt;$$\ce{[ H_{3}O^{+} ]= \frac{ \frac{10^{-14}}{\lvert mol~H^{+}_{HCl} - (mol~OH^{-}_{drop} \times n )\rvert}}{V_{init} + (V_{drop} \times n)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Here is the code:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;dropVolume=0.001#L&#xA;molOH=10**(-4)#mol&#xA;molH=10**(-2)#mol&#xA;intVolume=1#L&#xA;n = np.arange(0, 200, 0.1)&#xA;y=[]&#xA;for i in n:&#xA;    if (molH-molOH*i)&amp;gt;0:&#xA;        y.append(np.log10((intVolume+dropVolume*i)/(molH-molOH*i)))&#xA;    if  (molH-molOH*i)==0:&#xA;        y.append(7)&#xA;    if (molH-molOH*i)&amp;lt;0:&#xA;        y.append(np.log10((intVolume+dropVolume*i)/(((10**(-14))/np.absolute(molH-molOH*i)))))&#xA;&#xA;        &#xA;fig = plt.figure()&#xA;ax = fig.add_subplot(111)  &#xA;plt.plot(n,y)&#xA;plt.ylabel(r'pH (&lt;span class=&quot;math-container&quot;&gt;$-log[H_{3}O^{+}]$&lt;/span&gt;)') &#xA;plt.xlabel(r'&lt;span class=&quot;math-container&quot;&gt;$mL$&lt;/span&gt; of acid added')&#xA;plt.grid(color='red')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;With these conditions we obtain the usual titration curve:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QoioS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can easily see that derives directly from the previous example we can also plot the &lt;span class=&quot;math-container&quot;&gt;$[OH^{-}], ~ [H_3O^{+}]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2hdIO.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[H_3O^{+}]$&lt;/span&gt; simply goes lower down to &lt;span class=&quot;math-container&quot;&gt;$10^{-14}$&lt;/span&gt; at the equivalence point, and then continue to goes down while &lt;span class=&quot;math-container&quot;&gt;$[OH^{-}]$&lt;/span&gt; increase.&lt;/p&gt;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&lt;p&gt;So, in fact, the answer to your question is that relative to the Self-ionization of water equilibrium &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] [OH- ] = 10^{-14}}$&lt;/span&gt; at equivalence point you are switching between a solution with an excess of &lt;span class=&quot;math-container&quot;&gt;$\ce{[H3O+] }$&lt;/span&gt; to a solution with an excess of &lt;span class=&quot;math-container&quot;&gt;$\ce{[OH- ]}$&lt;/span&gt; and are these excesses that are determinant. Of course, if you plot the pH you have a steep rise if you plot &lt;span class=&quot;math-container&quot;&gt;$[H_3O^{+}]$&lt;/span&gt; you got only a little step. This is why the usage a logarithmic function (precisely a &lt;code&gt;p&lt;/code&gt; function) is more convenient even if maybe more tricky, otherwise, you can zoom a little bit with Python to the equivalence point, this gives you an idea of what I mean with &amp;quot;excess&amp;quot;...&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9Qg9g.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2020-10-25T07:08:55.023" LastActivityDate="2020-10-25T07:08:55.023" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9385" PostTypeId="1" CreationDate="2014-03-22T22:19:03.417" Score="3" ViewCount="7382" Body="&lt;p&gt;We haven't learned this in class yet, but after looking online, and through my textbook, I can't seem to find a definite answer. I am hopelessly and utterly lost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j7qq1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4396" LastEditorUserId="4945" LastEditDate="2020-06-03T19:11:50.350" LastActivityDate="2020-06-03T19:11:50.350" Title="How to convert an amide to an imine?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9387" PostTypeId="2" ParentId="9385" CreationDate="2014-03-22T23:44:56.893" Score="4" Body="&lt;p&gt;The problem can be solved using &lt;strong&gt;retrosythesis&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You look at the product and imagine how that was made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like looking at Lego construction, trying to identify the pieces - or looking at a pair of jeans, trying to identify how it was stitched together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you go back and do the same for the previous building blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final product apparently is an imine. I'm sure we have seen that before among one of the exercises you worked on - in that case it was an intramolecular reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ZSnS.png&quot; alt=&quot;retrosynthesis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imines are usually made from carbonyl compounds and primary amines under acidic conditions. This one is made from &lt;em&gt;propanal&lt;/em&gt; and &lt;em&gt;ethylamine&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks good here, I see $\ce{H2NCH2CH3,\ H3O+(cat)}$ among your reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the aldehyde then? The number of carbon atoms in the aldehyde and the amide (&lt;em&gt;propionamide&lt;/em&gt; is the starting material) are the same, just the (formal) oxidation of the carbonyl/carboxyl group are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think about a direct reduction of the amide with $\ce{LiAlH4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9wXA5.png&quot; alt=&quot;amide reduction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, that isn't it. Just two steps would not have been enough anyway ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we do remember that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;esters can be reduced to alkanols with $\ce{LiAlH4}$&lt;/li&gt;&#xA;&lt;li&gt;alkanols can be oxidized to aldehydes with PCC (pyridiniumchlorochromate)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All this is among the list of reagents. Nice, we know to make the propanal from a propionic ester!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to turn propionamide into a propionic ester then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct reaction with methanol under acidic conditions does not work, &lt;em&gt;but&lt;/em&gt; after alkylation of the amide with an alkyl halide, we should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets's summarize our five steps then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H57kd.png&quot; alt=&quot;full synthesis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-23T00:03:16.640" LastActivityDate="2014-03-23T00:03:16.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9388" PostTypeId="2" ParentId="9385" CreationDate="2014-03-23T01:45:55.717" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/08ND7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solved!&#xA;All it took was a little reading and understanding :) &lt;/p&gt;&#xA;" OwnerUserId="4396" LastActivityDate="2014-03-23T01:45:55.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9389" PostTypeId="1" AcceptedAnswerId="9428" CreationDate="2014-03-23T04:35:20.387" Score="5" ViewCount="337" Body="&lt;p&gt;I've been using QuteMol for generating 3D animated molecular images and I've encountered a problem; for some reason that's a complete mystery to me when I create a pdb file for etravirine (using MarvinSketch to create a sdf file and then using OpenBabel to convert the sdf to pdb and add implicit hydrogens) and load it I found the bromine atom was all by itself with no bond connecting the two and I would like a bond in the image. Any ideas? Thanks in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following image is the result I'm getting from QuteMol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3r99E.gif&quot; alt=&quot;The result I&amp;#39;m getting from QuteMol&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2014-03-24T15:46:36.340" Title="Halogen bonds in QuteMol" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9390" PostTypeId="2" ParentId="6229" CreationDate="2014-03-23T05:47:10.183" Score="0" Body="&lt;p&gt;It doesn't become pitch black, but it does become dull gray/dark gray. The reason is that even if metal surface is mirror bright, it is usually composed of multiple crystals with slightly different composition and orientations. Because of this differences, different faces of the microcrystals are etched with different speed, forming a complicated surface shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, fine metal powders usually have dark coloration. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-03-23T05:47:10.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9391" PostTypeId="1" AcceptedAnswerId="9397" CreationDate="2014-03-23T07:28:36.873" Score="6" ViewCount="1476" Body="&lt;p&gt;Well, we know that charge is &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_charge#Quantization&quot; rel=&quot;noreferrer&quot;&gt;quantized&lt;/a&gt;. Quoting wikipedia below,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And we know that there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_charge&quot; rel=&quot;noreferrer&quot;&gt;partial charge&lt;/a&gt;. Quoting wikipedia below,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A partial charge is a non-integer charge value when measured in elementary charge units.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGxWG.png&quot; alt=&quot;Polarity of chloromethane (left) and of the related Grignard compound with indication of the partial charge.&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Don't the above two things (&lt;em&gt;and both being from the same respectable source&lt;/em&gt;) contradict with each other?&lt;/p&gt;&#xA;&lt;p&gt;If yes, then which one is &lt;em&gt;correct&lt;/em&gt;? And if both are true, then in which conditions do they follow?&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-03-23T15:25:00.793" Title="Does partial charge violate the law of quantization of charge?" Tags="&lt;molecules&gt;&lt;quantum-chemistry&gt;&lt;formal-charge&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9392" PostTypeId="2" ParentId="9391" CreationDate="2014-03-23T09:31:47.260" Score="5" Body="&lt;p&gt;Both are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any isolated object has integral charge, as any elementary particle has integral charge. However, as we know, elementary particles are not truly located in one point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about stationary states in physics, we can assume, that particles inhabit some volume with special density function indicating how much of the particle in question is in any given unit of volume. When we consider isolated molecule, we can assume that it is in a stationary state. This means, that electrons may be viewed as electron clouds, and different parts of said cloud may be assigned to different atoms in the molecule. In case if total amount of electron density, assigned to specific atom is non-integral, we gain a partial charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever atomic partial charges are estimated, there is a significant arbitrariness in the process, as there is no unquestionable way to define, which parts of electron density assigned to any given atom, as there is not strict definition of atom borders (atomic radii are estimated using equilibrium positions in some circumstances, but in truth electron density is non-zero even far beyond that radii). Still, the concept is useful enough. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-03-23T09:31:47.260" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9393" PostTypeId="1" CreationDate="2014-03-23T09:46:04.330" Score="6" ViewCount="3963" Body="&lt;p&gt;When a room is not ventilated for a long time (days if people are present, weeks if not), humans perceive a change in air quality commonly described as &quot;stale&quot; or &quot;stuffy&quot; air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is going on here chemically? How is &quot;stale&quot; air different from normal air?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the difference cannot be the mere presence of biological exhalations from the inhabitants, as the phenomenon definitely also occurs when nobody lives in the building at all (though it takes a lot longer in that case).&lt;/p&gt;&#xA;" OwnerUserId="4920" LastActivityDate="2014-03-23T10:59:32.043" Title="What is &quot;stuffy&quot;/&quot;stale&quot; air composed of?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9394" PostTypeId="2" ParentId="9393" CreationDate="2014-03-23T10:59:32.043" Score="3" Body="&lt;p&gt;It may not be a complete answer but I can think of a few distinguishable features that may or may not be chemical reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;chemicals created by bacteria and organisms that are present in everyday households (mold, light activating some production from bacteria and micro-organisms, $NH_3, S^{2-}$ related compounds (they smell bad), etc). &lt;/li&gt;&#xA;&lt;li&gt;Compound this with the acknowledged lack of circulation and you have funky smelling air.&lt;/li&gt;&#xA;&lt;li&gt;high humidity: how one can &quot;feel&quot; &quot;stale, muggy&quot; air.&lt;/li&gt;&#xA;&lt;li&gt;high carbon dioxide concentration can make the air feel heavier (from interior design sources).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-03-23T10:59:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9395" PostTypeId="1" AcceptedAnswerId="9408" CreationDate="2014-03-23T12:31:12.977" Score="86" ViewCount="97722" Body="&lt;p&gt;Among my friends it is a sort of 'common wisdom' that you should throw away water after a couple of days if it was taken from the tap and stored in a bottle outside the fridge, because it has 'gone bad'.&#xA;First of all, the couple of days is not very well defined, which already makes me a bit suspicious. Second, I cannot think of anything in tap water that would make the water undrinkable after a couple of days already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone clarify this issue for me? Does tap water really 'go bad' after a couple of days outside the fridge? Why? &lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="21964" LastEditDate="2016-08-04T15:51:08.427" LastActivityDate="2016-08-04T15:51:08.427" Title="Does water really 'go bad' after a couple of days?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;biochemistry&gt;&lt;water-treatment&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="9396" PostTypeId="2" ParentId="9395" CreationDate="2014-03-23T12:57:36.980" Score="39" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I cannot think of anything in tap water that would make the water undrinkable after a couple of days already.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tagging the question with &lt;a href=&quot;/questions/tagged/biochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;biochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;biochemistry&lt;/a&gt; probably points in the right direction. The only effect I can think about is growth of anaerobic microbes, e.g. &lt;em&gt;Escherichia coli&lt;/em&gt;, causing diarrhea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;There's good chance to keep the water from 'going bad' and even 'turn it good' again when keeping the water in PET bottles in the sunlight, letting &lt;a href=&quot;http://www.cdc.gov/safewater/solardisinfection.html&quot;&gt;SODIS&lt;/a&gt; (solar disinfection) work for you.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-23T14:53:25.920" LastActivityDate="2014-03-23T14:53:25.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9397" PostTypeId="2" ParentId="9391" CreationDate="2014-03-23T13:34:28.857" Score="5" Body="&lt;p&gt;Partial charges are like resonance structures: they are a convenient notation which captures the most important aspects of a phenomenon, but they are not what is &quot;really&quot; going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say that a molecule contains partial charges, what we mean is that the electric field surrounding a bond is polarized &lt;em&gt;as if&lt;/em&gt; some fractional number of elementary charges had been displaced from one end to the other.  The phenomenon — the polarized electric field — is real, but the mechanism that produces it does not violate charge quantization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the electrons always have charge −1, and all of the protons always have charge +1.  But the electrons' &lt;em&gt;positions&lt;/em&gt; are indeterminate.  (The protons' positions are also indeterminate, but on a much smaller scale which is usually neglected in this scenario.)&#xA;A standard technique for calculating the electric field of the molecule is to take the nuclei as fixed and then &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow&quot;&gt;allow the valence electrons' wave functions to range over the entire molecule&lt;/a&gt;.  When you do this, you often discover that some bonds are &lt;em&gt;polar&lt;/em&gt;, with electrons (formally belonging to the atoms at both ends of the bond) more likely to be found at one end than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric field of such a bond is numerically the same as the electric field that would be produced if you could transfer a fraction of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Elementary_charge&quot; rel=&quot;nofollow&quot;&gt;elementary charge&lt;/a&gt; from one end of the bond to the other.  But that is not what has happened.  You can think of what is really going on in these terms: the electrons are moving rapidly back and forth along the bond, but they are &lt;em&gt;slightly more often&lt;/em&gt; at one end of the bond, so the average charge density at one end is greater, but not by a full elementary charge.  (This too is inaccurate, but is a standard way to bend quantum theory to make it more comfortable to intuition.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(N.B. this is the same as permeakra's answer, but in less technical language.)&lt;/p&gt;&#xA;" OwnerUserId="4921" LastEditorUserId="4921" LastEditDate="2014-03-23T15:25:00.793" LastActivityDate="2014-03-23T15:25:00.793" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9398" PostTypeId="2" ParentId="9391" CreationDate="2014-03-23T13:35:42.527" Score="2" Body="&lt;p&gt;It does violate that particular statement (and not quantization of charge). You just need to understand that charge is quantized mean that there exists a unit of charge which is indivisible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may say that charge on an object is always integral multiple of multiple of elementary charge, but it is wrong, as you have pointed out. It is just used to say that elementary charge is indivisible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, when an electrically neutral atom bonds chemically to another neutral atom that is more electronegative, its electrons are partially drawn away. This leaves the region about that atom's nucleus with a partial positive charge, and it creates a partial negative charge on the atom to which it is bonded. It is to be noted that elementary charge (electron in your case) is not divided.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Principle of quantization of charge can be stated as: Charge on any &lt;em&gt;isolated object&lt;/em&gt; is always integral multiple of elementary charge.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is to be noted that electron is still elementary particle. Proton and neutrons are not elementary. Quarks are now believed to be smallest indivisible unit of charge.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2014-03-23T13:48:00.323" LastActivityDate="2014-03-23T13:48:00.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9399" PostTypeId="1" AcceptedAnswerId="9400" CreationDate="2014-03-23T14:03:24.290" Score="2" ViewCount="2348" Body="&lt;p&gt;The answer is likely very obvious but I know pretty much nothing about this so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I make an alloy, using an alloy? That is, could I make an alloy of steel (in itself an alloy of iron and carbon) and silver?&lt;/p&gt;&#xA;" OwnerUserId="4922" LastEditorUserId="4734" LastEditDate="2014-03-23T15:10:44.890" LastActivityDate="2014-06-17T22:36:26.390" Title="can an alloy be used to make another alloy?" Tags="&lt;metal&gt;&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9400" PostTypeId="2" ParentId="9399" CreationDate="2014-03-23T14:19:43.187" Score="7" Body="&lt;p&gt;An alloy is purely a mixture of metals (and sometimes non-metals) which are non-chemically bonded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, you could mix steel and silver to create an alloy. There would be nothing stopping you from doing this, but I cannot say whether or not it would produce any desirable results. Mixing alloys with other materials is merely creating another alloy. Stainless steel is actually just regular steel combined with Chromium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alloy is usually created to increase certain properties of its constituents. For example, iron is mixed with carbon in order to create steel, which is far stronger and can be used to manufacture items such as girders. Or, in other cases, Chromium may  be added, if the steel is to be used in, say, cutlery.&lt;/p&gt;&#xA;" OwnerUserId="4923" LastActivityDate="2014-03-23T14:19:43.187" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9401" PostTypeId="1" CreationDate="2014-03-23T16:39:12.560" Score="5" ViewCount="6356" Body="&lt;p&gt;Why is $\ce{HClO4}$ more acidic than $\ce{HBrO4}$? But $\ce{HCl}$ is less acidic than $\ce{HBr}$? What determines the acidity? Is it the concentration of $\ce{H+}$? If so, how do these trends develop along the halogen group?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="3735" LastEditDate="2014-03-24T03:55:24.440" LastActivityDate="2020-12-20T04:53:26.697" Title="Halogen Oxyacid Trend (acid strength)" Tags="&lt;acid-base&gt;&lt;periodic-trends&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9402" PostTypeId="2" ParentId="2439" CreationDate="2014-03-23T16:46:18.383" Score="0" Body="&lt;p&gt;My guess would be that the yellow-brown color in your tapwater is mainly caused by small suspended particles of rust (i.e. hydrated &lt;a href=&quot;//en.wikipedia.org/wiki/Iron%28III%29_oxide&quot; rel=&quot;nofollow&quot;&gt;iron(III) oxides&lt;/a&gt;) from old iron pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in places with well-maintained plumbing and good drinking water, such rust contamination can often appear temporarily after maintenance work or some other interruption in the water supply dislodges built-up rust deposits from the pipes.  If you're getting rusty water all the time, it suggests either that the water distribution network in your building / town is in pretty poor shape, and/or that the raw water supply used to produce your tapwater is already quite rich in dissolved iron, which will precipitate out into rust when the water is aerated and chlorinated at the water processing plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas, I can't think of any simple &quot;household chemistry&quot; method to get rid of the rust suspension in a way that would be practical for drinking water.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In theory, if you simply filled a large plastic container with the rusty water and let it sit long enough, the rust should settle to the bottom.  However, if the particles are very fine (and they likely are), that may take a &lt;em&gt;long&lt;/em&gt; time.  You could speed the process up with a centrifuge, but that takes it way out of the &quot;easy and practical&quot; range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you just wanted to get rid of the visible coloration, you could add a chelating agent like &lt;a href=&quot;//en.wikipedia.org/wiki/Citric_acid&quot; rel=&quot;nofollow&quot;&gt;citric acid&lt;/a&gt;, &lt;a href=&quot;//en.wikipedia.org/wiki/Polyphosphate&quot; rel=&quot;nofollow&quot;&gt;polyphosphate&lt;/a&gt; or &lt;a href=&quot;//en.wikipedia.org/wiki/EDTA&quot; rel=&quot;nofollow&quot;&gt;EDTA&lt;/a&gt; to dissolve the rust.  However, this will leave you with a solution of e.g. ferric citrate, which, while indeed drinkable in principle, may not quite have the neutral flavor you'd expect of clean water.  (This could still be worth a try, if only to check if the color is indeed caused by rust; citric acid can be found in many stores, both as a food additive and as &quot;biodegradable rust / scale remover&quot;.  Do make sure it's pure, if you intend to drink the result, and don't overuse it.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most practical solution may simply be to get your own water filter.  You'll want to get one that's suitable for removing rust particles, or whatever it is that's actually coloring your water.  If this is indeed a common problem in your town, a local home improvement store can likely recommend a suitable filter system for you.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="283" LastActivityDate="2014-03-23T16:46:18.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9403" PostTypeId="1" AcceptedAnswerId="9413" CreationDate="2014-03-23T17:33:48.257" Score="2" ViewCount="173" Body="&lt;p&gt;I have to do an experiment where I will find out what the influence of the concentration on the rate of a homogenous reaction is.&lt;/p&gt;&#xA;&lt;p&gt;We are asked to create 2 solutions, respectively A and B:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{KI + Na2S2O3 + Starch + water}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O2 + H2SO4 + water}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As long as A and B aren't added together, the reaction cannot start.&lt;/p&gt;&#xA;&lt;p&gt;Now what I'm supposed to do is make 10 different A and B solutions and add these together and measure the time it takes for the solution to turn blue (Iodine + Starch as an indicator).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Now, my problem is that the following questions are also asked:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&amp;quot;Write down the different reactions that take place&amp;quot;&lt;/strong&gt;&lt;br /&gt;&#xA;This seems dubious to me but might not be.&lt;/p&gt;&#xA;&lt;p&gt;e.g. Does dissolving into ions also count as a reaction?&lt;br /&gt;&#xA;Does the Starch (for which I have no given chemical formula) react in solution A?&lt;br /&gt;&#xA;I've also read online that adding &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; creates a &amp;quot;Piranha&amp;quot; solution (&lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+ + HSO4- + O}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO5 + H2O}$&lt;/span&gt;, I've found both online but I don't know which is correct here.)&lt;/p&gt;&#xA;&lt;p&gt;In conclusion I just don't know which reactions I should give (Everything that happens in A and B isolated from eachother or everything that happens when I add A and B together? Either way, I'm also not sure what the reactions would be because, for instance, I don't have any certainty on what the chemical composition of the Starch is.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Is the reaction a multistep reaction or a single step reaction? Why? If it's a multistep reaction, write down the partial reactions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Assuming that adding B to A is &amp;quot;The&amp;quot; reaction, how do I find out if there are multiple steps or not? How do I find these steps?&lt;/p&gt;&#xA;&lt;p&gt;Given that this question is asked already makes me think that it is in fact a multi-step reaction.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="95133" LastEditDate="2021-05-04T06:36:45.867" LastActivityDate="2021-05-04T06:36:45.867" Title="The effect of concentration on the rate of a homogenous reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9404" PostTypeId="2" ParentId="9395" CreationDate="2014-03-23T17:36:12.063" Score="6" Body="&lt;p&gt;I am not an expert and its been a long time since I last studied chem and bio. The best answer will probably only come from a biologist. But out of interest, I will try to answer. Also, I suggest that you post this question on websites such as &lt;a href=&quot;http://www.askabiologist.org.uk/&quot; rel=&quot;noreferrer&quot;&gt;http://www.askabiologist.org.uk/&lt;/a&gt; . Its likely that you will get some &#xA;good answers there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the water in the bottle is thoroughly &quot;purified&quot; at the water plant , there has been absolutely no contamination (pipes, faucet etc.) on the way, and the bottle and handler are clean, then it is unlikely that water will go bad. Example that suggests contamination at faucet - &lt;a href=&quot;http://www.journalofhospitalinfection.com/article/S0195-6701(98)90295-X/abstract&quot; rel=&quot;noreferrer&quot;&gt;http://www.journalofhospitalinfection.com/article/S0195-6701(98)90295-X/abstract&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that most bottles are 100% air-tight. So, the next thing I would check is if there are any water-borne pathogens that can spread by air. If such pathogens exist, then is it possible for them to enter the bottle through small &quot;air spaces&quot; ? That seems unlikely to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if water is microbe free and can be kept that way by a bottle, then it should be good for a &quot;long&quot; time. &lt;/p&gt;&#xA;" OwnerUserId="4926" LastActivityDate="2014-03-23T17:36:12.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9405" PostTypeId="2" ParentId="2439" CreationDate="2014-03-23T18:19:16.350" Score="1" Body="&lt;p&gt;I believe, there are two possible options, that are not immediately harmful for a drinker&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;fine rust particles in water. I suspect it is not your case, but if it is, fine mechanical filter should do the job. To check, add rust-removing chemicals, such as oxalic or citric acid, to warm colored water. If the coloration will vanish, you probably can remove the rust with fine mechanical filter (beware, I have absolutely no idea how fine the filter may be in your case). The reason is bad condition of water pipe system and should be dealt with accordingly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;organic compounds, most likely tannin contamination, is very possible if your water source is swamp-fed. The contamination may be very durable, and, honestly, even being natural for local water, still may be harmful. To check, find bleaching agent based on any hypochlorite, and see if it will destroy/change the coloration. If it will, most likely the source is organic and may be slightly harmful if used for drinking for a long time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There should be also some government subdivision, concerned with water quality regulation. I believe, you will find more information if consult with them. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-03-23T20:48:59.537" LastActivityDate="2014-03-23T20:48:59.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9406" PostTypeId="1" CreationDate="2014-03-23T19:49:57.707" Score="1" ViewCount="1576" Body="&lt;p&gt;I was wondering what happens when you boil dry a solution of several salts for example a solution of KCl and NaNO3. Since the ions dissociate I was wondering if the ions would combine randomly upon being boiled out of solution making the original salts as well as KNO3 and NaCl? Does this happen? If not why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in a related question when you mix two solutions to form a precipitate in a double displacement reaction will the insoluble salt always form or are there situations where an insoluble salt can form but doesn't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2014-03-23T22:41:39.370" Title="Evaporation of a mixture of salts" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9407" PostTypeId="2" ParentId="5025" CreationDate="2014-03-23T20:18:36.533" Score="2" Body="&lt;p&gt;When you left steel wool in vinegar, iron reacted with acetic acid to produce ferrous acetate and HYDROGEN. The so produced NASCENT hydrogen is very strong reducing agent and it reduced the earlier formed brown ferric acetate (due to the presence of H2O2)  ---&gt; to ferrous acetate. This explains the loss of reddish brown colour.&lt;/p&gt;&#xA;" OwnerUserId="4931" LastActivityDate="2014-03-23T20:18:36.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9408" PostTypeId="2" ParentId="9395" CreationDate="2014-03-23T21:09:03.613" Score="49" Body="&lt;p&gt;First of all, it depends on how the tap water was treated before it was piped to your house. In most cases, the water was chlorinated to remove microorganisms. By the time the water arrives at your house, there is very little (if any) chlorine left in the water. When you fill you container, there is likely to be some microorganisms present (either in the container or in the water).&#xA;In a nutrient rich environment, you can see colonies within 3 days. For tap water, it will probably take 2 to 3 weeks. But that doesn't mean that the small amount of growth doesn't produce bad tasting compounds (acetic acid, urea, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW Nicolau Saker Neto, cold water dissolves more gas than hot water. Watch when you heat water on your stove. Before it boils, you will see gas bubbles that form on the bottom and go to the surface (dissolved gases) and bubbles that disappear while rising to the surface (water vapor).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-23T21:09:03.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9409" PostTypeId="2" ParentId="5025" CreationDate="2014-03-23T21:30:52.730" Score="3" Body="&lt;p&gt;Since your solution is acidic (from the acetic acid), you cannot have any carbonates; they react with the acid and carbon dioxide bubbles off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the brown solution is iron(III) acetate and the white precipitate is iron(II) acetate. The darkness of the solution indicates the concentration of iron(III) acetate.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-23T21:30:52.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9410" PostTypeId="1" CreationDate="2014-03-23T22:40:29.113" Score="4" ViewCount="1730" Body="&lt;p&gt;First, I would like to say that I understand the basic idea of disassociation and solubility. However, as I study intermolecular forces I feel like this basic Idea is being challenged. My understanding from basic chemistry is that ionic bonds are much stronger than covalent bonds, due to the actual exchange of electrons and the formation of cations and anions, causing an attractive force between the opposite charges to &quot;hold&quot; the compound together. When explaining solubility with IMF theory, however, I fail to understand how the IM forces present in water--dipole-dipole, H-bonds, and dispersion--are able to overcome that attraction between the cation and anion present in the ionic bond; because, I was under the impression that actual bonds are always stronger than IMF bonds. So, for instance, does it not infer that if, say, Sodium Chloride is placed into water, and the attraction of the h-bonds in water cause the NaCl to essentially break apart into ions, that the inherent attractions between cations and anions in an ionic compound are weaker than the attraction force of the h-bond water molecule?&lt;/p&gt;&#xA;" OwnerUserId="4936" LastActivityDate="2014-08-13T15:13:15.107" Title="Why Ionic compounds dissolve into water" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9411" PostTypeId="2" ParentId="9406" CreationDate="2014-03-23T22:41:39.370" Score="3" Body="&lt;p&gt;As the solvent evaporates the least soluble salts precipitate in succession.  This is used during seawater and salt lake evaporations to isolate values beyond NaCl. Overall, you will get a mixture of combinations of cations and anions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supersaturation is not uncommon.  Metastable solutions usually require nucleation sites to get over their initial activation energy for solids' crystallization.  High melting points (strong lattice forces) are difficult to supersaturate.  Benzil in xylenes remarkably supersaturates, and is stable for a week or more in silanized glassware and cooled very slowly within styrofoam and rags.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-23T22:41:39.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9412" PostTypeId="2" ParentId="7560" CreationDate="2014-03-23T22:41:47.903" Score="4" Body="&lt;p&gt;Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-23T22:41:47.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9413" PostTypeId="2" ParentId="9403" CreationDate="2014-03-23T22:48:44.020" Score="1" Body="&lt;p&gt;You could look up &quot;iodine clock reactions&quot; and get the straight skinny on the chemistries and their kinetics.  You might also look up &quot;Theory of Experimentation&quot; and see how confounded variables cannot be teased apart by varying them individually.  Instruct your instructor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_clock_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Iodine_clock_reaction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-23T22:48:44.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9414" PostTypeId="2" ParentId="7406" CreationDate="2014-03-23T23:03:47.073" Score="7" Body="&lt;p&gt;According to www.webelements.com, &quot;There is no one single or best structure for the periodic table...&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the table was split at lanthanum since it kept the elements of the &lt;em&gt;d&lt;/em&gt; orbitals together. You could argue that the table should be split with Sc, Y, La and Ac together, with the elements Ce through Lu and Th through Lr as the &lt;em&gt;f&lt;/em&gt; orbital elements. The next column would have Ti, Zr, Hf and Rf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, all elements have &lt;em&gt;f&lt;/em&gt; orbitals, below lanthanum they are not usually populated. Lanthanum is the first element that could have an electron in the &lt;em&gt;f&lt;/em&gt; orbital in the ground state. Since the transition energy of the electron between in the &lt;em&gt;d&lt;/em&gt; orbital and the &lt;em&gt;f&lt;/em&gt; orbital is quite low, there should be a significant number of electrons occupying the higher energy orbital, even at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-23T23:03:47.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9415" PostTypeId="2" ParentId="9401" CreationDate="2014-03-23T23:17:37.370" Score="1" Body="&lt;p&gt;Assuming you are talking about acids in water, and taking into account that the acids you mention are all regarded as &lt;em&gt;strong&lt;/em&gt; acids, I would focus on the relative stability of the corresponding conjugated bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br-}$ being larger than $\ce{Cl-}$ can better allocate the negative charge. Then, $\ce{HBr}$ ($pK_{a}$= −8.7) is stronger acid than $\ce{HCl}$ ($pK_{a}$=-6.3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand $\ce{ClO4-}$ is more stable than $\ce{BrO4-}$. Here, another factor plays a role and should explain the difference in stability: the charge dispersal associated to &lt;strong&gt;resonance&lt;/strong&gt;. Being the structure similar, we would expect the same effect: however, the resonance stabilization that characterizes such kind of ions has to be more effective for $\ce{ClO4-}$, due to a better overlap between $p$ orbitals ($2p$ of oxygen with $3p$ of chlorine, better than with $4p$ of bromine). As a consequence, $\ce{HClO4}$ ($pK_{a}$= −8) is stronger acid than $\ce{HBrO4}$ ($pK_{a}$= ? haven't found it, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_strength#Almost_strong_acids&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; $\ce{HBrO4}$ is said to be an &quot;almost strong acid&quot;, therefore not stronger than $\ce{H3O+}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These considerations tentatively explain the difference in &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_strength#Common_strong_acids&quot; rel=&quot;nofollow&quot;&gt;$pK_{a}$&lt;/a&gt; for such strong acids.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2014-03-23T23:17:37.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9416" PostTypeId="2" ParentId="9401" CreationDate="2014-03-24T01:01:51.620" Score="0" Body="&lt;p&gt;There are two different kinds of acids mentioned. First, binary acids such as $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, etc. The other acids mentioned were inorganic oxyacids or oxoacids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The most important factor in oxoacid strength is the ability to donate a proton which is influenced by electronegativity and bond polarity.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxoacids are usually designated with X as the central atom, and thus, their structure can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -\text{X}-\text{O}-\text{H} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X can be bonded to other things. i.e. hydrogens, more -OH groups, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If X was an alkali or alkaline earth metal, and thus electropositive, the corresponding compound would be a base. This is not an acid. The reason for this is the $X-O$ bond is extremely covalent, almost to the point where the strength is equivalent to bond in ionic strength. The oxygen atom thus has a large partial negative charge $\delta^- $ and the electron density accumulates on the $-O-H$ while the hydrogen, $\ce{H^+}$, binds tightly with the oxygen, preventing deprotonation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number of lone oxygens on $-X$ increases, the strength of the oxoacid increases as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from Principles of Modern Chemistry by Oxtoby et. al 7th Edition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the formula of these acids is written as $XO_nOH_m$, the corresponding acid strengths fall into distinct classes according to the value of n, the number of lone oxygen atoms  Each increase of 1 in n increases the acid ionization constant $K_a$ by a&#xA;  factor of about $10^5$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another way to examine the strength of acids is to look at the stability of their conjugate base which can be written as $XO_{n+1}(OH)^-_{m-1}$. As more oxygens are added on the central atom, the net negative charge is distributed more evenly around the base. As bases become more stable, they are less prone to accepting a proton. Remember that conjugate bases of strong acids are weak so intuitively, this makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put briefly, as $-X$ becomes more electronegative, electron density is pulled towards the central atom. As the electron density is pulled from the $-H$ bond to $-X-O$, the $O-H$ bond breaks easily, and a proton is produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we discuss organic acids and not binary nor inorganic acids, we not only have to talk about electronegativies, but steric hinderance, and resonance structures as well. If you would like, I can explain those as well.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-03-24T01:01:51.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9419" PostTypeId="1" CreationDate="2014-03-24T03:44:52.823" Score="5" ViewCount="5619" Body="&lt;p&gt;I'm using a sensor that measures $\ce{CO2}$ in ppm by volume aka atmosphere.&#xA;What I'm using it for is to measure $\ce{CO2}$ concentration in water which is measured in ppm by weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Ppmwco2} = \text{ppmvco2} \times 0.8317 \times 44.01 / 10000;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is quicker than $\text{co2}/1000000 \times 0.8317 \times 44.01 / 1000000$. 0.8317 accounts for Henry's Law and max ppmw will be 36.6 with 10,000 ppmv.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor floats on the waterline detecting any $\ce{CO2}$ bouncing out of the water. Do I even need to account for Henry's law? Is the math right?&lt;/p&gt;&#xA;" OwnerUserId="4943" LastEditorUserId="5026" LastEditDate="2018-05-18T01:31:55.773" LastActivityDate="2018-05-18T01:35:10.807" Title="How to convert CO2 ppm volume to ppm weight?" Tags="&lt;gas-laws&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9420" PostTypeId="2" ParentId="9410" CreationDate="2014-03-24T03:57:46.677" Score="1" Body="&lt;p&gt;An individual IMF bond is weaker than an ionic bond. But when a salt dissolves, at least 4 IMF bonds are formed for each ionic bond. This is why some salts (e.g. CaCl2) release heat when they dissolve. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-24T03:57:46.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9421" PostTypeId="2" ParentId="4992" CreationDate="2014-03-24T04:56:05.567" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://homepage.smc.edu/wissmann_paul/anatomy2textbook/1carbohydrates.html&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an explanation for alpha- and beta- glucose.&#xA;What it doesn't say is that the alpha-glucose can convert to beta-glucose and back again. Once the linkage is formed, it cannot change from alpha to beta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha linkage has the oxygen (on the aldehyde or ketone) below the ring and the beta has it above the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down the page, it shows beta-Maltose. The label beta (for the right most oxygen) is unimportant since the link has to be alpha for it to be maltose. A beta link would result in a molecule of &lt;a href=&quot;http://braukaiser.com/wiki/index.php?title=Carbohydrates#Maltose_and_Cellobiose&quot; rel=&quot;nofollow&quot;&gt;cellobiose&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-24T04:56:05.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9422" PostTypeId="2" ParentId="9317" CreationDate="2014-03-24T05:12:34.273" Score="4" Body="&lt;p&gt;If you extend the drawing in figure 1, you will discover that there are 5 Ga ions for 3 As ions. This would be suitable for a alloy. Since you are probably interested in a semiconductor, the arsenic needs to have a formal charge of -3. This means that the ratio of Ga to As will be 1:1.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-24T05:12:34.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9423" PostTypeId="2" ParentId="9403" CreationDate="2014-03-24T06:08:41.530" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I've also read online that adding &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; creates a &amp;quot;Piranha&amp;quot; solution [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is true - in principle. In reality, &lt;a href=&quot;http://en.wikipedia.org/wiki/Piranha_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Piranha solution&lt;/em&gt;&lt;/a&gt; is made from &lt;strong&gt;concentrated&lt;/strong&gt; sulfuric acid and &lt;strong&gt;30%&lt;/strong&gt; hydrogen peroxide.&lt;/p&gt;&#xA;&lt;p&gt;In the case of your experminent, you will most likely use diluted sulfuric acid and aqueous solutions of hydrogen peroxide with a much lower concentration. Consequently, the &lt;em&gt;Piranha effect&lt;/em&gt; will &lt;strong&gt;not&lt;/strong&gt; play a role in your experiment, neither is the oxidizing power of sulfuric acid relevant here.&lt;/p&gt;&#xA;&lt;p&gt;On a side side note, treating a sample with conc. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; is a method for the qualitative determination of iodide: Upon heating, one can observe the violet vapour of iodine.&lt;/p&gt;&#xA;&lt;p&gt;In your case, you need the sulfuric acid to keep the pH value low. Under these conditions, &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; acts as an oxidant:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O2 + 2 H+ + 2 e- -&amp;gt; 2 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Does dissolving into ions also count as a reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Does the Starch (for which I have no given chemical formula) react in solution A?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Starch is a mixture of two natural polymers (&lt;strong&gt;amylopectin&lt;/strong&gt; and &lt;strong&gt;amylose&lt;/strong&gt;), both made from D-glucose. Amylopectin doesn't play a role here, but it doesn't disturb either and it's just cheaper to let you work with the natural mixture.&lt;/p&gt;&#xA;&lt;p&gt;It's the helical amylose that is relevant here - and it acts as a &lt;strong&gt;reagent&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What actually happens in the reaction is the following:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 I- -&amp;gt; I2 + 2 e-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{I2 + I- -&amp;gt; I3-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The linear triodide ion forms a strongly coloured &lt;strong&gt;inclusion complex&lt;/strong&gt; with the amylose in the starch.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-27T08:01:42.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9424" PostTypeId="2" ParentId="9401" CreationDate="2014-03-24T06:55:51.483" Score="3" Body="&lt;p&gt;In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is &lt;strong&gt;irrelevant&lt;/strong&gt; at a certain point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/jo101409p&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ivo Leito&lt;/a&gt; and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2KQw.png&quot; alt=&quot;pentacyanopropene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this survey, 1,1,2,3,3-pentacyanopropene (&lt;strong&gt;1&lt;/strong&gt;), was found to be even more acidic with a $\mathrm{pK_a}$ of &lt;strong&gt;-15.0&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt; &#xA;It is important to note that these values are likely to be correct in their relative order. &lt;strong&gt;But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero&lt;/strong&gt;. This initially escaped my attention! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to &lt;a href=&quot;https://chemistry.stackexchange.com/users/4945/martin&quot;&gt;Martin&lt;/a&gt; for pointing out this important detail in his comment below!&#xA;&lt;hr&gt;&#xA;Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was synthesized by &lt;a href=&quot;http://dx.doi.org/10.1021/ic50072a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evan Appelman&lt;/a&gt; (even on a larger scale) by &lt;strong&gt;bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazingly, the author further states in a footnote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction is not smooth, and small explosions may take place in the vapor above the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-03-24T11:33:37.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9425" PostTypeId="1" CreationDate="2014-03-24T07:03:46.200" Score="4" ViewCount="7077" Body="&lt;p&gt;As everyone knows, the atomic orbital can be classified as $s, p_z, p_x, p_y, d_{z^2},d_{xz},d_{yz},d_{xy},d_{x^2-y^2}$ and so on. I want to know the meaning of $z^2,x^2-y^2$ and so on. Maybe this is a fundamental question, but I'm not familiar with chemistry.&#xA;&lt;HR/&gt;&#xA;I have some ideas about this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For example, we consider the $p$ orbital: $p_z$ is symmetric about $z$ axis; $p_x$ is symmetric about $x$ axis; $p_y$ is symmetric about $y$ axis. But what's the meaning of $z^2$ or $x^2 -y^2$?&lt;/li&gt;&#xA;&lt;li&gt;We consider the symmetry of atomic orbital with group theory. The change of orbital under the symmetry operation is the same as the orbital subscript.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4892" LastActivityDate="2014-03-24T23:51:12.287" Title="What does the subscript of atomic orbital mean?" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9427" PostTypeId="2" ParentId="9317" CreationDate="2014-03-24T07:21:53.890" Score="14" Body="&lt;p&gt;It is quite well established that stoichiometric gallium arsenide (CAS 1303-00-0) forms a zincblende analogous structure. It consists of two stacked face-centered cubic lattice systems (&lt;a href=&quot;http://www.ioffe.ru/SVA/NSM/Semicond/GaAs/basic.html&quot; rel=&quot;noreferrer&quot;&gt;Ioffe Physical Technical Institute database&lt;/a&gt;). See also Uncle Al's Answer. In other words, the arsenic forms the lattice and the gallium fills half of the tetrahedral positions (and vice versa). Therefore every gallium has four arsenic nearest neighbours.&lt;/p&gt;&#xA;&lt;p&gt;This does not necessarily mean that these are bonds. In this case however, the bonds are best described as covalent (having a somewhat ionic character).&#xA;The &lt;a href=&quot;http://global.britannica.com/EBchecked/topic/145105/crystal/51816/Covalent-bonds?anchor=ref506372&quot; rel=&quot;noreferrer&quot;&gt;Britannica Online Encyclopedia (Academic Edition)&lt;/a&gt; (You might need access provided through an educational institution) states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Besides the elemental semiconductors, such as silicon and germanium, some binary crystals are covalently bonded. Gallium has three electrons in the outer shell, while arsenic lacks three. Gallium arsenide (GaAs) could be formed as an insulator by transferring three electrons from gallium to arsenic; however, this does not occur. Instead, the bonding is more covalent, and gallium arsenide is a covalent semiconductor. The outer shells of the gallium atoms contribute three electrons, and those of the arsenic atoms contribute five, providing the eight electrons needed for four covalent bonds. The centres of the bonds are not at the midpoint between the ions but are shifted slightly toward the arsenic.&lt;/p&gt;&#xA;&lt;p&gt;Such bonding is typical of the III–V semiconductors — i.e., those consisting of one element from the third column of the periodic table and one from the fifth column. Elements from the third column (boron, aluminum, gallium, and indium) contribute three electrons, while the fifth-column elements (nitrogen, phosphorus, arsenic, and antimony) contribute five electrons. All III–V semiconductors are covalently bonded and typically have the zincblende structure with four neighbours per atom. Most common semiconductors favour this arrangement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is still a fluorishing field for researchers in the field of materials design semiconductors, &lt;a href=&quot;https://doi.org/10.1109/JQE.2014.2303075&quot; rel=&quot;noreferrer&quot;&gt;semiconductor lasers&lt;/a&gt;, computational physics/&lt;a href=&quot;https://doi.org/10.1103/PhysRevB.84.155204&quot; rel=&quot;noreferrer&quot;&gt;forcefield development&lt;/a&gt;, and more.&lt;/p&gt;&#xA;&lt;p&gt;The book &amp;quot;Electronic Devices and Circuit Theory&amp;quot; clearly gets it very wrong. I had a look into it and noticed they also argue with the Bohr model, which might not be a very good starting point at all. However, the focus of the book is clearly not solid state chemistry, and therefore it might fit the purpose.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-08-30T14:12:52.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9428" PostTypeId="2" ParentId="9389" CreationDate="2014-03-24T08:12:50.903" Score="2" Body="&lt;p&gt;I haven't used MarvinSketch in a while, but isn't it a 2D editor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generation of 3D coordinates from 2D files often means trouble. I tried that recently with a 2D BKChem drawing of a trypticene-based ligand for another answer and the result was useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I definitely recommend the use of &lt;strong&gt;Avogadro&lt;/strong&gt; as a free 3D editor. Please have a look at their &lt;a href=&quot;http://avogadro.cc&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; or the article in &lt;a href=&quot;http://dx.doi.org/10.1186/1758-2946-4-17&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Journal of Cheminformatics&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My typical workflow here is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;switch off auto-adding of hydrogen&lt;/li&gt;&#xA;&lt;li&gt;draw the skeleton with the right multiplicity of bonds, but using carbon only&lt;/li&gt;&#xA;&lt;li&gt;change the &quot;wrong&quot; carbons to heteroatoms&lt;/li&gt;&#xA;&lt;li&gt;let the program add the hydogen atoms&lt;/li&gt;&#xA;&lt;li&gt;run a force-field optimization&lt;/li&gt;&#xA;&lt;li&gt;manually adjust some dihedral angles&lt;/li&gt;&#xA;&lt;li&gt;run the force-field optimization again&lt;/li&gt;&#xA;&lt;li&gt;(optional) align a bond with one of the principal axis&lt;/li&gt;&#xA;&lt;li&gt;save as CML or MOL&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I put a MOL file of &lt;a href=&quot;http://pastebin.com/h48npafr&quot; rel=&quot;nofollow noreferrer&quot;&gt;etravirine&lt;/a&gt; created that way on pastebin. Please give QuteMol a try at it.&#xA;&lt;hr&gt;&#xA;I generated a PDB from the molfile mentioned above and ran it through QuteMol. The image below shows that the $\ce{C-Br}$ bond does exist in the PDB file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZYcC.png&quot; alt=&quot;QuteMol output&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Admittedly, checking the PDB file for the respective &lt;code&gt;CONNECT x y&lt;/code&gt; line would have been a (less flamboyant) proof too ;-)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-24T15:46:36.340" LastActivityDate="2014-03-24T15:46:36.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9430" PostTypeId="2" ParentId="9425" CreationDate="2014-03-24T08:45:33.003" Score="3" Body="&lt;p&gt;For each &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow&quot;&gt;azimutal quantum number&lt;/a&gt; $l$, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow&quot;&gt;magnetic quantum number&lt;/a&gt; $m$ ranges from $-l$ to $+l$. Those give you the number of atomic orbitals &quot;subsripts&quot; you should obtain.&#xA;Those are then expressed using &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;spherical harmonics&lt;/a&gt;. If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_spherical_harmonics&quot; rel=&quot;nofollow&quot;&gt;table of spherical harmonics&lt;/a&gt; at the $l=1$, you will find that they contain the cartesian axes as $x-y$, $z$ and $x+y$. Therefore it is advisable to form the linear combination out of them, so you get $x$, $y$, $z$. If you go to higher angular momentum, things get more complicated, but there is some tradition what linear combinations to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your second question, all this formulas are not some inevitable result you obtain by juggling the numbers. Contrary, it is carefully crafted in a way that the formulas are so simple.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-03-24T08:45:33.003" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="9433" PostTypeId="2" ParentId="9395" CreationDate="2014-03-24T11:49:32.327" Score="5" Body="&lt;p&gt;It depends on the water. If the water is already contaminated, then after a couple of days, there are more chances of water to go bad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the water is mineral, already purified, then it takes too many days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the other fact about the sunlight, if the water is directly faced with sunlight, then sunlight causes formation of the algae in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple words, you can drink a water, if it is a mineral water, then after couple of days, if it is a tap water, 2-3 days normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my individual opinion, I am not a doctor. You follow what you want. :)&lt;/p&gt;&#xA;" OwnerUserId="4952" LastActivityDate="2014-03-24T11:49:32.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9435" PostTypeId="2" ParentId="9395" CreationDate="2014-03-24T12:40:34.923" Score="13" Body="&lt;p&gt;A few lazy weeks of summer cottages, camping far from the beaten track or boat excursions in the archipelago: When summer comes, we seek to be closer to nature. Sometimes that means we may waive some of everyday life's certainties, like drinking water straight from the tap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill the drinking water of good quality, such as municipal drinking water in sterilized bottles or cans. Keep the water dark and as cool as possible. That keeps drinking water fresh for at least one week.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Smell and Taste&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Putting a limit on how long drinking water can be stored has more flavoring to it than health risks, says Torbjörn Lindberg, State inspector at NFA.&lt;/li&gt;&#xA;&lt;li&gt;The risk of getting sick of water is small in Sweden. But you should of course do not drink water that tastes or smells bad, is turbid or highly colored.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Get microorganisms in drinking water&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Good drinking water does not contain pathogenic microorganisms, such as viruses and bacteria. It is also a nutrient-poor environment that makes it difficult for such microorganisms to survive. They are simply ill-suited to the aquatic environment. If they are there from the beginning, it is therefore likely that they will disappear over time, and the risk of getting sick is reduced.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Drinking water can be bad&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Sometimes, bacteria, micro-sponges and algae that are adapted to the aquatic environment are found in the drinking water from the start. They can also multiply over time. It is completely natural and gives no health risks. However, drinking water can start to smell or taste badly, especially if kept long and hot, for instance in the sun. It can also be colored and can form a slimy layer on the inside of the container.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="3846" LastEditDate="2014-03-24T15:42:52.563" LastActivityDate="2014-03-24T15:42:52.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9436" PostTypeId="1" CreationDate="2014-03-24T13:33:46.243" Score="0" ViewCount="7397" Body="&lt;p&gt;Which of these solution have the lowest vapor pressure in 298 K? Please give a brief explanation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.01 m $\ce {CaCl2}$&lt;/li&gt;&#xA;&lt;li&gt;0.01 m $\ce {NaCl}$&lt;/li&gt;&#xA;&lt;li&gt;0.02 m $\ce {NaCl}$&lt;/li&gt;&#xA;&lt;li&gt;0.015 m $\ce {CaCl2}$&lt;/li&gt;&#xA;&lt;li&gt;0.01 m $\ce {Na2SO4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(This problem is taken from an Indonesian Chemistry Olympiad)&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-03-24T15:30:08.133" Title="Which of these solution have the lowest vapor pressure in 298 K?" Tags="&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2019-03-10T00:48:51.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="9437" PostTypeId="1" AcceptedAnswerId="9440" CreationDate="2014-03-24T14:16:19.163" Score="2" ViewCount="5913" Body="&lt;p&gt;Gold and Silver atoms interact with one another to form reasonably strong metallic bonds, but surprisingly they have almost no chemical interaction with atmosphere. How can we explain it based on the atomic structure of Gold and Silver. What is so unique about its chemical structure that is not present in other transition metals that makes them noble (non-reactive)?&lt;/p&gt;&#xA;" OwnerUserId="4941" LastEditorUserId="4923" LastEditDate="2014-03-24T23:04:40.423" LastActivityDate="2016-08-03T14:16:34.193" Title="Why Gold and Silver react minimally with atmosphere?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9438" PostTypeId="2" ParentId="9436" CreationDate="2014-03-24T14:51:44.963" Score="1" Body="&lt;p&gt;All examples are mixtures of not-volatile (here: salts, but it would be the same for a teaspoon of sugar) and volatile compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vapour pressure above these mixtures is proportional to the mole fraction of the volatile component. With other words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The higher the concentation of the salt, the lower the vapour pressure.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as user80551 pointed out in his comment, we should not look at it this way. Water isn't &quot;just the solvent&quot; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is the volatile compound in the mixture that matters! The higher the mole fraction of the volatile compound, the higher the vapour pressure!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With (again) other words: &lt;strong&gt;Raoult's Law&lt;/strong&gt; ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-24T15:30:08.133" LastActivityDate="2014-03-24T15:30:08.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9439" PostTypeId="2" ParentId="9436" CreationDate="2014-03-24T15:25:06.160" Score="0" Body="&lt;p&gt;Colligative properties.  The solution with the highest molality (note spelling), the largest number of particles/liter solvent (NOT solution) has the lowest vapor pressure.  All the electrolytes are strongly dissociated (salts of strong acids and bases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.03, 0.02, 0.04, 0.045, 0.03.  Why will sodium sulfate be the least completely dissociated of the salts?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-24T15:25:06.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9440" PostTypeId="2" ParentId="9437" CreationDate="2014-03-24T15:38:52.403" Score="4" Body="&lt;p&gt;Copper, silver, gold. What is the energy of the highest &quot;s&quot; orbital versus the highest &quot;d&quot; orbital?  Gold is a very heavy nucleus.  An &quot;s&quot; orbital has non-zero probability at its nucleus.  Special relativity effects make its &quot;s&quot; electrons more massive, bringing them into the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A silver atom is a soft Lewis acid.  It reacts well with large soft Lewis bases.  Silver quickly tarnishes with sulfide.  Gold is not chemically inert at all to even mild oxidation if the product cation is coordinated by ligands.  Salt water plus air will tarnish gold.  Dilute cyanide plus air will dissolve it (e.g., heap leaching of gold ore).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamics proposes, kinetics disposes.  Aluminum is tremendously reactive toward air and water.  Anything that permanently breaches its adherent oxide coating (e.g., abrasion plus amalgamation) sets it off.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-24T15:38:52.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9441" PostTypeId="1" CreationDate="2014-03-24T16:13:33.450" Score="0" ViewCount="1740" Body="&lt;p&gt;Once made by heating wood in the absence of air, methanol ($\ce{CH3OH}$) is now made by reacting carbon monoxide and hydrogen at high pressure: &#xA;$\ce{CO_{(g)} + 2 H2_{(g)} -&amp;gt; CH3OH_{(l)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many grams of $\ce{CH3OH}$ can be made from 30.0g of $\ce{CO}$ if it all reacts?&lt;/p&gt;&#xA;" OwnerUserId="4960" LastEditorUserId="418" LastEditDate="2014-03-24T16:16:08.583" LastActivityDate="2014-03-24T16:47:15.040" Title="How many grams of CH3OH can be made from 30.0g of CO if it all reacts?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-03-24T17:54:02.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="9442" PostTypeId="2" ParentId="9441" CreationDate="2014-03-24T16:47:15.040" Score="1" Body="&lt;p&gt;Moles of CO in equals moles of methanol out (at 100% yield), typically over copper-doped zinc oxide catalyst.  Do the arithmetic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One gram-mole of a chemical formula is [the magnitude of its atomic mass units] grams.  A nitrogen atom is 14.0067 amu.  A mole of nitrogen gas, $\ce{N2}$, is then 28.0134 grams.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-24T16:47:15.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9443" PostTypeId="2" ParentId="8099" CreationDate="2014-03-24T16:53:30.383" Score="3" Body="&lt;p&gt;Here's a simple explanation of why that happens:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the electron gets promoted to the higher level after absorbing one photon corresponding to the visible light. This makes you see the complementary color, as you already said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT when the electron comes back to the unexcited state, it does not do that in a single step! The electron comes back to the unexcited state taking smaller intermediate steps. Each of these smaller transitions corresponds to the emission of a low frequency photon belonging to the Infra Red region. This is outside the human visible range and thus we don't &quot;see&quot; these photons. But these Infra Red photons do heat up the solution though the change may not be measurable.&lt;/p&gt;&#xA;" OwnerUserId="487" LastActivityDate="2014-03-24T16:53:30.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9445" PostTypeId="2" ParentId="6917" CreationDate="2014-03-24T18:17:04.490" Score="2" Body="&lt;p&gt;The procedure is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid-base_titration&quot; rel=&quot;nofollow&quot;&gt;titration&lt;/a&gt;. It consists in adding a &lt;a href=&quot;http://en.wikipedia.org/wiki/PH_indicator&quot; rel=&quot;nofollow&quot;&gt;pH indicator&lt;/a&gt; (that changes color depending on pH) to the solution and adding measured quantities of base or acid until the indicator changes color and determining how much reacted from the amount used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative is to use a &lt;a href=&quot;http://en.wikipedia.org/wiki/PH_electrode&quot; rel=&quot;nofollow&quot;&gt;pH meter&lt;/a&gt; instead of an indicator specially in automated analysis.      &lt;/p&gt;&#xA;" OwnerUserId="1300" LastActivityDate="2014-03-24T18:17:04.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9446" PostTypeId="2" ParentId="7566" CreationDate="2014-03-24T20:30:06.290" Score="3" Body="&lt;p&gt;Disclaimer: I will be talking about the shear viscosity. I think that most of what I write is also applicable to extensional viscosity, although with different numerical constants, but I am not fully sure so I will limit myself to shear viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most theoretical models predict that the shear viscosity of both solutions will &lt;strong&gt;be equal&lt;/strong&gt;. The reason is that these models describe the viscosity of the suspension $\mu$ as the viscosity of the 'empty' liquid $\mu_0$ $+$ a correction term that depends on the &lt;strong&gt;volume concentration&lt;/strong&gt; of solids, $\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most famous one is the &lt;a href=&quot;http://dx.doi.org/10.1002/andp.19063240204&quot; rel=&quot;nofollow&quot;&gt;Einstein viscosity&lt;/a&gt; equation that simply reads:&#xA;$$\frac{\mu}{\mu_0}=1+2.5\phi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model works for low solid concentrations when inter-particle interactions can be neglected. Later &lt;a href=&quot;http://dx.doi.org/10.1017/S0022112077001062&quot; rel=&quot;nofollow&quot;&gt;Batchelor&lt;/a&gt; showed that including long-range hydrodynamic interactions an additional correction of $+6.2\phi^2$ should be included, but still only a dependence on the volume concentration is there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice summary of more models can be found in this paper: &lt;a href=&quot;http://www.lcpe.uni-sofia.bg/publications/2001/2001-06-KD.pdf&quot; rel=&quot;nofollow&quot;&gt;K.D Danov, &lt;em&gt;J. Colloid Interface Sci.&lt;/em&gt;, 235, 144–149 (2001)&lt;/a&gt;. That paper also includes a discussion that shows that the factor 2.5 should be dependent on the mobility of the particle. This would mean that it will be higher for smaller particles, so in your case that the solution with part B will be somewhat more viscosity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another overview of models, including models which have an increasing viscosity with increasing particle size can be found &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/466/2116/1201.full&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. So that would point to part A giving the more viscous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So to summarize:&lt;/strong&gt; to first order (i.e. low concentrations) the two viscosities will be the same. At higher orders you have competing effects of the particle radius that should increase the viscosity and the particle number that should also increase it. Which effect is dominant is probably hard to tell theoretically.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-03-24T20:30:06.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9447" PostTypeId="1" AcceptedAnswerId="9465" CreationDate="2014-03-24T20:47:44.983" Score="4" ViewCount="3329" Body="&lt;p&gt;I am currently studying computational chemistry and I am trying to understand the UV-VIS absorption of molecules. I know that UV-VIS absorption is electronic excitation and requires a specific energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why a molecule can not absorb a shorter wavelength as it has more energy? A slightly higher energy causes vibrational excitations which is then released through vibrational relaxation (releasing heat). Isn't it the case when a very shorter wavelength is used so that the electrons in the molecule are excited from more lower MOs (lower energy) to the higher ones? Isn't it still absorption? It is supposed to be by definition?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZHtPc.jpg&quot; alt=&quot;absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule can absorb a photon that has &quot;red&quot; amount of energy, it can also absorb &quot;green&quot; and &quot;blue&quot; amounts. But why can't it absorb anywhere between red and green or more basically isn't it supposed to absorb anything more than &quot;red&quot;? Just to be clear, by red I mean the energy difference between $n = 1$ and $n = 2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am already aware of the fact that, for example, any energy between $n = 2$ and $n = 3$ can not be emitted fully as some of them will be lost for rotational and vibrational transitions until the energy hits the $n = 2$ state but I am not expecting this to stop a molecule from absorbing that light.&lt;/p&gt;&#xA;" OwnerUserId="1060" LastEditorUserId="5026" LastEditDate="2018-06-25T05:55:23.750" LastActivityDate="2018-06-25T05:55:23.750" Title="A molecule absorbs light having a specific wavelength, why doesn't it absorb shorter wavelengths?" Tags="&lt;molecular-orbital-theory&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9448" PostTypeId="1" AcceptedAnswerId="9457" CreationDate="2014-03-24T23:34:48.077" Score="14" ViewCount="11879" Body="&lt;p&gt;What is the correct structure for the $\ce{NO2}$ compound (not ion)? I always thought it was like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HFeQW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the negative charge would be residing the on the highly electronegative oxygen, but was reading this &lt;a href=&quot;http://blog.wolframalpha.com/2013/10/07/introducing-wolframalphas-step-by-step-for-chemistry/&quot; rel=&quot;noreferrer&quot;&gt;blogpost&lt;/a&gt; by the guys at WolframAlpha, who state that contrary to popular belief, the correct structure is actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYVF6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They explain that it has to do with molecular orbital theory, but don't explain in depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain why this occurs?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T14:49:28.913" LastActivityDate="2019-07-14T04:03:15.347" Title="Structure of NO2 compound?" Tags="&lt;molecular-orbital-theory&gt;&lt;lewis-structure&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9449" PostTypeId="2" ParentId="9447" CreationDate="2014-03-24T23:39:36.540" Score="2" Body="&lt;p&gt;A cryogenic atom or molecule in vacuum is a narrow bandwidth oscillator (particle in a box).  If you do not hit the right note it will not resonantly electronically ring for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is polyatomic, warming it will introduce rotational (microwave) then vibrational (infrared) degrees of freedom, broadening lines and introducing new ones.  Put it in a fluid medium and the random jingle jangle of collisions broadens things further.  Put it in a solid and modes may be physically eliminated, e.g., whole molecule rotation.  Water's pale blue color is summed infrared vibronic modes, the only visible color caused by vibration absorption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher the frequency, the more localized the photon.  At very high energies photon collision is literally physical.  Things get collisionally knocked out rather than resonantly pumped up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heavy atomic nucleus has density ~$2.3{×10^{14} \ce{g/cm^3}}$.  Add a muon 206.768 times the mass of an electron into U-238.  It decays to orbit at a fair fraction of lightspeed just &lt;em&gt;inside&lt;/em&gt; the nuclear surface.  Selection rules for hadrons and leptons are different.  They may not see each other.  Triprismane, $\ce{C6H6}$, sits 90+ kcal/mole above benzene at room temp.  It won't fragment until increased temperature adds another 33 kcal/mole - and it doesn't crack to benzene.  Reaction path is defined by conservation of orbital symmetry.  Tetra-&lt;em&gt;tert&lt;/em&gt;-butyl tetrahedrane does not thermally fragment.  The insanely strained molecule is so symmetric that its lowest energy path is synchronous transformation into tetra-&lt;em&gt;tert&lt;/em&gt;-butyl cyclobutadiene' at 130°C!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSSr8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-24T23:39:36.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9450" PostTypeId="2" ParentId="5832" CreationDate="2014-03-25T01:27:18.920" Score="0" Body="&lt;p&gt;In short, yes, the electrons will change orbitals to fill shells (even if it is an ion). This is why the transition elements can have more than one ion.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-25T01:27:18.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9451" PostTypeId="2" ParentId="5287" CreationDate="2014-03-25T01:35:36.540" Score="3" Body="&lt;p&gt;I can't tell you a manufacturer, but I can tell you the purpose of the pointer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the height of the column of mercury changes, it changes the height of the mercury exposed to the atmosphere. Since the ruler is (usually) fixed in place, the level of the mercury needs to be adjusted to the zero on the ruler. The pointer shows you the zero point. You adjust the level of mercury so that it barely touches the point.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-25T01:35:36.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9452" PostTypeId="2" ParentId="5832" CreationDate="2014-03-25T01:47:05.983" Score="2" Body="&lt;p&gt;Check out the &lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/ionEnergy.html&quot; rel=&quot;nofollow&quot;&gt;NIST database&lt;/a&gt; for electronic configurations for atoms and ions in the gas phase. If you play around a bit, you will find that yes, some ionizations will not follow the expected trends from the &lt;em&gt;aufbau&lt;/em&gt; principle. For example, search for rhenium and compare the configurations between $\ce{Re^{11+}}$ and $\ce{Re^{17+}}$, and you'll see some of them are unexpected (and rather hard to explain). Osmium shows an anomaly going from $\ce{Os^{+}}$ to $\ce{Os^{2+}}$. There are other elements with ions containing &lt;em&gt;aufbau&lt;/em&gt; anomalies, but I can't remember many from the top of my head.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-03-25T01:54:31.393" LastActivityDate="2014-03-25T01:54:31.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9453" PostTypeId="2" ParentId="5229" CreationDate="2014-03-25T01:49:30.747" Score="1" Body="&lt;p&gt;There is a simple reason to have the purity greater than 100% for this compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the substance was exposed to a dry environment for several hours, a small amount of the  water of hydration could be lost, causing the calculation to have a higher purity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I believe that the purity is calculated from the perchlorate titration of the substance in acetic acid (I will need to check the USP monograph). There are many sources for error in the calculations: the weight of the L-Cysteine Hydrochloride Monohydrate, the volume of the perchlorate solution, the error from the standardization of the perchlorate solution, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the right conditions, you can easily get 101% for the purity. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-25T01:49:30.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9454" PostTypeId="1" CreationDate="2014-03-25T03:05:49.447" Score="11" ViewCount="2273" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{align}&#xA;\ce{Co^{3+}(aq) + e-  &amp;amp;-&amp;gt; Co^{2+}(aq)} &amp;amp;   E° &amp;amp;= \pu{+1.82 V} \\&#xA;\ce{Co^{2+}(aq) + 2e- &amp;amp;-&amp;gt;  Co(s)}      &amp;amp;   E° &amp;amp;= \pu{-0.28 V}&#xA;\end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;what is the standard reduction potential for $\ce{Co^{3+}(aq) +3e-  -&amp;gt; Co(s)}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I add up these two half-equations to get the desired half-equation, I get $E^\circ = \pu{+1.54 V}$. But the correct answer is $\pu{+ 0.42 V}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? &lt;/p&gt;&#xA;" OwnerUserId="4968" LastEditorUserId="16683" LastEditDate="2017-04-10T13:45:53.210" LastActivityDate="2017-04-10T22:28:58.283" Title="Deriving a reduction potential from two other reduction potentials" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9455" PostTypeId="1" CreationDate="2014-03-25T03:19:00.383" Score="2" ViewCount="10534" Body="&lt;p&gt;I am given a protein, BAPNA, and the initial concentration of the protein. The experiment involves reaction rates of varying protein concentrations. Each sample cuvette is inserted into a spectrometer, 100% transmittance is set, has the enzyme inserted, and then has transmittance measured every 20 s for 600 s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand absorbance is given as: $A = 2 - \log (T\%)$ and that's no issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I run into is trying to solve for the concentration or molar extinction constant. I understand that Beer-Lambert law, $A = \epsilon bc$  is used for this purpose. The only given concentration is the initial concentration. Assuming $b = 1$, we can solve for $\epsilon$ or $c$, but having 100% set in the beginning causes $A = 0$, meaning initially $\epsilon = 0$ if $c \neq 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where my calculations run into a dead end. I'm not given $\epsilon$ or concentration at any other point, but I'm supposed to be able to calculate the final concentration and maximum reaction rate. I've only a set of transmittance numbers over time, and that initial concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I am misunderstanding or not seeing?&lt;/p&gt;&#xA;" OwnerUserId="4967" LastEditorUserId="41328" LastEditDate="2017-10-11T12:49:48.280" LastActivityDate="2017-10-11T12:49:48.280" Title="Concentration from Absorbance vs Time" Tags="&lt;biochemistry&gt;&lt;analytical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9456" PostTypeId="2" ParentId="590" CreationDate="2014-03-25T03:44:27.597" Score="7" Body="&lt;p&gt;Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Antisymmetric exchange:&lt;/strong&gt;&#xA;At first I thought it was simply an &lt;a href=&quot;https://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;noreferrer&quot;&gt;exchange interaction&lt;/a&gt; where the wave function's sign is changed during exchange, now I don't think it's so simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antisymmetric exchange is also known as &lt;a href=&quot;http://sciencewise.info/definitions/Dzyaloshinskii-Moriya_interaction_by_George_Jackeli&quot; rel=&quot;noreferrer&quot;&gt;DM-interaction&lt;/a&gt; (for Dzyaloshinskii-Moriya). I don't know exactly what it is, here is the original paper citation - can't find it anywhere though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.E. Dzyaloshinskii, Thermodynamic theory of weak ferromagnetism in antiferromagnetic substances, Sov. Phys. JETP 5(6), 1259-1272 (1957)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sciencewise.info/definitions/Dzyaloshinskii-Moriya_interaction_by_George_Jackeli&quot; rel=&quot;noreferrer&quot;&gt;A definition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slideshow, which contains &lt;a href=&quot;http://magnetism.eu/esm/2009/slides/simon-slides-1.pdf&quot; rel=&quot;noreferrer&quot;&gt;some more info&lt;/a&gt; (slides 16-23)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;J-Strain:&lt;/strong&gt; Reading this, &quot;J-strain&quot; isn't the only &quot;strain&quot; that's referenced in this paper. Also referenced within is &quot;G-Strain&quot; which has a more &lt;a href=&quot;https://en.wikipedia.org/wiki/G-strain&quot; rel=&quot;noreferrer&quot;&gt;elucidating Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, maybe it's safe to say, like G-Strain, J-strain is the broadening of the &quot;distributed exchange coupling constants&quot; (a.k.a. J-values). Inferred from a line in the paper: &quot;Both proteins exhibit distributed J-values (J-strain)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What did J-strain have to do with the experiment?&lt;/em&gt; &quot;...The observed J-strain, which presumably represents a distribution of protein conformations “locked in” upon freezing the samples.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where does the broadening of the J-values (exchange coupling constants) a.k.a. J-strain come from?&lt;/em&gt; Well apparently there's this thing called &lt;a href=&quot;https://en.wikipedia.org/wiki/J-coupling&quot; rel=&quot;noreferrer&quot;&gt;J-coupling&lt;/a&gt; that seems relevant. J-coupling is also known as indirect dipole-dipole coupling, and gives info about molecule interconnectivity. J-coupling relates nuclei to local electrons (All this is from the J-coupling article). The broadening of the lines may come from anisotropic interactions, which affect nuclear spin energy levels. Sure enough, the paper says it can't rule out anisotropy. An Alternative source? Non-homogeneous regions in the sample, which could go back to the frozen/locked conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still think it'd be great if someone doing physics research in this field could elucidate some more though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-03-jcoupl.htm&quot; rel=&quot;noreferrer&quot;&gt;Another J-coupling Webpage (U. Wisc)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I should cite the original paper, &quot;Evidence for Antisymmetric Exchange in Cuboidal [3Fe−4S]+ Clusters&quot;. Linked in question.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-03-25T03:44:27.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9457" PostTypeId="2" ParentId="9448" CreationDate="2014-03-25T04:45:22.533" Score="5" Body="&lt;p&gt;Drawings of molecules are only representations to help us understand them, not the actual molecule itself. So both pictures you have there are correct; they both contribute to the actual structure of the $\ce{NO2}$ molecule. I'm guessing WolframAlpha chose the structure they did because it is the structure the most contributes to the actual structure of the molecule, with all formal charges being 0 like Nicolau said, it is the most 'favorable' of possible configurations that the molecule could be in.&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T14:49:53.623" LastActivityDate="2015-06-27T14:49:53.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9459" PostTypeId="1" AcceptedAnswerId="9461" CreationDate="2014-03-25T05:37:08.827" Score="14" ViewCount="2406" Body="&lt;p&gt;In a &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;noreferrer&quot;&gt;covalent bond&lt;/a&gt; between two &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms&quot; rel=&quot;noreferrer&quot;&gt;atoms&lt;/a&gt;, an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron&quot; rel=&quot;noreferrer&quot;&gt;electron&lt;/a&gt; from one of the either atom is shared by overlapping of their &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;noreferrer&quot;&gt;orbitals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&#xA;&lt;strong&gt;Why can't three atoms share an electron and overlap their orbitals?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="4734" LastEditDate="2014-04-12T07:40:01.180" LastActivityDate="2014-04-12T07:40:01.180" Title="Why only two atoms share an electron and not three?" Tags="&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9460" PostTypeId="2" ParentId="9455" CreationDate="2014-03-25T05:48:44.123" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The only given concentration is the initial concentration [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless there is a major misunderstanding on my side, you have all the data you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given&lt;/strong&gt; the initial concentration $c_0$ and &lt;strong&gt;knowing&lt;/strong&gt; the path length $d$ of your cuvette, you &lt;strong&gt;measure&lt;/strong&gt; the extinction $E$ at this point and use - as you have correctly stated in your question - the &lt;em&gt;Lambert-Beer law&lt;/em&gt; to &lt;strong&gt;calculate&lt;/strong&gt; the molar extinction coefficient $\epsilon$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \epsilon\cdot c\cdot d$$&#xA;$$\epsilon = \frac{E}{c\cdot d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Did I mention units? Units are your friend!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Write the &lt;em&gt;Lambert-Beer law&lt;/em&gt; &lt;strong&gt;with units&lt;/strong&gt;! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Percent (%) is &lt;strong&gt;NOT&lt;/strong&gt; a proper unit for the initial concentration of your protein.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but I'm supposed to be able to calculate the final concentration and maximum reaction rate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having determined $\epsilon$, you can now correlate at any point along your reaction the measured extinction with the actual concentration of your sample, including the final concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the slope of the curve of extinction $E\ $ &lt;em&gt;vs&lt;/em&gt; concentration $c$, the rate of your reaction can be obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the relation is not a straight line, &lt;em&gt;logarithms&lt;/em&gt; are your friend.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-25T09:07:57.970" LastActivityDate="2014-03-25T09:07:57.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9461" PostTypeId="2" ParentId="9459" CreationDate="2014-03-25T06:34:00.897" Score="13" Body="&lt;p&gt;The normal case for a covalent bond is indeed a &lt;strong&gt;2-centre 2-electron&lt;/strong&gt; bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are however cases, where 3 centres (= atoms) either share 2 or 4 electrons.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In organic chemistry, the most famous case for an &lt;em&gt;electron-deficient&lt;/em&gt; &lt;strong&gt;3-centre 2-electron&lt;/strong&gt; bond is the &lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/07/norbornyl-cation-nonclassical-structure-olah-herb-brown&quot; rel=&quot;nofollow noreferrer&quot;&gt;2-norbornyl cation&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;In inorganic chemistry, boranes with their &lt;a href=&quot;https://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane&quot;&gt;banana bonds&lt;/a&gt; are the role models.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hypervalent&lt;/em&gt; molecules with &lt;strong&gt;3-centre 4-electron&lt;/strong&gt; bonds, like &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4760/why-do-compounds-like-sf6-and-sf4-exist-but-sh6-and-sh4-dont/4762#4762&quot;&gt;$\ce{SF6}$&lt;/a&gt; or the bifluoride anion &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8549/existence-and-stability-of-bifluoride-ion&quot;&gt;$\ce{HF2-}$&lt;/a&gt; have been discussed here too.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-03-25T08:16:06.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9462" PostTypeId="2" ParentId="9454" CreationDate="2014-03-25T08:46:32.257" Score="14" Body="&lt;p&gt;That is because you cannot simply add the electrode potentials algebraically. In both cases, the electrode potentials have to be multiplied by $n$. What you can do instead is use Gibbs free energy change for each reaction which then can be added algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\circ = -nFE^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this you can get the $\Delta G^\circ$ for each reaction which can be then added algebraically in the same way you add the reactions. The resultant $\Delta G^\circ_\mathrm{net}$, on equating with $-nFE^\circ_\mathrm{net}$, we can get the value of $E^\circ_\mathrm{net}$, which should be your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this approach, since the value of $n$ for both the reactions are different, the relative weightage of the electrode potentials of each reaction in determining the resultant $E^\circ$ will differ and therefore the final $E^\circ$ will be different than what you get by just adding up the electrode potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{Co^3+ + e- &amp;amp;-&amp;gt; Co^2+} &amp;amp; E^\circ_1 &amp;amp;= \pu{+1.82 V} &amp;amp; \Delta G_1^\circ &amp;amp;= \pu{-175.6 kJ mol-1} \tag{1} \\&#xA;\ce{Co^2+ + 2e- &amp;amp;-&amp;gt; Co} &amp;amp; E^\circ_2 &amp;amp;= \pu{-0.28 V} &amp;amp; \Delta G_2^\circ &amp;amp;= \pu{+54.0 kJ mol-1} \tag{2} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\Delta G^\circ_\mathrm{net} &amp;amp;= \Delta G_1^\circ + \Delta G_2^\circ \\&#xA;&amp;amp;= \pu{-121.6 kJ mol-1} \\[3pt]&#xA;E^\circ_\mathrm{net} &amp;amp;= -\frac{\pu{-121.6 kJ mol-1}}{3F} \\&#xA;&amp;amp;= \pu{+0.42 V}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="16683" LastEditDate="2017-04-10T22:28:58.283" LastActivityDate="2017-04-10T22:28:58.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9463" PostTypeId="1" AcceptedAnswerId="9464" CreationDate="2014-03-25T09:31:43.557" Score="7" ViewCount="9793" Body="&lt;p&gt;This fact was written in my text book. I wonder why that should be so.&lt;/p&gt;&#xA;" OwnerUserId="4972" LastActivityDate="2014-03-25T19:04:25.653" Title="Why is it necessary to avoid even traces of moisture from a Grignard reagent?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9464" PostTypeId="2" ParentId="9463" CreationDate="2014-03-25T09:32:31.780" Score="5" Body="&lt;p&gt;Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{RMgX + H2O -&amp;gt; RH + Mg(OH)X}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, $\ce{R}$ is a hydrocarbon.&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="1431" LastEditDate="2014-03-25T11:22:32.870" LastActivityDate="2014-03-25T11:22:32.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9465" PostTypeId="2" ParentId="9447" CreationDate="2014-03-25T10:08:08.037" Score="5" Body="&lt;h1&gt;Classical approach&lt;/h1&gt;&#xA;&lt;p&gt;Absorption occurs when there are &amp;quot;oscillators&amp;quot; that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies.&lt;/p&gt;&#xA;&lt;p&gt;These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing their spins(NMR, ESR) orientation (microwave), configuration (infrared) nuclear configuration  (&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; rays) and finally electronic distribution (VIS, UV, X-Ray).&lt;/p&gt;&#xA;&lt;p&gt;Of course, not all the materials have oscillators that absorb in all the spectrum. For wavelength between 10 nm and 100 pm the oscillators, that mainly involve electrons in these regions, have to absorb energy between &lt;span class=&quot;math-container&quot;&gt;$10^{7}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10^{9}$&lt;/span&gt; J/mol without being ionized, in this case only the inner electrons of some atoms (almost always metals) absorbs these wavelengths see &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_absorption_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;XAS&lt;/a&gt;, so in our organic world seems that most compounds are X-Ray transparent but in fact there are a lot of inorganic compounds that absorb X-Ray.&lt;/p&gt;&#xA;&lt;p&gt;When radiation is absorbed there are many ways to release his energy(see figure 1). We can divide them into two great categories radiative relaxation and non-radiative relaxation. In fact radiative relaxation is most common when we are dealing with shorter wavelength because with more energy is possible to excite an electron to a higher level and this can sometimes re-emit radiation when it goes from a higher level to the ground level, this as you know, is called &lt;strong&gt;fluorescence&lt;/strong&gt; and is very common when you deal with UV, and X-Ray (see X-Ray Fluorescence spectroscopy). Eventually, there is also non-radiative relaxation process that are the main processes that occur from VIS (note that however is possible a VIS-NIR luminescence for some compounds) to the other spectral region with lower wavelength.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CtUSD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Quantum approach&lt;/h1&gt;&#xA;&lt;p&gt;Compound energy levels are &lt;strong&gt;quantized&lt;/strong&gt;, so compounds can absorb only some discrete energy values. These values are determinate by transitions that respect strictly &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;selection rules&lt;/a&gt;. If you know the energy of the transition, from Plank relation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E=\frac{ch}{\lambda}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;you can see that for every energy value there is a specific wavelength, so this means that compounds can absorb only at specific discrete wavelength. However if you see a UV-VIS spectrum you will see a continuous spectrum because roto-vibrational changes affect the energy levels of the compounds involved.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-06-25T05:50:35.623" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="9466" PostTypeId="1" CreationDate="2014-03-25T11:36:09.013" Score="3" ViewCount="853" Body="&lt;p&gt;My book says the &quot;angular momentum is always conserved in a transition&quot;. But if suppose we have a s electron then it's angular momentum is $\hbar\sqrt{l(l+1)}+\hbar\sqrt{s(s+1)}$ $\implies \hbar\frac{\sqrt{3}}{2}$ and suppose it transits to a p electron (which is in accordance with the selection rule as $\Delta l=1$) which has an angular momentum of           $\hbar\sqrt2 + \hbar\frac{\sqrt{3}}{2}$. As the angular momentum of a photon is $\hbar$ then it means (mathematically) that we would have to add $\sqrt2$photons to the atom but that's not possible. Then how is the angular momentum conserved?? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I can't see why the spin angular momentum of a photon is $\hbar$ as spin angular momentum is $\hbar\sqrt{s(s+1)}$ and as $s=1$ for a photon so it should be $\hbar\sqrt2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. the book I am using is Elements of Physical Chemistry by Peter Atkins and Julia de Paula.&lt;/p&gt;&#xA;" OwnerDisplayName="user4973" LastActivityDate="2014-04-24T13:15:28.590" Title="Conservation of angular momentum in electronic transition" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9467" PostTypeId="1" AcceptedAnswerId="9469" CreationDate="2014-03-25T11:36:19.390" Score="0" ViewCount="160" Body="&lt;p&gt;I just learned what chain reactions are, but I can't seem to wrap my head around them because it seems utterly useless and doesn't accomplish anything. At least that's how I understood it because the textbook doesn't have any real world applications for it. &#xA;So what I am asking is do chain reactions serve any purpose. If so, it would be great if you could support your answer with a real world example or application.&lt;/p&gt;&#xA;" OwnerUserId="4583" LastEditorUserId="73" LastEditDate="2014-03-25T11:47:50.597" LastActivityDate="2014-03-25T14:36:08.617" Title="What purpose do chain reactions serve?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9468" PostTypeId="2" ParentId="9466" CreationDate="2014-03-25T12:05:41.790" Score="2" Body="&lt;p&gt;I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.&#xA;&lt;hr/&gt;&#xA;I have some idea about your example. The electron transition is &quot;electron jumps from one energy level to another&quot;. So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.&lt;/p&gt;&#xA;" OwnerUserId="4892" LastEditorUserId="4892" LastEditDate="2014-03-25T12:40:55.663" LastActivityDate="2014-03-25T12:40:55.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9469" PostTypeId="2" ParentId="9467" CreationDate="2014-03-25T13:35:38.057" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] it would be great if you could support your answer with a real world example or application.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free radical polymerizations of monomers with $\ce{C=C}$ double bonds are not completely useless ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Polymethyl methacrylate (PMMA) is obtained by polymerization of methyl methacrylate (MMA). Think &lt;em&gt;acrylic glass&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Styrene can be polymerized to polystyrene (PS). Think &lt;em&gt;Styrodur&lt;/em&gt; and food containers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vinyl chloride is polymerized to polyvinyl chloride (PVC). Think pipes, insulation for electric cables, flooring.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-25T13:35:38.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9470" PostTypeId="2" ParentId="9467" CreationDate="2014-03-25T14:36:08.617" Score="1" Body="&lt;p&gt;Nuclear reactors and warheads, addition polymers (all hail Parylenes, too, and linear low density polyethylene, epoxy resins, truck liner beds, synthetic rubbers), genomes into proteomes, fire...to &quot;gray goo.&quot;  Now, some footnotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Living_polymerization&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Living_polymerization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONTROL.  A free radical polymerization or classical Ziegler-Natta zips until it dies.  Big molecular weight spread puts physical properties all over the map.  Living polymerization is an non-terminating polymerization.  The hot end(s) reversibly cap, allowing blocks to be assembled such that component pieces and the whole are hard by identical molecule to molecule.  This is called &quot;low dispersivity.&quot;  The polymer soles of your shoes are filled Kraton thermoplastic elastomers  (anionic living polymerization).  They are an identical short block of polystyrene, identical long block of disordered hydrocarbon, identical short block of polystyrene. It flows when hot, the polystyrene end segments being soluble in the melt.  It reversibly crosslinks as it cools, polystyrene domains coming out of solution and crystallizing.  There is no manufacturing waste.  Trimmings and rejects are remelted.  Physical properties are remarkably controllable and absolutely reproducible.  The polymer is inexpensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyethylene was a miracle, though an uncooperative one.  Dow's Insite catalysts changed the world with linear low density polyethylene.  The molecules now can be dialed into compositional anything, with molecular weight control.  Price/performance plummeted.  A plastic shopping bag weighs about five grams.  Less hair-thickness film has astounding strength. On the other end Dyneema and Spectra ultra-high molecular weight linear polyethylene, especially Plasma crystallized.  Much stronger than Kevlar/weight, does not hydrolyze, and it floats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chain reactions, their engineering and control, construct civilization on huge bulk scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra credit:  Why is there exactly one silicon atom in every molecule of Kraton triblock polymer regardless of molecular weight?  Where is it within the molecule?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-25T14:36:08.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9471" PostTypeId="2" ParentId="9463" CreationDate="2014-03-25T14:48:57.400" Score="5" Body="&lt;p&gt;Aside from decomposing as described, Grignard formation will not initiate given a trace of water.  There is a bulky lore of getting the things started even under carefully dried conditions.  Consider an oven-dried roundbottom containing absolute ether, Mg turnings, and a small portion of the total amount of  alkyl halide.  It routinely sits there doing nothing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crush a turning to expose fresh surface. Add a small iodine crystal to corrode the Mg.  Heat the thing, oh yeah, then scramble for the exotherm.  Making Grignards spawns neurotic behavior like lucky lab coats, flame drying everything, always adding a drop of freshly distilled $\ce{Me3SiCl}$ (always adding a drop of old $\ce{Me3SiCl}$), saving some of the last batch to set off the next.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-25T14:48:57.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9472" PostTypeId="2" ParentId="9448" CreationDate="2014-03-25T15:00:10.393" Score="2" Body="&lt;p&gt;The $^{16}\ce{O}$ nucleus is spin-0, the $^{14}\ce{N}$ nucleus is spin-1.  You can see the real world weighted fractional residence times of the unpaired spin with EPR.  $\pi$-bonding will complicate the spectrum with spin cross-coupling.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T14:51:49.767" LastActivityDate="2015-06-27T14:51:49.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9475" PostTypeId="2" ParentId="9321" CreationDate="2014-03-25T16:18:08.937" Score="3" Body="&lt;h1&gt;A practical tutorial with &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O5Yjv.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is not so easy to understand from the structure because polarity in fact depends on the arrangement of the groups in tridimensional space relative to one another and not simply on the groups present (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{trans-C2H2Cl2}$&lt;/span&gt; is not polar because the individual bond dipoles cancel one another, &lt;span class=&quot;math-container&quot;&gt;$\ce{cis-C2H2Cl2}$&lt;/span&gt; is polar) so you have to use a geometry optimization tool first. You can perform this optimization and make some assumptions from the model calculating the electrostatic potential with &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;. I've made this short tutorial.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw you molecule.&lt;/li&gt;&#xA;&lt;li&gt;Check it (doing a lot of calculations with the wrong molecule is not fun...)&lt;/li&gt;&#xA;&lt;li&gt;Start optimize geometry tool, wait until &lt;strong&gt;dE=0&lt;/strong&gt; or near 0&lt;/li&gt;&#xA;&lt;li&gt;Stop optimize geometry tool&lt;/li&gt;&#xA;&lt;li&gt;Go to &lt;strong&gt;Visualize&lt;/strong&gt; - &lt;strong&gt;Properties&lt;/strong&gt; - &lt;strong&gt;Molecule properties&lt;/strong&gt; read the polar momentum in Dalton&lt;/li&gt;&#xA;&lt;li&gt;If you want, you can create the potential surface from &lt;strong&gt;extension&lt;/strong&gt; - &lt;strong&gt;surface&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Eventually you can try to search for other conformers and repeat the process&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bEbqV.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;How accurate is this method?&lt;/h1&gt;&#xA;&lt;p&gt;From my attempts unfortunately not very. If you try with &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2Cl2}$&lt;/span&gt; you get about 1.086 D more for the cis conformer and 0.22 Dalton, more sometimes depending on the optimization, for the trans conformer after the research of other conformers.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-26T00:12:07.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9477" PostTypeId="1" CreationDate="2014-03-25T16:46:07.927" Score="3" ViewCount="166" Body="&lt;p&gt;This is a very simple and short question. I am looking for liquids that are low/non-volatile at 20$^\circ$C (at least substantially less volatile than water) which have a high contact angle with glass and oxidized silicon (in air) or with a coating of perfluoroalkanes (specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfluordecyltrichlorosilane&quot; rel=&quot;nofollow&quot;&gt;Perfluorodecyltrichlorosilane&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement on the contact angle is say $\theta&amp;gt;70^\circ$, i.e. the liquid should 'ball-up' a bit. The reason I am asking is that I cannot seem to find any (common) liquids that have both the low-volatility and the high contact angle $\theta$. Does anyone know some examples? Or, if no such thing exists, can someone explain what the fundamental reason is that these liquids don't exist?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-03-25T17:49:51.757" Title="Low-volatile liquids that 'ball-up' on siliconoxide or perfluoroalkanes" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9480" PostTypeId="2" ParentId="9477" CreationDate="2014-03-25T17:49:51.757" Score="2" Body="&lt;p&gt;Here it comes... mercury.  75-wt% Ga 25 wti%, mp. 15.5 C and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galinstan&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Galinstan&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in general wet glass.  Maybe ionic liquids, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Am. Chem. Soc. 129(13), 3804 (2007), DOI: 10.1021/ja070169d &lt;BR&gt;&#xA;Aust. J. Chem. 57(2) 113-155 (2004)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.researchgate.net/publication/233906460_Wetting_of_surfaces_with_ionic_liquids&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.researchgate.net/publication/233906460_Wetting_of_surfaces_with_ionic_liquids&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1016/j.jcis.2008.10.019&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://dx.doi.org/10.1016/j.jcis.2008.10.019&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://ura.unisa.edu.au/R/?func=dbin-jump-full&amp;amp;object_id=63927&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ura.unisa.edu.au/R/?func=dbin-jump-full&amp;amp;object_id=63927&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ppmp.pwr.wroc.pl/sorpa/file/eee7c6b05fb9497a9a59507ca0346479/download&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ppmp.pwr.wroc.pl/sorpa/file/eee7c6b05fb9497a9a59507ca0346479/download&lt;/a&gt; &lt;BR&gt;&#xA;[DMIM][Cl], 1,3-Dimethylimidazolium chloride, is reported 105.9 contact angle on glass.  That might be a typo - it's a huge number for the class of compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://english.cas.cn/ST/BR/br_progress/201106/t20110622_71442.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://english.cas.cn/ST/BR/br_progress/201106/t20110622_71442.shtml&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F67U3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-25T17:49:51.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9481" PostTypeId="2" ParentId="9463" CreationDate="2014-03-25T19:04:25.653" Score="0" Body="&lt;p&gt;The first answer doesn't really answer the question. It merely shows, correctly, what happens to water which is introduced into a Grignard solution already prepared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one hints at it. Water passifies the surface of the Mg (presumably coating it with MgO or Mg(OH)2) making it almost impossible to initiate the reaction. I have found that the techniques used in answer usually work.&lt;/p&gt;&#xA;" OwnerUserId="4978" LastActivityDate="2014-03-25T19:04:25.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9483" PostTypeId="1" AcceptedAnswerId="9486" CreationDate="2014-03-25T20:51:13.593" Score="9" ViewCount="1100" Body="&lt;p&gt;I will be doing a lab on Hydrogen Phosphate buffer system tomorrow and I have a few questions about the buffer system. I am not sure if my assumptions are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) What would happen to the rate of pH change if 5.0M HCl and NaOH were used instead of 0.50M?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that, since the pH of the buffer does not depend on the actual concentration of the buffer, but on the ratio of the two parts, therefore, the rate of pH should not expect to change. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Lets say I prepared a buffer at a different pH, how would the capacity of my buffer system change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that, the buffer capacity would not change upon the change of pH, is this correct?&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="4945" LastEditDate="2014-07-30T05:05:07.330" LastActivityDate="2014-07-30T05:05:07.330" Title="Questions about Buffer System (Lab)" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9484" PostTypeId="2" ParentId="5726" CreationDate="2014-03-25T21:06:08.383" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.ehs.neu.edu/hazardous_waste/fact_sheets/sodium_azide/&quot; rel=&quot;nofollow&quot;&gt;http://www.ehs.neu.edu/hazardous_waste/fact_sheets/sodium_azide/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Sodium azide reacts with heavy metals (such as silver, gold, lead, copper, brass, or solder in plumbing systems); and metal salts to form an accumulation of the highly explosive compounds such as lead azide and copper azide.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume reaction is fast.  How many micromoles of gold/cm^2 constitute your film?  How many micromoles of azide/volume are in your application?  Will the gold be corroded through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the gold there as a convenience or as a necessary surface?  A brief soak in dilute decanethiol solution plus rinse will cover the surface with a monomolecular layer of hydrocarbon, each chain being anchored by its sulfur.  The surface will be passivated and probably hydrophobic. 1-hydroxy-10-decanethiol will give a hydrophilic surface.  Is it worth the bother and expense to diddle with the gold surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold is selling for $(USD)42.15/g today.  How much gold is in your annual discard pile? When my school dumped a load of ancient undergrad combustion bombs I went through them.  They originally shipped each with a platinum crucible.  One remained.  Half a Troy ounce of platinum was a good hour's tax-free wage.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-25T21:06:08.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9486" PostTypeId="2" ParentId="9483" CreationDate="2014-03-25T22:46:13.903" Score="3" Body="&lt;p&gt;Let's consider two buffer solutions of the same kind ($pK_a=4.75)$, but of different concentrations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH3COOH}\,0.01\,\text{M}$ and $\ce{CH3COO-} 0.01\,\text{M}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CH3COOH}\,0.1\,\text{M}$ and $\ce{CH3COO-} 0.1\,\text{M}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The starting buffer volume is the same for both and equal to $100\,\text{mL}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, add a strong base $\ce{NaOH}\,0.05\,\text{M}$, few $\text{mL}$ only: see below $pH$ as a function of $\ce{NaOH}$ added volume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1xGBS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look how &quot;fast&quot; the $pH$ changes and you will conclude that higher concentrations buffers are more effective. Differences can indeed be appreciated from the graph: they witness the evidence that larger concentrations better &quot;buffer&quot; the addition of the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest you to do some calculations (e.g. for $1\,\text{mL}$ of  $\ce{NaOH}\,0.05\,M$ added) and see whether you get the same values of the graph (hopefully I have done them right...). Remember that the addition of a strong base will provoke a decrease in concentration of the acid, $\ce{CH3COOH}$ and a consequent increase of its conjugated base, $\ce{CH3COO-}$.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-25T23:20:04.750" LastActivityDate="2014-03-25T23:20:04.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9487" PostTypeId="1" CreationDate="2014-03-25T23:15:16.473" Score="0" ViewCount="320" Body="&lt;p&gt;I'm a student in highschool AP chemistry. I'm confused about a certain problem in my chem textbook. In summary, I am given a species and asked to identify its basicity as strong, weak, or negligible. I must then give its conjugate acid and identify likewise for its acidity. (in the context of Brønsted-Lowry theory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood all parts of the problem except for $\mbox{O}^{2-}$, to which I assigned the conjugate acid as $\mbox{OH}^{-}$. It was correct thus far. My guess that they were both weak was wrong, however. According to the solutions page, they're both strong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is $\mbox{O}^{2-}$ a strong base? How is $\mbox{OH}^{-}$ a strong acid? To my knowledge, neither are found in the canonical strong acid/base lists, and neither seem to satisfy the &quot;completely ionize in water&quot; definition.&lt;/p&gt;&#xA;" OwnerUserId="4918" LastEditorUserId="5026" LastEditDate="2018-03-04T07:42:53.170" LastActivityDate="2018-03-04T07:42:53.170" Title="Strength identification in conjugate acid-base pairing" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9488" PostTypeId="2" ParentId="9487" CreationDate="2014-03-25T23:56:42.973" Score="2" Body="&lt;p&gt;$\ce{H2O}$ is a weak acid.  Abstract a proton to give $\ce{HO^{-}}$, a strong base.  Abstract a proton from that to give $\ce{O^{2-}}$, a very strong base.  See a species as a whole, up and down for a property.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it with ammonia.  Feel free to wonder why $\ce{Li3N}$ is red.  Look what a very strong base can do,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Li3N + 2 H2 → LiNH2 + 2 LiH}$ (reversible at 270°C)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-25T23:56:42.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9489" PostTypeId="1" AcceptedAnswerId="9490" CreationDate="2014-03-26T02:56:33.113" Score="6" ViewCount="294" Body="&lt;p&gt;If I have a cup of milk and a cup of tea; to get them to a sweetness I like, I usually add less sugar to the milk as compared to the tea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that milk needs less sugar to be just as sweet as tea?&lt;/p&gt;&#xA;" OwnerUserId="4981" LastEditorUserId="5026" LastEditDate="2018-06-29T02:56:02.437" LastActivityDate="2018-06-29T02:56:02.437" Title="Why does sweetening milk need less sugar as compared to tea?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9490" PostTypeId="2" ParentId="9489" CreationDate="2014-03-26T03:57:41.593" Score="13" Body="&lt;p&gt;It's all in your taste buds. Lactose in milk is a form of sugar. It is naturally a little sweet. Tea is slightly bitter so it might seem you need to add more sugar to make it sweet.&lt;/p&gt;&#xA;" OwnerUserId="4943" LastActivityDate="2014-03-26T03:57:41.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9491" PostTypeId="1" AcceptedAnswerId="9494" CreationDate="2014-03-26T04:39:57.910" Score="8" ViewCount="50753" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A $\pu{100 mL}$ solution of $\ce{NaOH}$ has a $\mathrm{pH}$ of $13$. What volume of water in $\pu{mL}$ must be added to change the $\mathrm{pH}$ to $11$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Begin by calculating conc. of $[\ce{OH-}]$ for $\mathrm{pH}~13$.  &lt;/li&gt;&#xA;&lt;li&gt;Find the moles by multiplying concentration and volume ($\pu{0.1 L}$)&lt;/li&gt;&#xA;&lt;li&gt;Determine new concentration of $[\ce{OH-}]$ for $\mathrm{pH}~11$.&lt;/li&gt;&#xA;&lt;li&gt;Use $\text{amount}/\text{new concentration}$ to find the total volume.&lt;/li&gt;&#xA;&lt;li&gt;Use total volume ($\pu{100 mL}$) to find volume needed to be added.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The above doesn't produce the right answer, which is $\pu{9.900 mL}$. Can someone please fix my working out, or provide alternative steps to calculate this? &lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="41328" LastEditDate="2017-12-26T11:47:04.090" LastActivityDate="2019-10-13T09:46:47.403" Title="Volume required to dilute solution for a pH change" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9492" PostTypeId="1" AcceptedAnswerId="9496" CreationDate="2014-03-26T05:04:27.620" Score="3" ViewCount="8980" Body="&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using the following salts listed below, calculate the moles of acid and conjugate base needed to make $V = \pu{100 mL}$ of $c = \pu{0.50 M}$, $pH \in 6.5, 6.7, 6.8$ phosphate buffer solutions. ($\mathrm pK_\mathrm a(\ce{H2PO4}) = 6.64$)  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Molas masses of salts:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{KH2PO4} = 136.09 ~\pu{g/mol}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{NaH2PO\cdot H2O} = \pu{137.99 g/mol}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{Na2HPO4 \cdot 7H2O} = \pu{268.09 g/mol}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{K2HPO4} = \pu{174.2 g/mol}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried using the Henderson-Hasselbalch equation but I am stuck at the point where I have found the base-acid ratio and I need to find the moles of the acid and the moles of the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm pH = \mathrm pK_\mathrm a + \log\left(\frac{[\text{Conjugate base}]}{[\text{Acid}]}\right)\\&#xA;n = \pu{100 ml} \times \pu{0.5 M} =\pu{ 0.05 mol}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For pH 6.5:&#xA;$$&#xA;6.5 = 6.64 + \log\frac{[B]}{[A]}\\&#xA;\frac{[B]}{[A]} = 0.72&#xA;$$&#xA;Here I got the base-acid ratio as 0.72 but what should I do next? I tried looking up for solutions and a person suggests using $A + B = 0.05$ But how do I know what A and B are just by knowing the ratio is 0.72?&lt;/p&gt;&#xA;" OwnerUserId="4638" LastEditorUserId="5026" LastEditDate="2018-02-24T03:43:18.110" LastActivityDate="2018-02-24T16:31:05.413" Title="Calculate the moles of acid and conjugate base needed" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9493" PostTypeId="1" AcceptedAnswerId="9495" CreationDate="2014-03-26T05:20:13.753" Score="12" ViewCount="13095" Body="&lt;p&gt;Is it the same reason as to why transition metal complexes have colour?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="194" LastEditDate="2016-12-05T20:18:31.223" LastActivityDate="2021-03-04T10:59:53.757" Title="What is the origin of the colour of azo dyes?" Tags="&lt;aromatic-compounds&gt;&lt;transition-metals&gt;&lt;color&gt;&lt;photochemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9494" PostTypeId="2" ParentId="9491" CreationDate="2014-03-26T05:32:33.953" Score="4" Body="&lt;p&gt;Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start} &#xA;&amp;amp;= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\&#xA;V_\mathrm{end} &amp;amp;= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{pH}_\mathrm{start} &amp;amp;= 13 \\&#xA;c_\mathrm{start}(\ce{OH-}) &amp;amp;= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\&#xA;V_\mathrm{start} &amp;amp;= 0.1~\mathrm{L} \\&#xA;\ce{pH}_\mathrm{end} &amp;amp;=11 \\&#xA;c_\mathrm{end}(\ce{OH-}) &amp;amp;=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.&#xA;$$ &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2016-03-22T05:46:21.723" LastActivityDate="2016-03-22T05:46:21.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9495" PostTypeId="2" ParentId="9493" CreationDate="2014-03-26T05:40:20.630" Score="19" Body="&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, it is all about the absorption of light at specific wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TZ5vK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Azobenzene, the parent compound has an absorption maximum around $\lambda$= 430 nm in the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J667M.png&quot; alt=&quot;azo dyes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples are &lt;em&gt;Allura Red&lt;/em&gt; (&lt;strong&gt;1&lt;/strong&gt;), &lt;em&gt;Chrysoine Resorcinol&lt;/em&gt; (&lt;strong&gt;2&lt;/strong&gt;), &lt;em&gt;Janus Green B&lt;/em&gt; (&lt;strong&gt;3&lt;/strong&gt;) and &lt;em&gt;Direct Blue 1&lt;/em&gt; (&lt;strong&gt;4&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The colours of the molecules are chosen to resemble the colour in solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make the dyes soluble in water, either sulfonate groups are attached or the dye molecule is just one fat cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already mentioned by GM in another answer, +M substituents elongate the $\pi$ system (auxochromic effect).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-27T05:16:05.940" LastActivityDate="2014-03-27T05:16:05.940" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9496" PostTypeId="2" ParentId="9492" CreationDate="2014-03-26T06:58:50.080" Score="2" Body="&lt;p&gt;Since it is a buffer solution, the volume is constant ($V =\pu{0.1 L}$), hence&#xA;$$&#xA;\frac{c(b)}{c(a)} = \frac{n(b)}{n(a)}\overset{calc.}{=}0.72&#xA;$$&#xA; and with&#xA;$$&#xA;n_{\text{tot}}=n(b)+n(a) \overset{\text{def}.}{=}\pu{0.05 mol}\\&#xA;n(a)=\frac1{1.72}\cdot\pu{0.05 mol}\\&#xA;n(b)=\frac{0.72}{1.72}\cdot\pu{0.05 mol}&#xA;$$&#xA;Then $$m=M\cdot n$$ and you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-02-24T16:31:05.413" LastActivityDate="2018-02-24T16:31:05.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9497" PostTypeId="2" ParentId="9493" CreationDate="2014-03-26T07:54:29.283" Score="13" Body="&lt;h1&gt;What is the origin of colours?&lt;/h1&gt;&#xA;&lt;p&gt;Most of the colours that we perceive are originate by the selective absorption of some spectral bands and the reflection of the others wavelength, some times with the contribution of fluorescence from the absorption at a higher wavelength. If we exclude colour due to interference (e.g. some butterfly wings!) all the other phenomena are linked with the absorption of the photons. However, this absorption could be due to different causes and so we can differentiate between different absorption mechanisms.&lt;/p&gt;&#xA;&lt;h1&gt;Absorption of electromagnetic waves through bonding and anti-bonding &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; orbitals&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note: I suppose you are familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;MO&lt;/a&gt; theory&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;One very spread mechanism of absorption of electromagnetic waves is through electrons in pi and sigma orbitals. The absorption occurs at different wavelengths depending on the compound but however, most of the transitions occur normally in the UV region (400nm is the limit of VIS region). I have made this picture with TikZ from this &lt;a href=&quot;http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/uvvisab1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;site&lt;/a&gt; you can see that only &lt;span class=&quot;math-container&quot;&gt;$n \rightarrow \pi^{*}$&lt;/span&gt; is in the visible region. &lt;img src=&quot;https://i.stack.imgur.com/x2hN2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;How can azo dyes absorb light?&lt;/h1&gt;&#xA;&lt;p&gt;In the case of azo dyes, the Chromophore is &lt;span class=&quot;math-container&quot;&gt;$\ce{-N=N}-$&lt;/span&gt;. This group permits the absorption of light. However, Azobenzene has its absorption peak in UV region (it only absorbs a little bit of blue).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Qjbi1.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So how can azo dyes absorb light (of course for light we means 400-700 nm radiation) and so be colourful? One way to do that is to form a bigger &lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugation_(organic_chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjugated system&lt;/a&gt; to lower the energy, in this case delocalizing the electrons across the molecule. For azobenzene you can also add two auxochromes the hydroxyl group (&lt;span class=&quot;math-container&quot;&gt;$-OH$&lt;/span&gt;) and the amino group (&lt;span class=&quot;math-container&quot;&gt;$-NH_{2}$&lt;/span&gt;) these groups permit the formation of a charge-transfer complex &lt;a href=&quot;http://en.wikipedia.org/wiki/Charge-transfer_complex&quot; rel=&quot;nofollow noreferrer&quot;&gt;CT Complex&lt;/a&gt; that should be something like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KWBJP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Is it the same reason as to why transition metal complexes have colour?&lt;/h2&gt;&#xA;&lt;p&gt;I'm not sure that is right to say that is the same reason whereby transition metal complexes have colour. However, this is a tricky question because we have seen that there are many factors that cause the absorption so of course, a transition metal can be considered a conjugated system as said Dan S, most of the time, however, these interactions are called  ligand-to-metal charge-transfer (LMCT).&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-01-09T14:31:48.030" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="9498" PostTypeId="1" CreationDate="2014-03-26T08:56:08.953" Score="5" ViewCount="19084" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{MnO4-}$&lt;/span&gt; required to oxidize one mole of ferrous oxalate completely in acid medium will be:&lt;/p&gt;&#xA;&lt;p&gt;(a) 7.5 moles&lt;/p&gt;&#xA;&lt;p&gt;(b) 0.2 moles&lt;/p&gt;&#xA;&lt;p&gt;(c) 0.6 moles&lt;/p&gt;&#xA;&lt;p&gt;(d) 0.4 moles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In acidic medium, 1 mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{MnO_4^-}$&lt;/span&gt; accepts 5 moles of electrons. Oxalate is dibasic so will give 2 moles of electrons when oxidized to &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^2+ -&amp;gt; Fe^3+ + e-} $&lt;/span&gt; so total 3 moles of electrons.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{Amount of substance of } \ce{MnO_4^-}=\frac{3}{5}=0.6$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, the given answer is 0.4 moles indicating that only oxalate is considered to be oxidized.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Does  &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+} -&amp;gt; Fe^{3+} + e^-} $&lt;/span&gt;  actually occur in this reaction?&lt;/p&gt;&#xA;&lt;p&gt;What would be the net reaction when &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO_4}$&lt;/span&gt; is used(not just the ions)?&lt;/p&gt;&#xA;" OwnerUserId="1939" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-07-23T05:14:18.897" Title="How much potassium permanganate is needed to completely oxidize ferrous oxalate?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9499" PostTypeId="1" AcceptedAnswerId="9502" CreationDate="2014-03-26T08:57:16.993" Score="11" ViewCount="27317" Body="&lt;p&gt;I would like to ask what are the main differences between doxycycline HCl and hyclate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work in the pharmaceutical/veterinary raw materials industry and some clients are saying that they prefer hyclate because it is more concentrated than HCl, but I have yet to come on any articles confirming this. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="4982" LastEditorUserId="10185" LastEditDate="2015-10-02T16:24:31.780" LastActivityDate="2020-12-26T04:47:20.497" Title="What is the difference between doxycycline HCl and doxycycline hyclate?" Tags="&lt;organic-chemistry&gt;&lt;drugs&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9500" PostTypeId="2" ParentId="8498" CreationDate="2014-03-26T09:02:04.023" Score="6" Body="&lt;p&gt;Here is a neat experiment, but it unfortunately needs a welder to make the equipment, a metal sphere with a volume of about 100 mL. It should have a very small opening on top and a larger one on the bottom. It also needs feet to stand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/grgKY.png&quot; alt=&quot;experiment setup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal container should be capable of some pressure, because the gas will explode inside and you don't want to hurt anyone. Additional you need bucket or bowl filled with water and hydrogen gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, close the hole on the top of the container and place it in the bucket under water until it is completely full. Now you can fill the container with hydrogen gas until all water is gone from it. When you remove your container, make sure that the bottom hole stays on the bottom. As Hydrogen is much lighter than air, it will be trapped inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place the container on a stable, nonburnable table/ bench. Have a lighter ready. Now open the top hole and ignite the hydrogen gas. Move away from the experiment. After a while you will hear a phenomenal boom (make sure you have your mouth open). &lt;strong&gt;Do not try this in a small room.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick behind this is in the ratios. As Hydrogen will burn on top, air will be sucked in at the bottom. As long as the mixture has not the right ratio, it will not explode but just burn at the top. Once enough oxygen has entered it will ignite the whole thing and do a phenomenal bang. The experiment is pretty safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is part of this &lt;a href=&quot;http://www.chymiatrie.de/index.php/component/content/article/157-7lanaexperimentalvortrag&quot; rel=&quot;noreferrer&quot;&gt;lecture&lt;/a&gt;, which is in German unfortunately.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-03-26T09:02:04.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9501" PostTypeId="2" ParentId="4760" CreationDate="2014-03-26T09:07:30.723" Score="-3" Body="&lt;p&gt;$\ce{F}$ is much more electronegative compared to $\ce{H}$. Thus it causes a contraction in the $d$ orbital of $\ce{S}$. The $d$ orbital of $\ce{S}$ becomes much more energetically stabilised. It can then participate in bonding with the filled $p$ orbital of $\ce{F}$. For $\ce{H}$ this is not possible. The cost of hybridisation is much higher in the latter case. Thus $\ce{SH4}$ does not exist but $\ce{SF4}$ does.&lt;/p&gt;&#xA;" OwnerUserId="4983" LastEditorUserId="275" LastEditDate="2014-03-26T10:19:23.100" LastActivityDate="2014-03-26T10:19:23.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9502" PostTypeId="2" ParentId="9499" CreationDate="2014-03-26T09:18:42.063" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...]they prefer hyclate because it crystallizes more readily than HCl salt &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The principal component is doxycycline - in both cases. The &lt;em&gt;active&lt;/em&gt; component is a metabolite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doxycycline is an antibiotic and belongs to the &lt;em&gt;tetracyclines&lt;/em&gt; (= &lt;em&gt;four rings&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nen78.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prepare nice ions that can traverse membranes and pumps by the body easily, pharmaceutics containing amino groups or 'alkaloids' (see the red nitrogen atom in the structure), are often converted to their hydrochloride, citrate, oxalate, bromide etc. salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are salts with the (protonated) pharmaceutical as the cation and chloride as the anion. &lt;strong&gt;Salts are water soluble.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the notation for this is doxycycline$\ce{.HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystallisation is done in a solvent or a mixture of solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the doxycycline$\ce{.HCl}$ precipitates (=when it forms a solid), molecules of the solvent may get embedded, trapped, or &lt;strong&gt;entrained&lt;/strong&gt; in the crystal structure (but the crystals gets pulverized anyway and re-crystallized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is not just a wet solid! There would be little or no absorption in that case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly this happens when the &lt;em&gt;hyclate&lt;/em&gt; is formed during the crystallization of doxycycline hydrochloride in a mixture of water and ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hyclate&lt;/em&gt; is just an abbreviation for &lt;em&gt;hydrochloride hemiethanolate hemihydrate&lt;/em&gt;, which means that &lt;strong&gt;each molecule of doxycycline&lt;/strong&gt; comes with 1 $\ce{.HCl\ }$  &lt;strong&gt;and&lt;/strong&gt; 0.5 $\ce{.CH3CH2OH}$ and 0.5 $\ce{.H2O}$ in the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it is more concentrated (= more active component in one pill) only depends on the formulation of the respective pharmaceutical companies.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="20789" LastEditDate="2016-07-05T21:35:13.757" LastActivityDate="2016-07-05T21:35:13.757" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9503" PostTypeId="1" AcceptedAnswerId="9508" CreationDate="2014-03-26T10:22:53.493" Score="4" ViewCount="743" Body="&lt;p&gt;I was told that 'water' was the most common material in the universe and I disputed that claim. I was met with the argument that hydrogen and oxygen were the most common atoms in the world so water was as well. If this is true, then how come we can't find water anywhere in outer space? It must be the collection of 2 hydrogen atoms and 1 oxygen atom that is so rare then?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2014-03-26T14:48:44.463" Title="Most common material in the universe" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9504" PostTypeId="1" AcceptedAnswerId="9506" CreationDate="2014-03-26T11:40:46.197" Score="1" ViewCount="2485" Body="&lt;p&gt;I recently read about potassium aluminum sulfate that it has astringent property and has the ability to constrict body tissues and restrict blood flow. Since, for penis, the erection happens with the flow of the blood being directed to penis' blood vessels, does this effect of potassium aluminum sulfate cause an erection problem or any other effect on male/female sex organ?&lt;/p&gt;&#xA;" OwnerUserId="4985" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:54:29.293" LastActivityDate="2015-10-30T16:54:29.293" Title="Does potassium aluminum sulfate have any effect on sexual performance/sperm counts or anything relating to the functionality of the sexual organs?" Tags="&lt;pharmacology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9505" PostTypeId="2" ParentId="6950" CreationDate="2014-03-26T13:31:43.270" Score="1" Body="&lt;p&gt;In &lt;a href=&quot;http://dx.doi.org/10.1107/S0365110X67000453&quot; rel=&quot;nofollow&quot;&gt;Crystal structure of Sm5Ge4&lt;/a&gt;, the structure is described as &lt;em&gt;orthorhombic&lt;/em&gt; with $a = 7.75\,\mathring{A}$, $b = 14.94\,\mathring{A}$ and  $c = 7.84\,\mathring{A}$, having four $\ce{Sm5Ge4}$ units per unit cell. The authors further denote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The structure is conveniently described in terms of three layers which occur along the $b$ axis. Layer $G$ contains $\ce{Ge}$ atoms only; layer $S$ $\ce{Sm}$ atoms only Layer $C$ is a combination of $G$ and $S$, but with the roles of $\ce{Ge}$ and $\ce{Sm}$ interchanged. Layer $C$ serves as a mirror plane in $Pnma$, but only as a quasi-mirror plane in $Pn2_1a$. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Personally, I find it a bit irritating that this article from 1967, which seemingly is the original reference, is not directly cited on the Materials Project page.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-26T13:31:43.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9506" PostTypeId="2" ParentId="9504" CreationDate="2014-03-26T14:30:33.390" Score="0" Body="&lt;p&gt;Oral ingestion of simple aluminum compounds is not absorbed.  Topical genital application is definitely not recommended.  There is no obvious connection between aluminum and tumescence, one way or another  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soluble aluminum is toxic, likely cumulatively neurotoxic.  It is not absorbed from the GI tract unless complexed.  Aluminum hydroxide antacids as foaming tablets( citric acid plus sodium bicarbonate) gave aluminum citrate that is bioavailable.  Bad idea.  Pickles are made with alum to keep them ever so crunchy. The soluble aluminum is claimed not to be absorbed. Antiperspirant contains aluminum chlorhydroxide or glycinate to stopper sweat glands.  Topical application is not absorbed.  Exposing mucus membranes to soluble aluminum is definitely not recommended. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-26T14:30:33.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9508" PostTypeId="2" ParentId="9503" CreationDate="2014-03-26T14:48:44.463" Score="2" Body="&lt;p&gt;The Big Bang started with about 3/4 hydrogen atoms and 1/4 helium atoms.  Add a pinch of lithium and a whisper of boron.  Supernovae since then have released a lot of helium.  Heavier elements are still rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The most common [molecular] material in the universe&quot; must survive photolysis, radiolysis, and reaction.  Water is poor candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.astro.psu.edu/users/alex/astro497_7.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.astro.psu.edu/users/alex/astro497_7.pdf&lt;/a&gt; &lt;BR&gt;&#xA;What is out there.  CO is the marker. &lt;BR&gt;&#xA;&lt;a href=&quot;http://physics.nist.gov/cgi-bin/micro/table5/start.pl&quot; rel=&quot;nofollow&quot;&gt;http://physics.nist.gov/cgi-bin/micro/table5/start.pl&lt;/a&gt; &lt;BR&gt;&#xA;Molecules in space, isotopically and by &lt;BR&gt;&#xA;&lt;a href=&quot;http://science.gsfc.nasa.gov/691/cosmicice/interstellar.html&quot; rel=&quot;nofollow&quot;&gt;http://science.gsfc.nasa.gov/691/cosmicice/interstellar.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.astro.uni-koeln.de/cdms/molecules&quot; rel=&quot;nofollow&quot;&gt;http://www.astro.uni-koeln.de/cdms/molecules&lt;/a&gt; &lt;BR&gt;&#xA;atom count&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-26T14:48:44.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9509" PostTypeId="2" ParentId="9498" CreationDate="2014-03-26T15:16:43.057" Score="3" Body="&lt;p&gt;See &lt;a href=&quot;http://www2.bgsu.edu/departments/chem/faculty/endres/ch128/oxalate.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;green salt&quot; is $\ce{K3Fe(C2O4)3.3 H2O}$.&#xA;\begin{align}&#xA;\ce{MnO4- + 8H+ +5e- &amp;amp;-&amp;gt; Mn^2+ + 4 H2O}\\&#xA;\ce{C2O4^2- &amp;amp;-&amp;gt; 2 CO2 + 2e-}\\&#xA;\ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &amp;amp;-&amp;gt; 2 Mn^2+ + 8 H2O + 10 CO2 (g)}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One expects all the oxalate is oxidized to $\ce{CO2}$ and $\ce{Fe(II)}$ is oxidized to $\ce{Fe(III)}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4945" LastEditDate="2017-07-23T05:14:18.897" LastActivityDate="2017-07-23T05:14:18.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9510" PostTypeId="1" CreationDate="2014-03-26T17:48:59.990" Score="31" ViewCount="32511" Body="&lt;p&gt;According to molecular orbital theory, the bond order of $\ce{CO}$ is 3. When $\ce{CO+}$ is formed, the bond order decreases to 2.5, and thus the bond length should increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the bond length of $\ce{CO+}$ is found to be less than $\ce{CO}$. How is that so?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="16683" LastEditDate="2017-07-12T12:10:31.293" LastActivityDate="2017-07-12T12:10:31.293" Title="Why is the bond length of CO+ less than that of CO?" Tags="&lt;covalent-compounds&gt;&lt;molecular-orbital-theory&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="9511" PostTypeId="2" ParentId="9510" CreationDate="2014-03-26T20:00:46.760" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.meta-synthesis.com/webbook/39_diatomics/co.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.meta-synthesis.com/webbook/39_diatomics/co.jpg&lt;/a&gt; &lt;BR&gt;&#xA;Measured $\ce{CO}$ bond length is 1.128 Å, $\ce{CO^{+}}$ is 1.115 Å.  Given the following carbon monoxide MO diagram, from where is the electron removed to afford the cation with a shorter bond length?  $\ce{CO}$ has a dipole moment of 0.122 debyes, net negative charge on the carbon. One then suspects the electron is removed from the sigma anti-bonding orbital on carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w6qNx.jpg&quot; alt=&quot;carbon monoxide MO diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-26T20:00:46.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9512" PostTypeId="2" ParentId="9510" CreationDate="2014-03-26T20:02:56.647" Score="2" Body="&lt;p&gt;Molecular Orbital theory does NOT predict a longer bond. It correctly predicts a shorter bond. The HOMO for $\ce{CO}$ is a sigma antibonding orbital. When it is ionized, the bond order is increased to 3.5, and the bond length decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nWjgj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2014-03-26T20:02:56.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9513" PostTypeId="2" ParentId="9510" CreationDate="2014-03-26T20:46:24.810" Score="18" Body="&lt;p&gt;Unfortunately, the arguments presented by &lt;code&gt;buckminst&lt;/code&gt; and &lt;code&gt;Uncle Al&lt;/code&gt; aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in &lt;code&gt;buckminst&lt;/code&gt;'s diagramm, $\sigma_{3}$ in &lt;code&gt;Uncle Al&lt;/code&gt;'s diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the right symmetry (in this respect the MO scheme of &lt;code&gt;Uncle Al&lt;/code&gt; is better than the one of &lt;code&gt;buckminst&lt;/code&gt; because it indicates this mixing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the original problem still persists.&#xA;One possible explanation for the bond shortening after ionization is that the ionization leads to a shift of the $\ce{CO}$-electron-polarization (on ionization an electron is lost from the mostly $\ce{C}$-centered HOMO-$\sigma$ orbital and this leads to a formation of a positive partial charge on the $\ce{C}$ atom).&#xA;This strengthens the covalence of the $\ce{CO}$-bond and thus reduces the bond length (since the HOMO-$\sigma$ orbital is only slightly bonding there is not much bonding lost by taking away one of its electrons - the lost bonding is outweighed by the gain in covalency).&#xA;You can think of this strenghtening of the bond covalence in the following way: Two atomic orbitals can interact better (form stronger bonds) if their energies are close. Without the positive partial charge on $\ce{C}$ the AOs of $\ce{O}$ lie energetically a lot below the AOs of $\ce{C}$ (this is most pronounced for the $\ce{s}$ AOs which form the  HOMO-$\sigma*$ orbital). But with the positive partial charge on $\ce{C}$ the AOs of $\ce{C}$ are shifted down in energy and thus their energies are closer to the related AOs of $\ce{O}$ which leads to a stronger interaction when bonds are formed.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-07-13T23:16:29.927" LastActivityDate="2014-07-13T23:16:29.927" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9515" PostTypeId="2" ParentId="9493" CreationDate="2014-03-26T22:10:33.610" Score="2" Body="&lt;p&gt;Is it the same as transition metal colors? If you define this as &quot;an electron is promoted to a higher energy state&quot; and that photon absorption defines the color then the answer is yes.  If you define it as an absorption by d orbitals then the answer is no.  As others have indicated these dyes can absorb in the visible wavelengths due to the aromatic delocalized pi electrons.  These are electrons that are shared over many atoms.  In a transition metal you are simply promoting electrons in the d shell from one state to another.  The electrons aren't &quot;wandering&quot; over a bunch of other atoms.&lt;/p&gt;&#xA;" OwnerUserId="4989" LastActivityDate="2014-03-26T22:10:33.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9516" PostTypeId="1" AcceptedAnswerId="9518" CreationDate="2014-03-26T22:15:26.297" Score="2" ViewCount="351" Body="&lt;p&gt;I'm taking entry level collage chem. class, and we got to this topic that I don't really understand. I think I understood what an ion is (a atom whose amount of electrons is not equal to amount of protons, so it has positive or negative charge). But then polyatomic ions... Lets say peroxide; consists of two Oxygens, whose ions have charge of -2. Consequently, two of these anions are suppose to have charge of -4, but peroxide has charge of -2. How does this work? If polyatomic ions don't just sum the charge of ions they consist of, how else can I predict and understand their charge?&lt;/p&gt;&#xA;" OwnerUserId="4743" LastActivityDate="2014-03-26T23:54:16.157" Title="Charge of polyatomic ions doesn't equal the charge of ions it consists of?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9517" PostTypeId="1" AcceptedAnswerId="9522" CreationDate="2014-03-26T23:06:57.843" Score="17" ViewCount="2211" Body="&lt;p&gt;Free chlorine ($\ce{Cl}$) in the stratosphere can deplete ozone ($\ce{O3}$)  as follows:&#xA;$$\ce{Cl + O3 -&amp;gt; ClO + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chlorine atoms can then react with oxygen and return to the beginning of the cycle:&#xA;$$\ce{ClO + O -&amp;gt; Cl + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The EPA &lt;a href=&quot;http://www.epa.gov/ozone/science/process.html&quot;&gt;estimates&lt;/a&gt; the ozone-depleting potential of a single chlorine atom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the equations above, it seems that each chlorine atom returns to its original form. Why, after about 100,000 cycles, does it lose its ozone-depleting potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question, how is this estimated?&lt;/p&gt;&#xA;" OwnerUserId="4990" LastActivityDate="2017-12-26T12:09:59.350" Title="Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?" Tags="&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;atoms&gt;&lt;atmospheric-chemistry&gt;&lt;ozone&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9518" PostTypeId="2" ParentId="9516" CreationDate="2014-03-26T23:54:16.157" Score="2" Body="&lt;p&gt;H-O-O-H.  The bond in the middle is covalent.  Deprotonate to $\ce{{HOO}^{-}}$ (hydroperoxide) then $\ce^{^{-}{OO}^{-}}$, peroxide dianion, $\ce{O2^{2-}}$.  You learn it by doing it, having examples, learning what to look for, and gaining understanding.  Both oxygens are in the -1 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even good answers here have a legitimate diversity of explanation.  The world is complicated and dirty.  Unlike physicists (quantum gravitation, SUSY, dark matter), we can test our predictions with experiments.  Chemists are not too stinking proud to admit we were wrong, e.g., benzene displacements then palladium catalysis.  That is how we make real world progress, by discovering footnotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introductory inorganic does not have wiggle room.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-26T23:54:16.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9520" PostTypeId="1" AcceptedAnswerId="9521" CreationDate="2014-03-27T02:53:41.473" Score="3" ViewCount="287" Body="&lt;p&gt;Basically can someone explain why the molecularity cannot be determined from the rate law? &lt;/p&gt;&#xA;" OwnerUserId="4992" LastEditorUserId="4945" LastEditDate="2014-06-12T09:07:12.227" LastActivityDate="2014-06-12T09:07:12.227" Title="Why can't the molecularity be determined from the rate law?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9521" PostTypeId="2" ParentId="9520" CreationDate="2014-03-27T03:26:52.360" Score="4" Body="&lt;p&gt;The coefficients only correspond to the rate law coefficients if the reaction is elementary. Elementary reactions are ones where the reagent(s) combine through a mechanism implied by the reaction. These are simple. Many reactions aren't as simple and have multiple steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this &lt;a href=&quot;http://www.wiley.com/college/pratt/0471393878/student/review/kinetics/2_equations.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;, example two. The reaction is actually comprised of two reactions where the slow step determines the rate coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, you cannot just read the equation and get the coefficients. One can only determine whether a reaction is elementary (i.e. find its rate law coefficients) by experimentation.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-03-27T03:26:52.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9522" PostTypeId="2" ParentId="9517" CreationDate="2014-03-27T03:37:50.050" Score="17" Body="&lt;p&gt;Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a closed system, the catalytic cycle would continue until the ozone concentration is small enough, that the chance of chlorine atoms recombining is bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Nicolau's comment I decided to further look into the issue. Also qwersjc gave some valuable insights. For further reading I suggest, the following sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/&quot; rel=&quot;noreferrer&quot;&gt;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/&lt;/a&gt; (overview)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/rowland-lecture.pdf&quot; rel=&quot;noreferrer&quot;&gt;Nobel Lecture by S. Rowland&lt;/a&gt; (Free Access)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/v74-233&quot; rel=&quot;noreferrer&quot;&gt;Stratospheric Chlorine: a Possible Sink for Ozone. R. S. Stolarski, R. J. Cicerone; Canadian Journal of Chemistry, 1974, 52(8): 1610-1615.&lt;/a&gt; (Open access)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v249/n5460/abs/249810a0.html&quot; rel=&quot;noreferrer&quot;&gt;Original Paper by M. J. Molina and F. S. Rowland; Nature 249, 810 - 812 (28 June 1974)&lt;/a&gt; (Might not be free), &lt;a href=&quot;http://ozone.unep.org/pdf/stratopheric.pdf&quot; rel=&quot;noreferrer&quot;&gt;reprint on ozone.unep.org&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Ozon Cycle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ozone cycle is what protects us from quite dangerous UV-B radiation ($M$ is an inert Air molecule like $\ce{N2}$):&#xA;$$&#xA;\ce{O2 -&amp;gt;[h\nu] O + O} \\&#xA;\ce{O + O2 -&amp;gt;[&amp;lt;M&amp;gt;] O3}\\&#xA;\ce{O3 -&amp;gt;[h\nu] O + O2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Troublemakers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for chlorine atoms in the stratosphere is man made. Chlorofluorocarbons were very common in many products. These molecules are colourless and water-insoluble. Diffusion taking its course the end up in the stratosphere, where they may undergo dissociation.&#xA;$$\ce{CCl3F -&amp;gt;[h\nu] Cl + CCl2F}$$&#xA;So the main problem is, that free chlorine radicals are produced in the stratosphere. Also these CFCs have a lifetime in the stratosphere of up to 150 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some natural sources (of close to insignificant importance) that also produces free Chloride Radicals, i.e. volcanic eruptions. Those may transport Hydrochloride into the stratosphere, where it can react with the Hydroxide radical:&#xA;$$\ce{HCl + OH -&amp;gt; Cl + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Removal of Chlorine&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several steps involved, that lead to the removal of Chlorine from the stratosphere. Basically it will stay there for quite a while and cause significant damage. The stratosphere is very dry and water-soluble compounds, like HCl, will not be washed out quickly. In order to leave the stratosphere, the chlorine atoms or compounds therefore have to be near the troposphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most important reaction is with methane forming Hydrochloride, which may the be washed out by water (causing at least a part of acid rain nevertheless).&#xA;$$\ce{Cl + CH4 -&amp;gt; HCl + CH3}$$ The likeliness of this event is about 1000 times smaller than colliding with another ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However eventually, after about two years (&lt;a href=&quot;http://zebu.uoregon.edu/text/ozone&quot; rel=&quot;noreferrer&quot;&gt;a number I picked up on the way&lt;/a&gt;), a chlorine will leave the stratosphere. In the troposphere is more water present, so Hydrochloric acid will be formed and washed out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In our estimates, termination of the Cl--ClO chain results from a downward diffusion of the longer lived species in the chain (HCl, ClO) and eventual removal by tropospheric process. &lt;em&gt;From: Nature 249, 810-812 (28 June 1974)&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also a number of reactions that decrease the potential of Chlorine radicals.&#xA;$$\ce{ClO + NO2 -&amp;gt; ClONO2}\\&#xA;\ce{ClO + O2H -&amp;gt; ClOH + O2}\\&#xA;...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However these also form reservoir molecules that might diffuse back to the stratosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations behind this are rather complex and use a wide variety of empiric data along with complex models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is still a lot of research going on in this area, considering more molecules like $\ce{CO2}$ and other greenhouse gases, as well as global climate change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nobel lecture (1995) closes with the following remarks and we can only hope that this holds true:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, with the limitations just expressed, the threat of extensive further stratospheric ozone depletion during the 21st century appears to be under control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2016-04-20T11:57:41.380" LastActivityDate="2016-04-20T11:57:41.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9523" PostTypeId="1" CreationDate="2014-03-27T03:39:03.577" Score="5" ViewCount="1072" Body="&lt;p&gt;How would you draw the mechanism going from non-3-enoic acid to coconut aldehyde (5-pentyloxolan-2-one)? I know I added sulfuric acid and sodium carbonate, which I was confused how they worked into the mechanism. I am certain there is some type of internal reaction to create the five membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the only chemicals that could, in my opinion, be involved. The others are not needed in the mechanism. Pretty much confused on the first step of what happens when sulfuric acid is added. What does it protonate? How does this lead to an internal reaction and where does that reaction take place? Does the C-C double bond have anything to do with it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the lone pairs on the oxygen readily take the proton? But how would that affect the overall state of the compound in order to form that ring. I'm not seeing how that could be right.&lt;/p&gt;&#xA;" OwnerUserId="4994" LastEditorDisplayName="user7951" LastEditDate="2018-03-04T14:03:38.303" LastActivityDate="2018-03-04T14:03:38.303" Title="Coconut aldehyde mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9524" PostTypeId="2" ParentId="9523" CreationDate="2014-03-27T05:47:36.663" Score="5" Body="&lt;p&gt;I have the impression that you a very close to find the answer yourself! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will not give a full answer here, but some hints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] I know I added Sulfuric acid and sodium carbonate [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's assume that you did not add them simultaneously. Then, it seems likely that sulfuric acid &lt;em&gt;does the magic&lt;/em&gt; and sodium carbonate is added to neutralize the reaction mixture after your product has been formed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am certain there is some type of internal reaction to create the five membered ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good point! Let's &lt;strong&gt;draw&lt;/strong&gt; the starting material and the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6yfr.png&quot; alt=&quot;3-nonenoic acid&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-nonenoic acid is a carboxylic acid with a &lt;strong&gt;non-conjugated $\ce{C=C}$ double bond&lt;/strong&gt;. With other words, there are &lt;strong&gt;two functional groups&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think in the sulfuric acid! It's a strong acid while its anion is a rather weak nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XETkY.png&quot; alt=&quot;furanone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;Coconut aldehyde&lt;/em&gt;&quot; is  not an aldehyde at all, but a furanone, a &lt;strong&gt;five-membered lactone&lt;/strong&gt;. Think intramolecular ester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it conceivable that the $\ce{C=C}$ double bond is protonated?&lt;/li&gt;&#xA;&lt;li&gt;If so, what could possibly add to the resulting cation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-28T12:15:10.603" LastActivityDate="2014-03-28T12:15:10.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9525" PostTypeId="2" ParentId="6117" CreationDate="2014-03-27T11:40:07.170" Score="3" Body="&lt;p&gt;There's no further need to comment on the &lt;strong&gt;coffee maker&lt;/strong&gt;, but you might give the following cleaning method for the condenser and the rotation tube a try: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that you might want to first check whether this is in accordance with local safety regulations!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Take a small or medium-sized (max. 500 mL) 2-neck flask (NS29 socket connects to the rotavap, NS14.5 socket is for your thumb) and fill it with max. 250 mL acetone&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hook it up to the rotavap, support it with your hand&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;No spinning :D&lt;/li&gt;&#xA;&lt;li&gt;Start the vacuum pump&lt;/li&gt;&#xA;&lt;li&gt;Close the NS14.5 socket with your thumb&lt;/li&gt;&#xA;&lt;li&gt;Watch the acetone starting to boil under reduce pressure &lt;/li&gt;&#xA;&lt;li&gt;Rapidly remove your thumb from from the socket and let the acetone &quot;spit&quot; all through the rotavap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-27T11:40:07.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9526" PostTypeId="2" ParentId="9517" CreationDate="2014-03-27T13:49:52.290" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone_depletion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, Martin is correct; a chlorine molecule, being slightly heavier than oxygen, will eventually be transported out of the ozone layer back to the troposphere (taking approximately 2 years to do so). Additionally, presence of other atoms in the ozone layer have a small chance of destroying the radical chlorine, causing it to form $\ce{HCl}$ or $\ce{ClONO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most likely this estimate is based on some math crunching between the time-frame that the chlorine is in the ozone, the probabilities of chance reactions, and the rates/probabilities of these reactions.&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorUserId="41328" LastEditDate="2017-12-26T12:09:59.350" LastActivityDate="2017-12-26T12:09:59.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9528" PostTypeId="1" AcceptedAnswerId="9549" CreationDate="2014-03-27T16:38:45.307" Score="1" ViewCount="871" Body="&lt;p&gt;I am performing an experiment with DTZA and Ferrate. After a point of increasing the DTZA concentration too much the Ferrate takes longer (lower reaction rate) to react with the DTZA. Is this something reasonable or is it due to error? &lt;/p&gt;&#xA;" OwnerUserId="4830" LastEditorUserId="5026" LastEditDate="2018-02-19T12:59:42.420" LastActivityDate="2018-02-19T12:59:42.420" Title="Can an excessive reactant amount result in a longer reaction time?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9529" PostTypeId="1" CreationDate="2014-03-27T16:43:07.193" Score="1" ViewCount="8713" Body="&lt;p&gt;SiH4 and CH4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only intermolecular force they both have is London Dispersion forces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength of LDF is determined by molar mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;molar mass of SiH4 = 32.132&lt;/p&gt;&#xA;&#xA;&lt;p&gt;molar mass of CH4 = 48.42&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore shouldn't CH4 have a stronger LDF??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer says that SiH4 has a greater intermolecular force...why is that?&lt;/p&gt;&#xA;" OwnerUserId="4628" LastActivityDate="2017-04-02T21:49:29.123" Title="Determine which molecule has stronger intermolecular force?" Tags="&lt;structural-formula&gt;&lt;bond&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9530" PostTypeId="2" ParentId="9529" CreationDate="2014-03-27T17:19:23.613" Score="1" Body="&lt;p&gt;Look at the boiling points.  Interaction is based on geometry (surface area) and polarizability, among other things - and the phase: solid, liquid, vapor.  Molecular weight is not diagnostic.  SiH4 is &quot;fluffy&quot; versus methane being &quot;hard.&quot;  Compare the melting and boiling points of decane and isodecane  versus adamantane, all being $\ce{C_{10}}$.  Compare the melting and boiling points of &lt;em&gt;n&lt;/em&gt;-octane versus 2,2,3,3-tetramethylbutane, both $\ce{C8H_{18}}$.  Complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might check your MW for methane.  Bells and whistles should have gone off, $\ce{CH4}$ versus $\ce{SiH4}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-27T17:19:23.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9532" PostTypeId="2" ParentId="9529" CreationDate="2014-03-27T17:22:22.247" Score="0" Body="&lt;p&gt;The molar mass of SiH4 is ~32, however the molar mass of CH4 is ~16. This explains the textbook's answer. Uncle Al, however, provides a more detailed answer.&lt;/p&gt;&#xA;" OwnerUserId="4969" LastActivityDate="2014-03-27T17:22:22.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9533" PostTypeId="2" ParentId="6876" CreationDate="2014-03-27T18:51:57.407" Score="3" Body="&lt;p&gt;The reason for the ordering is that tertiary radicals have a lower energy (and are thus easier to form) than secondary radicals, which are in turn easier to form than primary radicals. This is because of a phenomenon called sigma conjugation - the weak overlap between the half filled orbital and the adjacent side group orbitals gives a slight lowering of energy.&#xA;This means that it is easier to remove a hydrogen to give a tertiary radical intermediate than to give a secondary radical intermediate and so on. &#xA;This is similar to &lt;a href=&quot;https://en.wikipedia.org/wiki/Markovnikov%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Markovnikov's rule&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason a tertiary product is not formed is that for butane there is no tertiary centre to begin with - nowhere that a Cl radical can react to give a tertiary butyl radical. Forming a tertiary product would require a rearrangement of the carbon chain.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-03-27T18:51:57.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9534" PostTypeId="2" ParentId="7467" CreationDate="2014-03-27T19:10:58.470" Score="4" Body="&lt;p&gt;Water at very high temperatures and pressures is corrosive (in the sense that it will react with a lot of common materials), but generally speaking pure water is less corrosive than water with impurities. Some power plants use purified water in their cooling systems for this reason.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-03-27T19:10:58.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9535" PostTypeId="1" AcceptedAnswerId="9539" CreationDate="2014-03-27T19:36:28.870" Score="3" ViewCount="390" Body="&lt;p&gt;I'm working on my chemistry lab report. The initial instructions were to test the hypothesis that the mass of products of a reaction will equal the mass of the reactants.&lt;/p&gt;&#xA;&lt;p&gt;Basically we where given a equation, our job was to balance and figure out the theoretical mass of reactants, and the theoretical mass of products.&lt;/p&gt;&#xA;&lt;p&gt;There we're a total of three questions. In the first one I got the following differential values, 0.2 and 0.1. This seemed to be the most logical for me since I knew the law of conservation of mass implied that mass cannot be created nor destroyed but could be transformed. I attributed this small differential value to this transformation theory.&lt;/p&gt;&#xA;&lt;p&gt;However, the next two questions I attempted still gave a very large differential value (40-120), which rejects my initial hypothesis.&lt;/p&gt;&#xA;&lt;p&gt;I believe I may have missed something in the second and third question, below are my attempted solutions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Initial Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4 + O2 -&amp;gt; CO2 + H2O}$&lt;/span&gt;&lt;br /&gt;&#xA;Balanced Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4 + 2O2 -&amp;gt; CO2 + 2H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccccccc}&#xA;\ce{CH4} &amp;amp; + &amp;amp; \ce{2O2} &amp;amp; \ce{-&amp;gt;} &amp;amp; \ce{CO2} &amp;amp; + &amp;amp; \ce{2H2O}\\&#xA;1(12+4) &amp;amp; + &amp;amp; 2(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 1(12+32) &amp;amp; + &amp;amp; 2(2+16)\\&#xA;1(16) &amp;amp; + &amp;amp; 2(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 1(44) &amp;amp; + &amp;amp; 2(18)\\&#xA;16 &amp;amp; + &amp;amp; 64 &amp;amp; \ce{-&amp;gt;} &amp;amp; 44 &amp;amp; + &amp;amp; 36\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Theoretical Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccc}&#xA;\ce{CH4}=\pu{16g} &amp;amp; &amp;amp; \ce{H2O}=\pu{36g}\\&#xA;+ &amp;amp; \ce{-&amp;gt;} &amp;amp; +\\&#xA;\ce{O2}=\pu{32g} &amp;amp; &amp;amp; \ce{CO2}=\pu{44g}\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Actual Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{43.8g}\\&#xA;\ce{H2O}=\pu{35.9g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Differential Value(Theoretical Calculation - Actual Calculation):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{0.2g}\\&#xA;\ce{H2O}=\pu{0.1g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Initial Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H6 + O2 -&amp;gt; CO2 + H2O}$&lt;/span&gt;&lt;br /&gt;&#xA;Balanced Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{2C2H6 + 7O2 -&amp;gt; 4CO2 + 6H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccccccc}&#xA;\ce{2C2H6} &amp;amp; + &amp;amp; \ce{7O2} &amp;amp; \ce{-&amp;gt;} &amp;amp; \ce{4CO2} &amp;amp; + &amp;amp; \ce{6H2O}\\&#xA;2(24+6) &amp;amp; + &amp;amp; 7(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 4(12+32) &amp;amp; + &amp;amp; 6(2+16)\\&#xA;2(30) &amp;amp; + &amp;amp; 7(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 4(44) &amp;amp; + &amp;amp; 6(18)\\&#xA;60 &amp;amp; + &amp;amp; 224 &amp;amp; \ce{-&amp;gt;} &amp;amp; 176 &amp;amp; + &amp;amp; 108\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Theoretical Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccc}&#xA;\ce{C2H6}=\pu{60g} &amp;amp; &amp;amp; \ce{H2O}=\pu{108g}\\&#xA;+ &amp;amp; \ce{-&amp;gt;} &amp;amp; +\\&#xA;\ce{O2}=\pu{224g} &amp;amp; &amp;amp; \ce{CO2}=\pu{176g}\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Actual Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{78.59g}\\&#xA;\ce{H2O}=\pu{48.2g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Differential Value(Theoretical Calculation - Actual Calculation):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{97.41g}\\&#xA;\ce{H2O}=\pu{48.2g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Initial Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H8 + O2 -&amp;gt; CO2 + H2O}$&lt;/span&gt;&lt;br /&gt;&#xA;Balanced Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H8 + 5O2 -&amp;gt; 3CO2 + 4H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccccccc}&#xA;\ce{C3H8} &amp;amp; + &amp;amp; \ce{5O2} &amp;amp; \ce{-&amp;gt;} &amp;amp; \ce{3CO2} &amp;amp; + &amp;amp; \ce{4H2O}\\&#xA;1(36+8) &amp;amp; + &amp;amp; 5(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 3(12+32) &amp;amp; + &amp;amp; 4(2+16)\\&#xA;1(44) &amp;amp; + &amp;amp; 5(32) &amp;amp; \ce{-&amp;gt;} &amp;amp; 3(44) &amp;amp; + &amp;amp; 4(18)\\&#xA;44 &amp;amp; + &amp;amp; 160 &amp;amp; \ce{-&amp;gt;} &amp;amp; 132 &amp;amp; + &amp;amp; 72\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Theoretical Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccc}&#xA;\ce{C3H8}=\pu{44g} &amp;amp; &amp;amp; \ce{H2O}=\pu{72g}\\&#xA;+ &amp;amp; \ce{-&amp;gt;} &amp;amp; +\\&#xA;\ce{O2}=\pu{160g} &amp;amp; &amp;amp; \ce{CO2}=\pu{132g}\\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Actual Calculation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{82.5g}\\&#xA;\ce{H2O}=\pu{45.0g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Differential Value(Theoretical Calculation - Actual Calculation):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$&#xA;\ce{CO2}=\pu{49.5g}\\&#xA;\ce{H2O}=\pu{27.0g}\\&#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Not sure if it matters, but we where given this program to enter our calculations to simulate the reactions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F54r0.png&quot; alt=&quot;Program for calculating mass of reactions&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4998" LastEditorUserId="106377" LastEditDate="2021-05-25T15:45:07.350" LastActivityDate="2021-05-25T15:45:07.350" Title="Figuring out the theoretical mass of reactants" Tags="&lt;everyday-chemistry&gt;&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9536" PostTypeId="2" ParentId="7467" CreationDate="2014-03-27T19:42:22.720" Score="3" Body="&lt;p&gt;If we define corrosivity as impurification of the water, ultrapure water eats nearly everything, including Pyrex.   Plumbing is usually polyvinylidene fluoride (PVDF, Kynar), polyperfluoroalkoxy (PFA), ethylene chlorotrifluoroethylene (ECTFE, Halar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stuff still grows in ultrapure water, slowly.  Chip fabs' water systems must be hard-UV irradiated, ozonized, or otherwise disinfected plus ultrafiltered against bio-particulate generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product tastes awful - bitter and flat.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-27T19:42:22.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9537" PostTypeId="1" CreationDate="2014-03-27T20:48:50.767" Score="2" ViewCount="2324" Body="&lt;p&gt;What role does heat play in the combustion process.Specifically, does the amount of heat  determine whether complete or incomplete combustion occurs, and is the amount of heat impacted by the amount of oxygen&lt;/p&gt;&#xA;" OwnerUserId="4583" LastEditorUserId="4583" LastEditDate="2014-03-27T20:54:36.487" LastActivityDate="2016-03-13T14:24:36.860" Title="Role of heat in combustion" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;combustion&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9538" PostTypeId="2" ParentId="9537" CreationDate="2014-03-27T21:38:32.510" Score="1" Body="&lt;p&gt;Indirectly yes, the amount of oxygen will have an impact on the amount of heat. This because, if there is enough oxygen, combustion will tend to be complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider methane combustion, complete and then incomplete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH4(g) + 2O2(g) -&amp;gt;CO2(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-891\,kJ/mol&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH4(g) + 3/2O2(g) -&amp;gt;CO(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-608\,kJ/mol&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of enthalpy, $\Delta H^{o}_{comb}$ ($25^{o}$) is larger for a complete combustion, as reported above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that more &lt;strong&gt;heat&lt;/strong&gt; (= enthalpy in conditions of constant pressure) is released if the combustion is complete.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-03-27T21:44:01.237" LastActivityDate="2014-03-27T21:44:01.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9539" PostTypeId="2" ParentId="9535" CreationDate="2014-03-27T21:48:50.860" Score="3" Body="&lt;p&gt;Try to answer to these questions, don't cheat, check the answer passing over the yellow box.&lt;/p&gt;&#xA;&lt;p&gt;I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have &lt;span class=&quot;math-container&quot;&gt;$\pu{100g}$&lt;/span&gt; of oxygen and &lt;span class=&quot;math-container&quot;&gt;$\pu{44g}$&lt;/span&gt; of propane.&lt;/p&gt;&#xA;&lt;p&gt;How many mole of propane and oxygen (&lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;) do you have?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Oxygen moles&lt;span class=&quot;math-container&quot;&gt;$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$&lt;/span&gt;,  Propane moles&lt;span class=&quot;math-container&quot;&gt;$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If this is your balanced equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C3H8 + 5O2 -&amp;gt; 3CO2 + 4H2O}$$&lt;/span&gt;&#xA;How many oxygen molecules do you need to react with one molecule of propane?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which is the ratio?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{5}$&lt;/span&gt; if you have &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; moles of oxygen you need &lt;span class=&quot;math-container&quot;&gt;$\frac{x}{5}$&lt;/span&gt; moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So how many moles of the products will you have?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$3\times0.6$&lt;/span&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$4\times0.6$&lt;/span&gt; moles of water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And so how many grams?&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{1.8mol}\times \pu{44g/mol}$&lt;/span&gt; for water, &lt;span class=&quot;math-container&quot;&gt;$\pu{2.4mol}\times \pu{18g/mol}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="106377" LastEditDate="2021-05-25T12:49:31.520" LastActivityDate="2021-05-25T12:49:31.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9540" PostTypeId="2" ParentId="9535" CreationDate="2014-03-27T22:59:23.933" Score="-1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 gram}$&lt;/span&gt; is (&lt;span class=&quot;math-container&quot;&gt;$\pu{E4 kg}$&lt;/span&gt;)(&lt;span class=&quot;math-container&quot;&gt;$\pu{299792458 m/s2}$&lt;/span&gt;) = &lt;span class=&quot;math-container&quot;&gt;$\pu{8.988E12 J}$&lt;/span&gt; or two kilotons nuclear yield equivalent.  Let's be conservative and say &amp;quot;experimental error.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Mass-energy is conserved because time is homogeneous throughout the universe.  Run this continuous symmetry through Noether's theorems to obtain the First Law of Thermodynamics.  However, time is &lt;em&gt;not&lt;/em&gt; locally homogeneous!  We live at the bottom of a gravitational potential well.  GPS atomic clocks run slower in orbit for special relativity (orbital velocity), but faster still for general relativity (higher in the potential well).  Therefore, don't transform &amp;quot;missing mass&amp;quot; except in physics labs.  Their party, their mess.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="95133" LastEditDate="2021-05-25T12:58:40.907" LastActivityDate="2021-05-25T12:58:40.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9541" PostTypeId="2" ParentId="9537" CreationDate="2014-03-27T23:34:05.160" Score="0" Body="&lt;p&gt;In contrast, the explosion of triacetone triperoxide (83% the pop of TNT) has hard by zero exotherm.  Ambient oxygen makes no difference.  Much of the acetone can be recovered and reused -  a near-zero carbon footprint, highly recyclable terrorist explosive.  Why do people complain?  J. Am. Chem. Soc. 127, 1146 (2005)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TATP, $\mathrm{Δ_{f}H°}$ = +151.4 kJ/mol &lt;BR&gt;&#xA;DOI: 10.1002/prep.201100100&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="24052" LastEditDate="2016-03-13T14:24:36.860" LastActivityDate="2016-03-13T14:24:36.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9542" PostTypeId="1" AcceptedAnswerId="9543" CreationDate="2014-03-27T23:36:17.550" Score="1" ViewCount="1006" Body="&lt;p&gt;I prepared some sodium aluminate by reacting a strong $NaOH$ solution with aluminium foil and filtering the result. I now have a clear solution and I'm not sure what it is. I stopped adding aluminium when the reaction rate showed down dramatically in a hot water bath. Basically what I'm asking is what are the properties of sodium aluminate since Wikipedia is severely lacking in information on sodium aluminate&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="2094" LastEditDate="2014-04-01T19:29:09.233" LastActivityDate="2014-04-01T19:29:09.233" Title="Testing a sodium aluminate solution" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9543" PostTypeId="2" ParentId="9542" CreationDate="2014-03-28T00:00:19.400" Score="0" Body="&lt;p&gt;There are multiple reactions going on simultaneously. Aluminum metal reacts with the oxygen in surrounding air in order to create a plating of aluminum (III) oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Al(s) + 3O2(l) -&amp;gt; 2Al2O3(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon addition of sodium hydroxide, the oxide layer reacts with the base, leaving the aluminum free to attack by water for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Al + 6 H2O -&amp;gt; 2 Al(OH)3 + 3 H2}$$&#xA;$$\ce{Al(OH)3 + NaOH -&amp;gt; Na+ + [Al(OH)4]-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, we can write the sum of the reactions as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Al(s) + 2NaOH(aq) + 6H2O(l) -&amp;gt; 2Na+(aq) + 2[Al(OH)4]- + 3H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the reaction proceeded slowly at first was due to the aluminum (III) oxide layer covering the aluminum metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this link, &lt;a href=&quot;http://www2.uni-siegen.de/~pci/versuche/english/v44-10.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www2.uni-siegen.de/~pci/versuche/english/v44-10.html&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This reaction is used in drain cleaners. They are mostly made out of strong alkalis, to which alumunim or zinc has been added. The alkalis break down organic residues chemically. In addition, the formation of hydrogen leads to a bubbling effect which adds an additional mechanical cleaning mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering the properties of aluminates, one can go to PubChem hosted by the NIH which has a lot of information for almost every single compound necessary as well as literature linked to the specific compound in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=14766&amp;amp;loc=ec_rcs&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=14766&amp;amp;loc=ec_rcs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/osala.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-29T03:12:15.810" LastActivityDate="2014-03-29T03:12:15.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9544" PostTypeId="2" ParentId="9537" CreationDate="2014-03-28T00:47:15.273" Score="5" Body="&lt;p&gt;Yes, &lt;strong&gt;temperature&lt;/strong&gt; and &lt;strong&gt;heat&lt;/strong&gt; do affect the &lt;strong&gt;spontaneity&lt;/strong&gt; and the &lt;strong&gt;equilibrium&lt;/strong&gt; of reactions. The term heat can be referred to enthalpy, which can be quite confusing in context, as I believe you are asking about raising the temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first take a look at the equation for the Gibb's Free Energy and raising temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibb's Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibb's Free energy describes how much &quot;free energy&quot; is available, or how much work is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, entropy, enthalpy are state functions, we can calculate them by using Hess's Law and using tabulated values for standard thermodynamic data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_{products} - \Sigma \Delta H^\circ_{reactants} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity's sake, I will use the combustion of methane that mannaia has so kindly provided and run the reaction at STP using heats of formation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H^\circ_{\text{comb of} \ \ce{CH_4}} = -890 \text{kJ} \cdot \text{mol}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H^\circ_{\text{f of} \ \ce{H_2O} \ \text{(l)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H^\circ_{\text{f of} \ \ce{CO_2} \ \text{(g)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4(g) + 2O2(g) -&amp;gt; CO2(g) + 2H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&#xA;$$ \Delta G^\circ  = -604.53 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -633.7 \text{kJ} \cdot \text{mol}^{-1} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the enthalpy of combustion for methane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$&#xA;$$ \Delta G^\circ  = -1569.34 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -1598.38 \text{kJ} \cdot \text{mol}^{-1} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would occur if we raise the T term? Obviously, the Gibbs Free Energy would become larger in magnitude and more negative, thus being more spontaneous. Inversely, if we lower the temperature to such a degree that $T\Delta S^\circ = \Delta H^\circ$, the reaction would not occur at all, and if $T\Delta S^\circ &amp;gt; \Delta H^\circ$, we would need to input energy in order for the reaction to proceed forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H&amp;lt;0$ and (-) and $\Delta S&amp;gt;0$ (+), the reaction is spontaneous at any temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H&amp;gt;0$ and (+) and $\Delta S&amp;lt;0$ (-), the reaction is NOT spontaneous at any temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H&amp;lt;0$ and (-) and $\Delta S&amp;lt;0$ (-), the reaction is spontaneous at low enough temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H&amp;gt;0$ and (+) and $\Delta S&amp;gt;0$ (+), the reaction is spontaneous at high enough temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to think of heat as a product or reactant. Combustion reactions are highly exothermic. Le Chatlier's principle includes raising the temperature as a stress. Alternatively, we can think about the equilibrium of this reaction to see how much product will be formed by looking at the magnitude of the equilibrium constant, $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly the van't Hoff isochore equation is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ln \frac{K_2}{K_1} = \frac{\Delta H}{R}(\frac{1}{T_1} - \frac{1}{T_2}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, an increase in temperature favors the formation of products if the reaction is endothermic. Inversely, a decrease in temperature favors the formation of products if the reaction is exothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, Uncle Al has stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ambient oxygen makes no difference&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The atmosphere is composed of about 70% nitrogen and 20% oxygen. If this reaction was to be observed in nature, there would be almost a limitless amount of oxygen for the reaction to rpoceed. Temperature does play a role in the stoichiometry of the reaction and thus whether or not the combustion is complete or incomplete. However, it is possible to limit one of the reactants as mannaia has done, specifically oxygen in a laboratory setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As related to the kinetic molecular theory of matter, increased temperature will increase the frequency of collisions of particles with sufficient energy to cleave bonds and reform them, driving the reaction further to the right. This is commonly done with the commercial production of ammonia using Van Haber process. Although this is thermodynamically unfavorable, when run at high temperatures, the rate of formation of ammonia is much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to say, an open environment, the temperature would increase the kinetic energy of the oxygen molecules, and so the particles would be spaced further apart. Describing this quantitatively... well, an experiment should be carried out. I would assume if one was so inclined to find if that would increase the equilibrium, rate, or completeness of combustion I'm sure they would find differences.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-05-04T20:08:38.447" LastActivityDate="2014-05-04T20:08:38.447" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9545" PostTypeId="1" AcceptedAnswerId="9547" CreationDate="2014-03-28T02:05:58.063" Score="2" ViewCount="1041" Body="&lt;p&gt;Quick question, in this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-Butenal + CH3MgBr  --&gt;  ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the grignard attack the double-bond or the aldehyde carbon?&#xA;Please explain this preference, thanks!&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorUserId="4969" LastEditDate="2014-03-28T17:51:28.203" LastActivityDate="2014-03-28T17:51:28.203" Title="Preference of grignard between aldehyde and double bond?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9546" PostTypeId="1" CreationDate="2014-03-28T04:03:35.220" Score="5" ViewCount="15376" Body="&lt;p&gt;In class we performed an experiment which I am unable to understand the results for&#xA;Why does the the amount of soot produced vary? What is the amount of soot  produced  dependent on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Experiment:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We had to have four test tubes upside down in a beaker of water. The test tubes were filled with water in them as well. We had to insert pieces of calcium carbide under the test tubes and have them as follows: one full of acetylene, one 3/4 full, one 1/2 full, one 1/10 full. Then remove the rubber stopper and bring a burning splint to the mouth of the test tube to ignite the acetylene gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Results:&lt;/em&gt;   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The full one: burned slow and produced a little bit of soot  &lt;/li&gt;&#xA;&lt;li&gt;3/4: burned fast and lots of soot  &lt;/li&gt;&#xA;&lt;li&gt;1/2: burned fast and lots of soot  &lt;/li&gt;&#xA;&lt;li&gt;1/10: burned fast and no soot   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4583" LastEditorUserId="4945" LastEditDate="2015-03-10T04:24:13.133" LastActivityDate="2019-10-29T00:03:55.700" Title="Incomplete Combustion products" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9547" PostTypeId="2" ParentId="9545" CreationDate="2014-03-28T05:01:25.547" Score="2" Body="&lt;p&gt;The Grignard will first attack the aldehyde carbon, giving you a secondary alcohol. Since the alcohol is strongly polarized, it would be difficult for the double bond to be attacked.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-28T05:01:25.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9548" PostTypeId="1" AcceptedAnswerId="9569" CreationDate="2014-03-28T05:59:28.223" Score="1" ViewCount="1525" Body="&lt;p&gt;Why is a sugar solution pre-treated with ammonium hydroxide before analyzing? And why is it specifically ammonium hydroxide instead of sodium hydroxide or calcium hydroxide? &lt;/p&gt;&#xA;" OwnerUserId="2433" LastActivityDate="2014-03-28T16:42:42.460" Title="Why is NH4OH used in Polarimetric analysis?" Tags="&lt;analytical-chemistry&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9549" PostTypeId="2" ParentId="9528" CreationDate="2014-03-28T07:00:29.060" Score="3" Body="&lt;p&gt;&lt;em&gt;Let me first say that I don't know this specific reaction, so I am not sure whether my answer holds for it, but let me answer the general question: can a high concentration slow down the reaction rate?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, yes it can. The mechanism by which this happens is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_inhibitor&quot; rel=&quot;nofollow&quot;&gt;inhibition&lt;/a&gt; and it is often (although not necessarily) connected to &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactions_on_surfaces&quot; rel=&quot;nofollow&quot;&gt;reactions on surfaces&lt;/a&gt;. I will explain inhibition in terms of a surface reaction here, but keep in mind that it can also happen in solution due to 'crowding', in which there is so much of 1 reactant that it basically 'cannot find' the other reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surface reactions the story is as follows. You will get a competition between adsorption on the surface and reaction on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Qualitatively&lt;/strong&gt;, if you have a large amount of one of the two reactants than many of the surface sites will have this reactant adsorbed to it meaning that some of the reactant on the surface has neighbors that it cannot react with (because it is the same molecule), this can slow down the reaction rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quantitatively&lt;/strong&gt;, the reaction order will be concentration dependent. A famous example of this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactions_on_surfaces#Langmuir.E2.80.93Hinshelwood_mechanism&quot; rel=&quot;nofollow&quot;&gt;Langmuir-Hinshelwood&lt;/a&gt; kinetics&#xA;$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_1 C_A+k_2 C_B)^2} $$ &#xA;where $C_s$ is the concentration of available adsorption sites and $C_A$ and $C_B$ are the concentrations of your reactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible ways in which this reaction rate equation can behave, depending on the values of the rate constants $k$, $k_1$ and $k_2$, but let me single out two cases that will explain a shift from increasing to decreasing reaction rate with increasing concentration:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;a very high concentration of $C_A$&lt;/strong&gt; will result in $k_1C_A&amp;gt;&amp;gt;&amp;gt;1, k_2C_B$. This allows us to neglect the $1$ and $k_2C_B$ terms in the denominator, turning the rate equation into this:&#xA;$$r =k C_s^2 \frac{k_2 C_B}{k_1 C_A} $$ &#xA;which indeed decreases with increasing concentration of $C_A$. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;a very low concentration of $C_A$&lt;/strong&gt; will result in $k_1C_A&amp;lt;&amp;lt;&amp;lt;1, k_2C_B$. Then the rate equation becomes:&#xA;$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_2 C_B)^2} $$ &#xA;which has an increasing rate with increasing concentration of $C_A$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So what you see here is that a change of $C_A$ from low to high concentration can change the reaction order in $C_A$ and therefore indeed result in an initial increase of the reaction rate with increasing $C_A$, but later a decrease if you increase it even further. When this crossover occurs depends on the values of the rate constants ($k$, $k_1$ and $k_2$) and the concentration of the other reactant.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-03-28T07:00:29.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9551" PostTypeId="2" ParentId="5727" CreationDate="2014-03-28T07:15:13.923" Score="3" Body="&lt;p&gt;Taking into account that gems often are doped oxides with rather high melting points, your project is quite a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A historical example for synthetic emerald is &lt;strong&gt;Igmerald&lt;/strong&gt;, produced by the &lt;em&gt;I.G. Farbenindustrie&lt;/em&gt; (hence the name) from 1929 until 1942 at the Bitterfeld facility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synthesis was performed at 800 °C in platinum crucibles, reacting $\ce{BeO}$ and $\ce{Al2O3}$ and $\ce{Li2CrO4}$ (as the colour-giving dopant). Lithium molybdate was used as a solvent and natural emeralds were used as seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucibles were covered with platinum lids having platinum tubes attached, through which more material was fed in during the crystallisation process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 18 growth periods over a year, prismatic synthetic emeralds with a length of 2-3 cm could be obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information giving above is taken from the article &lt;a href=&quot;http://www.ssef.ch/uploads/media/1998_Schmetzer_The_colour_of_Igmerald_-_I.G._Farbenindustrie_flux-grown_synthetic_emerald.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The colour of Igmerald: I.G. Farbenindustrie flux-grown synthetic emerald&lt;/strong&gt;&lt;/a&gt; by &lt;em&gt;Karl Schmetzer&lt;/em&gt; and &lt;em&gt;Lore Kiefert&lt;/em&gt;, hosted on the site of the &lt;a href=&quot;http://www.ssef.ch/home/&quot; rel=&quot;nofollow&quot;&gt;SSEF Schweizerisches Gemmologisches Institut&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-28T07:15:13.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9552" PostTypeId="1" AcceptedAnswerId="9554" CreationDate="2014-03-28T07:40:48.430" Score="10" ViewCount="48098" Body="&lt;p&gt;It says in my textbook that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + 2 O2 -&amp;gt; CO2 + 2 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems like a displacement reaction to me. Shouldn't the reaction be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + O2 -&amp;gt; CO2 + 2 H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\ce{4H}$ are displaced by $\ce{O2}$ (becoming $\ce{2 H2}$), which makes the reaction balanced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the reaction $\ce{CH4 + O2 -&amp;gt; CO2 + H2O}$, which is further balanced as my textbook says?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="865" LastEditDate="2015-11-04T18:32:37.290" LastActivityDate="2018-05-02T16:34:20.260" Title="CH4 + O2 =&gt; CO2 + 2H2, what's wrong with it?" Tags="&lt;reaction-mechanism&gt;&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9553" PostTypeId="1" CreationDate="2014-03-28T08:20:06.673" Score="0" ViewCount="580" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vldHr.png&quot; alt=&quot;Solubility/Purity Problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a chemistry final tomorrow and this question on my sample exam through me for a loop. I'm not entirely confident I can work out problems B and C, but those questions rely on the answer to A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could please especially explain how I'm supposed to read the solubility chart to understand which solvent is the best, that would be awesome.&lt;/p&gt;&#xA;" OwnerUserId="5002" LastEditorUserId="1362" LastEditDate="2014-03-31T05:41:15.697" LastActivityDate="2014-04-30T14:34:46.187" Title="Best solvent for crystallization + mass recovered from crystallization + purity of the product" Tags="&lt;homework&gt;&lt;solubility&gt;&lt;recrystallization&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9554" PostTypeId="2" ParentId="9552" CreationDate="2014-03-28T08:56:48.363" Score="19" Body="&lt;p&gt;The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in &quot;first&quot; step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH_4 + O_2 -&amp;gt; CO_2 + 2H_2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but then the hydrogen will be also combusted in reaction :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_2 + O_2 -&amp;gt; 2H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the overall reaction will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH_4 + 2O_2 -&amp;gt; CO_2 + 2H_2O}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWB4c.jpg&quot; alt=&quot;49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida, &lt;a href=&quot;http://enu.kz/repository/2011/AIAA-2011-94.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://enu.kz/repository/2011/AIAA-2011-94.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pXtzA.jpg&quot; alt=&quot;COMBUSTION AND FLAME 120:160–172 (2000)&quot;&gt;&#xA;Intermediate Species Proﬁles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptaﬂuoropropane: Comparison of Experimental Data with Kinetic Modeling, &lt;a href=&quot;http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="7475" LastEditDate="2015-11-04T17:34:19.400" LastActivityDate="2015-11-04T17:34:19.400" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9555" PostTypeId="1" AcceptedAnswerId="16251" CreationDate="2014-03-28T09:37:09.990" Score="5" ViewCount="1687" Body="&lt;p&gt;From Miessler's Inorganic Chemistry, 5th edition, there is this correlation diagram for $d^{2}$ in a octahedral field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDh2W.jpg&quot; alt=&quot;correlation diag&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left hand side, if it's not clear, it goes (from lowest to highest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{3}F, ^{1}D, ^{3}P, ^{1}G, ^{1}S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by Hund's rules mentioned even in the same chapter, one would expect the $^{3}P$ to show up right after the $^{3}F$ because of their higher spin multiplicities. And even after that, G should come before D. (the ordering for that being s,p,d,f,g in reverse for same spin multiplicities). I don't know of any reason for why the $^1D$ term is just above the $^3F$ term in this graph though. &lt;strong&gt;Is this an error or does anyone have an explanation?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="5017" LastEditDate="2014-09-12T23:23:40.797" LastActivityDate="2014-09-12T23:23:40.797" Title="Correlation Diagram Doesn't Match Selection Rules?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9556" PostTypeId="2" ParentId="8870" CreationDate="2014-03-28T10:15:44.580" Score="2" Body="&lt;p&gt;You can predict certain types of chemical reactions using formats.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Combination reaction: $\ce{A + B -&amp;gt; AB}$, for example $\ce{C + O2 -&amp;gt; CO2}$&lt;/li&gt;&#xA;&lt;li&gt;Displacement reaction: $\ce{AB + C -&amp;gt; AC + B}$, for example $\ce{H2SO4 + Zn -&amp;gt; ZnSO4 + H2 }$&lt;/li&gt;&#xA;&lt;li&gt;Double Displacement reaction: $\ce{AB + CD -&amp;gt; AD + BC}$ , for example $\ce{Pb(NO3)2 + 2KI -&amp;gt; PbI2 + 2KNO3}$&lt;/li&gt;&#xA;&lt;li&gt;Decomposition reaction: $\ce{ABC -&amp;gt;[\Delta T] A + B + C}$ [using heat], for example $\ce{CaCO3 -&amp;gt;[\Delta T] CaO + CO2}$ [using heat]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For redox reactions, I am still working&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="4945" LastEditDate="2014-04-28T08:37:27.067" LastActivityDate="2014-04-28T08:37:27.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9558" PostTypeId="2" ParentId="5631" CreationDate="2014-03-28T11:15:19.770" Score="4" Body="&#xA;&#xA;&lt;p&gt;I have never used the indigo framework, but there's more than one way to process SMILES strings in Python. &lt;a href=&quot;http://openbabel.org&quot; rel=&quot;nofollow&quot;&gt;Open Babel&lt;/a&gt; and &lt;a href=&quot;http://journal.chemistrycentral.com/content/2/1/5&quot; rel=&quot;nofollow&quot;&gt;Pybel&lt;/a&gt; might be an alternative:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import pybel&#xA;&#xA;# SMILES for flurazepam, taken from  &#xA;# http://chemspider.com/chemical-structure.3276&#xA;flurazepam = 'CCN(CC)CCN1C(=O)CN=C(C2=C1C=CC(=C2)Cl)C3=CC=CC=C3F'&#xA;&#xA;# SMILES for trimethylamine, taken from &#xA;# http://chemspider.com/chemical-structure.1114&#xA;trimethylamine = 'CN(C)C'&#xA;&#xA;# SMILES for nitromethane, taken from &#xA;# http://chemspider.com/chemical-structure.6135&#xA;nitromethane = 'C[N+](=O)[O-]'&#xA;&#xA;mol = pybel.readstring('smi', flurazepam)&#xA;&#xA;for atom in mol.atoms:&#xA;    print '{:&amp;lt;5} {:3} {:3} {:3}'.format(atom.type, atom.formalcharge, \&#xA;        atom.implicitvalence, atom.valence)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-28T11:15:19.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9560" PostTypeId="1" AcceptedAnswerId="9567" CreationDate="2014-03-28T13:39:50.247" Score="0" ViewCount="45" Body="&lt;p&gt;I have been looking for a paper that by using spectrophotometry method, we can determine rate of reaction of colourant. I could not find any paper. I was wondering if any body could help me.&lt;/p&gt;&#xA;" OwnerUserId="5006" LastEditorUserId="5006" LastEditDate="2014-03-28T14:18:14.613" LastActivityDate="2014-03-28T16:19:33.753" Title="Spectrophotometry for Colourant" Tags="&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9561" PostTypeId="1" AcceptedAnswerId="9562" CreationDate="2014-03-28T14:11:15.500" Score="16" ViewCount="8070" Body="&lt;p&gt;I'm not sure if this is a silly question, but I was sitting here with a cup full of cheezey poof balls thinking, &quot;My goodness, it's like an amazing cheesey delicious liquid - huge water molecules!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course my next thought was, &quot;Wait a minute - water has two hydrogen atoms bonded to an oxygen, so that's not quite right. They wouldn't be round like this.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I started thinking about the diagrams we see in chemistry textbooks, etc., and how the atoms are always pictured as round balls. How do we know this is accurate? Is it possible for the atoms to be configured in more complicated shapes (e.g. not solid, crystalline, some type of lattice)?&lt;/p&gt;&#xA;" OwnerUserId="5007" LastActivityDate="2017-01-31T21:15:40.993" Title="Are atoms really round?" Tags="&lt;atoms&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9562" PostTypeId="2" ParentId="9561" CreationDate="2014-03-28T14:39:08.613" Score="20" Body="&lt;p&gt;It depends how you define the surface of an atom. Atoms maintain no surface in the normal sense; only regions of space where you have a better chance of finding electrons. So in fact it is not correct to say they have a true shape at all.&lt;/p&gt;&#xA;&lt;h1&gt;Shapes of Atomic Orbitals&lt;/h1&gt;&#xA;&lt;p&gt;However if you plot the region of higher probability of finding electrons in an atom you can obtain something like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NWkFP.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;These are the shapes of the first five atomic orbitals: &lt;span class=&quot;math-container&quot;&gt;$1\mathrm{s}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2\mathrm{s}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2\mathrm{p}_x$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2\mathrm{p}_y$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$2\mathrm{p}_z$&lt;/span&gt; from &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. &lt;span class=&quot;math-container&quot;&gt;$1\mathrm{s}$&lt;/span&gt; orbital is sphere-shaped but other orbitals have more complex shapes so atoms with many electrons have orbitals very different from a sphere.&lt;/p&gt;&#xA;&lt;h1&gt;Shapes of Molecular Orbitals&lt;/h1&gt;&#xA;&lt;p&gt;Molecules have more electrons and so even more orbitals. They can have very strange &amp;quot;shapes&amp;quot;. I've calculated for you with GAMESS and Avogadro the LUMO 2&lt;span class=&quot;math-container&quot;&gt;$b_2$&lt;/span&gt; water's orbital that is one of the molecular orbitals of water. This is the result:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mdJ3Q.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Shapes of Atomic Constituents&lt;/h1&gt;&#xA;&lt;p&gt;For answering your question in the comment: In fact even protons, neutrons and electrons don't have a real shape due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality&quot; rel=&quot;noreferrer&quot;&gt;wave-particle duality&lt;/a&gt;. However we can assume in many cases that neutrons are particles (so we suppose a spherical symmetry) but the  de Broglie hypothesis states that they have also a wavelength!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-01-31T21:15:40.993" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9563" PostTypeId="2" ParentId="9561" CreationDate="2014-03-28T14:52:21.143" Score="2" Body="&lt;p&gt;If you can find a single atom in vacuum with a net dipole moment, its electron cloud is obviously not spherically symmetric.  Go across the periodic table's second row.  They are all $\ce{1s^2}$ $\ce{2s^2}$, so their atomic cores absent chemical combination and hybridization are first order spherically symmetric, filled s-orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B $\ce{2p^1}$, C $\ce{2p^2}$, N $\ce{2p^3}$, O $\ce{2p^4}$, F $\ce{2p^5}$, Ne $\ce{2p^6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three orthogonal 2p oribtals shaped like dumbbells: $\ce{2p_{x}}$, $\ce{2p_{y}}$, $\ce{2p_{z}}$.  Maximum multiplicity says they each fill before electron paring occurs,  One then suspects B, C, O, and F &lt;em&gt;single atoms&lt;/em&gt; in vacuum would have dipole moments.  If they superpose orbital hybridization absent chemical combination, symmetry says they do not have a dipole moment. Do they?  Google/Google Scholar are your friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1010/1010.2425.pdf&quot; rel=&quot;nofollow&quot;&gt;http://arxiv.org/ftp/arxiv/papers/1010/1010.2425.pdf&lt;/a&gt; &lt;BR&gt;&#xA;For alkali metals.  Is the vacuum phase spontaneously dimeric?&#xA;&lt;a href=&quot;http://home.physics.wisc.edu/~tgwalker/056.singlet_PRA.pdf&quot; rel=&quot;nofollow&quot;&gt;http://home.physics.wisc.edu/~tgwalker/056.singlet_PRA.pdf&lt;/a&gt; &lt;BR&gt;&#xA;Complicated.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-28T14:52:21.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9564" PostTypeId="2" ParentId="9552" CreationDate="2014-03-28T15:01:07.470" Score="5" Body="&lt;p&gt;Look at the enthalpies of formation of $\ce{CO}$, $\ce{CO2}$, and $\ce{H2O}$, then the net enthalpies of reaction.  In a deficiency of oxygen, and trimmed by the temperature and kinetics versus equilibrium conditions, are you more likely to obtain $\ce{CO}$ plus $\ce{H2O}$ than free hydrogen?  Given less oxygen, soot and water?  Note reaction energetics below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Water_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Water_gas&lt;/a&gt;  &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Syngas&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Syngas&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gasification&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Gasification&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.precision-combustion.com/High_Pressure_CPOX_methane.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.precision-combustion.com/High_Pressure_CPOX_methane.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7uKsL.jpg&quot; alt=&quot;enthalpies of formation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-03-28T15:09:27.610" LastActivityDate="2014-03-28T15:09:27.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9565" PostTypeId="2" ParentId="5727" CreationDate="2014-03-28T15:17:31.663" Score="1" Body="&lt;p&gt;You can purchase bulk raw or cut stones much cheaper than you can synthesize bulk raw,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.syntheticgems.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.syntheticgems.org/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, Dennis Elwell, &lt;em&gt;Man-Made Gemstones&lt;/em&gt;, 1979 or following. It includes Appendix 3, &quot;How to grow your own rubies,&quot; mentioning dopant that suppresses the otherwise platey crystal habit.  &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-28T15:17:31.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9566" PostTypeId="2" ParentId="9548" CreationDate="2014-03-28T15:27:16.227" Score="2" Body="&lt;p&gt;Do you see a potential corresponding to the sugars, or to their Schiff bases, or to said imines' trimers?  Perhaps ammonia gives you a nice pH buffer, as opposed to alkali hydroxides that will wildly swing pH and allow Cannizzaro disproportionation of reducing sugars' aldehyde groups at high pH. Ammonia does not reduce at the cathode.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-28T15:27:16.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9567" PostTypeId="2" ParentId="9560" CreationDate="2014-03-28T15:55:45.313" Score="2" Body="&lt;p&gt;I'm not sure I have answer to your question maybe, if you consider natural colorant too, these are the articles that go more close to what you want:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0003267003011541?np=y&quot; rel=&quot;nofollow&quot;&gt;Prediction of phenolic compounds in red wine fermentations by visible and near infrared spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0308814607010278&quot; rel=&quot;nofollow&quot;&gt;Spectroscopic, stability and radical-scavenging properties of a novel pigment from gardenia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to have a look a these too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eprints.iisc.ernet.in/1722/1/Combustion.pdf&quot; rel=&quot;nofollow&quot;&gt;Mimani, T., and Samrat Ghosh. &quot;Combustion synthesis of cobalt pigments: Blue and pink.&quot; Current Science 78.7 (2000): 892-896.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thaiceramicsociety.com/download/Comparison_of_blue_pigment_in_2_method.pdf&quot; rel=&quot;nofollow&quot;&gt;Bernardi, M. I. B., et al. &quot;Comparison of blue pigments prepared by two different methods.&quot; Journal of the European Ceramic Society 22.16 (2002): 2911-2919.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://old.icf.ro/PN_II/ECOPIG/CoxZn1-xal2O4.pdf&quot; rel=&quot;nofollow&quot;&gt;Visinescu, Diana, et al. &quot;The environmentally benign synthesis of nanosized Co&amp;lt; sub&gt; x Zn&amp;lt; sub&gt; 1− x Al&amp;lt; sub&gt; 2 O&amp;lt; sub&gt; 4 blue pigments.&quot; Dyes and Pigments 87.2 (2010): 125-131.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-03-28T16:19:33.753" LastActivityDate="2014-03-28T16:19:33.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9568" PostTypeId="2" ParentId="9552" CreationDate="2014-03-28T16:02:34.217" Score="3" Body="&lt;p&gt;There's nothing wrong. It's incomplete combustion.  It is one of a number of oxidizing reactions that can and will be occurring if you are burning this hydrocarbon in an oxygen-restricted environment. But the dominant reaction will be the one where the lowest energy state is reached. So you will likely wind up with one CO molecule, one water molecule and one hydrogen molecule instead.&lt;/p&gt;&#xA;" OwnerUserId="5009" LastActivityDate="2014-03-28T16:02:34.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9569" PostTypeId="2" ParentId="9548" CreationDate="2014-03-28T16:42:42.460" Score="1" Body="&lt;p&gt;If I'm not wrong, this is about polarimetry, not polarography. (No dripping mercury, but optical rotation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sugar is a hexose, the following species will exist in aqueous solution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;open form (a bit)&lt;/li&gt;&#xA;&lt;li&gt;furanose  (traces)&lt;/li&gt;&#xA;&lt;li&gt;pyranoses (major amount)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The pyranoses will adopt the $\alpha$- and the $\beta$-form. Both rotate polarized light by a different angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interconversion between these forms (&lt;em&gt;mutarotation&lt;/em&gt;) is facilitated by addition of acid or base. Thus, an equilibrium, resulting in a constant angle in polarimetry, is reached faster.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-28T16:42:42.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9570" PostTypeId="1" CreationDate="2014-03-28T18:51:07.413" Score="1" ViewCount="14147" Body="&lt;p&gt;I am given a blue solid W and it contains a tetraamine copper(II) complex and a $\ce{SO_4^{2-}}$. I added dilute sulfuric acid until the solution is pale blue; then added potassium iodide solution. A white precipitate forms ($\ce{CuI}$) in a brown solution ($\ce{I_2}$). However, I am confused about what reaction is actually going on.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="402" LastEditDate="2017-05-08T00:34:04.077" LastActivityDate="2017-05-08T03:41:30.020" Title="What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9571" PostTypeId="2" ParentId="9570" CreationDate="2014-03-28T20:06:07.203" Score="0" Body="&lt;p&gt;The formation of the tetraammine (II) copper sulfate ion complex is made through the addition of excess ammonia to copper (II) sulfate solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuSO4(aq) + 4NH3(aq) -&amp;gt; [Cu(NH3)4]^2+(aq) + SO4^{2-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add sulfuric acid in solution, we push the equilibrium to the left, as there will be an increase in the concentration of sulfate ions as sulfuric acid ionizes. As the equilibrium shifts to the left, there will be an increase in concentration of ammonia and copper (II) sulfate in solution according to Le Chatlier's Principle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Cu(NH3)4]2+(aq) + SO4^{2-}(aq) + H2SO4(aq) -&amp;gt; CuSO4(aq) + 4NH3(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add potassium iodide, an oxidation and reduction reaction takes place between the copper (II) sulfate solution and potassium iodide. Since the reaction is taking place in basic solution, we can balance the electrochemical reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KI (aq) + CuSO4(aq) -&amp;gt; CuI2(s) + K2SO4(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can see that solid white copper(II) iodide precipitates and potassium sulfate dissociates into its ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since copper (II) iodide is very unstable, it decomposes into copper(I) iodide and iodine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2CuI2 (s) -&amp;gt; 2CuI (s) + I2(s)}$$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="402" LastEditDate="2017-05-08T00:40:35.640" LastActivityDate="2017-05-08T00:40:35.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9572" PostTypeId="2" ParentId="9570" CreationDate="2014-03-28T20:16:42.267" Score="1" Body="&lt;p&gt;Copper(II) is a hard Lewis acid that would be octahedrally coordinated but for Jahn-Teller distortion.  Instead, it is square-planar coordinated with two distant weakly bound apical ligands.  The lovely deep blue color, hard Lewis base ammonia versus hard Lewis acid water coordination, illustrates the nephelauxetic effect.  What would obtain with triethylamine coordination of copper?  Cu(II) is substitution labile.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jahn%E2%80%93Teller_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jahn-Teller effect&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nephelauxetic_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nephelauxetic effect&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mjmorris.staff.shef.ac.uk/teaching/CHM1002/lect4b.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinetics (and thermodynamics) of substitution reactions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Adding acid protonates the ammonia, its lone pair now in a dative bond with a proton and unavailable for coordination.  Tetraamminecopper(II) (note the spelling of &quot;ammine&quot; when coordinated) becomes tetraaquocopper(II) with fast kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aqueous Cu(II) is a modest oxidizing agent; iodide is a modest reducing agent.  Cu(I) is a soft Lewis acid, iodide is a soft Lewis base, $\ce{CuI}$ is very insoluble (look up its $\ce{K_{sp}}$ -- see &lt;a href=&quot;https://web.archive.org/web/20140414131429/http://www.ktf-split.hr/periodni/en/abc/kpt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine reacts with excess iodide to give triiodide anion.  However, isolate the &quot;$\ce{CuI}$,&quot; wash, and titrate with thiosulfate.  $\ce{CuI}$ strongly adsorbs elemental iodine.  It is also modesty light sensitive.  Cyanide, being a pseudohalide, follows the same path to afford exceptionally insoluble CuCN plus cyanogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember who earned the &quot;A,&quot; or do your chair parade for better references, then understanding for the next question. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="11367" LastEditDate="2017-05-08T03:41:30.020" LastActivityDate="2017-05-08T03:41:30.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9573" PostTypeId="2" ParentId="9553" CreationDate="2014-03-28T20:50:59.140" Score="0" Body="&lt;p&gt;Is the double bond &lt;em&gt;cis&lt;/em&gt; ${Z}$ or &lt;em&gt;trans&lt;/em&gt; ${E}$?  (Terrible nomenclature.) What is the scent of the product?  What is it commonly called? Don't taste it. A tiny bit on your tongue will excavate a deep ulcer.  The ${E}$-isomer melts at -7.5 C.  How did you get the putative  ${Z}$-isomer, and how can it melt at least 40 C degrees higher than the ${E}$-isomer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two solvents are used, not five.  Different mixtures are used.  The technique is terrible!  You would dissolve contaminated product in warm ethanol (below product mp) with stirring, then titrate with water until the cloud point.  Add another drop of ethanol to clear.  Very slowly cool without stirring (cheesecloth wrap) to obtain large clean crystals.  Refrigerate for more crystals. Repeat a second time to maximize product yield.  Do not combine to two crops!  The second is not as pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the product is most likely a liquid, you would fractionally distill it under reduced pressure (aspirator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.st-andrews.ac.uk/labs/misc/nomo2.gif&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.st-andrews.ac.uk/labs/misc/nomo2.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-28T20:50:59.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9574" PostTypeId="1" AcceptedAnswerId="9576" CreationDate="2014-03-29T02:36:03.807" Score="2" ViewCount="30616" Body="&lt;p&gt;If $\ce{NaCl}$ is such a strongly bonded ionic compound, why it does it disassociate so readily in $\ce{H2O}$? One would think that because of the very strong ionic bond it would not disassociate it all. Further, why do other compounds like $\ce{AgCl}$ or $\ce{HF}$ not disassociate as readily even though they also have strong bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although others have marked this as duplicate, I prefer the wording of this question as it explicitly refers to NaCl has having a strong bond (which includes the search term I used to find this website). I have been writing a manuscript in which I have been attempting to answer this question (among others). The real answer becomes a bit complicated, but I will try to give a short answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my opinion that thermodynamic data is not always properly understood. The first thing we should understand is that the same thermodynamic change in energy should result whether the reaction were to take place in the gas phase (Born-Haber cycle) or in solution. This is a principle of conservation of energy, Hess's Law. In the Born-Haber Cycle, the sum of the energies for vaporization, ionization, and lattice energies are determined. If a hypothetical solution phase Born-Haber calculation were to be performed, the same amount of energy would exchange, except for the solvation energy, which is small. That is what the poster was asking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to think about the conditions of a solution phase Born-Haber cycle, then you would take the elements and convert them directly to solvent separated ions. What this should tell you is that you have just calculated the solution phase energy of a redox reaction of sodium and chlorine to give sodium and chloride ions. This bypasses the implication of sodium chloride having a strong affinity for its ions. Yes, it would take a lot of energy to separate sodium and chloride ions in the gas phase. That is why the lattice energy is so large, but if we did the reaction in solution, we will not have a gas phase lattice energy component in our calculations. Of course this seems plausible as the solution phases of HCl and NaOH are ionic. This is consistent with our expected properties of HCl to act as an acid (chloride having a low affinity for cations) and NaOH to act as a base (sodium having a low affinity for electrons). The sum is we should not expect NaCl to have a strong affinity (not none) for the ions of the opposite charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I think you will also find a good correlation of heterolytic solution phase bond strength with bond length (actually, ion gap). This is different from the homolytic bond energies you may find reported in many tables, including the poster's question, &quot;Why does NaCl dissolve in H2O despite its strong ionic bond?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PW&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="34709" LastEditDate="2016-09-10T19:42:05.220" LastActivityDate="2016-09-10T19:42:05.220" Title="Why does NaCl dissolve in H2O despite its strong ionic bond?" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2014-03-29T12:16:16.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="9575" PostTypeId="1" AcceptedAnswerId="9586" CreationDate="2014-03-29T02:42:50.637" Score="2" ViewCount="22697" Body="&lt;p&gt;Why does $\ce{Pb}$ have an oxidation number of $2^{+}$ while carbon and other elements in the same group have an oxidation number of $4^{+}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, why doesn't carbon also have a $2^{+}$ oxidation number, as can't it also give up its two electrons in the p-suborbital to have a full s-suborbital? Why does it prefer to gain four electrons instead to fill the p-suborbital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, why can't $\ce{Pb}$ gain four electrons just like carbon instead of giving up two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, doesn't this mean that $\ce{Pb}$ forms ionic bonds? Why doesn't carbon do the same?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="4945" LastEditDate="2015-03-19T06:50:39.120" LastActivityDate="2015-03-19T19:05:27.400" Title="Why does Pb normally have an oxidation number of +2?" Tags="&lt;oxidation-state&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9576" PostTypeId="2" ParentId="9574" CreationDate="2014-03-29T02:56:34.087" Score="3" Body="&lt;p&gt;The electrostatic forces between the positive cations and anions are strong. Coulomb's Law expresses the potential energy of two charges separated by a distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_p = \frac{Q_1Q_2}{4\pi \epsilon_0 r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may already know, water molecules stick to each other through dipole-dipole forces due to the permanent dipole that water faces due to the high electronegativity of the oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As ionic solids are added to water, water molecules proceed to surround each ion on the surface of the solid, forming a sphere of hydration. In the process, ions are separated from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\delta^-$ charge on the oxygen atoms of water are attracted to cations and inversely, repels the $\delta^+$ hydrogen atoms. Thus, for cations, the oxygens of water point inward, and for anions, the hydrogens face inward respectively. The most important thing is that the ion-dipole interactions and separation of ions with little change in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can relate the potential energy of the ions to the two partial charges of a polar molecule like water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_p \propto - \frac{|z|\mu}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z is the charge number of the ion and $\mu$ is the dipole moment of the polar molecule. Potential energy is lowered by the interaction between the solvent molecules and the ion. The $r^2$ term indicates that the interaction between ions and dipoles depends more on distance than the charges between two ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for hydration to occur, ion-dipole interactions must occur at the surface of the ion, and thus, ion-dipole interactions are strong for small, highly charged ions such as $\ce{Mg^{2+}}$, $\ce{Li^{2+}}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AgCl}$ is very slightly soluble in water and will not dissociate into it's ions. $\ce{HF}$ is a weak acid thus it does not deprotonate easily.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-30T23:43:21.937" LastActivityDate="2014-03-30T23:43:21.937" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9577" PostTypeId="1" AcceptedAnswerId="9590" CreationDate="2014-03-29T03:44:21.127" Score="2" ViewCount="14430" Body="&lt;p&gt;I am told that in a chemical equation the metal comes first and then the non metal. for example MgO, ZnSO&lt;sub&gt;4&lt;/sub&gt;,etc. But when both the elements are non metals or metals, the one with the lower atomic number would come first. for example:- H&lt;sub&gt;2&lt;/sub&gt;O,CO&lt;sub&gt;2&lt;/sub&gt;,HF,etc. But why is it that hydroxide is OH- and not HO-? Also why is methane CH&lt;sub&gt;4&lt;/sub&gt; and not H&lt;sub&gt;4&lt;/sub&gt;C?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastActivityDate="2014-03-29T14:14:20.113" Title="Why is it OH- and not HO-?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-03-29T14:46:01.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="9578" PostTypeId="2" ParentId="9575" CreationDate="2014-03-29T03:59:58.237" Score="0" Body="&lt;p&gt;[Carbon cannot shift the electrons from the S orbital to the P orbital because there is a significant difference in energy, unlike lead.] - oops, I was wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead also has a +4 oxidation state. Look at the chemistry of a lead acid battery. Lead doesn't form ionic bonds since the difference in electronegativity between lead and the anions is too small. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;NaCl - large difference in electronegativity&#xA;CO2 - small difference in electronegativity&#xA;PbO2 - small difference in electronegativity&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-03-29T13:48:35.840" LastActivityDate="2014-03-29T13:48:35.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9579" PostTypeId="1" CreationDate="2014-03-29T04:37:02.263" Score="3" ViewCount="367" Body="&lt;p&gt;Say I promote a molecule to a triplet state via repeated electronic excitation events.  Here, this is going to mean that an electron from the HOMO level, originally with a spin-UP and spin-DOWN electron, will be promoted to the LUMO level such that its spin is the same as the electron left-over in the HOMO level (we can then use the Pauli exclusion principle to partially understand that the triplet state will generally have a longer lifetime than the typically higher energy singlet state).  The classic example of a molecule in a triplet state is oxygen (though here, this is also the ground state, which is somewhat unusual): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/a1/MOO2a.svg/398px-MOO2a.svg.png&quot; alt=&quot;http://en.wikipedia.org/wiki/File:MOO2a.svg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to returning to the ground state via phosphorescence or a reverse intersystem crossing event, let's say that the triplet $T_1$ state molecule reacts with a reducing agent via an exothermic bond formation process, and that the reaction is reversible.  Consider that there are numerous instances where triplet state reactivity is higher than ground state reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following two things are unclear to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) When the reducing agent dissociates, is the molecule in the $T_1$ triplet state or the $G_0$ ground state?  If the molecule is still in the triplet state, can it enter the ground state while remaining bound to the reducing agent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Call the energy gap between the relevant $T_1$ triplet state and the $G_0$ ground state $E_1$.  Call the energy released into a surrounding thermal bath by the reducing agent reacting with the molecule in the $T_1$ triplet state $E_2$.  Must $||E_2|| &amp;gt; ||E_1||$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the pair of electrons donated by the reducing agent will necessarily set  $||E_2|| &amp;gt; ||E_1||$ and that the molecule will be left in the ground state upon dissociation of the reducing agent since the electrons will be reallocated to properly satisfy spin-pairing requirements.&lt;/p&gt;&#xA;" OwnerUserId="5018" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.643" LastActivityDate="2017-01-09T19:09:49.237" Title="A reversible exothermic reaction between a molecule in an excited triplet state and a reducing agent" Tags="&lt;energy&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9580" PostTypeId="1" CreationDate="2014-03-29T04:43:36.730" Score="-1" ViewCount="6825" Body="&lt;p&gt;Under the pressure of $\pu{340 atm}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO + 2H2 &amp;lt;=&amp;gt; CH3OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is its product during normal pressure of $\pu{1 atm}$? I think it would be a single displacement reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO + 2H2 &amp;lt;=&amp;gt; CH4 + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I'm not sure. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="41328" LastEditDate="2017-08-15T21:43:37.193" LastActivityDate="2017-08-15T21:44:45.767" Title="What is the product of CO + 2H2 under normal pressure?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9581" PostTypeId="1" CreationDate="2014-03-29T06:15:53.257" Score="13" ViewCount="2692" Body="&lt;p&gt;I know that cyclopropyl methyl carbocation is exceptionally stable compared to say, benzyl carbocation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I want to know how stable is cyclobutyl methyl carbocation compared to say, tertiary/secondary carbocations? Or, will it go rearrangement breaking the ring?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastActivityDate="2015-01-20T04:17:21.397" Title="Stability of cyclobutyl methyl carbocation" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9582" PostTypeId="2" ParentId="9579" CreationDate="2014-03-29T08:51:02.373" Score="2" Body="&lt;p&gt;We're looking at a case of PET (photoinduced electron transfer) with electron back transfer between an (electron) acceptor &lt;strong&gt;A&lt;/strong&gt; and a donor &lt;strong&gt;D&lt;/strong&gt; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For organic dyes, the ground state typically is a singlet state ($S_0)$, excitation of &lt;strong&gt;A&lt;/strong&gt; yields an excited singlet state (typically $S_1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A -&amp;gt;[h\nu]\  ^1A^{\ast}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quinone based dyes (or aza-BODIPY derivatives) often have a high triplet quantum yield, the following ISC (intersystem crossing) to a triplet state is efficient then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^1A^{\ast} -&amp;gt;[ISC]\  ^3A^{\ast}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the excited (triplet) state, electron transfer from a (ground state) donor may occur by filling the lower SOMO of the excited species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^3A^{\ast} + D -&amp;gt; A^{\bullet-} + D^{\bullet+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the acceptor radical anion still in a triplet state? No, since there no longer an electron pair with opposite spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: So, what is this good for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;: By electronic excitation, we managed to facilitate a redox reaction that would not have been possible in the ground state and generated a radical ion pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Radical ions are reactive intermediates.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spectroscopist, it can be favourable if the only reaction of the radical ion pair is  &lt;em&gt;electron back transfer&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A^{\bullet-} + D^{\bullet+} -&amp;gt; A + D}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This regenerates acceptor and donor in their former spin states, which - as mentioned before - are typically singlet states in the case of organic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, this means that no permanent bleaching occurs. You can excite the sample with laser pulses again and again to measure the transients in time-resolved uv-vis spectrosopy.&#xA;+&#xA;From a preparative perspective, this is the worst case: there is no product being formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have examined more productive cases of PET in the context of PET-initiated radical cyclizations of isoprenoid polyalkenes a while ago and published about these, e.g. in &lt;a href=&quot;http://dx.doi.org/10.1021/jp972027j&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Phys.Chem A&lt;/em&gt;&lt;/a&gt;, in &lt;a href=&quot;http://dx.doi.org/10.1039/A800322J&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J.Chem. Soc., Faraday Trans.&lt;/em&gt;&lt;/a&gt; and in &lt;a href=&quot;http://dx.doi.org/10.1351/pac199769010109&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Pure Appl. Chem.&lt;/em&gt;&lt;/a&gt;. (The last two should be freely accessible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7lpwb.png&quot; alt=&quot;PET cyclization of 1,1-dicarbonitrile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the course of these reactions, an alkene is oxidized to the corresponding radical cation by the excited state of an electron acceptor. The radical cation &#xA;traps a nucleophile (water, alkanol), immediately deprotonates and thus form a radical, which undergoes &lt;em&gt;5-exo-trig&lt;/em&gt; cyclization to a stabilized dicyanomethyl radical. The latter is reduced to its corresponding anion by $A^{\bullet-}$ and eventually protonated to yield the product (as a pair of separatable diastereomers).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-03-30T06:22:01.027" LastActivityDate="2014-03-30T06:22:01.027" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9583" PostTypeId="1" AcceptedAnswerId="9591" CreationDate="2014-03-29T08:53:36.890" Score="6" ViewCount="519" Body="&lt;p&gt;I want to do intramolecular aldol condensation of 7-oxooctanal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tzncl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tzncl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about the major product. Clearly, a 6-membered ring would be a major product, but there are 2 possibilities: taking hydrogen from $\ce{C^2}$ or $\ce{C^6}$. Which one would be preferred?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="41328" LastEditDate="2017-09-22T15:47:02.520" LastActivityDate="2017-09-22T15:47:55.703" Title="Aldol of 7-oxooctanal" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aldol-reaction&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9584" PostTypeId="2" ParentId="9580" CreationDate="2014-03-29T09:57:30.027" Score="1" Body="&lt;p&gt;Thanks a lot  Michiel I got the answer in the Fischer-Tropsch process. During normal atmospheric pressure the following reaction occurs :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO + 3H2 &amp;lt;=&amp;gt; CH4 + H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also could someone tell me the type of reaction this is? I think it is partially a displacement reaction for $\ce{O}$ is displaced by $\ce{H}$ and gives $\ce{CH4}$, but then $\ce{O}$ is left out and reacts with the remaining two hydrogen and gives $\ce{H2O}$. &lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="41328" LastEditDate="2017-08-15T21:44:45.767" LastActivityDate="2017-08-15T21:44:45.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9586" PostTypeId="2" ParentId="9575" CreationDate="2014-03-29T12:40:46.753" Score="3" Body="&lt;p&gt;&lt;em&gt;Why can't lead gain four electrons?&lt;/em&gt; The lower you go on the periodic table, the lesser you will find elements that have a more non-metallic character (so with this I mean elements that will have a lower tendency to gain electrons). Just visualize a lead atom with a very small nucleus and a large area surrounding the nucleus, filled with electrons. Now, the more you go away from the nucleus, the weaker the positively charged nucleus can 'pull' on the outer , negatively charged, valence electrons. Elements that haven't got a very tight grip on these valence electrons, will lose their valence electrons. However, elements with a small atomic radius (such as Fluorine) will have such a tight grip on its valence electrons, that it will even be able to snatch electrons from other atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the electron configuration of Radon is stable, because it has fully filled orbitals and it has a noble gas configuration, but you need to keep the nuclear forces in mind. This you can also see by looking at the electronegative value of lead : it is 1.87 (Pauling scale). Now compare it to carbon, which has 2.5. The lower you go in the same group, the lower the electronegativity goes and the more metallic character you will get. Elements with more metallic properties (such as lead) will mostly form ionic bonds and elements with more non-metallic properties (such as carbon) will mostly form covalent bonds. There exists a rule to predict (however it isn't always correct) if a compound will be covalently/ionically bonded : if the difference in electronegative value is higher than 1.66, then you have an ionic bond and if it is lower than 1.66 then you have a covalent bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Why does lead have an oxidation number of 2+ while carbon and other elements in the same group have an oxidation number of 4+?&lt;/em&gt; Just to correct you on this : lead &lt;strong&gt;does&lt;/strong&gt; have an oxidation state of 4+ and elements above such as carbon and tin also have 2+ as an oxidation state. Just look at the electron configuration : the outer $\ce{s^2 p^2}$ orbitals will be able to lose 2 or even 4 electrons, or even gain 4 electrons. In fact these are all possible oxidation states of the elements in the carbon-group:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Carbon 4, 3, 2, 1, 0, −1, −2, −3, −4&#xA;   Silicon 4, 3, 2, 1 -1, -2, -3, -4&#xA;   Germanium 4, 3, 2, 1, 0, -1, -2, -3, -4&#xA;   Tin  4, 3, 2, 1, -4&#xA;   Lead   4, 3, 2, 1&#xA;   (source : wikipedia)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Please do note the tendency to have more positive oxidation states, the lower you go in the group. (the reason why? see the beginning of the answer) Note : not all of these oxidation states are (very) stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also have a look at this picture:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ECe85.jpg&quot; alt=&quot;Size of atoms and ions/..&quot;&gt;&#xA;Notice the metallic (shown by red) character increases the lower you go on the periodic table.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorDisplayName="user2117" LastEditDate="2014-03-29T12:49:26.867" LastActivityDate="2014-03-29T12:49:26.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9588" PostTypeId="2" ParentId="9523" CreationDate="2014-03-29T12:54:24.863" Score="1" Body="&lt;p&gt;It is the double bond that gets protonated, and a secondary carbocation is formed. The carbonyl oxygen of the carboxyl group adds to this cation, forming the five-membered ring. This reaction is favorable because the resulting cation is more stable.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-03-29T12:54:24.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9589" PostTypeId="2" ParentId="9583" CreationDate="2014-03-29T14:02:22.790" Score="2" Body="&lt;p&gt;We expect the aldehyde to be more reactive toward nucleophilic attack for its smaller steric hindrance (kinetics).  The product ratio may vary with time, kinetics vs. thermodynamics at equilibrium.  Attack followed by elimination helps lock in the product.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-29T14:02:22.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9590" PostTypeId="2" ParentId="9577" CreationDate="2014-03-29T14:14:20.113" Score="3" Body="&lt;p&gt;OH- and HO- are both acceptable.  The charge is on the oxygen, of course.  One can argue cation then anion (salts) versus alphabetical (e.g., AuBe, which is brittle and weak rather than ductile, suggesting ionicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic formulas are C, H, then alphabetical.  Order may be altered to conform to structure (e.g., propane as $\ce{H3CCH2CH3}$) or convenience $\ce{H3C^{-}Li^{+}}$ (which is nowhere near the real structure of methyllithium, but is how it reacts).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-29T14:14:20.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9591" PostTypeId="2" ParentId="9583" CreationDate="2014-03-29T14:30:24.293" Score="4" Body="&lt;p&gt;The more stable enol would involve $\ce{C^6}$, with the aldol condensation you would get a six member ring with $\ce{-COCH3}$ attached to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the enol involving $\ce{C^2}$, the product would have a methyl and $\ce{-CHO}$ attached to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the distribution between the two products; it could be 51:49 or as much as 99:1. You should do the reaction to determine this. Factors that may effect the product ratio: temperature, solvent, reaction time, concentration, etc.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="41328" LastEditDate="2017-09-22T15:47:55.703" LastActivityDate="2017-09-22T15:47:55.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9592" PostTypeId="1" CreationDate="2014-03-29T15:08:52.040" Score="1" ViewCount="126" Body="&lt;p&gt;Suppose A(+)B(-) type fused salt is present in an electrolytic cell (not electrochemical cell). Let n electrons be entering the cell per second. Since each A(+) requires one electron to discharge, so n electrons will discharge n A(+) per second. Similarly, n B(-) are also discharged per second because the number of electrons entering and leaving the cell must be same. Now, each A(+) has unit positive charge and hence n A(+) means ne(+) charge flowing towards cathode. Also, n B(-) means ne(-) charge flowing towards anode or we can say ne(+) charge towards cathode. Does that mean total 2ne(+) charge is flowing towards cathode per second ? However, from above only ne(+) charge is entering the cell.&lt;/p&gt;&#xA;" OwnerUserId="5020" LastActivityDate="2014-03-29T15:08:52.040" Title="Discrepancy in current inside an electrolytic cell" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9594" PostTypeId="1" AcceptedAnswerId="9595" CreationDate="2014-03-29T20:47:03.040" Score="5" ViewCount="4342" Body="&lt;p&gt;Hi I have decided to give chemistry another shot (it didn't take so well way back in high school). My main interest is for reading up on brain chemistry and pharmacology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently have a textbook on inorganic chemistry [Introduction to Organic Chemistry 3rd edition: William Brown.Thomas Poon] that I picked up and am working my way through - well I have made it through chapter 1 at least [Covalent Bonding and Shapes of Molecules]. And am also making use of The UC Davis &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry&quot; rel=&quot;noreferrer&quot;&gt;(Organic) ChemWiki&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after finishing the chapter today I decided to take a break and play around with an experiment I sore on youtube &lt;a href=&quot;https://www.youtube.com/watch?v=9iyRUBvd260&quot; rel=&quot;noreferrer&quot;&gt;to make copper nanoparticles&lt;/a&gt; using copper sulphate and ascorbic acid. In the video he was pretty lenient with the measurements [1g sulphate, 5g ascorbic, 200ml water] and I wanted to scale the reaction up a bit so I had a go at trying to figure what the reaction was so I could get the true ratio... Obviously that duck wouldn't fly (I am only just starting on chapter 2 after all) so I just went ahead and multiplied his recipe by a factor of 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where I was going with that last paragraph is that even after I finish reading the organic chem textbook I'm not seeing how that's going to help me with something like $CuSO_4.5H_2O$ since i'ts got nothing to do with carbon. Should I put the organic book to the side and start of with something like the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry&quot; rel=&quot;noreferrer&quot;&gt;(Inorganic) ChemWiki&lt;/a&gt;? Is it necessary to go through both of them or would it be enough to learn one subject and fill the gaps as there encountered [this exercise is a personal mission - I don't envision any possible future where I become a full fledged working chemist]? Better yet are there any broader texts that teaches a unified chemistry that doesn't make artificial distinctions? Or should I go ahead and invent my own syllabus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.&lt;/p&gt;&#xA;" OwnerUserId="5023" LastActivityDate="2019-12-15T13:10:34.483" Title="Which branches of chemistry and in what order to learn (self study)?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9595" PostTypeId="2" ParentId="9594" CreationDate="2014-03-29T21:48:48.307" Score="1" Body="&lt;p&gt;&quot;&lt;em&gt;brain chemistry and pharmacology&lt;/em&gt;&quot;  Organic chemiisry, biochemistry; inorganic for foundation - pH, equilibrium, reduction and oxidation.  You'll need some physical chemistry, classical thermodynamics, to appreciate what should happen (thermodynamics) and what does happen (kinetics).  Diamond (the denser phase) is unstable versus graphite at room pressure, and calculably so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JwbXZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond is stable to 1000 C in air, 1500 C under hydrogen. CVD diamond is made in vacuum.  Know where Official Truth has footnotes.  You must be conversant with species, ions, molecules, reactions before you touch organic (which is all that plus wickedly structural).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.khanacademy.org/science/chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.khanacademy.org/science/chemistry&lt;/a&gt; &lt;BR&gt;&#xA;   Look at the Khan Academy.  Begin with inorganic chemistry. &lt;BR&gt;&#xA;&lt;a href=&quot;https://www.khanacademy.org/science/organic-chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.khanacademy.org/science/organic-chemistry&lt;/a&gt; &lt;BR&gt;&#xA;  And then organic.  Use LCAO to understand geometry, structure, hybridization, reactions.  LCAO is wholly ad hock and, point of fact, is wrong.  It sill works - splendidly.  Molecular orbitals are irrelevant until Woodward-Hoffmann.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint:  The folks who drool advocacy the loudest are clueless about the science, thereby not willfully lying.  Government is corrupt, ignorant, and mean.  Industry is all that plus desperate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to play with organic, buy HyperChem Lite.  Diddle the molecule together, add hydrogens (drop down menu), assign atom types (same menu), then minimize to at least ten times the recommended iterations plus much smaller acceptable error.  Learn how to see into stereograms (HyperChem genertes them, Display/Rendering).  2D representations get sloppy real fast,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5AC4Q.png&quot; alt=&quot;[6.6]chiralane enantiomers&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-29T21:48:48.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9596" PostTypeId="2" ParentId="9270" CreationDate="2014-03-29T23:54:11.153" Score="3" Body="&lt;p&gt;Although Nicolau Saker Neto's &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9242/what-are-some-good-examples-of-rate-equations-for-a-math-class/9244#9244&quot;&gt;answer&lt;/a&gt; provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First case: two steps process, the first being the slowest.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction between $\ce{H2O2}$ and $\ce{HBr}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{H2O2 + H+ +Br- &amp;amp;-&amp;gt; HOBr + H2O}  &amp;amp;&amp;amp; \text{slow}\\&#xA;\ce{HOBr + H+ +Br- &amp;amp;-&amp;gt; Br2 + H2O}   &amp;amp;&amp;amp; \text{fast}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of this reaction, determined by the slow step, equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is therefore a &lt;em&gt;third order&lt;/em&gt; reaction.  Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second case: two steps process, the first being the fastest.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction between $\ce{NO}$ and $\ce{O2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial equilibrium is rapidly established:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{2NO &amp;amp;&amp;lt;=&amp;gt; N2O2}      &amp;amp;&amp;amp; \text{fast}\\&#xA;\ce{N2O2 + O2 &amp;amp;-&amp;gt; 2NO2} &amp;amp;&amp;amp; \text{slow}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the rate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is therefore a &lt;em&gt;second order&lt;/em&gt; reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step (&quot;slow&quot; step), directly yields the order by which the concentration of that reactant is raised in the rate equation.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-07-28T03:08:53.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9598" PostTypeId="2" ParentId="9594" CreationDate="2014-03-30T01:32:04.333" Score="4" Body="&lt;p&gt;I agree with LDC3 that inorganic chemistry should definitely be your starting point; it builds foundations that apply to all of chemistry, which is why they teach it first; organic chemistry is just a branch, albeit a very large one, of chemistry that can also be very interesting once you know the fundamentals. Unfortunately using conventional textbooks etc, you will most likely be constricted to this order of learning, because of the level of these courses in the education system, where inorganic chemistry is very traditionally taught first, with organic chemistry only afterwards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any short-cuts would be greatly appreciated - I basically just want to get to a point where I can read a wikipedia page on neurotransmitter X, psychoactive Y, or food ingredient Z and be able to comprehend fully what is going on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if there are any shortcuts that you are seeking, Wikipedia can be pretty advanced because it is (usually) written by experts who know exactly what they're talking about; and especially on those high-level concepts, there are a lot of &quot;things&quot; (for lack of a better word) that you have to already know to understand fully what the big picture is. Trying to pick and choose would be maybe even more time-consuming than learning things in a progressive order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though this sounds pessimistic, learning chemistry can be rewarding if you work at it, not to mention you have this site and tons of other online resources to help you out along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Best of luck should you choose to continue, and I hope you do!&lt;/p&gt;&#xA;" OwnerUserId="4969" LastActivityDate="2014-03-30T01:32:04.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9599" PostTypeId="2" ParentId="9581" CreationDate="2014-03-30T06:41:29.937" Score="1" Body="&lt;p&gt;George Olah and coworkers have examined the &lt;a href=&quot;http://dx.doi.org/10.1021/ja9828962&quot; rel=&quot;nofollow&quot;&gt;cyclobutyldicyclopropylmethyl&lt;/a&gt; and the &lt;a href=&quot;http://dx.doi.org/10.1021/jo0701334&quot; rel=&quot;nofollow&quot;&gt;cyclobutylmethyl&lt;/a&gt; cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Yes, this is an awfully short answer and I can't really provide more information on the topic&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-30T06:41:29.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9600" PostTypeId="1" AcceptedAnswerId="9601" CreationDate="2014-03-30T12:50:08.877" Score="3" ViewCount="27643" Body="&lt;p&gt;In my textbook, a statement is given, as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boiling point of a liquid increases with increase in temperature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I was wondering, isn't this wrong? The boiling point of a liquid always remains the same, it doesn't change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, my question is:&#xA;&lt;strong&gt;Is the above statement wrong? If it is, then under what conditions does the boiling point of a liquid change?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="1362" LastEditDate="2014-03-30T12:52:18.670" LastActivityDate="2014-03-30T16:20:23.660" Title="Boiling point of a liquid increasing with temperature?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9601" PostTypeId="2" ParentId="9600" CreationDate="2014-03-30T13:12:27.110" Score="6" Body="&lt;p&gt;It sounds that they actually meant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boiling point of a liquid increases with increase in &lt;strike&gt;temperature&lt;/strike&gt; pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-30T13:12:27.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9602" PostTypeId="1" CreationDate="2014-03-30T13:19:03.250" Score="0" ViewCount="1936" Body="&lt;p&gt;In the equation for Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$, does $T$ refer to the temperature of the system or the surroundings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know we have to calculate Gibbs free energy of the system but the criterion for spontaneity says $\Delta S_\mathrm{total}$ should be greater than zero. When we relate it to Gibbs free energy to show that Gibbs free energy change should be always negative, we keep both system and surrounding temperature same with pressure also constant. Then how can the process occur? Wikipedia says that it is chemical potential that undergoes changes there but what about Gibbs free energy change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus, Gibbs free energy is most useful for thermochemical processes at constant temperature and pressure: both isothermal and isobaric. Such processes don't move on a &lt;em&gt;P—V&lt;/em&gt; diagram, such as phase change of a pure substance, which takes place at the saturation pressure and temperature. Chemical reactions, however, do undergo changes in chemical potential, which is a state function. Thus, thermodynamic processes are not confined to the two dimensional &lt;em&gt;P—V&lt;/em&gt; diagram. There is a third dimension for &lt;em&gt;n&lt;/em&gt;, the quantity of gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="13184" LastEditDate="2015-10-07T15:13:03.717" LastActivityDate="2015-10-07T15:21:04.373" Title="Which temperature is meant in the Gibbs free energy equation?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ClosedDate="2015-10-12T02:12:22.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="9603" PostTypeId="2" ParentId="9602" CreationDate="2014-03-30T13:49:30.943" Score="0" Body="&lt;p&gt;It depends on what you want to calculate. For the Standard Free Energy, $\Delta G°$, the temperature is 25°C. If you want to calculate the Free Energy at another temperature, $\Delta G$, it is usually provided. &#xA;&lt;a href=&quot;https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-03-30T13:49:30.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9604" PostTypeId="2" ParentId="9602" CreationDate="2014-03-30T14:33:00.157" Score="7" Body="&lt;p&gt;If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this must occur equally in both directions, otherwise a temperature gradient would settle between system and environment. We could describe thermal equilibrium as the result of &quot;a zero balance of rate of heat transfer between system and environment&quot;. That's what goes on during a phase transition (e.g. solid to liquid, where entropy of the system increases due to heat given by the environment, whereas temperature stays constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, applying the second law of thermodynamics to an ensemble system + environment, undergoing an infinitesimal transformation, we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm dS_\text{syst} + \mathrm dS_\text{env} \geq 0 \tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said above, the entropy of the environment changes if there's a  reversible heat exchange with the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm dS_\text{env} = \frac{\delta Q_{env}}{T}=-\frac{\delta Q_{syst}}{T} \tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Merging eq. $\text{(1)}$ and $\text{(2)}$, you write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm dS_\text{syst} - \frac{\delta Q_{syst}}{T} \geq 0 \tag3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the first law allows you to write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta Q - P\mathrm dV_\text{syst}=\mathrm dE_\text{syst} \tag4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\mathrm dS_\text{syst} - \left(\mathrm dE_\text{syst}+P\mathrm dV_\text{syst}\right) \geq 0  \tag5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now (and this is crucial), assuming both pressure and temperature constant, you end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm d{\left(E_\text{syst}+PV_\text{syst}-TS_\text{syst}\right)_{T,P}} \leq 0 \tag6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression within the brackets is a &lt;em&gt;state function&lt;/em&gt; of the only system and corresponds to the &lt;em&gt;free energy&lt;/em&gt;. Quoting &lt;a href=&quot;http://en.wikipedia.org/wiki/State_function&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering that $E+PV=H$, we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\mathrm dG_\text{syst}\right)_{T,P} = \left(\mathrm dH_\text{syst}-T\mathrm dS_\text{syst}\right)_{T,P} \leq 0 \tag7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, while equation $\text{(1)}$ is much more general, equation $\text{(7)}$ applies only to the situation where the work is expansion or contraction and for constant temperature and pression. Nonetheless, since for most situations of chemical interest, those conditions are satisfied, eq $\text{(7)}$ maintains a well deserved relevance and is very convenient to establish whether a process is spontaneous or not, within the limitation of constant temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A chemical reaction occurs down to the point where $\left(\Delta G\right)_{T, P}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{$a$A + $b$B &amp;lt;=&amp;gt; $l$L + $m$M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constant temperature and pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\Delta G\right)_{T, P}=l\mu_{\ce L} + m\mu_{\ce M} - a\mu_{\ce A} - b\mu_{\ce B}  \tag8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;being $\mu$ the chemical potential for each of the species involved.&#xA;The chemical potential relates to activity, according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\mu^\circ+RT\ln a \tag9$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation $\text{(8)}$ and $\text{(9)}$ shows that free energy varies with the composition (expressed as &lt;em&gt;activity&lt;/em&gt;) of our system: that's exactly what happens during a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proceeding further, you will end up with Van't Hoff isotherm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\Delta G\right)_{T, P}=\left(\Delta G^\circ\right)_{T}+RT\ln\frac{a_{\ce L}^{l}a_{\ce M}^{m}}{a_{\ce A}^{a}a_{\ce B}^{b}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activities vary until a chemical equilibrium is established, so that $\left(\Delta G\right)_{T, P}=0$.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorDisplayName="user7951" LastEditDate="2015-10-07T14:18:13.530" LastActivityDate="2015-10-07T14:18:13.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9605" PostTypeId="2" ParentId="9600" CreationDate="2014-03-30T14:54:19.720" Score="0" Body="&lt;p&gt;Given a pot of boiling water at one atmosphere pressure: Turn up the temperature of the burner.   More heat (energy) is transferred to the pot down the temperature gradient as the water continues to boil at 100 C.  Latent enthalpy of phase transition of one gram of water at 100 C to one gram of steam at 100 C is 540 cal/g.  Faster heat input means faster rate of boiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a 100 C boiling point only obtains at the surface.  The bottom of the pot has hydrostatic head and therefore a higher boiling point.  Sufficiently high temp at the bottom will get that boiling, too, resulting in faster heat transfer than mere convection.  Too high a temp will obtain the Leidenfrost effect, suddenly dropping heat transfer to near zero.  This is a tremendous problem in heat exchangers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.purdue.edu/BTPFL/BTPFL%20Publications/81.pdf&quot; rel=&quot;nofollow&quot;&gt;https://engineering.purdue.edu/BTPFL/BTPFL%20Publications/81.pdf&lt;/a&gt; &lt;BR&gt;&#xA;  Leidenfrost temperature versus liquid, surface, and surface composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The world in non-linear.  Do not believe engineers who think a line behaves as it travels to the right.  Oil storage tank fires fought with water accumulate heating water at the tank bottom, under the hydrostatic head of the tank.  The water superheats util it pushes the oil above it away, then it flashes to steam, expanding a thousand-fold, removing more hydrostatic head...resulting in a BLEVE.  The expanding cloud of oil mist and steam, if ignited, is a fuel-air explosion.  People get infrared thermal burns hundreds of meters distant.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-30T14:54:19.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9606" PostTypeId="2" ParentId="9575" CreationDate="2014-03-30T15:02:00.463" Score="2" Body="&lt;p&gt;The preference of $\ce{Pb}$ for the lower oxidation state is an example for the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Inert_pair_effect&quot; rel=&quot;nofollow&quot;&gt;inert pair effect&lt;/a&gt;. Due to relativistic contraction of the outermost $s$ orbital, its 2 electrons are closer to the nucleus and thus more tightly bound than the electrons in the $p$ orbitals. More energy is required to use the $s$ electrons for covalent or ionic bonds. Stable compounds with $\ce{Pb}$ in the oxidation state $\ce{+IV}$ are therefore mostly formed with very electronegative elements like $\ce{O}$ or $\ce{F}$.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2015-03-19T19:05:27.400" LastActivityDate="2015-03-19T19:05:27.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9607" PostTypeId="2" ParentId="9600" CreationDate="2014-03-30T16:14:13.963" Score="3" Body="&lt;p&gt;I agree with @Klaus Warzecha's answer but I would like to add a little more to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They probably mean to say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boiling point of a liquid increases with increase in pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since a liquid vaporises when molecules moves faster and farther apart but while increasing pressure, you are applying an external force o hold molecules closer to each other. Hence, molecules will require more energy to overcome this resistance thus this increases boiling point of liquid..!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boiling point of a liquid may also increase with decrease in volume of container.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think there is no need to explain this further.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-03-30T16:20:23.660" LastActivityDate="2014-03-30T16:20:23.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9608" PostTypeId="1" AcceptedAnswerId="9610" CreationDate="2014-03-30T19:05:58.213" Score="6" ViewCount="43413" Body="&lt;p&gt;When water is added to a weak acid like ethanoic acid, the number of ethanoic acid molecules that dissociate increases, however, pH increases (less acidic), too. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2019-05-23T13:45:41.667" LastActivityDate="2019-06-10T21:05:38.107" Title="How to rationalise that when we dilute a weak acid, the pH increases but also the dissociation increases?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9609" PostTypeId="2" ParentId="9608" CreationDate="2014-03-30T19:30:30.377" Score="3" Body="&lt;p&gt;Let's first examine the chemical reaction between our acids and bases. Ethanoic acid, commonly referred to as acetic acid reacts with our base, water. Water is amphiprotic, in that it can act as both a base and acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + H2O &amp;lt;=&amp;gt; CH3COO- + H3O+} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the acetate ion undergoes hydrolysis with water which now instead acts as an acid, reforming acetic acid as well has hydroxide ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COO^- + H2O &amp;lt;=&amp;gt; CH3COOH + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid is weak acid, however, the conjugate base, acetate, is a strong base/salt. Water acts as a strong base, however the hydronium ion is a weak acid, so we can except the pH to increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What many people often forget to realize is that once an acid/base reaction occurs, a salt is created along with water. However, a hydrolysis reaction between the salt and water can occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from Chemical Principles: The Quest for Insight by Atkins et. al on page 494 in the tenth edition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calcium acetate, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(CH3CO2)2 (aq)}$&lt;/span&gt; is used in medicine to treat patients with a kidney disease that results in high levels of phosphate ions in the blood. The calcium binds with the phosphates so that they can be excreted. If you are using calcium acetate for this purpose, it is important to know the pH of the solution to avoid complications in treatment. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Say we have a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.15 M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(CH3CO2)2 (aq)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the conjugate base of acetic acid is strong, we can expect the pH to increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our expression can be written as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COO^- + H2O &amp;lt;=&amp;gt; CH3COOH + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calcium is a spectator ion and thus can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_b = \ce{\frac{[CH3COOH][OH-]}{[CH3COO-]}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructing our RICE table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COO^- + H2O &amp;lt;=&amp;gt; CH3COOH + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline&#xA;&amp;amp;\ce{CH3COO^-} &amp;amp; \ce{H2O} &amp;amp; \ce{CH3COOH} &amp;amp; \ce{OH-}\\\hline&#xA;\text{Initial conc.} &amp;amp; 0.30 &amp;amp; - &amp;amp; 0 &amp;amp; 0 \\&#xA;\hline&#xA;\text{Change conc.} &amp;amp; -x &amp;amp; - &amp;amp; +x &amp;amp; +x\\&#xA;\hline&#xA;\text{End conc.} &amp;amp; 0.30 - x &amp;amp; - &amp;amp; x &amp;amp; x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;K_b &amp;amp;= 5.6\times10^{-10} = \frac{[x][x]}{[0.30-x]}\\&#xA;x &amp;amp;= 1.3 \times 10^{-5}\\&#xA;\text{pOH} &amp;amp;= -\log([\ce{OH-}])\\&#xA;\text{pH} &amp;amp;= 14 - \text{pOH} = 14 - 4.89 = 9.11&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="4945" LastEditDate="2019-05-23T13:57:55.990" LastActivityDate="2019-05-23T13:57:55.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9610" PostTypeId="2" ParentId="9608" CreationDate="2014-03-30T20:08:14.537" Score="7" Body="&lt;p&gt;It's simply due to dilution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's true that adding water increases dissociation (Gerard has done a nice effort in &lt;a href=&quot;https://chemistry.stackexchange.com/q/7648/4945&quot;&gt;Does the number of H+ ions in solution go up on dilution of a weak acid?&lt;/a&gt;), concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; inevitably decreases. As a consequence, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt;, defined as &lt;span class=&quot;math-container&quot;&gt;$-\log[\ce{H3O+}]$&lt;/span&gt;, increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt; acetic acid concentration &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M}$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H3O+]} = \sqrt{K_c} = \pu{4.22E-3M},$$&lt;/span&gt;&#xA;that corresponds to &lt;span class=&quot;math-container&quot;&gt;$\pu{4.22E-3mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;.&#xA;If you double the volume (&lt;span class=&quot;math-container&quot;&gt;$\pu{2 L}$&lt;/span&gt;),&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H3O+]}=\sqrt{K_c/2}=\pu{2.98E-3M},$$&lt;/span&gt;&#xA;that corresponds to &lt;span class=&quot;math-container&quot;&gt;$\pu{5.96E-3mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of substance is higher, but concentration is less: dilution does play a role here.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="4945" LastEditDate="2019-05-23T13:50:58.210" LastActivityDate="2019-05-23T13:50:58.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9611" PostTypeId="1" AcceptedAnswerId="9657" CreationDate="2014-03-30T20:30:48.007" Score="4" ViewCount="1057" Body="&lt;p&gt;So I've been trying to follow the derivation of the Goldman-Hodgkin-Katz equation on Wikipedia (&lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Goldman_equation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/w/index.php?title=Goldman_equation&lt;/a&gt; ) - I can follow it through after it sets up the initial model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm having trouble using the Einstein-Stokes relationship to derive the initial model in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the flux from diffusion down the chemical gradient, $-D_{A}\frac{d[A]}{dx}$. The negative sign is there because I'm imagining the ions being &quot;lost&quot; from the side I'm considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have the electrical field pushing ions the other way. This won't depend upon the rate of change of concentration because the electric field is uniform and so there's effectively a constant force on each of the particles. The flow will instead depend upon the constant velocity that these particles will be pushed at. So the flow will be $[A]v_d$ - the moles per unit volume being multiplied by the speed in the direction towards the membrane will give me the number of moles per unit area passing through the membrane per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all I need to do is work out $v_d$ I'm guessing. So I use the electrical mobility equation, $D_A=\frac{\mu_{q}k_BT}{q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ will be the charge, so $nzF$ (where $z$ is the valance, $nz$ gives the moles of electrons and multiplying by $F$ gives the charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_q$ will be $\frac{v_d}{E} = \frac{v_d L}{V_m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_B = \frac{R}{N_A}$ where $N_A$ is Avogadro's number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $D_A=v_d \frac{L}{V_m} \frac{RT}{zF} \frac{1}{N_A n}$, thus $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n=v_d$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this back in to get the ion flux, $D_A \frac{V_m}{L} \frac{zF}{RT} N_A n [A]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding my two fluxes together to get the total flux, I get: $j_a = -D_A(\frac{d[A]}{dx} - \frac{V_m}{L} \frac{zF}{RT} N_A n [A] )$ - this is not quite what the Wikipedia article has - a stray $n N_A$ has crept in, somehow!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me by pointing out where I'm making a mistake!&lt;/p&gt;&#xA;" OwnerUserId="5029" LastActivityDate="2014-04-01T20:32:43.080" Title="Deriving goldman equation" Tags="&lt;electrochemistry&gt;&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9612" PostTypeId="1" CreationDate="2014-03-31T01:11:01.760" Score="2" ViewCount="564" Body="&lt;p&gt;Was studying chemical bonding(Valence bond theory) but there came this topic called &lt;em&gt;wave equation&lt;/em&gt; but  have not studied quantum mechanics,so finding hard to understand these terms.Please help!!!&lt;/p&gt;&#xA;" OwnerUserId="4080" LastEditorUserId="4080" LastEditDate="2014-03-31T08:42:34.503" LastActivityDate="2014-04-30T14:50:08.193" Title="What is this wave function in Chemical bonding?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9613" PostTypeId="2" ParentId="9612" CreationDate="2014-03-31T01:44:27.280" Score="1" Body="&lt;p&gt;$\Psi$ is the wavefunction describing an electron's motion.  $\Psi^2$ is the probability density function for the electron. $\Psi^2$ is usually interpreted as the probability of finding the electron at a given set of coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case that $\Psi$ is a complex function (in that it contains imaginary numbers and not that it is complicated, which it usually is), $\Psi^2$ is defined as the product of $\Psi$ with its complex conjugate: $\Psi \Psi^*=|\Psi|^2$ so that the probability density function contains only real numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a simple case of $\Psi=a+bi$. The complex conjugate would be $\Psi^*=a-bi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi \Psi^* = (a+bi)(a-bi)=a^2+b^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the probability density function is &lt;em&gt;normalized&lt;/em&gt; so that its integral over all space is 1 (i.e. there is 100% chance of finding the electron in all of space). In the equation below, $A$ is the normalization constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1=A \int_o^{2\pi} \int_0^{2 \pi} \int_o^\infty \Psi \Psi^* dr d\theta \phi$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="275" LastEditDate="2014-04-30T10:39:20.623" LastActivityDate="2014-04-30T10:39:20.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9614" PostTypeId="2" ParentId="7642" CreationDate="2014-03-31T02:59:28.690" Score="2" Body="&lt;p&gt;After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Electron affinity (EA) refers to &quot;the energy released when an atom in the gas phase accepts an electron.&quot; (Burdge, &lt;em&gt;Atoms First&lt;/em&gt;, p.118)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower EA means that the process of accepting an electron is &lt;strong&gt;not&lt;/strong&gt; energetically favorable. In other words, lower EA = less likely to accept an electron. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Groups 4 and 8, in this case, already have filled subshells. Adding an electron would place it in a new subshell that is higher in energy. The EA would be lower for these groups because of the &quot;reluctance&quot; of the atom to accept an additional electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups 3 and 7 have the anomalously &lt;em&gt;high&lt;/em&gt; EAs, which might have been where the confusion was.&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="13339" LastEditDate="2015-05-18T08:04:32.590" LastActivityDate="2015-05-18T08:04:32.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9615" PostTypeId="1" CreationDate="2014-03-31T03:09:19.803" Score="1" ViewCount="542" Body="&lt;p&gt;I think that the first reaction would add into the ketone carbon, but I don't know where to go after that. &#xA;A step-by-step explanation of the answer would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KLRvf.png&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answer choices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0wDo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3747" LastActivityDate="2014-04-03T17:08:22.487" Title="Predict the product (organic chemistry)" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="21" ClosedDate="2015-11-09T06:06:36.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="9618" PostTypeId="2" ParentId="9615" CreationDate="2014-03-31T07:51:57.617" Score="8" Body="&lt;p&gt;Looks like a classic to me. But since this is homework, &lt;strike&gt;I will not provide a full reaction scheme&lt;/strike&gt; and only give some hints.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your starting material is a $\beta$-ketoester (1,3-dicarbonyl compound).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where is the most CH-acidic position in this molecule?&lt;/li&gt;&#xA;&lt;li&gt;Benzylchloride is an alkylating agent. What will happen when it reacts with the anion of your starting material?&lt;/li&gt;&#xA;&lt;li&gt;For extra credit: Why is sodium ethoxide in ethanol a reasonable choice when you want to isolate, purify and characterize the product? &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your starting material did have an ester group. Does it still have it after the first step? If so, what will happen to it in aqueous alkaline solution? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(&lt;em&gt;This step would not work with the saponification product of a simple ester.&lt;/em&gt;) &#xA;But you started with $\beta$-ketoester! Heating the product of step 2 in acidic media will let it loose some atoms. The mechanism may be drawn using a cyclic intermediate, in which the ketone has an assisting role.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dbZJH.png&quot; alt=&quot;beta-ketoester as synthon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deprotonation and subsequent benzylation of the $\beta$-ketoester occur at the most CH-acidic position, i.e beween the keto and the ester group. As a side note, benzylation at the methyl group adjacent to the carbonyl group is possible when using 2 equivalents of a strong base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using ethoxide rather than methoxide as a base makes sure that transesterification does not lead to a mixture of products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting benzylated $\beta$-ketoester is cleaved under alkaline conditions to the corresponding carboxylate. Under acidic conditions, this from a $\beta$-ketoacid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon heating, $\beta$-ketoacids undergo decarboxylation (= they loose $\ce{CO2}$) to form an enolate which tautomerizes to the final methyl ketone.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-03T17:08:22.487" LastActivityDate="2014-04-03T17:08:22.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9620" PostTypeId="1" CreationDate="2014-03-31T10:12:12.287" Score="2" ViewCount="268" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A tricarboxylic acid &lt;span class=&quot;math-container&quot;&gt;$\ce{H3A}$&lt;/span&gt; has three different types of &lt;span class=&quot;math-container&quot;&gt;$\text{p}K_\text{a}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$(\text{p}K_\text{a, 1} =2.0,$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\text{p}K_\text{a, 2}=5.0,$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\text{p}K_\text{a, 3}=8.0)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Identify the ratio of the three charged components(&lt;span class=&quot;math-container&quot;&gt;$\ce{H2A-}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\ce{HA^{2-}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{A^{3-}}$&lt;/span&gt;) in &lt;span class=&quot;math-container&quot;&gt;$0.2 ~\text{M}$&lt;/span&gt; solution of tricarboxylic acid &lt;span class=&quot;math-container&quot;&gt;$\ce{H3A}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.55 M}$&lt;/span&gt; of the strong base &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; is added into a solution of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.20 M}$&lt;/span&gt; tricarboxylic acid &lt;span class=&quot;math-container&quot;&gt;$(\ce{H3A})$&lt;/span&gt;. Calculate the final pH. (Assume that the addition of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; will not influence the volume of the solution.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I managed to get &lt;span class=&quot;math-container&quot;&gt;$\pu{0.04 M/L}$&lt;/span&gt; for the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2A-}$&lt;/span&gt; species but I do not know how to find the concentration of the acids.&lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="95133" LastEditDate="2021-05-13T12:47:58.420" LastActivityDate="2021-05-13T12:47:58.420" Title="Ratio of conjugate bases of a tricarboxylic acid" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9623" PostTypeId="1" AcceptedAnswerId="9628" CreationDate="2014-03-31T12:27:42.183" Score="13" ViewCount="6673" Body="&lt;p&gt;I've always been told in chemistry lessons that lone pairs repel more due to a greater charge density than bonding pairs. And that makes sense when steric effects are involved when bond orbitals overlap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this conflicts with what I have learnt in physics i.e. &lt;a href=&quot;https://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coulomb's law&lt;/a&gt;, which states that the only thing affecting the force between two charged bodies is the magnitude of the charge and the distance between them. Therefore it seems weird to me that a lone pair should &quot;repel&quot; more because of greater charge density, despite having the same charge magnitude as a bonding pair. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I neglecting or if I'm right, does that mean for large atoms (such as iodine, where electron pairs are further apart) the lone pairs and bonding pairs repel equally as per Coulomb's law?&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="41328" LastEditDate="2017-08-09T01:30:34.927" LastActivityDate="2021-02-14T18:37:38.353" Title="Correlation between Coulomb's law and VSEPR theory" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;orbitals&gt;&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9624" PostTypeId="2" ParentId="9583" CreationDate="2014-03-31T12:53:03.203" Score="6" Body="&lt;p&gt;If in doubt, draw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vxgqD.png&quot; alt=&quot;aldol condensation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deprotonation at C-6 under thermodynamic control results in the more stable enolate (&lt;strong&gt;1c&lt;/strong&gt;) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if that is your target molecule, addition of acetylene to cyclohexanone, followed by an acid-catalyzed rearrangement of the propargylic alcohol would be an alternative route.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-03-31T12:53:03.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9625" PostTypeId="2" ParentId="9623" CreationDate="2014-03-31T13:31:05.760" Score="1" Body="&lt;p&gt;The charge magnitude is the same in both cases; however, the sizes of the orbitals are different. The lone pair electrons are in the field of one nucleus, while the two electrons of a bonding pair are in the field of two nuclei. As a result, the orbital with the bonding pair tends to be smaller, and it requires less space than the more diffuse orbital of a lone pair. Therefore, the distance between bonding pair orbitals is larger. This reduces the repulsive force between them, because according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;nofollow&quot;&gt;Coulomb's law&lt;/a&gt;, the electrostatic force $F$ between two charges $q_1$ and $q_2$ is inversely proportional to the distance $r$ between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|F|=k_{e}\frac{|q_1q_2|}{r^2}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-12-02T17:53:00.537" LastActivityDate="2014-12-02T17:53:00.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9626" PostTypeId="2" ParentId="9623" CreationDate="2014-03-31T13:42:45.490" Score="0" Body="&lt;p&gt;Lone pair-Lone pair repulsion &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;$&lt;/span&gt; Lone pair-Bonding pair repulsion &lt;span class=&quot;math-container&quot;&gt;$&amp;gt;$&lt;/span&gt; Bonding pair-Bonding pair repulsion.&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jfENl.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Lone pairs are attached to same atom, they are free. As expected, they should repulse to greater extent. Bond pairs are attached to two different atoms, the heavy atoms attached to them might make them not repulse to greater extent. Now, lone pair-bond pair repulsion should be intermediate, as they have one lone pair and one bond pair.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-03-31T13:42:45.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9627" PostTypeId="1" AcceptedAnswerId="9644" CreationDate="2014-03-31T13:46:43.643" Score="0" ViewCount="24603" Body="&lt;p&gt;C + O = CO&lt;sub&gt;2&lt;/sub&gt; . This is because carbon has valency of 4 while oxygen has valency of 2. When they react the valencies are criss crossed which means we will get C&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;4&lt;/sub&gt; but this is simplified to CO&lt;sub&gt;2&lt;/sub&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my textbook say that C + O&lt;sub&gt;2&lt;/sub&gt; = CO&lt;sub&gt;2&lt;/sub&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is it possible that C + O&lt;sub&gt;2&lt;/sub&gt; and C + O both get the same product of CO&lt;sub&gt;2&lt;/sub&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can say that the reaction of C + O = CO&lt;sub&gt;2&lt;/sub&gt; is not balanced but the balanced chemical reaction would be C + 2O = CO&lt;sub&gt;2&lt;/sub&gt; not C + O&lt;sub&gt;2&lt;/sub&gt; = CO&lt;sub&gt;2&lt;/sub&gt; which is the reaction that my textbook says.&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="4945" LastEditDate="2014-04-01T09:49:47.870" LastActivityDate="2016-04-30T09:26:57.477" Title="C + O2 is equal to C + O, how is that possible" Tags="&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;&lt;combustion&gt;&lt;stoichiometry&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9628" PostTypeId="2" ParentId="9623" CreationDate="2014-03-31T15:28:21.233" Score="12" Body="&lt;p&gt;In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="41328" LastEditDate="2021-02-14T18:37:38.353" LastActivityDate="2021-02-14T18:37:38.353" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9629" PostTypeId="2" ParentId="9627" CreationDate="2014-03-31T19:29:28.190" Score="7" Body="&lt;p&gt;Let's first take a look at the allotropes of oxygen, and look into dioxygen more in depth.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Atomic oxygen ($\ce{O1}$, a free radical)&lt;/li&gt;&#xA;&lt;li&gt;Singlet oxygen ($\ce{O2}$), either of two metastable states of molecular oxygen&lt;/li&gt;&#xA;&lt;li&gt;Tetraoxygen ($\ce{O4}$), another metastable form&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From NASA, &lt;a href=&quot;http://www.nasa.gov/topics/technology/features/atomic_oxygen.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.nasa.gov/topics/technology/features/atomic_oxygen.html&lt;/a&gt;, regarding atomic oxygen:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atomic oxygen doesn't exist naturally for very long on the surface of Earth, as it is very reactive. But in space, where there is plenty of ultraviolet radiation, $\ce{O2}$ molecules are more easily broken apart to create atomic oxygen. The atmosphere in low Earth orbit is comprised of about 96% atomic oxygen. In the early days of NASA's space shuttle missions, the presence of atomic oxygen caused problems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dioxygen, or triplet oxygen, is the most commonly known allotrope of oxygen. It has the molecular formula $\ce{O2}$. Oxygen has 8 electrons with 2 in the 1s, 2 in the 2s, 4 in the 3p orbitals. Alternatively, there are 6 valence electrons. If there are another oxygen molecules, oxygen will pair up, form a double bond with bond order of two. &lt;strong&gt;In short, the potential energy of dioxygen is far less than that of atomic oxygen.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting aspect of oxygen is that it exhibits paramagnetism unlike $\ce{N2}$ and it can exist in two different electronic states called singlet oxygen. The picture of the molecular orbital (MO) diagram of oxygen makes this more clear:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BUw50.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MO diagrams above are for the singlet oxygen $a^1\Delta g$ excited state, the singlet oxygen $b^1\Sigma \text{g+}$ excited state, and the triplet ground state $X^3\Sigma \text{g-}$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may notice is a spin flip in the $b^1\Sigma \text{g+}$ excited state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition taken from Purdue University summarizes Hund's rule of maximum simplicity nicely: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two first diagrams are in violation of the 1.) spin-selection rule: spin-flips are foridden, and 2.) Laporte selection rule: trasnsitions between orbitals of the same parity are forbidden, where parity means symmetry with respect to inversion. There is a german notation, gerade - which refers to symmetric with respect to inversion and ungerade - antisymmetric with respect to inversion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to produce ozone. &lt;a href=&quot;https://en.wikipedia.org/wiki/Ozone#Production&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ozone#Production&lt;/a&gt;&#xA;Ozone is a triatomic molecule with 3 oxygens. It is much less stable than dioxygen and often breaks down into dioxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may have been confused with is Dalton's incorrect &quot;rule of greatest simplicity.&quot; Dalton was trying to resolve the issue of the correct ratio and number of atoms in respect to the chemical formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He assumed that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{H + O -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{H2 + O2 -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't until Avogadro and Gay-Lussac who stated the Law of Multiple Proportions and postulated the existence of diatomic molecules, that we can now resolve Dalton's incorrect hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When two elements form a series of compounds, the masses of one element that&#xA;  combine with a fixed mass of the other element are in the ratio of small integers&#xA;  to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-03-31T19:36:03.263" LastActivityDate="2014-03-31T19:36:03.263" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9630" PostTypeId="2" ParentId="9620" CreationDate="2014-03-31T20:27:03.367" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWunJ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's use citric acid as an example. Although the &lt;span class=&quot;math-container&quot;&gt;$pK_a$&lt;/span&gt; values of citric acid might not necessarily be the same as those above, it will be useful for demonstration.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H8O7 + H2O &amp;lt;=&amp;gt; C6H7O7- + H3O+ &amp;lt;=&amp;gt; C6H6O7^2- + H3O+ &amp;lt;=&amp;gt; C6H5O7^3- + H3O+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Throughout the polyprotic acid titration, two protons are donated. The way this problem is solved is you have to use successive &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; values and concentrations.&lt;/p&gt;&#xA;&lt;p&gt;Primary proton transfer equilibrium:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H8O7 + H2O &amp;lt;=&amp;gt; C6H7O7- + H3O+} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline&#xA;C_\mathrm{initial} &amp;amp; 0.20 &amp;amp; - &amp;amp; 0 &amp;amp; 0 \\&#xA;\hline&#xA;\Delta C &amp;amp; -x &amp;amp; - &amp;amp; +x &amp;amp; +x\\&#xA;\hline&#xA;C_\mathrm{final}&amp;amp; 0.20 - x &amp;amp; - &amp;amp; x &amp;amp; x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_\mathrm {a_1} = 0.01 = \frac{[\ce{C6H7O7-}][\ce{H3O+}]}{[\ce{C6H8O7}]} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_\mathrm{a1} = 0.01 = \frac{x^2}{0.20-x} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, we can't disregard the x in the denominator so we have to solve the quadratic:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x = 0.04 \ \mathrm{or} \ x = -0.05$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We disregard our negative root. From &lt;span class=&quot;math-container&quot;&gt;$[\ce{C6H7O7}]^- = [\ce{H3O+}]$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{C6H7O7}-] = \pu{0.04 mol L-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we use our second &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; to find the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H6O7^{2-}}$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm {a_2} &amp;lt;&amp;lt; K_\mathrm {a_1}$&lt;/span&gt;, we can assume that the hydronium ion concentration is unchanged by the second deportonation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H7O7- + H2O &amp;lt;=&amp;gt; C6H6O7^2- + H3O+} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline&#xA;C_\mathrm{initial} &amp;amp; 0.04 &amp;amp; - &amp;amp; 0 &amp;amp; 0.04 \\&#xA;\hline&#xA;\Delta C &amp;amp; -x &amp;amp; - &amp;amp; +x &amp;amp; +x\\&#xA;\hline&#xA;C_\mathrm{final} &amp;amp; 0.04 - x &amp;amp; - &amp;amp; x &amp;amp; 0.04 + x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_\mathrm {a_2} = .00001 = \frac{x(0.04 + x)}{0.04-x} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x = -0.04002 \ \mathrm{or} \ x = 9.995 \times 10^{-6} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Ignoring the negative root:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ [\ce{C6H6O7^2-}] = \pu{9.995 \times 10^{-6} mol L-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally, assuming that the concentration of hydronium ions and the conjugate base of citric acid after the 2nd deprotonation is unaffected by additional deprotonation, we can write up our equilibrium table again:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline\&#xA;C_\mathrm{initial}&amp;amp; 9.995*10^{-6} &amp;amp; - &amp;amp; 0 &amp;amp; 0.04 \\&#xA;\hline&#xA;\Delta C &amp;amp; -x &amp;amp; - &amp;amp; +x &amp;amp; +x\\&#xA;\hline&#xA;C_\mathrm{final} &amp;amp; 9.995*10^{-6} - x &amp;amp; - &amp;amp; x &amp;amp; 0.04 + x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_\mathrm {a_3} = .00000001 = \frac{x(0.04 + x)}{9.995 \cdot 10^{-6} -x} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x = -0.04002 \ \mathrm{or} \ x = 2.49875 \times 10^{-12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ [\ce{C6H5O7^3-}] = \pu{2.49875 \times 10^{-12} mol L-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="95133" LastEditDate="2021-05-13T12:46:35.280" LastActivityDate="2021-05-13T12:46:35.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9632" PostTypeId="1" AcceptedAnswerId="9638" CreationDate="2014-03-31T21:53:00.970" Score="1" ViewCount="270" Body="&lt;p&gt;As far as chemistry goes, there are laws or observations that can be useful to determine the perfect shape and disposition of the wood in a campfire ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example what chemistry says about the 2 most popular &quot;shapes&quot; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;teepee &lt;img src=&quot;https://i.stack.imgur.com/bq4Pn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;log cabin &lt;img src=&quot;https://i.stack.imgur.com/y8hhg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or even other variations such as the swedish stove ( 1 log with the top splitted in multiple segments )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties that I'm looking for:&#xA; - easy to start&#xA; - long lasting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties that I would like to have but I can discard:&#xA; - significant heat generation&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2014-04-01T02:20:11.200" Title="the perfect campfire" Tags="&lt;home-experiment&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9633" PostTypeId="2" ParentId="9627" CreationDate="2014-03-31T22:40:11.743" Score="5" Body="&lt;p&gt;The common products of carbonaceous materials' combustion are carbon monoxide and carbon dioxide. The products determine the equation.  The equation does not dictate the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon suboxide $\ce{C2O3}$ is known.   Benzene hexacarboxylic acid trianhydride is a carbon oxide.  Combustion generally outputs simple molecules in deep thermodynamic holes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-03-31T22:40:11.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9634" PostTypeId="1" CreationDate="2014-04-01T00:00:31.037" Score="-3" ViewCount="641" Body="&lt;p&gt;For my chemistry experiment, I was assigned to produce &lt;strong&gt;ammonium sulfide&lt;/strong&gt; and test the chemical properties of the compound. From my online research, I found out that I can mix household ammonia and match heads to produce Ammonia Sulfide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this will satisfy the purpose of my assignment, but I have a few concerns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what proportions I should mix these materials in order to produce an ideal solution. I am also unsure about the exact results of mixing ammonia and match heads, since match heads consists of more than just sulfer, but also potassium chlorate, antimony trisulfide, and some oxidizing agents. Will the end product potentially be dangerous? I plan to mix the two chemicals in a enclosed test tube, and seal it for a few days. &lt;/p&gt;&#xA;" OwnerUserId="5037" LastEditorUserId="16683" LastEditDate="2016-05-25T10:37:47.387" LastActivityDate="2016-05-25T10:37:47.387" Title="Is this reaction to produce ammonium sulfide safe?" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9635" PostTypeId="2" ParentId="9634" CreationDate="2014-04-01T00:20:37.817" Score="5" Body="&lt;p&gt;$$ \ce{H2S(aq) + 2 NH3(aq) -&amp;gt; (NH4)2S(aq)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon mixing ammonia with potassium chlorate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{KClO3(aq) + NH3(aq) -&amp;gt; NH4ClO3(aq)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Sb2S3 (aq) + NH3(aq) -&amp;gt; Sb(OH)3(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sb(III) reacts with aqueous ammonia to precipitate white Sb(OH)3. MSDS sheets say that antimony trisulfide is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Toxic to aquatic life with long lasting effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{Sb2S3(aq) + (NH4)2S(aq) -&amp;gt; (NH4)3SbS3 (aq)} $$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-04-01T11:47:57.073" LastActivityDate="2014-04-01T11:47:57.073" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9636" PostTypeId="1" CreationDate="2014-04-01T00:45:02.917" Score="2" ViewCount="4159" Body="&lt;p&gt;I can't seem to remember. I know that they all want to fill their valence shells, but, aren't these different?&lt;/p&gt;&#xA;" OwnerUserId="5038" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T16:59:10.033" LastActivityDate="2016-11-06T16:59:10.033" Title="Do lithium and beryllium want to lose electrons to be like helium?" Tags="&lt;physical-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9637" PostTypeId="2" ParentId="9636" CreationDate="2014-04-01T01:08:13.257" Score="5" Body="&lt;p&gt;Yes, they do. All alkali metals, in group 1 of the periodic table, display the tendency to lose electrons as well as group 2 alkaline earth metals. Instead of losing one electron like alkali metals, alkaline earth metals such as beryllium lose two!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loss of an electron to some other element, say a halogen, is very favorable as it gives the potential for alkali metals or alkaline earth metals to form a full valence shell or stable octet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it is very common to see ions such as $\ce{H+}$ as well as $\ce{Li^{+}}$ as well as $\ce{Be^{2+}}$. Thus, salts such as $\ce{NaCl}$ as well as other compounds such as hydroxides, $\ce{Li(OH)}$ are the product of these metals once they lose their electrons and react with another element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First ionization energies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oU9ak.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-04-01T12:31:27.163" LastActivityDate="2014-04-01T12:31:27.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9638" PostTypeId="2" ParentId="9632" CreationDate="2014-04-01T02:20:11.200" Score="2" Body="&lt;p&gt;From many years as a trained firefighter, I can tell you that there is certainly science involved. There are a number of very heavy calculations for calculating things like solid combustible burn time and heat release rates for combustible materials, which might possibly be useful to predict the perfect shape, fuel size and arrangement for a campfire, but are probably well beyond the scope of producing the perfect sausage sizzle. However, a number of key factors that influence fire behaviour, and which must be considered in building the best campfire include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size of fuels. For a given fuel type, the larger the fuel piece, the greater the resident burn time, but the greater the energy input required to obtain ignition. So small fuels will ignite and burn first, and provide energy through preheating to ignite larger fuels.&lt;/li&gt;&#xA;&lt;li&gt;Availability of fine fuels (typically those under 5mm diameter). For ignition to occur, you need to have fine fuels, especially in the absence of other freely burning ignition agents (fire starters, a cup full of petroleum etc). As heat rises, the fine fuels need to be placed at/near the base to preheat larger available fuels. Plenty of fine fuels is critical - you can't have too much.&lt;/li&gt;&#xA;&lt;li&gt;Available air capacity. Compact fuels that do not have sufficient air flow are poor propagators of fire. Oxygen is one of the critical elements of the fire triangle (or quadrilateral if you choose), and if not being positively injected (bellows etc), there must be sufficient air flow between fuels to allow drawing in of fresh oxygen supplies (As fire burns, the hot gases rise, and fresh oxygen is drawn in at the base, and must be able to freely disperse.&lt;/li&gt;&#xA;&lt;li&gt;Exposure to radiant and convection heat energy. Heat energy passes via radiant, convection and conductance pathways. Solid fuels need to be preheated to release the combustible gases, and this requires sufficent exposure to radiant and convection energy pathways, which is usually above the burning fuel source.&lt;/li&gt;&#xA;&lt;li&gt;Fuel moisture. Critical factor. Many fuels with greater than about 15% fuel moisture will have difficulty igniting; around 25% most fuels have difficulty in sustaining burn. Fine fuels exchange moisture with the atmosphere far more readily, and so weather conditions must form a real factor in your calculations (temperature, relative humidity, wind). On hotter, drier days, larger diameter fine fuels can be used for ignition. Drier fuels also burn hotter and provide greater energy output. Moist fuels release lower energy flames, taking longer to ignite other fuel.&lt;/li&gt;&#xA;&lt;li&gt;Resident burn time. Different materials (including different types of wood) have different resident burn times, and so your choice of fuels will be significant. Iron bark, for instance, has a greater resident burn time, and greater energy output than radiata pine. This derives from the density and composition of the fuels themselves. Fine fuels, of course, have a very short resident burn time, so it is important to have enough to ensure sufficient preheating of larger fuels. How much will depend on fuel moisture!&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surface area of fuels. Smooth, polished wood will take longer to ignite than rough sawn timber&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, you may have to stipulate some other limitations:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Availability of renewable fuels. Is you campfire fuels source static, or are you able to add more fuels as the campfire burns. This clearly influences how long it lasts, and heat generation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stability of design. As the finer fuels burn, there is a point of critical collapse, where the top fuels collapse on the sub-layers. This will influence a number of factors, including availability of air and exposure to preheating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My design for my favourite campfire is somewhat similar to your log cabin design. It is a crib design, which means it is layered, quite often with the bottom layer of fine fuels (tinder) sitting on a spaced grid of larger sticks to allow air flow underneath. The mix of large fuel and smaller fuels (kindling) is staggered to give stability and access to preheating. The overall shape gets smaller as the layers go up, so the final shape is somewhat square-pyramidal. Separation of fuels needs to be close, while still ensuring maximum air flow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-04-01T02:20:11.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9640" PostTypeId="2" ParentId="5025" CreationDate="2014-04-01T03:56:33.453" Score="1" Body="&lt;p&gt;I would like to add to the previous answers a Pourbaix Diagram, which demonstrates the transition mentioned between ferous and ferric ions as the electrochemical potential or redox properties of the solution change. &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Pourbaix_Diagram_of_Iron.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; This diagram suggests that the solution does indeed have a pH comparable to that of Vinegar which is pH=2.4 The H2O2 should not affect the pH, but rather the oxidation state (vertical axis below) from Fe+2 to Fe+3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G76rL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2015-07-22T17:02:40.427" LastActivityDate="2015-07-22T17:02:40.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9641" PostTypeId="1" CreationDate="2014-04-01T04:48:17.423" Score="2" ViewCount="1008" Body="&lt;p&gt;I found &lt;a href=&quot;http://chemistry.about.com/od/chemistryhowtoguide/a/Stink-Bomb-Recipe.htm&quot; rel=&quot;nofollow&quot;&gt;this recipe&lt;/a&gt; on how to make a stink bomb:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recipe basically says to mix ammonia and match heads to create ammonium sulfide. However, I am skeptical, because&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It seems that match heads have mostly phosphorus, and only a negligible amount of sulfur.&lt;/li&gt;&#xA;&lt;li&gt;I can't find any reference for synthesizing ammonium sulfide from ammonia and sulfur; rather, I found that it is synthesized from hydrogen sulfide and ammonia.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Would this recipe still work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it also make &lt;a href=&quot;http://en.wikipedia.org/wiki/Malodorant&quot; rel=&quot;nofollow&quot;&gt;malodorants&lt;/a&gt; other than ammonium sulfide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, would these other malodorants be safe? Or would it be better to use a different process to create a stink bomb, like using actual sulfur, or something  entirely different?&lt;/p&gt;&#xA;" OwnerUserId="5041" LastEditorUserId="4734" LastEditDate="2014-04-01T08:54:31.727" LastActivityDate="2014-04-03T09:35:12.733" Title="Would this recipe on stink bomb work?" Tags="&lt;synthesis&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9642" PostTypeId="1" AcceptedAnswerId="9659" CreationDate="2014-04-01T06:04:00.473" Score="5" ViewCount="21249" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Use activities to calculate the pH of each of the following solutions, being sure to use $\mathrm{\alpha}$ values and the extended Debye-Huckel equation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first one the concentration is $0.020 \mathrm{M}$ $\ce{NaCl}$&#xA;  with the answer being : $\mathrm{pH}$ = $6.99$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The second one the concentration is $0.10 \mathrm{M}$ $\ce{NaCl}$&#xA;  with the answer being : $\mathrm{pH}$ = $6.98$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thing is i have no idea how to find the activity coefficient of $[\ce{H}^+]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know ionic strength $u$ is calculated as &#xA;$\mathrm{\frac{1}{2}\left(\sum C_iZ_i^2\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how does $[\ce{H+}]$ work into the charge balance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the activity coefficient is calculated as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log\gamma=\frac{-0.51(z^2)(\sqrt{u})}{1+(\alpha(\frac{\sqrt{u}}{305})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I have is:&#xA;$$\ce{Na+ + Cl- + H+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that raises the question of how do I know what $[\ce{H+}]$ is and $[\ce{OH-}]$ for calculating the ionic strength? Am I approaching this wrong? Where do I start?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="5026" LastEditDate="2018-03-05T13:33:42.153" LastActivityDate="2018-03-05T13:33:42.153" Title="Find the pH of a sodium chloride solution using the extended Debye-Huckel equation" Tags="&lt;solubility&gt;&lt;analytical-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9643" PostTypeId="2" ParentId="9636" CreationDate="2014-04-01T09:02:39.220" Score="7" Body="&lt;p&gt;Yes, lithium wants to lose electrons to be like helium because &lt;strong&gt;full valence shells are more stable states&lt;/strong&gt; and all noble gases have full valence shells. So alkali metals lose one electron to achieve nearest noble gas configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;beryllium is exceptional because of its small atomic and ionic radii&lt;/strong&gt;. Due to this, its &lt;strong&gt;ionization potential is very high&lt;/strong&gt;. So it mostly forms covalent bonds.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T16:56:36.220" LastActivityDate="2016-11-06T16:56:36.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9644" PostTypeId="2" ParentId="9627" CreationDate="2014-04-01T09:16:35.370" Score="8" Body="&lt;p&gt;When thinking about chemical reactions it is very important to know which chemicals may react with each other. Jun-Goo Kwak already pointed out the nature of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick reminder: The ground state of elementary oxygen is the triplet biradical $\ce{o2}$, which is a gas. This is what we have on the surface of the earth. Carbon however comes in many  different forms in nature. The most popular and often is &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite&quot; rel=&quot;noreferrer&quot;&gt;graphite&lt;/a&gt;. Other forms include diamond, fullerenes and graphene.&#xA;At one point in your life you haw almost certainly come in contact with graphite: coal. As the principle repeating unit is carbon itself, its formula will be written as $\ce{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the binary combination of oxygen and carbon, there are also many different modifications. The most important of them are carbonmonooxide ($\ce{CO}$) and carbondioxide ($\ce{CO2}$). Like Uncle Al stated there are also suboxides known, which are usually byproducts of incomplete combustion (if not targeted explicitly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all that, if you burn coal, the following main reaction will happen (&lt;strong&gt;1&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C + O2 -&amp;gt; CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, given the right conditions (excess carbon) also carbonmonoxide  may be formed (sum of reaction, &lt;strong&gt;2&lt;/strong&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C + O2 -&amp;gt; 2CO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction itself will detour via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boudouard_reaction&quot; rel=&quot;noreferrer&quot;&gt;Boudouard reaction&lt;/a&gt;, which is very important in &lt;a href=&quot;http://en.wikipedia.org/wiki/Blast_furnace&quot; rel=&quot;noreferrer&quot;&gt;blast furnace&lt;/a&gt; processes. First forming carbondioxide via &lt;strong&gt;2&lt;/strong&gt; and then convertin excess carbon to carbonmonoxide via &lt;strong&gt;3&lt;/strong&gt;:&#xA;$$\ce{C + CO2 &amp;lt;=&amp;gt; 2CO}$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="3735" LastEditDate="2015-05-08T13:21:17.913" LastActivityDate="2015-05-08T13:21:17.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9645" PostTypeId="1" AcceptedAnswerId="9680" CreationDate="2014-04-01T10:10:46.807" Score="0" ViewCount="4471" Body="&lt;p&gt;I know, by nature, normally atoms are stable. But I wonder, if it is possible to compress an atom, say the Hydrogen atom, to infinite density by applying external force? &lt;em&gt;After all, electrons and protons would attract each other by electrostatic forces and nuclear forces are also there at least up-to a certain distance...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has it been done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I thought of it, was because I was trying to find the ratio of the amount vacuum to the amount of matter in the whole universe...&lt;/p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="4734" LastEditDate="2014-04-03T11:45:04.653" LastActivityDate="2017-12-20T17:34:47.383" Title="Is it possible to compress an atom to infinite density?" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;nuclear&gt;&lt;atomic-radius&gt;&lt;protons&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ClosedDate="2017-12-21T02:02:14.303" ContentLicense="CC BY-SA 3.0" />
  <row Id="9646" PostTypeId="2" ParentId="9055" CreationDate="2014-04-01T10:27:05.917" Score="0" Body="&lt;p&gt;when a thin lead sheet is hammered, its internal energy is increased. this is because when we are hammering it, we are providing energy to the system which results in the increase of the energy of the system. :-)&lt;/p&gt;&#xA;" OwnerUserId="5043" LastActivityDate="2014-04-01T10:27:05.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9647" PostTypeId="2" ParentId="9634" CreationDate="2014-04-01T10:49:16.800" Score="2" Body="&lt;p&gt;In fact there are many types of matches so you should be aware of what you are using. In home-made experiments the problem is that many times you don't know the composition or the concentration of your reactants.&lt;/p&gt;&#xA;&lt;p&gt;Take a look a &lt;a href=&quot;http://books.google.it/books?id=370UwG8CuNwC&amp;amp;pg=PR7&amp;amp;dq=match+composition&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=JIM6U5ibNKaR5ASd1oGQAQ&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q=match%20composition&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Preparatory Manual of Black Powder and Pyrotechnics&#xA;by Jared Ledgard&lt;/a&gt; for understand the composition of your matches.&lt;/p&gt;&#xA;&lt;p&gt;We can assume that you have 20-50% of potassium chlorate and if present 1-15% of sulfur. I don't know the weight of a match's head (if someone have a laboratory balance you could determine it) however I think surely it weights no more then 4 mg. So in 50 pics matches box you have max 0,2 g of heads matches. So 0,1 g of potassium chlorate and 0.06 g of sulfur.&lt;/p&gt;&#xA;&lt;h1&gt;Hydrogen sulfide&lt;/h1&gt;&#xA;&lt;p&gt;Hydrogen sulfide is toxic and so can be dangerous but assuming that all the sulfur became hydrogen sulfide you have &lt;span class=&quot;math-container&quot;&gt;$0.06g_{Sulfur}~/ 32 \frac{g}{mol}=0.001875~mol$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$H_{2}S$&lt;/span&gt; hence &lt;span class=&quot;math-container&quot;&gt;$$0.001875~mol \times 34.08\frac{g}{mol}= 0.0639 g$$&lt;/span&gt; of Hydrogen sulfide. So if you work in a room of &lt;span class=&quot;math-container&quot;&gt;$4~m \times 4~m \times 2~m $&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$ 32 m^{3}$&lt;/span&gt; you have 1.9 &lt;span class=&quot;math-container&quot;&gt;$\frac{mg}{m^{3}}$&lt;/span&gt; of hydrogen sulfide according to &lt;a href=&quot;http://www.epa.gov/ttn/atw/hlthef/chlorine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPA&lt;/a&gt; is above the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lowest-observed-adverse-effect_level&quot; rel=&quot;nofollow noreferrer&quot;&gt;LOAEL (HEC)&lt;/a&gt;: 1.9 &lt;span class=&quot;math-container&quot;&gt;$\frac{mg}{m^{3}}$&lt;/span&gt; dose. However I hope that you will to the experiment in a well ventilated ambient. In this condition I think that is safe, however I don't think is a good experiment to perform.&lt;/p&gt;&#xA;&lt;h1&gt;Chlorine&lt;/h1&gt;&#xA;&lt;p&gt;Assuming that the chlorine in potassium chlorate is release as chlorine with a 100% yield (I'm quite sure that this can't happen) you have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.1g_{KClO_{3}} / 122.5495 \frac{g}{mol}=0.00082 mol_{KClO_{3}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hence, 0.00041 mol of &lt;span class=&quot;math-container&quot;&gt;$Cl_{2}$&lt;/span&gt; and so if 70.9 is the molecular weight 0.029069 g of chlorine.&lt;/p&gt;&#xA;&lt;p&gt;So in a &lt;span class=&quot;math-container&quot;&gt;$32 m^{3}$&lt;/span&gt; &amp;quot;room&amp;quot; you have 0.90625 &lt;span class=&quot;math-container&quot;&gt;$\frac{mg}{m^{3}}$&lt;/span&gt; of chlorine according to &lt;a href=&quot;http://www.epa.gov/ttn/atw/hlthef/chlorine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPA&lt;/a&gt; is below the LOAEL 1.2&lt;span class=&quot;math-container&quot;&gt;$\frac{mg}{m^{3}}$&lt;/span&gt; dose.&lt;/p&gt;&#xA;&lt;p&gt;P.S. Up to my knowledge &lt;span class=&quot;math-container&quot;&gt;$\ce{Sb2S3}$&lt;/span&gt; is no more used.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-05-25T00:39:02.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9649" PostTypeId="2" ParentId="9641" CreationDate="2014-04-01T13:36:38.813" Score="1" Body="&lt;p&gt;Not all the matches have sulfur but only some types in &lt;a href=&quot;http://books.google.it/books?id=370UwG8CuNwC&amp;amp;pg=PR7&amp;amp;dq=match+composition&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=JIM6U5ibNKaR5ASd1oGQAQ&amp;amp;ved=0CDIQ6AEwAA#v=onepage&amp;amp;q=match%20composition&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Preparatory Manual of Black Powder and Pyrotechnics&#xA;by Jared Ledgard&lt;/a&gt; see page 191, you can find the best matches for your needs, classic matches are okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right, no reaction is reported between sulfur and ammonia. However if you put in a closed box a compound containing sulfur and a little bit of organic matter &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur-reducing_bacteria&quot; rel=&quot;nofollow&quot;&gt;Sulfur-reducing bacteria (SRB)&lt;/a&gt; reduce sulfur in sulfide anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{S + 2e- -&amp;gt; S^{2-}}$$&#xA;$S^{2-}$ is a base and reacts with water to produce hydrogen sulfide:&#xA;$$\ce{S^{2-} +2 H2O -&amp;gt; H2S + 2OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_{2}S$ is a conjugate acid of $S^{2-}$ so now can react with ammonia with an acid-base reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2S + NH3 &amp;lt;=&amp;gt; NH4SH}$$&#xA;or:&#xA;$$\ce{H2S + 2NH3 &amp;lt;=&amp;gt; (NH4)2S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course like many ammonium salts it can easily decompose into the two odorous gasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others odorous substances are  skatoles, mercaptans, indoles however I don't think that these are present in great quantities in this case. See this &lt;a href=&quot;http://www.alken-murray.com/H2SREM3.HTM&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; for further informations.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-04-03T09:35:12.733" LastActivityDate="2014-04-03T09:35:12.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9650" PostTypeId="1" CreationDate="2014-04-01T14:39:28.873" Score="3" ViewCount="813" Body="&lt;p&gt;I have some &lt;a href=&quot;http://en.wikipedia.org/wiki/Plastic_optical_fiber&quot; rel=&quot;nofollow&quot;&gt;plastic optical fiber&lt;/a&gt;, which consists of a thin core of PMMA (polymethyl methacrylate), covered by a fluorinated polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to heat a bundle of these fibers to ~130 degrees, in order to form a simple &lt;a href=&quot;http://i-fiberoptics.com/pof-image-guide.php&quot; rel=&quot;nofollow&quot;&gt;image guide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I won't start until I know this is safe. Can you help me find out what types of gases I may encounter by heating these materials? Anything known dangerous? What other safety precautions should I take?&lt;/p&gt;&#xA;" OwnerUserId="5045" LastActivityDate="2014-10-05T22:36:04.977" Title="Safety concerns - Heating PMMA (acrylic glass) and fluorinated polymers" Tags="&lt;safety&gt;&lt;polymers&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9651" PostTypeId="2" ParentId="9650" CreationDate="2014-04-01T14:54:33.947" Score="0" Body="&lt;p&gt;PMMA melts at 160 °C and ignites in air at 460 °C, forming CO and formaldehyde as toxic gases when burning. Since you are working with a much lower temperature, PMMA should be safe. What type of fluorinated polymer do you want to use?&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-01T14:54:33.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9652" PostTypeId="1" AcceptedAnswerId="9658" CreationDate="2014-04-01T15:08:53.897" Score="0" ViewCount="109" Body="&lt;p&gt;I know that the mass of Helium-4 is slightly less than the mass of 2x Deuterium it may be made of and that it is the reason fusion gives a lot of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where exactly is this missing mass taken from? Aren't those 2 protons, 2 neutrons and 2 electrons forming Helium-4 atom the same as they were in previous Deuterium atoms?&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2014-04-02T14:59:13.247" Title="What exactly happens to mass during deuterium fusion?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9653" PostTypeId="1" AcceptedAnswerId="9654" CreationDate="2014-04-01T15:27:22.390" Score="1" ViewCount="74" Body="&lt;p&gt;I want to use syringe pump to pump water at 0.82 ml/min for 30 min ? Is that possible ( I read some paper use that but don't know how ?)&lt;/p&gt;&#xA;" OwnerUserId="5047" LastEditorUserId="4945" LastEditDate="2015-10-23T17:16:52.637" LastActivityDate="2015-10-23T17:16:52.637" Title="Is there any way to use syringe pumpe in pumping flow" Tags="&lt;physical-chemistry&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9654" PostTypeId="2" ParentId="9653" CreationDate="2014-04-01T16:40:24.070" Score="0" Body="&lt;p&gt;This should be no problem when you use a syringe pump model with adjustable flow rate and sufficient volume (at least 30 ml).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-01T16:40:24.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9655" PostTypeId="2" ParentId="2660" CreationDate="2014-04-01T17:49:36.333" Score="9" Body="&lt;p&gt;The Madelung energy ordering rule gives the energy of the orbitals approximately:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \mathrm{1s &amp;lt; 2s &amp;lt; 2p &amp;lt; 3s &amp;lt; 3p &amp;lt; 4s &amp;lt; 3d &amp;lt; 4p &amp;lt; 5s &amp;lt; 4d &amp;lt; &#xA;5p &amp;lt; 6s &amp;lt; 4f &amp;lt; 5d &amp;lt; 6p &amp;lt; 7s &amp;lt; 5f &amp;lt; 6d &amp;lt; 7p}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;That would speak for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$&lt;/span&gt; if you follow the Aufbau principle (from German Aufbau = setup). But &lt;span class=&quot;math-container&quot;&gt;$\ce{Pt}$&lt;/span&gt; is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of &lt;span class=&quot;math-container&quot;&gt;$\ce{Pt}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.&lt;/p&gt;&#xA;&lt;p&gt;Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.&lt;/p&gt;&#xA;" OwnerUserId="5048" LastEditorUserId="95340" LastEditDate="2021-01-15T23:30:30.193" LastActivityDate="2021-01-15T23:30:30.193" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9656" PostTypeId="1" AcceptedAnswerId="9660" CreationDate="2014-04-01T18:17:08.170" Score="0" ViewCount="327" Body="&lt;p&gt;Testing for contamination of the ever so popular &lt;a href=&quot;http://en.wikipedia.org/wiki/Glyphosate&quot; rel=&quot;nofollow&quot;&gt;glyphosate&lt;/a&gt; based herbicide &quot;Roundup&quot; in water is a service that many labs provide, one of them is the &lt;a href=&quot;http://www.microbeinotech.com/home.aspx&quot; rel=&quot;nofollow&quot;&gt;Microbe Inotec Laboratory&lt;/a&gt; . They use an enzyme to check for levels of of contamination in ppb, which, I'm sure, is rather a complicated process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there any way to test for it more easily, without finding the actual amount of glyphosate but only checking whether it's present or not?&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2014-04-01T22:11:47.087" Title="Test for glyphosate in water, boolean style" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9657" PostTypeId="2" ParentId="9611" CreationDate="2014-04-01T19:31:07.917" Score="1" Body="&lt;p&gt;The mistake is in the charge $q$, that you have put in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;D_A=\frac{\mu_{q}k_BT}{q}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ is simply the particle charge (and not the whole charge, as you've unfortunately considered) and should then be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=\frac{zF}{N_{A}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $F$ the Faraday constant (charge / mole ), ${N_{A}}$ the Avogadro's number (number of particle / mole) and $z$ the charge valence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${N_{A}}$ will finally cancel out when you divide Boltzmann's constant $k_B$ by the the charge $q$, in the definition of diffusion coefficient. That will allow you to recover the Goldman equation found on &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Goldman_equation#Derivation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By checking all that through an &lt;em&gt;always useful&lt;/em&gt; dimension analysis for $D_A$ and using the defintions you have given in your question for ($\mu_q$ = $\frac{v_d}{E} = \frac{v_d L}{V_m}$ and $k_B = \frac{R}{N_A}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\frac{\mu_{q}k_BT}{q}=\frac{m/s\,\times m}{V}\times \frac{J/(mol\times K)\,\times K}{{N_A}\times C/mol}\times{N_A}=\frac{m^2}{s}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you find the correct units for a diffusion coefficient.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-04-01T20:32:43.080" LastActivityDate="2014-04-01T20:32:43.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9658" PostTypeId="2" ParentId="9652" CreationDate="2014-04-01T20:06:23.820" Score="2" Body="&lt;p&gt;Fusion of H-2 does NOT give He-4.  It gives ~50:50 H-3 (1.01 MeV) + p (3.02 MeV) or He-3 (0.82 meV) + n (2.05 MeV).  Either way, the missing &quot;mass&quot; is the mass equivalent of binding energy, E/c^2.  Binding energy is &lt;em&gt;negative&lt;/em&gt;.  A bound system masses less than its separated constituent parts.  You must input energy (that being mass-equivalent) to unbind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion converts weakly bound nuclei into more strongly bound nuclei.  The net difference in binding energies comes out as the kinetic energies of the product nuclei and particles.  Count neutrons and protons going in versus neutrons and protons coming out.  If one convetrs to the other, there must be elecron neutrinos or electron antineutrinos emitted to balance the books for energy, momentum and lepton generation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nuclear_fusion&lt;/a&gt; &lt;BR&gt;&#xA;   Middle&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-02T14:59:13.247" LastActivityDate="2014-04-02T14:59:13.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9659" PostTypeId="2" ParentId="9642" CreationDate="2014-04-01T21:10:34.223" Score="4" Body="&lt;p&gt;Let's first look at ionic strength:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the strength of 0.02 M NaCl solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient. &#xA;$$\textrm{activity} = a_i = \gamma_i[i]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a typical equilibrium expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B &amp;lt;=&amp;gt; C + D} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha$ is the size of the hydrated ion in pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the negative log and you will get pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In context with the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma_{\ce{H+}} = 1.1243166 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="21330" LastEditDate="2015-09-27T17:25:22.797" LastActivityDate="2015-09-27T17:25:22.797" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9660" PostTypeId="2" ParentId="9656" CreationDate="2014-04-01T22:11:47.087" Score="1" Body="&lt;p&gt;The proposed method above most likely uses ELISA (enzyme-linked immunoabsorbent assay). In most modern labs, ELISA is quite a standard procedure, however, each assay is easily worth a couple hundred dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method of detection is using HPLC (high-performance liquid chromatography), however those machines can run upwards of a couple hundred thousand dollars and each test is expensive as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The EPA has a protocol which they used to detect glyophosphate contamination in water:&#xA;&lt;a href=&quot;https://www.bucksci.com/catalogs/EPA%20Method%20-%20547%20-%20GLYPHOSATE%20IN%20DRINKING%20water%20by%20isocratic%20HPLC%20with%20flourecense%20detector.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.bucksci.com/catalogs/EPA%20Method%20-%20547%20-%20GLYPHOSATE%20IN%20DRINKING%20water%20by%20isocratic%20HPLC%20with%20flourecense%20detector.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What may be of interest to you is an article from the journal of the Royal Chemical Society in which they published an article where they used gold nanoparticles to detect presence of glyphosphate using the naked eye or UV light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2013/AY/c2ay26391b#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;http://pubs.rsc.org/en/content/articlelanding/2013/AY/c2ay26391b#!divAbstract&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-04-01T22:11:47.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9661" PostTypeId="2" ParentId="9650" CreationDate="2014-04-01T22:55:01.650" Score="1" Body="&lt;p&gt;One problem is bonding of the bundle.  The contact surface is the fluororesin cladding that allows more nearly perpendicular rays to be totally internally reflected.  If you nick the cladding you get a light leak.  Will the cladding fuse when heated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ligthly bond a bunch of cylinders you get air gaps where they touch.  Those will be a triangular grid off dark spots, 9.3% of the area of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Circle_packing&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Circle_packing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xaCuo.jpg&quot; alt=&quot;circle close packing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-01T22:55:01.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9665" PostTypeId="1" AcceptedAnswerId="9667" CreationDate="2014-04-02T03:14:27.923" Score="5" ViewCount="9515" Body="&lt;p&gt;Chemistry says that we are are we supposed to heat boiling flasks, and not Erlenmeyer flasks.  However, I can't find any information about this on Google.  Is there a particular reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just because we want to separate the purposes for our glassware, or is there an actual valid reason?&lt;/p&gt;&#xA;" OwnerUserId="5057" LastActivityDate="2020-08-19T13:25:34.350" Title="Why are we not supposed to heat Erlenmyer flasks?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;equipment&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9666" PostTypeId="2" ParentId="9665" CreationDate="2014-04-02T03:19:58.627" Score="2" Body="&lt;p&gt;Since most Erlenmeyer flasks are made with Pyrex glass, there should be no problem heating them up. I have been told not to heat volumetric flasks since the volume could change. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-02T03:19:58.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9667" PostTypeId="2" ParentId="9665" CreationDate="2014-04-02T03:34:42.370" Score="9" Body="&lt;p&gt;This really depends on what you mean by heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main uses for Erlenmeyer flasks is for recrystallization, which of course involves heating the container and contents. Typically this is only moderate heating, such as over a steam bath or with a hair dryer. Some manufacturers specify that their Erlenmeyer flasks are &lt;a href=&quot;http://www.duran-group.com/en/products-solutions/laboratory-glassware/products/boiling-flasks-and-general-laboratory-glassware.html&quot;&gt;ideal for heating applications&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, boiling flasks have more uniform and thicker walls, which allow for much greater heating limits, and a more uniform heating distribution. They are designed to handle more intense localised heating, and manage thermal and physical shock distribution. Erlenmeyer flasks usually have thinner walls and greater wall thickness variation, and are less able to handle sudden thermal changes, such as heating with a high energy source like a Bunsen, or heat gun. Using them in this way may well lead to cracking, fracturing or worse, and is unsafe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, although Erlenmeyer flasks can be heated, they are not specifically designed for high temperature, constant heating applications. For this, you should use a boiling flask. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-04-02T03:34:42.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9669" PostTypeId="2" ParentId="3942" CreationDate="2014-04-02T04:47:35.887" Score="27" Body="&lt;p&gt;In the conversion of primary alcohols to primary amines &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-CH2OH -&amp;gt; R-CH2NH2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;direct alkylation of ammonia normally is the last thing you want to do in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent further alkylation of the initial product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More promising strategies are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Gabriel&lt;/em&gt; synthesis: Hydroxy-halogen-exchange, followed by reaction of the primary halide with phthalimide and subsequent hydrazinolysis&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Mitsunobu&lt;/em&gt; reaction ($\ce{Ph3P, DEAD}$) of the primary alcohol using hydrazoic acid $\ce{HN3}$ as a nucleophile, followed by reduction of the alkyl azide &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An (almost-one-pot) combination of the &lt;em&gt;Appel&lt;/em&gt; and the &lt;em&gt;Staudinger&lt;/em&gt; reaction&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat the primary alcohol ($\ce{R-CH2OH}$) with $\ce{Ph3P}$  and bromotrichloromethane $\ce{CBrCl3}$ in benzene, then cool down to RT (room temperature) to yield $\ce{R-CH2Hal}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add $\ce{NaN3}$,tetrabutylammonium bromide and &lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-dimethylformamide and reflux again to obtain the azide $\ce{R-CH2N3}$, then cool down to RT, pour on water, extract with benzene and dry the organic layer&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add triethylphosphite $\ce{(EtO)3P}$ to the organic layer, stir at RT to yield the iminophosphorane $\ce{R-CH2N=P(OEt)3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hydrolyze the iminophosphorane by refluxing with hydrochloric acid, then neutralize to release the amine&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-04-22T19:21:04.620" LastActivityDate="2016-04-22T19:21:04.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9670" PostTypeId="1" AcceptedAnswerId="9705" CreationDate="2014-04-02T07:21:22.730" Score="6" ViewCount="4835" Body="&lt;p&gt;For a fluid flow experiment I am using DMSO (dimethylsulfoxide), because of its low volatility, reasonably high surface tension, low viscosity and relative safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the MSDS of DMSO I find that the liquid is hygroscopic, meaning it adsorbs moisture from the surrounding environment. Unfortunately, I cannot find anywhere how much water will be absorbed (mass%) or how quickly this will happen, is it saturated in minutes, hours, days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my question is twofold: (i) how much water will DMSO at maximum absorb and what is the rate at which this absorption occurs? (ii) with the absorbed water, how badly do the physical properties (viscosity, surface tension, density) of the (now) mixture change?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2016-03-24T21:23:11.700" Title="Hygroscopic behaviour of DMSO - how bad is it?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9671" PostTypeId="2" ParentId="9670" CreationDate="2014-04-02T08:04:58.503" Score="3" Body="&lt;p&gt;I only used DMSO in other contexts and I have no idea how fast the water uptake is or how it changes the physical properties, but &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;DMSO is miscible with water in any ratio&lt;/li&gt;&#xA;&lt;li&gt;it does not form an azeotrop with water&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The typical protocol for drying is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Keep over night over anhydrous $\ce{CaSO4}$ or powdered $\ce{BaO}$&lt;/li&gt;&#xA;&lt;li&gt;Decant off&lt;/li&gt;&#xA;&lt;li&gt;Distill over $\ce{CaH2}$ (about 10 g/liter) at reduced pressure (bp is around 75 °C at 16 hPa)&lt;/li&gt;&#xA;&lt;li&gt;Store over 4 A molecular sieve in a dark and tightly closed bottle&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-02T09:00:07.887" LastActivityDate="2014-04-02T09:00:07.887" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9672" PostTypeId="1" AcceptedAnswerId="9682" CreationDate="2014-04-02T08:38:44.410" Score="3" ViewCount="9072" Body="&lt;p&gt;If I test a pure solution of an acid chloride with a pH-meter, will I get a reading of pH 7?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I add water to it, I know $\ce{HCl}$ gas will form, but this gas will dissolve in water to give an acidic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main confusion is that we call ethanoyl chloride an &quot;acid&quot; chloride, but is its pure solution acidic?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2014-07-07T07:45:54.263" LastActivityDate="2015-11-08T14:50:36.240" Title="Are acid chlorides really acidic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9673" PostTypeId="1" CreationDate="2014-04-02T08:59:28.087" Score="1" ViewCount="867" Body="&lt;p&gt;Evaporating cold aqueous chromium(III) chloride produces violet crystals. However when hot aqueous chromium(III) chloride crystallizes, green crystals result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain the chemistry behind this... I mean both the crystals contain $Cr^{3+}$, so the difference in colour cannot be due to two different oxidation states. Also, they contain the same ligand $Cl^-$. Can someone please tell me where I am going wrong?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2014-04-02T09:16:39.303" Title="Different coloured chromium salts" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9674" PostTypeId="2" ParentId="9672" CreationDate="2014-04-02T09:07:23.927" Score="-6" Body="&lt;p&gt;Is there such a thing as a completely pure solution ? Would there not always be some water contamination and hence would you not always have some protons floating around ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as a thought experiment I imagine that in a solution of 'pure' ethanoyl chloride you would have a pH of 7 as there are no free protons moving around in solution.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2014-04-02T09:07:23.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9675" PostTypeId="2" ParentId="9673" CreationDate="2014-04-02T09:16:39.303" Score="1" Body="&lt;p&gt;Well, actually when we evaporate aqueous chromium(III) chloride, the hydrate of the crystal will be gone, so we get anhydrous chromium(III) chloride ($\ce {CrCl3}$). This type of anhydrous produce violet color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we crystallizes aqueous of the chromium(III) chloride, the hydrate won't go, so we get hydrous chromium(III) chloride (usually hexahydrate, like $\ce {CrCl3.6H2O}$). This crystal produce dark green color.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-04-02T09:16:39.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9676" PostTypeId="1" AcceptedAnswerId="9678" CreationDate="2014-04-02T09:59:54.623" Score="2" ViewCount="370" Body="&lt;p&gt;Given a mixture of: a saturated solution of a solute in a solvent together with excess of the solute compound as crystalline material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were to change the bulk composition of the mixture by removing some of the solution could this cause dissolution or precipitation of the solid in the remaining mixture, or the concentration of the remaining solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably one should expect some crystallisation in the portion of the solution that is removed from the mixture?&lt;/p&gt;&#xA;" OwnerUserId="5061" LastActivityDate="2014-04-02T10:24:41.197" Title="Does removing solution affect a supersaturated solution" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solvents&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9677" PostTypeId="2" ParentId="9676" CreationDate="2014-04-02T10:23:45.453" Score="2" Body="&lt;p&gt;When a solution is saturated, the remaining solid is in equilibrium with the solvent:&#xA;$$\ce{AB&amp;lt;=&amp;gt;[solv.]A+_{(solv)} + B-_{(solv)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that there is as much of your solid solved. If you take out some of the solution then the concentration remains the same, in your original flask as well as in the portion you withdraw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove only solvent, i.e. by heating your solution or diffusion over time. In this withdrawn portion crystals will form. The concentration will not change, only the amount of substance $n$ in your solution will change as the volume shrinks ($c=\frac{n}{V}$).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-02T10:23:45.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9678" PostTypeId="2" ParentId="9676" CreationDate="2014-04-02T10:24:41.197" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If one were to change the bulk composition of the mixture by removing some of the solution [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This does not change the composition of the solution above the solid. Since the concentration of the solute remains constant, I do not expect a change.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Presumably one should expect some crystallisation in the portion of the solution that is removed from the mixture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It still is a saturated solution. Unless you evaporate the solvent (= increase the concentration of the solute) or change the temperature, nothing will change.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-02T10:24:41.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9679" PostTypeId="1" CreationDate="2014-04-02T10:50:27.567" Score="13" ViewCount="10769" Body="&lt;p&gt;Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?&lt;/p&gt;&#xA;" OwnerUserId="3678" LastEditorUserId="16683" LastEditDate="2017-02-27T14:47:59.147" LastActivityDate="2021-02-22T21:56:53.017" Title="Why does having equivalent resonance structures give more stability?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9680" PostTypeId="2" ParentId="9645" CreationDate="2014-04-02T12:07:26.090" Score="5" Body="&lt;p&gt;We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to construct a 'Planck density' by computing $10^{-5}$ grams / $\text{(} 10^{-33} \text{centimeters)}^{3} = 10^{94}$ grams/cc. Many physicists think that this represents the highest density for any natural compression mechanism, because at higher densities, space-time itself transmutes into some type of quantum mechanical system for which 'density' is not a meaningful concept. So, if fundamental particles are truly without any structure larger than the Planck scale, then they can be packed together into a 'system' with a density of $10^{94}$ grams; the highest achievable density for energy/mass in this physical universe. Even within black holes, the famous 'Singularity' may not have infinite density, but merely a finite density near the Planck value. This is how 'quantum gravity' is thought to regulate, or eliminate real physical singularities; something which ordinary general relativity is not able to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I read it in a book Elegant Universe, by Brian Greene. This book has a whole seperate chapter for this.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2014-04-02T12:07:26.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9681" PostTypeId="2" ParentId="9679" CreationDate="2014-04-02T13:25:41.747" Score="-2" Body="&lt;p&gt;You can estimate stability by the number of resonance structures you can draw. For the acetate ion, there are 2 resonance structures, while for the protonated form (acetic acid) only one structure is possible, which favors deprotonation. For the phenolate ion, you can formulate 4 structures, with the negative charge delocalized into the benzene ring. For phenol, you can draw 4 structures as well, although 3 of them have a positive formal charge at the oxygen. This is why phenol is weakly acidic, and quite acidic for an alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The content of this post can also help understanding the different acidities of phenol and acetic acid: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6155/why-is-ethanoic-acid-more-acidic-than-phenol?rq=1&quot;&gt;Why is ethanoic acid more acidic than phenol?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-04-06T20:37:58.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9682" PostTypeId="2" ParentId="9672" CreationDate="2014-04-02T14:23:58.943" Score="6" Body="&lt;p&gt;An acid chloride is a covalent molecule.  It has no pH.  If you put a pH electrode into a reactive acid chloride you will dehydrate and acylate its hydrated glass surface ($\ce{#Si-OH}$) and damage its future response when properly used.  pH electrodes are stored sitting in buffer or saline solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An acid chloride is an oxo-acid whose acidic $\ce{-OH}$ has been replaced with $\ce{-Cl}$.  Hydrolysis restores the oxo-acid, displacement makes derivatives. Hydrochloric acid is &lt;strong&gt;not&lt;/strong&gt; an acid chloride.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="7475" LastEditDate="2015-11-08T14:44:46.850" LastActivityDate="2015-11-08T14:44:46.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9684" PostTypeId="2" ParentId="9665" CreationDate="2014-04-02T15:07:39.870" Score="4" Body="&lt;p&gt;The conical shape of an erlenmeyer flask creates a sharp angle between the bottom and sides of the flask.  Stress, be it from heating, cooling, mechanical shock, etc., is not as uniformly distributed across such an angular structure as it is in a spherically shaped flask.  Hence, the erlenmeyer flask is more prone to cracking under stress than a spherical flask.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-02T15:07:39.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9685" PostTypeId="2" ParentId="9665" CreationDate="2014-04-02T15:23:09.213" Score="5" Body="&lt;p&gt;In &lt;a href=&quot;http://books.google.it/books?id=k4jjq5s6vZsC&amp;amp;pg=PA231&amp;amp;dq=should+not+heat+Erlenmeyer+flasks&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=aSc8U9PQM8TI4AT6moDIAw&amp;amp;ved=0CFUQ6AEwBA#v=onepage&amp;amp;q=conical%20flask&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Essential Environmental Science: Methods &amp;amp; Techniques&#xA;by Simon Watts,Lyndsay Halliwell&lt;/a&gt; I've found this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7fLR2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This because his conical shape does not allow an uniform expansion and so the glass could break. Furthermore if you are heating substances that will leave a residue will be very hard to clean it.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-04-02T15:23:09.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9686" PostTypeId="1" CreationDate="2014-04-02T17:10:11.837" Score="7" ViewCount="2060" Body="&lt;p&gt;I am thinking about the possibility of making a hat or other clothing item that might show up funny on cameras, as they are more sensitive to infrared light than human eyes.&#xA;Therefore, I was wondering if there are any (non hazardous) materials that emit InfraRed light when they are exposed to visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these exist, and would they be suited for this kind of application?&lt;/p&gt;&#xA;" OwnerUserId="4839" LastEditorUserId="5017" LastEditDate="2014-09-29T14:37:34.847" LastActivityDate="2014-09-29T14:37:34.847" Title="Do materials exist that absorb Visible light and emit Near InfraRed light?" Tags="&lt;photochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9687" PostTypeId="2" ParentId="9686" CreationDate="2014-04-02T17:30:57.257" Score="4" Body="&lt;p&gt;Yes, there are some minerals that shows VIS-NIR luminescence here a couple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Synthetic ruby $\ce{Al2O3:Cr}$&lt;/li&gt;&#xA;&lt;li&gt;Kyanite ($\ce{Al2SiO5}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.viewsfromscience.com/documents/webpages/led_fluorescence_p2.html&quot; rel=&quot;nofollow&quot;&gt;Reference here&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated material is $\ce{LiSrPO4:Eu2+, Pr3+}$ you can read the full article in &lt;a href=&quot;http://www.opticsinfobase.org/view_article.cfm?gotourl=http%3A%2F%2Fwww%2Eopticsinfobase%2Eorg%2FDirectPDFAccess%2F6312A969%2DB317%2DEA47%2D868E052ED760D779%5F248882%2Foe%2D21%2D3%2D3161%2Epdf%3Fda%3D1%26id%3D248882%26seq%3D0%26mobile%3Dno&amp;amp;org=&quot; rel=&quot;nofollow&quot;&gt;11 February 2013 / Vol. 21, No. 3 / OPTICS EXPRESS 3161&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some pigments like egyptian blue or cadmium yellow (CI pigment yellow 37, that is cadmium sulfide) that were also used in the past. Egyptian blue (calcium copper silicate ($\ce{CaCuSi4O10}$))  is safe, other pigments with cadmium are more toxic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most spread pigment shows VIS-NIR luminescence: &lt;strong&gt;Chlorophyll&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look to &lt;a href=&quot;http://chsopensource.org/2013/04/29/infrared-fluorescence-photography-review-2/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; for a nice overview. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All the luminesce processes have not a high quantum yield, so to record them, most of the times, you have to avoid reflection. Tungsten, halogen lamps and sun emit a lot of NIR radiation so is possible that reflected light overhang the luminescence. When I performed NIR luminescence I used VIS LEDs and a long pass filter at 850 nm.  If you don't have result, first thing you have to try is to remove the ICF (IR-Cut off-Filter) filter from your camera so NIR light can reach the sensor more easily.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-04-02T19:04:14.797" LastActivityDate="2014-04-02T19:04:14.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9688" PostTypeId="2" ParentId="9686" CreationDate="2014-04-02T17:54:35.717" Score="4" Body="&lt;p&gt;NIR-emitting long-persistent luminescent nanoparticles have been prepared from $\ce{Cr(III)}$- and $\ce{Pr(III)}$-doped gallogermanates. According to the authors, these materials exhibit very long afterglows up to 360h (= 15 days). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://dx.doi.org/10.1038/nmat3173&quot; rel=&quot;nofollow&quot;&gt;DOI1&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1021/ja404243v&quot; rel=&quot;nofollow&quot;&gt;DOI2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-02T17:54:35.717" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9689" PostTypeId="2" ParentId="9686" CreationDate="2014-04-02T18:26:08.000" Score="4" Body="&lt;p&gt;CCD cameras typically have an IR-absorbing filter.  Short wavelength scattering (varies as inverse fourth power of wavelength) renders white cloth opaque, cellulose versus air refractive index.  If the scattering is quenched by wetting or avoided by IR-viewing, transparency results.  The champion visible-absorber, IR-fluorescent mateiral is chlorophyll.  Military camouflage extends into the IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IR-fluorescent dyes are common as biochemistry tags,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bruker.com/products/mr/nmr/avance-iii-hd/overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bruker.com/products/mr/nmr/avance-iii-hd/overview.html&lt;/a&gt; &lt;BR&gt; &#xA;&lt;a href=&quot;http://www.sigmaaldrich.com/life-science/cell-biology/antibodies/cf-dyes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sigmaaldrich.com/life-science/cell-biology/antibodies/cf-dyes.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8rNx2.png&quot; alt=&quot;IR fluorescent dyes&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-02T18:26:08.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9690" PostTypeId="2" ParentId="3959" CreationDate="2014-04-02T19:59:12.580" Score="2" Body="&lt;p&gt;Well, in fact, the &lt;strong&gt;alpha-proton of aldehyde&lt;/strong&gt; and other carbonyl/carboxylic acids are acidic, across a large range. The alpha-proton has a chance to be &lt;strong&gt;deprotonated&lt;/strong&gt; by organolithium and magnesium reagents and forms a so called &quot;&lt;strong&gt;enolate&lt;/strong&gt;&quot;, which undergoes &quot;&lt;strong&gt;aldol condensation&lt;/strong&gt;&quot; reaction with residual aldehyde. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some knowledge derived from practice but not covered in text book is, if alkyl lithium reagent is added to an enolizable aldehyde, the &lt;strong&gt;nucleophilic additon&lt;/strong&gt; and &lt;strong&gt;enolization-aldol condensation&lt;/strong&gt; of aldehyde would take place together, and the product will be a mixture of alcohol and aldol condensation in various ratio depending on substrates. Some aldehydes, e.g. phenylacetaldehyde, provides aldol condensation product in major due to their high enolizabization property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg/Na/Li alkoxides and amides also can deprotonate the alpha-proton of carbonyl as well as attacking certain certain carbonyls, like carboxylic ester/amide and nitrile. The carbonyl of aldehydes and ketone are not attacked.&lt;/p&gt;&#xA;" OwnerUserId="5064" LastActivityDate="2014-04-02T19:59:12.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9692" PostTypeId="1" CreationDate="2014-04-02T20:30:01.683" Score="3" ViewCount="1249" Body="&lt;p&gt;How can we identify bases and acids by their chemical formulas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example this reaction. I know that HNO3 is an acid and KOH is a base, but if I didn't know that how would I be able to identify that using the chemical formulas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HNO3(aq)	+	KOH(aq)	\rightarrow	KNO3(aq)	+	H2O(l)}$&lt;/p&gt;&#xA;" OwnerUserId="4583" LastEditorUserId="1362" LastEditDate="2014-04-02T20:58:11.400" LastActivityDate="2018-03-04T02:56:03.930" Title="Chemical Formulas of Acids and Bases" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2014-04-03T09:28:36.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="9693" PostTypeId="1" CreationDate="2014-04-02T20:37:13.117" Score="-1" ViewCount="10455" Body="&lt;p&gt;What type of double displacement reactions can form a gas. How can i determine if a double displacement reaction will form a gas?&lt;/p&gt;&#xA;" OwnerUserId="4583" LastActivityDate="2014-04-03T10:00:15.980" Title="Double Displacement Reactions that produce gas" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-04-03T17:09:34.667" ContentLicense="CC BY-SA 3.0" />
  <row Id="9694" PostTypeId="2" ParentId="9692" CreationDate="2014-04-02T20:53:43.950" Score="0" Body="&lt;p&gt;Look at The Periodic Table.  Metal (Group I and II) hydroxides are bases, non-metal &quot;hydroxides&quot; are acids (especially with high positive charges on the non-metal.)  Then we have amphoteric hydroxides like zinc and aluminum that afford zincates and aluminates in very alkaline media.  $\ce{Mn(OH)3}$ is a base but $\ce{HOMnO3}$ is a strong acid.  Look at core oxidation state, charge polarization, and oxophilicity of that ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arrhenius, Brønsted-Lowry, Lewis; hard, soft; solvo-cation, solvo-anion.  How do you classify $\ce{Al(OMe3)}$ that is a Brønsted base for methoxide and a Lewis acid for aluminum(III)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4}{^+}$ $\ce{NH3}$ $\ce{NH2}{^-}$ $\ce{NH}{^{2-}}$  $\ce{N}{^{3-}}$.  Ammonium is definitely the acid.  Nitride is definitely the base.  What are the species in-between?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-03-04T02:56:03.930" LastActivityDate="2018-03-04T02:56:03.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9695" PostTypeId="2" ParentId="9692" CreationDate="2014-04-02T20:57:52.610" Score="1" Body="&lt;p&gt;I'm going to assume you are in high school, thus talking about &lt;a href=&quot;http://en.wikipedia.org/wiki/Br%C3%B8nsted%E2%80%93Lowry_acid%E2%80%93base_theory&quot; rel=&quot;nofollow&quot;&gt;Brønsted acids/bases&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Brønsted acid is a compound that can donate a proton $\ce{H^+}$ whereas the base can accept a proton. For some chemicals it can be easily seen that they can donate a proton, like in the example of $\ce{HNO3}$ you give. There is obviously a proton in there so it makes sense that it is an acid. For chemicals with $\ce{O^-}$ in their structure (like the $\ce{KOH}$ in your example which dissociates into $\ce{K^+}$ and $\ce{OH^-}$) it is also clear that they can accept a proton and thus are obviously a base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the difficult part is when a molecule can do both: these are called amphiprotic substances, the most famous being water $\ce{H2O}$, but there are many more of these, think about $\ce{H2PO4^-}$, it both can accept and donate a proton. In these situations molecules can indeed be considered both a base and an acid and the way they behave depends on the other molecules they come into contact with. If the other chemical is a stronger acid, as measured by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_dissociation_constant&quot; rel=&quot;nofollow&quot;&gt;acid dissociation constant&lt;/a&gt; then the current molecule will act as base and vice versa.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="3735" LastEditDate="2014-04-02T22:16:04.660" LastActivityDate="2014-04-02T22:16:04.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9696" PostTypeId="2" ParentId="3959" CreationDate="2014-04-02T21:07:43.650" Score="1" Body="&lt;p&gt;Butyllithium is a powerful nucleophile much faster than it is a strong base.  Add two equivalents of dry N,N,N',N'-tetramethylethylenediamine to coordinate the lithium.  Now you have an extraordinary base (it will deprotonate THF at room temp) but a very poor nucleophile.  One then presumes that nucleophilic attack upon a carbonyl may be a four-center intermediate in which C-O-Li coordination sets up and directs attack of the alkyl anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BuLi plus Li &lt;em&gt;tert&lt;/em&gt;-butoxide is a strong enough base to initiate anionic polymerizations.  The alkyllithium cage polymer is disaggregated. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-02T21:07:43.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9698" PostTypeId="1" AcceptedAnswerId="9704" CreationDate="2014-04-02T21:56:17.520" Score="4" ViewCount="2695" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  How many milliliters of $0.202\ \mathrm M$ NaOH should be added to $25.0\ \mathrm{mL}$ of $0.0233\ \mathrm M$ salicylic acid to adjust the $\mathrm{pH}$ to $3.50$?&lt;br&gt;&#xA;  The answer from the text book is $2.2\ \mathrm{mL}$ NaOH&#xA;  &lt;strong&gt;My attempt&lt;/strong&gt;&#xA;  I believe the correct equation to use for this is the Henderson–Hasselbalch equation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH}=\mathrm pK_\mathrm a + \log\left(\frac{[\ce B]}{[\ce{BH}]}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm pK_\mathrm a = 2.972$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I set it up as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3.50=2.972+\log\left(\frac{x}{0.0233}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $x$ I get $0.0785$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where to go from here&#xA;since $x$ is the concentration I figure that if I multiply by $0.025$ (the volume) and then divide by the molarity ($0.202$) I would get the answer but that's not right.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2018-03-08T02:10:58.313" LastActivityDate="2018-03-08T02:10:58.313" Title="Adjusting the ph of salicylic acid using sodium hydroxide" Tags="&lt;equilibrium&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9699" PostTypeId="2" ParentId="9693" CreationDate="2014-04-02T22:15:24.233" Score="1" Body="&lt;p&gt;Reacting acids with carbonates will produce carbon dioxide gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3(s) + 2 HCl(aq) -&amp;gt; CaCl2(aq) + CO2(g) + H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With double replacement reactions you generally either get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) A neutralization reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl(aq) + NaOH(aq) -&amp;gt; NaCl (aq) + H2O(l)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Gas displacement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3(s) + 2 HCl(aq) -&amp;gt; CaCl2(aq) + CO2(g) + H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Precipitation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCl(aq) + AgNO3 (aq) -&amp;gt; NaNO3(aq) + AgCl(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is important to know your solubility rules.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-04-02T22:15:24.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9701" PostTypeId="1" AcceptedAnswerId="9702" CreationDate="2014-04-03T02:07:19.133" Score="1" ViewCount="374" Body="&lt;p&gt;In the reaction $\ce{E(g) &amp;lt;=&amp;gt; P(g)}$ at 25 °C, equilibrium is reached when the pressure of the product is 0.100 that of the reactant. What is $\Delta G$ in joules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\Delta G = RT\ln K$, my answer is $-8.31 \times 298 \times \ln (1.00) \ \text{J}$.&#xA;However, the answer is $-8.31 \times 298 \times \ln (0.100) \ \text{J}$. Wouldn't this mean that $\Delta G &amp;gt; 0$, therefore favoring the reactants? (As an aside, why would the reaction even be shifted to the left since it has a higher pressure?)&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorDisplayName="user7951" LastEditDate="2015-05-04T15:30:17.260" LastActivityDate="2015-05-04T15:30:17.260" Title="Why isn't ΔG equal to zero for this equilibrium reaction?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9702" PostTypeId="2" ParentId="9701" CreationDate="2014-04-03T03:52:19.063" Score="2" Body="&lt;p&gt;You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium.&#xA;And this quantity can very well be different from zero.&#xA;For non-standard conditions the Gibbs free energy of reaction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta G = \Delta G^{0} + RT \log K&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a derivation see &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/189&quot;&gt;this answer&lt;/a&gt; of mine.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-04-03T04:50:23.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9704" PostTypeId="2" ParentId="9698" CreationDate="2014-04-03T07:17:40.353" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; is the correct equation, but ask yourself what you are putting in there:&#xA;$$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $A^-$ and $HA$ occupy the same volume, you can rewrite:&#xA;$$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When adding Sodium Hydroxide consider the following equation&#xA;$$\ce{AH + NaOH &amp;lt;=&amp;gt; A- + H2O + Na+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you will find &#xA;$$ n(A^-) = n(\ce{NaOH})\\&#xA;   n(HA) = n_0(HA)-n(\ce{NaOH})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can substitute and solve for $n(\ce{NaOH})$:&#xA;$$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then divide by the concentration, yielding your volume.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intermediate results:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial amount of $HA$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;     $n_0(HA) = 0.5825\ \text{mmol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Amount of $\ce{NaOH}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final Volume:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-03T07:17:40.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9705" PostTypeId="2" ParentId="9670" CreationDate="2014-04-03T09:47:28.143" Score="9" Body="&lt;p&gt;Being curious about the hygroscopicity I have tried the experiment using pure, dried DMSO in our lab where the temperature was $21^\circ \text{C}$ and the relative humidity about $60\%$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the two images below. This is a DMSO droplet originally with $V=50\; \mu\text{L}$. You can clearly see that the volume of the droplet has increased substantially in 20 minutes from the different height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hBDYN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a crude approximation, assuming that both shapes are caps of a sphere, I calculated that the change in height from 100 px, to 112 px is equivalent to a volume increase of $15\%$, which is much too big for an accurate flow experiment. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-04-03T09:47:28.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9706" PostTypeId="2" ParentId="9693" CreationDate="2014-04-03T10:00:15.980" Score="1" Body="&lt;p&gt;Do redox reactions count? What about a some gas other than carbon dioxide then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition metal halides are reduced by $\ce{CN-}$ and form cyano complexes, while a part of the cyanide is oxidized and released as dicyan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 CuCl2 + 10 KCN -&amp;gt; 2 K3[Cu(CN)4] + (CN)2 ^  + 4 KCl}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AuCl3 + 4 KCN -&amp;gt;  K[Au(CN)2] + (CN)2 ^ + 3 KCl}$ &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-03T10:00:15.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9707" PostTypeId="2" ParentId="6876" CreationDate="2014-04-03T13:16:12.570" Score="1" Body="&lt;p&gt;To address your second question, fluorine and iodine are not considered for this reaction because the first is too reactive while the latter is too unreactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorine will react harshly with alkanes, producing an unselective mixture of fluorocarbons. The chain propagation steps of the reaction are so exothermic that it proceeds explosively, and special measures are necessary to control the reaction (e.g. low reactant concentrations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, iodine will react very slowly, if at all. Although the activation energy for the reaction (+151 kJ/mol) is the lowest of all halogens, the first chain propagation step (+142 kJ/mol) and the overall reaction enthalpy (+54 kJ/mol) are very endothermic, which disfavors radical iodination of alkanes (&lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/radikale/halogenierung_alkane.vlu.html&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-11-29T20:27:38.367" LastActivityDate="2014-11-29T20:27:38.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9708" PostTypeId="1" AcceptedAnswerId="9709" CreationDate="2014-04-03T14:07:30.660" Score="1" ViewCount="182" Body="&lt;p&gt;I have a problem to visualize the following assignment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum distance between $\ce{Rb^+}$ and $\ce{I^-}$ ions if radius of  $\ce{Rb^+}$ is ${1.49\cdot10^{-10}}~\mathrm{m}$ and radius of $\ce{I^-}$ is ${2.17\cdot10^{-10}}~\mathrm{m}$ if they are arranged in centered cube?&lt;/p&gt;&#xA;" OwnerUserId="5070" LastEditorUserId="4945" LastEditDate="2014-05-21T02:53:48.657" LastActivityDate="2014-05-21T13:58:07.123" Title="What is the minimum distance of Rubidium and Iod ions in a centered cube?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9709" PostTypeId="2" ParentId="9708" CreationDate="2014-04-03T14:27:12.550" Score="1" Body="&lt;p&gt;The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-21T13:58:07.123" LastActivityDate="2014-05-21T13:58:07.123" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9710" PostTypeId="1" CreationDate="2014-04-03T14:34:30.190" Score="-1" ViewCount="194" Body="&lt;p&gt;This problem is taken from the Indonesian Chemical Olympiad.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A sample containing barium peroxide was added to a glass full of sulfuric acid. It produced an insoluble product and its filtrate. The filtrate was reacted with potassium iodide in acidic conditions. The product was titrated with &lt;span class=&quot;math-container&quot;&gt;$11.8 \,\mathrm{mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$0.1 \,\mathrm{M}$&lt;/span&gt; sodium thiosulfate.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Write the reaction above.&lt;/li&gt;&#xA;&lt;li&gt;What is the mass of the barium peroxide in the sample?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Actually I know if the first reaction will make H2O2, but when this peroxide reacted with potassium iodide, I'm confused what is the product.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-03-29T08:52:16.900" Title="Indonesian Chemical Olympiad: Calculating mass from predicted reaction" Tags="&lt;redox&gt;&lt;stoichiometry&gt;&lt;titration&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9711" PostTypeId="2" ParentId="9710" CreationDate="2014-04-03T15:07:15.790" Score="3" Body="&lt;p&gt;In sulfuric acid, bariumperoxide generates hydrogen peroxide and barium sulfate (= precipitate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BaO2 + H2SO4 -&amp;gt; H2O2 + BaSO4 v}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under acidic conditions, iodide is oxidized to iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O2 + 2I- + 2H3O+ -&amp;gt; I2 + 4H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an excess of iodide, triodide is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{I- + I2 -&amp;gt; I3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of iodine (in triodide) is determined by titration with thiosulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{I3- + 2 S2O3^2- -&amp;gt; S4O6^2- + 3 I-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave the math for you :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2016-03-11T22:23:12.647" LastActivityDate="2016-03-11T22:23:12.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9713" PostTypeId="1" AcceptedAnswerId="9715" CreationDate="2014-04-03T16:03:30.760" Score="5" ViewCount="2782" Body="&lt;p&gt;Typical filter paper used in chemistry lab (e.g. Whatman #1) exhibits autofluorescence in near UV with excitation maximum at 365 nm and emission maximum at 440 nm. It was reported in literature&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0039914002005751&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; that fluorophores from filter paper can be extracted by PBS buffer in wide range of pH, and thus they are at least partially soluble in water. However, cellulose, the major component of filter paper, is not fluorescent in this wavelengths range and is insoluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the source of filter paper fluorescence?&lt;/strong&gt; I assume it comes from paper pretreatment and bleaching, but what is chemical identity of fluorescent compounds in it? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0039914002005751&quot; rel=&quot;nofollow&quot;&gt;Danielson, T. L. &lt;em&gt;et al.&lt;/em&gt;; Impact of filter paper on fluorescence measurements of buffered saline filtrates. &lt;em&gt;Talanta&lt;/em&gt; &lt;strong&gt;2003&lt;/strong&gt;, &lt;em&gt;59&lt;/em&gt;, 601–604.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="54" LastEditorUserId="4945" LastEditDate="2014-07-29T10:57:22.337" LastActivityDate="2018-07-03T13:35:08.460" Title="Why is filter paper fluorescent?" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9715" PostTypeId="2" ParentId="9713" CreationDate="2014-04-03T16:42:14.653" Score="9" Body="&lt;p&gt;In the production of white paper, the bleached pulp is often treated with optical brightening additives (OBA), that absorb in the uv and emit in the blue range. If I remember correctly, a typical example is 4,4'-diaminostilbene-2,2'-disulfonic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KUg9P.png&quot; alt=&quot;optical brightener&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an article from &lt;a href=&quot;http://dx.doi.org/10.2174/1876520300902010053&quot; rel=&quot;noreferrer&quot;&gt;The Open Textile journal&lt;/a&gt; (OA) in which the compound mentioned above is used as a starting material in the synthesis of another optical brightener for textiles.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-03T17:06:10.210" LastActivityDate="2014-04-03T17:06:10.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9716" PostTypeId="1" AcceptedAnswerId="9717" CreationDate="2014-04-03T17:31:22.240" Score="3" ViewCount="1467" Body="&lt;p&gt;I know what demethylation is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what &lt;strong&gt;mono-&lt;/strong&gt; means, the N is nitrogen of course, I know that from the line structure image in front of me if I could not guess it - and after quite a bit of google searching I gave up and came here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry is not my field but out of curiosity and for no practical purpose whatsoever I started learning a few things outside my field, and I came across this tiny little naming issue. I thought this is so simple I could easily find something, but apparently it is even MORE simple than I thought - meaning everybody uses the term but nobody bothers to explain it. :-(&lt;/p&gt;&#xA;" OwnerUserId="5022" LastEditorUserId="16683" LastEditDate="2015-10-09T15:53:11.597" LastActivityDate="2016-12-22T16:26:13.620" Title="Meaning of &quot;mono-N-demethylation&quot;" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9717" PostTypeId="2" ParentId="9716" CreationDate="2014-04-03T17:41:36.780" Score="7" Body="&lt;p&gt;Mono-&lt;em&gt;N&lt;/em&gt;-demethylation means the removal of a single (mono) methyl group from a nitrogen(&lt;em&gt;N&lt;/em&gt;)-containing functional group. For example, a reaction which transforms the alkylated amine group $\ce{-NHCH3}$ into $\ce{-NH2}$.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorDisplayName="user7951" LastEditDate="2016-12-22T16:26:13.620" LastActivityDate="2016-12-22T16:26:13.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9718" PostTypeId="2" ParentId="9716" CreationDate="2014-04-03T17:44:38.373" Score="4" Body="&lt;p&gt;You might know the demethylation from removing a methyl group from an oxygen atom. A methoxy group $\ce{-OCH3}$ is converted to a hydroxy group $\ce{-OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is correctly called a $\ce{O}$-demethylation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{N}$-demethylation is the same - for a nitrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;mono&lt;/em&gt; prefix denotes that only one (of several) methyl groups is removed from a nitrogen atom.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-03T17:44:38.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9719" PostTypeId="2" ParentId="4438" CreationDate="2014-04-03T18:09:05.120" Score="7" Body="&lt;p&gt;Three-body special cases can have closed solutions.  For $\ce{H2+}$, the heavy nuclei are assumed to be stationary while the electron travels. Confocal ellipsoidal coordinates then describe the system to first order, followed by perturbation treatments because the nuclei do move. As stated, there are no general analytical forms for the three-body problem overall.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="16683" LastEditDate="2016-11-12T20:01:37.373" LastActivityDate="2016-11-12T20:01:37.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9720" PostTypeId="2" ParentId="778" CreationDate="2014-04-03T18:27:21.163" Score="1" Body="&lt;p&gt;You have three &lt;em&gt;miscible&lt;/em&gt; colored liquids you want to separate quick, cheap, and easy at will? There must be two orthogonal properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Couple one chromophore to a superparamagnetic particle, e.g., a ferrofluid.  It is pulled out in a strong and strongly divergent magnetic field'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Couple the second chromophore to a strong acid, e.g, as a sulfonic acid.  It stays in water, certainly to pH 7 and more basic, resistant to immiscible solvent extraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Couple the third chromophore to a strong base, like guanidinium.  It stays in water until you go strongly alkaline, and then then be extracted into an immiscible solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't mind spending a year or ten in the lab, case (1) plus a temperature-driven critical phase change will do a pretty, polystyrene- and poly(vinyl methylene)-linked chromophores, &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/solid-solutions/printall.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.doitpoms.ac.uk/tlplib/solid-solutions/printall.php&lt;/a&gt; (bottom),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EcBlp.gif&quot; alt=&quot;inverse critical phase separation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-03T18:27:21.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9721" PostTypeId="1" CreationDate="2014-04-03T18:39:14.870" Score="4" ViewCount="4561" Body="&lt;p&gt;I'm performing a liquid-liquid extraction on a 1 l water sample for GC analysis. The procedure calls for three extractions with 50 ml of dichloromethane. Would one extraction with 150 ml produce a different result?&lt;/p&gt;&#xA;" OwnerUserId="316" LastEditorUserId="16683" LastEditDate="2018-04-15T21:07:05.753" LastActivityDate="2018-05-15T21:55:16.320" Title="Why should one perform an extraction with multiple small portions of solvent as opposed to one large portion?" Tags="&lt;experimental-chemistry&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9722" PostTypeId="2" ParentId="409" CreationDate="2014-04-03T18:39:34.510" Score="3" Body="&lt;p&gt;No biggie.  For aqueous ferrofluid, silanize the glass with Rain-X automotive windshield treatment.  If you want both a hydrophobic and oleophobic glass surface, fluoroalkylsilane surface coupling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GLASSCLAD 18 &lt;BR&gt;&#xA;AQUAPHILE AQ &lt;BR&gt;&#xA;AQUAPHOBE CF &lt;BR&gt;&#xA;AQUAPHOBE CM&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-03T18:39:34.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9724" PostTypeId="2" ParentId="9716" CreationDate="2014-04-03T19:28:54.790" Score="2" Body="&lt;p&gt;In organic chemistry, when you speak of mono-N-demethylation, normaly you are refering to an amine bearing 2 or more methyl groups. Let's say you have N,N-dimethylaniline and you want to get N-methylaniline, then you are talking about removing ONE methyl group thus mono-N-demethylation. For organic chemists out there... Yes I know you cant do that one easily and specifically but one keeps dreaming...&lt;/p&gt;&#xA;" OwnerUserId="5073" LastActivityDate="2014-04-03T19:28:54.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9725" PostTypeId="1" CreationDate="2014-04-04T00:10:18.003" Score="5" ViewCount="47321" Body="&lt;p&gt;I understand how the nitrite $\ce{NO2}$ ion can be formed and have a negative charge, but the nitrate $\ce{NO3}$ ion is confusing me. I made some simple drawings to try to explain what I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ho95f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the &quot;electron from outside&quot; that the oxygen receives, as indicated in the drawing, comes from the nitrogen atom? If so, then the N atom would be left with only 4 valence electrons instead of 5, thus being able to form another covalent bond with the extra oxygen, forming $\ce{NO3}$. Is this the correct line of thinking?&lt;/p&gt;&#xA;" OwnerUserId="5077" LastEditorUserId="402" LastEditDate="2016-12-12T02:56:28.397" LastActivityDate="2020-11-03T23:21:27.880" Title="How is the Nitrate Ion (NO3) formed?" Tags="&lt;lewis-structure&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9726" PostTypeId="2" ParentId="9725" CreationDate="2014-04-04T00:49:18.410" Score="3" Body="&lt;p&gt;An organiker would look at it this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJ819.gif&quot; alt=&quot;Nitrate resonance forms&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inorganiker might look at it as literal N(+5) with two double-bonded oxygens.  It does redox as N(+5). Coordination structures are varied,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Ki1F.gif&quot; alt=&quot;Nirate coordination&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-04T00:49:18.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9727" PostTypeId="1" CreationDate="2014-04-04T01:40:22.127" Score="0" ViewCount="2238" Body="&lt;p&gt;I am supposed to write the precipitation reaction for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(NH4)2C2O4 + AgNO3 -&amp;gt;[\ce{H2O}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my equation correct? I left out the Spector ions.&#xA;$$\ce{4(NH4) + 2 C2 +O4\rightarrow 4(NH4)2CO3}$$&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="1362" LastEditDate="2014-04-13T08:04:23.977" LastActivityDate="2014-04-13T08:04:23.977" Title="Which salt precipitates from a solution of ammonium oxalate and silver nitrate?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9728" PostTypeId="2" ParentId="9727" CreationDate="2014-04-04T05:35:32.510" Score="5" Body="&lt;p&gt;Your current solution is wrong and you don't seem to be familiar with the concepts involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's walk through this together and change that! &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You take solid ammonium oxalate $\ce{(NH4)2C2O4}$ and silver nitrate $\ce{AgNO3}$ and dissolve both &lt;strong&gt;salts&lt;/strong&gt; in water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The salts dissociate into their ions. Let's do that separately for both salts first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ (NH4)2C2O4 -&amp;gt;[\ce{H2O}] 2 NH4+_{(aq)} + C2O4^{2-}_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{AgNO3 -&amp;gt;[\ce{H2O}] Ag+_{(aq)} +  NO3^{-}_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are dissolving both salts in the the same beaker with water, your solution now contains all the ions of both equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget about the spectator ion fanciness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much more important to identify the chemical units (= the ions from which the salts are composed). Leave these ions intact and don't rip them apart into weird combinations of elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Precipitation will occur when the combination of these ions (chose a cation and an anion) yields a salt with a very low solubility in water. There are only two cations and anions here, so the number of possible combinations isn't too big. Look up the solubilities of the possible products to figure out which one will precipitate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not forget to balance the reaction!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-04T05:50:23.980" LastActivityDate="2014-04-04T05:50:23.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9729" PostTypeId="1" AcceptedAnswerId="9739" CreationDate="2014-04-04T06:55:36.583" Score="-1" ViewCount="1088" Body="&lt;p&gt;According to WolframAlpha, the change in enthalphy in the combustion of propane is -2220 kJ/mol. How do I calculate the activation energy required start the reaction in the first place? Do you use the same principles as when calculating delta H or am I missing something mind bogglingly obvious?&lt;/p&gt;&#xA;" OwnerUserId="5050" LastEditorUserId="44581" LastEditDate="2017-06-17T01:54:20.527" LastActivityDate="2017-06-17T01:54:20.527" Title="How much activation energy is required to combust propane?" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;enthalpy&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9730" PostTypeId="1" AcceptedAnswerId="9738" CreationDate="2014-04-04T07:15:04.983" Score="6" ViewCount="16229" Body="&lt;p&gt;My book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As single ions of a metal are not associated in the solid with single ions of a non metal, separate units of ionic compounds do not exist. It is, therefore, wrong to talk of a molecule of an ionic compound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know ionic solids exist in form of crystal lattice but why can't we isolate a single molecule of ionic compound?&lt;/p&gt;&#xA;" OwnerDisplayName="user4973" LastEditorUserId="7475" LastEditDate="2017-11-21T14:42:29.457" LastActivityDate="2017-11-21T14:42:29.457" Title="Why don't molecules of ionic compounds exist?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9731" PostTypeId="1" AcceptedAnswerId="9732" CreationDate="2014-04-04T08:39:42.397" Score="2" ViewCount="1020" Body="&lt;p&gt;If the absorbance of something is 0, what is the equilibrium constant? In a reaction, the absorbance is 0, so the concentration of the product is 0. The concentration of one of the reactants is also 0. What does this say about the equilibrium constant? You would have to divide by 0, suggesting that keq doesnt exist for this reaction.&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="418" LastEditDate="2015-04-18T05:32:54.840" LastActivityDate="2015-04-18T05:32:54.840" Title="If absorbance is 0, what is the equilibrium constant?" Tags="&lt;equilibrium&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9732" PostTypeId="2" ParentId="9731" CreationDate="2014-04-04T09:13:33.337" Score="1" Body="&lt;p&gt;Are we talking about uv-vis absorption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed that you are monitoring the right wavelength and that the extinction coefficients are not ridiculously small, you are outside the limits of experimental error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still do not see any absorption, this means that the species isn't there (anymore).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it isn't there anymore, it can't particiate in the equilibrium any longer - there is no measurable equilibrium anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still need to measure something in the case of an irreversible and fast reaction, your experimental setup needs to change: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might need to perform time-resolved  uv spectroscopy and use a stopped-flow setup for the controlled mixing of your reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have anything but a uv-vis spectrometer, a cuvette and a microliter syringe, do it the old way using the standard addition procedure&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prepare the cuvette with a solution of one reactant with a decent concentration &lt;/li&gt;&#xA;&lt;li&gt;Prepare a solution of the second reactant. This might be highly concentrated it you want to avoid dilution effects. If the reaction is fast, you can only measure the absorption of the first reactant and the product anyway&lt;/li&gt;&#xA;&lt;li&gt;Measure the initial absorption of the first reactant&lt;/li&gt;&#xA;&lt;li&gt;Add some volume of the second solution to your cuvette&lt;/li&gt;&#xA;&lt;li&gt;Measure the uv absorption of your first reactant. Do you see any change?   &lt;/li&gt;&#xA;&lt;li&gt;Can you measure the absorption of the product too?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Repeat steps 4-6.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-04T09:46:54.290" LastActivityDate="2014-04-04T09:46:54.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9733" PostTypeId="1" AcceptedAnswerId="9735" CreationDate="2014-04-04T13:17:34.940" Score="1" ViewCount="4875" Body="&lt;p&gt;Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.&lt;/p&gt;&#xA;" OwnerUserId="5050" LastActivityDate="2019-01-22T09:42:42.923" Title="Is photosynthesis endothermic and exothermic at the same time? ... somehow?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;synthesis&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9734" PostTypeId="2" ParentId="9733" CreationDate="2014-04-04T13:34:30.473" Score="0" Body="&lt;p&gt;The heat of formation is the difference in energy between the current state (in this case, glucose) and the state of the elements at 25 °C (carbon, oxygen, hydrogen, etc).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-04T13:34:30.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9735" PostTypeId="2" ParentId="9733" CreationDate="2014-04-04T13:39:18.917" Score="9" Body="&lt;p&gt;Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of  $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\ce{6CO2(g) + 6H2O(l) -&amp;gt; C6H12O6(s) + 6O2(g)}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that at $25\ \mathrm{^\circ C}$ and  $1\ \text{atm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you &quot;lose&quot; them, while you form glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorDisplayName="user7951" LastEditDate="2015-04-17T19:01:40.263" LastActivityDate="2015-04-17T19:01:40.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9736" PostTypeId="1" CreationDate="2014-04-04T13:50:02.107" Score="2" ViewCount="154" Body="&lt;p&gt;Is storing carbon dioxide in limestone a good idea? Wouldn't that lead to a reaction like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{C}\mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\to {\mathrm{C}\mathrm{O}_3}^{-2} + 2\mathrm{H}^+$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then the (unbalanced) reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Ca}\mathrm{C}\mathrm{O}_3 + \mathrm{H}^+\to \mathrm{C}\mathrm{O}_2 + \mathrm{Ca}(\mathrm{O}\mathrm{H})_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the the ground turns into a water caustic water solution. Is this right?&lt;/p&gt;&#xA;" OwnerUserId="5082" LastActivityDate="2014-04-06T21:36:41.367" Title="Carbon dioxide in lime ground" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9737" PostTypeId="2" ParentId="9736" CreationDate="2014-04-04T14:37:15.683" Score="4" Body="&lt;p&gt;Limestone is the natural sedimentary rock, e.g., the top of Mt. Everest that began as sea floor.  &quot;Storing&quot; $\ce{CO2}$ in limestone requires heating calcium carbonate white hot to dissociate it to lime, then scrubbing &quot;other&quot; $\ce{CO2}$ with the lime.  This is, to coin an adjective, smartless, since you also burned fuel to heat the limestone. Curing concrete imbibes and traps $\ce{CO2}$, but it was initially dissociated to release it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second equation is terrible. You obtain a strong base by adding acid to a salt.  $\ce{CaCO3}$ plus carbonic acid, $\ce{CO2}$ in water, makes water-soluble calcium bicarbonate $\ce{Ca(HCO3)2}$.  That is how limestone caves are dissolved out, and why karst topography is stark (surface erosion) and unstable (sinkholes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To vastly store carbon at a profit, plant forests, harvest, and build wood-frame buildings.  The photosynthesis bottleneck, carbon fixation, depends upon the slowest least efficient enzyme on Earth, RuBisCO.  A smart (at least, not corrupt) society then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Gene-gineers a better RuBisCO.  Ten times the productivity leaves it the worst enzyme on Earth.  Everybody would have a full belly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Gene-gineers a 10x fast-growing forest with huge lignin content in its wood.  It won't rot, it won't feed termites, it will be immensely strong.  If wood-frame buildings become unfashionable, bury the wood in deserts to store carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no problem so easily solved that social advocacy will not make it intractable, disastrous, and expensive.  The unabled maximize self-esteem to be credible - effectively selecting for the deluded, hence advocacy.  A society inundated with rules from religious and secular political classes ignoring productive ends will collapse.  USSR to Pol Pot, reality is what it appears to be - thermodynamics.  There is no cheating possible.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-04T14:37:15.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9738" PostTypeId="2" ParentId="9730" CreationDate="2014-04-04T14:52:41.283" Score="6" Body="&lt;p&gt;Ions are electrostatically bound.  One need only separate them with inegrated inverse squared distance work.  If you mix multiple ionic compounds in solvent, a melt, gas phase, the original paired associations are spontaneously scrambled.  Benzene plus xylene does not give you toluene.  &quot;Separating&quot; benzene is not electrostatic attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quaternized glycine, $\ce{^{+}N(CH3)3-CH2-C(=O)O^{-}}$, is betaine.  It is ionic, but the ions are covalently bound into a molecule.  Where are the discrete formula units in alumina?  You can draw lines to connect closest atoms, but those are not bonds. (Note fractional atom counts for atoms embedded in planes, edges, and corners.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZWjoo.gif&quot; alt=&quot;Alumina crystal structure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-04T14:52:41.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9739" PostTypeId="2" ParentId="9729" CreationDate="2014-04-04T15:00:28.583" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/fuels-ignition-temperatures-d_171.html&quot; rel=&quot;nofollow&quot;&gt;http://www.engineeringtoolbox.com/fuels-ignition-temperatures-d_171.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The autoignition temperature is an external variable.  It depends upon total pressure and oxygen partial pressure.  For paper it is famously 232.778 C (significant figures aside). &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-04T15:00:28.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9740" PostTypeId="2" ParentId="9679" CreationDate="2014-04-04T16:03:10.570" Score="15" Body="&lt;h1&gt;Preamble&lt;/h1&gt;&#xA;&lt;p&gt;One important thing to know is that what we call &amp;quot;resonance structure&amp;quot; is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term &lt;strong&gt;mesomeric structure&lt;/strong&gt; more appropriate (&lt;em&gt;meso-&lt;/em&gt; Greek mésos in the middle;  &lt;em&gt;-merism&lt;/em&gt; from Gk. merismos &amp;quot;dividing, partition&amp;quot;) because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a &amp;quot;normal steady&amp;quot; compound that simply has an electron distribution that cannot be described with a single chemical structure.&lt;/p&gt;&#xA;&lt;h2&gt;Why does having equivalent resonating structures give more stability?&lt;/h2&gt;&#xA;&lt;p&gt;The &lt;strong&gt;resonance energy&lt;/strong&gt; is due to &lt;strong&gt;electron delocalization&lt;/strong&gt;, so more the electrons are delocalized in the molecule higher is the &lt;strong&gt;resonance energy&lt;/strong&gt; ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid  like this:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/AqOM1.png&quot; height=&quot;100&quot;&gt;&#xA;&lt;p&gt;Here the electrons are equally distributed (delocalized) between the two oxygen atoms.&lt;/p&gt;&#xA;&lt;p&gt;However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the &lt;strong&gt;resonance energy&lt;/strong&gt; is lower and so the overall stability.&lt;/p&gt;&#xA;&lt;p&gt;In the case of &lt;strong&gt;phenolate ion&lt;/strong&gt; I think the first structure in the figure [2] gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately  I can't find the energy of the different resonance structures.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZALTa.png&quot; height=&quot;100&quot;&gt;&#xA;&lt;p&gt;Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall &lt;strong&gt;resonance energy&lt;/strong&gt; taking into account valency and electronegativity of the atoms involved.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="4945" LastEditDate="2021-02-21T18:06:43.717" LastActivityDate="2021-02-21T18:06:43.717" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="9742" PostTypeId="1" CreationDate="2014-04-04T20:06:42.743" Score="2" ViewCount="186" Body="&lt;p&gt;I have synthesised a compound which is basically a pyridine ring connected to a tetrazole. The procedure suggested to precipitate it at pH 1, but it did not precipitate. It then suggested to extract it with ethyl acetate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that it is not going to the organic layer but into the aqueous layer. I have tried both at pH 1 and 7 and it still does not move to the organic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas how to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="16683" LastEditDate="2017-03-22T23:20:20.313" LastActivityDate="2017-03-22T23:20:20.313" Title="Extraction of a pyridine tetrazole from the aqueous phase" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9743" PostTypeId="1" AcceptedAnswerId="18405" CreationDate="2014-04-04T20:15:48.303" Score="33" ViewCount="3035" Body="&lt;p&gt;&lt;strong&gt;THE STORY:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common example used to illustrate the limitations of restricted Hartree-Fock (RHF) theory is the H$_2$ dissociation energy ($D_e$) curves.  RHF enforces electrons to be paired into spin orbitals, $\chi$, or two spatial orbitals $\phi$ with the same set of spatial coordinates $\mathbf{r}$ but a different spin function (i.e. $\alpha$ or $\beta$ for spin-up $\uparrow$ and spin-down $\downarrow$, respectively).  Unrestricted Hartree-Fock theory gives each spatial orbital its own set of coordinates, allowing for $\phi_1(\alpha$) to have a different energy than $\phi_1(\beta)$, giving rise to a proper description of the H$_2$ dissociation energy (see the graph below).  I didn't include units in this graph but the energy is in kcal mol$^{-1}$.  The red curve gives the appropriate (and approximate) $D_e$.  The RHF formalism (blue curve) gives a $D_e$ that is three-times greater than the 'right answer'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGTtD.png&quot; alt=&quot;Dissociation of H$_2$ with the RHF and UHF formalisms using a minimal basis set&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corresponding electron configuration picture is shown below.  Here we clearly see RHF forcing the electrons to be paired at infinite distances, giving rise to an H$^-$H$^+$ description of the system which is clearly wrong.  (Note that this is a 'state' that included into the wave function by RHF).  UHF allows the electrons to reside in their own spatial orbitals giving rise to two doublets at infinite separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c7iu5.png&quot; alt=&quot;Electron configuration of H$_2$.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is where things get interesting.  In terms of the actual computations implemented to generate the curves in Figure &lt;a href=&quot;https://i.stack.imgur.com/oGTtD.png&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, the electrons adopt opposite spins.  However, let us imagine a theoretical scenario where two lone hydrogen atoms are floating around in space at infinite distance.  The corresponding electrons (which we will assume is not interacting with each other) can adopt any spin they want so both electrons can have a spin-up configuration.  Now imagine these two hydrogen atoms approaching each other.  For a covalent bond to form, the electrons will be paired into a spin-orbital which means that one electron must flip its spin according to the Pauli-Exclusion principle.  (We should note at this point that orbitals are merely mathematical functions whereas spin is an actual physical property).  This now leads us to...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;THE QUESTION:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an energy cost associated with flipping the spin of an electron such as &lt;em&gt;in the scenario I've described&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious follow up question (if the answer is yes to the first piece), would the H$_2$ dissociation energy curves be different for two H atoms each with spin up electrons and two H atoms with electrons of opposite spin (and how different would they be)?&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-04-07T14:30:55.023" LastActivityDate="2017-07-27T19:01:32.717" Title="Is there an energy cost associated with flipping the spin of an electron?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="31" FavoriteCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="9744" PostTypeId="2" ParentId="9742" CreationDate="2014-04-04T21:37:23.890" Score="1" Body="&lt;p&gt;Concentrate under reduced pressure on the rotovap. Add a miscible co-solvent that does not solubilize the zwitterion at its isoelectric point but dissolves starting materials, like THF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pyridine is basic, the terazole is acidic. Catch it on an ion exchange resin column in water, then displace it with a polyvalent counterion.  Possibly down neutral alumina in water to slow it and concentrate it.  You can run a microcolumn in a disposable pipette with a tiny ball of glass wool inside at the taper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Salt it out as a solid with saturated NaCl or potassium nitrate its isoelectric point. Precipitate it, perhaps as the tetraphenylborate or picrate (careful there).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-04T21:37:23.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9747" PostTypeId="1" CreationDate="2014-04-05T06:48:11.800" Score="3" ViewCount="1272" Body="&lt;p&gt;My textbook doesn't say anything about this but does say that on addition of organolithium compounds to carbon dioxide, they give ketones and not carboxylic acids like grignard reagents. This has been attributed to the greater nucleophilic character of Organolithium compounds compared to grignard reagents. I see that greater polarity of the C-Metal bond must relate to this according to many websites but I cannot understand why! Also, during formation of the RLi compounds using alkyl halides, what does the alkyl halide dissociate into? Please help! &lt;/p&gt;&#xA;" OwnerUserId="5089" LastActivityDate="2014-04-05T15:32:14.400" Title="Why are organolithium and organomagnesium compounds nucleophilic in nature?" Tags="&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9748" PostTypeId="2" ParentId="8481" CreationDate="2014-04-05T14:22:15.933" Score="1" Body="&lt;p&gt;Although this is indeed a bit more a math problem than chemistry, I think it is useful to answer it, because this is a commonly encountered problem in reaction kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, &lt;strong&gt;Yes&lt;/strong&gt; it is possible to get the rates with respect to each reactant separately, as long as you have sufficient experimental data. With sufficient I mean: $1+\text{number of reactants}$. This is because you don't only need the order for each reactant, but also the rate constant for the total reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The rate equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a rate equation will have the following structure:&#xA;$$ r = k [A]^\alpha [B]^\beta[C]^\gamma...\text{etc}$$&#xA;where with etc I mean that you can continue for as many reactants that you have. In your example you have 2 reactants so your rate equation in general will look like this:&#xA;$$ r = k [A]^\alpha [B]^\beta $$&#xA;To find the reaction order with respect to each reactant we need experimental data on $r$ for given $[A]%$ and $[B]$ like you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The fitting procedure&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to write a short program that takes an initial guess for $k$, $\alpha$ and $\beta$ and checks how good or bad that guess is by comparing the calculated rates $R$ with the experimental rates $r$ for each set of $[A]$ and $[B]$. The normal way to make this comparison is based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_squares&quot; rel=&quot;nofollow&quot;&gt;least-squares&lt;/a&gt;:&#xA;$$S = \sum_{i=1}^N (r_i-R_i)^2 $$&#xA;So you sum up the squared differences between the reaction rate you found experimentally and the calculated rates for all $N$ experimental datasets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the reaction orders and rate constant you want $S$ to be minimal, preferably 0 (although that normally doesn't happen with experimental data due to small measurement errors). Normally $S$ is not immediately optimal on your first guess for $k$, $\alpha$ and $\beta$, so you have to 'guess again'. You could indeed really guess again, but there are much smarter ways of determining the new guess, which will make sure your new guess is a &lt;em&gt;better&lt;/em&gt; one than the old one. A common technique for this is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Newton%27s_method&quot; rel=&quot;nofollow&quot;&gt;Newton's method&lt;/a&gt;. With the new guess you again check the value of $S$. You then repeat the cycle until you cannot find a lower value for $S$, which means that you have found the values of $k$, $\alpha$ and $\beta$ that are &lt;em&gt;optimal&lt;/em&gt; for this experimental dataset and thus should be the correct reaction orders for the reaction.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-04-05T14:22:15.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9749" PostTypeId="2" ParentId="9747" CreationDate="2014-04-05T14:32:02.317" Score="2" Body="&lt;p&gt;When Li reacts with an alkyl halide, a salt (lithium halide) is formed, whose stability is the primary thermodynamic driving force of the reaction: $\mathrm{2Li}+\mathrm{RCl} \rightarrow \mathrm{RLi}+\mathrm{LiCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the larger difference between the electonegativities of C and Li, the C-Li bond in organolithium compounds has more ionic character and C is more negatively polarized. The higher polarization makes R in RLi a harder nucleophile compared to RMgX, which explains why lithium organyls and Grignard reagents react differently when adding to certain carbonyl compounds.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-04-05T15:32:14.400" LastActivityDate="2014-04-05T15:32:14.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9750" PostTypeId="2" ParentId="5910" CreationDate="2014-04-05T14:39:20.240" Score="2" Body="&lt;p&gt;The assumptions made in the &lt;a href=&quot;http://aerosolschool.web.psi.ch/gothenburg_2008/Group_reviews/FloryHugginsTheory.pdf&quot; rel=&quot;nofollow&quot;&gt;Flory-Huggins theory&lt;/a&gt; are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quasi-solid lattice in the liquid&lt;/li&gt;&#xA;&lt;li&gt;Inter-changeability of segments (not necessarily the same as the polymer structure&#xA;units) of polymer and solvent molecules in the lattice&lt;/li&gt;&#xA;&lt;li&gt;Independence of lattice constants on composition (artificial)&lt;/li&gt;&#xA;&lt;li&gt;Polymer molecules are of same size&lt;/li&gt;&#xA;&lt;li&gt;Average concentration of polymer segments in cells adjacent to cells unoccupied&#xA;by the polymeric solute is taken to be equal to the overall average concentration&lt;/li&gt;&#xA;&lt;li&gt;The expected number of available positions for each successive segment is&#xA;overestimated in the formulation as the formula includes double counting of&#xA;segments, separated by 2 or more segments in the same chain, that will fall on the&#xA;same position twice&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In particular the assumption that polymer molecules are of the same size is a very unrealistic one, which was later removed in the &lt;a href=&quot;http://books.google.nl/books?id=OSAaRwBXGuEC&amp;amp;pg=PA202&amp;amp;lpg=PA202&amp;amp;dq=assumptions%20Flory-Krigbaum&amp;amp;source=bl&amp;amp;ots=wUMrb_5mzq&amp;amp;sig=6WqCD6fgk1y9o6FyF6pDPKq5uVg&amp;amp;hl=nl&amp;amp;sa=X&amp;amp;ei=oRRAU9yqBYvKOY-ygZgN&amp;amp;ved=0CDkQ6AEwAQ#v=onepage&amp;amp;q=assumptions%20Flory-Krigbaum&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Flory-Krigbaum theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-04-05T14:39:20.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9751" PostTypeId="2" ParentId="6640" CreationDate="2014-04-05T15:05:40.737" Score="2" Body="&lt;p&gt;Yes, that is completely correct. Let me summarize the approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the chemical reaction:&#xA;$$\ce{N2 + 3H2 \rightarrow 2 NH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-04-05T15:34:52.983" LastActivityDate="2014-04-05T15:34:52.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9754" PostTypeId="1" CreationDate="2014-04-05T17:05:07.493" Score="1" ViewCount="108" Body="&lt;p&gt;While studying about band theory of semiconductors, I observed that when the electrons were excited from the valence band to the conduction band, they left behind holes in the valence band. From my existing knowledge, I believe that the valence electrons alone occupy the valence band which tells me the valence band is negative. For the holes to exist, the valence band has to be neutral. So, why is the valence band neutral?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my reason for the valence band to be negative was wrong, I would like to know the reason.&lt;/p&gt;&#xA;" OwnerUserId="5096" LastActivityDate="2014-04-05T17:37:44.290" Title="Is the valence band neutral?" Tags="&lt;electrons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9755" PostTypeId="2" ParentId="9754" CreationDate="2014-04-05T17:37:44.290" Score="3" Body="&lt;p&gt;Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively.  If you continue this chain extension and wind up with a very large polyene, the energy difference between the all of the bonding (or antibonding) orbitals becomes small and instead of having discreet molecular orbitals, you can imagine them &quot;fuzzing&quot; out into a &quot;band&quot; of molecular orbitals.  All of the bonding orbitals would become what is termed the valence band, while the antibonding orbitals would comprise the conduction band.  Just as ethylene is neutral, so is the large polyene.  If you excite one of ethylene's pi electrons to the antibonding molecular orbital you have created an excited state which is still neutral.  Scientists working with the solid state would use different terms to describe this process.  They would say that the antibonding orbital contains 2 holes and the boding orbital 2 electrons. When the electron is promoted to the excited state a hole is also promoted (holes have lower energy the further they are from the nucleus) from the antibonding orbital to the boding orbital.  Just like the ground state, the excited state molecule is still neutral, the electron and hole are just in different orbitals from where they started.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-05T17:37:44.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9758" PostTypeId="1" AcceptedAnswerId="9830" CreationDate="2014-04-05T21:00:26.007" Score="5" ViewCount="407" Body="&lt;p&gt;Take the reaction:&#xA;$$ (1) \; \ce{A + B &amp;lt;=&amp;gt;X + Y}, \; K_1=\frac{[X][Y]}{[A][B]} $$&#xA;Now assume that X isomerizes to Z according to:&#xA;$$ (2) \; \ce{X &amp;lt;=&amp;gt; Z}, \; K_2=\frac{[Z]}{[X]} $$&#xA;We can then add the equations to obtain:&#xA;$$ (3) \; \ce{A + B &amp;lt;=&amp;gt; Z + Y}, \; K_3=K_1K_2=\frac{[Z][Y]}{[A][B]} $$&#xA;It's pretty easy to show that $ K_3=K_1K_2 $ when considering the ratio of products and reactants, but this makes less sense when considering the actual numeric value of $K_3$. The fact that $ K_3=K_1K_2 $ can be explained by the statement of multiple equilibria: all equilibria must be satisfied simultaneously. This is what I'm having trouble conceptualizing. Does the equilibrium constant in $(1)$ already take into account $(2)$, and vice versa? Equation $(3)$ will reach a state of maximum entropy in which $K_3$ applies. This must also occur when both $K_2$ and $K_3$ are satisfied? I can do the math all day, but I'm not quite getting the link between the three equilibrium constants.&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="275" LastEditDate="2014-04-06T10:51:27.813" LastActivityDate="2014-04-09T06:51:13.570" Title="Conceptualizing Multiple Equilibria" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9761" PostTypeId="1" AcceptedAnswerId="9783" CreationDate="2014-04-06T00:01:42.887" Score="5" ViewCount="9402" Body="&lt;p&gt;When viewing this on a shopping web site, one of the reviewers indicates &quot;This is about 50-60 [dollars] worth of pyrex glass and another 20 [dollars] worth of wood.&quot;  The company charges $499 for this setup, so I'd like to find out if there is any truth to this statement by identifying the parts and seeing how much they cost.  Image of the setup below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s41E6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5100" LastEditorUserId="2094" LastEditDate="2014-04-07T10:42:27.587" LastActivityDate="2018-11-26T13:36:45.087" Title="Yama Cold Brew Tower: what are these lab parts / components?" Tags="&lt;water&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9763" PostTypeId="1" CreationDate="2014-04-06T07:35:02.600" Score="6" ViewCount="557" Body="&lt;p&gt;Are there any chemistry spreadsheet-like software that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Generates 2D structure for the entered compounds (based on name or SMILES)&lt;/li&gt;&#xA;&lt;li&gt;Easy to search and create customized reports (such as similarity search)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;that preferably runs on linux and is not too expensive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of &lt;a href=&quot;http://www.molsoft.com/icm-chemist.html&quot; rel=&quot;noreferrer&quot;&gt;ICM Chemist&lt;/a&gt; and &lt;a href=&quot;http://www.chemaxon.com/products/jchem-for-office/&quot; rel=&quot;noreferrer&quot;&gt;JChem for Office&lt;/a&gt; (Windows), and I'm wondering if there are any other alternatives available.&lt;/p&gt;&#xA;" OwnerUserId="5101" LastActivityDate="2015-07-22T17:28:42.783" Title="Chemistry Spreadsheet Software?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9764" PostTypeId="1" CreationDate="2014-04-06T12:31:30.580" Score="1" ViewCount="243" Body="&lt;p&gt;I want to do some experiments on enantiomers. As it involves some calculations of potential energy surfaces, I would like to do this experiment on a simple enantiomer (For example $\ce{CHBrClF}$). But this is in liquid form I think. I need an enantiomer sample in the &lt;em&gt;powder&lt;/em&gt; form. Can anyone suggest a simple sample? Either one I can buy or can easily make in a chemical lab with the help of a chemist.&lt;/p&gt;&#xA;" OwnerUserId="2498" LastEditorUserId="402" LastEditDate="2017-04-04T04:46:37.643" LastActivityDate="2017-04-04T06:30:41.840" Title="Simple enantiomer in powder form?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;optical-properties&gt;&lt;chirality&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9766" PostTypeId="2" ParentId="9764" CreationDate="2014-04-06T12:54:21.160" Score="0" Body="&lt;p&gt;Sucrose, common table sugar, is a pure enantiomer that exists as a powder.  Many drugs are manufactured as a single enantiomer, Lipitor would be an example.  Any chemical supply company will sell numerous pure enantiomers; d- and l-carvone would be a common example.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-06T12:54:21.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9767" PostTypeId="2" ParentId="9764" CreationDate="2014-04-06T14:05:49.840" Score="2" Body="&lt;p&gt;If you value ease of acquisition more than structural simplicity, then chiral protein aminoacids are very cheap, easily obtained crystalline white powders, L-alanine being the simplest among them. Its structure is quite small, but its zwitterionic character could possibly introduce additional complexity depending on how you intend to use it. L-Malic acid is a slightly larger option that's also cheap and has no zwitterionic character. There are probably other examples of relatively small, chiral, biologically important compounds which are isolated as solids in large scale and have good enantiomeric purity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For maximum structural simplicity (such as a molecule with a single carbon atom that is also chiral), you will need some rather polar substituents in the structure to make sure that such a small molecule is a solid at room temperature, or some very heavy atoms such as iodine and bromine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: After a bit more searching, lactic acid may be the best combination of simplicity and low cost for an enantiomeric molecule which is a solid at ambient conditions. An enantiomerically pure sample is a white crystalline powder that melts at around 50°C.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-04-06T14:39:42.150" LastActivityDate="2014-04-06T14:39:42.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9768" PostTypeId="2" ParentId="9763" CreationDate="2014-04-06T14:13:44.643" Score="0" Body="&lt;p&gt;I use ACD/Labs Chemsketch in Windows. There is a version that runs on Linux.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Linux users can download a compatible version of ACD/ChemSketch&#xA;  Freeware that can be run with Code Weavers' CrossOver Office.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acdlabs.com/resources/freeware/chemsketch/&quot; rel=&quot;nofollow&quot;&gt;http://www.acdlabs.com/resources/freeware/chemsketch/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-06T14:13:44.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9769" PostTypeId="1" AcceptedAnswerId="9770" CreationDate="2014-04-06T14:24:41.317" Score="0" ViewCount="93" Body="&lt;p&gt;I'm certain that I misunderstand this law. It states that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = I_o e^{-kb} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and after rearranging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = ln{I_o}-lnI $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ stands for absorbance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does is mean that when the intensity of the exiting light is very low than the absorbance is extremely high?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) But from the graphs in my textbook, I've seen absorbance which was up to 1 or 1,2. Does it mean that most objects give little &quot;resistance&quot; to light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) My textbook says that molar absorption coefficient depicts the probability that photon will excite the energy states (not their words exactly :) ). Than why is f(c) = epsilon a constant function? Shouldn't more concentration increase the probability of change in energy states in some for example molecule?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5103" LastEditDate="2014-04-06T15:34:03.000" LastActivityDate="2014-04-06T15:34:03.000" Title="An issue with Bouger-Lambert Law" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9770" PostTypeId="2" ParentId="9769" CreationDate="2014-04-06T15:14:45.083" Score="1" Body="&lt;p&gt;1) Yes that is correct.  Generally, when the intensity of the light transmitted or reflected is low, then the absorbance is high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It is hard to say just what &quot;most&quot; means, but if we are speaking of visible light (remember absorbance is wavelength dependent), there are a LOT of colorless materials that do not absorb in the visible region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Yes, more molecules will increase the probability of light being absorbed by the solution, but as you noted, the absorption coefficient is a &quot;molar&quot; absorption coefficient.  Therefor when using the molar absorption coefficient, the absorbance is adjusted for the number of molecules in the solution.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-06T15:14:45.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9771" PostTypeId="2" ParentId="9764" CreationDate="2014-04-06T16:10:14.187" Score="3" Body="&lt;p&gt;Enantiomorphs are non-superposable mirror images (all signs of one coordinate axis reversed).  What is the maximum enantiomeric divergence possible?  If you have a  quantitatively negligible chiral divergence, there is nearly no quantitative difference to be realized in calculation or measurement.  BTW, the maximum energy difference between enantiomers cannot be more than a part-per-trillion difference/average.  If it were larger, there would be easily detectable enthalpy of fusion anomalies for calorimetry secondary standard benzil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chirality is geometry.  Geometric chirality is easily observed (no ${S_n}$ symmetries[A]) and easily calculated[B].  alpha-Quartz and gamma-glycine single crystals have $\text{CHI} = 1 \text{, COR} = 1 \text{, DSI} = 0$, maximum mathematical chirality[C], but it cannot be measured.  Solution optical rotations ignore atomic mass distribution[D].  Silver thiogallate, $\ce{AgGaS2}$ in &lt;strong&gt;achiral&lt;/strong&gt; space group ${I}$-${42d}$, rotates 522°/mm along [100] at 497.4 nm[E].  ${P3_121}$ quartz and ${P3_221}$ berlinite are both  levorotatory[F].  Achiral $\ce{PhCOCH3}$ impurity changes resolved $\ce{PhCH(OH)CH3}$ optical rotation[G].  CIP notation flips with composition at constant geometry (look at natural protein cysteine versus alanine, both L-amino acids but opposite CIP notations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you obtain $100\%$ optically pure maximally divergent enantiomers to your specs?  No problem!  Benzil, $\ce{(C6H5CO)2}$, is achiral when molten, gas phase, or dissolved. Solid state crystal lattice forces twist and stack benzil molecules into homochiral helices, space groups ${P3_121}$ (right-handed)or ${P3_221}$ (left-handed)[H].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6z5w.png&quot; alt=&quot;benzil crystal structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You obtain (racemic) benzil, modestly resolved (1 $\mathrm{S}$)-(-)-alpha pinene and (1 $\mathrm{R}$)-(+)-alpha pinene (both enantiomers), and mixed xylenes.  $20$ ml scintillation vial, foil-lined cap.  $100$ mg of benzil or so, $10$ ml resolved, alpha-pinene, heat to $100$ C. Drip in mixed xylenes until the molten benzil is just soluble, add another drop or two, very slowly cool undisturbed.  Only one space group crystallizes out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enantiomorphic space groups[I] are mathematically &lt;em&gt;perfectly&lt;/em&gt; divergent geometric chiralities.  Here is the calculation of benzil crystal lattice's atomic mass distribution from Petitjean's software, CHI = $1$ (rapidly asymptotic) COR = $1$ DSI = $0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HjtLX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a discrete molecule as opposed to a crystal lattice, resolved camphor comes in around $\text{CHI} = 0.5$.  Resolved ${D_3}$-trishomocubane is a much better geometric example and is easily synthesized in large quantities starting with benzoquionone and cyclopentadiene (Diels-Alder).  If you have a methyl group sticking up from its ${D_3}$-axis, calculated $\text{CHI} = 0.996$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Look at that methyl-bearing carbon!  It has no ${S_n}$ symmetries, for &lt;em&gt;R&lt;/em&gt; must reflect or invert as &lt;em&gt;S&lt;/em&gt;.  This chiral center's sense of chirality cannot be named.  Try doing it with CIP notation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9fnrP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uncle Al has been very naughty with geometric chirality, &quot;[6.6]Chiralane: a remarkably symmetric chiral molecule,&quot; Symmetry: Culture and Science 19(4) 307-316 (2008).  Note that the core carbon atom has four &lt;em&gt;rigorously identical&lt;/em&gt; substituents yet it is a maximally geometrically chiral center.  Chemical chiral nomenclature is fundamentally defective, for said chiral center (and the four to which it is attached) cannot be assigned a direction of handedness by any means, even in theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[A] &lt;a href=&quot;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&lt;/a&gt;  &lt;BR&gt;&#xA;  pp. 11-13.  &lt;BR&gt;&#xA;[B] J. Math. Phys. 40, 4587 (1999)  &lt;BR&gt; &#xA;  &lt;a href=&quot;http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM&lt;/a&gt;  &lt;BR&gt;&#xA;[C] &lt;a href=&quot;http://www.mazepath.com/uncleal/qzdense2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mazepath.com/uncleal/qzdense2.png&lt;/a&gt; for alpha-quartz  &lt;BR&gt;&#xA;  &lt;a href=&quot;http://www.mazepath.com/uncleal/glydense2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mazepath.com/uncleal/glydense2.png&lt;/a&gt; for gamma-glycine  &lt;BR&gt;&#xA;[D] &lt;a href=&quot;http://www.mazepath.com/uncleal/norbors.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mazepath.com/uncleal/norbors.gif&lt;/a&gt;  &lt;BR&gt;&#xA;    &lt;a href=&quot;http://www.mazepath.com/uncleal/norone.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mazepath.com/uncleal/norone.png&lt;/a&gt;  &lt;BR&gt;&#xA;[E] Appl. Cryst. 33, 126 (2000)  &lt;BR&gt;&#xA;[F] J. Appl. Crystallogr. 19, 108 (1986)   &lt;BR&gt;&#xA;[G] J. Org Chem. 38(10), 1870 (1973) &lt;BR&gt;&#xA;[H] Acta Cryst. B43 398 (1987) &lt;BR&gt;&#xA;[I] &lt;a href=&quot;http://elib.mi.sanu.ac.rs/files/journals/publ/69/7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://elib.mi.sanu.ac.rs/files/journals/publ/69/7.pdf&lt;/a&gt; Section 2: eleven pairs of enantiomorphic space groups; &lt;BR&gt;&#xA; &lt;a href=&quot;http://www.math.ru.nl/~souvi/papers/acta03.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.math.ru.nl/~souvi/papers/acta03.pdf&lt;/a&gt;  Section 3ff.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="41556" LastEditDate="2017-04-04T06:30:41.840" LastActivityDate="2017-04-04T06:30:41.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9773" PostTypeId="1" AcceptedAnswerId="15379" CreationDate="2014-04-06T17:08:01.467" Score="1" ViewCount="1540" Body="&lt;p&gt;I'm a little confused by this question, perhaps my understanding of enthalpy isn't as good as it should be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the compound $\ce{KNO3}$, which of the following reactions is used to determine the standard enthalpy of formation, $\Delta H^\circ_\mathrm f$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{K(s) + N (g) + O3(g) -&amp;gt; KNO3 (s)\\&#xA;2K + N2 + 2O3 -&amp;gt; 2KNO3\\&#xA;2K + N2 +3O2          -&amp;gt; 2KNO3\\&#xA;K  + 1/2 N2 + 3/2 O2  -&amp;gt; KNO3}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the $\Delta H^\circ_\mathrm f(\ce{KNO3})= \pu{−494.6 kJ/mol}$ from reading a table, I'm just confused about how do I get from that number to determine the formation of a compound.&lt;/p&gt;&#xA;" OwnerUserId="5104" LastEditorUserId="5026" LastEditDate="2018-02-25T03:00:02.307" LastActivityDate="2018-02-25T03:00:02.307" Title="Determine standard enthalpy of formation of salt given other reactions" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9774" PostTypeId="2" ParentId="9763" CreationDate="2014-04-06T17:26:24.143" Score="6" Body="&lt;p&gt;I can't comment on spreadsheet-like tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm frequently using &lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Open Babel&lt;/a&gt;, which is free (as in &lt;em&gt;free speech&lt;/em&gt; and &lt;em&gt;free beer&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not so much into using command-line tools or using programmes via their bindings to programming languages, a &lt;a href=&quot;http://openbabel.org/docs/2.3.1/GUI/GUI.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;GUI for Open Babel&lt;/a&gt; is available too, but I've never used it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Generates 2D structure for the entered compounds (based on name or SMILES)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Names might be tough in general, but &lt;a href=&quot;http://www.daylight.com/smiles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SMILES&lt;/a&gt; (Simplified Molecular Input Line Entry System) can be processed with Open Babel when the identifier is read from a file. Output to large number of chemical formats and image formats is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel -ismi stuff.smi -omol -O stuff.mol&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would convert a SMILES string to a mol file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting a collection of SMILES to MOL and writing them to a SDF file is not a big thing:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python&#xA;&#xA;import os&#xA;import pybel&#xA;&#xA;# dictionary of SMILES taken from chemspider.com&#xA;insecticides = {&#xA;    'parathion' : 'CCOP(=S)(OCC)OC1=CC=C(C=C1)[N+](=O)[O-]',&#xA;    'disulfaton' : 'CCOP(=S)(OCC)SCCSCC', &#xA;    'permethrin' : 'C1(C(C1C(=O)OCC2=CC(=CC=C2)OC3=CC=CC=C3)C=C(Cl)Cl)C',&#xA;    'furathiocarb' : 'CCCCOC(=O)N(C)SN(C)C(=O)OC1=CC=CC2=C1OC(C2)(C)C',&#xA;    'methomyl' : 'CC(=NOC(=O)NC)SC'&#xA;}&#xA;&#xA;# write dict items to multi-component SDF file&#xA;outdir = '/home/klaus/workspace/chemistry/stackexchange/similarity/'&#xA;outname = 'insecticides.sdf'&#xA;outpath = os.path.join(outdir, outname)&#xA;outfmt = 'sdf'&#xA;datafile = pybel.Outputfile(outfmt, outpath, overwrite = True)&#xA;&#xA;for name, smiles in insecticides.iteritems():&#xA;    mol = pybel.readstring('smi', smiles)&#xA;    mol.title = name&#xA;    datafile.write(mol)&#xA;&#xA;datafile.close()    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Easy to search and create customized reports (such as similarity search)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the substructure search, I used the &lt;a href=&quot;http://ligand.info/ligand_info_subset_5.sdf.gz&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drug-likeness NCI subset&lt;/a&gt; from &lt;a href=&quot;http://ligand.info/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ligand.info&lt;/a&gt;. The unpacked SDF file (445.2 MB) has 193,323 entries on 14,757,262 lines (&lt;code&gt;wc -l&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the SDF file, I first generated a binary fastsearch file using &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel data.sdf -ofs -O data.fs&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searches are based on SMILES and SMARTS patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CW78G.png&quot; alt=&quot;2-aminobenzothiyzole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that 2-aminobenzothiazole might be an interesting fragment to search for and with a bit of fiddling, I wrote the suitable SMARTS pattern: &lt;code&gt;c12ccccc1nc(N)s2&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in doubt about the correct representation, it is always a good idea to convert the pattern to an image file here: &lt;code&gt;obabel -:&quot;c12ccccc1nc(N)s2&quot; -opng -O pattern.png&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The search was performed using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel -ifs data.fs -s&quot;c12ccccc1nc(N)s2&quot; -osvg -O results.svg -xd -xc5 -xA -xC&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and generated a SVG table with 233 entries.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-08T14:08:49.933" LastActivityDate="2014-04-08T14:08:49.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9775" PostTypeId="2" ParentId="9736" CreationDate="2014-04-06T21:36:41.367" Score="3" Body="&lt;p&gt;Sincerely I'm not sure that carbon sequestration injecting &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; in the ground is a good idea. The risk is that possible leaks cause asphyxiation to the living creatures (humans included) in the surrounding &lt;a href=&quot;http://www.popularmechanics.com/science/environment/climate-change/are-carbon-sequestration-leaks-a-health-danger&quot; rel=&quot;nofollow noreferrer&quot;&gt;see this article&lt;/a&gt;, furthermore injecting gas can change the equilibria of the ecosystem or cause other hydrogeological instability.&lt;/p&gt;&#xA;&lt;p&gt;I strongly recommend you to read &lt;a href=&quot;http://sitemaker.umich.edu/sec005group04/negative_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, an excerpt of a natural &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; emission:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The year was 1986, the scene of the tragedy was Lake Nyos in Cameroon.&#xA;People were going about their daily lives when death struck; a large&#xA;amount of &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; exploded from Lake Nyos and more than 1,700 people died.&#xA;When medical examiners came upon the scene they found bodies sprawled&#xA;everywhere, many with clothes shredded and half-torn, a last attempt&#xA;to relieve the suffocation of asphyxiation.  Cattle had died, and&#xA;those who had survived had been a comatose state for hours.  Skin was&#xA;seared, bronchial tubes were burnt, and lungs filled with fluid.  It&#xA;was surely a quick way to die, but not something to be wished for.&#xA;The cause of the explosion was at first, but after multiple tests and&#xA;examinations it appears that Lake Nyos was connected to a volcanic&#xA;chamber that slowly released &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; into the lake.  After a large amount&#xA;accumulated, it erupted and rose to the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However there are others research more optimistic &lt;a href=&quot;http://www.uscsc.org/Files/Admin/Educational_Papers/Is_CCS_Safe.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper from 2009&lt;/a&gt; can give you a good review however nobody can predict all the possible implications of these injections, and the topic is still debated the conclusion of the article clarify this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“There is consensus that &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; storage in deep underground geologic has&#xA;great technological potential and  may be deployable on a widespread&#xA;basis. However, there are still substantial financial, institutional,&#xA;regulatory, and technical challenges that remain. To address these&#xA;challenges, multiple integrated &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt;  capture and storage system&#xA;projects are needed to prove out the technology. Also needed is an&#xA;array of  small, and intermediate and large-scale &lt;span class=&quot;math-container&quot;&gt;$CO_{2}$&lt;/span&gt; injection field&#xA;tests in diverse geologies to adequately  characterize and validate&#xA;the U.S. geologic resource” (Position Paper of the Coal Utilization&#xA;Research Council  on the Status of Technology for Carbon Capture and&#xA;Storage (CCS), 2008).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Back to the chemistry...&lt;/h1&gt;&#xA;&lt;p&gt;Your reaction (as stated Uncle Al) is wrong because:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaOH + CO2 -&amp;gt; CaCO3 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Has &lt;span class=&quot;math-container&quot;&gt;$K_{eq}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$4.60 \times 10^{19}$&lt;/span&gt;! So goes in the other side of your reaction. I think that you are being tricked by a typical reaction with an acid, like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaCO3 + 2HCl -&amp;gt; CaCl2 + 2H2O + CO2 ^}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However if you substitute HCl with carbonic acid you have a series of complex equilibria:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaCO3 + 2H2CO3 &amp;lt;=&amp;gt; Ca(HCO3)2 + H2O + CO2}$$&lt;/span&gt;&#xA;but:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + CO2 &amp;lt;=&amp;gt;H2CO3}$$&lt;/span&gt; However the ratio between carbonic acid and carabon dioxide is very low: &lt;span class=&quot;math-container&quot;&gt;$$\frac{[H_{2}CO_{3}]}{[CO_{2}]} ≈ 1.7×10^{−3}$$&lt;/span&gt;&#xA;In fact carbon sequestration normally take advantage of adsorption of gasses into the soils, but the carbonatation occur very slowly.&lt;/p&gt;&#xA;&lt;p&gt;By the way with carbon sequestration you will have a soil with a lower pH, more acid... and maybe sparkling tap water too! ;-)&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-04-06T21:36:41.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9778" PostTypeId="1" AcceptedAnswerId="9779" CreationDate="2014-04-07T03:24:13.000" Score="1" ViewCount="3240" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;There are two types of stationary phases used in thin layer chromatography. One stationary phase is made of aluminium oxide. Which of the following molecules would be expected to have the smallest &lt;span class=&quot;math-container&quot;&gt;$R_f$&lt;/span&gt; using aluminium oxide as the stationary phase on the glass slide?&lt;/p&gt;&#xA;&lt;p&gt;(a) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH2CH2COOH}$&lt;/span&gt;&lt;br /&gt;&#xA;(b) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH2CH2OH}$&lt;/span&gt;&lt;br /&gt;&#xA;(c) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH2NH2}$&lt;/span&gt;&lt;br /&gt;&#xA;(d) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH2CH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{AL2O3}$&lt;/span&gt; is a non-polar stationary phase, therefore if the molecule is also non-polar it would display high adsorption  to the stationary phase, and therefore a smaller &lt;span class=&quot;math-container&quot;&gt;$R_f$&lt;/span&gt; value.&lt;/p&gt;&#xA;&lt;p&gt;The answer is (a).&lt;/p&gt;&#xA;&lt;p&gt;Isn't (a) polar?&lt;/p&gt;&#xA;&lt;p&gt;Wouldn't this increase the &lt;span class=&quot;math-container&quot;&gt;$R_f$&lt;/span&gt; value, because it doesn't adsorb to the non-polar stationary phase? Can someone please explain, correct me?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="-1" LastEditorDisplayName="user10259" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-14T00:52:40.673" Title="Thin Layer Chromatography &amp; Stationary Phase" Tags="&lt;polarity&gt;&lt;adsorption&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9779" PostTypeId="2" ParentId="9778" CreationDate="2014-04-07T03:56:27.007" Score="1" Body="&lt;p&gt;Aluminum Oxide is more polar than Silica Gel and polar compounds do not migrate far from the starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://infohost.nmt.edu/~jaltig/TLC.pdf&quot; rel=&quot;nofollow&quot;&gt;Thin Layer Chromatographic Analyses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://courses.chem.psu.edu/chem36/Experiments/PDF%27s_for_techniques/TLC.pdf&quot; rel=&quot;nofollow&quot;&gt;7. Thin-Layer Chromatography&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-07T03:56:27.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9780" PostTypeId="1" CreationDate="2014-04-07T06:05:21.100" Score="1" ViewCount="48" Body="&lt;p&gt;When two small 2D carbon platelets are pushed into proximity, why do they release the edge attached molecule (maybe O) and then bond C to C?&lt;br&gt;&#xA;I think the answer I seek would address the relative difference in bond strengths, and what forces are involved to remove attached molecules, probably Oxygen.   .....&lt;br&gt;&#xA;BACKGROUND:   I experiment with electrostatic properties of conductive carbon films for oscillation and physical displacement applications. I use a mechanical 'tribological' method of creating 2D Carbon sheets.  In this case the method is: smearing powered graphite onto a smooth substrate, followed by agitation and polishing to a smooth highly conductive film.  The result is assumed to be hexagonal graphene film of several layers.&lt;/p&gt;&#xA;" OwnerUserId="5108" LastEditorUserId="544" LastEditDate="2019-01-16T14:30:24.983" LastActivityDate="2019-01-16T14:30:24.983" Title="Why does C=C bond replace C-O bond when two graphite platelets are rubbed against each other edge on edge?" Tags="&lt;valence-bond-theory&gt;&lt;carbon-allotropes&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9783" PostTypeId="2" ParentId="9761" CreationDate="2014-04-07T10:39:03.503" Score="5" Body="&lt;p&gt;I think that you can make your own cheaper Yama cold brew tower however what they are selling to you is the design! What you see in the picture is not chemistry glassware but if you want to make your own with chemistry glassware (this will cost more money because good chemistry glassware is purer and must have many expansive features that normal glassware don't have) I've written the analogous components in the figure. You need an open-top cylindrical dropping funnel with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Stopcock&quot; rel=&quot;nofollow noreferrer&quot;&gt;stopcock&lt;/a&gt; valve.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Qkj0E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.pltscientific.com/catalog/attachment/files/1352702523-reaction-vessels-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.alibaba.com/product-detail/Dropping-Funnels-_136154108.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for reference. You can use a normal funnel and attach to it a stopcock valve if you want. Then there is something similar to a Buchner funnel, I imagine that there is a filter in the bottom, may be included into the funnel, in this case, a 90-150 micron filter should be sufficient. See &lt;a href=&quot;http://www.piscolabglass.com/sintered-glass-buchner-funnel.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more info. Then there is a glass coil spiral, up to my knowledge this has no practical use is only beautiful and geeky but will cost a lot. Then you have a normal carafe.&#xA;(probably a piece of filter paper is placed over the grounded cafe to allow the diffusion of water in the whole volume of the cylinder and not only in the centre).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OIbxl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some &lt;a href=&quot;http://en.wikipedia.org/wiki/Filter_paper&quot; rel=&quot;nofollow noreferrer&quot;&gt;filter paper&lt;/a&gt; and a couple of half cut plastic bottles you can do your Yama tower with less of 5 $: you can make a little hole in the screw cap of your first bottle to make drip the first funnel, and use a piece of filter paper in the second.  If you want to use glassware don't use chemistry glassware you don't need it really, try to find normal glassware to adapt to your project and you can make your own cheaper and nice Yama tower. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2018-10-05T12:57:00.810" LastActivityDate="2018-10-05T12:57:00.810" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9785" PostTypeId="1" CreationDate="2014-04-07T14:11:59.790" Score="2" ViewCount="239" Body="&lt;p&gt;I've modelled the iso-surface of the highest occupied molecular orbital of a molecule with DFT in Gaussian.If I were to model the iso-surface of the molecular orbital of the singlet and triplet excited states, how would I do that?Using tdDFT, I've already calculated the excitation energy of both the triplet and singlet manifold.&lt;/p&gt;&#xA;" OwnerUserId="4829" LastActivityDate="2014-04-07T14:11:59.790" Title="How to model iso-surface of molecular orbitals of singlet and triplet excited states" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;photochemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9787" PostTypeId="1" AcceptedAnswerId="9789" CreationDate="2014-04-07T14:28:28.407" Score="2" ViewCount="134" Body="&lt;p&gt;I understand that most acids are weak acids, i.e. they do not fully disassociate in solution. However, certain acids, such as $\ce{H2SO4}$, do. I have been unable to find out what the defining characteristic is that differentiates strong acids from weak ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, HCl is strong, yet both HF and HBr are weak (even though they're based on halogens as well). Shouldn't HF be even stronger? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What characteristic makes only a select few acids fully disassociate (and hence belong to the strong acid group)?&lt;/p&gt;&#xA;" OwnerUserId="5114" LastEditorUserId="9961" LastEditDate="2015-11-13T19:42:45.243" LastActivityDate="2015-11-13T19:42:45.243" Title="Why do only certain acids disassociate in water?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9788" PostTypeId="2" ParentId="4302" CreationDate="2014-04-07T16:58:42.117" Score="0" Body="&lt;p&gt;For most of the metals have a look at &lt;a href=&quot;http://www.opticsinfobase.org/ao/abstract.cfm?id=26571&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical properties of the metals Al, Co, Cu, Au, Fe, Pb, Ni, Pd, Pt, Ag, Ti, and W in the infrared and far infrared Applied Optics, Vol. 22, Issue 7, pp. 1099-1119 (1983)&lt;/a&gt;. You will find something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5N72s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OCR software could help you to retrieve the data.You can use $k$ as an index that indicates the amount of absorption loss.&#xA;Free pdf of another version &lt;a href=&quot;https://www.google.it/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=7&amp;amp;ved=0CHEQFjAG&amp;amp;url=http%3A%2F%2Fwww.dtic.mil%2Fcgi-bin%2FGetTRDoc%3FAD%3DADA159736&amp;amp;ei=HNVCU4eeL8HWtQboyIHADQ&amp;amp;usg=AFQjCNFu_G7DKH5rOKQJuk5dAj1FP8MofA&amp;amp;sig2=GII3qkFJaZqaJ2ZNYlj-Gg&amp;amp;bvm=bv.64367178,d.Yms&amp;amp;cad=rja&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-07T16:58:42.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9789" PostTypeId="2" ParentId="9787" CreationDate="2014-04-07T17:23:29.043" Score="6" Body="&lt;p&gt;There are many factors that determine the strength of an acid:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;electronegativity of the non-metal atom&lt;/strong&gt;: an electronegative atom can attract more easily  the electrons of the hydrogen an so generate the proton.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;atomic radius&lt;/strong&gt; greater the radius of the conjugate base greater the acidity, this because the electrons are dispersed around the atom and there is a less attraction between the positive proton and the negative anion.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;stability of the conjugate base&lt;/strong&gt; if the conjugate base is stable due to some particular electronic configuration (e.g. stability due the resonance structure) the species are more prone to be dissociated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-07T17:23:29.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9790" PostTypeId="2" ParentId="5382" CreationDate="2014-04-07T18:22:48.833" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_point&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Triple_point&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a material's equilibrium melting and freezing points were non-identical, you would have a perpetual motion machine of the first kind.  Go head, trace the energy flows for temperature versus specific heats and latent enthalpies of transition.  Cycled divergent vapor pressures at the differing transition temperatures would power a perpetual wind generator.  Ya gotta think about these things. e.g., water below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0aBrg.gif&quot; alt=&quot;Water vapor pressures&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-07T19:33:14.413" LastActivityDate="2014-04-07T19:33:14.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9791" PostTypeId="2" ParentId="9778" CreationDate="2014-04-07T18:38:25.467" Score="0" Body="&lt;p&gt;&quot;&lt;em&gt;There are two types of stationary phases used in thin layer chromatography&lt;/em&gt;.&quot;  Sigma-Aldrich sells silica gel (unmodified, modified/bonded, chiral and high purity), aluminum oxide, cellulose (fibrous, microcrystalline), and polyamide 6 stationary phase TLC plates.  That barely scratches the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alumina is highly polar and highly basic unless modified.  Yellowed tetramethylenediamine run down a Dispopipette of dry Activity 0 basic alumina has everything bothersome ripped out of it within an inch or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carboxylic acid will ionize and salt bond, given an Rf of about zero.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-07T18:38:25.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9792" PostTypeId="1" CreationDate="2014-04-07T18:52:45.253" Score="9" ViewCount="2002" Body="&lt;p&gt;&lt;strong&gt;The Story:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction pathways can be difficult things to deal with using electronic structure methods.  Intrinsic reaction coordinates (IRCs) can be determined a number of ways (a discussion of these methods lie way outside the scope of this post).  Suffice to say, the reaction pathway is an 'energy' pathway of reaction, connecting reactants and products, usually through some intermediate and/or transition state (TS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of this question, imagine implementing some sort of IRC technique which begins with a previously resolved transition state (Figure &lt;a href=&quot;https://i.stack.imgur.com/Kor1Y.png&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;).  The computation follows the reaction coordinate in both the forward and reverse directions from the TS energetically downhill until it reaches a (for the sake of simplicity) a minimum energy point with features which satisfy the method used.  In general, the structure and energy of this final IRC geometry should be very representative of the structure which is connected to the TS via the pathway determined.  If you were to optimize this final geometry, you should get good agreement between the optimized structure and the IRC configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kor1Y.png&quot; alt=&quot;Hypothetical reaction pathway&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider this scenario.  An optimization was performed on the IRC resolved structure B and the energy relaxation was significant and the structure significantly changed, giving rise to a new (and lower) energy.  One could no longer (within the formalism of IRC methods) say that the optimized structure B lies on the reaction pathway, and therefore is not a representative configuration at point B on the reaction pathway associated with the TS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an example in the literature where this type of behavior occurs?  How does one 'deal' with a situation like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Information:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;That this scenario occurs with a few IRC procedures (not just one).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That the IRC computations were given a ridiculously large amount of&#xA;cycles and steps &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That various stepsizes were sampled in parallel IRC jobs with no&#xA;significant difference in the results&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I will add more as I think of them)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Potentially Helpful Imagery&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two somewhat complex molecules coming together to form a single covalent bond.  You run an IRC from the TS (which describes the bond formation) in the reverse direction and the intermolecular separation begins to increase.  This separation continues to increase until the IRC job terminates properly.  Now imagine running an optimization on these two (slightly complex) molecules which are separated by maybe 4.0 Å.  Obviously these molecules are going to interact because they are not infinitely separated.  You observe the (slightly complex) molecules moving toward each other during the optimization procedure (rotating, twisting, gyrating along the way) until you now have a minimum energy structure which is much lower in energy than the resulting IRC computation.  These two molecules interact via intermolecular interactions (no covalent bonds are formed between the molecules).  This optimized structure is analogous to a reactant complex (as opposed to separated reactants) and has an energy a few kcal mol$^{-1}$ lower than the corresponding IRC geometry.  How on earth does this fit in to the reaction pathway?&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="73" LastEditDate="2014-04-07T21:08:35.657" LastActivityDate="2014-04-15T18:56:02.340" Title="Reaction Pathways: What to do when IRC geometries are relaxed (optimized) and the energy changes significantly" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;reaction-coordinate&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9793" PostTypeId="1" CreationDate="2014-04-07T21:25:14.557" Score="3" ViewCount="326" Body="&lt;p&gt;I tried to google this but 'element' as a programming term complicated the search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a site or article that covers how one would find each element in nature, in its most abundant form. For example, what's the best natural source/form of Sulfur? Beryllium? Tugsten? etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5116" LastEditorUserId="7475" LastEditDate="2015-10-30T17:28:10.097" LastActivityDate="2015-10-30T17:28:10.097" Title="How to Find Each Element in Nature" Tags="&lt;everyday-chemistry&gt;&lt;reference-request&gt;&lt;elements&gt;" AnswerCount="5" CommentCount="1" ClosedDate="2020-06-25T09:28:20.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="9794" PostTypeId="2" ParentId="9793" CreationDate="2014-04-07T22:21:06.590" Score="4" Body="&lt;p&gt;The Wikipedia page for each element is probably the easiest place to start with. If you want some significantly more in-depth data, the USGS publishes &lt;a href=&quot;http://minerals.usgs.gov/minerals/pubs/mcs/index.html&quot; rel=&quot;nofollow&quot;&gt;free-access mineral commodity summaries&lt;/a&gt; on many (but not all) elements, including information about their origin, abundance and applications. Other good sources are books, especially descriptive inorganic chemistry books, such as a few of &lt;a href=&quot;http://johnemsley.com/book_home.html&quot; rel=&quot;nofollow&quot;&gt;John Emsley's publications&lt;/a&gt; including Nature's Building Blocks and The Elements, as well as many other college-level inorganic chemistry texts.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-04-07T22:21:06.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9795" PostTypeId="1" AcceptedAnswerId="9796" CreationDate="2014-04-07T23:47:30.637" Score="1" ViewCount="1436" Body="&lt;p&gt;$\ce{CH2CH2(g) + H2(g) -&amp;gt; CH3CH3(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this table, source: &lt;a href=&quot;http://www.kentchemistry.com/images/links/Kinetics/TB08_003.GIF&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.kentchemistry.com/images/links/Kinetics/TB08_003.GIF&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mTOSn.gif&quot; alt=&quot;http://www.kentchemistry.com/images/links/Kinetics/TB08_003.GIF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;C-C enthalpy = 348 kJ/mol&#xA;C-H enthalpy = 413 kJ/mol&#xA;H-H enthalpy = 436 kJ/mol&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way I see it, H-H has to break, C-C has to break, and then two C-H bonds have to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Units omitted) Therefore, should the answer be: $413 \cdot 2 - 348 - 436 = 42 \, \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then it has to be an endothermic reaction.  Is this correct, or am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="5057" LastEditorUserId="4945" LastEditDate="2014-04-08T07:12:23.273" LastActivityDate="2014-04-08T08:49:27.740" Title="What is the energy change of ethene reacting with hydrogen gas?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9796" PostTypeId="2" ParentId="9795" CreationDate="2014-04-08T01:26:35.017" Score="1" Body="&lt;p&gt;Your reasoning is correct, but your numbers are wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are not breaking a C-C bond, but changing a C=C bond to C-C. So you will also need the energy for C=C which is 614 kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2014-04-08T08:49:27.740" LastActivityDate="2014-04-08T08:49:27.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9797" PostTypeId="2" ParentId="9793" CreationDate="2014-04-08T01:33:02.143" Score="0" Body="&lt;p&gt;You should try &lt;a href=&quot;http://www.webelements.com/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;. You will need to click on each element to get the information you need.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-08T01:33:02.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9798" PostTypeId="1" AcceptedAnswerId="9799" CreationDate="2014-04-08T04:43:23.857" Score="4" ViewCount="790" Body="&lt;p&gt;Are there any approximate relationships between the (static) dielectric constant of a liquid and properties of the individual molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose I know the polarizability and dipole moment of a molecule, as well as it's density in the liquid phase. Can I estimate the dielectric constant of the liquid from these quantities (or other similar ones)?&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2021-04-02T10:21:40.870" Title="How to estimate the dielectric constant of a liquid?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;molecules&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9799" PostTypeId="2" ParentId="9798" CreationDate="2014-04-08T05:40:31.357" Score="5" Body="&lt;p&gt;Yes, such relations exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably there are two general theories for polar and non-polar fluids that connect the polarizability $P$ with the dielectric constant $\epsilon$ as reviewed in this article by &lt;a href=&quot;http://www.nist.gov/mml/acmd/fluids_modeling/upload/Harvey-Lemmon-50062.pdf&quot; rel=&quot;noreferrer&quot;&gt;Harvey and Lennon, International Journal of Thermophysics, Vol. 26, No. 1, January 2005&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Clausius-Mosotti equation for &lt;strong&gt;non-polar liquids&lt;/strong&gt; is:&#xA;$$P = \frac{\epsilon-1}{\epsilon +2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas an equation developed by Kirkwood for &lt;strong&gt;polar liquids&lt;/strong&gt; is:&#xA;$$ P = \frac{(\epsilon-1)(2\epsilon+1)}{9 \epsilon}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article then continues to explain that you can estimate the polarizability from theoretical relations involving the molar density $\rho$ and the dipole moment $\mu$, but in your case the two equations above are sufficient since you assume that you know $P$.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-04-08T05:40:31.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9800" PostTypeId="1" CreationDate="2014-04-08T07:14:57.347" Score="4" ViewCount="1043" Body="&lt;p&gt;I had read some where that in glasses the surface tension is high, but in a lota the surface tension is low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question how much time does the surface tension take to change? I mean that there was water in a glass but I put it into a lota in how much time will the surface tension of water change? Immediately?&#xA;This is a lota: &lt;img src=&quot;https://i.stack.imgur.com/e1qzX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4602" LastEditorUserId="73" LastEditDate="2014-04-10T09:18:54.993" LastActivityDate="2014-04-10T09:18:54.993" Title="Time taken in changing surface tension" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="9801" PostTypeId="1" CreationDate="2014-04-08T08:21:47.087" Score="3" ViewCount="1436" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The given graph of abundance of different samples of butanol with respect to time is-&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Jqjeh.png&quot; alt=&quot;Graph of abundance of different samples of butanol with time&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Textbook solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HZCpI.png&quot; alt=&quot;Calibration curve of butanol samples (given as answer)&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But why aren't the points like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(20,5);(40,10);(60,15);(a,7)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;after which, to plot the curve we can use &lt;span class=&quot;math-container&quot;&gt;$y=mx+c$&lt;/span&gt; and then find the value of &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="83758" LastEditorDisplayName="user8016" LastEditDate="2020-06-11T03:59:19.297" LastActivityDate="2020-06-11T03:59:19.297" Title="Plotting a calibration curve for Gas-Liquid Chromatography" Tags="&lt;analytical-chemistry&gt;&lt;concentration&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9802" PostTypeId="2" ParentId="9801" CreationDate="2014-04-08T08:35:25.670" Score="3" Body="&lt;p&gt;If I remember correctly, the amount of your compound isn't proportional to the peak height but to the peak area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the days, when plotters were just plotters (=no numerical integration of the peaks), people used to cut out the curves and weighted the paper pieces on a balance to determine the peak area.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-08T08:35:25.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9803" PostTypeId="1" AcceptedAnswerId="9807" CreationDate="2014-04-08T08:45:11.777" Score="16" ViewCount="4127" Body="&lt;p&gt;Knowing very little about the nature of water, wondering how it might behave at the centre of a planet or centre of an another massive gravitational body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could water take such pressures or might it break into separate hydrogen and oxygen to find something more accommodating for the pressure exerted, if such atmosphere would allow the volume even?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, bonus points: could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?&lt;/p&gt;&#xA;" OwnerUserId="5117" LastActivityDate="2021-04-04T06:54:13.177" Title="What happens to water under high pressures without possibility of escape?" Tags="&lt;thermodynamics&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9804" PostTypeId="2" ParentId="9793" CreationDate="2014-04-08T08:59:15.913" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://elements.vanderkrogt.net/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://elements.vanderkrogt.net/index.php&lt;/a&gt; is in my opinion the best site about how have been finden and discovered the elements.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zHIc8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-08T08:59:15.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9805" PostTypeId="2" ParentId="9793" CreationDate="2014-04-08T09:04:55.387" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.geokem.com/global-element-dist1.html#ge&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.geokem.com/global-element-dist1.html#ge&lt;/a&gt; is more specific for your needs,more concise, less history more about actual source of the element.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gxwbV.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-08T09:04:55.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9806" PostTypeId="1" CreationDate="2014-04-08T09:22:50.123" Score="2" ViewCount="5531" Body="&lt;p&gt;In air, how much $\ce{CO2}$ would be required to prevent a match to light or a butane lighter to work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a basement office that is unusable because of an apparent lack of oxygen, currently it is my guess that the oxygen is being displaced by $\ce{CO2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit research suggests that it would take a reduction to around 14% for a safety match to fizzle out without any flame. But my naive estimate suggests 7% $\ce{CO2}$, and at that level isn't $\ce{CO2}$ deadly? &#xA;And this isn't even accounting for the displacement of nitrogen, surely that would mean the levels of $\ce{CO2}$ would be even higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a 6% increase in $\ce{CO2}$ displace oxygen by only small percentage say 2% as it's only a smaller ratio of the atmosphere? Or would the presence of $\ce{CO2}$ contribute to the lack of ignition compounding the lack of oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/40562/what-would-cause-lack-of-oxygen-in-cellar&quot;&gt;See my related question on the DIY stack exchange&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;UPDATE&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;An insepector from the local council has installed a data-logger, whilst he was in the basement he took a reading that I've attached below. When this reading was taken the air felt poor, but by no means the worst it's been, a candle wouldn't stay alight. An breathing during conversation seemed laboured. It looks like the assumption of displacement doesn't look correct, there's more to it. &#xA;&lt;strong&gt;Here's the readings:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas - Reading (Normal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ 3.3% (0.04%) &lt;br&gt;&#xA;$\ce{O2}$ 17.5% (20.8%)&lt;br&gt;&#xA;&lt;strong&gt;Other&lt;/strong&gt; 79.2% (unchanged?)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9UVyQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5118" LastEditorDisplayName="user7951" LastEditDate="2018-07-31T17:17:19.773" LastActivityDate="2018-07-31T17:17:19.773" Title="How much Carbon Dioxide would be required to displace enough oxygen to prevent ignition?" Tags="&lt;combustion&gt;" AnswerCount="3" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="9807" PostTypeId="2" ParentId="9803" CreationDate="2014-04-08T09:28:53.977" Score="15" Body="&lt;h1&gt;What happens to water under high pressures without the possibility of escape?&lt;/h1&gt;&#xA;&lt;p&gt;It depends on what you mean for high pressure and on the temperature, however, the water phase diagram can help you to understand what will happen. This is from Wikipedia User Cmglee:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RpaIc.png&quot; alt=&quot;Edited image of User:Cmglee&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see that at high-pressure water assumes a solid form (ice): you will have ice X at 100 GPa, and what is labelled as hexagonal ice XI in the diagram but is actually &lt;a href=&quot;https://en.wikipedia.org/wiki/Superionic_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;superionic water or ice XVIII&lt;/a&gt; at 1 TPa (temperature range 0 to 650 K).  These sorts of ices have different lattice and internal energies.&lt;/p&gt;&#xA;&lt;h1&gt;What would happen to water as pressures increase towards infinity without the possibility of the water escaping confinement?&lt;/h1&gt;&#xA;&lt;p&gt;Unfortunately, chemists don't deal with &lt;strong&gt;infinity&lt;/strong&gt;, basically, however, there will be a point where the water molecule won't longer exist all the bonds will break.  After that probably you will have something called &lt;em&gt;electron-degenerate matter&lt;/em&gt; have a look &lt;a href=&quot;http://faculty.wcas.northwestern.edu/%7Einfocom/The%20Website/pressure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, here is all about quantum mechanics, Pauli exclusion principle and black holes so on more Physic S.E. stuff.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wbv88.png&quot; alt=&quot;Edited image of User:Cmglee&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Could triple points in water play a role in keeping temperatures low under high pressures via fluid thermodynamics?&lt;/h1&gt;&#xA;&lt;p&gt;You can also note that there are other triple points like (100 k, 62GPa) this however doesn't directly affect the properties of water but are the properties of water that determine where are the triple points. In a close system, kinetic energy is constant so I think that fluid thermodynamics doesn't matter in this case is more related to equilibria.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/ref.html#r81&quot; rel=&quot;nofollow noreferrer&quot;&gt;More info about water phase diagrams here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="17175" LastEditDate="2020-12-31T16:27:17.570" LastActivityDate="2020-12-31T16:27:17.570" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="9808" PostTypeId="2" ParentId="9792" CreationDate="2014-04-08T09:59:33.450" Score="3" Body="&lt;p&gt;The IRC algorithm usually follows the gradient from TS to product state B and by definition has to go always downhill in energy. In best case, it treats the Potential Energy Surface (PES) around the given point as quadratic and just integrates Newtonian equation of motion and does it until it reaches &lt;strong&gt;local&lt;/strong&gt; minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure is different from geometry optimization, which can use powerful methods to find energy minimum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Static picture, T = 0K&lt;/strong&gt; shows that the reaction will end in the &quot;upper&quot; B state. If you then ask for energy minimum, reorientation of any kind (intermolecular or conformational) takes place and you find &quot;lower&quot; B. Do not be decepted by your plot, the optimized B lies on different point of reaction coordinate (however you define it) than the IRC-type B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dynamic picture, T &gt; 0K&lt;/strong&gt; shows that IRC is not the only way to proceed from TS to product, although it is still important representative. Upon running molecular dynamics (MD) from TS, you will end up in the equilibrium state, probably well represented by &quot;lower&quot; B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be quite difficult to find exactly this type of behavior in literature, because such technical topics are quite often swept under the carpet and only indirect hints remain. Similar topics are discussed in &lt;a href=&quot;http://books.google.de/books/about/Molecular_Reaction_Dynamics.html?id=FVqyS31OM7sC&amp;amp;redir_esc=y&quot; rel=&quot;nofollow&quot;&gt;Levine: Molecular Reaction Dynamics&lt;/a&gt;. As a simple example take the E2 elimination from cyclohexane, where the leaving groups must be axial, so the reactive species is different from energy minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how to deal with it? It is up to you, where you set zero to your relative energy scale. It will depend on which energies you want to compare, if you propose some series of reactions.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-04-08T09:59:33.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9809" PostTypeId="1" AcceptedAnswerId="9810" CreationDate="2014-04-08T10:57:45.993" Score="0" ViewCount="2396" Body="&lt;p&gt;Branched chain molecules have lower boiling point than straight chain molecules because branched chain molecules have a smaller surface area where they can be attracted by other molecules. Then, why Isodecane (a branched chain molecule) has a higher boiling point than Decane (a straight chain molecule)?&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2015-11-16T08:04:11.647" Title="Why Isodecane (B.P.: 327 *C) has a higher boiling point than decane (B.P.: 174.1 *C)?" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9810" PostTypeId="2" ParentId="9809" CreationDate="2014-04-08T11:14:20.543" Score="2" Body="&lt;p&gt;If you mean 2-methylnonane with isodecane, your data is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-methylnonane has a boiling point around 166 °C, which would support the rule of thumb that you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-08T11:14:20.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9811" PostTypeId="2" ParentId="6944" CreationDate="2014-04-08T11:39:51.907" Score="1" Body="&lt;p&gt;Straight chain molecules have more places along its length where they can be attracted to other molecules, so there are more chances of London Dispersion Forces to be developed. Hence, they have stronger intermolecular forces as compared to the branched chain molecules which have a compact shape, therefore fewer spaces where they can be attracted to other molecules.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="402" LastEditDate="2017-04-02T20:21:52.357" LastActivityDate="2017-04-02T20:21:52.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9812" PostTypeId="2" ParentId="5239" CreationDate="2014-04-08T11:54:14.277" Score="1" Body="&lt;p&gt;In the case of alkanes, with increasing molecular mass, the number of atoms in a molecule also increase, which means an increase in the chain length of molecules. A molecule with a large chain length experiences stronger London Dispersion Forces. The reason is that longer molecules have more places where they can be attracted to other molecules. This is the reason why pentane (longer chain molecule) experiences stronger intermolecular forces of attraction than methane.&#xA;As alkanes are non-polar, therefore, they will only exhibit London Dispersion Forces.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="402" LastEditDate="2017-04-18T18:57:40.743" LastActivityDate="2017-04-18T18:57:40.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9813" PostTypeId="1" AcceptedAnswerId="12257" CreationDate="2014-04-08T12:27:20.853" Score="-1" ViewCount="11871" Body="&lt;p&gt;On one hand, covalent character increases the bond dissociation enthalpy (i.e., more energy is needed to be put in) and on the other it lowers the melting point of $\ce{Al2O3}$ to a value lower then would be expected with perfect ionic bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does it decrease or increase bond strength?&lt;/p&gt;&#xA;" OwnerDisplayName="user5110" LastEditorUserId="73" LastEditDate="2014-09-04T17:20:56.130" LastActivityDate="2017-09-24T12:39:01.760" Title="Does covalent character increase or decrease bond strength?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2017-09-24T15:25:21.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="9814" PostTypeId="2" ParentId="7520" CreationDate="2014-04-08T12:40:59.457" Score="15" Body="&lt;p&gt;Actually it is not necessary to dig deep into quantum mechanics. There are several reasons why noble gasses are stable (as gasses at room temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, there is the obvious full valence shell. &#xA;Trend in the periodic table make clear that the charge of the nucleus grows from left to right in every period. The attractive force towards the electrons therefore increases. (This holds also from top to bottom.) Electrons in lower orbitals ($n&amp;lt;\text{period}$) now shield the charge of the nucleus. This will be somewhat the same throughout the period. (The orbitals also shrink due to higher charge of the nucleus.) In noble gasses the valence shell is completely filled, providing quite a good shield for the next shell. Also the increase in the main quantum number means a significant increase in the energy level of the next orbital. This makes it unlikely, that a noble gas will accept another electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the high charge of the nucleus it is also not easy to remove an electron from an orbital. However, this is possible from (at least) Argon on downwards. For example: $\ce{HArF}$ is stable in a matrix at room temperature. It has a strong covalent $\sigma$ bond ($\ce{H\bond{-}Ar+}$) and a not quite as strong ionic bond ($\ce{[HAr]+\cdots F-}$). Krypton already does some fancy chemistry, that is stable at room temperature. As Uncle Al has pointed out Xenon is well known for its reactivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the sudden change? It is fairly intuitive, that the maximum electron density for each orbital with increasing main quantum number is also further away from the nucleus. That makes the valence shell very well polarizeable. Also the electrons are better shielded by the previous shells. However, the natural occurrence of these elements are as (homonuclear) gasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you bring these elements in close contact with each other it was found out, that they have a very small dissociation energy ($D_e&amp;lt;1~\text{meV}$). This is due to dispersion, and van-der-Waals Forces, which is the main interaction between these elements. However, in $\ce{He2}$ no binding mode was observed (instant dissociation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that still does not answer, why those elements are stable as gasses, as pointed out that there are in fact attractive forces. The reason is as simple as obvious: Entropy. If two noble gasses would form a molecule/adduct, the bond/association energy of this molecule would have to compensate for the loss of entropy (Two volume elements would become one, hence the gas would have to expand to cover the room previously occupied, which requires work energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation by tschoppi also covers, why from a MO point of view there could not be a bond in $\ce{He2}$. Go ahead and ask yourself the question if this would be true for $\ce{He3}$. We also know that orbital overlap is only one component to the truth. There are many other more. If it was not for such a nice thing like dispersion and other &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_Force&quot;&gt;weak chemical interactions&lt;/a&gt; we would not exist.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-08T12:40:59.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9815" PostTypeId="2" ParentId="9806" CreationDate="2014-04-08T12:53:18.513" Score="5" Body="&lt;p&gt;The parameter you are asking about is called the &lt;a href=&quot;http://www.processoperations.com/FireExplode/FE_Chp01.htm&quot;&gt;Minimum Oxygen Concentration&lt;/a&gt; (MOC), and is a measurement of the minimum amount of oxygen present in air to propagate a flame. The amount depends on the fuel being burned, and varies from about 4% for hydrogen to about 18% for some solid combustibles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider your example of butane, and round some numbers to make our calculations simple. &lt;a href=&quot;http://www.processoperations.com/FireExplode/FE_Tables/FE_Table02.htm&quot;&gt;Butane has a MOC of 12%.&lt;/a&gt; Normal oxygen content in air is 20%, with nitrogen and sundries making up the remaining 80%. Assume uniform mixing and displacement of gases is equal. In order to reduce the oxygen content to 12%, you need to introduce enough CO2 to make up 40% of the air content:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O2 + N2 = 100$  =&gt;20 + 80 = 100%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O2 + N2 + CO2 = 100$    =&gt; 12 + (12*80/20) + CO2 =100%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let us consider a couple of other very significant points. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/zoo00/zoo00755.htm&quot;&gt;An oxygen level of 12% is unsafe&lt;/a&gt;. Anything below 19.5% is unsafe. At levels of 12%, a person will have impaired judgement and coordination, and experience increased pulse rates. Prolonged exposure can have fatal consequences.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.ivhhn.org/index.php?option=com_content&amp;amp;view=article&amp;amp;id=84&quot;&gt;A CO2 level of 40% is fatal&lt;/a&gt;. Full stop. Levels above 7% can cause loss of consciousness and just 10% can be fatal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;CO2 is actually denser than air, so mixing of the main components is unlikely to be uniform. This means that higher CO2 concentrations will be found lower down in the room.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope you stay well clear of your basement until you have resolved this problem and have appropriate active ventilation installed, as well as possibly an oxygen monitor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the very least, get yourself a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_leak_detection#History&quot;&gt;canary&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-04-08T13:24:54.057" LastActivityDate="2014-04-08T13:24:54.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9816" PostTypeId="1" CreationDate="2014-04-08T13:28:32.423" Score="1" ViewCount="287" Body="&lt;p&gt;What exactly does the excitation spectrum tell us about a sample in Florescence Spectroscopy, what information does it tell us? I know that we will get an output of concentration vs intensity but that's it.&lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="418" LastEditDate="2014-04-09T08:59:55.407" LastActivityDate="2014-04-09T08:59:55.407" Title="Fluorescence Spectra, emission spectrum" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9817" PostTypeId="2" ParentId="9816" CreationDate="2014-04-08T13:40:23.620" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorescence_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;by analysing[sic] the different frequencies of light emitted in&#xA;  fluorescent spectroscopy, along with their relative intensities, the&#xA;  structure of the different vibrational levels can be determined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-08T13:40:23.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9818" PostTypeId="2" ParentId="9816" CreationDate="2014-04-08T14:17:22.840" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that we will get an output of concentration vs intensity but that's it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, that's not what you are getting. Typically, you run an emission spectrum first to determine the maximum of the fluorescence emission (or any other point of interest in the emission spectrum). In the excitation spectrum, you monitor this particular emission line and its intensity while varying the excitation wavelength.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-08T14:17:22.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9819" PostTypeId="1" AcceptedAnswerId="9820" CreationDate="2014-04-08T14:37:54.963" Score="6" ViewCount="21422" Body="&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I have very limited knowledge of chemistry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is the following: I have motorcycle part made from some sort of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_alloy&quot; rel=&quot;noreferrer&quot;&gt;Aluminium alloy&lt;/a&gt; - most probably 6005, 6061-T6 or 7005. And inside of this &lt;strong&gt;aluminium part&lt;/strong&gt; there is &lt;strong&gt;steel part&lt;/strong&gt; which is simply &lt;strong&gt;stuck&lt;/strong&gt; there and it's really hard to get it out physically without damaging the &lt;strong&gt;aluminium part&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of googling, I've found out several ways of doing that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;By putting &lt;strong&gt;aluminium part&lt;/strong&gt; with stuck &lt;strong&gt;steel part&lt;/strong&gt; into water and start adding &lt;strong&gt;concentrated nitric acid&lt;/strong&gt; until steel part starts to dissolve. But I've also read that &lt;strong&gt;dillute nitric acid&lt;/strong&gt; (which will occur after mixing with water) will also dissolve aluminium or I'm wrong here?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By using some sort of &lt;a href=&quot;http://en.wikipedia.org/wiki/Alum&quot; rel=&quot;noreferrer&quot;&gt;Alum&lt;/a&gt;: either by using &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_alum&quot; rel=&quot;noreferrer&quot;&gt;Potassium alum&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_aluminium_sulfate&quot; rel=&quot;noreferrer&quot;&gt;Ammonium alluminium sulfate&lt;/a&gt;. Some guy on the forum suggested to &lt;strong&gt;dissolve&lt;/strong&gt; above &lt;strong&gt;alum&lt;/strong&gt; in &lt;strong&gt;hot tap water&lt;/strong&gt; and then put &lt;strong&gt;aluminium&lt;/strong&gt; part &lt;strong&gt;with steel&lt;/strong&gt; part into resulting solution and let it stay there for a day. It should &lt;strong&gt;dissolve&lt;/strong&gt; or at least &lt;strong&gt;damage steel&lt;/strong&gt; part without affecting aluminium part which may &lt;strong&gt;ease physical removal&lt;/strong&gt; of steel part.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Could you please advise which of above solutions is better? And in case of using alum - which of them should I buy - potassium or ammonium one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help!&lt;/p&gt;&#xA;" OwnerUserId="5121" LastActivityDate="2018-03-28T04:20:32.813" Title="How to dissolve steel without affecting aluminium alloy" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;home-experiment&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9820" PostTypeId="2" ParentId="9819" CreationDate="2014-04-08T15:07:41.973" Score="4" Body="&lt;p&gt;The part of the steel insert that is stuck to the aluminum is not accessible to water.  Let's look at linear coefficients of thermal expansion,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;nofollow&quot;&gt;http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum is 22.2 ppm/K. Steels are smaller, the largest being 304 austentic at 17.3 ppm.  Ambient temp is around 24 C, a 400 F kitchen oven is around 204 C.  That should not be hot enough to lose temper in the alloys.  If the aluminum hole is an inch in diameter, worst case increasing the temp 180 K times 4.9 ppm/K net gives you 882 ppm or 22 microns of slack.  The steel insert pushes out.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-08T15:07:41.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9821" PostTypeId="1" AcceptedAnswerId="9822" CreationDate="2014-04-08T15:37:31.173" Score="9" ViewCount="12729" Body="&lt;p&gt;London dispersion forces (LDF) are present in all molecules, whether polar or non-polar. Molecules also exhibiting dipole-dipole interactions (in addition to the LDF) must have stronger forces of attraction than those molecules which exhibit only LDF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, why does tetrachloromethane (carbon tetrachloride), which is a non-polar molecule exhibiting only London dispersion forces, have a higher boiling point ($\pu{77 ^\circ C}$) than trichloromethane (chloroform) ($\pu{61 ^\circ C}$) which is a polar molecule, exhibiting dipole-dipole interactions?&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="5026" LastEditDate="2018-05-16T10:53:46.857" LastActivityDate="2018-05-16T10:53:46.857" Title="Why does tetrachloromethane have a higher boiling point than trichloromethane?" Tags="&lt;halides&gt;&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9822" PostTypeId="2" ParentId="9821" CreationDate="2014-04-08T15:46:16.633" Score="6" Body="&lt;p&gt;You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastActivityDate="2014-04-08T15:46:16.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9823" PostTypeId="1" CreationDate="2014-04-08T15:57:31.837" Score="5" ViewCount="69555" Body="&lt;p&gt;Is Diethyl ether (also known as ethyl ether) a polar molecule? What type of intermolecular forces dominate it? Dipole-Dipole Interactions, London Dispersion Forces or Hydrogen Bonding? Please Explain.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="14" LastEditDate="2014-04-08T16:04:22.670" LastActivityDate="2017-01-26T20:16:22.687" Title="What type of intermolecular forces will dominate Diethyl ether?" Tags="&lt;intermolecular-forces&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9824" PostTypeId="2" ParentId="6513" CreationDate="2014-04-08T16:47:42.920" Score="0" Body="&lt;p&gt;The valence electrons in the outermost orbit of an atom, get excited on availability of energy. They overcome the binding force to become free and move anywhere within the boundaries of the solid. Thus they contribute to conduction &lt;/p&gt;&#xA;" OwnerUserId="5123" LastActivityDate="2014-04-08T16:47:42.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9826" PostTypeId="2" ParentId="9819" CreationDate="2014-04-08T17:20:16.303" Score="1" Body="&lt;p&gt;&quot;&lt;em&gt;However, for more complex cases like broken steel screws inside aluminium parts which cannot be extracted&lt;/em&gt;&quot; Iron touching aluminum is a shorted electrolytic cell. The aluminum will preferentially erode in a wet corrosive environment. Hot concentrated nitric acid will passivate an aluminum surface by oxidation. Even if it preferentially goes after the steel, that is extremely hazardous - including inhalation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screw removal/extractor sets are tapered left-threaded hardened drill bits.  As they drill into the broken stub they unscrew it.  It still helps to heat the piece to equilibrium to disengage aluminum from steel.  You don't want galling as you force out the steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.harborfreight.com/12-piece-screw-extractor-set-40349.html&quot; rel=&quot;nofollow&quot;&gt;http://www.harborfreight.com/12-piece-screw-extractor-set-40349.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.sears.com/craftsman-3-pc-screw-out-174-damaged-screw/p-00952154000P&quot; rel=&quot;nofollow&quot;&gt;http://www.sears.com/craftsman-3-pc-screw-out-174-damaged-screw/p-00952154000P&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.liquidwrench.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.liquidwrench.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a drop of silicone oil, Kroil, WD-40, Liquid Wrench, or Mobil 1 0w20 before starting.  Silicone ends up everywhere.  This is a disaster in a high vacuum lab.  A machine shop is different.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-08T17:20:16.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9828" PostTypeId="1" AcceptedAnswerId="9960" CreationDate="2014-04-08T17:43:14.847" Score="1" ViewCount="262" Body="&lt;p&gt;In a book about electrochemistry the situation of a cell is described of which one half cell has an &quot;undefined thermodynamic potential&quot; as no redox couple exists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Pt/Cu^{2+}\ (0.01\ M),\ Cd^{2+}\ (0.01\ M),\ H_2SO_4\ (1\ M)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating the potential where reduction of Cu sets in by thermodynamic data when a potential is applied,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Cu^{2+} + 2e \rightleftharpoons Cu,\ +\ 0.340\ V\ vs. NHE\\&#xA;Cd^{2+} + 2e \rightleftharpoons Cu,\ -\ 0.4025\ V\ vs. NHE\\&#xA;O_{2+} + 4H^+ + 4e \rightleftharpoons H_2O,\ +\ 1.229\ V\ vs.\ NHE,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they argue that there is a rest potential range $0.340\ V ... 1.229\ V\ vs. NHE$ where no current is to be expected. Beyond these values current flow is expected. I don't understand this, because I understand $0.340\ V\ vs. NHE$ to be the standard potential of copper reduction under standard conditions, but the concentration of 0.01 M is not &quot;standard&quot; (= 1 M), right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I'm afraid I have a general lack of understanding a system where one species of a redox pair is missing, because argueing in terms of Nernst's equation for such systems always leads to infinitely large (or small) potentials. In the case of copper above Nernst is not applicable in my opinion (correct ?), but what does the general relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; \lim_{[R] \rightarrow 0} (E_0 +  \frac{RT}{nF} \ln\frac{[O]}{[R]})=\infty&#xA;$$&#xA;for the reaction $O+ne\rightleftharpoons R$ mean? And how can one figure out the rest potential range for a half cell on thermodynamic basis (neglecting kinetics)? What does the situation for copper looks like where [R] does not appear in Nernst's equation? When the solution in the book is right, why?&lt;/p&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2016-11-27T13:39:11.197" Title="Electrochemical half cell without redox couple" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9829" PostTypeId="1" AcceptedAnswerId="13475" CreationDate="2014-04-08T19:27:43.220" Score="11" ViewCount="449" Body="&lt;p&gt;It's hard to evaluate Environmental Tobacco Smoke lifecycle, it can certainly react, be absorbed by organisms, or be diluted in rain, and be emitted again from the ground.&lt;br&gt;&#xA;But in the end, I'm interested to understand (as a non expert) if the toxicity (cytotoxicity, genetoxicity) of the compounds decrease over time (especially how are they eliminated) in a global outdoor environment where we suppose emissions would stop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several resources list the components &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK53014/&quot;&gt;Chemistry and Toxicology of Cigarette Smoke&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/tx700275p&quot;&gt;Comparison between Marijuana and Cigarette Smoke&lt;/a&gt;, &lt;a href=&quot;http://ash.org/CAEPAProposal.pdf&quot;&gt;CAEPA&lt;/a&gt; (page6): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas and particle phase contain roughly the same constituents as explained in the first paper, some compounds are 1,3-butadiene, acrolein, ammonia, aromatic amines, benzene, CO, isoprene, nicotine, nitrosamines, PAHs (polycyclic aromatic hydrocarbons), pyridine, toluene, hydrogen cyanide...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be their half-life in outdoor environment?&lt;/p&gt;&#xA;" OwnerUserId="5126" LastEditorUserId="5126" LastEditDate="2020-12-27T19:20:31.980" LastActivityDate="2020-12-27T19:20:31.980" Title="How long burned tobacco/cigarette smoke remain toxic?" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="9830" PostTypeId="2" ParentId="9758" CreationDate="2014-04-08T21:06:10.523" Score="1" Body="&lt;p&gt;When there is a situation of multiple equilibria, all of them must be satisfied. If this is the case, the whole system will be at equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your example and as you report, the maths let you write that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;K_{3}=K_{1}\times K_{2} \quad (1)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This final equation, beyond being the mere mathematical combination of two initial equations, is obtained by considering that Gibbs free energy is a &lt;strong&gt;function of state&lt;/strong&gt;: iIt describes &quot;quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/State_function&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;). One may get to equilibrium via an unique straight step, or through several steps. But the final destination will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition allows us to write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\Delta G_{3}^{o}=\Delta G_{1}^{o} + \Delta G_{2}^{o}  \quad (2)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (2) can be converted into equation (1), by remembering that $&#xA;\Delta G^{o}=-RT\text{ln} K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can first of all say that $K_{1}$ and $K_{2}$ relate to the two &quot;pieces&quot; of the reaction path one has to cover to reach the overall equilibrium. Knowing them is essential, if one wants to obtain the quantities for all the chemical species, at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{3}$, on the other hand, relates to the &quot;shortcut&quot; (the unique &quot;ideal&quot; step connecting the initial and final states of a system) and is linked to $K_{1}$ and $K_{2}$, through equation (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the examples you provide and assuming initially 1 mole for $A$ and 1 mole for $B$, we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;K_{1}=\frac{(y-z)y}{(1-y)^2}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;K_{2}=\frac{z}{y-z}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could more generally write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;K_{1}=f(y,z)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;K_{2}=g(y,z)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two constants are different functions: but they both depend on the same parameters, $y$ and $z$, which are temperature and pressure (for gases) dependent. If temperature changes, both constants will accordingly change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the equilibrium quantities, at given temperature and pressure, we'll need the values for $K_{1}$ and $K_{2}$, since knowing only $K_{3}$ won't be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond all this, kinetics will affect the time needed to reach system equilibrium: but that's &lt;em&gt;another story&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="3871" LastEditDate="2014-04-09T06:51:13.570" LastActivityDate="2014-04-09T06:51:13.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9832" PostTypeId="2" ParentId="9806" CreationDate="2014-04-08T22:06:35.033" Score="3" Body="&lt;p&gt;As you know the atmospheric gasses concentration in dry air is very different, carbon dioxide normally is below 0.1%:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QIVSf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that &lt;a href=&quot;http://en.wikipedia.org/wiki/Limiting_oxygen_concentration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Limiting oxygen concentration&lt;/a&gt; for ignite the wood of your match is 17 % and that the ratio between the atmospheric gas is preserved you have to find the factor ($x$) that multiplied by the oxygen content give you 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$20.9646 \times~x = 17$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can multiply Argon%, Oxygen%, Carbon dioxide% etc. etc. for x and then find the % of the unknown gas subtracting to 100% the sum of all the gasses. In one formula:&#xA;$$100-((Ar\%\times x)+ (O_{2}\%\times x) + (CO_{2}\%\times x) \ldots)= Unknown~gas~\% $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tRdUI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the unknown gas could occupy 18.8% of the volume of your cellar. This means that if it is $CO_{2}$ you could have several problem: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within 1 minute of  exposure to 20-30% $CO_{2}$, unconsciousness and&#xA;  convulsions occur in humans. &lt;a href=&quot;http://www.blm.gov/pgdata/etc/medialib/blm/wy/information/NEPA/cfodocs/howell.Par.2800.File.dat/25apxC.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;blm.gov&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you calculate it with a LOC of 14% you have a $CO_{2}$ content of 33%, this content as stated long (credit to him) can kill you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mj3Yz.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA; However for what I've read in your question there are not enough evidence that you are dealing with carbon dioxide, what I think is sure is that your cellar is not a safe place to stay. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-08T22:06:35.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9833" PostTypeId="1" CreationDate="2014-04-08T23:42:59.053" Score="3" ViewCount="18391" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaCO3(s) &amp;lt;=&amp;gt; CaO(s) + CO2(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;How will the amount of each element change if the pressure is increased?&lt;/p&gt;&#xA;&lt;p&gt;My guess is that the equilibrium will shift to the left, because the right side has 1 gas molecule. I know my solution contradicts with Le Chatelier's principles, but there is no gas on the left. So how will the the amount of the elements on the right decrease and the solid on the left increase since pressure only has an effect on gases.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="95133" LastEditDate="2020-09-11T20:13:05.317" LastActivityDate="2020-09-11T20:20:54.260" Title="How does the equilibrium shift when the products are solid and gas when increasing pressure?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9834" PostTypeId="2" ParentId="9833" CreationDate="2014-04-09T00:39:50.547" Score="2" Body="&lt;p&gt;I don't think that your guess contradicts Le Chatelier's Principle, which states that introducing a change will prompt the reaction to oppose that change.&lt;/p&gt;&#xA;&lt;p&gt;Increasing the pressure will lead to a decrease in &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; as it converts back into the starting reactant, which will lower the pressure as the gas is converted into &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorUserId="95133" LastEditDate="2020-09-11T20:20:54.260" LastActivityDate="2020-09-11T20:20:54.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9836" PostTypeId="1" CreationDate="2014-04-09T01:47:38.430" Score="1" ViewCount="13084" Body="&lt;p&gt;We know that for strong acids with a $1:1$ mole ratio between all particles involved are suitable for use with the aforementioned formula for titration problems. I want to know the limits on this formula, that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider if we had $$\ce{H2SO4(aq) + 2KOH(aq) -&amp;gt; 2H2O(l) + K2SO4(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the moles of $\ce{H+}$ = moles of $\ce{~^{-}OH}$ in order to neutralize, how do we apply $M_aV_a=M_bV_b$ to this situation, if at all possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, does this formula work if we had weak acids? I would think not, because with weak acids we don't have the benefit of complete dissociation and thus not a $1:1 = \text{acid}:\ce{H+}$ ion ratio nor a $1:1 = \text{base}:\ce{~^{-}OH}$ ratio.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="4945" LastEditDate="2014-04-10T05:08:56.097" LastActivityDate="2018-06-13T08:16:08.217" Title="MaVa = MbVb - Limitations and Uses" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9838" PostTypeId="1" CreationDate="2014-04-09T03:30:15.583" Score="1" ViewCount="5527" Body="&lt;p&gt;Can someone help me find the mechanism for this reaction. I've been looking for it, but I &#xA;haven't found it. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5129" LastActivityDate="2014-04-09T16:46:44.810" Title="What's the mechanism for the oxidation of primary alcohol to carboxylic acid with HNO3?" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2014-04-10T12:37:30.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="9840" PostTypeId="1" CreationDate="2014-04-09T06:36:10.773" Score="1" ViewCount="45" Body="&lt;p&gt;lets say we increased the excitation wavelength higher than it should be, what effect would this have on the sensitivity on our analysis, or changing the emission wavelength to a higher value than it would be.&#xA;My guess is that the higher wavelength leads to a lower frequency and so lowers the sensitivity but I have no idea what changing the emission wavelength would do? &lt;/p&gt;&#xA;" OwnerUserId="1554" LastEditorUserId="418" LastEditDate="2014-04-09T09:00:21.030" LastActivityDate="2014-04-09T09:00:21.030" Title="Fluorescence spectrocopy, sensitivity" Tags="&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9841" PostTypeId="1" CreationDate="2014-04-09T10:54:28.440" Score="1" ViewCount="1245" Body="&lt;p&gt;I want to know the mechanism of this reaction. My intuition says $\ce{H-}$ will attack on $\ce{C}$ attached to $\ce{N}$ and hydrolysis afterwards would produce $\ce{-NH-OH}$. But the answer says that the amine will be formed.&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4945" LastEditDate="2014-05-22T12:11:38.207" LastActivityDate="2015-07-14T19:12:05.863" Title="Reduction of an Oxime with LiAlH4 ­­­­­­­­­" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2017-01-15T21:49:33.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="9842" PostTypeId="1" AcceptedAnswerId="9848" CreationDate="2014-04-09T11:43:39.620" Score="0" ViewCount="136" Body="&lt;p&gt;Wood countertops are frequently sealed, when a non-hardening, food-safe finish is desired, with some &lt;a href=&quot;http://www.howardproducts.com/prod-butcher-block.php&quot; rel=&quot;nofollow&quot;&gt;magic cocktail of waxes and oils&lt;/a&gt;. Typical components might include &lt;a href=&quot;http://en.wikipedia.org/wiki/Mineral_oil&quot; rel=&quot;nofollow&quot;&gt;mineral oil&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Beeswax&quot; rel=&quot;nofollow&quot;&gt;beeswax&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Carnauba_wax&quot; rel=&quot;nofollow&quot;&gt;carnauba wax&lt;/a&gt;, mixed in some unspecified ratio and then sold at premium prices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being content paying for a bottle of magic goo, I'm seeking a scientific understanding of how these products actually work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These products claim to (over time) build a coating of wax on the surface which is more durable than simply mineral oil would be. If the product is a mixture of oil and wax which is quite thin, how can it build a layer of wax on the surface of the wood with repeated applications? Does the wood selectively absorb the oils? When subsequent applications are made, are the thinner oils able to diffuse through whatever wax has built on the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the relevant concepts in play here, and in what directions might I research to gain a deeper understanding?&lt;/p&gt;&#xA;" OwnerUserId="5133" LastEditorUserId="5133" LastEditDate="2014-04-09T11:51:39.720" LastActivityDate="2014-04-09T14:42:50.357" Title="Wax and oil polish on wood" Tags="&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9843" PostTypeId="2" ParentId="9833" CreationDate="2014-04-09T13:33:11.203" Score="4" Body="&lt;p&gt;Consider the equilibrium expression for this reaction.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaCO3(s) &amp;lt;=&amp;gt; CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$&lt;/span&gt;&#xA;The concentrations of solids and liquids are constant.  They are the molar densities. Since &lt;span class=&quot;math-container&quot;&gt;$\ce{[CaO]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[CaCO3]}$&lt;/span&gt; don't change, they are moved to the left hand side and &amp;quot;folded into&amp;quot; the equilibrium constant.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;K_c^{'} &amp;amp;\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\&#xA;K_c &amp;amp;= \ce{[CO2]}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, as long as solid &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt; and solid &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt; are present along with &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; gas there will be an equilibrium.  Only changes to the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; will cause a shift in the equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;You asked how will the amounts change if the pressure is increased.  The pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is increased by either adding more &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; or by reducing the volume of the container.  Adding more &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; will increase the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; momentarily, which will shift the equilibrium to the left, using up some &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt; and making &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt;. The pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; can also be increased by reducing the volume of the container.  Again, the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is increased, which increases the reaction with &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt; to make additional &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt;.  The &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt; are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; affect the position of the equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt;.  The added gas won't affect the partial pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5134" LastEditorUserId="95133" LastEditDate="2020-09-11T20:20:09.907" LastActivityDate="2020-09-11T20:20:09.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9844" PostTypeId="2" ParentId="8265" CreationDate="2014-04-09T13:45:10.697" Score="6" Body="&lt;p&gt;Gallium metal may be obtained in the elemental form.  A classic demonstration is to hold gallium in your hand and watch it melt.&lt;br&gt;&#xA;&lt;a href=&quot;http://astounde.com/wp-content/uploads/2012/06/gallium.jpg&quot; rel=&quot;noreferrer&quot;&gt;http://astounde.com/wp-content/uploads/2012/06/gallium.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy from the heat lamp must be absorbed by the gallium, but if the IR light is reflected, then it may not be getting hot enough to melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that you have an alloy of gallium and some other metal and that the melting point is higher than for pure gallium.&lt;/p&gt;&#xA;" OwnerUserId="5134" LastActivityDate="2014-04-09T13:45:10.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9845" PostTypeId="2" ParentId="9172" CreationDate="2014-04-09T14:04:06.740" Score="1" Body="&lt;p&gt;The pKa of a base is just the pKa of the conjugated acid of that base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following reaction:&lt;br&gt;&#xA;$$\ce{B + H2O &amp;lt;=&amp;gt; HB+ + OH- } (1)$$&lt;br&gt;&#xA;you get the equilibrium constant: $Keq1 = \frac{(HB^{+})(OH^{-})}{(B)}$ with Kb = Keq1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the reaction with the conjugated acid HB+:&lt;br&gt;&#xA;$$\ce{HB+ + H2O &amp;lt;=&amp;gt;  B + H3O+}$$&lt;br&gt;&#xA;the equilibrium constant is then $Keq2 = \frac{(B)(H_{3}O^{+})}{(HB^{+})}$ which gives the corresponding Ka&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of the two gives:&lt;br&gt;&#xA;$$Kb * Ka = \ce{\frac{(HB+)(OH^{-})}{(B)}} \times \ce{\frac{(B)(H3O+)}{(HB+)}} = \ce{(OH^{-}).(H3O+)} = 10^{-14} = Kw$$&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="2094" LastEditDate="2014-04-09T17:18:45.610" LastActivityDate="2014-04-09T17:18:45.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9846" PostTypeId="2" ParentId="65970" CreationDate="2014-04-09T14:29:37.373" Score="3" Body="&lt;p&gt;The intermediate hydroxylamine is further deoxidatively reduced to the amine.  LAH is a blunt instrument; aluminum(III) is a potent oxophile.  Cf: ketone or aldehyde plus amine plus cyanoborohydride in mildly acidic buffer.  It only takes a tickle to reduce a Schiff base. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-09T14:29:37.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9847" PostTypeId="1" AcceptedAnswerId="27278" CreationDate="2014-04-09T14:36:08.157" Score="29" ViewCount="24716" Body="&lt;p&gt;Although the wikipedia page on Hydronium indicates a $\mathrm{p}K_\text{a}$ of −1.74, I noticed in the discussion of this page that the subject seems debated (cf. &lt;a href=&quot;http://en.wikipedia.org/wiki/Talk:Hydronium#pKa.3F&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Talk:Hydronium#pKa.3F&lt;/a&gt;) since alternative (apparently more rigorous) reasoning leads to $\mathrm{p}K_\text{a} = 0$ for $\ce{H3O+}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, many demonstrations lead to $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of 15.74 for $\ce{H2O}$ and thus, using $K_\text{w}=[\ce{H+}][\ce{OH-}]=10^{-14}$, give $\mathrm{p}K_\text{a}$ and $\mathrm{p}K_\text{b}$ of −1.74 for $\ce{H+}$ and $\ce{OH-}$ respectively. But those demonstrations seem to make use of the concentration $[\ce{H2O}]=55.56~\text{mol/L}$ instead of the activity $a_{\ce{H2O}}=1$ (for a solvent) in the equilibrium constant of the reactions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems to me that the $\mathrm{p}K_\text{a}$ of $\ce{H3O+}$ must be equal to $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this question still debated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS1:&lt;br&gt;&#xA;To add details about the question, the debate in the references mentioned in the discussion on the wikipedia page as well as in the link provided by Nicolau Saker Neto bellow seems to boil down to the coexistence of two different definitions for the equilibrium constant:&lt;br&gt;&#xA;- one based on the activities of the compounds&lt;br&gt;&#xA;- the other derived from the 'mass action law' and which thereby uses the molar fractions of the compounds&lt;br&gt;&#xA;Reading through the Callen (Thermodynamics and an Introduction to Thermostatistics) gives me the impression that the derivation from the mass action law of an equilibrium constant involving molar fractions is based on the assumption of an ideal fluid. However, it is precisely the departure from ideality that seems to justify the use of activities instead of molar fractions. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS2:&lt;br&gt;&#xA;Another point, which I have not looked in details yet, seems to be a debate about the possibility to apply the definition for $K_\text{a}$ to the water molecules $\ce{H2O}$ considering that they are not a highly diluted solute since they constitute the solvent. Is this a valid issue and should then the acidity of $\ce{H3O+}/\ce{H2O}$ be determined in another solvent than $\ce{H2O}$ to be compared with the other acids?&lt;/p&gt;&#xA;" OwnerUserId="5135" LastEditorUserId="5135" LastEditDate="2018-01-21T15:02:11.403" LastActivityDate="2021-01-20T08:38:23.017" Title="What is the pKa of the hydronium, or oxonium, ion (H3O+)?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9848" PostTypeId="2" ParentId="9842" CreationDate="2014-04-09T14:42:50.357" Score="2" Body="&lt;p&gt;Mineral oil serially penetrates the wood to render its surface hydrophobic.  Nothing eats mineral oil and it does not easily oxidize (go rancid).  It is also the solvent vehicle to transport beeswax (plasticizer, compatibilizer, internal soft sealant) and the hard wax surface sealant. Nothing polymerizes (re linseed or tung oil) or chemically bonds to the wood.  All three are FDA GRAS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like dissolves like.  While the surface will be sealed against polar materials and provide no food for bacteria and fungi of itself (and absence of water is a good biostat), it can imbibe fats and oils.  Butcher block surfaces and such are best made of dense oily woods that are naturally sealed against water.  Top them off.  Lignin is a fair antibacterial and antifungal agent.  Polyethylene cutting boards have higher sustained levels of viable bugs than do wood cutting boards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://faculty.vetmed.ucdavis.edu/faculty/docliver/Research/cuttingboard.htm&quot; rel=&quot;nofollow&quot;&gt;http://faculty.vetmed.ucdavis.edu/faculty/docliver/Research/cuttingboard.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-09T14:42:50.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9849" PostTypeId="2" ParentId="9838" CreationDate="2014-04-09T14:48:13.103" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://connection.ebscohost.com/c/articles/25511593/oxidation-nitric-acid-aliphatic-alcohols-diols-carboxylic-acids&quot; rel=&quot;nofollow&quot;&gt;http://connection.ebscohost.com/c/articles/25511593/oxidation-nitric-acid-aliphatic-alcohols-diols-carboxylic-acids&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pages.towson.edu/ladon/orgrxs/reagent/oxidizer.htm&quot; rel=&quot;nofollow&quot;&gt;http://pages.towson.edu/ladon/orgrxs/reagent/oxidizer.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a dangerous thing to do for the easy detonation of nitrate esters.  Activated and stabilized alcohols, e.g., benzoin to benzil, are reasonable.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-09T14:48:13.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9851" PostTypeId="1" AcceptedAnswerId="10584" CreationDate="2014-04-09T16:40:43.997" Score="3" ViewCount="3914" Body="&lt;p&gt;Why do certain solutes dissolved in water release heat and other solutes absorb heat?&lt;/p&gt;&#xA;" OwnerUserId="5138" LastActivityDate="2014-05-17T20:56:10.543" Title="Heat when dissolving solutes in water" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9852" PostTypeId="2" ParentId="9838" CreationDate="2014-04-09T16:46:44.810" Score="1" Body="&lt;p&gt;The reaction starts with the formation of an organic nitric acid ester (as these compounds are dangerously unstable, as Uncle Al already pointed out, I would not recommend to use nitric acid for this oxidation). The negatively charged nitrate oxygen abstracts an alpha proton from the alkyl group, and $\ce{HNO2}$ is eliminated, yielding the aldehyde. Nucleophilic attack of nitrate on the carbonyl group and subsequent elimination of another equivalent $\ce{HNO2}$ finally yields the carboxylate.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-09T16:46:44.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9853" PostTypeId="2" ParentId="4606" CreationDate="2014-04-09T17:26:11.650" Score="3" Body="&lt;p&gt;&lt;strong&gt;The self-accepted answer given here is wrong.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each line in a bond block of a MOL file represents &lt;strong&gt;one bond&lt;/strong&gt; between two atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line in a &lt;strong&gt;bond block&lt;/strong&gt; has the following compact format:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;111222tttsssxxxrrrccc&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With other words, each entry (=different number or letter) in such a line can take up to three spaces. If it needs less. e.g. when the atom number is smaller than 100, the data is right-aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;111&lt;/code&gt; is the number of the first atom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;222&lt;/code&gt; is the number of the second atom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These atom numbers refer to the position of the atom in the &lt;strong&gt;atom block&lt;/strong&gt; in the same file, where the cartesian coordinates and the atom type (element) are given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;ttt&lt;/code&gt; is the number decoding the bond type: &lt;code&gt;1=single&lt;/code&gt;, &lt;code&gt;2=double&lt;/code&gt;, &lt;code&gt;3=triple&lt;/code&gt;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;sss&lt;/code&gt; describes the stereochemistry, &lt;code&gt;1&lt;/code&gt; is a single bond pointing up, &lt;code&gt;6&lt;/code&gt; is a single bond pointing down, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter are used in the flat 2D MOL file of morphine cited in the question, e.g. for the only hydrogen atom in the file: a bridgehead H-atom.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-09T17:26:11.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9854" PostTypeId="2" ParentId="9851" CreationDate="2014-04-09T21:11:49.243" Score="1" Body="&lt;p&gt;Enthalpy of hydration is exothermic, Lewis acid-base neutralization.  Entropy of dissolution can be either positive or negative. Negative - ordering effect of ion on solvent is greater than the entropy increase of the crystal (highly ordered) lattice breaking down. Positive - increase in entropy because the solvent hydrogen bonding is disrupted.  Dissolving KOH is a very large exotherm,  Dissolving urea in water is a very large endotherm.  This is descriptive and qualitative,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/kosmos.html&quot; rel=&quot;nofollow&quot;&gt;http://www1.lsbu.ac.uk/water/kosmos.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We invite the physical chemists to put a quantitative mathematical model on it.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4222" LastEditDate="2014-05-17T20:56:10.543" LastActivityDate="2014-05-17T20:56:10.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9855" PostTypeId="1" CreationDate="2014-04-10T00:06:27.043" Score="1" ViewCount="59" Body="&lt;p&gt;I am doing a lab for school and I do not know what this question is asking. In this lab we are making electrochemical cells from different metals. &lt;/p&gt;&#xA;" OwnerUserId="5139" LastActivityDate="2014-09-08T22:17:58.147" Title="how to determine the relative differences in activity between metals" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9857" PostTypeId="1" AcceptedAnswerId="9861" CreationDate="2014-04-10T01:17:39.317" Score="13" ViewCount="6815" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A sample of &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2}$&lt;/span&gt; is placed in the flame of a Bunsen burner. The component of the substance which is responsible for the production of the flame colour is:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ca}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Why is the textbook answer &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca}$&lt;/span&gt; and not &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-20T08:20:31.907" Title="Flame test: Is the metal atom or the metal ion responsible for the flame colour?" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;ions&gt;&lt;color&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9860" PostTypeId="1" AcceptedAnswerId="9864" CreationDate="2014-04-10T04:32:20.687" Score="1" ViewCount="1774" Body="&lt;p&gt;Some references say that alkanes elute faster than alkenes, whereas in my textbook the higher volatility of alkenes means they elute faster than alkanes. This is for gas chromatography. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alkenes are more polar than alkanes due to more electron-withdrawing&#xA;  $\text{sp}^2$-hybrid orbitals (more s-character draws electrons closer to&#xA;  nucleus).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone please clarify - bit confused. &lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="189" LastEditDate="2014-04-10T05:10:59.820" LastActivityDate="2014-04-10T07:52:37.830" Title="Elution Rate - Alkanes VS Alkenes in Gas Chromatography" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9861" PostTypeId="2" ParentId="9857" CreationDate="2014-04-10T04:51:09.503" Score="13" Body="&lt;p&gt;The phenomenon that is witnessed during a flame test is an &quot;atomic emission&quot;.&#xA;This statement may seem inappropriate, since it is a solution of metal ions (and not atoms) that is tested. The reason for calling it atomic emission lies in the process occurring in the flame. One of the steps of the process is an atomization step. That is, the flame converts the metal ions into atoms. When a solution of sodium chloride is placed in a flame, for example, the solvent evaporates, leaving behind solid crystalline sodium chloride. This evaporation is then followed by the dissociation of the sodium chloride crystals into individual ground state atoms - a process that is termed atomization. Thus sodium atoms are actually present in the flame at this point rather than sodium ions, and the process of light emission actually involves these atoms rather than the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GCveC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-04-10T04:51:09.503" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9862" PostTypeId="2" ParentId="9836" CreationDate="2014-04-10T05:23:40.310" Score="0" Body="&lt;p&gt;Actually, you can. $\ce{H2SO4}$ liberates $\ce{2 H+}$ ions which is neutralized by the $\ce{OH^-}$ ions. The $K_\mathrm{a}$ for water is $10^{-14}$, this means that the concentration of those ions are very low compared to the $\ce{K^+}$ and the $\ce{SO4^{2-}}$ ions. You just need to use the stoichiometry of the equation properly to apply it to $M_aV_a=M_bV_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use weak acids and bases if you allow them time to react. Most of them will react in less than a second, but I seen a few that take several seconds. In fact, the best way to determine the concentration of sodium hydroxide solution (since it can absorb carbon dioxide from the air) is to weigh dried potassium hydrogen phthalate (a weak acid) to determine the concentration.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="24052" LastEditDate="2018-06-13T08:16:08.217" LastActivityDate="2018-06-13T08:16:08.217" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="9863" PostTypeId="2" ParentId="9800" CreationDate="2014-04-10T05:52:02.587" Score="2" Body="&lt;p&gt;Just as a quick answer, because apparently some people are in doubt so let me try to clean up the confusion.&lt;/p&gt;&#xA;&lt;h3&gt;The 'physics part'&lt;/h3&gt;&#xA;&lt;p&gt;Surface tension that people &lt;em&gt;normally&lt;/em&gt; talk about is the liquid-air surface tension, in this case water-air, and I am pretty sure this is the surface tension you are referring to in your question. &lt;strong&gt;This water-air surface tension is a property of the water and air and has absolutely nothing to do with the physical shape or material of the container&lt;/strong&gt;.&#xA;For completeness, I have to mention that the contact with the walls that have different chemical properties will change the shape of the water-air interface a bit, which you might misinterpret as a change in water-air surface tension but this is caused by the water-container and container-air surface tensions which indeed DO depend on the container.&lt;/p&gt;&#xA;&lt;h3&gt;The 'chemistry part'&lt;/h3&gt;&#xA;&lt;p&gt;What &lt;em&gt;could&lt;/em&gt; * change the water-air surface tension is the exchange of some material out of the container into the water or the other way around. This would mean that you are not really looking at the water-air surface tension anymore, but rather at the (water&lt;span class=&quot;math-container&quot;&gt;$+$&lt;/span&gt;some solute)-air surface tension. Which might indeed be different, typically lower.&lt;/p&gt;&#xA;&lt;p&gt;If this exchange is the mechanism then the time over which the surface tension will change is dependent on the kinetics of that process and without knowing what is being exchanged it is impossible to tell you what the time span will be, it could be seconds, it could be hours. Note that the explanation with stuff leaking out of or into the lota would mean that (assuming there will be equilibrium at some point) storing water in the lota and then drinking it from the glass would be the same as drinking it directly from the lota.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;*I'm not saying it does I have no clue whether this happens with a lota&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-04-10T05:59:46.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9864" PostTypeId="2" ParentId="9860" CreationDate="2014-04-10T07:52:37.830" Score="1" Body="&lt;p&gt;Chromatography in general is a separation method in which the partition equilibrium of a compound between two phases is permanently disturbed since one of the phases is changing/exchanged (mobile phase).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the mobile phase in gas chromatography is the carrier gas.&#xA;The stationary phase isn't just a hollow tube but a capillary covered with various differently functionalized materials, that may be rather unpolar (squalane), more polar (polyethylene glycol) or polar and shape-selective (cyclodextrines). Other very common materials are alkyl-substituted polysiloxanes, such as polydimethylsiloxane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the latter stationary phase, Alan Katritzky and coworkers published &lt;em&gt;QSPR: the correlation and quantitative prediction of chemical and physical properties from structure&lt;/em&gt;, &lt;em&gt;Chem. Soc. Rev.&lt;/em&gt;, &lt;strong&gt;1995&lt;/strong&gt;, &lt;em&gt;24&lt;/em&gt;, 279-287 (&lt;a href=&quot;http://dx.doi.org/10.1039/CS9952400279&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;, &lt;a href=&quot;http://www.ark.chem.ufl.edu/published_papers/pdf/377.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;). Here, the retention times for dodecane and 1-dodecene were given as 4.59 and 4.45 min, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ladislav Sojak published a freely available 35-page review &lt;em&gt;Separation and identification of isomeric hydrocarbons by capillary gas chromatography and hyphenated spectrometric techniques&lt;/em&gt; in &lt;em&gt;Petroleum &amp;amp; Coal&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;4(3)&lt;/em&gt;, 1-35 (&lt;a href=&quot;http://www.vurup.sk/sites/vurup.sk/archivedsite/www.vurup.sk/pc/vol46_2004/issue3/pdf/Sojak_1_35.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;). You might want to have a look at this article on the effect of having &lt;strong&gt;non-terminal&lt;/strong&gt; $\ce{C=C}$ double bonds and cis-trans isomers of longer alkenes, which therefore have significantly different shapes.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-10T07:52:37.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9866" PostTypeId="1" CreationDate="2014-04-10T10:08:08.457" Score="1" ViewCount="5387" Body="&lt;p&gt;Is there any trick to memorize the angles and sides of 7 crystal systems of crystalline solids?&lt;br&gt;&#xA;1. Cubic&lt;br&gt;&#xA;2. Tetragonal&lt;br&gt;&#xA;3. Orthorhombic&lt;br&gt;&#xA;4. Monoclinic&lt;br&gt;&#xA;5. Hexagonal&lt;br&gt;&#xA;6. Rhombohedral or Trigonal&lt;br&gt;&#xA;7. Triclinic&lt;/p&gt;&#xA;" OwnerUserId="5119" LastActivityDate="2014-06-09T16:58:45.890" Title="How to memorize 7 crystal systems?" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9867" PostTypeId="1" CreationDate="2014-04-10T12:28:51.203" Score="3" ViewCount="973" Body="&lt;p&gt;I am studying a reaction using electronic structure methods using various software packages (ORCA,GAMESS, G09) and have found two possible transition structures between product and reactant. Both have imaginary frequencies, but each has a different energy value, frequencies and geometries. Some software coincide in the found transition state while other differs, adding to the confusion.&#xA;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible to have multiple transition states, if so what is a good criteria to choose between candidate states? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Should I expect the frequency to be within certain range? (for example if the reaction is the movement of a H atom within the molecule)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there some quantitative value(s) (energy, vibration amplitude/frequency, vibration 3d direction vectors, 3d geometry, etc.) that can be used as a reasonable measure to choose the &quot;correct&quot; state? (That is to avoid visualizing the imaginary mode of vibration or computing the IRC).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4340" LastEditorUserId="4340" LastEditDate="2014-04-10T12:42:57.633" LastActivityDate="2014-04-10T12:42:57.633" Title="Are multiple transition states possible? Criteria to choose the correct one?" Tags="&lt;reaction-mechanism&gt;&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;reaction-coordinate&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9868" PostTypeId="1" CreationDate="2014-04-10T13:26:51.507" Score="8" ViewCount="1435" Body="&lt;p&gt;I need help identifying the following 5 substances; G, Y, T, L and J&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;G is a white solid which is insoluble in water. Consider the following reactions carried out on solid G.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt;&lt;br /&gt;&#xA;Excess dilute nitric acid is added to a small amount of G in a test-tube. A colourless and odourless gas Y is given off and it is noticed that Y extinguishes a lighted splint.&lt;br /&gt;&#xA;The solution remaining in the test tube is divided into two different test tubes, sample A and sample B.&lt;br /&gt;&#xA;&lt;strong&gt;Step 2:&lt;/strong&gt;&lt;br /&gt;&#xA;A few drops of potassium iodide solution are added to a sample A and a bright yellow precipitate L is observed.&lt;br /&gt;&#xA;&lt;strong&gt;Step 3:&lt;/strong&gt;&lt;br /&gt;&#xA;Sodium hydroxide solution is added drop wise to sample B. A dense white precipitate, T, is immediately observed. Precipitate T dissolves when excess sodium hydroxide solution is added to it.&lt;br /&gt;&#xA;&lt;strong&gt;Step 4:&lt;/strong&gt;&lt;br /&gt;&#xA;When some G is heated in a hard glass tube, gas Y which is colourless and odourless is given off. The solid J remaining in the hard glass tube changes colour on cooling from red-brown to yellow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;These are my solutions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G = &lt;span class=&quot;math-container&quot;&gt;$\ce{PbCO3}$&lt;/span&gt; because&lt;/li&gt;&#xA;&lt;li&gt;Y = &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; - colourless and odourless gas&lt;/li&gt;&#xA;&lt;li&gt;L = &lt;span class=&quot;math-container&quot;&gt;$\ce{PbI2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;T = &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb(OH)2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;J = &lt;span class=&quot;math-container&quot;&gt;$\ce{PbO2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Are they correct or not?&lt;/p&gt;&#xA;" OwnerUserId="5143" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-06-30T07:57:12.190" Title="Identifying 5 unknown substances" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9869" PostTypeId="2" ParentId="9868" CreationDate="2014-04-10T13:54:55.617" Score="11" Body="&lt;p&gt;If you don't have any idea:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with the hypothesis that your starting material is the salt of some metal.&lt;/li&gt;&#xA;&lt;li&gt;Some of the test are apparently useful to identify the cation, others tackle the anion.&lt;/li&gt;&#xA;&lt;li&gt;Is it likely that the flame-extinguishing gas comes from the cation?&lt;/li&gt;&#xA;&lt;li&gt;If it is not the cation, it must be the anion from which the gas is formed.&lt;/li&gt;&#xA;&lt;li&gt;Think in the main property of &lt;strong&gt;diluted&lt;/strong&gt; nitric &lt;strong&gt;acid&lt;/strong&gt;. Is it an oxidant in diluted form, or just something else?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I almost completely agree with the solution given in the question, except for the last step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three known lead oxides are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{PbO2}$: dark brown to black, strong oxidant&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Pb3O4}$: red-orange, used as a rust-proof primer (in German: &lt;em&gt;Bleimenninge&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{PbO}$: (normally) yellow&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When dark $\ce{PbO2}$ is heated, it splits off oxygen to form red $\ce{Pb3O4}$, when it's heated up more, more oxygen is released and yellow $\ce{PbO}$ is formed. Properly done in a quartz tube the colours of the German flag can be produced this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the puzzle, we are not starting with $\ce{PbO2}$ but with $\ce{PbCO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\ce{PbCO3}$(= &lt;strong&gt;G&lt;/strong&gt;) is heated, $\ce{CO2}$ (= &lt;strong&gt;Y&lt;/strong&gt;) is released. The questions mentions a &lt;em&gt;hard glass tube&lt;/em&gt;, which suggests that the tube has to withstand high temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that the reaction simply is $\ce{PbCO3 -&amp;gt;[\Delta] PbO + CO2 ^}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible explanation for the colour change would be that at high temperatures, red $\ce{\alpha-PbO}$ is initially formed, which converts to yellow $\ce{\beta-PbO}$ upon cooling.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-06-29T08:12:20.420" LastActivityDate="2018-06-29T08:12:20.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9870" PostTypeId="2" ParentId="9866" CreationDate="2014-04-10T15:40:39.827" Score="1" Body="&lt;p&gt;The first step is understanding what you have.  The seven crystal systems are derived from decreasing symmetry (lifting degeneracies) of the cubic system by altering lengths and angles.  Example: Cube volume is $\ce{a^3}$.  Parallelepiped volume with all symmetries explicit is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Parallelepiped&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Parallelepiped&lt;/a&gt;&lt;BR&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QxYI5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rockhounds.com/rockshop/xtal/index.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rockhounds.com/rockshop/xtal/index.shtml&lt;/a&gt; &lt;BR&gt;&#xA;Good reading about crystal systems and their expression.  &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Space_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Space_group&lt;/a&gt; &lt;BR&gt;&#xA;As usual, nomenclature is a mess. &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hermann%E2%80%93Mauguin_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Hermann%E2%80%93Mauguin_notation&lt;/a&gt; &lt;BR&gt;&#xA;That's a winner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystallographers often translate the unit cell within the lattice to make its contents  &quot;nice.&quot; The rhombohedral system is particularly nasty for choosing the unit cell itself.  65 Sohnke space groups of 230 overall contain chiral formula units.  Racemic screw axes make that interesting, ${2_1}$, ${4_2}$, ${6_3}$.  Within the 65, 11 enantiomorphic space groups - themselves chiral without regard to unit cell contents - allow the same substance to crystallize in two mirror-image space groups.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-10T15:40:39.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9871" PostTypeId="1" CreationDate="2014-04-10T16:13:32.753" Score="2" ViewCount="40173" Body="&lt;p&gt;I am trying to make  5N $\ce{HCl}$ solution. So far I understood that you need to get the molecular weight (36.5) and divide it by the number of $\ce{H+}$ that react in the acid-base reactions. The problem is that my $\ce{HCl}$ stock is 37 %. How should I take this into account?&lt;/p&gt;&#xA;" OwnerUserId="5145" LastEditorUserId="73" LastEditDate="2014-04-11T21:37:55.960" LastActivityDate="2015-07-30T15:44:18.713" Title="How do I prepare a 5N HCl solution?" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9872" PostTypeId="1" CreationDate="2014-04-10T17:47:13.737" Score="11" ViewCount="33623" Body="&lt;p&gt;Ever had a ballpoint pen that would all of a sudden hit a 'dead spot' on a sheet of paper and refuse to write? It's clearly not the pen, because when moved off to the corner it would draw perfect circles instantly. Move it back to the spot, and it stops working again. Usually I can get something blotchy if I write out the word a dozen or so times while pressing hard.  What's going on with the paper and ink here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some pages of my notebook are like this and it's very annoying. My best guess is that it got splashed and something about the paper changed, but my guess probably isn't a very good one since it doesn't have the signature crinkly waviness of wetted paper.  &lt;/p&gt;&#xA;" OwnerUserId="4068" LastActivityDate="2017-06-05T20:45:17.450" Title="What is going on when pen stops working on a section of paper?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9873" PostTypeId="2" ParentId="9872" CreationDate="2014-04-10T18:00:46.967" Score="14" Body="&lt;p&gt;If in doubt, try it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just spilled some olive oil on a sheet of paper and wiped it off again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result: The ballpoint pen does not write on the oily spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you &lt;em&gt;fat-finger&lt;/em&gt; the paper?&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment above confirmed the finding in the question but didn't give an explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, two hours later it is still not possible to write on the oil-soaked and more transparent parts of the paper with a ballpoint pen, while a felt-tip pen (here: an Edding, ther German equivalent to a Sharpie) leaves perfect lines on every part of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper is fibrous material with a porous structure - without that and the capillary effects coming with it, fountain pens wouldn't be able to write on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ballpoint pens, this probably isn't so important - the ink is very viscous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But paper also has a rough surface - and this might be crucial, as already suggested by &lt;em&gt;Lord Stryker&lt;/em&gt; in his comment. The rough paper surface provides friction, allows the tungsten carbide ball to &lt;strong&gt;roll&lt;/strong&gt;, take up ink and spread it on the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a smooth or slippery surface - and this seems to be the case for the oily regions on the paper - the tip of a ballpoint pen only slides but does not roll. As a result, no ink is transfered to the paper.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-10T21:08:08.167" LastActivityDate="2014-04-10T21:08:08.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9874" PostTypeId="1" CreationDate="2014-04-10T18:55:26.867" Score="0" ViewCount="89" Body="&lt;p&gt;The specific heat capacity of liquid water is $4.18 \, \text{J} \, \text{K}^{-1} \, \text{g}^{-1}$. Calculate the energy required to heat $1.0 \, \text{mol}$ of water from $298 \, \text{K}$ to $363 \, \text{K}$. &#xA;My question is: Do I have to change the moles of water into grams or should I sub in $1.0$ into the formula for specific heat capacity?&lt;/p&gt;&#xA;" OwnerUserId="5148" LastEditorUserId="4356" LastEditDate="2014-10-24T23:44:45.100" LastActivityDate="2015-03-31T19:29:19.057" Title="Which units do I use in calculating heat capacity?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2014-10-23T11:59:22.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="9875" PostTypeId="2" ParentId="9874" CreationDate="2014-04-10T19:13:51.210" Score="0" Body="&lt;p&gt;You have the operative value/g.  Have it act upon grams of sample.  Keep your units consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water boils at 373 K, so one presumes it is a reasonably closed container - or evaporation will skew the real world number, leveraged by surface area.  A mole of water is a little more than a tablespoon. Call it about 18 calories/K.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-11T00:20:28.137" LastActivityDate="2014-04-11T00:20:28.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9877" PostTypeId="2" ParentId="2384" CreationDate="2014-04-10T20:05:23.403" Score="0" Body="&lt;p&gt;This is almost all conjecture. Well as you mentioned many smoke molecules are phenolic. Pehnols combined with other materials are often used to create laminates for fabrics, an example is phenol-aldehyde fabric resins. &#xA;Also I believe many types of partial hydrocarbon burnings create aldehydes. Both are able to dissolve in water, and if there is any moisture in the couch it could cling to the water, when the water evaporates from one area you could have some kind of phenol aldehyde remnants sticking around. This would play  well into the surface area argument as well since water likes high surface area, and materials soluble in water moisture would then accumulate readily on high surface area objects.&lt;/p&gt;&#xA;" OwnerUserId="5149" LastActivityDate="2014-04-10T20:05:23.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9878" PostTypeId="1" CreationDate="2014-04-10T20:21:50.457" Score="1" ViewCount="113" Body="&lt;p&gt;As part of intern work I am running simulations on a group of compounds, but as it is I have little in the way of O-Chem experience and I wanted to be able to at least get some direction on naming these compounds (maybe some examples). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I've got the names of the functional groups down right, and I've recognized a few of the reaction types, but after that what do I do? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the other 2 reaction types? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I take the names of the products/reactants and the functional group names and stick em together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about when the functional group listed is bigger than the main group. (example R8 and ring opening)?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0UbSX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should have written F1-F8 but R1-R8 represent functional groups if I'm not mistakened.&#xA;Here are my best guesses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R1:alkane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R2:alkene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R3:alkyne&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R4:amine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R5:alcohol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R6:ketone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R7:amide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R8:benzene&lt;/p&gt;&#xA;" OwnerUserId="5149" LastActivityDate="2014-04-10T21:19:32.097" Title="Naming an ensemble of compounds" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9879" PostTypeId="2" ParentId="4482" CreationDate="2014-04-10T20:37:41.137" Score="3" Body="&lt;p&gt;Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta_{T}(A)+\Delta_{O}(B)&amp;gt;\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly &quot;prefers&quot; the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-04-13T14:42:37.183" LastActivityDate="2014-04-13T14:42:37.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9881" PostTypeId="2" ParentId="9878" CreationDate="2014-04-10T20:47:57.537" Score="2" Body="&lt;p&gt;Pericyclic ring opening, cyclobutene to butadiene (Woodward-Hoffmann rules). &lt;BR&gt;&#xA;Claisen rearrangement, vinyl ether to ketone (re bullvalene). &lt;BR&gt;&#xA;Diels-Alder addition. &lt;BR&gt;&#xA;Sn2 displacement; AgF will move that along (not a soft Lewis base solvent). &lt;BR&gt;&#xA;Alkene to expoxide is &lt;em&gt;not&lt;/em&gt; done with water.  Direct catalytic oxidation, base elimination from the cyanohydrin (first adding HOX), peracids. &lt;BR&gt;&#xA;Primary alcohol to aldehyde is Jones reagent (in water), Collins reagent (in anhydrous pyrdine, and much better. There is no product in the sludge), Swern oxidation (makes you unpopular downwind), fancier stuff. &lt;BR&gt;&#xA;Markovnikov addition of HCl to an alkene. &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Woodward-Hoffmann rules tell you what conditions are required (thermal vs. photochemical) and the associated stereochemistry (suprafacial vs. antarafacial) re conservation of orbital symmetry.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="418" LastEditDate="2014-04-10T21:19:32.097" LastActivityDate="2014-04-10T21:19:32.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9883" PostTypeId="2" ParentId="2384" CreationDate="2014-04-10T21:18:32.927" Score="1" Body="&lt;p&gt;Combined with the high surface area argument of @Paul J. Gans, I could conceive that since compounds like syringol have several polar and hydrogen bonding groups, they will cling much more readily to typical fabrics such as polyesters. Gasoline and natural gases, for example, have pretty distinctive smells but in my experience tend not to linger in fabrics the way smoke does and they are distinctly non-polar.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastActivityDate="2014-04-10T21:18:32.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9884" PostTypeId="2" ParentId="2384" CreationDate="2014-04-10T21:46:13.023" Score="7" Body="&lt;p&gt;My theory is that it's not the molecules that are mentioned that cling to the fabric but much larger (yet still tiny) particles of soot that cling to the fabric releasing the smelly molecules. this provides an explanation for how other smelly molecules do not linger on fabrics. In addition to that Syringol, as mentioned, is a solid at room temperature contributing to its clinging to the cloth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could prove for a interesting experiment, unfortunately i do not have a lab available at this time&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2014-04-10T21:46:13.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9885" PostTypeId="2" ParentId="9874" CreationDate="2014-04-10T22:52:57.423" Score="3" Body="&lt;p&gt;Yes, you would have to change it to grams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Units must be consistent. Specific heat is defined as the energy required to raise one gram of water by one kelvin. At atmospheric pressure in a closed container, the specific heat of water is 4.184 joules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One mole of water is 18.016 grams. So: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E =mc \Delta T $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m$ is the mass of the water, $c$ is the specific heat capacity of water, $\Delta T $ is the change in temperature, and $E$ is the energy required. Substitute in your numbers, and you shall get the right answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = (18.016\ \mathrm{g})(\frac{4.184\ \mathrm{J}}{\mathrm{g\ K}})(363\ \mathrm{K} - 293\ \mathrm{K})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = 5280\ \mathrm{J}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to use moles directly instead of converting to grams, you would have had to use the &lt;em&gt;molar heat capacity&lt;/em&gt;. Generally, people prefer to use specific heat. Engineers sometimes use the &lt;em&gt;volumetric heat capacity&lt;/em&gt; as well. &lt;/p&gt;&#xA;" OwnerUserId="5151" LastEditorDisplayName="user7951" LastEditDate="2015-03-31T19:29:19.057" LastActivityDate="2015-03-31T19:29:19.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9886" PostTypeId="1" AcceptedAnswerId="9887" CreationDate="2014-04-11T00:58:16.117" Score="2" ViewCount="4637" Body="&lt;p&gt;According to Modern Atomic Theory:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ratio in which the different atoms combine may be fixed and integral but may not always be simple.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The example given is that in sugar molecule the ratio of $\ce{C,H}$ and $\ce{O}$ atoms is $12:22:11$ &lt;em&gt;which is not simple&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, what I see here is that 12,22 and 11 are whole numbers, then why is this &lt;em&gt;not&lt;/em&gt; a simple ratio?&lt;/p&gt;&#xA;" OwnerUserId="5152" LastEditorUserId="5026" LastEditDate="2018-03-09T10:02:59.127" LastActivityDate="2018-03-09T10:04:45.933" Title="What is a simple ratio in which atoms combine?" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9887" PostTypeId="2" ParentId="9886" CreationDate="2014-04-11T01:09:59.477" Score="3" Body="&lt;p&gt;The ratio of C, H and O atoms of 12:22:11 is likely to be sucrose. Glucose has a simpler ratio of 1:2:1 (even though it has 6 carbons). The usual context for presenting the theory is with inorganic compounds with no more than a 2:3 or 3:5 ratio. With this limitation, they use the words simple ratio since the numbers are small and can easily be found. The ratio of 12:22:11 could be confused with another ratio if your calculations are off by 20%.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="5026" LastEditDate="2018-03-09T10:03:42.340" LastActivityDate="2018-03-09T10:03:42.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9888" PostTypeId="2" ParentId="9886" CreationDate="2014-04-11T01:40:12.753" Score="2" Body="&lt;p&gt;It is often not true within inorganic chemistry (isomorphous substitution, fractional valences; e.g., ruby, sapphire).  Discrete organics are overall exact single compositions.  Polymers are not.  Hydrocarbons have simple empirical formulas.  Add heteroatoms and all bets are off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hemoglobin is $\ce{C2952H4664O832N812S8Fe4}$. Every high molecular weight Kraton thermoplastic elastomer triblock polymer contains a single atom of silicon at its exact middle.  Plasma-drawn Spectra ultra-high molecular weight gel-crystallized polyethylene, some 20 times by weight stronger than steel cable, is $\ce{CH2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dicoronylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dicoronylene&lt;/a&gt;: $\ce{C48H20}$&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bacitracin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bacitracin&lt;/a&gt;: $\ce{C66H103N17O16S}$&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-03-09T10:04:45.933" LastActivityDate="2018-03-09T10:04:45.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9890" PostTypeId="1" AcceptedAnswerId="9904" CreationDate="2014-04-11T03:05:41.247" Score="0" ViewCount="1248" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L1N2x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the data booklet we must refer too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XtJLM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QHS6a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the exam paper solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mhu2W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont understand how they conclude this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Peak areas and the chemical shifts seem to be different for me. For example there is a peak area of 2 for the chemical shift (ppm) of 2.0. How can this be when the group is CH3-COOR... isn't the peak area 3 if there are 3 hydrogens? Can someone please help me clarify reading this table, and deducing structures. How do link the R groups and when they bold H, does that mean the H can be any number? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="418" LastEditDate="2014-04-11T20:07:05.587" LastActivityDate="2014-04-11T20:07:05.587" Title="1H-NMR and deducing structures" Tags="&lt;homework&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;mass-spectrometry&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9891" PostTypeId="2" ParentId="9890" CreationDate="2014-04-11T03:14:27.353" Score="0" Body="&lt;p&gt;For the chemical shift of 2.0, the person used the example of a methyl group attached to the carboxyl group, he didn't mean to imply that there are 3 hydrogens. In the structure he presented in B, you will see that there are only 2 hydrogens, not 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for the chemical shift of 4.1, he used ethyl (which has 2 hydrogens) instead of methyl (3 hydrogens).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-11T03:14:27.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9892" PostTypeId="1" AcceptedAnswerId="9893" CreationDate="2014-04-11T06:43:05.950" Score="0" ViewCount="6569" Body="&lt;p&gt;Here is the question and answer out of an exam paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Afp3s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I thought UV-Visible can also use radiation in the visible spectrum. Also when analyzing sodium chloride (a molecule), then UV-Visible would be more appropriate than AAS, because AAS would be used for just Na (Sodium). Would I be wrong to have said UV-Visible?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="2094" LastEditDate="2014-04-11T09:20:41.473" LastActivityDate="2014-04-11T09:25:04.573" Title="UV-Visible Spectroscopy in the analysis of sodium chloride in potato chips" Tags="&lt;homework&gt;&lt;electrons&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9893" PostTypeId="2" ParentId="9892" CreationDate="2014-04-11T09:12:18.307" Score="1" Body="&lt;p&gt;NaCl solution as you know is almost transparent in the VIS region, UV-VIS spectroscopy can be used to determinate concentration of soluble salt see &lt;a href=&quot;http://faculty.washington.edu/ghp/images/stories/pubPDF/2008-Chai_Zheng_Zhao-Spectroscopic_studies_of_solutes_in_aqueous_solution.pdf&quot; rel=&quot;nofollow&quot;&gt;J. Phys. Chem. A 2008, 112, 2242-2247&lt;/a&gt; however UV part of the spectra is used instead of VIS part, so this does not fit the restrictions of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely AAS has a greater sensibility and so is the best technique to use in this case.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-04-11T09:25:04.573" LastActivityDate="2014-04-11T09:25:04.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9894" PostTypeId="2" ParentId="9871" CreationDate="2014-04-11T10:44:32.457" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_chloride&quot;&gt;Hydrogen chloride&lt;/a&gt; in its standard state is a gas. We usually do not make solutions of $\ce{HCl}$ by the method of dissolving a known mass of the the solute in enough solvent to make the appropriate volume of the specific concentration since gaseous substances are difficult to weigh (and weight is one of the most direct ways we have to determine mass in the lab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrochloric acid (aqueous solutions of $\ce{HCl}$) is commercially available in several concentrations, usually $1\ \text{N}$, $0.1\ \text{N}$. and &quot;concentrated&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Concentrated&quot; simply means that you have in that bottle the highest concentration of $\ce{HCl}$ water that is stable. For $\ce{HCl}$, a concentrated solution is 36-38% by mass. If you know the density of the solution (should be on the bottle), you can convert % by mass into moles of $\ce{HCl}$ per liter of solution (you also need the molar mass of $\ce{HCl}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you had a solution that was 20% $\ce{HCl}$ by mass with a density of $1.098 \ \text{g/mL}$, you can determine molarity by converting mass of $\ce{HCl}$ into moles and mass of solution into liters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$20\% \ \ce{HCl} \ \text{by mass}=\frac{20 \text{ g }\ce{HCl}}{100\ \text{g solution}}$$&#xA;$$20 \text{ g }\ce{HCl} \times \frac{1 \text{ mol }\ce{HCl}}{36.46 \ \text{g }\ce{HCl}}=0.5485\ \text{mol }\ce{HCl}$$&#xA;$$100\ \text{g solution}\times \frac{1\ \text{mL solution}}{1.098\ \text{g solution}}\times \frac{1\text{ L}}{1000\ \text{mL}}=0.09107\ \text{L solution}$$&#xA;$$\frac{0.5485\ \text{mol }\ce{HCl}}{0.09107\ \text{L solution}}=6.023 \ \text{M}$$&#xA;However, other folks have the done the math for you. The Wikipedia article of hydrochloric acid lists the concentrations (and densities) for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrochloric_acid#Physical_properties&quot;&gt;several $\ce{HCl}$ solutions&lt;/a&gt; as a function of % by mass. This &lt;a href=&quot;https://www.erowid.org/archive/rhodium/chemistry/equipment/molarity.html&quot;&gt;listing of concentrated reagents&lt;/a&gt; lists $37.2\% \ \ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a concentration for $37.2\% \ \ce{HCl}$, you can proceed to work through the dilution process using $c_1 v_1 = c_2 v_2$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-11T10:44:32.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9895" PostTypeId="1" AcceptedAnswerId="9901" CreationDate="2014-04-11T10:44:44.937" Score="1" ViewCount="257" Body="&lt;p&gt;I am doing a project in which I fill a cylinder with a volume of $636cm^2$ (made of stainless steel) with molten salt: 60% Sodium Nitrate ($NaNO_3$) and 40% Potassium Nitrate($KNO_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The container might reach (worst case scenario) -35 degrees Celsius in a cold winter night and in the summer (with ideal conditions) up to 700 degrees Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find out if I should add a relief valve (like in old water boilers) or just seal it tight. This of course depends o how much the pressure increases between those 2 states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that molten salt  melts at 221 degrees Celcius and starts  freezing when it reaches 238 degrees Celcius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9oPnS.jpg&quot; alt=&quot;Molten Salt&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any relevant data, so I don't know how I could find the result for this. Any ideas or suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My worry is that it would blow up due to pressure or the lid would blow open...&lt;/p&gt;&#xA;" OwnerUserId="5156" LastEditorUserId="5156" LastEditDate="2014-04-11T10:50:41.900" LastActivityDate="2014-04-11T14:36:02.410" Title="Change in pressure or volume of molten salt" Tags="&lt;thermodynamics&gt;&lt;materials&gt;&lt;heat&gt;&lt;mixtures&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9896" PostTypeId="2" ParentId="2384" CreationDate="2014-04-11T11:52:00.620" Score="8" Body="&lt;h2&gt;&lt;strong&gt;Why is the smell of smoke so persistent, while many other strong smells are not?&lt;/strong&gt;&lt;/h2&gt;&#xA;&lt;p&gt;In fact this is a tricky question, because you can't make such a comparison. According to &lt;a href=&quot;http://goldbook.iupac.org/S05719.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC Gold book&lt;/a&gt; &lt;strong&gt;Smoke&lt;/strong&gt; is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An aerosol originating from combustion, thermal decomposition or&#xA;thermal evaporation. Its particles may be solid (magnesium oxide&#xA;smoke) or liquid (tobacco smoke).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it is a fluid colloidal system in which the &lt;em&gt;&amp;quot;the dispersed phase is a solid, a liquid or a mixture of both and the continuous phase is a gas&amp;quot;&lt;/em&gt; others strong smells that you smell during your every day life, are mainly already gasses. The interaction areosol-material an a gas-material is completely different.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Interaction between fabric and gasses&lt;/h2&gt;&#xA;&lt;p&gt;We can decompose the interaction in two different phenomena:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;gas trapped inside the little voids inside your fabric&lt;/li&gt;&#xA;&lt;li&gt;gas &lt;strong&gt;adsorbed&lt;/strong&gt; in the fibers surface&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The first is governed by &lt;a href=&quot;https://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion&quot; rel=&quot;noreferrer&quot;&gt;Fick law of diffusion&lt;/a&gt; this process reach the equilibrium very quickly e.g. if you enter in a room with a gas after a couple of minutes the concentration of gas inside the voids of your fabric will be the same as the outside concentration.&lt;/p&gt;&#xA;&lt;p&gt;The second is more hard to determine, however is more slower because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Adsorption&quot; rel=&quot;noreferrer&quot;&gt;adsoprtion&lt;/a&gt; occur if there are free sites where the molecule can be placed, and this is a competitive process between all the others gasses present in larger quantities in the atmosphere, that already occupy these sites.&lt;/p&gt;&#xA;&lt;p&gt;However if you stay a day in a room with a gas is possible the some gas molecule will be &lt;strong&gt;ad&lt;/strong&gt;-sorbed in your fabric. This interaction is more persistent, however I think that the amount of substance is quite negligible.&lt;/p&gt;&#xA;&lt;p&gt;So interaction between gasses and fabric is not very persistent.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Interaction between aerosol and fabric&lt;/h2&gt;&#xA;&lt;p&gt;This is completely different because colloidal suspensions are unstable. Aerosol are high energy system that tend to lower their energy forming bigger particles or bonds with other materials. So the deposition of these particles over the surface of others materials will be most of the times a spontaneous process. The high surface of the fabrics indeed make it easier.&lt;/p&gt;&#xA;&lt;p&gt;So what happen when a particulates/aerosols settles on the fabric? This is not  gas trapped inside the little voids of your fabric or gas adsorbed in the fibers surface is a mixture of &lt;em&gt;liquids&lt;/em&gt; and &lt;em&gt;solids&lt;/em&gt;: many big (relatively!) &amp;quot;clusters&amp;quot; of molecules spread all over the fabric, from the tiny pores to the fiber's ravines. What matter here is the vapor pressure of the substance  you got this equilibrium (for a liquid):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Molecule_{(l)} &amp;lt;=&amp;gt; Molecule_{(g)} }$$&lt;/span&gt; and the analogous with &lt;span class=&quot;math-container&quot;&gt;$_{s}$&lt;/span&gt; instead of &amp;quot;l&amp;quot; for the sublimation. For both of these equilibria only the molecules in the liquid-gas or solid-gas interface can evaporate\sublimate, so the cluster surface is very important!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;The case of Syringol&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/48mEx.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the case of Syringol you have a solid that melt at 50–57 °C. So the sublimation is the only path that allows you to smell it. For most solids the vapor pressure is very low so the equilibrium will keep going to the right for a long long time, releasing syringol, until the source end.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-04-11T11:52:00.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9897" PostTypeId="1" CreationDate="2014-04-11T13:04:57.723" Score="0" ViewCount="460" Body="&lt;p&gt;I'm going to be using a finings agent with a positive charge to remove negative trub from a solution (beer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to keep some chelated zinc in the solution - ideally zinc picolinate - so wondering what its charge is? I'm hoping it's neutral...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the answer, are there any zinc chelates that carry a neutral charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all so zinc can remain in the beer after fining to reduce the estrogenic effects of hops.&lt;/p&gt;&#xA;" OwnerUserId="5157" LastEditorUserId="4945" LastEditDate="2014-05-08T08:29:35.573" LastActivityDate="2014-05-08T08:29:35.573" Title="Zinc Picolinate - Positive, negative or neutral charge?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9899" PostTypeId="2" ParentId="9897" CreationDate="2014-04-11T13:40:49.973" Score="1" Body="&lt;p&gt;Zinc picolinate with two ligands per metal ion ($\ce{Zn(C6H4NO2)2}$) is a neutral chelate complex. Deprotonated picolinic acid has one negative charge at the carboxylate group ($\ce{NC5H4-COO^-}$). Two picolinate ligands thus compensate the two positive charges of $\ce{Zn^2+}$, forming a neutral complex in which they act as bidentate ligands. The nitrogen of the pyridine ring and the carboxylate coordinate the Zn center ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6QJmw.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://www.lookchem.com/Zinc-picolinate/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lookchem.com/Zinc-picolinate/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-04-11T16:24:22.950" LastActivityDate="2014-04-11T16:24:22.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9901" PostTypeId="2" ParentId="9895" CreationDate="2014-04-11T14:36:02.410" Score="2" Body="&lt;p&gt;Venting initiates from blockage - anywhere.  All your lines must be hot-traced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrate melts are powerful oxidizing agents.  If containment fails onto organics (e.g., wood) you've got a catastrophe.  Nitrate melts significantly shrink on cooling and solidifying, and they bond to polar surfaces (glass, metal).  If you have  a half-liter of solidified salt and heat the bottom of containment, the melt will expand against the plug of solid salt above it, possibly deforming or rupturing containment.  The very hot and oxidizing liquid shoots out.  Heat a stripe up the side to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrate melts thermally decompose.  Even if the average temp is OK, heating hot spots may give you oxidizing, corrosive, and toxic by inhalation nitrogen oxides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.springer.com/article/10.1361%2F10599490417542&quot; rel=&quot;nofollow&quot;&gt;http://link.springer.com/article/10.1361%2F10599490417542&lt;/a&gt; &lt;BR&gt;&#xA;&quot;Corrosion of stainless steels and carbon steel by molten mixtures of commercial nitrate salts&quot; &lt;BR&gt;&#xA;&lt;a href=&quot;http://prod.sandia.gov/techlib/access-control.cgi/2001/018518.pdf&quot; rel=&quot;nofollow&quot;&gt;http://prod.sandia.gov/techlib/access-control.cgi/2001/018518.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.electrochem.org/dl/ma/197/pdfs/0209.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.electrochem.org/dl/ma/197/pdfs/0209.pdf&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The articles say your hot end is out of the safety zone.  Welds.  Chloride.  Chloride corrosion, especially pitting, can be aggressive. Halide in general may destabilize nitrate.  Ammonium nitrate is rather tame.  Ammonium nitrate plus chloride contamination is Texas City, Texas 16 April 1947. 2300 tons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.energystoragenews.com/Molten%20Nitrate%20Salt%20for%20Solar%20Energy%20Storage.html&quot; rel=&quot;nofollow&quot;&gt;http://www.energystoragenews.com/Molten%20Nitrate%20Salt%20for%20Solar%20Energy%20Storage.html&lt;/a&gt; &lt;BR&gt;&#xA;1600 tonnes of molten nitrate eutectic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might put it behind grout-filled cement block or cinder block with #5 rebar weaving vertical and horizontal, and into the foundation, when you build the wall.  Have a large area vent direction.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-11T14:36:02.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9903" PostTypeId="2" ParentId="7437" CreationDate="2014-04-11T17:11:18.090" Score="1" Body="&lt;p&gt;26-60 femtoseconds depending upon impact angle for $\ce{F- + CH3Cl}$ according to &lt;a href=&quot;http://cg4.sm.chim.unifi.it/~cardini/papers/Martina1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mugnai et al.&lt;/a&gt; (Table IV.)&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-03-17T07:44:30.163" LastActivityDate="2018-03-17T07:44:30.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9904" PostTypeId="2" ParentId="9890" CreationDate="2014-04-11T17:18:49.783" Score="5" Body="&lt;p&gt;I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemical shift (ppm) and peak area (integral)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are in that environment, compared to the number of protons in all the other environments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the areas need to be considered as a ratio instead of absolute values is important: if the molecule was symmetric, with an ethyl group either side of a ketone e.g. C₅H₁₀O you would see a peak ratio of 3:2 (two environments), but there would actually be 6 protons in the first environment and 4 in the second. This is a bit like empirical vs molecular formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some tips for deducing structures&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Start with 'characteristic' shifts. These are usually the lowest and highest shifts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the 4.1 ppm shift is characteristic of ester protons, i.e. -(C=O)OC &lt;strong&gt;H&lt;/strong&gt;₂R (R can also be a proton too!) where, much like the data sheet above, bold is to tell you which protons are being referred to in a more complicated structure. The shift at 0.9 ppm is characteristic of protons which are attached to a standard saturated alkyl chain. This tells you that those protons aren't very near any oxygen, nitrogen or other electron withdrawing or electron donating groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Try and explain the other shifts using functional groups that you have identified from your characteristic shifts &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shift at 2.0 ppm is likely to be alkyl chain protons which are next to an electron withdrawing group, which conveniently you have already identified as an ester group by looking at your characteristic shifts! In this case, that means you can put the 0.9 ppm shifted protons at one end of your structure, because there is nowhere else to put them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Check your peak assignment makes sense when you look at your peak areas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to check is which end of the molecule the 0.9 ppm shifted protons go on. i.e. is the molecule methyl propanoate or ethyl ethanoate? To do this you can look at the peak areas. For this example it is quite easy to check because there are 3 protons at 4.1 ppm, so you know that you have a methyl ester group -(C=O)OC &lt;strong&gt;H&lt;/strong&gt;₃. Our R group from earlier is an H after all! The resulting structure is the one shown in the example: methyl propanoate.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-04-11T17:18:49.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9905" PostTypeId="2" ParentId="9419" CreationDate="2014-04-11T17:58:49.470" Score="3" Body="&lt;p&gt;You can't determine the concentration of CO2 in the water just from measuring the concentration in the air above the water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of CO2 is extremely dependent upon pH. See: &lt;a href=&quot;http://ion.chem.usu.edu/~sbialkow/Classes/3650/CO2%20Solubility/DissolvedCO2.html&quot; rel=&quot;nofollow&quot;&gt;http://ion.chem.usu.edu/~sbialkow/Classes/3650/CO2%20Solubility/DissolvedCO2.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no indication you are considering pH at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility of CO2 also depends upon temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you need to consider whether or not the phases are in equilibrium. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-04-11T17:58:49.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9907" PostTypeId="2" ParentId="9855" CreationDate="2014-04-11T19:17:07.057" Score="1" Body="&lt;p&gt;You have a standard pot of liquid electrolyte into which are dipped non-touching same-spaced paired strips of different clean metals.  The external circuit between the strips is a sensitive voltmeter (with extremely high electrical resistance - hard by zero current flow).  You measure the potential (noting direction!) between all combinations of paired strips, (n)(n-1)/2, and then draw certain conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are amalgams saturated with respect to excess solid metal interesting substrates for this experiment (e.g., aluminum)? How do you construct a standard hydrogen electrode?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-11T19:17:07.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9908" PostTypeId="1" CreationDate="2014-04-11T21:35:02.690" Score="1" ViewCount="18049" Body="&lt;p&gt;One equation to calculate a heat change is $q = c_p \cdot m \cdot\Delta T$. I know that $\Delta T$ can be in either degrees Celsius ($^\circ{}\mathrm{C}$) or kelvin ($\mathrm{K}$). However, one thing that confuses me is how the units cancel if $\Delta T$ is in degrees Celsius, since $c_p$ has units of $\mathrm{J\ g^{-1}\ K^{-1}}$. How does this work?&lt;/p&gt;&#xA;" OwnerUserId="5161" LastEditorDisplayName="user7951" LastEditDate="2016-10-10T06:55:15.427" LastActivityDate="2016-10-10T07:08:24.763" Title="Why can temperature changes have units of either degrees Celsius or kelvin?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9909" PostTypeId="2" ParentId="9908" CreationDate="2014-04-11T21:54:04.623" Score="3" Body="&lt;p&gt;The temperature in kelvin is equal to the temperature in degrees Celsius plus 273.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\Delta T$ will be the same whether the temperature was reported in $\mathrm{K}$ or $^\circ\mathrm{C}$.  As an example if $T_\mathrm{initial} = 0\ ^\circ\mathrm{C}\ (273\ \mathrm{K})$ and $T_\mathrm{final} = 100\ ^\circ\mathrm{C}\ (373\ \mathrm{K})$, $\Delta T$ is 100 degrees no matter whether you used $^\circ\mathrm{C}$ or $\mathrm{K}$ in your computation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2016-10-10T07:04:59.837" LastActivityDate="2016-10-10T07:04:59.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9910" PostTypeId="1" CreationDate="2014-04-12T02:28:23.233" Score="0" ViewCount="192" Body="&lt;p&gt;So, in mass spectrometry we have the following equation to describe a magnetic sector analyser,&#xA;$$\frac{m}{z}=\frac{B^2r^2e}{2V}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say that I have two particles with different mass/charge ratios,&#xA;$$\frac{m_1}{z_1} = 120.9,\quad \frac{m_2}{z_2} = 121.0,$$&#xA;and I have $B=2\text{T}$, $r=25\text{cm}$. For example, what accelerating voltage $V$ do I need to apply to separate these two particles? What criteria do I have to determine whether two particles are separated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** ** &lt;strong&gt;EDIT&lt;/strong&gt; ** **&#xA;I don't know where to start with this. So even though you don't generally answer homework-like questions on this forum, some pointers in the right direction would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="5164" LastEditorUserId="5164" LastEditDate="2014-04-12T02:40:35.597" LastActivityDate="2014-04-13T01:24:10.587" Title="Calculate accelerating voltage required to separate two particles in a magnetic sector analyser" Tags="&lt;analytical-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9911" PostTypeId="1" CreationDate="2014-04-12T02:41:38.350" Score="0" ViewCount="14362" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9y2C7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propan-1-ol : CH3CH2CH2OH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be broken up into:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CH3 + CH2CH2OH&lt;/li&gt;&#xA;&lt;li&gt;CH3CH2 + CH2OH&lt;/li&gt;&#xA;&lt;li&gt;CH3CH2CH2 + OH&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Base peak = most abundant fragment formed (highest peak)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do i know which fragment is the &quot;most abundant?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I know the fragment structure, i can work out the molar mass, and then deduce the m/e value on the assumption of a (+1 charge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the exam paper solution: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ed9ZQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2014-05-06T14:54:03.463" Title="How to calculate m/e value for the base peak in mass spectroscopy?" Tags="&lt;homework&gt;&lt;atoms&gt;&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9912" PostTypeId="1" AcceptedAnswerId="9913" CreationDate="2014-04-12T03:16:57.047" Score="0" ViewCount="3594" Body="&lt;p&gt;Ok, so the following question is given in my text book. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Of the molar conductance value of $\ce{Ca^2+}$ and $\ce{Cl^-}$ at infinite dilution are respectively $118.88\times10^{-4}$ and $77.33\times10^{-4}$ then that of $\ce{CaCl2}$ is  (all have same unit)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So what I did was I added the molar conductivity of $\ce{Ca^2+}$ and added $2\times$ molar conductivity of  $\ce{Cl^-}$. I got the answer. It was one of the options. So I basically wanted to know: Did I do it correctly and is my process correct? And if I did it wrong then what is the correct process?&lt;/p&gt;&#xA;" OwnerUserId="302" LastEditorUserId="13339" LastEditDate="2015-05-18T09:10:01.603" LastActivityDate="2017-01-05T07:19:21.690" Title="Calculating molar conductance of CaCl2" Tags="&lt;homework&gt;&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9913" PostTypeId="2" ParentId="9912" CreationDate="2014-04-12T03:42:13.847" Score="2" Body="&lt;p&gt;Yes, the process is correct, but do you understand why? &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-12T03:42:13.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9914" PostTypeId="1" CreationDate="2014-04-12T07:32:16.987" Score="2" ViewCount="187" Body="&lt;p&gt;I just read about &lt;a href=&quot;http://en.wikipedia.org/wiki/Alchemy&quot; rel=&quot;nofollow&quot; title=&quot;Alchemy&quot;&gt;Alchemy&lt;/a&gt;. And it seems to me that many of their principles are closely &lt;a href=&quot;http://en.wikipedia.org/wiki/Alchemy#Relation_to_the_science_of_chemistry&quot; rel=&quot;nofollow&quot;&gt;related&lt;/a&gt; to chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here is my doubt - &lt;strong&gt;Why is not possible to make that alchemist stone called &lt;a href=&quot;http://en.wikipedia.org/wiki/Philosopher%27s_stone&quot; rel=&quot;nofollow&quot;&gt;Philosopher's Stone&lt;/a&gt;? What was the misconception about the research of Philosopher's Stone and how does modern chemistry can prove, or at least consider, that actually it is a Dead-end research?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has any modern chemist ever tried to reopen the research on this topic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;p&gt;&#xA;&#xA;&lt;p&gt;&#xA;" OwnerUserId="4734" LastEditorUserId="73" LastEditDate="2014-05-09T06:32:25.360" LastActivityDate="2014-05-09T06:32:25.360" Title="What was the misconception about the search of Philosopher's Stone and how does modern chemistry deals with this topic?" Tags="&lt;physical-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="0" CommentCount="16" ClosedDate="2014-04-12T08:39:39.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="9915" PostTypeId="1" CreationDate="2014-04-12T07:43:55.547" Score="2" ViewCount="72" Body="&lt;p&gt;Humic substance consists of many, to my knowledge poorly understood, species of organic matter, mostly carbohydrates chained in intersting (and also poorly understood) ways. Humic substance is an endproduct of digestion of other organic matter, and is in itself known to be hardly biodegradable. Industrial processes to destroy humic substance or humic acid in wastewater involve UV and strong oxidizers. But how is humic substance destroyed in nature? Given that at least part of the carbon embedded in life ends up as humic substance at one point or other, there must be a natural pathway to destroy it, or several. What is it?&lt;/p&gt;&#xA;" OwnerUserId="102" LastActivityDate="2014-04-12T14:53:54.270" Title="How is Humic substance destroyed in nature?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9916" PostTypeId="1" AcceptedAnswerId="9917" CreationDate="2014-04-12T09:26:25.463" Score="2" ViewCount="113" Body="&lt;p&gt;It should be determined by some metal markers (meaning originating from traffic etc.) but I cannot find any source which would describe it. &#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5166" LastActivityDate="2019-03-19T19:22:10.620" Title="How to determine source of air pollution" Tags="&lt;environmental-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9917" PostTypeId="2" ParentId="9916" CreationDate="2014-04-12T13:05:15.983" Score="1" Body="&lt;p&gt;The main idea is the following: you will need multiple sampling points and record enough data to obtain a representative dataset. Then when drawing wind roses with high percentiles (let say P95 or above) you will be able to easily point out an unique source of a relevant and exclusive pollutant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, reality is much complex... And there are some adverse effects that you must address before pointing out the source with a reasonable confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you better be not so close to the source, in order to measure immission instead of emission. Then environment and local conditions will average out. And wind speed, wind direction and concentrations become meaningful. In this case high percentiles likely originate from your source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, your analyte must be relevant and if possible exclusive from your source. Measuring CO2 is not a good choice when monitoring traffic (it is stable, widespread thus global and not exclusively produced by traffic), NO is a better one because it is local, reactive and strongly related to high compression ratio thermal engine combustion source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, if there are multiple isotropic sources, you will get nothing useful using wind roses. And your problem becomes really tricky.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-04-12T13:05:15.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9918" PostTypeId="1" AcceptedAnswerId="9952" CreationDate="2014-04-12T13:55:16.167" Score="1" ViewCount="2090" Body="&lt;p&gt;Can anybody give me an example of an endothermic dissolution process, preferably one in which name of the substances involved are easy to remember. I have searched the Web thoroughly but could not find an example, only definitions. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="865" LastEditDate="2015-10-30T17:01:56.733" LastActivityDate="2015-10-30T17:01:56.733" Title="endothermic dissolution process" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9920" PostTypeId="2" ParentId="9916" CreationDate="2014-04-12T14:41:14.567" Score="1" Body="&lt;p&gt;All catalytic converters faintly emit heavy metal-containing microparticles continuously detached from the catalytic wash bonded upon their ceramic matrix.  Correlate your pollutant(s) with those particles and you establish proximity (though not causality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isotopic labeling of emissions (C-13 in fuel, O-17,18 in air) then expensively establishes causality.  You could use conceivably use inert taggants like $\ce{CF4}$, $\ce{SF6}$, $\ce{NF3}$ injected into the exhaust stream, but then Enviro-whiners will attack (Greenhouse Effect, Global Warming, Climate Change).  There is no postulated loss so small that social advocacy will not demand vastly greater real world losses to possibly prevent it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK98071/&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/books/NBK98071/&lt;/a&gt; &lt;BR&gt;&#xA;Particulates&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-12T14:41:14.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9922" PostTypeId="2" ParentId="9915" CreationDate="2014-04-12T14:53:54.270" Score="3" Body="&lt;p&gt;Sunlight is aggressive, as is microbiology.  Ozone, hydroxyl radicals (including Fenton chemistries with peroxides), air pollution in general eventually degrade and destroy near anything organic.  Examine sediments downstream of rivers, including outflows, especially for heavily colored waters.  Sampling vs. linear distance and depth will give you histories of degradative chemistries and biologies.  Note that phenolics are overall good microbiocides re lignin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terra preta soils demonstrate the survivability of charcoal.  Humic substances are more functionalized and more reactive.  One can start at the other end, too - weathering of coal.  Coal contains dispersed iron pyrite and other sulfide minerals that speed degradation.  Powdered coal heaps can self-ignite from cumulative oxidation heat.  Carbon dioxide, water, and elemental nitrogen are deep thermodynamic holes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-12T14:53:54.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9923" PostTypeId="2" ParentId="7605" CreationDate="2014-04-12T16:01:26.393" Score="12" Body="&lt;p&gt;$\ce{Hg^2+}$ has a high affintity to thiols and other sulfur functional groups, and therefore $\ce{HgCl2}$ forms a Lewis acid-base adduct with the dithiane. The cyclic sulfonium cation undergoes ring opening, and after nucleophilic attack by water, a second Hg-S bond under ring closure is formed. Deprotonation finally yields the deprotected carbonyl compund and the cyclic mercury dithiolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWoLZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="402" LastEditDate="2016-12-04T22:11:53.153" LastActivityDate="2016-12-04T22:11:53.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9924" PostTypeId="1" AcceptedAnswerId="9926" CreationDate="2014-04-12T16:03:32.863" Score="4" ViewCount="14389" Body="&lt;p&gt;I have a solution of water and oil mixed to as much as possible, since they won't actually mix. The supply is coming down a pipe at a relatively fast rate and I need to separate it so that the water goes down a pipe and the oil goes through another pipe.&lt;/p&gt;&#xA;&lt;p&gt;How would this be done, without waiting days for it to occur?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FM0Vs.png&quot; alt=&quot;Diagram of what I want.&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5171" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-08-16T21:41:15.350" Title="Instantly separate oil and water" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9926" PostTypeId="2" ParentId="9924" CreationDate="2014-04-12T22:06:48.550" Score="8" Body="&lt;p&gt;Oil-water separation is a big money industrial process, and there are a number of methods for successfully separating the two. So the answer to your question is best approached in two parts: a theoretical answer, and a practical answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, yes. A number of systems exist and the best are described below. There are off-the-shelf applications, as well as major capital infrastructure for major water treatment plants. Whether these solutions are for you will depend on your application, as well as budget. Hence, the practical answer may be different if you are looking for a cheap, backyard-employable device to solve the problem created by Granny filling the radiator through the sump oil inlet again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Industrial_wastewater_treatment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a very nice summary of water treatment, and includes a lot of information about the &lt;a href=&quot;http://en.wikipedia.org/wiki/API_oil-water_separator&quot; rel=&quot;nofollow noreferrer&quot;&gt;API (American Petroleum Institute) oil-water separator&lt;/a&gt;, which has been in effective use for decades. It is a very effective device, albeit costly, and can treat upwards of 50000 litres of water per hour. You can read about it in detail here, but I &lt;strike&gt;don't&lt;/strike&gt; &lt;em&gt;doubt&lt;/em&gt; it is what you are after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something more in the lines of what you describe is the &lt;a href=&quot;https://www.google.com/search?q=vertical%20gravity%20separator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vertical Gravity separator&lt;/a&gt;. It is a pressurized system which promotes separation through convection mixing as the liquid is pushed though a complex path. The differences in liquid densities encourages separation of the two components. VGS systems can treat up to 3000L per hour. A document describing the operation of this system can be found on a link to a &lt;a href=&quot;http://www.oilwaterseparator.com.au/products/vertical-gravity-separator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf document on this page&lt;/a&gt;. There are some problems associated with this type of device, in that the 'filter' mesh can get blocked if the material is very dirty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, any treated water is not 100% pure, and may need additional treatment depending on your requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXrni.jpg&quot; alt=&quot;Overview of VGS for oil-water separation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of interesting web resources where you can read more about oil-water separation methods. &lt;a href=&quot;http://www.pielkenrood.net/index.php?option=com_content&amp;amp;task=view&amp;amp;id=16&amp;amp;Itemid=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pielkenrood&lt;/a&gt; have an excellent page of various methods, past and present. &lt;a href=&quot;http://www.oil-water-separators.com/Blog/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This blog&lt;/a&gt; also has a number of useful pages, although it is a commercial site promoting their own preferred devices.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="4934" LastEditDate="2014-04-12T23:26:22.517" LastActivityDate="2014-04-12T23:26:22.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9927" PostTypeId="1" AcceptedAnswerId="9939" CreationDate="2014-04-12T23:33:50.953" Score="2" ViewCount="2698" Body="&lt;p&gt;I have a pretty basic question but the last time I took Chemistry was in 2007.  I am studying the Navy's Nuclear study guide for their interviews and one of the question I am faced with is below.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Determine the final pH and temperature when the following two solutions are mixed together in a &lt;span class=&quot;math-container&quot;&gt;$\pu{3L}$&lt;/span&gt; container.&lt;br /&gt;&#xA;Sol A: &lt;span class=&quot;math-container&quot;&gt;$\pu{2L}$&lt;/span&gt;, pH = 3, and &lt;span class=&quot;math-container&quot;&gt;$\pu{80 ^\circ F}$&lt;/span&gt;&lt;br /&gt;&#xA;Sol B: &lt;span class=&quot;math-container&quot;&gt;$\pu{1L}$&lt;/span&gt;, pH = 5, and &lt;span class=&quot;math-container&quot;&gt;$\pu{40 ^\circ F}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Would this be as simple as &lt;span class=&quot;math-container&quot;&gt;$(2/3) \cdot 3 + (1/3)\cdot5$&lt;/span&gt; for the pH and &lt;span class=&quot;math-container&quot;&gt;$(2/3)\cdot80 + (1/3)\cdot40$&lt;/span&gt; for the temperature?&lt;/p&gt;&#xA;" OwnerUserId="5173" LastEditorUserId="69830" LastEditDate="2021-05-25T05:37:13.557" LastActivityDate="2021-05-25T05:37:13.557" Title="Given the volumes: determine the pH and the final temperature of a mixture knowing only the initial pH and the temperature of the un-mixed components" Tags="&lt;solutions&gt;&lt;ph&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9928" PostTypeId="2" ParentId="9927" CreationDate="2014-04-12T23:54:01.780" Score="0" Body="&lt;p&gt;The temperature would be correct. And the pH would be correct for this exercise, but not in reality. Since most solutions contain weak acid and base, it is difficult to determine the pH unless you knew which buffers and quantities are in the system.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-12T23:54:01.780" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="9929" PostTypeId="1" AcceptedAnswerId="9934" CreationDate="2014-04-13T00:22:42.257" Score="3" ViewCount="1557" Body="&lt;p&gt;I working on the Navy study guide for their nuclear engineering programs and I am not a Chemist.  Thus, I have come here to try and develop a better understanding of the subject matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is pH important in materials selection?&lt;/p&gt;&#xA;" OwnerUserId="5173" LastEditorUserId="46306" LastEditDate="2018-03-02T15:41:50.723" LastActivityDate="2018-03-02T15:41:50.723" Title="pH and materials selction" Tags="&lt;physical-chemistry&gt;&lt;ph&gt;&lt;materials&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2018-03-03T04:47:47.653" ContentLicense="CC BY-SA 3.0" />
  <row Id="9930" PostTypeId="2" ParentId="9929" CreationDate="2014-04-13T00:45:28.890" Score="1" Body="&lt;p&gt;A few links that should shed some light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.uctm.edu/journal/j2013-4/8%20Tzaneva%20383-390.pdf&quot; rel=&quot;nofollow&quot;&gt;Effect of pH on the corrosion behaviour of high nitrogen stainless steel in chloride medium&#xA;&lt;/a&gt; i.e. ~seawater, the introduction should prove manageable even if some of the rest of the article gets overly technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia also has general articles on &lt;a href=&quot;http://en.wikipedia.org/wiki/Corrosion&quot; rel=&quot;nofollow&quot;&gt;corrosion&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/PH&quot; rel=&quot;nofollow&quot;&gt;pH&lt;/a&gt; that may provide some extra background, though I would suggest just using them to find key terms to combine in google searches to try and find an introductory level university course, they are typically more focused in their explanations.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-04-13T00:45:28.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9932" PostTypeId="2" ParentId="9929" CreationDate="2014-04-13T01:08:35.520" Score="1" Body="&lt;p&gt;Metals are overall are attacked by acid, especially plus air.  Aluminum and zinc are attacked by both acid and base.  Cooling and heating water, and especially boiler water, must be delicately adjusted for pH and solutes to avoid chemical and galvanic corrosion of metal containment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seawater is nasty stuff. Seawater plus air is nastier.  Acid seawater plus air is very bad indeed.  How big is your ship's paint locker?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could build things from Inconel 686 or Hasteloy C-2000. They would not corrode, not even in hot Green Death, hot Yellow Death, or any variation of seawater. Alas, they would be expensive, very heavy (dense), and allergenic to some people (the nickel).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-13T01:08:35.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9933" PostTypeId="2" ParentId="9910" CreationDate="2014-04-13T01:24:10.587" Score="1" Body="&lt;p&gt;Baseline separating C-12 from C-13 from C-14 requires accelerator mass spectrometry,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerator_mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Accelerator_mass_spectrometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example requires 100 times the finesse.  I'm not optimistic for reduction to practice vs. a paper solution.  Your target could be a charge-coupled device or a channel plate amplifier.  Observe the spatial separation of signals above noise.  A Faraday cup works differently.  Time-of-flight is not optimistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might diddle FT ion cyclotron resonance in a high-Q cavity.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-13T01:24:10.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9934" PostTypeId="2" ParentId="9929" CreationDate="2014-04-13T01:31:54.950" Score="9" Body="&lt;p&gt;The pH scale is merely a way of keeping track of the concentration a species (the hydrogen ion $\ce{H^+}$, $pH = -log\  a_{H^+} \simeq -log\  [H^+]$) in a medium (usually liquid, not necessarily aqueous). You can define arbitrary scales for other species, too. Analytical chemists are more familiar with scales such as pCl, with $pCl = -log\  a_{Cl^-} \simeq -log\  [Cl^-]$, and that's fine, but they're much rarer. The reason chemistry pays so much attention to hydrogen ions specifically is because the pH scale keeps track of what is arguably &lt;em&gt;the most reactive species in all of chemistry&lt;/em&gt; (one could make a case for alpha particles, i.e. bare helium-4 nuclei $\ce{^4_2 He^{2+}}$ instead, but hydrogen ions are nevertheless comparable in strength and far more common). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stop to think about it, a (non-solvated) hydrogen ion, $\ce{H^+_{(g)}}$ is a &lt;em&gt;bare proton&lt;/em&gt;. Let me explain why this is an important realization. In chemistry, we talk a lot about how electrons get shoved around when different molecules meet (also known as chemical reactions), and the exact way in the electrons shuffle about is dependent on complicated electromagnetism at the quantum level. However, to a simple but enlightening approximation, one can imagine how substances behave based on how much electrical charge they concentrate in a region; simply, concentrated charges will display stronger electromagnetic effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of chemistry, we deal with charges of order unity relative to the fundamental electrical charge ($e=1.6022 \times 10^{-19}\ C$) spread in the volume of a few atoms (of order $10^{-3}\  nm^3$). The resulting &lt;em&gt;charge density&lt;/em&gt; can be found by dividing the charge by the volume. Now here's the thing - the nucleus of an atom is &lt;strong&gt;tiny&lt;/strong&gt; compared to the size of the entire atom (with its electrons). A lone hydrogen atom has a radius of approximately $0.1\ nm$, but a lone hydrogen &lt;em&gt;nucleus&lt;/em&gt; has a radius of about $1\ fm = 0.000001\ nm$, a difference of five orders of magnitude. The volume difference is thus &lt;em&gt;fifteen&lt;/em&gt; orders of magnitude. Therefore, the bare nucleus which comprises a $\ce{H^+}$ ion has &lt;em&gt;a charge density 15 orders of magnitude greater&lt;/em&gt; than a similarly-charged atomic ion, for example $\ce{Li^+}$. In effect, this means that a bare $\ce{H^+}$ has such a huge positive charge density that it will tear opposite charges (electrons) from absolutely &lt;strong&gt;anything&lt;/strong&gt;. Helium and neon are notoriously unreactive, but both react quite happily (and rather exothermically) with gaseous $\ce{H^+}$ forming hydrohelium(+) and hydroneon(+) as the respective products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why doesn't water (or any ready source of $\ce{H^+}$ ions, i.e. protic substance) immediately dissolve absolutely everything it touches? It so happens that, in condensed phases, whatever molecules close to a hydrogen ion are so attracted that they effectively encase the hydrogen ions into &lt;em&gt;solvated hydrogen ions&lt;/em&gt;, forming a shell which protects everything else from the aggressive reactivity of the hydrogen ion. For example, in water, $\ce{H^+}$ is actually not a very correct description of reality. Rather, one finds $\ce{H(H_2O)_{n}^+}$ (often alluded to by formulae such as $\ce{H3O^+}$, $\ce{H5O2^+}$, $\ce{H9O4^+}$ and others), or more concisely (but less distinguishably) $\ce{H^+_{(aq)}}$. After being surrounded by water molecules, the hydrogen ion loses almost all of its reactivity, as the original charge is now effectively dispersed over several molecules, greatly dimishing the charge density of the solvated hydrogen ion compared to the non-solvated hydrogen ion. Even after the massive handicap, solvated hydrogen ions are &lt;em&gt;still&lt;/em&gt; quite reactive. Thus, it is important to keep track of their amount, to figure out if certain materials are capable of withstanding the onslaught.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all I said, it seems like it would be a good idea to minimize the amount of $\ce{H^+}$ in water, which can be done by going to high values in the pH scale. However, by doing so you begin to significantly increase the amount of hydroxide ions $\ce{OH^-}$ present, in order to obey the autodissociation reaction equilibrium of water. In some sense, hydrated $\ce{OH^-}$ ions in solution are about as reactive as solvated $\ce{H^+}$ (even though non-solvated $\ce{H^+}$ is much, much more reactive than non-solvated $\ce{OH^-}$). Thus, going to either end of the pH scale can be a bad idea. One could say that water solutions are least reactive* when their pH is close to 7 (including pure water), as one minimizes the sum of $\ce{H^+}$ and $\ce{OH^-}$ present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Least reactive in the specific sense that if a material resists acidic aqueous solutions or basic aqueous solutions, then it &lt;em&gt;must&lt;/em&gt; resist pure water and neutral aqueous solutions, assuming that whatever else anything else in the water incapable of altering the pH value is also unreactive; a pH 7 sodium chloride solution for example can be quite corrosive to metals, but not because of the water itself.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-04-13T01:43:38.200" LastActivityDate="2014-04-13T01:43:38.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9935" PostTypeId="1" AcceptedAnswerId="9936" CreationDate="2014-04-13T02:03:31.563" Score="2" ViewCount="37037" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/02Htf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;From the list below, circle the wavelength that would be the best setting for the UV-visible spectrophotometer used to determine the FITC concentration in the sample.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{array} {}&#xA;\mathrm{wavelengths:} &amp;amp; \pu{400 nm}&amp;amp; \pu{450 nm}&amp;amp; \pu{500 nm}&amp;amp;\pu{550 nm} &amp;amp;\pu{600 nm}&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The exam paper answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{500 nm}$&lt;/span&gt;, because we need the wavelength of &amp;quot;maximum absorbance.&amp;quot; But this graph shows two curves, and the absorbance curve is omitted.&lt;/p&gt;&#xA;&lt;p&gt;But what is the difference between an emission, absorbance, and an excitation spectrum? What are they measuring? How do you interpret/read these graphs - those with multiple curves.&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="95133" LastEditDate="2021-04-29T10:24:43.957" LastActivityDate="2021-04-29T10:24:43.957" Title="Difference between an absorbance, emission, and excitation spectrum for uv-visible spectroscopy" Tags="&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9936" PostTypeId="2" ParentId="9935" CreationDate="2014-04-13T02:17:56.237" Score="6" Body="&lt;p&gt;When you have an excitation spectrum and an emission spectrum, it is for fluorescence spectroscopy. To obtain these spectra, you hold either the emission wavelength or the excitation wavelength and scan the other to gather the information. The excitation spectrum shows at what wavelengths the solution uses to produce its fluorescence. The emission spectrum shows what wavelengths are given off from the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For absorbency spectrum, the solution has all the wavelengths shining on the solution, and you monitor the wavelengths of interest. The absorbency shows which wavelengths are absorbed by the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The excitation spectrum has the same peaks as the absorbency spectrum; although the absorbency spectrum may have more peaks.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-13T02:17:56.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9937" PostTypeId="1" CreationDate="2014-04-13T03:24:02.110" Score="1" ViewCount="487" Body="&lt;p&gt;According to Wikipedia, HF is miscible with water. But isn't hydrogen fluoride a gas at RT and a sensible pressure? What would happen if you just kept adding more HF gas to some water?&lt;/p&gt;&#xA;" OwnerUserId="3811" LastActivityDate="2014-05-01T12:17:25.557" Title="HF and water miscible?" Tags="&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9938" PostTypeId="2" ParentId="9937" CreationDate="2014-04-13T04:42:09.947" Score="4" Body="&lt;p&gt;Aqueous hydrofluoric acid of commerce is 48 wt-% HF, which averages tight to hydronium fluoride, $\ce{H3O^+ F^-}$.  You can snug 100% HF as $\ce{3 HF}$ being in equilibrium with $\ce{H2F^+ + {FHF}^{−}}$.  However, there is no concentration of HF that will separate into two liquid phases or a solid phase with solution at ambient temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrofluoric acid is exceptionally aggressive and lethal toward tissue by contact.  Effects can be delayed but not reversed.  Topical or infused calcium gluconate can ameliorate further damage.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-13T04:42:09.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9939" PostTypeId="2" ParentId="9927" CreationDate="2014-04-13T06:07:26.450" Score="5" Body="&lt;p&gt;&lt;strong&gt;For the temperature&lt;/strong&gt;, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this &lt;em&gt;enthalpy of mixing&lt;/em&gt; yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the pH&lt;/strong&gt;, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is &lt;em&gt;logarithmic&lt;/em&gt;, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (&lt;em&gt;I will make the assumption here that you have simple liquids, which are not buffered.&lt;/em&gt;) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the &lt;em&gt;amount&lt;/em&gt; of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Short discussion on the pH&lt;/em&gt;: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-04-13T16:28:08.140" LastActivityDate="2014-04-13T16:28:08.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9940" PostTypeId="1" CreationDate="2014-04-13T12:13:38.167" Score="1" ViewCount="1496" Body="&lt;p&gt;I'm working on a high school chemistry book and want to double check one of the homework assignments. The question is for students to write the empirical and molecular formula for 1,2-dichloroethane. &#xA;I can't find this online. The molecular formula is given as C2H4Cl2, but could this also be the empirical formula? I don't find any sites as listing the empirical formula as CH2Cl.&lt;/p&gt;&#xA;" OwnerUserId="5177" LastEditorUserId="865" LastEditDate="2021-05-22T18:58:33.523" LastActivityDate="2021-05-22T18:58:33.523" Title="Empirical and Molecular formula: Homework" Tags="&lt;nomenclature&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9941" PostTypeId="2" ParentId="9940" CreationDate="2014-04-13T12:56:34.247" Score="3" Body="&lt;p&gt;The molecular formula and empirical formula are right. As we known, empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the compound molecular formula is C2H4Cl2, so we divide it with 2, 4, and 2 GCD / greatest common divisor. The GCD is 2, so the empirical formula is CH2Cl.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-04-13T12:56:34.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9943" PostTypeId="1" CreationDate="2014-04-13T14:37:17.700" Score="1" ViewCount="243" Body="&lt;p&gt;I’m trying to do some quantum chemical calculations for the linear molecule $\ce{FeP2}$ in the gas phase as well as its crystal (&lt;a href=&quot;http://dx.doi.org/10.1088/0031-8949/4/3/010&quot; rel=&quot;nofollow&quot;&gt;orthorhombic symmetry like here&lt;/a&gt;). I am lead to believe that in both cases iron has an oxidation state of +6, and phosphorous has an oxidation state of −3. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are such oxidation states a reasonable assumption for this molecule?&lt;/li&gt;&#xA;&lt;li&gt;Would the molecule normally be a singlet (multiplicity of one)?&lt;/li&gt;&#xA;&lt;li&gt;What quantum chemistry methods would be appropriate for this system?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5178" LastEditorDisplayName="user7951" LastEditDate="2015-06-29T15:40:09.867" LastActivityDate="2015-06-29T15:40:09.867" Title="Bonding, multiplicity and quantum chemistry for FeP2" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9944" PostTypeId="2" ParentId="9937" CreationDate="2014-04-13T17:00:41.920" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But isn't hydrogen fluoride a gas at RT and a sensible pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, it depends on the room in question. $HF$ boils at 19.5 C, meaning one can check if it mixes well with chill water or at relatively hight pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would happen if you just kept adding more $HF$ gas to some water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Learn about binary diagrams (thermodynamic chemistry) and  find one for $HF/H_2O $ system&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-05-01T12:17:25.557" LastActivityDate="2014-05-01T12:17:25.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9945" PostTypeId="2" ParentId="9943" CreationDate="2014-04-13T18:30:51.193" Score="2" Body="&lt;p&gt;Admittedly it is vacuum phase, but are you certain about your oxidation state assignments?  &quot;nearly temperature independent susceptibility of -(8.8±0.7)centerdot10-6SI (at room temperature)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;doi:10.1016/j.jpcs.2013.04.015 &lt;BR&gt;&#xA;&lt;a href=&quot;https://sciborg.uwaterloo.ca/~lfnazar/publications/Electrochem_Solid-State_Lett_6_8_A162-A165_2003.pdf&quot; rel=&quot;nofollow&quot;&gt;https://sciborg.uwaterloo.ca/~lfnazar/publications/Electrochem_Solid-State_Lett_6_8_A162-A165_2003.pdf&lt;/a&gt;  &lt;BR&gt;&#xA;doi:10.1021/ja301173q  &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.staff.uni-bayreuth.de/~bt220235/refs/HPR_29_235.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.staff.uni-bayreuth.de/~bt220235/refs/HPR_29_235.pdf&lt;/a&gt;  &lt;BR&gt;&#xA;doi:10.1088/0031-8949/4/3/010 (given)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-13T18:30:51.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9951" PostTypeId="1" CreationDate="2014-04-14T02:58:11.437" Score="1" ViewCount="9365" Body="&lt;p&gt;How do I know if an atom has degenerate orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Understanding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that degenerate orbitals mean orbitals that have the same energy level for the same n. However, how do I distinguish between an atom that has degenerate orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Example of non-degenerate orbitals (&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/graphics/6_22.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l1wNW.gif&quot; alt=&quot;Source&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4830" LastEditorUserId="189" LastEditDate="2014-04-18T16:04:11.000" LastActivityDate="2014-04-19T12:24:15.587" Title="Degenerate Orbitals" Tags="&lt;homework&gt;&lt;atoms&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9952" PostTypeId="2" ParentId="9918" CreationDate="2014-04-14T04:32:10.793" Score="3" Body="&lt;p&gt;An example of an endothermic dissolution is dissolving potassium iodide in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: &lt;a href=&quot;http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/enthalpies%20of%20solution%20of%20electrolytes.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a table of Enthalpy of Solution for various compounds. Since this is the enthalpy of solution, some compounds will release energy when dissolve when the table indicates it would require energy to dissolve, and vice versa. If the number is negative, it is likely to be exothermic, and positive number are likely to be endothermic.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-04-15T01:40:07.420" LastActivityDate="2014-04-15T01:40:07.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9953" PostTypeId="1" CreationDate="2014-04-14T04:32:23.543" Score="2" ViewCount="129" Body="&lt;p&gt;Here       &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      oh = octahedral &#xA;      td = tetrahedral&#xA;      CN = co-ordination number&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My problem is that, that why it is not possible to have a void with CN = 5 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the definition of CN i.e. the number of nearest neighbors of a particle (here a void because we are talking about it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following figure if we treat silver sphere as the void then we say that it is a td void.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we place a sphere over this td void then we generate a void whose distance from center of void to the center of all the 5 spheres is equal. And so making the CN = 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So kindly help me in understanding that why a void with CN = 5 not possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bWisz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5175" LastActivityDate="2014-04-14T14:45:26.427" Title="Why there exist no void with co-ordination number 5" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9954" PostTypeId="2" ParentId="9951" CreationDate="2014-04-14T04:55:15.630" Score="4" Body="&lt;p&gt;If I understand your question correctly:&#xA;Isolated atoms always have degenerate orbitals: the p orbitals 3-times degenerate, the d orbital 5-times degenerate, and the interaction with a crystal field / ligand field/ presence of other atoms and their electronic occupation that split this degeneracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I misunderstood your question, please explain what you would like to distinguish from what?&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-14T04:55:15.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9956" PostTypeId="1" CreationDate="2014-04-14T09:44:07.613" Score="0" ViewCount="13864" Body="&lt;p&gt;I was asked to write a balanced chemical equation for the reaction between sulfuric acid and sodium carbonate. The balanced equation isn't the problem. The question asks to also show all the state symbols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the equations I find online have sodium carbonate as a solution, not a solid. I'm also aware there are two forms of the balanced equation:&#xA;$$\ce{H2SO4(aq) + Na2CO3(s) -&amp;gt; H2CO3 + Na2SO4 -&amp;gt; H2O + CO2 + Na2SO4}$$&#xA;So do I say the sodium sulfate is a solid or in solution? &lt;/p&gt;&#xA;" OwnerUserId="5177" LastEditorUserId="7448" LastEditDate="2017-04-08T16:19:24.387" LastActivityDate="2017-04-08T16:22:53.820" Title="Chemical Equation: Sulfuric acid and powdered sodium carbonate" Tags="&lt;solutions&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9957" PostTypeId="1" AcceptedAnswerId="10994" CreationDate="2014-04-14T10:30:58.420" Score="9" ViewCount="1835" Body="&lt;p&gt;I have a frequency calculation by Gaussian09. Now I would like to extract vibrational modes as a set of xyz coordinate files. I would like to manually recompute numerically the points along a transitional vector, since IRC calculations all fail. (I suspect it is no real saddle point, flat PES.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the formation of the Lewis-Acid-Base complex $\ce{H3N\bond{-}BF3}$ with a transition state given in $$\ce{NH3 + BF3 -&amp;gt; [H3N\cdots BF3]^{TS} -&amp;gt; H3N\bond{-}BF3}$$ where this procedure would equal a (rigid) scan of the $\ce{N\cdots B}$ bond. This would be easy to accomplish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get more difficult considering the pseudo vibration of ammonia, pyramidal to planar to pyramidal. Here it is not only one variable that has to be changed. (With dummy atoms it is still manageable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If molecules get more complicated, the manual specification of the bond parameters is too complicated and a source of many errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the days when some of the gui programs were not able to calculate the vibrations and showing them in that fashion it was common to produce a set of multiple xyz files, each of them a snapshot of the coordinates along a specified vibrational vector. (This is now done internally and you can look at fancy pictures.) One of those programs capable of doing so is &lt;a href=&quot;http://xvibs.sourceforge.net/&quot; rel=&quot;noreferrer&quot;&gt;XVibs&lt;/a&gt;. For various quantum chemistry packages they calculated the distorted geometries. Unfortunately that is now ancient. (It was possible with a Gaussian09 output.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Turbomole there is a tool distributed along called &lt;a href=&quot;http://www.turbomole-gmbh.com/manuals/version_6_5/Documentation_html/node9.html&quot; rel=&quot;noreferrer&quot;&gt;screwer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I have no access to that program package and I also do not want to change all the computations so far done to this software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lordstryker already posted in the comments, that it might be done by hand with chemcraft. I tried that and it is not very satisfying. One produces too many errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other programs/scripts that might be able to produce these files?&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-22T11:15:03.437" LastActivityDate="2016-11-30T08:03:24.723" Title="Animate molecular vibrations as multiple xyz coordinate files" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9958" PostTypeId="1" CreationDate="2014-04-14T10:48:29.850" Score="0" ViewCount="338" Body="&lt;p&gt;I'm doing revision questions and want to double check something. The opening information reads: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two chlorides of iron were prepared. One was prepared by reacting iron with dry chlorine&#xA;  gas. 4.50 g of iron reacted with chlorine gas to produce 13.01 g of the chloride.&#xA;  The other chloride was prepared by reacting iron with dry hydrogen chloride; 2.80 g of iron reacted with hydrogen chloride to produce 6.35 g of the chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The questions are then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.&lt;br&gt;&#xA;  (ii) Find the ratio of the different masses of chlorine that combined with 2.80 g of iron.&lt;br&gt;&#xA;  (iii) Which of the chemical combination laws is demonstrated using this data?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is that first question right? I would've thought they're asking to calculate the mass of chlorine, which you'd simply get by subtracting the mass of reactant iron from the mass of the end product.&lt;/p&gt;&#xA;" OwnerUserId="5177" LastEditorUserId="7448" LastEditDate="2017-04-08T16:24:33.530" LastActivityDate="2017-04-08T16:24:33.530" Title="Laws of Chemical Combination" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9960" PostTypeId="2" ParentId="9828" CreationDate="2014-04-14T11:04:36.143" Score="0" Body="&lt;p&gt;&lt;strong&gt;Nernst's equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nernst Equation can be derived from the Gibbs energy $\Delta G = nFE $. The free energy for a cell reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = \Delta G^\circ - RT \ln Q$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^\circ - \frac {RT}{nF} \ln Q$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O + $n$e -&amp;gt; R}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^\circ -  \frac{RT}{nF} \ln\frac{a_\mathrm R}{a_\mathrm O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_i$ is activity of species $i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Activity&lt;/strong&gt;&#xA;In a solution, the activity of $i$ is defined: $a_i=\gamma_i \frac {C_i}{C^0}$ where $C_i$ is the concentration of solute, $C^\circ$ is standard concentration ($1\ \mathrm M$), and $\gamma_i$ is unitless activity coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation can be simplified in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]}{[\ce{O}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^2+ + 2e- -&amp;gt; Cu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Nernst's equation will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E = E^{\circ'} -  \frac{RT}{nF} \ln\frac{[\ce{R}]=1}{[\ce{O}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because activity of solid (deposited copper) is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to calculate the potential of $0.01\ \mathrm M$ solution you will obtain the difference between standard concentration and the potential at $0.01\ \mathrm M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;E &amp;amp;= E^{\circ'} -  \frac{0.02559\ \mathrm V}{2} \ln\frac{1}{0.63}\\[6pt]&#xA;&amp;amp;= 0.34\ \mathrm V -  0.0059\ \mathrm V&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computed potential ($0.0059\ \mathrm V$) is lower than the standard potential and they are probably neglecting it. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorDisplayName="user7951" LastEditDate="2016-11-27T13:39:11.197" LastActivityDate="2016-11-27T13:39:11.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9961" PostTypeId="2" ParentId="9958" CreationDate="2014-04-14T11:34:42.207" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two chlorides of iron were prepared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do &lt;strong&gt;not&lt;/strong&gt; read that as: $\ce{FeCl_{n}}$ was prepared by two different methods, but as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reacting iron with dry chlorine gas yields $\ce{FeCl_{n}}$&lt;/li&gt;&#xA;&lt;li&gt;Reacting iron with hydrogen chloride yields $\ce{FeCl_{m}}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Apparently, iron is oxidized in both cases. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the oxidants (what is reduced)?&lt;/li&gt;&#xA;&lt;li&gt;Which of them is the stronger oxidant?&lt;/li&gt;&#xA;&lt;li&gt;What is the stoichiometry (what are &lt;strong&gt;n&lt;/strong&gt; and &lt;strong&gt;m&lt;/strong&gt;)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Could it be helpful to divide the amount of chloride produced by the amount of iron used? Would different numbers indicate which of the products has a higher chloride content? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(i) Calculate the mass of iron that reacted with 2.80 g of iron in each chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it possible that they meant the same as indicated above, i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the amount of iron chloride that would have been produced from 2.80 g (instead of 4.50 g) iron in the first experiment and compare with the outcome of the second experiment.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-14T11:34:42.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9962" PostTypeId="2" ParentId="9956" CreationDate="2014-04-14T11:37:35.967" Score="3" Body="&lt;p&gt;Try to follow this reasoning. (mouse over for answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which parameter is mandatory to determine if a substance precipitate or not?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Solubility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this parameter behave with sulphates?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Many sulphates are soluble in water. Generally the sulphates become less soluble as you go down the group. So sodium sulphate is surely soluble, confirm with Wikipedia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally think about others parameters that can have some sort of influence in this to improve your understanding of chemistry. So which parameters can affect this behavior? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; In fact as best practice you should always take in account the temperature and the amount of substance, and if the solution is saturated the excess will precipitate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the answer?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; (aq)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="7448" LastEditDate="2017-04-08T16:22:53.820" LastActivityDate="2017-04-08T16:22:53.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9963" PostTypeId="2" ParentId="9957" CreationDate="2014-04-14T11:49:48.843" Score="5" Body="&lt;p&gt;If I remember correctly, frequency data are in the &lt;code&gt;stdout&lt;/code&gt; clusterfuck usually named log file, as well as in the (formatted) checkpoint files. There shouldn't be any need to prepare a cube file for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If GaussView is not available, GM's comment has the answer: &lt;a href=&quot;http://avogadro.cc&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt; can visualize vibrations and I'm sure I've done this. Unfortunately, I don't have an output file at hand here.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-14T11:49:48.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9964" PostTypeId="1" AcceptedAnswerId="22450" CreationDate="2014-04-14T13:22:04.580" Score="3" ViewCount="6324" Body="&lt;p&gt;I fully understand and recognise the manner and which the series works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any references for the change in sigma donor capability moving up the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, comparing $\ce{Cl-}$ and $\ce{NH3}$. Which of the two would be more a more willing sigma donor ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have something to do with how diffuse the orbitals are ? &lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="7581" LastEditDate="2015-01-07T19:52:10.187" LastActivityDate="2015-01-07T19:52:10.187" Title="Spectrochemical Series - Sigma Donor Capability" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9965" PostTypeId="2" ParentId="9956" CreationDate="2014-04-14T14:33:44.460" Score="2" Body="&lt;p&gt;The answer depends on the reaction stoichiometry, reactant concentrations, and temperature.  As written and unless concentrated, $\ce{CO2}$ fizzes off and everything stays dissolved.  With excess acid you'll get sodium bisulfate.  Concentrated and cooled, you can precipitate sodium sulfate decahydrate (Glauber's salt), M.P. = $32.38 ~\rm^{\circ} C$.  That is an interesting M.P. standard and constant temperature bath during the phase transition.  There is also a heptahydrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/izDj0.jpg&quot; alt=&quot;Sodium sulfate phase diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="7448" LastEditDate="2017-04-08T16:21:22.873" LastActivityDate="2017-04-08T16:21:22.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9966" PostTypeId="2" ParentId="9953" CreationDate="2014-04-14T14:45:26.427" Score="3" Body="&lt;p&gt;Trigonal bipyramidal complexes are abundant.  However, if you want a periodic crystal, you must tesselate volume with translation - the unit cell of the crystal lattice.  A five-fold symmetric figure cannot do this, for the angles do not add up without gaps.  2-, 3-, and 6-fold symmetries are allowed, offering 230 unique 3-D space groups overall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow two kinds of unit cell, you can get approximate periodicity, as in quasicrystals and gas-water clathrates.  Both can have 5- or 10-fold symmetry overall and give sharp x-ray or neutron diffraction spots (the IUC definition of a crystal).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-14T14:45:26.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9967" PostTypeId="1" CreationDate="2014-04-14T16:41:57.790" Score="1" ViewCount="28" Body="&lt;p&gt;I have I microchannel test rig , I want to push a 150-300 ml/hr from water in my test , I would like to know which pump type i can use ? &lt;/p&gt;&#xA;" OwnerUserId="4788" LastActivityDate="2014-04-14T16:41:57.790" Title="To model early fluid flow which types of pumps should be used?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9968" PostTypeId="2" ParentId="9038" CreationDate="2014-04-14T21:21:07.700" Score="1" Body="&lt;p&gt;Fajan's rules for the prediction of bond character can be summarized in two points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Large cation with low positive charge + small anion $\rightarrow$ ionic&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Small cation with high positive charge + large anion $\rightarrow$ covalent&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1973/dt/dt9730000666#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://link.springer.com/chapter/10.1007%2F978-3-540-49850-6_53&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2016-10-29T14:52:37.133" LastActivityDate="2016-10-29T14:52:37.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9970" PostTypeId="1" CreationDate="2014-04-15T01:42:08.927" Score="2" ViewCount="13136" Body="&lt;p&gt;Can someone please tell me the chemical formula for just &quot;Hypo&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question came up in previous exams and I don't know what is the general /chemical formula for just &quot;Hypo&quot;. I can find the formula for hypo chlorite and hypo bromide, but I am unable to find formula for just &quot; Hypo&quot;&lt;/p&gt;&#xA;" OwnerUserId="5187" LastActivityDate="2014-04-15T09:51:00.303" Title="Formula for hypo" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="4" ClosedDate="2015-10-31T11:54:25.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="9971" PostTypeId="2" ParentId="9970" CreationDate="2014-04-15T01:50:13.437" Score="1" Body="&lt;p&gt;First, hypobromide does not exist, hypobromite does.&lt;/p&gt;&#xA;&lt;p&gt;Hypo- is not a formula per see. It is an indicator that the oxide has one less oxygen than the -ite oxide.&lt;/p&gt;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;chlorite ==&amp;gt; &lt;span class=&quot;math-container&quot;&gt;$ClO_2^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;hypochlorite ==&amp;gt; &lt;span class=&quot;math-container&quot;&gt;$ClO^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;bromite ==&amp;gt; &lt;span class=&quot;math-container&quot;&gt;$BrO_2^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;hypobromite ==&amp;gt; &lt;span class=&quot;math-container&quot;&gt;$BrO^-$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-04-15T01:50:13.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9972" PostTypeId="2" ParentId="9970" CreationDate="2014-04-15T01:57:34.977" Score="2" Body="&lt;p&gt;Back in the day when photographs were processed chemically rather than digitally, &quot;hypo&quot; was the chemical used to remove unexposed, undeveloped silver chloride, bromide and/or iodide from the emulsion to fix the image.  The chemical was a solution of sodium thiosulfate, $Na_2S_2O_3$.  The common name at the start of photography was sodium hyposulfate...&lt;/p&gt;&#xA;" OwnerUserId="3668" LastActivityDate="2014-04-15T01:57:34.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9974" PostTypeId="1" CreationDate="2014-04-15T04:47:26.877" Score="5" ViewCount="4670" Body="&lt;p&gt;There are many foodstuffs that are ionically conductive (for example, anything moist and salty). What about foods that have a high electronic conductivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of would be something which has been charred – presumably the char, being mostly carbon, would have a fairly high conductivity.&lt;/p&gt;&#xA;" OwnerUserId="955" LastActivityDate="2015-09-01T10:01:54.800" Title="Examples of electronically conductive foods?" Tags="&lt;materials&gt;&lt;food-chemistry&gt;&lt;conductivity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9977" PostTypeId="2" ParentId="9970" CreationDate="2014-04-15T07:25:27.107" Score="1" Body="&lt;p&gt;In chemistry, the prefixes &lt;em&gt;hyper&lt;/em&gt; and &lt;em&gt;hyper&lt;/em&gt; are often used to denote the number of oxygen atoms in oxoacids as compared to the &lt;em&gt;normal&lt;/em&gt; oxoacid. The prefixes are taken from Greek.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;hyper ($\upsilon\pi\epsilon\rho$) = &lt;strong&gt;above&lt;/strong&gt; in the sense of &lt;em&gt;more than&lt;/em&gt; or &lt;em&gt;super&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;hypo ($\upsilon\pi o$) = &lt;strong&gt;below&lt;/strong&gt; in the sense of &lt;em&gt;less than&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Consequently, there is no general formula for these prefixes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prefixes are used in medicine too, e.g. to refer to conditions resulting from vitamin levels above (&lt;em&gt;hypervitaminosis&lt;/em&gt;) or below (&lt;em&gt;hypovitaminosis&lt;/em&gt;) an average and healthy level.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-15T09:51:00.303" LastActivityDate="2014-04-15T09:51:00.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9978" PostTypeId="1" AcceptedAnswerId="10002" CreationDate="2014-04-15T10:42:15.190" Score="4" ViewCount="6575" Body="&lt;p&gt;The definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The abbreviation m/z is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although m is not the ionic mass nor is z a multiple or the elementary (electronic) charge, e. The abbreviation m/e is, therefore, not recommended. Thus, for example, for the ion $\ce{C7H7^2+}$, m/z equals 45.5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain the definition? I found it a bit difficult to comprehend. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is &quot;ionic mass&quot;?&lt;/li&gt;&#xA;&lt;li&gt;What do they mean by &quot;z a multiple or elementary charge&quot;?&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C7H7^2+}$, m/z equals 45.5 -&gt; what did they do here?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="7448" LastEditDate="2017-03-28T16:02:02.400" LastActivityDate="2017-03-28T16:02:02.400" Title="Could someone please explain the difference between m/z and m/e in mass spectroscopy" Tags="&lt;electronic-configuration&gt;&lt;electrons&gt;&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9980" PostTypeId="1" CreationDate="2014-04-15T11:07:30.800" Score="10" ViewCount="2834" Body="&lt;p&gt;We know that the Gibbs free energy is related to the equilibrium constant by the following equation:&#xA;$$\Delta_\mathrm{r}G^\circ=-RT\ln K$$&#xA;We also know the Van't Hoff equation:&#xA;$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left(R = 8.314\ \mathrm{J~mol^{-1}~K^{-1}}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what kind of $K$ do we use for these equations? Concentration quotient ($K_c$) or pressure quotient ($K_p$)?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="10185" LastEditDate="2015-06-18T16:08:15.887" LastActivityDate="2018-09-27T11:46:40.140" Title="What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9981" PostTypeId="2" ParentId="9974" CreationDate="2014-04-15T11:34:07.797" Score="2" Body="&lt;p&gt;Many foods conduct electricity. The charge carriers are ions, and the conductivity is a function of temperature and ion mobility. Liquids, therefore, are generally better conductors. The electrical properties of foods are important parameters to consider in food processing operations such as ohmic heating, microwave heating, pulsed field irradiation and induction heating. This has implications in a number of important areas, such as, the industrial microwave pasteurisation of egg-whites, and high temperature short exposure sterilisation of certain food products. A quick search of the Thomson Reuters Web Of Science database returns over 350 hits for &quot;electrical conductivity of foods&quot;. Indeed there are whole book chapters dedicated to the subject. See for example &lt;a href=&quot;http://www.crcnetbase.com/doi/abs/10.1201/9781420028805.ch10&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://link.springer.com/chapter/10.1007/978-3-540-34194-9_8&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what about some numbers then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductivity for some general materials at 20 °C are given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Resistivity_of_various_materials&quot; rel=&quot;nofollow&quot;&gt;this page in Wikipedia&lt;/a&gt;. (All values in S/m)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Copper  $6 \times 10^7$&lt;/li&gt;&#xA;&lt;li&gt;Sea Water $4.8$&lt;/li&gt;&#xA;&lt;li&gt;Drinking Water $10^{-4}$ to $10^{-2}$&lt;/li&gt;&#xA;&lt;li&gt;Deionized water $10^{-6}$&lt;/li&gt;&#xA;&lt;li&gt;Glass $10^{-15}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carbon $10^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following values for foods have been taken from a variety of published literature sources. Again units are S/m. Most temperatures are at or near 20 °C, where I can find the value stated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Beer 0.1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Coffee 0.18 &lt;/li&gt;&#xA;&lt;li&gt;Coffee with milk 0.36  &lt;/li&gt;&#xA;&lt;li&gt;Orange juice 0.36 &lt;/li&gt;&#xA;&lt;li&gt;Apple Juice 0.24  &lt;/li&gt;&#xA;&lt;li&gt;Cranberry Juice 0.09  &lt;/li&gt;&#xA;&lt;li&gt;Milk 0.5  &lt;/li&gt;&#xA;&lt;li&gt;Carrot Juice 1.3 &lt;/li&gt;&#xA;&lt;li&gt;Tomato juice 1.7  &lt;/li&gt;&#xA;&lt;li&gt;Beef 0.26  &lt;/li&gt;&#xA;&lt;li&gt;Chicken 0.19  &lt;/li&gt;&#xA;&lt;li&gt;Tomato paste 0.35-0.82 &lt;/li&gt;&#xA;&lt;li&gt;Gatorade 0.2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2462" LastEditorDisplayName="user7951" LastEditDate="2015-09-01T10:01:54.800" LastActivityDate="2015-09-01T10:01:54.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9982" PostTypeId="1" AcceptedAnswerId="9992" CreationDate="2014-04-15T12:22:08.703" Score="5" ViewCount="2198" Body="&lt;p&gt;If we have a long-chain hydrocarbon, such as decane, and we split it through thermal cracking (say in an industrial plant), we use high temperatures, and high pressures. Cracking produces smaller molecules - alkanes, and alkenes. It might look like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is cracked into products like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3-&amp;gt; 2CH2=CH2 + C3H6 + C8H18}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we use high pressure in this case? Le Chatelier's principle states that the equilibrium will shift to resist the change — in this case it would move the position of equilibrium to the left, not the right. In cracking we want maximum amount of products (ie. the right) — therefore surely a low pressure would shift the equilibrium that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is not dependent on molecules colliding — merely enough thermal energy has to be provided to enough molecules to cause them to split — homolytic fission — and become free radicals which react to form smaller molecules. Therefore a pressure requirement to increase the collisions doesn't seem necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then do we use high pressure when cracking hydrocarbons?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="30836" LastEditDate="2020-03-31T15:34:56.873" LastActivityDate="2020-03-31T15:34:56.873" Title="Why are high pressures used in cracking of long-chain hydrocarbons?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;hydrocarbons&gt;&lt;chemical-engineering&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9983" PostTypeId="1" CreationDate="2014-04-15T12:35:44.137" Score="4" ViewCount="504" Body="&lt;blockquote&gt;&#xA;  &lt;h3&gt;Passage-II&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3fdJR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3fdJR.png&quot; alt=&quot;1,3‐dibromocyclobutane, reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{&#xA;A -&amp;gt;[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})]&#xA;B -&amp;gt;[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}]&#xA;C -&amp;gt;[\text{dil.}~\ce{OH-}][Δ]&#xA;D [Mixture]&#xA;}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;31.&lt;/strong&gt; &quot;A&quot; gives&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) Lucas test&lt;br&gt;&#xA;  B) Silver mirror test&lt;br&gt;&#xA;  C) White precipitate with ammoniacal &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt;&lt;br&gt;&#xA;  D) Yellow precipitate with ammoniacal cuprous chloride&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;32.&lt;/strong&gt; How many isomeric products are existed in the mixture &quot;D&quot;?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) 2&lt;br&gt;&#xA;  B) 4&lt;br&gt;&#xA;  C) 6&lt;br&gt;&#xA;  D) 8 &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would be 'A' in the reaction? Any help would be deeply appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that alc. KOH does elimination, but I was confused whether it would eliminate one bromine each from two rings and create a bond between both the rings or it would simply eliminate both bromines from same ring and create two conjugated double bonds (cyclobut-1,3-diene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There is one more possibility: both the bromines get eliminated and a diagonal bond is formed in the ring (cyclobutane with any two diagonal carbons joined).&lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="41328" LastEditDate="2019-05-10T10:33:24.633" LastActivityDate="2019-05-10T10:33:24.633" Title="Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="9984" PostTypeId="2" ParentId="9978" CreationDate="2014-04-15T13:27:35.393" Score="3" Body="&lt;p&gt;$m$ is the atomic mass of the ion (in units ($u$); $1u=1.665402\times10^{-27} \mathrm{kg}$), $z$ is the number of its positive/negative charges. $z$ should not be used in interchange with $e$ because it is dimensionless, while $e$ has the unit $\mathrm{C}$ (Coulomb). For $\ce{[C7H7]^2+}$, $m/z$ is given by (the unit of $m$ is omitted):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(\ce{C7H7})/2=(7 \cdot 12+7 \cdot 1)/2=91/2=45.5$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="7448" LastEditDate="2017-03-28T15:16:59.517" LastActivityDate="2017-03-28T15:16:59.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9985" PostTypeId="1" AcceptedAnswerId="10028" CreationDate="2014-04-15T13:30:41.910" Score="5" ViewCount="5884" Body="&lt;p&gt;I absolutely love chemistry and I really want to take part in science fairs. I am also taking an AP chemistry class. Cleaner fuel and energy sources and air pollution (especially due to smog and vehicle exhaust) interest me. My initial idea was to make solar panels work along the lines of Chlorophyll, but that is a bit too advanced for my level. Where could I get inspiration for some science fair ideas or chemistry research topics (high school level)? &lt;/p&gt;&#xA;" OwnerUserId="4486" LastEditorUserId="4945" LastEditDate="2020-02-17T11:10:02.687" LastActivityDate="2020-02-17T11:10:02.687" Title="What are some good research topics for a high school student?" Tags="&lt;energy&gt;&lt;reference-request&gt;&lt;fuel&gt;&lt;green-chemistry&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9986" PostTypeId="2" ParentId="9983" CreationDate="2014-04-15T13:59:45.760" Score="1" Body="&lt;p&gt;When &lt;strong&gt;1,2&lt;/strong&gt;-dibromocyclobutane is heated with KOH, 2 moles of acetylene are formed, probably via a cyclobutadiene intermediate. &lt;strong&gt;1,3&lt;/strong&gt;-dibromocyclobutane reacts differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the final product is vinylacetylene (but-1-en-3-yne, $\ce{HC#C-CH=CH2})$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-15T13:59:45.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9987" PostTypeId="2" ParentId="9982" CreationDate="2014-04-15T14:33:39.457" Score="1" Body="&lt;p&gt;If you were a refinery, you'd throughput hundreds of tonnes/day from one reactor.  To keep process volumes manageable, high gas pressures and fast throughputs.  If you thermally cracked long chain alkanes, you would need red heat and get coking.  If you did it over acid catalyst, typically zeolite whose pores define product by fit, you'd get coking.  The way to do it is to reform under high pressure and modestly high temperature over noble metal catalyst-doped zeolite plus added hydrogen.  Coking is then deeply suppressed - &quot;Platformate.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethylene is better made otherwise; alpha-olefins overall are assembled catalytically.  You do not want straight-chain products.  You want a maximally branched ~$\ce{C8}$-cut for gasoline, with maximum octane number.  Linear ~$\ce{C10}$-cut and up is cheap diesel and kerosene, with maximum cetane number.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Government intervention tortures your access to liquid fuels.  The most recently built refinery is ancient.  Enviro-whinerism has carefully lowered the energy content/gallon.  &quot;Sustainability&quot; means you pay for it big time but don't have a right to receive it.  How is it bottom of the barrel diesel costs so much?  It's the law! written by idiots and crooks.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-15T14:33:39.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9989" PostTypeId="2" ParentId="9983" CreationDate="2014-04-15T16:09:32.907" Score="5" Body="&lt;p&gt;The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Csnn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-15T16:09:32.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9990" PostTypeId="2" ParentId="9237" CreationDate="2014-04-15T16:44:50.300" Score="1" Body="&lt;p&gt;There is a confusion here: &quot;molecular system&quot; is just a figure of speech, and there is no hard line between molecular systems and e.g. macroscopic solid bodies. It is not strictly about size of bonding type. If you use the definition of molecules as entities where the atoms are connected with covalent bonds then you will end up indeed with molecules like a chromosome, or a piece of silicon single crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, molecular systems are just a way to speak about systems which are smaller then lets say 5-10 nm, orbital picture is good to describe physical / chemical properties. The things that chemist are comfortable with. Polymers, DNA, proteins are treated generally as large molecules, because generally we look at smaller units where the molecular approach is useful. No one cares about the bangap of PE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solids are generally systems where the band picture is a much more useful description, and generally needs to be 100s of nanometers. A piece of silicon can be discussed a a molecule, yet, people are generally much more interested on questions like band structure, dopant effect etc, where the solid state approach is much more useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In between: we call them nanostructures, supramolecular systems, colloids, depending on who speaks about them and how to approach them. Again, it is not necessarily about the exact size, but what kind of theory you try to apply. There are gold nanoparticles of 1-3 nm size, which is much smaller then most of the enzymes and proteins. There are even nanoparticles with fixed number of atoms, i.e. well defined stoichiometry! Yet, I heard no one calling a gold nanoparticle molecular system, just because we prefer to look at them as a small piece of metal instead of a large molecule. If you prepare a silicon particle that size, most probalbly people will categorise it as nanosystem, and not as a large molecule, in spite the fact that the atoms are connected with covalent bonds. On the other hand, we prefer to call protein complexes to supramolecular systems, because the secondary interactions between building blocks is the defining feature we care about.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half-seriously: if you don't know if is is a molecular system, ask yourself, who would publish about this? If it is in the field of physics, than it is solid mater, maybe nano-system. If it sounds more like chemistry, then is it a molecular system, a colloid system (if it is something old and classic), or a supramolecular system. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T16:44:50.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9992" PostTypeId="2" ParentId="9982" CreationDate="2014-04-15T16:55:04.533" Score="4" Body="&lt;p&gt;First of all, cracking is not an equilibrium process, therefore La Chatelier-Brown principle doesn't apply! (irrelevant)&#xA;Also, the mechanism is much more complicated than this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In gas phase, radical reactions generally have extremely complex reaction networks. E.g. burning of methane with oxygen contains thousands if not more elementary reactions. I am pretty sure that gas phase cracking is not a pure first order reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In industry, cracking is generally done on heterogeneous catalyst, therefore not a gas phase reaction. Heterogeneous catalysis is pretty much controlled by the absorbed amount of molecules on the catalyst surface: high pressure, faster reactions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T16:55:04.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9993" PostTypeId="2" ParentId="9974" CreationDate="2014-04-15T16:59:16.243" Score="0" Body="&lt;p&gt;Char need not be complete.  It need only be sufficient to form a conducting network, and so is a legitimate answer for electronic conductivity of food.  An organic conductor generally pi-stacks and preferably has unpaired spins.  Look up the conductivity of melanins (if you like eating long pig or naked goat).  DNA is a pi-stack!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.drproctor.com/os/amorphous.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.drproctor.com/os/amorphous.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www1.ethz.ch/lbb/Education/Biosensors/DNA_conductivity.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www1.ethz.ch/lbb/Education/Biosensors/DNA_conductivity.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://mike.zwolak.org/publications/DNA_Electronics.pdf&quot; rel=&quot;nofollow&quot;&gt;http://mike.zwolak.org/publications/DNA_Electronics.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://physicsworld.com/cws/article/news/2007/dec/17/conduction-seen-in-dna-backbone&quot; rel=&quot;nofollow&quot;&gt;http://physicsworld.com/cws/article/news/2007/dec/17/conduction-seen-in-dna-backbone&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.physics.ucla.edu/research/biophysics/news/pdf/outlook.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.physics.ucla.edu/research/biophysics/news/pdf/outlook.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.ceesdekker.net/files/PW_DNA.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ceesdekker.net/files/PW_DNA.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-15T16:59:16.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9994" PostTypeId="2" ParentId="9980" CreationDate="2014-04-15T17:06:32.110" Score="-1" Body="&lt;p&gt;Strictly speaking these equations should contain some correction of activity (concentration quotients) or non-ideal behaviour of gases (pressure quotients). &#xA;If you don't need to deal with those and you can treat your system as ideal, the concentration of a gas and pressure are equivalent. Be careful with the units, though!&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T17:06:32.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9995" PostTypeId="2" ParentId="590" CreationDate="2014-04-15T17:30:43.890" Score="4" Body="&lt;p&gt;This is about the spin Hamiltonian of the system, and more like naming an empirical observation then giving an actual physical explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the most simple (regular) spin systems a Hamiltonian with H=JS1S2 where J is a scalar handles pretty well the behaviour os spin. Note, in this case there is no orientation preference for the coupled spin compared to any internal axis, e.g. ligand orientation or such, only tells if S=S1+S2 or S=S1-S2 is more stable (i.e. ferro or antiferromagnetic material). &#xA;Sometimes there is a preference of orientation of S1, S2.. compared to local molecular axis, eg. S1 prefers to align with the local z axis. You can models this with the zero-field splitting, i.e. you have a DS1^2 .. type of Hamiltonian in addition.&#xA;Now asymetric exchange is a contribution when the exchange interaction has a directional preference, i.e. JS1S2 with scalar J is not enough as it would be fully anisotropic. While in your system there is a magnetic frustration, the reason for asymmetric exchange is relativistic effects similar to zero-field splitting. There are low energy excited states that mix into the ground state due to strong spin-orbit coupling. Since iron(III) doesn't have such states, I guess it is coming from the extra electron of iron(II). But it is just an uneducated guess. I think it is nothing to do with the spin-strain (that exist with simple JSS coupling, too), it is just the strain makes it easier to observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that the spin-Hamiltonian model is just an empirical parametrisation of the behaviour of the spin (e.g. description of the EPR or in-field Mossbauer spectra). Frankly speaking, it is very rare to see someone making out any physical / chemical information out of it. I would read more of Munck/Bominaar group's later papers, as they often put some theoretical model (DFT and ligand field) behind their measurements. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T17:30:43.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9996" PostTypeId="2" ParentId="6720" CreationDate="2014-04-15T18:27:09.623" Score="5" Body="&lt;p&gt;In short:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;quantum chemistry, if you like academic chemistry, &lt;/li&gt;&#xA;&lt;li&gt;chemoinformatics /bioinformatics, if you like more industrial chemistry, pharma,&lt;/li&gt;&#xA;&lt;li&gt;simulations in nanotechnology, if you do not afraid of a little physics, and like interdisciplinar fields&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In long:&#xA;Quantum chemistry / molecular simulations is nice and interesting, but there are two main possible cons (depending on your taste):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;QC is generally very academy oriented. Though it is used in industry,&#xA;too, QC people are generally stick to academia or small scientific&#xA;software companies. Just google job advertisements to have a rough&#xA;idea, before you make any big decision. &lt;/li&gt;&#xA;&lt;li&gt;You may fell be late from the party. QC has many-many implementations, most of them is in&#xA;ancient FORTRAN dialects. There are even more unimplemented&#xA;algorithms, however you may find difficult to start a completely new&#xA;project from scratch, and you may find painful spagetthicoding in&#xA;FORTRAN if you join others.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You should note that in that respect molecular simulation people (molecular mechanics, MM based MD etc) are very similar to quantum chemistry folks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemoinformatics is a different field, and it generally requires different kind of math (much more machine learning / statistics oriented):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chemoinformatics is much less popular in academia, and the academic community is much smaller, on the other hand much more popular in industry. If you look for &quot;computational chemistry&quot; jobs, bigger part of the advertisements will be some flavour of chemoinfo. &lt;/li&gt;&#xA;&lt;li&gt;Generally it requires large (reliable) databases which can limit your research to certain fields. However, there are some innovative use of cheminfo, e.g. done by Grzybowski group at Notheastern digitalizing chemical synthetic databases. &lt;/li&gt;&#xA;&lt;li&gt;The field has an almost incestous relationship with pharma, so if it not your main interest, you may find most topics boring.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Combined fields: there is always interesting blends of  theory and calculation (and experiment), see e.g. the work of Christofer Wilmer about design of MOFs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations for nanotechnology should be similar to quantumchemistry, however it is an emerging field with slightly different culture, much less established codebases and much more diversity of theories. It is often a field where Physicists, Engineers and Chemists collide. It also mean you should have a better idea about what you do. It is essentially the Wild West. &#xA;In my experience, nanotech experimental groups are also much more willing to collaborate with computational / theoretical people than classical chemists and they have much more intimate relationship with machines and quantitative data than e.g. organic chemists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One word of caution&lt;/strong&gt;, though:&#xA;Scientific computational guys are smart and everything in many ways, but the majority of them are HORRIBLE programmers. By horrible programmers I mean people who never heard about version control, do not test their code properly, and willing to write totally undocumented, unreadable code in languages that most programmer have goosebumps. It is a field where even visualisation software are written in FORTRAN. &#xA;This can be a huge pro or a huge con, depending on your situation.  &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T18:27:09.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9997" PostTypeId="2" ParentId="9792" CreationDate="2014-04-15T18:56:02.340" Score="6" Body="&lt;p&gt;IRC is a gradient following algorithm that tries to follow the imaginary mode. Check the algorithm (several one implemented, so really depends on which you are talking about) and you will see that unlike in optimisation algorithms, there is a big emphasis on following the real gradient (and not just jump to the endpoint as fast as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works well around the TS where the gradient is large and has a well define direction. However getting closer to the optimised geometry, the requirements of following the exact gradient and having an effective optimization algorithm are contradicting. In analytical sense it may be not obvious, but when you have numeric gradients, small second derivatives (these are almost always very soft modes), the direction of the gradient may have big errors and you should keep a small stepsize to check it thoroughly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the situation you wrote down is more or less common at some extent in almost all the IRC calculations due to the numerical nature of the problem. Harder imaginary frequencies may help.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T18:56:02.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9998" PostTypeId="2" ParentId="472" CreationDate="2014-04-15T19:11:26.607" Score="0" Body="&lt;p&gt;You are talking about standard acid base reactions and formation of a salt (more covalent than ionic), so you can have an educated guess about the mechanism just thinking through what happens here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As(III) and As(V) are simple inorganic (weak) acids, where a proton jumps up and down from the O. In literature, you may find estimations about the frequency about this jump. If not I would just use pKs and similar data of e.g. phosphoric acid to have an idea about the magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetite surface is a (weak) base, so the OH can accept a proton somewhere. The formed water stays there as a water ligand on Fe, or something kicks it out and dissociate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both are weak, my guess would be that neither the unprotonated As acid nor the protonated magnetite surface is a dominant species. More probably the undissociated proton of the As-acids would attach to the OH of magnetite, and help to kick out the water while forming a salt. This salt is unlike NaCl or similar, and the Fe-O-As has a strong covalent character.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-15T19:11:26.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9999" PostTypeId="2" ParentId="7336" CreationDate="2014-04-15T19:24:53.110" Score="2" Body="&lt;p&gt;The position of substitution in EAS is determined by the electron-donating/-withdrawing effects of existing substituents. The effects can be inductive (the electronegativity of heteroatoms polarize the bonds), or they can be resonating (free electron pairs are donated to a double bond in the transition state). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. rule&lt;/strong&gt;; resonance effects overrule inductive effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a ring has both a resonating and an inductive substituent, or one substituent with both effects, the resonating effect will control the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. rule&lt;/strong&gt;; donating effects activate the ring (more electrons for the electrophile), withdrawing effects deactivate the ring (less electrons for the electrophile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong donating groups: -OH, -OR, -NH2, -NHR, -NR2, -SH, -SR, -F, -Cl, -Br, -I, -CH=CH-X (X is any donating group,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutral/weak donating groups: -CH3, -(CH2)R, -CHR2, -CR3 (R is any hydrocarbon), Ph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong withdrawing groups: -(C=O)R, -NO2, -SO3,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak withdrawing groups:  -(CH2)X (X is any atom more electronegative than carbon), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. rule&lt;/strong&gt;; activated rings favor para and ortho positions, deactivated rings favor meta position (para and ortho are deactivated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deactivated rings can still react, but much higher temperatures are required, and the reaction is much slower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. rule&lt;/strong&gt;; para is favored over ortho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the existing substituent is &quot;in the way&quot; of the attacking electrophile in ortho position.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;electrophilic&lt;/strong&gt; aromatic substitution the aromatic ring has a positive charge in the transition state (because an electrophile (E+) has &quot;borrowed&quot; to two electrons). The electrophile can, and will, attack all three possible positions, but this first step is reversible, and any charge on a carbon is highly unfavorable. Therefore, if the charge cannot be placed on in the bottom path in the example below, the electrophile will be ejected quickly, while the other two paths are stabilized long enough for a base (B-) to extract the proton, leaving the electrons used to bond with the proton to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vWEfg.png&quot; alt=&quot;EAS transition state example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;rule; Oxygen is inductivly withdrawing, but at the same time donating by resonanse. =&gt; donating.&lt;/li&gt;&#xA;&lt;li&gt;rule; Donating =&gt; activated ring.&lt;/li&gt;&#xA;&lt;li&gt;rule; Activated =&gt; ortho/para.&lt;/li&gt;&#xA;&lt;li&gt;rule; both ortho and para are available. =&gt; para favored&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Your example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Question (i), the -(C=O)R is a withdrawing group on the left ring, while the RO- group has two electron pairs that it can share with the right ring. ==&gt; para position on the right ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question (ii), the -NO2 on the left ring is deactivating for that ring, while -CH2R is weakly activating for both ring. ==&gt; para position on the right ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question (iii), the ring has a strong and a weak withdrawing group, and is thus highly deactivated. Also, as ssavec pointed out, NaI is no electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still let us assume that you had some strong electrophile; Both substituents deactivate their ortho positions, while having little effect on their meta positions. Since -NO2 is a stronger withdrawing group, it rules, and the reaction would happen meta to the -NO2 group (ortho to the -CH2Br).&lt;/p&gt;&#xA;" OwnerUserId="188" LastActivityDate="2014-04-15T19:24:53.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10000" PostTypeId="1" CreationDate="2014-04-15T20:06:01.857" Score="0" ViewCount="1193" Body="&lt;p&gt;Is there any kind of metal that melts very quickly with very low heat or temperature, and at the same time, becomes hard very quickly?&lt;/p&gt;&#xA;" OwnerUserId="5194" LastEditorUserId="16683" LastEditDate="2017-08-08T11:56:41.477" LastActivityDate="2017-08-08T12:06:52.280" Title="Is there a metal that melts and solidifies quickly?" Tags="&lt;metal&gt;&lt;melting-point&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10001" PostTypeId="2" ParentId="10000" CreationDate="2014-04-15T20:39:06.377" Score="4" Body="&lt;p&gt;Some bismuth alloys have rather low melting points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Wood's metal&lt;/em&gt; (50% bismuth, 25% lead, 12.5% cadmium and 12.5% tin) melts around 70 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Field's metal&lt;/em&gt; (51% indium, 32.5% bismuth and 16.5% tin) melts around 62 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alloy of 44.7% bismuth, 22.6% lead, 19.1% indium, 8.3% tin and 5.3% cadmium melts around 47 °C.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2017-08-08T12:06:52.280" LastActivityDate="2017-08-08T12:06:52.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10002" PostTypeId="2" ParentId="9978" CreationDate="2014-04-15T20:51:10.917" Score="1" Body="&lt;p&gt;There are no differences simply the second notation is not recommended by &lt;a href=&quot;http://goldbook.iupac.org/M03752.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC&lt;/a&gt;. The explanation is below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is ionic mass?&lt;/strong&gt; The ionic mass is the real mass of the ion that takes in account the weight of the electrons in the overall weight. However electrons are much more lighter (0.548 mDa) then nuclei,  so according to &lt;a href=&quot;http://goldbook.iupac.org/M03752.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC&lt;/a&gt; definition &lt;strong&gt;m&lt;/strong&gt; is intend to be A:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total number of heavy particles (protons and neutrons jointly called &gt; nucleons) in the atomic nucleus. Also called nucleon number. Symbol  m &#xA;  in mass spectrometry.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And not the ionic mass (that in fact would more accurate). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do they mean by &quot;z a multiple or elementary charge&quot;?&lt;/strong&gt; This is another approximation they suppose that we could use an elementary charge +1, +2, +3, -1, -2, -3 etc. etc. that is not a real quantification in Coulumb of the electrical charge of the ion as stated Jannis Andreska.&lt;br&gt;&#xA;&lt;strong&gt;$\ce{C7H7^2+}$, m/z equals 45.5 -&gt; what did they do here?&lt;/strong&gt; They simply divide &lt;strong&gt;m&lt;/strong&gt; (see definition above) by &lt;strong&gt;z&lt;/strong&gt; in this case (&lt;strong&gt;2+&lt;/strong&gt;=2).  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum:&lt;/strong&gt; This definition silently says that the m/z ratio is an approximation m is not the real weight of the ion (in fact is the weight only of the heavier particles and z is not the real charge of the ion) so it says: &lt;em&gt;&quot;don't use 'e' because this is an approximation we are not using the real charge nor the real weight, and the m/z ratio is is a dimensionless quantity!&quot;&lt;/em&gt;. I believe that in the future when even the lighter electrons, due the increasing accuracy of the instrument, will matter we will use $m_{i}/q_e$ where $m_i$ is the ionic mass and &lt;em&gt;$q_e$&lt;/em&gt; is the real charge... Only matter of time (and matter!)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.it/books?id=xc-nAAAAQBAJ&amp;amp;pg=PA11&amp;amp;dq=%22ionic+mass%22+mass+spectroscopy&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=4pRNU_fuK4PdtAa18YCoBQ&amp;amp;ved=0CGMQ6AEwBg#v=onepage&amp;amp;q=%22ionic%20mass%22%20mass%20spectroscopy&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Reference here!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-04-16T12:25:40.847" LastActivityDate="2014-04-16T12:25:40.847" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10003" PostTypeId="1" CreationDate="2014-04-15T20:55:57.900" Score="0" ViewCount="36" Body="&lt;p&gt;Is there a way to control the shape of metal? Suppose I have a fast melting metal, is there a way to make it take a specific shape? Of course I don't mean to like put it in a template to take the shape of a template, but rather to control the shape using a computer or something. Is something like this possible or done? If not possible, then why?&lt;/p&gt;&#xA;" OwnerUserId="5194" LastActivityDate="2014-04-15T21:44:27.963" Title="Controlling the shape of metal" Tags="&lt;computational-chemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10004" PostTypeId="2" ParentId="10003" CreationDate="2014-04-15T21:44:27.963" Score="4" Body="&lt;p&gt;Molten or solid at the end?  It can be shaped in an induction field (liquid)  or dimagnetically levitated (liquid to solid).  Magnetoforming, hydroforming.  Spin molten into into an oblate ellipsoid (solid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.capturedlightning.com/frames/shrinkergallery.html&quot; rel=&quot;nofollow&quot;&gt;http://www.capturedlightning.com/frames/shrinkergallery.html&lt;/a&gt; &lt;BR&gt;&#xA;Z-pinch and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Munroe effect for jets and planes (angle charges), and self-forging munitions.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-15T21:44:27.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10005" PostTypeId="2" ParentId="10000" CreationDate="2014-04-15T21:57:44.183" Score="5" Body="&lt;p&gt;The melting points of some eutectic mixtures are tabulated in an article by French.&lt;sup&gt;1&lt;/sup&gt; Some options for low-melting point metals/mixtures are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gallium-indium eutectic, mp 15.7 °C (&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/495425&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sigma-Aldrich catalogue&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Galinstan, mp –19 °C (&lt;a href=&quot;http://en.wikipedia.org/wiki/Galinstan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Various gallium-based Indalloy® alloys, mp between 7.6 and 29.78 °C (&lt;a href=&quot;http://www.indium.com/thermal-interface-materials/other/liquid-metal/&quot; rel=&quot;nofollow noreferrer&quot;&gt;corporate page&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;A general search for &quot;fusible alloys&quot; brings up several sources with many choices:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fusible_alloy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.rotometals.com/low-melt-fusible-alloys/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rotometals sales page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.belmontmetals.com/product/eutectic-alloys/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Belmont sales page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.boltonmetalproducts.com/Home_Page.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bolton Metal sales page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Reference&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;French, S. J. Melting Points of Eutectics. &lt;em&gt;Ind. Eng. Chem.&lt;/em&gt; &lt;strong&gt;1936,&lt;/strong&gt; &lt;em&gt;28&lt;/em&gt; (1), 111–113. &lt;a href=&quot;https://doi.org/10.1021/ie50313a026&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ie50313a026&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="16683" LastEditDate="2017-08-08T12:05:52.657" LastActivityDate="2017-08-08T12:05:52.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10006" PostTypeId="1" CreationDate="2014-04-15T22:53:22.253" Score="2" ViewCount="100" Body="&lt;p&gt;Given &lt;span class=&quot;math-container&quot;&gt;$K_\textrm{sp}(\ce{ZnS})= \pu{2.5E-22}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_\textrm{sp}(\ce{FeS})= \pu{3.7E-19}$&lt;/span&gt;, why would &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt; ions remain in solution but not &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^{2+}}$&lt;/span&gt; with acidification of the solution?&lt;/p&gt;&#xA;&lt;p&gt;I have little understanding of this, so I don't have a great attempt at the solution.&lt;/p&gt;&#xA;" OwnerUserId="5196" LastEditorUserId="69830" LastEditDate="2021-05-19T08:36:08.800" LastActivityDate="2021-05-19T09:58:43.327" Title="Why can acidification of solution lead to selective precipitation of ions such as Fe2+ and Zn2+?" Tags="&lt;acid-base&gt;&lt;solubility&gt;&lt;precipitation&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10008" PostTypeId="1" CreationDate="2014-04-16T00:12:15.203" Score="0" ViewCount="135" Body="&lt;p&gt;I understand that ate complexes can result from the addition of carbanions to neutral aluminum alkyls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the balanced equation for this reaction is/what the structure would be in solid state - like does it dimerize?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -&amp;gt; ?}$$&lt;/p&gt;&#xA;" OwnerUserId="5197" LastEditorUserId="16683" LastEditDate="2017-04-01T02:10:02.267" LastActivityDate="2017-04-01T02:10:02.267" Title="State of aggregation of &quot;ate&quot; complexes" Tags="&lt;solid-state-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10009" PostTypeId="1" AcceptedAnswerId="10010" CreationDate="2014-04-16T04:09:12.947" Score="6" ViewCount="3839" Body="&lt;p&gt;It has 5 valence electrons, but only loses 3 of them to make a +3 ion. Why does this occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it may have something to do with how losing 3 electrons leaves you with the p sublevel full, but I do not understand why Bismuth is okay with that.&lt;/p&gt;&#xA;" OwnerUserId="5200" LastActivityDate="2014-04-16T06:28:24.980" Title="Why does Bismuth only create a +3 ion instead of a +5 ion?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10010" PostTypeId="2" ParentId="10009" CreationDate="2014-04-16T04:30:15.113" Score="5" Body="&lt;p&gt;Isn't it just that +5 is a lot of (positive) charge for a single cation? Which other (transition) metals do form a $\ce{M^{5+}}$ cation? Just out of my mind, I can't think of any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, these high oxidation states exist in the form of oxo-anions ($\ce{CrO4-}$, $\ce{MnO4-}$, $\ce{VO4^{3-}}$, etc.) and bismuth is no exception here: $\ce{Bi(V)}$ exists as bismutate ($\ce{BiO3-})$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-16T06:28:24.980" LastActivityDate="2014-04-16T06:28:24.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10011" PostTypeId="2" ParentId="10008" CreationDate="2014-04-16T04:51:38.687" Score="2" Body="&lt;p&gt;The crystal structure of lithium tetraethylaluminate has been published (&lt;a href=&quot;http://dx.doi.org/10.1021/ic50016a021&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;)&#xA;and as far as I can tell from the abstract, there is no indication for a dimeric anion here. It might be save to assume that the balanced equation therefore can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -&amp;gt; 2 Al(CH2CH3)4^{\delta-}\,Li^{\delta+}}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-16T06:03:33.720" LastActivityDate="2014-04-16T06:03:33.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10012" PostTypeId="2" ParentId="10000" CreationDate="2014-04-16T06:55:38.627" Score="2" Body="&lt;p&gt;There is a game, which is a tradition in Germany around New Year, to predict good fortune for the upcoming year: &lt;a href=&quot;http://en.wikipedia.org/wiki/Molybdomancy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molybdomancy (german - Bleigießen)&lt;/a&gt; While this is unscientific jibber jabber, there is actually some real physics involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin has a (very) low melting point ($\approx \pu{230 ^\circ C}$) which can easily be achieved with a candle. If you pour the molten tin into water, it will immediately become solid again. (This also works with lead $\approx \pu{330 ^\circ C}$, but that is very toxic.)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-08-08T11:56:15.987" LastActivityDate="2017-08-08T11:56:15.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10013" PostTypeId="1" CreationDate="2014-04-16T08:40:46.200" Score="4" ViewCount="392" Body="&lt;p&gt;Why is the fulfilled electronic configuration of only $p$ orbital  is stable. I mean why $II-B $ group with fulfilled $d$ orbital,$II-A$ group with fulfilled $s$ orbital...are not stable. Why makes $p$ orbital so special?&lt;/p&gt;&#xA;" OwnerUserId="4698" LastEditorUserId="4734" LastEditDate="2014-04-19T11:59:54.773" LastActivityDate="2014-12-24T16:16:59.557" Title="Why is the 8-electron rule more important than the 2- or 18-electron rule" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10014" PostTypeId="1" CreationDate="2014-04-16T10:45:28.817" Score="1" ViewCount="341" Body="&lt;p&gt;Imagine a manometer where air pressure is $\pu{0.8 atm}$ and the gas pressure is more. We move it to sea level which the air pressure is $\pu{1 atm}$. How will the gas pressure change? (the temperature is constant)&lt;/p&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="5026" LastEditDate="2018-02-25T02:51:31.973" LastActivityDate="2018-02-25T02:51:31.973" Title="How will the pressure of a gas in manometer change if it goes to a place with higher pressure?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10015" PostTypeId="2" ParentId="10014" CreationDate="2014-04-16T12:15:51.167" Score="2" Body="&lt;p&gt;Assuming an ideal gas with T constant, $$ PV = nRT $$ the entire right-hand term is constant. Therefore$$ P \propto \frac{1}{V} $$ which is Boyle's law. Since the atmospheric pressure goes up, the liquid portion of the manometer will go down on the side open to the atmosphere, and it will go up on the side open to the gas. Since the volume of the gas goes down, its pressure must increase by Boyle's law.&lt;/p&gt;&#xA;" OwnerUserId="2368" LastActivityDate="2014-04-16T12:15:51.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10016" PostTypeId="1" CreationDate="2014-04-16T12:22:37.933" Score="3" ViewCount="786" Body="&lt;p&gt;I want to learn basic chemistry, so which textbook do you think I should pick? It should be something both concise and clear. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5204" LastActivityDate="2014-04-16T14:21:42.647" Title="Good Introduction to chemistry textbook" Tags="&lt;reference-request&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2014-04-17T05:11:46.587" ContentLicense="CC BY-SA 3.0" />
  <row Id="10017" PostTypeId="2" ParentId="10016" CreationDate="2014-04-16T12:26:28.027" Score="2" Body="&lt;p&gt;I would recommend &lt;strong&gt;&lt;em&gt;Zumdahl&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;Chemistry&lt;/strong&gt;. It discusses most of the basic topics, and contains lot of exercises. &lt;/p&gt;&#xA;" OwnerUserId="5167" LastActivityDate="2014-04-16T12:26:28.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10019" PostTypeId="2" ParentId="10016" CreationDate="2014-04-16T14:21:42.647" Score="2" Body="&lt;p&gt;You need the knowledge and a working mastery of it.  Choosing a good book requires already knowing what is useful and not useful to you.  Prime the pump,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.khanacademy.org/science/chemistry&quot; rel=&quot;nofollow&quot;&gt;https://www.khanacademy.org/science/chemistry&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-16T14:21:42.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10021" PostTypeId="1" AcceptedAnswerId="10026" CreationDate="2014-04-16T14:43:15.787" Score="9" ViewCount="46660" Body="&lt;p&gt;In your opinion, what would be the simplest way to balance any size of the chemical equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example here is an equation:  $$\ce{C12H26 + O2 -&amp;gt; CO2 + H2O}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw some ways by putting letters in front of each molecule, but I just don't get too much how to do this with this method! My teacher is always speaking about another method, but it simply doesn't feel logic/mathematics is enough for me.&lt;/p&gt;&#xA;" OwnerUserId="5206" LastEditorUserId="402" LastEditDate="2017-04-11T01:21:28.147" LastActivityDate="2018-09-28T11:51:32.220" Title="Simplest way to balance any chemical equation" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10022" PostTypeId="1" AcceptedAnswerId="20507" CreationDate="2014-04-16T14:59:02.633" Score="3" ViewCount="4438" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xGCJJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$&#xA;$$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$&#xA;$$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorDisplayName="user7951" LastEditDate="2016-11-11T11:09:29.730" LastActivityDate="2019-04-03T14:08:15.460" Title="Eutectic Phase Diagram and Lever Rule" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10023" PostTypeId="1" CreationDate="2014-04-16T15:22:49.710" Score="-2" ViewCount="614" Body="&lt;p&gt;$\Delta H$ for the reaction below is −126 kJ&#xA;$$\ce{2Na2O2(s) + 2 H2O(l) -&amp;gt;4 NaOH(s) + O2 (g)}$$&#xA;What is the amount of heat released by the reaction of 25.0 gram of $\ce{Na2O2}$ with water?&lt;/p&gt;&#xA;" OwnerUserId="5205" LastEditorDisplayName="user7951" LastEditDate="2015-04-29T19:43:49.477" LastActivityDate="2015-04-29T19:43:49.477" Title="Delta H of reaction and finding delta h of components?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-29T21:01:02.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="10024" PostTypeId="1" CreationDate="2014-04-16T16:00:37.837" Score="3" ViewCount="2903" Body="&lt;p&gt;I've been having a conceptual struggle with these two states for a while now, and I'd like to get it cleared up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound AB dissolves in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{AB(s) + H2O &amp;lt;-&amp;gt; A+(aq) + B- (aq) + H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that in the case of salts, there can be strong and weak electrolytes based on the degree of dissociation. For example, NaCl is a strong electrolyte because when dissolved in water it dissociates completely. However, in theory, there should be some salts that do not dissociate completely and thus are weak electrolytes, for example, I think, NaCH3CHOO, despite its high solubility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCH3COO(s) + H2O(l) &amp;lt;-&amp;gt; NaCH3COO(aq) + Na+(aq) + CH3COO- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example that confuses me. Assuming that CaO dissociates completely in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaO(s) + H2O(l) &amp;lt;-&amp;gt; Ca^{2+}(aq) +O^{2-} (aq) + H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O(2-)(aq) + H2O(l) --&gt; (irreversibly, I suppose because of the general formula for the combination of anhydride metal oxides with water) 2OH-(aq)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which then finally yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaO(s) + H2O(l) &amp;lt;-&amp;gt; Ca^{2+}(aq) + 2OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, here's the tricky part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ca^{2+}(aq) + 2OH- (aq) &amp;lt;-&amp;gt; Ca(OH)2(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't know whether this is true. Calcium hydroxide, according to solubility rules, is insoluble in water, but as we know, all compounds are at least a little bit soluble. The KSP may be very low, but it dissolves a bit. My confusion is in how we know if a precipitate forms or not without being given any molar concentration of Ca2+ or the KSP of Ca(OH)2. The latter is necessary to calculate QSP, and this is necessary to determine whether a precipitate actually forms. Without being given any other information in a school setting, is it correct to assume that all salts are strong electrolytes and that all low-KSP compounds form a precipitate in solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="2094" LastEditDate="2014-04-19T07:53:52.457" LastActivityDate="2016-03-13T17:50:43.107" Title="Aqueous vs. Solid" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10025" PostTypeId="2" ParentId="10021" CreationDate="2014-04-16T16:06:35.967" Score="6" Body="&lt;p&gt;Here is an example of the process I just used:&#xA;$$\ce{C12H26 + O2 -&amp;gt; CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Start with carbon (completely arbitrary choice).  Put $12$ on the right hand side:&#xA;$$\ce{C12H26 + O2 -&amp;gt; 12CO2 + H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carbons are balanced but oxygens are not.  Therefore, move on to $\ce{O}$.  Since we added $12$ to $\ce{O2}$ on the right hand side, do same for $\ce{O2}$ on left hand side.&#xA;$$\ce{C12H26 + 12O2 -&amp;gt; 12CO2 + H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Oxygen is still not balanced ($24$ vs. $25$... ouch).  What else can we do? Let's work with hydrogen for now (completely arbitrary choice).  We can return to oxygen later.  Put $13$ on right hand side to balance hydrogen. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C12H26 + 12O2 -&amp;gt; 12CO2 + 13H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Oxygen still not balanced (even though everything else is).  This is an even/odd situation.  The oxygens on the right hand side must be multiplied by an even number. So lets multiply $13$ by $2$ to do this.&#xA;$$\ce{C12H26 + 12O2 -&amp;gt; 12CO2 + 26H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now our hydrogens are not balanced.  Multiply left hand side by $2$:&#xA;$$\ce{2C12H26 + 12O2 -&amp;gt; 12CO2 + 26H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now our carbons are not balanced.  Let's fix that.&#xA;$\ce{2C12H26 + 12O2 -&gt; 24CO2 + 26H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Everything but oxygen is good, $24$ on left hand side and $74$ on right hand side. We can get $74$ on the left by changing our coefficient from $24$ to $37$ (since $37\times2 = 74$).&#xA;$$\ce{2C12H26 + 37O2 -&amp;gt; 24CO2 + 26H2O}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Check to see if the coefficients can be reduced (divided evenly by the same number to get corresponding whole numbers).  Since we have an odd number ($37$), we cannot.  Therefore, the equation is balanced in Step 7.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is an example of what I've just denoted as the 'Go-big then reduce' method (completely made this term up).  I find it to be the easiest way to balance an equation.  It may not be the most efficient but it will land you on the right answer (which, in a way, is efficient).&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="4945" LastEditDate="2017-06-12T10:16:59.910" LastActivityDate="2017-06-12T10:16:59.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10026" PostTypeId="2" ParentId="10021" CreationDate="2014-04-16T17:05:48.657" Score="16" Body="&lt;p&gt;You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your chemical equation contains 3 atomic species: &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;. This means that you need 3 equations to balance. First I write the chemical reaction as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{a C12H26 + b O2 -&amp;gt; c CO2 + d H2O}$$&lt;/span&gt; &#xA;Now I will write the 3 equalities for the 3 atomic species that we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;12a &amp;amp;= c \tag{for C}\\&#xA;26a &amp;amp;= 2d \tag{for H}\\&#xA;2b  &amp;amp;= 2c + d \tag{for O}&#xA;\end{align}&lt;/span&gt;&#xA;The numerical constants in front of &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; come from the number of atoms that are in the molecule for which they are included. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns (&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,&lt;sup&gt;1&lt;/sup&gt; just against convention. This means that we can simply set one of the unknowns to any value. Let's say &lt;span class=&quot;math-container&quot;&gt;$a = 1$&lt;/span&gt; for now and revisit this choice after solving the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$a = 1$&lt;/span&gt; it immediately follows that &lt;span class=&quot;math-container&quot;&gt;$c = 12$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d = 13$&lt;/span&gt;. From those two it then follows that &lt;span class=&quot;math-container&quot;&gt;$b = 18.5$&lt;/span&gt;. Our balanced reaction becomes:&lt;sup&gt;2&lt;/sup&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C12H26 + 18.5 O2 -&amp;gt; 12 CO2 + 13 H2O}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of &lt;span class=&quot;math-container&quot;&gt;$a = 1$&lt;/span&gt; and pick it such that all numbers will become integers. In this case you can easily see that this will happen for &lt;span class=&quot;math-container&quot;&gt;$a = 2$&lt;/span&gt; which then results in &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2C12H26 + 37 O2 -&amp;gt; 24 CO2 + 26 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for some reason it is not easy to see which value for &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the &lt;em&gt;greatest common divisor&lt;/em&gt; of the 4 numbers and divide by that to obtain the same equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Compare &lt;span class=&quot;math-container&quot;&gt;$10=10$&lt;/span&gt; to the multiplied version &lt;span class=&quot;math-container&quot;&gt;$30=30$&lt;/span&gt;, both are correct, but they are just different by a factor &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure whether you are familiar with &lt;em&gt;linear algebra&lt;/em&gt;, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="44114" LastEditDate="2018-09-28T11:51:32.220" LastActivityDate="2018-09-28T11:51:32.220" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10027" PostTypeId="1" CreationDate="2014-04-16T17:08:27.963" Score="-1" ViewCount="1534" Body="&lt;p&gt;How do I find the oxidation numbers for ester?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main issue is about finding the numbers for ester groups.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         /O\&#xA;     -    ||&#xA;R -- O -- C -- R'&#xA;     -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I want the oxidation numbers of those two O atoms and C atom.&lt;/p&gt;&#xA;" OwnerUserId="5208" LastEditorUserId="402" LastEditDate="2015-11-24T00:12:50.420" LastActivityDate="2015-11-24T00:12:50.420" Title="Oxidation numbers for esters?" Tags="&lt;organic-chemistry&gt;&lt;oxidation-state&gt;&lt;esters&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10028" PostTypeId="2" ParentId="9985" CreationDate="2014-04-16T18:05:32.933" Score="4" Body="&lt;p&gt;Green chemistry is a highly advanced topic. It ain't for the weak of heart, so it is probably not for below university level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally favor pyrotechnics as relatively safe &lt;strong&gt;iff every bit of safety rules is strictly followed&lt;/strong&gt; and stunningly spectacular topic that is pretty easy and cheap to do by own hands (most materials are inexpensive as well as most tools). However, it is &lt;strong&gt;extremely&lt;/strong&gt; time-consuming and has questionable legal status in some countries (and for a good reason: safety may be easily compromised and potential for disaster is high). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting topic is fluorescence. Many compounds shine under UV light, and there are families of such compounds that are not exhaustively studied and yet are relatively easy to make. However, a deep study is required, but it is probably advanced school level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With access to advanced materials/tools/labs more topics become available, but a full list of such tools is required to recommend anything.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-04-16T18:05:32.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10029" PostTypeId="2" ParentId="10024" CreationDate="2014-04-16T18:11:22.017" Score="2" Body="&lt;p&gt;$\ce{CaO}$ (lime) begins by reacting with water to $\ce{Ca(OH)2}$ (slaked lime) with a very large exotherm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ktf-split.hr/periodni/en/abc/kpt.html&quot; rel=&quot;nofollow&quot;&gt;http://www.ktf-split.hr/periodni/en/abc/kpt.html&lt;/a&gt; &lt;BR&gt;&#xA;  $\ce{Ca(OH)2, K_{sp}} = 5.02×10^{-6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not concerned about pH (via ion hydrolysis) and pH buffering, soluble salts are strongly dissociated.  As you say, depending on concentration, the answer can go either way.  If you are reacting it further (lime water as a $\ce{CO2}$ detector), precipitation of much less soluble $\ce{CaCO3}$ makes the intermediate step unimportant (perhaps aside from kinetics). Add more $\ce{CO2}$ to obtain soluble $\ce{Ca(HCO3)2}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="24052" LastEditDate="2016-03-13T14:47:05.803" LastActivityDate="2016-03-13T14:47:05.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10030" PostTypeId="1" AcceptedAnswerId="10053" CreationDate="2014-04-16T18:20:47.240" Score="2" ViewCount="362" Body="&lt;p&gt;For MALDI or ESI mass spectrometry of large macromolecules (e.g. in the $\approx 100$ kDa to $\approx 10 - 100$ MDa mass range), I've always how measurement error scales with particle mass.  Not having any hands on experience with mass spetrometry, and calling anything build in the last 5 - 10 years &quot;modern&quot;, are there any published tables of general rules of thumb for how measurement error scales with particle mass (the two must surely be positively correlated)?  If we're interested in measuring the mass of large intact nucleic acids or proteins, at what single-molecule particle masses does it make sense to use ESI or MALDI?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess that at higher mass ranges (e.g. in the MDa range) ESI might become preferable since it can produce multiply charged species, allowing for smaller $\frac{m}{z}$ ratios?&lt;/p&gt;&#xA;" OwnerUserId="5209" LastEditorUserId="5209" LastEditDate="2014-04-16T18:29:00.880" LastActivityDate="2014-04-18T10:10:31.590" Title="ESI &amp; MALDI Mass Spectrometry: particle mass vs. measurement error" Tags="&lt;mass-spectrometry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10031" PostTypeId="2" ParentId="10006" CreationDate="2014-04-16T18:21:17.263" Score="1" Body="&lt;p&gt;FeS has 1500 times the solubility of ZnS.  Given the slow addition of a small molar fraction of acid that solubilizes both Zn(II) and Fe(II), that fraction will retain at equilibrium.  It gets complicated if soluble polysufides exist. Iron gives pyrite and marcasite ($\ce{FeS2}$) via mackinawite ($\ce{FeS}$) and greigite ($\ce{Fe3S4}$), but they are insoluble.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-16T18:21:17.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10032" PostTypeId="2" ParentId="10027" CreationDate="2014-04-16T18:30:09.887" Score="3" Body="&lt;p&gt;$\ce{CO2}$ and derivatives (e.g., carbonates and their esters, urea, guanidine, $\ce{CCl4}$) are formally C(+4), fully oxidized.  Carboxylates are then C(+3) (but not a chloroformate ester, of course).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-16T18:30:09.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10033" PostTypeId="1" AcceptedAnswerId="10034" CreationDate="2014-04-16T18:46:06.667" Score="4" ViewCount="4552" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Find the bond angles as indicated, in the compounds given below, according to VSEPR theory-&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hINIj.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I got: &lt;span class=&quot;math-container&quot;&gt;$109.5^\circ$&lt;/span&gt;(6),&lt;span class=&quot;math-container&quot;&gt;$109.5^\circ$&lt;/span&gt;(7),&lt;span class=&quot;math-container&quot;&gt;$109.5^\circ$&lt;/span&gt;(8) for the compound (d). But the given answer says it is wrong. But I don't know why. Everything else I got them right.&lt;/p&gt;&#xA;" OwnerUserId="4896" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-04-03T11:46:54.307" Title="Mistake in finding bond angles in VSEPR theory" Tags="&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10034" PostTypeId="2" ParentId="10033" CreationDate="2014-04-16T19:03:44.737" Score="6" Body="&lt;p&gt;Looking at compound D, one of the first things we notice is the carbonyl group. The carbon atom in a carbonyl is $\ce{sp^2}$ hybridized, so angle 6 involves an $\ce{sp^2}$ hybridized carbon. In such cases the $\ce{H-C-O}$ bond angle is ~ 120 degrees.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to elaborate a bit more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you have four objects arranged in space about a central carbon atom (the $\ce{sp^3}$ hybridization case), the four bonds from carbon are going to arrange themselves spatially so that the electron - electron repulsions between the four bonds are minimized.  This is achieved by arranging the bonds such that they point towards the vertices of a tetrahedron.  In a case like $\ce{CH_4}$, where the 4 atoms bound to carbon are identical, the C-H bonds adopt the perfect tetrahedral angle of 109.5 degrees. In the case of $\ce{CH_2Cl_2}$ where the four bonds are not equivalent, the bond angles will differ somewhat from 109.5 degrees (can you guess which angle will be greater then 109.5 degrees and which angle smaller) - but still, the tetrahedral angle is a reasonable first approximation.  &lt;/li&gt;&#xA;&lt;li&gt;In the case of an $\ce{sp^2}$ hybridized carbon, the 3 sigma bonds will arrange themselves in a plane with an approximate angle of 120 degrees between the bonds.  This arrangement again minimizes the electron - electron repulsion between the 3 sigma bonds in this geometry.  &lt;/li&gt;&#xA;&lt;li&gt;Finally, what geometry will minimize electron - electron repulsion between the two sigma bonds in the $\ce{sp}$ hybridized case?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(mouse over for answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Yes, the two bonds will be in a plane 180 degrees from one another, just like the $\ce{H-C-C}$ bond angle in acetylene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Take home message:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{sp^3}$ hybridized carbon has angles ~109 degrees&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{sp^2}$ ~ 120 degrees&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and $\ce{sp}$ ~180 degrees. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-17T16:10:58.097" LastActivityDate="2015-03-17T16:10:58.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10035" PostTypeId="1" CreationDate="2014-04-16T22:11:25.820" Score="3" ViewCount="39" Body="&lt;p&gt;I am doing a senior project that synthesizes PbS quantum dots and our particles keep crashing out of solution. I was wondering if anyone can give me some advice or information on stabilizing the colloidal solution. The solvent is DI water. I am interested in the chemistry behind stabilization. like what needs to be added to lower the surface energies of particles? why it is added? and how it stabilizes the colloids?&lt;/p&gt;&#xA;" OwnerUserId="4505" LastActivityDate="2014-04-16T23:46:08.670" Title="Particle suspension and re-suspension of nano-particles" Tags="&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10036" PostTypeId="2" ParentId="10035" CreationDate="2014-04-16T23:46:08.670" Score="1" Body="&lt;p&gt;Try mercaptoacetate surface treatment to put a negative charge on their surfaces for repulsion around pH 8 or 9 basic solution.  Uncoordinated mercaptoacetate is oxidatively dimerized to the disulfide by air (oxygen).  The drug &lt;em&gt;S&lt;/em&gt;-mercaptocysteine as its salt (basic pH) should be more aggressive toward binding.  3-Mercapto-1-propanesulfonic acid will bind at lower pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dose must be low enough that you do not disaggregate your particles, especially over time.  You can permeable membrane desalt an excess.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-16T23:46:08.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10037" PostTypeId="1" AcceptedAnswerId="10039" CreationDate="2014-04-17T02:08:31.777" Score="6" ViewCount="24399" Body="&lt;p&gt;I am in high school and haven't had a formal chemistry course, however I am fascinated with biology and picked up a biology book more advanced than the one the school is using to read on my own. The book is: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073227390&quot; rel=&quot;nofollow noreferrer&quot;&gt;Biology by Peter Raven (2007) ISBN: 0073227390&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about something in the organic chemistry section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the convention that is used to number carbon atoms in for example, glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you don't know what I'm talking about, here is an example: a structural diagram of an a-glucose molecule courtesy of Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GmVI0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers next to the carbon atoms, I don't understand how they are ordered. Can someone please explain? I know this is an elementary question for most of you, but the help is appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5212" LastEditorDisplayName="user7951" LastEditDate="2018-06-20T17:49:57.107" LastActivityDate="2018-06-20T17:49:57.107" Title="What is the convention of numbering carbon atoms in organic molecules?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-10-16T22:02:11.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="10038" PostTypeId="2" ParentId="10027" CreationDate="2014-04-17T02:33:53.510" Score="1" Body="&lt;p&gt;By finding oxidation numbers a systematic way is to go from the element with the highest electronegativity to the lowest. Also always consider neighbouring atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I the presented case of an arbitrary ester $\ce{R\bond{-}O\bond{-}(C\bond{=}O)\bond{-}R'}$, this would be oxygen. Also the $\ce{R}$ moieties are of alkyl or aryl type, so they are bonded to a carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assign $-2$ to all Oxygens, assuming they will form a full valence shell (which is a requirement for oxidation states). Now consider that electrons between two carbons will be shared equally. Therefore you can ignore $\ce{R'}$, meaning you assign the total group a value of $0$. There is one Oxygen bonded to the carbon and to $\ce{R}$, it's effect of the oxidation number will be divided in both directions. Therefore the carboxyl carbon will &quot;feel&quot; only $-1$. Add up all the oxidation numbers around the carboxyl carbon ($-3$). Since it is a neutral compound the oxidation state of the carbon has to mirror its surroundings - the value is therefore $+3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Always keep in mind, that oxidation states are formal numbers only. They do not represent the actual charge of an atom. They will still give you an idea about the polarity of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-17T02:33:53.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10039" PostTypeId="2" ParentId="10037" CreationDate="2014-04-17T02:40:59.773" Score="4" Body="&lt;p&gt;Glucose occurs as the open chain form aldehyde that determines atom numbering, and as the epimeric cyclic hemiacetal(s) you provided,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cxFWA.gif&quot; alt=&quot;Glucose open and cyclic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;ether&quot; is a hemiacetal.  C-1's -OH can be up or down.  Either way, it opens and reversibly dehydrates to the aldehyde R-C(=O)H.  Number one atom in a chain is the end that bears the highest total mass.  Double bonds count double their other end atom.  The cyclic form's C=1 also has the largest summed mass attached, two oxygens plus a carbon and a hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-17T02:40:59.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10040" PostTypeId="1" CreationDate="2014-04-17T04:26:37.530" Score="3" ViewCount="742" Body="&lt;p&gt;Is it necessary that the valency of an element must have the same valency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eg. If 2 elements A and B combine with the same mass of C, is it necessary that they will have the same valency as they would have when combining AB?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="4734" LastEditDate="2014-04-19T21:36:54.817" LastActivityDate="2015-05-20T15:30:07.557" Title="Law of reciprocal proportion" Tags="&lt;atoms&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10041" PostTypeId="2" ParentId="10040" CreationDate="2014-04-17T05:31:44.537" Score="3" Body="&lt;p&gt;The following are ladders of oxidation states for two elements. Simple examination shows there is no hard and fast rule for valence and stoichiometry overall. Valence need not be integral, either, for example, &lt;a href=&quot;http://origin-ars.els-cdn.com/content/image/1-s2.0-S0301010405002764-fx1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Creutz-Taube ion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe(-2), $\ce{Na2[Fe(CO)4]}$, Collman's reagent &lt;BR&gt;&#xA;Fe(0), $\ce{Fe(CO)5}$ &lt;BR&gt;&#xA;Fe(0), $\ce{Fe3(CO)12}$ &lt;BR&gt;&#xA;Fe(+2), $\ce{FeCl2}$ &lt;BR&gt;&#xA;Fe(+3), $\ce{FeCl3}$ &lt;BR&gt;&#xA;Fe(+4), $\ce{K4[FeO4]}$ &lt;BR&gt;&#xA;Fe(+5), $\ce{K3[FeO4]}$ &lt;BR&gt;&#xA;Fe(+6), $\ce{K2[FeO4]}$ &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V(-2), $\ce{[Na(diglyme)2][V(CO)6]}$ &lt;BR&gt;&#xA;V(-1), $\ce{(Cp)V(CO)4}$ &lt;BR&gt;&#xA;V(0), $\ce{V(CO)6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VAJz.png&quot; alt=&quot;Vanadium oxidation states&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry seeks to establish rules.  Then the fun starts - breaking the rules.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="73" LastEditDate="2014-04-18T23:43:24.770" LastActivityDate="2014-04-18T23:43:24.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10042" PostTypeId="1" AcceptedAnswerId="10048" CreationDate="2014-04-17T06:22:27.343" Score="5" ViewCount="169" Body="&lt;p&gt;A proton obviously has no electronic energy, no vibrational and no rotational degrees of freedom. Therefore I think it is fair to assume, that $$H(\ce{H+})=\frac32\cdot \mathcal{R}\cdot T$$ holds for the translational degrees of freedom. I am looking for some quotable literature where they state that, or if that has not been stated anywhere, how to derive this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing quantum chemical calculations, i.e. gas phase. I define the &lt;a href=&quot;http://goldbook.iupac.org/P04907.html&quot; rel=&quot;nofollow&quot;&gt;proton affinities&lt;/a&gt; via &#xA;$$\text{PA}(\ce{M})=-\left(H(\ce{H+M})-\bigr[H(\ce{M})+H(\ce{H+})\bigr]\right)$$&#xA;for the reaction&#xA; $$\ce{H+ + M -&amp;gt; H+M}.$$ &#xA;This is respecting the (quite common) sign convention to show the energy released upon binding a proton. &#xA;I need a reliable source or at least a best-practice example(s) for $H(\ce{H+})$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-01-27T09:39:32.360" LastActivityDate="2015-01-27T09:39:32.360" Title="Quotable Literature for the Enthalpy of a Proton" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;enthalpy&gt;&lt;reference-request&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10043" PostTypeId="2" ParentId="8265" CreationDate="2014-04-17T08:06:03.390" Score="7" Body="&lt;p&gt;Having recently had an opportunity to play around with some gallium of my own, I have a few thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible you have a different metal than what you thought you had, but frankly, I don't think that's very likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes a surprising amount of absorbed heat to melt gallium. Putting a sizable chunk of gallium in $35~^\circ\mathrm C$ water doesn't cause it to melt immediately. Even boiling water doesn't melt it instantly. Because gallium is a metal, it conducts heat fairly well, so when heated gently, it won't melt until the temperature of the whole piece gets close to the melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by user5134, a shiny metal like gallium isn't going to pick up heat very quickly from a radiative source like a heat lamp. If you wait long enough, it'll definitely melt. If you want to melt the gallium relatively quickly, then you'll want to put it into physical contact with something warm. The standard method for melting solid gallium in a container is to just immerse the closed container in hot water (not boiling water, especially if the container is glass).&lt;/p&gt;&#xA;" OwnerUserId="5214" LastEditorUserId="11367" LastEditDate="2017-04-04T20:22:22.343" LastActivityDate="2017-04-04T20:22:22.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10044" PostTypeId="1" AcceptedAnswerId="10047" CreationDate="2014-04-17T09:14:26.677" Score="30" ViewCount="54342" Body="&lt;p&gt;I'm making some graphs and I have to label the axes. I want to be extra careful and put the units in even though the meaning of $\text{pH}$ is well known. But I have a problem (though a simple one): $\text{pH}$ is a minus logarithm (base 10) of concentration of hydrogen ions (or rather their activity). What is the unit then, is it $[-\log(\text{mol}/\text{L})]$? What should I write, could you help me?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="189" LastEditDate="2014-04-18T07:45:09.103" LastActivityDate="2014-04-18T14:03:10.813" Title="What is the unit of pH?" Tags="&lt;physical-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10045" PostTypeId="1" CreationDate="2014-04-17T09:18:54.870" Score="7" ViewCount="1333" Body="&lt;p&gt;Lets say we want the activity coefficient of $\ce{Cu^{2+}}$ ions in solution. Would we simply setup one half cell of a $\ce{Cu}$ electrode in a known concentration of $\ce{CuSO4}$ and then connect this to a Standard Hydrogen Electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the standard electrode potential of the reaction $\ce{Cu^{2+} + 2e-  -&amp;gt;  Cu}$  and the Nernst equation could we determine the activity coefficient of $\ce{Cu^{2+}}$ ions ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this all I would need to do ?&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="189" LastEditDate="2014-04-17T09:43:58.587" LastActivityDate="2014-07-21T05:46:49.620" Title="Activity Coefficients from Electrode Potentials" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10046" PostTypeId="2" ParentId="10044" CreationDate="2014-04-17T09:37:48.393" Score="12" Body="&lt;p&gt;Unless you have very good reason to do otherwise, treat pH as dimensionless. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-04-17T09:37:48.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10047" PostTypeId="2" ParentId="10044" CreationDate="2014-04-17T09:38:32.267" Score="37" Body="&lt;p&gt;The real definition of the $\text{pH}$ is not in terms of concentration but in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_%28chemistry%29&quot;&gt;activity&lt;/a&gt; of a proton,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \text{pH} = - \log a_{\ce{H+}} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the activity is a dimensionless quantity. You can think of the activity as a generalization of the mole fraction that takes into account deviations from the ideal behaviour in real solutions.&#xA;By introducing the (dimensionless) activity coefficient $\gamma_{\ce{H+}}$, which represents the effect of the deviations from the ideal behaviour on the concentration, you can link the activity to the concentration via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  a_{\ce{H+}} = \frac{\gamma_{\ce{H+}} c_{\ce{H+}}}{c^0} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c^0$ is the standard concentration of $1 \, \text{mol}/\text{L}$.&#xA;If you ignore the non-ideal contributions you can approximately express the $\text{pH}$ in terms of the normalized proton concentration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \text{pH} \approx - \log \frac{c_{\ce{H+}}}{c^0} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, there can be no logarithm of a quantity bearing a unit. If however you encounter such a case it is usually due to sloppy notation: either the argument of the logarithm is implicitly understood to be normalized and thus becomes unitless or the units in the logarithm's argument originate from using the mathematical properties of logarithms to divide the logarithm of a product which is by itself unitless into a sum of logarithms: $\log(a \cdot b) = \log(a) + \log(b)$.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-04-17T11:01:48.923" LastActivityDate="2014-04-17T11:01:48.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10048" PostTypeId="2" ParentId="10042" CreationDate="2014-04-17T10:16:12.917" Score="7" Body="&lt;p&gt;A quite complete procedure in the calculation of proton affinities from &lt;em&gt;ab initio&lt;/em&gt; calculations is referenced in an article by &lt;a href=&quot;http://dx.doi.org/10.1016/S0168-1176%2896%2904397-2&quot;&gt;H. Audier et al., &lt;em&gt;Int. J. Mass Spectrom.&lt;/em&gt; 1996&lt;/a&gt;. It goes much further in the procedure than only on the calculation of the enthalpy of a proton (including corrections due to the basis set superposition error and so on), but confirms that you are right in assuming that the enthalpy of a proton is $3/2 RT$.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2014-04-17T10:16:12.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10049" PostTypeId="2" ParentId="10023" CreationDate="2014-04-17T14:31:56.817" Score="2" Body="&lt;p&gt;Loosely put, enthalpy is the change in heat of a given reaction.  We see input of two moles of sodium peroxide to give a mole of oxygen in the balanced equation.  (What are the reduced and oxidized species?)  You have a mass of sodium peroxide input, you know how many moles that is, and you know the one equivalent of reaction requires two moles of sodium peroxide.  You then know the heat released.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-17T16:25:18.647" LastActivityDate="2014-04-17T16:25:18.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10052" PostTypeId="1" AcceptedAnswerId="10058" CreationDate="2014-04-18T08:42:25.970" Score="5" ViewCount="2821" Body="&lt;p&gt;The process of ironing of clothes removes (to a large extend) the wrinkles that where introduced in the clothes while drying. Obviously this has to be some physico-chemical process that is initiated/sustained by the heat of the iron and (perhaps) the pressure applied with the iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the actual process that is going on when ironing? Let's assume we have a cotton shirt, do the polymers only physically rearrange or do they actual undergo a chemical change? If it is the latter, does ironing your clothes often cause more rapid degradation?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="275" LastEditDate="2014-04-18T10:56:36.093" LastActivityDate="2016-05-01T10:00:56.087" Title="What are the physico-chemical processes at play in ironing of clothes?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;polymers&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10053" PostTypeId="2" ParentId="10030" CreationDate="2014-04-18T08:46:59.397" Score="3" Body="&lt;p&gt;The answer to your question is not an easy one, as multiple parameters need to be taken into account to give an answer. At the high masses that you quote, most of the litterature trends come to their limits in terms of extrapolation, as the major mass ranges in mass spectrometry covers an $m/z$ range 10 - 4 000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you indicate in your question, one should separate the question into two parts: how does mass accuracy change with high $m/z$ values and how does mass accuracy change with high mass values, considering a high number of charges, as can be achieved in ESI.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Dependence with ion charges&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I will first tackle the second part of the question, because this will be needed to answer the first part. If one considers a mass analyzer in which measurement accuracy ($\Delta m/z$) is directly proportional to the measured $m/z$ result, we have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Delta m/z}{m/z} = Constant = \frac{\Delta m}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore for such a mass analyzer there is no direct advantage in using multiply charged ions (as produced in ESI) compared to singly charged ions as produced in MALDI. From this point, it comes that on this criterium alone, one should consider if measurement accuracy increases faster or slower than a proportional increase to decide whether multiply charged ions come as an advantage or a disadvantage.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Dependence with mass analyzers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let us now consider a series of classic mass analyzers and consider how they would perform in terms of high mass measurement. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quadrupole filters and quadrupole ion traps&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I group these in the same subtopic, as when one is concerned about high masses they have fairly similar properties. Commercial instruments are usually limited to $m/z$ 4,000 (or 32,000 for the specially devised systems). So ESI is a requirement for high mass species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quadrupole mass filters are usually operated in the constant mass width mode ($\Delta m/z = C$, with $C$ on the order of 0.5 to 1.3 amu), which means that the mass error will be roughly only dependent on $z$ and that there is an advantage to work on the high end of the $m/z$ range, as the overall error will be lower than on the low end. One point worth notice though: if you consider a 2 MDa system at $m/z$ 1,000, it would mean $z = 2000$. Thus spacing between neighbouring charge states is of only 0.5, lower than quadrupole resolution, which means that a deconvolution of the charge state distribution will not be doable and that is will be globally not possible to determine the mass of the molecule with this technique.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Time-of-flight analyzers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These are probably the mostly used analyzers for high mass compounds, both in MALDI and ESI modes. In an ideal TOF, the mass accuracy would depend only on the digitizer (and thus scale as $1/\sqrt{m/z}$), but it also depends on the applied potentials and other factors leading to a mass accuracy that scales as $\sqrt{m/z}$. This still means that considering the above points, mass accuracy will be better achieved at low $z$ values. The mass range limit for TOF analyzers is in theory infinite, but it is in fact limited by the detector (see below).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Orbitrap mass analyzers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Technically an Orbitrap mass analyzer could benefit from working at high $m$, low $z$ conditions, as mass accuracy in an Orbitrap scales as $\sqrt(m/z)$. The instrument is nevertheless limited by the possibility to inject ions within the trap, which is currently limited to 4,000 (or maybe 8,000, I did not check the most recent developments in the field). Thus ESI is required for such high mass systems, but the highest mass accuracies would be achieved on the high end of the achievable mass range.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;FT-ICR mass analyzers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In FT-ICR mass analyzers, mass accuracy scales as $m/z$. Thus there is no effect of the charge on mass accuracy. Technically, FT-ICR instruments always include a quadrupole (trap or filter) between the ion source and the FT-ICR mass analyzer and the $m/z$ range is thus limited by the range of this filter (4,000 - 8,000).  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Other issues of high mass measurements&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;One of the difficulties that arises when using high mass ions is also ion detection: devices using a secondary electron emission (such as MCP's or conversion dynode) have efficiencies that depend on the linear velocity of the impacting ion. Therefore the standard detectors are not well adapted to high $m/z$ ions. Some detectors have been developped to circumvent this limitation, but this remains an issue for masses in the 2 - 10 MDa range that you quote in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other issue is ion desolvation: as size increases, adducts (with salt, solvent or matrix) tend to increase also, leading to changes in the total mass of the ions and a broadening of the peaks. Mass accuracy will also be reduced because there is no direct rule for removing the masses of these adducts from the measured mass to lead to the mass of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2014-04-18T10:10:31.590" LastActivityDate="2014-04-18T10:10:31.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10054" PostTypeId="1" AcceptedAnswerId="10055" CreationDate="2014-04-18T08:47:10.173" Score="4" ViewCount="1134" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A student is designing an experiment using paper chromatography. From previous results, she knows that the Rf values for two dyes that she wishes to investigate are 0.72 and 0.80.&lt;/p&gt;&#xA;&lt;p&gt;It is proposed that the dyes be run on the same piece of paper and, for various reasons, that the sports should have a separation of a at least 1cm at the end of the procedure.&lt;/p&gt;&#xA;&lt;p&gt;What is the minimum height of the chromatography paper that she should use in this experiment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can someone correct my working out. I don't know what the correct answer is.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;distance travelled by dye x = &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;distance travelled by dye y = &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;distance of solvent front = &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$x/p=0.72$&lt;/span&gt; therefore &lt;span class=&quot;math-container&quot;&gt;$x = p(0.72)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y/p=0.8$&lt;/span&gt;, therefore &lt;span class=&quot;math-container&quot;&gt;$y = p(0.8)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y-x=1$&lt;/span&gt;, therefore &lt;span class=&quot;math-container&quot;&gt;$y=1+x$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;new equation: &lt;span class=&quot;math-container&quot;&gt;$1+x=p(0.8)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$x=0(0.72)$&lt;/span&gt;, if we solve for &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; we get &lt;span class=&quot;math-container&quot;&gt;$12.5$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;so the piece of paper should be at-least 12.5 cm long.&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-27T02:49:20.977" Title="How do I find Rf values &amp; length size of paper in chromatography?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10055" PostTypeId="2" ParentId="10054" CreationDate="2014-04-18T09:06:58.953" Score="2" Body="&lt;p&gt;$$\Delta R_f = 0.08$$&#xA;Let the separation distance be $\Delta r$ = 1 cm and let $l$ be the length of the paper in cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta R_f = \frac{\Delta r\, \mathrm{cm}}{l\, \mathrm{cm}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $l$, obtain $l$ = 12.5 cm.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-18T09:51:30.827" LastActivityDate="2014-04-18T09:51:30.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10056" PostTypeId="2" ParentId="9978" CreationDate="2014-04-18T09:52:56.700" Score="1" Body="&lt;p&gt;One should note that the IUPAC definition for $m/z$ has changed in the latest recommendation (&lt;a href=&quot;http://www.iupac.org/publications/pac/85/7/1515/pdf/&quot; rel=&quot;nofollow&quot;&gt;Definitions of terms relating to mass spectrometry (IUPAC Recommendations 2013)&lt;/a&gt;) which helps clarifying the question asked above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of what the labeling of the x-axis in a mass spectrum is supposed to represent is listed as the most prominent &lt;strong&gt;Controversial term&lt;/strong&gt; in the introductory section of this definition document. Thus the question remains somewhat open. The definition of $m/z$ has slightly changed compared to the 1991 one cited in the above question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abbreviation representing the dimensionless quantity formed by dividing the ratio of the mass of an &lt;em&gt;ion&lt;/em&gt; to the &lt;em&gt;unified atomic mass unit&lt;/em&gt;, by its &lt;em&gt;charge number&lt;/em&gt; (regardless of sign). The abbreviation is written in italicized lowercase letters with no spaces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The major change from the 1991 definition (quoted in your question) is the replacement of the mass number by the actual ratio of mass to the unified atomic mass unit. This follows the deveolopment of high resolution instruments and of ESI, which means that the $m/z$ scale has become abundantly used for multiply charged ions and that one cannot approximate anymore the ion mass divided by $1 u$ as the total mass number of the atoms present in the ion.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2014-04-18T09:52:56.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10057" PostTypeId="1" CreationDate="2014-04-18T10:48:07.577" Score="3" ViewCount="12626" Body="&lt;p&gt;According to my book, metals want to &quot;lose&quot; electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?&lt;/p&gt;&#xA;" OwnerUserId="5225" LastEditorUserId="9961" LastEditDate="2017-07-14T15:01:25.357" LastActivityDate="2017-07-14T15:22:46.807" Title="If metals want to &quot;lose&quot; electrons, then why will copper ions take electrons from a zinc electrode?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10058" PostTypeId="2" ParentId="10052" CreationDate="2014-04-18T10:56:22.043" Score="8" Body="&lt;p&gt;Ironing of clothes is a physical change (if you believe that physical changes and chemical changes are not part of a single continuum of change).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most polymers (including the cellulose fibers that comprise cotton) have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass_transition&quot;&gt;glass transition&lt;/a&gt;. The glass transition is a reversible phase change from a rigid fixed solid amorphous &quot;glass&quot; to a solid phase where there is some mobility of the polymer chains, but not enough to be a liquid (this phase is sometimes called a &quot;rubber,&quot; &quot;plastic,&quot; or &quot;elastomer&quot; phase). The glass transition is sometimes used to refer to the phase change between an order crystalline (or semicrystalline) solid to a disordered amorphous solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the glass transition temperature, the solid material has some mobility - it can be shaped, molded, flattened, whatever. When it is cooled down again, it becomes rigid and holds its shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ironing raises the temperature of the fabric to above the glass transition. The action of ironing presses out the wrinkles in the now more flexible fabric. When the fabric cools, it retains its new wrinkle-free shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is used during ironing for one important purpose. I tried to find the glass transition temperature of cellulose, but am unable to do so. Presumably it is close to the temperature at which cellulose begins to burn. Water prevents the burning, but not just in the way you would expect. Water does absorb heat from the iron in order to keep the fabric from getting too hot. However, if this were the only thing water was doing, the fabric would not reach the glass transition and ironing would not work. Water also acts as a plasticizer, lowering the glass transition temperature. Most fabrics are hydrophilic (and cellulose as a carbohydrate is especially so). The water binds to the chains, swells the fibers, and changes the physical and chemical properties of the material.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-18T10:56:22.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10060" PostTypeId="1" AcceptedAnswerId="10061" CreationDate="2014-04-18T11:59:25.923" Score="0" ViewCount="10705" Body="&lt;p&gt;I'm having trouble in understanding how to identify chemical compounds as stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question below, &lt;strong&gt;D&lt;/strong&gt; is the correct answer according to the markscheme. I've drawn out the chemical structure but am unable to see what possible stereoisomers it can form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried visualising the possible isomers on Wolfram Alpha: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=CH3CH2CHOHCH3&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wolframalpha.com/input/?i=CH3CH2CHOHCH3&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two isomers which stand out to me, which I guess are the stereoisomers?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(S)-(+)-2-butanol&lt;/li&gt;&#xA;&lt;li&gt;(R)-2-butanol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, to me it looks like they are just rotated versions of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hCqaH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3886" LastActivityDate="2014-04-19T23:51:32.110" Title="Which compound can exist as stereoisomers?" Tags="&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10061" PostTypeId="2" ParentId="10060" CreationDate="2014-04-18T13:16:27.277" Score="3" Body="&lt;p&gt;&quot;D&quot; is the correct answer. 2-Butanol exists as two stereoisomers, R- and S-2-butanol, just as you described.  The &quot;2&quot; carbon in 2-butanol is a chiral carbon atom, it has 4 different groups attached to it (ethyl, methyl, hydroxyl and a hydrogen atom).  The R and S forms of 2-butanol cannot be rotated one into the other, they can only be interconverted by reflecting one in a mirror.  They are non-superimposable mirror images like your right and left hand. Build a model of 2-butanol and then build a second model that is the mirror image of the first - you'll find that you cannot intercovert them by rotation and one is not superimposable upon the other.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-18T13:16:27.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10064" PostTypeId="2" ParentId="10060" CreationDate="2014-04-18T14:39:11.020" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&quot; rel=&quot;nofollow&quot;&gt;http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf&lt;/a&gt;  &lt;BR&gt;&#xA;Everything you ever wanted to know about stereoisomerism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick answer:  Any reasonably tetrahedral atom bearing four different groups (including isotopic substitution and unbonded orbitals) is a chiral center.  However, pairs of enantiomeric chiral centers in a molecule can cancel (&lt;em&gt;meso&lt;/em&gt;-compounds.  Tartaric acid has three stereoisomers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical answer:  Any center (need not contain an atom - helicenes) that lacks all ${S_n}$ symmetries is a chiral center.  Any molecule that lacks all ${S_n}$ symmetries is chiral.  Consider methyl phenyl sulfoxide and that sulfur's lone pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rigorous answer (Michel Petitjohn, &lt;em&gt;J. Math. Phys.&lt;/em&gt; &lt;strong&gt;40&lt;/strong&gt;, 4587 (1999) et al.):  Any collection of anonymous unit mass countable points that sum to finite principle moments of inertia, in any number of dimensions greater than zero, that cannot be superposed upon its mirror image (one coordinate with all reversed signs) strictly by translations and rotations, is chiral.  An infinite number ($\ce{Aleph_0}$) of points in a hyperbolic tiling qualifies.  Petitjohn's QCM software quantitatively calculates geometric chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last one is important.  It is a recipe for creating an acyclic, undistorted tetrahedral carbon atom bearing four rigorously identical, freely rotating subtsituents that is a chiral center.  Planar $\ce{sp^2}$ carbons can be chiral.  It demonstrates that no organic nomenclature for chirality can be complete.  It demonstrates that at least one physics founding postulate describing vacuum symmetries can be &lt;em&gt;testably&lt;/em&gt; defective, measurably empirically falsified by chemical constructs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The philosophical 900 lb gorillas, &lt;BR&gt; &#xA;&lt;em&gt;Symmetry Through the Eyes of a Chemist&lt;/em&gt; Magdolna Hargittai, István Hargittai, 2009 &lt;BR&gt;&#xA;&lt;em&gt;Fearful Symmetry: Is God a Geometer?&lt;/em&gt;  Ian Stewart, Martin Golubitsky, 2011.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-18T16:03:39.167" LastActivityDate="2014-04-18T16:03:39.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10065" PostTypeId="1" AcceptedAnswerId="10068" CreationDate="2014-04-18T14:45:44.370" Score="3" ViewCount="278" Body="&lt;p&gt;I just wanted to check my understanding of this. Given the reaction,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (1) \; \ce{A-&amp;gt;B} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the corresponding rate law is $ v=k[A] $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (2) \; \ce{A &amp;lt;=&amp;gt; B} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with forward rate law: $ v_1=k_1[A] $ and reverse: $ v_2=k_2[B] $. This makes the corresponding rate equation $ v=v_1-v_2=k_1[A]-k_2[B] $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in the difference between the two. I've often seen reactions expressed in both ways. In $ (1) $, is the rate law assuming the reverse reaction not to occur (or at least to be completely negligible? That's why I drew it with $ \ce{-&amp;gt;} $ instead of $ \ce{&amp;lt;=&amp;gt;} $. Am I correct in my thinking? In actuality, is it always more correct to use the rate law of $ (2) $? The rate law for $ (1) $ could also correspond to initial reaction velocity, right? This is because there is no $ B $ present at the start of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, then the integrated rate equations will become substantially more complicated using rate laws of form $ (2) $. I can't use simple equations like $ [A]_t=[A]_0 e^{-kt} $, as this wouldn't take into account the reverse reaction.&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="2368" LastEditDate="2014-04-18T14:51:07.113" LastActivityDate="2015-11-09T15:53:07.947" Title="Simple Kinetics Clarification" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10066" PostTypeId="2" ParentId="10057" CreationDate="2014-04-18T14:52:33.897" Score="1" Body="&lt;p&gt;Reactive systems seek lowest free energy equilibrium.  In aqueous solution, Zn + Cu(II) is higher in energy than Zn(II) + Cu(O) re the electromotive series.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be manipulated.  In hard Lewis base water, soft Lewis acid Cu(I) is unstable to disproportionation to give hard Lewis acid Cu(II) and Cu(0).  In soft Lewis base acetonitrile, Cu(II) and Cu(0) are unstable to conproportionation to Cu(I).  That last is quite a reaction.  Anhydrous CuCl2 and copper powder (slight excess) are refluxed with stirring in MeCN. The solution turns a terrible opaque inhomogeneous khaki color, then flashes colorless and transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be interesting to try it with Fe(II) and Fe(O).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-18T14:52:33.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10067" PostTypeId="2" ParentId="10057" CreationDate="2014-04-18T15:46:22.673" Score="0" Body="&lt;p&gt;Metals want to lose electrons because the process requires very less energy. Some metals require more energy than others, e.g. consider Cu and Zn. Zn will require lesser amount of energy to lose electrons than Cu. This is easily seen by the greater Oxidation potential values of Zinc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a metal electrode is dipped in an electrolyte containing its own ion, then it may lose or gain electrons depending on whether it has a negative or a positive oxidation potential, again involving energy $(W=qV)$. Thus, Zn having a negative potential loses electrons and Cu having a positive potential gains electrons. Thus, a connected Zn and Cu system will allow passage of electrons from Zn to Cu.&lt;/p&gt;&#xA;" OwnerUserId="3863" LastActivityDate="2014-04-18T15:46:22.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10068" PostTypeId="2" ParentId="10065" CreationDate="2014-04-18T17:48:13.433" Score="3" Body="&lt;p&gt;Not all reactions can or should be represented with a double arrow.  The general-audience example is that you &quot;can't unbake a cake&quot;.  You are right that there are instances that we assume that k2 is negligible (relative to k1) and therefore a reaction is treated as if it is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sample plot of how the concentrations of A for the irreversible and reversible reactions compare for k1 = k2 = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uVx8l.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No surprises here; the reversible reaction results in an equilibrium concentration of A. We can compare the concentration of A in each of the reactions as a function of kr:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2aEJm.png&quot; alt=&quot;Mathematica graphics&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I've plotted Log[k2] on the x-axis and the ratio of A in the irreversible reaction to the reversible reaction.  The different lines are at various times (in seconds).  I've kept k1 = 1 and the initial concentration 1 as well.  The two equations predict the same concentration of A in the limit of very small k2, and by very small I suspect this means at least a factor of 10^4 or more (not shown on the graph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you are correct that the integrated rate laws get much more complicated for reversible reactions, the one you are considering is still &lt;a href=&quot;http://www.wolframalpha.com/input/?i=solve+a%27%5Bt%5D+%3D%3D+-k1+a%5Bt%5D+%2B+k2+%281-+a%5Bt%5D%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;solvable in closed form&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-04-18T17:48:13.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10072" PostTypeId="2" ParentId="10065" CreationDate="2014-04-18T19:39:27.710" Score="1" Body="&lt;p&gt;In $(1)$ the reaction is irreversible. For example, potassium chlorate decomposes on heating to form potassium chloride and oxygen:&lt;br&gt;&#xA;$$\ce{2KClO3 -&amp;gt; 2KCl + 3O2}$$&#xA;In this reaction, the products cannot combine to give back potassium chlorate (I think the reason would be because of reaction not being performed in closed container). In case of irreversible reactions, the reactions occur only in forward direction. In actuality, you can use rate equation of $(2)$ for the above reaction, as rate backward reaction is zero, you will end up with rate equation of $(1)$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity of reaction $(1)$ doesn't corresponds to velocity of reaction at time $t=0$. It is the general equation for velocity of the reaction.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a first order reaction $$\ce{A -&amp;gt; products}$$&lt;br&gt;&#xA;Velocity of the reaction is given by $\frac{\mathrm dx}{\mathrm dt}=k[\ce{A}]$. Here, $[\ce{A}]$ corresponds to concentration of reactant $\ce{A}$ at a particular time and it doesn't correspond to concentration of $\ce{A}$ at $t=0$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand it consider the following explanation. let the initial concentration of the reactant be $a\ \mathrm{mol\ dm^{-3}}$. Let $x\ \mathrm{mol\ dm^{-3}}$ decompose in $t$ seconds. The remaining concentration of the reactant is $a-x\ \mathrm{mol\ dm^{-3}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\therefore}$ Velocity of the reaction after a time 't' is given by $\frac{\mathrm dx}{\mathrm dt}=k[a-x]$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rate equations for reversible reactions will be different as given in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation#Stoichiometric_reaction_networks&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; (see the first order reaction part).&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorDisplayName="user7951" LastEditDate="2015-11-09T15:53:07.947" LastActivityDate="2015-11-09T15:53:07.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10073" PostTypeId="2" ParentId="10057" CreationDate="2014-04-18T20:01:01.747" Score="4" Body="&lt;p&gt;Typically, when one uses terms such as &quot;likes to lose&quot; or &quot;likes to gain&quot;, we should think of that phrase as &quot;likes to lose &lt;em&gt;relative to&lt;/em&gt;...&quot;.  In your case, when we think about electron configurations, it might be more helpful to think &quot;metals like to lose &lt;em&gt;rather than gain&lt;/em&gt; electrons to achieve a noble-gas-like configuration&quot;.  When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can explain the transfer of electrons by looking at the the metal activity series such as the one &lt;a href=&quot;http://www.mychemistrytutor.com/downloads/activity-series&quot; rel=&quot;nofollow noreferrer&quot;&gt;shown here&lt;/a&gt;.  The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode.  Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu.  These tables are based on standard reduction potentials: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^{2+} + 2 e^- -&amp;gt; Cu(s)}\ \tag{E = 0.34 V}$$&#xA;$$\ce{Zn^{2+} + 2 e^- -&amp;gt; Zn(s)}\ \tag{E = -0.76  V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^{2+} + Zn(s) -&amp;gt; Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="35555" LastEditDate="2017-07-14T15:22:46.807" LastActivityDate="2017-07-14T15:22:46.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10075" PostTypeId="2" ParentId="6609" CreationDate="2014-04-18T20:27:51.617" Score="2" Body="&lt;p&gt;The ratio of the radius of the cation to the radius of the anion is called radius ratio. The ratio of the radius of cation and the radius of anion i.e, radius ratio ($r_+/r_-$) plays an important role in determining the structures of ionic solids and coordination number of ions. The possible coordination numbers and structural arrangements of anions around  cations for different $r_+/r_-$ values are given below:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBY3e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2014-04-18T20:27:51.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10076" PostTypeId="2" ParentId="10024" CreationDate="2014-04-19T08:14:34.733" Score="1" Body="&lt;p&gt;Up to my knowledge the reaction mechanism of CaO and $H_{2}O$ is not well understood yet, however seems that for low density CaO,  $H_{2}O$ is absorbed over the CaO. The mechanism should be something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaO + 2H2O -&amp;gt; CaO\cdot2H2O -&amp;gt;Ca(OH)_2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reaction is wrong. &lt;a href=&quot;http://books.google.it/books?id=kBouzCwk32EC&amp;amp;pg=PA152&amp;amp;dq=CAO+h2o+mechanism&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=hChSU7_2M8rOtAbA3ICwBw&amp;amp;ved=0CHwQ6AEwCQ#v=onepage&amp;amp;q=CAO%20h2o%20mechanism&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Reference here&lt;/a&gt;.&#xA;The reaction is very favorite energetically so you don't have the equilibrium you mentioned: when $Ca(OH)_{2}$ is formed you will not have CaO back until you provide sufficient heat and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reaction is right:&#xA;$$\ce{ Ca(OH)2(s) &amp;lt;=&amp;gt; Ca^{2+}(aq) + 2OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of $Ca(OH)_{2}$ is 0.173 g/100 mL (20 °C). This cause the alkalinity of the solution. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How we know if a precipitate forms or not without being given any&#xA;  molar concentration of Ca2+ or the KSP of Ca(OH)2?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't know it: if you have less than 0.173 g/100 mL of calcium hydroxide you will not have any precipitate.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without being given any other information in a school setting, is it&#xA;  correct to assume that all salts are strong electrolytes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, some salts are absolutely not good electrolytes $CaCO_{3}$ is an example. You could say that most of the halogen salts are strong electrolytes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it correct to assume that all low-KSP compounds form a precipitate in solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, because if the solution is not saturated you will not have a precipitate, however is correct say that &lt;strong&gt;probably&lt;/strong&gt; a low-KSP compound will form a precipitate in solution.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-19T08:14:34.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10077" PostTypeId="1" CreationDate="2014-04-19T10:56:37.623" Score="3" ViewCount="731" Body="&lt;p&gt;You know how when you carry out a reaction with many compounds, when drawing the reactions you just draw a line at a right angle to the bond with a 'R' at the end (see image)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SqBiI.png&quot; alt=&quot;R&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can specify what group R is in a table (eg: $\ce{OMe}$). However, if instead of adding a substituent to a benzene ring, you have pyridine? How can you specify it?&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="16683" LastEditDate="2017-03-22T23:21:25.903" LastActivityDate="2017-03-22T23:21:25.903" Title="How can I specify pyridine as a substituent?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;aromatic-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10078" PostTypeId="1" CreationDate="2014-04-19T11:14:06.520" Score="3" ViewCount="198" Body="&lt;p&gt;The bonds are: (I have given pictures of what I mean in case there's any ambiguity that I leave open in these questions. Also so that if you want to number the R groups you've got a head-start)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Secondary amides&lt;/strong&gt;, I suspect I might know the answer, extensive. Example is the one in &lt;strong&gt;folic acid&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b6TNB.png&quot; alt=&quot;Secondary amides&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tertiary amides&lt;/strong&gt;. This is one that's a complete mystery to me. Example is the one in &lt;strong&gt;leucovorin&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jReTD.png&quot; alt=&quot;Tertiary amides&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sulfonamides&lt;/strong&gt;. Hydrolysis in this case, should lead a primary amine and a sulfonic acid. Or such is the hope. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AAlbp.png&quot; alt=&quot;Sulfonamide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Carboxylic esters&lt;/strong&gt;. To give a primary alcohol and a carboxylic acid. Basically the reverse of the &lt;em&gt;Fisher esterification&lt;/em&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HRed7.png&quot; alt=&quot;Carboxylic ester&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sulfonyl ester&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oR22m.png&quot; alt=&quot;Sulfonyl ester&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Secondary/tertiary amines&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R2BwT.png&quot; alt=&quot;Secondary/tertiary amines&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ether&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYQB9.png&quot; alt=&quot;Ether&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sulfyl ether&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytTfm.png&quot; alt=&quot;Sulfyl ether&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you're wondering how exactly is this question relevant? As it might sound pretty random for me to just ask this question but it's incredibly relevant to drugs when it comes to their tendency to be hydrolysed in the stomach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The simple fact is that I'm likely to come up with other questions so if it's possible for you to give me a guide for figuring out this stuff, or estimating it, please do share! If you do this and I can follow it, I will instantly accept your answer as the answer!&lt;/strong&gt; &lt;em&gt;Providing me with a &lt;strong&gt;decent ref&lt;/strong&gt; will also win you points.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2014-04-19T11:33:42.220" LastActivityDate="2014-04-19T11:33:42.220" Title="How much hydrolysis will likely occur of the following bonds at pH 1-4 in the time period of 1-3 hours?" Tags="&lt;ph&gt;&lt;chemical-biology&gt;&lt;drugs&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10079" PostTypeId="2" ParentId="10013" CreationDate="2014-04-19T12:16:57.573" Score="1" Body="&lt;p&gt;By the time the $p$ orbital is filled it is still the valence shell, you can see this by filling order e.g. $3s^2$ $3p^6$ $4s^2$ $3d^{10}$ ; You can see there isn't really an '18 electron rule' because there can never be 18 electrons in the valence shell and since chemistry is under most circumstances about the interaction of valence electrons then you can see that the $d$ shell electrons rarely have much to do with reactions because they are always 'one level down' from the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is a little unclear so I can't really say much more but I think you seem to be trying to ask why Neon or an analogous noble gas is so stable whilst Magnesium with a full $s$ orbital is not. Think about the difference in energy levels between going from one shell to another and you'll realise that an $\ce{Mg^2+}$ ion and Neon are very similar in terms of their stability because they are both in a low energy state.&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="4734" LastEditDate="2014-04-19T21:37:57.790" LastActivityDate="2014-04-19T21:37:57.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10080" PostTypeId="2" ParentId="9951" CreationDate="2014-04-19T12:24:15.587" Score="0" Body="&lt;p&gt;The ground state of any atom has degenerate sub-shell orbitals; however anything that increases the energy of one orbital more than the others, be it a steric effect or a field effect will break the degeneracy. There is only one s orbital so there isn't really anything for it to be degenerate with (ignoring hybridisation), p orbitals however since there are three of them AND they are mutually perpendicular there is opportunity for something to effect one p orbital more than the other two raising it's energy and breaking degeneracy. An example would be a magnetic field which running perpendicular to one orbital will have no effect but will increase the energy of the other two orbitals since they are mutually perpendicular. Another interesting example occurs in ligand field theory - &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Crystal_Field_Theory/High_Spin_and_Low_Spin_Complexes&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Crystal_Field_Theory/High_Spin_and_Low_Spin_Complexes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2014-04-19T12:24:15.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10081" PostTypeId="2" ParentId="10024" CreationDate="2014-04-19T12:40:07.000" Score="1" Body="&lt;p&gt;I think what's confusing you is how you know if an insoluble precipitate will form and the simple answer would be if an insoluble precipitate can form them it will form, and this is almost a direct consequence of le Chatelier's principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the reaction, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MgSO4(aq) + Na2CO3(aq) -&amp;gt; MgCO3(s) + Na2SO4(aq)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's actually happening is:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MgSO4(s) -&amp;gt; Mg^2+(aq) + SO4^{2-}(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2CO3(s) -&amp;gt; 2Na+(aq) + CO3^{2-}(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg^2+ (aq) + CO3^{2-}(aq) -&amp;gt; MgCO3(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Sodium and Sulphate ions in solution are spectators and do not 'react', they don't really do anything at all but you have a solution equivalent to that which you would have formed from the dissolution of Sodium Sulphate in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the poor format, I haven't figured out how to the proper chemical font yet and also you don't have to include water in your dissolution reactions, the involvement of water is implicit since your state symbols are changing&lt;/p&gt;&#xA;" OwnerUserId="4833" LastEditorUserId="5456" LastEditDate="2016-03-13T17:50:43.107" LastActivityDate="2016-03-13T17:50:43.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10082" PostTypeId="2" ParentId="152" CreationDate="2014-04-19T13:14:44.027" Score="7" Body="&lt;p&gt;Could not a simple way of rationalising this result be to use the formula for a classical Coulombic attraction between point charges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the radial distribution as shown, and the scaling of electrostatic force with r² rather than r, any time spent closer to the nucleus is vastly more important than time spent further away in terms of energy minimisation. This means that as radius increases (going left to right on the graph) what the lines do gets increasingly irrelevant unless they have exhibited very similar behaviour previously.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-04-19T13:14:44.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10083" PostTypeId="2" ParentId="10077" CreationDate="2014-04-19T13:39:30.053" Score="4" Body="&lt;p&gt;The issue, I think, is to specify how the pyridine is connected. This means you have to identify which carbon on the pyridine ring is connected. This picture shows the numbering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3onZL.png&quot; alt=&quot;pyridyl carbon numbering&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using your example the structure and names of the various possibilities would be (using connections on the 3-position of the benzonitrile ring as an example):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4pYBY.png&quot; alt=&quot;example structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can generalise from this example.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-04-19T13:39:30.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10084" PostTypeId="1" CreationDate="2014-04-19T15:49:47.957" Score="1" ViewCount="517" Body="&lt;p&gt;We know that chemical potential is defined as ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$. Here $a_i$ is the activity of $i^\text{th}$ component of solution. In case of gases instead of $a_i$ it is $f_i$ (fugacity). Activity is the activity coefficient multiplied by concentration or in case of fugacity it is the product of the fugacity coefficient multiplied by partial pressure of the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now this is not the only definition of Chemical potential. Chemical potential also means the partial molar Gibbs free energy when pressure and temperature of the system are constant. We know that &lt;strong&gt;change in&lt;/strong&gt; Gibbs free energy of a system at constant temperature and pressure also represents Non-PV work done by system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the term $RTln(a_i)$ can be correlated to Non-PV work done by system in order  for the two definitions to be equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NEW QUESTION AFTER READING YOUR ANSWER(S)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Philipp, Ok after reading all your answers I got now modified question that still remained.First of all we know K and Kc are related but kc can be modified by adding activity coefficients and standard concentrations to give equilibrium constant that should be equal to K.This requires the activity coefficients (That we used with Kc) to always follow relation with fugacity coefficients and RT terms which is not proved in answers there and that is what actually my question is asking.Sorry but I am reluctant here to write maths but I just say refer to the second equation from last in your &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5811/unit-of-the-equilibrium-constant-contradiction-of-bridgmans-theorem&quot;&gt;&lt;strong&gt;This Answer&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually earlier my question has title ways of expressing concentration in activity term &lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-04-23T05:06:10.440" Title="Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10086" PostTypeId="1" AcceptedAnswerId="10087" CreationDate="2014-04-19T17:42:41.843" Score="0" ViewCount="870" Body="&lt;p&gt;The only thing that is given is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The enthalpy of vaporization for $N_2(l)$ is 5.56 kJ/mol.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There aren't any examples similar to this problem in the textbook. I only know how to get energy using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q=mcΔT$ and $w=PΔV$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No clue what to do with the enthalpy of vaporization value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: Solved it by first getting number of moles ($n$) via $PV=nRT$ then multiplying moles by 5.56 kJ/mol. Moles cancel out, leaving only kJ, 1.75kJ to be precise.&lt;/p&gt;&#xA;" OwnerUserId="4686" LastEditorUserId="46306" LastEditDate="2018-12-07T19:06:46.810" LastActivityDate="2018-12-07T19:06:46.810" Title="How much heat (in J) is required to produce 2.0L of N2(g) at 77.36 K and 1.0 atm?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10087" PostTypeId="2" ParentId="10086" CreationDate="2014-04-19T17:52:09.537" Score="1" Body="&lt;p&gt;Since nitrogen's boiling point is 77.36 K , it appears that the question is asking how much heat is required to boil nitrogen to produce 2.0 L at 1 atm.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-19T17:52:09.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10088" PostTypeId="1" CreationDate="2014-04-19T18:00:50.903" Score="1" ViewCount="134" Body="&lt;p&gt;Double layer theories like that of Gouy-Chapman give the electrostatic potential $\ \phi\ $ in solution. For a symmetrical electrolyte and small potential drops $\ \phi\left(x=0\right)-\phi\left(x\rightarrow\infty\right)=\phi_0-0\ \text V=\phi_0$, the theory gives&#xA;$$&#xA;\phi=\phi_0\exp\left(-\kappa x\right)&#xA;$$&#xA;where $x$ is measured from the electrode/solution interface into the bulk solution.&#xA;As $\kappa$ is very large $\left(\sim10^{-9}\frac{1}{\text m}\right)$, the electric field&#xA;$$&#xA;E=-\frac{\text{d}\phi}{\text{d}x}=-\kappa\phi_0\exp\left(-\kappa x\right)&#xA;$$&#xA;falls to a constant value (0  V in this example) very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equations for the transport of ions in diluted solutions, the &lt;strong&gt;migration&lt;/strong&gt; flux of species $j$ is&#xA;$$&#xA;J_{\text {j,migr}}=-z_ju_jFc_j\frac{\text{d}\phi}{\text{d}x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the flux is proportional to $E$ it must be zero everywhere in solution according to Gouy-Chapman. But in every book about electrochemistry migration is said to contribute to the flux $J_j$ of species $j$ except when $c_j$ is much more smaller than the concentrations of other species. But why should there be migration in a macroscopic cell at all? In double layer theories ions gather near the electrode surface and compensate the charge on the metal, so there should be no electric field in solution. Isn't this what the equation for $\phi$ above says?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there two different types of &quot;electrostatic potentials&quot; in solution: One associated with the double layer theories and one associated with transport?&#xA;$$&#xA;\phi_{\text{Gouy-Chapman}}\neq\phi_{\text{Migration}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case: How are both potentials related to each other (and why do books always use the same symbol for both)?&lt;/p&gt;&#xA;" OwnerUserId="5122" LastActivityDate="2014-04-19T18:00:50.903" Title="What is the relationship between migration and the potential of double layer theory?" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10089" PostTypeId="2" ParentId="10060" CreationDate="2014-04-19T23:51:32.110" Score="0" Body="&lt;p&gt;To exhibit stereoisomerism, a compound must have a chiral centre i.e - a carbon atom with four different groups attached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is the only compound that meets this requirement.&lt;/p&gt;&#xA;" OwnerUserId="5237" LastActivityDate="2014-04-19T23:51:32.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10090" PostTypeId="2" ParentId="5055" CreationDate="2014-04-19T23:57:16.307" Score="3" Body="&lt;p&gt;Elemental carbon tends not to dissolve in solvents due to its very stable structure, but it will react with oxygen to form carbonates/carbonites, so the heating of the bowl in atmospheric oxygen (as mentioned in a previous comment) could be a possible solution.&lt;/p&gt;&#xA;" OwnerUserId="5237" LastActivityDate="2014-04-19T23:57:16.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10091" PostTypeId="1" AcceptedAnswerId="10095" CreationDate="2014-04-20T00:34:49.453" Score="1" ViewCount="1601" Body="&lt;p&gt;Why are there two possible bond angles for the nitrous acid molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After perusing Wikipedia I found that there are both cis and trans forms of the nitrous acid molecule. I also note that the cis form is sterically hindered. Could this be a reason for the two different bond angles - there are two different geometric isomers and in one case there are larger van der Waals repulsions, which in turn expands the bond angle? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:02:00.510" LastActivityDate="2018-03-04T02:02:00.510" Title="Bond angle of nitrous acid" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-08-25T09:16:20.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="10092" PostTypeId="1" AcceptedAnswerId="10144" CreationDate="2014-04-20T00:37:08.793" Score="1" ViewCount="2357" Body="&lt;p&gt;Given the molar heat capacity of a partition function as a function of temperature, how would one determine the partition function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the molar heat capacity = $\alpha T^2$ &lt;/p&gt;&#xA;" OwnerUserId="5238" LastActivityDate="2017-02-01T23:45:45.273" Title="Calculating the partition function based on molar heat capacity" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10093" PostTypeId="1" AcceptedAnswerId="10094" CreationDate="2014-04-20T00:45:39.430" Score="1" ViewCount="119" Body="&lt;p&gt;Is the end of a cigarette on fire when it is lit? Or is it smoldering? Is there a chemical difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="316" LastActivityDate="2014-05-20T01:43:16.623" Title="Is the end of a cigarette on fire when lit?" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;materials&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10094" PostTypeId="2" ParentId="10093" CreationDate="2014-04-20T01:07:24.330" Score="1" Body="&lt;p&gt;A lighted cigarette is burning, technically smoldering (no flame).  The stuff inside the paper bears only a passing resemblance to tobacco,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cigarette#Construction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cigarette#Construction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-20T01:07:24.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10095" PostTypeId="2" ParentId="10091" CreationDate="2014-04-20T01:22:39.857" Score="1" Body="&lt;p&gt;This ism the microwave structure of nitrous acid, 1971.  Your mileage may vary condensed phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI:10.1039/TF9716702179 (1971)&lt;BR&gt;&#xA;The structure of &lt;em&gt;cis&lt;/em&gt;-nitrous acid is found to be: O—H = 0.982, N—O(H)= 1.392, N—O = 1.185 Å, ∠ NOH = 104.0 and ∠ ONO = 113.6°, and for &lt;em&gt;trans&lt;/em&gt;-nitrous acid: O—H = 0.958, N—O(H)= 1.432, N—O = 1.170 Å, ∠ NOH = 102.1 and ∠ ONO = 110.7°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf&lt;/a&gt; &lt;BR&gt;&#xA;  2007 pape with more recent citations.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-20T01:22:39.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10096" PostTypeId="2" ParentId="5055" CreationDate="2014-04-20T01:49:15.320" Score="3" Body="&lt;p&gt;All these materials are hazardous by skin contact and inhalation.  The are toxic and evil.  One possibility is soaking overnight in auto supply store generic carburetor cleaner.  Another is overnight soak in automatic transmission fluid.  Gloves, goggles, disposable activated charcoal painter's mask.  If you get anything on a belt or your shoes, you throw them out.  Contaminated clothing must be quickly removed.  Wash skin with soap and water - twice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=bRiZOTTel34&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=bRiZOTTel34&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.break-free.com/?location=/products/index.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.break-free.com/?location=/products/index.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything works, wash the cleaned bowl in a good dishwashing detergent solution - twice.  There must be no residual product odor.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-20T01:49:15.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10097" PostTypeId="1" CreationDate="2014-04-20T02:20:15.957" Score="2" ViewCount="1580" Body="&lt;p&gt;$$\ce{4Fe + 3O2  -&amp;gt; 2Fe2O3(s) + 1625~$\mathrm{kJ}$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be endothermic because the energy is in the product. However, if the equation is viewed from the perspective of the reactant, it is endothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this endo- or exothermic?&lt;/p&gt;&#xA;" OwnerUserId="5239" LastEditorUserId="7448" LastEditDate="2017-04-08T16:34:23.740" LastActivityDate="2017-04-08T16:36:03.033" Title="Is this endo or exothermic?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10098" PostTypeId="1" CreationDate="2014-04-20T02:37:09.013" Score="5" ViewCount="79615" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the final temperature of water and iron if a &lt;span class=&quot;math-container&quot;&gt;$\pu{30 g}$&lt;/span&gt; piece of iron at &lt;span class=&quot;math-container&quot;&gt;$\pu{144 °C}$&lt;/span&gt; was dropped into a calorimeter with &lt;span class=&quot;math-container&quot;&gt;$\pu{40 g}$&lt;/span&gt; of water at &lt;span class=&quot;math-container&quot;&gt;$\pu{20 °C}$&lt;/span&gt;?&#xA;Specific heat of water is &lt;span class=&quot;math-container&quot;&gt;$\pu{4.184 J g-1 ^\circ C-1}$&lt;/span&gt;, and of iron is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.449 J g-1 ^\circ C-1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here is my work:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;Q &amp;amp;= mc\,\Delta T\\&#xA;Q_1 &amp;amp;= (\pu{30 g})(\pu{0.449 J g-1 ^\circ C-1})(x - \pu{144 ^\circ C}) \tag{Iron}\\&#xA;Q_2 &amp;amp;= (\pu{40 g})(\pu{4.184 J g-1 ^\circ C-1})(x - \pu{20 ^\circ C}) \tag{Water}\\&#xA;\text{Since,} \quad Q_1 &amp;amp;= -Q_2\\&#xA;13.47 (x-144) &amp;amp;= - (167.36)(x-20)\ \pu{J}\\&#xA;13.47x - 1939.68 &amp;amp;= -167.36x + 3347.20\\&#xA;180.83x &amp;amp;= \pu{5286.88 ^\circ C} \\&#xA;x &amp;amp;=  \pu{0.03420 ^\circ C}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;This is giving me an answer that isn't correct according to my book. What did I do wrong, and how can I fix it?&lt;/p&gt;&#xA;" OwnerUserId="5239" LastEditorDisplayName="user7951" LastEditDate="2020-09-19T09:02:38.230" LastActivityDate="2020-09-19T09:02:38.230" Title="How do I find the final temperature of a calorimeter using specific heat capacities?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;&lt;calorimetry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10100" PostTypeId="1" AcceptedAnswerId="10101" CreationDate="2014-04-20T02:53:32.880" Score="6" ViewCount="93" Body="&lt;p&gt;Is there an article, book, some account of the discovery that crude oil came from decayed biological material?&lt;/p&gt;&#xA;" OwnerUserId="316" LastEditorUserId="30836" LastEditDate="2019-04-13T22:09:22.673" LastActivityDate="2019-04-13T22:09:22.673" Title="When was oil discovered to be of biological origin?" Tags="&lt;reference-request&gt;&lt;history-of-chemistry&gt;&lt;geochemistry&gt;&lt;petrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10101" PostTypeId="2" ParentId="10100" CreationDate="2014-04-20T03:09:51.767" Score="7" Body="&lt;p&gt;If you look at a gas or liquid chromatogram of crude oil, you see a remarkable preponderance of (unbranched) even-numbered carbon compounds (re acetyl CoA homologation).  You also see traces of polycyclics that are net chiral - terpenoids, terpenes, pristane, phytane, cholestane, chlorins, and porphyrins.  Fischer-Tropsch gives a very different distribution and no net chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mikhailo V. Lomonosov, &quot;&lt;em&gt;Slovo o reshdenii metallov ot tryaseniya zemli&lt;/em&gt;&quot; &#xA;Proceedings of the Imperial Academy of Sciences, St. Petersburg, 1757.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Mikhail Tsvet was born in 1872)&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-20T03:09:51.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10102" PostTypeId="1" CreationDate="2014-04-20T08:34:00.870" Score="0" ViewCount="29" Body="&lt;p&gt;Imagine two companies X and Y. Both companies use very similar technical equipment to carry out a biotechnological process where a chemical reaction is catalyzed by an enzyme. Company X uses an enzyme with a reaction barrier of, say, 15 kcal/mol, while company Y uses a slightly different enzyme to catalyze the same reaction but this enzyme has an activation energy of only 12 kcal/mol.&#xA;The reaction energies in the two reactors are, obviously, the same since both companies do $\ce{A} \rightarrow \ce{B}$. Still, as I see it, company X has to input 3 kcal per mole of product more into activating the enzyme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does company Y therefore require 3 kcal less energy to produce 1 mole of product?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2014-04-20T09:16:25.703" Title="Can energy savings be estimated from activation energy differences?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10103" PostTypeId="2" ParentId="5055" CreationDate="2014-04-20T08:58:29.893" Score="0" Body="&lt;p&gt;The situation is not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your carbon is pure ( why would it be?) then burning down with flame would be the easiest and cleanest solution. Practically nothing solves pure carbon, except melt iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your carbon is not pure, but e.g. you burned some food on the surface, cooking with sodium-carbonate generally help. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-20T08:58:29.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10104" PostTypeId="2" ParentId="6417" CreationDate="2014-04-20T09:01:56.843" Score="1" Body="&lt;p&gt;I use the following to remember the spectrochemical series;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;I brought some cloth, nylon net from office, collect water near cyanomethane place, among each big pants, place no proper concentration or concern.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(&lt;span class=&quot;math-container&quot;&gt;$\ce{I, Br, S, Cl, NO_3, N_2, F, OH, CrO_2, H_2O, CH_3CN, py, NH_3, en, bpy, phen, NO_2, PPh3, CN, CO}$&lt;/span&gt; respectively.)&lt;/p&gt;&#xA;" OwnerUserId="5241" LastEditorUserId="65276" LastEditDate="2021-04-06T11:57:51.110" LastActivityDate="2021-04-06T11:57:51.110" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10105" PostTypeId="2" ParentId="10102" CreationDate="2014-04-20T09:16:25.703" Score="3" Body="&lt;p&gt;The statement of this questions mixes a number of issues, so there is, in my opinion, no definite answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, one must take care that activation energies will have an influence on the kinetics (rate constant) of the reaction that is catalyzed. In a very general point of view, the overall energetic requirement for a chemical reaction does not directly depend on the activation energy (for instance the equilibrium constant is not changed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus if two separate processes are considered, a different activation energy will mostly mean a reaction rate that will be slower, so overall a process that might take longer but not necessarily need a higher energy input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if one would add to this problem a constraint that would be that the reaction rates of bith processes should be similar, the main method to achieve such a result would be to change the temperature at which the process takes place. This is related to the activation energy through the Arrhenius equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = A e^{-\frac{E_{a}}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one assumes that the pre-exponential factor is the same for both processes (in effect meaning that the transition states in the catalyzed reaction have similar constraints) then one can reach some conclusions on both processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the reaction rates at constant temperature can be calculated. At 298 K, 3 kcal/mol difference would mean a ratio $k_{1}/k_{2}$ of about 150, so the process with lower activation energy would proceed 150 times faster than the other one. An other option would then be to increase the temperature of the slower process. In such a case, the temperature would depend on the ratio of the activation energies (and not the difference) as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{2} = \frac{E_{a2}}{E_{a1}} T_{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes the values of the qustion and considers a process that would take place at 298 K, one would need to heat at 373 K to reach the same rate constant. From this, one could consider the additional energy input required to heat the reaction medium up to this temperature.&lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2014-04-20T09:16:25.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10106" PostTypeId="1" CreationDate="2014-04-20T09:41:31.227" Score="18" ViewCount="100642" Body="&lt;p&gt;What causes the melting and boiling points of noble gases to rise when the atomic number increases? What role do the valence electrons play in this?&lt;/p&gt;&#xA;" OwnerUserId="5225" LastEditorUserId="9344" LastEditDate="2016-08-19T12:16:32.607" LastActivityDate="2018-05-14T21:06:51.100" Title="Why do the melting and boiling points of the noble gases increase when the atomic number increases?" Tags="&lt;inorganic-chemistry&gt;&lt;boiling-point&gt;&lt;melting-point&gt;&lt;noble-gases&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10108" PostTypeId="2" ParentId="10106" CreationDate="2014-04-20T13:00:56.067" Score="19" Body="&lt;p&gt;The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The van der Waals force increases with the increase in the size of the atom, and therefore, in general, the boiling and melting points increase from $\ce{He}$ to $\ce{Rn}$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helium boils at $-269\ \mathrm{^\circ C}$. Argon has larger mass than helium and have larger dispersion forces. Because of larger size the outer electrons are less tightly held in the larger atoms so that instantaneous dipoles are more easily induced resulting in greater interaction between argon atoms. Therefore, its boiling point ($-186\ \mathrm{^\circ C}$) is more than that of $\ce{He}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, because of increased dispersion forces, the boiling and melting points of monoatomic noble gases increase from helium to radon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k61Jv.gif&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For more data of melting and boiling points of noble gas &lt;em&gt;compounds&lt;/em&gt;, read this &lt;a href=&quot;http://books.google.co.in/books?id=6OcDAAAAMBAJ&amp;amp;pg=PA340&amp;amp;dq=melting+and+boiling+points+of+noble+gas+compounds&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=CuZTU5HoHcePrQe8loHwBA&amp;amp;ved=0CC0Q6AEwAA#v=onepage&amp;amp;q=melting%20and%20boiling%20points%20of%20noble%20gas%20compounds&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;page&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorDisplayName="user7951" LastEditDate="2015-06-19T17:41:18.377" LastActivityDate="2015-06-19T17:41:18.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10109" PostTypeId="2" ParentId="5055" CreationDate="2014-04-20T14:44:46.147" Score="2" Body="&lt;p&gt;When a cooking utensil used over a campfire becomes encrusted on the outside, it's because the contents of the utensil keep it relatively cool.  Simply heating an empty container (below its melting point!) should oxidize off the carbon-based material. (Probably creosote)&lt;/p&gt;&#xA;" OwnerUserId="3668" LastActivityDate="2014-04-20T14:44:46.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10110" PostTypeId="2" ParentId="10097" CreationDate="2014-04-20T16:10:28.777" Score="5" Body="&lt;p&gt;The reaction is exothermic left to right and endothermic right to left. Yes, as you noted, it's wise to note which way you are looking at the reaction, but generally, we look at reactions from left to right (because the arrow generally points left to right). For example, the following reaction is endothermic as written, i.e. very hot water (steam) is very corrosive and adding heat to water encourages further ionization of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O ⟶ H3O+ + HO-}$$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2017-04-08T16:35:11.540" LastActivityDate="2017-04-08T16:35:11.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10111" PostTypeId="2" ParentId="10106" CreationDate="2014-04-20T16:24:11.653" Score="7" Body="&lt;p&gt;Valence electrons have not much to do with this, as their outer shell is closed.&#xA;As the other answer mentioned, dispersion forces are the ones responsible for any interaction between these atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size dependence therefore is directly coming from the size dependence of dispersion forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very simplistic way, a random charge fluctuation can polarize the otherwise perfectly apolar atoms. This induced dipole moment is then responsible for the dispersion interactions.  The polarizibility of an atom increases (easier to polarize) if the atomic number increases, therefore the interactions in nobel gases will reflect this behavior. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-20T16:24:11.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10114" PostTypeId="1" CreationDate="2014-04-20T18:03:22.917" Score="4" ViewCount="19790" Body="&lt;p&gt;Mix &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrochloric_acid&quot; rel=&quot;nofollow&quot;&gt;$\ce{HCl}$&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;$\ce{NaOH}$&lt;/a&gt; together (reactants are at the same temperature) in a calorimeter. The temperature increases, and the book says it's an exothermic reaction. I don't get it. The thermometer is in the solution which means it measures the temperature of the solution. It's the system. The temperature of the system increases, this means that heat flows in. How could this be an exothermic reaction?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="4231" LastEditDate="2014-07-20T14:01:50.903" LastActivityDate="2015-09-07T17:18:12.450" Title="How to judge if a reaction is exothermic or endothermic" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10115" PostTypeId="1" CreationDate="2014-04-20T18:08:42.833" Score="14" ViewCount="5188" Body="&lt;p&gt;Rearranging the Eyring equation leads to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta^\ddagger S^\circ = R \ln{\frac{k \times h}{{k_\text{B}}{T}}}+\frac{\Delta^\ddagger H^\circ}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is the rate constant, $h$ is the Planck constant, $R$ is the universal gas constant, $k_\text{B}$ is the Boltzmann constant, $T$ is temperature, $H$ is enthalpy, and $S$ is entropy. The entropy term should ultimately be in units of $\mathrm{\frac{J}{mol\ K}}$. The right term of the equation will clearly yield the correct units. For the left term, the universal gas constant can be expressed in $\mathrm{\frac{J}{mol\ K}}$, which means that the log term should be dimensionless. For the bottom part of the fraction, the units evaluate to $\mathrm{J}$. For the top part, the Planck constant has units of $\mathrm{J\cdot s}$. To get the top part of the fraction to evaluate to $\mathrm{J}$, the rate constant would have to be in units of $\mathrm{s^{-1}}$. For first order reactions, this is indeed the case. For reactions of higher order, this will never work out to the correct units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me think that the Eyring equation is simply using the unitless values of these variables. This still leads to some ambiguity about which value of the rate constant to use. Take for instance a second-order reaction in which the rate constant is in units of $\mathrm{M^{-1}\ s^{-1}}$. If we rearrange the units to SI units ($\mathrm{m^3\ mol^{-1}\ s^{-1}}$), the value of the rate constant changes. So is the rate constant $k$ in the Eyring equation simply the &lt;em&gt;value&lt;/em&gt; of the rate constant when that rate constant is expressed in certain units? And if so, are the concentration units always expressed in molarity? I kept missing a homework problem because I was converting $k$ from molarity to SI units, which makes me believe this is indeed the case.&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T16:32:37.710" LastActivityDate="2021-05-03T14:30:27.330" Title="Rate Constant Units and Eyring Equation" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10116" PostTypeId="5" CreationDate="2014-04-20T18:57:51.487" Score="0" Body="&lt;p&gt;A fuel is any material that can be made to react with other substances so that it releases chemical or nuclear energy as heat or to be used for work. The concept was originally applied solely to those materials capable of releasing chemical energy but has since also been applied to other sources of heat energy such as nuclear energy (via nuclear fission and nuclear fusion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, heat energy released by reactions of fuels is converted into mechanical energy via a heat engine. Other times the heat itself is valued for warmth, cooking, or industrial processes, as well as the illumination that comes with combustion. Fuels are also used in the cells of organisms in a process known as cellular respiration, where organic molecules are oxidized to release usable energy. Hydrocarbons and related oxygen-containing molecules are by far the most common source of fuel used by humans, but other substances, including radioactive metals, are also utilized.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="39539" LastEditDate="2017-06-05T21:39:53.887" LastActivityDate="2017-06-05T21:39:53.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10117" PostTypeId="4" CreationDate="2014-04-20T18:57:51.487" Score="0" Body="Questions regarding fuels and the properties of fuels, like calorific value, octane number, mechanism of combustion, etc., should use this tag. " OwnerUserId="4934" LastEditorUserId="39539" LastEditDate="2017-06-05T21:38:49.237" LastActivityDate="2017-06-05T21:38:49.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10118" PostTypeId="2" ParentId="10114" CreationDate="2014-04-20T19:13:03.203" Score="2" Body="&lt;p&gt;HCl + NaOH give NaCl plus water.  Add up the free energies of formation of the species on each side and compare. Binding energies are negative.  The more strongly bound side is created with an exotherm.  The outside is irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-20T19:13:03.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10120" PostTypeId="1" AcceptedAnswerId="10130" CreationDate="2014-04-20T20:28:39.450" Score="0" ViewCount="6215" Body="&lt;p&gt;Using this equation, &#xA;$\ce{CaCl2(s) -&amp;gt; Ca^{2+}(aq) + 2Cl^{-}(aq)}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\Delta H = - 325.0 \; kJ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 15.0 grams of the solid, how much heat is released or gained in kilojoules? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q4mST.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5249" LastEditorUserId="5249" LastEditDate="2014-04-20T23:04:37.763" LastActivityDate="2014-04-21T04:39:28.150" Title="How much heat is gained/released based off of equation?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10122" PostTypeId="1" AcceptedAnswerId="10126" CreationDate="2014-04-21T00:29:52.843" Score="0" ViewCount="595" Body="&lt;p&gt;I am working on a straight-forward NMR graph as part of my test prep. I have a pretty limited understanding of NMR, but I thoroughly read the chapter leading up to this practice question and am having difficulty explaining all of the peaks that I am seeing. Let me show the question and my understanding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the chemical and the trace that results.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jznk5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this problem is that there are four groups of equivalent hydrogens, and I numbered them 1-4. The 1st group has 3 equivalent hydrogens, with two adjacent non-equivalent hydrogens (group #2). Using the n+1 rule, the H group #1 should create 2+1=3 peaks in a cluster. Same goes for H group #4, since there are two adjacent hydrogens, so another 3 peaks. When considering H-group #2, the three hydrogens from H-group #1 should be 3+1=4 peaks. Same goes for H-group #3, which has the three hydrogens from H-group #4 so 3+1=4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am expecting to see 2 groups of 3 peaks and 2 groups of 4 peaks. I see the two groups of 4 peaks, but not the 2 groups of three peaks. Instead, I see 1 group of 6 peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I supposed to believe that the 2 groups of 3 are just located so near it appears to be 6? Or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="1986" LastActivityDate="2014-04-21T02:54:54.837" Title="Unexplained peak cluster in proton NMR trace?" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10124" PostTypeId="2" ParentId="10122" CreationDate="2014-04-21T01:28:42.477" Score="2" Body="&lt;p&gt;You are correct. The chemical shift of methyl protons 1 and 4 are close to one another.  You still see the two expected triplets, they just have similar chemical shifts.  That's not too surprising since they are both methyl groups attached to a methylene group.  We could tell which triplet belongs to which methyl carbon by measuring and comparing the 2 sets of CH2-CH3 coupling constants. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-04-21T02:09:16.643" LastActivityDate="2014-04-21T02:09:16.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10125" PostTypeId="1" AcceptedAnswerId="10131" CreationDate="2014-04-21T02:49:17.437" Score="1" ViewCount="6245" Body="&lt;p&gt;When going through one of my chemistry textbooks, I saw that the electrolysis of aqueous lead nitrate led to oxygen being formed at the anode and lead being formed at the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in school, I was taught that only metals less reactive than hydrogen will form at the cathode.  This electrolysis contradicts what I learned, as lead is more reactive than hydrogen, and should remain as ions in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my question is: &lt;strong&gt;What forms at the cathode during electrolysis of aqueous lead nitrate and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5251" LastActivityDate="2021-01-21T21:04:54.683" Title="Electrolysis of aqueous lead nitrate" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10126" PostTypeId="2" ParentId="10122" CreationDate="2014-04-21T02:54:54.837" Score="2" Body="&lt;p&gt;I did a simulation of the peaks (I got the shifts wrong) for 60 MHz and 400 mHz. As you can see, the peaks overlap on the weaker magnet.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9PeqM.png&quot; alt=&quot;Ethyl Acetate 60 mHz&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5KV5U.png&quot; alt=&quot;Ethyl Acetate 400 mHz&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-21T02:54:54.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10127" PostTypeId="1" CreationDate="2014-04-21T03:17:44.433" Score="8" ViewCount="3099" Body="&lt;p&gt;How can I make $\ce{Cu2O}$ from copper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am seeing if electrolysis is a good route (in the context of kitchen style chemistry) to cuprous oxide, instead of the traditional heating copper and removing the cupric oxide, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Epsom salt electrolyte, it is very easy to make copper II hydroxide (cloudy ammonia soluble, according to the wiki). This can be heated to $\ce{CuO}$ easily enough, but I have no idea if this can be reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $\ce{CuO}$ be reduced easily? Or is there a simple way to produce (from copper) a copper (I) salt, from which a precipitate of $\ce{Cu2O}$ may be formed? What reducing agents could be used?&lt;/p&gt;&#xA;" OwnerUserId="5252" LastEditorUserId="4945" LastEditDate="2014-05-22T12:00:46.723" LastActivityDate="2020-05-09T05:46:20.727" Title="Possible methods for copper (II) reduction to copper (I)" Tags="&lt;redox&gt;&lt;home-experiment&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10130" PostTypeId="2" ParentId="10120" CreationDate="2014-04-21T04:39:28.150" Score="2" Body="&lt;p&gt;I'm going to assume that the question is presenting the given change in enthalpy as a molar enthalpy change (so as to say that -325.0 kJ are evolved per mole of CaCl2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under that assumption, answering the question is then a matter of using the a molar fraction of CaCl2 for the amount you have present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CaCl2 has a molar mass of 110.98 g/mol so with 15.0 g of CaCl2, you'd have roughly 0.1352 mol of CaCl2. 0.1352 mol * -325.0 kJ/mol = -43.94 kJ. The negative here means that the CaCl2 will release 43.94 kJ of heat. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastActivityDate="2014-04-21T04:39:28.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10131" PostTypeId="2" ParentId="10125" CreationDate="2014-04-21T05:01:14.680" Score="3" Body="&lt;p&gt;Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question more directly, it is actually lead that forms at the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastActivityDate="2014-04-21T05:01:14.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10132" PostTypeId="2" ParentId="10097" CreationDate="2014-04-21T05:19:31.980" Score="2" Body="&lt;p&gt;In the equation given, the energy is not actually &lt;strong&gt;in&lt;/strong&gt; the product, but is written &lt;strong&gt;as&lt;/strong&gt; a product, meaning that 1625 kJ of heat is evolved along with 2 moles of iron(II) oxide.  &lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="7448" LastEditDate="2017-04-08T16:36:03.033" LastActivityDate="2017-04-08T16:36:03.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10135" PostTypeId="1" CreationDate="2014-04-21T12:47:04.517" Score="1" ViewCount="928" Body="&lt;p&gt;I need to write balanced equations describing the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one mole of $\ce{Al2Me6}$ with two moles of water&lt;/li&gt;&#xA;&lt;li&gt;excess of $\ce{Al2Me6}$ with silicon dioxide&lt;/li&gt;&#xA;&lt;li&gt;excess of $\ce{Al2Me6}$ with tin(IV) chloride&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My answer so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a) $\ce{2H2O + Al2Me6 -&amp;gt; Al2Me4(H2O)2}$?&lt;/li&gt;&#xA;&lt;li&gt;b) $\ce{SiO2 + Al2Me6 -&amp;gt; SiMe4 + Al2Me2O2}$?&lt;/li&gt;&#xA;&lt;li&gt;c) $\ce{SnCl4 + Al2Me6 -&amp;gt; SnMe4 + Al2Me2Cl4}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="7448" LastEditDate="2017-04-08T16:38:09.873" LastActivityDate="2017-04-08T16:38:09.873" Title="Reactions involving trimethylaluminium Al2Me6" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-04-08T19:15:00.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="10136" PostTypeId="2" ParentId="5055" CreationDate="2014-04-21T12:59:24.580" Score="9" Body="&lt;p&gt;By far the safest solution is to physically abrade the surface until you have worn away everything that isn't copper. This is slow and requires some sort of grinding agent but has the upside that you are not likely to injure yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a chemical solution that will work but it requires access to some dangerous chemicals and is extremely hazardous even in a well-equipped lab. It is insane to try anywhere else as you won't have adequate safety equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to remove carbon black from anything is to use a mixture of concentrated sulfuric acid and hydrogen peroxide. This is often used (but also often highly discouraged because of the danger involved) in laboratories to clean sintered glass filters contaminated with things like activated carbon. In the case of a copper bowl, the solution will remove the carbon but will also etch the copper surface (it is sometimes used as an etchant to remove copper from printed circuit boards). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixture of sulfuric acid and hydrogen peroxide is sometimes known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Piranha_solution&quot; rel=&quot;noreferrer&quot;&gt;Piranha solution&lt;/a&gt; of reasons that are obvious when you have seen it work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not encourage anyone without good chemical safety training and access to very good safety equipment to work with the mixture. And even then they should work only with small amounts and in a fume hood behind a shield and while wearing eye protection, strong gloves, and a fireproof coat.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="402" LastEditDate="2017-04-09T18:33:05.940" LastActivityDate="2017-04-09T18:33:05.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10137" PostTypeId="1" CreationDate="2014-04-21T13:45:40.857" Score="5" ViewCount="317" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction of $\ce{(C5H5)Fe(CO)2Cl}$ with an equimolar amount of $\ce{Li+[C6H5]-}$ gives a product. Draw its structure and apply the 18 electron rule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a question I was given the molecular weight and percentages of each element in the compound. I worked out that there is $\ce{1 Fe}$, $\ce{13 C}$, $\ce{10 H}$ and $\ce{2 O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product seems to be an iron atom surrounded with ligands, $\ce{2 CO}$ ligands, $\ce{1 C5H5}$ and $\ce{1 [C6H5]-}$ ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 18 electron rule expands to 8 valence electrons from $\ce{Fe}$, 2 electrons from each $\ce{CO}$ ligand, 5 electrons from $\ce{C5H5}$ and 1 electron from the $\ce{[C6H5]-}$ anion: $8+4+5+1 = 18$. But what would the structure look like?  Especially how is the $\ce{[C6H5]-}$ anion connected?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="16683" LastEditDate="2017-04-01T16:17:20.383" LastActivityDate="2017-04-01T16:17:20.383" Title="Reaction of CpFe(CO)2Cl with phenyllithium" Tags="&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10138" PostTypeId="1" CreationDate="2014-04-21T14:01:22.317" Score="7" ViewCount="2309" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Under basic conditions a Michael addition reaction can be used to convert compound A into product C. Explain why no Robinson annulation cyclic product is observed?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Compound A&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/5L2dX.png&quot; alt=&quot;reactant A&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Product C&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/YMC1M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have worked out that the other compound involved in the Michael addition reaction would be compound B (below), but I don't know why no Robinson annulation reaction could occur after the Michael addition reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound B &#xA;&lt;img src=&quot;https://i.stack.imgur.com/zUkrO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="5026" LastEditDate="2018-02-26T02:05:20.777" LastActivityDate="2018-02-26T02:05:20.777" Title="Why does no Robinson Annulation occur after the Michael addition?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10139" PostTypeId="1" CreationDate="2014-04-21T14:11:29.973" Score="1" ViewCount="3075" Body="&lt;p&gt;Fe(η5-C5H5)2 does not react with hydrogen, but its nickel analogue, Ni(η5-C5H5)2 is readily hydrogenated to give Ni(η5-C5H5)(η3-C5H7). Use the 18 electron rule to explain this behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer so far...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe(η5-C5H5)2 18 electron rule = valence electrons from Fe is 8, electrons from C5H5 is 5 therefore 8 + 5 + 5 = 18 so it is stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ni(η5-C5H5)2 18 electron rule = valence electrons from Ni is 10, electrons from C5H5 is 5 therefore 10 + 5 + 5 = 20 so Ni is not in most stable form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ni(η5-C5H5)(η3-C5H7) 18 electron rule = valence electrons from  Ni is 10, electrons from C5H5 is 5.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how many electrons from (η3-C5H7) and how does this all explain why the iron complex doesn't react with hydrogen but the nickel one does? It is to do with the fact it has two too many electrons (20e) so it bonds with two protons from hydrogen? Also what does this symbol η mean? &lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="189" LastEditDate="2014-10-27T22:10:23.470" LastActivityDate="2014-10-27T22:10:23.470" Title="How can the 18 electron rule be used to explain that Fe(η5-C5H5)2 doesn't react with hydrogen but Ni(η5-C5H5)2 does?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;coordination-compounds&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10140" PostTypeId="2" ParentId="10138" CreationDate="2014-04-21T14:41:03.453" Score="4" Body="&lt;p&gt;As illustrated in the following figure, &lt;strong&gt;the Robinson annulation reaction involves two steps&lt;/strong&gt;.  The first step involves the &lt;strong&gt;Michael addition&lt;/strong&gt; of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below).  In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an &lt;strong&gt;intramolecular aldol condensation&lt;/strong&gt; to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mj6F7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mj6F7.png&quot; alt=&quot;Robinson annulation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often in Robinson annulations methyl vinyl ketone is used. &lt;strong&gt;It is important that  the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl&lt;/strong&gt; so that the enolate required in the second step can be formed. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Explain why no Robinson annulation cyclic product is observed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In your example&lt;/strong&gt;, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your &lt;strong&gt;Product C&lt;/strong&gt; can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-09-11T18:46:48.540" LastActivityDate="2016-09-11T18:46:48.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10141" PostTypeId="2" ParentId="10139" CreationDate="2014-04-21T14:52:59.190" Score="3" Body="&lt;p&gt;Hydrogenation of NiCp2 takes place on the ligand, not the metal, to reduce donor electron count, giving a low-energy closed-shell 18 electron count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nickelocene&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Nickelocene&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ferrocene&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Ferrocene&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hapticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Hapticity&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf&lt;/a&gt; &lt;BR&gt;&#xA;   p. 9, 10.  &quot;η&quot; is &quot;hapticity,&quot; the number of ligand atoms contributing pi-electron orbitals to the metal center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cp^-}$ is aromatic (draw the Frost diagram) when coordinated $\ce{η^5}$ on-face, or  $\ce{η^1}$ non-aromatic as a simple anion coordinated on-vertex.  Saturating one localized double bond leaves cyclic delocalized allyl anion coordinated $\ce{η^3}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{TiCp4}$, two Cp anions are face pentahapto, two are vertex monohapto.  If we take a proton or C-13 NMR, will we see discrete protons or carbons for the $\ce{η^1}$-Cp, or will allowed 1,5-shifts scramble all five protons?  All five carbons?  Will it be temperature dependent if it happens?  If it happens, will H and C have different kinetics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXZgm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-21T14:52:59.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10142" PostTypeId="2" ParentId="10137" CreationDate="2014-04-21T15:06:48.693" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://voh.chem.ucla.edu/vohtar/fall06/classes/277/pdf/18-Electron%20Rule.pdf&quot; rel=&quot;nofollow&quot;&gt;http://voh.chem.ucla.edu/vohtar/fall06/classes/277/pdf/18-Electron%20Rule.pdf&lt;/a&gt; &lt;BR&gt;&#xA;Summing ligand electron contribution. &lt;BR&gt;&#xA;DOI:10.1006/jmsp.2001.8470 &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.ccp14.ac.uk/ccp/web-mirrors/farrugia/~louis/chem2x/Organomet-2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ccp14.ac.uk/ccp/web-mirrors/farrugia/~louis/chem2x/Organomet-2.pdf&lt;/a&gt; &lt;BR&gt;&#xA; Document p. 34.  Note error in drawing the graphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct displacement gives $\ce{CpFe(CO)2Ph}$, with the phenyl being a simple anion sigma-bonded to the metal. If there are two carbons plus two oxygens in the product, implying the CO ligands are intact, your are done.  If some fancy stiff happened with those ligands and the phenyl anion, you must work it out or look it up.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-21T15:06:48.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10144" PostTypeId="2" ParentId="10092" CreationDate="2014-04-21T15:20:33.410" Score="1" Body="&lt;p&gt;All you have to do is to use some of the well known results from Statistical Thermodynamics:&#xA;The molar heat capacity at constant volume is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \bar{c}_V = \left( \frac{\partial \bar{U}}{ \partial T} \right)_V&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\bar{U}$ is the inner energy, $T$ is the temperature, and $V$ is the volume.&#xA;The molar inner energy in terms of the partition function $Z$ and the temperature is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \bar{U} = \frac{k_{\text{B}} T^{2}}{n} \frac{\partial \ln Z}{ \partial T} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the amount of substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you simply have to plug the equation for $\bar{U}$ into the equation for $\bar{c}_V$ and after using the chain rule of differentiation you get the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \bar{c}_V = 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} + \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you just have to use your formula for the molar heat capacity, $\bar{c}_V = \alpha T$, and solve the resulting differential equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} - \alpha T^2 &amp;amp;= 0  \\&#xA; \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{1}{T} \frac{\partial \ln Z}{ \partial T} - \frac{n \alpha}{k_{\text{B}}} &amp;amp;= 0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $\ln Z$.&#xA;If solving differential equations by hand is not your thing, you can simply charge &lt;a href=&quot;http://www.wolframalpha.com/input/?i=DSolve%5by%27%27%5bx%5d+%2B+%282%2Fx%29%2ay%27%5bx%5d+-+a+%3D%3D+0%2C+y%5bx%5d%2C+x%5d&quot; rel=&quot;nofollow&quot;&gt;Wolfram Alpha&lt;/a&gt; with the task and get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \ln Z &amp;amp;= \frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \\&#xA;  Z &amp;amp;= \exp\left(\frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_1$ and $C_2$ are constants of integration which can be fixed by enforcing boundary conditions for the temperature.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-04-21T15:20:33.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10146" PostTypeId="1" AcceptedAnswerId="10182" CreationDate="2014-04-21T15:53:42.103" Score="4" ViewCount="1876" Body="&lt;p&gt;I have the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{Mg(s) + 2HCl(aq) &amp;amp;-&amp;gt; MgCl2(aq) + H2(g)}\\&#xA;\ce{MgO(s) + 2HCl(aq) &amp;amp;-&amp;gt; MgCl2(aq) + H2O(aq)}\\&#xA;\ce{H2(g) + 1/2O2(g) &amp;amp;-&amp;gt; H2O(l)}\\&#xA;\ce{Mg(s) + 1/2O2(g) &amp;amp;-&amp;gt; MgO(s)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have $\Delta H$ in $\mathrm{kJ/mol}$ for the first three reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of this information, I need to find $\Delta H^\circ_\mathrm{f}{\left(\ce{MgO(s)}\right)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m pretty sure I have to use Hess’ law but I don’t know how to apply it here. I’ve tried using the second reaction as my “main” reaction and then I can use the third and fourth reaction as formation reactions for two of the compounds of the main reaction but after that I’m not sure how to proceed.&lt;/p&gt;&#xA;" OwnerUserId="4693" LastEditorUserId="95133" LastEditorDisplayName="user7951" LastEditDate="2021-04-30T11:30:15.237" LastActivityDate="2021-04-30T11:30:15.237" Title="Finding the enthalpy of formation of magnesium oxide using Hess' Law" Tags="&lt;physical-chemistry&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10147" PostTypeId="1" CreationDate="2014-04-21T18:08:18.473" Score="2" ViewCount="481" Body="&lt;p&gt;First, I would like to do a some form of extraction to only get the mercury, then followed by a titration to identify the concentration or the presence of mercury.&lt;/p&gt;&#xA;" OwnerUserId="5263" LastEditorUserId="73" LastEditDate="2014-04-23T14:58:23.383" LastActivityDate="2015-03-27T00:31:01.130" Title="How can I carry out a titration to identify mercury in wastewater?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10148" PostTypeId="1" CreationDate="2014-04-21T18:19:33.930" Score="2" ViewCount="4121" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are (is) the product formed when 1,1,1-trichloropropane is treated with aqueous potassium hydroxide?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the major product should be propionic acid, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="3863" LastEditorUserId="5026" LastEditDate="2018-03-04T02:52:30.143" LastActivityDate="2018-03-04T02:52:30.143" Title="Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10149" PostTypeId="2" ParentId="10148" CreationDate="2014-04-21T18:44:39.153" Score="1" Body="&lt;p&gt;$\ce{OH^-}$ could displace $\ce{Cl3C^-}$ to give ethanol plus chloroform plus tetrachloroethylene (via cage effect on dichorocarbene).  &lt;BR&gt;&#xA;$\ce{OH^-}$ could eliminate HCl to give 1,1-dichloro-1-propene, and on to 1-chloropropyne, and on to.... &lt;BR&gt;&#xA;$\ce{OH^-}$ could displace $\ce{Cl^-}$ to give the acid chloride hemicarbonyl chloride, then on to propionic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvent, concentration, and temperature are important for biasing reaction paths.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-22T15:58:33.720" LastActivityDate="2014-04-22T15:58:33.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10150" PostTypeId="2" ParentId="10148" CreationDate="2014-04-21T20:40:44.660" Score="1" Body="&lt;p&gt;The techinfo sheet for 1,1,1-trichloroethane says that major product of hydrolysis is 2,2-dichloroethylene, so I'd say that for 1,1,1-trichloropropane it would be 1,1-dichloropropene-1.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-04-21T20:40:44.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10153" PostTypeId="2" ParentId="10097" CreationDate="2014-04-21T23:52:48.680" Score="4" Body="&lt;p&gt;The reaction is &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Le_Chatelier%27s_Principle/Effect_Of_Temperature_On_Equilibrium_Composition/Exothermic_Versus_Endothermic_And_K&quot; rel=&quot;nofollow&quot;&gt;exothermic&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A system that releases heat to the surroundings is an exothermic reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Furthermore, forming a chemical bond releases energy and therefore is an exothermic process, such as when Iron and Oxygen bond to form Iron(II) Oxide. Sometimes it's not clear whether more bonds were formed or broken during a chemical reaction so we have to add up the energies to see, unless they are already given to us like in this example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, always consider the equation in the direction the arrow is pointing. The standard is to point from left to right, but sometimes it can point both ways simultaneously. In this case, the question should state which direction of the chemical reaction it wants you to consider.&lt;/p&gt;&#xA;" OwnerUserId="5266" LastActivityDate="2014-04-21T23:52:48.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10154" PostTypeId="1" AcceptedAnswerId="10156" CreationDate="2014-04-22T01:36:06.407" Score="3" ViewCount="14081" Body="&lt;p&gt;I have just been waken up by an alarm and found out that my neighbour did something terrible with his lighter. Luckily, there was no fire, only a plenty of smoke, and our flats were hermetically isolated, so that smoke didn't go to my flat. But, most probably because of my suspiciousness, I noticed some smell of burnt organic, which usually accompanies CO. I switched on ventilation and went out - just in case. There were no CO detectors in the house - only smoke detectors - my bad.&#xA;Now the adventure is over (I hope so, at least) and I have got an (im)practical question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any ways to detect CO using only &quot;kitchen chemistry&quot;?&lt;/p&gt;&#xA;" OwnerUserId="5267" LastActivityDate="2017-12-01T18:05:35.610" Title="Carbon monoxide detection without a detector" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10156" PostTypeId="2" ParentId="10154" CreationDate="2014-04-22T02:32:29.040" Score="2" Body="&lt;p&gt;Surf a civilian aircraft supply.  They sell a little absorbant white tab (chemical spot detector) containing $\ce{PdCl2}$.  When exposed to CO it goes black.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aircraftspruce.com/menus/ps/carbonmonoxidedetectors.html&quot; rel=&quot;nofollow&quot;&gt;http://www.aircraftspruce.com/menus/ps/carbonmonoxidedetectors.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canary.  Goldfish won't work - they can survive on anaerobic respiration,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aquaticcommons.org/3663/1/14P089.pdf&quot; rel=&quot;nofollow&quot;&gt;http://aquaticcommons.org/3663/1/14P089.pdf&lt;/a&gt; &lt;BR&gt;&#xA;   who needs powdered alcohol?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T02:32:29.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10157" PostTypeId="1" CreationDate="2014-04-22T02:47:28.930" Score="0" ViewCount="1220" Body="&lt;p&gt;$$\begin{align}\ce{2A + 2B &amp;amp;-&amp;gt; 2C }\\ \ce{A + B &amp;amp;-&amp;gt; C}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the equilibrium constant change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And why the rate of the first reaction square the rate of the second? I understand it mathematically, but logically and scientifically I can't understand it at all.   &lt;/p&gt;&#xA;" OwnerUserId="5268" LastEditorUserId="4945" LastEditDate="2014-06-12T09:03:06.270" LastActivityDate="2014-09-10T10:40:00.550" Title="Multiplication of the reaction and Equilibrium constant" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10158" PostTypeId="1" AcceptedAnswerId="13966" CreationDate="2014-04-22T04:13:20.183" Score="2" ViewCount="587" Body="&lt;p&gt;people I've been trying to adjust the bond angles in a protein data bank (PDB) file, but I'm a novice when it comes to computing so I would like to ask your assistance. I have a file for &lt;a href=&quot;https://en.wikipedia.org/wiki/Metamizole&quot; rel=&quot;nofollow noreferrer&quot;&gt;metamizole&lt;/a&gt;, but as has been &lt;a href=&quot;https://commons.wikimedia.org/wiki/User_talk:Fuse809#Nitrogen_geometry_in_File:Metazimole3Dan.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;pointed out&lt;/a&gt; by a user on Wikipedia the 90 degree bond angle in it isn't consistent with the trigonal planar geometry of the nitrogen atom in the molecule. Here is my PDB file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;COMPND    UNNAMED&#xA;AUTHOR    GENERATED BY OPEN BABEL 2.3.2&#xA;HETATM    1  C   LIG     1      -0.971  -5.437  -1.069  1.00  0.00           C  &#xA;HETATM    2  N   LIG     1      -1.931  -4.484  -0.575  1.00  0.00           N  &#xA;HETATM    3  C   LIG     1      -3.342  -4.804  -0.629  1.00  0.00           C  &#xA;HETATM    4  S   LIG     1      -4.562  -3.540  -0.090  1.00  0.00           S  &#xA;HETATM    5  O   LIG     1      -4.430  -2.149  -1.074  1.00  0.00           O  &#xA;HETATM    6  O   LIG     1      -4.326  -3.106   1.433  1.00  0.00           O  &#xA;HETATM    7  O   LIG     1      -6.044  -4.127  -0.251  1.00  0.00           O  &#xA;HETATM    8  C   LIG     1      -1.479  -3.307  -0.058  1.00  0.00           C  &#xA;HETATM    9  C   LIG     1      -1.218  -3.011   1.277  1.00  0.00           C  &#xA;HETATM   10  C   LIG     1      -1.353  -3.951   2.413  1.00  0.00           C  &#xA;HETATM   11  N   LIG     1      -0.863  -1.703   1.337  1.00  0.00           N  &#xA;HETATM   12  C   LIG     1      -0.731  -0.968   2.567  1.00  0.00           C  &#xA;HETATM   13  N   LIG     1      -0.775  -1.218   0.118  1.00  0.00           N  &#xA;HETATM   14  C   LIG     1      -1.185  -2.152  -0.766  1.00  0.00           C  &#xA;HETATM   15  O   LIG     1      -1.262  -2.037  -2.012  1.00  0.00           O  &#xA;HETATM   16  C   LIG     1      -0.309   0.020  -0.235  1.00  0.00           C  &#xA;HETATM   17  C   LIG     1      -1.201   0.998  -0.733  1.00  0.00           C  &#xA;HETATM   18  C   LIG     1      -0.737   2.291  -1.040  1.00  0.00           C  &#xA;HETATM   19  C   LIG     1       0.619   2.617  -0.856  1.00  0.00           C  &#xA;HETATM   20  C   LIG     1       1.520   1.640  -0.390  1.00  0.00           C  &#xA;HETATM   21  C   LIG     1       1.059   0.346  -0.078  1.00  0.00           C  &#xA;HETATM   22  H   LIG     1      -1.483  -6.301  -1.438  1.00  0.00           H  &#xA;HETATM   23  H   LIG     1      -0.312  -5.724  -0.276  1.00  0.00           H  &#xA;HETATM   24  H   LIG     1      -0.403  -4.993  -1.861  1.00  0.00           H  &#xA;HETATM   25  H   LIG     1      -3.492  -5.667  -0.014  1.00  0.00           H  &#xA;HETATM   26  H   LIG     1      -3.539  -4.923  -1.674  1.00  0.00           H  &#xA;HETATM   27  H   LIG     1      -3.538  -1.778  -0.991  1.00  0.00           H  &#xA;HETATM   28  H   LIG     1      -1.646  -4.913   2.048  1.00  0.00           H  &#xA;HETATM   29  H   LIG     1      -2.094  -3.586   3.092  1.00  0.00           H  &#xA;HETATM   30  H   LIG     1      -0.414  -4.034   2.920  1.00  0.00           H  &#xA;HETATM   31  H   LIG     1      -0.853  -1.635   3.395  1.00  0.00           H  &#xA;HETATM   32  H   LIG     1      -1.481  -0.206   2.610  1.00  0.00           H  &#xA;HETATM   33  H   LIG     1       0.238  -0.518   2.611  1.00  0.00           H  &#xA;HETATM   34  H   LIG     1      -2.196   0.763  -0.872  1.00  0.00           H  &#xA;HETATM   35  H   LIG     1      -1.394   3.001  -1.399  1.00  0.00           H  &#xA;HETATM   36  H   LIG     1       0.954   3.571  -1.064  1.00  0.00           H  &#xA;HETATM   37  H   LIG     1       2.519   1.873  -0.276  1.00  0.00           H  &#xA;HETATM   38  H   LIG     1       1.721  -0.367   0.265  1.00  0.00           H  &#xA;CONECT    1    2   22   23   24                                       &#xA;CONECT    1                                                           &#xA;CONECT    2    1    3    8                                            &#xA;CONECT    3    2    4   25   26                                       &#xA;CONECT    3                                                           &#xA;CONECT    4    3    5    6    7                                       &#xA;CONECT    4                                                           &#xA;CONECT    5    4   27                                                 &#xA;CONECT    6    4                                                      &#xA;CONECT    7    4                                                      &#xA;CONECT    8    2    9   14                                            &#xA;CONECT    9    8   10   11                                            &#xA;CONECT   10    9   28   29   30                                       &#xA;CONECT   10                                                           &#xA;CONECT   11    9   12   13                                            &#xA;CONECT   12   11   31   32   33                                       &#xA;CONECT   12                                                           &#xA;CONECT   13   11   14   16                                            &#xA;CONECT   14   13    8   15                                            &#xA;CONECT   15   14                                                      &#xA;CONECT   16   13   17   21                                            &#xA;CONECT   17   16   18   34                                            &#xA;CONECT   18   17   19   35                                            &#xA;CONECT   19   18   20   36                                            &#xA;CONECT   20   19   21   37                                            &#xA;CONECT   21   20   16   38                                            &#xA;CONECT   22    1                                                      &#xA;CONECT   23    1                                                      &#xA;CONECT   24    1                                                      &#xA;CONECT   25    3                                                      &#xA;CONECT   26    3                                                      &#xA;CONECT   27    5                                                      &#xA;CONECT   28   10                                                      &#xA;CONECT   29   10                                                      &#xA;CONECT   30   10                                                      &#xA;CONECT   31   12                                                      &#xA;CONECT   32   12                                                      &#xA;CONECT   33   12                                                      &#xA;CONECT   34   17                                                      &#xA;CONECT   35   18                                                      &#xA;CONECT   36   19                                                      &#xA;CONECT   37   20                                                      &#xA;CONECT   38   21                                                      &#xA;MASTER        0    0    0    0    0    0    0    0   38    0   38    0&#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you can also give me a general method of correcting this problem it will be much appreciated. Here is the gif file I get out of this PDB file from Jmol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wJCws.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I got from optimisation. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/5SwKi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2014-04-22T21:26:22.817" LastActivityDate="2014-07-04T11:05:50.757" Title="Bond angles in PDB" Tags="&lt;bond&gt;&lt;software&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10159" PostTypeId="1" AcceptedAnswerId="10161" CreationDate="2014-04-22T05:06:21.773" Score="9" ViewCount="2120" Body="&lt;p&gt;In the news recently is '&lt;a href=&quot;http://www.palcohol.com/home.html&quot; rel=&quot;nofollow&quot;&gt;Palcohol&lt;/a&gt;', powdered alcohol, that you add to water or existing liquids to make it alcohol. Presumably this would dissolve and break apart into alcohol and {some other compound}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could bond to ethyl alcohol in order to:&#xA;1) Create a solid, and&#xA;2) Not be disgusting as a side product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most things have some associated taste, $\ce{NaCl}$ is certainly out. If the 'other' components somehow combined to $\ce{H2O}$ or a gas like $\ce{O2}$ or $\ce{CO_2}$ that might remove any residual taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be in the list of available combinations with ethyl alcohol that are safe to consume, gaseous, or don't otherwise add weird tastes to the liquid?&lt;/p&gt;&#xA;" OwnerUserId="91" LastEditorUserId="15410" LastEditDate="2015-10-12T03:55:01.357" LastActivityDate="2015-10-12T03:55:01.357" Title="How could alcohol be made into a powder?" Tags="&lt;aqueous-solution&gt;&lt;solvents&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10160" PostTypeId="1" AcceptedAnswerId="10164" CreationDate="2014-04-22T05:15:25.823" Score="6" ViewCount="277" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Fugacity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Article on Fugacity&lt;/a&gt; contains statement &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The contribution of nonideality to the chemical potential of a real gas is equal to RT ln $\phi$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here $\phi$ is fugacity coefficient.I think it is wrong it should be fugacity coefficient multiplied by partial pressure of gas divided by standard pressure.So for the example of nitrogen gas at 100 atm it's chemical potential should be μ = μ$_{id}$ + RT ln 97.03 and not μ = μ$_{id}$ + RT ln 0.9703 as given in Wikipedia Article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused, who is wrong me or wikipedia ?&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="418" LastEditDate="2014-04-22T14:27:51.700" LastActivityDate="2014-04-22T14:27:51.700" Title="Mistake in Wikipedia article on fugacity?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10161" PostTypeId="2" ParentId="10159" CreationDate="2014-04-22T05:38:04.087" Score="10" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclodextrin&quot;&gt;Cyclodextrins&lt;/a&gt;, which comprise a class of cyclic carbohydrates, use their properties of being hydrophobic within their ring(s), yet hydrophilic outside. Alcohol can be put inside these rings. Then, the entire molecule is still capable of dissolving in water, releasing the encapsulated alcohol molecules. &lt;a href=&quot;http://animalnewyork.com/2014/powdered-alcohol-better-drinking-chemistry/&quot;&gt;(src)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.popsci.com/article/technology/how-make-powdered-booze-home&quot;&gt;popsci&lt;/a&gt; link has one way of making edible powdered alcohol. A nice casual step-by-step tutorial, actually...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still you want to maintain a balance between maximizing alcohol (or do you?) and having the powder &quot;feeling wet&quot; due to high absorption of alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;https://en.wikipedia.org/wiki/Alcohol_powder&quot;&gt;this&lt;/a&gt; may be a relevant Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-04-22T05:38:04.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10162" PostTypeId="1" AcceptedAnswerId="10179" CreationDate="2014-04-22T06:35:47.273" Score="1" ViewCount="81" Body="&lt;p&gt;What is happening &lt;a href=&quot;https://www.youtube.com/watch?v=kODW0qHgThU&quot; rel=&quot;nofollow&quot;&gt;in this video&lt;/a&gt;? Can you explain this technically as well as in layman's terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll summarize it. I see a technician moving fuming mixtures of presumably uranium and organic compounds on and off hot plates. Perchloric acid is mentioned, and a large bottle is present.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="2094" LastEditDate="2014-04-24T08:12:04.390" LastActivityDate="2014-04-24T08:12:04.390" Title="Explain isotopic uranium/perchloric acid digestion" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-07-29T12:01:56.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="10163" PostTypeId="1" CreationDate="2014-04-22T07:52:52.033" Score="2" ViewCount="6601" Body="&lt;p&gt;I'm working on solid propellant based microthrusters and I am using NCNG as the solid propellant. I want to know the reaction taking place and the mass fraction of the products obtained on combustion of NCNG propellant. &lt;/p&gt;&#xA;" OwnerUserId="5270" LastEditorUserId="189" LastEditDate="2014-04-22T09:32:06.297" LastActivityDate="2014-05-22T18:08:12.500" Title="Reaction involved in Combustion of Nitrocellulose-Nitroglycerine(NCNG) solid propellant?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10164" PostTypeId="2" ParentId="10160" CreationDate="2014-04-22T09:27:31.713" Score="8" Body="&lt;p&gt;I think you have misinterpreted the passage on Wikipedia a little. Of course, you are right that the chemical potential is given by the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \mu_{i} (p_{i}, T) &amp;amp;= \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \log a_{i} \\&#xA; &amp;amp;= \mu_{i}^{0} + RT \ln \left(\phi_{i} \frac{p_{i}}{p^0} \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the universal gas constant, $T$ is the temperature, and $\phi_{i}$, $p_{i}$, $a_{i}$, and $\mu_{i}$ are the fugacity coefficient, the partial pressure, the activity and the chemical potential of the $i$th component in the system, respectively.&#xA;But the passage you are quoting says that &lt;em&gt;the contribution of nonideality to the chemical potential of a real gas is equal to $RT \ln \phi$&lt;/em&gt; and in the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \mu = \mu_{\text{id}} + RT \ln \left(\phi \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{\text{id}}$ is not the same as $\mu^{0}$. Rather it is meant in the following way: For an ideal gas the fugacity coefficient is $1$ by definition and so the chemical potential of an ideal gas, $\mu_{\text{id}}$, is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \mu_{\text{id}} = \mu^{\text{id}}_{i} (p_{i}, T) = \mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use the mathematical property of logarithms that $\log(a \cdot b) = \log(a) + \log(b)$ you can rewrite the equation for the chemical potential of real gases to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \mu_{i} (p_{i}, T) &amp;amp;= \underbrace{\mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)}_{= \, \mu_{\text{id}}} + RT \ln \left(\phi_{i} \right) \\&#xA; &amp;amp;= \mu_{\text{id}} + RT \ln \left(\phi_{i} \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so you see, that the contribution of nonideality to the chemical potential of a real gas is indeed equal to $RT \ln \phi$ and you get to the equation from the Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-04-22T09:36:16.987" LastActivityDate="2014-04-22T09:36:16.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10165" PostTypeId="2" ParentId="10163" CreationDate="2014-04-22T10:46:06.543" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrocellulose&quot; rel=&quot;nofollow&quot;&gt;Nitrocellulose&lt;/a&gt; is complex stuff. It is generally considered to be the fully nitrated version of cellulose, in which every available alcohol functional group is converted to a nitrate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[C6H10O5]n +3HNO3 -&amp;gt; (C6H7O2(ONO2)3]n +3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principal it probably contains some small amount of partially and/or non-nitrated units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider nitrocellulose to be fully nitrated, then the combustion reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ 2C6H7N3O11 + 9/2O2 -&amp;gt; 12CO2 + 3N2 + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitroglycerin&quot; rel=&quot;nofollow&quot;&gt;nitroglycerin&lt;/a&gt; $(\ce{C3H5N3O9})$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C3H5N3O9  -&amp;gt; 6CO2 + 3N2 + 5H2O + 1/2O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will depend on the ratio of nitrocellulose to nitroglycerin what the overall $\ce{CO2}:&#xA;\ce{N2}:\ce{H2O}$ mole ratio is. I'll leave you to figure out the mass ratios using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_mass&quot; rel=&quot;nofollow&quot;&gt;molecular masses&lt;/a&gt; of the products.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-22T10:46:06.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10166" PostTypeId="1" AcceptedAnswerId="10177" CreationDate="2014-04-22T11:20:13.203" Score="2" ViewCount="368" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CrCl3(bpy)3}$&lt;/span&gt; (A) when heated forms (B) which has the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{CrCl3(bpy)2}$&lt;/span&gt; and has one free &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; per mole of (B). B can react with ethylene diamine (&lt;span class=&quot;math-container&quot;&gt;$\ce{en}$&lt;/span&gt;) to form (C) which has the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{CrCl3(bpy)2(en)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(B) can also be converted into (D) which also has the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{CrCl3(bpy)2}$&lt;/span&gt; but cannot react with &lt;span class=&quot;math-container&quot;&gt;$\ce{en}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Identify compounds A–D and draw their structures&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that B is &lt;em&gt;cis&lt;/em&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{[CrCl2(bpy)2]Cl}$&lt;/span&gt;&lt;br /&gt;&#xA;C is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cr(bpy)2(en)]Cl3}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\ce{en\bond{-&amp;gt;}metal}$&lt;/span&gt; bonds are where the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; ions in B are.&lt;br /&gt;&#xA;D is the &lt;em&gt;trans&lt;/em&gt; isomer of B.&lt;/p&gt;&#xA;&lt;p&gt;What I dont understand is how you're meant to know what A is. I assume it's &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cr(bpy)3]Cl3}$&lt;/span&gt; as a later part of the question asks about relative molar conductivity of A and B.&lt;/p&gt;&#xA;&lt;p&gt;Why would heating that cause a &lt;span class=&quot;math-container&quot;&gt;$\ce{bpy}$&lt;/span&gt; to leave the complex and two chlorides to replace it? Or is it &lt;span class=&quot;math-container&quot;&gt;$\ce{[CrCl2(bpy)2]Cl(bpy)}$&lt;/span&gt; in which case, why does heating it not form the more stable complex I've stated above?&lt;/p&gt;&#xA;" OwnerUserId="5272" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-11-06T01:30:07.693" Title="Heating CrCl3(bpy)3 — isomerism and ligand exchange" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10167" PostTypeId="1" AcceptedAnswerId="10176" CreationDate="2014-04-22T11:47:01.207" Score="0" ViewCount="194" Body="&lt;p&gt;I have designed a container which is heated up and has to heat up some tube-coils of water inside of it. This container is filled with a fluid that can help store and transfer the heat received from the container to the tube-coils.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan was to find some material that I can easily buy and which is not very expensive to fill the $636cm^3$ container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to have a high coefficient of heat transfer, high heat capacity and an operating temperature of at least the range of 10-400 degrees Celsius. I will also need to calculate how much the volume changes when it goes from around -35 to 500 degrees Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think a high density and if the material is non-toxic might be a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know any fluid that could be used for this or somewhere where I could find such information?&lt;/p&gt;&#xA;" OwnerUserId="5156" LastActivityDate="2016-05-09T01:36:58.993" Title="Generally available fluid with good heat transport properties" Tags="&lt;thermodynamics&gt;&lt;materials&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10171" PostTypeId="2" ParentId="10157" CreationDate="2014-04-22T13:02:51.420" Score="2" Body="&lt;p&gt;Rate will not be square in most of the cases and who said so? The equilibrium constant scientifically changes because Gibb's free energy is extensive quantity and it is related to equilibrium constant through relation that you might know $\Delta G=\Delta G^\circ +\mathcal{R}T\ln Q $. Where $Q$ is reaction quotient which at equilibrium becomes equilibrium constant and then it gives $\Delta G^\circ = -\mathcal{R}T\ln K$&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="4945" LastEditDate="2014-06-12T09:04:36.127" LastActivityDate="2014-06-12T09:04:36.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10172" PostTypeId="2" ParentId="10148" CreationDate="2014-04-22T13:40:32.920" Score="2" Body="&lt;p&gt;Figure: 1,1,1-trichloropropane     &lt;img src=&quot;https://i.stack.imgur.com/nzADU.jpg&quot; alt=&quot;1,1,1D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol $ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\color{green}{\checkmark}}$ It may be noted that generally &lt;em&gt;alcoholic&lt;/em&gt; $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while &lt;em&gt;aqueous&lt;/em&gt; $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, &lt;em&gt;aqueous&lt;/em&gt; $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above thought process, the most preferable reaction would be:&lt;br&gt;&#xA;$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) -&amp;gt;[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl} $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(OH)_3C-CH_2-CH_3 -&amp;gt; (OH)CO-CH_2-CH_3 + H_2O}$$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right, propionic acid along with water molecule is going to be the product. &lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="24052" LastEditDate="2016-03-11T22:25:16.937" LastActivityDate="2016-03-11T22:25:16.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10173" PostTypeId="1" AcceptedAnswerId="10174" CreationDate="2014-04-22T13:56:22.887" Score="1" ViewCount="924" Body="&lt;p&gt;I am a bit confused about the correct name of $\mathrm{LaMnO}_3$. Is it lanthanium manganate or lanthanum manganite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was assuming that since $\mathrm{SrTiO}_3$ is called strontium titanate, $\mathrm{LaMnO}_3$ would then be called lanthanium manganate, but I got confused when I heard a collegue referring to it as manganite and after reading a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanthanum_strontium_manganite&quot; rel=&quot;nofollow&quot;&gt;wikiepdia article&lt;/a&gt; on $\mathrm{La}_x\mathrm{Sr}_{1-x}\mathrm{MnO}_3$ where they also used the term manganite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct form, or are they simply interchangeable?&lt;/p&gt;&#xA;" OwnerUserId="5274" LastActivityDate="2014-04-22T14:32:44.880" Title="About the nomenclature: manganate or manganite?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10174" PostTypeId="2" ParentId="10173" CreationDate="2014-04-22T14:32:44.880" Score="0" Body="&lt;p&gt;Let's go down the ladder.  $\ce{KMnO4}$ is permanganate, Mn(VII).  $\ce{K2MnO4}$ is manganate, Mn(VI).  $\ce{K4MnO4}$ is manganite, Mn(IV).  $\ce{LaMnO3}$ is called both manganate and manganite, but Mn(III). In a proper world it should be lower than that manganite.  They are pedagogically both wrong if it is La(III).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dictionary.sensagent.com/potassium%20manganite/en-en/&quot; rel=&quot;nofollow&quot;&gt;http://dictionary.sensagent.com/potassium%20manganite/en-en/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to Cl(VII) and downward is not valid.  Drawing nomenclature parallels to titanium is poor, for Ti(IV) is the highest common oxidation state.  $\ce{SrO + TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is Ti(IV) on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic had the same problem. The highest valence carbocation was tricoordinate, hence a carbonium ion.  Magic Acid gave $\ce{CH5^+}$.  Carbonium ions were renamed carbenium ions to general confusion.  Then, $\ce{CH6^{2+}}$.  Organic dropped the ball with direction of optical rotation versus geometric left and right.  It was flat out stupid designating &lt;em&gt;E&lt;/em&gt; (&quot;&lt;em&gt;trans&lt;/em&gt;&quot;) and &lt;em&gt;Z&lt;/em&gt; (&quot;&lt;em&gt;cis&lt;/em&gt;&quot;) olefins.  The designation should have been abstract rather than language-originated.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T14:32:44.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10176" PostTypeId="2" ParentId="10167" CreationDate="2014-04-22T14:56:52.090" Score="3" Body="&lt;p&gt;400 C is beyond long term use of any organic heat transfer fluid.  Dow Corning boasts 427 C for one of its silicone fluids.  Don't bet on that.  Inorganic eutectic salts will long term decompose into corrosive volatiles.  Best bet for a red hot closed system is fusible alloys of indium, gallium, and bismuth - beware of metal boiling points.  Open systems will air oxidize.   Accommodate sealed inert gas fill (argon) with a metal bellows add-on, preferably on a pipe extension away from the hot zone.  Fill composition must be compatible with container composition.  Mercury is out for toxicity and boiling point, 357 C.  Thermal coefficient of linear expansion, solid vs. liquid alloy, must be looked up.  The increment is cubed for volume change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;500 C (red heat) will change the metallurgy of stainless steels over time.  They will harden (carbide precipitation), among other things.  Don't try to drill a hole or tap a thread in stainless that has been soaked at red heat.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T14:56:52.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10177" PostTypeId="2" ParentId="10166" CreationDate="2014-04-22T15:03:47.943" Score="1" Body="&lt;p&gt;It is Cr(III) not Co(III).  Cr(III) has reasonable substitution kinetics. Your structure for A is good. Heating drives off a mole of volatile bipy, then two chlorides inner sphere coordinate to take its place.  The third chloride remains outer sphere and readily exchangeable. Bidentate en comes in to displace two inner sphere chlorides outer sphere again.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T15:03:47.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10179" PostTypeId="2" ParentId="10162" CreationDate="2014-04-22T15:23:45.370" Score="5" Body="&lt;p&gt;If that is a perchlorate digestion, the technician needs more training.  The hood door must be lower despite the Official marker.  His cotton lab coat must NOT be exposed to vapors!  Sleeves should be &lt;em&gt;inside&lt;/em&gt; long gloves.  He has a bottle of perchloric acid next to a running hot plate!  I'd have a 5-gallon bucket of water nearby if the gloves ignite.  As perchlorate digestions occasionally explode, and he's doing a lot of them, I'd wear goggles under the face shield and wear Dyneema or Spectra body armor under the lab coat.  One might have some concern for hands (longer pair of tongs) and neck, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's apparently cooking off a sample of organics-contaminated uranium whatever for trace contamination or isotopic analysis.  Everything oxidizable will be gone, leaving uranium oxides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is NOT marked as a perchlorate fume hood!  Everything organic downwind of those fumes is a potential high explosive, and certainly a violently flammable material including its own oxidizer.  There will be massive corrosion of all metals.  You don't want to be standing, living, downwind of the exhaust stack.  OTOH, the operation is low cost/sample, and anybody off the street is a potential lab tech.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T15:23:45.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10180" PostTypeId="2" ParentId="10163" CreationDate="2014-04-22T17:26:19.107" Score="0" Body="&lt;p&gt;Energetic materials can emit a significant fraction of their carbon as CO.  Take advantage of excess local oxygen with more fuel, or push to CO2 with more oxidizer.  The lower the average molecular weight of the exhaust, the higher the Isp, all other things being equal.  US hydrogen/LOX engines run heavy on hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;re Cordite and Ballistite, grain size, and internal channel geometry if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mechanism of the Burning of Double-Base Propellants &lt;BR&gt;&#xA;DOI: 10.1021/j150480a015 &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.google.com/patents/US4072546&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/US4072546&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA425264&quot; rel=&quot;nofollow&quot;&gt;http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA425264&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Solid Propellant Chemistry, Combustion, and Motor Interior Ballistics&lt;/em&gt;, Volume 185. Vigor Yang, Thomas B. Brill, Wu-Zhen Ren, Paul Zarchanm, 2000,. p. 288 ff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double-base propellants (DB) give minimal smoke with medium-high performance, Isp ~ 235 s. Adding aluminum gives Isp ~ 250 s with visible smoke.  DB have 50-60% nitrocellulose and 30-40% nitroglycerin content.  High nitroglycerin for extrusion and casting, low for pressing.  DBs are boosted by adding dispersed micronized ammonium perchlorate.  Both nitrocellulose and nitroglycerin decompose to afford nitric acid.  This must be neutralized during long term storage, certainly if perchlorate is in there.  Beware of detonation under pressure, hence phlegmatizing agents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache%3amAvWdNxk28sJ%3awww.researchgate.net/profile/Jasmin_Terzic/publication/230776722_Influence_of_energetic_characteristics_of_double-base_propellants_on_internal-ballistics_parameters/file/79e415044a80253b9e.pdf%20&amp;amp;cd=5&amp;amp;hl=en&amp;amp;ct=clnk&amp;amp;gl=us&quot; rel=&quot;nofollow&quot;&gt;double-base propellant energetics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-22T17:26:19.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10181" PostTypeId="2" ParentId="10084" CreationDate="2014-04-22T19:33:34.003" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that chemical potential is defined as&#xA;  ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it.&#xA;The derivation of this equation can be found in &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/189&quot;&gt;this answer&lt;/a&gt; of mine.&#xA;The definition of the chemical potential is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that Gibbs free energy of a system at constant temperature and&#xA;  pressure also represents Non-PV work done by system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again this is not quite correct: It is the &lt;strong&gt;change in Gibbs free energy&lt;/strong&gt; of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now considering a general chemical reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C &amp;lt;=&amp;gt; \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the differential change in Gibbs free energy is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or introducing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extent_of_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;extend of reaction&lt;/a&gt;, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Extent_of_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;extend of reaction&lt;/a&gt; $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants.&#xA;Substituting the equation for the chemical potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \mu_{i} = \mu_{i}^{0} + RT \ln a_{i} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced.&#xA;This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-04-22T20:15:25.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10182" PostTypeId="2" ParentId="10146" CreationDate="2014-04-22T19:42:08.697" Score="2" Body="&lt;p&gt;You are trying to find out the enthalpy of formation.&#xA;The enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their standard state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation matches this, so that should be your 'main' or 'target' equation. From there you should be able to use Hess' law to calculate the enthalpy change of the formation of magnesium oxide.&lt;/p&gt;&#xA;" OwnerUserId="5237" LastActivityDate="2014-04-22T19:42:08.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10183" PostTypeId="2" ParentId="10158" CreationDate="2014-04-22T21:09:07.753" Score="2" Body="&lt;p&gt;Your interpretation of the remark on wikipedia is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that you provided a pdb file of the structure as drawn and &lt;strong&gt;without&lt;/strong&gt; any geometry optimization. As a result, N2, N3, and N13 are trigonal planar in your file, while there should in fact be pyramidalisation (there are lone pairs!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, &quot;the guy&quot; isn't writing about bond angles, but about the dihedral angle formed by C1-N2-C8-C9, which is indeed close to 90° in your structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criticism is justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your should run a geometry optimization on the structure before you submit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do that, open your pdb file in Avogadro, select MMFF94 as your force field and run a geometry optimization pressing &lt;kbd&gt;Alt&lt;/kbd&gt;&lt;kbd&gt;Ctrl&lt;/kbd&gt;&lt;kbd&gt;O&lt;/kbd&gt;. If necessary, fiddle with the dihedral in question and repeat the optimization.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-04-22T21:18:45.383" LastActivityDate="2014-04-22T21:18:45.383" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10184" PostTypeId="1" CreationDate="2014-04-19T10:57:21.787" Score="2" ViewCount="75" Body="&lt;p&gt;I know it's the proteins that somehow change their properties at some temparature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what kind of change, in terms of molecule-level structure?&lt;/p&gt;&#xA;" OwnerUserId="5078" OwnerDisplayName="Volker Siegel" LastActivityDate="2014-04-22T22:07:02.060" Title="How does the process of protein denaturatation work?" Tags="&lt;temperature&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10185" PostTypeId="2" ParentId="10184" CreationDate="2014-04-19T13:06:42.517" Score="2" Body="&lt;p&gt;Basically, the egg-white (just another protein structure) is made of long chains of amino acids. They're held in shape by weak bonds. When you heat the structure, the energy you put in, is enough to break those bonds, thereby destroy the structure. This process &lt;a href=&quot;https://en.wikipedia.org/wiki/Denaturation_%28biochemistry%29#Common_examples&quot; rel=&quot;nofollow&quot;&gt;has a name for it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the temperature is high enough, new covalent bonds can form between the amino acids, and the amino acids in some other protein and they become entangled into a much more complicated structure, which is completely different from the real one. As you cool, you get this finished structure as cooked egg...&lt;/p&gt;&#xA;" OwnerUserId="1414" OwnerDisplayName="Waffle's Crazy Peanut" LastActivityDate="2014-04-19T13:06:42.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10186" PostTypeId="1" CreationDate="2014-04-23T00:46:14.730" Score="6" ViewCount="315" Body="&lt;p&gt;First, I’m a high school student and I wonder why every textbook just tells you how to set up those equations and cancel them to get the entropy change but never tells you why you can cancel the same species on both sides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, for example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2B + 3H2O -&amp;gt; B2H6 + 3/2O2} \tag{  $\Delta H= 762~\mathrm{kJ}$}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2 + 1/2O2 -&amp;gt; H2O} \tag{  $\Delta H=-286~\mathrm{kJ}$}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the textbook tells us we should add up two reactions in order to get what we want (I didn’t post the whole question, just go with this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, why do we have to add the two whole reactions up instead of just change the $\ce{H2O}$ on the left side into $\ce{H2 + 1/2O2}$?  What I understand about this $\ce{H2 + 1/2 O2 -&amp;gt;H2O}$ is if we reverse it, we can say if $\ce{H2O}$ takes in $286~\mathrm{kJ}$ of energy, it will turn into $\ce{H2}$ and $\ce{1/2 O2}$. So why we can’t just add $286~\mathrm{kJ}$ to the equation? Why we still need to add the whole thing $\ce{H2 + 1/2 O2 -&amp;gt; H2O}$ to the other equation?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="5026" LastEditDate="2018-06-25T03:04:38.273" LastActivityDate="2018-06-25T03:04:58.423" Title="Calculation of the enthalpy change of reactions" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10187" PostTypeId="1" CreationDate="2014-04-23T01:44:56.103" Score="4" ViewCount="25948" Body="&lt;p&gt;At constant pressure $p=\text{const.}$, change of internal energy equals heat subtract work(PV). So, $q=\Delta E - pV$. So is the the change of enthalpy. So we say at constant pressure, the change of enthalpy equals the change of heat. But when the volume is constant $V = \text{const.}$, no work is done. $\Delta E= q$, $\Delta H = \Delta q$. So why not in this situation too?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="4945" LastEditDate="2014-04-23T05:03:29.070" LastActivityDate="2014-04-23T05:33:55.443" Title="at constant volume why the change of enthalpy doesn't equal the change of heat" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10188" PostTypeId="2" ParentId="10187" CreationDate="2014-04-23T03:54:18.403" Score="9" Body="&lt;p&gt;At constant volume, the pressure is not necessarily constant. The pressure will usually increase when the system is heated. This makes the $PV$ term not constant. For this reason, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \Delta (E + PV) $$&#xA;$$\Delta H = \Delta E + \Delta (PV) $$&#xA;$$\Delta H = \Delta E + V\Delta P $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now heat is just $\Delta E$, because there is no work. So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = q+ V \Delta P $$&lt;/p&gt;&#xA;" OwnerUserId="5277" LastEditorUserId="1362" LastEditDate="2014-04-23T05:33:55.443" LastActivityDate="2014-04-23T05:33:55.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10190" PostTypeId="1" AcceptedAnswerId="10202" CreationDate="2014-04-23T07:14:46.143" Score="1" ViewCount="28206" Body="&lt;p&gt;I know that it is possible to predict whether combinations of acids and bases will be acidic, basic, or neutral:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;weak acid (WA) and strong base (SB) reacts basic  &lt;/li&gt;&#xA;&lt;li&gt;strong acid (SA) and and weak base (WB) reacts acidic&lt;/li&gt;&#xA;&lt;li&gt;SA and SB reacts neutral&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How does a WA react with a WB? I've been told that it isn't neutral. How do you predict what it is and why is it not neutral?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="4945" LastEditDate="2014-04-24T02:17:15.623" LastActivityDate="2014-04-24T02:59:05.777" Title="Predicting the pH of a weak acid and weak base solution" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10191" PostTypeId="2" ParentId="7637" CreationDate="2014-04-23T11:01:46.827" Score="2" Body="&lt;p&gt;While the mechanism is not 100% certain (or wasn't when I worked on it some time ago), the current best guess is that it is similar to the well known &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzoin_condensation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Benzoin condensation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a2Pli.png&quot; alt=&quot;benzoin condensation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the following intermediate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UjpmU.png&quot; alt=&quot;benzoin first step&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied to the paraquat conditions this might go like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fuz8V.png&quot; alt=&quot;paraquat reaction mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutral quinone-like intermediate crystallises out of solution and can be separated and then oxidised (it is very oxygen sensitive) to the desired paraquat product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are partially guesswork, though, as it isn't the easiest reaction to investigate mechanistically.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-04-23T11:01:46.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10192" PostTypeId="1" CreationDate="2014-04-23T13:02:57.600" Score="6" ViewCount="146" Body="&lt;p&gt;How can I reduce Crystallisation temperature of seed oil to be used in winter as biodiesel?&#xA;The glycerol has been removed from the oil, and white spirit is used to reduce its viscosity.&lt;/p&gt;&#xA;" OwnerUserId="5280" LastEditorUserId="5456" LastEditDate="2014-05-24T13:40:39.623" LastActivityDate="2017-03-23T21:34:24.887" Title="How can I reduce Crystallization temperature of seed oil to be used in winter as biodiesel?" Tags="&lt;temperature&gt;&lt;fuel&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10193" PostTypeId="2" ParentId="10192" CreationDate="2014-04-23T14:21:19.197" Score="1" Body="&lt;p&gt;Waxing and pour point are typical specs for fuel. How cold is your winter? If it is very cold, and your tank is above ground, there is little to be done. Insulate the tank (metal-backed rock wool) and heat it.  Use a more unsaturated feedstock with no saturated components. That lowers its pour point and removes waxing.  Did you dry your fuel?  Unsaturated fuels oxidize and deposit varnish.  &quot;Biofuel&quot; is a good substrate for microbial growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enviro-whinerism, sustainability, social consciousness, are Luddite lies.  They assume thermodynamics is amenable to legislation and good feelings, and weather is always mild and cooperative.  None of that obtains in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pour_point&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pour_point&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.infineum.com/Pages/pourpoint.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.infineum.com/Pages/pourpoint.aspx&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.aftonchemical.com/Solutions/LubricantComponents/Pages/PourPointDepressants.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.aftonchemical.com/Solutions/LubricantComponents/Pages/PourPointDepressants.aspx&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.globalp.com/products/product.cfm?productID=32&quot; rel=&quot;nofollow&quot;&gt;http://www.globalp.com/products/product.cfm?productID=32&lt;/a&gt; &lt;BR&gt;&#xA;Chemicals&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-23T14:21:19.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10194" PostTypeId="1" CreationDate="2014-04-23T18:03:57.827" Score="3" ViewCount="3269" Body="&lt;p&gt;Which fragmentation is possible at this m/z value for phthalic acid diethyl ester? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried McLafferty but it didn't solve my problem. &lt;/p&gt;&#xA;" OwnerUserId="5255" LastEditorUserId="4945" LastEditDate="2014-10-22T15:02:45.013" LastActivityDate="2017-10-06T13:03:57.150" Title="Where does the characteristic peak for phtalic acid diethyl ester at m/z 149 originate?" Tags="&lt;organic-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10195" PostTypeId="2" ParentId="10194" CreationDate="2014-04-23T18:12:39.827" Score="0" Body="&lt;p&gt;Addition of a proton, then loss of $\ce{CO2Et}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-23T18:12:39.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10196" PostTypeId="2" ParentId="10194" CreationDate="2014-04-23T18:17:06.300" Score="3" Body="&lt;p&gt;The MW of diethyl phthalate is 222.  The peak at 149 corresponds to loss of a MW=73 fragment.  This would correspond to loss of one of the ester groups ($\ce{CO_2Et}$).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-23T18:17:06.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10197" PostTypeId="1" CreationDate="2014-04-23T21:32:37.950" Score="-2" ViewCount="6940" Body="&lt;p&gt;Please explain to me why $\ce{KSO4^-}$ doesn't exist as a molecular ion in (aqueous) solution. I couldn't find a reason why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed if you took one $\ce{SO4^{2-}}$ and one $\ce{K^+}$ ion, then only that one would form the molecular ion $\ce{KSO4^-}$.&lt;/p&gt;&#xA;" OwnerUserId="5284" LastEditorUserId="16683" LastEditDate="2018-08-27T17:26:38.010" LastActivityDate="2018-08-27T18:23:00.627" Title="Why can't a potassium sulfate anion KSO4(-) exist in solution?" Tags="&lt;solutions&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10198" PostTypeId="1" AcceptedAnswerId="10201" CreationDate="2014-04-23T22:00:12.167" Score="1" ViewCount="154" Body="&lt;p&gt;The salt that you can typically find in a kitchen is the $\ce{NaCl}$, isn't this salt supposed to act as an antiseptic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find a lot of recipes that suggest the addition of salt when preparing the mix for the leaven of the bread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the salt supposed to act against the formation and the culture of bacteria that are carrying forward the leaven?&lt;/p&gt;&#xA;" OwnerUserId="5036" LastEditorUserId="73" LastEditDate="2014-04-23T22:27:30.117" LastActivityDate="2014-04-24T02:16:37.113" Title="Why is it a good idea to put some salt in the bread dough?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10200" PostTypeId="2" ParentId="10197" CreationDate="2014-04-24T00:12:11.810" Score="2" Body="&lt;p&gt;In solution, $\ce{K2SO4}$ is going to fully dissociate into $\ce{2K^+}$ and $\ce{SO_4^2-}$, see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dissociation_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on dissociation&lt;/a&gt;.&lt;br&gt;&#xA;In solution, the ions will orient towards each other so as to functionally neutralize the charges, but you are still going to have two $\ce{K+}$ ions orienting towards each $\ce{SO4^2-}$ ion, unless there is something else in solution that is &quot;distracting&quot; one of the $\ce{K+}$ ions.&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4945" LastEditDate="2018-08-27T16:53:38.720" LastActivityDate="2018-08-27T16:53:38.720" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10201" PostTypeId="2" ParentId="10198" CreationDate="2014-04-24T02:16:37.113" Score="3" Body="&lt;p&gt;Common table salt is not necessarily an antiseptic in the conventional &quot;sterilizing&quot; sense. What gives it weakly antiseptic properties in biological context is its ability to absorb water from cells through osmosis, should the concentration of the salt outside the cell be too high, the cell would lose too much water through its membrane and die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is strongly dependent on concentration. For example, according to &lt;a href=&quot;http://chemistry.about.com/od/foodcookingchemistry/f/Why-Does-Salt-Work-As-A-Preservative.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;, a concentration upwards of 20% is required to kill bacteria; at lower concentrations the salt will only hinder growth, which is most likely the reason it is added (&lt;a href=&quot;https://cooking.stackexchange.com/questions/15526/amount-of-salt-in-bread?rq=1&quot;&gt;in moderation&lt;/a&gt;) to bread.&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:30.103" LastActivityDate="2014-04-24T02:16:37.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10202" PostTypeId="2" ParentId="10190" CreationDate="2014-04-24T02:59:05.777" Score="3" Body="&lt;p&gt;&lt;em&gt;Note: The expected $\text{pH}$ values refer to 1:1 mixtures as suggested in the question&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One major feature of weak acids and bases is that they do not dissociate completely in water, producing less free oxonium or hydroxide ions than strong acids and bases do (Brønsted). For example in Sodiumacetate, a weak base, there is an equilibrium where the reactions left to right and right to left occur at the same speed, hence cancelling each other. &#xA;\begin{aligned}\ce{&#xA;   2NaOAc_{(aq)} + H2O &amp;amp;&amp;lt;=&amp;gt; 2Na+_{(aq)} + {}^{-}OAc_{(aq)} + HOAc_{(aq)} + {}^{-}OH_{(aq)}} &amp;amp;(1)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same also applies to weak bases like ammonium hydroxide $\ce{NH4OH}$. Note, that ammonium itself is not a Brønsted base as it lacks to donate hydroxide ions.&#xA;\begin{aligned}\ce{&#xA;   NH4OH_{(aq)} &amp;amp;&amp;lt;=&amp;gt; NH4+_{(aq)} + {}^{-}OH_{(aq)}} &amp;amp;(2)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak acids can be expanded in the same way, example &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzoic_acid&quot; rel=&quot;nofollow&quot;&gt;benzoic acid&lt;/a&gt;:&#xA;\begin{aligned}\ce{&#xA;   Ph-COOH_{(aq)} + 2H2O&amp;amp;&amp;lt;=&amp;gt; Ph-COO^{-}_{(aq)} + H+3O_{(aq)}} &amp;amp;(3)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Lewis definition, weak acids are not able to complex the hydroxide ions of water's autoprotolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;   2H2O &amp;amp;&amp;lt;=&amp;gt; H+3O_{(aq)} + {}^{-}OH_{(aq)} &amp;amp;\\&#xA;   BF3_{(aq)} + 3H2O &amp;amp;&amp;lt;=&amp;gt; [HOBF3]^{-}_{(aq)} + 2H+3O_{(aq)} + {}^{-}OH_{(aq)} &amp;amp;(4)&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations are also in equilibrium, meaning all species are present in aqueous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Lewis definition ammonia $\ce{NH3}$ is a base and it reacts analogous to the above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;   2NH3_{(aq)} + 2 H2O &amp;amp;&amp;lt;=&amp;gt; NH4OH_{(aq)} + NH4+_{(aq)} + {}^{-}OH_{(aq)}} &amp;amp;(5)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the rate of dissociation is strongly dependent on the weakest bond. In $(1)$ that is $\ce{Na-O}$ and successively $\ce{H-O}$. Overall the reaction will produce excess hydroxide ions, hence the solution will have a $\text{pH}&amp;gt;7~(@25~^\circ{}\!C)$.&#xA;The same applies analogously to $(2)$ and $(5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $(3)$ the opposite is the case. The reaction will produce excess oxonium ions, hence having a $\text{pH}&amp;lt;7~(@25~^\circ{}\!C)$. The same applies to $(4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when adding a weak base to a weak acid, one has to evaluate the whole equilibrium:&#xA;\begin{aligned}\ce{&#xA;   NH4OH_{(aq)} + HOAc_{(aq)} + H2O &#xA;   &amp;lt;=&amp;gt;&amp;amp;\\ [(NH4)(OAC)]_{(aq)} + NH4+_{(aq)} &amp;amp;+ {}^{-}OH_{(aq)} +  {}^{-}OAc_{(aq)} + H+3O_{(aq)}&#xA;} &amp;amp;(6)&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is therefore very much dependent on the chemical properties of all compounds involved in the solution. These systems are also very well known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution&quot; rel=&quot;nofollow&quot;&gt;buffer solutions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a good start would be comparing $\text{p}K_a$ and $\text{p}K_b$ values of the involved components, giving you a hint whether it will be more acidic or more basic. But you also have to pay attention to the amount of substance in particular in these cases. Dilution also plays an important role in this kind of chemistry. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-24T02:59:05.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10203" PostTypeId="1" AcceptedAnswerId="10204" CreationDate="2014-04-24T03:35:36.843" Score="2" ViewCount="345" Body="&lt;p&gt;Why more attractive metals will be oxidized? Aren't they being reduced, because they attract and receive electrons from negative polyatomic ions? So they will be anodes, where oxidization occurs. Why anodes are negative in voltaic cells and positive in electrolytic cells? Same for cathodes.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-04-24T03:50:56.827" Title="Voltaic and electrolytic cells" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10204" PostTypeId="2" ParentId="10203" CreationDate="2014-04-24T03:50:56.827" Score="2" Body="&lt;p&gt;Metals want to lose electrons and become positively charged. When the metal loses the electron, it is called oxidation. The substance that gets the electron is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation could occur at the anode if it takes electrons from a metal (i.e. $Fe^{2+}$ to $Fe^{3+}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anodes are negative in voltaic cells since there are excessive electrons available to power the circuit. Cathodes give electrons away to the solution.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-24T03:50:56.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10205" PostTypeId="1" AcceptedAnswerId="10214" CreationDate="2014-04-24T04:31:08.113" Score="2" ViewCount="1353" Body="&lt;p&gt;Some ceramicists decorate pottery using soluble salts (usually chlorides and sulfates e.g. copper chloride, cobalt sulfate) which are used individually or mixed to produce different colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been exploring the use of these chemicals on porcelain.  From the MSDS of each chemical I've become aware of the safety concerns and the proper precautions to take when mixing, applying, and firing them in a kiln.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know much about chemistry and do not understand the reactions when mixing these chemicals.  Are there any chemicals which should absolutely not be mixed together?  For instance, the Wikipedia article on Potassium permanganate says &quot;Concentrated sulfuric acid reacts with $\ce{KMnO4}$ to give $\ce{Mn2O7}$, which can be explosive.&quot;  Does this mean I should not mix Potassium permanganate with a sulfate like Copper Sulfate?  In general, what happens when a sulfate is mixed with a chloride?  Are there any general safety guidelines you could offer?  Is there a fairly simple way to understand these interactions without the requirement of years of Chemistry study?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for my ignorance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some chemicals that ceramicists have used before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;gray&#xA;&lt;ul&gt;&#xA;&lt;li&gt;copper chloride (heavy application and heavy reduction can give pinks and reds)&lt;/li&gt;&#xA;&lt;li&gt;palladium chloride&lt;/li&gt;&#xA;&lt;li&gt;ruthenium chloride&lt;/li&gt;&#xA;&lt;li&gt;selenium (selenous acid, selenium toner)&lt;/li&gt;&#xA;&lt;li&gt;silver nitrate&lt;/li&gt;&#xA;&lt;li&gt;tellurium chloride&lt;/li&gt;&#xA;&lt;li&gt;vanadium (vanadyl sulfate, vanadium pentoxide)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;blue&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cobalt chloride&lt;/li&gt;&#xA;&lt;li&gt;molybdenum (molybdic acid)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;green&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ammonium chromate&lt;/li&gt;&#xA;&lt;li&gt;nickel chloride&lt;/li&gt;&#xA;&lt;li&gt;potassium dichromate&lt;/li&gt;&#xA;&lt;li&gt;sodium chromate&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;brown&#xA;&lt;ul&gt;&#xA;&lt;li&gt;iron chloride (iron chloride emits heat when mixed with water so the water    should be added gradually in small amounts)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;pink/purple/maroon&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gold chloride (1-5% solution, adding either cobalt, manganese or&lt;/li&gt;&#xA;&lt;li&gt;tellurium will give different shades)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;yellow&#xA;&lt;ul&gt;&#xA;&lt;li&gt;praseodymium chloride (very pale color)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;black&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cobalt chloride (50% solution) and iron chloride (100% solution)&lt;/li&gt;&#xA;&lt;li&gt;cobalt chloride (50% solution) and nickel chloride (50% solution)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5286" LastEditorUserId="1362" LastEditDate="2014-11-02T15:19:36.967" LastActivityDate="2018-01-04T13:33:57.957" Title="Are there any safety guidelines for mixing sulfate with chloride?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10207" PostTypeId="2" ParentId="6311" CreationDate="2014-04-24T07:43:14.340" Score="4" Body="&lt;p&gt;Hybridization is a concept, which is often misconceived. One has to understand, that it is a mathematical concept, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will be formed respecting steric and electronic interactions and possibly interactions with a medium or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hybridisation is therefore only a result of a certain geometric arrangement of atoms in a molecule. A linear combination of all available (atomic) orbitals will form molecular orbitals (MO). These can then be rearranged (recombined) in a certain way to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Localized_molecular_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;localised molecular orbitals&lt;/a&gt; (LMO). These can then be used to interpret a bonding situation in a simpler theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each LMO is expressed as a linear combination of the atomic orbitals, hence it is possible to calculate contributions of the atomic orbitals and describe these also as hybrid orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very simple case methane, the $\ce{C-H}$ bonds are formed from contributions of carbon's $\ce{2s, 2p}$ and hydrogen's $\ce{1s}$. For the carbon this may be interpreted as a $\ce{2sp^3}$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simple case of $\ce{{}^{-}CH3}$, as Uncle Al already pointed out is isoelectronic to ammonia, the bonding situation may be also described in terms of $\ce{2sp^3}$ orbitals. As well as ammonia (or amines in general) it may also undergo &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;nofollow noreferrer&quot;&gt;(nitrogen) inversion&lt;/a&gt;. &#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Carbanion_Structural_Formulae_V.1.svg/500px-Carbanion_Structural_Formulae_V.1.svg.png&quot; alt=&quot;carbanion inversion&quot;&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbanion&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this inversion scenario the hybridisation of the carbon (nitrogen) changes from $\ce{2sp^3}$ to $\ce{2sp^2}$ to $\ce{2sp^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With larger carbanions this becomes a little more complicated. As long as $\ce{R=Alkyl}$, hyper conjugation is not strong enough to prevent pyramidalisation. However, the inversion berrier is furthermore reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any of the $\ce{R}$ groups are different from alkyl substituents, then I would expect that the environment around the carbon becomes (more) planar, i.e. the contribution of carbon's $\ce{2s}$ orbital increases. That may have various reasons, but some kind of conjugation or hyperconjugation will be involved. When these $\ce{R}$ groups become pretty bulky, I would expect the same, due to steric interactions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please understand hybridisation as a tool to understand bonding situations. Therefore it is a result of a geometric alignment of atoms in a molecule. It is not the driving force to form a certain geometric constellation.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.660" LastActivityDate="2014-04-24T07:43:14.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10208" PostTypeId="1" AcceptedAnswerId="10209" CreationDate="2014-04-24T08:01:54.473" Score="3" ViewCount="157" Body="&lt;p&gt;I want to redo an experiment. They've used&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Nickel chloride: 40 g/L&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Nickel chloride can be bought in two forms: anhydrous ($\ce{NiCl2}$) and hydrate ($\ce{NiCl2.6H2O}$). The paper doesn't say which one they've used. Which one did they mean in the paper? In one liter of water should I dissolve 40 grams of $\ce{NiCl2}$ or 40 grams of $\ce{NiCl2.6H2O}$? Is there a convention?&lt;/p&gt;&#xA;" OwnerUserId="5287" LastEditorUserId="4945" LastEditDate="2014-04-24T08:53:51.037" LastActivityDate="2014-04-24T08:53:51.037" Title="metal salt solutions g/l means hydrate or anhydrous" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10209" PostTypeId="2" ParentId="10208" CreationDate="2014-04-24T08:12:26.250" Score="2" Body="&lt;p&gt;Unless the experimental part of the paper provides additional details, such as &lt;em&gt;equivalents&lt;/em&gt; or &lt;em&gt;moles&lt;/em&gt; or the colour of the nickel chloride used ($\ce{NiCl2*6H2O}$ is green, while the anhydrous chloride is yellowish) you're up to guessing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the experiment you're trying to repeat is conducted in aqueous solution, it is safe to assume that you can use the hexahydrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, gloves and proper waste management are a good idea here.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-24T08:12:26.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10210" PostTypeId="1" AcceptedAnswerId="13841" CreationDate="2014-04-24T10:29:24.337" Score="10" ViewCount="20456" Body="&lt;p&gt;A question on a past exam for a course I'm studying for asks: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the relation between Van der Waals forces and hydrophobic&#xA;  interactions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand, Van der Waals forces are just a name to summarize a bunch of little weak forces that only come into play when molecules are placed close to one other, based off &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8691/van-der-waals-nonspecific-interaction-definition&quot;&gt;this previous answer&lt;/a&gt; to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that hydrophobic interactions arise because non-polar components of a molecule are unable to form bonds with hydrogen in water, but disrupt the hydrogen, forcing them to make a cage-like structure (although I'm not clear on why), based off &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Hydrophobic_interactions&quot; rel=&quot;noreferrer&quot;&gt;this entry&lt;/a&gt; in the Chemistry wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the relation between these two forces that Van der Waals forces can be encouraged by hydrophic forces? Is there some other relation I'm missing here? Does it have something to do with the interactions denaturing at high temperatures?&lt;/p&gt;&#xA;" OwnerUserId="4622" LastEditorUserId="73" LastEditDate="2014-06-25T09:03:20.430" LastActivityDate="2019-02-02T17:31:32.027" Title="What's the relationship between Van der Waals forces and hydrophobic interactions?" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10211" PostTypeId="1" CreationDate="2014-04-24T11:55:21.913" Score="3" ViewCount="2863" Body="&lt;p&gt;What is the name of the molecule that has the greatest number of different kinds of atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, H2O is water has 2 different kinds of atoms which are H and O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, C6H12O6 is fructose and has 3 different kinds of atoms which are C, H and O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After warned to be a too broad question. I will rephrase it. The molecule must be stable and can be find in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the name of the molecule has the record?&lt;/p&gt;&#xA;" OwnerUserId="5288" LastEditorUserId="5288" LastEditDate="2014-04-25T13:43:51.883" LastActivityDate="2014-04-25T13:43:51.883" Title="what is the name of the molecule that has the greatest number of different atoms?" Tags="&lt;molecules&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2014-04-25T01:53:21.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="10213" PostTypeId="2" ParentId="10210" CreationDate="2014-04-24T14:21:06.263" Score="4" Body="&lt;p&gt;All sorts of dispersion force and polarization interactions are around the 5 kcal/mole of a hydrogen bond. Recent hype is the &quot;halogen bond&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cen.acs.org/articles/92/i1/Defining-New-Carbon-Bond.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cen.acs.org/articles/92/i1/Defining-New-Carbon-Bond.html&lt;/a&gt; &lt;BR&gt;&#xA;DOI:10.1021/cg500094b &lt;BR&gt;&#xA;DOI:10.1039/C3CP00054K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long before that, the curious co-crystallization of $\ce{CBr4}$ and $\ce{CPh4}$ was noted.  Tetra-(4-bromophenyl)methane gives the same structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI:10.1021/ja953373m &lt;BR&gt;&#xA;&lt;a href=&quot;http://galoppinigroup.com/wp-content/uploads/2012/06/Weak-Intermolecular.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://galoppinigroup.com/wp-content/uploads/2012/06/Weak-Intermolecular.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://ccaasmag.org/chem_2011/vol1/Darbost_halogenAryl.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ccaasmag.org/chem_2011/vol1/Darbost_halogenAryl.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clathrates with water (methane, $\ce{SF6}$, $\ce{CO2}$, xenon).  Benzil is interesting.  Pi-stacking and lattice forces make the solid state conformation very different from gas phase, solution, or molten,&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VBVDx.png&quot; alt=&quot;Benzil xtal structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in water, hydrophobic exclusion from the hydrogen-bonded polar continuum affords a lot of energy to play with, then diddle with the medium,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/kosmos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www1.lsbu.ac.uk/water/kosmos.html&lt;/a&gt; &lt;BR&gt;&#xA;kosmotropes and chaotropes&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-24T14:21:06.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10214" PostTypeId="2" ParentId="10205" CreationDate="2014-04-24T14:43:26.653" Score="4" Body="&lt;p&gt;Overall, always wear disposable gloves, goggles, and a dust mask.  Add a lab coat, smock, or apron so as not to contaminate your clothing.  Avoid skin contact, don't ingest and don't inhale (this includes the dust).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manganese heptoxide is a special case with concentrated sulfuric acid.  Avoid mixing strong oxidizers (permanganate, chromate/dichromate, perchlorate) with organics.  NEVER screw around with chlorate salts.  They are powerful oxidizers with very low kinetic barriers.  Finely divided metals (especially titanium) can go to white-hot burning in open air.  Titanium burns under air, carbon dioxide, nitrogen, water...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Palladium is a heavy metal, poison and tissue corrodant.  A microgram of tellurium by any contact gives &lt;a href=&quot;http://www.lenntech.com/periodic/elements/te.htm#Health%20effects%20of%20tellurium&quot; rel=&quot;nofollow&quot;&gt;&quot;tellurium breath&quot;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local contact with chromates can cause ulceration.  Neodymium gives you purples, cerium yellow to orange.  I seem to remember that cobalt blue (cobalt aluminate after firing) in tin oxide applied to white ware gives bubblegum purple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tin-glazing&quot; rel=&quot;nofollow&quot;&gt;Tin Glazing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold via the Purple of Cassius is very nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a pretty!  Suppose you mixed clay with hollow glass microbubbles, then fired.  Would the clay foam to an interesting surface or whole body?  This is the stuff, sold generally as lightweight resin filler, including hobby stores.  Most prices are outrageous when sold by volume.  Density of 3M Scotchlite K20 is 0.2 g/cm^3.  Goggles and dust mask.  Be careful or the stuff goes everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aircraftspruce.com/catalog/cmpages/bubbles.php&quot; rel=&quot;nofollow&quot;&gt;http://www.aircraftspruce.com/catalog/cmpages/bubbles.php&lt;/a&gt; &lt;BR&gt;&#xA;Good price. &lt;BR&gt;&#xA;&lt;a href=&quot;http://solutions.3m.com/wps/portal/3M/en_US/Energy-Advanced/Materials/Product/?N=5002440+7570097&amp;amp;rt=r3&quot; rel=&quot;nofollow&quot;&gt;http://solutions.3m.com/wps/portal/3M/en_US/Energy-Advanced/Materials/Product/?N=5002440+7570097&amp;amp;rt=r3&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5137" LastEditDate="2014-11-02T16:53:17.310" LastActivityDate="2014-11-02T16:53:17.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10215" PostTypeId="1" AcceptedAnswerId="10216" CreationDate="2014-04-24T18:16:58.167" Score="15" ViewCount="3749" Body="&lt;p&gt;This is a very basic question and I'm surprised it only just struck me. The nature of the bond in coordination complexes is a coordinate covalent bond. Only the ligand donates electrons for bond formation between the metal and ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that we never place a positive charge on the ligands? For example, in $\ce{[Pt(NH3)5Cl]Br3}$, $\ce{NH3}$ is attached to three Hydrogens and the lone pair is donated for forming the coordinate bond. So there &lt;em&gt;should&lt;/em&gt; be a positive charge on $\ce{NH3}$, right?&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="189" LastEditDate="2014-04-24T21:55:05.853" LastActivityDate="2017-06-17T15:52:20.010" Title="The bond in coordination complexes" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10216" PostTypeId="2" ParentId="10215" CreationDate="2014-04-24T20:09:27.067" Score="15" Body="&lt;p&gt;In a typical covalent bond, such as is found between carbon and hydrogen in&#xA;methane, each atom is considered to contribute one electron to the two-electron,&#xA;two-centre bond which is formed. The metal-ligand bond is often envisaged as a second type of covalent bond in which we still have a two-centre, two-electron bond, but where&#xA;both of the electrons come from one of the atoms or from a molecule. This type of&#xA;bond is known variously as a coordination, a dative covalent or a donor-acceptor&#xA;bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was recognized early on that the formality of &lt;em&gt;dative covalency&lt;/em&gt; in coordination&#xA;compounds presents some difficulties. Many inorganic compounds are conveniently&#xA;thought of as ionic salts in which there is an essentially complete charge separation&#xA;between cationic and anionic species. Compounds of the group 1 and group 2 metals&#xA;tend to be readily categorized as &lt;em&gt;ionic&lt;/em&gt;. Dissolution of a salt of a group 1 or a&#xA;group 2 metal results in the formation of solutions containing solvated cations and&#xA;anions - only in concentrated solutions are there significant cation-anion&#xA;interactions beyond simple ion-pairing. However, the ionic model does not appear&#xA;to be suitable for the description of the properties of many transition-metal&#xA;compounds. For example, the compound $\ce{K4[Fe(CN)6]}$ dissolves in water to give&#xA;solutions containing solvated potassium ions and the $\ce{[Fe(CN)6]^{4-}}$ ion, rather than&#xA;solvated potassium, iron(II) and cyanide ions. The interactions between the cyanide&#xA;and the iron(II) centre appear to result in longer lived species than result from&#xA;simple electrostatic interactions of the type observed in sodium chloride. This is, of&#xA;course, the sort of argument which led to the development of the description of&#xA;coordination compounds in terms of donor-acceptor interactions between the&#xA;ligands and the metal centre.&#xA;Let's have a look at the consequences of the formation of a donor-acceptor&#xA;bond in a little more detail. If the donor-acceptor bond is completely covalent,&#xA;then there is a net transfer of one unit of charge from the donor to the acceptor as&#xA;a direct consequence of the equal sharing of the electron pair between the two&#xA;centres. This result leaves a positive charge on the donor atom and a negative&#xA;charge on the acceptor atom. The limiting &lt;em&gt;ionic&lt;/em&gt; and &lt;em&gt;covalent&lt;/em&gt; descriptions of a&#xA;complex cation such as $\ce{[Fe(H2O)6]^{3+}}$ are shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pdzav.png&quot; height=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As commented above 'ionic' structures are not in accord with the&#xA;chemical properties of coordination compounds. However, there are also a number&#xA;of objections to the covalent description. The charge distribution is such that the&#xA;iron(III) centre bears a three minus charge, whereas the oxygen atoms of the water&#xA;ligands each bear a single positive charge. This would be unrealistic in view of the&#xA;electronegativities of these elements ($\ce{Fe}$, 1.8; $\ce{O}$, 3.5), which predict that the $\ce{Fe-O}$ bond should be polarized in the sense $\ce{Fe^{\delta +}-O^{\delta -}}$. The problem was addressed by Pauling, who recognized that, in reality, it was not appropriate to describe most bonds as being purely 'ionic' or purely 'covalent'.&#xA;He developed a description of the bonding between a metal and its ligands which included considerable ionic character in the metal-ligand bonds within a basically covalent regime. In order to determine the amount of &lt;em&gt;ionic&lt;/em&gt; character within a given metal-ligand bond, Pauling framed his electroneutrality principle. In this, he opined that the actual distribution of charges within a molecule is such that charges on any single atom are within the range -1 to +1.&#xA;It can be seen how this works by reconsidering the ion $\ce{[Fe(H2O)6]^{3+}}$. An &lt;em&gt;ionic&lt;/em&gt;&#xA;description results in a +3 charge on the metal centre, whilst a &lt;em&gt;covalent&lt;/em&gt; description&#xA;gives the metal a -3 charge. Now the electroneutrality principle suggests that the&#xA;&lt;em&gt;ideal&lt;/em&gt; charge on the metal centre is zero. That would be achieved if the iron centre&#xA;gains a total of three electrons from the six oxygen donor atoms; in other words, if&#xA;each oxygen loses one half of an electron rather than the whole electron which the&#xA;fully covalent model demands. Pauling describes this situation as 50%&#xA;covalent (or 50% ionic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jna4m.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pauling's model gives a good indication that the bonding situation in transition metal complexes is rather complex since the bonds are highly polarized (An 'accurate' view of the bonding in transition metal complexes can only be gained by more sophisticated methods such as the molecular orbital method).&#xA;Thus, Lewis structures are not particularly well suited to depict the bonding situation in coordination compounds because Lewis structures are intended for covalent compounds only.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2017-06-17T15:52:20.010" LastActivityDate="2017-06-17T15:52:20.010" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10218" PostTypeId="2" ParentId="10197" CreationDate="2014-04-24T22:32:15.187" Score="3" Body="&lt;p&gt;You &lt;em&gt;can&lt;/em&gt; make an aqueous solution of $\ce{KSO_4^{-}}$ (and no other cations), in the sense that you can make a regular $\ce{K2SO4}$ solution which lacks some potassium ions. By doing so, you are producing a solution with a net electric charge. The higher the net electric charge, the less stable the solution is, as electrons will start redistributing into the surroundings in order to correct the charge imbalance to a high precision. While a net charge is possible, it seems unlikely to be able to produce a solution which contains even close to a 1 µM excess of an ion, and I do not know if electrically imbalanced solutions can be prepared quantitatively. Take at look at the answers and comments to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6932/evaporating-the-water-of-a-solution-of-only-carbonate-ions-and-water-gives&quot;&gt;this previous question&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="16683" LastEditDate="2018-08-27T17:25:41.857" LastActivityDate="2018-08-27T17:25:41.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10219" PostTypeId="1" AcceptedAnswerId="10221" CreationDate="2014-04-25T00:33:48.403" Score="0" ViewCount="2153" Body="&lt;p&gt;What are the products of the following hydrolysis reaction? How do you know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4NO3 + H2O &amp;lt;=&amp;gt; ?}$$&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="41328" LastEditDate="2017-09-21T05:34:57.640" LastActivityDate="2017-09-21T05:36:36.767" Title="Determine products of NH4NO3 hydrolysis" Tags="&lt;inorganic-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2014-04-25T03:22:59.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="10220" PostTypeId="2" ParentId="10219" CreationDate="2014-04-25T00:41:10.833" Score="-1" Body="&lt;p&gt;The products will be $\ce{HNO3 + NH4OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because a salt hydrolysis reaction always produces an acid and a base. The $\ce{H+}$ in $\ce{H2O}$ will bond with $\ce{NO3-}$ to form $\ce{HNO3}$, a strong acid. The remaining $\ce{OH-}$ in $\ce{H2O}$ will bond with $\ce{NH4+}$ to form $\ce{NH4OH}$, a weak base.&lt;/p&gt;&#xA;" OwnerUserId="5292" LastEditorUserId="41328" LastEditDate="2017-09-21T05:36:36.767" LastActivityDate="2017-09-21T05:36:36.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10221" PostTypeId="2" ParentId="10219" CreationDate="2014-04-25T01:45:08.340" Score="3" Body="&lt;p&gt;It will never happen. You can not hydrolyze a salt to any strong acid (nitric acid in your case) or base because the reverse reaction is predominant.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-04-25T01:45:08.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10225" PostTypeId="2" ParentId="8866" CreationDate="2014-04-25T02:35:41.993" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.wiley-vch.de/books/sample/3527302409_c01.pdf&quot; rel=&quot;nofollow&quot;&gt;New Energetic Materials&lt;/a&gt;, table 1.2, a list of explosives ranges from 19-48 GPa in pressure achieved.  This is two orders of magnitude greater than the 60,000 psi figure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as maximum theoretical pressure, the question isn't really clear whether the maximum absolute pressure at which combustion can occur or the maximum change in pressure due to combustion is the value of interest.  The answer could be related to the maximum pressure at which chemical bonds exist.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-04-25T02:35:41.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10226" PostTypeId="1" CreationDate="2014-04-25T02:45:56.400" Score="4" ViewCount="133" Body="&lt;p&gt;At high enough pressure, atomic material no longer exists, for example under the conditions of a neutron star.  Is there a different maximal pressure above which covalent bonds no longer exist?&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-04-25T10:52:45.973" LastActivityDate="2015-01-26T17:56:26.693" Title="Maximum Pressure of Covalently Bonded Material" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10227" PostTypeId="1" CreationDate="2014-04-25T02:48:42.583" Score="2" ViewCount="1384" Body="&lt;p&gt;We use water to cool down some machines because water has high heat capacity, which means it takes more energy to increase 1 degree. Heat transfer occurs only when temperature is different. We cool down the machine because we don't want it to be too hot, but what does the hot mean? The temp or the heat? When we touch water and another liquid at the same temperature but contain different heat. We feel the heat or the temp?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastActivityDate="2014-04-25T16:09:52.990" Title="Which one really determines how people feel. Heat or temperature?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ClosedDate="2015-12-28T06:35:37.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="10228" PostTypeId="1" CreationDate="2014-04-25T02:53:18.957" Score="0" ViewCount="1497" Body="&lt;p&gt;Why is it that in titration a color change occurs initially but then the solution quickly reverts to its previous color?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much!&lt;/p&gt;&#xA;" OwnerUserId="4363" LastActivityDate="2014-04-25T03:21:21.617" Title="Titration color change duration" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-04-25T03:22:31.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="10229" PostTypeId="2" ParentId="10228" CreationDate="2014-04-25T03:21:21.617" Score="1" Body="&lt;p&gt;As an example, let's say you're titrating a strong acid with a strong base using phenolphthalein as an indicator. As you initially add the base to the acid, you may not see the color change occurring because the acid so readily neutralizes the added base. This is because of the large concentration of H+. As you approach a basic pH, however, concentration of H+ decreases and the acid is not able to neutralize the base as quickly. As the base disperses in the solution more slowly, the phenolphthalein will have time to react and show a slight color change. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastActivityDate="2014-04-25T03:21:21.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10230" PostTypeId="1" AcceptedAnswerId="10231" CreationDate="2014-04-25T04:11:49.573" Score="11" ViewCount="25723" Body="&lt;p&gt;The title says it all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this question when helping someone with their high-school chemistry homework. What are the reasons that someone would use an (from my point of view) obfuscated notation?&lt;/p&gt;&#xA;" OwnerUserId="4969" LastActivityDate="2019-11-16T15:17:19.777" Title="Why NH4OH? Why not NH3(aq)?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10231" PostTypeId="2" ParentId="10230" CreationDate="2014-04-25T05:02:38.353" Score="14" Body="&lt;p&gt;The lone pair at the nitrogen is very well suited to form hydrogen bonds. This bond is especially strong, since ammonia has a quite high proton affinity, it is a medium strong base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the autoprotolysis reactions in pure water and (liquid) ammonia:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2H2O &amp;amp;&amp;lt;=&amp;gt; H3O+ + OH-}\tag1\\&#xA;\ce{2NH3 &amp;amp;&amp;lt;=&amp;gt; NH4+ + NH2-}\tag2&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing these systems, the reaction of ammonia with water will yield&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;      &amp;amp;&amp;amp; \ce{2H2O + 2NH3 &amp;amp;&amp;lt;=&amp;gt; H3O+ + OH- + NH4+ + NH2-}\tag3 \\&#xA;&amp;amp; (~-~)&amp;amp; \ce{H3O+ + NH2- &amp;amp;-&amp;gt; H2O + NH3}\\\hline&#xA;      &amp;amp;&amp;amp; \ce{H2O + NH3 &amp;amp;&amp;lt;=&amp;gt; NH4+ + OH-}\tag4\label{eq:ammonia+water}&#xA;\end{align}&lt;/span&gt;&#xA;\eqref{eq:ammonia+water} is incorporated into a huge and strong network of hydrogen bonds. In this case the equilibrium will be very much pushed to the right side, meaning, that almost all ammonia molecules will be protonated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it is more appropriate to write:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{NH4OH (aq) &amp;amp;&amp;lt;=&amp;gt; NH4+ (aq) + OH- (aq)}\tag{4a}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is always advisable to look at the conditions of the whole reaction system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4OH}$&lt;/span&gt; is a pretty outdated notation, if you would want to consider one water entity explicitly, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3.H2O}$&lt;/span&gt; would be more appropriate. See the answer by Maurice for some more information: &lt;a href=&quot;https://chemistry.stackexchange.com/a/123839/4945&quot;&gt;Can acid-base reactions occur in a non-aqueous medium?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2019-11-16T15:15:20.990" LastActivityDate="2019-11-16T15:15:20.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10232" PostTypeId="1" AcceptedAnswerId="10813" CreationDate="2014-04-25T06:03:35.010" Score="5" ViewCount="11202" Body="&lt;p&gt;I'm confused by the following from &lt;a href=&quot;http://batteryuniversity.com/learn/article/lithium_based_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The cathode is a metal oxide and the anode consists of porous carbon. During discharge, the ions flow from the anode to the cathode through the electrolyte and separator; charge reverses the direction and the ions flow from the cathode to the anode.&lt;/p&gt;&#xA;&lt;p&gt;On discharge, the anode undergoes oxidation, or loss of electrons, and the cathode sees a reduction, or a gain of electrons. Charge reverses the movement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This says that the metal oxide electrode is always the cathode and the porous carbon electrode is always the anode. As far as I know, that designation should be correct on discharge but backwards on charge. The anode is always whichever electrode is doing the oxidation, and the cathode is the electrode doing the reduction.&lt;/p&gt;&#xA;&lt;p&gt;My naive understanding would be this:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The cathode is the electrode with the reduction half reaction. The anode is the electrode with the oxidation half reaction. This is true for both charge/discharge.&lt;/li&gt;&#xA;&lt;li&gt;The anode is the one that produces electrons and cathode receives electrons. This is true for both charge/discharge.&lt;/li&gt;&#xA;&lt;li&gt;When switching between charge/discharge, the redox reactions reverse, and the cathode/anode designations also switch to preserve cathode==reduction and anode==oxidation.&lt;/li&gt;&#xA;&lt;li&gt;During discharge, the battery functions as a galvanic cell, where the redox reaction produces electric energy, electrons go down their electrical gradient from the negative electrode to the positive electrode. The anode is the negative electrode, the cathode is the positive electrode.&lt;/li&gt;&#xA;&lt;li&gt;During charge, the battery functions as an electrolytic cell, where electric energy drives a nonspontaneous redox reaction, electrons go up their electrical gradient from the positive electrode to the negative electrode. The anode is the positive electrode, the cathode is the negative electrode.&lt;/li&gt;&#xA;&lt;li&gt;In a lithium ion battery, the positive electrode is the metal oxide and the negative electrode is the porous carbon. The anode/cathode designations switch depending on whether the battery is charging or discharging.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Please help me understand this.&lt;/p&gt;&#xA;" OwnerUserId="4309" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-03-04T08:43:47.510" Title="Cathode + Anode + Rechargeable battery" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10233" PostTypeId="1" CreationDate="2014-04-25T09:12:25.863" Score="1" ViewCount="450" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The total number of octahedral void(s) per atom present in a cubic closed pack structure is&lt;/p&gt;&#xA;&lt;p&gt;(&lt;strong&gt;a&lt;/strong&gt;) 1&lt;br /&gt;&#xA;(&lt;strong&gt;b&lt;/strong&gt;) 2&lt;br /&gt;&#xA;(&lt;strong&gt;c&lt;/strong&gt;) 3&lt;br /&gt;&#xA;(&lt;strong&gt;d&lt;/strong&gt;) 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I answered with (&lt;strong&gt;d&lt;/strong&gt;), but the key on internet says it is (&lt;strong&gt;a&lt;/strong&gt;). Please clarify the dispute.&lt;/p&gt;&#xA;" OwnerUserId="4103" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-12-16T15:56:41.347" Title="Total number of octahedral void(s) per atom in a cubic closed pack structure" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;crystallography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10234" PostTypeId="1" AcceptedAnswerId="10240" CreationDate="2014-04-25T09:42:23.093" Score="6" ViewCount="1919" Body="&lt;p&gt;Actually this is going to be a list of interconnected questions. Please bear with me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZDnA.jpg&quot; alt=&quot;Cyclodextrin structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia article on cyclodextrin (CD) says that it is hydrophobic from inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) How is it hydrophobic when there are so many hydroxyl groups on the interior surface ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) This &lt;a href=&quot;http://www.u.arizona.edu/~brusseau/images/research/remediation-hydroxines.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;&#xA;    labels the interir cavity as apolar rather than hydrophobic. I guess&#xA;    the apolarity is because of mutual cancellation of the dipole&#xA;    moments. &#xA;Am I right on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I don't understand how the cavity excludes water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are reports on trapping organic substances such as &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/app.12818/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;naphthalene&lt;/a&gt;, &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs10847-010-9776-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;toluene&lt;/a&gt; and &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2012/EM/c2em30223c#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;crude oil&lt;/a&gt; in the CD cavity. Another &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1387181104002859&quot; rel=&quot;nofollow noreferrer&quot;&gt;report&lt;/a&gt; says that nitrosamime is trapped in CD by interacting with the hydroxyls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) How does this trapping exactly occur ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Do the hydroxyls play a role in restricting the diffusion of the trapped organic molecules out of the cavity ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this, I started investigating because of a recent news about an alcoholic beverage in powder form- &lt;a href=&quot;http://www.palcohol.com/home.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;palcohol&lt;/a&gt;. Initially I didn't believe it but later I found that alcohol is absorbed in CD to make a &lt;em&gt;dry&lt;/em&gt; powder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) What prevents alcohol from adsorbing on the exterior surface and not make the powder sticky ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) Other studies say that there is a 1:1 ratio of trapping but palcohol claims that it is 58% alcohol by weight. Even &lt;a href=&quot;http://www.google.com/patents/US3795747&quot; rel=&quot;nofollow noreferrer&quot;&gt;this old patent&lt;/a&gt; says that CD absorbs a lot of alcohol. How does that happen ? In all it seems that CD behaves differently with different molecules.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2014-04-25T14:43:17.967" Title="Why is Cyclodextrin cavity hydrophobic?" Tags="&lt;adsorption&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10235" PostTypeId="1" AcceptedAnswerId="10266" CreationDate="2014-04-25T11:57:49.697" Score="0" ViewCount="150" Body="&lt;p&gt;I hope this doesn't sound too strange.&lt;br&gt;&#xA;I have plenty of cheap vodka and want to clean the A/C on my car.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought about boiling 1/2 litre of vodka in my car while I ran the A/C with recirculation on.&lt;br&gt;&#xA;The alcohol should get liquid within the air conditioner and desinfect and clean the vaporizer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just not sure if I create an explosive mixture inside the car this way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;David&lt;/p&gt;&#xA;" OwnerUserId="5296" LastActivityDate="2014-04-26T17:43:26.283" Title="Boilng vodka for A/C cleaning" Tags="&lt;boiling-point&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10236" PostTypeId="2" ParentId="10227" CreationDate="2014-04-25T12:14:16.900" Score="3" Body="&lt;p&gt;Regarding the last part &quot;Do we feel heat or the temp?&quot; - this is more physiological then chemical, but I want to share this anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We feel heat flux, i.e. how fast heat is removed or supplied to our skin. Since heat flux can be both advection in the surrounding medium (e.g. air) or conduction in the skin it depends on the situation which is dominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you can write this as &#xA;$$ Q=k A \Delta T$$&#xA;where $Q$ is the heat flux, $k$ is some transfer coefficient, $A$ is the exchange area and $\Delta T$ is the temperature difference. The value of $k$ depends on the mechanism of transfer e.g. for advection it scales with $\rho C_p v$ i.e. it depends on the heat capacity and flow velocity of the surrounding medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence of feeling heat flux is that we can only feel temperature &lt;em&gt;differences&lt;/em&gt; not absolute temperature and that advection gives rise to a phenomenon called 'wind chill' that causes us to feel colder when there is more wind, because the higher wind velocity increases the heat transfer coefficient.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-04-25T12:25:53.760" LastActivityDate="2014-04-25T12:25:53.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10238" PostTypeId="1" CreationDate="2014-04-25T13:57:05.647" Score="0" ViewCount="769" Body="&lt;p&gt;A cylinder contains 2 kg of steam at 200 kPa and 200 $^\circ$C. Now steam is compressed at constant pressure until its temperature rises to 250 $^\circ$C. What is the work done during this compression process?&lt;/p&gt;&#xA;" OwnerUserId="5297" LastEditorUserId="275" LastEditDate="2014-04-27T10:59:24.080" LastActivityDate="2014-04-27T11:41:27.547" Title="What is the compression work?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-04-28T01:03:33.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="10239" PostTypeId="2" ParentId="10226" CreationDate="2014-04-25T14:31:45.840" Score="2" Body="&lt;p&gt;With the twist of thumbscrew, a hand-held diamond anvil cell will exert 300 GPa (43,500,000 psi).  No chemical bonding problem there. If you stack stuff - planetary and stellar cores - it gets more interesting, depending on composition and temperature.  Temperature matters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_degeneracy_pressure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electron_degeneracy_pressure&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic fields exert pressure.  How much magnetic field diddles with chemical bonding to give you a pressure estimate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetar&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnetar&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_pressure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Magnetic_pressure&lt;/a&gt; &lt;BR&gt;&#xA;  10^10 teslas is 4.0×10^25 J/m^3 E/c^2 mass density more than 10^4 times that of lead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That'll put some hurt on chemical bonding.  Scale down to 347 kJ/mole-bond multiplied in a molar volume of diamond.  Atomizing diamond requires about 800 kJ/mol per atom against cohesive energy.  The strongest chemical bonding interaction is roughly 3 eV.  Then you can pontificate with a Lennard-Jones potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lennard-Jones_potential&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lennard-Jones_potential&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-25T14:31:45.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10240" PostTypeId="2" ParentId="10234" CreationDate="2014-04-25T14:43:17.967" Score="5" Body="&lt;p&gt;Cyclodextrin hydroxyls are the external surface (hydrophilic and hydrogen-bonding).  They circle the larger open end.  The carbon and hindered ether skeleton is internal (hydrophobic).  A cyclodextrin looks like a bonded hollow detergent micelle in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QOt1K.jpg&quot; alt=&quot;beta-cyclodextrin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you believe advertising hype, remember that underivatized cyclodextrins are nephrotoxic, crystallizing upon concentration, probably as the cholesterol endohedral complex.  Look it up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethanols insert their hydrocarbon tails into the center, then hydrogen bond their polar ends with each other and the rim. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-25T14:43:17.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10241" PostTypeId="1" CreationDate="2014-04-25T15:47:57.207" Score="-8" ViewCount="1213" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJUA2.png&quot; alt=&quot;Difficult Question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can Somebody please solve this chemistry question&#xA;It came as a part of a previous test and i wasn't able to find a way to use ny/nx since all my answers would be negative. It wasn't a homework problem.&lt;/p&gt;&#xA;" OwnerUserId="5299" LastEditorUserId="5299" LastEditDate="2014-04-26T12:54:27.457" LastActivityDate="2014-04-26T12:54:27.457" Title="Find the molecular formulas of these alkanas and alkenes" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-04-25T21:23:50.337" ContentLicense="CC BY-SA 3.0" />
  <row Id="10242" PostTypeId="2" ParentId="10186" CreationDate="2014-04-25T16:02:29.503" Score="2" Body="&lt;p&gt;You always need to add the equations since the break of bonds or the formation of bonds leads to an exchange in energy.&#xA;Here is a simple case &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{2 B + 3 H2O &amp;amp;-&amp;gt; B2H6 + 3/2 O2}\quad&amp;amp;&amp;amp;(\Delta H = 762\ \mathrm{kJ})\\&#xA;\ce{H2 + 1/2 O2 &amp;amp;-&amp;gt; H2O}\quad&amp;amp;&amp;amp;(\Delta H = −286\ \mathrm{kJ}) \times 3\\&#xA;\hline&#xA;\ce{3 H2 + $\boldsymbol{\ce{3/2 O2}}$ &amp;amp;-&amp;gt; $\boldsymbol{\ce{3H2O}}$}\quad&amp;amp;&amp;amp;(\Delta H = −858\ \mathrm{kJ})\\&#xA;\ce{2 B + $\boldsymbol{\ce{3 H2O}}$ &amp;amp;-&amp;gt; B2H6 + $\boldsymbol{\ce{3/2 O2}}$}\quad&amp;amp;&amp;amp;(\Delta H = 762\ \mathrm{kJ})&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the ones bold get canceled since the product is the reactant in the other equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_\mathrm{r}$ = sum of the $\Delta H$s so it’s equal to $-96\ \mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the second equation you were talking about one $\ce{H2O}$ molecule however in the first you are talking about three.&lt;/p&gt;&#xA;" OwnerUserId="5299" LastEditorUserId="5026" LastEditDate="2018-06-25T03:04:58.423" LastActivityDate="2018-06-25T03:04:58.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10243" PostTypeId="2" ParentId="10233" CreationDate="2014-04-25T16:03:18.377" Score="3" Body="&lt;p&gt;Finally got it, it was a misunderstanding. I ignored to read the word ‘&lt;em&gt;per atom&lt;/em&gt;’ in a hurry. So clearly the answer is 1.&lt;/p&gt;&#xA;" OwnerUserId="4103" LastEditorUserId="41328" LastEditDate="2019-12-16T15:56:41.347" LastActivityDate="2019-12-16T15:56:41.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10244" PostTypeId="2" ParentId="10227" CreationDate="2014-04-25T16:09:52.990" Score="1" Body="&lt;p&gt;Sensation of heat via the nervous system is complicated, as there are different kinds of proteins which respond to temperature and take part in thermoception.  The proteins themselves respond directly to temperature, but we don't sense the temperature-based conformation directly.  Rather the impact of this change on ion flux leads to a change in electrical potential across the cellular membrane, ultimately resulting in the 'firing' of neural impulses.  As such, higher heat transfer results in the sensation of more heat due to more rapid change in membrane potential even though the proteins are directly responding to absolute temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is further complicated by inhibitory neurons in the sensory processing system that 'accommodate' for ongoing signals by down-regulating a sensation when it occurs continuously.  This is why there is a qualitative difference between the feeling of initially stepping out of air conditioning into a hot day and how it feels after being out in the heat for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the feedback from the various sensors are further integrated to provide more complex sensory signals.  For instance, the sensation of intense (painful) heat can actually be created by suppressing the sensation of intense cold using cool stimuli.  This is the basis for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_grill_illusion&quot; rel=&quot;nofollow&quot;&gt;thermal grid illusion&lt;/a&gt; where contact with alternating warm and cool bars can create the illusion of painful heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, our sensation of hot and cold is not a simple direct relationship to heat or heat transfer, though it certainly correlates  in a useful way that helps us to avoid damage to our bodies under 'typical' conditions.&lt;/p&gt;&#xA;" OwnerUserId="5298" LastActivityDate="2014-04-25T16:09:52.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10245" PostTypeId="1" AcceptedAnswerId="11172" CreationDate="2014-04-25T17:32:30.997" Score="8" ViewCount="47032" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is weight of 1 mole of substance equal to atomic/molecular mass in grams? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me, it happens because mole has been defined in such a way. It is defined as the numbers of particles in $\pu{12 g}$ of $\ce{^{12}C}$. If it were $\pu{24 g}$ instead of $\pu{12 g}$, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am wrong somewhere (or maybe everywhere).&lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="41328" LastEditDate="2017-08-07T01:38:59.043" LastActivityDate="2019-03-02T10:59:58.533" Title="Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?" Tags="&lt;mole&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10246" PostTypeId="1" CreationDate="2014-04-25T17:53:05.270" Score="2" ViewCount="116" Body="&lt;p&gt;This is more of a meta-chemistry question, but, here goes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the structure of a polymer, how can I search online for which polymer it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I have an image like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/LW8mz2a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know which polymer specifically this is, but also a method for figuring this out in the future by myself :)&lt;/p&gt;&#xA;" OwnerUserId="5301" LastActivityDate="2014-04-25T22:47:28.277" Title="How do I lookup a polymer by its structure?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10247" PostTypeId="2" ParentId="10230" CreationDate="2014-04-25T18:35:34.230" Score="7" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH4OH}$&lt;/span&gt; is a terrible notation which makes no chemical sense, as there is no such a chemical species and therefore not a dissociation of a mythical species, but deprotonation of water what is behind the basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I guess textbooks like to use it for &quot;didactical&quot; purposes (means it is easier to tell the kids that &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4OH}$&lt;/span&gt; are bases then tell the chemistry behind it).&lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="4945" LastEditDate="2019-11-16T15:17:19.777" LastActivityDate="2019-11-16T15:17:19.777" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="10248" PostTypeId="2" ParentId="10245" CreationDate="2014-04-25T18:42:14.337" Score="1" Body="&lt;p&gt;Take the magnitude of formula mass and append &quot;grams&quot; to obtain the mass of one gram-mole.  If  you have a formula weight magnitude and append &quot;pounds,&quot; it is one pound-mole.  One gram-mole contains one Avogadro number of formula units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like a &quot;dozen eggs,&quot; a mole is simply a bigger basket.  It allows you to take microscopic descriptions to real world scales.  Moles are scaling factors for thermodynamics, as in molar volumes of gases. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-25T18:42:14.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10249" PostTypeId="1" CreationDate="2014-04-25T19:01:17.283" Score="11" ViewCount="30777" Body="&lt;p&gt;We have two solutions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Solution 1 is &lt;span class=&quot;math-container&quot;&gt;$\ce{HCOOH}$&lt;/span&gt;, its concentration is &lt;span class=&quot;math-container&quot;&gt;$c_1 = \pu{10^-2 mol/l}$&lt;/span&gt;, its volume is &lt;span class=&quot;math-container&quot;&gt;$V_1 = \pu{50 ml}$&lt;/span&gt;, and its &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}_1 = 2.9$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solution 2 is &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt;, its concentration is &lt;span class=&quot;math-container&quot;&gt;$c_2 = \pu{10^-2 mol/l}$&lt;/span&gt;, its volume is &lt;span class=&quot;math-container&quot;&gt;$V_2 = \pu{50 ml}$&lt;/span&gt;, and its &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}_2 = 3.4$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How would be the equation and the ICE table, and what is the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of the mixture of these two solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$&lt;/span&gt;, but it’s trivial and doesn't work.&lt;/p&gt;&#xA;" OwnerUserId="5302" LastEditorUserId="22102" LastEditDate="2020-03-12T07:13:52.613" LastActivityDate="2020-03-13T22:30:56.703" Title="How to determine the pH of a mixture of two weak acids?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="10250" PostTypeId="1" AcceptedAnswerId="10254" CreationDate="2014-04-25T19:44:41.093" Score="2" ViewCount="159" Body="&lt;p&gt;Let's  say I want to take a random plant , and get the chemical formulas of  the compounds he's made of. Is there an automated method for doing so  and how costly is it(roughly) ?&lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorUserId="4945" LastEditDate="2014-06-12T08:54:58.887" LastActivityDate="2014-06-13T01:20:03.767" Title="How do i detect the compounds a plant is made of, automatically?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10251" PostTypeId="2" ParentId="10241" CreationDate="2014-04-25T20:31:34.803" Score="4" Body="&lt;p&gt;Non-cyclic (or acyclic) alkanes have the general formula $\ce{C_{n}H_{2n + 2}}$, and corresponding alkenes have the general formula $\ce{C_{n}H_{2n}}$.  So in this problem let's say our alkane (x) is $\ce{C_{n}H_{2n + 2}}$, then our alkene (y) with one more carbon atom will be $\ce{C_{n + 1}H_{2(n + 1)}}$.  The molecular weight of the alkane will be $$\ce{MW(x) = (12*n) + (1*(2n +2 )) = 14n + 2}$$ and $$\ce{MW(y) = ((12*(n + 1)) + (1*(2n + 1 ))) = 14n + 14}$$  We also know that grams(x) = grams(y), and that moles(x)/moles (y) = 7/5.  The number of moles of a substance is equal to its weight in grams divided by its molecular weight.  Therefore $$\frac{N_x}{N_y}=7/5=\frac{G(x)/14n+2}{G(y)/14n+14}$$  But in this problem grams(x) = grams(y) so our equation simplifies to $$7/5=\frac{14n+14}{14n+2}$$ Solving for &quot;n&quot; gives you the number of carbon atoms in the alkane, and &quot;n+1&quot; will be the number of carbon atoms in the alkene.  I'll leave the mathematical solution and the naming of the two moleculesto you.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-25T20:31:34.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10252" PostTypeId="1" AcceptedAnswerId="10253" CreationDate="2014-04-25T20:51:26.780" Score="0" ViewCount="2015" Body="&lt;p&gt;The changing of the colour of $\ce{NO2}$ is because of a change in temperature, resp. heating and cooling?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="4945" LastEditDate="2014-06-12T09:01:46.387" LastActivityDate="2016-04-06T21:54:46.787" Title="Does Nitrogen dioxide change colour if we put it in a constant temperature of e.g. 100 kelvin?" Tags="&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10253" PostTypeId="2" ParentId="10252" CreationDate="2014-04-25T21:37:18.447" Score="6" Body="&lt;p&gt;The changing color of $\ce{NO_2}$ is because it exits in equilibrium with $\ce{N_2O_4}$ as shown in the following equation $$\ce{2 NO_2 &amp;lt;=&amp;gt; N_2O_4} + \mathrm{58~ kJ/m}$$ $\ce{NO_2}$ is a reddish-brown colored gas, while the $\ce{N_2O_4}$ is colorless.  Considering Le Chatelier's principle, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking heat out of the equilibrium will push it to the right&lt;/li&gt;&#xA;&lt;li&gt;Putting heat in, will push it to the left&lt;/li&gt;&#xA;&lt;li&gt;Increasing pressure will push it to the right&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As we shift the equilibrium we change the concentration of the components, and consequently the color of the gaseous mixture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to your question&lt;/strong&gt;, in the solid state (which is where it would be at 100°K), the equilibrium is shifted completely to the right.  So if we had a tube of gaseous $\ce{NO_2}$ and $\ce{N_2O_4}$ at room temperature it would have a distinct red-brown color.  If we then plunged it into a bath at 100°K, it would solidify to $\ce{N_2O_4}$ and appear colorless.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-19T01:09:32.230" LastActivityDate="2015-03-19T01:09:32.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10254" PostTypeId="2" ParentId="10250" CreationDate="2014-04-25T22:42:52.257" Score="7" Body="&lt;p&gt;HPLC-MS (high performance liquid chromatography- mass spectrometry), the latter with a suitable database for known small molecule plant components (alkaloids, terpenes, THC, etc.) would do it.  Cost would be three to low four figures/shot.  Saccharides and proteins are different and more expensive problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what you are asking.  Mere pokeberry contains astragalin, canthomicrol, caryophyllene, esculentic acid, jaligonic acid, lectins, oleanolic acid, 3-oxo-30-carbomethoxy-23-norolean-12-en-28-oic acid, 6-palmityl-Δ7-stigmasterol-Δ-D-glucoside, 6-palmytityl-α-spinasteryl-6-D-glucoside, phytolaccagenic acid, phytolaccagenin, phytolaccatoxin, phytolaccosides A,B,C,D,E,F,G (esculentoside E), pokeweed mitogen, tannin, triterpene alcohols, α-spinasterol, and α-spinasteryl-β-D-glucoside and a whole bunch more.  One injection is hardly a beginning.  If you do not know how to cook it, it kills you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uU6BW.jpg&quot; alt=&quot;Pokeberry&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-25T22:42:52.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10255" PostTypeId="2" ParentId="10246" CreationDate="2014-04-25T22:47:28.277" Score="4" Body="&lt;p&gt;That looks like bilirubin, from hemoglobin catabolized in the liver  Monomer, dimer..tetramer...oligomer...polymer...high polymer.  It is a tetrapyrrole, not even an oligomer.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-25T22:47:28.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10256" PostTypeId="2" ParentId="10249" CreationDate="2014-04-26T06:15:09.197" Score="-4" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$-\log[\ce{H}]=\text{p(H)}$&lt;/span&gt; you can calculate concentration of &lt;span class=&quot;math-container&quot;&gt;$[\ce{H}]$&lt;/span&gt; for each acid, then add the &lt;span class=&quot;math-container&quot;&gt;$[\ce{H}]$&lt;/span&gt; for two acids and calculate &lt;span class=&quot;math-container&quot;&gt;$\text{p(H)}$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$-\log[\ce{H}]$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5097" LastEditorUserId="4945" LastEditDate="2019-09-16T11:38:14.960" LastActivityDate="2019-09-16T11:38:14.960" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10257" PostTypeId="1" AcceptedAnswerId="10263" CreationDate="2014-04-26T07:23:42.493" Score="0" ViewCount="67" Body="&lt;p&gt;I am trying to simulate a chain of molecules which are planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please give me an example of the molecular chain which can stay as planar with it's minimum energy state is planar. A published study on the evolution of the geometry of the molecule will be highly helpful.&lt;/p&gt;&#xA;" OwnerUserId="5307" LastEditorUserId="4734" LastEditDate="2014-04-26T21:17:51.737" LastActivityDate="2014-04-26T21:17:51.737" Title="Modelling planar molecules and their evolutions" Tags="&lt;polymers&gt;&lt;structural-biology&gt;&lt;geochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10259" PostTypeId="1" CreationDate="2014-04-26T12:10:53.890" Score="3" ViewCount="7175" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which is more stable in aqueous solution: $\ce{Cr^3+}$ or $\ce{Mn^3+}$? Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My approach: $\ce{Cr^{3+}}$ should be more stable as the $\ce{3d}$ electrons will enter the $\mathrm{t_{2g}}$ orbitals. Due to CFSE (Crystal Field Stabilization Energy) of half filled $\mathrm{t_{2g}}$ orbitals $\ce{Cr^{3+}}$ should be most stable.&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="41328" LastEditDate="2017-10-05T13:50:16.440" LastActivityDate="2017-10-05T13:50:16.440" Title="Stability of transition metals in aqueous solution" Tags="&lt;aqueous-solution&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;&lt;stability&gt;&lt;crystal-field-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10260" PostTypeId="2" ParentId="7427" CreationDate="2014-04-26T12:27:07.027" Score="-2" Body="&lt;p&gt;Ionic bonds are stronger since they're between 2 charged particles and its because of the electrostatic force of attraction between the oppositively charged ions. They have higher melting and boiling points and require more energy than the covalent bonds break. Covalent bonds are as a result of Adams sharing electrons.  They can be easily broken into its primary structure as the atoms are close by to share the electron. &lt;/p&gt;&#xA;" OwnerUserId="5299" LastActivityDate="2014-04-26T12:27:07.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10261" PostTypeId="2" ParentId="10232" CreationDate="2014-04-26T12:41:38.533" Score="1" Body="&lt;p&gt;Okay, so your understanding is correct about rechargeable batteries after taking electrochemistry in class.However, electrons migrate from anode to cathode . The reaction at the anode produces electrons.&lt;/p&gt;&#xA;" OwnerUserId="5299" LastActivityDate="2014-04-26T12:41:38.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10262" PostTypeId="2" ParentId="8395" CreationDate="2014-04-26T12:49:32.580" Score="1" Body="&lt;p&gt;An electrochemical cell which causes external electric current flow can be created using any two different metals since metals differ in their tendency to lose electrons. Zinc more readily loses electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow through an external wire which leads from the zinc to the copper.As a zinc atom provides the electrons, it becomes a positive ion and goes into aqueous solution, decreasing the mass of the zinc electrode. On the copper side, the two electrons received allow it to convert a copper ion from solution into an uncharged copper atom which deposits on the copper electrode, increasing its mass.&#xA;Here the voltage of the battery means the potential difference between the two half cell, this is how the enode values are different . The bigger the difference in the enode value( its ability to lose or gain electrons) the bigger the voltage of the dry cell will be&lt;/p&gt;&#xA;" OwnerUserId="5299" LastActivityDate="2014-04-26T12:49:32.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10263" PostTypeId="2" ParentId="10257" CreationDate="2014-04-26T14:52:00.380" Score="1" Body="&lt;p&gt;One-dimensional Fabre and Bechgaard salts, TTF-TCNQ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4gOR2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.tamu.edu/rgroup/dunbar/Publications/2002-147.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chem.tamu.edu/rgroup/dunbar/Publications/2002-147.pdf&lt;/a&gt; &lt;BR&gt;&#xA;platinum blues&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI:10.1002/asia.200700370 &lt;BR&gt;&#xA;hexaphenylbenzene discotic liquid crystals &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.crcpress.com/product/isbn/9781439811436&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crcpress.com/product/isbn/9781439811436&lt;/a&gt; &lt;BR&gt;&#xA;&quot;Chemistry of Discotic Liquid Crystals: From Monomers to Polymers,&quot; Sandeep Kumar (2010)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.springer.com/chapter/10.1007/978-94-017-2319-0_16&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/chapter/10.1007/978-94-017-2319-0_16&lt;/a&gt; &lt;BR&gt;&#xA;middle &lt;BR&gt;&#xA;&lt;a href=&quot;http://physics.aps.org/synopsis-for/10.1103/PhysRevB.82.180401&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://physics.aps.org/synopsis-for/10.1103/PhysRevB.82.180401&lt;/a&gt; &lt;BR&gt;&#xA;  Organic magnets &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/cen/topstory/8201/8201notw8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/cen/topstory/8201/8201notw8.html&lt;/a&gt; &lt;BR&gt;&#xA;Ladderanes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qQILI.gif&quot; alt=&quot;Ladderanes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stacks &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.google.com/patents/WO2011017711A2?cl=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.google.com/patents/WO2011017711A2?cl=en&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.halcyon.com/nanojbl/Images/ConPoly3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.halcyon.com/nanojbl/Images/ConPoly3.gif&lt;/a&gt; &lt;BR&gt;&#xA;and a central hexacoordinate silicon with oxygen bridges&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-26T14:52:00.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10264" PostTypeId="1" AcceptedAnswerId="10265" CreationDate="2014-04-26T15:42:12.913" Score="0" ViewCount="880" Body="&lt;p&gt;I need to calculate the heat of combustion per mole and per kilogram of complete combustion of, say, X.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I write reaction : $\ce{X + O2 -&amp;gt; ... + Q}$&lt;/li&gt;&#xA;&lt;li&gt;Find Q using Hess law&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, what do I do after that?&lt;/p&gt;&#xA;" OwnerUserId="5310" LastEditorUserId="73" LastEditDate="2014-04-26T19:12:10.490" LastActivityDate="2014-04-26T19:12:10.490" Title="Heat of combustion per mole and per kilogram" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10265" PostTypeId="2" ParentId="10264" CreationDate="2014-04-26T15:58:35.920" Score="1" Body="&lt;p&gt;Q is the heat of combustion per mole. Next, you would use the formula weight to determine the heat of combustion per g (or kg).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-26T15:58:35.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10266" PostTypeId="2" ParentId="10235" CreationDate="2014-04-26T17:43:26.283" Score="1" Body="&lt;p&gt;Alcohol vapors can certainly form explosive mixtures with air, so filling the car with ethanol vapors is potentially dangerous. A source of ignition, for example a spark, could under the right circumstances ignite the vapors. If the fan motor of the AC system is a brushed motor it can create small sparks. So I would say there is a fire hazard.&lt;br/&gt; &#xA;But there are other problems with your suggested setup. The ethanol and water vapors would condense on every surface of your car's interior, the windows, the roof, the doors, etc. This would mean you loose a lot of alcohol, so you would need more. Ethanol might attack and start to dissolve some of the interior, glued parts may become loose. And of course the car would smell very strongly like vodka for a long time.&lt;br/&gt;&#xA;So I would suggest disassembling the dash cluster and whatever else you need to get to the AC system, and using the vodka in liquid form to clean it properly.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2014-04-26T17:43:26.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10267" PostTypeId="1" AcceptedAnswerId="13940" CreationDate="2014-04-26T19:13:38.280" Score="9" ViewCount="1749" Body="&lt;p&gt;What is the reason why oxalic acid has its second $\ce{-COOH}$ group flipped? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing it's something to do with steric repulsions between the $\ce{-COOH}$ groups but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="4511" LastEditorDisplayName="user10259" LastEditDate="2015-05-20T01:34:01.843" LastActivityDate="2015-05-20T01:34:01.843" Title="Oxalic acid structure" Tags="&lt;acid-base&gt;&lt;structural-formula&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10268" PostTypeId="2" ParentId="10267" CreationDate="2014-04-26T19:37:12.473" Score="0" Body="&lt;p&gt;The carbonyls' dipole moments minimize energy by pointing in opposite directions.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-26T19:37:12.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10269" PostTypeId="2" ParentId="10267" CreationDate="2014-04-26T20:44:06.830" Score="3" Body="&lt;p&gt;More structure stabilizing hydrogen bonds, both intra- and intermolecular, can be formed in the conformation where the carbonyl groups are anti-periplanar.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-04-26T20:44:06.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10271" PostTypeId="2" ParentId="10259" CreationDate="2014-04-26T22:59:53.367" Score="1" Body="&lt;p&gt;Cr(III) is common as dirt and stable in solution as any number of different salts.  Mn(III) is exotic, labile, and chemically reactive,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hach.com/asset-get.download.jsa?id=7639984472&quot; rel=&quot;nofollow&quot;&gt;http://www.hach.com/asset-get.download.jsa?id=7639984472&lt;/a&gt;‎ &lt;BR&gt;&#xA;DOI: 10.1039/AN9911600285 &lt;BR&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese%28III%29_acetate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Manganese(III)_acetate&lt;/a&gt; &lt;BR&gt;&#xA;DOI:10.1039/C3RA42480D &lt;BR&gt;&#xA;interesting free radical one-electron oxidation reagent&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-26T22:59:53.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10273" PostTypeId="1" CreationDate="2014-04-26T23:09:19.210" Score="8" ViewCount="11320" Body="&lt;p&gt;What do hardness and softness mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it something to do with the polarisability. &lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="9961" LastEditDate="2015-10-03T00:39:30.400" LastActivityDate="2017-01-28T17:38:14.463" Title="Hardness and softness are terms used to describe which aspect of acidic behaviour?" Tags="&lt;hsab&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10274" PostTypeId="2" ParentId="8365" CreationDate="2014-04-26T23:25:27.200" Score="3" Body="&lt;p&gt;Thermodynamics proposes, kinetics disposes.  The most stable form can have a high activation energy barrier surrounding it.  The medium matters.  Hard Lewis bases (water) stabilize high oxidation states (Cu(II), tungstate).  Soft Lewis bases (acetonitrile) stabilize low oxidation states (Cu(II); $\ce{W(CO)6}$ and &lt;a href=&quot;http://www.google.com/patents/EP1995347A1?cl=en&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/EP1995347A1?cl=en&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ti^{+4}}$ is the basement, especially with oxygen.  Now rationlalize titanium suboxide Magnéli phases.  They are spectacular as inert electrodes under extreme conditions, re Ebonex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI:10.1016/j.electacta.2010.05.011 &lt;BR&gt;&#xA;&lt;a href=&quot;http://eprints.soton.ac.uk/cgi/request_doc?eprintid=152857&amp;amp;_action_null=Request+a+copy&quot; rel=&quot;nofollow&quot;&gt;http://eprints.soton.ac.uk/cgi/request_doc?eprintid=152857&amp;amp;_action_null=Request+a+copy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-26T23:25:27.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10276" PostTypeId="2" ParentId="2776" CreationDate="2014-04-26T23:52:03.390" Score="1" Body="&lt;p&gt;If you need good control, put a spring-loaded or cantilever-loaded valve at the tip.  Rupture diaphragms typically have a needle point at the far end. The bubble deforms with pressure, contacting the point that then ruptures the diaphragm that is under tension.  It is much easier to control and reproduce deformation than it is to engineer catastrophic failure at maximum deformation.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-26T23:52:03.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10277" PostTypeId="2" ParentId="8424" CreationDate="2014-04-27T00:00:34.203" Score="0" Body="&lt;p&gt;The allyl anion is delocalized and planar.  You have a mirror plane through its middle carbon normal to the plane of the ring, or a ${C_2}$ axis through that carbon depending on how the other two carbons are canted.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-27T00:00:34.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10280" PostTypeId="1" CreationDate="2014-04-27T02:37:15.990" Score="0" ViewCount="52" Body="&lt;p&gt;This is the next step of my previous &lt;a href=&quot;https://chemistry.stackexchange.com/questions/10257/modelling-planar-molecules-and-their-evolutions&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to simulate a circular chain (ring) of molecules which are planar (two dimensional). What I do is take a planar unknot and embed it into a two dimensional lattice. Then I calculate energy of the unknot and try to evolve the unknot so that it reaches the canonical form. By definition it then has the minimum energy. To have a realistic definition of the energy I need to study real life planar unknots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please give me an example of the molecular chain which can stay as planar with it's minimum energy state is planar. A published study on the evolution of the geometry of the molecule will be highly helpful.&lt;/p&gt;&#xA;" OwnerUserId="5307" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-05-01T17:04:26.620" Title="Modelling planar ring molecules and their evolutions" Tags="&lt;polymers&gt;&lt;structural-biology&gt;&lt;geometrical-isomerism&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-01-23T06:49:55.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="10283" PostTypeId="1" CreationDate="2014-04-27T03:57:42.767" Score="2" ViewCount="1023" Body="&lt;p&gt;I am attempting to adapt a hydrothermal method for synthesizing $\ce{PbS}$ nanocrystals to my school that contains a mixture of cationic and anionic surfactants sodium dodecyl sulfate (.05 mmol, SDS) and cetyltrimethylammonium bromide (.05 mmol, CTAB) and ethylenediaminetetraacetic acid (.2 mmol, EDTA) as a chelating agent that diminishes the reactivity of lead ions in solution. The lead source within the method is Lead II Acetate and the sulfur source is an organosulfur compound with the formula $\ce{SC(NH2)2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The company that we bought our chemicals from sent us Lead subacetate and while I wait for them to send us the right one back I am wondering if using the appropriate ratio of moles of Lead subacetate which has 3 moles of lead per 1 mole of lead subacetate to the documented lead II acetate which contains 1 mole of lead would lead to lead sulfide $\ce{PbS}$ nanocrystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of subacetate contains two crystalline basic acetates containing hydroxide ions in addition to acetate ions. Does the subacetate create a conjugate acid that lowers the pH? &lt;/p&gt;&#xA;" OwnerUserId="4505" LastEditorUserId="402" LastEditDate="2017-04-01T01:48:38.087" LastActivityDate="2017-04-01T01:48:38.087" Title="What is the difference between the reactivity of Lead subacetate and Lead II acetate undergoing a hydrothermal process?" Tags="&lt;reaction-mechanism&gt;&lt;experimental-chemistry&gt;&lt;organometallic-compounds&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10285" PostTypeId="1" CreationDate="2014-04-27T05:31:51.677" Score="6" ViewCount="204" Body="&lt;p&gt;Suppose we have some fermented (optionally also distilled) fruit/fruit juice. The sample therefore contains water, sugars, ethanol etc. Which analysis method is the simplest to find out whether the toxic methanol is also present?&lt;/p&gt;&#xA;" OwnerUserId="5320" LastEditorUserId="4945" LastEditDate="2014-05-08T08:29:22.657" LastActivityDate="2016-03-01T22:39:22.747" Title="How to find trace methanol content in fermented fruit/distillate?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10286" PostTypeId="2" ParentId="10285" CreationDate="2014-04-27T08:37:27.040" Score="4" Body="&lt;p&gt;Methanol, formed by degradation of pectines from the skin of fruits, as well as higher alcohols is volatile and does not show significant absorption in the UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, gas chromatography is the method of choice to determine these components in the fermentation mix.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-04-27T08:37:27.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10288" PostTypeId="1" CreationDate="2014-04-27T08:57:28.230" Score="3" ViewCount="4284" Body="&lt;p&gt;A boron trihalide (&lt;span class=&quot;math-container&quot;&gt;$\ce {BX3}$&lt;/span&gt;) is a Lewis acid, because it can accept an electron pair, as per Gilbert N. Lewis' definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Arrange &lt;span class=&quot;math-container&quot;&gt;$\ce {BF3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce {BCl3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce {BBr3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {BI3}$&lt;/span&gt; from the weakest to the strongest acid properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="735" LastEditorUserId="83758" LastEditDate="2020-05-18T12:10:41.200" LastActivityDate="2020-05-18T12:10:41.200" Title="Boron trihalides acidity" Tags="&lt;acid-base&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10290" PostTypeId="1" CreationDate="2014-04-27T10:02:03.483" Score="1" ViewCount="270" Body="&lt;p&gt;If a ligand K[Ag(CN)&lt;sub&gt;2&lt;/sub&gt;] is dissociated.. Which one is correct dissociation equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K[Ag(CN)2] -&amp;gt; K+ + Ag+ + 2CN^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K[Ag(CN)2] -&amp;gt; K+ + [Ag(CN)2]^-}$$&lt;/p&gt;&#xA;" OwnerUserId="5324" LastEditorUserId="2328" LastEditDate="2014-04-27T12:11:09.083" LastActivityDate="2014-04-27T12:11:09.083" Title="Dissociation of ligand" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10291" PostTypeId="2" ParentId="10290" CreationDate="2014-04-27T10:33:31.240" Score="2" Body="&lt;p&gt;Three different types of dissociation can occur here, the two mentioned by the OP:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K[Ag(CN)2] &amp;lt;=&amp;gt; K+ + Ag+ +2CN-}$$&#xA;$$\ce{K[Ag(CN)2] &amp;lt;=&amp;gt; K+ + Ag(CN)2^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one where only one cyanide ligand dissociates from the silver:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K[Ag(CN)2] &amp;lt;=&amp;gt; K+ + AgCN + CN-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as Klaus points out in his comment dissociation $\ne$ ionization. Ionization is a chemical reaction that generates ions where there were none before. Consider the ionization reaction of ammonia $\ce{NH3}$ in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH3 + H2O &amp;lt;=&amp;gt; NH4+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this type of reaction is possible, although the cyanide anion is the conjugate base of a weak acid $(\ce{HCN}, \ \text{pK}_a =9.21)$. Thus, any liberated cyanide ion could react with water (assuming this is an aqueous solution!), though I would hardly call this an ionization (as there is no net increase in ions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CN- + H2O &amp;lt;=&amp;gt; HCN + OH-}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-27T10:33:31.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10292" PostTypeId="2" ParentId="10238" CreationDate="2014-04-27T11:41:27.547" Score="1" Body="&lt;p&gt;To calculate work at constant pressure, you need the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=-P \Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $w$ is work, $P$ is pressure and $\Delta V$ is the change in volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to solve some gas laws relations to get both $P$ and $\Delta V$ in terms of the single variable $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Algebraically, it will simplify to $P\Delta V = nR\Delta T$, assuming ideality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to assume a non-ideal gas (which is a good assumption for water at high pressure). If we use the van der Waals equation, things get dicey and calculus is needed. I'm hoping you don't need calculus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d[(P-\frac{n^2a}{V^2})(V-nb)]=d(nRT)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d(PV-\frac{n^2a}{V}-nbP+\frac{n^3ab}{V^2})=d(nRT)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PdV+VdP+\frac{n^2a}{V^2}dV-nbdP-\frac{2n^3ab}{V^3}dV=nRdT+RTdn$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $n$ and $P$ are constant $dP=0$ and $dn=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PdV+\frac{n^2a}{V^2}dV-\frac{2n^3ab}{V^3}dV=nRdT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{V_i}^{V_f}(P+\frac{n^2a}{V^2}-\frac{2n^3ab}{V^3})dV=\int_{T_i}^{T_f}nRdT$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-27T11:41:27.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10293" PostTypeId="1" AcceptedAnswerId="10296" CreationDate="2014-04-27T11:59:58.583" Score="1" ViewCount="71" Body="&lt;p&gt;I'm looking for a solvate that changes its solvation property when being dried.&#xA;One example of such a material is gypsum:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at temperatures up to 250°C it becomes γ-anhydrite which will slowly&#xA;hydrate to a hemihydrate. &lt;/li&gt;&#xA;&lt;li&gt;at temperatures above 250°C it becomes&#xA;β-anhydrite which will not hydrate again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However the temperature range is too high to use it as example in my thesis.&lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2014-08-25T17:02:51.627" Title="Searching for a solvate that changes its solvation property" Tags="&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10294" PostTypeId="2" ParentId="10245" CreationDate="2014-04-27T13:29:58.537" Score="1" Body="&lt;p&gt;It is about the definitions, but I suspect you are overcomplicating in thinking it isn't obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mole is an Avogadro number (6.022 x 10&lt;sup&gt;23&lt;/sup&gt;) of things (the &lt;em&gt;things&lt;/em&gt; could be anything but being a big number it is only a useful metric for atoms and molecules). Given this definition, if a molecule weighs the equivalent of a carbon atom then a mole of it will be 12g. A mole of hydrogen gas weighs about (about because it might have some deuterium) 2g as hydrogen gas consists of H&lt;sub&gt;2&lt;/sub&gt; molecules. &lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-04-27T13:29:58.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10295" PostTypeId="2" ParentId="10158" CreationDate="2014-04-27T14:37:35.203" Score="0" Body="&lt;p&gt;The nitrogen is trigonal planer with the lone pair orbitals aligned with the pi-bond orbitals on the 5-member ring. Also the benzene ring needs to be planer with the 5-member ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sx5Ru.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="14" LastEditDate="2014-05-27T18:09:39.000" LastActivityDate="2014-05-27T18:09:39.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10296" PostTypeId="2" ParentId="10293" CreationDate="2014-04-27T14:52:10.860" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/jres/086/jresv86n2p181_A1b.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nvlpubs.nist.gov/nistpubs/jres/086/jresv86n2p181_A1b.pdf&lt;/a&gt; &#xA;&lt;a href=&quot;http://www.phasediagram.dk/binary/magnesium_sulfate.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.phasediagram.dk/binary/magnesium_sulfate.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium sulfate hydrates,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://iopscience.iop.org/0022-3727/41/21/212002/fulltext/&quot; rel=&quot;nofollow&quot;&gt;http://iopscience.iop.org/0022-3727/41/21/212002/fulltext/&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.springerimages.com/Images/Chemistry/1-10.1007_s00269-008-0256-0-0&quot; rel=&quot;nofollow&quot;&gt;http://www.springerimages.com/Images/Chemistry/1-10.1007_s00269-008-0256-0-0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI:10.1209/0295-5075/102/28003 &lt;BR&gt;&#xA;&lt;a href=&quot;http://ej.iop.org/images/0295-5075/102/2/28003/Full/epl15377f5_online.jpg&quot; rel=&quot;nofollow&quot;&gt;http://ej.iop.org/images/0295-5075/102/2/28003/Full/epl15377f5_online.jpg&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.phys.tue.nl/nfcmr/PhD-Saidov-2012.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.tue.nl/nfcmr/PhD-Saidov-2012.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;====&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://patentimages.storage.googleapis.com/US7638109B2/US07638109-20091229-D00001.png&quot; rel=&quot;nofollow&quot;&gt;http://patentimages.storage.googleapis.com/US7638109B2/US07638109-20091229-D00001.png&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.genchem.com/properties.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.genchem.com/properties.asp&lt;/a&gt; &lt;BR&gt;&#xA;sodium carbonate&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-27T14:52:10.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10297" PostTypeId="1" CreationDate="2014-04-27T16:23:21.073" Score="0" ViewCount="1442" Body="&lt;p&gt;Is calculus and differential equations really that useful in physical chemistry? How does knowing that helps me in physical chemistry? And how much of calculus must I know in order to do physical chemistry topics such as thermodynamics? Because I am currently a chem major and am wondering what maths courses would be relevant. I currently completed single variable calculus and intend to go on to multivariate calculus as well as differential equations. I wonder how useful that'd be. &lt;/p&gt;&#xA;" OwnerUserId="5032" LastEditorUserId="1342" LastEditDate="2014-04-28T01:45:11.657" LastActivityDate="2014-04-28T01:45:11.657" Title="Calculus and physical chemistry" Tags="&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-10-23T17:45:01.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="10299" PostTypeId="2" ParentId="230" CreationDate="2014-04-27T17:51:42.627" Score="-1" Body="&lt;p&gt;Bond means force of attraction. Matter is made up of molecules and and atoms. The physical force of attraction which holds atoms and molecules in a matter is called physical bond. Vander waal's forces, coulombic forces are physical forces. The bond which is made due to the behaviour of atom to become stable is called chemical bond. The physical bond is due to property of particles called mass,electric charge(positive or negative).&lt;/p&gt;&#xA;" OwnerUserId="5326" LastActivityDate="2014-04-27T17:51:42.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10300" PostTypeId="2" ParentId="10297" CreationDate="2014-04-27T20:41:17.403" Score="5" Body="&lt;p&gt;Calculus is the basis of physical chemistry. Have a look to &lt;a href=&quot;https://en.wikipedia.org/wiki/Maxwell_relations&quot; rel=&quot;noreferrer&quot;&gt;Maxwell relations&lt;/a&gt;,&lt;a href=&quot;https://en.wikipedia.org/wiki/Clapeyron%27s_equation&quot; rel=&quot;noreferrer&quot;&gt;Clapeyron equation&lt;/a&gt;. How stated Uncle Al you should take a book of physical chemistry and look into, furthermore some books (see Atkins) have an appendix showing you the mathematical skills required to understand the content of the book. Here what you should know only for thermodynamics in my opinion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;algebra &lt;/li&gt;&#xA;&lt;li&gt;logarithms&lt;/li&gt;&#xA;&lt;li&gt;vectors&lt;/li&gt;&#xA;&lt;li&gt;calculus&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Partial derivative&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;differential equations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you go deep into the atoms more mathematical skills are required.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-27T20:41:17.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10301" PostTypeId="1" AcceptedAnswerId="10302" CreationDate="2014-04-27T20:44:20.137" Score="5" ViewCount="2073" Body="&lt;p&gt;Most health-food stores carry supplements supposedly containing iodine for people with iodine deficiency. I recently bought some kelp tablets that are supposed to have iodine but as a consumer how can I be sure that there is any iodine in them at all, let alone the claimed minimum daily requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know that I'm not just taking green-colored sawdust? Is there a cheap way to test a kelp tablet for its iodine content? If not are there testing labs that I could send some kelp to for testing that isn't going to cost me thousands of dollars?&lt;/p&gt;&#xA;" OwnerUserId="5328" LastEditorUserId="16683" LastEditDate="2017-04-09T19:19:36.270" LastActivityDate="2017-04-09T19:19:36.270" Title="Easy method of analysing iodine content in supplements?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10302" PostTypeId="2" ParentId="10301" CreationDate="2014-04-27T20:53:06.157" Score="11" Body="&lt;p&gt;Take some potato starch dissolved in water (you can easily get it from the water when you boil potatoes), and put a tablet into the water. If the water becomes bluish black, there is iodine in the tablet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there are testing labs that will determine the amount of iodine in the tablets which will cost somewhere around \$100 to \$200.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="402" LastEditDate="2017-04-09T18:34:25.337" LastActivityDate="2017-04-09T18:34:25.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10303" PostTypeId="2" ParentId="10297" CreationDate="2014-04-27T21:02:31.010" Score="1" Body="&lt;p&gt;My university required calculus (2 courses) and differential equations (1 course) to complete the chemistry degree.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-27T21:02:31.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10306" PostTypeId="2" ParentId="10301" CreationDate="2014-04-28T00:50:04.497" Score="2" Body="&lt;p&gt;Kelp is exceptionally good at scavenging the Periodic Table, as in Fukushima radwaste now bathing the US West Coast.  Iodized salt takes care of your iodine requirement, and generic multivitamins, and seafood and seaweed (sushi) overall.  You would need work at iodine deficiency unless you live in the Midwest with an idiosyncratic diet.  You can buy iodide pills at a pharmacy.  If a food supplement bottle does not have USP or NF in bold letters on its label, you should not put it into your mouth - no certified assay of contents, amounts, or bioavailability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faith-based engineering is rife with tests of faith.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-28T00:50:04.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10307" PostTypeId="1" CreationDate="2014-04-28T03:00:48.337" Score="0" ViewCount="238" Body="&lt;p&gt;If -2 is a oxidation number and 2- is the charge, then what is - by itself, is it a oxidation number or charge? As in $\ce{Ag+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also do you see $\ce{Cu + 2Ag+ -&amp;gt; Cu^{2+} + 2Ag}$ as a redox reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I don't see any increase or increase in the oxidation number of any element in this equation. All I see is charges&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="418" LastEditDate="2014-04-28T03:39:45.847" LastActivityDate="2014-04-28T03:39:45.847" Title="Confused with oxidation numbers" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10308" PostTypeId="2" ParentId="10307" CreationDate="2014-04-28T03:19:24.457" Score="1" Body="&lt;p&gt;In a chemical equation, the oxidation number is not provided, so any numbers are either the number of molecules, or the charge on the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the oxidation number, it is the number of electrons removed or added to the element to get to the stated charge. For example; $Cl^-$ has an oxidation number of -1; $Ca^{2+}$ has a oxidation number of +2; and in $SO_4^{2-}$, sulfur has an oxidation number of +6.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-28T03:19:24.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10309" PostTypeId="2" ParentId="10307" CreationDate="2014-04-28T03:34:31.870" Score="3" Body="&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC gold book&lt;/a&gt;, oxidation state is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;&lt;br/&gt;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's consider your reaction $Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag$ The superscripts represent charges. Reaction equations typically don't include oxidation states. In the reactants, $Cu$ has no charge and thus an oxidation state of 0 (rule 1). $Ag^+$ has a charge of +1 and thus an oxidation state of +1 (rule 2). In the products, $Cu^{2+}$ has a charge of +2 and thus an oxidation state of +2 (rule 2). $Ag$ has no charge and thus an oxidation state of 0 (rule 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In molecules, oxidation state can be thought of the charge of an atom when all bonds are treated as ionic. For example, in $H_2O$, Both $O-H$ bonds are polar and, since $O$ is more electronegative, all bonding electrons can be considered to belong to $O$. Thus, $O$ has 2 bonding pairs and 2 lone pairs for a total of 8. $O$ has 6 protons so it's oxidation state here is $6-8=-2$. Each $H$ has no electrons and 1 proton for an oxidation state of +1. The total oxidation state of the molecule is $2(+1)+-2=0$, which matches its charge.&lt;/p&gt;&#xA;" OwnerUserId="5220" LastActivityDate="2014-04-28T03:34:31.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10312" PostTypeId="1" CreationDate="2014-04-28T05:22:58.273" Score="2" ViewCount="2155" Body="&lt;p&gt;My teacher says buta-1,2-diene is symmetric and hence optically inactive but I think it's the opposite, because allenes with an even number of double bonds are optically active because they have no plane of symmetry. Which answer is correct?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="16683" LastEditDate="2016-09-07T16:25:52.593" LastActivityDate="2016-09-07T16:25:52.593" Title="Chirality of buta-1,2-diene" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10313" PostTypeId="2" ParentId="10312" CreationDate="2014-04-28T05:29:21.963" Score="1" Body="&lt;p&gt;2,3-butadiene is not a IUPAC name. It should be 1,2-butadiene. I think you are talking about 1,3-butadiene (natural rubber). In any case, neither of the molecules have a chiral center; therefore, no optical activity.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-28T05:29:21.963" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10314" PostTypeId="2" ParentId="10245" CreationDate="2014-04-28T07:05:08.530" Score="7" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;In short terms:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumption is correct. If the definition of mole would have been based on 24 grams of carbon-12, all molecular weights would double. This is not the case and I highly doubt, that this definition will ever change.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;In long terms:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very definition specific. One should not be confused with the several types of definitions. All these definitions are based on the carbon-12 isotope. Hence the &lt;a href=&quot;http://goldbook.iupac.org/A00497.html&quot;&gt;atomic mass (constant)&lt;/a&gt; has a specific value:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One twelfth of the mass of a carbon-12 atom in its nuclear and&#xA;  electronic ground state,  $m_u = 1.660 5402 10~\times~10^{−27}~\text{kg}$. It is equal&#xA;  to the unified atomic mass unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/A00497.html&quot;&gt;unified atomic mass unit&lt;/a&gt; is therefore a derived non-SI unit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Non-SI unit of mass (equal to the atomic mass constant), defined as&#xA;  one twelfth of the mass of a carbon-12 atom in its ground state and&#xA;  used to express masses of atomic particles,  $u\approx 1.660 5402 10~\times~10^{−27}~\text{kg}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The atomic weight (i.e. &lt;a href=&quot;http://goldbook.iupac.org/R05258.html&quot;&gt;relative atomic mass&lt;/a&gt;) is a ratio (=number) and therefore has no unit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ratio of the average mass of the atom to the unified atomic mass&#xA;  unit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also based on carbon-12 is the &lt;a href=&quot;http://goldbook.iupac.org/M03980.html&quot;&gt;definition of mole&lt;/a&gt;, which you stated correctly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SI base unit for the amount of substance (symbol:  mol). The mole is&#xA;  the amount of substance of a system which contains as many elementary&#xA;  entities as there are atoms in 0.012 kilogram of carbon-12. When the&#xA;  mole is used, the elementary entities must be specified and may be&#xA;  atoms, molecules, ions, electrons, other particles, or specified&#xA;  groups of such particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Derived from this is the &lt;a href=&quot;http://goldbook.iupac.org/A00543.html&quot;&gt;Avogadro constant&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fundamental physical constant (symbols: $L$,  $N_A$) representing the&#xA;  molar number of entities:  $L = 6.022 141 79 30 \times 10^{23}~\text{mol}^{−1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Therfore one can derive several other commonly used properties, such as molecular mass and molar mass. In a biannual publication &quot;&lt;a href=&quot;http://www.iupac.org/publications/pac/series/atomicweights/&quot;&gt;Atomic weights of the elements&lt;/a&gt;&quot; are published by IUPAC. In &lt;a href=&quot;http://www.iupac.org/publications/pac/85/5/1047/&quot;&gt;Pure Appl. Chem., 2013, Vol. 85, No. 5, pp. 1047-1078&lt;/a&gt; (or &lt;a href=&quot;http://dx.doi.org/10.1351/PAC-REP-13-03-02&quot;&gt;here&lt;/a&gt;) it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;atomic mass&lt;/strong&gt;, $m_\text{a}$, of an unbound neutral atom of&#xA;  carbon-12, $m_\text{a}(\ce{{}^{12}C})$, in its nuclear and electronic&#xA;  ground states is $12~u$ exactly, where $u$ is the unified atomic mass&#xA;  unit. The &lt;strong&gt;atomic weight&lt;/strong&gt; (also called the relative atomic mass) of&#xA;  isotope $^i\text{E}$ of element $\text{E}$, symbol&#xA;  $A_\text{r}(^i\text{E})$, in material $\text{P}$ is &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$A_\text{r}(^i\text{E})&#xA;&amp;gt; =\frac{m_\text{a}(^i\text{E})_\text{P}}{\frac{1}{12}m_\text{a}(\ce{{}^{12}C})}=\frac{m_\text{a}(^i\text{E})_\text{P}}{u}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus, the atomic mass of $\ce{{}^{12}C}$ is $12~u$, and the atomic&#xA;  weight of $\ce{{}^{12}C}$ is $12~$ exactly. All other atomic weight&#xA;  values are ratios to the $\ce{{}^{12}C}$ standard value and thus are&#xA;  dimensionless numbers. The atomic weight of element $\text{E}$,&#xA;  $A_\text{r}(\text{E})$, in a material $\text{P}$ is determined from&#xA;  the relation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$A_\text{r}(\text{E})_\text{P}=\sum\left[x(^i\text{E})_\text{P}\times A_\text{r}(^i\text{E})\right]$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where $x(^i\text{E})_\text{P}$ is the mole fraction of isotope&#xA;  $^i\text{E}$ in material $\text{P}$ (also called the isotopic&#xA;  abundance).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the standard atomic mass of carbon is (given in an interval)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\text{a}}(\ce{C})=A_\text{r}(\text{C})_\text{graphite}\times u = [12.0096, 12.0116]u$, with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u=\frac{1}{12}m_\text{a}(\ce{{}^{12}C})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the atomic mass of a single molecule is always an integer multiple of $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of carbon can then be defined as&#xA;\begin{aligned}&#xA;M(\ce{C})&amp;amp;=m_{\text{a}}(\ce{C})\times L\\&#xA;&amp;amp;=[12.0096, 12.0116]\times10^{-3}~\text{kg/ mol}\\&#xA;&amp;amp;\approx 12.01~\text{g/ mol}~\text{(4 s.f.)}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole relation becomes a little bit clearer when looking at bromine. There are two important Isotopes of bromine: $\ce{^{79}Br}$ and $\ce{^{81}Br}$ (&lt;a href=&quot;http://www.webelements.com/bromine/isotopes.html&quot;&gt;ref&lt;/a&gt;). So in elemental bromine ($\ce{Br2}$) the molecules may have three different masses $m_{\text{a}}(\ce{^{79}Br2})=158u$, $m_{\text{a}}(\ce{^{81}Br2})=162u$ and $m_{\text{a}}(\ce{^{79}Br^{81}Br})=160u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard atomic mass of bromine is $m_{\text{a}}(\ce{Br})= [79.901, 79.907]u$. &#xA;Therefore $M(\ce{Br})=[79.901, 79.907]\times 10^{-3}~\text{kg/ mol} \approx 79.90~\text{g/ mol}~\text{(4 s.f.)}$ (based on the reference value).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When calculating with molar masses $M$ one always has to keep in mind, that the used standard values are based on a (global) statistic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/A00297.html&quot;&gt;amount of substance&lt;/a&gt;, $n$, Also contains definition of: number of moles&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Base quantity in the system of quantities upon which SI is based. It&#xA;  is the number of elementary entities divided by the Avogadro constant.&#xA;  Since it is proportional to the number of entities, the&#xA;  proportionality constant being the reciprocal Avogadro constant and&#xA;  the same for all substances, it has to be treated almost identically&#xA;  with the number of entities. Thus the counted elementary entities must&#xA;  always be specified. The words 'of substance' may be replaced by the&#xA;  specification of the entity, for example: amount of chlorine atoms, &#xA;  $n_\ce{Cl}$, amount of chlorine molecules,  $n_{\ce{Cl2}}$. No&#xA;  specification of the entity might lead to ambiguities [amount of&#xA;  sulfur could stand for  $n_\ce{S}$,  $n_{\ce{S8}}$, etc.], but in many&#xA;  cases the implied entity is assumed to be known: for molecular&#xA;  compounds it is usually the molecule [e.g. amount of benzene usually&#xA;  means  $n_{\ce{C6H6}}$], for ionic compounds the simplest formula unit&#xA;  [e.g. amount of sodium chloride usually means $n_{\ce{NaCl}}$] and for&#xA;  metals the atom [e.g. amount of silver usually stands for &#xA;  $n_{\ce{Ag}}$]. In some derived quantities the words 'of substance'&#xA;  are also omitted, e.g. amount concentration, amount fraction. Thus in&#xA;  many cases the name of the base quantity is shortened to amount and to&#xA;  avoid possible confusion with the general meaning of the word the&#xA;  attribute chemical is added. The chemical amount is hence the&#xA;  alternative name for amount of substance. In the field of clinical&#xA;  chemistry the words 'of substance' should not be omitted and&#xA;  abbreviations such as substance concentration (for amount of substance&#xA;  concentration) and substance fraction are in use. The quantity had no&#xA;  name prior to 1969 and was simply referred to as the number of moles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/R05271.html&quot;&gt;relative molecular mass&lt;/a&gt;,  $M_r$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ratio of the mass of a molecule to the unified atomic mass unit.&#xA;  Sometimes called the molecular weight or relative molar mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/R05270.html&quot;&gt;relative molar mass&lt;/a&gt;&#xA;Molar mass divided by  $1~\text{g/ mol} (the latter is sometimes called the standard molar mass).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-04-28T07:05:08.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10315" PostTypeId="2" ParentId="10301" CreationDate="2014-04-28T09:12:12.780" Score="3" Body="&lt;p&gt;Just for reference there are at least other two methods, not so safe and easy to perform, but that can help you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;treat with a little bit of $\ce{HNO3}$ 1 ml of your dissolved sample add 0.5 ml of $\ce{AgNO3}$ 0.5 M. If $\ce{I-}$ is present, you should see a yellow precipitate: $\ce{AgI}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;more dangerous!&lt;/strong&gt;: put in your test tube 50 mg of your substance add 1 mL of concentrated sulfuric acid. Heat the test tube. Violet vapors will appear if $\ce{I-}$ is present.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="402" LastEditDate="2017-04-09T18:33:32.633" LastActivityDate="2017-04-09T18:33:32.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10316" PostTypeId="1" AcceptedAnswerId="10319" CreationDate="2014-04-28T10:11:04.790" Score="4" ViewCount="326" Body="&lt;p&gt;In cold countries, it is a common phenomena to add salt to melt or clear snow. However, only specific salts are being commercially. So my question is that why only sodium chloride and $\ce{CaCl2}$ are used for addition to snow, to melt/clear it?&lt;/p&gt;&#xA;" OwnerUserId="5335" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T02:38:48.550" LastActivityDate="2015-06-28T02:40:10.383" Title="Addition of NaCl and CaCl2 to snow" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10317" PostTypeId="2" ParentId="10316" CreationDate="2014-04-28T11:19:41.423" Score="2" Body="&lt;p&gt;Sodium chloride is used because it is cheaper than most of the others salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium Chloride $\ce{NaCl}$ / Industrial Salt : US $65-95 / Metric Ton&lt;/p&gt;&#xA;&#xA;&lt;p&gt;potassium chloride industrial grade: US $90-130 / Metric Ton&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prices taken from &lt;a href=&quot;http://www.alibaba.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.alibaba.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T02:40:10.383" LastActivityDate="2015-06-28T02:40:10.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10318" PostTypeId="2" ParentId="6932" CreationDate="2014-04-28T12:12:36.463" Score="6" Body="&lt;p&gt;I would love to add some quantum chemical bone to the stew we are trying to cook here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6932/evaporating-the-water-of-a-solution-of-only-carbonate-ions-and-water-gives/6936#comment10183_6932&quot;&gt;comments&lt;/a&gt;. Then &lt;a href=&quot;https://chemistry.stackexchange.com/a/6936/4945&quot;&gt;Rauru&lt;/a&gt; is also right that you will have to consider several equilibria:&#xA;\begin{aligned}\ce{&#xA;(1) &amp;amp;&amp;amp; 2H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + {}^{-}OH \\&#xA;(2) &amp;amp;&amp;amp; CO3^{2-} + H2O &amp;amp;&amp;lt;=&amp;gt; HCO3- + {}^{-}OH\\&#xA;(3) &amp;amp;&amp;amp; HCO3- + H2O &amp;amp;&amp;lt;=&amp;gt; H2CO3 + {}^{-}OH &#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $(3)$ seems a little far fetched, so we will ignore it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by &lt;a href=&quot;https://chemistry.stackexchange.com/a/6937/4945&quot;&gt;G M&lt;/a&gt;. So basically it all boils down to on question: &lt;strong&gt;Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yNPo6.jpg&quot; alt=&quot;homo of carbonate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6932/evaporating-the-water-of-a-solution-of-only-carbonate-ions-and-water-gives/6936#comment10152_6932&quot;&gt;comments&lt;/a&gt;.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: &quot;What happens to sulphate if you shoot it into space?&quot; The answer was given without any further explanation and I never bothered to ask: &quot;It looses an electron.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be&#xA;$$\ce{2CO3^{2-} -&amp;gt; 2CO2- + O2^{2-}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also collision with any particle may cause electrons to be transferred.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-04-28T12:12:36.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10319" PostTypeId="2" ParentId="10316" CreationDate="2014-04-28T13:49:34.303" Score="3" Body="&lt;p&gt;Salts will lower the melting point of snow or ice.&#xA;$\ce{NaCl}$ is chosen because it is cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited: Salt with more total ions is more effective. $\ce{CaCl_{2}}$ is the one of the most available salt with 3 total ions. $\ce{MgCl_{2}}$ is another option.&#xA;Some crude $\ce{NaCl}$ produced from sea water directly contain $\ce{Ca}$ and $\ce{Mg}$ salt and can be used to melt ice.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T02:39:48.417" LastActivityDate="2015-06-28T02:39:48.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10320" PostTypeId="1" AcceptedAnswerId="71272" CreationDate="2014-04-28T14:13:02.593" Score="24" ViewCount="26543" Body="&lt;p&gt;I just had to read some general descriptions of sodium chloride and it was always classified as odourless (e.g., by &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chloride&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). However, large amounts of table salts (around 1 kg) have a clearly perceptible odour to me. By &lt;em&gt;table salt&lt;/em&gt; I mean salt sold for the purpose of cooking – I have not been taking notes or making experiments, but as I had the experience on dozens of different occasions, I would guess that the salt does not need to be iodised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to describe the smell, I would first say &lt;em&gt;salty.&lt;/em&gt; It’s somewhat similar to the smell of the seaside, but the latter is much richer and has additional components. From the ten smell categories proposed &lt;a href=&quot;//dx.doi.org/10.1371/journal.pone.0073289&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, only &lt;em&gt;sweet&lt;/em&gt; and &lt;em&gt;chemical&lt;/em&gt; fit somehow. It also has something of &lt;a href=&quot;//en.wikipedia.org/wiki/Petrichor&quot; rel=&quot;noreferrer&quot;&gt;petrichor&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the classification as &lt;em&gt;odourless&lt;/em&gt; just an approximation or do I smell additional ingredients (such as iodide) or impurities that cause table salt not to be pure sodium chloride? In the latter case: What causes the odour?&lt;/p&gt;&#xA;" OwnerUserId="4669" LastEditorUserId="4669" LastEditDate="2017-03-26T08:03:55.283" LastActivityDate="2018-11-09T15:51:45.260" Title="Is sodium chloride really odourless? If yes, what do I smell?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10323" PostTypeId="1" CreationDate="2014-04-28T16:21:36.333" Score="3" ViewCount="677" Body="&lt;p&gt;I have a graph of a titration of a weak polyprotic acid with a strong base.&lt;/p&gt;&#xA;&lt;p&gt;I graphed &lt;span class=&quot;math-container&quot;&gt;$\mathrm {pH}$&lt;/span&gt; versus amount of strong base added.&lt;/p&gt;&#xA;&lt;p&gt;Now since the acid is polyprotic, how do I determine which equivalence point is the final one?&lt;/p&gt;&#xA;&lt;p&gt;Also not sure how to find the initial concentration of the weak acid? I know the volume I put in, as well as the volume of strong base at any given time.&lt;/p&gt;&#xA;" OwnerUserId="5336" LastEditorUserId="95133" LastEditDate="2021-05-25T12:47:53.770" LastActivityDate="2021-05-25T12:55:13.640" Title="Finding the equivalent point of a weak polyprotic acid when reacted with a strong base" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10324" PostTypeId="2" ParentId="10312" CreationDate="2014-04-28T16:38:18.973" Score="9" Body="&lt;p&gt;Let's see if a picture can help.  Newman projections can be handy in analyzing the stereochemistry of allenes.  On the left is a Newman projection of methyl allene (buta-1,2-diene).  There is a sigma plane of symmetry that is perpendicular to the screen and contains the 3 allenic carbons plus the methyl group.  Any molecule with a plane of symmetry is achiral.  On the other hand, look at the Newman projection to the right, 1,3-dimethyallene (penta-2,3-diene).  There is no plane symmetry in this case, only a $\ce{C_2}$ axis.  Draw the mirror image of the pictured allene and you will find that it is not superimposable on its mirror image. 1,3-dimethylallene is a chiral molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Azv86.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-04-28T20:58:57.807" LastActivityDate="2014-04-28T20:58:57.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10326" PostTypeId="2" ParentId="10288" CreationDate="2014-04-28T18:34:00.377" Score="5" Body="&lt;p&gt;The following order is experimentally found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{BF3}&amp;lt;\ce{BCl3}&amp;lt;\ce{BBr3}&amp;lt;\ce{BI3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stands in contrast to what is expected when the electronegativity of the halides is considered. A more electronegative halide should be able to stabilize the negative charge at &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; in the Lewis acid-base complex better, and this would suggest that &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt; would be the strongest Lewis acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the nature of the &lt;span class=&quot;math-container&quot;&gt;$\ce{B-X}$&lt;/span&gt; bond needs to be taken into account. The change of hybridization from &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{BX3}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{[BX4]-}$&lt;/span&gt; involves the loss of partial &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonding which results from the overlap of the &lt;span class=&quot;math-container&quot;&gt;$\ce{p_z}$&lt;/span&gt; orbitals of &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\ce{p_z}$&lt;/span&gt; orbitals of &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; are of similar size, because both elements are members of the same period. Therefore, they have the best overlap and form the strongest &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond. Because the reaction with a Lewis base would require some energy to break this bond, &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt; is a comparably weak Lewis acid. The larger &lt;span class=&quot;math-container&quot;&gt;$\ce{p_z}$&lt;/span&gt; orbitals of the heavier halides have increasingly less overlap with the smaller boron &lt;span class=&quot;math-container&quot;&gt;$\ce{p_z}$&lt;/span&gt; orbital. Therefore, the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonding is weaker and formation of the tetrahedral Lewis acid-base adduct via rehybridization becomes energetically more favorable.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="83758" LastEditDate="2020-05-18T09:11:30.297" LastActivityDate="2020-05-18T09:11:30.297" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10327" PostTypeId="2" ParentId="10320" CreationDate="2014-04-28T19:02:59.150" Score="13" Body="&lt;p&gt;Since odour is generally coming from vapour / gas form of a material, and NaCl has zero vapour pressure, it should be odourless.&#xA;However: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Moving around solid materials, dust, small particles can fly around that can be dissolved in the nose and interact with the odour sensors. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You almost always have something else together with salt. And if it is not pure, you cannot be sure what makes the odour, the salt or the impurity. Eg. sea salt often contains remains of algae. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2014-04-29T15:52:56.037" LastActivityDate="2014-04-29T15:52:56.037" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10328" PostTypeId="1" CreationDate="2014-04-28T21:07:45.497" Score="1" ViewCount="175" Body="&lt;p&gt;I have heard that mass spectrometry is very time consuming: why is this?  What is the process involved in performing a mass spectrometry experiment?  What steps are the most time consuming?&lt;/p&gt;&#xA;" OwnerUserId="5338" LastActivityDate="2014-04-29T17:46:03.050" Title="What is the process involved in performing a mass spectrometry experiment?" Tags="&lt;experimental-chemistry&gt;&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10330" PostTypeId="1" CreationDate="2014-04-29T01:01:24.667" Score="1" ViewCount="43" Body="&lt;p&gt;For example consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$aA + bB \rightarrow pP + nN$$ where $a, b, p$ and $n$ are moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant is $$ \frac{[P]^p [N]^n }{[A]^a [B]^b}$$ Where does this come from? I don't understand why anyone would multiply the concentrations of the products together and then divide it by the product of the reactants together. There doesn't seem to be anything in my prior experience to justify that. And why are the exponents the number of moles? What is the physical meaning of an equilibrium constant?&lt;/p&gt;&#xA;" OwnerUserId="5340" LastActivityDate="2014-04-29T01:01:24.667" Title="Why does the equilibrium law work mathematically and what does it really mean" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2014-04-29T12:39:43.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="10331" PostTypeId="1" AcceptedAnswerId="10345" CreationDate="2014-04-29T05:24:44.267" Score="1" ViewCount="791" Body="&lt;p&gt;Water molecules are described as an oxygen atom covalently (ignoring for the monent protonization/deprotonization) bonded to two hydrogen atoms with an angle of ~104° between the hydrogen atoms (like Mickey Mouse). I can understand this with multiple water molecules (the stacking would - for want of better descriptive powers - surround the oxygen's relative negative charge.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a single water molecule retain this configuration? It would seem that in space, it should look like CO&lt;sub&gt;2&lt;/sub&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5343" LastEditorUserId="91520" LastEditDate="2020-03-31T16:11:05.563" LastActivityDate="2020-03-31T16:11:05.563" Title="Configuration of a water molecule?" Tags="&lt;water&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10332" PostTypeId="2" ParentId="10328" CreationDate="2014-04-29T05:27:17.910" Score="3" Body="&lt;p&gt;This is hard to answer, because in my experience mass spectrometry isn't that time consuming at all. The experiment itself doesn't take that long. In this experiment the sample is ionized and charged fragments are accelerated in an electrical or magnetic field. Depending on their mass/charge ratio they will end up on specific locations in the detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process around the experiment itself is dependent on your specific lab, but typically the mass spectrometer is operated by only a few people that do all the samples of the entire department. Because of that, there will often be a queue which could be several days. Additionally, not all mass spectra can be digitally identified as coming from a certain chemical, especially for larger molecules there are often various options that require human interference for the identification. In our lab this human interpretation is left to the person owning the sample, but if that part is also done by the mass spectrometer operators (I know in some labs this is the case) then I can imagine that this last step will be time-limiting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: the experiment is short, the procedure around it (dependent on your specific lab) can take quite some time.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-04-29T05:27:17.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10333" PostTypeId="1" AcceptedAnswerId="10336" CreationDate="2014-04-29T07:54:09.150" Score="2" ViewCount="736" Body="&lt;p&gt;We have a 120 ml water sample with an unknown concentration of benzene. If the bottle size is 160 ml (with 40 ml of air) and benzene has reached equilibrium, what fraction of benzene is left in the water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are only given Hcc (Henry constant - Cair/Cwater) 0.225 and Log Kow (to which I believe is needed for a later question) 2.53. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to start. We do not know mole fraction so we cannot find partial pressure using Henry's law. Also this rules out Raoult's Law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also we can't use the Ideal gas law since we do not have a lot of the variables. I'm stumped. But we can make assumptions such as T for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="5344" LastActivityDate="2014-04-29T10:31:57.477" Title="Fraction of concentration of benzene in air water" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10334" PostTypeId="1" AcceptedAnswerId="10337" CreationDate="2014-04-29T08:03:19.383" Score="8" ViewCount="298" Body="&lt;p&gt;We are always told that to excite an electron from one state to a higher energy states, for example from the valance band to the conduction band, the energy must &lt;em&gt;equal&lt;/em&gt; the energy difference between the two energy states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conduction band consists of many allowed energy states, separated from the valance band by an energy gap. For the element Si, this energy gap is (at room temperature) about 1,1 eV. The range of energies for visible light is 1.8 eV to 3.1 eV. Since Si indeed conducts electricity at room temperature, is it then assumed that there is an allowed energy states with &lt;em&gt;exactly&lt;/em&gt; the same energy as the energy difference between the two states? The conduction band, or the valance band for that matter, is not continuous, but discrete. Or is it sufficient that the photon energy &quot;roughly equals&quot;, or is &quot;close enough&quot;? If so, how close is close enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my question is clearly stated.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastActivityDate="2014-04-29T12:11:32.163" Title="How strict is the &quot;to excite electrons the energy must equal the energy state difference&quot; fact?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;photochemistry&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10335" PostTypeId="1" CreationDate="2014-04-29T10:21:11.237" Score="1" ViewCount="357" Body="&lt;p&gt;Salt is very commonly added to ice even at room temperature to prevent it from melting. Can anyone explain how does it really work?&lt;/p&gt;&#xA;" OwnerUserId="5335" LastEditorUserId="4945" LastEditDate="2015-10-23T17:45:37.370" LastActivityDate="2015-10-23T17:45:37.370" Title="Addition of salt to ice" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10336" PostTypeId="2" ParentId="10333" CreationDate="2014-04-29T10:31:57.477" Score="3" Body="&lt;p&gt;You don't need the original concentration or any of the variables you are missing. &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry%27s_law&quot; rel=&quot;nofollow&quot;&gt;Henry's Law&lt;/a&gt; gives you everything you need. The usual formulation of Henry's Law &lt;strong&gt;at equilibrium&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{i,s}=k_{H,i,s}c_{i,s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{air}=c_{air}*V_{air}$$&#xA;$$n_{sol}=c_{sol}*V_{sol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-04-29T10:31:57.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10337" PostTypeId="2" ParentId="10334" CreationDate="2014-04-29T10:33:59.267" Score="4" Body="&lt;p&gt;In fact there are many factors to consider, so the answer to your question depends on what type of material you are analyzing. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First of all you should know that even if the absorption level are discrete the absorption and the emission of a material are never a single value but a peak. If you consider the absorption or the emission of a gas you can assume them discrete value (even is not completely true), and so you can call them &lt;strong&gt;lines&lt;/strong&gt; (see e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman_series&quot; rel=&quot;nofollow&quot;&gt;Lyman series&lt;/a&gt;), but if you condense the matter the interaction between the atoms/molecules broad the absorption/emission range and so you have &lt;strong&gt;peaks&lt;/strong&gt;. If you are talking about electrons you are dealing with UV-VIS spectra in this case if you look at solid or a liquid UV-VIS spectra you can see that you don't have a well define peak but a continuous absorption, in this case you have &lt;strong&gt;bands&lt;/strong&gt;. So it really depends of the state of the material you are analyzing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are other factors that determine the broadening of the absorption to multiple wavelengths. Form the previous explanation you can derive that more an atom is isolated, less interaction occurs, and the absorption is more limited to determinate wavelength, because you have similar &quot;oscillators&quot;. But if you increase &lt;strong&gt;temperature&lt;/strong&gt; and &lt;strong&gt;pressure&lt;/strong&gt; you increase the probability that some atoms/molecules interact, and have different energy levels. The effect is the same: you are broadening the absorption band, respectively you will have a _temperature and a &lt;em&gt;pressure broadening&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;So it seems that if you have a single atom this absorbs at a determinate wavelength. Even in this case is not true, the explanation is not so intuitive as the previous ones because you have to understand &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;Heisenberg uncertainty principle&lt;/a&gt; but roughly it states that you can't determine precisely the energy states of an atoms. This last broadening is called &lt;em&gt;Natural broadening&lt;/em&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For your specific case you should look at the the UV-VIS spectrum, you will not see a sudden step next to the absorption band but a gently curve. This means that there are less and less atoms absorbing in this region because statistically is less probable that atoms can be perturbed so much to absorb that wavelengths but there are still chances that some atoms absorbs even at wavelength far from the mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look to wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Spectral_line&quot; rel=&quot;nofollow&quot;&gt;spectral lines page&lt;/a&gt; for more infos.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-04-29T12:11:32.163" LastActivityDate="2014-04-29T12:11:32.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10338" PostTypeId="1" AcceptedAnswerId="10360" CreationDate="2014-04-29T11:15:20.237" Score="4" ViewCount="9827" Body="&lt;p&gt;How do I find the set of quantum numbers for a specific electron in an element? For example, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the set of quantum numbers for the 19th electron in chromium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1. From here on, how do I decide what would the principle quantum number be? &lt;/p&gt;&#xA;" OwnerUserId="5345" LastEditorUserId="73" LastEditDate="2014-04-29T23:16:26.610" LastActivityDate="2014-04-30T04:10:19.003" Title="What is the appropriate procedure for determining the quantum numbers of an electron?" Tags="&lt;electronic-configuration&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10339" PostTypeId="1" AcceptedAnswerId="10340" CreationDate="2014-04-29T12:30:49.740" Score="21" ViewCount="29743" Body="&lt;p&gt;I kind of get the idea of singlet and triplet states. But why are they called singlet and triplet (what is the single and what is the triple in these cases)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that I am missing something obvious!&lt;/p&gt;&#xA;" OwnerUserId="5274" LastActivityDate="2016-12-12T13:29:04.723" Title="Why is a singlet state called singlet and a triplet state called triplet?" Tags="&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10340" PostTypeId="2" ParentId="10339" CreationDate="2014-04-29T13:56:25.537" Score="17" Body="&lt;p&gt;The terms arose back in the early days of quantum physics when spectral lines that were expected to be singlets were actually observed to be more complex (doublets, triplets, etc.).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron can have a spin quantum number of $+\frac12$ or $-\frac12$.  For a system that exists as a singlet, all spins are paired and the total spin for the system is $S=0$.  If we have a single electron, $S=\frac12$.  If we have a triplet system with 2 unpaired electrons, $S=1$.  Also associated with these electrons/ systems is a spin-angular momentum vector, $L$.  Quantum mechanics tells us that $L$ is allowed to have $2S+1$ distinct values.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a singlet, $L = 2(0) + 1 = 1$     only one state exists (singlet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a doublet, $L = 2\left(\frac12\right) + 1 = 2$   two states exist (doublet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a triplet, $L = 2(1) + 1 = 3$     three states exist (triplet)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-04-29T16:33:48.993" LastActivityDate="2014-04-29T16:33:48.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10342" PostTypeId="2" ParentId="10335" CreationDate="2014-04-29T14:38:17.963" Score="1" Body="&lt;p&gt;Colligative properties, then freezing point depression.  The melting point of ice is lowered -1.86 C/molal of solute particles.  One molal is one mole of particles/kg of solvent (NOT solution).  Addition of salt lowers the melting point of water, causing the ice to melt, absorbing the latent enthalpy of fusion, until the system temp lowers to its equilibrium value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want your ice to stick around, thermally insulate it from the surrounding environment.  You must interdict conduction, radiation, and convection.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-04-29T16:33:34.777" LastActivityDate="2014-04-29T16:33:34.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10343" PostTypeId="2" ParentId="6450" CreationDate="2014-04-29T14:49:14.230" Score="4" Body="&lt;p&gt;No water!  $\ce{SbF5}$ is a tremendously strong Lewis acid: high oxidation state, fluorine electron withdrawal, and terrible mismatch of pi-orbital size inhibiting electron back donation.  Solvo-cation, solvo-anion.  $\ce{F^-}$ is a strong Lewis base.  $\ce{SbF6^-}$, given its large size, symmetry, charge delocalization, and lack of polarizabilty (those hard fluorines) is an exceptionally weak base and weak nucleophile.  The proton in $\ce{HSbF6}$ is naked and exceptionally acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is anhydrous, so there is no leveling effect making hydronium the most acidic  solvo-cation.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="9961" LastEditDate="2015-02-20T19:41:49.230" LastActivityDate="2015-02-20T19:41:49.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10345" PostTypeId="2" ParentId="10331" CreationDate="2014-04-29T15:03:48.363" Score="0" Body="&lt;p&gt;DOI:10.1063/1.1696094 &lt;BR&gt;&#xA;   You want the electron diffraction structure of water in vacuum.&#xA;DOI:10.1038/248405a0&#xA;   Liquid water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{H2O}$: r_g (OH) = 0.976±0.0030 Å, mean HOH angle α_g = 107.2°±3° &lt;BR&gt;&#xA;For $\ce{D2O}$: r_g (OD) = 0.970±0.0025 Å, mean DOD angle α_g = 104.2°±3°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the HOH angle slightly opens in vacuum as opposed to condensed phase.  That is 1965.  See if you can find a more recent study.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-29T15:03:48.363" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10346" PostTypeId="1" AcceptedAnswerId="10350" CreationDate="2014-04-29T15:31:18.797" Score="4" ViewCount="1421" Body="&lt;p&gt;This question was posed a couple years ago but not answered adequately. You can view the older thread &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1364/why-is-dialysis-tubing-impermeable-to-sucrose-when-it-should-retain-12-000-da/1366#1366&quot;&gt;here&lt;/a&gt;. The cellulose membrane commonly used in high school and university labs is widely available from suppliers such as Wards, &lt;a href=&quot;http://www.carolina.com/dialysis-tubing/dialysis-tubing-1-in-x-10-ft/684212.pr?&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carolina&lt;/a&gt;, Flinn, PASCO, etc. The specifications are either available online or were obtained by contacting the companies and in each case the molecular weight cut off (MWCO) was between 6,000 and 16,000 Da. Regardless of the specifications these products are used to conduct osmosis demonstrations by placing 0.2-1.0M sucrose on one side of the membrane and to create water potential. In addition to having conducted this exercise myself the results are validated in countless examples online. The mass/pressure results from osmosis and a benedicts test indicate sucrose is not moving across the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering Sucrose has a MW of ~342 Da the previous explanations seem insufficient;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The pore size of the membrane is variable. True, but according to specs at least 10% of the pores are larger than the specification and even with variation of 0-6,000 Da (worst case) it would still be like throwing a tennis ball through a garage door 10% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Interactions between solute molecules, shape, and solute-solvent causing a larger solvodynamic radius. Sucrose doesn't have any charges I'm aware of, it has a simple structure, and even covered in water molecules it would still be much smaller than the pores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions or alternative explanations?&lt;/p&gt;&#xA;" OwnerUserId="5349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-07-28T17:56:55.230" Title="Why is dialysis tubing impermeable to sucrose?" Tags="&lt;osmosis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10350" PostTypeId="2" ParentId="10346" CreationDate="2014-04-29T17:11:51.320" Score="3" Body="&lt;p&gt;Benedict's reagent detects reducing sugars (free aldehyde group) by reducing soluble blue Cu(II) to insoluble red Cu(I) oxide.  Sucrose aquoous solution is inert unless it is pre-hydrolyzed by heating with strong acid catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That certainly is a poser.  One possibility (SWAG, not fact) is regenerated crosslinked cellulose is overall not permeable to neutral solutes.  Its size selection is active toward dissociated solutes and proteins that are charged.  One can test this with anti-freeze.  The sodium fluorescene color should freely diffuse.  Tiny molecule ethylene glycol (EG) should be retained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.chemetrics.com/Glycol&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.chemetrics.com/Glycol&lt;/a&gt; &lt;BR&gt;&#xA;  Test kit and how it works:  Iodate cleavage  of 1,2-diol to aldehyde, then aldehyde detection with Purpald. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I3oFg.gif&quot; alt=&quot;Purpald aldehyde detection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could iodate cleave permeated EG (or propylene glycol, glycerin, etc.) then use Benedict's solution.  Serial dilution of standard EG solution for control and calibration.  One expects any dialysis membrane to leak somewhat.  Major passage of uncharged 1,2-dihydroxylic solutes smaller and larger than than sucrose is the SWAG's falsifying observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemsee-foods.com/antifreeze-detection&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemsee-foods.com/antifreeze-detection&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.chemworld.com/Ethylene-Glycol-Test-Kit-p/k-4815.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemworld.com/Ethylene-Glycol-Test-Kit-p/k-4815.htm&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.hach.com/glycol-visual-test-kit-model-eg-1/product?id=7640218461&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hach.com/glycol-visual-test-kit-model-eg-1/product?id=7640218461&lt;/a&gt; &lt;BR&gt;&#xA;  EG in food and fluids tests.   &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.astm.org/Standards/E202.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astm.org/Standards/E202.htm&lt;/a&gt; &lt;BR&gt;&#xA;  Testing standards&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-29T17:11:51.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10351" PostTypeId="2" ParentId="10339" CreationDate="2014-04-29T17:15:02.120" Score="10" Body="&lt;p&gt;A pair of electrons, being fermions, must have antisymmetric wave function, i.e. if $\psi(\xi_1,\xi_2)$ is a wavefunction describing the system, where $\xi_1$ are position and spin of electron 1 and $\xi_2$ is position and spin of electron 2, then $\psi(\xi_2,\xi_1)=-\psi(\xi_1,\xi_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first approximation, spin degree of freedom can be separated from orbital degrees of freedom, so that the wavefunction becomes $\chi(s_1,s_2)\phi(x_1,x_2)$, where $s_i$ is spin of $i$th electron, and $x_i$ is position of $i$th electron. Here $\chi$ is spin part of wavefunction, and $\phi$ is orbital part. To preserve total antisymmetry of the wavefunction, $\chi$ and $\phi$ can be either symmetric, or antisymmetric. If one is symmetric, the other must be antisymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spin of a single electron can be up $\uparrow$ or down $\downarrow$. I.e. simplest options for a two-electron system could be $\uparrow\uparrow$, $\downarrow\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$. But the latter two don't honour &lt;a href=&quot;http://en.wikipedia.org/wiki/Identical_particles&quot; rel=&quot;noreferrer&quot;&gt;indistinguishability&lt;/a&gt; of electrons. To correctly include indistinguishability of electrons, we should take symmetric and antisymmetric linear combinations of these spin states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have four options, split in two variants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Antisymmetric orbital and symmetric spin part of wavefunction&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\uparrow\uparrow$&lt;/li&gt;&#xA;&lt;li&gt;$\downarrow\downarrow$&lt;/li&gt;&#xA;&lt;li&gt;$\downarrow\uparrow+\uparrow\downarrow$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;b) Symmetric orbital and antisymmetric spin part&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\downarrow\uparrow-\uparrow\downarrow$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From here we can see that symmetric spin part of wavefunction gives rise to three different states — these are triplet states. If spin part of wavefunction is antisymmetric, there's only one such state — it's the singlet state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one makes spectroscopic measurements with not very high resolution, states with different spins but same orbitals will appear to have the same energies, so the spectral lines will appear the same. But if you put your system in magnetic field, you'll see that the spectral lines split according to spin multiplicities: spin-singlet states will remain single lines, while spin-triplets will split into three different spectral lines. This is the origin of such naming.&lt;/p&gt;&#xA;" OwnerUserId="2133" LastEditorUserId="2133" LastEditDate="2016-12-12T13:29:04.723" LastActivityDate="2016-12-12T13:29:04.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10353" PostTypeId="2" ParentId="10328" CreationDate="2014-04-29T17:46:03.050" Score="0" Body="&lt;p&gt;The problems are mainly in the pre-analytical steps, in most of the cases you can't analyze the sample as it is. You have do perform extraction, digestion and so on. Then if you want to have quantitative result you have to use standards. The calibration can be a quite difficult process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the time required depends from your sample and what and how you want achieve from your analysis.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-29T17:46:03.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10355" PostTypeId="2" ParentId="10335" CreationDate="2014-04-29T20:16:54.630" Score="5" Body="&lt;p&gt;Uncle Al is correct.  Colligative properties are properties of solution in which only the ratio between number of particles of solute and solvent are considered (often expressed in different forms, such as moles of solute per kilograms of solvent for molality); the identities of either are irrelevant.  So we could dissolve sodium chloride in water or magnesium bromide, but the general properties will be the same (though note that the magnitude of freezing point depression and similar properties will be slightly different depending on the substance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular for salt water, the point at which the solution will freeze is depressed.  To understand this, we must understand what occurs when pure water freezes.  In general terms, when a liquid freezes, its normally hectic and randomly moving particles form a stable, packed structure that we call a solid; a hexagonal structure in particular for water at standard conditions.  The specifics are more complicated, but this is the main idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you introduce a solute into the pure water, such as salt, the solute particles are getting in the way of the formation of the packed structure.  As a result, the temperature must be lowered slightly further in order for the solvent (i.e. water) molecules to properly overcome the solute molecules acting as obstacles for the formation of the solid structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when the solvent freezes, more often than not, the solute particles are often excluded, either pushed outside or trapped inside small pockets.  In the case of salt water, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;nofollow&quot;&gt;eutectic system&lt;/a&gt; is created, if you're interested in learning of them.  Salt water itself cannot freeze into salty-ice or otherwise.&lt;/p&gt;&#xA;" OwnerUserId="5352" LastEditorUserId="5352" LastEditDate="2014-04-29T23:43:25.917" LastActivityDate="2014-04-29T23:43:25.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10356" PostTypeId="2" ParentId="10135" CreationDate="2014-04-29T21:08:22.250" Score="3" Body="&lt;p&gt;For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-29T21:08:22.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10357" PostTypeId="2" ParentId="8308" CreationDate="2014-04-29T21:35:32.820" Score="7" Body="&lt;p&gt;All three carbon modifications indeed contain covalently bond carbon atoms. The $sp^2$-hybridized C atoms are connected by $\sigma$ and delocalized $\pi$ bonds in all cases. It is the geometric arrangement of the atoms that differentiates the modifications: Stacked planes in graphite, fullerenes: ball-shaped molecules (hence the name buckyball), and for carbon nanotubes, the name is quite self-explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single layer of the graphite structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QYQvi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\ce{C60}$ fullerene molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CaEMD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally some carbon nanotubes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BlnGl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2015-01-03T18:20:13.047" LastActivityDate="2015-01-03T18:20:13.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10358" PostTypeId="1" AcceptedAnswerId="10359" CreationDate="2014-04-29T22:52:06.763" Score="1" ViewCount="664" Body="&lt;p&gt;A common problem in shooting sports is precipitation of lead onto aluminum suppressor baffles.  (The lead presumably comes from both lead styphnate priming compound and from the base and bearing surface of lead bullets that vaporize during propulsion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lead precipitate is allowed to build up a contiguous layer it requires mechanical removal, which tends to damage and degrade the aluminum alloys used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common solution for ferrous baffles is to strip the lead with H2O2-boosted acetic acid, but that goes after aluminum almost as fast as lead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What practical electrochemical methods could be used to strip the solid lead from the aluminum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Bonus if you can propose a method that would also leave titanium baffles intact!)&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2014-04-30T01:24:15.397" Title="How to separate lead condensate from aluminum" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10359" PostTypeId="2" ParentId="10358" CreationDate="2014-04-30T01:24:15.397" Score="3" Body="&lt;p&gt;You could try hard anodizing Al and Ti (e.g., Calphalon cookware) to give them a non-delicate surface, and go on from there.  Black is a good gun component color.  CVD deposition of titanium boride, ion plating of titanium nitride or carbonitride gives exceptionally survivable hard surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.brycoat.com/coating-services/pvd-coatings/titanium-nitride-tin.html&quot; rel=&quot;nofollow&quot;&gt;http://www.brycoat.com/coating-services/pvd-coatings/titanium-nitride-tin.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.brycoat.com/coating-services/pvd-coatings/ticn.html&quot; rel=&quot;nofollow&quot;&gt;http://www.brycoat.com/coating-services/pvd-coatings/ticn.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among mostly cheap and harmless diddles, I'd try 1:2 &lt;strong&gt;mole ratio&lt;/strong&gt; of choline chloride  with urea or ethylene glycol (&quot;all natural,&quot; etc.).  Mix, the deep eutectic liquefies, soak.  Both eutectics conduct electricity, so you could make lead contamination the anode and plate it out on a closely spaced but not touching copper foil cathode. Test on contaminated waste Al or Ti.  Careful with the voltage - too high and you mobilize Al and Ti.  Avoid contact with skin pro forma, do not ingest, keep away from eyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try vacuum thermal deposition of a micron or so thickness of the Parylene family on the newly fabricated and cleaned metal - intrinsic, especially C, or F.  Parylene polymer surface coats survive the insides of excimer lasers, protecting the excitation electrodes.  If the lead still sticks, chemical cleaning should stop at the polyparylene film.  Or not.  Gotta test it.  You are worried about vapor deposition not abrasion, so an inert barrier film is reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google, parylene coating services&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For aluminum or titanium vs. lead in chemistry, I'd pull one out of left field to try on test coupons:  Rauchfuss' elixir: elemental sulfur dissolved in N-methylimdiazole (pyridine less so, very smelly, toxic, and don't get it on your skin or breathe the awful vapors).  It goes through polysulfide-forming metals like a buzz saw, giving soluble products.  I would expect aluminum to be inert; titanium is a toss-up.  It should hit lead hard, probably.  Test it small scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inorg. Chem. 30 (11): 2514 (1991) &lt;BR&gt; &#xA;DOI:10.1021/ic00011a011 &lt;BR&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/&lt;/a&gt; (resolve a DOI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last gasp:  Weapon Shield to clean before and after firing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentrated nitric acid (already a bad idea) passivates aluminum and eats lead.  Toxic and corrosive NOx fumes come off.  However...aluminum plus lead in galvanic contact make aluminum the sacrificial anode.  Homeland Severity does not like people possessing concentrated nitric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wear gloves and goggles, have ventilation.  None of this stuff is good for you.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-30T01:24:15.397" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10360" PostTypeId="2" ParentId="10338" CreationDate="2014-04-30T04:10:19.003" Score="3" Body="&lt;p&gt;1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of the 19th electron of chromium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the last shell is 4, therefore n=4,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;l=0(as there is only one orbital in the s sub shell) ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m=0 (since we're still in the s sub shell only) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and s=+1/2 or s=-1/2(arbitrarily assigned values)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note:+1/2 or -1/2 are used to distinguish between electrons as no two electrons can have the same electronic configuration, but here the question isn't that specific, so we need not bother. &lt;/p&gt;&#xA;" OwnerUserId="5345" LastActivityDate="2014-04-30T04:10:19.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10361" PostTypeId="1" AcceptedAnswerId="10362" CreationDate="2014-04-30T05:32:33.253" Score="-4" ViewCount="1312" Body="&lt;p&gt;What does it mean for an element to be reduced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean for an element to be oxidized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an oxidizing agent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a reducing agent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you identify oxidizing and reducing agents?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastActivityDate="2014-04-30T05:51:19.417" Title="Questions concerning redox reactions" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2014-05-01T05:08:49.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="10362" PostTypeId="2" ParentId="10361" CreationDate="2014-04-30T05:51:19.417" Score="2" Body="&lt;p&gt;My instructor told me to remember LEO and GER.&lt;br&gt;&#xA;LEO - loss of electron is oxidation&lt;br&gt;&#xA;GER - gain of electron is reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium metal wants to give up an electron. It looses an electron, so it is oxidized.&lt;br&gt;&#xA;Chlorine likes electrons. It gains electrons, so it is reduced.&lt;br&gt;&#xA;Sodium is a reducing agent since it donates an electron.&lt;br&gt;&#xA;Chlorine is an oxidizing agent since it will grab an electron.&lt;br&gt;&#xA;A reducing agent is oxidized and an oxidizing agent is reduced.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-04-30T05:51:19.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10363" PostTypeId="2" ParentId="472" CreationDate="2014-04-30T10:13:41.417" Score="8" Body="&lt;p&gt;Unfortunately there are no (or only very hard to find) studies to the mechanism of the actual sorption of arsenic (acids) to magnetite. Most literature does describe it as a simple Ligand exchange, although I find that hard to believe. From what I have read&lt;sup&gt;[1-4]&lt;/sup&gt; I will try to compose some hints on what may happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all lets have a look at arsenic acid. It is a medium strong acid and a quite well oxidising agent.&#xA;\begin{align}\ce{&#xA;H3AsO4 &amp;amp;\equiv AsO(OH)3\\&#xA;AsO(OH)3 + H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + AsO(O^- )(OH)2\\&#xA;AsO(OH)3 + 2H3+O + 2e- &amp;amp;&amp;lt;=&amp;gt; As(OH)3 + 3H2O&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In presence of any electron rich material and acidic environment it will therefore not be unusual that arsenous acid will be formed.&lt;sup&gt;[5]&lt;/sup&gt; We may therefore also focus on this, amphoteric molecule:&#xA;\begin{align}\ce{&#xA;H3AsO3 &amp;amp;\equiv As(OH)3\\&#xA;As(OH)3 + H3+O &amp;amp;&amp;lt;=&amp;gt; As+(OH)2 + 2H2O\\&#xA;As(OH)3 + {}^{-}OH &amp;amp;&amp;lt;=&amp;gt; AsO^{-}(OH)2 + H2O&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the pH of the solution one would expect to find many of different of these  species. Arsenic is also somewhat oxophil, meaning that the $\ce{As-O}$-bond is quite strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetite in aqueous solution reacts basic, also forming complex like structures or hydroxides, for example this fully hydrated cluster (ref. [2], stoichiometry omitted):&#xA;\begin{aligned}\ce{&#xA; Fe3O4 + $x$\, H2O &amp;amp;&amp;lt;=&amp;gt;  Fe2(OH)2(H2O)8^4+&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one has to pay attention, that the used ferric nano particles for sorpting arsen(o)ic acid(s) are far larger than the used model complex. However, we can see, that there are functional groups like $\ce{-OH}$ or $\ce{-OH2+}$ (or maybe even $\ce{-O-}$) covering the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this as prerequisites we can think about what might happen next. I generally would suggest to start with physisorption, i.e. the cluster and the acid molecules come in close range. Since we are dealing with contaminated water it is maybe safe to assume, that we can deal with soluble salts of arsenoic acid, hence focusing on $\ce{AsO3^{3-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the complexes come into close range hydrogen bonds will form, with &#xA;$&#xA;\newcommand{\nanoFe}{\bbox[lightgrey,3pt,border:1px solid black]{\ce{Fe}}}&#xA;\nanoFe$ being the nano particle:&#xA;$$\ce{\nanoFe-OH\cdots{}O-AsO2^{3-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the whole surface is covered with hydroxide groups, several of these hydrogen bonds may form. As we can see, with this we may now enter the realm of chemisorption. Once these complexes have formed, it is from an entropic point of view favoured to release smaller molecules, water in this case. Also the oxygen atoms bound to arsenic are now in close range to iron atoms. Since we are dealing with nano particle, there will be non-coordinated irons in proximity. This allows for a more permanent complet to form:&#xA;$$\ce{\nanoFe-O-AsO^3- -O\cdots{}HO-\nanoFe}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this monodentate complex we have an already stable complex of a bound arsenic to an iron. It is furthermore possible, and probably favoured, to form a bidentate complex with a somewhat similar mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us also consider a more basic scenario. In this case the surface of the magnetite will not be entirely covered with hydroxide atoms, but more freely accessible oxygen atoms will be present. Direct bond formation between $\ce{\nanoFe-O\cdots{}AsO3^{3-}}$ may occur. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After writing all this I also found a credible source that somehow proves my theory. In &amp;ldquo;Understanding Arsenate Reaction Kinetics with Ferric Hydroxides&amp;rdquo;&lt;sup&gt;[6]&lt;/sup&gt; they used a model system of $$\ce{HAsO4^2- + Fe2O10H15+ -&amp;gt; Fe2O10H15-HAsO4-}$$ to explain the mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this they describe that physisorption is barrierless, leading fast to the adduct that is $\Delta G^\circ_\mathrm{r}=-58~\mathrm{kJ/mol}$ more stable in energy. However, chemisorption and developing the equilibrium is a more lengthy process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the formed complexes become neutral, the activation barrier to release arsenate is very high, compared to adsorption. I will stop right here, please read the publication for further details.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Notes, references and further reading:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.gca.2006.05.012&quot; rel=&quot;noreferrer&quot;&gt;Dimitri A. Sverjensky, Keisuke Fukushi, &lt;em&gt;Geochimica et Cosmochimica Acta&lt;/em&gt; &lt;strong&gt;2006,&lt;/strong&gt; &lt;em&gt;70&lt;/em&gt; (15), 3778-3802.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/es050271f&quot; rel=&quot;noreferrer&quot;&gt;Nianliu Zhang , Paul Blowers , and James Farrell, &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt; &lt;strong&gt;2005,&lt;/strong&gt; &lt;em&gt;39&lt;/em&gt; (13), 4816–4822.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.degruyter.com/viewbooktoc/product/19732&quot; rel=&quot;noreferrer&quot;&gt;Arnold F Holleman; Egon Wiberg; Nils Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. Walter de Gruyter &amp;amp; Co., Berlin, New York, 2007.&lt;/a&gt; (german)&lt;br&gt;&#xA;&lt;a href=&quot;http://books.google.co.jp/books?id=Mtth5g59dEIC&amp;amp;dq=isbn:0123526515&quot; rel=&quot;noreferrer&quot;&gt;Egon Wiberg, Nils Wiberg: Inorganic Chemistry. Academic Press, 2001.&lt;/a&gt; (english)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/scitable/knowledge/library/introduction-to-the-sorption-of-chemical-constituents-59738640&quot; rel=&quot;noreferrer&quot;&gt;Thompson, A. &amp;amp; Goyne, K. W. (2012) Introduction to the Sorption of Chemical Constituents in Soils. &lt;em&gt;Nature Education Knowledge&lt;/em&gt; 3(6):15&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Heavier halogens ($\ce{X=I, Br, Cl}$) should do the job and they will most certainly be present in usual ground water (more or less acidic solution) :&#xA;$$\ce{AsO(OH)3 + 2H3+O + 2KX &amp;lt;=&amp;gt; As(OH)3 + 3H2O + X2 ^ + 2K+}$$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/es4013382&quot; rel=&quot;noreferrer&quot;&gt;James Farrell and Binod K. Chaudhary, &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt; &lt;strong&gt;2013,&lt;/strong&gt; &lt;em&gt;47&lt;/em&gt; (15), 8342–8347.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Also interesting: (a) &lt;a href=&quot;http://dx.doi.org/10.1021/es030309t&quot; rel=&quot;noreferrer&quot;&gt;Suvasis Dixit and Janet G. Hering, &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt; &lt;strong&gt;2003,&lt;/strong&gt; &lt;em&gt;37&lt;/em&gt; (18), 4182–4189&lt;/a&gt; (b) &lt;a href=&quot;http://dx.doi.org/10.1021/es051521v&quot; rel=&quot;noreferrer&quot;&gt;Dimitri A. Sverjensky and Keisuke Fukushi, &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt; &lt;strong&gt;2006,&lt;/strong&gt; &lt;em&gt;40&lt;/em&gt; (1), 263–271.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2016-03-15T06:53:49.910" LastActivityDate="2016-03-15T06:53:49.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10364" PostTypeId="2" ParentId="9553" CreationDate="2014-04-30T11:05:55.073" Score="2" Body="&lt;p&gt;Expanding on LDC3's comment a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information in the table (irrelevant of how good or bad the method proposed is) tells you what the saturation points are for solvent mixtures at given temperatures for each compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentally, you want to separate your desired product from your undesired product. In this case, you are trying to recrystallise the 3-phenylacrylaldehyde, leaving the benzaldehyde in solution. You know this because for all temperatures and solvent mixtures it has lower solubility (this cannot be assumed for a more complex solvent array).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need the minimum of hot solvent, here 100 degrees, to dissolve all of your crude compound, and you want to maximise the return of pure compound on cooling. This explains why you want the biggest ratio of hot to cold solubility as LDC3 indicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also want to minimise the amount of your starting material that crashes out with your product. So, calculate the same ratios for the starting material, but here you want to minimise the return on cooling, so you want the smallest ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now have a table of solubility ratios, columns being 'solvent mix', 'rP' (ratio of product solubilities) and 'rSM' (ratio of starting material solubilities). The largest value of rP/rSM gives you the best solvent mixture.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastEditorUserId="5075" LastEditDate="2014-04-30T14:34:46.187" LastActivityDate="2014-04-30T14:34:46.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10365" PostTypeId="1" CreationDate="2014-04-30T11:17:37.650" Score="1" ViewCount="400" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many ml of acidified $\ce{K2Cr2O7}$  solution, which includes 21.6 g of dichromate ions per Litre, are required for the oxidation of 44.4 g of  $\ce{CH3CH2CH(OH)CH3}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously 44.4 g of $\ce{CH3CH2CH(OH)CH3}$ means 0.6 moles. So from the stoichiometry of the the oxidation we find out that 0.2 moles of $\ce{K2Cr2O7}$ are needed. I have trouble finding the concentration of $\ce{K2Cr2O7}$ though and thats why I need help.&lt;/p&gt;&#xA;" OwnerUserId="5358" LastEditorUserId="69830" LastEditDate="2021-05-21T18:28:37.360" LastActivityDate="2021-05-21T18:28:37.360" Title="How to calculate the concentration of a K₂Cr₂O₇ solution?" Tags="&lt;organic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10367" PostTypeId="1" AcceptedAnswerId="10381" CreationDate="2014-04-30T12:22:11.170" Score="10" ViewCount="3754" Body="&lt;p&gt;Simple question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does household glue work? How does it attach to almost any other compound? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="73" LastEditDate="2014-04-30T21:12:16.150" LastActivityDate="2014-04-30T21:12:16.150" Title="How does glue work?" Tags="&lt;everyday-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10369" PostTypeId="2" ParentId="10365" CreationDate="2014-04-30T14:15:28.423" Score="1" Body="&lt;p&gt;First, you need a balanced equation for the oxidation.  The gives you the mole ratio of dichromate to substrate.  You know how many moles of substrate you have, you know how many moles of dichromate you need.  If you have 21.6 g of dichromate ions (as such)/liter, then you know how many moles/liter you have.  You know the number of moles you need, you know the number of moles/liter you have, you know the number of liters you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will be doing your Jones oxidation in acid, of course.  Collins reagent is a sweeter deal for workup, since MEK is 27.5 g/100 ml water-soluble.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-30T14:15:28.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10370" PostTypeId="2" ParentId="7443" CreationDate="2014-04-30T14:16:48.850" Score="1" Body="&lt;p&gt;In principle, this reaction should be possible. However, the intermediately formed 2-hydroxy-2-propyl radical is not stabilized by the OH substituent, as the alternative mesomeric structure with C=OH bond and unpaired electron at O would violate the octet rule. 2-bromo-2-propanol should still be the major product, but there are also a lot of possible side products like multiply brominated compounds. Other substituents which stabilize an adjacent radical (by conjugation, for example) will increase the selectivity of the reaction. The possibility of reaction of the bromine radicals with the functional groups and sensitivity of the substrate to the reaction conditions (UV radiation etc.) also need to be considered. For the selective halogenation of certain substrates, often more suitable reagents than $\ce{Br2}/ h\nu$ are available, for example, NBS/AIBN for allylic bromination.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-04-30T14:16:48.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10371" PostTypeId="2" ParentId="10367" CreationDate="2014-04-30T14:26:56.103" Score="6" Body="&lt;p&gt;There is glue (sets) and there is adhesive (stays sticky).  Glue is a fluid that wets both surfaces, then sets to a solid by solvent evaporation, reaction, or melt solidifying.  Solvent and melt glues are primarily mechanical bonds, especially for porous surfaces.  Reactive glues can be chemical bonds as well (resorcinol carpenters glue by oxidation, isocyanate Gorilla glue by nucleophilic attack, Super Glue by free radical grafting).  Solvent alone can weld plastic surfaces, such as cyclohexanone for vinyl (plasticized or filled PVC) or organic solvents in general for acrylic, polycarbonate, polystyrene, ABS, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failure of bonded surfaces can be through adhesion (interface separates) or cohesion (material separates).&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-30T14:26:56.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10373" PostTypeId="2" ParentId="9612" CreationDate="2014-04-30T14:50:08.193" Score="0" Body="&lt;p&gt;A theoretical chemist wants to mathematically describe electron density, energy, and directionality in space from first principles, &lt;em&gt;ab intio&lt;/em&gt;, pure math.  One can completely ignore atoms and do it with orthogonal polynomials and spherical harmonics.  This proves to be unworkable for needing unending curve fitting (like epicycles for solar system orbits before Copernicus; or SUSY, or dark matter) for things like hybridization, molecular orbitals, and solid state band structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One instead derives the math specifically from atomic, molecular, and sold state systems beginning with the Shroedinger equation via Hamiltonian or Lagrangian approaches, as quantum mechanics and its wavefunctions, thus wave equations.  At atomic and molecular scales, electrons being low mass and low energy, the de Broglie equation says there are no little pellets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing organic, you survive and prosper by returning to the 1930s with LCAO (linear combination of atomic orbitals - purely descriptive), plus the Woodward-Hoffmann rules as appropriate. LCAO is an empirical black box.  In goes the problem, crank it, out comes the correct answer.  If you are doing pharma, crude LCAO lock-and-key rapidly gives way to detailed calculations for the very many subtle ways electron densities interact.  Pharma requires fit &lt;em&gt;plus&lt;/em&gt; selectivity.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-30T14:50:08.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10374" PostTypeId="2" ParentId="6443" CreationDate="2014-04-30T16:01:58.160" Score="1" Body="&lt;p&gt;While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;111 g/100 mL at 20 °C&lt;/a&gt;). The reaction that occurs is a precipitation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -&amp;gt; Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-10-29T13:06:53.260" LastActivityDate="2014-10-29T13:06:53.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10375" PostTypeId="1" CreationDate="2014-04-30T17:12:47.920" Score="6" ViewCount="102324" Body="&lt;p&gt;Is there a complete list of all the half equations for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; - both oxidation and reduction, in acidic and alkaline conditions? I've looked on the internet but can't seem to find a list with all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are my first attempts: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2 + 2e^- -&amp;gt; 2OH^-}$&lt;/span&gt; (in alkaline conditions)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2 + 2H^+ + 2e^- -&amp;gt; 2H2O}$&lt;/span&gt; (in acidic conditions)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2 -&amp;gt; 2H^+ + O2 + 2e^-}$&lt;/span&gt; (in acidic conditions)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are these correct and are there any more?&lt;/p&gt;&#xA;" OwnerDisplayName="user5110" LastEditorUserId="402" LastEditDate="2019-05-03T22:38:15.320" LastActivityDate="2021-05-06T10:27:29.917" Title="Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10376" PostTypeId="1" CreationDate="2014-04-30T17:28:49.330" Score="2" ViewCount="242" Body="&lt;p&gt;I want to test a process that was documented in the literature using N-methylimidazole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not doing academic research, but rather testing a documented reaction for application to an industrial process, so I don't believe I need lab-grade chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look for this solvent online I can only find lab suppliers (e.g., Matrix Scientific, Acros Organics, Alfa Aeser) selling tiny quantities at very high prices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something unique to this chemical as a &quot;donor solvent&quot; that is not reproduced by any more common or inexpensive solvent?  I have seen it mentioned frequently in organic chemistry papers, but since I can't find industrial suppliers I assume it's peculiar to research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application I'm interested in is chemical stripping of lead using &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00061a035?journalCode=jacsat&quot; rel=&quot;nofollow&quot;&gt;Rauchfuss's Elixir&lt;/a&gt;, which consists of sulfur as an oxidant and a donor solvent.  I haven't been able to learn why they used methylimidazole as the solvent in the study.  Obviously I would prefer one more widely and/or cheaply available.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2015-12-07T16:02:34.443" LastActivityDate="2015-12-07T17:41:53.687" Title="Is N-methylimidazole an unusual solvent, or are there more common substitutes for it?" Tags="&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10378" PostTypeId="2" ParentId="10331" CreationDate="2014-04-30T18:17:29.893" Score="1" Body="&lt;p&gt;As shown by Uncle Al there is some (old) experimental data on the bond angle of water in this article: &lt;a href=&quot;http://dx.doi.org/10.1063/1.1696094&quot; rel=&quot;nofollow&quot;&gt;Shibata and Bartell &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;42&lt;/strong&gt;, 1147 (1965)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water in vacuum they find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{H2O}$: r_g (OH) = 0.976±0.0030 Å, mean HOH angle α_g = 107.2°±3°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there is &lt;em&gt;no statistically significant&lt;/em&gt; deviation from the value at atmospheric pressure.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-12-09T05:21:54.360" LastActivityDate="2014-12-09T05:21:54.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10379" PostTypeId="2" ParentId="10376" CreationDate="2014-04-30T18:19:02.843" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB1316726.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB1316726.htm&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.lookchem.com/cas-616/616-47-7.html&quot; rel=&quot;nofollow&quot;&gt;http://www.lookchem.com/cas-616/616-47-7.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.guidechem.com/cas-616/616-47-7.html&quot; rel=&quot;nofollow&quot;&gt;http://www.guidechem.com/cas-616/616-47-7.html&lt;/a&gt;  2000 RMB = $(USD)320 &lt;BR&gt; &#xA;&lt;a href=&quot;http://ahjinao.en.alibaba.com/product/595605676-209523668/Time_honored_Supplier_of_N_Methylimidazole_CAS_No_616_47_7_.html&quot; rel=&quot;nofollow&quot;&gt;http://ahjinao.en.alibaba.com/product/595605676-209523668/Time_honored_Supplier_of_N_Methylimidazole_CAS_No_616_47_7_.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.alibaba.com/countrysearch/CN/n--methylimidazole-supplier.html&quot; rel=&quot;nofollow&quot;&gt;http://www.alibaba.com/countrysearch/CN/n--methylimidazole-supplier.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.buyersguidechem.com/chemical_supplier/N-Methylimidazole.php&quot; rel=&quot;nofollow&quot;&gt;http://www.buyersguidechem.com/chemical_supplier/N-Methylimidazole.php&lt;/a&gt; &lt;BR&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Info and suppliers,  Note that a Chinese contract is often not the legal or philosophical equivalent of a Western contract.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-04-30T18:19:02.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10380" PostTypeId="2" ParentId="8308" CreationDate="2014-04-30T18:35:07.447" Score="2" Body="&lt;p&gt;Structure and bonding is interconnected, there is not such thing as &quot;only structural differences&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon has many allotropes, including graphite and diamond. You can say they just differ in structure, yet, this difference leads to a metallic sheet in the case of graphite, and an insulating 3D structure in the case of diamond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think about carbon nanotubes, fullerenes and graphene as somewhat modified  versions  od graphite with the same delocalized / metallic sp2 carbon sheets. However that the fact that you curve / roll the carbon sheet and you reconnect certain carbons e.g. to form a tube, you do actually modify the delocalized / metallic nature of the carbon sheet. Specifically, e.g. carbon nanotubes can be semi-conducting or metallic, depending on how you roll up the carbon sheet to a tube.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-04-30T18:35:07.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10381" PostTypeId="2" ParentId="10367" CreationDate="2014-04-30T19:41:43.183" Score="8" Body="&lt;p&gt;At a microscopic level most of the materials have many microscopic pores with different diameters and shapes, so in most of the cases any non solid material which can &lt;strong&gt;harden&lt;/strong&gt; and which has the ability to go inside the tiny pores can be a glue of a determinate strength simply hardening inside the pores. This is called &lt;strong&gt;mechanical adhesion&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the adhesion can be favored if some bonds between the molecules are formed so some glues are more effective for some sort of material. Generally van der Waals' force occurs spontaneously between compound with dipoles, this is called &lt;strong&gt;dispersive adhesion&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hydrogen bonds&lt;/strong&gt; can increase the strength of the glue. Covalent bond between glue and materials don't occurs often, I think that cyanoacrylate (super glue) probably form a covalent bond with the hydroxilated surface of the material (but nobody seems talk about this). In fact a real surface is always cover by other compounds so is very difficult that something interact directly with the surface, and this is why you should always clean (and in most of the cases dry) the surface before apply the glue. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-04-30T19:41:43.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10383" PostTypeId="1" CreationDate="2014-05-01T04:09:05.057" Score="0" ViewCount="2215" Body="&lt;p&gt;I don't understand why the mass of a molecule make it attract more to other polar molecules. &lt;/p&gt;&#xA;" OwnerUserId="4157" LastActivityDate="2014-05-01T06:26:50.617" Title="Why is mass proportional to the strength of a dipole-dipole attraction? meaning more mass more strength" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-05-01T10:59:48.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="10384" PostTypeId="2" ParentId="10383" CreationDate="2014-05-01T06:26:50.617" Score="0" Body="&lt;p&gt;Increasing mass increases the availability for molecules to attract through London dispersion forces (induced dipole-dipole). A longer chain has more atoms and hence more electrons and therefore experiences stronger bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say 'polar' I assume you mean permenant dipole-dipole and hydrogen bonding which is actually not affected by the 'mass' or length of a carbon chain in a molecule. However if you take ethanol and pentanol as an example they both have hydrogen intermolecular bonding on account of the -OH group but pentanol also experiences greater London dispersion forces than ethanol which results in its significantly higher melting and boiling points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to recap increase mass is not a direct causal link to charge. It is in fact more mass meaning more atoms meaning more electrons meaning induced dipoles leading to greater intermolecular forces. I hope that's clear (:&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2014-05-01T06:26:50.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10385" PostTypeId="1" AcceptedAnswerId="16450" CreationDate="2014-05-01T06:49:02.227" Score="4" ViewCount="2152" Body="&lt;p&gt;A problem in my Pchem textbook states that the electronic degeneracy of an oxygen atom is 5. My reasoning leads me to believe that the degeneracy is actually 6 (S = 1, L = 1, so (S+1)(L+1) = 6. This would indicate there are two unpaired electrons and a pair in the P orbitals... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I can reason a 5 degeneracy is if the ground state has a pair in the d-orbitals, which I do not believe is the case.&lt;/p&gt;&#xA;" OwnerUserId="5366" LastActivityDate="2014-09-18T11:03:15.740" Title="Why is the groundstate electronic degeneracy of monoatomic O(g) 5?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10386" PostTypeId="2" ParentId="10375" CreationDate="2014-05-01T07:38:23.237" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://books.google.it/books?id=cptVx_1zpawC&amp;amp;pg=PA198&amp;amp;dq=hydrogen+peroxide+electrochemistry&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ei=k_FhU_zvMoqFtQa1oIDIDg&amp;amp;ved=0CEsQ6AEwAA#v=onepage&amp;amp;q=hydrogen%20peroxide%20electrochemistry&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proceedings of the Symposium on Oxygen Electrochemistry&lt;/a&gt; and &lt;a href=&quot;https://doi.org/10.1016/S0043-1354(01)00235-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Water Research&lt;/em&gt; &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;36&lt;/em&gt; (1), 85-94&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{lcc}&#xA;\text{Formation}&amp;amp;:&amp;amp;~\ce{2H+ + O2 + 2e- &amp;amp;-&amp;gt;&amp;amp; H2O2} &amp;amp;\quad E^0=\pu{0.440V} \text{ vs. SCE}\\&#xA;\text{Reduction}&amp;amp;:&amp;amp; ~\ce{H2O2 + 2H+ + 2e- &amp;amp;-&amp;gt;&amp;amp; 2H2O} &amp;amp;\quad E^0=\pu{1.534V} \text{ vs. SCE}\\&#xA;\text{Direct water}&amp;amp;:&amp;amp; ~\ce{2H+ + 1/2O2 + 2e- &amp;amp;-&amp;gt;&amp;amp; H2O}\\&#xA;\text{Decomposition}&amp;amp;:&amp;amp;  ~\ce{H2O2 &amp;amp;-&amp;gt;&amp;amp; H2O + 1/2 O2}&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;span class=&quot;math-container&quot;&gt;$\text{SCE}$&lt;/span&gt; stands for &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturated_calomel_electrode&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturated Calomel Electrode&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another strange suggested reaction involves the anion &lt;span class=&quot;math-container&quot;&gt;$\ce{HO2-}$&lt;/span&gt; (&lt;a href=&quot;http://doi.org/10.1007/BF00938566&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. &lt;em&gt;Monatshefte für Chemie&lt;/em&gt; &lt;strong&gt;1952&lt;/strong&gt; &lt;em&gt;83(2)&lt;/em&gt; 422–39.&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Decomposition 2}:\ce{H2O2 + H2O- -&amp;gt; H2O + O2 + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At the cathode occur at the same time the hydrogen gas evolution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H+ + 2e- -&amp;gt; H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, at the anode, oxygen gas is evolved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$~\ce{2H2O -&amp;gt; 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2019-05-03T22:36:02.087" LastActivityDate="2019-05-03T22:36:02.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10387" PostTypeId="1" CreationDate="2014-05-01T07:44:22.567" Score="3" ViewCount="425" Body="&lt;p&gt;Consider a binary azeotropic solution of two volatile components $A, B$. I wanted to try and find out the required concentrations of $A, B$ in the solution in terms of their vapour pressures. I tried using Raoult's Law to do this, but the I got an unexpected answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the vapour pressure of $A, B$ at the current temperature be $P_A, P_B$, respectively. Let their concentrations be $\chi_A, \chi_B$ and their partial vapour pressures over the solution be $p_A, p_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by Raoult's Law, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_A = P_A\chi_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_B = P_B\chi_B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Dalton's Law of partial pressures, the mole fractions of $A, B$ in the vapor phase are given by $\frac{p_A}{p_A + p_B}$ and $\frac{p_B}{p_A + p_B}$ respectively. Since the solution is azeotropic, these two are equal to their mole fractions in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_A\chi_A}{P_A\chi_A + P_B\chi_B} = \chi_A \implies \frac{P_A}{P_B}(1-\chi_A) = \chi_B\tag1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_B}{P_A}(1-\chi_B) = \chi_A \tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $(1)$ in $(2)$, yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{P_B}{P_A}\left(1-\frac{P_A}{P_B}(1-\chi_A)\right) = \chi_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ = \frac{P_B}{P_A}-1+\chi_A = \chi_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\implies P_B = P_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are numerous azeotropes where this condition is not satisfied. How come? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="4945" LastEditDate="2014-05-01T13:11:01.940" LastActivityDate="2014-05-01T14:01:15.510" Title="Condition for formation of azeotropes" Tags="&lt;solutions&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10388" PostTypeId="5" CreationDate="2014-05-01T08:52:48.397" Score="0" Body="&lt;p&gt;Quantum chemistry is a subfield of &lt;a href=&quot;/questions/tagged/quantum-mechanics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;quantum-mechanics&amp;#39;&quot; rel=&quot;tag&quot;&gt;quantum-mechanics&lt;/a&gt;. Like its parent field, quantum chemistry focuses on physical phenomena that occur at an atomic scale, usually specifically aiming at chemical reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is closely related to &lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt; for which it provides the working equations and is therefore of more general nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roots of quantum theories go back to the early 20th century. Their postulation was a result of observed atomic-level phenomena that could not be suitably described by classical mechanics. The Pioneers of this field include Heisenberg and Schrödinger, that are nowadays very well known not only to chemists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, quantum theory has many applications in computing and a diversity of other areas of science. Modern applications include accurate description of chemical &lt;a href=&quot;/questions/tagged/bonding&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;bonding&amp;#39;&quot; rel=&quot;tag&quot;&gt;bonding&lt;/a&gt; as well as the development of &lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt;. It has therefore also contributed to our understanding of atoms, molecules, and reactions.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-04-08T01:05:37.647" LastActivityDate="2017-04-08T01:05:37.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10389" PostTypeId="4" CreationDate="2014-05-01T08:52:48.397" Score="0" Body="Quantum chemistry is a subfield of quantum mechanics. Like its parent field, quantum chemistry focuses on understanding physical phenomena occuring at the atomic scale. Quantum chemistry however is more focused on providing useful descriptions of electronic structure to aid in understanding chemical problems (e.g. reactions, spectra, dynamics, ...)." OwnerUserId="4945" LastEditorUserId="68944" LastEditDate="2018-12-23T00:02:56.147" LastActivityDate="2018-12-23T00:02:56.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10390" PostTypeId="1" AcceptedAnswerId="10392" CreationDate="2014-05-01T09:34:36.980" Score="10" ViewCount="3877" Body="&lt;p&gt;I have a chapter and in it there was a topic on the Rutherford experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is written that doubly charged helium particles or alpha particles were thrown towards a very thin gold foil and some of the alpha particles where bounced back because there were positive particles (protons) and positive and positive always repel each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this was the reason for the bouncing back of the alpha particles, then why didn't the electrons, which are much smaller than protons, get attracted from alpha particles?&#xA;Especially when the alpha particles are doubly charged.&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="4945" LastEditDate="2014-05-01T13:10:15.543" LastActivityDate="2019-11-05T09:42:33.700" Title="Why weren't the electrons attracted to doubly charged Helium particles in the Rutherford experiment?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;protons&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10391" PostTypeId="1" CreationDate="2014-05-01T10:02:16.240" Score="10" ViewCount="11672" Body="&lt;p&gt;Why does water volume decrease when salt is added? Our teacher asks us in the class but I don't find any strong reason.&lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="5026" LastEditDate="2018-02-25T07:37:18.637" LastActivityDate="2019-11-01T22:23:17.267" Title="Why does water volume decrease when salt is added?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ClosedDate="2019-11-02T05:40:49.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="10392" PostTypeId="2" ParentId="10390" CreationDate="2014-05-01T10:28:43.483" Score="10" Body="&lt;p&gt;The gold foil experiment conducted by Rutherford (or more appropriately by &lt;a href=&quot;http://en.wikipedia.org/wiki/Gold_foil_experiment&quot; rel=&quot;noreferrer&quot;&gt;Geiger and Marsden&lt;/a&gt; under Rutherford's direction) involves a stream of $\alpha$ particles (which are helium nuclei) bombarding a thin gold foil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why didn't some electrons go with the alpha particles because of electrostatic attraction? Some probably did. The experiment was equipped with a fluorescent zinc sulfide screen to detect alpha particles ($\ce{He^{2+}}$ ions). This screen might have also detected $\ce{He+}$ ions (but not necessarily in a way that was different than the detection of alpha particles), but probably not free electrons or helium atoms. The experiment was only set up to detect alpha particles, and so that's all that Rutherford, Geiger, and Marsden observed. We'll never know how many (if any) alpha particles picked up electrons in their experiment. We could, however, repeat the experiment, replacing the fluorescent screen with a mass analyze to detect at least the difference between $\ce{He^{2+}}$ and $\ce{He+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, we known that many alpha particles passed through (and some were deflected) unchanged. Why didn't these particles pick up electrons? The answer to this question is another question: What is holding those electrons in the gold foil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold atoms have 79 protons holding down those electrons. The first ionization energy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gold&quot; rel=&quot;noreferrer&quot;&gt;gold&lt;/a&gt; is 890.1 kJ/mol and the second is 1980 kJ/mol. However, these values are for isolated gold atoms. Gold (like all metals) is held together by &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bond&quot; rel=&quot;noreferrer&quot;&gt;metallic bonds&lt;/a&gt; with delocalized electrons occupying a valence band. Each electron is not being held by just one gold nucleus. Once the first few electrons are ejected or snatched by the alpha particles, the gold foil will have a net positive charge. It will be much harder to remove negatively charged particles from the positively charged foil.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-01T10:28:43.483" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10393" PostTypeId="2" ParentId="10391" CreationDate="2014-05-01T11:02:55.610" Score="10" Body="&lt;p&gt;What you are experiencing is the phenomenon of &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_molar_property&quot;&gt;partial molal properties&lt;/a&gt; - components of a mixture contribute unequally to the properties of the mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the components of a mixture do not contribute linearly to the properties of a mixture. A well-known example is the water-ethanol system, which has negative excess volume. When 5 mL of water and 95 mL of ethanol are mixed, the final volume is less than 100 mL. See the following graph constructed by &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Excess_Volume_Mixture_of_Ethanol_and_Water.png&quot;&gt;Wikipedia user Wilfried Cordes&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ivQrA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar behavior is occurring in your sodium chloride and water system. Now, as for why these phenomena occur, a lot of it has to do with intermolecular forces. For example, you could consider that the hydrogen-bonding and dipole-dipole interactions between water and ethanol are perhaps stronger than between water and water or ethanol and ethanol. Additionally, water molecules are smaller than ethanol molecules. There is also a thermodynamic consideration. This state of affairs must represent minimum free energy (maximum entropy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you relate the interactions of sodium and chloride ions with water to the observation of negative excess volume?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-01T11:02:55.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10394" PostTypeId="2" ParentId="10387" CreationDate="2014-05-01T14:01:15.510" Score="4" Body="&lt;p&gt;It is not that surprising that you got this answer: the existence of azeotropes is due to non-ideal behaviour of mixtures (see for instance &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope#Deviation_from_Raoult.27s_law&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page and the associated diagram&lt;/a&gt;). It is not surprising that when you impose an ideal behaviour of the solvent that follows Raoults law, you get some unexpected results. Basically, what your calculation leads to it that for ideal solvent mixture, the only situation in which the gas phase molar ratio can be equal to the solution molar ratio is when both solvent have equal vapour pressures. &lt;/p&gt;&#xA;" OwnerUserId="263" LastActivityDate="2014-05-01T14:01:15.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10395" PostTypeId="1" CreationDate="2014-05-01T14:12:30.820" Score="6" ViewCount="636" Body="&lt;p&gt;I am looking for a chemical test for the presence of gluten: however, having looked online for such a thing, all the results seem to be about 'home testing kits' (presumably for people with a gluten intolerance). Is there a simple chemical test which can be used for a school science project (meaning that the chemicals involved should be easily available), something similar to how iodine can be used to detect starch?&lt;/p&gt;&#xA;" OwnerUserId="5374" LastEditorUserId="16683" LastEditDate="2021-02-16T22:58:59.107" LastActivityDate="2021-02-17T08:56:55.170" Title="Is there a chemical test for gluten?" Tags="&lt;analytical-chemistry&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10396" PostTypeId="2" ParentId="10390" CreationDate="2014-05-01T14:25:38.567" Score="3" Body="&lt;p&gt;Consider radium decay alpha-particle energies: Ra-226 4.86 MeV, Ra-222 5.59 MeV. The first atomic ionization energy of gold is 9.22 eV.  We do not expect charge neutralization of the alpha-particle until it slows by collision. It will leave a short path of dense ionization as it transfers energy, continuously being stripped to a bare nucleus until it slows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum work function of bulk gold tops at 5.47 eV, giving the same expectation.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-01T14:25:38.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10397" PostTypeId="2" ParentId="6987" CreationDate="2014-05-01T15:06:18.130" Score="0" Body="&lt;p&gt;If you want to have a more comprehensive overview about your question, I would suggest you to read this book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/981024665X&quot; rel=&quot;nofollow&quot;&gt;The Nobel Prize: The First 100 Years&quot;&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this link &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/themes/physics/karlsso&quot; rel=&quot;nofollow&quot;&gt;The Nobel Prize in Physics 1901-2000&lt;/a&gt; you'll find one of the chapters for free with a reasonable explanation about why we treat our problems using different hierarchies or different approaches (bottom-up or top-bottom).&lt;/p&gt;&#xA;" OwnerUserId="5375" LastEditorUserId="5375" LastEditDate="2014-05-01T15:42:36.693" LastActivityDate="2014-05-01T15:42:36.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10399" PostTypeId="2" ParentId="6390" CreationDate="2014-05-01T16:13:13.640" Score="5" Body="&lt;p&gt;Regular test tube is not really a good choice for rapid cooling for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is often cheap glass,&lt;/li&gt;&#xA;&lt;li&gt;it has a relatively thin wall, and large space for the sample inside.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This later is important, because if you want to cool down something really quick, it should be relatively small, and have good contact with the wall. I assume you want to cool something really quick or else the question has not much point.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have you tried better quality tubes, e.g quarz, Pyrex? &lt;/li&gt;&#xA;&lt;li&gt;Have you tried transfer the sample to a cooled place, instead of cooling the whole thing down? E.g. rapid freeze quench techniques are often based on squirting the sample into a large volume of cooler liquid or on a cooled surface. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-05-01T16:13:13.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10400" PostTypeId="2" ParentId="10194" CreationDate="2014-05-01T16:54:34.217" Score="8" Body="&lt;p&gt;First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C84662&amp;amp;Units=SI&amp;amp;Mask=200#Mass-Spec&quot; rel=&quot;nofollow noreferrer&quot;&gt;this spectrum&lt;/a&gt;, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not aware of any complete study of phtalate ester fragmentation, but I did not carry a complete litterature search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below is a proposal for the mechanism leading to the formation of two possible structures for the $m/z$ 149 ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one consider that is fragments as do benzoate esters, through radical induced $\alpha$-bond cleavage leading to the loss of $\ce{CH3CH2O^.}$ followed by standard decarbonylation of the acylium ion. For phtalate esters, the question at this point is whether the structure will remain as &lt;strong&gt;I&lt;/strong&gt; or might undergo isomerisation to more stable structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FtdSK.png&quot; alt=&quot;Reaction mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have provided a few alternative structures: simple addition of the neighbouring carbonyl would lead to structure &lt;strong&gt;II&lt;/strong&gt; which although it has some constraints, allows a good stabilization of the positive charge both in the ester and by the aromatic ring. Other structures can be postulated, such as &lt;strong&gt;III&lt;/strong&gt; after addition of the ethanoyl oxygen to the aromatic cycle. This four membered ring can reopen to lead to the acylium structure &lt;strong&gt;IV&lt;/strong&gt; which can in turn undergo decarbonylation to form the ion $m/z$ 121, also present in the spectrum. So there is some indirect evidence that these structures might be involved in the fragmentation process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in order to explain the formation of ions at $m/z$ 105 and 77, one can also propose a last isomer for ion $m/z$ 149, structure &lt;strong&gt;V&lt;/strong&gt;, formed after 1,5 hydride transfer. This one can quite easily lead to loss of acetaldehyde, eventually followed by decarbonylation to lead to ion $m/z$ 77. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If phtalates present specific reactivity, the other option is that ion &lt;strong&gt;A&lt;/strong&gt; at $m/z$ 177 can cyclize to structure &lt;strong&gt;B&lt;/strong&gt; which is a very good candidate for ethene loss leading also to an ion at $m/z$ 149. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then becomes which of these two $m/z$ 149 is the actual ion. These two isobaric ions could be resolved by high resolution experiments, but after looking in publicly available databases, I did not find any high resolution spectra. Deuterium labelling of the ethyl groups could also prove handy, but once again I did not find any spectra of diethylphtalate labelled on the ethyl groups. (The usual diethylphtalate used in quantitation experiments seems to be the $d_{4}$ isotopomer, with deuterium on the aromatic ring.) Finally one can also have a look at isomers and related ions to see how they fragment. And indeed if one takes the dipropylphtalate ester, the ion at $m/z$ 149 is present (produced after propanoyl loss followed by propene loss) and there is no strong ion at $m/z 163$ which would tend to be a good indication of ethene loss for diethylphtalate. But, on the other hand, both the &lt;em&gt;meta&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; isomers of diethylphtalate have a much weaker ion at $m/z$ 149, which cannot be accounted so easily by an equivalent to intermediate &lt;strong&gt;B&lt;/strong&gt;. So I would probably favor the concurrent occurence of both pathways, with a ratio that still needs to be measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the answer above, it should have been quite immediate that a McLafferty rearrangement was not an option: since there are no nitrogen, the nitrogen rule allows to conclude directly that the odd mass of the ion at m/z 149 indicates that it is an even electron cation, as opposed to radical cations produced by McLafferty rearrangements. &lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="24052" LastEditDate="2017-10-06T13:03:57.150" LastActivityDate="2017-10-06T13:03:57.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10401" PostTypeId="2" ParentId="2376" CreationDate="2014-05-01T17:13:37.900" Score="2" Body="&lt;p&gt;Dichlorotriazenetrione, also known as dichloroisocyanuric acid, is slowly hydrolyzed in water to isocyanuric acid and hypochlorous acid, which serves as the oxidation agent. The N-Cl bond is easily hydrolyzed because of its relative weakness, which is also reflected in the stability of nitrogen halides. The slow hydrolysis releases a low but constant amount of oxidizers. $\ce{HOCl}$ itself oxidizes organic compounds, being reduced to $\ce{HCl}$ in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5PKz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution, $\ce{HOCl}$ is in equilibrium with its anhydride $\ce{Cl2O}$ and elemental chlorine, which also serve as oxidizers. Furthermore, it slowly decomposes to $\ce{HCl}$ and oxygen, so that a mixture of different oxidation agents is present in its solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + HOCl \rightleftharpoons Cl2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HOCl \rightleftharpoons Cl2O + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HOCl \rightarrow 2HCl + O2}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-01T17:13:37.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10404" PostTypeId="1" CreationDate="2014-05-01T18:16:03.530" Score="0" ViewCount="1564" Body="&lt;p&gt;I would like to know how I can clean a silk tie at the knot. I presume it got filthy, over time, due to &quot;fatty&quot; fingers. I am looking for answers either based on 1) good theory or 2) personal experience. I am &lt;em&gt;not&lt;/em&gt; looking for grandma's advice.&lt;/p&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="1430" LastEditDate="2014-05-02T04:10:50.590" LastActivityDate="2021-05-28T17:15:17.197" Title="How do I clean a silk tie at the knot?" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10405" PostTypeId="1" CreationDate="2014-05-01T18:23:42.860" Score="3" ViewCount="13508" Body="&lt;p&gt;My book says that-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In acidic water, the negative fatty acid ions become fatty acids. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, they did not include a chemical equation to demonstrate this, so I made my own:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C17H35COO- (aq) + HCl (aq) -&amp;gt; C17H35COOH (aq) + Cl- (aq)}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to this equation, the fatty acid ion is a base, since it gets a proton. So this must be wrong. Can anyone point out my mistakes (if any), and explain what happens when a negative fatty acid ion comes in contact with acidic water, and becomes a fatty acid?&lt;/p&gt;&#xA;" OwnerUserId="5225" LastEditorUserId="83758" LastEditDate="2020-05-30T08:44:50.443" LastActivityDate="2020-05-30T08:44:50.443" Title="What happens when C17H35COO-, a soap, reacts with an acidic solution?" Tags="&lt;acid-base&gt;&lt;fats&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10406" PostTypeId="2" ParentId="6987" CreationDate="2014-05-01T18:23:44.820" Score="0" Body="&lt;p&gt;&lt;em&gt;formula to perfectly describe the interactions&lt;/em&gt;  Calculating the bond angle of water, vacuum vs. condensed phase for three small atoms, has a huge literature offering no definitive answer.  Something as simple as the pendulum equation (ideal - in vacuo, no friction or other dissipation), is not a closed form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/math/6/4/c/64cd3db46ef18d401109423e81cdf6ce.png&quot; rel=&quot;nofollow&quot;&gt;http://upload.wikimedia.org/math/6/4/c/64cd3db46ef18d401109423e81cdf6ce.png&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo1.pdf&quot; rel=&quot;nofollow&quot;&gt;http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo1.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/mechan/Pendulo2.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://fy.chalmers.se/~f7xiz/TIF080/Period.pdf&quot; rel=&quot;nofollow&quot;&gt;http://fy.chalmers.se/~f7xiz/TIF080/Period.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.phys.uconn.edu/~hamilton/phys258/N/pend.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.phys.uconn.edu/~hamilton/phys258/N/pend.pdf&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.scielo.org.mx/pdf/rmfe/v54n1/v54n1a10.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.scielo.org.mx/pdf/rmfe/v54n1/v54n1a10.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HyperChem Lite for a very modest price has a lovely MM+ (semi-empirical) calculation engine.  I do blocks of 32500 iterations, update screen every 500 iterations, with 1e-020 limit until things look good enough, the method chokes, or the structure is ridiculous (pi bonds in high symmetry environments can go odd places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hyper.com/Products/HyperChemLite/FeaturesLite/tabid/375/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.hyper.com/Products/HyperChemLite/FeaturesLite/tabid/375/Default.aspx&lt;/a&gt; &lt;BR&gt;&#xA; &lt;em&gt;ab initio&lt;/em&gt; Hartree Fock/6-31G(d) is better if you have two weeks in a cluster.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-01T18:23:44.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10407" PostTypeId="2" ParentId="10405" CreationDate="2014-05-01T18:28:04.150" Score="4" Body="&lt;p&gt;Your equation is correct. Hydrochloric acid ($\ce{HCl}$) is a stronger acid than stearic acid ($\ce{C17H35COOH}$),i.e. it is dissociated into $\ce{H+}$ and $\ce{X-}$, the anionic rest, to a greater degree. Therefore, it protonates the stearate ion. One can also say that the conjugated base of stearic acid (the stearate ion) is a stronger base than chloride, the conjugated base of hydrochloric acid.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-01T18:34:41.743" LastActivityDate="2014-05-01T18:34:41.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10408" PostTypeId="1" AcceptedAnswerId="10411" CreationDate="2014-05-01T20:35:03.847" Score="2" ViewCount="1888" Body="&lt;p&gt;I came across a radical substitution which Methane reacts with $\ce{IBr}$ to form iodomethane and hydrogen bromide and many more products. I understand the whole process but I don't quite see how the $\ce{IBr}$ molecule will break via homolytic fission instead of heterolytic. Since Bromine is more electronegative than Iodine, wouldn't $\ce{Br}$ takes both electrons and form $\ce{Br-}$ ions?&lt;/p&gt;&#xA;" OwnerUserId="5377" LastEditorUserId="4945" LastEditDate="2014-05-02T07:53:57.060" LastActivityDate="2014-05-02T07:53:57.060" Title="UV radiation on IBr (Radical Substitution)" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10410" PostTypeId="1" AcceptedAnswerId="10418" CreationDate="2014-05-01T20:49:23.153" Score="3" ViewCount="1621" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The electrodes have different charges. One is positive ( the cathode ) while one is negative (anode). They will attract the particles in the solution which are charged. Hydrogen to the cathode while oxygen will go to the anode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been taught that water is a covalent bond. Also I have been taught that covalent bonds do no have a charge so how is it possible that this compound is charged if it's covalent?&lt;/p&gt;&#xA;" OwnerUserId="5378" LastActivityDate="2014-05-02T12:45:15.183" Title="How can water be electrolysed if it's not ionic compound?" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10411" PostTypeId="2" ParentId="10408" CreationDate="2014-05-01T21:19:53.343" Score="2" Body="&lt;p&gt;The presumed stability of $\ce{Br-}$ is overcompensated by the instability of $\ce{I+}$. While some polyatomic iodine cations like $\ce{I2+}$ exist, $\ce{I+}$ is found only to be stable in the gas phase and in the absence of a counterion. Due to the low difference of electronegativities (0,5) of Br and I, the I-Br bond is predominantly covalent, which also makes heterolysis unlikely. Homolytic cleavage of the I-Br bond is therefore energetically more favorable.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-01T21:27:05.450" LastActivityDate="2014-05-01T21:27:05.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10412" PostTypeId="1" CreationDate="2014-05-01T23:21:49.293" Score="1" ViewCount="1262" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2avTU.jpg&quot; alt=&quot;Reaction with Lucas Reagent&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the product(s) if this alcohol reacts with Lucas reagent? &#xA;Would Methyl shift happen ? I heard that in Groove's process, only primary and secondary alkyl chlorides are formed. &lt;/p&gt;&#xA;" OwnerUserId="5380" LastActivityDate="2015-03-20T15:21:37.377" Title="What is/are the product when this reacts with Lucas reagent?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10413" PostTypeId="2" ParentId="10412" CreationDate="2014-05-01T23:46:04.637" Score="1" Body="&lt;p&gt;A shift of the methyl group is likely, because this would yield a more stable tertiary carbocation. In concentrated hydrochloric acid, the double bond can also be protonated, producing a secondary carbocation which is captured by $\ce{Cl-}$. If both happens, the product would be 2,4-dichloro-4,5-dimethylhexane.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-01T23:46:04.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10414" PostTypeId="2" ParentId="10410" CreationDate="2014-05-02T00:42:11.730" Score="1" Body="&lt;p&gt;If you use NaCl as the aqueous electrolyte, you get sodium metal and chlorine. Sodium reacts with water to give hydrogen and hydroxide.  Chlorine eats the anode, dissolves in water, and disproportionates with hydroxide to give HOCl and chloride.  NaOH aqueous electrolyte gives hydrogen and oxygen.  Sodium carbonate aqueous electrolyte can be interesting at the cathode versus voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry tops off around 3 eV bonds.  A mere transistor battery gives you nine volts.  Products depend on electrolyte, concentration, voltage, current density, and electrode material (e.g., overvoltage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/21833/1/IJCB%2041B%282%29%20423-426.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nopr.niscair.res.in/bitstream/123456789/21833/1/IJCB%2041B(2)%20423-426.pdf&lt;/a&gt; &lt;BR&gt;&#xA;Electrochemistry can push things hard.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-02T00:42:11.730" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10415" PostTypeId="1" AcceptedAnswerId="15251" CreationDate="2014-05-02T05:16:48.277" Score="17" ViewCount="595" Body="&lt;p&gt;Below is replicated Question 1 from the Final Qualifying Exam of the Australian Chemistry Olympiad, 2004B, &lt;a href=&quot;http://www.chemteam.info/AChO/FQE/BF-2004.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of promotion-hybridization is quite successful at explaining why covalent molecules with a central atom from period 3 or higher (e.g. $\ce{SF6}$, $\ce{PF5}$, $\ce{I3}$) are stable even though they are hypervalent (i.e. they appear to &quot;exceed the octet rule&quot;). However, hybridization is inadequate in explaining the remarkable stability of molecules such as the so-called &quot;hyperlithiated&quot; carbon species such as $\ce{CLi6}$. Molecular orbital theory can be applied to $\ce{CLi6}$ (Reed and Weinhold, 1985) for an alternate explanation to when and why hypervalent molecules with central atom of period 2 can be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of all the different types of atomic orbitals (e.g. $s$, $p_x$, $d_{yz}$, etc) of a general central atom, six of these will participate in the forming of sigma bonds with the molecular orbitals of six ligands &lt;em&gt;in an octahedral geometry&lt;/em&gt;. The valence $s$ orbital is one of them. Note that the ligands approach the central atom along the three coordinate axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Which other five atomic orbitals can also do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we shall consider the molecular orbitals (MO) of the 6 $\ce{Li}$ $2s$ orbitals (in an octahedral geometry). The lowest energy MO is given below in figure 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWvVq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to quantitative calculations, the MO in figure 1 is the &lt;strong&gt;&lt;em&gt;only&lt;/em&gt;&lt;/strong&gt; MO lower in energy than the uncombined lithium $2s$ orbital; all other MOs are slightly higher in energy than uncombined $\ce{Li}$ $2s$ and can thus be considered as antibonding. (Note that this is generally true for 2 to 8 Li atoms arranged symmetrically around the central atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $s$ orbital will overlap with the MO given in figure 1. Exactly one of the other five MOs will have the right symmetry to overlap (and form a sigma bond) with each of the other five atomic orbitals of the central atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Hence draw the other five MOs obtained by linear combinations of the 6 $\ce{Li}$ $2s$ orbitals. Indicate their relative energies on an energy diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the central atom is carbon, three of the six MOs (of the 6 $\ce{Li}$ $2s$ orbitals) will not combine with the corresponding atomic orbital of carbon. One of these is the one drawn in figure 1. Because carbon is much more electronegative than lithium, the $2s$ orbital of carbon is too low in energy to mix with the MO in figure 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Which other two MOs will not combine when carbon is the central atom? Provide a rationale as to why this is so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) Hence, draw the molecular orbital energy diagram of $\ce{CLi6}$. Keep in mind the relative energies of carbon AOs and the six Li MOs (as carbon is more electronegative than lithium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) Using the energy diagram in part (d), explain the stability of $\ce{CLi6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) Propose an alternative to the &quot;octet rule&quot; when predicting stability of hyperlithiated species of period 2 atoms. Using this rule, identify possible neutral (i.e., not cations or anions) hyperlithiated species of nitrogen and oxygen that can exist due to the same stability reason as that of $\ce{CLi6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(g) Can $\ce{CH6}$ exist? If so, why so? If not, why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My response:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) These are the $p_x$, $p_y$, $p_z$, $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Assuming analogy with octahedral complexes, Googling provides &lt;a href=&quot;http://www.chem.tamu.edu/rgroup/dunbar/Teaching/CHEM362/Lecture_23-24.pdf&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;, pg. 10 of which appears to answer the question. Given that $E_d$ &amp;lt; $E_s$ &amp;lt; $E_p$, the $d$ orbitals represented would be found at the bottom and $p$ at the top of the energy diagram.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is an octahedral complex analogous to $\ce{CLi6}$?&lt;/li&gt;&#xA;&lt;li&gt;What is a linear combination of atomic orbitals?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(c) By logical reasoning, these would be the two similar $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals. I am uncertain regarding the rationale.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it because it is unfeasible for carbon's electrons to be in $3d$ orbitals?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(d) I really have no idea. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If three of the six possible MOs will not combine, does that mean that they are not present in the MO diagram?&lt;/li&gt;&#xA;&lt;li&gt;Since carbon is more electronegative than lithium, it should contribute more to the bonding orbitals, while lithium should contribute more to antibonding orbitals.&lt;/li&gt;&#xA;&lt;li&gt;How are the 10 electrons in $\ce{CLi6}$ distributed? In order to impart stability, there must be more electrons in bonding than in antibonding orbitals. Can we simply have three bonding MO orbitals with 2 electrons and three antibonding MO orbitals with 4 electrons between them?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(e) By logical reasoning, the larger number of electrons in bonding rather than in antibonding orbitals imparts stability to the molecule since bonding electrons are not entirely counteracted by antibonding electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) I really have no idea.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By guesswork, I would propose the decuplet rule, in which hyperlithiated species from period 2 are stable with 10 valence electrons, producing the species $\ce{NLi5}$ and $\ce{OLi4}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(g) By logical reasoning, yes: it satisfies the conditions of (f).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My gut says no.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given that my background in chemistry is limited to high school AP Chemistry, please curtail the material provided in your responses. Thanks for all your help in advance!&lt;/p&gt;&#xA;" OwnerUserId="5382" LastEditorUserId="16683" LastEditDate="2017-08-30T13:43:54.397" LastActivityDate="2017-08-30T13:43:54.397" Title="Hyperlithiated Carbon Species" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="8" ClosedDate="2017-08-30T13:41:46.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="10416" PostTypeId="1" CreationDate="2014-05-02T09:06:15.567" Score="3" ViewCount="396" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Glycine &lt;span class=&quot;math-container&quot;&gt;$(\ce{NH2CH2COOH})$&lt;/span&gt; is an amino acid with two &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_a$&lt;/span&gt; values &lt;span class=&quot;math-container&quot;&gt;$(\mathrm{p}K_{a,1}=2.0, \  \ce{COOH};\  \mathrm{p}K_{a,2}=10.0, \ \ce{NH2})$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(a) If &lt;span class=&quot;math-container&quot;&gt;$\pu{0.01 mol}$&lt;/span&gt; of this amino acid is used to prepare one &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt; solution. Calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H}$&lt;/span&gt; and the &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3+CH2COOH}$&lt;/span&gt; concentration in the solution.&lt;/p&gt;&#xA;&lt;p&gt;(b) If &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of the above glycine amino acid solution is mixed with &lt;span class=&quot;math-container&quot;&gt;$\pu{90 mL}$&lt;/span&gt; of basic buffer system. If the concentration of the &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2CH2COO-}$&lt;/span&gt; was identified to be &lt;span class=&quot;math-container&quot;&gt;$\pu{1.05 \cdot 10^{-4} M}$&lt;/span&gt; indicate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H}$&lt;/span&gt; of the glycine solution.&lt;/p&gt;&#xA;&lt;p&gt;(c) If &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of strong base &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.01 M}$&lt;/span&gt;) was added to solution in (a) what is the final &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;For part (a):&lt;/strong&gt;&lt;br /&gt;&#xA;I used &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]=\sqrt{10^{-10}\cdot0.01} =10^{-6}$&lt;/span&gt;.&lt;br /&gt;&#xA;Then I used &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H} =14-\log(10^{-6})$&lt;/span&gt; to get a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;.&lt;br /&gt;&#xA;While my &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3+CH2COOH}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{10^{-6} L}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;For part (b):&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I used the Henderson-Hasselbalch equation for a basic buffer &lt;span class=&quot;math-container&quot;&gt;$$\mathrm{p}\ce{H} =\mathrm{p}K_a     +\log\left(\frac{\ce{[NH2CH2COO- ]}}{\ce{[NH2CH2COOH]}}\right)$$&lt;/span&gt; I got a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$10.97$&lt;/span&gt;. But I'm   unsure of my answer.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;For part (c):&lt;/strong&gt;&#xA;I suspect that it is a basic buffer but I'm not sure which value to use. could anyone help me out with this.&lt;/p&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-06-04T14:18:52.437" Title="Amphoteric amino acid titration" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10417" PostTypeId="1" CreationDate="2014-05-02T11:39:44.697" Score="4" ViewCount="1263" Body="&lt;p&gt;I've been quite stunned during the last Winter Olympics in Sochi by the following fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperatures were quite high, sometimes 10° Celsius, and the organizers had to do something to prevent the snow from melting. I heard on the TV the commentators saying that they were discharging salt on the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However where I live (France), we have occasionnally snow in the winter, and to keep the roads clear for the vehicles or the sidewalks clear for the pedestrians, city services also discharge salt !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible to have these two different effects with the (seemingly) same component ?&lt;/p&gt;&#xA;" OwnerUserId="5384" LastActivityDate="2014-05-28T01:17:25.230" Title="Avoiding snow melting with salt" Tags="&lt;melting-point&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10418" PostTypeId="2" ParentId="10410" CreationDate="2014-05-02T12:32:01.660" Score="2" Body="&lt;p&gt;Pure water contains only small amounts of $\ce{H3O+}$ and $\ce{OH-}$ which are generated by autoprotolysis (the reaction given by Klaus Warzecha in his comment to your question). Reduction of $\ce{H3O+}$ at the cathode produces molecular hydrogen. At the anode, water is in turn oxidized, yielding molecular oxygen and $\ce{H3O+}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Cathode:}\ \ce{2H3O+ + 2e- \rightarrow H2 +2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Anode:}\ \ce{6H2O \rightarrow O2 + 4H3O+ + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Overall\ reaction:}\ \ce{2H2O \rightarrow 2H2 + O2}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-02T12:45:15.183" LastActivityDate="2014-05-02T12:45:15.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10419" PostTypeId="1" AcceptedAnswerId="10420" CreationDate="2014-05-02T15:19:33.410" Score="22" ViewCount="24284" Body="&lt;p&gt;Over the course of my studies, my lab coat has absorbed a non-negligible amount of different chemicals, some of them nasty organic polymers, some are more insoluble inorganic compounds like $\ce{Cr2O3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even after washing the lab coat some stains remain. Is there a way to get rid of those, or should I rather buy a new lab coat?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2018-04-11T04:53:56.757" Title="How can I clean my lab coat?" Tags="&lt;safety&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10420" PostTypeId="2" ParentId="10419" CreationDate="2014-05-02T16:06:37.773" Score="26" Body="&lt;p&gt;A lab coat will eventually get dirty, that's why it is a good idea to wear it. I had a lot of students complaining about the condition of their coat and they asked me if they could just wash it in the washing machine. My answer was always &lt;strong&gt;&lt;em&gt;no&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I remember - I am a theoretician now - a labcoat costs much less than a pair of trousers and a T-shirt. It should be renewed annually and disposed of according to the norms how you would dispose of an unknown chemical. You never know where the chemicals incorporated in your coat will end up, so dispose of them as safely as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some countries, it is required by the employer to provide equipment like that and taking care of either cleaning or disposing of it eventually. (In Germany that should be the case.) Wear it as a disposable layer of protection and please - for the safety of the environment and yourself - do never ever put it in a normal washing machine again.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an additional note, if you wash it in the same washer like your normal clothes, how can you assure that it will not end up in your everyday clothes? I can imagine, that having chromium whatever in my underwear would not make me feel alright.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2018-04-08T05:07:01.720" LastActivityDate="2018-04-08T05:07:01.720" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10421" PostTypeId="2" ParentId="10419" CreationDate="2014-05-02T16:07:03.750" Score="1" Body="&lt;p&gt;Use or discard.  There is no &quot;clean.&quot;  DuPont Teflon Advanced or an old can of perfluoroctanesulfonate Scotchguard for the new lab coat?  Classic Scotchguard, &#xA;&lt;img src=&quot;https://i.stack.imgur.com/qZ45p.jpg&quot; alt=&quot;classic Scotchguard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An eaten lab coat decorated with holes, stains, and blackened smears of stopcock grease along the sides is a badge of honor.  You keep a pressed new one aside for when hominims in suits visit (JAFOs).  Formal occasions demand the Black Lab Coat.  They cannot be purchased.  As with lightsabers, you must make your own.  Given the potential toxicity of aniline black, one recommends starting with a new 100% cotton lab coat washed (remove sizing) and double rinsed.  Then, Procion MX and/or Cibacon F reactive blacks (and lots of non-iodized salt). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.worlddyevariety.com/disperse-dyes/disperse-blue-354.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.worlddyevariety.com/disperse-dyes/disperse-blue-354.html&lt;/a&gt; &lt;BR&gt;&#xA;If you are a polyester sort of fellow, Foron Brilliant Blue disperse dyes.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="402" LastEditDate="2018-04-08T05:06:00.900" LastActivityDate="2018-04-08T05:06:00.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10422" PostTypeId="2" ParentId="10154" CreationDate="2014-05-02T18:24:08.607" Score="1" Body="&lt;p&gt;I have seen more than one website recommend a canary. If you'd prefer not to have a canary in the house I highly recommend investing in a CO detector... they save lives!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&#xA;If you would still not like to have a CO detector; I highly recommend knowing what the symptoms of CO poisoning are, if you experience these symptoms and believe they maybe caused by CO, get out of the house and call for help immediately! Flu-like symptoms without the fever, dizziness and headache. Here are two links, I hope you find their information helpful!&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&#xA;&lt;p&gt;&#xA;&lt;a href=&quot;http://www.nytimes.com/2004/01/18/realestate/your-home-detecting-carbon-monoxide.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nytimes.com/2004/01/18/realestate/your-home-detecting-carbon-monoxide.html&lt;/a&gt;&#xA;&lt;p&gt;&#xA;&lt;a href=&quot;http://www.health.state.mn.us/divs/eh/indoorair/co/&quot; rel=&quot;nofollow&quot;&gt;http://www.health.state.mn.us/divs/eh/indoorair/co/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5137" LastActivityDate="2014-05-02T18:24:08.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10423" PostTypeId="2" ParentId="10283" CreationDate="2014-05-02T18:31:08.267" Score="1" Body="&lt;p&gt;Compared to lead(II) acetate, lead subacetate contains additionally $\ce{2 Pb(OH)2}$. The hydroxide can either be neutralized by protons from EDTA or take part in the hydrolysis of thiourea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SC(NH2)2 + OH- \rightleftharpoons OC(NH2)2 + HS-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HS- + OH- \rightleftharpoons S^{2-} + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH of the solution should be higher unless all hydroxide is neutralized or consumed by the hydrolysis reaction.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-02T18:31:08.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10424" PostTypeId="1" AcceptedAnswerId="10425" CreationDate="2014-05-02T13:08:56.963" Score="7" ViewCount="2099" Body="&lt;p&gt;I have a bottle of Powerade and after probably around 20+ refills of water, the new water I put in still tastes/smells a little like Powerade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that happen?&lt;/p&gt;&#xA;" OwnerUserId="1312" OwnerDisplayName="mateoc" LastEditorUserId="73" LastEditDate="2014-05-02T19:55:21.660" LastActivityDate="2014-05-02T19:55:21.660" Title="Why does the water in a Powerade bottle tastes a little like Powerade after many refills?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10425" PostTypeId="2" ParentId="10424" CreationDate="2014-05-02T14:03:16.997" Score="18" Body="&lt;p&gt;Typically water bottles used in sports or biking are manufactured from high-density polyethylene (HDPE), which is essentially like a longer-chain version of paraffin wax. In short, it's basically a solid nonpolar hydrocarbon (with the exception of the Zeigler-Natta catalysts, which are only present in trace amounts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, a quick stroll through &lt;em&gt;Fenaroli's Handbook of Flavor Ingredients&lt;/em&gt; shows that many of the flavor compounds typically found in foods are either somewhat nonpolar, have nonpolar sections, or have (mostly unsubstituted) aromatic rings present. For example, ethyl valerate, one of the primary constituents of apple aroma, has a butyl chain, is fairly nonpolar, and is nearly insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it may be soluble in HDPE, and so it may infuse the bottle with the flavor agents so that the taste can remain for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, doing a double-blind test may be your best bet for checking to see if this is a real phenomenon, or your imagination. But I've personally noticed this as well, so it could very well be real.&lt;/p&gt;&#xA;" OwnerUserId="4987" OwnerDisplayName="DumpsterDoofus" LastActivityDate="2014-05-02T19:23:41.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10426" PostTypeId="2" ParentId="10419" CreationDate="2014-05-02T21:34:10.397" Score="8" Body="&lt;p&gt;You should &lt;em&gt;never&lt;/em&gt; wash your own lab coat. That defeats the whole purpose of lab coats: separating yourself from the potentially hazardous chemicals you work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your employer may have a program for laundering lab coats. My university has a program for safely and properly laundering lab coats. This way you can keep it reasonably clean until replacement.&lt;/p&gt;&#xA;" OwnerUserId="5387" LastActivityDate="2014-05-02T21:34:10.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10427" PostTypeId="2" ParentId="5515" CreationDate="2014-05-02T23:28:58.013" Score="1" Body="&lt;p&gt;Charge density is a measure of electrical charge per volume. It is independent from electronegativity, and increases with the number of charges of an ion and with decreasing radius, because a smaller ion has also a smaller volume. The number of charges for $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ is the same (+2), but $\ce{Mg^{2+}}$ has the smaller atomic radius. Its charge is therefore distributed over a smaller volume, and the charge density is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrated radius of an ion can somewhat differ from its atomic radius, because it is influenced by the charge density. If the ion charge is concentrated in a smaller volume, it exerts stronger attractive forces on the dipoles of water molecules, which results in the formation of a larger hydration shell. The smaller $\ce{Mg^{2+}}$ attracts more water dipoles and forms a larger hydration shell. Hence, its hydrated radius can be bigger than that of $\ce{Ca^{2+}}$, which has a larger atomic radius but a smaller hydrate shell.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-02T23:28:58.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10428" PostTypeId="1" AcceptedAnswerId="10429" CreationDate="2014-05-03T01:36:59.747" Score="7" ViewCount="43809" Body="&lt;p&gt;I want to do a density demonstration, and I want to inhale helium and sulfur hexafluoride as a fun demonstration. I am open to using any source to purchase from, and money is not an issue (but preferably as inexpensive as possible), but what concentration should I buy (what concentration is safe to inhale), and from where?&lt;/p&gt;&#xA;" OwnerUserId="5389" LastEditorUserId="5389" LastEditDate="2015-01-06T22:31:17.667" LastActivityDate="2017-04-09T18:29:42.630" Title="What concentration of sulfur hexafluoride is safe to inhale" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10429" PostTypeId="2" ParentId="10428" CreationDate="2014-05-03T05:17:03.823" Score="15" Body="&lt;p&gt;I'm afraid this answer doesn't address where you can get $\ce{SF6}$, as I don't know any suppliers myself (it would probably help to specify where you are, though). I just want to straighten out the safety details and make sure you (and anyone else who reads this in the future) know what you're doing. All chemistry experiments can take a turn for the worse if done without adequate preparation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure sulfur hexafluoride is a rather chemically inert gas (unlike the related but very toxic sulfur tetrafluoride $\ce{SF4}$ and disulfur decafluoride $\ce{S2F10}$), and it can be safely handled in regular situations. Decomposition of $\ce{SF6}$ usually requires aggressive conditions, &lt;a href=&quot;http://www.nature.com/nature/journal/v193/n4814/abs/193473a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;such as exposure to molten sodium metal at 200°C&lt;/a&gt;. There are some situations where $\ce{SF6}$ reacts at low temperatures, but that requires &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/134/10/10.1063/1.3560908?ver=pdfcov&quot; rel=&quot;nofollow noreferrer&quot;&gt;irradiation with deep-UV photons&lt;/a&gt; or &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic50018a033&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct contact with solvated electrons from alkali metals dissolved in ammonia&lt;/a&gt;. I don't suppose you'll have to worry about such situations. It also seems that $\ce{SF6}$ can decompose into a small amount of the other aforementioned toxic sulfur fluorides &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_hexafluoride#Dielectric_medium&quot; rel=&quot;nofollow noreferrer&quot;&gt;if it is exposed to electrical arcs&lt;/a&gt;, but since it is an electrically insulating gas, &lt;a href=&quot;https://www.youtube.com/watch?v=u19QfJWI1oQ#t=51s&quot; rel=&quot;nofollow noreferrer&quot;&gt;the electric fields required for arcing are even higher than in air&lt;/a&gt;, so this also shouldn't be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, both pure $\ce{SF6}$ and helium are very unlikely to cause chemical damage to the body, even indirectly. However, &lt;strong&gt;that doesn't imply they are completely safe&lt;/strong&gt;. Of course, they can still act as an asphyxiant simply by displacement of oxygen in the lungs (&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inert_gas_asphyxiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;make sure to read this!&lt;/a&gt;&lt;/strong&gt;). &lt;em&gt;Be aware that the brain is incapable of directly measuring the oxygen content in your blood&lt;/em&gt;, and instead measures oxygenation rather indirectly, by the amount of carbon dioxide. &lt;strong&gt;It is possible to faint and even die from hypoxia without any warning&lt;/strong&gt;. Additionally, since $\ce{SF6}$ is relatively heavy, it requires more effort from your diaphragm muscles to expel from your lungs, making $\ce{SF6}$ inhalation slightly more dangerous than helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a safety measure, &lt;em&gt;do not take consecutive breaths in the non-oxygenated gas and do not spend more than 5 seconds without inhaling air&lt;/em&gt;. While your lungs are devoid of oxygen, not only are they unable to transport oxygen into your blood, &lt;em&gt;the lungs actually work in reverse and help expel oxygen from your blood&lt;/em&gt;. After performing the trick, take deep breaths in regular air for a minute or two before trying it again, to be sure all the previously inhaled $\ce{SF6}$ or helium has exited your lungs and your oxygen supply has returned to normal. Hyperventilating before or after inhaling $\ce{SF6}$/helium is likely a bad idea, as it further messes with the balance of oxygen and carbon dioxide in the blood. You should avoid any possibility of becoming light-headed and fainting, especially if you decide to do the trick multiple times. Also be mindful that the gasses can accumulate in poorly ventilated spaces, with $\ce{SF6}$ being a particular problem as it will tend to stay close to the ground and disperse with more difficulty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going through the trouble to get $\ce{SF6}$, you should do the floating aluminium boat display, too. It's pretty cool! &lt;a href=&quot;https://www.youtube.com/watch?v=DzLX96VWTkc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's another video&lt;/a&gt; showing you can even pour the gas into the boat and make it sink.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-09T18:29:42.630" LastActivityDate="2017-04-09T18:29:42.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10431" PostTypeId="1" CreationDate="2014-05-03T11:06:19.447" Score="2" ViewCount="584" Body="&lt;p&gt;I already know that thymolphthalein is soluble in alcohols, very alkaline aqueous solutions, acetone, and propylene glycol methyl ether.  My questions are what other solvents dissolve thymolphthalein, and what additives increase solubility (for example, specific surfactants)?&lt;/p&gt;&#xA;" OwnerUserId="5394" LastActivityDate="2014-05-03T23:48:51.120" Title="Thymolphthalein solubility" Tags="&lt;acid-base&gt;&lt;solubility&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10432" PostTypeId="1" CreationDate="2014-05-03T11:10:39.110" Score="2" ViewCount="126" Body="&lt;p&gt;I have question about hetero-cycles: Why do we use (the most common hetero-cyclic elements) nitrogen, oxygen, sulfur as hetero compounds most of time?&lt;/p&gt;&#xA;" OwnerUserId="5393" LastEditorUserId="4356" LastEditDate="2014-10-24T23:44:55.963" LastActivityDate="2014-10-24T23:44:55.963" Title="What's unique about the most commonly used hetero-cyclic elements?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-10T03:25:28.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="10435" PostTypeId="2" ParentId="10432" CreationDate="2014-05-03T14:48:43.817" Score="3" Body="&lt;p&gt;A possible reason for this is that heterocycles with O, N or S are often easier to obtain and more stable than the corresponding compounds which contain other main group elements as heteroatoms. &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphole&quot; rel=&quot;nofollow&quot;&gt;Phosphole&lt;/a&gt;, for example, is less stable than furan, pyrrole and thiophene because of its lower aromaticity. &lt;a href=&quot;http://en.wikipedia.org/wiki/Borole&quot; rel=&quot;nofollow&quot;&gt;Borole&lt;/a&gt;, the structural analog with the lighter heteroatom B, has not yet been isolated in unsubstituted form. Ring strain and the stability of X-C and X-H bonds (X = heteroatom) also need to be considered. Bond lengths tend to increase with heavier heteroatoms. With decreasing electronegativity of X in relation to C, the X-C bond becomes less covalent and is more prone to be cleaved by hydrolysis, for example.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-03T14:48:43.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10439" PostTypeId="2" ParentId="7183" CreationDate="2014-05-03T16:26:11.907" Score="-1" Body="&lt;p&gt;You only need to know that at STP $V_m=24.4\:\mathrm{L\:mol^{-1}}$, and that $n$ (number of moles of of a compound or molecule) is equal to $\frac{V_{gas}}{V_m}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find limiting and excess reagents using stoichiometric ratios (in this case it is 1:1), therefore $\ce{H2}$ is limiting the limiting reagent.&#xA;Use stoichiometry to find amount of $\ce{Cl2}$ reacted, and then subtract the initial number of moles from the final number of moles.&lt;/p&gt;&#xA;" OwnerUserId="5299" LastEditorUserId="15410" LastEditDate="2015-05-19T06:29:24.083" LastActivityDate="2015-05-19T06:29:24.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10440" PostTypeId="2" ParentId="5027" CreationDate="2014-05-03T18:20:29.207" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mordant is a substance used to set dyes on fabrics or tissue&#xA;  sections by forming a coordination complex with the dye which then&#xA;  attaches to the fabric or tissue (&lt;a href=&quot;http://en.wikipedia.org/wiki/Mordant&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tannin is a biomolecule consisting of many carbohydrate and polyphenol residues with plenty of hydroxyl and carboxyl groups. Tannic acid is a common type of tannin:&lt;img src=&quot;https://i.stack.imgur.com/vew67.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ form colored complexes with these oxygen-containing functional groups. These iron-tannin complexes change the color of the wood. The color intensity increases with tannin content of the wood and can be increased by the addition of tannin-rich tea. Structures with a varying coordination sphere of iron are formed, some of which are shown here (sources: &lt;a href=&quot;http://dx.doi.org/10.5006/1.3294344&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; and &lt;a href=&quot;http://www.diva-portal.org/smash/get/diva2:29759/FULLTEXT01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O1rvj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-06T14:56:07.427" LastActivityDate="2014-05-06T14:56:07.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10441" PostTypeId="2" ParentId="10419" CreationDate="2014-05-03T19:40:31.507" Score="12" Body="&lt;p&gt;Lab coats, like other forms of &lt;a href=&quot;http://en.wikipedia.org/wiki/Personal_protective_equipment&quot; rel=&quot;nofollow noreferrer&quot;&gt;PPE&lt;/a&gt;, are supposed to provide a final barrier of protection, for when all other risk mitigation strategies have failed. They are not just an expensive form of a fancy cooking apron. If you are getting spills on your lab coat, you should take this as an opportunity to re-examine your laboratory techniques to try to establish why this is happening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contamination (including wiping hands) on lab coats should be treated at the time of incident, and contaminated lab coats should not be removed from the laboratory, unless in a contaminated laundry or waste container. Most workplaces will have a well-defined plan for laundering lab coats, and strict guidelines for how often this should be done. &lt;a href=&quot;https://www.google.com.au/?q=cleaning%20lab%20coats%20SOP#q=cleaning%20lab%20coats%20SOP&quot; rel=&quot;nofollow noreferrer&quot;&gt;You can search for just how common this is&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good lab coats can be expensive, but think of them a little like fire extinguishers. Most days in the lab, you shouldn't even need one. However, you want to make sure that when you DO need them, they will work properly and do the job they are designed for. Regular laundering of a non-contaminated lab coat will keep it in top condition, but once contaminated with hazardous chemicals that cannot be safely neutralized, they should then be replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question is: no, you should not attempt to launder your lab coat. Buy a new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a final 2 cents worth: Dirty lab coats are not a badge of honour; they are a sign of a sloppy chemist. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="402" LastEditDate="2018-04-08T05:06:30.920" LastActivityDate="2018-04-08T05:06:30.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10443" PostTypeId="2" ParentId="7578" CreationDate="2014-05-03T22:58:21.670" Score="2" Body="&lt;p&gt;Iron(III) acetate will be stable under these conditions. According to this &lt;a href=&quot;http://link.springer.com/article/10.1007/BF01981811&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, thermal decomposition to iron oxide only starts at 423 K (150 °C). Heating iron and acetic acid together with an oxidation agent (oxygen from air or, like in your case, $\ce{H2O2}$) is actually a suitable way to prepare iron acetate dyes (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_acetate&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-03T22:58:21.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10445" PostTypeId="2" ParentId="10431" CreationDate="2014-05-03T23:48:51.120" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6Bn8.png&quot; alt=&quot;Thymolphthaleine synthesis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One expects [poly(ethylene oxide)-co-(polypropylene oxide)-co-poly(ethylene oxide)] triblock surfactants will micellize it (e.g., Pluronics); possibly Tergitols.  It looks too bulky to fit inside even gamma-cyclodextrin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hang a suitable side chain off the phthalic anyhydride and the product will dissolve in pretty much anything.  What does pH mean outside water?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-03T23:48:51.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10446" PostTypeId="1" CreationDate="2014-05-04T05:02:53.603" Score="5" ViewCount="1697" Body="&lt;p&gt;The Wurtz reaction is $$\ce{R-X + Na -&amp;gt;[ether] R-R}$$ where $\ce{R}$ is alkyl. But if we take butylchloride with $\ce{Na}$, the product is n-octane and not 3,4-dimethylhexane. I am unable to understand the fact that rearrangement does not happen even when it's a free radical mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4133" LastEditorUserId="5026" LastEditDate="2018-04-15T11:51:59.277" LastActivityDate="2018-04-15T11:53:27.207" Title="Why does rearrangement not take place in Wurtz reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10447" PostTypeId="2" ParentId="10446" CreationDate="2014-05-04T06:39:36.807" Score="-3" Body="&lt;p&gt;You are making it too complicated, but it's really simple: $$\ce{Bu-Cl + 2Na + Cl-Bu -&amp;gt; Bu-Bu + 2NaCl}$$ where $\ce{Bu-{}}$ is the n-butyl group and $\ce{Bu-Bu}$ is n-octane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can understand wurtz's reaction much easier this way.&lt;/p&gt;&#xA;" OwnerUserId="5398" LastEditorUserId="5026" LastEditDate="2018-04-15T11:53:27.207" LastActivityDate="2018-04-15T11:53:27.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10450" PostTypeId="2" ParentId="6692" CreationDate="2014-05-04T10:10:01.953" Score="3" Body="&lt;p&gt;First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the mass of our solution:&#xA;$$ m = \rho \times V = 103.7~\mathrm{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):&#xA;$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total heat capacity of your system is given as&#xA;$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$&#xA;where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know all the quantities needed for the following equation:&#xA;$$ q = c_\text{tot}\Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We expand:&#xA;$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $c_\text{cal}$:&#xA;$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now you know the heat capacity of the calorimeter.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-05-04T10:10:01.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10451" PostTypeId="1" AcceptedAnswerId="10502" CreationDate="2014-05-04T10:10:05.223" Score="0" ViewCount="191" Body="&lt;p&gt;In a thought experiment, a friend proposed to me a machine capable of decomposing any object (e.g., a brick) into its molecules. I'm not a chemist or physicist, but my initial reaction was that this was certainly an environmental issue if used in a larger scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that right? Or is a molecule so basic that there would be no health concern from such a (theoretical) decomposition.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; @Aesin: &#xA;My thinking was:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;objects can't disappear; the brick will be breathed in, get into&#xA;water etc.&lt;/li&gt;&#xA;&lt;li&gt;if we start decomposing more advanced objects, we&#xA;get all kinds of harmful elements in the air (mercury etc.)&lt;/li&gt;&#xA;&lt;li&gt;possible radiation from decomposition&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for the purpose: Let's assume it would be used for general waste destruction. Let's assume energy is not a concern (thanks for mentioning it!)&lt;/p&gt;&#xA;" OwnerUserId="5401" LastEditorUserId="5401" LastEditDate="2014-05-05T07:00:34.140" LastActivityDate="2017-04-09T18:28:55.707" Title="Would decomposing an object into molecules be an environmental problem?" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2019-12-06T15:30:53.617" ContentLicense="CC BY-SA 3.0" />
  <row Id="10453" PostTypeId="1" CreationDate="2014-05-04T10:30:31.083" Score="6" ViewCount="1022" Body="&lt;p&gt;Reading Wikipedia on &quot;Beta Decay&quot; they list the equation of Carbon-14 decaying into N-14.&#xA;see equation &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot;&gt;https://en.wikipedia.org/wiki/Beta_decay&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume the 14 refers to total protons and neutrons.  The 6 in C, I believe means six protons.  In N, I believe the 7 refers to protons. Where did the seventh proton come from the decay of  C-14 into N-14?  I understand neutrons can decay into protons, but the math doesn't work for me.&lt;/p&gt;&#xA;" OwnerUserId="5404" LastEditorUserId="4325" LastEditDate="2016-01-13T15:45:41.063" LastActivityDate="2016-01-13T15:45:41.063" Title="Understanding Beta Decay" Tags="&lt;physical-chemistry&gt;&lt;nuclear&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10454" PostTypeId="2" ParentId="10446" CreationDate="2014-05-04T12:19:43.273" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wurtz Reaction&lt;/a&gt; mechanism involves radical anions. Neither radicals nor carbanions rearrange &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbocation_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;the same way that carbocations do&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The first step of the mechanism is a single electron transfer from &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{R-X}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Na + R-X -&amp;gt; Na+ +[R-X].^{-}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The alkyl halide radical anion fragments to a carbon radical and a halide anion:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[R-X].^{-}-&amp;gt;R. + X-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R-CH2-CH2. +.CH2-CH2-R-&amp;gt;R-CH=CH2 + CH3-CH2-R}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;... or coupling:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R-CH2-CH2. + .CH2-CH2-R -&amp;gt; R-CH2-CH2-CH2-CH2-R}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; is a good reducing agent, and that primary radical is readily reduced to an anion:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R. + Na -&amp;gt; R:^{-} + Na+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The carbanion then reacts with a molecule of alkyl halide in a S&lt;sub&gt;N&lt;/sub&gt;2 reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R:^{-} + X-R -&amp;gt; R-R + X-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The formation of the anion is actually problematic, especially if &lt;span class=&quot;math-container&quot;&gt;$\ce{R-X}$&lt;/span&gt; is secondary or tertiary, since E2 elimination will compete.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{(CH3)3C:^{-} + (CH3)3C-X -&amp;gt; (CH3)3C-H + (CH3)C=CH2 + X-}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-15T11:53:00.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10455" PostTypeId="2" ParentId="10453" CreationDate="2014-05-04T12:51:34.827" Score="12" Body="&lt;p&gt;For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{_{Z_1}^{A_1}X -&amp;gt; _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$&#xA;$$Z_1 = Z_2 + Z_3$$&#xA;$$A_1 = A_2 + A_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{_{Z}^{A}X -&amp;gt; _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrons decay into protons &lt;strong&gt;and electrons&lt;/strong&gt; (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{_0^1n -&amp;gt; _1^1p+ + _{0}^0e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{_6^{14}C -&amp;gt; _7^{14}N+ +e- -&amp;gt; _7^{14}N}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-04T12:51:34.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10456" PostTypeId="1" AcceptedAnswerId="10462" CreationDate="2014-05-04T13:01:50.033" Score="1" ViewCount="61" Body="&lt;p&gt;I have the following question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following would not be used to measure changes in atmospheric gases?&lt;/p&gt;&#xA;&lt;p&gt;i) Satellites&lt;/p&gt;&#xA;&lt;p&gt;ii) Mass Spectrometer&lt;/p&gt;&#xA;&lt;p&gt;iii) Digital Thermometer&lt;/p&gt;&#xA;&lt;p&gt;iv) Fractional Distillation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I assumed it would be fractional distillation since you cannot really measure anything with it? You only separate the fractions of different materials such as crude oil.&lt;/p&gt;&#xA;&lt;p&gt;However the answer was Digital Thermometer. I guess I can understand how this object would be useless at measuring atmospheric changes relating to gases however how would you detect atmospheric gases changes using fractional distillation?&lt;/p&gt;&#xA;" OwnerUserId="5378" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-05-04T18:49:45.983" Title="Atmospheric changes relating to gases" Tags="&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10457" PostTypeId="1" CreationDate="2014-05-04T13:22:10.227" Score="10" ViewCount="14821" Body="&lt;p&gt;For many salts there are solubility curves as a function of temperature that are smooth (don't have any kinks). Sodium sulfate, however, has a kink in the solubility-T curve at 30 degrees as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oBfxA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when the temperature is about 30 degree?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="1362" LastEditDate="2014-05-04T17:33:18.987" LastActivityDate="2016-04-17T08:58:45.433" Title="Why does sodium sulfate have an unusual solubility-temperature curve?" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10458" PostTypeId="2" ParentId="10457" CreationDate="2014-05-04T14:48:21.010" Score="10" Body="&lt;p&gt;&quot;Sodium sulfate&quot; is a mixture of hydrated phases in shifting equilibrium given temperature and concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9CNE5.gif&quot; alt=&quot;Na2SO4.(H2O)x phase diagram, 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uY5uj.jpg&quot; alt=&quot;Na2SO4.(H2O)x phase diagram, 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-04T14:48:21.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10461" PostTypeId="1" AcceptedAnswerId="10474" CreationDate="2014-05-04T18:02:28.897" Score="2" ViewCount="395" Body="&lt;p&gt;Working on a practice exam. The question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A volume of gas contains&lt;br&gt;&#xA;  0.25 moles of $\ce{CO2}$&lt;br&gt;&#xA;  0.30 moles of $\ce{N2}$&lt;br&gt;&#xA;  0.15 moles of $\ce{O2}$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the partial pressure of $\ce{O2}$ if the total pressure was $690\ \mathrm{mmHg}$ and the temperature was $25.0\ \mathrm{^\circ C}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know the formula is $p_a=\dfrac{n_aRT}{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_a = 0.25\ \mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R  = 0.0821\ \mathrm{l\ atm\ K^{−1}\ mol^{−1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = 298\ \mathrm K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't seem to figure out volume. I was reading somewhere that volume is directly proportional to The amount of moles, but maybe it has something to do with the total pressure. I'm a little lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of would be getting volume via the Ideal Gas Law as well. Would that work?&lt;/p&gt;&#xA;" OwnerUserId="5104" LastEditorDisplayName="user7951" LastEditDate="2018-05-22T19:50:52.583" LastActivityDate="2018-05-22T19:50:52.583" Title="Ideal gas law? implied volume?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10462" PostTypeId="2" ParentId="10456" CreationDate="2014-05-04T18:49:45.983" Score="1" Body="&lt;p&gt;Assuming 'change' means modification of composition, the answer is Digital Thermometer because even if temperature is linked to atmosphere composition it does not provide you any information about it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Satelite can embed LIDAR system which will provide you crude qualitative (spectrum) and quantitative (intensities) informations on atmosphere composition, but it requires complex mathematical processing and relies on weak hypothesis;&lt;/li&gt;&#xA;&lt;li&gt;Mass spectrography is probably the most accurate way because it is highly qualitative (retention time) and quantitative (ionization current);&lt;/li&gt;&#xA;&lt;li&gt;Distillation will allow you to qualitatively 'separate' (boiling point) and 'quantify' (mass flow rate) distinct components.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously temperature measurement is not able to provide any informations of this kind. Further more, temperature gradient in atmosphere is not only dominated by its composition, there is a lot of other phenomena involved in it.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-05-04T18:49:45.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10463" PostTypeId="2" ParentId="10453" CreationDate="2014-05-04T19:04:38.057" Score="3" Body="&lt;p&gt;Charge and energy are always conserved. Mass is a form of energy, and what we call 'mass number' - the sum of neutrons + protons - is a close approximation to mass (because electrons are so light, having just over a two-thousandth of the mass of a proton or a neutron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversion of a neutron into a proton when an electron is released takes care of charge conservation, because the charge on the proton is exactly the opposite of the charge on the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy needed to create the electron in beta decay comes from the reduced binding energy of the nucleons in N-14 compared with those in C-14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that make sense of the maths for you?&lt;/p&gt;&#xA;" OwnerUserId="5407" LastActivityDate="2014-05-04T19:04:38.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10464" PostTypeId="1" AcceptedAnswerId="10469" CreationDate="2014-05-04T21:04:38.167" Score="6" ViewCount="6614" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$55.0~\mathrm{mL}$ of $0.250~\mathrm{M}~\ce{NaOH}$ is used to titrate $35.0~\mathrm{ml}$ of $\ce{H2SO4}$. &#xA;  What is the molarity of $\ce{H2SO4}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the equation for this reaction is:&#xA;$$\ce{H2SO4 + 2NaOH &amp;lt;=&amp;gt; Na2SO4 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I have to use $$i_\mathrm a\cdot M_\mathrm a\cdot V_\mathrm a = i_\mathrm b\cdot M_\mathrm b\cdot V_\mathrm b$$ formula where $i$ is the number of ions, $M_\mathrm a$ and $M_\mathrm b$ are the molarities (of the acid and base, respectively), and $V_\mathrm a$ and $V_\mathrm b$ are the volumes of the acid and base (in milliliters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I starting doing this problem and got up to this point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i\cdot M_\mathrm a\cdot 35~\mathrm{ml} = 2\cdot 0.250~\mathrm{M}\cdot 55~\mathrm{ml}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: &#xA;What should I put for the number of ions on the left hand side? Is it three or two?&lt;/p&gt;&#xA;" OwnerUserId="5409" LastEditorUserId="5026" LastEditDate="2018-03-01T08:17:11.360" LastActivityDate="2020-08-07T13:13:41.907" Title="Titration of sulfuric acid with sodium hydroxide" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10465" PostTypeId="1" AcceptedAnswerId="10467" CreationDate="2014-05-04T21:15:04.737" Score="8" ViewCount="2904" Body="&lt;p&gt;A few drops of water can initiate a reaction between iodine and aluminium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=SKSU72-1ERc&quot;&gt;https://www.youtube.com/watch?v=SKSU72-1ERc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this happen, since iodine is only slightly soluble in water?&lt;/p&gt;&#xA;" OwnerUserId="5410" LastActivityDate="2017-03-07T03:49:17.327" Title="How can water catalyse a reaction between iodine and aluminium?" Tags="&lt;catalysis&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10466" PostTypeId="2" ParentId="10465" CreationDate="2014-05-04T21:29:05.643" Score="6" Body="&lt;p&gt;It is very difficult to have a solid react with another solid due to the low contact area and the inability of the molecules to obtain the activation energy. Even though there is little iodine dissolved in the water, the contact area is greatly increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the reaction starts, the heat of the reaction provides the activation energy and the gaseous iodine (which makes the contact area large).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-04T21:29:05.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10467" PostTypeId="2" ParentId="10465" CreationDate="2014-05-04T22:05:34.017" Score="12" Body="&lt;p&gt;A naked aluminum surface (freshly sanded) in air immediately grows about 4 nm of adherent aluminum oxide that passivates the surface.  Iodine slightly dissolves in water and disproportionates&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I2_{(s)} + H2O_{(l)} -&amp;gt; HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine is strongly solubilized by forming triiodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I2 + I- -&amp;gt; [I3]-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acids chew through the alumina surface barrier.  When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm&quot; rel=&quot;nofollow&quot;&gt;http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm&lt;/a&gt;&lt;br&gt;&#xA;$$\begin{alignat}{2}&#xA;&amp;amp;\ce{AlI3_{(s)}}\quad &amp;amp;&amp;amp;\Delta G_\text{f}^\circ = -300.8\ \mathrm{kJ/mol}\\ &#xA;&amp;amp;\ce{Al2O3_{(s)}}\quad &amp;amp;&amp;amp;\Delta G_\text{f}^\circ = -1582.4\ \mathrm{kJ/mol}\\&#xA;\end{alignat}$$&#xA;Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T16:35:49.367" LastActivityDate="2015-04-30T16:35:49.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10469" PostTypeId="2" ParentId="10464" CreationDate="2014-05-05T00:03:01.347" Score="6" Body="&lt;p&gt;That is one way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, this is a stoichiometric question where you are trying to find the molarity of an unknown polyprotic acid by titrating it with a strong base (woo-hoo!) of known molarity. I prefer using this method as it is easier to see the relationship between the units as well as use chain-ink conversion which most people are familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{M} = \frac{\text{mol}}{\text{L}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 55 \ \text{mL} \ \ce{ NaOH} * \frac{250 \ \text{mmol} \ \ce{NaOH}}{1000 \ \text{mL} \ \ce{NaOH}} * \frac{1 \ \text{mmol} \ \ce{H2SO4}}{2 \ \text{mmol} \ \ce{NaOH}} * \frac{1}{ 35 \ \text{mL} \ \ce{H2SO4}} = $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which equals &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0.19643 \ \frac{\text{mol}}{\text{L}} $$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2014-05-05T00:19:57.420" LastActivityDate="2014-05-05T00:19:57.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10471" PostTypeId="1" AcceptedAnswerId="10721" CreationDate="2014-05-05T08:00:35.070" Score="5" ViewCount="12015" Body="&lt;p&gt;What is the reaction mechanism of the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 + O3 -&amp;gt; H2O + 2O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried searching this on Google but am not getting good answers.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="402" LastEditDate="2017-05-02T23:26:26.987" LastActivityDate="2020-08-29T14:38:53.227" Title="Reaction between hydrogen peroxide and ozone" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10474" PostTypeId="2" ParentId="10461" CreationDate="2014-05-05T10:17:46.273" Score="6" Body="&lt;p&gt;You don't need the Ideal Gas Law. You can use it (and I'll show you how). The question asks for &lt;strong&gt;partial pressure&lt;/strong&gt;. You want &lt;a href=&quot;http://en.wikipedia.org/wiki/Dalton%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Dalton's Law of Partial Pressures&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dalton's Law&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dalton's Law states that the total pressure in a system is equal to the sum of the partial pressures of the components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_T=P_A + P_B + P_C + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant volume and temperature (which you have, even if you don't know the volume), the ideal gas law can be rearranged so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P}{n}=\frac{RT}{V}$$&#xA;$$\frac{P_T}{n_T}=\frac{P_A}{n_A}$$&#xA;$$P_A=\frac{n_A}{n_T}P_T$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ideal Gas Law&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find the volume of the system because you have the total moles $n_T$ and the total pressure $P_T$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\frac{n_T RT}{P_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then replace the number of moles with $n_A$ and solve for $P_A$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_A=\frac{n_A RT}{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach is basically equivalent to the previous except: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using the ideal gas law you have to do two calculation (volume and partial pressure), while you only need to do one with Dalton's Law&lt;/li&gt;&#xA;&lt;li&gt;Fewer math steps means less rounding, which means less error and higher precision.&lt;/li&gt;&#xA;&lt;li&gt;If you use Dalton's Law, you don't have to worry about the units of temperature or pressure. Temperature is not in the equation, and the pressures are a ratio. You can use whatever pressure unit you want without converting.&lt;/li&gt;&#xA;&lt;li&gt;You don't need to worry about $R$ and its units in the Dalton's Law approach.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solve problems as ratios whenever you have the chance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="61578" LastEditDate="2018-05-22T04:05:29.310" LastActivityDate="2018-05-22T04:05:29.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10476" PostTypeId="1" CreationDate="2014-05-05T14:52:26.037" Score="1" ViewCount="545" Body="&lt;p&gt;I know that fractional distillation is used to separate two liquids that are miscible in one another and have close boiling points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am really confused about vacuum and steam distillation - in which cases do we use these separation techniques? Can someone please give examples.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2014-07-29T11:46:45.020" LastActivityDate="2014-07-29T11:46:45.020" Title="Different techniques of distillations" Tags="&lt;experimental-chemistry&gt;&lt;purification&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10477" PostTypeId="1" CreationDate="2014-05-05T14:52:28.947" Score="19" ViewCount="9423" Body="&lt;p&gt;I recently added a potassium hydroxide, $\ce{KOH}$, pellet to a small droplet of hydrogen peroxide (30 %), $\ce{H2O2}$. You could clearly see the potassium hydroxide dissolving in the droplet when instantaneously the solutions began to fizz strongly (boil and/or release oxygen gas, I'm not sure) and release a lot of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what happened, but I've got a few ideas of what might have happened:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is some kind of reaction between the $\ce{KOH}$ and $\ce{H2O2}$.&lt;/li&gt;&#xA;&lt;li&gt;The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of  150.2 °C in which it also triggered the decomposition of $\ce{H2O2}$ into water and oxygen gas.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On Wikipedia I read:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;rate of decomposition increases with rising&lt;/strong&gt; temperature, concentration and &lt;strong&gt;pH&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So when $\ce{KOH}$ is dissolved, the solution has a high pH and this might trigger the decomposition of hydrogen peroxide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I did notice after a pH test that the plastic lid on which I had condensed some of the evaporated water had a very high pH, so this would most likely contain some dissolved potassium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case a reaction happened, what could that specific reaction be?&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="7448" LastEditDate="2017-04-08T16:53:12.900" LastActivityDate="2017-04-08T16:53:12.900" Title="What exactly happened when I added KOH to hydrogen peroxide?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10478" PostTypeId="1" AcceptedAnswerId="10483" CreationDate="2014-05-05T15:37:27.013" Score="13" ViewCount="70406" Body="&lt;p&gt;In the standard Periodic Table layout , all the elements up to 56 are in order i.e are in the same layout table. However, lanthanides and actinides are always shown separately from the layout like in this layout:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C4q1r.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C4q1r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason behind this structure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this standard layout or can I represent it like this too?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jqntf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqntf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5416" LastEditorUserId="10185" LastEditorDisplayName="user15489" LastEditDate="2015-09-18T18:00:50.227" LastActivityDate="2015-12-06T06:49:28.620" Title="Why lanthanides and actinides are shown separate from standard Periodic Table layout?" Tags="&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10479" PostTypeId="2" ParentId="10478" CreationDate="2014-05-05T15:59:02.800" Score="8" Body="&lt;p&gt;Because if you put lanthanides and actinides in to the Periodic Table like transition metals, the table will be way too wide.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorDisplayName="user15489" LastEditDate="2015-06-19T17:48:59.927" LastActivityDate="2015-06-19T17:48:59.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10480" PostTypeId="1" CreationDate="2014-05-05T16:01:09.643" Score="2" ViewCount="406" Body="&lt;p&gt;Where can I find atomic coordinates for guanosine / cytosine / thymine / adenine (either the ribonucleotide or deoxyribonucleotide bases) where the atoms are configured in the manner we would expect in the context of an idealized B-form or A-form nucleic acid duplex?  I'd like to calculate the angle between various bonds in these bases, and to play around with a few other things, but I'm having trouble finding proper atomic coordinate sets.&lt;/p&gt;&#xA;" OwnerUserId="5417" LastActivityDate="2014-07-31T16:27:09.977" Title="Atomic coordinates for DNA bases as they appear in a B-form or A-form duplex" Tags="&lt;biochemistry&gt;&lt;structural-biology&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10481" PostTypeId="2" ParentId="10476" CreationDate="2014-05-05T16:09:28.780" Score="1" Body="&lt;p&gt;Vacuum distillation is employed when the boiling point of the substance is too high (400 Celsius for example). Under reduced pressure, the boiling point will be reduced to the range that is easier to handle.&#xA;Steam distillation take advantage of azeotropic effect, water steam is bubbled into your liquid (aromatic compound for example) and carry out the mixture of water vapor and your liquid vapor. This technique can also avoid handling the high boiling point distillation because the azeotropic boiling point is usually lower than pure liquid. &lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-05T16:09:28.780" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10482" PostTypeId="2" ParentId="10476" CreationDate="2014-05-05T16:54:24.377" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.umsl.edu/~orglab/documents/distillation/Image169.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.umsl.edu/~orglab/documents/distillation/Image169.gif&lt;/a&gt; &lt;BR&gt;&#xA;   boiling point vs. pressure nomograph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The separation unit of a distillation (or countercurrent exchanges overall) is the theoretical plate (originally from petroleum distillation towers),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fch.upol.cz/skripta/fcc_and_zvem_english/rektif/rektif_teorie_clip_image002.jpg&quot; rel=&quot;nofollow&quot;&gt;http://fch.upol.cz/skripta/fcc_and_zvem_english/rektif/rektif_teorie_clip_image002.jpg&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.vaxasoftware.com/alen/MCTH_EN.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.vaxasoftware.com/alen/MCTH_EN.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each separation step must be at equilibrium.  Your basic lab rig is might be three theoretical plates at best, vapor vs. liquid in a hollow vertical column.  Efficient physical separation demands thin film countercurrent flow to maximize surface area/volume, hence packed columns then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.solvent--recycling.com/spinning_band_packed_column.html&quot; rel=&quot;nofollow&quot;&gt;http://www.solvent--recycling.com/spinning_band_packed_column.html&lt;/a&gt; &lt;BR&gt;&#xA;   spinning band distillation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wherein a tightly fitting teflon screw thread spinning like all get out jams fluid condensate downward against vapor flowing upward. Net output is a drop every now and again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural evolution of phase exchange is gas chromatography then liquid chromatography (much denser medium - look up the exchange maths). HPLC using monodisperse spheres with surface exchange only can have theoretical plate heights of a few microns, giving tens of thousands of theoretical plates in a rather short column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Supelco/Posters/1/T413105H.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Supelco/Posters/1/T413105H.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="4296" LastEditDate="2014-05-05T19:02:50.990" LastActivityDate="2014-05-05T19:02:50.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10483" PostTypeId="2" ParentId="10478" CreationDate="2014-05-05T17:58:04.147" Score="12" Body="&lt;p&gt;The Periodic Table arranges elements in blocks as each type of orbital fills with electrons - $s,p,d,f,g,h$.  Alkali metals and alkaline earths are $s$-block filling (but could be one $s$-block slot).  $p$-block six electrons to fill are trelides, tetralides, pnticides, chalcogenides, halides, inert gases (but could be one $p$-block slot).  Transition metal $d$-block is ten elements (but could be one $d$-block slot).  Filling the $f$-block are 14 elements, lanthanoids and actinoids.  That gets sloppy to print and the elements are (or at least were) overall obscure.  They get condensed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Representing the Periodic Table has become an an art form.  The plain vanilla variety is terse and useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alternative_periodic_tables&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Alternative_periodic_tables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorDisplayName="user15489" LastEditDate="2015-06-19T17:48:39.310" LastActivityDate="2015-06-19T17:48:39.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10484" PostTypeId="1" AcceptedAnswerId="10485" CreationDate="2014-05-05T21:24:28.737" Score="3" ViewCount="277" Body="&lt;p&gt;We have national elections coming up in two days, and like every time the election officials will mark people's thumbs with a dab of some purple solution that leaves a brown stain on the skin (and thumbnail) that takes a while to grow out. I'm pretty sure that's a potassium permanganate solution - that stuff behaves exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm not into electoral fraud (there are other controls in any case) but I dislike having that stain on my fingernail for a month or more afterward. Previously I've had partial success by rubbing butter into my skin, but the stain still gets through. I've also tried making some plain old sugar or even glucose available, hoping that it would get oxidized first, but I can't conclusively say that that worked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other tricks to avoid, or at least strongly mitigate, these stains? There's a window of about a minute or two where the purple solution sits on the cuticle while I'm unable to do anything about it.&lt;/p&gt;&#xA;" OwnerUserId="389" LastActivityDate="2019-09-29T21:32:29.797" Title="How can I protect my skin from KMnO4?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10485" PostTypeId="2" ParentId="10484" CreationDate="2014-05-05T22:04:21.950" Score="6" Body="&lt;p&gt;If it is permanganate, the brown $\ce{MnO2}$ stain will decolorize and solubilize in acidic reductant - lemon juice, perhaps boosted with a little crushed and dissolved Vitamin C.   I'd go for crystal/gentian violet or Disperse Red 9 as a marker, or a reactive Cibacron or Procion dye.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cthulhu for Fearless Leader!  Why vote for the lesser evil?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-05T22:04:21.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10486" PostTypeId="1" AcceptedAnswerId="10489" CreationDate="2014-05-05T23:23:09.883" Score="1" ViewCount="2400" Body="&lt;p&gt;There is a solid substance in a rigid cylinder with a piston placed on top such that the distance between the piston and the bottom of the cylinder is essentially zero, and then the substance decomposes and forms gas that expands and pushes the piston up against a constant external pressure to a maximum point. It then falls slowly as heat from the reaction is lost to the surroundings. Is it true that at the maximum point, the internal pressure is equal to the external pressure?&lt;/p&gt;&#xA;" OwnerUserId="5356" LastActivityDate="2014-05-06T00:14:08.823" Title="Internal and external pressure during a gas expansion" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10487" PostTypeId="1" AcceptedAnswerId="10490" CreationDate="2014-05-05T23:31:39.567" Score="3" ViewCount="281" Body="&lt;p&gt;If I take a compound like cyclopentadiene ($\ce{C5H6}$) and I substitute a nitrogen for a carbon at one of the $\ce{C=C}$ bonds (making a $\ce{C=N}$ bond), how does this typically change the hydrophobicity of the compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'm not really a chemist.  My thought here is that nitrogen is more electronegative than carbon, so this will polarize the bond, induce a dipole, and make the molecule more hydrophilic.  In general, it seems to me that mixing together atoms of differing electronegativity in alkane or alkene/etc. compounds should create dipoles that can interact with polar solvent molecules, improving hydrophilicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an additional thought, how would the nitrogen substitution effect stacking interactions between the substituted ring and a regular molecule of cyclopentadiene?  My guess here would be that hydrophobic effects would cause a more favorable pi-stacking interaction between two unsubstituted copies of cyclopentadiene relative to cyclopentadiene and the aforementioned and described nitrogen substituted version of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, by &quot;stacking interactions&quot; I mean the sort of $\ce{C=C}$ stacking interactions that lead to cycloaddition reactions, not that the two rings will lie on top of one-another (as in the case of something like a pair of benzene molecules).&lt;/p&gt;&#xA;" OwnerUserId="5420" LastEditorUserId="7448" LastEditDate="2017-04-08T16:57:34.913" LastActivityDate="2017-04-08T16:57:34.913" Title="Cyclopentadiene/etc. hydrophobicity after nitrogen atom substitution at a double bond?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10489" PostTypeId="2" ParentId="10486" CreationDate="2014-05-06T00:14:08.823" Score="1" Body="&lt;p&gt;If the frictionless massless piston is not moving, then internal and external pressures must be equal.  If the piston has mass, the internal pressure must be greater when it is stationary, to support the weight.  If there is static friction, all bets are off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the piston were a right cone rather than a right cylinder.  Would the excess surface area change the outcome, all other things being equal?&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-06T00:14:08.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10490" PostTypeId="2" ParentId="10487" CreationDate="2014-05-06T00:29:24.400" Score="2" Body="&lt;p&gt;The cyclic unsaturated imine (Schiff base) will be more hydrophilic for the nitrogen lone pair hydrogen bonding and being a Lewis base toward water.  You do know that CpH [4 + 2} Diels-Alder reaction reversibly thermally dimerizes, right?  The endo-dimer is an exo-flip then hydrogenation away from JP-10 cruise missile fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cyclic imine, ${2H}$-pyrrole is an isomer of pyrrole.  One expects it will be labile toward proton migration in base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.7330.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemspider.com/Chemical-Structure.7330.html&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.5256848.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemspider.com/Chemical-Structure.5256848.html&lt;/a&gt; &lt;BR&gt;&#xA;Click the Properties tab, etc. &lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-06T00:29:24.400" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10491" PostTypeId="1" CreationDate="2014-05-06T01:13:16.810" Score="0" ViewCount="277" Body="&lt;p&gt;I have a 14 karat gold ring with diamonds. I want to weaken the metal (to get the diamonds out without damaging them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I dip the ring in sulfuric acid, will I be able to dissolve the base metal in the alloy leaving the gold in place?&lt;/p&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="402" LastEditDate="2017-05-08T00:33:09.703" LastActivityDate="2017-05-08T00:33:09.703" Title="Removing diamonds from a 14 karat gold ring with acid" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10492" PostTypeId="2" ParentId="10477" CreationDate="2014-05-06T02:23:01.170" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;There is some kind of reaction between the $\ce{KOH}$ and $\ce{H_2O_2}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There was definitely a reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The $\ce{KOH}$ released a lot of energy when dissociating into the aqueous solution and then boiled the water, maybe reaching the temperature of 150.2 °C in which it also triggered the decomposition of $\ce{H_2O_2}$ into water and oxygen gas.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Water cannot reach a temperature above 100 °C unless it is pressurized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, $\ce{H_2O_2}$ can disassociate rather quickly under the right conditions. An enzyme called catalase, can break up $\ce{H_2O_2}$ at an astounding rate. If you add an amount you can barely see, like the size of a period at the end of a sentence, to 1 mL of 30% $\ce{H_2O_2}$ in a test tube, the oxygen production will push the liquid out of the tube.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2015-04-23T12:39:44.500" LastActivityDate="2015-04-23T12:39:44.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10493" PostTypeId="1" CreationDate="2014-05-06T08:05:07.237" Score="-2" ViewCount="149" Body="&lt;p&gt;How can I come to know how a catalyst affects a reaction when the reaction and the catalyst used is known.&lt;br&gt;&#xA;I searched the internet for an answer but could not find one.&lt;/p&gt;&#xA;" OwnerUserId="5423" LastActivityDate="2017-09-19T15:56:39.243" Title="How can one come to know how a catalyst affects a reaction when the reaction and the catalyst is known" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;catalysis&gt;&lt;enzymes&gt;" AnswerCount="4" CommentCount="1" ClosedDate="2017-09-16T02:15:11.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="10494" PostTypeId="2" ParentId="10493" CreationDate="2014-05-06T09:11:52.340" Score="1" Body="&lt;p&gt;There is no simple way to know it.&#xA;The mechanism in some of the cases of catalysis is known, therefore some predictions can be  made. Also there are many cases where the given catalytic reaction has a great practical significance, therefore it is thoughtlessly tested with different combinations of reagents, and the results are published. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-05-06T09:11:52.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10497" PostTypeId="2" ParentId="10451" CreationDate="2014-05-06T10:11:09.240" Score="3" Body="&lt;p&gt;Decomposing something into 'molecules' is immediately indicative of a non-chemistry background! To try to address your question a bit more directly, I can see a couple of ways of looking at the thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It decomposes everything to simplest possible component stable molecular structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of organic material, plastics etc, you will probably end up with a preponderance of various gases and water, as largely this disassembly is equivalent to complete combustion (CO₂, H₂O, NH₃) likely being the main resulting candidates. These would float out into the atmosphere unless captured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of inorganic material, e.g. a computer motherboard you will end up with many molecules of various metals, most of which would rapidly oxidise due to their very high surface area (being unimolecular), and various silicates and ceramic unit cells. All of these would be incredibly fine aerosols or dissolved in atmospheric water, similarly rapidly dispersing unless contained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results would likely include enhancing the greenhouse effect (on balance as some compounds will diminish it), inducing acid rain (sulfates, nitrates), causing algal blooms/biome damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had storage for different outputs then you would be producing several industrially useful compounds and if you partnered it with metal reclamation facilities to deal with all the metal oxides it might be a useful way of recycling, energy no object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It decomposes and sorts everything by chemical element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have a nice little storage container for each element (reasonable given you are sorting into the simplest molecule of a pure element), you would end up with various cylinders of oxygen, hydrogen, nitrogen, lumps of metal, lumps/powders of inorganic materials like phosphorus, sulfur, silicon. This would be potentially useful, but incredibly energetically expensive even beyond the separation system outlined by Uncle Al, so would be considerable only in the case of having essentially limitless free energy (many nuclear fusion plants/orbital solar collectors/dyson sphere type energy availability I suspect). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issues here would relate to various elements being highly reactive on contact with air (or rather the oxygen/water in it). Obvious examples include allotropes of phosphorus and the rare earth metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other issues unaddressed include fractional distillation and cracking of small molecule gases into smallest molecule gases, how to prevent reaction inside the vessel i.e. actually separate things rather than just provide them with huge amounts of energy to react with each other etc&lt;/p&gt;&#xA;" OwnerUserId="5075" LastEditorUserId="402" LastEditDate="2017-04-09T18:27:09.273" LastActivityDate="2017-04-09T18:27:09.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10498" PostTypeId="2" ParentId="10493" CreationDate="2014-05-06T10:20:38.063" Score="2" Body="&lt;p&gt;Empirically, you dump the catalyst to be tested into the reaction and determine the rate with/without for various catalytic loadings. This will tell you whether it increases or decreases the rate of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you are vaguely asking about a specific reaction and catalyst, you aren't going to get a specific answer without specifically mentioning them in your question. (I think that was enough specificity for now!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, you will probably want to search for university level teaching resources on transition metal or biological catalysts wherein you might find your answer. Higher level reading would be to use something like Web of Knowledge or Reaxys to find publications on your catalyst/reaction of question.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-05-06T10:20:38.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10499" PostTypeId="1" CreationDate="2014-05-06T13:44:30.603" Score="0" ViewCount="203" Body="&lt;p&gt;Reading in Atkin's Inorganic Chemistry book, I cam accross:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although an inversion and a twofold rotation may sometimes achieve the&#xA;  same effect, that is not the case in general and the two operations&#xA;  must be distinguished.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He does not give an example of an occasion where such cases are not the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will appreciate if someone can give an example of when inversion and twofold operation won't give the same result.&lt;/p&gt;&#xA;" OwnerUserId="5274" LastActivityDate="2014-05-06T14:21:28.773" Title="An example to illustrate the difference between an inversion operation and a twofold rotation" Tags="&lt;inorganic-chemistry&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10500" PostTypeId="2" ParentId="9911" CreationDate="2014-05-06T14:19:06.893" Score="3" Body="&lt;p&gt;I would say that the question is not quite well formulated and that the answer that you quote is even further misleading in giving the proper answer.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Finding the answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To start, let's consider fragmentation of primary alcohols, which is the question at hands here. First of all, can we rule out any of these answers?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Propan-1-ol undergoes &lt;strong&gt;complete&lt;/strong&gt; fragmentation in a mass spectrometer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sentence means that we can rule out the molecular ion $M^{\bullet +}$ as a base peak in the spectrum. Propan-1-ol has a monoisotopic peak at $m/z$ 60 so it rules out answer &lt;strong&gt;D&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply the standard fragmentation rules for alcohols: consider ionization on the oxygen lone pair, we have two possible $\alpha$-cleavage leading to $m/z$ 31 and $m/z$ 59. An additionnal side reaction could also be hydrogen transfer to the oxygen atom, followed by water loss. This would lead to an ion at $m/z$ 42. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ok3s.png&quot; alt=&quot;Ionized propanol fragmentation mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the options above, should we thus completely rule out $m/z$ 29? I think that in the spirit of those who asked the exam question, the answer is yes, and we should rule out answer &lt;strong&gt;A&lt;/strong&gt;. But see below...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now left with two options which can be explained by the fragmentation chemistry. Which should we choose? Technically, the problem here is that now we rely on abundances and not on the simple occurence of these fragments anymore. This is much more tricky, as there are no absolute rules. If we have a look at good textbooks, we find that loss of the largest radical is favored. Therefore, we have here to compare loss of $H^{\bullet}$ vs loss of $CH_{3}CH_{2}^{\bullet}$. Thus the answer is &lt;strong&gt;B&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Is the above answer complete?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The problem in such a question (and in the answer that is given below) is that it assumes that in mass spectrometry one could correctly predict fragment ion abundances based on very simple rules. And, as I tell my students, currently, even if you were able to map out the complete potential energy surface of the fragmentation pathway and do some RRKM calculations on the transition states, you would probably be off by a factor of 2 from the experimental results in terms of abundances. So asking for predicting the base peak (i.e. the highest abundance peak in a spectrum) can be quite wrong when two fragmentation channels are quite close to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at the &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C71238&amp;amp;Units=SI&amp;amp;Mask=200#Mass-Spec&quot; rel=&quot;nofollow noreferrer&quot;&gt;experimental electron ionization spectrum&lt;/a&gt;. Well OK, the base peak is indeed $m/z$ 31, so we are not too bad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's have a deeper look. Remember we crossed-out $m/z$ 29 as a possible base peak, and it happens to be the second most abundant peak in the spectrum, ahead of $m/z$ 42 and $m/z$ 59 which followed the basic rules! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m/z$ 29 could be either $HCO^{+}$ or $C_{2}H_{5}^{+}$. The former could be a secondary fragmentation of $m/z$ 31 after $H_{2}$ loss, whereas the latter could be produced as a heterolytic bond cleavage of the C-C bond. (In fact it is the same bond elongation leading to $m/z$ 31, except that they electron remains on the opposite side.) Energetically, these two channels differ by the difference in ionization energy between the $C_{2}H_{5}^{\bullet}$ and $^{\bullet}CH_{2}OH$ radicals, i.e. 8.11 eV &lt;em&gt;vs&lt;/em&gt; 7.56 eV. Although it amounts to almost 100 kJ/mol, with the fairly large energy distribution in electron ionization, this is still possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the second rule that we applied, is it always true that the largest radical if formed? This empirical rule works out quite well, even more so when the comparison is between a hydrogen radical (which is quite difficult to produce) and a carbon atom centered radical.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;And how about the exam paper solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here I am very highly surprised by what they seem to affirm. OK as we just saw above, $m/z$ 31 is a quite characteristic peak of primary alcohols. And loss of water is also characteristic of alcohols having side chains longer than 3 carbons. But does it mean that these should be the base peaks? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at the series of small primary alcohols:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Alcohol              m/z 31        Water loss           Other base peak&#xA;Ethanol              100            10&#xA;Propan-1-ol          100            15&#xA;Butan-1-ol            95           100&#xA;2-methylpropan-1-ol   38            10                  m/z 43&#xA;Pentan-1-ol           38            50                  m/z 42&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I do not see the rational behind this answer. It starts failing as soon as the small alcohol sizes reaches four carbon atoms. Of course there is a rational behind each of these base peak (for instance, in the case of 2-methylpropan-1-ol, the ionization energy of the isopropyl radical is of 7.37 eV, i.e. lower than that of $^{\bullet}CH_{2}OH$ and thus it becomes the most abundant ion, with exactly the same bond cleaved), but you have to go through the mechanisms, the thermochemistry and sometimes even the transition states to get to the global picture.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2014-05-06T14:54:03.463" LastActivityDate="2014-05-06T14:54:03.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10501" PostTypeId="2" ParentId="10499" CreationDate="2014-05-06T14:21:28.773" Score="1" Body="&lt;p&gt;If the body is chiral, a ${C_2}$-axis maintains chirality.  An inversion (${S_2}$, ${i}$, the product of ${C_2}$ and ${σ}$h) inverts chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/CHEM_3111_6170/Symmetry_and_group_theory.pdf&quot; rel=&quot;nofollow&quot;&gt;https://ww2.chemistry.gatech.edu/~wilkinson/Class_notes/CHEM_3111_6170/Symmetry_and_group_theory.pdf&lt;/a&gt; &lt;BR&gt;&#xA;Frame 5 &lt;BR&gt;&#xA;&lt;a href=&quot;http://alpha.chem.umb.edu/chemistry/ch612/documents/MolecularGeometryandPointGroups_002.pdf&quot; rel=&quot;nofollow&quot;&gt;http://alpha.chem.umb.edu/chemistry/ch612/documents/MolecularGeometryandPointGroups_002.pdf&lt;/a&gt; &lt;BR&gt;&#xA;More technical &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.b-u.ac.in/sde_book/msc_phychem.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.b-u.ac.in/sde_book/msc_phychem.pdf&lt;/a&gt; &lt;BR&gt;&#xA;The math&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-06T14:21:28.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10502" PostTypeId="2" ParentId="10451" CreationDate="2014-05-06T15:13:42.110" Score="3" Body="&lt;h1&gt;Molecular level&lt;/h1&gt;&#xA;&lt;p&gt;It depends on the molecules you obtain from the decomposition if you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_gasification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plasma gasification&lt;/a&gt; on mixed organic-inorganic materials you have syngas (&lt;span class=&quot;math-container&quot;&gt;$CO$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$H_2$&lt;/span&gt; ... ) and vitrify matter. Watch this &lt;a href=&quot;http://www.youtube.com/watch?v=gboSDLBV1ns&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; maybe a little bit misleading but can give you an idea. If you use a complete combustion with organic materials you will have mainly water, carbon dioxide, sulfur dioxide etc.,  depending on what you are burning. Indeed if you breath sulfur dioxide or nitrogen dioxide is not good for your health, and acid rain are a plague for forests. Incomplete combustion can lead to more hazard for the environment because you can have more complex molecules, sometimes very stable.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;objects can't disappear; the brick will be breathed in, get into water&#xA;etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Brick are composed mainly of silicon, not carbon, so you can't use combustion, at a very high temperature you can have silicon, a metal. Maybe some dangerous silane.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;if we start decomposing more advanced objects, we get all kinds of&#xA;harmful elements in the air (mercury etc.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Mercury is a liquid, not a gas so if you hypothetically discover a cheap technique to get it in the elemental form you will have a liquid.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;possible radiation from decomposition&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, you will not have radioactive elements so you don't have radioactivity ( &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;-ray) or other dangerous radiations.&lt;/p&gt;&#xA;&lt;h1&gt;Did you mean particulate?&lt;/h1&gt;&#xA;&lt;p&gt;I think that maybe what you mean if is a danger to powered materials to very tiny particles. &lt;a href=&quot;https://en.wikipedia.org/wiki/Particulates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Particulate&lt;/a&gt; is indeed  dangerous but in this case, you don't reach the molecular level. For your brick, this means that you have dust, something quite similar to what you have in your home.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-04-09T18:28:55.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10503" PostTypeId="2" ParentId="10491" CreationDate="2014-05-06T15:47:31.847" Score="3" Body="&lt;p&gt;I think that asking a jeweler is the best thing to do because gold is not something to throw away. However talking about the chemical concern, sulfuric acid can't oxidize gold, if the base metal (you should specify where and how it is) is iron theoretically you will be able to dissolve it with sulfuric acid:&#xA;$$\ce{Fe + H2SO4 -&amp;gt;FeSO4 + H2 ^}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something similar can happen with copper and tin but I think that temperature of the acid is a determinant parameter for these reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the base metal is inside the alloy is quite unlikely that you can affect it because gold is not a porous material, so you can only interact with the superficial layer.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="402" LastEditDate="2017-05-08T00:31:45.153" LastActivityDate="2017-05-08T00:31:45.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10505" PostTypeId="1" CreationDate="2014-05-06T22:44:00.680" Score="1" ViewCount="2761" Body="&lt;p&gt;I'm having trouble interpreting the results I got for my experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you should be well aware of, lithium and strontium both burn with crimson red when they are placed in a flame. Now, I can't decide which one it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I was told to do an experiment with the same salt where I had to add dilute hydrochloric acid and barium chloride. I had no visible change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this last test help me with identifying between strontium and lithium? What element is it?&lt;/p&gt;&#xA;" OwnerUserId="5427" LastEditorUserId="41328" LastEditDate="2020-04-24T05:48:01.460" LastActivityDate="2020-04-24T05:48:01.460" Title="Identification of lithium and strontium salts" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10506" PostTypeId="1" CreationDate="2014-05-07T00:47:23.850" Score="3" ViewCount="35642" Body="&lt;p&gt;I get increasingly nervous as I get my organic lab course reports. For, I often get asked whether the peak in the IR comes from an asymmetric or symmetric stretch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked at about ten books and turned google upside down. But I still can't answer the question in every case, though our molecules are quite simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you help me out with a table or something equivalent?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most urgent example is a $\ce{C-H}$ streching at $3006~ \mathrm{cm}^{-1}$ from p-tbutyl-(2-ethanol)-benzene (don't mind the nomenclature, please).&lt;/p&gt;&#xA;" OwnerUserId="5428" LastEditorUserId="2094" LastEditDate="2014-05-07T08:42:04.167" LastActivityDate="2016-08-27T15:24:22.873" Title="Distinguish between symmetric and asymmetric molecules with IR" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10507" PostTypeId="2" ParentId="10505" CreationDate="2014-05-07T01:38:53.573" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Solubility_table&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB6723042.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB6723042.htm&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&lt;/a&gt; &lt;BR&gt;&#xA;&lt;a href=&quot;http://www.chemicalbook.com/ProductChemicalPropertiesCB1184992_EN.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalbook.com/ProductChemicalPropertiesCB1184992_EN.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium chromate is listed as being exceptionally soluble in water.  Strontium chromate is listed as being insoluble.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-07T01:38:53.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10509" PostTypeId="1" CreationDate="2014-05-07T05:53:39.383" Score="24" ViewCount="18957" Body="&lt;p&gt;I have checked the internet and read quite a few books, but I still am not able to understand why vinylic and arylic carbocations are highly unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I found while surfing the internet is: For arylic carbocation, the carbon bearing postive charge cannot participate in resonance with adjacent carbons because it's p orbital is perpendicular to the other orbitals. But, I didn't really understand what this means.&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="5026" LastEditDate="2018-03-13T06:26:00.297" LastActivityDate="2021-05-29T07:06:46.113" Title="Why are vinylic and arylic carbocations highly unstable?" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10510" PostTypeId="2" ParentId="10493" CreationDate="2014-05-07T08:50:29.537" Score="0" Body="&lt;p&gt;At a basic level, a catalyst lowers the activation energy of a reaction, which makes it easier for the reaction to occur (&quot;speeding it up&quot;). The catalyst is not chemically consumed by the process. (Though corrosion, coking, and other side reactions can still deteriorate catalyst perofmance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details depend on the specific reaction and catalyst.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-05-07T08:50:29.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10511" PostTypeId="2" ParentId="10506" CreationDate="2014-05-07T09:48:01.373" Score="0" Body="&lt;p&gt;first of all I have no practical experience with this field (yet), but we had covered this in theory (and also I don't have chemistry in English). I hope that I can help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I recall, you have to first write the structural diagrams of the suspected molecules and compare them as it can be seen on the picture bellow &lt;img src=&quot;https://i.stack.imgur.com/bzN7H.jpg&quot; alt=&quot;picture 1&quot;&gt;&lt;br&gt;&#xA;(I had izopropanol and propanol in my book, but I think that you will get this from this). Then you search in tables for appropriate absorption frequencies and write them down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you check the experimental graph that you obtained in order to find out with what molecule you are dealing. So for example the corresponding graphs for these two substances are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fw587.jpg&quot; alt=&quot;enter image description here&quot;&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/tVv7c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5103" LastActivityDate="2014-05-07T09:48:01.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10512" PostTypeId="1" CreationDate="2014-05-07T14:12:03.140" Score="3" ViewCount="3393" Body="&lt;p&gt;Do you all know of any large graphs for the vibrational spectrums of HI, HBr, HF, and HCl? I need graphs that are large enough to see the spacing between the peaks in the spectra of HCl, HBr, and HI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, if you know of any raw data sets for any of the above, that would work. Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="5433" LastActivityDate="2014-05-08T10:51:28.740" Title="Vibrational Spectrums of HI, HBr, HF, and HCl" Tags="&lt;physical-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10514" PostTypeId="1" CreationDate="2014-05-07T14:49:16.163" Score="17" ViewCount="1124" Body="&lt;p&gt;I understand that boiling point is a colligative property for aqueous solutions, and that it empirically depends on the mole fraction of the solute, rather than the identity of the solute. I understand the van't Hoff factor / ionic strength and why it matters. I also get that an increase in entropy is the primary driving force for boiling point elevation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why does the influence of intermolecular attraction go out the window for solutions, as opposed to pure substances? Shouldn't the identity of the specific dissolved species - that is, their charge and their size, affect the intermolecular attractions in the solution? Shouldn't greater intermolecular attraction increase the boiling point by reducing vapor pressure, just as it does for pure substances? For example, neutral molecular solutes as opposed to strong electrolytes, or large ions like iodide vs smaller ones like fluoride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the effect simply too small to matter compared to the entropy factor, or am I missing something theoretically?&lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2015-09-19T23:53:59.703" Title="Why doesn't the boiling point of an aqueous solution depend on intermolecular attraction?" Tags="&lt;aqueous-solution&gt;&lt;boiling-point&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10515" PostTypeId="2" ParentId="6858" CreationDate="2014-05-07T16:09:05.707" Score="-3" Body="&lt;p&gt;This is because of the fact that in presence of light iodine and bromine absorbs light and excitation of the electron from the ground state to the excited state takes place which on returning back to the ground state radiates energy. This radiating energy comes under the visible region. Therefore they are co lured.&lt;/p&gt;&#xA;" OwnerUserId="5436" LastActivityDate="2014-05-07T16:09:05.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10516" PostTypeId="2" ParentId="10509" CreationDate="2014-05-07T16:20:41.993" Score="27" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital.  In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basis of the &lt;strong&gt;inductive effect&lt;/strong&gt;.  For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - &lt;strong&gt;the methyl group is inductively electron donating&lt;/strong&gt; to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon.  For a positive center to be attached to a more electronegative group is destabilizing.  Hence the vinylic cation is less stable than a typical alkyl cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things are even worse with the aryl carbocation on the right.  Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry.  A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$.  These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JdU5p.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JdU5p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-09-30T22:40:50.690" LastActivityDate="2015-09-30T22:40:50.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10517" PostTypeId="2" ParentId="10514" CreationDate="2014-05-07T18:14:00.507" Score="7" Body="&lt;p&gt;GOOD QUESTION! That is the nastiest p-chem question I've seen.  Atkins, &lt;em&gt;Physical Chemistry&lt;/em&gt; 2nd edition (the readable one), pp 228-232pp.  Boiling point elevation is presented as purely an entropy effect, given two assumptions:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The solute is non-volatile. &lt;BR&gt;&#xA;2) The non-volatile solute does not dissolve in the solid solvent. &lt;BR&gt;&#xA;Both assumptions are explored in Chapter 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solute increases the entropy of the solution but does not change the entropy of the vapor at a given P and T.  A higher T is then necessary to give the same entropy difference.  I am not satisfied that this wholly addresses a strongly interactive solute versus a non-interactive solute versus a strongly negatively interactive solute,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/kosmos.html&quot;&gt;http://www1.lsbu.ac.uk/water/kosmos.html&lt;/a&gt; &lt;BR&gt;&#xA;Kosmotropes and chaotropes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1429218126&quot;&gt;http://www.amazon.com/Physical-Chemistry-Peter-Atkins/dp/1429218126&lt;/a&gt; &lt;BR&gt;&#xA;Atkins 9th edition is current. &lt;BR&gt;&#xA;&quot;There is almost no chemical concept that Atkins can't make unnecessarily confusing&quot;&lt;BR&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0935702997&quot;&gt;http://www.amazon.com/Physical-Chemistry-A-Molecular-Approach/dp/0935702997/ref=zg_bs_16052591_2&lt;/a&gt; &lt;BR&gt;&#xA;Library - it's pricey. &lt;BR&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0073398179&quot;&gt;http://www.amazon.com/Thermodynamics-Engineering-Approach-Yunus-Cengel/dp/0073398179/ref=dp_ob_title_bk&lt;/a&gt; &lt;BR&gt;&#xA;Library - it's very pricey.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-07T18:14:00.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10518" PostTypeId="1" CreationDate="2014-05-07T18:17:41.967" Score="1" ViewCount="8619" Body="&lt;p&gt;Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not? And is there a reason why the products of reaction 2) are written separately, instead of &quot;2OH&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\ce{OH- + H2O &amp;lt;=&amp;gt; H2O + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\ce{O^{2-} + H2O -&amp;gt; OH- + OH-}$&lt;/p&gt;&#xA;" OwnerUserId="5225" LastEditorUserId="189" LastEditDate="2014-05-07T18:30:11.783" LastActivityDate="2018-03-04T02:57:10.600" Title="Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10519" PostTypeId="2" ParentId="10518" CreationDate="2014-05-07T19:50:23.593" Score="0" Body="&lt;p&gt;All reactions are reversible but some are more reversible than others. Some reactions are so little reversible that they are practically irreversible, like that reaction of yours between oxide anion and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all comes down to comparing the stabilities of reactants and products. The actual question is, why is oxide anion so unstable compared to hydroxide ions? One approach would be to think how would you make the oxide anion from a water molecule. Clearly, you can remove one hydrogen from a water molecule with a mild base but taking the second one would require a much stronger base. Take a look at the pkb value of hydroxide ion and visualize the equilibrium populations in your head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible approach would be to compare the electronic structures of the species. Oxide anion has twice the negative load of the hydroxide ion. Hydroxide ion has one hydrogen to relieve the stress.&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2014-05-07T19:50:23.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10520" PostTypeId="2" ParentId="10518" CreationDate="2014-05-07T21:13:53.693" Score="1" Body="&lt;p&gt;Hydroxide is the strongest base in water (leveling effect).  The first posted reaction is symmetric.  Given reactant and product sides of the equation, how do you tell them apart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxide is a much stronger base than hydroxide.  The second reaction spontaneously proceeds to the right (leveling effect).  If you look up the Brønsted basicities of oxide and hydroxide, you can write a simple thermodynamic ratio for their relative abundances at equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no reason not to write $\ce{2 OH^-}$ or $\ce{2 (OH^{-})}$.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastEditorUserId="5026" LastEditDate="2018-03-04T02:57:10.600" LastActivityDate="2018-03-04T02:57:10.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10521" PostTypeId="2" ParentId="10505" CreationDate="2014-05-07T21:19:31.860" Score="4" Body="&lt;p&gt;If your solution contains only &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{Sr^2+}$&lt;/span&gt;, you can differentiate them by adding a sulfate solution (for example, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2SO4}$&lt;/span&gt; or diluted sulfuric acid). If strontium is present, a white, fine-crystalline precipitate of &lt;span class=&quot;math-container&quot;&gt;$\ce{SrSO4}$&lt;/span&gt; will form. In the case of lithium, the solution will remain clear. This is because &lt;span class=&quot;math-container&quot;&gt;$\ce{SrSO4}$&lt;/span&gt; is much less soluble in water than &lt;span class=&quot;math-container&quot;&gt;$\ce{Li2SO4}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.0132 g}$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$\pu{34.8 g}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\pu{100 g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{20 ^\circ C}$&lt;/span&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding diluted &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl_{aq}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{BaCl2}$&lt;/span&gt; to a solution is commonly used as a test for &lt;span class=&quot;math-container&quot;&gt;$\ce{SO4^2-}$&lt;/span&gt; (if it is present, &lt;span class=&quot;math-container&quot;&gt;$\ce{BaSO4}$&lt;/span&gt; precipitates). Maybe this could help you to distinguish between &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Sr^2+}$&lt;/span&gt; given the different solubilities of their sulfates. If your unknown salt readily dissolves in water and &lt;span class=&quot;math-container&quot;&gt;$\ce{SO4^2-}$&lt;/span&gt; is found to be present in the solution, then it is likely that you have lithium sulfate. Strontium sulfate as unknown salt would not readily dissolve and therefore not give a positive test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if your solution can contain more ions besides &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{Sr^2+}$&lt;/span&gt;, I would suggest to try the separation with ammonium carbonate which is already mentioned in the comments to your question.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="61578" LastEditDate="2020-04-24T05:39:09.770" LastActivityDate="2020-04-24T05:39:09.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10522" PostTypeId="2" ParentId="10518" CreationDate="2014-05-07T21:19:55.373" Score="2" Body="&lt;p&gt;Correct, all reactions are reversible, just to different extents. Think about the combustion reaction in your vehicle. Is that reaction reversible? Sure, in that nothing is 100% irreversible. The reversing of the combustion just might take a while. But it's not reversible to any significant extent in the short and medium-term; i.e. you're not going to be re-capturing your combusted gasoline and be putting that back together as usable fuel for your vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, about the reaction of oxide ion with water - that reaction goes practically 100% because oxide ion is a very strong base. &lt;strong&gt;The strongest base supported in high concentration by a medium is the conjugate base of the medium.&lt;/strong&gt; (Leveling effect). For water, that would be hydroxide ion (remember that conjugate bases lose a proton). Oxide ion is the conjugate base of hydroxide ion. We would expect oxide ion to be an even stronger base than hydroxide ion because of oxide ion's even higher negative charge density - we're talking about a negative two charge here (versus negative one for hydroxide) and we're missing a hydrogen proton (which only further decreases the size of the oxide ion and further increases its negative charge density). If you think about this Coulombically, then of course oxide ion is going to be a stronger base than hydroxide ion - opposite charges attract (recall that water has two partial positive charges/dipoles stabilized by its hydrogen atoms). F=kq1q2/r^2 and one of the q's here is massive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also about how the products are written - it doesn't matter how the products are written. Either way you are creating two HO- (hydroxide ions). Write hydroxide with the negative next to the O (so either -OH or HO- are acceptable) because that's what you're doing - heterolytic clevage of the O-H bond in H-O-H, which (in this case) creates a negative formal charge that is mostly stabilized by the oxygen atom (NOT the hydrogen atom). Plus it's the &lt;em&gt;hydroxide ion&lt;/em&gt; not the oxyhydride ion (whatever the hell that is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, when writing -OH, make sure the &quot;-&quot; is clearly a superscript, lest people think you're talking about a hydroxyl group.  &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-07T21:25:35.903" LastActivityDate="2014-05-07T21:25:35.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10525" PostTypeId="1" CreationDate="2014-05-07T23:44:45.007" Score="-2" ViewCount="2638" Body="&lt;p&gt;Why are alkali and alkali earth metals indicative of a strong acid? Does anyone know why it is so? Or tell me what is indicative of strong acids?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="402" LastEditDate="2017-05-08T00:30:56.713" LastActivityDate="2017-05-08T00:30:56.713" Title="Why are alkali and alkali earth metals indicative of a strong acid?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="6" ClosedDate="2014-05-08T11:29:33.433" ContentLicense="CC BY-SA 3.0" />
  <row Id="10526" PostTypeId="1" AcceptedAnswerId="10541" CreationDate="2014-05-08T01:00:49.297" Score="2" ViewCount="10677" Body="&lt;p&gt;Given that hydrogen atoms that are attached to a compound must be ionizable for the compound to be an acid, what prevents a hydrogen from being ionized? Is it solubility. If the bond is nonviolent, water could not dissolve the compound, and so hydrogen atoms are not ionizable. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2018-03-04T02:58:55.520" Title="Ionizable hydrogen atoms for acids" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10527" PostTypeId="2" ParentId="10526" CreationDate="2014-05-08T01:17:16.803" Score="0" Body="&lt;p&gt;Ionization &lt;em&gt;here&lt;/em&gt; (in terms of acids) refers to heterolytic bond clevage, not homolytic bond clevage (which would create a hydrogen atom, complete with its one electron). Unfortunately in the Brønsted-Lowry definition of acids, we are dealing with hydrogen protons - i.e. hydrogen atoms stripped of their lone electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course we could take a whole hydrogen atom from a molecule, i.e. we could do this with water, and what we'd end up with is a hydrogen atom and a hydroxyl group. Unfortunately the lone hydrogen atom is rather unstable by itself - that's why at standard state conditions you find not the hydrogen radical but instead diatomic hydrogen. But then the bare hydrogen proton is rather unstable too in its own right; it's solvated in water actually (as the hydronium ion; it may even be solved by multiple waters). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:58:55.520" LastActivityDate="2018-03-04T02:58:55.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10528" PostTypeId="2" ParentId="10525" CreationDate="2014-05-08T01:27:57.377" Score="4" Body="&lt;p&gt;Alkali and alkali earth metals are not indicative of a strong acid; if anything, they're more indicative of a strong base, but this is one of those half-truths/Santa Claus/tooth fairy level &quot;truths&quot; that general chemistry textbooks tell students. Just because something has an &quot;-OH&quot; in it doesn't make it a base or an acid automatically. The textbook is trying to placate readers with a simple rule that is sometimes right so that when you go into the lab and see $\ce{NaOH}$ (sodium hydroxide), you recognize that's a base. $\ce{KOH}$, that's a base too. $\ce{Ca(OH)_2}$, that's a base too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about $\ce{(HO)_3PO}$? What about $\ce{(HO)_2SO_2}$? The two molecules I just mentioned are phosphoric acid and sulfuric &lt;strong&gt;&lt;em&gt;acid&lt;/em&gt;&lt;/strong&gt;, respectively (not phosphoric and sulfuric &lt;em&gt;base&lt;/em&gt;). The formulas are, in addition, written with respect to structure, to illustrate that both have $\ce{OH}$ groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about phenol? That's an organic molecule with an OH group hanging off. What about acetic acid, $\ce{H_3CCOOH}$? That has an OH group on the end hanging off (no, it has no O to O bonds). &lt;strong&gt;None&lt;/strong&gt; of these are bases (at least to any significant extent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One must realize that to be a base - an Arrhenius base (as it seems that your textbook is introducing you to - one must have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) An electropositive element. An electropositive element would be a group one or group two element and is able to stabilize positive charge no problemo. Example: Sodium, Na. We often find sodium as sodium ion: $\ce{Na^+}$. But do we find any of the &lt;em&gt;electronegative&lt;/em&gt; elements - say - fluorine - as $\ce{F+}$? Hell no. $\ce{F-}$, maybe. That's because those electronegative elements are better at stabilizing negative charge while electropositive elements are better at doing the opposite. Remember that chemistry is just one big free energy optimization game; in simpler terms - stability is paramount and is the underlying goal of chemical processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The electropositive element must be attached to an OH group. Through heterolytic bond clevage, the OH group will leave the electropositive group as the hydroxide ion, $\ce{HO^-}$. The heterolytic bond cleavage (in this case) leaves the electropositive element short of one electron but remember that's perfectly okay - electropositive elements are alright with a positive charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens with an Arrhenius base (water is implied)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KOH -&amp;gt;K^+ + HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what the Arrhenius theory might lead to to believe (which isn't true; and again, water is implied):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(HO)_2SO_2 -&amp;gt; 2HO^- + SO_2^2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, this is the proper (first) ionization of sulfuric acid. Remember that sulfur is much more electronegative than one of the group 1 or 2 elements. This, together with the &lt;em&gt;inductive&lt;/em&gt; effect of the attached oxygens allows the sulfur to stabilize a negative charge gained through heterolytic cleavage of the O-H bond. Note that I do not mention resonance as a factor (although textbooks might imply this as a factor) because sulfur does not generally access its d-orbitals in bonding, and hence, there is no double bond (and therefore a set of electrons to delocalize) in either sulfuric acid or the hydrogen sulfate ion. Both are best represented as charge-separated species with a +2 formal charge on the sulfur. In other molecules, however, resonance stabilization is a factor in their acidities - e.g. take the acetate ion when considering acetic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(HO)_2SO_2 + H2O -&amp;gt; H_3O^+ + HOSO_3^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in essence the aforementioned two factors (electropositive element and OH group) make an Arrhenius base. Unfortunately the Arrhenius definition of acids and bases is rather limited and if you study chemistry further, you will forget the Arrhenius definition and supplant it with the Bronsted-Lowry definition of acids and bases and the Lewis definition of acids and bases (at the least). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you are really looking for what is indicative of a strong base or acid, you just assess conjugate stability. A strong acid has a very stable conjugate base. A strong base has a very stable conjugate acid. A weak acid has a less stable conjugate base. A weak base has a less stable conjugate acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $\ce{HCl}$ is the poster child of strong acids. Its conjugate base is $\ce{Cl^-}$, which is basically unreactive in water solution, thereby making $\ce{HCl}$ a strong acid. Remember that IF $\ce{Cl^-}$ were more reactive, then the reverse reaction (un-dissociation, if you will) will be favored, and this isn't going for $\ce{HCl}$'s cause - which is losing its hydrogen proton and keeping it lost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H_2O -&amp;gt; H_3O^+ + Cl^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction doesn't go backward to any significant extent; i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl- + H_3O^+ -&amp;gt; not much of a reaction}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\ce{HCl}$ is a strong acid because it basically ionizes 100%; there is no reverse reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line for the OP: Take everything your general chemistry textbook has to offer with a grain of salt. Right now your textbook is helping you develop a hammer to take to chemistry problems - crude, but sometimes effective. Later in your studies you will progress to finer tools - the chisel and scalpel, if you will. Here are some common fairy tales offered by general chemistry textbooks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;$\ce{OH^-}$&quot; - the negative formal charge in the hydroxide ion is mainly stabilized by the more electronegative element (as common sense would dictate, and that element would be oxygen). So write $\ce{HO^-}$ and if your teacher asks you what the hell that is, smack him or her down with the science. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\ce{sp^3d^2}$ hybridization and other explanations of hypervalent molecules with d-orbital utilization. This myth started with the misreading of a chemical paper which made use of d-polarization functions (quantum mechanic stuff) as saying that hypervalent molecules access their d-orbitals on a regular basis. D-functions are not d-orbitals. Species such as sulfate ion are best represented with a +2 formal charge on the central atom. Linus Pauling was wrong about Vitamin C and he's wrong about bonding too. So is your textbook (likely, unless it was updated since 1990, the year in which a seminal paper was published which definitely refuted the use of d-orbitals in hypervalent molecules). Note that textbooks may be &quot;updated&quot; yearly but most of the time the only changes are 1) a new photo of a chemical molecule on the cover and 2) a new edition stamped onto the spine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The existence of the $\ce{H^+}$ proton in water. No. This is Coulombically (and comically) unlikely. The high positive charge density of the bare proton means that it is hydrated by water almost immediately in water, creating the hydronium ion. So in writing reactions, write $\ce{H_3O^+}$, not $\ce{H^+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The 90 degree bond angle. No, methane ($\ce{CH_4}$ does not have any 90 degree bond angles. 90 degree bond angles are actually relatively rare in chemistry. So no, a methane molecule does not look like a set of cardinal axes with H's stuck on the ends where North, East, South, and West are supposed to be. It's a tetrahedron; draw it like a tripod or draw it with projections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) $\ce{NaCl_(aq)}$ and other &quot;aqueous&quot; species which actually exist in their dissociated states in water. There is no unionized sodium chloride in water. Proof - put some table salt in water. Do you see anything in the water? No, all the crystals dissolved. So don't write $\ce{NaCl_(aq)}$. Your teacher may ask you to write the &lt;em&gt;net ionic&lt;/em&gt; equation for reactions; in my opinion, this is the &lt;em&gt;only&lt;/em&gt; correct equation one can right. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-08T03:01:42.070" LastActivityDate="2014-05-08T03:01:42.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10529" PostTypeId="2" ParentId="10525" CreationDate="2014-05-08T02:55:25.600" Score="3" Body="&lt;p&gt;I'd just like to add a small commentary to Dissenter's nice answer. Not only is it important to analyze the structure of the solute to determine its acidic or basic behavior, but also &lt;em&gt;the structure of the solvent&lt;/em&gt;. We are very used to thinking about aqueous solutions of acids and bases, but that's just a particular case, and limiting oneself to water hinders a deeper understanding of acidity and basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classic example is to consider nitric acid, $\ce{HNO3}$ as a solute. In liquid water as a solvent, nitric acid is a strong acid, and its behavior is represented by the formal reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3 + H2O -&amp;gt; NO_3^- + H3O^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if pure acetic acid $\ce{H3CCOOH}$ is used as a solvent (also known as glacial acetic acid), nitric acid is a &lt;em&gt;weak acid&lt;/em&gt;, behaving like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3 + H3CCOOH &amp;lt;=&amp;gt; NO_3^- + H3CCOOH_2^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the real fun is what happens if pure sulfuric acid $\ce{H2SO4}$ is used as a solvent. Here, nitric acid is a &lt;strong&gt;weak base&lt;/strong&gt;! Instead of the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3 + H2SO4 &amp;lt;=&amp;gt; NO_3^- + H3SO4^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, what happens in reality is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3 + H2SO4 &amp;lt;=&amp;gt; H_2NO_3^+ + HSO4^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The take-home is just that acidity and basicity are somewhat more fluid concepts than they might at first appear from Arrhenius theory, so avoid being too strict about classifying and memorizing what substances are acids, bases or neither.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-05-08T00:30:12.597" LastActivityDate="2017-05-08T00:30:12.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10530" PostTypeId="2" ParentId="10525" CreationDate="2014-05-08T02:59:47.460" Score="3" Body="&lt;p&gt;Exactly, and to expand on what Nicolau mentions, we can discuss the leveling effect. A chemist (can't remember who) found that some acids - e.g. hydrochloric acid and nitric acid - have the same strength in water. How can this be? They're clearly different, yet they have the same strength in water. This is because these strong acids are &quot;leveled&quot; in water solvent. Take-home:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The strongest acid that can exist in any solvent is the conjugate acid of the solvent. The strongest base that can exist in any solvent is the conjugate base of the solvent.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for water, the conjugate acid (remember that conjugate means inverse; acid means proton-losing, so inverse/opposite of proton losing is proton gaining) is the hydronium ion, which is formed precisely by tacking a proton to water (this is a Lewis acid/base or coordination chemistry reaction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the conjugate base of water is the hydroxide ion. So the strongest base that can exist in water solution is hydroxide ion - anything stronger will be leveled to hydroxide ion. Take the following leveling reactions as examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl_(l) + H_2O -&amp;gt; H_3O^+ +Cl^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here pure, liquid hydrochloric acid is leveled to its conjugate base and hydronium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S^2- + H_2O -&amp;gt;HS^- + HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, a strong base, sulfide ion, is being leveled to its conjugate acid and hydroxide ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I agree, don't try to memorize any list of strong acids or bases - such a list is just about useless and often the list only works for situations in which water is a solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also this discussion reminds me of a good quiz question my teacher once asked me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y15QC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one draws out the Lewis structure for the solvent, one finds that there is a lone pair on the nitrogen. This parallels ammonia, $\ce{H_3N}$, which also has a lone pair on the central nitrogen atom. The point was for us to draw the parallel between ammonia and liquid methylhydroxylamine. Upon realizing that, we can then realize that $\ce{H_2S}$ is a strong acid in this solvent because even in water solvent, $\ce{H_3N}$ is effectively protonated by $\ce{H_2S}$ - the two undergo a large extent reaction. One can ascertain this reaction quantitatively through calculating a complex K for the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_2S + H_3N &amp;lt;=&amp;gt;H_4N^+ + HS^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{above~reaction}=\frac{[HS^-][H_4N^+]}{[H_2S][H_3N]}=\frac{K_a(reactant~acid)}{K_a(product~acid)}=\frac{(1.1*10^{-7})}{(5.6*10^{-10})}=196=reaction~progresses~far~to~the~right$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or simply by comparing the $\ce{K_a}$ values of the reactant acid and the product acid. Off the top my head (I basically got the $\ce{K_a}$ values of all the acids memorized because of how much we studied B/L A/B chemistry), they are (also mentioned above in the calculation of the K for the reaction): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${K_a(H_4N^+) = 5.6 x 10^{-10}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${K_a(H_2S) = 1.1 x 10^{-7}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as we can see, we are going from a stronger acid (not strong acid) to a weaker acid. It makes sense that this equilibrium reaction would be a large extent reaction to the right; strong to weak; waterfalls; high to low, etc. Analogize as you see fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP: Let the chemistry talk to you; don't let the rules dictate everything because in chemistry, the &quot;rules&quot; are mostly illusions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: And further expanding on the fact that acidity and basicity are rather fluid concepts, consider this myth: &lt;em&gt;neutral water has a pH of 7.0&lt;/em&gt;. No. The only requirement for neutrality is $\ce{[H_3O^+]=[HO^-]}$ in a solution. Consider the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_2O + heat -&amp;gt;H_3O^+ +HO^-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is stoichiometric - i.e. hydronium and hydroxide ions are created in a 1:1 ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 25 degrees Celsius, the concentration of both ions produced in this &lt;em&gt;auto-ionization&lt;/em&gt; reaction is ${1.0 *10^{-7}}$ M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you heat up water, the reaction goes further to the right (Le Chatlier's principle - disturb a system at equilibrium and the system will try to return to equilibrium; in this case the system will try to consume the extra heat through further auto-ionization). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does this mean? It means that hot water (i.e. steam) is dangerous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot water is BOTH acidic and basic (i.e. corrosive) - and highly so. So don't come into contact with super-heated water/steam. Contact with skin may require a visit to your dermatologist or quite possibly your undertaker. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-08T16:50:46.047" LastActivityDate="2014-05-08T16:50:46.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10531" PostTypeId="1" AcceptedAnswerId="10538" CreationDate="2014-05-08T04:30:41.363" Score="3" ViewCount="2287" Body="&lt;p&gt;This is the procedure I carried out in class:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reduction of nitrobenzene by using tin and $\ce{HCl}$ acid&lt;/li&gt;&#xA;&lt;li&gt;After the completion of the reaction, $\ce{NaOH}$ added to neutralize excess acid and dissolve any $\ce{Sn(OH)2}$&lt;/li&gt;&#xA;&lt;li&gt;Steam distillation is carried out to obtain suspension of phenylamine&lt;/li&gt;&#xA;&lt;li&gt;Salting carried out where $\ce{NaCl}$ salt is added. Aqueous layer discarded&lt;/li&gt;&#xA;&lt;li&gt;Solvent extraction: dry ether added and mixture shaken vigorously. Aqueous layer discarded&lt;/li&gt;&#xA;&lt;li&gt;Distillation under reduced pressure carried out and solid phenylamine left in flask&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2595" LastEditorDisplayName="user7951" LastEditDate="2017-10-19T17:31:00.820" LastActivityDate="2017-10-19T17:31:00.820" Title="Preparation of phenylamine/aniline" Tags="&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10532" PostTypeId="1" AcceptedAnswerId="10533" CreationDate="2014-05-08T07:13:35.713" Score="2" ViewCount="1099" Body="&lt;p&gt;What's the process called where you can use electricity to separate the oxygen and hydrogen from  water? Is it electrolysis? Can it be applied to other liquids?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="4734" LastEditDate="2014-05-08T10:16:39.343" LastActivityDate="2014-05-09T10:33:33.120" Title="Which process is used to separate Hydrogen and Oxygen from Water?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10533" PostTypeId="2" ParentId="10532" CreationDate="2014-05-08T07:42:19.787" Score="2" Body="&lt;p&gt;Problem in your question: should be be O and &lt;strong&gt;H&lt;/strong&gt;, because $\ce{H2O}$ (water) is made of H (Hydrogen) and O (Oxygen). $\ce{CO2}$ is carbon dioxide, for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, electrolysis can be performed on many compounds, not just water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electrolytic reaction is where an electric current is used to drive an otherwise non-spontaneous chemical reaction. Splitting water is just one of such reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/183116/electrolysis&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="4734" LastEditDate="2014-05-08T10:17:06.200" LastActivityDate="2014-05-08T10:17:06.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10534" PostTypeId="5" CreationDate="2014-05-08T09:16:45.350" Score="0" Body="&lt;p&gt;The &lt;a href=&quot;https://goldbook.iupac.org/C01222.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;concentration&lt;/a&gt; characterizes the composition of a mixture with respect to the volume of the mixture. It is a group of four quantities: mass, amount, number and volume concentration. It is also usually used as the short form for amount (of substance) concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related and within the scope of this tag are normality, molality, mole fraction/ratio, mass fraction/ratio.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://goldbook.iupac.org/M03713.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass concentration&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass of a constituent divided by the volume of the mixture.&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $\rho, \gamma$&lt;/li&gt;&#xA;&lt;li&gt;SI Unit: $\rho_i = \frac{m_i}{V} = \mathrm{\frac{kg}{m^3}}$&lt;/li&gt;&#xA;&lt;li&gt;Common units: $\mathrm{\frac{g}{L}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://goldbook.iupac.org/A00295.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amount concentration&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Amount of a constituent divided by the volume of the mixture. For entities $B$ it is often denoted as $[B]$&lt;/li&gt;&#xA;&lt;li&gt;Synonyms amount-of-substance concentration, substance concentration, molar concentration, molarity&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $c$&lt;/li&gt;&#xA;&lt;li&gt;SI Unit: $\phi_i = \frac{n_i}{V} = \mathrm{\frac{mol}{m^3}}$&lt;/li&gt;&#xA;&lt;li&gt;Common units: $\mathrm{\frac{mol}{L}}=\mathrm{\frac{mmol}{mL}}=\mathrm{\frac{mol}{dm^3}}$ (Often denoted as $M$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://goldbook.iupac.org/N04260.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Number concentration&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Number of entities of a constituent in a mixture divided by the volume of the mixture.&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $C, n$&lt;/li&gt;&#xA;&lt;li&gt;SI Unit:$\phi_i = \frac{N_i}{V} = \mathrm{m^{-3}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://goldbook.iupac.org/V06643.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Volume concentration&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Volume of a constituent of a mixture divided by the sum of volumes of all constituents prior to mixing.&lt;/li&gt;&#xA;&lt;li&gt;Synonym: volume fraction&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $\phi$&lt;/li&gt;&#xA;&lt;li&gt;SI Umit: (dimensionless) $\phi_i = \frac{V_i}{V} = \mathrm{1}$&lt;/li&gt;&#xA;&lt;li&gt;Common Units: $\mathrm{\frac{L}{L}}$, or expressed as numbers $0.5$ or $50\%$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related &lt;a href=&quot;https://goldbook.iupac.org/E02192.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Normality&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.&lt;/li&gt;&#xA;&lt;li&gt;Examples:  $\ce{\frac12 H2SO4, \frac15 KMnO4, \frac13 Fe^{3+}}$&lt;/li&gt;&#xA;&lt;li&gt;Synonyms: equivalent concentration, equivalent entity &lt;/li&gt;&#xA;&lt;li&gt;IUPAC discourages the use of this&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related &lt;a href=&quot;https://goldbook.iupac.org/M03970.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molality&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Amount of entities of a solute divided by the mass of the solvent&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $m, b$&lt;/li&gt;&#xA;&lt;li&gt;SI Unit:  $b_i = \frac{n_i}{m_{\mathrm{solvent}}} = \mathrm{\frac{mol}{kg}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related &lt;a href=&quot;https://goldbook.iupac.org/A00296.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mole fraction&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Amount of a constituent divided by the total amount of all constituents in the mixture.&lt;/li&gt;&#xA;&lt;li&gt;Synonym: Amount fraction&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $x$ ($y$ for gaseous mixtures)&lt;/li&gt;&#xA;&lt;li&gt;SI Unit: (dimensionless) $x_i = \frac{n_i}{n} = \mathrm{1}$&lt;/li&gt;&#xA;&lt;li&gt;Common units: $\mathrm{\frac{mol}{mol}}$. parts per million/billion/trillion (ppm/b/t)&lt;/li&gt;&#xA;&lt;li&gt;Mole ratio: $r_i = \frac{n_i}{n - n_i}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related Number fraction&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of entities of one constituent divided by the total number of entities in the mixture.&lt;/li&gt;&#xA;&lt;li&gt;equal to mole fraction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related &lt;a href=&quot;https://goldbook.iupac.org/M03722.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass fraction&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass of a constituent divided by the total mass of all constituents in the mixture.&lt;/li&gt;&#xA;&lt;li&gt;Symbol: $w$&lt;/li&gt;&#xA;&lt;li&gt;SI Umit: (dimensionless) $w_i = \frac{m_i}{m} = \mathrm{1}$&lt;/li&gt;&#xA;&lt;li&gt;Common units: $\mathrm{\frac{kg}{kg}}$. parts per million/billion/trillion (ppm/b/t)&lt;/li&gt;&#xA;&lt;li&gt;Mass ratio: $\zeta_i = \frac{m_i}{m - m_i}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Further reading&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Concentration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://goldbook.iupac.org/F02494.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;fraction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://goldbook.iupac.org/M03948.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mixing ratio&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related Tags&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;/questions/tagged/molar&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molar&amp;#39;&quot; rel=&quot;tag&quot;&gt;molar&lt;/a&gt;&#xA;  &lt;a href=&quot;/questions/tagged/solutions&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solutions&amp;#39;&quot; rel=&quot;tag&quot;&gt;solutions&lt;/a&gt;&#xA;  &lt;a href=&quot;/questions/tagged/mixtures&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;mixtures&amp;#39;&quot; rel=&quot;tag&quot;&gt;mixtures&lt;/a&gt;&#xA;  &lt;a href=&quot;/questions/tagged/aqueous-solution&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;aqueous-solution&amp;#39;&quot; rel=&quot;tag&quot;&gt;aqueous-solution&lt;/a&gt;&#xA;  &lt;a href=&quot;/questions/tagged/solubility&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solubility&amp;#39;&quot; rel=&quot;tag&quot;&gt;solubility&lt;/a&gt;&#xA;  &lt;a href=&quot;/questions/tagged/solvents&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solvents&amp;#39;&quot; rel=&quot;tag&quot;&gt;solvents&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2017-06-25T13:03:07.963" LastActivityDate="2017-06-25T13:03:07.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10535" PostTypeId="4" CreationDate="2014-05-08T09:16:45.350" Score="0" Body="The concentration characterizes the composition of a mixture with respect to the volume of the mixture. There are mass, amount, number and volume concentrations. Related and within the scope of this tag are normality, molality, mole fraction/ratio, mass fraction/ratio." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-08T11:27:37.037" LastActivityDate="2014-05-08T11:27:37.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10536" PostTypeId="5" CreationDate="2014-05-08T09:45:16.753" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://goldbook.iupac.org/A00204.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alcohol&lt;/a&gt; refers to a class of organic compounds, which have a hydroxyl functional group bound to a saturated carbon atom (phenols are considered a separate class). It is often used as a synonym for ethyl-alcohol which is used in alcoholic beverages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://goldbook.iupac.org/SVG/A00204.png&quot; alt=&quot;generic structure of alcohol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several kinds of alcohols, depending on the moieties $\ce{R}$. Important classes include simple acyclic alcohols with the general formula $\ce{C_{n}H_{2n + 1}OH}$. The most famous one amongst this group is  ethyl-alcohol $\ce{H3C-CH2OH}$ which is used in beverages. It is therefore very common that alcohol refers to this particular molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hydroxylic group is connected with unsaturated carbon resulting compounds are called phenols, enols or ynols and have properties distinct from alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohols of any kind find various applications as antiseptics, fuels and preservatives. Especially in &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt; they are used as solvents. (See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-14T19:51:29.830" LastActivityDate="2018-03-14T19:51:29.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10537" PostTypeId="4" CreationDate="2014-05-08T09:45:16.753" Score="0" Body="For questions about saturated and unsaturated alcohols, their physical properties, their reactions, etc. For questions about phenol, use the [phenols] tag instead." OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-14T19:41:59.423" LastActivityDate="2018-03-14T19:41:59.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10538" PostTypeId="2" ParentId="10531" CreationDate="2014-05-08T10:32:39.603" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could have performed any procedure to remove the solvent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple distillation  - This would be slower than what you did and could overheat the undistilled aniline causing decomposition.&lt;/li&gt;&#xA;&lt;li&gt;Vacuum distillation - This is what you did, perhaps even on a special apparatus called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rotary_evaporator&quot; rel=&quot;nofollow&quot;&gt;rotary evaporator&lt;/a&gt;. This method is still a distillation, even if your product was not distilled. &lt;strong&gt;Something&lt;/strong&gt; was distilled.&lt;/li&gt;&#xA;&lt;li&gt;Evaporation on a hot plate (or steam bath)- &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Steam distillation is suitable because the aniline is already in an aqueous suspension. If do vacuum distillation, you would need to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Separate the aniline from the aqueous mixture (extract with ether)&lt;/li&gt;&#xA;&lt;li&gt;Remove the ether (simple distillation or reduced pressure distillation)&lt;/li&gt;&#xA;&lt;li&gt;Vacuum distil the aniline&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you look at what you did, it is the same steps in a different order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Distil the aniline &lt;/li&gt;&#xA;&lt;li&gt;Separate the aniline from the aqueous mixture (extract with ether)&lt;/li&gt;&#xA;&lt;li&gt;Remove the ether (simple distillation or reduced pressure&#xA;distillation)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Either is probably suitable. In the end, the steam distillation is easier and less hazardous than the full-blown vacuum distillation you would need.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-08T10:32:39.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10539" PostTypeId="2" ParentId="10512" CreationDate="2014-05-08T10:51:28.740" Score="2" Body="&lt;p&gt;The fine structure in the diatomic vibrational spectra is based on various population of the rotational levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best demonstration of isotopic effects in vibrational spectra is to switch hydrogen for deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of vibration (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law#Harmonic_oscillator&quot; rel=&quot;nofollow&quot;&gt;Hooke's Law&lt;/a&gt; adapted to a harmonic oscillator) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu = \frac{1}{2\pi}\sqrt{\dfrac{k}{\mu}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is the force constant of the oscillator and $\mu$ is the reduced mass: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu=\dfrac{M_1M_2}{M_1+ M_2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Switching $\ce{HCl}$ to $\ce{DCl}$ has a much larger effect than the two chlorine isotopes (35 and 37):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H^{35}Cl}, \ \mu\approx\frac{35}{36}=0.972$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H^{37}Cl}, \ \mu\approx\frac{37}{38}=0.974$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{D^{35}Cl}, \ \mu\approx\frac{70}{37}=1.892$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{D^{37}Cl}, \ \mu\approx\frac{74}{39}=1.898$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The difference between $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$ is in the third decimal place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a nice Google Image search for the &lt;a href=&quot;https://www.google.com/search?q=hcl%20and%20dcl%20ir&quot; rel=&quot;nofollow&quot;&gt;vibrational spectra of HCl and DCl&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-08T10:51:28.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10541" PostTypeId="2" ParentId="10526" CreationDate="2014-05-08T14:38:34.650" Score="2" Body="&lt;p&gt;Not all hydrogen atoms in a compound are equally acidic. The acidity of a hydrogen at a certain position in a molecule depends on its chemical environment, i.e. the solvent (if the compound is dissolved), which atoms and functional groups it is attached to, and how well the negative charge of the corresponding conjugated base is stabilized. A quantitative measure for the acidity of a proton is its $pK_a$ value which is defined as follows (&lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_dissociation_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pK_a=-\mathrm{log}_{10}K_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $K_a$ being the equilibrium constant of the dissociation reaction in water $\ce{HA + H2O \rightleftharpoons H3O+ + A-}$, in which the hydrogen is abstracted from the compound by heterolytic bond cleavage. $[\ce{H2O}]$ can be assumed to be constant (approx. 55 mol/l) and is therefore included in $K_a$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{a}=\frac{[A^{-}][H^{+}]}{[HA]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take formic acid ($\ce{HCOOH}$) as an example. As the name of the compound suggests, it has at least one acidic proton, and if all hydrogens would be equally acidic and dissociate from the molecule, the reaction would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCOOH + 2H2O \rightleftharpoons 2H3O+ + COO^{2-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this does not happen in aqueous solution, only one of the protons is acidic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCOOH + H2O \rightleftharpoons H3O+ + HCOO-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that both hydrogens are situated at different positions in the structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jVNJ4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a $pK_a$ of 3.77, the hydrogen of the carboxyl group is significantly more acidic than the hydrogen bonded to the carbon atom. The typical $pK_a$ of an aldehyde hydrogen (H bonded to the carbon of a carbonyl group) is in the range of 17 (&lt;a href=&quot;http://en.wikipedia.org/wiki/Aldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), so such a position cannot be deprotonated in water.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-12T15:13:34.677" LastActivityDate="2014-05-12T15:13:34.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10542" PostTypeId="5" CreationDate="2014-05-08T14:46:27.957" Score="0" Body="&lt;p&gt;Molar refers to a &lt;a href=&quot;/questions/tagged/concentration&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;concentration&amp;#39;&quot; rel=&quot;tag&quot;&gt;concentration&lt;/a&gt; given in mole per liter. It is often abbreviated with $M$. Mostly it is used in terms of x-molar solutions, e.g. one molar refers to a solution with a concentration of one mole per liter.&#xA;\begin{aligned}&#xA;       c_i        &amp;amp;=\frac{n_i}{V} \newline&#xA; \left[c_i\right] &amp;amp;=x\mathrm{\frac{mol}{L}} =x~M&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-08T15:15:27.307" LastActivityDate="2014-05-08T15:15:27.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10543" PostTypeId="4" CreationDate="2014-05-08T14:46:27.957" Score="0" Body="Molar refers to a concentration given in moles per liter. " OwnerUserId="4945" LastEditorUserId="73" LastEditDate="2014-05-08T15:15:40.677" LastActivityDate="2014-05-08T15:15:40.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10544" PostTypeId="2" ParentId="6499" CreationDate="2014-05-08T14:48:04.903" Score="1" Body="&lt;p&gt;&quot;&lt;em&gt;in excess of 2000 litres of water... per gram of dessicant [sic].&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the volume of one gram of desiccant? &lt;BR&gt;&#xA;2) What is the volume of 2000 liters of water? &lt;BR&gt;&#xA;3) What is the cognitive dissonance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = sin(1/x) in the interval (0,1).  Rotate about the x-axis. The volume of the cylinder is pi.  The surface area of the curve is infinite.  Wet that surface for a great storage reservoir.  If budget is tight, (1,0.0001) will hold the same volume.  Find the fallacy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ditto metal-oxide frameworks and gas adsorption to lower pressure of stored STP volume.  It not about the BET area, it's about the closed cycle - including adsorption and desorption enthalpies.&lt;/p&gt;&#xA;" OwnerUserId="4296" LastActivityDate="2014-05-08T14:48:04.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10545" PostTypeId="1" CreationDate="2014-05-08T15:32:44.803" Score="1" ViewCount="1074" Body="&lt;p&gt;How does carbonic acid cause acid rain when $K_b$ of bicarbonate is greater than $K_a$? Should it not create an alkaline solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = 4.8 \times 10^{-11}\ (mol/L)$. $K_b = 2.3 \times 10^{-8}\  (mol/L)$. These are the values for $\ce{HCO3-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers are from a school book that I read, but it's not in English. It's called &quot;Kjemi 1&quot; by Harald Brandt&lt;/p&gt;&#xA;" OwnerUserId="5225" LastEditorUserId="865" LastEditorDisplayName="user15489" LastEditDate="2021-05-22T21:14:27.190" LastActivityDate="2021-05-22T21:14:27.190" Title="How does carbonic acid cause acid rain when Kb of bicarbonate is greater than Ka?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10546" PostTypeId="2" ParentId="10545" CreationDate="2014-05-08T16:20:21.610" Score="2" Body="&lt;p&gt;What is the ${K_a}$  of carbonic acid? Note that sources differ in their ${K_a}$ values, and especially for carbonic acid, since there are two kinds - a pseudo-carbonic acid/hydrated carbon dioxide and the real thing (which exists in equilibrium with hydrated carbon dioxide but in a small concentration - about 4% of what what appears to be carbonic acid is true carbonic acid, with the rest simply being $\ce{H2O*CO_2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, I find that the ${K_a}$ of the mixed carbonic acid is about $4.2 \times 10^{-7}$, which is greater than $1.0  \times 10^{-7}$, and this implies that a solution of carbonic acid alone should be acidic no matter what. This suggests to me that your numbers are wrong; would you mind sharing your numbers and their source if possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I see that you have updated your numbers. Nonetheless, I believe that your ${K_a}$ for carbonic acid is wrong; that number looks suspiciously like the ${K_a}$ instead for hydrogen carbonate ion (or the bicarbonate ion). According to Wikipedia, the ${pKa}$ of carbonic acid, is 6.3 (and this is taking into account any aqueous carbon dioxide). Convert this to a ${K_a}$ value and we get about $5.0 \times 10^{-7}$. This is in-line with the value I obtained from a copy of Daniel C. Harris' &lt;em&gt;Qualitative Chemical Analysis&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: I think you've realized your mistake; as you say, the values are for $\ce{HCO_3^-}$, which is the hydrogen carbonate ion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T02:33:44.340" LastActivityDate="2015-06-28T02:33:44.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10547" PostTypeId="1" CreationDate="2014-05-08T17:26:42.010" Score="6" ViewCount="25869" Body="&lt;p&gt;Haloalkanes aren't very soluble in water because they can't form hydrogen bonds, and the energy required to break hydrogen bonds in water etc is higher than the energy released when new bonds between the haloalkane and water are formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as fluorine forms hydrogen bonds why couldn't a fluoroalkane form hydrogen bonds with water thus become more soluble in water?&lt;/p&gt;&#xA;" OwnerUserId="5445" LastEditorUserId="16683" LastEditDate="2017-04-06T13:34:56.333" LastActivityDate="2020-07-29T12:07:57.317" Title="The solubility of haloalkanes in water" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;halides&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10548" PostTypeId="2" ParentId="10547" CreationDate="2014-05-08T17:39:54.920" Score="3" Body="&lt;p&gt;I think the answer lies with the fact that alkanes are basically non-polar, even if they have halogen substituents. Water, on the other hand, is polar, so the solvent and solute are incompatible. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2017-04-25T20:48:11.370" LastActivityDate="2017-04-25T20:48:11.370" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10549" PostTypeId="2" ParentId="1" CreationDate="2014-05-08T18:00:40.150" Score="2" Body="&lt;p&gt;there is an important tool develop to give answer of this question which is effective nuclear charge, which is the effective nuclear charge experienced by an electron as an electron in outer shell will always be shielded by inner electrons so as we move from left to right electrons in outer shell increases where by electrons in inner shell remain same. Hence effective nuclear charge increases and forcing electrons to come closer.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-05-08T18:00:40.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10550" PostTypeId="2" ParentId="4404" CreationDate="2014-05-08T18:06:22.340" Score="1" Body="&lt;p&gt;first of all take one group a day.&#xA;right whatever you have to learn and then try to identify some relationship. Then reread the material and write down whatever you remember . Now look what you miss. &#xA;everyday take a new group and do the same.&#xA;Also write whatever you remember of the previous groups you have learned. &lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-05-08T18:06:22.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10551" PostTypeId="2" ParentId="10514" CreationDate="2014-05-08T22:08:37.543" Score="2" Body="&lt;p&gt;Perhaps the enthalpy of interaction between the solute and solvent does not change as the solvent evaporates (at least until the concentration of solute gets high enough that solute-solute interactions become important) so the entropy is the only factor that changes.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2014-05-08T22:08:37.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10552" PostTypeId="1" CreationDate="2014-05-08T22:52:19.360" Score="1" ViewCount="1389" Body="&lt;p&gt;I have to perform a chemical test to determine whether the solution is a secondary or a tertiary alcohol. How can I do this? Please explain the procedure and the apparatus.&lt;/p&gt;&#xA;" OwnerUserId="5427" LastEditorUserId="16683" LastEditDate="2018-10-24T18:25:35.427" LastActivityDate="2018-10-24T18:25:35.427" Title="What chemical test can differentiate secondary and tertiary alcohols?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;alcohols&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10553" PostTypeId="2" ParentId="10552" CreationDate="2014-05-08T23:18:30.413" Score="1" Body="&lt;p&gt;A secondary alcohol can be further oxidized to a ketone (&lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_of_secondary_alcohols_to_ketones&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;), while a tertiary alcohol cannot. For example, when you use dichromate as oxidation agent, you will observe a color change of the solution from orange to green. A tertiary alcohol will not react unless your oxidizer is too concentrated.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-08T23:18:30.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10554" PostTypeId="2" ParentId="10552" CreationDate="2014-05-08T23:20:38.530" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;Add sulfuric acid to a chilled aqueous solution of the alcohol and sodium nitrite.&lt;/li&gt;&#xA;&lt;li&gt;Record the UV spectrum.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tertiary alkyl nitrites show an absorption band at $\lambda$ = 400 nm, secondary and primary alkyl nitrites only below.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-05-08T23:20:38.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10555" PostTypeId="5" CreationDate="2014-05-09T05:13:11.190" Score="0" Body="&lt;p&gt;Protons are subatomic particles with the symbol $p$ or $p^{+}$. They have a positive electric charge of 1 elementary charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus of every element consists of protons and the number of these protons are referred to as its atomic number. There are also neutrons present in the nucleus. Both are usually referred to as nucleons. The main mass of an atom (or ion) comes from these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term proton is also often used for the positively charged Hydrogen ion $\ce{{}^{1}H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proton itself consists of two up quarks and one down quark and it is also a fermion. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-09T07:30:39.483" LastActivityDate="2014-05-09T07:30:39.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10556" PostTypeId="4" CreationDate="2014-05-09T05:13:11.190" Score="0" Body="Protons are subatomic particles with the symbol p or p+. They have a positive electric charge of 1 elementary charge. Questions referring to interactions of protons in solutions should rather use the acid-base tag." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-09T05:55:27.993" LastActivityDate="2014-05-09T05:55:27.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10557" PostTypeId="5" CreationDate="2014-05-09T06:20:17.750" Score="0" Body="&lt;p&gt;&lt;strong&gt;This general tag should &lt;em&gt;not&lt;/em&gt; be used anymore.&lt;/strong&gt; Other tags like  &lt;a href=&quot;/questions/tagged/stoichiometry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stoichiometry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stoichiometry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/acid-base&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;acid-base&amp;#39;&quot; rel=&quot;tag&quot;&gt;acid-base&lt;/a&gt;, ... should be used instead. This is an active burninate-request. See &lt;a href=&quot;https://chemistry.meta.stackexchange.com/q/3601/&quot;&gt;I have a terrible [reaction]!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2017-03-27T12:22:14.563" LastActivityDate="2017-03-27T12:22:14.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10558" PostTypeId="4" CreationDate="2014-05-09T06:20:17.750" Score="0" Body="DEPRECATED. Do not use this tag anymore. (See tag wiki and meta for more information.)&#xD;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2017-03-27T12:22:14.563" LastActivityDate="2017-03-27T12:22:14.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10559" PostTypeId="5" CreationDate="2014-05-09T07:06:19.733" Score="0" Body="&lt;p&gt;This tag should be applied to questions dealing with solutions of any kind. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://goldbook.iupac.org/S05746.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;solution&lt;/a&gt; is liquid or solid phase containing more than one substance when for convenience one (or more) substance(s), which is called the solvent, is treated differently from the other substances, which are called solutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a &lt;a href=&quot;http://goldbook.iupac.org/D01739.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;dilute solution&lt;/a&gt;. A superscript attached to the  $\infty$ symbol for a property of a solution denotes the property in the limit of infinite dilution. (from the IUPAC Gold Book)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution is also often referred to as a mixture of different compounds. The main characteristic is that it consists of only one homogeneous phase.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Examples of solutions:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gases in Liquids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carbon dioxide in water&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Liquid in Liquids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ethanol in water&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solids in Liquids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sodium chloride in water&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gases in Solids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrogen in Palladium&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Liquids in Solids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pentane/ Hexane in paraffin wax&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solids in Solids&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tin in Copper (known as bronze)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/I02934.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal dilute solution&lt;/a&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dilute solution in which the solute may be regarded as obeying Henry's law, so that all the solute activity coefficients may be approximated to 1.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/S05471.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;saturated solution&lt;/a&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A solution which has the same concentration of a solute as one that is in equilibrium with undissolved solute at specified values of the temperature and pressure.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/M03940.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mixed crystal (solid solution)&lt;/a&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A crystal containing a second constituent which fits into and is distributed in the lattice of the host crystal. (The use of 'solid solution' for amorphous materials is not recommended.)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related tags&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;/questions/tagged/solubility&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solubility&amp;#39;&quot; rel=&quot;tag&quot;&gt;solubility&lt;/a&gt; &lt;a href=&quot;/questions/tagged/solvents&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solvents&amp;#39;&quot; rel=&quot;tag&quot;&gt;solvents&lt;/a&gt; &lt;a href=&quot;/questions/tagged/aqueous-solution&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;aqueous-solution&amp;#39;&quot; rel=&quot;tag&quot;&gt;aqueous-solution&lt;/a&gt; &lt;a href=&quot;/questions/tagged/thermodynamics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;thermodynamics&amp;#39;&quot; rel=&quot;tag&quot;&gt;thermodynamics&lt;/a&gt; (&lt;a href=&quot;/questions/tagged/water&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;water&amp;#39;&quot; rel=&quot;tag&quot;&gt;water&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-04-08T01:51:43.813" LastActivityDate="2017-04-08T01:51:43.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10560" PostTypeId="4" CreationDate="2014-05-09T07:06:19.733" Score="0" Body="This tag should be applied to questions dealing with solutions of any kind. A solution is a liquid or solid phase containing more than one substance where, for convenience, one (or more) substance(s) called the solvent is treated differently from the other substances, which are called solutes. If water is the solvent, then the more appropriate 'aqueous-solution' tag should (also) be specified." OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-04-08T01:50:32.297" LastActivityDate="2017-04-08T01:50:32.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10562" PostTypeId="2" ParentId="915" CreationDate="2014-05-09T10:03:52.960" Score="1" Body="&lt;p&gt;The electronic configuration of carbon is $\ce{1s^2 2s^2 2p^2}$. So a carbon atom requires 4 electrons to complete its outermost orbital. The electronic configuration of  oxygen is $\ce{1s^2 2s^2 2p^4}$ so an oxygen atom requires 2 electrons to complete its octet configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus two oxygen atoms and 1 carbon atom form a double bond between them so more energy is required to break this bond. In this way $\ce{CO2}$ is formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For converting $\ce{CO2}$ back into carbon and oxygen atoms, $\ce{CO2}$ should be heated at almost $\pu{298 K}$ so carbon becomes gaseous and oxygen becomes part of air.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit from another deleted answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{CO2}$ can be produced by combining $\ce{C}$ and $\ce{O2}$ under the presence of air; then oppositely $\ce{CO2}$ can be separated to $\ce{C}$ and $\ce{O2}$ by heating $\ce{CO2}$ at almost that temperature which is impossible for $\ce{CO2}$ to absorb heat.&lt;/p&gt;&#xA;" OwnerUserId="5446" LastEditorUserId="16683" LastEditDate="2018-08-14T15:28:45.940" LastActivityDate="2018-08-14T15:28:45.940" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10566" PostTypeId="2" ParentId="5974" CreationDate="2014-05-09T11:28:40.253" Score="4" Body="&lt;p&gt;So if I understand the question correctly, you heated up water with some salt by performing electrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, you would also generate chlorine gas. However, the heat capacity of water is really large, so I doubt that you would even be capable of bringing water to a boil by just jolting current through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question: &lt;strong&gt;Yes, when performing electrolyzation of any solution you should apply the usual precautionary measures, such as working in an open space or under a fume hood.&lt;/strong&gt; Under no circumstances should it be possible for you to inhale the gases produced.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-05-09T11:28:40.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10567" PostTypeId="1" AcceptedAnswerId="10568" CreationDate="2014-05-09T14:01:32.873" Score="0" ViewCount="466" Body="&lt;p&gt;Are these structures equivalent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i.minus.com/jnojtCx6wYvqD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://i.minus.com/jnojtCx6wYvqD.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/diAXS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-08-19T19:57:03.007" LastActivityDate="2015-08-19T19:57:03.007" Title="Equivalent Cyclohexane Structures?" Tags="&lt;organic-chemistry&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10568" PostTypeId="2" ParentId="10567" CreationDate="2014-05-09T14:18:36.113" Score="2" Body="&lt;p&gt;Yes, the two structures you've drawn are the same; both substituents are in axial positions.  The two structures shown in the link you provided are not the same; they are conformational isomers.  The one on the left is the same as the one you've drawn with both substituents in the axial position.  The one on the right has both substituents in the equatorial position.  These two conformational isomers exist in equilibrium with the (e,e) isomer favored.  The two conformers interconvert by a cyclohexane ring flip.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-09T14:18:36.113" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10569" PostTypeId="1" AcceptedAnswerId="73239" CreationDate="2014-05-09T15:15:02.950" Score="36" ViewCount="3960" Body="&lt;p&gt;&lt;strong&gt;PREFACE:&lt;/strong&gt; I am no expert on this topic.  My questions at the bottom may be off base.  I have some experience with symmetry-adapted perturbation theory (SAPT) when it comes to analyzing intermolecular interactions.  The total interaction energy of a system can be quantified in four different energy components including exchange, induction, electrostatics, and dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural Bond Orbital (&lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_bond_orbital&quot; rel=&quot;noreferrer&quot;&gt;NBO&lt;/a&gt;) analyses translate the complex quantum-mechanical wavefunction into a more tangible Lewis-dot-like formalism (natural Lewis structure).  The introduction of the following series of equations has been largely influenced by the presentation of &lt;a href=&quot;http://www.wiley.com/WileyCDA/WileyTitle/productCd-1118119967.html&quot; rel=&quot;noreferrer&quot;&gt;Weinhold and Landis (2012)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important Terms:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lewis-type NBOs - filled 'donor' orbitals (e.g. $\sigma_{AB}, \sigma_{CD}, \cdots$)&lt;/li&gt;&#xA;&lt;li&gt;Non-Lewis-type NBOs - vacant 'acceptor' orbitals (e.g. $\sigma_{AB}^{\ast}, \sigma_{CD}^{\ast}, \cdots$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lewis and non-Lewis NBOs mix:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a wavefunction, $\Psi$, comprised of two components, a known natural Lewis structure wavefunction, $\Psi^{(L)}$, and a non-Lewis correction wavefunction, $\Psi^{(NL)}$, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi = \Psi^{(L)} + \Psi^{(NL)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural Lewis structure wavefunction is simply a fictitious representation of an exact physical system where resonance effects are non-existent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a corresponding Hamiltonian, $\hat{\mathcal{H}}$, which can also be treated as an additive scheme with corresponding components to each piece of the wavefunction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{\mathcal{H}} = \hat{\mathcal{H}}^{(L)} + \hat{\mathcal{H}}^{(NL)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these two equations and the form of the Schrödinger Equation, $\hat{\mathcal{H}}\Psi=E\Psi$, there must also exist a similar construction of the energy of the system, $E$, such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^{(L)} + E^{(NL)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three equations provide the foundation for a &quot;systematic 'perturbation theory' analysis&quot;, where $\hat{\mathcal{H}}^{(L)}$ is the unperturbed Hamiltonian with a corresponding (known) eigenfunction, $\Psi^{(L)}$, and energy eigenvalue $E^{(L)}$.  Note that $\Psi^{(L)}$ is not only known, but it is related to an 'idealized single reference picture' in which each Lewis-type NBO has exact double occupancy for closed shell systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with Hartree-Fock, the limit of a single-reference method can be reached by an infinite expansion of a basis ($\Omega)$ in the one-electron eigenvalue equation.  For the natural Lewis structure, this can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^{(0)} \Omega_i^{(L)} = \varepsilon_i^{(L)}\Omega_i^{(L)} \qquad \mathrm{where~~} i = 1,2,\cdots,n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon_i^{(L)}$ are orbital energies of Lewis-type NBOs.&#xA;Because $h^{(0)}$ also contains the non-Lewis contributions such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h^{(0)} \Omega_j^{(NL)} = \varepsilon_j^{(NL)}\Omega_j^{(NL)} \qquad \mathrm{where~~} j = n+1,\cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because natural Lewis structures contain zero resonance, the donor (Lewis-type NBOs) and acceptor (non-Lewis-type NBOs) orbitals do not interact due to their orthogonal nature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int \Omega_i^{(L)\ast}h^{(0)}\Omega_j^{(NL)}d\tau=0 \qquad \mathrm{for~all~~} i,j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a real-world $1e^-$ Hamiltonian operator, $F$ (i.e. Fock operator, Kohn-Sham operator, etc.), will have donor-acceptor interactions such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{ij} = \int \Omega_i^{(L)\ast}F\Omega_j^{(NL)}d\tau\neq0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leading to mixing (delocalization) which connects the non-Lewis type NBOs with Lewis-type NBOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second-Order Perturbative Treatment of Delocalizations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mixing of the donor and acceptor orbitals can be treated with second-order perturbation theory.  Even though $\Psi^{(L)}$ ignores interactions with non-Lewis acceptors, it contains more than 99% of the total electron density, $\rho_L$, offering a good starting point for an unperturbed wavefunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixing of donors and acceptors lead to an overall energy lowering (&quot;stabilization&quot;), a quantum mechanical phenomenon.  Consider two interacting orbitals, as schematized below.  One is doubly-occupied with a lone electron pair, $n$, and the other is an unoccupied antibonding orbital (here a $\pi^{\ast}$ orbital). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1in$ &lt;img src=&quot;https://i.stack.imgur.com/NVeyP.png&quot; alt=&quot;Donor-acceptor orbital mixing&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overlap of the lone electron pair, $n$, with the vacant antibonding orbital, $\pi^{\ast}$ causes an change in energy of the lower occupied orbital, $x$. This &lt;em&gt;change in energy&lt;/em&gt; between the non-mixing orbital with $n$ and the mixing orbital $x$ (denoted as $\Delta E$) is referred to as 'stabilization energy' of the electron delocalization.  This energy (in kcal mol$^{-1}$) is determined via the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E_{ij}^{(2)} = \frac{-q_i \left| F_{ij} \right|^2}{\left(\varepsilon_j^{(NL)}-\varepsilon_i^{(L)}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon_j^{(NL)}$ is the energy of the non-Lewis NBO (i.e. $\pi^{\ast}$),  $\varepsilon_i^{(L)}$ is the energy of the orbital occupied by $n$, and $q_i$ is the occupancy of the orbital ($q=2$ in the above figure).  The 'stabilization energy' $\Delta E_{ij}^{(2)}$ as determined by second-order perturbation treatments is commonly abbreviated as $E(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recap:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have introduced (albeit very poorly) the concept of a separable wavefunction into natural Lewis and non-Lewis type formalisms.  These two states mix, allowing for delocalization of electrons, a phenomenon which leads to lower energy states.  This energy lowering is commonly called 'stabilization energy'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical Significance?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, after such an introduction, I am still at a loss as to what all of this &lt;em&gt;really&lt;/em&gt; means.  As a theoretical chemist (I'm a chemist, theoretically speaking...), I am interested in the physical significance of this type of quantity [$E(2)$].  However, I am unsure how significant this type of value truly is.  Do these delocalizations (energy stabilizations) lead to an overall lower energy of the system?  Would a system with more of these delocalizations be expected to lie lower in energy than a comparable system without delocalizations?  Can that type of correlation be made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I just do not see any usefulness in computing these $E(2)$ values for any reason because, while it may tell you that particular donor-acceptor interactions lead to this energy lowering, it doesn't offer much more than that.  What am I missing here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR - What can I use $\mathbf{E(2)}$ values for?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="11367" LastEditDate="2017-04-24T01:34:28.793" LastActivityDate="2019-10-30T06:49:51.950" Title="Natural Bond Orbital analysis: Significance of stabilization energy determined by 2nd order perturbation" Tags="&lt;computational-chemistry&gt;&lt;orbitals&gt;&lt;theoretical-chemistry&gt;&lt;resonance&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10570" PostTypeId="2" ParentId="10506" CreationDate="2014-05-09T15:26:52.490" Score="3" Body="&lt;p&gt;Whether the stretch of a bond in a molecule is asymmetric or symmetric depends on the relative movement of the atoms connected by the bond. Lets take a $\ce{CH2}$ group as an example. If the two hydrogen atoms connected to the carbon move in the same direction during stretching, it s a symmetric stretch, if they are going in opposite directions, the stretch is asymmetric (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Infrared_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Q8aS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left: symmetric stretch, right: asymmetric stretch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bond vibration like stretching will only be IR-active (i.e. give a band in the IR spectrum) if it is accompanied by a change of dipole moment. In the case of the $\ce{C-H}$ bond, carbon is slightly negatively and hydrogen slightly positively polarized. A symmetric stretch increases and decreases the distance between these partial charges periodically, and therefore also changes the dipole moment, because the latter is dependent from the distance between charges (&lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). During an asymmetric stretch, the elongation of one $\ce{C-H}$ bond is compensated by the shortening of the other, so the overall change of dipole moment should be weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of 2-(4-&lt;em&gt;tert&lt;/em&gt;-butylphenyl)ethanol (the correct name for your compond), the band at 3006 $\mathrm{cm}^{-1}$ can either be assigned to the $\ce{CH2}$ groups of the substituted ethanol or to $\ce{C-H}$ bonds in the benzene ring. In the first case, you can distinguish between symmetrical and asymmetrical stretch (symmetrical seems more likely to me), while in the latter case, a single hydrogen is attached to a carbon, and a distinction of the stretching, like in the case of two substituents at the same carbon, cannot be made.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-09T17:18:57.393" LastActivityDate="2014-05-09T17:18:57.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10571" PostTypeId="1" AcceptedAnswerId="10575" CreationDate="2014-05-09T17:51:35.880" Score="10" ViewCount="309" Body="&lt;p&gt;I have tried for hours to calculate the commutator of angular momentum in the differential form, but I cannot get the correct answer. This is my first experience with actually checking if two operators commutes, so there may be some beginner's misunderstandings that causes the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from my textbook the correct answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\hat{L}_x,\hat{L}_y] = \frac{\hbar}{\mathrm{i}}\hat{L}_z = \frac{\hbar}{\mathrm{i}} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angular momenta are defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\&#xA;\hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\&#xA;\hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sets up the commutator expression, and writes it out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\hat{L}_x,\hat{L}_y] = \left[ y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} , z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right] &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I assume is a well-known algebraic rule for handling commutators, I get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right] - \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right] &#xA;- \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right] &#xA;+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was straight forward to find that the two middle terms commute, i.e. they equal zero. Hence, we are left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x}   \right]&#xA;+ \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}   \right] \tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon evaluating these terms, I let them act on a function $ \psi $. Looking at the first term:&#xA;First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on $ \psi $:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;\frac{\hbar}{\mathrm{i}} \left(y \frac{\partial \psi z}{\partial z} \frac{\partial \psi}{\partial x} - yz  \frac{\partial \psi }{\partial x}  \frac{\partial \psi}{\partial z}     \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The y in the second term above can be factored out from the differentials, because the differential is not done with respect to y. In the first term, however, the z must be differentiated along with psi by using the product rule. Doing this, and cancelling equal terms of opposite sign, yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\hbar}{\mathrm{i}} y \psi \frac{\partial \psi}{\partial x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating this calculation for the second term in equation $(1)$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\frac{\hbar}{\mathrm{i}} x \psi \frac{\partial \psi}{\partial y}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting into $(1)$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\hat{L}_x,\hat{L}_y] \psi = \frac{\hbar}{\mathrm{i}} \left(  y \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial y}  \right) \psi&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, comparing with the correct answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) My signs are opposite of what they should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) More importantly, I have a $ \psi $ in the differentials which should not be there. My question is, &quot;why, and how do I rid my answer of it?&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I have sufficiently showed what I have done. I have checked my notes several times, and I cannot get rid of those extra psi's in my answer.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="16683" LastEditDate="2018-03-02T14:04:28.517" LastActivityDate="2018-03-02T14:04:28.517" Title="Calculating Commutator of Differential Angular Momentum" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10572" PostTypeId="1" AcceptedAnswerId="10573" CreationDate="2014-05-09T20:53:51.127" Score="0" ViewCount="1032" Body="&lt;p&gt;Would the line structure of formaldehyde, H2COH, simply be a straight line with a OH on the end of the line? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="19383" LastEditDate="2015-10-09T18:12:41.623" LastActivityDate="2015-10-09T18:12:41.623" Title="Line structure of formaldehyde" Tags="&lt;organic-chemistry&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10573" PostTypeId="2" ParentId="10572" CreationDate="2014-05-09T21:09:38.770" Score="2" Body="&lt;p&gt;Formaldehyde, $\ce{H_2CO}$, has the following structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Q9wO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no $\ce{OH}$ group, just a central carbon atom with a single bond to each of two hydrogens and a double bond to oxygen.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-09T21:09:38.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10574" PostTypeId="2" ParentId="10572" CreationDate="2014-05-09T21:10:46.343" Score="1" Body="&lt;p&gt;When you omit the hydrogens, it would be a straight double line with an O at the end.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-09T21:10:46.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10575" PostTypeId="2" ParentId="10571" CreationDate="2014-05-09T21:18:07.260" Score="8" Body="&lt;p&gt;Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum_operator#Commutation_relations&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1].&#xA;So, substituting the differential operator expression for $\hat{L}_z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into the commutator equation you should get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your problem with the additional $\psi$:&#xA;Everything is fine until you let equation [1] act on a wavefunction $\psi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right]&#xA;+ \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \right) \psi&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations:&#xA;When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is incorrect. The correct way to do this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result.&#xA;So, in the case at hand this leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right] \psi &amp;amp;= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\&#xA;  &amp;amp;= -\hbar^{2} y \frac{\partial \psi}{\partial x}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \psi &amp;amp;= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\&#xA;  &amp;amp;= \hbar^{2} x \frac{\partial \psi}{\partial y}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you substitute these equations into the commutator equation you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;[\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is exactly the result you should expect.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-05-10T09:26:38.263" LastActivityDate="2014-05-10T09:26:38.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10576" PostTypeId="1" CreationDate="2014-05-09T21:43:17.873" Score="3" ViewCount="712" Body="&lt;p&gt;About two weeks ago I put some ordinary unleaded petrol into a plastic water bottle, as a temporary measure (long story, don't ask!). It has now formed a reddish liquid at the bottom of the bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottle was emptied prior to adding the petrol, but probably still contained a small amount of water when the fuel was poured in. The bottle was stored in ordinary room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume that the fuel has corroded/reacted with the plastic and left this substance, but I have looked to Stack Exchange in the hope of a more scientific answer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the red substance and cause of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5ath0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ath0l.jpg&quot; alt=&quot;The red substance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please note that the brownish 'slug-shape' is merely an artifact of the light refraction...&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5448" LastEditorUserId="3748" LastEditDate="2014-05-29T20:26:21.253" LastActivityDate="2014-07-28T20:56:55.120" Title="What is this red substance that has accumulated at the bottom of a plastic bottle full of petrol (see picture)?" Tags="&lt;plastics&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10577" PostTypeId="1" AcceptedAnswerId="10578" CreationDate="2014-05-10T01:27:43.927" Score="0" ViewCount="309" Body="&lt;p&gt;Are the following equivalent structures for 2,2-dimethylpropane?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yRiRd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think so, because my rendering simply rotates the cross 45 degrees! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, regarding 3,3-diethylpentane, are these both equally valid structures? My rendering seems to look way different but filling in the carbons and hydrogens, both appear correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kfywh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2016-12-23T22:13:44.403" LastActivityDate="2016-12-23T22:13:44.403" Title="Equivalent Line Angle Structures?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10578" PostTypeId="2" ParentId="10577" CreationDate="2014-05-10T01:48:35.040" Score="3" Body="&lt;p&gt;a) yes, those are equivalent structures of 2,2-dimethylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) yes, those are equivalent structures of 3,3-diethylpentane&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-10T01:48:35.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10579" PostTypeId="1" AcceptedAnswerId="10586" CreationDate="2014-05-10T05:04:49.843" Score="5" ViewCount="218" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is ammonia a nucleophile and hydrogen the electrophile in the Brønsted-Lowry acid-base reaction of acetic acid with ammonia?&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + NH3 &amp;lt;=&amp;gt; CH3COO- + NH4+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I say that ammonia is the nucleophile because I drew a curved arrow from the lone pair on the ammonia to the ionizable hydrogen on acetic acid, and I drew a curved arrow from the bonding pair in the hydroxyl group to the oxygen in the hydroxyl group of acetic acid. It seems that ammonia is the nucleophile because it is attacking the hydrogen proton (a bare proton) and that hydrogen is the electrophile because it likes the lone pair on ammonia.&lt;/p&gt;&#xA;&lt;p&gt;So if the above is right, this would make ammonia a Lewis base and the hydrogen proton a Lewis acid? Or do we say the entire acetic acid molecule is a Lewis acid?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="83758" LastEditDate="2020-09-07T08:28:01.313" LastActivityDate="2020-09-07T08:28:01.313" Title="Identifying nucleophiles and electrophiles in an acid-base reaction" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10580" PostTypeId="1" AcceptedAnswerId="10593" CreationDate="2014-05-10T05:33:42.830" Score="1" ViewCount="2596" Body="&lt;p&gt;I have a problem with this graphic, specifically the right hand side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a4nnv.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why a &quot;sextet&quot; atom (an atom with 6 valence electrons)? I thought only a select few elements (C, N, and O) actually follow the octet rule. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-10-16T17:42:07.580" LastActivityDate="2015-10-16T17:42:07.580" Title="Electron Pushing" Tags="&lt;organic-chemistry&gt;&lt;erratum&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10581" PostTypeId="2" ParentId="9836" CreationDate="2014-05-10T05:44:12.960" Score="1" Body="&lt;p&gt;Regarding the ionization extent of sulfuric acid solutions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{1)}$ The first ionization is complete at most molarities (until you start reaching extremely high molarities - i.e. $\text{18 M}$). So sulfuric acid is a strong acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{2)}$ The second ionization (i.e. the ionization of the hydrogen sulfate ion, $\ce{HOSO3^-}$, &lt;em&gt;can&lt;/em&gt; be complete at low starting concentrations of hydrogen sulfate ion. This is because $K_a(\ce{HOSO3^-}) = 1.2 \times 10^{-2}$ is rather sizable! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can generalize these two cases of sulfuric acid solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For very dilute sulfuric acid solutions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{H3O^+}]=2M_i - [\ce{HOSO3^-}] = [\ce{HOSO3^-}]+ 2\times[\ce{SO4^{2-}}]$, in which we generally take $[\ce{HOSO_3^-}]$ as negligible (i.e. = $\epsilon \to 0$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For non-dilute solutions of sulfuric acid, we only have these relationships:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H3O^+}]=M_i +[\ce{SO4^{2-}}] = [\ce{HOSO3^-}]+2[\ce{SO4^{2-}}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, the first relationship (one for very dilute solutions) follows from the second because in a very dilute solution, $[\ce{HOSO3^-}]\approx 0$ - i.e. it has fully ionized into $[\ce{SO4^{2-}}]$.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="15235" LastEditDate="2017-03-22T12:49:13.653" LastActivityDate="2017-03-22T12:49:13.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10582" PostTypeId="1" AcceptedAnswerId="10594" CreationDate="2014-05-10T06:06:21.383" Score="1" ViewCount="520" Body="&lt;p&gt;My book indicates that the following electron pushing is wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ejqM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what would the right electron pushing diagram be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following in mind. My rationale is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Electrons attack protons, not the other way around. This is just convention right? Is this just a conceit in chemistry; just a standardization? On a related note is this why one person described organic chemistry as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The foundation of every reaction in organic chemistry can be found in one statement: Nucleophile Attacks Electrophile.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than electrophile attacks nucleophile? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Water is the base here; water is the Lewis Base (nucleophile) so water will attack the hydrogen cation (which is a bare nucleus).&#xA;3) HCl is the acid here; H+ is the Lewis Acid (electrophile, so it's going to go toward the lone pairs on the water). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uGNON.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="12860" LastEditDate="2015-11-13T20:27:00.173" LastActivityDate="2015-11-13T20:27:00.173" Title="Correct Electron Pushing with HCl and Water" Tags="&lt;acid-base&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10584" PostTypeId="2" ParentId="9851" CreationDate="2014-05-10T08:14:17.337" Score="3" Body="&lt;p&gt;To understand this phenomenon first we must point out that there are very different solid substances and solvents. Different combinations of them lead to different interactions and properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway for the regular case - ionic or soluble compound in water - there is at least two steps to realize the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Break the solid structure (eg. the crystal lattice). &lt;strong&gt;This step generally absorbs energy&lt;/strong&gt; (endothermic, see lattice energy);&lt;/li&gt;&#xA;&lt;li&gt;Solvate the compound. Because there are interactions between matter, the solvent might organize itself around the compound (solvation shell). Those molecules involved are not any more available, they are tied to the compound, they are not any more free water. &lt;strong&gt;This step generally releases energy&lt;/strong&gt; (exothermic, see solvation energy). You can think about this step as a reaction of the solvent to recover form the intrusion by coating the intruder with one or several layers of its molecules. Then it 'feels' if there were only water. You will find details of the complex solvation process in good Physical Chemistry book.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Succeeding these two unit steps leads to dissolution. Solubility is the thermodynamic limit beyond that you cannot dissolve further molecule of a substance. Off course this is related with every substance present in the solution and available free water. Solubility are generally expressed for pure substance only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you realize your solution, speaking of heat exchanged with the environment, you experienced the balance between those two reactions. This is why the heat balance can lead to a global endothermic or exothermic reaction.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-05-10T10:33:39.350" LastActivityDate="2014-05-10T10:33:39.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10585" PostTypeId="1" AcceptedAnswerId="10587" CreationDate="2014-05-10T09:19:11.753" Score="1" ViewCount="1285" Body="&lt;p&gt;Acids are usually corrosive but what determines this? Is it the concentration of the acid or strength of the acid ($K_a$ value)?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2014-05-10T10:36:53.373" LastActivityDate="2014-05-10T10:36:53.373" Title="What are the factors that affect the corrosiveness of an acid?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10586" PostTypeId="2" ParentId="10579" CreationDate="2014-05-10T09:48:40.027" Score="5" Body="&lt;p&gt;In my opinion, your statements are both right, this is the &lt;a href=&quot;http://goldbook.iupac.org/L03508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC definition&lt;/a&gt; [1]:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Lewis acid&lt;/strong&gt;: A molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\underset{Lewis acid}{Me3B} + \underset{Lewis base}{:\!NH_\phantom{3}} -&amp;gt; \underset{Lewis adduct}{Me3\overset{-}{B}-\overset{+}{N}H3}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can see that the author says molecular entity (&lt;a href=&quot;http://goldbook.iupac.org/M03986.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity&lt;/a&gt; [1]) and the corresponding chemical species (a boarded concept, that in this case I interpret as the combination of different molecular entity), so I think that the exegesis of this definition is that both the acetic acid and the hydron can be called a Lewis acid. And in fact is what usually all the chemists use.&lt;/p&gt;&#xA;&lt;h3&gt;Reference&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;IUPAC. &lt;em&gt;Compendium of Chemical Terminology&lt;/em&gt;, 2nd ed. (the &amp;quot;Gold Book&amp;quot;). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (&lt;strong&gt;1997&lt;/strong&gt;). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. &lt;a href=&quot;https://doi.org/10.1351/goldbook&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1351/goldbook&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="41328" LastEditDate="2020-09-07T08:27:37.507" LastActivityDate="2020-09-07T08:27:37.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10587" PostTypeId="2" ParentId="10585" CreationDate="2014-05-10T10:02:18.463" Score="2" Body="&lt;p&gt;Corrosiveness of an acid is a more complex concept, of course the concentration of the acid and the strength of the acid are two main factors that determinate the success of the corrosion reaction. For example this is the Pourbaix diagram of iron both concentration and $K_a$ determine the pH of the solution so if this is higher then about 9 the metal is passivate and you can't have any corrosion. &lt;img src=&quot;https://i.stack.imgur.com/SCkal.png&quot; alt=&quot;Pourbaix diagram of Iron&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I add to these another important factor:  &lt;strong&gt;temperature&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some case the corrosiveness is due to a more complex interaction that is directly linked to the material you are dealing. One example is &lt;em&gt;acqua regia&lt;/em&gt; in this case the strong corrosiveness is due to the couple effect of two different acid that are able to de-passivate the metal allowing  an enhanced corrosive effect on noble metals. However acqua regia is not the best choice for an organic material in this case &lt;em&gt;Piranha Solution&lt;/em&gt; has a greater corrosive effect.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-05-10T10:13:08.920" LastActivityDate="2014-05-10T10:13:08.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10588" PostTypeId="2" ParentId="10576" CreationDate="2014-05-10T10:39:35.453" Score="1" Body="&lt;p&gt;Maybe could be some iron oxide or hydroxide from previous storage or ( if you have a iron tap) from it ore maybe directly from the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some fuel has red dye in it but I don't think this is the case, because I don't see how it can precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe it could be a residue of &lt;a href=&quot;https://en.wikipedia.org/wiki/Petroleum_resin&quot; rel=&quot;nofollow&quot;&gt;petroleum resin&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-05-10T10:39:35.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10589" PostTypeId="2" ParentId="2" CreationDate="2014-05-10T12:56:19.523" Score="-3" Body="&lt;p&gt;How does $\ce{NaCl}$ maintain its crystalline structure? It is a good question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because in crystalline structure of $\ce{NaCl}$  6 electrons of $\ce{Na+}$ ion are surrounded by 6 electrons of $\ce{Cl-}$ ion and 6 electrons of $\ce{Cl-}$ ion are surrounded by 6 electrons of $\ce{Na+}$. That's how $\ce{NaCl}$ maintains its crystalline structure. &lt;/p&gt;&#xA;" OwnerUserId="5446" LastEditorUserId="3846" LastEditDate="2014-05-10T14:32:12.830" LastActivityDate="2014-05-10T14:32:12.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10591" PostTypeId="2" ParentId="10582" CreationDate="2014-05-10T13:37:41.920" Score="1" Body="&lt;p&gt;1) Nucleophiles often have excess electron density while electrophiles are characterized by a lack thereof. The electrons flow in the direction of lower electron density (and not the other way around), because they are &quot;attracted&quot; by the positive partial or formal charge of the electrophile. This is the reason why the first scheme from your book is wrong, because it shows a nucleophile attacking another nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2), 3) and the reaction scheme you have drawn are correct.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-10T13:37:41.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10592" PostTypeId="2" ParentId="10580" CreationDate="2014-05-10T13:52:31.843" Score="1" Body="&lt;p&gt;A &quot;sextet&quot; atom has a lack of electrons and is eager to get 2 additional valence electrons to fill its octet. By a Lewis acid-base reaction (between ether and $\ce{BF3}$) or lone-pair donation from a neighboring atom (the mesomeric structures of the two carbocations), another bond with 2 electrons to the sextet atom is made and its octet is completed.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-10T13:52:31.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10593" PostTypeId="2" ParentId="10580" CreationDate="2014-05-10T13:58:51.537" Score="2" Body="&lt;p&gt;Where you find this picture? Very misleading. When sextet is mentioned, the better way is to say is lone pair acceptor or Lewis Acid. You simply need one empty molecular orbital on the atom to be able to do all the steps in the picture. In the case for common B, C, N and O they happen to be preferring sixtet. However, for heavier element, they are wrong. If you change the $BF_{3}$ to $FeCl_{3}$ in the picture, the reaction will be also true. But the iron center is not sextet, it is $3s^{2}3p^{6}3d^{5}$ in its outer electron layer.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-10T13:58:51.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10594" PostTypeId="2" ParentId="10582" CreationDate="2014-05-10T14:15:14.817" Score="1" Body="&lt;p&gt;If the electrons are to go somewhere, they should be attracted to that place. Consider the incorrect diagram you posted. The oxygen atom in water is partially negatively charged; the bonded electrons are drawn to the oxygen atom due to its high electronegativity. So it would not make much sense for the electrons in HCl to &quot;attack&quot; the oxygen atom, because the negative electrons would not be attracted the negative oxygen atom - in fact, they should be repelled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep in mind that protons are rather stationary, at least where chemical reactions are concerned. It is the electrons that move around and form bonds to whatever compound is willing. Since &quot;attack&quot; is a verb of action, it helps to think of something mobile to carry out that action. The protons do not attack; the electrons do. Even in &quot;electrophilic attacks&quot;, such as electrophilic addition, it is the electrons that move around and form bonds, even though the &quot;disturbance&quot; is provoked by something electrophilic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience in organic chemistry is much like your quotee's. Just find out where the electrons go, and draw the schemes.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastActivityDate="2014-05-10T14:15:14.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10595" PostTypeId="1" AcceptedAnswerId="10599" CreationDate="2014-05-10T14:58:47.543" Score="2" ViewCount="2196" Body="&lt;p&gt;I just got night contact lenses, and you are supposed to use a &quot;one-step&quot; cleaning solution to clean it over a period of six hours. At the bottom of the lenses case, there is a &quot;titanium plate neutralizer&quot;, which causes the liquid to bubble. What is this gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: it was platinum, not titanium&lt;/p&gt;&#xA;" OwnerUserId="5457" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:06:28.863" LastActivityDate="2015-10-30T17:06:28.863" Title="What is formed when hydrogen peroxide reacts with titanium?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10596" PostTypeId="1" CreationDate="2014-05-10T16:05:14.197" Score="0" ViewCount="872" Body="&lt;p&gt;For e.g .Due to rusting , iron becomes weak .So it should become weak only if the amount of iron is reduced. Even if we say the reduced iron is now the rust ,the amount of rust practically is very less comparing to iron.&lt;/p&gt;&#xA;" OwnerUserId="5321" LastActivityDate="2014-05-10T20:45:12.750" Title="Does corrosion follow the law of conservation of mass?" Tags="&lt;metal&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10597" PostTypeId="2" ParentId="5974" CreationDate="2014-05-10T16:06:36.460" Score="4" Body="&lt;p&gt;Your question is not very clear.&#xA;Did you just heat the salt water? Or you have put electrode to electrolysis water? What material is your electrode? Copper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me assume you are using copper electrode to electrolyze brine. Then you will have the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $$\ce{Cu - 2e -&amp;gt; Cu^2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anode: $$\ce{2H2O + 2e -&amp;gt; H2 + 2OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over all: &#xA;$$\ce{Cu + 2H2O -&amp;gt; Cu(OH)2 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when copper is your electrode, you will have no $\ce{Cl2}$ gas for sure. You will only breath in hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use graphite as the electrode, you will get overall reaction:&#xA;$$\ce{NaCl + 2H2O -&amp;gt; NaOH + H2 + Cl2}$$&#xA;It is possible to have the $\ce{Cl2}$ gas. However, if the amount of $\ce{Cl2}$ is not really big, it will be absorbed by water especially when it is basic.&#xA;$$\ce{Cl2 + 2OH- -&amp;gt; ClO- + Cl- + H2O}$$&#xA;And over all:&#xA;$$\ce{NaCl + H2O -&amp;gt; NaClO + H2}$$&#xA;No gas emitted. Electrolysis of water is actually one way to make bleach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is: it is &lt;strong&gt;safe&lt;/strong&gt; to inhale unless you are using extremely big battery.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="2093" LastEditDate="2014-05-10T16:12:54.320" LastActivityDate="2014-05-10T16:12:54.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10598" PostTypeId="2" ParentId="10596" CreationDate="2014-05-10T16:12:15.483" Score="1" Body="&lt;p&gt;The answer to the question in the title is yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron oxide flakes away from the iron so just looking at the iron would make you think that the mass has disappeared; it's actually on the surface below, unless someone has moved it.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-10T16:12:15.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10599" PostTypeId="2" ParentId="10595" CreationDate="2014-05-10T16:26:31.690" Score="1" Body="&lt;p&gt;Hydrogen peroxide will damage your eye tissue; that's why there are enzymes in our cells actually to catalyze the conversion of $HOOH$ to water and oxygen gas. The enzymes are actually contained in &lt;em&gt;peroxisomes&lt;/em&gt; (name is self-explanatory).* The neutralizer serves the same function of catalyzing the otherwise relatively slow decomposition of $HOOH$ into water and oxygen gas; hydrogen peroxide will decompose in the presence of light and that's why bottles of hydrogen perioxide are opaque - generally black or brown (but not transparent like bottles of &lt;strike&gt;which&lt;/strike&gt; &lt;em&gt;witch&lt;/em&gt; hazel or isopropyl alcohol). I have seen platinum used as a catalyst in contact cleaning solutions but I suppose titanium can be used as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit: peroxisomes actually create $HOOH$ upon catabolism of cell by-products but can also break down its own toxic byproduct. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4934" LastEditDate="2014-05-10T16:44:39.990" LastActivityDate="2014-05-10T16:44:39.990" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10600" PostTypeId="1" AcceptedAnswerId="10612" CreationDate="2014-05-10T16:35:42.010" Score="1" ViewCount="25" Body="&lt;p&gt;What type of probe or probe coating should one use to measure the temperature of a roughly 100c aqueous solution of cupric chloride?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stainless steel probe seems incorrect. Is a Teflon coated probe sufficient or is something else more appropriate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard way of protecting metal probes in corrosive liquids?&lt;/p&gt;&#xA;" OwnerUserId="5458" LastActivityDate="2014-05-10T22:10:40.053" Title="What is the correct temperature probe for a wet etch solution?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10601" PostTypeId="2" ParentId="6319" CreationDate="2014-05-10T16:55:36.037" Score="1" Body="&lt;p&gt;I am not sure if you are referring to carbanions or carbenes (I suspect the latter), because both contain a carbon with a lone pair, so I will give explanations on both in the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbanions are destabilized by the exact same effect that stabilizes carbocations and radicals: hyperconjugation. Donation of electron density from an adjacent $\sigma$ bond into the empty/half-filled $p_z$ orbital stabilizes the cation/radical, but not a carbanion center with its negative charge in the fully-filled $sp^3$ orbital lobe. Therefore, a primary carbanion is more stable than a secondary one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse is true for carbenes: Secondary carbenes are way more stable than primary ones. This is because carbenes are per se a very reactive species, and because of this property they are often used in organic synthesis as intermediates which are generated in situ. To obtain a reasonably stable carbene, the reactivity of the lone pair needs to be somewhat reduced. Two substituents at the carbene carbon are needed which stabilize the carbene either by steric hindrance (i.e. making the lone pair difficult to reach for other reactants) or mesomeric effects (for example, lone pair donors which alleviate electron lack of the &quot;sextet&quot; carbon, delocalization of the carbene lone pair into an aromatic system). The Arduengo carbene (named after its discoverer) is a good example for this. The stability of this compound is such that it can be isolated as a crystalline solid at room temperature. The carbene lone pair can be delocalized into the five-membered ring which then contains 6 $\pi$ electrons (2 double bonds + lone pair of uncharged nitrogen) and is aromatic. Bulky $\mathrm{R}^1$ groups like adamantyl further contribute to stability by shielding the lone pair from potential reactants (&lt;a href=&quot;http://en.wikipedia.org/wiki/Persistent_carbene&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XFf5O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arduengo carbenes are not the only persistent (especially stable) carbenes, and the understanding of their stability is still a subject of ongoing research, because structural elements (bulky substituents, aromaticity) which had been regarded as necessary for stability were often disproved as such by the synthesis of carbenes which did not contain them. The compounds do not need to be cyclic, as shown by &lt;a href=&quot;http://dx.doi.org/10.1002/anie.199611211&quot; rel=&quot;nofollow noreferrer&quot;&gt;bis(diisopropylamino) carbene&lt;/a&gt;, and even heteroatoms next to the carbene center do not need to be present, like in the case of &lt;a href=&quot;http://dx.doi.org/10.1126/science.1126675&quot; rel=&quot;nofollow noreferrer&quot;&gt;bis(diisopropylamino)cyclopropenylidene&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-10T16:55:36.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10602" PostTypeId="1" AcceptedAnswerId="10828" CreationDate="2014-05-10T17:11:16.080" Score="5" ViewCount="2211" Body="&lt;p&gt;Why is $\ce{Fe(OH_2)_6^{3+}}$ fairly acidic? This iron has six water molecules coordinated to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, water itself is a very weak acid. But when water is coordinated to iron, it becomes more acidic. I can think of a few reasons; can you think of more? Also are my reasons valid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Induction. The negative charge generated through loss of the hydrogen proton is stabilized by the coordinated water molecules. Plus the center of the square planar molecule is greatly positively charged (and therefore can handle the negative charge, despite the fact that iron itself is actually electropositive rather than electronegative - at least in iron's &lt;em&gt;ground state&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Size of the ferric ion. The ferric ion is rather big (it is, after all, able to coordinate six water molecules, as opposed to only four for some other metal ions). So whatever positive charge it has after being ionized once is spread out over a large surface area and that makes entire compound relatively stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Oxidation state of the central ferric ion - positive 3. This indicates a high degree of ionic character in the coordination bonds. Withdrawal of electron density from the water molecules makes the hydrogens even more positively charged and thus even more electrophilic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Positive charge density of the central ferric atom. Coulomb's law is in (significant) effect here. Positive and positive. Like charges repel. Kicking off a hydrogen proton - no problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2m3oY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-11-13T19:30:31.417" LastActivityDate="2015-11-13T19:30:31.417" Title="The Acidity of the Hydrated Iron (III) Ion" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10606" PostTypeId="1" CreationDate="2014-05-10T18:49:40.760" Score="4" ViewCount="3366" Body="&lt;p&gt;In internet resources, only $p_z$ orbitals are shown to overlap with $s$ orbitals to form sigma bond. But my teacher says that according to Valence Bond Theory, $s$ orbital overlaps with all three $p_x$, $p_y$ and $p_z$ orbitals to form sigma bonds. I am really confused about this. I heard that the concept changes when it comes to molecular orbital theory. Only similar orbitals (like $1s$-$1s$,$2p_x$-$2p_x$,$3p_y$-$3p_y$) form bonds but ($2s$-$2p_z$) do not.&#xA;Please clarify if there is any difference in concepts of overlapping between 'Molecular orbital theory' and 'Valence Bond Theory'.&lt;/p&gt;&#xA;" OwnerUserId="5459" LastEditorUserId="5016" LastEditDate="2014-05-10T23:36:32.517" LastActivityDate="2019-04-10T22:07:57.310" Title="Can a s orbital overlap with any p orbital to form a sigma bond?" Tags="&lt;valence-bond-theory&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10608" PostTypeId="2" ParentId="10606" CreationDate="2014-05-10T19:35:10.717" Score="7" Body="&lt;p&gt;$\ce{Px}$, $\ce{Py}$, and $\ce{Pz}$ orbitals can all overlap with an $\ce{S}$ orbital to form a bond.  It simply depends upon along which axis (X, Y or Z) the bond will be directed. These concepts of overlap do not change between Molecular Orbital theory and Valence Bond theory.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-10T19:44:21.277" LastActivityDate="2014-05-10T19:44:21.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10610" PostTypeId="2" ParentId="10606" CreationDate="2014-05-10T20:36:58.177" Score="0" Body="&lt;p&gt;Of course there is only one concept of bonding which never differ in either cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that here you have a doubt about the bonding between &lt;strong&gt;s&lt;/strong&gt; and &lt;strong&gt;p&lt;/strong&gt; orbital. First of all you should know that &lt;strong&gt;s&lt;/strong&gt; orbital can form bond between with &lt;strong&gt;p&lt;/strong&gt; orbital on any axis &lt;strong&gt;(x,y or z axis)&lt;/strong&gt; and all forms bond in &lt;strong&gt;sigma bond&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="23561" LastEditDate="2019-04-10T22:07:57.310" LastActivityDate="2019-04-10T22:07:57.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10611" PostTypeId="2" ParentId="10596" CreationDate="2014-05-10T20:45:12.750" Score="1" Body="&lt;p&gt;&lt;strong&gt;In chemistry we assume the total mass of reactants to be conserved in after reaction&lt;/strong&gt; (except in the nuclear reactions, of course). It is a fundamental law and hold true everywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in your question the Oxide layer formed above the iron makes the Ferrous Oxide which generally get removed, hence you think the mass had reduced.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2014-05-10T20:45:12.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10612" PostTypeId="2" ParentId="10600" CreationDate="2014-05-10T22:10:40.053" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polytetrafluoroethylene&quot; rel=&quot;nofollow&quot;&gt;Teflon&lt;/a&gt; is resistant against corrosive chemicals and stable up to temperatures of about 200 °C, so it is suitable for use in hot aqueous salt solutions. Using a coating or casing made of plastic which is unreactive and resistant against heat should be a good way to protect temperature probes.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-10T22:10:40.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10613" PostTypeId="1" CreationDate="2014-05-10T23:52:01.557" Score="2" ViewCount="870" Body="&lt;p&gt;&lt;a href=&quot;http://youtu.be/1f92vGOridg&quot; rel=&quot;nofollow&quot;&gt;http://youtu.be/1f92vGOridg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clicking on the link above will take you to a video with a diagram of a voltammogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through my own experiments I have managed to obtain a similar diagram. However, I am unclear on how to determine the overpotential from such a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the overpotential is simply the potential required over and above the amount required to get the reaction going thermodynamically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could, please be explicit about what points I should choose exactly to make my calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you ! &lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2016-04-24T07:16:16.260" Title="Cyclic Voltammogram - How to determine Overpotential" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10614" PostTypeId="1" AcceptedAnswerId="10615" CreationDate="2014-05-11T02:17:36.040" Score="0" ViewCount="2164" Body="&lt;p&gt;How does boric acid behave as an acid? Is it a Lewis acid or a Brønsted-Lowry acid? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:24:04.300" LastActivityDate="2018-03-03T11:24:04.300" Title="Mechanism of Acidity - Boric Acid" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ClosedDate="2014-05-11T20:10:16.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="10615" PostTypeId="2" ParentId="10614" CreationDate="2014-05-11T02:51:34.583" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boric_acid&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says boric acid is a Lewis acid. It later states that it is a tribasic Brønsted acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Brønsted acid is a compound that is able to provide $\ce{H^+}$ ions. A Lewis acid is a compound that can accept an electron pair; it would need to be electron poor. Some molecules have both characteristics and either feature could be used in explaining the reactivity of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tribasic means that there are 3 hydrogens that can be used. I'm believe that the author used the term basic acid to mean that the hydrogen is attached as a hydroxyl group instead of hydrogen bonded. Phosphorous acid is always written as $\ce{H3PO3}$, although it could be written as $\ce{P(OH)3}$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2016-03-13T18:24:56.707" LastActivityDate="2016-03-13T18:24:56.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10616" PostTypeId="1" CreationDate="2014-05-11T03:15:23.820" Score="3" ViewCount="1091" Body="&lt;p&gt;Stumbled across this interesting practice problem in my organic textbook. I have a few reasons for why $\ce{H4N+}$ is less acidic than $\ce{H3O+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Let’s assess conjugate stability first. Conjugates are $\ce{H3N}$ and $\ce{H2O}$. Let’s look at central atom oxidation state. Nitrogen in ammonia has an oxidation state of negative 3. Oxygen in water has an oxidation state of negative 2. The more negative oxidation state of nitrogen suggests it is more nucleophilic than oxygen (i.e. more reactive). The more stable the conjugate base, the stronger the acid. Water appears to be more stable through oxidation state analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Oxygen is more electronegative than nitrogen, so there is more electron withdrawal from the hydrogens in the hydronium ion. This gives the hydrogens in hydronium ion a higher partial positive charge, making them more electrophilic (and hence more reactive). This is the reason (the sole reason) my book offered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Size plays a role in the direct acidities of the two molecules. First, nitrogen is bigger than oxygen. Both hydronium and ammonium ion have a positive 1 formal charge. The bigger size of nitrogen decreases its positive charge density, making it less Coulombically favorable for ammonium ion to kick off a proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these reasons 1) valid and 2) are there any more reasons I may have overlooked? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T19:27:10.647" LastActivityDate="2015-05-11T19:27:10.647" Title="Rationalizing Acidities of Ammonium Ion and Hydronium Ion" Tags="&lt;acid-base&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10617" PostTypeId="1" AcceptedAnswerId="10646" CreationDate="2014-05-11T06:35:08.807" Score="0" ViewCount="932" Body="&lt;p&gt;I came across this definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When discussing nucleophilicity we’re specifically talking about donating a pair of electrons to an atom other than hydrogen (usually carbon).  When a species is donating a pair of electrons to a hydrogen (more specifically, a proton, H+) we call it a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make this drawing of mine wrong? I labeled the base here a nucleophile even though the nucleophile is lending n electron pair to the hydrogen. Are all bases nucleophiles?  Or should I stick with the definition above? (Forgive the forgotten arrow from the bond between oxygen and hydrogen which should be pointing at oxygen's valence environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AiMav.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5591" LastEditDate="2015-11-13T20:12:13.847" LastActivityDate="2015-11-13T20:12:13.847" Title="Definition of nucleophile" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10618" PostTypeId="2" ParentId="10617" CreationDate="2014-05-11T06:54:36.603" Score="0" Body="&lt;p&gt;It does not exclude a base of being a nucleophile. It is just that a specific label is provided for a pair of electrons being donated to the hydrogen ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are all bases nucleophiles? There are 2 definitions for a base, a lone pair doner and/or a hydrogen ion accepter. For many compounds, a base is also a nucleophile. I hesitate to say all bases are nucleophiles, but it may be true.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-11T06:54:36.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10619" PostTypeId="1" CreationDate="2014-05-11T10:23:03.030" Score="2" ViewCount="75" Body="&lt;p&gt;So I was thinking about corrosion and I came across this structure(see image)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think the inner metal/alloy that is corroding is steel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what do you think the outer layer (in white) is? perhaps the metal is just coated with paint to avoid corrosion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, what type of corrosion do you think its undergoing? General Corrosion, Galvanic corrosion, uniform corrosion and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OCLAg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5461" LastActivityDate="2014-05-11T14:26:23.550" Title="What type of material is this and how is it corroding?" Tags="&lt;metal&gt;&lt;materials&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10620" PostTypeId="1" CreationDate="2014-05-11T13:29:48.220" Score="5" ViewCount="1986" Body="&lt;p&gt;The original question was:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I find the number of structural and configurational isomers of a given compound?&#xA;  is there any formula?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a specific example, how would I find the number of structural and configurational isomers of a bromo compound $\ce{C5H9Br}$ formed by addition of $\ce{HBr}$ to 2-pentyne&lt;/p&gt;&#xA;" OwnerUserId="5345" LastEditorUserId="73" LastEditDate="2014-05-12T06:33:28.727" LastActivityDate="2018-02-22T14:53:06.073" Title="Is there a mathematical generalization for the number of isomers (structural and configurational) of a given compound?" Tags="&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10621" PostTypeId="1" AcceptedAnswerId="10626" CreationDate="2014-05-11T14:07:06.910" Score="6" ViewCount="3927" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is $\ce{Co(SCN)_{4}^{2-}}$ blue?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I simply can't predict the colors of such inorganic salts. I have learned that $\ce {Co^{2+}}$ salts are pink in color. Also, if the ligand is a strong field ligand ($\ce {SCN-}$ in this case), then the splitting is more and the color should shift to the red side...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my concept wrong?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="41556" LastEditDate="2018-05-27T19:30:39.780" LastActivityDate="2018-05-27T19:31:31.217" Title="Why is tetrathiocyanatocobaltate(II) blue?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10623" PostTypeId="2" ParentId="10619" CreationDate="2014-05-11T14:26:23.550" Score="2" Body="&lt;p&gt;Though a picture is not really enough information, approximate location at least and any simple tests on the materials shown would help a lot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its structural and ugly so its steel.&#xA;Its mostly likely painted steel. I'd say the paint flaked off so it has general uniform corrosion. Why it looks black and not the regular red-brown is probably combination of oxides and sulphides formed as the corrosion products. I don't see a hint that its galvanised steel.&#xA;Steel in exterior uses  rarely suffers any corrosion other than  general uniform corrosion.    &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-05-11T14:26:23.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10624" PostTypeId="1" CreationDate="2014-05-11T14:28:17.100" Score="3" ViewCount="2150" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do $\ce{NaHCO3}$ and $\ce{KHCO3}$ have different crystal structures?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We had this in a test today and this got me stumped...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do $\ce{NaHCO3}$ and $\ce{KHCO3}$ possess crystal structures?&lt;/li&gt;&#xA;&lt;li&gt;If they do, why should they have different crystal structures, because they are in the same group and must possess almost similar physical properties.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4945" LastEditDate="2014-06-11T07:28:25.190" LastActivityDate="2017-03-08T17:50:53.717" Title="Why do NaHCO₃ and KHCO₃ have different crystal structures?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10625" PostTypeId="1" AcceptedAnswerId="10627" CreationDate="2014-05-11T14:43:06.990" Score="2" ViewCount="252" Body="&lt;p&gt;Please consider 2-acetyl-1-pyrroline:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/2-Acetyl-1-pyrroline&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/2-Acetyl-1-pyrroline&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mlQaI.png&quot; width=&quot;86&quot; alt=&quot;1-(3,4-dihydro-2H-pyrrol-5-yl)ethan-1-one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the molecular geometry of the nitrogen lone pair?  Is the lone pair oriented in-plane with the ring?  How is this distinct from the orientation carbon-acetyl bond, which should presumably point out-of-plane with the ring?&lt;/p&gt;&#xA;" OwnerUserId="5463" LastEditorDisplayName="user7951" LastEditDate="2016-12-26T19:34:11.007" LastActivityDate="2016-12-26T19:34:11.007" Title="Nitrogen lone pair electrons on 2-acetyl-1-pyrroline" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10626" PostTypeId="2" ParentId="10621" CreationDate="2014-05-11T14:45:40.947" Score="13" Body="&lt;p&gt;Let us consider this &lt;em&gt;qualitatively&lt;/em&gt; with crystal field theory (CFT). Co is a transition metal with $d^7$ configuration. CFT says that depending on the geometry of the metal complex, the d-orbital energies will split into different energy levels. If the ligands (in this case, the four $\ce{SCN-}$ ligands) align with certain orbitals, then the electrons in these orbitals will feel an increased repulsion due to the closeness of the ligands, and increase in energy. The orbitals in-between the aligned axes, will not feel this repulsion, and will not increase in energy. The splitting diagram might look something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4eGH.jpg&quot; alt=&quot;Splitting diagram for a tetrahedral complex&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an equation that relates the (splitting)&#xA; energy to wavelength:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;\Delta E = \frac{hc}{\lambda} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, we see that if the splitting energy increases, wavelength will have to decrease, i.e. be shorter than before. This means a shift towards the violet region. As you say, $\ce{SCN-}$ is a strong ligand, meaning it leads to a rather high $\Delta E$. So we could say that $\ce{Co(SCN)_4^{2-}}$ is blue due to the high energy gap between the d-orbitals. Electrons are excited from one of the lower orbitals up to one of the higher orbitals. Upon de-exciting, light is emitted with an energy equal to $\frac{hc}{\lambda}$. This happens to be in the blue/violet region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weaker ligands would split the energy levels less, and the complex's color would therefore have a color closer to the red part of the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did this make it a little clearer?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="41556" LastEditDate="2018-05-27T19:31:31.217" LastActivityDate="2018-05-27T19:31:31.217" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10627" PostTypeId="2" ParentId="10625" CreationDate="2014-05-11T15:04:28.587" Score="3" Body="&lt;p&gt;The nitrogen lone pair is in plane with the ring because it is in one of the three $sp^2$ hybrid orbitals of N. The other two hybrid orbitals form $\sigma$ bonds to the adjacent carbons, and all three are in plane, with an angle of 120° between them. The $p_z$ orbital of N which is perpendicular to the ring plain forms the $\pi$ bond of the C=N double bond. The acetyl substituent should also lie in plane with the ring because this would allow energetically favorable conjugation between the two double bonds (C=N and C=O).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-11T15:04:28.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10629" PostTypeId="2" ParentId="10624" CreationDate="2014-05-11T15:48:30.637" Score="5" Body="&lt;p&gt;The reason why the crystal structure of $\ce{NaHCO3}$ is face centered and the crystal structure of $\ce{KHCO3}$ is body centered is due to the fact that the cation/anion ratio is different, meaning that certain types of hole are now too small for the cation to fit into them and as a result it must go into a different hole.  &lt;/p&gt;&#xA;" OwnerUserId="5464" LastEditorDisplayName="user2117" LastEditDate="2014-05-11T20:09:45.490" LastActivityDate="2014-05-11T20:09:45.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10630" PostTypeId="1" AcceptedAnswerId="10645" CreationDate="2014-05-11T16:11:11.260" Score="9" ViewCount="1349" Body="&lt;p&gt;I was trying to normalize the wave function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi (x) = \begin{cases} 0  &amp;amp; x&amp;lt;-b \\ A &amp;amp; -b \leq x \leq 3b \\ 0 &amp;amp; x&amp;gt;3b&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is done simply by evaluating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\limits_{-b}^{3b} | \psi (x) |^2 dx = 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = \pm \frac{\sqrt{b}}{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the normalized wave function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\psi (x) = \begin{cases} 0  &amp;amp; x&amp;lt;-b \\ \pm \frac{\sqrt{b}}{2} &amp;amp; -b \leq x \leq 3b \\ 0 &amp;amp; x&amp;gt;3b&#xA;\end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was quite straight forward... too straight forward for my liking. My question is twofold: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is my derivation above correct? And &#xA;2) How shall I deal with the ``$\pm$'' sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any &quot;physical significance&quot;, and can be dropped?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="41556" LastEditDate="2018-02-24T15:03:30.103" LastActivityDate="2018-02-24T15:04:15.397" Title="Normalization of the wavefunction φ(x) = A" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10631" PostTypeId="2" ParentId="7713" CreationDate="2014-05-11T17:25:18.593" Score="6" Body="&lt;p&gt;While we cannot be sure that $\ce{Fe(NH3)5}$ does not exist, the bonding between ligand and metal offers a good explanation for why it is expected to be less stable than $\ce{Fe(CO)5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In metal carbonyls like $\ce{Fe(CO)5}$, the bonding between CO and the metal consists of three single bonds, 1 $\sigma$ and 2 $\pi$ bonds. The $\sigma$ bond is formed by overlap of the carbon $sp$ hybrid orbital (which contains the lone pair) with an empty $d$ orbital of the metal. Electron density is donated from the orbital of CO into the metal orbital, so CO acts as a $\sigma$ donor. The two other $\pi$ bonds are formed by overlap of a filled metal $d$ orbital with a pair of $\pi^*$ molecular orbitals of CO. Electron density is donated from the metal orbital into the molecular orbitals of CO, so CO acts as a $\pi$ acceptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FLx1o.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\pi$ bonding to occur, the metal is required to have sufficient $d$ electrons, and thus preferably a lower oxidation state. This explains the relatively low oxidation state of iron (0) in $\ce{Fe(CO)5}$. The combination of one $\sigma$ and 2 $\pi$ bonds gives the M-CO bond a partial triple-bond character, while the C-O triple bond is weakened by the transfer of electron density into the $\pi^*$ orbitals (&lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_carbonyl&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/enBlB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of its ability to serve as a $\sigma$ donor and $\pi$ acceptor simultaneously, CO is a stronger ligand than $\ce{NH3}$ which can only act as a $\sigma$ donor with its lone pair in the nitrogen $sp^3$ hybrid orbital. $\ce{NH3}$ lacks a $\pi$ system and therefore cannot act as a $\pi$ acceptor. Because $\pi$ backbonding is crucial for stabilization of a metal center in low oxidation state, ammonia is unlikely to form stable ammin complexes which are analogous to the corresponding metal carbonyls.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-11T17:25:18.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10632" PostTypeId="1" AcceptedAnswerId="10648" CreationDate="2014-05-11T17:48:24.270" Score="0" ViewCount="1001" Body="&lt;p&gt;I have recently started reading about colour in organic molecules and come across conjugation of pi bonds. My question is pretty short... In transition metal ions I understand colour is caused by molecules jumping between the d sub orbitals as a result of ligand splitting them. The concept of excitation easily applies here. So what I want to know is. How do electrons get excited in a conjugated molecule?&lt;/p&gt;&#xA;" OwnerUserId="4833" LastActivityDate="2014-05-12T00:39:24.123" Title="How are electrons in conjugated molecules excited?" Tags="&lt;organic-chemistry&gt;&lt;electronic-configuration&gt;&lt;electrons&gt;&lt;color&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2014-05-12T12:08:11.650" ContentLicense="CC BY-SA 3.0" />
  <row Id="10633" PostTypeId="1" AcceptedAnswerId="10679" CreationDate="2014-05-11T18:17:42.657" Score="0" ViewCount="691" Body="&lt;p&gt;I have heard of of a special system which specially deals with naming of organic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we get names such as 2-methylpropan-1-ol.&#xA;I have tried googling it and found few links but their explanation was way above my head. I also asked my teacher about the same question and he said this is not required in your standard, but I really want to know about naming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Few Links that I found on Google were&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Another is wikihow very difficult, I need a simple explanation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="4945" LastEditDate="2014-08-17T07:21:46.330" LastActivityDate="2014-08-17T07:21:46.330" Title="How are Organic Compounds named?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-05-18T07:53:59.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="10634" PostTypeId="1" AcceptedAnswerId="10635" CreationDate="2014-05-11T18:53:54.187" Score="2" ViewCount="5890" Body="&lt;p&gt;I was a reading a book that a senior gave me, it said that redox is a type of reaction in which both reduction and oxidation take place. I know about oxidation but not reduction. After googling reduction reaction I found it's simply just the opposite. But then how can a reduction and oxidation take place together? It would be better if you put forth some examples.&lt;/p&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="7448" LastEditDate="2017-04-08T17:16:41.570" LastActivityDate="2017-04-08T17:16:41.570" Title="What is Redox reaction? Is it really possible?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ClosedDate="2017-04-08T20:50:52.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="10635" PostTypeId="2" ParentId="10634" CreationDate="2014-05-11T19:40:48.920" Score="5" Body="&lt;p&gt;I think you mean Redox, which is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Portmanteau&quot; rel=&quot;nofollow&quot;&gt;portmanteu&lt;/a&gt; of Reduction-Oxidation. An easy example of redox is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haber_process&quot; rel=&quot;nofollow&quot;&gt;Haber process&lt;/a&gt; (nitrogen and hydrogen form ammonia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2 + 3 H2 &amp;lt;=&amp;gt; 2 NH3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Haber process the nitrogen is reduced (the oxidation state changes from $0$ to $-3$), and the hydrogen is oxidized (the oxidation state changes from $0$ to $+1$). In that redox reaction, &lt;strong&gt;both oxidation and reduction take place&lt;/strong&gt;, as you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no oxidation without reduction and vice-versa. The electrons lost in oxidation have to go somewhere! If they wouldn't, then there's no reason for the reaction to occur in the first place.&lt;/p&gt;&#xA;" OwnerUserId="5390" LastEditorUserId="4945" LastEditDate="2014-05-12T08:31:19.937" LastActivityDate="2014-05-12T08:31:19.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10636" PostTypeId="2" ParentId="5748" CreationDate="2014-05-11T19:42:15.693" Score="7" Body="&lt;p&gt;We know that melting or freezing is an equilibrium process. The energy that is required to melt an ice cube will not contribute in elevating its temperature until all the solid water is molten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take two ice cubes and add salt to one of them, then put each of them at room temperature, both of the ice cubes will absorb energy from the surroundings, and this energy as we said will contribute in breaking down the bonds between water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube that salt has not been added to, has a melting point $0~\mathrm{^\circ C}$ and so if we measure its temperature during melting it will remain zero until all ice is molten. That ice cube to which we have added salt, the salt that is added lowers the melting and freezing points of water because it lowers the vapor pressure of water. This ice cube will absorb energy from the environment to help break bonds between water molecules. We know that the salt added will dissolve in the melted portion of the ice. This formed solution of salt will have a lowered freezing point, so the equilibrium between the solid phase and the aqueous phase will be shifted towards the liquid phase since such a solution will freeze at say $-2~\mathrm{^\circ C}$. Since both phases are close together, the ice will absorb energy from the salt solution and will reduce its temperature to the $-2~\mathrm{^\circ C}$ to maintain the equilibrium. When all ice is molten we end up with a salt solution that has got a temperature of say $-1.5~\mathrm{^\circ C}$. This is due to the solution being diluted now. After that, it will start absorbing heat from the room and reach zero and above. So, in conclusion that is how salt melts ice.&lt;/p&gt;&#xA;" OwnerUserId="5468" LastEditorUserId="-1" LastEditDate="2018-02-22T15:46:30.463" LastActivityDate="2018-02-22T15:46:30.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10637" PostTypeId="1" AcceptedAnswerId="10639" CreationDate="2014-05-11T19:51:11.360" Score="2" ViewCount="6185" Body="&lt;p&gt;I can't seem to find the answer in google, which is surprising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that ionic compounds dissolve into electrolytes. However, when non-ionic sacarose for example dissolves in water, does the sacarose crystals break into single atoms as well or does it stay as a whole molecule in water? What does exactly happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone =)&lt;/p&gt;&#xA;" OwnerUserId="5390" LastEditorUserId="5390" LastEditDate="2014-05-11T23:59:25.883" LastActivityDate="2014-05-11T23:59:25.883" Title="How do non-ionic compounds dissolve into water? Into molecules or atoms?" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10638" PostTypeId="2" ParentId="10614" CreationDate="2014-05-11T19:56:37.353" Score="2" Body="&lt;p&gt;Boric acid is a Lewis acid, therefore when in water it accepts the lone pair from oxygen and becomes $\ce{(OH)3B^{-}-OH2^{+}}$.&#xA;From this compound a proton can be donated and this is how boric acid behaves as a Brønsted acid.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="24052" LastEditDate="2016-03-13T14:54:53.517" LastActivityDate="2016-03-13T14:54:53.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10639" PostTypeId="2" ParentId="10637" CreationDate="2014-05-11T20:01:07.000" Score="3" Body="&lt;p&gt;The molecules remain as they are, as discrete particles, they do not break up into single atoms. What can happen to certain compounds when they are dissolved in water is a hydrolysis reaction. But even then, only certain bonds of the molecule are cleaved in a reaction with water, and the molecule does not spontaneously fall apart into the elements it consisted of, because there is no thermodynamic reason to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is also true for ionic compounds. Ammonium chloride ($\ce{NH4Cl}$), for example, consists of discrete $\ce{NH4+}$ and $\ce{Cl-}$ ions. $\ce{NH4+}$ is a molecular ion, and it remains as such in aqueous solution and does not break up into single nitrogen and hydrogen atoms/ions. However, the salt can hydrolyze in aqueous solution according to the following equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4Cl + H2O \rightleftharpoons NH3 + H3O^+ + Cl^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason why aqueous solutions of ammonium chloride are weakly acidic.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-11T20:22:01.383" LastActivityDate="2014-05-11T20:22:01.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10641" PostTypeId="1" AcceptedAnswerId="11216" CreationDate="2014-05-11T20:12:23.633" Score="12" ViewCount="605" Body="&lt;p&gt;What is the structure of &lt;span class=&quot;math-container&quot;&gt;$\ce{FeSO4 \cdot NO}$&lt;/span&gt; that is formed when &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; is passed through ferrous sulfate solution? If it is a brown ring complex then why does the complex sphere break upon heating to yield &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;? In general, when do the complex spheres break?&lt;/p&gt;&#xA;" OwnerUserId="5470" LastEditorUserId="5470" LastEditDate="2019-08-21T20:04:11.807" LastActivityDate="2019-08-21T20:04:11.807" Title="What is the Structure of FeSO₄ • NO?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10643" PostTypeId="2" ParentId="10630" CreationDate="2014-05-11T20:52:44.813" Score="5" Body="&lt;p&gt;Strictly speaking, both signs are valid solutions to the wave equation, so you probably should keep the $\pm$ in front of the wave function. If this is the only wave function you're working with, then you can use either sign and it won't matter: both signs will give the same probability density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the sign &lt;em&gt;would&lt;/em&gt; matter is if there is you need to have this wave function interact with some other wave function. In that case, the two wave functions will interfere, and the &lt;em&gt;relative&lt;/em&gt; phase between the two wave functions will determine how constructively or destructively they will interfere. The sign you apply to this wave function will partially determine that relative phase. In such a case, there will be some other information that will determine which sign is physical.&lt;/p&gt;&#xA;" OwnerUserId="323" LastActivityDate="2014-05-11T20:52:44.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10644" PostTypeId="2" ParentId="10632" CreationDate="2014-05-11T21:28:05.880" Score="1" Body="&lt;p&gt;To be able to show some color, the photons with wave length within visible light region have to be absorbed as the energy to make electron jump from lower orbital to higher orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of conjugated organic compound, electrons jumps between $\pi$ and $\pi^{*}$ molecular orbitals happen to be within the visible light region. The key is the energy difference, the conjugation system has to be big enough or the energy gap between $\pi$ and $\pi^{*}$ molecular orbitals is too big and the wave length will be too short to be visible.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-11T21:28:05.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10645" PostTypeId="2" ParentId="10630" CreationDate="2014-05-11T21:42:03.623" Score="6" Body="&lt;p&gt;Your derivation is mostly correct, but you made a small mistake when evaluating the integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$.&#xA;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &amp;amp;= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\&#xA;&amp;amp;\Rightarrow  A = \pm \frac{1}{2 \sqrt{b}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a &lt;strong&gt;constant&lt;/strong&gt; phase factor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\psi'=\mathrm e^{-\mathrm i\alpha}\psi,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation.&#xA;So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="41556" LastEditDate="2018-02-24T15:04:15.397" LastActivityDate="2018-02-24T15:04:15.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10646" PostTypeId="2" ParentId="10617" CreationDate="2014-05-11T22:00:24.213" Score="1" Body="&lt;p&gt;All nucleophiles are Lewis bases. They are just electron pair donor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A molecule or ion is called a &quot;base&quot; (or, “Brønsted base”) when it interact with proton, it is called a &quot;nucleophile&quot; when it interact with other atom centers, which can be carbon, nitrogen, sulfur and so on. So the row &quot;base&quot; and &quot;nucleophile&quot; may overlap. For example, $\ce{OH-}$ is a base when it take proton off an acid, and it can also be a nucleophile when it hydrolyzes ester (nucleophilic attack to carbonyl group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The broad nucleophilicity definition is how good the nucleophile interact the positive or partial positive atom center. When the atom center is proton, nucleophilicity is then basicity. Because proton center is hard and usually less steric hindred, a bulky and hard (for hardness $O$ &gt; $N$ &gt; $S$ center) electron donor will be likely to prefer proton over the softer and more hindred non-proton center. These electron donors are called &lt;strong&gt;Non-nucleophilic bases&lt;/strong&gt;. Potassium tert-Butoxide is a good example.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-11T22:00:24.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10647" PostTypeId="2" ParentId="8412" CreationDate="2014-05-11T23:29:42.193" Score="5" Body="&lt;p&gt;Hyperconjugation is the interaction between adjacent orbital (empty &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; for cation, &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; for alkene) with &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond, while common conjugation is between adjacent orbital and &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond.&lt;/p&gt;&#xA;&lt;p&gt;As one example, one reason why more alkyl substituted alkenes are more stable than less substituted ones is because more alkyl groups provide more hyperconjugation stabilization of the alkene. Why more alkyl substituted cations are more stable is also because the hyperconjugation stabilization from more adjacent alkyl group.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jVuYf.gif&quot; alt=&quot;Example for hyperconjugation&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="61578" LastEditDate="2021-03-06T23:14:45.410" LastActivityDate="2021-03-06T23:14:45.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10648" PostTypeId="2" ParentId="10632" CreationDate="2014-05-12T00:39:24.123" Score="3" Body="&lt;p&gt;This first picture shows what happens to the molecular orbitals of ethylene if we bring two ethylene molecules (the right and left sides of the molecular orbital vs. energy part of the diagram) close together and form 1,3 butadiene (center of the figure)&lt;img src=&quot;https://i.stack.imgur.com/BlqHF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next figure shows the result of adding a third ethylenic group to the molecule and form 1,3,5-hexatriene. Note how in each case the molecular orbitals combine (split) and generate a new set of molecular orbitals; and in each case the separation between the Highest Occupied Molecular Orbital (HOMO) and the Lowest Unoccupied Molecular Orbital (LUMO) decreases.  When light is absorbed an electron is promoted from the HOMO to the LUMO.  If we make the HOMO-LUMO separation small enough by extending conjugation, we can shift absorbtion into the visible region (longer wavelength, smaller HOMO-LUMO energy separation). We can add other groups too, like cyano, nitro, phenyl, carbonyl, etc. to extend the conjugation in our system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ovxjz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-12T00:39:24.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10650" PostTypeId="1" AcceptedAnswerId="10651" CreationDate="2014-05-12T01:28:13.703" Score="9" ViewCount="72627" Body="&lt;p&gt;This question is based on the Brønsted-Lowry model. I saw many answers to similar questions online. However none of them have given a detailed answer that is based on molecular structure (the answers all told the OP to rely on a reference table for a way to identify strong and weak acids).&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="73" LastEditDate="2014-05-12T06:10:34.070" LastActivityDate="2017-01-04T05:11:12.913" Title="Why do strong acids have weak conjugate bases?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10651" PostTypeId="2" ParentId="10650" CreationDate="2014-05-12T02:15:54.923" Score="12" Body="&lt;p&gt;All chemical processes are reversible at least to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping this in mind, a strong acid must have a weak conjugate &lt;em&gt;base&lt;/em&gt; (I think you mean strong acids have weak conjugate bases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Let's try to prove by contradiction. If a strong acid &lt;em&gt;had&lt;/em&gt; a &lt;em&gt;strong conjugate base&lt;/em&gt;, then the base would quickly re-associate itself with a hydrogen proton. This &lt;em&gt;reverses&lt;/em&gt; what the acid just did - dissociate itself from a hydrogen proton! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate this concept, consider hydrochloric acid - the oft-cited strong acid. $\ce{HCl}$ is a strong acid because in water solution it dissociates (nearly) completely into chloride ion and hydrogen protons (which are then solvated by water). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H_2O -&amp;gt;H_3O^+ + Cl^-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reaction of hydrochloric acid with water. Note the one-way arrow; the reaction is a one way street. $\ce{HCl}$ is the acid; $\ce{Cl^-}$ is the conjugate base; water is the &lt;em&gt;base&lt;/em&gt; and hydronium ion is the &lt;em&gt;conjugate acid&lt;/em&gt;. In other words, this is the generalized reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(strong)~acid + base -&amp;gt; conjugate ~acid + conjugate ~base}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bite-size pieces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(strong)~acid -&amp;gt; conjugate ~base}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{base -&amp;gt; conjugate ~acid }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the above two &quot;half-reactions&quot; together and you get a complete Brønsted acid/base reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So back to the dissociation of $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this reaction (the reverse reaction) happened to a significant extent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H_2O &amp;lt;- H_3O^+ + Cl^-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then that would be &lt;em&gt;antagonistic&lt;/em&gt; to the dissociation of $\ce{HCl}$! $\ce{HCl}$ would no longer be nearly completely deprotonated in water solution since the reverse reaction is significant! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we conclude that if one has a strong acid, then its conjugate base must be weak. The converse is also true; if an acid's conjugate base is weak, then the acid must be strong. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-01-04T05:11:12.913" LastActivityDate="2017-01-04T05:11:12.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10652" PostTypeId="2" ParentId="6033" CreationDate="2014-05-12T02:28:54.057" Score="0" Body="&lt;p&gt;Note that substances with mobile pi-electron systems can also conduct electricity. For example, graphite is a good conductor. Diamond, although made up of the exact same element (carbon), is not a conductor. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-12T02:28:54.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10653" PostTypeId="1" AcceptedAnswerId="10655" CreationDate="2014-05-12T06:08:54.520" Score="37" ViewCount="14873" Body="&lt;p&gt;In my exam, I was asked why cyclopropane could decolourise bromine water (indicating that it reacted with the bromine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I could guess was that it is related to the high angle strain in cyclopropane, as the C–C–C bond angle is $60^\circ$ instead of the required $109.5^\circ$. No book I have read mentions this reaction. What is the product formed, and why does it occur?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="16683" LastEditDate="2017-08-27T13:00:54.050" LastActivityDate="2017-08-27T13:15:43.083" Title="Why does cyclopropane react with bromine?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10655" PostTypeId="2" ParentId="10653" CreationDate="2014-05-12T07:20:41.900" Score="40" Body="&lt;p&gt;The following ring opening reaction will occour:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MpuWD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MpuWD.png&quot; alt=&quot;Reaction of cyclopropane with bromine to form 1,3-dibromopropane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ -  bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4SshJ.gif&quot; alt=&quot;orbital models&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very general approach can be derived from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Walsh_diagram&quot; rel=&quot;noreferrer&quot;&gt;Walsh diagram&lt;/a&gt;. &lt;a href=&quot;https://doi.org/10.1002/qua.560290513&quot; rel=&quot;noreferrer&quot;&gt;Schwarz &lt;em&gt;et.al.&lt;/em&gt;&lt;/a&gt; {&lt;a href=&quot;http://www.academia.edu/847307/Some_new_aspects_of_bonding_in_cyclopropane&quot; rel=&quot;noreferrer&quot;&gt;@academia.edu&lt;/a&gt;} and &lt;a href=&quot;https://doi.org/10.1021/ja01008a016&quot; rel=&quot;noreferrer&quot;&gt;Hoffmann&lt;/a&gt; {&lt;a href=&quot;http://www.roaldhoffmann.com/sites/all/files/30s_1.pdf&quot; rel=&quot;noreferrer&quot;&gt;@roaldhoffmann.com&lt;/a&gt;} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a &lt;a href=&quot;https://i.stack.imgur.com/9PyPo.jpg&quot; rel=&quot;noreferrer&quot;&gt;chart of all occupied molecular orbitals formed from valence orbitals and the LUMO&lt;/a&gt;. Here is a preview. Each orbital is viewed from three different angles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9PyPol.jpg&quot; alt=&quot;chart preview&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found &lt;a href=&quot;http://www.bluffton.edu/~bergerd/chem/walsh/cp.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy - as mentioned by &lt;a href=&quot;https://chemistry.stackexchange.com/a/10666/4945&quot;&gt;ron&lt;/a&gt; - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJV7am.jpg&quot; alt=&quot;total density&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/eladd/symbr2tt.html&quot; rel=&quot;noreferrer&quot;&gt;chemguide.co.uk&lt;/a&gt;. The essential part is the attack of the bromine at the HOMO(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yahBB.gif&quot; alt=&quot;attack of bromine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring opening reaction can be reversed by &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyclopropane#Synthesis&quot; rel=&quot;noreferrer&quot;&gt;adding sodium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when there are bromine radicals present (UV light) then substitution will occur:&#xA;\begin{aligned}\ce{&#xA;Br2 &amp;amp;-&amp;gt;[\ce{h\nu}] 2Br.\\&#xA;&amp;amp;+(CH2)3 -&amp;gt; (CH2)2(CHBr) + HBr&#xA;}\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-08-27T13:15:43.083" LastActivityDate="2017-08-27T13:15:43.083" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10656" PostTypeId="1" AcceptedAnswerId="10661" CreationDate="2014-05-12T08:06:58.690" Score="0" ViewCount="624" Body="&lt;p&gt;I surfed through the internet and visited a number of chemistry websites but only in vain. I couldn't find an answer.&lt;br&gt;&#xA;Please help! Your quick response will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="865" LastEditDate="2015-10-30T17:01:39.593" LastActivityDate="2015-10-30T17:01:39.593" Title="The unit of energy, 1 dm³ atm, is equal to how many calories?" Tags="&lt;everyday-chemistry&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10657" PostTypeId="5" CreationDate="2014-05-12T08:09:09.390" Score="0" Body="&lt;p&gt;A molecular orbital is a mathematical function that describes the wave-like behaviour of electrons in a molecule. It is used mainly used to calculate chemical and physical properties by ab initio and density functional theory methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/M03996.html&quot; rel=&quot;nofollow&quot;&gt;Molecular Orbital&lt;/a&gt; (from the IUPAC Goldbook):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A one-electron wavefunction describing an electron moving in the&#xA;  effective field provided by the nuclei and all other electrons of a&#xA;  molecular entity of more than one atom. Such molecular orbitals can be&#xA;  transformed in prescribed ways into component functions to give&#xA;  'localized molecular orbitals'. Molecular orbitals can also be&#xA;  described, in terms of the number of nuclei (or 'centres')&#xA;  encompassed, as two-centre, multi-centre, etc. molecular orbitals, and&#xA;  are often expressed as a linear combination of atomic orbitals. An&#xA;  orbital is usually depicted by sketching contours on which the&#xA;  wavefunction has a constant value (contour map) or by indicating&#xA;  schematically the envelope of the region of space in which there is an&#xA;  arbitrarily fixed high (say 96%) probability of finding the electron&#xA;  occupying the orbital, giving also the algebraic sign (+ or −) of the&#xA;  wavefunction in each part of that region.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-12T08:31:12.040" LastActivityDate="2014-05-12T08:31:12.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10658" PostTypeId="4" CreationDate="2014-05-12T08:09:09.390" Score="0" Body="A molecular orbital is a mathematical function that describes the wave-like behaviour of electrons in a molecule. It is used mainly used to calculate chemical and physical properties by ab initio and density functional theory methods." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-12T08:31:07.643" LastActivityDate="2014-05-12T08:31:07.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10659" PostTypeId="2" ParentId="10620" CreationDate="2014-05-12T08:25:08.877" Score="5" Body="&lt;p&gt;As I said in my comment, the mathematical way to solve this kind of problem relies on Graph Theory. This article (&lt;a href=&quot;http://dx.doi.org/10.1021/ci00047a033&quot; rel=&quot;nofollow&quot;&gt;Applications of Graph Theory in Chemistry&lt;/a&gt;, &lt;a href=&quot;http://www.tamug.edu/mars/no-show/19-JCICS-1985.pdf&quot; rel=&quot;nofollow&quot;&gt;preprint&lt;/a&gt;) reviews the problem and further more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you google 'isomer' and 'graph theory' you will find anything you want about this exciting topic. But for generic question, there is no generic answer. It is about building all possible graphs or find out inherent properties inside them. You cannot answer to this question unless it is well-defined and properly bounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have such question, please rewords it and post it on Mathematical Exchange. This should give you fine grained answer. If you just wanted to understand how it works, welcome to the fabulous world of Graph Theory, you will have to learn a bit to answer it by yourself.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-05-12T11:10:29.323" LastActivityDate="2014-05-12T11:10:29.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10660" PostTypeId="1" AcceptedAnswerId="10662" CreationDate="2014-05-12T08:49:50.017" Score="4" ViewCount="6491" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An organic compound contains 71.7% carbon, 6.7% hydrogen, 10.4% nitrogen and 11.8% oxygen by mass. Given that its vapour density is 67.5, determine its empirical formula and molecular formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the mass percentages, I have found out that the empirical formula is $\ce{C8H9NO}$. However, to find the molecular formula we usually require the molecular mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does &quot;vapour density&quot; mean, and how can I use it to calculate the molecular mass of the compound?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="16683" LastEditDate="2018-08-21T13:06:15.903" LastActivityDate="2018-08-21T13:06:15.903" Title="How can molecular mass be calculated from vapour density?" Tags="&lt;stoichiometry&gt;&lt;terminology&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10661" PostTypeId="2" ParentId="10656" CreationDate="2014-05-12T08:50:54.820" Score="5" Body="&lt;p&gt;$1\ \mathrm{dm^3} = 1\times 10^{-3}\ \mathrm{m}^3 $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ \mathrm{atm = 1013.25\ mbar = 101.325 kPa = 101325\ {kg}\cdot{}m^{-1}\cdot s^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ \mathrm{atm \cdot dm^3} = 1\times 10^{-3}\cdot 101325\ \mathrm{m^3 \cdot kg \cdot{}m^{-1}\cdot s^{-2}} = 101.325\ \mathrm{kg \cdot{}m^{2}\cdot s^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 1\ \mathrm{atm \cdot dm^3 = 101.325\ J = 101.325\cdot 0.2390\ cal= 24.217\ cal}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-05-12T08:50:54.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10662" PostTypeId="2" ParentId="10660" CreationDate="2014-05-12T09:49:02.440" Score="8" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Vapour_density&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the vapour density of a molecule is &quot;the density of a vapour in relation to that of hydrogen&quot;. The density of a gas, $\rho$, is proportional to its molecular mass, $M$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \frac{m}{V} \propto \frac{m}{n} = M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass of the gas, $V$ is the volume, and $n$ the amount of gas. Therefore, if we denote your compound by $\ce{X}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;67.5 &amp;amp;= \frac{\rho(\ce{X})}{\rho(\ce{H2})} \\&#xA;&amp;amp;= \frac{M(\ce{X})}{M(\ce{H2)}} \\&#xA;M(\ce{X}) &amp;amp;= 67.5 \cdot M(\ce{H2}) \\&#xA;&amp;amp;= 67.5 \cdot \pu{2 g mol-1} \\&#xA;&amp;amp;= \pu{135 g mol-1}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you can use this piece of information to find that the molecular formula is $\ce{C8H9NO}$ as well.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="16683" LastEditDate="2018-08-21T13:06:02.870" LastActivityDate="2018-08-21T13:06:02.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10663" PostTypeId="2" ParentId="10656" CreationDate="2014-05-12T09:55:19.823" Score="2" Body="&lt;p&gt;General Gas Constant, $\mathcal{R} = 0.0821\:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}}$&lt;br&gt;&#xA;General Gas Constant, $\mathcal{R} = 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;   \mathcal{R}               &amp;amp;: \mathcal{R}\\&#xA;   0.0821 \:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}}&#xA;                             &amp;amp;: 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}}  \\&#xA;   0.0821\:\mathrm{dm^3 atm} &amp;amp;: 1.989 \:\mathrm{cal}  \\&#xA;   1\:\mathrm{dm^3 atm}      &amp;amp;: (1.989\:\mathrm{cal}\cdot 1\:\mathrm{dm^3 atm})/ 0.0821\:\mathrm{dm^3 atm}$  \\&#xA;   1\:\mathrm{dm^3 atm}      &amp;amp;: 24.22\:\mathrm{cal}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="4945" LastEditDate="2014-06-11T07:39:09.647" LastActivityDate="2014-06-11T07:39:09.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10666" PostTypeId="2" ParentId="10653" CreationDate="2014-05-12T13:19:00.283" Score="23" Body="&lt;p&gt;The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°.  From this, and using Coulson's theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 + \lambda^2 \cos(114^\circ) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Now, using the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(which says that summing the &quot;s&quot; character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the C–C bond in cyclopropane is very high in p-character.  It is this high p-content that allows cyclopropane to behave in a similar fashion to an olefin in terms of stabilizing adjacent charge, absorbing bromine, etc. By the way, an x-ray study of a cyclopropane derivative&lt;sup&gt;1&lt;/sup&gt; shows significant electron density only exterior to the cyclopropane ring. There is &lt;strong&gt;no&lt;/strong&gt; significant electron density interior to the ring, consistent with this analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1): Hartman, A.; Hirshfeld, F. L. Structure of &lt;em&gt;cis&lt;/em&gt;-1,2,3-tricyanocyclopropane. &lt;em&gt;Acta Crystallogr.&lt;/em&gt; &lt;strong&gt;1966,&lt;/strong&gt; &lt;em&gt;20,&lt;/em&gt; 80–82. &lt;a href=&quot;https://doi.org/10.1107/S0365110X66000148&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1107/S0365110X66000148&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-08-27T13:07:10.867" LastActivityDate="2017-08-27T13:07:10.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10667" PostTypeId="1" AcceptedAnswerId="10671" CreationDate="2014-05-12T13:19:45.577" Score="2" ViewCount="208" Body="&lt;p&gt;I'm wondering if &lt;em&gt;any&lt;/em&gt; reaction pathway exists (not involving enzymes) to do the following in aqueous solvent at temperatures &amp;lt; ~100°C: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XA8UH.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, it's important however that the reaction is concerted, and that the W - X and Y - Z bonds are not broken until both the W-Y and X-Z bonds are formed.  Bonds here must be covalent, but otherwise can be of any type.  Any kind of catalyst can be used.  Of course, it would be nice to know of an efficient reaction with minimally exotic conditions or reagents (i.e. it would be nice to avoid heavy metals for oxidative-addition type reactions), but I won't push my luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does something like this exist?&lt;/p&gt;&#xA;" OwnerUserId="5475" LastEditorUserId="7448" LastEditDate="2017-04-08T17:20:36.633" LastActivityDate="2017-04-08T17:20:36.633" Title="Does anything like this covalent concerted bond exchange reaction exist?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2017-04-08T19:14:26.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="10668" PostTypeId="1" AcceptedAnswerId="10700" CreationDate="2014-05-12T13:29:04.630" Score="6" ViewCount="7881" Body="&lt;p&gt;A volumetric flask does not look like an Erlenmeyer flask, but has more of a pear shape. Is it just to distinguish more easily between them while they are on the shelf?&lt;/p&gt;&#xA;" OwnerUserId="5476" LastEditorUserId="4945" LastEditDate="2014-07-30T02:56:52.533" LastActivityDate="2019-07-10T13:30:42.707" Title="Why is a volumetric flask shaped the way it is?" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10669" PostTypeId="2" ParentId="10668" CreationDate="2014-05-12T13:48:07.977" Score="7" Body="&lt;p&gt;When the glassware was made by hand, the bulb at the bottom varied in size, so each flask had to be filled with the correct volume before the line could be etched on the glass.  A tube was used for the neck since the volume filling the tube would change at a consistent rate. So you have a bulb holding most of the liquid, and a tubular neck to indicate when you have the correct volume. The flask looks like a pear shaped container.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-12T13:48:07.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10670" PostTypeId="1" CreationDate="2014-05-12T14:05:25.450" Score="0" ViewCount="5100" Body="&lt;p&gt;I couldn't find a clear answer in my book and on the internet.&lt;br&gt;&#xA;Please help! Your quick response will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="5119" LastEditDate="2014-05-12T14:11:20.977" LastActivityDate="2014-05-12T14:42:44.383" Title="What is Inversion Temperature in terms of Joule Thomson Effect? Does Critical Temperature depend upon Inversion Temperature?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10671" PostTypeId="2" ParentId="10667" CreationDate="2014-05-12T14:14:52.873" Score="4" Body="&lt;p&gt;$$\ce{W=X + Y=Z -&amp;gt;[\mathrm{cat}] W=Y + X=Z}$$ is known as olefin metathesis, achieved using Grubbs/Hoyveda/Fürstner catalysts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water-soluble Ru catalysts have been prepared in the Grubbs group (&lt;a href=&quot;http://dx.doi.org/10.1021/ja058451c&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-05-12T15:10:28.233" LastActivityDate="2014-05-12T15:10:28.233" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10672" PostTypeId="2" ParentId="10667" CreationDate="2014-05-12T14:27:57.547" Score="3" Body="&lt;p&gt;And the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wittig_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wittig reaction&lt;/a&gt; works in a quite similar fashion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/89/Wittig_Reaktion.svg/500px-Wittig_Reaktion.svg.png&quot; alt=&quot;Wikipedia Wittig_reaction_scheme&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Phosphorous Oxygen Ylide Bond may has mesomeric forms: $$\ce{-{}^{+}P-C^{-}-&amp;lt;-&amp;gt;-P=C-{}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can again learn a lot for free at &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1979/wittig-facts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nobelprize.org&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.687" LastActivityDate="2016-07-14T23:31:43.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10673" PostTypeId="2" ParentId="10670" CreationDate="2014-05-12T14:35:08.410" Score="4" Body="&lt;p&gt;This is half-part answer to your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joule Thompson Inversion Temperature is the temperature when Joule Thomson coefficient $\mu_\mathrm{JT}$ is null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_\mathrm{JT} =  \left(\frac{\partial T}{\partial p}\right)_h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that the gas does not chill ($\mu_\mathrm{JT}&amp;gt;0$) or warm ($\mu_\mathrm{JT}&amp;lt;0$) when it experiences expansion. Pressurized Hydrogen has a negative Joule-Thomson coefficient, which must be taken in account if you plan to stock it and expand it, because it will significantly heat when expended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This factor is influenced by gas molecular interactions (repulsive and attractive). Therefore, for an Ideal Gas implies that $\mu_\mathrm{JT} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find detailed explanation in Atkins, Physical Chemistry, Chapter 2: First Law - The Joule-Thomson Effect.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-05-12T14:42:44.383" LastActivityDate="2014-05-12T14:42:44.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10674" PostTypeId="2" ParentId="10552" CreationDate="2014-05-12T14:47:43.300" Score="4" Body="&lt;p&gt;Or you can try the very simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Lucas%27_reagent&quot; rel=&quot;nofollow&quot;&gt;Lucas' test&lt;/a&gt; using zinc chloride and concentrated hydrochloric acid:&#xA;$$\ce{ROH + HCl -&amp;gt; RCl + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works best with tertiary alkohols, slow with secondary alcohols and not at all with primary alcohols.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-12T14:47:43.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10675" PostTypeId="1" CreationDate="2014-05-12T14:58:18.567" Score="1" ViewCount="988" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A sequence of reactions is shown below starting from an optically active compound P.&#xA;  $$\Large{\ce{\underset{\ce{C4H8O3}}{P}-&amp;gt;[SOCl2]\underset{\ce{C4H6Cl2O}}{Q}-&amp;gt;[C2H5NH2]\underset{\ce{C6H12ClNO}}{R}}}$$&#xA;  P does not react with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone. Q reacts very rapidly with one molar equivalent of ethylamine at low temperature to form R. Select the most appropriate statement(s).&#xA;  (A) Q as well as R are optically active.&#xA;  (B) Q on heating with an excess of ethylamine in the presence of a base gives a&#xA;  basic compound.&#xA;  (C) R is a basic compound and can react with an acid to form a salt.&#xA;  (D) An optically inactive isomer of P on heating can form compound S $(\ce{C4H6O2})$ which does not react with thionyl chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P is hydroxy butanoic acid. In Q the hydroxyl groups gets replaced by chlorine. I have confusion about the structure of R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel Option (A) to be correct. I am doubtful about (D). And have no idea about (B) and (C).&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="5026" LastEditDate="2018-02-24T02:47:37.540" LastActivityDate="2018-02-24T02:47:37.540" Title="What is true about this reaction involving hydroxy carboxlic acid?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10676" PostTypeId="2" ParentId="10675" CreationDate="2014-05-12T15:56:50.027" Score="1" Body="&lt;p&gt;I think your guess for P is correct, we just don't know whether it is 2- or 3-hydroxy-butanoic acid.  Reaction of thionyl chloride with the hydroxyl group in the alkyl chain can proceed with retention or inversion of configuration depending on reaction conditions, but the point is, if P is optically active, then Q will be optically active too (so A seems correct).  Reaction of Q, an acid chloride, with ethyl amine will produce the corresponding ethyl amide, R.  Since amides are basic (they can form salts with acids), I think B and C are true too.  An optically inactive isomer of P would be 4-hydroxy-butanoic acid.  On heating it would dehydrate and form a stable lactone (S) that would not react with thionyl chloride, so I think D is true also.  I just wonder if we're missing something that would allow us to determine if P is the 2- or 3- isomer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:  To be complete and accurate let's update.  Based on the comments below (particularly JA's) I was right about &quot;B&quot; (but for the wrong reason, I thought the amide would be basic enough, but it is not; B is true because, as JA points out, heating with excess amine will convert P into an amino amide (I've labeled it as X) which is basic).  I was wrong about C, R is not basic enough to form a salt.  I still don't know if P is 2- or 3-hydroxy-butanoic acid, but I've added a picture based on the 3-hydroxy isomer to try and summarize the chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4X2Wp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-12T20:19:04.123" LastActivityDate="2014-05-12T20:19:04.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10677" PostTypeId="1" CreationDate="2014-05-12T16:05:48.297" Score="5" ViewCount="7148" Body="&lt;p&gt;I want to do Wurtz reaction of 1-bromo-3-chlorocyclobutane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6yo1gm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6yo1gm.png&quot; alt=&quot;1-bromo-3-chlorocyclobutane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; suggests to do an intramolecular reaction. Instead, I remember reading somewhere that bromine is more reactive and hence we should do an intermolecular reaction, and attach two carbons to where $\ce{Br}$ was attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="5026" LastEditDate="2018-07-07T03:21:32.980" LastActivityDate="2019-12-02T04:18:35.943" Title="What product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10678" PostTypeId="1" AcceptedAnswerId="10702" CreationDate="2014-05-12T16:18:27.497" Score="4" ViewCount="954" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A compound &lt;span class=&quot;math-container&quot;&gt;$\ce X$&lt;/span&gt; of formula &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5NO2}$&lt;/span&gt; on treatment with &lt;span class=&quot;math-container&quot;&gt;$\ce {HNO2}$&lt;/span&gt; &#xA;   &lt;span class=&quot;math-container&quot;&gt;$(\ce{NaNO2}+\ce{HCl})$&lt;/span&gt; gives off an inactive gas &lt;span class=&quot;math-container&quot;&gt;$\ce Y$&lt;/span&gt; and forms &lt;span class=&quot;math-container&quot;&gt;$\ce Z$&lt;/span&gt;. When heated with soda lime, &lt;span class=&quot;math-container&quot;&gt;$\ce X$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$\ce{A(CH5N)}$&lt;/span&gt; soluble in acid. Find the true statements : &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce A$&lt;/span&gt; answers carbylamine test.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce X$&lt;/span&gt; is an amino acid.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce Y$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce Z$&lt;/span&gt; is a hydroxyl acid.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt :&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce X$&lt;/span&gt; is glycine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce A$&lt;/span&gt; is methyl amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce Y$&lt;/span&gt; is dinitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce Z$&lt;/span&gt; is 2-hydroxy ethanoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, correct options should be : &lt;span class=&quot;math-container&quot;&gt;$1,2,4$&lt;/span&gt;. But the answer given is &lt;span class=&quot;math-container&quot;&gt;$2,4$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me to understand if and where I went wrong?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="41686" LastEditDate="2020-03-18T15:06:45.097" LastActivityDate="2020-03-18T16:17:38.727" Title="Doesn't methyl amine give carbylamine test?" Tags="&lt;organic-chemistry&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10679" PostTypeId="2" ParentId="10633" CreationDate="2014-05-12T16:34:25.277" Score="2" Body="&lt;p&gt;I myself while searching and talking to my dad found the answer.Sounds weird, sorry for posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THE NAMES OF ORGANIC COMPOUNDS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Cracking the code&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A modern organic name is simply a code. Each part of the name gives you some useful information about the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the &quot;prop&quot; tells you that there aren't any carbon-carbon double bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting the carbon atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;code  no of carbons&#xA;meth    1&#xA;eth 2&#xA;prop    3&#xA;but 4&#xA;pent    5&#xA;hex 6&#xA;Types of carbon-carbon bonds&#xA;Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.&#xA;code    means&#xA;an  only carbon-carbon single bonds&#xA;en  contains a carbon-carbon double bond&#xA;For example, butane means four carbons in a chain with no double bond.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propene means three carbons in a chain with a double bond between two of the carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkyl groups&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds like methane, $\ce{CH4}$, and ethane, $\ce{CH3CH3}$, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A methyl group is $\ce{CH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ethyl group is $\ce{CH3CH2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These groups must, of course, always be attached to something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Types of compounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alkanes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1:  Write the structural formula for 2-methylpentane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any carbon-carbon double bonds? No - an tells you there aren't any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now draw this carbon skeleton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a methyl group on the number 2 carbon atom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had to name this yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any carbon-carbon double bonds? No - therefore pentane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particular end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.&#xA;If you want more detail goto: &lt;a href=&quot;http://www.chemguide.co.uk/basicorg/conventions/names.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/basicorg/conventions/names.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="275" LastEditDate="2014-05-14T10:22:38.220" LastActivityDate="2014-05-14T10:22:38.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10682" PostTypeId="1" AcceptedAnswerId="10687" CreationDate="2014-05-12T18:10:51.853" Score="1" ViewCount="175" Body="&lt;p&gt;I am studying this claim&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compensatory mechanism in respiratory alkalosis is decrease of SB i.e. bicarbonate ions ($\ce{HCO3^-}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are reactions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCO3- \rightarrow H+ + CO3^{2-}}$ (conjugate base)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCO3- + H+ \rightarrow H2CO3}$ (conjugate acid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but since the $pK_a$ for the second reaction is greater&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;decrease in $\ce{HCO3-}$ yields more acids &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so alleviates respiratory alkalosis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To claim such a thing, I think you must know that the $pK_a$ for the reaction is greater, so change in bicarbonate ions' content results in bigger change in the second reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does the decrease of bicarbonate ions lead to decreased respiratory alkalosis?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="292" LastEditorUserId="5016" LastEditDate="2014-05-12T19:22:04.260" LastActivityDate="2018-03-04T02:53:28.610" Title="Decrease of Bicarbonate to compensate Respiratory alkalosis" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10683" PostTypeId="1" CreationDate="2014-05-12T18:26:39.903" Score="8" ViewCount="19348" Body="&lt;p&gt;I'm having trouble (not alone, apparently) understanding the concept of Gibbs free energy, but I'm beginning to. What I still don't understand, however, is how Gibbs free energy of formation differs from heat- or enthalpy of formation. Is it more precise, somehow? given that entropy is accounted for in Gibbs free energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="5479" LastActivityDate="2014-05-12T21:49:29.577" Title="What is the difference between enthalpy of formation and Gibbs free energy of formation?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10684" PostTypeId="2" ParentId="10683" CreationDate="2014-05-12T18:52:30.990" Score="7" Body="&lt;p&gt;Gibbs free energy is not more &quot;precise.&quot; The two are &lt;em&gt;different&lt;/em&gt; but complementary. Enthalpy and Gibbs Free Energy indicate different things. Enthalpy can tell you about the relative stabilities of the products and reactants. Gibbs free energy however can tell you about whether a reaction is spontaneous (whether a reaction will occur) under a set of specified conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people will confuse exothermic - bond improving - reactions with spontaneous reactions. While it is true that a bond improving reaction is in itself quite &lt;em&gt;favorable&lt;/em&gt; - after all - all chemical processes tend toward greater stability and lower potential energy - we must remember that there is another dimension to the problem of reaching maximum stability - namely, &lt;em&gt;entropy&lt;/em&gt; (a measure of disorder). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that all physical processes tend toward &lt;em&gt;greater&lt;/em&gt; entropy (greater disorder) as well. Just as one's bedroom becomes more and more chaotic despite the pleas from Mom to clean one's room, physical processes tend toward chaos (disorder) as well. Gibbs free energy takes entropy into account with enthalpy and can tell us whether a process will actually go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NO_2(g) -&amp;gt; N_2O_4(g)}\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider only the forward reaction. The forward reaction is &lt;em&gt;spontaneous&lt;/em&gt; under standard state conditions and 298 K - i.e. we have a negative Gibbs free energy value. Don't believe me? Nitrogen dioxide is brown. Nitrogen tetraoxide however is colorless. Put some brown nitrogen dioxide into a flask and if you observe it, you'll notice that the brown color fades somewhat as the colorless nitrogen tetraoxide is made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enthalpy of reaction is definitely negative - the reaction is &lt;em&gt;exothermic&lt;/em&gt; because the reaction is a &lt;em&gt;dimerization reaction&lt;/em&gt; - we are forming a bond here (specifically, a N-N bond) by using two nitrogen dioxide molecules (which are actually radicals, and remember that radicals are highly unstable). Nitrogen tetraoxide however is not a radical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy of reaction however is negative as well - i.e. we have an &lt;em&gt;unfavorable&lt;/em&gt; change in entropy. Remember that $ΔS^o = S_{final} - S_{initial}$. In this case, it is clear that $S_{final} &amp;lt; S_{initial}$ because on the right hand side, we have one mole of gas. On the left hand (reactant) side, we have two moles of gas. There is definitely a lot more chaos (disorder) with 2 moles of gas molecules bumping around and whizzing around every single way imaginable. So we've gone from a disordered system on the left to a somewhat more ordered system on the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would describe this reaction as (at 298 K and std. state conditions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;em&gt;Spontaneous&lt;/em&gt;&#xA;2) &lt;em&gt;Exothermic&lt;/em&gt;&#xA;3) &lt;em&gt;Entropy lowering&lt;/em&gt;&#xA;4) &lt;em&gt;Exergonic&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-12T21:49:29.577" LastActivityDate="2014-05-12T21:49:29.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10685" PostTypeId="1" AcceptedAnswerId="10690" CreationDate="2014-05-12T18:54:06.770" Score="1" ViewCount="653" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The plot of solubility of some salts against temperature is given on the right. A salt like $\ce{NaCl}$ can be recovered fully, when dissolved in a solvent and the solvent is removed by evaporation. Choose the correct statement(s):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkdV9.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;No heat is absorbed or released when $\ce{NaCl}$ is dissolved in water.  &lt;/li&gt;&#xA;  &lt;li&gt;More heat is released when the ions of $\ce{Yb2(SO4)3}$ are solvated than that absorbed when the ions are separated.  &lt;/li&gt;&#xA;  &lt;li&gt;The heat of solution is negative for $\ce{H3BO3}$.  &lt;/li&gt;&#xA;  &lt;li&gt;When $\ce{AlCl3}$ is dissolved in water and the water is evaporated, solid $\ce{AlCl3}$ is recovered.  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt; (1) is clearly correct. (3) and (4) are clearly wrong. My confusion lies with (2).&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="7448" LastEditDate="2018-02-19T12:29:53.420" LastActivityDate="2018-02-19T12:29:53.420" Title="Solubility and heat release of ytterbium (III) sulfate in water" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10686" PostTypeId="1" CreationDate="2014-05-12T19:19:55.207" Score="2" ViewCount="243" Body="&lt;p&gt;And how do I find the diffusion coefficient from a Levich plot?&lt;/p&gt;&#xA;" OwnerUserId="5480" LastActivityDate="2014-08-11T15:07:08.297" Title="How does the diffusion coefficient of a species affect the limiting current for a RDE voltammagram?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10687" PostTypeId="2" ParentId="10682" CreationDate="2014-05-12T19:27:40.323" Score="1" Body="&lt;p&gt;Bicarbonate ion is amphiprotic, as you've recognized. It can react as both a Brønsted base and as a Brønsted acid in water solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = 4.8 * 10^{-11}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HOCO_2^- + H_2O &amp;lt;=&amp;gt; OH_3^+ + CO_3^{2-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_b = 2.1 * 10^{-4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HOCO_2^- + H_2O &amp;lt;=&amp;gt; HO^- + H_2CO_3}$ (not really true carbonic acid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through K value inspection, we can see that the bicarbonate ion preferentially reacts as a base in water solution. The magnitude of this difference is massive; $K_b$ is more than a million times greater than $K_a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by decreasing the concentration of bicarbonate ion, we can combat respiratory alkalosis. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:53:28.610" LastActivityDate="2018-03-04T02:53:28.610" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10688" PostTypeId="1" CreationDate="2014-05-12T19:31:08.847" Score="1" ViewCount="546" Body="&lt;p&gt;I know that the limiting current is proportional to the square root of the rotation speed, but what does the slope of the plot represent?&lt;/p&gt;&#xA;" OwnerUserId="5480" LastActivityDate="2014-05-13T12:11:55.410" Title="What does the slope of a Levich plot represent?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10689" PostTypeId="1" AcceptedAnswerId="10692" CreationDate="2014-05-12T19:47:06.967" Score="4" ViewCount="5678" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GUAe0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is NanoBalletDancer.&#xA;How many different carbon environments are there in NanoBalletDancer – i.e. how &#xA;many signals would be seen in total in the 13C NMR spectrum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I counted 8 signals from benzene ring carbons&#xA;4 from methyl group carbons&#xA;4 from triple bond carbons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the total amount of signals? Which did I miss?&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="848" LastEditDate="2014-05-12T23:50:35.073" LastActivityDate="2016-12-07T17:33:35.023" Title="How many signals would be seen in total in the 13C NMR spectrum" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10690" PostTypeId="2" ParentId="10685" CreationDate="2014-05-12T20:12:53.763" Score="1" Body="&lt;p&gt;When the solubility of a salt decreases with temperature, this indicates that the enthalpy of dissolution is exothermic, i.e. heat is released during this process. This is the case when the lattice energy, which is at least required to separate the ions by breaking up the crystal lattice, is smaller than the energy which is released when the ions are solvated. The solubility curve of ytterbium(III) sulfate therefore proves the statement (B).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-12T20:12:53.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10691" PostTypeId="2" ParentId="10689" CreationDate="2014-05-12T20:33:37.447" Score="6" Body="&lt;p&gt;See the given figure below, where each unique signal is given a specific color and a shape which is &#xA;unique to it. Carbons 6 and 8 produce the same signal, yet they are left purposely blank, since I ran out of colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting 2 and 15 makes it 22, however, since there isn't really much difference between the two apical methyl groups the answer should be 22.  I am guessing they took the apical 40,41 methyl groups to produce different signals thus making it 23.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvsZ4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5483" LastEditorUserId="73" LastEditDate="2014-05-13T05:20:34.197" LastActivityDate="2014-05-13T05:20:34.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10692" PostTypeId="2" ParentId="10689" CreationDate="2014-05-12T20:40:04.740" Score="8" Body="&lt;p&gt;3 different methyl signals: (1, 16) (40, 41) (27, 28, 29, 33, 34, 35)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 different methylene signals: (17) (2, 15) (3,14) (37, 39)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 aliphatic CH signal: (36)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 different aliphatic quaternary C signals: (26, 32) (38)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 different triple bond C signals: (24, 30) (25, 31) (4, 13) (5, 12)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 different aromatic carbons: (7) (10) (6, 8) (9, 11) (18) (21) (19, 23) (20, 22)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22 different signals &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looked at a model, methyls 40 and 41 are different; one is on the same side of the 6-membered ring as the benzene ring, the other is on the same side of the 6-membered ring as the hydrogen at C-36&lt;/p&gt;&#xA;&#xA;&lt;p&gt;23 different signals!&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-12T20:54:50.120" LastActivityDate="2014-05-12T20:54:50.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10694" PostTypeId="2" ParentId="10677" CreationDate="2014-05-12T21:06:28.547" Score="2" Body="&lt;p&gt;Br is a much better leaving group than Cl.First of all presuming that Cyclobutane is stabilized by substitutent groups is important since it wil gradually rearrange itself to buta diene.But since it is a theoretical question hence  we can say that it should add to the Br side since Br leaves more efficiently being larger in size than Cl and hence having less efficient orbital overlap with Carbon as well as a large electronic repulsion due to the large number of electrons it has is the reason behind this.And the major product will probably be a bicyclo [1.1.0] butane since it is the kinetic product.&#xA;so basically your answer-&#xA;&lt;img src=&quot;https://i.stack.imgur.com/olAT6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5483" LastActivityDate="2014-05-12T21:06:28.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10695" PostTypeId="1" AcceptedAnswerId="10696" CreationDate="2014-05-12T23:20:19.850" Score="3" ViewCount="2830" Body="&lt;p&gt;From ChemWiki by UC Davis:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a side note, it is important to note that $\ce{BF3}$ frequently bonds with a $\ce{F-}$ ion in order to form $\ce{BF4-}$ rather than staying as $\ce{BF3}$. This structure completes boron's octet and it is more common in nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wait, what? Boron has three electrons. Bonding with four fluoride ions gives it four electrons &amp;mdash; these are covalent bonds (shared electrons). That makes for a grand total of 7 valence electrons. Did the author mean $\ce{BF_5^2-}$? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2017-04-08T17:35:34.570" LastActivityDate="2017-04-08T17:35:34.570" Title="Why does boron trifluoride form boron tetrafluoride ion instead of boron pentafluoride ion?" Tags="&lt;bond&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10696" PostTypeId="2" ParentId="10695" CreationDate="2014-05-12T23:33:27.960" Score="6" Body="&lt;p&gt;As you point out, the $\ce{B}$ in $\ce{BF_3}$ is sharing a total of 6 electrons, 3 from the boron and 3 from the three fluorines.  Now add a fluoride anion that brings 2 more electrons to share.  This now makes for a total of 8 electrons around the boron.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-12T23:33:27.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10697" PostTypeId="2" ParentId="10115" CreationDate="2014-05-13T04:47:52.590" Score="11" Body="&lt;p&gt;Using Transition State Theory (&lt;a href=&quot;http://goldbook.iupac.org/T06470.html&quot; rel=&quot;noreferrer&quot;&gt;TST&lt;/a&gt;) one has to be aware of at least some of the many flaws of it. And may also often lead to wrong assumptions. However, it is a wonderful tool to describe basic kinetics in a theoretical manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important/ fundamental assumptions are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;separation of electron and nuclear movement (comparable to &lt;a href=&quot;http://goldbook.iupac.org/BT07008.html&quot; rel=&quot;noreferrer&quot;&gt;Born-Oppenheimer approximation&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;(all possible) transition states (TS) are in equilibrium with the reactants  $$\ce{A + B &amp;lt;=&amp;gt; [AB]^{\ddagger}}\tag1$$&lt;/li&gt;&#xA;&lt;li&gt;combined &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_statistics&quot; rel=&quot;noreferrer&quot;&gt;Boltzmann statistic&lt;/a&gt; for reactants and TS&lt;/li&gt;&#xA;&lt;li&gt;All molecules that reach the transition state from the reaction coordinate of the reactants will leave the TS to the reaction coordinate of the product(s) - and vice versa &#xA;\begin{align}&#xA;\ce{A + B &amp;lt;=&amp;gt; &amp;amp;[AB]^{\ddagger}-&amp;gt;P}\tag{2a}\\&#xA;\ce{P &amp;lt;=&amp;gt; &amp;amp;[AB]^{\ddagger}-&amp;gt; A + B}\tag{2b}&#xA;\end{align}&lt;/li&gt;&#xA;&lt;li&gt;at the TS the transition vector can be separated from all other motions and hence may be treated classical as &lt;strong&gt;translation&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words TST reduces a very complicated multidimensional problem to a one-dimensional translation. That implies that all treated molecules must be connected on a potential energy hypersurface with one vector. (Borrowing from the amazing &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/99/Rxn_coordinate_diagram_5.PNG/640px-Rxn_coordinate_diagram_5.PNG&quot; alt=&quot;reaction diagramm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence reactions that are described with TST are of first order by default&lt;/strong&gt;. One would have to make various adjustments to treat higher orders (and I would assume that not matching units would be of minor concern).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.707" LastActivityDate="2015-09-25T05:47:10.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10698" PostTypeId="5" CreationDate="2014-05-13T08:48:52.907" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/H02752.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat&lt;/a&gt; is a form to transfer energy from one body to another or from one body to its surroundings. It is closely connected to &lt;a href=&quot;/questions/tagged/thermodynamics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;thermodynamics&amp;#39;&quot; rel=&quot;tag&quot;&gt;thermodynamics&lt;/a&gt; (&lt;a href=&quot;/questions/tagged/physical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;physical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;physical-chemistry&lt;/a&gt;), &lt;a href=&quot;/questions/tagged/enthalpy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;enthalpy&amp;#39;&quot; rel=&quot;tag&quot;&gt;enthalpy&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/free-energy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;free-energy&amp;#39;&quot; rel=&quot;tag&quot;&gt;free-energy&lt;/a&gt;, change in the state of matter, &lt;a href=&quot;/questions/tagged/temperature&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;temperature&amp;#39;&quot; rel=&quot;tag&quot;&gt;temperature&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the transfer of heat, i.e. heating, may cause the temperature of a system to rise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating may also result in a change of the state of matter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;melting: $\ce{H2O_{(s)} (ice) -&amp;gt;[\Delta T] H2O_{(l)} (water)}$ (see also &lt;a href=&quot;/questions/tagged/melting-point&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;melting-point&amp;#39;&quot; rel=&quot;tag&quot;&gt;melting-point&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/sublimation&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;sublimation&amp;#39;&quot; rel=&quot;tag&quot;&gt;sublimation&lt;/a&gt;: $\ce{CO2_{(s)} -&amp;gt;[\Delta T] CO2_{(g)}}$&lt;/li&gt;&#xA;&lt;li&gt;boiling: $\ce{W_{(l)} -&amp;gt;[\Delta T] W_{(g)}}$ (see also &lt;a href=&quot;/questions/tagged/boiling-point&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;boiling-point&amp;#39;&quot; rel=&quot;tag&quot;&gt;boiling-point&lt;/a&gt;, here $\approx 6200~\mathrm{K}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The symbol of heat is $q$ or $Q$ and it is usually measured in joule $\mathrm{J}$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T13:17:42.140" LastActivityDate="2018-03-19T13:17:42.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10699" PostTypeId="4" CreationDate="2014-05-13T08:48:52.907" Score="0" Body="For questions about heat - a form of transfer of energy from one body to another or from one body to its surroundings. Also see these tags: [thermodynamics], [enthalpy], [energy], [temperature]." OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T13:17:25.180" LastActivityDate="2018-03-19T13:17:25.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10700" PostTypeId="2" ParentId="10668" CreationDate="2014-05-13T09:49:49.827" Score="10" Body="&lt;p&gt;A volumetric flask is designed to store a relatively great volume of liquid with great accuracy. If you want to reach a determinate volume of liquid in a large backer when you add liquid to it so the liquid height increases a little bit the volume increase a lot. What you measure with your eyes for determining the volume is the height of the liquid, so the error in the volume measure is directly linked with the error in measuring the liquid height. The relation between the increase of the height (&lt;span class=&quot;math-container&quot;&gt;$\Delta h$&lt;/span&gt;) and the volume for a cylinder is very simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V= \pi r^{2} \Delta h  $$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose that you make a systematic error &lt;span class=&quot;math-container&quot;&gt;$\Delta h$&lt;/span&gt; of 0.5 mm, that is quite a good esteem of average max human accuracy in measuring the liquid height in a cylinder with naked eyes. If you make this error in glassware with different radius you will change the accuracy of your measurements, according to the previous equation:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Fnw0n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you use ten times greater radius you will have a 100 times greater error!&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Why the body has not the conical shape of the Erlenmeyer?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This actually is not so important. Generally, the spherical surface more evenly distributes stress across its surface and is less prone to fractures. This is why we use &lt;a href=&quot;https://en.wikipedia.org/wiki/Round-bottom_flask&quot; rel=&quot;nofollow noreferrer&quot;&gt;round-bottom flasks&lt;/a&gt;. It doesn't seem there are a lot of differences in the production of both &lt;a href=&quot;https://www.youtube.com/watch?v=MMtRVuwO-HA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Erlenmeyer&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=jzubmTDebQc&quot; rel=&quot;nofollow noreferrer&quot;&gt;volumetric flask&lt;/a&gt;, so it's probably mainly due to a convention. In fact, you can find also volumetric flask with Erlenmeyer style. Like in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C10gK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C10gK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2019-07-10T13:30:42.707" LastActivityDate="2019-07-10T13:30:42.707" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10701" PostTypeId="1" AcceptedAnswerId="10703" CreationDate="2014-05-13T10:23:43.253" Score="16" ViewCount="84721" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following carbanions is the most stable?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cXfKx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cXfKx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think the answer is either &lt;strong&gt;a&lt;/strong&gt; or &lt;strong&gt;c&lt;/strong&gt;, as in &lt;strong&gt;d&lt;/strong&gt; there are no chances for the carbanion to have its charge become less. In &lt;strong&gt;b&lt;/strong&gt; I think there is a + inductive effect due to methyl group which gives it more negative charge.&lt;/p&gt;&#xA;&lt;p&gt;The ones left are &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;c&lt;/strong&gt;.  Nitrogen is an electronegative element so it might lessen the charge on the carbanion, but in both &lt;strong&gt;a&lt;/strong&gt; and &lt;strong&gt;c&lt;/strong&gt; the carbanion is same distance from it. I am unable to reason out it from here.&lt;/p&gt;&#xA;" OwnerUserId="5484" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-07-23T06:51:45.043" Title="How to determine the most stable carbanion?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10702" PostTypeId="2" ParentId="10678" CreationDate="2014-05-13T10:46:29.947" Score="3" Body="&lt;p&gt;You are indeed correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HOOC-CH2-NH2(X) -&amp;gt;[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HOOC-CH2-NH2(X) -&amp;gt;[soda lime] CH2-NH2(A)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;span class=&quot;math-container&quot;&gt;$ \ce{(A) + CHCl3 + 3KOH -&amp;gt;[Heat] CH3-NC + 3KCl + 3H2O}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methylamine &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbylamine_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;does give the carbylamine test&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So clearly, the answer given must be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct options are 1,2 and 4.&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="83758" LastEditDate="2020-03-18T16:17:38.727" LastActivityDate="2020-03-18T16:17:38.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10703" PostTypeId="2" ParentId="10701" CreationDate="2014-05-13T11:12:51.777" Score="9" Body="&lt;p&gt;Wikipedia says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The stability and reactivity of a carbanion is determined by several factors. These include&lt;/p&gt;&#xA;&lt;p&gt;1)The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;&lt;/p&gt;&#xA;&lt;p&gt;2)Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;&lt;/p&gt;&#xA;&lt;p&gt;3)The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can see that point 3) is not relevant here. We do not have a conjugated system, nor is there any chance for resonance.&lt;/p&gt;&#xA;&lt;p&gt;Keeping in mind point 1), the &lt;span class=&quot;math-container&quot;&gt;$\ce{N^+}$&lt;/span&gt; has a &lt;span class=&quot;math-container&quot;&gt;$\ce{-I}$&lt;/span&gt; effect, but inductive effect reduces over distance. The negative charges in options a), b) and d) are either in the meta or para position.&lt;/p&gt;&#xA;&lt;p&gt;Also, remember that &lt;span class=&quot;math-container&quot;&gt;$\ce{+I}$&lt;/span&gt; effect destabilises the carbanion.&lt;/p&gt;&#xA;&lt;p&gt;In c), the carbanion is primary, but in a), it is secondary. Primary carbanions are more stable due to lesser &lt;span class=&quot;math-container&quot;&gt;$\ce{+I}$&lt;/span&gt; effect.&lt;/p&gt;&#xA;&lt;p&gt;So your answer is c).&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-05-13T11:21:55.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10704" PostTypeId="2" ParentId="10701" CreationDate="2014-05-13T11:16:58.037" Score="2" Body="&lt;p&gt;In $a$ or $c$ $\ce N$ atom is equidistant from $-$ charge. Hence, $c$ is most stable as it has no $+I$ effect from other ends like $a$&lt;/p&gt;&#xA;" OwnerUserId="4597" LastActivityDate="2014-05-13T11:16:58.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10705" PostTypeId="2" ParentId="10688" CreationDate="2014-05-13T12:11:55.410" Score="2" Body="&lt;p&gt;if you mean the following plot:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zGGBU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Electrochemical Methods. fundamentals and applications, p. 341). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a bulk concentration is the same as the concentration at the surface of the electrode (the mass-transfer is very high) than the current will be independent on $\sqrt\omega$ and will be constant at the value $i_K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slope of the bold line represent deviation form Levich line and tend towards the limit $\lim_{\sqrt\omega \to \infty} i = i_K $. The higher the speed of rotation is, the less is the electrode reaction limited by mass-transfer effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Levich line (dashed; $\frac{i}{\sqrt\omega}$ ) is constant only when $i_K$ is much larger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ertbg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-05-13T12:11:55.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10706" PostTypeId="1" CreationDate="2014-05-13T12:25:32.037" Score="3" ViewCount="1629" Body="&lt;p&gt;Why is entropy not zero for an irreversible adiabatic process though q=0? Ideally entropy is a state function so entropy should be zero.&lt;/p&gt;&#xA;" OwnerUserId="5485" LastActivityDate="2015-10-19T15:04:32.410" Title="Entropy change in adiabatic process" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10707" PostTypeId="1" AcceptedAnswerId="10709" CreationDate="2014-05-13T13:05:09.067" Score="3" ViewCount="258" Body="&lt;p&gt;Can singlet oxygen facilitate a (2+2) cycloaddition reaction, for example between two alkenes, by reacting first with the pi-electrons in one alkene and then lowering the energy barrier for the second alkene to attack (given the instability of 1,2-dioxetane)?  Are there any examples of this sort of thing?&lt;/p&gt;&#xA;" OwnerUserId="5486" LastEditorUserId="5486" LastEditDate="2014-05-13T13:11:20.120" LastActivityDate="2014-05-13T14:35:29.440" Title="Can singlet oxygen act as a catalyst for a (2+2) cycloaddition reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10708" PostTypeId="2" ParentId="10686" CreationDate="2014-05-13T13:26:09.447" Score="1" Body="&lt;p&gt;I would recommend you take a look at the book Electrochemical Methods fundamentals and applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Koutecý-Levich equation (totally irreversible one-step, one-electron reaction):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Large \frac{1}{i}=\frac{1}{i_K}+\frac{1}{i_{l,c}}=\frac{1}{i_K}+\frac{1}{0.62nFAD^{2/3}\omega^{1/2}\nu^{-1/6}C^*_O} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher diffusion coefficient is, the higher is the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diffusion coefficient can be computed from the Koutecý-Levich equation if you know all other coefficients. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-05-13T13:34:41.913" LastActivityDate="2014-05-13T13:34:41.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10709" PostTypeId="2" ParentId="10707" CreationDate="2014-05-13T14:35:29.440" Score="3" Body="&lt;p&gt;I'm not aware of any such examples.  In the reactions of singlet oxygen with olefins only products from cycloaddition to form a dioxetane and \ or products from the ene reaction are observed.  There are no reports of olefin dimerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OhK0a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-13T14:35:29.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10710" PostTypeId="2" ParentId="10706" CreationDate="2014-05-13T16:24:05.617" Score="1" Body="&lt;p&gt;Because there is many way to produce work and there is always irreversibility associated with real production. This is why work and heat variation are not exact differential. The path (the way) used does matter. In your case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}U = - \delta W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adiabatic (no heat) irreversible (real) transformation produce entropy. Isentropic means adiabatic and reversible. Therefore you will produce work and irreversibilities (eg.: friction) which will account for internal energy variation.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="3879" LastEditDate="2014-05-13T16:38:26.147" LastActivityDate="2014-05-13T16:38:26.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10711" PostTypeId="1" AcceptedAnswerId="10715" CreationDate="2014-05-13T18:26:08.410" Score="2" ViewCount="1345" Body="&lt;p&gt;I was looking for some distillation equipment on Sigma Aldrich because I wanted to perform some vacuum distillation, when I came across &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/z102598?lang=en&amp;amp;region=NL&quot; rel=&quot;nofollow&quot;&gt;a vacuum-jacketed distilling column&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as far as I understood my high-school chemistry, I can perform vacuum distillation perfectly fine with a 'normal' column as long as there is an entry to which I can connect my vacuum source (in my case a cheap aspirator, but he it does the trick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I'm buying all this pretty expensive stuff, I want to make sure that I buy the right equipment. As far as I can see it, this vacuum-jacketed distilling column has nothing to do with vacuum distillation, it is simply a vacuum jacket around a normal column. But then, where it is used for and why, or am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:48:28.937" LastActivityDate="2015-10-30T16:48:28.937" Title="Vacuum-jacketed distilling column" Tags="&lt;equipment&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10713" PostTypeId="1" AcceptedAnswerId="10714" CreationDate="2014-05-13T19:31:30.850" Score="4" ViewCount="3033" Body="&lt;p&gt;I watched a Khan academy video on organic chemistry functional groups. It mentioned the amine and amide groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I see, an amide is an amine with a carbonyl group. However, the internet says that that assumption is wrong. I am guessing it has to do with the &quot;R&quot; groups on the amine functional group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the video, he used an example of 2 ethyls (2 carbon chains, not sure what they are called) attached to the nitrogen, and the third R group is just a hydrogen.&#xA;So what exactly is an R group, and are all amides also amines? Thanks in advance for the help.&lt;/p&gt;&#xA;" OwnerUserId="5488" LastEditorUserId="12860" LastEditDate="2015-10-18T22:01:38.923" LastActivityDate="2015-10-18T22:01:38.923" Title="Are all amides also amines?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10714" PostTypeId="2" ParentId="10713" CreationDate="2014-05-13T19:48:07.053" Score="12" Body="&lt;p&gt;An R group is used in structural formulae as a placeholder for a range of possible subtituents, e.g. hydrogen, alkyl chains or aryl groups etc. It can also be used as an abbreviation when R is a predefined rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An amide is a derivative of an organic acid where an $\ce{OH}$ group is substituted by $\ce{NR2}$ (with R = H or any organic rest, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Amide&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, for example). An &lt;a href=&quot;http://en.wikipedia.org/wiki/Amine&quot; rel=&quot;noreferrer&quot;&gt;amine&lt;/a&gt; is a base, and not formally derived from an acid, but from ammonia. It has the general formula $\ce{NR3}$, with R being H or any organic rest except X=O (X = C, S, etc.) or any other rest derived from an acid. So amides and amines are different classes of compounds. Here are examples for amides of a carboxylic acid, a sulfonic acid (sulfonamide), and an amine (from left to right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/blRMO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound from the video that you have described is diethyl amine. You can formally derive it from ammonia by substituting two hydrogens with ethyl groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXvf8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-05-13T21:48:54.203" LastActivityDate="2014-05-13T21:48:54.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10715" PostTypeId="2" ParentId="10711" CreationDate="2014-05-13T20:25:43.123" Score="7" Body="&lt;p&gt;The purpose of the vacuum jacket is to provide insulation of the column from cooling by ambient lab air.  You don't want to have to heat the &quot;pot&quot; (the flask containing the crude material you want to purify) any more than you have to, otherwise heat sensitive material in the pot may degrade and turn to tar.  Imagine if you cooled the exterior of your distillation column with cold water while performing the distillation.  You'd have to heat the pot a LOT more to get the material in the pot to finally reach the top of the column, then if you weren't cooling the column.  Air will have the same effect on your distillation, albeit to a smaller extent, than cold water.  So to avoid putting unnecessary extra heat into the pot, you vacuum jacket your column to preclude undesired cooling by ambient air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, a distillation column is different than a condensing column.  You don't want to cool a distillation column, you do want to cool a condensing column (or condenser).  Back in the day, very long columns were sometimes needed in the organic lab (5-10 feet) to separate closely boiling materials, they are still used in industrial applications.  If you had a 10-foot column that wasn't vacuum jacketed, you would turn most of your material to tar before it distilled over, because of the extra heat you had to supply to make up for the undesired cooling of the column by ambient air.  Eventually improvements like column packing materials (glass beads for example) and spinning bands (a metal helix rotating in the column) came along to increase column efficiency and allow chemists to be able to get good separation efficiencies (measured as &quot;height equivalent theoretical plates&quot;, HETP) without having to resort to long columns.  Still a vacuum jacketed column is preferred to lessen unnecessary heating.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-13T20:25:43.123" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10716" PostTypeId="1" AcceptedAnswerId="10720" CreationDate="2014-05-13T21:55:02.117" Score="7" ViewCount="250" Body="&lt;p&gt;According to my calculations, the reaction $\ce{H2O + O3 -&amp;gt; H2O2 + O2}$ should be favourable under standard conditions. However, I am told that the reaction is probably very slow in solution since ozone is poorly soluble in water and the reaction occurs through a radical mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water can be ignited in a fluorine atmosphere. Can water also be ignited in an ozone atmosphere? Can this be done at 1 atm?&lt;/p&gt;&#xA;" OwnerUserId="5490" LastEditorUserId="402" LastEditDate="2017-04-01T01:17:09.353" LastActivityDate="2017-04-01T01:17:09.353" Title="Can water be ignited in an ozone atmosphere?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;combustion&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10718" PostTypeId="2" ParentId="10713" CreationDate="2014-05-14T00:59:32.690" Score="6" Body="&lt;p&gt;While JA is formally correct, the description misses a crucial point, because it is a structural definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactivity characteristic of amines is their moderate nucleophilicity/basicity and low acidity.  This is a property of a lone pair of electrons on the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a carbonyl is introduced on an alpha-carbon, the electron withdrawing properties, both by induction and resonance, decreases the availability of the lone pair of electrons to engage in reactions as bases.  This leads to a strong decrease in nucleophilicity/basicity and a sharp increase in acidity, although amides are still fairly weak acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the difference between amines and amides is that amines are bases, and amides are not.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-05-14T00:59:32.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10719" PostTypeId="2" ParentId="10323" CreationDate="2014-05-14T01:28:20.083" Score="3" Body="&lt;p&gt;A simple way to determine whether an equivalence point in a titration is the last one is to figure out the volume of titrant added for the first titration point, then realize all subsequent equivalence points will be at exact integer multiples of this volume, under the assumption that the analyte contains only one acid (a mixture of acids won't behave the same). When an integer multiple of the first equivalence point volume fails to be an equivalence point itself, then there will be no further equivalence points.&lt;/p&gt;&#xA;&lt;p&gt;For example, let us imagine a generic polyprotic acid with formula &lt;span class=&quot;math-container&quot;&gt;$\ce{H_{k}A}$&lt;/span&gt;, being titrated with a strong monoprotic base such as &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;. Starting from the fully protonated acid, let us neutralize each ionizable hydrogen in steps. Let the acid  &lt;span class=&quot;math-container&quot;&gt;$\ce{H_{k}A}$&lt;/span&gt; being titrated have an initial concentration &lt;span class=&quot;math-container&quot;&gt;$c_{acid}$&lt;/span&gt; in a solution of volume &lt;span class=&quot;math-container&quot;&gt;$V_{acid}$&lt;/span&gt;, and let the titrant have a concentration &lt;span class=&quot;math-container&quot;&gt;$c_{base}$&lt;/span&gt; and a total volume added &lt;span class=&quot;math-container&quot;&gt;$V_{base}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;By definition, to reach the first equivalence point (&lt;span class=&quot;math-container&quot;&gt;$V_{base}=V_{eq\ 1}$&lt;/span&gt;), we consider the reaction which represents the neutralization of &lt;em&gt;only&lt;/em&gt; the first proton:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H_{k}A + OH- -&amp;gt; H_{k-1}A- + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each one mole of acid  &lt;span class=&quot;math-container&quot;&gt;$\ce{H_{k}A}$&lt;/span&gt; must surrender exactly one mole of hydrogen ions &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, each of which will react with exactly one mole of hydroxide ions &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;. Thus:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{released}\ \ce{H+}}=n_\mathrm {acid}=c_\mathrm {acid} \times V_\mathrm {acid}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{base}=c_\mathrm {base} \times V_\mathrm {eq\ 1} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{released}\ \ce{H^+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 1}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V_\mathrm {eq\ 1} = \frac{c_\mathrm {acid} \times V_\mathrm{acid}}{c_\mathrm {base}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, to reach the second equivalence point (&lt;span class=&quot;math-container&quot;&gt;$V_{base}=V_{eq\ 2}$&lt;/span&gt;), we consider the reaction which represents the neutralization of the first &lt;strong&gt;and&lt;/strong&gt; second protons:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H_{k}A + 2 OH- -&amp;gt; H_{k-2}A^{2-} + 2 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Each one mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{H_{k}A}$&lt;/span&gt; must surrender exactly &lt;strong&gt;two&lt;/strong&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, each of which will react with exactly one mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;. The relevant equations are now:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{released}\ \ce{H+}}=2\times n_\mathrm {acid}=2\times c_\mathrm {acid} \times V_\mathrm {acid}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{eq\ 2} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n_{\mathrm{released}\ \ce{H+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  2\times c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V_\mathrm {eq\ 2} = 2\times \frac{c_\mathrm {acid} \times V_\mathrm {acid}}{c_\mathrm{base}} = 2\times V_\mathrm {eq\ 1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice how the added volume to reach the second equivalence point is exactly twice that added to reach the first equivalence point. This pattern repeats, i.e. the third equivalence point will happen at &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm {base}=3\times V_\mathrm {eq\ 1}$&lt;/span&gt;, and the &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;-th equivalence point will happen at &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm {base}=k\times V_\mathrm {eq\ 1}$&lt;/span&gt;. &lt;em&gt;The equivalence points are equally spaced&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In short, if the titration of a pure polyprotic acid shows the first sudden pH swing after adding &lt;span class=&quot;math-container&quot;&gt;$\pu{15 mL}$&lt;/span&gt; of the base titrant, then &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm {eq\ 1}$&lt;/span&gt; equals 15 mL, and the only possible next equivalence points are at &lt;span class=&quot;math-container&quot;&gt;$\pu{30 mL}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{45 mL}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{60 mL}$&lt;/span&gt;, etc. If during the titration another sudden pH swing is observed at &lt;span class=&quot;math-container&quot;&gt;$\pu{30 mL}$&lt;/span&gt;, but not at &lt;span class=&quot;math-container&quot;&gt;$\pu{45 mL}$&lt;/span&gt;, then you can deduce that there is no third equivalence point, and hence the acid is diprotic.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="95133" LastEditDate="2021-05-25T12:55:13.640" LastActivityDate="2021-05-25T12:55:13.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10720" PostTypeId="2" ParentId="10716" CreationDate="2014-05-14T03:53:30.650" Score="8" Body="&lt;p&gt;We are basically talking about a derivation of the $\ce{H2/O2}$ system, which involves a radical mechanism. You maybe want to have a look at &lt;a href=&quot;https://chemistry.stackexchange.com/a/9617/4945&quot;&gt;this answer&lt;/a&gt; of mine going a bit more into detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction you are interested about is only a total sum of reactions and is most likely not to occur as other reactions will be favourable. Ozone itself is not a very stable compound and many other molecules may actively catalyse decomposition. There are various publications&lt;sup&gt;[1-3]&lt;/sup&gt; on the mechanism of this reaction:&#xA;\begin{align}\ce{&#xA;  M + O3 &amp;amp;&amp;lt;=&amp;gt; M + O2 + O\\&#xA;  O + O3 &amp;amp;&amp;lt;=&amp;gt; 2O2\tag1&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of $\ce{M}$ is in $\ce{O3, O2, CO2, N2, He, etc}$.&lt;sup&gt;[1]&lt;/sup&gt; Although not stated explicitly, there should basically no reason why water is not be able to accomplish the same. Hence the driving force of the total reaction would be entropy, giving the overall equation:&#xA;$$\ce{2 O3 -&amp;gt;[\ce{H2O}] 3 O2}\tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -&amp;gt; O3 + O2 + O}\tag3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In presence of Hydrogen donors ($\ce{RH}$ apparently any kind will do) a chain reaction will be initiated.&lt;sup&gt;[1]&lt;/sup&gt; $$\ce{O + RH -&amp;gt; OH + R}\tag4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main chain carriers will be $\ce{HO, HO2}$.&#xA;\begin{align}\ce{&#xA;  HO + O3 &amp;amp;-&amp;gt; HO2 + O2\\&#xA;  HO2 + O3 &amp;amp;-&amp;gt; HO + 2O2\tag5&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process will be terminated by this (quite unspecific) reaction. This may then also involve $\ce{H2O2}$, which &quot;is expected to be an excellent $\ce{H}$ donor.&quot;&lt;sup&gt;[1]&lt;/sup&gt;&#xA;$$\ce{2HO2 -&amp;gt;}\text{stable products}\tag6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O &amp;lt;=&amp;gt; H3+O + {}^{-}OH}\tag7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steahelin and Hoigné propose several hypothesis in which way ozone will be decomposed, but I will not go into absurd detail.&lt;sup&gt;[4]&lt;/sup&gt; The core assumptions involve $\ce{{}^{-}OH}$ as initiator of a radical chain.&#xA;\begin{align}\ce{&#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; .O2- + HO2. &amp;amp;  HO2. &amp;amp;&amp;lt;=&amp;gt; H+ + .O2-\\&#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; HO2- + O2   &amp;amp;  H2O2 &amp;amp;&amp;lt;=&amp;gt; HO2- + H+\\&#xA;                               &amp;amp;&amp;amp;  O3 + HO2- &amp;amp;-&amp;gt; .OH + .O2- + O2\\&#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; .O3- + .OH\tag8&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is basically followed by another set of radical chain reactions.&#xA;\begin{align}\ce{&#xA;  O3 + .O2- &amp;amp;-&amp;gt; .O3- + O2\\&#xA;  .O3- + H2O &amp;amp;-&amp;gt; .OH + {}^{-}OH + O2\tag9&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is all again resulting in $(2)$ again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary&lt;/strong&gt; the proposed reaction might occur, but will immediately be followed by another set of reactions resulting in the formation of biatomic oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.1743610&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sidney W. Benson, and Arthur E. Axworthy Jr., &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1957&lt;/strong&gt;, &lt;em&gt;26&lt;/em&gt;, 1718.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.1674697&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. V. Michael, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1971&lt;/strong&gt;, &lt;em&gt;54&lt;/em&gt;, 4455.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1084960/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oliver R. Wulf, and Richard C. Tolman, *Proc. Natl. Acad. Sci. USA, &lt;strong&gt;1927&lt;/strong&gt;, &lt;em&gt;13&lt;/em&gt; (5), 272–275.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/es00104a009&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johannes. Staehelin, and Juerg. Hoigne, &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt;, &lt;strong&gt;1982&lt;/strong&gt;, &lt;em&gt;16&lt;/em&gt; (10), 676–681.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-04-20T05:57:13.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10721" PostTypeId="2" ParentId="10471" CreationDate="2014-05-14T04:14:10.580" Score="5" Body="&lt;p&gt;This is basically an excerpt of the &lt;a href=&quot;https://chemistry.stackexchange.com/a/10720/4945&quot;&gt;answer&lt;/a&gt; I have given to another quite similar question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ben is quite right, neither the reaction nor the mechanism is simple. It involves several different radical chains all resulting in the decomposition of ozone.&lt;a href=&quot;http://dx.doi.org/10.1063/1.1743610&quot; rel=&quot;nofollow noreferrer&quot;&gt;[a]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -&amp;gt; O3 + O2 + O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In presence of Hydrogen donors like $\ce{H2O2}$, which &quot;is expected to be an excellent $\ce{H}$ donor.&quot;&lt;a href=&quot;http://dx.doi.org/10.1063/1.1743610&quot; rel=&quot;nofollow noreferrer&quot;&gt;[a]&lt;/a&gt; Those are forming the main chain carriers will be $\ce{HO, HO2}$.&#xA; Finally stopping at some point by recombination of these species.&#xA;\begin{aligned}\ce{&#xA;O + H2O2 &amp;amp;-&amp;gt; OH + HO2\\&#xA;  HO + O3 &amp;amp;-&amp;gt; HO2 + O2\\&#xA;  HO2 + O3 &amp;amp;-&amp;gt; HO + 2O2\\&#xA;  2HO2 &amp;amp;-&amp;gt;}\text{stable products}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O &amp;lt;=&amp;gt; H3+O + {}^{-}OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially important might also be protolysis of hydrogen peroxide. $$\ce{H2O2 + H2O &amp;lt;=&amp;gt; HO2- + H3+O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steahelin and Hoigné propose several hypotheses in which way ozone will be decomposed, but I will not go into absurd detail here.&lt;a href=&quot;http://dx.doi.org/10.1021/es00104a009&quot; rel=&quot;nofollow noreferrer&quot;&gt;[b]&lt;/a&gt; The core assumptions involve $\ce{{}^{-}OH}$ as the initiator of a radical chain. Followed by radical (ionic) species like $\ce{.O2-, .O3-, HO2., .OH}$ resulting again in a chain reaction. &#xA;\begin{aligned}\ce{&#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; .O2- + HO2.\\  &#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; HO2- + O2\\ &#xA;  O3 + {}^{-}OH &amp;amp;-&amp;gt; .O3- + .OH\\&#xA;  HO2. &amp;amp;&amp;lt;=&amp;gt; H+ + .O2-\\&#xA;  O3 + HO2- &amp;amp;-&amp;gt; .OH + .O2- + O2\\&#xA;  O3 + .O2- &amp;amp;-&amp;gt; .O3- + O2\\&#xA;  .O3- + H2O &amp;amp;-&amp;gt; .OH + {}^{-}OH + O2&#xA;}\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-02T23:28:38.090" LastActivityDate="2017-05-02T23:28:38.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10722" PostTypeId="1" AcceptedAnswerId="10739" CreationDate="2014-05-14T06:14:33.880" Score="23" ViewCount="1775" Body="&lt;p&gt;When there are $2$ hydroxy groups on same carbon, they tend to undergo dehydration ( of course with exceptions like chloral where $\ce H$ bonding prevails).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? We are breaking a sigma bond in order to get a pi bond between $\ce C$ and $\ce O$ ?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4231" LastEditDate="2015-03-16T13:35:38.090" LastActivityDate="2020-07-24T17:34:37.327" Title="What makes C=O more stable that C(OH)₂" Tags="&lt;organic-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10723" PostTypeId="1" CreationDate="2014-05-14T06:26:08.883" Score="0" ViewCount="1975" Body="&lt;p&gt;The rate of hydrogenation is faster for which alkene : cis or trans. I know it depends on steric hindrance on the carbon, but both have steric hindrance. Can someone please help&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="5591" LastEditDate="2015-10-09T17:29:16.880" LastActivityDate="2015-10-09T17:29:16.880" Title="Rate of hydrogenation" Tags="&lt;organic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10724" PostTypeId="2" ParentId="10723" CreationDate="2014-05-14T07:36:15.797" Score="1" Body="&lt;p&gt;Imagine that the mechanism for the addition of $\ce{H-Hal}$, $\ce{Hal-Hal}$, and $\ce{H-H}$ to an alkene are fundamentally different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, subsequent addition of the proton and (then) the halogen is assumed. In the second case, formation of a halonium ion is postulated as the first step, followed by addition of the halide (sort of backside attack to the existing $\ce{C-Hal}$ bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the latter case, the &lt;strong&gt;simultaneous&lt;/strong&gt; addition of both hydrogen atoms to the same face of the alkene is assumed. Isn't it likely under these conditions that a &lt;em&gt;cis&lt;/em&gt; alkene, which is less shielded on one side, is faster hydrogenated then?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-05-14T07:36:15.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10725" PostTypeId="5" CreationDate="2014-05-14T07:57:56.727" Score="0" Body="&lt;p&gt;Stoichiometry refers to the relationship between the amounts of substances of the reactants and the amount of substances of the formed products. It is closely related to &lt;a href=&quot;http://goldbook.iupac.org/C01034.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical reaction equations&lt;/a&gt;. As the total mass of the reactants has to equal the total mass of the products, it also reflects the law of conservation of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deriving from this it is possible to calculate the amount of product that will be formed, if the amount of reactants is known, and vice versa. A reaction can be classified as &lt;a href=&quot;http://goldbook.iupac.org/S06021.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;stoichiometric&lt;/a&gt; if it follows simple integer ratios, i.e. no excess of reactants or products.&#xA;$$\ce{NaCl_{(aq)} + AgNO3_{(aq)} -&amp;gt; Na_{(aq)}^{+} + NO3_{(aq)}^{-} + AgCl_{(s)} v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reactions are more complicated and involve several &lt;a href=&quot;http://goldbook.iupac.org/M03804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction paths or mechanisms&lt;/a&gt;, but the overall stoichiometry has to be preserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise IUPAC Definition of &lt;a href=&quot;http://goldbook.iupac.org/S06026.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stoichiometry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term refers to the relationship between the amounts of substances&#xA;  that react together in a particular chemical reaction, and the amounts&#xA;  of products that are formed. The general stoichiometric equation:&#xA;  $$\ce{a A + b B + \dots -&amp;gt; \dots + y Y + z Z}$$ provides the&#xA;  information that $\ce{a}$ moles of $\ce{A}$ reacts with $\ce{b}$ moles&#xA;  of $\ce{B}$ to produce $\ce{y}$ moles of $\ce{Y}$ and $\ce{z}$ moles&#xA;  of $\ce{Z}$. The stoichiometry of a reaction may be unknown or may be&#xA;  very complex. For example, the thermal decomposition of acetaldehyde&#xA;  yields mainly methane and carbon monoxide, but also a variety of minor&#xA;  products such as ethane, acetone, and diacetyl. The stoichiometric&#xA;  equation $$\ce{CH3CHO -&amp;gt; CH4 + CO}$$ is therefore only an approximate&#xA;  one. Even when the overall stoichiometry of a reaction is well&#xA;  defined, it may be time-dependent in that it varies during the course&#xA;  of a reaction. Thus if a reaction occurs by the mechanism  $\ce{A -&amp;gt; X&#xA;&amp;gt; -&amp;gt; Y}$, and $\ce{X}$ is formed in substantial amounts during the course of the process, the relationship between the amounts of&#xA;  $\ce{A}$, $\ce{X}$ and $\ce{Y}$ will vary with time, and no one&#xA;  stoichiometric equation can represent the reaction at all times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further reading:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/24245/43942&quot;&gt;Are matrices a good way to do stoichiometry?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="43942" LastEditDate="2017-05-26T06:13:02.123" LastActivityDate="2017-05-26T06:13:02.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10726" PostTypeId="4" CreationDate="2014-05-14T07:57:56.727" Score="0" Body="Stoichiometry refers to the relationship between the amounts of substances of the reactants and the amount of substances of the formed products. This tag should be used for specific questions about chemical reaction equations. " OwnerUserId="4945" LastEditorUserId="19616" LastEditDate="2017-04-21T19:15:32.153" LastActivityDate="2017-04-21T19:15:32.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10728" PostTypeId="5" CreationDate="2014-05-14T08:57:13.527" Score="0" Body="&lt;p&gt;A crystal structure is a unique arrangement of atoms, ions, or molecules in a crystalline solid (or liquid). It is a highly ordered three dimensional structure, that reflects the intrinsic nature of its components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be expressed as a &lt;a href=&quot;http://goldbook.iupac.org/U06562.html&quot; rel=&quot;nofollow&quot;&gt;unit cell&lt;/a&gt;, the smallest, regularly repeating arrangement of entities in a &lt;a href=&quot;/questions/tagged/materials&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;materials&amp;#39;&quot; rel=&quot;tag&quot;&gt;materials&lt;/a&gt;. It is generally a paralellepiped (in special cases a cuboid or cube) and is therefore composed of three base vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and its volume is therefore given through $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$. These vectors are usually called lattice parameters. The symmetry properties of a crystal are expressed as space group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crystal structure can be obtained by electron diffraction - a very common analytical method in &lt;a href=&quot;/questions/tagged/solid-state-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;solid-state-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;solid-state-chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-14T10:22:47.033" LastActivityDate="2014-05-14T10:22:47.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10729" PostTypeId="4" CreationDate="2014-05-14T08:57:13.527" Score="0" Body="A crystalline (or crystal) structure is a unique arrangement of atoms, ions, or molecules in a crystalline solid (or liquid). It is a highly ordered three dimensional structure, that reflects the intrinsic nature of its components. " OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-14T10:23:01.700" LastActivityDate="2014-05-14T10:23:01.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10730" PostTypeId="5" CreationDate="2014-05-14T09:22:21.513" Score="0" Body="&lt;p&gt;Material is a general term for a chemical substance or a mixture of substances with certain, well defined physical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials sciences, also known as materials engineering, is an interdisciplinary field (mostly between chemistry and physics) that is focused on designing materials with new properties, or new processes to produce materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common, prominent classes of materials are metals, glasses, ceramics, semiconductors, composites, and plastics. In recent years a main focus was &lt;a href=&quot;/questions/tagged/nanotechnology&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nanotechnology&amp;#39;&quot; rel=&quot;tag&quot;&gt;nanotechnology&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2018-10-14T18:51:39.423" LastActivityDate="2018-10-14T18:51:39.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10731" PostTypeId="4" CreationDate="2014-05-14T09:22:21.513" Score="0" Body="Material is a general term for a chemical substance or a mixture of substances with certain, well defined physical properties." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-14T09:39:43.837" LastActivityDate="2014-05-14T09:39:43.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10732" PostTypeId="1" AcceptedAnswerId="10733" CreationDate="2014-05-14T09:26:56.047" Score="7" ViewCount="3993" Body="&lt;p&gt;I am learning Chemistry now, and I want to take some notes that involve huge amount of organic chemistry on Mac. How can I do that? Pages does not seem support scientific use.&lt;/p&gt;&#xA;" OwnerUserId="1800" LastEditorUserId="14" LastEditDate="2014-05-14T14:03:58.107" LastActivityDate="2016-12-16T15:38:38.530" Title="How do I take chemistry notes on Mac?" Tags="&lt;software&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10733" PostTypeId="2" ParentId="10732" CreationDate="2014-05-14T09:41:47.153" Score="9" Body="&lt;p&gt;If you know LaTeX and you want to see what you are writing you can use &lt;a href=&quot;http://www.lyx.org/Download&quot; rel=&quot;nofollow noreferrer&quot;&gt;lyx&lt;/a&gt;, and some TeX package like &lt;a href=&quot;http://www.ctan.org/pkg/mhchem&quot; rel=&quot;nofollow noreferrer&quot;&gt;mhchem&lt;/a&gt; and &lt;a href=&quot;http://www.ctan.org/pkg/chemfig&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemfig&lt;/a&gt;. So you can write all the things you want and add comments. This is what I have done with my Mac just now:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NU1UX.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Here is the code, you can make these lines as a default template:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\documentclass[12pt,a4paper]{article}&#xA;\usepackage[utf8]{inputenc}&#xA;\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}&#xA;\usepackage{chemfig}&#xA;\usepackage{tikz}&#xA;\usepackage[version=4]{mhchem}&#xA;\setlewis{4pt}{0.8em}{black}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From here you have to write! The date is automatically added!&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\title{Introduction to chemistry notes}&#xA;\begin{document}&#xA;\maketitle&#xA;\section{mhchem}&#xA;You know that \ce{ CH4 + 2O2 -&amp;gt; CO2 + 2H2O}     this is a simple equation\ldots&#xA;Okay maybe you can do this with word, but if you are familiar with Chemistry S.E. you will be very fast with this notation.&#xA;&#xA;\ce{Zn^2+&#xA;&amp;lt;=&amp;gt;[\ce{+ 2OH-}][\ce{+ 2H+}]&#xA;$\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}$&#xA;&amp;lt;=&amp;gt;C[+2OH-][{+ 2H+}]&#xA;$\underset{\text{Hydroxozikat}}{\ce{[Zn(OH)4]^2-}}$&#xA;}&#xA;&#xA;&#xA;&#xA;\section{Chemfig}&#xA;\begin{figure}[h]&#xA;\setatomsep{1.8em}&#xA;\schemestart&#xA;\subscheme[90]{\chemfig{*6((-[@{a1}]H_2@{a2}\Lewis{26,N})=[@{r1}]-[@{r2}]=[@{b2}](-[@{b1}]N =[@{c1}] N -[@{c2}]*6(=[@{d1}]-[@{d2}] =[@{e1}](-[@{e2}] OH)-=-))-=-=)}\chemrel{&amp;lt;-&amp;gt;}&#xA;\chemmove[-stealth,shorten &amp;lt;=1pt, shorten &amp;gt;=1pt]{%&#xA;    \draw(a2)..controls +(90:5mm) and +(135:5mm)..(a1);% first arrow&#xA;    \draw(r1)..controls +(225:12mm) and +(300:12mm)..(r2);% ring arrow&#xA;    \draw(b2)..controls +(10:3mm)..(b1);% ring arrow &#xA;    \draw(c1)..controls +(340:8mm)..(c2);% ring arrow &#xA;    \draw(d1)..controls +(225:12mm) and +(300:12mm)..(d2);% ring arrow&#xA;    \draw(e1)..controls +(20:4mm)..(e2);% ring arrow  &#xA;    }        &#xA;\chemfig{*6((=[@{a1}]H_2@{a2}\chemabove{N}{\oplus})-[@{r1}]=[@{r2}]-[@{b2}](=[@{b1}]N-[@{c1}] N =[@{c2}]*6(-[@{d1}]=[@{d2}] -[@{e1}](=[@{e2}] \chemabove{O}{\ominus}H)-=-))-=--)}&#xA;}&#xA;\schemestop&#xA;\caption{This is a coniguated system\ldots}&#xA;\label{fig:con}&#xA;\end{figure}&#xA;Okay the lesson will end and in one hour you'll draw this figure only, maybe you should use an external software\ldots %what the professor means whit this last phrase?! &#xA;\end{document}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sincerely I think that is best to take handwrite notes because I think that writing and drawing chemical formula allow you to memorize better but of course this is only a thought. &lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="24052" LastEditDate="2016-12-16T10:53:42.020" LastActivityDate="2016-12-16T10:53:42.020" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10734" PostTypeId="5" CreationDate="2014-05-14T10:02:29.333" Score="0" Body="&lt;p&gt;Ozone is an allotrope of oxygen with the general formula $\ce{O3}$. It is hence also known as trioxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozone is a colourless or pale blue diamagnetic gas with a displeasing, painful smell, that is comparable to that of chlorine gas. It is a powerful oxidizing agent, which is also the reason for this smell. It is irritating to the respiratory system of mammals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bent polar molecule which's structure may be expressed with two mesomeric forms (from Wikipedia):&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d7/Ozon_Mesomerie.svg&quot; alt=&quot;mesomerie ozon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozone is closely connected to &lt;a href=&quot;/questions/tagged/atmospheric-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;atmospheric-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;atmospheric-chemistry&lt;/a&gt;, as the ozone layer protects the surface from quite dangerous UV-B radiation ($\ce{M}$ is an inert Air molecule like $\ce{N2}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 -&amp;gt;[h\nu] O + O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O + O2 -&amp;gt;[M] O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O3 -&amp;gt;[h\nu] O + O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In laboratory uses it may be synthesised by electrolysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3 H2O -&amp;gt; O3 + 6 H+ + 6 e^{−}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6 H+ + 6 e^{−} -&amp;gt; 3 H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 H2O -&amp;gt; O2 + 4 H+ + 4 e^{−}}$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-03T20:57:53.707" LastActivityDate="2017-05-03T20:57:53.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10735" PostTypeId="4" CreationDate="2014-05-14T10:02:29.333" Score="0" Body="Ozone is an allotrope of oxygen with the general formula O3. It is hence also known as trioxygen. This tag should be applied to questions, that are specifically about its structure, properties, and uses in chemistry." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-14T10:22:58.190" LastActivityDate="2014-05-14T10:22:58.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10736" PostTypeId="1" CreationDate="2014-05-14T10:33:50.020" Score="9" ViewCount="23370" Body="&lt;p&gt;In the nitration of compounds such as naphthalene and anthracene, how can you determine which would be the major product? I tried drawing resonance structures for the intermediate formed because stability of intermediate determines which product will be major, but this became too hard for me as there were too many resonance structures and I couldn't really compare them. Can someone please help?&lt;/p&gt;&#xA;" OwnerUserId="4474" LastEditorUserId="16683" LastEditDate="2018-08-03T13:56:49.830" LastActivityDate="2018-08-04T05:27:09.830" Title="Nitration of naphthalene and anthracene" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;carbocation&gt;&lt;regioselectivity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10737" PostTypeId="1" AcceptedAnswerId="10753" CreationDate="2014-05-14T11:01:38.293" Score="0" ViewCount="1108" Body="&lt;blockquote&gt;&#xA;Let us say that we have the following two reactions:  &lt;br&gt; &#xA;$$\ce{A + O2 -&amp;gt; B}$$ where $\ce A \text{ and } \ce B $ are different oxides of metal $\ce M$.&lt;br&gt;   &#xA;Weights of $\ce A$ and $\ce B$ are $\pu{5.72 g}$ and $\pu{6.36 g}$. &lt;br&gt; &#xA;$$\ce{B + H2-&amp;gt;M + H2O}$$  &lt;br&gt;&#xA;The weights of $\ce B$ and $\ce M$ are $\pu{4.77 g}$ and $3.81$ g, respectively. &lt;br&gt;  &#xA;&lt;/blockquote&gt;  &#xA;&#xA;&lt;p&gt;So I was trying to infer some information from the given data and this is what I did:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first reaction we can calculate the equivalent weight of $\ce A$ and $\ce B$ to be $\pu{71.5 g}$ and $\pu{79.5 g}$ respectively.&lt;br&gt;&#xA;In the second reaction we can calculate the equivalent weight of $\ce B$ and $\ce M$ to be $\pu{39.75 g}$ and $\pu{31.7 g}$ respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the equivalent weight of $\ce B$ in the two reactions, the equivalent weight of $\ce B$ in the first reaction is twice the equivalent weight in second reaction. What exactly does this mean?&lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="7448" LastEditDate="2017-04-08T17:50:18.147" LastActivityDate="2017-04-08T17:54:11.963" Title="Different Equivalent Weight for the same compound" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10738" PostTypeId="1" AcceptedAnswerId="10740" CreationDate="2014-05-14T12:30:41.710" Score="1" ViewCount="3199" Body="&lt;p&gt;I am sure that the answer is (b) because I am familiar with the other rings and I think they are showing resonance (and it is a single choice question so it was easy to guess)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to know why is (b) not showing resonance (because I think it can as far as I know) and if I am wrong about the answer then please correct me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XNFfm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5484" LastEditorUserId="5026" LastEditDate="2018-02-28T14:36:37.433" LastActivityDate="2018-02-28T14:37:29.870" Title="Resonance stabilized carbanion" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10739" PostTypeId="2" ParentId="10722" CreationDate="2014-05-14T12:46:19.473" Score="26" Body="&lt;p&gt;Bond stability is the answer.&lt;/p&gt;&#xA;&lt;p&gt;The energy required to break a &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\pu{85.5 kcal/m}$&lt;/span&gt;; the energy given off when a &lt;span class=&quot;math-container&quot;&gt;$\ce{C=O}$&lt;/span&gt; bond is formed is around &lt;span class=&quot;math-container&quot;&gt;$\pu{178 kcal/m}$&lt;/span&gt; (&lt;a href=&quot;http://www.cem.msu.edu/%7Ereusch/OrgPage/bndenrgy.htm#dissbe&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).  These bond energies suggest that a carbonyl will generally be a few &lt;span class=&quot;math-container&quot;&gt;$\pu{kcal/mol}$&lt;/span&gt; more stable than the corresponding &lt;em&gt;gem&lt;/em&gt;-diol &lt;span class=&quot;math-container&quot;&gt;$[(2 \cdot 85.5) - 178 = \pu{-7 kcal/mol}]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;[Entropic effects also play a role (see @Martins comment), but while bond enthalpies will change within a series of compounds the entropic effects should be similar and roughly cancel out when we compare compounds.]&lt;/p&gt;&#xA;&lt;p&gt;However, the &lt;span class=&quot;math-container&quot;&gt;$\pu{7 kcal/mol}$&lt;/span&gt; difference noted above isn't large, so there is a delicate balance in the equilibrium between a carbonyl compound and its corresponding &lt;em&gt;gem&lt;/em&gt;-diol. Since &lt;span class=&quot;math-container&quot;&gt;$\pu{1.4 kcal/mol}$&lt;/span&gt; will shift an equilibrium by roughly a power of 10, small factors like dipole moments, inductive effects, &lt;em&gt;etc&lt;/em&gt;., are enough to stabilize one side of the equilibrium more than the other and significantly shift the equilibrium from one side to the other.&lt;/p&gt;&#xA;&lt;p&gt;If water is present, then an equilibrium will be established between a carbonyl compound and its corresponding &lt;em&gt;gem&lt;/em&gt;-diol.  Below is an interesting comparison of acetone and formaldehyde.  A seemingly small change (changing &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt;) shifts the equilibrium from one side to the other (by 6 powers of 10!).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6SRyU.jpg&quot; alt=&quot;carbonyl-diol equilibria of Acetone and Formaldehyde&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is a link to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Geminal_diol&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; that has a few more interesting examples of carbonyl-diol equilibria.&lt;/p&gt;&#xA;&lt;p&gt;In the case of chloral (&lt;span class=&quot;math-container&quot;&gt;$\ce{CCl3CHO}$&lt;/span&gt;) it is the strong electron withdrawing inductive effect of the &lt;span class=&quot;math-container&quot;&gt;$\ce{CCl3}$&lt;/span&gt; group that destabilizes the already positively polarized carbonyl carbon driving the equilibrium to the diol side.&lt;/p&gt;&#xA;&lt;p&gt;As an aside, if you're not familiar with the history of chloral (or as it is often called, chloral hydrate) google &amp;quot;Mickey Finn&amp;quot; or &amp;quot;knock out drops.&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;As I've mentioned above a change of just a few &lt;span class=&quot;math-container&quot;&gt;$\pu{kcal/mol}$&lt;/span&gt; can markedly shift the equilibrium between a carbonyl and the corresponding &lt;em&gt;gem&lt;/em&gt;-diol. Lest the reader think that &lt;em&gt;gem&lt;/em&gt;-diols are a rarity, here are a number of examples of compounds that exist predominately in the &lt;em&gt;gem&lt;/em&gt;-diol form:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ninhydrin&quot; rel=&quot;noreferrer&quot;&gt;ninhydrin&lt;/a&gt; (because of the H-bond)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{R-(CO)-(CO)-(CO)-R'}$&lt;/span&gt; (middle ketone houses the &lt;em&gt;gem&lt;/em&gt;-diol; &lt;a href=&quot;https://chemistry.stackexchange.com/a/42796/5026&quot;&gt;see detailed explanation&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;1,1-dihydroxycyclopropane (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; carbon relieves ring strain)&lt;/li&gt;&#xA;&lt;li&gt;floral hydrate (similar to chloral hydrate)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2020-07-24T17:34:37.327" LastActivityDate="2020-07-24T17:34:37.327" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="10740" PostTypeId="2" ParentId="10738" CreationDate="2014-05-14T12:48:40.573" Score="2" Body="&lt;p&gt;(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates &lt;a href=&quot;http://en.wikipedia.org/wiki/Bredt%27s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bredt's rule&lt;/a&gt;. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-02-28T14:37:29.870" LastActivityDate="2018-02-28T14:37:29.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10741" PostTypeId="2" ParentId="10738" CreationDate="2014-05-14T13:01:49.047" Score="2" Body="&lt;p&gt;You are right about compound B.  Try to build a model of the resonance form for B where charge is delocalized to the adjacent carbonyl.  Here is a picture of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rl25C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note what is called a &quot;bridgehead double bond&quot; in the resonance form on the right.  Such a structure would be a violation of Bredt's Rule, which states that double bonds can't be placed in bridgehead positions if there would be a lot of ring strain as a result.  Again, try to build a model of the molecule on the right and see how strained that bridgehead double bond makes the system.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-14T13:01:49.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10742" PostTypeId="2" ParentId="10738" CreationDate="2014-05-14T13:03:45.000" Score="-1" Body="&lt;p&gt;For any cyclic compound to exhibit resonance, the following rules have to be obeyed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $(4n+2)\pi$ electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Conjugation. Conjugation is not possible for either b) or d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Planarity. Compound b) is not planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, compound b) violates Bredt's rule as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, neither b) nor d) can be resonance stabilised.&lt;/p&gt;&#xA;" OwnerUserId="2321" LastEditorUserId="2321" LastEditDate="2014-05-14T13:08:51.537" LastActivityDate="2014-05-14T13:08:51.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10743" PostTypeId="2" ParentId="10736" CreationDate="2014-05-14T13:45:32.973" Score="4" Body="&lt;p&gt;When you compare the stability of the carbocation intermediates of polycyclic aromatic compounds, you have to consider if the aromaticity of one of the rings is sacrificed in order to delocalize the positive charge. If this is the case, the intermediate is less stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For naphthalene, for example, substitution at C1 is favored, because the cation is stabilized by allylic resonance and the aromatic character of the second ring is maintained. After attack of the electrophile E at C2, the charge in the resulting cation is delocalized without retaining the aromaticity of the second ring (it no longer has 6 π electrons). Hence, this C2-substituted intermediate is less stable (&lt;a href=&quot;http://depts.washington.edu/chemcrs/bulkdisk/chem238A_win05/notes_12_pt2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NNaAj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNaAj.png&quot; alt=&quot;Cationic intermediates in C1 and C2 electrophilic substitution of naphthalene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make analogous considerations for anthracene, except that there is a third position for substitution, namely C9 in the middle ring. Delocalization of the charge on this position creates a cation with two intact benzene rings, which favors substitution at this position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XCJFa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XCJFa.png&quot; alt=&quot;C9 substitution in anthracene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="16683" LastEditDate="2018-08-04T05:27:03.293" LastActivityDate="2018-08-04T05:27:03.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10744" PostTypeId="2" ParentId="10736" CreationDate="2014-05-14T13:55:27.967" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qeSkJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qeSkJ.png&quot; alt=&quot;Nitration of naphthalene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To supplement Jannis Andreska's excellent answer, you can consider the total number of resonance structures available to both carbocation intermediates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitration at position 1 produces a carbocation that has 7 total resonance structures, 4 of which appear to preserve the aromaticity of the second ring. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M6yKp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6yKp.png&quot; alt=&quot;Resonance structures for intermediate in 1-nitration of naphthalene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitration at position 2 produces a carbocation that has 6 total resonance structures, 2 of which appear to preserve the aromaticity of the second ring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wOFrj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wOFrj.png&quot; alt=&quot;Resonance structures for intermediate in 2-nitration of naphthalene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same analysis for anthracene, and you will probably find that nitration at position 9 (on the middle ring) is favored.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2018-08-04T05:27:09.830" LastActivityDate="2018-08-04T05:27:09.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10745" PostTypeId="2" ParentId="10736" CreationDate="2014-05-14T13:57:52.883" Score="6" Body="&lt;p&gt;In the electrophilic substitution of polycyclic aromatics, when drawing resonance structures &lt;strong&gt;keeping as many benezene rings intact as possible is important&lt;/strong&gt;.  Note too that a naphthalene ring isn't as &quot;good&quot; as two separate benzene rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the structures below, electrophilic attack at the 1 position in naphthalene (top row of drawing) allows you to draw 2 resonance structures with a benzene ring remaining intact (4 if we count structures with the double bonds simply shifted in the intact benzene ring); attack at the 2 position (middle row in figure) only has 1 resonance structure (2 if we count structures with the double bonds simply shifted in the intact benzene ring) with a full benzene ring intact.  Hence, electrophilic attack at the 1-poisition is preferred in naphthalene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same considerations to anthracene, we see that only attack at the central ring position allows 2 full benzene rings to remain intact in the possible resonance structures. Consequently electrophilic attack at anthracene's 9-position is preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uxISX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-21T17:05:07.867" LastActivityDate="2015-05-21T17:05:07.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10746" PostTypeId="1" CreationDate="2014-05-14T14:24:39.053" Score="1" ViewCount="812" Body="&lt;p&gt;I am doing a project where I heat up a container (around 500ml) filed with castor oil at room temperature. I want to add a pressure/safety valve that opens up when the castor oil reaches around 300 degrees Celsius (boiling temperature). In order to get the right valve I have to know how much the pressure increases with in that process. How can I calculate this? And were can I find the required data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something else I need to consider?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Like I have experienced, it seems impossible to find the required data. What if I used mineral oil instead?&lt;/p&gt;&#xA;" OwnerUserId="5156" LastEditorUserId="5156" LastEditDate="2014-05-17T09:32:49.213" LastActivityDate="2014-05-17T09:32:49.213" Title="Taking Castor Oil from room temperature to 300 degrees Celsius" Tags="&lt;heat&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10748" PostTypeId="1" AcceptedAnswerId="10751" CreationDate="2014-05-14T15:40:21.437" Score="7" ViewCount="4229" Body="&lt;p&gt;While going through my chemistry text book I came through the following reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{RCH2OH + (O) -&amp;gt; RCHO + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above reaction why oxygen atom is enclosed in parenthesis ?&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="189" LastEditDate="2014-05-14T15:54:31.117" LastActivityDate="2016-03-21T20:49:32.800" Title="Why oxygen is enclosed in parenthesis in some chemical reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10749" PostTypeId="2" ParentId="10732" CreationDate="2014-05-14T15:42:44.867" Score="-5" Body="&lt;p&gt;Try Avogadro &lt;a href=&quot;http://avogadro.cc/wiki/Get_Avogadro&quot; rel=&quot;nofollow&quot;&gt;http://avogadro.cc/wiki/Get_Avogadro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an advanced molecule editor and visualizer designed for cross-platform use in computational chemistry, molecular modeling, bioinformatics, materials science, and related areas. It offers flexible high quality rendering and a powerful plugin architecture.&lt;/p&gt;&#xA;" OwnerUserId="5501" LastActivityDate="2014-05-14T15:42:44.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10750" PostTypeId="2" ParentId="10723" CreationDate="2014-05-14T15:46:58.373" Score="2" Body="&lt;p&gt;Are you asking about catalytic hydrogenation or non-catalytic hydrogenation (e.g. diimide)?  For the time being, let's assume that your question is about the catalytic process.  All the catalyst does is lower the activation energy required to reach the transition state (TS) and the TS will look about the same independent of which isomer you start with.  Therefor, ground state stability differences between the two olefins will determine which isomer reacts faster.  Since trans-olefins are generally more stable than their corresponding cis isomers, it will take more energy (the difference in stability between the cis and trans isomers) for the more stable (lower energy) trans isomer to reach the TS, and less energy for the cis isomer to reach the TS.  Consequently cis isomers react somewhat faster than their trans-counterparts during catalytic hydrogenation.  I suspect that the same considerations would apply to non-catalytic processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, remember that in the catalytic process, depending upon catalyst, temperature, solvent, pressure, etc.; many other things can be going on, such as cis-trans isomerization of the starting material, positional isomerization, etc.  So if the reaction conditions are not chosen carefully and 1) if cis-trans isomerization is faster than hydrogenation and 2) let's say the cis isomer reacts 100 times faster than the trans; then both isomers could appear to give the same rate of hydrogenation because all hydrogenation would proceed through the cis isomer.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-14T15:46:58.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10751" PostTypeId="2" ParentId="10748" CreationDate="2014-05-14T16:02:27.297" Score="9" Body="&lt;p&gt;It represents oxygen from an oxidising agent, as opposed to from diatomic oxygen. It's normally done in square brackets like this:&#xA;$$\ce{CH3CH2OH + 2[O] → CH3COOH + H2O}$$ &#xA;Is the reaction between ethanol and potassium manganate (VII) to form ethanoic acid. The oxygen in the reaction is from the permanganate, not from diatomic oxygen. It kind of simplifies the reaction, so you only see the reaction of what you want -the ethanol and oxygen - instead of the whole reaction with the manganate there. (Manganate is an oxidising agent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your reaction the alcohol:&#xA;$\ce{RCH2OH}$ is reduced by the oxygen to form an aldehyde and water, using oxygen from an oxidising agent - the [O].&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditDate="2014-05-14T16:35:54.907" LastActivityDate="2014-05-14T16:35:54.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10752" PostTypeId="1" AcceptedAnswerId="10758" CreationDate="2014-05-14T16:13:08.290" Score="5" ViewCount="185" Body="&lt;p&gt;I am working on a problem in the context of an infinite symmetric potential barrier around the origin, with barriers at &lt;span class=&quot;math-container&quot;&gt;$x=- \frac{a}{2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x=\frac{a}{2}$&lt;/span&gt;. The wave function is a symmetric triangle-function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\Psi (x) = \begin{cases} 0 &amp;amp; x &amp;lt; - \frac{a}{2} \\ \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right) &amp;amp; -\frac{a}{2} \leq x \leq \frac{a}{2} \\ 0 &amp;amp; x &amp;gt; \frac{a}{2} \end{cases}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must calculate the probability of measuring a particle in the above eigenstate in the ground state (when &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt;). The way to attack this problem is to calculate the coefficient &lt;span class=&quot;math-container&quot;&gt;$c_1$&lt;/span&gt;, and then square it. &lt;strong&gt;Here follows my (wrong) attempt.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;c_n = \int\limits_{-\infty}^{\infty} \psi_n^*(x)\Psi (x)\,\mathrm dx&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand from the above expression that the &lt;span class=&quot;math-container&quot;&gt;$\psi_n^* (x)$&lt;/span&gt; is just the &quot;regular&quot; solution to the particle in a box. Given this symmetric potential, and that &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt; (odd &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, even function), I must set&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\psi _1^* (x) = \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for &lt;span class=&quot;math-container&quot;&gt;$c_n$&lt;/span&gt; becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;c_n = \int\limits_{-a/2}^{a/2}  \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a} \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;c_n = \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be solved by partial integration, setting &lt;span class=&quot;math-container&quot;&gt;$u'=\cos \left( \frac{\pi x}{a} \right)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v= \left( \frac{a}{2} - \lvert x \rvert \right)$&lt;/span&gt;. Hence, &lt;span class=&quot;math-container&quot;&gt;$u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v'=-1$&lt;/span&gt;. The integral itself then becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx = \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right)\left( \frac{a}{2} - \lvert x \rvert \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) (-1)\,\mathrm dx&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cleaning up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mbox{Integral} = \frac{a^2}{2 \pi} \sin \left( \frac{\pi x}{a} \right) - \lvert x \rvert \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right) - \frac{a^2}{\pi ^2} \cos \left( \frac{\pi x}{a} \right)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just need to insert the limits. However, we see quite easily from the above expression that upon inserting &lt;span class=&quot;math-container&quot;&gt;$x=\pm \frac{a}{2}$&lt;/span&gt;, the last term vanishes, since &lt;span class=&quot;math-container&quot;&gt;$\cos ( \pm\pi / 2 = 0)$&lt;/span&gt;. The sine terms will equal &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; when the limit is positive and &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; when the limit is negative. Hence, we are left with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mbox{Integral} = \left[ \frac{a^2}{2 \pi} - \frac{a}{2} \frac{a}{\pi} \right] - \left[ \frac{a^2}{2 \pi}(-1) - \lvert \frac{-a}{2} \rvert \frac{a}{\pi}(-1)  \right] = \frac{a^2}{2 \pi} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the original expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;c_n = \frac{2 \sqrt{6}}{a^2} \cdot \frac{a^2}{\pi} = \frac{2 \sqrt{6}}{\pi} \approx 1.56&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you see the problem. My &lt;span class=&quot;math-container&quot;&gt;$c_n$&lt;/span&gt; is larger than one, which will give me a rather un-physical probability. Am I wrong in this derivation in my assumption for &lt;span class=&quot;math-container&quot;&gt;$\psi _n^* (x)$&lt;/span&gt;? If not, then where?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wolfram Alpha gives me that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mbox{Integral} = \frac{2a^2}{\pi ^2} \\&#xA;\Rightarrow c_1 = \frac{2 \sqrt{6}}{a^2} \cdot \frac{2a^2}{\pi ^2} = \frac{4 \sqrt{6}}{\pi ^2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this gives me a probability of &lt;span class=&quot;math-container&quot;&gt;$c_1^2 \approx 0.986$&lt;/span&gt;, which at least has physical meaning. Also, logic tells me that there is a very high probability of finding the particle in the ground state, so this further backs up WA's answer.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="402" LastEditDate="2019-05-20T23:58:01.127" LastActivityDate="2019-05-20T23:58:01.127" Title="Probability of measuring a particle in the ground state: having trouble with the integration" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10753" PostTypeId="2" ParentId="10737" CreationDate="2014-05-14T16:18:05.387" Score="3" Body="&lt;p&gt;Equivalent weights are considering a redox equation with one mole equivalent of electrons. But some redox reactions require more than one mole equivalent of electrons to be equilibrated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you have the second reaction that leads to a weight equivalent of $1/2$ of that for the first reaction, it means that the reaction 2 has consumed $1/2$ of  $B$ compared to reaction 1 for one mole equivalent of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You still cannot proceed to the molecular weights on these basis, but you can now try to consider that the most likely case is that the first one involved only one electron and the second one two electrons. Thus you can write that this is consistent with: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A \rightarrow B + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(equilibrated by $\ce{\frac 14 O_{2} + e- \rightarrow {1\over 2} O^2-}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the second equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B + 2 e- -&amp;gt; M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(equilibrated with $\ce{H2 + O^2- -&amp;gt; H2O + 2 e-}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One this is said, one should consider the molar weight of $\ce M$ to be twice the equivalent weight of $\ce M$ calculated from the second equation, leading to a molar weight of 63.5 g/mol. Quite unsurprisingly, this is the molar weight of copper, so $M = Cu$. And going backwards, $\ce{B = CuO}$ with copper in its second oxidation state. We can check this, as the molecular weight of CuO is 76.5 g/mol, exactly the equivalent weight calculated for reaction 1. And thus $\ce{A = Cu2O}$ with copper in its first oxidation state.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="7448" LastEditDate="2017-04-08T17:54:11.963" LastActivityDate="2017-04-08T17:54:11.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10754" PostTypeId="2" ParentId="10748" CreationDate="2014-05-14T16:55:50.433" Score="3" Body="&lt;p&gt;$\ce{(O)}$ means nascent oxygen. When you use an oxidising agent, fresh Oxygen is produced which is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Nascent_state_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;nascent&lt;/a&gt; oxygen.  &lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="5078" LastEditDate="2016-03-21T20:49:32.800" LastActivityDate="2016-03-21T20:49:32.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10755" PostTypeId="1" AcceptedAnswerId="10767" CreationDate="2014-05-14T17:19:36.527" Score="1" ViewCount="8828" Body="&lt;p&gt;Which of the following has highest solubility in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. $ Ca(OH)_2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. $ Fe(OH)_3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. $ Cr(OH)_3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. $ Zn(OH)_2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also please explain how can we predict their solubilities when we are not provided with their solubility products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that calcium hydroxide is sparingly soluble. So it cannot be the right option. I think Iron hydroxide is also soluble to a good extent. I haven't got any idea about option C. I think D may be the right answer.&lt;/p&gt;&#xA;" OwnerUserId="5119" LastEditorUserId="5119" LastEditDate="2014-05-14T19:33:25.327" LastActivityDate="2016-02-16T14:32:20.563" Title="Which of the following has highest solubility in water?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10756" PostTypeId="1" AcceptedAnswerId="10772" CreationDate="2014-05-14T17:36:22.053" Score="2" ViewCount="6440" Body="&lt;p&gt;I just read about non-metallic oxides in my textbook which says that non-metallic oxides are a compound formed of an oxide ion and a non-metal. It also said that they are acidic in nature. It gave examples of carbon dioxide and sulfur dioxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then wouldn't hydrogen oxide (water) also be a non-metallic oxide and hence, acidic in nature. But most textbooks say that water is neutral. So how's that possible?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastActivityDate="2015-11-05T03:28:59.600" Title="is water a non-metallic oxide? and hence acidic in nature?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10757" PostTypeId="1" CreationDate="2014-05-14T17:49:57.130" Score="-1" ViewCount="856" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDeHH.jpg&quot; alt=&quot;How to add a methyl group and an ethyl group to an alkyne with 2 terminal hydrogen separately.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to add a methyl group and an ethyl group to an alkyne with 2 terminal hydrogens separately?&lt;/p&gt;&#xA;" OwnerUserId="5502" LastEditorUserId="5591" LastEditDate="2015-10-09T17:29:52.177" LastActivityDate="2015-10-09T17:29:52.177" Title="How to do this reaction - adding 2 different groups to an alkyne with 2 terminal hydrogens?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-04-08T17:56:36.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="10758" PostTypeId="2" ParentId="10752" CreationDate="2014-05-14T17:59:51.700" Score="5" Body="&lt;p&gt;I think your tactics using partial integration from the start is the problem because $\lvert x \rvert$ is not differentiable at $x=0$ so working with its derivative is not quite kosher (and $v' = -1$ is certainly wrong since for $x &amp;lt; 0$ the gradient of $-\lvert x \rvert$ is $1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you go about it step by step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  c_1 &amp;amp;= \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right) dx \\&#xA;      &amp;amp;= \frac{2 \sqrt{6}}{a^2} \biggl( \frac{a}{2} \underbrace{\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{1} } - \underbrace{\int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{2} } \biggr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, integral $I_1$ is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  I_{1} &amp;amp;= \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx \\&#xA;   &amp;amp;= \left[ \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right)  \right]_{-a/2}^{a/2} = \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) + \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) = \frac{2a}{\pi}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For integral $I_2$ I suggest that you use that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \lvert x \rvert = \begin{cases} x &amp;amp;  x \geq 0 \\ -x &amp;amp; x \leq 0 \end{cases}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and split up the integral into two integrals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  I_{2} &amp;amp;= \int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx \\&#xA;   &amp;amp;= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now use integration by parts with $u'=\cos \left( \frac{\pi x}{a} \right)$ ($\Rightarrow u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $) and $v= x$ ($\Rightarrow v'= 1 $) to solve the integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \int x \cos \left( \frac{\pi x}{a} \right) dx &amp;amp;= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) dx \\&#xA;  &amp;amp;= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  I_{2} &amp;amp;= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \\&#xA;        &amp;amp;= \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{0}^{a/2} - \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{0} \\&#xA;        &amp;amp;= \frac{a^2}{2 \pi} - \frac{a^2}{\pi^2} - \frac{a^2}{\pi^2} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} - \frac{2 a^2}{\pi^2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the equation for $c_1$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  c_1 &amp;amp;= \frac{2 \sqrt{6}}{a^2} \left( \frac{a}{2} I_{1} - I_{2} \right) \\&#xA;      &amp;amp;= \frac{2 \sqrt{6}}{a^2} \left( \frac{a^{2}}{\pi} - \frac{a^2}{\pi} + \frac{2 a^2}{\pi^2} \right) \\&#xA;      &amp;amp;= \frac{4 \sqrt{6}}{\pi^2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that is indeed what Wolfram Alpha gives you.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-05-15T11:53:51.523" LastActivityDate="2014-05-15T11:53:51.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10759" PostTypeId="2" ParentId="10757" CreationDate="2014-05-14T18:03:28.627" Score="3" Body="&lt;p&gt;No catalytic hydrogenation. Doing a sequential alkylation of the starting acetylene using sodamide and liquid ammonia with the appropriate alkyl bromide will do the trick.  Be sure to only use one equivalent of the first alkylating agent.  It doesn't matter whether you use $\ce{MeBr}$ or $\ce{EtBr}$ first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pHEsZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-14T18:22:30.563" LastActivityDate="2014-05-14T18:22:30.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10760" PostTypeId="1" CreationDate="2014-05-14T18:19:28.580" Score="1" ViewCount="648" Body="&lt;p&gt;What is the maximum percentage of carbon allowed in stainless steel 202 so that it does not cause brittleness?&lt;/p&gt;&#xA;" OwnerUserId="5503" LastEditorUserId="7920" LastEditDate="2014-10-08T19:38:02.073" LastActivityDate="2019-08-19T21:57:51.967" Title="Percentage of carbon in stainless steel to avoid brittleness" Tags="&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10763" PostTypeId="1" AcceptedAnswerId="10793" CreationDate="2014-05-14T19:02:49.973" Score="3" ViewCount="20292" Body="&lt;p&gt;Octan-1-ol is polar due to the presence of the $\ce{-OH}$ group, but it is immiscible with water. How is it possible?&lt;/p&gt;&#xA;" OwnerUserId="5504" LastEditorUserId="4945" LastEditDate="2014-07-29T10:48:03.587" LastActivityDate="2014-07-29T10:48:03.587" Title="Why 1-octanol is immiscible with water?" Tags="&lt;water&gt;&lt;polarity&gt;&lt;mixtures&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10764" PostTypeId="2" ParentId="10763" CreationDate="2014-05-14T19:09:55.473" Score="1" Body="&lt;p&gt;It is not soluble because the head (OH) is slightly polar and the whole chain is fat. 1-octanol is used to assess partition coefficient in environmental literature because it mimics 'well' animal fat tissue absorption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{partition} = \frac{[X_\ce{octanol}]}{[X_\ce{water}]}$$&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-05-14T19:09:55.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10765" PostTypeId="2" ParentId="10763" CreationDate="2014-05-14T19:13:54.847" Score="4" Body="&lt;p&gt;Sometimes chemistry is a balancing act.  In the case of 1-octanol the hydroxyl group contributes 17 units towards the molecular weight, while the 8-carbon tail contributes 113 units.  In this case the molecule is more non-polar (hydrocarbon like) than polar (water like).  As you make the chain shorter and shorter you continually increase the solubility of the alcohol in water.  For the very short chains, methanol and ethanol, they are completely miscible in water.  Take a look at the chart here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.solubilityofthings.com/water/alcohols&quot; rel=&quot;nofollow&quot;&gt;chart&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-14T19:13:54.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10766" PostTypeId="2" ParentId="10732" CreationDate="2014-05-14T19:16:34.200" Score="1" Body="&lt;p&gt;My wife used ChemDraw when she was in grad school and teaching. &lt;a href=&quot;http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemDraw/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemDraw/Default.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5505" LastActivityDate="2014-05-14T19:16:34.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10767" PostTypeId="2" ParentId="10755" CreationDate="2014-05-14T19:53:57.983" Score="1" Body="&lt;p&gt;You can't predict solubility reasoning, you have to remember &lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/solrules.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;solubility rules&lt;/strong&gt;&lt;/a&gt; in your case this one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;6&quot;&gt;&#xA;  &lt;li&gt;Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble. &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in this case $\ce{Ca(OH)2}$ even if slightly soluble theoretically should be the right answer.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="9961" LastEditDate="2015-12-09T22:41:48.257" LastActivityDate="2015-12-09T22:41:48.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10768" PostTypeId="1" AcceptedAnswerId="10769" CreationDate="2014-05-14T20:25:56.837" Score="0" ViewCount="3626" Body="&lt;p&gt;When looking over an old organic midterm I came across this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gK70A.jpg&quot; alt=&quot;Question&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/Oep4c.jpg&quot; alt=&quot;Signals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see I got half of it right, but honestly I'm not sure how I got that one right or how you would answer this in general.  Looking back at this now it appears that the first signal was split into a doublet and each peak was then split into a triplet, which makes sense for proton &lt;code&gt;c&lt;/code&gt; (first split by &lt;code&gt;b&lt;/code&gt; and then by &lt;code&gt;d&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one seems trickier.  It looks as though the signal was split into a quartet, and then each peak into a triplet.  To me that looks like it should be &lt;code&gt;d&lt;/code&gt; - split into the quartet by &lt;code&gt;e&lt;/code&gt; and then split into the triplets by &lt;code&gt;c&lt;/code&gt; and &lt;code&gt;b&lt;/code&gt;.  So why is the answer &lt;code&gt;b&lt;/code&gt;?  These J values wouldn't have been supplied for now reason - is there a way I could have used them?  More generally, is it possible (and if so how) to identify the type of proton and the structure of a molecule if you just had access to the NMR and the J values like this?&lt;/p&gt;&#xA;" OwnerUserId="4148" LastActivityDate="2014-05-14T21:01:05.073" Title="Identifying protons in NMR using the J values" Tags="&lt;organic-chemistry&gt;&lt;hydrogen&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10769" PostTypeId="2" ParentId="10768" CreationDate="2014-05-14T20:45:19.250" Score="3" Body="&lt;p&gt;The proton(s) in question absorbs at 5.95 ppm, the olefinic region - so it cannot be D which is expected to absorb upfield in the aliphatic region.  B is the correct answer.  Proton B would be split by proton C into a doublet with a 15.5 Hz separation (the coupling constant $\ce{J_{BC}}$).  Each line in the doublet would be further split into another doublet with a J=8 Hz spacing by proton A.  So now we have 4 lines, a doublet of doublets.  Each of these four lines will be further split into a triplet with 1.5 Hz spacing by the two (hence the triplet) D protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chemical shift of a proton and the observed splitting (e.g. the coupling constants, or J values) can go a long way in helping to determine a structure, particularly in relatively simple molecules.  As molecules get more complicated you can do advanced experiments to help with the structure determination.  One example would be spin decoupling experiments.  For example, if we &quot;know&quot; which signal is due to the D protons, we could irradiate them with energy during our nmr experiment.  If they are coupled to proton B (they are), then the triplet in the B signal should diappear and we would just see our doublet of doublets.  This type of experiment confirms which protons are coupled to which other protons.  There are many other advanced nmr techniques that can be used to help in structure elucidation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-14T20:45:19.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10770" PostTypeId="1" AcceptedAnswerId="10776" CreationDate="2014-05-14T20:46:18.953" Score="13" ViewCount="532" Body="&lt;p&gt;My textbook (&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0534389511&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Chemistry, 2nd Ed. Hornback, Joseph&lt;/a&gt;) states that if a hydrogen is located near the center of an aromatic ring it will experience an (often quite sizable) upfield shift, as seen in these images (borrowed from aforementioned textbook):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RUjtc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RUjtc.png&quot; alt=&quot;Examples of molecules where a shielded proton is located directly above an aromatic ring&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically it all makes sense - the &lt;a href=&quot;https://en.wikipedia.org/wiki/Aromatic_ring_current&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring current of the aromatic system&lt;/a&gt; shields these protons rather strongly. But is there a way to estimate what the actual upfield shift will be? For example, if provided with structure 1 and asked to estimate the chemical shift of the red hydrogen, how could you do so? Or do you just run the NMR and find out? My book doesn't give any advice on how to figure this out, just that it happens.&lt;/p&gt;&#xA;" OwnerUserId="4148" LastEditorUserId="16683" LastEditDate="2019-01-11T21:46:23.150" LastActivityDate="2019-01-11T21:50:12.093" Title="Can the amount of shielding in proton NMR caused by ring currents be predicted?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10771" PostTypeId="2" ParentId="10768" CreationDate="2014-05-14T21:01:05.073" Score="2" Body="&lt;p&gt;You can read the coupling constant directly from the NMR spectrum. Typically you would use processing software for this which makes it much easier, but you can do it from the spectra you have available as long as you know the field strength of the spectrometer. Your scale is in ppm in this case, to convert it to Hz you have to divide by the spectrometer frequency in MHz, e.g. 0.1 ppm on a 300 MHz spectrometer are 30 Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've marked where you would measure the coupling in your first example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4rABvl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a nice example of the typical pattern you get for a proton on a double bond that couples once via the double bond and once via a single bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second example is more complicated, the most important thing you should recognize is that it is not a real quartet. The four peaks are all roughly at equal height, if this was a true quartet they should follow Pascal's triangle and be at a ratio of 1:3:3:1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there are more than two couplings involved, and if the couplings are identical or multiples of each other they can shift the peaks onto each other. In this case you have the big 15.5 Hz coupling, that signal is then split up again by the 6 Hz coupling which results in four roughly equally spaced signals that are again split up by a small 1.5 Hz coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example the absolute chemical shift is a rather big help, both signals are around 6 ppm which pretty much tells you that those two are the two protons directly at the double bond.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-05-14T21:01:05.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10772" PostTypeId="2" ParentId="10756" CreationDate="2014-05-14T22:07:10.453" Score="6" Body="&lt;p&gt;When textbooks say that water is &quot;neutral,&quot; textbooks mean that the water meets this condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[H_3O^+]} = [\ce{HO^-}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water can be neutral, in that its hydronium ion concentration equals its hydroxide ion concentration. Water can also be acidic if there is carbon dioxide dissolved in the water. Carbon dioxide can dissolve in water simply by letting the water come into contact with the atmosphere. Water can be basic too if you put a basic solute in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water itself can also &lt;em&gt;act&lt;/em&gt; as a weak &lt;em&gt;Brønsted&lt;/em&gt; acid by donating a proton. This however is not a favorable process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water can also act as a &lt;em&gt;Lewis&lt;/em&gt; acid because the oxygen in water withdraws electron density from the hydrogens, giving each hydrogen a partial positive charge. This in turns makes the hydrogens &lt;em&gt;electrophilic&lt;/em&gt; - fitting the definition of Lewis acidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what your textbook is getting at is Lewis acidity. Consider carbon dioxide. This doesn't fit the Brønsted definition of acidity because it has no hydrogen protons to donate. So consider the partial charges on the carbon dioxide molecule. There is a positive partial charge on the carbon because the oxygens are more electronegative than carbon and have withdrawn electron density from the carbon. This makes the carbon &lt;em&gt;electrophilic&lt;/em&gt;. And hence carbon dioxide as a whole can be considered a Lewis acid (but do understand that it's the carbon specifically in carbon dioxide acting as a Lewis acid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general, your textbook says that non-metallic oxides tend to be acidic (in the Lewis sense) because the oxygens withdraw electron density from the non-metal (remember that oxygen is one of the most electronegative elements). This in turns creates partial positive charge(s) and makes the centers of these partial positive charges electrophilic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you could say that non-metallic oxides are &lt;em&gt;precursors&lt;/em&gt; to &lt;em&gt;Brønsted&lt;/em&gt; acids; consider this Lewis acid/base reaction. Remember that the central conceit of Lewis acid/base chemistry (coordination chemistry) is electron flow; nucleophiles attack electrophiles (not the other way around; that would be the fatal conceit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO_3 + H_2O -&amp;gt; H_2SO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have sulfur with an electronegativity (EN) of 2.6. Oxygen has an electronegativity of 3.5. The difference is 0.9. The bond is clearly polar covalent. In addition, we have not one but &lt;em&gt;three&lt;/em&gt; oxygens isolating electron density from the sulfur. Therefore, the sulfur most definitely has a partial positive charge and the oxygens partial negative charges. We say the oxygens are &lt;em&gt;nucleophilic&lt;/em&gt;. And back to water, we have already concluded the hydrogen atoms in water are &lt;em&gt;electrophilic&lt;/em&gt; due to their positive partial chage. The sulfur trioxide will attack the hydrogens and form the sulfuric acid molecule - a strong &lt;em&gt;Brønsted&lt;/em&gt; acid (its first ionization is complete in water solution). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-27T22:08:24.617" LastActivityDate="2015-10-27T22:08:24.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10773" PostTypeId="2" ParentId="10770" CreationDate="2014-05-14T22:42:21.470" Score="2" Body="&lt;p&gt;Magnetic anisotropy effects are very difficult to calculate. As such I don't believe there is a table that help you calculate the value so an experiment or a similar example is required. A look at &lt;a href=&quot;http://www.organic-chemistry.org/books/reviews/3540418776.shtm&quot; rel=&quot;nofollow&quot;&gt;Pretsch&lt;/a&gt; shows only one example of this system (structure 1 in your book). Running one of the structures in a NMR &lt;a href=&quot;http://www.nmrdb.org/predictornew&quot; rel=&quot;nofollow&quot;&gt;predictor&lt;/a&gt; gives results that are quite far from the experimental values and obviously don't take this effect into consideration&lt;/p&gt;&#xA;" OwnerUserId="1055" LastActivityDate="2014-05-14T22:42:21.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10774" PostTypeId="1" AcceptedAnswerId="10779" CreationDate="2014-05-14T23:15:04.367" Score="3" ViewCount="3049" Body="&lt;p&gt;This question touches on whether sulfur is hypervalent (whether it expands its octet when bonding). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first picture shows both sulfurs with an expanded octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that in the top half of the below picture, we are supposed to end up with sulfuric acid, but we actually end up with sulfuric acid plus two lone pairs (if I drew the mechanism correctly). Did I? (I actually separated out the top part of the picture into two parts - first, hydrogen attacking oxygen; second, oxygen attacking sulfur - so the arrow pushing wouldn't get confusing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second picture follows updated chemical fact. According to Wikipedia, sulfur trioxide has one double bond and two dative single bonds. Sulfuric acid is also a charge separated species with four single bonds, not two single bonds and two double bonds. I seem to end up with the proper product here without any hand-waving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8wBbX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="11367" LastEditDate="2017-05-08T04:40:19.787" LastActivityDate="2018-02-23T17:33:09.780" Title="Mechanism of sulfur trioxide reaction with water to make sulfuric acid" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10775" PostTypeId="1" CreationDate="2014-05-14T23:31:45.393" Score="2" ViewCount="2994" Body="&lt;p&gt;Given a stick spectrum for an unknown diatomic molecule, how do you determine the rotational constants $B_\nu$ for both the excited and ground states given that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{j}=B_{\nu}j(j+1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{cc}&#xA;\text{Line position}\, \mathrm{(cm^{-1})} &amp;amp; \text{relative intensity} \\&#xA;\hline&#xA;27128.20 &amp;amp; 1.47 \\&#xA;27171.74 &amp;amp; 2.35 \\&#xA;27211.40 &amp;amp; 3.13 \\&#xA;27247.18 &amp;amp; 3.34 \\&#xA;27279.08 &amp;amp; 2.62 \\&#xA;27331.24 &amp;amp; 1.00 \\&#xA;27351.50 &amp;amp; 2.62 \\&#xA;27367.88 &amp;amp; 3.34 \\&#xA;27380.38 &amp;amp; 3.13 \\&#xA;27389.00 &amp;amp; 2.35 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s5Ohy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user5507" LastEditorUserId="41328" LastEditDate="2017-08-05T17:16:51.493" LastActivityDate="2017-08-05T17:16:51.493" Title="How to calculate the rotational constants for different states from a stick spectrum?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10776" PostTypeId="2" ParentId="10770" CreationDate="2014-05-14T23:41:40.110" Score="13" Body="&lt;p&gt;Johnson and Bovey first proposed a theory describing the ring current effect,&lt;sup&gt;[1]&lt;/sup&gt; and in 1972 Haigh and Mallion published their often-cited paper on ring current shielding.&lt;sup&gt;[2]&lt;/sup&gt; A later review by these authors in 1979 describes the mathematical prediction of ring currents very well.&lt;sup&gt;[3]&lt;/sup&gt; You may often see this graphic describing ring current contributions to chemical shifts in textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IoWmN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in answer to your question, it is possible &lt;em&gt;in theory&lt;/em&gt; to make chemical shift predictions for nuclei within the shielded regions of ring currents, however &lt;em&gt;in practice&lt;/em&gt;, it is very difficult and prone to large errors. Calculations require a very thorough knowledge of distances and angles between the aromatic plane and the nucleus of interest. And a reasonable computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very few chemical shift prediction tables in texts or literature give examples of upfield-shifted aromatics, and none that I'm aware of give enough tabulated data to extend an even ballpark prediction. This makes it very, very difficult to make a prediction unless you have experience with the particular type of molecules you are dealing with. For example, I work with one research group who makes quite good estimations of chemical shifts for the inner ring proton within porphyrin systems, but this has evolved over many years of experience of dealing with this particular class of molecule. They would really be at a loss trying to predict any of the shifts for the 4 compounds you have shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most desktop chemical shift predictors use an additive model based on combined chemical shift contributions, and do not do a full calculation of ring current properties. In fact, many desktop predictors give very poor estimations for these types of protons for this very reason, as they do not distinguish shielded and deshielded regions arising from ring current. For small organic molecules the variables dictating the ring current contribution are difficult to predict, especially for those signals experiencing increased shielding, such as overhanging methyl groups. They require knowledge of distance from the centre of the aromatic ring and angle from the plane. This information is often used qualitatively in spectral analysis to construct a 3-dimensional structure of a molecule, in explaining upfielded shifts of certain nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In proteins, NMR shift prediction models are more inclusive, as the molecule conformations are often very highly constrained and well predicted. Many shift predictors for protein NMR includes a ring current contribution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;References&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Johnson, C. E.; Bovey, F. A. Calculation of Nuclear Magnetic Resonance Spectra of Aromatic Hydrocarbons. &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;1958,&lt;/strong&gt; &lt;em&gt;29&lt;/em&gt; (5), 1012–1014. &lt;a href=&quot;https://doi.org/10.1063/1.1744645&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1063/1.1744645&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Haigh, C. W.; Mallion, R. B. New tables of ‘ring current’ shielding in proton magnetic resonance. &lt;em&gt;Org. Magn. Reson.&lt;/em&gt; &lt;strong&gt;1972,&lt;/strong&gt; &lt;em&gt;4&lt;/em&gt; (2), 203–228. &lt;a href=&quot;https://doi.org/10.1002/mrc.1270040203&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1002/mrc.1270040203&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Haigh, C.; Mallion, R. Ring current theories in nuclear magnetic resonance. &lt;em&gt;Prog. Nucl. Magn. Reson. Spectrosc.&lt;/em&gt; &lt;strong&gt;1979,&lt;/strong&gt; &lt;em&gt;13&lt;/em&gt; (4), 303–344. &lt;a href=&quot;https://doi.org/10.1016/0079-6565(79)80010-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/0079-6565(79)80010-2&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="16683" LastEditDate="2019-01-11T21:50:12.093" LastActivityDate="2019-01-11T21:50:12.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10777" PostTypeId="1" CreationDate="2014-05-15T00:45:31.690" Score="0" ViewCount="545" Body="&lt;p&gt;im doing pcb etching process (for those who don't know pcb is that thing with electronics on it lmao) using Ferric Chloride and it's taking forever &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there any chemicals i can find that will speed up this process (could it also be found at home, salt or acid for instance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="5509" LastActivityDate="2014-11-22T21:27:56.583" Title="how to speed up Ferric Chloride and copper reaction" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10778" PostTypeId="2" ParentId="10775" CreationDate="2014-05-15T01:14:42.227" Score="5" Body="&lt;p&gt;In your question, it's not clear whether the quantum number assignments for the transitions are known in advance, so for generality I'll assume that you don't know what they are, and help you find them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Combination-differences&lt;/em&gt; is the most common paper-based method to determine line assignments (for more detailed analysis, a computer-aided fitting program such as PGopher is best). From the line assignments, it's easy to find the $B$-values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotational energy levels are given by&#xA;$$F_\nu(J)\approx B_\nu J(J+1)+D_\nu [J(J+1)]^2$$&#xA;where $B_\nu$ and $D_\nu$ are the rotational constant and centrifugal distortion constant associated with the manifold $\nu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denoting the initial and final states by $1$ and $2$ respectively, the $P,R$ transition energies are given by &#xA;$$P(J)=F_2(J-1)-F_1(J)&#xA;\\R(J)=F_2(J+1)-F_1(J).$$&#xA;Combination-differences is can be done in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If your rovibronic transition originates from an initial manifold with well-documented spectral characteristics (including $B_1$), note that &#xA;$$R(J-1)-P(J+1)=2 (2 J+1) \left(B_1-2 D_1 \left(J^2+J+1\right)\right)\approx2B_1(2J+1)$$&#xA;when $D_1$ is small, and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. This allows you to number the peaks as to their identity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If instead the upper state has well-known $B_2$, then note that&#xA;$$R(J)-P(J)=2 (2 J+1) \left(B_2-2 D_2 \left(J^2+J+1\right)\right)\approx2 B_2 (2 J+1)$$&#xA;and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. Again, this allows you to number the peaks as to their identity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once the quantum numbers are assigned, finding both $B_1$ and $B_2$ is easy. For example, $R(J)-R(J-1)=2 \left(B_2 (J+1)-B_1 J\right)$, so if you measure it for two values of $J$, then you have two unknowns and two datapoints, so just solve for $B_1$ and $B_2$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fun Pictures&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Just to give a simple idea of how spectra are simulated, the following Mathematica code does a basic rotational spectrum assuming Boltzmann statistics at a reasonable temperature (hit &lt;code&gt;Cell&lt;/code&gt;&gt;&lt;code&gt;Convert To&lt;/code&gt;&gt;&lt;code&gt;StandardForm&lt;/code&gt; to get it to render the subscripts correctly):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Subscript[F, k_][J_] := &#xA;  Subscript[B, k] J (J + 1) - Subscript[D, k] J^2 (J + 1)^2;&#xA;R[J_] := Subscript[F, 2][J + 1] - Subscript[F, 1][J];&#xA;P[J_] := Subscript[F, 2][J - 1] - Subscript[F, 1][J];&#xA;ListPlot[Transpose[&#xA;  Table[{{R[J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}, {P[&#xA;       J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}} /. {Subscript[&#xA;      D, k_] :&amp;gt; 0, Subscript[B, 1] -&amp;gt; 1, Subscript[B, 2] -&amp;gt; 0.9}, {J, &#xA;    0, 15}], {2, 1, 3}], Filling -&amp;gt; Axis]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ARMD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4987" LastEditorUserId="4987" LastEditDate="2014-05-15T01:28:20.627" LastActivityDate="2014-05-15T01:28:20.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10779" PostTypeId="2" ParentId="10774" CreationDate="2014-05-15T02:38:18.670" Score="5" Body="&lt;p&gt;I've drawn a more correct mechanism for the reaction of dilute $\ce{SO3}$ with water in the liquid phase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SvpCZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SvpCZ.png&quot; alt=&quot;SO3 + water&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO3 (aq) + 3H2O (l) -&amp;gt; SO4^2- (aq) + 2H3O+ (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO3}$ is a strong electrophile, enough to react quickly with water, which is a relatively weak nucleophile. A water molecule is added to the structure, facilitated by the dislocation of a $\ce{S=O}$ bonding electron pair in $\ce{SO3}$ to the oxygen atom, forming the intermediate structure shown. Additional water molecules then quickly abstract hydrogen atoms from the intermediate (actually, the first abstraction is far easier than the second, so it might have been more proper to show each proton abstraction separately in sequence). The final product of the reaction is sulfate anions and hydronium cations ($\ce{H2SO4}$ is a strong acid in water, so it doesn't really exist as a molecule. The second proton isn't nearly as acidic as the first, but in dilute solution both are ionized). The formal global reaction is shown below the mechanism. Notice that at every point the formal charge on the central sulfur atom is exactly the same, zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with regards to our little discussion about the actual structure of $\ce{SO3}$. Classical valence bond theory, enhanced with the concept of hypervalency and VSEPR theory, attributes the structure shown, with three double bonds. &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_trioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Wikipedia article for $\ce{SO3}$&lt;/a&gt; mentions another structure, &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1997/FT/a700708f#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;based on modern valence bond theory computational calculations&lt;/a&gt;, with one double and two single bonds. This structure is quite interesting as it does not invoke hypervalency. Indeed, hypervalency is thought to be an old concept, and not really necessary in more modern theoretical treatments of chemical bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one opts to display the structure according to modern valence theory, &lt;em&gt;they would do well to make it explicit. Almost all structural formulae today are drawn in the context of classical valence bond theory&lt;/em&gt;. Certain assumptions from the classical theory do not carry over to the modern theory. For example, the authors of the article &quot;&lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1997/CS/cs9972600087#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modern valence bond theory&quot;, Chem. Soc. Rev., 1997,26, 87-100&lt;/a&gt; (&lt;a href=&quot;http://molphys.org/files/articulo_2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;free access mirror&lt;/a&gt;) find that the bonding in $\ce{N2O}$, described classically as a mixture of the canonical forms $\ce{^{-}N= N^{+}=O}$ and $\ce{N≡N^{+}-O^{-}}$, is best described as $\ce{N≡N=O}$. Even with five bonds, the central nitrogen atom has a calculated occupancy close to 7, i.e. it is not hypervalent. Since the structural construction rules of classical and modern valence bond theory aren't the same, the structures generated from both do not mesh and cannot be compared directly. The simplest thing to do, then, is to draw $\ce{SO3}$ according to classical valence bond theory, with three double bonds.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="31775" LastEditDate="2018-02-23T17:33:09.780" LastActivityDate="2018-02-23T17:33:09.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10780" PostTypeId="1" CreationDate="2014-05-15T03:01:36.077" Score="2" ViewCount="2281" Body="&lt;p&gt;While studying dipole moment in organic chemistry, I got stuck at a particular statement written in the book : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symmetrical molecules without lone pairs of electrons will have &lt;span class=&quot;math-container&quot;&gt;$\mu = 0.$&lt;/span&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main problem was, why would they have &lt;span class=&quot;math-container&quot;&gt;$\mu =0$&lt;/span&gt; ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of Symmetrical molecules are possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book gave the example of &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt;. How is this a symmetrical molecule? &lt;/p&gt;&#xA;" OwnerUserId="5512" LastEditorUserId="16683" LastEditDate="2019-08-26T23:44:44.170" LastActivityDate="2019-08-27T01:06:53.757" Title="Dipole moment of symmetrical molecules" Tags="&lt;molecular-structure&gt;&lt;symmetry&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10782" PostTypeId="1" AcceptedAnswerId="10791" CreationDate="2014-05-15T03:20:13.583" Score="0" ViewCount="206" Body="&lt;p&gt;Got a few issues with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RIfxE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Lewis Structure: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NX1GU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, we have five lone pairs on which a hydrogen proton can land and form the solvent's conjugate acid. Let's consider them left to right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Oxygen lone pair. Adding a hydrogen proton here would make the oxygen have a formal charge of +1. Not exactly ideal for an atom which is so electronegative (even more EN than nitrogen is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) sp3 nitrogen. This seems like a good site; it parallels that the ammonia molecule. However, we would be creating a +1 formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) sp2 nitrogen. We again would be creating a +1 formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) sp nitrogen. And finally, we would be creating a +1 formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the more s-character, the more able is the hybridized atom able to stabilize &lt;em&gt;negative&lt;/em&gt; charge. It isn't called &lt;em&gt;electronegativity&lt;/em&gt; for nothing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, is electronegativity inversely related to electropositivity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So should the sp nitrogen be &lt;em&gt;least&lt;/em&gt; able to stabilize a positive formal charge? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From a Lewis acid/base consideration, we would say that each of these lone pair sites are pretty nucleophilic. So which is the most nucleophilic. We need the strongest electron donating tendency. And that appears to be the sp3 nitrogen because in an sp3 nitrogen, the electrons are further from the nucleus on a time-average basis than the electrons in an sp2 or sp nitrogen. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-15T03:30:05.887" LastActivityDate="2014-05-15T09:13:16.120" Title="Strongest B/L Acid Which Can Exist (Leveling Effect)" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10784" PostTypeId="1" AcceptedAnswerId="10797" CreationDate="2014-05-15T03:35:54.083" Score="2" ViewCount="2768" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A container at &lt;span class=&quot;math-container&quot;&gt;$30~^\circ\pu{C}$&lt;/span&gt; contains &lt;span class=&quot;math-container&quot;&gt;$\ce{HNC_{(g)}}$&lt;/span&gt; reacting to form &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN_{(g)}}$&lt;/span&gt;. Knowing that the specific reaction constant is &lt;span class=&quot;math-container&quot;&gt;$4.4\cdot10^{-4}~\pu{s-1}$&lt;/span&gt; and that you initially have &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HNC}$&lt;/span&gt;: how many moles of each gas are in the container after one hour and a half?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm pretty sure that the reaction is reversible here. However, I have no idea how to calculate how many moles of each gas will there be after &lt;span class=&quot;math-container&quot;&gt;$\pu{1.5 hrs}$&lt;/span&gt; (even if it was not reversible).&lt;/p&gt;&#xA;&lt;p&gt;I have &lt;em&gt;many&lt;/em&gt; doubts:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Did they skip the concentration because there's only one reactant?&lt;/li&gt;&#xA;&lt;li&gt;Does &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; take into account the &lt;span class=&quot;math-container&quot;&gt;$30~^\circ\pu{C}$&lt;/span&gt;?&lt;/li&gt;&#xA;&lt;li&gt;What's the equation for &amp;quot;change of concentration over time&amp;quot; here?&lt;/li&gt;&#xA;&lt;li&gt;How can I figure out how many moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{HNC}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; the container will have after 90 minutes, ignoring the fact that it's a reversible reaction?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5506" LastEditorUserId="83758" LastEditDate="2020-06-17T16:27:43.667" LastActivityDate="2020-06-17T16:27:43.667" Title="How to calculate the concentration after a certain timespan from the specific reaction constant?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="10785" PostTypeId="2" ParentId="10780" CreationDate="2014-05-15T03:52:59.230" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; is a flat triangle molecule, so it has mirror symmetry. Since the dipoles cancel out, the molecule does not have a dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="16683" LastEditDate="2019-08-26T23:44:30.963" LastActivityDate="2019-08-26T23:44:30.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10786" PostTypeId="2" ParentId="10780" CreationDate="2014-05-15T03:57:39.713" Score="3" Body="&lt;p&gt;Yes, the statement in your book is very vague.  Molecules are often classified by the symmetry elements they contain.  For example, &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; contains a &lt;span class=&quot;math-container&quot;&gt;$C_3$&lt;/span&gt; axis perpendicular to the plane containing the &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; molecule.  &quot;&lt;span class=&quot;math-container&quot;&gt;$C_3$&lt;/span&gt;&quot; means that if you rotate the molecule around this axis by &lt;span class=&quot;math-container&quot;&gt;$360^\circ/3 = 120^\circ$&lt;/span&gt;, you'll get a molecule that is indistinguishable from the molecule you started with.  &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; also contains 3 &lt;span class=&quot;math-container&quot;&gt;$\sigma_\mathrm{v}$&lt;/span&gt; planes of symmetry.  These planes contain the &lt;span class=&quot;math-container&quot;&gt;$C_3$&lt;/span&gt; axis and one of the &lt;span class=&quot;math-container&quot;&gt;$\ce{B-H}$&lt;/span&gt; bonds.  &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; also contains one &lt;span class=&quot;math-container&quot;&gt;$\sigma_\mathrm{h}$&lt;/span&gt; symmetry plane that is perpendicular to the &lt;span class=&quot;math-container&quot;&gt;$C_3$&lt;/span&gt; axis and contains the &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; molecule.  In addition to these symmetry elements, a number of other symmetry elements exist as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that only molecules that belong to symmetry classes (point groups) &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C_n$&lt;/span&gt; (the molecule only contains a &lt;span class=&quot;math-container&quot;&gt;$C_n$&lt;/span&gt; axis&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C_{n\mathrm v}$&lt;/span&gt; (the molecule contains a &lt;span class=&quot;math-container&quot;&gt;$C_n$&lt;/span&gt; axis and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; total &lt;span class=&quot;math-container&quot;&gt;$\sigma_\mathrm{v}$&lt;/span&gt; planes)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C_{\mathrm s}$&lt;/span&gt; (the molecule only has a plane of symmetry that contains&#xA;the entire molecule)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;can have a permanent dipole moment.  If the molecule belongs to any other point group it cannot have a dipole moment.  Since &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; contains a &lt;span class=&quot;math-container&quot;&gt;$\sigma_\mathrm{h}$&lt;/span&gt; plane of symmetry (&lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; belongs to point group &lt;span class=&quot;math-container&quot;&gt;$D_{3\mathrm h}$&lt;/span&gt;) - it therefore does not belong to groups &lt;span class=&quot;math-container&quot;&gt;$C_n$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C_{n\mathrm v}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$C_{\mathrm s}$&lt;/span&gt; - and cannot have a dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2019-08-27T01:06:53.757" LastActivityDate="2019-08-27T01:06:53.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10789" PostTypeId="2" ParentId="5836" CreationDate="2014-05-15T06:59:38.120" Score="2" Body="&lt;p&gt;Since you've written your own code, you will probably be able to judge this best, but I'll give you some pointers as to where I'd start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I've gathered, you simulate a &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_dynamics#Canonical_ensemble_.28NVT.29&quot; rel=&quot;nofollow&quot;&gt;NVT ensemble&lt;/a&gt; of a pair of chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all you have to decide whether you want explicit representation of the solvent or not. &lt;em&gt;Hint: The latter is computationally far more expensive and I doubt that you have implemented all the optimizations that you'd find in larger MD software.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because this choice can drastically influence the diffusion properties of your chemicals. In a coarse-grained force field they might still be able to diffuse around pretty much unhindered, while when using explicit atomistic representation of the solvent, the properties may change to a larger degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I'd do it would be to check whether the two areas of the molecules that react with each other are close to one another, meaning that some distance criteria $(d&amp;lt;d_\text{rxn.})$ has to apply. When this criteria is met, you can &lt;strong&gt;make the reaction happen&lt;/strong&gt; by &lt;strong&gt;creating new bonds&lt;/strong&gt; and &lt;strong&gt;changing the atom types&lt;/strong&gt; that are bonded on-the-fly. It wouldn't be so tricky to implement this part, I guess. &lt;em&gt;Caveat: This will only work if you have a reaction whose mechanism is close to the proposed scheme. You might introduce nontrivial errors otherwise.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can model the &quot;real&quot; concentration curves of all the substances, &lt;strong&gt;and&lt;/strong&gt; simultaneously change your system to reflect the change in number of molecules. I wouldn't be surprised that you'd find that the pressure would slowly decrease as the reaction takes place (since you keep the volume constant).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-05-15T06:59:38.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10790" PostTypeId="1" CreationDate="2014-05-15T07:07:19.460" Score="1" ViewCount="1653" Body="&lt;p&gt;I'm extracting $\ce{N-(4-butoxyphenol)-N'-(4-dimethylaminophenyl)thiourea}$ from thiombutosine (anti-leprosy drug) tablets. I'm doing this extraction with hot ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why are 4 successive 10mL extractions better than a single 40mL extraction?&lt;/p&gt;&#xA;" OwnerUserId="5518" LastEditorUserId="4945" LastEditDate="2014-05-15T09:32:41.720" LastActivityDate="2014-05-15T09:32:41.720" Title="Why are multiple extractions better than a single extraction?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ClosedDate="2014-05-15T10:38:18.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="10791" PostTypeId="2" ParentId="10782" CreationDate="2014-05-15T07:48:45.763" Score="3" Body="&lt;p&gt;Your reasoning is not entirely correct, as you consider that the charge is only located on the atom which bears the proton upon protonation. But nearby atoms, bonds and molecules will also participate in the stabilization of the positive charge, through several factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the gas phase (so considering no solution phase effects), you should at least consider that, in addition to the instrinsic capability to bear a proton of the atom (which goes as the inverse of electronegativity, as you state in your discussion), one has also to consider the following factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Resonance effects (delocalization of the charge on other atoms, that may be less electronegative than the atom directly bearing the proton). In the molecule given above, this would lead to a stabilization of the $sp$ and $sp^{2}$ hybridized nitrogens, as there would be a partial positive charge on the neighbouring carbon atom, as: $$\ce{RCH=N+H-R' &amp;lt;-&amp;gt; RC+H-NH-R'} $$ or $$\ce{RC#N+H &amp;lt;-&amp;gt; RC^{+}=NH}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inductive effects (or if you prefer availability of electrons in nearby atoms that can be displaced towards the charged site). In your constrained textbook example, this would be hard to indicate: there are a lot of heteroatoms (nitrogens and oxygens) and electron withdrawing groups (CN) which are likely to remove electrons from the carbon atoms, and in the central part of the molecule, these are likely to contribute to destabilize the molecule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Intramolecular hydrogen bonds. Although sometimes neglected, these can contribute quite a lot to the stabilization of the charge when mutiple protonation sites involving heteroatoms are present. Just as a reminder, in the gas phase, a hydrogen bond in a protonated molecule can reach up to 120 kJ/mol. In your example they would be hard to predict without modelling the complete structure, but one could consider a possible H-bond between the protonated imine ($sp^{2}$ nitrogen) and the alcohol oxygen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then if we were to consider solution phase, the interactions between neighbouring molecules and the propensity to form hydrogen bonds with the protonated molecule will also play a role on the possible protonation sites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the above considerations, one should consider that the most likely stable forms would be protonation on either the nitrile group ($sp$ nitrogen) or the imine group ($sp^{2}$ nitrogen), as both allow to put a partial positive charge on the neighbouting carbon atom. The protonated nitrile group is a protonation on an $sp$ nitrogen, with charge delocalization on an $sp$ carbon which means that the charge stabilization will be less efficient than for an imonnium ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one considers purely gas phase consideration, one can use the proton affinity scale to rank these various contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look for proton affinities in the &lt;a href=&quot;http://webbook.nist.gov/chemistry/&quot; rel=&quot;nofollow&quot;&gt;NIST database&lt;/a&gt; we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;for a secondary amine $\ce{H3C-NH-CH3}$ 929 kJ/mol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for a nitrile $\ce{H3C-C#N}$ 779 kJ/mol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately little experimental data is available for imines. But some theoretical data is available in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja984077n&quot; rel=&quot;nofollow&quot;&gt;this article by Hammerum &lt;em&gt;et al.&lt;/em&gt;&lt;/a&gt;. If one considers the methylimine $\ce{Z-CH3-CH=N-CH3}$ proton affinity is calculated at 943 kJ/mol, higher than dimethylamine. But on the other hand if one considers $\ce{H2C=N-CH3}$, its proton affinity is of 880 (&lt;em&gt;exp&lt;/em&gt;) or 902 (&lt;em&gt;theo&lt;/em&gt;), lower than dimethylamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So inductive effects clearly play a very important role in the order between amine protonation and imine protonation in the gas phase. In a really constrained molecule as above, they are likely to play a major role, as is the possibility to form hydrogen bonds in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one now puts in the possible formation of intermolecular hydrogen bonds, protonated secondary amines can make two hydrogen bonds with two hydrogen bond acceptors, whereas protonated secondary imines can only make one hydrogen bond. This will likely tip the balance towards amines and would make answer &lt;strong&gt;(4)&lt;/strong&gt; most likely in solution, whereas I would be quite at loss to give an answer in the gas phase.&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="4945" LastEditDate="2014-05-15T09:13:16.120" LastActivityDate="2014-05-15T09:13:16.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10792" PostTypeId="1" CreationDate="2014-05-15T07:50:49.947" Score="20" ViewCount="454" Body="&lt;p&gt;From time to time I do little pro bono lessons for second grade in math, physics, and chemistry. Recently I came across this question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a container is $2~\mathrm{mol}$ of $\ce{SO3}$ (sulfur trioxide). This compound under certain conditions breaks down into $\ce{SO2}$ (sulfur dioxide) and $\ce{O2}$ (oxygen). At the same those two resulting compounds react together back to form $\ce{SO3}$. In equilibrium we have $1.52~\mathrm{mol}$ of $\ce{SO3}$.&#xA;  How much of remaining resulting compounds do we get?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer: It is obvious that $0.48~\mathrm{mol}$ of $\ce{SO3}$ broke down into $0.48~\mathrm{mol}$  of  $\ce{SO2}$ and $0.24~\mathrm{mol}$ of $\ce{O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem: A few of students cannot grasp how could we start with $2~\mathrm{mol}$ of some substance and end up with $2.24~\mathrm{mol}$ of other substances. I have tried to explain to them that $\mathrm{mol}$ is not a quantity as they are used to as in weight or volume, but that a $\mathrm{mol}$ is a number of molecules. I have tried to simplify an example using just 2 molecules of $\ce{SO3}$ and showed that the break down to 2 molecules $\ce{SO2}$ and 1 molecule of $\ce{O2}$. That was understandable. But as soon as we got back to moles: &quot;How can we get more than we started with?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to explain this? Any experience? To me, it seems so obvious.  &lt;/p&gt;&#xA;" OwnerUserId="5519" LastEditorUserId="402" LastEditDate="2017-05-08T00:25:51.183" LastActivityDate="2017-05-08T02:28:49.797" Title="How to explain the concept of mol and reactions involving mol quantities" Tags="&lt;equilibrium&gt;&lt;molecules&gt;&lt;mole&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10793" PostTypeId="2" ParentId="10763" CreationDate="2014-05-15T07:53:27.043" Score="6" Body="&lt;p&gt;I just thought I'd add a visualisation of this concept. Electrostatic maps show the electrostatic potential at a given point of the molecule. We know water is very polar, and this is evident from its electrostatic map:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bs0h4.jpg&quot; alt=&quot;Electrostatic potential map of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at octanol though, we see that the vast majority of the molecule is non-polar (evenly distributed electrostatic charge):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GsmeK.jpg&quot; alt=&quot;Electrostatic potential map of 1-octanol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydroxy-group of octanol is indeed hydrophilic (polar), but the majority of the molecule wants to stay in the non-polar phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules with strongly hydrophilic &quot;heads&quot; and non-polar tails have unique properties in aqueous solutions, including some very important for life. Take phospholipids. In water, they self assemble into the basic structure of the cell membrane, the so-called phospholipid bilayer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Qkz4.gif&quot; alt=&quot;phospholipid bilayer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how the phospholipids arrange themselves to have their polar portion in contact with water (hence &quot;hydrophilic&quot; - &quot;loving water&quot;), while their non-polar portion is in contact only with the non-polar portions of the surrounding phospholipids (hence &quot;hydrophobic&quot; - &quot;fearing water&quot;).&lt;/p&gt;&#xA;" OwnerUserId="1160" LastActivityDate="2014-05-15T07:53:27.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10794" PostTypeId="2" ParentId="10792" CreationDate="2014-05-15T08:11:35.973" Score="3" Body="&lt;p&gt;Say you have a piece of bread. Lets refer to that piece of bread as one large crumb. Then you split this large crumb into 10 smaller crumbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start: one large crumb &lt;/p&gt;&#xA;&#xA;&lt;p&gt;End: 10 smaller crumbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More advanced example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have a piece of bread, consisting of three different kinds of bread (three pieces of each type of bread) that you just squeezed together in your hand so that they would stick. Then, by some magical force, the large multicrumb is split into its constituents (the three different kinds of bread). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start: 1 large multicrumb&lt;/p&gt;&#xA;&#xA;&lt;p&gt;End: $\ce{3Crumb_{type 1} + 3Crumb_{type 2} + 3Crumb_{type 3}}$&lt;/p&gt;&#xA;" OwnerUserId="5334" LastActivityDate="2014-05-15T08:11:35.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10795" PostTypeId="2" ParentId="10792" CreationDate="2014-05-15T10:17:05.403" Score="5" Body="&lt;p&gt;This is indeed a advanced exercise so maybe first you should made more easier exercises. In this case the student need to know &lt;strong&gt;mole concept&lt;/strong&gt; and &lt;strong&gt;equilibrium concept&lt;/strong&gt; that are not directly linked, the problem here is more related to equilibrium concept. For the future he/she should begin to reasoning following these steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the balanced equation:&#xA;$$\ce{2SO3 &amp;lt;=&amp;gt; 2SO2 + O2}$$&#xA;Then for simplify you could write it so, this is more easier to understand, always ask &lt;strong&gt;Can I write it so?&lt;/strong&gt;:&#xA;$$\ce{SO3 &amp;lt;=&amp;gt; SO2 + \frac{1}{2}O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you should ask him: &lt;strong&gt;Which is the difference between this notation:&lt;/strong&gt; $\ce{SO3 -&amp;gt; SO2 + \frac{1}{2}O2}$ [eq. 1]  &lt;strong&gt;and this&lt;/strong&gt; $\ce{SO3 &amp;lt;=&amp;gt; SO2 + \frac{1}{2}O2}$ &lt;strong&gt;?&lt;/strong&gt; You should focus  that in the first reaction all the reagent are transformed in the products  in the second only a part. So you can show him what happen if this is not an equilibrium. For the first case you can write this table:&#xA;\begin{array}{|c|c|c|c|c|}\hline&#xA; &amp;amp;\mathrm{SO}_{3}  &amp;amp; \mathrm{SO}_{2} &amp;amp; \mathrm{O}_2 \\ \hline&#xA;\mathrm{Initial}              &amp;amp; 2\ \mathrm{mol}   &amp;amp;  0 &amp;amp;0    \\ \hline&#xA;\mathrm{Variation} &amp;amp; \small{(0-2=)}-~2 &amp;amp; (2-0=)+2&amp;amp;(1-0=)+1 \\ \hline&#xA;\mathrm{At\ the\ end}         &amp;amp; 0~\mathrm{mol}&amp;amp; 2~\mathrm{mol}&amp;amp; 1~\mathrm{mol}\\ \hline&#xA;\end{array}&#xA;(Maybe write the initial row and let the student write the &quot;At the end&quot; row)&#xA;At this point you can talk about equilibrium: (the variation row may lead to confusion so maybe you could write a 3 row table). You can tell him to fill something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|c|}\hline&#xA; &amp;amp;\mathrm{SO}_{3}  &amp;amp; \mathrm{SO}_{2} &amp;amp; \mathrm{O}_2 \\ \hline&#xA;\mathrm{Initial}               &amp;amp; 2\ \mathrm{mol}   &amp;amp;  0 &amp;amp;0    \\ \hline&#xA;\mathrm{Variation} &amp;amp; ? &amp;amp; ?&amp;amp;? \\ \hline&#xA;\mathrm{At\ the\ end\ (equilibrium)}   &amp;amp; 1.52\ \mathrm{mol}      &amp;amp; ? &amp;amp; ?\\ \hline&#xA;\end{array}&#xA;At the end:&#xA;\begin{array}{|c|c|c|c|c|}\hline&#xA; &amp;amp;\mathrm{SO}_{3}  &amp;amp; \mathrm{SO}_{2} &amp;amp; \mathrm{O}_2 \\ \hline&#xA;\mathrm{Initial}              &amp;amp; 2\ \mathrm{mol}&amp;amp;  0 &amp;amp;0    \\ \hline&#xA;\mathrm{Variation} &amp;amp; -~(2-1.52\ \mathrm{mol}) &amp;amp; +~(2-1.52\ \mathrm{mol}) &amp;amp; +~\frac{(2-1.52\ \mathrm{mol})}{2} \\ \hline&#xA;\mathrm{At\ the\ end\ (equilibrium)}           &amp;amp; 1.52\ \mathrm{mol} &amp;amp; 2-1.52\ \mathrm{mol} &amp;amp; \frac{(2-1.52\ \mathrm{mol})}{2}\\ \hline&#xA;\end{array}&#xA;So the core concept: &lt;strong&gt;you have a fixed number of molecules one part is transformed in products one part is still reagent&lt;/strong&gt;.&#xA;So starting form 2 mol of reagents only a part of its can be transform in products. The goal is simply find this part (with a simple subtraction) and calculate the products using eq. 1.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Could be useful use metaphors, some times I use dozen of eggs. This could be dangerous please note dozen can be compare to Avogadro's number but not to mole in fact a mole is not simply a number is a &lt;strong&gt;&quot;number of&quot;&lt;/strong&gt; because only refers to elementary entities  (may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles, so not only molecules but neither orange or ping pong balls). With these exercise maybe you should NOT use metaphors: one misconception you should avoid is that it is a static equilibrium, tell him that as $SO_{3}$ molecules decompose they are formed but at equilibrium the rate is constant. This is a dynamic equilibrium!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="14" LastEditDate="2014-09-05T13:41:21.090" LastActivityDate="2014-09-05T13:41:21.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10796" PostTypeId="2" ParentId="10792" CreationDate="2014-05-15T13:08:04.360" Score="2" Body="&lt;p&gt;Currency transactions or making change can be viewed as examples of an equilibrium process.  Maybe this example might help the students see your point.  For example, $$\ce{4Q &amp;lt;=&amp;gt; 2Q + 3D + 4N}$$ $$\ce{4Q &amp;lt;=&amp;gt; 3Q + 25P}$$ $$\ce{4Q &amp;lt;=&amp;gt; 1DB}$$ $$\ce{Q=quarter, D=dime, P=penny, DB=dollar bill}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-15T20:49:53.420" LastActivityDate="2014-05-15T20:49:53.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10797" PostTypeId="2" ParentId="10784" CreationDate="2014-05-15T13:58:09.883" Score="0" Body="&lt;p&gt;To answer this, you should know what is the concentration of HNC after 90 minutes. First, you should know the initial concentration of HNC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, look at the rate constant. It tell you if the reaction is a first order. So, you can apply the data in integrated first-order rate law equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nX5OK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget if you must change 90 minutes to seconds :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, now you find the concentration of HNC after 90 minutes. To figure out the moles of HCN after reaction, we know that the initial moles of HNC is the same as the moles of HNC after reaction plus the moles of HNC that changed to HCN. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after submit the initial moles of HNC and moles of HNC after reaction, you get the moles of HNC that changed to HCN. This is the same of moles of HCN that&#xA;present in the container.&lt;/p&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-05-15T13:58:09.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10798" PostTypeId="1" AcceptedAnswerId="10950" CreationDate="2014-05-15T14:52:04.680" Score="4" ViewCount="1288" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Consider &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O_{(g)} -&amp;gt; H2O_{(l)}}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{105^{\circ} C}$&lt;/span&gt;. Under these conditions it is false that:&lt;/p&gt;&#xA;&lt;p&gt;(1) &lt;span class=&quot;math-container&quot;&gt;$T\Delta S_r$&lt;/span&gt; is negative.&lt;br /&gt;&#xA;(2) Relative humidity is 100%.&lt;br /&gt;&#xA;(3) &lt;span class=&quot;math-container&quot;&gt;$|\Delta H_r|&amp;gt;|T\Delta S_r|$&lt;/span&gt;&lt;br /&gt;&#xA;(4) &lt;span class=&quot;math-container&quot;&gt;$P_{\ce{H2O_{(g)}}}&amp;gt;\pu{1 atm}$&lt;/span&gt;&lt;br /&gt;&#xA;(5) &lt;span class=&quot;math-container&quot;&gt;$|{\Delta G_f}_{\ce{H2O_{(g)}}}|=|{\Delta G_f}_{\ce{H2O_{(l)}}}|$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Been trying to understand this question. Here is my reasoning:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;True, because &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is positive &lt;span class=&quot;math-container&quot;&gt;$\pu{(105 + 273)K}$&lt;/span&gt; while &lt;span class=&quot;math-container&quot;&gt;$ΔS$&lt;/span&gt; should obviously be negative because the final entropy is lower than the initial entropy. We are after all going from water's gaseous state to its liquid state. There are fewer degrees of freedom in water's liquid state.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is an equilibrium between water vapor and water liquid, then by definition relative humidity is 100%. The air is &lt;em&gt;saturated&lt;/em&gt; with water vapor to the point where it can't even hold the water vapor and water vapor exists in a dynamic equilibrium with liquid water. From Wikipedia:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Saturated air is really just air which contains water vapor in equilibrium with a liquid water source.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;False, because &lt;span class=&quot;math-container&quot;&gt;$ΔG$&lt;/span&gt;, not &lt;span class=&quot;math-container&quot;&gt;$ΔG^{\circ}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$ΔG$&lt;/span&gt; for a system at equilibrium is always &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$ΔG^{\circ}$&lt;/span&gt;, not necessarily &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;. So we have &lt;span class=&quot;math-container&quot;&gt;$0 = ΔG = ΔH - TΔS$&lt;/span&gt;. This implies &lt;span class=&quot;math-container&quot;&gt;$ΔH = TΔS$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;True. For liquid water to boil, vapor pressure must equal atmospheric pressure. Similarly for water to condense, atmospheric pressure must equal vapor pressure. Also, vapor pressure is solely dependent on temperature of the pure liquid. The below statement is from UC Davis' ChemWiki. I would think that adding solute would change the vapor pressure (and yes it does, and it's discussed further down the page of the site).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;em&gt;Vapor pressures are dependent only on temperature and nothing else.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Taking into consideration that water boils at &lt;span class=&quot;math-container&quot;&gt;$\pu{100^{\circ}C}$&lt;/span&gt;, we may conclude that at &lt;span class=&quot;math-container&quot;&gt;$\pu{105^{\circ}C}$&lt;/span&gt;, the vapor pressure of liquid water is greater than &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;. And therefore, for this equilibrium to exist, both the vapor pressure of liquid water and the atmospheric pressure of gaseous water must be equal and thus greater than &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;This is true. This would make &lt;span class=&quot;math-container&quot;&gt;$ΔG = 0$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$ΔG = G_{\text{(products)}} - G_{\text{(reactants)}}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I would like to know if my reasoning is correct.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="106377" LastEditDate="2021-05-13T14:44:11.510" LastActivityDate="2021-05-13T14:44:11.510" Title="Water-water vapour equilibrium above the boiling point" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10799" PostTypeId="1" AcceptedAnswerId="10800" CreationDate="2014-05-15T15:13:13.730" Score="1" ViewCount="1065" Body="&lt;p&gt;I had a visit from an air conditioner repairman who brought with him a tank of refrigerant. He measured the system and determined that the pressure was about 20 psi too low. So he added refrigerant to the system until it was right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it came time to pay him, he said, &quot;well I had to raise your system by 20 psi so it's going to be [20 x price for 1 psi].&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The price was fine, but I am still wondering: would it have taken the same amount of gas to raise my system from 0 to 20 psi as it would to go from 80 to 100, say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in other words, is it absurd to charge by pressure delta instead of some more reliable measure of what was dispensed from the tank?&lt;/p&gt;&#xA;" OwnerUserId="5520" LastEditorUserId="5520" LastEditDate="2014-05-16T01:20:50.223" LastActivityDate="2015-04-06T08:53:57.093" Title="Can a volume of gas be measured by change in PSI?" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10800" PostTypeId="2" ParentId="10799" CreationDate="2014-05-15T15:20:11.250" Score="2" Body="&lt;p&gt;Volume is inversely proportional to pressure only when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the number of moles of gas is held constant. We can say this is true in your situation because the repairman measured the pressure in the tank and I don’t think a significant amount of gas disappeared from the tank in the time it took for him to measure the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The temperature is held constant. Again, I don’t think temperature dropped 20 kelvin in the time it took for him to measure the pressure in the tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This follows from the ideal gas law equation – a good approximation most of the time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PV=nRT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I say your repairman still has his wits about him and isn’t trying to rip anyone off.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-04-01T12:45:18.157" LastActivityDate="2015-04-01T12:45:18.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10801" PostTypeId="2" ParentId="10799" CreationDate="2014-05-15T18:10:34.777" Score="1" Body="&lt;p&gt;In fact $PV=nRT$ works better when you deal with gas at high temperature and low pressure. In your case it depends on the pressure of your conditioner and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Compressibility_factor&quot; rel=&quot;nofollow&quot;&gt;compressibility factor&lt;/a&gt; of your refrigerant. If you know the refrigerant and his compressibility factor you can evaluate it.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2015-04-06T08:53:57.093" LastActivityDate="2015-04-06T08:53:57.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10802" PostTypeId="1" AcceptedAnswerId="10803" CreationDate="2014-05-15T18:17:48.947" Score="2" ViewCount="626" Body="&lt;p&gt;Would it be valid to describe $N_2O_5$ isomers as cis or trans? As a matter of fact, does this molecule even have isomers, given that all its outer N-O bonds are resonance stabilized and therefore equal in length and strength? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for non-resonance stabilized compounds, would it be okay to describe say the N to O double bonds as cis or trans with respect to each other? I've just never heard of cis/trans being used this way. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-15T20:08:49.060" Title="Cis and Trans Isomerism in Resonance-Stabilized Compounds" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10803" PostTypeId="2" ParentId="10802" CreationDate="2014-05-15T20:08:49.060" Score="4" Body="&lt;p&gt;Thanks for the picture, now I understand your question. All of the atoms in $\ce{N2O5}$ don't need to lie in a plane as is shown in the picture in your link.  At room temperature there would be rapid rotation about the 2 bonds connecting the central oxygen to the two nitrogen atoms, so there is no &quot;fixed&quot; geometry from which you could use the cis and trans references.  For example, imagine the conformation with one $\ce{NO2}$ group in the plane of the screen and the other perpendicular to the screen, even if each $\ce{NO2}$ group had one $\ce{N=O}$ double bond and one $\ce{N-O}$ single bond, the terms cis and trans would still have no meaning in this conformation. Further, as you noticed, all 4 bonds between nitrogen and the exterior oxygens are equivalent due to resonance, so again terms like cis and trans would have no meaning.  Finally, I can't imagine a situation where an $\ce{N=O}$ double bond (e.g. where the oxygen is unsubstituted) could have cis-trans isomers because the 2 lone pairs on the oxygen are equivalent.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-15T20:08:49.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10805" PostTypeId="1" CreationDate="2014-05-15T20:11:13.440" Score="1" ViewCount="66" Body="&lt;p&gt;Consider a structure like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R - Cy - Cy - Ph - OR'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Cy represents a cyclohexane ring, Ph is a phenyl ring, and R, OR' is alkyl group and oxygen bonded to different alkyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2 linearly bonded fluorines on top of the phenyl ring are moved to either cyclohexane ring, do you get the same effect?&#xA;What &quot;effect&quot; means I'm not sure, but this is related to liquid crystals having  $\Delta\varepsilon=\varepsilon_{\parallel}-\varepsilon_{\perp}&amp;lt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial guess is that because cyclohexane rings have $\sigma\rightarrow\sigma^*$ transitions, resulting in lower birefringence $\Delta n=n_e-n_0$, the effect is the not the same. I'm also assuming the dipole will be different? This is an organic chemistry question but sort of physics related so i wasn't sure where to put this question.&lt;/p&gt;&#xA;" OwnerUserId="5522" LastEditorUserId="9961" LastEditDate="2015-10-09T16:52:59.010" LastActivityDate="2015-10-09T16:52:59.010" Title="Liquid crystal structure: Equivalent effect of alkoxy group?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;&lt;optical-properties&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10807" PostTypeId="1" AcceptedAnswerId="10822" CreationDate="2014-05-15T20:53:26.033" Score="1" ViewCount="1067" Body="&lt;p&gt;Are all polar molecules both nucleophilic and electrophilic, depending on which atom you are talking about? Take water for example. Oxygen bears the partial negative charge. Oxygen is nucleophilic. Because oxygen isolates electron density from the hydrogens, the hydrogens bear partial positive charges. So would the hydrogens be electrophilic? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-16T06:11:54.343" Title="Identifying Nucleophiles and Electrophiles" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10808" PostTypeId="1" AcceptedAnswerId="10809" CreationDate="2014-05-15T21:29:25.843" Score="2" ViewCount="64" Body="&lt;p&gt;I am trying to solve a question that asks to compare Q vs. K. K = 4.2, and Q = 1.0.&lt;/p&gt;&#xA;&lt;p&gt;Problem Details:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;2.0 mol of Samples of A, B, and C&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{A + B \rightleftharpoons 2C}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that the reaction equilibrium is to the right, but, from this information, how can I tell if it is far to the right or slightly to the right?&lt;/p&gt;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-05-16T20:01:35.467" Title="How to tell how far equilibrium is?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10809" PostTypeId="2" ParentId="10808" CreationDate="2014-05-15T21:33:47.243" Score="2" Body="&lt;p&gt;Try writing out the K expression. Remember that in general, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${K = \frac{[product]}{[reactant]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, ${K = \frac{[C]^2}{[A][B]}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in other words, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = \frac{[product]^2}{[reacant][reactant]} = \frac{[product]^2}{[reacant]^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I combined the $[reactant]$ terms because the equation you have, there is 1:1 ratio of both reactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take the square root of both sides: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{K} = \sqrt\frac{[product]^2}{[reactant]^2} = \frac{[product]}{[reactant]}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now just plug in your numbers to determine how far to the right the equilibrium lies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{K} = \sqrt{4.2} = \frac{[product]}{[reactant]} = 2.0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore (upon consideration of reaction stoichiometry), there is a 2:1:1 ratio of products to reactants at equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also plug in the Q (reaction quotient) value; remember that K is just a special Q value (K is the Q value when the system is at equilibrium):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{Q} = \sqrt{1.0} = \frac{[product]}{[reactant]} = 1.0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Q value indicates that the system currently lies away from the equilibrium mix of a 2:1:1 ratio between products and reactant gases. At Q = 1.0 we have a 1:1:1 mix of A, B, and C. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-16T20:01:35.467" LastActivityDate="2014-05-16T20:01:35.467" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10810" PostTypeId="1" CreationDate="2014-05-15T22:27:41.417" Score="-2" ViewCount="5203" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the $\mathrm{pH}$ of a $\pu{0.500 M}$ solution of $\ce{KC2H3O2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am preparing for an exam on equilibrium and I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How would I go about calculating this?&lt;/li&gt;&#xA;&lt;li&gt;Is the $\ce{H3O}$ in the middle of $\ce{KC2H3O2}$ one hydroxide ion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is what I have attempted to do: find the conjugate acid/bases, write out a balanced equation, then take the $-\log{[\ce{H+}]}$ of the conjugate acid. But I am a little stuck on the best course of action to find the conjugate acid/base.&lt;/p&gt;&#xA;" OwnerUserId="5523" LastEditorUserId="41328" LastEditDate="2017-12-26T12:07:05.897" LastActivityDate="2017-12-26T12:08:09.143" Title="How to tell the pH of HC2H3O2?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10812" PostTypeId="1" CreationDate="2014-05-15T23:39:56.287" Score="3" ViewCount="1072" Body="&lt;p&gt;I'm making herbal and medicinal mushroom tinctures and extracts.  Normally I'll use everclear but it isn't cheap and 95% alcohol.  Also I don't know what that other 5% is.  Is it just water?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can reduce alcohol to near 100% without a still?  I've heard poring it over epsom salts through a few layers of coffee filters will remove some of the non alcohol contents.  Is this true?&lt;/p&gt;&#xA;" OwnerUserId="2301" LastEditorUserId="4945" LastEditDate="2014-05-16T03:00:58.350" LastActivityDate="2014-05-16T04:42:35.500" Title="What is a good way for a laymen to purify alcohol to near 200 proof?" Tags="&lt;everyday-chemistry&gt;&lt;purification&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10813" PostTypeId="2" ParentId="10232" CreationDate="2014-05-15T23:40:12.107" Score="2" Body="&lt;p&gt;I have asked experts -&gt; I am right. The common convention is to label the physical electrodes as cathode/anode based on &lt;strong&gt;discharge&lt;/strong&gt; even though that will be backwards when the battery is charging. That is a poor, confusing, and illogical naming convention, but it is what it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more clear and logical naming convention would be to name the physical electrodes either positive or negative since those labels apply in both charge/discharge directions. Some people use this logical naming convention, but the illogical convention seems more prevalent.&lt;/p&gt;&#xA;" OwnerUserId="4309" LastActivityDate="2014-05-15T23:40:12.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10815" PostTypeId="2" ParentId="10810" CreationDate="2014-05-15T23:59:19.520" Score="2" Body="&lt;p&gt;You need the $K_\mathrm{a}$ value of the solute of interest. That would be $\ce{[C2H3O2]-}$ as $\ce{K+}$ has highly limited Brønsted-Lowry acid/base properties in water. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="41328" LastEditDate="2017-12-26T12:08:09.143" LastActivityDate="2017-12-26T12:08:09.143" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10816" PostTypeId="2" ParentId="10812" CreationDate="2014-05-16T00:36:59.413" Score="3" Body="&lt;p&gt;Distillation of ethanol yields 95% ethanol and 5% water. That is the best one can do unless you don't want to drink it afterwards (it will have a little benzene, a known carcinogen). The Epsom salts may remove some water, but I don't think it will remove all the water, not to mention that some of the Epson salt will dissolve into solution.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-16T00:36:59.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10818" PostTypeId="2" ParentId="10812" CreationDate="2014-05-16T03:48:32.747" Score="3" Body="&lt;p&gt;Compared to &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/459836?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;anhydrous ethanol&lt;/a&gt; with about 100 US\$/L -&lt;a href=&quot;http://www.winechateau.com/~~~Brand~Everclear~1.html&quot; rel=&quot;nofollow&quot;&gt;everclear&lt;/a&gt; seems to be a very nice and cheap substitute (30 US\$/L). It is perfectly safe to drink &lt;strong&gt;&lt;em&gt;diluted&lt;/em&gt;&lt;/strong&gt; so the remaining stuff should be almost only water. (Almost undetectable) Trace amounts of essential oils might be present, but as it is made from grain, very unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anhydrous &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00IOHYTAS&quot; rel=&quot;nofollow&quot;&gt;Magnesium Sulphate is about 6 US\$ for a quarter pound&lt;/a&gt; (about 110 g) and you might need to add all of it. As LDC3 already mentioned you might solve some of it, not sure if you want that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Coffee is also kind of expensive and in my humble opinion far too precious to use as a filter. Yet again you will solve some of its contents in your solution.&lt;/strike&gt; (In response to LCD3's comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use your potions for outside application only, you could try and search for perfume base alcohol, which could be a little cheaper. It is &lt;a href=&quot;http://en.wikipedia.org/wiki/Denatured_alcohol&quot; rel=&quot;nofollow&quot;&gt;denatured&lt;/a&gt;, so it is absolutely not safe to drink it. There is also industrial cleaning alcohol on the market - but here again you cannot be sure what exactly is in there. It is very hard and expensive to get rid of the additives in denatured alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main price of alcohol comes from taxes anyway. If it is intended for drinking, the in most countries spirits are heavily taxed (luxury goods). Working around that is no good idea. When it comes to your own safety and your potions users, do not by too cheap - consequences could be far more expensive. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-16T04:42:35.500" LastActivityDate="2014-05-16T04:42:35.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10819" PostTypeId="1" AcceptedAnswerId="10820" CreationDate="2014-05-16T04:07:51.490" Score="6" ViewCount="23141" Body="&lt;p&gt;We had this question in a test and I am not sure of the answer&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q.&lt;/strong&gt; Temperature changes don't effect the extent of inter and intra molecular hydrogen bonding. State true or false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that raising temperature would increase the kinetic energy of molecules and would thus lead to weakening of &lt;em&gt;inter&lt;/em&gt; molecular hydrogen bonds. But am not sure what will happen to &lt;em&gt;intra&lt;/em&gt; molecular hydrogen bonds...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For e.g. if ortho-&lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrophenol&quot; rel=&quot;nofollow&quot;&gt;nitrophenol&lt;/a&gt; is heated, will the &lt;em&gt;intra&lt;/em&gt; molecular hydrogen bonds between hydrogen of the hydroxyl group and oxygen of the nitro group break?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas/suggestions?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="4160" LastEditDate="2016-05-10T15:44:17.463" LastActivityDate="2019-01-25T02:34:55.197" Title="Effect of temperature on Hydrogen Bonding?" Tags="&lt;bond&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10820" PostTypeId="2" ParentId="10819" CreationDate="2014-05-16T04:17:29.950" Score="8" Body="&lt;p&gt;Yes. Temperature disrupts bond of all kinds. Heat up a protein hot enough and you can even disrupt its primary structure - the linear sequence of amino acids, and amino acids are held together through covalent bonds. This is why prion contaminated organisms and instruments must be heated to extremely, extremely high temperatures. The question is whether or not the bond will break; the question is simply what temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example nitrous acid, which exhibits two unique $\ce{O-N-O}$ bond angles. One of these bond angles (the smaller one) results from intramolecular hydrogen bonding between the hydrogen and the oxygen on the other end. This occurs because there free-rotation is permitted around the single bonds in the nitrous acid molecule. The energy difference between the two structures is calculated to be $0.45~\mathrm{kcal~mol^{-1}}$ or $1880~\mathrm{J~mol^{-1}}$. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-05-16T04:57:27.657" LastActivityDate="2014-05-16T04:57:27.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10821" PostTypeId="2" ParentId="10810" CreationDate="2014-05-16T04:35:05.267" Score="6" Body="&lt;p&gt;First of all let us introduce the given compound in a more bonding descriptive way: $\ce{H3C-COOK}$. This compound is commonly referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;potassium acetate&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/f5/Potassium_acetate_skeletal.png/150px-Potassium_acetate_skeletal.png&quot; alt=&quot;structure of potassium acetate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reacts with water in the following fashion:&#xA;$$\ce{H3C-COOK_{(aq)} + H2O &amp;lt;=&amp;gt; H3C-COOH_{(aq)} + K+_{(aq)} + {}^{-}OH_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_mass_action&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass action law&lt;/a&gt; can be applied in the following way:&#xA;$$K_c=\frac{\ce{[H3C-COOH]}\ce{[K+]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}\ce{[H2O]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can then rearrange to form the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_dissociation_constant#Bases_and_basicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;base dissociation constant&lt;/a&gt;:&#xA;$$K_b=\frac{\ce{[H3C-COOH]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{p}K_b$ for potassium should be around $4.7$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As dissenter pointed out, you do not need the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; to  calculate $\mathrm{p}\ce{H}$ values for a pure solution of potassium acetate. You can have &#xA;a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Weak_base#A_typical_pH_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; for an example calculation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For your second question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydroxide&lt;/a&gt; Ion: $\ce{{}^{-}OH}$ or $\ce{OH-}$ or $\ce{HO-}$ ...&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydronium&lt;/a&gt; Ion: $\ce{H3+O}$ or $\ce{H3O+}$ or $\ce{H+OH2}$ ...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.770" LastActivityDate="2014-05-16T04:52:01.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10822" PostTypeId="2" ParentId="10807" CreationDate="2014-05-16T06:11:54.343" Score="3" Body="&lt;p&gt;Nucleophilicity and Electrophilicity are no absolute measurable quantities - they are based on an per entry basis and usually in reference to some kind of reaction. So basically you are right in assuming that polar molecules have both properties. In the special case of water these properties also stem from $$\ce{H2O &amp;lt;=&amp;gt; {}^{-}OH + H3+O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let us have a look at some molecules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CO}$ pretty good nucleophile especially towards transition metals. It has quite a small dipole towards the carbon, but a massive overload of electron density due to the $\sigma$-type HOMO&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CO2}$ also quite polar bonds, but no (permanent) dipole - still quite a potent electrophile given the right nucleophile&lt;/li&gt;&#xA;&lt;li&gt;$\ce{SO3}$ has no permanent dipole (but a transitional one) although quite polar bonds. However, it is only a strong electrophile.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NH3}$ quite polar, hence a reasonably good nucleophile. But it is also almost always way too strong a base to act as an electrophile&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NH(Et)2}$ not so different from ammonia, quite polar, but yet a terrible nucleophile, because it is too big&lt;/li&gt;&#xA;&lt;li&gt;$\ce{BH3}$ not very polar (no permanent dipole) but quite a good electrophile&lt;/li&gt;&#xA;&lt;li&gt;$\ce{I-}$ a monopol, very polarisable, good nucleophile, good leaving group.&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The consideration of the strength of nucleophiles and electrophiles lies in many variables. It might be polarity, local electron density, molecular orbitals, polarisabilities. It is often highly dependent on the solvent and the other reactants. If you add $\ce{I-}$ to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Acyl_chloride&quot; rel=&quot;nofollow&quot;&gt;acyl chloride&lt;/a&gt; you will certainly have a substitution to some extend:&#xA;$$\ce{R-COCl + I- &amp;lt;=&amp;gt; R-COI + Cl-}$$&#xA;If you add hydroxide ions, then you will certainly end up with the acid:&#xA;$$\ce{R-COCl + I- + {}^{-}OH &amp;lt;=&amp;gt; R-COOH + I- + Cl-}$$&#xA;Look at it nucleophilicity or electrophilicity from a case to case basis.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-16T06:11:54.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10823" PostTypeId="2" ParentId="6774" CreationDate="2014-05-16T15:09:30.050" Score="5" Body="&lt;p&gt;According to this &lt;a href=&quot;http://books.google.de/books?id=1vnk6J8knKkC&amp;amp;pg=PA971&amp;amp;lpg=PA971&amp;amp;dq=Hall-Heroult%20process%20%2b%20graphite%20anode&amp;amp;source=bl&amp;amp;ots=WHYTrhn-89&amp;amp;sig=92pHA53RuJTk8hFxIprOJ0wzBgI&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=qhJ2U-K7L8qE4gTp1IBI&amp;amp;ved=0CH4Q6AEwCw#v=onepage&amp;amp;q=Hall-Heroult%20process%20%2b%20graphite%20anode&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;, graphite is used as anode material because carbon is easier to oxidize than $\ce{O^2-}$. The anodic oxidation of carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C + 2O^2- \rightarrow CO2 + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;proceeds more efficiently than the oxidation of $\ce{O^2-}$ at an inert anode&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2O^2- \rightarrow O2 + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C}$ is a much better reducing agent than $\ce{O^2-}$ whose oxidation state is not changed in the reaction with graphite. As a consequence, anodic current density and the deposition rate of aluminium at the cathode will increase.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-16T15:09:30.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10825" PostTypeId="1" CreationDate="2014-05-16T16:35:54.907" Score="4" ViewCount="401" Body="&lt;p&gt;I'm currently working on a chemistry project in which I need to convert ethyl methyl carbonate to ethyl formate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DCXeL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCXeL.png&quot; alt=&quot;ethyl methyl carbonate ‐&amp;gt; ethyl formate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have a simple background in organic chemistry (and I haven't seen anything about reaction mechanisms yet), but I was guessing water could be used in this reaction, to produce &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3OH}$&lt;/span&gt; and replace the &lt;span class=&quot;math-container&quot;&gt;$\ce{-OCH3}$&lt;/span&gt; group by an &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; group. Then the next step would be to remove the oxygen from the  &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; group. I've seen that compounds such as &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; can do this, but I'm looking for a catalyst to make the process more economically viable, in case the proposed synthetic path is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to perform the conversion, or even by following another synthetic path? I've got no idea whether this process is correct. I'm also trying to keep the process as low energy-consuming as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree it would be better to start from ethanol and formic acid, but I have to start with this molecule.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="4945" LastEditDate="2020-02-13T19:22:42.467" LastActivityDate="2020-02-13T19:28:12.370" Title="How to convert an organic carbonate to its formate ester?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="10827" PostTypeId="1" AcceptedAnswerId="10834" CreationDate="2014-05-16T17:05:26.713" Score="4" ViewCount="338" Body="&lt;p&gt;Here is how I saw the Hamiltonian being written in one Quantum Mechanics book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{H} = -\frac{\hbar^2}{2m_\mathrm{e}r^2} \left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin \theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right] - \frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, that's all very nice and complicated, but looking at the last term, and seeing the use of $r$ in there,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;creates a small doubt in me - what is this $r$ there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrostatics, this was simply the distance between the two charged particles, and that would mean that this was the distance between the proton and the electron, but isn't the position of the electron non-deterministic? Shouldn't that thing be a probability distribution of the distance instead? And why isn't it squared?&lt;/p&gt;&#xA;" OwnerUserId="5530" LastEditorUserId="16683" LastEditDate="2016-09-19T09:13:17.000" LastActivityDate="2016-09-19T09:13:17.000" Title="Potential term in the hydrogen atom Hamiltonian" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10828" PostTypeId="2" ParentId="10602" CreationDate="2014-05-16T17:07:54.110" Score="7" Body="&lt;p&gt;When a water molecule coordinates to $\ce{Fe^3+}$, one non-bonded $sp^3$ hybrid orbital (with one of the 2 lone pairs) of oxygen overlaps with an empty $d$ or $p$ orbital of $\ce{Fe^3+}$. Electron density is transferred from oxygen to iron, and due to the electronegativity of oxygen, the polarization of the $\ce{O-H}$ bond in the water ligand is increased (more electron density is &quot;pulled&quot; from $\ce{H}$ towards $\ce{O}$). The hydrogens are more positively polarized than in a free (uncoordinated) water molecule, and this increases their electrophilic character and acidity to the point that they are readily abstracted by solvent (non-coordinated) $\ce{H2O}$ in aqueous solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(H2O)6^3+ + H2O \rightleftharpoons  Fe(H2O)5(OH)^2+ + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(H2O)5(OH)^2+ + H2O \rightleftharpoons  Fe(H2O)4(OH)2^+ + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further deprotonation results in the neutral complex $\ce{Fe(H2O)3(OH)3}$, from which $\ce{Fe(OH)3}$ precipitates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aqueous solution of $\ce{Fe^3+}$ aquo complexes is fairly acidic, wth a pH of up to 1.5 depending on iron concentration. The first dissociation equilibrium lies mainly on the product side, which can also be seen in the color of the solution: Pure $\ce{Fe(H2O)6^3+}$ is pale violet and found in crystals of some hydrated $\ce{Fe(III)}$ salts, while aqueous solutions get their orange color from $\ce{Fe(H2O)5(OH)^2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have mentioned ion charge (oxidation state) and charge density (which is proportional to ion size) of the metal center as potential factors of influence. One can expect that with increased charge and/or charge density, the polarization of the water ligands also increases, and thus the aquo complex should be more acidic. This is true for many aquo complexes; however, there are also some deviations from this rule, like $\ce{Cr(H2O)6^2+}$ and $\ce{V(H2O)6^2+}$ (&lt;a href=&quot;http://www.chemguide.co.uk/inorganic/complexions/acidity.html&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="4231" LastEditDate="2015-03-03T23:11:09.530" LastActivityDate="2015-03-03T23:11:09.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10829" PostTypeId="1" AcceptedAnswerId="10848" CreationDate="2014-05-16T17:25:48.290" Score="1" ViewCount="845" Body="&lt;p&gt;I have a gold-plated yoyo bearing which needs cleaning. Most people clean regular non-gold-plated bearings with mineral spirits (naphtha), lighter fluid (hexane/heptane), and acetone, but these gold bearings are somewhat new and we don't have a consensus on a cleaning method. There's debate as to whether the gold plating influences spin time or if it's the precision manufacturing, but they do spin forever and are quite silent. Anyway, these bearings are $22, so constant replacement is an enormous barrier to using these outside of competitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've researched gold cleaning techniques, and all the results say to rinse with water. That's not a very good option since all yoyo bearings have tight spaces where dust gets trapped. To understand this particular bearing's sensitivity: I did try rinsing with tap water last night, and now it barely spins. It seems even the contaminants in the tap water were enough to block up this bearing. Also, while most bearings can be used with a very thin lubricant to prolong life and decrease noise, this one cannot. I could try distilled water, but we really need a precision solution.&lt;/p&gt;&#xA;" OwnerUserId="5528" LastActivityDate="2016-03-09T20:13:11.013" Title="Best way to clean a gold-plated yoyo bearing?" Tags="&lt;solvents&gt;&lt;cleaning&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10830" PostTypeId="1" CreationDate="2014-05-16T18:14:29.840" Score="2" ViewCount="6733" Body="&lt;p&gt;I'm building a home lab around &lt;em&gt;All Lab, No Lecture Illustrated Guide to Home Chemistry Experiments&lt;/em&gt; and I'd like to understand the chemistry that I'm doing a bit better. For example, here's an &lt;a href=&quot;http://www.homechemlab.com/iodine.html&quot; rel=&quot;nofollow&quot;&gt;iodine experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Weigh out 2.0 g of potassium iodide and transfer it to a test tube.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Add about 1.5 mL of distilled water to the test tube and swirl to dissolve the potassium iodide.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Add 1.5 mL of concentrated hydrochloric acid (or about 1.8 mL of hardware store muriatic acid) to the test tube and swirl to mix the&#xA;  solutions.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Add about 10 mL of drugstore 3% hydrogen peroxide. The solution immediately turns dark brown as the iodide ions are oxidized to&#xA;  elemental iodine, which precipitates out.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Swirl the test tube to suspend the iodine and pour the liquid through a funnel with a folded piece of filter paper to capture the&#xA;  iodine crystals.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Rinse the iodine crystals on the filter paper several times with a few mL of distilled water. The rinse solution appears brown from&#xA;  dissolved iodine, but iodine is not very soluble in water, so you're&#xA;  not losing much of your yield.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Spread out the filter paper on a watch glass or saucer and allow the crystals to dry thoroughly. Iodine gradually sublimates (passes&#xA;  directly from solid to gaseous form) at room temperature, so don't&#xA;  leave the crystals exposed to air any longer than necessary to dry&#xA;  them.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Once the crystals are dry, transfer them to a sealed storage bottle or vial.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's easy enough to follow and I know I have $$\ce{KI, H2O, HCl, H2O2}$$ with a result of $$\ce{I2\, +\, ???}$$ but then what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had HS Chemistry, skipped Chem in college :(, so I know how to balance equations, convert from g or ml to moles, etc, but where can I learn to translate an experiment into equations without having to ask others (Textbooks or online resources are ideal, going and taking chem at a local uni. is a distant second, since I work fulltime)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I'm sticking to inorganic chem, so that should simplify things a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently studying &lt;a href=&quot;https://www.edx.org/course/mitx/mitx-3-091x-introduction-solid-state-1751&quot; rel=&quot;nofollow&quot;&gt;EdX's Solid State Chemistry&lt;/a&gt;, so I hope that'll help some.&lt;/p&gt;&#xA;" OwnerUserId="5362" LastEditorUserId="418" LastEditDate="2015-01-08T08:40:47.830" LastActivityDate="2016-09-20T09:00:11.630" Title="Equation for the reaction of iodide to iodine using hydrogen peroxide under acidic conditions?" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10831" PostTypeId="1" AcceptedAnswerId="10832" CreationDate="2014-05-16T18:33:00.273" Score="3" ViewCount="2356" Body="&lt;p&gt;Most computer chips contain or are made of silicon. Why silicon? Why not copper or tellurium or some such?&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="5026" LastEditDate="2018-07-03T02:52:37.293" LastActivityDate="2018-07-03T02:52:37.293" Title="Why is silicon used in computer chips?" Tags="&lt;elements&gt;&lt;semiconductors&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10832" PostTypeId="2" ParentId="10831" CreationDate="2014-05-16T19:09:08.690" Score="6" Body="&lt;p&gt;Most electron components of computer chips and boards are &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiconductor_device&quot; rel=&quot;noreferrer&quot;&gt;semiconductor devices&lt;/a&gt; like &lt;a href=&quot;http://en.wikipedia.org/wiki/Transistor&quot; rel=&quot;noreferrer&quot;&gt;transistors&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Diode&quot; rel=&quot;noreferrer&quot;&gt;diodes&lt;/a&gt;.&#xA;Silicon is, from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiconductor_device&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on semiconductor devices&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By far, silicon (Si) is the most widely used material in semiconductor devices. Its combination of low raw material cost, relatively simple processing, and a useful temperature range make it currently the best compromise among the various competing materials&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Silicon is the &lt;a href=&quot;http://education.jlab.org/glossary/abund_ele.html&quot; rel=&quot;noreferrer&quot;&gt;second most common element in the earth's crust&lt;/a&gt; after oxygen, and present in a variety of &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicate_minerals&quot; rel=&quot;noreferrer&quot;&gt;silicate minerals&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Silicon#Electronic_grade&quot; rel=&quot;noreferrer&quot;&gt;Electronics grade silicon&lt;/a&gt; can be produced easily by molten salt electrolysis of these minerals followed by purification by zone refining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is a conductor, and therefor not suitable for semiconductor devices. Tellurium is &lt;a href=&quot;http://en.wikipedia.org/wiki/Tellurium#Occurrence&quot; rel=&quot;noreferrer&quot;&gt;one of the rarest elements&lt;/a&gt; in the earth's crust.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-16T19:09:08.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10833" PostTypeId="1" AcceptedAnswerId="14192" CreationDate="2014-05-16T19:17:04.463" Score="17" ViewCount="39550" Body="&lt;p&gt;I am confused about the migratory aptitude of various groups, as there are many different orders for the same given in different places, especially about -Ph and -H. I would like to know if someone could reliably tell the order. &lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5026" LastEditDate="2018-05-21T02:13:26.207" LastActivityDate="2018-05-21T02:13:26.207" Title="Migratory aptitude in pinacol-pinacolone rearrangement" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;carbocation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10834" PostTypeId="2" ParentId="10827" CreationDate="2014-05-16T19:42:59.163" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the $r$ there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$r$ is the same in the potential energy term of the Hamiltonian as in the kinetic energy term of the Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r$ is the distance from electron to the proton.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But isn't the position of the electron non-deterministic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the generally accepted Copenhagen interpretation of quantum mechanics, the position of the electron is not deterministic. However, expressing the kinetic and potential energy of the the electron at each position in space does not contradict this interpretation. It is only saying, for each point, if the electron is at this point, the energy is _.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shouldn't that thing be a probability distribution of the distance instead?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wave function solutions represent probability amplitude in the Copenhagen interpretation.  The square of the absolute value of the wavefunction represents a probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why isn't it squared?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Force would be inversely proportion to $r^2$, but potential is inversely proportional to $r$.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="194" LastEditDate="2016-09-18T21:23:33.957" LastActivityDate="2016-09-18T21:23:33.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10835" PostTypeId="1" AcceptedAnswerId="10836" CreationDate="2014-05-16T19:49:06.980" Score="4" ViewCount="10564" Body="&lt;p&gt;I got a problem with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcURe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we stuck with the IUPAC name and called the isopropyl group a 1-methylethyl group? That would definitely &lt;em&gt;reverse&lt;/em&gt; the alphabetical priorities; because the substituents here on the hydrocarbon are on equivalent positions, we must give the lower number to the substituent that comes first in the alphabet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's the rub! (No isopropyl alcohol pun!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isopropyl - &quot;I&quot; comes before &quot;M&quot; as in methyl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we name the isopropyl group systematically, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methylethyl comes after just methyl in the alphabet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So could the name of this also be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-06-10T11:16:00.443" LastActivityDate="2018-02-12T03:00:07.853" Title="Nomenclature - Common vs IUPAC Names and Alphabetical Priorities" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10836" PostTypeId="2" ParentId="10835" CreationDate="2014-05-16T20:07:03.967" Score="3" Body="&lt;p&gt;No, 2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane is not an acceptable name.  Like the rule you quoted says, start numbering at a bridgehead carbon and first pass through the largest ring (in this case the 5-membered ring) and number that ring so that a substituent will have the lowest number.  So this molecule would be named 2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane.  Usage of &quot;1-methylethyl&quot; is not acceptable.  If you weren't going to use the term &quot;isopropyl&quot;, you could call the fragment &quot;2-propyl&quot;. So you could write 2-(2-propyl)-7,7-dimethyl-bicyclo[3.2.0]heptane, but the term &quot;isopropyl&quot; is much more common usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: After reading Greg E's comment, I looked around the internet and found that &quot;1-methylethyl&quot; is both correct and the official IUPAC name.  Here is a paraphrasing of the relevant IUPAC rule...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These groups (those attached off the main chain) are always numbered so that the point of attachment to the rest of the molecule is C1, with the longest continuous chain beginning from the point of attachment taken as the parent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then all sites I visited would add something like this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic (or common, or trivial) names.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of these sites seemed to (strongly) favor the trivial name.  As one site put it, &quot;Common names are used because they are easier to say, shorter, and save paper&#xA;and ink in scientific publications when they have to be used repeatedly in manuscripts.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-17T15:03:33.213" LastActivityDate="2014-05-17T15:03:33.213" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10839" PostTypeId="2" ParentId="10829" CreationDate="2014-05-16T23:59:44.063" Score="0" Body="&lt;p&gt;You could use the same method they use to clean gold rings, gold-plated  watchbands, etc.  Get an ultrasonic cleaner and put in distilled water with a few drops (depending on the tank size) of dishwasher detergent.  This will remove oil, grease, contaminants, etc.  If you don't have an ultrasonic bath, just stop in at the jewelry store at the mall and 9 times out of 10 they'll be nice enough to clean your bearing in less than 5 minutes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-16T23:59:44.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10840" PostTypeId="1" CreationDate="2014-05-17T03:09:44.743" Score="3" ViewCount="735" Body="&lt;p&gt;For instance, if there is cup of water saturated with a salt to the point where any more added salt will just float around in the solution, are those undissolved salt molecules in equilibrium with the dissolved ones? For instance, do the undissolved ones become partially dissolved for a moment, while the dissolved ones swap places with them? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does this behave in such a manner that whatever salts were dissolved first to reach saturation are the ones that will stay dissolved, in spite of other added salts? &lt;/p&gt;&#xA;" OwnerUserId="5534" LastActivityDate="2014-05-17T11:25:03.767" Title="Do undissolved salts in saturated solvents enter equilibrium with the dissolved salts?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10841" PostTypeId="2" ParentId="10840" CreationDate="2014-05-17T03:34:10.920" Score="2" Body="&lt;p&gt;Yes. At the solution saturation point there is a dynamic equilibrium between the solid and solvated solutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the IUPAC Gold Book: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;saturated solution:&#xA;A solution which has the same concentration of a solute as one that is in equilibrium with undissolved solute at specified values of the temperature and pressure.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-17T03:40:48.140" LastActivityDate="2014-05-17T03:40:48.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10842" PostTypeId="2" ParentId="10840" CreationDate="2014-05-17T03:36:29.807" Score="4" Body="&lt;p&gt;Yes, they enter into a constant equilibrium. Just as we write equilibrium expressions for insoluble salts as equilibrium expressions, like $\ce{CaF2} \leftrightharpoons \ce{Ca}^{2+} + \ce{2F-}$, when you reach the saturation point, solute particles are in dynamic equilibrium between aqueous and solid phase.&lt;/p&gt;&#xA;" OwnerUserId="4455" LastActivityDate="2014-05-17T03:36:29.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10848" PostTypeId="2" ParentId="10829" CreationDate="2014-05-17T07:49:13.917" Score="4" Body="&lt;p&gt;In most of the case distilled water is enough, but you should be sure that there are no salts inside your object if you can perform a Conductometry test on the washing water follow this procedure: Wash the item inside a buck for 3 minutes with distilled water measure Conductometry, take another bucket with the same amount of water wash the item for 3 minutes, measure Conductometry repeat these steps until conducibliy drop down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a Conductometry test I've made while restoring an ancient metal object. You can see that I needed 5 washing to get good result. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/g1IJA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have high conductivity you should identify the salts with some spots test to understand where they came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have grease you can use a surfactant like &lt;a href=&quot;http://en.wikipedia.org/wiki/Polysorbate_20&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plysorbate 20&lt;/a&gt; 2% in distilled water not more!&#xA;Eventually if there to much salts and grease you can use: Thiourea solution 8-10% Sodium thiosulphate 30% EDTA 10 % (&lt;a href=&quot;http://www.historeum.org/ihawiki/index.php/Conservation_and_restoration_of_metals&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Remember to dry very carefully the object with a hair dryer, NEVER leave the object wet for more than a couple of minutes!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually you can use a protective like Soter or Zapon these normally cause in terms of colour change a $\Delta E$ of 2-4 so ask to your self if you really need it for a gold object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion avoid ultrasonic cleaning for composite materials! If you really care about the object ask for a metal restorer!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2015-03-15T12:56:02.117" LastActivityDate="2015-03-15T12:56:02.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10850" PostTypeId="1" AcceptedAnswerId="10852" CreationDate="2014-05-17T11:13:08.483" Score="7" ViewCount="5658" Body="&lt;p&gt;I've been studying organic chemistry and I think trans fats are actually more stable than cis fats. This is because the steric hindrance (for trans) would be lower and the molecule energy would be lower too (more stable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are unsaturated fats usually cis at room temperature if the steric hindrance is higher? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that because the room temperature provides enough energy to break the pi bond in the double covalent bond (for example) making the molecule rotate and have a isomer of higher energy? &lt;/p&gt;&#xA;" OwnerUserId="5540" LastActivityDate="2019-05-11T04:31:08.873" Title="Why unsaturated fats are usually cis?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;&lt;fats&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10851" PostTypeId="2" ParentId="10840" CreationDate="2014-05-17T11:25:03.767" Score="1" Body="&lt;p&gt;As others have said, they are in equilibrium. However there is a physical/mechanical (mass transfer)  barrier to this equilibrium. When a crystal is dissolving/being precipitated, all shrinkage/growth occurs at the surface of the crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the equilibrium state is reached, the exchange occurs only at the surface of the crystal. There is no change for the material in the body of the crystal (except by diffusion, which in crystals is an extremely slow process.) For a radioactive tracer to reach the centre of the crystals, complete crystals would have to dissolve and new ones be precipitated, which is only likely to happen if there is a gradient of conditions (for example a temperature gradient) between one part of the vessel and another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you have a layer of salt slurry with clear solution above it, the solid phase of the slurry part is only in equilibrium with the liquid phase in the slurry part. If you add a radioactive tracer to the clear solution, it will be quickly incorporated into the surface of the crystals in the top layer of the slurry, but take a long time to reach (by liquid diffusion) the bottom part of the slurry. And it may never make it into the middle of any of the crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS I am a chemist turned chemical engineer, and for quantitative design of dissolution/precipitation process, this distinction is important.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2014-05-17T11:25:03.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10852" PostTypeId="2" ParentId="10850" CreationDate="2014-05-17T11:39:16.293" Score="5" Body="&lt;p&gt;Nature produces cis fats exclusively (someone may prove me wrong with some weird example.) This is why unsaturated fats such as vegetable oil have lower melting point than saturated fats of the same MW (the cis double bonds lead to a more compact molecule which gets less tangled with its neighbours.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason nature produces cis bonds is due to the enzymatic mechanism in which these bonds are produced. Therefore this is down to reaction kinetics, and as often happens the product which is not the most thermodynamically stable of all possible products is produced. It just has to be more stable than the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When fat is artificially hydrogenated with a metal catalyst (for example nickel) to increase its viscosity/raise its melting point (to turn vegetable oil into margarine) some of the cis double bonds that escape being hydrogenated are able to change from cis to the more thermodynamically stable trans. These trans fats are unnatural and are therefore widely claimed to be unhealthy.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2014-05-17T11:39:16.293" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10853" PostTypeId="1" AcceptedAnswerId="10860" CreationDate="2014-05-17T14:21:33.010" Score="9" ViewCount="2230" Body="&lt;p&gt;I learnt that given a reaction: $$\ce{A -&amp;gt; B}$$ the enthalpy change is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By convention, we keep the minus sign in the equation, and keep all bond enthalpies positive, regardless of whether bonds are made or broken. In this case, the bonds in the reactants are broken, and the bonds in the products are formed. I was also given the equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are both $\Delta H$'s given by the above formulas equivalent?&lt;/p&gt;&#xA;" OwnerUserId="5542" LastEditorUserId="5026" LastEditDate="2018-05-16T03:19:23.917" LastActivityDate="2019-07-01T01:52:46.187" Title="Comparing formula for enthalpy change with bond dissociation energy and formation enthalpy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10855" PostTypeId="1" CreationDate="2014-05-17T14:25:08.547" Score="9" ViewCount="11476" Body="&lt;p&gt;I recently was in need of zinc for an experiment. So I resorted to obtain it from some dry cell. But I wondered if it was pure zinc or was it &lt;span class=&quot;math-container&quot;&gt;$\ce{ZnO2}$&lt;/span&gt; or any other compound of &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt;. It then occurred to me that I had knowledge of existence of pure &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; and its compounds, but how would its discovery have happened without the prior knowledge of its existence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did scientists discover the elements without knowing of their existence and unique properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?&lt;/p&gt;&#xA;" OwnerUserId="5550" OwnerDisplayName="karthikeyan" LastEditorUserId="41328" LastEditDate="2019-01-10T06:22:25.393" LastActivityDate="2019-01-10T06:22:25.393" Title="How were the elements discovered before modern technology?" Tags="&lt;analytical-chemistry&gt;&lt;elements&gt;&lt;history-of-chemistry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="10856" PostTypeId="2" ParentId="10855" CreationDate="2014-05-17T14:47:31.377" Score="1" Body="&lt;p&gt;Zinc doesn't naturally occur in elemental form.  Instead, it occurs as zinc sulfide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thought that zinc was first produced in India and China by roasting the zinc sulfide with a carbon reducing agent, possible dung.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc is a component of brass.  Zinc was first produced as this alloy with copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purer forms of zinc were &lt;a href=&quot;http://www.infinityfoundation.com/mandala/t_es/t_es_agraw_zinc_frameset.htm&quot; rel=&quot;nofollow&quot;&gt;first produced in Rajasthan by distillation&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5160" OwnerDisplayName="DavePhD" LastActivityDate="2014-05-17T14:47:31.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10857" PostTypeId="1" AcceptedAnswerId="10859" CreationDate="2014-05-17T16:04:23.550" Score="1" ViewCount="2203" Body="&lt;p&gt;I think I need some distilling (boiling) and collection flasks and a 3-way adapter with a thermometer.  I was thinking of using an open vacuum adaptor to join the lower end of the condenser to the receiving flask.  Does this sound ok?  And if so, what kind of condenser should I use?&lt;/p&gt;&#xA;" OwnerUserId="5545" LastEditorUserId="4148" LastEditDate="2014-05-17T16:48:50.780" LastActivityDate="2016-03-31T19:25:00.913" Title="What condenser can be used for simple distillation?" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10858" PostTypeId="2" ParentId="10855" CreationDate="2014-05-17T16:17:52.387" Score="3" Body="&lt;p&gt;Essentially, a continuous process of trying to find the purest forms of different materials, and then find patterns and distinctions in how they behave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: say you pick up a random rock. You can try to measure as many qualities as possible about it: colour, hardness, density, magnetic susceptibility, and so on. Then, try to make it into something else. Heat it, freeze it, attack it with acids, water, alkali, and see if it seems to behave like anything you already know about, or whether you can convert it into something that does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more methods you have, the more sure you can be that two different objects are composed of the same material, or different materials. Particular similarities can cause problems, for example it took a long time for people to manage to separate the lanthanides because they behave very similarly under many tests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, elements have also discovered after finding unexpected results -- for example, helium was discovered in the spectral lines of the sun's radiation before it was ever isolated or identified on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a common problem of trying to find the irreducible complexity behind an opaque emergent system -- in this case, trying to work back from the materials we have to their fundamental components. The more information you have, the easier it is to discern more elemental qualities and patterns, and the more other bits you can find.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-17T16:17:52.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10859" PostTypeId="2" ParentId="10857" CreationDate="2014-05-17T17:01:33.777" Score="1" Body="&lt;p&gt;That sounds good.  Here's a drawing of the set up.  I'd use a water-cooled condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hFAdW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-17T17:01:33.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10860" PostTypeId="2" ParentId="10853" CreationDate="2014-05-17T17:05:50.097" Score="8" Body="&lt;p&gt;Yes. The two methods are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try this with the dimerization of $\ce{NO_2}$ into $\ce{N_2O_4}$. In this reaction we form one $\ce{N-N}$ bond. Every other bond stays the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NO_2 &amp;lt;=&amp;gt; N_2O_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we calculate the enthalpy of the reaction through bond energies, we get the negative of the $\ce{N-N}$ bond energy. Why? Because we only formed &lt;em&gt;one&lt;/em&gt; bond in this entire reaction. Every other bond stays the same - we start with &lt;em&gt;four&lt;/em&gt; resonance stabilized $\ce{N-O}$ bonds and end with &lt;em&gt;four&lt;/em&gt; resonance stabilized $\ce{N-O}$ bonds. The net change in bonds is the formation of a $\ce{N-N}$ bond. Here is a picture of the process (note that this picture doesn't show any resonance stabilization but there is; the &quot;pi&quot; bonding electrons are delocalized despite what the picture implies). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TO54d.jpg&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond energy is a positive value; the negative of this value implies that the dimerization reaction is exothermic - which is obviously true! We are going from a radical species - nitrogen dioxide - into a species with no unpaired electrons. Radicals are reactive! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also calculate this using heats of formation for both nitrogen tetraoxide and nitrogen dioxide and we'd get the same enthalpy of reaction. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-05-16T03:20:25.550" LastActivityDate="2018-05-16T03:20:25.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10861" PostTypeId="2" ParentId="10853" CreationDate="2014-05-17T17:22:22.310" Score="6" Body="&lt;p&gt;Given the reaction $\ce{A -&amp;gt; B}$ you know that $\ce{A}$ and $\ce{B}$ must have the same atomic composition, i.e. they consist of the same number of atoms for each element, because if this wasn't the case there would be other reactants or products in the reaction equation (the reaction is some sort of rearrangement). So, $\ce{A}$ and $\ce{B}$ only differ in the way the atoms are linked together, i.e. some bonds that are present in $\ce{A}$ might not be present in $\ce{B}$ and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Hess' law, which states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps, leads you to the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \sum \Delta H_F \left( \text{products}\right)-\Delta H_F \left( \text{reactants}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta H_F$ is the enthalpy of formation. This $\Delta H_F$ is simply the energy difference between the target compound and the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta H_F = H_{\text{molecule}} - H_{\text{elements}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\Delta H_F$ (mostly) describes how much energy is gained by forming bonds in the molecule. Now, since $\ce{A}$ and $\ce{B}$ have the same atomic composition, i.e. they have the same $H_{\text{elements}}$, you see that the energy difference between them must come from the energy difference of the bonds in $\ce{A}$ and $\ce{B}$. The part of bonding energy that comes from bonds that both $\ce{A}$ and $\ce{B}$ have in common cancels out and what remains is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, both methods are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="41556" LastEditDate="2018-05-15T20:28:03.147" LastActivityDate="2018-05-15T20:28:03.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10862" PostTypeId="1" AcceptedAnswerId="10884" CreationDate="2014-05-17T18:36:05.073" Score="9" ViewCount="10828" Body="&lt;p&gt;Are molecules with bond orders less than one stable? My professor commented that they were &quot;barely alive&quot; but what does he mean, scientifically? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that molecules with fractional bond orders greater than one can exist - i.e. nitrogen oxide has a bond order of 2.5 (through MO calculations) and it exists just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are molecules with fractional bond orders below one just very unstable? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-05-05T06:30:26.067" LastActivityDate="2018-05-05T06:31:13.480" Title="Are molecules with bond orders less than one stable?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10863" PostTypeId="1" AcceptedAnswerId="14304" CreationDate="2014-05-17T20:15:12.383" Score="3" ViewCount="965" Body="&lt;p&gt;I have a Graham condenser and tested it out by distilling water from salt water then I washed it and now I can't get either the coils or the jacket completely dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice?  If there isn't a better way I was thinking of popping it in the oven for a while to see if that'd dry it out the rest of the way.&lt;/p&gt;&#xA;" OwnerUserId="5362" LastActivityDate="2014-07-13T08:06:54.473" Title="How do you dry a Graham condenser?" Tags="&lt;equipment&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10864" PostTypeId="2" ParentId="10857" CreationDate="2014-05-17T20:18:29.377" Score="0" Body="&lt;p&gt;That sounds fine. You only need the open vacuum adaptor if you're using a sealed distillation setup and you have cause to worry about the pressure while distilling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use a basic &lt;a href=&quot;http://www.sanjosescientific.com/catalog/4002.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;distillation apparatus&lt;/a&gt; with a Graham condenser and a Kjeldahl distillation bulb.  Currently, I skip the three-way adapter with the thermometer because I'm just doing basic distillations and while I can seal mine I generally don't because I'm not worried about my distillate evaporating .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XkKny.jpg&quot; alt=&quot;basic distillation apparatus&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5362" LastEditorUserId="5362" LastEditDate="2014-05-17T20:24:25.907" LastActivityDate="2014-05-17T20:24:25.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10865" PostTypeId="1" AcceptedAnswerId="10866" CreationDate="2014-05-18T01:50:00.047" Score="1" ViewCount="790" Body="&lt;p&gt;Is the flip of the cis-isomer wrong? Specifically the image on the bottom right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O4liB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I see is that on the cis-isomer in the first chair, we have an axial upward methyl group and an equatorial upward methyl group. When the ring flips we now have an equatorial upward methyl group (good) and a downward equatorial methyl group (bad). Shouldn't the second one be &lt;em&gt;axial&lt;/em&gt; &lt;em&gt;upward&lt;/em&gt;? From what I understand, ring flips shouldn't change the orientations of groups; just their axial/equatorial configurations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in general, would these two rings be equivalent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOcFO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="19616" LastEditDate="2015-10-16T17:58:25.320" LastActivityDate="2015-10-16T17:58:25.320" Title="Incorrect Ring Flip?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10866" PostTypeId="2" ParentId="10865" CreationDate="2014-05-18T02:24:37.957" Score="4" Body="&lt;p&gt;Yes, the cis isomer on the right should have the methyl group on C2 in the axial (up) position.  The two rings you've drawn at the bottom of your post are equivalent, they can be interconverted by a 180 degree rotation about an axis that bisects the C1-C2 bond and the C4-C5 bond.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-18T02:24:37.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10867" PostTypeId="1" AcceptedAnswerId="10868" CreationDate="2014-05-18T03:55:22.610" Score="6" ViewCount="2848" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:  I misquoted my teacher; he said something more along the lines of: in conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is &lt;em&gt;also&lt;/em&gt; $\ce{sp^2}$ hybridized. &lt;strong&gt;Would this be accurate?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My teacher said that if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is &lt;em&gt;also&lt;/em&gt; $\ce{sp^2}$ hybridized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why is this true?&#xA;2) Is this even true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing this has to do with physical limitations. For example, consider a carbonyl oxygen. Oxygen is $\ce{sp^2}$ hybridized. The carbon the oxygen is bound to has already &quot;used up&quot; two of its four bonds. The carbon can only form two more sigma bonds in addition to the one sigma bond it already has in the $\ce{C=O}$ double bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about &lt;strong&gt;&quot;hypervalent&quot;&lt;/strong&gt; (if there are such beasts) molecules? Here we come to a &lt;em&gt;self-consistency&lt;/em&gt; issue. My teacher still believes in &lt;em&gt;hypervalent&lt;/em&gt; molecules - i.e. he draws sulfuric acid with &lt;em&gt;six&lt;/em&gt; bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's consider sulfuric acid in its hypervalent form, with two double bonded $\ce{sp^2}$ oxygens. There are a total of six bonds and &lt;em&gt;four&lt;/em&gt; sigma bonds, implying $\ce{sp^3}$ hybridization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my revised set of questions is as listed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is my teacher being inconsistent with himself here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does $\ce{sp^2}$ hybridization still imply $\ce{sp^2}$ hybridization for linked atoms? Note that sulfuric acid does not disprove my teacher's statement &lt;em&gt;if&lt;/em&gt; we take sulfuric acid to have &lt;em&gt;four&lt;/em&gt; bonds, not &lt;em&gt;six&lt;/em&gt; (i.e. sulfuric acid is properly represented as charge separated due to the high ionic character of its $\ce{S-O}$ bonds. If I remember correctly, on a time-average basis, 1.7 electrons are in oxygen's valence and only 0.3 electrons are in sulfur's valence per bond.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-29T02:15:49.373" LastActivityDate="2014-05-29T07:08:02.270" Title="Why does sp2 hybridization imply adjacent sp2 hybridization?" Tags="&lt;hybridization&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10868" PostTypeId="2" ParentId="10867" CreationDate="2014-05-18T04:01:35.480" Score="0" Body="&lt;p&gt;Acetone has the oxygen and the carbon $sp^2$ hybridized. In carbon dioxide, the oxygens are $sp^2$, but the carbon is $sp$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;For $SO_4^{2-}$, the sulfur needs to utilize 2 of the d orbitals &lt;strike&gt;for the sigma bonds, so it would be spd^2 hybridized, and it uses the 2 p orbitals&lt;/strike&gt; for the pi bonds. The sulfur is $sp^3$ hybridized. (According to Wikipedia).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-05-18T04:35:09.673" LastActivityDate="2014-05-18T04:35:09.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10869" PostTypeId="1" AcceptedAnswerId="10904" CreationDate="2014-05-18T05:18:21.563" Score="4" ViewCount="17975" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The standard electrode potentials for the following redox couples are given as: &gt;$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$&lt;br&gt;&#xA;  What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt; $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -&amp;gt; Fe^{2+} + Fe}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="5026" LastEditDate="2018-02-27T01:56:09.760" LastActivityDate="2019-10-06T11:31:09.090" Title="Calculate the standard oxidation potential from given potentials" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10870" PostTypeId="1" AcceptedAnswerId="10880" CreationDate="2014-05-18T05:22:28.447" Score="5" ViewCount="9640" Body="&lt;p&gt;I was wondering why only one of the nitrogen lone pairs in imidazole is involved in the delocalized π-system. Specifically, I was wondering why it is that lone pair B in particular is delocalized, instead of lone pair A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aOWJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOWJu.png&quot; alt=&quot;Lone pairs in imidazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my conjecture about why B's lone pair is delocalized while A's lone pair cannot be.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If we delocalize A, we'd get extreme ring strain; we'd make the geometry around that nitrogen &lt;span class=&quot;math-container&quot;&gt;$\ce{sp}$&lt;/span&gt;. This implies a &lt;span class=&quot;math-container&quot;&gt;$120^\circ$&lt;/span&gt; bond angle. That's problematic. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Effective &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C0}$&lt;/span&gt;-orbital overlap (conjugation) only occurs when the p-lobes are planar — i.e. parallel with each other. If the nitrogen with the B lone pair kept its B lone pair, then that nitrogen would be &lt;em&gt;tetrahedral&lt;/em&gt; and that would distort the planarity of the molecule; everything else in the molecule appears to be &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We also run into formal charge issues; we get a positive formal charge on the &quot;A&quot; nitrogen and a negative formal charge on the B nitrogen. While this isn't a dealbreaker, it's not preferable, as there is separation of like formal charges. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Anything I miss? Any points I mangled? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-10-25T21:05:55.927" LastActivityDate="2018-10-25T21:12:04.143" Title="Why is only one lone pair in imidazole delocalised?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;heterocyclic-compounds&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10871" PostTypeId="1" AcceptedAnswerId="10877" CreationDate="2014-05-18T05:48:38.090" Score="2" ViewCount="1063" Body="&lt;p&gt;$\ce{MgSO4}$, on reaction with $\ce{NH4OH}$ and $\ce{Na2HPO4}$, forms a white crystalline precipitate.&#xA;Identify the precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: I guess the white crystalline precipitate is $\ce{Mg(OH)2}$, since it is white and insoluble in water. &lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="73" LastEditDate="2014-05-18T07:17:35.437" LastActivityDate="2015-03-25T15:14:48.067" Title="What is the precipitate of this reaction of magnesium?" Tags="&lt;inorganic-chemistry&gt;&lt;precipitation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10872" PostTypeId="2" ParentId="10870" CreationDate="2014-05-18T05:51:03.777" Score="4" Body="&lt;p&gt;You have a problem with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) If we conjugate A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $sp^2$. This implies a 180° bond angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the nitrogen is $sp^2$, you get 120° bond angles, not 180°. For 180° bond angles, you would need $sp$ hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, both A and B are conjugated. Both atoms are $sp^2$ hybridized and both have a p orbital used for pi bonding in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;Here is the bonding in &lt;a href=&quot;http://www.chemtube3d.com/Pyrrole-structure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyrrole&lt;/a&gt;. As you will notice, the nitrogen is $sp^2$ hybridized and the lone pair occupies the p orbital for pi bonding in the ring. In &lt;a href=&quot;http://www.chemtube3d.com/Pyridine-structure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyridine&lt;/a&gt;, the nitrogen is again $sp^2$ hybridized but the lone pair sticks out from the ring. These are the 2 nitrogen configurations in imidazole.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="418" LastEditDate="2017-01-15T07:00:26.727" LastActivityDate="2017-01-15T07:00:26.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10873" PostTypeId="2" ParentId="10869" CreationDate="2014-05-18T06:11:56.913" Score="-1" Body="&lt;p&gt;Your approach is a correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation: (I call the given two reduction half cell reactions as 1 and 2 respectively)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required half cell reaction $\ce{Fe^{3+} -&amp;gt; Fe}$ could be obtained by adding reaction 1 and 2 .Thus the net $\ce{E0}$ value would be the sum of $\ce{E0}$ values of reaction 1 and 2 (as both are reduction half cell reactions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence net $\ce{E0(Fe^{3+}/Fe^{2+})} = (x + y) \, \text{V}$&lt;/p&gt;&#xA;" OwnerUserId="5547" LastEditorUserId="189" LastEditDate="2014-05-18T12:54:41.643" LastActivityDate="2014-05-18T12:54:41.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10875" PostTypeId="1" AcceptedAnswerId="10882" CreationDate="2014-05-18T06:40:35.900" Score="1" ViewCount="1924" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Solid $\ce{NaHCO3}$ is heated to $90~^\circ\mathrm{C}$. At equilibrium the total pressure of the gases produced is $0.545~\mathrm{atm}$. Calculate $\Delta G^\circ$ at $90~^\circ\mathrm{C}$ for the reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NaHCO3(s) &amp;lt;=&amp;gt; Na2CO3(s) + H2O(g) + CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the $\Delta G^\circ = -\mathcal{R}T\cdot\ln(K)$ formula, I did the work below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\circ = -(8.31)(273.15 + 90)\cdot\ln(0.545)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gave me $1.83~\mathrm{kJ/mol}$ and therefore $3.66~\mathrm{kJ}$ for the entire reaction, $2~\mathrm{mol}$. Unfortunately, the answer is marked as $7.85~\mathrm{kJ}$, suggesting that $K=1$, why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't $K=(0.545)/1$ because of the gases present?&lt;/p&gt;&#xA;" OwnerUserId="5548" LastEditorUserId="4945" LastEditDate="2015-02-27T03:05:27.533" LastActivityDate="2015-02-27T03:05:27.533" Title="How to calculate the Gibbs free energy from the equilibrium constant for the decomposition of sodium bicarbonate?" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10876" PostTypeId="1" CreationDate="2014-05-18T07:21:28.650" Score="3" ViewCount="263" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The minimum pressure for the existence of liquid solution is?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$0.6\;\mathrm{atm}$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$0.8\;\mathrm{atm}$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$0.77\;\mathrm{atm}$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$0.9\;\mathrm{atm}$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although this seems to be an informative question, I want to know the process by which we get this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the critical pressure for a liquid solution is &lt;span class=&quot;math-container&quot;&gt;$p={ a\over {27b^2}}$&lt;/span&gt; but what is the value of a and b in that case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect an easy to understand answer for high-school level.&lt;/p&gt;&#xA;" OwnerUserId="5549" LastEditorUserId="83758" LastEditDate="2020-02-22T18:10:35.007" LastActivityDate="2020-02-22T18:10:35.007" Title="What is the minimum pressure for existence of a liquid solution?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2020-02-23T18:39:39.520" ContentLicense="CC BY-SA 4.0" />
  <row Id="10877" PostTypeId="2" ParentId="10871" CreationDate="2014-05-18T07:22:00.767" Score="2" Body="&lt;p&gt;Well, if we look at the possible compounds that can form here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg(OH)&lt;sub&gt;2&lt;/sub&gt; = Insoluble,&#xA;MgHPO&lt;sub&gt;4&lt;/sub&gt; = Soluble,&#xA;(NH&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; = Soluble,&#xA;(NH&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;HPO&lt;sub&gt;4&lt;/sub&gt; = Soluble,&#xA;Na&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt; = Soluble,&#xA;NaOH = Soluble&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it appears your guess is correct.&lt;/p&gt;&#xA;" OwnerUserId="5548" LastEditorUserId="13409" LastEditDate="2015-03-25T15:14:48.067" LastActivityDate="2015-03-25T15:14:48.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10879" PostTypeId="1" CreationDate="2014-05-18T09:28:54.350" Score="2" ViewCount="4096" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Among &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^2+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{V^2+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ni^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ti^2+}$&lt;/span&gt;, which cation is having the highest hydration energy to form the aqua complex &lt;span class=&quot;math-container&quot;&gt;$\ce{[M(H2O)6]^2+}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^2+}$&lt;/span&gt; should be the answer as it has the most stable electronic configuration.&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="41328" LastEditDate="2019-01-05T11:06:36.330" LastActivityDate="2019-01-05T11:07:38.240" Title="Which d-block cation has the maximum hydration enthalpy?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10880" PostTypeId="2" ParentId="10870" CreationDate="2014-05-18T10:36:42.727" Score="12" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vSVOA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSVOA.png&quot; alt=&quot;Structure of imidazole with locants&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any resonance effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect of orbital geometry is often invoked to explain the difference in basicity between &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyridine&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyridine&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrrole&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyrrole&lt;/a&gt;. In the former, the nitrogen lone pair is of type 3 and does not participate in resonance, and so it has a basicity comparable to regular amines and imines. However, in the latter, the lone pair is of type 1 and &lt;em&gt;does&lt;/em&gt; participate in the aromaticity of the ring. Protonation of the nitrogen atom would require localization of the electron pair on the nitrogen, breaking most of the resonance effect. This is energetically unfavourable, so pyrrole displays a much lower basicity than expected for amines or imines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pyridine#Molecular_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;This section&lt;/a&gt; on pyridine has a little more information on why the type A lone pair does not delocalize, including a helpful picture. Notice that the lone pair is in an &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbital parallel to the plane of the ring, while the π-bonded system is perpendicular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can make the following observations on your conjectures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I don't think this applies, actually. The fact that orbital overlap is disallowed in the first place precludes the possibility of even thinking what would happen to the geometry of the molecule if the electron pair in nitrogen 3 formed a π-bond; it just doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the molecule had a boron atom instead of carbon in position 2 (with an empty &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbital pointing in the same direction as the full &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbital in nitrogen 3), then I'm not sure what would happen. A more simple example of this type of structure would be 1,2-azaborine (of which only the &lt;a href=&quot;http://en.wikipedia.org/wiki/1,2-dihydro-1,2-azaborine&quot; rel=&quot;nofollow noreferrer&quot;&gt;1,2-dihydro derivative&lt;/a&gt; seems to have been isolated as of yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine canonical structures in which the electron pair on nitrogen links with the boron atom, forming an additional bond (analogous to &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzyne&lt;/a&gt;, which is isoelectronic with 1,2-azaborine). However, to form an extra N–B π-bond, then either the bond will be a &lt;span class=&quot;math-container&quot;&gt;$\pi(\mathrm{sp^2}\text{–}\mathrm{sp^2})$&lt;/span&gt; bond (which is unusual as far as I know), or both the nitrogen and boron atoms would have to partially rehybridize into a mix of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}$&lt;/span&gt; configurations to allow a &lt;span class=&quot;math-container&quot;&gt;$\pi(\text{p–p})$&lt;/span&gt; bond, which would likely imply an increase in bond angle and a huge increase in ring strain. Neither option seems particularly good, so I suspect such canonical structures would contribute little to the actual bonding picture in the boron-substituted ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The nature of the nitrogen atom which participates in resonance is actually something between &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt;. The concept of hybridization isn't very adept at explaining resonance naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not happy with how much my arguments rely on hybridization, but I don't know other effective means to explain what's going on.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="16683" LastEditDate="2018-10-25T21:12:04.143" LastActivityDate="2018-10-25T21:12:04.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10881" PostTypeId="2" ParentId="10876" CreationDate="2014-05-18T10:59:45.647" Score="1" Body="&lt;p&gt;The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_equation&quot; rel=&quot;nofollow&quot;&gt;van der Waals equation&lt;/a&gt; for nonideal gases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, these constants are empirically derived, and thus &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_constants_%28data_page%29&quot; rel=&quot;nofollow&quot;&gt;different for every substance&lt;/a&gt;, and so the critical pressure would be different for every substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-18T10:59:45.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10882" PostTypeId="2" ParentId="10875" CreationDate="2014-05-18T11:06:08.017" Score="3" Body="&lt;p&gt;For your reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=P_{\ce{CO2}}\cdot P_{\ce{H2O}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $K\ne0.545\text{ atm}$ because the &lt;strong&gt;sum&lt;/strong&gt; of the pressures is $0.545\text{ atm}$ and the equilibrium constant is the &lt;strong&gt;product&lt;/strong&gt; of the pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\ce{CO2}}+P_{\ce{H2O}}=0.545\text{ atm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you started with no gas, and the stoichiometric coefficients of both gases are equal, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\ce{CO2}}=P_{\ce{H2O}}=\dfrac{0.545\text{ atm}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the value of $K$ now?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-18T11:06:08.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10883" PostTypeId="2" ParentId="10879" CreationDate="2014-05-18T11:20:49.430" Score="6" Body="&lt;p&gt;Note that hydration energy is the energy released when water molecules are adsorbed on the surface of the cation. Hence the cation should have a high charge density, which means it should have a small size. So, the smaller, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ni}$&lt;/span&gt; has the smallest atomic radii compared to all other options due to greater effective nuclear charge, so it's ionic radii would also follow the same trend. Hence, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ni^2+}$&lt;/span&gt; would have greater hydration enthalpy.&lt;/p&gt;&#xA;" OwnerUserId="5549" LastEditorUserId="41328" LastEditDate="2019-01-05T11:07:38.240" LastActivityDate="2019-01-05T11:07:38.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10884" PostTypeId="2" ParentId="10862" CreationDate="2014-05-18T11:28:43.550" Score="9" Body="&lt;p&gt;Molecules possessing a bond of order below 1 can be perfectly stable, in the sense that their resulting molecular structure lies in an energetic potential well. Strictly speaking it is enough that at $T=\pu{0 K}$ and in the absence of any interactions with matter or fields, the molecule will not spontaneously disassemble. However, there is no need to go quite so far to protect such a molecule from decomposing; there are examples of species which are chemically important in regular laboratory conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other things being equal, it is true that species with bond orders below 1 are relatively unstable. This is mainly because the fractional bond is comparatively weak (requiring comparatively little activation energy to break, i.e. a smaller $E_\mathrm{a}$), and because in most cases the molecule can react with other substances in such a way as to form products with all covalent bonds of bond order 1 or higher (increasing the exergonicity of most reactions, i.e. a more negative $\Delta_\mathrm{r}G$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the kinetic barrier for reaction is comparatively low, and the thermodynamic drive for reaction is comparatively high, species with bond order below 1 tend to need some extra protection against the world to stay in one piece. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diborane&quot; rel=&quot;nofollow noreferrer&quot;&gt;diborane&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethylaluminium&quot; rel=&quot;nofollow noreferrer&quot;&gt;trimethylaluminium&lt;/a&gt; are compounds possessing bonds of order 0.5, and while being indefinitely stable when pure, they &lt;a href=&quot;https://www.youtube.com/watch?v=RfeaffoWajc&quot; rel=&quot;nofollow noreferrer&quot;&gt;spontaneously ignite in air&lt;/a&gt; from exposure to oxygen and moisture. As rightly pointed out in the comments, noble gas compounds require fractional bond orders, yet several compounds can be obtained and stored (especially xenon compounds), though they tend to be sensitive to moisture and heating. Boron is also responsible for &lt;a href=&quot;http://en.wikipedia.org/wiki/Boranes&quot; rel=&quot;nofollow noreferrer&quot;&gt;a very interesting class of compounds&lt;/a&gt; in which many boron atoms attach to  each other in cage-like structures with very complicated bonding, in which fractional order bonds are involved. Some of the larger and more symmetric structures can be relatively stable, especially with adequate substituents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space, there's really not much of anything around, so you might expect to find some molecules with fractional bond orders floating about. Indeed, one can find the &lt;a href=&quot;http://en.wikipedia.org/wiki/Trihydrogen_cation&quot; rel=&quot;nofollow noreferrer&quot;&gt;trihydrogen cation&lt;/a&gt;, which is actually one of the most common ions in the Universe!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="5026" LastEditDate="2018-05-05T06:31:13.480" LastActivityDate="2018-05-05T06:31:13.480" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10885" PostTypeId="1" CreationDate="2014-05-18T11:32:56.837" Score="2" ViewCount="366" Body="&lt;p&gt;I'm a enthusiastic and devoted amateur chemist (in disguise, I'm actually a mathematician), but lately I've been thinking to what extend it is safe to pursue my hobby. I'm after all an amateur with just high-school lab skills and some unnecessary math to back me up. I really love chemical experiments, but I don't want to get seriously injured. The point is that same very basic materials, like sulfuric acid or cyclohexylamine, already have a 3 on &lt;a href=&quot;http://en.wikipedia.org/wiki/NFPA_704#Blue&quot; rel=&quot;nofollow&quot;&gt;the NFPA 704 scale of health hazards&lt;/a&gt;, quoting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Short exposure could cause serious temporary or moderate residual injury (e.g. chlorine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course I read the MSDS of the chemicals I'm working with, wear a protective lab coat and tightly connected safety goggles, plus safety gloves, but I don't have nearly the safety equipment of a normal lab (fumehood, eye showers, sprinklers, etc.) To to concretize my question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Speaking for all amateur chemists out there: to what extend do you as a professional chemist consider amateurs like me to be safe. Can we handle level 3 or perhaps even level 4 chemicals, or should we stick with 2 (greatly reducing our potential experimenting terrain)?&lt;/li&gt;&#xA;&lt;li&gt;What measure can we (and should we) take to ensure our safety?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that the capabilities vary greatly between people, but there was not sufficient information regarding this topic (from reputable sources) for me to be found on the internet, so I though I post it here. I don't feel like this is a primarily opinion based question, as there is at least to some degree consensus that amateurs cannot (and are not allowed to!) safely handle, say for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Tabun_%28nerve_agent%29&quot; rel=&quot;nofollow&quot;&gt;tabun&lt;/a&gt; or potassium cyanide (just a ridiculous example to illustrate my point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can appreciate my concerns and that if this is not the right place for this type of question (I think it is, there is even a safety tag here) you cordially redirect me to an other more suited place and close this.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-05-21T08:26:44.343" LastActivityDate="2014-05-21T08:26:44.343" Title="Hazardous materials handled by amateur chemists" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="0" CommentCount="19" FavoriteCount="1" ClosedDate="2014-05-18T12:31:25.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="10886" PostTypeId="2" ParentId="10871" CreationDate="2014-05-18T12:08:36.900" Score="5" Body="&lt;p&gt;With $\ce{(NH4)2HPO4}$, $\ce{Mg^2+}$ forms a white, crystalline precipitate of &lt;a href=&quot;http://en.wikipedia.org/wiki/Struvite&quot; rel=&quot;noreferrer&quot;&gt;magnesium ammonium phosphate hexahydrate&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg^2+ + HPO4^2- + NH4^+ + OH- + 5H2O \rightleftharpoons Mg(NH4)PO4 \cdot 6H2O\downarrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very sensitive test for $\ce{Mg^2+}$; however, calcium, strontium, barium and other heavy metal ions also form precipitates of phosphates under these conditions. $\ce{Mg(NH4)PO4 \cdot 6H2O}$ can be distinguished by the form of its crystals when they are examined under a microscope. It forms rhombic crystals with prismatic shape, and is less microcrystalline than other earth alkali metal phosphates (sources: &lt;a href=&quot;http://books.google.de/books?id=CLCdzlwv3f8C&amp;amp;pg=PA86&amp;amp;lpg=PA86&amp;amp;dq=magnesium%20ammonium%20phosphate%20hexahydrate%20%2b%20qualitative%20test%20for%20magnesium&amp;amp;source=bl&amp;amp;ots=DuKIya4G0O&amp;amp;sig=e7nSA1oG382uladhClVw9USgqqI&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=bJ94U_W2CZOK4gSliYGoCg&amp;amp;ved=0CDwQ6AEwAQ#v=onepage&amp;amp;q=magnesium%20ammonium%20phosphate%20hexahydrate%20%2b%20qualitative%20test%20for%20magnesium&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://books.google.de/books/about/Lehrbuch_der_analytischen_und_pr%C3%A4parati.html?hl=de&amp;amp;id=nJkuAAAACAAJ&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-18T12:08:36.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10887" PostTypeId="1" CreationDate="2014-05-18T14:50:23.360" Score="0" ViewCount="1445" Body="&lt;p&gt;Combustion of 0.2000 g of vitamin C gives 0.2998 g $\ce{CO2}$ and 0.819 g of water. What is the empirical formula of vitamin C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated moles of products and found the mole ratio but it's not coming to be a whole number.&lt;/p&gt;&#xA;" OwnerUserId="5551" LastEditorUserId="5456" LastEditDate="2014-05-18T23:14:57.393" LastActivityDate="2014-05-18T23:14:57.393" Title="Calculating empirical formula of vitamin C" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10888" PostTypeId="1" AcceptedAnswerId="10889" CreationDate="2014-05-18T15:51:27.323" Score="2" ViewCount="2642" Body="&lt;p&gt;Somewhere along the line of researching why the $\ce{Cl-C-H}$ bond angle in methyl chloride is less than what is predicted ($\pu{109.5^\circ}$), I ran into a book which said that the electron withdrawing effects of the chlorine atom gave the $\ce{C-Cl}$ bond more &quot;p-character&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? Is he simply using this term to describe the elongated nature of the bond - i.e. s-orbitals are spherical and p-orbitals not spherical but rather elongated (at least as depicted by cartoon drawings in textbooks)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the author saying that the electron-withdrawing effects of chlorine cause the electrons to spend more time further away from the nucleus; hence the &quot;p-character&quot;? I understand that in a spherical shell, electrons are basically equidistant from the nucleus. But in an elongated, &quot;p-character&quot; shell, electrons can go further away from the nucleus. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:01:25.300" LastActivityDate="2018-03-04T02:01:25.300" Title="Bond character and bond angles" Tags="&lt;bond&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10889" PostTypeId="2" ParentId="10888" CreationDate="2014-05-18T16:20:55.470" Score="3" Body="&lt;p&gt;S orbitals are lower in energy than P orbitals.  Electrons prefer to be in as low an energy orbital as possible.  Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride).  The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond.  So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it?  Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them.  It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted).  The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-18T16:28:58.397" LastActivityDate="2014-05-18T16:28:58.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10890" PostTypeId="2" ParentId="10887" CreationDate="2014-05-18T17:03:04.983" Score="1" Body="&lt;p&gt;You are using the wrong number for the grams of water, it should be 0.0819.  Go back and redo your calculations with that change.  Now you should get the correct ratio of the elements in the molecular formula for ascorbic acid.  If you don't, report back and show your math.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-18T17:03:04.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10891" PostTypeId="2" ParentId="10887" CreationDate="2014-05-18T17:04:45.540" Score="0" Body="&lt;p&gt;When you get the mole ratios but they are not whole numbers, divide through by the smallest number. Say you get for some hypothetical compound the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${C_{2.5}H_{5}O_{7.5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case to get the empirical formula, just divide each subscript by 2.5, the smallest subscript. You will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${CH_{2}O_{3}}$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-18T17:04:45.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10892" PostTypeId="1" AcceptedAnswerId="10894" CreationDate="2014-05-18T17:49:29.717" Score="1" ViewCount="106" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At standard temperature and pressure, what volume in $\pu{mL}$ of $\ce{H2O}$ is produced in the following reaction?&#xA;  $$\ce{2 KOH(l)  +  H3PO4(l)  -&amp;gt;  K2HPO4(l)  + 2 H2O(l)}$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most questions I've found have more information than this so I'm confused as to how to do this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried multiplying $\pu{2 mol}$ of $\ce{H2O}$ by $\pu{22.4 L}$ and then converting it to $\pu{mL}$, but I've a feeling that isn't right. &lt;/p&gt;&#xA;" OwnerUserId="5552" LastEditorUserId="4945" LastEditDate="2018-02-21T09:33:28.383" LastActivityDate="2018-02-21T09:33:28.383" Title="How to determine the volume of a substance produced when only given the equation at STP?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10893" PostTypeId="1" AcceptedAnswerId="10895" CreationDate="2014-05-18T17:56:39.670" Score="1" ViewCount="138" Body="&lt;p&gt;With regard to a trigonal bipyramidal molecule, such as $\ce{PIBrClF_2}$, can there be any constituional isomers? I'm thinking no, because constitutional isomers differ in the connectivity of their atoms. The molecule only has single bonds. All are bound to a single central atom. There is no way for the connectivity of the atoms to change versus something such as ethanol, $\ce{C_2H_6O}$. I can think of two constitutional isomers off the bat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_3CCH_2OH}$ and $\ce{H_3COCH_3}$ in which the connectivity of the atoms differ; in the former, we have a hydroxyl group; in the latter, we are missing the hydroxyl group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\ce{PIBrClF_2}$ should not have any constitutional isomers but does have stereoisomers, correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-18T18:05:14.863" LastActivityDate="2014-05-18T18:05:14.863" Title="Constitutional Isomers or Stereoisomers?" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10894" PostTypeId="2" ParentId="10892" CreationDate="2014-05-18T18:00:55.987" Score="1" Body="&lt;p&gt;I think you erred in multiplying 2 moles of water by 22.4 L. There are 55.5 moles of water per liter of water. Keeping this relationship in mind, use dimensional analysis to find the moles of water produced in your reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're confusing the number 22.4 with the fact that 1 mol of gas at STP occupies approximately 22.4 L. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-18T18:00:55.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10895" PostTypeId="2" ParentId="10893" CreationDate="2014-05-18T18:04:19.577" Score="2" Body="&lt;p&gt;Correct, no constitutional isomers, but there are stereoisomers (different substituents in the apical and equatorial positions).  Just for the record, what is your fifth substituent?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-18T18:04:19.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10896" PostTypeId="1" AcceptedAnswerId="10897" CreationDate="2014-05-18T21:22:41.787" Score="1" ViewCount="1218" Body="&lt;p&gt;Are these all valid isomers for the formula $\ce{C_3H_6O}$? I drew the first five; my friend came up with the last three, which have double bonds. Are these all isomers or different? More importantly, are there any more? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in general, how can one come up with all the isomers for a molecule? I heard of graph theory, but I think that's a little bit too advanced for me. I guess the only way is to draw out as many as you can and name them to weed out redundant structures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1pC3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2021-05-22T21:37:08.410" LastActivityDate="2021-05-22T21:37:08.410" Title="Drawing Constitutional Isomers" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10897" PostTypeId="2" ParentId="10896" CreationDate="2014-05-18T21:46:33.957" Score="1" Body="&lt;p&gt;They are all constitutional isomers.  If you are counting enols (your friends #3, it is the enol of your #1), then include the enol of acetone as an additional constitutional isomer.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-18T21:46:33.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10898" PostTypeId="1" AcceptedAnswerId="10926" CreationDate="2014-05-18T22:02:50.150" Score="6" ViewCount="242" Body="&lt;p&gt;How would such a scientist be able to get the spectroscopy of chemicals?  Without access to modern computing or electronics how would it be possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any video link or reference link are much appreciated here.&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="4148" LastEditDate="2014-05-19T16:02:01.767" LastActivityDate="2014-05-19T16:18:01.037" Title="How could a scientist from the 1800s to the 1920s run spectroscopy of chemicals?" Tags="&lt;spectroscopy&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10899" PostTypeId="1" AcceptedAnswerId="10900" CreationDate="2014-05-18T22:20:15.103" Score="3" ViewCount="433" Body="&lt;p&gt;How were scientists able to produce these accurate structural diagrams long before scientists were able to capture images of the actual molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://www.sciencedaily.com/releases/2013/05/130530142007.htm&quot; rel=&quot;nofollow&quot;&gt;Scientists capture first images of molecules before and after reaction&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="4148" LastEditDate="2014-05-19T00:20:33.593" LastActivityDate="2014-05-19T00:20:33.593" Title="How could scientists produce accurate chemical structures before they could capture images of molecules?" Tags="&lt;structural-formula&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10900" PostTypeId="2" ParentId="10899" CreationDate="2014-05-18T22:24:53.627" Score="6" Body="&lt;p&gt;They ran chemical reactions on the molecule and identified the fragments.  Then like a jigsaw puzzle, using additional information derived from other chemical reactions, they pieced the molecule back together.  Some Nobel prizes were awarded for such structure determinations.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-18T22:24:53.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10901" PostTypeId="1" AcceptedAnswerId="10936" CreationDate="2014-05-18T22:57:29.673" Score="3" ViewCount="365" Body="&lt;p&gt;After consulting three organic chemistry texts (Wade, Klein, and Brown), only one actually clearly delineates a few guidelines for picking the most significant resonance structure. The others mention the same factors (maybe less one or two). So in sum, the important factors to consider are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A filled valance shell is preferable to a partially filled valance shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The more covalent bonds the better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Least separation of oppositely signed charges. I'm thinking this is a Coulombic effect. Some books imply formal charge through the term &quot;charge&quot; but what about partial charges (which are much more reflective of true charges than &lt;em&gt;formal&lt;/em&gt; charges. Which one should I really be looking at? Least separation of opposite &lt;em&gt;formal&lt;/em&gt; charges or opposite &lt;em&gt;partial&lt;/em&gt; charges? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Negative charges should generally be assigned to the more electronegative elements; the opposite is true for electropositive elements. Again, formal charge misleads - consider the ammonium ion (positive formal charge on nitrogen; negative partial charge on nitrogen). Again, one book implies &lt;em&gt;formal&lt;/em&gt; charge but really, isn't looking at &lt;em&gt;partial&lt;/em&gt; charges a bit more informative? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) &lt;strong&gt;Unmentioned factor in any of the three textbooks:&lt;/strong&gt; What about hybridization? If we have a molecule with two nitrogens, one $\ce{sp}$ hybridized and one $\ce{sp^2}$ hybridized but otherwise identical, shouldn't we also take into account &quot;orbital electronegativity&quot; - i.e. the ability of hybrid orbitals with higher s-character to better stabilize negative charge (and conversely higher p-character better stabilizes positive charge)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) &lt;strong&gt;Unmentioned factor in any of the three textbooks:&lt;/strong&gt; Induction? Shouldn't induction also play a role in resonance stabilization? Resonance delocalizes electron(s); &lt;s&gt;can't induction further stabilize electrons?&lt;/s&gt; &lt;em&gt;debunked&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-02T16:47:50.617" LastActivityDate="2015-10-02T16:47:50.617" Title="Orbital Electronegativity Considerations in Resonance Structures" Tags="&lt;resonance&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10902" PostTypeId="1" AcceptedAnswerId="10903" CreationDate="2014-05-18T23:25:34.480" Score="3" ViewCount="144" Body="&lt;p&gt;The fourth principle of Molecular Orbitals state that:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Molecular orbitals are best formed when composed of Atomic orbitals of like energies.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about how I should interpret this; which is the correct?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It means that the molecular orbitals that I need to consider are the ones formed by the same orbitals on each atom (like a pair of $2p$, two $3s$, etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It means that I need to evaluate the actual energy on each orbital using: $$E = \frac{-13.6 z^2}{n^2}$$ And then compare the energies of every orbital, to match each orbital on &lt;em&gt;Atom 1&lt;/em&gt; to the closest one on &lt;em&gt;Atom 2&lt;/em&gt;? Meaning that I could possibly have a $2s$ pairing with a $3p$, depending on both atoms? (Note that the formula gives different results for each &lt;em&gt;Atomic Orbital&lt;/em&gt; depending on the atom's $z$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I known that every possible orbital combination exists, but I'm asking about the ones that are more relevant and thus need to be considered.&lt;/p&gt;&#xA;" OwnerUserId="1809" LastActivityDate="2014-05-18T23:46:40.220" Title="Fourth principle of Molecular Orbitals" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10903" PostTypeId="2" ParentId="10902" CreationDate="2014-05-18T23:46:40.220" Score="3" Body="&lt;p&gt;Both of the statements have merit. When the orbitals are of the same type, the overlap is maximized. When the orbitals are nearly the same energy, the atoms are closer together (and give better overlap) than when the energy difference is larger. You don't really need to calculate the energies of the orbitals, just look at the size of the ions. The larger the ion, the higher the energy level of the bonding electrons.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-18T23:46:40.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10904" PostTypeId="2" ParentId="10869" CreationDate="2014-05-18T23:50:36.857" Score="4" Body="&lt;p&gt;You can write down the three half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA; \ce{Fe^{2+} + 2e- &amp;amp;-&amp;gt; Fe}      &amp;amp; E_1&amp;amp;= x\,\mathrm{V}\\&#xA; \ce{Fe^{3+} + e-  &amp;amp;-&amp;gt; Fe^{2+}} &amp;amp; E_2&amp;amp;= y\,\mathrm{V}\\&#xA; \ce{Fe^{3+} + 3e- &amp;amp;-&amp;gt; Fe}      &amp;amp; E_3&amp;amp;= z\,\mathrm{V}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of &lt;em&gt;volts&lt;/em&gt; is &lt;em&gt;joules per coulomb&lt;/em&gt;. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x$ joules per coulomb, or&lt;/li&gt;&#xA;&lt;li&gt;$96500x$ joules per mole of electrons, or&lt;/li&gt;&#xA;&lt;li&gt;$2\times 96500x$ joules per mole of reaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the same fashion, the second half reaction has the electrons carrying&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$y$ joules per coulomb,&lt;/li&gt;&#xA;&lt;li&gt;$96500y$ joules per mole of electrons,&lt;/li&gt;&#xA;&lt;li&gt;the same $96500y$ joules per mole of reaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The third half reaction is the sum of the first two, so its electrons carry&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2\times 96500x + 96500y$ joules per mole of reaction, which is&lt;/li&gt;&#xA;&lt;li&gt;$3\times 96500 z$ joules per mole of reaction, or&lt;/li&gt;&#xA;&lt;li&gt;$96500z$ joules per mole of electrons, or&lt;/li&gt;&#xA;&lt;li&gt;$z$ joules per coulomb.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.&lt;/p&gt;&#xA;" OwnerUserId="5277" LastEditorUserId="4945" LastEditDate="2014-05-19T03:16:28.807" LastActivityDate="2014-05-19T03:16:28.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10905" PostTypeId="2" ParentId="10876" CreationDate="2014-05-19T02:26:31.687" Score="-3" Body="&lt;p&gt;The minimum pressure for the existence of a liquid solution is just below 1 atm. Because at 1 atm pressure(vapour pressure) every liquid gets boiled off corresponding to the temperature at which it attains the atmospheric pressure as it's vapour pressure.&lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2014-05-19T02:26:31.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10906" PostTypeId="1" AcceptedAnswerId="10909" CreationDate="2014-05-19T04:33:35.977" Score="6" ViewCount="982" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbiLF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the above picture, why is the $\ce{C-C}$ $\sigma$ bond composed of two $\ce{sp^2}$ orbitals while the $\ce{C-H}$ $\sigma$ bonds composed of $\ce{sp^3}$ hybrid orbitals? I don't see why; I thought the geometry around the carbons was trigonal planar. Why does hybridization differ across atoms in the molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking the textbook authors made a mistake; yes, there are four $\ce{C-H}$ bonds, but this does not mean there is $\ce{sp^3}$ hybridization &lt;em&gt;because&lt;/em&gt; there are &lt;em&gt;two&lt;/em&gt; carbon atoms, &lt;em&gt;each&lt;/em&gt; with two $\ce{C-H}$ bonds and one $\ce{C-C}$ bond. Perhaps the authors mistakenly took there to be one carbon and four $\ce{C-H}$ bond (i.e. methane.) That would be $\ce{sp^3}$ hybridization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps I'm wrong altogether. Please enlighten me! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Saw this in another text: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yq9CS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="73" LastEditDate="2015-05-21T23:45:55.207" LastActivityDate="2015-05-21T23:45:55.207" Title="Is this textbook bonding picture of hybridization in ethene wrong?" Tags="&lt;hybridization&gt;&lt;erratum&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10907" PostTypeId="2" ParentId="10869" CreationDate="2014-05-19T04:45:21.410" Score="4" Body="&lt;p&gt;As correct as eflaschuks approach is, I found it rather confusing, because much text and few mathematics. So I am adding them here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let us write down the Nernst Equation for an arbitrary reaction $\ce{Ox + z\cdot e- &amp;lt;=&amp;gt; Red}$:&#xA;\begin{aligned}&#xA;  E=E^\ominus(\ce{Red/Ox}) - z^{-1}\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Red}}{\ce{Ox}}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our given system that boils down to the following reaction and their potentials:&#xA;\begin{aligned}&#xA;  \ce{Fe^{2+} + $2\cdot$ e- &amp;amp;-&amp;gt; Fe}&amp;amp; E_1 &amp;amp;= E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}&amp;amp; \text{(1)}\\&#xA;  \ce{Fe^{3+} + $1\cdot$ e- &amp;amp;-&amp;gt; Fe^{2+}}&amp;amp; E_2 &amp;amp;= E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}&amp;amp; \text{(2)}\\&#xA;  \ce{Fe^{3+} + $3\cdot$ e- &amp;amp;-&amp;gt; Fe}&amp;amp; E_3 &amp;amp;= E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}}&amp;amp; \text{(3)}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are looking for the standard potential $E^\ominus(\ce{Fe/Fe^{3+}})$, so you have to consider the potential $E_3$, which can be build from $E_1$ and $E_2$:&#xA;$$E_3 = a\cdot E_1 + b\cdot E_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $(1)$, $(2)$ and $(3)$ yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{multline}&#xA;E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\&#xA;a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+&#xA;b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)&#xA;\end{multline}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to obtain the standard potential the logarithmic terms have to cancel. Therfore we can split these equations:&#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;E^\ominus(\ce{Fe/Fe^{3+}}) &amp;amp;= &#xA;a\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +&#xA;b\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) \\&#xA;&amp;amp;&amp;amp;\frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &amp;amp;= &#xA;a\cdot\left(\frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) +&#xA;b\cdot\left(\frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)\\&#xA;\therefore&amp;amp;&amp;amp; \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &amp;amp;= &#xA;\frac{3a}{2} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}} +&#xA;      3b     \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the $\ce{Fe^{2+}}$ terms to cancel it is fairly obvious, that $$a=\frac23,\\ b=\frac13,$$ therfore resulting in &#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;E^\ominus(\ce{Fe/Fe^{3+}}) &amp;amp;= &#xA;\frac{2}{3}\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +&#xA;\frac{1}{3}\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}).&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also check that against experimental data (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;):&#xA;\begin{array}{llrlr}\hline&#xA;\text{halfreaction} &amp;amp;&amp;amp;&amp;amp;&amp;amp; \text{potential}\\\hline&#xA;  \ce{Fe^{2+} + $2\cdot$ e- &amp;amp;-&amp;gt; Fe}     &amp;amp; E_1 &amp;amp;=&amp;amp; -0.44\mathrm{V} \\&#xA;  \ce{Fe^{3+} + $1\cdot$ e- &amp;amp;-&amp;gt; Fe^{2+}}&amp;amp; E_2 &amp;amp;=&amp;amp; 0.77\mathrm{V}\\&#xA;  \ce{Fe^{3+} + $3\cdot$ e- &amp;amp;-&amp;gt; Fe}     &amp;amp; E_3 &amp;amp;=&amp;amp; -0.04\mathrm{V}\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="37585" LastEditDate="2016-11-18T12:29:54.953" LastActivityDate="2016-11-18T12:29:54.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10908" PostTypeId="1" AcceptedAnswerId="10917" CreationDate="2014-05-19T05:12:20.907" Score="12" ViewCount="7492" Body="&lt;p&gt;In formamide, the nitrogens appear to be $\ce{sp^3}$ hybridized, implying tetrahedral geometry. However, analysis shows that the molecule is actually very nearly planar with bond angles close to 120 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: as suggested by Martin and another poster, hybridization is a rough concept. So perhaps the hybridization of nitrogen upon further analysis should best be described as somewhere inbetween $\ce{sp^3}$ and $\ce{sp^2}$. This however would still necessitate planarity, correct? Pi bonds are formed through the above and below electron-pairing in p-orbitals; effective bonding is achieved when these p-orbitals are parallel with respect to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking this has to do with the partial double bond character in the molecule (also appears to be some ionic character to the molecule - likely due to electron-withdrawing effects of the nitrogen and the oxygen). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvFFa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard answer. However, would intramolecular hydrogen bonding also play a role? Couldn't there be a hydrogen bond between the peripheral hydrogen on the nitrogen and the oxygen? Couldn't this also help with achieving the 120 degree bond angles? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-19T05:19:40.787" LastActivityDate="2014-05-19T09:29:02.707" Title="Rationalizing the Planarity of Formamide" Tags="&lt;resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10909" PostTypeId="2" ParentId="10906" CreationDate="2014-05-19T05:16:14.490" Score="7" Body="&lt;p&gt;You are not mistaken. Your book is wrong. If one desperately wants to involve the hybridisation concept (which makes sense for carbon), then the description should be in accordance with the molecular geometry of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;ethene&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/3b/Ethylene-CRC-MW-dimensions-2D.png/320px-Ethylene-CRC-MW-dimensions-2D.png&quot; alt=&quot;ethene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the carbons can be described as $\ce{sp_{x}p_{y}=sp^2}$ hybridised, hence there are four $\ce{H_{s}-C_{sp^2}}$ $\sigma$ bonds, one $\ce{C_{sp^2}-C_{sp^2}}$ $\sigma$ bond and one $\ce{C_{p_{z}}-C_{p_{z}}}$ $\pi$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the concept of hybridisation is no law, but yet another tool to describe bonding. In case of methane, (ethane,) ethene and ethyne it is quite simple to come to certain $\ce{sp^{x}}$ hybrid orbitals where $\ce{x}\in\mathbb{N}$ - different molecule may lead to fractional $\ce{x}$. (Not a rule, just a guideline)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to Dissenters comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinate system is chosen arbitrarily to indicate the molecular plane to be $xy$, as indicated in the picture taken from the &lt;a href=&quot;http://de.wikipedia.org/wiki/Ethen&quot; rel=&quot;nofollow noreferrer&quot;&gt;german article of wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/32/Liaison_pi.svg/500px-Liaison_pi.svg.png&quot; alt=&quot;ethene plane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the π orbitals have to be perpendicular to this plane they only can be composed of $\ce{p_{z}}$ orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular plane can be chosen different, e.g. as the $xz$ plane, the the $\ce{sp^2}$ orbitals would be formed from $\ce{sp_{x}p_{z}}$ and the $\pi$ bond would consist of $\ce{p_{y}}$ orbitals.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.797" LastActivityDate="2014-05-19T05:30:37.777" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10910" PostTypeId="5" CreationDate="2014-05-19T06:09:41.237" Score="0" Body="&lt;p&gt;For questions, that are about scientific books, and/or scientific literature in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be used for reference recommendations or specific questions about a book or an article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be used in combination with &lt;a href=&quot;/questions/tagged/reference-request&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reference-request&amp;#39;&quot; rel=&quot;tag&quot;&gt;reference-request&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/textbook-erratum&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;textbook-erratum&amp;#39;&quot; rel=&quot;tag&quot;&gt;textbook-erratum&lt;/a&gt;, to narrow the field.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:16:55.187" LastActivityDate="2014-05-19T07:16:55.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10911" PostTypeId="4" CreationDate="2014-05-19T06:09:41.237" Score="0" Body="For questions, that are about scientific books, and/or scientific literature in general. Maybe especially useful in combination with reference-request or textbook-erratum." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:17:24.343" LastActivityDate="2014-05-19T07:17:24.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10912" PostTypeId="5" CreationDate="2014-05-19T06:37:05.493" Score="0" Body="&lt;p&gt;This tag is for questions looking for a single specific, or a small number of (citable) references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions of the form &quot;What paper/book reported the mechanism of X?&quot;, &quot;Where can I find the original publication of synthesis X?&quot;, &quot;Where can I find popular applications of catalyst X?&quot;, or &quot;Where is the crystal structure of X reported?&quot; are reference requests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valid answers should include a complete citation, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Author(s), Initial(s): Year of publication, Title, Edition, Publisher, Place of Publication. (ISBN, URL if available)&lt;/li&gt;&#xA;&lt;li&gt;Author(s), Initial(s), Journal name &lt;strong&gt;Year of publication&lt;/strong&gt; &lt;em&gt;Volume number&lt;/em&gt; (Issue or part number), first (and last page) numbers (Title, DOI, URL if availabile)&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:17:13.267" LastActivityDate="2014-05-19T07:17:13.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10913" PostTypeId="4" CreationDate="2014-05-19T06:37:05.493" Score="0" Body="This tag is for questions looking for a single specific, or a small number of (citable) references." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:17:17.027" LastActivityDate="2014-05-19T07:17:17.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10914" PostTypeId="1" AcceptedAnswerId="10921" CreationDate="2014-05-19T06:43:04.053" Score="0" ViewCount="1514" Body="&lt;p&gt;After studying ionic equilibrium, I was just making some wild questions and I was unable to calculate the $\ce{pH}$ of this solution. Assume, 1 mole of each was added to $1L$ solution of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH}+\ce{NH4OH&amp;lt;=&amp;gt;CH3COONH4}+\ce{H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction will happen till all $\ce{CH3COOH}$ is consumed by $\ce{NH4OH}$. More of $\ce{NH4OH} $ will be formed by Le Chatelier's principle. So, the answer must be same as that of a weak acid and weak base salt : $pH=\dfrac{1}{2}(pK_w+pK_a-pK_b)$.&#xA;But how will we calculate if there was $2$ mole of acetic acid?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4945" LastEditDate="2014-05-19T06:45:30.027" LastActivityDate="2014-05-19T13:30:10.250" Title="pH of a solution of acetic acid and ammonium acetate" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10915" PostTypeId="5" CreationDate="2014-05-19T06:43:17.433" Score="0" Body="&lt;p&gt;For questions that deal with a specific statement of a specified book. These questions should always include a complete reference:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Author(s), Initial(s): Year of publication, Title, Edition, Publisher, Place of Publication. (ISBN, URL if available) &lt;strong&gt;&lt;em&gt;Add page number, table/ figure/ scheme accordingly&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Author(s), Initial(s), Journal name &lt;strong&gt;Year of publication&lt;/strong&gt; &lt;em&gt;Volume number&lt;/em&gt; (Issue or part number), first (and last page) numbers (Title, DOI, URL if availabile) &lt;strong&gt;&lt;em&gt;Add page number (for longer publications e.g. reviews), table/ figure/ scheme accordingly&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;online resource with perma-link (e.g. wikipedia)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:16:45.960" LastActivityDate="2014-05-19T07:16:45.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10916" PostTypeId="4" CreationDate="2014-05-19T06:43:17.433" Score="0" Body="For questions that deal with a specific statement of a specified book. These questions should always include a complete reference." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-19T07:17:01.257" LastActivityDate="2014-05-19T07:17:01.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10917" PostTypeId="2" ParentId="10908" CreationDate="2014-05-19T09:29:02.707" Score="13" Body="&lt;p&gt;Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon is obviously $\ce{sp^2}$ hybridised (as this concept is very well applicable here), hence organising it's ligands in one plane with roughly $120^\circ$ angles. Naturally one would assume the nitrogen to be $\ce{sp^3}$ hybridised, which is the case for most amines. However, the inversion barrier for these molecules is (depending on the substituents) very low. $$\ce{[NH3]^{pyr-top} &amp;lt;=&amp;gt; [NH3]^{TS-plan} &amp;lt;=&amp;gt; [NH3]^{pyr-bot}}$$ For the nitrogen that means going from $\ce{sp^3}$ to $\ce{sp^2}$ and back again. You can now stabilise the intermediated structure with conjugation, and that is exactley the case here. In your chart, that refers to entry 2. This causes the nitrogen to most likely be of $\ce{sp^2}$ hybridisation and the lone pair of entry one would be in a $\ce{p}$ orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugation happens as stated in you chart by the overlap of that orbital with the antibonding $\pi^*~\ce{C-O}$ orbital. This causes the $\ce{N-C}$ bond order to increase, while the $\ce{C-O}$ BO has to decrease. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All those resonance structures are only descriptions of extreme states, the truth lies between them. The following scheme considers the most common ones and adds a third, that might explain delocalisation (in a non-traditional Lewis way) up to a certain visual point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/myaBa.png&quot; alt=&quot;resonance structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In molecular orbital theory you can form 3-centre orbitals from all molecules perpendicular &#xA;to the molecular plain. If you choose this plane to be $xy$ then the contributing orbitals will be $\ce{p_{$z$}}$. The following scheme may support this claim, the dipicted orbitals were obtained by a BP86/cc-PVTZ calculation. (The HOMO is an in-plane lone pair of oxygen.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXdrr.gif&quot; alt=&quot;orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there is certainly no intra molecular hydrogen bond ($d(\ce{O-H}\approx 2.57$ same level), there will also most certainly be an attraction between the $\ce{C-O}$ and $\ce{N-H}$ bond helping to stabilize the planarity. But I will not go into detail about that, because it would involve breaking away from the very handy hybridisation concept.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-19T09:29:02.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10918" PostTypeId="1" AcceptedAnswerId="10933" CreationDate="2014-05-19T10:28:23.523" Score="3" ViewCount="9856" Body="&lt;p&gt;I am having trouble with a problem involving a Semi-infinite potential square well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written down some notes that I added to the post. I have problems understanding the physical situation. See the figure in the notes. When the potential is just shifted upwards, so that the &lt;span class=&quot;math-container&quot;&gt;$V(x)=0$&lt;/span&gt; region is &lt;span class=&quot;math-container&quot;&gt;$0&amp;lt;x&amp;lt;a$&lt;/span&gt;, I have no trouble understanding the solution. However, I am just really confused about the negative potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you will see in the notes, I have interpreted Region I as a bound state (e.g. an electron bound to a proton). Region II, I believe, is the free state (e.g. &quot;bound&quot; electron in energy level &lt;span class=&quot;math-container&quot;&gt;$n=\infty$&lt;/span&gt;), since &lt;span class=&quot;math-container&quot;&gt;$V(x) = 0$&lt;/span&gt;.  Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I face some troubles with the boundary conditions, as one of my wave-functions simply disappear! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ocNm6.jpg&quot; alt=&quot;Page 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDqKI.jpg&quot; alt=&quot;Page 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MHMCs.jpg&quot; alt=&quot;Page 3&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2020-05-25T16:21:24.730" LastActivityDate="2020-05-25T16:21:24.730" Title="Semi-Infinite Potential Square Well: Negative Potential" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10920" PostTypeId="2" ParentId="10918" CreationDate="2014-05-19T12:08:54.267" Score="1" Body="&lt;p&gt;I believe there are two seperate mistakes, one in region I and one in region II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For region I, it looks like you made a sign error, and that you should have $-V_0 - E$ instead of $V_0 - E$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this may seen trival at first, the sign is important to the form of the solution to the differential equation (wavefunction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For region II, the way the diagram of the question is drawn, E is negative.  Assure that this fact is properly accounted for in the form of the solution.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-19T12:08:54.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10921" PostTypeId="2" ParentId="10914" CreationDate="2014-05-19T12:46:35.837" Score="2" Body="&lt;p&gt;You have some mistaken assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4OH}$ and $\ce{CH3COONH4}$ are not molecules that exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assuption &quot;all $\ce{CH3COOH}$ is consumed&quot; is incorrect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you say more $\ce{NH4OH}$ will be formed? How is this possibe? What can it be formed from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Review that &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson–Hasselbalch equation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-19T12:46:35.837" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="10924" PostTypeId="2" ParentId="10906" CreationDate="2014-05-19T13:58:40.690" Score="3" Body="&lt;p&gt;Martin has provided an excellent answer to your question.  And although I've arrived late at the party I'd like to comment on the hybridization in ethylene.  All of the introductory texts I've come across use the same bonding description as your text: the carbons in ethylene are hybridized such that we have 6 carbon $\ce{sp^2}$ bonds (4 $\ce{C-H}$, 2 $\ce{C-C}$) and 1 pi bond based on overlap between the unhybridized p porbital remaining on each carbon.  Actually, if we refine this analysis to take into account the known $\ce{H-C-H}$ angle of 117 degrees, we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bonding are $\ce{sp^{2.2}}$ hybridized and the carbon orbitals involved in the $\ce{C-C}$ bond are $\ce{sp^{1.6}}$ hybridized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to point out that, while staying within the MO framework where we only consider the s and p orbitals on carbon, there is an alternate way of describing the hybridization in ethylene - and these two descriptions predict the same physical reality, so they are equivalent descriptions.  The texts arbitrarily choose to hybridize only 2 of the carbon p orbitals along with the carbon s orbital.  It is just as valid to mix all 3 p orbitals with the s orbital.  If we constrain 2 of the resultant orbitals just as we did above (those that will bond to hydrogen) to produce an $\ce{H-C-H}$ angle of 117 degrees, then these two orbitals will be $\ce{sp^{2.2}}$ hybridized, just as the earlier analysis concluded.  The remaining two orbitals on carbon, however, can consequently be described as $\ce{sp^{4.3}}$ hybridized.  The picture that emerges from this alternate hybridization scheme is one of bent bonds - a 2-membered ring if you will!  Something to think about.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-19T13:58:40.690" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10925" PostTypeId="2" ParentId="10906" CreationDate="2014-05-19T15:10:06.340" Score="0" Body="&lt;p&gt;The bonding model commonly accepted is neither &quot;right&quot; or &quot;wrong&quot; - it is a model, i.e. an approximation of reality using simple concepts (sort of like describing movement of trains as being linear and at constant speed, and ignoring wobbles, curves, speeding up and slowing down). There is an alternate model, although not popular, that postulates $sp^3$ hybridization only. Try to search for &quot;banana bond theory&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5557" LastEditorUserId="5016" LastEditDate="2014-05-19T16:00:18.793" LastActivityDate="2014-05-19T16:00:18.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10926" PostTypeId="2" ParentId="10898" CreationDate="2014-05-19T15:25:19.887" Score="8" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zXkZo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light from a flame, the sun, or a star, was passed though a narrow slit, then directed through a prism (F) and observed.  The prism is rotatable about a vertical axis to adjust the wavelength range being observed.  Bright lines in the spectrum are emission transitions and dark lines are absorption transitions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further progress was made using &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffraction_grating&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffraction gratings&lt;/a&gt; instead of prisms and using photography instead of naked eye observation.  Photography permitted observation of infrared and ultraviolet lines.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By these techniques, known elements were detected in the sun and other stars, and helium was discovered in the sun before it was discovered on Earth, all before 1900.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information see &lt;a href=&quot;http://www.neafsolar.com/bb/spectroscopy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectroscopy History&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-05-19T16:18:01.037" LastActivityDate="2014-05-19T16:18:01.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10928" PostTypeId="1" AcceptedAnswerId="10930" CreationDate="2014-05-19T17:11:14.997" Score="1" ViewCount="88" Body="&lt;p&gt;Is visible spectroscopy the only non-electronic method of all the spectroscopy method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some of the spectroscopy method is possible being used because of electronics.&lt;/p&gt;&#xA;" OwnerUserId="370" LastActivityDate="2017-07-15T08:42:32.413" Title="Is visible spectroscopy the only non-electronic method of all the spectroscopy method?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10929" PostTypeId="1" AcceptedAnswerId="10932" CreationDate="2014-05-19T17:25:45.063" Score="1" ViewCount="3700" Body="&lt;p&gt;I don't think that cyclohexenone should have any resonance forms. Moving the double bond around would necessitate moving hydrogens around as well. For example, I know that benzene exhibits resonance (duh) and the second picture attached shows the pi bonding electrons shifting but doesn't show any hydrogens shifting positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TG324.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8yZT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, is there a easier way to determine if a resonance form of a cyclic compound is valid? Drawing out each individual atom and bond is time-consuming! Should I use Huckel's rule? Do all cyclic compounds which exhibit resonance follow Huckel's rule? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-19T17:31:22.883" LastActivityDate="2014-05-19T17:54:52.240" Title="Resonance in Cyclohexenone" Tags="&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10930" PostTypeId="2" ParentId="10928" CreationDate="2014-05-19T17:29:19.520" Score="3" Body="&lt;p&gt;No, Ultraviolet spectroscopy and Infra-red spectroscopy were practiced without any use of electricity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infra-red and ultraviolet light were both known by 1801.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultraviolet absorption lines had been observed by 1843.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="42741" LastEditDate="2017-07-15T08:42:32.413" LastActivityDate="2017-07-15T08:42:32.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10931" PostTypeId="2" ParentId="6956" CreationDate="2014-05-19T17:37:04.180" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chaotropic_agent&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chaotropic&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Kosmotropic&quot; rel=&quot;nofollow noreferrer&quot;&gt;kosmotropic&lt;/a&gt; effects of solutes can occur in any solvent in which interactions between solvent molecules lead to a more ordered structure, like a hydrogen bonding network in water. Such a network will also be present in alcohols like ethanol, although to a lesser degree, because the alcohol molecules have only one proton which can act as a hydrogen bond donor. The Hofmeister (or lyotropic) series for alcohols might still be similar to that for water (not identical), but it will be more different with increasing structural difference between the solvent molecule and water. In fact, the Hofmeister series for water is still not fully understood on a molecular basis, as it is described in this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/76/molecular-explanation-of-the-hofmeister-series&quot;&gt;post&lt;/a&gt;, and this makes it difficult to predict Hofmeister series for other polar, non-aqueous solvents.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-05-19T17:37:04.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10932" PostTypeId="2" ParentId="10929" CreationDate="2014-05-19T17:42:37.340" Score="6" Body="&lt;p&gt;The structure you've drawn on the left does have a resonance structure, but not the one you've drawn on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMi2F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing resonance structures out by hand gets quicker over time.  You find that you don't need to actually put in a &quot;C&quot; for carbon, or label the hydrogen atoms.  Huckel's rule generally applies to planar, cyclic molecules with continuous conjugation, although it has also be applied to transition states where an S or P orbital on a shifting C or H moiety completes a loop of continuous conjugation (for example, the 1,5 hydrogen shift in cyclopentadiene).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-19T17:54:52.240" LastActivityDate="2014-05-19T17:54:52.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10933" PostTypeId="2" ParentId="10918" CreationDate="2014-05-19T18:05:43.553" Score="4" Body="&lt;p&gt;Answering your quantum mechanics question seems to become a hobby of mine :). First let me say that there is nothing wrong with shifting the potential up so that $V_{0} = 0$. This choice would indeed simplify the problem as it gets rid of negative energy values and thus avoids the following point of confusion: when $E &amp;lt; 0$ the wave function of a particle in region $\text{II}$ is not a plane wave $\mathrm{e}^{\mathrm{i} k x}$ ($k \in \mathbb{R}$) like that of a free particle moving in vacuum but a decaying exponential $\mathrm{e}^{-\kappa x}$ (with $\kappa \in \mathbb{R}$). So, the choice $V_{0} = 0$ is advisable. Shifting the potential back later is always possible and might be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, of course the problem is solvable irrespective of the choice of $V_{0}$. I think your problem in this case is that you enforced an incorrect boundary condition. To avoid confusion I will start from the beginning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will only consider the case $E &amp;lt; 0$ in the following, so $E$ is always a negative number between $-V_{0}$ and $0$. For region $\text{II}$ you have to solve the Schroedinger Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &amp;amp;= E \psi_{\text{II}}(x) \\&#xA;  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &amp;amp;= - \frac{2mE}{\hbar^{2}} \psi_{\text{II}}(x) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here comes the first &quot;problem&quot; with the choice of $V_{0}$: In your solution you chose $\frac{2mE}{\hbar^{2}} = k^{2}$ as you would naturally do for a free particle. But since $E$ is negative $k$ will be a complex number (and not a real number as it would be for a free particle) and so this differential equation does not have an oscillatory solution $\psi_{\text{II}}(x) = A \sin(kx) + B \cos(kx)$ with $k \in \mathbb{R}$. Rather you get a decaying exponential function by doing the following: $- \frac{2mE}{\hbar^{2}}$ will be a positive number so you simply set $- \frac{2mE}{\hbar^{2}} = \kappa^{2}$ to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &amp;amp;= \kappa^{2} \psi_{\text{II}}(x) \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this differential equation has a solution of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{II}}(x) = A_{\text{II}} \mathrm{e}^{\kappa x} + B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you can enforce the boundary condition that the wavefunction $\psi_{\text{II}}$ must be square integrable over the whole interval $a &amp;lt; x &amp;lt; \infty$ (Note that you don't have to consider the interval $-\infty &amp;lt; x &amp;lt; \infty$ but only the interval on which $\psi_{\text{II}}$ is defined). Since $\lim_{x \to \infty} \mathrm{e}^{\kappa x} = \infty$ this mean that the prefactor $A_{\text{II}}$ must be equal to zero, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{II}}(x) = B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So much for region $\text{II}$. Let's move on to region $\text{I}$: Here you have to solve the Schroedinger Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} - V_{0} \psi_{\text{I}}(x)  &amp;amp;= E \psi_{\text{I}}(x) \\&#xA;  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &amp;amp;= - \frac{2m(E + V_{0})}{\hbar^{2}} \psi_{\text{I}}(x) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, care must be taken: Since $-V_{0} &amp;lt; E &amp;lt; 0$ the term $(E + V_{0})$ is positive and so $- \frac{2m(E + V_{0})}{\hbar^{2}}$ is a negative number and you can set $- \frac{2m(E + V_{0})}{\hbar^{2}} = -k^{2}$ which leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &amp;amp;= - k^{2} \psi_{\text{I}}(x) \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differential equation has a solution of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{I}}(x) = A_{\text{I}} \mathrm{e}^{\mathrm{i} k x} + B_{\text{I}} \mathrm{e}^{-\mathrm{i} k x} \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You could also formulate an equivalent solution in terms of $\sin$ and $\cos$ functions using Euler's formula if that is more to your liking. But I prefer plane waves ;))&#xA;This is the point where you made your main mistake: You discarded one of the terms in your $\psi_{\text{I}}$ wavefunction because it blew up for $x \to \infty$. But $\psi_{\text{I}}(x)$ is only defined on the interval $0 \leq x \leq a$ and within this interval both exponential terms are perfectly square-integrable and you can't discard any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you just have to remember that $\psi_{\text{III}}(x) = 0 \ \forall \ x$ and enforce the boundary conditions given by the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{I}}(0) &amp;amp;= \psi_{\text{III}}(0) \quad &amp;amp;&amp;amp;\Rightarrow \quad A_{\text{I}} \mathrm{e}^{0} + B_{\text{I}} \mathrm{e}^{0} = 0 \quad &amp;amp;&amp;amp;\Rightarrow \quad B_{\text{I}} = -A_{\text{I}} \\&#xA;  \psi_{\text{I}}(a) &amp;amp;= \psi_{\text{II}}(a) \quad &amp;amp;&amp;amp;\Rightarrow \quad A_{\text{I}} \mathrm{e}^{\mathrm{i} k a} - A_{\text{I}} \mathrm{e}^{-\mathrm{i} k a} = B_{\text{II}} \mathrm{e}^{-\kappa a} &amp;amp;&amp;amp; \\&#xA;  &amp;amp; &amp;amp;&amp;amp;\hphantom{\Rightarrow} \ \quad A_{\text{I}} \underbrace{\bigl(\mathrm{e}^{\mathrm{i} k a} - \mathrm{e}^{-\mathrm{i} k a} \bigr)}_{= \, 2 \mathrm{i} \sin(k a)} = B_{\text{II}} \mathrm{e}^{-\kappa a} \quad &amp;amp;&amp;amp;\Rightarrow \quad B_{\text{II}} = 2 \mathrm{i} \sin(k a) \mathrm{e}^{\kappa a} A_{\text{I}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning your comment: You can indeed enforce another boundary condition. The probability&#xA;flux is required to be continuous across any surface, since otherwise the surface would&#xA;contain sources or sinks. This entails the requirement that both $\psi$ and $\nabla \psi$ must be continuous. The continuity condition $\psi_{\text{I}}(a) = \psi_{\text{II}}(a)$ has already been enforced but the one for the gradient is still left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &amp;amp;= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above discussion we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{I}}(x) &amp;amp;= A_{\text{I}} \underbrace{\left(\mathrm{e}^{\mathrm{i} k x} - \mathrm{e}^{-\mathrm{i} k x} \right)}_{= \, 2 \mathrm{i} \sin(k x)} = 2 \mathrm{i} A_\text{I} \sin (kx) = C_\text{I} \sin(kx) &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \psi_{\text{II}}(x) &amp;amp;= \underbrace{2 \mathrm{i} A_{\text{I}} \sin(k a)}_{= \, \psi_{\text{I}}(x=a)} \, \mathrm{e}^{\kappa a} \mathrm{e}^{- \kappa x} = \psi_{\text{I}}(x \! = \! a) \, \mathrm{e}^{- \kappa (x - a)} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $C_\text{I} = 2 \mathrm{i} A_{\text{I}}$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &amp;amp;= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \\&#xA;  k C_\text{I} \cos(ka) &amp;amp;= - \kappa C_\text{I} \sin(ka) \\&#xA;  \kappa &amp;amp;= - \frac{k}{\tan(k a)}  \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this equation looks innocent enough it is not easily solved for $E$ and you will need the assistance of something like Mathematica for that.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-05-21T09:57:08.490" LastActivityDate="2014-05-21T09:57:08.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10934" PostTypeId="2" ParentId="7106" CreationDate="2014-05-19T19:32:31.593" Score="3" Body="&lt;p&gt;The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, &lt;a href=&quot;http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stability_constants_of_complexes&quot; rel=&quot;nofollow&quot;&gt;stability constants&lt;/a&gt; of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-19T19:32:31.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10935" PostTypeId="2" ParentId="4021" CreationDate="2014-05-19T19:56:53.813" Score="2" Body="&lt;p&gt;I'm not sure you did do the experiment wrong. I performed this experiment and it closely matched this &lt;a href=&quot;https://www.youtube.com/watch?v=sZaPcO4hzVU&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; which had a light yellow liquid and foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my results&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMEsV.jpg&quot; alt=&quot;After first filtration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I filtered it a couple more times to remove more foam. Then added the aluminum foil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V65nD.jpg&quot; alt=&quot;After aluminum foil&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I got at the end before I filtered the liquid out at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end I was able to extract 6.2 grams of bismuth powder from 67 tablets of pepto (a bit less than the theoretical ~10 grams I was hoping to get).&lt;/p&gt;&#xA;" OwnerUserId="5362" LastActivityDate="2014-05-19T19:56:53.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10936" PostTypeId="2" ParentId="10901" CreationDate="2014-05-19T20:28:05.130" Score="4" Body="&lt;p&gt;As far as hybridization goes, the hybridization of all of the atoms must be the same in all resonance structures. In considering resonance structures, none of the nuclei should be moved in space. Since hybridization is designed to explain the arrangement of nuclei/electrons in space, hybridization cannot change between resonance structures, otherwise nuclei may be forced to occupy new positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, this means that certain atoms may appear to be sp3 hybridized but with consideration of resonance structures actually should be treated as sp2. This often means that a lone pair is not in an sp3 orbital as expected, but an unhybridized p-orbital.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-05-19T20:28:05.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10938" PostTypeId="1" CreationDate="2014-05-19T21:19:15.337" Score="0" ViewCount="78" Body="&lt;p&gt;I haven't taken a chemistry class in several years, and I have honestly been searching the web for the past hour and can't find anything. I know this is something basic we learned back in high school chemistry but maybe I am not searching for the right term...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula? I know how to calculate distillation speed based on wattage, and the vaporization constant of the molecule, but cannot find any particular formula that can calculate how that number changes under reduced pressure. &lt;/p&gt;&#xA;" OwnerUserId="5562" LastActivityDate="2017-01-16T11:28:18.670" Title="What is the formula for heat of vaporization under reduced pressure?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10939" PostTypeId="1" AcceptedAnswerId="10940" CreationDate="2014-05-19T23:23:50.050" Score="6" ViewCount="26151" Body="&lt;p&gt;&lt;a href=&quot;http://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2000-local-olympiad-exam.pdf&quot; rel=&quot;noreferrer&quot;&gt;Link to exam&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Which metal reacts most vigorously with water?&lt;/p&gt;&#xA;&lt;p&gt;(A) Ca (B) K (C) Mg (D) Na&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The given answer is B and the reasoning is that is is the most electropositive since it the furthest left and down according to periodic trends.&lt;/p&gt;&#xA;&lt;p&gt;However, isn't &lt;span class=&quot;math-container&quot;&gt;$\ce{K}$&lt;/span&gt; the conjugate acid from a strong base -- &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt;? So isn't the conjugate base of a strong acid quite weak? So why would &lt;span class=&quot;math-container&quot;&gt;$\ce{K}$&lt;/span&gt; react strongly?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-10-23T05:58:12.003" Title="Why does potassium react the most vigorously with water?" Tags="&lt;acid-base&gt;&lt;kinetics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10940" PostTypeId="2" ParentId="10939" CreationDate="2014-05-20T00:28:35.823" Score="4" Body="&lt;p&gt;I might be wrong here, but I believe that the question asks about the metals in their atomic (non ionic) form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for a $\ce{KOH}$ bond to form, the K must be ionized (since the $\ce{KOH}$ bond is ionic). This means that when $\ce{KOH}$ dissolves in water it'll form the $\ce{K}^+$ electrolyte, which is not the same thing as a pure $\ce{K}$ atom with its 19th electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\ce{K}$ atom, then, since it's the most electropositive of that list, is the one to react &quot;more vigorously&quot; (which I guess it refers to more speed/energy) with water (a polar molecule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helped you. I'm sorry if I'm wrong!&lt;/p&gt;&#xA;" OwnerUserId="5390" LastEditorUserId="-1" LastEditDate="2014-05-20T16:47:49.033" LastActivityDate="2014-05-20T16:47:49.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10941" PostTypeId="1" AcceptedAnswerId="10942" CreationDate="2014-05-20T01:45:59.953" Score="3" ViewCount="6728" Body="&lt;p&gt;I understand what causes elements to be conductive or nonconductive. But what causes a substance say rubber to be nonconductive at a molecular level?&lt;/p&gt;&#xA;" OwnerUserId="316" LastActivityDate="2014-05-20T17:38:46.110" Title="What attributes cause a substance to be nonconductive or conductive" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10942" PostTypeId="2" ParentId="10941" CreationDate="2014-05-20T02:02:46.967" Score="1" Body="&lt;p&gt;Conductivity requires mobile electrons. Metals tend to have loosely held &lt;em&gt;valence&lt;/em&gt; electrons so metals are generally conductive. Metals have loosely held valence electrons for two reasons: metals tend not to be very electronegative (able to stabilize electron density) and metals tend to be somewhat big atoms by nature of many metals having a d-group of electrons. D electrons are further from the nucleus on a time average basis than s electrons or p electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, conductors do not have to be metals. Consider graphite. Graphite is $\ce{sp^2}$ hybridized. Graphite is made solely of carbon atoms - and carbon isn't a metal! The mobile pi electrons in graphite due to resonance allows graphite to be a good conductor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also note that conductivity does not always imply mobile electrons; solutions can be conductive too and this is due to mobile &lt;em&gt;electrolytes&lt;/em&gt; (ions). For example, you will find that deionized water is completely unconductive. However, regular tap water is generally conductive due to the presence of ions (e.g. chloride ions from chlorination). Gatorade is also very conductive because it is especially formulated to contain a lot of electrolytes. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-20T17:38:46.110" LastActivityDate="2014-05-20T17:38:46.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10943" PostTypeId="2" ParentId="10939" CreationDate="2014-05-20T02:10:01.543" Score="14" Body="&lt;p&gt;I interpret &quot;reacts most vigorously&quot; to mean which metal reacts the fastest.  This is a question of kinetics, not thermodynamics (e.g. the answer is not based on the enthalpy of the reaction).  Kinetic rates are determined by the height of the energy barrier that needs to be surmounted, we need to determine and compare the activation energies in this process for the various metals.  The activation energy in this process is determined by the following two steps: $$\ce{M(s) -&amp;gt; M(g)}$$ $$\ce{M(g) -&amp;gt; M^+(g) + e^-}$$ $$\ce{or}$$ $$\ce{M(g) -&amp;gt; M^{+2}(g) + 2 e^-}$$ depending on whether we are analyzing a group I or group II metal.  The following link provides a nice analysis and tabulation of the energies required for these two steps with the group I metals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/s-Block_Elements/Group__1%3a_The_Alkali_Metals/Properties_of_the_Group_1_Elements/Reactions_of_Group_1_Elements_with_Water&quot;&gt;Group I link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this link does the same for the Group II metals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/s-Block_Elements/Group__2_Elements%3a_The_Alkaline_Earth_Metals/Chemistry_of_the_Group_2_Elements/Reactions_of_Group_2_Elements_with_Water&quot;&gt;Group II link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll summarize their findings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal ...   Activation Energy (kJ/mol)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K   ...         +508&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Na  ...        +603&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mg  ...       +2200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ca  ...       +1950&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CONCLUSION: Potassium has the lowest activation energy and should react the fastest.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-05T16:18:48.530" LastActivityDate="2014-07-05T16:18:48.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10944" PostTypeId="1" AcceptedAnswerId="10946" CreationDate="2014-05-20T02:35:03.327" Score="16" ViewCount="8298" Body="&lt;p&gt;I have never seen anything other than a set of rules like these when textbooks present how to assign oxidation numbers. Such as these: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YlxeX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we keep in mind that oxidation numbers are simply imaginary numbers which suppose all bonding to be ionic - i.e. not electron sharing - and if we keep in mind simply &lt;em&gt;relative&lt;/em&gt; electronegativity, we can easily work out the oxidation state of &lt;em&gt;any&lt;/em&gt; element in &lt;em&gt;any&lt;/em&gt; compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take water. Bonding order: $\ce{H-O-H}$. Oxygen has two lone pairs. Oxygen is more electronegative than hydrogen. So we suppose that oxygen takes both electrons in both bonding pairs. Oxygen has 8 electrons. Its valence is 6. 6-8 is -2; oxygen has two more electrons assigned to it than what it has in its valence, so naturally its oxidation state is &lt;em&gt;negative&lt;/em&gt; 2. No rules needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now hydrogen peroxide. $\ce{H-O-O-H}$. Here we have an oxygen-oxygen bond so in this case, neither element wins the electronegativity battle. Electrons are split between the oxygen and the oxygen. However, since the oxygens are much more electronegative than the hydrogens, we assign both electrons in both the $\ce{O-H}$ bonding pairs to oxygen. Oxygen has 7 electrons assigned to it; oxygen has a valence of 6; oxidation state: -1. No need to memorize exceptions as stated in the table above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could go on. However, I am curious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Has anyone been taught to assign oxidation states this way?&#xA;2) If not, what do you think of this method? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-07-18T07:52:28.250" LastActivityDate="2019-06-18T08:58:04.100" Title="Electronegativity Considerations in Assigning Oxidation States" Tags="&lt;oxidation-state&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10945" PostTypeId="2" ParentId="10939" CreationDate="2014-05-20T05:25:10.067" Score="10" Body="&lt;p&gt;Please read ron's post before continuing here (too long for a comment). Its reasoning is very good and makes a lot of sense. The activation barrier is the defining component for how vigorously components react. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His presented reaction may cover the initial activation barrier, but subsequent reactions also have to be taken into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall reaction that will occur in the case of sodium can be formulated like this:&#xA;\begin{aligned}&#xA;  \ce{2Na(s) + 2H2O(l) &amp;amp;-&amp;gt; 2Na+(aq) + H2(g) ^ + 2{}^{-}OH(aq)}&amp;amp;&#xA;  \Delta H &amp;amp;= -184~\mathrm{kJ/ mol}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the start of the reaction has to be considered as the atomisation energy of the metal, after that the ionisation energy, both of these reactions need energy. Energy is released to the system by the hydration of the ion. For sodium this boils down to&#xA;\begin{aligned}&#xA;  \ce{Na(s)  &amp;amp;-&amp;gt; Na(g)}             &amp;amp; \Delta H &amp;amp;= +109~\mathrm{kJ/ mol}\\&#xA;  \ce{Na(g)  &amp;amp;-&amp;gt; Na+(g) + e-}       &amp;amp; \Delta H &amp;amp;= +494~\mathrm{kJ/ mol}\\&#xA;  \ce{Na+(g) &amp;amp;-&amp;gt; Na+(aq)}           &amp;amp; \Delta H &amp;amp;= -406~\mathrm{kJ/ mol}\\[2ex]\hline&#xA;  \ce{Na(s)  &amp;amp;-&amp;gt;[\ce{H2O}] Na+(aq)} &amp;amp; \Delta H &amp;amp;= +197~\mathrm{kJ/ mol}.\\\hline\hline&#xA;\end{aligned} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why would this reaction happen, when there is no energy released? The cause of this are the secondary and tertiary reactions, that then again fuel the initial reactions. So the key question here is, what happens to the electron? &#xA;\begin{aligned}&#xA;\ce{2H2O + 2e- &amp;amp;&amp;lt;=&amp;gt;  H2 ^ + 2{}^{-}OH}\\&#xA;\end{aligned}&#xA;While I cannot find a value for this reaction, however I strongly suspect it to be negative. It more certainly gives us a hint on what to consider else, i.e. the hydration of hydroxide ions. (&lt;a href=&quot;http://dx.doi.org/10.1021/ed054p540&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;)&#xA;\begin{aligned}&#xA;  \ce{{}^{-}OH &amp;amp;-&amp;gt; {}^{-}OH(aq)} &amp;amp; \Delta H &amp;amp;= -460~\mathrm{kJ/mol}&#xA;\end{aligned}&#xA;So this releases a lot of energy to the system, helping to atomize and ionize the sodium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tertiary reaction also has to be considered. Because of the produced heat, hydrogen will immediately burn when coming in contact with air. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;)&#xA;\begin{aligned}&#xA;  \ce{H2(g)  + 1/2 O2(g) &amp;amp;-&amp;gt;  H2O(g)} &amp;amp; \Delta H &amp;amp;= -242~\mathrm{kJ/mol}\\&#xA;  \ce{2H2(g) +     O2(g) &amp;amp;-&amp;gt; 2H2O(g)} &amp;amp; \Delta H &amp;amp;= -484~\mathrm{kJ/mol}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all these reactions successfully overcome the activation barrier once the reaction started. This also then has the effect of multiplying the initial reaction, causing a chain reaction and hence reacting even more furiously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch &lt;a href=&quot;http://youtu.be/ODf_sPexS2Q&quot; rel=&quot;nofollow&quot;&gt;Sodium burn in water&lt;/a&gt; and at one point you see the hydrogen ignite and how vigorous this reaction is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to all other metals, but as ron stated the kinetics depend on the activation barrier.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-10-23T05:58:12.003" LastActivityDate="2015-10-23T05:58:12.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10946" PostTypeId="2" ParentId="10944" CreationDate="2014-05-20T06:28:05.820" Score="19" Body="&lt;p&gt;In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronegativity battle scheme is most helpful for all kinds of compounds since it is one of the most generic ways to derive oxidation states. The table represents just a cheat sheet that might be very helpful in the beginning. If you are spending most of your time with chemistry, this table will be present as some sort of muscle memory - usually referred to as chemical intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC &lt;a href=&quot;http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gold book&lt;/a&gt; now defines oxidation state as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;oxidation state&lt;/strong&gt;&lt;br&gt;&#xA;  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:&lt;br&gt;&#xA;  &lt;em&gt;OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The underlying principle is that the ionic sign in an &lt;span class=&quot;math-container&quot;&gt;$\ce{AB}$&lt;/span&gt; molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear &lt;span class=&quot;math-container&quot;&gt;$\ce{AA}$&lt;/span&gt; bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/bzfzQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bzfzQ.png&quot; alt=&quot;O04365-1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/bHvBW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bHvBW.png&quot; alt=&quot;O04365-2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;NOTE:&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;em&gt;Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;When used in chemical nomenclature as a symbol, the OS value is in roman numerals.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating &lt;span class=&quot;math-container&quot;&gt;$\text{OS} = +1$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$−2$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; yields correct OS in oxides, hydroxides, and acids like &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;; with coverage extended to &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Source:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2014,&lt;/strong&gt; &lt;em&gt;86&lt;/em&gt; (6), 1017–1081 &lt;a href=&quot;https://doi.org/10.1515/pac-2013-0505&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1515/pac-2013-0505&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2016,&lt;/strong&gt; &lt;em&gt;88&lt;/em&gt; (8), 831–839 &lt;a href=&quot;https://doi.org/10.1515/pac-2015-1204&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1515/pac-2015-1204&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from &lt;a href=&quot;https://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;old.iupac.org&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I recommend having a look at the sources cited above, they contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to 2016 the IUPAC defined oxidation states in their &lt;a href=&quot;https://web.archive.org/web/20141102154648/http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gold book (via the Internet Archive)&lt;/a&gt; differently. I am including this definition, since it will be still present in prominent text books and it is what lead to the postulates given in the table of the original question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;oxidation state&lt;/strong&gt; (deprecated, pre-2016 definition)&#xA;  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;the oxidation state of a free element (uncombined element) is zero; &lt;/li&gt;&#xA;  &lt;li&gt;for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;&lt;/li&gt;&#xA;  &lt;li&gt;hydrogen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and oxygen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt; when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; in hydrides of active metals, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{LiH}$&lt;/span&gt;, and oxygen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; in peroxides, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt;;&lt;/li&gt;&#xA;  &lt;li&gt;the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{S8}$&lt;/span&gt; (elementary sulfur), &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; are, respectively: &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$04, $&lt;/span&gt;+4&lt;span class=&quot;math-container&quot;&gt;$, $&lt;/span&gt;+6&lt;span class=&quot;math-container&quot;&gt;$ and $&lt;/span&gt;+6$. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What we see is, that the table in the original question actually reflects some of these rules. However, this set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions. With only these rules it is impossible do determine the oxidation states for &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt; and many, if not most, compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in &lt;a href=&quot;http://dx.doi.org/10.1021/ed1005213&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanded Definition of the Oxidation State&lt;/a&gt;, which resembles the currently used definition quite well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So he basically came to the same conclusion as Dissenter in the question. Another statement from this article is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. &lt;a href=&quot;http://store.doverpublications.com/0486656225.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Republished by Courier Dover Publications, 2012&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2019-06-18T08:58:04.100" LastActivityDate="2019-06-18T08:58:04.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10947" PostTypeId="5" CreationDate="2014-05-20T06:41:53.780" Score="0" Body="&lt;p&gt;In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC &lt;a href=&quot;http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gold book&lt;/a&gt; now defines oxidation state as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;oxidation state&lt;/strong&gt;&lt;br&gt;&#xA;  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:&lt;br&gt;&#xA;  &lt;em&gt;OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The underlying principle is that the ionic sign in an &lt;span class=&quot;math-container&quot;&gt;$\ce{AB}$&lt;/span&gt; molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear &lt;span class=&quot;math-container&quot;&gt;$\ce{AA}$&lt;/span&gt; bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/bzfzQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bzfzQ.png&quot; alt=&quot;O04365-1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/bHvBW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bHvBW.png&quot; alt=&quot;O04365-2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;NOTE:&lt;/em&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;em&gt;Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;When used in chemical nomenclature as a symbol, the OS value is in roman numerals.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;em&gt;At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating &lt;span class=&quot;math-container&quot;&gt;$\text{OS} = +1$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$−2$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; yields correct OS in oxides, hydroxides, and acids like &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;; with coverage extended to &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.&lt;/em&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Source:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2014,&lt;/strong&gt; &lt;em&gt;86&lt;/em&gt; (6), 1017–1081 &lt;a href=&quot;https://doi.org/10.1515/pac-2013-0505&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1515/pac-2013-0505&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2016,&lt;/strong&gt; &lt;em&gt;88&lt;/em&gt; (8), 831–839 &lt;a href=&quot;https://doi.org/10.1515/pac-2015-1204&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1515/pac-2015-1204&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from &lt;a href=&quot;https://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;old.iupac.org&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sources cited above contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to 2016 the IUPAC defined oxidation states in their &lt;a href=&quot;https://web.archive.org/web/20141102154648/http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gold book (via the Internet Archive)&lt;/a&gt; differently. This definition will likely be still present in prominent (older) text books.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;oxidation state&lt;/strong&gt; (deprecated, pre-2016 definition)&#xA;  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;the oxidation state of a free element (uncombined element) is zero; &lt;/li&gt;&#xA;  &lt;li&gt;for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;&lt;/li&gt;&#xA;  &lt;li&gt;hydrogen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; and oxygen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt; when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; in hydrides of active metals, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{LiH}$&lt;/span&gt;, and oxygen has an oxidation state of &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; in peroxides, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt;;&lt;/li&gt;&#xA;  &lt;li&gt;the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{S8}$&lt;/span&gt; (elementary sulfur), &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; are, respectively: &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$04, $&lt;/span&gt;+4&lt;span class=&quot;math-container&quot;&gt;$, $&lt;/span&gt;+6&lt;span class=&quot;math-container&quot;&gt;$ and $&lt;/span&gt;+6$. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions, which eventually lead to the change. With only these rules it is impossible do determine the oxidation states for &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt; and many, if not most, compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in &lt;a href=&quot;http://dx.doi.org/10.1021/ed1005213&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanded Definition of the Oxidation State&lt;/a&gt;, which resembles the currently used definition quite well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another statement from this article is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. &lt;a href=&quot;http://store.doverpublications.com/0486656225.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Republished by Courier Dover Publications, 2012&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2019-06-18T09:02:08.617" LastActivityDate="2019-06-18T09:02:08.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10948" PostTypeId="4" CreationDate="2014-05-20T06:41:53.780" Score="0" Body="Oxidation state is a concept, which assigns hypothetical charges to atoms in compounds. This tag should be applied to questions which relate to that concept or to oxidation states in atoms of compounds." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-20T09:42:51.003" LastActivityDate="2014-05-20T09:42:51.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10949" PostTypeId="1" CreationDate="2014-05-20T07:37:36.647" Score="2" ViewCount="85" Body="&lt;p&gt;Is there any difference between the terms specific conductivity and conductance. If yes, please explain.&lt;/p&gt;&#xA;" OwnerUserId="5564" LastActivityDate="2014-07-19T17:30:43.523" Title="Confusion regarding specific conductivity" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-07-31T13:50:51.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="10950" PostTypeId="2" ParentId="10798" CreationDate="2014-05-20T07:46:27.523" Score="5" Body="&lt;p&gt;All the arguments are perfect except the fourth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement is true, but the premise that the vapour pressure of water must equal the atmospheric pressure is incorrect. A phase equilibrium between the gaseous and the liquid phase can be established at a particular temperature without the liquid boiling. Boiling is a special case of this equilibrium when the atmospheric pressure is entirely constituted by water vapour in the vicinity of the liquid surface and hence the vapour pressure can no longer rise at that pressure. In normal cases, as is possible in this case also, vapour pressure of water is just one component of the atmospheric pressure and the liquid need not be boiling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more accurate sentence would be $P_{atm}\geq P_{vap}$. This would still mean that both vapour pressure and the atmospheric are higher than 1 atm but &lt;em&gt;not necessarily&lt;/em&gt; equal.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-05-20T07:46:27.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10951" PostTypeId="1" CreationDate="2014-05-20T12:54:17.403" Score="1" ViewCount="91" Body="&lt;p&gt;I am perparing for a presentation on the application of TiO2 nanotubes in cement which is a recent topic in cement chemistry. Do you happen to have any papers (or know of any accessible on Internet) on the topic from the recent years that you can share with me? Or even if you have any information/experience to share, I would be grateful. &lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorDisplayName="user7951" LastEditDate="2015-06-20T08:56:41.843" LastActivityDate="2015-06-20T08:56:41.843" Title="TiO2 nanotube application in cement" Tags="&lt;inorganic-chemistry&gt;&lt;materials&gt;&lt;reference-request&gt;&lt;nanotechnology&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2018-10-18T20:36:44.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="10952" PostTypeId="1" AcceptedAnswerId="10953" CreationDate="2014-05-20T13:33:23.470" Score="1" ViewCount="1472" Body="&lt;p&gt;What is the colour of elemental Ag? I have doubt whether it is black or silver in colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In Tollens' test, we get a silvery precipitate of elemental silver - hence silver in colour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have read somewhere and also in the following links as a reference that it was not mere speculation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.in/books?id=AFY2gKwh4AsC&amp;amp;pg=PA553&amp;amp;lpg=PA553&amp;amp;dq=elemental+silver+precipitate&amp;amp;source=bl&amp;amp;ots=nzfsQcBAm6&amp;amp;sig=jTCQk7-k0HzAByI25ESI6OSclJs&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=S1d7U6aXIY-OuASHp4HIDQ&amp;amp;redir_esc=y#v=onepage&amp;amp;q=elemental%20silver%20precipitate&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.in/books?id=AFY2gKwh4AsC&amp;amp;pg=PA553&amp;amp;lpg=PA553&amp;amp;dq=elemental+silver+precipitate&amp;amp;source=bl&amp;amp;ots=nzfsQcBAm6&amp;amp;sig=jTCQk7-k0HzAByI25ESI6OSclJs&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=S1d7U6aXIY-OuASHp4HIDQ&amp;amp;redir_esc=y#v=onepage&amp;amp;q=elemental%20silver%20precipitate&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(towards the end)&#xA;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.co.in/books?id=cZL_-hyw98sC&amp;amp;pg=PA675&amp;amp;lpg=PA675&amp;amp;dq=elemental+silver+precipitate&amp;amp;source=bl&amp;amp;ots=DDhdhikKhU&amp;amp;sig=zIq-qQM7VI4i5enohZVi3oIagHw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=S1d7U6aXIY-OuASHp4HIDQ&amp;amp;redir_esc=y#v=onepage&amp;amp;q=elemental%20silver%20precipitate&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.co.in/books?id=cZL_-hyw98sC&amp;amp;pg=PA675&amp;amp;lpg=PA675&amp;amp;dq=elemental+silver+precipitate&amp;amp;source=bl&amp;amp;ots=DDhdhikKhU&amp;amp;sig=zIq-qQM7VI4i5enohZVi3oIagHw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=S1d7U6aXIY-OuASHp4HIDQ&amp;amp;redir_esc=y#v=onepage&amp;amp;q=elemental%20silver%20precipitate&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please confirm this. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="5429" LastEditorUserId="5016" LastEditDate="2014-05-20T14:40:19.780" LastActivityDate="2014-05-20T16:28:26.777" Title="Elemental Silver Colour" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10953" PostTypeId="2" ParentId="10952" CreationDate="2014-05-20T15:33:31.383" Score="3" Body="&lt;p&gt;Silver:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bLQXv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver sulfide precipitate is black.  You are pobably confusing a precipitate in a test for silver, with actual silver.  The precipitate in a test will be a compound of silver, such as silver sulphide which is black.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nanophase metal particles are optically different from bulk in many cases, and can be black, but this doesn't seem to be what the question is referring to.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-05-20T16:20:43.990" LastActivityDate="2014-05-20T16:20:43.990" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10954" PostTypeId="2" ParentId="10952" CreationDate="2014-05-20T16:28:26.777" Score="2" Body="&lt;p&gt;(this is just to illustrate with an image, even though the answer was already formulated by DavePhD )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what you define as colour - a silver bar will be metallic-silver, however silver powder may seem gray  (or even black in water) (see image: &lt;a href=&quot;http://www.nbchangjing.com/upload/2009031714521679.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nbchangjing.com/upload/2009031714521679.jpg&lt;/a&gt; ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently performed this silver-compound reaction : $\ce{Zn + 2AgNO3 -&amp;gt; Zn(NO3)2 + 2Ag}$ After a minute you could clearly see silver 'growing' from the zinc wire that I added to the silver nitrate solution. The silver was powdery black, however when I 'flattened' the silver powder, I had a silver-metallic surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even found an image of a similar reaction involving copper wire : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VVT3w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You clearly see that there is a grayish-black silver powder forming on the wire, yet this doesn't have the same colour as a silver coin, for example.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-05-20T16:28:26.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10956" PostTypeId="2" ParentId="10949" CreationDate="2014-05-20T17:01:39.817" Score="1" Body="&lt;p&gt;Let's take a wire inside cable for example. That wire's &lt;strong&gt;conductance&lt;/strong&gt; is just the inverse of the resistance this cable is making when electricity passes through it. This depends on things like the length, what is made of, the maximum/minimum width of certain areas of the wire, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the &lt;strong&gt;conductivity&lt;/strong&gt; of that wire is a direct property of the material it's made of. If the wire is made of copper for example, it'll have (at the same temperature) the same &lt;em&gt;conductivity&lt;/em&gt; as any copper wire in the world (because they are all made of copper), but not the same &lt;em&gt;conductance&lt;/em&gt; (because they are not physically equal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps, I'm sorry if I was wrong!&lt;/p&gt;&#xA;" OwnerUserId="5390" LastActivityDate="2014-05-20T17:01:39.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10958" PostTypeId="1" AcceptedAnswerId="10961" CreationDate="2014-05-20T17:59:14.397" Score="4" ViewCount="197" Body="&lt;p&gt;This is the gas-phase structure of hypofluorous acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVkLj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid-phase structure however has a bond angle of 101 degrees. Why is there a difference in the solid and gas phase bond angles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this due to packing in the solid-phase? In the solid-phase the molecules should be much closer together than in the gas-phase, so I expect more intermolecular forces and interactions. However, what exactly would be responsible for the angle expansion in the solid phase? From one book I read that bond lengths often differ among the phases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what exactly is behind the bond angle expansion in the solid phase? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-20T18:16:42.433" Title="Bond Angles of Hypofluorous Acid - Gas and Solid Phases" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10959" PostTypeId="1" CreationDate="2014-05-20T18:03:47.940" Score="11" ViewCount="1179" Body="&lt;p&gt;Are there well developed reaction mechanisms in inorganic chemistry like those in organic chemistry? If yes, please suggest an easy to follow reference.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="7653" LastEditDate="2015-01-18T06:13:53.380" LastActivityDate="2019-12-29T16:47:35.380" Title="Inorganic chemistry reaction mechanisms" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10960" PostTypeId="1" CreationDate="2014-05-20T18:07:18.730" Score="1" ViewCount="140" Body="&lt;p&gt;If I have the amount of the molecule in moles, how do I calculate the mass of an individual molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the molecule was water?&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="5456" LastEditDate="2014-05-20T21:42:52.543" LastActivityDate="2014-07-20T05:08:33.797" Title="Calculating the mass in g of a molecule" Tags="&lt;molecules&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2014-07-20T14:17:47.287" ContentLicense="CC BY-SA 3.0" />
  <row Id="10961" PostTypeId="2" ParentId="10958" CreationDate="2014-05-20T18:16:42.433" Score="4" Body="&lt;p&gt;It is due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;hydrogen bonding&lt;/a&gt; in the solid but not gas phase, in analogy to &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/water.html&quot; rel=&quot;nofollow&quot;&gt;what occurs for the water bond angle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow&quot;&gt;VSPER&lt;/a&gt; point of view, the oxygen atom of one molecule would form a hydrogen bond with the hydrogen of another molecule, decreasing the repulsion of the lone pairs to be closer to that of the covalently bonded pairs, so the geometry would be closer to tetraherdral in the solid phase than in the gas phase.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-20T18:16:42.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10962" PostTypeId="2" ParentId="10960" CreationDate="2014-05-20T18:25:01.180" Score="2" Body="&lt;p&gt;The &quot;amount of molecules in moles&quot;, such as 7 moles, provides no information concerning the mass of an individual molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is like asking, &quot;if I have the amount of eggs in dozens, how do I calculate the mass of an egg?&quot; &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-20T18:25:01.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10964" PostTypeId="2" ParentId="10960" CreationDate="2014-05-20T19:05:02.977" Score="2" Body="&lt;p&gt;It sounds like you're trying to calculate the mass of a water molecule.  Water has a molecular weight of $18\:\mathrm{g/mol}$.  In a mole of water there will be Avogadro's number of molecules ($6.02\cdot 10^{23}\:\mathrm{molecules/mol}$).  Therefore, $$\text{mass of a water molecule} =  \frac{18}{6.02 \cdot 10^{23}} = 2.99 \cdot 10^{-23}\:\mathrm{g/molecule}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-07-20T05:08:33.797" LastActivityDate="2014-07-20T05:08:33.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10965" PostTypeId="1" AcceptedAnswerId="10973" CreationDate="2014-05-20T20:49:17.690" Score="2" ViewCount="901" Body="&lt;p&gt;Trying to identify a mystery compound, here are some of my notes, I'm trying to get a second opinion on what it might be!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Observations&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Appears as a white powder, some chunks are formed, sticking to side of tube slightly&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Solubility&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;3 drops of distilled water, all of substance dissolved&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Flame Test&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Red Flame&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sodium Hydroxide Test&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A white ppt was formed when mystery liquid was combined with sodium hydroxide&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Concentrated sulfuric acid test&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A yellow colored gas formed when 4 drops of sulfuric acid were added with substance in test tube&lt;/p&gt;&#xA;&lt;p&gt;My thoughts based on these results narrow the cation down to lithium and strontium. After performing a precipitate test again with my two knowns (lithium, strontium)strontium was the only one to form a ppt. So I'm almost 100% sure that my cation is strontium.&lt;/p&gt;&#xA;&lt;p&gt;I have the anion narrowed down to bromide, nitrate, and nitrite. It has to be one of those three! I'm leaning towards bromide as the anion, just interested on what you guys think from my notes.&lt;/p&gt;&#xA;&lt;p&gt;Here are the options again, with my choice in bold:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Cation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lithium&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Strontium&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Anion&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bromide &lt;strong&gt;?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Nitrate&lt;/li&gt;&#xA;&lt;li&gt;Nitrite&lt;/li&gt;&#xA;&lt;li&gt;Iodine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&lt;p&gt;I have a book with 16 possible cations, and 11 possible anions. Based on my results and the information given in the book, I've narrowed it down to 2 cations, lithium, and strontium, and 3 possibly 4 anions. I'm almost certain the cation is strontium, based on the fact that it formed the same white ppt. that the mystery substance did when mixed with sodium hydroxide. Lithium did not. As for the anion, I'm not sure, I'm leaning towards bromide, but was looking for a second opinion. I've been doing some research on how bromide reacts with sulfuric acid, and the results online have been almost identical to what happened with the unknown substance, however in my option book it says bromide and nitrate have very similar reactions.&lt;/p&gt;&#xA;&lt;p&gt;If you have any more thoughts on how I could identify to anion, please let me know!&lt;/p&gt;&#xA;" OwnerUserId="5569" LastEditorUserId="69830" LastEditDate="2021-05-18T05:24:44.240" LastActivityDate="2021-05-18T05:54:39.193" Title="How can I check if this mystery compound is strontium bromide?" Tags="&lt;solubility&gt;&lt;heat&gt;&lt;identification&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10967" PostTypeId="1" AcceptedAnswerId="10968" CreationDate="2014-05-21T01:07:09.957" Score="0" ViewCount="691" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGlch.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this graphic in &lt;em&gt;Organic Chemistry&lt;/em&gt; by Klein (2nd edition too!) and this has to be wrong, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror images should have the double bond on the far side of the ring in both images, right? I am sure, unless my visualization skills have totally failed me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UG2h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the result when I flip the entire image using an online mirror effect creator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8mG32.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-21T01:21:26.920" Title="Enantiomer of Amphetamine" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10968" PostTypeId="2" ParentId="10967" CreationDate="2014-05-21T01:21:26.920" Score="6" Body="&lt;p&gt;The graphic from the book is correct, those two molecules are mirror images.  Same for your drawing, those are the same two mirror images as in the book.  Remember that the electrons in the aromatic ring are delocalized.  The double bonds in the aromatic ring are not fixed.  Where you choose to draw the double bonds in aromatic rings is your choice and does not play any role in determining if molecules are enantiomers.  Bottom line, the following three structures all represent the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6Jtg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-21T01:21:26.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10972" PostTypeId="1" CreationDate="2014-05-21T07:36:10.190" Score="2" ViewCount="82" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gJ4KU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture shows one of the step of heme synthesis. In this step, two propanoate groups of Coproporhyrinogen III are decarboxylated to from protoporphyrinogen IX. The enzyme used is oxidase. Oxidase enzymes catalyze the removal of hydrogen in the presence of oxygen to from water or hydrogen peroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?&lt;/p&gt;&#xA;" OwnerUserId="1343" LastEditorUserId="1343" LastEditDate="2014-05-21T11:18:29.730" LastActivityDate="2014-07-20T14:41:26.970" Title="Protoporphirinogen IX formation in Heme synthesis" Tags="&lt;enzymes&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10973" PostTypeId="2" ParentId="10965" CreationDate="2014-05-21T12:53:21.707" Score="4" Body="&lt;p&gt;Relying on the fact that the compound is almost certainly a strontium compound, I'll look at the reactions with sulfuric acid of the specific anions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SrBr2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SrBr2 + H2SO4 -&amp;gt;  2HBr + SrSO4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Both compounds &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{SrSO4}$&lt;/span&gt; are white or colourless. BUT! &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_bromide&quot; rel=&quot;nofollow noreferrer&quot;&gt;concentrated sulfuric acid oxidizes &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; to bromine gas:&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 HBr + H2SO4 → Br2 + SO2 + 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And bromine gas has a yellow/red colour. So your compound is &lt;strong&gt;strontium bromide&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Sr(NO2)2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://strontium.atomistry.com/strontium_nitrite.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is a yellow compound&lt;/a&gt;. As you mentioned, the compound was white, so it isn't strontium nitrite.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Sr(NO3)2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Sr(NO3)2 + H2SO4 -&amp;gt;   2HNO3 + SrSO4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Both compounds &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{SrSO4}$&lt;/span&gt; are white or colorless. After looking some more stuff up, it turns out that sulfuric acid and nitric acid don't react (at room temperature) to produce any yellow/brown &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; gas.&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastEditorUserId="17368" LastEditorDisplayName="user2117" LastEditDate="2021-05-18T05:54:39.193" LastActivityDate="2021-05-18T05:54:39.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10974" PostTypeId="2" ParentId="10972" CreationDate="2014-05-21T13:42:02.097" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, note that there are 4 more hydrogen atoms in the reactant than in the product (one from each carboxylic acid group and one where each double bond forms).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at physiological pH, the two protons on the carboxyl groups will not really be there.  The equation in the question also ignores the 4 electrons transferred.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other two protons, a 5’-deoxyadenosyl radical abstracts a hydrogen atom (proton + electron) from each propionate side chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://udel.edu/~forbescr/Coproporphyrinogen.htm&quot; rel=&quot;nofollow&quot;&gt;Coproporphyrinogen III Oxidase Enzyme&lt;/a&gt; and references cited therein for more information. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-21T13:42:02.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10975" PostTypeId="1" AcceptedAnswerId="10983" CreationDate="2014-05-21T13:56:54.537" Score="6" ViewCount="186" Body="&lt;p&gt;Following up &lt;a href=&quot;https://chemistry.stackexchange.com/questions/10885/hazardous-materials-handled-by-amateur-chemists&quot;&gt;on a somewhat too broad question&lt;/a&gt; about amateur chemist safety, this will be the first to address some of the issues discussed there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often use the &lt;a href=&quot;http://cameochemicals.noaa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NOAA MSDS database&lt;/a&gt; (material safety data sheet) to investigate the chemicals I'm working with (they also have a nice reactivity predictor). As an example, let us examine the compound cyclohexylamine (I chose this as leading example for these followup question, because of the &lt;a href=&quot;http://www.gezondheidsraad.nl/sites/default/files/0015021OSH.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;availability of extensive research&lt;/a&gt; in my country)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f0qNS.png&quot; alt=&quot;Cyclohexylamine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sometimes used as a plasticizer or catalyst. As an amateur I would carefully go through the MSDS and notice for example that it is flammable, caustic compound, scoring a 3 on flammability and health in the &lt;a href=&quot;http://en.wikipedia.org/wiki/NFPA_704&quot; rel=&quot;nofollow noreferrer&quot;&gt;NFPA 704&lt;/a&gt; hazard diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an amateur hobby chemist without all comfort of a basic lab (aside from safety goggles, gloves, lab coat, fire blanket), how would I judge if I can safely handle this compound with regard to its flammability, aside from conforming to the &lt;em&gt;Protective Clothing&lt;/em&gt; section?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I could imagine that flammable liquids with high vapor pressure (11 mm Hg for cyclohexylamine according to NOAA, I guess this is measured at a standard temperature as vapor pressure varies with temperature) or low auto ignition points (560 ° F) or low boiling points (274.1 °F) are unsafe for amateurs. Are there empirical (or theoretical) recommendation considering vapor pressure and ignition points, i.e., starting at what value will these factors become too dangerous for the compound to be handled in a relatively amateurish lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I understand that the answers will be partly subjective, as for all answers regarding safety, but please try to give an objective as possible answer, preferably referring to external sources)&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2019-05-10T19:12:03.280" Title="Assessing fire hazard of compounds" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10976" PostTypeId="1" AcceptedAnswerId="10990" CreationDate="2014-05-21T16:05:39.223" Score="8" ViewCount="9422" Body="&lt;p&gt;I am fairly sure the first diagram I drew for carbon dioxide is wrong in terms of showing π bonding. This is because we use a π orbital twice, which isn't possible. The second diagram corrects this by realizing there are two unused p orbitals on the carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6HDjX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is what I got for allene. Again we have an central $\ce{sp}$ carbon and this implies that two p-orbitals are free to form π bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nzifX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="42741" LastEditDate="2017-05-19T04:47:55.383" LastActivityDate="2017-05-19T04:47:55.383" Title="π Bonding in Carbon Dioxide" Tags="&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;lewis-structure&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10977" PostTypeId="2" ParentId="10976" CreationDate="2014-05-21T16:23:22.163" Score="2" Body="&lt;p&gt;Your second drawing is correct.   The bonding scheme in carbon dioxide is the same as that in allene and ketene in that the two pi systems are orthogonal.  Draw or build a model of allene and notice what effect the orthogonality has on the two planes the hydrogens lie in. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-21T17:10:25.737" LastActivityDate="2014-05-21T17:10:25.737" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10978" PostTypeId="2" ParentId="915" CreationDate="2014-05-21T17:31:27.137" Score="7" Body="&lt;p&gt;&lt;strike&gt;Use&lt;/strike&gt; Exploit the fact that burning magnesium continues to burn in atmosphere of carbon dioxide.&#xA;$${\ce{Mg}}+\color{\red}{\ce{CO2}}\to \ce{MgO}+\color{\red}{\ce{C}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can electrolyse your mixture : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MgO}+\ce{H2O}\to \ce{Mg(OH)2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4OH-}\to\color{\red}{\ce{ O2}}+\ce{2H2O}+\ce{4e-}$$&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4597" LastEditDate="2014-05-22T13:13:10.453" LastActivityDate="2014-05-22T13:13:10.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10979" PostTypeId="1" AcceptedAnswerId="10981" CreationDate="2014-05-21T17:40:06.100" Score="3" ViewCount="692" Body="&lt;p&gt;Would this be proper resonance structures? I know there's a lone pair on the top carbon. Having a carboanion isn't favorable as opposed to a negative formal charge on the more electronegative carbon so I flipped a bonding pair to oxygen's valence. I then proceeded to move the pi electrons inside the ring as well. This seems to satisfy all octet requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UfYRZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XFUrz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second resonance diagram I saw the negative charge on the oxygen and decided to take one of its lone pairs and make it a bonding pair so that oxygen has 0 formal charge. This in turn necessitates pushing around a few pi electron pairs inside the ring. Same thing for the third one; I tried giving nitrogen a positive formal charge. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4231" LastEditDate="2014-05-21T21:26:54.570" LastActivityDate="2014-05-21T21:26:54.570" Title="Proper Resonance Structures" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10980" PostTypeId="2" ParentId="10975" CreationDate="2014-05-21T17:45:25.463" Score="0" Body="&lt;p&gt;Here is an illustrating link: &lt;a href=&quot;http://www2.galcit.caltech.edu/EDL/public/flammability/USBM-680.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www2.galcit.caltech.edu/EDL/public/flammability/USBM-680.pdf&lt;/a&gt; I hope it has a good analysis of what you want.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="1887" LastEditDate="2014-05-21T21:08:14.300" LastActivityDate="2014-05-21T21:08:14.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10981" PostTypeId="2" ParentId="10979" CreationDate="2014-05-21T20:02:06.553" Score="3" Body="&lt;p&gt;Yes, they are all valid resonance structures.  In the first series, there is a third resonance structure you could draw with the negative charge alpha to the carbonyl carbon. In the second series, the phenoxide anion, you can draw two more structures.  One with the negative charge in the other ortho position and one with the negative charge in the para position.  This significant resonance stabilization of the phenoxide anion explains why phenol is such a strong acid.  In your third series there is also a third structure with the positive charge situated alpha to the nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-21T20:02:06.553" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10982" PostTypeId="1" CreationDate="2014-05-21T22:46:22.433" Score="2" ViewCount="105" Body="&lt;p&gt;For example, say you react ethyl carbamate with methyl-$\ce{LiCu}$; would two methyl groups displace both the amide and the ester, creating an isopropyl alcohol?&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="7448" LastEditDate="2015-10-09T15:44:21.773" LastActivityDate="2015-10-09T15:44:21.773" Title="Would a carbamate be attacked twice by an alkyl lithium cuprate reagent?" Tags="&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10983" PostTypeId="2" ParentId="10975" CreationDate="2014-05-21T23:34:02.190" Score="3" Body="&lt;p&gt;I guess I don't completely agree with the premise of your question. Gasoline is highly flammable, even potentially explosive, yet &quot;amateurs&quot; deal with it constantly. The key is that it is &lt;em&gt;stored and used properly&lt;/em&gt;, from the pump to your gas tank. It becomes less safe if you pump it into a gas can, but gas cans also have certain requirements for storage safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the issue may be what sorts of flammables would an amateur have a high chance of storing or using inappropriately. For example, any &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrophoricity&quot; rel=&quot;nofollow&quot;&gt;pyrophoric&lt;/a&gt;&lt;/em&gt; compound would be a good candidate to avoid, as most require a way to exclude air and moisture, which often requires specialist equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvents with low flash points and high vapor pressure should also be treated with respect, but rubbing alcohol (isopropanol) or acetone might qualify for that designation. Storage in airtight fireproof containers in cool and dry places is often sufficient. Better yet, use a flammables cabinet. This simple step is something an amateur might skip for cost, but it may be the most sensible precaution. I have flammables cabinets, and I use them for ordinary hardware solvents along with experimental chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, work safely! No flames, as they say in organic chemistry. Use hotplates or heating mantles rather than burners. If you must use flames for some reason, put away all flammables and keep them a safe distance. Pay careful attention to any spills and attend to them immediately before resuming work. And perhaps key for amateurs, small quantities! No need for an amateur to have gallons of flammables on hand. Even professional labs are trying to reduce quantities in the name of green chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: My go-to source for using and disposing of hazardous chemicals is &lt;em&gt;Hazardous Laboratory Chemicals Disposal Guide&lt;/em&gt; by Margaret-Ann Armour. In addition to disposal techniques, each chemical has an extensive incompatibility section about what combinations can make a chemical even more hazardous, and all these entries have literature references. I use this book literally every time I do an experiment. It deals with much more than flammables, but flammables are definitely present.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2014-05-21T23:39:56.100" LastActivityDate="2014-05-21T23:39:56.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10984" PostTypeId="1" CreationDate="2014-05-22T00:28:52.967" Score="4" ViewCount="567" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lRHsp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lRHsp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there are at most seven consecutive carbon atom, and four alkyl groups attached to the chain of carbon, I would name it &lt;strong&gt;2,4-tetramethyl heptane&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="41328" LastEditDate="2017-08-06T20:13:12.447" LastActivityDate="2017-08-06T20:13:12.447" Title="How would you name this organic compound?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10985" PostTypeId="2" ParentId="10984" CreationDate="2014-05-22T00:44:20.827" Score="12" Body="&lt;p&gt;Yes, it's tetramethyl heptane, where are the 4 methyls? 2,4,4,6-tetramethylheptane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-22T00:44:20.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10986" PostTypeId="2" ParentId="10984" CreationDate="2014-05-22T00:44:52.283" Score="9" Body="&lt;p&gt;You are nearly right. You forgot to include that there is also a methyl group on carbon atom 6 and the name should also reflect that there are actually 2 methyl groups on carbon 4, so it's 2,4,4,6-tetramethylheptane.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-05-22T00:44:52.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10987" PostTypeId="2" ParentId="10982" CreationDate="2014-05-22T00:51:30.093" Score="4" Body="&lt;p&gt;I don't think so, carbonyls don't typically react with Gilman's reagent (dimethyllithium cuprate) unless the carbonyl happens to be $\ce{{\alpha}-{\beta} }$ unsaturated. So I would expect no reaction under normal reaction conditions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-22T00:51:30.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10988" PostTypeId="1" CreationDate="2014-05-22T01:17:13.637" Score="3" ViewCount="1055" Body="&lt;p&gt;The title says it all. I've been researching as well but all I could come close was solid state ion selective electrodes. Could someone please help me out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be two methods(or one) to measure the concentrations. &lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorUserId="4945" LastEditDate="2014-05-22T08:12:00.770" LastActivityDate="2014-09-25T09:49:35.017" Title="Determine Cu+2 and SO4-2 concentration using Potentiometry?" Tags="&lt;analytical-chemistry&gt;&lt;titration&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10989" PostTypeId="1" CreationDate="2014-05-22T07:49:38.367" Score="1" ViewCount="1223" Body="&lt;p&gt;Time and time again textbooks and websites state how wonderful it is that the deviations of Hydration Enthalpies from a linear relationship can be accounted for by the CFSE parameter for the particular complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what is the mechanistic explanation in terms of water and the metal ion interacting ? Why does CFSE lead to much more negative hydration enthalpies ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : A simplie overview of what I'm talking about : &lt;a href=&quot;http://sst-web.tees.ac.uk/external/u0000892/Web%20thing/CoordChem/CFSE5.htm&quot; rel=&quot;nofollow&quot;&gt;http://sst-web.tees.ac.uk/external/u0000892/Web%20thing/CoordChem/CFSE5.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="1549" LastEditDate="2014-05-22T08:56:48.333" LastActivityDate="2014-05-22T08:56:48.333" Title="CFSE/LFSE Explains Hydration Enthalpy" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10990" PostTypeId="2" ParentId="10976" CreationDate="2014-05-22T07:53:07.117" Score="14" Body="&lt;p&gt;In terms of a simple bonding concept the answer of ron is correct, however due to symmetry constraints it is not possible to distinguish the two $\pi$ systems. The reason for this is, that these $\pi$ orbitals are degenerated and perpendicular to each other.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Carbon dioxide&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The molecule is very symmetric and one of only a few with the point group $D_{\infty h}$. That also imposes constraints on how the spatial distribution looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the molecular axis ($\ce{O-C-O}$) is a rotational axis of infinite fold, $C_\infty$, as the molecule is linear. Perpendicular to that axis there is a horizontal mirror plane $\sigma_h$ passing through the centre of mass, i.e. the carbon. Additionally there is an infinite number of two-fold symmetry axis $C_2$ lying in that plane, which is the criterium for $D_{xh}$ point group. In line with the main axis there is also an infinite number of vertical mirror planes $\sigma_v$ (which is implicit due to the horizontal mirror plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of convenience we can define an upper and a lower half of the molecule (taking the cartesian $z$ axis as the main symmetry axis). These two halfs are indistinguishable, hence their orbitals must reflect that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a look at the $\pi$-orbitals, clearly shows that (&lt;a href=&quot;https://i.stack.imgur.com/O6VSc.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O6VSc.jpg&quot; alt=&quot;co2 pi orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your first drawing was not too wrong after all. The only thing you forgot is, that there is an identical orbital perpendicular to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ is also a nice example to see the boundaries of the hybridisation concept. In this example, due to symmetry, all atoms have to be $\ce{sp}$ hybridised. And that matches all the $\sigma$ orbitals (&lt;a href=&quot;https://i.stack.imgur.com/6PnPN.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PnPN.jpg&quot; alt=&quot;co2 sigma orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Allene&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The allene on the other hand is not linear and has a much lower symmetry. There is the main symmetry $C_2$ axis (for convenience) along the $\ce{C-C-C}$ axis, accompanied be two $\sigma_v$ mirror planes. There are also two diagonal symmetry planes $\sigma_d$ right between them. Perpendicular to the main symmetry axis we have two more $C_2$ axis (also perpendicular to each other). The point group is therefore $D_{2d}$ and has a lot less symmetry constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the orbitals are actually (surprisingly) quite similar though. We can define $\sigma$ and $\pi$ orbitals accordingly. The biggest distinction is still in the $\pi$-orbitals (&lt;a href=&quot;https://i.stack.imgur.com/lK5lH.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lK5lH.jpg&quot; alt=&quot;allene pi orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lack of symmetry and therefore less constraints make is possible, that the orbitals combine in a different manner. The top and the lower half of the molecule does not have to be the same anymore, hence the orbitals are more localised to one bond in particular. A notably thing is, that hydrogens take part in $\pi$ bonding (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;noreferrer&quot;&gt;Hyperconjugation&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sigma orbitals are quite the same as in $\ce{CO2}$ (&lt;a href=&quot;https://i.stack.imgur.com/A7IwA.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A7IwA.jpg&quot; alt=&quot;allene sigma orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the traditional bonding picture of localised orbitals is quite valid here. The central carbon is still being $\ce{sp}$ hybridised and the moieties are $\ce{sp^2}$ hybridised.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ketene is a combination of the two above, with even lower symmetry ($C_{2v}$). Therefore the conventional bonding picture should hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Oxygen it is usually always good to check, if hybridisation is a good concept to describe the bond. When bonded to more than one partner, the answer is usually yes. If less, then there is a good chance, that hyperconjugative effects outweigh the information gained from the hybridisation concept.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Additional fun note&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that the carbon suboxide ($\ce{C3O2}$) is not too different, but here again you will be surprised, as it has a very shallow bending potential. (&lt;a href=&quot;http://dx.doi.org/10.1002/1521-3749%28200101%29627:1%3C73::AID-ZAAC73%3E3.0.CO;2-A&quot; rel=&quot;noreferrer&quot;&gt;crystal structure&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1246/bcsj.43.2776&quot; rel=&quot;noreferrer&quot;&gt;vibrations&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-22T07:53:07.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10991" PostTypeId="1" CreationDate="2014-05-22T08:57:25.613" Score="18" ViewCount="3163" Body="&lt;p&gt;I've always wondered why the N-glycosidic bond, which connects the adenine or guanine purine bases to their sugars, was the easiest to cleave via acid hydrolysis, or why purines have the lowest ionization potentials?  What is it about the rings / resonance / etc. of purine vs. pyrimidine bases in nucleic acid polymers that makes this so?  For example, according to (Lindahl, Nature 362, pp. 709–715 (1993), &lt;a href=&quot;http://www.nature.com/nature/journal/v362/n6422/abs/362709a0.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/nature/journal/v362/n6422/abs/362709a0.html&lt;/a&gt;), at least in the context of deoxyribonucleic acid (DNA) at pH ~ 7 (probably the rate is much slower in RNA under the same conditions due to the 2'-OH) the rate of depyrimidation is $\approx 20$ fold slower than the rate of depurination.  Note that the actual base rate of depurination is $\approx 4 \times 10^{-9} \space \mathrm s^{-1}$ at 70 °C and pH ~ 7.4 (Lindahl &amp;amp; Nyberg, Biochemistry 11(19), pp. 3610–3618 (1972), &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/bi00769a018&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/bi00769a018&lt;/a&gt;), implying a base rate for depyrimidation of $\approx 2^{-10} \space \mathrm s^{-1}$ under the same conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess, I'd say that the reason has to do with purine bases (or the local context of the N9-glycosidic bond on purine bases) being less electrophilic, which is why I mention generally lower ionization energies for purines vs. pyrimidines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting from: &lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/o74-139#.U32-3vmSyDZ&quot; rel=&quot;nofollow&quot;&gt;http://www.nrcresearchpress.com/doi/abs/10.1139/o74-139#.U32-3vmSyDZ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Contrary to previous suggestions, pyrimidine as well as purine deoxynucleosides are subject to base cleavage, the relative ease of thermolysis being 2′-deoxyguanosine &gt; 2′-deoxyadenosine &gt; 2′-deoxycytidine &gt; 2′-deoxyuridine &gt; thymidine.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this actually seems to match the relative (experimentally measured) ionization energies of the bases --- Guanine (7.77 eV) &amp;lt; Adenine (8.26 eV) &amp;lt; Cytosine (8.68 eV) &amp;lt; Thymine (8.87 eV): Orlov, V. M., Smirnov, A. N., Varshavsky, Y. M. Ionization potentials and electron-donor ability of nucleic acid babes and their analogues. Tetra Lett. 1976;17(48):4377–4378.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my guess regarding the lower electrophilicity of purine bases is true, it's not clear to me, however, why this is the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update / Thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the four bases (where uracil is just missing the C5 methyl group on thymine) - (&lt;a href=&quot;http://img.sparknotes.com/figures/7/749a4182b7527e44d289a612e420f40c/dna_bases.gif&quot; rel=&quot;nofollow&quot;&gt;http://img.sparknotes.com/figures/7/749a4182b7527e44d289a612e420f40c/dna_bases.gif&lt;/a&gt;) - we can notice that the N9 nitrogen connecting purines to their sugars is only ever adjacent to carbon atoms with single- or double-bonds to nitrogen.  However, the N1 nitrogen connecting pyrimidines to their sugars is adjacent to a C2 carbon double-bonded with an oxygen atom.  Given that oxygen is more electronegative than nitrogen (oxygen has an electronegativity of ~3.44 vs. nitrogen's electronegativity of ~3.04) could this be a hint as to why depyrimidation is much less common than depurination?  Are there pyrimidine base analogues with a less electronegative group attached to the C2 carbon, and do depyrimidation rates here match or exceed depurination rates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could, of course, be completely off.  For example, maybe N7 protonation on purines presents a route to depurination, since guanine having a greater depurination rate than adenine (or lower ionization energy than adenine) is consistent with a decrease in the N7 nitrogen's basicity.  Maybe pyrimidines are much less prone to destabilizing protonation.  Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My objective here: I'd like to know the &quot;most significant&quot; factor for (and hopefully most general reason for) the greater observed rates of depurination vs. depyrimidation.  For example, perhaps the aforementioned oxygen on pyrimidines decreases depyrimidation rates by a factor of two or so (this isn't meant to be an actual guess), but what explains the $\approx 20$ fold difference noted by (Lindahl, 1993)?&lt;/p&gt;&#xA;" OwnerUserId="5581" LastEditorDisplayName="user7951" LastEditDate="2016-04-11T19:20:38.553" LastActivityDate="2017-03-10T07:36:21.567" Title="Why does depurination happen at a higher pH (or under less harsh conditions) than depyrimidation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;ph&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10992" PostTypeId="1" AcceptedAnswerId="10996" CreationDate="2014-05-22T09:00:19.703" Score="10" ViewCount="2093" Body="&lt;p&gt;If you have $\ce{CH3-\color{\red}{CH2}-CH2Br}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $\ce{CH2}$ (in red) would have a peak splitting of $12$ because $(n+1)(m+1)$&#xA;where $n$ for the $\ce{CH3}$ is $(3+1=4)$ and m for $\ce{CH2}$ is $(2+1=3)$ so its $4\times3=12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wczl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the general case, a signal will be split into $(n + 1) \times (m + 1)$&#xA;  peaks for an $\ce{H}$ atom that is coupled to a set of $n~\ce{H}$ atoms with one&#xA;  coupling constant, and to a set of $m~\ce{H}$ atoms with another coupling&#xA;  constant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can we determine differences in the coupling constant by just looking at the structure? Otherwise how do I know when to apply $n+1$ OR $(n+1)(m+1)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the other textbook it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3-\color{\red}{CH2}-CH2-COOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{CH2}$ (in red) has a peak splitting of $6$ because $(n+1)$ where $n=5$, $(5+1=6)$ and the $n$ is the $3~\ce{H}$ on $\ce{CH3}$ plus the $2~\ce{H}$ on $\ce{CH2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wxPAT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct method? Or are both right? and why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively symmetric adjacent hydrogen groups: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3–\color{\red}{CH2}–CH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the $\ce{CH2}$ split to 4 peaks due to symmetric adjacent hydrogen environments? Due to $n+1$, where $n=3$ so $(3+1=4)$.  I read that it should be split into 7 peaks - but where are they getting 7 from - is it $n+1$, where $n=6$ so $(6+1=7)$. For symmetric adjacent groups do you allow $n$ to be the sum of all hydrogens adjacent/or not? &lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="5026" LastEditDate="2018-02-26T05:38:44.573" LastActivityDate="2019-02-19T22:50:30.240" Title="HNMR - Peak Splitting" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10994" PostTypeId="2" ParentId="9957" CreationDate="2014-05-22T11:24:34.720" Score="3" Body="&lt;p&gt;As of today I decided to add an answer to this problem to declare it as unresolved. The answer of Klaus only answers the part of visualisation, which is not what I want or need. There are at least ten different programs that can do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While turbomole comes with the built-in functionality of screwer, other program packages might not have that. The mentioned XVibs is more than outdated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution LordStryker added in the comments, doing it with the ChemCraft software is workable but highly ineffective and may introduce many errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will certainly come back, whenever I find a program capable of this, but until further notice: &lt;strong&gt;*There is no such thing.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-22T11:24:34.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10996" PostTypeId="2" ParentId="10992" CreationDate="2014-05-22T12:08:43.337" Score="6" Body="&lt;p&gt;Both are correct to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first book predicting 12 lines is being very theoretical.  In principle with narrow enough line widths 12 lines would be observed.  However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j78ol.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j78ol.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch13/H-nprbr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;ucalgary.ca&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Credit: Univ. Calgary&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="15751" LastEditDate="2019-02-19T22:50:30.240" LastActivityDate="2019-02-19T22:50:30.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10997" PostTypeId="1" AcceptedAnswerId="10998" CreationDate="2014-05-22T13:07:52.220" Score="8" ViewCount="187" Body="&lt;p&gt;I faced this reaction which can be balanced in $2$ distinct ways which are &lt;strong&gt;not&lt;/strong&gt; multiples of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\ce{6XeF4 + 12H2O &amp;amp;-&amp;gt; 2XeO3 + 24HF + 4Xe + 3O2}\tag{1}\\[0.4em]&#xA;\ce{4XeF4 + 8H2O &amp;amp;-&amp;gt; 2XeO3 + 16HF + 2Xe + O2}\tag{2}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reaction is as it is given in the books and the second reaction was balanced by me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sense that only a mechanism can resolve this. I found a similar &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20131018233353AAm1iee&quot; rel=&quot;nofollow noreferrer&quot;&gt;question&lt;/a&gt; on googling, but the answer is unsatisfactory. (It's Yahoo Answers. What can you expect?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found two independent reactions &lt;a href=&quot;http://books.google.co.in/books?id=Mtth5g59dEIC&amp;amp;pg=PA397&amp;amp;lpg=PA397&amp;amp;dq=xef4%20hydrolysis%20mechanism&amp;amp;source=bl&amp;amp;ots=tEUIlzJ3UC&amp;amp;sig=rndfUriYptOI5gj0ZcaTQyZxoIg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=7_V9U_mEE8bqrAfdyIGYBA&amp;amp;redir_esc=y#v=onepage&amp;amp;q=xef4%20hydrolysis%20mechanism&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; which on adding produce equation $(2)$. I am confused as to what exactly happens. &lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="7475" LastEditDate="2017-04-04T12:57:24.023" LastActivityDate="2017-04-04T12:57:24.023" Title="How come the same reaction can be balanced in two distinct ways?" Tags="&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;stoichiometry&gt;&lt;hydrolysis&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10998" PostTypeId="2" ParentId="10997" CreationDate="2014-05-22T13:48:16.610" Score="4" Body="&lt;p&gt;As per Nicolau's suggestion, : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{(1-2)\times\color\red{4}\}-2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Xenon_trioxide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; also says that $\ce{XeO3}$ is unstable. &lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4597" LastEditDate="2014-05-23T03:12:52.370" LastActivityDate="2014-05-23T03:12:52.370" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10999" PostTypeId="1" CreationDate="2014-05-22T13:55:35.070" Score="5" ViewCount="1793" Body="&lt;p&gt;I was wondering why specific number of water molecules are attached to salts like $\ce{CuSO4.5H2O}$? Searching gave me a satisfying picture : &#xA;&lt;img src=&quot;https://i.stack.imgur.com/JDg0Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think $\ce{FeSO4.7H2O}$&#xA; should have a similar structure with $6$ water molecules in coordination sphere. But what about $\ce{MgSO4.7H2O}$? Or $\ce{Na2CO3.10H2O}$? How do we explain that?&lt;/p&gt;&#xA;" OwnerUserId="4597" LastEditorUserId="4945" LastEditDate="2014-08-25T07:41:23.767" LastActivityDate="2014-08-25T07:41:23.767" Title="How can we rationalise the amount of crystal water in coordination complexes?" Tags="&lt;bond&gt;&lt;water&gt;&lt;molecules&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11000" PostTypeId="1" CreationDate="2014-05-22T14:05:47.967" Score="4" ViewCount="12270" Body="&lt;p&gt;The question is(See image):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cZ4sA.jpg&quot; alt=&quot;Hydrolysis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that initially a cyanohydrin will be formed. On reaction with $\ce{H2SO4}$, the cyanohydrin will give a carboxylic acid/an amide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's my doubt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the medium is highly concentrated acid medium(95% $\ce {H2SO4}$), the quantity of water in the medium is less, which means lesser of the nucleophile(water in this case). Hence the cyanohydrin should undergo partial hydrolysis to give an amide(because of lesser water!).  Therefore, according to me the answer should be (D).(Had it been dilute acid medium I would have gone with the carboxylic acid product.) However, in this case my answer is not correct...:(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is incorrect about my thinking?&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="30836" LastEditDate="2020-04-15T17:20:02.860" LastActivityDate="2020-04-15T17:20:02.860" Title="Hydrolysis of cyanohydrin" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11001" PostTypeId="2" ParentId="11000" CreationDate="2014-05-22T15:01:31.143" Score="4" Body="&lt;p&gt;So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat).  I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted to the corresponding carboxylic acid; keep in mind that an equilibrium will initially exist between the amide and the carboxylic acid, but everytime the acid forms from the amide, ammonia is released and escapes from the system making the reaction irreversible, pushing this reaction further and further to the carboxylic acid side, ultimately only the carboxylic acid will be present.), and 2) acid catalyzed elimination of water from the alcohol portion of the molecule.  Taken together I would expect the final product to be the various cis-trans isomers of &quot;A&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-22T15:01:31.143" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11002" PostTypeId="2" ParentId="10999" CreationDate="2014-05-22T15:25:12.850" Score="3" Body="&lt;p&gt;&lt;em&gt;I have given two answers: one regarding the hydration of metal cations in solution and one regarding salts of metal cations and their hydration extents.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When in solution, the answer lies with size.&lt;/strong&gt; Think of a BB gun pellet and a basketball. How many basketballs can you stick to the sides of a BB gun pellet? 2, maximum. That's why some metal cations are hydrated with fewer waters than others; these metal cations tend to be the smaller ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it is not $\ce{FeSO4}$ that is hydrated but rather just $\ce{Fe^{2+}}$. Iron sulfate is soluble in water so there shouldn't be unionized iron sulfate in solution; in addition, if the iron sulfate were unionized, it would be hard to coordinate water molecules to the metal cation - it's charge density is diluted, so to speak, by the presence of the doubly negative sulfate anion. So the correct formula should be $\ce{[Fe(OH_{2})_6]^{2+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When not in solution - i.e about the salt examples you posted: outside of water these salts can coordinate that many water molecules likely because there is no water to hydrolyze the metal-sulfate bond.&lt;/strong&gt; By having the sulfate part of the molecule stick around, this likely facilitates more hydrogen bonding with water molecules - note that the sulfate anion is polarized by the water molecules; the negative charge density resides overwhelmingly on the oxygen molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vChlT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the above picture of ferric sulfate. Note that there are four oxygens with partial negative charges on the sulfate ion. &lt;em&gt;At least&lt;/em&gt; some of these partial negative charges facilitate hydrogen bonding in the solid-state. Also note the charge on the iron atom; this positive charge density also facilitates hydrogen bonding to water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, we can consider calcium sulfate dihydrate - gyspum, drywall, sheet rock - the stuff your walls are likely made of. From here we clearly see that two of the oxygens on the sulfate anion are available for association with water molecules. Hence the existence of $\ce{CaSO_4*2H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QZL51.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-22T15:45:35.353" LastActivityDate="2014-05-22T15:45:35.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11003" PostTypeId="2" ParentId="10999" CreationDate="2014-05-22T15:41:42.113" Score="3" Body="&lt;p&gt;For Magnesium Sulphate heptahydrate, again 6 waters coordinate the $\ce{Mg^{2+}}$. The 7th water is hydrogen bonded, but not coordinated to the $\ce{Mg^{2+}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystal structure is published as &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1973/DT/DT9730000816#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;Refinement of the Crystal Structure of Magnesium Sulphate Heptahydrate (Epsomite) by Neutron Diffraction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium Sulphate hexahydrate is also known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of (Di)Sodium Carbonate Decahydrate, again 6 waters coordinate each sodium, but 2 of the waters are shared between the to sodium ions, as two octahedra with a shared edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystal structure is published as &lt;a href=&quot;http://scripts.iucr.org/cgi-bin/paper?S0567740869006236&quot; rel=&quot;nofollow&quot;&gt;Crystal structure of $\ce{Na2CO3.10H2O}$&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="1499" LastEditDate="2014-05-24T10:31:17.240" LastActivityDate="2014-05-24T10:31:17.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11004" PostTypeId="1" AcceptedAnswerId="11013" CreationDate="2014-05-22T15:59:56.500" Score="21" ViewCount="32322" Body="&lt;p&gt;I can't find an answer to that question. I was told that in B3LYP, more variables implemented in the method are empirical, but I can't find anywhere if it's true, and I'm sure it's not the only difference.&#xA;I'm new to calculations, and it seems so hard and confusing, so please help me.&lt;/p&gt;&#xA;" OwnerUserId="5586" LastEditorUserId="4945" LastEditDate="2014-05-23T02:05:17.897" LastActivityDate="2014-09-26T19:24:53.180" Title="What's the difference between PBE and B3LYP methods?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="11005" PostTypeId="1" AcceptedAnswerId="11006" CreationDate="2014-05-22T17:29:24.110" Score="7" ViewCount="1425" Body="&lt;p&gt;I think there's a basic idea I'm missing about 3 Dimensional representation of molecules. I think it may be about converting to Fischer representation. Consider the following two products of bromination of an alkene, the first one being for cis-form and the 2nd one for trans-form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRCv1.png&quot; alt=&quot;3d representation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the question may sound stupid, but it's crucial to get the basics. Why is the product II) the meso form and the products I) are the optically active forms. By what I know, the solid wedge goes towards the observer, so in product II), shouldn't the Bromine substituent(or maybe the $CH_3$ substituent) and $H$ atom be on the 'same side', making it the one optically active?&lt;/p&gt;&#xA;" OwnerUserId="3863" LastEditorUserId="305" LastEditDate="2014-05-22T23:43:03.087" LastActivityDate="2017-10-26T02:36:02.057" Title="What is there in this 3D representation of organic molecules I fail to grasp?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;stereochemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11006" PostTypeId="2" ParentId="11005" CreationDate="2014-05-22T18:31:25.797" Score="14" Body="&lt;p&gt;You have drawn the compounds correctly, and yes &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a solid wedge means the bond is coming out of the plane of the screen towards you&lt;/li&gt;&#xA;&lt;li&gt;a dashed wedge means the bond is going behind the plane of the screen away from you&lt;/li&gt;&#xA;&lt;li&gt;a solid line means the bond lies in the plane of the screen&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a Newman projection of your molecules, sometimes Newman projections can make it easier to &quot;see&quot; things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvGqJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Build a model (I did) or flip the drawings around, there is no way you can interconvert the isomers labeled &quot;S,S' and &quot;R,R&quot;; they are chiral enantiomers.  The meso compound and its mirror image are easily converted into one another (try rotating the molecule about a $\ce{C_2}$ axis that lies in the plane of the screen, 2) is perpendicular to the mirror plane [the dashed line in my drawing] and 3) bisects the $\ce{C_2-C_3}$ bond in the molecule); they are indeed representations of the same molecule.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-22T18:31:25.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11007" PostTypeId="1" AcceptedAnswerId="11011" CreationDate="2014-05-22T18:49:06.267" Score="3" ViewCount="4383" Body="&lt;p&gt;Consider nitric oxide, which has a real bond order of 2.5, contrary to what this UC Davis rendering might imply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KjHIC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to visualize a molecule with a fractional bond order? What exactly is &quot;half a bond&quot;? The electron spends half its time participating in bonding and the other half of the time as a lone electron on nitrogen? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5456" LastEditDate="2014-05-22T20:37:39.113" LastActivityDate="2015-11-30T20:24:00.677" Title="Fractional Bond Order - How to Visualize?" Tags="&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11008" PostTypeId="1" AcceptedAnswerId="11012" CreationDate="2014-05-22T20:19:09.853" Score="1" ViewCount="167" Body="&lt;p&gt;A rule of thumb in chemistry is that a reaction doubles in rate for every 10 °C increase in temperature. Considering the impact of a 10 °C increase in global temperature, would the same rule apply to kinetic weather processes – for example, would the average wind speed around the world double?&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorDisplayName="user7951" LastEditDate="2015-03-20T20:00:27.677" LastActivityDate="2015-03-20T20:00:27.677" Title="Global warming kinetics" Tags="&lt;kinetics&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11009" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2014-05-22T20:24:07.523" Score="6" ViewCount="151" Body="&lt;p&gt;I was watching &lt;a href=&quot;https://www.youtube.com/watch?v=BXCfBl4rmh0&amp;amp;index=17&amp;amp;list=PL7A1F4CF36C085DE1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic Video's chlorine video&lt;/a&gt;, which discusses chlorine's ferocity in stripping electrons whenever possible, and the professors mentioned its consequent use in &lt;a href=&quot;https://en.wikipedia.org/wiki/World_War_I&quot; rel=&quot;nofollow noreferrer&quot;&gt;WWI&lt;/a&gt; as a chemical agent. This got me thinking, if I were in that situation (say Syria if they decided to go old-school or a chemical leak), what could I do to protect myself against gaseous chlorine using the chemistry of everyday materials around me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine a wet towel won't help much to prevent the chlorine from getting in my lungs. Are there any easily obtainable substances I could safely breath through that would neutralize the chlorine? Or are there any (realistic) MacGyver solutions to quickly building a semi-functional gas mask? Just kinda curious what could be done.&lt;/p&gt;&#xA;" OwnerUserId="5587" LastEditorUserId="41328" LastEditDate="2019-04-23T01:33:05.190" LastActivityDate="2019-04-23T01:33:05.190" Title="MacGyvered chlorine gas protection" Tags="&lt;safety&gt;&lt;halides&gt;&lt;chemical-weapons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11010" PostTypeId="2" ParentId="10992" CreationDate="2014-05-22T21:34:58.680" Score="3" Body="&lt;p&gt;The second text is merely reporting the observed number of peaks, not predicting the splitting pattern you would obtain from this molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is not a case of one being an exact rule, and one being an approximate rule. It is a case of one text describing how to determine the expected splitting pattern (which would be a quartet of triplets, or 12 peaks) and the other simply reporting how many peaks are observed (six peaks). If you understand the method explained in the first text, you can very easily rationalise why the expected 12 peak qt appears as a 6 line signal; if you merely follow the observations described in the second text, you will fail to understand complicated splitting patterns that provide valuable information about molecular structure in 'real' molecules. (No disrespect to C3H7X text book examples intended). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these systems would commonly be referred to as a A3M2X2 system, representing $\ce{CH3CH2CH2X}$. To predict the splitting of the M2 group, simply consider the effect of each coupled group in turn. The A3 group will cause the M2 protons group to split into a quartet. Each of those quartet signals is then split due to coupling to the X2 protons, so each quartet signal becomes a triplet. this gives you the 12 lines, as so elegantly described in the first text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you then consider that the coupling constants between M and A, (J&lt;sub&gt;MA&lt;/sub&gt;) and  M and X (J&lt;sub&gt;MX&lt;/sub&gt;) are roughly equivalent (let's say roughly 7.5Hz), then you get overlap of some of the 12 lines, and the resulting spectrum appears, roughly, as 6 lines. On a lower field instrument, or with poor resolution, this may be indeed what you see (first spectrum below, simulated at 200MHz). However at high field with good resolution (without having to resolution enhance the data), you can clearly see that there are in fact 12 lines, and the coupling constants are not quite equal (second spectrum, simulated at 800MHz). In fact, coupling constants are closer to J&lt;sub&gt;MA&lt;/sub&gt;=7.8, and J&lt;sub&gt;MX&lt;/sub&gt;=7.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y7okZ.png&quot; alt=&quot;Low field 1H spectrum of A3M2X2 spin system&quot;&gt;&#xA;&lt;em&gt;Simulated Low Field Spectrum of A3M2X2 Spin System (Simulated at 200MHz, 0.5Hz linewidth)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rq97D.png&quot; alt=&quot;High field 1H spectrum of A3M2X2 spin system&quot;&gt;&#xA;&lt;em&gt;Simulated High Field Spectrum of A3M2X2 Spin System (Simulated at 800MHz, 0.3Hz linewidth)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an added note, in both molecules, $\ce{CH3-CH2-CH2OH}$ and $\ce{CH3CH2CH2Br}$, the central $\ce{CH2}$ is prochiral. A prochiral centre is defined as one that will become chiral if a single substituent is changed. Prochirality is a very important property to consider, as it is very common for protons at a prochiral centre to be chemically and magnetically non-equivalent if there is a chiral centre elsewhere in the molecule. This often gives rise to methylene protons of ethyl sidechains appearing as non-equivalent groups in NMR.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="5026" LastEditDate="2018-02-26T05:39:58.107" LastActivityDate="2018-02-26T05:39:58.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11011" PostTypeId="2" ParentId="11007" CreationDate="2014-05-22T22:17:00.140" Score="6" Body="&lt;p&gt;Maybe some examples will help.  In the picture below, I've drawn some resonance structures for butadiene and acrolein.  What do they tell us?  They suggest that the central &quot;single&quot; bond should have some double bond character, it's more than a single bond, but less than a double bond.  By the same token, the &quot;double&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lKWPR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bonds at the ends of the molecules should both have some single bond character.  These are examples of fractional bond order.  The real world consequences of fractional bond order is a higher barrier to rotation about the central &quot;single&quot; bond in these two molecules than we might have expected, and a lower barrier to rotation in the pi bonds than we might have expected.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-22T22:17:00.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11012" PostTypeId="2" ParentId="11008" CreationDate="2014-05-22T22:56:33.873" Score="5" Body="&lt;p&gt;That is a good rule of thumb.  It is based on the Arrhenius equation $$k = A {\operatorname{e}^{\frac{-{E_{\text{act}}}}{RT}}}$$ You'll find that the &quot;temperature doubles every 10 degrees&quot; rule  only applies if the following assumptions are valid&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The reaction must obey the Arrhenius equation&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it must have an entropic pre-term based on collision probability&lt;/li&gt;&#xA;&lt;li&gt;the reaction must be an activated process (e.g. it must have an activation energy)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Arrhenius equation works well for simple unimolecular reactions like the ring opening of cyclobutene to buta-1,3-diene.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you'll need to ask a weather physicist or a meteorologist if weather can be treated as an activated process.  My guess is that the answer is &quot;no&quot;, but that's just a guess.  It seems to me that there is no minimum energy required (activation energy) for the rain to fall or for the wind to blow. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-03-20T19:58:28.143" LastActivityDate="2015-03-20T19:58:28.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11013" PostTypeId="2" ParentId="11004" CreationDate="2014-05-22T23:58:00.030" Score="28" Body="&lt;h2&gt;PBE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The PBE functional${}^{[1]}$ belongs to the class of generalized gradient approximation (GGA) functionals for the exchange-correlation energy $E_{\mathrm{xc}}$. Considering that the dependence $E_{\mathrm{xc}}[\rho]$ may be non-local, i.e. $E_{\mathrm{xc}}$ may depend on the density $\rho$ at a given point (locality), but also on $\rho$ nearby (non-locality) the assumption made by the local spin density approximation${}^{[2,3]}$ (LSDA) that $E_{\mathrm{xc}}[\rho]$ is strictly local leaves a lot room for improvement. But still LSDA is a good starting point since it is a rather simple efficient model that gives good results. The GGA functionals improve on LSDA in a way that's pretty much in the spirit of a Taylor series with LSDA as a starting point: When we are at a point, what happens further off depends not only on $\rho$ at that point, but also the gradient of $\rho$ at the point, etc. (much as in a Taylor series: if you want to go further you may need not only the gradient, but also the Laplacian, etc.). So, these functionals are typically constructed by adding gradient corrections to the LSDA functionals. But that is not a trivial thing to do and doing it &quot;naively&quot; leads to wrong results. Therefore, some GGA functionals are parametrized by fitting experimental data. But there are also some GGA functionals that achieve the inclusion of gradient correction without introducing experimentally fitted parameters which makes them valid for a wide range of systems. One example of such a &lt;strong&gt;parameter-free&lt;/strong&gt; GGA functional is the one developped by Perdew, Burke and Ernzerhof (PBE). It is known for its general applicability and gives rather accurate results for a wide range of systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Side note:&lt;/em&gt; GGA functionals are frequently termed non-local&#xA;functionals in the literature. This is a somewhat misleading and actually sloppy terminology that should be avoided and since my above statements might be misread in that direction I want to make one thing perfectly clear: All GGA functionals are perfectly&#xA;local in the mathematical sense: the value of the functional at a point $\vec{r}$ depends only on information about the density $\rho(\vec{r})$, its gradient $\nabla\rho(\vec{r})$, and possibly other information at this very point and is absolutely independent of properties of $\rho(\vec{r}^{\, \prime})$  at points $\vec{r}^{\, \prime} \neq \vec{r}$. Calling these functionals 'non-local' is only motivated by the fact that these functionals go beyond the 'local' density approximation and of course the observation that knowledge of&#xA;the gradients is the first step towards accounting for the inhomogeneity of the real density.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;B3LYP&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The B3LYP approach${}^{[4,5,6]}$ belongs to the hybrid approximations for the exchange–hybrid correlation functional. The approximation is famous, because it gives very good results and, therefore, is extremely popular. The distinguishing feature of such hybrid approximations is that they mix in a certain amount of the exact Hartree–Fock exchange energy into the exchange and correlation obtained from other functionals. There is actually some justification for that: At the lower limit ($\lambda = 0$, where $\lambda$ is the coupling constant) of the so called coupling constant integration the exchange-correlation hole is equal to the exact exchange hole. This observation led Becke${}^{[7, 8]}$ to conclude that a fraction of exact exchange should be mixed with GGA exchange and correlation. The simplest such hybrid functional is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;E^{\mathrm{hyb}}_{\mathrm{xc}} = a E_{\mathrm{x}}^{\mathrm{exact}} + (1 − a) E^{\mathrm{GGA}}_{\mathrm{x}} + E^{\mathrm{GGA}}_{\mathrm{c}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the constant $a$ can be fitted empirically or estimated theoretically ${}^{[9, 10, 11]}$&#xA;as $a \approx \frac{1}{4}$ for molecules.&#xA;So far so good, but it smells like a witch's brew for the B3LYP exchange–correlation potential $E_{\mathrm{xc}}$: &lt;em&gt;take the exchange–correlation energy from the LSDA&#xA;method, add a pinch (20%) of the difference between the Hartree–Fock exchange energy &#xA;$E^{\mathrm{KS}}_{\mathrm{x}}$ (well, in fact, this is Kohn–Sham exchange energy, because the Slater determinant wave function, used to calculate it, is the Kohn–Sham determinant, not the Hartree–Fock one) and the LSDA $E^{\mathrm{LSDA}}_{\mathrm{x}}$. Then, mix well 72% of Becke exchange potential${}^{[12]}$ $E^{\mathrm{B88}}_{\mathrm{x}}$ which includes the 1988 correction, then strew in 81% of the Lee–Yang–Parr correlation potential${}^{[13]}$ $E^{\mathrm{LYP}}_{\mathrm{c}}$. You will like this homeopathic magic potion most if you conclude by putting in 19% of the Vosko–Wilk–Nusair potential$^{[14]}$ $E^{\mathrm{VWN}}_{\mathrm{c}}$:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  E^{\mathrm{B3LYP}}_{\mathrm{xc}} = E^{\mathrm{LSDA}}_{\mathrm{xc}} + 0.2 (E^{\mathrm{KS}}_{\mathrm{x}} - E^{\mathrm{LSDA}}_{\mathrm{x}}) + 0.72 E^{\mathrm{B88}}_{\mathrm{x}} + 0.81 E^{\mathrm{LYP}}_{\mathrm{c}} + 0.19 E^{\mathrm{VWN}}_{\mathrm{c}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can see B3LYP contains lots of empirical parameters but that's fine as long as you only want the results and don't want to know why this is working :)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;${}^{1}$ J. P. Perdew, K. Burke, M. Ernzerhof, &quot;Generalized Gradient Approximation&#xA;Made Simple&quot;, &lt;em&gt;Phys. Rev. Lett.&lt;/em&gt; &lt;strong&gt;1996&lt;/strong&gt;, &lt;em&gt;77&lt;/em&gt;, 3865; J. P. Perdew, K. Burke,&#xA;M. Ernzerhof, &quot;Erratum to Generalized Gradient Approximation Made Simple&quot;, &lt;em&gt;Phys. Rev. Lett.&lt;/em&gt; &lt;strong&gt;1997&lt;/strong&gt;, &lt;em&gt;78&lt;/em&gt;, 1396 (E). (&lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.77.3865&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{2}$ W. Kohn, L. J. Sham, &lt;em&gt;Phys. Rev.&lt;/em&gt; &lt;strong&gt;1965&lt;/strong&gt;, &lt;em&gt;140&lt;/em&gt;, A1133. (&lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.140.A1133&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{3}$ U. von Barth, L. Hedin, &quot;A local exchange-correlation potential for the spin polarized case&quot;, &lt;em&gt;J. Phys. C: Solid State Phys.&lt;/em&gt; &lt;strong&gt;1972&lt;/strong&gt;, &lt;em&gt;5&lt;/em&gt;, 1629–1642. (&lt;a href=&quot;http://iopscience.iop.org/0022-3719/5/13/012/&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{4}$ A. D. Becke, &quot;Density-functional thermochemistry. III. The role of exact exchange&quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;1993&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 5648-5652. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/98/7/10.1063/1.464913&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{5}$ P. J. Stephens, F. J. Devlin, C. F. Chabalowski, M. J. Frisch, &quot;Ab Initio Calculation of Vibrational Absorption and Circular Dichroism Spectra Using Density Functional Force Fields&quot;, &lt;em&gt;J. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 11623–11627. (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100096a001&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{6}$ K. Kim, K. D. Jordan, &quot;Comparison of Density Functional and MP2 Calculations on the Water Monomer and Dimer&quot;, &lt;em&gt;J. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 10089–10094. (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100091a024&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{7}$ A. D. Becke, &quot;A new mixing of Hartree–Fock and local density‐functional theories &quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1993&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 1372. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/98/2/10.1063/1.464304&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{8}$ A. D. Becke, &quot;Density‐functional thermochemistry. III. The role of exact exchange&quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1993&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 5648. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/98/7/10.1063/1.464913&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{9}$ J. P. Perdew, M. Ernzerhof, and K. Burke, &quot;Rationale for mixing exact exchange with density functional approximations&quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1996&lt;/strong&gt; &lt;em&gt;105&lt;/em&gt;, 9982. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/105/22/10.1063/1.472933&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{10}$ C. Adamo and V. Barone, &quot;Toward reliable density functional methods without adjustable parameters: The PBE0 model&quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt; &lt;em&gt;110&lt;/em&gt;, 6158. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/110/13/10.1063/1.478522&quot;&gt;Link&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{11}$ M. Ernzerhof and G. E. Scuseria, &quot;Assessment of the Perdew–Burke–Ernzerhof exchange-correlation functional&quot;, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;110&lt;/em&gt;, 5029. (&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/110/11/10.1063/1.478401&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{12}$ A. D. Becke. &quot;Density-functional exchange-energy approximation with correct asymptotic behaviour&quot;, &lt;em&gt;Phys. Rev. A&lt;/em&gt; &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;38&lt;/em&gt;, 3098-3100. (&lt;a href=&quot;http://journals.aps.org/pra/abstract/10.1103/PhysRevA.38.3098&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{13}$ C. Lee, W. Yang, R. G. Parr, &quot;Development of the Colle-Salvetti correlation-energy formula into a functional of the electron density&quot;, &lt;em&gt;Phys. Rev. B&lt;/em&gt;, &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;37&lt;/em&gt;, 785-789. (&lt;a href=&quot;http://journals.aps.org/prb/abstract/10.1103/PhysRevB.37.785&quot;&gt;Link&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{14}$ S. Vosko, L. Wilk, M. Nusair, &quot;Accurate spin-dependent electron-liquid correlation energies for local spin density calculations: a critical analysis&quot;, &lt;em&gt;Can. J. Phys.&lt;/em&gt; &lt;strong&gt;1980&lt;/strong&gt;, &lt;em&gt;58&lt;/em&gt;, 1200-1211. (&lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/p80-159&quot;&gt;Link&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-26T19:24:53.180" LastActivityDate="2014-09-26T19:24:53.180" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11014" PostTypeId="2" ParentId="11009" CreationDate="2014-05-22T23:58:40.990" Score="9" Body="&lt;p&gt;Interestingly, the British had such a solution in WWI: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypo_helmet&quot;&gt;Hypo Helmet&lt;/a&gt;. This was simply a bag soaked in sodium thiosulfate solution and worn over the head.  Thiosulfate reduces chlorine (and halogens generally) to chloride (halide) ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 S2O3^{2−} + Cl2 → S4O6^{2−} + 2 Cl^{−}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the article notes, this was not the best solution, but I think it fits your criteria for rough and ready protection.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-05-22T23:58:40.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11015" PostTypeId="1" AcceptedAnswerId="11017" CreationDate="2014-05-23T03:07:51.900" Score="4" ViewCount="1646" Body="&lt;p&gt;Would these be valid conformations of butane? I tried matching these up with the group but my drawings all seem rotated versus the book.&lt;/p&gt;&#xA;&lt;p&gt;I'm fairly sure my anti conformer is correct, along with my gauche conformer. I'm not too sure about the two eclipsed conformers; from what I understand, there are two eclipsed conformations - one with strong torsional strain resulting from the methyl group interacting with the other methyl group, and the other resulting from the methyl groups interacting with the hydrogens, right?&lt;/p&gt;&#xA;&lt;p&gt;Also, can we consider the dihedral angle to be the angle between the methyl groups?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3y0Wk.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/htsT8.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Another problem is that what I've drawn doesn't seem to match with the book's energy diagram too. I'd expect the anti-conformer to be the lowest in energy, but I'd also expect a difference in E for the two eclipsed conformations. One should have more torsional strain from placing the two methyl groups next to each other and the other, less, because the methyl groups are eclipsing the comparatively small hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-10-02T16:36:41.977" Title="Conformations of butane" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11016" PostTypeId="1" AcceptedAnswerId="11018" CreationDate="2014-05-23T03:16:08.597" Score="2" ViewCount="151840" Body="&lt;p&gt;We were told to rinse the buret with the base so that the concentration won't be messed up, but when it comes to the flask that is going to hold the acid, there is no need to rinse, because it would not affect the result if it has water in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? Wouldn't the acid be diluted a little bit and that throws off the accuracy?&lt;/p&gt;&#xA;" OwnerUserId="5138" LastEditorUserId="4945" LastEditDate="2019-05-21T21:32:46.390" LastActivityDate="2019-05-21T23:00:17.980" Title="How come water would not affect the results of a titration?" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;titration&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11017" PostTypeId="2" ParentId="11015" CreationDate="2014-05-23T03:19:38.407" Score="5" Body="&lt;p&gt;You have got them correct. Just your diagrams in eclipsed should have very close bonds. In your third diagram, one $\ce H$ is a bit far for another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alkane_stereochemistry&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; provides a nice description for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram too  : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVKKQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct.&lt;/p&gt;&#xA;" OwnerUserId="4597" LastActivityDate="2014-05-23T03:19:38.407" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11018" PostTypeId="2" ParentId="11016" CreationDate="2014-05-23T03:23:17.997" Score="4" Body="&lt;p&gt;The volume and concentration of base equals the number of moles of base. This is use to determine the number of moles of acid in the flask. Since you are interested in the number of moles of acid in the flask, why would adding water (which doesn't have any acid) change the number of moles of acid?&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-23T03:23:17.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11019" PostTypeId="1" AcceptedAnswerId="11025" CreationDate="2014-05-23T04:51:29.693" Score="0" ViewCount="704" Body="&lt;p&gt;My teacher insists that we can push electrons either from a bond to an atom's valence, or vice versa. Bond to lone pair; lone pair to bond, he insists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what about bond to bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a pedagogical reason he might be avoiding bond-to-bond movement? This is a legal move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does he simply not see bond-to-bond movement as a separate entity; could bond-to-bond electron pushing be construed as simply a bond to lone pair and lone pair to new bond move? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5591" LastEditDate="2015-10-09T17:52:14.180" LastActivityDate="2015-10-09T17:52:14.180" Title="Resonance - Legal Moves" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11020" PostTypeId="2" ParentId="11015" CreationDate="2014-05-23T07:46:25.523" Score="6" Body="&lt;p&gt;Some years ago I did a (relaxed) rotational scan of n-butane for the purpose of showing that fatty acids are all-trans. That is also why I found the graph in Awesome's answer some kind of disturbing - it should form a sinus like wave. I include this graph here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vxEzx.jpg&quot; alt=&quot;energy diagram of the rotation of n-butane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited out the error noticed by ron. Thanks again.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-05-27T10:59:11.733" LastActivityDate="2014-05-27T10:59:11.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11021" PostTypeId="1" CreationDate="2014-05-23T08:11:05.330" Score="2" ViewCount="3791" Body="&lt;p&gt;We are doing an experiment and would like to be able to identify zinc/aluminium/ magnesium . Would a copper sulphate solution work ? I hear that it turns black on contact with zinc, stays clear on contact with aluminium and fizzes on contact with magnesium.&lt;/p&gt;&#xA;" OwnerUserId="5590" LastEditorUserId="4945" LastEditDate="2014-07-29T11:44:20.480" LastActivityDate="2014-07-29T11:44:20.480" Title="Identification of zinc/aluminium/magnesium with copper sulphate solution" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11022" PostTypeId="2" ParentId="11019" CreationDate="2014-05-23T08:37:05.047" Score="3" Body="&lt;p&gt;Resonance is a concept that originates from the incapability of the original Lewis structure to describe delocalised systems. The formulation of these resonance structures are a representation of extreme states that are possible. The true bonding pattern lies between all possible resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider benzene, where we, among others, consider two major resonance structures. In fact the electron structure is well known to be a delocalised $\pi$ system, therefore better described by the right figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XdjtZ.png&quot; alt=&quot;resonance structures of benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, electrons appear to be &quot;pushed&quot; from one bond to another. This is not the case, as electrons are occupying one (or more degenerate) orbitals the whole time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when it comes to resonance structures, there are no clear rules, what is allowed and what is not allowed, however it has to be a possible representation of the actual bonding pattern. Resonance is a mere concept of describing a very complicated bonding pattern in simpler terms.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-23T08:37:05.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11025" PostTypeId="2" ParentId="11019" CreationDate="2014-05-23T15:38:47.823" Score="2" Body="&lt;p&gt;Pedagogically speaking, many classes first introduce the concept of Lewis structures before resonance theory.  The lone pairs are very &quot;visible&quot; in Lewis structures and so moving lone pairs before moving bonding electrons around is often a normal progression.  But this is a good question to ask your teacher, it could lead to an interesting conversation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your second question, I've never thought of bond-to-bond electron pushing as bond-to-lone-pair-and-lone-pair-to-new-bond electron pushing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there really is just one rule to drawing resonance structures: You can't change the position of the atoms in resonance structures.  For example in the figure below, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gclpl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top line shows 3 resonance structures for nitromethane.  The structure on the right probably doesn't contribute much to the overall description of nitromethane because of the unusually long $\ce{O-O}$ bond length.  In the second row, the $\ce{O-O}$ bond length has been shortened to a reasonable length - but nuclei have been moved - so this is not a resonance structure of the nitromethane drawn to its left, but rather this is an equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the don't move atoms rule, everything else is fair game.  Look at the resonance structures drawn for methane on the bottom line.  They are both valid, but using our chemical &quot;intuition&quot; we would just say that the structure on the right is of such high energy that it doesn't contribute much towards the overall description of methane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-23T15:38:47.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11026" PostTypeId="2" ParentId="11021" CreationDate="2014-05-23T15:54:51.537" Score="2" Body="&lt;p&gt;If you mean identification of the metals in their elemental form, then yes, it might be possible to distinguish between them with an aqueous $\ce{CuSO4}$ solution due to the different reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc in $\ce{CuSO4}$ will be oxidized to colorless $\ce{Zn^2+}$, while $\ce{Cu^2+}$ is reduced to elemental copper, because copper is the more noble metal. The $\ce{Cu}$ precipitate is normally not black, although it can appear dark in the background of the blue solution. It can also form a layer on the piece of zinc, and in the course of the reaction, the initially blue solution will become more and more colorless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + CuSO4 \rightarrow Cu\downarrow + ZnSO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium does not react at all because it is protected by its native oxide layer which is stable in this solution. Magnesium, however, is oxidized under evolution of $\ce{H2}$ gas, which fizzes off from the solution because it is poorly soluble in water. The reaction will gradually become slower when insoluble $\ce{Mg(OH)2}$ passivates the metal surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + 2 H2O \rightarrow Mg(OH)2 + H2\uparrow}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-05-23T15:54:51.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11027" PostTypeId="1" AcceptedAnswerId="11031" CreationDate="2014-05-23T17:19:16.130" Score="1" ViewCount="5841" Body="&lt;p&gt;I've been unable to find an energy diagram for propane either in my book or online via Google. So I resorted to making my own. I suspect that propane isn't covered because it's essentially the same deal as ethane - we can have one staggered conformation and one eclipsed conformation (only in propane it's a methyl eclipsing a hydrogen, not hydrogens eclipsing each other). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I drew the Newman projections and an energy diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dpbOz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="30836" LastEditDate="2020-04-05T13:02:23.957" LastActivityDate="2020-04-05T13:02:23.957" Title="Energy Diagram for Propane" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;hydrocarbons&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11028" PostTypeId="1" CreationDate="2014-05-23T17:45:22.177" Score="11" ViewCount="1072" Body="&lt;p&gt;It is well known that the A in both DNA and RNA stands for acid, but where is the acid in chemical formula for the compound, and it is classified so based on what acid-base theory? Like Arrhenius, Brønsted-Lowry or Lewis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to better explain my question, when I say DNA I am essencially thinking about nucleotides linked together, am I missing something in the structure that contains my answer or this is the way to go?&lt;/p&gt;&#xA;" OwnerUserId="5595" LastEditorUserId="5026" LastEditDate="2018-03-03T11:22:49.533" LastActivityDate="2019-01-04T04:10:38.847" Title="Where is the acid in DNA/RNA?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11029" PostTypeId="2" ParentId="11028" CreationDate="2014-05-23T17:52:57.487" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Just to better explain my question, when I say DNA I am essencially thinking about nucleotides linked together, am I missing something in the structure that contains my answer or this is the way to go?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you do. *NA consist of nucleotides, linked by phosphodiester fragments. This leaves one hydroxyl of the phosphate fragment intact, so it can act as an acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look closely: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aRNA_chemical_structure.GIF&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7PCEs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a $\ce{P-O-}$ fragment, which means that there is a cation somewhere nearby. So, *NA are Brønsted acids, usually occuring as salts. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="3748" LastEditDate="2014-05-24T10:14:17.667" LastActivityDate="2014-05-24T10:14:17.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11030" PostTypeId="2" ParentId="11028" CreationDate="2014-05-23T17:53:25.190" Score="10" Body="&lt;p&gt;The phosphate ester groups which connect the nucleotides contain one acidic proton at their OH group, and two at the end of each strand. As you can see &lt;a href=&quot;http://en.wikipedia.org/wiki/File:DNA_chemical_structure.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://commons.wikimedia.org/wiki/Category:DNA_chemical_structures#mediaviewer/File:PhosphodiesterBond_of_DNA.PNG&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and in the picture below (with the groups marked turquoise), most of these groups are deprotonated depending on pH. They act as Brønsted-Lowry acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FSWte.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-03-03T11:23:03.947" LastActivityDate="2018-03-03T11:23:03.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11031" PostTypeId="2" ParentId="11027" CreationDate="2014-05-23T18:10:46.153" Score="3" Body="&lt;p&gt;Your diagram is correct.  Here is a &lt;a href=&quot;http://www.uwplatt.edu/~sundin/353/image/l960925f.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to an on-line energy diagram for propane and here is the diagram itself&lt;img src=&quot;https://i.stack.imgur.com/rthtc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-23T18:10:46.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11032" PostTypeId="1" CreationDate="2014-05-23T18:35:16.607" Score="3" ViewCount="702" Body="&lt;p&gt;I bought a 2 pH meters several months back from Amazon for roughly $10 a piece.  I couldn't tell you how they work, but here's the link on Amazon:  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0087LOWIA&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/gp/product/B0087LOWIA/ref=wms_ohs_product?ie=UTF8&amp;amp;psc=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also got a neutral buffer solution for calibration.  It worked fine initially but I'm thinking it's off.  It's now being used for pH testing in the salt water of an floatation tank (147 gallons of water + 750 lbs pharmaceutical grade epsom salts and 50 pounds of dead sea salt).  One meter shows a pH of 6 and another of the same kind at about 8.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does the salt effect the accuracy of the meter?  Are the meters inherently bad, or do they need periodic recalibration?  I didn't want to keep buying litmus paper.&lt;/p&gt;&#xA;" OwnerUserId="5597" LastActivityDate="2014-07-24T23:58:50.303" Title="Accuracy of pH meters in MgSO4 solution" Tags="&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11033" PostTypeId="1" AcceptedAnswerId="11037" CreationDate="2014-05-23T18:51:51.597" Score="0" ViewCount="4135" Body="&lt;p&gt;So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why are the $\ce{S-O}$ and $\ce{S-OH}$ bonds not equivalent then? Why do their lengths differ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect this has something to do with the addition of the hydrogen on the oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else could be a cause for the discrepancy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because my teacher challenged the fact that sulfur might not be hypervalent on the basis that the bond lengths differ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In retrospect, I think that he:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is placing too much weight on Lewis Structures - which are just rough renditions of what chemicals really are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The $\ce{S-O}$ bonds and $\ce{S-OH}$ bonds may differ in length not because the $\ce{S-O}$ bonds are necessarily double bonded but also because the $\ce{S-OH}$ bonds are different in that S is bound to a hydroxyl group rather than a single oxygen. Also can't there be very minor resonance contributors involving 5-6 bonds on sulfur? It's not that sulfur absolutely can't access its d-orbitals but rather it doesn't to any great extent. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5456" LastEditDate="2014-05-24T13:40:02.317" LastActivityDate="2014-05-24T13:40:02.317" Title="Sulfuric Acid - Bond Lengths and Charge Separation" Tags="&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11034" PostTypeId="2" ParentId="11032" CreationDate="2014-05-23T19:01:57.487" Score="2" Body="&lt;p&gt;The answer is yes, ionic force does influence real pH signal. Because pH meter is sensitive to activities (instead of concentration of H+ only) of ions that modifying it junction potential you must take in account ionic force when it is strength enough to significantly impact activities. There is a lot of limitation with pH-meter: above pH&gt;12 or bellow pH&amp;lt;2 they are not precise at all (see for acidic and basic errors). Or if they are used in different solvents than water. Organic chemists should tell you what pH could mean in hexane. Assessing pH is a complex task that must be addressed with the care it deserves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, you should find the user-guide of your pH meter on internet. I am sure the manufacturer will provide limitation and use tips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a detailed chapter on pHmetry in Skoog and West Analytical Chemistry. It may help you to deeper understand those phenomena.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="74" LastEditDate="2014-05-25T23:31:58.623" LastActivityDate="2014-05-25T23:31:58.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11035" PostTypeId="1" CreationDate="2014-05-23T19:36:47.517" Score="5" ViewCount="4564" Body="&lt;p&gt;If I have aqueous sulfuric acid solution as electrolyte and the anode electrode is made out of copper and the cathode electrode is made out of graphite, what are the two reactions at the respective electrodes and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\text{Anode:}  &amp;amp;&amp;amp; \ce{Cu &amp;amp;-&amp;gt; Cu^2+ + 2e-}\\&#xA;\text{Cathode:}&amp;amp;&amp;amp; \ce{2e- + 2H+ &amp;amp;-&amp;gt; H2 (g)}\\\hline&#xA;\text{Total:}  &amp;amp;&amp;amp; \ce{2H+ + Cu &amp;amp;-&amp;gt; H2 + Cu^2+}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should be the reaction, but why? Why isn't sulfuric acid participating? Why does that specific reaction happen?&lt;/p&gt;&#xA;" OwnerUserId="5599" LastEditorUserId="4945" LastEditDate="2020-02-26T12:56:46.413" LastActivityDate="2020-02-26T12:56:46.413" Title="Why doesn't sulfate get electrolyzed in sulfuric acid electrolysis?" Tags="&lt;acid-base&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11036" PostTypeId="1" AcceptedAnswerId="11138" CreationDate="2014-05-23T19:40:46.367" Score="11" ViewCount="4228" Body="&lt;p&gt;Consider the reversible unimolecular reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A &amp;lt;=&amp;gt;[k_1][k_2] B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the forward reaction is often considerably more thermodynamically favourable than the reverse reaction, and therefore the relationship $k_1 \gg k_2$ holds between the rate constants. The rate constants are in the same units, and so it is possible to write this relationship. A similar example would be a reaction that is bimolecular on both sides, $\ce{A + B &amp;lt;=&amp;gt; C + D}$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider this reversible reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B &amp;lt;=&amp;gt;[k_1][k_2] C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the reaction proceeds at rate $k_1 [\ce{A}][\ce{B}]$ in the forward direction and $k_2 [\ce{C}]$ in reverse. If the overall reaction rate has units $M s^{-1}$, then $k_1$ necessarily has units $M^{-1} s^{-1}$ and $k_2$ has units $s^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; For this second case, we can no longer impose that $k_1 \gg k_2$, due to difference in units, but can we say &lt;em&gt;anything&lt;/em&gt; about their relationship? Does setting $k_1$ to some value constrain the choice of $k_2$ in any way?&lt;/p&gt;&#xA;" OwnerUserId="5598" LastEditorUserId="30836" LastEditDate="2017-10-14T12:47:09.607" LastActivityDate="2017-10-14T12:47:09.607" Title="Relationship between magnitudes of forward and reverse kinetic rate constants" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11037" PostTypeId="2" ParentId="11033" CreationDate="2014-05-23T19:53:46.913" Score="0" Body="&lt;p&gt;Good questions, let's step through them one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. &quot;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the resonance structures below.  The one on the left with sulfur hypervalent is the only uncharged resonance structure.  I suspect that that counts for something.  Here is a &lt;a href=&quot;http://www.uwosh.edu/faculty_staff/gutow/Lewis_Tutorial/H2SO4_8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; that says it is the most important of the structures (don't get confused when you view the link, they use equilibrium arrows (wrong) in their diagram instead of  resonance arrows (correct).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e7woJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &quot;why are the S-O and S-OH bonds not equivalent then? Why do their lengths differ?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very fundamental question.  My answer is, they are not equivalent because they are different.  Just as you wouldn't expect the same $\ce{C-O}$ bond length in $\ce{CH_3-O-CH_3}$ and $\ce{CH3-OH}$, you wouldn't expect the same bond lengths in an $\ce{S-O}$ bond and an $\ce{S-OH}$ bond.  In fact, since these bonds are not superimposable one on the other by a symmetry operation, we can say that they MUST be different (i.e. have different lengths).  As one of my favorite professors used to say, now if were to actually measure the $\ce{S-O}$ and $\ce{S-OH}$ bond lengths and our measurements gave rise to the same numeric bond length out to 10 decimal places, we could still step back and say that both we and God know that the lengths are not the same, we just need to measure out to more decimal places!  Let me say the converse, the only bonds that can be identical are those that can be interconverted one with another through a symmetry operation.  For example in chloroform, $\ce{CHCl3}$, there is a $\ce{C_3}$ axis containing the $\ce{C-H}$ bond.  Rotation about this axis interconverts the 3 $\ce{C-Cl}$ bonds, therefor they are equivalent bonds.  Their bond lengths MUST be the same.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-23T19:53:46.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11039" PostTypeId="1" AcceptedAnswerId="11053" CreationDate="2014-05-24T03:33:56.913" Score="3" ViewCount="145" Body="&lt;p&gt;What are some incapabilities that the earlier spectroscopy (Visible, Ultraviolet and Infra-red) have when compared to the more modern method?&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="4734" LastEditDate="2014-05-24T10:29:31.950" LastActivityDate="2014-05-24T20:05:18.707" Title="What are some incapabilities that the early spectroscopy have when compared to the more modern method?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11040" PostTypeId="1" AcceptedAnswerId="11043" CreationDate="2014-05-24T04:09:52.333" Score="5" ViewCount="2135" Body="&lt;p&gt;In a reaction (as given in my textbook),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-CH=CH2 + (CH3-CO)2O  -&amp;gt; R-CH2=CH-COCH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an alkene reacts with acetic anhydride to give methyl alkenyl ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure of how this proceeds, a search on google displays no links to any relevant matter, and I have not successfully found this reaction in any of the other books I possess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, it should have been an addition reaction (characteristic of alkenes) across the double bond. But this isn't the case here. So, what exactly is the reaction mechanism?&lt;/p&gt;&#xA;" OwnerUserId="5429" LastEditorUserId="5026" LastEditDate="2018-04-04T14:53:09.297" LastActivityDate="2018-04-04T14:53:20.063" Title="Reaction mechanism of a terminal alkene with an anhydride" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11042" PostTypeId="1" CreationDate="2014-05-24T10:27:18.790" Score="4" ViewCount="2249" Body="&lt;p&gt;I read some articles and journals about activated carbon and found a problem about how the adsorbate could bind to the activated carbon. Some said it is because of intramolecular forces (van der Waals) and some said it is because of chemisorption. If chemisorption plays a role, what would the bonding be?&#xA;And if the bonding is just caused by intramolecular forces, what is the active site of the activated carbon? An answer or illustration for better understanding is greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5605" LastEditorUserId="5016" LastEditDate="2014-05-24T16:29:42.543" LastActivityDate="2016-11-24T15:49:58.947" Title="How does activated carbon bind the adsorbate?" Tags="&lt;bond&gt;&lt;adsorption&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11043" PostTypeId="2" ParentId="11040" CreationDate="2014-05-24T12:50:23.750" Score="7" Body="&lt;p&gt;This is the  Friedel-Crafts acylation reaction of alkene. Catalyzed by Lewis acid. Addition-elimination mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AjyOs.gif&quot; alt=&quot;Mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="5026" LastEditDate="2018-04-04T14:53:20.063" LastActivityDate="2018-04-04T14:53:20.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11044" PostTypeId="2" ParentId="5220" CreationDate="2014-05-24T13:48:25.510" Score="1" Body="&lt;p&gt;It's pretty easy. Put one of your bottles in the freezer when at home and let the water turn into ice. Then wrap an old cotton t-shirt around the bottle so that it makes a thick insulating layer around it. Then fasten the cloth with a safety pin. This would prevent the ice from being melted quickly and provide you with cool water for almost a day. With a little enthusiasm you can make it so that you just have to change bottle through pulling old ones and pushing new ones into the cloth without necessarily opening the cloth.&lt;/p&gt;&#xA;" OwnerUserId="5607" LastActivityDate="2014-05-24T13:48:25.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11046" PostTypeId="1" AcceptedAnswerId="11056" CreationDate="2014-05-24T17:00:09.150" Score="4" ViewCount="2192" Body="&lt;p&gt;I found this build-plan recycling on the internet (I believe it was originally published in a book written by the infamous clandestine chemist Hobart Huson) which seemed interesting to me. I do not have the money to pay for a professional fume hood, nor do I have the space to fit such thing. I have to switch the function of my work workplace sometimes because of lack of space (it is also used for woodworking, another hobby of me), so a simple demountable fume hood would be ideal. As said, I found this one with a simple build description, but I was wondering about its effectiveness and safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1ZbX.png&quot; alt=&quot;Demountable fume hood&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown on the picture, the fume hood consists of a demountable frame, a thick plastic sheet covering the frame that is attached to a tube, and a simple exhaust system connected to an other tube that is placed outside the building, so that the fumes can diffuse in the atmosphere. The following four points concern me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?&lt;/li&gt;&#xA;&lt;li&gt;Is this plastic sheet a potential danger in case of fire? Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?&lt;/li&gt;&#xA;&lt;li&gt;Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? Or is reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/Material_safety_data_sheet&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSDS&lt;/a&gt; (i.e. the &lt;em&gt;environment&lt;/em&gt; section) for the chemicals being used enough?&lt;/li&gt;&#xA;&lt;li&gt;There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Take note of the fact that we are talking about amateur chemistry here. There are some nasty fumes that need to be removed, like chlorine for example, but no dense fogs of hydrogen cyanide or phosgene.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-05-24T17:24:08.793" LastActivityDate="2014-05-25T06:34:43.677" Title="Demountable simple fume hood" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11047" PostTypeId="2" ParentId="11046" CreationDate="2014-05-24T18:51:17.490" Score="6" Body="&lt;p&gt;Since the chemicals will not be in contact with the plastic for very long, very little will diffuse through the plastic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are concerned about fire, you could wrap the outside (not the inside) with aluminum foil; this should keep the fire contained. Have a container of sand handy in case the fire is extremely hot and the aluminum starts to burn. In fact, sand would be handy for most fires since it will decrease the amount of oxygen getting to the flames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you have a high air flow and the volatiles are coming from a small area, there should be no problem. &lt;em&gt;Added:&lt;/em&gt; The concentration of the volatiles will not be high enough for a spark to ignite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't really comment about the environmental effect unless you provide more detail. If the MSDS doesn't mention a hazardous exposure from the fumes, then you probably don't need a fume hood. You can get some activated charcoal (from a pet store) and pass the fumes through the charcoal, but I don't know how effective that would be.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-05-24T21:28:27.460" LastActivityDate="2014-05-24T21:28:27.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11048" PostTypeId="1" AcceptedAnswerId="11052" CreationDate="2014-05-24T18:52:53.257" Score="34" ViewCount="116406" Body="&lt;p&gt;In mineralogy class, I was taught that metallic and ionic bonds are weaker than covalent bonds and that's why quartz and diamond have such a high hardness value. However, in organic chemistry class, I learned that covalent bonds are weaker than metallic and ionic bonds, thus organic substances have a much lower melting point than that of metals and ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I getting wrong? Are ionic and metallic bonds weaker than covalent bonds or not?&lt;/p&gt;&#xA;" OwnerUserId="3955" LastEditorUserId="9961" LastEditDate="2015-06-10T15:37:45.843" LastActivityDate="2018-05-31T02:34:14.730" Title="Are metallic/ionic bonds weaker than covalent bonds?" Tags="&lt;bond&gt;&lt;metal&gt;&lt;crystal-structure&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="11049" PostTypeId="2" ParentId="11048" CreationDate="2014-05-24T19:16:30.113" Score="2" Body="&lt;p&gt;Ionic and metallic bonds are weaker than covalent bonds. This is correct, it is why covalent crystal is much harder than ionic and metallic crystal/polycrystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second statement is wrong because firstly melting point is not proportional to the strength of chemical bond. There are more factors such as flexibility of molecules. Boiling point is more proportional instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More important, the inter-particle forces to be compared among the organic compound vs ionic vs metallic compound is &lt;strong&gt;NOT&lt;/strong&gt; among &lt;strong&gt;&lt;em&gt;covalent bond&lt;/em&gt;&lt;/strong&gt; vs ionic bond vs metallic bond. It is among &lt;strong&gt;&lt;em&gt;intermolecular force&lt;/em&gt;&lt;/strong&gt; (dipole-dipole, H-bond, Van der waals) vs ionic bond vs metallic bond. And the first one is much weaker than the second and third for sure. So the boiling point of organic compounds is much lower.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-24T19:16:30.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11050" PostTypeId="2" ParentId="11042" CreationDate="2014-05-24T19:22:41.203" Score="1" Body="&lt;p&gt;Chemisorption means chemical reaction occurs during absorption process, and any bond type other than intermolecular forces (dipole-dipole, H-bond, Van Der Waals) is possible. Activated carbon absorb using intermolecular forces, and I do not think it is chemisorption. The absorption is strong because the surface area is huge, not because of any real chemical bonds.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-05-24T19:22:41.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11051" PostTypeId="2" ParentId="11048" CreationDate="2014-05-24T19:28:35.580" Score="7" Body="&lt;p&gt;What you learned in your mineralogy class was correct; bond strength decrease in the following order covalent &gt; ionic &gt; metallic.  The reasoning for this is as follows.  In covalent bonds such as those in methane and oxygen, the valence electrons are shared between the atoms involved in the bond and they (the electrons) spend most of their time in the region between the nuclei involved in the bond; this makes for a strong bond.  In ionic materials such as sodium chloride, the electrons are donated from one (the electropositive) atom to the other (the electronegative) atom in order for the atoms to achieve a filled shell structure.  The ionic atoms are attracted to one another through electrostatic attraction and the crystal lattices that are formed.  The bonds formed through electrostatic attraction are not as strong as those formed from covalent sharing of electrons.  Finally, in metals the outermost electrons are donated or &quot;pooled&quot; in the band structure that exists in metals.  The electrons are free to travel great distances (hence the conductivity of metals) and serve as a glue to hold all of the positively charged metal nuclei together.  So in the case of metals, there are no significant metal-metal bonds and these bonds are therefor the weakest.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-24T19:28:35.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11052" PostTypeId="2" ParentId="11048" CreationDate="2014-05-24T19:29:06.357" Score="19" Body="&lt;p&gt;Quartz and diamond are stronger substances because their molecules form network covalent structures. These structures form a lattice-like structure, much the same as ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This molecular network is also the reason that diamond and quartz form a crystalline structures, just like you'd see in ionic substances such as NaCl. Some other structures you might want to look into are Graphite and Graphene, which are both allotropes of carbon (allotropes are, simply put, different molecular arrangements of an element). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The network structure combines to make the substance stronger than normal covalent bonded substances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, substances with standard covalent bonds seem to be weaker than those with ionic bonds because the ionic bonds tend to form a lattice structure, that makes them much stronger. You can see this in the fact that the boiling points of ionic salts are much higher than that of a covalent substance like water. However, when covalent bonds form network covalent structures, atoms combine to form a singular macromolecule that is much stronger than singular covalent bonds. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="5234" LastEditDate="2014-05-24T19:38:48.877" LastActivityDate="2014-05-24T19:38:48.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11053" PostTypeId="2" ParentId="11039" CreationDate="2014-05-24T20:05:18.707" Score="4" Body="&lt;p&gt;Some of the things that come to mind are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Resolution&lt;/strong&gt;  Modern equipment has much higher resolution than older equipment.  Optical component surface smoothness and uniformity, optical component alignment and electrical component stability all permit the user to see more detail and fine structure in the spectrum, which assists in analysis&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;More stable light sources&lt;/strong&gt;  Again resulting in more detailed and more reproducible spectra&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Improved monochromaticity of the light sources&lt;/strong&gt;  Further improving spectral resolution&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Improved sample cells&lt;/strong&gt;  Polished glass surfaces provide for better signal to noise making it easier to discern the signal&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Signal processing software&lt;/strong&gt;  Such as Fourier Transform methods which increase the signal to noise ratio making it easier to discern the true peaks in a spectrum&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Other computer and software enhancements&lt;/strong&gt;  That can be applied to sample handling and spectrum processing.  They dramatically reduce the time it takes to record and process a spectrum&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-24T20:05:18.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11054" PostTypeId="2" ParentId="11046" CreationDate="2014-05-24T21:03:08.637" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Actually, you should care more about the integrity of the tube&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this plastic sheet a potential danger in case of fire? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends. But care more about inflammable and unmeltable working surface (metal or ceramic). Some old institutions work just fine with fume hoods using wood frames.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PVC is self-extinguishing plastic that is extremely hard to lit up. However, when burned, it produce toxic vapours, and is a spark risk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Care about your car. It affects environment more than most other fumes you can produce.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The engine must not touch the air sucked of, as there is a spark in common electric engines.The fan itself must not touch anything to not produce sparks. Everything should be oiled.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-05-24T21:03:08.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11055" PostTypeId="2" ParentId="10273" CreationDate="2014-05-24T21:40:26.877" Score="24" Body="&lt;p&gt;The terms softness and hardness are used in &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot;&gt;HSAB theory&lt;/a&gt; to distinguish Lewis acids and bases by charge density, polarizability, electronegativity, and, in the case of molecular compounds, HOMO and LUMO energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Charge_density&quot;&gt;Charge density&lt;/a&gt; is inversely proportional to atomic/ionic radius and proportional to oxidation state (in case of a neutral atom) or charge (in case of an ion). Hence charge density increases with decreasing radius and increasing oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizability&quot;&gt;Polarizability&lt;/a&gt; is the tendency of the electron shell of an atom/molecule to become deformed in an electric field, which can originate e.g. from the charge of an adjacent ion or the partial charge of a dipole. An electron cloud of an atom or ion is more easily deformed with increasing volume, as it is more diffuse, and its outer electrons tend to be more loosely bound. Therefore, larger atoms and molecules tend to be more polarizable, and higher polarizability is often accompanied by lower charge density or oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of molecules, when a Lewis acid reacts with a Lewis base, electrons from the highest occupied molecular orbital (HOMO) of the Lewis base are transferred to the lowest unoccupied molecular orbital (LUMO) of the Lewis acid. How well this transfer of electron density works depends on the relative energies of the acid LUMO and the base HOMO. The stability of the reaction product will increase with the energy difference between its HOMO and LUMO (&lt;a href=&quot;http://www.chem.tamu.edu/rgroup/dunbar/Teaching/CHEM362/Lecture_30-31.pdf&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HSAB theory predicts that hard Lewis bases react preferentially with hard acids, and that likewise soft bases form stronger bonds with soft acids. For example, $\ce{Li+}$ and $\ce{BF3}$ are hard acids and $\ce{F-}$ and $\ce{NH3}$ are hard bases due to their small size and high charge density/low polarizability of the ions. $\ce{BF3}$ therefore readily forms Lewis acid-base adducts with fluoride and ammonia, and $\ce{LiF}$ is less soluble in water than $\ce{LiCl}$ and $\ce{LiBr}$. $\ce{Ag+}$ is an example for a soft acid (large ion with low charge), and the halide ions become increasingly softer bases with increasing ionic radius. Silver therefore forms stronger bonds with heavier halides like $\ce{I-}$ and $\ce{Br-}$, and this also explains the decreasing solubility of the silver halides in water, which is classified as a hard solvent. In the case of soft acids and bases, the difference of electronegativity tends to be smaller, and the bonds have more covalent character (e.g. silver iodide) than those between hard acids and bases (e.g. lithium fluoride) with larger difference in electronegativity (&lt;a href=&quot;http://de.scribd.com/doc/24498037/Hard-Soft-Acid-Base-HSAB-Theory&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shift of reaction equilibria can also be predicted with this concept, like in the following example (&lt;a href=&quot;http://www.chem.tamu.edu/rgroup/dunbar/Teaching/CHEM362/Lecture_30-31.pdf&quot;&gt;reference&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HgO + H2S \rightleftharpoons HgS + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equilibrium will be shifted to the right side, because $\ce{Hg^2+}$ as a soft acid prefers to bond to the soft base $\ce{S^2-}$. Likewise, oxygen as a hard base forms stronger bonds with hydrogen (hard acid).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-07-13T15:24:52.020" LastActivityDate="2014-07-13T15:24:52.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11056" PostTypeId="2" ParentId="11046" CreationDate="2014-05-24T22:19:51.607" Score="5" Body="&lt;p&gt;If you do a little reading about fume cupboards, and in particular, benchtop extraction hoods, you will discover that they are constructed from materials not too dissimilar to your design; consider the difference being amateur use versus professional use. To address your four main concerns:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Provided the plastic &lt;em&gt;is&lt;/em&gt; thick enough. So, certainly thicker than &lt;a href=&quot;http://en.wikipedia.org/wiki/Plastic_wrap&quot; rel=&quot;nofollow&quot;&gt;cling wrap&lt;/a&gt;, and I'd be happy using the types of material common to decent &lt;a href=&quot;http://www.greenhousemegastore.com/category/portable-greenhouses&quot; rel=&quot;nofollow&quot;&gt;portable greenhouses&lt;/a&gt;. It needs to be robust enough to withstand the extraction forces from your exhaust fan. If your exhaust is working correctly, any chemicals will not be in contact with the plastic long enough to be a concern. Be sure you run a post-purge extraction once you have finished your day's chemistry to remove all traces of chemical fumes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this plastic sheet a potential danger in case of fire? Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most fume hoods are made from combustible material. I have, in my 20+ year career as a research chemist, seen many fume hoods destroyed by fire. This includes hoods made from wood, various forms of plastic, even those with metal frames and plastic components. If you have a fire, yes the plastic sheet will burn, and depending on the material, it may burn quickly or slowly - you will have to judge that. This, then, will increase the volume of your fire, and the amount of combustible material. So you will need to manage that in your workshop layout. So, yes, this is a hazard, but one that you should be able to plan for, and adequately implement a risk mitigation strategy. But the real concerns are not whether the plastic sheet would be a danger &lt;em&gt;if&lt;/em&gt; there is fire, but (a)will the plastic sheet &lt;em&gt;increase&lt;/em&gt; the risk of fire, and (b)what effect will your exhaust fan have in the event of fire.  If you are doing chemistry that involves refluxing, as your picture suggests, you need to make sure that the size of the hood is big enough that you are not heating your plastic, and that the air flow is sufficient to extract any hot air from the roof of your hood. You also need to consider that fires in fume hoods are generally more active than fires on open bench areas because they are actively fanned and have a fresh air source driven into the fire zone because of your extraction fan. A fire blanket in your workshop is not a sufficient measure against the type of fire you might have doing chemistry, Buy yourself at least one &lt;a href=&quot;http://en.wikipedia.org/wiki/Fire_extinguisher#Dry_chemical&quot; rel=&quot;nofollow&quot;&gt;dry chemical extinguisher&lt;/a&gt;, which are sold in all major retailers that sell fire blankets. If you are using chlorine, then you should &lt;strong&gt;NOT&lt;/strong&gt; use the &lt;a href=&quot;http://en.wikipedia.org/wiki/ABC_dry_chemical&quot; rel=&quot;nofollow&quot;&gt;ABC type dry extinguishers&lt;/a&gt; that use ammonium based material, as they will react violently with chlorine. You should use a BC type, which is based on sodium bicarbonate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? Or is reading the MSDS (i.e. the environment section) for the chemicals being used enough?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you should be worried. Reading the MSDS is not enough. You need to read it, understand it, and then take appropriate measures to mitigate any hazardous risk to the local environment. If you are extracting chlorine gas, where is it going? Apart from being pulled through your pump system, which will be corroded pretty badly after only a few uses, what is directly outside at your pump exhaust? If you position your exhaust just outside a window of your building, what local wind channelling exists to push that exhaust straight into the next window of your, or your neighbour's house? You could consider installing a &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_filtering&quot; rel=&quot;nofollow&quot;&gt;carbon filter&lt;/a&gt; like the types used on water systems, as these are not overly expensive, but for many commercial systems, no scrubbing or filtration occurs, but the key issue is where is your exhaust going? So, understand your design, consider your risks, and plan your exhaust appropriately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that all building codes (certainly all the ones I have ever encountered) will specify that any motor used in an area containing flammable vapours (and dust, if you need to consider that for your woodworking) should be of the type approved for such an environment. What dust extraction do you use for your woodworking? If you get an arc from the pump motor, you will very likely get ignition. Again, this is something I &lt;em&gt;have&lt;/em&gt; seen in research laboratories for all sorts of non-approved electrical equipment, including pumps, air heaters, fridges. So, if you are exhausting flammable vapours, this is a real risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final consideration you haven't really mentioned is the exhaust pump itself. You need to make sure that it is of suitable size to provide correct laminar flow into the front of the hood, around all your equipment, into the exhaust outlet, and pushed back through the exhaust port after the pump. You should test this using &lt;a href=&quot;http://www.screwfix.com/p/smoke-pellets-5g-pack-of-10/64647&quot; rel=&quot;nofollow&quot;&gt;a smoke pellet&lt;/a&gt;, or something similar. the larger the opening of your hood, the larger the exhaust fan needs to be, which is why you should always lower the sash hood on fume cupboards when doing chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-05-25T06:34:43.677" LastActivityDate="2014-05-25T06:34:43.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11057" PostTypeId="2" ParentId="8424" CreationDate="2014-05-24T22:36:33.563" Score="1" Body="&lt;p&gt;Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion.  Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about.  So here's a picture of the reaction we're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkQNd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is a picture of the 5 molecular orbitals for the pentadienyl system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LuscN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons.  That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO).  The HOMO is so important because it is the orbital that will control our process.  If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other.  If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap.  The type of rotation that leads to bonding overlap in this example is termed &quot;disrotatory&quot;.  It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen.  This is the symmetry element that is preserved during the ring closure.  The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-25T02:11:08.527" LastActivityDate="2014-05-25T02:11:08.527" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11059" PostTypeId="1" CreationDate="2014-05-25T12:09:44.213" Score="1" ViewCount="1899" Body="&lt;p&gt;The solubility product of a salt having same molecular formula as $\ce{Ca_3(PO_4)_2}$ is $1.08\cdot10^{-28}$. Then find the number of moles of ions present in the saturated solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: I found out the solubility of the salt i.e.: $x=10^{-6}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means $x$ moles of salt are present in a litre. As 1 mole of the salt yields 5 moles of ions I thought the answer would be $5x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help required.&lt;/p&gt;&#xA;" OwnerUserId="5226" LastEditorUserId="865" LastEditDate="2021-05-22T21:03:35.193" LastActivityDate="2021-05-22T21:03:35.193" Title="How to calculate the number of moles present in a litre of saturated solution?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11060" PostTypeId="1" CreationDate="2014-05-25T14:48:05.503" Score="2" ViewCount="974" Body="&lt;p&gt;Before I begin, I think it's worth mentioning that I know little about chemistry and even less about knives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this, having seen many people talk about how patinas work and the best ways of applying them and which ones are best etc., It seemed to me like people just assumed that they work, presumably because they heard someone else do it, and them from someone else and so forth but that doesn't always make it true. If anyone has any idea If there is any reasoning behind this and if there any flaws behind the logic in it that would be greatly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't exactly want to ruin or change my new knife for no reason. I assume people might think that this works similar to the thin oxide coating on aluminium, but my guess is that the corrosion acts like it would on any metal and slowly &quot;eat away&quot; at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, simply, does it work?&lt;/p&gt;&#xA;" OwnerUserId="5616" LastEditorUserId="73" LastEditDate="2014-05-25T21:11:37.217" LastActivityDate="2014-05-25T21:11:37.217" Title="Does a patina on a knife prevent rust?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11061" PostTypeId="2" ParentId="11059" CreationDate="2014-05-25T16:27:50.863" Score="2" Body="&lt;p&gt;For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon dissolution of one unit of calcium phosphate, we get five units of ions &lt;em&gt;total&lt;/em&gt;, true, but the solubility product is defined as follows (from the IUPAC &lt;em&gt;Gold Book&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PO_{4}^{3-} + H_2O -&amp;gt;HOPO_3^{2-} + HO^-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, based on the IUPAC definition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $\ce{x}$ to be &lt;em&gt;one&lt;/em&gt; unit of calcium ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the above yields $x = 10^{-6}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;strong&gt;with respect to your original question&lt;/strong&gt;, the number of moles of calcium phosphate present &lt;strong&gt;initially&lt;/strong&gt; can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-25T16:58:36.030" LastActivityDate="2014-05-25T16:58:36.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11062" PostTypeId="2" ParentId="11060" CreationDate="2014-05-25T16:31:12.343" Score="1" Body="&lt;p&gt;In this case, the use of the word patina is a little misleading. Patina refers to the oxidation film on bronze or aged polish. I assume that you mean that the knife is coated with a substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, when steel oxidizes, the iron oxide that forms does not stick well and will flake off the surface. This would not be useful for a knife since the edge will become dull quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if the knife is made with stainless steel, it does not need to be coated to prevent rust, only keep reasonably dry. Some stainless steel will actually rust since it was not properly formulated. Very few knifes are made of other metals since stainless steel is reasonably cheap.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-25T16:31:12.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11063" PostTypeId="1" AcceptedAnswerId="11064" CreationDate="2014-05-25T18:18:32.207" Score="0" ViewCount="67" Body="&lt;p&gt;If I have a glass of water, and the vapor pressure is the point of equilibrium where the liquid molecules are becoming gases and the gas molecules are colliding and becoming liquid at the same rate, does that mean none of the liquid is lost? I feel as if I am misunderstanding this concept because when I leave my cup of water on the table and come back to it later on I usually notice that there's less water in the cup because it evaporated. &lt;/p&gt;&#xA;" OwnerUserId="5617" LastActivityDate="2014-05-25T18:28:13.523" Title="Clarification on vapor pressure" Tags="&lt;equilibrium&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11064" PostTypeId="2" ParentId="11063" CreationDate="2014-05-25T18:28:13.523" Score="2" Body="&lt;p&gt;Yes, the vapor pressure is the point of &lt;em&gt;equilibrium&lt;/em&gt; between the liquid and the vapor. But air &lt;em&gt;moves&lt;/em&gt; about so the liquid never saturates the air and an equilibrium is never reached. So more and more water evaporates until it is all gone.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-25T18:28:13.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11065" PostTypeId="1" CreationDate="2014-05-25T18:33:50.293" Score="5" ViewCount="115" Body="&lt;p&gt;I would be interested to know people's opinion on the following matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, when the far-from equilibrium behaviour of a chemical reaction system is analysed mathematically (in papers or text books), the system is placed in a gradient between a &lt;strong&gt;fixed&lt;/strong&gt; non-zero concentration of &quot;resource&quot; species, and a &lt;strong&gt;fixed&lt;/strong&gt; concentration of &quot;waste&quot; species (which is often zero). The other species in the reaction system are designated as &quot;intermediates&quot;, and it is their behaviour which is of prime interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the oscillating Lotka-Volterra reaction system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R + X \rightarrow 2X$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X + Y \rightarrow 2Y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y \rightarrow W$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Species $R$ and $W$ are the fixed concentration resource and waste species respectively, and species $X$ and $Y$ are the (oscillating) intermediates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears as if these reaction systems are being studied in a physical scenario which could not be realised in the lab, a &lt;em&gt;magical reaction tank&lt;/em&gt; if you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two reasons most often stated as to why resource and waste are held at constant concentrations, whilst the intermediates can vary, are that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; The reaction takes place in a large reservoir, where resource is abundant and where any waste produced by the reaction quickly diffuses away into the vastness &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; The reaction has resource species quickly &quot;pumped in&quot; by some external agent, and likewise waste species quickly &quot;pumped away&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problems with these explanations are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If in explanation 1, a very large reservoir is keeping resource and waste species at constant concentration, then if follows that the intermediate species can never reach any appreciable concentration, for they too exist in the same large volume. In other words, if the waste diffuses away, then why don't the intermediates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, in explanation 2, if a pumping process is artificially maintaining resource and waste concentrations, it is hard to see how this pumping process is being &quot;micro selective&quot; in completely removing the waste compounds, but leaving the intermediate compounds totally unaffected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More physically realistic models of chemical reactions can be made by also explicitly including the &lt;em&gt;reactor&lt;/em&gt; in the model - for example by considering the reaction as taking place in semi-permeable compartments or in bulk in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor_model&quot; rel=&quot;nofollow&quot;&gt;CSTR&lt;/a&gt;. However, additionally modelling the reactor often &lt;strong&gt;changes&lt;/strong&gt; the mathematical properties for the &quot;bare&quot; reaction reported in the literature (for example, in a CSTR, the intermediates as well as the waste are removed).&lt;/p&gt;&#xA;" OwnerUserId="5598" LastEditorUserId="10185" LastEditDate="2015-09-27T09:31:52.330" LastActivityDate="2015-09-27T09:31:52.330" Title="Magical reaction tanks" Tags="&lt;kinetics&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11067" PostTypeId="1" AcceptedAnswerId="11068" CreationDate="2014-05-25T20:40:55.287" Score="1" ViewCount="109" Body="&lt;p&gt;Suppose you have the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NH4SCN(s) + Ba(OH)2(s) -&amp;gt;Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You initiate this reaction using 15.22g of the $\ce{NH4SCN}$ (and excess of the barium hydroxide) in a boiling tube resting in 50g of water. The temperature rises by 11 degrees. Therefore energy = $4.2 \times 11 \times 50 = 2.31kJ$ and moles = $\frac{15.22}{76.1} = 0.2mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would then assume that it would just be $\frac{2.31}{0.2} = 11.55kJmol^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the actual answer is 23.1 (double my answer) which I am guessing is because there are 2 moles of $\ce{2NH4SCN}$. I do not understand why this is the case, because you are using 0.2 moles and the energy raised is 2.31 so why the multiplication by 2?&lt;/p&gt;&#xA;" OwnerUserId="5619" LastActivityDate="2014-05-25T22:11:11.430" Title="Enthalpy change of reaction" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11068" PostTypeId="2" ParentId="11067" CreationDate="2014-05-25T20:51:00.363" Score="0" Body="&lt;p&gt;The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="189" LastEditDate="2014-05-25T22:11:11.430" LastActivityDate="2014-05-25T22:11:11.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11069" PostTypeId="1" AcceptedAnswerId="11071" CreationDate="2014-05-25T21:11:34.390" Score="17" ViewCount="19018" Body="&lt;p&gt;The electronic configuration for $\ce{Cu}$, adjusted for Hund's rule, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{Ar}] 3d^{10} 4s^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, shouldn't $\ce{Cu}$ have an oxidation number of $+1$? whereby it gives off its outermost 4s electron? Why can it also have a $+2$ charge?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="189" LastEditDate="2014-05-25T22:09:59.590" LastActivityDate="2017-05-15T02:22:07.363" Title="Why can Cu have an oxidation number of +2?" Tags="&lt;electronic-configuration&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11070" PostTypeId="1" AcceptedAnswerId="11077" CreationDate="2014-05-25T22:27:47.270" Score="1" ViewCount="2513" Body="&lt;p&gt;I'm watching a video, and the speaker states that the molality formula is better than the molarity formula because it is more constant, since molarity deals with volume and molality deals with mass. If someone adds more solvent than the volume changes, so the answer in molarity changes as well. But if more solvent is added, doesn't the mass increase as well?&lt;/p&gt;&#xA;" OwnerUserId="5617" LastEditorUserId="4945" LastEditDate="2014-05-26T06:41:47.127" LastActivityDate="2014-05-26T17:59:27.367" Title="Increasing mass" Tags="&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11071" PostTypeId="2" ParentId="11069" CreationDate="2014-05-25T22:50:07.867" Score="22" Body="&lt;p&gt;Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$.  By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced.  While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the energy gained through bond formation and lattice energy (or hydration energy) then $\ce{Cu^{+2}}$ compounds will form.  In fact $\ce{Cu^{+2}}$ is the most common oxidation state of copper so the energetics must generally work out that the energy gained by forming more than one bond to copper and gaining additional lattice (or hydration) stabilization, more than offsets the energy cost of removing that second electron.  Removal of additional electrons can also occur with copper to form $\ce{Cu^{+3}}$ and $\ce{Cu^{+4}}$ compounds, for example, $\ce{K3CuF6}$ and $\ce{Cs2CuF6}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-25T03:04:04.827" LastActivityDate="2014-09-25T03:04:04.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11074" PostTypeId="1" AcceptedAnswerId="11081" CreationDate="2014-05-26T02:15:38.220" Score="2" ViewCount="83" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YeDU0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this formula derived? What does it mean; what is the use of this formula? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that at infinite dilution, ionization extent should be very high because of Le Chatlier's principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not seeing how this formula is derived and its significance in actual chemistry. I tried rewriting the formula as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} \cdot 100\%}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this isn't elucidating where the formula came from or its significance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could also rewrite the formula as this for the hypothetical acid, $\ce{HA}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} = \frac{\frac{[H_3O^+][A^-]}{[HA]}}{1.0 \cdot 10^{-7}+\frac{[H_3O^+][A^-]}{[HA]}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I think the trouble I'm having is defining maximum ionization extent.&lt;/strong&gt; For example, I know that in general ionization extent is defined as (where x is the concentration of the derived species - i.e. $\ce{[A^{-}]}$ for the hypothetical acid $\ce{HA}$ and $\ce{M_{i}}$ the initial molarity of the hypothetical acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{\frac{x}{M_{i}}}=\text{ionization extent}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling that the $1.0*10^{-7}$ term refers to the concentration of one of the products of the auto-ionization of pure water at 25 degrees Celsius. This makes sense because at infinite dilution the concentration of solute should be negligible (=0). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon further thought ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/talJ5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="24052" LastEditDate="2016-03-13T19:39:17.060" LastActivityDate="2016-03-13T19:39:17.060" Title="Max Ionization Extent" Tags="&lt;acid-base&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11076" PostTypeId="1" AcceptedAnswerId="11168" CreationDate="2014-05-26T03:02:11.190" Score="3" ViewCount="241" Body="&lt;p&gt;What is some advance reference books specifically about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas with explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to find nothing really very interesting for myself to learn general theory of alloy while searching the google books because these books is too wordy in detail but too few formula and laboratory skill to support it.&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="4945" LastEditDate="2014-05-26T06:43:39.060" LastActivityDate="2014-05-29T17:33:42.597" Title="What is some advance reference books about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas" Tags="&lt;materials&gt;&lt;reference-request&gt;&lt;alloy&gt;&lt;books&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11077" PostTypeId="2" ParentId="11070" CreationDate="2014-05-26T06:22:05.900" Score="3" Body="&lt;p&gt;I think someone may have made a slip of the tongue there: if you add more &lt;strong&gt;&lt;em&gt;solute&lt;/em&gt;&lt;/strong&gt;, the volume of the solution will change, without changing the mass of solvent present. If you just add more solvent, of course you'll change both the molarity and the molality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a bit of a thought experiment to illustrate this: I have some toluene and I'd like to dissolve a whole bunch of tri-octylphosphine sulfur into it (don't worry if you don't know what these are, just remember that toluene is the solvent and TOPS is the solute).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I want to make a 1 molal solution. The definition of one molal is a mole of solute in a kilogram of &lt;strong&gt;solvent&lt;/strong&gt;. I measure out one mole of TOPS and dissolve it in 1 kg of toluene. Done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to make a 1 molar solution. The definition of one molar is a mole of solute in a liter of &lt;strong&gt;solution&lt;/strong&gt;. I might be tempted to measure out one liter of toluene, dissolve the TOPS in it, and call it a day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is because one mole TOPS takes up a certain amount of volume. A rough estimate is that it takes up about a third of a liter. My solution is now one mole of TOPS in &lt;strong&gt;1.33 liters of solution&lt;/strong&gt;, making it 0.75 molar instead of 1 molar!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can fix this by starting out with the TOPS, then slowly adding toluene until the total volume of everything is 1L. This will give me a 1 molar solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can extend this idea a little further to see that adding solute to a solution makes its molarity a &lt;strong&gt;mess&lt;/strong&gt; to calculate unless you pretend that the solution volume doesn't change. The molality, meanwhile, is easy, because the mass of solvent hasn't changed a whit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As Martin points out, molality has an additional benefit in that solutions naturally expand as they heat. If we heat a solution up, its volume will increase, and its molarity will change. The molality, however, will not, since the mass of solvent does not depend on temperature.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="2087" LastEditDate="2014-05-26T17:59:27.367" LastActivityDate="2014-05-26T17:59:27.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11078" PostTypeId="1" CreationDate="2014-05-26T06:27:38.947" Score="1" ViewCount="223" Body="&lt;p&gt;I'm told that in the normal cubic system there are $\frac18$ of an atom at each corner and these atoms' radius are such that $a = 2r$ ($a$ is the lattice constant, $r$ radius of an atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bbc (body centered cubic system) they say the atoms in the diagonals are touching, and as a consequence the atoms at each corner are not big enough to satisfy the above relation $a = 2r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? Couldn't it be like the normal cubic system but stick an atom in the center to fill up the remaining space? How do we know that they touch along the cube's diagonal and not along the edges?&lt;/p&gt;&#xA;" OwnerUserId="3955" LastEditorUserId="4945" LastEditDate="2014-05-26T06:40:33.280" LastActivityDate="2014-05-28T09:07:46.463" Title="How do we know the size of atoms in body centered cubic system?" Tags="&lt;crystal-structure&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11080" PostTypeId="1" CreationDate="2014-05-26T08:52:23.633" Score="1" ViewCount="420" Body="&lt;p&gt;How can a phase that appears at high temperature be thermodynamically stable at room temperature? For example, $\alpha$-alumina is a phase that is formed at temperatures above $1000^\circ C$. How is the same phase stable at room temperature?&lt;/p&gt;&#xA;" OwnerUserId="5624" LastEditorUserId="5633" LastEditDate="2014-05-30T20:47:49.503" LastActivityDate="2014-05-30T20:47:49.503" Title="How can a phase be thermodynamically stable at room temperature if it only appears at high temperature?" Tags="&lt;thermodynamics&gt;&lt;stability&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11081" PostTypeId="2" ParentId="11074" CreationDate="2014-05-26T10:29:26.650" Score="1" Body="&lt;p&gt;For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis.&#xA;\begin{aligned}\ce{&#xA;(1)&amp;amp;&amp;amp;HA + H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + A^{-}\\&#xA;(2)&amp;amp;&amp;amp;2H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + {}^{-}OH \\\hline&#xA;(3)&amp;amp;&amp;amp;HA + 3H2O &amp;amp;&amp;lt;=&amp;gt; 2H3+O + A^{-} + {}^{-}OH&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degree of dissociation is usually defined as&#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;\alpha &amp;amp;= \frac{c(\ce{A^{-}})}{c_0}&amp;amp;\text{with}&amp;amp;&amp;amp;c_0&amp;amp;=c(\ce{A^{-}})+c(\ce{HA})\\&#xA;\therefore&amp;amp;&amp;amp;\alpha &amp;amp;= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&amp;amp;\text{with}&amp;amp;&amp;amp;&#xA;c(\ce{A^{-}})&amp;amp;=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\&#xA;(4)\therefore&amp;amp;&amp;amp;\alpha &amp;amp;= \frac{K_a}{K_a + c(\ce{H3+O})}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as&#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;K_w&amp;amp;=c(\ce{H3+O})c(\ce{{}^{-}OH})\\&#xA;(5)\therefore&amp;amp;&amp;amp;c(\ce{H3+O})&amp;amp;=\sqrt{K_w}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $25~^\circ{}C$ you will find your given equation:&#xA;\begin{aligned}&#xA;\alpha[\%] &amp;amp;= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\%&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-26T10:29:26.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11082" PostTypeId="1" CreationDate="2014-05-26T13:34:18.370" Score="4" ViewCount="1638" Body="&lt;p&gt;I am building a greywater harvesting system. It is a 3-stage system: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 1st stage breaks down soap and detergents. &lt;/li&gt;&#xA;&lt;li&gt;The 2nd stage filters the water. &lt;/li&gt;&#xA;&lt;li&gt;The 3rd stage pumps water out of system. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the best way to break down soap and detergents in a timely manner with as few mechanical parts as possible?&lt;/p&gt;&#xA;" OwnerUserId="5625" LastEditorUserId="30836" LastEditDate="2017-10-08T16:57:45.440" LastActivityDate="2017-10-08T16:57:45.440" Title="How to break down soap and detergents?" Tags="&lt;reaction-mechanism&gt;&lt;water&gt;&lt;aqueous-solution&gt;&lt;home-experiment&gt;&lt;solubility&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11083" PostTypeId="2" ParentId="10249" CreationDate="2014-05-26T13:37:19.823" Score="6" Body="&lt;p&gt;Given the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;[\ce{CH3COOH}]     &amp;amp;= \pu{1 * 10^-2 mol L-1}\\&#xA;[\ce{HCOOH}]       &amp;amp;= \pu{1 * 10^-2 mol L-1}\\&#xA;K_{z-\ce{CH3COOH}} &amp;amp;= \pu{1.8 * 10^-5 mol L-1}\\&#xA;K_{z-\ce{HCOOH}}   &amp;amp;= \pu{1.8 * 10^-4 mol L-1}\\&#xA;[\ce{H3O+}]_1      &amp;amp;= \pu{1.3 * 10^-3 mol L-1}\\&#xA;[\ce{H3O+}]_2      &amp;amp;= \pu{4.0 * 10^-4 mol L-1}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing that we could write&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HCOOH + CH3COOH + 2H2O -&amp;gt; HCOO- + CH3COO- + 2H3O+},$$&lt;/span&gt;&#xA;we can derive a equilibrium&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \frac{[\ce{HCOO-}][\ce{CH3COO-}][\ce{H3O+}]^{2}}&#xA;    {[\ce{HCOOH}][\ce{CH3COOH}]}&#xA;  &amp;amp;= \frac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]}&#xA;    \frac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]}\\&#xA;  &amp;amp;= K_{z-\ce{HCOOH}}K_{z-\ce{CH3COOH}}.&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we still need to know the original concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCOOH}$&lt;/span&gt;, we can calculate&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{CH3COO-}] = \frac{[\ce{CH3COOH}]}{[\ce{H3O+}]_2} K_{z-\ce{CH3COOH}} = \pu{4.5000 * 10^-4 mol L-1}.$$&lt;/span&gt;&#xA;For &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCOO-}]$&lt;/span&gt; we get &lt;span class=&quot;math-container&quot;&gt;$\pu{1.3846 * 10^-3 mol L-1}$&lt;/span&gt; (keeping some significance). &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  n_{\ce{HCOOH}} &amp;amp;= \pu{1.1385 * 10^-2 mol L-1} &#xA;    &amp;amp;&amp;amp;\text{(total)}\\&#xA;  n_{\ce{CH3COOH}} &amp;amp;= \pu{5.225 * 10^-3 mol L-1} &#xA;    &amp;amp;&amp;amp; \text{(total)}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; (neglecting the already present &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; of water)&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  K_{z-\ce{HCOOH}} &amp;amp;= \frac{[\ce{HCOO-}][\ce{H3O+}]}{[\ce{HCOOH}]}\\ &#xA;  &amp;amp;= \frac{x(x+y)}{(0.5n_{\ce{HCOOH}}-x)}&#xA;\end{align}&lt;/span&gt;&#xA;gives &#xA;&lt;span class=&quot;math-container&quot;&gt;$$y = -0.00018 - x + \frac{1.02465 \cdot 10^{-6}}{x},$$&lt;/span&gt; &#xA;using &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; in&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  K_{z-\ce{CH3COOH}} &amp;amp;= \frac{[\ce{CH3COO-}][\ce{H3O+}]}{[\ce{CH3COOH}]}\\&#xA;  &amp;amp;= \frac{y(x+y)}{(0.5n_{\ce{CH3COOH}}-y)}&#xA;\end{align}&lt;/span&gt;&#xA;gives &#xA;&lt;span class=&quot;math-container&quot;&gt;$$x = 0.000884322 \vee x = 0.00690291$$&lt;/span&gt; &#xA;(there are more solutions, but these are the only real positive ones). From this we can calculate &#xA;&lt;span class=&quot;math-container&quot;&gt;$$y = 0.0000943623 \vee y = -0.00693447.$$&lt;/span&gt; &#xA;&lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; should be positive, thus &lt;span class=&quot;math-container&quot;&gt;$x = 0.000884322$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y=0.0000943623$&lt;/span&gt;. This gives us a &lt;span class=&quot;math-container&quot;&gt;$pH$&lt;/span&gt; of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH} = -\log(0.000884322+0.0000943623) \approx 3.01$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which seems to me as correct (it will be slightly more acidic than when adding the amounts of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2020-03-05T12:35:00.713" LastActivityDate="2020-03-05T12:35:00.713" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="11084" PostTypeId="1" CreationDate="2014-05-26T14:31:30.037" Score="1" ViewCount="627" Body="&lt;p&gt;For $\ce{HCl(aq) -&amp;gt; H^{+}(aq) + Cl^{-}(aq)}$, if $\ce{HCl}$ has a concentration of 1 molar then $\ce{H+}$ and $\ce{Cl-}$ also have 1 molar. But for $\ce{HF(aq) &amp;lt;=&amp;gt; H^{+}(aq) + F^{-}(aq)}$, if $\ce{HF}$ has a concentration of 1.5 molar, then why doesn't $\ce{H+}$ and $\ce{F-}$ have 1.5 molar as well?&lt;/p&gt;&#xA;" OwnerUserId="5617" LastEditorUserId="189" LastEditDate="2014-05-26T14:51:39.380" LastActivityDate="2014-07-07T07:56:49.223" Title="Molality and dissociation" Tags="&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11085" PostTypeId="2" ParentId="267" CreationDate="2014-05-26T14:37:08.120" Score="2" Body="&lt;p&gt;@F'x: With the 3 reactions of Ball's system, there is no way that you can assign free energy values to species $A$, $B$, $C$ such that the combined free energy of the reactant species is always greater than the combined free energy of the product species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Try it on Wolfram Alpha:&#xA;solve (A+B) &gt; 2A, (B+C) &gt; 2B, (C+A) &gt; 2C &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, from a thermodynamics point of view, I would say that the Ball reaction system is invalid.&lt;/p&gt;&#xA;" OwnerUserId="5598" LastEditorUserId="5598" LastEditDate="2014-06-05T12:38:16.917" LastActivityDate="2014-06-05T12:38:16.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11086" PostTypeId="2" ParentId="11084" CreationDate="2014-05-26T14:51:07.223" Score="4" Body="&lt;p&gt;HF isn't a strong acid; its dissociation isn't complete in water, unlike HCl's dissociation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason HF isn't a strong acid while HCl is a strong acid has to do with charge density of the conjugate base. Charge density controls the reactivity of the conjugate base; charge is what governs chemistry and physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find that the $\ce{F^-}$ anion is fairly reactive relative to $\ce{Cl^-}$, which, in water, is essentially inert. Both are halogens; both have the same magnitude of charge; the difference is that the fluoride anion is smaller than the chloride anion. Therefore, the real negative charge density on the $\ce{F^-}$ anion is greater than that of the chloride anion, and this effective negative charge density is what drives the reactivity of the fluoride anion. In other words, the below reactions occurs to significant extent due to the concentrated negative charge on the fluoride anion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{F^{-} + H_2O &amp;lt;=&amp;gt; HF +HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This above reaction is antagonistic to the dissociation of $\ce{HF}$ in water because the above reaction makes more $\ce{HF}$! Note that $\ce{K_{b}(F^{-})}$ $=6.6\cdot 10^{-4}$ while $\ce{K_{b}(Cl^{-})}$ $=8.0\cdot 10^{-21}$! So there you have it; empirical evidence that chloride ion is a very unreactive species in water solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find that with the binary halogen acids, as you go down group 7, the acids get stronger. For example, $\ce{HBr}$ is an even stronger acid than the already strong acd, $\ce{HCl}$. This again can be rationalized using effective negative charge density; $\ce{Br^-}$ has the same magnitude of negative charge as $\ce{Cl^-}$ but in the bromide anion the negative charge is spread out over an even larger area, making bromide ion all the less reactive than chloride ion and hydrobromic acid stronger than hydrochloric acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your questions and gives you some insight into acids. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-07-07T07:56:49.223" LastActivityDate="2014-07-07T07:56:49.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11087" PostTypeId="1" AcceptedAnswerId="11090" CreationDate="2014-05-26T16:46:59.073" Score="4" ViewCount="140" Body="&lt;p&gt;I have been given task to write possible reaction equations, determine oxidizer, reducer, write corresponding electron-equations and calculate electromotive force (I do not know if that is what you call it in English) of reactions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Iron reaction with diluted sulfuric acid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aluminium reaction with solution of sodium hydroxide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Could you be so kind and help me with this task? &lt;/p&gt;&#xA;" OwnerUserId="5627" LastEditorUserId="7448" LastEditDate="2015-06-30T22:02:02.010" LastActivityDate="2015-06-30T22:02:02.010" Title="How to prove that reaction is possible" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-07-01T12:19:07.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="11088" PostTypeId="2" ParentId="2547" CreationDate="2014-05-26T17:00:52.573" Score="-1" Body="&lt;p&gt;$\ce{ZnSO4}$ is colorless because it has completely filled d orbitals while $\ce{FeSO4}$ is green because its d orbitals are incompletely filled.&lt;/p&gt;&#xA;" OwnerUserId="5628" LastEditorUserId="4231" LastEditDate="2014-05-26T17:24:50.540" LastActivityDate="2014-05-26T17:24:50.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11089" PostTypeId="2" ParentId="11084" CreationDate="2014-05-26T17:20:17.397" Score="7" Body="&lt;p&gt;The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$.  Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{pK_a}$ does measure the relative strength of an acid, look at the following equilibrium&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + HX&amp;lt;=&amp;gt;H3O^+ + X^-}$$&#xA;$$\ce{so~~~K_{eq}=\frac{[H3O^+][X^{-}]}{[H2O][HX]}}$$ but in dilute solutions the concentration of water stays constant so we can rewrite this equation as&#xA;$$\ce{K_{a}=\frac{[H3O^+][X^{-}]}{[HX]}}$$ and finally, $$\ce{pK_{a}~=~-Log_{10}K_{a}}$$ Now we can see that the more highly dissociated our acid $\ce{HX}$, the higher the concentrations of $\ce{H3O^+}$ and $\ce{X^{-}}$, and in turn the larger the value for $\ce{K_{a}}$, and the more negative the value of $\ce{pK_{a}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: highly dissociated acids have large negative $\ce{pK_{a}}$ values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is a Table from &lt;a href=&quot;http://www.sparknotes.com/chemistry/acidsbases/fundamentals/section2.rhtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;SparkNotes&lt;/a&gt;.  It compares the $\ce{pK_{a}s}$ (degree of ionization or strength) of the various haloacids.  You can see that as we move down the Table from $\ce{HF}$ to $\ce{HI}$ the $\ce{pK_{a}}$ becomes much more negative; our acids become much more dissociated in water.  So we see that $\ce{HF}$ is not very dissociated ($\ce{K_{a}=10^{-3.1}}$, equilibrium lies to the left). On the other hand, 1 mole of $\ce{HCl}$ will produce about 1 mole of $\ce{H^+}$ ($\ce{K_{a}=10^{+7}}$, equilibrium lies far to the right).  Note too that the $\ce{pK_{a}s}$ of these acids correlate with their bond strength which makes sense.  As the $\ce{H-X}$ bond gets weaker, it becomes easier for the acid to ionize (dissociate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NsqRk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-26T18:56:28.810" LastActivityDate="2014-05-26T18:56:28.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11090" PostTypeId="2" ParentId="11087" CreationDate="2014-05-26T18:15:38.240" Score="4" Body="&lt;p&gt;The ionic equation for this reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe + 2H+ ~(+\;SO4^2- ) -&amp;gt; Fe^2+ + H2 ~(+\;SO4^2- )}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As two half-reactions, it is seen that the iron is oxidized:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe -&amp;gt; Fe^2+ + 2e- }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reductant (iron) transfers electrons to hydrogen, and is thus itself oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the hydrogen is reduced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+ + 2e- -&amp;gt; H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidant (hydrogen) removes electrons from iron, and is thus itself reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the reduction potential you have to solve the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E = E_\text{acceptor} - E_\text{donor} = E_{\ce{2H+/H2}}- E_{\ce{Fe/Fe^2+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E = 0\ \mathrm{V} - (-0.41\ \mathrm{V}) = 0.41\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions runs voluntarily if $\Delta E &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the same for b).&lt;/p&gt;&#xA;" OwnerUserId="5614" LastEditorDisplayName="user7951" LastEditDate="2015-06-30T18:58:25.730" LastActivityDate="2015-06-30T18:58:25.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11091" PostTypeId="1" AcceptedAnswerId="14656" CreationDate="2014-05-26T20:28:36.130" Score="3" ViewCount="1127" Body="&lt;p&gt;I'm wondering if there is any way to create adenine, thymine, cytosine, or guanine from chemicals available to anyone, using basic lab equipment.&lt;/p&gt;&#xA;" OwnerUserId="2630" LastActivityDate="2015-11-20T07:14:29.627" Title="Is it possible to synthesis adenine, thymine, cytosine, or guanine at home?" Tags="&lt;home-experiment&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11092" PostTypeId="1" AcceptedAnswerId="11101" CreationDate="2014-05-26T20:53:14.233" Score="5" ViewCount="12213" Body="&lt;p&gt;Why is it that the Lewis structure of $\ce{Cl_2SO}$ has a total of $24$ valence electrons? I thought that the number of valence electrons should be $2\times7 + 1\times6 + 1\times6 = 26$.&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="4945" LastEditDate="2014-05-27T01:46:02.817" LastActivityDate="2016-03-26T23:00:20.020" Title="Lewis structure of Cl2SO" Tags="&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11093" PostTypeId="1" CreationDate="2014-05-26T23:01:33.107" Score="0" ViewCount="3158" Body="&lt;p&gt;As we all know ammonium chloride is a acid. And what explains this is its salt hydrolysis. It breaks down into ammonium and chloride ions. Ammonium react with water to produce hydronium or hydrogen ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then why can't chlorine ions react with water to produce hydroxide ions?&#xA;Is there any way other than using an indicator to know, whether a compound is acidic or basic? We know the second reaction can't happen, because we can tell the compound is acidic by using an indicator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really stops the second reaction from happening?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="97862" LastEditDate="2021-05-24T15:09:32.420" LastActivityDate="2021-05-24T15:09:32.420" Title="Hydrolysis of ammonium cloride" Tags="&lt;acid-base&gt;&lt;hydrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11094" PostTypeId="2" ParentId="11093" CreationDate="2014-05-26T23:16:55.330" Score="0" Body="&lt;p&gt;Hydrogen chloride is a strong acid, which means that when HCl is mixed into water, it completely disassociates into $H^+$ and $Cl^-$. Chloride ion will only be found in solution; not as a salt (although it may precipitate). Ammonia is a weak base, so it will exists as both ammonia and ammonium ion in solution. When you have a salt like ammonium chloride, you know the chloride is unaffected, but the ammonium ion will distribute into ammonia and ammonium ion, producing $H^+$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-26T23:16:55.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11097" PostTypeId="2" ParentId="11036" CreationDate="2014-05-27T00:51:56.753" Score="1" Body="&lt;p&gt;The reaction rates can be related to each other through the equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is important to note that we are making two key assumptions in order to do this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The solution is nearly &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_solution&quot; rel=&quot;nofollow&quot;&gt;ideal&lt;/a&gt;&lt;/em&gt; - in layman's terms, that means the concentration of one substance does not depend on any of the others. When that is true, we can use an equilibrium constant defined in terms of concentrations, $K_{eq} = \frac{[C]}{[A][B]}$, instead of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow&quot;&gt;activities&lt;/a&gt; (effective concentrations).&lt;/li&gt;&#xA;&lt;li&gt;The rate laws are in fact given by $k_1[A][B]$ and $k_2[C]$ - this may be true for elementary reactions, but you cannot in the general case determine the reaction rates solely from a reaction equation. You must find them empirically, or have a good idea of what the underlying mechanism is (and even then you might not get the correct rate law.) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming that these two conditions hold, we can then relate the two rate constants to each other through the equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, we know that the rate of change in one direction is equal to the rate of change in the other direction. In other words, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_1[A][B] = k2[C]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This  means that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{k_1}{k_2} = \frac{[C]}{[A][B]}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{eq} = \frac{[C]}{[A][B]}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{eq} = \frac{k_1}{k_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say that a reaction is more thermodynamically favorable in the forward direction, we are also saying that $K_{eq}$ is large (relatively speaking) and that $k_1 &amp;gt;&amp;gt; k_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right to be suspicious, since it doesn't seem like we should be able to compare two numbers with different units. The key here is that whether we explicitly look at the units or not, they are still there in the equilibrium constant. We are comparing the magnitudes alone, usually in an effort to reduce the complexity of an equation by assuming that some terms are approximately equal to zero.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-05-28T18:14:12.570" LastActivityDate="2014-05-28T18:14:12.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11098" PostTypeId="1" AcceptedAnswerId="11099" CreationDate="2014-05-27T01:17:20.847" Score="4" ViewCount="1026" Body="&lt;p&gt;I have an empty aluminum Ether bottle (Shipping name diethyl Ether) that I'd like to use as a large drinking bottle as this is thick aluminum. Is it dangerous? Here is some of the info on the bottle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemical Composition: (C-2 H-5)2-O f.w.74.12.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Name:  Ether Purified (For Fat Extraction)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Other: Mallinckrodt (Brand?)&#xA;   5.51 lb&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0884&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Diethyl Ether 60-29-7&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Ethyl Alcohol 64-17-5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5634" LastEditorUserId="4945" LastEditDate="2014-05-27T01:43:34.333" LastActivityDate="2014-05-27T21:03:35.577" Title="Drinking water out of a 24 year old Ether Bottle" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11099" PostTypeId="2" ParentId="11098" CreationDate="2014-05-27T01:21:03.080" Score="6" Body="&lt;p&gt;Mallinckrodt is the chemical manufacturer. The ethanol is added to the ether to prevent peroxides from forming in the ether (like an antioxidant). If you wash and rinse the container thoroughly a few times, then it should be fine.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-27T01:21:03.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11101" PostTypeId="2" ParentId="11092" CreationDate="2014-05-27T01:53:13.467" Score="5" Body="&lt;p&gt;Sulfur is sometimes described as a hypervalent atom, or one that has more than eight valence electrons. This is possible to some extent; sulfur can indeed utilize its d-orbitals for bonding. However, recent quantum mechanical calculations suggest that the extent of this utilization is negligible. Therefore, your book is most likely representing the $\ce{Cl_2SO}$ molecule as a &quot;charge-separated&quot; molecule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Upon further thought this still makes no sense. Charge is conserved and thionyl chloride should always have 26 electrons (no matter how they are distributed). So perhaps, due to the high degree of ionic character in the $\ce{S-Cl}$ and $\ce{S-O}$ bonds, the missing two electrons are &quot;dispersed&quot; among these ligands. (This is speculation). Perhaps your book is genuinely in error. Perhaps they did not count a lone pair on the central sulfur atom or otherwise miscounted the number of electrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: Here's a picture of two possible Lewis structures of thionyl chloride; as we can see, the book agrees that the thionyl chloride molecule is better represented as a &quot;charge-separated&quot; molecule. Nonetheless, all the electrons are present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EjV4F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 3: Here's a picture of thionyl chloride which makes it seem as if the molecule only has 24 valence electrons. NB: the lone pair is still there and its presence is implied through the lack of any indication of a non-zero formal charge on the sulfur atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q7UYu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-27T03:30:17.450" LastActivityDate="2014-05-27T03:30:17.450" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11102" PostTypeId="2" ParentId="11093" CreationDate="2014-05-27T02:09:09.847" Score="3" Body="&lt;p&gt;Chloride ion does not react with the water to any significant extent. To understand this, we must first accept that all chemical reactions are equilibriums. This means that all chemical reactions are reversible to some extent. Some reactions aren't very reversible at all (at least not in the short or medium term) - such as the combustion reaction in your car's gasoline engine. But if you give it a few million years, the carbon dioxide released by your car's combustion process will be fixed (biological term) by plants and these plants might eventually turn into fossil fuels again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've established that all chemical reactions are equilibriums. Now we can consider the case of the chloride ion and why this reaction does not happen, as you say, or &lt;em&gt;does not happen to a significant extent&lt;/em&gt;, as I say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl^- + H_3^+O -&amp;gt; HCl + HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that if the above reaction happened to any significant extent, then this would make HCl a weak acid. Strong acids are defined to dissociate nearly completely in water. HCl is a strong acid by any account. Think of it as a waterfall. This reaction represents going down a waterfall (this reaction is large extent; heck, it's practically 100%):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H_2O -&amp;gt; Cl^- + H_3^+O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the above reaction is the going down part of the water fall, its reverse reaction must represent the going up part of the waterfall. Going up the waterfall is hard no matter how you slice it; the only difference is whether you're trying to go up a small waterfall or Niagara falls. The reaction of hydrochloric acid with water is akin to going down Niagara falls - this is a favorable reaction. This is basically a one-way street. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, chloride ion can't affect solution pH; if chloride ion could somehow react again to form more hydrochloric acid then we'd violate the definition of a strong acid - nearly complete dissociation. This is what makes hydrochloric acid a strong acid! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is empirically true. Chloride ion can't even be protonated effectively by the strongest acid in a solution of hydrochloric acid, which is hydronium ion, so there is no way that you can expect chloride ion to be protonated by water (which is only about a trillion times weaker than hydronium ion as an acid). Thus, your solution of ammonium chloride has a pH controlled not by the chloride ion but by the ammonium ion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-27T05:15:13.620" LastActivityDate="2014-05-27T05:15:13.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11104" PostTypeId="1" AcceptedAnswerId="11107" CreationDate="2014-05-27T07:28:18.363" Score="12" ViewCount="4151" Body="&lt;p&gt;A table does not sublimate, and nor does a spoon. Ice does, however. What is the fundamental difference?&lt;/p&gt;&#xA;" OwnerUserId="5636" LastEditorUserId="4923" LastEditDate="2014-05-28T08:57:07.200" LastActivityDate="2014-05-28T09:12:07.883" Title="Why doesn't a table sublimate, while ice does?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11105" PostTypeId="1" AcceptedAnswerId="11110" CreationDate="2014-05-27T07:47:03.293" Score="6" ViewCount="8790" Body="&lt;p&gt;For a $1^{\text{st}}$ order reaction $ A \rightarrow B$, the corresponding rate law is: $$\text{rate} = k[A]^{1}$$ and the integrated rate law for this is $$ \ln \left(\frac{A}{A-x}\right) = kt$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going by this equation, every first-order reaction must get completed only at &lt;em&gt;infinity&lt;/em&gt;. Is this true? If not, then what is the correct explanation?&lt;/p&gt;&#xA;" OwnerUserId="4995" LastEditorUserId="16683" LastEditDate="2017-04-26T23:18:08.773" LastActivityDate="2017-04-26T23:19:49.643" Title="Does a first order reaction really get completed only at infinity?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11106" PostTypeId="2" ParentId="11104" CreationDate="2014-05-27T08:02:47.427" Score="7" Body="&lt;p&gt;&amp;lt;&amp;lt; &lt;strong&gt;This answer was posted before the question was completely rephrased and originally asked &quot;why does a table not evaporate&quot;&lt;/strong&gt; &gt;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who said a spoon doesn't evaporate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice is a solid below its melting point ($0~^\circ\mathrm{C}$) but above its melting point it goes through a &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_transition&quot; rel=&quot;nofollow&quot;&gt;phase transition&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Melting_point&quot; rel=&quot;nofollow&quot;&gt;melts&lt;/a&gt; to become a liquid (so it's liquid at room temp). Then when it's heated above room temperature to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_point&quot; rel=&quot;nofollow&quot;&gt;boiling point&lt;/a&gt; ($100~^\circ\mathrm{C}$), it goes through another &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_transition&quot; rel=&quot;nofollow&quot;&gt;phase transition&lt;/a&gt; and it evaporates to become gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;Iron&lt;/a&gt; (assuming an iron spoon here to keep it simple) is a solid (at room temperature) then when it's heated above room temperature to its melting point (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;$1538~^\circ\mathrm{C}$&lt;/a&gt;) it becomes a liquid.  Then when it's heated above its melting point to its boiling point (&lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;$\approx2861~^\circ\mathrm{C}$&lt;/a&gt;), it evaporates.&lt;/p&gt;&#xA;" OwnerUserId="5637" LastEditorUserId="5637" LastEditDate="2014-05-28T09:12:07.883" LastActivityDate="2014-05-28T09:12:07.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11107" PostTypeId="2" ParentId="11104" CreationDate="2014-05-27T08:18:10.990" Score="18" Body="&lt;p&gt;This relates to the difference between how ice, iron spoons and wooden tables stick together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In wood, most or all of the bonds between the individual atoms and fibre units are covalent; making them very strong. Similarly, a spoon is an alloy of iron, nickel, chrome and carbon, also a very coherent structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ice, on the other hand, the units of the crystal is held together with comparatively weak hydrogen bonds, meaning that not a lot of energy is required for a surface molecule of water to escape, in a process called sublimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, once a water molecule escapes the ice, it is, given a relatively constant temperature, very unlikely to rejoin the body of ice; meaning that especially in open air and windy conditions, ice will sublimate at sub-zero temperatures, although as the temperature of the ice drops, this effect becomes steadily less noticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; It is worth noting that given enough time, a table or a spoon &lt;em&gt;can&lt;/em&gt; indeed sublimate as well, at any temperature, but the odds against it are utterly staggering.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="5638" LastEditDate="2014-05-28T08:32:12.197" LastActivityDate="2014-05-28T08:32:12.197" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11108" PostTypeId="1" AcceptedAnswerId="11115" CreationDate="2014-05-27T08:56:52.057" Score="-2" ViewCount="2211" Body="&lt;p&gt;Before $\ce{sp^3}$ hybridisation, does the $\ce{C}$ -atom get excited to $\ce{1s^2~2s^1~2p^3}$ state, as it happens before $\ce{sp^2}$ hybridisation?&#xA;What is the energy of the $\ce{sp^3}$ hybridised orbital?&lt;/p&gt;&#xA;" OwnerUserId="5639" LastEditorUserId="4945" LastEditDate="2014-05-27T09:47:00.943" LastActivityDate="2014-05-27T15:09:19.780" Title="sp3 hybridisation in carbon" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11109" PostTypeId="2" ParentId="11108" CreationDate="2014-05-27T09:30:26.893" Score="0" Body="&lt;p&gt;sp3 hybridisation is what we call the formation of hybrid atomic orbitals, by the combination of an s orbital and 3 p orbitals on the carbon atom. I think you understand this. However, like in a dative bond, even though one of p orbitals is empty before the carbon becomes hybridized, it can still combine with the other atomic orbitals, to create new hybrid atomic orbitals. So no, the atom doesn't have to get excited to 1s2 2s1 2p3 before hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of thinking about sp3 hybridization, a more advanced view, is that they are merely a nice simple way of describing the bonding in atoms. If we were to use the molecular orbital approach we would have to keep going back to our computers to work out the MO's. This isn't really needed, unless we want exact energies and a detailed description of the bonding. In the case of sp3 hybridization, say in methane, the carbon s orbital interacts with ALL of the hydrogen orbitals to form a MO. Then the carbon p-orbitals interact with the hydrogen 1s orbitals. Here the case is that several AO are interacting to form one MO. Then the sp3 model can just be said to be a simplification of a more complex interaction of AOs. This hasn't directly answered your question, yet. The carbon atom doesn't have to be excited to the 1s2 2s1 2p3 state to interact like described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next question. The energy of the sp3 hybridised orbital is not one, but 4 orbitals. The energy cannot really be found because they don't really exist. After they bond, however, the energy changes based on the bond, as they form a new MO, so it may be more useful to know the energy of each MO in the individual case.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditDate="2014-05-27T13:20:58.033" LastActivityDate="2014-05-27T13:20:58.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11110" PostTypeId="2" ParentId="11105" CreationDate="2014-05-27T09:37:01.367" Score="4" Body="&lt;p&gt;As far as I can see, the expression you have come up with is not entirely correct. &#xA;If we assume that the reaction is $\ce{A -&amp;gt; B}$ (i.e. saying that it is approximately irreversible, or that $k_\mathrm{reverse} \ll k_\mathrm{forward}$), we have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{rate} = \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separating: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\mathrm{d}[\ce{A}]}{[\ce{A}]} = -k\,\mathrm{d}t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integration gives: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln\frac{[\ce{A}]}{[\ce{A}]_0} = -kt $$&#xA;$$ [\ce{A}] = [\ce{A}]_0 \mathrm e^{-kt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $[\ce{A}]_0$ is the initial concentration of $\ce{A}$. It is correct that mathematically, $[\ce{A}] &amp;gt; 0$ for all $t&amp;lt;\infty$, from the last equation. In practise, however, the amount of $[\ce{A}]$ after a certain time interval $t$ will be negligibly small. How long it takes depends on $k$; if $k$ is large, the time for the reaction to go essentially to completion will be short. &lt;/p&gt;&#xA;" OwnerUserId="475" LastEditorUserId="16683" LastEditDate="2017-04-26T23:19:49.643" LastActivityDate="2017-04-26T23:19:49.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11111" PostTypeId="2" ParentId="11080" CreationDate="2014-05-27T11:15:05.600" Score="0" Body="&lt;p&gt;As far I have read- alpha-Alumina is stable at room temperature. They just sinter alumina powder at 1000C so the mostly physical reaction happens quickly.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-05-27T11:15:05.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11113" PostTypeId="2" ParentId="11105" CreationDate="2014-05-27T12:54:45.137" Score="3" Body="&lt;p&gt;To add to Kjetil's answer, the expression given by the OP and the expression derived by Kjetil are (almost) equivalent. If we define the OP's $A$ as $[A]_0$, the original concentration of $A$ and the OP's $A-x$ as $[A]_t$, the concentration of $A$ at time $t$, then we rearrange and take the natural log of both sides...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[A]_t=[A]_0e^{-kt}\ \ \ \ \ (1)$$&#xA;$$\ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=\ln{\left(e^{-kt}\right)}\ \ \ \ \ (2)$$&#xA;$$ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=-kt\ \ \ \ \ (3)$$&#xA;$$ln{\left(\dfrac{[A]_0}{[A]_t}\right)}=kt\ \ \ \ \ (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental differences between the original equation and the one with the natural logarithms is that we have to make some assumptions. Our derivation produces an undefined term if $[A]_0=0$ (equation 2), so we must assume that $[A]_0&amp;gt;0$, which is logical. We also must assume $[A]_t\ne 0$ (equation 4). Only equation 1 works with $[A]_t=0$, which as was pointed out is only technically true at $t=\infty$ (or $[A]_0=0$).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-05-27T12:54:45.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11114" PostTypeId="1" CreationDate="2014-05-27T13:53:17.200" Score="2" ViewCount="105" Body="&lt;p&gt;I want to perform a molecular dynamics simulation with alpha-quartz as my mineral surface. I want to study adsorption of water on alpha-quartz. The most dominant error that I came up with is the problem with my initial physical configuration. In other words error is: System is too far from equilibrium. I think the major problem is related to chemical structure of alpha-quartz slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the experimental data for unit cell of alpha quartz and then replicate it for making a slab of it. My question is: Is this structure stressed in some way? My cell is periodic in 3 directions and some of oxygen atoms are not in the right place! Can it be the problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to replicate unit cell of alpha quartz for obtaining a reasonable structure? Where can I find a right data for unit cell of alpha quartz?&lt;/p&gt;&#xA;" OwnerUserId="5573" LastActivityDate="2014-05-27T13:53:17.200" Title="Making a reasonable slab of alpha quartz" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11115" PostTypeId="2" ParentId="11108" CreationDate="2014-05-27T15:09:19.780" Score="1" Body="&lt;p&gt;To follow up on Swedish Architect's answer (and Martin's comment):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of an orbital is a poorly defined concept, but what can be measured is the energy it takes to excite an electron from a molecular orbital (or an atomic orbital in some cases-- but that is not relevant for hybridization).  In the case of an sp3 hybridized carbon one may infer something about the 'energy' of the orbital from the bond dissociation energy of of the bond formed by such an orbital (as previously mentioned this is a fairly imprecise use of language, but the concept does have some instructional value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reference (&lt;a href=&quot;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&lt;/a&gt;) provides some of these.  In particular on p4.46 of that reference in the chart, CH3CH3 (sp3) is listed at 410 kJ/mol, while CH2CH2 (sp2) is listed at 427 kJ/mol, and HCCH (sp) is listed at 523 kJ/mol.  this is the amount of energy required to abstract a hydrogen atom from that systems, which is a reflect of how 'strong' the bond is, which is a reflection of how low in energy the 'orbital' is.  In essence, the higher the bond dissociation energy, the lower the energy of the corresponding orbital is.  Since all of the molecules in question here from combinations of the hydrogen orbital with the carbon orbital, as a crude approximation, you can assign a relative energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be advised, this analysis really only has meaning for trying to teach the concept.  The actual molecular orbital includes contributions from all atoms in a molecule, and the fact that the ethane has more hydrogens than ethene or ethyne is perturbing this analysis slightly.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-05-27T15:09:19.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11116" PostTypeId="1" CreationDate="2014-05-27T15:19:50.067" Score="2" ViewCount="100" Body="&lt;p&gt;I'm a master student working on networks analysis in general. A network is something that has nodes and there are links between the nodes. Nodes and links could have attributes. An evolving network is one that changes overtime (new nodes and links are added..etc). An example of that is Facebook. Nodes are users and links represent the friendship relationship. Users have attributes (gender, age ..etc). A Facebook network as you know is an example of a social network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that so many people studied traditional evolving networks like social networks, the web, or transportation networks. Currently I'm looking for novel examples of evolving networks to study them. So I thought there might be some examples in &lt;em&gt;chemistry&lt;/em&gt; that could represent some kind of an evolving network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question: Can you give me examples in &lt;em&gt;chemistry&lt;/em&gt; for evolving networks?&lt;/p&gt;&#xA;" OwnerUserId="5194" LastEditorUserId="4945" LastEditDate="2017-03-10T07:56:15.577" LastActivityDate="2017-03-10T07:56:15.577" Title="Examples of evolving networks in chemistry" Tags="&lt;theoretical-chemistry&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2014-05-28T00:19:38.187" ContentLicense="CC BY-SA 3.0" />
  <row Id="11117" PostTypeId="2" ParentId="11104" CreationDate="2014-05-27T18:56:45.210" Score="8" Body="&lt;p&gt;One term I'm missing so far in the discussion is the vapor pressure.&lt;/p&gt;&#xA;&lt;p&gt;You may say that the water vapor pressure above ice is far higher than the iron vapor pressure above iron  (at &amp;quot;normal&amp;quot; temperatures), or the vapor pressure of the various substances forming the table over that table.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;See &lt;a href=&quot;http://www.its.caltech.edu/%7Eatomic/snowcrystals/ice/ice.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for the water vapor pressure. I take from the &lt;a href=&quot;http://www.its.caltech.edu/%7Eatomic/snowcrystals/ice/Vapor2.gif&quot; rel=&quot;noreferrer&quot;&gt;diagram&lt;/a&gt; that at -20 °C, the water vapor pressure above ice is approximately 1 mbar &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 1 hPa, and at slightly below 0°C, about 7 mbar.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressures_of_the_elements_(data_page)&quot; rel=&quot;noreferrer&quot;&gt;this table in Wikipedia&lt;/a&gt;, you need to heat iron well above the melting point to reach such iron vapor pressures. They also cite a formula for the iron vapor pressure above solid iron:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\log (P/\mathrm{Pa}) = 12.106 - 21723 / (T/\mathrm{K}) + 0.4536~\log (T/\mathrm{K}) - 0.5846 (T/\mathrm{K})^{-3}$&lt;/span&gt;&#xA;with the specification that this could be used from 298 K upwards.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Plugging in 298 K, I get &lt;span class=&quot;math-container&quot;&gt;$p_{Fe} \approx 10^{-60}$&lt;/span&gt; Pa.&lt;/p&gt;&#xA;&lt;p&gt;This is so low that lots of other processes will lead to more substantial loss of iron from the spoon via the gas phase (e.g. according to &lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB5444364.htm&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; has a vapor pressure of 1 mmHg &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 133 hPa at 194 °C).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="160" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-05-27T18:56:45.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11118" PostTypeId="1" CreationDate="2014-05-27T20:02:59.303" Score="1" ViewCount="591" Body="&lt;p&gt;My wife and I have had an ongoing argument about various cleaning agents, and I wanted to get a Chemists' view on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we are doing laundry, she will generally clean using the cold water setting and a standard amount of detergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do laundry, I will use the warm water setting with a standard amount of detergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the clothes come out (seemingly) clean. Using cold water saves energy from the heating process, which has been her excuse for doing such a dastardly thing. Her argument against using warm water is that it will damage/shrink the cotton. While I don't disagree on that point, I am curious if the use of warm water causes the detergent to work better or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, there are various soaps that &lt;a href=&quot;http://www.nytimes.com/2011/09/17/business/cold-water-detergents-get-a-chilly-reception.html?pagewanted=all&amp;amp;_r=0&quot; rel=&quot;nofollow&quot;&gt;work&lt;/a&gt; best in cold water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the actual mechanism at work here? Why do some soaps work better in cold (or hot) temperatures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things a bit better, let's remove some more variables. We use this detergent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx&quot; rel=&quot;nofollow&quot;&gt;Tide Pods&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I can't find information on the normal temperatures those machines operate at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also who is right? (I bet it's her)&lt;/p&gt;&#xA;" OwnerUserId="5643" LastEditorUserId="5643" LastEditDate="2014-05-27T20:45:13.987" LastActivityDate="2014-05-27T20:56:40.253" Title="Do various soaps work better when dissolved/used in different water temperatures?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11119" PostTypeId="2" ParentId="11118" CreationDate="2014-05-27T20:37:38.097" Score="2" Body="&lt;p&gt;To start, soaps and detergents have two different chemical makeups, meaning that they act different under different conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general soaps, here is an explanation of how it works:&#xA;&lt;a href=&quot;http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general chemistry note, as the temperature of a solution increases, the ease of a chemical to change (whether it is to react or change shape) increases.  This means that colder temperatures will allow a chemical to remain in the same state.  If the soap depends on a specific conformation/shape to act then colder temperatures may be preferable.  However, if a reaction needs to occur in order for the soap to act some heat may be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Tide Pods, it states that it &quot;dissolves quickly in hot &amp;amp; cold.&quot;&#xA;&lt;a href=&quot;http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx&quot; rel=&quot;nofollow&quot;&gt;http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical makeup of Tide Pods is composed primarily of salt compounds and enzymes.  Both salts and enzymes tend to operate better at higher temperatures.  So as long as you don't pass the denaturing temperature of the enzyme, higher temperatures would be preferred.&lt;/p&gt;&#xA;" OwnerUserId="5644" LastEditorUserId="5644" LastEditDate="2014-05-27T20:56:40.253" LastActivityDate="2014-05-27T20:56:40.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11120" PostTypeId="2" ParentId="7260" CreationDate="2014-05-27T20:58:36.840" Score="3" Body="&lt;p&gt;When you find a minimum in the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow&quot;&gt;thermodynamic potential energy&lt;/a&gt;&lt;/strong&gt; of a system, you will have found an equilibrium point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is the &lt;em&gt;thermodynamic potential&lt;/em&gt; part. Depending on the constraints of the system (i.e. which thermodynamic state variables are held constant) the thermodynamic potential energy function will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a system of a fixed number of particles at constant temperature and pressure (we would call it NPT), the &lt;strong&gt;Gibbs free energy&lt;/strong&gt; is the thermodynamic potential, and it will be minimized at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For NVT (constant volume and temperature) the &lt;strong&gt;Helmholtz free energy&lt;/strong&gt; is the thermodynamic potential, and that is what will be minimized at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a system consisting of a rock dropped from a height above the ground, gravitational potential energy is the thermodynamic potential (I am abusing the language a little bit here) and it will be minimized at equilibrium (when the rock hits the ground).&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-27T20:58:36.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11121" PostTypeId="2" ParentId="11098" CreationDate="2014-05-27T21:03:35.577" Score="5" Body="&lt;p&gt;I'm sure the bottle itself will be safe to use, but I would strongly recommend that you remove or destroy the label. Mislabeled items sometimes lead to unanticipated consequences. Admittedly, that also takes some of the fun out of it.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-05-27T21:03:35.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11122" PostTypeId="2" ParentId="11116" CreationDate="2014-05-27T21:05:36.060" Score="1" Body="&lt;p&gt;If you are looking for a biochemistry answer to this question, there are many cases in the cell structure of any living organism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the skin cells are mapped in a multi-layer system with many physical links (like gap junctions, tight junctions, etc).  Old cells are removed from the linked system and new ones added continuously.&lt;/p&gt;&#xA;" OwnerUserId="5644" LastActivityDate="2014-05-27T21:05:36.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11123" PostTypeId="2" ParentId="11116" CreationDate="2014-05-27T21:30:02.103" Score="2" Body="&lt;h3&gt;Synthetic Pathways&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;... is the only example I can think of right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far back as 1971¹ people have been positing the possibility of an AI to search for novel synthesis paths, but only a couple of years ago a decent one was developed² which could interpret literature and create novel sets of steps for particular reactions -- for example, given a particular reactant and product, produce a set of steps which would produce one from the other. The particular focus of the paper for the 2012 one was to produce single-pot (multi-step) procedures to replace awkward separated multi-step syntheses, but in principle the procedure is generalisable to any path.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sridharan, N.S., An Application of Artificial Intelligence to Organic Chemical Synthesis, Doctoral Thesis, State University of New York at Stony Brook, New York, July 1971.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/anie.201202155&quot; rel=&quot;nofollow&quot;&gt;C M Gothard et al, Angew. Chem., Int. Ed., 2012, 51, 7922 (DOI: 10.1002/anie.201202155)&lt;/a&gt; but there's a &lt;a href=&quot;http://www.rsc.org/chemistryworld/2012/08/automatic-chemist&quot; rel=&quot;nofollow&quot;&gt;good write-up by Philip Ball in Chemistry World, August 2012 issue&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-05-27T21:30:02.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11124" PostTypeId="1" CreationDate="2014-05-28T00:00:52.203" Score="-1" ViewCount="7739" Body="&lt;p&gt;There are two choices. One is Nitric acid, and the other is phosphoric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution of each type of these acids has a concentration of 0.1 mole. I answer would be phosphoric acid, because each molecule contains more than 1 hydrogen ion. Thus it would be more conductive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does strong/ weak affect conductivity? &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="12860" LastEditDate="2015-11-13T20:19:29.240" LastActivityDate="2015-11-13T20:19:29.240" Title="Which is more conductive?" Tags="&lt;acid-base&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11125" PostTypeId="2" ParentId="11124" CreationDate="2014-05-28T00:54:03.873" Score="2" Body="&lt;p&gt;Whether the acid is a strong or weak acid does affect its conductivity.  Weaker acids are less dissociated and generate fewer ions in the equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid $\ce{K_{a}=24}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid $\ce{K_{a}=7.1 x 10^{-3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://depts.washington.edu/eooptic/links/acidstrength.html&quot; rel=&quot;nofollow&quot;&gt;link to Ka reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid is more dissociated than phosphoric acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrical Conductivity κ in mS/cm at 1 Mass Percent Concentration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrical%20conductivity%20of%20aqueous%20solutions.pdf&quot; rel=&quot;nofollow&quot;&gt;link to conductivity reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid ~ 56.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid ~ 10.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid is more dissociated and more conductive than phosphoric acid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-28T01:14:34.457" LastActivityDate="2014-05-28T01:14:34.457" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11126" PostTypeId="2" ParentId="10417" CreationDate="2014-05-28T01:17:25.230" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://gizmodo.com/the-great-and-secret-salt-heist-that-saved-the-sochi-ol-1524359889&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; and &lt;a href=&quot;http://www.nytimes.com/2014/02/17/sports/olympics/emergency-shipment-of-salt-averts-olympic-crisis.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; (posted by @Jun-Goo Kwak in the comments above) say that the salt was used to lower the melting point of the snow on the surface, which would cause it melt. It would then re-freeze (I am assuming either at night or due to being in contact with snow beneath it), forming an icy layer on the surface. I don't know anything about Alpine skiing, but it seems that an icy surface is better than a slushy surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ice would also form an insulating layer on the surface, which might help keep more snow from melting. Ice is a better conductor of heat than snow is (snow contains a lot of air, which is a terrible thermal conductor), so it is unlikely that it would act as an effective conductive insulator. However, it would shield the sub-surface snow from convective heat transfer caused by the movement of air. This is how &lt;a href=&quot;http://people.howstuffworks.com/igloo.htm&quot; rel=&quot;nofollow&quot;&gt;igloos&lt;/a&gt; and &lt;a href=&quot;http://nsidc.org/cryosphere/snow/science/characteristics.html&quot; rel=&quot;nofollow&quot;&gt;snow caves&lt;/a&gt; work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could also be an effect due to the phase transition of melting - when ice melts, it is an endothermic transition, which means it absorbs heat. Some of that heat comes from the air, but some of it will also come from the snow directly below the surface. The result could be that the snow melts more slowly than it would if it were not covered by a layer of ice. The reverse of this effect is the reason that &lt;a href=&quot;http://extension.arizona.edu/sites/extension.arizona.edu/files/pubs/az1222.pdf&quot; rel=&quot;nofollow&quot;&gt;spraying an orange crop with water right before a hard freeze can save the crop&lt;/a&gt; (pg. 3) - as the water freezes, it releases heat (freezing is exothermic) and some of that heat goes into the orange trees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hard to guess which effect is most important (insulation vs. endothermic heat transfer), but I would say that both should be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, in both cases the idea is to melt snow and/or ice, but in the case of roads, we want the ice to melt immediately and flow away, so we use lots of salt. In the case of Alpine ski events, we want a little snow to melt and then re-freeze, so we use a little salt.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-28T01:17:25.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11127" PostTypeId="2" ParentId="6172" CreationDate="2014-05-28T01:30:03.933" Score="2" Body="&lt;p&gt;Neon exists as single atoms in its elemental form. It is a noble gas, and as @jonsca said, that means (for various reasons) that it is extremely difficult for it to form bonds (I don't know of any neon compounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the confusion may be coming from a missing definition: what is a molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;molecule&lt;/strong&gt; is two or more atoms joined by covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since neon exists as solitary atoms, it is not a molecular element.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-28T01:30:03.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11128" PostTypeId="1" CreationDate="2014-05-28T03:00:48.550" Score="0" ViewCount="821" Body="&lt;p&gt;I am looking for a list of classes of synthetic organic chemicals. I'm not very strong in chemistry. The only information I could find are from &lt;a href=&quot;http://www.nmenv.state.nm.us/dwb/contaminants/SOCs.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; which seem to describe SOC's are like pesticides or fuel additives, but I don't think you can call those classes.&lt;/p&gt;&#xA;" OwnerUserId="5651" LastEditorUserId="4734" LastEditDate="2014-05-28T06:13:29.643" LastActivityDate="2014-05-28T06:13:29.643" Title="Types of organic chemicals?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-06-03T12:14:59.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="11129" PostTypeId="1" AcceptedAnswerId="11132" CreationDate="2014-05-28T05:10:51.810" Score="1" ViewCount="5060" Body="&lt;p&gt;Here is the overall formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COONa  + HCl &amp;lt;=&amp;gt; NaCl + CH3COOH}$$&#xA;(sodium acetate + hydrochloric acid $\ce{&amp;lt;=&amp;gt;}$sodium chloride + acetic acid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way does the above reaction naturally proceed? I've tried Googling this, but different sources seem to suggest &lt;em&gt;contradictory&lt;/em&gt; answers, and I can't really tell what's correct.&lt;/p&gt;&#xA;" OwnerUserId="5588" LastEditorUserId="33991" LastEditDate="2017-04-26T19:17:34.603" LastActivityDate="2017-04-26T19:17:34.603" Title="Which way does this reaction between NaCl and acetic acid naturally proceed?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11130" PostTypeId="2" ParentId="11128" CreationDate="2014-05-28T05:32:39.397" Score="2" Body="&lt;p&gt;There are a &lt;em&gt;lot&lt;/em&gt; of organic chemicals&lt;sup&gt;[citation needed]&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the incomprehensibly vast number of chemical species and variation between them, almost every classification system is domain-specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of pure chemistry, the best you can do is usually to organize them according to what functional groups are in play; this will usually reveal something about the fundamental chemical interactions available to the species in question, but this is not necessarily useful or enlightening:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many applications, such as as antifreeze, fuel or a solvent, methanol and ethanol are (largely) interchangeable: They both have short carbon chains attached to a single alcohol group (-OH), their boiling points, dipole moments, chemical energy and other properties are similar, and they mostly react in the same way on a chemical level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other applications, however, they are as  night and day: In biology, methanol is a potent poison, and ethanol is a mostly harmless intoxicant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic chemistry is, in short, notoriously resilient to generally useful classification systems.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastActivityDate="2014-05-28T05:32:39.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11131" PostTypeId="2" ParentId="11129" CreationDate="2014-05-28T05:36:31.633" Score="2" Body="&lt;p&gt;As denoted by the equilibrium arrows, the reaction proceeds in both directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, the net direction depends on the concentrations of the chemical species (Le Chatelier's principle). For example, if there's a lot of acetic acid, the reaction will go left. If there's lots of acetate, the reaction will go right (&quot;lots&quot; is a relative term, see the equilibrium constant for acetic acid: $K_a = 1.8\times10^{-5}$). The net reaction will proceed in this direction until the equilibrium concentrations are achieved. However, keep in mind that even at this stage the reaction still occurs in both directions, but there will be a net direction (one direction is favoured over the other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once equilibrium occurs, there is no net direction but, again, the reaction still occurs in both directions (at the same rate now). &lt;/p&gt;&#xA;" OwnerUserId="5220" LastEditorUserId="4945" LastEditDate="2014-05-28T06:06:06.700" LastActivityDate="2014-05-28T06:06:06.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11132" PostTypeId="2" ParentId="11129" CreationDate="2014-05-28T06:33:04.303" Score="4" Body="&lt;p&gt;&lt;em&gt;Mostly&lt;/em&gt; the reaction will proceed from left to right, unless you have ridiculous amount of acetic acid. Hydrochloric acid is much stronger, so it suppresses dissociation of acetic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is only part of the truth. In fact, in water solutions many compounds (partially) dissociates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HCl = H^+ + Cl^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is strongly endothermic in vacuum, but subsequent interaction of ions with water makes it mostly energy neutral or even highly exothermic for some compounds. This means, that in you case the solution would actually contain $H^+$, $OH^-$ (very little), $CH_3COO^-$, $CH_3COOH$, $HCl$ (almost none, may be ignored in most cases), $Na^+$. The actual reaction would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CH_3COO^- + H^+ = CH_3COOH$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual direction of the reaction would depend on initial concentraion and guarded by equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The keywords for googling/textbook reading are 'electrolitic dissociation', 'ionic reaction', 'chemical equilibrium', 'equilibrium constant'. Generally, the procedure of determining which concentration of which ions the system will have requires solution of system of equations, including equilibrium constants, electroneutrality equation and material balance equations. The system often may have no analytic solutions, in which case it is usually approximated by simpler system, dropping some elements of sums in the system.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-05-28T06:33:04.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11134" PostTypeId="1" AcceptedAnswerId="20038" CreationDate="2014-05-28T09:03:29.330" Score="5" ViewCount="538" Body="&lt;p&gt;Henry's Law is for a gas dissolved in a liquid and states P = kC where P is the partial pressure of the gas above the solution and C is the concentration of the gas dissolved in the liquid. The k is a &quot;constant&quot;, that is, it is independent of P, C and time but depends on the type of gas and the type of liquid. According to Pickover, page 184 (reference below), Henry's Law assumes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A1.  the gas/liquid system has reached equilibrium, and&#xA;A2.  the gas does not chemically react with the liquid, and&#xA;A3.  the concentration of gas in the liquid is low, and&#xA;A4.  the gas pressure is low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful for any help in answering the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1.  Is the list of assumptions, A1 to A4, correct and complete? I have no reason to doubt it but I would like to confirm it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2.  Some authors (e.g. Dickinson E.J., Road Materials and Pavement Design, 2000, 1(3), 255-280) have applied Henry's Law to the diffusion of oxygen in air at atmospheric pressure into bitumen which is a semi-solid at room temperature. Is there any experimental evidence that Henry's Law can be extended to such a system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference&#xA;Pickover, C., &quot;Archimedes to Hawking&quot;, Oxford University Press 2008&lt;/p&gt;&#xA;" OwnerUserId="5613" LastActivityDate="2019-12-03T11:17:46.123" Title="Validity of Henry's Law for an Oxygen/Bitumen System?" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11135" PostTypeId="2" ParentId="11078" CreationDate="2014-05-28T09:07:46.463" Score="1" Body="&lt;p&gt;The easiest answer in my opinion; try it! If each atom has the same radius, it is simply not possible to fit one of them into the center of the cubic system (known as a cubic hole). It is, however, possible if the atom is sufficiently small. &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_chloride#Crystal_structure&quot; rel=&quot;nofollow&quot;&gt;CsCl&lt;/a&gt; is an example of such a compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can show this geometrically (as is done &lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/c123/bcc.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), but you can also convince yourself that it must be the case using equally-sized tennis balls or something similar to try it out in practise. &lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2014-05-28T09:07:46.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11136" PostTypeId="2" ParentId="9908" CreationDate="2014-05-28T09:44:00.173" Score="3" Body="&lt;p&gt;Elaborating Ron’s answer through a concrete example, we start by looking at the Celsius scale&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_\text{C1} = 25\ \mathrm{^\circ C} ,~ T_\text{C2} = 100\ \mathrm{^\circ C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta T_\text{C} = T_\text{C2} - T_\text{C1} = 100\ \mathrm{^\circ C} - 25\ \mathrm{^\circ C} = 75\ \mathrm{^\circ C} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar fashion, using the Kelvin scale&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_\text{K1} = 298.15\ \mathrm K = (273.15 + 25)\ \mathrm{K},~ T_\text{K2} = 373.15 = (273.15 + 100)\ \mathrm{K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta T_\text{K} = T_\text{K2} - T_\text{K1} = (273.15 + 100)\ \mathrm{K} - (273.15 + 25)\ \mathrm{K} = 75\ \mathrm{K} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the constant offset of 273.15 between the Celsius scale and the Kelvin scale cancels, and the difference is exactly the same. The &lt;em&gt;units&lt;/em&gt; remain, but the &lt;em&gt;numerical value&lt;/em&gt; of the difference is the same.&lt;/p&gt;&#xA;" OwnerUserId="475" LastEditorDisplayName="user7951" LastEditDate="2015-03-31T19:20:29.767" LastActivityDate="2015-03-31T19:20:29.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11138" PostTypeId="2" ParentId="11036" CreationDate="2014-05-28T11:30:31.847" Score="3" Body="&lt;p&gt;A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic &lt;a href=&quot;http://goldbook.iupac.org/C01023.html&quot; rel=&quot;nofollow&quot;&gt;equilibrium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\Delta G_r &amp;amp;= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\&#xA;\Delta G_r^\circ &amp;amp;= − \mathcal{R}T \cdot \ln K&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot; rel=&quot;nofollow&quot;&gt;equilibrium constant&lt;/a&gt; may be derived from the &lt;a href=&quot;http://goldbook.iupac.org/M03725.html&quot; rel=&quot;nofollow&quot;&gt;mass-action law&lt;/a&gt; and should be defined as a product of activities (for the standard state). For $\ce{A + B &amp;lt;=&amp;gt; C}$ this will lead to&#xA;$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\gamma\frac{c}{c^{\circ}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.&#xA;$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.&#xA;$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-28T11:30:31.847" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11139" PostTypeId="1" CreationDate="2014-05-28T13:46:57.173" Score="0" ViewCount="476" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ksp of CaSO4 is 7.10*10^-5&lt;/li&gt;&#xA;&lt;li&gt;Molar mass of Na2SO4 is 142.04g/mol &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1 - figure out molar solubility of CaSO4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;√7.10*10^-5 = 8.43*10^-3M &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2 - convert Na2SO4 from mass to moles &lt;/p&gt;&#xA;&#xA;&lt;p&gt;105mg = 0.105g &#xA;0.105/142.04 = 7.39*10^-4 mol &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3 - figure out moles of CaSO4 produced &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we need 1 mol of Ca2+ &amp;amp; 1 mol of Na2SO4 to form 1 mol of CaSO4 &#xA;And we have less Na2SO4 than Ca2+ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CaSO4 produced = 7.39*10^-4 mol &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4 - figure out molarity of CaSO4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.39*10^-4mol / 100ml = 7.39*10^-3 / L = 7.39*10^-3M &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5 - figure out if any CaSO4 precipitates &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.43*10^-3M &gt; 7.39*10^-3M &#xA;Therefore NO CaSO4 precipitates &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The answer in the book is a Yes, can anyone please explain to me where I have gone wrong? &#xA;Thanks!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="5661" LastEditorUserId="5661" LastEditDate="2014-05-28T14:58:57.213" LastActivityDate="2014-05-28T14:58:57.213" Title="A sample of tap water is found to be 0.025 M in Ca2+. If 105 mg of Na2SO4 is added to 100.0 mL of the tap water, will any CaSO4 precipitate?" Tags="&lt;equilibrium&gt;&lt;solubility&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11140" PostTypeId="2" ParentId="8870" CreationDate="2014-05-28T14:18:20.213" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is there a general way to predict products of a chemical reaction, given only the atomic masses of masses of all elements involved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Unforturnately, no.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If there's no method to do so, are we supposed to memorize all chemical reactions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There is some amount of reactions one have to memorize individually. However, there are classes of reactions that are quite broad and have uniform rules to check if some reactions happens. Two most common of such classes are redox reactions in  solutions, guarded by electrode potentials, and ionic reactions in solutions, guarded by compound solubility expressed as solubility product.&lt;/p&gt;&#xA;&lt;p&gt;In your example&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CuFeS2 + O2 -&amp;gt;}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is an example of oxidation of sulfides. Most sulfides can be oxidized at elevated temperature by oxygen into stable oxides, in this case (if there is excess oxygen): &lt;span class=&quot;math-container&quot;&gt;$$\ce{CuFeS2 + O2 -&amp;gt; CuO + Fe2O3 + SO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{K2CrO4 + H2SO4 -&amp;gt;}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{K2Cr2O7 + H2SO4 -&amp;gt;}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Both reactions are ionic and guarded by relative acidity of chromic and sulfuric acids and led to formation of chromic anhydride.&lt;/p&gt;&#xA;&lt;p&gt;The remaining reaction is unlikely to proceed at significant rate, as the oxide considered is kinetically inert. However, if it would react, the products would be potassium chromate (VI) and later ferrate (VI).&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-06-09T16:38:39.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11141" PostTypeId="2" ParentId="11139" CreationDate="2014-05-28T14:51:48.510" Score="1" Body="&lt;p&gt;$K_{sp} =[Ca^{2+}][SO_4^{2-}] = 7.10 \times 10^{-5} M^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.025M[SO_4^{2-}]  = 7.10 \times 10^{-5} M^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[SO_4^{2-}] = 0.00284M$ (this is the maximum soluble concentration of $SO_4^{2-}$ in the presence of $0.025M Ca^{2+}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 0.000739 moles of $SO_4^{2-}$ are added to 100 mL of water, corresponding to a concentration of $0.00739M SO_4^{2-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precipitation occurs because 0.00739M is greater than 0.00284M.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-28T14:51:48.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11142" PostTypeId="1" AcceptedAnswerId="11143" CreationDate="2014-05-28T15:45:07.697" Score="2" ViewCount="2625" Body="&lt;p&gt;I'm in class right now and my teacher put this up on the board as a representation of 1,2-dichloroethane. He also projected a model. I'm thinking ... his board drawing is wrong. Shouldn't the chlorines also be on dashes too? There's no way the chlorines can be in the same plane as the carbon atoms; the model makes it too obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ly4jH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R9bf6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AA6gr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2015-10-16T17:13:52.263" LastActivityDate="2015-10-16T17:13:52.263" Title="Wedge Dash Diagram of 1,2-Dichloroethane" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11143" PostTypeId="2" ParentId="11142" CreationDate="2014-05-28T15:57:11.090" Score="4" Body="&lt;p&gt;The teacher's drawing is correct.  Of course there are other possible conformers that could be drawn, but there is nothing wrong with the one that is pictured.   Basic rule:  a dashed wedge indicates a bond projecting &lt;strong&gt;behind&lt;/strong&gt; the plane of the screen, a bold wedge indicates a bond projecting out &lt;strong&gt;in front&lt;/strong&gt; of the plane of the screen, and a simple solid line indicates a bond existing &lt;strong&gt;in&lt;/strong&gt; the plane of the screen.  Also note that this drawing notation can be used for any molecule, not just carbon systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AHoe8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-28T19:39:50.930" LastActivityDate="2014-05-28T19:39:50.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11144" PostTypeId="1" AcceptedAnswerId="11147" CreationDate="2014-05-28T17:27:21.927" Score="8" ViewCount="15546" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$5\:\mathrm{g}$ of an unknown salt are dissolved in $325\:\mathrm{g}$ of water. Both the water and the salt are initially the same temperature. The water's temperature falls by $11.4\:\mathrm{^\circ{}C}$. Explain how it is possible for the salt and water to change temperature even though both substances are initially at the same temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $$q = mc \Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $q, m,$ and $\Delta T$ aren't changed by this action, this process must result in the raising of water's specific heat. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="5026" LastEditDate="2018-02-25T09:26:40.173" LastActivityDate="2018-03-13T20:27:47.477" Title="Thermodynamics adding salt to water changes the temperature" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11146" PostTypeId="2" ParentId="11144" CreationDate="2014-05-28T18:29:55.690" Score="2" Body="&lt;p&gt;You are starting on the right track, but the specific heat of water won't change (at least not by a measurable amount).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that the flow of heat is defined by $q=mC{\Delta}T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you also know that energy is conserved. So if heat is flowing out of the water, where is it going?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like a homework question (if it is, please tag it as one!), so I am reluctant to just give you the answer. I will give you a hint though - the problem states that the salt &lt;em&gt;dissolves&lt;/em&gt; - I think that will be important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Dissenter said, dissolution of any salt (ionic compound) can be either exothermic or endothermic, depending on the salt and on the solution. In this case, it must be endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this because ${\Delta}T &amp;lt; 0$, which means $q &amp;lt; 0$ &lt;em&gt;for the water.&lt;/em&gt; The heat must have gone somewhere - the only place it could have gone (assuming that it wasn't just lost to the environment) is into the molecules as stored energy (enthalpy). Since the heat transferred from the water is negative (exothermic), that means the enthalpy change for dissolving the salt must be positive (endothermic).&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5026" LastEditDate="2018-02-25T09:29:04.063" LastActivityDate="2018-02-25T09:29:04.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11147" PostTypeId="2" ParentId="11144" CreationDate="2014-05-28T19:38:49.563" Score="14" Body="&lt;p&gt;The keyword is that the salt dissolves. Dissolution entails &lt;em&gt;at least&lt;/em&gt; two steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Overcoming solvent-solvent interactions and bonds. An extreme example: your dinner plate doesn't dissolve in your kitchen table. One reason is that there is an extremely high energy barrier to overcoming the hypothetical solvent's (in this case, the kitchen table's) intermolecular forces. You'd have to take an ax to the table to overcome the strong intermolecular forces that hold the molecules of your dinner table together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand breaking these solvent-solvent &quot;bonds&quot; or intermolecular forces in water is easy; you can jump into a swimming pool just fine. I wouldn't suggest jumping into a table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as this step's suggests, this is an energy intensive process or an endothermic process. Energy must be consumed to break these solvent-solvent bonds or interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Breaking solute-solute interactions and bonds. This however isn't &lt;em&gt;required&lt;/em&gt;. For example, some unionized $\ce{NaCl}$ may be solvated by water. You'll find this more true with some of the less soluble salts. But most of the sodium chloride you toss in water will have dissolved (we'll get to why later), so what is solvated generally isn't $\ce{NaCl_{(s)}}$ or $\ce{NaCl_{(aq)}}$ (this is somewhat misleading terminology) but rather $\ce{Na^+}$ and $\ce{Cl^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This again is an &lt;em&gt;endothermic&lt;/em&gt; process as bonds are being broken.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Solute-solvent stabilization (solvation). This steps entails the formation of solute-solvent intermolecular forces. For example, the sodium ion, with its positive charge, may form a hydration shell of water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an exothermic process, as &quot;bonds&quot; or more accurately, stabilizing intermolecular forces are formed. &lt;img src=&quot;https://i.stack.imgur.com/s54lx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couple all three of these processes together and you have the actual dissolution process of $\ce{NaCl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's interesting about $\ce{NaCl}$ is that even with the third, exothermic step, the dissolution (which comprises all three steps) is &lt;em&gt;still&lt;/em&gt; a tiny bit endothermic. So that's what the question is premised on - the addition of sodium chloride to water kicks off a &lt;em&gt;spontaneous&lt;/em&gt; process which is &lt;em&gt;endothermic&lt;/em&gt;. So one might ask why might this &lt;em&gt;endothermic&lt;/em&gt; process still occur. The answer lies with entropy; the dissolution of sodium chloride is not enthalpically favorable (dissolution &lt;em&gt;consumes&lt;/em&gt; energy - specifically - from the water) - yet the dissolution is &lt;em&gt;entropically&lt;/em&gt; favorable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that all processes tend toward greater disorder over time, and the dissolution of sodium chloride is a perfect way to increase disorder; we go from unionized sodium chloride to two ions! The two ions are definitely going to have a lot more degrees of freedom than a single, unified molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCl_{(s)} + H_2O -&amp;gt;Na^+_{(aq)} + Cl^{-}_{(aq)}}$$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="15235" LastEditDate="2017-03-22T12:53:05.823" LastActivityDate="2017-03-22T12:53:05.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11148" PostTypeId="1" CreationDate="2014-05-28T20:39:01.150" Score="13" ViewCount="262" Body="&lt;p&gt;I tried to make different crystals in the cubic system by stacking cubes in a 3D software. However, I need to tell the software where to skip rows or stop stacking the unit cells in a specific direction else it will only make a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this work in nature? How does the fluorite &quot;&lt;em&gt;know&lt;/em&gt;&quot; where to skip rows in order to make the octahedron crystal? Does it have something to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Bravais_lattice&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bravais lattices&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Coordination_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;coordination numbers&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I thought that the unit cell needs to look like the final shape, but I think I'm wrong there.&lt;/p&gt;&#xA;" OwnerUserId="3955" LastEditorUserId="41328" LastEditDate="2017-08-09T01:22:00.073" LastActivityDate="2017-10-01T14:13:28.830" Title="How does the crystal know where to stop stacking the unit cells?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;crystallography&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11151" PostTypeId="1" AcceptedAnswerId="11154" CreationDate="2014-05-29T01:52:43.170" Score="3" ViewCount="1931" Body="&lt;p&gt;Would the formal charge of terminal atoms differ from the FC of the non-terminal atoms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you had a sheet of graphite; inside the network covalent solid all the carbons are bound to one another, but the terminal atoms are obviously missing some C-C bonds since they are terminal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D2XvO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their formal charges would differ, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the consequence of the differing charges? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-30T05:27:46.607" Title="Formal Charge of Terminal Atoms" Tags="&lt;formal-charge&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11152" PostTypeId="2" ParentId="11151" CreationDate="2014-05-29T02:11:19.050" Score="3" Body="&lt;p&gt;Yes, the formal charge at such atoms would be different.  Such atoms, typically found at the at the edge of some crystalline or regular solid, are known as &quot;defects&quot; in the solid.  Because they have an incomplete outer shell and lack a stable bonding configuration (bonding vacancies), they usually exist as a radical, anion or cation.  They define an area of extremely high reactivity on the solid's surface and mark the spot where reactions often begin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: references to charges at defect sites&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-inst.eecs.berkeley.edu/~ee143/sp06/lectures/Defects_tutorial.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; covers inorganic systems, see p.5 for some nice pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.purdue.edu/quantum/files/Nanotechnology_Romaneh.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; covers graphene systems, from the &quot;Conclusions&quot; section; &quot;We have performed scanning gate microscopy on &lt;strong&gt;graphene&lt;/strong&gt; and &lt;strong&gt;observed such charge inhomogeneity&lt;/strong&gt; and puddles due to extrinsic doping from &lt;strong&gt;possible sources&lt;/strong&gt; such as metal contacts, &lt;strong&gt;graphene edges&lt;/strong&gt; and surface residues.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5084" LastEditDate="2014-05-30T05:27:46.607" LastActivityDate="2014-05-30T05:27:46.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11153" PostTypeId="2" ParentId="10867" CreationDate="2014-05-29T02:31:07.983" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely not, here are a few examples&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJ0RM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-05-29T07:08:02.270" LastActivityDate="2014-05-29T07:08:02.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11154" PostTypeId="2" ParentId="11151" CreationDate="2014-05-29T02:31:47.493" Score="1" Body="&lt;p&gt;The terminal carbons &lt;em&gt;may&lt;/em&gt; or &lt;em&gt;may not&lt;/em&gt; be missing some bonds. In any case, unless there is a charge on the surface/edge, the formal charge will always be zero.&lt;/p&gt;&#xA;&lt;p&gt;Let's look at a single graphene ring to make it easier to see:&lt;/p&gt;&#xA;&lt;h1&gt;Free radicals&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKmc5.png&quot; alt=&quot;A graphene ring with single electrons on each carbon&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this first case, there are no atoms attached, and it is pure carbon. However, for the molecule to be electrically neutral, there must be a total of 6*4 valence electrons: 24. If we only have alternating single and double bonds, that gives us 18, so the other 6 have to go somewhere. In this structure I have drawn them as free radicals (although it is extremely unlikely they would be for long). In this case, you can see that the formal charge on each carbon is still zero:&lt;/p&gt;&#xA;&lt;p&gt;# valence electrons - [&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt; # bonding electrons + #non-bonding electrons] = formal charge&lt;/p&gt;&#xA;&lt;p&gt;4 - (&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt;*6 + 1) = 0&lt;/p&gt;&#xA;&lt;h1&gt;Triple Bonds&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OlUiE.png&quot; alt=&quot;A graphene ring with triple bonds&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The second image is more likely if we don't allow any other elements to attach. Since the alternating double bonds are are really resonance structures formed by the overlap of p-orbitals, those lone electrons can do the same and form another &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-bond, resulting in a triple bond. This idea is support by this &lt;a href=&quot;http://www.nature.com/ncomms/2014/140113/ncomms4040/full/ncomms4040.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;, which shows that under vacuum, they have observed that graphene sheets can form triple bonds along the edge. Examining each carbon, we again see that the formal charge is zero:&lt;/p&gt;&#xA;&lt;p&gt;4 - (&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt;8 + 0) = 0&lt;/p&gt;&#xA;&lt;h1&gt;'Other Stuff'&lt;/h1&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rflTA.png&quot; alt=&quot;Benzene&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The third picture is what we commonly assume happens - something (in this case hydrogen) is attached. This would again give a total of four bonds, resulting in a formal charge of zero for each carbon.&lt;/p&gt;&#xA;&lt;p&gt;4 - (&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt;8 + 0) = 0&lt;/p&gt;&#xA;&lt;p&gt;So the answer to your first question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Their formal charges would differ, correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is most likely &amp;quot;no&amp;quot; unless the surface is charged.&lt;/p&gt;&#xA;&lt;p&gt;However, your second question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is the consequence of the differing charges?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;is a good one, considering that regardless of formal charge, the terminal atoms will obviously have a different &lt;em&gt;character&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;How exactly it will change depends on the specifics, but in general, we can say that they are &amp;quot;more reactive.&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-05-29T15:22:56.310" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11155" PostTypeId="1" CreationDate="2014-05-29T03:10:46.680" Score="4" ViewCount="2396" Body="&lt;p&gt;I'm a student working in a high school laboratory that is looking to pursue research into perovskite cells for next year, but I have some solubility and safety questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good solvent (please be specific) of copper(I) iodide?, copper(I) thiocyanate?  Silver iodide?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to deposit these materials out of solution onto the substrate.  Christians et al. in 2014 deposited copper(I) iodide from a 1:39 chlorobenzene/di-&lt;em&gt;n&lt;/em&gt;-propyl sulfide solution, but my mentors have rejected chlorobenzene as a substance with high environmental toxicity and difficult disposal procedures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the solvent for any of the three compounds would dissolve the compound well, and then evaporate off with heating, while having no to low toxicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I have received approval for purchasing methylammonium iodide and lead(II) iodide which I will dissolve in gamma-butyrolactone in 1:1 molar.  There are no safety sheets for the product, methylammonium lead iodide, so my mentors wanted me to see if there were any harmful properties that would come about from making this compound, ie: is there anything more dangerous or something more to watch for from methylammonium lead iodide than methylammonium iodide or lead(II) iodide?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Article: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja411014k&quot; rel=&quot;nofollow&quot;&gt;An Inorganic Hole Conductor for Organo-Lead Halide Perovskite Solar Cells. Improved Hole Conductivity with Copper Iodide&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5672" LastEditorUserId="4945" LastEditDate="2014-05-29T07:29:14.003" LastActivityDate="2015-07-21T13:21:33.587" Title="Solvent for copper(I) iodide, copper(I) thiocyanate, silver iodide for HTM deposition; safety for methylammonium lead iodide" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;safety&gt;&lt;solvents&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11157" PostTypeId="2" ParentId="10867" CreationDate="2014-05-29T05:15:37.033" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be seen as a general rule of thumb, but it is not accurate. The defining feature for &lt;a href=&quot;http://goldbook.iupac.org/C01267.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjugated&lt;/a&gt; systems is that they have (local) planarity. This is necessary in order to form &lt;a href=&quot;http://goldbook.iupac.org/D01583.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;delocalised&lt;/a&gt; $\pi$ orbitals. The &lt;a href=&quot;http://goldbook.iupac.org/S05434.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\pi$&lt;/a&gt; orbitals are antisymmetric to that defining plane, hence have to be formed at least by $\ce{p}$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all $\ce{sp^2}$ orbitals are in a plane, there are available $\ce{p}$ orbitals perpendicular to that plane for $\pi$ orbitals. The best example for this is benzene and on a larger scale graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same concept can be applied to $\ce{sp}$ orbitals. The best example might be $\ce{HC#C-C#CH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therefore a combination of both. A fantastic example for this is phenylacetylene (BP86/def-SVP only orbitals of out of plane $\pi$ symmetry no 27, 26, 24, 21).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhiEv.jpg&quot; alt=&quot;pi orbitals of phenylacetylene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even much more fantastic example is the $D_{2h}$ symmetric $\ce{HC#C-CH=CH-C#CH}$ (BP86/cc-pVTZ only orbitals of out of plane $\pi$ symmetry no 20, 17, 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hRb0C.jpg&quot; alt=&quot;pi orbitals of (HCC-CH)2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-05-29T05:15:37.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11160" PostTypeId="1" AcceptedAnswerId="11161" CreationDate="2014-05-29T11:07:27.327" Score="6" ViewCount="110981" Body="&lt;p&gt;I know that the lower the pH the faster fermentation occurs. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="5603" LastActivityDate="2019-12-18T14:08:49.363" Title="Why does pH affect fermentation?" Tags="&lt;ph&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11161" PostTypeId="2" ParentId="11160" CreationDate="2014-05-29T13:04:15.583" Score="7" Body="&lt;p&gt;pH affects the shape of proteins. In the case of fermentation a collection of enzymes is responsible for the metabolic processes that occur. - an enzyme is a protein which performs a metabolic process. For example sucrase is an enzyme which breaks sucrose down into fructose and glucose. They are kind of like organic catalysts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proteins are made of amino acids strung together in a long line, based off DNA. The amino acids bond up into a long polymer. Then cross linking occurs between the R groups on the amino acids. This makes the unique shape of the protein, because the cross-linking turns the protein into a 3-d shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amino acids may be acidic amino acids, or basic amino acids, due to the R group on the amino acids. If the pH is increased, this affects the shape of proteins, by disrupting the bonds in the protein. In the case of fermentation, you say the rate increases when it get's more acidic - when the pH is lower. This is because the organisms - the yeast - producing the enzymes to ferment glucose, have adapted to acidic conditions. This means that through natural selection the enzymes today have been selected because they work best in acidic conditions, which bend the protein into the correct shape to allow fermentation to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read: &lt;a href=&quot;http://www.icr.org/article/glycolysis-alcoholic-fermentation/&quot; rel=&quot;noreferrer&quot;&gt;http://www.icr.org/article/glycolysis-alcoholic-fermentation/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this article it says that 10 enzymes turn glucose in pyruvic acid:&lt;a href=&quot;http://en.wikipedia.org/wiki/File%3aPyruvic-acid-2D-skeletal.png&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/File:Pyruvic-acid-2D-skeletal.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then two enzymes turn this into ethanol and CO2. The last two enzymes function best in acidic conditions, presumably because they have evolved to work best in the acidic conditions made by the pyruvic cid. However all fermentation stops after the pH drops below about 4.2. The optimum pH is about 4.8 - 5.0. After this we can assume that the pH bends the proteins out of shape too much, and the protein is said to have been denatured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to conclude... The rate increases because enzymes are all suited to specific conditions, dictated by natural selection - the conditions they have evolved around - and in the case of fermentation the last 2 enzymes - turning pyruvic acid to ethanol and CO2 - work best in acidic conditions, as they have evolved to those conditions. But increase the pH too much... and the protein denatures, and fermentation stops completely.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2014-05-29T13:04:15.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11162" PostTypeId="2" ParentId="11160" CreationDate="2014-05-29T13:15:35.440" Score="3" Body="&lt;p&gt;I don't know that it's correct to say that lower pH increases fermentation rate in all cases. It is true for pH ranges that are close to neutral, but there is a point where increasing acidity will kill the yeast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general though, yeasts and molds are more tolerant of acidic conditions than they are of basic conditions, down to a pH of about 5. Bacteria, on the other hand, tend to not be very tolerant of acidic conditions (at least, not the bacteria commonly found in places where we are trying to encourage fermentation.) This &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1087585/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; compares the growth and fermentation rates of bacteria to yeast under different conditions, and finds that under mildly acidic conditions (pH of 5.5 to 5.0) the bacterial growth rate was slowed and the yeast growth rate was maximized. Since bacteria competes with yeast for food, the ideal pH is the one that kills the most bacteria without slowing the yeast growth rate down. A pH of 5 is about the same as a cup of black coffee, and is about one hundred times less acidic than orange juice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC294626/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; which describes another interesting effect - the bacteria that was used in the first experiment actually starts producing ethanol as a fermentation product under acidic conditions, instead of just lactose. I suspect that growth rates are a bigger factor, but it is an interesting complication.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-05-29T17:28:04.133" LastActivityDate="2014-05-29T17:28:04.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11163" PostTypeId="1" CreationDate="2014-05-29T15:41:30.267" Score="1" ViewCount="643" Body="&lt;p&gt;One example of this is $\ce{HCl}$ and $\ce{KOH}$. Why would the acid not just give the hydrogen ion in $\ce{HCl}$ to form $\ce{Cl-}$ and $\ce{KOH2+}$? &lt;/p&gt;&#xA;" OwnerUserId="5675" LastEditorUserId="46306" LastEditDate="2018-12-07T19:07:59.263" LastActivityDate="2018-12-07T19:07:59.263" Title="Why are acids characterized by giving hydrogen and bases characterized by receiving it when acid and base reactions are double displacement reactions?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11164" PostTypeId="2" ParentId="11163" CreationDate="2014-05-29T15:57:48.433" Score="1" Body="&lt;p&gt;$\ce{KOH}$ does not exist as a molecule.  In water it is completely dissociated as $\ce{K+}$ and $\ce{OH-}$.  It does not make sense to write $\ce{KOH_2^+}$ for this reason.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorDisplayName="user15489" LastEditDate="2015-05-17T19:34:24.220" LastActivityDate="2015-05-17T19:34:24.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11165" PostTypeId="2" ParentId="11065" CreationDate="2014-05-29T16:44:40.163" Score="1" Body="&lt;p&gt;I would say it's worth investigating. My suspicion is that the answer won't change much, but the way to find out is to try it both ways and see how the assumptions affect the answer under various conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a CSTR, for this particular reaction a batch reactor might be more appropriate, since that is the easiest setup for a lab. In that case, to model the reaction all we have to do is &lt;em&gt;not&lt;/em&gt; assume that some of the rates of change are zero, and see how that changes the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately for us, this &lt;a href=&quot;http://www.idea.wsu.edu/OscilChem/&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; has already done the heavy lifting and even has a &lt;a href=&quot;http://www.idea.wsu.edu/OscilChem/ltk_ans.htm&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt; in a mathcad file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't looked at the solution closely enough to determine if they are making the approximations you mentioned (I don't have mathcad installed), or if they are just directly integrating all four equations. They only show graphs for the concentrations of X and Y. However, it probably wouldn't be too hard to modify in either direction to compare the results under different sets of assumptions.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-29T16:44:40.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11166" PostTypeId="2" ParentId="11163" CreationDate="2014-05-29T17:10:13.830" Score="1" Body="&lt;p&gt;This is a good question. First, we must be cognizant of what actually exists. If we don't know what actually exists then any chemistry we attempt to do is just a shot in the dark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are talking about &lt;em&gt;solutions&lt;/em&gt;, then $\ce{HCl}$ doesn't exist; there isn't much to any unionized hydrochloric acid in the system. Thus it is misleading to write $\ce{HCl_{(aq)}}$, at least for introductory chemistry students, as one might take this to mean there is unionized hydrochloric acid that is solvated by water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually exists in solutions of strong acids (apart from the counterion) is at the &lt;em&gt;bare minimum&lt;/em&gt; the hydronium ion, $\ce{H_3^+O}$. The bare hydrogen proton, $\ce{H^+}$ &lt;em&gt;does not&lt;/em&gt; exist by itself; it's too reactive. Recent studies have also suggested that the $\ce{H_5^+O_2}$ molecule exists in acidic solution in equal quantity as $\ce{H_3^+O}$. And of course other &lt;em&gt;waters of hydration&lt;/em&gt; may exist due to the sheer positive charge density of a bare proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3UzUm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of $\ce{KOH}$, this is a very soluble salt, and unless we have a saturated solution of $\ce{KOH}$, we generally won't have any unionized $\ce{KOH}$. What we do have however in solution is $\ce{K^+}$ and $\ce{HO^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I suggest that before trying to predict the products of any chemical reactions, write out a complete list of what you have in your system - a &lt;em&gt;principle system inventory&lt;/em&gt;. Ignore the very minor species such as the higher waters of hydration, such as the $\ce{H_9^+O_4}$ molecule - this likely won't be a big deal as far as introductory-level calculations go, and scientists have yet to even fully characterize the solvation of the hydrogen proton in water themselves. Only upon realizing what's really in the system can we do accurate chemistry. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, if we were dealing with pure, liquid $\ce{HCl}$ and solid $\ce{KOH}$, then what we have is simply $\ce{KOH}$ and $\ce{HCl}$. There is nothing to dissolve the potassium hydroxide and no medium for the pure hydrochloric acid to completely ionize in. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Speaking specifically of why $\ce{KOH2+}$ doesn't exist or isn't stable: Note that its components are the $\ce{K^+}$ and $\ce{HO^-}$ ions. Note the opposite charges on these two ions. Remember Coulomb's law about the attraction between electrically charged particles. The opposite charges give $\ce{KOH}$ a substantial lattice energy, allowing it to exist as an ionic solid. If you were to protonate the $\ce{HO^-}$ part of the molecule, however, you'd rob the $\ce{HO^-}$ part of its negative charge. This substantially diminishes the lattice energy of the solid, so the solid wouldn't be stable anymore. Not to mention that this protonation also creates water, and $\ce{KOH}$ is soluble in water. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-29T17:21:45.860" LastActivityDate="2014-05-29T17:21:45.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11167" PostTypeId="2" ParentId="11163" CreationDate="2014-05-29T17:23:08.160" Score="0" Body="&lt;p&gt;You have the right idea - many simple acid/base reactions (using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Arrhenius_definition&quot; rel=&quot;nofollow&quot;&gt;Arrhenius definition&lt;/a&gt;) follow the same pattern as a double-displacement reaction in aqueous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalized pattern for a &lt;strong&gt;double displacement&lt;/strong&gt; reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AB + CD -&amp;gt; AD + CB}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, ionic double displacement reactions are driven by the formation of a precipitate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AB_{(aq)} + CD_{(aq)} -&amp;gt; AD_{(aq)} + CB_{(s)}v}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, the compund $\ce{CB}$ is &lt;strong&gt;insoluble&lt;/strong&gt; in water, and so it &lt;strong&gt;precipitates&lt;/strong&gt; out of solution (indicated by the arrow pointing downward). The reaction goes forward because the $\ce{C}$ ions and $\ce{B}$ ions leave the solution as soon as they meet each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an Arrhenius acid/base reaction, we see a similar pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA_{(aq)} + BOH_{(aq)} -&amp;gt; BA_{(aq)} + H_2O_{(l)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the formation of &lt;strong&gt;water&lt;/strong&gt; drives the reaction - as protons meet hydroxide ions, they form water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the pattern is the same, which is good! However, pay careful attention to the details of the pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA}$ corresponds to $\ce{HCl}$, and $\ce{BOH}$ corresponds to $\ce{KOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, $\ce{A=Cl}$ and $\ce{B=K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting those into the pattern above, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl_{(aq)} + KOH_{(aq)} -&amp;gt; KCl_{(aq)} + H_2O_{(l)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helps to always think about these as ions when in aqueous solution (as @DavePhD mentioned). For this reason, these reaction patterns are typically taught along with solubility and dissociation of ionic compounds in aqueous solution. If you get in the habit of mentally (or explicitly on paper) dissociating all &lt;strong&gt;strong electrolytes&lt;/strong&gt;, the patterns are much easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;strong&gt;ionic equation&lt;/strong&gt; for the general precipitation reaction pattern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A^+_{(aq)} +B^{-}_{(aq)} + C^+_{(aq)} +D^{-}_{(aq)} -&amp;gt; A^+_{(aq)} +D^{-}_{(aq)} + CB_{(s)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the &lt;strong&gt;net ionic&lt;/strong&gt; equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B^{-}_{(aq)} + C^+_{(aq)} -&amp;gt; CB_{(s)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For acid/base reactions (only when both are &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_strength&quot; rel=&quot;nofollow&quot;&gt;strong&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+_{(aq)} +A^{-}_{(aq)} + B^+_{(aq)} +OH^{-}_{(aq)} -&amp;gt; B^+_{(aq)} +A^{-}_{(aq)} + H_2O_{(l)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+_{(aq)} + OH^{-}_{(aq)} -&amp;gt; H_2O_{(l)}}$&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="3846" LastEditDate="2015-05-17T19:33:43.723" LastActivityDate="2015-05-17T19:33:43.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11168" PostTypeId="2" ParentId="11076" CreationDate="2014-05-29T17:33:42.597" Score="3" Body="&lt;p&gt;The classic would be Porter and Easterling's &quot;Phase Transformations in Metals and Alloys&quot;, apparently still available. This should give you a good basis for understanding the thermodynamics of binary (and higher) systems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a deeper look at how computational thermodynamics is done, like CALPHAD, the Lukas, Fries, and Sundman book &quot;Computational Thermodynamics&quot; might do, but that has zero historical background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also try asking this in Physics using the materials-science tag.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-05-29T17:33:42.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11169" PostTypeId="1" AcceptedAnswerId="11171" CreationDate="2014-05-29T17:42:13.173" Score="23" ViewCount="5375" Body="&lt;p&gt;I have asked my teacher, as she was introducing the concept of mole to us, why &lt;em&gt;that&lt;/em&gt; number was chosen, instead of more convenient one. She told me that it came from the definition of the mole, that is the number of atoms in 12g of Carbon 12. When I asked why that definition was chosen, she answered me that my question wasn't really a question, and I understood from her stare that asking &lt;em&gt;why?&lt;/em&gt; twice in a row was not appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I certainly hope that this definition wasn't chosen randomly and that there is a reason for it being it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gave it some thought and I came to believe that it may be for convenience after all. There are 12 nucleons in a carbon atom and if we say that the electrons mass can be neglected, then we get that a mole is the number of nucleon it takes to get one gram of them. However, can we really neglect the mass of electrons, when the atoms get bigger? And if so, why do the element's mass per mole do very rarely end up being integers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me think there may be another reason, or some things I do not understand. And if there are no reason, then I guess I might just as well be doing religious studies..!&lt;/p&gt;&#xA;" OwnerUserId="5678" LastEditorUserId="5678" LastEditDate="2014-05-29T18:39:04.970" LastActivityDate="2019-01-18T21:24:29.323" Title="Why is the definition of the mole as it is?" Tags="&lt;mole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11171" PostTypeId="2" ParentId="11169" CreationDate="2014-05-29T18:12:08.183" Score="25" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the definition of the moles as it is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a rather arbitary definition that the mole is the number of atoms in 12g of carbon 12.  This has not always been the definition.  For example, prior to 1960, the definition was based upon oxygen rather than carbon-12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first standard was based upon 1 gram of hydrogen.  Later, the standard was changed to 16 grams of oxygen being a mole, for convenience because oxygen formed compounds with many other elements.  Eventually it was realized that elements including oxygen have different isotopes of different mass, and the mass of a particular isotope is a more specific and measurable standard than the mass of natural abundance oxygen.  12 grams of carbon-12 matches the old standard of 16 grams of natural abundance oxygen more closely than choosing 16 grams of oxygen 16.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;can we really neglect the mass of electrons, when the atoms get bigger?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mass of a proton or neutron is about 1836 times that of an electron. So depending upon the ratio of neutrons to protons, electrons are at most 1/1836 of the mass of a neutral atom.  If accuracy of more than 1 part in 1836 is desired, it is important to consider the electrons, regardless of whether atoms are big or small.    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why do the element's mass per mole do very rarely end up being integers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No element other than carbon-12 will exactly be an interger.  This is because masses of atoms depend upon number of protons, neutrons, electrons and &lt;a href=&quot;http://en.wikipedia.org/wiki/Binding_energy&quot;&gt;binding energy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, considering that carbon-12 has an equal number of protons and neutrons (and electrons), other isotopes with a 1:1 proton/neutron ratio would have essentially integral atomic masses except for binding energy.  For example cadmium 112 is 111.90 instead of exactly 112 because binding energy is slightly more than 0.1 amu, partially offest by a higher number of neutrons than protons, the neutron mass being slightly greater than the combined mass of an electron and proton.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by Matt Black, the natural abundance atomic masses further deviate because they are weighted averages of the masses of isotopes of a given element, weighted by their abundance on Earth.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-05-30T12:21:00.373" LastActivityDate="2014-05-30T12:21:00.373" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11172" PostTypeId="2" ParentId="10245" CreationDate="2014-05-29T18:54:24.390" Score="8" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why is weight of 1 mole of substance equal to atomic/molecular mass in grams?&lt;/p&gt;&#xA;&lt;p&gt;According to me, it happens because mole has been defined in such a way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes! That is correct.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is defined as the numbers of particles in 12 g of C12. If it were 24 g, instead of 12 g, then the weight of 1 mole of substance would equal 2 times the atomic/molecular mass in grams.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Also correct, assuming that the definition of &lt;strong&gt;unified atomic mass units&lt;/strong&gt; (amu) remained the same. @Martin's answer is correct, but we can also arrive at the same conclusion using a simple dimensional analysis approach.&lt;/p&gt;&#xA;&lt;p&gt;First we need the definition of an amu:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{1 atom}~ \ce{^12C} = \pu{12 amu}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now take the real definition of a mole:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{1 mol}~ \ce{^12C} = \pu{12 g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, divide the first equation by the second:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{12 g}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Cross-multiply and reduce:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What this tells us is that the &lt;strong&gt;ratio&lt;/strong&gt; of g/mol to amu/atom is exactly one - and we made sure it would work out that way by carefully choosing how we defined moles and amus.&lt;/p&gt;&#xA;&lt;p&gt;Since the masses of &lt;em&gt;all&lt;/em&gt; elements are defined in terms of amu, which are ultimately based on the mass relative to carbon-12, this ratio holds for all any atom or molecule.&lt;/p&gt;&#xA;&lt;p&gt;Let's take it further and put in your hypothetical doubling of the number of grams of carbon-12 per mole:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\pu{1 atom}~ \ce{^12C}}{\pu{1 mol}~ \ce{^12C}} = \frac{\pu{12 amu}}{\pu{24 g}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now when we cross-multiply and reduce, we get:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2~\frac{\pu{g}}{\pu{mol}~ \ce{^12C}} = 1~\frac{\pu{amu}}{\pu{atom}~ \ce{^12C}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since we kept the definition of an amu the same, but changed that of the mole, what we see is that 1 mole of carbon-12 using these units would have to weigh 24 g (the &lt;em&gt;molar mass&lt;/em&gt;), even though its &lt;em&gt;atomic mass&lt;/em&gt; would still be 12.&lt;/p&gt;&#xA;&lt;p&gt;That's a lot harder to keep track of mentally, and so it is a good thing for us that the definitions of amus and moles were chosen so carefully!&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-03-09T17:09:02.200" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11173" PostTypeId="1" AcceptedAnswerId="11175" CreationDate="2014-05-29T19:18:43.257" Score="12" ViewCount="3737" Body="&lt;p&gt;The bond enthalpy associated with a $\ce{O=O}$ double bond is equal to 495 kJ/mol. Does that mean that adding enough kinetic energy in the form of heat will eventually cause the bonds to break and create monoatomic oxygen?&lt;/p&gt;&#xA;" OwnerUserId="5679" LastEditorDisplayName="user7951" LastEditDate="2015-05-13T18:47:54.917" LastActivityDate="2016-12-04T17:27:08.440" Title="Will heating diatomic oxygen enough break the O=O bonds?" Tags="&lt;bond&gt;&lt;kinetics&gt;&lt;enthalpy&gt;" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11174" PostTypeId="2" ParentId="11173" CreationDate="2014-05-29T19:25:37.377" Score="2" Body="&lt;p&gt;Yes. Adding enough thermal energy to anything will disrupt the bonds. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-05-29T19:25:37.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11175" PostTypeId="2" ParentId="11173" CreationDate="2014-05-29T19:28:10.230" Score="11" Body="&lt;p&gt;Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 -&amp;gt;C[energy]\ 2O}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-25T16:57:12.077" LastActivityDate="2015-04-25T16:57:12.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11176" PostTypeId="1" AcceptedAnswerId="11182" CreationDate="2014-05-29T19:38:04.890" Score="2" ViewCount="124" Body="&lt;p&gt;I was taught to balance redox equations with acid/base considerations. Instead of arbitrarily adding $\ce{H^+}$ and $\ce{HO^-}$ ions and memorizing separate rules for acidic and basic solutions, I was taught simply to keep these four things in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Charge must be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Mass is similarly conserved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Do what's thermodynamically more favorable. For example, in strongly basic solution, $\ce{O^2-}$ oxygen ions come from not hydronium ion (because that doesn't exist in significant quantity in basic solution) and not from water (because that requires heterolytic clevage of &lt;strong&gt;2&lt;/strong&gt; $\ce{H-O}$ bonds but instead $\ce{HO^-}$ because that only requires clevage of only &lt;strong&gt;one&lt;/strong&gt; $\ce{H-O}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Strongest base reacts with strongest acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, here's a gem from Chem Wiki by UC Davis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l88S8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How come I've never seen my professor's method mentioned &lt;em&gt;anywhere&lt;/em&gt;? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How were you taught?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Which way do you think is more advantageous? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5456" LastEditDate="2014-05-29T21:29:40.503" LastActivityDate="2014-05-29T21:29:40.503" Title="Balancing Redox Equations" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11177" PostTypeId="2" ParentId="11173" CreationDate="2014-05-29T19:38:33.980" Score="1" Body="&lt;p&gt;In principle yes, but it will need much more than 495 kJ/mol for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond enthalpy is based on (experimental) enthalpy of formation. It does not take into account the enthalpy reaction barrier of the $\ce{O2}$ dissociation. &lt;/p&gt;&#xA;" OwnerUserId="445" LastActivityDate="2014-05-29T19:38:33.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11179" PostTypeId="1" AcceptedAnswerId="11181" CreationDate="2014-05-29T19:46:02.657" Score="2" ViewCount="95" Body="&lt;p&gt;What is the $\ce{K_{a}}$ value for $\ce{H^+}$. I understand that the hydrogen proton doesn't stick around in solution by itself for very long. Nonetheless, does it have a  $\ce{K_{a}}$ value? Can it have a  $\ce{K_{a}}$? I think it &lt;em&gt;can&lt;/em&gt; have one but quantifying it might be hard. Its  $\ce{K_{a}}$ value I think would correspond to this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+ + H_2O -&amp;gt;H_3^+O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this doesn't look like a conventional acid/base reaction; we don't have the conjugate base of $\ce{H^+}$ here, unless the definition of a conjugate base isn't what I think it is (acid missing one proton). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does the above mentioned $\ce{K_{a}}$ value exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is my definition of a conjugate base too narrow? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5456" LastEditDate="2014-05-29T20:26:15.637" LastActivityDate="2014-05-30T03:56:50.547" Title="What's the Acid Ionization Constant of the Hydrogen Proton?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11180" PostTypeId="2" ParentId="11173" CreationDate="2014-05-29T19:48:52.777" Score="6" Body="&lt;p&gt;Yes. For quantitative information see &lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS-20.pdf&quot;&gt;Gas Phase Reaction Kinetics of Neutral Oxygen Species&lt;/a&gt;, particularly section 2.1, especially table 4, which gives the equillibrium constant $K_D$ as a function of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_D = \frac{[O]^2}{[O_2]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_D$ exceeds 1 above ~1000K, in units of particles per cubic centimeter.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-05-29T19:48:52.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11181" PostTypeId="2" ParentId="11179" CreationDate="2014-05-29T20:13:56.877" Score="2" Body="&lt;p&gt;The equillibrium between a proton plus water molecule and a hydronium ion is better considered using the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_affinity&quot; rel=&quot;nofollow&quot;&gt;proton affinity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{H^+(g) &amp;amp;-&amp;gt; H^+(aq)} \\&#xA;\Delta E &amp;amp;= −1530~\mathrm{kJ/mol}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could roughly convert to an equillibrium constant using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = -RT \ln(K)$$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2014-05-30T03:56:50.547" LastActivityDate="2014-05-30T03:56:50.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11182" PostTypeId="2" ParentId="11176" CreationDate="2014-05-29T20:39:44.963" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is entirely possible that no one has published that approach! It is certainly not common in introductory chemistry textbooks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) How were you taught?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was taught (and currently teach) the &quot;standard&quot; algorithm you found on the Chem Wiki.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Which way do you think is more advantageous?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on your goal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your goal is to make sure students understand all of the steps and have a good intuition for what is physically happening, then your professor's method makes a lot of sense. This is because you need to already have a clear understanding of what things like &quot;thermodynamically favorable&quot; and &quot;relative acid strength&quot; really mean in order for that approach to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if your goal is to make sure students have the functional ability to work through a particular type of problem, regardless of how well they grasp the underlying concepts, then the algorithmic method makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, if your goal is to solve a problem quickly without having to think too hard about details, the algorithmic method also makes more sense.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-29T20:39:44.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11183" PostTypeId="2" ParentId="11179" CreationDate="2014-05-29T20:46:08.230" Score="2" Body="&lt;p&gt;I don't think $\ce{K_{a}}$ has meaning in that context.  $\ce{K_{a}}$ is defined as the equilibrium constant for the following dissociation reaction $$\ce{HA{&amp;lt;=&amp;gt;}H^+ + A^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+}$ can't dissociate; that is, there is no &quot;$\ce{A}$&quot; in the case your considering.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-05-29T21:20:20.867" LastActivityDate="2014-05-29T21:20:20.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11185" PostTypeId="1" CreationDate="2014-05-29T21:27:11.500" Score="2" ViewCount="249" Body="&lt;p&gt;Where could we find a good description of graph theory applied to the general problem of isomer enumeration and chemical graph theory?&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="4945" LastEditDate="2015-10-23T16:57:04.947" LastActivityDate="2015-10-23T16:57:04.947" Title="Graph theory, isomer enumeration and chemical graph theory" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11186" PostTypeId="2" ParentId="11169" CreationDate="2014-05-29T22:38:28.590" Score="8" Body="&lt;h1&gt;Why is the definition of the moles as it is?&lt;/h1&gt;&#xA;&lt;p&gt;Here what say &lt;a href=&quot;http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf&quot; rel=&quot;noreferrer&quot;&gt;The International System of Units (SI)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;“Atomic  weights” were originally referred to the atomic weight of&#xA;oxygen, by general  agreement taken as 16. But whereas physicists&#xA;separated the isotopes in a mass  spectrometer and attributed the&#xA;value 16 to one of the isotopes of oxygen, chemists  attributed the&#xA;same value to the (slightly variable) mixture of isotopes 16, 17 and&#xA;18,  which was for them the naturally occurring element oxygen.&#xA;Finally an agreement  between the International Union of Pure and&#xA;Applied Physics (IUPAP) and the  International Union of Pure and&#xA;Applied Chemistry (IUPAC) brought this duality to  an end in 1959/60.&#xA;Physicists and chemists have ever since agreed to assign the value&#xA;12, exactly, to the so-called atomic weight of the isotope of carbon&#xA;with mass number  12 (carbon 12, 12C), correctly called the relative&#xA;atomic mass Ar( 12C). The unified  scale thus obtained gives the&#xA;relative atomic and molecular masses, also known as the  atomic and&#xA;molecular weights, respectively&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iupac.org/publications/pac/1992/pdf/6410x1535.pdf&quot; rel=&quot;noreferrer&quot;&gt;Atomic Weight' - The Name,&#xA;Its History, Definition, And Units&#xA;Prepared For Publication By&#xA;P. De Bievre' And H. S. Peiser&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The unit could be fixed arbitrarily. By choice, it was linked  to 12C&#xA;and the macroscopic mass unit, the kilogram of the SI. Nevertheless,&#xA;one should note from the  definition below that the magnitude of that&#xA;link factor of NA does not affect the amount of substance of  any&#xA;entity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h1&gt;Can we really neglect the mass of electrons when the atoms get bigger?&lt;/h1&gt;&#xA;&lt;p&gt;I would say yes, Neutrons mass is 1.00866491588(49) u for a proton is 1.007276466879(91) u while the electrons at rest have a mass of 0.000548579909 u so even if we take the heaviest atom Oganesson that has 118 electrons we will have as a very approximate weight of the electrons 0.06473242927025999 u.&lt;/p&gt;&#xA;&lt;h1&gt;Why do the element's mass per mole do very rarely end up being integers?&lt;/h1&gt;&#xA;&lt;p&gt;At the root, there is the fact that neither the atomic mass unit &lt;code&gt;u&lt;/code&gt; is an integer actually it's 1.660539040(20)×10−27 kg. The kilogram was defined long before the studies of sub-atomic particles so it wasn't designed to be a perfect multiple of the atomic mass unit. However, even if we decide to adjust the kilogram so that the atomic mass unit is 1.00000000000×10−27 kg we will eventually end up with non-integer masses, this is due mainly to two factors:&lt;/p&gt;&#xA;&lt;h2&gt;Nuclear binding energy&lt;/h2&gt;&#xA;&lt;p&gt;The &amp;quot;mass defect&amp;quot; that is related to the fact that: &lt;span class=&quot;math-container&quot;&gt;$$E=mc^{2}$$&lt;/span&gt;&#xA;So the mass is linked to the energy, different elements have different energy due to different &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy&quot; rel=&quot;noreferrer&quot;&gt;Nuclear binding energy&lt;/a&gt; and so different mass!&lt;/p&gt;&#xA;&lt;h2&gt;Neutrons heavier than protons&lt;/h2&gt;&#xA;&lt;p&gt;Neutrons weight 1.00866491588(49) u while protons 1.007276466879(91) u so elements with more protons are heavier.&lt;/p&gt;&#xA;&lt;h2&gt;Isotopic abundance? Not really...&lt;/h2&gt;&#xA;&lt;p&gt;Natural abundance (NA) is not at the root of the problem, your observation is valid not only for Standard atomic weight (Ar, standard,) that take into account the isotopic abundance but also for the weight of single isotopes. The periodic table below shows you that also element without isotopes (blue background) have non integer mass. Of course, if you consider the standard atomic weight of an element ( that is the result of a weighted average so you might already expect that is a non integer... ) the isotopic abundance have a greater impact on the result, but if you consider the atomic mass as defined by &lt;a href=&quot;https://goldbook.iupac.org/html/A/A00496.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC&lt;/a&gt; so you are analyzing a single isotope the root of the problem are the reasons described above.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YPmT1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YPmT1.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Holden, N., Coplen, T., Böhlke, J., et al. (2018). IUPAC Periodic&#xA;Table of the Elements and Isotopes (IPTEI) for the Education Community&#xA;(IUPAC Technical Report). Pure and Applied Chemistry, 90(12), pp.&#xA;1833-2092. Retrieved 18 Jan. 2019, from doi:10.1515/pac-2015-0703&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-01-18T21:24:29.323" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="11187" PostTypeId="2" ParentId="8838" CreationDate="2014-05-29T23:56:51.690" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;With no knowledge of chemistry, is it realistic to undertake a biochemistry class?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say that taking a biochemistry course without &lt;em&gt;any&lt;/em&gt; background in chemistry or biology would be very &lt;strong&gt;difficult&lt;/strong&gt;, but it would be &lt;strong&gt;possible&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The course description lists the following prerequisites: &quot;Basic knowledge in general and organic chemistry&quot;. How important are these subjects in biochemistry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The info page link that you gave would not work for me, so I can't tell how in-depth this course would be, but I know that the undergraduate biochemistry courses I have experience with expect you to be familiar with (at a minimum):&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Chemistry&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bonding theory&lt;/li&gt;&#xA;&lt;li&gt;Molecular structure&lt;/li&gt;&#xA;&lt;li&gt;Nomenclature of organic molecules&lt;/li&gt;&#xA;&lt;li&gt;Polarity and its relation to hydrophobicity/hydrophilicity&lt;/li&gt;&#xA;&lt;li&gt;Basic organic reaction chemistry (particularly Lewis acid/base and functional group chemistry)&lt;/li&gt;&#xA;&lt;li&gt;Basic reaction kinetics&lt;/li&gt;&#xA;&lt;li&gt;Basic thermodynamics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Biology&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Macromolecules (proteins, carbohydrates, fats)&lt;/li&gt;&#xA;&lt;li&gt;Cell structures&lt;/li&gt;&#xA;&lt;li&gt;Basic anatomy/physiology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these are fundamental to just about every concept in biochemistry, and they will probably expect you to already be familiar with if not good at them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are some good references (books) for self-learning these subjects? Is Stryer's Biochemistry a good book for self-learning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That I can't answer. I can tell you that &lt;a href=&quot;http://amzn.com/0470570954&quot; rel=&quot;nofollow&quot;&gt;Voet &amp;amp; Voet's &quot;Biochemistry&quot;&lt;/a&gt; is a good biochem book, but I am not sure if it would be suitable for self-learning. I don't have experience with Stryer's book. Voet &amp;amp; Voet also have a &lt;a href=&quot;http://amzn.com/0470547847&quot; rel=&quot;nofollow&quot;&gt;&quot;Fundamentals of Biochemistry&quot;&lt;/a&gt; book that might be better - but again, I don't have any experience with that one and can't compare to Stryer. In my personal experience, I have found that reading as many different books as possible is the best way to self-study, but your results may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give you some suggestions for general chem and organic books:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;General Chemistry&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://amzn.com/0321696727&quot; rel=&quot;nofollow&quot;&gt;Brown, Lemay, and Bursten's &quot;Chemistry: The Central Science&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://amzn.com/0321809246&quot; rel=&quot;nofollow&quot;&gt;Tro's &quot;Chemistry: A Molecular Approach&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are both very easy to read and clearly explain all of the fundamental concepts, with lots of example problems worked out in detail.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Organic Chemistry&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://amzn.com/0321768418&quot; rel=&quot;nofollow&quot;&gt;Wade's &quot;Organic Chemistry&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is the only organic chem book I have much experience with, but I remember it being very easy to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For biology textbooks, I'm afraid I can't be much help.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are biochemistry and biostatistics related at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only in the sense that biostatistics is statistical methods applied to problems found in biology, and many problems found in biology are related to biochemistry. You would certainly want to learn some basic biochemistry if you were going to work on bioinformatics problems, for example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there many applications of statistics in biochemistry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are - particularly in areas like design of experiments, genomics, proteomics, and drug screening, and other various assays.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-29T23:56:51.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11188" PostTypeId="2" ParentId="8820" CreationDate="2014-05-30T00:02:45.583" Score="1" Body="&lt;p&gt;If you are assuming a pseudo-steady state condition (where the bulk of the liquid is in thermal equilibrium with itself for the timescale of interest), and you want to study the bulk molecules of the liquid, then you want to use the temperature of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are going to be studying the &lt;strong&gt;surface&lt;/strong&gt; of the liquid, using a Maxwell-Boltzmann distribution is much more questionable - it definitely isn't strictly correct, at least on time scales of any substantial length, but you might be able to get away with it for rough approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you gave some more details as to the specific application, I could probably give you a better answer.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-30T00:02:45.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11189" PostTypeId="2" ParentId="9107" CreationDate="2014-05-30T00:06:49.157" Score="1" Body="&lt;p&gt;Just above the line, in the two-phase field where $\epsilon$ is in equilibrium with the liquid, the chemical potentials of both Cu and Sn are constant (very likely not equal to each other mind you).  This is a consequence of the tie-line on the Gibbs free energy diagram - check out a book on binary alloy thermodynamics such as Porter and Easterling, Phase Transformations in Metals and Alloys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the peritectic point itself, the tie line connects all three phases, $\epsilon$, $\eta$ and the liquid, so the chemical potentials for each element are equal. Dropping the temperature just a smidge leads to two separate two-phase regions, one of $\epsilon$ with $\eta$, and one with $\eta$ and liquid.  The chemical potential of Sn (or Cu) is different in each of those two-phase regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, Porter and Easterling would be a good reference.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-05-30T00:06:49.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11190" PostTypeId="2" ParentId="5091" CreationDate="2014-05-30T01:07:23.207" Score="3" Body="&lt;p&gt;I can't tell you for sure what &lt;em&gt;is&lt;/em&gt; happening, but I can tell you what &lt;em&gt;could be&lt;/em&gt; happening.&lt;/p&gt;&#xA;&lt;h1&gt;Physical Adsorption&lt;/h1&gt;&#xA;&lt;p&gt;It is possible that the proteins are not &lt;em&gt;chemically&lt;/em&gt; adsorbed, but rather &lt;em&gt;physically&lt;/em&gt; adsorbed. The article you cited implies that this is the case.&lt;/p&gt;&#xA;&lt;p&gt;Many proteins in the presence of nano-scale hydrophobic &amp;quot;features&amp;quot; will partially denature, and the hydrophobic core will physically adsorb (attach via van der Waals forces). This effect is amplified if water is present, since the hydrophilic parts of the protein will be stabilized in water, allowing more of the hydrophobic parts to attach to the carbon feature.&lt;/p&gt;&#xA;&lt;p&gt;Now you are probably wondering, &amp;quot;Why doesn't it attach to the PEG surface as well, since that is hydrophobic?&amp;quot; There are two possible reasons.&lt;/p&gt;&#xA;&lt;h2&gt;PEG is not as hydrophobic as you think&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hvwtZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The PEG chains have oxygen in each monomer, which would reduce the non-polar character of the chains. It is possible that the difference is enough to make adsorption of non-polar amino acids on PEG thermodynamically unfavorable, while on pure-carbon structures it would be favorable.&lt;/p&gt;&#xA;&lt;h2&gt;Shape matters&lt;/h2&gt;&#xA;&lt;p&gt;Another possibility is that the actual shape of the features makes the difference. In order for a protein to denature and &amp;quot;attach&amp;quot; to carbon feature via hydrophobic interactions, it has to first expose the hydrophobic groups in the core. For a flat surface, the amount of initial denaturization (maybe not a real word) would need to be higher than for a &amp;quot;sharp&amp;quot; feature. Sort of like how a needle can slip into your skin without doing much damage. So, the smaller the characteristic length of the features, the more likely the protein would adsorb. There are also protein-protein effects on the surface, which are again reduced by smaller-scale features. In this scenario, you could think of the proteins as kinetically trapped in solution when near a regular PEG surface, while the activation energy for adsorption to a small feature is smaller.&lt;/p&gt;&#xA;&lt;p&gt;I can think of two ways to test this idea:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use different proteins on the same type of surface features - the ones with less exposed hydrophobic groups should adsorb less easily&lt;/li&gt;&#xA;&lt;li&gt;Use different sizes of features - there should be a feature size that gives a maximum adsorption for a given protein.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h2&gt;Chemical Bonding&lt;/h2&gt;&#xA;&lt;p&gt;The other possibility is chemical bonding between the carbon features and the protein. It is unlikely that this will happen if the carbon features are really pure carbon (graphene, for example). However, if they are creating them by ionizing PEG with an electron beam, then it is highly likely that they are &lt;strong&gt;functionalized&lt;/strong&gt; with &lt;em&gt;something&lt;/em&gt;. Carbonyl groups would be my guess, but depending on the conditions there could be nitro groups as well. In this case, there is a possibility that hydrogen bonds would form, and if the surface is reactive enough, chemical bonds might form as well.&lt;/p&gt;&#xA;&lt;p&gt;To test this idea, I would do the same thing as before - only this time, if we are right:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Proteins with more exposed hydrophobic groups should be &lt;strong&gt;less likely&lt;/strong&gt; to adsorb&lt;/li&gt;&#xA;&lt;li&gt;Feature size should not matter at all&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;If this turned out to be the case, you could then systematically test polypeptides made up of different amino acids to verify it and determine whether hydrogen bonding or covalent bonding was more likely.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-05-30T01:07:23.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11193" PostTypeId="2" ParentId="8542" CreationDate="2014-05-30T06:27:47.370" Score="4" Body="&lt;p&gt;In the oxidation of aldehyde, the hydrogen attached to carbonyl compound is replaced by an $\ce{-OH}$ group resulting in the formation of carboxylic compound. In case of ketones, bulkier alkyl groups ($\ce{-R}$) should be replaced by an $\ce{-OH}$ group to form carboxylic compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that hydrogen abstraction is much easier than that of bulkier alkyl groups. Likewise, after the oxidation of the aldehyde, the carbon chain containing the $\ce{C=O}$ group remains same but due to loss of $\ce{-R}$ group during oxidation, the length of the carbon chain decreases in ketones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEXXX.png&quot; alt=&quot;schematics of the reactions&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4945" LastEditDate="2014-05-30T08:13:17.297" LastActivityDate="2014-05-30T08:13:17.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11194" PostTypeId="2" ParentId="7918" CreationDate="2014-05-30T07:01:36.803" Score="1" Body="&lt;p&gt;Fuel cells don't convert water into H and O.  The generation of hydorgen gas is done at some point prior to being introduced into the cell (as in the case of hydrogen combustion engines).  The first stage of the fuel cell uses a catalyst, like platinum, to split hydrogen gas (H2) into atomic hydrogen.  Since the catalyst is a conductor and is sandwiched next to a proton-permeable membrane (proton exchange membrane), the electrons from the hydrogen are also separated and conducted to the final stage of the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fuel cell itself was generating hydrogen from H2O, the system as a whole would generate no net energy--all of the energy obtained from the hydrogen reaction at the end would need to go back into separating the H2O in the first place.&lt;/p&gt;&#xA;" OwnerUserId="5687" LastActivityDate="2014-05-30T07:01:36.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11195" PostTypeId="1" CreationDate="2014-05-30T07:19:24.747" Score="2" ViewCount="105" Body="&lt;p&gt;I am trying to prepare some iron(III) naphthenate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the raw materials are : ferric(III) chloride, and naphthenate acid,  NaOH, toluene, deionized water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything worked well in lab scale: final iron(III) naphthenate dark red in color and soluble in toluene was obtained easily; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I encountered problem when scaling it up: it seems there are undesired side reactions, the finish goods looked like black precipitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that because naphthenate acid was oxidised by Iron(III)? or is there any other side reaction when being scaled up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I first dissolve &lt;code&gt;NaOH&lt;/code&gt; and &lt;code&gt;FeCl3&lt;/code&gt; into water separately;&lt;/li&gt;&#xA;&lt;li&gt;add &lt;code&gt;NaOH&lt;/code&gt; aqueous solution into naphthenate acid while stirring at about 60°C;&lt;/li&gt;&#xA;&lt;li&gt;add &lt;code&gt;FeCl3&lt;/code&gt; aqueous solution into the soldium naphthenate solution slowly;&lt;/li&gt;&#xA;&lt;li&gt;keep reaction temperature at about 80°C for a while and remove the water;&lt;/li&gt;&#xA;&lt;li&gt;reflux distillation to obtain finish goods (the maximum temperature might be high)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="3793" LastEditDate="2014-05-30T10:59:13.030" LastActivityDate="2014-05-30T10:59:13.030" Title="What are the possible side reactions when preparing iron(III) naphthenate?" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11196" PostTypeId="2" ParentId="11155" CreationDate="2014-05-30T11:45:40.953" Score="2" Body="&lt;p&gt;Tell your mentor that to be concerned about chlorobenzene toxicity on perovskites is riddiculous. The most toxic thing in perovskites is the lead iodide...chlorobenzene´s toxicity is much lower than that. If he is so worried about toxicity, he should not work on perovskites.&lt;/p&gt;&#xA;" OwnerUserId="5690" LastActivityDate="2014-05-30T11:45:40.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11197" PostTypeId="2" ParentId="11080" CreationDate="2014-05-30T12:21:29.107" Score="1" Body="&lt;p&gt;$\alpha$-alumina is thermodynamically stable at room temperature, and also at high temperature (around $1,200-1,400 \space ^{\circ}C$) - however, alpha alumina is not what is typically formed when alumina is made from the various aluminum ores (&lt;a href=&quot;http://ref.scielo.org/qprp3k&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;). Typically, one of the several other crystal structures will form, with different particle sizes and amounts of water (hydration). These and all other phases that are found along the way to $\alpha$-alumina are called &lt;strong&gt;transition aluminas&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://alteo-alumina.com/en/transition-aluminas&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; explains what happens upon heating. In general, first the crystal is dehydrated, which causes it to form a different, dehydrated phase. Then, as the temperature continues to rise, the alumina crystal structure rearranges into more compact structures, until eventually, near the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sintering&quot; rel=&quot;nofollow&quot;&gt;sintering point&lt;/a&gt;, the atoms have sufficient kinetic energy that they can &quot;find&quot; the most thermodynamically stable configuration - $\alpha$-alumina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the transition alumina phases are kinetically trapped in structures that are &lt;em&gt;not&lt;/em&gt; the most thermodynamically stable, but heating allows them to get out of that trap and into the most favorable state.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-05-30T12:21:29.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11198" PostTypeId="1" CreationDate="2014-05-30T12:49:17.083" Score="2" ViewCount="513" Body="&lt;p&gt;&lt;strong&gt;Scenario&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you randomly spread in a cubic space of volume $m^3$ (or in a sphere of volume $V_s$, as you prefer), $n_A$ particles $A$ and $n_B$ particles $B$. Particles $A$ are spheres of volume $V_A$ and are not moving. Particles $B$ have no volume (they are points) and are moving linearly (reflecting at the boundaries of the space (cube or sphere)) at a constant speed $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that when a particle $B$ meet a particle $A$ (collision), we count a collision but the particle $B$ just go through the particle $A$. Whatever is the model that is easier to calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What function describes the probability that there is $x$ collisions in a time interval $\Delta t$. Or what is the expected the number of collisions?&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="5691" LastEditDate="2017-08-02T05:19:13.293" LastActivityDate="2017-08-02T05:19:13.293" Title="Number of collisions of particles in a box (or a sphere)" Tags="&lt;physical-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11199" PostTypeId="2" ParentId="11173" CreationDate="2014-05-30T14:51:40.737" Score="0" Body="&lt;p&gt;It sure will. But as soon as the gas cools down it will just re-form into O&lt;sub&gt;2&lt;/sub&gt; unless you give it something else to react with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to get it to react with fluorine at around 700 degrees by heating O&lt;sub&gt;2&lt;/sub&gt; and F&lt;sub&gt;2&lt;/sub&gt; through it. You now have monatomic O and F ready to party, the fluorine in particular &quot;&lt;a href=&quot;http://pipeline.corante.com/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;losing it's gentle and forgiving nature&lt;/a&gt;&quot; when reduced to single life. After cooling (to 90K), you will have a flask full of FOOF. That's assuming you still have an intact lab (hint: please do not ever try this, for your sake or the sake of others in your vicinity &lt;em&gt;--Ed.&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5692" LastEditorUserId="16683" LastEditDate="2016-12-04T17:27:08.440" LastActivityDate="2016-12-04T17:27:08.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11200" PostTypeId="2" ParentId="10939" CreationDate="2014-05-30T15:01:11.073" Score="5" Body="&lt;p&gt;The other answers explain things in terms of the activation energy barrier, but they don't explain &lt;strong&gt;why&lt;/strong&gt; the activation energy barrier is lowest for K. I will attempt to do that here.&lt;/p&gt;&#xA;&lt;p&gt;When a metal reacts with water, several things must happen (not necessarily in this order):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One or more electrons have to be removed from the metal&lt;/li&gt;&#xA;&lt;li&gt;One or more &lt;span class=&quot;math-container&quot;&gt;$O-H$&lt;/span&gt; bonds have to break on water molecules (one per electron)&lt;/li&gt;&#xA;&lt;li&gt;The electron(s) must be &amp;quot;picked up&amp;quot; by the &lt;span class=&quot;math-container&quot;&gt;$OH$&lt;/span&gt; to become &lt;span class=&quot;math-container&quot;&gt;$OH^-$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;As @ron said, the &amp;quot;vigorousness&amp;quot; of this process depends on the &lt;strong&gt;kinetics&lt;/strong&gt; of the reaction, which are in turn determined by the activation energy of the rate-limiting step.&lt;/p&gt;&#xA;&lt;p&gt;In the links that he provided (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/s-Block_Elements/Group__1:_The_Alkali_Metals/Properties_of_the_Group_1_Elements/Reactions_of_Group_1_Elements_with_Water&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/s-Block_Elements/Group__2_Elements:_The_Alkaline_Earth_Metals/Chemistry_of_the_Group_2_Elements/Reactions_of_Group_2_Elements_with_Water&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;), they show that the activation energy is lowest for K, which means it would react most vigorously. (note - these are not &lt;em&gt;real&lt;/em&gt; activation energies - the atoms are not actually atomized to the gas phase, and the electrons are not removed under vacuum to infinite distance, but they still give us a good &lt;em&gt;relative&lt;/em&gt; estimate)&lt;/p&gt;&#xA;&lt;p&gt;However, knowing that the kinetics of the reaction depends on the activation energy won't help you if you don't happen to have a chart of activation energies handy. On an exam, you certainly won't, and it is good to develop an intuition for these sorts of things anyhow. And so, you might ask:&lt;/p&gt;&#xA;&lt;h1&gt;Why is the activation energy lower?&lt;/h1&gt;&#xA;&lt;p&gt;Looking at the links, you can see that the activation energies are dominated by the ionization energy. In other words, when it takes more energy to remove electrons, the reaction slows down.&lt;/p&gt;&#xA;&lt;p&gt;Why should K have the lowest ionization energy? First we need to look at the equation that governs the potential energy between charges:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E=\frac{{\kappa}Q_1Q_2}{d}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here &lt;em&gt;E&lt;/em&gt; is electrostatic potential energy, &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Coulomb's constant&lt;/a&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Q_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q_2$&lt;/span&gt; are charges.&lt;/p&gt;&#xA;&lt;p&gt;You can see from this that if the charges have the opposite sign, then energy will be become more negative as they move closer together (inversely proportional to distance). As either of the charges increase, the energy will become more negative as well (directly proportional to charge). In other words, big charges are harder to pull apart, and charges that are close together are harder to pull apart than charges that are far apart. As a rough analogy, imagine pulling apart two magnets - it is more difficult when they are close together.&lt;/p&gt;&#xA;&lt;p&gt;Now compare K to the other group I element (Na). K is further down the column. Since atomic radius increases with the row number (due to the fact that orbital radius increases with principle quantum number &lt;strong&gt;n&lt;/strong&gt;), the valence electron on K is further from the nucleus (on average) than it is on Na. This means that it is easier to remove a valence electron from K. As an analogy, imagine trying to move a rock up a hill. You have to do less work (put in less energy) if the rock is already halfway up the hill. Since the electrons are further from the nucleus in K, they are already &amp;quot;further up the hill.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;What about group II elements? As you progress from left to right, the atomic radius &lt;em&gt;decreases&lt;/em&gt; - this is because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;noreferrer&quot;&gt;effective nuclear charge&lt;/a&gt; seen by the valence electrons increases. The combination of increased charge and decreased distance leads to a steeper &amp;quot;hill&amp;quot; that must be overcome. On top of that, now we have to remove &lt;em&gt;two&lt;/em&gt; electrons for the reaction to work, which is roughly twice as difficult (technically we don't &lt;em&gt;have&lt;/em&gt; to remove both, but the overall reaction thermodynamics won't work out if we don't). As a result, group I elements will &lt;em&gt;always&lt;/em&gt; react more vigorously with water than group II will.&lt;/p&gt;&#xA;&lt;p&gt;I think that it helps to imagine reactions as sort of a &amp;quot;fight&amp;quot; between elements over electrons. Who will win the fight and how it will go is determined by the number of protons in the nucleus and the electron orbital configuration. This approach is useful because it applies to more advanced chemistry as well.&lt;/p&gt;&#xA;&lt;p&gt;For some demonstrations of alkali metal reactions with water, check out this video:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.open.edu/openlearn/science-maths-technology/science/chemistry/alkali-metals&quot; rel=&quot;noreferrer&quot;&gt;Alkali Metals&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-05-30T15:01:11.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11201" PostTypeId="2" ParentId="11042" CreationDate="2014-05-30T15:02:48.327" Score="3" Body="&lt;p&gt;Activated carbon works by enlarging the surface area of elemental carbon. Activated carbon can reach surface areas of several hundred square meters to several thousand square meters per gram. Also elemental carbon is fairly non reactive, non toxic and cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDci9.jpg&quot; alt=&quot;Activated carbon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen in the picture, activated carbon forms a micro structure with incredibly small pores, so that many molecules can be adsorbed (the forces involved decrease with a factor $1/r^{6}$ with distance). The adsorption itself is an application of the London dispersion force (one of the Van der Waals forces).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe chemisorption has anything to do with this.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2016-11-24T15:49:58.947" LastActivityDate="2016-11-24T15:49:58.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11202" PostTypeId="1" AcceptedAnswerId="11203" CreationDate="2014-05-30T16:53:33.673" Score="2" ViewCount="2882" Body="&lt;p&gt;Is it preferable to balance redox equations with  $\ce{H+}$ or $\ce{H3O+}$? I think the latter, because first, the bare hydrogen proton doesn't exist in water solution. So why write down something we know not to exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because in this problem, if we balance using both species, $\ce{H^+}$ and $\ce{H3O+}$, we achieve two different equations. Mass is conserved in both equations as well as charge; all the differs is one coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are the two equations equivalent in that they are still describing the same reaction? I think so, because sometimes I see textbook authors write these two equations for $K_\mathrm{w}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O &amp;lt;=&amp;gt; H+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we were actually going to run an experiment and needed exact quantities, we would probably prefer the second reaction both above and below because the coefficients in these reactions better represent the chemistry of the system right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;em&gt;which of the following is &quot;better&quot;?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO_{3}^- + 6I^- + 6H^+ \leftrightharpoons 3I_2 + Cl^- + 3H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO_{3}^- + 6I^- + 6H_3^+O \leftrightharpoons 3I_2 + Cl^- + 9H_2O}$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2016-10-22T15:03:41.643" LastActivityDate="2016-10-22T15:03:41.643" Title="Balancing redox equations in acidic media: to use H+ or H3O+?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11203" PostTypeId="2" ParentId="11202" CreationDate="2014-05-30T17:07:21.543" Score="6" Body="&lt;p&gt;This is a question of &quot;style&quot; so there is no correct answer.  In fact, just as $\ce{H^+}$ doesn't really exist in solution, neither does $\ce{H3O+}$ exist in solution.  In reality higher hydrates of $\ce{H^+}$ are what exist in solution. See the first paragraph in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;.  Again, both $\ce{H^+}$ and $\ce{H3O+}$ are just shorthand notations for reality, choose the style that you like best.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-30T17:07:21.543" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11204" PostTypeId="2" ParentId="11198" CreationDate="2014-05-30T17:26:48.833" Score="0" Body="&lt;p&gt;With assumption that $n_A$ and $n_B$, i.e., the number of particles $A$ and $B$, are large and are in random motion, then the probability of a particle $B$ to be found within the volume of particles $A$ is $$\frac{n_A V_A}{V_s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will hold for every particle $B$. So, when each particle moves $s \Delta t$ distance, the number of collisions it will undergo is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{s\Delta t\cdot n_A V_A}{V_s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n_B$ particles the number of collisions is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{n_B \cdot s \Delta t \cdot n_A V_A}{V_s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With my assumption at top, it will either approach 1 or 0.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="16683" LastEditDate="2017-08-02T04:34:35.217" LastActivityDate="2017-08-02T04:34:35.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11205" PostTypeId="1" AcceptedAnswerId="11207" CreationDate="2014-05-30T17:28:32.597" Score="0" ViewCount="65" Body="&lt;p&gt;How come the oxygens &lt;em&gt;transferred&lt;/em&gt; in redox reactions are always the $\ce{O^{2-}}$anion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have this set of rules, and the rules are implicitly referring to the $\ce{O^{2-}}$  anion, a potent base (otherwise the rules wouldn't make sense; for example, it makes sense that a base in acidic solution is protonated to water, and that a strong base in basic solution is leveled to hydroxide ion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how come, when electrons are transferred, such as in this unbalanced reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO_{3}^- + 6I^- \leftrightharpoons 3I_2 + Cl^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{O^{2-}}$ anion is formed (if only to be consumed again)? I remember my prof would say that specifically &lt;strong&gt;3&lt;/strong&gt; $\ce{O^{2-}}$ anions &quot;disappear&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not the $\ce{O^{-}}$ anion or simply $\ce{O}$? Does this have to do with stability? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bSRpX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2015-05-19T07:31:44.210" LastActivityDate="2015-05-19T07:31:44.210" Title="What is the origin of the transferred oxygendianion in redox reactions?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11206" PostTypeId="2" ParentId="11134" CreationDate="2014-05-30T18:05:35.880" Score="0" Body="&lt;p&gt;the gas/liquid system has reached equilibrium,beacause pressure is defined only when gas is in equilibrium otherwise gas will not have same pressure throughout gas.A3. the concentration of gas in the liquid is low, and A4. the gas pressure is low.because gas will deviate from ideal nature and there will be more collisions which will deviate it from following Henry's LAw.A2,the gas does not chemically react with the liquid because laws in chemistry that relates physical properties assumes that the interaction between particles is zero or negligible which is not possible in chemical reaction&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-05-30T18:05:35.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11207" PostTypeId="2" ParentId="11205" CreationDate="2014-05-30T18:37:34.573" Score="2" Body="&lt;p&gt;I think you are misunderstanding the meaning of &lt;a href=&quot;http://en.wikipedia.org/wiki/Superoxide&quot; rel=&quot;nofollow&quot;&gt;superoxide&lt;/a&gt; and the rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superoxide is $\ce{O_2^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superoxide is NOT $\ce{O^{2-}}$ as you have written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are referring to hydroxide which is $\ce{OH^-}$, not superoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason hydroxide is used to balance equations in basic aqueous solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_2O \leftrightharpoons H^+ + OH^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is hydroxide present in aqueous solution, but not superoxide, peroxide, hydroxyl radial, etc. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-05-30T19:17:53.757" LastActivityDate="2014-05-30T19:17:53.757" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11208" PostTypeId="2" ParentId="11173" CreationDate="2014-05-30T19:19:36.667" Score="0" Body="&lt;p&gt;Heating it even further will rip all the electrons away from the nuclei themselves to create a plasma state of free floating positive nuclei and electrons whereby, if enough energy was put into the system, the oxygen nuclei would fuse &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen-burning_process&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Oxygen-burning_process&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5693" LastActivityDate="2014-05-30T19:19:36.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11212" PostTypeId="1" AcceptedAnswerId="17048" CreationDate="2014-05-31T02:12:55.353" Score="4" ViewCount="4559" Body="&lt;p&gt;A leak in the air conditioning system of an office building releases $12~\mathrm{kg}$ of $\ce{CHF2Cl}$ per month. If the leak continues, how many kilograms of $\ce{Cl}$ will be emitted into the atmosphere each year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working out is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;12~\mathrm{kg} &amp;amp;= \mathrm{CHF}_2\mathrm{Cl} \\ &#xA;\text{Cl mass in compound} &amp;amp;= \frac{35.45}{86.448}\times100\% = 41\% \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$41\%$ of $12~\mathrm{kg}$ is $4.92~\mathrm{kg}$, so I multiplied it by $12$ to get the annual result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;4.92~\mathrm{kg} \times 12(\text{months}) = 59.04~\mathrm{kg\ year}^{-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have access to the answer so I'm not sure whether or not my answer is correct. Also, this process seems a little to simple to be correct. I say this because the question prior to it was very difficult. Can someone please tell me if I am right, and if I'm not, where did I go wrong?  &lt;/p&gt;&#xA;" OwnerUserId="4685" LastEditorUserId="4945" LastEditDate="2015-02-26T07:34:32.147" LastActivityDate="2015-02-26T07:34:32.147" Title="How to calculate the amount of chlorine emitted yearly by a leak in the air conditioning system?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11213" PostTypeId="1" CreationDate="2014-05-31T03:58:11.053" Score="3" ViewCount="2868" Body="&lt;p&gt;I have failed in permanently suspending cocoa powder into water.  I have tried heat (212'f) and pressure (40 psi) into a sugar water solution (50/50).  Also, powdered milk and evaporated milk have failed to yield a permanent suspension.  There was another method of Irish moss to coagulate the whole thing together, but that failed as well.  I need the bonding/ emulsifying agent to be edible, delicious and shelf stable.  I am a low level cook and do not have the chemistry knowledge to do this myself. &lt;br/&gt;&lt;br/&gt;&#xA;Thank you for reading.&lt;/p&gt;&#xA;" OwnerUserId="5699" LastActivityDate="2017-02-11T17:14:10.630" Title="Permanent suspension of cocoa powder into water" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11215" PostTypeId="1" CreationDate="2014-05-31T07:38:48.720" Score="-1" ViewCount="79" Body="&lt;p&gt;I have a question about the refrigerant's pressure in the inlet of an evaporator. Why should the sat-temp of the refrigerant be below the required temp of the cooled compartment ? Shouldn't the refrigerant's temp at the exit of the evaporator be equal to that of the cooled compartment ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks all. &lt;/p&gt;&#xA;" OwnerUserId="5700" LastEditorUserId="15477" LastEditDate="2017-06-26T01:08:30.630" LastActivityDate="2017-06-26T01:08:30.630" Title="Why should the sat-temp of refrigerant be below the required temp of the cooled compartment?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11216" PostTypeId="2" ParentId="10641" CreationDate="2014-05-31T10:34:14.073" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulphate solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The structure is octahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fe ion is at the center of the octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five water molecules and the NO molecule occupy the vertices of the octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfate is a separate spectator ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall charge of the iron coordination complex is 2+.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is brown ring complex, why does the complex sphere break on heating to give $\ce{NO}$? In general, when do the complex spheres break ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Brown ring&quot; refers to the macroscopic appearance of the product in a tube.  This is unrelated to the chemical structure.  The binding of NO to Fe is an exothermic reaction.  In accordance with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot;&gt;Van't Hoff Equation&lt;/a&gt;, the equillibrium constant shifts in favor of the reverse reaction as temperature increases.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For detailed infomation see &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ic010628q&quot;&gt;Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of Nitric Oxide to Aquated Iron(II)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, for many years it was thought the iron was reduced to Fe(I), the NO being oxidized to NO+; however, according to the above reference, the current thinking is that the iron is oxidized to Fe(III), and the NO is reduced to NO-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the complex is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-02T18:03:29.307" LastActivityDate="2014-06-02T18:03:29.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11217" PostTypeId="1" CreationDate="2014-05-31T13:16:11.760" Score="5" ViewCount="21811" Body="&lt;p&gt;This was a statement in my book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution of a weak acid cannot be titrated with a weak base using an indicator to find&#xA;the end-point because &lt;strong&gt;the pH change is too gradual close to the equivalence point.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please elaborate the bold text.&lt;/p&gt;&#xA;" OwnerUserId="2595" LastActivityDate="2017-04-07T00:17:25.103" Title="No indicator is suitable when titrating weak acid with weak alkali" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11218" PostTypeId="2" ParentId="10976" CreationDate="2014-05-31T15:02:25.570" Score="6" Body="&lt;p&gt;I wish to synthesize a new answer by combining Martin and ron's answers.  I think the real issue is to ensure that the cylindrical symmetry of the electron density of $\ce{CO_2}$ is preserved: that is, $\ce{CO_2}$ should be like a &quot;rod&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below, I have drawn two &quot;resonance structures&quot; (valence bond configurations) that must be combined in order to generate the cylindrical symmetry of the electron density, while maximizing the conceptual relationship with resonance structures.  I have chosen to use a) unhybridized C and b) two ${sp}^2$ fragments for the O's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small missive: Of course, multiconfigurational MO theory and multiconfigurational VB theory are equivalent.  However, certain problems can be describe most economically in either MO theory or VB theory.  This is a case where the electronic structure is probably easiest in MO theory, since it requires only one configuration to have cylindrical symmetry (+1 for Martin's answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your question is about &quot;resonance structures&quot; for $\ce{CO_2}$ (+1 for ron's answer), but the constraints on the density mandate the use of the admixture of two resonance structures.  If we use only one resonance structure, it is easy to see from the lone pairs that the density would not be cylindrically symmetric!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please note that I am not sure about the sign of the combination, it may be - since we are assuming a singlet state.  And maybe I am overthinking it. Perhaps it should be a double-headed arrow for resonance structures.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mtGtO.png&quot; alt=&quot;CO2ResonanceStructures&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-05-31T16:00:12.187" LastActivityDate="2014-05-31T16:00:12.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11219" PostTypeId="1" CreationDate="2014-05-31T15:14:53.247" Score="3" ViewCount="6460" Body="&lt;p&gt;From a chemistry point of view I would like to learn about possible hazards of fog machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such devices, also called “foggers”, create vapor which appears like fog or smoke. It is used in discos, parties, etc.&#xA;It works by heating “fog juices” which are made of glycerine and distilled/de-ionized water mix or propylene glycol and distilled water mix.&#xA;I don't know the temperature these mixes are heated by the machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know glycerine and propylene glycol are not hazardous, but I would like to know if heating them can pose any risk. I've talked with someone who works in this industry and he couldn't tell if there has been any study to show if the fog produced by these foggers is harmful or not.&#xA;So even though these devices have existed for years I would still like to know if they can be harmful and if they should be avoided when possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you can help as chemist is by telling me if vaporizing glycerine and distilled water mix at certain temperature can cause any change to the supposedly harmful substance and make it dangerous or not.&lt;/p&gt;&#xA;" OwnerUserId="5702" LastEditorUserId="4945" LastEditDate="2015-12-14T09:24:21.430" LastActivityDate="2015-12-14T09:24:21.430" Title="What are possible hazards of fog machines?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;heat&gt;&lt;environmental-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11220" PostTypeId="2" ParentId="11173" CreationDate="2014-05-31T15:39:02.967" Score="4" Body="&lt;p&gt;It is important to point out that heat is a bulk (statistical mechanical) concept while bond strength is largely quantum mechanical.  If I am not mistaken, heat/temperature can be defined using pictures like molecules striking the edges of a container, or perhaps from energy/entropy relationships that involve Avagadro's number of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question, this means that given a large enough thermal energy, other molecules of $\ce{O2}$ will likely collide with an $\ce{O_2}$ molecule in question -- thereby bringing their nuclear kinetic energy to bear, and perhaps causing a dissociation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that an isolated molecule of $\ce{O_2}$ (or anything for that matter) will spontaneously rupture because of heat, because here heat is not a well-defined quantum mechanical concept, as it is mediated by many moving atoms.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-06-01T01:09:27.973" LastActivityDate="2014-06-01T01:09:27.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11221" PostTypeId="1" AcceptedAnswerId="11224" CreationDate="2014-05-31T15:41:49.937" Score="3" ViewCount="3733" Body="&lt;p&gt;This is the question from the text book: &lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/v4gQ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I did, since it worked with all the previous questions (similar to this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\frac{4}{3}\pi(7*10^8)^3*1.4}{1.008} * 6.022*10^{23}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equals $1.2 *10^{51}$, but the answer is $1.2*10^{57}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I done wrong? Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="4685" LastActivityDate="2015-12-10T20:14:35.570" Title="Finding out the number of atoms in the sun." Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11222" PostTypeId="1" CreationDate="2014-05-31T15:53:18.627" Score="3" ViewCount="235" Body="&lt;p&gt;I believe the molecule $\ce{CH3CH2CH2Cl}$ to be saturated, since there are only single bonds between the carbon atoms, but my workbook says it's unsaturated. Why?&lt;/p&gt;&#xA;" OwnerUserId="5705" LastEditorUserId="10185" LastEditDate="2015-10-02T16:56:29.747" LastActivityDate="2015-10-02T16:56:29.747" Title="Is CH₃CH₂CH₂Cl saturated or unsaturated?" Tags="&lt;organic-chemistry&gt;&lt;erratum&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11224" PostTypeId="2" ParentId="11221" CreationDate="2014-05-31T16:01:28.673" Score="10" Body="&lt;p&gt;The problem is that you did not include units with your numbers. If you had, you would have easily seen that you are multiplying $m^3$ and $cm^{−3}$ and coming up with 1 instead of 1000000.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-05-31T16:01:28.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12222" PostTypeId="2" ParentId="11222" CreationDate="2014-05-31T16:33:38.743" Score="2" Body="&lt;p&gt;The workbook must be wrong, it is a saturated molecule.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-31T16:33:38.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12223" PostTypeId="2" ParentId="11222" CreationDate="2014-05-31T16:50:50.220" Score="2" Body="&lt;p&gt;You are correct. Your workbook might be wrong. It is definitely a saturated molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one quoting wikipedia text must see that the article is talking about saturated hydrocarbon from end of line one , so wikipedia is not wrong it's our interpretation of sentece is wrong.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5424" LastEditDate="2014-07-02T20:25:38.013" LastActivityDate="2014-07-02T20:25:38.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12224" PostTypeId="2" ParentId="11217" CreationDate="2014-05-31T16:58:40.517" Score="6" Body="&lt;p&gt;An indicator works when the solution's pH is in its range. It's not necessarily the equivalence point.  The point is called the end point where it actually changes color. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the difference in volume is small between the equivalence and end points, then the indicator can be used.  Since a gradual change implies more volume must be added to obtain the same pH change in other acid base titration, the difference between equivalence point and end point(if you use indicator) will be more and therefore indicator will not be suitable for it.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="73" LastEditDate="2014-07-31T13:45:11.783" LastActivityDate="2014-07-31T13:45:11.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12225" PostTypeId="1" AcceptedAnswerId="12228" CreationDate="2014-05-31T17:15:09.580" Score="2" ViewCount="435" Body="&lt;p&gt;I am asked to find the enthalpy change of formation of the following:&#xA;$$\ce{N2 + 1/2O2 -&amp;gt; N2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given the following enthalpies of reaction:&#xA;\begin{align}&#xA;\ce{C + N2O &amp;amp;-&amp;gt; CO + N2} &amp;amp;\Delta H_f =-193 \tag{1} \\&#xA;\ce{C + 1/2O2 &amp;amp;-&amp;gt; CO} &amp;amp;\Delta H_f=-111 \tag{2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate enthalpy change of formation of the nitrous oxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it would just be adding the two enthalpies together because I substitute the second equation into the first one, but this is wrong.&lt;/p&gt;&#xA;" OwnerUserId="5619" LastEditorUserId="41328" LastEditDate="2017-08-11T08:58:03.103" LastActivityDate="2017-08-11T09:06:14.427" Title="Hess Cycle, determining enthalpy change of formation" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12226" PostTypeId="2" ParentId="12225" CreationDate="2014-05-31T17:26:57.930" Score="2" Body="&lt;p&gt;Try adding, subtracting, multiplying by real numbers and reversing the equations as necessary to get the desired equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, since your &lt;em&gt;target&lt;/em&gt; equation has $\ce{N_2}$ on the reactant side, I would flip the first given equation so that diatomic nitrogen ends up on the reactant side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then note that if you have the same molecule on both sides of the equation in equal quantities, you may remove it from the equation since these species will not affect the &lt;em&gt;enthalpy&lt;/em&gt; of the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note) Catalysts are similarly not consumed by a chemical reaction, yet they definitely affect the rate (kinetics) of the reaction. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-05-31T17:51:32.217" LastActivityDate="2014-05-31T17:51:32.217" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12227" PostTypeId="2" ParentId="12225" CreationDate="2014-05-31T17:45:59.173" Score="4" Body="&lt;p&gt;If you start by reversing equation &lt;strong&gt;1&lt;/strong&gt;, you get $$\ce{CO + N2 -&amp;gt;C + N2O~~ {\Delta}H_{f}=+193}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, add this result with equation &lt;strong&gt;2&lt;/strong&gt; and you get $$\ce{CO~+N2~+C~+1/2O2-&amp;gt;C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2-&amp;gt;N2O~~{\Delta}H_{f}=82}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-05-31T17:45:59.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12228" PostTypeId="2" ParentId="12225" CreationDate="2014-05-31T18:34:09.567" Score="3" Body="&lt;p&gt;I took the first equation away from the second equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lll}&#xA;  &amp;amp;\ce{C + 1/2O2 &amp;amp;-&amp;gt; CO} &amp;amp;\Delta H_f &amp;amp;= -111 \\&#xA;- &amp;amp;\ce{C + N2O &amp;amp;-&amp;gt; CO + N2} &amp;amp;\Delta H_f &amp;amp;= -193 \\&#xA;\hline&#xA;= &amp;amp;\ce{1/2O2 - N2O &amp;amp;-&amp;gt; - N2} &amp;amp;\Delta H_f &amp;amp;= 82 \\&#xA;= &amp;amp;\ce{N2 + 1/2O2 &amp;amp;-&amp;gt; N2O }&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="5619" LastEditorUserId="41328" LastEditDate="2017-08-11T09:06:14.427" LastActivityDate="2017-08-11T09:06:14.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12229" PostTypeId="1" CreationDate="2014-05-31T22:21:21.320" Score="2" ViewCount="71" Body="&lt;p&gt;I understand that all polyurethane (PU) materials off gas VOCs. I tried finding the % composition of the gases off-gassed by PU, but I am not able to get satisfactory answer. On the same note, why are these gases off-gassed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="5706" LastActivityDate="2014-05-31T22:21:21.320" Title="What is the chemical composition of VOC's off-gassed from polyurethane?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12230" PostTypeId="2" ParentId="9650" CreationDate="2014-06-01T02:12:34.607" Score="1" Body="&lt;p&gt;(you've probably done this by now, but...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply some common sense to the quantity you are working with. Are you making this on an industrial scale, or one small item to use in your lab? If the latter, don't worry about it. Unless something massively toxic is expected to come out (VX, for example) just don't go out of your way to breathe the products of the process. Even arsenic won't hurt you in small quantities.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-06-01T02:12:34.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12231" PostTypeId="2" ParentId="11219" CreationDate="2014-06-01T02:24:27.303" Score="1" Body="&lt;p&gt;Given that these machines have been used &lt;strong&gt;&lt;em&gt;extensively&lt;/em&gt;&lt;/strong&gt; in the entertainment industry for decades, and one (Rosco Fogger) even has an Academy Award, common sense says that they aren't dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breathing in copious quantities of the fog causes irritation, but that applies to a lot of things. Dry ice fog[1] is basically water vapour - fill the room with it and you die from asphyxiation from the CO&lt;sub&gt;2&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;looks better in many cases as it's &quot;fog&quot; and not &quot;smoke&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-06-01T02:24:27.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12232" PostTypeId="1" AcceptedAnswerId="12251" CreationDate="2014-06-01T06:07:18.440" Score="4" ViewCount="5683" Body="&lt;p&gt;Which is more acidic between methanal ($\ce{HCHO}$) and ethanal ($\ce{CH3CHO}$).&#xA;Please explain using General organic chemistry basic concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Effort:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw the stability of the conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be $\ce{HCO-}$ and $\ce{CH2^{(-)}CHO}$.&#xA;Due to resonance I can say that second one is more stabilized and thus a better acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But answer says that methanal is more acidic than ethanal.&lt;/p&gt;&#xA;" OwnerUserId="5708" LastEditorUserId="5708" LastEditDate="2014-06-02T02:43:18.860" LastActivityDate="2014-06-02T02:43:18.860" Title="Acidity of aldehydes" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12233" PostTypeId="1" CreationDate="2014-06-01T06:12:13.777" Score="8" ViewCount="3546" Body="&lt;p&gt;Aluminum mercury alloy is used as a chemical reagent to reduce compounds, such as the reduction of imines to amines,in this case mercury destroy aluminium oxide layer which protect aluminium to react with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alloy of aluminium and gallium which similar to above alloy make the aluminium more reactive by preventing it from forming an oxide layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases mercury and gallium plays a same role of destroying oxide layer on aluminium to react with water but why can't gallium aluminium alloy reduce imines to amines like mercury aluminium alloy ?&lt;/p&gt;&#xA;" OwnerUserId="5709" LastEditorUserId="4231" LastEditDate="2014-06-01T18:53:31.267" LastActivityDate="2017-12-15T18:40:37.350" Title="What is difference between aluminium mercury alloy and gallium aluminium alloy?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12234" PostTypeId="1" AcceptedAnswerId="12291" CreationDate="2014-06-01T07:12:11.840" Score="-2" ViewCount="2667" Body="&lt;p&gt;I know that $\ce{RCH=NH}$ is an imine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it's structure?&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="4734" LastEditDate="2014-06-01T18:02:29.017" LastActivityDate="2014-06-03T06:55:29.633" Title="What is the structure of RCH=NH?" Tags="&lt;organic-chemistry&gt;" AnswerCount="3" CommentCount="2" ClosedDate="2014-06-03T07:32:09.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="12235" PostTypeId="2" ParentId="6979" CreationDate="2014-06-01T07:23:58.677" Score="4" Body="&lt;p&gt;A few things come to mind: 1) my understanding is that (ab|cd) in a gaussian basis is exactly solved for arbitrary &quot;angular momenta&quot; using recurrence relations such as the Obara-Saika method. The O-S method also has ways of generating the requisite derivatives with respect to coefficients, without which it is difficult to minimize the energy or do anything practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) On the question of alternative basis sets to the Gaussian, one simply needs a complete basis set (of which there are many), helps to be orthogonal (almost as many), has to describe the boundary conditions of vanishing at infinity or being periodic with unit cell (fewer), hopefully is cusp-like near the nuclei (very few), and integration has to be tractable without resorting to quadrature, since there are four indices (each with 3 dimensions) but also millions of these integrals to compute -- and that leaves you with the Gaussian.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-06-02T05:38:27.333" LastActivityDate="2014-06-02T05:38:27.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12236" PostTypeId="1" AcceptedAnswerId="12239" CreationDate="2014-06-01T08:17:57.600" Score="1" ViewCount="528" Body="&lt;p&gt;I know that $\ce{NaHCO3}$ is sodium bicarbonate, but According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonate&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemistry, a carbonate is a salt of carbonic acid, characterized by the presence of the carbonate ion, $\ce{CO3^{2-}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaHCO3}$ also has the $\ce{CO3^{2-}}$ group, but is it a carbonate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, are bicarbonates a subset of carbonates?&lt;/p&gt;&#xA;" OwnerUserId="5710" LastEditorUserId="4945" LastEditDate="2014-06-01T12:46:57.437" LastActivityDate="2014-06-01T15:39:17.790" Title="Is NaHCO₃ a carbonate?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12238" PostTypeId="1" AcceptedAnswerId="12240" CreationDate="2014-06-01T12:59:06.603" Score="2" ViewCount="184" Body="&lt;p&gt;It is known that binding (aka association) constants are in fact unitless, as has been discussed &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1137/why-are-equilibrium-constants-unitless&quot;&gt;here&lt;/a&gt; already. However, I'm not a chemist and am confused about when one should or should not use units when working with association constants. &lt;a href=&quot;http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;One source&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$K_\text{eq}$ for a reaction with unequal numbers of reactants and products is &lt;strong&gt;always&lt;/strong&gt; given with units, even in published papers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But why is that? Why not always use unitless values? Is there something inherently wrong with never using units for binding constants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this example. Let $\ce{A}$ bind to $\ce{X}$ as an n-mer (e.g. $\ce{A}$ can be a transcriptional activator binding to gene promoter). This results in active state, $\ce{X_{A}}$ (e.g. a state leading to gene transcription):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{X + nA -&amp;gt;[k_{\text{on}}][k_{\text{off}}] X_{A}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Association constant: $K_\text{A} = \frac{k_\text{on}}{k_\text{off}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming equilibrium and law of mass action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{A} \cdot \text{X} \cdot \mathrm{A^n} = \mathrm{X_A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the fractional occupancy (active states to all states ratio) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\mathrm{\frac{X_A}{X+X_A}} = \frac{K_\text{A}\cdot \mathrm{A^n}}{1 + K_\text{A}\cdot \mathrm{A^n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $y$ must be unitless (i.e. it can be interpreted as a probability), for this particular equation, $K_\text{A}$ must have units of $\mathrm{M^{-n}}$ (in general, $\text{concentration}^{-n}$) for the equation to work out, correct (assuming concentration of $\ce{A}$ has units $\mathrm{M}$)? So in this particular case, using unitless $K_\text{A}$ seems wrong, but is it really, or is there something I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's extend this equation so that it includes a half-saturation constant $h$, i.e. concentration of $\ce{A}$ required for $y=0.5$ (50% activation). If I'm doing this correctly, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{K_\text{A}\cdot \mathrm{A^n}}{K_\text{A}\cdot h^n + K_\text{A}\cdot \mathrm{A^n}} = \frac{\mathrm{A^n}}{h^n + \mathrm{A^n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is equivalent to the &lt;a href=&quot;http://www.springerreference.com/docs/html/chapterdbid/376584.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill equation&lt;/a&gt; for an activator. This generalized equation, unlike the previous one, works just fine regardless of whether or not $K_\text{A}$ is unitless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my understanding of this correct, and does the choice of unitless vs non-unitless binding constant indeed depend on the formulation of a specific equation?&lt;/p&gt;&#xA;" OwnerUserId="5630" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-09-27T17:26:05.730" Title="Should binding constants be unitless when deriving fractional occupancy equations from reactions?" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12239" PostTypeId="2" ParentId="12236" CreationDate="2014-06-01T13:17:47.563" Score="4" Body="&lt;p&gt;Carbonates are salts and esters of carbonic acid $\ce{H2CO3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In generic nomenclature you may therefore consider a hydrogen carbonate also a subset of all carbonates. They are sometimes also referred to as primary carbonates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another considerable point is, that the structural subunit $\ce{O=C(-O-)2}$is still conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a stricter sense a hydrogencarbonate is only all forms involving $\ce{HCO3-}$, so it appears to be not a generic carbonate, but a class of its own. This refers to its one $\ce{O-H }$ bond and therefore it being a different structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another point of view should always consider aqueous solution equilibrium.&#xA;\begin{aligned}&#xA;\ce{HCO3- + H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + \color{\red}{CO3^{2-}}}\\&#xA;\ce{HCO3- + H2O &amp;amp;&amp;lt;=&amp;gt; H2CO3 + {}^{-}OH}&#xA;\end{aligned} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that carbonate ions will be present to a certain extend in solution. The chemistry will therefore also to some extend resemble the chemistry of the carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical properties of bicarbonate compound might differ from the pure carbonates - incorporation of water, solubility, ... .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another page, in organic chemistry the compound class &lt;strong&gt;carbonates&lt;/strong&gt; refers to structures of the type $\ce{\color{\red}{R}-{CO3}-\color{\navy}{R'}}$, where $\ce{\color{\red}{R},\color{\navy}{R'}$\in$ Alkyl, Aryl}$ - hence esters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it all up, since a more stringent definition of carbonate is not present to me, I would always consider a bicarbonate a carbonate.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-01T13:17:47.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12240" PostTypeId="2" ParentId="12238" CreationDate="2014-06-01T14:40:34.207" Score="4" Body="&lt;p&gt;The equilibrium constant does not need to be unitless, because it is depending on its definition. See &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot; rel=&quot;nofollow&quot;&gt;goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Equilibrium Constant&lt;/em&gt;&lt;/strong&gt;&#xA;  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The standard equilibrium constant is always unitless, as it is defined differently (&lt;a href=&quot;http://goldbook.iupac.org/S05915.html&quot; rel=&quot;nofollow&quot;&gt;goldbook&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;Standard Equilibrium Constant&lt;/em&gt;&lt;/strong&gt; $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)&#xA;  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your example both would work if you straighten your definitions, involving concentrations&#xA;\begin{aligned}&#xA;K_A \cdot c(\ce{X}) \cdot c^n(\ce{A}) &amp;amp;= c(\ce{X_{A}}),\\&#xA;y=\frac{c(\ce{X_A})}{c(\ce{X})+c(\ce{X_{A}})} &amp;amp;= \frac{K_A\cdot c^n(\ce{A})}{1 + K_A\cdot c^n(\ce{A})},&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or activities&#xA;\begin{aligned}&#xA;K^\circ_A \cdot a(\ce{X}) \cdot a^n(\ce{A}) &amp;amp;= a(\ce{X_{A}}),\\&#xA;y=\frac{a(\ce{X_A})}{a(\ce{X})+a(\ce{X_{A}})} &amp;amp;= \frac{K^\circ_A\cdot a^n(\ce{A})}{1 + K^\circ_A\cdot a^n(\ce{A})}.&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases $y$ remains unitless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In experimental chemistry it is much easier to observe concentrations instead of activities. Therefore to fist named constant is more often in use. A straightforward use of the thermodynamical equilibrium constant might sometimes prove a little tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasonable dilutions one can simply substitute &lt;a href=&quot;http://goldbook.iupac.org/A00115.html&quot; rel=&quot;nofollow&quot;&gt;activity&lt;/a&gt; with a unitless concentration as given through&#xA;$$a(\ce{Y})=\gamma\frac{c(\ce{Y})}{c^\circ},$$&#xA;with $\gamma\approx1$ for $c(\ce{Y})\to0$ and $c^\circ=1\:\mathrm{mol/L}$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-01T14:40:34.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12241" PostTypeId="2" ParentId="12236" CreationDate="2014-06-01T15:38:01.337" Score="1" Body="&lt;p&gt;I would consider it a carbonate. Chemistry of sodium bicarbonate, or sodium hydrogen carbonate, or simply baking soda, is very similar to that of just sodium carbonate, or $\ce{Na_2CO_3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding a strong acid to either $\ce{Na_2CO_3}$ or $\ce{NaHCO_3}$ will effect bubbling. Why? The acid protonates the carbonate or bicarbonate and forms the unstable $\ce{H_2CO_3}$, which is generally referred to as &lt;em&gt;carbonic acid&lt;/em&gt;, but is in fact a mixture of carbonic acid and the much less potent acid, &lt;em&gt;hydrated carbon dioxide&lt;/em&gt;. $\ce{H_2CO_3}$ decomposes rapidly to water and carbon dioxide. Hence, in the lab, if you wished to test for the present of carbonate in a salt, you could try adding a strong acid and look for bubbling. Copious bubbling is an indication of the presence of a carbonate or a carbonate derivative such as $\ce{HOCO_2^-}$ ($\ce{HCO3^-}$). To verify which carbonate you may simply perform a pH test as the $\ce{CO_3^2-}$ anion is a much strong base than the $\ce{HOCO_2^-}$ ion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-01T15:38:01.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12242" PostTypeId="2" ParentId="12234" CreationDate="2014-06-01T16:05:44.227" Score="3" Body="&lt;p&gt;The general structure of an imine contains a carbon-nitrogen double bound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u91Yg.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can add hydrogen atoms or organic groups instead of $R^1$, $R^2$ or $R^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of your formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RVHqM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2014-06-01T16:05:44.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12243" PostTypeId="2" ParentId="11219" CreationDate="2014-06-01T16:08:00.947" Score="1" Body="&lt;p&gt;Extreme heating of glycerine produces acrolein, which is toxic and highly irritating (I'd say it should be also carcinogenic, but was unable to find any suggestions for it with quick googling).  However, the heating should be extreme or an acidic catalyst is needed. Extreme heating of 1,2 propylene glycol will produce propanal, that is also toxic, but not that much as acrolein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free glycerol and 1,2 propylene glycol are &lt;em&gt;slightly&lt;/em&gt; toxic, but their LD50 are high. (&gt;10 g/ kg), so nothing special. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-01T16:08:00.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12244" PostTypeId="2" ParentId="12232" CreationDate="2014-06-01T16:17:15.493" Score="1" Body="&lt;p&gt;The answer lies with induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen is better able to stabilize the residual negative charge on the conjugate base of methanal by virtue of residing closer to the negative formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the conjugate base of ethanal note that the electronegative oxygen is one carbon removed from the negative formal charge while on the conjugate base of methanal, the oxygen is on the same carbon with the negative formal charge. Thus, the oxygen in methanal will be better able to withdraw electron density from the spot where heterolytic bond clevage occurred. This follows from Coulomb's Law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${F=k\frac{q_1q_2}{r^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see from Coulomb's law, force and distance are related in an inverse-square fashion. In other words, increasing the distance of the oxygen nucleus from the site of the negative charge &lt;em&gt;dramatically&lt;/em&gt; decreases the electron-withdrawing effects of oxygen. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-01T16:46:51.600" LastActivityDate="2014-06-01T16:46:51.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12246" PostTypeId="2" ParentId="12234" CreationDate="2014-06-01T17:01:09.473" Score="2" Body="&lt;p&gt;The structure of an imine is analogous to that of an olefin.  There is a central $\ce{C-N}$ sigma bond along with a pi bond just as in an olefin. In the figure below, the olefin has four substituents attached to the two carbons forming the carbon-carbon double bond.&lt;img src=&quot;https://i.stack.imgur.com/OZxxA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All 6 of these atoms or groups lie in the same plane, the plane of the screen.  Similarly in an imine, all of the groups attached to the central $\ce{C=N}$ double bond also lie in the plane of the screen, along with the carbon and nitrogen atoms themselves.  Note that while the olefin had four substituents ($\ce{R_1 - R_4}$), the imine has only 3 substituents.  The fourth substituent position in the imine is occupied by a lone pair of electrons on the imine nitrogen.  This gives rise to an important difference between olefins and imines.  Appropriately substituted olefins can exist as &lt;em&gt;cis-trans&lt;/em&gt; pairs.  It typically takes energies in excess of 60 kcal/mol to break the pi bond portion of a carbon-carbon double bond and cause &lt;em&gt;cis-trans&lt;/em&gt; isomerization.  The barrier for interconversion (called &lt;em&gt;syn-anti&lt;/em&gt; isomerization in imines) in an imine is much lower, on the order of 20 kcal/mol, so rotation about the imine double bond occurs much more readily than in the olefin case.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-01T17:01:09.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12247" PostTypeId="2" ParentId="11215" CreationDate="2014-06-01T18:04:13.743" Score="2" Body="&lt;p&gt;In order for heat transfer to occur, there must be a temperature gradient. If the temperature of the refrigerant was equal to the temperature of the cooled compartment, no heat would flow from the cooled compartment into the refrigerant to drive the phase change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason that convective flow over the evaporator coils is important - it increases the rate of heat transfer by keeping the fluid moving so that the temperature right next to the evaporator coil is always close to the average compartment temperature. A larger ${\Delta}T$ leads to a faster rate of heat flow.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-01T18:04:13.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12248" PostTypeId="1" CreationDate="2014-06-01T18:15:47.973" Score="1" ViewCount="5685" Body="&lt;p&gt;Is it possible to crystallize vegetable glycerin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw a myth about it online and was wondering if it is possible. &lt;/p&gt;&#xA;" OwnerUserId="5715" LastEditorUserId="7475" LastEditDate="2015-10-16T17:11:58.793" LastActivityDate="2015-10-16T17:11:58.793" Title="Crystallization of Glycerin" Tags="&lt;organic-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12249" PostTypeId="2" ParentId="11213" CreationDate="2014-06-01T18:41:23.437" Score="3" Body="&lt;p&gt;Sounds like you need a food-grade surfactant. Maybe you could try &lt;a href=&quot;http://en.wikipedia.org/wiki/Lecithin&quot; rel=&quot;nofollow&quot;&gt;lecithin&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that cocoa powder is mostly made of stuff that is not soluble in water (hydrophobic). So, to create the emulsion, you will need to &quot;package&quot; the powder molecules in something they are soluble in. You could either use a surfactant directly, or build up in layers using fat molecules and proteins, with a surfactant around the whole package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lecithin would be a good starting choice since it is FDA approved and used in lots of products already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add lecithin in gradually larger amounts until the suspension worked (you will have to try different heat/pressure and emulsification treatments - consider high-pressure &lt;a href=&quot;http://en.wikipedia.org/wiki/Homogenization_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;homogenization&lt;/a&gt; as well.)&lt;/li&gt;&#xA;&lt;li&gt;Add some whole milk (we are looking for the fat molecules and proteins) and see if you can find an acceptable milk/lecithin ratio that tasted good and that was still stable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If that doesn't work, your other option is to modify the cocoa powder granule surface so that it is hydrophilic. That means you will need to add polar functional groups. One way to do that is with an acid (citric, for example), but you will add a sour taste which might be hard to get rid of. There is a &lt;a href=&quot;http://www.google.com/patents/US8372456&quot; rel=&quot;nofollow&quot;&gt;patent&lt;/a&gt; for an approach using enzymes, but I don't know which one they used (it just refers to &quot;starch degrading,&quot; &quot;protein degrading,&quot; and &quot;cell-wall degrading&quot; enzymes), or how well it works.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-01T18:41:23.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12250" PostTypeId="1" AcceptedAnswerId="12252" CreationDate="2014-06-01T18:50:13.827" Score="10" ViewCount="3821" Body="&lt;p&gt;Acet* indicates Ethyl, but does not offer two Carbon atoms, but three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetaldehyde is Ethanal, Acetic acid is Ethanoic acid, but Aceton is Propanon (yes, I'm aware that there is no Ethanon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is Acetone called Acetone?&lt;/p&gt;&#xA;" OwnerUserId="625" LastActivityDate="2015-06-16T21:02:42.003" Title="Why is it ACETone?" Tags="&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12251" PostTypeId="2" ParentId="12232" CreationDate="2014-06-01T18:51:38.727" Score="3" Body="&lt;p&gt;Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data.  If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively.  A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZonf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if someone asked, &quot;would pulling the same proton off from ethanal be easier or harder?&quot;, you could answer &quot;harder&quot; because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon.  But it turns out that when ethanal is treated with base a &lt;strong&gt;different&lt;/strong&gt; proton is removed!  The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed.  Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-02T00:30:53.347" LastActivityDate="2014-06-02T00:30:53.347" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12252" PostTypeId="2" ParentId="12250" CreationDate="2014-06-01T19:08:23.533" Score="8" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://www.jstor.org/discover/10.2307/27757220?uid=3739568&amp;amp;uid=2&amp;amp;uid=4&amp;amp;uid=3739256&amp;amp;sid=21103819510541&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to the first page of a book entitled, &quot;The History of Acetone, 1600-1850&quot; by Mel Gorman.  The author points out that acetone was known in the Middle Ages and was frequently produced by heating dry lead acetate.  I suspect that the &quot;acetate&quot; (or whatever the Latin, French or German term was) root stuck and then it was just modified a bit - to acetone - to make it distinct.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-02T13:33:11.600" LastActivityDate="2014-06-02T13:33:11.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12253" PostTypeId="1" CreationDate="2014-06-01T19:14:22.010" Score="2" ViewCount="162" Body="&lt;p&gt;I was looking up some ion channel blockers. What does (+) or (-) below mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tocris.com/dispprod.php?ItemId=2028#.U4t67vnoTy6&quot; rel=&quot;nofollow&quot;&gt;(+)-Bicuculline&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tocris.com/dispprod.php?ItemId=159237#.U4t6cfnoTy4&quot; rel=&quot;nofollow&quot;&gt;(-)-Bicuculline methiodide&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5716" LastActivityDate="2014-06-02T02:27:45.603" Title="(+)-Bicuculline, what does (+) mean?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12254" PostTypeId="2" ParentId="12233" CreationDate="2014-06-01T19:28:17.457" Score="2" Body="&lt;p&gt;We can rationalize the differences in the reactivity as follows.  Gallium has a higher ionization energy compared to mercury.  During a redox reaction the metal is oxidized and it loses an electron, therefore how &quot;easy&quot; it is for a metal to lose its electon is iimportant. Since gallium is harder to inonze than mercurey, it implies that it is &quot;easier&quot; for mercury to lose its electron compared to Gallium, making mercury a better stronger reducing agent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may come back and add some quantitative information later, but i'm going to go eat now. I think looking in a table of standard reduction potentials would be reasonable. &lt;/p&gt;&#xA;" OwnerUserId="5717" LastActivityDate="2014-06-01T19:28:17.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12255" PostTypeId="2" ParentId="12253" CreationDate="2014-06-01T19:32:12.783" Score="5" Body="&lt;p&gt;Bicuculline is a chiral molecule.  That means that its mirror images are non-superimposable.  These non-superimposable images are termed &quot;enantiomers.&quot;  When you see a &quot;(+)&quot; or &quot;(-)&quot; in front of a chemical name it tells you that only one the enantiomers is present (e.g. it is not a racemic mixture of the enantiomers).  Chiral molecules have the ability to rotate the plane of a polarized light beam that passes through it. Enantiomers will rotate the plane of polarization in exactly equal amounts (same magnitude) but in opposite directions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dextrorotary designated as (+), indicates clockwise rotation (to the right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Levorotary designated as (-), indicates anti-clockwise rotation (to the left)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-02T02:27:45.603" LastActivityDate="2014-06-02T02:27:45.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12256" PostTypeId="1" CreationDate="2014-06-01T20:43:58.063" Score="2" ViewCount="1770" Body="&lt;p&gt;Using these equations:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA;\text{(Fe2):}&amp;amp;&amp;amp;\ce{ Fe + CuSO4 &amp;amp;-&amp;gt; FeSO4 + Cu}\\&#xA;\text{(Fe3):}&amp;amp;&amp;amp;\ce{ 2Fe + 3CuSO4 &amp;amp;-&amp;gt; Fe2(SO4)3 + 3Cu}&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Beginning with &lt;span class=&quot;math-container&quot;&gt;$\pu{0.78 g}$&lt;/span&gt; of iron, the theoretical yields of copper are &lt;span class=&quot;math-container&quot;&gt;$\pu{0.89 g}$&lt;/span&gt; (with &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\pu{1.33 g}$&lt;/span&gt; (with &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{3+}}$&lt;/span&gt;), and the actual yield was &lt;span class=&quot;math-container&quot;&gt;$\pu{0.78 g}$&lt;/span&gt; of copper. I know to find of the ratio of the moles of iron used to the moles of copper formed (88% for &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt;, 58% for &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{3+}}$&lt;/span&gt;), and so the clear choice would be that this reaction formed &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt; ions.&lt;/p&gt;&#xA;&lt;p&gt;But my teacher said this would form &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{3+}}$&lt;/span&gt; ions, but why is that? Shouldn't it have formed &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt;? The reaction formed &lt;span class=&quot;math-container&quot;&gt;$\pu{0.12 mol}$&lt;/span&gt; of copper, compared to the &lt;span class=&quot;math-container&quot;&gt;$\pu{0.14 mol}$&lt;/span&gt; in the theoretical yield, and the other &lt;span class=&quot;math-container&quot;&gt;$\pu{0.02 mol}$&lt;/span&gt; could have just been lost somewhere along the reaction process.&lt;/p&gt;&#xA;" OwnerUserId="5718" LastEditorUserId="83758" LastEditDate="2020-09-15T14:30:01.870" LastActivityDate="2020-09-15T14:30:01.870" Title="Which ion of iron is produced in a reaction between iron and copper(II) sulfate?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12257" PostTypeId="2" ParentId="9813" CreationDate="2014-06-01T21:39:06.350" Score="0" Body="&lt;p&gt;In complex terms we can't ever really describe something as been fully ionic or fully covalent but more of a mix of the two. For example, a HCl molecule will have a small amount of ionic bonding character even though we define it as covalent bonding. This is besides the point though... in this case, we generally assume ionic bonds are stronger than covalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increased the covalent bonding character in a species you are likely to decrease the ionic bonding character resulting in a lower melting point as forces of attraction are now weaker. However, we do see additional bonding occur in species such as vulcanised rubber where you get cross-linking between carbon atoms on adjacent chains. This then obviously increases the melting point as you have not reduced any ionic bonding character&lt;/p&gt;&#xA;" OwnerUserId="5719" LastActivityDate="2014-06-01T21:39:06.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12258" PostTypeId="1" AcceptedAnswerId="12260" CreationDate="2014-06-01T22:31:44.003" Score="1" ViewCount="3449" Body="&lt;p&gt;I tried drawing out (R)-1,1,2-trimethylcyclohexane but I can't seem to figure it out. I think I&quot;m getting hung on assigning priorities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've labeled the chiral carbon with an asterisk. I've also drawn out all the groups it is attached to. I gave the hydrogen the lowest priority (4) and I'm giving the methyl carbon a priority of 3. I am giving the other carbons priorities of 1 and 2 (as can be seen in the picture). However, this doesn't seem to make any sense; I get a molecule that I can neither assign S nor R to since the numbers are out of order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWzbb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is this really an image of (R)-1,1,2-trimethylcyclohexane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vY4Bh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4231" LastEditDate="2014-06-02T00:24:52.617" LastActivityDate="2014-06-02T01:48:45.847" Title="Drawing (R)-1,1,2-trimethylcyclohexane" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12259" PostTypeId="1" CreationDate="2014-06-01T23:53:24.030" Score="3" ViewCount="2391" Body="&lt;p&gt;I am trying to dissolve copper(I) iodide and copper(I) thiocyanate in ammonia.  I am quite sure that they both dissolve in liquid ammonia, and are insoluble in ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to purchase a 2.0M solution of ammonia in ethanol, like the one &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/search?interface=All&amp;amp;term=ammonia&amp;amp;N=0&amp;amp;mode=match%20partialmax&amp;amp;focus=product&amp;amp;lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, would the CuI and CuSCN still dissolve in the ammonia solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are there any other solvents that you know of for these materials?&lt;/p&gt;&#xA;" OwnerUserId="5672" LastActivityDate="2014-07-31T03:20:33.383" Title="CuI, CuSCN Solubility in Ammonia Solution" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12260" PostTypeId="2" ParentId="12258" CreationDate="2014-06-01T23:53:56.280" Score="4" Body="&lt;p&gt;You've correctly identified the chiral center in your molecule (the carbon with the asterisk), that is the first step.  When doing work with chiral centers, it is very important to use solid lines to denote bonds in the plane of the screen, bold wedges to denote bonds coming from the plane of the screen towards the viewer and dashed wedges to denote bonds from the plane of the screen away from the viewer.  Drawing all 4 bonds as simple solid lines, as you did, will often lead to confusion.  Look at the figures below, both enantiomers of 1,1,2-trimethylcyclohexane are shown.  The next step in assigning R/S configuration is to rank the 4 groups around the chiral carbon by molecular weight, with &quot;1&quot; being assigned to the highest molecular weight group, and &quot;4&quot; to the lowest (you also did this correctly).  Now, grab the &quot;4&quot; group in your hand so that groups &quot;1&quot; through &quot;3&quot; form a circular pattern between your eye and your hand.  If moving from 1-&gt;2-&gt;3 involves clockwise movement, then this is the &quot;R&quot; configuration.  If the movement is counterclockwise, then it is the &quot;S&quot; compound you are observing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/seTTB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/sterism3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; for a nice picture of what I'm describing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot determine the R/S configuration of the chiral carbon in the compound you've drawn on paper at the top of your post because it is unclear which group 3 or 4 is projecting towards the viewer and which is projecting away from the viewer.  Finally, note that my &quot;R&quot; drawing matches the &quot;R&quot; drawing at the bottom of your post.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-02T01:48:45.847" LastActivityDate="2014-06-02T01:48:45.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12261" PostTypeId="2" ParentId="9813" CreationDate="2014-06-02T04:06:47.037" Score="0" Body="&lt;p&gt;There was a previous post concerning strength differences between ionic bonds &lt;a href=&quot;https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds-i-was-taught-the-opposite/11052#11052&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bond strength decrease in the following order covalent &gt; ionic &gt; metallic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Substances with covalent bonds tend to appear weaker than ionically bonded substances (i.e. exhibiting lower boiling points) because ionically bonded substances often form lattice structures from their constituent ions, creating a much more resilient substance, if you will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the difference in bond strengths, however, when covalent substances form lattice-like structures as well. These network covalent structures are far &quot;stronger&quot; than normal covalent structures because, like ionic substances, they have formed molecular networks that make them stronger. Some examples of these network covalent structures are carbon in the form of diamond or graphene and silicon dioxide (quartz). You will notice that quartz and diamond both exhibit a crystalline structure similar to an ionic substance like NaCl, which is due to their network covalent structures. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-06-02T04:06:47.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12265" PostTypeId="1" CreationDate="2014-06-02T06:30:54.177" Score="6" ViewCount="472" Body="&lt;p&gt;The question is about emergence and reductionism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except for a few exceptions, such as dark matter, most phenomena in Nature could be reproduced from the standard model and general relativity with great accuracy. Up to a Planck scale, there is no inconsistency between general relativity and quantum theory by means of effective theory. However, at different level of complexity, a new principle may emerge. One example is the theory of evolution in Biology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any similar emerged master theory in Chemistry that comparable with evolution in Biology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the emerged master theory means:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is not just a brute-force &lt;em&gt;ab initio&lt;/em&gt; computation (emergence). Here I have excluded explicitly correlated Gaussian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory is universally valid. Not like 8- and/or 18-electron principle and/or VSEPR and/or periodic table, we know there are many exceptions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory should generate experimentally testable predictions, hopefully quantitative. Here I have excluded Bader's QTAIM at its current form. Atom in molecule and types of chemical bonds are anyway not observables. One may see some correlation between Laplacian at bond critical point with other molecular properties, and do some fitting or hand-waving argument. Such fitted relation could be useful. However, it is a fitting, not a derivation from its principle. I do not regard it as a prediction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory should not be discovered in condensed matter physics first. Here I have excluded spontaneous symmetry breaking.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I do not expect the statement like, there is no master principle in chemistry is the master principle in chemistry.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2629" LastEditorUserId="30836" LastEditDate="2018-11-17T15:54:04.057" LastActivityDate="2018-11-17T15:54:04.057" Title="Is there any emerged master principle in chemistry, like the theory of evolution in biology?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="12266" PostTypeId="2" ParentId="12265" CreationDate="2014-06-02T08:17:49.907" Score="9" Body="&lt;p&gt;Quantum Mechanics and Statistical Mechanics are the &quot;master theories&quot; of chemistry. Prediction can be had from difference of descriptions determined from these rigorously defined and realizable scientific concepts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relativistic contributions can be seen especially in heavier elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In the spirit of your question, perhaps you are looking for something that has evolved from chemistry, that may not have been directly apparent in the physics community.  I would propose that the concept of a &lt;strong&gt;&lt;em&gt;functional group&lt;/em&gt;&lt;/strong&gt;--a rather predictably-behaving collection of atoms--is a master principle.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-06-03T11:48:21.663" LastActivityDate="2014-06-03T11:48:21.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12267" PostTypeId="1" CreationDate="2014-06-02T12:27:00.833" Score="1" ViewCount="35" Body="&lt;p&gt;I'm doing some research on heat pipes &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_pipe&quot; rel=&quot;nofollow&quot; title=&quot;check here for wiki&quot;&gt;(check here for wiki)&lt;/a&gt;. One of the things I'm planning to do is try some different fluids in the pipes. Water (commonly used) has a freezing point that's too high, so I'm looking into fluids that don't freeze up in winter and disrupt the equilibrium in the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've randomly picked ethanol so far, but I'm looking for suggestions from someone more knowledgeable then me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly interested in vertical heat pipes that don't need a wick like &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/a/a3/Heat_Pipe_Mechanism.png&quot; rel=&quot;nofollow&quot;&gt;here on wikipedia&lt;/a&gt;, but rely on gravity to get the drops of condensed fluid back to the warmer area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential things that could matter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Viscosity of the fluid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bubble forming at boiling point (bubbles can form an isolating barrier perhaps?), is there a fluid that boils in vacuum with very small bubbles?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surface tension&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All suggestions are very welcome, please explain why you think a certain fluid is a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Johan Rensink&lt;/p&gt;&#xA;" OwnerUserId="5726" LastActivityDate="2014-06-02T12:27:00.833" Title="Suggestions for research into different fluids in heat pipes" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12270" PostTypeId="1" AcceptedAnswerId="12271" CreationDate="2014-06-02T16:56:24.833" Score="1" ViewCount="1798" Body="&lt;p&gt;I don't understand why this molecule is assigned S instead of R. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZYqU4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H is assigned a priority of 4; the methyl group on the chiral carbon gets 3; the left carbon gets 1 and the right carbon gets 2. The rotation should therefore be clockwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess what I have an issue with is assigning priorities in cases like these: what does moving away from the chirality center mean? Do I choose the H? The C? The other C? &lt;img src=&quot;https://i.stack.imgur.com/HHejM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we just ignore the attached hydrogens? It seems like the hydrogen on the carbon left of the chirality center is ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WBOB9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uX8DS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the above case, why are we comparing an O with a C? The first atom to the left of the chirality center is a C, not an O. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yI0WH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the last picture I give my configuration for the molecule in the first picture. Why doesn't the upper right carbon have a priority of 2? Double bonded carbon counts twice; the rest are hydrogens. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-02T17:29:54.893" LastActivityDate="2015-01-26T08:12:30.327" Title="R,S Configuration" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12271" PostTypeId="2" ParentId="12270" CreationDate="2014-06-02T17:31:11.220" Score="2" Body="&lt;p&gt;The right carbon gets a &amp;quot;1&amp;quot;, so the compound is &amp;quot;S&amp;quot;.  Attached hydrogens are not ignored.  Here is a &lt;a href=&quot;http://www.chem.purdue.edu/courses/chm25500/handouts/chiral_center_priority_rules.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a nice set of rules.  See rule #3 for handling double\triple bonds, that's where you made an error. It says, &amp;quot;Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar “phantom” atoms by single bonds. Note: “phantom” atoms are bonded to no other atoms&amp;quot;.  See the page for a clearer picture.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/laKOQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the case of the diol you show near the bottom of your post, the 2 carbons attached to the chiral center are equal, so we next look at, and compare, all of the substituents attached directly to them.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-06-02T17:31:11.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12272" PostTypeId="2" ParentId="12270" CreationDate="2014-06-02T17:32:26.953" Score="1" Body="&lt;p&gt;As you move away from the chiral center along two chains you are comparing, for each atom in the chain consider the highest priority atom it is bonded to first.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See step 1, parts 2 and 3 in this reference: &lt;a href=&quot;http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For double and triple bonds you have to add ficticous atoms before you evaluate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See step 1, part 4 in the above reference.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-02T17:32:26.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12273" PostTypeId="1" CreationDate="2014-06-02T18:00:24.323" Score="18" ViewCount="434" Body="&lt;p&gt;How does Lewis acidity correlate with bactericidal activity? I read on Wikipedia that the two are positively correlated but no explanation was given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with the movement of electrons - i.e. a current? I'm guessing that having a current passed through you isn't good for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose also that the interactions (i.e. coordination bonds) of metal (ions) with bacterial components can interfere with cellular processes. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-06-05T11:41:51.413" LastActivityDate="2016-04-14T05:44:35.700" Title="Lewis Acidity and Bacteria" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12274" PostTypeId="2" ParentId="7183" CreationDate="2014-06-02T18:04:31.997" Score="0" Body="&lt;p&gt;At STP, 1 mol of any gas occupies 22.4L. Using this knowledge, you can then convert each volume of gas to moles and then determine the limiting reactant using the 2:1 molar ratio of product to reactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for part a, the amount of HCl produced would be dependent on the limiting reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part c could then be answered by subtracting the number of moles of limiting reactant from the excess reactant. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastActivityDate="2014-06-02T18:04:31.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12275" PostTypeId="1" AcceptedAnswerId="12276" CreationDate="2014-06-02T18:56:24.190" Score="0" ViewCount="57" Body="&lt;p&gt;I'm having trouble assigning priorities for this molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried assigning priorities with the rules for doubly bonded atoms in mind. I know that the lowest priority group should be going into the page (not on a wedge) but I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned as the true configuration of the molecule. Problem is, I get R. Answer key says S. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mktm4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-02T23:00:40.783" Title="Configuration, chirality" Tags="&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12276" PostTypeId="2" ParentId="12275" CreationDate="2014-06-02T20:38:16.987" Score="2" Body="&lt;p&gt;Your drawing is correct and yes, &quot;I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned&quot; this is true too.  In your drawing 1-&gt;2-&gt;3 is a clockwise rotation (therefore &quot;R&quot; &lt;strong&gt;but&lt;/strong&gt;), but the lowest priority item (4) is coming out towards you.  Therefor, the R/S assignment needs to be reversed (from &quot;R&quot;) to &quot;S&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5084" LastEditDate="2014-06-02T23:00:40.783" LastActivityDate="2014-06-02T23:00:40.783" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12277" PostTypeId="1" CreationDate="2014-06-02T20:56:01.493" Score="-1" ViewCount="999" Body="&lt;p&gt;From my understanding NMR spectroscopy works by analyzing the resonance frequency of each structure around a hydrogen or carbon. Since every structure is unique and has unique resonance frequency it is easy to identify them. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other nuclei besides hydrogen and carbon 13 could be detected using NMR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is H1 NMR the same as saying NMR?&lt;/p&gt;&#xA;" OwnerUserId="4830" LastEditorUserId="4945" LastEditDate="2019-01-22T13:04:50.340" LastActivityDate="2019-01-22T16:38:10.337" Title="What other nuclei besides hydrogen and carbon 13 could be detected using NMR and is H1-NMR synonymus to NMR?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12278" PostTypeId="1" AcceptedAnswerId="12283" CreationDate="2014-06-02T21:01:22.887" Score="5" ViewCount="27520" Body="&lt;p&gt;Take as example &lt;a href=&quot;http://njcmr.njit.edu/distils/lab/aspirins/nap4.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; &#xA;or even better &lt;a href=&quot;http://wwwchem.csustan.edu/consumer/aspirincons/aspirincons.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wash the crystals with 2-3 ml of chilled water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does the aspirin need to be cooled down? And why do we wash the aspirin with chilled water?&lt;/p&gt;&#xA;" OwnerUserId="2389" LastEditorUserId="4945" LastEditDate="2014-10-23T10:27:46.447" LastActivityDate="2014-10-23T10:27:46.447" Title="Why is crystalline aspirin washed with chilled water in the purification process?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12279" PostTypeId="2" ParentId="12277" CreationDate="2014-06-02T21:19:10.143" Score="7" Body="&lt;p&gt;The term &lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt; NMR is more specific than the term &quot;NMR&quot; in that it is referring only to NMR experiments involving protons.  It is usually relatively straight forward to identify which atom in a molecule is responsible for a certain NMR signal.  There is also a lot of other information that can be extracted from an NMR signal by doing secondary NMR experiments that can help with peak assignment.  That said, some molecules are so large and complex that signal assignment can become problematical. As to which other nuclei can be detected by NMR, any isotope with an odd number of protons and/or neutrons will have a magnetic moment and be detectable by the NMR experiment.  Here is a listing of some nuclei with non-zero spin that have been studied by NMR (&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_magnetic_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;see this Wikipedia article&lt;/a&gt;):&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^2H}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^6Li}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^10B}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^11B}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^13C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^14N}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^15N}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^17O}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^19F}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^23Na}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^29Si}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^31P}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^35Cl}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^113Cd}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^129Xe}$&lt;/span&gt;,  &lt;span class=&quot;math-container&quot;&gt;$\ce{^195Pt}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2019-01-22T16:38:10.337" LastActivityDate="2019-01-22T16:38:10.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12280" PostTypeId="1" AcceptedAnswerId="12282" CreationDate="2014-06-02T22:33:43.387" Score="6" ViewCount="1936" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;&#xA;  $$\ce{2NO(g) + H_2(g) &amp;lt;=&amp;gt; N_{2}O(g) + H_{2}O} + \pu{351kJ}$$&#xA;  If the temperature is increased in the reaction above, which of the following occurs?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. The concentration of $\ce{H_{2}O}$ increases&lt;br&gt;&#xA;  B. The concentration of $\ce{N_{2}O}$ decreases  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My teacher said the answer is B;&#xA;however, I thought that since the reverse reaction is endothermic, adding more heat would produce more $\ce{2NO}$ and $\ce{H_2}$ which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.&#xA;Did my teacher make a mistake? If he did not can you please explain why he is correct.&lt;/p&gt;&#xA;" OwnerUserId="4157" LastEditorUserId="5026" LastEditDate="2018-03-04T07:37:50.513" LastActivityDate="2018-03-04T07:37:50.513" Title="Effect of temperature and Le Chatelier's principle" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12281" PostTypeId="2" ParentId="12280" CreationDate="2014-06-02T22:52:36.910" Score="1" Body="&lt;p&gt;I'm not sure if you could actually predict what would happen if you raise the temperature of the reaction &lt;em&gt;system&lt;/em&gt;. That would &lt;em&gt;change&lt;/em&gt; the equilibrium constant &lt;em&gt;K&lt;/em&gt; for this equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you don't know what the concentration/pressure of each system component is, there is the possibly that K could be adjusted so that it equals &lt;em&gt;Q&lt;/em&gt; when you raise the temperature for the &quot;reaction.&quot; That would make both answers &lt;em&gt;incorrect&lt;/em&gt; since at equilibrium there is no net change in system composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell your teacher that. He's being ambiguous by not specifying if we're raising the temperature of the reaction system, or whether we're simply heating up the reactants or what. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example illustrates why your teacher is ambiguous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{flour~+~milk~+~eggs -&amp;gt;cake}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you heat up the entire reaction system, then you'd have a warm cake, and a warm bag of flour, some warm milk, and some warm eggs. Or maybe you'd have a warm cake. It depends on how much heat you add. Here's one ambiguity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add heat to the reactants only - i.e. you bake the combination of reactants - then you'd get cake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add heat to the product only, then you'd get a warm cake. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regardless of however crudely your teacher might have phrased the question, you've thought too much into the question. True, adding heat to the product side will encourage the reverse reaction, but you don't know to what extent and it's best not to speculate what might happen also as a result. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-02T22:58:37.310" LastActivityDate="2014-06-02T22:58:37.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12282" PostTypeId="2" ParentId="12280" CreationDate="2014-06-02T23:13:57.653" Score="4" Body="&lt;p&gt;Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -&amp;gt; N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$&#xA;So the reaction to the right is exothermic, it gives off heat.  That tells us the products are more stable than the reactants and at equilibrium $\mathrm{\frac{[Products]}{[Reactants]}~&amp;gt;1}$.  Here we could guess that if we heat the reaction (put heat back into the reaction), we will tend to level out the equilibrium thereby pushing the reaction back to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To examine this guess let's work an example.  You were given an enthalpy of reaction, but for a &quot;back of the envelope&quot; calculation, let's run a comparison and assume that the change in entropy will be roughly the same for the two conditions (temperatures) we're going to compare.  With this assumption, we can let $$\Delta G= -RT\cdot\ln K_\text{eq}= -351\ \mathrm{kJ/mol}$$  We know $\Delta G$, $R=8.31\times10^{-3}$ and let's say the data is given for room temperature, so $T=298\ \mathrm K$.  Solving for $K_\text{eq}$ we find $K_\text{eq}=3.6\times10^{61}$.  Now let's let our reaction equilibrate $100^\circ$ above room temperature (now $T=398\ \mathrm K$).  $\Delta G$ and $R$ are fixed, so lets resolve the equation and we now find $K_\text{eq}=1.2\times10^{46}$, indeed heat has pushed our reaction to the left.  The teacher was right.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:43:20.270" LastActivityDate="2016-03-12T00:43:20.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12283" PostTypeId="2" ParentId="12278" CreationDate="2014-06-02T23:43:32.750" Score="7" Body="&lt;p&gt;Because aspirin is less soluble in cold water. In fact, aspirin is not very soluble in water at all, which is why you are supposed to take it with lots of water. You are not &quot;chilling the aspirin&quot;, you are preventing it from dissolving too much so that you have a better yield of your product. You wash it to remove impurities that &lt;em&gt;do&lt;/em&gt; dissolve in water.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-06-02T23:43:32.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12284" PostTypeId="1" CreationDate="2014-06-03T00:30:40.950" Score="2" ViewCount="711" Body="&lt;p&gt;Assume these 2 reactions. I must write the Lewis structures of the products.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$$\ce{C6H5COOH + CH3CH2OH -&amp;gt;[\ce{H+}] ?}$$&#xA;(comment: A benzene ring + $\ce{CO_2H}$ on one end, I don't know its name)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$$\ce{CH3CH2OH + (CH3CO)2O -&amp;gt;[\ce{H+}]?}$$ (comment: acetic anhydride)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &lt;strong&gt;first&lt;/strong&gt; reaction I will get acetic anhydride (&lt;a href=&quot;http://www.chem.ucla.edu/harding/IGOC/A/acetic_anhydride01.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2PjI.jpg&quot; alt=&quot;acetic anhydride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &lt;strong&gt;second&lt;/strong&gt; reaction I will get aspirin and acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the above correct?&lt;/li&gt;&#xA;&lt;li&gt;Is there a connection between the two reactions? Is it an intermediate step of some sort? What is the importance between the two reactions?&lt;/li&gt;&#xA;&lt;li&gt;Why is $\ce{H+}$ added?&lt;/li&gt;&#xA;&lt;li&gt;What is the benzene ring and the $\ce{CO2H}$ called? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4830" LastEditorUserId="5026" LastEditDate="2018-02-24T02:42:06.970" LastActivityDate="2018-02-24T02:42:06.970" Title="What products will the following reactions yield?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12285" PostTypeId="2" ParentId="12284" CreationDate="2014-06-03T00:42:47.147" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No, I hesitate to give the correct products since this is homework, but maybe some of what's below will help&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, the two reactions are both esterifications; converting an acid, an acid anhydride, an acid chloride, etc., to the corresponding ester&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proton is a catalyst, it is not consumed in the reaction.  It protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;benzoic acid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-06-03T05:18:41.073" LastActivityDate="2014-06-03T05:18:41.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12286" PostTypeId="2" ParentId="12277" CreationDate="2014-06-03T01:19:08.790" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Since every structure is unique and has unique resonance frequency it is easy to identify them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, in NMR we do not deal with structures, but instead nuclei. Scroll &lt;strong&gt;all the way to the bottom&lt;/strong&gt; to see a list of some NMR active nuclei on the website of Hans J. Reich (University of Wisconsin):&#xA;&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/07-multi-01-nuclear.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Summary of Nuclear Properties&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll note that some are listed as having low sensitivity which is related to certain properties which we won't go into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing about NMR, from an organic chemistry point of view at least, is that carbon has only one NMR active nuclei, &lt;span class=&quot;math-container&quot;&gt;$\ce{^{13}C}$&lt;/span&gt;, that additionally has very low abundance &lt;span class=&quot;math-container&quot;&gt;$(\approx1\%)$&lt;/span&gt;. &#xA;Hence, when you perform a &lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt; NMR experiment on an organic molecule like ethanol, for example, you are unlikely to see carbon nuclei interacting and causing splitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask how we obtain &lt;span class=&quot;math-container&quot;&gt;$\ce{^{13}C}$&lt;/span&gt; NMR spectra? Well, it is possible, just not easy. Beyond having very low abundance it also possesses a long relaxation time which essentially means that when you excite a &lt;span class=&quot;math-container&quot;&gt;$\ce{^{13}C}$&lt;/span&gt; nuclei there is a long period of time that elapses before it relaxes back to the ground state and is capable of being excited again to obtain a clean spectra.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="4945" LastEditDate="2019-01-22T16:19:28.427" LastActivityDate="2019-01-22T16:19:28.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12287" PostTypeId="2" ParentId="12265" CreationDate="2014-06-03T01:29:05.477" Score="10" Body="&lt;p&gt;You may be asking for something quite a bit deeper, and it's rather hard to strictly adhere to the requirements in the question, but after Eric's interesting edit showing some outside-the-box thinking, I went along those lines and came up with some options. I'll let you decide which (if any) make the cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two good candidates for master principles in chemistry may be &lt;em&gt;acid-base theory&lt;/em&gt; and &lt;em&gt;reduction-oxidation theory&lt;/em&gt;. I think ultimately they're just somewhat coarse measures of the relative positions between HOMOs and LUMOs for different species, but after a couple centuries of research and some tabulation, we've come up with general top-level rules for understanding the chemical behaviour of large classes of compounds, with significant predictive power, without necessarily resorting to complicated calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;periodic table&lt;/em&gt; itself could also work; different elements possess certain chemical similarities which allows them to be grouped with one another. Mendeleev famously used this to predict a handful of new elements, such as germanium (though he did also predict some other elements which never came to be).  A related principle which is simple but of great importance is that elements are entirely characterized by the number of protons in their nuclei. Every single atom in the Universe with 79 protons in its nucleus is a gold atom, no matter how many neutrons and electrons are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I wonder if you'd count &lt;em&gt;materialism&lt;/em&gt; as a master principle which emerged in chemistry; every material in the Universe is made of atoms (not counting dark matter). If two materials are the same, then they must be made of the same types and proportions of atoms in the same configurations. If two materials are different, then either the atoms are different, or they are present in different proportions, or they are present in different configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I myself would actually have included the octet rule as a master principle. It has many holes and limited applicability, but it is uncanny that such a simple concept works &lt;em&gt;at all&lt;/em&gt;. I find it a surprising way to effectively skip tons of gritty quantum mechanical calculations, helping chemistry education to begin in school around ages 14-15.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-06-03T01:29:05.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12288" PostTypeId="1" AcceptedAnswerId="12289" CreationDate="2014-06-03T03:04:37.637" Score="1" ViewCount="585" Body="&lt;p&gt;Which path do I take when we have a branched chiral center? None of the textbooks I've consulted adequately covers this or even provides an example problem in which there was branching. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jiAx2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Chiral carbon with the following attachments H, an carboxylic acid group, a methyl group, and then the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) H gets the lowest priority. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The methyl group gets the second lowest priority since it has just one carbon and then hydrogens; the other paths have either more carbons or oxygens! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Between the carboxylic acid group and the path along the ring, we have an oxygen and a hydrogen going for the carboxylic acid group. But which path do I take? After comparing the carbon in the carboxylic acid group to the carbon on the ring, do I compare the double-bonded oxygen to the next carbon on the ring or the single-bonded oxygen? How does one choose in cases of branching such as these? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-06-03T04:42:56.853" LastActivityDate="2015-05-18T09:13:54.267" Title="R, S Configuration and Branching - Which Path?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12289" PostTypeId="2" ParentId="12288" CreationDate="2014-06-03T03:25:50.697" Score="3" Body="&lt;p&gt;No,  you have a carbon figuratively attached to 3 oxygen atoms vs.  a carbon figuratively attached to 3 carbon atoms; the former has higher priority. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-03T03:25:50.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12290" PostTypeId="2" ParentId="12288" CreationDate="2014-06-03T06:42:43.470" Score="1" Body="&lt;p&gt;convert it to &lt;b&gt;fischer projection&lt;/b&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PlPAs.png&quot; alt=&quot;fischer projection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see it is (&lt;b&gt;S&lt;/b&gt;)&lt;br&gt;&#xA;&lt;b&gt;Edit: Added Explanation&lt;/b&gt;&#xA;Going according to atomic numbers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atomic no.1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1.$\ce{H}$- lowest priority.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atomic no. 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2.$\ce{CH3}$ - next priority.Has $\ce{H}[1]$&lt;br&gt;&#xA;3.$\ce{C6H4..}$ - next priority. Has $\ce{C}[6]$&lt;br&gt;&#xA;4.$\ce{COOH}$ - next priority.Has $\ce{O}[8]$  &lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-06-09T03:38:39.870" LastActivityDate="2014-06-09T03:38:39.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12291" PostTypeId="2" ParentId="12234" CreationDate="2014-06-03T06:55:29.633" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOnS8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is one of the isomers.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-03T06:55:29.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12293" PostTypeId="2" ParentId="12277" CreationDate="2014-06-03T10:42:20.923" Score="4" Body="&lt;p&gt;It's not always easy to interpret NMR spectra. Different nuclei can show the same chemical shift, leading to overlapping signals. NMR signals are also not infinitely sharp, overlapping signals are rather common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are also more advanced experiments you can use to solve these issues, if you have enough of your sample and enough time on a spectrometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;H NMR is the most common type because &lt;sup&gt;1&lt;/sup&gt;H is almost 100% abundant, and hydrogens are present in most organic compounds. &lt;sup&gt;13&lt;/sup&gt;C and &lt;sup&gt;15&lt;/sup&gt;N are also used often because they are common in organic chemistry and especially in biology. Those isotopes are rather rare in nature though, this makes experiments involving them much, much less sensitive than proton NMR. If you want to perform more complicated experiments with these nuclei you have to isotope-label your sample, which is routinely done for proteins, but extremely expensive or even impossible for small organic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;19&lt;/sup&gt;F is another isotope that works well for NMR, it is also around 100% abundant and therefore very sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can do NMR on any spin 1/2 nucleus. Nuclei with higher spins is also possible, but that gets more difficult.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-06-03T10:42:20.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12294" PostTypeId="2" ParentId="12248" CreationDate="2014-06-03T12:12:38.340" Score="1" Body="&lt;p&gt;yes it is possible, for example by adding a seed crystal under the right conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the reference for further information: &lt;a href=&quot;http://www.google.com/patents/US2233606&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/US2233606&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-03T12:12:38.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12295" PostTypeId="2" ParentId="12250" CreationDate="2014-06-03T13:32:48.500" Score="4" Body="&lt;p&gt;According to Fownes' Elementary Chemistry 12th Edition (pp. 676ff), ketones are obtained from the dry distillation of calcium salts of the fatty acids. So calcium acetate yields acetone, calcium propionate yields &quot;propione&quot; (diethyl ketone), etc. Lead salts also are stated to work.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-06-03T13:32:48.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12296" PostTypeId="1" AcceptedAnswerId="12297" CreationDate="2014-06-03T14:24:26.763" Score="1" ViewCount="7207" Body="&lt;p&gt;I've always been told that carbon bonds with four hydrogens and that nitrogen bonds with three hydrogens, like in benzene and hexazine, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the name of this type of bonding and how does one come up with the number?&lt;/p&gt;&#xA;" OwnerUserId="4612" LastEditorUserId="4612" LastEditDate="2020-03-03T19:45:20.853" LastActivityDate="2020-03-03T19:45:20.853" Title="How do I find out the number of bonds an atom must link to?" Tags="&lt;bond&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12297" PostTypeId="2" ParentId="12296" CreationDate="2014-06-03T15:01:16.803" Score="1" Body="&lt;p&gt;It's named a covalent bond. The amount of hydrogen atoms that can be bond (or any other atom) can be  calculated most of the time using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;nofollow&quot;&gt;octet rule&lt;/a&gt;, that states that any atom will try to get the electron configuration of a noble gas (i.e. 8 electrons in the outer shell). The electrons from the covalent bond count for both atoms that take part in the bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example consider $\ce{N}$ that has five electrons in its outer shell, and $\ce{H}$ that has just one. $\ce{N}$ wants three more electrons and can get them from covalent bond with hydrogen. $\ce{H}$ want one more electron (this is an exception) in its outer shell to fill it up, so will kindly receive them from nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-06-03T15:01:16.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12298" PostTypeId="2" ParentId="12280" CreationDate="2014-06-03T15:13:02.137" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a frequent misconception about Le Chatelier's principle.  Pressure only changes the equillbrium if pressure changes in a way that changes concentrations, for example by changing volume or adding more of a reactant or product.  If pressure changes in a way that doesn't change concentrations, for example by adding a noble gas to the mixture, or in this case changing temperature, equillibrium is unchanged.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When temperature changes, the equillibrium constant itself changes, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;nofollow&quot;&gt;Van't Hoff Equation&lt;/a&gt;.  For exothermic reactions, increasing temperature favors the reverse reaction.  For endothermic reactions, increasing temperature favors the forward reaction.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5084" LastEditDate="2014-06-03T16:47:50.780" LastActivityDate="2014-06-03T16:47:50.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12299" PostTypeId="2" ParentId="12296" CreationDate="2014-06-03T15:22:37.207" Score="1" Body="&lt;p&gt;Atoms, especially those in the first 2 or 3 rows of the Periodic Table, try to attain a filled outer shell of electrons.  This gives them an inert gas electron configuration and a special stability.  They can achieve this filled outer shell of electrons in one of two ways, they can donate or accept electrons (ionic bonding) or share electrons (covalent bonding).  An example of ionic bonding would be sodium with an electronic configuration of $\ce{1s^2 2s^2 2p^6 3s^1}$.  By donating (losing) one electron it attains the inert gas electronic configuration of neon ($\ce{1s^2 2s^2 2p^6}$).  On the other hand, atoms like carbon ($\ce{1s^2 2s^2 2p^2}$) and nitrogen ($\ce{1s^2 2s^2 2p^3}$) would have to gain or lose too many electrons to achieve an inert gas configuration, so they typically form covalent bonds by sharing electrons.  In order for carbon to achieve the neon configuration it would need to share 4 more electrons.  That is why carbon typically forms &lt;strong&gt;4&lt;/strong&gt; covalent bonds.  In the case of methane, $\ce{CH4}$, each hydrogen atom brings one electron to share with the carbon atom and in turn each hydrogen shares one electron from carbon.  By sharing electrons like this, each hydrogen achieves a stable helium configuration and carbon achieves a stable neon electron configuration.  In the case of nitrogen, it only needs to share 3 more electrons in order to achieve the neon configuration.  It can do this by forming &lt;strong&gt;3&lt;/strong&gt; covalent bonds.  If we look at ammonia, $\ce{NH3}$, we can apply the same reasoning we applied to methane.  By sharing electrons with 3 hydrogen atoms, both the nitrogen and hydrogen atoms can achieve inert gas electronic configurations.  Simply stated, atoms try to form as many bonds as necessary, be they ionic or covalent, in order to achieve an inert gas configuration.  While there are many exceptions to this approach, especially for heavier atoms, it does seem to work reasonably well for the lighter elements.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-03T15:22:37.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12300" PostTypeId="1" CreationDate="2014-06-03T17:01:31.613" Score="2" ViewCount="113" Body="&lt;p&gt;If you have some sample in solution (of a pure R substance), wouldn't the individual molecules all have a random orientation in space? Or do the photons emitted into the solution cause all of the molecules to align? Or is it just that, statistically, a given percentage of the particles will have an orientation which corresponds to the polarized light source. &lt;/p&gt;&#xA;" OwnerUserId="5733" LastEditorUserId="4231" LastEditDate="2015-10-09T17:37:52.473" LastActivityDate="2015-10-09T17:37:52.473" Title="How does polarimetry account for the orientation of the molecules?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12301" PostTypeId="1" AcceptedAnswerId="12303" CreationDate="2014-06-03T17:35:53.973" Score="0" ViewCount="188" Body="&lt;p&gt;I was searching for a sol gel method for deposition of carbon nanotube thin films and there aren't too many articles on that.  Does anybody know the reason?  Is it because Carbon nanotubes are difficult to disperse in many solvents or?  Does anyone know if it is possible to effectively use a sol gel method to make a CNT thin film?&lt;/p&gt;&#xA;" OwnerUserId="3730" LastActivityDate="2014-06-03T18:08:03.390" Title="Is there a good sol gel method for carbon nanotube thin film deposition?" Tags="&lt;solubility&gt;&lt;solvents&gt;&lt;nanotechnology&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12302" PostTypeId="2" ParentId="12300" CreationDate="2014-06-03T17:43:15.233" Score="3" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; The light beam interacts with all of the randomly aligned molecules in its path.  There are roughly $\ce{6 x 10^{23}}$ molecules in a mole of material. If we were to dissolve a micro mole of this compound in 10 ml (1 deciliter is a typical polarimetry sample size) of solvent we would have $\ce{6 x 10^{17}}$ molecules in our sample cell.  Since the molecules are randomly oriented, there are enough molecules so that the polarized light beam will sample all possible orientations of the molecule with respect to the light beam many, many times over.  This allows for measurements to be made in an accurate and statistically reproducible manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Detailed answer:&lt;/strong&gt; A beam of light has associated with it an electric and magnetic field.  In simple polarimetry experiments it is the electric field (or vector) associated with the light beam that is of interest.  Let's say the beam is propagating in the Z direction, then the propagating electric vector lies in an infinite number of planes that contain the Z axis.  When we plane-polarize the electric vector associated with the light beam, we remove all of the planes that the electric vector resides in except for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next important point is to realize that a plane polarized electric vector associated with a beam of light can also be represented by (is equivalent to) two circularly polarized beams (or vectors), rotating in opposite directions (clockwise and counter-clockwise), each one-half the intensity of the plane polarized electric vector.  What we wind up with is two helices, one right-handed and one left-handed, traveling through the sample in our polarimeter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say our polarimeter sample contains a significant concentration of chiral molecules in solution.  Although the chiral molecules are randomly aligned with respect to the light beam, each chiral molecule will interact with the right-handed or clockwise rotating helical electric vector differently than it will interact with the left-handed or counter-clockwise rotating helical electric vector.  Our sample must contain a large number of molecules so that, despite all of the different molecular alignments between our molecules and the beam, when averaged over the entire sample, the difference in interaction of the right- and left-handed helices with our chiral sample will be repeatable.  It is not hard to have a significant concentration of molecules because a reasonably long sample path length is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unequal (dissymmetric) interaction between our chiral sample and the right- and left-handed helices results in a rotation of the plane of polarization of the emerging light beam from its original position - and this angle of rotation is what the polarimeter measures. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-04T18:06:09.770" LastActivityDate="2014-06-04T18:06:09.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12303" PostTypeId="2" ParentId="12301" CreationDate="2014-06-03T18:08:03.390" Score="0" Body="&lt;p&gt;I think it may depend on what you mean by &quot;effectively.&quot; I imagine that it would be difficult to get a thin film of &lt;em&gt;oriented&lt;/em&gt; CNTs using a sol-gel method, since the gel would necessarily have randomly oriented tubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other potential issue is making the sol-gel in the first place. Unless they are functionalized, carbon nanotubes won't gel in hydrophilic solvents. It might be possible to create gels in hydrophobic solvents, but a quick search didn't turn anything like that up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.ipme.ru/e-journals/RAMS/no_13614/05_13614_fu.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; has a brief description of a typical process - you can see that the idea is to make the sol-gel with something that will gel in aqueous solution first (silica, for example), and then add the nanotubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want &lt;em&gt;pure&lt;/em&gt; CNT thin films, that would probably be a lot more difficult.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-03T18:08:03.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12304" PostTypeId="1" AcceptedAnswerId="12362" CreationDate="2014-06-03T19:14:35.677" Score="1" ViewCount="317" Body="&lt;p&gt;At Wikipedia a &lt;a href=&quot;http://en.wikipedia.org/wiki/Monomer&quot; rel=&quot;nofollow&quot;&gt;monomer&lt;/a&gt; is defined to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a molecule that may bind chemically to other molecules to form a&#xA;  polymer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymer&quot; rel=&quot;nofollow&quot;&gt;polymer&lt;/a&gt; is defined to be&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a molecule composed of many monomers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These definitions are obviously circular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there non-circular definitions of monomers and polymers?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5735" LastActivityDate="2014-06-05T03:58:50.730" Title="Monomers and polymers" Tags="&lt;polymers&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12306" PostTypeId="2" ParentId="12304" CreationDate="2014-06-03T20:40:53.077" Score="1" Body="&lt;p&gt;The term &quot;monomer&quot; only has meaning in the context of discussing polymerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polymer is a molecular chain where the links of the chain are repeated identical or a least similar units.  Examples where the units are idential would be polypropylene, polyvinylchloride and polysiloxane.  Example where the units are similar but not necessarily identical would be DNA and proteins.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A monomer is a molecule that can react to form a single unit of a polymer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, propylene is a monomer of poylpropylene and amino acids are monomers of a protein.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-03T21:47:36.820" LastActivityDate="2014-06-03T21:47:36.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12307" PostTypeId="1" CreationDate="2014-06-03T20:56:42.870" Score="0" ViewCount="151" Body="&lt;p&gt;How does formylation, of the sort shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zJ3t8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zJ3t8.png&quot; alt=&quot;formylation diagram&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://en.wikipedia.org/wiki/Formylation&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;) affect the hydrophobicity of the target protein?  Can we say anything general about adding a formyl group to a nitrogen?&lt;/p&gt;&#xA;" OwnerUserId="5737" LastEditorUserId="31313" LastEditDate="2017-04-01T02:24:16.590" LastActivityDate="2017-04-01T02:24:16.590" Title="How does formylation of a nitrogen change the hydrophobicity of a target protein?" Tags="&lt;organic-chemistry&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12308" PostTypeId="1" CreationDate="2014-06-03T21:11:50.620" Score="1" ViewCount="33" Body="&lt;p&gt;I have been given a protein of unknown concentration and have been asked to make a set of standards using a protein with a known concentration so we can quantify roughly via dot blot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know how to work out the calculation part. I would like to learn both mass and molar.For example, the information I have for my standard protein is it is at a concentration of 30mg/ml and 550uM. I want to put, for example, 1ul dots of this protein on the membrane for my blot. I want several dilutions of my standard, for example ranging from 20pg to 100ng. How do I work it out to get varied concentrations from the stock to my 1ul volumes on my blot?&lt;/p&gt;&#xA;" OwnerUserId="5738" LastActivityDate="2014-06-03T21:11:50.620" Title="How do I go about calculating protein concentration to make a set of protein standards for protein quantification via dot blot?" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12309" PostTypeId="1" AcceptedAnswerId="12316" CreationDate="2014-06-03T21:13:12.797" Score="3" ViewCount="3409" Body="&lt;p&gt;In my teacher's power-point states that we must consider that inductive effect in $CH_{3}COOH$ exists because of the electron deficiency in $C$ (partial positive charge $\delta^{+}$) as a result of the electronegativity of the double bonded $O$. As you may notice in the figure &lt;a href=&quot;https://i.stack.imgur.com/xNA7d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; although the single bonded $O$ is more electronegative than $C$, the electron density of the single bonded $O$ moves to $C$ in order to &quot;neutralize&quot; the $\delta^{+}$, leaving $H$ less attached to the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we consider that double bonded $O$ establishes that $\delta^{+}$ on our $C$ (and creates an electron withdrawing inductive effect $-I$) and not the single bonded $O$? Is this because of the $\pi$ double bond? Can you tell me why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xNA7d.png&quot; alt=&quot;asd&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure &lt;a href=&quot;https://i.stack.imgur.com/xNA7d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/strong&gt;: Inductive effect created by the electron deficiency of $C$. As we can see the $H$ is now less attached to our molecule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now supose we have the following carboxylic acids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$CH_{3}COOH$, &lt;em&gt;pka&lt;/em&gt; = $4.80$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$FCH_{2}COOH$, &lt;em&gt;pka&lt;/em&gt; = $2.66$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_{2}^{-}CCH_{2}COOH$, &lt;em&gt;pka&lt;/em&gt; = $5.69$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$O_{2}^{-}C(CH_{2})_{4}COOH$, &lt;em&gt;pka&lt;/em&gt; = $5.41$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me it makes sense that $FCH_{2}COOH$ is more acidic than $CH_{3}COOH$, because $F$ is an electronegative element and he creates a partial negative charge moving de electron density to him, leaving our $H$ in the carboxyl group less attached and easier to remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the fact $O_{2}^{-}CCH_{2}COOH$ is less acidic than $O_{2}^{-}C(CH_{2})_{4}COOH$ for me it's less intuitive. In my teacher's power-points states that $O_{2}^{-}$ linked to an alkyl chain will create an electron donating inductive effect $+I$. Can you explain to me why? Isn't there a withdrawing resonance effect $-R$? If $O$ is an electronegative element why doesn't it create an $-I$ effect instead of $+I$? Beacause if that was the case adding carbons $(CH_{2})_{4}$ in the chain would reduce the the $-I$ effect and the $O_{2}^{-}C(CH_{2})_{4}COOH$ would be less acidic than $O_{2}^{-}CCH_{2}COOH$, and that isn't the case.&lt;/p&gt;&#xA;" OwnerUserId="5540" LastEditorUserId="4945" LastEditDate="2014-06-05T11:50:14.983" LastActivityDate="2014-06-05T11:50:14.983" Title="Inductive effect" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12310" PostTypeId="2" ParentId="12309" CreationDate="2014-06-03T22:05:20.810" Score="1" Body="&lt;p&gt;I think both oxygens will have an inductive effect on the bonded carbon, but since the single bonded oxygen is also attached to a hydrogen, its inductive effect is decreased relative to that of the double-bonded oxygen. You can think of both oxygens as having an inductive effect, but the net inductive effect is definitely away from the carbon and in the direction of the double-bonded oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also did you mean to write negatively charged diatomic oxygen? I think you meant to write $\ce{O^{2-}}$. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-03T22:37:17.930" LastActivityDate="2014-06-03T22:37:17.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12311" PostTypeId="2" ParentId="12309" CreationDate="2014-06-03T22:27:33.580" Score="1" Body="&lt;p&gt;The last pka is for the negative ion? Isn't it obvious, that in that case leaving hydrogen cation has to overcome attraction from two negative charges instead of one and have troubles with it?&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-03T22:27:33.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12312" PostTypeId="2" ParentId="12288" CreationDate="2014-06-03T22:29:34.453" Score="3" Body="&lt;p&gt;As @ron mentioned $\ce{-CO_{2}H}$ has higher priority so the numeration given by @ADG in the fischer projection is incorrect because he assumes that the ring has higher priority than $\ce{-CO_{2}H}$. You can also think in the ring priority as $(C,C,C)$ and $\ce{-CO_{2}H}$ priority as $(C,O,O)$. As you may notice$(C,O,O)$ has higher priority than $(C,C,C)$ (the total atomic number of the substituent group in the first case is higher).&lt;br&gt;&#xA;See the following figure: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SNzeO.jpg&quot; alt=&quot;&quot;&gt;&lt;br&gt;&#xA;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Absolute_Configuration,_R-S_Sequence_Rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case I don't think you need to continue the path, but if that's the case you can see how it works in the figure.&lt;/p&gt;&#xA;" OwnerUserId="5540" LastEditorUserId="13339" LastEditDate="2015-05-18T09:13:54.267" LastActivityDate="2015-05-18T09:13:54.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12313" PostTypeId="1" AcceptedAnswerId="12324" CreationDate="2014-06-03T22:44:16.887" Score="1" ViewCount="254" Body="&lt;p&gt;I'm trying to apply conjugate theory, which I can apply very well to acids/bases, to redox. Can you verify my logic, which I have broken down below? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I know that solid sodium is a strong reducer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I know this because solid sodium when placed in water reacts violently, forming, among other products, sodium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Reducers are oxidized. Reducers lose electrons. Correct; $\ce{Na^{+}}$ is the product of placing solid sodium in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) So in this half-reaction: $\ce{Na&amp;lt;=&amp;gt;Na^{+} +e^-}$, sodium is the reducer. $\ce{Na^{+}}$ is the &lt;em&gt;conjugate&lt;/em&gt; oxidizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Oxidizers are reduced. Oxidizers &lt;em&gt;gain&lt;/em&gt; electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Because sodium is a strong reducer, which means it has a strong tendency to be oxidized, or a strong tendency to lose electrons, its &lt;em&gt;conjugate&lt;/em&gt; oxidizer &lt;em&gt;must&lt;/em&gt; be weak. Oxidizers are reduced; oxidizers gain electrons. The potential for $\ce{Na^{+}}$ to gain the electron that $\ce{Na}$ just lost must be small, or $\ce{Na}$ would not be a good reducer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Therefore, a strong reducer's conjugate oxidizer must be weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) Likewise, a strong oxidizer's conjugate reducer must be weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;9) Likewise, a weak reducer's conjugate oxidizer must be strong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;10) Likewise, a weak oxidizer's conjugate reducer must be strong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if anyone could point me to a good reference on gaining &lt;em&gt;insight&lt;/em&gt; into redox, tha would be great! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while we're on the topic of redox, what's the mechanism for this reaction? Are both bonds cleaved homolytically? I see how the nucleophilic chlorines attack the electrophilic hydrogens but how's the bond broken? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_2 + Cl_2 -&amp;gt;2HCl}$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-03T22:59:48.660" LastActivityDate="2014-06-04T04:01:26.410" Title="Conjugate Theory and Redox" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12314" PostTypeId="2" ParentId="12307" CreationDate="2014-06-03T23:00:51.493" Score="1" Body="&lt;p&gt;Formylmethionine is used as the start amino acid for a protein chain in certain organisms (the prokaryotes.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the alpha nitrogen is formylated, it cannot be in the middle of the chain. This is never seen in real proteins, partly because there would be two carbonyl groups attached to the nitrogen, which would have a very strong electron withdrawing effect and be quite unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/N-Formylmethionine&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/N-Formylmethionine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other organisms (the eukaryotes, including the animals and humans) use methionine as the start amino acid of the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Genetic_code&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Genetic_code&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there can only ever be one N-formylated methionine in a real protein (and it is often removed after protein synthesis in a step called post-translational modification) the effect on hydrophobicity will not be very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand, you were to take an existing protein and formylate the -(CH2)4-NH2 side chains on all the lysine amino acid residues, I would expect a slight increase in hydrophobicity in neutral solution. In acid solution I would expect the difference to be more marked, as the formylated NH2 groups would lose their ability to become ionised by accepting a proton from the acid.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2014-06-03T23:00:51.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12315" PostTypeId="1" AcceptedAnswerId="12371" CreationDate="2014-06-03T23:06:54.943" Score="5" ViewCount="2566" Body="&lt;p&gt;What is the mechanism for the electrolysis of water? I understand that hydrogen and oxygen gas are made, but how exactly does this happen when electrons are passed through water? A diagram would be very useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this have anything to do with electronegativities? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-06T18:04:59.537" LastActivityDate="2020-08-16T15:12:56.287" Title="Electrolysis of Water" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12316" PostTypeId="2" ParentId="12309" CreationDate="2014-06-03T23:25:27.840" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do we consider that double bonded O establishes that δ+ on our C&#xA;  (and creates an electron withdrawing inductive effect −I) and not the&#xA;  single bonded O?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the resonance structures for acetic acid in the top line of the following drawing.  We can see that the middle resonance structure takes the pi electrons involved in the $\ce{C=O}$ double bond and moves them to oxygen, creating a positive charge at the carbonyl carbon.  In the third, or left-most resonance structure we see that the &quot;single-bonded oxygen&quot; is also involved in resonance delocalization, but it does not create any positive charge at the carbonyl carbon.  Aside from resonance factors, both oxygens will create partial positive charge at the carbonyl carbon through inductive effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwCrx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question about the relative $\ce{pK_{a}s}$ of these acetic acid derivatives let's proceed as follows.  First, let's consider resonance effects of the substituent ($\ce{X}$) to be negligible.  This is because 1) the substituent is not directly attached to the carbonyl group or the resulting carboxylate anion, 2) it turns out there is a node at $\ce{C2}$ in the carboxylate anion (zeroth order approximation), so the $\ce{X}$ group is disconnected in a resonance sense.  So we need only consider the inductive effect of $\ce{X}$ upon both the reactant and the product and this should allow us to estimate how the equilibrium involving acid ionization will shift.  $\ce{X=H}$ is our baseline and we will compare the other two compounds against it.  When $\ce{X=F}$ the starting acid ($\ce{A}$) will be inductively destabilized due to the electron withdrawing inductive effect of fluorine.  On the other hand, the inductive effect of fluorine will serve to stabilize the resultant carboxylate anion ($\ce{B}$).  These two effects (destabilization of the reactant and stabilization of the product) will act in concert to push our equilibrium to the right, making fluoroacetic acid a stronger acid than acetic acid.  In the case of $\ce{X=CO2^-}$, the inductive electron donating effect of the $\ce{-CH2CO2^-}$ group should be stronger than the $\ce{CH2-H}$ group in acetic acid, so our reactant should be more stabilized.  On the other hand, the inductively electron donating $\ce{-CH2CO2^-}$ should destabilize the already electron rich carboxylate anion ($\ce{B}$).  Here both inductive effects again work in concert, but now to stabilize the reactant and destabilize the product making this derivative less acidic than acetic acid.  In addition to this inductive effect there is also an electrostatic effect.  When $\ce{X=CO2^-}$, we will have two negative charges in $\ce{B}$.  This electrostatic effect further destabilizes $\ce{B}$.  The combination of inductive and electrostatic effects in this last case will make this acid much less acidic than acetic acid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-03T23:57:21.960" LastActivityDate="2014-06-03T23:57:21.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12318" PostTypeId="1" AcceptedAnswerId="12319" CreationDate="2014-06-04T02:05:30.020" Score="2" ViewCount="195" Body="&lt;p&gt;In other words, are these two structures equivalent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tsOjE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tsOjE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2017-04-06T23:16:49.967" LastActivityDate="2017-04-06T23:36:10.997" Title="Does the arrangement of atoms around a chiral carbon matter?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12319" PostTypeId="2" ParentId="12318" CreationDate="2014-06-04T02:20:27.633" Score="3" Body="&lt;p&gt;Yes, they are both (S)-2-chlorobutane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-04T02:20:27.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12320" PostTypeId="1" AcceptedAnswerId="12354" CreationDate="2014-06-04T02:38:52.857" Score="0" ViewCount="4339" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/llwim.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that to correctly assign R,S priorities I have to rotate the above figure until the H is projecting into the page (away from me). However, I don't know how to rotate the H and the rest of the molecule when the H is in the plane of the page. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I imagine tilting the molecule back I get a see-saw shaped molecule ... obviously incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZGlV2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZGlV2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2017-04-06T23:18:02.547" LastActivityDate="2019-01-21T20:23:41.473" Title="How to rotate a chiral carbon" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12321" PostTypeId="1" CreationDate="2014-06-04T03:24:50.607" Score="4" ViewCount="2264" Body="&lt;p&gt;What is amine inversion, and how does it affect basicity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, why is &lt;span class=&quot;math-container&quot;&gt;$\ce{Et3N}$&lt;/span&gt; less basic than (2.2.2.)cyclohexyl-amine despite the +I from ethyl groups? The answer to the question is option 3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5fKL3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5fKL3.png&quot; alt=&quot;compounds&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  Compare the basic strengths of the compounds given:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;(i) &gt; (ii) &gt; (iii)&lt;/li&gt;&#xA;  &lt;li&gt;(ii) &gt; (i) &gt; (iii)&lt;/li&gt;&#xA;  &lt;li&gt;(ii) &gt; (iii) &gt; (i)&lt;/li&gt;&#xA;  &lt;li&gt;(iii) &gt; (ii) &gt; (i)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5708" LastEditorUserId="4945" LastEditDate="2020-02-19T16:51:38.733" LastActivityDate="2020-02-19T16:51:38.733" Title="How does amine inversion affect basicity?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12322" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2014-06-04T03:30:42.063" Score="2" ViewCount="10853" Body="&lt;p&gt;What is the order of basicity between $\ce{Me2NH}$, $\ce{Me3N}$, $\ce{MeNH2}$ in a protic solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has something to do with field effect which I could not understand and could not find anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain?&lt;/p&gt;&#xA;" OwnerUserId="5708" LastEditorUserId="7448" LastEditDate="2015-10-16T18:06:36.887" LastActivityDate="2015-10-16T18:06:36.887" Title="Field Effect and it's effect on basicity on amines" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;solvents&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12323" PostTypeId="1" AcceptedAnswerId="12356" CreationDate="2014-06-04T03:34:02.117" Score="2" ViewCount="2116" Body="&lt;p&gt;What are the complete steps of reaction from sodium benzoate and ascorbic acid to benzene? I can't seem to visualize it, and, much less, find anything online regarding the actual steps of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally: how does it make use of, say Cu(II) or Fe(III) to complete the reaction? It seems that the reaction can only proceed with these present.&lt;/p&gt;&#xA;" OwnerUserId="5740" LastEditorUserId="41328" LastEditDate="2017-08-05T22:06:14.867" LastActivityDate="2017-08-05T22:06:14.867" Title="What is the complete mechanism of reaction for sodium benzoate and ascorbic acid?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12324" PostTypeId="2" ParentId="12313" CreationDate="2014-06-04T03:45:20.467" Score="2" Body="&lt;p&gt;Your question covers two different topics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Redox properties can be well described with redox potentials attributed to a given reaction. It is essentially the same concept that you are talking about: we assign a number (conveniently it is an actual measurable potential) to an electrode reaction which includes an oxidized and reduced form of the species we are talking about. This number, the redox potential, can show you how strong oxidizer/reducer is the corresponding corresponding form, and just as you guess, strong oxidizers generally have a weak reducer pair. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\ce{HCl}$ formation reaction is generally a radical chain reaction, triggered e.g. by light, and not a redox reaction. The key here is to split the $\ce{Cl2}$ (e.g photo-chemically), then the obtained free radicals induce a chain-reaction with elementary steps like &#xA;$\ce{Cl + H2 -&amp;gt; HCl + H}$ or &#xA;$\ce{H + Cl2 -&amp;gt; HCl + Cl}$ &#xA;The two gas mixed is more or less inert in total darkness. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="4945" LastEditDate="2014-06-04T04:01:26.410" LastActivityDate="2014-06-04T04:01:26.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12325" PostTypeId="2" ParentId="12322" CreationDate="2014-06-04T04:11:20.547" Score="1" Body="&lt;p&gt;The basicity of amines has to do not only with the degree of substitution with electron-releasing alkyl groups but also factors such as steric hindrance and degree of solvation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronegative nitrogen is able to isolate electron density from the alkyl groups. So having more electron-releasing alkyl groups to stabilize the positive formal charge created by the abstraction of a proton from the protic solvent might seem like a good thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand steric hindrance due to bulky alkyl groups reduces basicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, having more alkyl groups reduces solvation due to hydrogen &quot;bonding.&quot; Consider ammonia, $\ce{NH_3}$. This molecule can form four hydrogen bonds. On the other hand, a tertiary amine that has been protonated can form only &quot;good&quot; hydrogen bond; one might argue that the hydrogens on the alkyl groups can form hydrogen &quot;bonds&quot;, but these are obviously going to be weaker &quot;bonds&quot; compared to the ones among ammonia molecules.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-04T04:17:02.430" LastActivityDate="2014-06-04T04:17:02.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12326" PostTypeId="1" AcceptedAnswerId="15310" CreationDate="2014-06-04T07:05:30.863" Score="4" ViewCount="203" Body="&lt;p&gt;In preparation for the Birch reduction, I am trying to prepare an electride salt, &lt;span class=&quot;math-container&quot;&gt;$\ce{[Na(NH3)6]+, e-}.$&lt;/span&gt; This salt should be brilliant blue, as far as I know, but I am getting a black sludge when adding sodium to the liquid anhydrous ammonia at −60 °C or so.&lt;/p&gt;&#xA;&lt;p&gt;Is this because the salt is missing the electron?&#xA;How do I remedy this?&lt;/p&gt;&#xA;" OwnerUserId="5741" LastEditorUserId="41328" LastEditDate="2020-11-19T21:15:58.153" LastActivityDate="2020-11-19T21:15:58.153" Title="Electride salt preparation" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;salt&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12327" PostTypeId="1" CreationDate="2014-06-04T07:11:36.267" Score="0" ViewCount="404" Body="&lt;p&gt;I came across this question in a test:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The enol form is the ________ form for 3-hydroxy pent-2-enoate?  (major/minor)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My argument is that the enol form must be more stable because of hydrogen bonding in the enol form and subsequent six membered ring formation (chelation). However the keto form has no such stabilization factor. Hence according to me the enol form must be more stable. But sadly, that is the wrong answer. What's going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure for reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xvlWw.png&quot; alt=&quot;enol form&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4160" LastEditorUserId="7448" LastEditDate="2015-10-16T16:42:48.263" LastActivityDate="2016-02-07T12:20:43.607" Title="Major form of methyl 3-hydroxy pent-2-enoate?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12328" PostTypeId="2" ParentId="12327" CreationDate="2014-06-04T07:41:08.093" Score="1" Body="&lt;p&gt;quick googling shows that for a very similar compound ethyl acetoacetate enol and ketoforms are in equilibrium with 10% of enol form. However, this equilibrium is influenced by solvent significantly. Actually, the percentage shows that the difference in stability between these two forms is veeery little. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference may be attributed to the difference in stability of the bonds, i.e.   C=O + C-H is better then C=C + C-O + O-H. However, internal hydrogen bonding DOES exist and participates in stability of enol form. For example, for hexan-1,3-one, the percentage of enol form is much lower. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-04T07:41:08.093" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12331" PostTypeId="1" AcceptedAnswerId="12347" CreationDate="2014-06-04T14:34:21.137" Score="1" ViewCount="119" Body="&lt;p&gt;I have read about &quot;electroflocculation&quot; that uses &lt;em&gt;less&lt;/em&gt; flocculant/alum, but is there a method – no matter how expensive or impractical – that uses &lt;em&gt;no&lt;/em&gt; flocculant at all?&lt;/p&gt;&#xA;" OwnerUserId="5679" LastEditorUserId="21964" LastEditDate="2016-08-17T17:49:19.317" LastActivityDate="2020-08-20T07:04:00.040" Title="Water purification procedure that does not involve a flocculant of any kind?" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;colloids&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12332" PostTypeId="1" CreationDate="2014-06-04T14:48:18.517" Score="4" ViewCount="5913" Body="&lt;p&gt;I read in a book that &quot;Sometimes, both the electrodes dip in the same electrolyte solution &amp;amp; in such cases we do not require a salt bridge&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="5743" LastEditorUserId="73" LastEditDate="2014-06-04T15:44:45.410" LastActivityDate="2015-05-09T16:44:52.663" Title="Is there always a need for a salt bridge to let a galvanic cell work continuously?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12334" PostTypeId="2" ParentId="12318" CreationDate="2014-06-04T15:08:26.773" Score="0" Body="&lt;p&gt;The difference between these structures is a rotation of the molecule by 60 degrees. You could build a model to convince yourself that they are the same. The arrangement of the atoms around the chiral center &lt;em&gt;does&lt;/em&gt; matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The utility of the R/S convention is that you can unambiguously decide whether the two compounds are the same or different, which can be difficult depending on the orientation of the compound.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="402" LastEditDate="2017-04-06T23:36:10.997" LastActivityDate="2017-04-06T23:36:10.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12335" PostTypeId="1" AcceptedAnswerId="12336" CreationDate="2014-06-04T16:42:18.713" Score="2" ViewCount="1522" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Do52y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above molecule, I know that the longest carbon chain has 7 carbons, so the backbone is heptane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two substituents are at the 4 position on the heptane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the substituents is a methyl group. That much is obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other appears to be a complex substituent and is a 1-chloroethyl group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, would the name be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4,4-(1-chloroethyl)methylheptane? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the issue is what happens when we have &lt;em&gt;different&lt;/em&gt; substituents with the same loctant. I've looked in one book and at all its practice problems, and this scenario &lt;em&gt;never&lt;/em&gt; occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that 4-methylheptane is a real compound and that its name is &lt;em&gt;correct&lt;/em&gt;. So I am leaning toward 4,4-(1-chloroethyl)methylheptane as being a correct name; the complex substituent is correctly named; both locatants are present; everything is alphabetized; my only issue is whether I am missing some convention (e.g. should there be a dash somewhere, a comma somewhere, or what?) &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-08-17T07:26:01.280" LastActivityDate="2014-08-17T07:26:01.280" Title="Nomenclature with Complex Substituents" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12336" PostTypeId="2" ParentId="12335" CreationDate="2014-06-04T16:49:35.757" Score="6" Body="&lt;p&gt;Close. But each (different type of) group added to the heptane backbone is numbered separately: 4-(1-chloroethyl)-4-methylheptane. Also, you have an asymmetric carbon, so the chirality needs to be specified (I will leave that to you).&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-06-04T16:49:35.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12337" PostTypeId="1" AcceptedAnswerId="12338" CreationDate="2014-06-04T17:05:24.977" Score="1" ViewCount="8676" Body="&lt;p&gt;Consider (R)-3-bromo-1,1-dimethylcyclohexane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to clarify how I assign R,S priorities in a cyclic compound. My book says to compare the atoms along the &quot;path of higher priority.&quot; The problem is in a cyclic compound, if you go far along this path of higher priority without finding a point of difference, you may collide (picture below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that because the path taken initially results in no difference, we can still assign the top path a higher priority than the lower path because the top path has two methyl groups. My book says that if the paths are identical then choose the one with more of the same atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZaHPY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tqIP5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SRss0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-04T17:40:44.530" LastActivityDate="2014-06-04T17:40:44.530" Title="R,S in Cyclic Compounds - Path of Highest Priority?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12338" PostTypeId="2" ParentId="12337" CreationDate="2014-06-04T17:16:21.070" Score="3" Body="&lt;p&gt;If you &quot;collide&quot; without finding some difference, then the molecule is not chiral.  In the example at hand, it appears that you have analyzed it correctly.  First, you look at the 2 carbons adjacent to the chiral center.  They are both $\ce{CH_{2}{'}s}$; no difference, therefore move out to the next carbon on each side of the ring.  Here we find a $\ce{C-CH_{2}-C}$ on one side and a $\ce{C-CMe_{2}-C}$ on the other side.  The latter has higher priority than the former.  Stop at this point, no need to go further, and complete the &quot;1-2-3-4&quot; prioritization.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-04T17:16:21.070" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12339" PostTypeId="1" AcceptedAnswerId="12342" CreationDate="2014-06-04T18:07:05.070" Score="2" ViewCount="1204" Body="&lt;p&gt;According to this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.youtube.com/watch?v=ns4Py96Bkn8&amp;amp;NR=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.youtube.com/watch?v=ns4Py96Bkn8&amp;amp;NR=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cgWJt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is this true? (You might recall I asked a similar question about $\ce{sp^2}$ hybridization implying $\ce{sp^2}$ hybridization for all directly linked atoms a while back, but I couldn't remember my prof's exact wording). I &lt;em&gt;think&lt;/em&gt; this YouTube video is saying the same thing my prof said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does this rule imply that both the oxygens in the below diagram are $\ce{sp^2}$ hybridized? Each is connected to an $\ce{sp^2}$ carbon atom. Each has lone pair(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/owZLy.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Even if we didn't know that the rule existed, we should still know that both oxygens are $\ce{sp^2}$ hybridized because of the following resonance possibility, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6zoOe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ETA: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above is true then this doesn't make sense: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pi5WS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-21T14:09:35.367" LastActivityDate="2014-06-21T14:09:35.367" Title="Hybridization in Light of Resonance" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;hybridization&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12341" PostTypeId="1" AcceptedAnswerId="17167" CreationDate="2014-06-04T18:19:07.180" Score="3" ViewCount="417" Body="&lt;p&gt;If we add an ionic solute to water, will the specific heat of the water be increased or decreased? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the specific heat is a measure of how much energy we must add to water for the average kinetic energy of the water molecules to go up by a certain unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand that hydrogen bonding is what limits the kinetic energy of the water molecules. This accounts for water's high specific heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when adding an ionic solute to water, such as $\ce{KI}$, we disrupt the hydrogen bonding. So we may expect the specific heat of a water solution to be lower than that of pure water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there are ion-dipole interactions created, correct? These will act to at least partially counteract the specific heat depression, correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-06-05T11:46:06.120" LastActivityDate="2014-10-07T05:10:51.357" Title="Specific Heat of Water Solution" Tags="&lt;physical-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12342" PostTypeId="2" ParentId="12339" CreationDate="2014-06-04T20:03:56.667" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Is this true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's more or less true in the cases were hybridization methods \ techniques work.  For example, when the hybridization of water is discussed it is often said that the central oxygen is more or less $\ce{sp^3}$ hybridized and therefor the two lone pairs on oxygen are in (roughly) $\ce{sp^3}$ orbitals.  Current thinking based on photoelectron spectral data suggests that, instead of this traditional view, one lone pair appears to be in an orbital that is predominantly $\ce{P}$ and the other in an orbital that is predominantly $\ce{S}$.  So hybridization models don't always give the correct answer.  I'm not sure what the hybridization of an atom with a lone pair that's attached to a double bond really is.  But it is probably OK to say what the teacher said in an introductory class in order to keep things simple for a while.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Does this rule imply that both the oxygens in the below diagram are&#xA;  sp2 hybridized?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the traditional hybridization approach is applicable here (e.g. is the ester oxygen behaving like water or is it &quot;traditionally&quot; $\ce{sp^3}$ hybridized?) then yes, both oxygens are roughly $\ce{sp^2}$ hybridized.  Remember, hybridization is not a step function.  There are hybridization indices between 2 and 3.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Even if we didn't know that the rule existed, we should still know&#xA;  that both oxygens are sp2 hybridized because of the following&#xA;  resonance possibility, correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some data on esters suggests that the $\ce{C-O-C}$ angle is around 115 degrees.  Somewhat between 109 degrees ($\ce{sp^3}$) and 120 degrees ($\ce{sp^2}$).  And maybe an intermediate hybridization is what we should expect based on the resonance structures for an ester.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-04T20:03:56.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12343" PostTypeId="2" ParentId="8282" CreationDate="2014-06-04T20:46:33.677" Score="-2" Body="&lt;p&gt;Refrigeration condenses the CO2 and allows for a &quot;softer&quot; bubble feel in the mouth.  &lt;br/&gt;&#xA;&lt;br/&gt;As for shelf life of soda, it should not effect a &quot;canned&quot; soda one way or another. &lt;br/&gt;&lt;/p&gt;&#xA;" OwnerUserId="5699" LastActivityDate="2014-06-04T20:46:33.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12344" PostTypeId="1" CreationDate="2014-06-04T21:00:59.860" Score="-1" ViewCount="1914" Body="&lt;p&gt;What's a natural substitute for baking power or baking soda? I am doing a project where I need a raising agent in liquid form. Naturally, I can't use anything carbonated so baking powder/soda are out of question.&lt;/p&gt;&#xA;" OwnerUserId="5747" LastEditorUserId="73" LastEditDate="2014-06-04T21:07:59.767" LastActivityDate="2014-07-05T08:27:45.737" Title="What's a natural substitute for baking powder or baking soda?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-03-06T02:48:30.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="12345" PostTypeId="1" AcceptedAnswerId="12346" CreationDate="2014-06-04T22:01:19.223" Score="1" ViewCount="839" Body="&lt;p&gt;How do I tell which one is cis/trans? I know the definitions: cis is same side; trans is across; but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What am I comparing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How do I compare/visualize? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fmxjH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Das8V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-04T22:43:04.220" LastActivityDate="2014-06-04T22:43:04.220" Title="Cis/Trans Isomerism in Cycloalkenes" Tags="&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12346" PostTypeId="2" ParentId="12345" CreationDate="2014-06-04T22:39:26.367" Score="5" Body="&lt;p&gt;In simple molecules like what you've drawn above, see if the hydrogens are on the same side of the double bond or not.  If they are on the same side, then the compound is &lt;em&gt;cis&lt;/em&gt;; if the hydrogens are on opposite sides, then the compound is &lt;em&gt;trans&lt;/em&gt;.  In more complex cases, prioritize the four substituents on the double just as you've been doing with &quot;R\S&quot; nomenclature.  If the two highest priority groups are on the same side then the compound is &lt;em&gt;cis&lt;/em&gt;, if they are on opposite sides, then the compound is &lt;em&gt;trans&lt;/em&gt;.  The terms &quot;&lt;em&gt;cis&lt;/em&gt;&quot; and &quot;&lt;em&gt;trans&lt;/em&gt;&quot; are still used, but especially in the more complex cases they have been replaced with the first letters of the German words &quot;entgegen&quot; (E, opposite) and &quot;zusammen&quot; (Z, together).  So cis-2-butene could also be called Z-2-butene.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-04T22:39:26.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12347" PostTypeId="2" ParentId="12331" CreationDate="2014-06-05T00:04:56.463" Score="1" Body="&lt;p&gt;This of course depends on the scale and intent of the purification/treatment process, but flocculation is but one step in the water treatment process. So, I'm not sure whether you are asking if you can purify water without flocculation, or if you can get flocculation/aggregation to occur without the addition of floccing agents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest answer to water purification, on a small scale, is filtration and distillation. Flocculation serves to remove solid particles from the water which contribute to turbidity, but which may also include pathogens that are of real concern to human safety. On larger scales (household to community), incredibly effective methods for removing these types of contaminants include various forms of filtration. Some of the first successful water treatment plants on a large scale used sand filtration, and fast and slow sand filters are still used widely today. There are many commercially available drinking water filters for the removal of sediment, microbes and chemicals which are highly effective. Of course, on an industrial scale, filters of this nature would require frequent changing, and this is where sediment removal via flocculation is most effective in reducing costs. You could Google 'water purification in developing worlds' for a range of water purification methods that do not involve flocculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, inducing flocculation without the addition of flocculant again depends on nature of the contaminant you wish to remove. The biofuel industries are pursuing this goal, developing methods of harvesting valuable microalgae stocks. Autoflocculation (or bioflocculation) of microalgae is known to occur as a function of pH, which can be controlled by restricting CO2 levels. Again a Google search on 'microalgae autoflocculation' will give you many hits. Another method is the light induced autoflocculation of microalgae. There is an interesting post on &lt;a href=&quot;http://www.researchgate.net/post/What_are_the_best_methods_to_induce_microalgae_flocculation_without_chemical_contamination&quot; rel=&quot;nofollow&quot;&gt;ResearchGate&lt;/a&gt; that may be of some interest to you if you are focused on flocculation of microalgae (if not, then ignore!)&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-06-05T00:04:56.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12348" PostTypeId="1" AcceptedAnswerId="12352" CreationDate="2014-06-05T01:17:59.517" Score="8" ViewCount="2743" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\implies 14 - \log(1.0001\times10^{-7})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\implies \mathrm{pH} \approx 7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense?&lt;/p&gt;&#xA;" OwnerDisplayName="user4603" LastEditorUserId="5026" LastEditDate="2018-02-26T05:22:45.087" LastActivityDate="2018-02-26T05:22:45.087" Title="Calculating the pH of a dilluted aqueous KOH solution" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12349" PostTypeId="2" ParentId="7244" CreationDate="2014-06-05T01:24:29.967" Score="4" Body="&lt;p&gt;The simplest answer in thermodynamic terms is &quot;a reversible process is one in which the entropy generation term is zero.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-05T01:24:29.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12350" PostTypeId="2" ParentId="12344" CreationDate="2014-06-05T01:38:12.980" Score="0" Body="&lt;p&gt;Carbonate is very natural.  Wood ash has a significant amount of potassium and sodium carbonate.  You can see this by collecting white powdery wood ash and adding vinegar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock formations like limestone naturally contain carbonate, although limestone is calcium carbonate.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-05T01:38:12.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12351" PostTypeId="2" ParentId="12322" CreationDate="2014-06-05T01:52:25.847" Score="0" Body="&lt;p&gt;In case of Ethyl in protic solvent the basicity is:&#xA;$$\ce{Et2NH&amp;gt;Et3N&amp;gt;EtNH2}$$&#xA;This is a result of two opposing factors, Inductive Effect and Solvation.In case of other alkyl groups than Ethyl:&#xA;$$\ce{R2NH&amp;gt;RNH2&amp;gt;R3N}$$&#xA;And it is as the following when Inductiive effect becomes dominant in gaseous state:&#xA;$$\ce{R3N&amp;gt;R2NH&amp;gt;RNH2}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Field Effect(or I may say inductive effect)  as stated on Wikipedia&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The 'Inductive Effect' is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule.&lt;br&gt;&#xA;The permanent dipole induced in one bond by another is called inductive effect.&#xA;The electron cloud in a $\sigma$-bond between two unlike atoms is not uniform and is slightly displaced towards the more electronegative of the two atoms. This causes a permanent state of bond polarization, where the more electronegative atom has a slight negative charge ($\delta^{–}$) and the other atom has a slight positive charge ($\delta^{+}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the electronegative atom is then joined to a chain of atoms, usually carbon, the positive charge is relayed to the other atoms in the chain. This is the electron-withdrawing inductive effect, also known as the -I effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some groups, such as the alkyl group, are less electron-withdrawing than hydrogen and are therefore considered as electron-releasing. This is electron releasing character and is indicated by the +I effect.&#xA;In short, alkyl groups tend to give electrons, leading to induction effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The $\ce{R}$ groups are $\ce{+I}$ that give $\ce{N}$ a more electron density and hence $\ce N$ becomes more basic and we can thus say that $\ce{R3NH}$ would be most basic as it has most alkyl groups and thus the basicity in gaseous state can be predicted, but in solution the water molecule tend to solvate amines and decrease its basicity, the solvation depends, here, mainly on size as difference between $\ce{-R}$ and $\ce{-H}$ groups and thus the two orders of basicity arise for alkyl amines.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-06-05T04:21:03.487" LastActivityDate="2014-06-05T04:21:03.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12352" PostTypeId="2" ParentId="12348" CreationDate="2014-06-05T01:53:34.530" Score="6" Body="&lt;p&gt;Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-05T03:09:32.783" LastActivityDate="2014-06-05T03:09:32.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12353" PostTypeId="1" AcceptedAnswerId="12355" CreationDate="2014-06-05T02:01:53.683" Score="0" ViewCount="1682" Body="&lt;p&gt;How do I draw a double Newman projection for cyclohexane? Which bond(s) do I sight down? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UWXt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, if we have 1-methylcyclohexane, we sight C1-C2 and C5-C4, as shown in my attempt below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PxDri.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be correct? I wish to achieve level 4! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-05T02:07:07.433" LastActivityDate="2014-06-05T02:13:45.583" Title="Double Newman Projections" Tags="&lt;organic-chemistry&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12354" PostTypeId="2" ParentId="12320" CreationDate="2014-06-05T02:08:03.400" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qHsM5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHsM5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine viewing it from right side:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H is upwards and CH3CH2 is downwards.&lt;/li&gt;&#xA;&lt;li&gt;OH is leftwards and CH3 rightwards.&lt;/li&gt;&#xA;&lt;li&gt;H and CH3CH2 are backwards(from right) so they must be on vertical line of Fischer projection and similarly for OH and CH3.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternatively:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Put the dotted wedge CH3 in any vertical position let it be bottom.&lt;/li&gt;&#xA;&lt;li&gt;Then imagine all other groups to be planar, we see in clockwise order:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CH3CH2 , H , OH&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Draw them similarly on the Fischer projection in clockwise sense.!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wp4FW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wp4FW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Note that both are (S)&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="1782" LastEditDate="2017-04-06T23:32:53.843" LastActivityDate="2017-04-06T23:32:53.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12355" PostTypeId="2" ParentId="12353" CreationDate="2014-06-05T02:13:45.583" Score="1" Body="&lt;p&gt;Note that -Cl and -CH(CH3)2 are on equatorial positions(The slant lines suggest that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the -H is on axial position(The perfectly vertical line suggest that)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is not just a coincidence, rather it is intentional. I think it would suffice further.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-05T02:13:45.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12356" PostTypeId="2" ParentId="12323" CreationDate="2014-06-05T02:22:45.490" Score="6" Body="&lt;p&gt;From &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jf00029a001&quot; rel=&quot;nofollow&quot;&gt;Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Cu^{2+} + H_2Asc \rightarrow Cu^+ + HAsc^.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Cu^+ + O_2 \rightarrow Cu^{2+} + O_2^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {2O_2^- + 2H^+ \rightarrow H_2O_2 + O_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Cu^+ + H_2O_2 \rightarrow Cu^{2+} + OH^- + OH^.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the hydroxyl radical reacts with benzonic acid as opposed to benzoate.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-19T02:59:03.603" LastActivityDate="2014-06-19T02:59:03.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12357" PostTypeId="2" ParentId="12348" CreationDate="2014-06-05T02:40:27.193" Score="3" Body="&lt;p&gt;Your result is correct but I don't agree with your method. Your method would fail under certain circumstances. Which circumstances you might ask? When the pH of the solution is &lt;em&gt;actually&lt;/em&gt; between 6.8 and 7.2. In these cases, your assumption that water auto-ionizes to create exactly $\ce{[HO^-]} = 1.0\times10^{-7} M $ will have significant figure implications). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About your assumption: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[HO^-]} = 1.0\times10^{-7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;em&gt;not&lt;/em&gt; true when you add any amount of acid or base to water because the addition of either (or both) will disrupt the equilibrium that exists in water (referred to as $\ce{K_{w}}$, the auto-protolysis of water, or the auto-ionization of water):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_2O &amp;lt;=&amp;gt; HO^- + H_3^+O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_{w}=[HO^-][H_3^+O]}=1.0\times10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All&lt;/em&gt; you know about your system is that $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$. This is in fact true for &lt;em&gt;any&lt;/em&gt; water solution; as we can see from the auto-ionization reaction above, there exists a 1:1 ratio between $\ce{[HO^-]}$ and $\ce{[H_3^+O]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also know that because $\ce{KOH}$ dissociates completely in water solution, $\ce{[HO^-]_{KOH}} = 1.0 \times 10^{-11} M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the &lt;em&gt;correct&lt;/em&gt; way to solve this system would be using $\ce{K_{w}}$. Work below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_{w}=\sum{[HO^-]\sum[H_3^+O]}}=1.0\times10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{=([HO^-]_{water}}+\ce{[HO^-]_{KOH})([H_3^+O]_{water})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$, we can make this substitution: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $x=\ce{[HO^-]_{water} = [H_3^+O]_{water}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{=(x}+\ce{[HO^-]_{KOH})(x)} = 1.0\times10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can solve for x from here. Keep in mind what x equals! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-05T02:59:10.940" LastActivityDate="2014-06-05T02:59:10.940" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12359" PostTypeId="1" AcceptedAnswerId="12360" CreationDate="2014-06-05T03:41:31.017" Score="0" ViewCount="4126" Body="&lt;p&gt;This does not compute for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qwRvR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would the four-carbon butyl group have a lower relative priority than the 3 carbon isopropyl group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the carbons one by one between the isopropyl and the butyl group, we'd run out of carbons in the isopropyl group first! How then can isopropyl possibly have a higher priority than the butyl group? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2016-12-06T04:47:50.563" Title="Isopropyl and Butyl Groups - Relative Priorities" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12360" PostTypeId="2" ParentId="12359" CreationDate="2014-06-05T03:45:50.590" Score="2" Body="&lt;p&gt;Comparing IsoPropyl and n-Butyl&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On first position both have C.&lt;/li&gt;&#xA;&lt;li&gt;On second position IsoPropyl has two C's whereas n-Butyl has only one C.&lt;/li&gt;&#xA;&lt;li&gt;Thus IsoPropyl has higher priority and we do not see the further chain.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-05T03:45:50.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12361" PostTypeId="2" ParentId="12359" CreationDate="2014-06-05T03:50:18.780" Score="1" Body="&lt;p&gt;If two groups have the same first atom, then compare the second atom from the chiral center.  If there are multiple second atoms, then compare them in order of atomic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from &lt;a href=&quot;http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html&lt;/a&gt;, step 1, part 2&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-05T03:50:18.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12362" PostTypeId="2" ParentId="12304" CreationDate="2014-06-05T03:58:50.730" Score="1" Body="&lt;p&gt;Let's have a look at what IUPAC say; they would be a reliable source to get definitions of chemical terminology from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/M04017.html&quot; rel=&quot;nofollow&quot;&gt;Monomer&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A substance composed of monomer molecules.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/M04019.html&quot; rel=&quot;nofollow&quot;&gt;Monomer molecules&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A molecule which can undergo polymerization thereby contributing&#xA;  constitutional units to the essential structure of a macromolecule. &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/P04735.html&quot; rel=&quot;nofollow&quot;&gt;Polymer&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A substance composed of macromolecules.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/M03667.html&quot; rel=&quot;nofollow&quot;&gt;Macromolecule&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A molecule of high relative molecular mass, the structure of which&#xA;  essentially comprises the multiple repetition of units derived,&#xA;  actually or conceptually, from molecules of low relative molecular&#xA;  mass. &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These definitions are consistent with the latest &lt;a href=&quot;http://www.iupac.org/nc/home/projects/project-db/project-details.html?tx_wfqbe_pi1%5Bproject_nr%5D=2002-048-1-400&quot; rel=&quot;nofollow&quot;&gt;IUPAC Compendium of Polymer Terminology and Nomenclature, IUPAC Recommendations 2008&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, historically, the definition of a polymer has changed somewhat. Up until about 80 years ago, polymers were seen as molecules comprised of similar empirical formula - benzene, for example, was considered a polymer of acetylene. It was Staudinger's work in the 1920's (contributing to his Nobel Prize) that defined polymers as macromolecules.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-06-05T03:58:50.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12365" PostTypeId="2" ParentId="12321" CreationDate="2014-06-05T07:14:20.183" Score="4" Body="&lt;p&gt;Amine inversion is also known as nitrogen inversion. You can find more here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nitrogen_inversion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has nothing to do with relative basicity of amines, that is defined by 1) energy of particular N-H  bonding (increases with +I effect) and 2) energy of ammonium cation hydration. (decreases with alkyl group number and their size). The relative balance of said factors is fragile and cannot be easily predicted without reference to experimental data except cases when amine has lone pair involved into bonding interactions (aniline, pyrrole, amides) or connected with extremely strong electron acceptors. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-06-05T08:49:49.960" LastActivityDate="2014-06-05T08:49:49.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12366" PostTypeId="2" ParentId="12344" CreationDate="2014-06-05T07:34:16.110" Score="2" Body="&lt;p&gt;I never heard about chemical rising agent in liquid form. You can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Monocalcium_phosphate&quot; rel=&quot;nofollow noreferrer&quot;&gt;monocalcium phosphate&lt;/a&gt; that was an ancient leavening agent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6pPv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe you should follow the advice of Dissenter and use yogurt, &lt;a href=&quot;http://en.wikibooks.org/wiki/Cookbook:Sourdough_Starter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sourdough Starter&lt;/a&gt; or &lt;a href=&quot;http://en.wikibooks.org/wiki/Cookbook:Kefir&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kefir&lt;/a&gt; all these produce carbon dioxide through fermentation, and are more &quot;natural&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at &lt;a href=&quot;http://en.wikibooks.org/wiki/Cookbook:Leavening_Agent&quot; rel=&quot;nofollow noreferrer&quot;&gt;this wikibook&lt;/a&gt; for a list of leavening agent.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-05T07:34:16.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12367" PostTypeId="1" AcceptedAnswerId="12392" CreationDate="2014-06-05T08:36:46.513" Score="2" ViewCount="316" Body="&lt;p&gt;If I know the chemical shift tensor (from ab initio methods), $\sigma$, for each atom in my structure, is it possible to construct an NMR spectra for my system? If yes, how and what mathematical implementations do I need?&lt;/p&gt;&#xA;" OwnerUserId="5274" LastActivityDate="2015-01-23T20:08:28.400" Title="How to construct NMR spectra from chemical shift tensors?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12369" PostTypeId="2" ParentId="12315" CreationDate="2014-06-05T09:35:52.797" Score="2" Body="&lt;p&gt;In pure water at the negatively charged cathode, a reduction reaction takes place, with electrons ($\ce{e^{−}}$) from the cathode being given to hydrogen cations to form hydrogen gas (the half reaction balanced with acid):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction at cathode: $\ce{2 H+(aq) + 2e^{−} -&amp;gt; H2(g)}$&#xA;At the positively charged anode, an oxidation reaction occurs, generating oxygen gas and giving electrons to the anode to complete the circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation at anode: $\ce{2 H2O(l) -&amp;gt; O2(g) + 4 H+(aq) + 4e^{−}}$&#xA;The same half reactions can also be balanced with base as listed below. Not all half reactions must be balanced with acid or base. Many do, like the oxidation or reduction of water listed here. To add half reactions they must both be balanced with either acid or base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode (reduction): $\ce{2 H2O(l) + 2e^{−} -&amp;gt; H2(g) + 2 {}^{-}OH(aq)}$&#xA;Anode (oxidation): $\ce{4 {}^{-}OH (aq) -&amp;gt; O2(g) + 2 H2O(l) + 4 e^{-}}$&#xA;Combining either half reaction pair yields the same overall decomposition of water into oxygen and hydrogen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall reaction: $\ce{2 H2O(l) -&amp;gt; 2 H2(g) + O2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electrolysis_of_water&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="4945" LastEditDate="2014-08-13T16:04:36.343" LastActivityDate="2014-08-13T16:04:36.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12371" PostTypeId="2" ParentId="12315" CreationDate="2014-06-05T11:21:49.657" Score="6" Body="&lt;p&gt;Nobody has given an answer yet, so I'll give it a crack.&lt;/p&gt;&#xA;&lt;p&gt;In electrolysis, as you probably know, there are two electrodes – the cathode (negative) and the anode (positive). We can deal with what happens at each separately, then tie them together.&lt;/p&gt;&#xA;&lt;p&gt;At the cathode:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2O + 2e- -&amp;gt; H2 + 2OH-}$$&lt;/span&gt;&#xA;You're interested in the mechanism of these reactions. We know that water exists in solution in an equilibrium:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&lt;/span&gt;&#xA;At the cathode, electrons are supplied to the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions, reducing them:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H3O+ + e- -&amp;gt; H2O + H* }$$&lt;/span&gt;&#xA;With many reactions happening at once, the &lt;span class=&quot;math-container&quot;&gt;$\ce{H*}$&lt;/span&gt; radicals react together, forming &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;. By removing the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions, it shifts the equilibrium of water to the right, to oppose the change – removing &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; ions and increasing the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H2O &amp;lt;=&amp;gt;&amp;gt; H3O+ + OH-}$$&lt;/span&gt;&#xA;Because the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions doesn't change, the local solution around the cathode becomes basic – this can be done in a school lab. Add indicator. Electrolyse. Observe the color change near the cathode.&lt;/p&gt;&#xA;&lt;p&gt;At the anode:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2O -&amp;gt; O2 + 4H+ + 4e-}$$&lt;/span&gt;&#xA;Here I suspect the mechanism is a bit similar:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&lt;/span&gt;&#xA;We take the &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; and oxidise it:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{OH- -&amp;gt; OH* + e-}$$&lt;/span&gt;&#xA;This results in a radical – the hydroyxl radical – being formed. The steps from here, to get to water being formed are probably quite complicated as a mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorDisplayName="user7951" LastEditDate="2020-08-16T15:12:56.287" LastActivityDate="2020-08-16T15:12:56.287" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="12372" PostTypeId="1" AcceptedAnswerId="12377" CreationDate="2014-06-05T13:01:25.730" Score="3" ViewCount="3160" Body="&lt;p&gt;Why is silicon oxide giant covalent structure? Why it doesn't bond like carbon dioxide?&lt;/p&gt;&#xA;" OwnerUserId="5755" LastEditorUserId="23561" LastEditDate="2016-07-25T16:11:38.140" LastActivityDate="2021-03-06T14:06:13.763" Title="Why is silicon oxide giant covalent structure?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12373" PostTypeId="2" ParentId="12331" CreationDate="2014-06-05T13:10:25.553" Score="0" Body="&lt;p&gt;Sure.  We could imagine, for instance, that a strong electrical field (anode/cathode pair) and a special stirring, would cause the migration of all non-H2O molecules to nucleate together and/or adhere to walls and be removed from solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flocculants just start this process going, and are used in context to other processes, as @long mentions.&lt;/p&gt;&#xA;" OwnerUserId="296" LastActivityDate="2014-06-05T13:10:25.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12374" PostTypeId="1" AcceptedAnswerId="12379" CreationDate="2014-06-05T13:16:03.127" Score="1" ViewCount="3460" Body="&lt;p&gt;I am confused, should it be 2 or 4?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I consider $\ce{BeCl2}$ as a chain structure the coordination number is 4, and in the gas phase it is 2. &lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="73" LastEditDate="2014-08-31T12:37:44.217" LastActivityDate="2014-08-31T12:37:44.217" Title="What is the coordination number for BeCl2?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12375" PostTypeId="2" ParentId="537" CreationDate="2014-06-05T14:18:36.527" Score="11" Body="&lt;p&gt;You can write it however you want. Check this out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VcByH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the presence of formulas such as $\ce{(HO)_2SO_2}$ for sulfuric acid. This information sheet is the same formula sheet I used when I was taught. Serves a good pedagogical purpose - emphasizing structure - and its relation to reactivity, stability, etc. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-05T14:18:36.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12376" PostTypeId="2" ParentId="12374" CreationDate="2014-06-05T14:40:07.433" Score="1" Body="&lt;p&gt;Beryllium is the most electronegative of the Group 1 and Group 2 elements (excluding hydrogen) and so rather than lose 2 electrons (and become $\ce{Be^{+2}}$ - a lot of charge on such a small atom) it can often bond covalently.  Just like with carbon, the covalent bonding will involve mixing of beryllium's $\ce{2s}$ and $\ce{2p}$ orbitals.  The coordination number for Beryllium can vary depending upon how the orbitals mix and the what hybridization results.  In the gas phase the $\ce{BeCl2}$ monomer has a linear structure and the coordination number is 2.  This is because the molecule is $\ce{sp}$ hybridized with two empty $\ce{p}$ orbitals.  $\ce{sp}$ hybridization with two empty $\ce{p}$ orbitals requires bonding to two groups (coordination number = 2) in a linear arrangement. In the solid phase, where other $\ce{BeCl2}$ molecules are close, the chlorine atom from one $\ce{BeCl2}$ molecule can also form a dative bond to an adjacent beryllium atom.  In this case the molecules can rehybridize to $\ce{sp^3}$ and form 4 bonds (coordination number = 4) arranged along the vertices of a tetrahedron.  In this case the formerly empty $\ce{p}$ orbitals have become $\ce{sp^3}$ orbitals and share the lone pair of electrons contributed from a chlorine atom on an adjacent $\ce{BeCl2}$ molecule.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-05T20:23:47.830" LastActivityDate="2014-06-05T20:23:47.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12377" PostTypeId="2" ParentId="12372" CreationDate="2014-06-05T14:55:56.130" Score="3" Body="&lt;p&gt;Carbon dioxide is a molecule with double bonds.&lt;/p&gt;&#xA;&lt;p&gt;Elements of 3+ rows are not happy with double bonds, they prefer single bonds, even in compounds like &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; 'double bonds' are actually dative single bonds.&lt;/p&gt;&#xA;&lt;p&gt;The reason for it is that pi-orbitals for 3+ rows are very diffuse, making overall overlapping with p-orbitals of 2-rows extremely ineffective&lt;/p&gt;&#xA;&lt;p&gt;For this reason, &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; forms structure with only singular Si-O bonds. It happens so that for it the only way to form one is to make 3-d infinite structure.&lt;/p&gt;&#xA;&lt;p&gt;For similar reason &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O5}$&lt;/span&gt; is exactly like this in gas phase while &lt;span class=&quot;math-container&quot;&gt;$\ce{P2O5}$&lt;/span&gt; forms tetrahedral &lt;span class=&quot;math-container&quot;&gt;$\ce{P4O10}$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3}$&lt;/span&gt; loves to form trimers and chains.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="9961" LastEditDate="2021-03-06T14:06:13.763" LastActivityDate="2021-03-06T14:06:13.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12378" PostTypeId="1" AcceptedAnswerId="12388" CreationDate="2014-06-05T15:17:03.507" Score="-1" ViewCount="332" Body="&lt;p&gt;I came across this statement and I don't understand what it meant by it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Square planar geometry is favoured by ligands that can form pi bonds by accepting electrons from metal atoms or ions.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is this possible? I mean a ligand, itself, donates electrons. How is it supposed to be accepting electrons?&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="9961" LastEditDate="2016-12-04T02:17:40.653" LastActivityDate="2016-12-04T02:17:40.653" Title="How is square planar geometry possible in metal complexes?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12379" PostTypeId="2" ParentId="12374" CreationDate="2014-06-05T15:28:03.193" Score="2" Body="&lt;p&gt;Coordination number is determined by compound. It is normal for an element to have several different CN. Preferable CN can change for an element depending on the atom's neighbors. For example, Fe has preferrable CN=6 for fluorine anions, but CN=4 for chloride anions. When heated or under extreme pressure preferable CN may change as well. For example, $SiO_2$ has high-pressure modification with 6-coordinated silicon and 3-coordinated oxygen, while at normal pressure they are 4- and 2-coordinated respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of $Be$ it prefers, just like all elements of the 2nd row, to have CN=4, but in many cases is forced to live with smaller numbers. BTW, in gas phase, $BeCl_2$ exists in equilibrium between monomer with Be's CN=2 and dimer with Be's CN=3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case CN for Be &lt;em&gt;in cristalline&lt;/em&gt; $BeCl_2$ is 4, while in gas phase it is 3 and 2 for dimere and monomer respectively. There is no 'one CN for Be atom right for all Be compounds.'&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-05T15:28:03.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12382" PostTypeId="1" AcceptedAnswerId="12413" CreationDate="2014-06-05T18:11:13.420" Score="0" ViewCount="212" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the combustion of 0.214g of sulfur provided exactly 0.428g of sulfur dioxide. After a second try, 0.782g of SO2 was obtained. What mass of sulfur was used?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The combustion of lithium forms lithium oxide. Combustion involving a 1.451g sample of oxygen results in obtaining 2.710g of Li2O. What mass of lithium was involved in this reaction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Natural argon has three isotopes: Ar-36 (35.967u), Ar-38 (37.962u), and Ar-40 (39.962u). if the isotopic abundance of Ar-36 is 0.006%, find the isotopic abundance of the other two&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After pulling my hair out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I still have no idea how to do this question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I balanced the equation $4Li + O_{2} -&amp;gt; 2Li_{2}O$ then found the moles of both oxygen and lithium oxide, which was 0.090688 mols and 0.090333 mols. Used the smaller number of moles, set up the ratios $\frac{4Li}{2Li_{2}O} = \frac{x}{0.090333mols}$&lt;br&gt;&#xA;x = 0.1807 mols, which equals to 1.2647g of Li&lt;br&gt;&#xA;Is this correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948&lt;br&gt;&#xA;0.00215802 + 37.962x + 39.962 - 39.962x = 39.948&lt;br&gt;&#xA;-2x = -0.01615802&#xA;x = 0.00807901&lt;br&gt;&#xA;which mean Ar-38 is 0.8%, but the answer clearly says it is 0.588%&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5138" LastEditorUserId="5138" LastEditDate="2014-06-06T01:07:40.413" LastActivityDate="2014-06-06T18:17:50.000" Title="Questions about combustion and isotopic abundance" Tags="&lt;combustion&gt;&lt;stoichiometry&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12383" PostTypeId="1" AcceptedAnswerId="12408" CreationDate="2014-06-05T18:22:37.890" Score="1" ViewCount="540" Body="&lt;p&gt;Do carboxylic acids (or any compound with a carbonyl group) count as a branch chained organic compound due to the C=O? Or does it need a carbon - carbon branch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xefo2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5619" LastEditorUserId="4945" LastEditDate="2014-06-12T08:29:38.343" LastActivityDate="2014-06-12T08:29:38.343" Title="Does the carboxyl group count as a branched chain" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12385" PostTypeId="1" CreationDate="2014-06-05T19:05:04.663" Score="-1" ViewCount="1381" Body="&lt;p&gt;${H}_2O(s)\leftrightharpoons{H}_2O(l)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) the mass of ice must equal to the mass of water&#xA;B) the mass water and the mass of ice each remain constant&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2016-03-12T00:43:31.523" Title="What is true about this equation, at equilibrium, and why?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2020-05-29T12:46:33.670" ContentLicense="CC BY-SA 3.0" />
  <row Id="12386" PostTypeId="2" ParentId="12385" CreationDate="2014-06-05T19:30:35.130" Score="2" Body="&lt;p&gt;At equilibrium, the amount of reactant being converted to product is equal to the amount of product being converted to reactant.  Therefor, the concentration of both the reactant and product are no longer changing and the ratio of these no longer changing concentrations can be represented as a fixed number or constant - the equilibrium constant.  $$\mathrm{K_{eq}=\frac{[Product]}{[Reactant]}}$$ where the brackets denote concentrations.  In the case at hand, a liquid-solid equilibrium, the expression becomes (thanks to DavePhD for pointing this out, see his UC Davis link in the comments if you'd like more info.) $$\ce{K_{eq}=[H2O~ (l)]}$$  This equation tells us that at equilibrium, the concentration of liquid water is a constant, it is not changing.  If the concentration of liquid water is not changing, then the concentration of ice cannot be changing.  This does not say that the concentration (or mass, since concentration is related to mass divided by molecular weight per unit volume) of liquid water must equal the concentration or mass of ice; it just says that the concentrations or masses of liquid water and ice must be constant.  Hence &quot;B&quot; is the correct choice.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:43:31.523" LastActivityDate="2016-03-12T00:43:31.523" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12387" PostTypeId="2" ParentId="12385" CreationDate="2014-06-05T19:31:40.820" Score="1" Body="&lt;p&gt;If the ice and liquid water are in equillibrium, and no heat is added to or removed from the system, B) the mass of water and the mass of ice each remain constant&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-05T19:31:40.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12388" PostTypeId="2" ParentId="12378" CreationDate="2014-06-05T19:54:26.993" Score="3" Body="&lt;p&gt;Maybe I'll break your world apart, but back-donation is actually pretty common. There are actually systems where it dominates, like antimony/naphthalene adducts. Back-donation is what participates to extreme stability of transition metal carbonyls. The key is, that not only transition metal has vacant orbitals, but the ligand have too, and in some cases even low-energy ones, like $CO$. Such orbitals can accept electron density from filled orbitals of the central atom, say $d_{z^2}$ in square planar complexes. Orbitals for back-donation are usually $\pi^*$ orbitals, and the interaction commonly found in phosphines, alkenes and arenes, especially if the have substituents with significant $-I$ effect &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-05T19:54:26.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12389" PostTypeId="2" ParentId="374" CreationDate="2014-06-05T20:02:46.720" Score="36" Body="&lt;p&gt;According &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ic010628q&quot; rel=&quot;noreferrer&quot;&gt;Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of&#xA;Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction&#xA;Revisited&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many years it was thought that iron was reduced to $\ce{Fe^{I}}$ and $\ce{NO}$ oxidized to $\ce{NO+}$, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin $\ce{Fe^{III}}$ ($S=5/2$) antiferromagnetically couples with $\ce{NO-}$ ($S=1$) for an observed spin of $S=3/2$.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7475" LastEditDate="2016-08-28T15:20:23.947" LastActivityDate="2016-08-28T15:20:23.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12390" PostTypeId="1" AcceptedAnswerId="12395" CreationDate="2014-06-05T20:35:10.440" Score="3" ViewCount="288" Body="&lt;p&gt;Technetium is probably the most awkward element of the periodic table for me. It seems to me that Technetium is an exception, in the sense that it doesn't have any stable isotopes, despite having a relatively low atomic number when compared to other unstable elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought, maybe there is something special about it. Some exceptional property that causes it to be so unstable with such a low atomic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me. What is it that makes Technetium so unstable?&lt;/p&gt;&#xA;" OwnerUserId="5760" LastActivityDate="2016-08-08T11:58:17.567" Title="Why is Technetium unstable?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;transition-metals&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12391" PostTypeId="1" CreationDate="2014-06-05T20:52:51.493" Score="0" ViewCount="1062" Body="&lt;p&gt;I know there are many more terms out there like polymerization, condensation, etc. But the important thing is whether they can be differentiated or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Esterification, which I just learned, needs to have a dehydrating agent when involving bad leaving group, such as $\ce{-OH}$.They are typically Brønsted-Lowry acids. Correct me if I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does heterolysis work in the case of $\ce{CH3COOC2H5}$. [the reverse of the esterification of ethanoic acid and ethanol]&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="5456" LastEditDate="2014-06-07T15:14:59.363" LastActivityDate="2014-06-07T15:36:21.143" Title="Esterification and dehydration synthesis" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12392" PostTypeId="2" ParentId="12367" CreationDate="2014-06-05T21:11:43.543" Score="3" Body="&lt;p&gt;Yes, it is possible. It is done often. The solution depends on the dynamics and orientation(s) of the spins in the ensemble (as will be shown bellow). I will assume you want to calculate the spectrum in frequency space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\omega_{cs} = \sigma_{zz}^{LF}(-\gamma \vec{B}_{0})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can hide a few variables and rewrite this in a much more useful way as:&#xA;$$&#xA;\omega_{CS} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{LF} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{PAS} =(\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{RF}&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\vec{b_0} = \frac{\vec{B_0}}{B_0}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above allows us to calculate the chemical shift in many reference frames. The frame where $\omega$ is diagonal (the PAS) is often convenient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\omega_{CS} = \omega_{xx}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{yy}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{zz}^{PAS}(\vec{b}^{PAS}_{0,x})^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can define $\theta$ as the angle between $Z^{PAS}$ and $b_0$ and $\phi$ as the angle between $X^{PAS}$ and the x-component of the $b_0$ field. This lets us write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\omega_{CS} = \omega_{xx}^{PAS} \cos^2\phi \sin^2\theta + \omega_{yy}\sin^2\phi \sin^2\theta+ \omega_{zz}^{PAS}\cos^2\theta&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the main equation you will need to calculate a chemical shift spectrum in frequency space. Numerically you will sum over all the angles in your ensemble. If your sample is a powder you could use angles sampled from a Gaussian distribution on a sphere for example. Typically people then apply convolution with a Lorentzian function (to account for relaxation and smooth the numerical artifacts). If your sample is a solution and the molecular reorientation time is fast compared to the inverse of the anisotropy, everything but the isotropic component will be averaged away. To put this in context (and to get to some possibly familiar equations) it will be useful to introduce some convention. In the Haeberlen-convention:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\delta_{zz} - \delta_{iso}| \ge |\delta_{xx} - \delta_{iso}| \ge |\delta_{yy} - \delta_{iso}|$$&#xA;$$\delta_{iso} = \frac{1}{3}(\delta_{xx} + \delta_{yy}+ \delta_{zz})$$&#xA;$$\delta = \delta_{zz} =\delta_{iso}$$&#xA;$$\eta = (\delta_{yy}-\delta_{xx})/\delta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\delta$ is called the anisotropy and $\eta$ is called asymmetry. Graphically:&lt;img src=&quot;https://i.stack.imgur.com/1Oll9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this convention and some geometry you can get to:&#xA;$$&#xA;\omega _{CS} = \frac{\delta}{2} (3\cos ^2\theta -1 + \eta \sin^2 \theta \cos 2\phi) + \delta_{iso}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;em&gt;ab initio&lt;/em&gt; chemical shift calculations normally are reported in shielding values, you will probably want to convert to shift values using a sutable reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &#xA;Multidimensional Solid-State NMR and Polymers, by Klaus Schmidt-Rohr and Hans Wolfgang Spiess, chapter 2&lt;/p&gt;&#xA;" OwnerUserId="5759" LastEditorUserId="5759" LastEditDate="2015-01-23T20:08:28.400" LastActivityDate="2015-01-23T20:08:28.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12394" PostTypeId="1" AcceptedAnswerId="12400" CreationDate="2014-06-05T22:27:33.387" Score="6" ViewCount="6195" Body="&lt;p&gt;For example, silver chloride, silver sulfide, and silver oxide are three exceedingly stable silver salts in water solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that silver forms such stable salts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have to do with charge density of the component ions? This doesn't seem to hold water because we know that silver oxide is much more soluble than silver sulfide, and the oxide anion is obviously a lot smaller than the sulfide ion, so the oxide ion should have a much higher negative charge density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So simple charge density rationalizations go out the window. I've Googled, I've looked in books, etc., but I can't find an explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, can a Lewis acid/base model be used to rationalize the stability of the silver compounds? For example, the sulfur, by nature of being less electronegative than oxygen, is a better Lewis base/electron pair donor/nucleophile; the sulfide ion is more willing to &quot;donate&quot; its electrons. Also the sulfide ion should be bigger and has empty &lt;em&gt;3d&lt;/em&gt; valence orbitals which may or may not be accessible to a great extent. Nonetheless, metals, by nature of possessing a large number of valence &lt;em&gt;d-electrons&lt;/em&gt;, may be able to increase the stability of the sulfide ion by populating its d-valence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Lewis acid/base model seems to get a lot of mileage. I also know that silver is a good Lewis acid, as are many metal cations, because of the poor shielding afforded by the &lt;em&gt;d&lt;/em&gt; electrons. Also, the silver cation has an empty 5s orbital. Thus, we can form a sulfide-to-cation sigma coordinate covalent bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may also form a pi-type coordinate covalent bond as the far-away d-electron density &quot;drifts&quot; to empty 3d valence orbitals on the sulfide anion, making the sulfide anion now more nucleophilic than ever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if my rationalizations are incorrect or if there are better explanations of silver salt stability. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-05T22:41:59.460" LastActivityDate="2014-06-05T23:31:16.100" Title="Why are many silver salts insoluble?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12395" PostTypeId="2" ParentId="12390" CreationDate="2014-06-05T22:28:13.303" Score="4" Body="&lt;p&gt;If you google &quot;why is technetium unstable&quot; you'll see that this question has been around for a while.  Over on physics.stackexchange, it is discussed in &lt;a href=&quot;https://physics.stackexchange.com/questions/40960/why-is-technetium-unstable&quot;&gt;Why is technetium unstable?&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read some of what others have posted on this topic and the short answer appears to be: because we have such a poor understanding of the atomic nucleus, there is no answer to this question at this time.  Again, the link above would be a good place to start.  One of the answers (3 upvotes) claims to be an answer.  The selected answer (10 upvotes) says there is no answer.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2016-08-08T11:58:17.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12397" PostTypeId="1" AcceptedAnswerId="12401" CreationDate="2014-06-05T22:44:51.627" Score="6" ViewCount="1385" Body="&lt;p&gt;There was a section about atomic orbitals in my organic chemistry textbook that I did not quite understand. First the author explained a bit about the Schrödinger equation, which I understood to be (vaguely, they did not go to any detail, even omitting the formula all together) a equation that describes the likeliness of positions of an electron in 3d space. $\Psi$ (i.e. the Schrödinger equation) can be solved for different quantum numbers that respectively represent the different orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This made me wonder what a picture (like below) of an orbital actually depicts, as it are chances that are described by $\Psi$, not boundaries. My best guess was that such picture represents an area that has a high chance, say 90% or more, of containing the electron (how they determine the shape, I don't know, but my best guess again here is that they only take those parts of the real orbital that add a significant percentage to the total chance of the electron actually being in the depicted orbital).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are all these assumptions right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rpljG.png&quot; alt=&quot;Some orbitals&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-06-06T07:29:51.510" LastActivityDate="2014-06-06T07:29:51.510" Title="Meaning of depiction of atomic orbitals" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12398" PostTypeId="1" CreationDate="2014-06-05T23:17:02.013" Score="2" ViewCount="1273" Body="&lt;p&gt;For a school project, I am in a group tasked with calculating Avogadro's Number in multiple manners.  Other than the Millikan Oil Drop Experiment, we know of no ways to compute it.  A Google search yielded no methods that could be of use (either complex beyond the high school level, or through using a mole of materials, which defeats the point of deriving the number).  What methods can we (intelligent high schoolers on a limited budget) use to compute Avogadro's Number? &lt;/p&gt;&#xA;" OwnerUserId="5762" LastActivityDate="2014-06-07T04:45:37.387" Title="Calculating Avogadro's Number" Tags="&lt;atoms&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12400" PostTypeId="2" ParentId="12394" CreationDate="2014-06-05T23:31:16.100" Score="5" Body="&lt;p&gt;This is just an unproven guess on my part.  Whether compounds are soluble or not depends on 1) how strong the bonding (including lattice) forces are that hold the compound together (e.g. how much energy do we have to put in to liberate the ions, and 2) how much energy is recovered through solvation of the resulting ions.  I don't see anything odd about the ionic radius of the silver ion so I'm guessing that the energy of solvation of the resultant ions would be fairly normal; so it is probably not factor &quot;2&quot; that makes most silver salts insoluble.  Here is a &lt;a href=&quot;http://ibchem.com/IB/ibnotes/full/ene_htm/15.2.htm&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to an article that suggests that bonding in silver compounds may have a covalent component, it says, for silver salts &quot;the structure has some contribution from covalent bonding. This appears to be bonding the structure together more tightly, as in a giant covalent structure.&quot;  The words &quot;more tightly&quot; catch my eye.  This seems to suggest that factor &quot;1&quot; may be in play in that silver salts may have increased bond\lattice energies.  If true, this would help explain the insolubility of many silver salts.  Finally I'll add that some of the soluble silver salts ($\ce{AgF, AgNO3, Ag(CH3CO2)}$) seem to involve anions that would be expected to be strongly hydrated.  The extra energy generated through solvation would further assist in solubilizing these silver salts.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-05T23:31:16.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12401" PostTypeId="2" ParentId="12397" CreationDate="2014-06-06T00:13:14.007" Score="8" Body="&lt;p&gt;Without going into a long discussion about what the wave function is, I will try to briefly answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For atoms, the (approximate) solution to $\Psi$ is the product of two functions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the radial components (distance from the nucleus) &lt;/li&gt;&#xA;&lt;li&gt;the angular/spherical components (behavior going around the nucleus, like latitude and longitude)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) Since the radial components of each component extend quite far, decaying to zero only at infinity, it would be very difficult to make a plot of the 3D object on a 2D surface that does not cover up the features that would be nearer the nucleus (further from eye), without using varying degrees of transparency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spherical component is usually solved analytically using complex numbers (i.e. a+b*i) but they can be transformed into real values by linear combinations. So, instead of two &quot;channels&quot; for real/imaginary, we can use only one &quot;channel&quot; (real) for each function, 1s, 3dxy, etc.  Often, these are colored so that if the sign of the real function is positive, it is e.g. red, otherwise, e.g. blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second feature of these types of graphics is called the &quot;isosurface&quot;--which means, plot only if the value of the function is equal to a chosen value.  Sometimes the chosen value is taken to be &quot;90% of the corresponding electron density contribution from this atomic orbital is enclosed&quot; (which, b.t.w. is $\Psi^* \Psi$ or simply $\Psi^2$.  Though $\Psi$ may be real/complex/negative/positive valued (hence need all sorts of colors and &quot;channels&quot; to represent it) the electron density is &lt;em&gt;positive&lt;/em&gt; and &lt;em&gt;real&lt;/em&gt;.  This is very important, as complex quantities are not observable, but are rather mathematical constructs (vehicles) for arriving at things that are actually observable, such as the electron density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But practically speaking, in order to see certain features like the doughnut in $dz^2$, people play a little fast and loose with precisely what the value of the isosurface is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shading on the surfaces that you have provided could be misinterpreted, because the colors do not correspond in any way to the value of the function being plotted, but rather seem to be an aid to the volumetric rendering of an isosurface that has already been set to either the absolute value or the square of the atomic orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that in your center drawing, which adds all of the individual atomic orbitals, that you are beginning to form a sphere.  It had better, as atoms are, in fact, spherical!${}^{\dagger}$  The formal name of the functions are spherical harmonics, and the &lt;strong&gt;&lt;em&gt;set&lt;/em&gt;&lt;/strong&gt; of e.g. $p_x$, $p_y$, $p_z$ or the five $d$'s &lt;strong&gt;&lt;em&gt;taken together&lt;/em&gt;&lt;/strong&gt; form a spherical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}^{\dagger}$ assuming the nuclear quadrupole is zero&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-06-06T00:41:50.640" LastActivityDate="2014-06-06T00:41:50.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12403" PostTypeId="2" ParentId="4903" CreationDate="2014-06-06T00:32:08.917" Score="0" Body="&lt;p&gt;Any energy that belong to orbitals with non-zero angular moment (ie. anything, except s orbitals) will split in magnetic field. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-06-06T00:32:08.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12404" PostTypeId="1" AcceptedAnswerId="12415" CreationDate="2014-06-06T02:08:16.687" Score="25" ViewCount="38794" Body="&lt;p&gt;Our teachers made us accept (without any explanation) that diamond conducts heat better than graphite. What is the reason behind this (alleged) fact?&lt;/p&gt;&#xA;" OwnerUserId="5763" LastEditorUserId="30836" LastEditDate="2019-04-19T12:18:50.683" LastActivityDate="2019-04-19T12:18:50.683" Title="Why does diamond conduct heat better than graphite?" Tags="&lt;heat&gt;&lt;materials&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="12405" PostTypeId="2" ParentId="12397" CreationDate="2014-06-06T04:25:57.347" Score="-2" Body="&lt;p&gt;What I understand of orbitals is simply that ELECTRONS can not be pin-pointed accurately, because they are in continous motion and that too not in a fixed circular path (as Bohr had depicted), so it is virtually impossible to know where an electron is in an atom. So scientist found a way to counter this problem by providing another model - THE QUANTUM MECHANICAL MODEL of the atom, where instead of electrons residing in orbits, it was stated that electrons have a possibility that it may reside in a particular region - what we call as orbitals. It is quite intresting to note that the probality of finding an electron never becomes 0, which means an atom in USA can have its electrons in Australia (atleast in theory).&lt;/p&gt;&#xA;" OwnerUserId="5763" LastEditorUserId="5763" LastEditDate="2014-06-06T04:40:08.243" LastActivityDate="2014-06-06T04:40:08.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12406" PostTypeId="1" AcceptedAnswerId="12409" CreationDate="2014-06-06T10:47:19.700" Score="1" ViewCount="1153" Body="&lt;p&gt;In the Arrhenius equation do Arrhenius constant and activation energy depend on concentration ? (Their units do)&lt;/p&gt;&#xA;" OwnerUserId="5767" LastActivityDate="2014-06-06T13:36:50.763" Title="Variation of Arrhenius equation with concentration" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12407" PostTypeId="1" CreationDate="2014-06-06T10:52:32.573" Score="8" ViewCount="1300" Body="&lt;p&gt;I should like to know roughly how the hormone melotonin is &lt;em&gt;synthetically&lt;/em&gt; made for the pharmaceutical industry (i.e. not how it is made in living beings). I can't seem to find any reference to its manufacture on the web, it is seen to have a fairly simple structure on its Wiki page and biologically it is synthesised from the forerunner amino acid tryptophan. So I believe it is likely to be readily synthesised for use in pharmaceuticals without the use of animal products, but I cannot confirm this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, a close relative of mine has been prescribed melotonin and I want to confirm that this substance is not made from animal products, as I live in a country where the authorities in their great wisdom have deregulated the food and meat processing industries. Astoundingly, they did this at exactly the same time that a UK royal commission recommended &lt;em&gt;re-regulation of the same&lt;/em&gt; to prevent contamination of food and pharmaceuticals by the Creutzfeld-Jacob prion.&lt;/p&gt;&#xA;" OwnerUserId="5766" LastEditorUserId="5084" LastEditDate="2014-06-06T14:58:26.457" LastActivityDate="2017-07-05T11:41:42.950" Title="How is Melatonin Synthesised?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;synthesis&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12408" PostTypeId="2" ParentId="12383" CreationDate="2014-06-06T11:26:22.377" Score="2" Body="&lt;p&gt;In organic chemistry, the terminology of &lt;em&gt;branched&lt;/em&gt; vs. &lt;em&gt;unbranched&lt;/em&gt; is generally used to refer to the structure of the hydrocarbon skeleton of a molecule. Hence, heteroatomic functional groups are usually not relevant. The structure in your question is therefore &lt;em&gt;unbranched&lt;/em&gt;. For an authoritative reference, see the relevant sections of the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_5.htm&quot; rel=&quot;nofollow&quot;&gt;IUPAC blue book&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-06T11:26:22.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12409" PostTypeId="2" ParentId="12406" CreationDate="2014-06-06T11:45:55.960" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/Arrhenius_Equation&quot; rel=&quot;nofollow&quot;&gt;Arrhenius equation&lt;/a&gt; gives you a rate constant $k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions&quot; rel=&quot;nofollow&quot;&gt;second order reaction&lt;/a&gt;, $k$ has units of $1/Ms$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/First-Order_Reactions&quot; rel=&quot;nofollow&quot;&gt;first order reaction&lt;/a&gt;, $k$ has units of $1/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arrhenius prefactor $A$ has the same units as $k$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate, as opposed to the rate constant will depend upon concentration. For the first order reaction, the rate constant is multiplied by one concentration to get the rate, and for a second order reaction the rate constant is multiplied by two concentrations to get the rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k$ and $A$ do depend upon concentration, that would imply that the reaction is not accurately modeled by the rate constant (for example not really first order when you are modeling as first order or not really second order when you are modeling as second order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as activation energy $E_a$, the equation can be written in terms of energy per mole divided by $RT$ or energy per molecule divided by $k_BT$ (where $k_B$ is Boltzmann constant).  $E_a$ does not depend upon concentration.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-06T13:36:50.763" LastActivityDate="2014-06-06T13:36:50.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12411" PostTypeId="1" CreationDate="2014-06-06T17:12:32.077" Score="0" ViewCount="1204" Body="&lt;p&gt;I know that optical activity means polarisation of light but what is so special in that that it changes the properties of the substances significantly&lt;/p&gt;&#xA;" OwnerUserId="5767" LastActivityDate="2014-07-08T05:18:44.623" Title="Effect of optical activity" Tags="&lt;optical-properties&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12412" PostTypeId="1" AcceptedAnswerId="12420" CreationDate="2014-06-06T17:51:18.757" Score="2" ViewCount="3174" Body="&lt;p&gt;I am dealing with the iodine clock reaction, which has the half-reactions:&#xA;$$\begin{alignat}{2}\ce{H2O2 (aq) + 2I- (aq) &amp;amp;-&amp;gt; I2 (s) + 2 OH- (aq)} &amp;amp;&amp;amp;\quad (\text{slow}) \\ \ce{I2  (s) + 2S2O3^2- (aq) &amp;amp;-&amp;gt; 2I- (aq) + S4O6^2- (aq)} &amp;amp;&amp;amp;\quad  (\text{fast}) \end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have to find the rate law. What I did was 4 solutions – one with standard amounts, another with double the $\ce{I-}$, one with double the $\ce{H2O2}$, and one with double the $\ce{S2O3^2-}$. I recorded the amount of time for the ppt ($\ce{I2}$) to appear. Now I have to calculate the rate law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, I am not quite sure how. Usually you take the &quot;slow&quot; step and set up an equation based on that, like $\text{rate}=\ce{[H2O2]$^x$ [I- ]$^y$}$. However, the reaction is not just dependent on those chemicals, it's also dependent on the $\ce{S2O3^2-}$. When it was doubled, the reaction was significantly slower (for obvious reasons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I set up the rate law for this reaction? Or, if I can't, what else can I do with my data?&lt;/p&gt;&#xA;" OwnerUserId="3944" LastEditorDisplayName="user7951" LastEditDate="2015-10-01T23:17:44.830" LastActivityDate="2015-10-01T23:17:44.830" Title="Rate law of the iodine clock reaction" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12413" PostTypeId="2" ParentId="12382" CreationDate="2014-06-06T18:07:45.633" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\frac{0.214g}{0.428g} = \frac {x}{0.782g}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your answer is ok, but on an exam you want to solve quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.710g - 1.451g = ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;wasn't that easier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is slightly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.00006, x, 0.99994-x&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-06T18:17:50.000" LastActivityDate="2014-06-06T18:17:50.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12415" PostTypeId="2" ParentId="12404" CreationDate="2014-06-06T19:35:44.410" Score="27" Body="&lt;p&gt;Diamond is one of the best thermal conductors known, in fact diamond is a better thermal conductor than many metals (thermal conductivity (W/m-K): aluminum=237, copper=401, diamond=895).  The carbon atoms in diamond are $\ce{sp^3}$ hybridized and every carbon is bonded to 4 other carbon atoms located at the vertices of a tetrahedron.  Hence the bonding in diamond is a uniform, continuous 3-dimensional network of $\ce{C-C}$ single (sigma) bonds.  Graphite on the other hand is formed from $\ce{sp^2}$ hybridized carbon atoms that form a continuous 2-dimensional sigma and pi bonding network.  This 2-dimensional network forms sheets of graphite, but there is little connection between the sheets, in fact, the sheet-sheet separation is a whopping ~3.4 angstroms.  This might lead us to suspect that heat conduction in the 2-dimensional sheet of graphite would be superior to diamond, but that heat conduction between graphite sheets would be very low.  This is, in fact, an accurate description of thermal conduction in graphite.  Thermal conductivity parallel to the graphite sheets=1950, but thermal conduction perpendicular to the sheet=5.7.  Therefor, when we consider thermal conduction over all possible directions (anisotropic) diamond would be superior to graphite.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-06T19:35:44.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12417" PostTypeId="1" AcceptedAnswerId="12419" CreationDate="2014-06-06T22:52:25.630" Score="18" ViewCount="31881" Body="&lt;p&gt;For most Diels–Alder reactions, the major product is &lt;em&gt;endo&lt;/em&gt; because there are favourable interactions between the newly forming pi bond and the electron withdrawing groups of the dienophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the reaction between furan and maleic anhydride an exception? From what I understand, the carbonyl groups of the anhydride are electron withdrawing.&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="16683" LastEditDate="2017-09-21T22:34:48.913" LastActivityDate="2017-09-21T22:55:50.970" Title="Why is the exo product formed in the Diels–Alder reaction between furan and maleic anhydride?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;pericyclic&gt;&lt;stereoselectivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12418" PostTypeId="1" AcceptedAnswerId="12426" CreationDate="2014-06-06T23:06:13.310" Score="3" ViewCount="589" Body="&lt;p&gt;If I had some NaCl dissolved in water, it is conductive, so if I were to put two electrodes in the container (powered), what is causing the flow of electrons?&#xA;Do the NaCl ionic bonds break and the ionic atoms collect at the oppositely charged electrode (how I'd assume electroplating works) with all of the atoms themselves moving to carry current, or are electrons loose in a &quot;sea&quot; like a metallic bond, with some of the atoms being pulled along with the flow?&lt;/p&gt;&#xA;" OwnerUserId="5772" LastActivityDate="2014-06-07T14:54:54.120" Title="Behavior of NaCl in water with electrodes" Tags="&lt;ions&gt;&lt;electrons&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12419" PostTypeId="2" ParentId="12417" CreationDate="2014-06-06T23:27:15.770" Score="20" Body="&lt;p&gt;Great question!  It turns out that the rate of formation of the &quot;expected&quot; &lt;em&gt;endo&lt;/em&gt; product is actually ~500 times faster than the rate of formation of the &lt;em&gt;exo&lt;/em&gt; product. However, the Diels–Alder is a reversible reaction. In this case, the &lt;em&gt;exo&lt;/em&gt; product is thermodynamically favored over the &lt;em&gt;endo&lt;/em&gt; product by about $\pu{1.9 kcal/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, even though the predicted &lt;em&gt;endo&lt;/em&gt; product is the kinetically favored product (and formed first), thermodynamics eventually takes over and the more stable &lt;em&gt;exo&lt;/em&gt; adduct is what you wind up with. More details are given in &lt;a href=&quot;https://doi.org/10.1021/jo00397a031&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;1978,&lt;/strong&gt; &lt;em&gt;43&lt;/em&gt; (3), 518.&lt;/a&gt; What a nice example of competing kinetic and thermodynamic pathways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QIN2S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QIN2S.png&quot; alt=&quot;Exo vs endo Diels–Alder reaction between furan and maleic anhydride&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-09-21T22:55:50.970" LastActivityDate="2017-09-21T22:55:50.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12420" PostTypeId="2" ParentId="12412" CreationDate="2014-06-07T01:37:54.313" Score="3" Body="&lt;p&gt;IMO, the only reaction that you are studying is the first reaction:  the conversion of $I^-$ to $I_2$ by hydrogen peroxide.  The thiosulphate reaction is simply a dramatic way to indicate when a particular amount of iodine has been produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction was &lt;strong&gt;not&lt;/strong&gt; slower when you doubled the amount of thiosulphate.  Twice as much iodine needed to be produced (and reconverted to iodide ion) before the thiosulphate was consumed and the first &lt;strong&gt;permanent&lt;/strong&gt; iodine changed the starch indicator blue.  &lt;/p&gt;&#xA;" OwnerUserId="3668" LastActivityDate="2014-06-07T01:37:54.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12423" PostTypeId="2" ParentId="12398" CreationDate="2014-06-07T02:18:20.470" Score="2" Body="&lt;p&gt;Placing a drop of oleic acid on the surface of water contained in a pan is a relatively straightforward experiment that has been widely used to determine Avogadro's number.  Here are two links that provide good background on a) how to run the experiment, b) how to analyze the data, and 3) the science behind the experiment.  The two links are just variations on the same experiment, but if you read them both it will really enhance your understanding of the experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.phys.ufl.edu/courses/phy1033/fall09/molecule.pdf&quot; rel=&quot;nofollow&quot;&gt;link 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lahc.edu/classes/chemistry/arias/Exp%203%20-%20AvogadroF11.pdf&quot; rel=&quot;nofollow&quot;&gt;link 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-07T02:18:20.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12424" PostTypeId="2" ParentId="12398" CreationDate="2014-06-07T04:45:37.387" Score="0" Body="&lt;p&gt;Here is a crude method of performing the experiment:-  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take two copper electrodes, weigh them, and dip them in $\ce{CuSO4}$ solution  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a constant current source or a fixed-EMF battery.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Connect an ammeter in series with the battery and the electrolytic setup&lt;/li&gt;&#xA;&lt;li&gt;Note down the time of closing the switch. Pass a current for a fixed time and also note down the current magnitude. &lt;/li&gt;&#xA;&lt;li&gt;After say some time $t$ of passing the current $I$, open the switch and remove the cathode and weigh it.  &lt;/li&gt;&#xA;&lt;li&gt;Say the increase in mass of the cathode is $m$. The equivalent weight of copper is $63.54/2$, hence the equivalents of copper deposited is $2m/63.5$. &lt;/li&gt;&#xA;&lt;li&gt;The amount of electricity passed is $It$. This must be equal to one $2m/63.5$ equivalents of electrons. Since the charge on one electron is $\approx 1.6\times 10^{-19}=e$, the number of electrons that passed is $It/e%$. Therefore the number of electrons in one equivalent or one mole, (which is the avogadro's number) will be $(It/e)/(2m/63.5)$.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You will have to make appropriate arrangements for accurate electrolysis to ensure correct weight measurement.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastActivityDate="2014-06-07T04:45:37.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12425" PostTypeId="1" CreationDate="2014-06-07T05:31:39.553" Score="2" ViewCount="800" Body="&lt;p&gt;How does the path of light become visible due to scattering of light? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me it only indicates the presence of colloidal particles, but my book gives the example of a cinema hall projector and says that you could see the path of light due to its scattering. Nevertheless, it's possible to see the lights path without the scattering.&lt;/p&gt;&#xA;" OwnerUserId="5774" LastEditorUserId="3846" LastEditDate="2014-06-07T15:07:12.513" LastActivityDate="2014-06-08T16:09:17.543" Title="Scattering of light" Tags="&lt;colloids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12426" PostTypeId="2" ParentId="12418" CreationDate="2014-06-07T06:12:37.193" Score="3" Body="&lt;p&gt;$\ce{NaCl}$ once in water dissociates, producing $\ce{Na+}$ and $\ce{Cl-}$ ions. Note: &lt;strong&gt;this process is possible only because immediately after dissociation ions are solvated by water, i.e. water molecules bind to ions&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ul5L3.png&quot; alt=&quot;http://en.wikipedia.org/wiki/Solvation#mediaviewer/File:Na%2BH2O.svg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is releasing energy and increasing effective size of ion, decreasing ion charge density. Once relatively free ions are produced, they can move between electrodes, transferring charge between electrodes. On electrodes ions may discharge taking/giving up electrons, and this is what usually happens with $\ce{Cl-}$ ion. However, if this is not an option, as the expected product is no stable enough in water, water molecule may react instead, producing another ion and either hydrogen or oxygen. This is what happens with $\ce{Na+}$: $\ce{Na}$ is too active to be stable in contact with water, so instead of sodium, hydrogen and $\ce{{}^{-}OH}$ are produced instead, resulting in  net reaction (assuming cold solution) $\ce{2NaCl + 2 H2O &amp;lt;=&amp;gt; 2 NaOH + Cl2 + H2}$ with chlorine and hydrogen released on different electrodes. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2014-06-07T14:54:54.120" LastActivityDate="2014-06-07T14:54:54.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12427" PostTypeId="1" AcceptedAnswerId="14877" CreationDate="2014-06-07T09:23:12.597" Score="13" ViewCount="9746" Body="&lt;p&gt;In an experiment, I set up a cell with lead nitrate (w/ lead electrode) and zinc sulfate (w/ zinc electrode), with a salt bridge containing potassium nitrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observed that by increasing the number of salt bridges, the voltage measured increased. However, given what I know about electrochemical cells, I can't work out why. I initially thought it would have no effect on the voltage/potential difference.&lt;/p&gt;&#xA;" OwnerUserId="5542" LastEditorUserId="5456" LastEditDate="2014-06-07T15:01:36.967" LastActivityDate="2019-11-14T10:05:29.950" Title="Why does increasing number of salt bridges increase voltage of electrochemical battery?" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12429" PostTypeId="1" AcceptedAnswerId="12431" CreationDate="2014-06-07T10:35:02.150" Score="4" ViewCount="3828" Body="&lt;p&gt;My question stands:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to oxidize toluene to benzoic acid using PCC?&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="33991" LastEditDate="2017-04-27T08:01:22.070" LastActivityDate="2017-04-27T08:01:22.070" Title="Can toluene be oxidized to benzoic acid using Pyridinium Chlorochromate(PCC)?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12431" PostTypeId="2" ParentId="12429" CreationDate="2014-06-07T12:27:31.457" Score="8" Body="&lt;p&gt;PCC is generally regarded as a milder oxidizing agent, used to effect selective oxidation of alcohols to aldehydes (from $1^{\circ}$ alcohols) or ketones (from $2^{\circ}$ alcohols). While over-oxidation to a carboxylic acid from a primary alcohol is definitely possible, PCC should not react with a simple unfunctionalized benzylic carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, oxidation at the benzylic position is easily achieved with more aggressive oxidizing agents. In particular, chromic acid (generated &lt;em&gt;in situ&lt;/em&gt; from chromium salts in aqueous sulfuric acid, as in the Jones oxidation) as well as aqueous potassium permanganate will fully oxidize virtually any benzylic carbon all the way to the carboxylic acid, providing the benzylic carbon has at least one hydrogen. Of course, the rest of the chain is completely cleaved in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Out of curiosity, I consulted my copy of March's &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt; to determine if any other common reagents were able to effect the same benzylic oxidation. Nitric acid is listed along with the two methods I described above. Naturally, a survey of various more exotic methods is also given. Intriguing was the use of sodium hypochlorite in acetonitrile, or NBS in aqueous sodium hydroxide under photochemical conditions, for the specific complete oxidation of aryl methyl groups.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-06-07T15:24:42.130" LastActivityDate="2014-06-07T15:24:42.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12432" PostTypeId="2" ParentId="12425" CreationDate="2014-06-07T15:13:02.697" Score="2" Body="&lt;p&gt;Light can only be seen when it enters your eye, so it somehow has to find a way between the light source and your cornea. Since light travels in &lt;a href=&quot;http://en.wikipedia.org/wiki/Gravitational_lens&quot; rel=&quot;nofollow&quot;&gt;more or less&lt;/a&gt; straight lines, it somehow has to change direction if you aren't looking into the source of said light directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is called scattering. It's what makes the path of light visible, because at every step on this path, the light gets scattered. This of course leads to a loss of intensity at the other end of the light path compared to a non-scattering medium.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-06-07T15:13:02.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12433" PostTypeId="2" ParentId="12391" CreationDate="2014-06-07T15:36:21.143" Score="3" Body="&lt;p&gt;The reverse of esterification is called &quot;hydrolysis.&quot;  The hydrolysis reaction can be performed under acidic or basic conditions.  This &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-3-1.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; shows the mechanisms involved for both the acidic and basic hydrolysis of an ester.  Each step in the reaction is explained in detail and accompanied by a picture. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-07T15:36:21.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12434" PostTypeId="2" ParentId="12411" CreationDate="2014-06-07T18:43:46.557" Score="0" Body="&lt;p&gt;Optical activity is the rotation of the plane of polarized light. It does not change the substances at all; rather, it is a property of the substance itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's special because compounds that rotate the plane of light can either rotate it to the right (+) or to the left (-) and that's measurable! Enantiomers can be differentiated in this way, since they will rotate the light by exactly the same amount but in opposite directions.&lt;/p&gt;&#xA;" OwnerUserId="5779" LastActivityDate="2014-06-07T18:43:46.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12436" PostTypeId="1" AcceptedAnswerId="13445" CreationDate="2014-06-08T03:41:07.800" Score="0" ViewCount="1338" Body="&lt;p&gt;I have three questions about waters of hydration, the first two of which I'm really stuck on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A sample of cobalt(II) nitrate hydrate was heated to remove all the water of hydration. The hydrate was found to be $65.96\%$ oxygen. Calculate the number of water molecules associated with each formula unit of cobalt(II) nitrate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Epsom salts is $\ce{MgSO4.xH2O}$. The hydrate was found to contain $71.4\%$ oxygen. Calculate the  value of $x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Zinc nitrate $\ce{Zn(NO3)2.xH2O}$ contains $21.98\%$ zinc by mass. What is the value of $x$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have absolutely no idea for the first two questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the third:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume $\pu{100 g}$ of material&lt;/li&gt;&#xA;&lt;li&gt;$\pu{21.98 g}$ of Zn = $\pu{0.3284 mol}$ of $\ce{Zn}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I couldn't get any further than this.&lt;/p&gt;&#xA;" OwnerUserId="5138" LastEditorUserId="11367" LastEditDate="2017-03-28T15:40:49.037" LastActivityDate="2017-03-28T15:40:49.037" Title="How to calculate the number of waters of hydration?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12437" PostTypeId="1" AcceptedAnswerId="12441" CreationDate="2014-06-08T03:51:31.047" Score="1" ViewCount="8543" Body="&lt;p&gt;Why does a buffer work most effectively when the concentration of salt and acid is 1:1?&lt;/p&gt;&#xA;" OwnerUserId="2595" LastEditorUserId="4945" LastEditDate="2014-06-08T11:14:21.017" LastActivityDate="2014-06-09T12:58:16.250" Title="Why is a buffer solution most effective when concentrations of salt and acid are the same?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12439" PostTypeId="2" ParentId="12437" CreationDate="2014-06-08T03:58:24.900" Score="1" Body="&lt;p&gt;It will take an equal amount of acid or base to change the pH of the buffer the same amount. If the buffer system was established with 1:2, it would be the same as a 1:1 system after adding an equivalent of a base (or acid).&lt;/p&gt;&#xA;&lt;p&gt;Added&lt;br&gt;&#xA;Support: Just after equation 1 on &lt;a href=&quot;http://www.science.smith.edu/departments/Biochem/Biochem_353/Buffer_Theory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Since a buffer is most effective when about half is in the acid form (Buffer H+) and half in the base form (Buffer)...&lt;/p&gt;&#xA;&lt;p&gt;The buffer concentration must be sufficient to maintain the pH within acceptable limits with the changes in proton concentration expected to occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-06-08T13:18:27.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12441" PostTypeId="2" ParentId="12437" CreationDate="2014-06-08T11:29:07.267" Score="4" Body="&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch-Equation&lt;/a&gt;:&#xA;$$\mathrm{p}\ce{H}=\mathrm{p}{K_a}+\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}$$&#xA;and buffer capacity:&#xA;$$\phi=\frac{\text{number of moles of acid/base added to 1L solution}}{\Delta\mathrm{p}\ce{H}}$$&#xA;where:&#xA;$$\Delta \mathrm{p}\ce{H} = \left|\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}\right|$$&#xA;it is minimum when $$[\ce{salt}]=[\ce{acid}]$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-06-09T03:47:42.467" LastActivityDate="2014-06-09T03:47:42.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12442" PostTypeId="2" ParentId="10547" CreationDate="2014-06-08T11:34:04.783" Score="-2" Body="&lt;p&gt;The haloalkanes are only very slightly soluble in water. In order for a&#xA;haloalkane to dissolve in water, energy is required to overcome the&#xA;attractions between the haloalkane molecules and break the hydrogen&#xA;bonds between water molecules. Less energy is released when new&#xA;attractions are set up between the haloalkane and the water molecules&#xA;as these are not as strong as the original hydrogen bonds in water. As a&#xA;result, the solubility of haloalkanes in water is low. However, haloalkanes&#xA;tend to dissolve in organic solvents because the new intermolecular&#xA;attractions between haloalkanes and solvent molecules have much the&#xA;same strength as the ones being broken in the separate haloalkane and&#xA;solvent molecules.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-08T11:34:04.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12445" PostTypeId="1" AcceptedAnswerId="12446" CreationDate="2014-06-08T14:11:48.977" Score="2" ViewCount="1506" Body="&lt;p&gt;What is the meaning of the 1,1,1,2 part in the compound 1,1,1,2-Tetrafluoroethane, a refrigerant commonly known as R-134a, some info:&lt;/p&gt;&#xA;&lt;p&gt;Formula: &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_2FCF_3}$&lt;/span&gt;&lt;br /&gt;&#xA;Structure:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vzs7l.png&quot; alt=&quot;Structure of 1,1,1,2-Tetrafluoroethane&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-06-09T22:49:08.850" Title="Meaning of 1,1,1,2 in 1,1,1,2-Tetrafluoroethane &amp; similar" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12446" PostTypeId="2" ParentId="12445" CreationDate="2014-06-08T14:18:10.330" Score="3" Body="&lt;p&gt;The numbers indicate which carbons the substituents - the fluorine atoms - are on. The root of the molecular name is &quot;tetrafluoro,&quot; indicating the presence of four fluorine substituents. So now where are they? That's what the numbers tell us; the &quot;1,1,1&quot; part indicates that there are three fluorine substituents on the first carbon, and &quot;2&quot; tells us there is one more fluorine substituent on the second carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the first carbon and which is the second carbon? The numbering scheme should be selected such that it gives the lowest number at the first point of difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, we can write &quot;1,1,1,2&quot; to indicate the presence of the four fluorine substituents rather than &quot;1,2,2,2.&quot; &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-09T22:49:08.850" LastActivityDate="2014-06-09T22:49:08.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12447" PostTypeId="2" ParentId="12445" CreationDate="2014-06-08T15:12:34.097" Score="1" Body="&lt;p&gt;1,1,1,2 in tetrafluro ethane[HFC 134a] is a position of fluorine in ethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOk6a.png&quot; alt=&quot; ethane structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is ethane structure in which 3 hydrogen of first carbon (1,1,1) and 1 hydrogen of second carbon (2) will be replaced by fluorine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHM6D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1,1,2 tetrafluro ethane is used instead of chlorofluorocarbon [CFC] as it is less hazardous to environment as compare to CFC&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-06-09T10:01:05.163" LastActivityDate="2014-06-09T10:01:05.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12448" PostTypeId="2" ParentId="12425" CreationDate="2014-06-08T16:09:17.543" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLh6l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of scattering what you are talking about is generically called &lt;a href=&quot;http://en.wikipedia.org/wiki/Tyndall_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tyndall effect&lt;/a&gt;. You can &quot;see&quot; the path of the light because some of the photons from the source are scattered in others direction so so they can reach your eyes. But in fact what you see are the photons that don't travel in straight trajectory, so not the beam itself but the path of the beam due to the photons that change direction due the effects of particles suspended in the medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually you can't see the light path without these effects...&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-08T16:09:17.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12449" PostTypeId="1" AcceptedAnswerId="13444" CreationDate="2014-06-08T16:48:50.790" Score="2" ViewCount="1395" Body="&lt;p&gt;I've been making some ice cream without a ice cream maker, using salt, ice, mixer and steel canister since I can't get a ice cream machine around here and its fun. It works pretty good but it seems slower than a simple ice cream machine that uses a freezer bowl. And the freezer wall volume seems smaller than the amount of ice I use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will using a frozen brine (if I can even freeze it) or liquid cool faster than &#xA;using frozen ice and then adding salt to melt it and lower the temperature? whats the difference? Is temperature all the matter? (liquid brine at -10c vs ice at 0)&#xA;any other factors that can contribute to the time it takes to freeze the mixture?&#xA;Any idea what coolant those bowls use? (seems to freeze at freezer temps).&lt;/p&gt;&#xA;" OwnerUserId="5788" LastActivityDate="2014-06-08T17:44:58.703" Title="Will ice cream freeze faster with frozen brine or adding salt to ice?" Tags="&lt;thermodynamics&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13444" PostTypeId="2" ParentId="12449" CreationDate="2014-06-08T17:44:58.703" Score="2" Body="&lt;p&gt;If the brine melts at the same temperature you reach with the ice and salt, then it should take the same time to freeze the ice cream since the freezing time depends on heat transfer. The benefit of frozen brine is that there is more heat capacity in the frozen brine than in ice-salt cooled water, so the ice-salt water may become warmer than the frozen brine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be that there is poor heat transfer from the cream to the ice so it takes longer than you think. Also, if you are the one turning the crank, time seems to stretch out since your muscles get tired and yet the ice cream is still not frozen.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-08T17:44:58.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13445" PostTypeId="2" ParentId="12436" CreationDate="2014-06-08T18:07:19.697" Score="2" Body="&lt;p&gt;All of the questions involve the same reasoning, so let's look at question #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2016-03-11T21:42:26.180" LastActivityDate="2016-03-11T21:42:26.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13448" PostTypeId="1" CreationDate="2014-06-08T20:15:19.227" Score="2" ViewCount="12624" Body="&lt;p&gt;In Chemistry class, we had to combine $\ce{K2Cr2O7}$ with 2-Methylpropan-2-ol and &lt;em&gt;heat&lt;/em&gt; it for the oxidation to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do we have to &lt;em&gt;heat&lt;/em&gt; the mixture when, for example, butanol, can be oxidized via use of the &lt;em&gt;same&lt;/em&gt; reagent but &lt;em&gt;without&lt;/em&gt; heating it?&lt;/p&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="33991" LastEditDate="2017-04-27T08:08:55.653" LastActivityDate="2017-04-27T08:08:55.653" Title="Oxidizing 2-Methylpropan-2-ol using potassium dichromate. Why is heating required?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13450" PostTypeId="1" AcceptedAnswerId="13453" CreationDate="2014-06-08T21:02:59.993" Score="5" ViewCount="3180" Body="&lt;p&gt;On skin permeability, according to Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Human skin has a low permeability; that is, most foreign substances are&#xA;unable to penetrate and diffuse through the skin. Skin's outermost&#xA;layer, the stratum corneum, is an effective barrier to most inorganic&#xA;nanosized particles.[22][23]&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is little to be found on exactly what chemicals can permeate the unbroken skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper &lt;a href=&quot;http://oem.bmj.com/content/61/4/376.full&quot; rel=&quot;noreferrer&quot;&gt;http://oem.bmj.com/content/61/4/376.full&lt;/a&gt; states the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Many chemicals can cross the unbroken skin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;However, the highly lipophilic nature of most solvents can also result&#xA;in dermal uptake when deposited on the skin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Wikipedia's statement incorrect? What chemicals can permeate the skin, which ones cannot, and why?&lt;/p&gt;&#xA;" OwnerUserId="6790" LastActivityDate="2014-06-09T11:57:00.597" Title="Skin permeability, for all chemicals or just some?" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13451" PostTypeId="2" ParentId="13448" CreationDate="2014-06-08T21:12:12.880" Score="4" Body="&lt;p&gt;Tertiary alcohols (such as 2-methyl-2-propanol, in your specific case) cannot be directly oxidized by $\ce{Cr(VI)}$ salts like $\ce{K2Cr2O7}$ because they have no $\alpha$-hydrogen. On the other hand, a primary alcohol (such as 1-butanol) will be readily oxidized, even under mild conditions and with reagents much gentler than $\ce{K2Cr2O7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With heating and sufficiently acidic conditions, a tertiary alcohol can be eliminated to form an alkene, which can then be oxidatively cleaved to yield two molecules with carbonyl groups. In the case of 2-methyl-2-propanol, one would presumably obtain acetone and either formaldehyde or formic acid, depending on how far the oxidation is permitted to go. (Though, under conditions vigorous enough to effect the oxidative cleavage of the alkene, it's highly unlikely the aldehyde could be isolated. I would also note that carbon dioxide is a product of oxidative cleavage of terminal alkenes when $\ce{KMnO4}$ is the oxidizer, but whether this occurs with $\ce{Cr(VI)}$ salts is unknown to me.) This is really the only reaction I can imagine to have plausibly occurred, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge, $\ce{K2Cr2O7}$ is not very efficient for this purpose, and the procedure itself strikes me as pretty odd if the oxidative cleavage described above is the intended reaction. It may be that no reaction was intended to happen, since oxidation reactions of this type can be used as chemical tests to differentiate alcohols by degree of substitution.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-08T21:12:12.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13452" PostTypeId="1" AcceptedAnswerId="13476" CreationDate="2014-06-08T21:19:43.133" Score="2" ViewCount="1055" Body="&lt;p&gt;My daughter has a chemistry exam on Tuesday. She is doing quite well with her revision except for the following table which she is having difficulty remembering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an easy way to remember this? Or perhaps if someone could explain why these exceptions are insoluble that will help her instead?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PXOjX.jpg&quot; alt=&quot;Insoluble salts&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1714" LastActivityDate="2014-06-09T15:42:42.900" Title="Memorising exceptions for soluble salts" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13453" PostTypeId="2" ParentId="13450" CreationDate="2014-06-08T21:32:27.943" Score="4" Body="&lt;p&gt;&quot;Skin's outermost layer, the stratum corneum, is an effective barrier to most &lt;strong&gt;inorganic&lt;/strong&gt; nanosized particles.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But skin is quite permeable to many &lt;strong&gt;organic&lt;/strong&gt; compounds. DMSO is one example, and it is warned that DMSO is a good solvent for other compounds that could be more toxic than DMSO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example: we all (should) know that mercury is toxic. However, it's not going to hurt you much if you roll it around in your hand for a brief period (don't breathe in, though!). But organic mercury compounds are terribly toxic and skin permeable, including the infamous &lt;em&gt;dimethyl mercury&lt;/em&gt; which killed a famous researcher of toxins.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-06-08T21:32:27.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13457" PostTypeId="2" ParentId="10706" CreationDate="2014-06-08T23:53:32.583" Score="2" Body="&lt;p&gt;The 2nd law is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d S \ge \displaystyle \frac{d q}{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the best you can do is zero for a reversible process.   Or if you don't like inequalities, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d S = \displaystyle \frac{d q}{T} + d S_{irr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d S_{irr}$ is the entropy production due to the irreversible processes within the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a intuitive in depth discussion in Prigogine's book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern Thermodynamics: From Heat Engines to Dissipative Structures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=6qFxngEACAAJ&amp;amp;dq=Modern+Thermodynamics%3a+From+Heat+Engines+to+Dissipative+Structures&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=IveUU8HvH8KVqAbDvILoBA&amp;amp;ved=0CCcQ6AEwAA&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=6qFxngEACAAJ&amp;amp;dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=IveUU8HvH8KVqAbDvILoBA&amp;amp;ved=0CCcQ6AEwAA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6793" LastActivityDate="2014-06-08T23:53:32.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13460" PostTypeId="2" ParentId="7184" CreationDate="2014-06-09T00:38:40.913" Score="0" Body="&lt;p&gt;For (a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{\ce{H2}} = \frac{p V}{R T} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times (22.0\ \text{K} + 273.15\ \text{K})} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times 293.2\ \text{K}} = 0.06251389252288155\ \text{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the correct number of significant figures &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{\ce{H2}} = 0.06251\ \text{mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the balanced reaction tells you that 1 mole of $\ce{Mg}$ makes 1 mole of $\ce{H2}$, so $n_{\ce{Mg}} = 0.06251\ \text{mol}$.  To get the mass of $\ce{Mg}$ use the atomic weight&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\ce{Mg}} = n_{\ce{Mg}} \times M_{\ce{Mg}} = 0.06251\ \text{mol} \times M_{\ce{Mg}} = 1.519400157768636\ \text{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, don’t forget we only have four significant figures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\ce{Mg}} = 1.519$ g&lt;/p&gt;&#xA;" OwnerUserId="6793" LastEditorDisplayName="user7951" LastEditDate="2015-05-10T20:26:29.060" LastActivityDate="2015-05-10T20:26:29.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13461" PostTypeId="2" ParentId="7183" CreationDate="2014-06-09T00:57:51.483" Score="2" Body="&lt;p&gt;Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure.   That means your volumes can be used &quot;as if&quot; they were moles in the chemical reaction stoichiometry.   So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl.  So, you get 20.0 L of HCl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (c) the excess reagent is the H$_2$.  Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over.   Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{H_2} = 0.22$ mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Review check out:&#xA;gas laws: &lt;a href=&quot;http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases&quot; rel=&quot;nofollow&quot;&gt;http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://itunes.apple.com/us/app/physycalc/id644105995&quot; rel=&quot;nofollow&quot;&gt;https://itunes.apple.com/us/app/physycalc/id644105995&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6793" LastActivityDate="2014-06-09T00:57:51.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13463" PostTypeId="1" AcceptedAnswerId="13505" CreationDate="2014-06-09T05:48:42.367" Score="6" ViewCount="20893" Body="&lt;p&gt;From what I understand generally Coke is carbonated to roughly 6.2 g per liter in bottles and cans, or 3.1 volumes of CO2.  I was watching this video because I was curious about how fountain soda machines worked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=RRHNt67xbMs&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=RRHNt67xbMs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the end of the video they suggest that the soda is carbonated to around 5 volumes of carbonation.  Which as I understand would be somewhere in the realm of 11 g of CO2 per liter, almost twice that of the normal soda.  I imagine there are a couple other factors involved, like they are dispensing more CO2 than is actually dissolved in the soda as a consequence of trying to instantly carbonate.  According to everything I know about carbonation I know from beer making they are carbonating almost 20 times higher than a forced beer carbonation technique in home brewing.  Can anyone help me better understand this?  I would think that the carbonation level difference between bottled and fountain soda would be something highly controlled by beverage companies for the sake of product consistency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some additional reading and found that the correct maintenance levels on soda fountains with chilled water to roughly 40 degrees F is 75 psi if that helps.  For 3.1 volumes of carbonation as I understand equilibrium would be just like 5 psig.  I know I missing something big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only assume that the rate of carbonation loss is inversely exponential as the carbonation approaches closer to equilibrium for the environmental pressure.  Would 5 volumes rapidly decarbonate to 2-3 volumes but not get much lower very quickly?&lt;/p&gt;&#xA;" OwnerUserId="6794" LastEditorUserId="4325" LastEditDate="2016-02-22T05:13:11.413" LastActivityDate="2019-01-05T02:11:32.107" Title="How much carbonation is in one liter of commercial fountain Coca-Cola" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;gas-laws&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13464" PostTypeId="1" CreationDate="2014-06-09T08:34:44.353" Score="0" ViewCount="1523" Body="&lt;p&gt;Can the size of molecule oxygen reduce smaller? If yes, how is that possible? Is it related to proton and electron surrounding the nucleus?&lt;/p&gt;&#xA;" OwnerUserId="6795" LastEditorUserId="1499" LastEditDate="2014-06-09T11:45:56.643" LastActivityDate="2014-06-10T17:22:34.633" Title="Reduce sizing of molecule - Oxygen" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13465" PostTypeId="1" CreationDate="2014-06-09T09:37:12.590" Score="1" ViewCount="109" Body="&lt;p&gt;How does charge come on dispersion medium in colloidal sols ? (It is said that in electro-osmosis flow of dispersion medium takes because it is also electrically charged, but why ? )&lt;/p&gt;&#xA;" OwnerUserId="5774" LastActivityDate="2014-06-09T09:37:12.590" Title="Charges in colloidal dispersion" Tags="&lt;colloids&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13467" PostTypeId="1" AcceptedAnswerId="13473" CreationDate="2014-06-09T10:20:38.003" Score="0" ViewCount="345" Body="&lt;p&gt;If we look at the right-hand-side table here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hammett_equation&lt;/a&gt;, we notice something particular.  It appears that the free energy of activation (for example, measured in terms of the acidicity of a carboxyl group on a benzene ring), and with respect to substituents in the para- position on the benzene ring, increases as one moves down column 17 on the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this happening?  Shouldn't we expect the opposite to be true?  Consider that, as we travel down column 17 corresponding to the halogens, electronegativity decreases and the probability of resonance contributions via pi-bonding should increase.&lt;/p&gt;&#xA;" OwnerUserId="6797" LastEditorUserId="6797" LastEditDate="2014-06-09T10:41:53.490" LastActivityDate="2014-06-09T13:02:25.567" Title="Halogens and the free energy of activation in a resonant benzene system" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13468" PostTypeId="2" ParentId="13464" CreationDate="2014-06-09T11:35:00.413" Score="1" Body="&lt;p&gt;Surprisingly enough, the answer is &lt;strong&gt;yes&lt;/strong&gt;, though it would be extremely difficult in practice to reduce the size of a molecule as large as $\ce{O_2}$ by the method I will describe here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I must be clear that &quot;shrinking&quot; as seen in many science-fiction stories is completely impossible. Size is not some arbitrarily determined parameter which you can change with a ray gun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though the concept of size is a little fuzzy at atomic/subatomic scales, we know it is the valence electrons who are responsible for determining the distance scales in atoms and molecules (since they're the farthest from the nuclei). After solving the Schrödinger equation for a system, one finds that the electrons are concentrated in regions around the nuclei called &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital#mediaviewer/File%3aElectron_orbitals.svg&quot; rel=&quot;nofollow&quot;&gt;orbitals&lt;/a&gt;. A meaningful definition of size for these orbitals implies that they are around 100 picometres or more across, conveniently described as twice the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_radius&quot; rel=&quot;nofollow&quot;&gt;Bohr radius&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, this would be the limit for any atom or molecule. However, nature has a rather curious quirk. We have discovered that all atoms in the Universe involved in chemistry are made of electrons, protons and neutrons (the latter two being themselves made of different combinations of up quarks and down quarks). However, there are more subatomic particles out there, and there is nothing to stop &lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_atom&quot; rel=&quot;nofollow&quot;&gt;(exotic) atoms&lt;/a&gt; being made out of these other subatomic particles, as long as they are combined in the right way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics doesn't yet understand why, but the fundamental subatomic particles in the Universe come in &lt;em&gt;three generations&lt;/em&gt;. That is, there are groups of particles which behave exactly the same (have the same spin, electric charge, among other parameters), but differ only in their mass. In particular, electrons are just one type in a group of three types of particles called &lt;a href=&quot;http://en.wikipedia.org/wiki/Lepton&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;leptons&lt;/em&gt;&lt;/a&gt;. The other types are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon&quot; rel=&quot;nofollow&quot;&gt;muons&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Tau_%28particle%29&quot; rel=&quot;nofollow&quot;&gt;taus&lt;/a&gt; (or tauons). Again, all leptons share exactly the same properties, except for mass; the muon is approximately 200 times heavier than the electron, while the tau is around 3500 times heavier. We don't naturally see atoms containing these heavier cousins of the electron because they decay very quickly and are hence much rarer. Even though the muon is one of the most stable subatomic particles, its mean lifetime is a mere $2.2\ \mu s$, and the tau is much more unstable still. In the case of the muon, however, it is long-lived enough to produce it artificially and get some chemistry done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one were to mix muons with normal atomic nuclei in an appropriate chamber, then for a fleeting moment it would be possible to make atoms and molecules where all the electrons are replaced with muons. The resulting atoms/molecules would have &lt;em&gt;muon orbitals&lt;/em&gt; instead of electron orbitals. By solving the Schrödinger equation for &lt;em&gt;muons instead of electrons&lt;/em&gt; (changing $m_e$ for $m_{\mu}\approx200\ m_e$), the Bohr radius becomes around 200 times smaller and as a direct consequence the muon orbitals are smaller by approximately the same factor. Thus, an oxygen atom where &lt;em&gt;all electrons have been replaced by muons&lt;/em&gt; will be about 1/200th the size, but will display almost exactly the same chemistry; two such exotic atoms would be expected to combine exactly as normal oxygen atoms do, creating an $\ce{O_2}$ molecule with sixteen muons instead of electrons, with an expected bond length of around $0.6\ pm$ instead of the usual $121\ pm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may all seem theoretical, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon#Muonic_atoms&quot; rel=&quot;nofollow&quot;&gt;muonic atoms&lt;/a&gt; have already been made in the laboratory before. Given the difficulties in producing and manipulating muons in their short lifespan, the only atoms produced as of yet seem to be muonic hydrogen (where the single electron is replaced by a muon) and quite recently &lt;a href=&quot;http://www.sciencemag.org/content/331/6016/448.short&quot; rel=&quot;nofollow&quot;&gt;muonic helium&lt;/a&gt; where only one of two electrons was replaced with a muon. In the latter case, the mix of electrons and muons makes the atom behave chemically quite differently from regular helium or doubly-muonic helium. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-06-09T11:35:00.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13469" PostTypeId="2" ParentId="13450" CreationDate="2014-06-09T11:47:46.540" Score="2" Body="&lt;p&gt;There are many phsico-chemical properties of the penetrant that determine the skin permeability of the substance. I cite this list from Kanerva, Lasse, ed. Handbook of occupational dermatology. Springer, 2000.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Polarity&lt;/li&gt;&#xA;&lt;li&gt;Volatility&lt;/li&gt;&#xA;&lt;li&gt;Solubility&lt;/li&gt;&#xA;&lt;li&gt;Concentration&lt;/li&gt;&#xA;&lt;li&gt;Molecular weight and particle size&lt;/li&gt;&#xA;&lt;li&gt;Electrostatic charge&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course it also depends on the site of application and how is the skin at the moment of the application however:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in general, non polar, lipid-soluble substances traverse skin more readily than do ionic species. Substances that penetrate skin easily include lipd-soluble endogenous substances (hormones, vitamins D and K) and a number of xeonobiotic compounds. Common examples of these are  phenol. nicotine, and strychnine. Some military poisons, such as the nerve gas sarin permeate the skin very readily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Manahan, Stanley E. Toxicological chemistry and biochemistry. CRC Press, 2002. p.121&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-06-09T11:57:00.597" LastActivityDate="2014-06-09T11:57:00.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13470" PostTypeId="2" ParentId="13464" CreationDate="2014-06-09T12:19:37.790" Score="0" Body="&lt;p&gt;If you could eliminate the Pauli exclusion principle then the size of all atoms and molecules would collapse.    Because electrons cannot have the same four quantum numbers in an atom, $n$, $l$, $m_l$, and $m_s$, the electrons in multi-electron atoms and molecules are forced to occupy the higher energy and larger radius orbitals.   More electrons: bigger atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about Pauli exclusion principle here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grandinetti.org/Teaching/Chem121/Lectures/QuantumNumbers&quot; rel=&quot;nofollow&quot;&gt;http://www.grandinetti.org/Teaching/Chem121/Lectures/QuantumNumbers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and atomic radii trends here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grandinetti.org/Teaching/Chem121/Lectures/MultiElectronAtoms&quot; rel=&quot;nofollow&quot;&gt;http://www.grandinetti.org/Teaching/Chem121/Lectures/MultiElectronAtoms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grandinetti.org/Teaching/Chem121/Lectures/AtomicRadii&quot; rel=&quot;nofollow&quot;&gt;http://www.grandinetti.org/Teaching/Chem121/Lectures/AtomicRadii&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6793" LastActivityDate="2014-06-09T12:19:37.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13472" PostTypeId="2" ParentId="12437" CreationDate="2014-06-09T12:58:16.250" Score="0" Body="&lt;p&gt;Just look at a titration plot of a weak acid or weak base.   When you are half way to the equivalence point the rate of change (slope) of pH vs mL titrant is near zero.  That point is when you are at 1:1 for salt:acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrIyd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6793" LastActivityDate="2014-06-09T12:58:16.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13473" PostTypeId="2" ParentId="13467" CreationDate="2014-06-09T13:02:25.567" Score="0" Body="&lt;p&gt;Resonance between a carbon and a halogen atom decreases as we move down the series $\ce{F, Cl, Br, I}$.  This is because overlap is most effective between orbitals of the same size.  Therefor, carbon-fluorine overlap, both involving $\ce{2p-2p}$ orbitals should be much stronger than carbon $\ce{2p}$ orbital overlap with the $\ce{3p, 4p, etc_{.}}$ orbitals in the larger halogens.  Consequently carbon-halogen resonance becomes less significant as we move down the halogen series.  If we now consider that 1) the resonance and inductive effects for the &lt;em&gt;para&lt;/em&gt;-halogens are in opposition to each other (tend to counteract each other) and 2) the resonance effect decreases as we move down the halogen series, then the order of halogens in the Table you mentioned makes sense.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-09T13:02:25.567" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13475" PostTypeId="2" ParentId="9829" CreationDate="2014-06-09T13:35:59.500" Score="10" Body="&lt;h1&gt;Toxicity of tobacco combustion products in different phases&lt;/h1&gt;&#xA;&lt;p&gt;For answering your question I think is better split it in two. First thing you need to understand that toxicity depends on &lt;strong&gt;the method of administation&lt;/strong&gt; of the toxic agent. Tobacco smoke is a physical state of the tobacco combustion products. It is an aerosol originating from combustion where its particles are liquid, gas and solid (tar). Quoting &lt;a href=&quot;http://journal.chemistrycentral.com/content/6/1/98&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry Central Journal 2012, 6:98&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Cigarette smoke is a complex and dynamic aerosol consisting of at least 5,600 chemicals and toxicants found across two phases, the particulate (tar) and vapour phase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(I would like to add the gas phase...) so when the product of the combustion of the tobacco are in this physical state you have to look at a different &lt;strong&gt;method of administration&lt;/strong&gt; in this case vapour inhalation, Dust and Mist Inhalation and gas inhalation these are the most potentially dangerous method of administration. It seems that after about 40 minutes the liquid and solid part of the aerosol deposit (of course this depends on the condition, T, wind etc. etc.) so is probable that after this period of time the toxicity of tobacco combustion products decrease considerably. However this is not true for all the combustion products. For example: carbon monoxide is recognize tobacco combustion product and it is a gas so it wont deposit after this time and it will be &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/004565359090098E&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;removed by reactions with OH radicals (85%), by soils (10%), and by diffusion into the stratosphere&amp;quot;&lt;/a&gt; very slowly.&#xA;&lt;strong&gt;Benzene&lt;/strong&gt; is a good example of how the toxicity depends on the method of administration. LC50 inhalation value of benzene is 10000 ppm TFLo Dermal is 0.92 mL/kg so it is a big difference if you wait that the areosol deposit or if you inhale it.&lt;/p&gt;&#xA;&lt;p&gt;If you are interested in rate of decomposition of benzene it seems that biodegradation of benzene can reach 0.95% at day according to &lt;a href=&quot;http://info.ngwa.org/gwol/pdf/892950286.PDF?pagewanted=all&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chiang&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Regarding &lt;strong&gt;1-3 butadiene&lt;/strong&gt; according to &lt;a href=&quot;http://www.epa.gov/ttnamti1/files/ambient/passive/carbo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;William A. McClenny and Donald Whitaker&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;i.e., a 10-6 lifetime risk level for cancer due to inhalation exposure&#xA;of 0.03 &lt;span class=&quot;math-container&quot;&gt;$g/m^{3}$&lt;/span&gt; (12.4 pptv at 20 °C and 1 atm pressure).1 This compound&#xA;is very reactive in the ambient atmosphere with a short&#xA;atmosphericlifetime, estimated to be 2–3 hr.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The half-life of &lt;strong&gt;acrolein&lt;/strong&gt; is &lt;a href=&quot;http://cat.inist.fr/?aModele=afficheN&amp;amp;cpsidt=22225462&quot; rel=&quot;nofollow noreferrer&quot;&gt;14.4 hr&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;You can also find a very complete EPA report &lt;a href=&quot;https://www.google.it/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;ved=0CFEQFjAF&amp;amp;url=http%3A%2F%2Foaspub.epa.gov%2Feims%2Feimscomm.getfile%3Fp_download_id%3D36309&amp;amp;ei=VbuaU7LEH6mc0QWR3IGIDQ&amp;amp;usg=AFQjCNFIRnUw8Zqv3fhcqQX77zO-Z9Z8fA&amp;amp;sig2=er_VegOtv-kKgKVJ9kCUdA&amp;amp;bvm=bv.68911936,d.bGQ&amp;amp;cad=rja&quot; rel=&quot;nofollow noreferrer&quot;&gt;EPA/600/P-98/001F&#xA;October 2002&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;However determine the half-life of all the compounds is a very complex task because is related to how the ecosystem respond to these compounds so to the actual capability to decompose them through different mechanism is strictly related with thousands of factors.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-07-11T18:48:22.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13476" PostTypeId="2" ParentId="13452" CreationDate="2014-06-09T15:42:42.900" Score="2" Body="&lt;p&gt;Understanding why it happens unfortunately will not help her to remember. This is the classic case where you have to find a dirty way to remember them. Maybe you can make some practical experiment I remember most of them with a practical application of the salts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of online resource regarding Mnemonic Tricks to remember solubility rules, these are two video on youtubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=AsCLuLS-yZY&quot; rel=&quot;nofollow&quot;&gt;Memonic tricks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=slf8w2DHH9Y&quot; rel=&quot;nofollow&quot;&gt;Song!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting the song:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is solubility it's so crazy, don't be a fool and follow these rules!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-09T15:42:42.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13477" PostTypeId="1" AcceptedAnswerId="13494" CreationDate="2014-06-09T16:18:26.617" Score="3" ViewCount="340" Body="&lt;p&gt;I saw an ad on tv last night for Mitchum deodorant that showed big text on the screen that said &quot;OXYGEN eliminates all odors!&quot; or something similar. Their website indeed claims &lt;a href=&quot;http://www.mitchum.com/OurTechnology/OurTechnology.aspx&quot; rel=&quot;nofollow&quot;&gt;&quot;Our exclusive formula releases pure oxygen, a powerful odor fighter...&quot;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does oxygen fight odors, and how can this aspect of the deodorant be more effective than the abundance of pure O$_2$ in the air?&lt;/p&gt;&#xA;" OwnerUserId="4185" LastActivityDate="2014-06-10T00:36:33.963" Title="How does oxygen fight odors? (Deodorant ad)" Tags="&lt;redox&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13478" PostTypeId="1" CreationDate="2014-06-09T16:22:51.440" Score="1" ViewCount="4179" Body="&lt;p&gt;I am trying to detect formalin in different types of food. Formalin $(\ce{CH2(OH)2})$is an aqueous solution of formaldehyde, which is $\ce{CH2O}$ or $\ce{H2CO}$. I used many chemicals to do this, but every time I fail. Please help me to do this. I got some results in my test by using NFPA 704 and ChEMBL, but that is not enough to save all. I would like to make a simple and easy food formalin detector that can save all the people. Please help me or give me some direction.&lt;/p&gt;&#xA;" OwnerUserId="6800" LastEditorUserId="275" LastEditDate="2014-06-09T18:49:29.023" LastActivityDate="2015-04-05T16:21:57.107" Title="How can i detect Formalin in food easily" Tags="&lt;biochemistry&gt;&lt;home-experiment&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13479" PostTypeId="1" AcceptedAnswerId="13481" CreationDate="2014-06-09T17:28:20.047" Score="1" ViewCount="141" Body="&lt;p&gt;I'm not a chemist, and nor do I study chemistry so please try to be gentle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glucose has the molecular formula $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ and the empirical formula $\mathrm{C}\mathrm{H}_{2}\mathrm{O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the empirical formula and working the other way, am I guaranteed that $\mathrm{C}_n\mathrm{H}_{2n}\mathrm{O}_n$, where $n$ is a positive whole number, are all going to be well-define molecular formulae?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples are Formaldehyde ($n=1$), Acetic Acid ($n=2$) and Ribose ($n=5$). As a starting point, does $n=3$ and $n=4$ make sense, i.e. $\mathrm{C}_3\mathrm{H}_{6}\mathrm{O}_3$ and $\mathrm{C}_4\mathrm{H}_{8}\mathrm{O}_4$ respectively?&lt;/p&gt;&#xA;" OwnerUserId="6801" LastEditorUserId="4945" LastEditDate="2015-10-23T17:59:26.593" LastActivityDate="2015-10-23T17:59:26.593" Title="Empirical and Molecular Formulae" Tags="&lt;physical-chemistry&gt;&lt;isomers&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13480" PostTypeId="1" AcceptedAnswerId="13495" CreationDate="2014-06-09T17:50:51.357" Score="1" ViewCount="121" Body="&lt;p&gt;If I had a beaker of water (i.e. many molecules), at exactly 0 °C and at atmospheric pressure, what phase would be encountered?&lt;br&gt;&#xA;And would the answer to this also apply to other substances at their freezing/melting point, and out of interest, would the same effect be observed at their other &quot;phase intercepts&quot; too?&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorDisplayName="user7951" LastEditDate="2015-03-23T18:29:39.363" LastActivityDate="2015-03-23T18:29:55.413" Title="Behaviour of water at exactly 0 °C" Tags="&lt;water&gt;&lt;phase&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13481" PostTypeId="2" ParentId="13479" CreationDate="2014-06-09T18:06:44.937" Score="2" Body="&lt;p&gt;Yes they make sense but the same molecular formula can represent different compounds, for the first case &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H6O3}$&lt;/span&gt; may rappresent:&lt;/p&gt;&#xA;&lt;p&gt;Glyceraldehyde:&lt;img src=&quot;https://i.stack.imgur.com/rjuSM.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;or lactic acid (and its optical isomers):&lt;img src=&quot;https://i.stack.imgur.com/ua9l7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C4H8O4}$&lt;/span&gt; leads to even more possibilities see the three different &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrose&quot; rel=&quot;nofollow noreferrer&quot;&gt;tetrose&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EoPTk.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Generally these compounds may be created when the sum of the formal charges of each atom in the molecular formula is equal to zero. Oxygen, hydrogen normally have formal charge of -2 and -1 carbon +4 so:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-2\times n + (-1 \times n \times 2) + (+4 \times n)=-4\times n +4\times n=0$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-06-09T18:06:44.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13482" PostTypeId="1" CreationDate="2014-06-09T19:15:29.777" Score="-1" ViewCount="117" Body="&lt;p&gt;I understand how liquids can become solids by decreasing temperature. To my understanding this is because the molecules slow down and are able to form bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm having problems understanding how we can create a solid out of liquid by beating it, such as the case with cream becoming butter if you beat it long enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is going on in such a process?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2014-06-10T02:44:13.683" Title="From liquid to solid by beating" Tags="&lt;everyday-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13483" PostTypeId="1" CreationDate="2014-06-09T19:35:24.520" Score="13" ViewCount="5741" Body="&lt;p&gt;Can you please explain the difference between the bond enthalpies in sulfurtrioxide $\ce{SO3}$ and the sulfite anion $\ce{SO3^{2-}}$?&lt;/p&gt;&#xA;" OwnerUserId="6803" LastEditorUserId="16683" LastEditDate="2018-03-02T10:40:32.453" LastActivityDate="2018-03-02T10:40:32.453" Title="Is the S–O bond strength in sulfur trioxide or sulfite anion larger?" Tags="&lt;bond&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13485" PostTypeId="1" AcceptedAnswerId="13492" CreationDate="2014-06-09T21:04:39.493" Score="1" ViewCount="19073" Body="&lt;p&gt;Why is a C-Br bond considered polar? From electronegativity considerations, both carbon and bromine have very similar electronegativities - 2.5 and 2.8 - respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, I am told that such a bond would be polar. We generally classify bond between atoms with EN differences &amp;lt; 0.5 as non-polar; why would C-Br be an exception? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-17T19:50:58.337" LastActivityDate="2014-08-17T19:50:58.337" Title="Why is the C-Br bond considered polar?" Tags="&lt;bond&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13486" PostTypeId="2" ParentId="13483" CreationDate="2014-06-09T21:18:17.107" Score="2" Body="&lt;p&gt;My hypothesis is that the bond enthalpy of the S-O bond in the sulfate anion is smaller than that of the S-O bond enthalpy in sulfur trioxide. I would say that the S-O bond strength lies with atomic size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sulfate anion, we have the electronegative oxygen atoms stabilizing the bulk of the negative 2 formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bearing in mind that anions are generally bigger than their respective neutral counterparts, and that cations are smaller than their respective neutral counterparts. Why? This lies with electron-electron shielding and $\ce{Z_{eff}}$, or effective nuclear charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having more electrons in a valence, such as in the case of the oxygens in the sulfate anion as opposed to the oxygens in sulfur dioxide, decreases $\ce{Z_{eff}}$. This in turn allows the electrons to be further from the nucleus, and this obviously increases size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increased size of the oxygens in the sulfate anion in turn lengthen the S-O bonds, and this in turn decreases the bond enthalpy of the S-O bonds in the sulfate anion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-09T21:18:17.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13487" PostTypeId="1" AcceptedAnswerId="13490" CreationDate="2014-06-09T22:19:10.807" Score="0" ViewCount="542" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/flS52.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the question mean by how many electron pairs are in 2p orbitals? How does one tell? Given that both the oxygen and the nitrogen are $\ce{sp^2}$ hybridized, wouldn't there be &lt;em&gt;two&lt;/em&gt; 2p orbitals? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ow8Mu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5591" LastEditDate="2015-10-09T18:06:20.467" LastActivityDate="2015-10-09T18:06:20.467" Title="2p Orbital Occupancy" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13488" PostTypeId="2" ParentId="13483" CreationDate="2014-06-09T22:23:41.460" Score="1" Body="&lt;p&gt;Sulfur Trioxide is typically represented with Sulfur double bonded to each of three oxygens.  In this context, the oxygens have full octets, while the Sulfur has an 'expansion'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the oxygens are full (and sulfur is pretty close to 'full'), to add electrons to this molecule requires the placement of electrons into 'antibonding orbitals'.  These orbitals would be pi-antibonding orbitals, weakening the S-O bonds somewhat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be advised, it is difficult to speak about the enthalpy of either bond(sigma or pi) within a double bond.  Typically the enthalpy of a double bond is compared to a single bond and the difference is assigned to the 'pi-bond'.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-06-09T22:23:41.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13489" PostTypeId="1" AcceptedAnswerId="128394" CreationDate="2014-06-09T22:40:23.920" Score="3" ViewCount="2910" Body="&lt;p&gt;Is there any single element (e.g. possibly carbon) that can be eaten, which has a nutritional value such that it provides calories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found on Wikipedia (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dietary_element&quot; rel=&quot;nofollow noreferrer&quot;&gt;dietary element&lt;/a&gt;) that there are certain elements that our bodies require.  I suppose oxygen could count as such a nutritional element, although we breathe it instead of eat it.&lt;/p&gt;&#xA;" OwnerUserId="6806" LastEditorUserId="402" LastEditDate="2020-02-28T21:44:59.503" LastActivityDate="2020-03-17T05:38:29.120" Title="Does any element have nutritional (caloric) value?" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;heat&gt;&lt;food-chemistry&gt;&lt;elements&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13490" PostTypeId="2" ParentId="13487" CreationDate="2014-06-09T22:44:56.010" Score="2" Body="&lt;p&gt;In the resonance structure drawn at the top of your image, the $\ce{p}$ orbital on carbon and the $\ce{p}$ orbital on oxygen share a bonding pair of electrons, a lone pair of electrons reside in a nitrogen $\ce{p}$ orbital, two lone pairs reside in two oxygen $\ce{sp^2}$ orbitals.  So for this case, the most commonly drawn representation, there is one lone pair in a $\ce{p}$ orbital.  In your middle resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital, and another lone pair in the nitrogen $\ce{p}$ orbital; hence, 2 lone pairs in two different $\ce{p}$ orbitals.  In your bottom resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital and a shared bonding electron pair in the $\ce{C-N}$ $\ce{pi}$ bond; therefor, one lone pair in a $\ce{p}$ orbital for this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a poorly worded question.  Since the real molecule is represented by a combination of the 3 resonance structures, the correct answer would be &quot;somewhere between one and two lone pairs in $\ce{p}$ orbitals.&quot;  Adding the detail contained in the above paragraph would probably help the grader understand your explanation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-09T22:44:56.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13491" PostTypeId="2" ParentId="7033" CreationDate="2014-06-09T23:04:16.110" Score="4" Body="&lt;p&gt;Charcoal is a good absorbent because its quite selective and cheap. Its high surface area means nothing if it doesn't bond to the components that you want absorbed and not a lot else- witness a household sponge which has high surface area and mercury- the mercury just slides off.  Charcoal is used in the gold mining/processing industry. The gold(and other metals) is oxidised to form cyanide complexes. The charcoal mostly only absorbs the cyanide complexes not all the other rubbish and water. Charcoal wouldn't be much good for the water purification industry if it absorbed water. Charcoal absorbation processes are a multitudinous- its thousands of papers of primary metallurgy(gold).   &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-06-09T23:04:16.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13492" PostTypeId="2" ParentId="13485" CreationDate="2014-06-09T23:21:26.120" Score="4" Body="&lt;p&gt;I poked around on the internet and found a lot of sites use the Pauling electronegativity scale (&lt;a href=&quot;http://periodictable.com/Properties/A/Electronegativity.al.html&quot; rel=&quot;nofollow&quot;&gt;link 1&lt;/a&gt;, &lt;a href=&quot;http://www.tutor-homework.com/Chemistry_Help/electronegativity_table/electronegativity.html&quot; rel=&quot;nofollow&quot;&gt;link 2&lt;/a&gt;).  For carbon the value is 2.55 and for bromine 2.96.  Not quite a 0.5 difference, but closer - and this &quot;rule&quot; is somewhat arbitrary to begin with.  Also the dipole moment of bromobenzene is around 1.7 D (&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/22998499&quot; rel=&quot;nofollow&quot;&gt;link 3&lt;/a&gt;), certainly not insignificant.  I too would have said bromobenzene is polar.  It is not strongly polar, whether it is &quot;a little&quot; polar, &quot;mildly&quot; polar, &quot;somewhat&quot; polar, etc. seems to be the question - but that's just semantics.  Maybe the reason many people consider the $\ce{C-Br}$ bond to be polar is because it has a very observable effect on electrophilic aromatic substitution reaction rates.  Specifically, why should the relative rate for electrophilic aromatic substitution at the &lt;em&gt;meta&lt;/em&gt; position in bromobenzene be slower that the rate in benzene itself unless you invoke the inductive effects caused by a polar $\ce{C-Br}$ bond?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-10T00:27:00.647" LastActivityDate="2014-06-10T00:27:00.647" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13493" PostTypeId="2" ParentId="13489" CreationDate="2014-06-09T23:30:18.897" Score="5" Body="&lt;p&gt;Interesting question! You may get a better answer, but I believe the answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calories are units of heat and represent energy available to the body. The form of fuels that can be (biochemically) &quot;burned&quot; by the body are hydrocarbons; i.e. molecules of carbon and water, like sugars. Carbon alone I do not believe can work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other single element fuel is hydrogen, and again I do not think this is biologically available to the body. By the way, oxygen is required as the oxidizer in any combustion reaction, and we definitely need oxygen (from breathing) to convert the food we eat to energy. But oxygen is not a fuel: it is the oxidizer that is reduced during combustion.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="402" LastEditDate="2020-02-28T22:05:06.397" LastActivityDate="2020-02-28T22:05:06.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13494" PostTypeId="2" ParentId="13477" CreationDate="2014-06-10T00:36:33.963" Score="7" Body="&lt;p&gt;In my estimation, this is basically marketing nonsense. If you examine the list of ingredients, the primary active agent is invariably some aluminum or aluminum-zirconium salt, which act as anti-perspirants primarily by the mechanical actions of shrinking pores (due to their astringency) and clogging ducts in the apocrine glands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Listed among the inactive ingredients is hydrogen peroxide, $\ce{H2O2}$. At sufficient concentrations, $\ce{H2O2}$ can be a potent oxidizing agent, and probably the source of the oxygen the ad refers to. It might indeed be able to neutralize some odors by oxidizing the molecules which cause them, thereby transforming those molecules into odorless ones. However, as it's listed among the inactive ingredients, it's almost certainly present in very low concentration. This is to be expected, since prolonged skin contact will invariably cause chemical burns unless the mixture is very dilute. Furthermore, much of the odor of sweat is the result of metabolic byproducts from skin-dwelling bacteria, chief among them certain short-chain carboxylic acids that are already maximally oxidized. The only malodorous substances in sweat that can be neutralized by oxidation, as far as I can tell, are certain amines and thiols. That said, I doubt the concentration of $\ce{H2O2}$ is sufficient to be of any use here. $\ce{H2O2}$ does also have anti-septic properties, so perhaps its purpose is bacteriostatic, but there I'm just speculating. Ultimately, I'm uncertain of its purpose in the formulation, but I'm extremely dubious that it helps substantially in eliminating odor.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-10T00:36:33.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13495" PostTypeId="2" ParentId="13480" CreationDate="2014-06-10T01:32:15.713" Score="2" Body="&lt;p&gt;It depends on how you achieved water at 0 °C. If you cooled water down to 0 °C, then it will remain water, unless you remove more heat, at which point it will start to freeze. Water needs to be at 0 °C to freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you warm ice up from below freezing to 0 °C, then it will remain ice, unless you add more heat to melt the ice. Ice needs to be at 0 °C to melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be the same for all substances.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorDisplayName="user7951" LastEditDate="2015-03-23T18:29:55.413" LastActivityDate="2015-03-23T18:29:55.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13496" PostTypeId="2" ParentId="13482" CreationDate="2014-06-10T01:46:05.653" Score="4" Body="&lt;p&gt;Butter is at least 80% milk fat, 16% water and about 3% milk solids (such as proteins and sugars). The milk fat is normally a solid at room temperature. To make butter, cream is separated from the milk and pasteurized. The cream is then churned (beaten) to separate out buttermilk (mostly water) from the milk fat (butter). So, when butter is made, a solid is separated from a solution (actually an emulsion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing I can think of where a liquid is transformed into a solid is beaten egg whites. In this case, air is mixed into the liquid to form an emulsion. It appears solid (but is still a liquid) since the liquid surrounds many air bubbles (which don't pop since they are very small). When you cook the beaten egg whites, then the protein in the egg white are denatured and become a solid.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-10T01:46:05.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13497" PostTypeId="1" AcceptedAnswerId="13503" CreationDate="2014-06-10T01:48:04.700" Score="0" ViewCount="2447" Body="&lt;p&gt;In transition metals, is the shell with the highest energy considered the valence shell? For example, in copper the electronic configuration is ${[Ar]\text{ } 3d^{10}\text{ } 4s^{1}}$. However, in accordance with the afbau principle the $3d$ subshell actually has higher energy than the $4s$ one. Does this mean the $3d$ subshell is considered the valence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, why is gold not considered to be a reactive substance. It's electronic configuration is not stable: $[Xe]\text{ } {4f}^{14} {5d}^{10} 6s^1$, as it has an unfilled $6s$ orbital. So why is it not highly reactive like some other transition metals?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2016-01-01T21:43:40.263" Title="Valence Shell for Transition Metals" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13498" PostTypeId="2" ParentId="13489" CreationDate="2014-06-10T02:29:58.060" Score="4" Body="&lt;p&gt;Caloric value is basically stored energy.  Plant-based foods create this stored energy through the absorption of sunlight, and in some extreme situations, I imagine there may be plants that generate energy from thermal vents and such.  They then store that energy in a molecular combination that you can think of as compressing a spring, giving it potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for animals, whereas rather than sunlight that energy is created from eating either plant or other animal life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass inherently has energy only at the mass-energy equivalent level, and biologically we are not capable of accessing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the conservation of energy, if the element you are putting in your body is not part of a molecular &quot;energy spring&quot; as per the analogy above, it will not spontaneously posses energy just because you are ingesting it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as Carbon, this is one of the great misunderstandings about fuel.  We burn Hydrocarbons as a fuel source.  Wood originally, then coal, and now we have jet fuel.  They are basically all the same thing, with decreasing Carbon to hydrogen ratios.  The energy stored is actually being released from the hydrogen bonds.  The ideal goal is for us to be using carbon-less hydrocarbons as fuel sources as the only burned waste is water as it combusts with the air, no CO2 providing the highest weight to fuel ratio (no carbon along for the ride reduces the weight) and doesn't pollute.&lt;/p&gt;&#xA;" OwnerUserId="6794" LastEditorUserId="402" LastEditDate="2020-02-28T22:04:20.983" LastActivityDate="2020-02-28T22:04:20.983" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13499" PostTypeId="1" AcceptedAnswerId="13501" CreationDate="2014-06-10T02:29:58.337" Score="3" ViewCount="1600" Body="&lt;p&gt;I am trying to write a balanced equation for the above mentioned reaction but having some trouble figuring it out. So far, I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{([C6H4N3O2]+ Cl^{-}) + H2SO4 + H2O -&amp;gt; C6H5NO3 + N2 + ?}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I &quot;incorporate&quot; sulfuric acid and water into the reaction? Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="11367" LastEditDate="2017-04-28T13:27:41.733" LastActivityDate="2017-04-28T13:27:41.733" Title="Balanced equation for synthesis of p-nitrophenol from p-nitrodiaznoium chloride salt and sulfuric acid/water?" Tags="&lt;organic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13500" PostTypeId="2" ParentId="13482" CreationDate="2014-06-10T02:44:13.683" Score="2" Body="&lt;p&gt;There's also Non-Newtonian Shear Thickening Fluids.  However I cannot personally explain how beating them turns them solid (temporarily).  But as best I understand, nobody is 100% sure so I'm not too embarrassed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shear_thickening&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Shear_thickening&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Newtonian_fluid&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Non-Newtonian_fluid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However LDC3s explaination of butter is likely a more accurate answer to your question, and its not actually a change of state from liquid to solid as a result of impact force, but rather water being removed from a material that is its own solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of taking the water out of salt water.&lt;/p&gt;&#xA;" OwnerUserId="6794" LastActivityDate="2014-06-10T02:44:13.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13501" PostTypeId="2" ParentId="13499" CreationDate="2014-06-10T03:02:02.293" Score="4" Body="&lt;p&gt;The sulfuric acid is not consumed in the reaction pathway to product, so there's no reason for it to appear in the equation for the reaction.  Therefor, the overall equation for the reaction is $$\ce{([C6H4N3O2]+ Cl^{-}) + H2O -&amp;gt; C6H5NO3 + N2 + HCl}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-10T13:13:12.060" LastActivityDate="2014-06-10T13:13:12.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13502" PostTypeId="2" ParentId="13499" CreationDate="2014-06-10T03:05:55.753" Score="4" Body="&lt;p&gt;The water is the source of the hydroxyl ($\ce{-OH}$) group in the phenol, which accounts for one oxygen and one hydrogen. The other hydrogen is ultimately abstracted, probably by a second molecule of water to form $\ce{H3O+}$. The simplest way of representing this, however, is to simply write $\ce{HCl}$ on the product side, though it's tacitly understood that it's unlikely to actually exist undissociated in any significant quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the $\ce{H2SO4}$, I'm fairly certain its only purpose is to maintain the acidity of the solution, with the aim of avoiding azo-coupling side reactions. (Although, with a para-substituted deactivating nitro group, this is probably not much of a concern.)&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-10T03:05:55.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13503" PostTypeId="2" ParentId="13497" CreationDate="2014-06-10T03:36:41.837" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Nope, as wiki tells us, only outermost electron shell is considered&#xA;valence shell, though d-electrons for transition metals are &#xA;considered as valence electrons. No logic here, only tradition.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;as you can see, there are three electron subshells above obviously inert Xe core. 5d-electons, that most of the time are far from the nuclei because of the form of their orbitals, feel dramatic increase in effective core charge through the third d-row. For example, for Ti effective charge, keeping its d-electrons is 4. For gold it is 11, so the electrons are well-held. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For s-electrons the form of the orbital puts very significant part of electronic density near the nuclei. This means, that for heavy elements (staring from as far as Ga) a so-known 'inert electron pair effect' is observed: in comparison with p- and d-electrons, due the form of the s-orbital, s-electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. As effect, Bi is highly unstable in valence state +5, perbromates are much stronger oxidizing agent than perchlorates, but same is not true for bromates and chlorates, and Ag, Au and Hg have very tightly bound s-electons, leading to relative inertness of said metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Au and Ag, however, strongly bound s-electrons participate in extremely powerful (due to relative small size of s-orbitals) s-s interactions, while for Hg only relatively weak s-p and d-p interactions are possible because of filled d- and s- subshells. Thus, metallic Ag and especially Au in metallic state have s-electrons involved into effective covalent interactions, leading to relatively inert nature. Beware, however, that they are far from unreactive: Ag have high affinity to sulfur, and both Ag and Au dissolve in cyanide solutions when oxygen is present, not saying about such agressive reagents as chlorine and bromine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, Au, Ag and Cu share one trait with Ia (I main subgroup) elements. Due to only spherical s-orbitals involvement in atom bonding in metal state, all three metals have high plasticity. Other d-elements have significant d-d bonding in metal states, and d-orbitals are not spherically symmetrical, leading to much less plastic behavior.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2016-01-01T21:43:40.263" LastActivityDate="2016-01-01T21:43:40.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13504" PostTypeId="1" CreationDate="2014-06-10T03:37:38.270" Score="1" ViewCount="304" Body="&lt;p&gt;How to form Schiff's Base at both the ends of the thiosemicarbazide?&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH2}$&lt;/span&gt; of hydrazine part reacts easily to form Schiff's base i.e. imine. But &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2}$&lt;/span&gt; attached to &lt;span class=&quot;math-container&quot;&gt;$\ce{C=S}$&lt;/span&gt; does not. How to do that?&lt;br /&gt;&#xA;One approach is by using NCS but that forms imine at &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2}$&lt;/span&gt; attached to &lt;span class=&quot;math-container&quot;&gt;$\ce{C=S}$&lt;/span&gt; only and not at the other end.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZZ9oN.png&quot; alt=&quot;thiosemicarbazide&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5255" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-07-29T09:28:25.707" Title="Reactivity of thiosemicarbazide" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13505" PostTypeId="2" ParentId="13463" CreationDate="2014-06-10T03:58:19.353" Score="4" Body="&lt;p&gt;If the soda is not distributed, the rate of loss of $\ce{CO2}$ would be an exponential decay. So changing from 5 volumes to 3 volumes is faster than from 3 to 1 volumes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A can of soda with 3 volumes of $\ce{CO2}$ has about 15 psi of extra pressure inside the can. If it was increased to 25 psi (5 volumes, like the fountain sodas), the can would need to be thicker to ensure it doesn't rupture, which would increase the cost of the can. There is a balance of the amount of $\ce{CO2}$ and the cost for the drink.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-06-10T04:06:11.140" LastActivityDate="2014-06-10T04:06:11.140" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13506" PostTypeId="1" AcceptedAnswerId="13508" CreationDate="2014-06-10T05:13:57.390" Score="2" ViewCount="7076" Body="&lt;p&gt;The question says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A proton is accelerated to one tenth of the velocity of light. If it's velocity can be measured with a precision &lt;span class=&quot;math-container&quot;&gt;$\pm 1\%$&lt;/span&gt;. What must be its uncertainty in position?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v=0.1 \times c =3\cdot 10^7\ \pu{m/s}\\&#xA; \Delta(v)=\frac{1}{100}\\&#xA; m= 1.6\cdot10^{-27}\ \pu{kg}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then I directly substituted these values into the formula:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta(v)\cdot\Delta(\text{position})\cdot \text{mass}=\frac h{4\pi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To get the uncertainty in position, however, the answer I got was approx. &lt;span class=&quot;math-container&quot;&gt;$3.5 \cdot 10^{-6}\ \pu{m}$&lt;/span&gt; which is way too different from the correct answer:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0.5\cdot10^{-13}\ \pu{m}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can anyone please explain me how to solve this question?&lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="83758" LastEditDate="2020-06-12T20:05:43.700" LastActivityDate="2020-06-12T20:05:43.700" Title="Find the uncertainty in position" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13508" PostTypeId="2" ParentId="13506" CreationDate="2014-06-10T07:28:26.317" Score="4" Body="&lt;p&gt;The question tells you&#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;v&amp;amp;=3\cdot10^7\:\mathrm{m/s}\pm1\%\\&#xA;&amp;amp;&amp;amp;\Delta v &amp;amp;= |(v+1\%v) - (v-1\%v)|\\&#xA;\implies&amp;amp;&amp;amp;\Delta v &amp;amp;= \frac{2}{100}v\\&#xA;\therefore&amp;amp;&amp;amp;\Delta v&amp;amp;= 2\cdot3\cdot10^5\:\mathrm{m/s}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&#xA;\begin{aligned}&#xA;&amp;amp;&amp;amp;\Delta x\cdot\Delta p &amp;amp;\geq \frac{\hbar}{2}\\&#xA;\implies&amp;amp;&amp;amp;\Delta x\cdot\Delta (v\cdot m) &amp;amp;\geq \frac{\hbar}{2}\\&#xA;\implies&amp;amp;&amp;amp;\Delta x&amp;amp;\geq \frac{\hbar}{2\cdot\Delta v\cdot m }\\&#xA;\therefore&amp;amp;&amp;amp;\Delta x &amp;amp;\gtrapprox\frac{1.05\cdot10^{−34}\:\mathrm{J\cdot s}}{2\cdot6\cdot10^5\:\mathrm{m/s}\cdot1.6\cdot10^{−27}\:\mathrm{kg}}\\&#xA;&amp;amp;&amp;amp;\Delta x&amp;amp;\gtrapprox5.5\cdot10^{-14}\:\mathrm{m}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-06-10T07:33:32.190" LastActivityDate="2014-06-10T07:33:32.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13510" PostTypeId="2" ParentId="13464" CreationDate="2014-06-10T17:22:34.633" Score="0" Body="&lt;p&gt;This is going to sound simplistic relative to the previous 2 answers, but what about putting a crystal of solid oxygen into an anvil at a temperature below its solidification point, and compressing it. Phase diagrams (e.g. &lt;a href=&quot;http://commons.wikimedia.org/wiki/File%3aPhase_diagram_of_solid_oxygen.svg&quot; rel=&quot;nofollow&quot;&gt;http://commons.wikimedia.org/wiki/File:Phase_diagram_of_solid_oxygen.svg&lt;/a&gt;) show that oxygen exists in different crystal forms at different temperatures and pressures. It becomes superconducting at low temperatures and high pressures.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-06-10T17:22:34.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13511" PostTypeId="1" AcceptedAnswerId="13512" CreationDate="2014-06-10T18:00:10.420" Score="3" ViewCount="97" Body="&lt;p&gt;I was wondering that since acids dissociate in water to make a lot of  $\ce{H+}$ cations. Is it possible that introducing a supply of electrons via an electric current, would create $\ce{H2}$ gas?  &lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="4945" LastEditDate="2014-06-11T07:19:56.870" LastActivityDate="2014-06-11T07:19:56.870" Title="Since aqueous acids release Hydrogen cations into the solution, will an electric current create Hydrogen gas?" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13512" PostTypeId="2" ParentId="13511" CreationDate="2014-06-10T21:17:48.507" Score="4" Body="&lt;p&gt;Yes.  In fact the first electrolysis was done with acidified water.  Pure water is only slightly ionized and is a poor conductor.  Adding ions to water increases its conductivity and speeds up the electrolysis process.  So by adding a source of protons to pure water the following reaction will occur at the cathode $$\ce{2H+ (aq) + 2e^{-}-&amp;gt;H2(g)}$$  Of course a parallel reaction will occur at the anode $$\ce{4OH- -&amp;gt;2H2O + O2(g) + 4e-}$$ otherwise a current wouldn't flow and the electrolysis would not continue.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-10T21:17:48.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13513" PostTypeId="1" AcceptedAnswerId="13515" CreationDate="2014-06-10T22:53:22.190" Score="4" ViewCount="3182" Body="&lt;p&gt;Why are unpaired electrons especially reactive? Does pairing electrons decrease the reactivity of the electrons? But then doesn't forcing two electrons into the same orbital cost energy? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-10T23:21:45.420" Title="Why are radicals unstable?" Tags="&lt;bond&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13514" PostTypeId="1" AcceptedAnswerId="13518" CreationDate="2014-06-10T23:10:57.767" Score="5" ViewCount="397" Body="&lt;p&gt;We know that water auto-ionizes (to a rather small extent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is which solvent would have the fewest ions (or at least fewer ions than water). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that perhaps some kind of hydrocarbon, as these are both weaker acids and weaker bases than water and thus even less reactive. I've seen computers submerged in oils (as well as distilled water) due to their low conductivities. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2017-09-18T03:58:09.670" Title="What solvent would have the fewest ions?" Tags="&lt;solutions&gt;&lt;conductivity&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13515" PostTypeId="2" ParentId="13513" CreationDate="2014-06-10T23:21:45.420" Score="6" Body="&lt;p&gt;Free radicals are usually very reactive.  If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form.  Yes, it does cost energy to bring those electrons close together.  But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually more than enough energy to offset the energy cost of bringing those electrons close enough to form a new bond.  Because of the exothermicity of bond formation the reaction profile usually has a small activation energy.  Translation: a small activation energy means a  fast reaction.  A caveat, at the outset I said that &quot;free radicals are usually reactive&quot;.  There are many known free radicals that are quite stable.  This is usually due to steric factors (the reactive free radical center is surrounded by bulky groups and just not sterically accessible for reaction) or electronic factors (some free radicals exist in very large delocalized systems, hence the spin density at any one atom in the system is so small that reaction is unlikely).  One further point, most of what I've said relates to free radicals, which was the title of your question, however you did mention &quot;unpaired electrons&quot; in the beginning of your post.  Molecules with unpaired electrons (note the plural, electrons) are different than free radicals with (usually) just one unpaired electron.  Some molecules like $\ce{O2}$ are quite stable even though they have two unpaired electrons.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-10T23:21:45.420" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13516" PostTypeId="1" CreationDate="2014-06-11T00:05:06.463" Score="0" ViewCount="515" Body="&lt;p&gt;The hydrolysis of salt tells you whether a salt is acidic or basic. But it does not tell you how acidic or basic a salt is. Is there a way to determine this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that taking of the weak base part of a salt( presuppose that it is acidic) and see how many hydronium ion it could produce. So the more hydronium ions, the more acidic this salt is?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="87609" LastEditDate="2020-11-19T17:37:25.883" LastActivityDate="2020-11-19T17:37:25.883" Title="Hydrolysis of salt" Tags="&lt;acid-base&gt;&lt;hydrolysis&gt;&lt;salt&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13517" PostTypeId="2" ParentId="13514" CreationDate="2014-06-11T00:05:07.230" Score="8" Body="&lt;p&gt;2-Methylpropane has a $\mathrm{p}K_\mathrm{a}$ of 53 (&lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm{p}K_\mathrm{a}$ table&lt;/a&gt;).  That sounds pretty non-ionic.  Can't think of any common all-carbon (no substituents of any kind) liquid compounds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2017-09-18T03:58:09.670" LastActivityDate="2017-09-18T03:58:09.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13518" PostTypeId="2" ParentId="13514" CreationDate="2014-06-11T01:32:43.673" Score="5" Body="&lt;p&gt;Ron's answer is a good option, however &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid-base_reaction_theories#Solvent_system_definition&quot; rel=&quot;nofollow noreferrer&quot;&gt;autoionization is not necessarily equivalent to self-protonation&lt;/a&gt;. There may be some solvents where autoionization is even more suppressed than any self-protonation reaction. Perhaps the liquid with the lowest tendency to ionize may be something composed of molecules with extremely strong bonds, for example liquid nitrogen or liquid carbon monoxide. Of course, in Ron's acid table there already are several compounds which would be expected to ionize less than once in even reasonably macroscopic volumes of liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As Greg points out in the comments, why not go all out and consider noble gasses. The only plausible ionization reaction for the pure liquid elements is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ng(l) &amp;lt;=&amp;gt; Ng+_{(Ng)} + e-_{(Ng)}}\ ,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{Ng}$ indicates a noble gas and the $\ce{(Ng)}$ subscript indicates solvation by the noble gas atoms. The electron doesn't attach to anything, as $\ce{Ng-}$ would actually be unbound (no barrier to dissociation, or at least this is true for non-ultraheavy noble gasses such as ununoctium and beyond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that noble gas atoms likely make for an extremely poor solvent, we can get a very rough measure of the equilibrium constant by assuming the solvation is nil. Thus, the reaction would be equivalent to the gas-phase ionization of the atoms. Assuming the entropy change is a small factor relative to the very large ionization enthalpies, then the reaction $\Delta G_\mathrm{ion}^\circ$ is simply equal to the ionization energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For helium, this equals a massive $\pu{2372.3 kJ mol-1}$. Applying $\Delta G^\circ = -RT\ln K_\mathrm{ion}$, then for liquid helium at its boiling point under atmospheric pressure ($\pu{4.2 K}$), the ionization constant is as low as $10^{-29500}$ (though most of the reason the exponent is tiny is because of the really low temperature; for comparative purposes the value at room temperature would be around $10^{-400}$). A more accurate calculation would drive this value up quite a bit, but it would still be many, many orders of magnitude more difficult to ionize than anything else.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="41328" LastEditDate="2017-09-18T03:57:09.273" LastActivityDate="2017-09-18T03:57:09.273" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13519" PostTypeId="1" CreationDate="2014-06-11T04:16:00.040" Score="3" ViewCount="110" Body="&lt;p&gt;Does anyone know of a step by step guide to building a microbial fuel cell? Could someone here on Chem SE possibly provide one? I am looking for something like &lt;a href=&quot;http://www.engr.psu.edu/ce/enve/logan/bioenergy/pdf/Cathode_093008.pdf&quot; rel=&quot;nofollow&quot;&gt;THIS&lt;/a&gt; but for the whole fuel cell and not just the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Descriptive papers on the topic would be helpful also. &lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="4363" LastEditDate="2014-06-11T04:39:58.600" LastActivityDate="2014-10-16T04:40:36.323" Title="Microbial Fuel Cell" Tags="&lt;fuel&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13520" PostTypeId="2" ParentId="13516" CreationDate="2014-06-11T04:36:10.047" Score="2" Body="&lt;p&gt;&lt;strong&gt;Steps for determining acidity or basicity of a salt by hydrolysis:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write the neutralization reaction in reverse (salt and water makes base and acid).&lt;/li&gt;&#xA;&lt;li&gt;Break up the salt into its ions.&lt;/li&gt;&#xA;&lt;li&gt;To write the base, add an $\ce{OH-}$ to the cation (the positive ion) for each + charge.&lt;/li&gt;&#xA;&lt;li&gt;To write the acid, Add $\ce{H+}$ to the anion (the negative ion) for each negative charge.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you for a strong base, the salt is &lt;strong&gt;basic&lt;/strong&gt;.  $\ce{(NaOH, KOH, LiOH, Mg(OH)_2,  ...)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you form a strong acid it is &lt;strong&gt;acidic&lt;/strong&gt;.  $\ce{(HCl, HNO_3, HClO_4, HBr,...)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you form both a strong acid and a strong base it is &lt;strong&gt;neutral&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EXAMPLE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH_4Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4Cl + H_2O &amp;lt;=&amp;gt; NH4OH + HCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{HCl}$ is a strong acid it breaks up and yields $\ce{H+}$, the salt is &lt;strong&gt;acidic&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2014-06-11T10:26:18.020" LastActivityDate="2014-06-11T10:26:18.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13521" PostTypeId="1" AcceptedAnswerId="14758" CreationDate="2014-06-11T06:01:25.380" Score="1" ViewCount="320" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In which of the following chemical compounds are the building blocks molecules?&lt;/p&gt;&#xA;&lt;p&gt;(A) &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt;&lt;br /&gt;&#xA;(B) &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt;&lt;br /&gt;&#xA;(C) &lt;span class=&quot;math-container&quot;&gt;$\ce{KNO3}$&lt;/span&gt;&lt;br /&gt;&#xA;(D) &lt;span class=&quot;math-container&quot;&gt;$\ce{NaNO3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(B) is the correct answer, but I have no idea why. Could you briefly explain why this is the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="6812" LastEditorUserId="69830" LastEditDate="2021-05-22T07:30:52.697" LastActivityDate="2021-05-22T07:30:52.697" Title="In which of the following chemical compounds are the building blocks molecules?" Tags="&lt;molecules&gt;&lt;covalent-compounds&gt;&lt;salt&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="13523" PostTypeId="1" CreationDate="2014-06-11T09:47:55.437" Score="0" ViewCount="163" Body="&lt;p&gt;Tetrahydrofolate:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i6sx2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess (6S) means some sixth &lt;em&gt;S&lt;/em&gt;-center and 5 some other kind of numbering.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2014-06-11T21:08:55.867" Title="What do '(6S)' and '5' mean in (6S)-5-formyltetrahydrofolate?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13524" PostTypeId="1" AcceptedAnswerId="13526" CreationDate="2014-06-11T10:28:39.233" Score="0" ViewCount="83" Body="&lt;p&gt;$m(\ce{FeSO4. $x$~H2O})=20.0\:\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N(\ce{H2O})=3.03 \cdot10^{23}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$x=?$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1.)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{aligned}&#xA;   m (\ce{H2O})&amp;amp;= \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{N_{A}}\\&#xA;   &amp;amp;= \frac{3.03 \cdot10^{23} \cdot 18.02$\:\mathrm{g/mol}}{6.02 \cdot 10^{23}/\:\mathrm{mol}}\\&#xA;   &amp;amp;= 9.06987\:\mathrm{g}&#xA;  \end{aligned}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2.)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{aligned}&#xA;   m(\ce{FeSO4})&amp;amp;= m(\ce{FeSO4} \cdot x \ce{H2O}) - m(\ce{H2O})\\&#xA;   &amp;amp;= 20.00\:\mathrm{g} - 9.06987\:\mathrm{g}\\&#xA;   &amp;amp;= 10.9301\:\mathrm{g}&#xA;  \end{aligned} &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any suggestions what to do next?&lt;/p&gt;&#xA;" OwnerUserId="6812" LastEditorUserId="4945" LastEditDate="2014-06-11T11:26:24.703" LastActivityDate="2014-06-11T11:26:24.703" Title="Any suggestions how to solve this?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13526" PostTypeId="2" ParentId="13524" CreationDate="2014-06-11T11:19:14.590" Score="0" Body="&lt;p&gt;Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculate $n(\ce{H2O})$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{H2O})\approx0.5\:\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x\approx 6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-11T11:19:14.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13527" PostTypeId="2" ParentId="13523" CreationDate="2014-06-11T13:15:24.283" Score="2" Body="&lt;p&gt;I've redrawn a truncated version of the title molecule below with the &quot;5-formyl&quot; group in place.  I've also inserted small numbers so you can see how the base system is numbered.  So &quot;5-formyl&quot; means that a formyl group ($\ce{-CHO}$) is attached to the &quot;5&quot; position.  The 6-position has 4 different substituents attached to it so it is a chiral carbon atom.  The R-S nomenclature system is used to describe the 2 mirror image possibilities at chiral centers.  The enantiomer you've drawn is the &quot;6S&quot; stereoisomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIshA.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-11T21:08:55.867" LastActivityDate="2014-06-11T21:08:55.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13532" PostTypeId="2" ParentId="374" CreationDate="2014-06-11T14:13:52.093" Score="3" Body="&lt;p&gt;Oxidation State of &quot;Fe&quot; in brown ring complex depends on the binding mode of NO ligand to Iron (whether is it &quot;bent&quot; or &quot;linear&quot;). Unless we specify the IR frequencies of NO ligand for different modes, we can not say the oxidation state of &quot;Fe&quot;. Depending on the IR stretching  frequencies of NO ligand which is bound to the Iron, it can be NO+, NO- or simply NO.&lt;/p&gt;&#xA;" OwnerUserId="6817" LastActivityDate="2014-06-11T14:13:52.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13533" PostTypeId="1" CreationDate="2014-06-11T15:34:39.970" Score="2" ViewCount="676" Body="&lt;p&gt;Why doesn't alkaline earth metals lose only one electron when they are ionized ? I know that magnesium atoms like to have the electronic configuration of neon but I don't understand the reason that we can't find $\ce{Mg+}$ atoms .&lt;/p&gt;&#xA;" OwnerUserId="6819" LastEditorUserId="4734" LastEditDate="2014-06-16T11:18:18.037" LastActivityDate="2014-06-16T11:18:18.037" Title="Why don't alkaline earth metals lose only one electron when they are ionized?" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13535" PostTypeId="2" ParentId="13533" CreationDate="2014-06-11T17:36:25.233" Score="3" Body="&lt;p&gt;Magnesium can definitely exist as $\ce{Mg^{+}}$. Hence the availability of first ionization energy data for magnesium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for other metals; there are only &lt;em&gt;uncommon&lt;/em&gt; valences, such as $\ce{Na^{3+}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had enough energy, you could make $\ce{Na^{4+}}$ if you wanted to. It would just cost, according to the table below, 9543 + 6912 + 4562.4 + 495.8 kJ/mol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZiILX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your question is why isn't $\ce{Mg^{+}}$ more commonly encountered. That has to do with stability. Losing two electrons and becoming isoelectronic with a noble gas is more stable than having an unpaired electron. An unpaired electron is available for bond formation, and bond formation tends to be exothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to read up on the +1 valence of magnesium, such as in $\ce{MgH}$, you can read more here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bernath.uwaterloo.ca/media/24.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bernath.uwaterloo.ca/media/24.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-16T04:14:41.367" LastActivityDate="2014-06-16T04:14:41.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13536" PostTypeId="1" AcceptedAnswerId="13583" CreationDate="2014-06-11T17:45:36.060" Score="8" ViewCount="6428" Body="&lt;p&gt;Is bond formation &quot;strictly&quot; exothermic? The IUPAC definition of exothermic doesn't make any reference to bond formation. However, I have seen the aforementioned statement before - that bond formation is &quot;strictly&quot; exothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the answer is &quot;it depends&quot; and that &quot;it depends&quot; at least on how accurately bonds are characterized. For example, the first reaction below is endothermic, while the second one is exothermic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NO -&amp;gt; ONNO}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NO_2 -&amp;gt; O_2NNO_2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis structure analysis for the first reaction at best misleads because one might suspect the nitric oxide molecule as having a bond order of 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TmVbG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the lone electron is actually delocalized. So the actual $\ce{N-O}$ bond order should be 2.5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VIaNp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\ce{ONNO}$ molecule however there is no delocalization of electrons and the bond order of the $\ce{N-O}$ bond is 2. So the $\ce{N-O}$ bond is weakened, and even the formation of the $\ce{N-N}$ bond cannot compensate for this weakening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the above example reminds me of certain unstable homonuclear species. Such as $\ce{Ne_2}$, which according to MO theory has a bond order of 0 - i.e. there is no bond. So I suppose that bond formation is not always exothermic. On the other hand, if the bond order is 0, is a bond really &quot;formed&quot;? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-11T17:56:13.737" LastActivityDate="2017-12-29T02:03:28.297" Title="Is Bond Formation &quot;Strictly&quot; Exothermic?" Tags="&lt;bond&gt;&lt;energy&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13537" PostTypeId="1" AcceptedAnswerId="13574" CreationDate="2014-06-11T17:59:51.050" Score="0" ViewCount="160" Body="&lt;p&gt;Does the hyponitrite molecule exist? I know the hyponitrite ion exists (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyponitrite&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hyponitrite&lt;/a&gt;) but what about the molecule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found no reference to the molecular form on Wikipedia. But it does seem that such a molecule could at least come together through the dimerization of nitric oxide. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-12T20:00:21.833" Title="Existence of Hyponitrite Molecule" Tags="&lt;bond&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13540" PostTypeId="1" CreationDate="2014-06-11T21:57:34.327" Score="2" ViewCount="41" Body="&lt;p&gt;Does evaporation of non-water liquids into the atmosphere vary with atmospheric humidity?&lt;/p&gt;&#xA;" OwnerUserId="6822" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:12:16.183" LastActivityDate="2015-10-23T17:12:16.183" Title="Atmospheric Humidity Effects on Non-water Evaporation" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13541" PostTypeId="1" CreationDate="2014-06-11T22:24:51.743" Score="4" ViewCount="17211" Body="&lt;p&gt;I am a physicist, not a chemist. I'm trying to get a basic understanding of the reactions taking place in a battery using a saltwater electrolyte with copper and zinc terminals. I'm writing a general science level article about it. I have found that the reactions taking place at the terminals are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA; \text{Zinc terminal:}&amp;amp;&amp;amp;\ce{Zn(s) &amp;amp;-&amp;gt; Zn^{2+}(aq) + 2e^-} &amp;amp;(\Delta G = -0.76\:\mathrm{V})\\&#xA; \text{Copper Terminal:}&amp;amp;&amp;amp;\ce{Cu^{2+}(aq) + 2e^- &amp;amp;-&amp;gt; Cu(s)} &amp;amp;(\Delta G = -0.34\:\mathrm{V})&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I understand the first reaction in terms zinc leaving the terminal and going into solution kind of as if it were dissolved $\ce{ZnCl2}$ (does that make sense?). But I need help understanding the second reaction. I've inserved a copper terminal into the solution. But not copper ions. Am I to understand it like this: when I put the copper terminal in the solution, copper ions immediately drift off, perhaps for the same reason the zinc ions do, but then deposit back onto the terminal in the presence of the electrons released when the zinc goes into solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like an explanation somewhere between that it just happens and full on redox theory, if that is possible.&lt;/p&gt;&#xA;" OwnerUserId="6821" LastEditorUserId="4945" LastEditDate="2014-06-12T04:54:43.650" LastActivityDate="2017-01-25T11:20:09.060" Title="Explanation for the reactions in a saltwater battery with zinc and copper electrodes" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13542" PostTypeId="1" AcceptedAnswerId="13543" CreationDate="2014-06-11T22:39:56.260" Score="3" ViewCount="250" Body="&lt;p&gt;I am working on a chemistry assignment and can’t figure out part of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.2 mol}$&lt;/span&gt; of a compound with a molecular weight of &lt;span class=&quot;math-container&quot;&gt;$82$&lt;/span&gt; contain &lt;span class=&quot;math-container&quot;&gt;$\pu{9.6 g}$&lt;/span&gt; of carbon, &lt;span class=&quot;math-container&quot;&gt;$\pu{1.2 g}$&lt;/span&gt; of hydrogen and the rest is nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is asking for the empirical and molecular formula which I can solve, but how do you determine the amount of nitrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I assumed a &lt;span class=&quot;math-container&quot;&gt;$\pu{100 g}$&lt;/span&gt; sample but that didn't get me anywhere. The empirical formula answer is supposed to be &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H3N}$&lt;/span&gt; but I couldn’t achieve that. Unless the book is wrong.&lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="4945" LastEditDate="2020-02-17T16:05:22.440" LastActivityDate="2020-02-17T22:11:53.030" Title="How to determine the molecular formula based on molecular weight, amount of substance, and partial masses of some, but not all contained elements?" Tags="&lt;organic-chemistry&gt;&lt;stoichiometry&gt;&lt;elemental-analysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13543" PostTypeId="2" ParentId="13542" CreationDate="2014-06-11T22:48:46.943" Score="5" Body="&lt;p&gt;If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T12:32:09.650" LastActivityDate="2015-06-17T12:32:09.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13544" PostTypeId="2" ParentId="6337" CreationDate="2014-06-11T22:50:33.667" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;${\Delta G}$ and ${\Delta H}$ are normally expressed on a per mole basis.  Your assertion that ${\Delta G}$ is related to concentration is not precise.  It is related to the ratio of concentrations, and as such the incongruity with ${\Delta H}$ is not troubling.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you found a fuel cell with a gain in entropy, according to your equation you would have more than 100% efficiency.  I seriously doubt such a situation exists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-06-11T22:50:33.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13545" PostTypeId="1" CreationDate="2014-06-11T23:17:30.727" Score="1" ViewCount="11203" Body="&lt;p&gt;I've got this question and I cannot figure out how to get the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the number of moles of Phosphorus in $15.95\:\mathrm{g}$ of tetraphosphorus decaoxide. ($\ce{P4O10}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried doing it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar Mass of (element) = Number of atoms * Relative Atomic Mass = ### grams/mole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of Compound in Grams = $15.95\:\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(\ce{P4}) = 4(30.97376) = 123.89504 \:\mathrm{g/mol}$&lt;br&gt;&#xA;$M(\ce{O10}) = 10(15.9994) = 159.994 \:\mathrm{g/mol}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(\ce{P4O10}) = 123.89504\:\mathrm{g/mol} + 159.994\:\mathrm{g/mol} = 283.88904 \:\mathrm{g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{P}: (123.89504 \:\mathrm{g/mol})/(283.88904 \:\mathrm{g/mol}) = 0.4364207 \cdot (100\% = 43.64\% \ce{P})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$15.95\:\mathrm{g} \cdot 0.4364207 = 6.9565\:\mathrm{g} \ce{P}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{P} = 6.9565\:\mathrm{g}/ (123.89504\:\mathrm{g/mol}) = 0.056148 \:\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer in the back of the book is $0.2247 \:\mathrm{mol}$&lt;/p&gt;&#xA;" OwnerUserId="6825" LastEditorUserId="5026" LastEditDate="2018-03-04T02:50:16.380" LastActivityDate="2018-03-04T02:51:00.163" Title="Calculate the number of moles of phosphorus in 15.95g of tetraphosphorus decaoxide" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13546" PostTypeId="2" ParentId="13545" CreationDate="2014-06-11T23:28:38.827" Score="2" Body="&lt;p&gt;You can go about it more directly as follows.  You have $(15.95/283.889)=\pu{0.0562 moles}$ of $\ce{P4O10}$.  From the molecular formula you see that for every mole of $\ce{P4O10}$ you have 4 moles of phosphorous.  Therefor you have $4 \times 0.0562 = \pu{0.2247 moles}$ of phosphorous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your error involved converting moles of $\ce{P4}$ to moles of $\ce{P}$.  If you multiply your answer by 4 you get the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-04T02:51:00.163" LastActivityDate="2018-03-04T02:51:00.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13547" PostTypeId="1" CreationDate="2014-06-12T01:05:31.863" Score="5" ViewCount="4378" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/guc34.gif&quot; alt=&quot;C NMR for Benzaldehyde&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this image is correct, why is the Carbon-3 more deshielded than Carbon-2 and Carbon-1? From what I understand, deshielding is the greatest when the Carbon atom is closer to electron withdrawer and it appears Carbon-3 is the farthest.&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="4945" LastEditDate="2015-02-16T11:49:13.700" LastActivityDate="2015-02-16T11:49:13.700" Title="Why is the para carbon in Benzaldehyde NMR more deshielded than the ortho and meta carbons?" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13548" PostTypeId="2" ParentId="13547" CreationDate="2014-06-12T01:30:29.577" Score="5" Body="&lt;p&gt;Resonance effects come into play along with the inductive effects.  I've drawn a few of the resonance structures for benzaldehyde below, notice how resonance effects place some positive charge on carbons 2, 3 and 5 which results in their deshielding.  I might have guessed that the combination of resonance and inductive effects would place C2 below C3, but that's not the case.  At least C5 is at lowest field and C2 and C3 are close and both are below C1. The interplay of inductive and resonance effects is complex and estimating the magnitudes of these effects can get tricky.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fK7LM.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-12T01:30:29.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13549" PostTypeId="1" AcceptedAnswerId="13550" CreationDate="2014-06-12T01:41:53.623" Score="0" ViewCount="188" Body="&lt;p&gt;What happens when strongly basic water is electrolyzed? I know that the electrolysis of water in strongly acidic solution involves hydrogen protons; will these protons still be involved in strongly basic solution? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-12T02:07:33.567" Title="Electrolysis of Strongly Basic Water" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13550" PostTypeId="2" ParentId="13549" CreationDate="2014-06-12T02:07:33.567" Score="2" Body="&lt;p&gt;In a basic solution $\ce{H2O}$ will act as the $\ce{H}$-source for the hydrogen evolution reaction. The reaction will look something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \ce{2H2O + 2e- -&amp;gt; 2OH- + H2}&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-06-12T02:07:33.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13551" PostTypeId="1" AcceptedAnswerId="13552" CreationDate="2014-06-12T02:12:21.157" Score="2" ViewCount="457" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If, in 2 M $\ce{HONO2}$, 2 mols $\ce{Al (s)}$ are oxidized to $\ce{Al(OH2)6^3+ (aq)}$, mols of reduction product hydrogen is&lt;br&gt;&#xA;  (&lt;strong&gt;a&lt;/strong&gt;) two;&lt;br&gt;&#xA;  (&lt;strong&gt;b&lt;/strong&gt;) three;&lt;br&gt;&#xA;  (&lt;strong&gt;c&lt;/strong&gt;) four;&lt;br&gt;&#xA;  (&lt;strong&gt;d&lt;/strong&gt;) five;&lt;br&gt;&#xA;  (&lt;strong&gt;e&lt;/strong&gt;) six.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the above question, how would one solve it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be valid to simply say that I know that 6 electrons are being transferred because we have this skeleton equation (balanced for charge): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al -&amp;gt; Al(OH2)6^3+ + 6 e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And given that 6 electrons are being transferred and that hydrogen is a reduction product, we must obtain hydrogen from some source in this system. The source would probably be hydronium ion, since hydronium ion is less stable than water. Plus abstracting protons from water to form hydrogen gas wouldn't make much sense in strongly acidic solution (we'd be forming hydroxide ions as well)! And knowing the following reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H+ + 2 e- -&amp;gt; H2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we just say since we know that 6 electrons are being moved, and that hydrogen gas is a product, we know that the immediately above reaction is occurring, and according to its stoichiometry, 2 moles of electrons yields 2 moles of hydrogen gas? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2017-10-01T14:50:51.027" LastActivityDate="2017-10-01T14:53:25.513" Title="Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13552" PostTypeId="2" ParentId="13551" CreationDate="2014-06-12T02:40:07.437" Score="5" Body="&lt;p&gt;You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.)&#xA;\begin{aligned}\ce{&#xA;(1)&amp;amp;&amp;amp; Al + 6H2O &amp;amp;-&amp;gt; Al(OH2)_{6}^{3+} + 3e^{-}&amp;amp;|\cdot2\\&#xA;+ (2)&amp;amp;&amp;amp; 2H3O+ + 2e^{-} &amp;amp;-&amp;gt; H2 ^ + 2H2O &amp;amp;|\cdot3\\\hline&#xA;(3)&amp;amp;&amp;amp; 2Al + 12H2O + 6H3O+ &amp;amp;-&amp;gt; 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\&#xA;(4)\equiv(3)&amp;amp;&amp;amp; 2Al + 6H2O + 6H3O+ &amp;amp;-&amp;gt; 2Al(OH2)_{6}^{3+} + 3H2 ^ \\&#xA;(5)\equiv(3)&amp;amp;&amp;amp; Al + 3H2O + 3H3O+ &amp;amp;-&amp;gt; Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.&lt;br&gt;&#xA;Hence, three mole of Hydrogen gas are formed by two mole of Aluminium. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit upon Dissenters follow up question &lt;em&gt;&quot;Why exactly would hydronium ion be the source?&quot;&lt;/em&gt;&lt;br&gt;&#xA;It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="7475" LastEditDate="2017-10-01T14:53:25.513" LastActivityDate="2017-10-01T14:53:25.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13553" PostTypeId="2" ParentId="13551" CreationDate="2014-06-12T02:40:45.850" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;2 moles of electrons yields 2 moles of hydrogen gas&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doesn't your second equation say that 2 moles of electrons yields 1 mole of $\ce{H2}$?  Following your reasoning, wouldn't 2 moles of $\ce{Al}$ yield 12 moles of electrons, which in turn would yield 6 moles of $\ce{H2}$?  In any case, the balanced equation is $$\ce{12H2O + 6HNO3 + 2Al -&amp;gt; 2Al(H2O)_6^{+3} + 3H2 + 6NO3^-}$$ so 2 moles of $\ce{Al}$ would produce 3 moles of $\ce{H2}$ (answer B).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-12T02:40:45.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13554" PostTypeId="2" ParentId="13541" CreationDate="2014-06-12T03:31:28.730" Score="2" Body="&lt;p&gt;The key point you are missing is that the thermodynamic values you report are for standard conditions, meaning that there would also be 1 M $\ce{Zn^2+}$ and 1 M $\ce{Cu^2+}$ in the solutions.  The Nernst equation can be used to determine the non-standard condition potential (in your case, sticking a copper wire into a solution with effectively no copper ions.  In practice, this would be very hard to do, as adventitious ions (and other reactions not considered) will contribute to the cell potential when the intended ion concentrations are so low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept can be used to determine the concentration of copper ions in solution, and may be of interest in explaining the phenomenon to a general audience.  Two $\ce{Cu}$/$\ce{Cu^2+}$ half-cells with the &lt;em&gt;same&lt;/em&gt; concentration of copper ions will result in a cell potential of zero; however, if one of the cell concentrations is known (for argument's sake, say 1 M) then the &quot;unknown&quot; ion concentration could be found using the Nernst equation.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorDisplayName="user7951" LastEditDate="2015-11-12T00:26:43.917" LastActivityDate="2015-11-12T00:26:43.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13555" PostTypeId="1" AcceptedAnswerId="13560" CreationDate="2014-06-12T03:38:33.323" Score="4" ViewCount="500" Body="&lt;p&gt;I have looked at several sources including &lt;a href=&quot;https://www.khanacademy.org/science/chemistry/acids-and-bases/v/buffers-and-hendersen-hasselbalch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Khan Academy&lt;/a&gt;, &lt;a href=&quot;http://chemcollective.org/activities/tutorials/buffers/buffers3&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemCollective&lt;/a&gt; and the answer to a similar question:&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/9183/can-someone-please-explain-buffers-to-me&quot;&gt;Can someone please explain buffers to me?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not clear whether buffers can only protect against the opposite species. I.e. a weak acid can only protect against the pH change caused by strong bases, or if it can also protect against the pH change caused by strong acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have $\ce {H_2CO_3_{(aq)} + H_2O_{(l)}&amp;lt;=&amp;gt;H_CO^{-}_3_{(aq)} +  H_3O^+_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we add a strong acid $\ce{HCl + H_2O-&amp;gt; H_3O^+ + Cl^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $x$ amount of protons in the solution and for every one of the weak acid's conjugate base, there is a proton. So, to me it appears that there is no way for the weak acid to act as a buffer for a strong acid. Is this correct or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-02-08T20:15:50.973" Title="Do buffers only work for species of the opposite variety? (I.e. weak acids buffer strong bases and vice versa)" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13556" PostTypeId="5" CreationDate="2014-06-12T03:45:58.623" Score="0" Body="&lt;p&gt;Gloves are a layer of protection in terms of laboratory safety. This tag should be specified if this is a question about particular gloves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various types of gloves used in laboratories. The most common are &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubber_glove&quot; rel=&quot;nofollow&quot;&gt;disposable rubber gloves&lt;/a&gt; for short time protection. They are often made from &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrile_rubber&quot; rel=&quot;nofollow&quot;&gt;nitrile rubber&lt;/a&gt;, which are more resistant than plain &lt;a href=&quot;http://en.wikipedia.org/wiki/Latex&quot; rel=&quot;nofollow&quot;&gt;latex&lt;/a&gt; gloves, which are usually only protecting against dirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various multi use gloves necessary in experimental chemistry. Examples include heat protection and acid/base protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related tag: &lt;a href=&quot;/questions/tagged/safety&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;safety&amp;#39;&quot; rel=&quot;tag&quot;&gt;safety&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-06-12T07:34:39.240" LastActivityDate="2014-06-12T07:34:39.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13557" PostTypeId="4" CreationDate="2014-06-12T03:45:58.623" Score="0" Body="Gloves are a layer of protection in terms of laboratory safety. This tag should be specified if this is a question about particular gloves." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-06-12T07:34:57.547" LastActivityDate="2014-06-12T07:34:57.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13558" PostTypeId="2" ParentId="13555" CreationDate="2014-06-12T03:58:38.633" Score="3" Body="&lt;p&gt;If you just add straight carbonic acid to water, the $\mathrm{p}\ce{H}$ will drop as it is ionized (as represented by your equation). Since it is a weak acid it won't ionize completely and you'll be left with carbonic acid, bicarbonate and hydronium in solution at equilibrium. Now, if you increase the concentration of $\ce{H+}$ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left, as protons react with bicarbonate to produce carbonic acid). Because the new protons are not free in solution, the pH will not drastically change until the bicarbonate is exhausted. Thus the solution is buffered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, the acid dissociation constant, which is unchanging for a specific species by definition, is given by $K_a=\frac{[\ce{HCO3^{-}}][\ce{H+}]}{[\ce{H2CO3}]}$. If you add more $\ce{H+}$, the ratio of $\ce{HCO3^-}$ to $\ce{H2CO3}$ must decrease for the system to regain equilibrium &lt;/p&gt;&#xA;" OwnerUserId="5220" LastEditorUserId="4945" LastEditDate="2014-06-12T06:12:47.577" LastActivityDate="2014-06-12T06:12:47.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13559" PostTypeId="2" ParentId="13541" CreationDate="2014-06-12T05:15:41.513" Score="5" Body="&lt;p&gt;This answer deals with the situation, when but electrodes are in the same salt water solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming your salt solution contains sodium chloride&#xA;$$\ce{NaCl (s) -&amp;gt;[\ce{H2O}] Na+ (aq) + Cl- (aq)},$$&#xA;then your reaction at the zinc terminal will certainly be&#xA;$$\ce{Zn (s) + 2Cl- -&amp;gt; ZnCl2 (aq) + 2e-}.$$&#xA;On the copper terminal water will be electrolysed&#xA;$$\ce{H2O + 2e- -&amp;gt;[\ce{Cu}] 1/2 H2 ^ + {}^{-}OH},$$&#xA;or using autoprotolysis&#xA;\begin{aligned}\ce{&#xA;2H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + {}^{-}OH\\&#xA;H3+O + e- &amp;amp;-&amp;gt;[\ce{Cu}] 1/2 H2 ^ + H2O&#xA;}.\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will happen since sodium has a much lower electron affinity than water/ hydronium&#xA;\begin{aligned}&#xA;\ce{Na+ + e- &amp;amp;-&amp;gt; Na} &amp;amp;E^\circ&amp;amp;=−2.71\:\mathrm{V}\\&#xA;\ce{2H2O + 2e- &amp;amp;-&amp;gt; H2 + 2 {}^{-}OH}&amp;amp;E^\circ&amp;amp;=−0.83\:\mathrm{V}\\&#xA;\ce{2H+ + 2e- &amp;amp;-&amp;gt; H2}&amp;amp;E^\circ&amp;amp;=0\:\mathrm{V}.&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what salt solution you are using, different cations might be reduced. You can find a collection of values at &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a battery (&lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;nofollow&quot;&gt;Galvanic cell&lt;/a&gt;) there are usually coupled half cells, and in this case this would probably be $\ce{Zn|Zn^2+||Cu^2+|Cu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, that the Zinc terminal is in a zinc containing solution and the copper terminal is in a copper containing solution. In this case your assumption is correct &#xA;\begin{aligned}&#xA; \text{Zinc terminal:}&amp;amp;&amp;amp;\ce{Zn(s) + ZnCl2 (aq)&amp;amp;&amp;lt;=&amp;gt; 2Zn^{2+}(aq) + 2Cl- + 2e^-} \\&#xA; \text{Copper Terminal:}&amp;amp;&amp;amp;\ce{CuCl2(aq) + Cu (s) + 2e^- &amp;amp;&amp;lt;=&amp;gt; 2Cu(s) + 2Cl-} &#xA;\end{aligned}&#xA;There has to be a salt bridge to neutralise charge differences (ion flow).&lt;br&gt;&#xA;Note that in that case zinc ions will go into solution, while copper ions will be removed from the solution. If the concentration of copper drops close to zero, then no current can be measured. The galvanic cell can be charged by &quot;reversing&quot; the process, apply opposite current. Then zinc ions will be removed from solution and copper will go into solution.&lt;br&gt;&#xA;In reality, this is a equilibrium process.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2016-01-22T03:47:20.400" LastActivityDate="2016-01-22T03:47:20.400" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13560" PostTypeId="2" ParentId="13555" CreationDate="2014-06-12T05:33:17.140" Score="4" Body="&lt;p&gt;Think of it in terms of the equilibrium constant: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$&#xA;Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q\; =\; \frac{9\cdot 2}{1} = 18$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $Q &amp;gt; K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q\; =\; \frac{8\cdot 1}{2} = 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and because $Q = K$ the reaction is now at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="2142" LastEditDate="2015-02-08T20:15:50.973" LastActivityDate="2015-02-08T20:15:50.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13562" PostTypeId="1" CreationDate="2014-06-12T09:31:04.693" Score="8" ViewCount="1841" Body="&lt;p&gt;Compounds with triple bonds generally seem to be unstable. $\ce{HCN}$ and $\ce{C2H2}$ are high-energy, relatively short-lived molecules that will readily polymerise or react with other organic molecules. My naïve mental picture is that they don't really want to have a triple bond because the geometry makes it awkward, so they'll do anything to transfer an electron somewhere else and turn it into a double or a single bond instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\ce{N2}$ seems to be an exception. By my naïve reasoning, when an $\ce{N2}$ meets an organic molecule it should be eager to shrug off its triple bond and join the party. But in fact this doesn't happen, and nitrogen fixing organisms have to do quite a bit of work to get nitrogen to participate in organic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what is it about $\ce{N2}$ that makes its triple bond so energetically favourable while $\ce{C#C}$ and $\ce{C#N}$ bonds are so unfavourable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this question isn't about reactivity with $\ce{O2}$ but rather with organic molecules. I'm interested in why it's so difficult for $\ce{N2}$ to react with organic molecules in general, given that other molecules with triple bonds seem to react with them very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Information to guide the scope of answers: I'm coming at this from a physics background, so thermodynamic concepts can be taken as understood, but I quickly get lost when it comes to electrons and orbitals and so on, hence the rather basic question.)&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="148" LastEditDate="2014-06-13T05:29:10.937" LastActivityDate="2017-09-04T20:22:55.347" Title="Why is N₂ stable but HCN and C₂H₂ unstable?" Tags="&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13563" PostTypeId="1" CreationDate="2014-06-12T12:16:40.490" Score="2" ViewCount="840" Body="&lt;p&gt;The amount of water(g) that air can accommodate depends on temperature. The absolute humidity then is defined as the percentage of water to air and decreases when the temperature decreases. The relative humidity is the ratio between the current and the maximum humidity, which increases with decreasing temperature. The temperature at which the relative humidity reaches 100% is called the dew point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, why can air keep more water when it is warmer?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="1235" LastEditDate="2014-06-12T14:13:50.920" LastActivityDate="2015-03-20T05:43:34.613" Title="Why can air accommodate more water when the air has a higher temperature?" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13564" PostTypeId="2" ParentId="13563" CreationDate="2014-06-12T12:56:51.413" Score="4" Body="&lt;p&gt;For a substance to be a gas, the kinetic energy of the molecules that make it up must be higher than the electrostatic potential energies that hold it together. When we talk about air &quot;holding&quot; water we are really talking about the amount of water that would be in the vapor phase relative to the amount that would be liquid at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since temperature is a measure of the average kinetic energy of molecules, as the temperature increases, the fraction of water that can exist as vapor increases. Since air is already a gas under any conditions where water vapor can exist, it looks like the air &quot;holds&quot; more water as temperature increases.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-12T12:56:51.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13565" PostTypeId="2" ParentId="13562" CreationDate="2014-06-12T12:56:53.607" Score="4" Body="&lt;p&gt;I think this question can be answered reasonably well with thermodynamic considerations.  When we say something is stable, it is (or should be) assumed that the compound does not engage in a chemical reaction.  It is important when we discuss the stability of compounds to focus not only on the compound itself but &lt;em&gt;what the products of the reaction are&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three triple bonds you mention ($\ce{N#N}$, $\ce{C#N}$ and $\ce{C#C}$) all have very high bond dissociation energies (from various sources such as &lt;a href=&quot;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; they are $\approx950, 940\text{ and }960\:\mathrm{kJ/mol}$) so all of these triple bonds require a large amount of energy to break.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when we think about the stability of acetylene, I suspect most people are thinking about its stability towards ignition (which has &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/je60013a037&quot; rel=&quot;nofollow&quot;&gt;been studied&lt;/a&gt;, by the way).  That gives us a reaction we can explore, namely the combustion of acetylene in the presence of oxygen to produce carbon dioxide and water.  &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%22C2H2+%2B+O2+-%3E+CO2+%2B+H2O%22&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; can calculate the thermodynamics of this reaction, giving a result of $-2600\:\mathrm{kJ/mol}$; clearly a very exothermic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the combustion of nitrogen in the presence of oxygen, which could form nitrogen dioxide.  Again, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%22N2+%2B+O2+-%3ENO2%22&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha&lt;/a&gt; (Note, if you happen to play around with WA, make sure you put your reaction in quotation marks or it won't be interpreted properly) gives us a result of +66 kJ/mol or an endothermic reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do we make of these results?  Focusing on the &lt;em&gt;products&lt;/em&gt; we see that for the acetylene combustion, there is a large amount of energy &lt;strong&gt;released&lt;/strong&gt; when creating the products.  The same is not true for the products of nitrogen combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, when considering stability, we must think about we mean by the word, since that meaning will lead to a chemical reaction, and the tools we use to describe chemical reactions help us quantify that stability.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="4945" LastEditDate="2014-06-13T06:16:15.500" LastActivityDate="2014-06-13T06:16:15.500" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13566" PostTypeId="1" AcceptedAnswerId="13567" CreationDate="2014-06-12T13:12:29.203" Score="1" ViewCount="2276" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/joOZ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to assign the peaks here to dimethylphenylphatalate but all the benzene rings are throwing me off. All I can get is that the further most signal is the carbonyl carbon (168.71) and the one prior to that (143.27) might be the ester carbon (from the methyl). Can someone help?&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="16683" LastEditDate="2016-10-22T15:19:20.073" LastActivityDate="2016-10-22T15:19:20.073" Title="Carbon-13 NMR of dimethyl 3,4,5,6-tetraphenylphthalate?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13567" PostTypeId="2" ParentId="13566" CreationDate="2014-06-12T15:38:50.233" Score="4" Body="&lt;p&gt;Here's a numbered drawing of the title compound to help us keep track of things.  First thing we notice is that there are 13 different carbon atoms in the structure and we see 13 signals in the spectrum, that's a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sSeUV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the easy ones.  The upfield peak at 52.28 is in the expected position for C1, the ester methyl carbon.  Down at the other end of the spectrum, the peak at 168.71 is in the expected position for an ester carbonyl carbon (C2).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the region for aromatic carbons we see 5 low intensity peaks between 143 and 132.  These belong to the (unprotonated) quaternary carbons 3, 4, 5, 6, 10.  Using &lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_84-66-2_13CNMR.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;diethyl phthalate&lt;/a&gt; as a model compound, I would guess that the peak at 132.19 is from C3.  The two peaks at 138.73 and 138.60 have the same height and are closely spaced suggesting very similar environments, I might guess them to be C6 and C10, just not sure which is which.  Noting that the peak for the carbon corresponding to our C5 is downfield of the peak corresponding to our C4 in the model compound, I would expect the same trend to hold here, just with each of the peaks shifted downfield due to attachment of a phenyl group.  So I would assign the peak at 143.27 to C5 and the peak at 139.29 to C4.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the protonated aromatic carbons for us to consider.  The 2 peaks at 125.97 and 126.83 are half the intensity of the other protonated aromatic signals, therefor assign them to C9 and C13 as there are only half as many of these carbons as compared to the other protonated aromatic carbons. It's not clear which peak belongs to C9 and which to C13.  This leaves us with the 4 signals due to C7, C8, C11 and C12.  Using the same resonance structure - delocalization arguments I made in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13547/benzaldehyde-nmr-question&quot;&gt;your benzaldehyde cnmr question&lt;/a&gt; from yesterday, I would expect C7 and C11 to be downfield from C8 and C12.  Therefor, I would assign C7 and C11 to the peaks at 129.71 and 130.88, not knowing which is which.  C8 and C12 then belong to the signals at 126.88 and 127.41, not knowing which belongs to which. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there is some guess work in here and if it was an important project I'd do some further experiments and look for some additional model compounds in order to confirm these assignments and remove the discrepancies, but this is a good start. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-06-12T15:38:50.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13568" PostTypeId="1" AcceptedAnswerId="13572" CreationDate="2014-06-12T16:20:34.407" Score="4" ViewCount="661" Body="&lt;p&gt;The process I have seen in class to activate a carbonyl function is acetalisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the steps involved is the following : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4YizS.png&quot; alt=&quot;http://i.imgur.com/WA0W7bC.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering, is that a $\ce{S_{N}1}$ or a $\ce{S_{N}2}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS : drawing molecules with paint is awful, anyone knows a good free molecules drawer ?)&lt;/p&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="4945" LastEditDate="2014-06-13T04:47:06.533" LastActivityDate="2014-06-13T04:47:06.533" Title="Does this activation of carbonyls usually involve a SN1 or a SN2?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13569" PostTypeId="1" CreationDate="2014-06-12T16:36:27.060" Score="9" ViewCount="762" Body="&lt;p&gt;I'm supposed to write a balanced equation for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Iron(s) + copper(II) sulfate(aq) -&amp;gt; ?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use Iron(II)/ferrous I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe + CuSO4 -&amp;gt; FeSO4 + Cu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if I use Iron(III)/ferric,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe + CuSO4 -&amp;gt; Fe2(SO4)3 + Cu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Before balancing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to determine which one I am supposed to use?&lt;/p&gt;&#xA;" OwnerUserId="6826" LastEditorUserId="33991" LastEditDate="2017-04-28T07:10:39.103" LastActivityDate="2017-04-28T07:10:39.103" Title="How do I determine whether it is the *ferrous* or *ferric* ion that is implied in this displacement reaction?" Tags="&lt;electrochemistry&gt;&lt;ions&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13570" PostTypeId="2" ParentId="13568" CreationDate="2014-06-12T16:45:25.203" Score="2" Body="&lt;p&gt;This is an SN1-like process, the intermediate is an alyoxonium ion (R1R2C=O+R), which, having a double-bond resonance form is a somewhat better intermediate than a free carbocation, as in the tradition SN1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is too much steric hinderance for this to be a SN2 reaction with a reasonable reaction rate.  That's a definite no-go.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2014-06-12T16:45:25.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13572" PostTypeId="2" ParentId="13568" CreationDate="2014-06-12T18:31:25.233" Score="5" Body="&lt;p&gt;It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction.  It is commonly referred to as an &quot;addition-elimination&quot; reaction.  You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate.  Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction.  This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl.  Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously.  Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product.  You &quot;add&quot; a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you &quot;eliminate&quot; one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uaFRg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another, more specific drawing for the case at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTD8U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-12T20:55:11.730" LastActivityDate="2014-06-12T20:55:11.730" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="13573" PostTypeId="1" AcceptedAnswerId="13577" CreationDate="2014-06-12T18:32:17.463" Score="1" ViewCount="187" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxa0p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding this question, how can one solve this without access to thermodynamic data? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose we should keep the following information in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Double bonds are not double the strength of single bonds. Same goes for triple bonds. Reason has to do with orbital overlap with regard to sigma and pi bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In order of increasing stability: $\ce{O_2, O_3, O}$. We know this because diatomic oxygen is what usually exists in large quantities at ground level; ozone is a pollutant because it is quite reactive and can thus harm tissues; we don't usually hear of monoatomic oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Cannot be the answer. Going from ozone to diatomic oxygen is exothermic. At least part of this reaction is exothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Definitely not the answer. Going from ozone to diatomic oxygen exclusively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) This is endothermic, but perhaps not as endothermic as creating monoatomic oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Looks pretty endothermic; we are splitting oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Definitely not the answer, this is fundamentally exothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we definitely pick out the right answer? What reasoning would you employ? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-13T02:10:14.967" Title="Endothermicity of Reaction" Tags="&lt;bond&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13574" PostTypeId="2" ParentId="13537" CreationDate="2014-06-12T20:00:21.833" Score="2" Body="&lt;p&gt;Yes, see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00755a006&quot; rel=&quot;nofollow&quot;&gt;Structural and Stablity of Nitric Oxide Dimer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just don't call it &quot;hyponitrite molecule&quot;, it might get offended.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-12T20:00:21.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13575" PostTypeId="2" ParentId="13536" CreationDate="2014-06-12T20:40:26.447" Score="5" Body="&lt;p&gt;At very high pressure, say $&amp;gt;\pu{100 GPa}$, diatomic molecules such as $\ce{H2}$ and $\ce{O2}$ are no longer favorable relative to monoatomics, even at low temperature where Gibbs free energy $G$ is dominated by enthalpy $H$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.nature.com/nmat/journal/v10/n12/fig_tab/nmat3189_F2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phase Diagram of Hydrogen&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the cores of Jupiter and Saturn are monoatomic hydrogen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diatomic bond formation would be endothermic under such conditions.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2017-12-29T02:03:28.297" LastActivityDate="2017-12-29T02:03:28.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13576" PostTypeId="2" ParentId="10250" CreationDate="2014-06-13T01:20:03.767" Score="3" Body="&lt;p&gt;In general, no it is not possible to catalog &lt;strong&gt;all&lt;/strong&gt; of the compounds a plant is made of automatically. Uncle Al's answer is as close as you could get, but even that is far, far away from getting every single molecule (there would be tens of thousands, at least). &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_chromatography%E2%80%93mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;HPLC-MS&lt;/a&gt; is good first step and would get you most of the small molecular weight substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For larger biomolecules, you would have to use &lt;a href=&quot;http://en.wikipedia.org/wiki/SDS_PAGE&quot; rel=&quot;nofollow&quot;&gt;gel electrophoresis&lt;/a&gt;, which is not very &quot;automated.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are a long way from cheap, automated separations of biological mixtures, but it is one of the major research goals of the pharmacology, molecular biology, and medical fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK22410/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; describes many of the currently available methods for protein purification, which is one of the most difficult problems in the field of separations.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-13T01:20:03.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13577" PostTypeId="2" ParentId="13573" CreationDate="2014-06-13T02:10:14.967" Score="4" Body="&lt;p&gt;I believe that you would have to know at least these things to start with:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;All enthalpies of bond dissociation are endothermic&lt;/li&gt;&#xA;&lt;li&gt;Complete dissociation of triple bonds is more endothermic than double bonds, which is more endothermic than single bonds (when compared for the same elements).&lt;/li&gt;&#xA;&lt;li&gt;The difference between going from a triple bond to a double bond is smaller than double to single, and that is smaller than going from single to complete dissociation (you referenced this in your question)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can summmarize these facts with the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fvldt.png&quot; alt=&quot;relative bond dissociation enthalpies for oxygen-oxygen bonds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here D(O-O) represents the bond dissociation enthalpy of an oxygen-oxygen single bond. The numbers inside of circles represent &quot;steps&quot; between each successively stronger bonds, and are there just for easier reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, let's look at the options again, and count how many of each type of bond are broken and formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ a) \space O_3(g) \rightarrow O_2(g) + O(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozone has a resonance structure, which means that we need to consider the bonds as being somewhere between a single and double bond in strength. We could imagine them being about halfway up step 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broken: 2 x single/double resonance bonds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formed: 1 x double bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we are looking at an overall reaction that would be slightly more endothermic than the breaking of a single bond. To make that easier to see, think about taking these energetic &quot;steps&quot; using the figure above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(+1.5, +1.5) - breaking the two resonance bonds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( -1, -2) - forming the double bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ b) \space  2O_3(g) \rightarrow 3O_2(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we are breaking two sets of resonance bonds, but forming 3 double bonds. So our steps look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(+1.5, +1.5, +1.5, +1.5) - breaking six resonance bonds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(-1, -2, -1, -2, -1, -2) - forming three double bonds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three -1, -2 steps, each of which is larger than a +1.5 step. However, they are not much larger, and the total of the difference may or may not be greater than the fourth 1.5 step. This means we can't say for sure whether it is exo- or endo-thermic, but we can at least say it would be less endothermic than breaking a single bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ c) \space  3O_2(g) \rightarrow 2O_3(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are doing the opposite of (b). We still don't know the sign, but we know the magnitude is the same as (b), which was less than (a), so either way it can't be (b) or (c).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ d) \space  O_2(g) \rightarrow 2O(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we are breaking a double bond, and forming nothing. This is the clear winner so far, with steps +1, +2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ e) \space  O_3(g) + O(g) \rightarrow 2O_2(g) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the final case, we break two resonance bonds, and form two double bonds. Our steps are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;+1.5, +1.5, -1, -2, -1, -2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the formation of each double bond releases more energy than the dissociation of each resonance bond, this is clearly exothermic, as you stated in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking all of this together, we have to choose (d) as the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-13T02:10:14.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13579" PostTypeId="1" AcceptedAnswerId="13762" CreationDate="2014-06-13T03:51:17.087" Score="2" ViewCount="4284" Body="&lt;p&gt;Why does calcium carbonate, when heated strongly, decompose into calcium oxide and carbon dioxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism? Can someone illustrate the means by which this happens? What bonds/associations are broken? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9W5s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that one of the calcium ion has a strong intermolecular association with one of the partially negative oxygens, and this changes the bond character of the carbonate ion. Upon applying heat, this bond remains while carbon dioxide is off-gassed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the positive charge on the calcium ion isolates the electron density to the nearest oxygen atom on the carbonate ion. This isolation of electron density away from the carbon and toward the calcium weakens the oxygen-carbon bond while strengthening the calcium ion-oxygen intermolecular association force. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-06-13T06:05:32.480" LastActivityDate="2014-06-20T02:07:30.220" Title="Why does calcium carbonate decompose into calcium oxide?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;reaction-mechanism&gt;&lt;bond&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13580" PostTypeId="1" CreationDate="2014-06-13T06:49:13.340" Score="0" ViewCount="612" Body="&lt;p&gt;Is it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and are there multiple types of ORR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for context: &lt;a href=&quot;http://www.nature.com/srep/2014/140122/srep03821/full/srep03821.html&quot; rel=&quot;nofollow&quot;&gt;http://www.nature.com/srep/2014/140122/srep03821/full/srep03821.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1235" LastActivityDate="2014-06-13T07:58:58.100" Title="What exactly is the &quot;oxygen reduction reaction&quot;?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13581" PostTypeId="1" CreationDate="2014-06-13T07:05:36.143" Score="1" ViewCount="320" Body="&lt;p&gt;Are they thought to continuously pop in and out of existence at various points inside the orbital defined by probabilities or do they follow definite paths that are made fuzzy by the Heisenberg uncertainty principle?&#xA;I sometimes find myself as thinking of the electron cloud as a gas with varying charge density corresponding to the shape of the atomic orbital. &lt;/p&gt;&#xA;" OwnerUserId="6830" LastEditorUserId="5456" LastEditDate="2014-06-13T08:45:28.893" LastActivityDate="2018-06-27T05:49:08.447" Title="I am trying to picture how electrons move around in atomic orbitals" Tags="&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13582" PostTypeId="2" ParentId="13580" CreationDate="2014-06-13T07:58:58.100" Score="2" Body="&lt;p&gt;According to your &lt;a href=&quot;http://dx.doi.org/10.1038/srep03821&quot; rel=&quot;nofollow noreferrer&quot;&gt;cited paper&lt;/a&gt; the reaction they care most about is $$\ce{O2 + 4H+ + 4e- -&amp;gt; 2H2O}$$ as you are suggesting. This should also be the total reaction and the most interesting one for applications, e.g. in fuel cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various suggestions for the mechanistics of the Oxygen Reduction Reaction. A quite recent review was published by &lt;a href=&quot;http://dx.doi.org/10.1155/2012/491604&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nagappan Ramaswamy and Sanjeev Mukerjee&lt;/a&gt;. A very early general scheme can be found at &lt;a href=&quot;http://electronicstructure.wikidot.com/oxygen-reduction-reaction-on-platinum-using-dft&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronicstructure.wikidot.com&lt;/a&gt;, where they cited &lt;a href=&quot;http://dx.doi.org/10.1016/S0022-0728%2876%2980250-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halina S. Wroblowa &lt;em&gt;et.al.&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://electronicstructure.wdfiles.com/local--files/oxygen-reduction-reaction-on-platinum-using-dft/wroblowa.PNG&quot; alt=&quot;wroblowa scheme&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.860" LastActivityDate="2014-06-13T07:58:58.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13583" PostTypeId="2" ParentId="13536" CreationDate="2014-06-13T08:47:26.377" Score="13" Body="&lt;p&gt;Bond formation is alway strictly &lt;a href=&quot;http://goldbook.iupac.org/E02269.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;exothermic&lt;/a&gt; in the sense of the change of enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;exothermic reaction&lt;/strong&gt; A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A bond can only exist, if it needs energy to break it, i.e. the &lt;a href=&quot;http://goldbook.iupac.org/B00699.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;bond dissociation energy&lt;/a&gt; is always positive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;bond-dissociation energy, $D$&lt;/strong&gt; The enthalpy (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -&amp;gt; .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This has absolutely nothing to do with a reaction being exothermic/endothermic or exergonic/endergonic, because this is defined by the rearrangements of bonds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding noble gas diatomics, it is quite clear from MO-Theory, that there is no bond. However, even these non bonded elements have a non-zero dissociation energy. Please refer to &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/7520/4945&quot;&gt;Why are noble gases stable&lt;/a&gt;&quot; and to answers and comments within.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-06-13T09:21:55.367" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13584" PostTypeId="1" CreationDate="2014-06-13T09:13:38.850" Score="11" ViewCount="12459" Body="&lt;p&gt;What is the reason for anomalous expansion of water? Why doesn’t it simply expand on heating or contract on cooling? Why it shows anomalous behavior at 4 degrees Celsius? Why not on 4.6 or 10 (or whatever) degrees Celsius?&lt;/p&gt;&#xA;" OwnerUserId="6831" LastEditorDisplayName="user7951" LastEditDate="2015-06-19T18:20:37.880" LastActivityDate="2020-10-05T02:35:14.130" Title="What is the reason for anomalous expansion of water?" Tags="&lt;water&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13586" PostTypeId="2" ParentId="13581" CreationDate="2014-06-13T13:28:28.460" Score="7" Body="&lt;p&gt;I think you miss an important notion to get a better understanding of the behavior of the electrons: &lt;a href=&quot;http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality&quot; rel=&quot;nofollow noreferrer&quot;&gt;wave-particle duality&lt;/a&gt;. I believe that a better visualization takes into account that you have to deal even with a wave. From what you say it seems that your picture is shifted only to the particle-side,  keep meditating for a half an hour trying to visualize the electron as a wave. For me is a very interesting mind experiment. Then move to the orbital concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Lund University researches took a picture of a moving electron. This image maybe could be helpful for your meditation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xeHcE.jpg&quot; alt=&quot;captured electron movement&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lund University. &quot;Electron Gets Film Debut In First-ever Video Of Its Kind.&quot; ScienceDaily. ScienceDaily, 25 February 2008. &lt;a href=&quot;http://www.sciencedaily.com/releases/2008/02/080222095358.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.sciencedaily.com/releases/2008/02/080222095358.htm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="4945" LastEditDate="2018-06-26T08:55:40.800" LastActivityDate="2018-06-26T08:55:40.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13587" PostTypeId="2" ParentId="13581" CreationDate="2014-06-13T14:02:47.273" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;interpretation of quantum mechanics&lt;/a&gt; is an unresolved issue in physics and philosophy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I sometimes find myself as thinking of the electron cloud as a gas with varying charge density corresponding to the shape of the atomic orbital. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was Schrodinger's original interpretation when he developed/postulated the Schrodinger equation, that the wave function should be interpreted as charge density.  However, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Copenhagen_interpretation&quot; rel=&quot;nofollow&quot;&gt;Copenhagen interpretation&lt;/a&gt; soon became most popular. The Copenhagen interpretation associates the wavefunction with a probability distribution rather than a charge distribution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are they thought to continuously pop in and out of existence at various points inside the orbital defined by probabilities or do they follow definite paths that are made fuzzy by the Heisenberg uncertainty principle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The electrons of atoms do not pop in and out of existence.  However, &lt;a href=&quot;http://en.wikipedia.org/wiki/Virtual_particle&quot; rel=&quot;nofollow&quot;&gt;virtual electrons&lt;/a&gt; pop in and out of existence and can affect the energy levels of atoms slightly, for example contributing to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lamb_shift&quot; rel=&quot;nofollow&quot;&gt;Lamb shift&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Copenhagen interpretation, the electrons do not follow definite paths; however, there are other interpretations which have not been disproven, such as &lt;a href=&quot;http://plato.stanford.edu/entries/qm-bohm/&quot; rel=&quot;nofollow&quot;&gt;Bohmian Mechanics&lt;/a&gt; where &lt;a href=&quot;http://arxiv.org/abs/quant-ph/0404175&quot; rel=&quot;nofollow&quot;&gt;definite paths are followed&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-13T14:02:47.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13588" PostTypeId="2" ParentId="13584" CreationDate="2014-06-13T14:19:05.807" Score="7" Body="&lt;p&gt;So, the first question to ask would be: how &quot;anomalous&quot; is change in the thermal expansion coefficient?  It turns out that for most real materials the coefficient of thermal expansion is not a constant, and takes on both positive (expansion) and negative (contraction with increasing temperature) values.  For silicon as an example, see &lt;a href=&quot;http://www.ioffe.ru/SVA/NSM/Semicond/Si/thermal.html.&quot; rel=&quot;nofollow&quot;&gt;ioffe&lt;/a&gt;,  Above room temperature Si has a positive (but not constant) CTE, below about 100K silicon's CTE is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question is why would this occur?  Taking a liquid or a solid (ignore gases for the moment), an individual atom (or molecule) sits in some kind of potential well which is what is causing the material to be bound together. At absolute zero, these atoms sit at the very bottom of the potential well.  As the temperature increases, heat is added and the atom starts rattling around in the potential well, and is able to explore more of the well.  There is no particularly good reason to believe that the shape of the well, across the entire temperature range of existence of the phase, will be such that a single linear coefficient will describe where the atom will most likely sit in the well at a given temperature. On top of that in a solid is the equilibrium concentration of point defects that is dependent on temperature.  As an example, Google for Simmons and Balluffi and their experiments combining x-ray measurements of the lattice parameter (where the atoms sit in the potential well) and thermal expansion (change in length of a heated bar) to determine the point defect concentrations vs temperature for various solids.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5456" LastEditDate="2014-10-30T04:44:15.393" LastActivityDate="2014-10-30T04:44:15.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13589" PostTypeId="1" CreationDate="2014-06-13T14:54:03.193" Score="3" ViewCount="10943" Body="&lt;p&gt;So I had an organic chemistry lab where we had to dissolve p-nitrophenol in water, dilute $\ce{NaHCO3}$, dilute $\ce{NaOH}$, and dilute $\ce{HCl}$ to test for solubility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The p-nitrophenol dissolved in everything but water. I just want to make sure if these results are theoretically correct. From what I understand, the nitrophenol should have also dissolved in water because its a polar substance and it shouldn't have dissolved in $\ce{HCl}$ because I think the Hydrogen in the phenol is itself acidic due to electron withdrawing effects of the benzene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain what should happen? &lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="5456" LastEditDate="2014-06-13T15:03:13.003" LastActivityDate="2017-09-11T04:26:25.870" Title="Solubility of p-nitrophenol in various substances?" Tags="&lt;organic-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13590" PostTypeId="2" ParentId="13584" CreationDate="2014-06-13T15:07:20.033" Score="5" Body="&lt;p&gt;The question &quot;Why exactly 4.6 grad&quot; has no simple answer. The question 'wtf happens with water' may have reasonably simple answer. I think about it this way: water has surprisingly low-density crystal structure with lot of holes, meaning that many defects putting a molecule into such hole would result in increase of density. Thus, the actual form of density-temperature dependency for water is result of two opposing forces: various defects, that increase its density and thermal expansion that decrease its density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, is not the only way for such dependency with extremum to occur. Another reason for negative thermal expansion coefficient would be fact, that with thermal oscillation increase and increase of mean distance between atoms, the position between mean position of some atoms may decrease. A simplistic model would be this simple physical experiment: &lt;a href=&quot;http://youtu.be/rpnoq9hz1gw&quot;&gt;http://youtu.be/rpnoq9hz1gw&lt;/a&gt; with increased speed of rotation of the chain, mean distance between its rings becomes less and less, but their mean position are slowly drageed to one point, and if chain is rotated fast enough, its vertiacal size becomes a bit more than size of one ring. IRL similar mechanism is the reason for some pulled polymers to contract at elevated temperatures. It is possible for some compounds to have negative thermal expansion for similar reason, though I can't give an example at the moment.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-13T15:07:20.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13591" PostTypeId="1" CreationDate="2014-06-13T15:21:02.500" Score="17" ViewCount="7555" Body="&lt;p&gt;Recently the question was asked, &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/13536/is-bond-formation-strictly-exothermic&quot;&gt;Is bond formation “strictly” exothermic?&lt;/a&gt;&quot;  As a corollary to that question, what is the relationship between entropy and bond formation?  Does the entropy of the bond formation reaction always decrease when a chemical bond is formed?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-26T02:02:41.790" LastActivityDate="2018-03-26T02:02:41.790" Title="Can entropy increase during bond formation?" Tags="&lt;thermodynamics&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13592" PostTypeId="1" AcceptedAnswerId="13684" CreationDate="2014-06-13T16:24:14.723" Score="9" ViewCount="754" Body="&lt;p&gt;I am testing my small api library that generates lots of chemical files such as MOL and SDF. I was wondering if there is a software that can show me the pictures of chemical structures inside each file so that I don't have to open each of these separately. would prefer something free for Windows and Mac.&lt;/p&gt;&#xA;" OwnerUserId="6834" LastEditorUserId="4945" LastEditDate="2015-12-14T09:47:56.593" LastActivityDate="2018-07-06T07:59:58.283" Title="What free software exists to view molecule pictures within a file explorer?" Tags="&lt;software&gt;&lt;molecular-structure&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13593" PostTypeId="1" AcceptedAnswerId="13596" CreationDate="2014-06-13T17:02:35.583" Score="3" ViewCount="13010" Body="&lt;p&gt;Why is the $\ce{sp^2}$ hybridized nitrogen atom in guanidine the most basic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gYnIC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that as a general guideline, $\ce{sp^3}$ hybridized atoms are more basic than $\ce{sp^2}$. Bases are electron pair donors; having the electrons further away from the nucleus makes donation easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, protonating the double bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule. Protonating the other nitrogens does not allow for the same degree of resonance stabilization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, in one textbook, I found mention of a +M effect - what is the +M effect? And while we're on that, what's +/- I? I understand that I stands for induction, but what's +I? What's positive induction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more note, since the double bonded nitrogen is $\ce{sp^2}$ hybridized, does that mean it has a greater negative charge density than the other nitrogens by nature of having greater s-character? Could this + charge density be giving the double bonded nitrogen more nucleophilic character? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y9f2z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="33991" LastEditDate="2016-12-09T18:56:12.660" LastActivityDate="2016-12-09T18:56:12.660" Title="Protonation of Guanidine" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13594" PostTypeId="1" AcceptedAnswerId="13597" CreationDate="2014-06-13T17:03:59.770" Score="4" ViewCount="9704" Body="&lt;p&gt;When the diazonium ion couples with dimethylaniline, why is the para position favored?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FVCzp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVCzp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this only due to steric factors or do electronic factors (such as ortho-para directivity) play a role here too? In particular, I suspect ortho-para electronic factors might be significant because some of the resonance structures have a cation directly next to the electron-donating group.&lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="11367" LastEditDate="2017-05-03T02:35:31.780" LastActivityDate="2017-05-03T02:35:31.780" Title="Why does diazo coupling preferentially occur at the para position?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13595" PostTypeId="2" ParentId="13593" CreationDate="2014-06-13T18:53:20.710" Score="3" Body="&lt;p&gt;The basicity of the double-bonded nitrogen is chiefly a result of the delocalization of charge permitted by the two single-bonded nitrogens with allylic lone pairs, as the resonance contributors you've drawn suggest. Notice that, upon protonation, the three nitrogens become chemically equivalent, with the positive charge shared equally among all three. If either of the single-bonded nitrogens is protonated, the resulting cation does not benefit from charge delocalization, which is a comparatively less stable arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the basicity is a function of the stability of the resulting guanidinium cation, it's a thermodynamic phenomenon. However, it may be worth mentioning that kinetic factors also likely favor protonation at the double-bonded nitrogen, since the lone pairs on the other two nitrogens are somewhat delocalized. This is the meaning of the $+M$ notation, which refers to the amine groups being electron-donating by the mesomeric effect (basically synonymous with resonance), while $-M$ would indicate groups which are electron-withdrawing. This is an effect active within the $\pi$-electron system of a molecule, which is probably better understood when considered from the standpoint of MO theory. The concept of resonance/mesomerism is a convenient abstraction that usually works well qualitatively to approximate the electron distribution in molecular orbitals (especially frontier orbitals). In reality, MO theory probably provides a more accurate rendering of the electron distribution, with $\pi$ orbitals that span multiple atoms. I say this to emphasize that resonance &lt;em&gt;is not&lt;/em&gt; a physical process, but just a very expedient way of representing electron delocalization in $\pi$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similary, $+I$ indicates a group being electron-donating by induction, while $-I$ denotes groups that are electron-withdrawing via induction. In essence, an electronegative group creates bond polarization, which can act to draw electron density away from an atom, while an electropositive group can do the opposite. This effect operates principally through the $\sigma$-bond skeleton of a molecule, and weakens as distance from the polarized group increases.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-13T18:53:20.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13596" PostTypeId="2" ParentId="13593" CreationDate="2014-06-13T19:45:10.130" Score="1" Body="&lt;p&gt;This may be another example of kinetics &lt;em&gt;vs&lt;/em&gt;. thermodynamics. It is generally the case that $\ce{sp^3}$ hybridized lone pairs are more basic than $\ce{sp^2}$ hybridized lone pairs because the more s-character in an orbital, the more stable (unreactive) the lone pair.  However in guanidine, ground state resonance structures (see picture below) tend to make the $\ce{sp^3}$ nitrogens less basic.  Whether they are actually more or less basic than the $\ce{sp^2}$ nitrogen I don't know.  But let me start by assuming that the $\ce{sp^3}$ nitrogens are more basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To whatever extent the $\ce{sp^3}$ hybridized nitrogens are more basic, they will protonate more rapidly than the $\ce{sp^2}$ nitrogen.  However, due to resonance stabilization, the product formed by protonation of the $\ce{sp^2}$ nitrogen is thermodynamically more stable than the product formed upon protonation of the $\ce{sp^3}$ nitrogen.  All of these protonations are simple equilibria between starting material and protonated product.  In this case with guanidine there are two protonation equilibria occurring in parallel.  Even though protonation of the $\ce{sp^3}$ nitrogen is kinetically favored due to its increased basicity, the product we observe is from protonation of the $\ce{sp^2}$ nitrogen due to its greater thermodynamic stability.  Because both of these processes are in equilibria, thermodynamics controls the outcome.  If the $\ce{sp^3}$ nitrogens are actually less basic than the $\ce{sp^2}$ nitrogen, then kinetics and thermodynamics favor the same outcome, kinetic protonation of the $\ce{sp^2}$ nitrogen and formation of its thermodynamically favored products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rcSHw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-13T21:52:53.290" LastActivityDate="2014-06-13T21:52:53.290" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13597" PostTypeId="2" ParentId="13594" CreationDate="2014-06-13T20:17:19.003" Score="5" Body="&lt;p&gt;Both steric and electronic factors play a role in this reaction.  Resonance structures such as those shown below favor electrophilic attack (the diazonium ion is an electrophile) at the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions. Since there are twice as many &lt;em&gt;ortho&lt;/em&gt; positions as &lt;em&gt;para&lt;/em&gt; positions in N,N-dimethylaniline, on a purely statistical basis we might expect a 67:33 &lt;em&gt;ortho/para&lt;/em&gt; product distribution if only electronic effects controlled the reaction.  Since diazonium coupling usually produces a majority of the &lt;em&gt;para&lt;/em&gt; coupled product, steric factors must be playing a role as well.  The somewhat bulky dimethylamino group does make it more difficult for an electrophile to approach the &lt;em&gt;ortho&lt;/em&gt; position and the diazonium ion is relatively large.  These two steric factors combine to reduce the amount of the &lt;em&gt;ortho&lt;/em&gt; isomer and the &lt;em&gt;para&lt;/em&gt; isomer predominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TFq5w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-13T20:17:19.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13600" PostTypeId="1" AcceptedAnswerId="13601" CreationDate="2014-06-13T23:09:19.977" Score="1" ViewCount="1175" Body="&lt;p&gt;For simple acid/base reactions - i.e. those involving inorganic compounds - does activation energy ever play in reactions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or can we ignore kinetics entirely and just consider thermodynamic factors such as conjugate stability? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be very interested if there were some book or article examining inorganic molecules' acid-base behavior from a kinetic standpoint. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-13T23:33:13.500" Title="Activation Energy in Acid/Base reactions" Tags="&lt;acid-base&gt;&lt;thermodynamics&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13601" PostTypeId="2" ParentId="13600" CreationDate="2014-06-13T23:33:13.500" Score="3" Body="&lt;p&gt;Proton transfer, especially in aqueous media, has an activation barrier that is easily surmounted at room temperature or higher.  Such reactions are said to be diffusion controlled (the reaction happens at the rate of proton diffusion through the media) or spontaneous.  In such cases thermodynamics will control the process.   That said, I'm sure there are cases where the base the proton is transferring to, is, for example, surrounded by bulky ligands substantially increasing the activation energy for the process.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-13T23:33:13.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13602" PostTypeId="1" CreationDate="2014-06-14T01:26:53.517" Score="6" ViewCount="361" Body="&lt;p&gt;Just as &lt;span class=&quot;math-container&quot;&gt;$\ce{HArF}$&lt;/span&gt; was synthesized...Could something like &lt;span class=&quot;math-container&quot;&gt;$\ce{Li-Ar-Li}$&lt;/span&gt; exist with AXE geometry &lt;span class=&quot;math-container&quot;&gt;$\ce{AX2E3}$&lt;/span&gt;? Any other argon compound proposals out there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that there are some posible excimers like &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAr}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF}$&lt;/span&gt; (maybe &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF2}$&lt;/span&gt;). Can they be considered covalent compounds, even if explained via excited dimers/excimers or the OM theory (recall the technical name for lithium-doped argon matrices in the chemical literature)? &lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="17175" LastEditDate="2020-05-25T23:20:16.137" LastActivityDate="2020-05-25T23:20:16.137" Title="Which argon compounds exist, supported by experiment or computation?" Tags="&lt;noble-gases&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13603" PostTypeId="1" AcceptedAnswerId="13604" CreationDate="2014-06-14T07:08:31.530" Score="2" ViewCount="1573" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Are the following two compounds isomers?&lt;/p&gt;&#xA;&lt;p&gt;(A) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_{3}-CO-CH_{2}-CH_{2}-CH_{2}-CH_{2}-OH}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(B) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-COOH}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Although they have the same molecular formula (&lt;span class=&quot;math-container&quot;&gt;$\ce{C_{6}H_{12}O_{2}})$&lt;/span&gt;, I think that this two compounds are not isomers.&lt;/p&gt;&#xA;" OwnerUserId="6812" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-08T19:19:28.320" Title="Are 6-hydroxyhexan-2-one and hexanoic acid isomers?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13604" PostTypeId="2" ParentId="13603" CreationDate="2014-06-14T08:09:50.860" Score="5" Body="&lt;p&gt;Isomers are compounds having same molecular formula, but a different structural formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As they satisfy the above condition, they are isomers and they are structural isomers. Structural isomers have a few variety based on connectivity of constituent atoms. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chain : which has a different arrangement of carbon molecules&lt;/li&gt;&#xA;&lt;li&gt;Position : different position of same functional group&lt;/li&gt;&#xA;&lt;li&gt;Functional : different functional group for the same atoms&lt;/li&gt;&#xA;&lt;li&gt;the others (tautomerism, metamerism and ring-chain isomerism)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are functional isomers as they have different functional groups. First one has a ketone and an alcoholic group, while the second one has a carboxylic acid functional group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will come to see that it is easier to recognise structural isomers compared to the rest (stereoisomerism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When speaking about isomerism, a picture is worth a thousand words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQRXm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQRXm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6839" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-08T15:45:19.283" LastActivityDate="2018-03-08T15:45:19.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13605" PostTypeId="2" ParentId="13592" CreationDate="2014-06-14T08:34:57.053" Score="4" Body="&lt;p&gt;What about &lt;a href=&quot;http://jmol.sourceforge.net&quot; rel=&quot;nofollow&quot;&gt;JMol&lt;/a&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It opens .Mol and a slew of other formats in Java which is fairly widely compatible. Open-source too. &lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="5456" LastEditDate="2014-10-09T04:45:30.973" LastActivityDate="2014-10-09T04:45:30.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13606" PostTypeId="1" CreationDate="2014-06-14T11:03:59.977" Score="1" ViewCount="181" Body="&lt;p&gt;The electronic conductance of a metal depends on its density.Does this mean that if we increase the density of the metal,its electronic conductance will increase?How does it happen?&lt;/p&gt;&#xA;" OwnerUserId="5775" LastActivityDate="2014-10-12T16:44:38.803" Title="how does electronic conductance of a metal change with density?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13607" PostTypeId="1" AcceptedAnswerId="13640" CreationDate="2014-06-14T11:39:29.350" Score="17" ViewCount="91050" Body="&lt;p&gt;I know that, solubility of alkaline earth metal hydroxides increases down the group and solubility of alkaline earth metal sulfates decreases down the group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to understand the reason for these trends.&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="5026" LastEditDate="2018-07-10T09:44:49.180" LastActivityDate="2021-05-17T10:16:30.487" Title="Why does the solubility of Group II hydroxides increase and the solubility of sulphates decrease down the group?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;alkaline-earth-metals&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="13608" PostTypeId="1" CreationDate="2014-06-14T11:45:16.427" Score="1" ViewCount="2109" Body="&lt;p&gt;When $\ce{NCl3}$ is dissolved in water, $\ce{NH3}$ and $\ce{HOCl}$ is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please explain me whether formed solution is acid or basic or neutral? thank you! &lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="4734" LastEditDate="2014-06-15T10:11:59.507" LastActivityDate="2014-06-15T10:11:59.507" Title="NCl₃ + H₂O → NH₃ + HOCl , the resulting solution is acidic, basic or neutral?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13609" PostTypeId="1" AcceptedAnswerId="13611" CreationDate="2014-06-14T13:03:44.123" Score="10" ViewCount="1870" Body="&lt;p&gt;Why is it that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{p}K_\mathrm{a}(\ce{HF}) &amp;lt; \mathrm{p}K_\mathrm{a}(\ce{HCl}) &amp;lt; \mathrm{p}K_\mathrm{a}(\ce{HBr}) &amp;lt; \mathrm{p}K_\mathrm{a}(\ce{HI}),$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;even though the electronegativity decreases down the group? The more electronegative the atom accompanying hydrogen, the lower the energy of the σ* bond. The lower the energy of the σ* bond, the easier it is for nucleophiles to attach to it (i.e. hydrogen is more easily removed from the compound, thus making it more acidic).&lt;/p&gt;&#xA;&lt;p&gt;Obviously, something is wrong with this reasoning. I could guess it is because hydrogen binds to different &lt;span class=&quot;math-container&quot;&gt;$\mathrm s$&lt;/span&gt; orbitals with the different halogens. For &lt;span class=&quot;math-container&quot;&gt;$\ce{F},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Br},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{I}$&lt;/span&gt; these are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2sp^3},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3sp^3},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4sp^3},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5sp^3},$&lt;/span&gt; respectively.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="41328" LastEditDate="2021-05-14T18:45:08.090" LastActivityDate="2021-05-14T18:45:08.090" Title="Hydrogen halide acidity" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;halides&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13610" PostTypeId="2" ParentId="13606" CreationDate="2014-06-14T14:21:29.773" Score="1" Body="&lt;p&gt;You cannot change the density of a metal without also changing other properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brass is a mixture of several elements. It can have different densities depending on the mixture of the elements. I haven't researched it, but the electronic conductance would likely be different for each type of brass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a source for your statement?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The electronic conductance of a metal depends on its density.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-14T14:21:29.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13611" PostTypeId="2" ParentId="13609" CreationDate="2014-06-14T15:19:27.973" Score="8" Body="&lt;p&gt;Yes, you are on the right track, let's look at the situation in more detail.  In the haloacid equilibrium $$\ce{HX &amp;lt;=&amp;gt; H+ + X-}$$ anything that stabilizes HX will push the equilibrium to the left and make the $\mathrm{p}K_\mathrm{a}$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make the $\mathrm{p}K_\mathrm{a}$ less positive. Here's a table that summarizes some of the haloacid data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{cccc}&#xA;\text{X} &amp;amp; \text{H-X bond strength / kJ mol}^{-1} &amp;amp; \text{Electronegativity of X} &amp;amp; \mathrm{p}K_\mathrm{a}(\ce{HX}) \\&#xA;\hline&#xA;\ce{F} &amp;amp; 565 &amp;amp; 4.0 &amp;amp; 3.1 \\&#xA;\ce{Cl} &amp;amp; 427 &amp;amp; 3.0 &amp;amp; -7.0 \\&#xA;\ce{Br} &amp;amp; 363 &amp;amp; 2.8 &amp;amp; -9.0 \\&#xA;\ce{I} &amp;amp; 295 &amp;amp; 2.5 &amp;amp; -11.0&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 main factors that influence this equilibrium: bond strength, electronegativity and polarizabilty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &quot;Bond Strength&quot; column we see that the $\ce{HX}$ bond strength decreases as we move down the column. This is because, as you noted, overlap between the hydrogen $\mathrm{1s}$ orbital and the halogen orbital is most effective with the fluorine $\mathrm{2sp^3}$ orbital (in fact, overlap is so good in the fluorine case that the bond has a significant covalent nature) and the effectiveness of the overlap decreases as we move down the column. Therefore, $\ce{HX}$ bond strengths should make the $\mathrm{p}K_\mathrm{a}$ decrease (less positive, more negative) as we move down the halogen column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your analysis of electronegativity is correct. The more electronegative the halogen, the more stable the halide anion. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to increase (more positive) as we move down the halogen column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to decrease (more negative) as we move down the halogen column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have two factors, bond strength and polarizability, that work together to push the $\mathrm{p}K_\mathrm{a}$ lower as we move down the halogen column; and one factor, electronegativity, that works in the opposite direction. Based on the observed $\mathrm{p}K_\mathrm{a}$'s, the first two factors win out over the electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I like your $\sigma$, $\sigma^*$ reasoning. Here's how I think of it in those terms.  As you bring two atomic orbitals closer together, they will split and form two new molecular orbitals. The degree of splitting is dependent upon the overlap of the two atomic orbitals.  Overlap is a measure of electron density between the atoms.  An electronegative atom will share its electron less readily decreasing the splitting.  If there is a strong bond, then the electrons are well shared and the splitting will be larger.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-03-26T18:23:18.150" LastActivityDate="2016-03-26T18:23:18.150" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="13612" PostTypeId="2" ParentId="13608" CreationDate="2014-06-14T16:13:00.470" Score="6" Body="&lt;p&gt;Here's our initial reaction, $$\ce{NCl3 + 3H2O -&amp;gt; NH3 + 3HOCl}$$ $\ce{NH3}$ is a weak base and $\ce{HOCl}$ is a weak acid, but there is more of the acid, so at this point the solution would be mildly acidic. However $\ce{NH3}$ and $\ce{HOCl}$ are known to react as follows $$\ce{NH3 + HOCl-&amp;gt;NH2Cl + H2O}$$ and in aqueous solution the resultant chloramine can decompose as shown in the following equation $$\ce{3 NH2Cl ⟶ N2 + NH4Cl + 2 HCl}$$  Clearly, an acid is generated in this last reaction.  So although the initial products are neutral, eventually the solution will be strongly acidic.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-14T16:20:44.677" LastActivityDate="2014-06-14T16:20:44.677" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="13613" PostTypeId="1" CreationDate="2014-06-14T16:28:11.050" Score="3" ViewCount="5664" Body="&lt;p&gt;I think I know what these terms mean:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;each row on the periodic table has a new shell of electrons, &lt;/li&gt;&#xA;&lt;li&gt;a subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.,&lt;/li&gt;&#xA;&lt;li&gt;and energy levels are the same as shells, &lt;/li&gt;&#xA;&lt;li&gt;whilst orbitals are the same as subshells. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct or totally wrong?&lt;/p&gt;&#xA;" OwnerDisplayName="user5110" LastEditorUserId="16683" LastEditDate="2019-01-25T00:35:57.470" LastActivityDate="2019-02-12T13:23:41.790" Title="What are similarities and differences among shells, orbitals, subshells, and energy levels?" Tags="&lt;orbitals&gt;&lt;theoretical-chemistry&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13614" PostTypeId="2" ParentId="13613" CreationDate="2014-06-14T16:45:20.127" Score="1" Body="&lt;p&gt;Shells are the major grouping of the orbitals (1, 2, 3, 4, etc). Subshells are the groupings inside each shell (s, p, d, f, etc). Energy levels correspond to the orbitals and vary with the element and the number of electrons in the atom (&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{3+}}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2019-01-24T18:40:01.017" LastActivityDate="2019-01-24T18:40:01.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13615" PostTypeId="1" AcceptedAnswerId="13617" CreationDate="2014-06-14T17:39:34.403" Score="1" ViewCount="2889" Body="&lt;p&gt;Less electronegative atom is the central atom. $\ce{N}$ is more electronegative than $\ce{Cl}$ as $\ce{Cl}$ has larger atomic radius. So why in $\ce{NCl3}$, $\ce{N}$ is the central atom?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="275" LastEditDate="2014-12-02T14:17:50.867" LastActivityDate="2015-10-25T02:33:17.763" Title="Determining central atom in NCl3" Tags="&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13616" PostTypeId="2" ParentId="13608" CreationDate="2014-06-14T17:50:35.580" Score="0" Body="&lt;p&gt;It is less but acidic since in HOCl, OH bond dissociates to form proton.Though NH3 is a lewis base,it is not  strong enough to neutralize the acidic HOCl.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastActivityDate="2014-06-14T17:50:35.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13617" PostTypeId="2" ParentId="13615" CreationDate="2014-06-14T18:39:36.877" Score="4" Body="&lt;p&gt;Cl is slightly more electronegative according to the tables I've seen: &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electronegativity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I wouldn't say it's an absolute rule that the less electronegative atom is central.  For example, $\ce{NH3}$ has the N central and $\ce{Cl_2O}$ has the O central.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-14T18:44:51.410" LastActivityDate="2014-06-14T18:44:51.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13618" PostTypeId="2" ParentId="13613" CreationDate="2014-06-14T19:05:12.430" Score="4" Body="&lt;p&gt;There is an arcane lore of quantum mechanics, that says that movement of electron in a hydrogen-like atom (with exactly one electron) is fully specified by four numbers: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Principal quantum number (&lt;em&gt;n&lt;/em&gt;) (1, 2, 3 ...) - specifies a shell&lt;/li&gt;&#xA;&lt;li&gt;Azimuthal quantum number(&lt;em&gt;ℓ&lt;/em&gt;) (1 .. &lt;em&gt;n&lt;/em&gt;) - specifies a subshell within the shell&lt;/li&gt;&#xA;&lt;li&gt;Magnetic quantum number (&lt;em&gt;m&lt;/em&gt;) (-&lt;em&gt;ℓ&lt;/em&gt; .. 0 .. &lt;em&gt;ℓ&lt;/em&gt;) - specifies an orbital within the subshell&lt;/li&gt;&#xA;&lt;li&gt;Spin quantum number (&lt;em&gt;s&lt;/em&gt;) (+/- 0.5) specifies one of two possible positions of electron on the orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While movement of electrons in non-hydrogen-like atoms is somewhat different, the form of movement, i.e. orbitals, are mostly same, so it is generally said, that electrons in a multielectron atom may be assigned the same numbers (with a side-note that electrons cannot be distinguished) and no two electrons can have all four numbers the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Energy level' is a completely different concept: it is a group of states possible for some quantum chemical system and having the same energy. This term may (randomly) pop-up in discussions where the electron is viewed as a singular particle with other parts of the system being 'framework'. While somewhat working, such considerations are incorrect and should better be avoided.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2019-01-24T18:39:22.220" LastActivityDate="2019-01-24T18:39:22.220" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="13619" PostTypeId="1" AcceptedAnswerId="13733" CreationDate="2014-06-14T20:44:25.190" Score="4" ViewCount="596" Body="&lt;p&gt;I am pharmacy student and would like to work in the field of drug design. We learn to work with the programs but thats not enough for me. I want to learn the theoretical principles of computational chemistry and develop software themselves, or at least parts of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a few questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I start with the chemistry part?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solving the Schrödinger equation&lt;/li&gt;&#xA;&lt;li&gt;Hartree–Fock method&lt;/li&gt;&#xA;&lt;li&gt;Density functional theory&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which mathematical principles do I need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a good starting point, especially in terms of mathematics and hope you can help me.&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2014-12-08T10:45:44.610" Title="Theoretical principles of molecular simulations" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13620" PostTypeId="1" CreationDate="2014-06-14T20:49:33.240" Score="0" ViewCount="301" Body="&lt;p&gt;For example like HCl, why isn't it a hydrogen bond?&lt;/p&gt;&#xA;" OwnerUserId="6840" LastEditorUserId="9961" LastEditDate="2015-10-11T22:39:18.263" LastActivityDate="2015-10-11T22:39:18.263" Title="How would you know if ... is H-bond or permanent dipole or induced dipole?" Tags="&lt;intermolecular-forces&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13621" PostTypeId="2" ParentId="13620" CreationDate="2014-06-14T20:56:23.867" Score="2" Body="&lt;p&gt;Hydrogen bonding is defined as an electrostatic attraction between a hydrogen atom bearing a partial positive charge in a molecule and an electronegative element of another molecule. The partial positive charge is the result of a dipole, caused by an electronegative element, &lt;em&gt;such as&lt;/em&gt; F, O, or N. The other electronegative element to which the partially positive hydrogen is attracted to is also &lt;em&gt;usually&lt;/em&gt; F, O, or N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the IUPAC definition; note that it does &lt;em&gt;not&lt;/em&gt; exclude the possibility of hydrogen bonding in molecules such as HCl: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A form of association between an electronegative atom and a hydrogen&#xA;  atom attached to a second, relatively electronegative atom. It is best&#xA;  considered as an electrostatic interaction, heightened by the small&#xA;  size of hydrogen, which permits proximity of the interacting dipoles&#xA;  or charges. Both electronegative atoms are usually (but not&#xA;  necessarily) from the first row of the Periodic Table, i.e. N, O or F.&#xA;  Hydrogen bonds may be inter-molecular or intramolecular.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/H02899.html&quot; rel=&quot;nofollow&quot;&gt;http://goldbook.iupac.org/H02899.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-14T20:56:23.867" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13623" PostTypeId="1" AcceptedAnswerId="13631" CreationDate="2014-06-14T22:16:40.377" Score="1" ViewCount="198" Body="&lt;p&gt;My freshwater fish tank has a pH of 5.8 and a KH value in excess of 40 or 720+ppm while still maintaining a water hardness of 20 ppm or GH&amp;lt; 1 now I am a math/phsyics student but I have taken both first year chem class at university so I expect to follow most basic ideas pertaining to chemistry but this doesn't make a lot of sense. Most water with a high alkalinity also has a basic pH around 8-8.5 normally I'd say that a high KH value is simply saying I have an acidic solution that is very resistant to a pH change i.e. a large buffering capacity but my water is very soft to the point of almost not having anything in it ( GH&amp;lt;1) or less than 20ppm of minerals how can water be so acidic have such a large buffering capacity and have nothing in it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the reason why I am asking here is all the aquarium forums i have tried can't seem to figure out what's going on in my tank)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer poseing an intresting possibility my tank has Turtles in who obviously are advanced enough to excreet urea instead of amonium i Titrated out a reaction and Got KH value of 28 and GH value of 4.5 i also added several air stones ( my buddie has a bio chem degree and figured it could of been too much $CO_{2}$ we managed to get the Ph up to ~6.4 and drop the KH value ot be more closely in line with accpetable parameters ( a value of 8 still oddly high compared to the GH value of 4.5)&lt;/p&gt;&#xA;" OwnerUserId="6841" LastEditorUserId="1362" LastEditDate="2014-11-02T15:12:26.933" LastActivityDate="2014-11-02T15:12:26.933" Title="How can water with a very low hardness have substantial buffer capacity?" Tags="&lt;everyday-chemistry&gt;&lt;ph&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13624" PostTypeId="1" AcceptedAnswerId="13637" CreationDate="2014-06-14T22:59:54.427" Score="1" ViewCount="73" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l7G8s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the above massaging of various K expressions valid? Something in me says &quot;Hess's Law.&quot; But then doesn't that deal with enthalpy? What's the true reason? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-15T15:27:42.997" Title="Why May K Expressions Be Manipulated As Such?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13625" PostTypeId="1" AcceptedAnswerId="13627" CreationDate="2014-06-15T00:33:48.673" Score="6" ViewCount="6839" Body="&lt;p&gt;$\ce{Fe}$, $\ce{Mg}$, $\ce{Ni}$, $\ce{Pb}$, $\ce{Sn}$, and $\ce{Zn}$ all react when they (in solid form) are submerged in an acid solution with the presence of a strong acid like $\ce{HCl}$, but silver and copper do not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have examined the electronegativity values and both $\ce{Ag}$'s and $\ce{Cu}$'s electronegativity value is only different from $\ce{Ni}$ by $\mathrm{0.1}$ on the Pauling scale. I don't see a pattern; the only thing I see which is common between copper and silver is that they both may have an oxidation number of +1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is it that $\ce{Ag}$ and $\ce{Cu}$ are immune to the H cations in the acidic solution?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="305" LastEditDate="2014-06-15T04:47:33.263" LastActivityDate="2017-07-06T15:15:54.587" Title="Why is it that neither silver nor copper react with a strong acid?" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13626" PostTypeId="1" AcceptedAnswerId="13632" CreationDate="2014-06-15T01:05:53.203" Score="2" ViewCount="340" Body="&lt;p&gt;For the reaction of silver and hydrochloric acid to form silver chloride: would the following be the correct way to couple the below two half reactions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Ag(s) + 2Cl^- -&amp;gt;2AgCl(s) + 2e^-} ~~E^0 =-0.222 ~V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2e^- + 2H_3^+O -&amp;gt; H_2 + H_2O}{ ~~E^0 =0.000~ V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_3^+O +2Ag(s) + 2Cl^- -&amp;gt; 2AgCl(s) + H_2 + H_2O}{ ~~E^0 =-0.222~ V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implying that the reaction is non-spontaneous as we have a &quot;negative potential&quot;? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-15T11:10:48.297" Title="Calculating Redox Potential; Coupled Reactions" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13627" PostTypeId="2" ParentId="13625" CreationDate="2014-06-15T02:31:54.557" Score="6" Body="&lt;p&gt;This will involve some degree of hand-waving. Let me first limit the analysis to the transition metals so that more effective comparisons can be made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction between a metal and an acid requires the metal to be oxidized, i.e. lose electrons to the acid, creating positively-charged metal ions in solution. The general idea is that as you go along the transition metals in a period from left to right, there is an increase in &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective nuclear charge&lt;/a&gt; felt by the valence $ns$ electrons (due to the additional $(n-1)d$ subshell electrons poorly shielding the $ns$ electrons from the nucleus) making the valence electrons harder to remove. This trend can be seen to some extent in the increase in ionization energies of the elements, for example, and also in the standard reduction potentials for metal cations of same charge (they become more positive, in other words making oxidation and hence attack by acid more difficult). Thus, metals at the left of the transition metals tend to be more easily attacked by acids than the metals at the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a rather interesting exception though: zinc strongly breaks the trend, being far easier to oxidize than copper, the element before it, even though zinc has a much higher ionization energy, clearly indicating its electrons are held tighter (cadmium also exhibits this anomaly, to a milder extent). What gives? The thing is that the tendency for a metal to oxidize is the result of several combined factors, one of which is also how strongly bound the solid metal is in the first place. If a metal contains atoms which are strongly bonded to each other, then oxidation tends to be more unfavourable as it would require these bonds to break. When comparing copper and zinc, it is clear that the latter has far weaker metallic bonding in the solid (their melting/boiling points are $1360\ \rm{K}$ / $2835\ \rm{K}$ and $695\ \rm{K}$ / $1180\ \rm{K}$, respectively). Thus, if you include the energy necessary to separate the atoms from the solid before ionizing them, it turns out that the process is easier for zinc than copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing different rows of the transition metals is a bit less clear, in part because not all elements have ions of the same charge which can be directly compared. The general tendency is that the metals become tougher to oxidize as you go down the rows. For the transition metals in the sixth and seventh period this is probably a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanthanide_contraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;lanthanide/actinide contraction&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/search?q=relativistic%20effects&quot;&gt;relativistic effects&lt;/a&gt;, which decrease the energy of the valence $ns$ orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for some other elements. Magnesium is very reactive towards acids because it is both a metal containing relatively weak bonds and because its ionization energy is comparatively low, being an alkaline earth element. For tin, the metallic bonding is not weak, but its ionization energy is not too high, so it will oxidize in acids, though not as strongly. Lead is somewhat less reactive to acids, possibly because its oxidation tends to stop at +2 instead of tin's +4, creating a significantly &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;softer cation&lt;/a&gt; which isn't as well solvated by water.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-06-15T02:31:54.557" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13628" PostTypeId="1" AcceptedAnswerId="13652" CreationDate="2014-06-15T03:55:09.293" Score="5" ViewCount="6170" Body="&lt;p&gt;I was reviewing my grad Thermodynamics textbook to go over a concept that I never learned well: the Lattice model.  One set of equations&lt;sup&gt;1&lt;/sup&gt; has:&#xA;\begin{align}&#xA;zN_1 &amp;amp;= 2N_{11} + N_{12}\\&#xA;zN_2 &amp;amp;= 2N_{22} + N_{12}\\&#xA;\end{align}&#xA;where $z$ is the coordination number, &lt;em&gt;or&lt;/em&gt; the number of nearest neighbors (defined twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My trouble is a simple thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pure solution of $1$ &lt;/li&gt;&#xA;&lt;li&gt;$z = 10$&lt;/li&gt;&#xA;&lt;li&gt;$N = 100$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So then the first equation becomes $10{\times}100=2{\times}N_{11}+0$, so $N_{11}=500$. Five-hundred nearest pairs, right?  But if I actually had a solution of 100 species 1 each with 10 neighbors, then I'd expect (ignoring boundaries) to have about $100{\times}10=1000$ pairs of $1-1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to figure out what is meant by $N_{11}$.  Could someone explain it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; J. Prausnitz, R. Lichtenthaler, and E. Gomes de Azevdeo. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd Edition, p.331, Eq.7-68.&lt;/p&gt;&#xA;" OwnerUserId="6843" LastEditorUserId="4945" LastEditDate="2017-02-14T06:23:30.797" LastActivityDate="2017-02-14T07:04:53.183" Title="In the lattice model, what's a nearest neighbor?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13630" PostTypeId="1" AcceptedAnswerId="13635" CreationDate="2014-06-15T10:00:27.980" Score="4" ViewCount="16246" Body="&lt;p&gt;Ozone is a resonating structure, is it because of its resonance and bent structure that it is a polar molecule? Or Is it because of the formal charges whereby the formal charges determine the direction of the dipole moment. If yes, how does this work with resonating structures. Initially, I thought that resonating structures automatically are polar, however, the azide ion is a clear counter-example. I am curious, how one deals with such a case, should we take each resonating structure by itself and consider the superposition in the end. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;Ultimately, my question is how can you explain ozone's polarity without formal charges because on a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_polarity&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;, they explain it using formal charges. If that's the case, shouldn't resonance be considered. If yes, how?&lt;/p&gt;&#xA;" OwnerUserId="6846" LastEditorUserId="4945" LastEditDate="2014-06-16T09:44:50.877" LastActivityDate="2014-06-16T10:50:00.643" Title="Does resonance affect the polarity of the molecule?" Tags="&lt;resonance&gt;&lt;polarity&gt;&lt;formal-charge&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13631" PostTypeId="2" ParentId="13623" CreationDate="2014-06-15T10:55:09.023" Score="1" Body="&lt;p&gt;Distilled water may have pH 5-6 because it dissolves significant amount of carbon dioxide. Carbon dioxide in water is in equilibrium &#xA;$$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3 &amp;lt;=&amp;gt; H+ + HCO3-}.$$ Since amount of dissolved carbon dioxide is much larger than formed acid, the solution acts as buffer. &#xA;Tap/sweet water usually contains fair amount of K, Na, Ca, and Mg cations, that traps significant amount of carbon dioxide forming $\ce{HCO3- /CO3^{2-}}$ buffer, that is alkaline. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2014-06-15T14:19:48.917" LastActivityDate="2014-06-15T14:19:48.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13632" PostTypeId="2" ParentId="13626" CreationDate="2014-06-15T11:10:48.297" Score="2" Body="&lt;p&gt;$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As two half-reactions, it is seen that the silver is oxidized:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2~Ag \rightarrow 2~Ag^{+} +2~e^- ~~ E^{\circ}= 0.80\,V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the hydrogen is reduced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2~H^+ + 2~e^- \rightarrow H_2 ~~ E^{\circ}= 0.0\,V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the reduction potential you have to solve the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E = E_{acceptor} - E_{donor} = E_{2H^{+}/H_2}- E_{Ag/Ag^{+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E =  0\,V - (0.80\,V) = -0.80\,V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions doesn't run voluntarily if $\Delta E &amp;lt; 0$&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2014-06-15T11:10:48.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13633" PostTypeId="2" ParentId="13603" CreationDate="2014-06-15T12:09:34.837" Score="5" Body="&lt;p&gt;They are certainly isomers since they have same molecular formula but differ in structural formula and their chemical properties are also different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isomers can be classified into structural isomer/constitutional isomer and stereo-isomer. The above compounds are structural isomers and in depth, they are functional isomers which is sub division of structural isomers in which the isomers differ in functional groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound A has two functional groups ketone and alcoholic groups (priority order ketone&gt; alcoholic group) and compound B has carboxylic acid as functional group and they have same molecular formula.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorDisplayName="user15489" LastEditDate="2015-06-24T10:45:00.087" LastActivityDate="2015-06-24T10:45:00.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13634" PostTypeId="2" ParentId="13630" CreationDate="2014-06-15T13:53:08.593" Score="5" Body="&lt;p&gt;Yes, ozone has a dipole moment because it is bent and has resonance structures that involve formal charge separation.  Resonance can affect the polarity and dipole moment of a molecule, but only if at least one of the resonance structures involves formal charge separation, and even then there are cases where the overall molecule will not have a dipole moment.  Some examples will help.  Let's start with benzene. Resonance structures can be drawn for benzene, but none of them involve charge separation, therefore benzene is not polar and does not have a dipole moment. So just being able to draw resonance structures does not mean the molecule must be polar and have a dipole moment.  Next let's examine cyclohexanone.  In this case two resonance structures can be drawn and one of them involves charge separation.  Therefor, we would expect this molecule to be polar and have a dipole moment. For 1,4-cyclohexanedione we can also draw resonance structures with formal charge separation, however in this case the symmetry of the molecule causes the local polarization at each carbonyl to cancel and overall, the molecule has no dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aq5e4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5084" LastEditDate="2014-06-15T16:54:13.073" LastActivityDate="2014-06-15T16:54:13.073" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13635" PostTypeId="2" ParentId="13630" CreationDate="2014-06-15T14:14:07.983" Score="7" Body="&lt;p&gt;When talking about &lt;a href=&quot;http://goldbook.iupac.org/R05326.html&quot; rel=&quot;noreferrer&quot;&gt;resonance&lt;/a&gt; it is important to know that this is only a concept to visualise certain bonding features. It is very important to understand, that the actual bonding situation is a mixture of all possible resonance structures. The Molecule itself exists at all times as this mixture, hence neither of these resonance representations can be observed. One can regard it also as a superposition of these structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structures had to be introduced into the &lt;a href=&quot;http://goldbook.iupac.org/L03513.html&quot; rel=&quot;noreferrer&quot;&gt;Lewis formula&lt;/a&gt; context, when it was obvious, that one formula can no longer describe a bonding pattern remotely accurate. While the Lewis formalism is great for many things, as it keeps chemistry visualisable and simple, it unfortunately has its boundaries. Only for molecules with very simple bonding patterns it is somehow possible to guess some physical properties (qualitatively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formal charges are also only a concept, or a result from the limited depiction of the Lewis representation - hence the name formal. They do not usually represent any physical charge. However, sometimes, the physical charge will coincide with the formal charge. Therefore it is usually not possible to derive any actual dipole moment from them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge separation within molecules depend of the geometric arrangements of the nuclei and the electrons around them, i.e. the electron density. It is a physical property which can be observed and also predicted by means of quantum chemical considerations. Molecular Orbital Theory is a very fundamental concept which allows to derive these properties (at least on a qualitative level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Upon the extension of the question and some comments:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Thomij already stated, the formal charges are only for bookkeeping. They always have an integer value. A partial charge, however, is a physical observable, resulting from the difference from the electron density and the field of the nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases the overall formal charge has to add up to the overall physical charge, which can also be calculated from all partial charges. These partial charges can also be calculated from quantum chemical calculations, since you are approximately solving for the optimal electron density at a certain nuclear arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is absolutely no possibility to predict a dipole moment without knowing the electron density (or wavefunction) at a given nuclear geometry. A Lewis formula can be generated from these facts, showing the simplest but still well matching bond description. This representation has to sometimes include formal charges. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_polarity&quot; rel=&quot;noreferrer&quot;&gt;linked wikipedia page&lt;/a&gt; does not employ this concept to explain the polarity, they rather use partial charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial charges in ozone result from a superposition of the molecular orbitals (&lt;a href=&quot;https://i.stack.imgur.com/YC1nn.jpg&quot; rel=&quot;noreferrer&quot;&gt;click here to enlarge&lt;/a&gt;). These are the valence orbitals obtained from BP86/cc-VPDZ:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YC1nn.jpg&quot; alt=&quot;valence orbitals of ozone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially the HOMO (very right) and the HOMO-3 (4th from right) and HOMO -6 (4th from left) contribute massively to a slight charge separation in ozone.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-06-16T10:50:00.643" LastActivityDate="2014-06-16T10:50:00.643" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13636" PostTypeId="2" ParentId="13623" CreationDate="2014-06-15T14:35:14.633" Score="3" Body="&lt;p&gt;In an aquarium, there could be several non-mineral buffers. The most likely one is a phosphate buffer. There would also be carboxylic acid (amino acids) and probably urea, as well as carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-15T14:35:14.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13637" PostTypeId="2" ParentId="13624" CreationDate="2014-06-15T15:27:42.997" Score="2" Body="&lt;p&gt;Your intuition is correct. The reason equilibrium constants from different reactions can be strung together and multiplied comes from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess%27_law#Extension_to_free_energy_and_entropy&quot; rel=&quot;nofollow&quot;&gt;a generalization of Hess' law which covers Gibbs' free energy changes&lt;/a&gt; (apparently also called &lt;a href=&quot;http://en.wikipedia.org/wiki/Bordwell_thermodynamic_cycle&quot; rel=&quot;nofollow&quot;&gt;Bordwell thermodynamic cycles&lt;/a&gt;). Enthalpy is a thermodynamic function with a special property; it is a &lt;em&gt;state function&lt;/em&gt;. That is, enthalpy changes in cycles are determined solely by the initial and end states of a system, and not the trajectory in which the initial state travels to the end state (as happens for functions such as work $W$ and heat $q$). Any state function can be manipulated in the way enthalpy is used for Hess diagrams. Thus, by adding reactions, you can add their respective reaction free energy changes $\Delta_rG$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us now take two reactions, with free energy changes equal to $\Delta_rG_1$ and $\Delta_rG_2$. There is a relationship between a reaction's free energy change and its reaction quotient $Q$, namely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rG = \Delta_rG^o + RT\ ln\ Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, where $^o$ indicates standard conditions. For reactions 1 and 2, we therefore have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rG_1 = \Delta_rG^o_1 + RT\ ln\ Q_1$$&#xA;$$\Delta_rG_2 = \Delta_rG^o_2 + RT\ ln\ Q_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the reactions, we obtain a reaction 3 where the same formula applies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\color{\red}{\Delta_rG_3} = \color{\navy}{\Delta_rG^o_3} + \color{\green}{RT\ ln\ Q_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since we can add the free energy equations for reactions 1 and 2, reaction 3 is also represented by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_rG_1 + \Delta_rG_2 = \Delta_rG^o_1 + \Delta_rG^o_2 + RT\ ln\ Q_1 + RT\ ln\ Q_2$$&#xA;$$\color{\red}{\Delta_rG_1 + \Delta_rG_2} = \color{\navy}{\Delta_rG^o_1 + \Delta_rG^o_2} + \color{\green}{RT\ ln\ Q_1 Q_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, we have $Q_3 = Q_1 \times Q_2$. Considering the equilibrium conditions of all reactions ($\Delta_rG = 0$), then the reaction quotients are defined as the equilibrium constants $K$, and similarly one finds $K_3 = K_1 \times K_2$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-06-15T15:27:42.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13638" PostTypeId="2" ParentId="2881" CreationDate="2014-06-15T16:49:43.100" Score="4" Body="&lt;p&gt;Kinetic stability refer to a pre equilibrium conditions. For example the equilibrium constant K may be very large meaning the formation of the product is favorable, but the condition is such that the reaction does not go (or goes extremely slowly) &#xA;Thermodynamic stability refer to a system in which products formation is not favourable. For example we have larger reaction quotient Q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction can be thermodynamically unstable but kinetically stable because the condition are such that the rate of the reaction is zero.  A mixture of coal and oxygen are kinetically stable because the condition for combustion are not obtained until the coal is heated to its ignition temperature. The system is thermodynamically unstable because at equilibrium the coal oxygen reaction favors the formation of $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="6847" LastEditorUserId="4734" LastEditDate="2014-06-18T22:57:59.410" LastActivityDate="2014-06-18T22:57:59.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13639" PostTypeId="1" AcceptedAnswerId="14663" CreationDate="2014-06-15T18:25:51.667" Score="3" ViewCount="380" Body="&lt;p&gt;Consider the reaction &lt;strong&gt;100ml 10M&lt;/strong&gt; $\ce{NaOH_{(aq)}} +$ &lt;strong&gt;100ml 10M&lt;/strong&gt; $\ce{NaHSO3_{(aq)} +}$ &lt;strong&gt;100mL 0.01M&lt;/strong&gt; $\ce{KMnO4_{(aq)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture posted here, the left columns give information on the Molarity of species present &lt;strong&gt;before&lt;/strong&gt; 100mL 0.01 M $\ce{KMnO4}$ is added. The right column is &lt;strong&gt;after&lt;/strong&gt; $\ce{KMnO4}$ is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H3aID.jpg&quot; alt=&quot;Redox Reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the resulting solution, $\ce{Mn}$ is oxidized to various states: $\ce{Mn^2+ Mn^4+ Mn^6+ Mn^7+}$ (where $\ce{Mn^7+}$ is the original form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know, why do different oxidation states of $\ce{Mn}$ appear? How can we predict the concentration of oxidation states?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="3926" LastEditDate="2014-08-04T04:56:09.977" LastActivityDate="2014-08-04T04:56:09.977" Title="In a redox reaction, why does more than one oxidation state of an element form?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13640" PostTypeId="2" ParentId="13607" CreationDate="2014-06-15T18:46:30.720" Score="5" Body="&lt;p&gt;A few weeks ago I would have given the hard soft acid base theory (HSAB) explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard this &quot;trend&quot; explained in this way:  Hydroxide is hard and sulfate is soft.  So, the harder ions are, the more soluble with sulfate and less with hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About a week ago I was playing around with some DH, lattice energy, and Ksp data and I found that barium seems to go against the &quot;trend&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found these posts that show that many of the &quot;trends&quot; we are taught in chem class aren't as trendy as we are told. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at this:  &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/group2/solubility.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemguide.co.uk/inorganic/group2/solubility.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a homework question or a curiosity question?  If the former, the &quot;trend&quot; isn't that trendy and the teacher should read the links above.  If the latter, do you know that barium salts tend to go against the &quot;trend&quot;?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="4945" LastEditDate="2014-07-16T09:19:52.313" LastActivityDate="2014-07-16T09:19:52.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13641" PostTypeId="2" ParentId="13533" CreationDate="2014-06-15T18:59:47.230" Score="1" Body="&lt;p&gt;How is the magnesium being ionized?  If it is reacting with oxygen or another oxidizer, look at the Born-Haber cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnesium losing electrons is endothermic.  Oxygen gaining one electron is exothermic but the 2nd electron affinity of oxygen is endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg^{2+}}$ and oxide making magnesium oxide is exothermic.  Removing the second electron from magnesium and adding the second electron to oxygen is &quot;paid for&quot; by the increase in lattice energy due to having 2 doubly charged ions instead of 2 singly charged ions.  &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="4734" LastEditDate="2014-06-16T06:55:34.240" LastActivityDate="2014-06-16T06:55:34.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13642" PostTypeId="1" AcceptedAnswerId="13644" CreationDate="2014-06-15T20:11:53.020" Score="1" ViewCount="232" Body="&lt;p&gt;What is the extent of say, this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_2CO_3 + HOCO_2^- &amp;lt;=&amp;gt; HOCO_2^- + H_2CO_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K for the reaction appears to equal 1, and given the stoichiometry of the reaction, this implies that the reaction reaches an extent of 50% at equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is this actually true?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="33991" LastEditDate="2017-04-28T07:23:47.547" LastActivityDate="2017-04-28T07:23:47.547" Title="What is the extent of a reaction between an acid and its conjugate base?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13643" PostTypeId="2" ParentId="13642" CreationDate="2014-06-15T20:55:51.007" Score="3" Body="&lt;p&gt;Thermodynamics only applies to statistical behaviour over macroscopic quantities, rather than individual proton shuffles. When considered on this scale and with this level of abstraction, no reaction has taken place and so this isn't a meaningful question.&lt;/p&gt;&#xA;" OwnerUserId="39" LastActivityDate="2014-06-15T20:55:51.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13644" PostTypeId="2" ParentId="13642" CreationDate="2014-06-15T23:07:52.123" Score="9" Body="&lt;p&gt;The reaction you've shown is termed an &quot;identity reaction&quot;, the products are the same as the reactants so $\ce{K_{eq}}$ &lt;em&gt;must&lt;/em&gt; equal one.  In the particular case you've drawn the reaction involves a simple proton transfer and will likely be diffusion controlled.  Other examples of identity reactions are shown below.  That they are actually occurring can be demonstrated using isotopically labeled substrates.  The last one is one of my favorites, the degenerate rearrangement of bullvalene into itself via a Cope rearrangement.  My drawing of bullvalene doesn't do it justice, just google bullvalene and prepare to be fascinated.  If you number each of the ten carbon atoms in bullvalene you'll find that there are over 1.2 million possible arrangements.  Sequential Cope rearrangements will convert each of these arragments into every other possible rearrangement.  Connectivity in bullvalene is lost, each carbon atom is rapidly bound to every other carbon atom.  At a bit over 100 °C, both the proton and carbon-13 NMR of bullvalene show only 1 sharp peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w34Hm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-10-18T20:22:26.650" LastActivityDate="2015-10-18T20:22:26.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13645" PostTypeId="2" ParentId="9980" CreationDate="2014-06-16T05:50:57.347" Score="5" Body="&lt;p&gt;The first equation actually contains the definition of the &lt;a href=&quot;https://www.goldbook.iupac.org/html/S/S05915.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard equilibrium constant&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$&lt;/span&gt;&#xA;With this definition the equilibrium constant is dimensionless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under standard conditions the van't Hoff equation is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$&lt;/span&gt;&#xA;and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =&#xA;  \frac{{\ - \Delta H^\circ }}{R}&#xA;  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the ordinary &lt;a href=&quot;https://www.goldbook.iupac.org/html/E/E02177.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilibrium constant&lt;/a&gt; may be defined in various forms:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_x = \prod_B x_B^{\nu_B}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably one of the best representations for the standard equilibrium constant involves relative &lt;a href=&quot;https://www.goldbook.iupac.org/html/A/A00115.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;activities&lt;/a&gt;, for an arbitrary reaction, &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\nu_{A}A + \nu_{B}B -&amp;gt; \nu_{C}C + \nu_{D}D},$$&lt;/span&gt;&#xA;this resolves in &#xA;&lt;span class=&quot;math-container&quot;&gt;$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration is connected to the activity via&#xA;&lt;span class=&quot;math-container&quot;&gt;$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$&lt;/span&gt;&#xA;where the standard concentration is &lt;span class=&quot;math-container&quot;&gt;$c^\circ = \pu{1 mol/L}$&lt;/span&gt;. &#xA;At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$&lt;/span&gt;&#xA;See also a very detailed answer of &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/4945&quot;&gt;Philipp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial pressure is connected to the activity via&#xA;&lt;span class=&quot;math-container&quot;&gt;$$a(\ce{A}) = &#xA;\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$&lt;/span&gt;&#xA;with the fugacity &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; and the fugacity coefficient &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; and the fraction occupied by the gas &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, the total pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, as well as the standard pressure &lt;span class=&quot;math-container&quot;&gt;$p^\circ=\pu{1 bar}$&lt;/span&gt; or traditional use of &lt;span class=&quot;math-container&quot;&gt;$p^\circ=\pu{1 atm}$&lt;/span&gt;.&#xA;For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure &lt;span class=&quot;math-container&quot;&gt;$p(\ce{A})$&lt;/span&gt;, since&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &amp;amp;=1, &amp;amp;&#xA;p(\ce{A}) &amp;amp;= y_{\ce{A}}\cdot{}p.&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course concentrations and partial pressures are connected via the ideal gas&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA;pV\ &amp;amp;=nRT\\&#xA;p\ &amp;amp;\propto \text{const.} \cdot c,&#xA;\end{aligned}&lt;/span&gt;&#xA;and therefore it is valid to write:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_c\propto \text{const.} \cdot K_p.$$&lt;/span&gt;&#xA;It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of &lt;span class=&quot;math-container&quot;&gt;$(\mathcal{R}T)^{\sum{}\nu}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2018-09-27T11:46:40.140" LastActivityDate="2018-09-27T11:46:40.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13646" PostTypeId="1" AcceptedAnswerId="13668" CreationDate="2014-06-16T06:50:27.820" Score="6" ViewCount="9184" Body="&lt;p&gt;This may be in relation to the $\ce{-O-H}$ group present in cellulose, but I would like to have a more specific answer.&lt;/p&gt;&#xA;" OwnerUserId="6851" LastEditorUserId="9961" LastEditDate="2015-10-30T17:05:47.063" LastActivityDate="2020-11-24T02:39:56.770" Title="Why do cotton clothes take a longer time to dry as compared to synthetic clothes?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13647" PostTypeId="2" ParentId="13533" CreationDate="2014-06-16T07:03:27.320" Score="1" Body="&lt;p&gt;$\ce{Mg+}$ is not commonly encountered because compounds with $\ce{Mg^{2+}}$ are more stable. This does not mean that $\ce{Mg^{2+}}$ is more stable on its own, in gas phase even $\ce{Mg+}$ is less energetically favorable than $\ce{Mg^0}$. However, when interacting with other atoms, stabilization due electrostatic interaction is much higher for $\ce{Mg^{2+}}$ than for $\ce{Mg+}$ because of much smaller radius and double charge of the former. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, at least some compounds of $\ce{Mg(+1)}$ are known, even if in exotic conditions, like $\ce{MgCl}$ is known in gas phase.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2014-06-16T07:31:04.970" LastActivityDate="2014-06-16T07:31:04.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13648" PostTypeId="1" AcceptedAnswerId="13649" CreationDate="2014-06-16T11:43:10.043" Score="0" ViewCount="966" Body="&lt;p&gt;I have a series of questions similar to the following. I want to make sure I do this correct before I do the others. I am in need of a hint.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Given molarity of 6M NaOH with d=1.4g/mL, calculate, percent, mole fraction, and molality.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So in the above question, I have the molarity, but how am I supposed to get liters of solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured it was 6M per 1L of water, but then what am I supposed to do with the density? Is that the density of the water with the NaOH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in a sub part it gives me the mole fraction as 0.1 with a density of 1.3g/mL. Why does the question give me that number? Is it possible to figure out percent, molality, and molarity based on mole fraction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for you help. &lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorDisplayName="user7951" LastEditDate="2015-09-27T00:17:05.613" LastActivityDate="2015-09-27T00:17:05.613" Title="Molarity to Percent, Mole Fraction, Molality" Tags="&lt;homework&gt;&lt;concentration&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13649" PostTypeId="2" ParentId="13648" CreationDate="2014-06-16T13:18:40.660" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I figured it was 6M per 1L of water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is incorrect you have 6 mol per 1L of water. M(stands for mol/L) not for mol!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to tackle this problem?&lt;/strong&gt;&#xA;Check water density, is 1 $g/cm^{2}$. So yours is the density of the solution. Then understand what the problem is asking what is percent? What is mole fraction? What is molality? Are you able to answer these questions? If not go back to the text book.&#xA;Then is only a matter of simple algebra and units matching. And remember if you have weight and molecular weight you can calculate the number of mole!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose you have a liter of this solution &lt;strong&gt;what's the weight of NaOH?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; molecular weight ($\frac{g}{mol}$) of NaOH $\times$ number of moles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now what's the &lt;strong&gt;Overall weight&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Overall weight=density $\times$ volume (remember to use the right units)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Percent what? In chemistry there are a lot of percent, you mean percentage by weight (wt%) I imagine. Try to figure out from your self:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;$ = \frac{NaOH weight}{Overall weight}\times100 $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mole fraction require mole of both, we start from the weight, so whats the weight of water? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; you got the overall weight and the weight of NaOH&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now calculate the moles of water, and it is simple to find the mole fraction, and molality. Make these steps in the opposite way and you will find the answer to your last question.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-16T13:18:40.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13651" PostTypeId="2" ParentId="13602" CreationDate="2014-06-16T14:13:26.927" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp311901a&quot;&gt;ArBeO and ArBeS&lt;/a&gt; are experimentally known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.201203306/abstract&quot;&gt;ArAuF, ArAgF and ArCuF&lt;/a&gt; are also experimentally known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CUOAr$_n$ and related complexes &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/12617681&quot;&gt;are known&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp409941v&quot;&gt;ArBeNCN and ArBeNBO&lt;/a&gt; are predicted by theory.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-06-16T14:38:22.100" LastActivityDate="2014-06-16T14:38:22.100" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="13652" PostTypeId="2" ParentId="13628" CreationDate="2014-06-16T16:07:29.283" Score="4" Body="&lt;p&gt;A nearest neighbour in general terms is literally that: Find the closest atom of any given element, that is your nearest neighbour distance for that element in the lattice. The number of nearest neighbours for that element is the number of atoms that are at this distance from your starting atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{11}$ is the &lt;strong&gt;number of nearest neighbour pairs of atoms of element 1&lt;/strong&gt;, and $N_{12}$ is the number of nearest neighbour pairs between any atom of 1 and any atom of 2 etc. For the $N_{11}$ or $N_{22}$ case, you have to consider that each atom of 1 or 2 in the equation has a contribution to $zN$, hence the term being multiplied by 2 when viewed from the perspective of element 1 or 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your thought experiment, your expected value neglects edges (you can't just neglect edges!) so will be too high by a good margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dissertation of Alexander Zhou&lt;sup&gt;1&lt;/sup&gt; covers these equations (p.34) and explains them slightly differently (and more clearly I think) than de With's textbook.&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alexander Zhou. Explicit Non-Random Contribution in Lattice Fluid Equation of State. Explizite nicht zufällige Wahrscheinlichkeitsverteilungen in gitterbasierten Zustandsgleichungen. Doktor Ingenieur, Technische Fakultät der Universität Erlangen-Nürnberg, Nürnberg, 2013. See &lt;a href=&quot;https://opus4.kobv.de/opus4-fau/frontdoor/index/index/year/2013/docId/3063&quot; rel=&quot;nofollow noreferrer&quot;&gt;UB Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gijsbertus de With. &lt;em&gt;Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications&lt;/em&gt;; John Wiley &amp;amp; Sons: 2013. See &lt;a href=&quot;https://books.google.co.uk/books?id=7VIbAAAAQBAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Books&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5075" LastEditorUserId="4945" LastEditDate="2017-02-14T07:04:53.183" LastActivityDate="2017-02-14T07:04:53.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13653" PostTypeId="1" CreationDate="2014-06-16T19:45:56.563" Score="3" ViewCount="340" Body="&lt;p&gt;So I had these two reactions that confused me .. I tried the standard way of balancing it (Putting oxygen number everywhere, +1,+2,-2 etc.) and then go on with it but it didn't work &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anyways here are the 2 reactions:&#xA;$$\ce{3 Cu + 8HNO3-&amp;gt; 3Cu(NO3)2 + 2NO + 4H2O}$$&#xA;$$\ce{3Ag + 4HNO3 -&amp;gt; 3AgNO3 + NO + 2H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've no idea how do you get those coefficients (numbers).. Please, is there an easy way to do it (perhaps it has to do with electrons added/given? ) and if so - where do I start? &lt;/p&gt;&#xA;" OwnerUserId="6856" LastEditorUserId="275" LastEditDate="2014-06-16T20:16:28.530" LastActivityDate="2014-06-19T15:43:14.763" Title="Balancing redox equation" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13654" PostTypeId="2" ParentId="13646" CreationDate="2014-06-16T20:39:00.123" Score="0" Body="&lt;p&gt;Cotton is very absorbent so the reason it takes longer to dry is physical not chemical. It has capillaries which is why it is so absorbent.&lt;/p&gt;&#xA;" OwnerUserId="6857" LastActivityDate="2014-06-16T20:39:00.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13655" PostTypeId="1" CreationDate="2014-06-16T22:34:45.257" Score="1" ViewCount="175" Body="&lt;p&gt;I have seen certain metals such as Aluminium used in a situation where it is in a powdered form and &quot;compacted&quot; into a shape under intense pressure (in a mould), and it ends up looking mostly like solid metal. It is also heated afterwards, I assume to form a complete metallic bond.&lt;br/&gt;&#xA;I have a few questions on this:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How are oxide layers on the powder affected when it is compacted? Eg. Can the $Al_2O_3$ layer, or at least the O atoms from them, be &quot;inside&quot; the moulded object when it is compacted?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Metals in a powdery form aren't as reflective as in their &quot;solid&quot; form (also please let me know the right word for that). Why is this?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Are most metals in their powdered form pure, eg. Not like a salt, but a single element?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;I know gold powder, obtained from chloroauric acid, becomes gold in its metal form when melted down. It looks brownish and takes up far more volume than the gold it contains. Is this pure Au or a compound/salt?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Also, in 4. with the gold powder, it actually forms rather large clumps from the chloroauric acid. How is this possible assuming it's a pure element?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thought I better get all the questions done in one go. Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="5772" LastEditDate="2014-06-17T20:58:01.883" LastActivityDate="2014-12-23T17:18:19.203" Title="Properties of powdered metals" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13656" PostTypeId="2" ParentId="13653" CreationDate="2014-06-16T22:46:37.073" Score="3" Body="&lt;p&gt;There are a few basic principles that are useful to keep in mind when you want to balance a redox equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) OILRIG - Oxidation Is Loss, Reduction Is Gain (of electrons)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the basics of the first equation:  $$\ce{Cu + HNO3-&amp;gt; Cu(NO3)2} + ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left hand side we have Cu(0) metal and N(V), on the right hand side we have Cu(II) ion and N(V), so from OILRIG we know that the copper is being oxidised... but we know from OILRIG we have to put the electrons from the Cu(0) somewhere...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Hydrogen ions are always +1, and oxygen ions are always (there are a few exceptions but this is a good general rule) -2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... So the only thing that can change is the nitrogen. The nitrogen must be being reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen actually has a large number of possible oxidation states, taking any value between -3 and +5, so how do we decide which one it goes to in this case?  This depends on the temperature and the concentration of the acid. We can probably assume room temperature and bench concentration of acid (otherwise conditions need to be specified), and that the nitrogen isn't going to gain 5 electrons and become nitrogen gas, which means that we are going to form an oxide. Of the oxides, the most thermally stable is NO (nitrogen is in oxidation state +2 here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that look like now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu + HNO3-&amp;gt; Cu(NO3)2 + NO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To balance the electron transfers, we need 3 Cu(O) going to Cu(II) to donate 6 electrons for 2 N(V) to go to N(II). Balancing for electrons and nitrogens, we now have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Cu + 8HNO3-&amp;gt; 3Cu(NO3)2 + 2NO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, now we need to look at balancing oxygens and hydrogens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Use water if you can to balance oxygen and hydrogen. If not, start looking at hydroxide or hydrogen ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we need to somehow account for 8 hydrogens and 8x3 - 3x3x2 + 2 = 4 oxygens... That's 4 water molecules, so we finally have the answer you provided at the start!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Cu + 8HNO3-&amp;gt; 3Cu(NO3)2 + 2NO + 4H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="5075" LastEditorUserId="5075" LastEditDate="2014-06-19T15:43:14.763" LastActivityDate="2014-06-19T15:43:14.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13657" PostTypeId="1" CreationDate="2014-06-16T22:53:26.107" Score="25" ViewCount="36526" Body="&lt;p&gt;I'm looking for an explanation of the bonding in the phosphate (PO&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;3−&lt;/sup&gt;) ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dXYQl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Image courtesy of &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Phosphat-Ion.svg&quot; rel=&quot;noreferrer&quot; title=&quot;Wikipedia&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus (&lt;sub&gt;15&lt;/sub&gt;P) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1s&lt;sup&gt;2&lt;/sup&gt;&lt;br /&gt;&#xA;2s&lt;sup&gt;2&lt;/sup&gt; 2p&lt;sup&gt;6&lt;/sup&gt;&lt;br /&gt;&#xA;3s&lt;sup&gt;2&lt;/sup&gt; 3p&lt;sup&gt;3&lt;/sup&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the Phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, why are the oxygens shown as carrying a negative charge?&lt;/p&gt;&#xA;" OwnerUserId="7147" LastActivityDate="2018-10-07T22:50:24.163" Title="Bonding in the phosphate ion" Tags="&lt;bond&gt;&lt;ions&gt;&lt;covalent-compounds&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13659" PostTypeId="1" AcceptedAnswerId="13663" CreationDate="2014-06-17T02:29:24.723" Score="6" ViewCount="1221" Body="&lt;p&gt;In converting glucose from its acyclic (open-chain) form to either cyclic form (alpha or beta) why is the C5 hydroxyl group used instead of the bottom-most hydroxyl group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, are these structures considered isomers? All sources I've seen only say &quot;form&quot; or &quot;representation.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3747" LastEditorUserId="33991" LastEditDate="2017-04-28T07:35:34.950" LastActivityDate="2020-06-25T08:14:59.867" Title="Why does the cyclization of open-chain glucose occur via the C5 hydroxyl group?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13660" PostTypeId="2" ParentId="13659" CreationDate="2014-06-17T02:34:23.500" Score="3" Body="&lt;p&gt;The cyclic form of glucose uses the C5 oxygen for bonding since it forms a 6 member ring (more stable then either 5 or 7 member ring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are isomers since they have the same chemical formula.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-06-17T05:12:02.037" LastActivityDate="2014-06-17T05:12:02.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13661" PostTypeId="2" ParentId="13657" CreationDate="2014-06-17T03:39:08.413" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Phosphorus (&lt;span class=&quot;math-container&quot;&gt;$\ce{_{15}P}$&lt;/span&gt;) - being the fifteenth element - has fifteen&#xA;  electrons, five valence electrons and the following electron&#xA;  configuration:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{1s^2,~2s^2,~2p^6,~3s^2,~3p^3}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very observant! What happens is that phosphorus is hybridized in the phosphate ion. Hybridization allows the central phosphate atom to form more bonds than you might expect from its &lt;em&gt;ground&lt;/em&gt; state electron configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the framework of hybridization, one may look at the bonds in the phosphate anion as having a great deal of ionic character. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronegativity of phosphorus is 2.19; the electronegativity of oxygen is 3.44. The difference is 1.25 - or rather sizable. As a result, electron density is largely withdrawn from the central phosphorus. Hence the negative charges you see on the three oxygen molecules in your picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other depictions, which &lt;em&gt;reject&lt;/em&gt; a large degree of d-orbital utilization in hybridization, might give a negative formal charge to &lt;em&gt;all&lt;/em&gt; the oxygens and a positive formal charge to the central phosphorus by eschewing the &lt;span class=&quot;math-container&quot;&gt;$\ce{P=O}$&lt;/span&gt; double bond. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Apart from the oxygen with a double-bond, we have three oxygens&#xA;  forming single bonds. Does this not leave them one electron short?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, why are the oxygens shown as carrying a negative charge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could draw double bonds for all the oxygens but there aren't that many valence electrons to be distributed. The structure shown above is a compromise between number of valence electrons and electronegativity considerations - we wouldn't want a surfeit of electrons around the central phosphorus since that's not as thermodynamically favorable as having the more electronegative oxygen support these electrons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the sum of the formal charges on an atom must add up to the overall charge on the atom. Phosphate anion is triply negative charged; note how there are three oxygens bearing a single negative charge. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="23561" LastEditDate="2018-10-07T22:50:24.163" LastActivityDate="2018-10-07T22:50:24.163" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13662" PostTypeId="1" CreationDate="2014-06-17T05:15:55.707" Score="7" ViewCount="1402" Body="&lt;p&gt;Recently I've been going through an exercise of &lt;strong&gt;Chemical Kinetics and Nuclear Chemistry&lt;/strong&gt; and right now I am stuck on one question that was previously asked in &lt;a href=&quot;http://en.wikipedia.org/wiki/Joint_Entrance_Examination#JEE_Advanced&quot; rel=&quot;noreferrer&quot;&gt;JEE(Advanced)-2013&lt;/a&gt;. The question says,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the reaction,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{P + Q -&amp;gt; R + S}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The time taken for &lt;span class=&quot;math-container&quot;&gt;$75\%$&lt;/span&gt; reaction of &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; is twice that time taken for &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; reaction of &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt;. The concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt; varies with reaction time as shown in the figure. What is the overall order of the reaction?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3WQKo.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And here was my approach to the question,&lt;/p&gt;&#xA;&lt;p&gt;In the question different data have been given for different reactants (i.e. &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt;). For &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; it has been given that it take &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; double the time to react &lt;span class=&quot;math-container&quot;&gt;$75\%$&lt;/span&gt; than that from &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt;. So assuming the time taken for &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; reaction to be &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; and that of &lt;span class=&quot;math-container&quot;&gt;$75\%$&lt;/span&gt; to be &lt;span class=&quot;math-container&quot;&gt;$2t$&lt;/span&gt;, we can easily find that order of reaction w.r.t P is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. As,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{t_{\frac12}}{t_{\frac34}}=\frac{\log 2}{\log 4}=\frac12$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Further w.r.t &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; there is a graph given which is something like in form &lt;span class=&quot;math-container&quot;&gt;$\ce{[Q]=[Q_o]-mt}$&lt;/span&gt;. So comparing with the equation of zero order reaction which is like &lt;span class=&quot;math-container&quot;&gt;$\ce{[A]=[A_o]-kt}$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$\ce{[A]}$&lt;/span&gt; is the concentration of reactant and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is rate constant), we can conclude that order of reaction w.r.t to &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So I concluded that overall order of reaction will be&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;order of reaction of &lt;span class=&quot;math-container&quot;&gt;$P +$&lt;/span&gt; order of reaction of &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\implies 1+0=1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So was this the right approach from my side or there is any kind of error in any step?&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-12T18:05:01.763" Title="Finding overall order of reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13663" PostTypeId="2" ParentId="13659" CreationDate="2014-06-17T05:25:46.630" Score="5" Body="&lt;p&gt;Due to the internal &lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_strain&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring strain&lt;/a&gt; the best possible conformation is a six-membered heterocycle in this case called &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyranose&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyranose&lt;/a&gt; . A five-membered heterocycle may in rare cases also be formed, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Furanose&quot; rel=&quot;nofollow noreferrer&quot;&gt;furanose&lt;/a&gt;. (From left to right: α-D-Glucopyranose, β-D-Glucopyranose, α-D-Glucofuranose, β-D-Glucofuranose; courtesy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose#Cyclic_forms&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Alpha-D-Glucopyranose.svg/100px-Alpha-D-Glucopyranose.svg.png&quot; alt=&quot;α-D-Glucopyranose&quot;&gt;  &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/82/Beta-D-Glucopyranose.svg/100px-Beta-D-Glucopyranose.svg.png&quot; alt=&quot;β-D-Glucopyranose&quot;&gt;  &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/f4/Alpha-D-Glucofuranose.svg/100px-Alpha-D-Glucofuranose.svg.png&quot; alt=&quot;α-D-Glucofuranose&quot;&gt;  &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Beta-D-Glucofuranose.svg/100px-Beta-D-Glucofuranose.svg.png&quot; alt=&quot;β-D-Glucofuranose&quot;&gt;&lt;br&gt;&#xA;Including the linear chain form they are all &lt;a href=&quot;http://goldbook.iupac.org/I03289.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;isomers&lt;/a&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Isomer&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;), as they have the same chemical formula. More specifically the relationship between $\ce{linear &amp;lt;-&amp;gt; pyranose &amp;lt;-&amp;gt; furanose}$ is called constitutional or structural isomer.&lt;br&gt;&#xA;The $\ce{\alpha-D &amp;lt;-&amp;gt; \beta-D}$ forms are diastereomers, or more specific diastereomeric conformers.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:44.917" LastActivityDate="2014-06-17T06:58:11.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13666" PostTypeId="2" ParentId="13657" CreationDate="2014-06-17T11:23:17.920" Score="27" Body="&lt;p&gt;First of all, let me state the obvious: &lt;strong&gt;&lt;em&gt;Phosphorus is awesome.&lt;/em&gt;&lt;/strong&gt;&#xA;After we got that out of the way we can focus on why. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are many different modifications of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorus&quot; rel=&quot;noreferrer&quot;&gt;phosphorus&lt;/a&gt; in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -&amp;gt; P_{red} -&amp;gt; P_{violet} -&amp;gt; P_{black}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples:&#xA;$$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$&#xA;While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphine&quot; rel=&quot;noreferrer&quot;&gt;phosphane&lt;/a&gt; $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method.&#xA;\begin{array}{rlrr}\hline&#xA;       &amp;amp;             &amp;amp;\mathrm{\%P (hyb)}  &amp;amp;\mathrm{\%O (hyb)}\\\hline&#xA;3\times&amp;amp;\ce{Bd(O-P)} &amp;amp; 24 (\ce{sp^3}) &amp;amp; 76 (\mathrm{sp^{2.3}})\\&#xA;       &amp;amp;\ce{Lp(O)}   &amp;amp;                &amp;amp; 100 (\mathrm{sp^{0.4}})\\&#xA;2\times&amp;amp;\ce{Lp(O)}   &amp;amp;                &amp;amp; 100 (\ce{p})\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cZ0Tg.png&quot; alt=&quot;phosphate lewis structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rZrE0.jpg&quot; alt=&quot;phosphate nbo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9MLqb.jpg&quot; alt=&quot;canonical orbitals of phosphane&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-07-03T15:17:17.127" LastActivityDate="2017-07-03T15:17:17.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13667" PostTypeId="1" CreationDate="2014-06-17T13:55:06.027" Score="22" ViewCount="24939" Body="&lt;p&gt;Many metals have relatively high melting and boiling points, but mercury has relatively low melting and boiling points. What are the possible reasons for this?&lt;/p&gt;&#xA;" OwnerUserId="6864" LastEditorUserId="15410" LastEditDate="2015-05-17T06:31:35.497" LastActivityDate="2019-11-14T18:08:35.747" Title="Why does mercury have low melting and boiling points?" Tags="&lt;inorganic-chemistry&gt;&lt;relativistic&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13668" PostTypeId="2" ParentId="13646" CreationDate="2014-06-17T14:10:16.043" Score="6" Body="&lt;p&gt;Cotton fiber is primarily &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot; rel=&quot;noreferrer&quot;&gt;cellulose&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ly2lg.png&quot; alt=&quot;Cellulose&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red atoms are oxygen, and the white atoms are hydrogen. Whenever you have hydrogen attached to something extremely &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;electronegative&lt;/a&gt; (such as oxygen), there is a potential for &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bonding&quot; rel=&quot;noreferrer&quot;&gt;hydrogen bonding&lt;/a&gt; to occur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSyTX.png&quot; alt=&quot;Hydrogen Bonding in Water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the electronegative oxygen atom draws electron density away from the more electropositive hydrogen atom, which results in a very polar bond. The positive charge on one molecule attracts the negative charge on another, and a very strong physical bond forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare cellulose with something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Nylon&quot; rel=&quot;noreferrer&quot;&gt;nylon&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TcLzH.png&quot; alt=&quot;Nylon&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water can hydrogen bond to cellulose, it &quot;sticks&quot; much better - we say that it is &lt;strong&gt;hydrophilic&lt;/strong&gt;. Since it can't form hydrogen bonds to nylon, we say nylon is more &lt;strong&gt;hydrophobic&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for most synthetic polymer fibers, and as a result, cotton absorbs water much more readily than other fibers, and tends to &quot;hang on&quot; to the molecules for longer - which means cotton fibers take longer to dry.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-17T14:10:16.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13669" PostTypeId="2" ParentId="9399" CreationDate="2014-06-17T14:13:00.120" Score="1" Body="&lt;p&gt;Silver and Iron are almost immiscible- 1% in the solid and liquid phase.  Steel  which is iron plus  perhaps only 0.03% carbon and some manganese. You aren't going to able to mix much steel into silver or much silver to steel as a liquids or after it freezes. Any more than 1% of either and it will just separate.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.e-journals.in/pdf/v2ns1/s153-s156.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.e-journals.in/pdf/v2ns1/s153-s156.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="2581" LastEditDate="2014-06-17T22:36:26.390" LastActivityDate="2014-06-17T22:36:26.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13670" PostTypeId="2" ParentId="5543" CreationDate="2014-06-17T14:20:06.263" Score="1" Body="&lt;p&gt;Copper doesn't need a protective layer for atmospheric corrosion- its inherently inert. It might not be a patina rather sulphide pollution from sea air caused simple corrosion and so it won't affected in other environments. Yes rust is very weakly protective- only slightly physically protective as long as its not broken but rust is rather weak and non-adhesive.   &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-06-17T14:20:06.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13671" PostTypeId="2" ParentId="13667" CreationDate="2014-06-17T14:48:54.027" Score="27" Body="&lt;p&gt;Mercury is different!  It is not as reactive as its neighbors in the Periodic Table, it doesn't conduct heat and electricity as well as other metals, and it is a liquid unlike other metals.  The electronic structure of mercury is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Xe] 4f^{14} 5d^{10} 6s^2}$&lt;/span&gt;, so the first thing we notice is that all of its orbitals are full, there are no unpaired electrons in mercury.  This helps explain mercury's reluctance to react with other materials or form bonds with other mercury atoms.  Most metal atoms share their outer electrons with other metal atoms, in fact, all of these &quot;shared&quot; metal electrons exist as a diffuse &quot;sea&quot; of electrons.  It is the extended nature of this sharing that makes metals good conductors of heat and electricity.  The fact that the metal electrons are shared provides a strong bonding interaction between metal atoms that gives metals their solid structure.  None of this applies to mercury.  With its filled shell electron configuration it is very reluctant to form bonds even with other mercury atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One additional, interesting effect makes mercury even more reluctant to share its electrons.  Special relativity suggests that for larger nuclei (and mercury with atomic number 80 is large enough for the effect to apply) electrons will begin to travel closer to the speed of light.  In turn, their mass will increase and as mass increases, orbital radius decreases.  The electron is closer to the nucleus, more attracted to the nucleus and less available for bonding.  Here is a nice &lt;a href=&quot;http://blogs.scientificamerican.com/the-curious-wavefunction/2013/07/31/what-does-mercury-being-liquid-at-room-temperature-have-to-do-with-einsteins-theory-of-relativity/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt; that explains this relativistic effect and how it applies to mercury in plain English and a bit more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two effects, filled outer electron shell and contraction of the outer orbital closer to the nucleus, combine to make mercury reluctant to form bonds, even with other mercury atoms.  It is this lack of electron bonding between mercury atoms that makes it melt and boil at such low temperatures.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2019-11-14T18:08:35.747" LastActivityDate="2019-11-14T18:08:35.747" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13672" PostTypeId="2" ParentId="857" CreationDate="2014-06-17T15:52:35.677" Score="0" Body="&lt;p&gt;I suggest you to read carefully this sheet &lt;a href=&quot;http://nj.gov/health/eoh/rtkweb/documents/fs/2845.pdf&quot; rel=&quot;nofollow&quot;&gt;nj.gov&lt;/a&gt; I'm quoting the main infos:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;WAYS OF REDUCING EXPOSURE&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Where possible, enclose operations and use local exhaust ventilation at the site of chemical release. If local exhaust ventilation or&#xA;  enclosure is not used, respirators should be worn.&lt;/li&gt;&#xA;  &lt;li&gt;Wear protective work clothing.&lt;/li&gt;&#xA;  &lt;li&gt;Wash thoroughly immediately after exposure to Trimethyltin Chloride and at the end of the workshift.&lt;/li&gt;&#xA;  &lt;li&gt;Post hazard and warning information in the work area. In addition, as part of an ongoing education and training effort, communicate all&#xA;  information on the health and safety hazards of Trimethyltin Chloride&#xA;  to potentially exposed workers.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;WORKPLACE CONTROLS AND PRACTICES&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Where possible, automatically transfer Trimethyltin Chloride from drums or other storage containers to process containers. Good WORK&#xA;  PRACTICES can help to reduce hazardous exposures. The following work&#xA;  practices are recommended:&lt;/li&gt;&#xA;  &lt;li&gt;Workers whose clothing has been contaminated by Trimethyltin Chloride should change into clean clothing promptly.&lt;/li&gt;&#xA;  &lt;li&gt;Do not take contaminated work clothes home. Family members could be exposed.&lt;/li&gt;&#xA;  &lt;li&gt;Contaminated work clothes should be laundered by individuals who have been informed of the hazards of exposure to Trimethyltin&#xA;  Chloride.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-17T15:52:35.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13673" PostTypeId="1" AcceptedAnswerId="13676" CreationDate="2014-06-17T18:46:41.283" Score="2" ViewCount="15120" Body="&lt;p&gt;According to my book, equivalent weight is defined as the ‘number of parts by weight’ of an element that interacts with 1.008 parts by weight of hydrogen. It is dimensionless while gram equivalent weight has gram as units. But in &lt;a href=&quot;https://en.wikipedia.org/wiki/Equivalent_weight&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Equivalent_weight&lt;/a&gt;, equivalent weight and gram equivalent weight are same. I’ve two questions: Which one is correct, my book or Wikipedia? And what is ‘number of parts by weight’?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:02:12.880" LastActivityDate="2019-04-01T20:59:37.840" Title="What is defined by part by weight in the definition of equivalent weight?" Tags="&lt;physical-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13674" PostTypeId="2" ParentId="5543" CreationDate="2014-06-17T18:51:14.140" Score="7" Body="&lt;p&gt;The previous answers doesn't take into account the environment: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They were operated in conditions of intense humidity/heat and were placed near the &lt;strong&gt;sea&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Near the sea means that there are certainly a lot of $\ce{Cl-}$ ions in your metal. Chloride ions are a very dangerous species for metals conservation they tends to depassivate the metal forming complex ions. In the case of copper occurs this series of reactions:&#xA;$$\ce{Cu^{+} + Cl^{-} &amp;lt;=&amp;gt; CuCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^{+} + 2Cl^{-} &amp;lt;=&amp;gt; CuCl2- + Cl- &amp;lt;=&amp;gt; CuCl3^{2-} + Cl- &amp;lt;=&amp;gt;CuCl4^{3-} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That lead to different copper corrosion product. (e.g. atacamite $\ce{Cu2(OH)3Cl}$, malachite $\ce{CuCO3.Cu(OH)2}$ (Did you see some blue-green pigment?), Cupric hydroxide $\ce{Cu(OH)2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all &lt;strong&gt;dynamic equilibria&lt;/strong&gt; so your copper with dry-wet, hot-cold cycle will continuously object of degradation until chloride ions are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use distilled water with cotton to clean the salts deposit, dry quickly where you clean.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.benthamsciencepublisher.com/open/tocorrj/articles/V002/71TOCORRJ.pdf&quot; rel=&quot;nofollow&quot;&gt;The Open Corrosion Journal, 2009, 2, 71-76&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0010938X02002573&quot; rel=&quot;nofollow&quot;&gt;Corrosion Science 46 (2004) 109–135&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2094" LastEditorDisplayName="user467" LastEditDate="2015-07-18T21:29:40.110" LastActivityDate="2015-07-18T21:29:40.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13675" PostTypeId="2" ParentId="13657" CreationDate="2014-06-17T20:11:01.233" Score="3" Body="&lt;p&gt;Thank you very much, both of you, for some very detailed replies. I'm going to try to synthesise them into a single answer, which I hope will be accurate. Here is how I understand it, armed with the new knowledge contained within the posts above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the phosphate ion we have a central Phosphorus atom, with five valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is bonded to four oxygen atoms, which have six valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five P electrons plus 4 times 6 O electrons gives 29 electrons. We also have a charge of 3- indicating that we need to take into account 3 additional electrons, giving us 32 electrons overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could model this situation by proposing that the P atom exists at the centre, with four O atoms bonded to it. One O is double bonded to the P and the other three Os are single bonded, giving 10 electrons to the P (which it is happy with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the double-bonded O we would have 2 lone pairs of electrons (makes 4) and a further 4 electrons from the double bond. Oxygen has 6 valence electrons and so the 2 lone pairs are its own electrons and its other 2 valence electrons form the double bond with two of the P electrons. The double-bonded O has eight electrons and is happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of each single-bonded O we would have 3 lone pairs (presumably the oxygen's own six valence electrons?) plus an extra electron (presumably from elsewhere? &lt;strong&gt;I don't really understand this&lt;/strong&gt;) which participates in a single bond with a P electron. This gives us eight electrons, which the oxygen is happy with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the model shown in the wikipedia drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dXYQl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we draw the Lewis structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2EIn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see all of this put together and we can also see that, while the double-bonded O is neutral, having six electrons assigned to it, the three single bonded O atoms have seven electrons assigned and therefore carry a -1 charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no reason, however, why we should select the top O atom as the one to have a double-bond. This is not at all what we find in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can build a simplified model of reality, where the double-bond (and its electrons) is exchanged between each O atom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kbm6m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And where there is no double-bond at all:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cZ0Tg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this, however, is really nothing more than an extreme simplification. The reality is a hybrid. It is possible to propose that we have four sigma bonds with four sp&lt;sup&gt;3&lt;/sup&gt; hybridised orbitals from the phosphorus and a fifth delocalised pi bond with the d orbital of the phosphorus. This gives our P 10 electrons and each O 8 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the reality seems to be that, in fact, the d orbital is not involved at all and instead we have four sigma bonds formed by sp&lt;sup&gt;3&lt;/sup&gt; hybridised orbitals from the phosphorus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's what I just don't understand. When I drew the Lewis structure, I added three extra electrons from somewhere. This would make sense if, as martin says, each oxygen has three lone pairs, we can imagine a sea of the five P valence electrons and these extra three electrons spread evenly with two forming each sigma bond. This gives each oxygen eight electrons and it would give the P eight electrons too. The higher electronegativity of Oxygen means that the O atoms attract the electrons more strongly and each atom receives a negative charge, whereas the P attracts them less strongly and receives a positive charge. I'm really not sure I've understood the arrangement of electrons though, I would like to know how the electrons are apportioned in the ion, where they came from and how the overall charge comes about.&lt;/p&gt;&#xA;" OwnerUserId="7147" LastActivityDate="2014-06-17T20:11:01.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13676" PostTypeId="2" ParentId="13673" CreationDate="2014-06-17T20:51:23.123" Score="2" Body="&lt;p&gt;The term &quot;parts by weight&quot; is how you would specify in a recipe the relative proportions of the ingredients. So if the recipe called for one part by weight of flour and two parts by weight of sugar, you could use 1 ounce of flour and two ounces of sugar (to make a small batch) or 1/2 pound of flour and 1 pound of sugar to make a large batch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;recipe&quot; for water calls for exactly 1.008 parts by weight of hydrogen and 8.000 parts by weight of oxygen. If you use too much of either ingredient, there will be an excess of it after the reaction is run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Parts by volume&quot; would apply if you were combining liquids and measuring them by volume rather than by weight.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-06-17T20:51:23.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13677" PostTypeId="1" AcceptedAnswerId="13681" CreationDate="2014-06-17T21:36:56.447" Score="0" ViewCount="2068" Body="&lt;p&gt;Just a little question, I was taught that the electronegativity of potassium[K] is 1, but I recently discovered most internet resources say 0.82. &lt;a href=&quot;http://www.lenntech.com/periodic-chart-elements/electronegativity.htm&quot; rel=&quot;nofollow&quot;&gt;Some even say both&lt;/a&gt;. &lt;strong&gt;What is the actual electronegativity of potassium, 0.82 or 1?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5567" LastEditorUserId="7448" LastEditDate="2016-08-14T22:03:03.640" LastActivityDate="2016-08-15T18:03:21.880" Title="Electronegativity of Potassium (Pauling)" Tags="&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13678" PostTypeId="1" AcceptedAnswerId="13689" CreationDate="2014-06-17T21:41:20.397" Score="0" ViewCount="250" Body="&lt;p&gt;I have a sheet of thin unperforated teflon that I need to perforate by tomorrow for a lab I work in. None of my coworkers know how to perforate the sheet. Does anyone know how I can do this using basic tools from my house?&lt;/p&gt;&#xA;" OwnerUserId="4363" LastEditorUserId="4945" LastEditDate="2014-06-18T07:15:53.277" LastActivityDate="2014-06-18T07:15:53.277" Title="How do you perforate a teflon sheet?" Tags="&lt;materials&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13679" PostTypeId="1" CreationDate="2014-06-17T21:44:37.200" Score="2" ViewCount="953" Body="&lt;p&gt;I'm trying to figure out the relationship between the molecular weight distribution of a polyethylene plastic and its strength and heat sealability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The molecular weight distribution is described as being narrow, which apparently is helpful in terms of strength and heat sealability).&lt;/p&gt;&#xA;" OwnerUserId="6866" LastEditorUserId="4945" LastEditDate="2014-06-18T06:44:44.580" LastActivityDate="2015-01-28T09:35:52.377" Title="What is relationship between molecular weight distribution of PE and its strength?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;materials&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13680" PostTypeId="1" AcceptedAnswerId="13682" CreationDate="2014-06-17T21:50:54.997" Score="1" ViewCount="253" Body="&lt;p&gt;I was told that dd-water will superheat when heated up in a microwave because of a lack of ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Doesn't water auto-ionize? I asked this; the answer was it auto-ionizes &quot;too slowly&quot; in dd-water to prevent superheating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) This begs the question: what are the kinetics of the auto-ionization of water? What's the rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) This also begs the question - what is superheating? What does ions have anything to do with superheating? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-17T22:27:24.613" Title="Double Deionized Water and Superheating" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13681" PostTypeId="2" ParentId="13677" CreationDate="2014-06-17T22:03:03.340" Score="5" Body="&lt;p&gt;It depends which electronegativity scale you choose to use.  There are a lot of different scales for comparing electronegativities of the elements.  I've included a &lt;a href=&quot;http://www.meta-synthesis.com/webbook/36_eneg/electroneg.html&quot; rel=&quot;nofollow&quot;&gt;list&lt;/a&gt; of 11 of them below if you're interested.  In any case here are the electronegativities reported for potassium using 5 different electronegativity scales.  All values are in &quot;Pauling units&quot; so that the numbers can be compared.  The Pauling scale is the most commonly used, so 0.82 would be a reasonable estimate for the electronegativity of potassium.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pauling 0.82&lt;/li&gt;&#xA;&lt;li&gt;Sanderson 0.45&lt;/li&gt;&#xA;&lt;li&gt;Allred Rochow 0.91&lt;/li&gt;&#xA;&lt;li&gt;Mulliken-Jaffe 0.73&lt;/li&gt;&#xA;&lt;li&gt;Allen 0.734&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the Pauling scale, electronegativities range from around 0.7 to 3.98 (fluorine), so as expected, potassium is at the electropositive end.  None of these scales report a value of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;List of 11 Different Electronegativity Scales&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pauling Scale: (1932)  Obtains values by thermochemical methods&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mulliken Relation: (1934)  Defines a relation that depends upon the&#xA;orbital characteristics of an atom in a molecule. Mulliken&#xA;electronegativity is the numerical average of the ionisation&#xA;potential and electron affinity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Gordy Scale: (1946)  Defines electronegativity in terms of the&lt;br&gt;&#xA;effective nuclear charge and the covalent radius. (Zeff)e/r         &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Walsh Scale: (1951)  Relates electronegativity to stretching force&lt;br&gt;&#xA;constants of the bonds of an atom to a hydrogen atom.&lt;/li&gt;&#xA;&lt;li&gt;Huggins Scale: (1953) Alternative to Pauling's thermochemical&lt;br&gt;&#xA;procedure&lt;/li&gt;&#xA;&lt;li&gt;Sanderson Scale: (1955)  The ratio of the average electron density of&#xA;an atom to that of a hypothetical &quot;inert&quot; atom having the same number of electrons. This ratio is a measure of the relative compactness of the atom&lt;/li&gt;&#xA;&lt;li&gt;Allred-Rochow Scale: (1958)  Defines electronegativity in terms of the effective nuclear charge and covalent radius. Like the Gordy scale but uses (Zeff)e/r2&lt;/li&gt;&#xA;&lt;li&gt;Jaffe Scale: (1962)  Uses the electronegativity of orbitals rather&lt;br&gt;&#xA;than atoms to develop group electronegativities for molecular fragments (&lt;em&gt;e.g.&lt;/em&gt; $\ce{CH3}$ vs $\ce{CF3}$) that take into account the charge of a group, the effects of substituents, and the hybridization of the bonding orbital&lt;/li&gt;&#xA;&lt;li&gt;Phillips Scale: (1968)  Defines electronegativity in terms of the&lt;br&gt;&#xA;dielectric properties of atoms in a given valence state&lt;/li&gt;&#xA;&lt;li&gt;Martynov &amp;amp; Batsanov Scale: (1980) Obtained by averaging the&lt;br&gt;&#xA;successive ionisation energies of an element's valence electrons&lt;/li&gt;&#xA;&lt;li&gt;Allen CE Scale: (1992) Configuration energy (CE), the average&lt;br&gt;&#xA;one-electron valence shell energy of the ground-state free atom, is&lt;br&gt;&#xA;used to quantify metal-covalent-ionic bonding&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-08-15T18:03:21.880" LastActivityDate="2016-08-15T18:03:21.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13682" PostTypeId="2" ParentId="13680" CreationDate="2014-06-17T22:27:24.613" Score="4" Body="&lt;p&gt;This has a lot to do with nucleation.  Just like crystallization and initial rain drop formation, for boiling to occur an initial site within the solution for a bubble of vapor to form on is required. So if you have very clean water that has been deaerated in a highly polished vessel there is an increased chance for superheating to occur due to the reduction in potential nucleation sites.  Tap water can superheat too, it is just less likely to do so because it typically contains more impurities that can serve as nucleation sites. This why boiling chips are commonly used in lab distillations.  They provide a site for vapor nucleation and usually prevent superheating of the material to be distilled.  Introducing other sites for nucleation, such as dissolved bubbles, solid contaminants, rough spots on the wall of the container all serve to lessen the probability of superheating.  I'm not sure what ions would have to do with the superheating process other than they are often representative of microscopic solid impurities that can provide that surface for the initial nucleation event.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-17T22:27:24.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13684" PostTypeId="2" ParentId="13592" CreationDate="2014-06-18T02:41:03.680" Score="10" Body="&lt;p&gt;I found this tool named &lt;a href=&quot;http://abhimanyusirohi.github.io/ThumbFish&quot; rel=&quot;nofollow noreferrer&quot;&gt;thumbfish&lt;/a&gt; (Windows only). It can display thumbnails for variety of formats and is free&lt;/p&gt;&#xA;" OwnerUserId="6834" LastEditorUserId="5026" LastEditDate="2018-07-06T07:59:58.283" LastActivityDate="2018-07-06T07:59:58.283" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13685" PostTypeId="1" CreationDate="2014-06-18T02:56:56.397" Score="1" ViewCount="1163" Body="&lt;p&gt;What really is a radical? For example, when $\ce{Cl2}$ receives enough energy to dissociate the bond and turn into $\ce{ 2Cl.}$, the two chlorine atoms each carry one of the electrons that form the bond. What is the difference between the $\ce{Cl}$ radical and the original $\ce{Cl}$? They have exactly the same number of electrons. And how do they act as a chain carrier?&lt;/p&gt;&#xA;" OwnerUserId="4591" LastEditorUserId="4945" LastEditDate="2014-06-18T04:54:56.477" LastActivityDate="2016-03-11T22:20:35.493" Title="What are radicals and how do they act in radical chain reactions?" Tags="&lt;reaction-mechanism&gt;&lt;radicals&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13686" PostTypeId="2" ParentId="13685" CreationDate="2014-06-18T03:39:17.950" Score="1" Body="&lt;p&gt;A radical is a species with an unpaired electron. The chlorine molecule has a covalent bond joining the 2 chlorine atoms, there are no free or unpaired electrons. Because of this, the chlorine molecule is a stable species and energy must be put into the system to break the bond. If you break that bond you now have a chlorine atom with an unpaired electron. Such a radical with the unpaired electron is a very reactive species because of the unpaired electron. When such a radical reacts with a double bond a new free radical is generated at the other end of the double bond and chain propagation continues. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-18T12:28:20.967" LastActivityDate="2014-06-18T12:28:20.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13687" PostTypeId="2" ParentId="6321" CreationDate="2014-06-18T05:05:02.830" Score="1" Body="&lt;p&gt;As it is already pointed out in the comments, what you are learning for is hybridisation and some basic MO theory. Do not go too far into the MO direction, because you will get into somewhere completely different place, and you will hard time to make the connections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So aftre you reviewed hybridization: Once you can identify in a molecule the sp1, sp2 and sp3 carbons, you should be able to orient the hybrid orbitals. For example in the case of sp2 carbon of the lower image, it is in the HHCC plane. In case of sp1 there are two, for sp2 one un-hybridized p orbitals. These unhybridized orbitals should be perpendicular to the hybridized ones: perpendicular to the axis of sp1 or the plane of sp2. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-06-18T05:05:02.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13688" PostTypeId="1" AcceptedAnswerId="16603" CreationDate="2014-06-18T06:28:00.183" Score="11" ViewCount="845" Body="&lt;p&gt;I originally posted this on EE.SE but might be better here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.batteryshowdown.com/results-lo.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery showdown&lt;/a&gt; compared the mAh and mWh capacity of different AA batteries discharged at 200mA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the relative capacities of each battery type (within the same chemistry type) be the same at much lower current draw, e.g. 1mA? If not, why not? Are there things battery manufacturers do to 'tune' batteries to a certain discharge range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I've noted that as current goes down alkaline batteries approach lithium batteries in capacity. Is there a current below which alkaline batteries have more than lithium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from battery showdown:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ldxZY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6869" LastActivityDate="2014-09-22T18:35:41.450" Title="Is greater relative AA battery capacity at high currents indicative of greater capacity at low currents?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13689" PostTypeId="2" ParentId="13678" CreationDate="2014-06-18T07:03:02.840" Score="1" Body="&lt;p&gt;You should contact a specialized company, however you should also provide more infos about the diameter and the number of the holes you want. This is an excerpt from &lt;a href=&quot;http://www.fedtech.com/perforated-sheets.html&quot; rel=&quot;nofollow&quot;&gt;FedTech&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;laser cutting is the method of choice for small perforation, as laser&#xA;  cutting has a smaller, tighter cutting tolerance and thus has the&#xA;  capability to create tiny, intricate perforations. However, if you&#xA;  prefer, abrasive waterjet cutting can be used as well but for larger&#xA;  perforation (.030&quot; and up, respectively). If you prefer to use&#xA;  non-abrasive waterjet cutting, you can create perforations up to .004&quot;&#xA;  in diameter, however this form of waterjet cutting is restricted in&#xA;  cutting certain types of material. With non-abrasive waterjet cutting,&#xA;  you can cut up to about 12&quot; thick from the following: plastic, rubber,&#xA;  foam, composites, silicone, foam, felt, cork, Teflon (PTFE), neoprene,&#xA;  and other soft materials&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-06-18T07:03:02.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13692" PostTypeId="1" AcceptedAnswerId="13693" CreationDate="2014-06-18T12:55:30.633" Score="1" ViewCount="23642" Body="&lt;p&gt;I don't know why my solutions manual says these are enantiomers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvOw0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fisher projections I drew suggest that these are identical compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SC6gu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-19T08:05:17.950" Title="Enantiomers or Identical Compounds?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13693" PostTypeId="2" ParentId="13692" CreationDate="2014-06-18T13:22:02.377" Score="4" Body="&lt;p&gt;The solutions manual is correct, they are enantiomers.  The compound on the left is (R)-2-chloro-(S)-3-butanol and the compound on the right is (S)-2-chloro-(R)-3-butanol.  Try to use the &quot;R\S&quot; nomenclature system on the 2 chiral carbons, that should help figuring out whether they are identical [(R,R and R,R), (S,S and S,S), (R,S and R,S), (S,R and S,R)] enantiomers [(R,R and S,S), (R,S and S,R), (S,R and R,S)] or diastereomers [(R,R and R,S), (R,R and S,R), (S,S and S,R), (S,S and R,S)].&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-18T13:30:07.207" LastActivityDate="2014-06-18T13:30:07.207" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13694" PostTypeId="2" ParentId="6244" CreationDate="2014-06-18T13:37:24.750" Score="0" Body="&lt;p&gt;Pi bonds involve sideways overlap while sigma bond involves head on or axial overlap. Axial overlaps have higher degree of overlapping than sideways overlap.&lt;/p&gt;&#xA;" OwnerUserId="6872" LastEditorUserId="4734" LastEditDate="2014-06-18T22:57:52.243" LastActivityDate="2014-06-18T22:57:52.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13695" PostTypeId="2" ParentId="13685" CreationDate="2014-06-18T13:42:17.497" Score="4" Body="&lt;p&gt;free radical is a molecule with one or more unpaired electrons. Despite this definition, d-element compounds with unpaired electrons on d-element, like $\ce{CrCl3.6H2O}$ are not usually considered as free radicals because their unpaired electrons are relatively unreactive. Free radicals are especially reactive, because their reactions usually go through activated complex with relatively small energy. The reasons behind it are not that easy do describe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general most reaction with free radicals in organic chemistry falls into two categories: one, where free radical reacts with exposed hydrogen/heteroatom and one where it reacts with exposed multiple bond. In both cases, orbitals of interests are three orbitals of the radical in consideration and two atoms of the organic molecules. MO theory says, that such interaction results in three MOs: bonding, nonbonding and antibonding. With three electrons occupying this orbitals, the energy of the system is mostly same as for the radical and the molecule separated, so the activation energy of the reaction is near zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of nucleophilic/electrophilic reactions in , the interaction in activated complex are similar, with two electrons on bonding and zero or two electrons on nonbonding orbitals. However, in case of aliphatic derivatives, the reaction usually goes throw sterically hindered state with 5-coordinated carbon  and/or requires generation of hi-energy particles that are not preserved after the molecule reacts. In cases when hi-energy intermediate may be preserved (for example, ionic polymerization), and/or reaction goes through sterically unhindered intermediates (for example, nucleophilic substitution in acyl halides), and/or one of the reagents easily produces active intermediates (for example, nitrating mixture in aromatic substitution) reactions without free radicals as intermediates may proceed swiftly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of chain reaction for aliphatic radical substitution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2 + CH4 -&amp;gt; CH3Cl + HCl}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;in case if $\ce{CH4}$ is presented in huge excess, resulting in reduced rate of multichlorination, the reaction proceeds as following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chain initiations: $\ce{Cl2 -&amp;gt; 2.Cl}$&lt;/li&gt;&#xA;&lt;li&gt;Chain growth 1: $\ce{.Cl + CH4 -&amp;gt; HCl + .CH3}$ &lt;/li&gt;&#xA;&lt;li&gt;Chain growth g 2: $\ce{.CH3 + Cl2 -&amp;gt; CH3Cl + .Cl}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In case of ionic polymerization&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chain initiation: $\ce{H2SO4 + CH2=CH-Ph -&amp;gt; HSO4- + H-CH_2-\stackrel{+}{C}H-Ph}$&lt;/li&gt;&#xA;&lt;li&gt;Chain growth: $\ce{H-(CH2-CHPh-)_{n}-CH2-\stackrel{+}{C}H-Ph + CH2=CH-Ph -&amp;gt;} $  $\ce{H-(CH2-CHPh-)_{n + 1}-CH2-\overset{+}{C}H-Ph}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both reactions proceed smoothly at room temperature if source of active particles is available (UV-light for breaking $\ce{Cl-Cl}$ bond for the former and concentrated sulfuric acid for the latter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare nucleophilic substitution $\ce{KOH + CH3I -&amp;gt; CH3OH + KI}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{KOH -&amp;gt; K+ + OH-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{OH- + CH3-I -&amp;gt; (HO-CH3-I)- -&amp;gt; HO-CH3 + I-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{I- + K+ -&amp;gt; KI}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you see, the reaction proceeds through sterically hindered state with 5-coordinated carbon. Compare to acyl halide case $\ce{KOH + CH3COCl -&amp;gt; CH3COOH + KCl}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{KOH -&amp;gt; K+ + OH-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{OH- + CH3-(C=O)-Cl -&amp;gt; CH3-C(O^{-})(OH)Cl -&amp;gt; CH3-COOH + Cl-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cl- + K+ -&amp;gt; KCl}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;in this case the reaction proceeds through sterically unhindered state with 4-coordinated carbon, so acetylchloride reacts with water at room temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR : free radicals are particles with unpaired electron, readily available for covalent interactions, they reacts with molecules with easy because they are ones of big family of 'active particles' with low energy of activation of many reactions, and, unlike many other active species, they usually regenerate themselves in reaction, allowing chain reactions to proceed. Other active particles or plainly active substrates also exists, so free radicals are not special. The fact of presence of unpaired electron does not necessary make the particle an active free radical, many relatively inactive free radicals exists, like $\ce{NO}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading: MO theory (case of 3-atomic interactions) and kinetics (case of chain reactions) parts of your favorite advance chemistry textbook, parts on reaction mechanisms and active particles of your favorite advanced organic chemistry textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite above answer based exclusively on organic chemistry examples, everything stated above is true for inorganic chemistry. However the set of possible kinds of elementary acts, active particles and interactions is much wider than for basic organic chemistry, leading to many special and unique cases and consequently to much less predictable reactions in complicated cases.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="24052" LastEditDate="2016-03-11T22:20:35.493" LastActivityDate="2016-03-11T22:20:35.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13696" PostTypeId="1" AcceptedAnswerId="13699" CreationDate="2014-06-18T14:33:50.020" Score="1" ViewCount="1369" Body="&lt;p&gt;I have a couple of basic questions on stereochemistry that confused me. Please help me understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why does a molecule become achiral if it has a plane of symmetry even though it has steriocentes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) When working with Fischer Projections, they say you can rotate a Fischer projection 180 degrees, but not by 90 degrees. I don't understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, can you please suggest me a good resource wherein I can learn stereochemistry (not to very advanced level)?&lt;/p&gt;&#xA;" OwnerUserId="4995" LastActivityDate="2018-09-13T17:31:29.030" Title="Some questions on stereochemistry" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13697" PostTypeId="2" ParentId="2881" CreationDate="2014-06-18T14:43:57.620" Score="52" Body="&lt;p&gt;To understand the difference between kinetic and thermodynamic stability, you first have to understand &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy_surface&quot; rel=&quot;noreferrer&quot;&gt;potential energy surfaces&lt;/a&gt;&lt;/strong&gt;, and how they are related to the state of a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A potential energy surface is a representation of the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;noreferrer&quot;&gt;potential energy&lt;/a&gt;&lt;/strong&gt; of a system as a function of one or more of the other dimensions of a system. Most commonly, the other dimensions are spatial. Potential energy surfaces for chemical systems are usually very complex and hard to draw and visualize. Fortunately, we can make life easier by starting with simple 2-d models, and then extend that understanding to the generalized N-d case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we will start with the easiest type of potential energy to understand: gravitational potential energy. This is easy for us because we live on Earth and are affected by it every day. We have developed an intuitive sense that things tend to move from higher places to lower places, if given the opportunity. For example, if I show you this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C94wb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can guess that the rock is eventually going to roll downhill, and eventually come to rest at the bottom of the valley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you also intuitively know that it is not going to move unless something moves it. In other words, it needs some &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_energy&quot; rel=&quot;noreferrer&quot;&gt;kinetic energy&lt;/a&gt;&lt;/strong&gt; to get going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could make it even harder for the rock to get moving by changing the surface a little bit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xk3ys.png&quot; alt=&quot;2d potential surface with local minimum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it is really obvious that the rock isn't going anywhere until it gains enough kinetic energy to overcome the little hill between the valley it is in, and the deeper valley to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call the first valley a &lt;strong&gt;local minimum&lt;/strong&gt; in the potential energy surface. In mathematical terms, this means that the first derivative of potential energy with respect to position is zero: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm dE}{\mathrm dx} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the second derivative is positive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm d^2E}{\mathrm dx^2} \gt 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the slope is zero and the shape is concave up (or convex).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper valley to the right is the &lt;strong&gt;global minimum&lt;/strong&gt; (at least as far as we can tell). It has the same mathematical properties, but the &lt;em&gt;magnitude&lt;/em&gt; of the energy is lower – the valley is deeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put all of this together, (and can tolerate a little &lt;a href=&quot;http://www.merriam-webster.com/dictionary/anthropomorphize&quot; rel=&quot;noreferrer&quot;&gt;anthropomorphization&lt;/a&gt;) you could say that the rock &lt;em&gt;wants&lt;/em&gt; to get to the global minimum, but whether or not it &lt;em&gt;can&lt;/em&gt; get there is determined by the amount of kinetic energy it has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It needs at least enough kinetic energy to overcome all of the local maxima along the path between its current local minimum and the global minimum.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't have enough kinetic energy to move out of its current position, we say that it is &lt;em&gt;kinetically stable&lt;/em&gt; or &lt;em&gt;kinetically trapped.&lt;/em&gt; If it has reached the global minimum, we say it is &lt;em&gt;thermodynamically stable.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply this concept to chemical systems, we have to change the potential energy that we use to describe the system. Gravitational potential energy is too weak to play much of a role at the molecular level. For large systems of reacting molecules, we instead look at one of several &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;noreferrer&quot;&gt;thermodynamic potential energies&lt;/a&gt;.&lt;/strong&gt; The one we choose depends on which &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/State_variable&quot; rel=&quot;noreferrer&quot;&gt;state variables&lt;/a&gt;&lt;/strong&gt; are constant. For macroscopic chemical reactions, there is usually a constant number of particles, constant temperature, and either constant pressure or volume (NPT or NVT), and so we use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Gibbs Free Energy&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; for NPT systems) or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Helmholtz_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Helmholtz Free Energy&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; for NVT systems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these is a thermodynamic potential under the appropriate conditions, which means that it does the same thing that gravitational potential energy does: it allows us to predict where the system will go, if it gets the opportunity to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For kinetic energy, we don't have to change much - the main difference between the kinetic energy of a rock on a hill and the kinetic energy of a large collection of molecules is how we measure it. For single particles, we can measure it using the velocity, but for large groups of molecules, we have to measure it using temperature. In other words, increasing the temperature increases the kinetic energy of all molecules in a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can describe the thermodynamic potential energy of a system in different states, we can figure out whether a transition between two states is thermodynamically favorable – we can calculate whether the potential energy would increase, decrease, or stay the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at all accessible states and decide that the one we are in has the lowest thermodynamic potential energy, then we are in a &lt;strong&gt;thermodynamically stable&lt;/strong&gt; state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example using methane gas, we can look at Gibbs free energy for the reactants and products and decide that the products are more thermodynamically stable than the reactants, and therefore methane gas in the presence of oxygen at 1 atm and 298 K is thermodynamically &lt;em&gt;unstable&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you would have to wait a very long time for methane to react without some outside help. The reason is that the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state&quot; rel=&quot;noreferrer&quot;&gt;transition states&lt;/a&gt;&lt;/strong&gt; along the lowest-energy reaction path have a much higher thermodynamic potential energy than the average kinetic energy of the reactants. The reactants are &lt;em&gt;kinetically trapped&lt;/em&gt; - or stable just because they are stuck in a local minimum. The minimum amount of energy that you would need to provide in the form of heat (a lit match) to overcome that barrier is called the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy&quot; rel=&quot;noreferrer&quot;&gt;activation energy&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can apply this to lots of other systems as well. One of the most famous and still extensively researched examples is &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass&quot; rel=&quot;noreferrer&quot;&gt;glasses&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glasses are interesting because they are examples of kinetic stability in physical phases. Usually, phase changes are governed by thermodynamic stability. In glassy solids, the molecules would have a lower potential energy if they were arranged in a crystalline structure, but because they don't have the energy needed to get out of the local minimum, they are &quot;stuck&quot; with a liquid-like disordered structure, even though the phase is a solid.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorDisplayName="user7951" LastEditDate="2019-07-27T18:47:01.933" LastActivityDate="2019-07-27T18:47:01.933" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13698" PostTypeId="2" ParentId="13696" CreationDate="2014-06-18T15:12:04.180" Score="2" Body="&lt;p&gt;If you rotate the substituents on a Fischer projection by 90 degrees you change the stereochemistry of each substituent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The up/down positions on a Fischer projection indicate substituents projected &lt;strong&gt;away&lt;/strong&gt; from you. The left and right positions indicate substituents projected &lt;strong&gt;toward&lt;/strong&gt; you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rotated tetrahedral molecule to help you see what I mean by the substituents are projected away/toward you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4P4tC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can represent toward using a solid wedge and away using a hatched wedge: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kEqI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-18T19:02:11.427" LastActivityDate="2014-06-18T19:02:11.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13699" PostTypeId="2" ParentId="13696" CreationDate="2014-06-18T17:25:24.850" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Why does a molecule become achiral if it has a plane of symmetry&#xA;  even though it has stereocenters?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If a plane of symmetry bisects a molecule such that one half of the bisected molecule is the mirror image of the other half of the bisected molecule, then the entire molecule &lt;em&gt;must&lt;/em&gt; be superimposable on its mirror image, it must be achiral.  Said differently, if mirror symmetry exists within a molecule, then the mirror image of the entire molecule will be superimposable upon the original drawing of the entire molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a molecule ($\ce{A}$) with 2 chiral groups ($\ce{G_{R}}$ and ($\ce{G_{S}}$).  There is a plane of symmetry that bisects $\ce{A}$, it is perpendicular to the plane of the screen and contains the central carbon atom and the 2 methyl groups.  Now, when we draw the mirror image of $\ce{A}$ (let's call it ($\ce{\bar{A}}$), we find that it is superimposable upon the original drawing of $\ce{A}$ by a simple 180-degree rotation.&#xA;Even though $\ce{A}$ contains 2 chiral substituents, the molecule as a whole is achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mk3Oz.jpg&quot; alt=&quot;Illustration depicting a molecule, A, as superimposable on its mirror image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41556" LastEditDate="2018-09-13T17:31:29.030" LastActivityDate="2018-09-13T17:31:29.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13700" PostTypeId="1" AcceptedAnswerId="13707" CreationDate="2014-06-18T22:21:14.173" Score="1" ViewCount="2012" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mFlfU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xRENg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help me walk through the logic here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) HF is a weaker acid than HCl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) This is as far as I can understand. How does sigma star affect the acidities of the binary halogen acids? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If a covalent bond was formed, you would expect a decreasing acidity&#xA;  down the periodic table of the halogens, because the energy of the σ∗&#xA;  would increase forming a less attractive LUMO (lowest unoccupied&#xA;  molecular orbital), which would result in less hydrogen atoms being&#xA;  pulled of the halide (i.e. lower acidity, higher pKa).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are some questions about the above excerpt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Covalent bond formed between what? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why would the sigma star orbital be populated if the hydrogen were removed and a bond formed in place of it? Or is a bond going to be formed with the hydrogen still present? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://usflearn.instructure.com/courses/986898/files/31117039&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://usflearn.instructure.com/courses/986898/files/31117039&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LuuMN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="12860" LastEditDate="2015-11-13T19:22:35.083" LastActivityDate="2015-11-13T19:22:35.083" Title="Rationalizing Strength of Binary Halogen Acids with MO Theory" Tags="&lt;acid-base&gt;&lt;molecular-orbital-theory&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13701" PostTypeId="1" AcceptedAnswerId="13729" CreationDate="2014-06-18T22:25:53.727" Score="2" ViewCount="4915" Body="&lt;p&gt;From EN alone we might expect $\ce{HCl}$ to be more polar but a book says that the bent structure of $\ce{H2S}$ gives rise to its polarity - which is also apparently more than $\ce{HCl}$. Why? Does this have to do with the lone pairs? Lone pairs supply electron density, right? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T00:21:16.357" LastActivityDate="2015-06-28T00:22:02.910" Title="Why is H2S more polar than HCl?" Tags="&lt;bond&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13702" PostTypeId="2" ParentId="13692" CreationDate="2014-06-19T00:51:27.713" Score="2" Body="&lt;p&gt;In the following figure, everything on the left is identical and everything on the right is identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5uJB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Fischer_Projections&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; for how to draw Fischer Projections.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-19T00:51:27.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13703" PostTypeId="1" AcceptedAnswerId="13705" CreationDate="2014-06-19T00:54:00.760" Score="2" ViewCount="1177" Body="&lt;p&gt;Like &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Simple_distillation_apparatus.svg&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/9c/Simple_distillation_apparatus.svg/500px-Simple_distillation_apparatus.svg.png&quot; alt=&quot;distillation aparatus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two medium sized glass flasks. I understand I need a T-shaped adapter from the first flask that will connect the thermometer and the condenser. So, I also need the condenser and a L-type adapter that will connect the last flask with the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the condenser &lt;em&gt;really&lt;/em&gt; necessary for a completely beginner set up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I really need the adapters? I'm worried they will not fit. How high are the costs for these parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any replies. I know some theoretical chemistry stuff, but with this glassware I've got no idea what I'm doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thanks guys, I accidentally found a very nice chemical equipment store and a nice lady there sold me all the parts. I only didn't get the thermometer because it was way too expensive (and I can make something similar myself). It cost 80 eur. Flask (1l), adapter, condenser, another adapter, mount for the flask and rubber pipes. The thermometer was extra 35.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="402" LastEditDate="2015-11-30T02:52:38.193" LastActivityDate="2015-11-30T02:52:38.193" Title="How do I make a simple home distillation device?" Tags="&lt;home-experiment&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13704" PostTypeId="1" AcceptedAnswerId="13715" CreationDate="2014-06-19T01:47:27.780" Score="1" ViewCount="340" Body="&lt;p&gt;What is the fire point of sucrose with an average particle diameter of 60 micrometres? I can't seem to find any documentation on this, online.&lt;/p&gt;&#xA;" OwnerUserId="5740" LastEditorUserId="4945" LastEditDate="2014-07-29T10:00:33.207" LastActivityDate="2014-07-29T10:00:33.207" Title="What is the fire point of sucrose with an average particle diameter of 60 micrometres?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13705" PostTypeId="2" ParentId="13703" CreationDate="2014-06-19T02:20:35.050" Score="1" Body="&lt;p&gt;This depends on what you need this setup to accomplish, as the amount of solutions that this setup can manipulate vary greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic consensus is that if you would want an &lt;em&gt;optimal&lt;/em&gt; apparatus you would need:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Round Bottom Flask&lt;/li&gt;&#xA;&lt;li&gt;Thermometer&lt;/li&gt;&#xA;&lt;li&gt;A Column&lt;/li&gt;&#xA;&lt;li&gt;Condenser (with Water In/Water Out nozzles)&lt;/li&gt;&#xA;&lt;li&gt;Collection Flask&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6861" LastEditorUserId="17952" LastEditDate="2015-11-30T01:35:55.477" LastActivityDate="2015-11-30T01:35:55.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13707" PostTypeId="2" ParentId="13700" CreationDate="2014-06-19T02:51:10.900" Score="4" Body="&lt;p&gt;Some initial comments on the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Bonding in the $\ce{HX}$ series is a varying mix of ionic and covalent bonding.  $\ce{HF}$ has the largest covalent bonding component and $\ce{HI}$ the smallest.  This trend goes against what electronegativities alone would suggest, but is consistent with what overlap integrals (e.g. overlap between a $\ce{1s}$ orbital and a $\ce{2sp^3}$ orbital is better than overlap between a $\ce{1s}$ orbital and a $\ce{5sp^3}$ orbital) would predict.  Therefor, the latter effect appears to predominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) An MO diagram depicts only the covalent portion of the bonding situation.  Therefor it is unlikely to accurately predict acid strength over the entire $\ce{HX}$ series where the (covalent\ionic) contribution to bonding is varying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The MO diagram should probably have hybridized the $\ce{Cl}$ AO's prior to mixing them with the $\ce{H}$ AO.  That is, I'd guess that the covalent portion of the $\ce{HX}$ bond is between a $\ce{H}$ $\ce{1s}$ orbital and a $\ce{Cl}$ $\ce{3sp^3}$ orbital, not a $\ce{Cl}$ $\ce{3p}$ orbital.  This could be assessed by looking at the photoelectron spectrum of $\ce{HCl}$ and analyzing it (e.g. how many chlorine lone pair signals?) in terms of hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) $\ce{\sigma^{\ast}}$ orbitals don't really need to be brought in to the discussion since they are unoccupied throughout the $\ce{HX}$ dissociation process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Increased $\ce{HX}$ acidity correlates inversely with $\ce{HX}$ bond strength. $\ce{HF}$ with the best overlap and strongest bond is the weakest acid; $\ce{HI}$ with the poorest overlap and weakest bond is the strongest acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQb3w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Response to OP update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand just what % ionic means in the Table you posted.  When bonding is discussed, usually 3 types of bonding are presented (see this &lt;a href=&quot;http://facstaff.gpc.edu/~proessle/LECTURE/CHEM%201211%20Lectures/Unit%203/Bonding.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; for example) 1) covalent bonding where electrons are roughly equally shared, 2) polar bonding where we have a covalent bond with unequal electron sharing and 3) ionic bonding where electrons are not shared.  For $\ce{HX}$ you can draw 2 resonance structures, one covalent and one ionic that both contribute to the true picture of the covalent bond in $\ce{HX}$.  Is the % ionic in the Table referring to the polarization in the $\ce{HX}$ covalent bond, or is referring to a truly ionic bond?  How does the increasing covalency suggested by the Table as you move down the $\ce{HX}$ series square with decreased bond strength down the series?  Since bromobenzene and methyl bromide have dipole moments, the equation used to generate the % ionicity in the Table could also generate a percent ionicity for these compounds, but they are certainly not &quot;ionic&quot; compounds.  I suspect the Table is measuring the polarization of a covalent bond and not referring to the ionic character found in a salt.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-19T21:59:12.870" LastActivityDate="2014-06-19T21:59:12.870" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13708" PostTypeId="1" CreationDate="2014-06-19T03:10:05.917" Score="2" ViewCount="1519" Body="&lt;p&gt;I have the following IR Spectrum that I am trying to interpret:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Rsp9Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I am thinking this compound have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a primary amine due to two peaks at 3624 and 3533.&lt;/li&gt;&#xA;&lt;li&gt;The sharp peak 3416 could indicate an alkyne(sp C-H).&lt;/li&gt;&#xA;&lt;li&gt;And the other sharp peak at 1716 is an aldehyde or ketone.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this analysis correct ? And what have I missed ? Thanks for any help !&lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="6833" LastEditDate="2014-06-19T11:53:36.770" LastActivityDate="2014-08-02T15:12:58.950" Title="Interpretation of the following IR?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13709" PostTypeId="1" AcceptedAnswerId="13731" CreationDate="2014-06-19T03:30:36.043" Score="1" ViewCount="558" Body="&lt;p&gt;The problem statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lets say you have $1 L$ of $2 M$ methane and the same amount of chlorine. Lets also say that both are liquids since those are most likely to react. Now the only way they can both be liquids at STP is if the temperature is as cold as an antarctic winter so this is not aqueous. Gases more often bump the wrong way and solids don't react unless it is oxidization or dissolving.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now the initiation step is forming the first molecule of $\ce{HCl}$ and Methyl.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now the methyl and chlorine atom really want to react and form chloromethane&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Relevant equations&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{CH4 + Cl2 -&amp;gt; HCl + CH3Cl}$&lt;br&gt;&#xA;  (this continues up to tetrachloromethane)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{2 CH3Cl -&amp;gt; Cl2 + C2H6}$&lt;br&gt;&#xA;  (this can continue for much longer than the previous one can)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The attempt at a solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 CH4 + 2 Cl2 -&amp;gt; 2 HCL + 2 CH3 + 2 Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Cl + 2 CH3 -&amp;gt; 2 CH3Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 CH3Cl + 2 CH3Cl -&amp;gt; 1 C2H6 + 1 Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 HCl -&amp;gt; 2 H2 + 2 Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This obviously can't happen because then we have more chlorine than we started out with. Why? Well that $1 M\; \ce{Cl2}$ from ethane plus $2 M\; \ce{Cl2}$ from $\ce{HCl}$ is $ 3 M\; $$\ce{Cl2}$ and we started with $2 M \ce{Cl2}$. Just like the number of each element the molarity has to be balanced. What I find confusing is figuring out the molarity of each compound at each step of the process not the compounds &lt;em&gt;per se&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="33991" LastEditDate="2017-05-04T20:26:22.333" LastActivityDate="2017-05-04T20:26:22.333" Title="Free Radical Reaction of Methane and Chlorine" Tags="&lt;organic-chemistry&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13710" PostTypeId="2" ParentId="6244" CreationDate="2014-06-19T03:49:31.200" Score="1" Body="&lt;p&gt;As stated previously, it is due to the head-on overlap of sigma bonds and the lateral overlap of pi-bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller overlap of pi bonds also explains why double and triple bonds basically exist only for 2nd row elements (C,N,O especially) and not for higher row elements. A C=C bond has a length of 133 pm. A Si-Si bond length is at around 186 pm, therefore the contribution of the pi-integral overlap is almost negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inability of Silicium  to form strong pi bonds is also one of the answer why life on earth is carbon based and not silicium based, since the richness of organic chemistry is in part due to the ability of carbon to form strong double and triple bonds.&lt;/p&gt;&#xA;" OwnerDisplayName="user6876" LastActivityDate="2014-06-19T03:49:31.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13711" PostTypeId="2" ParentId="4429" CreationDate="2014-06-19T04:03:53.560" Score="5" Body="&lt;p&gt;It is mainly due to &lt;strong&gt;kinetic motion&lt;/strong&gt; in molecules &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of molecules follows the Boltzmann distribution. Therefore, at room temperature molecules are constantly moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intermolecular interaction&lt;/strong&gt;, such as intermolecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intramolecular interactions&lt;/strong&gt; on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit &lt;strong&gt;strain&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user6876" LastEditorUserId="51553" LastEditDate="2018-04-05T19:32:41.057" LastActivityDate="2018-04-05T19:32:41.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13713" PostTypeId="1" AcceptedAnswerId="13718" CreationDate="2014-06-19T05:49:53.143" Score="2" ViewCount="1016" Body="&lt;p&gt;I recently looked up hydrogen sulfate ($\ce{HSO4-}$) to see how it's put together, and found this image of methyl hydrogen sulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0zst9.png&quot; alt=&quot;lewis structure of methylsulfate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd assume that hydrogen sulfate looks the same, but without the methyl group attached. However I'm confused, how can it be constructed like that? This shows the sulfur having a total of 12 valence electrons, but shouldn't the maximum be 8?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might be asking something that'll be obvious to me later on in chemistry, but this (and other groups that act as atoms) appear early on. I probably don't &lt;em&gt;need&lt;/em&gt; to know it yet but I'd prefer to know how it works if I can as opposed to just accepting it.&lt;/p&gt;&#xA;" OwnerUserId="2662" LastEditorUserId="402" LastEditDate="2017-05-08T00:25:11.070" LastActivityDate="2017-05-08T00:25:11.070" Title="Why is hydrogen sulfate put together as it is?" Tags="&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13714" PostTypeId="1" CreationDate="2014-06-19T06:05:04.453" Score="0" ViewCount="3627" Body="&lt;p&gt;This question is related to the toasting of bread. When we toast a slice of bread, the starch in it breaks down into dextrin. I want to know the balanced chemical equation for the process. &lt;/p&gt;&#xA;" OwnerUserId="6879" LastEditorUserId="16683" LastEditDate="2017-04-04T21:26:31.813" LastActivityDate="2020-05-25T16:29:01.823" Title="What the balanced chemical equation for the conversion of starch to dextrin by dry heating?" Tags="&lt;stoichiometry&gt;&lt;food-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13715" PostTypeId="2" ParentId="13704" CreationDate="2014-06-19T06:08:20.787" Score="1" Body="&lt;p&gt;I managed to find the answer in the MSDS of sugar given by the US Department of Labour, after some searching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the MSDS found &lt;a href=&quot;http://www.unitedsugars.com/assets/pdf/quality/MSDS_Powdered.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; there is an ignition temperature (which I imagine to be the same as the fire point) of 350 C, with a minimum explosion concentration of 0.045 g/l in air. There also seems to be a possible ignition from a dramatic increase in pressure (mentioned in the MSDS to be 5,000 psi/sec), though no further dangers are expressed.&lt;/p&gt;&#xA;" OwnerUserId="5740" LastActivityDate="2014-06-19T06:08:20.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13716" PostTypeId="2" ParentId="13701" CreationDate="2014-06-19T06:33:25.943" Score="1" Body="&lt;p&gt;Clearly, dipole of $\ce{HCl}$ (&lt;strong&gt;1.05D&lt;/strong&gt;) is more than $\ce{H2S}$ (&lt;strong&gt;0.97D&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be an error in your book or your book's author might have a different hypothetical theory about Dipole Moment. It might also be possible that the author is quantifying the dipole in different perspective (Not quite sure about what it can be).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://forums.studentdoctor.net/threads/calling-all-chem-pros-unite-the-clans.891541/&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt;. The guys over there had quite a lot discussion on the very same topic. Everyone came up with same result, Dipole of $\ce{HCl}$ &gt; $\ce{H2S}$ in every possible perspective.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T00:22:02.910" LastActivityDate="2015-06-28T00:22:02.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13717" PostTypeId="2" ParentId="1238" CreationDate="2014-06-19T06:53:57.380" Score="4" Body="&lt;p&gt;Salt + ice takes the temperature down from 0 °C ice to −19 °C salt water, and can be used to make more ice (or ice cream!). So you have an ice-generating engine powered by salt and water, and as much cooling as you might desire (down to −19 °C). You could retrieve the salt later by evaporation. This is obvious, but I’m not aware of any device that does this, so maybe there are obstacles in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to know how to get down to −80 °C so I can make dry ice.&lt;/p&gt;&#xA;" OwnerUserId="6880" LastEditorUserId="6880" LastEditDate="2018-01-30T04:42:49.607" LastActivityDate="2018-01-30T04:42:49.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13718" PostTypeId="2" ParentId="13713" CreationDate="2014-06-19T06:56:13.730" Score="-1" Body="&lt;p&gt;I think it's time for you to know about Octet Expansion rule. The molecules that expand their octet are known as &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypervalent Molecules&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Normally, atoms in group 2 have 8 electrons and can thus form 4 bonds.  This is mainly because they have s and p orbitals. However, in elements of group 3, they have extra orbitals called d orbitals that can be used to form more bonds.  when forming extra bonds, the atom promotes electrons into the d orbitals and allows the atom to from more than 4 bonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Same thing happens in the case of sulfur. Sulfur belongs to 3rd period and therefore it has a vacant 3d orbital and thus can accommodate more than 8 electrons in its valence cell.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="402" LastEditDate="2017-05-08T00:24:57.120" LastActivityDate="2017-05-08T00:24:57.120" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13723" PostTypeId="5" CreationDate="2014-06-19T08:15:46.840" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-06-19T08:15:46.840" LastActivityDate="2014-06-19T08:15:46.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13724" PostTypeId="4" CreationDate="2014-06-19T08:15:46.840" Score="0" Body="Fourier transform infrared spectroscopy is a technique which is used to obtain an infrared spectrum of a solid, liquid or gas." OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2014-06-19T12:16:15.070" LastActivityDate="2014-06-19T12:16:15.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13726" PostTypeId="2" ParentId="7439" CreationDate="2014-06-19T10:56:48.630" Score="6" Body="&lt;p&gt;I think you can make coffee with the machine shown in the picture.&#xA;This is my analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fa50M.png&quot; alt=&quot;breaking bad coffee machine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6884" LastActivityDate="2014-06-19T10:56:48.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13727" PostTypeId="1" AcceptedAnswerId="15020" CreationDate="2014-06-19T11:10:23.027" Score="2" ViewCount="220" Body="&lt;p&gt;I am using one quantum chemistry package where I am supposed to assign a number of active orbitals I want to be included in the FCI (full configuration interaction) calculation. For example if I have  ${H_2}$ molecule (singlet) and I use $cc-pV6Z$ basis set of Dunning, what would be the maximum number of active orbitals in the FCI? Can somebody provide me insight into this problem?&lt;/p&gt;&#xA;" OwnerUserId="6885" LastEditorUserId="5456" LastEditDate="2014-06-19T15:18:27.927" LastActivityDate="2014-08-08T17:13:05.513" Title="Number of orbitals in full configuration interaction" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13728" PostTypeId="1" CreationDate="2014-06-19T11:27:04.293" Score="2" ViewCount="273" Body="&lt;p&gt;This is an extremely basic question about biochemistry from someone who doesn't know much about organic chemistry. I'm aware of how organisms store energy ATP and release it by dephosphorylating it into ADP. My question is about the role that the adenosine part of the ATP molecule plays in this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, it seems as if the triphosphate group could be attached to any old thing, and you'd get the same energy by dephosphorylating it. But as far as I'm aware, it's only ATP that's used for this purpose in biology, and not any other triphosphate. Why is this? Is the fact that the triphosphate is attached to an adenosine just a &quot;frozen accident&quot; like the genetic code, or does it play an important role in the phosphorylation/dephosphorylation reactions themselves?&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2014-07-19T19:31:32.843" Title="What does the adenosine do in ATP?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13729" PostTypeId="2" ParentId="13701" CreationDate="2014-06-19T11:47:18.477" Score="3" Body="&lt;p&gt;I have some additional comments. As mentioned previously, by the most straightforward measure of polarity for small molecules (net molecular dipole moment), $\ce{HCl}$ is undoubtedly more polar than $\ce{H2S}$. If some other measure of polarity is being considered, then it needs to be stated for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the authors of the book and some of its readers are taking the higher boiling point of $\ce{H2S}$ compared to $\ce{HCl}$ to mean the former is a more polar molecule. In fact, this behaviour holds for &lt;a href=&quot;https://i.stack.imgur.com/f0HwH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;all of the group 15/16/17 molecular hydrides&lt;/a&gt;. However, this is only an indirect measure of polarity, and there are other confounding factors. In this particular case, I believe the reason $\ce{H2S}$ has a higher boiling point does not have to do with the strength of each intermolecular interaction, but the average number of intermolecular interactions which can be held at once, compared to its neighbouring hydrides ($\ce{PH3}$ and $\ce{HCl}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hand-waving explanation is that the group 16 hydrides have a better &quot;balance&quot; of partially charged atoms compared to the group 15 and 17 hydrides in the same period. For example, looking at the hydrogen bonded molecules of the second period and considering an optimal hydrogen bonded network, $\ce{NH3}$ is capable of being the donor atom for three H-bonds and being the acceptor atom for only one H-bond. In $\ce{HF}$, the situation is inverted; now there are three H-bond acceptor sites, and only one H-bond donor. $\ce{H2O}$, however, is a convenient match of being a &lt;a href=&quot;https://i.stack.imgur.com/OIP8c.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;H-bond double donor and double acceptor&lt;/a&gt;. Thus, water performs more hydrogen bonds per molecule on average than either ammonia or hydrogen fluoride, and has a significantly higher boiling point, even though the actual hydrogen bonds are arguably strongest for $\ce{HF}$. The logic would be similar for the group 15/16/17 hydrides of the third period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I'd like to clarify the role of non-bonding pairs in the dipole moment of a molecule. Electrons in pure atomic orbitals are symmetric with respect to the nucleus and produce no net dipole. In a hybridization picture, one might say that &lt;a href=&quot;http://booksite.elsevier.com/9780444594365/downloads/16755_10040.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-hybridized $s$ and $p$ orbitals possess no dipole moment&lt;/a&gt;, while hybrid orbitals are &lt;em&gt;not&lt;/em&gt; symmetric and &lt;em&gt;do&lt;/em&gt; possess a dipole moment. In the case of the second period element hydrides, the lone pairs in hybridized $sp^3$ orbitals would provide a net dipole moment. However, for the hydrides in the third period and below, the molecular geometries are consistent with no hybridization (bond angles close to 90°), and therefore the lone pairs would be present in effectively $s$ and $p$ atomic orbitals, with no effect on a molecule's polarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more accurate picture without invoking hybridization, the molecular orbitals which contain the lone pairs are almost the same as the central atom's atomic orbitals in the heavier group 15/16/17 hydrides (and hence produce no net dipole), while for the second period hydrides the molecular orbitals are not as symmetric and hence actually produce a dipole.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-06-19T11:47:18.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13730" PostTypeId="1" AcceptedAnswerId="13732" CreationDate="2014-06-19T13:33:06.707" Score="1" ViewCount="108" Body="&lt;p&gt;I got tripped up by this question; I thought these were all enantiomers when in fact they were all meso compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one prevent this from happening? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zuoe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one rotate the Fischer projection, for example, in the first problem; I understand that one has to hold one group in place, but then what? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="72973" LastEditDate="2019-10-19T19:15:50.537" LastActivityDate="2019-10-19T19:15:50.537" Title="Stereochemistry of meso compounds" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13731" PostTypeId="2" ParentId="13709" CreationDate="2014-06-19T13:40:49.717" Score="1" Body="&lt;p&gt;OK, let's focus on the stoichiometry of your last 4 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 CH4 + 2 Cl2 -&amp;gt; 2 HCL + 2 CH3 + 2 Cl~~~~}$  this looks OK&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Cl + 2 CH3 -&amp;gt; 2 CH3Cl~~~~~~~~~~~}$ this looks OK&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 CH3Cl + 2 CH3Cl -&amp;gt; 1 C2H6 + 1 Cl2~~~~}$ this should be $\ce{1 CH3Cl + 1 CH3Cl -&amp;gt; 1 C2H6 + 1 Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 HCl -&amp;gt; 2 H2 + 2 Cl2~~~~~~~}$ this should be $\ce{2 HCl -&amp;gt; 1 H2 + 1 Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I agree with all of your chemical reactions, but at least now you don't wind up with more chlorine then you started with.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-19T13:40:49.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13732" PostTypeId="2" ParentId="13730" CreationDate="2014-06-19T13:52:19.567" Score="1" Body="&lt;p&gt;To rotate on a single bond in a Fischer Projection, all 3 groups move in the same direction. $\frac {1} {2\ \ \ 3} \rightarrow \frac {2} {3\ \ \ 1}$. See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Meso_Compounds&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; for more information. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-19T13:52:19.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13733" PostTypeId="2" ParentId="13619" CreationDate="2014-06-19T13:57:13.363" Score="10" Body="&lt;p&gt;Computational chemistry is a synthesis of many different subjects, each of which is very complex on its own. So, if you really want to learn the theoretical basis, you should be prepared to put in a lot of time and effort. I don't want to discourage you, I just want you to have realistic expectations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamentals of computational chemistry are physical chemistry and computer science. The fundamentals of those are math, physics, and logic. So, depending on what your background is, you might have to start with math and physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Math:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Calculus&lt;/li&gt;&#xA;&lt;li&gt;Vector Calculus&lt;/li&gt;&#xA;&lt;li&gt;Differential Equations&lt;/li&gt;&#xA;&lt;li&gt;Statistics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Classical Dynamics&lt;/li&gt;&#xA;&lt;li&gt;Classical Electrodynamics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you have taken those courses (if you haven't already) you can start with physical chemistry. You could conceivably teach yourself those subjects, but it would be very difficult - a free online course would be a good option if you can't take a university course. I believe Khan academy covers all of these subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Physical Chemistry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Statistical Mechanics&lt;/li&gt;&#xA;&lt;li&gt;Quantum Mechanics&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I always liked Atkin's &lt;a href=&quot;http://amzn.com/1429218126&quot;&gt;physical chemistry&lt;/a&gt; book (although I don't know if the newer editions are still good). McQuarrie's &lt;a href=&quot;http://amzn.com/1891389505&quot;&gt;quantum chemistry&lt;/a&gt; and &lt;a href=&quot;http://amzn.com/1891389157&quot;&gt;statistical mechanics&lt;/a&gt; are very good as well and are classics in the field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where do I start with the chemistry part?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;solving the Schrödinger equation&lt;/li&gt;&#xA;  &lt;li&gt;Hartree–Fock method&lt;/li&gt;&#xA;  &lt;li&gt;Density functional theory&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics in physical chemistry will cover solutions to Schrodinger's equation for all of the cases where a closed-form solution exists. This will give you an understanding of the basic principles, so you should start with that. To solve &quot;real&quot; systems, you need more advanced approximation techniques, and Hartree-Fock and DFT will get you started. That is a good order to study them in, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For computer science, a lot depends on which software you will be using or what languages you want to use. Most of the core, highly optimized algorithms are written in fortran or C. Some go as far as assembly for the most important parts. Most of the software I have seen (non-core) is written in fortran or C++. My experience is a little out of date (I last worked in the field about 7 years ago) so that may no longer be true. At any rate, you will need at least a basic programming course in the language you want to use (C would be good to start with, since you will really learn the fundamentals). I would also take some object-oriented courses, so courses involving more advanced C++ and java would be good as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key concepts (beyond programming and architecture) that you will need are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stochastic methods (Monte Carlo)&lt;/li&gt;&#xA;&lt;li&gt;Linear Algebra&lt;/li&gt;&#xA;&lt;li&gt;Numerical integration&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With all of the above, you will have a good set of base skills and knowledge on which you can build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am only familiar with one book that might be suitable for beginners in computational chemistry, so perhaps others can give better suggestions. The one I used was &lt;a href=&quot;http://amzn.com/0582382106&quot;&gt;Molecular Modelling&lt;/a&gt; by Leach. It gives a good overview of common techniques, but doesn't go too in-depth. However, it was last updated in 2001 - so it may be out of date. The state of the art moves pretty quickly in terms of detailed application, although the fundamentals tend to stay pretty consistent over time. Another classic text which covers the basics in some depth is &lt;a href=&quot;http://amzn.com/0198556454&quot;&gt;Allen &amp;amp; Tildesley's Computer Simulation of Liquids&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these books focus more on the modeling aspects of molecules as opposed to chemical reactions, so if you are more interested in simulating reactions as opposed to statistical mechanics or protein-ligand interactions, you would need different books.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-19T13:57:13.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13734" PostTypeId="1" AcceptedAnswerId="13814" CreationDate="2014-06-19T15:07:27.943" Score="2" ViewCount="3685" Body="&lt;p&gt;My prof told me that we can compare Lewis base strength based on electron donation to the $\ce{H+}$ proton - i.e sigma bond formation with the $\ce{H+}$ proton - and proceeded to conclude that $\ce{S^{2-}}$ is a weaker Lewis base than $\ce{O^{2-}}$, since the former is a weaker base than the latter. He then proceeded to generalize that the oxide ion is always a stronger Lewis base than the sulfide ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a valid method of comparing the strength of Lewis bases? I have two issues:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Isn't Lewis acid/base strength quantified using kinetics, not thermodynamics? Don't stronger LAs react faster, and weaker LAs react slower? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Even if this comparison were valid, would the comparison still be valid outside of water solution? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does this mean that this question is invalid? No medium is specified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VIO1n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1r8yx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;I think he knows what he's talking about. In explaining the &lt;strong&gt;much&lt;/strong&gt; greater stability of silver sulfide ($K_{\mathrm{sp}} \approx 6 \cdot 10^{-51}$) vs. $\ce{Ag2O}$ ($K_{\mathrm{sp}} = 3.5 \cdot 10^{-16}$), he notes that charge density does not explain this disparity. In fact, charge density would imply that the oxide salt should be more stable. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For instance, although $\ce{O^2-}$ is a stronger B/L base than $\ce{S^2-}$, $\ce{S^2-}$ can be (and usually is) a stronger Lewis base than $\ce{O^2-}$ since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with with the Lewis base interacts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The text goes on to mention other factors such as having a metal cation with a surfeit of valence d electrons, and how that can populate the empty valence 3d orbitals of $\ce{S^2-}$. (S has 3d orbitals?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense; a good nucleophile has less stabilized electrons - i.e. electrons that are further from the nucleus. It still seems that the questions posted above put B/L acidity ahead as the main deciding factor for sigma-donor strength. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-02-28T16:03:02.553" LastActivityDate="2018-02-28T16:04:07.520" Title="Lewis base strength and the hydrogen proton" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13735" PostTypeId="2" ParentId="13619" CreationDate="2014-06-19T15:32:47.403" Score="4" Body="&lt;p&gt;Disclaimer: I don't have significant experience in the field, but I did touch it a little.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to learn the theoretical principles of computational chemistry and develop software themselves, or at least parts of it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The idea is noble, but extremely naive.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would like to work in the field of drug design.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lucky for you, you don't actually need any advanced math staff to work in this specific field. Sure, it may be handy, but not necessary. Most drug-related simulations may be performed using molecular mechanics/molecular dynamics approach, that is quasi-classical in nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;general programming notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the most common language in scientific community is, unfortunately, fortran. It is a very old language with huge amount of inherent problems. I advice to learn programming in some more sane language first and then move to fortran/c/c++. Freely available SICP (structure and interpretation of computer programs) based on scheme is a good choice. Sure, it uses quite esoteric scheme, but it does discourage bad habits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;minimal list, in order&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;classical newtonian physics (forces, acceleration, speed - all in vector form, their relations and electrostatic interactions)&lt;/li&gt;&#xA;&lt;li&gt;theory of programming: big-O notation, algorithmics, space-sorting, indexes in space etc. Specifically, MM/MD needs a way to estimate total forces in many-body system with electrostatic interaction where thousands and millions of bodies may be present. This requires effective algorithm to discard small interactions, which in turn requires effective algorithm producing list of close pairs.&lt;/li&gt;&#xA;&lt;li&gt;optional: genetic algorithms. &lt;/li&gt;&#xA;&lt;li&gt;optional: machine learning. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;after learning that, it is time to start to read actual articles/reviews. Areas of interests are QSAR (relies on machine learning) and docking (relies on MM/MD approach to 'fit' molecule to receptor.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in case you still have resolve to code something, it is time to actually code. take some existing package and tune it. &lt;/li&gt;&#xA;&lt;li&gt;in case you want to work with huge systems or hi-performance computing, learn&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;theory of parallel programming: threading libraries, synchronisation primitives&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;common parallelization technologies: mpi, openmp.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;GPGPU : OpenCL or Cuda&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;maximal list, in order. Overkill for drug design. In addition for minimal list&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;linear algebra.&lt;/li&gt;&#xA;&lt;li&gt;analytical geometry&lt;/li&gt;&#xA;&lt;li&gt;advanced calculus (multidimensional integrals, partial derivatives and related stuff)&lt;/li&gt;&#xA;&lt;li&gt;differential equations&lt;/li&gt;&#xA;&lt;li&gt;function approximation using linear combination of other functions &lt;/li&gt;&#xA;&lt;li&gt;quantum mechanics, nonrelativistic case.&lt;/li&gt;&#xA;&lt;li&gt;any general quantum chemistry book - qualitative theory of bonding, hartree-fock equations, black magic of MP2 and DFT &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;up to this point the topics are useful for any chemical student. Specialized staff follows.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;general programming&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;algorithmics&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;big-O notation&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;numerical errors, estimating them.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;IEEE floating-point format, use and limitations&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;numerical methods, specifically&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;fast fourier transform&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;multidimensional integration&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;iterative diagonalization methods&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;at this point it becomes possible to code something simple. In case hi-performance computing is of interest (it is a separate area, that is not chemistry related per se)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;parallel programming, &lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;see minimal list&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;distributed storage and processing&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-06-19T15:39:50.930" LastActivityDate="2014-06-19T15:39:50.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13736" PostTypeId="1" AcceptedAnswerId="13741" CreationDate="2014-06-19T16:49:39.980" Score="2" ViewCount="497" Body="&lt;p&gt;A trivial research in atoms and their basic theories led me to this term: &quot;isotones&quot;&lt;br&gt;&#xA;Nuclides sharing the same number of neutrons but different atomic numbers.&lt;br&gt;&#xA;A simple request or shall I say, quest:&lt;br&gt;&#xA;please indicate me to any feasible similarities existing between isotones and please, DO NOT use jargon.&lt;/p&gt;&#xA;" OwnerDisplayName="user6887" LastActivityDate="2014-06-22T21:53:38.783" Title="Do isotones share any similarities?" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13737" PostTypeId="2" ParentId="13736" CreationDate="2014-06-19T17:42:08.750" Score="2" Body="&lt;p&gt;I suspect that having the same number of neutrons may mean something at the nuclear chemistry and nuclear physics level, but in terms of general chemistry, including inorganic, organic and biochemistry, it's all about the number of electrons and protons.  So I don't think that isotones would show any differences in terms of (non-nuclear) reaction chemistry.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-19T17:42:08.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13738" PostTypeId="2" ParentId="13734" CreationDate="2014-06-19T18:10:11.743" Score="3" Body="&lt;p&gt;One of the keys in the definition of Lewis acids and Lewis bases is that they react to form a Lewis adduct.  To compare the relative strengths of different Lewis bases, you would allow them to react with the same Lewis acid, form the required adduct and measure the enthalpy of formation of the adduct. The adduct with the largest enthalpy of formation would contain the strongest Lewis base.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) So Lewis acid and base strength is about thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You could run the reaction in any solvent (or gas phase) as long as all of the Lewis acids or bases you want to compare are able to form an adduct in that solvent.  There is no reason why the order of a series of enthalpies of bond formation has to remain constant when you move from solvent to solvent, so your relative strengths could well change from one solvent to another.  The same applies to oxide \ sulphide relative Lewis base strengths if you substitute some other Lewis acid in place of the proton.  Again the relative ordering might well change.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-19T18:10:11.743" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13739" PostTypeId="2" ParentId="13728" CreationDate="2014-06-19T19:26:11.077" Score="1" Body="&lt;p&gt;The adenosine part is needed to make DNA and RNA.  It is logical for the triphospate to be attached to ADP, GDP, UDP or CDP since they are needed to make DNA and RNA.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some instances, GTP is used instead of ATP.  In fact the &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krebs cycle&lt;/a&gt; does involve phosphorylation of GDP.   GTP is used in the formation of microtubes by &lt;a href=&quot;http://en.wikipedia.org/wiki/Tubulin&quot; rel=&quot;nofollow noreferrer&quot;&gt;tubulin&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen some give the rate of hydrolysis being greater for ATP than GTP as a reason for ATP being more utilized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a related question here: &lt;a href=&quot;https://biology.stackexchange.com/questions/11286/why-is-atp-the-preferred-choice-for-energy-carriers&quot;&gt;https://biology.stackexchange.com/questions/11286/why-is-atp-the-preferred-choice-for-energy-carriers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:14.547" LastActivityDate="2014-06-19T19:26:11.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13740" PostTypeId="2" ParentId="13736" CreationDate="2014-06-19T19:47:22.357" Score="1" Body="&lt;p&gt;In chemistry, if an atom has a closed valence electron shell, as do the noble gases, it is particularly stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is a &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/shell.html&quot; rel=&quot;nofollow&quot;&gt;shell model&lt;/a&gt; of the nucleus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain numbers of neutrons (and also protons) are referred to as magic numbers: 2,8,20,28,50,82,126, all even numbers, because nuclei with these numbers of neutrons (or protons) are particularly stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The similarity between isotones is that stability of a nucleus is somewhat correlated with the number of neutrons.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-19T19:47:22.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13741" PostTypeId="2" ParentId="13736" CreationDate="2014-06-19T20:16:57.210" Score="3" Body="&lt;p&gt;Chemistry is determined entirely by electron interactions. Since the number of electrons in an atom is determined by the number of protons in the nucleus, the atomic number (number of protons) is what defines the chemical behavior of a given element. This means that in terms of chemical properties, isotones would have no similarities unless they happened to be in the same group (column) of the periodic table, but that would be a coincidence and would have nothing to do with the number of neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As DavePhD and ron said, the nuclear stability could be similar, since nuclear stability has a lot to do with the total number of nucleons (neutrons + protons) as well as the relative proportion of protons to neutrons. The wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotone&quot; rel=&quot;nofollow&quot;&gt;isotones&lt;/a&gt; discusses stability in a little more detail.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-06-22T21:53:38.783" LastActivityDate="2014-06-22T21:53:38.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13742" PostTypeId="2" ParentId="10114" CreationDate="2014-06-19T20:36:50.190" Score="3" Body="&lt;p&gt;This is one of the most commonly misunderstood principles in thermochemistry - if you can wrap your mind around it, you are most of the way there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is, there is an easy way to do that. In any physics problem (thermochemistry is really physics) you have to be very careful with how you define the system and the surroundings. This is important for two reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It lets you write energy balance equations across the system boundaries (so we can use math to solve the problem)&lt;/li&gt;&#xA;&lt;li&gt;By choosing the right system boundaries, we can make the math much, much easier.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since thermochemistry at this level always involves the same types of problems, chemists have done the hard part for you and already selected the best system boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The system is whatever is in the reaction equation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it. The entire rest of the universe is the &quot;surroundings.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking back at your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The thermometer is in the solution which means it measures the temp of the solution. It's the system. The temp of a system increases means heat flows in, how could this be an exothermic reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the reaction molecules are the only thing in the system, then that means the water, the calorimeter, the themometer, you, me, the room - everything else - counts as surroundings. Now, the water did heat up. We know this because the thermometer absorbed some heat and showed a temperature increase. Since they are all part of the &lt;em&gt;surroundings,&lt;/em&gt; the heat must have come from the &lt;em&gt;system.&lt;/em&gt; Since heat left the system, the sign of $q_{reaction}$ is negative, and the reaction is &lt;strong&gt;exothermic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a teacher, I have found that getting system boundaries and the signs right is one of the biggest problems students have. If you focus on getting that concept down early - and really understanding &lt;em&gt;why&lt;/em&gt; they are that way, you are well on your way.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-19T20:36:50.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13743" PostTypeId="1" CreationDate="2014-06-19T20:37:13.520" Score="11" ViewCount="111183" Body="&lt;p&gt;In laboratory setting, is there a difference between glycerin and glycerol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some conflicting info on this topic.&lt;/p&gt;&#xA;" OwnerUserId="6888" LastActivityDate="2015-10-21T17:56:59.713" Title="is there any difference between glycerin and glycerol?" Tags="&lt;nomenclature&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13744" PostTypeId="2" ParentId="13743" CreationDate="2014-06-19T20:44:29.777" Score="1" Body="&lt;p&gt;Glycerin and glycerol are both names for the same molecule. However, depending on where you are getting the glycerol from, it could be more or less pure.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-19T20:44:29.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13745" PostTypeId="2" ParentId="13743" CreationDate="2014-06-19T20:45:12.320" Score="10" Body="&lt;p&gt;Nope, there is no chemical difference between glycerol, glycerin or glycerine.  All 3 names refer to the same compound, propane-1,2,3-triol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EGILr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-19T20:45:12.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13746" PostTypeId="2" ParentId="13743" CreationDate="2014-06-19T20:45:14.473" Score="1" Body="&lt;p&gt;As far as I know, glycerin and glycerol both refer to the same compound: propantriol.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-19T20:45:14.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13747" PostTypeId="1" AcceptedAnswerId="13750" CreationDate="2014-06-19T20:53:31.727" Score="3" ViewCount="284" Body="&lt;p&gt;If all chemical processes are equilibria at least to some extent, then where does the energy to overcome the activation energy come from? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5396" LastEditDate="2014-06-20T06:35:06.590" LastActivityDate="2014-06-20T06:35:06.590" Title="If all chemical processes are equilibria at least to some extent, where does the activation energy come from?" Tags="&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13748" PostTypeId="2" ParentId="13714" CreationDate="2014-06-19T21:05:04.137" Score="5" Body="&lt;p&gt;Let's take a look at the structure of both molecules. First, &lt;a href=&quot;http://en.wikipedia.org/wiki/Starch&quot; rel=&quot;nofollow noreferrer&quot;&gt;starch&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rr0uf.png&quot; alt=&quot;Starch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dextrin&quot; rel=&quot;nofollow noreferrer&quot;&gt;dextrin&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/crRdr.png&quot; alt=&quot;Dextrin - shorter than starch witht he same structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see the difference? They are the same molecule (both are &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;nofollow noreferrer&quot;&gt;glucose&lt;/a&gt; polymers), the only difference is the molecular weight, or the length of the polymer chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starch, by convention, is 300-600 monomers long. Dextrin is something less than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the number of dextrin molecules you get will depend on the polydispersity (size distribution), average chain length, and extent of the reaction, there is no way to write a balanced reaction equation for a real reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could assume that a starch molecule of a given length breaks into dextrin molecules of a given length, and write an equation. It would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H11O5[C6H10O5]_{m}C6H11O6 + (\frac{$m-n$}{2})H2O-&amp;gt;} \frac{(m+2)}{(n+2)}\ce{C6H11O5[C6H10O5]_{n}C6H11O6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would only work for one length of starch and dextrin (in reality there could be 10 - 1000) and would also only be true if $m &amp;gt; n$, and $m$ is evenly divisible by $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the question you specify dry heat, but there is no way to balance the equation without including water (unless you use hydrogen and oxygen gas). I suspect that even under dry heat conditions, there is moisture on the bread, and it is consumed in the reaction. This could be part of the reason that toasting drys bread.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="16683" LastEditDate="2017-04-04T21:26:15.540" LastActivityDate="2017-04-04T21:26:15.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13749" PostTypeId="2" ParentId="13747" CreationDate="2014-06-19T21:39:04.930" Score="1" Body="&lt;p&gt;Maybe this is another case where semantics enters in.  It seems that if a reaction coordinate connects two species, then over time an equilibrium will be achieved between them.  &quot;Over time&quot; could mean microseconds or millennia.  The kinetic energy distribution of a  collection of molecules is described by a Boltzmann distribution.  Therefor even at room temperature, there will be some molecules with enough energy to surmount a high barrier along the reaction coordinate.  If there is a high barrier connecting two molecules, then it will just take longer for equilibrium to be achieved.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-19T21:39:04.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13750" PostTypeId="2" ParentId="13747" CreationDate="2014-06-19T22:10:28.350" Score="4" Body="&lt;p&gt;I don't think the statement that all processes are in equilibrium to some extent is true. By definition, dynamic equilibrium means that the rate of change in one direction is equal to the rate of change in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly not true for all &lt;strong&gt;processes,&lt;/strong&gt; otherwise there would not be a process - nothing would appear to change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for any given &lt;strong&gt;reaction equation&lt;/strong&gt; that is known to occur, we can say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;for a reaction at equilibrium, neither the concentration of reactants nor the concentration of products will be zero&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that may be what you meant in your question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, then let's look at your example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not an equilibrium process. However, the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na(g)-&amp;gt;Na(g)+ + e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could be &lt;strong&gt;at equilibrium.&lt;/strong&gt; In other words, the ratio of reactants to products could be such that the Gibbs free energy is at a minimum for this temperature and pressure. A more mathematical way to say that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = K_{eq}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q=\frac{[Na^+]}{[Na]}$ and $K_{eq}$ is the equilibrium constant under these conditions. (Brackets mean &quot;partial pressure&quot; - not concentration here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under normal lab conditions (1 atm, 298 K), this is not thermodynamically favorable. The reaction quotient heavily favors the reactants, or another way to say it is $K_{eq}$ is really close to zero. So close you may not be able to measure it. The activation energy is extremely high (equal to the ionization energy), and so any electrons that are removed would have to get that energy from somewhere - which was your next question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an idealized, isolated system, you only have one source of energy for things like that: kinetic. Thermal motion of the atoms could conceivably put enough energy into a single atom that an electron could be &quot;kicked off&quot; and result in an ion. However - the odds of that happening are extremely low, given that the distribution of speeds would have a vanishingly small, although still non-zero, probability of occurring at the magnitude needed to overcome the activation energy barrier. Since the reaction is thermodynamically unfavorable, the probability that the electron would jump right back on is very high - much higher than it would be for leaving in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you have $6.022 \cdot 10^{23}$ &lt;em&gt;opportunities&lt;/em&gt; each femtosecond, very low odds can still work out. But the total amount that would be ionized at any one time is still very, very low. Possibly so low that you would have to wait for an essentially infinite amount of time to see a measurable amount of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;em&gt;real&lt;/em&gt; system, a much larger fraction would be ionized (and this may or may not be reflected in $K_{eq}$, depending on how it is measured). The reason is that there is always ionizing radiation present on earth, and there will be more or less depending on where you are located. Statistical thermodynamics doesn't take this into account, so any prediction of the relative amounts of reactants and products would not be correct. This is kind of a unique example in that respect - most equilibrium reactions, even with very slow kinetics, deal with systems where the thermal energy of the system has a much bigger effect than radiation.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-06-20T00:40:49.227" LastActivityDate="2014-06-20T00:40:49.227" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13751" PostTypeId="1" AcceptedAnswerId="13753" CreationDate="2014-06-19T23:00:30.550" Score="0" ViewCount="8234" Body="&lt;p&gt;here is a question that i don't know how it is solved :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sample that is 75 % chloride by mass is dissolved in water and treated with an excess of AgNO3. If the mass of the AgCl precipitate that forms is 2.013 g, what was the mass of the original sample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you give me the the proper way of solving it &lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-06-19T23:48:59.213" Title="what's the mass of the original sample (g)?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13752" PostTypeId="1" AcceptedAnswerId="13759" CreationDate="2014-06-19T23:37:25.293" Score="1" ViewCount="623" Body="&lt;p&gt;the question says :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two elements, A and B, combine to form two binary compounds. In the first&#xA;compound, 3.5 g of A combines with 8.00 g of B. In the second compound,&#xA;5.0 g of A combines with 17.1 g of B. If the formula of the first compound is&#xA;AB2, then the formula of the second compound would be ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't even make the first step :(&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-06-20T01:10:08.473" Title="Use the given information to write the formula of this compound" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13753" PostTypeId="2" ParentId="13751" CreationDate="2014-06-19T23:48:59.213" Score="1" Body="&lt;p&gt;It is first of all important to recognise that all the chloride in the original sample will precipitate upon treatment with excess AgNO3. This will give you a starting point, which is a measurable amount of AgCl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular mass of AgCl = (107.87+35.45)=143.32g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the proportion of chloride in AgCl = (35.45/143.32) = 0.2473&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore in 2.013g of AgCl, there is (0.2473 x 2.013) = 0.4978g of chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 0.4978g of chloride constitutes 75% of the mass of the original sample, then the amount of original sample is (0.4978 / 0.75) = 0.6637g&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-06-19T23:48:59.213" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13754" PostTypeId="1" AcceptedAnswerId="13757" CreationDate="2014-06-20T00:09:42.660" Score="1" ViewCount="1467" Body="&lt;p&gt;Why aren't these enantiomers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you place a mirror behind the compound and look into the mirror, you get the second drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqpFW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-20T00:46:25.163" Title="Flipping a molecule vs. mirror behind compound" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13755" PostTypeId="1" CreationDate="2014-06-20T00:37:32.883" Score="0" ViewCount="721" Body="&lt;p&gt;here is a question says&#xA;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NDYNM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that atom's number is found by =  $6.022*10^{23} * mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how this formula is used here is what i dont know&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what's the $idea$ of this question?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-06-20T01:02:25.243" Title="what's the ratio of atoms of these two isotopes?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13756" PostTypeId="1" CreationDate="2014-06-20T00:41:31.223" Score="2" ViewCount="2057" Body="&lt;p&gt;Two questions: what reaction forms between ammonium nitrate, black iron oxide, and sucrose; and a story related question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the whole story:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was playing around a few months ago and stumbled upon a mixture that generated something similar to Black Snakes (firework). &#xA;I recently decided to demo this mixture to someone, but I could not find my notes about the experiment, so I decided to try to replicate the mixture from what I remembered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am certain that the mixture contained ammonium nitrate, black iron oxide, and some sort of fuel (probably sucrose, impure in the form of powdered sugar). &#xA;Combining these three ingredients (one part by volume, whatever particle sizes I have), and igniting with a lighter, the mixture does form black snaking ash, and turns red and crumbles as it cools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1: So, what kind of reaction has taken place between the black iron oxide, ammonium nitrate, and sucrose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I leave out any one of the ingredients, the ash snake does not form, so all three compounds are definitely involved. If we leave out either the ammonium nitrate or sucrose, little happens with the application of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://drive.google.com/file/d/0Byd6ngUnOQEeSTVMRWFhanBXWXM/edit?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;Here's a picture after the fact.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't had a very good batch since I lost my notes, so maybe I am missing an ingredient, or maybe I just haven't got the right ratios.&#xA;Q2: Any Suggestions?^&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I have been having more success with this mixture (by volume):&#xA;1 part ammonium nitrate&#xA;1 part potassium nitrate&#xA;2 parts black iron oxide&#xA;2 parts sucrose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mix contents and melt over medium indirect heat (skillet is fine); once it is all melted, immediately remove from heat and form into pellet as it cools. The mixture will reduce in volume by several times.&#xA;I suggest doing this in small batches since it can self ignite if you are not careful, or do not heat it at the right rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ammonium nitrate is omitted or substituted by potassium nitrate, the ash does not expand or cohere. So the ammonium nitrate is doing something more than just providing oxygen here.&#xA;The reduction of iron oxide seems to make the ash less monolithic, and moderate reductions in iron oxide forms snakes that fall apart quickly during production.&lt;/p&gt;&#xA;" OwnerUserId="6890" LastEditorUserId="6890" LastEditDate="2014-06-21T02:11:47.487" LastActivityDate="2014-07-21T05:51:16.043" Title="React these iron oxide, ammonium nitrate, and sucrose" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13757" PostTypeId="2" ParentId="13754" CreationDate="2014-06-20T00:46:25.163" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you place a mirror behind the compound and look into the mirror,&#xA;  you get the second drawing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;If you place a mirror behind compound 1 and look into the mirror&quot; - If by that you mean, generate the mirror image of compound 1, well OK, but that does not produce compound 2. It generates the molecule labeled &quot;mirror image&quot; in the following diagram, and it is a non-superimposable mirror image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F6P4Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-20T00:46:25.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13758" PostTypeId="2" ParentId="13755" CreationDate="2014-06-20T00:55:24.210" Score="1" Body="&lt;p&gt;You need more information.  They probably expect you to look in a book and see that the atomic weight (which is a weighted average over all isotopes) of $\ce{Tl}$ is 204.383.  Now if x is the amount of $\ce{^203Tl}$ with atomic weight 202.97 and (1-x) is....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that get you started?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(x*202_{.}97) + ((1-x)*204_{.}97)~=~?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you get to the answer from here?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-20T01:02:25.243" LastActivityDate="2014-06-20T01:02:25.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13759" PostTypeId="2" ParentId="13752" CreationDate="2014-06-20T01:10:08.473" Score="2" Body="&lt;p&gt;For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14.  This means a molecule A&lt;sub&gt;n&lt;/sub&gt;B&lt;sub&gt;m&lt;/sub&gt; will give a mass ratio for A:B of n:1.14xm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a molecule AB, for every 1g of A, you will have 1.14g of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a molecule AB&lt;sub&gt;2&lt;/sub&gt;, for every 1g of A, you will have 2.28g of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a molecule A&lt;sub&gt;2&lt;/sub&gt;B&lt;sub&gt;3&lt;/sub&gt;, for every 1g of A, you will have (1.14x3/2) 1.71g of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;etc etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for 5.0g of A, in order to have 17.1g of B, you will have 1.14 x (m/n) = 17.1/5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; m/n = 17.1 / 5 / 1.14 = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means m/n=3, or AB3 (alternatively it could also be A2B6)&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-06-20T01:10:08.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13760" PostTypeId="1" AcceptedAnswerId="13761" CreationDate="2014-06-20T01:25:10.677" Score="2" ViewCount="295" Body="&lt;p&gt;The key tells me these are enantiomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HxMvV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I however don't think they are; their chiral carbons' configurations aren't inverted - i.e. only one chiral carbon's R/S configuration changes! This suggests diastereomers to me! Plus the Sawhorse projections I draw don't seem to be mirror images. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwKVQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kGImZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-20T01:49:44.490" LastActivityDate="2017-08-13T10:23:00.807" Title="Enantiomers or Diastereomers?" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13761" PostTypeId="2" ParentId="13760" CreationDate="2014-06-20T01:34:03.727" Score="4" Body="&lt;p&gt;They are diastereomers.  Both of the &quot;front&quot; chiral carbons (the one with the bromine attached) are &quot;S&quot;.  Rotate the one on the right 120 degrees CCW and it is identical to the one on the left.  But the &quot;back&quot; carbon (the one with the hydroxyl attached) is &quot;S&quot; on the left figure and &quot;R&quot; on the right figure - diastereomers.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-20T01:34:03.727" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="13762" PostTypeId="2" ParentId="13579" CreationDate="2014-06-20T02:07:30.220" Score="4" Body="&lt;p&gt;Some quick googling did not turn up a definitive reaction mechanism, and most of the papers I did find seemed to instead be focused on the overall reaction rate and controlling factors - probably because this is such an industrially important reaction and (at least according to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009250968870022&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;) gas transport is the rate-limiting factor - this is common for heterogeneous reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not too hard to propose a &lt;em&gt;plausible&lt;/em&gt; mechanism. I would suggest the same mechanism you did - what is likely happening is that the oxygen near one calcium ion can pull some electron density away from the carbon, leaving $\ce{CO2}$ which diffuses away as a gas. The calcium ion stabilizes the new oxygen anion and forms a new, different lattice structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Martin points out, the reason that this process is thermodynamically favorable is that the products have a higher entropy. The standard Gibbs free energy of reaction (using data from &lt;a href=&quot;http://amzn.com/B00AWIEPCE&quot; rel=&quot;nofollow&quot;&gt;Tro's Chemistry&lt;/a&gt;) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\Delta G &amp;amp;= \Delta H - T\Delta S \\&#xA;&amp;amp;= 179.2\space kJ/mol - T(0.1602\space J/mol*K)&#xA;\end{align}&#xA;$$&#xA;Note that $\Delta H$ is positive, but so is $\Delta S$.&#xA;At 298 K (room temp) $\Delta G = 131.3 \space kJ/mol$ - the reaction is not favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium temperature (ignoring temperature corrections) should be around 1120 K.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-20T02:07:30.220" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13763" PostTypeId="2" ParentId="13747" CreationDate="2014-06-20T03:14:48.550" Score="1" Body="&lt;p&gt;I think there are a couple conceptual problems here you should consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It is not true that all reactions are equilibrium in the normal sense of equilibrium: just burn any complex organic compound, and now try to re-make the material from carbon-dioxide and water. In other words: entropy does mater. Technically any elementary reaction can go both direction, but this is a statement about microscopic processes, and you should careful about it when you try to conclusions about ensembles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are talking about reactions and equilibrium in a macroscopic sense, while you are asking about microscopic mechanism (transfer of activation energy). In microscopic scale there is a fluctuation of the energy what a molecule can get from its environment. The kinetic energy, vibration energies etc of molecules are not a single value, rather a distribution of values, so there is a non-zero probability that two molecules collide with enough energy or enough energy is stored in a vibration mode so the reaction can go.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-06-20T03:14:48.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13764" PostTypeId="1" AcceptedAnswerId="13774" CreationDate="2014-06-20T03:21:25.987" Score="8" ViewCount="130531" Body="&lt;p&gt;Do we have that $\pu{1 g/mol} = \pu{1 amu}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we have, for the mass of an atom of carbon 12, call it $m(\ce{^12C})$, that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(\ce{^12C}) = \pu{12 amu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and furthermore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pu{1 mol} \cdot m(\ce{^12C}) = \pu{12 g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(\ce{^12C}) = \pu{12 amu} = \pu{12 g/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally we get that $\pu{1 g/mol} = \pu{1 amu}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my chemistry teacher is telling me that those are two completely different things and that I am confused between the mass per atom and the mass per $6.022\cdot10^{23}$ atoms. I can't understand how, and this is really bugging me, so help is very appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this requires the mole to be a number (or a &quot;constant&quot;), which may be where I'm wrong.&lt;/p&gt;&#xA;" OwnerUserId="5678" LastEditorUserId="24052" LastEditDate="2018-03-14T13:33:47.723" LastActivityDate="2020-05-04T14:55:30.150" Title="amu and g/mol relation" Tags="&lt;mole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13766" PostTypeId="1" CreationDate="2014-06-20T06:24:03.843" Score="2" ViewCount="16268" Body="&lt;p&gt;I am learning about structure of atom, in which i saw right from J.J. Thomsom atomic model to modern nuclear atomic model all are spherical in shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen how different discoveries help to develop the atomic model but i have't came through a single discovery which proves that atoms are spherical in shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Why atoms are spherical in shape?   &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2018-03-07T03:14:33.840" Title="Why are atom spherical in shape ?" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13767" PostTypeId="2" ParentId="13756" CreationDate="2014-06-20T08:01:53.423" Score="2" Body="&lt;p&gt;Disclaimer: what I offer here is an educated guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4NO3}$ is known to decompose quite easily, especially if proper catalyst is present. Thus, one of the reactions is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH_4NO_3 = N_2 + 2H_2O + [O]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;generated oxygen may be used to oxidize sucrose or black iron oxide. I doubt that $\ce{N2O}$ is formed, since it is quite strong oxidizer and decomposes quite easily, especially if redox catalyst is present&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C_12H_22O_11 + 24[O] = 12 CO_2 + 11 H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Fe3O4 +[O] = 3 Fe2O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you said, the snake is black, so the second reaction does not occur&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the final process is dehydration of sugar, producing carbon (actually, the carbon may contain significant amount of hydrogen and oxygen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C_12H_22O_11 = 12 C + 11 H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the same time numerous other processes, producing gaseous reducing organics, happens, so I advise to use a fume hood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cooled, reducing atmosphere from sugar decomposition vanishes, and black iron oxide reacts with air&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4Fe3O4 +O2 = 6 Fe2O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since black iron oxide acts as a catalyst and does not melt on reaction conditions, it is good to use as fine one as you can get and maybe it may be used in smaller amounts. Since ammonium nitrate acts as main energy source here, it is likely that it can be used in larger amounts. Sucrose forms the body of the 'snake', so it may be used in smaller amount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'snake' consists of (partially) carbonized/caramelized sucrose binding other ashes together and leftovers from iron oxide. The expansion of volume results from gases emitted in the processes of decomposition of ammonium nitrate and decomposition of sucrose, though decomposition of ammonium nitrate is probably the main source here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, thanks for recipe, I think it would come handy for me this year.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-06-21T05:43:20.430" LastActivityDate="2014-06-21T05:43:20.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13768" PostTypeId="1" CreationDate="2014-06-20T09:26:28.353" Score="3" ViewCount="104" Body="&lt;p&gt;I got this reaction and I can't figure out the end products&#xA;$$\ce{SO2 + S + CaO + Na2SO4 -&amp;gt; Na2S2O3 +  ~?}$$&#xA;I am guessing some calcium sulfur salt should form, but which one?&lt;/p&gt;&#xA;" OwnerUserId="6893" LastEditorUserId="5026" LastEditDate="2018-07-18T03:07:38.023" LastActivityDate="2018-07-18T03:07:38.023" Title="What are the other end products in sulfur oxoacid formation?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13769" PostTypeId="1" AcceptedAnswerId="13776" CreationDate="2014-06-20T11:43:38.020" Score="1" ViewCount="1624" Body="&lt;p&gt;A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water.&#xA;What is the total number of atoms in the compound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what's the idea of this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what i know is that atoms number is found by multiplying the Avogadro number by the mol's number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;??&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-06-20T16:52:01.030" Title="A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13771" PostTypeId="2" ParentId="13764" CreationDate="2014-06-20T14:00:26.963" Score="1" Body="&lt;p&gt;You need to be more careful with your units. The erroneous result is that you are equating a value in amu (a measure of mass, like grams) with a value in grams per mole (an invariant property of an element or compound, regardless of the amount you have). &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-06-20T14:00:26.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13772" PostTypeId="2" ParentId="13766" CreationDate="2014-06-20T15:28:14.597" Score="3" Body="&lt;p&gt;I think you should think this question a little through:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The easy answer is that atoms are not spherical in shape. Sphere is a solid object, atoms to not have well defined boundaries with &quot;solid cutoffs&quot;. Atoms are more like clouds consisting distributions of electrons, neutrons and protons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Atoms, per definition, are electrons + a single nucleus, ie. the huge positive charge of a point-like object (nucleus) pulls together the whole thing. That result is that electrons approximately move in a central field. Such central field generates spherical symmetry of the electronic states, and that is why you have e.g. spherical harmonics when you describe the atomic orbitals. Check the solution of the Schrodinger equation how it leads to states, and how wave-functions with only radial dependence (i.e. sphere-like) appears.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;+++++ Incorrect part of the answer was deleted based on the comments of DavePhD. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2015-01-05T17:04:21.820" LastActivityDate="2015-01-05T17:04:21.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13773" PostTypeId="1" CreationDate="2014-06-20T15:47:36.310" Score="2" ViewCount="2507" Body="&lt;p&gt;I tried to get the formula of potassium ammonium sulfate. What I got is $\ce{K(NH4)2SO4}$ but I don't think this the right formula. I searched on the internet and found an answer ($\ce{KNH4SO4}$), but I don't think it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the correct formula of potassium ammonium sulfate?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="194" LastEditDate="2018-02-17T16:08:40.777" LastActivityDate="2018-02-17T16:46:37.697" Title="What is the formula of potassium ammonium sulfate?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13774" PostTypeId="2" ParentId="13764" CreationDate="2014-06-20T16:26:13.640" Score="16" Body="&lt;p&gt;You are correct, but to make it a little more clear you can include the assumed &quot;atom&quot; in the denominator of amu:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;m_{\ce{C}^{12}} &amp;amp;= \pu{12amu atom^-1} \\ \\&#xA;m_{\ce{C}^{12}} &amp;amp;= \pu{12g mol^-1} \\ \\&#xA;\pu{12amu atom^-1} &amp;amp;= \pu{12g mol^-1} \\ \\&#xA;\pu{1amu atom^-1} &amp;amp;= \pu{1g mol^-1}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the &lt;strong&gt;ratio of amu/atom is the same as the ratio of g/mol.&lt;/strong&gt; The definitions of amu and moles were intentionally chosen to make that happen (I'm surprised your teacher didn't explain this, actually). This allows us to easily relate masses at the atomic scale to masses at the macroscopic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check this, look at the mass of an amu when converted to grams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{1amu}= \pu{1.6605E-24 g}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now divide one gram by one mole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{1g mol^-1}= \frac{\pu{1 g}}{\pu{6.022E23 atom}} = \pu{1.6605E-24 g atom^-1}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the same number! Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{1g mol^-1}= \pu{ 1 amu atom^-1}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2018-10-03T11:27:20.343" LastActivityDate="2018-10-03T11:27:20.343" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="13776" PostTypeId="2" ParentId="13769" CreationDate="2014-06-20T16:52:01.030" Score="1" Body="&lt;p&gt;You have to calculate the amount of substance with the mass of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can simply multiply all values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$  &lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2014-06-20T16:52:01.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13777" PostTypeId="2" ParentId="13730" CreationDate="2014-06-20T17:02:36.910" Score="0" Body="&lt;p&gt;Exchanging two groups in fischer projection changes it to its enantiomer and further exchange reverts it back.After exchanging as follows gives meso compound with internal plane of symmetry:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that these are for the compunds on the left. &#xA;  Do similiarly for others on right &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;a) $\ce{CH3&amp;lt;-&amp;gt;H}$ THEN $\ce{H&amp;lt;-&amp;gt;OH}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)Rotate nearer $C$ by $\pi/3$ rad. anticlock wise.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c)Imagine rotation about $C_2-C_3$ by $\pi$ rad., the $OH$ in the right back of the plane comes in front projecting out towards us hanging down and thus we have a vertical plane of symmetry.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-20T17:02:36.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13778" PostTypeId="2" ParentId="13773" CreationDate="2014-06-20T17:03:34.330" Score="7" Body="&lt;p&gt;$\ce{K(NH_4)SO_4}$ is the correct formula. You have two unipositive cations and a single dinegative anion. The total charge must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yn3Cx.gif&quot; alt=&quot;structural formula of the ionic compound&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5614" LastEditorUserId="5026" LastEditDate="2018-02-17T16:46:37.697" LastActivityDate="2018-02-17T16:46:37.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13779" PostTypeId="1" AcceptedAnswerId="14456" CreationDate="2014-06-20T17:24:33.433" Score="12" ViewCount="2212" Body="&lt;p&gt;Using the rigid rotor approximation to the level energies, and such other appropriate assumptions, we can approximate rotational partition function, &lt;span class=&quot;math-container&quot;&gt;$Q_{\mathrm{rot}}$&lt;/span&gt;, of linear molecules as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{eqnarray}&#xA;  Q_{\mathrm{rot}} &amp;amp; = &amp;amp; \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{E_J}{kT}\right)}\\&#xA;  &amp;amp; \simeq &amp;amp; \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{hB_0J(J+1)}{kT}\right)} \\&#xA;\end{eqnarray}&lt;/span&gt;&#xA;In the book, &lt;a href=&quot;http://app.knovel.com/web/toc.v/cid:kpMMSE0001/viewerType:toc/root_slug:microwave-molecular-spectra/b-off-set:0/b-cat-name:Chemistry%20&amp;amp;%20Chemical%20Engineering/b-cat-slug:chemistry-chemical-engineering/b-cat-id:213/b-sub-cat-id:89/b-topic-name:Analytical%20Chemistry/b-topic-slug:analytical-chemistry/b-off-set:0/b-order-by:name/b-sort-by:ascending/b-filter-by:all-content&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microwave Molecular Spectra&lt;/a&gt; (Gordy &amp;amp; Cook, 1984), Chapter 3, Equation 3.64, the authors made approximation as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_{rot}\simeq\frac{kT}{hB_0}+\frac{1}{3}+\frac{1}{15}\left(\frac{hB_0}{kT}\right)+\frac{4}{315}\left(\frac{hB_0}{kT}\right)^2$$&lt;/span&gt;&#xA;Also, there is another approximation made by McDowell, R. S. 1988, Journal of Chemical Physics, 88, 356:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_{rot}\simeq\frac{kT}{hB_0}\exp\left(\frac{hB_0}{3kT}\right)$$&lt;/span&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/AnhGk.png&quot; alt=&quot;rotational partition fuction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can see from the picture that &lt;span class=&quot;math-container&quot;&gt;$Q_{\mathrm{rot}}\simeq\frac{kT}{hB_0}+\frac{1}{3}$&lt;/span&gt; is also a good approximation. Then I can guess how to derive McDowell's approximation, but I cannot derive Gordy &amp;amp; Cook's approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have got are as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$$n\exp(-x(n+1)n)+\int_{(n+1)n}^{(m+1)m}{\exp(-xy)}\,\mathrm dy+(m+1)\exp(-x(m+1)m) \\&#xA;\leq\sum_{J=n}^{m}{(2J+1)\exp(-x(J+1)J)} \\&#xA;\leq\sum_{i=n^2}^{m^2+2m}{\exp(-xi)}$$&lt;/span&gt; hence, &lt;span class=&quot;math-container&quot;&gt;$$\int_{0}^{\infty}{\exp(-xy)}\,\mathrm dy=\frac{1}{x} \\ &#xA;\leq\sum_{J=0}^{\infty}{(2J+1)\exp(-x(J+1)J)} \\ &#xA;\leq\sum_{i=0}^{\infty}{\exp(-xi)}=\frac{1}{1-\exp(-x)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., &lt;span class=&quot;math-container&quot;&gt;$$\frac{kT}{hB_0}&amp;lt;Q_{\mathrm{rot}}&amp;lt;\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the latter one &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1/%281-e%5E%28-x%29%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;approximates&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{1-\exp\left(-\frac{hB_0}{kT}\right)}\approx \frac{kT}{hB_0}+\frac{1}{2}+\frac{1}{12}\frac{hB_0}{kT}-\frac{1}{720}\left(\frac{hB_0}{kT}\right)^3$&lt;/span&gt; when &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is rather high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyone knowing how to derive Gordy &amp;amp; Cook's approximation? Please tell me and I appreciate it very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, another little question, are such approximations still useful as nowadays computers are greatly powerful?&lt;/p&gt;&#xA;" OwnerUserId="6892" LastEditorDisplayName="user7951" LastEditDate="2019-01-26T15:14:32.600" LastActivityDate="2019-01-26T15:14:32.600" Title="How to make approximation of rotational partition function of diatomic linear molecules?" Tags="&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13780" PostTypeId="1" CreationDate="2014-06-20T20:37:16.857" Score="3" ViewCount="1566" Body="&lt;p&gt;Can Argon hybridize orbitals and/or form covalent-like/ionic-like compounds? &#xA;Is there any study of that? I would be happy to read concrete references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of molecular geometries for argon compounds are known or hypothesized at the current time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some peculiarity to apply something like VSEPR or molecular orbital theory to theoretical or experimentally known argon compounds?&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="4945" LastEditDate="2015-02-24T12:58:14.087" LastActivityDate="2016-07-20T23:15:39.857" Title="Are ther known compounds of Argon? What is their molecular geometry and hybridisation?" Tags="&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;reference-request&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13781" PostTypeId="1" AcceptedAnswerId="13782" CreationDate="2014-06-20T21:13:51.190" Score="5" ViewCount="1642" Body="&lt;p&gt;Lets say you have a buffered solution containing $0.50~\mathrm{M}$ acetic acid ($K_a = 1.8 \cdot 10^{-5}$) and $0.50~\mathrm{M}$ sodium acetate, and you want to calculate the $\mathrm{p}\ce{H}$ of this solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done in two ways: using the standard equilibrium chart (initial, change, equilibrium), or using the Henderson Hasselbalch equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do these two give the same result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Henderson Hasselbalch equation is derived starting with the equilibrium equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = \frac{\ce{[H+][CH3COO^{-}]}}{\ce{[CH3COOH]}}$ and then the $\ce{[H+]}$ is found, and then the $\log_{10}$ is taken. However, doesn't this assume that the system is already at equilibrium and that the concentrations of $\ce{[CH3COO^{-}]}$ and $\ce{[CH3COOH]}$ are the concentrations at equilibrium? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that is clearly not the case as shown by the ICE chart. However, both give the correct pH. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="22102" LastEditDate="2020-11-19T08:53:21.047" LastActivityDate="2020-11-19T08:53:21.047" Title="How to calculate the pH of a buffered solution with Henderson Hasselbalch?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;henderson-hasselbalch-equation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13782" PostTypeId="2" ParentId="13781" CreationDate="2014-06-20T22:16:14.443" Score="2" Body="&lt;p&gt;The main equilibrium in such a solution is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH +CH3COO^- &amp;lt;=&amp;gt;CH3COOH + CH3COO^-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH +H_2O &amp;lt;=&amp;gt;CH3COO^- + H_3^+O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, doesn't this assume that the system is already at equilibrium&#xA;  and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$&#xA;  are the concentrations at equilibrium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH +CH3COO^- &amp;lt;=&amp;gt;CH3COOH + CH3COO^-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have these equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH + H2O- &amp;lt;=&amp;gt;H_3^+O + CH3COO^-}$&#xA;$\ce{H2O + CH3COO^- &amp;lt;=&amp;gt;CH3COOH + HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution &lt;em&gt;electro&lt;/em&gt; neutrality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass conservation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution electroneutrality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-20T22:28:35.827" LastActivityDate="2014-06-20T22:28:35.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13783" PostTypeId="1" AcceptedAnswerId="14495" CreationDate="2014-06-20T22:20:51.517" Score="5" ViewCount="3097" Body="&lt;p&gt;I'm told that when sodium chlorite $\ce{(NaClO2)}$ is mixed with aqueous acid, chlorine dioxide ($\ce{ClO2}$) is produced and remains mostly dissolved in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the strength of the acid (concentration of acid and/or the type of acid) affects this reaction. I would also like to know what happens (afterwards) to the $\ce{ClO2}$ over time. For example, if undisturbed, does the $\ce{ClO2}$ enter the air, convert into another compound, or something else?&lt;/p&gt;&#xA;" OwnerUserId="6895" LastEditorUserId="7475" LastEditDate="2017-09-08T10:39:36.503" LastActivityDate="2017-09-08T10:39:36.503" Title="What happens when a solution of sodium chlorite is acidified?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13784" PostTypeId="1" AcceptedAnswerId="14623" CreationDate="2014-06-20T22:32:57.917" Score="4" ViewCount="1926" Body="&lt;p&gt;I keep reading that solutions are electrically neutral, but I've never read an explanation. I've read after the fact justifications, such as one problem which asked me to calculate the force of repulsion between two beakers with a tiny, tiny bit of excess charge (it ended up being over 10,000 pounds). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are solutions electrically neutral? Not interested in after the fact explanations like the above example. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-25T19:08:59.007" Title="Why are solutions electrically neutral?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13786" PostTypeId="1" CreationDate="2014-06-21T10:50:20.120" Score="0" ViewCount="2896" Body="&lt;p&gt;The solubility product of $\ce{AgBr}$ is $7.7 \cdot 10^{-13}\:\mathrm{mol^2/L^2}$.&#xA;What was the initial concentration of $\ce{AgNO3}$ solution, if the precipitation of $\ce{AgBr}$ appears after the addition of $20\:\mathrm{mL}$ of a $0.001$ molar solution of $\ce{NaBr}$ to $500\:\mathrm{mL}$ of the $\ce{AgNO3}$ solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the solution as $0.054\:\mathrm{M}$. I'm confused with the procedure. This is what I have done.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Precipitate occurs at $K_{sp}=Q$ and $Q=[\ce{Ag+}][\ce{Br-}]$&lt;/li&gt;&#xA;&lt;li&gt;$[\ce{Ag+}] = [\text{(Vol of $\ce{AgNO3}$)}\cdot\text{Molarity} ] / \text{Total Volume of the mixture}$&lt;/li&gt;&#xA;&lt;li&gt;Similarly for $\ce{Br-}$&lt;/li&gt;&#xA;&lt;li&gt;$[\ce{Ag+}][\ce{Br-}] = 2\cdot10^5$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The answer I got it $0.054\:\mathrm{M}$ ? Is that correct ?&lt;/p&gt;&#xA;" OwnerUserId="6899" LastEditorUserId="4945" LastEditDate="2014-06-21T17:33:48.933" LastActivityDate="2014-07-28T13:37:28.537" Title="Finding the initial concentration" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13787" PostTypeId="1" AcceptedAnswerId="13791" CreationDate="2014-06-21T12:27:03.333" Score="2" ViewCount="996" Body="&lt;p&gt;I understand that the strength of HF differs significantly depending on the concentration. I understand the basic theory behind this - homoassociation - and stabilization of the highly negative charge dense fluoride anion through hydrogen &quot;bonds&quot;/intermolecular attractions with undissociated solute $\ce{HF}$ molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am unclear on the Wikipedia explanation of homoassociation:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Homoassociation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Homoassociation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the equations - I cannot tell if the authors meant to refer to pure, liquid $\ce{HF}$ or a solution of HF. I'm thinking that the authors are referring to liquid $\ce{HF}$ because auto-ionization of HF is mentioned. I'm thinking that auto-ionization would only be significant in liquid $\ce{HF}$. Or in a concentrated solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, regarding the second equation - the authors make it look like the fluoride anion is forming an actual, &lt;em&gt;intramolecular&lt;/em&gt;, bond with the $\ce{HF}$ molecule. I think the authors are implying the formation of an &lt;em&gt;intermolecular&lt;/em&gt; hydrogen bond, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the bifluoride anion seems to be a real molecular entity (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bifluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Bifluoride&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ltIj2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only muddies the water further. I.e. which of the following best represents how the fluoride anion is stabilized, causing concentrated solutions of $\ce{HF}$ to ionize further than one would expect from $\ce{K_a}$ values? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HF}$ ... $\ce{F^-}$ (hydrogen bonding) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HF_2^-}$ (formation of bifluoride anion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any further links to resources about homoassociation would be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: Also, can't the fluoride anion form hydrogen bonds with water? The fact that concentrated HF is stronger than dilute HF than what mere $\ce{Ka}$ values suggest implies that H-bonding with water isn't as effective as H-bonding with HF molecules. Why? Is water less polar? On the other hand water can form a maximum of &lt;em&gt;two&lt;/em&gt; hydrogen bonds with the fluoride anion, while a HF molecule can only form a maximum of &lt;em&gt;one&lt;/em&gt; hydrogen bond with a fluoride anion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-21T14:34:41.637" LastActivityDate="2014-06-21T16:32:44.047" Title="Homoassociation and Acid Strength" Tags="&lt;acid-base&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13789" PostTypeId="1" AcceptedAnswerId="13790" CreationDate="2014-06-21T13:02:46.677" Score="6" ViewCount="1046" Body="&lt;p&gt;What is the systematic name of carbonyl sulfide? My professor commented that the systematic name was &quot;very odd&quot; so he didn't bother to mention it. Is it really that odd? And what is it, actually?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the name &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.9644.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Thioxomethanone&quot;&lt;/a&gt;. Is that really odd? From my limited knowledge, I've seen the root &quot;thiol&quot; before in describing something with a sulfur in it. So at least the root doesn't seem weird. I also see &quot;oxo.&quot; There's an O in OCS. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-07-24T11:21:57.210" LastActivityDate="2018-07-24T11:58:55.870" Title="Systematic name of OCS (COS)" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13790" PostTypeId="2" ParentId="13789" CreationDate="2014-06-21T13:22:03.430" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonyl_sulfide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; lists the IUPAC name as &quot;carbon oxide sulfide.&quot;  Odd? Sounds kind of logical to me.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-21T13:22:03.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13791" PostTypeId="2" ParentId="13787" CreationDate="2014-06-21T16:24:43.393" Score="4" Body="&lt;p&gt;I think this &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_fluoride&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on hydrogen fluoride&lt;/a&gt; will answer your questions.  It discusses the acidity of $\ce{HF}$ from dilute to concentrated aqueous solution, and then pure $\ce{HF}$.  They present four equilibria to describe these situations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{H2O + HF &amp;amp;~&amp;lt;=&amp;gt; [H3O^+ {.} F^{-}]&amp;amp;&amp;amp;(I)}\\&#xA;\ce{[H3O^+ {.} F^{-}] &amp;amp;~&amp;lt;=&amp;gt; H3O^+ + F^{-}&amp;amp;&amp;amp;(II)}\\&#xA;\ce{[H3O^+ {.} F^{-}] + HF &amp;amp;~&amp;lt;=&amp;gt; H3O^+ + HF2^{-}&amp;amp;&amp;amp;(III)}\\&#xA;\ce{3HF &amp;amp;~&amp;lt;=&amp;gt; H2F^+ + HF2^{-}&amp;amp;&amp;amp;(IV)}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dilute solution equilibrium (I) lies far to the right and the second equibrium (II) far to the left.  So even though the $\ce{HF}$ is extensively dissociated, the resultant hydrogen bonded ion pair is very stable and reluctant to undergo significant dissociation.  Hence the solution is weakly acidic.  As the concentration of $\ce{HF}$ is increased homodissociation comes into play [equilibrium (III)] producing the hydrogen difluoride ion as a discrete entity which is stabilized through intermolecular hydrogen bonding.  The increase in $\ce{H3O+}$ makes the resultant solution very acidic.  In pure $\ce{HF}$, autoionization occurs producing an exteremly acidic solution.  Page 4 of &lt;a href=&quot;http://www.chem.purdue.edu/chemsafety/Equip/HFfacts10.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; shows a graph of HF acidity from near zero to 50% concentration.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-06-21T16:32:44.047" LastActivityDate="2014-06-21T16:32:44.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13792" PostTypeId="1" CreationDate="2014-06-21T16:48:41.887" Score="1" ViewCount="76" Body="&lt;p&gt;Is there a possible theoretical effect of mechanical vibration on the stability of &#xA;organic molecules.  E.g. some medicines degrade if they are not stored in a fridge, is there a theoretical foundation that suggests that low frequency vibrations might have the same effect?&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="5354" LastEditDate="2015-09-11T12:56:32.883" LastActivityDate="2015-09-11T12:56:32.883" Title="Is there a possible effect of mechanical vibration on the stability of organic molecules?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;stability&gt;&lt;mechanochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13793" PostTypeId="2" ParentId="13792" CreationDate="2014-06-21T17:37:29.767" Score="3" Body="&lt;p&gt;What do you mean by &quot;low frequency vibrations&quot;?  The term &quot;extemely low frequency&quot; is used to describe earthquake type frequencies (3-30 Hz, give or take), whereas &quot;low frequency&quot; is used to describe frequencies in the 30kHz - 300 kHz range.  If you mean &quot;low frequency&quot; as defined in the previous sentence, then yes, that frequency range of vibration is known to have chemical effects.  The process involves the formation and collapse of bubbles (cavitation) in the solution.  High temperatures and pressures can be generated during the bubble formation and collapse process.  The typical lab apparatus can homogenize samples, produce catalytic effects by cleaning metal surfaces or increasing phase transfer, disrupt cells, and break some (weak?) chemical bonds.  This &lt;a href=&quot;http://www.europeansocietyofsonochemistry.com/introduction/&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; reports,&quot;Ligand-metal bond cleavage in transition metal complexes to give coordinatively unsaturated species or modified complexes as well as complete strip off of ligands to produce amorphous metals disrupture of the solvent structure altering the solvation of reactants Sonolysis of molecules (homolytic fragmentation to radicals, rupture of polymers, generation of excited states, cell disrupture).&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-21T17:37:29.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13794" PostTypeId="2" ParentId="6901" CreationDate="2014-06-21T18:02:00.507" Score="11" Body="&lt;p&gt;The electron being gained by fluorine would be taken in to a much smaller 2p orbital and requires more electron coupling energy than that of much larger 3p orbital of chlorine. Therefore, energy released during the electron gaining process of fluorine is less than that of chlorine.&lt;/p&gt;&#xA;" OwnerUserId="6905" LastEditorUserId="13339" LastEditDate="2015-05-18T07:38:11.467" LastActivityDate="2015-05-18T07:38:11.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13795" PostTypeId="1" AcceptedAnswerId="13798" CreationDate="2014-06-21T23:42:41.623" Score="1" ViewCount="3487" Body="&lt;p&gt;I will be given a binary mixture containing two unknown compounds(the mixture could be either liquid or solid- I don't know yet). The plan of the experiment is to identify the two compounds. So my primary concern is to separate the binary mixture and purify the individual compounds(then the subsequent classification tests/NMR/IR analysis should be straightforward).  I am thinking of doing an acid-base extraction but don't know how to since I don't know any physical properties of the compounds. Any suggestions would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="6833" LastEditDate="2014-06-21T23:56:51.620" LastActivityDate="2014-06-22T00:42:26.580" Title="Separation of Binary mixture?" Tags="&lt;acid-base&gt;&lt;analytical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13796" PostTypeId="1" AcceptedAnswerId="14559" CreationDate="2014-06-22T00:16:47.800" Score="2" ViewCount="1963" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3bwy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the central chlorine atom is best described as $\ce{sp^3}$ hybridized. This makes sense; the molecule is pyramidal. Also, due to the molecule's pyramidal geometry, this rules out the possibility of &lt;em&gt;effective&lt;/em&gt; resonance due to less overlap between the  $\ce{sp^3}$ lobes, which are angled, as opposed to the parallel pi lobes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JmClE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains quite well the difference in acidity between nitric acid and chloric acid; nitric acid is stronger than chloric acid. At first it seems difficult to differentiate between the acidities of both; the nitrate anion also has three peripheral oxygen atoms, which provide a similar (negative) inductive effect by delocalizing electron density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/28oPK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to electronegativity, the chlorate anion seems like it should be more &lt;em&gt;slightly&lt;/em&gt; stable than the nitrate anion, implying that chloric acid is stronger than nitric acid; chlorine is more electronegative than nitrogen (if only slightly so - difference is 0.12 on the Pauling scale).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation state analysis suggests that the two should be similar in acidity; both the chlorine and the nitrogen have oxidation states of +5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems that the best rationalization of why nitric acid is stronger than chloric acid is the &quot;better&quot; (overlapping pi orbitals) resonance form of the nitrate anion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the above rationalization, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How, exactly, do parallel pi orbitals enable &quot;better&quot; resonance? I get the principle of maximum overlap. If I had to guess, I would say that more overlap enables greater distribution of electrons among the relevant orbitals ... is there a more concise and scientific way of putting this? Is this even correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Are parallel pi orbitals necessary for resonance? I got the impression it is from organic class. Or are parallel pi orbitals only necessary for a special form of resonance - conjugation? Is the picture on the bottom referring to conjugation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Are there any other ways to rationalize the difference in strength between these two acids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BBqw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-28T22:18:51.830" LastActivityDate="2014-07-28T22:18:51.830" Title="Nitric acid vs. chloric acid strength" Tags="&lt;hybridization&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13797" PostTypeId="1" AcceptedAnswerId="13805" CreationDate="2014-06-22T00:37:49.283" Score="1" ViewCount="118" Body="&lt;p&gt;The energy level of a hydrogen atom in its ground state is $-13.6\ \mathrm{eV}$. &#xA;How is this value calculated and how can we calculate the same for different orbits?&lt;/p&gt;&#xA;" OwnerUserId="6902" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T13:21:46.873" LastActivityDate="2015-05-01T13:21:46.873" Title="Hydrogen atom energy level" Tags="&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-01T16:06:36.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="13798" PostTypeId="2" ParentId="13795" CreationDate="2014-06-22T00:42:26.580" Score="3" Body="&lt;p&gt;&lt;strong&gt;N.B.&lt;/strong&gt;: the advice below mostly applies to organic compounds. If you're dealing with inorganic salts, or solutions thereof, the approach will be very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a solid mixture, I would take an IR first. That should allow you to rule out potentially useless chemical tests by identifying the functional groups present in your mixture without wasting any of the sample. This would also give you some indication whether acid/base extraction is likely to be of any use. After that, it seems logical to test solubility in a wide range of solvents and at different temperatures. If you can find one that dissolves your mixture only partially at low temperature, but completely at high temperature, then you can likely separate the components immediately by recrystallization. Judicious use of a centrifuge can help if the crystals form a fine suspension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a liquid mixture, I would again take an IR initially. In some cases, one component of the mixture can be chemically separated, and IR results would again be useful in determining the viability of that approach. Ketones and aldehydes, for example, will react to form semicarbazones that can be recrystallized and then identified by melting point. As far as physical methods of separation, standard are distillation and fractional freezing.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-22T00:42:26.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13799" PostTypeId="1" AcceptedAnswerId="13804" CreationDate="2014-06-22T02:28:45.170" Score="0" ViewCount="17996" Body="&lt;p&gt;According to the dipole moment's definition, as polarity increases the distance between the centres of positive charge &amp;amp; negative charge increases.In the ionic bond ,as the polarity is maximum, &lt;strong&gt;distance between the two charged centres is maximum&lt;/strong&gt;.Is this concept correct? If not what is the relationship between distance and polarity. I m not getting any intuition.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="5531" LastEditDate="2014-06-22T12:49:30.890" LastActivityDate="2014-06-22T13:05:55.633" Title="Does the distance increase as the polarity increases?" Tags="&lt;physical-chemistry&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13800" PostTypeId="2" ParentId="13799" CreationDate="2014-06-22T02:43:32.897" Score="1" Body="&lt;p&gt;I'm not sure if the guideline actually works out in empirical observations. There are at least two factors affecting bond length - atomic radii - and polarity of the bond. A more polar bond will have a less even distribution of electrons. The electron density will generally be closer to the more electronegative atom. This isolation of electrons &lt;em&gt;may&lt;/em&gt; have the effect of &quot;stretching&quot; the bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following two molecules, both of which are linear and polar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Plycd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hd3gR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From polarity considerations alone, we might expect the $\ce{H-F}$ bond to be longer than the $\ce{H-Cl}$ bond because fluorine is more electronegative than chlorine. However, chlorine is a bigger atom than fluorine by nature of having additional sublevels of electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, consider the below data: bond length actually increases with decreasing polarity (quantified through dipole moments). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluoromethane Bond length 136pm, Bond strength 108 kcal/mol, Dipole moment (D) 1.85&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromomethane Bond length 178pm, Bond strength 84 kcal/mol, Dipole moment (D) 1.87&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloromethane Bond length 193pm, Bond strength 70 kcal/mol, Dipole moment (D) 1.81&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodomethane Bond length 214pm, Bond strength 56 kcal/mol, Dipole moment (D) 1.62&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unfortunate thing about your guideline is that I thought that it would work at least across periods - in which atomic size stays &lt;em&gt;relatively&lt;/em&gt; constant. However, the facts do not fit the case. Consider $\ce{H-N}$ and $\ce{H-O}$. Oxygen is more electronegative than nitrogen; we'd expect the $\ce{H-O}$ bond to be more polar, and thus longer than the $\ce{H-N}$, at least according to your guideline. The opposite is actually true; the $\ce{H-O}$ bond is actually &lt;em&gt;shorter&lt;/em&gt; than the $\ce{H-N}$ bond. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-22T03:00:35.413" LastActivityDate="2014-06-22T03:00:35.413" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13801" PostTypeId="1" CreationDate="2014-06-22T07:42:58.283" Score="0" ViewCount="1784" Body="&lt;p&gt;Clay brick is made by heating them so their clay structure changes. What i understand is ultimately all materials have one or another base minerals in it particularly here this brick still it remains in some other form. So after process of heat, those minerals or nutrients can remain in such condition a plant can absorb? Can a brick dust be a good source of nutrients for plants to grow?&lt;/p&gt;&#xA;" OwnerUserId="6910" LastEditorUserId="9961" LastEditDate="2015-10-23T18:22:26.423" LastActivityDate="2015-10-23T18:22:26.423" Title="Can a clay brick waste in a soil be a source of minerals for plants?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13802" PostTypeId="1" CreationDate="2014-06-22T10:26:02.263" Score="3" ViewCount="1038" Body="&lt;p&gt;One day I took aluminium foil(sanitary brand,  don't know if you've  heard it)  then I casually burned it over  a gas lighter then it burned and gave off Grey fumes and the foil became a dull Grey where it had burnt.  I mistakenly may have inhaled a bit of fumes.  I had a slight pain in my right nostril the whole day. How is that?  Is that dangerous? &lt;/p&gt;&#xA;" OwnerUserId="6911" LastEditorUserId="4945" LastEditDate="2014-06-22T10:27:39.483" LastActivityDate="2014-06-23T11:34:42.073" Title="I heard aluminium doesn't produce fumes when burnt but the opposite happened, how?" Tags="&lt;safety&gt;&lt;metal&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13803" PostTypeId="2" ParentId="13802" CreationDate="2014-06-22T12:35:28.357" Score="4" Body="&lt;p&gt;It doesn't. Consumer aluminium foil often is a thin film on thicker polymer film, it is much more durable this way. The most common polymer for films is likely polyethylene, and when heated it produces nasty mix of pyrolysis and oxidation products. While not extremely toxic, they are definitely irritating and unhealthy. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-22T12:35:28.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13804" PostTypeId="2" ParentId="13799" CreationDate="2014-06-22T12:39:35.033" Score="1" Body="&lt;p&gt;Actually dipole moment is:&#xA;$$\huge \vec\mu=q.\vec d$$&#xA;where $q$ is the magnitue of charge on any particle and $\vec d$ is the distance between them;&#xA;This must imply that if charge increases or distance between the charges increases, dipole moment increases; The larger the dipole moment, the greater the polarity&lt;img src=&quot;https://i.stack.imgur.com/qb0wq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually when the more electronegative elemets pulls electrons, the partial charge, $q\le e$, increases and in case of 100% polar compund, $q=e$ whereas in 100% covalent compunds, $q=0$; so, you can see dipole moment, and polarity is maximum for ioninc compounds.But if two atoms have greater bond length, surely dipole moment and polarity increases.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-06-22T13:05:55.633" LastActivityDate="2014-06-22T13:05:55.633" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13805" PostTypeId="2" ParentId="13797" CreationDate="2014-06-22T12:56:15.887" Score="1" Body="&lt;p&gt;From the Bohr theory:&#xA;$$\large E=-\frac{Z^2e^4m}{8\epsilon_0^2n^2h^2}=-13.6\times\frac{Z^2}{n^2}\ \mathrm{eV}$$&#xA;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|}&#xA;&amp;amp; Z&amp;amp;\text{atomic number}&amp;amp;-\\&#xA;&amp;amp; e&amp;amp;\text{charge of electron}&amp;amp;1.60 \times 10^{-19}\ \mathrm{C}\\&#xA;&amp;amp; m&amp;amp;\text{mass of electron}&amp;amp;9.10 \times 10^{-31}\ \mathrm{kg}\\&#xA;&amp;amp; \epsilon_0&amp;amp;\text{electric permittivity}&amp;amp; 8.85 \times 10^{−12}\ \mathrm{F/m}\\&#xA;&amp;amp; n&amp;amp;\text{principal quantum number}&amp;amp;-\\&#xA;&amp;amp; h&amp;amp;\text{Planck's constant}&amp;amp;6.62 \times 10^{-34}\ \mathrm{m^2\ kg/s}\\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can put values of $Z$ and $n$ for any shell of a hydrogenic species like $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, $\ce{B^4+}$…&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T13:21:41.620" LastActivityDate="2015-05-01T13:21:41.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13806" PostTypeId="1" AcceptedAnswerId="13807" CreationDate="2014-06-22T14:06:10.877" Score="5" ViewCount="16250" Body="&lt;p&gt;Is the number of hyperconjugation structures for a hydrocarbon with number of 'alpha' hydrogens 'a' equal to 'a' or 'a+1'?&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="7448" LastEditDate="2015-10-16T16:50:52.207" LastActivityDate="2019-12-27T15:16:55.900" Title="Number of hyperconjugation structures" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;hyperconjugation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13807" PostTypeId="2" ParentId="13806" CreationDate="2014-06-22T14:26:44.100" Score="4" Body="&lt;p&gt;Let's use the n-propyl carbocation as an example.  It has &lt;strong&gt;two&lt;/strong&gt; hydrogens alpha to the cationic carbon and &lt;strong&gt;two&lt;/strong&gt; resonance structures involving hyperconjugation of those alpha hydrogens can be drawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PW7j8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-22T14:26:44.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13808" PostTypeId="1" CreationDate="2014-06-22T15:26:35.523" Score="1" ViewCount="168" Body="&lt;p&gt;I have a steam distillation setup with the flask for boiling the açaí powder with distilled water. Are you able to extract the essential oil from the boiling material?&lt;/p&gt;&#xA;" OwnerUserId="6914" LastEditorUserId="402" LastEditDate="2017-04-02T06:30:10.420" LastActivityDate="2017-04-02T06:30:34.003" Title="Steam distillation of açaí powder" Tags="&lt;home-experiment&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13809" PostTypeId="1" CreationDate="2014-06-22T15:36:25.510" Score="2" ViewCount="511" Body="&lt;p&gt;I have discerned these factors affecting orbital overlap:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Symmetry&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Type of orbital&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Nodes along the internuclear axis&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$n$ quantum number of the atomic orbitals - the number of shells of electrons&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Size of the atoms&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Geometry of molecule&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Angle of overlap&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With respect to the binary halogen acids, I still have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand how one might argue that the binary halogen acids become &lt;em&gt;more&lt;/em&gt; polar as one goes down group 7, in that electron density isn't very well shared; overlap is poor because of the small 1s orbital of hydrogen and the increasingly larger hybridized orbitals of the halogens. But this is half of what constitutes ionic character. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. what about the electron on hydrogen? As the halogens get bigger, electronegativity decreases. How can the electron from hydrogen be isolated toward the halogen as EN decreases?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="194" LastEditDate="2016-12-05T19:31:19.033" LastActivityDate="2016-12-05T19:31:19.033" Title="Effects of orbital overlap and electronegativity on polarity of binary halogen acids" Tags="&lt;bond&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13810" PostTypeId="1" AcceptedAnswerId="13835" CreationDate="2014-06-22T15:54:24.790" Score="6" ViewCount="18437" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following are in the correct order of their ionization energies? (multi-answer question)  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{O &amp;gt; S &amp;gt; S- &amp;gt;O-}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{F &amp;gt; F- &amp;gt; Cl- &amp;gt; Cl}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{O &amp;gt; O- &amp;gt; S- &amp;gt; S}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{F &amp;gt; Cl &amp;gt; Cl- &amp;gt; F-}$  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are 1 and 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is ionization energy of $\ce{S-}$ greater than $\ce{O-}$? The sulfur ion has an extra shell than an oxygen ion, so the atomic size of sulfur should be greater than oxygen. Therefore, the ionization energy of an oxygen ion should be more than a sulfur ion because the electrons in oxygen are closer to the nucleus when compared to sulfur. But why is the reverse true?&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="5026" LastEditDate="2018-07-16T02:26:32.317" LastActivityDate="2018-07-16T02:29:02.363" Title="Why is the ionization energy of sulfur anion more than that of oxygen anion?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13811" PostTypeId="1" AcceptedAnswerId="13812" CreationDate="2014-06-22T16:00:10.660" Score="0" ViewCount="5807" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the given compounds is most acidic? &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I solve this question, and how do I explain it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the oxidation state of carbon in both. The oxidation state in &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is higher, therefore it is more electronegative, so therefore &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; may be more acidic.&lt;br&gt;&#xA;For &lt;span class=&quot;math-container&quot;&gt;$\ce{CO }$&lt;/span&gt; I don't know about the charge so that I can use it to understand if it would give electron or accept electrons, so that I can apply Lewis acid/base theory.&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="4945" LastEditDate="2019-01-18T13:56:18.843" LastActivityDate="2019-01-18T13:56:18.843" Title="Is carbon dioxide more acidic than carbon monoxide?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13812" PostTypeId="2" ParentId="13811" CreationDate="2014-06-22T16:50:50.110" Score="5" Body="&lt;p&gt;Carbon dioxide is a much stronger Lewis acid than carbon monoxide. In carbon dioxide, the carbon has a much higher oxidation state (formally, $+4$ vs. the $+2$ of carbon monoxide), and the two oxygen atoms heavily withdraw electron density from the carbon atom, leaving a very large partial positive charge ($\delta^{+}$). This is rigorously confirmed by electrostatic potential maps and calculated orbital coefficients. As a consequence, the electron-deficient carbon of $\ce{CO2}$ is highly electrophilic, and reacts with a variety of nucleophiles (e.g., Grignard reagents, organolithium reagents, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon monoxide, on the other hand, actually contains substantial electron density at the carbon atom (in spite of the bonds being generally polarized towards the oxygen atom), which possesses a fairly localized and active lone pair. Consequently, a few notable exceptions aside, carbon monoxide mostly functions as a Lewis base.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-22T16:50:50.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13813" PostTypeId="1" AcceptedAnswerId="13826" CreationDate="2014-06-22T17:34:41.100" Score="1" ViewCount="958" Body="&lt;p&gt;Are there any solids with hydronium ion in its crystal structure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why not? I understand there are hydroxide-containing solids and gels such as aluminium hydroxide but why not hydronium ions? Does this have to do with the positive charge of hydronium ion? I could see the positive charge of hydronium ion excluding the possibility of forming any salts/solids with the group 1 and 2 metals. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="72973" LastEditDate="2020-02-20T14:05:16.807" LastActivityDate="2020-02-20T16:24:00.473" Title="Solid salts of hydronium ion" Tags="&lt;acid-base&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13814" PostTypeId="2" ParentId="13734" CreationDate="2014-06-22T17:48:32.163" Score="3" Body="&lt;p&gt;1) There is no, actually, a definitive, wide acknowledged measure of Lewis base/acid strength. For Brønsted acids/bases there is $\mathrm pK_\mathrm a$/$\mathrm pK_\mathrm b$ , which is thermodynamical (and actually depends on solvent, so there is no absolute scale of acid strength). [1] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping what said above in mind, a qualitative measure of Lewis base strength would be its ability to donate its lone pair. Obviously, it grows with negative charge of the particle and lowering its electronegativity, there is nothing new with it. It should grow with shrinking of negative ion, also understandable. The most common Lewis acid is proton, so comparing  strength of conjugated acid we can roughly estimate order of strength of Lewis base. [2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Quantitatively - no. Qualitatively - yes, it would in most cases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] On a side not, in some cases there IS a detectable difference between thermodynamical and kinetic Brønsted acidity. For details see 'sterically hindered bases' in context of organic synthesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] This, however, comes with a little twist: a Lewis base working as a strong base for one element may work as a weaker base for another element. This depends on relative size of donating and receiving orbitals size/energy. Some food for thoughts may be found in solubility table, comparing solubility of, for example, Be, Ba, Hg and Pb salts. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5026" LastEditDate="2018-02-28T16:04:07.520" LastActivityDate="2018-02-28T16:04:07.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13815" PostTypeId="1" AcceptedAnswerId="13883" CreationDate="2014-06-22T17:53:57.263" Score="2" ViewCount="104" Body="&lt;p&gt;What is the concentration of hydrochloric acid needed to have a solution with the activity of the hydrogen proton = 1? I read 1.18 M somewhere but I am unable to find a corroborating source. Also, how can the activity of the hydrogen proton = 1 when it doesn't exist by itself, suggesting that its activity should = 0 (since it can't act as an independent particle)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also how would one get to this result? Does one use a high-grade pH meter? Don't analytical-level pH meters measure activities rather than concentrations of hydronium ion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-24T20:50:52.907" Title="Activities of Solutes - especially acids and bases" Tags="&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13816" PostTypeId="1" AcceptedAnswerId="13817" CreationDate="2014-06-22T18:13:43.100" Score="1" ViewCount="7764" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;em&gt;Acetylation (adding an acetyl group) and phosphorylation (adding a phosphate group) make the histones more negatively charged because acetyl and phosphoryl groups are negative. They are &quot;glass is half empty&quot; molecules. By making histones more negatively charged, their grip on DNA will be much looser because DNA is also negatively charged. Similar charges (negative and negative) repel one another&lt;/em&gt;&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know oxygen and carbon have different electronegativities but that generates a partial negative charge and a partial &lt;strong&gt;positive&lt;/strong&gt; charge. So why do we consider the negative charge prevails over the positive charge in order to histones &quot;repel&quot; DNA?&lt;/p&gt;&#xA;" OwnerUserId="5540" LastActivityDate="2014-06-22T18:25:22.020" Title="Why does acetyl have negative charge?" Tags="&lt;organic-chemistry&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13817" PostTypeId="2" ParentId="13816" CreationDate="2014-06-22T18:25:22.020" Score="4" Body="&lt;p&gt;The acetyl group itself does not have any significant charge. But the residue that gets acetylated is lysine in this case, which has a positive charge. The acetylation results in a neutral amide, removing the positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HBUUZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Lysine_acetylation.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Wikimedia Commons&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd add that the effect of histone acetylation is likely more subtle than just the charge effect. This kind of posttranslational modification often affects the protein structure in some way, or is recognized by other proteins interacting with that site.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-06-22T18:25:22.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13818" PostTypeId="1" AcceptedAnswerId="13824" CreationDate="2014-06-22T18:32:46.270" Score="4" ViewCount="903" Body="&lt;p&gt;Can you reflux without any problems/dangers with a tilted condenser (e.g. because you don't have enough space in your fumehood)? I would say there isn't a problem, because both the surface area of the refluxed liquid and the length of the condenser (and thus the length of the trajectory of the water flowing through the condenser) remain unchanged. Is this thought correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my current setup&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOWgl.jpg&quot; alt=&quot;My setup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you should use countercurrent exchange right?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-07-17T10:33:50.653" LastActivityDate="2016-02-22T21:45:56.917" Title="Refluxing with tilted condenser" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13819" PostTypeId="1" AcceptedAnswerId="13821" CreationDate="2014-06-22T19:23:42.590" Score="11" ViewCount="22259" Body="&lt;p&gt;My book says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the number of electrons lost must equal the number gained, the half-reactions must be multiplied by integers as necessary to achieve the balanced equation. However, the value of e° is not changed when a half-reaction is multiplied by an integer. Since a standard reduction potential is an intensive property (it does not depend on how many times the reaction occurs), the potential is not multiplied by the integer required to balance the cell reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I do not understand why the reduction potential is unaffected by the number of mols. If you have a greater number of mols of the metal, doesn't that allow for a greater potential difference?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2017-11-20T00:50:26.130" Title="Why is standard reduction potential an intensive property?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;reduction-potential&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13820" PostTypeId="2" ParentId="13808" CreationDate="2014-06-22T19:27:09.477" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/A%C3%A7a%C3%AD_oil&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki on açaí oil&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Açaí  oil is obtained by cold pressing.  Steam distillation may be able to extract compounds like vanillic acid but may not get the anthocyanins or the triglycerides.  Many of the health benefits are from the healthy triglycerides and the anthocyanins.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="402" LastEditDate="2017-04-02T06:30:34.003" LastActivityDate="2017-04-02T06:30:34.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13821" PostTypeId="2" ParentId="13819" CreationDate="2014-06-22T19:29:28.593" Score="13" Body="&lt;p&gt;Look at the unit volt. 1 volt = 1 joule per coulomb. If the number of moles is doubled, the coulombs are doubled, as is the number of joules. Think of the volt as the driving force behind an individual electron in an electrochemical system.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorDisplayName="user7951" LastEditDate="2015-06-18T17:49:20.587" LastActivityDate="2015-06-18T17:49:20.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13822" PostTypeId="2" ParentId="13818" CreationDate="2014-06-22T19:36:25.200" Score="0" Body="&lt;p&gt;I can think about three problems in the case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Glass is fragile, so if any significant load falls onto flask's or condenser's necks, it is a risk. Thus, the condenser should be light and have separate support, and the flask should have a second support for bottom or have a very small load (say, if it would be more than 100 g, I would hesitate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) even round-bottom flasks does not necessarily have a perfectly spherical bottom, so fast stirring is absolutely out of question, and even gentle stirring for medium loads is not good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If highly tilted, Allihn condenser may trap significant amount of liquid in its spheres.  Dimroth and Inland Revenue condenser may trap some (little) amount of solvent on their coils too. Dimroth condenser MAY be used for refluxing even if should not ideally, and for solvents with high BP (say, toluene) should be OK. In fact, in case if violent reaction may occur it is PREFERABLE to use condenser with low internals resistance to straight float so the stream of boiling reaction mixture runned through the condenser in did not lift and smash it or teared the flask apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, if no gas exhaust is expected, solvent, reactants and products do not  have low bp and high toxicity I would prefer to reflux outside of the fume hood rather then tilted. It is OK. Even if some gas exhaust is expected, I would shrug of nonviolent non-toxic gas exhaust. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-06-22T19:43:18.867" LastActivityDate="2014-06-22T19:43:18.867" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13823" PostTypeId="1" CreationDate="2014-06-22T20:11:11.413" Score="1" ViewCount="30" Body="&lt;p&gt;I'm trying to simulate effects of carbonic anhydrase on cellular pH as caused by changes in bicarbonate. I assume a bicarbonate homeostatic mechanisms which replenishes intracellular concentrations from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 + H2O &amp;lt;=&amp;gt; H+ + HCO3- }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One paper I’ve found states the assumptions and rates, but doesn’t parameterize the differential (difference) relation between the substrates and products. The paper merely states that given the time scales I’m considering, it’s ok to assume $p(\ce{CO2})$ is constant, since carbon dioxide’s diffusion is fast in aqueous media. Secondly, the article states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The equilibrium constant of the reaction was $10^{-6.35}\ \mathrm{M}$ and the rate constant for $\ce{CO2}$ hydration was $10^6\ \mathrm{s^{-1}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do should I think of this reaction, and more importantly how do I simulate it?&lt;/p&gt;&#xA;" OwnerUserId="6908" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T16:27:58.260" LastActivityDate="2015-06-17T16:27:58.260" Title="Simulating enzymatic change rates of Carbonic anhydrase" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13824" PostTypeId="2" ParentId="13818" CreationDate="2014-06-22T20:37:58.103" Score="2" Body="&lt;p&gt;As long as everything is clamped and balanced - so that there isn't excessive weight on the joint connecting the flask to the condenser - the system should be OK.  After all, many rotary evaporators have the flask plus condenser tilted at an angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LTeDx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-22T20:37:58.103" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13826" PostTypeId="2" ParentId="13813" CreationDate="2014-06-22T21:09:44.360" Score="5" Body="&lt;p&gt;There is a section in this &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium#Solid_hydronium_salts&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia link&lt;/a&gt; entitled, &quot;Solid hydronium salts.&quot;  It says that any acid with an ionization constant of 10^9 or higher may form stable hydronium ion salts.  Examples they give include hydronium perchlorate and carborane-based hydronium ion salts.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-22T21:09:44.360" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13827" PostTypeId="1" CreationDate="2014-06-22T22:35:42.707" Score="4" ViewCount="17327" Body="&lt;p&gt;My question is about what it is - at the subatomic level - that determines the charge on atoms which participate in covalent bonding in reality. I understand that formal charge is just that - it's a formality, which gives human beings a very useful tool for modelling reality and making predictions. I, though, would like to know what is responsible for the charge we find in covalently bonded molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain. If we picture an ionic bond as a highly polar covalent bond, we can see that in an atom like NaCl we have the state of affairs where the highly electronegative chlorine atom has &quot;pulled away&quot; an electron from the sodium atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q21D9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Orbitals omitted for ease of drawing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately this means that Na, with its 11 protons has just 10 electrons, which gives it a positive charge, while Cl, with its 17 protons, now has 18 electrons, giving it a negative charge. Electrostatic forces between the two nuclei result in the bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it might be more accurate to say that the Na nucleus simply ceases to be significantly affected by its former 3s&lt;sup&gt;1&lt;/sup&gt; electron. As a result the 11 protons in the Na nucleus, which result in a +11 charge, are countered by only 10 electrons, giving a +1 charge overall. Meanwhile, the 17 protons of the Cl nucleus (+17 charge) experience the negative charge of 18 electrons (-18) giving a -1 charge overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare this with the situation we find in the carbonate ion. (&lt;a href=&quot;http://www.chem.ucla.edu/harding/tutorials/resonance/draw_res_str.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;) For the purposes of my question I won't worry too much about resonance, I don't mind too much whether we have the accurate -2/3 charge, but let's keep it simple and just look at one of the contributing structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FZX2A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Orbitals again omitted for ease of drawing. Orange indicates an additional electron from some other source. Oxygen has only eight electrons ordinarily, but one will be gained when forming this ion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we have covalent bonding. In the case of the singly bonded oxygens, we have 8 protons and 10 electrons. This is two more negatively charged electrons than positively charged protons. This, though, is not ionic bonding, it is covalent bonding and the polarisation is not such that the electron contributed by carbon dissociates completely from the Carbon. Instead it is &quot;shared&quot; between the two. Nevertheless, the &quot;sharing&quot; is not equal and oxygen attracts the electrons more strongly than the carbon, so we have a -1 charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, though, is - why is it that the charge works out like this? Why is it that when you have oxygen (8 protons) (charge of +8) and a filled outer shell (8 electrons plus buried inner core gives -10 (charge of -1 on each electron)) you have a charge of -1 not -2, is it simply the covalent bond, is it simply the fact that the two electrons in the sigma bond are sufficiently far enough away in reality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then why should it be that the doubly bonded oxygen, which also has ten electrons surrounding it, has no charge at all? Meanwhile, the carbon atom in the  middle is surrounded by 10 electrons also, whereas it has only 6 protons and yet carries no charge at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively the answer seems obvious. The bonds are covalent and the electrons are &quot;shared.&quot; So, in the double bond (which, of course, has a different nature to the single bond) one must assume that the electrons are positioned in such away that they are balanced by the protons of the oxygen nuclei. There is therefore no charge on this atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Carbon atom we surely again understand that the right balance of charges and distance from the nucleus is arrived at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall we have 32 electrons and 30 protons so the overall -2 charge on the whole atom makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate question I've been trying to lead up to is what is it that dictates the distribution of charge. What is the difference - at the subatomic level - between two ionically bonded atoms, a covalently bonded molecule like the CO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;2-&lt;/sup&gt; ion which carries a charge, and a covalently bonded molecule like H&lt;sub&gt;2&lt;/sub&gt;O - which is also polar (hence we can have hydrogen bonding) but isn't usually assigned a charge. On what basis do we assign charge?&lt;/p&gt;&#xA;" OwnerUserId="7147" LastActivityDate="2014-06-24T00:11:44.430" Title="How is the charge of covalently bonded atoms determined?" Tags="&lt;bond&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13828" PostTypeId="1" CreationDate="2014-06-23T00:38:36.117" Score="3" ViewCount="1247" Body="&lt;p&gt;Hydralazine is a drug used to treat heart disease. It is $59.99\%$ carbon, $5.03\%$ hydrogen, and $34.98\%$ nitrogen and has a molecular mass of $160.178$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the molecular formula for hydralazine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did:&#xA;\begin{align}&#xA;59.99~\ce{C}/12.011&amp;amp;=4.994\\&#xA;5.03~\ce{H}/1.01&amp;amp;=4.9801\\&#xA;34.98~\ce{N}/14.007&amp;amp;=2.4973.&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then divided all $3$ by $2.4973$, and got 2 carbon, 2 hydrogen, and 1 nitrogen atoms i.e. $\ce{C_2H_2N}$. The online homework said that the stoichiometry was wrong. I don't know what I did wrong.&lt;/p&gt;&#xA;" OwnerUserId="6918" LastEditorUserId="5026" LastEditDate="2018-02-21T10:38:32.587" LastActivityDate="2018-02-21T10:38:32.587" Title="How to determine the molecular formula of a compound given its molecular mass and mass percentage contribution of the involved elements?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13829" PostTypeId="2" ParentId="13796" CreationDate="2014-06-23T04:50:42.757" Score="1" Body="&lt;p&gt;First, we must see $pK_a's$, too see which factors which we must account:&#xA;$$\begin{array}{c|c|}&#xA;&amp;amp;\ce{HNO3}&amp;amp;-1.4\\&#xA;&amp;amp;\ce{HClO3}&amp;amp;-1\\&#xA;\end{array}$$&#xA;The strength of a Bronsted-Lowry acid depends upon the extent to which the proton can be separated from it and transferred to a base.Removing the proton involves breaking a bond tp the proton, and it involves making the conjugate base more electrically negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relationship between structure and acidity can be seen through factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hybridisation&lt;/li&gt;&#xA;&lt;li&gt;Inductive Effects(Electronegativity)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is also an effect of the solvent on acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's compare both acids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c}&#xA;&amp;amp;acid&amp;amp;EN&amp;amp;\text{Hybridisation}\\&#xA;&amp;amp;\ce{HNO3}&amp;amp;N\equiv3.0^*&amp;amp;sp^2&amp;amp;\\&#xA;&amp;amp;\ce{HClO3}&amp;amp;Cl\equiv3.0^*&amp;amp;sp^3&amp;amp;\\&#xA;\end{array}$$&#xA;$^*$may vary a little&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Hybridisation suggests N to be more acidic as %s character is more and thus the negative charge on O will be better stabilised; But Solvation is different due to size. Both have almost same electronegativity.In water essentialy both ionize to 100%.So you must &lt;strong&gt;here only&lt;/strong&gt; cosider solvation to be major factor.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Parallel pi-orbitals are necessary for resonance; consider pyridine where there is no resonance of N's lone pair as N is $sp^2$ which makes its lone pair in the plane of benzene(type) ring and thus this leaves a lone pair of electrons available on nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-07-26T05:16:34.367" LastActivityDate="2014-07-26T05:16:34.367" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13830" PostTypeId="2" ParentId="13828" CreationDate="2014-06-23T04:59:33.993" Score="4" Body="&lt;p&gt;The percentages are based on the molar mass. You have to multiply the percentages with the molar mass and divide by the mass of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{C}:&amp;amp;&amp;amp; \frac{0.5999 \cdot 160.178}{12.011} &amp;amp;\approx 8\\&#xA;\ce{H}:&amp;amp;&amp;amp; \frac{0.0503 \cdot 160.178}{1.01} &amp;amp;\approx 8\\&#xA;\ce{N}:&amp;amp;&amp;amp; \frac{0.3498 \cdot 160.178}{14.007} &amp;amp;\approx 4\\&#xA;       &amp;amp;&amp;amp; \implies \ce{C8H8N4}\\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="5614" LastEditorUserId="4945" LastEditDate="2014-12-02T09:02:34.390" LastActivityDate="2014-12-02T09:02:34.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13831" PostTypeId="2" ParentId="13810" CreationDate="2014-06-23T05:34:31.720" Score="0" Body="&lt;p&gt;Write the electronic configuration for all atoms:&#xA;$$\begin{array}{lll}\hline&#xA;\text{atom}&amp;amp;\text{before ionization}&amp;amp;\text{after ionization}\\\hline&#xA;\ce{O}    &amp;amp;1s^22s^22p^4             &amp;amp;1s^22s^22p^3(^*)\\\hline&#xA;\ce{O-}   &amp;amp;1s^22s^22p^5             &amp;amp;1s^22s^22p^4\\\hline&#xA;\ce{S}    &amp;amp;1s^22s^22p^63s^23p^4     &amp;amp;1s^22s^22p^63s^23p^3(^*)\\\hline&#xA;\ce{S-}   &amp;amp;1s^22s^22p^63s^23p^5     &amp;amp;1s^22s^22p^63s^23p^4\\\hline&#xA;\ce{F}    &amp;amp;1s^22s^22p^5             &amp;amp;1s^22s^22p^4\\\hline&#xA;\ce{F-}   &amp;amp;1s^22s^22p^6(^*)         &amp;amp;1s^22s^22p^5\\\hline&#xA;\ce{Cl}   &amp;amp;1s^22s^22p^63s^23p^5     &amp;amp;1s^22s^22p^63s^23p^4 \\\hline&#xA;\ce{Cl-}  &amp;amp;1s^22s^22p^63s^23p^6(^*) &amp;amp;1s^22s^22p^63s^23p^5 \\\hline&#xA;\end{array}$$&#xA;$(^*)$:Stable Configurations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{O^-}$ and $\ce{S^-}$ go to stable configurations  $\ce{S,O&amp;gt;S^- ,O^- }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since size of $\ce{S}$ is larger than $\ce{O}$ $\implies$ $\ce{S &amp;lt; O } $.&#xA;Assuming IE of $S^-$ reverse of EA of $S$ and $O^-$ reverse of EA of O, $O^-&amp;lt;s^-$ because S has large size to accomodate electron and less repulsion than small O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[My incorrect order $\ce{S&amp;lt;O &amp;lt; S- &amp;lt; O- }$ was actually $\ce{S&amp;lt;O &amp;lt; S+ &amp;lt; O+ }$]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\ce{O^- &amp;lt; S^- &amp;lt; S &amp;lt; O }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply same concept to others; Remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a)Nuclear Charge&lt;/li&gt;&#xA;&lt;li&gt;b)Shielding&lt;/li&gt;&#xA;&lt;li&gt;c)Atomic radii&lt;/li&gt;&#xA;&lt;li&gt;Penetration Effect[s&gt;p&gt;d&gt;f]&lt;/li&gt;&#xA;&lt;li&gt;d)Stability of fully and halfly filled orbitals.&lt;/li&gt;&#xA;&lt;li&gt;Successive Ionisations are always harder starting from neutral compound.$IE_1\ll IE_2\lll IE_3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a rule of Thumb:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$IE\propto \frac{(a)(d)}{(b)(c)} $, &lt;strong&gt;very roughly&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-06-23T15:56:18.567" LastActivityDate="2014-06-23T15:56:18.567" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13834" PostTypeId="1" AcceptedAnswerId="13838" CreationDate="2014-06-23T07:33:53.810" Score="2" ViewCount="6713" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Calculate the electronegativity of fluorine from the following data:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\text{BDE}(\ce{H-H}) &amp;amp;= \pu{104.2 kcal/mol} \\&#xA;\text{BDE}(\ce{F-F}) &amp;amp;= \pu{36.6 kcal/mol} \\&#xA;\text{BDE}(\ce{H-F}) &amp;amp;= \pu{134.6 kcal/mol} \\&#xA;\chi_\ce{H} &amp;amp;= 2.1&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I thought that electronegativity should be related to effective nuclear charge, as it is a measure of how strongly atoms &amp;quot;pull&amp;quot; on bond pairs.&lt;/p&gt;&#xA;&lt;p&gt;Is there a formula to calculate the electronegativity from bond dissociation energies, and what is the rationale behind it?&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-10-28T22:19:49.120" Title="Calculating the electronegativity of an element from bond dissociation energies" Tags="&lt;inorganic-chemistry&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13835" PostTypeId="2" ParentId="13810" CreationDate="2014-06-23T09:09:42.950" Score="7" Body="&lt;p&gt;The ionisation energy of an anion is the same as the electron affinity of the element. $$\text{IE}(\ce{A-}) = \text{EA}(\ce{A})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning is not wrong and one might expect exactly this behaviour. Another point for this reasoning is the substantial lower electronegativity of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur&quot; rel=&quot;nofollow noreferrer&quot;&gt;sulfur&lt;/a&gt; ($2.6$) versus &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxygen&lt;/a&gt; ($3.4$).  However, experimental findings contradict this reasoning.&#xA;$$\begin{array}{lll}&#xA;&amp;amp;&amp;amp;/\mathrm{kJ\cdot{}mol^{-1}}\\\hline&#xA;\text{IE}\ce{(O)}  &amp;amp;        &amp;amp;=1314\\&#xA;\text{IE}\ce{(S)}  &amp;amp;        &amp;amp;=1000\\&#xA;\text{IE}\ce{(S− )} &amp;amp;=\text{EA}\ce{(S)} &amp;amp;=200\\&#xA;\text{IE}\ce{(O− )} &amp;amp;=\text{EA}\ce{(O)} &amp;amp;=141\\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Wikipedia articles on &lt;a href=&quot;http://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionisation energies&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron affinities&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it all boils down to electron-electron repulsion. Oxygen ($152~\mathrm{pm}$) is much smaller than sulfur ($180~\mathrm{pm}$) (&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;van der Waals radius, source&lt;/a&gt;). This goes hand in hand with the smaller orbitals (lower main quantum number for the valence shell)  The electrons can therefore only occupy lesser space and repulse each other more. This is mainly the reason, why sulfur has a greater electron affinity than oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reasoning may be applied to fluorine and chlorine.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-07-16T02:29:02.363" LastActivityDate="2018-07-16T02:29:02.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13837" PostTypeId="2" ParentId="13685" CreationDate="2014-06-23T13:40:28.523" Score="0" Body="&lt;p&gt;The difference between the Cl2 and the Cl (chlorine and chloride respectively) is that the in the chlorine &lt;strong&gt;MOLECULE&lt;/strong&gt; the chlorine atoms are covalently bonded to each other thus filling all the occupied orbitals as one 3p electron from each Chlorine (2 in total) is shared between the two. As a result, one can take each chlorine atom's electron configuration to be 1s2, 2s2, 2p6, 3s2, 3p6. (identical to the unreactive noble gas Argon) When the covalent bond is broken, typically using UV light, it is broken by homolytic fission which means that one of the electrons from the two used in the covalent bond goes to each chlorine atom leaving them uncharged. However, if one takes a look at the electron configuration of the atom now - 1s2, 2s2, 2p6, 3s2, 3p5 - it's clear that the 3p orbital is incomplete (one p orbital is only occupied by one electron). This renders the Chlorine &lt;strong&gt;atom&lt;/strong&gt; a Chlorine radical; there is no distinction between a chlorine atom and a chlorine radical. This also explains why chlorine is diatomic (is mostly present as a Cl2 molecule) with a single covalent bond. This is similar the same as all group 7 elements. Also similar to elements like oxygen which are di-radical as there are two unpaired electrons in the the 3p subshell. This explains why oxygen molecules have a double bond  and are also diatomic. I know &quot;&lt;strong&gt;two&lt;/strong&gt; un**paired** electrons&quot; sounds counter intuitive but the p sub-shells are filled in such a way as to avoid pairing the electrons in an orbital until absolutely necessasry (i.e once three electrons are already in the p sub-shell). N2 (Nitrogen) follows the pattern, forming triple bonds in a diatomic molecule to pair electrons in orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radicals are extremely reactive as a result. They're chain carriers in reactions called radical substitution reactions (search &quot;radical substitution chemguide&quot; on google). An example is the production of halogenoalkanes from alkanes which are useful for synthesis of alcohols and amines. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-23T13:40:28.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13838" PostTypeId="2" ParentId="13834" CreationDate="2014-06-23T14:25:07.117" Score="4" Body="&lt;p&gt;Considering the data you have been given, the equation you need is the one that defines the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Pauling_electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pauling electronegativities&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left|\chi_\ce{A} - \chi_\ce{B}\right|=\left(\text{eV}\right)^{-1/2}\sqrt{E_\mathrm d(\ce{A-B})-\frac{E_\mathrm d(\ce{A-A})+E_\mathrm d{\ce{(B-B)}}}{2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equation, &lt;span class=&quot;math-container&quot;&gt;$\left(\text{eV}\right)^{-1/2}$&lt;/span&gt; is in the equation to negate the units of the bond dissociation energies. This means that you need to convert all your bond dissociation energies from &lt;span class=&quot;math-container&quot;&gt;$\pu{kcal/mol}$&lt;/span&gt; into &lt;span class=&quot;math-container&quot;&gt;$\pu{eV}$&lt;/span&gt;. You can use the BDEs in units of &lt;span class=&quot;math-container&quot;&gt;$\pu{kcal/mol}$&lt;/span&gt;, but then the formula will require a constant in front to adjust for the difference between units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{1 eV}$&lt;/span&gt; is equal to &lt;span class=&quot;math-container&quot;&gt;$\pu{1.602 \times 10^-19 J}$&lt;/span&gt;, so you need to be able to convert from &lt;span class=&quot;math-container&quot;&gt;$\pu{kcal/mol}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{J/mol}$&lt;/span&gt;, then divide through by Avogadro's constant &lt;span class=&quot;math-container&quot;&gt;$N_\mathrm A = \pu{6.022 \times 10^19 mol-1}$&lt;/span&gt;. There is a &lt;a href=&quot;http://www.colby.edu/chemistry/PChem/Hartree.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;handy website&lt;/a&gt; which can help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;E_\mathrm d(\ce{H-H}) &amp;amp;= \pu{4.519 eV} \\&#xA;E_\mathrm d(\ce{F-F}) &amp;amp;= \pu{1.59 eV} \\&#xA;E_\mathrm d(\ce{H-F}) &amp;amp;= \pu{5.837 eV}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging the information you have been given, we find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\left|2.1 - \chi_\ce{F}\right| &amp;amp;= \sqrt{5.837 - \frac{4.519+1.59}{2}} \\&#xA;&amp;amp;= 1.668&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you two possibilities for &lt;span class=&quot;math-container&quot;&gt;$\chi_\ce{F}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$0.4$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$3.8$&lt;/span&gt;, depending on whether the quantity &lt;span class=&quot;math-container&quot;&gt;$(2.1 - \chi_\ce{F})$&lt;/span&gt; is positive or negative. This is where we need to use some knowledge of chemistry: fluorine is more electronegative than hydrogen, so we should have &lt;span class=&quot;math-container&quot;&gt;$\chi_\ce{F} &amp;gt; \chi_\ce{H}$&lt;/span&gt;, and consequently we take the solution &lt;span class=&quot;math-container&quot;&gt;$\chi_\ce{F} = 3.8$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why should bond energies be related to electronegativity? The idea that Pauling had when he devised this equation was that in a covalent bond between two of the same atom, there is no electronegativity difference and hence no ionic contribution to the bond strength. Hence, the &quot;average covalent contribution&quot; of each atom to a bond could be measured by the mean of the two bond strengths, i.e. &lt;span class=&quot;math-container&quot;&gt;$[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when there is a difference in electronegativity, the bond strength has an ionic contribution. The larger the difference in electronegativity &lt;span class=&quot;math-container&quot;&gt;$\left|\chi_\ce{A} - \chi_\ce{B}\right|$&lt;/span&gt;, the larger the ionic contribution. This ionic contribution could be roughly measured by subtracting the covalent contribution, &lt;span class=&quot;math-container&quot;&gt;$[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$&lt;/span&gt;, from the &lt;span class=&quot;math-container&quot;&gt;$\ce{A-B}$&lt;/span&gt; bond strength.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2018-10-28T00:17:15.750" LastActivityDate="2018-10-28T00:17:15.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13839" PostTypeId="1" CreationDate="2014-06-23T16:20:28.910" Score="0" ViewCount="3948" Body="&lt;p&gt;There is a specific order for filling electron for transition metals, but I don't know if it applies to other elements too?&lt;img src=&quot;https://i.stack.imgur.com/DQLqf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some thing to take note of in this picture is the order of 3d and 4s. My belief is: 3d has a higher energy level than 4s, and but 4s is still the outermost shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that transition metal would form colorful ions is that there are empty or half filled d orbitals, so when excited electrons drop back to the d orbitals, they emit energy in form of light. I am a bit confused about this.( it is from a book) what electron is it referring to? Is it the d electron or electron of lower energy level orbitals, as for instance 3p?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-06-23T19:30:47.510" Title="Transition metal order of filling electron" Tags="&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13840" PostTypeId="1" CreationDate="2014-06-23T16:46:31.947" Score="0" ViewCount="230" Body="&lt;p&gt;A professor I have gave the class this question, can anyone help? This is the first part of the question. (Also, my professor is from Bangladesh...so yea...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the year 2395, and the world is in chaos. Two nations have conquered the world, USA and Bangladesh. You have been hired by Bangladesh's government to monitor acid baths used to remove oxidation on aerospace metallic material desperately needed for the new   F-95 fighter jet engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acid bath is made up of three components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25-30% HNO3&#xA;4-5% HF&#xA;Balance H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, a spy has broken into your facility and added large amounts of Chlorine to the acid bath. As the lead chemist, it is your job to determine how much Chlorine was added to the acid bath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During sample collection, it is uncertain what any of the concentrations are, however once a sample is obtained you may manipulate it with any method you wish. Devise a method to determine the concentration of chlorine in the solution.&lt;/p&gt;&#xA;" OwnerUserId="6926" LastEditorUserId="4945" LastEditDate="2014-10-28T09:07:30.677" LastActivityDate="2014-10-28T09:07:30.677" Title="How can I find out the concentration of chlorine in an acid bath?" Tags="&lt;reaction-mechanism&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13841" PostTypeId="2" ParentId="10210" CreationDate="2014-06-23T17:15:56.380" Score="10" Body="&lt;p&gt;Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. This may be attractive or repulsive, depending on the dipoles involved.&lt;/li&gt;&#xA;&lt;li&gt;Deybe force - This is an effect caused between a molecule with is polar, and one that is not. Because the one that is polar affects the electrons on the second (non-polar) molecule, it creates an attractive force between the two molecules.&lt;/li&gt;&#xA;&lt;li&gt;London dispersal effect - This is a force that acts between two non-polar molecules/atoms. Because the electrons around each molecule/atom repel each other, it creates a redistribution of charge, inducing an instantaneous dipole moment. This is the force that acts in liquid noble gases, to make them liquid, or between methane molecules in liquid form.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hydrophobic forces are caused because molecules like ethane and other hydrocarbons - petrol etc. and oils like sunflower oil etc. are non-polar, and therefore don't &quot;like&quot; being dissolved in a polar solvent like water. In water, there are hydrogen bonds between the individual molecules, which is why water is a liquid - the van der Waals forces are far too small in water to bind the water as a liquid at room temperature. If we introduce non-polar substances into the water - polar solvent - it upsets these hydrogen bonds, and creates an increase in enthalpy because of this. Therefore the lowest energy state is for the hydrophobic hydrocarbons to separate themselves from the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system. To counter this decrease in entropy, there must be some decrease in enthalpy. Because the hydrophobe disrupts the hydrogen bonding in water, when they separate, the hydrogen bonding then causes an decrease in enthalpy, because of the favorable interactions. This is also where the van der Waals forces come in. They are favourable interactions, so cause a decrease in enthalpy, and this helps the separation of the hydrophobe and the water into two separate phases, because it makes the separation more energetically favourable. They are not necessary, but make separation more favourable, because the enthalpy change is greater for the separation process. This makes it more favourable.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="4222" LastEditDate="2014-06-25T08:11:03.230" LastActivityDate="2014-06-25T08:11:03.230" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13842" PostTypeId="2" ParentId="13839" CreationDate="2014-06-23T17:22:52.813" Score="1" Body="&lt;p&gt;This is the order of electrons filling for all elements but typically this diagram is only used for transition elements because when writing electron configurations for any other element its simple and intuitive but transition metals are somewhat irregular so a helpful diagram like that is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its transition metal compounds or complexes that are coloured not the metal on its own. Its coloured because when it forms compounds or complexes the orbitals in the 3d subshell aren't the same energy as they would be if it wasn't part of a compound or complex. The electrons make transitions between these different energy 3d orbitals when they absorb photons of visible light. The colour of the compound will be the complementary colour of the colour absorbed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3d orbital is typically seen as the higher energy orbital but when a transition metal is ionized the 4s electrons are taken first.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-23T17:22:52.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13846" PostTypeId="2" ParentId="13839" CreationDate="2014-06-23T19:30:47.510" Score="1" Body="&lt;p&gt;In response to your comments on my last answer: the term valence electrons isn't really used for compounds - especially ionic compounds. The valence electrons of the element will be the 4s electrons or electron in the case of Chromium and Copper. It's not so much a change in electron configuration its more about the energy or the orbitals. Each 4s orbital will have a different energy level. The electrons make transitions between these orbitals - this is why it is key that the transition metal ion has a partially filled d sub-shell (so there's room for the electrons to go to and from). It's not about the colour it emits, its about the colour it absorbs when visible light is passed through it. A good website to explain is &quot;&lt;a href=&quot;http://www.chemguide.co.uk/inorganic/complexions/colour.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/inorganic/complexions/colour.html&lt;/a&gt;&quot;. It talks about complexes but its the same idea for compounds. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-23T19:30:47.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13847" PostTypeId="2" ParentId="13840" CreationDate="2014-06-23T19:35:08.673" Score="0" Body="&lt;p&gt;I would suggest some form of multi-step redox titration. You'll need values for electrode potentials though.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-23T19:35:08.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13848" PostTypeId="2" ParentId="13840" CreationDate="2014-06-23T20:01:14.183" Score="1" Body="&lt;p&gt;The addition of chlorine to the bath will produce hydrochloric acid.  If the chloride ion concentration in the bath sample could be determined, then you could back-calculate to the amount of hydrochloric acid added.  The standard method to determine the amount of chloride ion in an aqueous sample is to precipitate it by adding silver nitrate solution followed by collecting the precipitate by filtration and then drying and weighing it.  You could use that approach here because silver nitrate and silver fluoride (which would be created from silver nitrate and the hydrofluoric acid) are soluble* in water, so only the silver chloride would precipitate if silver nitrate were added to your acid bath sample.  &lt;a href=&quot;http://www.ux1.eiu.edu/~cfjpb/teaching/quant/labs/experiment1.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's a link&lt;/a&gt; to an actual gravimetric procedure for determining chloride ion content.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Note: silver fluoride is soluble to the extent of 1.8 kg/L, therefor silver fluoride should not precipitate upon addition of silver nitrate to your chlorine-contaminated sample because the hydrofluoric acid content of the bath was only 4-5%.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-23T21:33:35.630" LastActivityDate="2014-06-23T21:33:35.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13850" PostTypeId="1" CreationDate="2014-06-23T22:03:27.193" Score="1" ViewCount="173" Body="&lt;p&gt;A very typical example of a compound of peroxide is $\ce{H2O2}$. When like the title says, will such compounds form, despite usually oxygen will have a oxidation number of 2-?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="7448" LastEditDate="2015-10-16T16:48:56.593" LastActivityDate="2015-10-16T16:48:56.593" Title="When will peroxide form?" Tags="&lt;covalent-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13851" PostTypeId="1" CreationDate="2014-06-23T22:29:52.300" Score="0" ViewCount="4942" Body="&lt;p&gt;When non metal react they form molecule, and whether the bond is polar or not, the elements in the molecular compound will not have charges what so ever, because they are not ionic. But can non metals be the only that is involved Ina redox reaction ?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-06-24T03:34:26.920" Title="Can redox reactions involve only non metals?" Tags="&lt;redox&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13852" PostTypeId="1" CreationDate="2014-06-23T22:41:28.593" Score="4" ViewCount="102" Body="&lt;p&gt;For a research project, I am trying to synthesize lead (II) sulfide quantum dots using the method of &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2010/CE/b917833c#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;Li &lt;em&gt;et al.&lt;/em&gt;&lt;/a&gt;.  I am injecting sulfur dissolved in 1-octadecene into a mixture of $\ce{PbO}$, oleic acid, oleylamine, and 1-octadecene at $210^{\circ}\ce{C}$, waiting 5 minutes, and quenching the reaction in an ice bath. The oleic acid : $\ce{PbO}$ molar ratio is 10 : 1 which, I think (roughly extrapolating from the data presented in the paper), should result in uniform sized quantum dots with absorption peak of about 900 nm. As described in the paper, the solution I'm working with turns black almost instantly, suggesting nanocrystal nucleation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, when I characterize them with UV-Vis-NIR (after precipitation, centrifugation, and dispersion in hexane), instead of getting a clear absorption peak, I get a weird wavelength-vs-absorbance graph that shoots up abruptly at 300 nm and then gradually curves downward, with no indication of a peak anywhere near 900 nm. Why is this, and what are possible solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it might have something to do with how I'm extrapolating the trend for oleic acid : $\ce{PbO}$ ratio, or perhaps how I'm precipitating/centrifuging/redissolving the QDs (though I'm not sure how this would affect the size or spectra). Or maybe some other factor I haven't thought of that affects the nucleation rate. Any ideas? &lt;/p&gt;&#xA;" OwnerUserId="6929" LastEditorUserId="14" LastEditDate="2015-06-17T15:57:22.120" LastActivityDate="2015-06-17T15:57:22.120" Title="Synthesizing monodisperse PbS quantum dots" Tags="&lt;experimental-chemistry&gt;&lt;spectroscopy&gt;&lt;solutions&gt;&lt;nanotechnology&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13853" PostTypeId="2" ParentId="13850" CreationDate="2014-06-23T23:14:40.723" Score="4" Body="&lt;p&gt;From Greenwood and Earnshaw's excellent book &lt;em&gt;Chemistry of the Elements&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the metals are burned in a free supply of air the predominant product depends on the mtal: $\ce{Li}$ forms the oxide $\ce{Li2O}$ (plus some $\ce{Li2O2}$), $\ce{Na}$ forms the peroxide $\ce{Na2O2}$ (plus some $\ce{Na2O}$) [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, formation of inorganic peroxide salts in this fashion doesn't appear to be especially favorable for other metals, at least by comparison to the variety of other oxides possible. Sodium's peculiar preference seems anomalous to me, especially given that the lattice energy for the oxide is higher (albeit slightly). This may be a function of the availability of oxygen, which is roughly $21\%$ in air, and is certainly temperature-dependent as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for organic substances, formation of peroxide linkages is possible from a variety of functional groups and using various reagents. Ordinary $\ce{O2}$ will form peroxide bonds with various substrates via radical mechanisms, especially under photochemical conditions. Carboxylic acids react with $\ce{H2O2}$ to form peroxyacids. These are just two examples that come to mind, and numerous others exist.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-06-23T23:23:33.757" LastActivityDate="2014-06-23T23:23:33.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13854" PostTypeId="2" ParentId="13851" CreationDate="2014-06-23T23:28:36.130" Score="3" Body="&lt;p&gt;A redox reaction is any reaction in which the oxidation states of the reactants change. There are certainly reactions involving metals which fit that classification. A relatable example would be the rusting of iron (one of the possible reactions involved in rusting):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe + O_2 + 2H_2O -&amp;gt; 2Fe(OH)_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here iron is oxidized and its oxidation state changes from 0 to +2.&lt;/p&gt;&#xA;" OwnerUserId="5220" LastActivityDate="2014-06-23T23:28:36.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13855" PostTypeId="1" AcceptedAnswerId="13859" CreationDate="2014-06-24T00:02:30.343" Score="12" ViewCount="73510" Body="&lt;p&gt;Of my belief, sulfuric acid might be a little stronger than hydrochloric acid. Because even though they both are strong acids, one mole of sulfuric acid produces two times as much $\ce{H+}$ as one mole of hydrochloric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="16683" LastEditDate="2016-09-01T06:12:25.137" LastActivityDate="2017-05-10T16:38:49.007" Title="Is sulfuric acid or hydrochloric acid stronger?" Tags="&lt;acid-base&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13856" PostTypeId="2" ParentId="13827" CreationDate="2014-06-24T00:11:44.430" Score="6" Body="&lt;p&gt;As you point out, formal charge is a &quot;bookkeeping&quot; method of assigning electric charges to atoms in a way that lets us easily approximate where the electron density is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry is made up of layers of successively less detailed approximations. The concept of &quot;formal charge&quot; is near the top of the stack - very little detail is involved. It is only slightly more detailed than Lewis structures, in that we are at least &lt;strong&gt;trying&lt;/strong&gt; to acknowledge that covalent bonds don't really share electrons equally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, we will look at the next higher and next lower level of detail, to see why formal charges are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we draw a Lewis structure, we imply that each line is made up of two electrons that share a molecular orbital that exists somewhere in the space between the two atoms that it connects. Without knowing any better, we would assume that each electron can be wherever it &quot;needs&quot; to be to fill the valence shell of an atom. So, in the case of $\ce{CO3^2-}$, the carbon has four bonds and no lone pairs, giving it a full octet. The oxygen atoms have either three lone pairs and a single bond, or two lone pairs and a double bond, again giving each a full octet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6WGI.png&quot; alt=&quot;Valence count in carbonate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we counted each electron twice for the purpose of checking the valence configuration. It's as if we are saying &quot;each atom can &lt;em&gt;borrow&lt;/em&gt; electrons from the other whenever it needs them.&quot; We are effectively ignoring any tendency of atoms to pull electrons towards each other - they float between atoms as needed, as if maintaining a full octet were the most important job in the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we look at the formal charges. The rules for determining formal charge say that we count the number of valence electrons an atom normally has (without bonds), then subtract the electrons we assign to it. We assign to an atom all of the electrons that are in lone pairs, plus one half of the bonding electrons. In other words, instead of saying that each atom gets the electrons whenever it wants them, we are saying &quot;each atom in the bond divides the electrons equally between themselves.&quot; When you do this, you find that some atoms have &quot;extra&quot; electrons (more than they would normally have in the valence shell) and some have too few - thus, a &lt;strong&gt;formal charge&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EsepG.png&quot; alt=&quot;Formal charge in carbonate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This starts to give us a slightly more correct picture of how electrons are distributed inside of a molecule, but it doesn't really tell the whole story. Notice that we didn't care about what type of atom it was when we split up the electrons. We only looked at that to find the &quot;natural&quot; number of electrons it would have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next layer of detail involves &lt;strong&gt;electronegativity&lt;/strong&gt;. In any covalent bond between two different elements, one element &quot;likes&quot; electrons more. The reasons for this are complex, and you have to go down another couple of layers to get to the reason. For now, just know that it is related to quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we know the relative electronegativities of elements, we can start to account for this tendency as well. We do this when assigning &lt;strong&gt;oxidation numbers&lt;/strong&gt;. With oxidation numbers, you assign to each atom all of its loan pair electrons. Then, for each bond, you give all electrons in the bond to the more electronegative atom. Then we subtract the total from the number of valence electrons the atom normally has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rdZxR.png&quot; alt=&quot;Oxidation numbers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice what we are doing. First we said &quot;all electrons are shared equally all the time.&quot; Then we said &quot;Divide the electrons equally.&quot; Finally we said &quot;the most electronegative atom gets all of the electrons.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these is actually true, but each gives a quick approximation that is useful for a particular type of problem. To get a real picture of what is happening to the electron density, you have to do detailed quantum mechanical calculations. They are really hard to do! We typically use computers to solve them. But, if we know what the limitations are, we can use quick approximations to get an idea of where the electrons are at that is good enough to, for example, make reasonable predictions in organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Valence count - assume all electrons are shared equally&lt;/li&gt;&#xA;&lt;li&gt;Formal charge - assign one half of bonding electrons to each atom&lt;/li&gt;&#xA;&lt;li&gt;Oxidation number - assign all bonding electrons to the most electronegative atom&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-24T00:11:44.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13857" PostTypeId="2" ParentId="13851" CreationDate="2014-06-24T00:18:34.673" Score="5" Body="&lt;p&gt;A redox reaction is one in which oxidation state of some atom changes, and such reactions need not involve metals, nor any ionic species. Oxidation state is the idealized charge that a given atom would have if all its bonds are treated as completely ionic. Thus, strictly for the purpose of calculating oxidation state, the electrons of a covalent bond are typically assigned to the more electronegative atom. (I'd note, however, that IUPAC provides a different set of rules for calculating oxidation number.) Hence, in order for a redox reaction to occur, there &lt;em&gt;does not&lt;/em&gt; need to be a complete transfer of electrons (in the sense of ionization). Examples of redox reactions involving molecular compounds are numerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the reaction $\ce{N2 + 3H2 &amp;lt;=&amp;gt; 2NH3}$. This is the Haber process for the production of ammonia. Here nitrogen is reduced from an oxidation state of $0$ to $-3$, while hydrogen is oxidized from $0$ to $+1$. Again, nitrogen being the more electronegative atom, is assigned the electrons of the bonds. This idealization is useful in that it provides some rough idea of electron distribution, and hence has some utility in predicting reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Also, the statement &lt;em&gt;&quot;[...] the elements in the molecular compound will not have charges what so ever [...]&quot;&lt;/em&gt; is inaccurate. There is a concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;nofollow&quot;&gt;formal charge&lt;/a&gt; that applies to neutral molecules as well as ions.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-06-24T02:54:37.260" LastActivityDate="2014-06-24T02:54:37.260" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13858" PostTypeId="2" ParentId="13855" CreationDate="2014-06-24T00:48:17.320" Score="8" Body="&lt;p&gt;Hydrochloric acid is the stronger of the two.  It has a $\ce{pK_{a}}$ around -6.3 while the $\ce{pK_{a}}$ of sulfuric acid is only around -3.  Here are the chemical equations for the dissociations of the two acids.  You can see that the second dissociation in the case of sulfuric acid is not very extensive and doesn't contribute much to the acidity of sulfuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{H2SO4 + H2O ~&amp;amp;&amp;lt;=&amp;gt;~ H3O+ + HSO4^{−}&amp;amp;&amp;amp;pK_{a}~=~-3}\\&#xA;\ce{HSO4- + H2O ~&amp;amp;&amp;lt;=&amp;gt;~ H3O+ + SO4^{−2}&amp;amp;&amp;amp;pK_{a}~=~2}\\&#xA;\ce{HCl + H2O ~&amp;amp;&amp;lt;=&amp;gt;~ H3O^+ + Cl^{-}&amp;amp;&amp;amp;pK_{a}~=~$-6.3$}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-06-24T02:44:44.157" LastActivityDate="2014-06-24T02:44:44.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13859" PostTypeId="2" ParentId="13855" CreationDate="2014-06-24T01:01:12.407" Score="11" Body="&lt;p&gt;When using $\mathrm{p}K_\mathrm{a}$, one typically does not consider multiple dissociations for polyprotic acids, as the acidity of the conjugate base ($\ce{HSO4-}$ in the case of sulfuric acid) can and should be measured (or calculated) separately. Using that metric, $\ce{HCl}$ is stronger (at least per Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_strength&quot; rel=&quot;nofollow&quot;&gt;acid strength&lt;/a&gt; page). One needs to keep in mind that $\mathrm{p}K_\mathrm{a}$ is solvent-dependent, and that the values typically given are for relatively dilute solutions. Acidity can sometimes vary widely with concentration ($\ce{HF}$ being a notorious example). It should also be noted that experimentally measuring acidity for very strong acids is actually quite difficult for numerous reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you evaluate acid strength by, say, the $\mathrm{pH}$ of the resulting solution, then you'd need to take subsequent dissociations into account. Using the crude approximation that an aqueous &quot;strong acid&quot; dissociates completely, then &lt;em&gt;any&lt;/em&gt; diprotic &quot;strong acid&quot; is going to be stronger than a monoprotic one for equal concentrations (assuming a non-zero $K_\mathrm{a}$ for the second dissociation). Of course, this is not accurate, and a proper calculation would take into account not only $K_\mathrm{a}$ values for all dissociations, but also water's auto-ionization. I suspect that $\ce{HCl}$ would still be stronger, given that the $K_\mathrm{a}$ for the dissociation of $\ce{HSO4-}$ is fairly small, but one would need to actually crunch the numbers to be sure.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="16683" LastEditDate="2016-09-01T06:06:13.383" LastActivityDate="2016-09-01T06:06:13.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13860" PostTypeId="2" ParentId="13855" CreationDate="2014-06-24T03:00:48.380" Score="3" Body="&lt;p&gt;Although sulfuric acid has two acidic protons, only one of those protons dissociates completely in solution.  After the first proton has dissociated, leaving $\ce{HSO4-}$, that species is now a weak acid as well as a weak base.  As weak acids do not dissociate completely in solution, the remaining acidic proton is now &quot;less&quot; acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A strong acid has a weak conjugate base. The $\ce{Cl-}$ ion created in solution by the dissociation of $\ce{HCl}$ is a terrible base, making hydrochloric acid very strong. The hydrogen sulfate anion is a weak base also, but it is a &lt;em&gt;better&lt;/em&gt; base then a $\ce{Cl-}$ ion, adding to its weakened acidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $K_\mathrm{a}$ values are a nice measure of acidity, only &lt;em&gt;weak&lt;/em&gt; acids have a $K_\mathrm{a}$ value associated with them.  Both hydrochloric acid and sulfuric acid are strong acids. In fact, because the $K_\mathrm{a}$ of sulfuric acid is so high, the eleventh edition (newest printed) textbook, Chang's &quot;Chemistry&quot;, used in College chemistry, simply states it as &quot;extremely high&quot;, assigning no numerical value to it. Likewise, the hydrogen sulfate anion is listed as having a $K_\mathrm{b}$ as &quot;extremely low&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6930" LastEditorUserId="16683" LastEditDate="2016-09-01T06:01:09.520" LastActivityDate="2016-09-01T06:01:09.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13861" PostTypeId="2" ParentId="13851" CreationDate="2014-06-24T03:34:26.920" Score="3" Body="&lt;p&gt;Yes. Apart from Greg's answer there are many more examples.&#xA;But since you asked specifically about a redox reaction that may produce ions, I am glad to provide an example. This reaction is very common chemistry, as it is used in public pools. It is the redox dissolution of chlorine in water (coupled with an acid-base reaction). It is also a &lt;a href=&quot;http://en.wikipedia.org/wiki/Disproportionation&quot; rel=&quot;nofollow&quot;&gt;disproportionation&lt;/a&gt; reaction.&#xA;\begin{array}{lcll}&#xA;\ce{1/2 Cl2 + e-  &amp;amp;-&amp;gt; &amp;amp; Cl-}                &amp;amp;\text{reduction}\\&#xA;\ce{1/2 Cl2 + H2O &amp;amp;-&amp;gt; &amp;amp; HClO + H+ + e-}     &amp;amp;\text{oxidation}\\\hline&#xA;\ce{Cl2 + H2O     &amp;amp;-&amp;gt; &amp;amp; HClO + HCl}         &amp;amp;\text{redox}\\\hline&#xA;\ce{HClO + H2O    &amp;amp;&amp;lt;=&amp;gt;&amp;amp; H3+O + ClO-}        &amp;amp;\text{acid-base}\\&#xA;\ce{HCl + H2O     &amp;amp;&amp;lt;=&amp;gt;&amp;amp; H3+O + Cl-}         &amp;amp;\text{acid-base}\\\hline&#xA;\ce{\overset{\color{\green}{0}}{Cl2} + 3H2O &#xA;                  &amp;amp;&amp;lt;=&amp;gt;&amp;amp; 2H3+O + \overset{\color{\green}{+I}}{Cl}O- &#xA;                              + \overset{\color{\green}{-I}}{Cl- } }&#xA;                                            &amp;amp;\text{total}\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the concentration and the $\ce{pH}$ of the solution, the extent of ionisation will vary.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-24T03:34:26.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13862" PostTypeId="2" ParentId="13768" CreationDate="2014-06-24T04:37:27.840" Score="0" Body="&lt;p&gt;First thing that came to my mind was calcium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unbalanced equation would be$$\ce{SO2 +S +CaO +NaSO4 -&amp;gt;Na2S2O3 + CaSO4}$$&lt;/p&gt;&#xA;" OwnerUserId="6930" LastEditorUserId="73" LastEditDate="2014-06-29T23:44:22.257" LastActivityDate="2014-06-29T23:44:22.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13864" PostTypeId="2" ParentId="13768" CreationDate="2014-06-24T07:50:33.003" Score="4" Body="&lt;p&gt;I am not one hundred percent certain, but I will try to give you some guidelines. There is the well known &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_dioxide#As_an_air_pollutant&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction of sulfur dioxide with calcium oxide&lt;/a&gt;.&#xA;$$\ce{SO2 + CaO -&amp;gt; CaSO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under aerobic conditions, this would further react with oxygen to form sulfate, since we have excess sulfur here I think it is fair to assume it will react to form thiosulfate.&#xA;$$\ce{1/8S8 + CaSO3 -&amp;gt; CaS2O3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding sodium sulfate is probably just a way to control/ buffer the pH during the reaction, I fail to see the actual relevance. But judging from the side reaction&#xA;$$\ce{SO2 + 1/8S8 (s) + 3H2O &amp;lt;=&amp;gt; S2O3^{2−} + 2 H3+O}$$&#xA;it is actually necessary to have a buffer system based on sulfate.&#xA;$$\ce{Na2SO4 + H3+O &amp;lt;=&amp;gt; 2Na+ + HSO4- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any reaction of sulfur with sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total I would write your equation like this:&#xA;$$\begin{multline}\ce{SO2 (g) + S (s) + CaO  (s) + Na2SO4 (s) -&amp;gt;[\ce{H2O}]\\ 2Na+ (aq) + Ca^{2+} (aq) + S2O3^{2-} (aq) +  SO4^{2-} (aq)}\end{multline}$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-08T00:24:11.697" LastActivityDate="2017-05-08T00:24:11.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13866" PostTypeId="1" AcceptedAnswerId="13869" CreationDate="2014-06-24T08:38:36.190" Score="7" ViewCount="2971" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWYnd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my understanding, by bonding to the free radical species, say by one of the double bonds, a new free radical on the $\beta$-carotene develops.&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorUserId="4945" LastEditDate="2014-07-17T10:28:31.863" LastActivityDate="2017-09-10T20:43:48.740" Title="How does neutralization of free radicals by $\beta$-carotene work?" Tags="&lt;reaction-mechanism&gt;&lt;radicals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13867" PostTypeId="2" ParentId="13784" CreationDate="2014-06-24T08:56:04.370" Score="-1" Body="&lt;p&gt;Because the solute is neutral when its solid so it must me neutral when it's aqueous.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-24T08:56:04.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13868" PostTypeId="2" ParentId="13866" CreationDate="2014-06-24T09:13:07.180" Score="5" Body="&lt;p&gt;The extended, conjugated series of alternating double-bonds in $\beta$-carotene comprises a system of highly delocalized $\pi$ electrons. If a radical forms by hydrogen abstraction at one of the two terminal allylic positions, that lone electron is delocalized over the entire $\pi$ electron system, and thereby stabilized. Similarly, radical addition across any of the double bonds will result in a highly stabilized radical (though addition to one of the internal double bonds will actually disrupt the conjugation, but is nevertheless quite possibile for reasons of sterics and the greater lability of $\pi$-bonds generally). The stability of the resulting radical drastically lowers the energy of activation for the reaction, hence making $\beta$-carotene so liable to react. In essence, its antioxidant activity is (at least in large measure) a direct consequence of the &lt;em&gt;ease&lt;/em&gt; with which it is itself sacrificially oxidized. The same factors are responsible for, e.g., the propensity of unsaturated fats to spoil more readily than saturated ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, as is typical of radical reactions, the resulting $\beta$-carotene radical, while reasonably stable, can still go on to propagate the radical chain reaction. There are various other antioxidants which form highly stable end products after initial radical formation, and thereby terminate the chain reaction.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-24T09:13:07.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13869" PostTypeId="2" ParentId="13866" CreationDate="2014-06-24T09:16:20.693" Score="11" Body="&lt;p&gt;$\beta$-Carotene is a highly delocalised $\pi$ system. &#xA;According to &lt;a href=&quot;http://www.fasebj.org/content/9/15/1551.long&quot;&gt;George Britton&lt;/a&gt; there are numerous mechanisms to achieve this. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Oxidation&lt;/strong&gt;&lt;br&gt;&#xA;Oxidising radicals can remove an electron from the carotene molecule.&#xA;$$\ce{CAR + .R+ -&amp;gt; .CAR+ + :\!R}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Reduction&lt;/strong&gt;&lt;br&gt;&#xA;Analogous it can accept an additional electron.&#xA;$$\ce{CAR + .R- -&amp;gt; .CAR- + R}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Hydrogen Abstraction&lt;/strong&gt;&lt;br&gt;&#xA;It is possible to abstract a hydrogen atom from a saturated carbon in allylic position.&#xA;$$\ce{H-CAR + .R -&amp;gt; .CAR + HR}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Addition&lt;/strong&gt;&lt;br&gt;&#xA;You are correct in assuming, that a free radical may form a bond with that system. This is most common for hydroxyl radicals.&#xA;$$\ce{CAR + .OH -&amp;gt; .CAR-OH}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You are also correct in assuming, that another radical will always be created. However, the single electron will be very delocalised (resonance stabilised) throughout the whole system and not nearly as reactive as the free radical before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining molecule can then of course attract more free radicals to form other singlet species, which the body can digest or excrete. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-24T09:16:20.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13871" PostTypeId="1" CreationDate="2014-06-24T10:36:12.767" Score="1" ViewCount="131" Body="&lt;p&gt;If we put a piece of coal in open air we don't get any energy of it. But if we put fire to it in open air we get energy. Explain it with help of Gibbs free energy. &lt;/p&gt;&#xA;" OwnerUserId="6938" LastEditorUserId="41328" LastEditDate="2019-03-13T11:46:26.070" LastActivityDate="2019-03-13T11:46:26.070" Title="How is it thermodynamically possible to obtain energy from a piece of coal?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13872" PostTypeId="1" CreationDate="2014-06-24T12:22:15.847" Score="0" ViewCount="58" Body="&lt;p&gt;I just bought a new hair clipper and I was about to start shaving my head when I noticed that it was almost at 0%. It has a nickel metal hydride battery. &#xA;The manual says I have to load it for 24h hours before using it the first time or whenever it has been unused for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started calculating. 24 hours are 1440 minutes. Dividing it by 100 makes it 14.4 minutes to gain 1%. To gain 20% it'll take 288 minutes (more than 4 hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I started to load it like an hour ago and it has almost 40% so far. That's not 4 hours. So it'll be at 100% in a few hours. I think in the evening it's done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the manual emphasized that it should be 24 hours ! &lt;strong&gt;So I wondered why do they want me to load it that long even when the percent-bar shows me it's fully loaded and ready to use?&lt;/strong&gt; I also looked up some reviews where the customers wrote that the battery's lifetime is very short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it cause a shorter battery-lifetime if I use it for like 10 up to 20 minutes in the evening (in about 6 hours)? Or more in general, is there something about the intrinsic chemistry of the battery which requires (in the most cases) so long before we use it the first time? Is that even necessary? &lt;/p&gt;&#xA;" OwnerUserId="2441" LastEditorUserId="865" LastEditDate="2015-10-30T17:09:08.587" LastActivityDate="2015-10-30T17:09:08.587" Title="Loading nickel metal hydride battery for the first time" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13873" PostTypeId="2" ParentId="8717" CreationDate="2014-06-24T12:23:25.200" Score="7" Body="&lt;p&gt;Inert pair effect is mostly shown by the 15-17th group elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the oxidation state reduces by 2 for elements below ($\ce{As}$, $\ce{Sb}$), which is more stable than the other oxidation states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is the inertness of the inner $s$ electrons due to poor shielding.&lt;/p&gt;&#xA;" OwnerUserId="6939" LastEditorUserId="4734" LastEditDate="2014-06-24T17:23:49.983" LastActivityDate="2014-06-24T17:23:49.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13874" PostTypeId="1" AcceptedAnswerId="13877" CreationDate="2014-06-24T12:29:16.250" Score="4" ViewCount="6040" Body="&lt;p&gt;When performing electrolysis at home, I usually used salt water as electrolyte. The problem is, that this method also produces chlorine and chlorine is aggressive to both humans and electrical equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of creating &lt;em&gt;pure&lt;/em&gt; hydrogen and oxygen by using distilled water. I was taught that water is not conductive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since water has &quot;&lt;em&gt;positive&lt;/em&gt;&quot; and &quot;&lt;em&gt;negative&lt;/em&gt;&quot; part (which allows for it to dissolve polar bonds, doesn't it?). Couldn't I just tear the molecules apart if I use electrical field strong enough? If I can, how high the voltage must be?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2018-06-17T12:42:55.397" Title="Can I perform electrolysis without electrolyte? (with distilled water)" Tags="&lt;bond&gt;&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13877" PostTypeId="2" ParentId="13874" CreationDate="2014-06-24T17:05:34.863" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I can, how high the voltage must be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not possible at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For obtaining pure hydrogen and oxygen it is still required to use an electrolyte.  However, there is some amount of them that are not electrolyzed, allowing to electrolyze only water. I'd try sodium/potassium hydroxide or sulfate. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-06-24T17:05:34.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13878" PostTypeId="2" ParentId="13801" CreationDate="2014-06-24T17:30:27.787" Score="2" Body="&lt;p&gt;In general clay is not a good &lt;em&gt;source&lt;/em&gt; of nutrients. &lt;a href=&quot;http://en.wikipedia.org/wiki/Clay&quot; rel=&quot;nofollow&quot;&gt;Clay&lt;/a&gt; is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicate_minerals#Phyllosilicates&quot; rel=&quot;nofollow&quot;&gt;class of hydrated alumina-silicates&lt;/a&gt; containing small amounts of metals. The alumina and silica are not plant &lt;a href=&quot;http://www.ncagr.gov/cyber/kidswrld/plant/nutrient.htm&quot; rel=&quot;nofollow&quot;&gt;nutrients&lt;/a&gt;, but some of the metals can be, in particular: potassium, magnesium, calcium, and iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these, only potassium is needed in large amounts by plants, and conversely, is fairly uncommon in clays. Calcium and iron are much more common, but are needed by plants in very small quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any mineral nutrient in any case I can think of, other sources would be better than clay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in the specific case where there was a mineral deficiency in the soil, and you didn't have another option available, clay &lt;em&gt;could&lt;/em&gt; provide that nutrient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, clay can help plants in another manner. This &lt;a href=&quot;http://www.ncagr.gov/agronomi/pdffiles/sfn13.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; describes the ability of clay soil to attract and hold positive ions (metals that can be absorbed by plants are positive ions) due to the high surface area and negative surface charge of the clay particles. Since leaching can be a problem in nutrient-poor soils, having a soil component that &quot;holds on&quot; to those nutrients can be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not brick dust would be helpful to plant growth would depend on the specific case, in particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The type of plant(s)&lt;/li&gt;&#xA;&lt;li&gt;The composition of the rest of the soil&lt;/li&gt;&#xA;&lt;li&gt;The type of clay used to make the brick&lt;/li&gt;&#xA;&lt;li&gt;The climate conditions (amount of rainfall, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-24T17:30:27.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13879" PostTypeId="1" AcceptedAnswerId="13884" CreationDate="2014-06-24T19:11:56.287" Score="1" ViewCount="2785" Body="&lt;p&gt;I have some sodium carbonate. But it's over 20 years old and was not kept in tight bottle. I wonder if this compound absorbs $\ce{H2O}$ and $\ce{CO2}$ from air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I'd need to &lt;a href=&quot;http://naturesnurtureblog.com/2012/05/08/ttt-turn-baking-soda-into-washing-soda/&quot; rel=&quot;nofollow&quot;&gt;burn it a little&lt;/a&gt; to turn it back in case the answer was yes.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2014-06-25T01:57:27.147" LastActivityDate="2014-06-25T01:57:27.147" Title="Does washing soda (sodium carbonate) turn into baking soda (sodium bicarbonate) over time?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13880" PostTypeId="2" ParentId="13871" CreationDate="2014-06-24T19:29:00.567" Score="2" Body="&lt;p&gt;Coal in open air can and does sponstaneously combust.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1877705813012289&quot; rel=&quot;nofollow&quot;&gt;Spontaneous Heating and Fire in Coal Mines&lt;/a&gt; and &lt;a href=&quot;http://daily.sightline.org/2012/04/11/coals-spontaneous-combustion-problem/&quot; rel=&quot;nofollow&quot;&gt;Coal’s Spontaneous Combustion Problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low temperature the change in Gibbs free energy of coal oxidation is still strongly negative, just like at high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at low temperature, the reaction rate is low.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, if the rate at which heat is dissipated is less that the rate at which heat is produced, temperature increases and the reaction rate increases until there is appreciable combusiton. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-24T19:29:00.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13881" PostTypeId="1" AcceptedAnswerId="13897" CreationDate="2014-06-24T19:45:25.080" Score="10" ViewCount="4853" Body="&lt;p&gt;I had used spare pieces of metal to perform electrolysis. They all had the disadvantage that they corrode when used as anode - and some of the oxides are toxic (copper, &lt;a href=&quot;http://antique-engines.com/stainless-steel-electrodes.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromium&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found that carbon, which can be harvested from pencils and old batteries does not corrode contrary to my expectations (I'd expect it to produce $\ce{CO2}$). But soldering on carbon is completely impossible. Also, I need some bigger electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are there any metals that do not corrode during electrolysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using sodium carbonate as electrolyte.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="41328" LastEditDate="2017-08-07T07:42:01.490" LastActivityDate="2020-10-22T22:17:47.577" Title="Which electrodes do not corrode at all?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;metal&gt;&lt;electrolysis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13882" PostTypeId="1" CreationDate="2014-06-24T20:05:15.877" Score="1" ViewCount="287" Body="&lt;p&gt;Here's the problem in the simplest way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a closed box containing methanol adsorbed by activated carbon. The box is made from copper. A pipe is connected to that box, which has condenser on the other end to return the methanol back to the liquid state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The box is painted black and put in the sun to absorb heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I have collected are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;initial and final pressure&lt;/li&gt;&#xA;&lt;li&gt;initial and final temperature&lt;/li&gt;&#xA;&lt;li&gt;0.26 kg of methanol was condensed in the other end&lt;/li&gt;&#xA;&lt;li&gt;0.3744 kg of methanol remained in the tank&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can we calculate enthalpy change in the box (heat absorbed by the box from the sun)?&lt;/p&gt;&#xA;" OwnerUserId="6944" LastEditorUserId="4945" LastEditDate="2014-06-25T02:07:51.450" LastActivityDate="2014-06-25T02:07:51.450" Title="How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13883" PostTypeId="2" ParentId="13815" CreationDate="2014-06-24T20:09:21.843" Score="1" Body="&lt;p&gt;For almost the last 100 years, the definition of pH has been -log of hydrogen ion activity.  Prior to 1920, the definition was -log of hydrogen ion concentration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen ion activity or concentration does not refer to bare protons, but solvated protons, as $\ce{H_3O^+, H_9O_4+}$ etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://link.springer.com/article/10.1023%2FB%3AJOSL.0000035359.00943.14#&quot; rel=&quot;nofollow&quot;&gt;Determination of Hydrogen Single Ion Activity Coefficients in Aqueous HCl Solutions at 25°C&lt;/a&gt; for authoritative data.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-06-24T20:09:21.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13884" PostTypeId="2" ParentId="13879" CreationDate="2014-06-24T21:05:08.237" Score="3" Body="&lt;p&gt;Industrially, sodium bicarbonate is produced by reacting sodium carbonate with water and carbon dioxide.  However, under normal household conditions sodium carbonate simply reacts with water or moisture in the air (it's hygroscopic) to form a hydrate - it doesn not form sodium bicarbonate. The water of hydration can be removed by mild heating (~100 C).  At high temperatures (&gt;850 C) sodium carbonate can decompose according to the following equation.$$\ce{Na2CO3(s) -&amp;gt; CO2(g) + Na2O(s)}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-25T00:07:06.877" LastActivityDate="2014-06-25T00:07:06.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13885" PostTypeId="1" CreationDate="2014-06-24T21:25:06.067" Score="0" ViewCount="3985" Body="&lt;p&gt;I've got a problem in determining the amount of $\ce{CO2}$ in soda water. Basically, I have boiled some purified water to remove the dissolved gas.  By using a SodaStream machine, I made carbonated water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, I have tried to determine the amount of carbon dioxide through titration with phenolphthalein and also $c(\ce{NaOH})=0.1~\mathrm{mol/L}$ sodium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I am not sure if this is the correct method to determine the amount of dissolved $\ce{CO2}$ in water.&lt;/p&gt;&#xA;" OwnerUserId="6945" LastEditorUserId="73" LastEditDate="2014-06-25T08:58:58.260" LastActivityDate="2018-01-29T14:51:00.493" Title="How to determine amount of CO₂ in soda water" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13886" PostTypeId="1" AcceptedAnswerId="13887" CreationDate="2014-06-24T22:32:28.270" Score="8" ViewCount="4635" Body="&lt;p&gt;According to my textbook, element $\ce{X}$ can oxidize element $\ce{Y}$ if $\ce{X}$ was lower in the activity series.&#xA;Lead is lower than Tin in the activity series table in the textbook, but the following question contradict that rule and I can't figure out why:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the reaction $\ce{Sn(s) + Pb(NO3)2(aq) -&amp;gt; Sn(NO3)2(aq) + Pb(s)}$ occur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book reveals at the end that the answer is no, but doesn't explain why.&lt;/p&gt;&#xA;" OwnerUserId="6947" LastEditorUserId="41328" LastEditDate="2019-06-11T05:15:18.520" LastActivityDate="2019-06-12T17:52:12.397" Title="Can lead(II) oxidize tin metal in aqueous solution?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;aqueous-solution&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13887" PostTypeId="2" ParentId="13886" CreationDate="2014-06-24T23:12:12.740" Score="8" Body="&lt;p&gt;Are you certain that your statement from the textbook is correct? The cation of X will oxidize element Y if element Y is &lt;strong&gt;more active&lt;/strong&gt; than X, which is usually labeled (may be towards the top or the bottom of the list). For reference, see the activity series at &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, where the more active metals are above the less active metals in the list.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to use the activity series:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, what happens when iron metal is placed into an aqueous solution of copper (II) nitrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write the potential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe(s) + 3Cu^{2+} -&amp;gt; 2Fe^{3+} + 3Cu(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since iron is more active than copper, iron will reduce copper cations (and copper cations will oxidize iron metal). The iron will dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if copper metal is placed into an aqueous solution of iron (III) nitrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since copper is less active than ion, copper metal will not reduce iron cations. No reaction occurs and the copper will not dissolve.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Your question&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lead is less active than tin in all activity series that I have seen. Lead metal should reduce tin cations. This reaction should work. It is possible that the textbook/solutions manual has a typo. I find them in my books all the time.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-06-24T23:12:12.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13888" PostTypeId="2" ParentId="13882" CreationDate="2014-06-24T23:28:21.537" Score="2" Body="&lt;p&gt;Enthalpy ($H$) is defined as, $H=U+PV$ where $U$ is the internal energy and $P$ is pressure and $V$ is volume. The change in enthalpy is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \Delta U + \Delta(PV)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\Delta U$ is equal to the sum of the heat transferred $q$ and the work done $w=-P\Delta V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant $V$ (which is what you have in a closed system), $\Delta V =0$ (and thus $w=0$) and $\Delta(PV)=V\Delta P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = q+V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can determine $q$ based on the amount of methanol that vaporized using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_of_vaporization#Other_common_substances&quot; rel=&quot;nofollow&quot;&gt;heat of vaporization of methanol&lt;/a&gt;. We can assume that the majority of heat absorbed was used to distil methanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_{vap}=m\cdot c_{vap}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we determine the volume of the system? Or... if the change in pressure is very small, perhaps that term can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-06-24T23:28:21.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13889" PostTypeId="1" AcceptedAnswerId="13890" CreationDate="2014-06-25T00:20:40.670" Score="0" ViewCount="1822" Body="&lt;p&gt;It is to my understanding that ΔG can be positive or negative given whether the reactant is exothermic or endothermic.  How would you go about measuring the energy (in joules) of a reaction using the heat of the reaction? &lt;/p&gt;&#xA;" OwnerUserId="2197" LastEditorUserId="73" LastEditDate="2014-06-25T00:21:44.400" LastActivityDate="2014-06-25T01:26:23.703" Title="How exactly do you measure the free energy of a reaction?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13890" PostTypeId="2" ParentId="13889" CreationDate="2014-06-25T00:28:13.847" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How would you go about measuring the energy (in joules) of a reaction&#xA;  using the heat of the reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can measure the heat of a reaction (the enthalpy of reaction, $\ce{\Delta H}$) using a calorimeter, but the enthalpy of the reaction is only one part of the free energy of the reaction ($\ce{\Delta G}$). $\ce{\Delta G}$ is related to $\ce{\Delta H}$ through the following equation $$\ce{\Delta G = \Delta H} - T \Delta S$$ where $\ce{T}$ is the temperature and $\ce{\Delta S}$ is the change in entropy for the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use a calorimeter you put your reactants into the calorimeter and let the reaction proceed.  The calorimeter is thermally isolated from its surroundings, so any heat given off (exothermic) or absorbed (endothermic) during the course of the reaction can be accurately measured.  Knowing the moles of reactant you put into the calorimeter and the heat change in joules (using the measured $\ce{\Delta T}$ and the known heat capacity of the calorimeter) you can determine $\ce{\Delta H}$ in joules/mole.  Here is a &lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/c120/calorimetry.html&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; that provides more detail about calorimeters along with some problems that are worked out in detail.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-25T01:26:23.703" LastActivityDate="2014-06-25T01:26:23.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13891" PostTypeId="1" CreationDate="2014-06-23T23:42:00.017" Score="2" ViewCount="22198" Body="&lt;p&gt;In this test, four pieces of magnesium fillings (ribbon) are added to the ethanolic extract followed by a few drops of concentrated hydrochloric acid. A reddish colour indicates the presence of flavonoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the basic chemistry of the test?  What are the chemical reactions taking place in the test that lead to the formation of red precipitate, and also, why have we used magnesium and concentrated HCl in the test?&lt;/p&gt;&#xA;" OwnerUserId="19583" OwnerDisplayName="katherinebridges" LastEditorUserId="73" LastEditDate="2014-06-25T04:02:24.910" LastActivityDate="2014-06-25T14:41:18.083" Title="Basic chemistry of the Shinoda test for flavonoids?" Tags="&lt;biochemistry&gt;&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13893" PostTypeId="1" CreationDate="2014-06-25T05:46:19.660" Score="4" ViewCount="130" Body="&lt;p&gt;I've been struggling with this for a while, and I can't seem to solve the problem. Alcohol is made in the fermentation process in beer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sugars formed during mashing (maltose and maltotriose) will undergo hydrolysis and form multiple molecules of the simple sugar glucose. This glucose will then be fermented by the yeast under anaerobic conditions into ethyl alcohol and carbon dioxide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand that perfectly. My problem is to figure out the reaction schemes for the reactions, for each step. Can anyone help me with that?&lt;/p&gt;&#xA;" OwnerUserId="1642" LastEditorUserId="7475" LastEditDate="2017-11-11T14:42:33.083" LastActivityDate="2017-11-11T14:56:20.373" Title="Fermentation in beer (reaction schemes)?" Tags="&lt;reaction-mechanism&gt;&lt;biochemistry&gt;&lt;alcohols&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13894" PostTypeId="2" ParentId="13891" CreationDate="2014-06-25T05:50:46.230" Score="4" Body="&lt;p&gt;I found &lt;a href=&quot;http://books.google.com/books?id=pfcRQQnVQ78C&amp;amp;pg=PA27&amp;amp;lpg=PA27#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;one source&lt;/a&gt; indicating that the reaction involves the conversion of the flavonoid to the corresponding anthocyanidin. The structure of that resulting anthocyanidin contains an extended system of conjugated $\pi$-electrons, which constitute a &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromophore&quot; rel=&quot;nofollow&quot;&gt;chromophore&lt;/a&gt;. As it happens, the electrons of molecules containing conjugated $\pi$ systems have energy level transitions corresponding to absorbance in the visible (and also ultraviolet) region of the electromagnetic spectrum. This accounts for their perceived color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the precise mechanism involved, I was not able to find any sources whatsoever. It's obvious that the flavonoid is reduced, with the magnesium metal being the reducing agent. It could be the process occurs by single electron transfer (SET) with radical and anionic intermediates, or via a cyclic transition state and carbenoid intermediates with coordination of the magnesium, or something else entirely. In any case, the purpose of the acid is likely to activate the carbonyl of the flavonoid to reduction by protonating it, and providing a source of $\ce{H+}$ ions for the subsequent formation of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I find it unlikely that a definitive mechanism for this reduction has been determined, if only because the mechanisms of many similar reductions involving metals and mineral acids widely used in synthetic organic chemistry have yet to be entirely resolved.)&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-25T05:50:46.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13895" PostTypeId="2" ParentId="13881" CreationDate="2014-06-25T06:33:19.303" Score="7" Body="&lt;p&gt;you do not need to solder a graphite electrode to connect it. It is possible to connect mechanically (screws, wires etc.). It is possible to buy some bigger pieces here: &lt;a href=&quot;http://www.penta-edm.cz/text-grafit/&quot;&gt;Penta trading&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question: Platinum is really hard to corrode. It is commonly used platinum coated titanium mesh, but it is quite expensive. link: &lt;a href=&quot;http://www.alibaba.com/showroom/platinum-coated-titanium-mesh.html&quot;&gt;Alibaba&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2014-06-25T06:33:19.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13896" PostTypeId="2" ParentId="13713" CreationDate="2014-06-25T06:58:55.617" Score="6" Body="&lt;p&gt;Hypervalency is an obsolete concept - it was used to explain bonding situations of molecules, that seem to exceed the octet rule. It was accomplished by the help of $\ce{spd}$ hybrid orbitals. It is now known, that for almost all molecules that were described this way the contribution of $\ce{d}$ orbitals is negligible (&amp;lt;1%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focussing on the question at hand, the bonding situation of the core $\ce{SO4^{2-}}$ anion is isoelectronic to the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13657/bonding-in-the-phosphate-ion/13666#13666&quot;&gt;phosphate anion $\ce{PO4^{3-}}$&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the methylsulfate this situation may be extended. A qualitative analysis of the natural charges (BP86/cc-pVDZ) reveals no difference. All of the oxygen atoms carry a negative charge, while sulfur is highly positive charged. Further more, NBO analysis confirms that the terminal oxygens have three lone pairs, while the other oxygens have two lone pairs. A more accurate Lewis structure would represent that fact and only use single bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lp5rO.jpg&quot; alt=&quot;charges&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/23q0Z.jpg&quot; alt=&quot;lewis structure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-06-25T06:58:55.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13897" PostTypeId="2" ParentId="13881" CreationDate="2014-06-25T08:09:03.583" Score="9" Body="&lt;p&gt;This highly depends on what you are electrolyzing. When using solutions that are not very acidic, oxide-coated metals sometimes work, like lead-oxide electrodes, that can be harvested from old lead accumulators (attention, lead is toxic, wear lab coat and gloves when working with them. Lead accumulators contain ~20% sulfuric acid, so wear gloves as well and unassemble them in an acid-proof working space)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search the web for recipes of perchlorate preparation by electrolysis at home, you will find extensive discussions on the topic of anode materials and making oxidation-resistant electrodes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Platinum is really hard to corrode, but even it corrodes in some cases.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5402" LastEditDate="2015-10-02T06:51:02.417" LastActivityDate="2015-10-02T06:51:02.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13898" PostTypeId="2" ParentId="13872" CreationDate="2014-06-25T09:32:58.683" Score="1" Body="&lt;p&gt;Battery on 0% usually still has some voltage, and battery on 100% usually can be overcharged. However, for many types of batteries it is not healthy, leading to premature degradation, the exact 0% and 100% charges as well as initial load procedure and storage requirements are marked at point that happens to prevent battery degradation and increase its capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm not aware of what exact process is intended to happen in Ni-hydride battery, I'm pretty aware that preparation procedure for battery MAY be required. For example, during initial cycles of lead accumulator, its capacity increases. The reason is in its chemistry, as it actually works on lead oxides transformations and the cycles increase layer of oxides on electrodes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Ni-hydride cell includes electrode with half-reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ni(OH)2 + OH- = NiO(OH) + H2O + e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess, that during initial charge the manufacturer expects that layer of nickel hydroxide is formed on positive electrode.  But this is just a guess, you'll have to search specific info on this particular type of battery to be sure. Ni-hydride batteries are resistant to overcharge at low currents, so if you use appropriate smart charger, say, shipped with the batteries, the overcharge is safe, so, since the manufacturer requires it, you should follow the manual. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-06-25T10:28:06.687" LastActivityDate="2014-06-25T10:28:06.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13899" PostTypeId="1" AcceptedAnswerId="14724" CreationDate="2014-06-25T10:26:22.540" Score="6" ViewCount="4880" Body="&lt;p&gt;Recently at school, I did some experiments on corrosion. Everything mostly went as expected, but the result in one of the test was quite baffling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This test had an iron nail in water with some of the nail protruding the surface. The tip of the nail was then covered by oil so that it was completely submerged. It was left as is for 4 weeks. At the end of this time, there was a bit of rusting (as I had expected), but most of it was right at the surface of the water, right where it and the oil meet. Is there any reason why this would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;I had a beaker with the nail in it, leaning upright against the wall of the beaker. I put about 200 ml of water in, then about 50–100 ml of oil. The water didn't completely cover it, but the oil did. After 4 weeks, there was rust, mostly situated around (but below) the oil/water border. As for why it was done, I don't know. My lab partner decided to test this. I think it was kind of testing how much it would rust when starved of oxygen. &lt;/p&gt;&#xA;" OwnerUserId="6951" LastEditorDisplayName="user7951" LastEditDate="2016-10-07T10:05:51.983" LastActivityDate="2016-10-07T10:05:51.983" Title="Rusting of an Iron nail at oil water boundary surface" Tags="&lt;redox&gt;&lt;water&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13900" PostTypeId="1" CreationDate="2014-06-25T11:16:25.650" Score="1" ViewCount="1507" Body="&lt;p&gt;This was a question on the mt exam two days ago :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chemist added 40.0 mL of an $NaOH$ solution to 90.0 mL of 0.400 M $HCl$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The solution was then treated with an excess of $\text {nickel(II) nitrate}$, resulting in the formation of 1.06 of $Ni(OH)_2$ precipitate. Determine the concentration of the original $NaOH$ solution.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I didn't know what to do with this question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew that I should get the number of mol of NaOH to get the solution's concentration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I  wrote first the equation that describes the first part of the reaction :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$NaOH +  HCl \rightarrow NaCl + H2O $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I wrote the second part of the reaction :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what will be the the other products ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stopped here and I guessed an answer.&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="6889" LastEditDate="2014-06-25T20:22:40.770" LastActivityDate="2014-07-25T21:15:10.580" Title="Determine the concentration of the original NaOH solution" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13901" PostTypeId="1" AcceptedAnswerId="13903" CreationDate="2014-06-25T11:57:29.313" Score="1" ViewCount="8327" Body="&lt;p&gt;Sodium hydroxide dissolved in water is a very strong base. I think this could mean it could de-oxidate the metal oxides during re-ox reaction. Basically, I ask whether the following happens:&#xA;$$\ce{NaOH + Fe2O3 -&amp;gt; Fe + something}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that it does reduce carbon dioxide producing sodium carbonate and water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NaOH + CO2 -&amp;gt; Na2CO3 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that work with metals? If yes, which ones?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-06-25T12:08:08.633" LastActivityDate="2018-01-09T12:17:27.720" Title="Can I use sodium hydroxide (NaOH) to remove oxides from metals?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2015-02-21T19:45:49.647" ContentLicense="CC BY-SA 3.0" />
  <row Id="13902" PostTypeId="1" CreationDate="2014-06-25T12:30:11.213" Score="1" ViewCount="3147" Body="&lt;p&gt;I want to avoid anything poisonous during electroplating. I can't avoid copper oxide though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have dissolved $\ce{Na2CO3}$ in water and I'm trying to electroplate aluminium with copper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SH2sU.jpg&quot; alt=&quot;photo of the electrolysis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see I'm doing it with no actual chemical equipment. Actually I'm trying to discover ways that can be done at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the water, I can see copper hydroxide ($\ce{Cu(OH)2}$) and copper carbonate ($\ce{CuCO3}$), both formed from the copper anode. But the aluminium heatsink does not seem to have any copper on it. Will it take time or am I just wasting copper here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not gonna work, what &lt;em&gt;is&lt;/em&gt; gonna work? Could I use vinegar?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-06-27T09:53:47.387" LastActivityDate="2020-08-15T15:18:46.440" Title="Can I perform electroplating with sodium carbonate and copper?" Tags="&lt;redox&gt;&lt;home-experiment&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13903" PostTypeId="2" ParentId="13901" CreationDate="2014-06-25T13:01:13.113" Score="-1" Body="&lt;p&gt;The short answer is no. Sodium Hydroxide is a soluble base (alkali) and Iron (III) oxide is a base too. An acid will react with metal oxides forming a salt and water. For example, if a sample of Iron(III) oxide was reacted with HCl, Iron chloride and water would form. It's possible to extract the pure Iron by electrolysis. Although, I am not sure that Iron Chloride is soluble so another acid may be more suitable or electrolysis could be performed using molten iron chloride.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-25T13:01:13.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13904" PostTypeId="2" ParentId="13900" CreationDate="2014-06-25T13:09:30.220" Score="1" Body="&lt;p&gt;I am going to need a unit for the 1.06. The question needs you to figure out the amount in moles of Nickel Hydroxide. Then multiply this by two as two moles of NaOH are required to form one mole of the Nickel precipitate. Then this is the number of moles that was surplus in the reaction with HCl (NaOH is added in excess - the key to the question). Find the number of moles of HCl using the volume and concentration. Add this number to the number of moles that were used to form the Nickel precipitate and then find the concentration using the moles and the volume.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-06-25T13:09:30.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13905" PostTypeId="2" ParentId="13891" CreationDate="2014-06-25T13:22:35.617" Score="4" Body="&lt;p&gt;Here is a diagram of the Clemmensen reduction.  The metal supplies the electrons and the hydrochloric acid supplies the protons to achieve the carbonyl reduction.  The mechanism of reactions that occur on metal surfaces is poorly understood, but &lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/clemmensen-reduction.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is a link&lt;/a&gt; to the current thinking on the mechanism for the Clemmensen reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlSWc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Shinoda test is similar to the Clemmensen reduction in that a carbonyl group is reduced.  However, in this case, a reductive-elimination occurs.  Due to the stability (extended conjugation) of the anthocyanidin product, instead of the reduction going all the way the to the corresponding methylene compound, the hydroxyl-like intermediate undergoes elimination to create the stable anthocyanidin.  In the Shinoda test, magnesium is used in place of the zinc-mercury amalgam.  The Shinoda test is a test for the presence of flavones.  If they are present in the test sample then they are reduced to anthocyanidins under the Shinoda reaction conditions as shown below.  The conjugation in flavinoid compounds produces a yellow color, while the extended conjugation in the resultant anthocyanidin shifts the color further out to the red region of the visible spectrum.  The dramatic change in color makes this a simple visual test for the presence of flavones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4MhKr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-25T14:41:18.083" LastActivityDate="2014-06-25T14:41:18.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13906" PostTypeId="1" AcceptedAnswerId="13914" CreationDate="2014-06-25T14:40:34.390" Score="2" ViewCount="3234" Body="&lt;p&gt;Why is it that fluorine has a &lt;em&gt;much&lt;/em&gt; greater reduction potential than chlorine, despite the fact fluorine has a lower electron affinity than chlorine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that fluorine has a lower electron affinity than chlorine due to fluorine's compact size and higher &lt;em&gt;negative&lt;/em&gt; charge density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what makes fluorine a good oxidizing agent - one better than chlorine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that the reduction potentials are referring to diatomic fluorine rather than elemental fluorine? And similarly diatomic chlorine rather than elemental chlorine? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-25T18:04:23.057" Title="Reduction potential and electron affinity" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13907" PostTypeId="1" AcceptedAnswerId="13915" CreationDate="2014-06-25T15:09:32.973" Score="23" ViewCount="49863" Body="&lt;p&gt;Dipole moment is the degree of polarity, i.e. the seperation of positive and negative charges. But I am not getting the intuition why and how lone pairs affect the polarity and dipole moment. I cannot connect the  &lt;strong&gt;separation of positive and negative charges&lt;/strong&gt;  definition with lone pair.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="4945" LastEditorDisplayName="user5764" LastEditDate="2017-10-10T08:18:39.767" LastActivityDate="2019-10-27T06:14:19.547" Title="How does a lone pair of a central atom affect the dipole moment?" Tags="&lt;physical-chemistry&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13908" PostTypeId="1" AcceptedAnswerId="13910" CreationDate="2014-06-25T16:11:59.200" Score="4" ViewCount="2629" Body="&lt;p&gt;I read that tungsten is 1.8 times more dense than lead. What other elements are denser than lead but safe to handle.  I want something that will fit under the keys of my computer keyboard so my pinky doesn't have to press so hard on the key.&lt;/p&gt;&#xA;" OwnerUserId="6952" LastActivityDate="2014-06-25T16:37:30.350" Title="Something denser than lead but with lead like properties?" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13909" PostTypeId="2" ParentId="13907" CreationDate="2014-06-25T16:21:04.693" Score="0" Body="&lt;p&gt;Dipole moment can be defined as the product of magnitude of charges and the distance of separation between the charges. Dipole moment may refer to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Electric dipole moment, the measure of the electrical polarity of a system of charges&lt;/li&gt;&#xA;  &lt;li&gt;Transition dipole moment, the electrical dipole moment in quantum mechanics&lt;/li&gt;&#xA;  &lt;li&gt;Molecular dipole moment, the electric dipole moment of a molecule.&lt;/li&gt;&#xA;  &lt;li&gt;Bond dipole moment, the measure of polarity of a chemical bond&lt;/li&gt;&#xA;  &lt;li&gt;Electron electric dipole moment, the measure of the charge distribution within an electron&lt;/li&gt;&#xA;  &lt;li&gt;Magnetic dipole moment, the measure of the magnetic polarity of a system of charges&lt;/li&gt;&#xA;  &lt;li&gt;Electron magnetic dipole moment&lt;/li&gt;&#xA;  &lt;li&gt;Nuclear magnetic moment, the magnetic moment of an atomic nucleus&lt;/li&gt;&#xA;  &lt;li&gt;Topological dipole moment, the measure of the topological defect charge distribution&lt;/li&gt;&#xA;  &lt;li&gt;The first order term (or the second term) of the multipole expansion of a function&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;(from &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole_moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The dipole moment is actually affected by the presence of a lone pair of electrons because the electrons on the central atom can cause shielding effect as the inner orbital electrons does for the outer-orbitals.&#xA;It can cause hindrances by charge imbalances which could result in imbalance of charges and which could destabilise the atom due to decrease in dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a molecule of carbon dioxide where two oxygens are present at each side of carbon: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O8q4S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the formal charge on carbon is zero and the same formal charge is there on both the oxygens. Coming to the topic of dipole moments you should consider the lone pair of electrons on oxygen. The lone pair is generally resided on the other side of the bond [as shown in figure]. These electrons are a part of oxygen and these makes oxygen more electronegative because of which the dipole moment increases ans hence the bond becomes stronger.&lt;br&gt;&#xA;In most of the cases the central atom in any compound of oxygen is more electropositive atom and electropositive atoms do not contain lone pairs. You can just imagine there are lone pairs on the central atom carbon which could generate some partial negative character in them and hence the bond between them must not be as strong as the bond between an electropositive and electronegative atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all is governed by dipole moment only.&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="4945" LastEditDate="2017-10-10T10:50:56.313" LastActivityDate="2017-10-10T10:50:56.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13910" PostTypeId="2" ParentId="13908" CreationDate="2014-06-25T16:37:30.350" Score="6" Body="&lt;p&gt;Here's a &lt;a href=&quot;http://www.lenntech.com/periodic-chart-elements/density.htm&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to a listing of the elements in the periodic table sorted by density.  The density of lead is 11.35 g/cm^3.  If you wanted higher density, no toxicity and to really let your pinky live in style, then gold with a density of 19.32 would be a possibility.  Other relatively available elements that are denser than lead and without toxicity concerns include palladium (12.02), rhodium (12.41), tungsten (19.35) and platinum (21.45). If you're willing to be a bit less dense than lead, then you can reduce price by using copper (8.96), molybdenum (10.22) or silver (10.5).  Alloys of some of these elements are also available at high density, but the cost (due to fabrication costs) is likely to be as high or higher than that the element itself.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-25T16:37:30.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13911" PostTypeId="1" AcceptedAnswerId="13917" CreationDate="2014-06-25T17:38:18.793" Score="10" ViewCount="43416" Body="&lt;p&gt;How should the water flow in your condenser? From the flask to the top of the condenser, or from the top of the condenser back to the flask? I could imagine both: flask to top will fill the entire condenser with water, because of the resistance of gravity, enlarging cooling area, or top to flask, because this will generate counter current exchange increasing heat flow, but then it is almost impossible to get a decent cooling surface without extreme water pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture suggests the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kDE6X.jpg&quot; alt=&quot;Refluxing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-07-17T10:30:07.510" LastActivityDate="2017-03-18T19:44:14.810" Title="Water flow in condenser" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13912" PostTypeId="2" ParentId="13911" CreationDate="2014-06-25T17:51:35.723" Score="7" Body="&lt;p&gt;Water should always enter from the bottom of a condenser (the end closest to the flask) and exit from the top of the condenser.  Doing it this way always &lt;em&gt;ensures&lt;/em&gt; that your condenser will be full of cooling water.  If you set it up the opposite way, and if for some reason the rate of water flow into the condenser decreased and became slower than the rate of water exiting the condenser, then the condenser would not remain filled with water.  If the reaction were unattended, this could cause problems.  BTW, here is a &lt;a href=&quot;http://www.chem.ualberta.ca/~orglabs/Techniques%20Extra%20Info/Distillation.html#same&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; you might enjoy.  It contains a lot of lab tips, including &quot;how to hook up the water tubing to a condenser&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5084" LastEditDate="2015-09-28T06:09:02.060" LastActivityDate="2015-09-28T06:09:02.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13913" PostTypeId="2" ParentId="13901" CreationDate="2014-06-25T17:52:27.527" Score="0" Body="&lt;p&gt;Your second reaction isn't a redox reaction. When carbon dioxide is dissolved in water, two equilibriums are forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray}&#xA;    \ce{H_2CO_3 + H_2O &amp;amp;&amp;lt;=&amp;gt; HCO_3^- + H_3O^+} \\&#xA;    \ce{HCO_3^- + H_2O &amp;amp;&amp;lt;=&amp;gt; CO_3^{2-} + H_3O^+}&#xA;  \end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that water also is in equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H_2O &amp;lt;=&amp;gt; OH^- + H_3O^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $\ce{NaOH}$ will — according to Le Chatelier's principle, the equilibriums are disturbed, and more carbonate ($\ce{CO_3^{2-}}$) is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in a nice redox reaction involving iron(III) oxide, aluminium is much better choice.&lt;/p&gt;&#xA;" OwnerUserId="396" LastEditorUserId="7475" LastEditDate="2018-01-09T12:17:27.720" LastActivityDate="2018-01-09T12:17:27.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13914" PostTypeId="2" ParentId="13906" CreationDate="2014-06-25T18:04:23.057" Score="2" Body="&lt;p&gt;Oxidising power , i.e the ability to accept eelectrons for any element depends on its electron gain enthalpy as also 2 other factors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Hydration energy - greater the better, which basically means that water molecules easily overcome attractive forces in the solute particles to break chemical bonds and that easy separation of solvent molecules to accommodate the solute occurs in a solution and that formation of new attractive interactions between solute and solvent particles is not that common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rem: sum of delta H 1, 2, and 3 is the overall enthalpy of the solution process. Rem also that H1, H2 are positive, H3 are negative...........read this up if you have forgotten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to its small size Fluorine's hydration energy is very high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Enthalpy of dissociation (i.e. bond dissociation energy) - lower the better&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluoride has a greater electron-electron repulsion among the lone pairs in the small sized F2 molecule where they are much closer to each other than in case of Cl2, hence the enthalpy of dissociation of F2 is lower than Cl2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two results in fluorine to have a greater positive electrode potential than chlorine, which results in greater oxidising power for F2.&lt;/p&gt;&#xA;" OwnerUserId="6953" LastActivityDate="2014-06-25T18:04:23.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13915" PostTypeId="2" ParentId="13907" CreationDate="2014-06-25T18:39:22.730" Score="21" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does lone pair of a central atom affect the dipole moment?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no single answer to your question, let me explain.  Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons.  This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond.  Next we must understand the directionality of the lone pair of electrons.  Consider the two molecules pictured below, ammonia and phosphine.  The molecules appear to be very similar, they are in the same column in the Periodic Table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VQ7YS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron.  You can see that in this case (as shown by the arrows, the &quot;arrowhead&quot; end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment.  Next, let's examine phosphine.  The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals.  You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will &lt;strong&gt;not&lt;/strong&gt; make a contribution to the overall molecular dipole moment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density.  But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-25T18:39:22.730" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13916" PostTypeId="1" CreationDate="2014-06-25T18:41:39.863" Score="1" ViewCount="1669" Body="&lt;p&gt;I need to make a solution that has a 15% (wt/wt%) HCl acid concentration and a 6% (wt/wt%) NaI concentration. I'm not sure yet how much of this solution I will be required to make so a general recipie would be awesome.  I would also like to know how to make the concentrations wt/vol% as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a bottle of Muriatic Acid. Its label says the contents are HCl acid, 20 degree Baume (31.45% by weight). The product's MSDS sheet says its 65-75% water and 25-35% HCl, and its specific gravity is 1.16 (1.16 g/ml).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Sodium Iodide (a brick of it). Its label says 99+%, pure, anhydrous.&lt;/p&gt;&#xA;" OwnerUserId="6956" LastEditorUserId="6956" LastEditDate="2014-06-26T23:43:22.840" LastActivityDate="2014-08-29T20:59:45.583" Title="Making a solution with two solutes" Tags="&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13917" PostTypeId="2" ParentId="13911" CreationDate="2014-06-25T19:00:13.540" Score="11" Body="&lt;p&gt;In general the cooling water is attached according to the countercurrent exchange mechanism to maximize the efficiency. Of course this doesn't work well for the Liebig condenser you pictured, but you generally wouldn't use this kind of condenser for reflux anyway. The most common condenser for reflux is the Dimroth condenser which can be used for countercurrent exchange without any problems as the entry and exit of the water are both at the top of the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image from &lt;a href=&quot;http://www.ioc-praktikum.de/methoden/skript/Arbeitsmethoden.pdf&quot; rel=&quot;noreferrer&quot;&gt;Arbeitsmethoden in der organischen Chemie (page 43)&lt;/a&gt; that shows where to attach the cooling water for different condenser types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VwzB5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-06-25T19:00:13.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13918" PostTypeId="1" AcceptedAnswerId="13920" CreationDate="2014-06-25T19:03:52.003" Score="12" ViewCount="57471" Body="&lt;p&gt;Upon treatment with Tollens' reagent (ammoniacal silver(I) nitrate), aldehydes are oxidised to carboxylic acid, and silver(I) is reduced to silver metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a mechanism for the this reaction online, but the only thing I can find is the balanced equation. Can someone propose or help me find the mechanism?&lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="16683" LastEditDate="2017-09-07T16:49:34.713" LastActivityDate="2017-09-07T17:04:14.120" Title="Mechanism for reaction of Tollens' reagent with aldehydes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;analytical-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13919" PostTypeId="2" ParentId="13918" CreationDate="2014-06-25T19:47:28.567" Score="1" Body="&lt;p&gt;The balanced equation can tell you a lot. This is the balanced equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-CHO + 2[Ag(NH3)2]+ + 3HO- -&amp;gt; R-COO- + 2Ag + 2H2O + 4NH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's check out what's happening:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) First, the aldehyde is being oxidized. Specifically, the carbon in the aldehyde is being oxidized; it is losing electrons to silver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation half reaction: $\ce{R-CHO -&amp;gt; R-COO- + 2e^- + H+}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation states of C: $~~~~~~+1~~~~~~~~~~~~~+3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Mass isn't balanced; we need a source of $\ce{O^2-}$ (negative 2 oxidation state oxygen atoms). Given that this reaction is happening with a solution of silver ammonia, it makes thermodynamic sense that your best source of $\ce{O^2-}$ oxygen is the hydroxide anion: $\ce{HO^-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is a &lt;em&gt;possible&lt;/em&gt; source but it's pricier to &lt;em&gt;heterolytically&lt;/em&gt; cleave two $\ce{H-O}$ bonds as opposed to just one $\ce{H-O}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically favorable: $\ce{H-O^- -&amp;gt; H^+ +O^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not so thermodynamically favorable: $\ce{H-O-H -&amp;gt; 2H^+ +O^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The freeing of one $\ce{O^2-}$ yields the hydrogen proton, $\ce{H+}$. This is an acid; it will react immediately with the strongest base in the system, which would be the hydroxide anion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that a hydrogen proton is freed from the aldehyde. The hydrogen atom is freed &lt;em&gt;heterolytically&lt;/em&gt; as to yield 2 electrons, which go to the silver ammonia complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-CHO -&amp;gt; R-COO- + H+ + 2e^-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two protons require two additional hydroxide anions to react with; this is why we use a total of three hydroxide anions and this yields two water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+ + 2HO- -&amp;gt; 2H2O}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) The above explains the hard stuff. The reduction of silver is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Ag(NH3)_2]+ + e^- -&amp;gt; Ag + 2NH3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The silver in the silver ammonia complex ion has an oxidation state of +1; gaining an electron allows solid silver to precipitate out, giving a positive Tollens test for aldehydes. The ammonia remains unchanged. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-25T20:18:34.853" LastActivityDate="2014-06-25T20:18:34.853" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13920" PostTypeId="2" ParentId="13918" CreationDate="2014-06-25T20:17:22.290" Score="17" Body="&lt;p&gt;Here are the two half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{[Ag(NH3)2]+ + e- &amp;amp;-&amp;gt; Ag^0 + 2NH3} \\&#xA;\ce{RCHO + 3OH- &amp;amp;-&amp;gt; RCO2- + 2H2O + 2e-}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which together yield the overall reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -&amp;gt; 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the process and the $\ce{Ag^+}$ is reduced. The resultant oxidized aldehyde (now a radical cation) reacts with hydroxide to form a tetrahedral intermediate. A gem-diol like intermediate is formed via a hydrogen shift, which then continues on to the final carboxylate anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U8DCz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U8DCz.png&quot; alt=&quot;reaction mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-09-07T16:59:35.700" LastActivityDate="2017-09-07T16:59:35.700" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13922" PostTypeId="2" ParentId="13918" CreationDate="2014-06-25T20:41:42.590" Score="7" Body="&lt;p&gt;A Google search for &lt;em&gt;&quot;Tollens' test mechanism&quot;&lt;/em&gt; gave me a link to this article: &lt;a href=&quot;https://doi.org/10.3184/174751911X13206824040536&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Chem. Res.&lt;/em&gt; &lt;strong&gt;2011,&lt;/strong&gt; &lt;em&gt;35&lt;/em&gt; (12), 675–677&lt;/a&gt;. A quick glance at the text of the article leads to the conclusion that the currently proposed mechanism is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{R-CHO + H2O &amp;amp;-&amp;gt; R-CH(OH)2} \\&#xA;\ce{R-CH(OH)2 + Ag+ &amp;amp;-&amp;gt; R-C^.(OH)2 + H+ + Ag^0} \\&#xA;\ce{R-C^.(OH)2 + Ag+ &amp;amp;-&amp;gt; R-COOH + H+ + Ag^0}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In strongly alkaline solution (pH &gt; 10) the mechanism changes to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{R-CHO + OH &amp;amp;-&amp;gt; R-CH(OH)O-} \\&#xA;\ce{R-CH(OH)O- + Ag+ + OH- &amp;amp;-&amp;gt; R-C^.(OH)O- + H2O + Ag^0} \\&#xA;\ce{R-C^.(OH)O- + Ag+ + OH- &amp;amp;-&amp;gt; R-COO- + H2O + Ag^0}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems that there is no direct evidence for these exact mechanisms, though they seem believable. That is normal, though; proving any mechanism is a long and tedious task that can take decades of dedicated studies, so we have to live with it. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2017-09-07T17:04:14.120" LastActivityDate="2017-09-07T17:04:14.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13923" PostTypeId="1" AcceptedAnswerId="13936" CreationDate="2014-06-25T22:05:25.200" Score="3" ViewCount="21999" Body="&lt;p&gt;What is a possible mechanism for the Beilstein test for halogens ? I can't seem to find a arrow pushing mechanism for this reaction. &lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="4945" LastEditDate="2014-06-26T06:07:15.493" LastActivityDate="2014-06-26T15:04:58.437" Title="Mechanism for the Beilstein test?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13924" PostTypeId="1" CreationDate="2014-06-25T22:51:31.077" Score="1" ViewCount="112" Body="&lt;p&gt;We use micron-grade hexagonal boron nitride (hBN) to coat the surface of bullets that are jacketed in gilding metal.  This coating is mechanically embedded/plated onto the surface in order to increase the lubricity of the bullets as they are propelled down the rifled bore of a gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can capture fired bullets in a water tank.  There is some question as to whether any of the hBN survives the temperature, pressure, and friction involved in shooting the bullet through a firearm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we determine whether the hBN coating survives the firing process?  Or better yet, can we characterize the quantity that remains on the bullet, or any part of it, after firing and recovery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Destructive methods of analysis are fine.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2017-07-31T12:51:45.063" Title="How can I determine/quantify the presence of boron nitride on a metal surface?" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;metallurgy&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13925" PostTypeId="5" CreationDate="2014-06-25T22:56:20.923" Score="0" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Alloy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An alloy is a mixture of chemical elements, which forms an impure substance (admixture) that retains the characteristics of a metal. An alloy is distinct from an impure metal in that, with an alloy, the added elements are well controlled to produce desirable properties, while impure metals such as wrought iron are less controlled, but are often considered useful. Alloys are made by mixing two or more elements, at least one of which is a metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-14T19:42:25.890" LastActivityDate="2018-03-14T19:42:25.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13926" PostTypeId="4" CreationDate="2014-06-25T22:56:20.923" Score="0" Body="For questions about alloys." OwnerUserId="5354" LastEditorUserId="5026" LastEditDate="2018-03-14T19:41:32.633" LastActivityDate="2018-03-14T19:41:32.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13927" PostTypeId="5" CreationDate="2014-06-25T23:18:08.260" Score="0" Body="&lt;p&gt;Metallurgy, in a highly simplified manner, has several crucial steps. These are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Crushing of the ore by powdering or grinding to increase its surface area&lt;/li&gt;&#xA;&lt;li&gt;Concentration/Benefaction/Dressing (removal of impurities) by froth floatation/levigation/magnetic separation/etc.&lt;/li&gt;&#xA;&lt;li&gt;Roasting or calcination (where appropriate) to form a metal oxide&lt;/li&gt;&#xA;&lt;li&gt;Reduction of the pure metal oxide to metal: electrolytically, by reducing agent, self-reduction, or by heating&lt;/li&gt;&#xA;&lt;li&gt;Purification of the metal obtained in step 4&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Several metals like iron or copper also involve formation and removal of &quot;matrix&quot; (infusible impurities) removed as fusible &quot;slag&quot; by addition of acidic or basic &quot;flux&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-19T04:45:16.427" LastActivityDate="2018-03-19T04:45:16.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13928" PostTypeId="4" CreationDate="2014-06-25T23:18:08.260" Score="0" Body="Chemical and physical characteristics of metals and alloys." OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2014-06-25T23:54:13.497" LastActivityDate="2014-06-25T23:54:13.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13929" PostTypeId="1" AcceptedAnswerId="13931" CreationDate="2014-06-25T23:37:06.420" Score="1" ViewCount="828" Body="&lt;p&gt;How does one determine the formula for two non metal elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such as Nitrogen and Oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I look at the charges with Nitrogen having -3 and Oxygen having -2. How would I get to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you build a formula based on those two elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it equal $\ce{N2O5}$         ?&lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="4945" LastEditDate="2014-06-26T01:39:05.120" LastActivityDate="2014-06-26T02:36:27.033" Title="Binary Molecular Compounds of Nonmetals" Tags="&lt;covalent-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13930" PostTypeId="2" ParentId="13929" CreationDate="2014-06-26T01:02:10.537" Score="2" Body="&lt;p&gt;If you're asking how can you determine a molecular formula when simply given two elements, then there are no hard and fast rules, especially with nitrogen. The only rule I can think of is that molecules, as opposed to ions, are electrically neutral. This still leaves you with plenty of possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen is a versatile element and can adopt a variety of oxidation states. So you'll find nitrogen in a variety of compounds in various proportions. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-26T02:36:27.033" LastActivityDate="2014-06-26T02:36:27.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13931" PostTypeId="2" ParentId="13929" CreationDate="2014-06-26T01:07:55.003" Score="5" Body="&lt;p&gt;Nonmetals form covalent bonds with other nonmetals, not ionic bonds. This means you can't use ionic bonding rules to figure out what the structure will be. In fact, there is no way to predict the structure - all you could do is propose Lewis structures for various combinations to predict what it &lt;em&gt;could&lt;/em&gt; be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also why there is a special set of nomenclature rules for molecular compounds - it's not enough to just give the names of the elements, you also have to specify the number of each atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, if all you are given is N and O, you would have no way of predicting the formula. It could be any one of maybe half a dozen compounds (maybe more). To end up with&#xA;$\ce{N2O5}$, you would need to know the &lt;em&gt;name&lt;/em&gt; of the compound first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For binary molecular compounds, the naming rules are pretty easy. First, you need to know the prefixes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1: mono-&lt;/li&gt;&#xA;&lt;li&gt;2: di-&lt;/li&gt;&#xA;&lt;li&gt;3: tri-&lt;/li&gt;&#xA;&lt;li&gt;4: tetra-&lt;/li&gt;&#xA;&lt;li&gt;5: penta-&lt;/li&gt;&#xA;&lt;li&gt;6: hexa-&lt;/li&gt;&#xA;&lt;li&gt;7: hepta-&lt;/li&gt;&#xA;&lt;li&gt;8: octa-&lt;/li&gt;&#xA;&lt;li&gt;9: nona-&lt;/li&gt;&#xA;&lt;li&gt;10: deca-&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then, for each element (starting with the least electronegative one - the one furthest to the left on the periodic table) you just use the prefix in front of the element name. $\ce{N2O5}$ would be called &lt;strong&gt;dinitrogen pentoxide&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two small details. The first is that we don't use &quot;mono&quot; when there is only one atom of the first element. This means that $\ce{CO}$ is &lt;strong&gt;carbon monoxide&lt;/strong&gt; rather than monocarbon monoxide. The second is that when an element starts with a vowel and the prefix ends in a vowel, we drop the prefix's ending vowel. This means that we say &lt;strong&gt;pentoxide&lt;/strong&gt; instead of pentaoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the formula given the name, you just reverse this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO}$ - nitrogen monoxide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N2O}$ - dinitrogen monoxide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... you get the idea.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-06-26T01:07:55.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13932" PostTypeId="1" CreationDate="2014-06-26T01:12:00.107" Score="4" ViewCount="14405" Body="&lt;p&gt;I've been studying photosynthesis and something made me think about how chlorophyll itself is made in plants.How exactly is it formed ?&lt;/p&gt;&#xA;" OwnerUserId="6957" LastActivityDate="2020-05-10T15:14:11.387" Title="How is chlorophyll made in plants?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13933" PostTypeId="1" AcceptedAnswerId="13941" CreationDate="2014-06-26T02:43:24.410" Score="1" ViewCount="81" Body="&lt;p&gt;I have a question that gives two concentrations and asks for the mass of HCl formed by the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two concentrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;250mL of 4.00M $\ce{H2SO4}$, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;250mL of 1.00M $\ce{NaCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the balanced reaction equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4 + 2NaCl -&amp;gt;Na2SO4 + 2HCL}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how do I add those two concentrations? I assume the mixture will be at least 500mL, but how do I add the molarity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="186" LastEditDate="2015-10-02T15:50:10.717" LastActivityDate="2015-10-02T15:50:10.717" Title="Adding concentrations to obtain mass" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13934" PostTypeId="2" ParentId="10259" CreationDate="2014-06-26T02:47:31.813" Score="4" Body="&lt;p&gt;Both elements are stable under certain conditions in various oxidation states. In &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium&quot; rel=&quot;nofollow&quot;&gt;chromium&lt;/a&gt; they range from -2 to +6 and in &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese&quot; rel=&quot;nofollow&quot;&gt;manganese&lt;/a&gt; from 0 to +7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some of them are more common than others, especially in aqueous solution. In &lt;a href=&quot;http://www.webelements.com/chromium/compounds.html&quot; rel=&quot;nofollow&quot;&gt;chromium the most common&lt;/a&gt; are +3 and +6. As a &quot;free&quot; ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one electron in the d shell. If you dissolve chromium(III)chloride in water you will get a complex formation.&#xA;$$\ce{CrCl3 + 6H2O &amp;lt;=&amp;gt; [Cr(H2O)6]^{3+} + 3Cl-}$$&#xA;Assuming an octahedral ligand alignment in &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_field_theory&quot; rel=&quot;nofollow&quot;&gt;crystal field theory&lt;/a&gt;, you are correct in assuming, that the $t_{2g}$ orbitals will be half filled, resulting in a more stable configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://www.webelements.com/manganese/compounds.html&quot; rel=&quot;nofollow&quot;&gt;manganese the most common oxidation states&lt;/a&gt; are +2, +4 and +7. If you dissolve manganese(III)chloride, then you will also form a aqueous complex, which will slowly transform to manganese(III)hydroxide and precipitate.&#xA;$$\ce{MnCl3 + 6H2O &amp;lt;=&amp;gt; [Mn(H2O)6]^{3+} + 3Cl- &amp;lt;=&amp;gt; Mn(OH)3 v + 3H3+O + 3Cl-}$$&#xA;The resulting hydroxide may be oxidised by air to manganese(IV)oxide hydroxide.&#xA;$$\ce{2\overset{\color{\green}{+III}}{Mn}(OH)3 + \overset{\color{\green}{\pm0}}{O2} -&amp;gt; 2\overset{\color{\green}{+IV}}{Mn}O(OH)2 + 2{}^{-}OH}$$&#xA;Over time this will convert into manganese dioxide. &#xA;$$\ce{MnO(OH)2 -&amp;gt; MnO2 + H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-26T02:47:31.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13935" PostTypeId="2" ParentId="13933" CreationDate="2014-06-26T03:39:48.283" Score="0" Body="&lt;p&gt;You're right; you can't directly add the two concentrations since these concentrations are specific to a certain volume of solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to recalculate concentrations of each system component in a combined solution, you need to find the moles of each system component in this combined solution, and then divide by the &lt;em&gt;total solution volume&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-06-26T03:39:48.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13936" PostTypeId="2" ParentId="13923" CreationDate="2014-06-26T04:15:12.220" Score="3" Body="&lt;p&gt;When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Cu(s) + O2(g) -&amp;gt; 2CuO(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when you mix it with the halide and heat it, the higher reduction potential makes it displace the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example with sodium chloride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuO(s) + 2NaCl(s) -&amp;gt; CuCl2(g) + Na2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green.&lt;/p&gt;&#xA;" OwnerUserId="6907" LastEditorUserId="6907" LastEditDate="2014-06-26T15:04:58.437" LastActivityDate="2014-06-26T15:04:58.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13937" PostTypeId="1" AcceptedAnswerId="13946" CreationDate="2014-06-26T05:12:01.603" Score="3" ViewCount="1328" Body="&lt;p&gt;I managed to separate a binary mixture(in solid form) of two unknown substances into two individual solid compounds. For the separation, I used an acid-base extraction with dichloromethane, NaOH, and after separation HCl.  My next step is to purify the two separated solids using recrystallization(this is the technique mandated by my instructor for solid purification) but I am not sure which solvents to use. Can anyone suggest how I should proceed ? Thanks a lot !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I was thinking since I used NaOH and HCl on my aqueous layer, NaCl would have formed. Since compound doesn't dissolve in water, is it okay to just put the compound in water, let NaCl dissolve and vacuum filter the solid(which presumably should be pure solid) ? &lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="6833" LastEditDate="2014-06-27T13:53:47.790" LastActivityDate="2014-06-27T15:18:04.770" Title="Recrystallization of two unknown compounds?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;purification&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13938" PostTypeId="2" ParentId="10825" CreationDate="2014-06-26T05:19:29.337" Score="2" Body="&lt;p&gt;First use &lt;span class=&quot;math-container&quot;&gt;$\ce{EtO-/EtOH}$&lt;/span&gt;, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or another way is to hydrolyse it in basic medium to get &lt;span class=&quot;math-container&quot;&gt;$\ce{H2CO3}$&lt;/span&gt; and then evaporate &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2020-02-13T19:28:12.370" LastActivityDate="2020-02-13T19:28:12.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13939" PostTypeId="1" CreationDate="2014-06-26T05:32:36.757" Score="0" ViewCount="319" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists: &#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{HCO3- &amp;amp;&amp;lt;=&amp;gt; H+ + CO3^{2-}}&amp;amp;&#xA;\frac{\ce{[CO3^{2-}]}}{\ce{[HCO3-]}}  &amp;amp;= 0.958&#xA;\end{align}&lt;/span&gt;&#xA;  When &lt;span class=&quot;math-container&quot;&gt;$\pu{10 cm^3}$&lt;/span&gt; of lake water were titrated with  &lt;span class=&quot;math-container&quot;&gt;$\pu{0.2 mol/dm^3}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{22 cm^3}$&lt;/span&gt; of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:&#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \ce{H+ + HCO3- &amp;amp;-&amp;gt; H2O + CO2}\\&#xA;  \ce{2H+ + CO3^{2-} &amp;amp;-&amp;gt; H2O + CO2}\\&#xA;\end{align}&lt;/span&gt;&#xA;  Calculate the total amount of substance of acid used, and thus, by using the ratio quoted, calculate &lt;span class=&quot;math-container&quot;&gt;$\ce{[CO3^{2-}]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[HCO3-]}$&lt;/span&gt; in the lake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was able to find the amount of substance of acid &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, which is &lt;span class=&quot;math-container&quot;&gt;$\pu{4.4E-3 mol}.$&lt;/span&gt; How do I go on?&lt;/p&gt;&#xA;" OwnerUserId="6958" LastEditorUserId="4945" LastEditDate="2018-10-25T17:49:58.297" LastActivityDate="2018-10-25T18:00:56.377" Title="How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2016-10-31T13:01:16.637" ContentLicense="CC BY-SA 4.0" />
  <row Id="13940" PostTypeId="2" ParentId="10267" CreationDate="2014-06-26T05:38:08.097" Score="11" Body="&lt;p&gt;Like ron already stated, the reason for that arrangement is the possibility to form more intermolecular hydrogen bonds. For oxalic acid there are &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02578605&quot; rel=&quot;noreferrer&quot;&gt;two modifications&lt;/a&gt; which depict this quite clearly.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RYOf5.png&quot; alt=&quot;solid state modifications of oxalic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From charge considerations itself it is not obvious, that the periplanar alignment should be better. However one can see, that there might be a little less repulsion. Given here are natural partial charges from NBO/BP86/cc-pVDZ.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xCqT4.jpg&quot; alt=&quot;charges of oxalic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotational scan reveals that the rotation barrier is just around $7~\mathrm{kJ/mol}$ and the difference between these are just around $2~\mathrm{kJ/mol}$ (BP86/cc-pVDZ). The dihedral angle of $0^\circ$ refers to the hydroxyl moieties being on the same side.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YCquU.jpg&quot; alt=&quot;rotational scan of oxalic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barrier results mainly from breaking the partial $\pi$ bond between the two carbon centres.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KmjNA.gif&quot; alt=&quot;animation of the rotational barrier&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{O...H}$ intramolecular distance is around $230~\mathrm{pm}$ and in gas phase and higher temperatures hydrogen tunneling should be also considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution, exchange rates between water and oxalic acid are very fast, so that all oxygen atoms should be equal - there is no difference observable at this time scale.&#xA;\begin{aligned}\ce{&#xA;(COOH)2 + 2H2O &amp;amp;&amp;lt;=&amp;gt; HOOC-COO- + H3+O + H2O\\&#xA;               &amp;amp;&amp;lt;=&amp;gt; (COO^{-})2 + 2H3+O\\&#xA;}\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-26T05:38:08.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13941" PostTypeId="2" ParentId="13933" CreationDate="2014-06-26T06:24:59.463" Score="2" Body="&lt;p&gt;Your working equation is correct.&#xA;$$\ce{H2SO4 + 2NaCl -&amp;gt;Na2SO4 + 2HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the amount of substance of protons of each solution via $n=c\cdot V$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the limiting agent?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Chlorine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many moles of hydrogen chloride can only be formed?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{HCl})=0.25~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Calculate the mass of hydrogen chloride via $m = n\cdot M$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-06-26T06:24:59.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13942" PostTypeId="2" ParentId="10416" CreationDate="2014-06-26T09:21:26.807" Score="3" Body="&lt;p&gt;As in aqueous solution:&#xA;$$\ce{NH3+CH2COOH_{[b]}&amp;lt;=&amp;gt; NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} &amp;lt;=&amp;gt; NH2CH2COO-_{[a]} + H+_{[x]}}$$ &#xA;So,&#xA;$$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$&#xA;&lt;code&gt;Notice that a=b as the solution is always neutral; also z is a neutral compund&lt;/code&gt;&lt;br&gt;&#xA;As, &#xA;$$[H^+]=10^{-6}\implies p^H=6$$&#xA;$$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You should have done $p^H=-log[H^+]$, your [b] is correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For part (b), yes, using Henderson-Hasslebalch equation:&#xA;$$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH,&#xA;$$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}&amp;lt;=&amp;gt; 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$&#xA;In millimoles:&#xA;$$\begin{array}{c|c|c|c|c}&#xA;&amp;amp;entity&amp;amp;b&amp;amp;k&amp;amp;a&amp;amp;y\\\hline&#xA;&amp;amp;initial&amp;amp;10&amp;amp;.1&amp;amp;0&amp;amp;0\\\hline&#xA;&amp;amp;final&amp;amp;9.95&amp;amp;0&amp;amp;.1&amp;amp;.1+110(original)\\\hline&#xA;\end{array}$$&#xA;Now you can similiarly apply agin the Henderson-Hasslebalch equation:&#xA;$$p_H=12+log(.1/9.95)\approx 10.01$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note that I am treating the compund as a monoprotic acid of&lt;/strong&gt; $pK_a =10(-NH_3^+)+2(-COOH)=12$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-06-26T10:03:08.303" LastActivityDate="2014-06-26T10:03:08.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13944" PostTypeId="1" CreationDate="2014-06-26T12:25:22.247" Score="3" ViewCount="1499" Body="&lt;p&gt;What is the difference when it comes to separating Carvacrol and Thymol on HPPLC with reverse phase - C18 column?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carvacrol elutes first, is that because it is ortho?&lt;/p&gt;&#xA;" OwnerUserId="6960" LastEditorUserId="4945" LastEditDate="2014-06-27T03:15:35.457" LastActivityDate="2014-06-27T03:15:35.457" Title="What's the difference between Carvacrol and Thymol?" Tags="&lt;analytical-chemistry&gt;&lt;aromatic-compounds&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13945" PostTypeId="2" ParentId="13944" CreationDate="2014-06-26T13:16:27.937" Score="4" Body="&lt;p&gt;On a reverse phase C18 column  (which itself is hydrophobic, it is an octadecyl hydrocarbon chain bound to the support material) the more hydrophobic the molecule, the more strongly it will bind to the stationary phase [like dissolves (or in this case binds to) like], and the longer it will take to elute with an organic solvent. Below are the structures of carvacrol and thymol.  The more exposed the hydroxyl group, the less hydrophobic the molecule.  The hydroxyl group in carvacrol (&lt;em&gt;ortho&lt;/em&gt; to a methyl group) is more exposed than the hydroxyl group in thymol (&lt;em&gt;ortho&lt;/em&gt; to an isopropyl group) because of the relative steric sizes of a methyl group &lt;em&gt;vs&lt;/em&gt;. an isopropyl group. Therefor thymol is more hydrophobic than carvacrol.  Consequently we would expect thymol to bind to the column more strongly than carvacrol, and therfor we would expect the carvacrol to elute first which matches your observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B7L23.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-27T00:14:28.960" LastActivityDate="2014-06-27T00:14:28.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13946" PostTypeId="2" ParentId="13937" CreationDate="2014-06-26T13:23:54.717" Score="3" Body="&lt;p&gt;Choosing a recrystallization solvent is as much an art as a science, even when you know what the substance is. Without knowing, it is a matter of making educated guesses. You want a solvent where it is much more soluble in the hot solvent than in the cold. Given that this is a lab exercise, we can assume that these are common substances. So, the one that dissolves in base is likely a carboxylic acid--perhaps benzoic acid or something close. That substance dissolves about 10 times as much in hot water as in cold - so I would try water first.  If it is too insoluble, then you could go to methanol, ethanol, etc., moving toward more nonpolar solvents until you get it to dissolve when heated. Then, hopefully it will crystallize out on cooling.  For the other substance, it is obviously soluble in dichloromethane and is not an acid. Perhaps it is non-polar. I would start with ethanol, and see how that works - if you need to go to a more non-polar solvent, toluene might be a good choice. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5577" LastEditDate="2014-06-27T15:18:04.770" LastActivityDate="2014-06-27T15:18:04.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13947" PostTypeId="1" CreationDate="2014-06-26T13:26:55.797" Score="0" ViewCount="2129" Body="&lt;p&gt;Choose the right answers in the following statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiling point of a liquid substance is higher as the Van der Waals forces (in/between) the molecules of the substance are (lower/higher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the Van der Waals forces are lower when the boiling point is lower. Second, do you say: the forces &lt;em&gt;in&lt;/em&gt; the molecules of a substance or the forces &lt;em&gt;between&lt;/em&gt; the molecules of a substance?&lt;/p&gt;&#xA;" OwnerUserId="6962" LastEditorUserId="6962" LastEditDate="2014-06-26T13:38:56.390" LastActivityDate="2014-06-26T15:50:58.667" Title="Relationship between Van der Waals forces and the boiling point" Tags="&lt;homework&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13948" PostTypeId="2" ParentId="13947" CreationDate="2014-06-26T13:44:59.570" Score="1" Body="&lt;p&gt;We talk about the forces between molecules.  The Van der Waals force represents the summation of all attractive and repulsive forces between molecules, other than those due to electrostatic charges or actual bonds.  This summation most often turns out to be a weak attractive force.  Typically the magnitude of the Van der Waals force is small compared to electrostatic or bonding forces and (again typically) only apply at relatively short range.  So they usually serve to weakly hold molecules together.  The stronger the Van der Waals force, the more strongly molecules will be held together and the more energy will be needed to be put into the system to break them apart from each other.  In the case of boiling a liquid, you are really talking about breaking molecules apart from one another.  So in the case of boiling, the weaker the Van der Waals attraction between molecules, less energy will needed to break them apart from each other and allow them to boil.&lt;/p&gt;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to a Wikipedia article&lt;/a&gt; on Van der Waals forces if you'd like to read more about them.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-06-26T15:50:58.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13949" PostTypeId="1" AcceptedAnswerId="13950" CreationDate="2014-06-26T17:43:14.653" Score="11" ViewCount="4782" Body="&lt;p&gt;There are many compounds in which the stability of a molecule is not governed by the presence of octet configuration in central atom. In most of the cases the central atom is generally sulfur or phosphorus. In these cases either the central atom has 6 electrons, or sometimes the electron count exceeds 8 and 10 or 12 electrons in its valence electron shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the concept of octet somewhat incorrect. My question is how can they increase their octet to such an extent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain this in relation with sulfur trioxide?&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="9961" LastEditDate="2020-06-12T14:22:48.267" LastActivityDate="2020-06-12T14:22:48.267" Title="Does sulfur and phosphorus expand their octet?" Tags="&lt;bond&gt;&lt;hybridization&gt;&lt;covalent-compounds&gt;&lt;oxidation-state&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="13950" PostTypeId="2" ParentId="13949" CreationDate="2014-06-26T17:45:45.377" Score="16" Body="&lt;p&gt;Whether sulfur or phosphorous actually expand their octet is contested within the chemistry community. Another term for this octet expansion is &quot;hypervalency.&quot; You can find many works of research regarding hypervalency. The consensus, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;according to Wikipedia&lt;/a&gt;, is that both &lt;em&gt;can&lt;/em&gt; expand their octets, but not to a significant extent. In other words, d-orbital utilization is low at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding sulfur trioxide, &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1997/FT/a700708f#!divAbstract&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; may be of interest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We find no evidence to support notions of p π –d π back-donation from oxygen to sulfur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: $\ce{C, N, O}$, and $\ce{F}$. And note that for carbon, there are well known exceptions in organic chemistry - carbocations (which have less than an octet of electrons). Carbocations however tend to exist mainly as reactive intermediates. Therefore, a more nuanced statement might be $\ce{C, N, O}$, and $\ce{F}$, when involved in neutral and thermodynamically stable compounds, &lt;em&gt;generally&lt;/em&gt; have 8 valence electrons. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;this makes the concept of octet somewhat incorrect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-27T13:30:37.470" LastActivityDate="2018-03-27T13:30:37.470" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13951" PostTypeId="1" AcceptedAnswerId="13953" CreationDate="2014-06-26T17:56:33.183" Score="2" ViewCount="2585" Body="&lt;p&gt;Whenever using copper as anode, I end up with cyan dust that doesn't dissolve in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q8yLY.jpg&quot; alt=&quot;mysterious copper oxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's that (presumably) oxide? Is it poisonous? Can I dissolve it in something to perform electroplating with it?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-06-26T19:07:00.033" Title="What copper oxide is being created upon electrolysis?" Tags="&lt;home-experiment&gt;&lt;electrolysis&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13952" PostTypeId="1" CreationDate="2014-06-26T18:04:14.980" Score="1" ViewCount="98" Body="&lt;p&gt;I was just thinking - plants are mostly acid. Even grass is quite acid. Could you create battery cell from squashed plants (after cutting them in pieces)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much effective would it be? Could you compare it to solar panels? Would you need to replace electrodes often? &lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-11-23T12:53:12.773" Title="Can I use harvested plants to gain electric power?" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13953" PostTypeId="2" ParentId="13951" CreationDate="2014-06-26T19:07:00.033" Score="3" Body="&lt;p&gt;It appears to be copper hydroxide or copper carbonate.  Copper oxide tends to be black/grey/brown.  Copper hydroxide is blue and copper carbonate is green.  Copper carbonate forms when the hydroxide is exposed to CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like to dissolve it, try an acidic solution.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-06-26T19:07:00.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13954" PostTypeId="1" CreationDate="2014-06-26T19:09:28.157" Score="1" ViewCount="1047" Body="&lt;p&gt;What is 'information depth' in XPS and what does determine it? Also how does it depend on the 'take-off angle'? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2014-07-27T15:46:57.893" Title="Information depth in x-ray photoelectron spectroscopy (XPS)" Tags="&lt;spectroscopy&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13957" PostTypeId="1" AcceptedAnswerId="13962" CreationDate="2014-06-26T21:27:37.280" Score="6" ViewCount="1970" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the two reduction half-reactions for the following fuel cell, determine the standard cell potential &lt;span class=&quot;math-container&quot;&gt;$E^\circ_\mathrm{cell}$&lt;/span&gt;,&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH4 (g) + 2 O2 (g) + 2 OH- (aq) -&amp;gt; CO3^2- (aq) + 3 H2O}.$$&lt;/span&gt;&#xA;  Reduction half-reactions:&#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CO3^2- (aq) + 7 H2O (l) + 8 e- &amp;amp;-&amp;gt; CH4 (g) + 10 OH- (aq)}&amp;amp;&#xA;E^\circ &amp;amp;= \pu{-0.1 V}\\&#xA;\ce{O2 (g) + 2 H2O (l) + 4 e- &amp;amp;-&amp;gt; 4 OH- (aq)}&amp;amp;&#xA;E^\circ &amp;amp;= \pu{+0.55 V}&#xA;\end{align}&lt;/span&gt;&#xA;  Answers: (a) &lt;span class=&quot;math-container&quot;&gt;$\pu{+0.45 V}$&lt;/span&gt; (b) &lt;span class=&quot;math-container&quot;&gt;$\pu{-0.45 V}$&lt;/span&gt; (c) &lt;span class=&quot;math-container&quot;&gt;$\pu{-0.65 V}$&lt;/span&gt; (d) &lt;span class=&quot;math-container&quot;&gt;$\pu{+0.65 V}$&lt;/span&gt;&#xA;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that to apply Hess' Law to this equation the top equation sign would be reversed and the outcome would be &lt;span class=&quot;math-container&quot;&gt;$\pu{+0.65 V}$&lt;/span&gt;. Often the system produces incorrect answers. I was looking through my textbook to double check this, and there is an instance where it says &#xA;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode},$$&lt;/span&gt;&#xA;but I believe that rule does not fit this instance. Could somebody validate that this answer provided is indeed correct or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to clarify that my answer was (d) and the system provided that the correct answer was (a).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I sent an email to my teacher to contest the question using an example from the textbook. Given that Martin has provided me with an answer concurrent with the &lt;span class=&quot;math-container&quot;&gt;$\pu{+0.45 V}$&lt;/span&gt;, then there must be a mistake in the text or I am misinterpreting something. I would really like to understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first my teacher told me that after flipping the equation you still need to subtract the anode from the cathode. So I responded like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to reference the E Cell example from the text. In the text it explains that &lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode}.$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The example given in the book is identical to the one provided in the exam:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA; (1): &amp;amp;&amp;amp;\ce{3 Ag+(aq) + Fe (s)   &amp;amp;-&amp;gt; 3 Ag (s) + Fe^{3+} (aq)}\\&#xA; (2): &amp;amp;&amp;amp;\ce{3 Ag+(aq) + 3 e-     &amp;amp;-&amp;gt; 3 Ag (s)} &amp;amp; E^\circ &amp;amp;= 0.80~\mathrm{V}\\&#xA; (3): &amp;amp;&amp;amp;\ce{Fe^{3+} (aq) + 3 e- &amp;amp;-&amp;gt; 3 Fe (s)} &amp;amp; E^\circ &amp;amp;= -0.04~\mathrm{V}\\&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By using the same logic that I have applied to the exam question the &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{cell}$&lt;/span&gt; was found to be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.84 V}$&lt;/span&gt;. This is because the sign on the &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; value of the &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt; half reaction was reversed like I did to the top equation (which consequently is the cathode).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As the given equations line up with the primary equation and signs are flipped whenever the equations are flipped, then it is not important to decipher which is the anode and which is the cathode. This is also in accordance with Hess' Law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That being said, I am confused whether the textbook example was incorrect, the exam question was incorrect, or my understanding is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="3926" LastEditDate="2020-01-29T17:45:16.630" LastActivityDate="2020-01-29T17:45:16.630" Title="How to calculate the potential via Hess's law? Is my textbook wrong about this?" Tags="&lt;electrochemistry&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13958" PostTypeId="1" AcceptedAnswerId="13959" CreationDate="2014-06-27T00:21:58.973" Score="2" ViewCount="3576" Body="&lt;p&gt;What are the blocks of the periodic table? What is the purpose of naming the elements per block group? What do they signify?&lt;/p&gt;&#xA;" OwnerUserId="6967" LastActivityDate="2014-06-27T00:37:31.007" Title="Block on the Periodic Table?" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13959" PostTypeId="2" ParentId="13958" CreationDate="2014-06-27T00:32:16.417" Score="4" Body="&lt;p&gt;Below is a picture of the 4 blocks in the Periodic Table.  The elements in a block all use the same orbital to hold their valence or outer shell electrons.  For example, all of the elements in block 1 have their valence electrons in an $\ce{s}$ orbital, those in block 2 have their valence electrons in a $\ce{p}$ orbital, block 3 the $\ce{d}$ orbitals and block 4 the $\ce{f}$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9Pmg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-27T00:37:31.007" LastActivityDate="2014-06-27T00:37:31.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13960" PostTypeId="1" AcceptedAnswerId="13961" CreationDate="2014-06-27T01:18:59.643" Score="8" ViewCount="8638" Body="&lt;p&gt;Is there a difference between these two molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so how does one tell? What is the correct nomenclature for cycloalkenes? I know that the double bond is assumed to be between C1 and C2 but why can't we number a cyclohexene to get 2,3? Is it because a lower sum is not preferred but simply lower individual numbers if possible? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-06-08T09:02:29.913" LastActivityDate="2016-02-17T12:02:04.730" Title="1,6-dimethylcyclohexene and 2,3-dimethylcyclohexene" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13961" PostTypeId="2" ParentId="13960" CreationDate="2014-06-27T01:47:53.420" Score="10" Body="&lt;p&gt;The IUPAC rule is to have the first branch to have the lowest number, even if it gives you 1,6-dimethylcyclohexene.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-27T01:47:53.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13962" PostTypeId="2" ParentId="13957" CreationDate="2014-06-27T03:57:36.577" Score="5" Body="&lt;p&gt;You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;(A)E^\circ = -0.10~\mathrm{V}&amp;amp;&amp;amp;\ce{CO3^{2-} + 7H2O + 8e- &amp;amp;-&amp;gt; CH4 + 10 {}^{-}OH}  \\&#xA;(B)E^\circ = +0.55~\mathrm{V}&amp;amp;&amp;amp;\ce{O2 + 2H2O + 4e- &amp;amp;-&amp;gt; 4 {}^{-}OH}  \\\hline&#xA;\color{\red}{-1}\cdot(A) + 2\cdot(B)&#xA;   &amp;amp;&amp;amp;\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&amp;amp;-&amp;gt; CO3^{2-} + 7H2O + 4 {}^{-}OH}\\&#xA;\implies&#xA;   &amp;amp;&amp;amp;\ce{CH4 + 2O2 + 2{}^{-}OH &amp;amp;-&amp;gt; CO3^{2-} + 3H2O}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;nofollow&quot;&gt;galvanic cell&lt;/a&gt; of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is&#xA;\begin{aligned}&#xA;\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~&#xA;               &amp;amp;= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\&#xA;               &amp;amp;= E^\circ(\ce{O2|{}^{-}OH}) &#xA;                  -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\&#xA;               &amp;amp;= +0.55~\mathrm{V}          &#xA;                  -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\&#xA;\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~&#xA;               &amp;amp;= +0.45~\mathrm{V}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just to address the second edit of the OP and to clarify:&#xA;\begin{aligned}&#xA; (1): &amp;amp;&amp;amp;\ce{3 Ag+(aq) + Fe (s)   &amp;amp;-&amp;gt; 3 Ag (s) + Fe^{3+} (aq)}\\&#xA; (2): &amp;amp;&amp;amp;\ce{3 Ag+(aq) + 3 e-     &amp;amp;-&amp;gt; 3 Ag (s)} &amp;amp; E^\circ &amp;amp;= 0.80~\mathrm{V}\\&#xA; (3): &amp;amp;&amp;amp;\ce{Fe^{3+} (aq) + 3 e- &amp;amp;-&amp;gt; 3 Fe (s)} &amp;amp; E^\circ &amp;amp;= -0.04~\mathrm{V}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-02-27T06:14:35.377" LastActivityDate="2015-02-27T06:14:35.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13963" PostTypeId="1" AcceptedAnswerId="13964" CreationDate="2014-06-27T04:14:57.677" Score="6" ViewCount="19848" Body="&lt;p&gt;The standard reduction potential of diatomic oxygen in acidic conditions is +1.23 volts. However, the standard reduction potential of diatomic oxygen in basic conditions is only +0.40 volts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nc1vr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the reduction of diatomic oxygen by acidic media more favored than the reduction of oxygen by water that is basic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of a few reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) We are likely going to be forming negative two oxidation state oxygen anions when we oxidize diatomic oxygen. Oxygen adopts an oxidation state of negative two except in peroxides, which aren't very stable, and I don't think the oxygen radical will be very stable in solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can someone provide a less fluffy reason that we're going to be forming negative two oxidation state oxygen atoms?&lt;/strong&gt; I'd imagine that the electronegativity and electron affinity of oxygen are two reasons it might favor the negative two oxidation state over the negative one oxidation state in solution. I.e. why doesn't this occur to a significant extent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 + 2e^- -&amp;gt; 2O^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 + 2e^- -&amp;gt; HOOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i.e. it won't react to a large extent with either hydronium or hydroxide ion. Plus the oxygen in water has a -2 oxidation state. So if we look at our skeleton half reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 + 4e^- -&amp;gt; 2H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We realize that we need a source of protons. And the proton source in acidic medium is likely hydronium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in basic medium, the proton source is water. I'd imagine that water is a poorer proton source than hydronium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would this account for the difference in standard reduction potential of oxygen in media that vary in pH? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that voltage is an extension of K through the Nernst equation; is what I'm seeing here basically that the protonation of a base ($\ce{O^2-}$) is more favored in acidic rather than basic media? Could there also be kinetic factors? In other words, why is this statement true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPkFr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What's the mechanism for the reduction of diatomic oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Why is the reduction of protons at high pH less favored than the reduction of protons at low pH? My first guess: you don't find as many protons in basic solutions. But then, protons don't exist by themselves anyway. So what can the proton source be? I'm guessing that in basic pH, there doesn't exist much hydronium ion. So the only two sources of protons are water and hydroxide anion. Heterolytically cleaving the hydroxide anion to form the highly basic oxide anion doesn't seem thermodynamically favorable. Cleaving water to form hydroxide anions ... are we running into a Le Chatlier's Principle effect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: Or should I be considering proton activities? Is the activity of the proton less in basic solution than in acidic solution? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-27T05:22:56.257" LastActivityDate="2014-06-27T05:22:56.257" Title="Why is the water reduction of oxygen favored in acidic conditions?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13964" PostTypeId="2" ParentId="13963" CreationDate="2014-06-27T05:16:38.077" Score="3" Body="&lt;p&gt;If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-27T05:16:38.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13965" PostTypeId="1" AcceptedAnswerId="13968" CreationDate="2014-06-27T10:21:01.240" Score="1" ViewCount="14700" Body="&lt;p&gt;Do I really need glass bottles if the acids and NaOH is so diluted? I thought plastics dont really react with acids. And I already have them in plastic bottles (probably a different kind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="43942" LastEditDate="2017-05-08T05:11:04.560" LastActivityDate="2017-10-22T07:14:44.967" Title="Is storing diluted (~1M) sulfuric acid, HCl, NaOH, etc. in plastic PET bottles a bad idea?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13966" PostTypeId="2" ParentId="10158" CreationDate="2014-06-27T10:38:31.000" Score="4" Body="&lt;p&gt;The crystal structure of metamizole can be found in &lt;a href=&quot;http://dx.doi.org/10.1107/S0567740879004271&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acta Cryst. (1979). B35, 612-615&lt;/a&gt;. It is also available via the &lt;a href=&quot;http://www.ccdc.cam.ac.uk/pages/Home.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;​Cambridge Crystallographic Data Centre (CCDC)&lt;/a&gt; database, entering the DOI: 10.1107/S0567740879004271.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SQSB3.png&quot; alt=&quot;metamizole&quot;&gt;&lt;br&gt;&#xA;There are essentially two conformations. The phenyl ring is rotated of plane of the pyrazolone moiety by about $42.5/44.5^\circ$. The nitrogen(4) bound to the phenyl moiety is slightly out of plane of the surrounding atoms ($-0.17/-0.18~\mathring{\mathrm{A}}$), so is the neighbouring nitrogen(5) ($0.39~\mathring{\mathrm{A}}$). The methylaminomethane group is also pyramidalised [nitrogen(9)], as can be seen by the dihedral angles with respect to the pyrazolone ring,  $\ce{C(3)-C(2)-N(9)-C(12)}$ and $\ce{C(3)-C(2)-N(9)-C(12)}$, being $89.3/104.6^\circ$ and $-43/-42.2^\circ$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a stick model screenshot of this structure, hydrogen atoms are not included in the files. (grey: carbon; red: oxygen; blue-ish: nitrogen; yellow: sulfur; purple: sodium)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6duFw.png&quot; alt=&quot;crystal structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A calculation (BP86/cc-pVTZ in gas phase at $O~\mathrm{K}$) based on one of the modifications in the crystal structure reveals the same picture.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LygFA.jpg&quot; alt=&quot;gas phase optimisation of metamizole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for completeness the xyz coordinates of the optimised compound.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  38&#xA;BVP86/cc-PVDZ scf done: -1369.314862&#xA; C     0.634952    10.049072     4.900129&#xA; C    -0.038364     9.052812     5.625610&#xA; C    -0.843603     9.404086     6.721280&#xA; C    -0.967888    10.746974     7.078721&#xA; C    -0.317360    11.745954     6.347432&#xA; C     0.482122    11.389915     5.258069&#xA; N     0.088512     7.682548     5.283073&#xA; N     0.524862     7.332867     3.979484&#xA; C     1.016111     6.033439     4.052906&#xA; C     1.037214     5.613638     5.356904&#xA; C     0.456058     6.652593     6.201473&#xA; C    -0.418749     7.686282     2.908454&#xA; C     1.415171     5.281438     2.830834&#xA; N     1.587824     4.410206     5.852441&#xA; C     0.695266     3.536091     6.539780&#xA; S     0.466632     1.963130     5.547604&#xA; O     1.708338     1.174668     5.559744&#xA; O     0.313685     6.699922     7.422867&#xA; C     2.964673     4.465074     6.365693&#xA; O    -0.210611     2.273661     4.290449&#xA; O    -0.634313     1.159020     6.505778&#xA; H    -0.141454     0.406631     6.889481&#xA; H    -1.345054     8.622837     7.287776&#xA; H    -1.589369    11.013825     7.934553&#xA; H    -0.429995    12.793897     6.626878&#xA; H     1.005549    12.158525     4.687515&#xA; H     1.289838     9.765864     4.076298&#xA; H    -1.364716     7.127508     2.998793&#xA; H     0.053315     7.462436     1.945681&#xA; H    -0.625790     8.761093     2.953974&#xA; H     0.547989     5.057675     2.191339&#xA; H     2.145811     5.841566     2.228931&#xA; H     1.854760     4.324136     3.132627&#xA; H    -0.315695     3.946934     6.648080&#xA; H     1.067111     3.196537     7.517244&#xA; H     3.601787     4.973633     5.631670&#xA; H     3.344097     3.442734     6.490780&#xA; H     3.030898     5.004553     7.327684&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Upon request, I will also provide a pdb file. There are probably a million ways for easy conversion of &lt;code&gt;*.xyz&lt;/code&gt; files to &lt;code&gt;*.pdb&lt;/code&gt; files. For example on linux with &lt;a href=&quot;http://openbabel.org/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;openbabel&lt;/a&gt; via&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PROMPT&amp;gt; babel &amp;lt;filename&amp;gt;.xyz -opdb &amp;lt;filename&amp;gt;.pdb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Much more easier, because with graphical user interface: &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt; (because it has openbabel built in).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;open the &lt;code&gt;*.xyz&lt;/code&gt; via file - open or &lt;kbd&gt;ctrl&lt;/kbd&gt;+&lt;kbd&gt;o&lt;/kbd&gt;&lt;/li&gt;&#xA;&lt;li&gt;save the file as &lt;code&gt;*.pdb&lt;/code&gt; via file - save as or &lt;kbd&gt;ctrl&lt;/kbd&gt;+&lt;kbd&gt;shift&lt;/kbd&gt;+&lt;kbd&gt;s&lt;/kbd&gt;&lt;br&gt;&#xA;make sure you select as filetype &quot;common molecule formats&quot; and enter the ending&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And as a result you will get this beautiful file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;COMPND    METAMIZOLE scf done: -1369.314862 &#xA;AUTHOR    GENERATED BY OPEN BABEL 2.3.0&#xA;HETATM    1  C   LIG     1       0.635  10.049   4.900  1.00  0.00           C  &#xA;HETATM    2  C   LIG     1      -0.038   9.053   5.626  1.00  0.00           C  &#xA;HETATM    3  C   LIG     1      -0.844   9.404   6.721  1.00  0.00           C  &#xA;HETATM    4  C   LIG     1      -0.968  10.747   7.079  1.00  0.00           C  &#xA;HETATM    5  C   LIG     1      -0.317  11.746   6.347  1.00  0.00           C  &#xA;HETATM    6  C   LIG     1       0.482  11.390   5.258  1.00  0.00           C  &#xA;HETATM    7  N   LIG     1       0.089   7.683   5.283  1.00  0.00           N  &#xA;HETATM    8  N   LIG     1       0.525   7.333   3.979  1.00  0.00           N  &#xA;HETATM    9  C   LIG     1       1.016   6.033   4.053  1.00  0.00           C  &#xA;HETATM   10  C   LIG     1       1.037   5.614   5.357  1.00  0.00           C  &#xA;HETATM   11  C   LIG     1       0.456   6.653   6.201  1.00  0.00           C  &#xA;HETATM   12  C   LIG     1      -0.419   7.686   2.908  1.00  0.00           C  &#xA;HETATM   13  C   LIG     1       1.415   5.281   2.831  1.00  0.00           C  &#xA;HETATM   14  N   LIG     1       1.588   4.410   5.852  1.00  0.00           N  &#xA;HETATM   15  C   LIG     1       0.695   3.536   6.540  1.00  0.00           C  &#xA;HETATM   16  S   LIG     1       0.467   1.963   5.548  1.00  0.00           S  &#xA;HETATM   17  O   LIG     1       1.708   1.175   5.560  1.00  0.00           O  &#xA;HETATM   18  O   LIG     1       0.314   6.700   7.423  1.00  0.00           O  &#xA;HETATM   19  C   LIG     1       2.965   4.465   6.366  1.00  0.00           C  &#xA;HETATM   20  O   LIG     1      -0.211   2.274   4.290  1.00  0.00           O  &#xA;HETATM   21  O   LIG     1      -0.634   1.159   6.506  1.00  0.00           O  &#xA;HETATM   22  H   LIG     1      -0.141   0.407   6.889  1.00  0.00           H  &#xA;HETATM   23  H   LIG     1      -1.345   8.623   7.288  1.00  0.00           H  &#xA;HETATM   24  H   LIG     1      -1.589  11.014   7.935  1.00  0.00           H  &#xA;HETATM   25  H   LIG     1      -0.430  12.794   6.627  1.00  0.00           H  &#xA;HETATM   26  H   LIG     1       1.006  12.159   4.688  1.00  0.00           H  &#xA;HETATM   27  H   LIG     1       1.290   9.766   4.076  1.00  0.00           H  &#xA;HETATM   28  H   LIG     1      -1.365   7.128   2.999  1.00  0.00           H  &#xA;HETATM   29  H   LIG     1       0.053   7.462   1.946  1.00  0.00           H  &#xA;HETATM   30  H   LIG     1      -0.626   8.761   2.954  1.00  0.00           H  &#xA;HETATM   31  H   LIG     1       0.548   5.058   2.191  1.00  0.00           H  &#xA;HETATM   32  H   LIG     1       2.146   5.842   2.229  1.00  0.00           H  &#xA;HETATM   33  H   LIG     1       1.855   4.324   3.133  1.00  0.00           H  &#xA;HETATM   34  H   LIG     1      -0.316   3.947   6.648  1.00  0.00           H  &#xA;HETATM   35  H   LIG     1       1.067   3.197   7.517  1.00  0.00           H  &#xA;HETATM   36  H   LIG     1       3.602   4.974   5.632  1.00  0.00           H  &#xA;HETATM   37  H   LIG     1       3.344   3.443   6.491  1.00  0.00           H  &#xA;HETATM   38  H   LIG     1       3.031   5.005   7.328  1.00  0.00           H  &#xA;CONECT    1   27    6    2                                            &#xA;CONECT    2    1    7    3                                            &#xA;CONECT    3    2    4   23                                            &#xA;CONECT    4    5    3   24                                            &#xA;CONECT    5    6   25    4                                            &#xA;CONECT    6   26    1    5                                            &#xA;CONECT    7    8    2   11                                            &#xA;CONECT    8   12    9    7                                            &#xA;CONECT    9   13    8   10                                            &#xA;CONECT   10    9   14   11                                            &#xA;CONECT   11    7   10   18                                            &#xA;CONECT   12   29   30   28    8                                       &#xA;CONECT   12                                                           &#xA;CONECT   13   31   32   33    9                                       &#xA;CONECT   13                                                           &#xA;CONECT   14   10   19   15                                            &#xA;CONECT   15   16   14   34   35                                       &#xA;CONECT   15                                                           &#xA;CONECT   16   20   17   21   15                                       &#xA;CONECT   16                                                           &#xA;CONECT   17   16                                                      &#xA;CONECT   18   11                                                      &#xA;CONECT   19   36   14   37   38                                       &#xA;CONECT   19                                                           &#xA;CONECT   20   16                                                      &#xA;CONECT   21   16   22                                                 &#xA;CONECT   22   21                                                      &#xA;CONECT   23    3                                                      &#xA;CONECT   24    4                                                      &#xA;CONECT   25    5                                                      &#xA;CONECT   26    6                                                      &#xA;CONECT   27    1                                                      &#xA;CONECT   28   12                                                      &#xA;CONECT   29   12                                                      &#xA;CONECT   30   12                                                      &#xA;CONECT   31   13                                                      &#xA;CONECT   32   13                                                      &#xA;CONECT   33   13                                                      &#xA;CONECT   34   15                                                      &#xA;CONECT   35   15                                                      &#xA;CONECT   36   19                                                      &#xA;CONECT   37   19                                                      &#xA;CONECT   38   19                                                      &#xA;MASTER        0    0    0    0    0    0    0    0   38    0   38    0&#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you do not see any double bonds (or $\pi$ bonds as they should correctly be referred as) then you messed up your programs settings. (This is probably valid for any common molecular geometry file)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-07-04T11:05:50.757" LastActivityDate="2014-07-04T11:05:50.757" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13967" PostTypeId="2" ParentId="13902" CreationDate="2014-06-27T10:59:38.937" Score="1" Body="&lt;p&gt;You can try the electrodeposition in a solution prepared by mixing baking soda with vinegar and the copper sulfate. Wear glove and protection glasses. The aluminium must be cleaned before plating. Please let me know the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acidic copper plating on aluminium will cause evolution of hydrogen, so less acidic solution is better. On the other hand you need to have the solution conductive and the baking soda could help.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2014-06-27T11:52:32.400" LastActivityDate="2014-06-27T11:52:32.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13968" PostTypeId="2" ParentId="13965" CreationDate="2014-06-27T13:53:12.413" Score="3" Body="&lt;p&gt;Here you can find thorough and clear instructions about using plastic bottles for acids and bases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.corning.com/lifesciences/emea/en/technical_resources/product_guid/bottles/material.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.corning.com/lifesciences/emea/en/technical_resources/product_guid/bottles/material.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common rule is to not store strong acids and bases (like sulfuric and hydrochloric acids, and NaOH) in plastic bottles if you are not sure about the type of plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polystyrene bottles are OK for strong alkali.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a PDF document on this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://csmedia2.corning.com/lifesciences/media/pdf/PSUG_Storage_Bottles_CLS_GL_025.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://csmedia2.corning.com/lifesciences/media/pdf/PSUG_Storage_Bottles_CLS_GL_025.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6972" LastEditorUserId="14617" LastEditDate="2017-10-22T07:14:44.967" LastActivityDate="2017-10-22T07:14:44.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13969" PostTypeId="2" ParentId="13954" CreationDate="2014-06-27T14:58:29.387" Score="2" Body="&lt;p&gt;Do you understand the basics of XPS?  The Wikipedia article is pretty clear. In summary, a (usually monochromatic) x-ray source irradiates a sample.  The x-ray can be absorbed by electrons in the sample, and can provide enough energy for a core electron to be set free with an energy equal to the incident x-ray energy minus its binding energy. Collecting the electrons and measuring their energy gives a plot of counts vs electron energy.  Lower electron energy indicates an electron coming from deeper in the atomic shell. Given knowledge of atomic binding levels, you can associate specific peaks in the spectrum with elements (with chemical effects on binding energies giving additional information). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - what is the &quot;information depth&quot;? A slightly odd term, I've generally heard/used 'detection depth'.  Two questions to ponder - (1) how far will x-rays of a few keV penetrate into a solid (if an atom doesn't see any x-rays, it won't emit any electrons), and (2) how far can a ~100eV-1000eV electron travel in a solid before being scattered (and thus losing the useful info).  Not surprisingly, (2) dominates for XPS, and is often called the 'escape depth'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the 'take-off angle' - again, the interesting part is where your electron detector is placed relative to the sample, and how far the electrons have to travel through the sample to head out in the direction of the detector - the 'escape depth' will vary with exit angle (take-off angle) since the amount of material the electron has to go through is a simple trig function of the exit angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-06-27T14:58:29.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13971" PostTypeId="1" CreationDate="2014-06-27T15:48:51.703" Score="3" ViewCount="890" Body="&lt;p&gt;I have already attempted to interpret some of the peaks. Please tell me if they are correct/incorrect and anything that I have missed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Es1xu.jpg&quot; alt=&quot;Unknown #49&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WFx3C.jpg&quot; alt=&quot;Unknown #50&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="69830" LastEditDate="2021-05-05T04:49:41.577" LastActivityDate="2021-05-05T06:34:54.890" Title="Interpreting Infrared Spectroscopy (IR) Spectra" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="13975" PostTypeId="1" AcceptedAnswerId="13976" CreationDate="2014-06-27T18:19:39.437" Score="3" ViewCount="113" Body="&lt;p&gt;Potassium is known to be the most freely available mobile element.&lt;br&gt;&#xA;I am wondering why that is the case?  Is the answer hidden in its atomic structure or its monovalent cationic existance?&lt;/p&gt;&#xA;" OwnerUserId="6954" LastEditorUserId="1362" LastEditDate="2014-06-29T07:55:44.867" LastActivityDate="2014-07-01T11:20:37.387" Title="why is potassium the most free element in most of plants" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13976" PostTypeId="2" ParentId="13975" CreationDate="2014-06-27T18:29:49.920" Score="1" Body="&lt;p&gt;There are many functions related to potassium some of which are shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium regulates the opening and closing of the stomata by a potassium ion pump. Since stomata are important in water regulation, potassium reduces water loss from the leaves and increases drought tolerance. Potassium deficiency may cause necrosis or interveinal chlorosis. K+ is highly mobile and can aid in balancing the anion charges within the plant. Potassium helps in fruit colouration, shape and also increases its brix. Hence, quality fruits are produced in Potassium rich soils. It also has high solubility in water and leaches out of rocky or sandy soils. This water solubility can result in potassium deficiency. Potassium serves as an activator of enzymes used in photosynthesis and respiration[4] Potassium is used to build cellulose and aids in photosynthesis by the formation of a chlorophyll precursor. Potassium deficiency may result in higher risk of pathogens, wilting, chlorosis, brown spotting, and higher chances of damage from frost and heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the answer to your question is the small size of potassium when compared to all the other cations in plants &lt;/p&gt;&#xA;&#xA;&lt;p&gt;another reason for this is its association with the chloride ion .&#xA;chloride and potassium ions will be playing follow the master game in which if any cation takes chloride ion near it then potassium will also follow chlorine and the same is the case with potassium whenever any other anion attracts it this makes potassium highly mobile.  source &lt;a href=&quot;http://en.wikipedia.org/wiki/Plant_nutrition#Potassium&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hope you got your answer. if any doubts left feel free to comment&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="6953" LastEditDate="2014-07-01T11:20:37.387" LastActivityDate="2014-07-01T11:20:37.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13977" PostTypeId="1" AcceptedAnswerId="13979" CreationDate="2014-06-27T18:53:10.510" Score="1" ViewCount="49" Body="&lt;p&gt;I have read in bio chemistry that calcium plays an important role in the cell division but the in depth information is'nt available in the textbook and I cannot memorize any fact without a supporting fact I just want to know what is its role in the process of cell division&lt;/p&gt;&#xA;" OwnerUserId="6953" LastActivityDate="2014-06-27T18:58:06.180" Title="in which way calcium promotes cell division" Tags="&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13978" PostTypeId="1" CreationDate="2014-06-27T18:56:58.030" Score="1" ViewCount="511" Body="&lt;p&gt;I've tried to research about the issue of health concerns about enamel coating of cooking utensils (i.e. dutch ovens) and failed. I should prepare a small report about the health concerns if they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help to answer that question? Any reliable sources would be greatly appreciated, especially scientific articles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4548" LastActivityDate="2014-06-27T20:37:19.167" Title="Heating of Enamel coated cooking utensils and health hazards" Tags="&lt;heat&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13979" PostTypeId="2" ParentId="13977" CreationDate="2014-06-27T18:58:06.180" Score="1" Body="&lt;p&gt;see its simple all you need to know is that calcium is responsible for the formation of middle lamellum which is a key structure in the cell wall formation if only karyokinesis occurs and cytokinesis fails to happen then the condition of resultant cell would become multinucleated which most plants do not have &lt;/p&gt;&#xA;" OwnerUserId="6954" LastActivityDate="2014-06-27T18:58:06.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13980" PostTypeId="2" ParentId="13978" CreationDate="2014-06-27T20:37:19.167" Score="2" Body="&lt;p&gt;I unfortunately don't have time to dig through these articles in great detail, but they should help get you started. Feel free to comment with more questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall impression I got from my brief search is that there is definitely a potential health hazard depending on how the coloring was made, but in most modern countries, there are  regulations in place to prevent those kinds of problems. Enough about what I think though, let's get into the references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the coating on most cookware is &lt;a href=&quot;https://en.wikipedia.org/wiki/Cast-iron_cookware#Enameled_cast_iron&quot; rel=&quot;nofollow&quot;&gt;vitreous enamel&lt;/a&gt;, which is made by heating a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vitreous_enamel&quot; rel=&quot;nofollow&quot;&gt;mixture of glass powder and pigment&lt;/a&gt; to a very high temperature and letting it melt onto a surface (much like glazing in pottery).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass itself is pretty much harmless, so the main danger comes from the added pigments. Here, we're stuck, because it really does depend on what was used to make the pigment. If it's a zinc pigment, you're probably fine. If it was cadmium based, you might be kinda screwed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found references to several articles which describe experiments trying to find out what the leach rate of metals from the glass into a solution is (usually by boiling in vinegar), but I can't hunt down the fulltexts right now, aside from the one open-access article from the NIH. Here's what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3127121/pdf/ijms-12-02336.pdf&quot; rel=&quot;nofollow&quot;&gt;Full text:Lead, Cadmium and Cobalt (Pb, Cd, and Co) Leaching of Glass-Clay Containers by pH Effect of Food&lt;/a&gt; &#xA;While this study was on clay containers, it should theoretically apply to glass-on-metal as well. Keep in mind that the containers seem to have been acquired from local craftsmen in Mexico, so the results won't necessarily apply to stuff you find on the supermarket shelves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/6863183&quot; rel=&quot;nofollow&quot;&gt;Abstract: Hot leaching of ceramic and enameled cookware: collaborative study.&lt;/a&gt;&#xA;This is what you're looking for, but I can't find the full article, and the authors seem to have been more interested in establishing a method than finding the health effects. If you can find the full paper and cross-reference their results with acceptable limits of exposure (maybe from the EPA), that'd probably be your best bet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second link also has other articles on similar problems. If you can't find your information from these two articles, maybe you can find it by following the links on PubMed. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2014-06-27T20:37:19.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13981" PostTypeId="2" ParentId="13786" CreationDate="2014-06-27T22:25:41.013" Score="1" Body="&lt;p&gt;The key is to get the concentration of bromide ions and use that value in the Solubility Equation as defined in step 1 to get $\ce{[Ag^+]}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{sp} = [Br^-][Ag^+]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis and procedure is fine, except the product in step 4 it's a little bit big. Check the algebra reorder there. The answer I get is $2\cdot 10^{-8}~\mathrm{M}$. I would comment but I'm new in Chemistry Beta and can't do that. Hope that helps,&lt;/p&gt;&#xA;" OwnerUserId="5683" LastEditorUserId="4945" LastEditDate="2014-07-28T04:48:41.883" LastActivityDate="2014-07-28T04:48:41.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13982" PostTypeId="2" ParentId="13786" CreationDate="2014-06-28T03:48:12.300" Score="0" Body="&lt;p&gt;The way you posted your calculation is confusing. You should be clear on what you want in your statement.&lt;/p&gt;&#xA;&lt;p&gt;First, find the number of moles of &lt;span class=&quot;math-container&quot;&gt;$Br^-$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\#\ moles\ Br^-=0.020L\cdot 0.001 M$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\#\ moles\ Br^-=2\cdot 10^{-5} moles$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now find the concentration of &lt;span class=&quot;math-container&quot;&gt;$Ag^+$&lt;/span&gt; in the 520 mL solution,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_{sp}=[Ag^+][Br^-]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=\frac {K_{sp}}{[Br^-]}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{\frac {2\cdot 10^{-5} moles}{0.520L}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=\frac {7.7⋅10^{−13}mol^2/L^2}{3.84\cdot 10^{-5} moles/L}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=2.00\cdot 10^{-8} moles/L$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now find the concentration of &lt;span class=&quot;math-container&quot;&gt;$AgNO_3$&lt;/span&gt; of the initial solution&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=2.00\cdot 10^{-8} moles/L\cdot \frac {0.520 L}{0.500 L}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[Ag^+]=2.10\cdot 10^{-8} moles/L$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the concentration of &lt;span class=&quot;math-container&quot;&gt;$AgNO_3$&lt;/span&gt; of the initial solution is &lt;span class=&quot;math-container&quot;&gt;$2.10\cdot 10^{-8} moles/L$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-07-28T13:37:28.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13983" PostTypeId="1" AcceptedAnswerId="13999" CreationDate="2014-06-28T05:33:54.527" Score="1" ViewCount="160" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xCee1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the kinetic validity of the following way of balancing the above redox reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some searching of the literature; I've never come across oxide anions being mentioned in redox reaction mechanisms. Also, I read something about how two-electron reduction is impossible in a single step - something about spin pairing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm fairly sure the following method is a good &lt;em&gt;thermodynamic&lt;/em&gt; method of thinking about the reaction, but what about its kinetic validity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that the below method isn't kinetically valid, but neither was it meant to be kinetically valid, and neither are the traditional methods of balancing redox reactions kinetically valid. Traditional methods call for balancing &quot;O&quot; and &quot;H.&quot; But we aren't dealing with 0 oxidation state oxygens and hydrogens, are we? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Here is the thought process that my professor would take in solving this problem:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Skeleton reaction. Note that sulfur is being reduced from a +6 oxidation state to a +4 oxidation state. This suggests that the solid zinc loses two electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + HSO4^- -&amp;gt;SO_2 +Zn^2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Note that &lt;em&gt;two&lt;/em&gt; &lt;strong&gt;negative two oxidation state oxygens&lt;/strong&gt; are &quot;disappearing&quot; from left to right. In other words, what happened to the oxygens in the hydrogen sulfate ion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + HSO4^- -&amp;gt;SO_2 +Zn^2+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4~ negative ~two~ oxidation~ state ~oxygens -&amp;gt; 2 ~negative ~2~ oxidation~ state~ oxygens }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The special thing about these oxygens is that they're negative two oxidation state oxygens. This suggests that these oxygens are the oxide anion. The oxide anion is highly basic in solution. Therefore, the oxide anions must have reacted with some system component, which caused them to &quot;disappear.&quot; This would be the strongest &lt;strong&gt;acid&lt;/strong&gt; in the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HSO4^- -&amp;gt;SO_2 + 2O^2- (unstable)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2O^2- + 4H3O^+ -&amp;gt;6H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Also note that a hydrogen proton disappears from left to right in the skeleton equation. Because the proton is highly acidic, it is attacked by the strongest base in the solution - water (since we're in an acidic solution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HSO4^- -&amp;gt;SO_2 + 2O^2- + H^+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H^+ + H2O -&amp;gt; H3O^+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Final balanced equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + HSO4^- + 4H3O+ + H2O -&amp;gt;SO_2 +Zn^2+ +6H2O + H3O+}$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-06-28T18:30:58.250" LastActivityDate="2014-06-28T18:43:44.870" Title="Kinetic validity; redox reactions" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13984" PostTypeId="1" AcceptedAnswerId="13990" CreationDate="2014-06-28T07:13:56.463" Score="8" ViewCount="1209" Body="&lt;p&gt;What would be the product formed when 1-bromo-3-chlorocyclobutane with two equivalents of metallic sodium in ether? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there would be a free radical formation and then it will form a compound whose bond line structure would look like a square with a diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is such an unstable compound being formed as there is a lot of torsional strain?&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="5026" LastEditDate="2018-02-25T09:04:59.837" LastActivityDate="2018-02-25T09:04:59.837" Title="Why does the Wurtz reaction form such an unstable compound with torsional strain?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13987" PostTypeId="2" ParentId="13984" CreationDate="2014-06-28T12:00:24.813" Score="5" Body="&lt;p&gt;Considering the torsional strain; the maximum relief will be in the bottom most molecule[&lt;strong&gt;Note that there may be other possible products from various combinations&lt;/strong&gt;]; but from top to bottom as you can see, you require more and more molecules and thus there will be a huge loss of entropy; but as$$\Delta G=\Delta H-T\Delta S$$&#xA;;If the tempertaure is not low, so as to reduce effect of $\Delta S$, we can thus expect without a loss of entropy and by formation of an intramolecular product; it is favorable.&lt;strong&gt;Note that generally this is the case with intramolecular reaction favourability&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what you mention &lt;strong&gt;&quot;I know that there would be a free radical formation and ...&quot;&lt;/strong&gt;; this is not true, or I must say partially true, because wurtz reaction can be explained both on the basis of radical and bimolecular elimination mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddbmh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-06-28T12:00:24.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13988" PostTypeId="1" CreationDate="2014-06-28T13:38:39.243" Score="4" ViewCount="14250" Body="&lt;p&gt;Why is the alpha hydrogen of a acetoacetic ester less acidic than the alpha hydrogen of 1,3 diketone ? I was under the impression that acetoacetic ester has more resonance structures so it more acidic. Please explain. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/oC6l7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="6833" LastEditDate="2014-06-28T13:49:38.383" LastActivityDate="2018-02-14T17:03:04.007" Title="Which alpha Hydrogens are more acidic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13989" PostTypeId="1" CreationDate="2014-06-28T13:48:18.333" Score="3" ViewCount="133" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;White solid &lt;strong&gt;A&lt;/strong&gt; water insoluble cleaved hot to solid gas &lt;strong&gt;B&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt;. &lt;strong&gt;B&lt;/strong&gt; is insoluble in water and in $\ce{HCl}$ but soluble to $\ce{HNO3}$ the react gives solution &lt;strong&gt;D&lt;/strong&gt; and &lt;strong&gt;E&lt;/strong&gt; colorless gas which becomes red on contact with air. The solution &lt;strong&gt;D&lt;/strong&gt; gave white precipitate &lt;strong&gt;F&lt;/strong&gt; with $\ce{HCl}$. The gas &lt;strong&gt;C&lt;/strong&gt; is reacted by heating with magnesium &lt;strong&gt;G&lt;/strong&gt; yielding a white solid which reacts with water yielding a white solid and the gas &lt;strong&gt;J&lt;/strong&gt; characteristic suffocating odor. &lt;strong&gt;J&lt;/strong&gt; is a basic , &lt;strong&gt;H&lt;/strong&gt; is soluble in dilute $\ce{H2SO4}$. The gas &lt;strong&gt;J&lt;/strong&gt; oxidized to &lt;strong&gt;C&lt;/strong&gt; while the gas under specific conditions ($\ce{Pt}$ catalyst and $800^\circ\mathrm{C}$) is oxidized to the gas &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;br&gt;&#xA;&lt;strong&gt;C&lt;/strong&gt;: $\ce{N2}$,&lt;br&gt;&#xA;&lt;strong&gt;G&lt;/strong&gt;: $\ce{Mg3N2}$,&lt;br&gt;&#xA;&lt;strong&gt;H&lt;/strong&gt;: $\ce{Mg(OH)2}$,&lt;br&gt;&#xA;&lt;strong&gt;J&lt;/strong&gt;: $\ce{NH3}$,&lt;br&gt;&#xA;&lt;strong&gt;E&lt;/strong&gt;: $\ce{NO}$&#xA;\begin{aligned}\ce{ &#xA;N2 + 3Mg &amp;amp;-&amp;gt; Mg3N2\\&#xA; Mg3N2 + 6H2O &amp;amp;-&amp;gt; 3Mg(OH)2 + 2NH3\\&#xA; Mg(OH)2 + H2SO4 &amp;amp;&amp;lt;=&amp;gt; 2 H2O + MgSO4\\&#xA; 4 NH3 + 5 O2 &amp;amp;-&amp;gt;[\ce{Pt~$800^\circ\mathrm{C}$}] 6 H2O + 4 NO\\&#xA; 4 NH3 + 3 O2 &amp;amp;-&amp;gt; 6 H2O + 2 N2\\&#xA;}\end{aligned} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot really find anything else.&lt;/p&gt;&#xA;" OwnerUserId="6977" LastEditorUserId="5026" LastEditDate="2018-07-24T01:50:27.867" LastActivityDate="2018-07-24T01:51:16.590" Title="Identification of unknowns with varying solubilities and states of matter" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13990" PostTypeId="2" ParentId="13984" CreationDate="2014-06-28T13:58:07.857" Score="8" Body="&lt;p&gt;In the Wurtz coupling reaction, an alkyl halide first reacts with a metal in a one-electron transfer step to produce an alkyl radical which quickly reacts in a second electron transfer with the metal to form a carbanion.  In the final step of the reaction, the carbanion displaces a halogen from a carbon-halogen bond by a $\ce{S_{N}2}$ process to form a carbon-carbon bond.  In the case of 1-bromo-3-chlorocyclobutane two reaction pathways are possible, the carbanion could react intramolecularly to form bicyclobutane or intermolecularly to from a cyclobutane dimer.  Considering the equation $$\ce{\Delta G = \Delta H - T \Delta S}$$ we would say that even though the $\ce{\Delta H}$ term is less favorable for bicyclobutane formation due to the increased ring strain, the change in entropy required to bring two separate molecules together to form the cyclobutane dimer overwhelms the enthalpic term and drives the reaction to proceed by the entropically more favorable intramolecular process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the cyclobutyl carbanion drawn in the figure below.  The carbanion is ideally set up for an intramolecular $\ce{S_{N}2}$ reaction (backside displacement) on the carbon containing the chlorine diagonally situated across from it.  Entropically, this is extremely favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, because 1) the carbanion and the chloride to be displaced are in the same molecule, 2) because they are so well positioned for the $\ce{S_{N}2}$ displacement reaction to occur and 3) because entropy strongly favors the unimolecular process (bicyclobutane formation) over the bimolecular process (cyclobutane dimer formation), bicyclobutane formation readily occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCOhP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-28T13:58:07.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13991" PostTypeId="2" ParentId="13988" CreationDate="2014-06-28T14:16:13.383" Score="1" Body="&lt;p&gt;When you remove the α-hydrogen, you get the following structures:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/t4odM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the carbonyl of the ester doesn't contribute (as much) to the stabilization since it already has some stabilization with the ester oxygen. Also the dicarbonyl tends to be flat due to Van der Waals forces. The ester group has the ability to rotate, which would inhibit stabilization.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-06-28T14:23:04.863" LastActivityDate="2014-06-28T14:23:04.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13992" PostTypeId="2" ParentId="13988" CreationDate="2014-06-28T14:18:19.303" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was under the impression that acetoacetic ester has more resonance structures so &lt;strong&gt;it more acidic.&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's actually wrong. Deprotonated acetoacetic ester has structures where electrons are donated from ester oxygen into conjugated system. This, while stabilizing system as a whole, lessens effect of electron donation from deprotonated carbon. Think about it this way: carbonyl oxygen wants electrons. While in acetylacetone electrons may be harvested from deprotonated carbon only and in acetylacetic ester they may be harvested from both deprotonated carbon and ester oxygen, stabilization effect in the first case is higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is well known, that stabilization effect on acidity of nearby $CH$ grows in row&#xA;$COO^- &amp;lt; -CO(NH_2) &amp;lt; -COOR &amp;lt; -COR &amp;lt; -CHO &amp;lt; -NO_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more electron deficient the conjugated system is, the more stabilized carbanion becomes, thus the less resonance structures with transfer of the charge onto carbonyl oxygen the group has, the more it stabilize carbanion. Thus, $\alpha-CH$-acidity in carbonyl compounds correlates with electrophilic activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that negative charge from deprotonated carbon must be delocalized as much as possible, which is especially easy with strongly electron-accepting groups. In case the compound already has an electron pair conjugated with carbonyl group,  this is more problematic.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2018-02-14T17:03:04.007" LastActivityDate="2018-02-14T17:03:04.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13993" PostTypeId="2" ParentId="13988" CreationDate="2014-06-28T14:36:58.200" Score="3" Body="&lt;p&gt;Here is a picture of the basic reaction we are talking about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5XxYf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that we wind up placing a negative charge alpha to the carbonyl carbon.  When we examine how substituents in various positions will affect the acidity of protons alpha to a carbonyl group, we must look at how both the inductive and resonance effects exerted by the substituent will stabilize or destabilize a nearby negative charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ULW38.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above chart we see that by attaching a $\ce{CF3}$ to a carbonyl we make the alpha-hydrogens very acidic. This is because the $\ce{CF3}$ group is inductively extremely electron withdrawing and this effect acts to stabilize the negative charge alpha to the carbonyl.  A methyl group is electron donating both inductively ($\ce{sp^2 - sp^3}$ bond) and &lt;em&gt;via&lt;/em&gt; resonance (hyperconjugation), so the alpha-protons in this case become less acidic.  Attaching an $\ce{OR}$ group to the carbonyl will stabilize the negative charge inductively since the oxygen is electron withdrawing, but we can now also draw resonance structures involving the $\ce{OR}$ and carbonyl groups that are electron donating to the carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHK7x.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonance effect is larger and outweighs the inductive effect.  The net result of donating electrons and moving them closer to the negative charge destabilizes the adjacent negative charge.  Hence the ester is less acidic than the previous compounds.  Applying the same reasoning to an amino group leads us to conclude that here the stabilizing inductive effect is even further decreased and the destabilizing resonance effect increased, both because nitrogen is less electronegative than oxygen.  These same arguments can be applied to simple carbonyl compounds as well as the 1,3-dicarbonyl compounds used here as examples.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-28T16:22:48.883" LastActivityDate="2014-06-28T16:22:48.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13994" PostTypeId="1" AcceptedAnswerId="13995" CreationDate="2014-06-28T15:48:06.230" Score="1" ViewCount="1249" Body="&lt;p&gt;I have been trying to solve an exercise on nuclear chemistry, but I was not sure whether I should ask this on physics or chemistry but I think that if chemistry had an exercise based on this then chemistry experts might be having the way to overcome this problem. So here is what my nuclear chemistry question says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One gram atom of $ _{79}^{198}\mathrm {Au} $ (&lt;strong&gt;having a half life of 65&#xA;  hours&lt;/strong&gt;) decays by $\beta$-emission to produce stable nuclide of Hg.&#xA;  How much Hg will be present after 260 hours?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thing which is making me confused is $\beta$ decay. I think I have a formula for this question but that $\beta$ thing is creating a problem. I would like to make that equation public. Here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_t=N_0\cdot\exp(-\lambda\cdot t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\lambda$ is the radioactive decay constant&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$N_0$ is initial number of nuclides present&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$N_t$ is the co. of parent nuclides present at time $t$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="670" LastEditorUserId="10185" LastEditDate="2016-05-21T15:44:57.063" LastActivityDate="2016-05-21T15:44:57.063" Title="Final amount of Hg present after $\beta$ decay" Tags="&lt;analytical-chemistry&gt;&lt;nuclear&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13995" PostTypeId="2" ParentId="13994" CreationDate="2014-06-28T16:50:36.687" Score="3" Body="&lt;p&gt;$$\ce{^198Au -&amp;gt; ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your equation we (of course) get the same result. &#xA;$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction) &#xA;$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find&#xA;$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-28T17:17:39.537" LastActivityDate="2014-06-28T17:17:39.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13996" PostTypeId="1" CreationDate="2014-06-28T17:15:44.287" Score="2" ViewCount="153" Body="&lt;p&gt;I have an assignment to calculate the concentration from a sample. The assignment is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You have 100 µL of RNA. You take 10 µL from the sample and put it in 990 µL of water in the cuvette. The OD is 0.05. Calculate the concentration of RNA in the sample. Calculate how many µL you have to take to make a solution of 100 ng.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus far I have succeeded to calculate the concentration from the OD. It is 2 µg/µL. I used the formula that I found on the internet :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$OD\times \text{Standard coeff.}(\pu{40 µg/ml})\times \text{Sample dilution}$$&lt;/span&gt;&#xA;but I cannot figure out how from that I can calculate the solution with 100 ng, because there is not mentioned the final volume for 100 ng, or how to calculate how much I have to take µL from the cuvette. I do not understand where to begin from.&lt;/p&gt;&#xA;&lt;p&gt;I tried with the formula &lt;span class=&quot;math-container&quot;&gt;$C_1V_1=C_2V_2$&lt;/span&gt;, but I could not figure that out.&lt;/p&gt;&#xA;" OwnerUserId="6978" LastEditorUserId="69830" LastEditDate="2021-05-07T06:31:17.657" LastActivityDate="2021-05-07T06:50:39.143" Title="How do I find a concentration from a sample tested on a spectrophotometer?" Tags="&lt;solutions&gt;&lt;concentration&gt;&lt;spectrophotometry&gt;&lt;dna-rna&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13997" PostTypeId="1" AcceptedAnswerId="13998" CreationDate="2014-06-28T17:48:01.940" Score="1" ViewCount="84" Body="&lt;p&gt;However I am thinking of this question to be damn easy but still I'll not say as I am not getting it,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two radio nuclei $A$ and $B$. $A$ is $\alpha$ emitter and&#xA;  $B$ is $\beta$ emitter and their decay constant are in ratio $1:2$.&#xA;  What should be the number of atoms of $A$ and $B$ at time $t=0$, so&#xA;  that probability of getting $\alpha$ and $\beta$ particles are same at&#xA;  $t=0$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So here how can the decay constant be dependent on number of atoms of $A$ and $B$ so that $\alpha$ and $\beta$ particle are in same ratio? I really need a right approach for this one.&lt;/p&gt;&#xA;" OwnerUserId="670" LastActivityDate="2014-06-28T18:10:50.227" Title="Find ratio of number of atoms of $A$ and $B$" Tags="&lt;analytical-chemistry&gt;&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13998" PostTypeId="2" ParentId="13997" CreationDate="2014-06-28T18:10:50.227" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two radio nuclei A and B. A is α emitter and B is β &#xA;  emitter and their decay constant are in ratio 1:2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So B decays twice as fast as A.  If we want the actual number of decay events from A to equal the actual number of decay events from B at t=0, then we must have twice as much A present as B at t=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given that $\ce{k_{B} = 2k_{A}}$. For the decay events at T=0 to be equal for A and B, then $$\ce{[B]_{o} * k_{B}~=~[A]_{o} * k_{A}}$$&#xA;$$\ce{[B]_{o} * 2k_{A}~=~[A]_{o} * k_{A}}$$&#xA;or $$\ce{\frac{2}{1}~=~\frac{[A]_{o}}{[B]_{o}}}$$&#xA;At t=0, the concentration of A must be twice the concentration of B in order for equal decay events to occur from A and B&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-28T18:10:50.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13999" PostTypeId="2" ParentId="13983" CreationDate="2014-06-28T18:43:44.870" Score="1" Body="&lt;p&gt;When you balance the equation for a redox reaction, the process of balancing it or the balanced equation itself say nothing about kinetics.  That said, redox reactions, as the name implies, deal with electron transfer.  Unless you are dealing with unusual molecules (bulky ligands for example) or unusual conditions (solids, glasses, very low temperature, etc.) electron transfer reactions are close to diffusion controlled, so kinetics is usually not an issue and is often glossed over in such reactions.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-28T18:43:44.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14000" PostTypeId="1" AcceptedAnswerId="14003" CreationDate="2014-06-28T18:54:38.337" Score="5" ViewCount="704" Body="&lt;p&gt;Is it possible for a nuclear reaction to occur simultaneously with a chemical reaction involving bond formation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest an example of this, please? &lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="16683" LastEditDate="2017-09-07T17:50:50.133" LastActivityDate="2017-09-07T18:10:13.813" Title="Can nuclear fission or fusion occur together with bond formation in a reaction?" Tags="&lt;nuclear-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14001" PostTypeId="2" ParentId="13996" CreationDate="2014-06-28T19:14:01.327" Score="1" Body="&lt;p&gt;The question being asked is, how many uL of your solution do you need to withdraw in order for that volume to contain 100 ng of the RNA.  If there are 2 ug/uL in the RNA solution, that is equivalent to 2000 ngm/uL.  So if you want 100 ng of RNA, you will need to withdraw &#xA;$$\ce{\frac{2000 ngm}{1 uL}~=~ \frac{100 ngm}{x ~uL}}$$&#xA;solving for x yields an answer of 1/20 uL or 0.05 uL.  0.05 uL of the RNA solution will contain 100 ng of the RNA.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-28T19:14:01.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14002" PostTypeId="2" ParentId="14000" CreationDate="2014-06-28T19:39:02.297" Score="1" Body="&lt;p&gt;To start with, fusion can be eliminated because of the intense pressure and high temperature needed. Chemicals would not exist since the electrons have been liberated from the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an atom undergoes fission, the release of energy would probably force the other nearby atoms away. Pitchblende is composed mostly of $\ce{UO2}$, and some $\ce{U3O8}$. $\ce{^238U}$ eventually decays to $\ce{^226Ra}$. The oxide formed would be $\ce{RaO}$. I don't know what happens to the freed electrons and oxygen ions, though.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="16683" LastEditDate="2017-09-07T17:26:19.960" LastActivityDate="2017-09-07T17:26:19.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14003" PostTypeId="2" ParentId="14000" CreationDate="2014-06-28T20:06:12.017" Score="4" Body="&lt;p&gt;I can't think of any examples of a nuclear reaction leading to a chemical reaction with bond formation.  But you got me curious and I started wondering if the reverse process, i.e. chemical bond formation leading to a nuclear reaction might be possible. I couldn't think of any traditional bond formation reactions that would fall into this category.  The closest I could come is the following reaction.&lt;br&gt;&#xA;$$\ce{^{2}_{1}H + ^{3}_{1}H -&amp;gt; ^{5}_{2}He -&amp;gt; ^{4}_{2}He + ^{1}_{0}n + Energy}$$&#xA;Once formed, the $\ce{^5He}$ is consumed by a nuclear decay process - there's half of what we were looking for. When we produce $\ce{^5He}$ from deuterium and trtium atoms, the two electrons that were once orbiting those separate atoms are now both orbiting a common atom - not a traditional chemical bond, but those two electrons are now occupying the same $\ce{1s}$ orbital and obeying the Pauli exclusion principle just as electrons in molecular orbitals do; and their once independent trajectories in space has been dramatically altered to where they are now orbiting a common nucleus.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-28T21:08:56.917" LastActivityDate="2014-06-28T21:08:56.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14004" PostTypeId="2" ParentId="8754" CreationDate="2014-06-28T22:09:00.960" Score="2" Body="&lt;p&gt;Your best bet for this will be to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation&quot; rel=&quot;nofollow&quot;&gt;Clausius-Clapeyron&lt;/a&gt; relation. This is an exponential function that describes the vapor pressure of a substance as a function of temperature, and uses the enthalpy of vaporization (latent heat) and a constant as parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant is dependent on the particular substance, but there is a way to eliminate it from the equation if you can get one data point: the vapor pressure at a specific temperature. If you can do that, then you can use the two-point form to predict vapor pressure at another temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln\frac{p_2}{p_1} = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This version only has one parameter - the enthalpy of vaporization ($\Delta H_{vap}$). $p_1$ and $T_1$ are the vapor pressure and temperature at point 1 (an arbitrary point), and $p_2$ and $T_2$ are the vapor pressure and temperature at a different arbitrary point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pure substance, if you know the enthalpy of vaporization and the boiling point, you an calculate the vapor pressure at any other temperature. That is because the boiling point is the temperature at which the vapor pressure equals one atmosphere. In other words, you would set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1 = 1 \space \mathrm{atm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1 = T_b$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve for $p_2$. The equation would be:&#xA;$p_2 = p_1\exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{T_2}-\frac{1}{T_1})) \tag{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you use this equation, make sure you first convert the temperatures to &lt;a href=&quot;http://en.wikipedia.org/wiki/Kelvin&quot; rel=&quot;nofollow&quot;&gt;Kelvin&lt;/a&gt;. This only works with an absolute temperature scale. Pressure can be in whatever units you want, as long as both pressures have the same units. In other words, convert 1 atm to whatever units you want to use for vapor pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data you listed didn't have the enthalpy of vaporization for either component, but they are available from &lt;a href=&quot;http://webbook.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;NIST&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_{vap} = 91.7 \space \mathrm{\frac{kJ}{mol}}$ - &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C56815&amp;amp;Mask=4&quot; rel=&quot;nofollow&quot;&gt;glycerol&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_{vap} = 71.2 \space \mathrm{\frac{kJ}{mol}}$ - &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C57556&amp;amp;Mask=4#Thermo-Phase&quot; rel=&quot;nofollow&quot;&gt;ppg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, NIST lists different values for the boiling point than you have. I would double check those before plugging numbers in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you don't have a pure substance; you have a mixture. Depending on how non-ideal it is (which basically means how much the molecules of one component will affect the other components) you might have to account for that if you want to get very accurate results. On the other hand, if you just want an estimate and are planning to experimentally adjust the power output from there, you can get a good starting point by assuming that the mixture and the vapor are both ideal. Under those conditions, the total vapor pressure of the mixture will be equal to the sum of the partial pressures...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_t = p_{gly} + p_{ppg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and you can use the Clausius Clapeyron equation from above for each component individually to find the respective vapor pressures. Then you can add them together if you want the total vapor pressure. However, you may be more interested in the vapor pressure of each component individually, since (from what I have heard) glycerol is more responsible for the &quot;cloud&quot; that forms, while propylene glycol is the better organic solvent. It would be helpful to know how much vapor of each component a given mixture would produce at a given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how you can predict vapor pressure from temperature. To relate this to the amount of heat you need, you will need to do another step. Temperature is a function of heat supplied and the specific heat of the substance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = mC\Delta T \tag{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q$ is the amount of heat transferred, $m$ is mass, and $C$ is the specific heat. You know the specific heat of the mixture, and the mass is an independent variable since you will control the composition and amount of the mixture. If you assume that volumes and heat capacities are additive, that mixing is fast, and that heat transfer is fast (at least locally), you can estimate the temperature (or rather, the maximum temperature) by connecting this to the power supplied by your device. Power in watts is the same as Joules/second. Therefore, you can multiply the power output by the time to get a total (maximum) amount of heat transfer (in Joules):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = P*t \tag{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can combine equations $(3)$ and $(2)$ and solve for the operating temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T=\frac{Pt}{mC}+T_i \tag{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T_i$ is the starting temperature of the mixture (before you apply power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can substitute this back into $(1)$ to come up with an estimate of the vapor pressure of each component as a function of the mass of the liquid, the power applied, and the length of time that it is applied for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{vap} = (1 \space \mathrm{atm}) \exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{\frac{Pt}{mC}+T_i}-\frac{1}{T_b})) \tag{5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real vapor pressure will likely be lower, but I think this would get you to a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things to remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$m$ refers to the mass of each component that is in close proximity to the heating element.&lt;/li&gt;&#xA;&lt;li&gt;$Delta H_{vap}$ is in kJ/mol - make sure that the $R$ you use has the right units.&lt;/li&gt;&#xA;&lt;li&gt;$T$ must be in Kelvin (both of them)&lt;/li&gt;&#xA;&lt;li&gt;The units of $P$, $m$, and $C$ have to cancel out to give K.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to get a (rough) upper estimate of the mass that will be vaporized, you could assume an ideal gas and use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m=\frac{pVM}{RT}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is the volume of the vaporizer chamber, and $M$ is the molar mass of the component.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-06-28T22:16:12.863" LastActivityDate="2014-06-28T22:16:12.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14005" PostTypeId="1" AcceptedAnswerId="14048" CreationDate="2014-06-28T22:44:55.453" Score="4" ViewCount="310" Body="&lt;p&gt;While I am &lt;em&gt;not&lt;/em&gt; asking the question here, a recent &lt;a href=&quot;https://biology.stackexchange.com/questions/17790/diffusion-of-hypotonic-solution-into-solution-with-glucose-and-starch/17802#17802&quot;&gt;question concerning osmosis&lt;/a&gt; at the Bio SE caught my interest. In a nutshell that question was, given isotonic solutions of sugar inside and outside an immersed semipermeable bag, what will happen if a second solute--too big to escape through the membrane-- is added to the bag? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question has an SAT flavor but I think it should be possible to answer it with a fairly straightforward DIY procedure. So my question here is whether the outline I have in mind would work, or whether something better can be devised.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Procedure: Make an n-molar solution of dextrose. Make an (n+x)-molar solution of dextrose [n] and a second [x] solute S (sucrose? starch? urea?--something that will not pass through the membrane). Fill the bag with the dextrose/S solution. Immerse the bag in the dextrose solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Osmosis alone will draw water into the bag. The question to be answered by the procedure is whether there will also be a net diffusion of dextrose out of the bag so that the final concentration of dextrose is higher outside the bag than inside, i.e., higher than the initial n-molarity.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to one internet source, dextrose will rotate plane-polarized light about $30^o$ clockwise. If my null hypothesis is that the n-molar solution and the post-osmosis/diffusion solution outside the bag have the same molarity (i.e., that there is no net movement of dextrose out of the bag), and if I am able to construct an apparatus to measure optical rotation consistently in my kitchen, could such an apparatus be sensitive enough to reject $H_o$ if sugar in fact diffuses out of the bag? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest a good practical way to set this up? Suggestions for types of membranes, necessary polarizing/light equipment, second solute S, etc. all welcome. Also, if there is a simpler/better way of going about this that would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sucrose reportedly rotates light more (perhaps $40^o$) but it may be too large to pass through some commercially available membranes. Fructose only rotates light about $10^o$ so I don't think it can be used here. Could sucrose be used as second solute?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermodynamics tag is added since I think diffusion is properly an aspect of thermodynamics (see Fermi, Thermodynamics, ch. VII). Please feel free to edit if I am mistaken.  &lt;/p&gt;&#xA;" OwnerUserId="5752" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:14.547" LastActivityDate="2016-12-08T21:21:22.487" Title="How to create a simple do-it-yourself procedure to measure diffusion?" Tags="&lt;thermodynamics&gt;&lt;home-experiment&gt;&lt;experimental-chemistry&gt;&lt;osmosis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14006" PostTypeId="1" AcceptedAnswerId="14024" CreationDate="2014-06-28T23:50:19.533" Score="9" ViewCount="1518" Body="&lt;p&gt;After seeing &lt;a href=&quot;https://www.youtube.com/watch?v=qorl6rKLmRs&quot; rel=&quot;noreferrer&quot;&gt;this youtube video&lt;/a&gt;, I was left to wonder why to use a $30^\circ$ angle between the flask and the pipette when emptying it. I can imagine it is important to touch the flask with your pipette when emptying, because the liquid prefers being attached to a solid surface due to capillary forces. When just emptying it from above the flask droplets would remain at the tip of the pipette. But why the $30^\circ$ angle? Is this an important standard or just arbitrary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aFRnL.png&quot; alt=&quot;Using a pipette under a 30° angle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="7475" LastEditDate="2015-12-14T02:24:53.930" LastActivityDate="2018-08-20T06:51:27.230" Title="Why pipetting with a 30° angle?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14007" PostTypeId="1" CreationDate="2014-06-29T00:28:30.520" Score="2" ViewCount="178" Body="&lt;p&gt;See the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dG8aF.png&quot; alt=&quot;Polymer Crystallinity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is A because of the hydrogen bond in the side group, thus the degree of secondary bonding between parallel chain segments is higher.  Is my thinking correct?&lt;/p&gt;&#xA;" OwnerUserId="6979" LastEditorUserId="9961" LastEditDate="2015-10-11T22:39:35.310" LastActivityDate="2018-08-20T05:58:17.283" Title="Polymer Crystallinity" Tags="&lt;crystal-structure&gt;&lt;polymers&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14008" PostTypeId="1" AcceptedAnswerId="14010" CreationDate="2014-06-29T02:53:21.433" Score="3" ViewCount="6745" Body="&lt;p&gt;I have seen my textbooks always give formula and description of hybridization of central atom of a molecule. But I wonder will other atoms present (not Hydrogen) go in hybridization or only the central atom only uses hybrid orbital and other use pure orbital?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="42741" LastEditDate="2017-05-12T03:09:55.300" LastActivityDate="2017-05-12T03:09:55.300" Title="Does hybridization take place in other atoms apart from central atom in a molecule?" Tags="&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14009" PostTypeId="2" ParentId="14008" CreationDate="2014-06-29T03:20:11.273" Score="0" Body="&lt;p&gt;Hybridization is the rearrangement of the orbitals predicted by the Schrodinger Equations to coincide with the observed orbital arrangements. So any atom with electrons in the S and P orbitals (lithium and heavier elements) can have hybridization rearrangements.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-06-29T03:20:11.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14010" PostTypeId="2" ParentId="14008" CreationDate="2014-06-29T03:52:07.103" Score="1" Body="&lt;p&gt;Yes, all atoms with p or d orbitals can hybridize in a molecule, not just the central atom.  Take acetone for example, the carbonyl carbon is $\ce{sp^2}$ hybridized and the carbonyl oxygen is also $\ce{sp^2}$ hybridized.  Another example is methyl chloride, the methyl carbon is $\ce{sp^3}$ hybridized and the chlorine is also $\ce{sp^3}$ hybridized.  As a final example consider dimethyl ether, each of the carbon atoms is $\ce{sp^3}$ hybridized and the central oxygen is also $\ce{sp^3}$ hybridized.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-29T03:52:07.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14011" PostTypeId="1" CreationDate="2014-06-29T05:31:37.833" Score="2" ViewCount="811" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.mhhe.com/physsci/chemistry/chang7/ssg/chap19_4sg.html&quot; rel=&quot;nofollow&quot;&gt;this Mc Graw-Hill link&lt;/a&gt; during a spontaneous redox reaction Gibbs Free Enthalpy must decrease and at the same time the change in the Gibbs Free Enthalpy is the maximum electric work that the reaction can do:&#xA;$$\Delta G = W_\text{max}=-n F E_\text{cell}$$ &#xA;in which $F$ stands for the Faraday constant, the amount of charge carried by one mole of electrons, and $n$ represents the number of moles of electrons to be transferred from the reducing agent to the oxidizing agent&#xA;according to the balanced equation, and $E_\text{cell}$ is the electromotive force of the cell, which equals $V_\text{cathod} - V_\text{anode}$, or alternatively, $E_\text{oxidizer} - E_\text{reducer}$ wherein $E$ here stands for the reduction potential. As $\Delta G$ must be nagative in the spontanous direction so $E_\text{cell}$ must be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the first and second law we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;&amp;amp;\delta Q-\delta W=\mathrm dU=T\mathrm dS-p\mathrm dV+\sum_i\mu_i\mathrm dN_i\\&#xA;&amp;amp;\Rightarrow\;\delta W-p\mathrm dV+\sum_i\mu_i\mathrm dN_i=\delta Q-T\mathrm dS\le0\qquad\because\text{2nd law}\\&#xA;&amp;amp;\Rightarrow\;\delta W-p\mathrm dV+\mathrm dG\bigr|_{p,T}\le0\\&#xA;&amp;amp;\Rightarrow\;\delta W-p\mathrm dV\le-\mathrm dG\bigr|_{p,T}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\delta W=p\mathrm dV$ then $\mathrm dG\le 0$, that is, the condition $\mathrm dG\le 0$ coincide with the 2nd law only id the only work done by the system is the pressure work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if $\delta W=p\mathrm dV+\text{other works}$, then $\text{other works}\le-\mathrm dG\bigr|_{p,T}$. This means the change in the &quot;minus Gibbs function&quot; is the maximum work attainable from the system beside the pressure work. This extra work can be positive or negative, in the form of electric work or friction work or etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here I have some problems with the content of the link provided at the beginning of my question, as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First of all, the maximum work that a system can do on its surrounding equals minus the change of the Gibbs Free Enthalpy, so the equality $\Delta G=W_\text{max}$ doesn't hold!?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If $\Delta G=W_\text{max}$ is valid, that is, I mean if the work attained is the maximum possible, then the process must be assumed as reversible , but if the process is reversible thenwe will need $\Delta G=0$ and not $\Delta G&amp;lt;0$, so that the process should be at equilibrium and spontaneous in no direction!?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That the Gibbs Free Energy must decrease in a closed system in the constant temperature and pressure is true only when the only type of work which is done is in the form of pressure work in expansion of the system's volume, not if other works like friction or electric works are also available, so that again using at the same time $\Delta G=W_\text{max}$ and $\Delta G\le0$ is misleading!?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The electric work is an internal work for a system containing the redox reaction and shouldn't enter the above formulation at all, as by work in the 1st law formulated above we mean only those work-based energy transport that occur at the boundaries of the system. So the electric force can never enter this formulation unless we change the control system under study, isn't it right?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And the main question. if I was wrong in analyzing the content of the provided link (which is very probable) where is my mistakes and if I was correct then how can I obtain the same result (which is certainly correct), that &quot;in a spontaneous redox reaction we should have $E_\text{oxidizer}-E_\text{reducer}\ge0$&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6980" LastEditorDisplayName="user7951" LastEditDate="2016-07-22T15:58:10.390" LastActivityDate="2016-07-22T15:58:10.390" Title="On Spontaneity of the Redox Reactions" Tags="&lt;thermodynamics&gt;&lt;redox&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14012" PostTypeId="1" AcceptedAnswerId="16837" CreationDate="2014-06-29T05:58:49.793" Score="6" ViewCount="176" Body="&lt;p&gt;Let $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ be a general reaction whose progress during time interval $dt$ is measured by $d\zeta$, so the amount of reactants consumed and products generated in mole would be $d N_i=\nu_i d\zeta$, wherein, $\nu_i$ would take negative values for the reactants being consumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From on the other hand we already know from the second law of thermodynamics that the reaction will occur in the direction in which $\sum_i \mu_i dN_i \le 0$, wherein, $\mu_i=\frac{\partial G}{\partial N_i}\bigr|_{p,T,N_j\;(j\ne i)}$ are the chemical potentials, so that we would have:&#xA;$$\left(\sum_i \mu_i \nu_i\right)\; d\zeta \le 0$$&#xA;The equality implies reversibility and should denote the equilibrium (&lt;em&gt;am I right?&lt;/em&gt;). At the equilibrium it is clear that the progression of the reaction should vanish and we should have $d\zeta=0$, so that the term $\left(\sum_i \mu_i \nu_i\right)$ should be free to gain any positive, zero or negative finite value. However, this is not what Guggenheim has written in his Thermodynamics book. He has first assumed in a given direction the reaction progresses, so that in that direction $d\zeta&amp;gt;0$, then has canceled out this positive quantity from the inequality and achieved: $\left(\sum_i \mu_i \nu_i\right) \le 0$, and eventually concluded that at the equilibrium the equality holds and we should have $\left(\sum_i \mu_i \nu_i\right) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how is it justified to be true when we already know at the equilibrium $d\zeta&amp;gt;0$ doesn\t hold and we should instead write: $d\zeta=0$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for bearing with me, I'm new to chemistry.&lt;/p&gt;&#xA;" OwnerUserId="6980" LastEditorUserId="1434" LastEditDate="2014-07-14T09:28:21.077" LastActivityDate="2014-09-28T16:45:13.397" Title="How can we justify setting the affinity $\sum_i \mu_i \nu_i$ equal to zero at chemical equilibrium?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14013" PostTypeId="2" ParentId="13939" CreationDate="2014-06-29T08:27:38.653" Score="-2" Body="&lt;p&gt;I would do it like follows: By using the ratio of 0.958 you can simply state the ratio of amount of substances. Then by using the balanced equations you can find the ratio of amount of substances of HCl required for each. Having both the ratio and total amount, you can get the the HCl for each; and then you can find the the required concentraion of those two ions from the balanced equation and total volume.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2018-10-25T18:00:56.377" LastActivityDate="2018-10-25T18:00:56.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14014" PostTypeId="1" AcceptedAnswerId="14020" CreationDate="2014-06-29T10:55:04.540" Score="2" ViewCount="1560" Body="&lt;p&gt;Unlike ammonia, the lone pair of phosphorous does not add to the dipole moment of phosphine. My textbook says this is due to the presence of lone pair of phosphorous in $\ce{PH3}$ in an s-orbital. I couldn't understand my book's explanation. How can the orbital of lone pair determine its contribution to the dipole moment? &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="194" LastEditDate="2017-05-12T03:06:25.117" LastActivityDate="2017-05-12T03:06:25.117" Title="Why does the lone pair of phosphine not contribute to its dipole moment?" Tags="&lt;physical-chemistry&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14015" PostTypeId="2" ParentId="14011" CreationDate="2014-06-29T11:08:00.313" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done &lt;strong&gt;by the system on the surroundings&lt;/strong&gt; is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_%28physics%29#Work_by_a_gas&quot; rel=&quot;nofollow&quot;&gt;pressure-volume work&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=-p\Delta V$$&#xA;$$\delta W=-pdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV&amp;gt;0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=\delta Q + \delta W =TdS - pdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=\delta Q - \delta W = TdS -pdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your &lt;strong&gt;chemistry&lt;/strong&gt; textbook is correct. An equivalent definition in physics would be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_{\text{max}}=nFE_{\text{cell}}$$&#xA;$$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = 0$ is the definition of &lt;strong&gt;equilibrium&lt;/strong&gt; not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 3:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum &lt;strong&gt;possible&lt;/strong&gt; electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 4:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 5: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that your initial equation is correct, then for a spontaneous reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G &amp;lt; 0$$&#xA;$$\therefore -nFE_\text{cell} &amp;lt; 0$$&#xA;$$nFE_\text{cell} &amp;gt; 0$$&#xA;$$E_\text{cell} &amp;gt; 0$$&#xA;$$E_\text{red} - E_\text{ox} &amp;gt; 0$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-06-29T11:08:00.313" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14017" PostTypeId="1" AcceptedAnswerId="14019" CreationDate="2014-06-29T13:48:55.413" Score="3" ViewCount="4894" Body="&lt;p&gt;Both delocalize the added proton. Is it because the bond lengths in amidines are the same, or perhaps because oxygen is more electronegative than nitrogen and thus is more reluctant at keeping the proton? But how can that explane the &lt;strong&gt;huge&lt;/strong&gt; difference in basicity between amidines and amides (resp $12$/$13$ versus $-1$/$0$ of $\mathrm{p}K_\mathrm{aH}$)?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="7475" LastEditDate="2015-10-09T16:49:38.500" LastActivityDate="2015-10-09T16:49:38.500" Title="Why are amidines so much more basic than amides?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14019" PostTypeId="2" ParentId="14017" CreationDate="2014-06-29T14:59:15.363" Score="3" Body="&lt;p&gt;Here is a diagram comparing what happens when we protonate an amide and an amidine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jYLC2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare the protonated products in the two reactions.  In the case of the amidine we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product.  In the case of the protonated amide the resonance structure with the double bond to oxygen places a positive charge on oxygen.  Therefor this resonance structure is a less significant contributor than the corresponding structure for the protonated amidine where the positive charge resides on the less electronegative nitrogen.  Since we can draw more significant resonance structures for the amidine case, the protonated amidine is more stable than the protonated amide.  This difference in protonated product stabilities explains why amidines are more basic than amides.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-29T19:53:39.250" LastActivityDate="2014-06-29T19:53:39.250" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14020" PostTypeId="2" ParentId="14014" CreationDate="2014-06-29T15:50:50.973" Score="1" Body="&lt;p&gt;Let's see if we can use the following picture to help clarify things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CcdIM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still &quot;point&quot; towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't &quot;point&quot; in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-29T15:50:50.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14021" PostTypeId="1" CreationDate="2014-06-29T16:03:58.847" Score="5" ViewCount="2959" Body="&lt;p&gt;Is there any database or book that has an index of organic reactions where the mechanisms are written out, complete with curly arrows? It would really help if I could look up a reaction and see a proposed mechanism with curly arrows.&lt;/p&gt;&#xA;" OwnerUserId="5679" LastActivityDate="2014-06-30T03:16:06.347" Title="Database with organic reaction mechanisms (curly arrows)" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14022" PostTypeId="1" AcceptedAnswerId="14023" CreationDate="2014-06-29T17:10:15.370" Score="8" ViewCount="5849" Body="&lt;p&gt;I don't know much about polymers but recently teflon made me get addicted to study their properties and nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the teflon polymer contains fluorine which is one of the most reactive elements but in the polymer state it is totally stable and moreover, it is applied as a coating to kitchen-wares and many utensils which are exposed to very high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did a reactive element like fluorine got such an exciting stability?&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="73" LastEditDate="2014-06-30T01:43:44.127" LastActivityDate="2014-06-30T01:43:44.127" Title="What makes teflon such a strong material?" Tags="&lt;polymers&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14023" PostTypeId="2" ParentId="14022" CreationDate="2014-06-29T19:40:48.463" Score="12" Body="&lt;p&gt;Teflon (PTFE, polytetrafluoroethylene) is prepared by the polymerization of tetrafluoroethylene as shown below.  If we took polyethylene and replaced all of the hydrogens with fluorines, we would have PTFE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NfldO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a Table that compares the strengths of some single bonds involving a methyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}\hline&#xA;Experimental ~Bond~ Enthalpies for~ CH_3 - X~ (kcal/mole)\\ \hline&#xA;\end{array}&#xA;\begin{array}{|c|c|c|}\hline&#xA;I &amp;amp; 57.6    \\ \hline&#xA;Br &amp;amp; 72.1    \\ \hline&#xA;Cl &amp;amp; 83.7    \\ \hline&#xA;CH_{3} &amp;amp; 90.1    \\ \hline&#xA;H &amp;amp; 104.9    \\ \hline&#xA;F &amp;amp; 115.0    \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{C-F}$ bond is one of the strongest bonds to carbon known.  As you add additional fluorine atoms to the carbon, the bond strength increases even further.  While the $\ce{C-F}$ bond is a covalent bond, it has considerable ionic character making the carbon positively polarized and the fluorine negatively polarized.  This bond polarization results in a strong electrostatic attraction between carbon and fluorine which accounts for the high bond strength observed.  It is this very high bond strength that makes fluorocarbons so stable (unreactive).  There are not many conditions that can supply enough energy to break this very strong bond.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-06-29T20:02:21.300" LastActivityDate="2014-06-29T20:02:21.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14024" PostTypeId="2" ParentId="14006" CreationDate="2014-06-29T19:52:22.613" Score="9" Body="&lt;p&gt;The angle of 30 degrees is not so important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to make sure that the pipette is straight and the liquid runs along inside of the glass. You should not hold the pipette directly into the liquid. The drop in the tip has been included in the calibration. At the end you just have to wipe the pipette at the glass edge.&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2014-06-29T19:52:22.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14025" PostTypeId="2" ParentId="14021" CreationDate="2014-06-30T01:03:04.690" Score="0" Body="&lt;p&gt;This is the closest I've seen to what you're asking for; some of the reactions have curly arrows. &lt;a href=&quot;http://www.chemthes.com/&quot; rel=&quot;nofollow&quot;&gt;The Chemical Thesaurus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6907" LastActivityDate="2014-06-30T01:03:04.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14026" PostTypeId="2" ParentId="14021" CreationDate="2014-06-30T01:55:23.740" Score="6" Body="&lt;p&gt;As far as free online resources go, I've found the following useful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.organic-chemistry.org/reactions.htm&quot;&gt;Organic Reactions (organic-chemistry.org)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.name-reaction.com&quot;&gt;Name-reaction.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_organic_reactions&quot;&gt;List of organic reactions (en.wikipedia.org)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would also strongly recommend acquiring a copy of &lt;em&gt;March's Advanced Organic Chemistry&lt;/em&gt; at some point. I think it's fair to say that it's widely regarded as one of the most essential reference works in the field of organic chemistry. It's concise, more comprehensive than one would've thought possible, beautifully organized, and contains thousands of primary literature references. While not every reaction is presented with an accompanying mechanism, many are, and useful exposition is typically provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another book you might find useful is &lt;em&gt;Strategic Applications of Named Reactions in Organic Synthesis&lt;/em&gt;. More than a mere reference text, it includes an overview describing important applications for the reactions it covers, examples of uses in practical syntheses, as well as comprehensive stepwise mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-06-30T01:55:23.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14027" PostTypeId="2" ParentId="14021" CreationDate="2014-06-30T03:16:06.347" Score="0" Body="&lt;p&gt;The answer to your question depends on what type of reactions you are looking for. If you are looking for simple 'name reactions' such as aldol condensation reaction, or Grignard reaction, then the links suggested by Greg would suffice. On the other hand, if you are looking for very specific reactions or reactions published in recent literature, then you must search in a dedicated chemistry database such as Scifinder. You may check the website here: &lt;a href=&quot;http://www.cas.org/products/scifinder/what-s-new-in-scifinder&quot; rel=&quot;nofollow&quot;&gt;http://www.cas.org/products/scifinder/what-s-new-in-scifinder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, scifinder is a paid database, and your academic institution will need to have an active subscribtion to this service. Most universities in the US and the Europe have subscription to Scifinder. You will not be access this service from your home, as it is IP restricted.&lt;/p&gt;&#xA;" OwnerUserId="6986" LastActivityDate="2014-06-30T03:16:06.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14028" PostTypeId="1" CreationDate="2014-06-30T06:12:18.200" Score="1" ViewCount="6150" Body="&lt;p&gt;In my textbook it says in the reaction $\ce{AlCl3 + Cl- -&amp;gt; AlCl4-}$, Aluminium Chloride ($\ce{AlCl3}$) has six valence electrons and accepts a lone pair from $\ce{Cl-}$, but wouldn't that mean the aluminium atom covalently bonded with the three chlorine atoms to get six electrons (which is wrong as it's ionic)? &lt;/p&gt;&#xA;" OwnerUserId="6988" LastEditorUserId="4945" LastEditDate="2014-06-30T06:22:35.503" LastActivityDate="2014-06-30T09:41:26.707" Title="Why is Aluminium Chloride a Lewis Acid if it's an ionic compound?" Tags="&lt;acid-base&gt;&lt;bond&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14029" PostTypeId="2" ParentId="14028" CreationDate="2014-06-30T07:56:10.870" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;if it's an ionic compound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, it is debatable. $\ce{Al^{3+}}$ is a very small cation and its formation is strongly unfavored. In fact, this cation, just like many others, may exist only in solvates, like $\ce{[Al(OH2)6]^{3+}}$, or unchemical environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular case of anhydrous $\ce{AlCl3}$ the molecule is a strong Lewis acid, strong enough to coordinate with lone pair of another similar molecule, so it forms either dimer $\ce{Al2Cl6}$ with two chlorine bridges, donating one pair to both Al ions, or polymer, with Al atoms exhibiting octahedral coordination. Unlike many ionic compounds, melted $\ce{AlCl3}$ has poor conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, does no apply to $\ce{AlCl3.6H2O}$ which is, again, a ionic compound $\ce{[Al(OH2)6]^{3+} (Cl^{-})3}$ and does not exhibit any Lewis acidity (though it does exhibit strong Brønsted acidity, i.e. tendency for hydrolysis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said, however, I have to clarify: ionicity of compound does not prevent particular ion from being Lewis acid/base. In fact, cations often are excellent Lewis acids. The only thing needed for a particle to be a Lewis acid is presence of low  vacant orbital, and sometimes presence of such orbital is not derivable from octet rule. For example, $\ce{SnCl4}$ is Lewis acid, though in it $\ce{Sn}$ has full octet. The orbital in question is describable in terms of MO theory which I strongly encourage to study. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most often considered Lewis acid is hydrogen cation. In fact, this Lewis acid is so strong, that it never occurs free, it is always bound to Lewis base (again, except unchemical environment). Another common Lewis acid is $\ce{Na^+}$. This cation is quite weak Lewis acid, but it is sterically accessible, and usually is paired with anion, which is usually a Lewis base. This said, despite mostly ionic nature of $\ce{NaCl}$, there is considerable covalent bonding in it. (of course, on the other hand there is such anion as $\ce{BPh4^-}$ which is not a lewis base of considerable strength, so its salts are almost purely ionic in nature)&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2014-06-30T09:41:26.707" LastActivityDate="2014-06-30T09:41:26.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14030" PostTypeId="1" AcceptedAnswerId="14031" CreationDate="2014-06-30T08:32:24.000" Score="3" ViewCount="260" Body="&lt;p&gt;I just read an experiment in which mercury has to be heated and in the same experiment there was a note warning not to inhale &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_%28element%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;mercury&lt;/a&gt; vapors as they are very harmful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't get the point that why are mercury vapors harmful.&lt;/p&gt;&#xA;" OwnerUserId="6954" LastEditorUserId="43942" LastEditDate="2017-05-07T13:42:06.763" LastActivityDate="2017-05-07T13:42:06.763" Title="What makes mercury vapors so harmful?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14031" PostTypeId="2" ParentId="14030" CreationDate="2014-06-30T08:41:11.227" Score="2" Body="&lt;p&gt;Mercury poisoning (also known as hydrargyria or mercurialism) is type of metal poisoning and a medical condition caused by exposure to mercury or its compounds. Mercury (chemical symbol $\ce{Hg}$) is a heavy metal occurring in several forms, all of which can produce toxic effects in high enough doses. Its zero oxidation state $\ce{Hg^0}$ exists as vapour or as liquid metal, its mercurous state $\ce{Hg2^{2+}}$ exists as inorganic salts, and its mercuric state $\ce{Hg^{2+}}$ may form either inorganic salts or organo-mercury compounds; the three groups vary in effects. Toxic effects include damage to the brain, kidneys and lungs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury poisoning can result in several diseases, including &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrodynia&quot; rel=&quot;nofollow noreferrer&quot;&gt;acrodynia&lt;/a&gt; (pink disease) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Minamata_disease&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minamata&lt;/a&gt; disease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_poisoning&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="16683" LastEditDate="2017-05-07T10:25:20.693" LastActivityDate="2017-05-07T10:25:20.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14032" PostTypeId="2" ParentId="14006" CreationDate="2014-06-30T12:54:20.933" Score="3" Body="&lt;p&gt;&lt;strong&gt;The main advantage is that you don't risk sprinkling your solution outside the beaker, so choose the angle you are more comfortable with to obtain this goal. Sometimes you don't see little drops spreading around!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a good practice but if you make multiple pipetting using the same pipette you could contaminate your reagent touching the beaker, so maybe sometimes emptying the pipette slowly, quite near to the solution, is the best thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some pipettes must be completely empty when you use them, while others take into account that some amount of solution will remain inside. Check what type yours is but generally touching the beaker's side does not change the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is more important to use gloves and cut your nails (the men in the photo seems to ignore it!).&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="2094" LastEditDate="2018-08-20T06:51:27.230" LastActivityDate="2018-08-20T06:51:27.230" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14034" PostTypeId="1" AcceptedAnswerId="14043" CreationDate="2014-06-30T13:52:15.097" Score="5" ViewCount="9390" Body="&lt;p&gt;What is the difference between backbonding and hyperconjugation and conjugation? &#xA;I mean the basic differences as all are concerned with donation of electrons. What is the most striking effect of each that differentiate them from others.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5424" LastEditDate="2016-11-02T19:26:31.120" LastActivityDate="2020-05-15T04:23:30.550" Title="Difference between backbonding and hyperconjugation and conjugation" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ClosedDate="2014-07-03T02:49:43.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="14035" PostTypeId="2" ParentId="7424" CreationDate="2014-06-30T14:46:13.173" Score="6" Body="&lt;p&gt;These authors are comparing &lt;strong&gt;relative&lt;/strong&gt; energies. Look at what they set as their reference... the &lt;em&gt;products&lt;/em&gt;. They could have easily made &quot;starting materials&quot; as their 0.0 reference point and stated that the products of the acyl chloride reaction is lower in energy relative to the anhydride reaction mechanism.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-06-30T14:46:13.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14036" PostTypeId="1" CreationDate="2014-06-30T16:41:01.040" Score="3" ViewCount="2962" Body="&lt;p&gt;How does one &quot;convert&quot; between an equilibrium constant calculated from the Gibbs free energy for a dissociation reaction and experimentally determined dissociation constants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta G^\circ=-RT\log(K_{eq})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I understand the derivation of this equation and why the equilibrium constant should be unitless if it's defined in terms of activities.&lt;br&gt;&#xA;But I have a binding reaction that has an estimated free energy change of $-32~\mathrm{kcal/mol}$, which yields a dissociation constant on the order of $10^{-24}$ at $298~\mathrm{K}$.&lt;br&gt;&#xA;Is it correct to just multiply the equilibrium constant by the standard concentration (1 molar) to get a more typical dissociation constant with units? A yoctomolar dissociation constant seems kind of strange to me, but I'm not sure where I've made a mistake.&lt;/p&gt;&#xA;" OwnerUserId="6992" LastEditorUserId="4945" LastEditDate="2014-07-03T03:16:15.527" LastActivityDate="2014-08-02T04:05:48.690" Title="Units for dissociation constant and relationship to Gibbs Free Energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14037" PostTypeId="1" CreationDate="2014-06-30T16:44:35.183" Score="2" ViewCount="291" Body="&lt;p&gt;A friend of mine told me that she used the hypotenuse of the pythagorean theorem to compute the work in graph below and the to compute the coefficient of efficiecy in this thermodynamics triangle cycle. What do you tnink about? &lt;img src=&quot;https://i.stack.imgur.com/Xq5jD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2014-06-30T17:00:27.867" LastActivityDate="2014-08-15T11:46:34.713" Title="Can I use the hypotenuse of the pythagorean theorem to compute the work?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14038" PostTypeId="1" AcceptedAnswerId="14039" CreationDate="2014-06-30T17:06:30.613" Score="3" ViewCount="5554" Body="&lt;p&gt;The CRC Elecrochemical Series and the solution to our exercise suggest that the oxidation states of borohydride are -V for Boron and +I for Hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since H (2.2) is more electronegativ than B (2.0), I would have expected the oxidation states to be +III for B and -I for H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to our exercise lists this (WARNING: it turned out that the solution was wrong):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Omazz.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the CRC Elecrochemical Series lists this (also suggesting -V for B):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RZHuZ.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="2116" LastEditDate="2014-07-03T13:18:52.357" LastActivityDate="2014-07-03T13:18:52.357" Title="What are the oxidation states in Borohydride (BH4-)" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14039" PostTypeId="2" ParentId="14038" CreationDate="2014-06-30T17:21:50.673" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would have expected the oxidation states to be +III for B and -I for&#xA;  H.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're right, boron is in the (+3) oxidation state in these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-06-30T17:21:50.673" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14040" PostTypeId="1" CreationDate="2014-06-30T17:35:41.893" Score="6" ViewCount="135" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This question has been re-asked on &lt;a href=&quot;https://hsm.stackexchange.com/&quot;&gt;History of Science and Mathematics&lt;/a&gt;:&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://hsm.stackexchange.com/q/3581&quot;&gt;Who is Hræðraford, the “learned clerk” “writing in modern chemical Latin”?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Truesdell, C. &lt;em&gt;Rational Thermodynamics&lt;/em&gt;. New York, NY: Springer New York, 1984. &lt;a href=&quot;http://dx.doi.org/10.1007/978-1-4612-5206-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://dx.doi.org/10.1007/978-1-4612-5206-1&lt;/a&gt;, p. 41 says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;… in the numerous papers and books whose titles and topics include &quot;irreversible&quot;—the learned clerk HRÆðRAFORD, writing in modern chemical Latin, has recently called their field &quot;thermodynamice inreparabilis&quot;—…&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who is this &quot;learned clerk&quot; Hræðraford?&lt;/p&gt;&#xA;" OwnerUserId="6994" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:33.357" LastActivityDate="2016-03-28T07:55:35.400" Title="Who is Hræðraford, the &quot;learned clerk&quot; &quot;writing in modern chemical Latin&quot;?" Tags="&lt;thermodynamics&gt;&lt;history-of-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ClosedDate="2016-03-25T18:37:15.840" ContentLicense="CC BY-SA 3.0" />
  <row Id="14041" PostTypeId="1" AcceptedAnswerId="14045" CreationDate="2014-06-30T18:17:17.453" Score="3" ViewCount="1260" Body="&lt;p&gt;Molten salt is pretty hot, 801 C. With such heat and the corrosive nature of salt, I am wondering what type of metal alloy or other material hybrid would be appropriate to handle and work with molten salt?&lt;/p&gt;&#xA;" OwnerUserId="6995" LastEditorUserId="4231" LastEditDate="2014-06-30T18:55:55.273" LastActivityDate="2014-07-01T01:34:19.620" Title="What kind of metal alloys can resist corrosion from molten salt?" Tags="&lt;alloy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14042" PostTypeId="2" ParentId="14041" CreationDate="2014-06-30T18:24:34.470" Score="2" Body="&lt;p&gt;I think you should use tungsten as it has the highest and best melting point among all the stable metallic elements its melting point is nearly &lt;a href=&quot;http://en.wikipedia.org/wiki/Melting_point&quot; rel=&quot;nofollow&quot;&gt;$3700~\mathrm{K}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More over its &lt;a href=&quot;http://www.infomine.com/investment/metal-prices/ferro-tungsten/&quot; rel=&quot;nofollow&quot;&gt;price&lt;/a&gt; is also considerably lower than other metals with higher melting points and if you think of corrosive nature, then I would recommend you to use silver but it becomes very &lt;a href=&quot;http://www.postalbullion.com/silver-price-per-grams-week-gbp-5.aspx&quot; rel=&quot;nofollow&quot;&gt;costly&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to comment for further clarification. I will definitely help you.&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="4945" LastEditDate="2014-07-01T01:34:19.620" LastActivityDate="2014-07-01T01:34:19.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14043" PostTypeId="2" ParentId="14034" CreationDate="2014-06-30T18:46:14.660" Score="13" Body="&lt;p&gt;OK, let's step through these terms and let's add &amp;quot;cross-conjugation&amp;quot; and &amp;quot;homoconjugation&amp;quot; for good measure to round out the series.&lt;/p&gt;&#xA;&lt;h3&gt;Backbonding:&lt;/h3&gt;&#xA;&lt;p&gt;This term is most often used to describe a type of bonding that occurs in inorganic compounds.  Metals bonding with carbon monoxide provide good examples of backbonding. The carbon atom in carbon monoxide has a lone pair of electrons that can be used to form a sigma bond with a metal.  Because carbon monoxide has low-lying &lt;span class=&quot;math-container&quot;&gt;$\pi^{\ast}$&lt;/span&gt; orbitals, it can accept electrons &lt;strong&gt;back&lt;/strong&gt; from the metal and further strengthen the bonding between the metal and the carbon monoxide ligand.  This process of &amp;quot;accepting electrons &lt;strong&gt;back&lt;/strong&gt; from the metal&amp;quot; is termed backbonding. Here is a good picture of the &lt;span class=&quot;math-container&quot;&gt;$\ce{\sigma}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}-\pi^{\ast}$&lt;/span&gt; (backbonding) bonding interactions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/atHRh.png&quot; alt=&quot;Pi and sigma type interactions&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It happens most often when 1) the metal is electron rich, 2) the ligand has low-lying &lt;span class=&quot;math-container&quot;&gt;$\pi^{\ast}$&lt;/span&gt; orbitals and 3) the metal has available a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; orbital with the proper symmetry to overlap constructively with the carbon monoxide &lt;span class=&quot;math-container&quot;&gt;$\pi^{\ast}$&lt;/span&gt; orbital.  Here is a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Organometallic_Chemistry/Ligands/Carbon_Monoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a great, one page article on backbonding if you'd like to read more.&lt;/p&gt;&#xA;&lt;h3&gt;Hyperconjugation:&lt;/h3&gt;&#xA;&lt;p&gt;This term is most commonly used in discussions about resonance.  Hyperconjugation is a stabilizing interaction that results from the interaction of the electrons in a &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;-bond (usually &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt;) with an adjacent (empty or filled) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbital. This interaction creates an extended molecular orbital that increases the stability of the system.  Most typically it involves the breaking of a carbon-hydrogen bond and then using what were the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonding electrons to stabilize an adjacent charge. In resonance structures, we can't move nuclei, but we can move electrons. For example, below is a diagram showing resonance structures that involve hyperconjugation that serve to stabilize the n-propyl carbocation. Hyperconjugation has been used as part of the explanation for the carbocation order of stability being tertiary &amp;gt; secondary &amp;gt; primary.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oCOlm.jpg&quot; alt=&quot;Example for Hyperconjugation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Recently it has been suggested that hyperconjugation, rather than steric effects, is what stabilizes the staggered conformation of ethane (very interesting!).    &lt;a href=&quot;https://www.princeton.edu/chemistry/macmillan/group-meetings/hyperconjugation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link&lt;/a&gt; to a fairly extensive, but very well written article on hyperconjugation, it covers the effects of hyperconjugation on resonance structures and conformational stability.&lt;/p&gt;&#xA;&lt;h3&gt;Conjugation:&lt;/h3&gt;&#xA;&lt;p&gt;This term is used to describe a system of continuous, overlapping p-orbitals where electron delocalization can occur.  Examples of conjugated systems include the 1,3-butadiene, acrolein, benzene, beta-carotene, etc.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0GZDL.jpg&quot; alt=&quot;Conjugate systems&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oWYnd.png&quot; alt=&quot;beta-Carotene&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Cross-conjugation:&lt;/h3&gt;&#xA;&lt;p&gt;Sometimes you'll hear the term &amp;quot;&lt;strong&gt;cross-conjugation&lt;/strong&gt;&amp;quot; used.  In this case, the continuity of the overlap between the p-orbitals does not follow a simple linear pattern as seen in the examples of conjugation pictured above.  Benzophenone and 3-methylene-1,4-pentadiene are examples of molecules with cross-conjugation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2KLvo.png&quot; alt=&quot;Benzophenone&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl8E2.jpg&quot; alt=&quot;2-Vinyl-1,3-butadiene&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Homoconjugation:&lt;/h3&gt;&#xA;&lt;p&gt;Sometimes &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbitals that are not directly conjugated (&lt;em&gt;e.g.&lt;/em&gt;, that are not directly connected, there is a &amp;quot;spacer&amp;quot; in between them) can still interact or overlap in a &amp;quot;through space&amp;quot; fashion.  The solvolysis of the &lt;em&gt;exo&lt;/em&gt;- and &lt;em&gt;endo&lt;/em&gt;-norbornenyl chlorides shown below provides a striking example. The &lt;em&gt;exo&lt;/em&gt;-isomer generates the corresponding carbocation thousands of times faster than its &lt;em&gt;endo&lt;/em&gt;-isomer. This difference is attributed to homoconjugation. The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbitals on the remote norbornenyl double bond can interact through-space with the developing &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbital that is being formed on the carbon where the exo-chloride is leaving from, dramatically stabilizing the formation of the resultant carbocation. This interaction is geometrically impossible for the &lt;em&gt;endo&lt;/em&gt;-isomer.  Again note, there is no direct conjugation between the remote double bond and the carbon atom where the carbocation will form, it is entirely a through space interaction that generates a &lt;em&gt;pseudo&lt;/em&gt;-allyl system.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jGBfy.jpg&quot; alt=&quot;Exo- and edo-nobonene systems&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-05-15T04:19:12.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14044" PostTypeId="2" ParentId="14034" CreationDate="2014-06-30T18:47:51.450" Score="4" Body="&lt;p&gt;The most basic difference lies in the kind of orbitals involved in all. In hyperconjugation, there occurs an interaction between sigma orbitals on one side and completely or partially filled orbitals on the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conjugation, alternating single and double bonds are present so as to create a region of overlapping &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals having in between them bridges of sigma bonds. Electrons are said to be delocalised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Backbonding can be best described as &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt;-coordinate bond, wherein the excess negative charge is donated to the central atom having vacant orbitals (&lt;em&gt;e.g.&lt;/em&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5255" LastEditorUserId="61578" LastEditDate="2020-05-15T04:23:30.550" LastActivityDate="2020-05-15T04:23:30.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14045" PostTypeId="2" ParentId="14041" CreationDate="2014-06-30T19:34:52.583" Score="8" Body="&lt;p&gt;George Lai's &lt;em&gt;High-Temperature Corrosion and Materials Applications&lt;/em&gt; (excerpts available on &lt;a href=&quot;http://books.google.ca/books?id=tfDwOe7xWeQC&amp;amp;pg=PA409&amp;amp;dq=george+lai+molten+salt+corrosion&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=b7qxU_nOAYKqyATn-IGQAw&amp;amp;ved=0CC0Q6AEwAA#v=onepage&amp;amp;q=george%20lai%20molten%20salt%20corrosion&amp;amp;f=false&quot;&gt;Google Books&lt;/a&gt;) is a good source. Very often for molten salts, Hastelloys are used; they're special-purpose alloys developed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hastelloy#Hastelloy&quot;&gt;Haynes International&lt;/a&gt;. I can't find a reference, but I believe Oak Ridge National Laboratories used Hastelloys for their molten salt reactors (which had fluoride salts, very reactive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're working with sodium chloride (little potassium), according to Lai, &lt;a href=&quot;http://www.hightempmetals.com/techdata/hitempHaynes188data.php&quot;&gt;Hastelloy-188&lt;/a&gt; seems to be the best bet, although many others would work as well. One reality of chemical engineering is that there will always be corrosion, it's just a matter of keeping it at an acceptable rate for the application.&lt;/p&gt;&#xA;" OwnerUserId="6907" LastActivityDate="2014-06-30T19:34:52.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14046" PostTypeId="1" AcceptedAnswerId="14186" CreationDate="2014-07-01T01:49:55.653" Score="7" ViewCount="737" Body="&lt;p&gt;I live in Southern California and sometimes hike in areas that can be overgrown with a plant called &lt;a href=&quot;http://en.wikipedia.org/wiki/Turricula_%28plant%29&quot;&gt;purple poodle-dog bush&lt;/a&gt;. As I found out the first time I went crashing through a stand of the stuff, it's a horrific skin irritant, much worse than poison oak. It has microscopic hairs that get stuck in your skin and act like syringes. The hairs have chemicals on them that are skin irritants. &lt;a href=&quot;http://en.wikipedia.org/wiki/Turricula_%28plant%29#Skin_irritant&quot;&gt;Wikipedia says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This contact dermatitis is due to prenylated phenolics exuded by hairs (glandular trichomes) of the plant. The principal irritants are derivatives of farnesyl hydroquinone and 3-farnesyl-P-hydroxybenzoic acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not a chemist, and all the nomenclature is Greek to me. My question is this. When I hike in an area where it will be hard to avoid the bush, I try to wear long sleeves and change my clothes immediately afterwards. Will throwing my clothes in the washing machine remove these chemicals and/or break them down into harmless ones? What if my shoes aren't machine washable? Will the chemicals break down naturally within a certain period of time? Would other methods such as exposure to sunlight help? Are these chemicals soluble in water, so that simple rinsing would help? How about making sure they aren't lurking in the upholstery of my car?&lt;/p&gt;&#xA;" OwnerUserId="6999" LastActivityDate="2014-07-08T01:50:07.647" Title="Removing prenylated phenolics from clothing, shoes, and upholstery" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14047" PostTypeId="1" AcceptedAnswerId="14051" CreationDate="2014-07-01T14:43:56.493" Score="3" ViewCount="1968" Body="&lt;p&gt;I would like to test for fluoride levels in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any methods or test kits available to test the levels of fluoride in home water supply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;https://uk.answers.yahoo.com/question/index?qid=20110824093408AA5tigC&quot; rel=&quot;nofollow&quot;&gt;Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="478" LastEditorUserId="478" LastEditDate="2014-07-01T14:52:31.893" LastActivityDate="2016-03-13T19:40:12.317" Title="How to test for fluoride levels in drinking water?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14048" PostTypeId="2" ParentId="14005" CreationDate="2014-07-01T14:50:19.317" Score="3" Body="&lt;p&gt;There are way more sensitive ways to measure dextrose (glucose) concentration than polarimetry. For example, there are reducing sugar tests like the Dinitrosalicylic Acid test or (better) the 2,2'-Biconchoninic acid test. Just measure the dextrose concentrations inside and outside of the bag. I have protocols for these tests if you want; it involves getting ahold of the reagents and a spectrophotometer (unless you just want to do it qualitatively, look at the color and see if they're different), but presumably that would be no more difficult than getting hold of a polarimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the second solute S goes, you want something that's a lot larger than glucose. I'm only familiar with dialysis tubing that separates small molecules from proteins, so I would go with that. Cheap regenerated cellulose dialysis tubing has a molecular weight cutoff of 10,000, so use glucose (MW 180) and cheap (and soluble) Bovine Serum Albumin (MW 66,000). You always want something at least five times less than or five times greater than the cutoff for these tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful, there are reducing tests for sugars and reducing tests for proteins, they often use the same techniques but in different conditions, so with one protocol they'll be very sensitive for one and much less sensitive for the other. As long as you have blanks, it shouldn't be a problem, but make sure you're following the right protocol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you really want to be fancy, there's a fluorimetric test that is extremely specific to glucose, using glucose oxidase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you want to be DIY, just use a diabetes glucose meter and strips.&lt;/p&gt;&#xA;" OwnerUserId="6907" LastActivityDate="2014-07-01T14:50:19.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14049" PostTypeId="1" AcceptedAnswerId="14098" CreationDate="2014-07-01T14:52:06.273" Score="2" ViewCount="954" Body="&lt;p&gt;I would like to remove fluoride from my water supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any methods, any test kits or water distillation units to remove fluoride from drinking water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://uk.answers.yahoo.com/question/index?qid=20110824093408AA5tigC&quot; rel=&quot;nofollow&quot;&gt;Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tunemind.net/fluorides.htm&quot; rel=&quot;nofollow&quot;&gt;A simple and easy way to neutralize fluorides in drinking water&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.trueactivist.com/scientists-discover-new-technique-to-remove-fluoride-from-drinking-water/&quot; rel=&quot;nofollow&quot;&gt;Scientists Discover New Technique to Remove Fluoride from Drinking Water&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Read More: &lt;a href=&quot;http://www.trueactivist.com/scientists-discover-new-technique-to-remove-fluoride-from-drinking-water/&quot; rel=&quot;nofollow&quot;&gt;http://www.trueactivist.com/scientists-discover-new-technique-to-remove-fluoride-from-drinking-water/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="478" LastEditorUserId="21964" LastEditDate="2016-07-25T16:04:33.343" LastActivityDate="2016-12-12T23:21:23.310" Title="What are the methods of fluoride removal from drinking water?" Tags="&lt;water&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14051" PostTypeId="2" ParentId="14047" CreationDate="2014-07-01T15:40:53.817" Score="6" Body="&lt;p&gt;A good reliable set-up is pricey.  &lt;a href=&quot;http://www.hach.com/intellical-isef121-fluoride-ion-selective-electrode-ise-1-meter-cable/product?id=7640513803&quot; rel=&quot;nofollow&quot;&gt;Here's a link&lt;/a&gt; to a reliable, quality probe for fluoride ion detection in water, it costs \$740  (US).  &lt;a href=&quot;http://www.hach.com/hq30d-portable-ph-conductivity-optical-dissolved-oxygen-do-orp-and-ise-multi-parameter-meter/product?id=7640494072&quot; rel=&quot;nofollow&quot;&gt;The meter&lt;/a&gt; that it plugs into sells for \$790 (US) if you get the less expensive hand-held device.  There also manuals and instructions on how to actually use the gear, at the website as well.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-13T19:40:12.317" LastActivityDate="2016-03-13T19:40:12.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14052" PostTypeId="1" AcceptedAnswerId="14472" CreationDate="2014-07-01T18:55:14.363" Score="4" ViewCount="449" Body="&lt;p&gt;I have noticed that, when exposed to water for a few days, the steel in my wagon turns into rust. After several more weeks, the rust that formed is dissolved in the water, and once I pour the water out there isn't any rust where there previously was rust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is unusual because iron oxides and iron hydroxides (the components of rust) don't normally dissolve in water, and so the water just keeps being reduced and the iron keeps being oxidized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can it apparently happen such that steel turns to rust, then back to steel with no rust, in the presence of water if rust does not normally dissolve in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens in a cycle like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;steel + water = rust -&gt; rust + water = steel + rusty water&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="10185" LastEditDate="2015-08-05T16:24:53.310" LastActivityDate="2015-08-05T16:24:53.310" Title="Rust forming and being washed off by water" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14053" PostTypeId="1" AcceptedAnswerId="14062" CreationDate="2014-07-01T19:30:18.060" Score="2" ViewCount="252" Body="&lt;p&gt;I use a solar still that holds 182 cubic feet of water. Here is how it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an insulator on both sides of the metal bowl. On the outside it covers the whole thing. on the inside it covers the sides but not the bottom. This collects the rain and also heats up the rainwater during sunny days so that the evaporation rate is faster. I have the pipe nearest the metal bowl the largest it can be while the metal bowl still collects rain and gets heated up by the sun. The water and any gases pass through there and into the condenser. I have this cold enough to condense water into a liquid but not so cold that the water turns solid and the $\ce{CO2}$, $\ce{O2}$, $\ce{N2}$ etc. turn liquid. Last but not least there is a spigot at the very end and a bucket below it. This separates atmospheric gases from liquid water with $\ce{CO2}$ being the main concern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also dust and pathogens from the air in rainwater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust most likely will stay behind and just make the bowl dusty. The pathogens get killed by the heat that there still is when all the water has been evaporated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say the maximum loss with this system is from 1 pint or so to 1 gallon or maybe more depending on the water level in the bowl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I think there is a way to calculate maximum and minimum loss(neither of these is 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I calculate maximum and minimum loss based on the depth of the water(because obviously at different depths there is a different amount of loss) and the size of everything(which by the way the pipe slowly tapers off to a certain size and so you have to take pressure into account as well as volume)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also how could I calculate the pressure in the pipe based on the fact that it tapers off and starts at the maximum size possible with the bowl still collecting rain and being heated by the sun?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lrlb8.jpg&quot; alt=&quot;Solar Still&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="4945" LastEditDate="2014-07-02T03:08:07.063" LastActivityDate="2020-02-13T02:07:06.667" Title="Solar still and how it works" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14054" PostTypeId="2" ParentId="11222" CreationDate="2014-07-01T19:53:37.860" Score="0" Body="&lt;p&gt;I refer you to definition 3) via Wikipedia:&#xA;         &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturation_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Saturation_%28chemistry%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up: a saturated hydrocarbon must have no double bonds, no triple bonds, and no rings (only single bonds), and each carbon is attached to two hydrogen atoms, apart from the end carbons, which are attached to three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound that you're given is not saturated since one of the end carbon atoms has a Cl attached to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, you can think of saturated as meaning that the carbons that make up the chain have to have as many hydrogens as possible attached; the chain is 'saturated' (or completely full) with hydrogen atoms.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;    Turns out I had a brain misfire: it's absolutely not a hydrocarbon (which would only contain carbon and hydrogen). The very first sentence of definition 3) that I referenced means that your workbook is wrong (and the other replies are correct): &lt;strong&gt;it's a saturated compound&lt;/strong&gt;. My mistake - I erroneously conflated the first two sentences of 3) from Wikipedia while having a simultaneous brain fart.&lt;/p&gt;&#xA;" OwnerUserId="7001" LastEditorUserId="7001" LastEditDate="2014-07-02T22:26:21.057" LastActivityDate="2014-07-02T22:26:21.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14055" PostTypeId="1" AcceptedAnswerId="14064" CreationDate="2014-07-01T20:29:11.550" Score="3" ViewCount="3639" Body="&lt;p&gt;My friend was very nervous when he noticed that I'm performing electrolysis in a room that we've just been residing. (it was under open window to let $\ce{H2}$ out) After a sharp argument, he made me shut it down, though I still thought that it was safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using $\ce{Na2CO3}$ as electrolyte and as far as I know, it doesn't get electrolysed. I use normal pipe water. Such water surely contains some minerals however in really small amount. My friend pointed out that it also contains chlorine and that's what he's afraid of. He would disregard my notice that he's drinking this water often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using copper electrodes, producing copper hydroxide and copper carbonate. I failed to find out whether these compounds are hazardous. They're in solid state anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've said you can see that I'm very sure that the reaction is not hazardous. But I'm not chemist. There could be something I didn't think of in which case I've been poisoning myself for quite a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you make me sure or warn me? &lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-07-02T12:30:32.883" Title="What dangerous gases could electrolysis with water and sodium carbonate produce?" Tags="&lt;safety&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14056" PostTypeId="2" ParentId="14055" CreationDate="2014-07-01T22:32:36.570" Score="0" Body="&lt;p&gt;Copper salts are toxic, which means the water you use would not be healthy after a while and be careful on disposing of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minerals you have are probably in small amounts, and you can check this measuring the conductivity of the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make the gases pass through a filter of CaCO3 that will absorb any chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the answers to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5974/is-it-dangerous-to-inhale-the-steam-coming-from-electrolysis-of-salt-water-solut&quot;&gt;this question&lt;/a&gt; for details on what is happening.&lt;/p&gt;&#xA;" OwnerUserId="1300" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-02T12:30:32.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14057" PostTypeId="1" AcceptedAnswerId="14580" CreationDate="2014-07-01T23:11:48.847" Score="23" ViewCount="759" Body="&lt;p&gt;Bullvalene (tricyclo[3.3.2.0&lt;sup&gt;2,8&lt;/sup&gt;]deca-3,6,9-triene) is a fluxional molecule able to interconvert any two carbon atoms through a series of degenerate Cope rearrangements (for more information, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Bullvalene#Stereodynamics&quot; rel=&quot;noreferrer&quot;&gt;the Wikipedia article on bullvalene&lt;/a&gt;).  At slightly above room temperature bullvalene produces a single line in its &lt;sup&gt;1&lt;/sup&gt;H-NMR spectrum.  Considering fluorobullvalene, there are 4 possible isomers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1-fluorotricyclo[3.3.2.0&lt;sup&gt;2,8&lt;/sup&gt;]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent cyclopropyl carbons&lt;/li&gt;&#xA;&lt;li&gt;3-fluorotricyclo[3.3.2.0&lt;sup&gt;2,8&lt;/sup&gt;]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons alpha to the cyclopropane ring&lt;/li&gt;&#xA;&lt;li&gt;4-fluorotricyclo[3.3.2.0&lt;sup&gt;2,8&lt;/sup&gt;]deca-3,6,9-triene, i.e. the fluorine could be attached to any of the 3 equivalent olefinic carbons beta to the cyclopropane ring&lt;/li&gt;&#xA;&lt;li&gt;5-fluorotricyclo[3.3.2.0&lt;sup&gt;2,8&lt;/sup&gt;]deca-3,6,9-triene, i.e. the fluorine could be attached to the unique tris-allylic carbon&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UNuFJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNuFJ.png&quot; alt=&quot;Isomers of fluorobullvalene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, at room temperature or thereabouts, only a single line is observed in the &lt;sup&gt;1&lt;/sup&gt;H, &lt;sup&gt;13&lt;/sup&gt;C and &lt;sup&gt;19&lt;/sup&gt;F-NMR spectra.  I remember once reading that about 80% of the fluorine is in the tris-allylic position (&lt;strong&gt;4&lt;/strong&gt;), most of the remainder is attached to the two olefinic carbons (&lt;strong&gt;2&lt;/strong&gt; and &lt;strong&gt;3&lt;/strong&gt;) and a very small amount is attached at the cyclopropyl position (&lt;strong&gt;1&lt;/strong&gt;). This was often cited as an example to support the argument that fluorine prefers to form bonds to carbon when the involved carbon orbital has high p-content (low s-content).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what the isomer distribution is?  Those 4 numbers and a reference would be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2017-08-29T18:01:58.580" LastActivityDate="2017-08-29T19:45:25.637" Title="What is the isomer distribution in monosubstituted fluorobullvalene?" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14058" PostTypeId="1" AcceptedAnswerId="14059" CreationDate="2014-07-02T00:36:24.740" Score="27" ViewCount="42858" Body="&lt;p&gt;Take for instance the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2(g) + I2(s) &amp;lt;=&amp;gt; 2HI(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant would not include the solid &lt;span class=&quot;math-container&quot;&gt;$\ce{I2}$&lt;/span&gt;, but why is this? I have read that its concentration is a constant, but what does that mean? If you increase the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{I2}$&lt;/span&gt; the reaction should shift towards the right so logically it does appear to play a part in the equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For liquids, I can understand why you don't include it when you have something like water which (usually) does not play a part in the reaction, but why do you still not include when the liquid takes part in the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="41328" LastEditDate="2019-02-27T07:25:05.077" LastActivityDate="2021-04-13T21:05:56.503" Title="Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="14059" PostTypeId="2" ParentId="14058" CreationDate="2014-07-02T03:49:47.770" Score="15" Body="&lt;p&gt;It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot; rel=&quot;noreferrer&quot;&gt;goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Equilibrium Constant&lt;/strong&gt;&lt;br&gt;&#xA;  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The standard equilibrium constant is always unitless, as it is defined differently (&lt;a href=&quot;http://goldbook.iupac.org/S05915.html&quot; rel=&quot;noreferrer&quot;&gt;goldbook&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Standard Equilibrium Constant&lt;/strong&gt; $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)&#xA;  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $R$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reaction $$\ce{H2 (g) + I2 (s) &amp;lt;=&amp;gt; 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities&#xA;$$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$&#xA;with $$a=\frac{f}{p^\circ}.$$&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;noreferrer&quot;&gt;activity&lt;/a&gt; for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore&#xA;$$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant &lt;strike&gt;and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible)&lt;/strike&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-02-24T11:16:46.067" LastActivityDate="2018-02-24T11:16:46.067" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14060" PostTypeId="1" CreationDate="2014-07-02T03:57:17.400" Score="4" ViewCount="3174" Body="&lt;p&gt;I need some help identifying the following unknown compound. Any suggestions or hints regarding the possible compound would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I am thinking the broad peak on the IR is an alcohol that somehow mixed with possible &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3-CH}$&lt;/span&gt; bonds. And the peak around &lt;span class=&quot;math-container&quot;&gt;$\pu{1600 cm-1}$&lt;/span&gt; is I think a &lt;span class=&quot;math-container&quot;&gt;$\ce{sp-C-C}$&lt;/span&gt; bond. The compound should have four distinct groups of protons. Beyond this basic analysis, I couldn't really come up with a compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KL7QM.jpg&quot; alt=&quot;NMR (8-6 ppm)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have updated (07/09/14) the question with a newer NMR (below) and an IR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EOl19.jpg&quot; alt=&quot;newer NMR (7.6-7.0 ppm)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DQ0xl.jpg&quot; alt=&quot;IR&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other additional info I worked out in the lab:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;look: white powder like solid.&lt;/li&gt;&#xA;&lt;li&gt;melting point: &lt;span class=&quot;math-container&quot;&gt;$\pu{116.3-119.0 ^\circ C}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solubility:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water - insoluble&lt;/li&gt;&#xA;&lt;li&gt;NaOH - insoluble&lt;/li&gt;&#xA;&lt;li&gt;Sulfuric Acid - soluble&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Classification Tests:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water insoluble phenol test - positive; a dark red precipitate formed. &lt;/li&gt;&#xA;&lt;li&gt;Chromic acid test - positive; green precipitate formed. &lt;/li&gt;&#xA;&lt;li&gt;Bromine test for unsaturation - positive; bromine discoloration happened until about &lt;span class=&quot;math-container&quot;&gt;$\pu{2 ml}$&lt;/span&gt; of bromine was added. White precipitate formed. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is my first time doing these tests so please don't consider their results too strictly. My Instructor said the IR is a much safer way to determine functional groups. &lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="4945" LastEditDate="2020-02-19T14:00:49.263" LastActivityDate="2020-02-19T14:00:49.263" Title="How to identify an unknown compound with spectroscopic data?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14061" PostTypeId="1" AcceptedAnswerId="14253" CreationDate="2014-07-02T04:23:43.000" Score="5" ViewCount="1706" Body="&lt;p&gt;I have the following IR and NMR of an unknown substance. Can someone tell me what is a possible structure of the compound or point me in the correct direction. So far I am thinking of a benzene ring based compound with an OH group. Something like 2-phenylethanol but not sure. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/ycc0E.jpg&quot; alt=&quot;NMR data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yr4Hs.jpg&quot; alt=&quot;IR data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;melting point: &lt;span class=&quot;math-container&quot;&gt;$\pu{116 ^\circ C}$&lt;/span&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Composition: white powdery solid.&lt;/li&gt;&#xA;&lt;li&gt;Soluble in &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NaHCO3}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on the IR, the broad peak at &lt;span class=&quot;math-container&quot;&gt;$\pu{3424 cm-1}$&lt;/span&gt; represents an alcohol while the 4 humps in the region &lt;span class=&quot;math-container&quot;&gt;$\pu{2100-2600 cm-1}$&lt;/span&gt; is indicative of an aromatic ring (most likely a benzene ring). The medium peak around &lt;span class=&quot;math-container&quot;&gt;$\pu{1660 cm-1}$&lt;/span&gt; is most likely a carbon-carbon double bond. NMR is where I get stuck; I know there should be 6 distinct types of protons and since most of the peaks are around &lt;span class=&quot;math-container&quot;&gt;$\pu{7-8 ppm}$&lt;/span&gt;, it is most likely the benzene ring protons, but I am not sure how to calculate J values and splitting as it was done in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14060/&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="4945" LastEditDate="2020-02-22T13:12:48.330" LastActivityDate="2020-02-22T13:12:48.330" Title="How do I find out what benzene containing compound this is with IR and NMR?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14062" PostTypeId="2" ParentId="14053" CreationDate="2014-07-02T04:33:25.870" Score="2" Body="&lt;p&gt;There is no way to answer your question accurately for this setup. It will depend on the temperature of the water in the bowl, the exposed surface area, the convective flow rate of air near the surface, and the relative humidity of the air - none of which are constant or even very predictable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet is to set it up and measure the amount before and after distillation. The difference would be your loss.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-02T04:33:25.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14063" PostTypeId="2" ParentId="14060" CreationDate="2014-07-02T06:10:43.453" Score="7" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Spectroscopy/Infrared_Spectroscopy&quot; rel=&quot;nofollow&quot;&gt;typical IR bands&lt;/a&gt; (&lt;a href=&quot;http://www.chem.ucla.edu/~webspectra/irtable.html&quot; rel=&quot;nofollow&quot;&gt;other source&lt;/a&gt;) and identify them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; - 3424 phenolic $\ce{OH}$ stretch&lt;br&gt;&#xA; - 1640 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch&lt;br&gt;&#xA; - 1264 aromatic $\ce{C-O}$ stretch&lt;br&gt;&#xA; -  747 aromatic $\ce{C-H}$ bending&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have a look at you spectrum, identify your residual solvent signal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;  7.15 is chloroform $\ce{CDCl3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have a look at the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Reference_Tables/Typical_values_for_1H-NMR_chemical_shifts&quot; rel=&quot;nofollow&quot;&gt;typical $\ce{{}^{1}H-NMR}$ shifts table&lt;/a&gt; and identify the protons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; These are aromatic protons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the summary of your first conclusions?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It's an aromatic compound with an hydroxyl group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How many protons?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; multiple of 5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, hence $\nu_0=500~\mathrm{MHz}$.)&lt;br&gt;&#xA;Example: Doublet at 7.55 ppm:&#xA;$$J=(7.5615-7.5461)\cdot500~\mathrm{MHz}=7.7~\mathrm{Hz}$$&#xA;Solutions:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7.55 doublet, $2\ce{H}$, $J=7.7~\mathrm{Hz}$&lt;br&gt;&#xA; 7.40 doublet of doublet (fake triplet), $2\ce{H}$, $J=7.6~\mathrm{Hz}$, $J=7.5~\mathrm{Hz}$&lt;br&gt;&#xA; The last one is tricky because there is one peak not picked&lt;br&gt;&#xA; It should be at 7.30 a doublet of doublet, $1\ce{H}$,  ${}^3J=7.7~\mathrm{Hz}$, $J=7.6~\mathrm{Hz}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Disclaimer:&lt;br&gt;&#xA;Please read long's answer and follow his thoughts. They make a lot of sense and hint that the compound we are looking at is not the one I am proposing. But due to my lack of current knowledge, I cannot make another suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine all you see and you will get a clue which substance it is.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 7.55 ortho&lt;br&gt;&#xA; 7.40 meta&lt;br&gt;&#xA; 7.30 para&lt;br&gt;&#xA; Phenol.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-07-06T12:35:27.677" LastActivityDate="2014-07-06T12:35:27.677" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="14064" PostTypeId="2" ParentId="14055" CreationDate="2014-07-02T07:59:24.203" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My friend pointed out that it also contains chlorine and that's what he's afraid of. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The amount is laughable at. However, using distilled (deionized) water is recommended for other reasons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I failed to find out whether these compounds are hazardous. They're in solid state anyway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are reasonably harmless. Of course, eating them is not recommended, but copper compounds have low toxity. Move to graphite or platinum electrodes is still recommended.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-02T07:59:24.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14065" PostTypeId="1" AcceptedAnswerId="14113" CreationDate="2014-07-02T11:35:35.563" Score="2" ViewCount="233" Body="&lt;p&gt;After research I found that enzymes are responsible for metabolism, specifically catabolism and anabolism. This in turn creates the necessary chemicals in your body. The only question left is: what are enzymes? I asked in Biology and the answer led me to enzymes being created by ribosomes. Ribosomes by RNA. RNA by the four chemicals, adenine, guanine, cytosine, and uracil. &lt;em&gt;Okay&lt;/em&gt; so, my question is: what chemical elements are those four chemicals comprised of? &lt;/p&gt;&#xA;" OwnerUserId="7002" LastEditorUserId="73" LastEditDate="2014-07-04T06:22:56.333" LastActivityDate="2014-07-04T08:12:33.283" Title="Human Body and RNA/DNA" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14066" PostTypeId="1" AcceptedAnswerId="14193" CreationDate="2014-07-02T12:13:18.067" Score="23" ViewCount="618" Body="&lt;p&gt;I am currently testing a piece of software that generates the InChI for a given structure. I also want to test error situations as end users will also deliver the mol files that will be used as input for the software.&lt;/p&gt;&#xA;&lt;p&gt;I have found this description about &lt;a href=&quot;http://www.inchi-trust.org/technical-faq/#4.15&quot; rel=&quot;noreferrer&quot;&gt;what the current version of InChI cannot represent&lt;/a&gt; (Technical FAQ section 4.15 at inchi-trust.org)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;InChI currently does not support the representation of:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Polymers&lt;/li&gt;&#xA;&lt;li&gt;Complex organometallics&lt;/li&gt;&#xA;&lt;li&gt;Markush structures&lt;/li&gt;&#xA;&lt;li&gt;Mixtures&lt;/li&gt;&#xA;&lt;li&gt;Conformers&lt;/li&gt;&#xA;&lt;li&gt;Excited state and spin isomers&lt;/li&gt;&#xA;&lt;li&gt;Non-local stereochemistry/chirality&lt;/li&gt;&#xA;&lt;li&gt;Topological isomers&lt;/li&gt;&#xA;&lt;li&gt;Cluster molecules&lt;/li&gt;&#xA;&lt;li&gt;Polymorphs&lt;/li&gt;&#xA;&lt;li&gt;Unspecific isotopic enrichment&lt;/li&gt;&#xA;&lt;li&gt;Reactions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Also, InChI is not suitable for very large compounds; technically, InChI input may not contain more than 1023 atoms.&lt;/p&gt;&#xA;&lt;p&gt;Please note that there are IUPAC InChI subcommittee working groups currently addressing some of these matters. Details of these efforts can be found at: &lt;a href=&quot;http://iupac.org/web/ins/802&quot; rel=&quot;noreferrer&quot;&gt;http://iupac.org/web/ins/802&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Unfortunately I am lacking the chemical background to generate a suitable structure as molfile that meets the described features to produce an error as an InChI cannot be generated for it.&lt;/p&gt;&#xA;" OwnerUserId="4719" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-10-20T03:42:37.810" Title="Structure that breaks InChI" Tags="&lt;computational-chemistry&gt;&lt;software&gt;&lt;notation&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14067" PostTypeId="1" AcceptedAnswerId="14069" CreationDate="2014-07-02T13:06:31.523" Score="1" ViewCount="86" Body="&lt;p&gt;Can anyone find any data on the entropies of solvation for these two anions in water solution? I can't find any.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2016-11-20T20:08:46.623" LastActivityDate="2016-11-20T20:08:46.623" Title="Entropies of Solvation for Chlorate and Nitrate Anions" Tags="&lt;solutions&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14068" PostTypeId="2" ParentId="14036" CreationDate="2014-07-02T13:33:39.707" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;free energy change of -32 kcal/mol, which yields something on the order of 10^-24 at 298K. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Seems the main mistake is a sign error.  Also &quot;log&quot; must mean &quot;natural logarthim&quot; the way the equation in the question is written to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the free energy change is negative, the equillbrium constant is $&amp;gt;1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should be $10^{\frac{32\frac{kcal}{mol}}{2.303\times0.00199\frac{kcal}{Kmol}\times{298K}}} = 10^{23}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-07-02T13:33:39.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14069" PostTypeId="2" ParentId="14067" CreationDate="2014-07-02T15:39:35.430" Score="4" Body="&lt;p&gt;Look at the &lt;a href=&quot;http://books.google.com/books?id=JoNUVLrNg14C&amp;amp;pg=PA63&amp;amp;lpg=PA63&amp;amp;dq=Entropies%20of%20Solvation%20for%20Chlorate%20and%20Nitrate%20Anions&amp;amp;source=bl&amp;amp;ots=hHBaWdVYeh&amp;amp;sig=2OXVRCGEe351WJNwleMRPg5jdh4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=tCG0U66-CdCkyASyoYDAAQ&amp;amp;ved=0CEEQ6AEwBg#v=onepage&amp;amp;q=Entropies%20of%20Solvation%20for%20Chlorate%20and%20Nitrate%20Anions&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;graph on p. 65&lt;/a&gt;, it puts the entropy of solvation (298 K) for chlorate and nitrate at about 25 and 21 kJ/mol respectively.  &lt;a href=&quot;http://books.google.com/books?id=frE5AAAAIAAJ&amp;amp;pg=PA118&amp;amp;lpg=PA118&amp;amp;dq=Entropies%20of%20Solvation%20for%20Chlorate%20and%20Nitrate%20Anions&amp;amp;source=bl&amp;amp;ots=3qXhfKQPhg&amp;amp;sig=aWWbaQ2hiUyjiMEgMbHKp_CYqQM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=tCG0U66-CdCkyASyoYDAAQ&amp;amp;ved=0CDoQ6AEwAw#v=onepage&amp;amp;q=Entropies%20of%20Solvation%20for%20Chlorate%20and%20Nitrate%20Anions&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;This reference&lt;/a&gt; suggests that the entropy term for nitrate is 116 J/(K-mol), multiplying that by 298 K yields 34.6 kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-02T15:39:35.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14070" PostTypeId="1" CreationDate="2014-07-02T16:54:54.183" Score="3" ViewCount="294" Body="&lt;p&gt;Is there any general ordering or rule for heteronuclear molecules like $\ce{AX_2}$, $\ce{AX_{n}}$, and their molecular bonding/antibonding configurations ($\sigma/\pi$ orbitals) accordingly to some order in the Molecular Orbital Theory (MOT)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the molecular orbital configurations and their order depend on the scheme and the linear combination of atoms (homo and heteronuclear, di and polyatomic) or is there some &quot;standard ordering&quot; like the aufbau principle $\ce{1s~2s~2p~3s~3p~4s~3d~4p~5s~\dots}$ for atomic orbitals in atoms? Or are molecular configurations a totally dependent on the molecule type and geometry? That is, I want to know if there is some (general) rule for molecular orbitals (ordering and energies) just like the aufbau principle for atomic orbitals.&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="4945" LastEditDate="2014-07-03T02:39:58.210" LastActivityDate="2014-07-03T02:39:58.210" Title="Molecular orbitals and &quot;generalized&quot; aufbau principle for heteronuclear molecular configurations" Tags="&lt;molecules&gt;&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14071" PostTypeId="1" AcceptedAnswerId="14072" CreationDate="2014-07-02T17:33:26.760" Score="2" ViewCount="24029" Body="&lt;p&gt;There is a question, that says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What total energy (in $\mathrm{kJ}$) is contained in $1.0~\mathrm{mol}$ of photons, all with a frequency of $2.75 \cdot 10^{14}~\mathrm{Hz}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer was:&lt;br&gt;&#xA;The energy of photon in hydrogen atom is given by the formula $E=h\nu$, where $h$ is Planck constant and $\nu$ is the frequency. After that I got an answer different from the correct answer which is $110~\mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain why?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="69830" LastEditDate="2021-05-08T12:31:58.173" LastActivityDate="2021-05-08T12:31:58.173" Title="What total energy (in kJ) is contained in 1.0 mol of photons, all with a frequency of 2.75 • 10¹⁴ Hz?" Tags="&lt;thermodynamics&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14072" PostTypeId="2" ParentId="14071" CreationDate="2014-07-02T18:37:39.220" Score="9" Body="&lt;p&gt;The formula $E=h\nu$ is for the energy of &lt;strong&gt;one&lt;/strong&gt; photon. You have a mole of photons. You need to use a slightly modified form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=Nh\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N$ is the number of photons, in this case &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;N&amp;amp;=n\cdot N_\mathrm A\\[6pt]&#xA;&amp;amp;=1\ \mathrm{mol}\times6.02\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]&#xA;&amp;amp;=6.02\times10^{23}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you are being asked to report energy in &lt;strong&gt;kilojoules&lt;/strong&gt; not kilojoules per mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use $E=nN_\mathrm A h \nu$, I get the correct answer you cite of $110~\mathrm{kJ}$. Do you?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorDisplayName="user7951" LastEditDate="2016-11-10T14:48:39.397" LastActivityDate="2016-11-10T14:48:39.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14073" PostTypeId="1" AcceptedAnswerId="14130" CreationDate="2014-07-02T18:38:23.023" Score="6" ViewCount="335" Body="&lt;p&gt;I am trying to solve problem 2 from the 2009 International Chemistry Olympiad (IChO), which is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\ce{H2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\ce{H2}$ rapidly desorbs. This question examines two kinetic models for $\ce{H2}$ formation on the surface of a dust particle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In both models, the rate constant for adsorption of $\ce{H}$ atoms onto the surface of dust particles is $k_\mathrm a = \pu{1.4*10^-5 cm3 s-1}$. The typical number density of $\ce{H}$ atoms (number of H atoms per unit volume) in interstellar space is $[\ce{H}] = \pu{10 cm-3}$. (You may treat numbers of surface adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Desorption of $\ce{H}$ atoms is first order with respect to the number of adsorbed atoms. The rate constant for the desorption step is $k_\mathrm d = \pu{1.9 * 10^-3 s-1}$. Assuming that only adsorption and desorption take place, calculate the steady-state number, $N$, of $\ce{H}$ atoms on the surface of a dust particle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, what do they mean by steady-state number here? I tried searching online but I could not find anything. The solution they give has the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_\mathrm d N − k_\mathrm a[\ce{H}] = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which they solve for $N$. I know this has something to do with steady state, but in this case why must the rate of absorption be equal to the rate of desorption? Is it because you are considering a single $\ce{H}$ atom which is considered to be an intermediate? Also, why is it $k_\mathrm dN$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qmnaB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the above, how did they come up with those rates? Why does absorption have $\ce{[H]}x_m$ but desorption have $\ce{m}x_m$? Finally, for the reaction rate, where did they get $m(m-1)$? I think the $\frac{1}{2}$ comes from the fact that the rate of production of $\ce{H2}$ is half the rate of absorption/desorption, is that right?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From these, they calculate the rate of change of $x_0$, $x_1$, and $x_2$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\begin{align}&#xA;\frac{\mathrm dx_0}{\mathrm dt} &amp;amp;= -k_\mathrm a[\ce{H}]x_0 + k_\mathrm d x_1 + k_\mathrm r x_2 \\&#xA;\frac{\mathrm dx_1}{\mathrm dt} &amp;amp;= k_\mathrm a[\ce{H}]x_0 - (k_\mathrm a [\ce{H}] + k_\mathrm d) x_1 + 2k_\mathrm dx_2 \\&#xA;\frac{\mathrm dx_2}{\mathrm dt} &amp;amp;= k_\mathrm a[\ce{H}]x_1 - (2k_\mathrm d + k_\mathrm r)x_2&#xA;\end{align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I sort of understand how they got the value for rate of change of $x_0$ as it is the rate of production minus the rate of consumption, and desorption from state 1 and reaction produce state 0, while absorption consumes it. However, why does the $k_\mathrm a$ term come with $[\ce{H}]$ while the other terms do not? What happened to the $m x_m$ that was mentioned above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, how did they get the rate of change for $x_1$ and $x_2$? &lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="16683" LastEditDate="2018-03-25T09:14:47.583" LastActivityDate="2018-03-25T09:34:15.537" Title="Kinetics of interstellar chemistry: Applying steady state to formation of H₂" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14075" PostTypeId="2" ParentId="14070" CreationDate="2014-07-03T02:11:41.067" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do the molecular orbital configurations and their order depends on the scheme and the linear combination of atoms (homo and heteronuclear, di and poliatomic) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not only orbitals arising depends on atom configuration, but also on atomic orbitals available (s,p,d) and their ordering depends on geometry (i.e. size) and relative energy of atomic orbitals available. Unfortunately, there is no simple rule here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lucky enough, there is some amount of free computational packages you can use to calculate and then visualize molecular orbitals. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-03T02:11:41.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14076" PostTypeId="2" ParentId="367" CreationDate="2014-07-03T04:42:38.007" Score="3" Body="&lt;p&gt;You can functionalize silica to add polymer &quot;brushes&quot; to the surface. You can also buy pre-made functionalized silica, although the downside to that is it isn't always easy to get the manufacturer to tell you exactly what is on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't really going to increase the attractive interactions between the silica and the polystyrene, however. Polystyrene has a bulky aromatic ring hanging off of each monomer, and no sites that can participate in hydrogen bonding. When you add hydrophobic brushes to the surface of the silica, what you are doing is limiting the ability of the silica to aggregate due to hydrophilic interactions with itself. As a result, given enough kinetic energy (temperature for melt casting, or sonication for solvent casting) you can get the particles to disperse. Then, if you solidify the polymer quickly enough, they will be kinetically trapped in a dispersed state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what your goal is, this could be enough for your application. However, if you want to get the particles to strongly attract to the polystyrene, you aren't going to be able to get much farther than van der Waals interactions. That's a limitation of the polystyrene, not the silica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other particles (gold, or carbon nanotubes, for example) will be more hydrophobic and easier to disperse without functionalizing, but you still are not going to get very strong attractive forces between the particles and the polymer.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-03T04:42:38.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14077" PostTypeId="1" CreationDate="2014-07-03T04:58:50.313" Score="3" ViewCount="830" Body="&lt;p&gt;I saw the following plot of the Arrhenius Equation &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;on Wikipedia (cc-by Zivilverteidigung and Matthias M.)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gg98E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gg98E.png&quot; alt=&quot;NO2 Arrhenius k against T.svg (cc-by Zivilverteidigung and Matthias M.)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a concave upwards graph, but according to the equation &#xA;&lt;span class=&quot;math-container&quot;&gt;$$k= A \mathrm e^\frac{-E_\mathrm a}{RT},$$&lt;/span&gt;&#xA; it is evident that the graph is of the form &#xA;&lt;span class=&quot;math-container&quot;&gt;$$y= \mathrm e^\frac{-1}{x}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QVgyY.png&quot; alt=&quot;graph of e^(-1/x)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this should have been the graph (please disregard the negative side of the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis). It is a concave downwards curve. Why is there a discrepancy? &#xA;This maybe also suitable in Mathematics.se, but I have a gut feeling that the solution to this lies in Chemistry and not maths.&lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorDisplayName="user7951" LastEditDate="2019-01-15T20:32:54.357" LastActivityDate="2019-01-15T20:32:54.357" Title="Plot of Arrhenius Equation with temperature" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14078" PostTypeId="2" ParentId="14061" CreationDate="2014-07-03T07:28:10.093" Score="3" Body="&lt;p&gt;I tried to solve it, but I horribly failed. But as I put some effort in it, I still share this with you. Maybe you can work it out eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit that this spectrum is bad to work with, it does not contain an internal standard and the solvent signal is not marked. It is possible, that it is integrated. If this is an educational exercise, your supervisor clearly did a very bad, a horrible job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signals are way too small to see a splitting pattern, it is very hard to even put a number to the signals. The aromatic area should at least have been enlarged. If you have another possibility, you do never ever scale the integrals to an aromatic proton, especially not the one that mixes with other signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following you did already: Have a look at &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Spectroscopy/Infrared_Spectroscopy&quot; rel=&quot;nofollow&quot;&gt;typical IR bands&lt;/a&gt; (&lt;a href=&quot;http://www.chem.ucla.edu/~webspectra/irtable.html&quot; rel=&quot;nofollow&quot;&gt;other source&lt;/a&gt;) and identify them.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3424 phenolic $\ce{OH}$ stretch  &lt;/li&gt;&#xA;&lt;li&gt;3054 aromatic $\ce{C-H}$ stretch  &lt;/li&gt;&#xA;&lt;li&gt;around 1600 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch  &lt;/li&gt;&#xA;&lt;li&gt;1265 aromatic $\ce{C-O}$ stretch  could be ether alkoxy stretch  &lt;/li&gt;&#xA;&lt;li&gt;737 and 704 aromatic $\ce{C-H}$ bending, hints towards meta substituted phenyl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now you could go ahead and identify the residual solvent signal.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;7.14 should be chloroform $\ce{CDCl3}$&lt;/li&gt;&#xA;&lt;li&gt;7.28 is also a good candidate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It can also be none of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a look at the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Reference_Tables/Typical_values_for_1H-NMR_chemical_shifts&quot; rel=&quot;nofollow&quot;&gt;typical $\ce{{}^{1}H-NMR}$ shifts table&lt;/a&gt; you can identify some of the protons. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;7-8.5ppm These are aromatic protons  &lt;/li&gt;&#xA;&lt;li&gt;2.20ppm allylic/benzylic proton, could also still be $\ce{sp^3-C}$, it is very low field. As it is a singlet it has no neighbour carbons with a hydrogen, but it integrates to only one, or maybe two - it is hard to tell.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the IR you identified a phenolic $\ce{OH}$ and = there should be a signal around 5ppm, which is not there. So I am very puzzled about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the splitting pattern is completely random to guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, as it says bruker500, hence $\nu_0=500~\mathrm{MHz}$.)&lt;br&gt;&#xA;Example: Doublet at 8.15 ppm:&#xA;$$J=(8.1560-8.1402)\cdot500~\mathrm{MHz}=7.9~\mathrm{Hz}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8.15 doublet, $2\ce{H}$, $J=7.9~\mathrm{Hz}$  &lt;/li&gt;&#xA;&lt;li&gt;7.65 multiplet, $1\ce{H}$, $J=1.4~\mathrm{Hz}$, $J=5.8~\mathrm{Hz}$, $J=6.5~\mathrm{Hz}$  &lt;/li&gt;&#xA;&lt;li&gt;7.52 multiplet, $3\ce{H}$, $J=8.3~\mathrm{Hz}$, $J=7.0~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$, $J=7.8~\mathrm{Hz}$&lt;/li&gt;&#xA;&lt;li&gt;7.39 doublet of doublet or triplet, $1\ce{H}$, $J=7.5~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$&lt;/li&gt;&#xA;&lt;li&gt;7.28 singlet, $1\ce{H}$ - or solvent&lt;/li&gt;&#xA;&lt;li&gt;7.14 singlet, $1\ce{H}$ - or solvent&lt;/li&gt;&#xA;&lt;li&gt;2.20 singlet, $1\ce{H}$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am very sorry I could not be of any more help. A $\ce{{}^{13}C-NMR}$ would certainly have been helpful at least.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-03T07:28:10.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14079" PostTypeId="1" CreationDate="2014-07-03T08:10:04.040" Score="3" ViewCount="2093" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can good quality crystal meth really be made as in &lt;em&gt;Breaking Bad&lt;/em&gt;?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a pharmaceutical drug really be reverse-engineered?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="30836" LastEditDate="2020-04-27T09:13:53.580" LastActivityDate="2020-04-27T09:13:53.580" Title="Crystal meth in Breaking Bad" Tags="&lt;synthesis&gt;&lt;drugs&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ClosedDate="2014-07-03T11:40:27.727" ContentLicense="CC BY-SA 4.0" />
  <row Id="14081" PostTypeId="2" ParentId="14079" CreationDate="2014-07-03T09:54:55.703" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can good quality crystal meth really be made and sold as in Breaking Bad?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, most procedures are accurate, although the whole methylamine is so hard to get idea in Breaking Bad is nonsense. It is DEA List I chemical, but can be easily manufactured and no sane kingpin would go steal it from a cargo train (although it makes a good scene). The real problem is getting the P2P (phenyl-2-propanone, or phenylacetone) which the authors of breaking bad seem to ignore (probably on purpose, so that stupid kids won't try and make it at home).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a pharmaceutical drug really be reverse engineered?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reversed engineered? Assuming you are talking about the transition of pseudoephedrine into methamphetamine, this is just a chemical reduction, following this scheme (if this kid wants to make meth, he would have found this procedure anyway, I'm not giving out any special information that would help anyone manufacture methamphetamine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOYnj.png&quot; alt=&quot;Methamphetamine from pseudoephedrine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-20T20:54:56.513" LastActivityDate="2015-02-20T20:54:56.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14082" PostTypeId="1" AcceptedAnswerId="14084" CreationDate="2014-07-03T11:58:08.603" Score="3" ViewCount="6197" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{25.95 g}$&lt;/span&gt; sample of methanol at &lt;span class=&quot;math-container&quot;&gt;$\pu{35.6 ^\circ C}$&lt;/span&gt; is added to a &lt;span class=&quot;math-container&quot;&gt;$\pu{38.65 g}$&lt;/span&gt; sample of ethanol at &lt;span class=&quot;math-container&quot;&gt;$\pu{24.7 ^\circ C}$&lt;/span&gt; in a constant pressure calorimeter. If the final temperature of the combined liquids is &lt;span class=&quot;math-container&quot;&gt;$\pu{28.5 ^\circ C}$&lt;/span&gt; and heat capacity of the calorimeter is &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3 J/^\circ C}$&lt;/span&gt; determine the specific heat of methanol. The heat capacity of ethanol is &lt;span class=&quot;math-container&quot;&gt;$\pu{2.46 J/(g * ^\circ C)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  q_\mathrm{methanol} &amp;amp;= -q_\mathrm{ethanol}\\&#xA;  msT &amp;amp;= -msT\\&#xA;  \pu{25.95 g} \cdot  (\pu{28.5 ^\circ C} - \pu{35.6 ^\circ C}) \cdot  s&#xA;    &amp;amp;= -\pu{38.65 g} \cdot  (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})&#xA;       \cdot  \pu{2.46 J/(g* ^\circ C)}\\&#xA;\therefore s &amp;amp;= \pu{1.96 J/(g * ^\circ C)}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The correct answer, however, is &lt;span class=&quot;math-container&quot;&gt;$\pu{2.36 J/(g * ^\circ C)}$&lt;/span&gt;.&#xA;I think that there is a mistake in my solution, because the heat capacity of the calorimeter is given, and I didn't use it.&lt;/p&gt;&#xA;&lt;p&gt;Could anyone explain where I have been wrong?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="4945" LastEditDate="2021-01-06T11:27:20.883" LastActivityDate="2021-01-06T11:35:44.860" Title="How to determine the specific heat of methanol from calorimeter data?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14083" PostTypeId="1" AcceptedAnswerId="14088" CreationDate="2014-07-03T13:18:15.300" Score="5" ViewCount="1408" Body="&lt;p&gt;Is there any rule why a crystal made of Si and C is named silicon carbide SiC, but never carbon silicate CSi? Both silicon and carbon are in the same group, so share many of the atomic properties, so why the asymmetry in naming? Are all such compounds named like &amp;lt;HigherPeriod&gt;&amp;nbsp;&amp;lt;LowerPeriod&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2133" LastActivityDate="2014-07-03T18:32:59.003" Title="Why isn't SiC called carbon silicate?" Tags="&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14084" PostTypeId="2" ParentId="14082" CreationDate="2014-07-03T13:27:14.483" Score="3" Body="&lt;p&gt;You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at &lt;span class=&quot;math-container&quot;&gt;$\pu{24.7 ^\circ C}$&lt;/span&gt; before the methanol was added and then it rose &lt;span class=&quot;math-container&quot;&gt;$\pu{3.8 ^\circ C}$&lt;/span&gt; along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3 J/^\circ C}$&lt;/span&gt; then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $&lt;/span&gt; &lt;em&gt;&amp;quot;Joules absorbed by the calorimeter&amp;quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Then we simply apply the heat lost equals heat gained concept:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;\pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\&#xA;  = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C}) &#xA;    \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C} &#xA;    \cdot \pu{3.8 ^\circ C}&#xA;\end{multline}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;After rearranging, we get &lt;span class=&quot;math-container&quot;&gt;$S = \pu{2.36 J/(g * ^\circ C)}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="4945" LastEditDate="2021-01-06T11:35:44.860" LastActivityDate="2021-01-06T11:35:44.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14085" PostTypeId="1" AcceptedAnswerId="14428" CreationDate="2014-07-03T13:36:38.153" Score="2" ViewCount="1569" Body="&lt;p&gt;I am trying to grasp how internal standardization works in principle, and I am having some trouble wrapping my head around it. It is not the choosing of an IS I think about, orwhen it might benefit the analysis, but just the underlying principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I add a known amount of the IS to every sample: unknown samples, blank samples, calibration samples, and quality control samples. Each of these samples contain equal amounts of the internal standard (or as &quot;equal&quot; as our pipettes may give). Considering the calibration, when the calibration samples are analyzed, I see two peaks in the chromatogram: the analyte and the IS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I ran calibration samples of 0, 1, 2, 5, 10, and 20 ppm, each one consiting 20 ppm internal standard. The analyte peaks will grow for each calibration sample, but the IS peaks remain more or less stationary. I construct the calibration curve by plotting the ratio of analyte peak area to IS peak area (or height), as a function of the analyte conentration in the calibration samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that there is a constant ratio involved, one that is used to correct analyte concentrations for loss of analyte, or varying sample introduction system. If the IS peak is lower, then the analyte concentration will be corrected upwards. This apparently is because this ratio I mentioned is constant, so when reducing the denominator, the nominator must increase. Mathematically, this is trivial; however, what is this mystical constant, and what is its value? Is that arbitrary? Can I choose this constant's value myself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite confused. Hopefully someone here can help my out!&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2014-07-17T10:26:59.267" LastActivityDate="2014-07-18T19:03:34.977" Title="The Principle of Internal Standardization (in the context of ion chromatography)" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14087" PostTypeId="1" AcceptedAnswerId="14100" CreationDate="2014-07-03T15:03:30.170" Score="38" ViewCount="29925" Body="&lt;p&gt;$\ce{N}$ &amp;amp; $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference? &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="16683" LastEditDate="2017-07-03T15:27:24.340" LastActivityDate="2017-07-03T15:35:32.957" Title="Why is the bond angle H-P-H smaller than H-N-H?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;periodic-trends&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="14088" PostTypeId="2" ParentId="14083" CreationDate="2014-07-03T15:25:42.040" Score="3" Body="&lt;p&gt;As per the rules of inorganic chemistry nomenclature the element of the compound which is more electronegative is to be written at the end of molecular formula and its name is to be written after all the elements which have lower electronegativity are written hence the given compound is pronounced as silicon-carbide and not carbon-silicide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a more detailed view about electronegativity of elements can be found in this picture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDFjW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6953" LastActivityDate="2014-07-03T15:25:42.040" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14089" PostTypeId="2" ParentId="14077" CreationDate="2014-07-03T15:43:09.457" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I believe that this should have been the graph...(please omit the&#xA;  negative side of x axis), so concave downwards. Why is there a&#xA;  discrepancy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your analysis and your graph are correct.  Let's examine why the discrepancy arises.  I've drawn two graphs below.  The top one is similar to your graph, only I'm examining a smaller range on the x-axis.  My lower graph examines an even smaller range and it is a range close to x=0.  You can see that this lower graph appears to be &quot;concave-up&quot; just like the Wikipedia graph you referenced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/haBpE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a third graph where I changed the numerator from -1 to -100, notice how it moves the curve further to the right along the x-axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrzbL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending upon 1) what numbers we put in the numerator and denominator of our exponential equation and 2) depending upon how we scale our axes, we can dramatically change how our graph appears.  There will always be a relatively flat part to this curve located close to x=0, then there will be a concave-up section, followed by a concave-down section, and finally another relatively flat section further out on the x-axis.  You happened to choose a large range (up to x=60-70) for the x-axis of your graph.  This compressed your graph, so it did not show the first two parts of the curve (the relatively flat part located close to x=0 and the concave-up section) clearly.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-05T19:51:34.870" LastActivityDate="2014-07-05T19:51:34.870" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14090" PostTypeId="1" CreationDate="2014-07-03T15:44:22.973" Score="0" ViewCount="2743" Body="&lt;p&gt;Acid is a substance that releases $\ce{H+}$ ions when mixed with water But if it releases $\ce{H+}$ ions the  acid will ultimately not exist. For e.g we take the acid $\ce{H2SO4}$. If we add water then it becomes $$\ce{H2SO4 + H2O -&amp;gt; H+ + SO4^{-2} + H2O}$$ So the bond between hydrogen and sulphate will be broken therefore the acid will not exist, but we see that diluted acids are possible. Why?&lt;/p&gt;&#xA;" OwnerUserId="5321" LastEditorUserId="4945" LastEditDate="2014-07-04T10:18:23.030" LastActivityDate="2014-07-06T16:37:22.327" Title="Why does an acid exist when dissolved in water?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14091" PostTypeId="2" ParentId="14083" CreationDate="2014-07-03T15:55:39.657" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why a crystal made of Si and C is named silicon carbide SiC, but never&#xA;  carbon silicate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To add to agha rehan abbas' and permeakra's comments, the &quot;ate&quot; suffix typically implies the presence of oxygen (carbonate, silicate, phosphate, etc.).  Since $\ce{SiC}$ does not contain oxygen, you would not use the &quot;ate&quot; suffix in this case.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-03T15:55:39.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14092" PostTypeId="2" ParentId="14083" CreationDate="2014-07-03T16:05:52.080" Score="3" Body="&lt;p&gt;Silicates are salts of silicic acids, $\ce{xH2O * ySiO2}$ and $\ce{SiC}$ is not one. The proper term is silicide. It is &lt;strong&gt;usually&lt;/strong&gt; assumed, that s/p element upper or righter from considered one is more electronegative, and &quot;-ide&quot; nomenclature assumes that second element has negative charge and is more electronegative. Of course, there are exceptions. For example, $\ce{RhB}$ is still written as is, despite $\ce{Rh}$ having higher electronegativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the ordering of elements in formulas and naming of compounds are subjects of tradition. For example, despite having higher negativity, nitrogen in the formula of ammonia precedes  hydrogen: $\ce{NH_3}$, while for water hydrogen precedes oxygen $\ce{H2O}$, and both may be called 'hydrides' in some cases. It may be viewed as another rule: in case of a isolated particle central atom of the molecule first, substituents later (composite substituents written in curly braces), but acidic hydrogens first. In brutto-formula indexes symbols of elements AFAIK are ordered alphabetically except that carbon and hydrogen are first two.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-07-03T18:32:59.003" LastActivityDate="2014-07-03T18:32:59.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14093" PostTypeId="1" CreationDate="2014-07-03T16:41:48.023" Score="1" ViewCount="41212" Body="&lt;p&gt;I want to understand this concept in detail. I am getting confused with increase in temperature backward reaction takes place or forward?? And how?? Proper explanation is needed.&lt;/p&gt;&#xA;" OwnerUserId="7010" LastActivityDate="2014-07-03T17:26:43.477" Title="How does temperature affect equilibrium, according to Le Chatelier's principle??" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2014-07-04T13:55:00.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="14094" PostTypeId="2" ParentId="14090" CreationDate="2014-07-03T16:57:37.070" Score="-1" Body="&lt;p&gt;Yes, exactly; only ions can't not exist. Water molecule has dipole nature. When acid is mixed with water, then because of ion-dipole interaction which is stronger than dipole -dipole interaction b/w water molecules and ion-ion interaction b/w acid molecules, opposite ends of water dipoles surround the ions of acid ( means the dipole getting delta + charge will attract negative ion of acid and vica versa). So, now ions get hydrated as process of hydration takes place. They are still free entities, but surrounded by water dipoles. So, the solution of acid and water exists.&lt;/p&gt;&#xA;" OwnerUserId="7010" LastActivityDate="2014-07-03T16:57:37.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14095" PostTypeId="2" ParentId="14087" CreationDate="2014-07-03T17:24:59.227" Score="-1" Body="&lt;p&gt;Since Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.&lt;/p&gt;&#xA;" OwnerUserId="1452" LastActivityDate="2014-07-03T17:24:59.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14096" PostTypeId="2" ParentId="14093" CreationDate="2014-07-03T17:26:43.477" Score="3" Body="&lt;p&gt;In a simple equilibrium we have either $$\ce{A + heat &amp;lt;=&amp;gt; B}$$ where the forward reaction is endothermic and the back reaction exothermic, or $$\ce{A &amp;lt;=&amp;gt; B + heat}$$ where the forward reaction is exothermic and the back reaction endothermic.  So in a simple equilibrium, the reaction in one direction will be exothermic and the reaction in the other direction will be endothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the temperature of an equilibrium is increased, the rate of both the forward and back reaction is increased.   However the rate of the endothermic reaction (the reaction that absorbs the added energy) is increased more.  So in the first reaction shown above, adding heat will shift the reaction to the right; in the second reaction heat will push it to the left.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason behind this behavior is that both the forward and back reaction can be described by the Arrhenius equation $$\ce{rate=Ae^{-E_{act}/RT}}$$  and the rate with the larger activation energy (the endothermic reaction) will always be more sensitive to changes in temperature. Therefor, the endothermic reaction will increase faster with increasing temperature than the exothermic reaction will.  Put some numbers into the equation and convince yourself.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-03T17:26:43.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14097" PostTypeId="2" ParentId="12250" CreationDate="2014-07-03T17:30:01.400" Score="3" Body="&lt;p&gt;The naming is done on the basis of the number of carbons present in the parent chain.The first member of each group is &quot;form-&quot;. However we do not have a &quot;formone&quot; as ketonic groups are non-terminal.The $\ce{CH3CO-}$ groups are called &quot;acetyl&quot; groups(as they already contain 2 atoms of carbon) and thus the very first compound that it can form with a methyl group is propanone or acetone.&#xA;However these are old-fashioned names and do not really need much intervention.The IUPAC nomenclature has logic and rightly names the compound &quot;propanone&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7000" LastEditorDisplayName="user15489" LastEditDate="2015-06-16T21:02:42.003" LastActivityDate="2015-06-16T21:02:42.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14098" PostTypeId="2" ParentId="14049" CreationDate="2014-07-03T17:48:52.953" Score="5" Body="&lt;p&gt;There are many filters which &lt;strong&gt;do not&lt;/strong&gt; remove fluoride from potable water. Boiling or freezing does not help either. Reverse osmosis filtration may be a solution. However they are expensive ($100+).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activated alumina defluoridation filters are used in areas where contamination from fluoride is common. These can reduce fluoride levels from 0.5 ppm to 0.1 ppm, but these require frequent replacement and are not a solution for domestic treatment. Distillation filters can be purchased to remove fluoride.&lt;/p&gt;&#xA;" OwnerUserId="7000" LastEditorUserId="9961" LastEditDate="2016-12-12T23:21:23.310" LastActivityDate="2016-12-12T23:21:23.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14099" PostTypeId="1" CreationDate="2014-07-03T18:39:19.173" Score="0" ViewCount="206" Body="&lt;p&gt;Suppose two compounds with, say, $\mathrm{p}K_{\mathrm{a}1}=2.8$ and $\mathrm{p}K_{\mathrm{a}2}=3.2$, and formulae $\ce{C_{$a$}H_{$b$}O_{$c$}}=\ce{C_{$p$}O_{$q$}(OH)_{$r$}}$ and $\ce{C_{$d$}H_{$e$}O_{$f$}}=\ce{C_{$u$}O_{$v$}(OH)_{$w$}}$, respectively, with given $abcdef$, $pqruvw$. Can we explain the above $\mathrm{p}K_\mathrm{a}$ with (ONLY) the knowledge of the given numbers of atoms of $\ce{C}$, $\ce{H}$ and $\ce{O}$ using the theory of Usanovich, and no knowledge of the internal chemical structure of the chemical bonds inside the two molecules?&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T18:56:46.693" LastActivityDate="2015-06-17T18:56:46.693" Title="pKa explanation through empirical formula?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14100" PostTypeId="2" ParentId="14087" CreationDate="2014-07-03T18:49:41.120" Score="39" Body="&lt;p&gt;&lt;strong&gt;Starting point:&lt;/strong&gt; 2s orbitals are lower in energy than 2p orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are two choices for the central atom:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;remain unhybridized:&lt;/strong&gt; [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;substituents will be arranged closer together and destabilizing steric interactions will be increased&lt;/li&gt;&#xA;&lt;li&gt;due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy&lt;/li&gt;&#xA;&lt;li&gt;the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;hybridize:&lt;/strong&gt; [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart&lt;/li&gt;&#xA;&lt;li&gt;due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy&lt;/li&gt;&#xA;&lt;li&gt;the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been &quot;saved&quot; can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to our problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-07-03T15:35:32.957" LastActivityDate="2017-07-03T15:35:32.957" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14101" PostTypeId="2" ParentId="14099" CreationDate="2014-07-03T21:23:28.287" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we explain the above pka with (ONLY) the knowledge of the given&#xA;  numbers of atoms of C, H and O using the theory of Usanovich&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, nobody's theory (including &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Usanovich_definition&quot;&gt;Usanovich's&lt;/a&gt;) can explain $\ce{pK_{a}}$ differences based on molecular formula coefficients.  This is best illustrated by the observation that even compounds with the same coefficients (same values for &quot;abcdef&quot;) can have different $\ce{pK_{a}}$ values.  For example, $\ce{C3H6O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{&amp;amp;acetone &amp;amp;&amp;amp;pK_{a}=20}\\&#xA;\ce{&amp;amp;allyl~ alcohol &amp;amp;&amp;amp;pK_{a}=$15.5$}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-03T21:23:28.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14102" PostTypeId="1" CreationDate="2014-07-04T00:03:24.793" Score="0" ViewCount="11663" Body="&lt;p&gt;we have a metal surface with electrons in a specific energy state say X ,now we shine light on it of specific wavelength and intensity.&#xA;now what if i double the intensity of light , what would happen to the number of electrons emitted and the average energy of the electrons ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;my guess is that since intensity is doubled, then number of photons emitted also double and the energy also doubles .... but i am not really sure about it&lt;/p&gt;&#xA;" OwnerUserId="7006" LastActivityDate="2014-07-04T02:59:43.160" Title="light Intensity and its effects on energy and emitted electron on metal surface" Tags="&lt;spectroscopy&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14103" PostTypeId="2" ParentId="14102" CreationDate="2014-07-04T00:23:54.327" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what if i double the intensity of light , what would happen to the&#xA;  number of electrons emitted and the average energy of the electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is known as the &quot;Photoelectric Effect&quot; and is described by the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{E_{max}=h\nu- WF_{M}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{E_{max}}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\ce{WF_{M}}$ is the workfunction for the particular metal. Your guess is part right, doubling the intensity of the incoming beam doubles the number of electrons ejected from the metal surface, but their kinetic energy distribution remains unchanged.  The maximum kinetic energy of an emitted electron is still determined by the above equation and therefor still capped by $\ce{E_{max}}$ and the kinetic energies of all electrons emitted are still distributed from 0 to $\ce{E_{max}}$.  If you'd like more information, &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_States_of_Atoms_and_Molecules/2._Foundations_of_Quantum_Mechanics/Photoelectric_Effect&quot; rel=&quot;nofollow&quot;&gt;here's a good, concise reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-04T00:51:23.583" LastActivityDate="2014-07-04T00:51:23.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14104" PostTypeId="1" AcceptedAnswerId="14112" CreationDate="2014-07-04T01:10:32.820" Score="3" ViewCount="217" Body="&lt;p&gt;$\ce{NF4+}$ ion reacts with water as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NF4+ + H2O -&amp;gt; NF3 + H2F+ + {1\over 2}O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does $\ce{NF4+}$ function as an acid or base in this reaction, and why?&lt;/p&gt;&#xA;" OwnerUserId="6988" LastEditorUserId="11367" LastEditDate="2017-04-08T16:13:51.260" LastActivityDate="2017-04-08T16:15:41.553" Title="Does nitrogen tetrafluoride act as an acid or a base in its reaction with water?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14105" PostTypeId="2" ParentId="14102" CreationDate="2014-07-04T02:59:43.160" Score="0" Body="&lt;p&gt;The intensity of emitted light varies directly with the number of photoelectrons emitted. If the intensity of light is varied, keeping the frequency of the source and the accelerating potential constant, the resulting photo-current produced is found to increase,i.e. the photo-current vs intensity of incident light curve is linear having a constant slope. Since the number of photoelectrons emitted per second is directly proportional to photo current produced, therefore it is also &lt;strong&gt;proportional to the intensity of incident light.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The above equation that has been stated in the previous answer can also be written as:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E(max)=eV, where e is the basic electronic charge and V represents the stopping potential. Now it has been found that the stopping potential is dependent on the frequency of incident light and is independent of its intensity. This implies that the &lt;strong&gt;maximum kinetic energy of emitted photo-electrons is independent of intensity&lt;/strong&gt; too[since, it remains proportional to stooping potential].       &lt;/p&gt;&#xA;" OwnerUserId="7000" LastActivityDate="2014-07-04T02:59:43.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14106" PostTypeId="1" CreationDate="2014-07-04T03:36:54.623" Score="1" ViewCount="254" Body="&lt;p&gt;Which acids are created by the combination of a gas and water?  I know that carbonic acid is created by combining carbon dioxide and water, but are there others?  Is there a specific name for this kind of acid?&lt;/p&gt;&#xA;" OwnerUserId="6895" LastActivityDate="2014-07-04T04:01:22.503" Title="Which acids are created by the combination of a gas and water?" Tags="&lt;acid-base&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14107" PostTypeId="2" ParentId="14106" CreationDate="2014-07-04T04:01:22.503" Score="1" Body="&lt;p&gt;I don't think there is a specific name for these acids. But I can think of at least 4 more. The gases are $HCl,\ SO_2,\ SO_3,$ and $NO_2$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-04T04:01:22.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14109" PostTypeId="1" CreationDate="2014-07-04T04:11:41.173" Score="1" ViewCount="480" Body="&lt;p&gt;Recently, I was introduced to a method of desalinization called electrodialysis in which a current was applied to salt water in such a way that the positive and negative were separated and drawn through membranes, leaving less salty water in between.  how do the membranes facilitate a one-way flow of ions?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this method sounds a lot like the electrolysis.  How is electrolysis prevented or suppressed?&lt;/p&gt;&#xA;" OwnerUserId="6895" LastEditorUserId="6895" LastEditDate="2014-07-07T11:57:26.963" LastActivityDate="2014-07-07T11:57:26.963" Title="What is electrodialysis and is electrolysis prevented or suppressed?" Tags="&lt;water&gt;&lt;ions&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14110" PostTypeId="1" CreationDate="2014-07-04T04:17:52.757" Score="2" ViewCount="55" Body="&lt;p&gt;When you mix ice cream and a soft drink (like coke,)  a reaction happens that creates a thick foam.  I am assuming the foam id composed of carbon dioxide gas and milk curd, which is produced by the milk reacting with an acid.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the milk curd produced by the carbonic acid or the phosphoric acid?&#xA;Is the carbon dioxide released by the added sugar, or another ingredient?&#xA;Are there other important processes going on in there?&lt;/p&gt;&#xA;" OwnerUserId="6895" LastEditorUserId="73" LastEditDate="2014-07-04T06:20:15.550" LastActivityDate="2014-07-04T06:20:15.550" Title="ice cream float - what causes it?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2014-07-04T18:16:04.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="14111" PostTypeId="1" AcceptedAnswerId="14399" CreationDate="2014-07-04T05:48:50.640" Score="9" ViewCount="653" Body="&lt;p&gt;Simply put, why does adding acid to the cathode up the cell voltage? From what I understand, acids are proton donors. Do these positive protons attract electrons toward the cathode compartment of the electrochemical cell? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-17T08:22:14.787" Title="Why do acids favor the oxidizer? And bases the reducer?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14112" PostTypeId="2" ParentId="14104" CreationDate="2014-07-04T07:26:16.330" Score="4" Body="&lt;p&gt;&lt;strong&gt;Neither.&lt;/strong&gt; The reaction is NOT one of ion exchange; it is a redox reaction. Thus, $\ce{NF4+}$ works as electron acceptor here, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NF4+ + 2e- -&amp;gt; NF3 + F-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction mechanism is unclear, it is possible that some of its stages may have acid-base character (for example, $\ce{F-}$ sometimes acts as a base), but $\ce{NF4+}$ certainly isn't acting primarily as base or acid here. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="11367" LastEditDate="2017-04-08T16:15:41.553" LastActivityDate="2017-04-08T16:15:41.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14113" PostTypeId="2" ParentId="14065" CreationDate="2014-07-04T08:12:33.283" Score="3" Body="&lt;p&gt;A quick search would tell you what the 4 nucleobases for RNA are made of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uracil:&lt;/strong&gt; 4 Carbon 4 Hydrogen 2 Nitrogen 2 Oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TdwdY.png&quot; alt=&quot;u&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Adenine:&lt;/strong&gt; 5 Carbon 5 Hydrogen 5 Nitrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/db/Adenine.svg/120px-Adenine.svg.png&quot; alt=&quot;a&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Guanine:&lt;/strong&gt; 5 Carbon 5 Hydrogen 5 Nitrogen 1 Oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Guanin.svg/200px-Guanin.svg.png&quot; alt=&quot;g&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cytosine:&lt;/strong&gt; 4 Carbon 5 Hydrogen 3 Nitrogen 1 Oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/10/Cytosine_chemical_structure.png/120px-Cytosine_chemical_structure.png&quot; alt=&quot;c&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-07-04T08:12:33.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14115" PostTypeId="2" ParentId="14087" CreationDate="2014-07-04T10:04:14.667" Score="13" Body="&lt;p&gt;First note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves unnatural or anomalous. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you compare nitrogen with phosphorus, you will note, that the former is much smaller than the latter, i.e. van der Waals radii $r(\ce{N})=155~\mathrm{pm};\ r(\ce{P})=180~\mathrm{pm}$ (ref. &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;), covalent radii $r(\ce{N})=71~\mathrm{pm};\ r(\ce{P})=107~\mathrm{pm}$ (ref. &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_radius&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;). Therefore also the orbitals in nitrogen are smaller, and $\ce{s}$ and $\ce{p}$ orbitals will occupy more of the same space than in phosphorus. As a result the $\ce{N-H}$ bond distance will naturally also be shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lone pair is usually most stable in an orbital that has high $\ce{s}$ character. Bonds will most likely be formed with the higher lying $\ce{p}$ orbitals. The orientation of these towards each other is exactly $90^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ammonia this would lead to very close $\ce{H\cdots{}H}$ contacts, which are repulsive and therefore the hydrogen atoms are pushed away from each other. This is possible since in the second period the $\ce{s-p}$ splitting is still very small and the nitrogen $\ce{s}$ orbital is accessible for the hydrogen atoms. This will ultimately result in mixing $\ce{s}$ and $\ce{p}$ orbitals for nitrogen in the respective molecular orbitals. This phenomenon can be referred to as hybridisation - the linear combination of orbitals from the same atom. This term is therefore somewhat independent from its most common usage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also very important to know, that the molecular wavefunction of a molecule has to reflect its overall symmetry. In this case it is $C_{3v}$, which means there is a threefold rotational axis and three vertical mirror planes (the axis is element of these planes). This gives also rise to degenerate orbitals. A canonical orbital picture has to reflect this property (BP86/cc-pVDZ; valence orbitals are ordered with increasing energy from left to right).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GRjGc.jpg&quot; alt=&quot;canonical valence orbitals of ammonia&quot;&gt;&lt;br&gt;&#xA;Note that the lowest lying valence molecular orbital is formed only from $\ce{s}$ orbitals (There is one additional $\ce{1s^2-N}$ core orbital.)&lt;br&gt;&#xA;Now Natural Bond Orbital (NBO) Theory can be used to transform these delocalised molecular orbitals to a more common and familiar bonding picture, making use of atomic hybrid orbitals. This method is called localising orbitals, but it has the expense of losing the energy eigenvalue that may be assigned to canonical orbitals (NBO@BP86/cc-pVDZ; valence NBO cannot be ordered by energy levels).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q5Gpy.jpg&quot; alt=&quot;nbo valence orbitals of ammonia&quot;&gt;&lt;br&gt;&#xA;In this theory you will find three equivalent $\ce{N-H}$ bonds, that are composed of $32\%~\ce{1s-H}$ and $68\%~\ce{s^{$0.87$}p^3-N}\approx\ce{sp^3-N}$ orbitals. Note that the lone pair orbital at nitrogen has a slightly higher $\ce{s}$ orbital contribution, i.e. $\ce{s^{1.42}p^3-N}\approx\ce{sp^3-N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the thermodynamically most favoured angle is found to be $107^\circ$ due to a compromise between optimal orbital overlap and least internuclear repulsion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The canonical bonding picture in phosphine is very similar to ammonia, only the orbitals are larger.  Even in this case it would be wrong to assume, that there is no hybridisation present at all. However, the biggest contribution to the molecular orbitals stems from the $\ce{p}$ orbitals at phosphorus.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gdU0O.jpg&quot; alt=&quot;canonical valence orbitals of phosphine&quot;&gt;&lt;br&gt;&#xA;Applying the localisation scheme, one end up with a different bonding picture. Here are three equal $\ce{P-H}$ bonds that are composed of $48\%~\ce{1s-H}$ and $52\%~\ce{s^{$0.5$}p^3-P}$ orbitals. The lone pair at phosphorus is composed of $57\%\ce{s} + 43\%\ce{p}$ orbitals.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SDnXh.jpg&quot; alt=&quot;valence nbo orbitals of phosphine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One can see the difference of the molecules also in their &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;noreferrer&quot;&gt;inversion barrier&lt;/a&gt;, while for ammonia the inversion is readily available at room temperature, $\Delta E \approx 6~\mathrm{kcal/mol}$, it is very slow for phosphine, $\Delta E \approx 40~\mathrm{kcal/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mostly due to the fact, that the nitrogen hydrogen bonds have already a significant $\ce{s}$ orbital contribution, which can be easily increase, to form the planar molecule with formally $\ce{sp^2}$ hybrids.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-07-03T15:28:34.990" LastActivityDate="2017-07-03T15:28:34.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14116" PostTypeId="1" AcceptedAnswerId="14138" CreationDate="2014-07-04T11:43:39.283" Score="1" ViewCount="107" Body="&lt;p&gt;As part of a simulation I am developing I have an acid in aqueous solution represented as a &lt;em&gt;concentration field&lt;/em&gt; of mass density (units &lt;code&gt;g/L&lt;/code&gt;) at various points in space (denoted $[S]$). I want to calculate the pH of the solution. I am attempting to use the method for weak acids outlined &lt;a href=&quot;https://en.wikipedia.org/wiki/PH#Weak_acids_and_bases&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but am struggling with the dimensionality of the variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[H]^2 + K_a [H] - K_a C_a = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the quadratic equation, and knowing that $K_a$ has no units, it implies that $[H]$ and $C_a$ are also dimensionless. Does this mean that $[H]$ is in fact the &lt;em&gt;activity&lt;/em&gt; of the the solution, and not the $H^+$ concentration? And is the relationship between $C_a$ and $[S]$ $$C_a = \gamma \frac{[S]}{M(S)c^\ominus},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is the &lt;em&gt;activity coefficient&lt;/em&gt;, $M(S)$ is the &lt;em&gt;molar mass&lt;/em&gt; of the acid, and $c^\ominus = 1$M is the &lt;em&gt;standard amount concentration&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Followup question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to the quadratic equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[H] = \frac{-K_a \pm \sqrt{K_a^2 + 4K_a C_a}}{2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus $[S] = 0 \implies [H] = \frac{-K_a \pm K_a}{2} = -K_a$ or $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $pH = -\log_{10}\left([H]\right)$ and $K_a &amp;gt; 0$  there is no real solution, but logic would dictate that $pH = 7$ in this case. Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="7016" LastEditorUserId="7016" LastEditDate="2014-07-04T12:40:18.607" LastActivityDate="2014-07-05T14:53:13.740" Title="Reconciling the units of pH calculation" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14117" PostTypeId="1" AcceptedAnswerId="14119" CreationDate="2014-07-04T11:52:46.323" Score="6" ViewCount="788" Body="&lt;p&gt;Does the hydronium ion donate a proton to water? If the collision of 2 molecules of water can cause a proton to be transferred, does hydronium 'pass' a proton to water? &lt;/p&gt;&#xA;" OwnerUserId="7015" LastEditorUserId="31313" LastEditDate="2017-04-01T02:15:08.947" LastActivityDate="2018-03-22T14:04:49.277" Title="Does H3O+ donate a H+ to Water?" Tags="&lt;acid-base&gt;&lt;protons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14118" PostTypeId="2" ParentId="14117" CreationDate="2014-07-04T13:18:28.873" Score="4" Body="&lt;p&gt;Yes, it is actually the main reason for superior mobility of hydrogen ion in water solutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see the link: &lt;a href=&quot;http://en.wikipedia.org/wiki/Grotthuss_mechanism&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Grotthuss_mechanism&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-04T13:18:28.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14119" PostTypeId="2" ParentId="14117" CreationDate="2014-07-04T14:51:44.820" Score="10" Body="&lt;p&gt;$\ce{H3O^+}$ is really just a &quot;shorthand notation&quot; for what really occurs in solution.  Typically higher hydrates of the proton, particularly $\ce{H5O2+}$ the &quot;Zundel&quot; cation and $\ce{H9O4+}$ the &quot;Eigen&quot; cation are thought to be the main players in the proton transfer process.  As permeakra has pointed out, Grotthuss was one of the early pioneers studying this process and his name is still used to describe the mechanism.  The mechanism is still an active area of research with considerable controversy remaining over the details of the mechanism.  &lt;a href=&quot;http://www1.lsbu.ac.uk/water/grotthuss.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; provides some diagrams of the general process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujjX1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, just as electrons are said to use a &quot;hopping&quot; transport mechanism, the same has been said of proton transport.  The making and breaking of hydrogen bonds, accompanied by the twisting and turning of protons completes the process.  This proton hopping mechanism between these larger water aggregates is thought to be a key step at room temperature.  However proton tunneling plays a role as well, particularly as the temperature is decreased.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-03-22T14:04:49.277" LastActivityDate="2018-03-22T14:04:49.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14121" PostTypeId="1" AcceptedAnswerId="14124" CreationDate="2014-07-04T21:30:00.007" Score="11" ViewCount="3318" Body="&lt;p&gt;Recently I was reading through the owner's manual of a washing machine; notably, the warnings in the front of the booklet. One warning stood out to me; it said that if the hot water in your house has not been run in a few weeks $\ce{H2}$ can build up in your lines. The warning of course stated that this can only occur under certain circumstances so I'm assuming it is quite rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Okay, so my question is, under what conditions does hydrogen gas build up in a hot water line?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="7475" LastEditDate="2015-10-30T16:37:11.973" LastActivityDate="2020-10-28T16:22:07.373" Title="How does hydrogen gas build up in hot water lines?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;water&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14122" PostTypeId="1" AcceptedAnswerId="14128" CreationDate="2014-07-04T21:50:06.347" Score="0" ViewCount="152" Body="&lt;p&gt;What are the chemical reactions and/or chemical processes that take place during SFE(supercritical fluid extraction)? Namely supercritical $\ce{CO2}$.&lt;/p&gt;&#xA;" OwnerUserId="7002" LastEditorUserId="2094" LastEditDate="2014-07-04T23:03:47.933" LastActivityDate="2014-07-05T00:08:46.820" Title="Chemical reactions in SFE?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14123" PostTypeId="1" CreationDate="2014-07-04T22:07:07.077" Score="2" ViewCount="1609" Body="&lt;p&gt;The basic + - electrolyte set up. Modern batteries and the like. How exactly do they work?&lt;/p&gt;&#xA;" OwnerUserId="7002" LastEditorUserId="4945" LastEditDate="2014-07-24T03:30:19.233" LastActivityDate="2014-07-24T16:31:07.043" Title="How does a simple alkaline battery work?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14124" PostTypeId="2" ParentId="14121" CreationDate="2014-07-04T22:30:12.877" Score="8" Body="&lt;p&gt;I think that it is probably due to a reaction between an acid and a metal. These reactions can lead to the formation of hydrogen following this generic forumula:&#xA;$$\ce{M + 2H+ -&amp;gt; M^{2+} + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example iron could be a source of electrons. If you look at the corrosion of iron: $$\ce{Fe-&amp;gt;Fe^{2+} + 2e^{-}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low pH (notice that there is a correlation between pH and temperature) hydrogen gas is formed:&#xA;$$\ce{2H^{+} +2e- -&amp;gt; 2H2}$$&lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-07-04T22:30:12.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14125" PostTypeId="2" ParentId="14123" CreationDate="2014-07-04T22:41:56.887" Score="0" Body="&lt;p&gt;The chemistry of alkaline batteries involve the oxidation of zinc and the reduction of manganese dioxide. The description of a cell can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_battery&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-04T22:41:56.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14126" PostTypeId="2" ParentId="14090" CreationDate="2014-07-04T22:57:33.147" Score="3" Body="&lt;p&gt;Your doubts are legitimates, but there are some aspects that you should take in account:&lt;/p&gt;&#xA;&lt;h1&gt;1-Not all acid are fully dissociated&lt;/h1&gt;&#xA;&lt;p&gt;Not all acids fully dissociate, so if you take acetic acid you have in fact an equilibrium with some acetic acid molecules not dissociated. So your question holds only for strong acid like sulfuric acid you mentioned.&lt;/p&gt;&#xA;&lt;h1&gt;2-You are using pseudo-Arrhenius definition, why don't you use Lewis acid definition?&lt;/h1&gt;&#xA;&lt;p&gt;If you think the acid as a 'substance that that releases &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt;' this is still an acid even if it dissociate, and even here you should notice that acidity is linked with &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt;. However if you want to dig deeper there are many definitions of acid if you use Lewis acid theory and so you think the acid as a specie that accepts electrons you can easily see that in fact that the acidity is linked mainly with &lt;span class=&quot;math-container&quot;&gt;$H^{+}$&lt;/span&gt;!&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-07-06T16:37:22.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14127" PostTypeId="2" ParentId="14122" CreationDate="2014-07-04T23:12:59.270" Score="1" Body="&lt;p&gt;Ideally no reactions should occur during an SF extraction. In all the extractions a solvent is used to extract different substances from a matrix. In SFE the solvent is a supercritical fluid, in most of the cases is used $CO_{2}$ but not always. The chemicals process related to the extraction is mainly the solubilization of the target substance in the SF depending on the properties of the target substance and the SF (e.g. polar or non polar). &lt;/p&gt;&#xA;" OwnerUserId="2094" LastActivityDate="2014-07-04T23:12:59.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14128" PostTypeId="2" ParentId="14122" CreationDate="2014-07-05T00:08:46.820" Score="2" Body="&lt;p&gt;Supercritical fluid technology is used primarily to enhance the chemical extraction process.  The solubility of a compound in a supercritical solvent increases as the system pressure increases at constant temperature. This is due to the highly pressurized supercritical solvent disrupting the intermolecular attractive forces between the molecules of the compound to be extracted, and replacing those forces with increased attractive forces between the compound and the supercritical solvent.  Carbon dioxide is a very common material used for supercritical extraction.  It is cheap, not flammable, has low toxicity and, after the extraction is complete you simply return the system to STP, allow the carbon dioxide to evaporate and you are left with only the extracted material.  Some examples of commercial uses for supercritical carbon dioxide include caffeine extraction from coffee beans, sterilization of surgical equipment, fracking and dry cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical reactions involving supercritical carbon dioxide are now being intensely studied.  One reaction that has been commercialized involves reaction of supercritical carbon dioxide with hardened, alkaline cements. A harder cement containing carbonate groups throughout the entire volume of cement is formed along with water (environmentally friendly process).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-05T00:08:46.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14130" PostTypeId="2" ParentId="14073" CreationDate="2014-07-05T02:16:39.423" Score="9" Body="&lt;p&gt;OK, let's step through your questions one at a time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First off, what do they mean by steady-state number here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Intermediates formed in reactions usually exist in low concentration.  The steady-state assumption assumes that the concentration of this (low-concentration) intermediate will not change appreciably over the course of the reaction once steady state has been established.  The rate of its formation is assumed to be equal to the rate of its consumption.  Making the steady state assumption allows us to recast the previous statement as a mathematical equation as we'll see below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this case why must the rate of absorption be equal to the rate of desorption&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Just like in a chemical equilibrium where it takes a finite time for the equilibrium to establish itself, it takes a finite time before the steady state is established. Once there is a finite, non-zero concentration of the intermediate &lt;em&gt;and&lt;/em&gt; we have arrived at the steady state, &lt;em&gt;then&lt;/em&gt; the rate of absorption and desorption will be equal and the non-zero concentration of the intermediate(s) will remain constant – a steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the steady-state assumption to the problem at hand allows us to state that for all of the intermediates (dust particles with 1 or 2 hydrogen atoms on them), the rate of hydrogen atom adsorption on an intermediate must be equal to the rate of hydrogen atom desorption from that intermediate. This can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_\mathrm a [\ce{H}] = k_\mathrm d N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then rearranged and solved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;N &amp;amp;= \frac{k_\mathrm a}{k_\mathrm d}[\ce{H}] \\&#xA;&amp;amp;= \frac{\pu{1.4 * 10^-5 cm3 s-1}}{\pu{1.9 * 10^-3 s-1}} \cdot \pu{10 cm^-3} \\&#xA;&amp;amp;= \pu{7.4 * 10^-2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is it $k_\mathrm d N$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because $k_\mathrm d N$ represents the &lt;strong&gt;total&lt;/strong&gt; rate of desorption of hydrogen atoms from dust particles, the right side of our verbal equation. It represents desorption from $x_1$ and $x_2$ - the only two species that can desorb, and it is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_\mathrm d N = k_\mathrm d x_1 + 2k_\mathrm d x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the factor of &quot;2&quot; arising because there are 2 hydrogen atoms on each dust particle in $x_2$, and both are viable candidates for desorption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is the same rate, that's what $k_\mathrm a [\ce{H}] = k_\mathrm d N$ means.  Rates are always a positive number, we are multiplying a rate constant times a concentration, both positive numbers.  However, when we write an equation and insert the rate expression into it, we can insert a minus sign before the rate expression if the rate expression represents a pathway that is causing the concentration of a product, reactant or intermediate to decrease.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How did they come up with those rates? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adsorption rate: $r_\mathrm a = \sum_m k_\mathrm a [\ce{H}] x_m = k_\mathrm a [\ce{H}]x_0 + k_\mathrm a [\ce{H}]x_1$&lt;br&gt;&#xA;This is just the typical kinetic expression – adsorption rate constant multiplied by the concentration of hydrogen atoms available to adsorb times the concentration of the various dust particles that can adsorb a hydrogen atom, which are only $x_0$ and $x_1$.&lt;/li&gt;&#xA;&lt;li&gt;Desorption rate: $r_\mathrm d = \sum_m k_\mathrm d m x_m = (k_\mathrm d \cdot 0 \cdot x_0) + (k_\mathrm d \cdot 1 \cdot x_1) + (k_\mathrm d \cdot 2 \cdot x_2)$&lt;br&gt;&#xA;Here $m$ is a concentration term analogous to $[\ce{H}]$.  It represents the number of hydrogen atoms on a dust particle, and can be 0, 1 or 2.  Note how it &quot;removes&quot; $x_0$ from the desorption equation as it should since a particle with no hydrogen atoms cannot desorb.  Notice too how it adjusts $x_2$ for the fact that there are two adsorbed hydrogen atoms in this case, so desorption is statistically twice as likely to happen.  It differs from $[\ce{H}]$ in its units: $[\ce{H}]$ is in $\pu{atoms/cm3}$ while &quot;m&quot; is in units of atoms/dust particle. &lt;/li&gt;&#xA;&lt;li&gt;Reaction rate: $r_\mathrm r = \left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right) = k_\mathrm r x_2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, after your explanation I understood how they got the absorption and desorption rates, but I'm still confused on how they got the reaction rate $\left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right)$. Could you please elaborate further on where all the numbers came from?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note we have a factor of $1/2$ because it takes 2 hydrogen atoms to produce 1 hydrogen molecule.  The $2$ and the $1$ come from the concentration terms $m(m-1)$.  In the &quot;Reaction&quot; case we have to combine two hydrogen atoms.  It is a bimolecular reaction and just like in the bimolecular reaction $\ce{A + B -&amp;gt; C}$ where the rate is proportional to $[\ce{A}][\ce{B}]$, here the analogous concentrations are $m(m-1)$.  Initially the concentration is $m=2$, but after that first hydrogen atom is &quot;selected&quot; the remaining concentration from which to pick our second hydrogen atom has been decreased to $m-1$.  What we're saying is that when we pick the first hydrogen we have two atoms to choose from, when we pick the second hydrogen atom we have one atom to choose from. If we had 5 hydrogen atoms on a particle and were still going to make 2 picks the equation for our two picks would become $5(5-1)=20$.  We would expect this case to proceed $(20/2)=10$ times faster than the case with only 2 atoms on the dust particle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 ways to form $\ce{H2}$ in our problem. Label the hydrogen atoms A and B, then we can either pick &lt;em&gt;via&lt;/em&gt; an AB or BA sequence. $m(m-1)$ is exactly analogous to $[\ce{H}][\ce{H}]$, it's just not a square because there is a significant difference between $2\cdot 2$ and $2\cdot 1$. When dealing with a mole of atoms ($N = 6 \cdot 10^{23}$, or Avogadro's number) and they react with one another in a bimolecular reaction, there is no need to write $(6\cdot 10^{23})(6\cdot 10^{23} - 1)$, but in our case with such small numbers the difference matters.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does absorption have $[\ce{H}]x_m$ but desorption have $mx_m$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As noted above, $m$ is a concentration term analogous to $[\ce{H}]$, but with different units.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, for the reaction rate, where did they get $m(m−1)$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why does the $k_\mathrm a$ term have $[\ce{H}]$ while the other terms do not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other terms use $m$ which, as explained above, is analogous to $[\ce{H}]$ in that they both reflect concentrations, $[\ce{H}]$ for hydrogen atoms in space and $m$ for hydrogen atoms on dust particles; they use different units as explained above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happened to the $mx_m$ that was mentioned above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the full rate expressions have been written out, they have been replaced with integers, 0, 1, or 2, representing the concentration of hydrogen atoms on a dust particle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further, how did they derive the rate of change for $x_1$ and $x_2$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rate of change of $x_1$ is the rate of formation minus the rate of destruction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of formation is given by $k_\mathrm a[\ce{H}]x_0$ (the rate of a bare dust particle picking up a hydrogen atom) plus $2k_\mathrm d x_2$ (the rate of a dust particle with two hydrogen atoms losing one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of destruction is given by $k_\mathrm a[\ce{H}]x_1 + k_\mathrm d x_1$ (the rate of a dust particle with one hydrogen atom either picking up one more, or losing it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the time being, I'll leave the rate of change of $x_2$ as an exercise for you to test yourself on what you've learned.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-03-25T09:34:15.537" LastActivityDate="2018-03-25T09:34:15.537" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14131" PostTypeId="1" CreationDate="2014-07-05T08:01:16.067" Score="8" ViewCount="998" Body="&lt;p&gt;In introductory chemistry books, courses, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many &quot;definitions&quot; contain words like: typically, most, often, likely, generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read since questions on this site mentioning different definitions of same concept in different books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just read about a concept where the explanation contained smth like &quot;some other chemists believe it should be the other way&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are there many non-rigorous definitions in introductory chemistry?&lt;/p&gt;&#xA;" OwnerUserId="7022" LastEditorUserId="73" LastEditDate="2014-09-04T17:18:11.780" LastActivityDate="2014-09-05T03:45:05.247" Title="Why is introductory chemistry so non-rigorous?" Tags="&lt;terminology&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14132" PostTypeId="1" AcceptedAnswerId="14134" CreationDate="2014-07-05T08:33:00.390" Score="4" ViewCount="1923" Body="&lt;p&gt;Question: Why is ionic lattice energy inversely proportional to the radius of the atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most heterogeneous covalent molecules are polar to some extent. The degree of polarity, or the dipole moment, depends on the difference in electronegativity difference between the two atoms. The larger the dipole moment, the higher the ionic character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know: Electronegativity decreases as you go downward in a group, however, the size increases, usually, as you go downwards in a group. Thus, ionic character will increase upon going downward, but the ionic lattice energy will decrease? This seems contradictory. Is this true? And if it is, why is it so?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastActivityDate="2015-04-30T13:59:08.007" Title="Chemical bonding - lattice energy" Tags="&lt;bond&gt;&lt;energy&gt;&lt;crystal-structure&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14133" PostTypeId="2" ParentId="14131" CreationDate="2014-07-05T10:56:55.127" Score="9" Body="&lt;p&gt;I like this merriam-webster definition of rigorous:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;done carefully and with a lot of attention to detail&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemistry is an experimental science there are a lot of patterns that work fine in most of the cases but not in all (e.g. the Octet rule) these are called &lt;strong&gt;rules of thumb&lt;/strong&gt;. Chemistry is however a rigorous science, simply there are exceptions that don't follow clear patterns or easy simplifications so they are harder to predict or to comprehend for a student in an introductory text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in all the science (physics, mathematics etc. etc.) there are some topics that are still debated, this don't mean that introductory chemistry is non-rigorous.&#xA;I suggest you to read some books about history of chemistry (e.g. Asmiov, A Short History of Chemistry), you will find that chemistry has been for most of his history (and it is in part now days) a trial and error research, followed by the attempt to find a theory, a pattern or a sense to the observations done, this doesn't mean that there isn't a rigorous approach but simply that reality don't fit the scheme that we propose so easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think maybe most of the text books don't emphasize in this aspect of chemistry, but most of the reasoning is &lt;a href=&quot;http://en.wikipedia.org/wiki/A_priori_and_a_posteriori&quot;&gt;&lt;em&gt;a posteriori&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2094" LastEditorUserId="957" LastEditDate="2014-07-05T19:27:22.003" LastActivityDate="2014-07-05T19:27:22.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14134" PostTypeId="2" ParentId="14132" CreationDate="2014-07-05T11:29:03.337" Score="7" Body="&lt;p&gt;Considering your doubt, from &lt;strong&gt;&lt;em&gt;Born-Lande&lt;/em&gt;&lt;/strong&gt; equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$N_\text{A}$ = &lt;em&gt;&lt;strong&gt;Avogadro constant&lt;/strong&gt;&lt;/em&gt;;&lt;br&gt;&#xA;      $M$ = &lt;em&gt;&lt;strong&gt;Madelung&lt;/strong&gt;&lt;/em&gt; constant, relating to the geometry of the crystal;&lt;br&gt;&#xA;      $z^+$ = charge number of cation&lt;br&gt;&#xA;      $z^-$ = charge number of anion&lt;br&gt;&#xA;      $e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$&lt;br&gt;&#xA;      $ε_0$ = permittivity of free space&lt;br&gt;&#xA;      $4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$&lt;br&gt;&#xA;      $r_0$ = distance to closest ion&lt;br&gt;&#xA;      $n$ = &lt;em&gt;&lt;strong&gt;Born&lt;/strong&gt;&lt;/em&gt; exponent, typically a number between 5 and 12, determined experimentally &#xA;      by measuring the compressibility of the solid, or derived theoretically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And what you are assuming the radius is in fact the distance between the two atoms.&lt;br&gt;&#xA;As the distance &lt;strong&gt;increases&lt;/strong&gt;, the attraction &lt;strong&gt;decreases&lt;/strong&gt; from &lt;strong&gt;&lt;em&gt;Coulomb's law&lt;/em&gt;&lt;/strong&gt; :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thus there is &lt;strong&gt;less&lt;/strong&gt; energy involved in breaking the lattice, so lattice energy &lt;strong&gt;decreases&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As you move down, the size &lt;strong&gt;increases&lt;/strong&gt; and the valence electron in the outermost shell experiences &lt;strong&gt;lesser&lt;/strong&gt; repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ &lt;strong&gt;increases&lt;/strong&gt;, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also as charges $z^+$, $z^-$ &lt;strong&gt;increases&lt;/strong&gt;, the magnitude of lattice energy does also &lt;strong&gt;increase&lt;/strong&gt; as can be seen clearly from the &lt;em&gt;&lt;strong&gt;Born-Lande&lt;/strong&gt;&lt;/em&gt; equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{r|c|l}&#xA;&amp;amp;&amp;amp;\text{Ionic Solid}&amp;amp;\text{Lattice Energy (kJ/mol)}\\\hline&#xA;&amp;amp;\text{Variation of}&amp;amp;\ce{NaF}&amp;amp;926\\&#xA;&amp;amp;r_0 \text{ i.e. distance}&amp;amp;\ce{NaCl}&amp;amp;786\\&#xA;&amp;amp;\text{between ions}&amp;amp;\ce{NaBr}&amp;amp;752\\&#xA;&amp;amp;&amp;amp;\ce{NaI}&amp;amp;702\\\hline&#xA;&amp;amp;\text{Variation of}&amp;amp;\ce{NaF}&amp;amp;926\\&#xA;&amp;amp;z^+,z^- \text{ i.e. charge}&amp;amp;\ce{MgO}&amp;amp;3800\\&#xA;&amp;amp;\text{on ions}&amp;amp;\ce{Al2O3}&amp;amp;15900\\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T13:59:08.007" LastActivityDate="2015-04-30T13:59:08.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14136" PostTypeId="2" ParentId="14116" CreationDate="2014-07-05T14:03:12.893" Score="0" Body="&lt;p&gt;I suggest you consider the following:&#xA;$$\ce{HA-&amp;gt;H+ +A-}\\&#xA;\begin{array}{c|c|}\hline&#xA;&amp;amp;&amp;amp;HA&amp;amp;H^+&amp;amp;A^-\\\hline&#xA;&amp;amp;\text{initial}&amp;amp;C&amp;amp;0&amp;amp;0\\\hline&#xA;&amp;amp;\text{final}&amp;amp;C(1-\alpha)&amp;amp;C\alpha&amp;amp;C\alpha\\\hline&#xA;\end{array}$$&#xA;where $C$ is initial concentration and $\alpha$ ($\le1$) is dissociation constant of the acid.&#xA;In equilibrium:&#xA;$$K_a=\frac{[H^+][A^-]}{[HA]}=\frac{C\alpha^2}{1-\alpha}$$&#xA;As $\alpha\ll1$, &#xA;$$K_a\approx C\alpha^2\implies\alpha=\sqrt{\frac{K_a}C}\implies[H^+]=\sqrt{K_aC}$$&#xA;$$\LARGE pH=-\frac12\log{(K_aC)}\;\small\text{for weak acids}$$ &lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4934" LastEditDate="2014-07-05T14:53:13.740" LastActivityDate="2014-07-05T14:53:13.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14137" PostTypeId="2" ParentId="14077" CreationDate="2014-07-05T14:41:30.023" Score="0" Body="&lt;p&gt;The two problems are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You must have an &quot;power-intercept&quot; with inverse-x as&#xA;  $$\ y(x)=e^{a-(b/x)}$$&#xA;  Important but not necessary, as to provide a limiting value $e^a$ to $y$ at $\infty$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also what you are worried about the derivatives is simple:&#xA;  $$y'(x)=\frac{b}{x^2}e^{a-(b/x)}$$&#xA;  $$y''(x)=\frac{b(b-2x)}{x^4}e^{a-(b/x)}$$with second derivate a root at $x=\frac b2$&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An example graph with two resolutions:&#xA;$$\text{legend: }[\color{red}{y(x)}\;,\color{purple}{y'(x)}\;,\color{black}{y''(x)}]\;\text{ with }a=2.5,b=2.2$$&#xA;So the derivative changes the sign at $1.1$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vfk4G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xeYaB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-07-05T14:41:30.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14138" PostTypeId="2" ParentId="14116" CreationDate="2014-07-05T14:44:29.317" Score="2" Body="&lt;p&gt;$K_a$ is not unitless (in this case). Some equilibrium constants may have their units totally cancel, but the acidity constant has units of concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_a = \dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\implies\dfrac{\frac{\text{mol}}{\text{L}}\cdot\frac{\text{mol}}{\text{L}}}{\frac{\text{mol}}{\text{L}}}=\frac{\text{mol}}{\text{L}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used moles per liter (molarity) as it is a common concentration unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $K_a$ and $[\ce{H+}]$ both have concentration units, as does $C_a$, so your initial equation $[\ce{H+}]^2 + K_a [\ce{H+}]+ K_a C_a =0$ is valid (at least for units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you recast $K_a$ and $C_a$ in terms of activities, then you need to replace $[\ce{H+}]$ with $\alpha_{\ce{H+}}$ (which is unitless).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-05T14:44:29.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14139" PostTypeId="2" ParentId="14131" CreationDate="2014-07-05T14:57:59.113" Score="9" Body="&lt;p&gt;The problem you are encountering has two origins:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Historically, chemistry has accumulated many models for chemical behavior that work most of the time. Some of these models are based on completely invalid theory (VSEPR), but they provide a set of rules that work. Some are based on decent theory (the octet rule, hybridization), but generate inconsistencies because the underlying theory is incomplete. Most of these models work in most cases, so they remain because they are easy to use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Higher level theories/models exist that contain less inconsistencies and fewer exceptions, but the maths are &lt;strong&gt;hard&lt;/strong&gt;. At the simplest, these models require multivariable calculus, linear algebra, and/or differential equations. At the more complex, we are talking about a computer spending 30 minutes (or more) iterating through the matrix approximations to many-electron wavefunctions (for which there are still no exact solutions) to find the best self-consistent minimum-energy solution. We're talking about problems in theoretical chemistry that used to consume one person's PhD thesis to just derive one proof and complete one calculation (or a small group of related calculations).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The field of chemistry has chosen to be more accessible to beginners by allowing the continuation of such imperfect models. There are a number of reasons for this. First, it allows students to progress through the more qualitative chemistry courses while they work up to the math requirements. Second, it allows students of other sciences (especially biology and medicine) to develop a conceptual understanding of chemistry (which they need far more than the calculus-based understanding).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2014-07-05T19:23:46.160" LastActivityDate="2014-07-05T19:23:46.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14140" PostTypeId="1" AcceptedAnswerId="14143" CreationDate="2014-07-05T16:25:01.190" Score="2" ViewCount="180" Body="&lt;p&gt;The methyl radical, $\ce{.CH3}$, is a strong oxidant and a strong reductant according to Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so? Any additional reasons to the ones below? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that an oxidant itself is reduced; an oxidant helps something else lose electrons, so the oxidant must be the one gaining electrons. Given that the methyl radical has an unpaired electron, it is quite reasonable that it would be amenable to having an extra electron to pair up with its lone electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that a reductant itself is oxidized; a reductant helps something else gain electrons. The carbon in the methyl radical has an oxidation state of -3, and I know that carbon can adopt a variety of higher oxidation states, such as +3 in acetic acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-05T19:18:10.067" Title="Redox properties of methyl radical" Tags="&lt;redox&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14141" PostTypeId="1" AcceptedAnswerId="14142" CreationDate="2014-07-05T18:33:55.683" Score="13" ViewCount="10564" Body="&lt;p&gt;Doing a first year chem class.&lt;br&gt;&#xA;Just read through the molecular naming of compounds and now I'm confused as to why $\ce{CaCl2}$ is called calcium chloride and not calcium dichloride?  &lt;/p&gt;&#xA;" OwnerUserId="7029" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2016-10-01T06:22:01.820" LastActivityDate="2017-05-19T22:51:16.527" Title="Why is CaCl2 called calcium chloride?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;ionic-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14142" PostTypeId="2" ParentId="14141" CreationDate="2014-07-05T18:41:25.367" Score="9" Body="&lt;p&gt;Elements from the first 2 columns only have 1 oxidation state, so any compound they form, it is easy to determine the number of anions.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-05T18:41:25.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14143" PostTypeId="2" ParentId="14140" CreationDate="2014-07-05T19:18:10.067" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is this so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is a &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html&quot; rel=&quot;nofollow&quot;&gt;tabulation&lt;/a&gt; of the ionization potentials (IP) and electron affinities (EA) for some alkali metals and halogens (kJ/mol):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|c|c|c|}\hline&#xA;&amp;amp; IP &amp;amp; EA &amp;amp; &amp;amp;&amp;amp; IP &amp;amp; EA          \\ \hline&#xA;F &amp;amp; 1681 &amp;amp; 331 &amp;amp;&amp;amp; Li &amp;amp; 520 &amp;amp; 60  \\ \hline&#xA;Cl &amp;amp; 1251 &amp;amp; 352 &amp;amp;&amp;amp; Na &amp;amp; 496 &amp;amp; 53 \\ \hline&#xA;Br &amp;amp; 1140 &amp;amp; 328 &amp;amp;&amp;amp; K &amp;amp; 419 &amp;amp; 49  \\ \hline&#xA;I &amp;amp; 1008 &amp;amp; 298 &amp;amp;&amp;amp; Rb &amp;amp; 403 &amp;amp; 47  \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reducing agent readily donates electrons and would be expected to have a low IP.  Sodium would be expected to act as a good reducing agent and chlorine as a poor one, and, indeed the IP of sodium is much lower than that of chlorine.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good oxidizing agent readily accepts electrons and would be expected to have a large EA.  Chlorine would be expected to act as a good oxidizing agent and sodium as a poor one, and, indeed, the EA of chlorine is much higher than that of chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C2229074&amp;amp;Mask=20&quot; rel=&quot;nofollow&quot;&gt;This NIST webpage&lt;/a&gt; tabulates experimental IP's and EA's for the methyl radical.  The IP is given as (9.84 eV) 945 kJ/mol and the EA appears to be less than (0.6 eV) 58 kJ/mol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare the IP and EA of the methyl radical to values in the above table for strong oxidizing and reducing agents.  The IP of the methyl radical is significantly larger than those reported for the alkali metals and a bit less, but relatively close to the IP's for all of the halogens except fluorine. This comparison suggests that methyl radical is, at best, a weak reducing agent. The EA of methyl radical is significantly less than those reported for the halogens and right around those reported for the alkali metals. This comparison suggests that methyl radical is, at best, a very weak oxidizing agent.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-05T19:18:10.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14144" PostTypeId="2" ParentId="14141" CreationDate="2014-07-05T19:26:40.437" Score="5" Body="&lt;p&gt;Some names are just used out of habit. &quot;Calcium chloride&quot; has been the common name for so long, that it has simply become common usage.  According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_chloride&quot;&gt;this Wikipedia article&lt;/a&gt;, calcium dichloride is an acceptable &quot;other&quot; name for the compound.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-05T19:26:40.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14145" PostTypeId="1" AcceptedAnswerId="14146" CreationDate="2014-07-05T20:17:59.777" Score="1" ViewCount="2145" Body="&lt;p&gt;What is Markovnikov's rule? How can we use it to predict the products of additions to alkenes? And is it always obeyed, or are there exceptions to this rule?&lt;/p&gt;&#xA;" OwnerUserId="7030" LastEditorUserId="16683" LastEditDate="2018-10-19T00:22:10.983" LastActivityDate="2018-10-19T00:29:51.760" Title="What is Markovnikov's rule?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;&lt;regioselectivity&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14146" PostTypeId="2" ParentId="14145" CreationDate="2014-07-05T20:57:36.513" Score="8" Body="&lt;p&gt;Markovnikov's rule is used to predict how (for example) HX, where X is a halide, adds to an unsymmetrical olefin.  As the diagram below shows, Markovnikov addition follows the rule that the &quot;H&quot; will add to the least substituted carbon while the &quot;X&quot; will add to the most substituted carbon.  This is due to the fact that these reactions proceed by protonaton of the olefin to yield a carbocation.  Protonation at the least substituted carbon yields the most stable carbocation (in this case a secondary carbocation).  Finally the carbocation reacts with &lt;span class=&quot;math-container&quot;&gt;$\ce{X^-}$&lt;/span&gt; to yield the final product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TLRwB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TLRwB.png&quot; alt=&quot;Examples of Markovnikov and anti-Markovnikov additions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases the Markovnikov rule is not followed.  Such additions to olefins are termed &quot;anti-Markovnikov&quot;.  These reactions typically proceed through a radical intermediate rather than a carbocation.  For example, addition to the olefin might be initiated by the addition of a bromine radical to generate the most stable radical intermediate, which would place the radical center at the most highly substituted carbon. Addition of a hydrogen radical completes the process and generates the final product which has the reverse addition pattern as compared to the Markovnikov addition product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-10-19T00:29:51.760" LastActivityDate="2018-10-19T00:29:51.760" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14147" PostTypeId="1" CreationDate="2014-07-06T02:41:59.823" Score="3" ViewCount="352" Body="&lt;p&gt;I came from a birthday party and someone gave me a sundae. I'm already full so I just decided to eat it at home. After arriving at home the sundae melted. So I left it in the freezer overnight. When I looked at my sundae it just formed ice crystals. It didn't went back to its original form. Why is that?. Why is it when I do this to ordinary ice cream it goes back to it's original form?&lt;/p&gt;&#xA;" OwnerUserId="7034" LastEditorUserId="4945" LastEditDate="2014-07-06T06:15:48.413" LastActivityDate="2014-07-06T06:15:48.413" Title="Why does ice cream not go back to its original form upon refreezing?" Tags="&lt;thermodynamics&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14148" PostTypeId="1" CreationDate="2014-07-06T05:07:04.850" Score="1" ViewCount="277" Body="&lt;p&gt;While sleeping on an air mattress yesterday a thought occurred to me. I had to re-inflate the air mattress since it was getting rather low and after it had a good stiffness to it, I went to sleep. The next morning I noticed that a lot of the pressure that I put into the air mattress had left and I remembered Le Chatelier's principle which says that gases in a closed system will convert to the least number of molecules possible if the pressure is increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is my thought:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Breathing into the air mattress had added moles of gas to a closed system (the air mattress).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lying on the air mattress increased the pressure of the system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given Le Chatelier's Principle, the gases should have transitioned to the lowest number of moles to relieve the pressure introduced on the system. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The thing I'm not sure about is if there are any equilibrium reactions for the gases in our atmosphere that will actually do this. Considering the makeup of our atmosphere, does Le Chatelier's Principle apply to the air inside an air mattress?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastActivityDate="2014-07-06T05:11:20.620" Title="Does Le Chatelier's principle apply to the gas inside an air mattress?" Tags="&lt;equilibrium&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14149" PostTypeId="2" ParentId="14148" CreationDate="2014-07-06T05:11:20.620" Score="4" Body="&lt;p&gt;No, the air we breathe is about 78% nitrogen, 21% oxygen, 0.5% carbon dioxide and other gases. Nitrogen doesn't react with anything at room temperature and there is nothing to burn inside the air mattress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What probably happened was that there is a very small hole that the air leaked out. About the only way to find the hole would be to submerge the air mattress in a pool.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-06T05:11:20.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14150" PostTypeId="1" AcceptedAnswerId="14156" CreationDate="2014-07-06T05:25:29.150" Score="10" ViewCount="16748" Body="&lt;p&gt;I understand the principle behind an eraseable pencil. Graphite is highly slippery, you rub it onto the paper, and an eraser can rub it off again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had friends at school with 'eraseable ink pens'. The ink was 'special' and the eraser was 'special' and it could erase the pen in the same way a regular eraser could erase pencil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is - &lt;strong&gt;how do eraseable pens work?&lt;/strong&gt; (Ie what substances are involved and what are their properties?)&lt;/p&gt;&#xA;" OwnerUserId="4478" LastEditorUserId="4945" LastEditDate="2014-07-06T06:10:47.643" LastActivityDate="2020-07-14T11:10:21.987" Title="How do erasable pens work?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14151" PostTypeId="2" ParentId="14147" CreationDate="2014-07-06T05:49:38.263" Score="7" Body="&lt;p&gt;Ice cream is a frozen emulsion mixed with air. When the ice cream melted, the air escaped. When you re-froze it, it formed a solid mass.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-06T05:49:38.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14152" PostTypeId="1" CreationDate="2014-07-06T07:02:56.843" Score="10" ViewCount="3208" Body="&lt;p&gt;I am in Papua New Guinea at the moment, where rain water harvesting is the most common way of getting water.&#xA;When washing hands, the soap feels hard to rinse off, and my skin feels &quot;greasy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read around a bit &lt;a href=&quot;http://www.madsci.org/posts/archives/jul2001/996090332.Ch.r.html&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://chemistry.about.com/od/howthingsworkfaqs/a/softwaterrinse.htm&quot;&gt;here&lt;/a&gt; and I think I understand what is said: essentially, sodium stearate in the soap separates into sodium and stearate ions, which then precipitate with other ions (like Calcium, into calcium stearate) and get rinsed away. But in very soft water, there are sodium ions rather than calcium ones, so this reaction is slowed down a lot, resulting in the greasy feeling.&#xA;As it turns out, I have access to hard ground water at the same time, and the difference is very obvious: with hard water, soap is rinsed off instantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question now: I assume rain water has in fact almost no ions in it (I'm thinking Sodium ions mostly), and I would therefore assume that the Sodium stearate dissolves very easily, but has no replacement ion to precipitate with, therefore sticks to the skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the right explanation or are there other factors at play in the explanation about that greasy feeling?&lt;/p&gt;&#xA;" OwnerUserId="7035" LastActivityDate="2014-07-06T07:36:09.343" Title="Soap and rain water, why does it not rinse off?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;surfactants&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14153" PostTypeId="2" ParentId="14152" CreationDate="2014-07-06T07:36:09.343" Score="6" Body="&lt;p&gt;Yes, this is a good explanation. Rain water is almost pure water, it lacks of bivalent ions such as $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ which helps soap to rinse of. You will get the same effect when a water softener is installed on water distribution.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastActivityDate="2014-07-06T07:36:09.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14154" PostTypeId="2" ParentId="14060" CreationDate="2014-07-06T09:21:04.793" Score="3" Body="&lt;p&gt;This answer aims to build on the general approach that Martin has provided, which overall makes a reasonable summation based on the data provided. IR is not really my specialty, but there is some more information that we can get out of the NMR data which should be helpful, and more reliable (in my opinion) than the IR data. Unfortunately, I am away away from my office for the next week, so cannot provide immediate references to support some statements here, so you'll have to take some things on face value. Let's begin with an overall summary of what data we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A partial 1H NMR spectrum, with only some of the peaks integrated. This is an expanded region of what we can assume to be a 500MHz (based on the export path). The linewidths are broad, and there is no clear source to allow confirmation of correct calibration. We therefore need to make two assessments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The calibration is incorrect, and the peak at 7.15, which has no integration, is in fact the residual CHCl3, and all chemical shifts need to adjust downfield (0.09-0.11 depending on what value for CHCl3 in CDCl3 you use; I use 7.26. For simplicity, let's adjust the chemical shifts downfield by +0.1ppm&lt;/li&gt;&#xA;&lt;li&gt;The calibration is correct, in which case the peak at 7.15 cannot be discounted, and should therefore have its integral determined. &#xA;A full display NMR spectrum would be very useful here to look for underlying exchange broadened proton signals. This might occur anywhere from about 2-15ppm, and may be very broad such that they appear as a hump in the baseline, but even in CDCl3, we should see them, and &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An IR spectrum which looks to have been run at pretty low concentration. I did not see your original IR spectrum, and wonder why you needed to redo it. What would be nice to know is whether the ratio of intensities for your absorbance peaks are the same for both IR data sets; particularly did the ratio of the broad stretch at 3422 change with respect to absorbances at 3019, 763 and 692?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your sample is a solid, as you mention in one of your comments. This would be a useful peice of information to have from the start. It is soluble in dichloromethane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So let's now start with collating information from the data provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IR Spectrum.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the significant signal is the broad peak at 3422, and this is textbook-indicative of an O-H stretch. It's probably a little too high to consider a N-H group of any sort. So, it could be an alcohol or an acid, but we have no C=O peak, so it leaves us with an -OH group.&#xA;We have absorbances at 3019, 763 and 692; all indicative of an aromatic. More specifically, 763 and 692 are indicative of a mono-substituted benzene ring.&#xA;As I say though, IR is not really my thing, and that's about all I can get from this spectrum. My biggest concern is the reliability of the OH peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NMR Spectrum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IhcbZ.png&quot; alt=&quot;Mono-substituted benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.39(2H, dd, H3) and 7.55(2H, d, H2). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and &amp;lt;1 for para. Looking at the H2 signal at 7.55, we can use our knowledge of coupling constants to determine the frequency of the spectrometer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.5615-7.5461 = 0.0154 x Spect.Freq ~ 7.5Hz =&gt; 487MHz, so close enough to 500MHz, and confirms our suspicions that it is a 500MHz, as the export path suggests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can calculate an accurate ortho coupling for H2-H3 to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.5615-7.5461 = 0.0154 x 500 = 7.7Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, mono-substituted benzene rings have been extensively studied and are very well understood; chemical shift data has been widely tabulated, and forms the basis for many chemical shift prediction algorithms. Starting with the benzene chemical shift (7.34ppm) as a basis, it is possible to use the shifts of each group to infer some information about the type of substituent.  The first thing to look for with this type of system is the order of H2 versus H3 (versus naked benzene). An electron-donating group increases shielding, and the ortho proton (H2) is typically found upfield of the meta proton (H3). Electron withdrawing groups decrease shielding, and H2 typically experiences a downfield shift from benzene, and usually resonates downfield from the meta (H3) proton. For the system you have, H2 is downfield of H3, and this is indicative of an electron-withdrawing group.  This is a very strong argument against this system being phenol. Phenol has its H2 protons upfield of H3. This is also what is so confusing about the IR spectrum you have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's now consider the possible structure for this unknown compound you have. Let's make the assumption that, as a homework/tutorial problem, this is going to be a fairly simple molecule, with a pretty common substituent. Remember we have two scenarios to consider for our NMR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 1 (corrected for CHCl3 at 7.26ppm): the substituents come at H2 (+0.21), H3 (+0.05), H4 (-0.04). Looking at &lt;a href=&quot;http://link.springer.com/book/10.1007%2F978-3-540-93810-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pretsch, Buhlmann and Badertscher&lt;/a&gt;, this matches incredibly well for the substituent being a phenyl group [H2 (+0.22), H3 (+0.06), H4 (-0.04)]. This would give the structure biphenyl, a white solid, which has a reported H2-H3 coupling of 7.71Hz. Very strong evidence by NMR, but is not supported by -OH stretch in IR data, although all other IR data is in agreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario 2 (spectrum already correctly calibrated): If we assume that the spectrum is correctly calibrated, then the CHCl3 residual peak comes under the H4 signal - probably could be the sharp peak which is the second peak from the right in this group. This means that the peak at 7.15 needs to be considered. By eye, its integral is roughly 1. A singlet of chemical shift of 7.15 is typical of a bis-halide, and so we could consider &amp;alpha;,&amp;alpha;-dichlorotoluene or &amp;alpha;,&amp;alpha;-dibromotoluene. Both are sufficiently electron withdrawing to give H2 downfield of H3, and  However, the former is definitely a liquid at room temp, and I suspect the latter is also. Therefore, not strong candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to have seen the original IR spectrum, and the full NMR spectrum to have confidence in any prediction. However, if I were just shown the NMR data, I would have confidence in predicting the structure as biphenyl. I hope you can provide the real solution to this eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rikdq.png&quot; alt=&quot;Possible structures&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-07-06T09:21:04.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14156" PostTypeId="2" ParentId="14150" CreationDate="2014-07-06T10:11:35.977" Score="9" Body="&lt;p&gt;Instead of a typical oil/dye ink,  erasable pens use a compound closer to rubber cement, which acts like a liquid when a bit of pressure is exerted. &lt;a href=&quot;http://pilotpen.us/categories/frixion-erasable-ink-pens/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A bit of history and a link to patents can be found &lt;a href=&quot;http://home.howstuffworks.com/pen5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;EDIT: Be sure to check out the other answer here about thermochromic ink too.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="3704" LastEditDate="2020-07-14T11:10:21.987" LastActivityDate="2020-07-14T11:10:21.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14157" PostTypeId="2" ParentId="14090" CreationDate="2014-07-06T11:10:58.310" Score="0" Body="&lt;p&gt;As a user mentioned, not all acids dissociate that way. Sometimes, there is also the production of Hydronium ions.&#xA;Refer : &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/water.php&quot; rel=&quot;nofollow&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/water.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O~(l) &amp;lt;=&amp;gt; H3+O~(aq) + {}^{-}OH~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydronium ions react with the sulphate radical, forming an acid like entity. Hence, the acidity is not lost.&lt;/p&gt;&#xA;" OwnerUserId="7037" LastEditorUserId="4945" LastEditDate="2014-07-06T12:39:49.063" LastActivityDate="2014-07-06T12:39:49.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14158" PostTypeId="2" ParentId="13881" CreationDate="2014-07-06T11:14:56.770" Score="1" Body="&lt;p&gt;Other than platinum, you can use Gold electrodes, but those are pretty expensive. So unless you use &lt;a href=&quot;https://en.wikipedia.org/wiki/Aqua_Regia&quot; rel=&quot;nofollow&quot;&gt;Aqua Regia&lt;/a&gt; as electrolyte, the corroding is negligible.&lt;/p&gt;&#xA;" OwnerUserId="7037" LastEditorUserId="1677" LastEditDate="2014-07-07T18:46:31.067" LastActivityDate="2014-07-07T18:46:31.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14159" PostTypeId="1" CreationDate="2014-07-06T16:39:28.973" Score="0" ViewCount="6191" Body="&lt;p&gt;My teacher told yes that salt is paramagnetic and it is attracted by magnet slightly. But when I tried that at home it did not work.&lt;/p&gt;&#xA;" OwnerUserId="7039" LastEditorUserId="9961" LastEditDate="2015-09-17T18:35:08.047" LastActivityDate="2019-01-15T22:58:13.520" Title="Is NaCl paramagnetic?" Tags="&lt;inorganic-chemistry&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14160" PostTypeId="2" ParentId="14141" CreationDate="2014-07-06T16:42:53.723" Score="-3" Body="&lt;p&gt;Calcium chloride is an ionic compound so cation is named first followed by anion. Do not get confused with carbon dioxide and sulfur dioxide. They are not ionic compounds.&lt;/p&gt;&#xA;" OwnerUserId="6839" LastEditorUserId="402" LastEditDate="2017-05-19T22:51:16.527" LastActivityDate="2017-05-19T22:51:16.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14161" PostTypeId="2" ParentId="14159" CreationDate="2014-07-06T17:02:10.193" Score="6" Body="&lt;p&gt;Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins. &#xA;The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-06T17:02:10.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14162" PostTypeId="1" AcceptedAnswerId="14163" CreationDate="2014-07-06T17:38:43.303" Score="5" ViewCount="1029" Body="&lt;p&gt;I had a chemistry lab where we had to titrate three solutions with &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;: a buffer, a diluted buffer and deionized water. My partner and I didn't label the charts we made, so now I have to figure out which one is which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see is that from the Henderson-Hasselbalch equation:&#xA; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}\ce{H} = \mathrm{p}K_a + \log\left(\ce{[A^{-}]/[HA]}\right)$&lt;/span&gt;,&#xA; the diluted buffer should have a larger slope at the beginning then the original buffer solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is my logic correct, and how do I differentiate between the original buffer and the deionized water?&lt;/p&gt;&#xA;" OwnerUserId="7040" LastEditorUserId="83758" LastEditDate="2020-02-20T21:06:45.827" LastActivityDate="2020-02-20T21:06:45.827" Title="Differentiating between Titration, Buffer, diluted buffer, Deionized H₂O" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="14163" PostTypeId="2" ParentId="14162" CreationDate="2014-07-06T18:22:37.123" Score="2" Body="&lt;p&gt;What is the expected pH of deionized water and what would the pH be after adding a few drops of acid? For both buffer solutions, there is an equivalence point, how would you use that to tell them apart? Usually deionized water is slightly acidic. I kind of mentioned that water doesn't have an equivalence point. Is there a graph without one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of a buffer? A diluted buffer would have the same characteristic, but in a smaller amount. Water would be missing this characteristic. You should be able to figure out which graph is which now.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-06T18:22:37.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14164" PostTypeId="1" AcceptedAnswerId="14167" CreationDate="2014-07-06T18:23:40.233" Score="0" ViewCount="529" Body="&lt;p&gt;I just want to know, what was the procedure through which the value of one mole of a substance was considered as a an aggregate of $6.023\cdot10^{23}$ particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, how do we calculate the volume occupied by one mole of gas?&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="4945" LastEditDate="2014-07-07T07:21:45.277" LastActivityDate="2014-07-07T07:21:45.277" Title="How can we say that one mole of a substance contains avogadro number of particles?" Tags="&lt;physical-chemistry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14165" PostTypeId="1" AcceptedAnswerId="14168" CreationDate="2014-07-06T18:54:08.933" Score="24" ViewCount="11595" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The “real way” boric acid behaves as an acid in water is first by acting as a Lewis acid (bottom reaction, in black).  In fairness to the undergrads, I’d bet &lt;strong&gt;&gt;25% of graduate students around the country would not be able to correctly explain the acidity of boric acid.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How are we supposed to know that boric acid heterolytically cleaves water as to leave a proton behind? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this something that is supposed to be immediately obvious? Or do we just have to read about experiments and then realize that this is how boric acid behaves as an acid - as exclusively a Lewis acid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other structurally analogous acids such as sulfuric acid also have a electrophilic center atom because of the electron-withdrawing oxygens, but these acids do not behave the way that boric acid does; these acids simply donate protons - they don't chop up water molecules to create protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a hint that $\ce{B(OH)_3}$ behaves this way found in the fact that boron can form either three or four bonds with elements? Should this clue us off to the possibility of $\ce{B(OH)_4^-}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is another hint found in the fact that boric acid is trigonal planar with an unused p-orbital? Sulfuric acid on the other hand doesn't have any empty orbitals (well, it might have empty d-orbitals which may or may not be accessible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MfJSk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfJSk.gif&quot; alt=&quot;boric acid acid-base mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T07:41:31.687" LastActivityDate="2020-08-17T16:39:48.337" Title="The acidity of boric acid" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14166" PostTypeId="2" ParentId="14165" CreationDate="2014-07-06T19:15:03.170" Score="4" Body="&lt;p&gt;At least according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Boric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wkipedia article&lt;/a&gt; (look in the &quot;Properties&quot; section) there is some disagreement as to the basis for boric acid's acidic properties.  One set of experiments seems to favor the Lewis acid interpretation shown above in black, while the other camp suggests that the data supports the view that boric acid is both a Lewis acid and a Brønsted acid (both the blue and the black mechanisms shown above are operative). &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-04T07:41:01.187" LastActivityDate="2018-03-04T07:41:01.187" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14167" PostTypeId="2" ParentId="14164" CreationDate="2014-07-06T19:35:15.830" Score="5" Body="&lt;p&gt;There's a nice &lt;a href=&quot;http://www.scientificamerican.com/article/how-was-avogadros-number/&quot;&gt;Scientific American article&lt;/a&gt; that provides an interesting historical perspective on Avogadro's Number.  They go on to point out that one of the early accurate determinations was based on the first accurate measurement of the charge on an electron.  The charge on a mole of electrons (1 Faraday) had been known for some time.  Dividing the latter by the former yielded Avogadro's Number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the volume of 1 mole of gas at STP, imagine the following experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) weigh out 1 mole of oxygen gas (32 gm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) fill a balloon with this weight of gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) attach a pressure gauge to the balloon and expand or contract the balloon's volume until the gauge reads 1 atmosphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Immerse the balloon in a water tank and measure the volume of water displaced - 22.4 liters&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-06T19:35:15.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14168" PostTypeId="2" ParentId="14165" CreationDate="2014-07-06T20:19:35.790" Score="15" Body="&lt;p&gt;I think your suppositions and observations are basically correct. On the basis of general principles, it is certainly possible to imagine that such a mechanism might be at play. If one considers the structure of boric acid, it should be readily apparent that the boron atom &lt;em&gt;could&lt;/em&gt; be quite electrophilic and hence act as a Lewis acid: it possesses a formally vacant $p$-orbital; it is bound to highly electronegative oxygen atoms, which withdraw electron density and create a partial positive $(\delta^+)$ charge on it (though this effect may be mitigated by the oxygen lone pairs, which can donate electron density into the aforementioned $p$-orbital); and finally, it has a planar geometry that makes it quite sterically accessible to approaching nucleophiles. Such a mechanism is also lent plausibility by the well-known propensity of various other boron compounds for forming Lewis adducts and being highly susceptible to hydrolysis (where applicable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of that having been said, I don't think it's fair to claim that such a mechanism is &quot;obviously&quot; correct for the reasons given above. Such arguments are strictly qualitative, and while they're very suggestive of the possibility, they are in no sense definitive. The only way to even approach something vaguely approximating certainty is to perform experiments (perhaps in concert with theoretical computations), and even there considerable difficulty exists in reaching firm conclusions. The confirmed presence of $\ce{B(OH)4-}$ under one particular set of conditions is not sufficient proof that the mechanism involving boron as a Lewis acid is always operative, nor does it rule out other pathways. (Note that I'm not necessarily disputing that mechanism, but only pointing out that extrapolating too widely from specific experimental results has certain liabilities.) As ron points out in his answer, the issue apparently isn't settled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I would also add that there are arguments to be made against such a mechanism. There's likely some degree of $\pi$-bonding between the oxygen lone pairs and the boron, which would be disrupted by the pyramidalization of the molecule upon formation of the intermediate Lewis adduct resulting from nucleophilic attack by $\ce{H2O}$. The associated energy penalty might be relevant to the reaction kinetics under some conditions. My instinct is that this is &lt;em&gt;probably&lt;/em&gt; not significant, but it may be a factor worth considering. Of course, this is all speculative on my part, though I think it further demonstrates that qualitative reasoning in chemistry is rarely clear-cut, and there's an awful risk of falling prey to fuzzy thinking. Hence, rigorous experimentation and calculation are vital.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="11367" LastEditDate="2017-04-08T16:20:23.823" LastActivityDate="2017-04-08T16:20:23.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14169" PostTypeId="1" CreationDate="2014-07-06T22:07:48.923" Score="5" ViewCount="383" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Ten $\mathrm{kg}$ of steam at $500~\mathrm{bar}$ is expanded at constant pressure until its volume increases to seven times its initial value of $0.01~\mathrm{m^3}$.&lt;br&gt;&#xA;  (a) Calculate the initial and final temperature of the steam&lt;br&gt;&#xA;  (b) Calculate the heat that must be supplied to carry out the process&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An attempt at solving this problem:&lt;br&gt;&#xA;$$\Delta H + \Delta E_k + \Delta E_p = Q - W_S\ \ \text{(open system)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking out unnecessary terms leaves us with&lt;br&gt;&#xA;\begin{aligned}&#xA;\Delta H &amp;amp;= Q\\ &#xA;500~\mathrm{bar} &amp;amp;= 5\cdot10^7~\mathrm{Pa}\\&#xA;\Delta H &amp;amp;= H_f - H_i\\&#xA;H_i &amp;amp;= (5\cdot10^7~\mathrm{Pa})\cdot(0.01~\mathrm{m^3}) = 5\cdot10^5~\mathrm{J}\\&#xA;H_f &amp;amp;= (5\cdot10^7~\mathrm{Pa})\cdot(0.07~\mathrm{m^3}) = 3.5\cdot10^6~\mathrm{J}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to kJ/kg (10 kg of steam)&#xA;\begin{aligned}&#xA;\hat{H}_i &amp;amp;= 50~\mathrm{kJ/kg}\\&#xA;\hat{H}_f &amp;amp;= 350~\mathrm{kJ/kg}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's where I stopped. I tried looking up the values of $\hat{H}_i$ and $\hat{H}_f$ in steam tables to interpolate the temperatures, but the values are too low. Since the pressure is too high, I also know the ideal gas law $(PV=nRT)$ won't apply to this scenario. Can someone point me in the right direction? &lt;/p&gt;&#xA;" OwnerUserId="7041" LastEditorUserId="4945" LastEditDate="2019-05-20T13:33:08.223" LastActivityDate="2019-05-20T13:33:08.223" Title="How to calculate initial and final temperatures during supercritical steam expansion?" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14170" PostTypeId="1" AcceptedAnswerId="14171" CreationDate="2014-07-07T04:48:44.817" Score="10" ViewCount="2323" Body="&lt;p&gt;Why isn't neptunium used in nuclear reactors in nuclear power plants?  Uranium is, and plutonium is. But neptunium isn't and it is in the middle of them.  Is it like it is too hard to make it do fission?  Can someone please kindly explain it to me?&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="17175" LastEditDate="2018-04-25T22:57:59.643" LastActivityDate="2018-04-25T22:57:59.643" Title="Why isn't neptunium used in nuclear reactors in nuclear power plants?" Tags="&lt;physical-chemistry&gt;&lt;elements&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14171" PostTypeId="2" ParentId="14170" CreationDate="2014-07-07T05:27:25.030" Score="10" Body="&lt;p&gt;There are several reasons that neptunium may not be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next would be the likelihood of fission when struck with a neutron. If it is the same or less than uranium, I would go with uranium. It would have to be substantially more than uranium to offset the cost of refinement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is the energy output per gram. I don't think it would be much greater than uranium, probably less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isotopes of neptunium that have long half-lives are $\ce{{}^{236}Np}$ and $\ce{{}^{237}Np}$. These are not produced in great abundance from $\ce{{}^{235}U}$ or $\ce{{}^{238}U}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make plutonium from uranium, the uranium must absorb neutrons without fission. This is much more likely to happen then to make $\ce{{}^{236}Np}$ or $\ce{{}^{237}Np}$&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-07-07T06:05:37.013" LastActivityDate="2014-07-07T06:05:37.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14172" PostTypeId="1" CreationDate="2014-07-07T06:13:24.960" Score="6" ViewCount="458" Body="&lt;p&gt;I understand the concept of a reactive species on a sort of vague and intuitive level: a reactive species is one that has lots of reactions that it can participate in. These tend to have large kinetic constants, be thermodynamically downhill and often have products that are also reactive, leading to knock-on effects. But reactivity isn't the same as energy, and it's possible to have highly reactive species with relatively low free energies and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in whether there's any kind of formal theory of reactivity, and/or a quantitative definition of the concept. For example, we're often told that free radicals are reactive because they have an unpaired electron and they &quot;want&quot; to fill their outer shell. This makes perfect sense on an intuitive level, but I'd like to know if it can be taken further. Is there some way we can write down an equation that will tell us how reactive a species will be in a given environment, or is there some way we can measure a species' reactivity empirically and use it to make predictions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I'm not looking for quantum-level predictions, although I will look into the theories mentioned in Sam and Greg's answers. Rather, I'm looking for something a bit more heuristic that will help me get a handle on what reactivity really &lt;em&gt;is&lt;/em&gt;. In particular, what does it really mean to say that one species is more or less reactive than another? Can we put a number to it and say that a species has a reactivity of 8.2 on some meaningful scale?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ron's answer is a start. It surely must be the case the reactivity relates to the energy landscape. But ron's answer deals with a one-dimensional landscape, corresponding to a single reaction. Reactivity is a concept that involves many reactions, and so a satisfying theory of reactivity would have to involve the properties of a massively multidimensional landscape. If there is anything that has been written about it from that point of view I would be very happy to read it.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="148" LastEditDate="2014-07-08T08:28:33.877" LastActivityDate="2014-07-21T14:40:01.670" Title="What is reactivity really, and can it be quantified?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14173" PostTypeId="1" CreationDate="2014-07-07T07:59:58.197" Score="2" ViewCount="8629" Body="&lt;p&gt;From &quot;Smarter than a 5th Grader&quot;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the simplest structural unit of an element or compound?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;molecule, &lt;/li&gt;&#xA;  &lt;li&gt;atom, &lt;/li&gt;&#xA;  &lt;li&gt;neutron, &lt;/li&gt;&#xA;  &lt;li&gt;proton&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For covalent compounds, the answer would definitely be molecule. But I'm unclear as to how to show that the answer for ionic compounds or elements is &quot;molecule&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://sites.google.com/site/rccchem2a/chapter-4-chemical-bonding-the-i&quot; rel=&quot;nofollow&quot;&gt;https://sites.google.com/site/rccchem2a/chapter-4-chemical-bonding-the-i&lt;/a&gt; states that ionic compounds: &quot;Do not have molecules as basic structural unit. Instead extended array of positively and negatively charged particles called ions.&quot; Is the term 'structural unit' defined across Science as only for covalent compounds?&lt;/p&gt;&#xA;" OwnerUserId="7046" LastEditorUserId="4945" LastEditDate="2014-07-07T12:20:48.877" LastActivityDate="2017-05-19T22:50:48.360" Title="What is the simplest structural unit of an element or compound?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14174" PostTypeId="1" AcceptedAnswerId="14177" CreationDate="2014-07-07T11:00:03.823" Score="14" ViewCount="9052" Body="&lt;p&gt;Since sodium chloride is sodium and chlorine atoms bonded as a lattice and there are no discrete molecules, doesn't that mean in gas state, sodium chloride is simply sodium and chlorine atoms separate from each other, resulting in a mixture of sodium gas and chlorine gas? &lt;/p&gt;&#xA;" OwnerUserId="6988" LastEditorUserId="6988" LastEditDate="2014-07-07T12:15:57.767" LastActivityDate="2014-07-07T14:10:37.210" Title="What is Sodium Chloride like in gas state?" Tags="&lt;physical-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14175" PostTypeId="2" ParentId="14173" CreationDate="2014-07-07T13:32:26.723" Score="2" Body="&lt;p&gt;For a solid lattice (metal, ionic solid, crystal) I would say the answer is &lt;strong&gt;unit cell&lt;/strong&gt;, for a non-charged compound  that isn't a lattice (including solution/gas phase) I would say &lt;strong&gt;molecule&lt;/strong&gt;, and for a charged one I would say &lt;strong&gt;ion-pair / ternary ion-associates / quaternary-ion associates&lt;/strong&gt; for two / three / four ion clusters. For polymers, I would refer to the basic unit(s) being the component monomer(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, as has been identified in the comments, an issue with allotropes, Carbon (Diamond, Graphite etc) can be encompassed within 'unit cell', but for elements like phosphorus and sulfur, it's a little more complicated. There I would contend that the basic unit of e.g. P&lt;sub&gt;4&lt;/sub&gt; (white Phosphorus) is the molecule, and for the rest, I would return to referring to the unit cell, or 'monomer' in the case of amorphous networks.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastEditorUserId="402" LastEditDate="2017-05-19T22:50:48.360" LastActivityDate="2017-05-19T22:50:48.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14176" PostTypeId="2" ParentId="14172" CreationDate="2014-07-07T13:45:46.773" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;Density Functional Theory&lt;/a&gt; has been used in the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1420065432&quot; rel=&quot;nofollow&quot;&gt;manner&lt;/a&gt; you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.ch.ic.ac.uk/harrison/Teaching/DFT_NATO.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for an introduction to the topic.&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-07-07T13:45:46.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14177" PostTypeId="2" ParentId="14174" CreationDate="2014-07-07T14:10:37.210" Score="16" Body="&lt;p&gt;No, in the gas phase sodium chloride exists as a monomer (the sodium chloride molecule) along with its dimer $\ce{Na2Cl2}$.  The dimer has a roughly rectangular shape and is quite floppy with chlorines located diagonally across from each other.  The dimer makes up about 27% of the mix.  All of the bond lengths, etc. can be found in &lt;a href=&quot;https://www.era.lib.ed.ac.uk/bitstream/1842/2601/2/McCaffrey%20PD%20thesis%2007.pdf&quot; rel=&quot;noreferrer&quot;&gt;this thesis&lt;/a&gt;.  Go to the end and you'll see the full paper that appeared in JACS with all of the data.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-07T14:10:37.210" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14178" PostTypeId="2" ParentId="14172" CreationDate="2014-07-07T14:45:04.277" Score="3" Body="&lt;p&gt;I think the term &quot;reactivity&quot; is often used to denote kinetic lability, a proclivity to react.  Let's look at the two reaction coordinates shown below.  They both represent the interconversion of molecules A and B through intermediate I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7J6da.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the potential well that I exists in is much deeper in the drawing on the left than the drawing on the right.  Therefor the activation energy for I to escape will be much larger for the intermediate on the left than the intermediate on the right.  The intermediate on the right would be said to be more reactive (have a shorter lifetime) than the intermediate on the left, because it requires less energy for it to escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experimentalist can infer the depth of the well through experiments.  Through such experiments the experimentalist learns that free radicals, highly strained hydrocarbons, carbenes, etc. exist in potential wells that must be very shallow because they react so readily.  Using this empirical data the experimentalist can then infer how similar compounds are likely to react. A computational chemist can infer the same thing by calculating the profile of the reaction coordinate for the various reactions under investigation by the experimentalist, or as is becoming increasingly common the computational chemist can predict what should be and let the experimentalist confirm those predictions.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-07T14:45:04.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14179" PostTypeId="1" AcceptedAnswerId="14189" CreationDate="2014-07-07T16:50:09.840" Score="9" ViewCount="1092" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XokvF.png&quot; alt=&quot;enter image description here&quot;&gt;What does this question mean by generally octahedral? Can this molecule adopt other shapes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I have my suspicions about the Br-S-F bond angle being a perfect 180 degrees. Wouldn't vdW repulsions distort the bond angle and make it ~180 degrees? The way I see it, if we place the bromine axial, wouldn't its sheer size push down on the fluorines in the equatorial plane and thus distort the 180 degree bond angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, now that I think of it, we only have one bromine pushing down on all the fluorines with an equal force and the net force should cancel out, making the 180 degree bond angle a theoretically perfect 180 degree bond angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any ways to verify this other than jumping through hoops in one's mind? I tried making Marvin Sketch render the molecule but it wasn't useful (or I wasn't using it correctly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, another choice I was doubting is choice 1. Are you sure there are 8 F-S-F bond angles &amp;lt; 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles &amp;lt; 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-08T06:15:14.630" LastActivityDate="2014-07-09T01:50:04.310" Title="&quot;Generally Octahedral&quot; - Meaning?" Tags="&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14180" PostTypeId="2" ParentId="14172" CreationDate="2014-07-07T18:25:57.150" Score="1" Body="&lt;p&gt;Depending on what kind of theory you are looking for, the maximum hardness principle and HSAB theory actually tells a lot about what species should be reactive and with whom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unrelated to density functional theory, but formulated in a rather chemically intuitive way. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-07T18:25:57.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14182" PostTypeId="2" ParentId="14179" CreationDate="2014-07-07T19:06:12.173" Score="8" Body="&lt;p&gt;I believe the writer is using the term &quot;generally octahedral&quot; much like when I say an orbital is &quot;roughly $\ce{sp^2}$ hybridized.&quot;  Few compounds are perfect octahedrons, few orbitals are exactly $\ce{sp^2}$ hybridized.  If a molecule's structure looks generally like an octahedron, but it isn't perfectly octahedral, then we can say that it is &quot;generally octahedral&quot; or better yet, a distorted octahedron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful way to determine if something is a perfect octahedron, tetrahedron, $\ce{sp^2}$ orbital, whatever, is through the use of symmetry. If any two atoms can be interconverted by rotation about an axis of symmetry or reflection through a plane of symmetry, then those atoms &lt;strong&gt;must&lt;/strong&gt; be equivalent.  If two atoms cannot be interconverted by such symmetry operations than they &lt;strong&gt;must&lt;/strong&gt; be non-equivalent.  Point groups and the symmetry elements they contain can make for an entire chemistry course, but the essentials can be learned quickly.  Here is a &lt;a href=&quot;http://csi.chemie.tu-darmstadt.de/ak/immel/script/redirect.cgi?filename=http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index7.html&quot; rel=&quot;noreferrer&quot;&gt;useful reference&lt;/a&gt; (one of many, you can google &quot;molecular symmetry, point groups&quot; to find something more to your liking) on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use the following drawing as a starting point to discuss the geometry of $\ce{BrSF_5}$.  We will use the term &quot;axial&quot; or &quot;apical&quot; to describe the F and Br atoms at the top and bottom of the drawing respectively, while the remaining 4 fluorines are termed &quot;equatorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uiGtj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying symmetry to $\ce{BrSF_5}$ tells us the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is a 4-fold rotational axis passing through the Br-S-F(axial) atoms&lt;/li&gt;&#xA;&lt;li&gt;There are 4 planes of symmetry containing this axis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rotational axis interconverts the 4 equatorial fluorine atoms, therefor they are equivalent as are the 4 F-S-F angles which must be (360/4) 90 degrees. Further, at least as I've drawn it here, the same rotational axis interconverts the bromine with itself (it sounds silly, but think about it) and the axial fluorine with itself.  Therefor, the Br-S-F(axial) angle must be 180 degrees. There is no plane of symmetry containing the S and 4 equatorial fluorines (because reflection through such a plane does not interconvert what's above the plane (F) with what's below the plane (Br).  This is a very important observation.  It tells us that the Br-S-F(equatorial) angle cannot be exactly 90 degrees.  It might be 90.001 or 85.278 degrees, but it can't be exactly 90 degrees - because of the lack of symmetry.  Symmetry can only tell us if things are the same or different, it can't tell us if different things differ by a lot or a little, we need to use our other chemical skills to do that.  So if the angle is not 90 degrees, then the S and 4 equatorial fluorines do not lie in a common plane; the 4 equatorial fluorines must be canted towards either the axial fluorine or the axial bromine.  I don't know which of those two options is preferred, but I would guess that the energy difference between these two forms is small and the barrier to their interconversion is small, so I might expect rapid interconversion between these 2 slightly distorted octahedral forms at room temperature.  At absolute zero, or perhaps in a crystal, only one of these two forms would prevail.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-09T01:50:04.310" LastActivityDate="2014-07-09T01:50:04.310" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14184" PostTypeId="1" AcceptedAnswerId="14185" CreationDate="2014-07-07T21:35:48.117" Score="1" ViewCount="9514" Body="&lt;p&gt;Please answer with as much detail as possible. I can't find the answer anywhere else and I need it for an essay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any websites I have found have conflicting answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One said that they don't actually burn underwater, rather than separate hydrogen from oxygen and burn at the surface. Another actually said that they do burn underwater because they take the oxygen source from the water. I do not have a specific definition for &quot;burn&quot; (the question itself is my assignment- write an essay answering that question!), so I know about as much as you do concerning the wording of the assignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that alkali and alkaline earth metals are generally very reactive with water, and usually result in exothermic reactions. &lt;/p&gt;&#xA;" OwnerUserId="7052" LastEditorUserId="865" LastEditDate="2020-04-12T16:33:46.787" LastActivityDate="2020-04-12T16:44:58.257" Title="How can metals like magnesium and sodium burn underwater?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;combustion&gt;" AnswerCount="3" CommentCount="2" ClosedDate="2020-04-13T08:56:47.910" ContentLicense="CC BY-SA 4.0" />
  <row Id="14185" PostTypeId="2" ParentId="14184" CreationDate="2014-07-07T22:11:07.637" Score="5" Body="&lt;p&gt;Your last sentence was the key.  Metals like sodium and magesium react with water according to the following equations&#xA;$$\ce{2Na + 2H2O -&amp;gt; 2NaOH +H2 + heat}$$&#xA;$$\ce{Mg + 2H2O -&amp;gt; Mg(OH)2 +H2 + heat}$$&#xA;Both reactions produce hydrogen gas and a lot of heat.  As the hydrogen gas bubbles out of the water it mixes with atmospheric oxygen. The intense heat ignites the hydrogen-oxygen mixture and the mixture continues to burn (possibly explosively depending upon the quantities involved) as long as hydrogen continues to be generated.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-08T02:41:16.573" LastActivityDate="2014-07-08T02:41:16.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14186" PostTypeId="2" ParentId="14046" CreationDate="2014-07-08T01:50:07.647" Score="3" Body="&lt;p&gt;As a starting point, I shall assume that the irritant compounds are freely exposed to the environment (which is not necessarily the case, as I mention further on). As such, one of the conceivably simplest ways to remove them is to wash them off, i.e. disperse them in a solvent and discard the liquid mixture. For several reasons (abundance, extensive solvation ability, non-toxicity, etc.) water is a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, how well would pure water work at dispersing the prenylated phenolics? Probably not too great. The representative parent compounds of the irritants, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroquinone&quot; rel=&quot;nofollow&quot;&gt;hydroquinone&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/4-Hydroxybenzoic_acid&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;p&lt;/em&gt;-hydroxybenzoic acid&lt;/a&gt; have modest solubilities in water (59 g and 5 g per 1 L, respectively), and the addition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Farnesyl&quot; rel=&quot;nofollow&quot;&gt;farnesyl chains&lt;/a&gt; in the compounds would quite likely cause a severe reduction in water solubility, unless the farnesyl chains are highly substituted with hydrophilic groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there is no substance that is &lt;em&gt;entirely&lt;/em&gt; insoluble in any solvent, though it is very common for certain solutes to require impractical amounts of solvent to disperse completely. It could well be that a washing machine's load of water (of order 100 L) is actually capable of dissolving the farnesylated toxins, especially because natural products are generally present in rather small quantities. However, there are some techniques which can possibly help the process. I list a few in the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use hot water - The majority of substances become more soluble in hot solvents because dissolution very often increases the entropy of a system ($\Delta S&amp;gt;0$), and hence from the Gibbs' free energy relation $\Delta G=\Delta H - T \Delta S$, an increase in temperature leads to a larger decrease in free energy, making the process more favourable. Of course, this only really works well if the water is hot throughout the soaking/washing process; adding hot water and letting it cool off will just cause part of the solute to precipitate out of solution again.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a surfactant (detergent) - Since the offending compounds potentially have large hydrophobic chains, they can conceivably removed by the same process we use to wash away oils and fats. An amphipatic molecule contains a hydrophobic part (such as the hydrocarbon chain of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_dodecylbenzenesulfonate&quot; rel=&quot;nofollow&quot;&gt;sodium dodecylbenzenesulphonate&lt;/a&gt;) responsible for interacting with the oils &lt;em&gt;and&lt;/em&gt; a hydrophilic part (in SDBS, the sulphonate group), which is well solvated by water. Thus, the surfactant bridges the interaction between oil and water, allowing effective dispersion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a basic solution - In this specific case, the undesirable compounds have some acidic character. In particular, the hydrogen atom in the -COOH (carboxyl) group of the &lt;em&gt;p&lt;/em&gt;-hydroxybenzoic acid derivatives are about as acidic as acetic acid. The -OH (hydroxyl) groups in both &lt;em&gt;p&lt;/em&gt;-hydroxybenzoic acid and hydroquinone are also slightly acidic, though weaker by some 5-6 orders of magnitude. Thus, mixing in a basic solution may allow an acid-base reaction to take place, forming salts of the organic compounds with significantly increased solubility in water. Many washing powders/liquids are already slightly alkaline, but it is also easy to prepare basic solutions separately. A cheap and very safe base would be sodium carbonate, $\ce{Na2CO3}$, though a few tens of grams may be required to prepare a decent volume of basic medium for soaking. Sodium bicarbonate, $\ce{NaHCO3}$, also works though it is a weaker base and would require a higher amount to reach a given pH value. Caustic soda, $\ce{NaOH}$, is yet another cheap option, but care must be taken to not add too much, as it is a strong base (in contrast to the previous two). I don't know how resistant different fabrics are but, very approximately, any more than a couple of grams of lye per litre of water risks partially hydrolysing the fabric and weakening its structural integrity. Use at your own risk!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a bleaching agent (chlorinated or non-chlorinated) - Bleaches work by partially oxidizing organic matter, breaking organic molecules into smaller chunks and making each piece more hydrophilic (due to added oxygenated functionalities). Even if the compounds aren't removed entirely, the reaction products may be much less harmful.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Miscellaneous - You suggest exposure to sunlight. Light is certainly capable of decomposing organic substances, especially UV light, but there's no guarantee that the exposure required to eliminate the irritants won't also break down fabric and bleach dyes. Waiting is also probably not very efficient, as there is no guarantee the irritant compounds are unstable enough to undergo spontaneous decomposition in ambient conditions at a significant rate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, all this discussion is of limited use if the initial assumption (freely accessible irritants) is incorrect. It could be that the compounds are stored in cavities in the plant hair's surface or interior, and the hairs themselves are insoluble and impermeable to water, effectively locking the toxins away. In this case, the compounds will likely be removed only slowly, as the hairs are progressively worn out by abrasion and other processes. Thus a physical removal method may be preferable. I have heard that adhesives (sticky tape, waxing strips, etc.) can be successfully used to peel out painful hairs. This is also a useful option for surfaces which cannot be washed easily (such as car upholstery).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I am not sure how this will all work out. It's been a few days since I wrote my comments, so perhaps you've had time to test something out? If so, I would be curious to hear some feedback!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-07-08T01:50:07.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14187" PostTypeId="2" ParentId="14184" CreationDate="2014-07-08T02:35:30.607" Score="-2" Body="&lt;p&gt;I specialize in environmental problems, namely waste-water. From what I can remember, sodium and magnesium react with water, very violently. Sodium is actually used in nuclear submarines to detect steam/water leaks. Cesium, as far as I know, is stored in some form of oil to keep it from reacting with moisture. I was in the 5th grade when my science teacher showed me a demonstration of sodium and how oxygen and hydrogen implode to form water - just like a rocket-ship engine works.&lt;/p&gt;&#xA;" OwnerUserId="7053" LastEditorUserId="7054" LastEditDate="2014-07-08T08:08:00.187" LastActivityDate="2014-07-08T08:08:00.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14188" PostTypeId="2" ParentId="12411" CreationDate="2014-07-08T05:18:44.623" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL;DR version&lt;/strong&gt;: Chirality is a basic property of a molecule, while optical activity is a bulk physical property possessed by samples of chiral compounds. Specific rotation is an intrinsic property of chiral compounds and a measure of optical activity under standardized conditions. Optical rotation is not responsible for the disparate behavior of enantiomers of biologically active compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chirality is an intrinsic property of a molecule (or, indeed, of certain geometric structures, more generally). A chiral molecule is one that is not superposable on its mirror image molecule (its enantiomer), meaning that no series of rotations or translations will make the two geometrically identical in three-dimensional space. One of the consequences of chirality is that chiral molecules &lt;em&gt;in bulk&lt;/em&gt; will measurably rotate plane-polarized light by a specific amount and in a specific direction. (The distinction here is crucial, since &lt;em&gt;any molecule[s]&lt;/em&gt; will in general cause some rotation, but the random distribution and spatial orientation of the large number of molecules in any given sample will produce no observable net rotation if the molecules are achiral. The rotation caused by one molecule is likely to be negated by another identical molecule in an opposite spatial orientation.) The key point is that chirality is an intrinsic property of a molecule, while optical activity is a bulk physical property of said molecules gathered &lt;em&gt;en masse&lt;/em&gt;. Specific rotation, on the other hand, can be viewed as an intrinsic property, since it is defined for a specific sample under an exact set of conditions (light wavelength, path length, concentration [for samples in solution], density [for pure liquid samples], temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optical activity is not the cause of any changes in the properties of chiral compounds, nor is it responsible for the sometimes wildly different biological effects of certain enantiomers (which I presume may be what you're actually referring to, forgive me if I'm wrong). The enantiomers of chiral molecules can often differ widely in their biological effects due to the basic nature of proteins, which are the constituent biological molecules that most if not all biologically active substances target in one way or another. Proteins are constructed from amino acids, which often themselves contain multiple chiral centers, and these proteins go on to fold into very specific and complex conformations. Enzyme active sites and cell receptors (among various other biological structures) are mostly comprised of proteins with very specific three-dimensional structures. As such, they interact significantly only with molecules that have the appropriate shape. Hence, enantiomers of chiral substances, which differ in their three-dimensional shape, will exhibit variable degrees and types of interactions with proteins, and can hence impact biological targets in disparate ways. This is largely what makes chirality so important to chemists, and partly explains why so much of the most respected (not to mention economically important) cutting-edge work recently (at least in organic chemistry) has revolved around enantioselective or enantiospecific synthesis techniques.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-08T05:18:44.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14189" PostTypeId="2" ParentId="14179" CreationDate="2014-07-08T06:06:56.193" Score="10" Body="&lt;p&gt;The term &quot;generally octahedral&quot; refers to a distorted octahedral arrangement of the ligands referring to the central atom (as ron already stated). In this case the distortion is due to the massively different bond lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to ron's answer I did a little calculation at BP86/cc-pVDZ. I did not use any symmetry constraints at first and the final geometry deviated from ideal &lt;a href=&quot;http://goldbook.iupac.org/P04703.html&quot; rel=&quot;noreferrer&quot;&gt;$C_\mathrm{4v}$&lt;/a&gt; geometry only slightly due to rounding errors. After symmetry constraint reoptimisation, the following parameters may be found:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrl}&#xA;\text{parameter}                &amp;amp; \text{value}&amp;amp;\\\hline&#xA;\mathbf{d}(\ce{S-Br})           &amp;amp; 2.039 &amp;amp;\mathrm{\mathring{A}}\\&#xA;\mathbf{d}(\ce{S-F_{ax}})       &amp;amp; 1.638 &amp;amp;\mathrm{\mathring{A}}\\&#xA;\mathbf{d}(\ce{S-F_{eq}})       &amp;amp; 1.650 &amp;amp;\mathrm{\mathring{A}}\\\hline&#xA;\angle(\ce{F_{ax}-S-Br})        &amp;amp; 180.0 &amp;amp;{}^{\circ}\\&#xA;\angle(\ce{F_{eq}-S-Br})        &amp;amp;  90.7 &amp;amp;{}^{\circ}\\&#xA;\angle(\ce{F_{ax}-S-F_{eq}})    &amp;amp;  89.3 &amp;amp;{}^{\circ}\\&#xA;\angle(\ce{F_{eq}-S-F'_{eq}})   &amp;amp;  90.0 &amp;amp;{}^{\circ}\\&#xA;\angle(\ce{F_{eq}-S-F''_{eq}})  &amp;amp; 178.7 &amp;amp;{}^{\circ}\\\hline&#xA;\angle_\mathrm{D}(\ce{F_{eq}-S-F'_{eq}-F''_{eq}})  &#xA;                                &amp;amp; 178.7 &amp;amp;{}^{\circ}\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eh9Ic.jpg&quot; alt=&quot;geometrical parameters of BrSF5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was unable to determine a interconversion barrier related to the $\angle(\ce{F_{eq}-S-Br})$ bond angle change. I was also not able to find a different stable isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solid state it is most likely, that the geometric parameters will change due to packing effects and/or more internal entropy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Addressing the edit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are you sure there are 8 F-S-F bond angles &amp;lt; 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles &amp;lt; 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the bromine pushes down the fluorines of the equatorial plane resulting in $\angle(\ce{F_{ax}-S-F_{eq}})=89.3^{\circ}$ bond angle. Due to this, the sulfur is out of plane with the equatorial fluorine atoms. Since the planes expanded by $\ce{Br-S-F_{eq}}$ and $\ce{Br-S-F'_{eq}}$ have to be at $90^\circ$ due to symmetry constraints, this is creating also a slightly smaller angle than $90^\circ$ for the equatorial fluorines. In this very case it can be found to be $\angle(\ce{F_{eq}-S-F'_{eq}})=89.992^{\circ}$, which was rounded by the program to fit the one decimal place restriction. The application of this is just trigonometry and you can derive it as a simple thought experiment - if you like.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-07-08T06:43:58.880" LastActivityDate="2014-07-08T06:43:58.880" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14191" PostTypeId="2" ParentId="14184" CreationDate="2014-07-08T07:18:10.950" Score="0" Body="&lt;p&gt;To my knowledge, neither of these metals &quot;burn&quot; underwater.  The word burn generally suggests that something reacts with oxygen.  As others have stated above, these very active metals will react with water by displacing hydrogen as H&lt;sub&gt;2&lt;/sub&gt;.  The hydrogen definitely will react with oxygen assuming that enough heat is evolved to start the reaction (very likely for sodium, less so for magnesium). &lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2014-07-08T07:18:10.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14192" PostTypeId="2" ParentId="10833" CreationDate="2014-07-08T10:11:47.843" Score="10" Body="&lt;p&gt;According to my FIITJEE (a coaching institute in India) textbook, the relative order of migratory aptitude of groups in pinacol-pinacolone rearrangement is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H &amp;gt; Ph &amp;gt; Me_3C &amp;gt; MeCH_2 &amp;gt; Me}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does not seem to be a direct correlation between the mass and the migratory aptitude of these groups.&lt;/p&gt;&#xA;" OwnerUserId="5487" LastEditorUserId="5026" LastEditDate="2018-05-21T01:53:04.363" LastActivityDate="2018-05-21T01:53:04.363" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14193" PostTypeId="2" ParentId="14066" CreationDate="2014-07-08T11:34:45.283" Score="17" Body="&lt;p&gt;Okay, let's tackle at least one problem here. Consider the rotation of 1,2-dichloroethane (BP86/cc-pVDZ):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TCW7d.png&quot; alt=&quot;rotation of dichloroethane&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These conformational changes can be further rationalised:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;C&lt;/strong&gt; and &lt;strong&gt;C'&lt;/strong&gt; are the same conformation, since these are mirror images. &lt;/li&gt;&#xA;&lt;li&gt;The same applies to &lt;strong&gt;B&lt;/strong&gt; and &lt;strong&gt;B'&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt; are local minima and can be referred to as &lt;em&gt;conformes&lt;/em&gt; in the above given way&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;B&lt;/strong&gt; and &lt;strong&gt;D&lt;/strong&gt; are transition states - they are not stable.&lt;br&gt;&#xA;If you consider the rotation &lt;strong&gt;A&lt;/strong&gt; to &lt;strong&gt;C&lt;/strong&gt; as a reaction, this state would mark the maximum energy needed for the conversion. I believe with &lt;em&gt;reactions&lt;/em&gt; it is meant, that an accurate description of the transition state by InChI is not possible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the term non-local chirality I suggest you look at &lt;a href=&quot;http://de.wikipedia.org/wiki/BINAP&quot; rel=&quot;nofollow noreferrer&quot;&gt;BINAP&lt;/a&gt; or &lt;a href=&quot;http://de.wikipedia.org/wiki/BINOL&quot; rel=&quot;nofollow noreferrer&quot;&gt;BINOL&lt;/a&gt;.&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/d/d6/BINAP_Enantiomers_Structural_Formulae_V.1.svg/350px-BINAP_Enantiomers_Structural_Formulae_V.1.svg.png&quot; alt=&quot;structure of BINAP&quot;&gt;&lt;br&gt;&#xA;In this case I have used a slightly abridged version of BINAP for educational purposes, where $\ce{Ph -&amp;gt; Me}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UyAMJ.jpg&quot; alt=&quot;R-BINAP&quot;&gt;  &lt;img src=&quot;https://i.stack.imgur.com/waqGx.jpg&quot; alt=&quot;S-BINAP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other wonderful examples of non-local chirality include &lt;a href=&quot;http://en.wikipedia.org/wiki/Helicene&quot; rel=&quot;nofollow noreferrer&quot;&gt;helicenes&lt;/a&gt;, which have helical chirality. Here is one example of a fancy &lt;a href=&quot;http://dx.doi.org/10.1107/S0108270102023739&quot; rel=&quot;nofollow noreferrer&quot;&gt;azonia derivative of hexahelicene&lt;/a&gt; (&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.89294.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;parent compound and links&lt;/a&gt;). The structure is available at the &lt;a href=&quot;http://www.ccdc.cam.ac.uk/pages/Home.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cambridge Crystallographic Data Centre (CCDC) Database&lt;/a&gt;. I removed the anion and converted it to xyz coordinates which I include in the appendix. Here are the structures, left &lt;em&gt;P&lt;/em&gt;-chirality, right &lt;em&gt;M&lt;/em&gt;-chirality:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lfhuZ.gif&quot; alt=&quot;scheme of 8a-azonia[6]helicene&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aH98d.jpg&quot; alt=&quot;P-8a-azonia[6]helicene&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IAvn9.jpg&quot; alt=&quot;M-8a-azonia[6]helicene&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another account of non/local chirality, there is the planar chirality, as can be seen in &lt;em&gt;trans&lt;/em&gt;-&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclooctene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclooctene&lt;/a&gt; (the bottom two). For further reading I suggest &lt;a href=&quot;http://dx.doi.org/10.1021/jo202176j&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neuenschwander &lt;em&gt;et.al.&lt;/em&gt; &quot;The Conformations of Cyclooctene: Consequences for Epoxidation Chemistry.&quot;&lt;/a&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/7/77/Cycooctene_Isomers_Structural_Formulae.png/298px-Cycooctene_Isomers_Structural_Formulae.png&quot; alt=&quot;Structures of cyclooctene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Markush_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Markush Structures&lt;/a&gt; are a kind of summary structure for certain compound classes. A very simple example would be a monosubstituted biphenyl system, where the upper structure is a representation of all three below. However, it is noteworthy, that each instance of a Markush structure can be well explained by InChI.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YE0gp.png&quot; alt=&quot;biphenyl markush structure&quot;&gt;&lt;br&gt;&#xA;More complicated structures are often used in patents and in other scientific publications, usually specifically referring to a very general procedure or mechanism. I found one on wikipedia commons, that could well represent a million molecules:&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/9e/Markush_structure.png&quot; alt=&quot;markush structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And another way to break the InChI are &lt;a href=&quot;http://goldbook.iupac.org/CT06769.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;clusters&lt;/a&gt;. Although the definition is not very specific, the $\ce{[KO{}^{\mathit{t}}Bu]_4}$ tetramer (&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_tert-butoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;, &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.63266.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemspider&lt;/a&gt;) certainly contains one. While the monomer can be described via InChI, the tetra- or Polymer should fail. It was published by &lt;a href=&quot;http://dx.doi.org/10.1016%2FS0277-5387%2800%2980154-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chisholm &lt;em&gt;et.al.&lt;/em&gt; in Polyhedron&lt;/a&gt; and the &lt;code&gt;*.cif&lt;/code&gt; file can be obtained from CCDB. I cleaned it up a little bit and included the &lt;code&gt;*.xyz&lt;/code&gt; in the appendix. You can quite nicely see the $\ce{K4}$ tetrahedral cluster in the middle (hydrogens omitted for clarity). &#xA;&lt;img src=&quot;https://i.stack.imgur.com/YoBsv.jpg&quot; alt=&quot;molecular structure of the potassium butoxide tetramer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Last but not least, I'd like to cover the matter of complicated organometallic molecules. As such I'd like to introduce the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grubbs%27_catalyst&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grubbs' catalysts&lt;/a&gt;. &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/2005/grubbs-facts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grubbs&lt;/a&gt; obtained the &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/2005/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nobel Prize in Chemistry 2005&lt;/a&gt; for his research in conjunction with Yves Chauvin and Richard R. Schrock. One of the key publications was &lt;a href=&quot;http://dx.doi.org/10.1021/ja00036a053&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Ring-opening metathesis polymerization (ROMP) of norbornene by a Group VIII carbene complex in protic media&quot;&lt;/a&gt;. The crystal structure can also be obtained at CCDC. Another example stems from &quot;&lt;a href=&quot;http://dx.doi.org/10.1021/ja952676d&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthesis and Applications of $\ce{RuCl2(=CHR')(PR3)2}$:  The Influence of the Alkylidene Moiety on Metathesis Activity&lt;/a&gt;&quot; and is also available via CCDC. This structure can also be found in the appendix. It is generally known as a 1st generation Grubbs catalyst (Hydrogens omitted for clarity).&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/1f/Grubbs_Catalyst_1st_Generation.svg/200px-Grubbs_Catalyst_1st_Generation.svg.png&quot; alt=&quot;scheme of Grubbs I&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KCcIa.jpg&quot; alt=&quot;molecular structure of Grubbs I&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Appendix&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The following section contains the nuclear coordinates in &lt;code&gt;*.xyz&lt;/code&gt;(XMol) file format and angstroms as unit. (If not denoted otherwise, structures are obtained at BP86/cc-pVDZ.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,2-dichloroethane &lt;strong&gt;A&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C        0.000000000      0.000000000      0.000000000&#xA;C        0.000000000      0.000000000      1.521307000&#xA;Cl       1.723641000      0.000000000     -0.593797000&#xA;H       -0.486731000     -0.904743000     -0.410753000&#xA;H       -0.486731000      0.904743000     -0.410753000&#xA;Cl      -1.723641000      0.000000000      2.115104000&#xA;H        0.486731000     -0.904743000      1.932060000&#xA;H        0.486731000      0.904743000      1.932060000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1,2-dichloroethane &lt;strong&gt;B&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C        0.030242000      0.085780000     -0.007620000&#xA;C       -0.030242000      0.085780000      1.528927000&#xA;Cl       1.518312000     -0.758527000     -0.631286000&#xA;H       -0.837191000     -0.442465000     -0.444148000&#xA;H        0.070555000      1.115212000     -0.412109000&#xA;Cl      -1.518312000     -0.758527000      2.152593000&#xA;H        0.837191000     -0.442465000      1.965455000&#xA;H       -0.070555000      1.115212000      1.933416000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1,2-dichloroethane &lt;strong&gt;C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C        0.066377000      0.127741000      0.004558000&#xA;C       -0.066377000      0.127741000      1.516749000&#xA;Cl       1.083384000     -1.241726000     -0.615213000&#xA;H       -0.919786000      0.050600000     -0.492443000&#xA;H        0.568277000      1.063384000     -0.317636000&#xA;Cl      -1.083384000     -1.241726000      2.136520000&#xA;H        0.919786000      0.050600000      2.013750000&#xA;H       -0.568277000      1.063384000      1.838943000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;1,2-dichloroethane &lt;strong&gt;D&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C        0.164887000      0.141535000     -0.000240000&#xA;C       -0.164887000      0.141535000      1.521547000&#xA;Cl       0.337146000     -1.473219000     -0.795160000&#xA;H       -0.633932000      0.666019000     -0.558099000&#xA;H        1.123149000      0.665665000     -0.177529000&#xA;Cl      -0.337146000     -1.473219000      2.316468000&#xA;H        0.633932000      0.666019000      2.079406000&#xA;H       -1.123149000      0.665665000      1.698836000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;R&lt;/em&gt;-BINAP&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C       -0.086423000     -0.055821000     -0.031546000&#xA;C       -0.090885000     -0.047699000      1.480085000&#xA;C        1.148913000     -0.158515000      2.237313000&#xA;C        1.064686000     -0.139089000      3.676951000&#xA;C       -0.131097000     -0.014450000      4.357670000&#xA;C       -1.357362000      0.116010000      3.652970000&#xA;C       -1.345263000      0.107319000      2.211098000&#xA;P        2.768926000     -0.321256000      1.451350000&#xA;H        1.988600000     -0.233144000      4.265654000&#xA;H       -0.135982000     -0.008251000      5.459401000&#xA;C       -2.590782000      0.269421000      4.370415000&#xA;C       -2.623021000      0.305506000      1.559779000&#xA;C        0.437392000      1.073619000     -0.788457000&#xA;C        0.410661000      0.991852000     -2.228054000&#xA;C       -0.079670000     -0.105834000     -2.908992000&#xA;C       -0.579162000     -1.233570000     -2.204534000&#xA;C       -0.579999000     -1.219208000     -0.762676000&#xA;P        1.107602000      2.557133000     -0.001640000&#xA;H        0.790955000      1.839345000     -2.816335000&#xA;H       -0.077540000     -0.112552000     -4.010726000&#xA;C       -1.063317000     -2.378355000     -2.921905000&#xA;C       -1.047368000     -2.424579000     -0.111239000&#xA;C       -3.792005000      0.431032000      3.705083000&#xA;H       -2.558632000      0.263827000      5.471798000&#xA;H       -4.735477000      0.552772000      4.258523000&#xA;C       -3.797353000      0.457087000      2.279022000&#xA;H       -2.662047000      0.340912000      0.466043000&#xA;H       -4.750761000      0.602046000      1.748556000&#xA;C       -1.503463000     -3.517255000     -0.830391000&#xA;H       -1.036094000     -2.475958000      0.982488000&#xA;H       -1.854657000     -4.415344000     -0.299810000&#xA;C       -1.523650000     -3.499569000     -2.256520000&#xA;H       -1.052652000     -2.347534000     -4.023279000&#xA;H       -1.890198000     -4.377463000     -2.809877000&#xA;C        4.061075000     -0.509159000      2.730645000&#xA;C        2.844045000     -1.846744000      0.446017000&#xA;C        1.585856000      3.772639000     -1.280334000&#xA;C       -0.173490000      3.383258000      1.008127000&#xA;H        0.254332000      4.289560000      1.485098000&#xA;H       -1.067700000      3.703930000      0.418133000&#xA;H       -0.532498000      2.717241000      1.815764000&#xA;H        1.997887000      4.670081000     -0.777298000&#xA;H        2.377005000      3.362349000     -1.938832000&#xA;H        0.749130000      4.118465000     -1.941170000&#xA;H        5.044249000     -0.601334000      2.227532000&#xA;H        4.100503000      0.383637000      3.385840000&#xA;H        3.944572000     -1.404262000      3.395364000&#xA;H        3.842692000     -1.931312000     -0.030903000&#xA;H        2.674387000     -2.779210000      1.039384000&#xA;H        2.087489000     -1.826871000     -0.361580000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;S&lt;/em&gt;-BINAP&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C       -1.383946000     -3.448543000     -2.256996000&#xA;C       -1.065989000     -2.266144000     -2.910129000&#xA;C       -0.652376000     -1.110469000     -2.179491000&#xA;C       -0.563545000     -1.177237000     -0.738280000&#xA;C       -0.899026000     -2.413534000     -0.098700000&#xA;C       -1.299620000     -3.519323000     -0.838209000&#xA;C       -0.154302000     -0.008671000      0.004167000&#xA;C        0.166467000      1.182512000     -0.673969000&#xA;C        0.072046000      1.221328000     -2.098779000&#xA;C       -0.323252000      0.113168000     -2.831301000&#xA;C       -0.061735000     -0.093981000      1.507135000&#xA;C       -1.110430000      0.372113000      2.322674000&#xA;C       -0.989013000      0.267385000      3.742090000&#xA;C        0.132224000     -0.289631000      4.336601000&#xA;C        1.210137000     -0.777916000      3.542877000&#xA;C        1.113248000     -0.682564000      2.103804000&#xA;C        2.206553000     -1.174312000      1.320794000&#xA;C        3.329975000     -1.726929000      1.922868000&#xA;C        3.419059000     -1.820539000      3.340084000&#xA;C        2.377998000     -1.353835000      4.130094000&#xA;P       -2.590653000      1.186493000      1.502738000&#xA;C       -3.750658000     -0.293139000      1.402088000&#xA;P        0.636241000      2.692004000      0.339668000&#xA;C        2.511918000      2.555623000      0.253510000&#xA;C       -3.385046000      2.079953000      2.947654000&#xA;C        0.388676000      4.078351000     -0.898992000&#xA;H       -1.802841000      0.635802000      4.383638000&#xA;H        0.202504000     -0.360978000      5.434128000&#xA;H        0.319305000      2.150696000     -2.632672000&#xA;H       -0.386992000      0.166196000     -3.930250000&#xA;H        2.436377000     -1.419019000      5.228476000&#xA;H        4.313241000     -2.262708000      3.805535000&#xA;H        2.146133000     -1.110484000      0.224695000&#xA;H        4.157224000     -2.097887000      1.298347000&#xA;H       -0.834575000     -2.478936000      0.997042000&#xA;H       -1.552289000     -4.458551000     -0.322477000&#xA;H       -1.128190000     -2.199481000     -4.008215000&#xA;H       -1.700498000     -4.331984000     -2.832408000&#xA;H        2.950535000      3.473538000      0.698290000&#xA;H        2.842563000      1.687658000      0.858736000&#xA;H        2.884117000      2.439555000     -0.786611000&#xA;H        0.678948000      5.023670000     -0.395042000&#xA;H        0.990357000      3.982867000     -1.828352000&#xA;H       -0.686746000      4.150262000     -1.160750000&#xA;H       -4.302108000      2.573768000      2.564446000&#xA;H       -3.673500000      1.421270000      3.794842000&#xA;H       -2.700063000      2.872570000      3.312044000&#xA;H       -4.742112000      0.063129000      1.051704000&#xA;H       -3.359638000     -1.014277000      0.656570000&#xA;H       -3.871382000     -0.807607000      2.379049000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;M&lt;/em&gt;-8a-azonia[6]helicene&lt;/strong&gt; (from crystal structure)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N        3.565811000      0.259443000     13.724558000&#xA;C        7.445232000     -1.196589000     12.564039000&#xA;C        8.629416000     -1.762757000     12.151335000&#xA;C        8.885324000     -1.978555000     10.783139000&#xA;C        7.914184000     -1.694865000      9.864763000&#xA;C        6.665024000     -1.171130000     10.265448000&#xA;C        5.584378000     -1.088690000      9.330104000&#xA;C        4.339207000     -0.800151000      9.738213000&#xA;C        4.088732000     -0.406137000     11.083608000&#xA;C        2.763235000     -0.187914000     11.543327000&#xA;C        2.522579000      0.029096000     12.830042000&#xA;C        3.246403000      0.443720000     15.064121000&#xA;C        4.198045000      0.692252000     15.970654000&#xA;C        5.531025000      0.961394000     15.546639000&#xA;C        6.527506000      1.327523000     16.488699000&#xA;C        7.769629000      1.675468000     16.086777000&#xA;C        8.044342000      1.864594000     14.697195000&#xA;C        9.259232000      2.459858000     14.281133000&#xA;C        9.466841000      2.806590000     12.982045000&#xA;C        8.445715000      2.608977000     12.035743000&#xA;C        7.265466000      1.985829000     12.396483000&#xA;C        7.061105000      1.553020000     13.720492000&#xA;C        5.833567000      0.912900000     14.172787000&#xA;C        4.884651000      0.303088000     13.284635000&#xA;C        5.151668000     -0.288539000     12.002691000&#xA;C        6.444932000     -0.841371000     11.632053000&#xA;H        7.296301000     -1.066868000     13.466331000&#xA;H        9.304643000     -2.036748000     12.840647000&#xA;H        9.760406000     -2.388329000     10.496986000&#xA;H        8.038683000     -1.964007000      8.943383000&#xA;H        5.780686000     -1.333585000      8.395468000&#xA;H        3.604979000     -0.909262000      9.084780000&#xA;H        2.029733000     -0.315211000     10.922946000&#xA;H        1.672696000      0.060618000     13.273677000&#xA;H        2.290439000      0.327335000     15.253240000&#xA;H        3.954366000      0.824398000     16.914659000&#xA;H        6.255785000      1.333585000     17.427224000&#xA;H        8.447606000      1.891266000     16.702563000&#xA;H        9.925491000      2.630800000     14.988120000&#xA;H       10.286033000      3.200604000     12.619714000&#xA;H        8.559741000      2.909640000     11.135042000&#xA;H        6.570317000      1.891266000     11.711236000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;M&lt;/em&gt;-8a-azonia[6]helicene&lt;/strong&gt; (from BP86/STO-3G)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N        0.241946000      3.510759000     13.746203000&#xA;C       -1.207087000      7.492952000     12.512135000&#xA;C       -1.801472000      8.692275000     12.089683000&#xA;C       -2.102492000      8.912745000     10.707738000&#xA;C       -1.853647000      7.899258000      9.772330000&#xA;C       -1.278056000      6.643678000     10.182732000&#xA;C       -1.146616000      5.553783000      9.229249000&#xA;C       -0.770948000      4.283969000      9.646130000&#xA;C       -0.392477000      4.045979000     11.023979000&#xA;C       -0.161103000      2.688046000     11.484550000&#xA;C        0.042219000      2.428893000     12.821687000&#xA;C        0.400500000      3.183323000     15.136378000&#xA;C        0.639392000      4.176721000     16.059802000&#xA;C        0.949043000      5.527644000     15.625469000&#xA;C        1.361398000      6.519481000     16.597330000&#xA;C        1.808770000      7.766658000     16.182869000&#xA;C        1.981223000      8.066019000     14.770425000&#xA;C        2.627560000      9.287588000     14.362758000&#xA;C        2.913363000      9.530579000     13.012425000&#xA;C        2.578886000      8.546652000     12.028110000&#xA;C        1.916409000      7.363991000     12.392255000&#xA;C        1.565834000      7.097902000     13.763105000&#xA;C        0.907348000      5.851096000     14.211296000&#xA;C        0.281492000      4.885356000     13.292158000&#xA;C       -0.309154000      5.144896000     11.968134000&#xA;C       -0.894790000      6.445456000     11.575101000&#xA;H       -1.007504000      7.339193000     13.585554000&#xA;H       -2.048406000      9.472285000     12.834293000&#xA;H       -2.558542000      9.868716000     10.388465000&#xA;H       -2.123797000      8.035037000      8.708340000&#xA;H       -1.422274000      5.738729000      8.172836000&#xA;H       -0.763274000      3.430969000      8.943112000&#xA;H       -0.251900000      1.843110000     10.776492000&#xA;H        0.087281000      1.420761000     13.274365000&#xA;H        0.295196000      2.107892000     15.372408000&#xA;H        0.697253000      3.916258000     17.133249000&#xA;H        1.320711000      6.254708000     17.669775000&#xA;H        2.110372000      8.530075000     16.926174000&#xA;H        2.921715000     10.016221000     15.141242000&#xA;H        3.423497000     10.463796000     12.708003000&#xA;H        2.853573000      8.716110000     10.969950000&#xA;H        1.691114000      6.611596000     11.618453000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\ce{KO{}^{\mathit{t}}Bu}$ tetramer&lt;/strong&gt; (from crystal structure)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;K       -1.309381000     -1.309381000     -1.309381000&#xA;O       -1.313567000      1.313567000     -1.313567000&#xA;C       -2.107232000      2.107232000     -2.107232000&#xA;C       -3.502008000      2.298114000     -1.478495000&#xA;H       -3.842748000      1.465100000     -1.465100000&#xA;H       -4.060420000      2.821364000     -2.059512000&#xA;H       -3.516240000      2.821364000     -0.619528000&#xA;O       -1.313567000     -1.313567000      1.313567000&#xA;O        1.313567000     -1.313567000     -1.313567000&#xA;K       -1.309381000      1.309381000      1.309381000&#xA;K        1.309381000      1.309381000     -1.309381000&#xA;C       -1.478495000      3.502008000     -2.298114000&#xA;C       -2.298114000      1.478495000     -3.502008000&#xA;K        1.309381000     -1.309381000      1.309381000&#xA;C       -2.107232000     -2.107232000      2.107232000&#xA;C        2.107232000     -2.107232000     -2.107232000&#xA;O        1.313567000      1.313567000      1.313567000&#xA;H       -1.465100000      3.842748000     -1.465100000&#xA;H       -2.059512000      4.060420000     -2.821364000&#xA;H       -0.619528000      3.516240000     -2.821364000&#xA;H       -1.465100000      1.465100000     -3.842748000&#xA;H       -2.821364000      2.059512000     -4.060420000&#xA;H       -2.821364000      0.619528000     -3.516240000&#xA;C       -1.478495000     -3.502008000      2.298114000&#xA;C       -3.502008000     -2.298114000      1.478495000&#xA;C       -2.298114000     -1.478495000      3.502008000&#xA;C        3.502008000     -1.478495000     -2.298114000&#xA;C        1.478495000     -2.298114000     -3.502008000&#xA;C        2.298114000     -3.502008000     -1.478495000&#xA;C        2.107232000      2.107232000      2.107232000&#xA;H       -1.465100000     -3.842748000      1.465100000&#xA;H       -2.059512000     -4.060420000      2.821364000&#xA;H       -0.619528000     -3.516240000      2.821364000&#xA;H       -3.842748000     -1.465100000      1.465100000&#xA;H       -4.060420000     -2.821364000      2.059512000&#xA;H       -3.516240000     -2.821364000      0.619528000&#xA;H       -1.465100000     -1.465100000      3.842748000&#xA;H       -2.821364000     -2.059512000      4.060420000&#xA;H       -2.821364000     -0.619528000      3.516240000&#xA;H        1.465100000     -1.465100000     -3.842748000&#xA;H        3.842748000     -1.465100000     -1.465100000&#xA;H        4.060420000     -2.059512000     -2.821364000&#xA;H        3.516240000     -0.619528000     -2.821364000&#xA;H        2.059512000     -2.821364000     -4.060420000&#xA;H        0.619528000     -2.821364000     -3.516240000&#xA;H        1.465100000     -3.842748000     -1.465100000&#xA;H        2.821364000     -4.060420000     -2.059512000&#xA;H        2.821364000     -3.516240000     -0.619528000&#xA;C        1.478495000      2.298114000      3.502008000&#xA;C        3.502008000      1.478495000      2.298114000&#xA;C        2.298114000      3.502008000      1.478495000&#xA;H        1.465100000      1.465100000      3.842748000&#xA;H        2.059512000      2.821364000      4.060420000&#xA;H        0.619528000      2.821364000      3.516240000&#xA;H        3.842748000      1.465100000      1.465100000&#xA;H        4.060420000      2.059512000      2.821364000&#xA;H        3.516240000      0.619528000      2.821364000&#xA;H        1.465100000      3.842748000      1.465100000&#xA;H        2.821364000      4.060420000      2.059512000&#xA;H        2.821364000      3.516240000      0.619528000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Grubbs I catalyst&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ru      1.923022    1.782458    4.279841&#xA;Cl      3.508375    2.054525    6.061616&#xA;P       0.620071    3.495015    5.336792&#xA;C       2.967497    2.881791    3.239789&#xA;Cl      0.101984    1.197998    2.838894&#xA;P       2.855052   -0.401312    3.741075&#xA;C       2.954465    3.343954    1.839333&#xA;Cl      3.310061    4.678711   -2.473972&#xA;C       1.880873    3.257547    0.960743&#xA;C       1.973325    3.667349   -0.352875&#xA;C       3.185197    4.168660   -0.812833&#xA;C       4.252303    4.292067    0.012884&#xA;C       4.145603    3.890748    1.341393&#xA;C      -0.637087    4.207002    4.175769&#xA;C      -1.979932    4.694567    4.692950&#xA;C      -2.932151    4.983356    3.538285&#xA;C      -2.339331    5.946324    2.517809&#xA;C      -0.957803    5.505405    2.056512&#xA;C      -0.041546    5.252113    3.235271&#xA;C       1.483967    4.856669    6.241316&#xA;C       0.518279    5.817396    6.934517&#xA;C       1.256971    6.738015    7.899778&#xA;C       2.346867    7.495106    7.161401&#xA;C       3.302557    6.547168    6.465690&#xA;C       2.565538    5.630984    5.497585&#xA;C      -0.333586    2.698072    6.719178&#xA;C      -1.014762    1.418485    6.241149&#xA;C      -1.859827    0.796208    7.358000&#xA;C      -1.043744    0.536540    8.601177&#xA;C      -0.361483    1.805588    9.075860&#xA;C       0.493782    2.427234    7.979254&#xA;C       2.628515   -0.975242    1.991593&#xA;C       3.236690   -0.059043    0.937820&#xA;C       2.687448   -0.416553   -0.435697&#xA;C       2.971677   -1.856587   -0.778868&#xA;C       2.466883   -2.799363    0.290130&#xA;C       2.999176   -2.423218    1.675026&#xA;C       4.622054   -0.540862    4.248552&#xA;C       5.523290    0.453308    3.517538&#xA;C       6.887074    0.518241    4.175936&#xA;C       7.549339   -0.842629    4.233326&#xA;C       6.649166   -1.859830    4.903939&#xA;C       5.271941   -1.927013    4.232992&#xA;C       1.995357   -1.737789    4.704830&#xA;C       2.010202   -1.452357    6.212537&#xA;C       1.292567   -2.535071    7.008639&#xA;C      -0.120796   -2.764155    6.507353&#xA;C      -0.110733   -3.109699    5.031603&#xA;C       0.567738   -2.010334    4.218268&#xA;H       3.716836    3.154747    3.670969&#xA;H       1.120535    2.917815    1.276640&#xA;H       1.189061    3.632874   -0.962082&#xA;H       5.017596    4.645026   -0.262690&#xA;H       4.792512    4.003484    1.897392&#xA;H      -0.815388    3.399122    3.615754&#xA;H      -1.902150    5.510235    5.223685&#xA;H      -2.378700    3.974732    5.228704&#xA;H      -3.726326    5.342882    3.821555&#xA;H      -3.181986    4.184074    3.125510&#xA;H      -2.984577    6.066660    1.748479&#xA;H      -2.250423    6.781889    2.909669&#xA;H      -1.054116    4.674447    1.537657&#xA;H      -0.576082    6.145292    1.467384&#xA;H       0.723372    4.965838    2.939787&#xA;H       0.125317    6.138803    3.761321&#xA;H       1.921040    4.375890    6.908583&#xA;H       1.676706    6.235725    8.636983&#xA;H      -0.145301    5.369077    7.433963&#xA;H       0.091950    6.422548    6.224250&#xA;H       0.676388    7.347621    8.295653&#xA;H       2.791319    8.070082    7.723424&#xA;H       1.896364    8.099751    6.522077&#xA;H       3.914233    7.048872    6.008409&#xA;H       3.698100    6.058287    7.055823&#xA;H       3.116098    5.021349    5.094850&#xA;H       2.158660    6.180195    4.786983&#xA;H      -1.006024    3.336226    6.935353&#xA;H      -1.559257    1.598311    5.456258&#xA;H      -0.325083    0.765200    6.003389&#xA;H      -2.608220    1.404611    7.524315&#xA;H      -2.214635    0.029761    7.044111&#xA;H      -0.418035   -0.117788    8.397718&#xA;H      -1.627822    0.182306    9.277813&#xA;H       0.176244    1.645756    9.829964&#xA;H      -1.053680    2.460437    9.333028&#xA;H       0.877440    3.225032    8.287287&#xA;H       1.235327    1.881817    7.763580&#xA;H       1.720632   -0.869158    1.853889&#xA;H       4.184273   -0.146643    0.965428&#xA;H       3.012517    0.820454    1.136093&#xA;H       1.847900   -0.200343   -0.409930&#xA;H       3.096540    0.151437   -1.100956&#xA;H       3.904789   -1.982340   -0.878422&#xA;H       2.584028   -2.105280   -1.617970&#xA;H       2.763571   -3.703670    0.105411&#xA;H       1.556168   -2.778368    0.317905&#xA;H       4.607903   -0.216537    5.166797&#xA;H       3.990062   -2.477871    1.686571&#xA;H       2.642313   -3.025155    2.339113&#xA;H       5.156005    1.289122    3.512016&#xA;H       5.590006    0.129866    2.598457&#xA;H       6.737696    0.838429    5.032942&#xA;H       7.431919    1.106030    3.739569&#xA;H       8.390108   -0.830548    4.679900&#xA;H       7.729962   -1.148361    3.309561&#xA;H       6.514020   -1.632968    5.797587&#xA;H       7.041226   -2.668142    4.887375&#xA;H       4.662063   -2.600521    4.676553&#xA;H       5.418404   -2.225128    3.326293&#xA;H       2.516573   -2.545817    4.572816&#xA;H       1.564734   -0.621934    6.319621&#xA;H       2.866067   -1.356878    6.528769&#xA;H       1.774146   -3.375710    6.968817&#xA;H       1.285356   -2.270001    7.937592&#xA;H      -0.502310   -3.449358    7.037418&#xA;H      -0.746615   -1.928105    6.670990&#xA;H       0.328488   -3.937948    4.917492&#xA;H      -1.013735   -3.235955    4.696631&#xA;H       0.558651   -2.294513    3.264385&#xA;H       0.034321   -1.214512    4.228140&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.140" LastActivityDate="2014-07-11T08:54:08.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14194" PostTypeId="2" ParentId="6759" CreationDate="2014-07-08T11:46:10.573" Score="2" Body="&lt;p&gt;These are &quot;strike anywhere&quot; matches &lt;a href=&quot;http://www.industrialrev.com/strike-anywhere-matches.html&quot; rel=&quot;nofollow&quot;&gt;http://www.industrialrev.com/strike-anywhere-matches.html&lt;/a&gt;    They need to be stored securely in a fireproof container, and you should never shake or drop the box, or store them near flammable goods. They have special haz-mat shipping requirements too. Oh, and matches of both kinds can be carcinogenic.&lt;/p&gt;&#xA;" OwnerUserId="7058" LastActivityDate="2014-07-08T11:46:10.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14198" PostTypeId="1" AcceptedAnswerId="14199" CreationDate="2014-07-08T14:08:26.290" Score="1" ViewCount="459" Body="&lt;p&gt;I was wondering the weight of 1 liter of ethylene at 2000 bar and 310 Kelvin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to just multiply the weight at 1 bar with 2000 but I don't think i got the correct answer because the answer I got was just below 2.4 kg per liter which seems way too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated it with constant temperature.&lt;/p&gt;&#xA;" OwnerUserId="7062" LastEditorUserId="4945" LastEditDate="2014-07-09T03:02:31.723" LastActivityDate="2014-07-09T03:02:31.723" Title="What is the mass of 1 Liter of ethylene at 2000 bar pressure?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14199" PostTypeId="2" ParentId="14198" CreationDate="2014-07-08T14:40:27.103" Score="3" Body="&lt;p&gt;If we treat it as an ideal gas, then&#xA;\begin{aligned}&#xA;  PV &amp;amp;= n\mathcal{R}T \\ \mathcal{R} &amp;amp;=0.08314~\mathrm{L~bar~K^{-1}~mol^{-1}}\\&#xA;  n &amp;amp;=\frac{PV}{\mathcal{R}T}=\frac{2000\cdot1}{0.08314\cdot310}=77.6~\mathrm{mol}\\&#xA;  M(\ce{ethylene}) &amp;amp;=28~\mathrm{g~mol^{-1}}\\&#xA;  \text{mass of ethylene} &amp;amp;= (28\cdot77.6) = 2.17~\mathrm{kg}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA; Using the link&lt;sup&gt;[a]&lt;/sup&gt; in the comment I find that the density of ethylene at $310~\mathrm{K}$ and $1800~\mathrm{bar}$ is $\varrho = 20.385~\mathrm{mol/L}$, and $\varrho = 21.065~\mathrm{mol/L}$ at $2200~\mathrm{bar}$.  Averaging these two numbers gives an estimated density of $\varrho = 20.725~\mathrm{mol/L}$ at $310~\mathrm{K}$ and $2000~\mathrm{bar}$.  So 1 liter of ethylene under these conditions would contain &#xA;\begin{aligned}&#xA;  \varrho\cdot M_{\ce{C2H4}}\cdot V &amp;amp; ~= m_{\ce{C2H4}}\\&#xA;  20.725~\mathrm{\frac{mol}{L}} \cdot 28~\mathrm{\frac{g}{mol}}\cdot 1~\mathrm{L} &amp;amp; ~\approx 580~\mathrm{g}.\\&#xA;\end{aligned}&#xA;  Is that more in line with your expectations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[a] Majid Jahangiri, Richard T Jacobsen, Richard B. Stewart and Robert D. McCarty: Thermodynamic Properties of Ethylene from the Freezing Line to 450 K at Pressures to 260 MPa. &lt;a href=&quot;http://dx.doi.org/10.1063/1.555753&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Phys. Chem. Ref. Data&lt;/em&gt; &lt;strong&gt;1986&lt;/strong&gt;, &lt;em&gt;15 (2)&lt;/em&gt;, 593&lt;/a&gt;. (DOI:10.1063/1.555753) (&lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd294.pdf&quot; rel=&quot;nofollow&quot;&gt;mirror at www.nist.gov&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-07-09T02:54:31.860" LastActivityDate="2014-07-09T02:54:31.860" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14200" PostTypeId="1" AcceptedAnswerId="14203" CreationDate="2014-07-08T15:35:35.367" Score="2" ViewCount="10985" Body="&lt;p&gt;I am looking for any liquid that does not dissolve in water with higher density and does not react with water, one that is readily available.&lt;/p&gt;&#xA;" OwnerUserId="7063" LastEditorUserId="73" LastEditDate="2014-07-08T21:45:30.623" LastActivityDate="2017-03-01T15:10:07.520" Title="Is there a readily available liquid that's more dense than water and insoluble?" Tags="&lt;water&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ClosedDate="2017-03-02T13:55:34.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="14202" PostTypeId="1" CreationDate="2014-07-08T16:54:47.080" Score="2" ViewCount="1511" Body="&lt;p&gt;Radical Anions are hundreds if not thousands of times more reactive than nucleophiles (right?). A radical anion is just a molecule with an unpaired electron. Where as a normal radical is missing an electron and an atom. If an acidic organic compound is deprotonated to form a nucleophile by removing a proton, then is there not an unpaired electron availible? (which makes the compound nucleophilic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't nucleophiles as reactive as radical anions, since they both have &quot;free electrons&quot;? What is the functional difference?&lt;/p&gt;&#xA;" OwnerUserId="7064" LastActivityDate="2014-07-08T18:20:03.657" Title="What is the functional difference between a radical anion and a nucleophile?" Tags="&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14203" PostTypeId="2" ParentId="14200" CreationDate="2014-07-08T17:08:09.687" Score="4" Body="&lt;p&gt;Trichloroethylene is a common chemical solvent that has reasonably low toxicity, it has density ~1.5 g/ml. However, it slowly becomes restricted. Tetrachloroethylene also may be considered. Dichlormethan has reasonably low toxicity, but its BP is too low. I would not use the liquids mentioned without fume hood on regular basis. One-time exposure is probably OK, but repeating exposure may lead to cancer.  Chloroform may be considered, but I recommend to avoid it, it has accumulating liver toxicity. Tetrachloromethane should be avoided, it also has undesired toxicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury MAY be considered, it is toxic if one is repeatedly exposed to vapours, but has very low vapour pressure and is reasonably cheap. But it has annoying tendency to produce small drops that are very hard to collect and move into smallest holes and accumulates in the system, so it clearly requires dedicated  workplace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquids mentioned above are reasonably cheap, so if they are not restricted in your home country, you should be able to allow yourself one. Unfortunately, that's all ideas I got. Well, there are some more, but they either has unreasonable cost, or are too toxic. Bromine, for example, is toxic and, though not mixing with water, partially dissolves in it.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-08T17:08:09.687" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14204" PostTypeId="2" ParentId="14202" CreationDate="2014-07-08T18:20:03.657" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Radical Anions are hundreds if not thousands of times more reactive&#xA;  than nucleophiles (right?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Radical anions are very reactive, but some nucleophiles (such as $\ce{CH3^{-}}$) are also very reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A radical anion is a molecule that has had an electron added to it.  For example, treating naphthalene with sodium in liquid ammonia transfers an electron from sodium to naphthalene forming the naphthalene radical anion, the electron is unpaired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A radical arises from the homolytic breaking of a bond.  This process leaves an unpaired electron with each of the two molecular fragments.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an acidic organic compound is deprotonated to form a nucleophile by&#xA;  removing a proton, then is there not an unpaired electron availible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Removal of a proton $\ce{H^+}$ (just a proton, no electrons) would leave 2 paired electrons (and a negative charge) on the remainder of the original molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleophiles come in a lot of flavors.  They can be neutral, like the nitrogen in an amine, or they can be negatively charged.  If negatively charged, the nucleophile could be a radical anion or a carbanion.  So radical anions just belong to a sub-class of nucleophiles.  It's probably fair to say that carbanions and radical anions are usually better nucleophiles than a neutral nucleophile.  Comparing the nucleophilicity of a carbanion and a radical anion is harder to generalize since they are both so reactive, but the carbanion with 2 electrons would probably be more nucleophilic than the radical anion with one electron..&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-08T18:20:03.657" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14207" PostTypeId="1" CreationDate="2014-07-08T21:09:17.413" Score="0" ViewCount="670" Body="&lt;p&gt;On wiki, linoleic acid has the code number &lt;strong&gt;FA 18:2 cis,cis-9,12&lt;/strong&gt;. But then what exactly is FA 18:2 undifferentiated?&lt;/p&gt;&#xA;" OwnerUserId="7066" LastEditorUserId="16683" LastEditDate="2015-10-16T16:46:18.813" LastActivityDate="2015-10-16T16:46:18.813" Title="Is FA 18:2 undifferentiated the same as linoleic acid?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14208" PostTypeId="2" ParentId="14207" CreationDate="2014-07-08T22:07:01.143" Score="3" Body="&lt;p&gt;It's part of an old nomenclature used in the fatty acid industry.  FA 18:2 translates to &lt;strong&gt;F&lt;/strong&gt; atty &lt;strong&gt;A&lt;/strong&gt; cid with &lt;strong&gt;18&lt;/strong&gt; carbon atoms and &lt;strong&gt;2&lt;/strong&gt; double bonds.  You can read more about these &quot;Lipid numbers&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Fatty_acid#Nomenclature&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  So to answer your question, there are number of fatty acids with 18 carbons and 2 double bonds, linoleic acid is just one of them.  FA 18:2 includes linoleic acid, but is not limited to just linoleic acid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-08T22:48:03.873" LastActivityDate="2014-07-08T22:48:03.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14209" PostTypeId="1" AcceptedAnswerId="14211" CreationDate="2014-07-09T06:12:35.183" Score="5" ViewCount="204" Body="&lt;p&gt;For example, let us that say we have a question where we have to use the ideal gas law and that the moles and temperature are constant. Then the gas law can also be written as $$nRT=PV$$ However, according to my solutions manual, this can also be written as $$nRT=P_1V_1=P_2V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how are we justified in rearranging the equation like this?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="194" LastEditDate="2018-02-16T14:15:26.227" LastActivityDate="2018-02-16T14:15:26.227" Title="Rearrangement of a gas law equation" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14210" PostTypeId="1" CreationDate="2014-07-09T06:45:33.547" Score="1" ViewCount="1324" Body="&lt;p&gt;I would like to know the highest density material between 10-17 g/mol range that is less than mercury and does not react with mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also would I like to know the highest density metal that is in solid form naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which ones are easily available and are highly dense?&lt;/p&gt;&#xA;" OwnerUserId="7063" LastEditorUserId="73" LastEditDate="2014-07-09T18:11:29.640" LastActivityDate="2014-07-09T18:11:29.640" Title="Which metals with high density do not react with mercury?" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14211" PostTypeId="2" ParentId="14209" CreationDate="2014-07-09T07:39:35.563" Score="9" Body="&lt;p&gt;If one rearranges the ideal gas law equation, you can obtain the following (assuming $n$ and $T$ are non-zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{PV}{nT} = R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ is a constant, and there are in fact infinitely many possible sets of values $(P, V, n, T)$ that satisfy the equation. Let $(P_1, V_1, n_1, T_1)$ denote one such set, and let $(P_2, V_2, n_2, T_2)$ denote a second one. Then, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_1V_1}{n_1T_1} = R$$&#xA;$$\frac{P_2V_2}{n_2T_2} = R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by the transitive property:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can let $n$ and $T$ be constant as well (as in your specific example), but the principal is the same and leads to an analogous result, since any product and/or ratio of constants will itself be a constant.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-09T07:39:35.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14212" PostTypeId="1" AcceptedAnswerId="14213" CreationDate="2014-07-09T12:18:12.453" Score="2" ViewCount="958" Body="&lt;p&gt;Increasing the boiling point of water by adding salts or other compounds would be very useful when working with a water bath while refluxing, because with pure water you cannot reach temperatures over 70°C (at that point too much evaporates from the bath). I know that simple table salt would work, but are there other and better compounds/salts available?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="177" LastEditDate="2014-07-09T18:48:29.653" LastActivityDate="2014-07-09T18:48:29.653" Title="What salts (or other substances) are available for increasing water's boiling point?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14213" PostTypeId="2" ParentId="14212" CreationDate="2014-07-09T13:16:38.067" Score="5" Body="&lt;p&gt;For relatively dilute solutions of non-volatile solutes, the following equation (which assumes ideal behavior) can be used to predict boiling point elevation with reasonable accuracy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta T_b = K_b \cdot b \cdot i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $K_b$ is the solvent-specific ebullioscopic constant, $b$ is the molality of the solute, and $i$ is the so-called van 't Hoff factor, which accounts for the extent of solute dissociation. For example, if one were to dissolve $\ce{NaCl}$ in water and assume total dissociation, $i = 2$, since two ions are produced in solution. It follows that, &lt;em&gt;ceteris paribus&lt;/em&gt;, a salt that dissociates to yield a greater number of ions will produce a greater boiling point elevation. In practice, the magnitude of boiling point elevation is capped by solubility and deviation from ideal behavior at concentration extremes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water also forms negative (i.e., higher-boiling) azeotropes with various substances. Addition of a miscible higher-boiling liquid compound can also effect elevation of the boiling point (assuming the mixture does not form a positive azeotrope), a colligative property similar to the above case for salts. Various polyols (e.g., glycerol, propylene glycol, etc.) are commonly used for such applications.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-09T13:24:15.133" LastActivityDate="2014-07-09T13:24:15.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14214" PostTypeId="2" ParentId="14210" CreationDate="2014-07-09T13:21:58.410" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://periodictable.com/Properties/A/Density.v.html&quot; rel=&quot;nofollow&quot;&gt;Here is a link&lt;/a&gt; to a chart of all the elements in the Periodic table and their densities.  Iridium at $22.65~\mathrm{g/cm^3}$ is the densest of all the elements with Osmium a close second at $22.61~\mathrm{g/cm^3}$.  Sigma-Aldrich sells both of these metals (Iridium \$126.50 for 0.5 gram; Osmium \$195 for 1 gram).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the highest density material between 10 17 range that is less than&#xA;  mercury and does not react with mercury&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mercury has a density of $13.534~\mathrm{g/cm^3}$.  It sounds like you're asking for something with a density right below mercury that doesn't react with mercury. Rhodium at $12.45~\mathrm{g/cm^3}$ is probably the most readily available element with a density less than mercury.  As to rhodium's reaction with mercury, &lt;a href=&quot;http://link.springer.com/article/10.1023%2FA%3A1010174412608#page-1&quot; rel=&quot;nofollow&quot;&gt;an alloy is formed at 175 C&lt;/a&gt;.  If you're working at room temperature, especially for shorter periods of time (like hours) you'd probably be OK, but it really depends on your specific application.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-07-09T13:39:40.070" LastActivityDate="2014-07-09T13:39:40.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14215" PostTypeId="1" AcceptedAnswerId="14216" CreationDate="2014-07-09T13:25:13.180" Score="12" ViewCount="642" Body="&lt;p&gt;This is a follow up on &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/14179/4945&quot;&gt;Generally Octahedral - Meaning?&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Dissenter asked in the comments to my answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not sure why the $\ce{F_{eq}-S-F'_{eq}}$ bond angles would be compressed, even if the $\ce{S}$ were out of the plane. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since ron's answer already satisfies the original question about $\ce{BrSF5}$ and I did not want to clutter my answer even further, I am asking it here in a &lt;a href=&quot;http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/&quot;&gt;Q&amp;amp;A&lt;/a&gt; style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does the pyramidalization effect the bonding angles in a distorted octahedral environment?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-05-25T04:43:49.960" Title="How does the pyramidalization effect the bonding angles in a distorted octahedral environment?" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14216" PostTypeId="2" ParentId="14215" CreationDate="2014-07-09T13:25:13.180" Score="11" Body="&lt;p&gt;So we are again talking about $\ce{BrSF5}$ but this post extends to all generally octahedral complexes/ molecules and even quadratic planar geometries. The main requirement is, that the molecule has a $C_{\mathrm{4v}}$ point group. (It may be extended to all non-linear $C_{n\mathrm{v}}\ (n&amp;gt;1)$ point groups.) First of all reconsider the molecular shape again.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eh9Icm.jpg&quot; alt=&quot;BrSF5&quot;&gt;&lt;br&gt;&#xA;Now admittedly the degree of pyramidalization is more than minimal, with less than $1^\circ$ deviation from the plane. However, this has a slight effect on the equatorial $\angle(\ce{F-S-F'})$ bond angle, purely to symmetric reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic trigonometric considerations are enough to understand this phenomena. Consider the following scheme, where the peak of the pyramid is represented by $\mathbf{\ce{Y}}=\ce{S}$ and the base is span by $\mathbf{\ce{X_{eq}}}=\ce{F_{eq}}$, $\mathbf{\ce{X'_{eq}}}=\ce{F'_{eq}}$, $\mathbf{\ce{X''_{eq}}}=\ce{F''_{eq}}$ and for completeness $\mathbf{\ce{X'''_{eq}}}=\ce{F'''_{eq}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z02Yx.png&quot; alt=&quot;pyramid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have some basic quantities, that we can address, but in the end we can derive the connection of the two angles without caring about the other quantities, only knowing the point group.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$a$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F'_{eq}})$ distance&lt;/li&gt;&#xA;&lt;li&gt;$d$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F''_{eq}})$ distance&lt;/li&gt;&#xA;&lt;li&gt;$s$ is the $\mathbf{d}(\ce{S\cdots{}F_{eq}})$ distance&lt;/li&gt;&#xA;&lt;li&gt;$h$ is the height of the pyramid and it refers to the out of plane shift of $\ce{S}$ compared to the $\ce{F_{eq}\cdots{}F'_{eq}\cdots{}F''_{eq}\cdots{}F'''_{eq}}$ plane&lt;/li&gt;&#xA;&lt;li&gt;$h'$ is another virtual quantity, the angle bisector of $\angle(\ce{F_{eq}-S-F'_{eq}})$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F''_{eq}})$&lt;/li&gt;&#xA;&lt;li&gt;$\beta$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F'_{eq}})$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;noreferrer&quot;&gt;Pythagoras's Theorem&lt;/a&gt; we can easily derive the following connection (square in the scheme):&#xA;\begin{align}&#xA;             &amp;amp;&amp;amp; d^2 &amp;amp;= a^2 + a^2\\&#xA;  \therefore &amp;amp;&amp;amp; d   &amp;amp;= \sqrt{2}\cdot a\tag{1}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we are going to use some &lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonometric_functions&quot; rel=&quot;noreferrer&quot;&gt;trigonometric functions&lt;/a&gt; for the angles (look at the triangles):&#xA;\begin{align}&#xA;             &amp;amp;&amp;amp; \sin\left(\frac{\alpha}{2}\right) &#xA;                    &amp;amp;= \frac{\frac12d}{s}\\&#xA;  \therefore &amp;amp;&amp;amp; s   &amp;amp;= \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}\tag{2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;             &amp;amp;&amp;amp; \sin\left(\frac{\beta}{2}\right) &#xA;                    &amp;amp;= \frac{\frac12a}{s}\\&#xA;  \therefore &amp;amp;&amp;amp; s   &amp;amp;= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\tag{3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's derive the connection between the angles by equalizing $(2)$ and $(3)$:&#xA;\begin{align}&#xA;             &amp;amp;&amp;amp; \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)} &#xA;                     &amp;amp;= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\\&#xA;  \therefore &amp;amp;&amp;amp; \sin\left(\frac{\beta}{2}\right)&#xA;                     &amp;amp;= \frac{a}{d}\sin\left(\frac{\alpha}{2}\right)\\&#xA;  \text{with (1)}\implies &amp;amp;&amp;amp;&#xA;                \sin\left(\frac{\beta}{2}\right)&#xA;                     &amp;amp;= \frac{1}{\sqrt{2}}\sin\left(\frac{\alpha}{2}\right)\\&#xA;  \equiv     &amp;amp;&amp;amp; \sqrt{2} &amp;amp;= \frac{\sin\left(\frac{\alpha}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}\tag{4}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can see, that the degree of pyramidalization is codependent on the equatorial bond angles. When the pyramidalization increases, $\alpha\to0$, then the equatorial bond angle has to decrease, $\beta\to0$, i.e. a constant ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case of $\alpha = 180^\circ$ this simplifies to $\beta=90^\circ$. In all other cases you can derive the degree of pyramidalization from the equatorial bond angles and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transformation of $(4)$ yields a function that displays the equatorial bond angle in dependence of the pyramidalization, with $\alpha\in~]0^\circ;180^\circ]$ or $\alpha\in~]0;\pi]$.&#xA;$$\beta = 2\cdot\arcsin\left[\frac{1}{\sqrt{2}}\cdot\sin\left(\frac{\alpha}{2}\right)\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PKRNP.jpg&quot; alt=&quot;graph of angle dependency&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Addendum&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The out of plane height can also be derived from Pythagoras's theorem (look at right triangle):&#xA;\begin{aligned}&#xA;                 &amp;amp;&amp;amp; h^2 &amp;amp;= s^2 + \left(\frac{d}{2}\right)^2\\&#xA;  \therefore     &amp;amp;&amp;amp; h   &amp;amp;= \sqrt{s^2 + \left(\frac{d}{2}\right)^2}\\&#xA;  \text{with (1)}&amp;amp;&amp;amp; h &amp;amp;= \sqrt{s^2 + \frac{1}{2}a^2}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is usually a value that is given for domed complexes, so it is nice to know how to find it.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-05-25T04:43:49.960" LastActivityDate="2015-05-25T04:43:49.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14217" PostTypeId="2" ParentId="14212" CreationDate="2014-07-09T13:31:10.820" Score="4" Body="&lt;p&gt;We always used (silicone) oil baths in the lab for refluxing and distilling.  It offers a much wider temperature range than water.  Here's two links that describe &lt;a href=&quot;http://www.chem.wisc.edu/areas/organic/orglab/tech/oilbath.htm&quot; rel=&quot;nofollow&quot;&gt;oil bath usage&lt;/a&gt; and &lt;a href=&quot;http://www.nap.edu/openbook.php?record_id=4911&amp;amp;page=117&quot; rel=&quot;nofollow&quot;&gt;lab reaction heating in general&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-09T13:31:10.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14218" PostTypeId="1" CreationDate="2014-07-09T14:26:50.173" Score="5" ViewCount="787" Body="&lt;p&gt;It is an easily defensible argument that this is a medical question rather than a chemistry one, however, I feel that it leans ever so slightly towards chemistry, and so I posted it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;new&quot; OTC stimulant being pushed around that shall remain unnamed so that this isn't flagged as an advertisement or anything.  In looking at the list of ingredients, I don't see anything that would act as a stimulant anywhere in that list.  Having said that, I'm not an expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I came to the experts.  ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the ingredients according to the bottle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z4lXJ.jpg&quot; alt=&quot;Ingredients&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just to clarify, the heart of this question is &quot;Which ingredient is the stimulant?&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7073" LastEditorUserId="73" LastEditDate="2014-07-09T17:32:41.663" LastActivityDate="2014-07-13T00:39:29.150" Title="What is the active ingredient in this &quot;stimulant?&quot;" Tags="&lt;biochemistry&gt;&lt;medicinal-chemistry&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14220" PostTypeId="2" ParentId="14218" CreationDate="2014-07-09T16:56:29.763" Score="2" Body="&lt;p&gt;If you google Schizandrol A ( 1,2,3,10,11,12-hexamethoxy-6,7-dimethyl-5,6,7,8-tetrahydrodibenzo[a,c]cycloocten-6-ol, &lt;a href=&quot;http://journals.iucr.org/e/issues/2008/08/00/bt2738/bt2738bdy.html&quot; rel=&quot;nofollow&quot;&gt;here's a link&lt;/a&gt; to a picture of the structure) you find that it is an herb capable of curing just about anything.  If you poke around a little deeper, you find two repetitive themes 1) it increases focus and 2) synergistically enhances the effect of other chemicals ingested along with it.  You also note this herb has caught the attention of the bodybuilding community.  The lifters use terms like &quot;pump and rush&quot;, &quot;being on another planet&quot;, etc.  Apparently it can take focus to a whole new level. From what I read it looks like it's always ingested with other stuff so it's hard to separate out what's what.  My guess is that all the ingredients in the bottle come into play and the effect they have together, particularly with the Schizandrol A present, is much greater than any ingredient alone.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-09T16:56:29.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14221" PostTypeId="1" CreationDate="2014-07-09T17:00:28.657" Score="3" ViewCount="4820" Body="&lt;p&gt;I am &lt;a href=&quot;https://earthscience.stackexchange.com/q/2223/641&quot;&gt;trying to determine&lt;/a&gt; what the derived partial pressure would be for CO2 in the global atmosphere from historical inconstant temperatures and CO2 concentrations of the global atmosphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aG30w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to do this?  If so, how?&lt;/p&gt;&#xA;" OwnerDisplayName="user7075" LastEditorUserId="-1" LastEditorDisplayName="user7075" LastEditDate="2017-04-13T12:50:29.617" LastActivityDate="2014-08-20T03:42:57.327" Title="Henry's Law with varying temperature" Tags="&lt;gas-laws&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14222" PostTypeId="1" CreationDate="2014-07-09T17:13:45.570" Score="9" ViewCount="13855" Body="&lt;p&gt;I learnt about intramolecular hydrogen bonding today, which occurs between molecules such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrophenol&quot; rel=&quot;nofollow noreferrer&quot;&gt;ortho-nitrophenol&lt;/a&gt;. What I was told is that in case of intramolecular bonding, the molecules separate from each other, opposite of what happens during intermolecular hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why is this. Why would attraction within a molecule cause separation of the molecules? There is obviously some intermolecular hydrogen bonding present in this case, but I understand that this might be negligible. Still, why should the molecules spread away from each other?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastEditorUserId="4945" LastEditDate="2019-04-04T14:15:47.803" LastActivityDate="2021-03-15T17:39:52.347" Title="How does intramolecular hydrogen bonding cause the molecules to be separated from each other?" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14223" PostTypeId="1" AcceptedAnswerId="14224" CreationDate="2014-07-09T17:23:01.163" Score="2" ViewCount="1389" Body="&lt;p&gt;For the equation $E_k=\frac{1}{2}mu^2$ what does the $m$ variable represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say mass but the textbook does not say explicitly that it is anything.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="30836" LastEditDate="2017-10-14T19:13:30.500" LastActivityDate="2017-10-14T19:13:30.500" Title="What are the units for m in the average kinetic energy formula?" Tags="&lt;kinetic-theory-of-gases&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14224" PostTypeId="2" ParentId="14223" CreationDate="2014-07-09T17:33:29.540" Score="6" Body="&lt;p&gt;$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-09T17:53:23.980" LastActivityDate="2014-07-09T17:53:23.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14225" PostTypeId="2" ParentId="14222" CreationDate="2014-07-09T19:57:56.640" Score="10" Body="&lt;p&gt;Molecules with &lt;strong&gt;inter&lt;/strong&gt;-molecular hydrogen bonds tend to associate with one another, while molecules with &lt;strong&gt;intra&lt;/strong&gt;-molecular hydrogen bonds tend to associate with themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-nitrophenol are often used as examples of molecules with &lt;strong&gt;intra&lt;/strong&gt;- and &lt;strong&gt;inter&lt;/strong&gt;-molecular hydrogen bonds respectively (see picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n3JML.jpg&quot; alt=&quot;ortho/para-nitrophenol and their hydrogen bonds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of their molecular structures the &lt;em&gt;ortho&lt;/em&gt; isomer can form an intramolecular hydrogen bond with itself, whereas the &lt;em&gt;para&lt;/em&gt; isomer can only hydrogen bond with another molecule of &lt;em&gt;p&lt;/em&gt;-nitophenol.  One would say that the &lt;em&gt;para&lt;/em&gt; isomer associates with other &lt;em&gt;p&lt;/em&gt;-nitrophenol molecules through hydrogen bonding, whereas the &lt;em&gt;ortho&lt;/em&gt; isomer associates with itself through hydrogen bonding.  Because the &lt;em&gt;para&lt;/em&gt; isomer associates with other molecules of &lt;em&gt;p&lt;/em&gt;-nitrophenol through hydrogen bonding, it takes more energy to break up this association and cause the molecule to melt or boil.  On the other hand, the &lt;em&gt;ortho&lt;/em&gt; isomer melts and boils at lower temperatures because there are fewer intermolecular forces (such as the hydrogen bonding) holding the &lt;em&gt;ortho&lt;/em&gt; molecules together. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2019-04-04T14:22:48.613" LastActivityDate="2019-04-04T14:22:48.613" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="14226" PostTypeId="2" ParentId="14060" CreationDate="2014-07-09T23:38:15.513" Score="6" Body="&lt;p&gt;The data given is this question has changed sufficiently from the original post that it warrants a new answer to the latest (and much more reliable) data, although should be taken in conjunction with Martin's response, and Long's original response. Let's first look at the IR data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IR Spectrum:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much better spectrum, and, compared to the original spectrum, serves as a very good example of how easy it is to introduce unwanted water into a sample. Peaks around 3020, 763 and 692 indicate an aromatic. The 763 and 692 are indicative of a mono-substituted benzene ring. There are also some clear aromatic overtones between 1950-1750. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NMR Spectrum:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still only a partial NMR spectrum, and one that we can only assume is correctly calibrated. Linewidths are better, and significantly, we see a small coupling from the peak at 7.15ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.41(2H, , H3) and 7.56(2H, d, H2). Long-range (small couplings) are ignored. We also have a peak at 7.15 (1H, d, H5). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and &amp;lt;1 for para. Shifted from benzene (7.34ppm), the substituents come at H2 (+0.22), H3 (+0.07), H4 (-0.04). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak at 7.15 could be one of few things. A bis-halide is a possibility, but more likely is an alkene. No real evidence in the IR supports an alkene (although my IR knowledge is limited). However, a &lt;strong&gt;symmetrical&lt;/strong&gt; alkene may be a possibility, as the main -C=C- absorbance would be IR inactive, although Raman active. This gives us two possibilities that come to immediate mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cis&lt;/em&gt;-stilbene (&lt;em&gt;cis&lt;/em&gt;-1,2-Diphenylethylene) m.p. 5-6oC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1eGjd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additive prediction for the aromatics and alkene proton can be done using many basic NMR texts. I'm using Pretsch, Buhlmann and Badertscher for this case (as it's the only one I have with me at the moment):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H2 : 7.34+0    = 7.33&lt;/li&gt;&#xA;&lt;li&gt;H3 : 7.34-0.07 = 7.27&lt;/li&gt;&#xA;&lt;li&gt;H4 : 7.34-0.15 = 7.19&lt;/li&gt;&#xA;&lt;li&gt;H5 : 5.25+1.38(Ar&lt;sub&gt;gem&lt;/sub&gt;)-0.07(Ar&lt;sub&gt;trans&lt;/sub&gt;) = 6.56&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;trans&lt;/em&gt;-stilbene (&lt;em&gt;trans&lt;/em&gt;-1,2-Diphenylethylene) m.p. 123-125oC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMhmN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H2 : 7.34+0.16 = 7.50&lt;/li&gt;&#xA;&lt;li&gt;H3 : 7.34+0    = 7.34&lt;/li&gt;&#xA;&lt;li&gt;H4 : 7.34-0.15 = 7.19&lt;/li&gt;&#xA;&lt;li&gt;H5 : 5.25+1.38(Ar&lt;sub&gt;gem&lt;/sub&gt;)+0.36(Ar&lt;sub&gt;cis&lt;/sub&gt;) = 6.99&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As this is clearly a problem session for you to make a final determination of structure, I will leave it for you to make that judgement. however, as a final point, this whole exercise, I hope, serves to demonstrate the importance of preparing good quality samples in order to collect high quality and reliable data. I hope you can see the difference between your original data and final data. This is a message that I continually preach to all those researchers I work with, both at an undergraduate level and postgraduate. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-07-09T23:38:15.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14228" PostTypeId="2" ParentId="14223" CreationDate="2014-07-10T03:02:02.167" Score="4" Body="&lt;p&gt;$$E_k = \frac{1}{2}mu^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&#xA;&lt;strong&gt;$m$ is molar mass (unit - kg/mol)&lt;/strong&gt;, and&#xA;$u$ is root mean square speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to wikipedia,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The molar mass of atoms of an element is given by the atomic mass of the element multiplied by the molar mass constant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Molar_mass&quot; rel=&quot;nofollow&quot;&gt;this page (from wikipedia)&lt;/a&gt; for complete explanation.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2014-07-24T03:25:45.210" LastActivityDate="2014-07-24T03:25:45.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14230" PostTypeId="1" AcceptedAnswerId="14252" CreationDate="2014-07-10T09:36:10.747" Score="3" ViewCount="92" Body="&lt;p&gt;I was wondering why FCKWs can destroy the ozone layer.&#xA;It seems like these gases have a high molecular weight in comparison to &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;.&#xA;How can these gasses come up to the outer layer of our atmosphere? In particular, how long (on average) would it take to get to such an altitude (through the diffusion of such a particle)?&lt;/p&gt;&#xA;&lt;p&gt;I mean, they are basically several times heavier than all the other components of our atmosphere combined. Hence, it shouldn't be very energetically stable for them to float on the outer layers of space on top of other lighter particles.&lt;/p&gt;&#xA;" OwnerUserId="7082" LastEditorUserId="102629" LastEditDate="2021-01-08T11:14:00.017" LastActivityDate="2021-01-08T11:14:00.017" Title="How does Dichlorofluoromethane get into the stratosphere?" Tags="&lt;thermodynamics&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14231" PostTypeId="1" CreationDate="2014-07-10T10:15:47.693" Score="1" ViewCount="296" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming pure $\ce{2s}$ and $\ce{2p}$ orbitals of carbon are used in forming $\ce{CH4}$ molecule, which of the following statements is false? (single choice question)&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Three $\ce{C-H}$ bonds will be at right angles&lt;/li&gt;&#xA;  &lt;li&gt;One $\ce{C-H}$ bond will be weaker than the other three $\ce{C-H}$ bonds&lt;/li&gt;&#xA;  &lt;li&gt;The shape of the molecule will be tetrahedral&lt;/li&gt;&#xA;  &lt;li&gt;The angle of $\ce{C-H}$ bond formed by $\ce{s-s}$ overlapping will be uncertain with respect to other three bonds.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me, the second option is false because $\ce{1s-2s}$ overlapping is stronger than $\ce{1s-2p}$. I just randomly found this question and I don't know the correct answer. &lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="865" LastEditDate="2021-05-22T20:58:39.970" LastActivityDate="2021-05-22T20:58:39.970" Title="Determining covalent bond strengths without hybridisation" Tags="&lt;bond&gt;&lt;molecular-structure&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14232" PostTypeId="2" ParentId="13483" CreationDate="2014-07-10T11:38:16.853" Score="22" Body="&lt;p&gt;Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-&lt;a href=&quot;http://goldbook.iupac.org/A00441.html&quot; rel=&quot;noreferrer&quot;&gt;aromatic&lt;/a&gt;, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following valence molecular orbitals were derived at BP86/cc-pVTZ, Energy level decreases from top to bottom (&lt;a href=&quot;https://i.stack.imgur.com/rR7zr.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rR7zrl.jpg&quot; alt=&quot;valence molecular orbitals of SO3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding picture is usually trivialised as each $\ce{S-O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=-\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three $\pi$ bonding orbitals, resulting in electron density deficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the $\ce{S-O}$ bond in sulfurtrioxide is still stronger and shorter, than in the sulfite anion. (BO means bond order)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrr}&#xA;              &amp;amp; \mathbf{d}(\ce{S-O}) &amp;amp; \text{BO}(\ce{S-O})\\\hline&#xA;\ce{SO3}      &amp;amp; 1.458                &amp;amp; 1.4                \\&#xA;\ce{SO3^{2-}} &amp;amp; 1.578                &amp;amp; 1.1                \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for this is, that upon reduction of sulfurtrioxide to sulfite, $$\ce{SO3 + 2e- &amp;lt;=&amp;gt; SO3^{2-}},$$ you will fill the LUMO. As already stated, this orbital is antibonding with respect to the $\ce{S-O}$ bonds. This is energetically unfavourable for the molecule, therefore the molecule distorts to get rid of the symmetry constraints, i.e. it adapts the lower $C_\mathrm{3v}$ point group. This results in pyramidalization and a different bonding pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will now no longer find any $\pi$ type bonding orbitals. More accurately this molecule is best described consisting of three equal $\ce{S-O}$ covalent single bonds, and three lone pairs at each oxygen. This results in a charge at each oxygen of $q=-1$ and at the sulfur of $q=1$ (which does not fit the traditional Lewis type bonding description.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also reflected in the molecular orbitals (BP86/cc-pVTZ, energy decreases from top to bottom, &lt;a href=&quot;https://i.stack.imgur.com/scO47.jpg&quot; rel=&quot;noreferrer&quot;&gt;enlarge&lt;/a&gt;).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/scO47l.jpg&quot; alt=&quot;valence molecular orbitals of $\ce{SO3^{2-}}$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also very important to understand that the latter molecule would not be stable in gas phase, since it is too highly charged. Without compensation of solvent and/ or counter ions, this molecule will most certainly fragmentate. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-10T11:38:16.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14233" PostTypeId="2" ParentId="14231" CreationDate="2014-07-10T12:07:17.860" Score="5" Body="&lt;p&gt;As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisation&lt;sup&gt;note&lt;/sup&gt; is forbidden - which is wrong, $\text{[very]}_n$ wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. &lt;strong&gt;Answer 3 would be wrong.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. &lt;strong&gt;Orbitals are not rigid.&lt;/strong&gt; They are quite flexible in the end.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;br&gt;&#xA;Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-10T12:07:17.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14235" PostTypeId="1" AcceptedAnswerId="14271" CreationDate="2014-07-10T12:34:45.347" Score="0" ViewCount="2467" Body="&lt;p&gt;If the quantum numbers of an ion are given, how can we identify this ion by its quantum numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;n = 3; $l$ = 2; $m_l$ = 0; m$_s$ = + ½&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I have so far: $n$ is 3 and $l$ is 2 which corresponds to $d$, so the last shell is $3d1$, therefore we have 21 $e^-$ and the answer will be $\ce{Sc}$ ion (from the periodic table) and the answer I got is $\ce{Mn^{2+}}$.   Is this the correct approach?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="73" LastEditDate="2014-07-10T22:18:37.097" LastActivityDate="2014-07-11T20:45:57.480" Title="How to identify ions by their quantum number?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14239" PostTypeId="1" CreationDate="2014-07-10T16:10:43.133" Score="-3" ViewCount="5108" Body="&lt;p&gt;Why is the unit joule is written as &quot;joule&quot; and &quot;J&quot;, I mean what does &quot;J&quot; mean and what does &quot;joule&quot; mean.&lt;/p&gt;&#xA;" OwnerUserId="7088" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T20:30:47.953" LastActivityDate="2016-11-06T20:30:47.953" Title="Why is it written as &quot;joule&quot; and &quot;J&quot;?" Tags="&lt;terminology&gt;&lt;units&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2016-11-06T22:54:26.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="14240" PostTypeId="2" ParentId="14239" CreationDate="2014-07-10T16:25:24.263" Score="6" Body="&lt;p&gt;In SI, every base units and many derived units have an official name and an official symbol. For example &quot;meter&quot;-&quot;m&quot;, &quot;second&quot;-&quot;s&quot; or &quot;joule&quot;-&quot;J&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are grammatical rules how to use them, but more or less they are interchangeable. If you wonder about the capitalization: in English, the unit names are common nouns and not capitalized (&quot;joule&quot;) even if they are derived from a person's name (Joule) or the symbol is a capital letter (&quot;J&quot;).&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-10T16:25:24.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14241" PostTypeId="1" AcceptedAnswerId="14245" CreationDate="2014-07-10T18:57:17.250" Score="6" ViewCount="903" Body="&lt;p&gt;I particulary enjoy performing chemical experiments without buying any chemicals for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was always sad about the fact that one glowstick lights out, you just throw it into trash can (I like all means of recycling). So I was wondering if I could (pick one or more):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Recycle the chemicals in it to make it glow again&lt;/li&gt;&#xA;&lt;li&gt;Use the chemicals for other interesting (and maybe even useful) experiment&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="33991" LastEditDate="2016-11-23T11:59:53.430" LastActivityDate="2016-11-23T11:59:53.430" Title="Can I somehow 'reuse' the chemicals in glow sticks?" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14243" PostTypeId="2" ParentId="14241" CreationDate="2014-07-10T20:34:52.573" Score="0" Body="&lt;p&gt;IIRC the reaction is an oxidation.  Perhaps it could be reduced.  I believe that hair permanent may be a reducing agent.  It would be interesting to see if permanent would work but only done in a safe ventilated place.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-10T20:34:52.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14244" PostTypeId="1" CreationDate="2014-07-10T22:09:06.997" Score="4" ViewCount="29410" Body="&lt;p&gt;I am doing some adsorption tests on bio materials and would like to test the data with Langmuir and Freundlich adsorption isotherm models. Assume these values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;adsorbent(g).   initial con (µg/L)  final con.(µg/L)&#xA;0.1                   1                0.65&#xA;0.1                   5                2.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(Sample volume is 20ml. In both cases the dose of adsorbent was the same (0.1g) and in the first initial concentration is 1ppb and, after adsorption, is 0.65ppb.  In the second initial concentration is 5ppb and, after adsorption, is 2.1ppb.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these data how can I use Langmuir and Freundlich isotherms models to get the constants and check whether my adsorption data fit the model or not.  (If any other parameter is needed you can assume some reasonable value.)&lt;/p&gt;&#xA;" OwnerUserId="7091" LastEditorUserId="5354" LastEditDate="2015-05-01T20:12:55.390" LastActivityDate="2017-07-13T07:30:06.987" Title="How to determine Langmuir adsorption isotherm constants and use the models" Tags="&lt;adsorption&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14245" PostTypeId="2" ParentId="14241" CreationDate="2014-07-10T23:17:20.890" Score="5" Body="&lt;p&gt;Here is a generalized version of what's going on chemically.  The key steps are dioxetane formation (the highly strained 4-membered ring) and its subsequent decomposition and energy transfer to a fluorescent dye, which then decays back to its ground state by emitting light.  By using different fluorescers, you can change the wavelength (color) of the light emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZKYFZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each company that manufactures glowsticks has their own patent(s) and proprietary process, hence the variation from product to product.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So I was wondering if I could (pick one or more):&#xA;  1.Recycle the chemicals in it to make it glow again&#xA;  2.Use the chemicals for other interesting (and maybe even useful) experiment&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) Since the starting oxalate ester and concentrated hydrogen peroxide are consumed in the reaction, to make it glow again you would need to put in a new batch of both the oxalate ester and concentrated hydrogen peroxide.  You probably wouldn't want to open the glowstick because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there might be some concentrated hydrogen peroxide remaining (if I were the manufacturer, I would want this cheap reagent present in a slight excess to make sure all of the ester is consumed)&lt;/li&gt;&#xA;&lt;li&gt;many of the commonly used fluorescers are irritants, both externally and internally&lt;/li&gt;&#xA;&lt;li&gt;phthalate esters are commonly used as solvents and they pose health risks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) Even after all of the starting materials are used up, the fluorescer still remains in solution.  Exposing the glowstick to a black light will still cause it to fluoresce.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-10T23:17:20.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14246" PostTypeId="1" AcceptedAnswerId="14248" CreationDate="2014-07-11T00:42:35.490" Score="5" ViewCount="2475" Body="&lt;p&gt;I am new to the organic chemistry world.&#xA;I was delving for the exact answer to my question, and I've found some results:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In wikipedia (french), is was mentioned that an organic molecule contains Hydrogen, so that $\ce{CO2}$ is a non-organic molecule, but wait, what about $\ce{H2O}$? It's almost said a non-organic molecule and it does contain Hydrogen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And on other websites, I've read that an organic molecule is supposed to have 'Carbon' in its components, else, it will be called 'a non-organic molecule', the result that is an obvious contradictory to the first one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I've deducted from all my results is, that the organic molecule must contain Hydrogen or Carbon, and I'm not pretty sure of my deduction.&lt;/p&gt;&#xA;" OwnerUserId="7093" LastEditorUserId="7475" LastEditDate="2016-10-06T19:50:00.417" LastActivityDate="2016-10-06T19:50:00.417" Title="What does 'organic/non-organic molecule' means exactly?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;molecules&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="1" ClosedDate="2016-10-06T21:55:39.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="14247" PostTypeId="1" AcceptedAnswerId="14249" CreationDate="2014-07-11T00:49:01.410" Score="7" ViewCount="2508" Body="&lt;p&gt;Is there a scientific explanation to why p orbitals are shaped like two balloons, etc. I think it has got to do with electron repulsions. Wikipedia says they are 'characterised by unique values of quantum numbers', of which I don't understand. &lt;/p&gt;&#xA;" OwnerUserId="6988" LastActivityDate="2014-07-11T17:37:38.770" Title="Why do atomic orbitals have their unique shapes?" Tags="&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14248" PostTypeId="2" ParentId="14246" CreationDate="2014-07-11T01:25:33.607" Score="5" Body="&lt;p&gt;There are many different answers to the question, &quot;What is an organic compound&quot;.  Here are a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/1organic/organic.html&quot;&gt;Pudue&lt;/a&gt; - &quot;compounds that contain both carbon and hydrogen&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.britannica.com/EBchecked/topic/431954/organic-compound&quot;&gt;Britanica&lt;/a&gt; - compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_compound&quot;&gt;Wikipedia&lt;/a&gt; - any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon. For historical reasons discussed below, a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon (such as CO and CO2), and cyanides are considered inorganic&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Chemical_Compounds/Nomenclature_of_Organic_Compounds&quot;&gt;UC Davis&lt;/a&gt;- Any compound containing carbon atom(s) is classified as an organic compound&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Trying to pull the best from each of these while still keeping it simple, perhaps a good definition of an organic compound would be - any covalent compound containing, at least, carbon and hydrogen.  Everything else would be inorganic.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-11T01:25:33.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14249" PostTypeId="2" ParentId="14247" CreationDate="2014-07-11T01:36:45.193" Score="5" Body="&lt;p&gt;The pictures that we typically use to represent orbitals are really just graphs showing the 3-dimensional probability of finding an electron (that occupies the orbital being examined) in space.  $\ce{\Psi}$ is the wavefunction that describes the electron with a set of specific quantum numbers.  $\ce{\Psi^2}$, gives rise to the electron density distribution (the shape of an orbital) plotted in the graph.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-11T01:36:45.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14250" PostTypeId="2" ParentId="14246" CreationDate="2014-07-11T02:30:55.317" Score="2" Body="&lt;p&gt;For organic compounds, I like the &quot;carbon skeleton&quot; definition. There is some base carbon structure (linear, branched, cyclic) that we attach other structures to (functional groups). Hydrogens I think of as the &quot;endcaps&quot; of organic chemistry: if there is no interesting functional group to attach, you default to a hydrogen. This definition emphasizes the fact that chains of carbons (the skeleton) don't seem to have limits as to size. But we can also have very small carbon skeletons (e.g. $\ce{CH_4}$). So this is how I think about organic molecules.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-07-11T02:30:55.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14251" PostTypeId="2" ParentId="14246" CreationDate="2014-07-11T03:15:24.230" Score="1" Body="&lt;p&gt;Organic\non-organic molecules - is archaic term, which created in that time, when scientists thought, organics cannot be created in lab, only in living organism.&#xA;Now, &quot;Organic&quot; mean &quot;hydrocarbons&quot;, and &quot;compound of hydrocarbons&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7094" LastActivityDate="2014-07-11T03:15:24.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14252" PostTypeId="2" ParentId="14230" CreationDate="2014-07-11T03:35:23.720" Score="4" Body="&lt;p&gt;This is one of the more &lt;a href=&quot;http://www.esrl.noaa.gov/csd/assessments/ozone/1998/faq1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;frequently asked questions about ozone&lt;/a&gt; depletion. I will try to give a short summary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, it is important to know that Chlorofluorocarbons (CFCs) are extremely stable at room temperature and pressure. It is therefore only natural, that in the lower atmosphere these gases are enriched, therefore they have plenty of time getting in higher altitudes via diffusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earths atmosphere is in constant motion, you can experience that on a macroscopic level via different winds. The sun heats up the earth's surface, causing all sorts of liquids to be evaporated. While evaporating water, these small molecules may form little conglomerates, taking other molecules up to higher heights. This will happen to dust, so it will also happen to CFCs. There will be local areas (or volumes, layers) of the air in different heights of the atmosphere that have different densities, concentrations and composition. As a result they will also have different air pressures. You can experience this on a plane at higher altitudes. Assuming that all reactions tend to react towards equilibrium, there will be forces acting between these areas. In a consequence there will be also upward forces acting on the CFC containing gas mixture. Slowly these molecules will move upwards to the stratosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this process is heavily dependent on many different factors and may well require years. I think it is actually very hard to find any accurate numbers for this process. My research on the subject turned up nothing about that. In most of the cases it was just stated, that these gasses are there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/9517/4945&quot;&gt;Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://www.nature.com/nature/focus/ozonehole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;collection of nature articles&lt;/a&gt;. Some are free access.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ozonewatch.gsfc.nasa.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ozonewatch @ NASA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ozzyozone.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ozzyozone.org/&lt;/a&gt; (for children, in english, french, spanish)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.scienceinschool.org/2010/issue17/ozone&quot; rel=&quot;nofollow noreferrer&quot;&gt;A hole in the sky&lt;/a&gt; @ Science in school (different languages available)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-07-11T03:35:23.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14253" PostTypeId="2" ParentId="14061" CreationDate="2014-07-11T05:54:52.480" Score="11" Body="&lt;p&gt;This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's start with IR&lt;/strong&gt; (not my strong point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first and second IR spectra are so different that it is hard to be confident about which one is correct. Are they collected in different media (one a solution and one a mull perhaps)? The first looks very low concentration&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;~3420 is clearly an -OH stretch. Probably too high to consider a N-H group of any sort. So, it could be an alcohol or an acid. &lt;/li&gt;&#xA;&lt;li&gt;1682 a C=O group. Probably in the acid region, likely aromatic acid, conjugated or intramolecular H-bonded. Possibly amide, but not convincing based on other peaks. &lt;/li&gt;&#xA;&lt;li&gt;There are lots of other peaks, most of which could mean many things, but I do see 749 and 708 which support the NMR notion of a monosubstituted benzene. I usually think that IR is only useful to corroborate peaks of a known structure, but then I'm an NMR advocate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NMR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the first and second spectra are very different; the obvious concern is that they are of different batches of sample - were the first and second analyses taken from the same sample, or from a different batch. In any case, let us look at the second, which is only a partial spectrum, and omits some possibly very significant information, especially in the region above 10ppm, where I would like to see a broad acid peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXuLW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aromatic protons shown exhibit a classic mono-substituted benzene ring pattern, labelled above. There is a strong downfield shift of the ortho (H2) protons, consistent with an electron withdrawing group substitution (such as -(C=O)R). Ortho coupling for the H2 proton is difficult to determine accurately as there are 3 peaks picked. This may be a digitizer point average, but considering the average of each peak, and the coupling from the first spectrum, we get a ortho coupling of 7.2-7.9. Not overly useful, but less than 8, so we can exclude ethers and halides. Couplings are calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8.1651-8.1507) x 500 = 7.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using additive analysis for monosubstituted benzene rings, we can determine the following shifts from benzene:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;H2 (8.162) 7.34 + &lt;strong&gt;0.82&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;H3 (7.514) 7.34 + &lt;strong&gt;0.17&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;H4 (7.651) 7.34 + &lt;strong&gt;0.31&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Looking at the published tabulated data for substituent effects on monosubstituted benzene rings (I'm using &lt;a href=&quot;http://link.springer.com/book/10.1007%2F978-3-540-93810-1&quot; rel=&quot;noreferrer&quot;&gt;Pretsch, Buhlmann and Badertscher&lt;/a&gt;, but lots of other sources are available), we find one possible candidate that matches closely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Benzoic Acid H2(+0.79), H3(+0.14), H4(0.28)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AIGsv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caveat that I would add to this is that the data presented is not the best quality, and the solution presented is consistent with the data only (not definitive). The key concern is the missing -OH peak in the NMR spectrum. Unless this sample was run in D2O, or washed with D2O, I would expect to see an acid peak around about 12 or so. Depending on the amount of water in the sample, this may be shifted somewhat and broadened, but should still be visible, even if a broadened hump in the baseline. Additional NMR experiments could be useful to confirm the structure.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-07-11T05:54:52.480" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14255" PostTypeId="1" CreationDate="2014-07-11T09:23:58.167" Score="5" ViewCount="17380" Body="&lt;p&gt;Why is the molar enthalpy of vaporization of a substance larger than its molar enthalpy of fusion (at constant pressure); for example, in the case of ice and water.&lt;/p&gt;&#xA;" OwnerUserId="7098" LastEditorUserId="4231" LastEditDate="2014-07-11T15:27:43.933" LastActivityDate="2018-08-14T21:25:48.310" Title="Why is the enthalpy of vaporization greater than the enthalpy of fusion?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14258" PostTypeId="2" ParentId="14247" CreationDate="2014-07-11T15:06:17.993" Score="3" Body="&lt;p&gt;The orbitals are solutions (also called wave functions) to the (time-independent) Schrödinger equation. The solutions are related to Bessel functions and Legendre polynomials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawing of orbitals as you often find in text books is in a sense the Copenhagen interpretation of quantum mechanics. In this interpretation, the wave functions (actually, the square of the wave function) are seen as probability distributions of where to find the electrons with 90 % probability (the cutoff might be different from 90 %).&lt;/p&gt;&#xA;" OwnerUserId="396" LastActivityDate="2014-07-11T15:06:17.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14259" PostTypeId="2" ParentId="14255" CreationDate="2014-07-11T15:24:16.920" Score="0" Body="&lt;p&gt;Ice is less dense than water, that's why ice floats on water.  The lower density of ice means that the average distance between water molecules in ice is greater than the average distance between water molecules in the liquid state.  Because of the greater distance between water molecules in the solid compared to the liquid, molecule - molecule interactions (such as van der Waals and dipole interactions, as well as hydrogen bonding) will be less in the solid than the liquid.  So while we need to put energy into ice to disrupt the lattice structure and break the attractive interactions, this energy is offset to some degree by the even stronger attractive forces that exist in the liquid, since the attractive forces are actually greater in the liquid than the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the gas phase the molecules are far enough apart that attractive forces between molecules are minimal. Therefor, when we go from liquid to gas we must put in a lot of energy to break all of the strong attractive forces that exist in the liquid without any offset because of the lack of significant attractive forces in the gas.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-11T15:24:16.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14260" PostTypeId="1" CreationDate="2014-07-11T16:12:13.597" Score="4" ViewCount="192" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The work done during a reversible compression - if $V_\mathrm{i} = V_1$ (initial) and $V_\mathrm{f} = \frac{1}{6}V_1$ (final) - of an ideal gas is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose $1.8 p_1 V_1$. Is it right or should it be $-1.8 p_1 V_1$?&lt;/p&gt;&#xA;" OwnerUserId="7101" LastEditorUserId="4945" LastEditDate="2018-02-21T10:41:10.787" LastActivityDate="2018-02-21T10:43:03.033" Title="Sign of the work done during a reversible compression" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14261" PostTypeId="2" ParentId="14247" CreationDate="2014-07-11T17:37:38.770" Score="3" Body="&lt;p&gt;I like to think of an atom as a 3-dimensional version of a vibrating string. If you pluck a guitar string while lightly touching the exact center point, it will vibrate in the first harmonic - make a sound one octave higher (twice the frequency). That vibrational mode is kind of like a 2p orbital in that there is zero amplitude at the exact center.  You get the same behavior in 2 dimensions--and more complex vibrational modes--with a vibrating drum head. Put some iron filings on a drum head and then use different frequencies of sound to stimulate vibration. Some of the vibrational modes look a lot like pictures of the orbitals.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-07-11T17:37:38.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14262" PostTypeId="2" ParentId="14260" CreationDate="2014-07-11T18:13:01.103" Score="3" Body="&lt;p&gt;PV work is defined as&#xA;$$w = -p\Delta V,$$&#xA;where $p$ is pressure and $\Delta V$ is $V_\mathrm{final} - V_\mathrm{initial}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has this form because in chemistry, work is defined from the perspective of the &lt;em&gt;system&lt;/em&gt; - if it is compressed, we say that we did work &lt;strong&gt;on&lt;/strong&gt; the system and so the sign of work is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, $V_\mathrm{f}$ is less than $V_\mathrm{i}$, so $\Delta V &amp;lt; 0$. If $\Delta V$ is negative, then $w$ must be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means your answer should have a positive sign. However, I get a different number from $1.8$ - you should double-check that.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="4945" LastEditDate="2018-02-21T10:43:03.033" LastActivityDate="2018-02-21T10:43:03.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14263" PostTypeId="1" CreationDate="2014-07-11T18:49:53.097" Score="3" ViewCount="203" Body="&lt;p&gt;Suppose one could generate an enzyme with arbitrary conditions at the active site. What conditions would then be necessary to cause carbon from dissolved $\ce{CO2}$ in low $pH$ water to combine with the free hydrogen and make methane?&lt;/p&gt;&#xA;" OwnerUserId="7104" LastEditorUserId="7104" LastEditDate="2014-07-12T03:23:07.113" LastActivityDate="2020-04-21T19:22:43.443" Title="Conversion of CO2 to methane" Tags="&lt;acid-base&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14264" PostTypeId="2" ParentId="14221" CreationDate="2014-07-11T18:58:45.633" Score="1" Body="&lt;p&gt;You currently have numbers in ppm (of ${CO_2}$); ppm values for gas tend to be equivalent to ${\mu mol/mol}$ units. Can't you just use that, the the definition of partial pressure, and the atmospheric pressure and be set (though it varies by altitude and a little over time too)? &lt;a href=&quot;http://forum.onlineconversion.com/showthread.php?t=11529&quot; rel=&quot;nofollow noreferrer&quot;&gt;Some forum&lt;/a&gt; has a similar homework-level question. Not sure if I'm missing something with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take it you wanted to use Henry's Law to relate concentration to partial pressure. typically that's used in the realm of vapor-liquid systems; not sure how it fares in a mixture of many gases that I'd qualify the atmosphere as. See things to note: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Dissolving_Gases_In_Liquids%2C_Henry&amp;#39;s_Law&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Chemwiki)&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at a (pretty crude IMHO) pressure vs altitude graph of the atmosphere &lt;a href=&quot;http://en.wikipedia.org/wiki/Atmospheric_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Wikipedia: Atmospheric pressure)&lt;/a&gt;: you can see the pressure even within the troposphere ranges from ~12kPa all the way to ~100kPa (sea level: ~101 kPa). The troposphere has 80% of the atmosphere's mass and you want the whole atmosphere anyways, so just take 101kPa as atmospheric pressure? I also assume atmosphere's mass is constant over the span of 400 million years - no clue on whether I can neglect losses and gains from weather, the biosphere, and space over this time span. Then you will have a bunch of carbon dioxide partial pressures over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/9d/Comparison_US_standard_atmosphere_1962.svg/512px-Comparison_US_standard_atmosphere_1962.svg.png&quot; alt=&quot;stdatm&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On short-term constancy of the atmosphere's mass, see &lt;a href=&quot;https://physics.stackexchange.com/questions/65923/is-the-total-mass-of-earths-atmosphere-essentially-constant-in-time&quot;&gt;SE.Physics&lt;/a&gt;. The mean even changes by season. It's comparatively small comparing 1.5E+15 kg to the total of 5.15E+18 kg.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-07-11T18:58:45.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14265" PostTypeId="2" ParentId="14221" CreationDate="2014-07-11T18:59:31.120" Score="4" Body="&lt;p&gt;The short answer is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henry's law works for small concentrations of ideal mixtures at equilibrium. &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry&amp;#39;s_law&quot; rel=&quot;nofollow&quot;&gt;Henry's Law constant&lt;/a&gt; varies with temperature according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;nofollow&quot;&gt;Van't Hoff equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;k(T)=k(T^{\circ})exp\left[-C\left(\frac{1}{T}-\frac{1}{T\circ}\right)\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $C$ is a constant related to the enthalpy of solvation for each gas, and $^{\circ}$ represents the standard state, T = 298 K and 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using standard state data, you could get an estimate of the partial pressure at a given temperature. However, two key assumptions that are implicit to Henry's law and the Van't Hoff equation do not hold for global atmospheric gas composition:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The atmosphere and the oceans are not at equilibrium (not even close when you consider atmospheric layers and convection)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CO2}$ in seawater is not an ideal mixture (very non-ideal since $\ce{CO2}$ reacts with water).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is one of the many reasons that climate modeling is difficult - the oceans act as huge carbon sinks with a very complex and not-very-stable relationship between dissolved $\ce{CO2}$ content, temperature, composition of rock on the sea floor, convection currents in the atmosphere and the ocean, amount of shellfish (no kidding), and probably lots of other factors (I'm not a climate expert).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net result is that while Henry's law can get you a quick idea of how much $\ce{CO2}$ could be dissolved at a given temperature under pseudo-equilibrium conditions (or what the corresponding partial pressure would be for a given concentration in water), it isn't enough to get you anywhere close to a good prediction of what the global atmospheric composition will be, particularly for $\ce{CO2}$ (due to its reactivity with water and incorporation into biomass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you really just want to convert $\ce{CO2}$ concentration in ppm to a partial pressure, you can do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ in air is &lt;em&gt;close enough&lt;/em&gt; to ideal that we can use the ideal gas law. For an ideal gas, the partial pressure is the same as whatever the pressure &lt;em&gt;would&lt;/em&gt; be if it were in a container by itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_1 = \frac{n_1RT}{V}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find V for a given mass of air, find the mass of $\ce{CO2}$ from the concentration, then convert to moles and plug it into the above.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-11T18:59:31.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14267" PostTypeId="1" AcceptedAnswerId="14278" CreationDate="2014-07-10T20:27:03.517" Score="63" ViewCount="4692" Body="&lt;p&gt;I was making pasta, and I noticed the pasta boiling over. I thought about it some more, and I realized I had no idea why this was happening. When the lid is on, the foam rises. When the lid is off, the foam dies back. Clearly there's some surfactant at work allowing bubbles to form, and the bubbles are mostly full of steam. If you remove the lid, they cool and condense down to a tiny size. &lt;a href=&quot;http://gizmodo.com/the-science-of-why-your-past-boils-over-1498537569&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; is honestly just not specific enough. Why is the starch important?(It also happens with potatoes) How does a polysaccharide, a hydrophilic molecule, become a surfactant(if indeed it does)? Is it phospholipids from the cells that became pasta? Do starch molecules form a polymer and trap steam underneath like a balloon? Is it some kind of hybrid where bubbles are stabilized by viscosity? What's going on here?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://cooking.stackexchange.com/questions/7909/what-causes-water-to-boil-over&quot;&gt;This&lt;/a&gt; question on seasoned advice tries to tackle it but gets surface tension wrong(I think) so that makes me a little wary of it. Surfactants like soap &lt;em&gt;reduce&lt;/em&gt; the surface tension, allowing bubbles to form by letting water molecules spread out into thin films. They also keep your alveoli in your lungs from collapsing. If starches increased the surface tension, wouldn't that reduce the likelihood of forming bubbles by increasing the energetic penalty? High surface tension materials act like mercury.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20091228190520AAcoZUw&quot; rel=&quot;noreferrer&quot;&gt;This answer&lt;/a&gt; is all over the internet but doesn't make any sense to me. You need &lt;em&gt;less&lt;/em&gt; surface tension for bubbles, surely?&lt;/p&gt;&#xA;&lt;p&gt;&lt;del&gt;If it's just phospholipids acting as detergent, could I 'boil over' pasta in cold water with a whisk?&lt;/del&gt; (I tried this, you can't.)&lt;/p&gt;&#xA;&lt;p&gt;I considered putting this on seasoned advice, but I'm not interested in how to prevent boil-over(or really any practical results), and the supplied answer there lacks scientific rigor.&lt;/p&gt;&#xA;" OwnerUserId="7103" OwnerDisplayName="Jeremy Kemball" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-09-15T19:22:40.640" Title="Why does pasta really boil over?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14268" PostTypeId="2" ParentId="14267" CreationDate="2014-07-10T21:14:06.593" Score="8" Body="&lt;p&gt;This isn't really a biology question I don't think...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a chemist I can tell you that the rough surface of the pasta does tend to make it produce boiling over.  The same can be produced with stones or wood or nearly any rough surface.  Chemists use &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_chip&quot;&gt;little chips of alumina or some other mineral&lt;/a&gt; to prevent over boiling in fact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a sealed soda bottle you can see the same effect.  Its the same scenario - in this case CO2 rather than steam is ready to come out of supersaturated liquid, but won't unless there is a Bubbles are forming at the plastic surface inside when you open the cap. If you see a stream of bubbles forming within the liquid they are always nucleating from a speck of dust or some other tiny bit of solid floating around.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.newscientist.com/article/dn14114-science-of-mentosdiet-coke-explosions-explained.html&quot;&gt;surface of mentos candy is particularly good&lt;/a&gt; at getting bubbles to form.  I'd bet that if you dropped mentos into boiling water, at least until the mentos dissolved you'd see a bit more bubbling than pasta.  I don't advise you to do this, its just a thought experiment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;there is something you can do to prevent pasta from boiling over&lt;/em&gt;!  Sorry for including this, but cooks are always using too much energy when cooking and its wasteful. When the water starts boiling  you can turn down the heat and put a lid on the pot.  Usually putting the burner on the lowest setting is enough.  This lowers the amount of steam that is supercritical in the water and fewer steam bubbles will form.  This saves energy too, even though the temperature of the water is the same (boiling water is always about 100 deg C), it lowers the amount of supercritical steam in the water. &lt;/p&gt;&#xA;" OwnerUserId="2387" OwnerDisplayName="shigeta" LastActivityDate="2014-07-10T21:20:55.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14269" PostTypeId="1" AcceptedAnswerId="14274" CreationDate="2014-07-11T20:01:30.677" Score="2" ViewCount="12243" Body="&lt;p&gt;My inorganic chemistry is at a basic level, so my question is, if I heat alumina ($\ce{Al2O3}$), to its melting point around 2000 C, will I wind up with pure Al?  I think the oxygen will release from the $\ce{Al2O3}$ when the temperature approaches 2000 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought 5 kilograms from a friend who works at a refinery near my town (he works in logistics and does not have a chemical background) with 1.75$ of almost pure alumina ~99.95%, and I want to know if I melt that white powder will I obtain 2.5kg of 99.95% aluminium?&lt;/p&gt;&#xA;" OwnerUserId="7106" LastEditorUserId="73" LastEditDate="2014-08-31T12:36:54.243" LastActivityDate="2015-01-09T07:36:25.613" Title="Will heating alumina (Al2O3) to its melting point yield pure Al?" Tags="&lt;inorganic-chemistry&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14270" PostTypeId="1" AcceptedAnswerId="14272" CreationDate="2014-07-11T20:02:21.447" Score="2" ViewCount="616" Body="&lt;p&gt;This is the question from the textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ozone O3 can be prepared in the laboratory by passing an electrical discharge through oxygen gas. Assume that an evacuated steel vessel  with a volume of 10L is filled with 32 ATM of O2 at 25 degrees Celsius and an electrical discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 30.64 ATM at 25 degrees Celsius .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions manual subtracts 3x from the initial pressure of O2 to determine the final pressure of O2 after the electrical discharge(32-3x=final pressure of O2). X is the amount of pressure lost (32-30.63=1.36=x). My question is, how are we justified in using the stiochiometric ratios to determine the final pressure. In other words, how is (32-3x) justified when finding the final pressure of O2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my question is not clear. Please let me know and I'll try my best to rephrase it.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastActivityDate="2014-07-11T21:42:17.640" Title="What is the relationship between pressure and stoichiometry ratios" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14271" PostTypeId="2" ParentId="14235" CreationDate="2014-07-11T20:45:57.480" Score="2" Body="&lt;p&gt;A pattern-based approach: Remember that the values for ${m_l}$ are centered about 0. This means for a d-orbital, where ${l=2}$, the values of ${m_l}$ = ${(-2, -1, 0, +1, +2)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for ${l=2}$ and ${m_l=0}$ and ${m_s=+1/2}$, you should arrive at ${d^3}$. The orbitals each get an electron for all the positive spins first, then are filled completely with the negative spin ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you &quot;visit&quot; ${m_l}$ twice (in rising order: -2, -1, 0, +1, +2, repeat). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the first d-electron (↑ _ _ _ _) such as ${Sc}$ has the configuration ${m_l=-2, m_s=+1/2.}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3rd electron (↑ ↑ ↑ _ _) would have ${m_l=0, m_s=+1/2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 5-electron configuration (↑ ↑ ↑ ↑ ↑) would have ${m_l=+2, m_s=+1/2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on the right-hand side of the d-group, the 6th electron (↑↓ ↑ ↑ ↑ ↑) would have ${m_l=-2, m_s=-1/2}$, and then the magnetic quantum numbers rise to +2 again at ${d10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, ${Mn^{+2}}$ is ${d^{3}}$ (5 to start, then subtract 2 because of its +2 charge) which matches the quantum numbers you specified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an 8 electron configuration because of the spin specified. That'd be (↑↓ ↑↓ ↑↓ ↑ ↑).&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-07-11T20:45:57.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14272" PostTypeId="2" ParentId="14270" CreationDate="2014-07-11T21:42:17.640" Score="1" Body="&lt;p&gt;The actual reaction being discussed is $$\ce{3O2 -&amp;gt; 2O3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know from the gas law that $$\ce{PV=nRT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the current experiment the only things changing are the pressure and the number of moles; the volume and temperature remaining constant.  In this case the gas law simplifies to $$\ce{\Delta P = \Delta n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor, we know the change in moles for the overall reaction (-3+2=-1) is $\ce{\Delta n}$ and by the above stoichiometry the change in the moles of oxygen will be three times this amount or $\ce{3\Delta n}$ or $\ce{3  \Delta P}$ or $\ce{3x}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-11T21:42:17.640" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14273" PostTypeId="1" AcceptedAnswerId="14275" CreationDate="2014-07-11T23:58:14.523" Score="1" ViewCount="15413" Body="&lt;p&gt;If place a can of pressurized liquid butane (such as a lighter refill canister) into a freezer to get it below its boiling point of $-1\ \mathrm{^\circ C}$,&#xA;and then release it into a container kept at a freezing temperature, will it remain in the liquid state without boiling off (until it reaches its boiling point), or will it boil off immediately?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a cold metal container kept at a stable temperature of $-7\ \mathrm{^\circ C}$, but when I released room temperature butane into it, it boiled off instantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did that happen, and what will happen when it is released from the can at a temperature at or below $-1\ \mathrm{^\circ C}$?&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="865" LastEditDate="2015-08-05T15:51:34.303" LastActivityDate="2015-08-05T15:51:34.303" Title="Liquefying butane in a freezer" Tags="&lt;boiling-point&gt;&lt;phase&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14274" PostTypeId="2" ParentId="14269" CreationDate="2014-07-12T00:22:41.913" Score="11" Body="&lt;p&gt;You've figured out one half of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hall-Heroult_process&quot; rel=&quot;noreferrer&quot;&gt;Hall-Heroult process&lt;/a&gt;. Just melting alumina on its own won't liberate the oxygen. The reason is difficult to explain without getting into chemical thermodynamics - one way to think about it is that the lowest energy state is one where oxygen is bound to alumina - where the electrons from alumina have been &quot;taken&quot; by the oxygen. In diatomic oxygen, the electrons are &quot;shared&quot; (the bonds are covalent) and so the total energy is higher. This is a very brief and not-very-detailed explanation, but the end result is that oxygen and aluminum, if given the opportunity, would rather form a bond than not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add thermal energy, unless it changes that balance (it doesn't, at least not until you get way past the melting point), all you are doing is making it &lt;em&gt;more&lt;/em&gt; possible for the reaction to happen. If you kept going, you could eventually add enough thermal energy to completely dissociate the alumina, at which point (depending on how you did it) you might be able to remove the oxygen and recover the aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However - the amount of energy it would take to do that is extremely large. Instead, it turns out to be much easier to simply force oxygen to give its electrons back to aluminum using an electric potential. When you do that, you &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis&quot; rel=&quot;noreferrer&quot;&gt;electrolyze&lt;/a&gt; the compound and create diatomic oxygen at the anode and aluminum at the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hall-Heroult process takes this another step to make this work at an even lower temperature (1000 C) by dissolving the alumina in molten &lt;a href=&quot;http://en.wikipedia.org/wiki/Cryolite&quot; rel=&quot;noreferrer&quot;&gt;cryolite&lt;/a&gt;. As a result, it is one of the most (maybe &lt;em&gt;the&lt;/em&gt; most) efficient industrial process for producing aluminum, and is responsible for the widespread availability and low cost of aluminum in modern times (and also for the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcoa&quot; rel=&quot;noreferrer&quot;&gt;Alcoa&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea about using solar energy isn't bad - although you might be surprised at the amount of energy you need to produce aluminum - both to melt it, and to electrolyze it. One thing you might look into is using solar thermal heating (with a collector, for example) to melt the alumina/cryolite mixture, and then use photovoltaics just for the electrolysis step. That would be quite a bit cheaper than using photovoltaic electricity alone to run an electric heater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.alcoa.com/global/en/about_alcoa/pdf/Smeltingpaper.pdf&quot; rel=&quot;noreferrer&quot;&gt;Industrially&lt;/a&gt;, the cryolite mixture is kept molten by passing a large electric current through it. The electricity is usually produced by power plants, since the amount required is so large (15 kWh/kg according to Alcoa, and that's just for the electrolysis step!). Although coal and gas power isn't renewable, the advantages are a relatively low cost per unit of power (which might not stay true in the next few decades) and on-demand power generation (likely to always be a big advantage over solar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also nuclear, hydroelectric, and geothermal power - all of which provide large amounts of on-demand power very inexpensively, once the plants are built.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-31T20:05:34.353" LastActivityDate="2014-08-31T20:05:34.353" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14275" PostTypeId="2" ParentId="14273" CreationDate="2014-07-12T00:43:51.437" Score="5" Body="&lt;p&gt;Since the room temperature butane didn't instantly cool when it hit the container, it was still above the boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a large freezer that was roughly at equilibrium at a temperature below $-1^\circ C$, and you put both the pressurized container and the empty container in there and let them reach equilibrium, then when you opened the pressurized container it would not boil and you could conceivably pour it into the other container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, butane has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Butane_(data_page)&quot; rel=&quot;noreferrer&quot;&gt;large vapor pressure&lt;/a&gt; (the enthalpy of vaporization is relatively small). This means that even at low temperatures, it will evaporate very quickly. This is especially true because the atmosphere does not &lt;em&gt;normally&lt;/em&gt; have any butane in it - so even a small partial pressure combined with convective transport in the air will cause it to evaporate even faster. If you imagine pouring some rubbing alcohol out on a table and then putting a fan next to it, it is the same idea. Butane molecules randomly have enough energy to escape the liquid, and air currents carry them away. Since there is no butane vapor nearby to establish an equilibrium with the liquid, more butane evaporates to take its place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net result of this is that even if you did the experiment in a freezer and made sure everything was below $-1^\circ C$ (which includes you wearing well-insulated gloves and not breathing on it), it might still evaporate so fast that it would &lt;strong&gt;look&lt;/strong&gt; as if it were boiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipedia, here is a plot of vapor pressure vs T for butane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiling point is the point where the vapor pressure line exceeds atmospheric pressure. Atmospheric pressure is usually around 760 mmHg. This is a log plot for the P axis, which means one large grid step is ten times larger than the previous one. From this graph, you can see that at $-30^\circ C$, the vapor pressure is still ~200 mmHg. This means that roughly 26% of the air &lt;em&gt;at equilibrium&lt;/em&gt; would be butane. Unless you have a very tiny freezer, that is a lot. This means that you might have to wear a gas mask and get a 55 gallon drum of butane before you could really do any &quot;pouring&quot; in the traditional sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $-100^\circ C$ you are down to 1 mmHg vapor pressure. At this point I would say that even a small amount (like what is in a lighter) would stay liquid long enough that you could pour it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PFzT6.png&quot; alt=&quot;Log plot of vapor pressure of butane vs temperature&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-12T00:43:51.437" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14276" PostTypeId="1" AcceptedAnswerId="14277" CreationDate="2014-07-12T01:41:02.423" Score="1" ViewCount="220" Body="&lt;p&gt;I have a small Butane blowtorch, and use it to heat an Aluminium plate. The plate is thick so takes 20 seconds or so to really start heating up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as I apply the heat, droplets of some liquid start &quot;condensing&quot; on the surface of the metal around the general area of the flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liquid is colorless and odorless and is at room temperature, note this happens when the metal is still at room temperature as it hasn't heated up much yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this liquid exactly and why/how does it form? I suspect it could be a $C_{4}H_{10} + O_{2}$ reactant but not sure.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="5772" LastEditDate="2014-07-12T01:56:03.097" LastActivityDate="2014-07-12T02:10:24.783" Title="Butane blowtorch condensing liquid on heating surface" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14277" PostTypeId="2" ParentId="14276" CreationDate="2014-07-12T01:54:30.257" Score="2" Body="&lt;p&gt;This is a guess - but I think it might be water formed as a byproduct of butane combustion precipitating on the plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2C4H10(g) + 13O2(g) -&amp;gt; 8CO2(g) + 10H2O(g)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum has very high thermal conductivity, so it is possible that the plate is rapidly absorbing heat from the water vapor, causing it to condense. If this stops happening right around the point when the aluminum temperature reaches 100 C, that would be evidence to support this hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-12T02:10:24.783" LastActivityDate="2014-07-12T02:10:24.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14278" PostTypeId="2" ParentId="14267" CreationDate="2014-07-12T03:01:06.380" Score="52" Body="&lt;p&gt;The starch forms a loosely bonded network that traps water vapor and air into a foamy mass, which expands rapidly as it heats up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Starch&quot;&gt;Starch&lt;/a&gt; is made of glucose polymers (amylopectin is one of them, shown here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hede9.png&quot; alt=&quot;Amylopectin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the chains are branched, some are linear, but they all have $\ce{-OH}$ groups which can form &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot;&gt;hydrogen bonds&lt;/a&gt; with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's follow some starch molecules through the process and see what happens. In the beginning, the starch is dehydrated and tightly compacted - the chains are lined up in nice orderly structures with no water or air between them, maximizing the hydrogen bonds between starch polymers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7hRE.png&quot; alt=&quot;Nicely aligned starch polymers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the water heats up (or as you let the pasta soak), water molecules begin to &quot;invade&quot; the tightly packed polymer chains, forming their own hydrogen bonds with the starch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8rc9n.png&quot; alt=&quot;Water molecules invading the polymer structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soon, the polymer chains are completely surrounded by water, and are free to move in solution (they have dissolved):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zjXq3.png&quot; alt=&quot;Polymers have dispersed in solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the water/starch solution is not completely uniform. In the middle of the pot of water, the concentration of starch is low compared to water. There are lots and lots of water molecules available to surround the starch chains and to keep them apart. Near the surface, when the water is boiling, the water molecules escape as vapor. This means that near the surface, the &lt;strong&gt;local concentration&lt;/strong&gt; of starch increases. It increases so much as the water continues to boil, that the starch can collapse back in on itself and hydrogen bond to other starch molecules again. However, this time the orderly structure is broken and there is too much thermal motion to line up. Instead, they form a loosely packed network of molecules connected by hydrogen bonds and surrounding little pockets of water and air (bubbles):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2XFTT.png&quot; alt=&quot;Starch network&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This network is very weak, but it is strong enough to temporarily trap the air as it expands due to heating - thus, the bubbles puff up and a rapidly growing foam forms. Since they are very weak, it doesn't take much to disrupt them. Some oil in the water will inhibit the bubbles from breaking the surface as easily, and a wooden spoon across the top will break the network mechanically as soon as it touches it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many biomolecules will form these types of networks under different conditions. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Gelatin&quot;&gt;gelatin&lt;/a&gt; is a protein (amino acid polymer) that will form elastic hydrogen-bonded networks in hot water. As the gelatin-water mixture cools, the gel solidifies, trapping the water inside to form what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sol-gel&quot;&gt;sol-gel&lt;/a&gt;, or more specifically, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogel#Hydrogels&quot;&gt;hydrogel&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gluten in wheat is another example, although in this case the bonds are &lt;a href=&quot;http://en.wikipedia.org/wiki/Disulfide_bond&quot;&gt;disulfide&lt;/a&gt; bonds. Gluten networks are stronger than hydrogen-bonded polysaccharide networks, and are responsible for the elasticity of bread (and of pasta).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DISCLAIMER:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pictures are not remotely to scale, starch is usually several hundred glucose monomers long, and the relative size of the molecules and atoms isn't shown.&lt;/li&gt;&#xA;&lt;li&gt;there aren't nearly enough water molecules - in reality there would be too many to be able to see the polymer (1,000's).&lt;/li&gt;&#xA;&lt;li&gt;the starch molecules aren't &quot;twisty&quot; enough or showing things like branching - the real network structure and conformations in solution would be much more complicated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But, hopefully you get the idea!&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-13T18:10:14.677" LastActivityDate="2014-07-13T18:10:14.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14279" PostTypeId="2" ParentId="14263" CreationDate="2014-07-12T03:42:42.710" Score="1" Body="&lt;p&gt;Let's take a close look at what the reaction entails. You want to convert &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_4}$&lt;/span&gt; using &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt;. It is possible that the hydrogen ion will co-ordinate with &lt;span class=&quot;math-container&quot;&gt;$\ce{CO_2}$&lt;/span&gt; to produce &lt;span class=&quot;math-container&quot;&gt;$\ce{[CO_2H]^+}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately it is very unlikely to have a second hydrogen ion attaching to the same oxygen to produce &lt;span class=&quot;math-container&quot;&gt;$\ce{[CO_2H_2]^{2+}}$&lt;/span&gt;. Now for the impossible step of water leaving and producing &lt;span class=&quot;math-container&quot;&gt;$\ce{[CO]^{2+}}$&lt;/span&gt;. This step would require a large amount of energy and &lt;span class=&quot;math-container&quot;&gt;$\ce{[CO]^{2+}}$&lt;/span&gt; would be extremely reactive. Since it is highly charged, you would not have any hydrogen ions approaching the moiety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, any further progress towards &lt;span class=&quot;math-container&quot;&gt;$\ce{CH_4}$&lt;/span&gt; is fantasy. To stabilize the reaction, you need to add electrons to the carbon (which you don't have a source).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="30836" LastEditDate="2020-04-21T19:22:43.443" LastActivityDate="2020-04-21T19:22:43.443" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14280" PostTypeId="2" ParentId="14255" CreationDate="2014-07-12T08:26:22.263" Score="0" Body="&lt;p&gt;Enthalpy is state function which is defined as $H = U + pV$, it includes internal energy of the system and energy to create the system volume (boundary work). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal energy may be decomposed in several terms: oscillation around a equilibrium centre, vibrational states (stretching and bending). Generally speaking (there might be counter example for some exotic material or at temperature close to $0\ \mathrm{K}$), internal energy increase with temperature because the partition of accessible vibrational states increases (higher states activate with temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the $pV$ term, you first must consider that it cannot be negative. Therefore $H \geq U$. Now you just have to consider that, generally, $\Delta V_\mathrm{fus} &amp;lt; \Delta V_\mathrm{vap}$. For a given amount of matter, volume difference when melting is small before volume difference when vaporizing. This inequality holds, even for water where $\Delta V_\mathrm{fus} &amp;lt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining those two arguments, you can explain why enthalpy of vaporization is many order of magnitude higher than enthalpy of fusion.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorDisplayName="user7951" LastEditDate="2018-08-14T21:25:48.310" LastActivityDate="2018-08-14T21:25:48.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14281" PostTypeId="1" CreationDate="2014-07-12T08:39:45.167" Score="2" ViewCount="118" Body="&lt;p&gt;I would like to know what precautions should be taken when handling liquids like trichloroethylene, tetrachloroethylene, dichlormethane, and mercury. Of these, which is most toxic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen people handle mercury in rooms without fume hoods. I don't have access to a fume hood, so is it adequate protection if I wear a suitable mask when handling mercury?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am interested in the conditions in which mercury vapor is formed. Why is it formed when its gaseous temperature is high?&lt;/p&gt;&#xA;" OwnerUserId="7063" LastEditorUserId="865" LastEditDate="2015-08-07T13:24:45.867" LastActivityDate="2015-08-07T13:24:45.867" Title="Instructions for handling toxic liquids" Tags="&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14282" PostTypeId="2" ParentId="14281" CreationDate="2014-07-12T11:29:47.207" Score="1" Body="&lt;p&gt;Each liquid produces some vapours. When you leave wet clothes, they eventually dry. Same here. It is true, that vapour pressure for mercury at room temperature is extremely low, but since mercury accumulates and has quite high toxicity, it is still a trouble. Same for other liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury must be kept in sealed container and/or under layer of water. Any works with it must occur in large pan, so no tiniest drop of mercury escaped. Escaped drops must be collected using thin copper plate, washed in concentrated nitric acid. All solutions of mercury compounds and remaining mercury should be utilized according to local procedures (there is an agency for it at my city). Metallic mercury may be transformed into salts using $\ce{FeCl3}$ concentrated solution. While not removing the mercury, this transforms it into compounds that does not produce vapours. However, in case of mercury leak a special agency must be informed to perform all necessary procedures. Consult local regulations on the cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dichloromethane has regrettably low boiling point. It must be kept away from heat sources and in case of high temperatures it must be cooled or kept in sealed container. It evaporates swiftly on open air, carrying away much heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the liquids should be used under fume hood. In case one is unavailable, an acceptable alternative is open air as long as there is a gentle wind. The chlorinated hidrocarbons mentioned has strong odor, strong enough to make handling them without good ventilation impossible. None of the liquids should contact skin. It is not immediately fatal, but still. In case the contact occurred, the chloroalkanes will evaporate quite quickly, though washing skin with soap and water is highly encouraged. Again, one-time breathing-in of the vapours is not fatal with this compounds, but repeated breathing-in guarantees health problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good thing: none of the liquids is flammable, so flammability is not a trouble here. Unfortunately, heating them at presence of air is still highly unrecommended as well as continuous exposure to sunlight: it may produce phosgene. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-07-12T14:18:18.043" LastActivityDate="2014-07-12T14:18:18.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14284" PostTypeId="1" AcceptedAnswerId="14336" CreationDate="2014-07-12T15:04:59.633" Score="2" ViewCount="142" Body="&lt;p&gt;I still distinctly remember reading about how Lewis acidity/basicity was characterized using kinetics. The reason given was that because formation constants are so large, these reactions are not equilibrium reactions and completion reactions. So we just characterize strength using the rate of reaction/formation. This at least makes partial sense; formation constants are huge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have found what I read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many reactions of nucleophiles are not reversible. A bond forms, a bond breaks, and that’s the end of the reaction. The problem with this from a measurement standpoint is that we often can’t determine an equilibrium constant for a reaction. And if we can’t do that, then we can’t develop a reactivity scale based on equilibria.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If we can’t measure equilibria, then what do we do? Well, we use the next best measurement available: to measure reaction rates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) Does the quoted text make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Am I interpreting it correctly or out of context? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Reference to previously answered question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13734/lewis-base-strength-and-the-hydrogen-proton&quot;&gt;Lewis base strength and the hydrogen proton&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-07-17T10:48:51.527" Title="Formation Constants; Lewis Acidity" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14285" PostTypeId="1" AcceptedAnswerId="14290" CreationDate="2014-07-12T15:10:30.357" Score="0" ViewCount="4432" Body="&lt;p&gt;here is a problem :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TjNSJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did &quot; as shown in the picture below &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rewrote the information given in terms of reactants and products ( equations )// in Part 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I used the Hess law to obtain the desired result //in part 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in part 3 // I added and cancelled the terms that appear in both sides &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I left with $O_{(g)}$ !! and the enthalpy is 1192 which is inconsistent with the correct value 1442 !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/diDNf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-07-12T18:23:04.433" Title="Calculate the second ionization energy for the Mg from the following data?" Tags="&lt;homework&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14286" PostTypeId="1" CreationDate="2014-07-12T15:34:03.263" Score="1" ViewCount="52630" Body="&lt;p&gt;If we are given 5 molecules without their electronegativity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BrF5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PBr5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CCl4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XeF2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XeF4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to determine that one of them is polar or not ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know by using the difference in electronegativity we can determine the polarity&#xA;But again, electronegativity is not given  ?? /:&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2014-07-12T17:48:35.800" Title="without using electronegativity, which one of these molecules is polar?" Tags="&lt;polarity&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2014-07-13T17:03:31.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="14287" PostTypeId="2" ParentId="14286" CreationDate="2014-07-12T16:11:32.530" Score="4" Body="&lt;p&gt;Use molecular geometries. Some geometries are asymmetrical, so they necessarily are polar. Other molecules are symmetrical, so even though they contain elements of great differences in electronegativities, these molecules are non-polar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider carbon dioxide - which has the elements carbon and oxygen in its composition. The EN of carbon is 2.5; the EN of oxygen is 3.5. Clearly, there is a big difference in electronegativity, which suggests that the oxygens are isolating electron density away from the carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iR2o8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see in the above picture, the two permanent dipoles cancel each other out since they are equal in magnitude and opposite in direction. So carbon dioxide might have elements of great difference in EN, but is still non-polar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now apply this knowledge to your problem. To determine molecular geometries, understand that both electrons and attached atoms influence geometry. Try looking up and applying the AXE method. I will leave my answer at here for now, since you have not shown much work in tackling this problem. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-12T16:16:57.487" LastActivityDate="2014-07-12T16:16:57.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14288" PostTypeId="2" ParentId="14286" CreationDate="2014-07-12T16:34:01.477" Score="5" Body="&lt;p&gt;When we say something is &lt;strong&gt;polar&lt;/strong&gt;, it has a very specific meaning. It means that there is a spatial separation of charges, or a &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;noreferrer&quot;&gt;dipole moment&lt;/a&gt;&lt;/strong&gt;. In the simplest case, this is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mu = Qr&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mu$ is the dipole moment, $Q$ is the charge on each pole (equal and opposite) and $r$ is the distance between the charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything with charges can have a dipole moment, and therefore be polar. In chemistry, we typically worry about whether or not two things are polar:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bonds&lt;/li&gt;&#xA;&lt;li&gt;Molecules&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;Bond Polarity&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Chemists are concerned with the polarity of bonds because polarity affects the &lt;strong&gt;character&lt;/strong&gt; of the bond - the more polar the bond is, the more it behaves like an ionic bond. The less polar, the more it behaves like a covalent bond. This has implications for everything from naming rules to reactivity, but all you need to know for this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;All&lt;/em&gt; bonds between two atoms are polar, unless the atoms are identical. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This is including whatever else the atoms may be bonded to). The only question after that is &quot;how polar is the bond,&quot; and that is where tables of electronegativity come in. Using an arbitrarily defined scale, we can measure the relative polarity of bonds by comparing the difference in electronegativity between the two elements.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Molecule Polarity&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For molecules, polarity isn't quite so simple. We know that all bonds between dissimilar atoms are polar, but in a molecule, sometimes the dipole moments add up to form a net dipole moment of zero. This is a little bit difficult to explain if you don't have a mathematics and physics background that includes vectors and summations of forces/moments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do, then when I say something like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The net dipole moment for a molecule is equal to the sum of dipole moments over each bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It will make sense to you. If you don't, then you have to get a little more creative. I tell my students to imagine that there are ropes connecting each outer molecule to the central molecule, along each bond. Then imagine pulling each rope towards the side that is more electronegative (has a higher electron density). The more polar the &lt;em&gt;bond&lt;/em&gt;, the harder you pull on the rope. Then imagine whether the molecule moves. If the forces balance, it stays put, so the net dipole moment is zero and it is not polar. If it does move, there was a net dipole moment, so it is polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_structures&quot; rel=&quot;noreferrer&quot;&gt;Lewis structures&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR&quot; rel=&quot;noreferrer&quot;&gt;VSEPR&lt;/a&gt; theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - &lt;em&gt;whether or not you know the individual atom electronegativity&lt;/em&gt;. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at each one, using wikipedia's geometric pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bromine_pentafluoride&quot; rel=&quot;noreferrer&quot;&gt;$\ce{BrF5}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9l3Wr.png&quot; alt=&quot;BrF5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there is a lone pair at the bottom (this is a &lt;strong&gt;square pyramidal&lt;/strong&gt; geometry). The net dipole moment will be pointing &quot;up&quot;, which makes this a polar molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://de.wikipedia.org/wiki/Phosphorpentabromid&quot; rel=&quot;noreferrer&quot;&gt;$\ce{PBr5}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0vTzj.png&quot; alt=&quot;PBr5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one has a &lt;strong&gt;trigonal bipyramidal&lt;/strong&gt; geometry, with each bond symmetrically opposed to each of the others. If you add up the vectors, they result in a net dipole moment of zero. Therefore, it is non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As ron points out in the comments below, this is not really a binary molecular compound, and in reality forms an &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorus_pentabromide&quot; rel=&quot;noreferrer&quot;&gt;ionic crystal structure&lt;/a&gt; with $\ce{PBr4+}$ and $\ce{Br-}$. I don't think the author of the question intended you to worry about that, but if they did, the answer is still &quot;non polar,&quot; although we wouldn't really call it a &quot;molecule&quot; any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_tetrachloride&quot; rel=&quot;noreferrer&quot;&gt;$\ce{CCl4}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eZThu.png&quot; alt=&quot;CCl4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon tetrachloride has a &lt;strong&gt;tetrahedral&lt;/strong&gt; geometry, and all the dipole moment vectors cancel. Therefore, it is non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/XeF2&quot; rel=&quot;noreferrer&quot;&gt;$\ce{XeF2}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uf5Ks.png&quot; alt=&quot;XeF2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is &lt;strong&gt;linear&lt;/strong&gt; - no net dipole moment, so it is non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/XeF4&quot; rel=&quot;noreferrer&quot;&gt;$\ce{XeF4}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxfjc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could only find the ball-and-stick model for this one. It is &lt;strong&gt;square planar&lt;/strong&gt; - there is a lone pair of electrons (not shown) on each &quot;face&quot; of the square on Xe. There is no net dipole moment, so it is also non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of all of these, the only polar molecule is $\ce{BrF5}$ (the first one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To review the steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw the Lewis structure&lt;/li&gt;&#xA;&lt;li&gt;Figure out the geometry (using VSEPR theory)&lt;/li&gt;&#xA;&lt;li&gt;Visualize or draw the geometry&lt;/li&gt;&#xA;&lt;li&gt;Find the net dipole moment (you don't have to actually do calculations if you can visualize it)&lt;/li&gt;&#xA;&lt;li&gt;If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-12T17:48:35.800" LastActivityDate="2014-07-12T17:48:35.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14289" PostTypeId="1" AcceptedAnswerId="14292" CreationDate="2014-07-12T16:44:20.020" Score="0" ViewCount="16290" Body="&lt;p&gt;In my chemistry book, they say that an approach to making a buffer solution of $\mathrm{pH}=5.09$ is by adding an appropriate amount of strong base $(0.052\ \mathrm{mol}\ \ce{NaOH})$ to $0.300\ \mathrm{l}$ of $0.025\ \mathrm{mol/l}\ \ce{CH3COOH}$. But, I haven’t been able to figure out how they got those numbers. If I am not mistaken the relevant equation is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COOH +OH- &amp;lt;=&amp;gt;CH3COO^- + H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the weak acid and conjugate base are respectively $\ce{CH3COOH}$ and $\ce{CH3COO^-}$. My concern is that this equation above goes nearly to completion $(K=1.8\times10^9)$ so I don’t see how this can even be a buffer solution. If not, how do they get the values? I would appreciate any help. &lt;/p&gt;&#xA;" OwnerUserId="6846" LastEditorDisplayName="user7951" LastEditDate="2015-04-01T09:03:47.670" LastActivityDate="2017-09-21T13:12:47.550" Title="How to create a buffer solution from NaOH and CH₃COOH?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14290" PostTypeId="2" ParentId="14285" CreationDate="2014-07-12T16:44:45.833" Score="1" Body="&lt;p&gt;You forgot that you only need 1/2 mole of $\ce{O2}$. Since the product is $\ce{MgO}$, the third equation in step one should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{1/2O2(g) -&amp;gt; O(g)} \space \space \rm{249.35 ~kJ/~mol}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I divided the bond dissociation energy of $\ce{O2}$ by two as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very common mistake in Born-Haber type problems - the way to catch it early is to write out the balanced reaction first, then adjust the coefficients of each step accordingly.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-12T17:39:46.040" LastActivityDate="2014-07-12T17:39:46.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14291" PostTypeId="1" AcceptedAnswerId="14295" CreationDate="2014-07-12T17:03:59.193" Score="2" ViewCount="984" Body="&lt;p&gt;Why is reduction defined as the gain of hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because hydrogen has a very low electronegativity and easily donates its electrons to other elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand reduction is formally defined as a decrease in oxidation state; oxidation state assignment involves supposing that all bonding is ionic and that electrons are therefore parceled to the most electronegative atom in a bond. And if something gains H - an atom of exceedingly low EN - then that element is reduced because that element will likely be more EN than H and therefore &quot;take&quot; the hydrogen atom's electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/akgmU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-12T23:08:05.630" LastActivityDate="2014-07-13T02:24:57.943" Title="Reduction and Oxidation - Alternate definitions" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14292" PostTypeId="2" ParentId="14289" CreationDate="2014-07-12T17:04:48.370" Score="4" Body="&lt;p&gt;Try considering reaction stoichiometry. You know the extent of the reaction. Now consider stoichiometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If these are really the numbers your textbook gave you, then the textbook is wrong; you cannot create a buffer system with these numbers specifically. I suspect that you didn't transcribe the numbers from your textbook correctly. In any case the general idea, however, is correct - you can indeed form a buffer solution using sodium hydroxide and a solution of acetic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to do to realize this is to know what a buffer is - a buffer has significant concentrations of both an acid and its conjugate base (or base and its conjugate acid) - and consider reaction stoichiometry. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-12T17:04:48.370" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14293" PostTypeId="2" ParentId="14285" CreationDate="2014-07-12T17:13:22.877" Score="2" Body="&lt;p&gt;Thanks for showing your work.  We're given that &#xA;$$\ce{\Delta H_{f} (MgO (s))~ =~ -601$.$8 ~kJ/mol}$$&#xA;So any balanced equation we write that has MgO as the product will have this same heat of formation.  An equation that produces MgO and allows us to solve the problem is&#xA;$$\ce{Mg^{+2}(g) + O^{-2}(g) -&amp;gt; MgO (g)}$$&#xA;The heat of formation of $\ce{O^{-2}}$ is given by&#xA;$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2 the ~O2~ bond~ energy~ (only~ half~ because~ we~ only~ need~ 1~ oxygen~ atom) + the ~first ~ionization~ potential~ for~oxygen + the~second~ionization~ potential~ for~ oxygen}$$&#xA;or&#xA;$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2(498.7)~-~141+780~ = ~$888.4$~ kJ/mol}$$&#xA;Similarly the heat of formation of $\ce{Mg^{+2}}$ is given by&#xA;$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~ \Delta H_{sublimation}Mg(s) + the ~first ~ionization~ potential~ for~Mg + the~second~ionization~ potential~ for~ Mg}$$&#xA;or&#xA;$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~130 + $738.1$ + IP_2(Mg)}$$&#xA;Before solving for $\ce{IP_2(Mg)}$ there is one more adjustment to make.  Everything we've done so far has been for the gas phase except for our starting heat of formation for $\ce{MgO}$ which was for the solid, so let's adjust this heat of formation to its gas phase value.&#xA;$$\ce{\Delta H_{f} (MgO (g))~ =~ \Delta H_{f} (MgO (s)) + lattice~ energy~ MgO }$$&#xA;or&#xA;$$\ce{\Delta H_{f} (MgO (g))~ =~ $-601.8$+ 3800 ~=~$3198.2$~ kJ/mol}$$&#xA;Now, inserting all of these values into our equation we get,&#xA;$$\ce{130 + $738.1$ + IP_2(Mg) + $888.4$~=~$3198.2$}$$&#xA;or&#xA;$$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-12T18:23:04.433" LastActivityDate="2014-07-12T18:23:04.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14294" PostTypeId="2" ParentId="14255" CreationDate="2014-07-12T17:25:11.150" Score="8" Body="&lt;p&gt;Enthalpies of phase changes are fundamentally connected to the electrostatic potential energies between molecules. The first thing you need to know is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is an attractive force between all molecules at long(ish) distances, and a repelling force at short distances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you make a graph of potential energy vs. distance between two molecules, it will look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F6kjv.png&quot; alt=&quot;Potential energy between two particles&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the y-axis represents electrostatic potential energy, the x-axis is radial separation (distance between the centers), and the spheres are &quot;molecules.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a potential energy curve, you can imagine the system as if it were the surface of the earth, and gravity was the potential. In other words, the white molecule &quot;wants&quot; to roll down the valley until it sits next to the gray molecule. If it were any closer than just touching, it would have to climb up another very steep hill. If you try to pull them away, again you have to climb a hill (although it isn't as tall or steep). The result is that unless there is enough &lt;strong&gt;kinetic energy&lt;/strong&gt; for the molecules to move apart, they tend to stick together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the potential energy function between any two types of molecules will be different, but it will always have the same basic shape. What will change is the &quot;steepness,&quot; width, and depth of the valley (or &quot;potential energy well&quot;), and the slope of the infinitely long &quot;hill&quot; to the right of the well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are talking about relative enthalpies of fusion and vaporization for a given system, we don't have to worry about how this changes for different molecules. We just have to think about what it means to vaporize or melt something, in the context of the spatial separation or relativity of molecules, and how that relates to the shape of this surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's think about what happens when you add heat to a system of molecules (positive enthalpy change). &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat&quot; rel=&quot;noreferrer&quot;&gt;Heat&lt;/a&gt; is a transfer of thermal energy between a hot substance and a cold one. It is defined by a change in temperature, which means that when you add heat to something, its temperature increases (this might be common sense, but in thermodynamics it is important to be very specific). The main thing we need to know about this is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Temperature is a measure of the average kinetic energy of all molecules in a system&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, as the temperature increases, the average kinetic energy (the speed) of the molecules increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go back to the potential energy diagram between two molecules. You know that energy is conserved, and so ignoring losses due to friction (there won't be any for molecules) the potential energy that can be &lt;em&gt;gained&lt;/em&gt; by a particle is equal to the kinetic energy it started with. In other words, if the particle is at the bottom of the well and has no kinetic energy, it is not going anywhere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZE9wL.png&quot; alt=&quot;Schematic of potential energy between two particles in a solid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it literally has no kinetic energy, we are at absolute zero, and this is an ideal crystal (a solid). Real substances in the real world always have &lt;em&gt;some&lt;/em&gt; thermal energy, so the molecules are always sort of &quot;wiggling&quot; around at the bottom of their potential energy wells, even in a solid material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how much kinetic energy do you need to melt the material? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a liquid, molecules are free to move but stay close together&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means you need enough energy to let the molecules climb up the well at least a little bit, so that they can slide around each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we draw a &quot;liquid&quot; line approximating how much energy that would take, it might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/arzmv.png&quot; alt=&quot;Potential energy between two particles in a liquid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red line shows the average kinetic energy needed for the particles to pull apart just a little - enough that they can &quot;slide&quot; around each other - but not so much that there is any significant space between them. The height of this line compared to the bottom of the well (times Avogadro's number) is the &lt;strong&gt;enthalpy of fusion&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we want to vaporize the substance? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a gas, the molecules are free to move and are very far apart&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the kinetic energy increases, eventually there is enough that the molecules can actually fly apart (their radial separation can approach infinity). That line might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZWZs.png&quot; alt=&quot;Potential Energy for the gas phase&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have drawn the line a little bit shy of the &quot;zero&quot; point - where the average molecule would get to infinite distance - because kinetic energies follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution&quot; rel=&quot;noreferrer&quot;&gt;statistical distribution&lt;/a&gt;, which means that some are higher than average, some are lower, and right around this point is where enough molecules would be able to vaporize that we would call it a phase transition. Depending on the particular substance, the line might be higher or lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the height of this line compared to the bottom of the well (times Avogadro's number) is the &lt;strong&gt;enthalpy of vaporization&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, it's a lot higher up. The reason is that for melting, the molecules just need enough energy to &quot;slide&quot; around each other, while for vaporization, they need enough energy to completely escape the well. This means that the enthalpy of vaporization is always going to be higher than the enthalpy of fusion.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-12T19:15:15.940" LastActivityDate="2014-07-12T19:15:15.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14295" PostTypeId="2" ParentId="14291" CreationDate="2014-07-12T23:06:46.063" Score="2" Body="&lt;p&gt;The &quot;official&quot; definition I've always used is that reduction is the gain of electrons. IUPAC seems to use the &lt;a href=&quot;http://goldbook.iupac.org/R05222.html&quot; rel=&quot;nofollow&quot;&gt;same definition&lt;/a&gt;. It looks like organic chemistry books often use the addition of oxygen to identify oxidation, and the addition of hydrogen or the removal of oxygen to identify reduction. This is in line with the historical definitions (at least &lt;a href=&quot;http://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow&quot;&gt;according to wikipedia&lt;/a&gt;) that defined reduction in terms of the change in mass on heating (if oxygen was lost, the weight would be reduced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using these shorthand rules, I think it's important to distinguish between the purposes of the definitions. The more formal &quot;official&quot; (according to IUPAC) definition is intended to get as close to the underlying physics as possible. The less formal &quot;shortcut&quot; definition used here is intended to quickly allow chemists to identify electron transfer just by looking at the structural formulas. In fact, in my copy of Wade's Organic Chemistry (4th ed) it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most organic chemists habitually use the following simple rules, based on the change in formula of the structure:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It then goes on to define redox reactions in terms of gain or loss of oxygen and hydrogen. This implies that the author doesn't intend it to be read as a formal definition - just a shortcut you can use in a pinch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a case of &quot;it works well enough&quot; being used to make life easy - but it is important to remember that reduction isn't really being &lt;strong&gt;defined&lt;/strong&gt; as gain of hydrogen in the formal scientific sense. It is still defined as a gain of electrons - it just so happens that in organic reactions, it is very difficult to add electrons to something without also adding hydrogen, and so the shortcut works (due to electronegativity differences between O, C, N, P, and H, as you pointed out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a redox reaction where this rule won't work, but looking at electrons will:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Li(s) + 1/2H2(g) -&amp;gt; LiH(s)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are adding hydrogen to lithium, and the oxidation state of Li goes from 0 to +1 - it is oxidized, not reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for an example of an organic reduction that does not involve the addition of hydrogen, but haven't been able to find one - which is probably why this shortcut rule works.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-13T02:24:57.943" LastActivityDate="2014-07-13T02:24:57.943" CommentCount="5" CommunityOwnedDate="2014-07-13T19:34:20.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="14296" PostTypeId="1" CreationDate="2014-07-12T23:50:14.903" Score="-1" ViewCount="65" Body="&lt;p&gt;When the plastic of water bottles is exposed to light some atoms come off from it and blend with the water, so my question is which are the chemicals and its effects&lt;/p&gt;&#xA;" OwnerUserId="7113" LastActivityDate="2014-07-13T08:12:44.587" Title="Dangerous chemicals from plastic" Tags="&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14297" PostTypeId="2" ParentId="14218" CreationDate="2014-07-13T00:39:29.150" Score="1" Body="&lt;p&gt;There actually appear to be multiple stimulants in this formulation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1,3,7-trimethylxanthine is one systematic chemical name for caffeine.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Synephrine&quot; rel=&quot;nofollow&quot;&gt;Synephrine&lt;/a&gt; is a sympathomimetic drug, which is a common ingredient in various OTC medicines, where it finds use as, e.g., a nasal decongestant due to the vasoconstriction produced as a side-effect of adrenergic activation. If you examine its molecular structure, you'll see that it's closely related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenethylamine&quot; rel=&quot;nofollow&quot;&gt;phenethylamine&lt;/a&gt;, and many (if not most) phenethylamine derivatives and structural analogs exhibit stimulant effects. Notable examples include: amphetamine, phentermine, clenbuterol, phenelzine, and various others. These substituted phenethylamines are an important class of drugs and exert a diverse variety of actions; some are stimulants, others anorectics, others decongestants, yet others antidepressants, etc.&lt;/li&gt;&#xA;&lt;li&gt;N,N-dimethyl-4-hydroxyphenylethylamine was formerly unfamiliar to me, but it appears to be a structural analog of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tyramine&quot; rel=&quot;nofollow&quot;&gt;tyramine&lt;/a&gt;, which is a naturally occurring amine present in a variety of foodstuffs. Tyramine is known to have mild stimulant effects itself, and is notorious for being potentially lethal in high doses if combined with an MAOI due to its hypertensive (blood-pressure raising) action.&lt;/li&gt;&#xA;&lt;li&gt;Beta-phenethylamine is an endogenous neurotransmitter within the human nervous system. It exerts a stimulant effect, but it typically has no effect when orally ingested because it is rapidly metabolized before it can be absorbed. Note that this is the same compound as plain &quot;phenethylamine&quot; (described above).&lt;/li&gt;&#xA;&lt;li&gt;Schizandrol A was also unfamiliar to me. There is at least &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/1678991&quot; rel=&quot;nofollow&quot;&gt;one animal study&lt;/a&gt; indicating possible CNS effects, though how this translates to humans is unclear, and I wasn't able to find any clinical studies or reputable sources reporting on effects on human subjects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In my estimation, the stimulant effects are probably prominently caused by the caffeine content of the formulation, with synephrine having secondary importance. Beta-phenethylamine in particular is unlikely to have any activity, due to its near total metabolism prior to reaching systemic circulation. As for N,N-dimethyl-4-hydroxyphenylethylamine and Schizandrol A, there doesn't appear to be enough literature on human effects to draw any conclusions, so their inclusion strikes me as dubious. They may or may not be significant in their action.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-13T00:39:29.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14298" PostTypeId="1" AcceptedAnswerId="14306" CreationDate="2014-07-13T01:15:55.400" Score="12" ViewCount="84896" Body="&lt;p&gt;I have three factors in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) vdW repulsions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Bond length &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Electron-electron repulsions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that vdW repulsions between ligand atoms push the atoms apart. And bond length obviously influences bond angle; if the bonds were very long, then the bond angle can shrink without bringing the ligand atoms too close together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other factors or are these it? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2017-04-18T21:09:19.260" Title="What affects bond angles?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14299" PostTypeId="1" CreationDate="2014-07-13T01:16:06.850" Score="6" ViewCount="618" Body="&lt;p&gt;I have recently found out that the white/yellowish &quot;sweat&quot; stains one  occasionally finds on t-shirts might be aluminium chlorohydrate polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While talk about aluminium and antiperspirants is all over the press and there are lots of DYI-instructions for removing them, they all sound quite pseudoscientific.  I find it strange, for instance, that some guides recommend both vinegar and/or baking soda as a remedy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas what else I could use to dissolve such stains without also dissolving much of the pigments?&lt;/p&gt;&#xA;" OwnerUserId="3974" LastEditorUserId="73" LastEditDate="2014-07-31T13:46:36.097" LastActivityDate="2021-05-02T20:40:33.940" Title="What is an effective fabric-friendly solvent for aluminium chlorohydrate?" Tags="&lt;polymers&gt;&lt;solvents&gt;&lt;color&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14301" PostTypeId="2" ParentId="14299" CreationDate="2014-07-13T03:10:22.053" Score="2" Body="&lt;p&gt;Have you tried baking soda or washing soda?  It may sound pseudoscientific but many people have had success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more successful is using baking soda as a deodorant, avoiding the aluminum altogether.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-13T03:10:22.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14302" PostTypeId="1" AcceptedAnswerId="14303" CreationDate="2014-07-13T03:11:14.207" Score="3" ViewCount="16460" Body="&lt;p&gt;I understand that that energy is required to overcome intermolecular forces holding solute particles together in the crystal, but I don't understand how this relates to lattice energies and respective solubilities in water.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="11367" LastEditDate="2016-11-21T19:31:57.920" LastActivityDate="2017-12-26T05:58:46.193" Title="Why do ionic substances with higher lattice energies tend to be less soluble in water than substances with lower lattice energies?" Tags="&lt;thermodynamics&gt;&lt;solutions&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14303" PostTypeId="2" ParentId="14302" CreationDate="2014-07-13T03:20:11.753" Score="10" Body="&lt;p&gt;When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy of solvation of the ions (gain) the compound won't dissociate, and it will be insoluble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large grey area where lattice energy and solvation energy are close, and for those compounds temperature has a big influence on solubility.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-13T03:20:11.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14304" PostTypeId="2" ParentId="10863" CreationDate="2014-07-13T06:25:43.987" Score="3" Body="&lt;p&gt;There are two ways I can think of to get rid of the remaining water:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As per LDC3's answer, flush the system with ethanol which will hopefully not have the problem of slow dry time.&lt;/li&gt;&#xA;&lt;li&gt;Put the glassware in an oven, which will also nicely sterilize the system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7116" LastEditorUserId="305" LastEditDate="2014-07-13T08:06:54.473" LastActivityDate="2014-07-13T08:06:54.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14305" PostTypeId="1" AcceptedAnswerId="15052" CreationDate="2014-07-13T07:18:06.243" Score="1" ViewCount="8447" Body="&lt;p&gt;Could there be some kind of process, a kind of &quot;retrograde electrolysis&quot; where hydrogen gas (or in some other form) and oxygen gas (or in some there form) are mixed with required energy to combine to form water as a byproduct?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="73" LastEditDate="2014-07-13T21:47:00.380" LastActivityDate="2014-08-09T16:31:29.540" Title="Can water be produced by mixing hydrogen gas and oxygen?" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14306" PostTypeId="2" ParentId="14298" CreationDate="2014-07-13T07:46:16.600" Score="7" Body="&lt;p&gt;Bond angles mainly depend on the following three factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hybridization: Bond angle depends on the state of hybridization of the central atom&lt;br&gt;&#xA;Hybridization: $\ce{sp^3}$, Bond angle: $109^\circ$, Example: $\ce{CH4}$&lt;br&gt;&#xA;Hybridization: $\ce{sp^2}$, Bond angle: $120^\circ$, Example: $\ce{BCl3}$&lt;br&gt;&#xA;Hybridization: $\ce{sp}$,   Bond angle: $180^\circ$, Example: $\ce{BeCl2}$&lt;br&gt;&#xA;Generally s- character increase in the hybrid bond, the bond angle increases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Good Side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triple bonds repel other bonding-electrons more strongly than double bonds.&lt;br&gt;&#xA;Double bonds repel other bonding-electrons more strongly than single bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heres a nice read into the different components (&lt;a href=&quot;http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text7/Tx73/tx73.html&quot; rel=&quot;noreferrer&quot;&gt;extra&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7112" LastEditorUserId="4945" LastEditDate="2014-07-14T11:12:33.433" LastActivityDate="2014-07-14T11:12:33.433" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14307" PostTypeId="2" ParentId="14239" CreationDate="2014-07-13T07:54:25.197" Score="3" Body="&lt;p&gt;The joule , symbol &quot;$\mathrm{J}$&quot;, is a derived unit of energy, work, or amount of heat in the SI  Units. It is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or $\mathrm{N\cdot m}$), or in passing an electric current of one ampere through a resistance of one ohm for one second. It is named after the English physicist James Prescott Joule (1818–1889).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms firstly of base SI units and then in terms of other SI units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm J  = {}\rm \frac{kg \cdot m^2}{s^2} = N \cdot m = \rm Pa \cdot m^3={}\rm W \cdot s = C \cdot V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{kg}$ is the kilogram, $\mathrm{m}$ is the metre, $\mathrm{s}$ is the second, $\mathrm{N}$ is the newton, $\mathrm{Pa}$ is the pascal, $\mathrm{W}$ is the watt, $\mathrm{C}$ is the coulomb, and $\mathrm{V}$ is the volt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One joule can also be defined as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The work required to move an electric charge of one coulomb through&#xA;an electrical potential difference of one volt, or one '&quot;coulomb&#xA;volt&quot; ($\mathrm{C\cdot V}$). This relationship can be used to define the volt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The work required to produce one watt of power for one second, or one&#xA;&quot;watt second&quot; ($\mathrm{W\cdot s}$) (compare kilowatt hour). This relationship can be&#xA;used to define the watt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Source: Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="7112" LastEditorUserId="4945" LastEditDate="2014-07-24T03:23:09.690" LastActivityDate="2014-07-24T03:23:09.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14308" PostTypeId="2" ParentId="14223" CreationDate="2014-07-13T08:03:51.857" Score="1" Body="&lt;p&gt;Because this is Chemistry, I'll assume you're dealing with some form or shape of Kinetic-Molecular Theory. So, with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_k = \frac{1}{2}mv^2$$ $E_k$ is the &#xA;average kinetic energy, m is the molecular mass and $v^2$&#xA; is the average of the squares of the molecular speeds. (if you're dealing with one particle, its just the velocity of said particle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gases at &#xA;the same temperature have the same average kinetic energy, so, on average, molecules with a lower speed have a higher &#xA;mass. If the gas is pure, all molecules will have the same mass, but at any instant the direction of movement will be &#xA;different for each. &lt;/p&gt;&#xA;" OwnerUserId="7112" LastActivityDate="2014-07-13T08:03:51.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14309" PostTypeId="2" ParentId="14296" CreationDate="2014-07-13T08:12:44.587" Score="1" Body="&lt;p&gt;It all depends on the polymer used in making the plastic bottle. I have heard that some of these chemicals you refer to are carcinogens. I won't pretend to know the formula nor the name of these carcinogens. However, the polymer must ve, to some extent, photodegradable given that the breakdown that you speak of occurs after being subjected to light. This would indicate that the polymer chain has a carbon oxygen double bond as part of the main &quot;backbone&quot; of the polymer and will have been formed by condensation polymerization. This may help to indicate the type or chemical that you're asking about.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-07-13T08:12:44.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14311" PostTypeId="2" ParentId="4877" CreationDate="2014-07-13T09:31:44.607" Score="14" Body="&lt;p&gt;Here in Mexico, a guy was arrested a couple years ago for &quot;dissolving&quot; more than 300 bodies killed by the cartels.&#xA;They found 55-gallon drums around town with a sludge in it, usually with a note to warn other. &#xA;His recipe?&#xA;A 55-gallon drum, several bags of lye, add body, fill with h20, and build fire beneath drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took about 24hrs I believe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I got caught up In the comments. And I'm New to this and all (bulletin boards?)&#xA;Like the case in Australia with sealed drums w/ HCL, in a bank vault, it Actually did more to preserve the bodies. I am no chemist, but it is my understanding, that when acids burn the skin the results are the same for both Fire and chemicals burns both rapidly dehydrate the body, destroying it. But is in a sealed container, that water can only go to the lid. Stopping the process. &#xA;I bet any acid will destroy a body.&#xA;When you're in the bath or pool too long you get those wrinkles, I've heard of people with wet feet in boots for weeks, remove their boots, and literally stripping the flesh off their feet to the bone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is how long will it take like the method used in here in Mexico and another post, boiling seems to be an important component. And again I'm guessing (and vegetarian), the heat tenderizes the meat, helping the acid along, but also the rolling of the water stirs it. If you were to put a piece of meat in acid that was stagnant, the acid might react, dissolve some meat, as the dissolved meat dilutes the acid depending on the makeup. It could settle or form a layer of meat/acid that sits on top of the meat, protecting it. Think layered shots. Or a unflushed end toilet that has had time to settle. &#xA;Hydrofluoric acid will eat thru skin, google it. One of the reasons the burns are so bad is @ 50%, it can take 8 hours to realize you were burned.&#xA;So that would be a slow process, and fluoride, as we know, preserves teeth, by extension bones. And for hydrofluoric acid burns, the standard treatment is calcium.&#xA;So it would be very slow, very likely stop, possible preserve the bones, making the resistant to rotting and harden them, but as your, we'll my fingers have less meat than most parts of my body, it would get to bone faster, ossicle dissolving some calcium, then render all the acid useless. &#xA;Again I am not a chemist but have self-studied chemistry for years, I would be very interested to hear anyone else's opinions on my determinations!&lt;/p&gt;&#xA;" OwnerUserId="7118" LastEditorUserId="402" LastEditDate="2017-04-08T22:25:51.550" LastActivityDate="2017-04-08T22:25:51.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14312" PostTypeId="1" AcceptedAnswerId="18346" CreationDate="2014-07-13T10:53:49.997" Score="1" ViewCount="1742" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If an element $\ce{X}$ forms the highest oxide of the formula $\ce{XO3}$, then it belongs to group:&lt;br&gt;&#xA;  &lt;strong&gt;A&lt;/strong&gt;) 14&lt;br&gt;&#xA;  &lt;strong&gt;B&lt;/strong&gt;) 15&lt;br&gt;&#xA;  &lt;strong&gt;C&lt;/strong&gt;) 16&lt;br&gt;&#xA;  &lt;strong&gt;D&lt;/strong&gt;) 17  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is an oxidation number of an element related to its group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried comparing $\ce{X}$ with $\ce{N}$-, $\ce{S}$-, $\ce{As}$-compounds, but all 3 come in the form $\ce{XO3}$ ($\ce{NO3}$, $\ce{SO3}$, $\ce{AsO3}$), so I can't guess the group number.&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="41328" LastEditDate="2017-08-15T21:35:19.743" LastActivityDate="2017-08-15T21:35:19.743" Title="How is the oxidation number related to the group number?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14313" PostTypeId="2" ParentId="14305" CreationDate="2014-07-13T12:27:16.800" Score="11" Body="&lt;p&gt;Indeed, it's called combustion! When you burn hydrogen and oxygen gas, the reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2_{(g)} + O2_{(g)} -&amp;gt; 2H2O_{(g)} + energy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you could see the former U.S. space shuttle as a large machine that turns hydrogen and oxygen into water. (Note: the space shuttle used liquid hydrogen and oxygen, but otherwise the principle and formula is exactly the same)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Gibbs free energy of the forward reaction is negative ($-242\textrm{ kJ/mol}$), you don't need to add energy to the system, the reaction is spontaneous (mind you, you generally need a spark to ignite the combustion because there's a &lt;em&gt;kinetic&lt;/em&gt; barrier to the reaction called the energy of activation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse reaction, water to hydrogen and oxygen, has a positive Gibbs free energy, so you need to add energy in the form of electricity to the system to make the reaction happen.&lt;/p&gt;&#xA;" OwnerUserId="6907" LastEditorUserId="6907" LastEditDate="2014-08-09T13:02:11.973" LastActivityDate="2014-08-09T13:02:11.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14314" PostTypeId="1" AcceptedAnswerId="14466" CreationDate="2014-07-13T15:33:34.550" Score="9" ViewCount="5729" Body="&lt;p&gt;I have created a solution that should contain hydroxide ions. Therefore, the pH &gt; 7, but is there any way to roughly estimate the pH in absence of any indicator? Maybe using current, or checking whether it dissolves certain chemicals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please mind that I am limited in equipment and so complex science is not feasible for me.&lt;/p&gt;&#xA;" OwnerUserId="3743" OwnerDisplayName="user209347" LastEditorUserId="73" LastEditDate="2014-07-13T21:42:15.263" LastActivityDate="2019-06-13T23:58:27.523" Title="Is there a means to measure pH without indicators?" Tags="&lt;home-experiment&gt;&lt;ph&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14315" PostTypeId="1" AcceptedAnswerId="14326" CreationDate="2014-07-13T19:09:36.207" Score="2" ViewCount="7696" Body="&lt;p&gt;It would make sense that when the benzene rings are fused they keep their pi bonds and form a polycyclic alkyne. However this is not the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis is that because monocyclic alkynes at 8C or above are stable polycyclic alkynes would be even more stable and more stable with smaller rings than monocyclic alkynes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However with 7 benzene rings fused in a hexagonal tiling type fashion the center ring has all single bonds like cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that fusing benzene or really any aromatic compound does not form polycyclic alkynes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0TFWP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A resonance structure of this would be a polycyclic alkyne along with just the regular pi bond delocalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bicyclo[4.4.0]deca-1,3,6,8,tetraene-4-yne is what I am thinking about. I am moving just the pi bonds in the resonance structure to get this. I am not moving actual carbon nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a picture of this but hopefully from the name you can figure it out.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2014-07-26T15:42:49.053" LastActivityDate="2014-07-26T16:34:01.283" Title="Why does fusing benzene rings not produce polycyclic alkynes?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14316" PostTypeId="1" AcceptedAnswerId="14317" CreationDate="2014-07-13T19:43:55.847" Score="1" ViewCount="5822" Body="&lt;p&gt;I've got two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What sort of compound can dissolve grass/leaves fastest without leaving &quot;much&quot; trace?&lt;/li&gt;&#xA;&lt;li&gt;What sort of enclosure is required for this to be doen safely and without damaging the enclosure?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am doing a small experiment and would like to try dissolving few leaves within a glass or something of the sort.&lt;/p&gt;&#xA;" OwnerUserId="7027" LastActivityDate="2014-07-13T20:44:59.817" Title="How to dissolve grass/leaves without affecting enclosure?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14317" PostTypeId="2" ParentId="14316" CreationDate="2014-07-13T20:44:59.817" Score="1" Body="&lt;p&gt;Plant matter is mostly cellulose which is insoluble in most solvents so it is unlikely you can &quot;dissolve&quot; the material.  You can react the material.  Two possibilities are nitric acid and sulfuric acid but there are a number of safety precautions you need to follow.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-13T20:44:59.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14318" PostTypeId="2" ParentId="14315" CreationDate="2014-07-13T21:01:40.187" Score="5" Body="&lt;p&gt;The ideal bond angles for alkynes are $180^{\circ}$, which is necessary for optimal overlap of atomic orbitals combining to produce mutually perpendicular $\pi$-type molecular orbitals. Any significant deviation from this is normally inherently destabilizing. Rings, by the very nature of their geometry, often incur additional strain due to their conformational inflexibility and possible demands for non-ideal bond angles (relative to acyclic molecules, that is). Introducing $\pi$-bonds into rings can potentially compound this problem, since they force at least local planarity and shorter bond lengths. These factors all contribute to the immense instability of cyclic alkynes in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you mention stable cycloalkynes, keep in mind that stability is a relative term in this context. While those with a sufficient number of carbons can be produced and isolated under specific experimental conditions, they are not thermodynamically stable relative to their open-chain equivalents, or to most of their various possible structural isomers. In some cases, they can, however, be made &lt;em&gt;kinetically&lt;/em&gt; stable by the addition of flanking $\alpha$-substituents which inhibit reaction by their steric bulk&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Syntheses of cycloalkynes below cyclooctyne are often elaborate, and frequently proceed through extremely reactive intermediates that decompose by specific reaction pathways so as to carefully avoid other products. Benzene (and the various polycyclic aromatic hydrocarbons, for that matter), on the other hand, is an extremely stable molecule which participates in quite a limited collection of reactions. Whatever particular reaction you're suggesting would almost certainly be hugely uphill in energy (i.e., yields a significantly &lt;em&gt;less stable&lt;/em&gt; product), and hence very thermodynamically disfavored; it would also certainly have immense kinetic barriers as well (since any such reaction pathway implies at least temporary loss of aromaticity in the transition state[s] and intermediate[s]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant reading might include &lt;em&gt;Dehydrobenzene and Cycloalkynes&lt;/em&gt; by Reinhard W. Hoffmann. There is extensive discussion of the reactivity of cycloalkynes, and numerous examples of spontaneous dimerizations, isomerizations, etc. Evidently, isomerizations to cycloallenes and cyclobutadiene dimers are common, which is a profound testament to the instability of cycloalkynes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;1. For example, 3,3,7,7-tetramethylcycloheptyne, a good description of which (and synthetic scheme for) is given in &lt;em&gt;Strained Hydrocarbons&lt;/em&gt; by Dodziuk.&lt;/sub&gt; &lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-13T21:18:03.180" LastActivityDate="2014-07-13T21:18:03.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14319" PostTypeId="1" AcceptedAnswerId="14324" CreationDate="2014-07-13T21:34:14.123" Score="2" ViewCount="2172" Body="&lt;p&gt;Preface: I've read &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9463&quot;&gt;this question&lt;/a&gt; about water preventing an initiation from occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why it's important to slowly add the alkyl halide to the dry ether and magnesium for &lt;strong&gt;exotherm control&lt;/strong&gt;. But why is the first addition called an initiation and what's its purpose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just to confirm that the conditions are ready to continue slowly adding the rest of the reagent? I know it's a free-radical reaction, maybe that's where the term &quot;initiation&quot; comes from. But I've seen the reaction described as &quot;self-sustaining&quot; after the initiation, but it's not a chain reaction. Perhaps that's referring to a predictable reaction proceeding forward instead of stalling then blowing up your glassware.&lt;/p&gt;&#xA;" OwnerUserId="7122" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-07-14T15:59:42.590" Title="What's the purpose of a Grignard reagent initiation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14320" PostTypeId="1" AcceptedAnswerId="14323" CreationDate="2014-07-13T21:57:28.010" Score="1" ViewCount="1183" Body="&lt;p&gt;I know electrolyzing water separates it into hydrogen and oxygen.  However, I am curious, if ice is electrolyzed, could it explode?  My thought process is that oxygen and hydrogen bubbles would build up in the ice, and continued electrolysis would ignite the gasses (forming water again).  However, since it is contained within the ice, it could ignite explosively.  Is this what would indeed happen?  One way to try this might be using a tazer or bug zapper on ice.  I believe this could be dangerous, and appropriate precautions would need to be taken.&lt;/p&gt;&#xA;" OwnerUserId="6806" LastEditorUserId="6806" LastEditDate="2014-07-13T22:04:10.467" LastActivityDate="2014-07-13T22:31:03.090" Title="Can electrolyzing ice cause an explosion?" Tags="&lt;water&gt;&lt;combustion&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14321" PostTypeId="1" CreationDate="2014-07-13T21:58:47.263" Score="3" ViewCount="4944" Body="&lt;p&gt;I was just wondering if there is any difference between supersecondary structure and motifs. As well, is a protein domain an independently folded region of a protein that has a particular function associated with it - that may or may not contain one or multiple motifs.&lt;/p&gt;&#xA;" OwnerUserId="7123" LastEditorUserId="73" LastEditDate="2014-07-14T00:20:26.197" LastActivityDate="2018-10-25T19:22:04.863" Title="What is the difference between protein domains, motifs, and supersecondary structures?" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14322" PostTypeId="2" ParentId="14319" CreationDate="2014-07-13T22:23:30.943" Score="0" Body="&lt;p&gt;First, you are mixing magnesium metal with an alkyl halide. The product of this reaction is $\ce{R-Mg-I}$, where $\ce{R}$ is the alkyl group and $\ce{I}$ is the halide. It is called an initiation reaction because you first need to make the reactive compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you add the alcohol to produce the product. Mixing an alkyl halide and an alcohol together will not yield the product you want.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2014-07-14T07:08:49.800" LastActivityDate="2014-07-14T07:08:49.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14323" PostTypeId="2" ParentId="14320" CreationDate="2014-07-13T22:31:03.090" Score="6" Body="&lt;p&gt;Bulk electrolysis of ice would likely not get very far. Electrolysis works in liquids because the liquid is able to transport an ion to complete the circuit. In the case of water electrolysis, water is needed to transport protons (or hydroxide, but we'll stick to just one thing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half reactions for &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water#Equations&quot;&gt;electrolysis of water&lt;/a&gt; are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation at anode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O (l) -&amp;gt; O2(g) + 2H+(aq) + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons travel through the wires and the protons through solution to the cathode where the following happens:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction at cathode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H+(aq) + 2e- -&amp;gt; H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrolysis requires ion mobility. In &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice_Ih&quot;&gt;solid water&lt;/a&gt; this mobility is severely decreased. All water molecules are locked in a rigid hydrogen bond network. A proton can propagate slowly through this network by proton exchange, but it will be slow. Thus electrolysis will be slowed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, electrolysis only occurs at the surface of the electrodes, which must provide a means for the gas to escape (whether they are on the surface of the ice or driven into the ice, gaps will exist for gas to escape).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-13T22:31:03.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14324" PostTypeId="2" ParentId="14319" CreationDate="2014-07-13T23:13:57.593" Score="3" Body="&lt;p&gt;The first step in the Grignard reaction involves transfer of an electron from the surface of the metal to the alkyl halide to form a radical anion.  The radical anion then disproportionates into a free radical and halide anion.&#xA;\begin{aligned}&#xA;\ce{R−X + Mg &amp;amp;-&amp;gt; R−X^{•−} + Mg^{•+}}\\&#xA;\ce{R−X^{•−} &amp;amp;-&amp;gt; R^{•} + X^{−}}\\&#xA;\ce{R^{•} + Mg^{•+} &amp;amp;-&amp;gt; RMg^{+}}\\&#xA;\ce{RMg^{+} + X^{−} &amp;amp;-&amp;gt; RMgX}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you take the magnesium out of the jar, the surface has a thin oxidation layer on it that prevents the initial electron transfer from occurring.  We used to take one piece of magnesium and scratch it good with a knife to expose a clean, fresh magnesium surface.  We would place this piece of magnesium in the flask (dry ether and the rest of the magnesium already added) and drip in a few drops of the alkyl halide.  Without fail, within a minute or two, mild bubbling would be noticed (local ether boiling), which became vigorous as the initiation phase of the reaction &quot;cleaned&quot; the surface of other pieces of oxidized magnesium.  After the vigorous bubbling started to subside we would drip in the remaining alkyl halide plus ether solution at a rate to maintain ether reflux.  So the reaction does involve a free radical but the &lt;em&gt;initiation phase&lt;/em&gt; really has more to do with removing the oxidation layer from all of the magnesium and allowing that initial electron transfer step to occur.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2014-07-14T07:07:12.317" LastActivityDate="2014-07-14T07:07:12.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14326" PostTypeId="2" ParentId="14315" CreationDate="2014-07-13T23:50:42.150" Score="4" Body="&lt;p&gt;If a polycyclic aromatic compound were to ring open, an allene would be the initial, symmetry-allowed, ring-opened product instead of an alkyne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uj2bx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've built a Dreiding model of the bis-allene product derived from naphthalene and it does not appear to be particularly strained.  I suspect the reason we do not see this process occurring is because the starting polycyclic aromatic is highly (resonance) stabilized, whereas the bis-allene product, while not strained, has no resonance stabilization due to the lack of pi-overlap in the cyclic structure due to the orthogonality imposed by the allene units.  In other words, the activation energy would be too high and the reaction would not proceed.  However, if we were to pass naphthalene through a tube furnace at very high temperature (this has been done), it would be interesting to see if any of the resultant products could be explained on the basis of the bis-allene intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;  Here's the bicyco[4.4.0] skeleton.  Using these numbers tell me where you want the 4 double and 1 triple bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qtmfy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, so this (the structure on the right) is what you mean!  There are 2 issues with it&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In this structure the bridgehead carbons each have 5 bonds, but&#xA;&lt;strong&gt;carbon can only have 4 bonds&lt;/strong&gt; (at least in compounds such as these) making this an unrealistic structure.  &lt;/li&gt;&#xA;&lt;li&gt;This structure has 12 pi electrons, whereas the starting naphthalene only had 10.  Resonance structures involve the movement of electrons, there is no way to go from 10 pi electrons to 12 pi electrons while keeping all of the sigma bonds intact, so this is not a resonance structure for naphthalene.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-26T16:34:01.283" LastActivityDate="2014-07-26T16:34:01.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14327" PostTypeId="1" CreationDate="2014-07-14T00:28:15.877" Score="4" ViewCount="1263" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$1.000~\mathrm{g}$&lt;/span&gt; sample of ethanol, &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5OH}$&lt;/span&gt;, was burned in a bomb calorimeter whose heat capacity had been determined to be &lt;span class=&quot;math-container&quot;&gt;$2.71~\mathrm{kJ/^\circ C}$&lt;/span&gt;. The temperature of &lt;span class=&quot;math-container&quot;&gt;$3.000~\mathrm{kg}$&lt;/span&gt; of water rose from &lt;span class=&quot;math-container&quot;&gt;$24.284~^\circ\mathrm{C}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$26.225~^\circ\mathrm{C}$&lt;/span&gt;. Determine the &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; for the reaction in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{kJ/mol}$&lt;/span&gt; of &#xA;  ethanol? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I said :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA;q_\mathrm{cal} &amp;amp;= C_\mathrm{cal}\cdot T\\&#xA;        &amp;amp;= (2.71\ \mathrm{kJ/^\circ C})(1.94\ \mathrm{^\circ C})\\&#xA;        &amp;amp;= 5.26~\mathrm{kJ}\\&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the &lt;span class=&quot;math-container&quot;&gt;$q_\mathrm{r}=-5.26\ \mathrm{kJ}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then calculated the number of mol by dividing the mass g over molar mass and I got &lt;span class=&quot;math-container&quot;&gt;$0.02171~\mathrm{mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I divided &lt;span class=&quot;math-container&quot;&gt;$-5.26\ \mathrm{kJ}$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$0.02171\ \mathrm{mol}$&lt;/span&gt; and I got &lt;span class=&quot;math-container&quot;&gt;$= -242.3~\mathrm{kJ/mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the correct value is &lt;span class=&quot;math-container&quot;&gt;$-1365~\mathrm{kJ/mol}$&lt;/span&gt;. What's wrong?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorDisplayName="user7951" LastEditDate="2018-12-10T17:42:32.353" LastActivityDate="2018-12-10T18:02:42.940" Title="Determine the ΔH for the reaction in kJ/mol of ethanol?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14328" PostTypeId="1" CreationDate="2014-07-14T03:11:39.697" Score="1" ViewCount="2994" Body="&lt;p&gt;I am trying to explain why some solids are more soluble in water than others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the energy released from the formation of bonds between the solid solute molecules and water molecules compensate the energy needed to break the intermolecular bonds between water molecules and the intermolecular bonds between the solute molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I'm not sure if I have used the correct terminology, for example I have never used intermolecular 'bonds', I've only used intermolecular forces before.&lt;/p&gt;&#xA;" OwnerUserId="6988" LastActivityDate="2014-07-14T14:22:14.310" Title="What is the difference between a bond, force, and interaction?" Tags="&lt;homework&gt;&lt;bond&gt;&lt;solubility&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14330" PostTypeId="1" AcceptedAnswerId="14341" CreationDate="2014-07-14T06:10:01.703" Score="5" ViewCount="1199" Body="&lt;p&gt;What will be the wavelength of an X-ray, if we apply a potential difference of, say &lt;span class=&quot;math-container&quot;&gt;$\pu{20 kV}$&lt;/span&gt; across the ends of the X-ray tube?&lt;/p&gt;&#xA;&lt;p&gt;Alright, I started with the modified de-Broglie equation-&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Lambda=\frac{h}{\sqrt{meV}}$$&lt;/span&gt;&#xA;I pinned the values of Planck constant, electronic charge and potential difference but can't guess what would be the mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; here.&lt;/p&gt;&#xA;&lt;p&gt;The answer comes &lt;span class=&quot;math-container&quot;&gt;$\pu{0.62*10^-10m}$&lt;/span&gt; if solved perfectly. Can anybody confirm?&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="83758" LastEditDate="2020-09-08T21:01:26.810" LastActivityDate="2020-09-25T12:15:20.717" Title="Wavelength of X-rays" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14331" PostTypeId="1" CreationDate="2014-07-14T08:53:39.937" Score="11" ViewCount="20235" Body="&lt;p&gt;My book says:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxygen is absorbed in colorless alkaline solution of pyrogallol and turns dark brown?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd like to know why and how.&lt;/p&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="5026" LastEditDate="2018-02-26T05:17:54.510" LastActivityDate="2018-02-26T05:17:54.510" Title="Why is oxygen absorbed in a colorless alkaline solution of pyrogallol and why does it turn dark brown?" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14334" PostTypeId="1" CreationDate="2014-07-14T12:58:11.630" Score="1" ViewCount="196" Body="&lt;p&gt;I need to produce paper bags. Is there any simple receipt for making paper glue so that brown (packaging) paper will stick to itself?&lt;/p&gt;&#xA;" OwnerUserId="7128" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:19:28.843" LastActivityDate="2015-10-30T17:19:28.843" Title="Recipe for paper glue" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14336" PostTypeId="2" ParentId="14284" CreationDate="2014-07-14T13:55:28.770" Score="4" Body="&lt;p&gt;I think I got this figured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acidity and basicity are thermodynamic properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleophilicity/electrophilicity are kinetic properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis acids are electrophiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis bases are nucleophiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use electrophile/nucleophile in describing Lewis acids/bases. But care should be taken to ensure we don't conflate kinetics with thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I finally got this issue resolved. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-14T13:55:28.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14337" PostTypeId="2" ParentId="14331" CreationDate="2014-07-14T13:59:39.957" Score="11" Body="&lt;p&gt;Pyrogallol is the common name for 1,2,3-trihydroxybenzene.  Hydroxy groups are strong activavtors of aromatic systems.  By the time 3 hydroxyl groups have been added to a benzene nucleus you have a very reactive molecule, reactive enough to react with oxygen in the air.  The reaction is very complex and free radicals are thought to be involved.  Free radical reactions involving activated aromatic rings often lead to colored, high molecular weight products that are not easy to characterize.  &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02906706#page-1&quot; rel=&quot;noreferrer&quot;&gt;Here's a link&lt;/a&gt; to a paper that suggests the initial products are probably the &lt;em&gt;ortho&lt;/em&gt;-quinone and 2-pyrone pictured below.  Certainly the quinone will be colored, and as it reacts further to produce the higher molecular weight products, they will be colored too.  &lt;a href=&quot;http://books.google.com/books?id=2oJgU1-40psC&amp;amp;pg=PA537&amp;amp;lpg=PA537&amp;amp;dq=why%20does%20pyrogallol%20absorb%20oxygen&amp;amp;source=bl&amp;amp;ots=FQ9frY-sNH&amp;amp;sig=5tUDZC3M1Lnb3ly9qLL4mw1HHWw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=otjDU-vCO4SNyAS2zIKoCg&amp;amp;ved=0CEQQ6AEwBjgK#v=onepage&amp;amp;q=why%20does%20pyrogallol%20absorb%20oxygen&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Here's another link&lt;/a&gt; to a very old paper suggesting that some of the higher molecular weight products contain at least 20 carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZIuna.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced equation describing the initial radical oxidation of pyrogallol to the quinone can be written as&#xA;$$\ce{2~ pyrogallol + O_2 -&amp;gt; 2 ~o-quinone + 2H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-14T13:59:39.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14338" PostTypeId="1" AcceptedAnswerId="14340" CreationDate="2014-07-14T14:17:58.057" Score="3" ViewCount="307" Body="&lt;p&gt;Is there any particular reason why this mechanism is used in the solution manual rather than an acid-catalyzed hydration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8bhCx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of having the double bond attack a hydronium ion, which would create a carbocation. This would then create a place for water to attack. Finally, the water would be attacked by water to reform hydronium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why wouldn't this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have anything to do with the fused ring? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-14T14:47:00.897" Title="Acid-Catalyzed Hydration" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14339" PostTypeId="2" ParentId="14328" CreationDate="2014-07-14T14:22:14.310" Score="1" Body="&lt;p&gt;Here is how I think of the situation.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The term &quot;bond&quot; is generally reserved to describe an attractive, stabilizing interaction between atoms or groups in the same molecule or in different molecules, that has a magnitude of at least several kcal/mol.  Everything from hydrogen bonds at the low end up to full chemical bonds at the high magnitude end are included.&lt;/li&gt;&#xA;&lt;li&gt;The terms &quot;force&quot; and &quot;interaction&quot; can be used to describe any attractive or repulsive situation, of any magnitude, between atoms or molecules. So these terms would be used to describe the van der Waals &quot;interaction&quot;, where the &quot;forces&quot; involved are much less than those involved in bonding.  However, the terms &quot;force&quot; and &quot;bonding&quot; can also be used to describe situations involving larger magnitude interactions such as, &quot;bringing the atoms together led to a bonding interaction.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&quot;Force&quot; and &quot;interaction&quot; can be used to describe any situation - attractive, repulsive, high energy, low energy; whereas &quot;bonding&quot; is typically reserved to describe stabilizing, higher energy interactions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-14T14:22:14.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14340" PostTypeId="2" ParentId="14338" CreationDate="2014-07-14T14:33:48.017" Score="4" Body="&lt;p&gt;Acid catalyzed hydration probably would work, but might well be accompanied by other materials derived from the carbocation intermediate, such as rearrangement and elimination products, which would make this route less desirable.  The mercuric acetate route does not involve free carbocations, so such side products will not contaminate the desired product. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-14T14:47:00.897" LastActivityDate="2014-07-14T14:47:00.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14341" PostTypeId="2" ParentId="14330" CreationDate="2014-07-14T16:00:41.147" Score="3" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\lambda=\frac h{\sqrt{meV}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E=mc^2$$&lt;/span&gt;&#xA;Also &lt;span class=&quot;math-container&quot;&gt;$$E=\frac{hc}\lambda$$&lt;/span&gt;&#xA;Therefore,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$mc^2=\frac{hc}\lambda$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$mc=\frac h\lambda$$&lt;/span&gt;&#xA;Hence &lt;span class=&quot;math-container&quot;&gt;$$m=\frac h{\lambda c}$$&lt;/span&gt;&#xA;Plugging the value of &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; from here we get an equation with all known constants. Solving for &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt;, we can get the result as &lt;span class=&quot;math-container&quot;&gt;$0.62\times10^{-10}\ \mathrm m$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorDisplayName="user7951" LastEditDate="2020-09-25T12:15:20.717" LastActivityDate="2020-09-25T12:15:20.717" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14342" PostTypeId="2" ParentId="14321" CreationDate="2014-07-14T16:56:53.910" Score="1" Body="&lt;p&gt;A few well-defined searches can quickly resolve questions like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: is any difference between supersecondary structure and motifs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motifs and supersecondary structures are synonymous. &lt;a href=&quot;http://www.cryst.bbk.ac.uk/PPS95/course/6_super_sec/super.html&quot; rel=&quot;nofollow&quot;&gt;src&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: is a protein domain an independently folded region of a protein that has a particular function associated with it? (that may or may not contain one or multiple motifs?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Each domain forms a compact three-dimensional structure and often can be independently stable and folded. Many proteins consist of several structural domains.&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Structural_domain&quot; rel=&quot;nofollow&quot;&gt;Wiki: Structural Domain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Several motifs pack together to form compact, local, semi-independent units called domains.&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Structural_domain#Tertiary_structure_of_domains&quot; rel=&quot;nofollow&quot;&gt;Wiki: Protein Domain&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-07-14T16:56:53.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14343" PostTypeId="1" AcceptedAnswerId="14349" CreationDate="2014-07-14T17:04:58.353" Score="1" ViewCount="10493" Body="&lt;p&gt;I have been thinking about my future goals and I have recently decided to go for Biomedical Engineering. sites recommend to start in Chemical Engineering because its a good pathway for biomedical engineering. is this true? please share what you think and from your perspective, thanks&lt;/p&gt;&#xA;" OwnerUserId="7130" LastActivityDate="2014-07-15T00:57:07.247" Title="Is Chemical Engineering a good pathway to Biomedical Engineering for graduate school?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-07-23T00:56:40.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="14344" PostTypeId="2" ParentId="14321" CreationDate="2014-07-14T17:17:01.093" Score="2" Body="&lt;p&gt;All supersecondary structures are motifs, but not all motifs are supersecondary structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motifs, in a biological sense, are very similar to fashion motifs. They are patterns that repeat in a lot of different places. Supersecondary structures are motifs that are made of several secondary structures. Some motifs are much smaller (pockets for holding ions are 3-4 amino acids long), or involve no secondary structure interactions (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Short_linear_motif&quot; rel=&quot;nofollow noreferrer&quot;&gt;SLM&lt;/a&gt;s). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second half of your question, a protein domain (depending on who you ask) is any region or regions of a protein that do something, regardless of how it is folded. Very often, domains can fold correctly on their own. Generally, the required protease/functional assay to determine whether a possible domain is really technically a domain is not usually done. For example, the kinase insert domain found in some tyrosine kinase receptors is considered a domain, despite being a) quite small b) unable to perform its function in isolation (not entirely fair, as its function is to &lt;a href=&quot;http://www.jbc.org/content/274/10/6453.long&quot; rel=&quot;nofollow noreferrer&quot;&gt;phosphorylate the rest of the receptor&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum: protein domains have a couple of competing definitions, so you cannot assume that something referred to as a 'protein domain' is capable of performing its function when separated from the rest of the protein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Domains can be of nearly any size (due to their functional and not structural classification), but very small and very large ones are rare. They can be made of any number of motifs, or none at all.&lt;/p&gt;&#xA;" OwnerUserId="7103" LastEditorUserId="9961" LastEditDate="2018-10-25T19:22:04.863" LastActivityDate="2018-10-25T19:22:04.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14345" PostTypeId="1" CreationDate="2014-07-14T18:00:43.340" Score="3" ViewCount="99" Body="&lt;p&gt;I am learning about base excision repair mechanisms. I understand the main idea how glycosylases are working, but I am trying to understand this in much more detail. I found a paper where they have written: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Monofunctional glycosylases catalyze the single-step displacement of damaged base via attack of an activated water molecule on c1' of the substrate &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure what an &quot;activated water molecule&quot; is, how, for example, an aspartate residue is making a water molecule 'activated', and how this &quot;activated water molecule&quot; is removing the damaged base (hydrolysis N-glycosyl bond). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0959440X04000041&quot; rel=&quot;nofollow&quot;&gt;J Christopher Fromme, Anirban Banerjee, Gregory L Verdine: DNA glycosylase recognition and catalysis.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7132" LastEditorUserId="4945" LastEditDate="2014-07-16T11:27:01.453" LastActivityDate="2014-07-16T11:27:01.453" Title="What is the mechanism of monofunctional glycosylase catalysis?" Tags="&lt;reaction-mechanism&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-07-16T19:20:14.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="14347" PostTypeId="1" AcceptedAnswerId="14350" CreationDate="2014-07-14T23:27:17.513" Score="3" ViewCount="971" Body="&lt;p&gt;1) Which Lewis structure is preferential of the two below? EN considerations suggest the first structure. Size/charge density suggests the second one; the second structure would likely be more stable because if we think of the halide series of binary acids - the bromide anion is more stable than the chloride anion - and hence hydrobromic acid is a stronger acid than hydrochloric acid - the sheer size of the bromide ion is able to effectively dilute its charge and overcome its lack in electronegativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TC7Bp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I am told that peripheral atoms should have octets. Why is this so? What is the physical reasoning behind this? Does this have anything to do with minimizing the overall potential energy of the molecule - i.e. the peripheral atoms are perhaps more subject to attack than internal atoms (due to steric factors) - so molecules try to minimize such a possibility by completing the octet for peripheral atoms? On the other hand, the octet rule is more of a guideline rather than a rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is the bond-angle of the Cl-Br-Cl bond? I am told that the bond angle is a perfect 180 degrees. What about the first structure? If we consider the first structure a resonance contributor, wouldn't that lead to a distorted/non-linear bond angle? Or would that still lead to a perfect 180 degree bond angle, because to me, it seems that all the lone pairs on the bromine can participate in resonance (if not all at once). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-15T00:04:48.047" LastActivityDate="2014-07-15T01:40:36.963" Title="Lewis Structures - Peripheral Octets" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14348" PostTypeId="1" AcceptedAnswerId="14351" CreationDate="2014-07-15T00:33:59.703" Score="11" ViewCount="90385" Body="&lt;p&gt;I have a chemistry test coming up and I might need to know the charge that goes with the different ions like $\ce{SO4}$ has $-2$, $\ce{NO2}$ is $-1$ and  $\ce{PO4}$ is $-3$.  Is there an easy way to remember this by looking at the periodic table or something like remembering that the second row elements are all going to be $\ce{something-O3}$  and the next row is all $\ce{something-O4}$ for the -ate endings (except $\ce{Cl}$ which is the odd man out). It will make writing out the different acid combos easier, if I can remember the charge difference that needs offsetting by the hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just trying to lay down some associative triggers for myself.&lt;/p&gt;&#xA;" OwnerUserId="7029" LastEditorUserId="5026" LastEditDate="2018-02-25T07:38:24.570" LastActivityDate="2018-02-25T07:43:45.170" Title="Is there an easy way to remember charges on ions?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14349" PostTypeId="2" ParentId="14343" CreationDate="2014-07-15T00:57:07.247" Score="5" Body="&lt;p&gt;If you want to get into a biomedical engineering program in graduate school, chemical engineering is not a bad option. You would of course probably have better luck if you majored in biomedical as an undergrad as well, since your coursework would more directly prepare you for those programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things to note though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a (big) difference between bio-&lt;strong&gt;medical&lt;/strong&gt; engineering and bio-&lt;strong&gt;chemical&lt;/strong&gt; engineering. Biomedical engineering is more closely related to pre-med and materials science than to biochemistry. That said, bio chem courses probably wouldn't &lt;em&gt;hurt&lt;/em&gt; you, but if you have the option, taking real biomedical engineering courses as electives (or as a minor) would be better. For bio-&lt;strong&gt;chemical&lt;/strong&gt; engineering, these days regular chem E programs almost always have at least an elective component in biochem engineering, and often times departments are called &quot;Chemical and Biological Engineering&quot; due to the similarities. In other words, chemical engineering undergrad degrees are very well suited (and possibly over-prepare) you for biochemical engineering graduate work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing to know is: the real secret to getting into a good graduate program is to get research experience as an undergrad in the field that you want to work in. This demonstrates to the acceptance committee that you meet the two main requirements PIs have for grad students:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You are interested in the subject.&lt;/li&gt;&#xA;&lt;li&gt;You have the aptitude to actually accomplish something. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A chemical engineering degree demonstrates aptitude on its own, and as @rch mentioned, is a pretty good &quot;fall back&quot; plan in case you change your mind later (although chem E is a tough major if it's just a fall back plan - I would not recommend it unless you really like chemistry and physics and could imagine working in a chemical plant, since that is likely what you will do if you &quot;fall back&quot; on it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is going to be that lots of other people will be able to demonstrate aptitude, with degrees in biomedical engineering, materials engineering, chemical engineering (lots of them get into biomedical later), pre-med, and other stuff that is just as well suited. In addition to aptitude, a lot of them (if not most of them) will also have research experience through undergrad projects or internships. To be competitive with them, you will want some as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strike&gt;best&lt;/strike&gt; easiest way to get research experience as an undergrad is to find a professor who is working on something you are interested in, and volunteer. Many times they will have paid positions specifically for undergrads, but if you can put in volunteer hours and are reasonably good at communicating, managing your time, and getting things done, most professors would be happy to put you to work for free. If you can manage to get a publication before you finish your undergrad degree, you will have an excellent shot at getting into the school of your choice, provided your GRE scores and grades are good (another reason to carefully consider chem E - not an easy major).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is internships, although a research internship carries more weight in grad school circles than a regular old &quot;gets the coffee and fills out spreadsheets&quot; internship. They are much harder to get, but usually pay pretty well, so it is definitely worth trying for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point I am trying to make here is - major in (almost) whatever you want, but if you want to go to grad school in a particular field, &lt;strong&gt;get research experience in that field as an undergrad.&lt;/strong&gt; Go for paid work first, but be prepared to volunteer if you have to, at least for a semester. You can quickly turn a volunteer position into a paid one if you do good work. Work experience also goes a long way toward getting you a job if you don't end up going to grad school. I'd put your odds of getting a good job with a biomedical engineering undergrad degree &lt;em&gt;with&lt;/em&gt; experience at higher than getting a job with a chemical engineering degree &lt;em&gt;without&lt;/em&gt; experience, especially these days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just so you don't think I'm trying to talk you out of a Chem E degree, Chem E was my major (BS, MS, and PhD) and I really liked it. If I didn't like chemistry and physics so much though, I probably wouldn't have made it through.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-15T00:57:07.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14350" PostTypeId="2" ParentId="14347" CreationDate="2014-07-15T01:18:40.130" Score="5" Body="&lt;p&gt;Quoting Greenwood &amp;amp; Earnshaw's &lt;em&gt;Chemistry of the Elements&lt;/em&gt;, discussing the triatomic polyhalide anions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[I]nteratomic distances [...] are individually always substantially greater than for the corresponding diatomic interhalogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the following data table does not contain an actual value for the bond length, though it does indicate that the two $\ce{Br-Cl}$ bonds are of equal length (which we probably could've safely assumed from symmetry considerations and the energetic degeneracy of the orbitals of the two chlorine atoms, if nothing else), and that the $\ce{Cl-Br-Cl}$ bond angle is (approximately) $180^{\circ}$. Bond length can often be correlated to both bond strength and bond multiplicity, especially when comparing compounds of the same atoms. Since the bonds in $\ce{BrCl2-}$ are longer than those in $\ce{BrCl}$, it's reasonable to infer that they are also weaker, and do not have greater multiplicity (i.e., there is no &lt;em&gt;additional&lt;/em&gt; $\pi$-bonding). I would further reason that any $\pi$-orbital overlap would have to be very poor (judging by the mismatch in both size and energy of the $3p$ and $4p$ orbitals of chlorine and bromine, respectively), which makes $\pi$-bonding unlikely. (For the sake of completeness, I would note that I could be mistaken here, since $4p$ orbitals are somewhat smaller and lower in energy than expected, given the poor shielding of the nucleus by $3d$ electrons. That said, I don't think the magnitude of this effect is large enough to make a major difference here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A full quantitative molecular orbital theory treatment of the molecule would require computation, and I wasn't able to find many useful literature references. However, as the molecule is relatively simple, we can probably make reasonable progress with even a very naive analysis. On the valence level(s), we have a total of $22$ electrons in $12$ twelve atomic orbitals, all $s$ or $p$. These will combine to form twelve molecular orbitals, eleven of which are going to be doubly occupied. Note that for any bonding MO generated, an anti-bonding MO will also be generated. Hence, even if we had six bonding MOs (and, in reality, some would almost certainly be non-bonding), we would have five anti-bonding MOs occupied as well. This would yield an overall bond order of $1$, indicating a bond order of $\frac{1}{2}$ for each individual bond. By analogy to similar (but more common and hence extensively studied) polyhalogen anions (e.g., $\ce{I3-}$), this conclusion seems justified. $\ce{I3-}$, and numerous similar isoelectronic molecules, are often described using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_four-electron_bond&quot; rel=&quot;nofollow&quot;&gt;three-center four-electron (3c4e) bond&lt;/a&gt;, and I think this model is probably applicable to $\ce{BrCl2-}$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot of all this is that, while it's true that you can draw certain valid resonance contributors, they may not accurately reflect the structure and electron density of the actual molecule. The structure with the negative formal charge on the bromine is best, which agrees with your intuition regarding its size and polarizability being able to stabilize the charge. Due to the disparity in electronegativity, however, the bonds are nevertheless polarized towards the chlorine atoms, so the formal charge alone doesn't fully describe the electron density correctly.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-15T01:40:36.963" LastActivityDate="2014-07-15T01:40:36.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14351" PostTypeId="2" ParentId="14348" CreationDate="2014-07-15T01:25:19.750" Score="10" Body="&lt;p&gt;I don't believe there is any simple way to determine what they are except by just plain memorization. I have given some tricks below that have helped me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I write down my poly-atomic ions, I generally put them in similar groups in a certain order. From least amount of oxygen to most amount oxygen. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;hypo&lt;/strong&gt; (name of ion) &lt;strong&gt;ite&lt;/strong&gt; -&gt; least oxygen&lt;/li&gt;&#xA;&lt;li&gt;(name of ion) &lt;strong&gt;ite&lt;/strong&gt; -&gt; one more oxygen than above&lt;/li&gt;&#xA;&lt;li&gt;(name of ion) &lt;strong&gt;ate&lt;/strong&gt; -&gt; one more oxygen than 'ite'&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;per&lt;/strong&gt; (name of ion) &lt;strong&gt;ate&lt;/strong&gt; -&gt; one more oxygen than 'ate'&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Understanding the above helps most with 'ite' and 'ate'. 'Ate' generally being 1 more than 'ite' for oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I list them from least charge to most charge. I order them this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitr &lt;strong&gt;ite&lt;/strong&gt; $\ce{NO2-}$&lt;/li&gt;&#xA;&lt;li&gt;Nitr &lt;strong&gt;ate&lt;/strong&gt; $\ce{NO3-}$&lt;/li&gt;&#xA;&lt;li&gt;Sulf &lt;strong&gt;ite&lt;/strong&gt; $\ce{SO3^2-}$&lt;/li&gt;&#xA;&lt;li&gt;Sulf &lt;strong&gt;ate&lt;/strong&gt; $\ce{SO4^2-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can see that 'ate' has one more oxygen, and that I have listed them in charge order. Sometimes looking at the trends in oxygen and charge will help you memorize. I have had to remember them by writing them down several times and now I am comfortable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example: Hypochlorite is &#xA;$\ce{ClO^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, chlorite should be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\ce{ClO2-}$, because it has one more oxygen, and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chlorate should be?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\ce{ClO3-}$, because it has one more oxygen than the above, and&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO4-}$, will be called?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;Per&lt;/strong&gt; chlor &lt;strong&gt;ate&lt;/strong&gt;, because it has the maximum amount of oxygen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above is not a hard and fast rule. One exception I can think of is manganate and permanganate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manganate is $\ce{MnO4^2-}$ but permanganate is $\ce{MnO4^-}$; same amount of oxygen, just a different charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to figure out any of the above from a periodic table didn't make a difference to me because my professor tests us without it. Yikes!&lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="5026" LastEditDate="2018-02-25T07:40:41.763" LastActivityDate="2018-02-25T07:40:41.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14352" PostTypeId="1" CreationDate="2014-07-15T05:30:02.590" Score="2" ViewCount="763" Body="&lt;p&gt;How do I represent a chemical structure in a matrix format in Matlab? Can I input a chemical structure into Matlab using the S.M.I.L.E.S notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to specify the connectivity of different atoms with others and type of bonds between them. I have to write the structure in matrix form, so that it can be utilized in some linear programming. For example, if one atom is connected to another atom I have to store 1 in matrix otherwise 0 (binary variables).&#xA;I need to develop back end code for my purpose to design some molecules using this approach. As of now I am dealing with 2D structures.&lt;/p&gt;&#xA;" OwnerUserId="7134" LastEditorUserId="4945" LastEditDate="2014-07-16T11:20:51.027" LastActivityDate="2014-07-16T20:01:03.897" Title="How can a chemical structure be stored in a matrix?" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14356" PostTypeId="1" AcceptedAnswerId="14360" CreationDate="2014-07-15T08:47:32.490" Score="9" ViewCount="2050" Body="&lt;p&gt;I just read about van der Waals radius and had some questions about it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Would it be correct to say that in general, van der Waals radius decreases as we move from left to right in the periodic table?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The van der Waals radius of the noble gas in a period is larger than the radius of halogen that precedes it. Why is it so? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the difference between van der Waals radius for monoatomic (say neon) and non-monoatomic (say fluorine) gases?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="4945" LastEditDate="2014-07-16T02:15:00.683" LastActivityDate="2017-07-06T04:20:20.510" Title="Van der Waals Radius clarifications" Tags="&lt;intermolecular-forces&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14357" PostTypeId="1" CreationDate="2014-07-15T14:12:55.787" Score="5" ViewCount="17613" Body="&lt;p&gt;What is the material of which the transparent part of CD cases are made of? It is nicely rigid and really lightweight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8ei4.jpg&quot; alt=&quot;cd case&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7136" LastEditorUserId="4945" LastEditDate="2014-07-16T10:58:08.367" LastActivityDate="2017-04-04T12:25:08.133" Title="What is the plastic material of the CD cases?" Tags="&lt;materials&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14360" PostTypeId="2" ParentId="14356" CreationDate="2014-07-15T16:23:34.257" Score="9" Body="&lt;p&gt;Let me start by providing a &lt;a href=&quot;http://periodictable.com/Properties/A/VanDerWaalsRadius.bt.html&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt; to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes, that is a general trend, but as you can see there are many exceptions.  The trend exists for the same reason that covalent radii generally shrink as we move from left to right across the periodic table, every time we add another proton and electron to an atom, the electrons do not effectively screen the outermost electrons from the positive nucleus so they become drawn in closer to the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;&amp;amp; 3. There are a lot of different ways to measure the van der&#xA;Waals radius of an atom, but in recent times the preferred method is&#xA;to examine the crystal structure of a molecule.  Using $\ce{F2}$ as&#xA;an example, the crystal would be composed of fluorine molecules.  if&#xA;we measure the crystallographic distance between two adjacent, but&#xA;non-bonded fluorine atoms and divide by two, we would have a single&#xA;value that could be used to estimate the van der Waals radius for&#xA;fluorine (we would probably want to measure a number of similar&#xA;values for fluorine from other molecules, maybe using $\ce{CH3F,&#xA;NF3, ClF,}$ etc., for example, and average them together). Our&#xA;crystal might look something like the drawing below.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qqG3U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because diatomic and polyatomic molecules are polarizable, the bonded fluorine atoms will likely be slightly polarized when packed in the crystal in order to stabilize (lower the energy) the overall system.  This polarization will cause non-bonded fluorines to draw a bit closer to each other then they would if there were no polarization.  A crystal of a noble gas (let's say neon) will just be a collection of monatomic neon atoms, no bonds, no polarization.  Therefor there is no &quot;attractive&quot; force to draw adjacent neon atoms closer together like there was in the case of fluorine.  So even before we make our measurements we might expect the van der Waals radius for neon to be larger than that for fluorine, which turns out to be the case.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-15T16:23:34.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14361" PostTypeId="1" AcceptedAnswerId="14370" CreationDate="2014-07-15T16:32:59.713" Score="12" ViewCount="20109" Body="&lt;p&gt;I would like to know why water doesn't ever become like syrup before it freezes when many other liquids, such as hydrocarbons, seem to have a more gradual change from becoming more viscous and then turning solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the more viscous a material is, the stronger the inter-molecular forces are. When the liquid is heated up, the molecules have more kinetic energy and so they can more easily break free of these forces--the liquid will become less viscous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't water have the same spectrum of viscosity as other materials? It seems like it is either solid, or it has very low viscosity. &lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="5633" LastEditDate="2014-07-16T00:04:36.097" LastActivityDate="2015-11-12T15:13:22.623" Title="Why doesn't the viscosity of water change much with temperature like it does for other substances?" Tags="&lt;intermolecular-forces&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14362" PostTypeId="1" AcceptedAnswerId="14367" CreationDate="2014-07-15T16:42:44.817" Score="3" ViewCount="69" Body="&lt;p&gt;I am looking for normal resolution IR-spectra of: $\ce{O2}$ and $\ce{N2}$ at high concentrations and; common minor gas in atmosphere: $\ce{H2O}$, $\ce{CO}$, and $\ce{CO2}$ at environmental concentrations. I have found some reference on internet (mostly from Green House Effect articles), but I cannot found it on a free, accurate and reliable source even on the free part of NIST databases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find good references for such spectra?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="5456" LastEditDate="2014-07-15T18:08:04.167" LastActivityDate="2014-07-15T18:50:14.770" Title="Atmopsheric Gas IR-Spectra" Tags="&lt;spectroscopy&gt;&lt;reference-request&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14363" PostTypeId="2" ParentId="14357" CreationDate="2014-07-15T16:44:23.453" Score="4" Body="&lt;p&gt;These are called slimline jewel cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_disc_packaging#Slimline_jewel_case&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical Disc Packaging&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bulk of slimline cases are made with translucent or transparent &lt;strong&gt;polystyrene&lt;/strong&gt;, and are available in multiple colors. A stronger alternative is made from semi-opaque, semi-flexible &lt;strong&gt;polypropylene&lt;/strong&gt; which is strong enough to protect the disc, but flexible enough not to break easily. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polystyrene#mediaviewer/File:Caja_de_CD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an image&lt;/a&gt; of a polystyrene case to compare to yours:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ezLGO.jpg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="4945" LastEditDate="2014-07-16T11:02:41.903" LastActivityDate="2014-07-16T11:02:41.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14364" PostTypeId="2" ParentId="14357" CreationDate="2014-07-15T16:44:25.893" Score="3" Body="&lt;p&gt;CD jewel cases are made from injection molded polystyrene.  Clam-shell alternatives use polypropylene.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-15T16:44:25.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14365" PostTypeId="1" CreationDate="2014-07-15T16:56:51.613" Score="9" ViewCount="87593" Body="&lt;p&gt;Question: The dipole moment of $\ce{HBr}$ is $2.60 \times 10^{-30}$ and the interatomic spacing is $1.41$. What is the percentage ionic character of $\ce{HBr}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know is that the percentage ionic character is observed dipole moment divided by calculated dipole moment $\times\ 100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculated dipole moment is $\textrm{charge on electron} \times \textrm{radius of molecule}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how would I convert interatomic spacing into radius?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastEditorUserId="305" LastEditDate="2014-07-15T18:22:45.930" LastActivityDate="2018-09-25T10:08:56.693" Title="Dipole moment - calculation of percentage ionic character" Tags="&lt;ionic-compounds&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14366" PostTypeId="2" ParentId="14365" CreationDate="2014-07-15T18:01:12.010" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The calculated dipole moment is charge on electron * radius of&#xA;  molecule&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's off a bit, the calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{aligned}&#xA;\mu_{100\%} &amp;amp;= \mathrm{electron ~charge \times interatomic ~spacing}\\&#xA;\mu_{100\%} &amp;amp;= (1.6 \times 10^{-19})\ \mathrm C \times (1.41 \times 10^{-10})\ \mathrm m\\&#xA;\mu_{100\%} &amp;amp;= 2.26 \times 10^{-29}\ \mathrm{C\ m}\\&#xA;\end{aligned}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you point out,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\% ~ionic~ character} = \frac{[\mu_\mathrm{obs}]}{[\mu_{100\ \%}]} = \frac{[2.6 \times 10^{-30}]}{[2.26 \times 10^{-29}]} = 0.12 = 12\ \%$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2018-09-25T10:08:56.693" LastActivityDate="2018-09-25T10:08:56.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14367" PostTypeId="2" ParentId="14362" CreationDate="2014-07-15T18:50:14.770" Score="4" Body="&lt;p&gt;Completely symmetrical diatomic molecules like $\ce{O2}$ and $\ce{N2}$ are not IR-active because they have a zero dipole moment, even when oscillating within their one vibrational mode. Hence, there is no change in dipole moment possible, so these molecules will not be detectable by IR. The other molecules you mention, however, are IR-active, and I was able to find spectra of all three in the NIST database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other spectroscopic methods that can detect IR-inactive molecules, however. Raman spectroscopy may be applicable, and I'm dimly aware of sophisticated methods based on absorptions in electronically excited states, but I don't know nearly enough about the subject to offer any concrete information on those techniques.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-15T18:50:14.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14368" PostTypeId="1" AcceptedAnswerId="14369" CreationDate="2014-07-15T19:34:36.767" Score="4" ViewCount="287" Body="&lt;p&gt;It is in pellets. And the box says main ingredient $\ce{NaOH}$ at least $30\%$ (I hope it's more). The box also says &quot;aluminium catalyst&quot;. My goal was to take these pellets and dissolve them in water so that I would make a $1\textrm{M}$ solution of $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set everything up and most of it refused to dissolve in the solution. And the solubility for $\ce{NaOH}$ is pretty high. So I added some heat.This made almost everything slowly dissolve. Another thing was these small aluminium bits seemingly reacting and going up and down in the solution. I'm not sure if there was any gas released, but the solution definitely was not still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I filtered it to remove the small aluminium balls. Any ideas If I have a relatively pure solution of $\ce{NaOH}$, or is it full of some other drain cleaner components. It is transparent-whiteish-little-yellowish. Filter kinda yellowish too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could these other components be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="305" LastEditDate="2014-07-16T03:40:31.420" LastActivityDate="2014-07-16T03:40:31.420" Title="Any possible additives to a drain cleaner I use as a source of NaOH?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14369" PostTypeId="2" ParentId="14368" CreationDate="2014-07-15T20:18:31.630" Score="5" Body="&lt;p&gt;The product you've described contains the following ingredients (per the &lt;a href=&quot;http://www.globalcosmed.eu/pl/index/html/id:4/cat:78/brand:87/pid:224/Kret_jedyny_oryginalny&quot; rel=&quot;noreferrer&quot;&gt;ingredient data sheet&lt;/a&gt; on the manufacturing company website):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sodium hydroxide&lt;/li&gt;&#xA;&lt;li&gt;Sodium chloride&lt;/li&gt;&#xA;&lt;li&gt;Aluminium&lt;/li&gt;&#xA;&lt;li&gt;Mineral oil&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One possibility is that the aluminium is there to provide a foaming action, as its oxidation in the presence of aqueous sodium hydroxide liberates $\ce{H2}$ gas, per the following reaction (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_aluminate#Reaction_of_aluminium_metal_and_alkali&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Al + 2NaOH + 2H2O -&amp;gt; 2NaAlO2 + 3H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aluminium and sodium hydroxide would need to be kept separate and relatively anhydrous in the bottle to avoid reaction, so I wonder if the purpose of the mineral oil (which is hydrophobic) is to act as a moisture barrier and prevent mixing. If this is indeed the reaction that occurs, then the hydrated aluminium salts present are going to be at least somewhat soluble in solution, so I think it's improbable that what you have is pure sodium hydroxide. That said, if the quantity of aluminium is indeed small, this might not be a problem for your purposes.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-15T20:18:31.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14370" PostTypeId="2" ParentId="14361" CreationDate="2014-07-15T20:33:50.147" Score="19" Body="&lt;p&gt;This is something I spent a lot of time thinking about during my PhD, so let me see if I can explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are talking about is the relationship between macroscopically observable properties (like viscosity), the &lt;strong&gt;relaxation time spectrum&lt;/strong&gt; (which is a way of looking at the time scales involved in molecular movements), and the structure of the molecules themselves. It's kind of complicated! But, I think I can explain it in way that will make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that as we increase the temperature, molecules have more kinetic energy and can more easily break free of the forces holding them together. I have some other answers &lt;a href=&quot;https://physics.stackexchange.com/questions/114525/how-does-hot-water-clean-better-than-cold-water/114600#114600&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14255/why-is-the-enthalpy-of-vaporization-greater-than-the-enthalpy-of-fusion/14294#14294&quot;&gt;here&lt;/a&gt; that explain this in more detail. With something like viscosity, however, there is more to the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you need to know is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Viscosity is a measure of momentum transfer in a fluid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, it is a way of quantifying the effect that movement in one part of a fluid has on the parts immediately adjacent to it. If the movements are highly coupled (meaning that if one layer moves, the layers next to it move almost as much), the viscosity is high. If not, the visocosity is low. &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a pretty good explanation of this on the macroscopic scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what causes a fluid to have a viscosity? Since you know about intermolecular &lt;em&gt;attractive&lt;/em&gt; forces, this is an obvious candidate. And indeed, the stronger the intermolecular forces are in a fluid, the higher the viscosity tends to be. However, there is another factor involved - intermolecular &lt;strong&gt;repulsive&lt;/strong&gt; forces (sometimes called &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_effects&quot; rel=&quot;noreferrer&quot;&gt;stearic effects&lt;/a&gt;&lt;/strong&gt; or &lt;strong&gt;excluded volume interactions&lt;/strong&gt;). The idea is that if one molecule wants to move in a dense fluid, the other molecules have to get out of the way first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OA0i5.png&quot; alt=&quot;A small molecule moves through a crowd.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an analogy, imagine moving through the crowd at a packed concert or club. In order for you to move, someone else needs to move, and for them to move, someone else has to move, and so on. You can be polite and wait patiently for them to move, and eventually someone will. Or you could be assertive and ask, and move faster. If you are really impatient, you could barrel through people, knocking them out of the way (and hope you don't get beat up). Your impatience in a crowd is analogous to shear stress in a fluid. At lower temperatures, the molecules are closer together (since their kinetic energy isn't high enough to move very far out of the potential energy wells of their neighbors), and they move more slowly in general. This means that the time it takes for neighbors to move goes up, and as a result, viscosity goes up too - if you force them to move, they will knock others out of the way, transferring momentum through the fluid. At higher temperatures molecules move much more frequently, and so it is relatively easy to &quot;drag&quot; some through the fluid - they don't have to knock other molecules out of the way, because by the time they get there, they have moved on their own. The viscosity is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for simple molecules, the line between &quot;can't move&quot; and &quot;moves easily&quot; is pretty distinct. Once the molecules are mobile enough that excluded volume interactions are overcome, the material melts, and further changes to viscosity are small and mostly due to intermolecular attractive forces. For example, here is a plot of the &lt;a href=&quot;http://www.engineeringtoolbox.com/water-dynamic-kinematic-viscosity-d_596.html&quot; rel=&quot;noreferrer&quot;&gt;viscosity of water vs. temperature&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tsirf.png&quot; alt=&quot;Viscosity of water as a function of temperature.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, it decreases smoothly as temperature increases. Below the freezing point, the viscosity is harder to define (although large masses of &lt;a href=&quot;http://www.iceandclimate.nbi.ku.dk/research/flowofice/modelling_ice_flow/&quot; rel=&quot;noreferrer&quot;&gt;ice do flow&lt;/a&gt;), but the curve is discontinuous and it &quot;jumps up&quot; to some higher value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about complex molecules? Let's link the monomers together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3BMyE.png&quot; alt=&quot;It&amp;#39;s harder for dimers to move.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it becomes much harder for the molecules to move. At a given temperature, you have to wait about the same amount of time for a single monomer to move, but it takes much longer for a pair to move together. Going back to the crowded room analogy - imagine trying to move through a crowd in which everybody - including you - was handcuffed in pairs to someone else. It would be a lot more difficult! In general, it turns out that the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stress_relaxation&quot; rel=&quot;noreferrer&quot;&gt;relaxation time&lt;/a&gt;&lt;/strong&gt; for linear polymers increases as (approximately) the square of the chain length (for short lengths below the entanglement length). This means that if you double the number of monomers in a polymer chain, the average time it takes for them to move out of the way goes up 4 times. This is true up to a certain length of chain called the &lt;strong&gt;entanglement length&lt;/strong&gt; (which depends on the polymer), at which point the likelihood of coordinated movement of the whole chain becomes so low that each polymer is sort of &quot;trapped&quot; in a tube made of other polymers. The analogy often used to explain this is a nest of snakes - they have to slither around each other. As a result, this type of motion was named &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reptation&quot; rel=&quot;noreferrer&quot;&gt;reptation&lt;/a&gt;&lt;/strong&gt; - after reptiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing is that for shorter lengths (below the entanglement length) each length scale of the chain moves essentially independently of the rest. This means that in a longer chain, on short time scales the shorter lengths act as though they were in a low-viscosity fluid - they move easily and motion is not correlated with their neighbors. For longer length scales, the chain is essentially &quot;frozen&quot; - it can't move until its neighbors move - and so it acts more like a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjc6s.png&quot; alt=&quot;Chain motion is much slower&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that short lengths of chain can &quot;wiggle&quot; on short time scales, but longer lengths move more slowly. This gives us a &lt;strong&gt;relaxation time spectrum&lt;/strong&gt; - it means that if you apply very fast changes (high shear rates) the molecules don't have time to get out of the way, and so the viscosity is high. If you increase the temperature, things move faster, and so at the same shear rate, the viscosity will be lower (more of the chain can move out of the way). The same is true at lower shear rates for the same temperature - the shorter length scales can move out of the way, and so it becomes more liquid - the viscosity decreases. On the longest time scales (lowest shear rates, coldest temperatures) the entire length of the chain can move, and so you see the lowest viscosity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Shear_thickening&quot; rel=&quot;noreferrer&quot;&gt;shear-thickening&lt;/a&gt;, and is commonly seen in linear polymer melts or concentrated solutions. It is an example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscoelasticity&quot; rel=&quot;noreferrer&quot;&gt;viscoelastic&lt;/a&gt; behavior, which means that depending on the conditions (temperature or time scale of interaction) the material will exhibit a range of liquid-like (viscous) or solid-like (elastic) behavior. An example commonly used to demonstrate this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Silly_Putty&quot; rel=&quot;noreferrer&quot;&gt;silly putty&lt;/a&gt; (although silly putty is not a pure linear polymer.) On short time scales, the polymers don't have time to relax completely and so it behaves like an elastic solid. This is why you can bounce a ball of silly putty on a table. At longer time scales, the polymers can fully relax, and so the silly putty flows. This is why if you let a lump of it sit out for a while, it will form a puddle. Silly putty is unique in that the relaxation time spectrum ranges from elastic to viscous over time scales that are commonly observed at room temperature, so you can see these effects in high contrast very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other types of non-Newtonian behavior (which just means viscosity is not constant), but they are all fundamentally connected to the &lt;strong&gt;relaxation time spectrum&lt;/strong&gt; of the fluid, which is directly related to the molecular structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Water has a narrow viscosity spectrum (as a function of temperature) because it has a narrow stress relaxation time spectrum.&lt;/li&gt;&#xA;&lt;li&gt;It has a narrow stress relaxation time spectrum because it is a relatively simple, small molecule.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-15T23:59:34.700" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14371" PostTypeId="1" AcceptedAnswerId="14380" CreationDate="2014-07-15T21:22:11.360" Score="6" ViewCount="390" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The density at &lt;span class=&quot;math-container&quot;&gt;$20^\circ\mathrm{C}$&lt;/span&gt; of a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.500 M}$&lt;/span&gt; solution of acetic acid in water is &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0042 g/mL}$&lt;/span&gt;. What is the molality of the solution? The molar mass of acetic acid is &lt;span class=&quot;math-container&quot;&gt;$M(\ce{CH3CO2H})=\pu{60.05 g/mol}$&lt;/span&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;The density at &lt;span class=&quot;math-container&quot;&gt;$20^\circ\mathrm{C}$&lt;/span&gt; of a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.258 M}$&lt;/span&gt; solution of glucose in water is&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0173 g/mL}$&lt;/span&gt;, and the molar mass of glucose is &lt;span class=&quot;math-container&quot;&gt;$M(\ce{Glc}) = \pu{180.2 g/mol}$&lt;/span&gt;. What is the molarity of the solution?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In question one we can assume a 1 liter solution and in question two we can assume a 100 gram sample. I understand these assumptions are needed to find other information, but when are we justified in making these assumptions?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="83758" LastEditDate="2020-03-17T21:05:57.213" LastActivityDate="2020-03-17T21:05:57.213" Title="When are we justified in making assumptions in questions" Tags="&lt;solutions&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14372" PostTypeId="2" ParentId="10045" CreationDate="2014-07-15T21:56:14.590" Score="2" Body="&lt;p&gt;Sort of 'yes', sort of 'no'.  Consider the following points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Activity coefficients are notorious for changing with $\mathrm{p}\ce{H}$, ionic strength, and concentration (among other things).  You might measure the the activity coefficient of $\ce{Cu^{2+}}$ in a $0.7~\mathrm{M}$ $\ce{CuSO4}$ solution and get one value, but try to measure the same thing in a low $\mathrm{p}\ce{H}$ and $1~\mathrm{\mu M}$ $\ce{CuCl2}$ solution and you may get a completely different value.  In practice, people usually end up measuring activity coefficients across a range of conditions that are relevant to their specific use-case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Standard Hydrogen Electrode is a theoretical device; and actually constructing something similar is difficult.  In practice, you would just want to use a commercially available double-junction silver-silver chloride or saturated calomel electrode. The potentials of those two are well-known and fairly stable. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See Chapter 2 of &quot;Electrochemical Methods&quot; 2nd Ed, by Bard and Faulkner for more information.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditDate="2014-07-21T05:46:49.620" LastActivityDate="2014-07-21T05:46:49.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14373" PostTypeId="1" AcceptedAnswerId="14377" CreationDate="2014-07-15T22:30:42.017" Score="13" ViewCount="135100" Body="&lt;p&gt;I know water boils at below room temperature in a vacuum, and want to know if there's an equation to calculate the melting or boiling points of elements or compounds at a given pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go at attempting this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason I ask is because I'd like to know if Gallium, a metal which melts at $29.77^\circ C$ at atmospheric pressure would be in a liquid state at room temperature if stored under a vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5772" LastActivityDate="2016-03-28T05:39:17.297" Title="How to calculate melting/boiling points at different pressures" Tags="&lt;melting-point&gt;&lt;vapor-pressure&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14374" PostTypeId="2" ParentId="14373" CreationDate="2014-07-15T23:00:13.860" Score="2" Body="&lt;p&gt;There is no general equation for melting points and boiling points that applies to all elements.  However, the melting point for gallium has been studied in considerable detail because gallium's melting point has been adopted as a calibration standard.  Isotech, a calibration company posted the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;2.3 The temperature of the melting point equilibrium of pure gallium depends upon the ambient pressure of the surrounding environment, and varies with atmospheric pressure expressed upon the surface of the gallium column as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t C = 29.7646 +/- (-0.002 C per std. atmos.) Eq. 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The negative sign of the coefficient is a result of the solidus expansion (liquidus contraction).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.isotech.co.uk/files/document_library_file-31.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to the above formula (see page 3).  Perhaps you could contact Isotech to get further clarification on their formula.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-15T23:00:13.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14375" PostTypeId="1" AcceptedAnswerId="14378" CreationDate="2014-07-15T23:45:56.320" Score="2" ViewCount="62" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Also good against fungi, amoeba, and enveloped viruses,[7] quats are believed to act by disrupting the cell membrane. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quaternary ammonium cleaners have long-alkyl substituent chains. Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have anything do with increasing the ability of the cleaners to penetrate the phospholipid bilayer, which has its non-polar portion oriented outward and its polar phosphate backbone oriented inside the cell? Does the alkyl portion dissolve into the lipid bilayer and wreck havoc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the exact mechanism? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-16T01:40:12.177" Title="Quaternary ammonium anti-microbials" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14376" PostTypeId="1" CreationDate="2014-07-15T23:51:27.647" Score="2" ViewCount="8644" Body="&lt;p&gt;Does thermal decomposition of metal oxides always produce oxygen?&lt;/p&gt;&#xA;" OwnerUserId="7141" LastEditorUserId="11367" LastEditDate="2017-04-08T16:24:43.970" LastActivityDate="2017-04-08T16:34:55.227" Title="Oxygen production from thermal decomposition of metal oxides" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14377" PostTypeId="2" ParentId="14373" CreationDate="2014-07-16T00:55:46.010" Score="13" Body="&lt;p&gt;I don't believe there is an equation that you can use for &lt;strong&gt;melting points&lt;/strong&gt; of a general substance as a function of pressure (since the melting phase transition has a lot to do with the geometry of the molecule and the structure of the solid), but there is one for the &lt;strong&gt;boiling point&lt;/strong&gt; of any pure substance when you are not near the critical point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liquid-vapor transition follows a very specific curve on the pressure-temperature plane of a PVT diagram. This is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation&quot; rel=&quot;noreferrer&quot;&gt;Clausius-Clapeyron&lt;/a&gt; relation, which at temperatures and pressures that are not close to the critical point, can be approximated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ln\frac{P_2}{P_1} = \frac{-\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the two-point form, which lets you predict the pressure of the vapor phase at a given temperature if you know the temperature and pressure at another point, and the enthalpy of vaporization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this to predict boiling points as a function of pressure by recognizing that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The boiling point is the temperature at which the vapor pressure at equilibrium exceeds atmospheric pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that if you know the boiling point of a substance at 1 atm (the normal boiling point), and you know $\Delta H_{vap}$ for that substance, you can predict the boiling point at another pressure by plugging in values for $P_1$ and $T_1$ (the normal boiling point), $P_2$ (the pressure you are interested in), and $\Delta H_{vap}$, the enthalpy of vaporization (also sometimes called the &lt;strong&gt;latent heat&lt;/strong&gt;), then solving for $T_2$ (the new boiling point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said, this only works when you aren't near the critical point, but for most applications that's not a concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific case, especially since you are probably more interested in the liquid-solid transition at this temperature, you are better off looking at a phase diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one for gallium (&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Phase_diagram_of_gallium_(1975).png&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;) (&lt;a href=&quot;http://www.osti.gov/scitech/servlets/purl/4010212&quot; rel=&quot;noreferrer&quot;&gt;original source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1oVI7.png&quot; alt=&quot;Phase diagram of gallium&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid-liquid line has a negative slope at low pressure (as ron pointed out) which means the melting temperature &lt;em&gt;increases&lt;/em&gt; as pressure decreases. Depending on how high the vacuum is, and on what temperature &quot;room temperature&quot; is, you might be in the solid phase - but the slope is steep, so even at 0 atm the melting point is still right around 300 K.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-16T13:13:28.843" LastActivityDate="2014-07-16T13:13:28.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14378" PostTypeId="2" ParentId="14375" CreationDate="2014-07-16T01:40:12.177" Score="3" Body="&lt;p&gt;A phospholipid has a charged, hydrophilic phosphate head and a neutral, hydrocarbon hydrophobic tail.  In a phospholipid bilayer, two phospholipids combine tail-to-tail such that the two outer, or exposed, surfaces are comprised of the polar phosphate heads.  The bilayer can function as a cell membrane; it can keep what is in &quot;in&quot; and what is out &quot;out&quot;.  It can also regulate critical cell functions such as pH and ion balance through the incorporation of proteins in the bilayer that can serve as ion pumps.  Clearly, any disruption to the bilayer could allow the cell contents to leak out and bad stuff to leak in, which would lead to either cell inactivation or cell death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of action of quaternary salts on lipid bilayers is still an active area of investigation, but here is a simplified overview of what is thought to be happening. .  The first step is thought to involve the electrostatic interaction of the positively charged ammonium ion with the exposed, negatively charged bilayer surface, followed by incorporation of the hydrocarbon chain of the ammonium ion into the bilayer. The electrostatic interaction affects membrane surface pressure which, in turn, disrupts cell function.  Shorter chain lengths (n=12-14) are more effective against gram-positive bacteria, while longer  chain lengths (n=14-16) are more effective against gram-negative bacteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://etd.uovs.ac.za/ETD-db/theses/available/etd-08152012-120125/unrestricted/JansenAC.pdf&quot; rel=&quot;nofollow&quot;&gt;Here's a reference to a recent thesis&lt;/a&gt; (2012) that studies the resistance of bacteria to quaternary ammonium compounds.  Section 1.2 (pages 2-5) provides a clear, concise summary of the latest thinking on the subject.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-16T01:40:12.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14380" PostTypeId="2" ParentId="14371" CreationDate="2014-07-16T04:54:41.453" Score="9" Body="&lt;p&gt;You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous solution, the solute particles are distributed uniformly among the particles of solvent, and the two form a constant ratio. If I split the sample in half, the solute particles are halved, but so are the solvent particles, so the &lt;em&gt;ratio&lt;/em&gt; remains the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, there's no need to assume any particular quantities to solve this problem; instead, you can use algebraic substitutions using the known values given and common definitions to solve for either molarity or molality directly. Let $M = \frac{m_{solute}}{n_{solute}}$ be the molar mass of the solute, $\rho = \frac{m_{solution}}{V_{solution}}$ be the mass density of the solution, $c = \frac{n_{solute}}{V_{solution}}$ be the molarity, and $b = \frac{n_{solute}}{m_{solvent}}$ be the molality, and note that $m_{solvent} = m_{solution} - m_{solute}$. Then, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;b\ &amp;amp; = \frac{n_{solute}}{m_{solution} - m_{solute}}\\&#xA; &amp;amp; = \frac{n_{solute}}{\rho \cdot V_{solution} - M \cdot n_{solute}}\\&#xA; &amp;amp; = \frac{n_{solute}}{\rho \cdot \frac{n_{solute}}{c} - M \cdot n_{solute}}\\&#xA; &amp;amp; = \frac{c}{\rho - M \cdot c}\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can rearrange this equation to solve for $c$ (molarity). Note, of course, that units need to match, but it's trivial to convert between grams and kilograms, mililiters and liters, etc.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-16T05:12:07.870" LastActivityDate="2014-07-16T05:12:07.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14383" PostTypeId="1" AcceptedAnswerId="14401" CreationDate="2014-07-16T12:49:45.703" Score="2" ViewCount="2259" Body="&lt;p&gt;For instance, I have a chemical reaction like :&#xA;$$\ce{Fe^{2+} + $0.5$O2 + H^{+} -&amp;gt;Fe^{3+} + $0.5$ H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I would like to use Action-Mass law, but I need to have equilibrium constant for this reaction. And this for many other reaction.&#xA;That's why I'm wondering if there exist a nice database with this information ?&lt;/p&gt;&#xA;" OwnerUserId="7146" LastEditorUserId="275" LastEditDate="2014-07-17T10:19:58.130" LastActivityDate="2018-12-05T02:06:44.673" Title="Is there a database which give reaction constant?" Tags="&lt;equilibrium&gt;&lt;redox&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14384" PostTypeId="1" AcceptedAnswerId="14387" CreationDate="2014-07-16T14:23:08.697" Score="4" ViewCount="1697" Body="&lt;p&gt;If I dissolve $\ce{K_2HPO_4}$ and $\ce{CaHPO_4}$ in water (two different reactions), the latter shows better solubility. Is it because $\ce{Ca^{2+}}$ has ha higher valence of $2$ than $\ce{K^+}$, which has only $1$? Or has it something to do with the lattice enthalpy?&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="4945" LastEditDate="2014-07-17T07:10:47.497" LastActivityDate="2014-07-17T07:10:47.497" Title="Which is more soluble in water and why: K₂HPO₄ or CaHPO₄?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14385" PostTypeId="1" CreationDate="2014-07-16T14:47:31.630" Score="3" ViewCount="2489" Body="&lt;p&gt;What is the Schrödinger wave equation?&lt;br&gt;&#xA;What are &lt;strong&gt;radial probability functions&lt;/strong&gt;, nodes-&lt;strong&gt;angular and radial&lt;/strong&gt; and &lt;strong&gt;electron cloud&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really confused with all these terms and all my book seems to offer me is a set of very complex words and definitions. I am relatively new to atomic structure and related things, so a brief answer in simple words would really be helpful! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also would like to know more about &lt;strong&gt;Hamiltonian&lt;/strong&gt; and &lt;strong&gt;Laplacian&lt;/strong&gt; operators and related subjects.&lt;/p&gt;&#xA;" OwnerUserId="7148" LastEditorUserId="4945" LastEditDate="2014-07-17T08:26:57.763" LastActivityDate="2014-09-08T12:57:52.800" Title="Schrödinger equation and wave function" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14386" PostTypeId="2" ParentId="14111" CreationDate="2014-07-16T15:24:21.867" Score="1" Body="&lt;p&gt;I donn't know what a &quot;detailed canonical answer&quot; is, but intutitively, I can appreciate that it is easier to take an electron away from (i.e. oxidize) something with a negative charge than something that is electrically neutral. That is because it costs energy to separate charged particles and it produces energy to reunite them. So many examples - a thiol is stable to air in the absence of a base, but in the presence of base, air will oxidize it to a disulfide. Same with phenol - stable in air but with base present, you soon get colored condensation products.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-07-16T15:24:21.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14387" PostTypeId="2" ParentId="14384" CreationDate="2014-07-16T15:34:40.670" Score="5" Body="&lt;p&gt;The charge and size of an ion influences both lattice stability and stability of the hydrated ion.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The less stable the lattice the more soluble the ion (less energy required to disrupt the lattice).  The more highly charged and smaller the ion is, the larger are the electrostatic repulsions within the lattice which will act to destabilize the lattice.&lt;/li&gt;&#xA;&lt;li&gt;The more stable the solvated ion, the more soluble the ion.  The higher the charge on the ion and the smaller its radius, the more stable the solvated ion.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The size of the potassium ion is around 152 pm, the size of the calcium ion is around 114 pm.  So the calcium ion is both the smaller and the more highly charged of the two.  Therefor we would expect the calcium ion to have the less stable lattice and be more highly solvated (stabilized) when in solution.  Both of these factors would lead us to expect $\ce{CaHPO4}$ to be more soluble than $\ce{K2HPO4}$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-16T15:34:40.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14390" PostTypeId="1" CreationDate="2014-07-16T17:46:24.097" Score="17" ViewCount="19126" Body="&lt;p&gt;I guess the ester is a weaker nucleophile because it does have an additional oxygen atom, unlike the ketone, that is pulling electrons from the C-O double bond towards the carbon atom (this happens vice versa too of course with the ester oxygen). This lowers the ionic character of the C-O double bond (smaller $\delta^{+}$ and $\delta^{-}$ charge on the carbon and oxygen atoms) and thus increases the energy of the C-O $\pi^{*}$ bond, making it less nucleophilic.   Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-07-17T10:35:16.930" LastActivityDate="2017-08-23T14:24:03.700" Title="Why is a ketone more nucleophilic than an ester?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14391" PostTypeId="2" ParentId="14390" CreationDate="2014-07-16T18:31:07.687" Score="17" Body="&lt;p&gt;The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester.  Look at the resonance structures drawn below.  Both the ketone and ester have a resonance structure that places positive charge on the carbonyl carbon, however you can also draw a third resonance structure for the ester that removes some of the positive charge from the carbonyl carbon and places it on the ester oxygen. This makes the ester carbonyl carbon less positive (more negative, more nucleophilic, less prone to attack by a nucleophile) than the ketone carbonyl carbon (more positive, less nucleophilic, more prone to attack by a nucleophile).  A nucleophile has electrons it would like to share and will react preferentially with centers of low electron density (more positively charged).  Therefor a nucleophile will react faster with a ketone carbonyl than an ester carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lRQpX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I believe Martin's approach (basically frontier MO) is fundamentally correct.  I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;acyl halide &gt; acid anhydride &gt; aldehyde &gt; ketone &gt; ester ~ carboxylic acid &gt; amide &gt; carboxylate ion&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My argument based on how resonance affects the charge density at the carbonyl carbon is one approach used to explain (&quot;teach&quot;) the above order of reactivity.  See for example page 4 on &lt;a href=&quot;http://www.chem.ucla.edu/harding/notes/notes_14D_C=Ofun01.pdf&quot; rel=&quot;noreferrer&quot;&gt;these&lt;/a&gt; UCLA notes.  I was just trying to keep it simple.&lt;/li&gt;&#xA;&lt;li&gt;The issue I have with Martin's approach is, what should the reader do next week when he wants to know if an acid chloride reacts faster or slower than an ester.  While the resonance approach may not be rigorous, it does yield the correct order of reactivity and allows the user to make predictions while only using pencil and paper.&lt;/li&gt;&#xA;&lt;li&gt;I am considering further editing of this post to use the resonance approach to explain differences in ground state stabilization of the carbonyl series.  Then, since they all go (more or less) to the same tetrahedral intermediate, one could explain the reactivity order based on differences in Eact.  This is another textbook method used to explain the carbonyl reactivity pattern.&lt;/li&gt;&#xA;&lt;li&gt;Finally, what is the &quot;correct&quot; answer?  Is it the real truth as nature knows it, but in a form that is difficult for the common individual to use and extend; or is it a rough approximation of the truth that can be easily used?  I think the answer to that question depends upon who is asking the question, who is the audience. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-18T02:35:09.447" LastActivityDate="2014-07-18T02:35:09.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14392" PostTypeId="1" AcceptedAnswerId="14492" CreationDate="2014-07-16T18:57:56.987" Score="3" ViewCount="2939" Body="&lt;p&gt;I am working with a company to help improve their reaction process where they use SDA 2B Alcohol (Karl Fischer titration has it at 99.66% EtOH, 0.3375% toluene). Specifically, I am attempting to determine what theoretical %wt or %vol of water is pulled from atmospheric air into pure ethanol given an infinite amount of time? Does it ever reach an equilibrium state and, if so, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut guess is ~95.5%wt EtOH/4.5%wt H2O, the azeotropic concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume constant relative humidity and room temperature. The ethanol is stored in a closed tote (~220 gallons). Initially, the tote is full of as-shipped ethanol (negligible air volume). However, half the ethanol is removed as part of a batch run. So, really, what I'm trying to get at is what happens to the leftover ethanol that is now sitting in a closed tote with ambient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if additional details will help answer the question. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7150" LastEditorUserId="7150" LastEditDate="2014-07-17T00:19:36.853" LastActivityDate="2014-07-26T04:57:01.197" Title="Ethanol and water hygroscopic equilibrium concentration" Tags="&lt;equilibrium&gt;&lt;water&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14393" PostTypeId="2" ParentId="14352" CreationDate="2014-07-16T20:01:03.897" Score="2" Body="&lt;p&gt;Are the use of MATLAB (or OCTAVE) and a self-designed molecular representation mandatory for you project? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Established 2D- and 3D representations for molecules beyond a connectivity matrix, i.e. with coordinates, (partial) charges, annotation of isotopes, etc. exist, such as in the form of the &lt;a href=&quot;http://www.xml-cml.org/&quot; rel=&quot;nofollow&quot;&gt;Chemical Markup Language (CML)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, well-documented toolkits, frameworks and modules are available to operate on these data are available for C++, Java, Perl, Python, and Ruby.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-07-16T20:01:03.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14394" PostTypeId="2" ParentId="459" CreationDate="2014-07-17T00:16:45.903" Score="2" Body="&lt;p&gt;Graphite is a semimetal, and no more mambojumbo than with metals, but it doesn't do anything with the problem. &#xA;Silicon/silica can be melted, purified, doped, can single crystals grown, can oxidize the surface, making the silicon semiconductor an silica insulator. By &quot;can&quot; I mean that engineeers know well tested methods, optimized for half a century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite cannot be reasonably melted, cannot grown to single crystals, cannot doped with similar techniques, highly anisotropic, so cutting maybe an issue, and when you make thin layers, it is hard to oxidize in a controlled way to make an insulator out of it. The resulted oxides are semiconductors, and there is a large variety of local structures that can appear with different electronic properties. By cannot I mean that no one really though about these problems until the last decade, so there might be equivalent solutions with e.g graphene, nanotube or other nanocarbon structures. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-17T00:16:45.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14395" PostTypeId="1" AcceptedAnswerId="14396" CreationDate="2014-07-17T02:25:42.813" Score="3" ViewCount="229" Body="&lt;p&gt;Let's say I want to find the oxidation numbers of the atoms in a chemical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: In &lt;span class=&quot;math-container&quot;&gt;$\ce{H3AsO3}$&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen has a &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt; charge. They're &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of them, so its &lt;span class=&quot;math-container&quot;&gt;$-6$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen has a &lt;span class=&quot;math-container&quot;&gt;$+1$&lt;/span&gt; charge. They're &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; of them, so its &lt;span class=&quot;math-container&quot;&gt;$+3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So arsenic must have a charge of &lt;span class=&quot;math-container&quot;&gt;$+3$&lt;/span&gt; for all the charges to add up to be &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;? Correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my professor keeps telling me to keep the positive oxidation numbers away from each other. I thought it only applied when drawing Lewis structures or resonance structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that a rule when writing a formula? &lt;/p&gt;&#xA;" OwnerUserId="6823" LastEditorUserId="83758" LastEditDate="2020-03-27T07:50:10.983" LastActivityDate="2020-03-27T07:50:10.983" Title="Can two positive ions be next to each other in a formula?" Tags="&lt;oxidation-state&gt;&lt;chemical-formula&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14396" PostTypeId="2" ParentId="14395" CreationDate="2014-07-17T02:46:12.613" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can two positive ions be next to each other in a formula?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there are many examples of this, $\ce{HNO3}$ for instance&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So Arsenic must have a charge of 3+ for all the charges to add up to&#xA;  be 0? Correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that's correct&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, my professor keeps telling me to keep the positive oxidation&#xA;  numbers away from each other. Is that a rule when writing a formula?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not a rule that I'm familiar with.  Here is a &lt;a href=&quot;https://scilearn.sydney.edu.au/fychemistry/iChem/inorganic_nomenclature.shtml&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to a nice, concise listing of the rules for writing formulas for inorganic compounds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought it only applied when drawing Lewis structures or resonance&#xA;  structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it certainly applies in those two cases&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-17T02:46:12.613" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14397" PostTypeId="1" AcceptedAnswerId="14403" CreationDate="2014-07-17T07:36:39.703" Score="7" ViewCount="5031" Body="&lt;p&gt;What is the maximum possible density of water at room temperature?&#xA;The density of the water in the Dead Sea at 37°C is 1.24. Would it be possible to increase the density using salts we add to our food, or possibly by using different salts than are present in the sea?&lt;/p&gt;&#xA;" OwnerUserId="7063" LastEditorUserId="42741" LastEditDate="2017-06-29T06:31:42.840" LastActivityDate="2017-06-29T06:31:42.840" Title="How can the density of water be maximised?" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;density&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14398" PostTypeId="2" ParentId="14390" CreationDate="2014-07-17T07:55:55.300" Score="15" Body="&lt;p&gt;Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case and the explanation lies within the molecular orbitals. The LUMO of the ketone has a slightly lower ($-1.57~\mathrm{eV}$) energy than that of the ester ($-0.87~\mathrm{eV}$). More important, the coefficient in this orbital is larger at the carbonyl carbon in the ketone compared to the ester. And this orbital will be attacked by the nucleophile, which usually has a high lying HOMO. So the LUMO of the ketone will be more accessible compared to the ester (smaller energy difference). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important fact is, that the ester is much more rigid than the ketone as it has one low lying MO that spans the $\ce{O=C-O}$ moiety (MO20), which is obviously not present (as it cannot be) in the ketone (MO18). You can also see, that the $\pi$ contribution of carbon in the ketone is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2n7RP.jpg&quot; alt=&quot;orbitals&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-17T07:55:55.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14399" PostTypeId="2" ParentId="14111" CreationDate="2014-07-17T08:22:14.787" Score="4" Body="&lt;p&gt;The effect that can be observed by adding acids to a solution to increase conductivity is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grotthuss_mechanism&quot; rel=&quot;nofollow&quot;&gt;Grotthuss mechanism&lt;/a&gt;. Protons have a much higher mobility in solution ($3.62\cdot10^{-3}~\mathrm{cm^2~V^{-1}~s^{-1}}$) than any other cation, basically allowing it to tunnel through a solution. (A similar effect can be noted for hydroxyl anions.)&#xA;Basically this allows a much faster charge exchange through hydrogen bond networks, also resulting in a higher cell voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But then again I am not really sure I understood your question correctly.)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-17T08:22:14.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14401" PostTypeId="2" ParentId="14383" CreationDate="2014-07-17T10:48:55.413" Score="3" Body="&lt;p&gt;The information you are looking for is out there, but sometimes you might have to work a little for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example reaction you post is a redox reaction. If we split the redox reaction into its two half-reactions, we can look up their reduction potentials (easier to find sometimes than equilibrium constants) and use the Nernst equation to calculate the equilibrium constant. I use &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_standard_electrode_potentials&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for my list of standard potentials, but only because the values are identical to those in textbooks that I own. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxidation half-reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{2+} -&amp;gt; Fe^{3+} + e-} \ \ \ E^\circ_{ox}= -0.77 \ \text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reduction half-reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 4H+ + 4e- -&amp;gt; 2H2O} \ \ \ E^\circ_{red} = +1.229 \ \text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Full balanced redox reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Fe^{2+} + O2 +4H+ -&amp;gt; 4Fe^{3+} + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cell potential&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_{cell}=E^\circ_{ox} + E^\circ_{red} = -0.77 \ \text{V} + 1.229 \ \text{V} = 0.459 \ \text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow&quot;&gt;Nernst Equation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ_{cell}=\frac{RT}{zF}\ln{K_{eq}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_{eq}=e^{\dfrac{E^\circ_{cell}zF}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation, $z$ is the number of electrons transferred, $T$ is the temperature in kelvins, $R$ is the ideal gas constant, and $F$ is the Faraday constant. At 0 Celsius (273.15 K):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq}=e^{\dfrac{0.459 \ \text{V} \cdot 4\text{ mol}\cdot 96.49\times 10^3 \frac{\text{C}}{\text{mol}}}{ 8.314\ \frac{\text{J}}{\text{K}\cdot \text{mol}}\cdot 273.15\ \text{K} }}=7.565\times 10^{33}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can look up the enthalpy change or free energy change for the reaction (many of which are available at &lt;a href=&quot;http://webbook.nist.gov/chemistry&quot; rel=&quot;nofollow&quot;&gt;webbook.nist.gov/chemistry&lt;/a&gt;, then you can use a variant of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Development_from_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;van 't Hoff equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_r G^\circ = -RT\ln{K_{eq}}$$&#xA;$$K_{eq}=e^{\dfrac{-\Delta_R G^\circ}{RT}}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-17T10:48:55.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14402" PostTypeId="2" ParentId="14397" CreationDate="2014-07-17T13:45:18.780" Score="3" Body="&lt;p&gt;I think they use epsom salt (sodium sulfate) in the commercial isolation tanks designed to allow a person to float in water (i.e. salt solution) and meditate without having any tactile input (a phenomenon of the 70s, but I think there still are some around).&lt;br&gt;&#xA;The more salt added to the water, the higher the density. Also, some salts are denser than others - so a heavy metal salt like cesium chloride would give a heavier water solution than say sodium chloride.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="4945" LastEditDate="2014-07-18T02:23:14.697" LastActivityDate="2014-07-18T02:23:14.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14403" PostTypeId="2" ParentId="14397" CreationDate="2014-07-17T14:51:22.470" Score="11" Body="&lt;p&gt;Clerici solution is a solution of equal parts of thallium formate and thallium malonate in water, density is 4.25 g/ml,     Thoulets solution is solution of potassium-mercury iodide in water with density 3.2 g/ml. &lt;strong&gt;Don't even think of using them.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium polytungstate $\ce{Na6[H2W12O40]}$ is a relatively safe compound and allows densities up to 3.1 g/ml. $\ce{ZnCl2}$ allows densities up to ~ 1.9 g/ml, but has acidic solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, common nontoxic salts usually either are not dense enough or not soluble enough to make really heavy liquids, but densities around 1.1-1.2 g/ml are quite easy to achieve even with common salt. Note, however, that concentrated salt solutions 'siphons' water from human body through damaged skin and mucous membranes, and to some extent even through undamaged skin, which may lead to desiccation, trauma of mucous membranes and very unpleasant sensations.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-17T14:51:22.470" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14404" PostTypeId="1" CreationDate="2014-07-17T15:34:54.780" Score="2" ViewCount="109" Body="&lt;p&gt;What is a molecular system in general terms?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the solution of sugar in water be called a molecular system, if we were not interested in sugar-water interaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today Jean Marie Lehn told me that bulk water is classical supramolecular system, but I cannot understand why.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="73" LastEditDate="2014-09-04T17:14:53.650" LastActivityDate="2014-09-04T17:16:34.117" Title="What is molecular system in general terms? What is a supramolecular system?" Tags="&lt;molecules&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14407" PostTypeId="1" CreationDate="2014-07-17T20:32:11.890" Score="4" ViewCount="992" Body="&lt;p&gt;I read detailed explanations of what oxidation numbers/ states represent and it boils down simply to this: If it's positive, it shows how many electrons the element lost, and in the case of a negative oxidation number, it shows how many electrons the element gained. That I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is, in Phosphorus pentachloride $\ce{PCl5}$, phosphorus forms 5 single bonds with chlorine atoms and 'gains' $5$ more electrons to accommodate a total of $10$ electrons. Shouldn't its oxidation number be $-5$ instead of $+5$?&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="4945" LastEditDate="2014-07-18T02:30:14.550" LastActivityDate="2014-07-18T02:30:14.550" Title="Oxidation numbers of phosphorus pentachloride" Tags="&lt;inorganic-chemistry&gt;&lt;molecules&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14408" PostTypeId="2" ParentId="14407" CreationDate="2014-07-17T21:01:25.960" Score="3" Body="&lt;p&gt;When considering the oxidation state of covalent compounds, we need to pretend they are ionic. When we do so, we &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_state#Using_electronegativity&quot; rel=&quot;nofollow&quot;&gt;assign all of the electrons&lt;/a&gt; to the &lt;strong&gt;more electronegative&lt;/strong&gt; element. The electronegativity of phosphorous using &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Pauling_electronegativity&quot; rel=&quot;nofollow&quot;&gt;the Pauling Scale&lt;/a&gt; is 2.19 and the electronegativity of chlorine is 3.16. Since chlorine is the more electronegative element, we assign all electrons in the bonds to the chlorine atoms. Thus $\ce{PCl5}$ is treated as the fictitious ionic compound comprised of $\ce{P^{5+}}$ ions and $\ce{Cl-}$ ions.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-17T21:01:25.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14409" PostTypeId="1" AcceptedAnswerId="14410" CreationDate="2014-07-17T21:09:32.823" Score="1" ViewCount="2441" Body="&lt;p&gt;My solution manual implied that we can have a racemic mixture of a meso compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to rotate the molecule to make it symmetrical. Sure, the bromination in the nucleophilic solvent gives you an anti product, but from my understanding, if you can rotate the molecule so that it is symmetrical, the molecule can be meso. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking no because meso compounds by definition have no enantniomers and a racemic mixture has both enantiomers. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-18T01:27:25.393" LastActivityDate="2014-07-18T01:27:25.393" Title="Racemic mixture of meso compound" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14410" PostTypeId="2" ParentId="14409" CreationDate="2014-07-17T21:24:49.713" Score="2" Body="&lt;p&gt;You are correct and you enumerated the reasons why.  A racemic mixture is a mixture of &lt;em&gt;d&lt;/em&gt; and &lt;em&gt;l&lt;/em&gt; enantiomers in equal proportion.  A meso compound has no &lt;em&gt;d&lt;/em&gt; or &lt;em&gt;l&lt;/em&gt; isomers, it is achiral.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-17T21:24:49.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14411" PostTypeId="2" ParentId="13519" CreationDate="2014-07-17T23:55:46.573" Score="0" Body="&lt;p&gt;Here this might help&#xA;&lt;a href=&quot;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p026.shtml#materials&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p026.shtml#materials&lt;/a&gt;&#xA;But you might want to use different electrodes.&lt;/p&gt;&#xA;" OwnerUserId="7141" LastActivityDate="2014-07-17T23:55:46.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14412" PostTypeId="1" CreationDate="2014-07-17T23:59:57.340" Score="4" ViewCount="391" Body="&lt;p&gt;I am doing a science fair experiment about Microbial Fuel Cells and was wondering how exactly do they work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the basics, that they work similarly to a normal Fuel Cell through oxidation and reduction but am still a little unclear on the full details. &lt;/p&gt;&#xA;" OwnerUserId="7141" LastActivityDate="2014-10-03T18:34:47.533" Title="How does a Microbial Fuel Cell work?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14413" PostTypeId="1" CreationDate="2014-07-18T02:09:40.830" Score="8" ViewCount="1131" Body="&lt;p&gt;On the potential energy surface, if you find a local maximum and calculate its frequencies in Gaussian or something like that, will you get all negative frequencies or all positive frequencies?&#xA;I know that at a local minimum the frequencies are all positive and that at a saddle point there are both negative and positive frequencies, but what about a maximum point?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastActivityDate="2014-07-18T03:28:42.973" Title="how are the frequencies at a local maximum of PES like?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14414" PostTypeId="2" ParentId="14413" CreationDate="2014-07-18T03:11:43.977" Score="8" Body="&lt;p&gt;The potential energy surface (PES) is a very complex thing and it is hard to be understood (if we can understand it at all). With quantum chemical approaches we can map out this surface. We differentiate between positive, real and imaginary (usually shown as negative) frequencies. There are no negative frequencies. What you may encounter analyzing the PSE, are one of the following extreme points.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Local/Global Minimum&lt;/strong&gt;&lt;br&gt;&#xA;All derivatives have to be positive, the number of imaginary frequencies (NImag) is zero $i=0$. This corresponds to an observable state of the system/ molecule, i.e. it can be trapped and analysed with spectroscopy. Exciting any of these frequencies, the energy of the molecule will increase. But you already knew that.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Transition State&lt;/strong&gt;&lt;br&gt;&#xA;A transition state is a first order saddle point on the PES. The Nimag is exactly one, $i=1$. It connects two local minima with each other. Try to imagine a molecule, like ammonia. This can undergo inversion, i.e. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;noreferrer&quot;&gt;nitrogen inversion&lt;/a&gt;. The two pyramidal forms are local minima, which are connected via a planar form (nitrogen is $\ce{sp^2}$ hybridised), which is the transition state. At that point one of the frequencies is imaginary, which means, that it corresponds to lowering the energy while exciting this frequency. All other vibrations would still increase the energy of a molecule.&lt;br&gt;&#xA;You cannot observe a state like this via common spectroscopic methods, but it still has a physical meaning. There are various theories in use to describe reactions of molecules. The most popular is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;noreferrer&quot;&gt;Transition State Theory&lt;/a&gt;, which was developed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry_Eyring&quot; rel=&quot;noreferrer&quot;&gt;H. Eyring&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Michael_Polanyi&quot; rel=&quot;noreferrer&quot;&gt;M. Polanyi&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Second Order Saddle point&lt;/strong&gt;&lt;br&gt;&#xA;It formally corresponds to a transition between transition states. Its NImag is $i=2$, so there are two vibrations that lower the energy. It usually has no physical meaning, as the molecule would probably decompose before reaching this state.&lt;br&gt;&#xA;If one of these vibrations correspond to a rotation of a moiety, e.g. a methyl group, and is therefore close to zero, it corresponds to the so called rotational space and might be within range of the transition state.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;$n$-th Order Saddle point&lt;/strong&gt;&lt;br&gt;&#xA;It is like the above a transition between two $(n-1)$-th order saddle points and has even less physical meaning. You sometimes encounter these saddle points as artefacts of the used calculation model. It is strongly advised to not use these states for any explanation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Maximum&lt;/strong&gt;&lt;br&gt;&#xA;This is one cannot be found. The PES is usually like a well. The borders of this would be total fragmentation of the molecule. Imagine the homolytic dissociation of dihydrogen (a 2D PES). The maximum value of the PES corresponds to superimposed atoms and its energy value is at infinity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An additional note. A transition state always connects two local minima. And vice versa, two local minima will always be connected via a series of transition states and local minima. It does not make much sense, looking at any other points than these, as they can most likely not interpreted in a physical meaningful way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comment by Ath&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What I mean is not global maximum, but local maximum, which doesn't necessarily represent bond breaking...The local maximum I'm interested in is actually the rotation of part of a molecule. I wonder whether the frequencies are all positive or all negative at that point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no maxima on the PES, since the maximum energy corresponds to superimposed atoms. A rotation of moieties is a transition state, or a first order saddle point, and has exactly one imaginary mode.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-07-18T03:28:42.973" LastActivityDate="2014-07-18T03:28:42.973" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14415" PostTypeId="1" AcceptedAnswerId="14416" CreationDate="2014-07-18T07:24:15.693" Score="4" ViewCount="919" Body="&lt;p&gt;I have got the homework to balance the following redox reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cr(OH)3- + HO2- -&amp;gt; Cr2O7^{-2} + OH- }$(Basic medium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To balance this reaction first thing we need to do is to find the oxidation number of each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My efforts:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found out oxidation number of each element on product side.. But it is bit confusing in reactant side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any help me in finding the oxidation number of elements on reactant side ?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="51553" LastEditDate="2018-08-17T13:30:14.517" LastActivityDate="2018-08-17T13:30:14.517" Title="Finding the Oxidation number" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14416" PostTypeId="2" ParentId="14415" CreationDate="2014-07-18T08:15:00.553" Score="6" Body="&lt;p&gt;About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: &lt;a href=&quot;https://chemistry.stackexchange.com/q/10944/4945&quot;&gt;Electronegativity Considerations in Assigning Oxidation States&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen, it shall have the oxidation state $-2$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What about the other atoms?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$&lt;br&gt;&#xA;Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen gets $+1$ (see above)&lt;br&gt;&#xA; Charge has to remain, i.e. $-1$&#xA; Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the hydroxyl ion is straightforward.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Hydrogen gets $+1$ (see above)&lt;br&gt;&#xA; Oxygen gets $-2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at dichromate and apply the above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Oxygen gets $-2$, most electronegative&lt;br&gt;&#xA; Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In total that will give you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -&amp;gt; \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-18T08:15:00.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14417" PostTypeId="1" AcceptedAnswerId="19288" CreationDate="2014-07-18T09:38:44.780" Score="22" ViewCount="38828" Body="&lt;p&gt;Here are the MO schemes of $\ce{N2}$ (left) and $\ce{O2}$ (right).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSj3G.gif&quot; height=&quot;325&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9sfpZ.gif&quot; height=&quot;325&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the $\sigma$-MO formed by the $p$ AOs energetically above the $\pi$-MO for $\ce{N2}$ but not for $\ce{O2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be explained this way:&lt;br&gt;&#xA;In second period elements with more than a half filled $\ce{p}$ orbitals, the energy pattern of the MOs is regular.&lt;br&gt;&#xA;In second period elements with less than or half filled $\ce{p}$ orbitals, as the $\ce{s}$ and $\ce{p}$ atomic orbitals have similar energies, so the formed $\ce{\sigma_{g}}$ MO would have similar energy as $\ce{\sigma_{u}}$, but because of electron cloud repulsion it has a big increase in energy, hence it is above the $\pi$ orbitals? And this phenomenom is known as sp mixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I wrong somewhere? Is my reasoning corret?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:17.090" LastActivityDate="2019-02-06T15:53:23.727" Title="What is the origin of the differences between the MO schemes of O₂ and N₂?" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14421" PostTypeId="1" AcceptedAnswerId="14429" CreationDate="2014-07-18T14:30:23.810" Score="15" ViewCount="12552" Body="&lt;p&gt;Consider a neutral atom that has $n$ protons and $n$ electrons. Is it possible to remove a proton from the atom via some technique such as bombardment with another particle or applying an appropriate amount of energy to the system?&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="7475" LastEditDate="2016-10-02T22:41:14.583" LastActivityDate="2019-10-18T10:17:51.400" Title="Can a proton be ejected from an atom?" Tags="&lt;protons&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14422" PostTypeId="2" ParentId="14421" CreationDate="2014-07-18T15:04:20.683" Score="6" Body="&lt;p&gt;I don't have much knowledge of Nuclear Chemistry but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is it possible that by any method like bombarding other particle or providing high energy can we remove proton from atom ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The best example that I can think of right now is &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Alpha_decay&quot; rel=&quot;noreferrer&quot;&gt;alpha decay&lt;/a&gt;. An alpha particle consists of two protons and two neutrons bound together into aparticle identical to a helium nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also what we do in nuclear fission/fusion [atom bomb etc.]. &lt;strong&gt;But I think adding or removing just one proton might be very hard.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2014-07-18T15:04:20.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14423" PostTypeId="1" AcceptedAnswerId="14430" CreationDate="2014-07-18T15:49:06.873" Score="2" ViewCount="7022" Body="&lt;p&gt;How does the electron from the 2s orbital &quot;jump&quot; to the 2p, thus leaving 4 unpaired electrons to form four covalent bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the octet rule not apply to carbon? If it does, how can it bond with only two electrons? It would lack two more.&lt;/p&gt;&#xA;" OwnerUserId="7163" LastEditorUserId="7163" LastEditDate="2014-07-18T15:57:41.197" LastActivityDate="2014-07-18T18:48:27.417" Title="Why is carbon tetravalent?" Tags="&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14424" PostTypeId="1" AcceptedAnswerId="14426" CreationDate="2014-07-18T16:10:39.277" Score="1" ViewCount="31271" Body="&lt;p&gt;I'm watching some (really awesome) chemistry courses, and I think I have a fairly decent handle on acids and bases (or maybe not, and I'm fooling myself). But the teacher just showed a table of weak/strong acids. $\ce{HSO4-}$  was one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, why hydrogen sulfate &lt;em&gt;ion&lt;/em&gt; instead of just hydrogen sulfate? Why is $\ce{HSO4-}$ an acid and $\ce{HSO4}$ apparently isn't? This was the only ion on the table. Or does $\ce{HSO4}$ simply not exist?&lt;/p&gt;&#xA;" OwnerUserId="1974" LastEditorUserId="73" LastEditDate="2014-07-25T15:47:33.690" LastActivityDate="2014-07-25T21:22:45.387" Title="Why is HSO4- shown as an example of a weak acid instead of HSO4?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14425" PostTypeId="2" ParentId="14423" CreationDate="2014-07-18T16:28:48.167" Score="4" Body="&lt;p&gt;Look up &quot;orbital hybridization&quot;.  It says that atomic orbitals can be combined to make new orbitals for bonding purposes.  Carbon can take 1 s and 3 p orbitals to make 4 sp3 orbitals that can make 4 bonds.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-18T16:28:48.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14426" PostTypeId="2" ParentId="14424" CreationDate="2014-07-18T16:30:16.487" Score="5" Body="&lt;p&gt;The bisulfate molecule doesn't exist in solution. I can only find references to the hydrogen sulfate molecule as part of salts, which implies that hydrogen sulfate is usually found as an ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bisulfate (hydrogen sulfate) ion is indeed an acid. It is a relatively strong weak acid too, with a $\ce{K_a}$ value of $1.2 * 10^{-2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. This makes the hydrogen more partially positive and thus more reactive (more likely to be taken by a base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pSRyT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-25T21:22:45.387" LastActivityDate="2014-07-25T21:22:45.387" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14427" PostTypeId="2" ParentId="14424" CreationDate="2014-07-18T16:31:11.797" Score="4" Body="&lt;p&gt;$\ce{HSO4}$ does not exist.  There is $\ce{H2SO4}$ and $\ce{HSO4-}$  Try drawing the Lewis structure for sulfate ion.  You'll see that there are two oxygens with formal charges.  They can be left as is or they can have $\ce{H}$ bonded to them.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="73" LastEditDate="2014-07-25T15:49:13.913" LastActivityDate="2014-07-25T15:49:13.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14428" PostTypeId="2" ParentId="14085" CreationDate="2014-07-18T17:09:37.103" Score="2" Body="&lt;p&gt;This ratio has nothing mystical, and will not work in every case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's first comment on the reasons to use an internal standard. As you mention, when you inject multiple times solutions containing the same concentration of IS: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but the IS peaks remain &lt;strong&gt;more or less&lt;/strong&gt; stationary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More or less&lt;/em&gt; is a major point here: this means that repeatability of the measurement process is an issue, when multiple samples with the same concentration lead to a variation in the response. If the accuracy required for the measurement is such that these small variations might not be an issue, there is no need for an IS. In chrmatography measurement, sources of variability include repeatability in the volume injected, detector sensitivity. If there is some sample preparation steps involved, loss of analyte (in transfer from one tube to the other, in the filtration steps etc...) can also lead to variations from one analysis to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how will an IS solve this variability issue? The basic point is the assulmption that these external variability issues will affect both the analyte and the IS in a similar way. So one has to look at the sources of errors. For instance, let's consider an error in the measurement of a sample volume (let's call it $\epsilon$ and assume it is close to 1) . We have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{analyte} = V_{real} c_{analyte} = \epsilon V_{set} c_{analyte}$$&#xA;$$m_{IS} = V_{real} c_{IS} = \epsilon V_{set} c_{IS}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the measurement system accurately measures $m_{analyte}$ and $m_{IS}$, it is quite obvious that by dividing both equations we can remove this variability parameter and we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{analyte/IS} = {m_{analyte} \over{m_{IS}}} = {c_{analyte} \over c_{IS}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the point were the ratio comes out: taking a ratio allows to remove the variability parameter ($\epsilon$) in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a bit too simplistic, and a measurement procedure is not as simple. The general concept will still be to try to factor out the variability parameter as a proportion of the measured value, and then to take a ratio which will remove this variability from the measured values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the IS is correctly chosen so that it behaves as the analyte through the entire analytical procedure (both in terms of physico-chemical properties, as well as in the response of the detector), and if the chromatographic properties as well as the detector are such that peak area is proportional to concentration for the analyte as for the IS, then one can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{analyte} = \epsilon(analyte) R_{method}(analyte) c_{analyte}$$&#xA;$$A_{IS} =  \epsilon(IS) R_{method}(IS) c_{IS}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A_{analyte}$ and $A_{IS}$ are the measured peak areas for both analyte ans IS, $R_{method}(x)$ is the method response factor, which is dependent on each compound. In general the variability factor $\epsilon$ should depend on each compound, but under the above assumption of similar properties, one can assume that $\epsilon(analyte) \approx \epsilon(IS) \approx \epsilon$. Without IS, one would determine $R_{method}(analyte)$ through a calibration procedure. With an IS, one takes the ratio and the above equation can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{area} = {A_{analyte} \over A_{IS}} = {R_{method}(analyte) \over R_{method}(IS)} {c_{analyte} \over c_{IS}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_{analyte/IS} = {R_{method}(analyte) \over R_{method}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the ratio of the response factors of the method for analyte and IS. This is probably the ratio that is referred to. And it is not mystical at all, it is obtained through a standard calibration procedure with the IS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caution word to conclude this answer: I have tried to be quite precise in the answer. I have often times heard some people state that the use of an, internal standard is a general solution for analytical issues. Thus I want to insist on a warning: the solution is good as long as the assumptions stated above hold. If any one of these fails, then the whole analytical procedure will also fail. And of course, the use of an IS will only protect from errors after addition of the IS, and not from variability in the addition of the IS (you mention pipetting it in the analyte solution, which will limit the repeatability to that achieved by the pipettes used).&lt;/p&gt;&#xA;" OwnerUserId="263" LastEditorUserId="263" LastEditDate="2014-07-18T19:03:34.977" LastActivityDate="2014-07-18T19:03:34.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14429" PostTypeId="2" ParentId="14421" CreationDate="2014-07-18T18:12:14.380" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to remove a proton from the atom via some technique&#xA;  such as bombardment with another particle or applying an appropriate&#xA;  amount of energy to the system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this occurs in nature and can also be done in the laboratory.  In nature the process is known as radioactive decay.  Listed below are two decay processes that will remove a proton(s)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;alpha decay: an atom ejects a helium nucleus (2 protons and 2 neutrons) &#xA;$$\ce{^239_94Pu -&amp;gt; ^4_2He + ^235_92U}$$&lt;/li&gt;&#xA;&lt;li&gt;beta-plus decay: a positron is emitted and a proton is converted to a neutron.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the (nuclear) lab, a process known as phototransmutation can be employed.  If a nucleus is irradiated with high-energy gamma rays (high-energy photons) it can absorb the energy and change to another element by ejecting a proton and\or neutron&#xA;$$\ce{^25_12Mg -&amp;gt;C[{h\nu}]\   ^1_1H + ^24_11Na}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make a rough estimate of the energy required as follows:&#xA;$$\mathrm{24.98583692\,amu~ \longrightarrow 1.00727647\,amu + 23.99096278\,amu}$$&#xA;comparing the starting and final mass, we find that 0.01240233 amu has been gained in the process.  Since 1 amu = 931.5 MeV, the gamma ray photon would have to have an energy of at least 11.6 MeV.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-13T13:53:10.680" LastActivityDate="2016-03-13T13:53:10.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14430" PostTypeId="2" ParentId="14423" CreationDate="2014-07-18T18:48:27.417" Score="3" Body="&lt;p&gt;The octet rule does apply to carbon.  In order for carbon to form a complete octet, it must either gain 4 electrons, lose 4 electrons or form 4 covalent bonds.  Placing 4 units of charge on such a small nucleus would be exceedingly difficult, so carbon typically binds in a covalent manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine an experiment where we place carbon atoms in a container of hydrogen.  This becomes a chemical reaction, just like any other.  $\ce{CH4}$ will be the final product because we need 4 bonds around the central carbon atom in order for the carbon to have an octet.  As the carbon and hydrogen begin to react, a chemist would say that the atoms begin to move along a reaction coordinate that connects starting materials with products.  As the atoms move along this trajectory they explore many possible configurations.  For example, carbon could form 4 bonds to hydrogen using its unhybridized atomic orbitals.  If this pathway produced $\ce{CH4}$ it would have 2 C(2p)-H(1s) bonds and 2 C(2s)-H(1s) bonds.  The two C-H bonds formed from the carbon p orbitals would have an H-C-H angle around 90 degrees.  The 2 bonds formed using the carbon 2s orbitals would be rather floppy as they could travel over the surface of an s orbital sphere.  The most stable configuration would probably involve these 2 C(2s)-H(1s) bonds arranging themselves as far away from the other 2 hydrogens as possible.  So we'd probably wind up with a distorted tetrahedron with one H-C-H angle around 90 degrees and another H-C-H angle much wider than the 109 degree tetrahedral angle.  Note that the 2 hydrogens involved in the H-C-H 90 degree angle would be very close together resulting in a destabilizing steric interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now consider a second possible path along our reaction coordinate.  In this case let's mix our carbon atomic orbitals (the 2s and 2p orbitals) into a new set of atomic orbitals.  We will &quot;raise&quot; the energies of the two 2s electrons and &quot;lower&quot; the energies of the two 2p electrons such that the 4 new orbitals we generate will have the same energy as the average energy of our initial s and p orbitals.  Equal mixing of the s and p orbitals results in the formation of 4 equivalent new atomic orbitals, they are often referred to as $\ce{sp^3}$ hybridized orbitals.  These 4 equivalent orbitals point towards the vertices of a tetrahedron.  Reaction with hydrogen will produce a $\ce{CH4}$ molecule that has 4 equivalent C(sp3)-H(1s) bonds and all of the resulting bond lengths and angles will be equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of carbon, the reaction coordinate follows this last pathway.  Why? Because it produces the lowest energy (most stable) version of $\ce{CH4}$ possible.  This lowest energy version of $\ce{CH4}$ does not suffer from unfavorable H-H steric repulsions due to small (90 degree) H-C-H angles.  All of the C-H bonds in this stable version of $\ce{CH4}$ also have stronger (more stable) bonds.  The C(sp3)-H bonds are much more directed and have better overlap that the diffuse, non-directed C(2s)-H bonds we discussed earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we move from carbon to phosphorous, we find that in the molecule $\ce{PH3}$ the bond angles are close to 90 degrees and the central phosphorous uses its original atomic orbitals (pure P in this case) to form bonds to hydrogen - no further mixing of orbitals occurs in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that nature is smart.  Nature knows how to mix (or not mix) atomic orbitals in order to produce the lowest energy compound possible, and it's method of achieving this is to travel along a reaction coordinate sampling all possible pathways and then letting energetics drive the reaction along the lowest energy pathway.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-18T18:48:27.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14431" PostTypeId="1" CreationDate="2014-07-18T19:39:27.543" Score="2" ViewCount="415" Body="&lt;p&gt;Every day we use chemical reactions to perform mechanical work by creating either heat (e.g. in a piston engine) or electricity (batteries). I know muscles perform work by taking stored chemical energy and converting it directly into mechanical force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than in organic chemistry, are there any examples of chemical reactions that we use to perform mechanical work without the intermediate step of harnessing the heat or electrical potential created by the reaction?&lt;/p&gt;&#xA;" OwnerUserId="7164" LastActivityDate="2014-07-18T19:39:27.543" Title="Are there practical chemical reactions that do mechanical work without heat or electricity?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14432" PostTypeId="2" ParentId="14421" CreationDate="2014-07-18T19:51:51.063" Score="11" Body="&lt;p&gt;There are actually a (very small) number of nuclei that exhibit spontaneous &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_emission&quot; rel=&quot;noreferrer&quot;&gt;proton emission decays&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more common are photon, neutron or alpha initiated proton emission reaction such as the &lt;a href=&quot;http://arxiv.org/abs/1010.4958&quot; rel=&quot;noreferrer&quot;&gt;the alpha initiated reaction on Sulfur discussed in this pre-print&lt;/a&gt;. Moderate energy alphas incident on light nuclei generate many interesting final states a fact that has been instrumental in figuring out how the nucleus works (which is admittedly usually framed as a physics rather than chemistry question).&lt;/p&gt;&#xA;" OwnerUserId="508" LastEditorUserId="508" LastEditDate="2016-10-03T17:16:51.540" LastActivityDate="2016-10-03T17:16:51.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14433" PostTypeId="2" ParentId="14421" CreationDate="2014-07-18T20:15:00.430" Score="8" Body="&lt;p&gt;That's more a nuclear physics question. But I will write some anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is a good amount of naturally radioactive nuclei, that emits alpha particle, i.e. $\ce{^4He}$ nucleus. Than, there is a less common $\beta^+$ decay, when nucleus emits positron, transforming one proton into neutron in the process (and emitting neutrino). Next, some nuclei captures electron (usually from lowest electron shell), emitting neutrino and again, transforming one proton into neutron. Proton emission is known, but extremely rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, there is some amount of nuclear reactions, when an incoming hi-energy particle merges with nucleus, than the resulting excited nucleus emits some other particles. This reactions are extremely specific, each pair of nucleus and particle has their own reactions routes, etc, etc. Example of proton-emitting reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\ ^3He + D = ^4He +^p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, heavy nuclei after capturing a neutron, undergoes $\beta$-decay, transforming into $\alpha$-radioactive nuclei. The net result would be losing a proton and maybe a neutron or two. But this an awfully specific for few nuclei. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR : no, there is no general technique, but there are some techniques (very costly), than work in some specific cases. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2016-02-06T08:31:14.723" LastActivityDate="2016-02-06T08:31:14.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14434" PostTypeId="1" AcceptedAnswerId="14448" CreationDate="2014-07-18T21:48:55.767" Score="1" ViewCount="1300" Body="&lt;p&gt;How many ligands does Fe has in heme ?&#xA;I am not sure how many are in heme and how many in cytochrome C.&#xA;Its about iron complexes.&lt;/p&gt;&#xA;" OwnerUserId="6977" LastEditorUserId="4945" LastEditDate="2014-07-19T14:00:58.157" LastActivityDate="2014-07-19T15:46:47.853" Title="How many ligands in Heme?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14435" PostTypeId="1" CreationDate="2014-07-18T21:49:39.400" Score="1" ViewCount="241" Body="&lt;p&gt;Is it possible to incorrectly predict that two bond angles are different from van der Waals repulsions alone? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that because bond lengths differ when different atoms are involved that we could, at least theoretically, have two identical bond angles involving different molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say we are considering the bond angles of molecule $\ce{AXY}$ and $\ce{BZA}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;X and Y are highly electron dense and also big. The first suggests a big bond angle; the second suggests a long bond length, which in turn suggests the possibility of a smaller bond angle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider $\ce{BZA}$. Z and A are not electron dense but also large. The first suggests a small bond angle. The second suggests a long bond length and thus a smaller bond angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it still possible for the two to have identical bond angles? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-18T21:49:39.400" Title="Identical bond angles" Tags="&lt;bond&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2018-03-26T16:26:39.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="14436" PostTypeId="1" AcceptedAnswerId="14480" CreationDate="2014-07-18T22:10:05.040" Score="4" ViewCount="207" Body="&lt;p&gt;I know that if I have two compounds in equilibrium and I know $\mathrm{\Delta G}$, then I can use $$\mathrm{\Delta G=-RTlnK}$$&#xA;to determine the concentrations of the two components at equilibrium.  I have 4 compounds that are all in equilibrium with one another&#xA;\begin{aligned}&#xA;\mathrm{\Delta G(1,4)} ~&amp;amp;= ~~~3.41~ \mathrm{kcal/mol}\\&#xA;\mathrm{\Delta G(3,1)}~&amp;amp;=-2.21~ \mathrm{kcal/mol}\\&#xA;\mathrm{\Delta G(2,3)}~&amp;amp;=-0.46~ \mathrm{kcal/mol}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I applied the equation sequentially and found that &#xA;\begin{aligned}%&#xA;[1]~&amp;amp;=~0.34\%\\&#xA;[2]~&amp;amp;=~29.1\%\\&#xA;[3]~&amp;amp;=~13.5\%\\&#xA;[4]~&amp;amp;=~57\%\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are those the correct concentrations? Is there an &quot;easy&quot; way to do it?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-04T15:18:15.240" LastActivityDate="2018-03-04T15:18:15.240" Title="4 compounds in equilibrium with one another - determine their equilibrium concentrations" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14437" PostTypeId="1" CreationDate="2014-07-18T22:44:16.540" Score="3" ViewCount="38785" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A $\pu{4.305 g}$ sample of a nonelectrolyte is dissolved in $\pu{105 g}$ of water. The solution freezes at $\pu{-1.23^\circ C}$. Calculate the molar mass of the solute. $k_f$ for water is $\pu{1.86 C/m}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1.23 = 1.86 \times \frac{x}{1.05},$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x = y / 4.305$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.66 = \frac{x}{1.05}$$&#xA;$$x = 0.69$$&#xA;$$0.69 = y / 4.305$$&#xA;$$y = \pu{2.989 g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I know this is not correct.&lt;/p&gt;&#xA;" OwnerUserId="7165" LastEditorUserId="69830" LastEditDate="2021-05-25T06:41:06.070" LastActivityDate="2021-05-25T06:41:06.070" Title="Calculating the molar mass of a solute from freezing point" Tags="&lt;solubility&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14438" PostTypeId="1" CreationDate="2014-07-19T00:14:47.620" Score="1" ViewCount="1001" Body="&lt;p&gt;A question that appeared on my last chemistry exam was :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following has greater entropy A) An atom &#xA;  B) A&#xA;  macromolecule&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question doesn't specify anything else(i.e. type/size of atom or macromolecule). The macromolecules we learned class were proteins, sugars, nucleic acids(DNA/RNA) or lipids. What is the correct answer and why(hypothetical calculation, if possible) ?&lt;/p&gt;&#xA;" OwnerUserId="6833" LastEditorUserId="6833" LastEditDate="2014-07-19T17:22:50.703" LastActivityDate="2014-07-19T18:25:56.007" Title="Greater entropy: an atom or macromolecule?" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;molecules&gt;&lt;atoms&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14439" PostTypeId="1" CreationDate="2014-07-19T00:19:42.253" Score="2" ViewCount="70158" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;We have a solution &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt; (acetic acid) with &lt;span class=&quot;math-container&quot;&gt;$c=0.02~\mathrm{mol/L}$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm a(\ce{CH3COOH}) = 1.8\cdot10^{-5}$&lt;/span&gt;. Calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of this solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;All I know is that it is a weak acid and that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH}= \mathrm{p}K_\mathrm a + \log\left(\frac{B}{A}\right).$$&lt;/span&gt;&#xA;I know &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; but there is no information about &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7166" LastEditorUserId="95133" LastEditDate="2020-09-12T11:48:00.267" LastActivityDate="2020-09-12T11:48:00.267" Title="Calculate the pH of acetic acid for a 0.02 M solution" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14440" PostTypeId="2" ParentId="14437" CreationDate="2014-07-19T01:19:34.063" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Freezing_point_depression&quot; rel=&quot;nofollow noreferrer&quot;&gt;freezing point depression&lt;/a&gt; is based on the molal concentration (moles of solute per kg of solvent).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta T_f = -k_f \cdot m$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You know the freezing point depression of the solution and the cryoscopic constant, so you can calculate the molality:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m = -\dfrac{\Delta T_f}{k_f}=-\dfrac{\pu{-1.23 ^\circ C}}{1.86 \frac{\textrm{kg}\cdot ^\circ \text{C}}{\textrm{mol}}}=\pu{0.66 molal}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since molality is the number of moles of solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_\textrm{solute}= \pu{0.66 m} \cdot \pu{ 0.105 kg}= \ ...$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Divide your mass of solute by the moles of solute and you will get the molar mass (and not 2.989 g).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="69830" LastEditDate="2021-05-25T06:40:48.807" LastActivityDate="2021-05-25T06:40:48.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14443" PostTypeId="2" ParentId="14438" CreationDate="2014-07-19T02:41:38.767" Score="5" Body="&lt;p&gt;There are many equations that describe the entropy of a system, one is &#xA;$$\ce{S = k~ ln ~W}$$&#xA;where S is the entropy of the system, k is Boltzmann's constant and W is the number of states a system can exist in.  If we take an atom, it can only move in 3 directions x, y and z - that's it.  If we take a small molecule like ethane, it can also move in these same 3 dimensions but it can also rotate about any one of the atoms, the various bonds can stretch and compress and the various bond angles can expand and collapse.  Clearly, there are a lot more states that ethane can exist in as compared to a single atom, clearly W is larger for ethane than a single atom, clearly S is larger for ethane than a single atom.  As we increase the complexity of the molecule, W and S increase exponentially.  A macromolecule will have a much larger entropy than an atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; (based on @Gdgames Gamers comment below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy is a measure of the disorder in a system.  It will always be true that the entropy of substance in the gas phase will always be greater than the entropy of the same substance in the liquid phase, which, in turn, will be always greater than the entropy of the same substance in the solid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the person who wrote the exam is assuming that the atom is in the gas phase and the macromolecule is in the solid phase and that based on the phase difference alone the atom has a larger entropy than the macromolecule.  I don't think this is always correct.  First off, the temperature is not specified, nor the state of the macromolecule (is it in solution?).  The macromolecule, depending upon the temperature and state, may or may not translate and rotate through space, but it will vibrate (even at 0K).  I'm just not sure how to compare the entropy contribution from the potential 3n-6 vibrational modes in the macromolecule to the entropy from the 3 translational modes of the atom.  &lt;strong&gt;Very poor question by the examiner&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-19T18:25:56.007" LastActivityDate="2014-07-19T18:25:56.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14444" PostTypeId="1" CreationDate="2014-07-19T06:27:19.027" Score="1" ViewCount="186" Body="&lt;p&gt;What are some examples of solutions deviating significantly from ideal behavior, and, by extension, showing pHs significantly different from those predicted by usage of concentrations instead of activity coefficients in K expressions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Wikipedia lists two examples but what other examples are there? Do all dilute solutions behave ideally or mostly ideally? Or are there major exceptions in which the use of concentrations instead of activities fails? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there an easy way to incorporate activities into the calculation of pH for complex systems? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-19T06:27:19.027" Title="Deviation from ideal solutions - activities and concentrations" Tags="&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14445" PostTypeId="1" CreationDate="2014-07-19T06:40:45.023" Score="-1" ViewCount="8678" Body="&lt;p&gt;As I recall a Lewis acid is a substance that can accept an electron pair from a donor and oxygen seems to be able to do that. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2 + O -&amp;gt; H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is oxygen a Lewis acid?&lt;/p&gt;&#xA;" OwnerUserId="7141" LastEditorUserId="43942" LastEditDate="2017-05-07T19:16:46.393" LastActivityDate="2017-05-07T19:16:46.393" Title="Is oxygen a Lewis acid?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14447" PostTypeId="2" ParentId="14445" CreationDate="2014-07-19T07:16:27.147" Score="3" Body="&lt;p&gt;Usually the terms Lewis acid and Lewis base are defined within the context of a specific chemical reaction. So in a reaction, if there is lewis acid there must be a lewis base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lone-pair acceptors are lewis acids where as lone-pair donors are lewis bases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some compounds, such as $\ce{H2O}$, are both Lewis acids and Lewis bases, because they can either accept a pair of electrons or donate a pair of electrons, depending upon the reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In reaction $\ce{H2 + O -&amp;gt; H2O}$, you can say that oxygen is lewis acid but that would be contradictory since hydrogen doesn't have a lone pair to donate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Oxygen acts as a lewis base by donating a lone pair to hydrogen ion when Acids are dissolved in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more clear on Lewis Acid/Base concept, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-07-19T07:25:59.947" LastActivityDate="2014-07-19T07:25:59.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14448" PostTypeId="2" ParentId="14434" CreationDate="2014-07-19T08:22:21.947" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Heme&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heme:&lt;/a&gt; The Iron molecule is bound to 4 Nitrogen atoms, but take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Porphyrin&quot; rel=&quot;nofollow noreferrer&quot;&gt;porphyrin&lt;/a&gt;. All the nitrogens are part of one actual structure. You can visualize the iron molecule in the center of the larger compound's structure, replacing the two Hydrogen atoms in N-H:&lt;/p&gt;&#xA;&lt;p&gt;The large structure surrounding and coordinating with Fe is also known as a &lt;strong&gt;polydentate ligand&lt;/strong&gt; since it bonds to Iron at more than one site. So there is just one ligand on Fe.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Porphyrin.svg/300px-Porphyrin.svg.png&quot; alt=&quot;porphyrin&quot; /&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/4d/Heme.svg/375px-Heme.svg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Cytochrome C: is a heme protein, so you may expect the same answer: A single polydentate ligand binds to Iron. &lt;a href=&quot;http://www.proteopedia.org/wiki/index.php/Cytochrome_c&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; has a PDB Model of it in a plugin.&lt;/p&gt;&#xA;&lt;p&gt;As one ligand coordinates to four sites in this case, other ligands typically attach (e.g. oxygen binding in hemoglobin) to fill the octahedral group (up to 6 ligands attached to Fe the coordinate center).&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-07-19T15:46:47.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14449" PostTypeId="1" AcceptedAnswerId="14519" CreationDate="2014-07-19T11:15:05.853" Score="11" ViewCount="17269" Body="&lt;p&gt;I am an amateur, with very little professional chemistry knowledge. I was doing a demonstration / experiment with my 6 year old child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We took very warm normal tap water, and dissolved into it a large pile of normal table salt. I didn't measure anything particularly, because I was just showing the basic concept of 'dissolving'. That's all. My kid was amazed to see the white pile of salt disappear visibly, but upon tasting the water understood what had happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we left the bowl up on a high shelf for 2 weeks, to let the water evaporate. This morning we took a look, and were rewarded by a dry bowl full of beautiful perfectly square salt crystals, many of them between 4-6mm in width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the crystals were perfect squares did not surprise me, as I have read that sodium and chlorine atoms form a cube shape at an atomic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I could not answer the question about what the 'X' pattern marking was on each square. I have tried to attach photographs to show what I mean. The 'normal' picture is blurry, but if you look you can faintly see an 'X' on each crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other 2 pictures are each at 20x, using the Natural History Museum Pocket Microscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what explains those 'X' shapes? If possible I need to translate the explanation into 6-year-old talk :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L14Cp.jpg&quot; alt=&quot;2 salt crystals&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LzuV0.jpg&quot; alt=&quot;Salt crystal at 20x&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/AdEwi.jpg&quot; alt=&quot;Salt crystal at 20x&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q4DMo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q4DMo.jpg&quot; alt=&quot;Here&amp;#39;s a closer look if anyone is interested&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7171" LastEditorUserId="24434" LastEditDate="2016-08-06T07:45:33.490" LastActivityDate="2018-06-25T11:38:32.990" Title="If salt crystals make a cube shape, what is this 'X' marking?" Tags="&lt;everyday-chemistry&gt;&lt;crystal-structure&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14450" PostTypeId="1" CreationDate="2014-07-19T12:48:10.420" Score="3" ViewCount="57" Body="&lt;p&gt;There's been a lot written (and debunked) about alleged biological effects from RF radiation, even when it's too low powered to cause heating effects or ionization. Is there any chemical reason to believe it's possible for an effect to exist? &lt;/p&gt;&#xA;" OwnerUserId="7173" LastActivityDate="2014-07-19T19:01:54.183" Title="Other than heating, is there any evidence that normal RF radiation (say, microwave range) can affect chemical reactions?" Tags="&lt;physical-chemistry&gt;&lt;reaction-control&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14451" PostTypeId="1" AcceptedAnswerId="14453" CreationDate="2014-07-19T12:52:54.560" Score="12" ViewCount="2071" Body="&lt;p&gt;Is $\ce{BCl5}$ possible? I saw it  in book. As far as we can see there is no way  to get 5 chlorine atoms with boron.&lt;/p&gt;&#xA;" OwnerUserId="7174" LastEditorUserId="73" LastEditDate="2014-07-29T08:53:51.843" LastActivityDate="2014-07-29T08:53:51.843" Title="Is BCl₅ possible?" Tags="&lt;crystal-structure&gt;&lt;covalent-compounds&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14452" PostTypeId="1" CreationDate="2014-07-19T12:57:13.963" Score="1" ViewCount="329" Body="&lt;p&gt;I have difficulty understanding the use of gauche correction in Bensen group increment method to calculate the heat of formation. As an example, 4 times gauche correction (0.8 kcal / mol) is considered in the calculation in the hypothetical equilibrium between the following two isomers. How is that figure decided?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vqsn9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2014-07-19T13:56:21.263" Title="Gauche correction in Bensen group increment" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14453" PostTypeId="2" ParentId="14451" CreationDate="2014-07-19T13:36:46.080" Score="14" Body="&lt;p&gt;Boron pentachloride is likely not stable except perhaps in extreme conditions, such as under very high pressures. Even then it may be possible that a description such as $\ce{[BCl4^{-}]Cl^+}$ containing a tetrahedral boron anion could turn out to be more accurate than any hypercoordinate structure (a boron atom surrounded by more than four ligand atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, pentacoordinate and hexacoordinate boron are not unheard of. Indeed they are common in &lt;a href=&quot;http://en.wikipedia.org/wiki/Boranes#Cluster_types&quot; rel=&quot;noreferrer&quot;&gt;borane clusters&lt;/a&gt;, though this type of structure has unusual bonding and perhaps is best left in its own category. Even without considering clusters, it is still possible to create hypercoordinate boron atoms by exploring steric demands of certain ligands. Examples can be found in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00331a064&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;, among several others. A simple diagram exemplifying how to force hypercoordination in boron is given &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo1024656&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;; notice how a four-coordinate structure requires rotation of the C-C bond between the left and middle aromatic rings, decreasing the extent of $\pi$-conjugation between rings and causing that particular molecular geometry to be less stable relative to the five-coordinate structure. Hexacoordinate carbon has also been prepared with &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja710423d&quot; rel=&quot;noreferrer&quot;&gt;even higher steric forcing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, it should be mentioned that &quot;hypercoordinate&quot; is the preferred term instead of &quot;hypervalent&quot;. Not only is hypercoordination more general (all hypervalent compounds are hypercoordinate, while not all hypercoordinate compounds need be hypervalent), but hypervalency usually implies some form of octet expansion, which is nowadays not generally believed to be a correct description of bonding. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-07-19T23:12:23.830" LastActivityDate="2014-07-19T23:12:23.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14454" PostTypeId="2" ParentId="14452" CreationDate="2014-07-19T13:56:21.263" Score="2" Body="&lt;p&gt;Here is how I do it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sight down each C-C bond along the backbone of the molecule and draw a Newman projection&lt;/li&gt;&#xA;&lt;li&gt;Look at all 3 possible conformations in the Newman projection and pick the conformation with the fewest gauche interactions&lt;/li&gt;&#xA;&lt;li&gt;Count the number of gauche interactions&lt;/li&gt;&#xA;&lt;li&gt;Repeat with the next C-C bond along the backbone&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is a drawing of the 3 Newman projections in 3,4-dimethylhexane that contain gauche interactions (there are no gauche interactions along the $\ce{C1-C2}$ or $\ce{C5-C6}$ bonds.  You can see that there are &lt;strong&gt;4&lt;/strong&gt; gauche interactions in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytv3m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-19T13:56:21.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14455" PostTypeId="2" ParentId="14123" CreationDate="2014-07-19T16:29:43.303" Score="2" Body="&lt;p&gt;An alkaline battery (or more precise alkaline manganese cell) consists of a zinc anode, a $\gamma\!-\!\ce{MnO2}$ cathode and $\ce{KOH}$ (in paste form) electrolyte. Zinc is the negative electrode and $\ce{MnO2}$ the positive electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During battery discharge, zinc at the anode is oxidized and dissolves in the electrolyte as hydroxozincate. When the electrolyte is saturated with zincate, zinc oxide starts to precipitate. With continued discharge and decreasing $\ce{OH-}$ concentration, zinc hydroxide is formed, which can slowly dehydrate to zinc oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;Zn + 4OH- &amp;amp;-&amp;gt;~ [Zn(OH)4]^{2-} + 2e- \\&#xA;[Zn(OH)4]^{2-} &amp;amp;-&amp;gt;~ ZnO + 2OH- + H2O\\&#xA;Zn + 2OH- &amp;amp;-&amp;gt;~ Zn(OH)2 + 2e- \\&#xA;Zn(OH)2 &amp;amp;-&amp;gt;~ ZnO + H2O\\&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode, $\ce{MnO2}$ is in turn reduced, first to $\ce{MnO(OH)}$. When the battery is discharged slowly, this can be further reduced to manganese(II) hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}\ce{&#xA;MnO2 + H2O + e- &amp;amp;-&amp;gt;~ MnO(OH) + OH- \\&#xA;MnO(OH) + H2O + e- &amp;amp;-&amp;gt;~ Mn(OH)2 + OH- \\&#xA;}\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall reaction, when only considering the first reduction step at the cathode, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + 2MnO2 + 2H2O + 2OH- -&amp;gt;~ [Zn(OH)4]^{2-} + 2MnO(OH)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During discharge, zinc is irreversibly oxidized and water is consumed. As a consequence, the battery is &quot;dead&quot; when the electrolyte is dried out and/or all zinc has dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_battery&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://de.wikipedia.org/wiki/Alkali-Mangan-Zelle&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-07-24T16:31:07.043" LastActivityDate="2014-07-24T16:31:07.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14456" PostTypeId="2" ParentId="13779" CreationDate="2014-07-19T18:03:22.667" Score="11" Body="&lt;p&gt;Sorry for the late answer. I just discovered this question.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Preludium&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Firstly, I think you might have an error or some non-standard notation in your formula for $Q_{\mathrm{rot}}$. The rotational energy levels are given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  E_{\mathrm{rot}} = \frac{\hbar^2}{2 I} J ( J + 1 ) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $I$ is the moment of inertia.&#xA;Then the rotational constant $B_{0} = \frac{\hbar^2}{2 I}$ (in units of energy) is introduced, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  E_{\mathrm{rot}} = B_{0} J ( J + 1 ) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{B_0 J(J+1)}{kT}\right)} \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas your exponent contains another $h$ which should not be there. At least that's the standard notation I am used to. For convenience I'll also introduce the rotational temperature $\Theta_{\mathrm{rot}} = \frac{B_{0}}{k}$, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right)} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Main Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ok, now that that's sorted out, let's start the real work. It is actually a pretty standard approximation that is used here: They use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula&quot; rel=&quot;noreferrer&quot;&gt;Euler-Maclaurin formula&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \sum_{n = a}^{b} f(n) &amp;amp;= \int_{a}^{b} f(n) \, \mathrm{d} n + \frac{1}{2} \bigl( f(b) + f(a) \bigr) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} \Bigl( f^{(2i - 1)}(a) - f^{(2i - 1)}(b) \Bigr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f^{(k)}(a)$ is the $k$th derivative of $f$ evaluated at $a$. The $b_{2i}$s are the absolute values of the even Bernoulli numbers, $b_{2} = \frac{1}{6}$, $b_{4} = \frac{1}{30}$, $b_{6} = \frac{1}{42}$, $...$.&#xA;It provides a way to approximate the summation of a function over a discrete variable by an analogous integration over a continuous variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case at hand, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  f(J) = (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the summation limits are $a=0$ and $b=\infty$ and where $f(\infty) = f^{\prime}(\infty) = \ldots = 0$ this formula boils down to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  Q_{\mathrm{rot}} = \sum_{J = 0}^{\infty} f(J) &amp;amp;= \int_{0}^{\infty} f(J) \, \mathrm{d} J + \frac{1}{2} f(0) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} f^{(2i - 1)}(0)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is what is usually called the classical or the high-temperature limit. It can be easily calculated by making the substitution $K = J(J + 1)$ $\Rightarrow$ $\mathrm{d}J = \frac{\mathrm{d}K}{2J + 1}$. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \int_{0}^{\infty} f(J) \, \mathrm{d} J &amp;amp;= \int_{0}^{\infty} (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \, \mathrm{d} J \\&#xA;  &amp;amp;= \int_{0}^{\infty} \exp\left(-\frac{\Theta_{\mathrm{rot}} K}{T}\right) \, \mathrm{d} K \\&#xA;  &amp;amp;= \frac{T}{\Theta_{\mathrm{rot}}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, only the derivatives $f^{(k)}(0)$ need to be sorted out. Those that contain terms below the third order in $\frac{\Theta_{\mathrm{rot}}}{T}$ can be calculated to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  f(0) &amp;amp;= 1 \\&#xA;  f^{(1)}(0) &amp;amp;= 2 - \frac{\Theta_{\mathrm{rot}}}{T} \\&#xA;  f^{(3)}(0) &amp;amp;= - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \\&#xA;  f^{(5)}(0) &amp;amp;= 120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing this by hand is very tedious. Best you use something like Mathematica. Plugging all this into the Euler-MacLaurin formula and dropping all terms above second order in $\frac{\Theta_{\mathrm{rot}}}{T}$ you get exactly the result of Gordy &amp;amp; Cook&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  Q_{\mathrm{rot}} &amp;amp;\approx \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} \cdot 1 + (-1) \frac{1}{6 \cdot 2!} \left( 2 - \frac{\Theta_{\mathrm{rot}}}{T} \right) \\&#xA; &amp;amp;\quad + (-1)^{2} \frac{1}{30 \cdot 4!} \left( - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\&#xA;  &amp;amp;\quad + (-1)^{3} \frac{1}{42 \cdot 6!} \left(120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \right) \\&#xA;  &amp;amp;= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} - \frac{1}{6} + \frac{1}{12} \frac{\Theta_{\mathrm{rot}}}{T} - \frac{1}{60} \frac{\Theta_{\mathrm{rot}}}{T} \\&#xA;  &amp;amp;\quad + \frac{1}{60} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \frac{1}{252} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\&#xA;  &amp;amp;= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note: If you have a homonuclear diatomic molecule you need to introduce the so called symmetry number $\sigma$ which represents the number of indistinguishable orientations a molecule can have. It originates from symmetry constrains: for homonuclear diatomic molecules there is a two-fold axis of symmetry perpendicular to the internuclear axis which gives rise to two indistinguishable orientations, so the above formula for $Q_{\mathrm{rot}}$ overcounts the number of available quantum states&#xA;by a factor of two because of the inherent indistinguishability of the nuclear pair. Thus, $Q_{\mathrm{rot}}$ needs to be divided by $\sigma = 2$. For heteronuclear diatomic molecules you simply have $\sigma = 1$.&#xA;With the introduction of $\sigma$ the equation for $Q_{\mathrm{rot}}$ becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  Q_{\mathrm{rot}} &amp;amp;\approx \frac{1}{\sigma} \left( \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \right) \\&#xA;  &amp;amp;= \frac{T}{\sigma \Theta_{\mathrm{rot}}} \left(1  + \frac{1}{3} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{1}{15} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} \right) \right)&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-07-20T15:51:13.417" LastActivityDate="2014-07-20T15:51:13.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14458" PostTypeId="2" ParentId="14439" CreationDate="2014-07-19T19:16:26.327" Score="8" Body="&lt;p&gt;This type of problem is best solved using the ICE (Initial Change Equilibrium) method.&lt;/p&gt;&#xA;&lt;p&gt;First, write down the chemical equation and the law of mass action for this reaction.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH &amp;lt;=&amp;gt; H+ + CH3COO-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Second, determine the initial concentration of each of the three species. Make a little table to keep everything straight:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lc} \hline&#xA;&amp;amp;\ce{CH3COO-} &amp;amp;\ce{&amp;lt;=&amp;gt;}&amp;amp;\ce{H+} &amp;amp;+&amp;amp;\ce{CH3COO-} \\ \hline&#xA;\text{I}&amp;amp; \pu{0.02 M} &amp;amp;&amp;amp;\pu{0 M} &amp;amp;&amp;amp;\pu{0 M} \\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At equilibrium, some of the acetic acid will be ionized. How much? We don't know right now, so let's have that be a variable; &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. At equilibrium, the concentration of acetic acid will have decreased by &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. How much will the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and acetate anion increase? Since the stoichiometric coefficient of all species is 1, then those concentrations will increase by &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. Let's add that to our table. Let's also write an expression for the equilibrium concentrations in terms of the initial concentrations and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lc} \hline&#xA;&amp;amp;\ce{CH3COO-} &amp;amp;\ce{&amp;lt;=&amp;gt;}&amp;amp;\ce{H+} &amp;amp;+&amp;amp;\ce{CH3COO-} \\ \hline&#xA;\text{I}&amp;amp; \pu{0.02 M} &amp;amp;&amp;amp;\pu{0 M} &amp;amp;&amp;amp;\pu{0 M} \\ &#xA;\text{C}&amp;amp; -x &amp;amp;&amp;amp;+x  &amp;amp;&amp;amp;+x \\ &#xA;\text{E}&amp;amp; 0.02-x  &amp;amp;&amp;amp;x &amp;amp;&amp;amp;x \\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, plug these expressions into the law of mass action:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{(x)(x)}{(0.02-x)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As long as &lt;span class=&quot;math-container&quot;&gt;$x\ne 0.02$&lt;/span&gt;, which is very unlikely at equilibrium, we have a solvable quadratic equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_a=1.8\times 10^{-5}= \dfrac{(x)(x)}{(0.02-x)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$3.6\times 10^{-7}-1.8\times 10^{-5}x-x^2=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Some texts will encourage you to make the simplification &lt;span class=&quot;math-container&quot;&gt;$0.02-x \approx 0.02$&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$x&amp;lt;&amp;lt;0.02$&lt;/span&gt;, however, WolframAlpha can solve &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1.8e-5%20%3Dx%5E2%2F%28.02-x%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;the original equation&lt;/a&gt; (and so can you).&lt;/p&gt;&#xA;&lt;p&gt;The equation has two roots: &lt;span class=&quot;math-container&quot;&gt;$x=5.91\times 10^{-4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x=-6.09\times 10^{-4}$&lt;/span&gt;. The second one is nonsense (we cannot have negative concentrations).&lt;/p&gt;&#xA;&lt;p&gt;If you make the assumption that &lt;span class=&quot;math-container&quot;&gt;$0.02−x \approx 0.02$&lt;/span&gt;, then the positive root is &lt;span class=&quot;math-container&quot;&gt;$x=6.0 \times 10^{−4}$&lt;/span&gt;, not a big difference given the significant figures. If you knew your initial concentration more precisely, this assumption would be invalid.&lt;/p&gt;&#xA;&lt;p&gt;Now, we know what &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is. What do we still need to know to determine pH? How is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; related to pH?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="95133" LastEditDate="2020-09-12T11:34:41.320" LastActivityDate="2020-09-12T11:34:41.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14459" PostTypeId="1" CreationDate="2014-07-19T21:26:30.437" Score="0" ViewCount="2913" Body="&lt;p&gt;I thought the answer was obvious, but endless google searching has revealed nothing but speculation and people who have completely conflicting opinions. Some people say $\ce{CO2}$ absorbs no heat, others say it can cool things extremely rapidly. Anyone who has ever had compressed $\ce{CO2}$ sprayed on themselves or on hands knows it can cause very rapid drop in temperature on what ever it hits very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I have a really hot mass of liquid, would spraying compressed $\ce{CO2}$ on it be an effective means of freezing/solidying the mass? - Assuming the liquid is well above room temperature (100+) and turns to a solid after a 20-30 degree drop in temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why the $\ce{CO2}$ is cold coming out of the tank, as it is decompressed, but is it effective at absorbing heat?&lt;/p&gt;&#xA;" OwnerUserId="7175" LastEditorUserId="39539" LastEditDate="2017-05-29T04:38:12.883" LastActivityDate="2020-12-07T13:49:52.720" Title="Can compressed CO2 sprayed onto a hot mass/liquid be an effective means to cool it?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14460" PostTypeId="1" AcceptedAnswerId="14461" CreationDate="2014-07-19T21:56:19.160" Score="3" ViewCount="1581" Body="&lt;p&gt;Here's a &quot;simple&quot; bond diagram of 1,1,1,2-Tetrafluoroethane which I believe is a Lewis structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/1,1,1,2-Tetrafluoroethane.svg/100px-1,1,1,2-Tetrafluoroethane.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/a4/1,1,1,2-Tetrafluoroethane.svg/100px-1,1,1,2-Tetrafluoroethane.svg.png&quot; alt=&quot;1,1,1,2-Tetrafluoroethane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, that's easy enough - four bonds from each carbon atom, and each fluorine atom having one bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some more &quot;complicated&quot; bonds which I have yet to do in my high school chemistry class, so I don't understand how to interpret them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's caffeine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/Caffeine-2D-skeletal.svg/150px-Caffeine-2D-skeletal.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/Caffeine-2D-skeletal.svg/150px-Caffeine-2D-skeletal.svg.png&quot; alt=&quot;Caffeine&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/Caffeine_(1)_3D_spacefill.png/150px-Caffeine_(1)_3D_spacefill.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/0/0e/Caffeine_(1)_3D_spacefill.png/150px-Caffeine_(1)_3D_spacefill.png&quot; alt=&quot;Caffeine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And chloroform:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u7ZZL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u7ZZL.png&quot; alt=&quot;Chloroform&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Chloroform_3D.svg/100px-Chloroform_3D.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Chloroform_3D.svg/100px-Chloroform_3D.svg.png&quot; alt=&quot;Chloroform&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go on interpreting these diagrams? Or is there missing information? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image sources: Wikimedia&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="15751" LastEditDate="2019-07-19T18:58:05.913" LastActivityDate="2019-07-19T18:58:05.913" Title="Explanation of certain chemical bond diagrams" Tags="&lt;structural-formula&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14461" PostTypeId="2" ParentId="14460" CreationDate="2014-07-19T21:58:50.587" Score="5" Body="&lt;p&gt;The second structure you posted is a line angle diagram. In this diagram, methyl groups (-CH3) are implied using sticks with nothing at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kXRjA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon atoms are also implied; carbons exist at every &quot;corner&quot; of the polygon which doesn't have an element explicitly identified. Hydrogens &lt;em&gt;attached to carbons&lt;/em&gt; are also implied. So at the intersection of two lines, if there are no other attachments and no charge, you can expect there to be two hydrogens (carbon is generally tetravalent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/weIxP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third structure you posted is a wedge-dash diagram. Wedges depict substituents angled toward you; dashes are substituents pointing away from you; the lines are coplanar. The advantage of this diagram is that it shows the 3D nature of the molecule. Your first picture of the fluoro-carbon compound implies 90 degree bond angles. However, you should realize that the carbon atoms are sp3 hybridized and thus should not have 90 degree bond angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JPxgg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nuw8k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading and exploration of other depictions of molecular structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/&quot; rel=&quot;noreferrer&quot;&gt;http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-19T22:04:48.700" LastActivityDate="2014-07-19T22:04:48.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14463" PostTypeId="1" AcceptedAnswerId="14464" CreationDate="2014-07-19T22:59:53.320" Score="1" ViewCount="1271" Body="&lt;p&gt;What is the relationship between electric potential and conductivity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, gold has a more negative electric potential than silver (-1.100 vs -0.800), but silver is a better conductor.&lt;/p&gt;&#xA;" OwnerUserId="743" LastActivityDate="2014-07-19T23:05:42.900" Title="Electric conductivity vs potential" Tags="&lt;conductivity&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14464" PostTypeId="2" ParentId="14463" CreationDate="2014-07-19T23:05:42.900" Score="2" Body="&lt;p&gt;I would not expect to find any relationship.  The potential you are referring to involves breaking metallic bonds and ionizing the resulting atoms while conductivity involves electrons moving through the solid metal.  Copper's is -.34 and it is between gold and silver in conductivity.  Aluminum's is 1.66 and it is #4 behind gold in conductivity.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-19T23:05:42.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14465" PostTypeId="2" ParentId="14459" CreationDate="2014-07-19T23:10:43.940" Score="5" Body="&lt;p&gt;Celebrity chef Alton Brown produces ice cream using a device called the Jet Cream.  He sprays $\ce{CO2}$ from a fire extinguisher at a stream of chocolate milk, instantly producing ice cream. The only difference in your scenario and his is the starting temperature of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that $\ce{CO2}$ absorbs no heat is absurd.  If it is colder than what it touches, it should absorb heat.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="39539" LastEditDate="2017-05-29T04:39:29.870" LastActivityDate="2017-05-29T04:39:29.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14466" PostTypeId="2" ParentId="14314" CreationDate="2014-07-20T00:34:53.230" Score="5" Body="&lt;p&gt;Have you tried with red or black tea? It turns lighter in acidic media and turns dark in basic media. I tried once adding the pulp juice of one lemon to a cup of tea and it lighted up (mainly for acetic acid), then I added NaOH-based salt and turned very black and opaque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe that suits for your solution.&lt;/p&gt;&#xA;" OwnerUserId="5683" LastActivityDate="2014-07-20T00:34:53.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14467" PostTypeId="1" CreationDate="2014-07-20T00:36:14.260" Score="4" ViewCount="2477" Body="&lt;p&gt;A Grignard reagant was prepared from an alkyl halide and magnesium chips, then reacted with a ketone. The resulting solution has been stirring for a few days at low temperature (0-5°C). It's now time to transfer the reaction solution slowly to a vessel containing an acid quench. My problem is the unreacted magnesium chips (smaller than when we started, of course) have settled, and are plugging the tubing and stopper valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What solvents could I use to flush the magnesium through, breaking it up, and possibly dissolve it? This can't be added to the reaction solution (the Drain-O method), but only through the tubing itself. Obviously water is a bad idea. Any insights?&lt;/p&gt;&#xA;" OwnerUserId="7122" LastEditorUserId="16683" LastEditDate="2017-04-03T15:23:55.670" LastActivityDate="2017-04-03T15:23:55.670" Title="Removing unreacted Mg metal in a Grignard reaction" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;metal&gt;&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14468" PostTypeId="2" ParentId="14467" CreationDate="2014-07-20T02:20:15.760" Score="2" Body="&lt;p&gt;Pure water would certainly be a bad idea, but a 1% solution of 1 M hydrochloric acid in ether would probably work without a vigorous reaction.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-20T02:20:15.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14472" PostTypeId="2" ParentId="14052" CreationDate="2014-07-20T07:50:35.177" Score="4" Body="&lt;p&gt;Rust appears because iron is oxidized in presence of oxygen (it is a oxydo-reduction reaction, meaning an electron transfer occurs between species), and water helps this phenomenon occur. Iron loses 3 electrons to the favour of the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In presence of water, oxygen can oxidize iron by the mean of hydroxyl group. It leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe(OH)_3 \leftrightharpoons Fe_2O_3\cdot 3H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where water is not the oxidizer, rather it is a mediator for oxidation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydroxides have low solubility, so water can only dissolve a small fraction of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, poured water can abrade rust (via mechanical stress) and remove it of from the bulk metal, leaving the surface clean for producing more rust. This is how huge rods of metal can vanish in a few months in &lt;em&gt;corrosive&lt;/em&gt; environmental conditions (such as wet air or water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preventing rust requires the use of specific alloys and surface finishing. You can also add a galvanic electrode (generally made of $\ce{Zn}$, a so-called sacrificial anode) to protect submerged metallic structures such as the hull of a boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in your example water is both a mediator of oxidation and mechanical abrasion agent.&lt;/p&gt;&#xA;" OwnerUserId="3879" LastEditorUserId="865" LastEditDate="2015-08-05T15:39:12.483" LastActivityDate="2015-08-05T15:39:12.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14473" PostTypeId="1" AcceptedAnswerId="14477" CreationDate="2014-07-20T07:56:34.137" Score="18" ViewCount="50443" Body="&lt;p&gt;Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry?&lt;/p&gt;&#xA;" OwnerUserId="7181" LastEditorDisplayName="user7951" LastEditDate="2016-12-05T12:20:56.857" LastActivityDate="2017-08-22T13:50:16.390" Title="How much can the pH change through dilution?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14475" PostTypeId="1" CreationDate="2014-07-20T08:32:23.503" Score="9" ViewCount="1376" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the final temperature and the change in internal energy when $\pu{500 J}$ of energy is transferred as heat to $\pu{0.900 mol}$ $\ce{O2(g)}$ at $\pu{298 K}$ and $\pu{1.00 atm}$ at constant pressure. Treat the gas as ideal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this? For example, does having constant pressure automatically mean there will be a temperature change and if so, why? What about volume? Why do we not need to know the volume for this equation? Basically, I want to understand what's going on conceptually rather than memorize a few rules. Any help would be much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="7182" LastEditorUserId="5026" LastEditDate="2018-02-19T15:09:31.780" LastActivityDate="2018-02-19T15:09:31.780" Title="On the process of heating a gas at constant pressure" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14476" PostTypeId="2" ParentId="14473" CreationDate="2014-07-20T10:27:04.713" Score="10" Body="&lt;p&gt;You forget that pure water already contains $\ce{H_3O^{+}}$ ($\ce{Kw}=\ce{[H_3O+][OH^{-}]}=14$,). If you have an acidic solution, you will be able to reach the neutral point ($\ce{pH} = 7$) after enough dilution, but it will never be alkaline. You can see this as similar to a limit or infinite sum $\lim_{x\rightarrow \infty } \frac{x}{x+1}=1$, but when you look at finite $x$ it will never actually be 1.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-07-20T10:27:04.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14477" PostTypeId="2" ParentId="14473" CreationDate="2014-07-20T10:46:30.363" Score="26" Body="&lt;p&gt;Water undergoes &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot;&gt;autoionization&lt;/a&gt;, i.e., it reacts as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + OH-}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-20T12:01:15.543" LastActivityDate="2014-07-20T12:01:15.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14478" PostTypeId="1" AcceptedAnswerId="14481" CreationDate="2014-07-20T12:38:27.147" Score="6" ViewCount="2572" Body="&lt;p&gt;Ok, most people including me know that formal charge is just a book-keeping tool. I know how to calculate it, apply it to Lewis structures etc etc. But I am confused in its very basic definition which says - &quot;&lt;strong&gt;&lt;em&gt;Formal charge is the charge on an atom if the electronegativity differences are ignored&quot;&lt;/em&gt;&lt;/strong&gt; . If the electronegativity difference is ignored, then how can an atom have charge ? I know this is not a place to ask such a basic question , but still I am very confused about this miniscule definition.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastActivityDate="2018-08-19T18:22:36.520" Title="The definition of formal Charge" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;atoms&gt;&lt;lewis-structure&gt;&lt;formal-charge&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14479" PostTypeId="1" AcceptedAnswerId="14497" CreationDate="2014-07-20T13:01:32.260" Score="2" ViewCount="887" Body="&lt;p&gt;Just to explain this question in a better way just think about the Rutherford's experiment(the alpha particle bombarded on Gold foil) be conducted using a electron gun in place of the Alpha one.. So what will happen when electron is targeted on a gold foil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either the atom rather the electrons of Ag atom become excited and therefore emit some sort of waves(if yes,then what kind of waves?), or in return electrons are ejected from the atom along with some waves,  or it may be possible that the electron pass by the gold foil without any effect...&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="4945" LastEditDate="2014-07-20T13:56:02.110" LastActivityDate="2014-07-21T16:09:14.243" Title="Bombarding atoms with electron gun?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14480" PostTypeId="2" ParentId="14436" CreationDate="2014-07-20T13:05:13.190" Score="5" Body="&lt;p&gt;In case anyone else travels this road...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\mathrm{\Delta G~=~-RTlnK_{eq}}$ and the 3 $\mathrm{\Delta G's}$ supplied above, we find that&#xA;\begin{aligned}&#xA;\mathrm{K^{1/4}_{eq}}~&amp;amp;=~0.003\\&#xA;\mathrm{K^{3/1}_{eq}}~&amp;amp;=~40.749\\&#xA;\mathrm{K^{2/3}_{eq}}~&amp;amp;=~2.163\\&#xA;\end{aligned}&#xA;We also know that the concentrations of the 4 species must total to 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${[1] + [2] + [3] + [4] = 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have the following 4 equations and 4 unknowns&#xA;\begin{aligned}%&#xA;[1]~ &amp;amp;=~~~ 0.003 ~[4]\\&#xA;[3]~ &amp;amp;=~ 40.749 ~[1]\\&#xA;[2]~ &amp;amp;=~~~ 2.163 ~[3]\\&#xA;[1] + [2] + [3] + [4]~ &amp;amp;=~ 1\\&#xA;\end{aligned}&#xA;solving for the 4 concentrations yields&#xA;\begin{aligned}%&#xA;[1]~ &amp;amp;=~~~ 0.23\%\\&#xA;[2]~ &amp;amp;=~ 20.27\%\\&#xA;[3]~ &amp;amp;=~~~ 9.37\%\\&#xA;[4]~ &amp;amp;=~ 70.13\%\\&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-13T19:39:04.247" LastActivityDate="2016-03-13T19:39:04.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14481" PostTypeId="2" ParentId="14478" CreationDate="2014-07-20T13:38:21.310" Score="5" Body="&lt;p&gt;A good description appears in the book, &quot;&lt;a href=&quot;http://books.google.com/books?id=xta7AgAAQBAJ&amp;amp;pg=PA381&amp;amp;lpg=PA381&amp;amp;dq=Formal%20charge&amp;amp;source=bl&amp;amp;ots=dq_xjUjh6e&amp;amp;sig=QpLrl2OdXQTRQaghEec5F2o9GK0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=-cHLU5TQAcb-oQTS3oCQBg&amp;amp;ved=0CCUQ6AEwAQ#v=onepage&amp;amp;q=Formal%20charge&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry A Molecular Approach&lt;/a&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the molecule of hydrogen fluoride, we know that it has a dipole moment, and fluoride is slightly negative. Ignoring this information (electronegativity difference) and we share the bonding electrons equally, the formal charge can be calculated, and both of them are determined to be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, we know that fluoride has the electron from the hydrogen most of the time (due to electronegativity difference). We say the hydrogen has a partial positive charge and the fluoride has a partial negative charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, it is apparent that the formal charge is different from the charge on the atoms in a molecule.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="9961" LastEditDate="2018-08-19T18:22:36.520" LastActivityDate="2018-08-19T18:22:36.520" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14482" PostTypeId="1" AcceptedAnswerId="81655" CreationDate="2014-07-20T14:09:27.940" Score="39" ViewCount="53792" Body="&lt;p&gt;I was told in my organic chemistry course that $\text{S}_\text{N}1$ and $\text{S}_\text{N}2$ reactions did not occur at $\text{sp}^2$ centres. When I asked why, I was not given a satisfactory explanation. For $\text{S}_\text{N}2$ it was suggested that the reaction could not proceed with inversion of configuration, as this would disrupt the orbital overlap causing the $\pi$ bond. I couldn't find any explanation in Clayden's &lt;em&gt;Organic Chemistry&lt;/em&gt; (2nd ed.), or even a mention of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please explain why? I have attached some curly arrow diagrams that seem reasonable enough to me, although I realise that being able to draw a reaction mechanism does not mean it is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iXFLU.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iXFLU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7185" LastEditorUserId="16683" LastEditDate="2018-10-25T23:48:47.533" LastActivityDate="2018-10-25T23:48:47.533" Title="Why do SN1 and SN2 reactions not occur at sp2 centres?" Tags="&lt;organic-chemistry&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="14483" PostTypeId="2" ParentId="14482" CreationDate="2014-07-20T15:05:28.793" Score="29" Body="&lt;p&gt;Sometimes, especially in introductory courses the instructor will try to keep things &quot;focused&quot; in order to promote learning.  Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, and therefore somewhat less common. Consider when a nucleophile reacts with a carbonyl compound, the nucleophile attacks the carbonyl carbon atom in an $\ce{S_{N}2}$ manner.  The electrons in the C-O $\pi$&amp;ndash;bond can be considered as the leaving group and a tetrahedral intermediate is formed with a negative charge on oxygen. It is harder to do this with a carbon-carbon double bond (energetically more costly) because you would wind up with a negative charge on carbon (instead of oxygen), which is energetically less desirable (because of the relative electronegativities of carbon and oxygen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TbdD7.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the Michael addition reaction, the 1,4-addition of a nucleophile to the carbon-carbon double bond in an $\ce{\alpha-\beta}$ unsaturated carbonyl system, this could be viewed as an $\ce{S_{N}2}$ attack on a carbon-carbon double bond, but again, it is favored (lower in energy) because you create an intermediate with a negative charge on oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBypJ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S_{N}1}$ reactions at $\ce{sp^2}$ carbon are well documented.  Solvolysis of vinyl halides in very acidic media is an example.  The resultant vinylic carbocations are actually stable enough to be observed using nmr spectroscopy.  The picture below helps explain why this reaction is so much more difficult (energetically more costly) than the more common solvolysis of an alkyl halide.  In the solvolysis of the alkyl halide we produce a traditional carbocation with an empty p orbital.  In the solvolysis of the vinyl halide we produce a carbocation with the positive charge residing in an $\ce{sp^2}$ orbital. Placing positive charge in an $\ce{sp^2}$ orbital is a higher energy situation compared to placing it in a p orbital (electrons prefer to be in orbitals with higher s density, it stabilizes them because the more s character in an orbital the lower its energy; conversely, in the absence of electrons, an orbital prefers to have high p character and mix the remaining s character into other bonding orbitals that do contain electrons in order to lower their energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LhtRL.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7448" LastEditDate="2015-11-20T12:43:25.970" LastActivityDate="2015-11-20T12:43:25.970" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14484" PostTypeId="1" AcceptedAnswerId="14486" CreationDate="2014-07-20T17:16:20.890" Score="1" ViewCount="94" Body="&lt;p&gt;Can somebody help me in identifying the name of $\ce{SnMg2O4}$?&lt;br&gt;&#xA;Other examples are $\ce{ZnAl2O4}$ (zinc aluminate), $\ce{ZnIn2S4}$ (zinc thioindate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please kindly provide the reference if possible.&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="41328" LastEditDate="2017-09-24T14:42:42.613" LastActivityDate="2017-09-24T14:42:42.613" Title="Finding the chemical name of SnMg₂O₄" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14485" PostTypeId="1" AcceptedAnswerId="14503" CreationDate="2014-07-20T17:25:29.490" Score="2" ViewCount="28600" Body="&lt;p&gt;Transition elements can form ions with different charges. Why can't elements other than transition elements form ions with different charges? If it is a Fluoride ion, why does it have to be an anion of charge -1? Why can't it have a -2 charge, or more? &lt;/p&gt;&#xA;" OwnerUserId="7186" LastActivityDate="2014-07-21T17:38:29.060" Title="Why does a Fluoride ion only have a -1 charge and not a -2 charge or more?" Tags="&lt;ionic-compounds&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14486" PostTypeId="2" ParentId="14484" CreationDate="2014-07-20T17:27:19.923" Score="3" Body="&lt;p&gt;$\ce{Mg2SnO4}$ Magnesium Orthostannate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MgSnO3}$ Magnesium Metastannate&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-20T17:27:19.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14487" PostTypeId="2" ParentId="14485" CreationDate="2014-07-20T17:45:06.923" Score="0" Body="&lt;p&gt;The fluoride anion can have a negative two charge. It's not impossible to add two electrons to the fluoride anion. But at that point it would be rather unstable as the fluoride anion with a negative 1 formal charge is isoelectronic with a noble gas - neon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, consider the high charge density of the fluoride anion. Tacking on an extra electron to fluorine is OK, but not as favorable relative to the less charge dense chlorine atom. Fluorine actually has a lower electron affinity than chlorine despite fluorine's higher electronegativity. Tacking on two extra electrons would be unfavorable. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-07-20T17:45:06.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14488" PostTypeId="2" ParentId="14485" CreationDate="2014-07-20T17:47:31.610" Score="2" Body="&lt;p&gt;The electronic structure of a fluorine atom is $\ce{1s^2 2s^2 2p^5}$.  There is a strong driving force for atoms to attain an octet (achieve an inert gas configuration) due to the extra stability associated with a filled shell of electrons.  In order to complete its octet and achieve the neon inert gas configuration ($\ce{1s^2 2s^2 2p^6}$), fluorine must gain 1 electron and become the fluoride anion ($\ce{F^{-}}$).  If it gains two electrons and becomes $\ce{F^{-2}}$, or loses 1 electron and becomes $\ce{F^{+}}$, it won't have an octet in its outer shell - it won't be anywhere near as stable as if it just gains one electron.  The same reasoning explains why sodium, for example, prefers to lose only one electron. Transition elements have larger electronic configurations and may need to gain or lose 3 or 4 electrons to achieve an inert gas configuration.  It requires a lot of energy to add or remove 3 or 4 electrons, so adding or removing 1 or 2 electrons, while not achieving an inert gas configuration, becomes an acceptable alternative because of the lower energy required to remove fewer electrons.  In these cases, oxidation states of (plus or minus) 1, 2, 3 or 4 become feasible.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-20T17:47:31.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14490" PostTypeId="1" AcceptedAnswerId="14494" CreationDate="2014-07-20T18:03:12.023" Score="4" ViewCount="53345" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Complete and balance the following molecular equation (in aqueous solution); include phase labels. Then, write the net ionic equation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$~\ce{NH_3 + HNO_3 -&amp;gt; ?}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I thought that the acid &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; would just give its hydrogen to &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; and make the resulting reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH_3 + HNO_3 -&amp;gt; HNH_3 + NO_3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However the correct answer is &lt;span class=&quot;math-container&quot;&gt;$\ce{NH_3 + HNO_3 -&amp;gt; NH_4NO_3}$&lt;/span&gt;. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2019-10-05T06:41:27.933" Title="Neutralization reaction between ammonia and nitric acid" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14491" PostTypeId="1" CreationDate="2014-07-20T18:08:59.640" Score="0" ViewCount="251" Body="&lt;p&gt;I would like to know what is valence molecular connectivity index of 3rd order cluster. Please provide an example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="402" LastEditDate="2016-11-05T18:29:44.770" LastActivityDate="2016-11-05T18:29:44.770" Title="Molecular connectivity index" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14492" PostTypeId="2" ParentId="14392" CreationDate="2014-07-20T18:14:26.333" Score="4" Body="&lt;p&gt;You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air.  There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote.  Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water.  That's 0.01%.  Whether this much water ruins your alcohol or not, you will know better than anyone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time you open the tote, you're potentially letting in 40g of water.&#xA;I say &quot;potentially&quot; because, since the gas content is not instanty exchanged, much less water than that might go in.  But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-20T18:19:58.983" LastActivityDate="2014-07-20T18:19:58.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14493" PostTypeId="1" AcceptedAnswerId="14496" CreationDate="2014-07-20T18:45:03.333" Score="6" ViewCount="606" Body="&lt;p&gt;Today I was pipetting some cyclohexylamine into smaller bottles so that I don't have to use the liter flask every time I need a little bit of catalyst or plasticizer. I pipetted several 10-20 mL portions in the open air on a regular table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I lack experience I'm extra precautious with what I do. From the MSDS's of cyclohexylamine I found that 10 mL will kill an adult male, which if you look at is isn't very much, less than a swig of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I'm wearing a lab coat, safety goggles and chemical resistant gloves, but a few things concerned me (or at least made me wonder if it was the right/safe way to handle). According to &lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9923628&quot; rel=&quot;nofollow&quot;&gt;the MSDS of ScienceLab&lt;/a&gt; the odor threshold is 26 ppm, while &lt;a href=&quot;http://cameochemicals.noaa.gov/chemical/496&quot; rel=&quot;nofollow&quot;&gt;according to CAMEO&lt;/a&gt;, the PAC-3 for cyclohexylamine is 30 ppm, which Wikipedia defines as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PAC-3 : Life-threatening health effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I opened the bottle the amine fish odor was very strongly noticeable, I however found no irritating effect on my eyes or lungs, although according to the documents provided above I was inhaling a lethal concentration. I'm quite confused here. More easily put as a question: as long as I don't feel irritation of my lungs or eyes, am I safe? And if I do, is there enough time to clean up (i.e. the most essential things, like shutting off electricity and closing open jars of chemicals) and leave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I accidentally touched a drop of cyclohexylamine (from the pipette) when I was cleaning up, with no gloves on. Although cyclohexylamine is listed as highly corrosive, I did not feel any irritation at all. Then I rinsed my pipette 3x with distilled water and put the waste in a jerrycan, but I still haven't figured out if I can pour the diluted mixture of water and cyclohexylamine down the drain, without poisoning myself or the neighborhood. The MSDS's keep quiet about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone clear some confusions up for me. I will be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="865" LastEditDate="2015-10-16T17:53:24.637" LastActivityDate="2015-10-16T17:53:24.637" Title="Safety when handling cyclohexylamine in non-industrial amounts" Tags="&lt;safety&gt;&lt;home-experiment&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14494" PostTypeId="2" ParentId="14490" CreationDate="2014-07-20T19:02:32.770" Score="12" Body="&lt;p&gt;Your answer is very close to the answer given, except for the following two tidbits (the first being more significant).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An acid-base reaction is not the exchange of a hydrogen atom &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;. It is the exchange of a hydrogen ion (or proton) &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;. Thus your answer should be: &lt;span class=&quot;math-container&quot;&gt;$$\ce{NH3(aq) +HNO3(aq) -&amp;gt; NH4+(aq) + NO3-(aq)}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The given answer combines the two ions produced into a single compound. &lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4+(aq) + NO3-(aq) -&amp;gt;NH4NO3(aq)}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The result of #2 would be reasonable if you had not been told that the reaction was occurring in aqueous solution. In aqueous solution, the products should have been those of #1, since ammonium nitrate is freely soluble in water.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="23561" LastEditDate="2019-02-28T01:08:47.197" LastActivityDate="2019-02-28T01:08:47.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14495" PostTypeId="2" ParentId="13783" CreationDate="2014-07-20T19:36:07.313" Score="4" Body="&lt;p&gt;When acid is added to sodium chlorite in solution (note: $\ce{ClO2-}$ is the chlorite ion), chlorous acid ($\ce{HClO2}$) is transiently formed which then goes on to decompose. However, you need hydrochloric acid, not just any acid, because chloride ion ($\ce{Cl-}$) itself acts as a catalyst for the decomposition. What happens is that chlorous acid disproportionates; that is, some of it gets oxidized while the rest gets reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, out of five $\ce{Cl}$ atoms in chlorous acid (oxidation state +3), one will end up as chloride in the oxidation state $-1$, and four will end up as $\ce{ClO2}$ in the oxidation state +4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{5HClO2 -&amp;gt; 4ClO2 + Cl- + H+ + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much any concentration will do. Calculating the proportions is easy if you know a bit of stoichiometry: you need 5 moles of chlorite for 4 moles of hydrochloric acid. The reaction proceeds to completion, and you get 4 moles of $\ce{ClO2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{ClO2}$ remains in solution and is a good bleaching agent; see &lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorine_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. It can undergo further disproportionation but if the solutions are kept in the dark, this process is slow.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2017-09-07T20:49:50.683" LastActivityDate="2017-09-07T20:49:50.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14496" PostTypeId="2" ParentId="14493" CreationDate="2014-07-20T20:28:15.037" Score="8" Body="&lt;p&gt;I once did some work with some small chain alkyl amines and they were bad news.  They would actually burn you if some got on your skin.  I always used a respirator and worked in a hood when using them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;as long as I don't feel irritation of my lungs or eyes, am I safe?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this is a one time occurrence, given the small quantities, I'd say the odds are way in your favor, but they call it &quot;odds&quot; for a reason.  My advice would be not to do it again without a respirator and hood where the exhaust is treated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if I do, is there enough time to clean up (i.e. the most essential&#xA;  things, like shutting off electricity and closing open jars of&#xA;  chemicals) and leave?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ask an industrial hygiene professional, I'm sure the answer would be &quot;no&quot;.  If it were me and I just began to notice some very mild sensation, I think I'd hold my breath and try to shut things down.  I know it doesn't make sense, but that's what I'd do.  If I noticed anything beyond the onset of very minor discomfort, I'd clear out fast and call the fire department.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally I accidentally touched a drop of cyclohexylamine (from the&#xA;  pipette) when I was cleaning up, with no gloves on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd keep washing it every so often during the day and I'd put some burn cream on it in between the washings.  I'd also take a good shower.  No harm in overreacting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;still haven't figured out if I can pour the diluted mixture of water&#xA;  and cyclohexylamine down the drain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm almost positive that the answer is no.  Besides you and the neighborhood, it might react with the pipes or what's stuck in the pipes and create longer term toxicity problems when you're in that room at some later date.   Where I live, we have hazardous chemical drop-off centers (mainly for paint, oil and household chemicals, but used for everything chemical).  I'd pay the relatively small amount they charge and drop it off there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final note, even when using the right protective gear, some people become &quot;sensitized&quot; to amines.  This means that future exposure at even lower levels to a wide array of amines can be more dangerous.  Also, I'd try to open up the exposed room to as much outside air as possible, use a room fan too if you have one.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-20T21:05:47.553" LastActivityDate="2014-07-20T21:05:47.553" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14497" PostTypeId="2" ParentId="14479" CreationDate="2014-07-20T20:43:48.333" Score="2" Body="&lt;p&gt;Well, I think, in principle, depending on the energy of electrons, different events can happen, but in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_diffraction#History&quot; rel=&quot;nofollow&quot;&gt;classical experiment&lt;/a&gt; electrons will diffract from the spacing within crystal lattice of gold in the foil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. The experiment is actually very well known in the history of science due to the following reason (quoted from &lt;a href=&quot;http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourquan/thomps.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is interesting to recall that G.P. Thompson, who shared the 1937&#xA;  Nobel Prize with Davisson for these experiments which proved that&#xA;  electrons are waves, is the son of J.J. Thompson who received the&#xA;  Nobel Prize in 1906 for proving that cathode rays were actually&#xA;  particles - electrons! And the amazing thing is that they were both&#xA;  right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-07-20T20:43:48.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14498" PostTypeId="1" CreationDate="2014-07-20T20:54:38.220" Score="3" ViewCount="5190" Body="&lt;p&gt;What kind of glass can I purchase that will melt in a Bunsen burner flame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to treat glass such that the melting point is lowered? I tried adding some sodium hydroxide but the glass shattered when heated. &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="23561" LastEditDate="2019-03-14T13:59:37.643" LastActivityDate="2019-03-14T13:59:37.643" Title="Glass with a low melting point?" Tags="&lt;home-experiment&gt;&lt;materials&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14499" PostTypeId="2" ParentId="10166" CreationDate="2014-07-20T23:34:10.633" Score="0" Body="&lt;p&gt;Nicholas, Cr(III) is generally 6-coordinate and both bipy and en are bidentate.  So 3 bipy in (A) connect to the metal to form the cation, leaving 3 $\ce{Cl-}$ in the outer sphere.  Then in (B) one bipy leaves and its two positions are occupied by $\ce{Cl}$, and as you say the configuration must be cis (both because a bidentate ligand departed and because another can easily come in).  Isomerization of (B) to (D), the trans isomer, probably occurs slowly, and (D) cannot accept en because the $\ce{Cl}$ sites are opposite instead of adjacent.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-20T23:34:10.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14500" PostTypeId="2" ParentId="14485" CreationDate="2014-07-21T00:00:53.710" Score="1" Body="&lt;p&gt;Some main group elements do form multiple ions, however, analogues to the transition elements are less common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_ion&quot; rel=&quot;nofollow&quot;&gt;Hydrogen&lt;/a&gt; - Hydrogen forms both the $\ce{H+}$ ion (the proton) and the $\ce{H-}$ ion (the hydride)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_chloride&quot; rel=&quot;nofollow&quot;&gt;Tin&lt;/a&gt; forms the $\ce{Sn^{2+}}$ and $\ce{Sn^{4+}}$ cations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-21T00:00:53.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14501" PostTypeId="2" ParentId="14493" CreationDate="2014-07-21T00:11:03.947" Score="6" Body="&lt;p&gt;There are some general points that need to be made about this question, and some will not directly answer your question. They may, hopefully be useful for future generations of home-chemists seeking answers on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, all of these very practical questions should have been asked well before you carried out this process. Hindsight is a wonderful gift. Prior to any chemical handling process, all industrial and research chemists will carry out an extensive Hazard Identification and Risk Assessment. &lt;strong&gt;Prior&lt;/strong&gt; to taking possession of this chemical you should be aware of how to store, handle and dispose of your chemicals. Disposal management includes spill control and safe disposal of unwanted waste materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, spilling some material on yourself during cleanup indicates that you were not sufficiently protected. Gloves, glasses, labcoats and other protective equipment need to be used during cleanup as well. Accidents can happen at any stage, and you need to be prepared for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, if you are using a hazardous chemical in an environment where you can so readily smell it suggests to me that you have inadequate ventilation in your work area. You should address this immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the first Risk Control measure the home chemist (any chemist!) needs to consider is subsitution. Is this really the only suitable reagent for you to use, or can you substitute it for another, less harmful, chemical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexylamine is a nasty chemical. It has a GHS signal word DANGER, which should immediately make any handler seek further information prior to using it. It is classed as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Category 3 flammable liquid, &lt;/li&gt;&#xA;&lt;li&gt;Category 2 Oral Acute Toxicity, &lt;/li&gt;&#xA;&lt;li&gt;Cat 3 Dermal Acute Toxicity, &lt;/li&gt;&#xA;&lt;li&gt;Cat 1 Skin corrosion, &lt;/li&gt;&#xA;&lt;li&gt;Cat 1 Serious Eye Damage and &lt;/li&gt;&#xA;&lt;li&gt;Cat 2 Reproductive Toxicity. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes, Cyclohexylamine is a nasty chemical. Now, on to your questions.......&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;as long as I don't feel irritation of my lungs or eyes, am I safe?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, no, no, good god, no. You will be subject to the dangerous effects of any chemical as long as you are exposed to them. Many effects will not be instantaneous; many can take hours or days (or even weeks) to manifest symptoms. I recall a case in the mid-90's, as an example, where a postgrad student died after cleaning up a spill of acrylonitrile in a lab. Symptoms of distress only became apparent several hours after the incident, and the person was hospitalized and died within 24 hours. Acrylonitrile essentially dissolves the tissues of the lungs. Not a nice way to die. Your lungs and eyes are the most sensitive areas of your body, and are likely to show irritation first. Consider, however, that your skin (especially your face) may have a much greater area of exposure, and this may be a pathway for contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PAC guidelines you ask about relate to extended periods of exposure. It does not mean that if you inhale 30ppm of a particular chemical you will experience life-threatening health effects immediately. Perhaps a more relevant and practical exposure guide is the Acute Exposure Guideline Levels (listed in your first reference), which shows that extended exposure at 30ppm for an hour could lead to life-threatening health effects or death. Or 10 minutes at 38ppm. &lt;strong&gt;The take home message from this is that if you can smell it, you are working in a dangerous environment.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I accidentally touched a drop of cyclohexylamine (from the pipette) when I was cleaning up, with no gloves on&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly, your previous statement &quot;Of course I'm wearing a lab coat, safety goggles and chemical resistant gloves&quot; was not quite correct. Personal safety equipment should be the first thing to go on, and the last thing to come off. I hope you treated the exposed area according to the MSDS first aid advice. Like other adverse chemical affects, corrosive properties do not need to be instantaneous. They can take a long time to manifest symptoms; HF, for instance, does not produce inflammation or burns until many hours after exposure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I still haven't figured out if I can pour the diluted mixture of water and cyclohexylamine down the drain, without poisoning myself or the neighborhood.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many SDS do not explicitly describe safe disposal considerations, and you should probably seek a more comprehensive source. Chemical suppliers should now be moving toward the GHS system for chemical management, and the SDS from these companies should include (albeit brief and generic) disposal considerations. However, this is a chemical which is toxic to the environment and should be disposed of correctly. Most local Council or Municipal Waste Management Facilities will have chemical collection days, or specific chemical drop-off centres to receive these types of items. Speak to your local government authority, and seek advice from them. Occasionally, for some chemicals, you can return any unused portions to the place of sale. Normally, these types of chemicals are burnt in a chemical incinerator with afterburner and scrubber. You should collect waste mixtures in an appropriate container and store safely until ready for disposal. For small volumes of waste, you may consider using the spill control guidelines, which may offer a more practical approach to providing a safer chemical to store and dispose of.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-07-21T00:11:03.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14502" PostTypeId="1" AcceptedAnswerId="14527" CreationDate="2014-07-21T02:00:59.893" Score="2" ViewCount="314" Body="&lt;p&gt;My Main purpose is using quantum chemistry for simulation of some reactions. Actually we are working on making some new polymers and I need to simulate some molecular properties of these polymers after adding some molecules to them. These include bond breaking and bond making as well as evaluation of optical properties.&lt;br&gt;&#xA;That's why I started studying quantum chemistry models. I started with the Schödinger Mechanical Model and then Hartree Fock Theory, but as everyone knows these models are not that efficient for molecules larger than lithium. Then I started learning DFT and got interested in semi-empirical models, but could not learn them yet. Finally I am absolutely confused about which method should I select.&lt;br&gt;&#xA;This is my question: As I have not enough time to learn all those quantum chemistry models/techniques, which method is currently best option if I just want to learn ONE method. (I have access to some super-computers in our university.) &lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="4945" LastEditDate="2015-12-14T09:33:01.050" LastActivityDate="2017-03-29T04:20:50.663" Title="How to choose the level of theory for modelling reactions of polymers?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14503" PostTypeId="2" ParentId="14485" CreationDate="2014-07-21T04:46:38.507" Score="5" Body="&lt;p&gt;The octet rule is usually used to explain the charges that atoms in groups I, II, VI, and VII have in ionic compounds, by saying something like &quot;noble gases are stable, therefore having 8 valence electrons must be stable for some reason.&quot; This is true, but understanding &lt;em&gt;why&lt;/em&gt; noble gas configurations are stable is important, and without knowing that, it does seem like magic that some atoms &lt;em&gt;don't&lt;/em&gt; adopt a noble gas configuration in ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The octet rule is a useful way to quickly predict ionic charges and write Lewis structures, but it's not a magical law that atoms follow. They don't &quot;want&quot; or &quot;try&quot; to obtain specific electronic structures, although we often talk like they do. What is really important is the total energy of the electron configuration. The electron configuration with the lowest energy is the most stable one.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Stability&quot; really means &quot;lowest energy&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The total energy of the electron configuration is governed by two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The energy of the &lt;strong&gt;electron orbitals&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;effective nuclear charge&lt;/strong&gt; that is &quot;seen&quot; by the electrons&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The orbital energies determine the order in which atomic orbitals will be populated by electrons in the ground state - in other words, which orbitals the electrons will exist in for a given atom. We know from empirical evidence and from quantum mechanical calculations that the order goes like this for most elements (not for some transition metals and heavier elements):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h9kW0.png&quot; alt=&quot;Electron orbital energy levels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons will occupy these orbitals from the &quot;ground up&quot; (called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aufbau principle&lt;/a&gt;). We know, therefore, that for 4th-period elements, the $3d$ sub-shell will fill before the $4p$ sub-shell will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic orbital diagram indirectly tells us something else. The principle quantum number, $n$ (which for the valence shell is equal to the row or period number), roughly corresponds to the distance from the nucleus of the orbital's highest electron density region. In other words, any sub-shell with $n=4$ will have most of its electron density further away from the nucleus than a sub-shell with $n=3$. This means that $3d$ electrons are &lt;em&gt;closer to the nucleus&lt;/em&gt; than $4s$ electrons, even though $3d$ electrons initially have  &lt;em&gt;higher energy.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The distance of the electron from the nucleus is roughly proportional to $n$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the electrons have a negative charge, and the nucleus is positively charged, there is an electrostatic potential energy between the electrons and the nucleus. If the electrons were uniformly distributed throughout the atom, we would expect that the nuclear charge &quot;felt&quot; by each electron would be equal to the number of protons - the atomic number. However, because of the &quot;rules&quot; of quantum mechanics (the allowed solutions to Schrodinger's equation), electrons are &lt;strong&gt;not&lt;/strong&gt; uniformly distributed - they occupy orbitals in a specific arrangement. Since some electrons are closer to the nucleus (lower $n$) than others, this means that the nuclear charge is partially screened by the inner electrons, and the outer electrons experience an &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective nuclear charge&lt;/a&gt;&lt;/strong&gt; that is somewhat less than the full nuclear charge. This diagram from wikipedia demonstrates the idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ky3Wa.png&quot; alt=&quot;Effective nuclear charge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this one shows the idea using the Bohr model (note that this is a schematic - not an accurate picture of what electrons &quot;look&quot; like in orbitals - only the relative distances from the nucleus is right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOAjM.png&quot; alt=&quot;Effective nuclear charge schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the valence electrons experience a large effective nuclear charge, they are harder to remove (the energy cost for removing them increases), and when the effective nuclear charge is lower, they are easier to remove. The actual effective nuclear charge is very difficult to accurately calculate, but it is very easy to estimate - all you need to do is take the number of protons, then subtract the number of &quot;screening&quot; electrons. Basically, we are assuming that each electron between the valence shell and the nucleus cancels out one proton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Z_{eff} = Z - S&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z_{eff}$ is the effective nuclear charge, $Z$ is the actual nuclear charge, and $S$ is the screening constant, here assumed to equal the number of &quot;core&quot; electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find $S$, all you have to do is count the number of electrons with $n$ less than the valence shell (highest $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking fluorine as an example, the electron configuration is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^5&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that $Z=9$ and $S=2$, giving an effective nuclear charge of +7. Compare that to lithium: $3-2=+1$. The valence electrons on fluorine are roughly 7 times harder to remove than the valence electron on lithium. In fact, any element to the left of fluorine has a lower effective nuclear charge for the valence shell, and will give up electrons more easily than fluorine will. In other words, it's not so much that fluorine won't give up electrons, it's that nothing can take them from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we add an electron? For $\ce{F-}$, the electron configuration becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^6&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{eff}$ is still +7, since the number of core screening electrons hasn't changed. This means that it is energetically favorable for fluorine to pick up an electron, since the additional electron would still &quot;see&quot; a large positive nuclear charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we add another, to make $\ce{F^{2-}}$? Now the configuration looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^63s^1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{eff}$ is now $9 - 10 = -1$ - note that the number of screening electrons jumped to 10 because the &quot;new&quot; valence shell has $n=3$. In other words, &lt;em&gt;because of the allowed orbital configurations&lt;/em&gt; the new electron would &quot;see&quot; a &lt;em&gt;negative&lt;/em&gt; charge on the nucleus - it would require energy to keep it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work through this process for all of the &quot;fixed charge&quot; elements, you find similar things - the effective nuclear charge is large for elements that gain electrons, until they gain too many (an octet), at which point it becomes negative. For fixed-charge metals, something else happens. Let's look at magnesium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg}$: $12-10=+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg^1+}$: $12-10=+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg^2+}$: $12-2=+10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you remove two electrons, the effective nuclear charge jumps to +10 for the remaining valence electrons. The same is true for all the group I and II metals - once you get to the noble gas &quot;core&quot; - the octect - the effective nuclear charge seen by the valence electrons jumps way up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why &quot;fixed charge&quot; elements always have the same charge in ionic compounds, but what about transition metals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at chromium. The electron configuration looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^63s^23p^64s^23d^4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which electrons are the valence electrons? The ones with the highest $n$ - the 4s electrons. The effective nuclear charge they &quot;see&quot; is $24 - 22 = +2$ - note that the 4 electrons in the $3d$ sub-shell are counted as screening, since they have $n=3$ which is less than 4, putting them &lt;em&gt;closer&lt;/em&gt; to the nucleus than the valence shell, despite their initially higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we remove an electron, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^63s^23p^64s^13d^4&#xA;$$&#xA;(This isn't exactly correct, there would really be no $4s$ electron and 5 $3d$ electrons, but the reason is complicated and it doesn't matter to this explanation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we removed a &lt;em&gt;valence electron&lt;/em&gt; - a $4s$ electron. Now $Z_{eff} = 24 - 22 = +2$ - it hasn't changed. We can remove one more to make $\ce{Cr^{2+}}$ and get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^63s^23p^63d^4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the definition of &quot;valence&quot; becomes a little less clear - we know that the highest $n$ value corresponds to the valence shell, but what about $l$, the orbital shape? It turns out that d orbitals are larger than s and p orbitals, and so the s and p electrons end up screening the nucleus from the d electrons. This means that d electrons would be removed next, and the effective nuclear charge &quot;seen&quot; by them is $24 - 18 = +6$. This is a larger charge, but not so large that other elements (notably oxygen and the halogens) can't still pull the d electrons away. This gives us a range of chromium charges (oxidation states) from +1 to +6. For $Cr^{6+}$, the electron configuration would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1s^22s^22p^63s^23p^6&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the effective nuclear charge is $24 - 10 =  +14$ - much too strong for anything else to &quot;win&quot; against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the rules of quantum mechanics determine the relative size of atomic orbitals and the order in which they are filled, which in turn determines the effective nuclear charge that valence electrons &quot;see&quot; for any given neutral atom or ion. The combination of the two leads to constraints on the charges that atomic ions can have, with the practical result being that group I, II, VI, and VII elements adopt a &quot;filled octet&quot; or &quot;noble gas&quot; configuration, while the other groups have more flexibility.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-07-21T17:38:29.060" LastActivityDate="2014-07-21T17:38:29.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14504" PostTypeId="1" AcceptedAnswerId="14506" CreationDate="2014-07-21T05:43:24.430" Score="5" ViewCount="573" Body="&lt;p&gt;Why is cetyl alcohol bent at the 9th position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOUiQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7189" LastEditorUserId="1974" LastEditDate="2014-07-21T16:53:03.500" LastActivityDate="2014-07-21T16:53:03.500" Title="Shape Of Cetyl Alcohol" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14505" PostTypeId="2" ParentId="14504" CreationDate="2014-07-21T06:04:22.447" Score="3" Body="&lt;p&gt;It is not.  This flat representation is schematic.  In fact the molecule can wobble a lot, and no C-C bond is more likely to be the site of a sharp bend than any other.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-21T06:04:22.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14506" PostTypeId="2" ParentId="14504" CreationDate="2014-07-21T06:05:48.307" Score="9" Body="&lt;p&gt;This is simply how the chemist chose to draw the molecule in that particular instance, and cetyl alcohol is formally regarded as a straight-chain molecule. Cetyl alcohol has all single bonds and no rings, so it is not structurally locked into any particular &lt;a href=&quot;http://en.wikipedia.org/wiki/Conformational_isomerism&quot;&gt;conformation&lt;/a&gt; or geometry, and every bond is free to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very simple saturated molecules like this, where the conformation is not in any way fixed (under typical conditions), the chemist is free to represent the molecule as one of any of its possible conformers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the most stable (i.e., lowest energy) conformer should statistically predominate, and this is the one with all groups &lt;a href=&quot;http://en.wikipedia.org/wiki/Staggered_conformation&quot;&gt;staggered&lt;/a&gt; and the largest groups antiperiplanar (q.v., the &lt;a href=&quot;http://goldbook.iupac.org/T06406.html&quot;&gt;IUPAC Gold Book entry on torsion angles&lt;/a&gt; for details), the conventional explanation being that this minimizes &lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_effects#Types_of_steric_effects&quot;&gt;steric&lt;/a&gt; hindrance/strain. The skeletal formula corresponding to this minimum-energy conformer is the simple straight-chain representation (seen &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=2682&quot;&gt;here&lt;/a&gt;, for example).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-21T07:18:07.440" LastActivityDate="2014-07-21T07:18:07.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14507" PostTypeId="1" CreationDate="2014-07-21T10:45:26.943" Score="5" ViewCount="558" Body="&lt;p&gt;I've been trying to understand why different alcohols have different effects on the brain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methanol is obviously poisonous, apparently because it's converted to formic acid. Because the body seems to cut organic substances in two's, any odd-numbered alcohol would seem to yield methanol, and therefore be poisonous. But why is 1-butanol (according to certain sites) more toxic than ethanol? Does it have a different effect on the receptors in the brain? And if so, why? Same with other simple even-numbered alcohols: why are their effects different from ethanol?&lt;/p&gt;&#xA;" OwnerUserId="7192" LastActivityDate="2014-07-27T04:41:25.423" Title="Why do different alcohols have different effects on the brain?" Tags="&lt;organic-chemistry&gt;&lt;toxicity&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14508" PostTypeId="1" AcceptedAnswerId="14521" CreationDate="2014-07-21T10:53:55.953" Score="2" ViewCount="1038" Body="&lt;p&gt;I would like to know what min and max values of densities are achievable with water solutions.&lt;br&gt;&#xA;There are some conditions:&lt;br&gt;&#xA;1. Normal or close to normal temperature;&lt;br&gt;&#xA;2. Non-radioactive solutions/components;&lt;br&gt;&#xA;3. Acids are good as long as they can be safely contained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="7193" LastActivityDate="2014-07-21T16:28:32.283" Title="High/low density water solution values" Tags="&lt;water&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14509" PostTypeId="2" ParentId="14485" CreationDate="2014-07-21T11:31:10.697" Score="0" Body="&lt;p&gt;Here's an easy argument to complement the others, which doesn't directly involve octets and electron shells. Compare the isoelectronic species  $\ce{Na^0}$, $\ce{Ne^{-}}$ and $\ce{F^{2-}}$, which all have the same $1s^22s^22p^63s^1$ fundamental configuration. Notice that going from left to right in the list, you are gradually removing a proton from the nucleus, making the valence electrons more poorly bound to the atom (less positive charge in the nucleus to pull the electrons inwards). Elemental metallic sodium is already very reactive because it has a comparatively loose valence electron (a low ionization energy). Removing two protons from the nucleus to generate $\ce{F^{2-}}$ would create a species &lt;strong&gt;much&lt;/strong&gt; more reactive than even metallic sodium(!), and it would be expected to immediately lose at least one electron in any conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In truth, $\ce{F^{2-}}$ is too unstable to be formed in the first place. Again, comparing the isoelectronic species in the list above, we look to $\ce{Ne^{-}}$. Neon actually already has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity#Electron_affinities_of_the_elements&quot; rel=&quot;nofollow&quot;&gt;electron affinity&lt;/a&gt; that is practically zero or even endergonic. In other words, the reaction $\ce{Ne^0_{(g)} + e^{-} -&amp;gt; Ne^{-}_{(g)}}$ already has very little drive to the right, if not actually being driven to the left. This suggests the reaction $\ce{F^{-}_{(g)} + e^{-} -&amp;gt; F^{2-}_{(g)}}$ (fluorine's second electron affinity) is strongly endergonic, and will not ever happen to a significant extent.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-07-21T11:31:10.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14510" PostTypeId="1" AcceptedAnswerId="14516" CreationDate="2014-07-21T11:45:04.907" Score="0" ViewCount="18934" Body="&lt;p&gt;Question is simple.. If we take an atom of any element and then supply heat energy to it then what will happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought is that in the beginning, energy (quanta; due to excitement of electron and then moving back to ground state) should be  evolved and then electron should be emitted... Am I right?&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="7448" LastEditDate="2015-07-03T11:58:32.067" LastActivityDate="2015-07-03T11:58:32.067" Title="What happens when we heat an atom?" Tags="&lt;atoms&gt;&lt;quantum-chemistry&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14511" PostTypeId="2" ParentId="14510" CreationDate="2014-07-21T12:54:58.270" Score="4" Body="&lt;p&gt;Well, strictly speaking, you can not heat an individual atom, because heat is a &lt;em&gt;macroscopic&lt;/em&gt; phenomenon: it is a mechanism of energy transfer from one macroscopic body to another one. And the same is true for the related notion of temperature.&#xA;But, both temperature and heat can be explained at &lt;em&gt;microscopic&lt;/em&gt; level as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particles that constitute bulk matter are in constant random motion and temperature is nothing but a measure of the mean &lt;em&gt;kinetic energy&lt;/em&gt; of particles random motion. Heat then can be understood as macroscopic manifestation of transfers of this kinetic energy of random motion from a statistically significant number of particles of one body to particles of another body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you can heat a body (not an individual atom), i.e. you can transfer energy to it by means of heat, and while the body is heated, the mean kinetic energy of all particles (atoms, nuclei, electrons) will rise, i.e. it temperature will rise.&#xA;The kinetic energy of an electron acquired then can be partly redistributed with its potential energy when electron ``jumps'' into higher orbital. After excitation the atom may return to the ground state or a lower excited state, by emitting a photon with a characteristic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, of course, in principle, we could add so much energy to a body be heating, that electrons will indeed leave atoms.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-07-21T12:54:58.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14512" PostTypeId="1" CreationDate="2014-07-21T12:58:08.377" Score="1" ViewCount="139" Body="&lt;p&gt;Could anyone please help to find the height of the potential barrier between silver and copper or $\ce{Ag-Cu}$. I am working on a piezo-resistive sensor based on copper fabric and silver electrodes. I was wondering if I can validate my experimental results with the model given by Zhang, X.-W., et al. &lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemistry expert and have a little knowledge on the height of the potential barrier. It would be highly appreciated if anyone could help me with providing the value for the height of the requested potential barrier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Zhang, X.-W., et al., Time dependence of piezoresistance for the conductor-filled polymer composites. Journal of Polymer Science Part B: Polymer Physics, 2000. 38(21): p. 2739-2749.&lt;/p&gt;&#xA;" OwnerUserId="7194" LastEditorUserId="10185" LastEditDate="2015-07-03T19:39:48.743" LastActivityDate="2015-07-03T19:39:48.743" Title="Height of potential barrier" Tags="&lt;electrochemistry&gt;&lt;reference-request&gt;&lt;silver&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14514" PostTypeId="2" ParentId="13589" CreationDate="2014-07-21T13:35:26.577" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/4-Nitrophenol&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;p&lt;/em&gt;-Nitrophenol&lt;/a&gt; is actually modestly soluble in water at neutral pH (16 g/l at 25 °C) because of its weak acidity ($p\mathrm{K_a}=7.16$ at 22 °C) and the formation of intermolecular hydrogen bonds between the phenolic OH group and water molecules. The latter would also explain why the compound dissolves in diluted $\ce{HCl_{aq}}$ despite being protonated at lower pH values. The importance of intermolecular hydrogen bonds is also demonstrated by the significantly lower solubility of the &lt;em&gt;ortho&lt;/em&gt; isomer, which forms intramolecular hydrogen bonds and therefore appears to be less polar to the solvent (&lt;a href=&quot;http://www.google.de/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=18&amp;amp;ved=0CFMQFjAHOAo&amp;amp;url=http%3A%2F%2Fchemistry.csudh.edu%2Ffaculty%2Fjim%2Fphenols.ppt&amp;amp;ei=zRTNU7q0PPT44QSEt4DYAw&amp;amp;usg=AFQjCNGGlPgxnL_gh7UZoFbESUub5m0DBQ&amp;amp;bvm=bv.71198958,d.bGE&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4v2L1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In basic solutions, &lt;em&gt;p&lt;/em&gt;-nitrophenol is deprotonated to &lt;em&gt;p&lt;/em&gt;-nitrophenolate. The anion is stabilized by delocalization of the negative charge into the benzene ring and the electron-withdrawing nitro group (only two of the possible resonance structures are shown):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2MIt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-07-21T13:35:26.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14515" PostTypeId="1" CreationDate="2014-07-21T13:57:00.800" Score="3" ViewCount="911" Body="&lt;p&gt;I found this &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3263647/&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; that says nicotine was found in major toothpaste brands in India. Rather than playing blame game, wouldn't it be easier to test ourselves ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to test for presence of Tobacco/nicotine in a product ?&lt;/p&gt;&#xA;" OwnerUserId="7195" LastEditorUserId="7448" LastEditDate="2015-10-09T15:50:28.003" LastActivityDate="2015-10-09T15:50:28.003" Title="How to test for presence of Tobacco/nicotine in a product?" Tags="&lt;toxicity&gt;&lt;identification&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14516" PostTypeId="2" ParentId="14510" CreationDate="2014-07-21T14:00:56.073" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If we take an atom of any element and then supply heat energy to it&#xA;  then what will happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A single atom can move in three dimensions (x, y, z) through space, it cannot rotate about an external axis like a diatomic molecule can, nor can it vibrate.  As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space.  If you supply enough energy, then the electrons surrounding the atom will absorb that energy and undergo electronic transitions from the ground state to various excited states.  This process can be diagrammed using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_potential&quot; rel=&quot;noreferrer&quot;&gt;Morse potential diagram&lt;/a&gt; , I've drawn a simple one below.  The diagram is meant to show promotion of an electron from the ground electronic state to higher and higher electronic levels as more energy is applied to the system.  Note that as the electron is promoted to higher energy levels, the average separation distance between it and the nucleus increases.  If enough energy is supplied, an electron will dissociate from the atom and we will be left with an ion.  Continue to supply even more energy and more electrons will be removed from the atom. Pumping even more energy into the system will ultimately lead to transformations in the nucleus of the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QR84r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-21T14:09:47.250" LastActivityDate="2014-07-21T14:09:47.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14519" PostTypeId="2" ParentId="14449" CreationDate="2014-07-21T15:30:13.733" Score="12" Body="&lt;p&gt;When salt (NaCl) crystals are grown under normal conditions, the edges of the cubes usually grow faster than the faces (because the edges have more contact with the saturated salt solution than the faces). So the crystals grow with a square-pyramidal indentation on each face. The &quot;X&quot; you observe is your view of the edges of that inverted pyramid. On the other hand, if the crystals are grown really slowly under controlled conditions, this does not occur and you get a more or less perfect cube.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-07-21T15:30:13.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14520" PostTypeId="2" ParentId="14479" CreationDate="2014-07-21T16:09:14.243" Score="2" Body="&lt;p&gt;Multiple interactions can happen.&lt;br&gt;&#xA;First, lets look at the &quot;simple&quot; case of an electron interacting with a single atom.  There are three options.  (1) the electron does not interact with the (neutral) atom - it misses completely and keeps on going.  (2) the electron may interact with the electron cloud of the atom.  This may lead to a scattering event, kicking a bound electron into a higher energy level, or knocking it off the atom entirely (generating the secondary electrons used for imaging in an SEM).  The atom can then relax, possibly releasing an x-ray or an Auger electron, both of which are used for composition analysis of materials.  The incident electron is deflected from the original path because of the scattering event. (3) the electron may interact with the nucleus and undergo Rutherford scattering (yup, same formula applies). &#xA;The more complex case of an electron impinging on a crystal adds the periodic potential of the crystal.  All of the events listed above apply, but in addition the periodic potential allows for diffraction, where the electron scatters off the Bloch waves of the crystal.  In a TEM diffraction can be imaged from the transmitted beam, in an SEM it can be imaged from the back-scattered beam (EBSD).&#xA;So, pretty much everything you mentioned can and does happen.  All are used for various measurements in materials science and physics. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-07-21T16:09:14.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14521" PostTypeId="2" ParentId="14508" CreationDate="2014-07-21T16:16:18.897" Score="1" Body="&lt;p&gt;Highest known is about 4.25 (all&#xA;densities in g/cm${}^3$) at room temperature:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Clerici_solution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Clerici_solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're willing to allow heating it comes close to 5 because the solubility of the salts increases with temperature:&#xA;&lt;a href=&quot;http://books.google.com/books?id=kaa2qeFRXmUC&amp;amp;pg=PA99&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=kaa2qeFRXmUC&amp;amp;pg=PA99&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite runner-up is the Thoulet solution mentioned at&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_tetraiodomercurate(II)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Potassium_tetraiodomercurate(II)&lt;/a&gt; but it&#xA;only reaches around 3.20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lowest would depend on what you mean by &quot;water solution&quot;. If you mix a&#xA;drop of water with a gallon of a very light liquid such as diethyl&#xA;ether (which is not very miscible with water but does dissolve some&#xA;-- hence the need to dry it in the lab over sodium or molecular&#xA;sieves), then the answer is the density of the liquid (ether = 0.73).&#xA;If water has to be the majority component I'd expect the&#xA;minimum to be very close to the density of water, since mixtures tend to decrease a bit in volume. (The density of water is 1.0 at 4${}^\circ$, close to 1.0 at room temperature, 0.94 near the boiling point).&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-21T16:28:32.283" LastActivityDate="2014-07-21T16:28:32.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14522" PostTypeId="2" ParentId="14449" CreationDate="2014-07-21T16:36:53.330" Score="11" Body="&lt;p&gt;Let's look at structure of $\ce{NaCl}$ ideal cubic crystal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ts8F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, atoms on the faces have 5 neighbors, atoms on the edges has four neighbors, and atoms on the vertices of the cube have 3 neighbors. This means, that a) edges and vertices pulls new atoms into crystal structure stronger and b) they 'collect' new atoms from larger volume, than faces. On the other hand, addition of atoms to an edge forms 'step' on the crystal surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NKKAn.png&quot; alt=&quot;steps on crystal surface&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A position near the step has three potential neighbors, so faces grows mostly on steps, and at exceptional speed. Given that, the fastest growing faces usually have 'spiraling' step&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MWCwG.png&quot; alt=&quot;spiraling step&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of unequal growth are expressed most at significantly oversaturated solutions, otherwise 'steps' travels fast enough over crystal surface to make a surface nearly flat. On the other hand, in case of highly oversaturated solution crystals grow mostly near verticies, edges and surface defects like 'spiraling step', leading to formation of so-known 'dendrites'. The most known example is snowflake (though perfect symmetry of snowflakes is a mystery on its own), the most suited for demonstrations example is growth of ammonium chloride crystals during natural evaporation of concentrated solution.  When the crystal in consideration grows on edges, it may form edges in the form of inverted pyramid, just like in your case. Growth of inverted pyramids is also wide known, and, may lead to quite &lt;a href=&quot;http://en.wikipedia.org/wiki/Bismuth#mediaviewer/File:Bismuth_crystals_and_1cm3_cube.jpg&quot; rel=&quot;noreferrer&quot;&gt;bizarre forms&lt;/a&gt; . In your case. the 'X' is, indeed, an edge of inverted pyramid on the crystal surface.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-21T16:36:53.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14523" PostTypeId="1" CreationDate="2014-07-21T16:42:47.207" Score="3" ViewCount="7363" Body="&lt;p&gt;In an electrochemical cell, the anode undergoes oxidation and the cathode undergoes reduction. This means that the anode &lt;em&gt;loses&lt;/em&gt; electrons while the cathode &lt;em&gt;gains&lt;/em&gt; electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the anode is negatively-charged while the cathode is positively-charged. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="7198" LastEditDate="2014-07-28T11:37:05.053" LastActivityDate="2015-07-14T06:54:29.887" Title="Electrochemical cell polarity and electrodes" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14524" PostTypeId="1" AcceptedAnswerId="14526" CreationDate="2014-07-21T17:33:15.513" Score="1" ViewCount="786" Body="&lt;p&gt;My prof noted that if the two components of any ionic compound undergo a large extent reaction with each other, then the substance CANNOT exist as a pure substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider ammonium hydroxide. This is simply ammonia (at least in the solution phase). Can we have a solid, crystalline ammonium hydroxide lattice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is thermodynamically valid. What about kinetically? Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any such examples? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what do you think of the below question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/siXxE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-21T21:18:05.047" LastActivityDate="2014-07-21T21:18:05.047" Title="Kinetic and Thermodynamic Stability" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14525" PostTypeId="2" ParentId="14523" CreationDate="2014-07-21T17:57:47.967" Score="5" Body="&lt;p&gt;The electrons which are &quot;given up&quot; by the oxidized species at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Anode&quot; rel=&quot;noreferrer&quot;&gt;anode&lt;/a&gt; flow from the anode through the electric circuit to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathode&quot; rel=&quot;noreferrer&quot;&gt;cathode&lt;/a&gt;, where they are &quot;needed&quot; for the reduction reaction. The oxidation creates an excess of electrons (negative charge) at the anode, which is compensated by the consumption of electrons at the electron-deficient cathode (positive charge) for the reduction reaction. In summary, the flow of electrons in the electrochemical cell is from anode (electron source) to cathode (electron drain).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-07-21T18:09:39.453" LastActivityDate="2014-07-21T18:09:39.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14526" PostTypeId="2" ParentId="14524" CreationDate="2014-07-21T18:32:04.740" Score="7" Body="&lt;p&gt;Chemistry is a field with far too many exceptions and edge cases to be making sweeping categorical statements. There are numerous examples of ionic compounds that are isolable and stable, the components of which will spontaneously decompose &lt;em&gt;under certain conditions&lt;/em&gt; (this is where reaction kinetics is relevant). Two that readily come to mind are ammonium nitrate ($\ce{NH4NO3}$) and ammonium bicarbonate ($\ce{NH4HCO3}$). Both of those can be isolated as pure salts, and both also can and will chemically decompose when circumstances are favorable (the former by multiple possible pathways depending on conditions).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't just theoretically possible, it's ubiquitous, it's the norm. If it weren't, human life couldn't exist. Combustion reactions are probably the most obvious and mundane example; just about every organic compound imaginable is usually more thermodynamically stable as a mixture of $\ce{CO2}$, $\ce{H2O}$, $\ce{N2}$, etc., but the kinetic barrier to the reaction (along with physical hindrances and the availability of an oxidizer, e.g., $\ce{O2}$) prevents this from occurring spontaneously without some trigger and the appropriate environmental conditions.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-21T18:38:59.460" LastActivityDate="2014-07-21T18:38:59.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14527" PostTypeId="2" ParentId="14502" CreationDate="2014-07-21T19:37:40.253" Score="9" Body="&lt;p&gt;The question is by far &lt;em&gt;not&lt;/em&gt; an elementary one. Short answer is: unfortunately, there is no silver bullet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a molecule of around 65 atoms I would go with DFT as the methodology which provides probably the best quality/cost ratio for molecules of this size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to choose functional and basis set, and this is a tricky business. But to start from:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do &lt;em&gt;not&lt;/em&gt; blindly use B3LYP. &lt;/li&gt;&#xA;&lt;li&gt;Try to &lt;em&gt;search literature&lt;/em&gt; for the same types of calculations (energetics of ring opening reactions, optical properties) for similar compounds. Maybe someone already did the so-called benchmark calculations, i.e. compared on how good different functionals in combination with different basis sets are on reproducing the properties (say, specific rotation) either measured experimentally or calculated at some high level of theory (say, CCSD(T) with reasonably sized basis set).&lt;/li&gt;&#xA;&lt;li&gt;Do your own &lt;em&gt;benchmark calculations&lt;/em&gt; with different functionals and basis sets. First, benchmark different functionals with the same basis set (say, Dunning correlation-consistent basis sets, cc-pVTZ or better cc-pVQZ), vary the functional, compare results with experimental values, choose the best one. Once method is chosen, try to decrease and also increase (if possible) the basis set to see the effect of basis set size (with the above mentioned bases it is quite simple: if cc-pVTZ was used to benchmark different functionals, try the best functional in combination with cc-pVDZ and cc-pVQZ). Choose the smallest basis set that reproduces experimental results with desired accuracy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S. You could also ask Google, there are some resources of information &lt;a href=&quot;http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/quantum-practical.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.chem.mun.ca/tcms/archive/gauldtut.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;there&lt;/a&gt; on how to choose method and basis set, though, as I said, this question is not elementary, so only general advice can be given. &lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="194" LastEditDate="2017-03-29T04:20:50.663" LastActivityDate="2017-03-29T04:20:50.663" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14528" PostTypeId="1" AcceptedAnswerId="14537" CreationDate="2014-07-21T22:11:06.270" Score="5" ViewCount="1512" Body="&lt;p&gt;The energy of molecule orbitals increases with more nodal planes. W1 (in the attached picture) has no nodal plane. I'd like to know how to draw the nodal planes in cyclopropane molecule orbitals but do not know how. I found no picture in the web with the nodal planes of cyclopropane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many nodal planes have W3, W4 and W5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rKM7E.jpg&quot; alt=&quot;expected nodal planes in cyclopropane in the picture from walsh orbitals, picture modificated from www.enu.kz/repository/repository2014/orbital-interaction.pdf page 85&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="305" LastEditDate="2014-07-22T00:17:39.163" LastActivityDate="2014-07-22T08:50:08.583" Title="Counting Nodal Planes in cyclopropane" Tags="&lt;energy&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14529" PostTypeId="2" ParentId="14528" CreationDate="2014-07-22T02:21:09.213" Score="1" Body="&lt;p&gt;Instead of nodal planes think of nodal surfaces.  For W1 there is one nodal surface, a cylinder that looks like a circle seen from above (the view in the book).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing with the hints.  All the combinations that come from orbitals that have radial symmetry (the ones where the bulges point toward the center and away from the center) have the same cylinder as a nodal surface.  If you draw the circle representing this cylinder seen from above, you'll basically have completed the nodal surface diagram for W1, W5, W6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of nodal surfaces helps build intuition, but it only goes so far.  For example, it doesn't have to be true that more nodal surfaces equals higher energy. &lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-22T06:51:14.050" LastActivityDate="2014-07-22T06:51:14.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14530" PostTypeId="1" CreationDate="2014-07-22T03:30:16.380" Score="1" ViewCount="829" Body="&lt;p&gt;Reaction was carried out in DMF, and TLC shows that the reaction has completed. But how to separate reaction mixture i.e. product from DMF?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;Steglich Esterification was carried out and the product is soluble in DMF. –  user 24 (Jul 22 at 7:13)&lt;/p&gt;&#xA;" OwnerUserId="5255" LastEditorUserId="4945" LastEditDate="2014-07-29T09:20:08.267" LastActivityDate="2014-07-29T09:20:08.267" Title="How to crystallize out reaction mixture from DMF?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;purification&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14531" PostTypeId="1" AcceptedAnswerId="14532" CreationDate="2014-07-21T18:36:54.747" Score="7" ViewCount="3725" Body="&lt;p&gt;I have heard that washing one's hands with vinegar is effective in neutralizing the strong bases in cement and is more effective than water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how cement burns and what bases are involved in this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried googling &quot;cement burns&quot; and found websites explaining the importance of washing one's hands with vinegar as a preventative measure. I have tried searching PubMed but I've found mostly case reports that do not address how chemical burns actually happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows of some good papers or books that cover the topic of chemical burns in greater detail than what I've found; please let me know.  &lt;/p&gt;&#xA;" OwnerUserId="7209" OwnerDisplayName="Slavatron" LastEditorUserId="73" LastEditDate="2014-07-22T10:02:37.450" LastActivityDate="2014-12-28T01:37:30.250" Title="Does vinegar neutralize the strongly basic compounds in cement?" Tags="&lt;biochemistry&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14532" PostTypeId="2" ParentId="14531" CreationDate="2014-07-21T20:33:21.557" Score="8" Body="&lt;p&gt;If you look for &lt;a href=&quot;http://en.wikipedia.org/wiki/Portland_cement#Safety_issues&quot; rel=&quot;nofollow&quot;&gt;Portland Cement safety issues&lt;/a&gt;, it is stated that cement is highly alkaline so it can cause alkaline burn and presumably since &lt;a href=&quot;http://en.wikipedia.org/wiki/Vinegar&quot; rel=&quot;nofollow&quot;&gt;vinegar&lt;/a&gt; is acetic acid, it neutralises the alkaline effects of cement as tested &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/12711953&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and the burns you suffered are probably related to alkaline burn. You can look at this article section 6.5 for some details on &lt;a href=&quot;http://www.prevor.com/FR/sante/RisqueChimique/diphoterine/publications/media/Burns_juil2009_Chemical%20burns_pathophysiology.pdf&quot; rel=&quot;nofollow&quot;&gt;alkaline burns&lt;/a&gt;, which was suggested by @Cornelius or perhaps the mentioned chapter in this book (Harchelroad FP, Rottinghaus DM. Chemical burns. In: Tintinalli JE, Kelen GD, Stapczynski JS, Ma OJ, Cline DM, eds. Emergency Medicine: A Comprehensive Study Guide. 6th ed. New York, NY: McGraw-Hill; 2004: chap 200.).&lt;/p&gt;&#xA;" OwnerUserId="7204" OwnerDisplayName="Bez" LastEditorUserId="4945" LastEditDate="2014-07-22T11:43:07.443" LastActivityDate="2014-07-22T11:43:07.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14533" PostTypeId="1" CreationDate="2014-07-22T05:25:44.270" Score="6" ViewCount="29633" Body="&lt;p&gt;I tried to draw the Lewis structure of $\ce{HCOOH}$ but I didn't get the correct representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the picture below, at no. 1, I put $\ce{C}$ as the central atom because it has less electronegativity than $\ce{O}$ ($\ce{H}$ cannot be a central atom). I drew the structure, but it is wrong because it violates the octet rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I came up with another structure as shown at no. 2,&#xA;and it seemed correct to me, but when I searched on Google for the correct structure, I found out that I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IvzDx.jpg&quot; alt=&quot;suggested Lewis structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the structure that I drew in no. 2  wrong, even though the formal charges are zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I figure out the correct order of atoms in such molecules like the one above. What I mean is how do I know that the $\ce{C}$ will have $\ce{H}$ on the left and one $\ce{O}$ up and $\ce{O, H}$ on the  right side?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="194" LastEditDate="2018-03-03T14:39:26.457" LastActivityDate="2018-03-03T14:39:26.457" Title="What is the proper Lewis structure for HCOOH?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14534" PostTypeId="2" ParentId="14533" CreationDate="2014-07-22T06:30:05.530" Score="4" Body="&lt;p&gt;Formal charges don't tell the whole story.  What atoms are connected to what is the first thing you need to consider. Note that $\ce{HCOOH}$ gives you a hint that the $\ce{OH}$ goes together, rather than the 2nd $\ce{H}$ attaching to $\ce{C}$. (Though the linear formula is still misleading in that both $\ce{O}$ atoms are attached to $\ce{C}$, which seems not to be the case in the formula).&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorDisplayName="user15489" LastEditDate="2015-05-23T09:14:42.833" LastActivityDate="2015-05-23T09:14:42.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14535" PostTypeId="1" AcceptedAnswerId="14536" CreationDate="2014-07-22T06:42:26.633" Score="4" ViewCount="398" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Balance the following reaction:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{CuS + SO4^{2-} -&amp;gt; CuO + SO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below, the number mentioned in the parenthesis is the oxidation number of sulfur.&lt;br&gt;&#xA;I have found out that the oxidation half-reactions and reduction half-reactions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)}     &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Oxidation half reaction}\\&#xA;\ce{^{(6)}SO4^{2-} &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Reduction half reaction}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, what about &lt;span class=&quot;math-container&quot;&gt;$\ce{CuO}$&lt;/span&gt; in the product side? I have also learned that we need to balance the spectator ions before charges if the spectator ion is other than oxygen and hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, there is no &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2019-05-21T11:48:19.590" LastActivityDate="2019-05-21T11:57:07.747" Title="What do the half-reactions need to be for balancing a redox reaction in an acidic medium?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="15" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14536" PostTypeId="2" ParentId="14535" CreationDate="2014-07-22T08:41:18.630" Score="3" Body="&lt;p&gt;Below, the number mentioned in the parenthesis is the oxidation number of sulfur.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Writing down half reactions:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)}     &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Oxidation}\\&#xA;\ce{^{(6)}SO4^{2-} &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Reduction}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Adding electrons in both reactions:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)}           &amp;amp;-&amp;gt; ^{(4)}SO2 + 6e-}&amp;amp;&amp;amp; \text{Oxidation}\\&#xA;\ce{^{(6)}SO4^{2-} + 2e- &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Reduction}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Balancing &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)}&amp;amp;-&amp;gt; ^{(4)}SO2 + 6e- + Cu^{+2}}&amp;amp;&amp;amp; \text{Oxidation}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Balancing charges of both half reactions:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)}&amp;amp;-&amp;gt; ^{(4)}SO2 + 6e- +CuO + 6H+}&amp;amp;&amp;amp; \text{Oxidation}\\&#xA;\ce{^{(6)}SO4^{2-} + 2e- +4H+ &amp;amp;-&amp;gt; ^{(4)}SO2}&amp;amp;&amp;amp; \text{Reduction}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Balancing oxygen and hydrogen:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CuS^{(-2)} + 3H2O&amp;amp;-&amp;gt; ^{(4)}SO2 + 6e- +CuO + 6H+}&amp;amp;&amp;amp; \text{Oxidation}\\&#xA;\ce{^{(6)}SO4^{2-} + 2e- +4H+ &amp;amp;-&amp;gt; ^{(4)}SO2 + 2H2O}&amp;amp;&amp;amp; \text{Reduction}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CuS + 3SO4^{2-}  + 6H+ -&amp;gt; 4SO2 +CuO + 3H2O}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Thanks to Silvio Levy for helping me to solve this problem.)&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2019-05-21T11:57:07.747" LastActivityDate="2019-05-21T11:57:07.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14537" PostTypeId="2" ParentId="14528" CreationDate="2014-07-22T08:50:08.583" Score="7" Body="&lt;p&gt;The assessment of nodal planes is not as trivial as it might seem at first, because they are usually not planes at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitals can be explained in terms of symmetry operations. There are three different operations and each of them has a unique element (in cartesian coordinates):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotation: Axis&lt;/li&gt;&#xA;&lt;li&gt;Mirror: Plane&lt;/li&gt;&#xA;&lt;li&gt;Inversion: Point&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When we talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen type atomic orbitals&lt;/a&gt;, we usually refer to their cartesian form. It is already fairly obvious for the $\ce{2s}$, that there is no nodal plane, while there are certainly nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The representations of $\ce{s}$ orbitals are always trivial with respect to the total symmetry of the external field (= Point group). That means any symmetry operation, which is given through the external field will be matched by the $\ce{s}$ orbital. Another term for this is totally symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The representation of $\ce{p}$ orbitals are always antisymmetric with respect to inversion in any given external field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The representation of $\ce{d}$ orbitals is always antisymmetric to two mirror planes. That excludes of course the $\ce{d_{z^2}}$ orbital, which is not a cartesian, but a spherical representation. (Note: In cartesian space there are 6 $\ce{d}$ orbitals. As &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum numbers&lt;/a&gt; only allow 5 spherical orbitals, they have to be transformed. See for example at &lt;a href=&quot;http://www.chemissian.com/ch5&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemissian.com&lt;/a&gt; or &lt;a href=&quot;http://www.chem.wayne.edu/schlegel/Pub_folder/174.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schlegel and Frisch&lt;/a&gt;.) Early basis sets for quantum chemical calculations used the cartesian representation, because it wa easier to code and calculate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Due to the geometric arrangement of nuclei in a molecule, the external field is given through the point group of the molecule. Every molecular orbital can be expressed through symmetry operations of that point group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclopropane has $D_\mathrm{3h}$ symmetry. And every molecular orbital has to respect that. I have already written about orbitals these orbitals in &lt;a href=&quot;https://chemistry.stackexchange.com/q/10653/4945&quot;&gt;Why does cyclopropane give bromine water test?&lt;/a&gt;&#xA;However, I would like to shed some more light on this, because I believe the picture you found is not quite correct.&lt;br&gt;&#xA;The following picture shows the depicted orbitals from your source, as well as the LUMO.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zWA0Hl.png&quot; alt=&quot;orbs of cyclopropane&quot;&gt;&lt;br&gt;&#xA;Where W1 corresponds to HOMO-2, W2 and W3 correspond to both HOMO, W4 corresponds to LUMO+3, finally W5 and W6 correspond to LUMO+2.&lt;br&gt;&#xA;The ordering of these orbitals is taken from a BP86/cc-pVTZ calculation of Gaussian09. A methodology that I believe gives fairly accurate results for molecular orbitals. (It is however noteworthy, that unoccupied orbitals are constructed from occupied orbitals. That means, that their physical meaning is limited.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a closer look at the orbitals. The HOMO-2 is of $A_1'$ symmetry and therefore behaves like an $\ce{s}$ orbital. There is no antibonding nodal surface in this orbital. That means at any bonding axis of the ring structure the sign of the wavefunction does not change. There is however a nodal surface that does not affect bonding, as was already pointed out in some comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HOMOs are of $E'$ symmetry and therefore behave like $\ce{p_x}$ and $\ce{p_y}$ orbitals. These orbitals have one antibonding nodal surface. In a result of symmetry restrictions these nodal surfaces are perpendicular to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also true for the LUMO+2s. In addition to this there is another antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LUMO again is of $A_1'$ symmetry and behaves like a $\ce{s}$ orbital. There is again no antibonding nodal surface with respect to the $\ce{C-C}$ bonds, but the main electron density would be outside of any bond and there is an antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LUMO+3 is of $A_2'$ symmetry and does therefore not behave like any other atomic orbital. Here we find three antibonding nodal surfaces with respect to the $\ce{C-C}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding situation in cyclopropane is very complicated and can be explained in many different ways. It is not as obvious, that the molecule is a $\sigma$ aromate, having in plane $\pi$ orbitals (HOMO). The explanation of its molecular orbitals were first composed in terms of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Walsh_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Walsh&lt;/a&gt; like behaviour on the basis of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_H%C3%BCckel_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;extended Hückel calculation&lt;/a&gt; with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)#Minimal_basis_sets&quot; rel=&quot;nofollow noreferrer&quot;&gt;minimum basis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you disregard the $\ce{C-H}$ bonds and only allow three carbon $\ce{sp^2}$ and three carbon $\ce{p}$ orbitals, then you can only arrive at the boning picture that was depicted in your source (and many other publications), as you can only have six molecular orbitals formed from these. There are, however, many more atomic orbitals that mix into the final wavefunction. First of all it is necessary to state, that the used $\ce{sp^2}$ orbitals are not of the ideal $\frac13\ce{s}+\frac23\ce{p}$ composition (see more about that in the linked question and the answer of ron therein).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now finally for the discussion about the correlation of nodal planes and increasing energy. It is a very common statement, that is missing it's most important limitation. It is only completely true for orbitals of the same symmetry. (This is also a necessary requirement.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the energy of the $\ce{s}$ orbital series increase with the main quantum number, as you add one more nodal surface per one increment in $n$: $\ce{1s-&amp;gt;0; 2s-&amp;gt;1; 3s-&amp;gt;2;...}$. The same applies to the $\ce{p}$ orbital series: $\ce{2p-&amp;gt;1; 3p-&amp;gt;2; ...}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this statement became popular, when talking about aromatic systems. In the framework of &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hückel's molecular orbital method&lt;/a&gt;, this must be true, since all regarded orbitals are of the same symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is usually true, that an orbital with fewer nodal planes is more stable than another with more, but this is more a gut feeling, than actual, factual science. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-07-22T08:50:08.583" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14538" PostTypeId="2" ParentId="14533" CreationDate="2014-07-22T09:19:18.390" Score="4" Body="&lt;p&gt;From a sum formula like $\ce{HCOOH}$ it is usually not possible to construct the Lewis structure. In this case, there are some obvious points, why your structures cannot be correct.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the oxygens only has an electron sextett. As oxygen is the most electronegative element in this compound, this is just not possible. But you saw this yourself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In these structures the carbon has only an electron sextett. Since it is much more electronegative than hydrogen, this can also be not correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in praxis sum formulas should always be accompanied with a name, or a schematic drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you try to formulate a Lewis structure based on the sum formula $\ce{C2H6O}$, you will end up with &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethyl_ether&quot; rel=&quot;nofollow&quot;&gt;dimethyl ether&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethanol&quot; rel=&quot;nofollow&quot;&gt;ethyl alcohol/ ethanol&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-22T09:19:18.390" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14539" PostTypeId="1" CreationDate="2014-07-22T09:36:38.540" Score="1" ViewCount="841" Body="&lt;p&gt;From Sc, to Mn, the number of oxidation states increases from one (Sc) to seven (Mn). The explanation for this is because the unpaired 3d electrons can be lost along with the 4s electrons during bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should this be the case? Wouldn't it be easier to lose an electron when it is paired because of interelectronic repulsion in the orbital? Could someone explain this? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7198" LastActivityDate="2014-07-23T17:44:10.437" Title="Explanation for increase in the number of oxidation states for transition elements from Sc to Mn of the 4th period" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14540" PostTypeId="1" CreationDate="2014-07-22T12:43:24.067" Score="1" ViewCount="1866" Body="&lt;p&gt;Looking to remove the hydrazine byproduct left after the mitsunobu reaction. Using DIAD and polymer bound triphenyl phosphine. The reduced diad is coeluting with my product on silica. My thoughts for getting rid of it are trying an alumina column, acidic wash (I have an N-Boc so it can't be too acidic), or switching to a different azodicarboxylate. &lt;/p&gt;&#xA;" OwnerUserId="7205" LastEditorUserId="1878" LastEditDate="2014-07-22T13:02:17.390" LastActivityDate="2014-07-27T23:21:29.243" Title="Removing the hydrazine byproduct from the mitsunobu reaction" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14542" PostTypeId="2" ParentId="14533" CreationDate="2014-07-22T14:39:02.603" Score="9" Body="&lt;p&gt;With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtOk6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have  involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y39wf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming &lt;em&gt;four&lt;/em&gt; bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above, in addition to keeping in mind that a carboxylic acid is a &lt;em&gt;Bronsted/Lowry&lt;/em&gt; acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-14T15:05:30.090" LastActivityDate="2014-08-14T15:05:30.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14543" PostTypeId="2" ParentId="14531" CreationDate="2014-07-22T16:28:42.843" Score="0" Body="&lt;p&gt;Injury to your skin can occur while the skin is in contact with the cement--the more prolonged the contact, the greater the degree of skin damage. As soon as you wash off the cement (using either vinegar or water--they will be equally effective) you will stop additional damage, but the vinegar will not reverse the skin damage that has already occurred. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-07-22T16:28:42.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14544" PostTypeId="1" AcceptedAnswerId="14546" CreationDate="2014-07-22T16:46:48.463" Score="3" ViewCount="162" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qhq2b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the nickel complex exhibiting the strongest pi-bonding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, charge density plays a role. Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I note that some have positive overall charges; could this play a role? Metal cation &amp;lt;-&gt; NO+ repulsion? So we might want a metal anion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also would size play a role? Size goes hand in hand with charge density. And what about &quot;backbonding&quot;?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2015-04-02T05:55:34.293" LastActivityDate="2015-04-02T05:55:34.293" Title="Carbonyl/Nitrosonium Complex Stability" Tags="&lt;inorganic-chemistry&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14545" PostTypeId="1" CreationDate="2014-07-22T17:37:42.270" Score="2" ViewCount="203" Body="&lt;p&gt;I am not a chemist! My chemistry classes from my molecular biology undergrad are far behind me. But I think I created a salt, and I don't know how. Here is my mystery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently got some water in my car, and wanted to prevent mold growth in my carpet mats. I brought the mats inside, vacuumed off, and completely soaked them in household vinegar. Then I placed the mats near a running dehumidifier (where the warm air comes out in the back). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;White crust appeared on the carpet. At first I thought my dehumidifier was spitting something out, but the white substance was not on any other nearby surface--only on the car carpet. Eventually as the further parts of the carpet dried, they also showed the white crust. It covers pretty much all of the carpet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this, in fact, a salt? If so, what basic substance could have been in my car carpet mat that would cause this reaction? Thanks everyone!&lt;/p&gt;&#xA;" OwnerUserId="7208" LastEditorUserId="73" LastEditDate="2014-07-22T17:38:59.337" LastActivityDate="2014-07-23T20:51:42.583" Title="How could household vinegar create a salt on car carpet?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14546" PostTypeId="2" ParentId="14544" CreationDate="2014-07-22T18:22:30.980" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ahem. Nope. $\pi$-bonding in $\ce{CO/NO+}$ complexes is from back-donation, from occupied metal d-orbitals onto unoccupied $\pi^*$ orbitals of the ligand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks like homework question, so no further details provided.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-22T18:22:30.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14547" PostTypeId="1" CreationDate="2014-07-22T20:26:13.713" Score="5" ViewCount="1270" Body="&lt;p&gt;I can't find the answer to this question on this SE website, and I apologize for the repetition if it has been answered before. It is my understanding that compound formation has only been observed for the noble gases argon, xenon, and krypton, while no compounds have ever been observed for neon and helium, and that this might have something to do with both the decreased ionization potential of larger noble gas atoms and the high value of electronegativity of the halogen (usually fluorine) that bonds with the noble gas atoms to form compounds. So, my question is why aren't compounds formed from argon, xenon, and krypton bonding with chlorine, bromine, and iodine as prevalent as those formed from those noble gases and fluorine? Could the smaller size of the fluorine atom be a reason for its apparently higher reactivity?&#xA;-John&lt;/p&gt;&#xA;" OwnerUserId="7210" LastEditorUserId="7210" LastEditDate="2014-07-22T20:39:11.540" LastActivityDate="2014-07-23T01:39:17.473" Title="Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;atoms&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14548" PostTypeId="1" AcceptedAnswerId="14551" CreationDate="2014-07-22T20:59:10.780" Score="6" ViewCount="1418" Body="&lt;p&gt;I love soup, even a thick soup for dinner, yet &lt;em&gt;many&lt;/em&gt; soup products have 400 to 1000 mg of sodium in them. Is there a simple chemical procedure any consumer could do to take the sodium out of the product without 'hurting' the flavor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one takes some, say, Campbell's soup and heats it to boiling, then freezes it, then heats it again, then freezes it etc., would this extract some sodium?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="73" LastEditDate="2014-07-24T01:54:14.200" LastActivityDate="2020-08-15T05:21:08.297" Title="Is it possible to remove salt from food without changing the flavor?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14549" PostTypeId="2" ParentId="8949" CreationDate="2014-07-22T21:22:22.153" Score="5" Body="&lt;p&gt;I think your confusion arose because of an error in the translation of the text you linked (the discussion on p. 359) and that this text was unclear about which equilibrium it was referring to. In that figure, the equilibrium that the authors are presumably referring to is the (planar) carbonyl- (tetrahedral) hemiacetal equilibrium (as described on page 358). So the equilibrium that the text refers to isn't the &quot;proton transfer equilibrium,&quot; but the planar-tetrahedral equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9uAcO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the scheme you posted with your question assumed that the formation of a hemiacetal (or hemiketal) is irreversible, as indicated by a unidirectional arrow. This reaction should be treated with an bidirectional equilibrium arrow as shown in p. 359 of the document that you linked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, the steric argument is reasonable and the reaction as a whole analogous to the more well-known relative hydration equilibria of aldehydes and ketones (which is related the general notion of aldehydes being more electrophilic than ketones). &lt;/p&gt;&#xA;" OwnerUserId="7211" LastEditorUserId="7211" LastEditDate="2014-07-23T16:49:32.187" LastActivityDate="2014-07-23T16:49:32.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14550" PostTypeId="2" ParentId="14533" CreationDate="2014-07-22T21:42:01.157" Score="2" Body="&lt;p&gt;The physical reason why answer 2 is wrong is because it represents a class of compounds that is very reactive (a carbene) and if the compounds in (2) actually existed, they would probably rearrange very quickly to the carboxylic acid isomers. The reason the rearrangement would occur is because of bond strengths (i.e. the energy it takes to cleave a chemical bond homolytically)--carbon-oxygen bonds are stonger than oxygen-oxygen bonds. See &lt;a href=&quot;http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; for rough numbers on this property &lt;/p&gt;&#xA;" OwnerUserId="7211" LastActivityDate="2014-07-22T21:42:01.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14551" PostTypeId="2" ParentId="14548" CreationDate="2014-07-22T21:51:25.743" Score="9" Body="&lt;p&gt;Short answer: no, there is no easy (or even moderately hard) procedure, and certainly not boiling and freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that water dissolves sodium ions better than just about anything else, so almost everything will come out of solution before the sodium does. You're much better off with a product that starts with less salt to begin with.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-22T21:57:36.870" LastActivityDate="2014-07-22T21:57:36.870" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14552" PostTypeId="2" ParentId="14547" CreationDate="2014-07-23T01:39:17.473" Score="4" Body="&lt;p&gt;You know that the bigger the noble gas atom, the easier it is to ionize it. By the same token it is also easier for a Xe atom than for a He atom to share some electron density in a covalent bond.  That is, Xe is more reactive (a very little bit) than He (not at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For halogens, the tendency is in some sense the opposite: The smaller the halogen atom the more eager it is to gain electron density -- that is, the more electronegative it is.  Fluorine is so electronegative that it can persuade Xe, Kr and even Ar (with difficulty) to give up some electron density to form a covalent bond.  Cl is not quite so electronegative, and Br and I even less.  Basically you need the least &quot;noble&quot; of the noble gases and the greediest halogen of all to even have a chance at a bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can justify this handwaving with thermodynamic calculations and molecular orbital arguments (though the latter are often handwavy too).&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-23T01:39:17.473" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14553" PostTypeId="1" AcceptedAnswerId="14813" CreationDate="2014-07-23T04:10:05.330" Score="6" ViewCount="12659" Body="&lt;p&gt;I am trying to understand this problem and its corresponding solution (complete solution not shown)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Frfxa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First off, from the above one can see that the equilibrium constants when $\ce{H2ASO4-}$ donates and receives a proton are as follows:&#xA;\begin{aligned}&#xA;\ce{H2AsO4 &amp;amp;&amp;lt;=&amp;gt; HAsO4^{2-} + H+} &amp;amp; K_2 &amp;amp;= 1.1 \cdot 10^{−7}\\&#xA;\ce{H2AsO4- + H+ &amp;amp;&amp;lt;=&amp;gt; H3AsO4}    &amp;amp; K_{-1} &amp;amp;= 1.6 \cdot 10^{2}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if this is an amphoteric substance as the solution claims, shouldn't both $K_a$ values be comparable? It seems logical that if it is able to act as both an acid and a base, the equilibrium constants should at least be similar to each other. Instead, one is several order of magnitudes larger than the other? Why can this be true and $\ce{H2ASO4-}$ still be an amphoteric substance?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Next, how did they arrive at the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We approximately have the equality relation: $$\ce{[H3AsO4]} = \ce{[HAsO4^{2−}]}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the evidence that shows this to be true? Isn't the $K_a$ value of the basic reaction ($1.6 \times 10^2$) much larger than that of the acidic reaction? Why should the concentrations be equal?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, let's say you are calculating the pH for the diprotic substance $\ce{H2S}$, which undergoes the following diassociations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2S &amp;lt;=&amp;gt; H+ + HS-}$$&#xA;$$\ce{HS- &amp;lt;=&amp;gt; H+ + S^{2-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the pH you would assume that the first reaction is dominant, and that the effects of the second reaction consuming $\ce{HS-}$ are minimal, and then calculate the pH from there — right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why can't you treat $\ce{HS-}$ as an amphoteric substance (because it can both gain and lose a proton) and then calculate the pH as you would for an amphoteric substance (as seen in the image at the beginning)?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="2142" LastEditDate="2014-07-28T00:28:58.810" LastActivityDate="2014-08-02T09:58:05.463" Title="Calculating pH of diprotic and amphoteric solutions" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14554" PostTypeId="2" ParentId="14553" CreationDate="2014-07-23T04:38:15.263" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First, I have read that $\ce{H2AsO4-}$ is a amphiprotic substances in&#xA;  that the dominant reactivity is $$\ce{2H2AsO4- &amp;lt;=&amp;gt; H3AsO4 + HAsO4^{2-}}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, this also seems to be a polyprotic substance. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two terms - amphiprotic and polyprotic - aren't mutually exclusive. Amphiprotic substances can simply behave as both Bronsted-Lowry acids and bases in that they can both give up and release hydrogen protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyprotic acids have multiple protons to donate. For example, phosphoric acid is triprotic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often times amphiprotic substances will be polyprotic. By nature of being able to lose multiple protons, these substances are &lt;em&gt;also&lt;/em&gt; able to &lt;em&gt;gain&lt;/em&gt; multiple protons. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is one $K_a$ value so much larger than the other? If this were a&#xA;  amphoteric substance shouldn't the $K_a$ values for both be&#xA;  comparable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good observation. This lies with Coulomb's law. Think about it. A neutral molecule kicks off a hydrogen proton. Now you have the first conjugate base. The first conjugate base is negatively charged. It also attempts to kick off a proton. Which do you think is easier? Clearly the first process is. Remember that oppositely charged particles attract. This explains why $K_{a1}$ values for polyprotic molecular acids are generally significantly larger than corresponding $K_{a2}$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because the $K_a$ values differ by several orders of magnitude doesn't mean the ion/molecule can't be amphiprotic. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Next, I researched how to calculate the $\mathrm{pH}$ value for&#xA;  amphiprotic substances, and I found this page that details how to&#xA;  do so, and the end result is that the $\mathrm{pH} = \sqrt{K_1 K_2}$.&#xA;  Why is the $\mathrm{pH}$ independent of the concentrations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, amphiprotic substances don't have $\mathrm{pH}$ values by themselves. In solution, then yes. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further, I also found another page that gets to the same result in a&#xA;  different manner. They start by saying:&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;We approximately have the equality relation: $$\ce{[H3AsO4]} = \ce{[HAsO4^{2−}]}$$&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the validity of the quoted statement: just note the main reaction stoichiometry. And the magnitude of the main reaction's K. The main K can be found through dividing the Ka of the reactant acid by the Ka of the product acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{H2AsO4^{−} + H2AsO4^{−} &amp;amp;&amp;lt;=&amp;gt; H3AsO4 + HAsO4^2-}\\&#xA;K_\mathrm{rxn} &amp;amp;= \frac{K_a(\ce{H2AsO4^{−}})}{K_a(\ce{H3AsO4})}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Next, why can't this method of calculating pH in amphoteric acids be&#xA;  used for diprotic acids? I found another page that details how to do&#xA;  so but even though the reactions are essentially similar to those for&#xA;  amphiprotic acids, the end result is different. Why does the same line&#xA;  of reasoning as used for deriving the formula for amphoteric&#xA;  substances fail for diprotic/polyprotic acids?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For diprotic acid, just keep $K$ value magnitudes in mind. This statement applies to diprotic and polyprotic acids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{a2}(\ce{H3AsO4}) = \ce{[HAsO4^{2−}]}$$&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-07-23T08:49:21.580" LastActivityDate="2014-07-23T08:49:21.580" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14555" PostTypeId="1" CreationDate="2014-07-23T05:02:00.303" Score="5" ViewCount="612" Body="&lt;p&gt;Suppose we have a redox reaction and we want to see in what direction the reaction is spontaneous. We do this by comparing the standard reduction potentials of the sub-reactions available in the main reaction and say if &#xA;$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}&amp;gt;0$$&#xA; then the reaction is spontaneous in the direction mentioned otherwise in the reverse direction. The same result can be obtained by comparing the standard oxidation potentials. However, as every spontaneity has its root in the second law of thermodynamics I am eager to see how the spontaneity of a redox equation in a given direction can be checked based on thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically how can we derive the condition &#xA;$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}&amp;gt;0$$&#xA; for spontaneity of a reaction from the fundamental thermodynamics laws?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To clarify the question better let me add that I already know that the second law for a common chemical reaction yields into the condition $$dG \le 0\,,$$ my question is about to find the condition &#xA;$$E^\circ =E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}&amp;gt;0$$ from that inequality. Also let me add further that I even already know that the electric work equals $-n F E^\circ_{\text{cell}}$ but I cannot justify for myself why $\Delta G$ is equal to this work. (&lt;strong&gt;Note.&lt;/strong&gt; the electric work is an internal work not crossing the boundaries of the system, it is not the work of the system, and if it was then at best we had $\Delta G=W_{\text{max}}$ but the inequality $\Delta G\le 0$ no longer hold.)&lt;/p&gt;&#xA;" OwnerUserId="6980" LastEditorUserId="402" LastEditDate="2016-11-05T18:17:25.503" LastActivityDate="2016-11-05T18:17:25.503" Title="The spontaneity of a redox reaction" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;entropy&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14557" PostTypeId="1" AcceptedAnswerId="16411" CreationDate="2014-07-23T05:58:46.750" Score="5" ViewCount="203" Body="&lt;p&gt;Water clusters made up of &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_cluster&quot; rel=&quot;nofollow&quot;&gt;various numbers of water molecules&lt;/a&gt; have been described.&lt;br&gt;&#xA;Are all of these structures theoretical?&lt;br&gt;&#xA;What experimental evidence supports any of these structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the free energy of a particular water cluster (per number of molecules in the cluster) vary from one thermodynamically stable arrangement to another? Is there a general trend, e.g. larger clusters are more stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In response to Martin's initial comment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious about water clusters in the context of receptor-ligand interactions (for example, in protein receptors, enzymes or synthetic host cavities). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cases where the cavities of these macromolecules is small (for example, a small enzyme active site) what kinds of water clusters are likely to fill this void in the absence of ligand? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put more broadly, does the structure of the water that is filling the cavity destabilize the receptor in such a way that ligand binding is enhanced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Considering a hypothetical receptor that is known to bind ligands (whose sizes range between 4 and 20 non-hydrogen atoms) in water, what are likely structures for the encapsulated water cluster that will be displaced by the ligands upon binding?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7211" LastEditorUserId="7211" LastEditDate="2014-07-25T19:55:03.013" LastActivityDate="2014-09-17T14:42:43.637" Title="What stable arrangements of hydrogen-bonded water exist in the range of 4-20 water molecules?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;experimental-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14558" PostTypeId="2" ParentId="7683" CreationDate="2014-07-23T08:02:54.953" Score="43" Body="&lt;p&gt;I would like to back up Klaus' answer with some &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;noreferrer&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt; (QTAIM) results, based on a DF-BP86/def2-SVP calculation. Note that these are results, obtained without the consideration of solvation or condensed phases. I believe they still prove a valid point in the case of electronic structure theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I revisited this question in order to answer another, similar &lt;a href=&quot;https://chemistry.stackexchange.com/q/15412/4945&quot;&gt;question&lt;/a&gt;. While putting more effort into this, I realised, that the here treated structures are actually transition states. This does not mean, that the addressed issues are invalidated. Even if the only exist for very short moments, they still exist and have to be considered.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;o&lt;/em&gt;-methylaniline you can clearly see the suggested intramolecular $\ce{H}$ bond. The distance $\mathbf{d}(\ce{N-H})=239.3~\mathrm{pm}$ is only little shorter than the sum of the van der Waals radii, $\mathbf{r}(\ce{N})=155~\mathrm{pm}$, $\mathbf{r}(\ce{H})=110~\mathrm{pm}$, but neglecting it is also wrong. Even if this interaction does only exist for very short periods of time, it still means, that it stabilises this state. It will however not be the dominant feature.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IY7wr.png&quot; alt=&quot;QTAIM for o-methylaniline&quot;&gt;&lt;br&gt;&#xA;(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho&amp;lt;0$, dashed blue lines indicate charge accumulation  $\nabla^2\rho&amp;gt;0$,&#xA;Red spheres are bond critical points, purple spheres are ring critical points, black lines are bond paths, red lines are zero flux surfaces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steric effects are usually electronic or dispersive effects in disguise, hence they also refer to an intramolecular hydrogen bond. The average $\ce{N-H}$ bond is only about $\mathbf{d}_\text{av.}(\ce{N-H})\approx99-105~\mathrm{pm}$.&lt;br&gt;&#xA;The solvation point made by user4604 should still be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This interaction has to lower proton affinity and or lewis acid affinity and therefore decreases also basicity. Not so much because its electronic effects stabilise one particular conformation, but especially, making the lone pair unavailable for short periods of time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can also analyse this for &lt;em&gt;o&lt;/em&gt;-methylbenzoic acid and here the effect changes direction.&lt;br&gt;&#xA;In benzoic acid there is already some intramolecular hydrogen bonding from the ortho hydrogens, one of these is still present in the substituted case. The distance $\mathbf{d}(\ce{O-H_{o'}})=226.2~\mathrm{pm}$ is only little shorter than the sum of the van der Waals radii, $\mathbf{r}(\ce{O})=151~\mathrm{pm}$, $\mathbf{r}(\ce{H})=110~\mathrm{pm}$, but neglecting it would still be wrong wrong.&lt;br&gt;&#xA;The distance $\mathbf{d}(\ce{O-H_{Me}})=210.8~\mathrm{pm}$ is significantly shorter than the sum of the van der Waals radii. You can again see the interaction via a bond path, and ring critical points. In the optimised geometry, the methyl moiety is slightly rotated, giving rise to two equidistant interactions. The rotation of this group can be considered as a free rotation at room temperature.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PBkn4.png&quot; alt=&quot;QTAIM for o-methylbenzoic acid&quot;&gt;&lt;br&gt;&#xA;It is noteworthy, that the bond critical point of the $\ce{O-H_{acid}}$ is almost at the position of the proton, which indicates also that most of the electron density already belongs to the oxygen.&lt;br&gt;&#xA;It is also obvious, that the charge concentration in this bond is significantly lower than in any other $\ce{E-X}$ bond, which can indicate a weak bond.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-01-05T14:56:41.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14559" PostTypeId="2" ParentId="13796" CreationDate="2014-07-23T10:00:07.747" Score="3" Body="&lt;p&gt;The two ions you are comparing to each other are very different. In $\ce{ClO3-}$ there are two more electrons than in $\ce{NO3-}$. This is also the reason why the former pyramidalizes, while the latter does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In nitrate the negative charge can be delocalised over the whole molecule, since there is one molecular orbital with $\pi$ symmetry. One basic requirement for these orbitals is, that they are antisymmetric with respect to a mirror plane, that includes the bonding axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This MO scheme would also be thinkable for the chlorate ion, but here you have two more electrons, so you have to fill an antibonding orbital. The charge can no longer be delocalised as well in this molecule. The conclusion is, that the symmetry of the whole molecule is lowered and the $\pi$ orbitals vanish. A lone pair is always more stable in an orbital with high $\ce{s}$ contribution, which leaves only the $\ce{p}$ orbitals of chlorine for bonding with oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance is a model which is necessary to describe bonding situations, that cannot be described by one Lewis structure alone. It is therefore more a rule of thumb than any evidence. $\pi$ bonds (or orbitals) are not necessary for valid resonance structures. They are however necessary for ([negative] hyper-)conjugation.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-23T10:00:07.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14560" PostTypeId="2" ParentId="2511" CreationDate="2014-07-23T10:17:59.020" Score="11" Body="&lt;p&gt;To complete the previous answers: according to this article &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/17579381&quot; rel=&quot;noreferrer&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/17579381&lt;/a&gt; NaCl exists in the gas phase both as a monomer (NaCl), about 73% of atoms at 943K, and as a dimer where the four atoms are arranged in a wobbly square (the remaining 27%).  Traces of Na${}_2$ and Cl${}_2$ are also observed.  I expect that as the temperature goes up the proportion of the dimer drops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the boiling point of NaCl is  1,686 K at 1atm the study must have been conducted at low pressure or with the NaCl vapor mixed in with other gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article also says (GED = gas-phase electron diffraction) : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The determination of the structure of sodium chloride vapor&#xA;by GED dates back as far as 1937, when it was studied by&#xA;heating salt to around 1000 K. As a consequence of the visual&#xA;analysis methods in use at the time, only the predominant and&#xA;expected monomer species was detected. It was not until the&#xA;1950s that the existence of associated species such as dimers&#xA;(Na2Cl2) were discovered using mass spectrometry. Microwave&#xA;spectroscopy has provided detailed information on the structure&#xA;of the monomer, but the lack of a permanent dipole moment&#xA;makes this impossible for the dimer.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2015-09-04T06:47:34.660" LastActivityDate="2015-09-04T06:47:34.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14562" PostTypeId="1" CreationDate="2014-07-23T11:35:59.867" Score="7" ViewCount="30094" Body="&lt;p&gt;Sucrose decomposes at 186 °C. It doesn’t melt but when it reaches decomposition temperature, sucrose decomposes into fructose and glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After cooling and hardening it is no longer sucrose. Am I right?&lt;/p&gt;&#xA;" OwnerUserId="4012" LastEditorDisplayName="user7951" LastEditDate="2016-12-25T01:26:12.430" LastActivityDate="2020-10-09T19:41:54.817" Title="Why do other sugars melt whereas sucrose decomposes?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14564" PostTypeId="2" ParentId="14562" CreationDate="2014-07-23T15:02:48.667" Score="7" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcIk7.gif&quot; alt=&quot;dehydration process&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are doing when you heat a sugar like sucrose is you are dehydrating it. The crystalline structure of sucrose breaks down and the molecules decompose into glucose and fructose and then lose water and then become isomers and polymerize to form caramel, a red-orange colored solid at room temperature. However there are several other products other than just fructose and glucose.Caramelization is poorly understood so I cannot give you a direct answer stating all of the products formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this at home, just put table sugar in a pot and heat it up. The same type of process is used in candy making.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.food-info.net/uk/colour/caramel.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorUserId="15751" LastEditDate="2018-12-30T19:08:03.603" LastActivityDate="2018-12-30T19:08:03.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14565" PostTypeId="2" ParentId="14562" CreationDate="2014-07-23T15:09:48.773" Score="13" Body="&lt;p&gt;You're mostly right.  Sucrose is a disaccharide composed of glucose and fructose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJl4d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a crystalline material.  Heating sucrose results in a &lt;a href=&quot;http://lib.tkk.fi/Diss/2010/isbn9789526032825/article1.pdf&quot; rel=&quot;noreferrer&quot;&gt;complex thermal process&lt;/a&gt; that involves both melting and decomposition.  Depending how fast you heat it, you will observe different things.  Basically, sucrose will start to decompose when it starts to melt.  If there is some water present during the heating process, then a reaction known as sucrose inversion will occur and the sucrose will break down into its components, one molecule of glucose and one molecule of fructose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if dry sucrose is heated to its melting \ decomposition point, then more complex things happen.  Significant amounts of glucose are initially detected.  Continued heating leads to &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Caramel#Chemistry&quot; rel=&quot;noreferrer&quot;&gt;caramelization&lt;/a&gt;&quot;.  Caramelization is a condensation process where water is eliminated from sucrose; the initial loss of water is followed by isomerization and polymerization.  &lt;a href=&quot;http://www.internetchemistry.com/news/2012/apr12/caramel-chemical-composition.html&quot; rel=&quot;noreferrer&quot;&gt;Over 4,000 products&lt;/a&gt; are thought to be formed during caramelization.  When this complex mixture is cooled, it is no longer sucrose.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-23T15:09:48.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14566" PostTypeId="1" CreationDate="2014-07-23T15:16:20.893" Score="3" ViewCount="715" Body="&lt;p&gt;I'm curious to know how basicity and acidity arises when a salt or covalent compound is dissolved in water.&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="73" LastEditDate="2014-07-23T22:20:53.070" LastActivityDate="2018-03-04T02:54:30.793" Title="What does the hydrolysis of a salt tell us about its basic/acidic character?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14567" PostTypeId="2" ParentId="14566" CreationDate="2014-07-23T15:40:57.223" Score="4" Body="&lt;p&gt;Think of the salt in terms of its components: for example, sodium sulfide can be thought of as derived from sodium hydroxide (a strong base) and hydrogen sulfide (a weak acid) - resulting in a net basic character. When you dissolve sodium sulfide in water, the sulfide ion partially reacts with water to form hydrosulfide ion and hydrogen sulfide, plus hydroxide ion. Aluminum chloride, on the other hand, is derived from aluminum hydroxide (a weak base) and hydrogen chloride (a strong acid), so it is an acidic salt. Sodium chloride, from sodium hydroxide and hydrogen chloride, is a neutral salt (fully ionized). Knowing the dissociation constants of the corresponding acid and base of a salt, one can calculate how acidic or basic the salt is.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-07-23T15:40:57.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14568" PostTypeId="2" ParentId="14566" CreationDate="2014-07-23T16:29:13.557" Score="3" Body="&lt;p&gt;Some salts may have basic Brønsted/Lowry anions but are extremely stable in water. This belies the true nature of the salt's anion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider silver sulfide as an example. This salt has a Ksp value of about ${6*10^{-51}}$. Despite the fact that the sulfide anion is a highly basic Brønsted/Lowry base, a solution of silver sulfide will actually exhibit a pH of 7.0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that salts do not have to be exclusively Brønsted-Lowry bases/acids. Some salts may also exhibit Lewis acidity/basicity in water. This is another way in which basicity and acidity may arise when a salt or covalent compound is dissolved in water. Consider boric acid. This isn't a salt, but it is a covalent compound. It is an acid, but some argue it is not an acid in the sense of being a proton donor. The central boron instead utilizes its empty 2p orbital to heterolytically cleave water molecules as to form H+ in solution. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:54:22.570" LastActivityDate="2018-03-04T02:54:22.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14569" PostTypeId="2" ParentId="14566" CreationDate="2014-07-23T16:42:46.340" Score="1" Body="&lt;p&gt;This question is directly related to the inter relatedness of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Br%C3%B8nsted%E2%80%93Lowry_acid%E2%80%93base_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brønsted&lt;/a&gt; acidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical metal salts (let's say MCl3) undergo hydrolysis (presumably in an aqueous solution) as Lewis basic water (which donates electrons to the acid in this context) coordinates (binds) to the Lewis acidic metal (which accepts electrons from the Lewis base). After, before or during coordination (it's hard to tell which without extensive experiment), water will give up one or two protons (i.e. hydrogen atom minus an electron), thereby generating a metal oxo/hydroxide (in this case, e.g. MO2H) or metal hydroxide (e.g. M(OH)3), either of which will be accompanied by acid in solution in the form of 3 eq of HCl (which would be completely dissociated into H+ and Cl-).&lt;/p&gt;&#xA;" OwnerUserId="7211" LastEditorUserId="5026" LastEditDate="2018-03-04T02:54:30.793" LastActivityDate="2018-03-04T02:54:30.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14570" PostTypeId="2" ParentId="14539" CreationDate="2014-07-23T17:44:10.437" Score="1" Body="&lt;p&gt;The oxidation state of a transition metal is an accounting device.  It's not the number of &quot;lost electrons&quot;.  In potassium permanganate, which has Mn (VII), the manganese atom shares its seven 3d electrons with the ligand (oxygen) in what are essentially covalent bonds, and the &quot;positive charge&quot; associated with the metal is not very different than what it is for ionic calcium compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &quot;paired electrons are less likely to be involved in bonding than unpaired electrons&quot; -- both paired and unpaired electrons in a transition metal can participate in bonding, provided they are in the outermost shell.  The paired electrons in inner shells are too deep in energy to take part in bonding.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-23T17:44:10.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14571" PostTypeId="2" ParentId="7683" CreationDate="2014-07-23T17:46:16.767" Score="5" Body="&lt;p&gt;User4604 is correct with the solvation argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, the acidities and basicities of these complexes are usually measured in water, where the free energy of hydrating a charged species can be very high (i.e. it is favorable to dissolve charged molecules). If nonpolar substituents are placed near the charged moiety, it will prevent that charged group from being effectively interacting with and becoming solvated by water. In doing so, the nonpolar substituent will influence the pKa (basicity and acidity) of the charged group. Whether the nonpolar group makes the moiety more basic or acidic depends on specific case (2-methyl benzoic acid will be more basic/less acidic than 4-methyl benzoic acid; however 2-methyl aniline will be less basic/more acidic than 4-methyl aniline). &lt;/p&gt;&#xA;" OwnerUserId="7211" LastActivityDate="2014-07-23T17:46:16.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14572" PostTypeId="1" CreationDate="2014-07-23T17:48:13.387" Score="2" ViewCount="445" Body="&lt;p&gt;I've been playing around with filtration methods, and got my hands on a casting machine for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first mission was to find out how porosity can affect filtration. To see this for myself, I just ran some titratable solvent through the membrane I produced, and analyzed the difference in concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking through some textbooks, I found a neat formula for calculating porosity, but there's one thing that's been bugging me. The formula requires the density of the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't something that I can just use the Archimedes Principle to get. I need to look at the wet weight and dry weight of the membrane or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dilemma right now is I can't find a single resource that lets me measure the density of a membrane without requiring either some pricey machine or extremely esoteric methods that's beyond my abilities. I'm really at a loss right now for this particular method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's &lt;a href=&quot;http://membrane.ces.utexas.edu/files/2012/07/Water-Group-Analytical-and-Characterization-Capabilities.pdf&quot; rel=&quot;nofollow&quot;&gt;this equipment&lt;/a&gt; here that's usable, but that looks like one high-end son of a gun. I really don't want to throw any more money than I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I measure the density of a membrane with tools available in the home?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7218" LastEditorUserId="4945" LastEditDate="2014-07-24T02:48:40.990" LastActivityDate="2014-07-24T02:48:40.990" Title="How do I measure the density of a cellulose acetate membrane?" Tags="&lt;physical-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14573" PostTypeId="2" ParentId="14545" CreationDate="2014-07-23T20:51:42.583" Score="2" Body="&lt;p&gt;I once had a humidifier and I ran it with tap water.  I soon noticed a build up of white dust on the carpet right in front of the humidifier.  I suspect your situation might be related to what caused my white dust.  Tap water contains salts (just like table salt), the harder the water the more salts it contains.  They are present in low concentration so you don't normally see them.  In my case, as more and more tap water evaporated it led to a visible build up of these salts.  In your case, as the water on the carpet dried, it left behind the same salts.  Maybe you really soaked your carpet with a lot of water, maybe your water is very hard, and finally, maybe salts were actually put into the carpet during their manufacturing process - and maybe washing them with vinegar (dilute acetic acid) solubilized the salts and brought them to the surface, making them visible upon drying.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-23T20:51:42.583" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14574" PostTypeId="2" ParentId="7427" CreationDate="2014-07-23T21:32:57.133" Score="2" Body="&lt;p&gt;Covalent and ionic bonds are best thought of as opposite ends of a spectrum, where electrons are shared evenly (as in the case of a symmetric molecule like N2) or unevenly (like HF). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, bonds with more ionic character are generally considered stronger due to the significant electostatic contribution to bonding (the cationic and anionic components experience attraction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle is the core concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Pauling_electronegativity&quot; rel=&quot;nofollow&quot;&gt;Pauling Electronegativities&lt;/a&gt;, where the degree to which electrons in a bond are &quot;attracted&quot; to one atom over the other (that is, the degree to which a bond is ionic) is ultimately derived from experimental bond dissociation energies (i.e. bond strengths).&lt;/p&gt;&#xA;" OwnerUserId="7211" LastActivityDate="2014-07-23T21:32:57.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14575" PostTypeId="1" CreationDate="2014-07-24T01:55:18.193" Score="5" ViewCount="16122" Body="&lt;p&gt;A ligand is a neutral molecule or negative ion with at least one lone pair of electrons which forms a dative covalent bond with a complex ion that has a vacant, energetically accessible orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is said that in terms of ligand strength, or 'binding power', the cyanide ion is stronger than ammonia, which is in turn stronger than water, which is in turn stronger than the chloride ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines the strength of a ligand?&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="7198" LastEditDate="2014-07-24T03:45:28.867" LastActivityDate="2014-07-24T04:47:45.573" Title="What makes a ligand stronger than another?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14576" PostTypeId="1" AcceptedAnswerId="14577" CreationDate="2014-07-24T02:36:24.193" Score="1" ViewCount="670" Body="&lt;p&gt;I bought a low salt Ketchup from Heinz and it says it uses ‘ALSOSALT’ substitute (potassium chloride). It says it does have 5 mg of sodium. Is potassium chloride (with possible additives) a good alternative to sodium chloride and does it not cause problems for the heart or general health?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorDisplayName="user7951" LastEditDate="2015-05-17T12:57:22.167" LastActivityDate="2015-05-17T12:57:22.167" Title="Are salt substitutes good alternatives to sodium chloride?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14577" PostTypeId="2" ParentId="14576" CreationDate="2014-07-24T02:44:52.493" Score="2" Body="&lt;p&gt;Small quantities of potassium are fine. Windsor Salt of Canada makes a 50% blend of sodium chloride and potassium chloride called Half Salt. You could use it like regular salt, although it would be good to decrease the amount you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nerve impairment and muscle function are affected from too much potassium.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-24T02:44:52.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14578" PostTypeId="1" CreationDate="2014-07-24T03:30:10.747" Score="33" ViewCount="8434" Body="&lt;p&gt;Consider all the MOs of some isolated molecule. (It could be a single&#xA;atom too; I'll use MO to refer to AOs as well.) Number them in&#xA;increasing order of the number of nodes (node = surface where the wave&#xA;function has zero density). Orbitals with the same number of nodes can&#xA;be numbered in any order. Now you have a sequence of orbitals $O_1,&#xA;O_2, ...$. Let their respective energies be $E_1, E_2, ...$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be &quot;common knowledge&quot; that $E_n \le E_{n+1}$ for any such&#xA;system and any $n$. As Martin put it to me yesterday, &quot;&lt;a href=&quot;http://chat.stackexchange.com/rooms/15903/counting-nodal-planes-in-cyclopropane&quot;&gt;An orbital&#xA;with 47 nodes can never be lower in energy than one with only 46&lt;/a&gt;.&quot; (Follow up on &lt;a href=&quot;https://chemistry.stackexchange.com/q/14528/4945&quot;&gt;Counting Nodal Planes in cyclopropane&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For various reasons, given below, I think this cannot be true in&#xA;general, and I'd really like to know under what conditions it is&#xA;&lt;em&gt;known&lt;/em&gt; to be true. &quot;Known&quot; here can mean either a rigorous statement&#xA;with a reference to a proof (a trivial example would be: It is true&#xA;for a one-electron atom; we can calculate the energies exactly) or a&#xA;precise statement with empirical justification (something like &quot;no&#xA;counterexamples are known for class X of molecules&quot; - again with a&#xA;reference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Important&lt;/em&gt;: Please, I'm not looking for an explanation that re-states&#xA;the rule in some equivalent or even looser fashion (&quot;more nodes means&#xA;the orbital is larger and less dense, therefore it must be higher in&#xA;energy&quot;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why do I think the statement can't be always true? Well, a calcium&#xA;atom has a filled 4s orbital and empty 3d orbitals. &lt;em&gt;If this does not&#xA;count as a counterexample, please explain what notion of orbital&#xA;energy the statement applies to.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I'm happy to believe that two MOs must satisfy the rule if&#xA;they have &quot;comparable&quot; sets of nodes (say, constant $n_x$ and $n_y$ in the&#xA;example I'm about to discuss), but I'd like to understand what is&#xA;meant by &quot;comparable&quot; in general. In a molecule with no symmetry, are&#xA;there comparable MOs at all? If so, how do we know if two given MOs&#xA;are comparable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of a much simpler system, the 3D rectangular box, may&#xA;also be relevant.  The energy levels for such a box are of course&#xA;$$&#xA;\frac{\hbar^2\pi^2}{2m}\biggl(&#xA;\frac{n_x^2}{L_x^2}+&#xA;\frac{n_y^2}{L_y^2}+&#xA;\frac{n_z^2}{L_z^2}&#xA;\biggr),&#xA;$$&#xA;where $n_x$, $n_y$ and $n_z$ are one more (or one less, if you count&#xA;the walls) than the number of nodal planes in the respective&#xA;direction. If we take $L_x = L_y = 1$ and $L_z = 0.1$ (say), the&#xA;wavefunction with $n_x=5$, $n_y=1$, $n_z =1$ has energy $5^2+ 1^2 +&#xA;(1/0.1)^2 = 126$ and 4 nodes (or 10, if you count the walls), while&#xA;the wavefunction for $n_x=1$, $n_y=1$, $n_z=2$ has energy $402$ and 1&#xA;node (or 7). So clearly the rule is not true here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted that molecules are not boxes, but this does show that&#xA;arguments based simply on the number of sign changes are not rigorous,&#xA;so they don't answer my question.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-09-26T02:23:10.773" Title="When is it true that more nodes equals higher energy?" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="14579" PostTypeId="2" ParentId="14575" CreationDate="2014-07-24T04:35:56.130" Score="2" Body="&lt;p&gt;One definite factor is the ability of the ligand to stabilize its lone pair. You observe that the cyanide ion is a better ligand than the nitrogen ligand. Let's think about this. The lone pair that reacts on the cyanide ion is the lone pair on the carbon atom. Not the lone pair on the nitrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon is less electronegative than nitrogen. Thus carbon is less able to stabilize its valence electron density, which in turns enhances the reactivity of its valence electron density through increased negative charge density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same with sulfide and oxide anions. Sulfur is less electronegative than oxygen, and thus less able to stabilize valence electron density. Hence the &lt;em&gt;much&lt;/em&gt; greater stability of silver sulfide as opposed to silver oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other competing factors include charge density; the Lewis acid the ligand is coordinating to, medium (solvent), etc. For example the oxide anion is more basic than the sulfide anion in water solvent; i.e. the oxide anion forms a stronger bond with the hydrogen proton in water. But we just saw that silver sulfide is much more stable than silver oxide. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-24T04:41:13.840" LastActivityDate="2014-07-24T04:41:13.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14580" PostTypeId="2" ParentId="14057" CreationDate="2014-07-24T04:45:21.107" Score="19" Body="&lt;p&gt;Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth &lt;em&gt;et al.&lt;/em&gt; Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz &lt;em&gt;et al&lt;/em&gt;., and these included some rather nice low temperature &lt;sup&gt;19&lt;/sup&gt;F and &lt;sup&gt;13&lt;/sup&gt;C NMR studies in both solution and solid state (and make it much quicker and easier for me to read).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assign the following isomers 1–4, as is done in the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NJl3w.png&quot; alt=&quot;Isomers of fluorobullvalene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oth &lt;em&gt;et al.&lt;/em&gt;&lt;sup&gt;[1]&lt;/sup&gt; studied fluorobullvalene at −25 °C, and found no evidence by &lt;sup&gt;19&lt;/sup&gt;F of isomer 1. The equilibrium populations of isomers 4:3:2 was 78%:7%:15%. However, they could not distinguish between isomers 3 and 2, so it could be 78:15:7. In any case, predominantly isomer 4. They do state, however, that isomer 1 must exist [&quot;Das C-Isomer muss also durchlaufen werden.&quot;], as the Cope rearrangement mechanism for isomer 4 to go to isomer 3 must go through isomer 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luz &lt;em&gt;et al.&lt;/em&gt;&lt;sup&gt;[2]&lt;/sup&gt; studied the solution state Cope rearrangement of a number of monosubstituted bullvalenes.&#xA;At −30 °C, they identify the ratio of isomers 4:3:2 to be 87%:3.4%:9.6%. Again no evidence of isomer 1 by &lt;sup&gt;19&lt;/sup&gt;F at this temperature. However, at −55 °C, they observe a peak in the &lt;sup&gt;19&lt;/sup&gt;F spectrum that they ascribe to isomer 1, with an intensity of 0.2% relative to isomer 4. This peak broadens beyond detection at −40 °C. The ratios of isomers at −55 °C, then, was found to be 1.0:0.018:0.065:0.002 for isomers 4:3:2:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just out of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the other monsubstituted bullvalenes all have different preferred isomer distributions in solution. In the solid state, all monosubstituted bullvalenes studied by Luz crystallize as well-ordered single isomer crystals. Fluorobullvalene&lt;sup&gt;[3]&lt;/sup&gt; crystallizes exclusively as isomer 4. Cyanobullvalene, bullvalenecarboxylic acid&lt;sup&gt;[4]&lt;/sup&gt;, and (ethylthio)bullvalene&lt;sup&gt;[5]&lt;/sup&gt; crystallize exclusively as isomer 3. Bromo- and iodobullvalene&lt;sup&gt;[6]&lt;/sup&gt; crystallize entirely as isomer 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful references for you:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Oth, J. F. M.; Merényi, R.; Röttele, H.; Schröder, G. Fluor-, chlor- und jodbullvalen. Tetrahedron Lett. 1968, 9 (36), 3941–3946. &lt;a href=&quot;https://doi.org/10.1016/S0040-4039(00)72372-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/S0040-4039(00)72372-0&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Poupko, R.; Zimmermann, H.; Müller, K.; Luz, Z. Dynamic NMR Investigation of the Cope Rearrangement in Solutions of Monosubstituted Bullvalenes. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;118&lt;/em&gt; (34), 7995–8005. &lt;a href=&quot;https://doi.org/10.1021/ja954004t&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja954004t&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 1. Carbon-13 NMR and X-ray Crystallography of Fluorobullvalene. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;118&lt;/em&gt; (34), 8006–8014. &lt;a href=&quot;https://doi.org/10.1021/ja954005l&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja954005l&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 2. Carbon-13 NMR and X-ray Crystallography of Cyanobullvalene and Bullvalenecarboxylic acid. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;118&lt;/em&gt; (34), 8015–8023. &lt;a href=&quot;https://doi.org/10.1021/ja954006d&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja954006d&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Luger, P.; Roth, K. X-Ray, n.m.r., and theoretical studies of the structures of (ethylthio)bullvalene. &lt;em&gt;J. Chem. Soc., Perkin Trans. 2 (1972-1999)&lt;/em&gt; &lt;strong&gt;1989,&lt;/strong&gt; 649–655. &lt;a href=&quot;https://doi.org/10.1039/P29890000649&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1039/P29890000649&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Luz, Z.; Olivier, L.; Poupko, R.; Müller, K.; Krieger, C.; Zimmermann, H. Bond Shift Rearrangement of Chloro-, Bromo-, and Iodobullvalene in the Solid State and in Solution. A Carbon-13 and Proton NMR Study. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1998,&lt;/strong&gt; &lt;em&gt;120&lt;/em&gt; (22), 5526–5538. &lt;a href=&quot;https://doi.org/10.1021/ja9728029&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja9728029&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="16683" LastEditDate="2017-08-29T10:05:47.880" LastActivityDate="2017-08-29T10:05:47.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14582" PostTypeId="1" AcceptedAnswerId="14612" CreationDate="2014-07-24T05:06:02.320" Score="13" ViewCount="4409" Body="&lt;p&gt;Recently I was watching a video on &lt;em&gt;Valence Bond Theory&lt;/em&gt;, on the &lt;em&gt;MIT Open-Course&lt;/em&gt; youtube channel. The teacher there said that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For hybridization to occur, electron promotion from fully filled orbitals to empty ones is a must.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But here in a book it is specifically mentioned that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electron promotion, though common, but is not '&lt;em&gt;a must&lt;/em&gt;' for hybridization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, it is clearly written that &lt;strong&gt;&lt;em&gt;sometimes&lt;/strong&gt; even the filled orbitals take part in hybridization.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried finding some examples of such a case on the web, but could not. It would be a great help if someone tells me even one example where the fully filled orbitals also take part in hybridization, with an explanation why it is so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link for the 41 min long video is &lt;a href=&quot;http://www.youtube.com/watch?v=3AVSORIJJJY&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="7448" LastEditorDisplayName="user2619" LastEditDate="2015-07-03T11:08:01.210" LastActivityDate="2019-10-30T04:12:18.173" Title="Do filled Orbitals also hybridize?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="12" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14583" PostTypeId="2" ParentId="14439" CreationDate="2014-07-24T05:11:14.707" Score="3" Body="&lt;p&gt;@bennorris method works, but I think my method may save some time.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lc} \hline&#xA;&amp;amp;\ce{CH3COO-} &amp;amp;\ce{&amp;lt;=&amp;gt;}&amp;amp;\ce{H+} &amp;amp;+&amp;amp;\ce{CH3COO-} \\ \hline&#xA;\text{I}&amp;amp; \pu{0.02 M} &amp;amp;&amp;amp;\pu{0 M} &amp;amp;&amp;amp;\pu{0 M} \\ &#xA;\text{C}&amp;amp; -x &amp;amp;&amp;amp;+x  &amp;amp;&amp;amp;+x \\ &#xA;\text{E}&amp;amp; 0.02-x  &amp;amp;&amp;amp;x &amp;amp;&amp;amp;x \\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I keep the acetic acid concentration to be the same because I assume the x value will be really small( you can prove it later)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_a = 1.8\times 10^{-5}=\dfrac{[\ce{H+}][\ce{C2H3O2-}]}{[\ce{HC2H3O2}]} = \dfrac{(x)(x)}{(0.02)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x=6.0 \times 10^{-4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can prove the assumption by dividing your x value with your initial concentration of acetic acid&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \dfrac{[\pu{6\times 10^{-4} M}]}{[\pu{0.02 M}]} =0.03 =3\% &amp;lt; 5\%$$&lt;/span&gt;&#xA;If the percentage you get is smaller than five percent, your assumption will be correct.&#xA;This method will save you some time as you don't have to deal withe the quadratic equation and give you a approximate answer&lt;/p&gt;&#xA;" OwnerUserId="7221" LastEditorUserId="95133" LastEditDate="2020-09-12T11:46:31.737" LastActivityDate="2020-09-12T11:46:31.737" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14584" PostTypeId="1" CreationDate="2014-07-24T05:26:26.497" Score="2" ViewCount="162" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$1.00~\mathrm{L}$&lt;/span&gt; gas at &lt;span class=&quot;math-container&quot;&gt;$100~^\circ\mathrm{C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$500~\mathrm{Torr}$&lt;/span&gt; contains &lt;span class=&quot;math-container&quot;&gt;$70.0\ \%~\ce{He}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$30.0\ \%~\ce{Ne}$&lt;/span&gt; by mass. What is the partial pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{He}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is &lt;span class=&quot;math-container&quot;&gt;$350~\mathrm{Torr}$&lt;/span&gt; whereas the correct answer is &lt;span class=&quot;math-container&quot;&gt;$421~\mathrm{Torr}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution (&lt;a href=&quot;https://i.stack.imgur.com/m3hZG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;source; commas &lt;code&gt;,&lt;/code&gt; should be points &lt;code&gt;.&lt;/code&gt;; I used to write it this way.&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{aligned}&#xA;  n_\mathrm{total} &amp;amp;= \frac{pV}{RT} = \frac{(1\ \mathrm l)(0.652\ \mathrm{atm})}{(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}=0.0215~\mathrm{mol}\\&#xA;  70\ \%~\ce{He} &amp;amp;\implies (70\ \%)(0.0215\ \mathrm{mol}) = 0.0151~\mathrm{mol}~\ce{He}\\&#xA;  p_\ce{He} &amp;amp;= \frac{nRT}{V}\\&#xA;            &amp;amp;= \frac{(0.0151\ \mathrm{mol})(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}{1\ \mathrm l} = 0.460~\mathrm{atm}\end{aligned}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.460~\mathrm{atm}\cdot\frac{101325~\mathrm{Pa}}{1~\mathrm{atm}}\cdot\frac{1~\mathrm{Torr}}{133.322~\mathrm{Pa}}\approx 350~\mathrm{Torr}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorDisplayName="user7951" LastEditDate="2019-05-30T13:44:02.297" LastActivityDate="2019-05-30T13:45:21.620" Title="A 1.00 L gas at 100 °C and 500 Torr contains 70.0 % He and 30.0 % Ne by mass. What is the partial pressure of He?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14586" PostTypeId="2" ParentId="14584" CreationDate="2014-07-24T08:07:26.997" Score="4" Body="&lt;p&gt;You're getting a different answer because partial pressure is proportional to &#xA;the fraction of each gas &lt;em&gt;in moles&lt;/em&gt;, rather than by weight. Why is that?&#xA;Looking at &lt;span class=&quot;math-container&quot;&gt;$pV=nRT$&lt;/span&gt;, you can see that the ratio between partial&#xA;pressures is the same as the ratio between number of moles (since &lt;span class=&quot;math-container&quot;&gt;$V,&#xA;R, T$&lt;/span&gt; are the same for both gases). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you respect proportions,&#xA;your answer will come out the same whether you consider &lt;span class=&quot;math-container&quot;&gt;$1.00\ \mathrm{L}$&lt;/span&gt; of the mix&#xA;or some unspecified number of liters. This lets you avoid a messy&#xA;calculation with &lt;span class=&quot;math-container&quot;&gt;$pV/RT$&lt;/span&gt;.&#xA;You may as well assume that there are 70 g of helium and 30 g of&#xA;neon. Find how many moles of each gas that represents. The partial&#xA;pressures of each gas are in the same proportion as the number of&#xA;moles. With this and the total pressure, you'll get the answer.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorDisplayName="user7951" LastEditDate="2019-05-30T13:45:21.620" LastActivityDate="2019-05-30T13:45:21.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14587" PostTypeId="2" ParentId="14404" CreationDate="2014-07-24T08:26:56.773" Score="5" Body="&lt;p&gt;Molecular chemistry is based predominantly on interactions through covalent bonds. It looks at molecules as individual entities, and is limited in its scope of accomodating intermolecular interactions. Your solution of sugar in water could easily be studied as a molecular system, if all you are interested in is the sugar molecule itself. Subtle differences in molecular conformation of the sugar in different solvent systems, for instance, would simply be written off as 'solvent effects'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supramolecular_chemistry&quot; rel=&quot;nofollow&quot;&gt;Supramolecular chemistry&lt;/a&gt; is based on intermolecular interactions. Typically, these are based on weak interactions (non-covalent forces), and include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ion-ion interactions&lt;/li&gt;&#xA;&lt;li&gt;ion-dipole interactions&lt;/li&gt;&#xA;&lt;li&gt;dipole-dipole interactions&lt;/li&gt;&#xA;&lt;li&gt;hydrogen bonding&lt;/li&gt;&#xA;&lt;li&gt;van der Waals forces&lt;/li&gt;&#xA;&lt;li&gt;pi stacking&lt;/li&gt;&#xA;&lt;li&gt;hydrophobic interactions&lt;/li&gt;&#xA;&lt;li&gt;and probably some others&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is through these types of interactions of that we can explore phenomena such as host-guest association, self-assembly, metal-ligand chelation, micelles, and even the structure of DNA, as all of these systems rely on the weak intermolecular interactions outlined above. You could also study your aqueous sugar system as a supramolecular system, and here you could consider what predominant intermolecular interactions are at work, and how they affect the overall chemistry of the bulk system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is supposed to exist as &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_cluster&quot; rel=&quot;nofollow&quot;&gt;water clusters&lt;/a&gt;, weak oligomeric and polymeric forms of $\ce{(H2O)_{n}}$ &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="73" LastEditDate="2014-09-04T17:16:34.117" LastActivityDate="2014-09-04T17:16:34.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14588" PostTypeId="1" AcceptedAnswerId="14591" CreationDate="2014-07-24T09:00:08.333" Score="4" ViewCount="4375" Body="&lt;p&gt;My teacher said that &lt;em&gt;propan-1-ol&lt;/em&gt; can also be named &lt;em&gt;propanol&lt;/em&gt; in the latest IUPAC recommendations. I doubt this because Wikipedia only mentions &lt;strong&gt;propan-1-ol&lt;/strong&gt; as the IUPAC name (&lt;em&gt;propanol&lt;/em&gt; being relegated to the &quot;other names&quot; section).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1-Propanol&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/xqnM3YD.png&quot; alt=&quot;im&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &quot;propanol&quot; an acceptable name?&lt;/p&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="16683" LastEditDate="2017-09-02T13:21:13.433" LastActivityDate="2017-09-09T11:02:15.447" Title="Is &quot;propanol&quot; a correct name in IUPAC nomenclature?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14589" PostTypeId="1" CreationDate="2014-07-24T10:02:25.990" Score="0" ViewCount="490" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5frqm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the theory, but am struggling to answer the following discussion questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i)&lt;br&gt;&#xA;Why isn't this represented by: $\ce{Fe^{3+} (aq) + 3 SCN- (aq) &amp;lt;=&amp;gt;  Fe(SCN)3 (aq)}$&lt;br&gt;&#xA;Why do we dilute the solution with water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii)&lt;br&gt;&#xA;I don't understand what the question is asking, can someone please explain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii)&lt;br&gt;&#xA;I understand this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv)&lt;br&gt;&#xA;I am unsure, can someone suggest a reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(v)&lt;br&gt;&#xA;I understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(vi)&lt;br&gt;&#xA;Constant volume of container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDITIONAL QUESTIONS:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I thought ALL equilibrium reactions needed to be in a closed vessel - so if this experiment was conducted in an open test tube, wouldn't this be an error? How would this affect the accuracy of the results - would it be under/over estimated and why?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="7920" LastEditDate="2014-11-14T04:15:52.437" LastActivityDate="2014-11-14T04:15:52.437" Title="What is the appropriate way of writing the Iron/thiocyanate equilibrium?" Tags="&lt;homework&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14590" PostTypeId="1" AcceptedAnswerId="14621" CreationDate="2014-07-24T10:14:30.163" Score="0" ViewCount="891" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VhU3J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I2 and H2 began at the same concentration - then do we start at the same point? ... or is it acceptable to start at different concentrations... &lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2014-07-30T15:10:36.047" Title="Equilibrium Graphs" Tags="&lt;homework&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14591" PostTypeId="2" ParentId="14588" CreationDate="2014-07-24T10:15:59.163" Score="6" Body="&lt;p&gt;The last IUPAC nomenclature recommendations were published in 1993. The current draft documents (release date uncertain) are far more lenient in their acceptance of semi-systematic naming of molecules. Please make sure you understand the differences between the &lt;a href=&quot;http://en.wikipedia.org/wiki/Preferred_IUPAC_name&quot;&gt;Preferred IUPAC Name&lt;/a&gt; (PIN), the &lt;a href=&quot;http://en.wikipedia.org/wiki/Retained_name&quot;&gt;retained name&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Trivial_name&quot;&gt;trivial name&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propanol is not be the &lt;em&gt;preferred&lt;/em&gt; IUPAC name, but significantly, neither is it listed in the tables of &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_671.htm&quot;&gt;Trivial and semisystematic names retained for naming organic compounds.&lt;/a&gt; It is, however, a widely accepted name for this chemical, and you are far more likely to encounter propanol in industry than propan-1-ol. Nobody uses propan-1-ol in the real world, even if it is the preferred IUPAC systematic name.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-07-24T10:15:59.163" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14593" PostTypeId="1" AcceptedAnswerId="14777" CreationDate="2014-07-24T12:09:01.823" Score="4" ViewCount="3322" Body="&lt;p&gt;The atomic radius of Sc is $\pu{162pm}$, Ti is $\pu{147pm}$, $\ce{Fe}$ is $\pu{126pm}$, and that of $\ce{Co}$ is $\pu{125pm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic configuration of $\ce{Fe}$ is $\ce{[Ar] 3d^6 4s^2}$, and that of $\ce{Co}$ is $\ce{[Ar] 3d^7 4s^2}$.  The difference in atomic number, and hence the difference in the number of 3d electrons, is 1. So, due to screening, the extra charge is &quot;canceled&quot; and they have nearly the same radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in atomic number and 3d electron count between $\ce{Sc}$ and $\ce{Ti}$ is also the same — 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know why there is a considerable difference between the radii of Sc and Ti but not between the radii of Fe and Co. &lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="16683" LastEditDate="2018-08-31T16:45:33.347" LastActivityDate="2018-08-31T16:45:56.993" Title="Atomic radii of Sc, Ti, Fe, and Co" Tags="&lt;atoms&gt;&lt;periodic-trends&gt;&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14594" PostTypeId="1" AcceptedAnswerId="14601" CreationDate="2014-07-24T13:56:26.783" Score="2" ViewCount="25836" Body="&lt;p&gt;Given a set of reactants, what aspects should I consider when predicting possible products?&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="194" LastEditDate="2017-03-29T04:24:20.400" LastActivityDate="2017-03-29T05:10:17.127" Title="How to predict the products of a organic chemical reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2017-03-29T05:10:41.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="14595" PostTypeId="1" AcceptedAnswerId="14661" CreationDate="2014-07-24T15:31:29.003" Score="4" ViewCount="1883" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3 g}$&lt;/span&gt; mixture of oxygen and argon is found to occupy a volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{16.2 L}$&lt;/span&gt; when measured at &lt;span class=&quot;math-container&quot;&gt;$\pu{675.9 mmHg}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{43.4 ^\circ{}C}$&lt;/span&gt;. What is the partial pressure of oxygen in this mixture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{436 mmHg}$&lt;/span&gt;, but I didn't manage to get to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it is found by the formula &#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_i=X_i\cdot P_\text{total},$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$P_i$&lt;/span&gt; is the partial pressure and &lt;span class=&quot;math-container&quot;&gt;$X_i$&lt;/span&gt; is the mole fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_\text{total}$&lt;/span&gt; is given to find &lt;span class=&quot;math-container&quot;&gt;$X_i$&lt;/span&gt;. I should divide the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; by the total amount of substances of the mixture &lt;span class=&quot;math-container&quot;&gt;$n_{\ce{O2}}/n_\text{total}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n_\text{total}$&lt;/span&gt; is easy to find with &lt;span class=&quot;math-container&quot;&gt;$n=PV/RT$&lt;/span&gt;,&#xA;but my problem is, that I can't find the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;. The mass in the problem is given for the mixture, so how to extract only the mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;, or in other words, how to get the amount of substance for &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="4945" LastEditDate="2020-02-13T11:23:05.533" LastActivityDate="2020-02-13T17:22:11.703" Title="How to calculate the partial pressure of oxygen in this mixture of oxygen and argon?" Tags="&lt;gas-laws&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14596" PostTypeId="2" ParentId="6109" CreationDate="2014-07-24T15:51:58.870" Score="4" Body="&lt;p&gt;You  can find  the answer in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M. Randić &#xA;    Aromaticity of Polycyclic Conjugated Hydrocarbons &#xA;    Chem. Rev. 103 (2003) 3449-3605.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on the kind of conjugated circuits within the set of Kekule valence structures.&#xA;If all conjugated circuits are of size 4n+2 (as the are in pyrene and all benzenoid hydrocarbons and a number of nonbenzenoid systems (like azulene and other) molecular is aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If molecule has 4n+2 and 4n conjugated circuits it is at best partially aromatic (which includes also buckminsterfullerene C60).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has only 4n conjugated circuis (as cyclooctatetrane) it is not aromatic. &lt;/p&gt;&#xA;" OwnerUserId="7223" LastActivityDate="2014-07-24T15:51:58.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14598" PostTypeId="1" CreationDate="2014-07-24T16:07:53.553" Score="2" ViewCount="1820" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The acid-base indicator methyl red has a &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-5}$&lt;/span&gt;. Its acidic form is red while its alkaline form is yellow. If methyl red is added in a colorless solution with a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 7$&lt;/span&gt;, the color will be :&lt;/p&gt;&#xA;&lt;p&gt;a) Pink&lt;/p&gt;&#xA;&lt;p&gt;b) Red&lt;/p&gt;&#xA;&lt;p&gt;c) Orange&lt;/p&gt;&#xA;&lt;p&gt;d) Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Seeing the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$1 \times 10^{-5}$&lt;/span&gt;, I concluded that it is a weak acid. Thus I think that when it is added to a neutral solution (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}=7$&lt;/span&gt;), the resultant solution must be weakly acidic and thus its color must be red.&lt;/p&gt;&#xA;&lt;p&gt;However, contrary to my logic, the answer given in my book is option&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;d) Yellow&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is there a flaw in my argument?&lt;/p&gt;&#xA;" OwnerUserId="4854" LastEditorUserId="83758" LastEditDate="2020-06-12T20:05:26.217" LastActivityDate="2020-06-12T20:05:26.217" Title="Find the color of the solution" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14599" PostTypeId="2" ParentId="14598" CreationDate="2014-07-24T16:28:54.540" Score="2" Body="&lt;p&gt;A pH indicator is generally a weak acid and often work in a fashion similar to buffers in that most of the time both the acid and its conjugate base exist in significant concentration. From the Ka of the indicator you should therefore be able to find the &quot;halfway&quot; point of the indicator or the middle of the range of pH values over which the indicator operates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about how indicators work. Indicators work simply by mixing colors. If I have an indicator which is black at pH = 0.0 and white at pH = 2.0, then I know that the protonated form of the indicator must be black-colored, and that the deprotonated form of the indicator must be white colored. This should make sense. As pH increase, hydroxide ion molarity increases, and hydroxide ion deprotonates other molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, at pH = 1.0, we should have a 50/50 mix of black and white molecules. In other words, the indicator shows a gray color at pH = 1.0. At pH = 1.0 we also have a buffer solution - we have the acid and conjugate base - and specifically in a 1:1 ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By application of the Henderson-Hasselbach equation we will realize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{pH = pK_{a} + log\frac{base}{conjugate~acid}}$ simplifies to ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{pH = pK_{a}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you take the problem from here. Don't forget to apply common-sense considerations. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-24T16:37:08.340" LastActivityDate="2014-07-24T16:37:08.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14600" PostTypeId="2" ParentId="14598" CreationDate="2014-07-24T16:37:31.747" Score="2" Body="&lt;p&gt;It's an indicator problem, so trying to balance the pH seems a bit off, since you typically add only a few drops of indicator to determine a solution's pH. Ideally I'd imagine you'd want the indicator to interfere as little as possible with whatever's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, different indicators have different ranges. (They don't have to change color at exactly ${pH=7}$.) You're given that the acidic color is red, and the basic color is yellow. The ${K_a}$ helps to determine this switching point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding simple indicators, if the solution's pH is lower than the indicator's ${pK_a}$ you'll see the acidic color. Vice versa for the basic color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ${pK_a}$ that corresponds to ${K_a=1*10^{-5}}$ is simply ${5}$. As ${7 &amp;gt; 5}$, you'd expect the alkaline color to appear (in this case, yellow)&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-07-24T16:37:31.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14601" PostTypeId="2" ParentId="14594" CreationDate="2014-07-24T16:41:18.233" Score="4" Body="&lt;p&gt;You need to build a working knowledge of known reaction mechanisms that include the functional groups of your reactants. Put another way, (1) identify the functional groups present in your reactants and (2) check known reaction mechanisms that proceed by reaction of these functional groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, the only way to be good at this is to study and memorize organic reactions. There's a number of resources to help with this (e.g. online flashcards, and I really like &lt;a href=&quot;http://chemistrybydesign.oia.arizona.edu/&quot; rel=&quot;nofollow&quot;&gt;Chemistry by Design&lt;/a&gt;, an app that you can use on a smartphone).&lt;/p&gt;&#xA;" OwnerUserId="7211" LastActivityDate="2014-07-24T16:41:18.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14605" PostTypeId="2" ParentId="14594" CreationDate="2014-07-24T19:51:31.727" Score="3" Body="&lt;p&gt;Sharpen up your general chem skills.  One skill is recognizing redox reactions.  Many organic reactions are redox.  In problems where you have the reactants and products and are prompted to give the reagents you can recognize reductions and oxidations.  For the problems where you have the reactant and conditions you can more easily recognize oxidizers and reducers.  You can work on relative acidity and identify the most acidic proton in a molecule.  You can recognize the different types of carbon in a molecule and remember which are more reactive to which conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep a notebook where you put in 5-10 reactions a day and memorize them.  You probably don't want to start today but you wish you started a month ago and a month from now you'll wish you started today.  Just learning name reactions would be a good start.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-07-24T19:51:31.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14606" PostTypeId="2" ParentId="14589" CreationDate="2014-07-24T19:52:10.617" Score="2" Body="&lt;p&gt;The second representation is more accurate because in water there can be no ``naked'' $\ce{Fe^3+}$ ion; as it goes into solution, iron combines with water to form an ion consisting of&#xA;a $\ce{Fe}$ atom surrounded by six molecules of water. It is the oxygens that form bonds with the iron atom, in a roughly octahedral arrangement: two oxygen&#xA;atoms in each coordinate direction, one on either side. The reaction&#xA;consists of the replacement of &lt;em&gt;one&lt;/em&gt; of these coordinated water&#xA;molecules by the $\ce{SCN-}$ ion; now the iron has bonds to 5 oxygens&#xA;and 1 nitrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of $\ce{H2O}$ to the formula doesn't mean we are&#xA;diluting the solution with water.  It just means we're letting the&#xA;water that's  present anyway play a role in the equation, to reflect reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not clear to me what they are asking in (ii).  The colors change&#xA;dramatically when you mix $\ce{[Fe(H2O)6]^3+}$ (pale yellow) with $\ce{SCN-}$&#xA;(colorless); the result is deep red. Perhaps some context will help.  Read the discussion of the reaction in the text to see what they say about color intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iv) Again, what is the context?  If the text is about the spectrospic&#xA;measuring of the equilibrium constant, low&#xA;concentrations ensure that the reation obeys the ideal law.  In&#xA;concentrated solutions the law doesn't work quite right unless you&#xA;replace concentrations by activities, something that is only&#xA;indirectly measurable (it's whatever makes the formula work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(vi) Again, we need to know the context.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No, not all equilibrium reactions need to be in a closed vessel. An&#xA;aqueous reaction that only takes a short time (so that for example&#xA;water won't evaporate appreciably, messing up the concentration) is&#xA;just fine in a test tube.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-24T19:59:01.180" LastActivityDate="2014-07-24T19:59:01.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14607" PostTypeId="2" ParentId="14594" CreationDate="2014-07-24T20:34:06.120" Score="1" Body="&lt;p&gt;Short answer: You can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: for simple substrates and simple reagents there is a big stock of reaction protocols with more-or-less known general mechanisms, that can give a good guess of resulting product. However, in case of complex substrates or tricky reagents (yet unknown) strange transformations may occur. They can get ad-hoc explanations, but.. that's not satisfactory enough. A really good book, considering key intermediates and most important reactions with mechanisms is March's Organic Chemistry (that's probably one of the best books on the subject).&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-24T20:34:06.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14608" PostTypeId="2" ParentId="14475" CreationDate="2014-07-24T23:56:55.053" Score="4" Body="&lt;p&gt;OK, let's go one question at a time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does having constant pressure automatically mean there will be a temperature change and if so, why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite.  &lt;em&gt;Adding energy&lt;/em&gt; means there will likely be a temperature change. What&#xA;constant pressure means is that it will be easy to calculate this change, because you can use the value of the &lt;em&gt;constant-pressure molar heat capacity&lt;/em&gt;, $C_p$, which is the same for all diatomic ideal gases.  It is also independent of temperature, volume and pressure.  You will find it in your book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar heat capacity tells you how much energy you need to transfer to one mole of the gas to raise the temperature by one degree.  You now have everything you need to know by how much the temperature changes, and therefore the final temperature. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What about volume? Why do we not need to know the volume for this equation? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually you do know the volume, or can easily figure it out.  And though the volume is not important for finding the temperature change, for the reason explained above, it's helpful in finding the change in internal energy. Why?&#xA;The change in internal energy is the energy you put in (500 J) minus whatever energy the gas spends expanding.  Since the pressure is constant the energy spent expanding is simply the product of the pressure by the increase in volume.  You can calculate the increase in volume because you know the pressure, number of moles and initial/final temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a bit of meta advice.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am trying to understand how to approach this question without simply plugging the numbers into an equation. Can someone please explain their thought process while solving this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't make clear what, if anything, you know (either conceptually or equation-wise) and what, if anything, you've tried. &#xA;You'll get help faster if you show you've done your part.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Basically, I want to understand what's going on conceptually rather than memorize a few rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Understanding what's going on conceptually goes hand-in-hand with learning the rules.  You can't really do one without the other.  (Well, you can memorize the rules without understanding the concepts, but you won't know which ones to apply.)&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-25T00:15:05.573" LastActivityDate="2014-07-25T00:15:05.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14610" PostTypeId="1" CreationDate="2014-07-25T00:24:38.907" Score="1" ViewCount="115" Body="&lt;p&gt;What would be the product if baking soda and common salt and vinegar are added together?&lt;/p&gt;&#xA;" OwnerUserId="5698" LastEditorUserId="4945" LastEditDate="2014-07-25T10:56:58.767" LastActivityDate="2014-07-25T10:56:58.767" Title="What is the chemistry of this household cleaning recipe?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14611" PostTypeId="2" ParentId="14610" CreationDate="2014-07-25T01:01:48.947" Score="5" Body="&lt;p&gt;Baking soda is sodium bicarbonate, and vinegar is dilute acetic acid. Common salt (table salt) is sodium chloride. Sodium bicarbonate will react with acetic acid to form sodium acetate (in solution), carbon dioxide gas, and water. Sodium chloride will not participate in the reaction. The reaction equation would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{NaHCO3(s) + HCH3COO(aq) -&amp;gt; NaCH3COO(aq) + CO2(g) + H2O(l)}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-25T01:01:48.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14612" PostTypeId="2" ParentId="14582" CreationDate="2014-07-25T01:02:26.427" Score="4" Body="&lt;p&gt;You mean like ammonia? If the filled orbital did not partake in hybridization, then the molecule would be flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water would be linear if the filled orbitals did not partake in hybridization.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-25T01:02:26.427" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14613" PostTypeId="2" ParentId="14582" CreationDate="2014-07-25T01:48:36.033" Score="11" Body="&lt;p&gt;First off, hybridization is a concept chemists developed to help explain reality (their observations).  Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A &quot;rule&quot; on hybridization: hybridization occurs in response to a bonding interaction.  Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form &quot;directed&quot; bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce.  Finally, energetically speaking, hybridization is a net neutral process.  When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs.  So if some electrons are &quot;promoted&quot;, then some are lowered in energy as well&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals.  The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon.  In the case of phosphine ($\ce{PH_3}$) the situation is reversed.  The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes.  The central phosphorous atom remains unhybridized.  In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I like the concept of hybridization and find it quite useful.  But you have to be careful where to apply it.  Hybridization is best when it is used to explain reality after the fact.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-25T01:48:36.033" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14614" PostTypeId="1" CreationDate="2014-07-25T02:12:29.517" Score="3" ViewCount="399" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the following half-reactions:&#xA;   &lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;    \ce{Ce^{4+} + e^{−} &amp;amp;-&amp;gt; Ce^{3+}} &amp;amp; E° &amp;amp;= \pu{1.72 V}\\&#xA;    \ce{Fe^{3+} + e^{−} &amp;amp;-&amp;gt; Fe^{2+}} &amp;amp; E° &amp;amp;= \pu{0.771 V}\\&#xA; \end{align}&lt;/span&gt;&lt;br&gt;&#xA;   A solution is prepared by mixing &lt;span class=&quot;math-container&quot;&gt;$\pu{3.0 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.30 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\pu{2.0 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.12 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ce^{4+}}$&lt;/span&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)? Calculate &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ce^{4+}]}$&lt;/span&gt; in the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt at solving the problem (which was incorrect):&lt;/strong&gt;&lt;br&gt;&#xA;First to find the potential (relative to SHE):&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;E&amp;amp;=E^\circ-0.05916\cdot\log(\ce{Fe^{2+}/Fe^{3+}})\\ &#xA;0.771&amp;amp;=E^\circ-0.05916\cdot\log((3\cdot0.3)/(3\cdot0.3 + 2\cdot0.12))\\&#xA;E^\circ&amp;amp;=\pu{0.7649 V}\\&#xA;\end{align}&lt;/span&gt;&#xA;I don't think this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&#xA;I am not sure how to find &lt;span class=&quot;math-container&quot;&gt;$E^\circ$&lt;/span&gt; cerium or how to find &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ce^{4+}]}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7228" LastEditorUserId="83758" LastEditDate="2020-06-06T11:07:34.217" LastActivityDate="2020-06-06T11:07:34.217" Title="How to find the potential of a mixed Iron/ Cer solution?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14615" PostTypeId="2" ParentId="14614" CreationDate="2014-07-25T04:30:12.853" Score="3" Body="&lt;p&gt;First it helps to have an idea of what will happen. Since 1.72 V is a&#xA;lot more than 0.771 V (a difference of 1V in chemistry goes a&#xA;long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much&#xA;better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of&#xA;enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$.&#xA;Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$&#xA;will be reduced to $\ce{Ce^{3+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$&#xA;left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look&#xA;at the number of moles of each ion that went into the reaction at the&#xA;moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts&#xA;with one unit of $\ce{Ce^{4+}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So now you've figured out which is the limiting reagent (the one that gets&#xA;consumed almost entirely). Let's suppose for the sake of argument it's&#xA;the $\ce{Fe^{2+}}$. Because you know how many&#xA;moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know&#xA;how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can&#xA;find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the&#xA;equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$,&#xA;because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's&#xA;all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same&#xA;logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and&#xA;$\ce{Ce^{3+}}$ in the equilibrium mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you have 3 concentrations out of 4.  Finding the 4th requires&#xA;knowing the equilibrium constant, which you can calculate from the&#xA;difference in potentials. (Write the combined full reaction based on the half-reactions.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find that the 4th concentration is&#xA;very small, justifying the claim that one of the reagents&#xA;was almost entirely consumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're with me so far, please update the question with these&#xA;results, and if you still need help with finding the potential at that&#xA;point, we can take it from there...&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-25T07:01:36.210" LastActivityDate="2014-07-25T07:01:36.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14616" PostTypeId="2" ParentId="13784" CreationDate="2014-07-25T06:16:14.027" Score="5" Body="&lt;p&gt;It is more acurate to say that they aren't very far from being electrically neutral. If you take a charged sphere and drop it into a beaker of some electrolyte sitting on an insulating counter, the result is a charged solution, albeit slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in spite of your amusing ceiling analogy (it made me chuckle), the argument about electrostatic attraction and repulsion does justify why solutions (and solids) can't be too far from being electrically neutral.  The flaw in the analogy is that humans don't need ceilings to exist; but you can' t have two charged beakers without a force between them. If each has a charge of 1 C -- which is a mere $10^{-5}$ times the charge of a mole of electrons -- and they're sitting 1m apart the force between them is about a million tons -- clearly this situation won't last very long!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-25T06:16:14.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14617" PostTypeId="1" CreationDate="2014-07-25T06:30:33.053" Score="9" ViewCount="571" Body="&lt;p&gt;I am currently calculating some reactions and I would like to express the ratio of the products in per cents. I am using the Boltzmann statistic in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The probability $p_j$ of a particle $j$ to be in a state with the energy $\varepsilon_j$ is given as $\text{(1a,a')}$. The statement $\text{(1b)}$ must therefore hold, whereas $\text{(1c)}$ also follows.&#xA;\begin{align}&#xA;\text{(a)}  &amp;amp;&amp;amp; p_j        &amp;amp;= \frac{N_j}{N}\\&#xA;\text{(a')} &amp;amp;&amp;amp; \%P_j      &amp;amp;= \frac{N_j}{N}\cdot100\%\\&#xA;\text{(b)}  &amp;amp;&amp;amp; 1          &amp;amp;= \sum_j p_j\\&#xA;\text{(c)}  &amp;amp;&amp;amp; N          &amp;amp;= \sum_j N_j\\\tag1&#xA;\end{align}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For any system with $k$ particles, that are in thermal equilibrium, the probability to find particle $j$ can be expressed via the Boltzmann statistics $(2)$, where $Z$ is the canonical partition function, the inverse temperature is $\beta = (\mathcal{k}_\mathrm{B}\cdot T)^{-1}$ and the degeneracy coefficient is $g_j$.&#xA;$$p_j=g_j\cdot \frac1Z\cdot \exp\left\{-\beta\varepsilon_j\right\}\tag2$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can assume that a state with the energy $\varepsilon_0$ is the ground state, therefore no other state can have lower energy. I can therefore assign the ground state as the baseline and switch to relative energies, as demonstrated in in $(3)$.&#xA;\begin{align}&#xA;\text{def.}&amp;amp;&amp;amp; \varepsilon_0 &amp;amp;= 0.0\\&#xA;           &amp;amp;&amp;amp; \varepsilon_j &amp;amp;= \varepsilon_0 + \Delta\varepsilon_j' \\&#xA;\therefore &amp;amp;&amp;amp; \varepsilon_j &amp;amp;= \Delta\varepsilon_j'\\\tag3&#xA;\end{align}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a consequence I can express the canonical partition function in terms of the ground state probability. &#xA;\begin{align}&#xA;            &amp;amp;&amp;amp; p_0 &amp;amp;= g_0\cdot \frac1Z\cdot \exp\left\{-\beta\varepsilon_0\right\}\\&#xA; \therefore &amp;amp;&amp;amp; p_0 &amp;amp;= g_0\cdot \frac1Z\\&#xA; \equiv     &amp;amp;&amp;amp; Z   &amp;amp;= \frac{g_0}{p_0}\\\tag4&#xA;\end{align}&#xA;The canonical partition function is also defined in $(5)$, but I am quite sure this one would not lead me to my desired goal.&#xA;$$Z = \sum_l^\infty g_l\cdot\exp\left\{-\beta\varepsilon_l\right\}\tag5$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can express any probability in terms of the ground state probability following from $(2-4)$.&#xA;\begin{align}&#xA;   &amp;amp;&amp;amp; p_j &amp;amp;= \frac{g_j}{g_0}\cdot p_0\cdot \exp\left\{-\beta\Delta\varepsilon_j'\right\}\\\tag6&#xA;\end{align}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a last step I will switch to macroscopic numbers, as my quantum chemical calculations give me $\Delta{}G$ values. &#xA;\begin{align}&#xA; \text{def.} &amp;amp;&amp;amp; \Delta{}G_j &amp;amp;= \mathcal{N}_\mathrm{A}\cdot \Delta\varepsilon_j'\\&#xA;   &amp;amp;&amp;amp; p_j &amp;amp;= \frac{g_j}{g_0}\cdot p_0\cdot &#xA;                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\\tag7&#xA;\end{align}&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;Finally the questions:&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do I obtain $p_0$ in a way, that I can use it in spreadsheet software like libreOffice calc, so that I satisfy $\text{(1b)}$? If I insert this equation, I would have to solve it iteratively, as I have the variable to solve for on both sides. I could then go ahead and use $Z$ directly.&#xA;\begin{align}&#xA;  p_0 &amp;amp;= 1 - p_j -\sum_{i\neq0,j} p_i\\&#xA;  \color{\red}{p_j} &#xA;      &amp;amp;= \frac{g_j}{g_0}\cdot \left(1 - \color{\red}{p_j} &#xA;                                 -\sum_{i\neq0,j} p_i\right)\cdot &#xA;                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\&#xA;\end{align}&lt;/li&gt;&#xA;&lt;li&gt;Am I making correct assumptions along the way?&lt;/li&gt;&#xA;&lt;li&gt;Could the following work? Assign an arbitrary probability to $p_0$, to obtain an arbitrary value for $[p_j]^\ddagger$. What would be reasonable? I picked $0.25$ as an example. Can I then normalize to get the total percentages?&#xA;\begin{align}&#xA; \text{def.}  &amp;amp;&amp;amp; p_0   &amp;amp;= [p_0]^\ddagger\ =\ 0.25\\&#xA;              &amp;amp;&amp;amp; [p_j]^\ddagger &amp;amp;= \frac{g_j}{g_0}\cdot \color{\green}{0.25}\cdot &#xA;                \exp\left\{-\frac{\Delta{}G_j}{\mathcal{R}\cdot{}T}\right\}\\&#xA; \text{Norm.} &amp;amp;&amp;amp; \%P_j &amp;amp;= \frac{[p_j]^\ddagger}{\sum_i [p_i]^\ddagger}\cdot 100\%&#xA;\end{align}&#xA;This would maybe break down the iterative procedure to a two step process. Is this procedure really equivalent to $\text{(1a')}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Closely related: &lt;a href=&quot;https://chemistry.stackexchange.com/q/14436/4945&quot;&gt;4 Compounds in Equilibrium with One Another - Determine Their Equilibrium Concentrations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-08-21T10:07:59.897" Title="How to obtain product ratio from energy differences via Boltzmann statistics" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14618" PostTypeId="1" CreationDate="2014-07-25T11:37:22.657" Score="25" ViewCount="56087" Body="&lt;p&gt;This is immediately following ron's answer from &lt;a href=&quot;https://chemistry.stackexchange.com/q/14390/4945&quot;&gt;Why is a ketone more nucleophilic than an ester?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most simplest questions you can ask, how can you rationalise the order of reactivity towards nucleophiles, which is given as&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/l0VqP.png&quot; alt=&quot;order of carbonyl compound versus nucleophiles&quot;&gt;&lt;br&gt;&#xA;acyl halide &gt; acid anhydride &gt; aldehyde &gt; ketone &gt; ester ~ carboxylic acid &gt; amide &gt; carboxylate ion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be many concepts out there, that explain this order, the above linked question could only halfway satisfactory answer it in the kind of generality I am looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am looking here for &lt;strong&gt;any&lt;/strong&gt; model that can be used to explain this reactivity.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I will also later on post an answer to this question using quantum chemical calculations, trying to generalise from example. The key points that will be included are stated in ron's answer linked above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I like Martin's posts, they make me think - and that is always a good&#xA;  thing. Here is where my thinking has taken me:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;I believe Martin's approach (basically frontier MO) is fundamentally&#xA;  correct. I would be more convinced that his actual numbers are&#xA;  meaningful if he were to calculate 1) the carbonyl carbon LUMO&#xA;  coefficient and 2) the HOMO-LUMO separation for the following series&#xA;  of compounds in order to see if his calculated values correlated with&#xA;  this order of reactivity of carbonyl compounds towards nucleophilic&#xA;  attack (we could use water as the nucleophile) - but Martin, I think&#xA;  such calculations might be a poor use of your time and resources.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I agree with most of this, I do strongly disagree with the last statement. Which I will hopefully be able to prove soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime I welcome everyone to share their thoughts on the matter.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-08-18T11:08:04.880" Title="Rationalising the order of reactivity of carbonyl compounds towards nucleophiles" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="14619" PostTypeId="2" ParentId="14057" CreationDate="2014-07-25T13:18:39.417" Score="15" Body="&lt;p&gt;Let me begin by thanking Martin and long for supplying the key reference (&lt;a href=&quot;https://doi.org/10.1021/ja954004t&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;118&lt;/em&gt; (34), 7995–8005&lt;/a&gt;).  Importantly, long also provided the isomer ratio I was asking for; according to him, the ratio was estimated using the numbers at bottom of page 7997 (with the 0.002 added).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free energy differences between the isomers of fluorobullvalene are reported in this paper as  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\Delta G(1,4) &amp;amp;= \pu{+3.41 kcal/mol} \\&#xA;\Delta G(3,1) &amp;amp;= \pu{-2.21 kcal/mol} \\&#xA;\Delta G(2,3) &amp;amp;= \pu{-0.46 kcal/mol} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram showing the 4 possible isomers of fluorobullvalene in equilibrium, as well as a simple example of the symmetry-allowed, 6-pi electron Cope rearrangement that occurs in bullvalene and interconverts the isomers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jaPvU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jaPvU.png&quot; alt=&quot;Fluorobullvalene isomers and Cope rearrangementmechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://chemistry.stackexchange.com/a/14480/16683&quot;&gt;this answer&lt;/a&gt;, the equilibrium concentrations of the four isomers were determined to be as follows. These numbers differ somewhat from long's numbers - which were an approximation long made based on nmr peak intensities - but appear to match those shown in Fig. 4 in the paper at 300 K:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;[1]~ &amp;amp;=~~~ 0.23\% \\&#xA;[2]~ &amp;amp;=~ 20.27\% \\&#xA;[3]~ &amp;amp;=~~~ 9.37\% \\&#xA;[4]~ &amp;amp;=~ 70.13\% \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question left to be answered: &lt;strong&gt;why does the isomer distribution fall out like this, and, in particular, why such an avoidance of the cyclopropyl position by fluorine?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we work towards an answer to this question we will discuss bonds that are $\ce{sp^{i}}$ hybridized.  “i” is referred to as the hybridization index.  The hybridization index is such that the ratio $1:i$ is the ratio of the s:p atomic orbitals mixed to form the hybrid orbital.  For example in an $\mathrm{sp^3}$ bond, $i=3$, and the ratio of s:p atomic orbital content in the bond is $1:3$.  Further, we can see from this that the percentage s character in a bond is given as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{s-character} = \frac{1}{1+i} \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is $1/(1+3) = 25\%$ in the $\mathrm{sp^3}$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of ways to estimate the percent s-character in a bond.  One way is through use of the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$^1\!J_\ce{C-H} = \pu{500 Hz} \cdot \text{s-character in the bond} \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $^1\!J_\ce{C-H}$ is the one-bond C–H coupling constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $\ce{^12C}$ does not have a magnetic moment, therefore it cannot interact with a proton spin and split it. $\ce{^13C}$ is a spin-½ nucleus, it can interact with a proton spin and cause signal splitting in the carbon and proton nmr spectra.  $\ce{^13C}$ only occurs at 1.1% natural abundance, but one can still measure the resultant C–H splitting (e.g. the coupling constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, consider the polarization in a $\ce{C-F}$ bond. It will look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c91y9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c91y9.png&quot; alt=&quot;Polarity of C–F bond&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the relative electronegativities of carbon and fluorine, electron density will shift away from carbon towards fluorine. To understand the effect of this electron shift, let's compare the hybridization in methane ($\ce{CH4}$) and methyl fluoride ($\ce{CH3-F}$).  In methane, the carbon orbital used to form the C–H bond is exactly $\mathrm{sp^3}$ hybridized – this is demanded by the symmetry of the molecule.  Using equation (1) we know that the percent s-character in the atomic orbital used by carbon to interact with and form a molecular orbital with hydrogen is 25% in this case.  From equation (2) we would predict that $^1\!J_\ce{C-H}$ in methane would be (500*0.25) Hz, or 125 Hz.  Indeed, the experimentally determined coupling constant for methane is measured as 125 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In methyl fluoride &lt;a href=&quot;http://books.google.com/books?id=2rxFRgp57_0C&amp;amp;pg=PA190&amp;amp;lpg=PA190&amp;amp;dq=methyl%20fluoride%20Jc-H&amp;amp;source=bl&amp;amp;ots=zzOFhnyvOT&amp;amp;sig=G2feDNmsn5Dsl71AWQw0LAsPkOI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=CszOU5O8BMWhyATbh4HYCw&amp;amp;ved=0CEcQ6AEwBA#v=onepage&amp;amp;q=methyl%20fluoride%20Jc-H&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;the experimentally observed $^1\!J_\ce{C-H}$ coupling constant is 149 Hz&lt;/a&gt;.  Using equation (2) we find that there is 149/500 = 0.30 or 30% s-character in the carbon orbital used to form the C–H bond.  Using equation 1, this is an $\mathrm{sp^{2.36}}$ bond.  There are 3 of these C–H bonds, so a total of 90% of the available carbon 2s orbital is used in forming them.  This leaves 10% of the carbon 2s orbital remaining and available for use in the C–F bond. Using equation 1 we find that i=9 and this carbon orbital is $\mathrm{sp^9}$ hybridized – very nearly pure p!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What a difference in bond hybridization between methane and methyl fluoride and it teaches an important lesson. Since the carbon 2s orbital is lower in energy than the carbon 2p orbital, the more s-character there is in a bond, the lower the energy of the electrons in that bond. If the electron density in an orbital is low (like in the carbon portion of the C–F bond in methyl fluoride), then nature doesn’t mix much s-character into that orbital and, instead, uses that s-character to stabilize orbitals that do have significant electron density.  Hence, the carbon orbitals used to form the C–F bonds in methyl fluoride only have 10% s-character because the electron density remaining around carbon in that bond is so low. Instead, the s-character is used in the 3 C–H bonds which do contain significant electron density around carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we apply this type of analysis to fluorobullvalene, let’s make some predictions based on what we’ve learned so far.  Clearly, fluorine will preferentially bond with carbon atoms that have low s-character in the bonding orbital.  The $^1\!J_\ce{C-H}$ coupling constant in cyclopropane is 161 Hz.  Using equations (2) and (1), we find that the C–H bond in cyclopropane has 32% s-character and is $\mathrm{sp^{2.1}}$ hybridized. In bullvalene itself, carbon (4) in the tris-allylic position looks like a $\ce{sp^{3}}$ hybridized carbon whereas the olefinic carbons are probably $\ce{sp^{2}}$ (more or less).  So right off the bat, we would expect more fluorine at position 4, than at either of the olefinic positions or the cyclopropyl position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant $^1\!J_\ce{C-H}$ coupling constants in the parent molecule bullvalene are presented in the table below, along with the percent s-character derived from equation (2), the hybridization index $i$ from equation (1), the value of $\mathrm{sp}^i$, and the percentage of the isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|c|c|c|c|c|}\hline&#xA;\text{Carbon} &amp;amp; ^1\!J_\ce{C-H}\text{ / Hz}  &amp;amp; \text{% s-character} &amp;amp; i &amp;amp; \mathrm{sp}^i &amp;amp; \text{% isomer} \\ \hline&#xA;1 &amp;amp; 169.0 &amp;amp; 33.8 &amp;amp; 1.96 &amp;amp; \mathrm{sp}^{1.96} &amp;amp; 0.23 \\ \hline&#xA;2 &amp;amp; 160.8 &amp;amp; 32.2 &amp;amp; 2.11 &amp;amp; \mathrm{sp}^{2.11} &amp;amp; 20.27 \\ \hline&#xA;3 &amp;amp; 166.1 &amp;amp; 33.2 &amp;amp; 2.01 &amp;amp; \mathrm{sp}^{2.01} &amp;amp; 9.37 \\ \hline&#xA;4 &amp;amp; 133.9 &amp;amp; 26.8 &amp;amp; 2.73 &amp;amp; \mathrm{sp}^{2.73} &amp;amp; 70.13 \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We immediately see that the amount of fluorine at a given position changes with the percentage s-character in the bond. Just as we expected, the higher the s-character, the lower the amount of that fluorinated isomer. In fact, if we measure the correlation coefficient between the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position, a correlation coefficient of –0.9982 is obtained (if we use long's suggested isomer distribution [0.18%, 5.99%, 1.66%, 92.17%] the correlation coefficient is –0.9875).  A value of 1 would be a perfect correlation, a value of –1 would be a perfect inverse correlation. In the case at hand, the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position demonstrate a near-perfect inverse correlation. An inverse correlation means that as the percentage s-character increases, the isomer with fluorine attached at this position decreases and the high value for the coefficient indicates that the movement of the two values is highly correlated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bullvalene provided an interesting test case where we could measure fluorine’s affinity to bond to different types of carbon. From this example we can see that fluorine strongly prefers to form bonds to carbon that utilize carbon atomic orbitals with low s-character.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2017-08-29T19:45:25.637" LastActivityDate="2017-08-29T19:45:25.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14620" PostTypeId="2" ParentId="14617" CreationDate="2014-07-25T16:33:16.813" Score="7" Body="&lt;p&gt;There are two aspects to the problem: theory and practice.  In theory you have a complete answer in the form of the partition function (5).  If you know all the energy levels and their degeneracies, you can at least in principle carry out the sum, and you've solved the problem.  One case where this is practical is where you have a (small) number $n$ of particles with spin $\frac12$ interacting with each other; there are $2^n$ configurations, each with potentially a different energy, but you can still carry out an explicit sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That example probably does not help  in your case; in chemistry you usually have infinitely many energy levels.  As you point out in the comments, you can cut off the sum at some point. This may still not give you a practical solution, either because the sum up to the cutoff is still too hard to calculate, or because the cutoff error is unacceptably high (say, if levels are closely spaced and the sum in (5)  converges very slowly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But assuming there is a reasonable cutoff, the method may very well be practical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind here are the answers to your questions at the end:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I obtain  $p_0$   in a way, that I can use it in spreadsheet software like libreOffice calc, so that I satisfy (1b) ? If I insert this equation, I would have to solve it iteratively, as I have the variable to solve for on both sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;iterative&quot; aspect is a red herring and the difficulty of having $p_0$ on both sides is illusory, since the $p_i$ are all proportional to $p_0$, and the proportionality factors are known and independent of $p_0$.  You can simply factor out $p_0$, which gives $p_0$ multiplied by the inverse of the partition function; then you equate this expression to 1 and (if you can calculate the partition function) you are done.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I making correct assumptions along the way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, I think you've stated the theory and the assumptions quite accurately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the following work? Assign an arbitrary probability to $p_0$, to obtain an arbitrary value for  $[p_j]^\ddagger$. What would be reasonable? I picked  0.25 as an example. Can I then normalize to get the total percentages?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that would work; the formulas you wrote under this item are correct.  You can save yourself the trouble of choosing a number like  0.25, as  this number cancels out of the fraction  $\frac{[p_j]^\ddagger}{\sum_i [p_i]^\ddagger}$ -- notice that it appears as a factor in the numerator and also in each summand in the denominator, where it can be factored out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last remark: if you have access to symbolic algebra software like Matlab, Maple or Mathematica, you may be able to do the calculation more conveniently than with libreOffice calc. If you're very lucky, you may even find the exact solution to your infinite sum!    (Not all sums can be solved exactly, of course; still it might be worth a try, and you can sometimes guide the software.  It helps to be familiar with generating functions and summation techniques as explained for example in the book &lt;em&gt;Concrete Mathematics&lt;/em&gt;, by Graham, Knuth and Patashnik.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-25T16:33:16.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14621" PostTypeId="2" ParentId="14590" CreationDate="2014-07-25T18:03:22.790" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What if I2 and H2 began at the same concentration - then do we start at the same point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes (and in textbook problems involving graphs this situation would be avoided for the reason pointed out in brinnb's comment)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is it acceptable to start at different concentrations... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also yes.  No law against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the graphs are part of a textbook problem. Are you able to figure out which graph can occur?&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-30T15:10:36.047" LastActivityDate="2014-07-30T15:10:36.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14622" PostTypeId="2" ParentId="14582" CreationDate="2014-07-25T18:55:42.807" Score="6" Body="&lt;p&gt;With hybridisation, &lt;strong&gt;&lt;em&gt;you only hybridise the number of orbitals you need i.e. that are filled.&lt;/em&gt;&lt;/strong&gt; So in methane, with four bonds, you have four electron pairs which need to fit somewhere. So you need four orbitals. Therefore we say that three 2p orbitals &lt;em&gt;hybridise&lt;/em&gt; with the 2s orbital, making four sp&lt;sup&gt;3&lt;/sup&gt; hybrids (the superscripts denoting the number of each pure orbitals involved) each with 1 electron in them. Note that we had to invoke the idea of hybridisation here, since the electron configuration of carbon is 1s&lt;sup&gt;2&lt;/sup&gt;2s&lt;sup&gt;2&lt;/sup&gt;2p&lt;sup&gt;2&lt;/sup&gt;, and this would only give us two bonds. We know experimentally that this isn't true (in other words, methane exists!) and so have to conjure a trick to work around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, let us look at the CH&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; cation. This has three bonds, and so three electron pairs. This time we need to invoke the idea of hybridisation to give us three orbitals. So we say that two 2p orbitals hybridise with the 2s orbital, to give three sp&lt;sup&gt;2&lt;/sup&gt; orbitals. This leaves us with one p-orbital that we didn't hybridise, which would be a non-bonding orbital (in fact, empty), and so this lies perpendicular to the plane of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another example, to clarify further, consider ammonia. Now ammonia has three bonds, but it also has a lone pair. We &lt;em&gt;count&lt;/em&gt; the lone pairs in our hybridisation scheme. Therefore we also need to hybridise to obtain four orbitals, like in methane, despite only having three bonds. Thus ammonia is also sp&lt;sup&gt;3&lt;/sup&gt;, mixing three 2p and one 2s orbital. Note however that in contrast to methane, three of these hybrid orbitals are singly occupied, and one is completely filled (the one which contains the lone pair). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can look at cases by which orbitals do not hybridise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For heavy atoms such as lead and bismuth (the same groups as carbon and nitrogen respectively) hybridisation does not occur. This is referred to as the &lt;em&gt;inert pair effect&lt;/em&gt;. In this case &lt;em&gt;inert pair&lt;/em&gt; refers to the unwillingness of the 6s&lt;sup&gt;2&lt;/sup&gt; orbital to undergo hybridisation. This is because heavy atom self-self bonds i.e. Pb-Pb are incredibly weak. We can imagine that some energy is expended in the hybridisation process because we are mixing s orbitals with p-orbitals of the same principal quantum number, of which the p are always higher in energy (said to be non-degenerate). So the hybrid orbitals are higher in energy than the pure s, and ever so slightly lower than the pure p (but not by much, as the p-character of an sp&lt;sup&gt;3&lt;/sup&gt; hybrid is 75%). So, the energy gained in forming a heavy atom bond such as Pb-Pb isn't enough to warrant hybridisation, since hybridisation requires energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I've used anthropogenic words to explain this i.e. (we hybridise as opposed to the atoms hybridise). This is deliberate. It's unclear whether hybridisation really occurs or not; indeed more sophisticated theories can be used to explain the phenomena. Hybridisation is more of a caveat used to explain something we couldn't readily explain with the previous level of theory. It's perhaps telling that we only invoke hybridisation when we need to. It's a characteristic of a bad theory when that happens. The idea is invoked readily in chemistry however, so you would do well to familiarize yourself with the terminology.If you're interested more in this topic, look up &lt;em&gt;molecular orbital theory&lt;/em&gt; for a more nuanced explanation. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-07-27T13:47:24.107" LastActivityDate="2014-07-27T13:47:24.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14623" PostTypeId="2" ParentId="13784" CreationDate="2014-07-25T19:08:59.007" Score="3" Body="&lt;p&gt;I think there is a misunderstanding here about the difference between &lt;strong&gt;explanations&lt;/strong&gt; and &lt;strong&gt;observations&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that two beakers of aqueous solution do not have a measurable force between them is an &lt;strong&gt;observation&lt;/strong&gt; that proves the solution has no (large) charge. Calculating a hypothetical force due to a charged solution and finding that it is much larger than what we observe is not an after-the-fact justification, it is a falsification of a hypothesis by an empirical observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;explanation&lt;/strong&gt; is that no matter what the ionic compound that you started with was, it was electrically neutral. Therefore, the solution must still be neutral.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-25T19:08:59.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14624" PostTypeId="2" ParentId="14424" CreationDate="2014-07-25T19:22:59.473" Score="5" Body="&lt;p&gt;I notice that there is no mention of &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid#Polyprotic_acids&quot;&gt;polyprotic acids&lt;/a&gt; in the other answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfuric_acid&quot;&gt;sulfuric acid&lt;/a&gt;) is diprotic - meaning it has two &quot;detachable&quot; protons which can come off in aqueous solution. The first one is strongly acidic - the $Ka = 2.4 × 10^6$ - this means the odds of finding a non-dissociated $\ce{H2SO4}$ molecule in solution are something like 1 out of 2.4 million.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the second proton is &lt;em&gt;weakly acidic&lt;/em&gt; - the $Ka = 1.0 × 10^{-2}$. This means that there will mostly be $\ce{HSO4-}$ instead of $\ce{SO4^2-}$ in solution (roughly 100 times as much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers are correct in saying that $\ce{HSO4}$ does not exist in solution - but it is important to realize that we list the bisulfate ion ($\ce{HSO4-}$) as a weak acid because it is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Conjugate_base&quot;&gt;conjugate base&lt;/a&gt; of sulfuric acid, and is in turn the conjugate acid of the sulfate ion. This becomes very important when you start to study acid/base equilibria.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-25T19:22:59.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14625" PostTypeId="1" CreationDate="2014-07-25T19:57:03.243" Score="0" ViewCount="32423" Body="&lt;p&gt;The symbols :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T : temperature //&#xA;V : volume //&#xA;P : pressure //&#xA;n : number of mole &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what it is clearly known about the relationship between one another is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P is inversely proportional to V ( from Boyle's law : PV=constant )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V is directly proportional to T ( from Charles's law : V/T=constant )&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;but the other ones I am not sure about them, So i might say :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P is inversely proportional to T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is inversely proportional to n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p is directly proportional to n&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastActivityDate="2016-03-12T00:43:14.850" Title="The relationships between n , T , P , V for a gas?" Tags="&lt;gas-laws&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14626" PostTypeId="2" ParentId="14625" CreationDate="2014-07-25T20:15:02.127" Score="2" Body="&lt;p&gt;All statements like &quot;P is inversely proportional to V&quot; should be qualified with &quot;other variables being equal&quot;. With this qualification you can read off all the proportionalities from $pV=NRT$.  Any two variables on the same side are inversely proportional; any two on opposite sides are directly proportional.  So $p$ is directly proportional to $T$.  Your other assertions are correct.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-25T20:15:02.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14627" PostTypeId="2" ParentId="14618" CreationDate="2014-07-25T22:38:38.810" Score="17" Body="&lt;p&gt;Here is my &quot;old school&quot; explanation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left.  As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well representing the tetrahedral intermediate.  $\ce{E_{act}}$ is the activation energy required to pass over the transition state.  To whatever extent a substituent might stabilize the starting carbonyl compound, the energy of the carbonyl compound will be lowered, and consequently $\ce{E_{act}}$ will increase.  The central carbon (it was the carbonyl carbon) in the tetrahedral intermediate is fully saturated (just $\ce{sp^3}$ bonds) and without charge, so the substituent X cannot interact through resonance to stabilize the tetrahedral intermediate.  In other words, while substituent X can stabilize or destabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OWtmp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are the resonance structures we can draw for our carbonyl compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M1UsZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If structure III contributes significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structure III contributes little.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by comparing the ester to the amide.  In both cases structure III involves a 2p-2p pi bond, but the resultant positive charge is placed on oxygen when we start with the ester and on nitrogen when we start with the amide.  Clearly, for electronegativity reasons, the amide case is better with the positive charge on the less electronegative amide nitrogen.  Our prediction: the amide is more stabilized than the ester and therefor should be the slower reacting of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ester &amp;gt; amide}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the carboxylate anion, resonance structure III is equivalent to resonance structure I (note: there is no positive charge in structure III for the carboxylate case).  Since they are equivalent, they carry high weight stabilizing and in describing the carboxylate anion.  The carboxylate anion is so stabilized by resonance structures I and III that it is basically unreactive to nucleophiles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ester &amp;gt; amide &amp;gt;&amp;gt; carboxylate}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's really no difference between the ester and carboxylic acid in terms of resonance structure III contribution so we would expect them to have similar reactivity towards a nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ester \sim acid &amp;gt; amide &amp;gt;&amp;gt; carboxylate}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare the ester to the anhydride.  Resonance structure III is equally effective for both compounds, BUT, there are two carbonyls in the anhydride for the oxygen to interact with.  Therefor, the oxygen will only be half as effective at stabilizing any one anhydride carbonyl as it was for the ester carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{anhydride &amp;gt; ester \sim acid &amp;gt; amide &amp;gt;&amp;gt; carboxylate}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the acid chloride, resonance structure III involves a 2p-3p pi bond, not very effective overlap due to the size difference (also chlorine is electronegative so placing a positive charge on it is not desirable, but the poor pi overlap is what limits the contribution from structure III in the acid chloride case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{acid~ chloride &amp;gt; anhydride &amp;gt; ester \sim acid &amp;gt; amide &amp;gt;&amp;gt; carboxylate}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ketone, our resonance structure III has a slightly different look, it involves hyperconjugation as shown in the following figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WBo6P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are 3 hydrogens, there are three such hyperconjugated structures,  They are not super significant resonance structures, but there are 3 of them, they do place the positive charge on a proton, so they do count for some stabilization.  To be honest, I would have guessed that it creates more stabilization than resonance structure III in the acid chloride case, but I wouldn't have known where it fits between the acid chloride and the ester.  Wherever I put it, I'd put the aldehyde to its left.  There is no resonance structure III for the aldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wind up with &#xA;$$\ce{acid~ chloride &amp;gt; anhydride &amp;gt; ester \sim acid &amp;gt; amide &amp;gt;&amp;gt; carboxylate}$$&#xA;and $\ce{aldehyde &amp;gt; ketone}$ fitting in somewhere between the acid chloride and the ester&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-26T15:45:20.720" LastActivityDate="2014-07-26T15:45:20.720" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14628" PostTypeId="2" ParentId="14625" CreationDate="2014-07-25T23:03:09.410" Score="1" Body="&lt;p&gt;Just remember the general gas law&#xA;$$\mathrm{PV=nRT}$$&#xA;or rearranging terms we see that&#xA;$$\mathrm{\frac{[PV]}{[nT]}=R}$$&#xA;where R is a constant.  Therefore,&#xA;$$\mathrm{\frac{[P_1V_1]}{[n_1T_1]}=\frac{[P_2V_2]}{[n_2T_2]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this very general set of ratios relates all of the variables in state (1) to the same variables in state (2).  If volume and number of moles don't change, then the equation reduces to&#xA;$$\mathrm{\frac{[P_1]}{[T_1]}=\frac{[P_2]}{[T_2]}}$$&#xA;and so on and so forth.  All of the proportionalities are clearly layed out in one general equation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:43:14.850" LastActivityDate="2016-03-12T00:43:14.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14629" PostTypeId="1" AcceptedAnswerId="14632" CreationDate="2014-07-25T23:51:39.687" Score="2" ViewCount="1137" Body="&lt;p&gt;What are the conditions for the evaporation of water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for the boiling of water, the vapor pressure of water must equal the partial pressure of water vapor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about for evaporation? Is it the same condition above in addition to the requirement that the relative humidity cannot be 100%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we have boiling when the relative humidity is 100%? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:16:48.177" LastActivityDate="2015-10-23T17:16:48.177" Title="Evaporation of water" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;water&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14630" PostTypeId="1" AcceptedAnswerId="14740" CreationDate="2014-07-25T23:52:59.293" Score="16" ViewCount="1798" Body="&lt;p&gt;I just started learning DFT and now I am totally confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I want to use B3LYP:&#xA;\begin{align}&#xA;v_s\left(\textbf{r}\right) &amp;amp;= v_\text{ext}\left(\textbf{r}\right) + \int d^3r\frac{n\left( \textbf{r}\right) }{\left|\textbf{r} - \textbf{r}^\prime\right|} + v_{\text{XC}}\left[ n \right]\left(\textbf{r}\right)\\&#xA;v_{\text{XC}}\left( r \right) &amp;amp;= \frac{\delta E_\text{XC}}{\delta n \left(\textbf{r}\right) }\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that we use B3LYP for approximation of $v_{\text{XC}}$ - but I have no idea about how - and then we use some basis sets to run an SCF calculation to minimize the energy.&#xA;I downloaded a bunch of articles and bought some books and read them, but all of them give very detailed discussion about all the basics (&lt;em&gt;e.g.&lt;/em&gt; what a functional is, or Kohn and Sham theorem, etc.), but no one talks about how to use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a reading material or video that explains all the calculation, step by step, for example, some material that shows all the calculations for a CO molecule, from the beginning to the end of the SCF calculation.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="4945" LastEditDate="2014-07-28T04:05:02.017" LastActivityDate="2017-08-17T06:50:17.830" Title="Are there any full worked examples of DFT calculations?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="3" CommentCount="16" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14631" PostTypeId="2" ParentId="14629" CreationDate="2014-07-26T02:18:52.157" Score="1" Body="&lt;p&gt;Water evaporates (in the sense that more water leaves a body of liquid or solid water than comes back to it) whenever the partial pressure of water vapor in contact with the liquid or solid water is less than the vapor pressure of water at the given tempearture.  This is the same as saying that the relative humidity is less than 100%, by the definition of relative humidity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we have boiling when the relative humidity is 100%? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can.  Suppose you have a flask of water with a narrow neck and surrounded by thermal insulation.  If you heat the water to the boiling point you will soon displace all the air, so both the liquid and gas phases are 100% water, and the water will continue to boil as long as you keep supplying heat. &lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-27T00:58:48.337" LastActivityDate="2014-07-27T00:58:48.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14632" PostTypeId="2" ParentId="14629" CreationDate="2014-07-26T02:23:47.840" Score="3" Body="&lt;p&gt;Evaporation is a surface phenomenon, boiling is a bulk phenomenon.  Evaporation will always be occurring for any liquid with an exposed surface.  Water molecules in a liquid sample have a distribution of kinetic energies.  It takes less work for a surface water molecule to break free from its neighbors, then for a water molecule in the bulk to escape, since the former has fewer neighbors (no molecules on top).  Those molecules with enough kinetic energy to break away from their neighbors will escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water to boil its vapor pressure must equal the atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only requirement for evaporation to occur is an exposed liquid surface.  I don't think relative humidity affects the boiling point, but I suspect it would affect the rate at which a liquid will evaporate from an open container.  Evaporation is always occurring, but to notice a change in the amount of material in an open system, more material must leave the surface then returns to the surface and the rate of return will be affected by the composition (water content) of the air above the open sample.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-26T02:23:47.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14633" PostTypeId="2" ParentId="4486" CreationDate="2014-07-26T03:05:26.833" Score="4" Body="&lt;p&gt;Sounds like you're referring to &lt;a href=&quot;http://federalist.wordpress.com/2008/06/16/plasma-waste-conversion/&quot; rel=&quot;nofollow&quot;&gt;plasma waste conversion&lt;/a&gt;.  If you heat matter to 4-figure temperatures all molecular bonds are broken and you have a gaseous mass of dissociated atoms.  It is theoretically possible to extract the constituent elements from the plasma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to preserve a particular element then the cooling of the plasma has to be controlled so that the element does not come into contact with others with which it naturally reacts until it is below the threshold temperature for those reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun bonuses: Not only can you recover specific elements from homogeneous waste streams, but plasma waste reactors can also run with an energy surplus created by the &quot;burning&quot; of energy-rich inputs.  They can also be designed to produce synthetic gas from organic feedstocks.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2014-07-26T03:13:28.353" LastActivityDate="2014-07-26T03:13:28.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14634" PostTypeId="2" ParentId="14595" CreationDate="2014-07-26T03:25:58.247" Score="2" Body="&lt;p&gt;Another visualization:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3 g}$&lt;/span&gt; of pure oxygen;  use the ideal gas law to find the volume &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume &lt;span class=&quot;math-container&quot;&gt;$\pu{19.3 g}$&lt;/span&gt; of pure argon;  use the ideal gas law to find the volume &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt;, smaller than &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; since argon is heavier than oxygen&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hopefully, the actual volume will be somewhere in between &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How far along the gap from &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; is the actual volume?&#xA;That's how far you are from zero oxygen to 100% oxygen.&#xA;With this fraction and the total pressure, the partial pressure is simple.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3668" LastEditorUserId="4945" LastEditDate="2020-02-13T13:14:19.600" LastActivityDate="2020-02-13T13:14:19.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14639" PostTypeId="1" CreationDate="2014-07-26T06:15:47.617" Score="10" ViewCount="202" Body="&lt;p&gt;I have recently read about the Z topological index found by Hosoya. The papers/sources The Z index is the sum of the coefficients of the matching polynomials (no. of ways to choose an independent edge from the set of edges in the graph - if we let each carbon atom be a vertex and each bond between a pair of carbon atoms is an edge.). The paper states that in particular for alkanes or saturated hydrocarbons, there is a correlation found between the Z index (which measures the extent of branching - in what way??) and the boiling point (bp) of these alkanes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the bp of alkanes increases with its size (due to strong force of attractions). However, I am confused about how the Z index measures the extent of branching, and any underlying reasons that the Z index will correlate with the bp and other physiochemical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers I have read so far fail to enlighten me on these questions above. Please help. &#xA;Thank you. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some ideas I have so far: &#xA;  Since Z index measures branching (or at least correlates to branching since more compacted graph will have a smaller sum of matching coefficient?), and branching correlates to the bp of alkanes (due to less surface area sometimes if you have more branching), perhaps thats why the Z index correlates to bp... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6944/why-do-branched-chain-compounds-have-lower-boiling-points-than-the-corresponding?rq=1.&quot;&gt;Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7236" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2016-07-20T00:26:44.473" Title="Hosoya Z Index and Correlation with Boiling Point" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14640" PostTypeId="1" CreationDate="2014-07-26T11:49:22.040" Score="2" ViewCount="86" Body="&lt;p&gt;Are there any known safety issues if you mix: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.scjprofessional.co.uk/File.ashx?id=82727&amp;amp;cid=220&quot; rel=&quot;nofollow&quot;&gt;Mr.Muscle glass cleaner&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.gt85.co.uk/wp-content/uploads/2013/02/Safety_Data_Sheet-GT85_Maintenance_Spray-SDS-19-11-2012.pdf&quot; rel=&quot;nofollow&quot;&gt;GT85 spray lubricant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7237" LastEditorUserId="73" LastEditDate="2014-07-27T02:05:58.930" LastActivityDate="2014-07-27T02:05:58.930" Title="Are there any hazardous products that would come from mixing glass cleaner with lubricant?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14641" PostTypeId="1" AcceptedAnswerId="14645" CreationDate="2014-07-26T12:38:23.490" Score="14" ViewCount="25470" Body="&lt;p&gt;I understand that polarity corresponds to an electronegativity difference and that the larger the electronegativity difference, the more polar the bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have read that carbon dioxide is nonpolar. This doesn't make sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In carbonyl compounds, the carbonyl carbon is partially positive and thus carbonyls are polar. $\ce{CO2}$ is simply two such units bonded together, so wouldn't it be that you would have an even more positive carbon and two partially negative oxygens causing $\ce{CO2}$ to be overall polar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that, $\ce{CO2}$ dissolves to a much larger extent in water compared to $\ce{O2}$, despite both molecules supposedly being nonpolar.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="16683" LastEditDate="2017-09-24T10:03:45.963" LastActivityDate="2017-09-24T10:03:45.963" Title="Why is carbon dioxide nonpolar?" Tags="&lt;inorganic-chemistry&gt;&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14642" PostTypeId="2" ParentId="14640" CreationDate="2014-07-26T12:49:18.460" Score="3" Body="&lt;p&gt;Nothing compromising safety will happen. Now, these are two completely different cleaning agents based on alcohol/water and organic-fuel respectively. Most likely will turn white  because these substances are immiscible. Therefore, it will not be a good idea to apply a mix for you cleaning tasks.  This will be  specially so  over wood or any rough porous surface likely as they will absorb the substances resulting in unappealing white or smear white spots.  Application over polished metal surfaces will be ok.&lt;/p&gt;&#xA;" OwnerUserId="5179" LastActivityDate="2014-07-26T12:49:18.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14643" PostTypeId="2" ParentId="14641" CreationDate="2014-07-26T13:21:46.847" Score="4" Body="&lt;p&gt;It's like they are at tug-of-war with each other, and they both have the same strength; it's not going anywhere. Since the oxygen is more electronegative than carbon, it has a partial negative charge and the carbon has a partial positive charge (twice of the oxygen since there are 2 oxygen).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-26T13:21:46.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14644" PostTypeId="2" ParentId="14641" CreationDate="2014-07-26T13:42:11.643" Score="11" Body="&lt;p&gt;&lt;strong&gt;In understanding molecular polarity you need to take the &lt;em&gt;whole&lt;/em&gt; structure into account.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning is correct as far as the &lt;em&gt;parts&lt;/em&gt; of the molecule are concerned. The individual bonds are polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, a &lt;em&gt;molecule&lt;/em&gt; can only be polar if it has a net dipole moment (that is, the charges don't balance out in direction across the whole molecule). So CO is polar as the polarity of the carbon-oxygen bond is unbalanced and the distribution of partial charge on the bond affects the whole molecule (giving the whole molecule a dipole moment). But CO&lt;sub&gt;2&lt;/sub&gt; is a &lt;em&gt;linear&lt;/em&gt; molecule and the partial dipoles of the two bonds are in exactly opposite directions. Hence they exactly balance out giving a &lt;em&gt;molecule&lt;/em&gt; which is non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If CO&lt;sub&gt;2&lt;/sub&gt; were not linear (like SO&lt;sub&gt;2&lt;/sub&gt; which is bent) it would be a polar molecule.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-07-26T13:42:11.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14645" PostTypeId="2" ParentId="14641" CreationDate="2014-07-26T13:42:48.560" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;so wouldn't it be that you&#xA;  would have an even more positive carbon and 2 partially negative&#xA;  oxygens &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, your analysis is correct to this point.  A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment.  However the molecule itself is linear and the two bond dipole moments are oriented 180 degrees with respect to one another and cancel each other out, so overall the molecule does not have a dipole moment and is non-polar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kzRfP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.&#xA;$$\ce{CO2(aq) + H2O &amp;lt;=&amp;gt; H2CO3(aq)}$$&#xA;This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-26T15:33:03.150" LastActivityDate="2014-07-26T15:33:03.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14646" PostTypeId="2" ParentId="14641" CreationDate="2014-07-26T14:09:46.710" Score="9" Body="&lt;p&gt;The formula for the net dipole moment $\vec{\mu_{net}}$ of an &lt;em&gt;overall neutral&lt;/em&gt; system of $n$ charged point particles is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\vec{\mu_{net}} = \sum\limits_{i}^{n} q_i\vec{r_i}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q_i$ is the charge of the $i$th point particle and $\vec{r_i}$ is the position vector for said particle; each individual dipole moment points from a negative charge to a positive charge. Carbon dioxide is a linear and symmetrical molecule, meaning that in the ground state the bonds between each respective oxygen atom and the carbon atom have the same lengths, and each oxygen atom bears an identical (partial) negative charge. All of this ultimately means that the two individual dipole moments between the carbon and oxygens one and two, respectively, perfectly oppose each other geometrically and have the same magnitude, i.e., $q\vec{r_1} = -q\vec{r_2}$, hence the vector sum $q\vec{r_1} + q\vec{r_2} = \mathbf{0}$.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-07-26T14:09:46.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14647" PostTypeId="1" AcceptedAnswerId="14649" CreationDate="2014-07-26T15:51:03.540" Score="6" ViewCount="30172" Body="&lt;p&gt;Why is it that $\ce{CO2}$ is considerably more soluble in water than $\ce{O2}$ is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ is nonpolar but dissolves in water which like $\ce{CO2}$ being nonpolar, doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the bond polarity and that the bond polarity is more important in solubility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not than why is $\ce{CO2}$ more soluble in water than $\ce{O2}$.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="275" LastEditDate="2014-07-26T18:25:18.443" LastActivityDate="2016-08-04T21:36:26.947" Title="If CO2 is nonpolar how come much more dissolves in water than O2?" Tags="&lt;solubility&gt;&lt;polarity&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14648" PostTypeId="2" ParentId="14647" CreationDate="2014-07-26T16:01:53.543" Score="2" Body="&lt;p&gt;That is because $\ce{CO2}$ will react with water in equilibrium to form $\ce{H2CO3}$, which is an acid and thus will dissociate to form ions that can easily fit in the polar solvent (i.e. water). $\ce{O2}$ on the other hand does not react with water because it is not a very good electrophile in comparison to $\ce{CO2}$. The $\pi$-bond is weak in $\ce{CO2}$ (i.e. high in energy), partly because the $\ce{C=O}$ bond is polarized and because of delocalisation in remaining $\pi$-bond of $\ce{CO2}$. $\ce{O2}$ does not have that luxury.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-07-26T16:21:47.720" LastActivityDate="2014-07-26T16:21:47.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14649" PostTypeId="2" ParentId="14647" CreationDate="2014-07-26T16:15:27.877" Score="11" Body="&lt;p&gt;Taken from my answer to your &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14641/why-is-o-c-o-nonpolar&quot;&gt;original question&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$.  Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds.  Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility.  Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid.&#xA;$$\ce{CO2(aq) + H2O &amp;lt;=&amp;gt; H2CO3(aq)}$$&#xA;This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-26T16:15:27.877" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14650" PostTypeId="2" ParentId="14647" CreationDate="2014-07-26T16:27:08.043" Score="1" Body="&lt;p&gt;Though the whole molecule has a net dipole moment of zero due to its geometry, the C=O bond has a dipole moment. Compared to the O=O bond which has no dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="7240" LastActivityDate="2014-07-26T16:27:08.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14651" PostTypeId="1" AcceptedAnswerId="14653" CreationDate="2014-07-26T21:32:07.937" Score="3" ViewCount="369" Body="&lt;p&gt;I understand that the molecule $\ce{IBrF2}$ is T-shaped, with three ligand atoms and two non-bonding pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how many unique bond angles can this molecule exhibit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor argues for the existence of only 2 unique bond angles in this molecule, with the somewhat $&amp;lt; 180^\circ$ axial $\ce{F-I-F}$ bond angle and the two equivalent and somewhat $&amp;lt; 90^\circ$ $\ce{F-I-Br}$ bond angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, can't we also arrange this molecule as to have this axial arrangement of atoms: $\ce{Br-I-F}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor argues that this arrangement is &lt;strong&gt;impossible&lt;/strong&gt; due to van der Waals repulsions. I suppose this makes sense; the two lone-pairs $\ce{&amp;lt;-&amp;gt;}$ large bromine atom repulsions would make this shape unfavorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impossible? I'm not sure about that. Dammit, nothing in chemistry is impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think? I'm arguing for the existence of &lt;em&gt;either&lt;/em&gt; two or three bond angles because we can either have a $\ce{F-I-F}$ axial arrangement or a $\ce{Br-I-F}$ axial arrangement. Plus, axial bonds are longer than equatorial bonds. So that would &lt;em&gt;lower&lt;/em&gt; the energy of the $\ce{Br-I-F}$ axial conformer. Side note: &lt;em&gt;why are axial bonds longer than equatorial bonds?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rn2Cp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-07-27T07:19:00.707" LastActivityDate="2014-07-27T07:19:00.707" Title="IBrF2 - van der Waals repulsions - and number of unique bond angles" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;theoretical-chemistry&gt;&lt;conformers&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14652" PostTypeId="2" ParentId="14334" CreationDate="2014-07-26T22:41:57.250" Score="2" Body="&lt;p&gt;There are a few different ways of making your own glue, here are four different versions you can try :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple Glue Ingredients:&#xA;1/2 cup flour&#xA;1/3 cup water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper Mache Glue Ingredients:&#xA;1 cup flour&#xA;1/3 cup sugar&#xA;1-1/2 cup water with 1 tsp white vinegar mixed in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cornstarch Paste Ingredients:&#xA;1-1/2 cups cold water&#xA;2 tablespoons cornstarch&#xA;2 tablespoons corn syrup&#xA;1 teaspoon white vinegar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple No-Cook Flour Paste:&#xA;1 cup flour&#xA;1/2 cup water&#xA;1/8 teaspoon salt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from &lt;a href=&quot;http://www.wikihow.com/Make-Glue&quot; rel=&quot;nofollow&quot;&gt;http://www.wikihow.com/Make-Glue&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7244" LastActivityDate="2014-07-26T22:41:57.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14653" PostTypeId="2" ParentId="14651" CreationDate="2014-07-27T02:14:27.163" Score="4" Body="&lt;p&gt;I suspect the molecule $\ce{IBrF2}$ does not exist, and is purely conjectural (theoretical) at this point.  If it ever had been synthesized or observed, it would have made the cover of Science or Nature.  &quot;First neutral ternary interhalogen compound found!&quot; (A handful of anionic  ternary interhalogen species are known.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my hunch is that, if the molecule did exist, the conformation your professor describes would be favored because of symmetry. Having the same atom in (roughly) opposite directions maximizes the amount of coupling because the opposite orbitals have same energy and can combine better than if it were two different atoms.  Hence greater stabilization.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Side note: why are axial bonds longer than equatorial bonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true in binary compounds like $\ce{PCl5}$, and apparently a sufficient explanation is that each axial chlorine wants to avoid 3 other chlorines at 90$^\circ$ angles, while the equatorial chlorines have 2 at 90$^\circ$ but the other 2 are further away anyway, at 120$^\circ$.  In other words, if you just want to maximize the &lt;em&gt;average&lt;/em&gt; Cl-Cl distance, subject to the constraint that the P-Cl distance is reasonable, you're better off with a slightly elongated bipyramid instead of one inscribed in a sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hypothetical $\ce{IBrF2}$ of course the axial I-F bond would be shorter than the &quot;equatorial&quot; I-Br bond, since Br is a bigger atom.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-27T03:18:36.967" LastActivityDate="2014-07-27T03:18:36.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14654" PostTypeId="2" ParentId="7438" CreationDate="2014-07-27T04:26:20.943" Score="1" Body="&lt;p&gt;A large part is the agriculture behind the fabric and what the industry and customer demands. Cotton and bamboo are easy to farm because we have a lot of experience and techniques for raising and harvesting and processing the plants and fibers ( especially cotton, which is available as GMO round-up ready to simplify weed control, and centuries of breeding for better quality and quantity of fiber ). Additionally, the textile industry has machines that are calibrated for cotton, other plant fibers may not work as well and re-designing the machines would be expensive. Customer demand is important, and customers are used to cotton clothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as chemical reasons go I don't have a lot of specifics. However, different plants have different compositions of fibers. Though all plant fibers are mostly cellulose/hemicellulose, lignin is also important, and would make for woodier, stiffer, tougher fibers ( though bamboo gets pretty woody, I don't know much about bamboo textiles ). Other plant fibers may contain pigments that would have to be bleached out before it could dyed with other colors. Length of fibers is also important. Shorter fibers would be harder to work into threads and textiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And keep in mind that the textile industry is not limited to cotton and bamboo. For centuries, cotton was a luxury material because picking the seeds out of the cotton boll was very hard work, cotton was far more expensive than silk. The widespread adaption of the cotton gin ( and slavery ) made cotton cheap and easily available. Prior to this, fabric would have been made from flax to make linin, or from hemp. Flax and hemp are still commonly grown for fabric. ( and we haven't even mentioned animal sources of fiber such as silk or wool ) &lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-27T04:26:20.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14655" PostTypeId="2" ParentId="14507" CreationDate="2014-07-27T04:41:25.423" Score="1" Body="&lt;p&gt;As the chain length of the alcohol increases, it becomes more hydrophobic. This would make it easier for them to insert into membranes and hydrophobic pockets of proteins. This could interfere with the protein function. Hydrophobic molecules also tend to be harder to eliminate than shorter molecules and can accumulate in fatty tissues, including the brain. Longer chains can also be metabolized into a greater number of products simply because there is more to metabolize. And sometimes seemingly innocuous compounds can be metabolized into very toxic substances, such as n-Hexane, which can be converted to a molecule with an epoxide on each end. The epoxides can cross-link proteins, particularly in neuron axons.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-27T04:41:25.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14656" PostTypeId="2" ParentId="11091" CreationDate="2014-07-27T04:54:56.243" Score="4" Body="&lt;p&gt;Based on jonsca's comment, extract DNA from something. The protocol will involve detergent and isopropol alcohol. The details are available in several places and is fairly simple, but you won't get much DNA per onion, so be ready to prepare a lot of them. To do the degradation, there are several things you could try. If your detergent contains nucleases ( many contain enzymes but do a poor job of labeling which enzymes ) you could mix that in and react it at 37C for a while. Can also try high pH solutions, if you happen to have NaOH around. If you do the DNA extraction differently, you can get the RNA instead of DNA. RNA is very unstable at high pH, and RNAse enzymes are EVERYWHERE. So the RNA should degrade into nucleotides pretty readily if you just use sloppy lab technique. When I extract RNA in the lab, I use phenol/chloroform to pull the proteins and genomic DNA away from the RNA, which stays in the aqueous layer. You probably don't have access to those at home, but cooking oils may be hydrophobic enough to pull it off. Also, it helps to have a centrifuge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't ask me how to purify the nucleotides from your solution and ask me even less about how to verify what you have at home. Verification would require something like a UV-Vis or HPLC or NMR. Good luck getting those at home.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-27T04:54:56.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14657" PostTypeId="2" ParentId="7227" CreationDate="2014-07-27T05:03:09.970" Score="3" Body="&lt;p&gt;I did some bioconjugation with NHS and maleimide biotins to proteins, peptides, and oligosaccharides. I removed the free biotin using spin filters or dialysis. Just be sure to use a MWCO well below the mass of your protein. I even used prep scale HPLC to remove biotin from the oligosaccharide. The HABA kit works well, I think I used that same one to quantify peptide - avidin linking. But try to keep your biotin protected from light and oxygen, it oxidizes fairly easily, I was getting +16 and +32 masses on LC-MS and couldn't explain it until I found some 50 year old paper on biotin sulfoxide. The oxidized biotin doesn't bind avidin nearly as well as normal biotin.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-27T05:03:09.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14658" PostTypeId="2" ParentId="10480" CreationDate="2014-07-27T05:22:29.427" Score="1" Body="&lt;p&gt;I played around with a program called GROMACS about 4 years ago and don't remember much, but I do remember doing a tutorial simulating a DNA double helix and part of it included making the DNA by giving it a sequence and running a script. It can probably make A or B form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This was a long time ago and I forgot. It wasn't GROMACS, it was Amber. Look here &lt;a href=&quot;http://ambermd.org/tutorials/&quot; rel=&quot;nofollow&quot;&gt;http://ambermd.org/tutorials/&lt;/a&gt; Tutorial B1 looks like the same tutorial I did with the DNA.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="137" LastEditDate="2014-07-31T15:21:14.860" LastActivityDate="2014-07-31T15:21:14.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14659" PostTypeId="2" ParentId="7227" CreationDate="2014-07-27T05:43:32.893" Score="0" Body="&lt;p&gt;You could precipitate the protein with alcohol (or acetone) (of course it will probably denature the protein), centrifuge the mixture and redissolve the protein to determine the amount of biotin bound to the protein.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-27T05:43:32.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14660" PostTypeId="1" AcceptedAnswerId="14662" CreationDate="2014-07-27T05:45:18.487" Score="15" ViewCount="444" Body="&lt;p&gt;The title is &lt;em&gt;not&lt;/em&gt; a reference to a Jason Derulo song. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How is &lt;em&gt;change&lt;/em&gt; in entropy measured, experimentally? I've Googled this for a bit and I've found all sorts of mathematical equations such as delta S = q/T as a way to get a grasp on entropy. However, I want to know how entropy is actually determined in the lab (as opposed to on paper). I understand that the units on entropy are in terms of energy/temperature. So if I had to guess I would suppose that some sort of calorimetry is used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the best way to teach entropy (at an introductory level)? I know that textbooks like to say it's a &quot;measure of disorder.&quot; On the other hand I have lots of websites telling me that entropy is best NOT described at a measure of disorder. So what is it really, and what do you think about the term &quot;disorder&quot; as a synonym? My professor refers to it as the degrees of freedom in a system; he also notes that longer carbon chains tend to have higher entropy values due to a greater ability to &quot;wiggle&quot; along their carbon chains. What do you think of this? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Discarding the archaic idea of &quot;disorder&quot; in regard to entropy is essential. It just doesn't make scientific sense in the 21st century and its apparent convenience often is flat-out misleading. As of November 2005, fifteen first-year college texts have deleted “entropy is disorder” although a few still retain references to energy “becoming disorderly”. (This latter description is meaningless&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="19616" LastEditDate="2015-10-23T16:54:52.547" LastActivityDate="2020-03-17T18:17:35.857" Title="Entropy - &quot;Wiggle&quot;?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14661" PostTypeId="2" ParentId="14595" CreationDate="2014-07-27T06:32:51.657" Score="4" Body="&lt;p&gt;How is the mixture composed? Find expressions for mass, volume and amount of substance.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag1\label1&#xA;  V_\text{total} &amp;amp;= V_{\ce{O2}} + V_{\ce{Ar}}\\&#xA;\tag3\label2&#xA;  n_\text{total} &amp;amp;= n_{\ce{O2}} + n_{\ce{Ar}}\\&#xA;\tag3\label3&#xA;  m_\text{total} &amp;amp;= m_{\ce{O2}} + m_{\ce{Ar}}&#xA;\end{align}&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transform &lt;span class=&quot;math-container&quot;&gt;$V_\text{total}$&lt;/span&gt; with the ideal gas and solve for &lt;span class=&quot;math-container&quot;&gt;$n_{\ce{Ar}}$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \tag4\label4&#xA;  V_\text{total} &amp;amp;= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\&#xA;  \tag5\label5&#xA;  n_{\ce{Ar}}  &amp;amp;= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation (&lt;span class=&quot;math-container&quot;&gt;$m = nM$&lt;/span&gt;), then plug in \eqref{5}:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \tag3&#xA;  m_\text{total} &amp;amp;= m_{\ce{O2}} + m_{\ce{Ar}}\\&#xA;  \tag6\label6&#xA;  m_\text{total} &amp;amp;= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\&#xA;  \tag7\label7&#xA;  m_\text{total} &amp;amp;= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}}&#xA;\end{align}&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is only one variable left, which you do not know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \tag8\label8&#xA;  m_\text{total} &amp;amp;= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\&#xA;  \tag9\label9&#xA;  n_{\ce{O2}} &amp;amp;= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the following data (It would be by far better to only use SI [derived] units):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;M_\ce{Ar} &amp;amp;= \pu{39.9 g//mol}\\&#xA;M_\ce{O2} &amp;amp;= \pu{32.0 g//mol}\\&#xA;R &amp;amp;= \pu{8.314 L kPa //K mol}\\&#xA;\pu{1 mmHg} &amp;amp;= \pu{133.322 Pa}\\&#xA;p_\text{total} &amp;amp;\approx \pu{90 kPa}\\&#xA;T &amp;amp;= \pu{316.5 K}\\&#xA;m_\text{total} &amp;amp;= \pu{19.3 g}\\&#xA;V_\text{total} &amp;amp;= \pu{16.2 L}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into \eqref{9}:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \tag9&#xA;  n_{\ce{O2}} &amp;amp;= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\&#xA;  n_{\ce{O2}} &amp;amp;= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\&#xA;  n_{\ce{O2}} &amp;amp;= \pu{0.36 mol}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ideal gas with the data provided yields:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  n_\text{total} &amp;amp;=\frac{pV}{RT}\\&#xA;  n_\text{total} &amp;amp;=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\&#xA;  n_\text{total} &amp;amp;=\pu{0.55 mol}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the mole fraction is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the partial pressure is therefore &lt;span class=&quot;math-container&quot;&gt;$\pu{57.7 kPa}$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$\pu{433 mmHg}$&lt;/span&gt; which is close enough to the solution, given that I rounded the &lt;a href=&quot;https://chat.stackexchange.com/transcript/message/53529625#53529625&quot;&gt;medieval witchcraft units&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2020-02-13T17:22:11.703" LastActivityDate="2020-02-13T17:22:11.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14662" PostTypeId="2" ParentId="14660" CreationDate="2014-07-27T07:58:40.067" Score="8" Body="&lt;p&gt;You can't measure entropy directly, any more than you can measure interatomic distances.  You measure other quantities -- for instance often you can measure energy gain/loss and temperature, and then you integrate $dS=dE/T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to explain it?  One of the best expositions I know is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0195008286&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;The Second Law&lt;/em&gt;&lt;/a&gt; by Henry A. Bent.  It is full of insightful examples, lays the ground carefully and avoids woolly talk, unlike many other thermodynamics books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some elementary but valid comments about the link between information-theoretical entropy and thermodynamical entropy can be found here: &#xA;&lt;a href=&quot;http://www.madsci.org/posts/archives/1998-10/909712896.Ph.r.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.madsci.org/posts/archives/1998-10/909712896.Ph.r.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Greg reminds us that energy and enthalpy are not directly measurable either.  That's why I was careful to write energy &lt;em&gt;gain/loss&lt;/em&gt;, which are somewhat more accessible to measurement; but of course even when doing a calorimetric experiment you're not measuring heat per se but other quantities: how much gas you burned or how intense the current was and how long it was left on. And even some of &lt;em&gt;these&lt;/em&gt; quantities again are indirectly measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree 100% that &quot;entropy is not more abstract than energy&quot; -- they are all abstractions.&#xA;Ultimately, however, I do think the measurement of thermodynamic entropy is one step more indirect than that of energy, if only because it involves the notion of temperature, which (if you think about it carefully) is subtle indeed. &lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-27T16:04:30.440" LastActivityDate="2014-07-27T16:04:30.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14663" PostTypeId="2" ParentId="13639" CreationDate="2014-07-27T09:29:46.817" Score="3" Body="&lt;p&gt;They all appear because they are all finite amount of energy apart, but note that some concentrations are many orders of magnitude greater than others, and the concentration of $\ce{MnO4-}$ is so low that the table shows a zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Predicting the concentration requires being able to read tables of potentials. In standard tables like &lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm&quot; rel=&quot;nofollow&quot;&gt;http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm&lt;/a&gt;  you will find equations such as $$\ce{MnO2{}+4H+ +4e-  \rightleftharpoons Mn + 2H2O}\quad (1.23\mathrm{V})$$ and $$\ce{Mn^{2+}{}+2e^-  \rightleftharpoons Mn}\quad (-1.185\mathrm{V})$$ (the numbers in parentheses are the standard reduction potentials, $E^\circ$, always measured in volts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this and the Nernst equation you can relate the concentrations of any set of species that form a balanced reaction.  In a complicated mix like this you don't know a priori the concentration of any species, only the certain sums (for instance, for all sulfur-containing ions) but you still can, perhaps iteratively, solve the corresponding set of algebraic equations and find the individual concentrations.  &lt;em&gt;All&lt;/em&gt; the equilibrium equations have to be satisfied simultaneously; if they aren't the system is not in equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In this answer I've used the word &quot;concentration&quot; but the equations written with concentrations are only approximately satisfied; to account for second-order effects the notion of &quot;activity&quot; was introduced, and while it is an essentially empirical notion, it's still useful if the dependence of activity on concentration for a particular problem has been tabulated.)&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-27T09:29:46.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14664" PostTypeId="1" AcceptedAnswerId="14665" CreationDate="2014-07-27T13:38:00.990" Score="5" ViewCount="5003" Body="&lt;p&gt;Is the following statement always true? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The more $\ce{-OH}$ functional groups in the molecule, the more is its viscosity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is true, because it is known that weak intermolecular forces lead to lower viscosities and strong intermolecular forces lead to higher viscosities &#xA;and because the more $\ce{OH}$ functional groups in the molecule will give rise to a molecule having stronger molecular forces. As the case in glycerol ($\ce{C3H8O3}$) and water ($\ce{H2O}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the statement must be true. What do you think?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="5026" LastEditDate="2018-03-03T15:38:30.243" LastActivityDate="2018-03-03T15:38:30.243" Title="Is viscosity proportional to the number of hydroxyl groups?" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14665" PostTypeId="2" ParentId="14664" CreationDate="2014-07-27T14:39:40.287" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the following statement always true?  The more −OH  functional&#xA;  groups in the molecule, the more is its viscosity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's hard to find anything that is &quot;always&quot; true in chemistry, but I'd bet that within a series of molecules where the only variable is the number of $\ce{OH}$ groups, your statement is generally true, so I basically agree with your position.  Your analysis and examples are also good. As you pointed out, it is about intermolecular forces.  Specifically, molecules with hydroxyl groups can form &lt;strong&gt;intermolecular hydrogen bonds&lt;/strong&gt; (see first picture below).  These hydrogen bonds cause the molecules to &quot;stick&quot; together and act as if they had a higher molecular weight.  Sugars have many hydroxyl groups that give rise to many intermolecular hydrogen bonds and cause sugars to flow in a slow, syrupy manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pwxCx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;strong&gt;comparative series of alcohols all involving a 3 carbon chain&lt;/strong&gt;.  When the number of hydroxyl groups remains the same and we just vary their position on the chain the viscosity hardly changes.  On the other hand, &lt;strong&gt;note how the viscosity increases markedly each time we add another hydroxyl group&lt;/strong&gt; to the chain.  It appears that the total number of hydroxyl groups is what really matters, so your statement does appear to be generally true as long as the number of hydroxyl groups is the only variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5mcQV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-07T17:40:29.787" LastActivityDate="2015-04-07T17:40:29.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14666" PostTypeId="2" ParentId="14664" CreationDate="2014-07-27T15:09:20.907" Score="11" Body="&lt;p&gt;It is a reasonable rule of thumb, but &lt;strong&gt;certainly not always true&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare for example the viscosities of Dodecane ($\ce{C12H26}$, $\mu=1.374 \ce{\; mPa\cdot s}$, no $\ce{OH}$ groups) and Ethanol ($\ce{C2H5OH}$, $\mu=1.095 \ce{\; mPa\cdot s}$, 1 $\ce{OH}$ group) (&lt;a href=&quot;http://www.engineeringtoolbox.com/absolute-viscosity-liquids-d_1259.html&quot;&gt;source&lt;/a&gt;). There you can see that ethanol has a higher number of $\ce{OH}$-groups, but a lower viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I do think you are on to something regarding the strenght of intermolecular interactions, it is just not purely the number of $\ce{OH}$-groups that matters for that. The length of the carbon chain for example (like in the example I mention above) also plays a role.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-07-27T15:09:20.907" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14667" PostTypeId="1" CreationDate="2014-07-27T16:02:51.130" Score="3" ViewCount="22362" Body="&lt;p&gt;I have read mannaia's &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9270/rate-and-order-confusion&quot;&gt;answer&lt;/a&gt; on the question of 'rate order and confusion', as well as Nicolau Saker Neto's &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9242/what-are-some-good-examples-of-rate-equations-for-a-math-class/9244#9244&quot;&gt;answer&lt;/a&gt; to a related question. They have both been very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, the equilibrium is actually derived from the rate constants of a reaction. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction, hence in a hypothetical reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{pA + qB &amp;lt;=&amp;gt; rC + sD}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the rate equations $Rate_{fwd} = k_{fwd}[A]^{a}[B]^{b}$ and $Rate_{bck} = k_{bck}[C]^c[D]^d$, with $a, b, c,$ and $d$ not related to $p, q, r$ and $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium is derived as such:&#xA;\begin{align}&#xA;  R_{fwd} &amp;amp;= R_{bck}\\&#xA;  k_{fwd}[A]^{a}[B]^{b}&amp;amp;=k_{bck}[C]^c[D]^d\\&#xA;  \frac{k_{fwd}}{k_{bck}} &amp;amp;= \frac{[C]^c[D]^d}{[A]^a[B]^b}\\&#xA;  \therefore k&amp;amp;=\frac {k_{fwd}} {k_{bck}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the derivation, and the fact that the stoichiometric coefficients of the reactants in the rate-determining step (RDS) of the reaction is equal to the order of the reactants in the rate equation, is it correct to state that the reaction equations given as the equilibrium equation is actually the rate-determining step of two reactions?&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-28T04:50:04.783" Title="Relationship between rate equation and equilibrium constant" Tags="&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;kinetics&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14669" PostTypeId="1" AcceptedAnswerId="14670" CreationDate="2014-07-27T17:11:03.280" Score="3" ViewCount="817" Body="&lt;p&gt;What would be the effect of putting a solution in a strong electric field? Would the reactions in the solution be affected? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking yes – lots (if not all) – chemistry is dependent on charge (partial/real charges) and by applying an electric field, we'd be affecting the very property of the molecules/ions that drives reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this been researched? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:49:15.597" LastActivityDate="2015-10-23T16:49:15.597" Title="Electric field effect on reactions" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14670" PostTypeId="2" ParentId="14669" CreationDate="2014-07-27T18:11:48.660" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would the reactions in the solution be affected?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they can be.  &lt;a href=&quot;http://www.scs.illinois.edu/burke/files/group_meetings/13.10.05%20Electric%20Fields%20SGB.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; are some nice examples.  Electric fields also exert a significant effect on &lt;a href=&quot;http://web.stanford.edu/group/boxer/papers/paper94.pdf&quot; rel=&quot;nofollow&quot;&gt;electron transfer reactions&lt;/a&gt;.  Basically you would expect charged molecules to be more affected than polar molecules, which should be more affected than non-polar, but polarizable molecules.  It's just a question of how big a field you can produce and how polarized and long-lived are the various entities (starting materials, intermediates, etc.) that exist along the reaction coordinate.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-27T18:11:48.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14671" PostTypeId="1" AcceptedAnswerId="14689" CreationDate="2014-07-27T18:32:30.373" Score="6" ViewCount="183" Body="&lt;p&gt;Amongst many models, including the valence bond model (VB) or the molecular orbital (MO) model, which are the ones with best predictive power? (e.g. the MO is thought to predict spectroscopic properties better than the VB)&lt;/p&gt;&#xA;" OwnerUserId="7251" LastEditorUserId="10185" LastEditDate="2015-09-27T09:29:51.770" LastActivityDate="2015-09-27T09:29:51.770" Title="Electronic model with highest prediction rate" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14672" PostTypeId="2" ParentId="6246" CreationDate="2014-07-27T19:50:29.013" Score="3" Body="&lt;p&gt;Even if no impurities are present in your copper sulfate,  your water might be slightly basic. (See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5125/what-chemicals-are-in-tap-water-that-gives-it-a-slightly-basic-ph-level&quot;&gt;this question&lt;/a&gt;.) Also water typically has dissolved $\ce{CO2}$, which in neutral solution means some $\ce{HCO3-}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By lowering the pH throught he addition of dilute sulfuric acid you ensure that the concentration of $\ce{OH-}$ and $\ce{HCO3-}$ is extremely low. Either of these ions, if present in significant amounts, will cause some coprecipitation (since both  copper hydroxide and copper carbonates are much less  soluble than copper sulfate) and as a result you wouldn't get nice big crystals of pure $\ce{CuSO4\cdot 5H2O}$.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-07-27T20:14:43.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14673" PostTypeId="1" AcceptedAnswerId="14674" CreationDate="2014-07-27T20:13:37.067" Score="4" ViewCount="3740" Body="&lt;p&gt;I was asked to draw the lewis-dot structure for ClO and I drew this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   ..  ..&#xA;  :Cl-  O.&#xA;   ..  ..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The correct answer is apparently this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   .   ..&#xA;  :Cl-  O:&#xA;   ..  ..&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How would I know that? In the first example, all the formal charges are zero, but in the second, Oxygen has -1 and Chlorine has +1.&lt;/p&gt;&#xA;" OwnerUserId="1169" LastEditorUserId="1169" LastEditDate="2014-07-27T20:59:25.703" LastActivityDate="2019-06-18T16:02:20.980" Title="Why is the Chlorine missing an electron in Chlorine Monoxide?" Tags="&lt;lewis-structure&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14674" PostTypeId="2" ParentId="14673" CreationDate="2014-07-27T20:21:19.863" Score="5" Body="&lt;p&gt;In fact the unpaired electron belongs neither to Cl nor to O, but to a molecular orbital spread over both atoms [but not equally: see below].  The Lewis structure is simply not a complete representation of what is going on.  (And for what it's worth wikipedia gives the same Lewis structure you drew: &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine_monoxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Chlorine_monoxide&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Actually it seems the highest occupied molecular orbital (where the unpaired electron is) is mostly localized on the oxygen atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MO Diagram (taken from &lt;a href=&quot;https://www.chm.uri.edu/weuler/chm401/studyaids/molecularstructurepracticeproblems_answers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zgVsd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zgVsd.jpg&quot; alt=&quot;MO diagram for ClO-&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, your structure is more correct than the one in the answer key.  However the justification involves molecular orbital arguments outside the scope of a general chemistry course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another edit:&lt;/strong&gt; I too like your argument about formal charges and would accept it as a justification that the unpaired electron belongs to O rather than to Cl.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="69977" LastEditDate="2019-06-18T16:02:20.980" LastActivityDate="2019-06-18T16:02:20.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14675" PostTypeId="1" AcceptedAnswerId="14679" CreationDate="2014-07-27T20:31:46.843" Score="3" ViewCount="209" Body="&lt;p&gt;I use Sodium Persulfate to etch copper on a PCB process. I have been using glass heater on the etching tank but the glass is too fragile; it breaks too easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that titanium heater is used commercially (in aquariums as well). I don't know much about chemistry, could anyone tell me if it is okay to use titanium with Sodium Persulfate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium Persulfate to water ratio is 1:10 and the maximum temperature is about 50 C. &lt;/p&gt;&#xA;" OwnerUserId="1940" LastEditorUserId="1940" LastEditDate="2014-07-28T20:08:29.763" LastActivityDate="2014-07-28T20:08:29.763" Title="can sodium persulfate etch or damage a titanium heater?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14676" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2014-07-27T21:01:39.420" Score="5" ViewCount="15309" Body="&lt;p&gt;Why is $\ce{B2}$ paramagnetic? It's just 2 borons with a triple bond between them. Aren't all the electrons paired?&lt;/p&gt;&#xA;" OwnerUserId="1169" LastEditorUserId="16683" LastEditDate="2018-08-17T07:21:24.923" LastActivityDate="2018-08-17T07:21:24.923" Title="Why is diboron (B2) paramagnetic?" Tags="&lt;molecular-orbital-theory&gt;&lt;electronic-configuration&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14677" PostTypeId="2" ParentId="14676" CreationDate="2014-07-27T21:23:52.480" Score="7" Body="&lt;p&gt;For the same reason $\ce{O2}$ is paramagnetic: degeneracy.&lt;br&gt;&#xA;The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$.  The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're right to think that this is unexpected:  why are the $\pi(\mathrm{2p})$ orbitals  lower than the $\sigma(\mathrm{2p})$? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Matt Law's lecture notes for the expected and observed ordering:&#xA;&lt;a href=&quot;https://web.archive.org/web/20160519100538/https://www.chem.uci.edu/~lawm/10-9.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MO Diagrams for Diatomic Molecules; Chapter 5; Friday, October 9, 2015; PDF via the Internet Archive&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="4945" LastEditDate="2018-08-16T13:33:00.630" LastActivityDate="2018-08-16T13:33:00.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14678" PostTypeId="2" ParentId="14540" CreationDate="2014-07-27T23:21:29.243" Score="3" Body="&lt;p&gt;Acidic wash is not going to help. As you said you can try a DIAD alternative for example &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/680850?lang=en&amp;amp;region=GB&quot; rel=&quot;nofollow&quot;&gt;DCAD&lt;/a&gt; can give good results (sry no advertising intended here). There is also the option of taking your contaminated product to the next step(s) and hope you can get rid of the impurity there. I would take option A if I have to make a bunch of similar analogues by a Mitsunobu and B otherwise for not wasting too much time&lt;/p&gt;&#xA;" OwnerUserId="7252" LastActivityDate="2014-07-27T23:21:29.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14679" PostTypeId="2" ParentId="14675" CreationDate="2014-07-27T23:30:00.380" Score="4" Body="&lt;p&gt;If the surface of the immersed element is all titanium, the answer appears to be that it won't get corroded unless the temperature is much higher than 50$^\circ$C; see &lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=1336&quot; rel=&quot;nofollow&quot;&gt;http://www.azom.com/article.aspx?ArticleID=1336&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case is changed if there are joints or parts made of something else, as in this illustration: &lt;a href=&quot;http://www.saltwatereddies.com/content-popup_image/pID-2199/popup_image.html&quot; rel=&quot;nofollow&quot;&gt;http://www.saltwatereddies.com/content-popup_image/pID-2199/popup_image.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that would not be so safe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if you've been breaking glass heating elements right and left, you have little to lose by trying the titanium one...&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-27T23:30:00.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14680" PostTypeId="1" CreationDate="2014-07-28T00:56:56.127" Score="7" ViewCount="218" Body="&lt;p&gt;I asked a question here &lt;a href=&quot;https://photo.stackexchange.com/questions/53048/how-can-walter-mitty-examine-the-negatives-of-photos-in-the-bright-sunny-day&quot;&gt;How can Walter Mitty examine the negatives of photos in the bright sunny day?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commonly known, there are three steps using chemical baths to bring out the images of negatives:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;developer to bring out the image, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;stop bath&quot; to halt the process, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;fixer&quot; to make the image permanent. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in details what are the Chemical procedure/compounds/reaction for the negative's developer/stop bath/fixer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wish the answer is in depth. &lt;/p&gt;&#xA;" OwnerUserId="3981" LastEditorUserId="23561" LastEditDate="2019-03-26T20:48:00.387" LastActivityDate="2019-03-26T20:50:02.367" Title="Chemistry of Walter Mitty's negative developer/stop bath/fixer?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14681" PostTypeId="2" ParentId="14680" CreationDate="2014-07-28T01:22:04.163" Score="4" Body="&lt;p&gt;basics here: &lt;a href=&quot;http://science.csustan.edu/nhuy/chem1002/photoexp.htm&quot; rel=&quot;nofollow&quot;&gt;http://science.csustan.edu/nhuy/chem1002/photoexp.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In depth: there are many good books about the chemistry of photography, starting with this 120-year old classic:&#xA;&lt;a href=&quot;https://archive.org/details/chemistryphotog02meldgoog&quot; rel=&quot;nofollow&quot;&gt;https://archive.org/details/chemistryphotog02meldgoog&lt;/a&gt;&#xA;‎&lt;p&gt;&#xA;Here is another that's fully available online and recent enough to cover every major aspect of the subject:&#xA;&lt;a href=&quot;https://openlibrary.org/books/OL5953553M/Photographic_chemistry_in_black-and-white_and_color_photography&quot; rel=&quot;nofollow&quot;&gt;https://openlibrary.org/books/OL5953553M/Photographic_chemistry_in_black-and-white_and_color_photography&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy more recent books, especially for developments in color.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-28T01:22:04.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14682" PostTypeId="1" CreationDate="2014-07-28T01:44:53.567" Score="0" ViewCount="169" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming standard conditions, answer the following questions.&#xA;  (yes/no)&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{Sn^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{Cr}$ metal capable of reducing $\ce{Fe^{2+}(aq)}$?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{VO^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{H2(g)}$ capable of reducing $\ce{Ag+ (aq)}$? &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Is $\ce{Cr^{3+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;yes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;yes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;no&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;yes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;yes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;no&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="7101" LastEditorUserId="5026" LastEditDate="2018-02-25T02:44:30.920" LastActivityDate="2018-02-25T02:44:30.920" Title="How do I find out if ions are oxidised/ reduced based on half cell potentials?" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2018-02-26T07:44:33.613" ContentLicense="CC BY-SA 3.0" />
  <row Id="14683" PostTypeId="2" ParentId="14667" CreationDate="2014-07-28T01:57:54.967" Score="3" Body="&lt;p&gt;At equilibrium it does not matter if the reaction has multiple steps; your equilibrium constant only has to involve the initial reactants and final products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in an S$_\mathrm{N}$1 reaction, obviously the RDS is the departure of the leaving group, but you don't have to (and usually are not in a position to) know the concentration of the intermediate carbocation in order to calculate the equilibrium of the overall reaction.  You can simply write an  equilibrium equation for the initial reactants and final products: $$K=\frac{[\ce{LG}][\ce{product}]}{[\ce{nucleophile}][\ce{substrate}]}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course finding $K$ may be not be easy, even if you have access to good thermodynamic data for the reactants and products, because $\Delta G_0$ (and therefore $K$) depends on solvation, temperature and other conditions. But given identical conditions, $K$  is constant over a wide range of concentrations, so you might be able to find it in the literature or measure it with a single experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this assumes that the RDS is fast enough that you're able to reach  equilibrium before the solvent evaporates, or your reactants and products decompose via other pathways, or you die of boredom!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can also write an equilibrium equation for the RDS alone:&#xA;$$&#xA;K'=\frac{[\ce{LG}][\ce{carbocation}]}{[\ce{substrate}]}&#xA;$$&#xA;But since you seldom know the concentration of the carbocation, this equation isn't as useful in practice (it is of course extremely useful for mechanistic studies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6243/how-is-it-that-the-equilibrium-constant-does-not-depend-on-mechanism&quot;&gt;This page&lt;/a&gt; has a more complete and very nice answer!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-07-28T04:50:04.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14684" PostTypeId="2" ParentId="13786" CreationDate="2014-07-28T05:07:07.957" Score="3" Body="&lt;p&gt;This is a titration problem for quantitatively determining the concentration of a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What reaction does occur?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\ce{AgNO3 (aq) + NaBr (aq) &amp;lt;=&amp;gt; AgBr v + Na+ (aq) + NO3- (aq)}$&lt;br&gt;&#xA; or essentially $\ce{Ag+ + Br- &amp;lt;=&amp;gt; AgBr v}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is the solubility product important? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can you tell about the equilibrium state at the point when the first precipitate falls?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The solubility product is matched, hence $\ce{[Ag^+][Br^-]}&amp;lt;K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the amount of bromine ions added to the solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What can you tell about the concentrations in the final mixture?&lt;br&gt;&#xA;First, what is the concentration of bromide ions in this mixture?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$&lt;br&gt;&#xA; $c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second, what can you tell about the concentration of silver ions in the final mixture?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the number of moles of silver ions in the final mixture?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the initial concentration of the silver nitrate solution?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="7187" LastEditDate="2014-07-28T08:25:56.367" LastActivityDate="2014-07-28T08:25:56.367" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14686" PostTypeId="2" ParentId="14641" CreationDate="2014-07-28T07:36:43.557" Score="2" Body="&lt;p&gt;The lone pairs on both oxygen atoms cancel each others effect, as simple as that. And the structure is symmetrical from all aspects, which again leads to zero dipole moment. Plus there is no net negative charge or lone pair of Carbon as well, therefore the compound cannot be polar. REMEMBER polarity is a vector.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastActivityDate="2014-07-28T07:36:43.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14687" PostTypeId="2" ParentId="14682" CreationDate="2014-07-28T08:06:07.697" Score="5" Body="&lt;p&gt;Look at this table of half-reactions: &lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm&quot; rel=&quot;nofollow&quot;&gt;http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll explain how it contradicts your answer to the first question; you can then rework your answers and update your question with justifications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One entry of the table says $$ \ce{Cr^{3+}(aq) + e- -&amp;gt; Cr^{2+} (aq)} \quad –0.424.$$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $-0.424$ is what's called the standard potential, expressed in volts.  It is a measure of how favorable the half-reaction is (compared with a standard half-reaction).  The higher the number, the more favored the products will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have to find a half-reaction involving $\ce{Fe^{2+}}$.  Since you want Fe to reduce something, it will itself be oxidized.  The table shows no half-reaction where $\ce{Fe^{2+}}$ is oxidized, but that's because by convention only &lt;em&gt;reduction&lt;/em&gt; half-reactions are shown; to get the opposite, &lt;em&gt;oxidation&lt;/em&gt;, half-reactions, just swap the terms and negate the potential.  That is, take this line from the table:&#xA;$$&#xA;\ce{Fe^{3+}(aq) + e^{–}  -&amp;gt; Fe^{2+}(aq)	} \quad 0.771&#xA;$$&#xA;and turn it into &#xA;$$&#xA; \ce{Fe^{2+}(aq) -&amp;gt; Fe^{3+}(aq) + e^{–}} \quad {-}0.771&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add this, term by term, to the Cr half-reaction. The result is&#xA;$$&#xA;\ce{Fe^{2+} + Cr^{3+} -&amp;gt; Fe^{3+} + Cr^{2+}} \quad {-}1.195&#xA;$$&#xA;If the potential in the final reaction is negative, the reaction is impossible (more precisely, equilibrium lies to the left --- in the case very far to the left, becase 1 volt goes a long way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the basics.  There is a subtlety involved when the number of electrons exchanged is not the same in the two half-reactions; you'll find it in your textbook.  You could try answering the other 5 questions now and if you get stuck, tell us precisely where.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-28T14:56:01.757" LastActivityDate="2014-07-28T14:56:01.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14688" PostTypeId="2" ParentId="7953" CreationDate="2014-07-28T08:13:18.263" Score="2" Body="&lt;p&gt;The answer is no, because as you remark $\ce{F2}$ will react with water and produce $\ce{HF}$.  So it is not the case that $\ce{F2}$ passes through unreacted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm puzzled by this sentence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you clarify what you mean by &quot;it produces data&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-28T08:13:18.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14689" PostTypeId="2" ParentId="14671" CreationDate="2014-07-28T08:24:49.580" Score="8" Body="&lt;p&gt;The so called gold standard of quantum chemical calculations of ground state energies and properties is usually considered to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupled_cluster&quot;&gt;Coupled Cluster&lt;/a&gt; Singles Doubles (with perturbative) Triples as a method. This method is based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Ab_initio&quot;&gt;&lt;em&gt;ab initio&lt;/em&gt;&lt;/a&gt; molecular orbital theory and it is size consistent, but not variational. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot;&gt;basis set&lt;/a&gt; of at least a triple zeta quality with additional valence and polarisation functions should be employed, good examples for this are Dunning's (augmented) correlation consistent basis sets, e.g. (aug-)cc-pV{T,Q,5,6}Z, or Weigend's def2-basis set family, def2-TZVPP, def2-QZVP, but there are various others. (If you like to have some citations, drop a note in the comments.) This level of theory is usually denoted as CCSD(T)/TZVP, where TZVP is a placeholder for one of the various basis sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle CC theory is exact at the complete basis set limit, but in practise, this is not solvable. Another theory that is in principle exact, because they become equivalent, is Full &lt;a href=&quot;http://en.wikipedia.org/wiki/Configuration_interaction&quot;&gt;Configuration Interaction&lt;/a&gt;, F-CI. The truncated variants CISD, CISD(T), CISDT, ... are variational, but not size consistent. For chemical systems it is much more important, that the calculations are size consistent, therefore these methods lost their importance. An attempt to solve this was to introduce &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadratic_configuration_interaction&quot;&gt;Quadratic Configuration Interaction&lt;/a&gt;, QCI, but this is almost the same as coupled cluster theory, so it was not successful either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The importance of &lt;a href=&quot;http://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)&quot;&gt;Many Body Perturbation Theory&lt;/a&gt;, MBPT, of high orders, like &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B8ller%E2%80%93Plesset_perturbation_theory&quot;&gt;Møller Plesset Perturbation Theory&lt;/a&gt; of fourth order, MP4, has very much decreased because of it inefficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some cases, which cannot be accurately described by these models. Those are cases, where the ground state cannot be well defined by one determinante, the initial guess wave function obtained from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock&quot;&gt;Hartree-Fock&lt;/a&gt;, HF. Here you have to use a different description. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Multi-configurational_self-consistent_field&quot;&gt;Multi Configuration&lt;/a&gt;, MC, or Multi Reference, MR, ansatz is necessary. The qualitative equivalent to HF is &lt;a href=&quot;http://en.wikipedia.org/wiki/Complete_active_space&quot;&gt;Complete Active Space&lt;/a&gt; Self Consistent Field, CASSCF. Based on this you can use Perturbation Theory to include more correlation effects. The best possible description so far (judging accuracy versus computational cost) is second order perturbation theory on a CASSCF wave function, for short CASPT2. It is also very reliable for spectroscopic calculations. There are also various developments on &lt;a href=&quot;http://en.wikipedia.org/wiki/Multireference_configuration_interaction&quot;&gt;Multi Reference Configuration Interaction&lt;/a&gt; as another theory. (I have limited expertise on that field.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of treating chemical systems is &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot;&gt;Valence Bond theory&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Modern_valence_bond_theory&quot;&gt;with the application to modern computational aspects&lt;/a&gt;. Here you propose the mixing of various Lewis like structures, to obtain the ground state energy and properties. The formalism can get quite complicated, maybe this is why its performance for excited states properties is not very accurate at the moment. It is still a very active field of research, by a few scientific groups. (I am not one of them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, affordable spectroscopic calculations have still huge errors and sometimes the numbers can vary quite a bit. Currently most used is a variety of methods derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Time-dependent_density_functional_theory&quot;&gt;Time Dependent Density Functional Theory&lt;/a&gt;, TDDFT. An &lt;em&gt;ab initio&lt;/em&gt; method of great importance in this field is Configuration Interaction Singles. Also fairly well implemented are approximate coupled cluster techniques, e.g. CC2. The struggle for finding an accurate, robust and computationally cheap method are one of the main goals of modern quantum chemistry. (I can apply these methods fairly well, but I have no deeper insight how they work.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extended systems like surfaces and solids there are various approaches. Since they get very demanding very soon, the most commonly used practical approaches are based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot;&gt;plain or hybrid density functionals&lt;/a&gt;. They often also employ plane wave basis sets with periodic boundary condition. (Unfortunately I am not an expert on that field either.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there is no definite answer to your question and the development of new and efficient methods is still a very active field. Depending on what type of description you want or need, there are many possibilities. The most advanced models are obviously for ground state energies and properties, since they have been researched for about a century now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you, I apologise for the chatty answer. If something remains unclear, drop me a note in the comments.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-28T08:24:49.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14691" PostTypeId="1" AcceptedAnswerId="14693" CreationDate="2014-07-28T13:22:03.883" Score="2" ViewCount="310" Body="&lt;p&gt;Does anyone know which specific azo compounds are used in DVD-R/CD-R recordable media?&lt;/p&gt;&#xA;" OwnerUserId="7260" LastEditorUserId="73" LastEditDate="2014-07-29T08:29:20.207" LastActivityDate="2014-07-29T08:29:20.207" Title="Which azo compounds are found in DVD-R and CD-R media?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14692" PostTypeId="1" AcceptedAnswerId="14694" CreationDate="2014-07-28T14:14:11.947" Score="4" ViewCount="1613" Body="&lt;p&gt;Every kid who has been to a science fair knows that you can make a battery with a lemon, a potato, a penny and a nail. I've never done this and I'm not entirely clear on how it works beyond a hazy recollection of undergraduate-level chemistry class; but I gather that you do not need to use a lemon and potato as your power source but can in fact use any number of substances willing to donate or accept electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is this: are there any home-made batteries that can produce sufficient electrical energy that its cheaper to make them yourself rather than plug your appliance into the wall or use store-bought batteries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that there are a number of variables here that might make a significant difference in whether its cheaper to use established electricity sources or create your own so for arguments sake lets say that I am trying to:&#xA;- charge a cell phone/laptop&#xA;- run a microwave long enough to make popcorn&#xA;- keep a refrigerator cold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets also consider the dual possibilities that I: &#xA;- live in a city apartment and have to buy all my fuels and materials &#xA;- live on my own far where I can grow as much produce as I want to devote to making batteries&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are other good fuel sources that could be a logistic variable; let me know in the comments so I can edit the question accordingly. &lt;/p&gt;&#xA;" OwnerUserId="7209" LastActivityDate="2017-08-28T14:46:59.007" Title="Are homemade batteries economically practical?" Tags="&lt;everyday-chemistry&gt;&lt;fuel&gt;&lt;electricity&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14693" PostTypeId="2" ParentId="14691" CreationDate="2014-07-28T14:15:06.560" Score="5" Body="&lt;p&gt;Each company in the DVD business has their own proprietary twist on the actual dye they use and they don't go out of their way to reveal the structure of the molecule they use.  Also the actual dye used changes fairly rapidly (several years or so) as technology shifts (e.g. higher density, Blu-ray, etc.).  That said, metal-azo dyes are popular today.  A Mitsubishi patent from a few years ago (WO 2006/104196 A1 (10/03/2006)) says &quot;The recording layer contains a pyridone azo organic dye having a 6-hydroxy-2-pyridone structure as a coupler component and an isoxazole, 1,2,4-triazole or pyrazole as a diazo component, coordinated to a metal, such as nickel, cobalt, iron, zinc, copper or manganese.&quot; Here is the structure of 2-pyridone (6-hydroxy would be attached on the other side of the nitrogen) and pyrazole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NOY1v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A0Dtd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://www.google.com/patents/EP1570004A2?cl=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to a Ciba patent&lt;/a&gt; that has more detail. And here is a general structure a 2-pyridone-based metal-azo dye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5I0r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-28T18:40:37.740" LastActivityDate="2014-07-28T18:40:37.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14694" PostTypeId="2" ParentId="14692" CreationDate="2014-07-28T15:25:51.493" Score="11" Body="&lt;p&gt;No: trying to catch up with 100+ years of technological optimizations in your kitchen won't be an economical enterprise.  Store-brought batteries are standardized and relatively inexpensive.  Power from an outlet is even less expensive (by one or two orders of magnitude).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,&#xA;in many places you can save on electricity by going solar. Even if you can't install solar panels on your roof because you don't own a house, you could buy solar chargers for small appliances if you have a sunny corner in your apartment...&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-28T21:29:26.033" LastActivityDate="2014-07-28T21:29:26.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14695" PostTypeId="1" AcceptedAnswerId="14696" CreationDate="2014-07-28T15:28:35.120" Score="12" ViewCount="13816" Body="&lt;p&gt;Many times I have seen the catalyst as &lt;a href=&quot;https://en.wikipedia.org/wiki/Raney_nickel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raney nickel&lt;/a&gt; instead of nickel catalyst. Raney nickel catalyst is developed by &lt;a href=&quot;https://en.wikipedia.org/wiki/Murray_Raney&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murray Raney&lt;/a&gt;, and I think &lt;strong&gt;Raney nickel catalyst&lt;/strong&gt; and &lt;strong&gt;nickel catalyst&lt;/strong&gt; are two different things, because various sources use both names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they differ from each other?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="41328" LastEditDate="2017-08-09T01:48:35.943" LastActivityDate="2017-08-09T01:48:35.943" Title="Difference between nickel catalyst and Raney nickel catalyst" Tags="&lt;reaction-mechanism&gt;&lt;nomenclature&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14696" PostTypeId="2" ParentId="14695" CreationDate="2014-07-28T16:05:31.083" Score="12" Body="&lt;p&gt;Yes, the Raney nickel catalyst is different from simple, fine nickel powder.  Raney nickel is prepared in two steps.  First, the nickel is alloyed with aluminum and then in second step much of the aluminum is removed through a leaching process with sodium hydroxide. As the aluminum is removed two things happen.  The resulting nickel-aluminum alloy becomes very textured and porous with extremely high surface area (sponge-like).  Also hydrogen gas is liberated during the leaching process and some of it remains adsorbed on the alloy surface.  This is the active catalyst that is ready for use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Al + 2 NaOH + 6 H2O -&amp;gt; 2 Na[Al(OH)4] + 3 H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to read more about the initial alloying process or the activation (leaching) step, see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Raney_nickel&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, Murray Raney discovered this catalyst (US1628190) in the late 1920's and he was the sole assignee.  I'm guessing that W. R. Grace bought the patent rights from him since they hold the trademark on use of the term &quot;Raney nickel&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user15489" LastEditDate="2015-06-22T18:08:29.813" LastActivityDate="2015-06-22T18:08:29.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14697" PostTypeId="1" CreationDate="2014-07-28T16:16:36.493" Score="2" ViewCount="124" Body="&lt;p&gt;Does there exist a mathematical model of isomerism? (a model which is capable of predicting, for given circumstances, the re-arrangement that is going to occur, e.g. Demjanov, Wagner-Meerwein etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;understand how a geometrical structure of a molecule changes for isomers&lt;/li&gt;&#xA;&lt;li&gt;understand under what conditions certain geometrical structure of one molecule is transformed into a different geometrical structure of an isomer&lt;/li&gt;&#xA;&lt;li&gt;which geometrical isomer structures are impossible to exist&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7251" LastEditorUserId="7251" LastEditDate="2014-07-31T08:35:49.623" LastActivityDate="2014-07-31T08:35:49.623" Title="Mathematical model of isomerism" Tags="&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14698" PostTypeId="2" ParentId="4961" CreationDate="2014-07-28T16:48:22.573" Score="1" Body="&lt;p&gt;Here is a &lt;a href=&quot;http://www.arkat-usa.org/get-file/20028/&quot; rel=&quot;nofollow&quot;&gt;link to a detailed reference on HMF&lt;/a&gt;.  In section 3.3 (page 32) they show a mechanism for the acid catalyzed degradation of HMF.  As you noted, they do propose two pathways for decomposition, one leads to polymeric materials, the other leads to 2,5-dioxo-3-hexenal, which then can react further to produce 4-oxopentanoic acid. The work they describe was done by Horvat (Horvat, J.; Klaič, B.; Metelko, B.; Sunjic, V. Tetrahedron Lett. 1985, 26, 2111).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to reproduce their drawing, but follow the link and look for Scheme 17.  Most of the steps involve protonation of either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a double bond to form a resonance stabilized carbocation followed by addition of water and elimination of a proton to yield an alcohol&lt;/li&gt;&#xA;&lt;li&gt;a hydroxyl group followed by elimination of water to yield a resonance stabilized carbocation followed by elimination of a proton to yield a double bond&lt;/li&gt;&#xA;&lt;li&gt;or protonation of the furan oxygen followed by ring opening to generate a protonated, resonance stabilized carbonyl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you have any specific questions on the mechanism, just post back here.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-28T19:42:19.940" LastActivityDate="2014-07-28T19:42:19.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14699" PostTypeId="2" ParentId="4894" CreationDate="2014-07-28T17:27:58.860" Score="1" Body="&lt;p&gt;I would suggest adding anhydrous $\ce{MgSO4}$ to your distillate, like Nicolau Saker Neto already mentioned in his comments. Mix the suspension thoroughly (e.g. by stirring) and look how long the salt remains dispersed. When it clumps up and settles quickly due to the formation of hydrates, there is still some water in your mixture, and you need to repeat the procedure until the added $\ce{MgSO4}$ remains dispersed as fine particles for a longer time (it should look similar to a snow storm). After the salt has settled, you can recover the dried oil with a pipette. Because &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_sulfate&quot; rel=&quot;nofollow&quot;&gt;magnesium sulfate&lt;/a&gt; is not listed as toxic or environmentally hazardous, it can be considered a &quot;greener&quot; alternative to, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper(II)_sulfate&quot; rel=&quot;nofollow&quot;&gt;copper(II) sulfate&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Diethyl_ether&quot; rel=&quot;nofollow&quot;&gt;diethyl ether&lt;/a&gt; (the latter is additionally highly flammable).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-07-28T17:27:58.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14700" PostTypeId="1" CreationDate="2014-07-28T17:56:43.900" Score="2" ViewCount="361" Body="&lt;p&gt;Why is energy released during the process of adsorption? How is this energy change observed and how is it calculated?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="305" LastEditDate="2014-07-28T19:01:54.877" LastActivityDate="2020-08-17T13:58:29.747" Title="Adsorption on surface of adsorbent" Tags="&lt;adsorption&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14701" PostTypeId="1" CreationDate="2014-07-28T18:40:37.907" Score="1" ViewCount="91" Body="&lt;p&gt;What invariants related to geometrical structures that describe molecular geometry are used in chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation: Here, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14671/&quot;&gt;Electronic model with highest prediction rate&lt;/a&gt;…, one can learn about electronic models. Now, I want to learn which invariants are used in all major models used (and why).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of invariant: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14701/invariants-in-chemistry-used-for-model-selection&quot;&gt;Invariants in chemistry used for model selection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of invariants in physics of ideal gas (invariance of the geometry of ideal gas with respect to Legendre transformations): &lt;a href=&quot;http://arxiv.org/pdf/0811.0222v1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arxiv.org/pdf/0811.0222v1.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7251" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-09-27T09:28:07.567" Title="Invariants in chemistry used for model selection" Tags="&lt;molecules&gt;&lt;theoretical-chemistry&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14702" PostTypeId="2" ParentId="9823" CreationDate="2014-07-28T18:57:02.620" Score="6" Body="&lt;p&gt;As you have already figured out, diethyl ether is a small dipole due to the electronegativity difference between carbon and oxygen, which can also be seen in this &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Diethyl_Ether_Electron_Rendering.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron density plot&lt;/a&gt;. As both dipole moment vectors point roughly in the same direction, they do not cancel each other out like in the case of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14641/why-is-o-c-o-nonpolar&quot;&gt;carbon dioxide&lt;/a&gt;, so the molecule has a net dipole moment and is therefore (weakly) polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2DGNj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;London dispersion forces&lt;/a&gt;, which result from short-lived dipoles induced by fluctuations in the electron shell of molecules, are also present. They are responsible for the weak interactions between the alkyl chains, like in other alkanes and non-polar molecules. The permanent dipoles can also interact with each other, in the form of intermolecular attractive interactions between centers with $\delta+$ and $\delta-$ partial charge (and likewise repulsion between centers of equal partial charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonding between C-H groups and oxygens, although possible in principle (see comments to your question) and, for example, found in &lt;a href=&quot;http://dx.doi.org/10.1074%2Fjbc.M204514200&quot; rel=&quot;nofollow noreferrer&quot;&gt;protein structures&lt;/a&gt;, can be neglected in this case due to the low number of donors and the weakness of the interaction (&amp;lt; 17 kJ/mol). Nevertheless, the ether oxygen can act as a hydrogen bond acceptor in combination with better hydrogen bond donors like OH groups, which explains why diethyl ether is sparingly soluble in water despite its low polarity (69 g/l at 20 °C (&lt;a href=&quot;http://en.wikipedia.org/wiki/Diethyl_ether&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)). On the other hand, the absence of hydrogen bonding between ether molecules becomes apparent in the much lower boiling point of diethyl ether (34.6 °C, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diethyl_ether&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) compared to water.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-07-28T18:57:02.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14703" PostTypeId="1" AcceptedAnswerId="14707" CreationDate="2014-07-28T19:47:45.297" Score="4" ViewCount="6876" Body="&lt;p&gt;I am interested in learning organic chemistry in order to solidify my knowledge of molecular biology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are a lot of options for organic chemistry textbooks, and I plan on simply sampling a few in order to figure out which ones are best, but I thought I would also ask for some advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I am self-learning, I have some specific items on my textbook wishlist:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The textbook should be &lt;em&gt;insightful&lt;/em&gt; - I don't simply want facts presented, I want to see how the facts connect together into a larger whole. Usually, teachers fulfill this task, but in this case, I will need some help from the textbook.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The textbook should have a good section on learning IUPAC nomenclature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't want organic chemistry to be an experience in advanced memorization for me. The previous point helps in addressing this issue, but I would like to reiterate it. I would like to have a textbook that helps me understand why each class of reactions is important (what can one do, that the other can't?) in a chemist (or an organism's) toolbox - I don't want to be memorizing reactions, I want to be putting them into my toolbox.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With such requirements in mind, which textbooks would you recommend?&lt;/p&gt;&#xA;" OwnerUserId="7264" LastActivityDate="2014-07-28T20:59:41.430" Title="Organic chemistry textbook for self-learning?" Tags="&lt;organic-chemistry&gt;&lt;books&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2016-11-04T18:41:54.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="14704" PostTypeId="1" AcceptedAnswerId="14718" CreationDate="2014-07-28T20:06:08.287" Score="14" ViewCount="3157" Body="&lt;p&gt;After a discussion about the usefulness of hydrogen fuel for cars with a friend of mine, I wondered what the reaction mechanism for combustion of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; was. I only study (well almost) organic chemistry, so I have no idea if the mechanisms from organic chemistry can be applied at all to non-organic molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2\ce{H2 + O2 -&amp;gt; 2 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most textbooks this would be described as a redox-reaction with the two half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2\ce{H+ + 2 e- -&amp;gt; H2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{O2 + 4 H+ + 4 e- -&amp;gt; 2 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I wonder what the real mechanism is. Based on my organic thinking I thought the following (it's probably completely wrong, but I think it is important to state because someone can correct me much better if I express my thinking than if he has no idea where the misconception started):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature is high enough a long pair of oxygen could break the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-H}$&lt;/span&gt; (the lone pair will be donated to the &lt;span class=&quot;math-container&quot;&gt;$\sigma^{*}$&lt;/span&gt; bond) bond and form &lt;span class=&quot;math-container&quot;&gt;$\ce{2 H-}$&lt;/span&gt; (each H gets one electron because of their equal electronegativity). &lt;span class=&quot;math-container&quot;&gt;$\ce{H-}$&lt;/span&gt; is very unstable and one will immediately react with the empty oxygen orbital, the other will break the &lt;span class=&quot;math-container&quot;&gt;$\ce{O=O}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond. The electrons in the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond will go to the positively charged oxygen atom (the one that donated its electrons in the first step). Now one oxygen atom is neutral and the other has a positive charge (because the electrons of the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond went for the other oxygen), where the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-}$&lt;/span&gt; will attack forming hydrogen peroxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{HO-OH}$&lt;/span&gt;). The previous process will now repeat, but instead, the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; bond between the oxygen atoms will be broken, resulting in two water molecules, &lt;span class=&quot;math-container&quot;&gt;$\ce{2 H2O}$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="402" LastEditDate="2019-05-29T22:53:48.043" LastActivityDate="2019-05-31T10:25:59.060" Title="Reaction mechanism of combustion of hydrogen" Tags="&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;hydrogen&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14705" PostTypeId="2" ParentId="14700" CreationDate="2014-07-28T20:37:59.827" Score="2" Body="&lt;p&gt;Under normal conditions, atoms will bond to one another and produce a molecule, only if the bond formation process is exothermic (e.g. it releases energy) - this translates into the products (the molecule) being thermodynamically more stable that the starting material (the atoms).  The same is true for the adsorption process.  A gas will adsorb on a surface only if this configuration is more stable than a gas and a free surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically when a gas is adsorbed on a surface, something akin to bond formation between the surface and the gas takes place and we wind up with a chemisorbed species.  There are many ways to study this process.  One simple way is to measure the amount of gas admitted to the system containing the surface to be studied and measuring the change in pressure over time.  When done in a calorimeter, the heat given of during the process can also be measured.  Combining these two measurements(change in moles of free gas and energy released) allows one to understand the basic energetics of the process. The &quot;isoteric method&quot; is one experimental procedure for accomplishing this in a laboratory.  It relies on measuring moles of gas &lt;em&gt;vs&lt;/em&gt;. the pressure in the system (an isotherm) at various temperatures.  From these measurements the enthalpy of adsorption can be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study of the adsorption of a gas on a surface is complicated.  There is not a surface with a monolayer of gas adsorbed on it with free gas above this.  Rather there is a surface and bulk gas and in between the two is a multilayer structure of adsorbed gas where the gas concentration varies throughout.  &lt;a href=&quot;http://thesis.library.caltech.edu/5574/1/Purewal_Thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;Here is a link&lt;/a&gt; to a recent Cal Tech thesis on the characterization of hydrogen gas adsorption onto various surfaces; take a look at section 1.5 and the beginning of chapter 3 for a nice overview of the adsorption process and the interfaces involved. The rest of chapter 3 is a discussion of the thermodynamics and various experimental techniques used to study the process.  It sounds like you might also be interested in performing calculations on the adsorption process.  If so, section 5.8 delves into computational methods used on one of the surfaces studied.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-28T20:43:36.383" LastActivityDate="2014-07-28T20:43:36.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14706" PostTypeId="1" AcceptedAnswerId="14709" CreationDate="2014-07-28T20:46:50.297" Score="1" ViewCount="570" Body="&lt;p&gt;The &lt;a href=&quot;http://ambermd.org/&quot; rel=&quot;nofollow&quot;&gt;Amber&lt;/a&gt; force field parameters describe hydrogen atoms in various states:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HC 1.008                             H aliph. bond. to C without electrwd.group&#xA;H1 1.008                             H aliph. bond. to C with 1 electrwd. group&#xA;H2 1.008                             H aliph. bond. to C with 2 electrwd.groups&#xA;H3 1.008                             H aliph. bond. to C with 3 eletrwd.groups&#xA;HA 1.008                             H arom. bond. to C without elctrwd. groups&#xA;H4 1.008                             H arom. bond. to C with 1 electrwd. group&#xA;H5 1.008                             H arom. bond. to C with 2 electrwd. groups&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I cannot figure what they mean by 'electrwd group' and it doesn't specify anywhere in the documentation. I'm hoping it will be glaringly obvious to a chemist...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7265" LastActivityDate="2014-07-28T21:17:24.397" Title="&quot;Electrwd&quot; group?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14707" PostTypeId="2" ParentId="14703" CreationDate="2014-07-28T20:53:58.767" Score="6" Body="&lt;p&gt;I'll provide a rundown of the general organic chemistry textbooks I've studied in detail:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Organic Chemistry&lt;/em&gt;, David R. Klein: A very solid introductory text. Emphasis is on expediency and development of intuition over rigorous theoretical correctness. The coverage of molecular orbital theory is extremely scant, and classic valence bond theory (with associated concept of orbital hybridization) is the major explanatory paradigm. This is probably the most accessible text for students with a minimal background, but it's also somewhat superficial.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Organic Chemistry&lt;/em&gt;, Jonathan Clayden, et al.: This is a very thorough text, with a slightly wider array of reactions discussed than Klein's textbook (most notably in the coverage of organometallic reagents). Again, the MO treatment is lacking, but it's invoked more frequently than in Klein. This book tends to delve quite deeply into the mechanisms underlying specific reactions, making it rather lengthy. There's also some superficial discussion of HSAB theory, which is mostly (if not completely) absent in Klein's book.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Advanced Organic Chemistry&lt;/em&gt;, Carey &amp;amp; Sundberg. This is something of a seminal classic, but it's not for the faint of heart or for those uninitiated in the fundamentals. Mechanisms are elucidated in great detail (primarily via MO theory, with results from both &lt;em&gt;ab initio&lt;/em&gt; and semi-empirical methods), and the exposition of pericyclic reactions is notably excellent. It also contains a great deal of illuminating content on conformational analysis. This is definitely not appropriate for a first textbook, but I would consider it essential reading for advanced undergrads and above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Ultimately, either of Klein's or Clayden's textbooks should satisfy all of your criteria, but I strongly recommend Carey &amp;amp; Sundberg for subsequent study, if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond those, I've found Fleming's &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions&lt;/em&gt; to be a very good introduction to qualitative MO theory, but it's not a general-purpose text. March's &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt; is the ultimate, all-encompassing reference manual, and I'd consider it the &lt;em&gt;sine qua non&lt;/em&gt; of any good organic chemistry library. As an aside, McQuarrie and Atkins have both published very good physical chemistry textbooks, which in many ways have been more enlightening than any textbook specific to organic chemistry. I'm also a fan of Oxtoby's &lt;em&gt;Principles of Modern Chemistry&lt;/em&gt; as a general chemistry textbook; its mathematical and physical rigor is incredibly admirable, even if it does sometimes lack severely in concision. I've also heard unanimous praise for Housecroft &amp;amp; Sharpe's &lt;em&gt;Inorganic Chemistry&lt;/em&gt;, but I've yet to read more than a few sections of it.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-28T20:59:41.430" LastActivityDate="2014-07-28T20:59:41.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14708" PostTypeId="2" ParentId="14692" CreationDate="2014-07-28T21:00:34.930" Score="3" Body="&lt;p&gt;In general, think of batteries (including chemical nonrechargeable ones) as tools for storing and transporting energy, not as an energy source.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to create energy through &lt;em&gt;chemistry&lt;/em&gt;, then the economically efficient way is the way everyone else does it, through exothermic oxidation - burn stuff and use it to power a generator.&lt;/p&gt;&#xA;" OwnerUserId="4928" LastActivityDate="2014-07-28T21:00:34.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14709" PostTypeId="2" ParentId="14706" CreationDate="2014-07-28T21:17:24.397" Score="2" Body="&lt;p&gt;electron-withdrawing (e.g. F in CF3COOH)&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-28T21:17:24.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14710" PostTypeId="2" ParentId="14555" CreationDate="2014-07-28T22:53:40.063" Score="6" Body="&lt;p&gt;The cell potential is related to the change in Gibbs free energy by this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta G^\circ= -nFE^{\circ}_\text{cell}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_and_Thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;UC Davis chem wiki&lt;/a&gt; has a brief explanation of the connection between cell potential, the reaction quotient (via the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nernst Equation&lt;/a&gt;), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0321809246&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tro's Chemistry&lt;/a&gt; has a derivation of all of these relations, but to summarize the first relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w = -qE^{\circ}_\text{cell}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;q = nF&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $n$ is the number of moles of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can substitute this back into the work equation to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;w = -nFE^{\circ}_\text{cell}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta G^\circ = w = -nFE^{\circ}_\text{cell}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that &quot;things move downhill&quot; – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorDisplayName="user7951" LastEditDate="2016-11-05T17:55:11.603" LastActivityDate="2016-11-05T17:55:11.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14711" PostTypeId="1" CreationDate="2014-07-29T01:03:42.250" Score="10" ViewCount="7232" Body="&lt;p&gt;If I choose GIAO HF/STO-3G, is the default solvent $\ce{CD3Cl}$? Is the default frequency for $\ce{^1H-NMR}$ 500 MHz? What about the frequency for $\ce{^13C-NMR}$?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastEditorUserId="4231" LastEditDate="2014-07-29T01:08:52.720" LastActivityDate="2020-12-22T02:56:52.043" Title="When using Gaussian to calculate NMR, what's the default solvent and frequency?" Tags="&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14712" PostTypeId="2" ParentId="14711" CreationDate="2014-07-29T01:16:29.620" Score="13" Body="&lt;p&gt;Performing calculations with the &quot;NMR&quot; keyword in Gaussian gives the magnetic shielding tensor in ppm and the spin-spin coupling in Hz. These numbers are independent of instrumental parameters like the frequency of the H-NMR. It doesn't contain any correction for any kind of solvent effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One remark: HF/STO-3G is a terrible, terrible low level of theory for ANY kind of calculations.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-29T01:16:29.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14713" PostTypeId="2" ParentId="14630" CreationDate="2014-07-29T01:34:35.090" Score="3" Body="&lt;p&gt;Let me be the elephant here and answering your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually perform calculations technically speaking you don't really need all this stuff. You need them to able to read papers, design scientifically sound research, understand what the input parameters mean and discuss your results, but not for the manual part of the labor. There are several + 1 commercial and free software are available which does these things for you as those are mentioned in the very useful comments. If you find an unscratched itch to add some energy term that is not in the software and you don't know how, it is kind of a sure sign you don't know what you are doing. It may sound harsh, but the truth is that you need to understand much more about the calculations before you can just add terms here and there, and at that level you will be able to use the developer version of those programs anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a funny side effect of education: generally QChem is taught by people who develop code, who talk all day about HF theory and numerical details, but mostly used by students who does not need to develop code, but use already existing programs. You come out from the lecture, learnt all those equations, you think you need to use somewhere, but the cold truth is that you have to open a window an click-click here and there or open a textfile and put some cryptic switches. At manual level, QChem calculations are much more about know-how and experience than writing equations. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-29T01:34:35.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14714" PostTypeId="1" AcceptedAnswerId="14716" CreationDate="2014-07-29T03:19:02.820" Score="0" ViewCount="283" Body="&lt;p&gt;If your home lost power without warning; are homemade batteries (such as the lemon/potato variety described in this question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14692/are-homemade-batteries-economically-practical&quot;&gt;Are homemade batteries economically practical?&lt;/a&gt;) a practical alternative for powering your appliances while you wait for the electrician or what have you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise? &lt;/p&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-07-29T05:03:54.743" Title="Are homemade batteries a practical alternative during a blackout?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;fuel&gt;&lt;electricity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14715" PostTypeId="2" ParentId="14714" CreationDate="2014-07-29T03:38:54.293" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... are homemade batteries ... a practical alternative for powering your appliances ...?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, those type of batteries don't have enough energy to power more than a small light bulb or a transistor radio. If fact the light bulb probably won't stay lit for more than 10 minutes. An LED would be much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need at least 30 car batteries to power your home for a significant amount of time; this is why most people going off the grid have a significant infrastructure cost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other than the alkaline batteries you buy in stores, not that I can think of.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-29T03:38:54.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14716" PostTypeId="2" ParentId="14714" CreationDate="2014-07-29T05:03:54.743" Score="3" Body="&lt;p&gt;Short answer:&#xA;Home made batteries are expensive, easy to fail, easy to spoil, potentially harmful to the environment and you need really a lot of them to get out meaningful amount of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&#xA;Your question is pretty vague, it is not clear what you mean by &quot;usable electricity&quot;.&#xA;If you need light, it is easier (cheaper, reliable) to use candles or oil lamps.&#xA;If you need electricity to drive electric devices, than you most probably need a predetermined voltage, which will be difficult to reliably produce.&#xA;If you need a substantial amount of electricity to drive light and several electric devices, it is very hard to do it with any kind of battery-like equipment: just think it over how many batteries you need the produce a kW electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Batteries per definition contains corrosive in one way or another, and DIY level technology risks that corrosion occurs much less predictable way. You have to figure out how long a battery can be stored reliably before you get everything soaked in rotten citrus juices.&lt;/li&gt;&#xA;&lt;li&gt;Since you most probably use metals in this battery, it is a good question how poisonous your new and used battery will be and if you can trash it without major environmental damage. It is not nice to spill out large amount of dissolved copper or lead in the garden.&lt;/li&gt;&#xA;&lt;li&gt;If you do any kind of electricity work, and you are not good at it, especially with potentially corroded equipment, you may have a lot of problem with contact errors and other reliability issues. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-29T05:03:54.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14717" PostTypeId="2" ParentId="14711" CreationDate="2014-07-29T06:09:32.177" Score="17" Body="&lt;p&gt;While Greg's answer describes sufficiently what the program does, I would like to extend a little on his remark. (This should have been a comment, but it was too long.)&lt;/p&gt;&#xA;&lt;p&gt;The &lt;code&gt;HF/STO-3G&lt;/code&gt; level of theory is what you call the minimised &lt;em&gt;ab initio&lt;/em&gt; approach. It is unreliable in many ways, but it is fast. However, even geometries of well known and/or calculated molecules may be wrong. In short, it fails more often than it is correct. The problems start with HF, since it does not describe correlation sufficiently. (It only describes Pauli correlation, but this is exact.) The next problem is the basis set: It substantially lacks basis functions. It is called the minimal basis for a reason. Using this setup for NMR calculations is worse than guessing.&lt;/p&gt;&#xA;&lt;p&gt;The first thing to do would be to increase the &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/m_basis_sets.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;basis set&lt;/a&gt;. Maybe move away from Pople basis all together. A few basis set families come to mind: Dunning's augmented family, &lt;code&gt;aug-cc-pVDZ&lt;/code&gt; and larger, Ahlrichs and Weigend's def2 family, &lt;code&gt;def2-SVP&lt;/code&gt; and larger, and many others, see &lt;a href=&quot;https://www.basissetexchange.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;EMSL&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Calculating NMR tensors is always an accuracy versus time tradeoff. And it might be complicated to find the best (in the sense of most efficient) setup. The best way is to start a geometry optimisation on quite a low level. For reasonably large molecules a semi-empirical model like &lt;code&gt;pm6&lt;/code&gt; can be used to scan for configurational space.&lt;/p&gt;&#xA;&lt;p&gt;Increase the level of theory to pure DFT. You can use density fitting here to speed up your calculation: e.g. &lt;code&gt;#P BP86/Def2SVP/W06 DenFit&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Validate your geometries with a higher basis set, e.g. &lt;code&gt;#P BP86/Def2TZVPP/W06 DenFit&lt;/code&gt; and &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_dft.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;different functionals&lt;/a&gt;, e.g. &lt;code&gt;TPSS&lt;/code&gt;, &lt;code&gt;PBE1PBE&lt;/code&gt;, &lt;code&gt;M06&lt;/code&gt;, etc.&lt;/p&gt;&#xA;&lt;p&gt;Validate your geometries with the dispersion correction of Grimme, e.g. &lt;code&gt;BP86 EmpiricalDispersion=GD3BJ&lt;/code&gt;, &lt;code&gt;PBE1PBE EmpiricalDispersion=GD3BJ&lt;/code&gt;, &lt;code&gt;B2PLYPD3&lt;/code&gt;, etc.&lt;/p&gt;&#xA;&lt;p&gt;Apply &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_scrf.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;solvent corrections&lt;/a&gt; for your system, e.g. &lt;code&gt;SCRF=(Solvent=Chloroform)&lt;/code&gt;. Make sure to run single point calculations first and try optimisations later.&lt;/p&gt;&#xA;&lt;p&gt;If you have fully converged geometries at DFT level, you might want to check basic properties with &lt;em&gt;ab initio&lt;/em&gt; techniques. Most appropriate would be &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_mp.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;perturbation theory&lt;/a&gt;, e.g. &lt;code&gt;MP2&lt;/code&gt;.&lt;/p&gt;&#xA;&lt;p&gt;After you have done all the above you can go ahead and calculate &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_nmr.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NMR&lt;/a&gt; tensors. It may be combined with solvent approximations. It requires a very well converged structure, a local minimum on the potential energy hyper surface. (Check that with a frequency calculation, e.g. &lt;code&gt;FREQ&lt;/code&gt;.) You can set it up like this sample input:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;%oldchk=filenameOfPreviousOptimization.chk&#xA;%chk=newCheckpoitfile.chk&#xA;#P MP2/def2TZVPP&#xA;SCRF(solvent=chloroform)&#xA;NMR(GIAO)&#xA;geom=AllCheck guess=read&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Be advised that that computing NMR data with only one method can lead to many misinterpretations. It is very important to verify your structures and tensor on a variety of methods. If you cannot afford MP2 with a large basis, then you should at least check 4 or 5 DFT methods. It is also important to know, that including solvent effects is still something where the theory is not exact. If you have a system that changes a lot with the solvent, then you should consider explicit treatment. Then you might want to dive into &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_mm.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular modelling&lt;/a&gt;, molecular dynamics, MM/QM, &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_oniom.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ONIOM&lt;/a&gt;, etc. pp.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="41556" LastEditDate="2020-12-22T02:56:52.043" LastActivityDate="2020-12-22T02:56:52.043" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="14718" PostTypeId="2" ParentId="14704" CreationDate="2014-07-29T10:14:34.167" Score="18" Body="&lt;p&gt;This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2 + O2 -&amp;gt; 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2 &amp;lt;=&amp;gt;[\Delta T] 2 H.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy).&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H. + O2 -&amp;gt; HO. + O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting oxygen radicals can again react with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; to form more hydrogen radicals. This reaction is also endothermic.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{O + H2 -&amp;gt; HO. + H.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydroxyl radicals can also react with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net result of those equations leads to a slightly exothermic sum, with high potential:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{3H2 + O2 -&amp;gt; 2H2O + 2H.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scheme will eventually end when the concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2,H2}$&lt;/span&gt; will become lower forcing the excess radicals to react with each other. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HO. + H. -&amp;gt; H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H. + O2 -&amp;gt; HO2.}\\&#xA;\ce{2HO2. -&amp;gt; H2O2 +O2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also happening but not as important: &lt;span class=&quot;math-container&quot;&gt;$\ce{HO. + HO. &amp;lt;=&amp;gt; H2O2}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;. peroxide is also easily cleaved again, or reacts with each other:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2O2 -&amp;gt; 2H2O + O2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all results in the main product water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/8498/4945&quot;&gt;How explosive is hydrogen gas?&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.439114&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980)&lt;/a&gt; most comprehensive (theoretical) study on the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2-O2}$&lt;/span&gt; system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.degruyter.com/view/product/19732&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiberg, Nils: Lehrbuch der Anorganischen Chemie&lt;/a&gt; Kap. 3.2 (german) &lt;em&gt;The mechanism is in part taken from this source.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf&lt;/a&gt; &lt;em&gt;Thanks to Silvio Levy&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2019-05-29T22:52:43.977" LastActivityDate="2019-05-29T22:52:43.977" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="14720" PostTypeId="1" AcceptedAnswerId="14723" CreationDate="2014-07-29T14:12:32.577" Score="2" ViewCount="509" Body="&lt;p&gt;Below is the structure of Ace-Ala-Nme. I have labelled just four of the carbon atoms (1-4 in blue). I need to identify (i) the hybridization state, and (ii) the number of electron withdrawing groups for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/INRpT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the amine (-NH) groups are electron donating while the carbonyls (-CO) are electron withdrawing. I therefore suspect that all four labelled carbon atoms are sp3 with 0 withdrawing groups. But this is assuming that the donating/withdrawing tendency of the amine and carbonyls cancel each other out. Can anyone confirm this to be the case?&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditDate="2014-07-29T15:32:09.470" LastActivityDate="2014-07-29T17:18:28.357" Title="Electronic configuration of Ace-Ala-Nme" Tags="&lt;organic-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14721" PostTypeId="1" AcceptedAnswerId="14722" CreationDate="2014-07-29T14:29:52.513" Score="2" ViewCount="449" Body="&lt;p&gt;If the leaving of the leaving group creates a secondary carbocation, and a tertiary carbocation has the possibility of being formed through a 1,2 methyl shift:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Will the methyl shift? I'm guessing the answer is an emphatic yes. Tertiary carbocations are more stable than secondary carbocations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Will there still be some secondary carbocation left over? I.e. will there be both major and minor products? I'm guessing yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because I was watching a video on E1 elimination and the narrator ignored the possibility of a shift, and I was wondering if this was because he forgot or because it was irrelevant. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-10-16T16:53:36.910" LastActivityDate="2015-10-16T16:53:36.910" Title="Rearrangements; Carbocations; E1 Mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14722" PostTypeId="2" ParentId="14721" CreationDate="2014-07-29T15:18:50.490" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Will the methyl shift? I'm guessing the answer is an emphatic yes.&#xA;  Tertiary carbocations are more stable than secondary carbocations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, if that is the only way to generate a tertiary carbocation and the C-Methyl bond can more or less line up with the carbocation p orbital, then the methyl would likely shift.  If there are other alkyl groups attached to the carbon initially bearing the methyl group, they could shift too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Will there still be some secondary carbocation left over? I.e. will&#xA;  there be both major and minor products? I'm guessing yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An equilibrium exists between the various carbocations.  At 300 K, an energy difference of 1.4 kcal/mol results in a power of ten difference in the equilibrium (e.g. from 50:50 to 90:10).  So it really depends on the energy difference between the various carbocations involved, as well as the rate they are captured or proceed to the next step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if there is a hydrogen on the carbon adjacent to the carbocation center, then elimination to create a double bond becomes a competing pathway,&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-29T15:18:50.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14723" PostTypeId="2" ParentId="14720" CreationDate="2014-07-29T15:32:28.477" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;1: sp3 with 1 electron withdrawing group&lt;/p&gt;&#xA;&lt;p&gt;2: sp3 with 2 electron withdrawing groups&lt;/p&gt;&#xA;&lt;p&gt;3: sp3 with 0 electron withdrawing groups&lt;/p&gt;&#xA;&lt;p&gt;4: sp3 with 1 electron withdrawing group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I would say your earlier analysis was correct.  If you look at the resonance structures that can be drawn for an amide linkage, you find there is considerable positive charge both on carbon and nitrogen.  This should serve to make them both electron withdrawing, relative to an attached carbon, in your structure.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ne4cc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; To clarify this answer with regards to some of the comments posted above, I'd like to add that typically when an amino nitrogen is situated on a double bond or aromatic ring it can donate electrons through resonance.  In fact, that is what it is doing in the resonance structure on the right - it is donating its lone pair of electrons via resonance.  However, the amino group is also electron withdrawing in an inductive sense relative to carbon.  Further the &lt;span class=&quot;math-container&quot;&gt;$\ce{-NH3^+}$&lt;/span&gt; group can no longer interact via resonance and is simply a strong electron withdrawing group.  The resonance structure on the right is analogous to the &lt;span class=&quot;math-container&quot;&gt;$\ce{-NH3^+}$&lt;/span&gt; situation and since it is a significant contributor to the overall description of the amide (&lt;em&gt;cf&lt;/em&gt;. there is significant double bond character in the amide &lt;span class=&quot;math-container&quot;&gt;$\ce{C-N}$&lt;/span&gt; bond) it acts inductively as a strong electron withdrawing group for the attached &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; carbon.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-07-29T17:18:28.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14724" PostTypeId="2" ParentId="13899" CreationDate="2014-07-29T16:52:07.293" Score="3" Body="&lt;p&gt;According to your description, this looks like &lt;a href=&quot;http://www.engineersedge.com/corrosion/localized_corrosion.htm&quot; rel=&quot;nofollow&quot;&gt;localized corrosion&lt;/a&gt;. Due to differences in the availability of oxygen in the solution, the two half-reactions of the redox process take place on different areas of the surface of the corroding metal. The oxygen reduction reaction preferably takes place on the nail surface in the medium with higher concentration of dissolved $\ce{O2}$ (water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 4e- + 2H2O -&amp;gt; 4 OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal dissolution (oxidation) is then limited to the more oxygen-starved surface area, which is presumably right at the the phase border, where consequently the deposit of corrosion products is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{Fe &amp;amp;-&amp;gt;~ Fe^2+ + 2e-}\\&#xA;\ce{Fe^2+ + 2OH- &amp;amp;-&amp;gt;~ Fe(OH)2 \rightleftharpoons FeO + H2O}\\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these two half-reactions are coupled, the amount of corrosion products formed depends on the oxygen concentration. When there is sufficiently dissolved oxygen available, $\ce{Fe^2+}$ is easily oxidized by to $\ce{Fe^3+}$, and thus the final product, which is commonly called &lt;a href=&quot;http://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow&quot;&gt;rust&lt;/a&gt;, is a mixture of more or less hydrated iron(III) oxides and hydroxides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^3+ + 3OH- -&amp;gt; Fe(OH)3 \rightleftharpoons FeO(OH) + H2O}$$&#xA;$$\ce{2FeO(OH) \rightleftharpoons Fe2O3 + H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-07-29T16:59:13.117" LastActivityDate="2014-07-29T16:59:13.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14725" PostTypeId="2" ParentId="6174" CreationDate="2014-07-29T18:29:30.487" Score="2" Body="&lt;p&gt;Just as an experienced synthetic chemist can make an educated guess about how compound A and compound B will react, an experienced photochemist can usually do the same.  There are many, many well-known photochemical reactions that are mechanistically understood and sometimes even used in synthetic procedures, some commercially.  A few such reactions are the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Norrish Type I reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQRyl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Norrish Type II reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IoUpV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paterno-Buchi reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g8xP3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;di-pi-methane reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhkyP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;photochemical induced halogenation, olefin dimerization and so many more.  So yes, an experienced photochemist can make reasonable predictions about which chemicals will react and what the possible products might be.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-29T18:29:30.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14726" PostTypeId="1" CreationDate="2014-07-29T18:34:21.467" Score="2" ViewCount="3857" Body="&lt;p&gt;Suppose you have a container with a drain at the bottom full of frozen H2O which you leave a room temperature to melt. Will the ice melt faster if you drain the water as it melts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect that the water, being warmer than the ice, will speed the melting of the ice. However, I would also expect that draining the water would remove allow more air to enter the system and heat the ice faster. Which explanation is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question: if the container is a cooler to fill with warm beverages, will the beverages cool faster in a wet or dry ice-filled cooler?&lt;/p&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="7209" LastEditDate="2014-07-29T18:44:09.553" LastActivityDate="2019-02-11T21:49:22.910" Title="Does ice melt faster when submerged in water?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14727" PostTypeId="2" ParentId="14726" CreationDate="2014-07-29T18:45:40.540" Score="6" Body="&lt;p&gt;Assuming the water and air are the same temperature (and that the air is mostly stationary) then the ice will melt quicker in the water. This is due to water having a higher &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_conductivity&quot; rel=&quot;nofollow&quot;&gt;thermal conductivity&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;heat capacity&lt;/a&gt;. In fact, this is easy to test: place a cocktail stick through an ice cube and lay it on top of a glass filled right to the top with (room temperature) water. The submerged half will melt quicker than that on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second question: A warm object will cool quicker in icy water than in just ice for the same reason above but also because of increased surface area contact. However, icy water will be warmer than just ice. So the item will potentially be able to achieve a lower temperature in just ice (slow and steady wins the race).&lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditDate="2014-07-29T19:23:11.627" LastActivityDate="2014-07-29T19:23:11.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14728" PostTypeId="1" CreationDate="2014-07-29T20:27:24.847" Score="6" ViewCount="3664" Body="&lt;p&gt;There is a drink that claims to be sweetened with stevia, but I think I taste Sucralose in there even though it is not listed on the ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easy way I could test for the presence of Sucralose to confirm my hunch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about Aspartame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By easy, I mean not needing access to something like a gas chromatograph. &lt;/p&gt;&#xA;" OwnerUserId="60418" OwnerDisplayName="bigjosh" LastEditorUserId="9961" LastEditDate="2015-10-23T18:08:38.890" LastActivityDate="2018-02-08T19:04:47.520" Title="How to detect if a beverage has Sucralose or aspartame?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14729" PostTypeId="2" ParentId="14728" CreationDate="2014-07-29T21:06:26.753" Score="3" Body="&lt;p&gt;I think it would be not an easy task. Aspartam does not contain any special groups, just a carboxyl, amide, phenyl... Nothing that you would not find in the beverage ingredients. In case of sucralose maybe you can search for some tests for chloroalkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the reaction between silver nitrate and certain organic halides (those compounds containing chlorine, bromine, or iodine) results in the formation of a silver halide precipitate as a positive test for organic halides.&quot; (&lt;a href=&quot;http://www.britannica.com/EBchecked/topic/108529/chemical-analysis/80788/Classical-qualitative-analysis&quot; rel=&quot;nofollow&quot;&gt;http://www.britannica.com/EBchecked/topic/108529/chemical-analysis/80788/Classical-qualitative-analysis&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in your case I would expect the product to contain free chloride ions and therefore giving false positive result.&lt;/p&gt;&#xA;" OwnerUserId="7272" LastActivityDate="2014-07-29T21:06:26.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14730" PostTypeId="1" AcceptedAnswerId="14756" CreationDate="2014-07-29T21:13:38.010" Score="1" ViewCount="1546" Body="&lt;p&gt;My professor showed us a sample calculation of the Ka value for a strong acid. He used standard state thermodynamic data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an issue with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Wouldn't this give us an understated Ka value? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Standard state conditions start with 1 molar solutions of solute. So in writing the reaction equation of an acid with water and by using that equation to find the delta G std. of reaction and then using the relationship between delta G and K to find K, wouldn't we be running into a sort of &quot;common ion&quot; effect issue? There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions. Finding the K of an acid dissolving in water that has a pH of 0.0 doesn't give us a Ka value (or does it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT to address Martin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what he did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HNO_3 + H2O -&amp;gt;H_3O^+ +NO_3^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find ${\Delta G^o_r}$ for above reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use ${\Delta G^o_r=-RTlnK}$ to solve for K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Is this K that you find a $\ce{K_a}$ value? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-07-30T12:47:29.230" LastActivityDate="2014-07-30T19:24:04.103" Title="Calculating Ka values from std. state thermodynamic data" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14731" PostTypeId="2" ParentId="6564" CreationDate="2014-07-29T21:14:36.667" Score="3" Body="&lt;p&gt;This is an old question, but in case anyone comes down this road...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aren't alpha halo carbonyl compounds extremely fast and reactive via $\ce{S_{N}2}$&#xA;  (the fastest i guess)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think so. In an $\ce{S_{N}2}$ reaction a nucleophile is attacking a positively polarized carbon atom.  The more positive charge on the carbon, the more the nucleophile is attracted.  Look at structure I below, it shows one of the resonance structures for a carbonyl that leads to a polarized carbonyl with positive charge on the carbonyl carbon.  Now to whatever degree the adjacent C-Cl bond is polarized with positive charge on carbon, we wind up with an unfavorable electrostatic situation of adjacent positive charges.  Said differently, the presence of a carbonyl group suppresses polarization of the adjacent C-Cl bond, which will make the C-Cl carbon less prone to nucleophilic attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Da9ij.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if the alpha carbon(to which the halide is attached) is tertiary,&#xA;  there should be a high chance of an elimination product via E 2  (leading to the &#xA;  formation of alpha beta unsaturated carbonyl), right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you want to remove hydrogen D in the diagram below.  Hydrogen D is much less acidic than hyrdrogen A because removal of the latter results in a resonance stabilized carbanion.  The difference in acidity between hydrogens A and D is so large that effectively only hydrogen A will be removed by a nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two arguments effectively eliminate pathways C and D from consideration.  Pathway B is simply an equilibrium that goes nowhere, so it has no real effect on the outcome of the reaction. This leaves pathway A, the Favorskii reaction, as the only viable alternative.  Pathway A, besides being the only game in town, is actually favored by two factors, 1) as noted above, hydrogen A is acidic and it's removal produces a resonance stabilized carbanion, and 2) completion of the Favorskii reaction is strongly favored by entropic factors since the carbanion and the C-Cl carbon are contained in the same molecule and held relatively close together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L8CJi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-29T21:14:36.667" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14732" PostTypeId="2" ParentId="14728" CreationDate="2014-07-29T22:05:54.493" Score="10" Body="&lt;p&gt;Here's the structure of sucralose&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9gCsj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is the structure of aspartame&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWKhx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are very different one from the other and should be easily separable.  You can do a &quot;poor man's&quot; separation using paper chromatography.  Aldrich sells both of these materials, but I'm sure you could find them much cheaper at a local Health or Health Food store.  Get the samples and some filter paper.  You'll need some solvents, a tank (preferably glass) with a lid, some filter paper coated with a fluorescent material (as an easier, but more expensive alternative, you could always buy some TLC plates from a chemical supply house) and a black light.  Put one of the solvents in the tank so that it is 1/8 - 1/4&quot; deep.  Put the top on the tank and let it equilibrate. Make dilute solutions of your known samples using the same solvent that's in the tank.  Using the tip of a needle or a toothpick place a small spot of each of the dissolved materials on the filter paper, put the spots about 1&quot; apart and high enough above the bottom edge of the filter paper so that they will be above the solvent level when you place the filter paper in the tank.  Give the spots a moment to dry and then put the spotted paper in the tank.  Watch as the solvent level climbs up the filter paper (or tlc plate) and remove the filter paper when the solvent is about 3/4 of the way up the paper or plate.  Let the paper or plate dry and then examine it with the black light.  The paper should fluoresce under black light except where the two spots have migrated up to.  The spots cover the fluorescent material and will appear as dark spots under the black light.  If you got a good separation between the two spots stop here and repeat the experiment putting a spot of your diluted drink on a new piece of filter paper along with spots of the two known materials.  Rerun it in the tank and see if there are any spots in your drink that match either of the spots from the two known compounds.  If there is no match, then sucralose or aspartame are not in your drink.  If there is a match, then whichever spots matched might be in your drink - more advance experiments will be needed to further analyze the situation.  If you didn't get a good separation between the spots for the two known compounds to begin with, start over and repeat with different solvents until you do get a separation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-29T22:05:54.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14733" PostTypeId="1" AcceptedAnswerId="14734" CreationDate="2014-07-29T22:43:19.607" Score="3" ViewCount="6207" Body="&lt;p&gt;If a very minute quantity spilled acid (e.g. $\ce{H2SO4}$, though the question would apply to other acids as well) is cleaned up with water (e.g. yielding a $0.0001~\mathrm{M}$ solution), and some of the water (dilute acid solution) is left behind, will the resulting solution become more concentrated as the water evaporates?  Would there be any danger that a drop of an initially-harmless concentration might become a visually-unnoticeable quantity of acid that was strong enough to pose a contact hazard?  Would wiping an area down with a moderately-dilute solution of a weak base such as sodium bicarbonate be an appropriate safety precaution, or would it be unnecessary if any acid droplets that might exist on a surface have been thoroughly diluted?&lt;/p&gt;&#xA;" OwnerUserId="7274" LastEditorUserId="4945" LastEditDate="2014-07-30T02:35:22.817" LastActivityDate="2015-03-22T07:27:05.687" Title="Can very dilute acids exposed to air become more concentrated as water evaporates?" Tags="&lt;acid-base&gt;&lt;safety&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14734" PostTypeId="2" ParentId="14733" CreationDate="2014-07-29T22:59:06.107" Score="3" Body="&lt;p&gt;Sulfuric acid (and other acids) can be concentrated through evaporation of water.  It would be a good idea to wash the surface down with dilute, basic solution.  That said, if I'm reading your post correctly, your droplets are already 0.1M.  That is still a strong acid and care should be taken in its clean up.  I was once working with a 0.01M solution of sulfuric acid and when cleaning the flask a few drops of the dilute solution splattered onto my foot, I didn't even notice it at the time.  An hour or so later my foot started to burn.  When I looked, my sock was gone in the affected area.  If you're dealing with anything more than drop or two use a spill kit and let someone know what you are doing.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-29T22:59:06.107" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14735" PostTypeId="2" ParentId="14728" CreationDate="2014-07-30T00:58:38.327" Score="0" Body="&lt;p&gt;Stevia is an extract from a plant and contains the sweeteners stevioside and rebaudioside. Both of these compounds are very sweet (about 300 times and 400 times sweeter than sucrose), but stevioside has a bitter after taste. It is very likely that what you think is sucralose is rabaudioside, since it leaves a very sweet after taste.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-30T00:58:38.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14736" PostTypeId="2" ParentId="14728" CreationDate="2014-07-30T01:32:06.143" Score="1" Body="&lt;p&gt;A trick my mother uses is to weigh it, but this only works in some very select cases and tends not to be conclusive in all but the most extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because some modern soft drinks are up to 1g sugar per 10mL volume, they tend to be a fair bit denser than water. Since aspartame is much, much sweeter than sugar, you usually only need a tiny bit (according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Aspartame#Intake&quot; rel=&quot;nofollow&quot;&gt;Wikipedia, only 180mg per 12 fl oz&lt;/a&gt;) and it doesn't increase the density as much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this only has a hope of working if you're comparing very sweet beverages (e.g. Mountain Dew and its &lt;a href=&quot;http://media.lanecc.edu/users/snydert/225Lectures/04B/thumbnails/10mtndewfacts.jpg&quot; rel=&quot;nofollow&quot;&gt;0.13g/mL of sugar&lt;/a&gt;), and isn't terribly conclusive on iced teas or lightly sweetened beverages. Still, it's an interesting technique that doesn't rely on trying to chemically separate the two substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you actually do try this, make sure that both beverages are completely flat (decarbonated) and at the same temperature.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2014-07-30T01:32:06.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14737" PostTypeId="1" CreationDate="2014-07-30T02:04:30.470" Score="2" ViewCount="509" Body="&lt;p&gt;In order to turn sea water into drinkable water, virtually all of the salt must be removed.  By definition, salts are ionic compounds.  I am assuming that there are other non-ionic compounds devolved in the sea water, must these compounds be removes as well before water becomes safe to drink?  In general, do non-ionic compounds present the same dehydration dangers as salts when consumed in drinking water?  Why or why not?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this question, lets not consider other toxic effects besides dehydration.&lt;/p&gt;&#xA;" OwnerUserId="6895" LastEditorUserId="9961" LastEditDate="2016-07-25T17:09:29.337" LastActivityDate="2016-07-25T17:09:29.337" Title="Are non-ionic compounds generally as dehydrating as ionic compounds in drinking water?" Tags="&lt;ionic-compounds&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14738" PostTypeId="1" AcceptedAnswerId="14747" CreationDate="2014-07-30T02:13:35.610" Score="3" ViewCount="2318" Body="&lt;p&gt;In Infrared spectroscopy, stretches are easily understandable. But, how do I visualize (or conceptually understand) the difference between dihedral (torsions), out of plane bends, coplanar bends, and perpendicular bends? Can you give examples?&lt;/p&gt;&#xA;" OwnerUserId="6986" LastEditorUserId="40220" LastEditDate="2017-03-05T20:28:41.377" LastActivityDate="2017-03-05T20:28:41.377" Title="Difference between torsion, out of plane, coplanar and perpendicular bends?" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;spectroscopy&gt;&lt;molecular-structure&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14739" PostTypeId="2" ParentId="14737" CreationDate="2014-07-30T02:22:59.470" Score="3" Body="&lt;p&gt;Actually, the salt in seawater doesn't kill you directly. It dehydrates you and you die of thirst. If you mixed 1 part of seawater with 2 parts of fresh water, it would not dehydrate you, but you would need to eliminate the excess salt so you would need to drink a lot of this salty water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many living things in seawater that can kill you. Red Tide (excuse me, harmful algal bloom) is an algae common to warm water that produces a toxin. Very little of this toxin is needed to kill a person.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-07-30T02:22:59.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14740" PostTypeId="2" ParentId="14630" CreationDate="2014-07-30T03:16:16.843" Score="6" Body="&lt;p&gt;The XC potential for DFT actually consists of two terms: $V_x$ and $V_c$. Depending on which XC functional you choose, the exchange part is either computed exactly (using Hartree-Fock) or using fitted parameters or in a combination of both. In the case of B3LYP the exchange is 20% Hartree + 8% Slater + 72% Becke88. The correlation functional consists of three (3) different terms (LYP) in a different form. If you take a different functional, say M06-2X, it has different functional type (54% Hartree exchange, 200+ parameters!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to the next part: You use a basis set only to compute the kinetic energy and exchange terms in the system. Let me walk you through a HF routine first: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian consists of four terms(nuclear kinetic energy is zero because of Born-Oppenheimer Approximation) : $T_{elec} + V_{nn} + V_{en} + V_{ee}$. The nuclear-nuclear potential energy is constant (because they are 'clamped'). The first and third terms are fairly trivial, and are computed with relative ease. The electron-electron potential term consists of Pauli's exchange term, and the coulomb repulsion term. They are the computationally expensive part of the HF calculation, and the eigenvalues (energies) are solved self-consistently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when you obtain the HF energy of a system, it is obtained variationally with respect to all the energy components in an interacting system of electrons and nuclei. Coming to DFT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term you mentioned $V_{ext}$ is the external potential, which is usually the nuclear potential for an unperturbed system. The second term is equivalent to Coulomb repulsion. The third term is evaluated using the XC potential of your choice(every DFT method differs only at this part). You evaluate the energies for each and every orbital using the effective potential $V_s$ which you mentioned. Note: &lt;strong&gt;You evaluate energies for every orbital separately.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because of the KS theorem that equates the density created by non-interacting particles to be the same as that of interacting particles. So, we solve Kohn-Sham equation instead of Schrodinger equation, under a constraint that the sum of square of all the orbitals gives the density:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho(\mathbf r)=\sum_i^N |\phi_{i}(\mathbf r)|^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the energy computed using DFT is minimum with respect to a non-interacting system of electrons. This is the biggest assumption of DFT, which is quite valid too. As a result, DFT provides a good account of correlation energy at a very minimal computational expense (HF accounts for ZERO correlation)!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a computational chemistry course, where DFT was a small part, so I am not able to explain more in detail. I used Prof. Kieron Burke's &lt;a href=&quot;http://www.chem.uci.edu/~kieron/dftold2/materials/bookABCDFT/gamma/g1.pdf&quot; rel=&quot;noreferrer&quot;&gt;ABC of DFT&lt;/a&gt; book, which he has uploaded online in his site for free use. It is a good source to start. There is also a video lecture of Prof. David Sherrill on Youtube: &lt;a href=&quot;https://www.youtube.com/watch?v=5orzn-XA29M&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=5orzn-XA29M&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I missed the last question: For any molecule:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute the kinetic energy (using the KE functional)&lt;/li&gt;&#xA;&lt;li&gt;Compute the external potential ($V_{ext}$)&lt;/li&gt;&#xA;&lt;li&gt;Compute the Coulomb repulsion using an initial guess density ($n(r)$)&lt;/li&gt;&#xA;&lt;li&gt;Compute the XC potential from the XC functional&lt;/li&gt;&#xA;&lt;li&gt;Use Kohn Sham equations to find out orbital energies (under the constraint that the density ($n(r)$) is equal to the density of interacting particles (electrons). In this case, the density must be equal to the total number of electrons.&lt;/li&gt;&#xA;&lt;li&gt;Using the newly obtained density, recompute the effective potential ($V_s$)&lt;/li&gt;&#xA;&lt;li&gt;Perform until self-consistency is achieved. Then total energy is the sum of occupied orbital energies times the occupancy number.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="6986" LastEditorUserId="6986" LastEditDate="2014-07-30T20:26:00.467" LastActivityDate="2014-07-30T20:26:00.467" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14742" PostTypeId="2" ParentId="6564" CreationDate="2014-07-30T07:10:00.653" Score="3" Body="&lt;p&gt;In extension to ron's answer, there are two reasons, why I think a $\ce{S_{N}2}$ reaction is unlikely to occur.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The LUMO, where the nucleophile will attack, of 2-Chlorocyclohexanone is the antibonding orbital of the carbonyl bond. Also the computed natural charges show a much more polarised carbonyl carbon (DF-BP86/dfe2-SVP).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zKfL6m.jpg&quot; alt=&quot;lumo&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wYhSPm.jpg&quot; alt=&quot;charge&quot;&gt;&lt;br&gt;&#xA;On the german wikipedia page I found a proposed mechanism, that would make use of the LUMO instead of the acidity of the $\alpha$ hydrogen for the rearrangement. I would suspect, that it heavily depends on the substrate and the base, which one is favoured.&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Faworski-Umlagerung_3a.svg/800px-Faworski-Umlagerung_3a.svg.png&quot; alt=&quot;reaction mechanism&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;The most stable conformation of this compound is where the chlorine is in equatorial position. The antibonding $\ce{C-Cl}$ bond orbital is sterically shielded by the ring framework. The molecule has to flip the ring to make it accessible. The energy difference is about $1.5~\mathrm{kcal\cdot{}mol^{-1}}$ (DF-BP86/def2-SVP).&lt;br&gt;&#xA;I do not have the time to calculate the activation barrier, but I think it is fair to assume an unfavoured equilibrium for the $\ce{S_{N}2}$ state, at room temperature about $1:9$. This would make this reaction not very likely.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.267" LastActivityDate="2014-07-30T07:10:00.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14747" PostTypeId="2" ParentId="14738" CreationDate="2014-07-30T14:08:26.750" Score="4" Body="&lt;p&gt;A molecule with N atoms has 3N degrees of freedom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 translational modes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if it is linear, it will have 2 rotational modes; if non-linear 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These leaves 3N-6 (3N-5 if linear) vibrational modes.  Vibrational modes are further subdivided into stretching, bending or torsional motions. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stretching motions can be symmetric or asymmetric&lt;/li&gt;&#xA;&lt;li&gt;Bending motions are often referred to as scissoring, twisting, wagging, and rocking&lt;/li&gt;&#xA;&lt;li&gt;Torsional motions refer to rotations about bonds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This diagram illustrates these various stretching and bending vibrations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ODjNu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the geometric relationship between the transition moment for a specific vibrational mode and the symmetry axis for the mode, vibrational modes may be further classified as parallel or perpendicular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://mackenzie.chem.ox.ac.uk/teaching/Molecular%20Vibrational%20Spectroscopy.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a 31 page article (slide show) that has a lot of background and pictures on this subject. Section 4.11 (page 19) might be particularly interesting to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final example that you might be interested in, formaldehyde has 4 atoms and is non-linear so it will have 6 vibrational modes.  These are all pictured on page 2 in this &lt;a href=&quot;http://szerves.chem.elte.hu/oktatas/ea/Perczel/IR.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-30T14:08:26.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14749" PostTypeId="1" AcceptedAnswerId="14755" CreationDate="2014-07-30T15:14:41.783" Score="14" ViewCount="30224" Body="&lt;p&gt;Can anyone explain the reaction mechanism of the two reactions given below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are the conversion of nitrobenzene to aniline in catalytic ($\ce{H2/Pd}$ in ethanol) and acidic ($\ce{Sn/HCl}$ or $\ce{Fe/HCl}$) mediums respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ar-NO2 -&amp;gt;[\ce{H2/Pd}][\ce{EtOH}] Ar-NH2}$$&#xA;&lt;br&gt;&#xA;$$\ce{Ar-NO2 -&amp;gt;[\ce{Sn + HCl} \text{ or } \ce{Fe + HCl}] Ar-NH2}$$&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="23561" LastEditDate="2018-08-16T00:19:34.683" LastActivityDate="2018-09-04T06:29:13.200" Title="Reaction mechanism for reduction of nitrobenzene to aniline in catalytic and acidic media" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14752" PostTypeId="2" ParentId="9055" CreationDate="2014-07-30T16:36:14.333" Score="2" Body="&lt;p&gt;When you hammer a sheet of lead, you are adding &lt;em&gt;energy in form of work&lt;/em&gt; into it, increasing it's internal energy. If this experiment were to be done in complete vacuum, it'd retain most of the energy, however if this experiment were to be on air, it'd lose a lot of energy in the form of sound.&lt;/p&gt;&#xA;" OwnerUserId="7281" LastActivityDate="2014-07-30T16:36:14.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14753" PostTypeId="2" ParentId="14737" CreationDate="2014-07-30T16:38:11.653" Score="2" Body="&lt;p&gt;Yes, non-ionic but solvable compounds, if present in high enough concentration in water, can dehydrate human. For example, sugars, if present in high concentrations, dehydrate living organisms, and this is the reason why honey can be stored for years without significant loss in quality. Sugar sirup is a well known preservative, traditionally used in slavic countries to preserve fruits and berries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, seawater contains very little amount of non-ionic compounds, and most of them belongs to living things of various nature and, to some extent, their wastes. Moreover, traditions schemes of desalination filters off this impurities as well. This is actually a problem, as perfectly clean water is not healthy to consume, so generally it is preferable to add some carefully balanced amount of salts to such water.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-30T16:38:11.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14755" PostTypeId="2" ParentId="14749" CreationDate="2014-07-30T17:05:10.653" Score="17" Body="&lt;p&gt;The first reaction is called a catalytic reduction.  Here is a picture of how it works with an olefin like ethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F6wNi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form.  The same things happen with nitrobenzene.  As the following diagram shows, &lt;a href=&quot;http://books.google.com/books?id=i-Xk7OTU6AIC&amp;amp;pg=PA84&amp;amp;lpg=PA84&amp;amp;dq=catalytic%20reduction%20of%20nitro%20group,%20mechanism&amp;amp;source=bl&amp;amp;ots=dpCFX13QQ5&amp;amp;sig=epAE0y1ukjZRTQJtZ2MSJpANOJ0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=Ph3ZU-3ENoOwyASu_4C4DQ&amp;amp;ved=0CHgQ6AEwDQ#v=onepage&amp;amp;q=catalytic%20reduction%20of%20nitro%20group%2C%20mechanism&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;the reduction passes through&lt;/a&gt; the nitroso and hydroxylamine intermediates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KTH7I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism for the Sn\HCl reduction involves reductive electron transfer of the nitro group while the tin is being oxidized. This diagram shows the electron transfers involved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4KkKY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is the mechanistic scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/du4gM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-07-30T17:05:10.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14756" PostTypeId="2" ParentId="14730" CreationDate="2014-07-30T17:55:45.010" Score="1" Body="&lt;p&gt;The definition of $K_a$ for $\ce{HNO3}$ is $$\frac{[\ce{H3O+}][\ce{NO3-}]}{[\ce{HNO3}]}$$ (concentrations in mol/L), which is precisely the $K$ that you get from the $\Delta G_r^0 = -RT\ln K$ formula, assuming you use the right free energies (which is not a trivial matter since they must take into account the interaction of each species with surrounding water molecules, an effect known as solvation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be wondering why the concentration of water does not appear in the equation for $K_a$ (or more generally the equation for $K$ in a reaction in aqueous solution).  Because that concentration remains essentially unchanged throughout, it is kept out of the defining expression for $K$, for simplicity. From the point of view of $G^0$ this is correct since the standard state to which the $G^0$ of water refers is pure liquid water, while the standard state for the other three species is 1M concentration. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're correct that for specifying the $G^0$ of $\ce{H3O+}$ the standard state is 1M. But the standard state of $\ce{H3O+}$ is not where the reaction takes place. The same $K$ will be found whether you add the $\ce{HNO3}$ to pure water or to a solution of $\ce{HCl}$ of pH zero -- the final concentrations of $\ce{H+}$, $\ce{NO3-}$ and $\ce{HNO3}$ will be different in the two cases, but $K$ will be the same.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You can find a good discussion of the &quot;standard state&quot; for ionic species in solution here: &#xA;&lt;a href=&quot;http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set10/lecture10.doc&quot; rel=&quot;nofollow&quot;&gt;http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set10/lecture10.doc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll quote a couple of paragraphs.  You can see from this discussion that the calculation of these standard free energies is based exactly on the equilibrium constants, so the formula your professor used works almost by definition!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As always, we can talk about the Gibbs free energy, enthalpy and entropy of such reactions. However, if we try to break these reactions down into the Gibbs energy, enthalpy and entropy of formation of each species in the reaction, we end up defining things like the Gibbs energy of a sodium cation in solution. Yet there is no way to generate just a sodium cation in solution (that would violate charge balance). Thus, we can only determine the Gibbs energy of formation of pairs of ions. This does not cause any real physical chemistry problems, but it causes a book keeping problem: how do we write down the Gibbs energy of formation of a Sodium cation in a book so that we can use it in any reaction where sodium cations are formed?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The answer is that we come up with a sort of standard ion, $\ce{H+}$ (really something more like $\ce{ H3O+}$) and we define the free energy of formation of all other ions relative to this one. For example, I can measure the Gibbs energy change when I dissolve HCl in water to form $\ce{H+}$ and $\ce{Cl-}$.&#xA;  $$\ce{HCl}\text{(g)}\to\ce{H+}\text{(aq)}+\ce{Cl-}\text{(aq)}&#xA; \quad\Delta G^0{}_1&#xA;$$&#xA;  I can now just define the Gibbs energy of formation of $\ce{H+}$ as zero and then determine the relative Gibbs energy of formation of $\ce{Cl-}$.&#xA;  $$\begin{align}&amp;amp;&#xA;\Delta G^0{}_1=\Delta G^0_{\mathrm{F}}(\ce{H+})+\Delta G^0_{\mathrm{F}}(\ce{Cl-})-\Delta G^0_{\mathrm{F}}(\ce{HCl})&#xA;\\&amp;amp;&#xA;\Delta G^0_{\mathrm{F}}(\ce{H+})=0\quad\text{(definition)}\\&amp;amp;&#xA;\Delta G^0_{\mathrm{F}}(\ce{Cl-})= \Delta G^0{}_1+\Delta G^0_{\mathrm{F}}(\ce{HCl})&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-07-30T19:24:04.103" LastActivityDate="2014-07-30T19:24:04.103" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="14757" PostTypeId="2" ParentId="8140" CreationDate="2014-07-30T20:11:16.463" Score="6" Body="&lt;p&gt;If you're trying to modify the source code &lt;strong&gt;and&lt;/strong&gt; run DFT jobs, then you basically have only three options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;PSI4 (very powerful, written in both C and Fortran)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NWChem (very powerful and fast)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GAMESS (kind of fast, fully written in FORTRAN)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you would like to play with codes and tweak it as you wish, not worrying about the speed and optimization level, then you must try &lt;strong&gt;PyQuante&lt;/strong&gt;. They recently added DFT and MP2 codes in it, so you it will be easy for you to play with. Needless to say, it runs slowly because the top layer is written in Python, while the inner layers are written in C and FORTRAN. I guess you will love playing with it!&lt;/p&gt;&#xA;" OwnerUserId="6986" LastActivityDate="2014-07-30T20:11:16.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14758" PostTypeId="2" ParentId="13521" CreationDate="2014-07-30T20:18:52.783" Score="3" Body="&lt;p&gt;Answer B is correct because nitric acid is the only compound whose building blocks satisfy the &lt;a href=&quot;http://goldbook.iupac.org/M04002.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC Gold Book definition&lt;/a&gt; of a molecule, a neutral (uncharged) entity consisting of at least 2 or more atoms. The building blocks of all other given compounds are atomic ($\ce{K+}$, $\ce{Na+}$ and $\ce{Ag+}$) or molecular ions ($\ce{NO3-}$), which by their nature do not fulfill the requirement of being electrically neutral. Per definition, they are &lt;a href=&quot;http://goldbook.iupac.org/M03986.html&quot; rel=&quot;nofollow&quot;&gt;molecular entities&lt;/a&gt;, i.e. distinct particles which can be identified as separate entities. This term is more general and also includes charged species like ions.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-07-30T20:18:52.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14760" PostTypeId="1" AcceptedAnswerId="14763" CreationDate="2014-07-30T20:50:17.623" Score="6" ViewCount="3457" Body="&lt;p&gt;I'm confused by this picture I found on &lt;a href=&quot;https://chemistry.stackexchange.com/q/20009/23561&quot;&gt;masterorganicchemistry.com&lt;/a&gt;—a very popular organic chemistry website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0m9i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that the product is a product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, can't there be another product? Specifically, can't the methyl group by the bromine have an &lt;em&gt;anti&lt;/em&gt; and &lt;em&gt;beta&lt;/em&gt; hydrogen relative to the bromine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So shouldn't there be two products?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, this one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tug56.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="23561" LastEditDate="2018-08-16T00:16:22.057" LastActivityDate="2018-08-16T00:16:22.057" Title="E2 Elimination; Anti-periplanar beta hydrogen" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;cyclohexane&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14761" PostTypeId="2" ParentId="14760" CreationDate="2014-07-30T21:30:51.410" Score="2" Body="&lt;p&gt;Yes that should be correct. Your product is less favoured but still likely. In fact I guess, ( I am not near a database to confirm that but you could try it) in reality you get the E1 product too. So a &quot;nice&quot; mixture of at least 3 products&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-07-30T21:30:51.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14762" PostTypeId="2" ParentId="14760" CreationDate="2014-07-30T21:52:21.027" Score="5" Body="&lt;p&gt;If the elimination occurs via an ${\mathrm{E_2}}$ mechanism, the other endocyclic double bond (i.e., the one between the two methyl bearing carbons) cannot be formed, as the $\ce{C_2}$ $\beta$-hydrogen will not be anti-periplanar to the $\ce{Br-}$ leaving group. That said, the other possible elimination product you've identified, with the exocyclic double bond, is entirely possible. However, this would be the Hofmann product, as it contains the less-substituted double bond. The major product of an elimination reaction is typically the Zaitsev product, which contains the most-substituted double bond. The Zaitsev product is the thermodynamic one (as the more-substituted alkene is typically more stable), while the Hofmann product is the kinetic one (as it results from a base abstracting a more sterically accessible or otherwise more acidic hydrogen). The Hofmann product can be obtained if a sterically bulky base is used, as this raises the energy of the transition state due to steric hindrance between the approaching base and the substrate molecule, and puts the reaction under kinetic control (assuming reaction temperature is controlled such that the $E_a$ for the thermodynamic pathway is prohibitively high). $\ce{NaOEt}$ is not bulky, so we expect the Zaitsev product (as given in the reaction scheme you cite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an $\mathrm{E_1}$ mechanism occurs, a tertiary carbocation intermediate is generated, and the double-bond is likely to form between the two methyl bearing carbons, as this would give the maximally stable tetrasubstituted alkene, and the requirement for anti-periplanarity is lifted. That said, $\mathrm{E_1}$ reactions are not favorable given a high concentration of strong base, so I would expect $\ce{E_2}$ to predominate, assuming at least a stoichiometric amount of $\ce{NaOEt}$ is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that, with cyclohexanes, the issue is slightly more complex, as the anti-periplanarity requirement of the $\mathrm{E_2}$ mechanism requires the leaving group to be axial, which in turn necessitates a ring flip that has an associated energy barrier. Hence, in theory, the reaction with a cyclic molecule may be more likely to be kinetically controlled under a wider range of conditions.)&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-07-30T22:06:52.873" LastActivityDate="2014-07-30T22:06:52.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14763" PostTypeId="2" ParentId="14760" CreationDate="2014-07-30T22:33:04.153" Score="7" Body="&lt;p&gt;A reasonable chemist would be on the lookout for two products, especially given that E2 elimination from 1-bromo-1-methylcyclohexane produces some methylenecyclohexane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dDWLK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the transition state for the E2 elimination the pi bond is already partially developed.  This means that the relative thermodynamic stabilities of the products will play a role in determining the product distribution.  Since more highly substituted olefins are thermodynamically more stable than less substituted olefins, one would predict that the product with the double bond internal to the ring would predominate over the analogue with the external double bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/leQ9w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction temperature, base size, steric constraints around the various hydrogens that might be involved and the numbers of the various hydrogens all factor into the final product ratio, but these effects are usually secondary unless you change circumstances so much that the reaction mechanism changes and shifts away from E2.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-07-30T22:44:26.700" LastActivityDate="2014-07-30T22:44:26.700" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14764" PostTypeId="1" AcceptedAnswerId="14776" CreationDate="2014-07-30T23:35:56.740" Score="6" ViewCount="7729" Body="&lt;p&gt;I know this undergoes an $\ce{S_{N}1}$ reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mSGQ9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ethanol is the solvent. Weak nucleophile. Neutral nucleophile. This points toward $\ce{S_{N}1/E1}$ (competing pathways). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The leaving group is on a tertiary carbon. This precludes backside attack, $\ce{S_{N}2}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The professor labeled this reaction as $\ce{S_{N}1}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question: where's the substitution part of this reaction? I don't see any substitution. Is there supposed to be no reaction here? From what he wrote it looks like a bromine left as bromide ion and a hydrogen appeared out of nowhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My expected transition state is a tertiary carbocation, which will not rearrange, and then the nucleophilic oxygen part of the ethanol molecule attacks the carbocation. Bromide ion deprotonates the oxonium ion, and I get an ether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who's right? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-07-31T03:12:13.117" LastActivityDate="2015-06-09T16:44:11.840" Title="SN1 reaction of ethanol at tertiary alkyl halide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14765" PostTypeId="1" CreationDate="2014-07-31T00:07:50.157" Score="1" ViewCount="253" Body="&lt;p&gt;$\ce{2A_{(s)} + xB_{(g)} &amp;lt;=&amp;gt; 3C_{(g)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{K_p}=19.3, \mathrm{K_c}=0.45, \mathrm T=523\pu K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work $\Delta n$ to be $1$, So would $x$ be $1$ or $2$? I'm not sure if the coefficient is part of $\Delta n$.&lt;/p&gt;&#xA;" OwnerUserId="7285" LastEditorUserId="23561" LastEditDate="2018-08-16T00:13:29.877" LastActivityDate="2018-08-16T00:13:29.877" Title="Finding the value of stoichiometric coefficient x" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14766" PostTypeId="1" AcceptedAnswerId="14769" CreationDate="2014-07-31T00:23:19.010" Score="9" ViewCount="3386" Body="&lt;p&gt;I recently came across a practice problem in my textbook asking me to name a few compounds using -ic and -ous endings. The exact wording is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Write the name of each of the following ionic substances, using &lt;em&gt;-ous&lt;/em&gt; and &lt;em&gt;-ic&lt;/em&gt; endings to indicate the charge of the cation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one was $\ce{CoCl2}$. However, cobalt has more than just two oxidization states (3, 2, 0, and -1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it is one of the states in the middle, how do I decide whether to use &lt;em&gt;-ous&lt;/em&gt; or &lt;em&gt;-ic&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7286" LastEditorUserId="23561" LastEditDate="2018-08-17T00:23:50.603" LastActivityDate="2018-08-17T00:23:50.603" Title="-ic -ous nomenclature" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14767" PostTypeId="1" AcceptedAnswerId="14770" CreationDate="2014-07-31T00:25:58.547" Score="4" ViewCount="9875" Body="&lt;p&gt;Does steric hindrance affect Brønsted basicity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that basicity is a thermodynamic quality. Not a kinetic factor. So it doesn't matter if it takes a million years for the proton to reach the site of basicity; time is nothing in thermodynamics. We only care about the initial and final state. So this suggests to me that steric hindrance does NOT affect Brønsted basicity &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the hydrogen proton is quite small. So I wouldn't expect it to be particularly sterically hindered by anything (unless we're in solution and the proton is heavily solvated). Still, once the proton reaches the site of basicity, it's reached it, and if I understand thermo correctly, thermo doesn't deal with anything in the middle of the energy graph. Only the points on the two ends.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TUgNw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also it would be nice if one could comment on the difficulty of this problem in general. Just curious. Personally I think it's a fascinating problem.&lt;/strong&gt; (On the other hand I'm not the one actually doing the problem). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:13:35.373" LastActivityDate="2018-03-03T11:13:35.373" Title="Steric Hindrance and Brønsted Basicity" Tags="&lt;acid-base&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14768" PostTypeId="2" ParentId="14765" CreationDate="2014-07-31T00:28:02.347" Score="3" Body="&lt;p&gt;The relationship between $K_p$ and $K_c$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p=K_c(RT)^{\Delta n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta n$ is the change in the number of moles of gas molecules over the course of the reaction. The only way you could know this change is from the stoichiometric coefficients. The $\Delta n$ term derives from the laws of mass action in which all partial pressures and/or concentrations are raised to the stoichiometric coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NO(g) + O2(g) &amp;lt;=&amp;gt; 2NO2(g)}$$&#xA;$$K_c = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$&#xA;$$K_p = \dfrac{P^2_{\ce{NO2}}}{P^2_{\ce{NO}}P_{\ce{O2}}}$$&#xA;$$P_x = \dfrac{n_x}{V_x}RT=[\ce{X}]RT$$&#xA;$$K_p = \dfrac{([\ce{NO2}]RT)^2}{([\ce{NO}]RT)^2([\ce{O2}]RT)}$$&#xA;$$K_p = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}\dfrac{(RT)^2}{(RT)^3}=K_c(RT)^{-1}$$&#xA;$$\Delta n = -1 =2-(2+1)$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2014-07-31T02:29:20.150" LastActivityDate="2014-07-31T02:29:20.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14769" PostTypeId="2" ParentId="14766" CreationDate="2014-07-31T00:36:13.970" Score="11" Body="&lt;p&gt;Cobalt may possibly have all those other oxidation states (and others too) for this nomenclature system, we only care about the two &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobalt#Compounds&quot;&gt;most common oxidation states&lt;/a&gt; in ionic compounds: $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$. Cobalt(II) compounds would thus be named  &lt;em&gt;cobaltous&lt;/em&gt; and cobalt(III) compounds would be &lt;em&gt;cobaltic&lt;/em&gt;. Which is yours?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-07-31T00:36:13.970" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14770" PostTypeId="2" ParentId="14767" CreationDate="2014-07-31T01:01:33.527" Score="6" Body="&lt;p&gt;Anything that stabilize one of the species in an acid-base equilibrium influences the free energy difference between the reactants and products, which alters the equilibrium constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA &amp;lt;=&amp;gt; H+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_r G^\circ = -RT\ln K_{eq}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You forget the solvent. Acidity and basicity depend on the interaction between the acid/base and the solvent. While steric hindrance will decrease the rate solvent access, it more importantly also decreases &lt;strong&gt;solvation&lt;/strong&gt; of the acidic or basic site. More steric hindrance prevents additional solvent molecules from reaching the site to stabilize an ionic conjugate acid or base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the difference in $\mathrm{p}K_a$ between methanol, ethanol, 1-butanol, and &lt;em&gt;tert&lt;/em&gt;-butanol. These are equilibrium constants, so they most certainly derived from &lt;strong&gt;thermodynamic&lt;/strong&gt; phenomena!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methanol&quot; rel=&quot;nofollow&quot;&gt;Methanol&lt;/a&gt; $\ce{CH3OH}$ has a $\mathrm{p}K_a = 15.5$ (in water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ethanol&quot; rel=&quot;nofollow&quot;&gt;Ethanol&lt;/a&gt; $\ce{CH3CH2OH}$ has $\mathrm{p}K_a = 15.9$ (in water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1-butanol&quot; rel=&quot;nofollow&quot;&gt;1-Butanol&lt;/a&gt; $\ce{CH3CH2CH2CH2OH}$ has $\mathrm{p}K_a = 16.1$ (in water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tert-Butyl_alcohol&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;t&lt;/em&gt;-Butanol&lt;/a&gt; $\ce{(CH3)3COH}$ has $\mathrm{p}K_a = 17$ (in water) (I've also seen 18 as its $\mathrm{p}K_a$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between methanol, ethanol, and 1-butanol (all primary alcohols) is small, while &lt;em&gt;t&lt;/em&gt;-butanol has a larger pKa. If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in methanol, ethanol, and 1-butanol, more water molecules can solvate the alkoxide anion conjugate base, stabilizing that conjugate base more than the conjugate base of &lt;em&gt;tert&lt;/em&gt;-butanol, which due to sterics cannot have as much solvation of the anion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, gas phase acidities, in which there is no solvent, should have no solvent and solvation effects, and thus no (or at least reduced) steric effects.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2014-07-31T02:44:38.407" LastActivityDate="2014-07-31T02:44:38.407" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14771" PostTypeId="1" CreationDate="2014-07-31T01:02:39.537" Score="11" ViewCount="57939" Body="&lt;p&gt;So apparently, &lt;a href=&quot;http://ntr.oxfordjournals.org/content/early/2014/05/14/ntr.ntu078.full&quot;&gt;e-cigarettes produce formaldehyde/acetaldehyde&lt;/a&gt; because they vaporize propylene glycol/glycerin. What is the chemical reaction through which propylene glycol/glycerin forms formaldehyde/acetaldehyde? Why does propylene glycol form more carbonylated products than glycerol?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="402" LastEditDate="2017-05-16T02:01:40.300" LastActivityDate="2017-05-16T02:01:40.300" Title="How does propylene glycol/glycerin form formaldehyde/acetaldehyde when vaporized?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14772" PostTypeId="2" ParentId="14765" CreationDate="2014-07-31T02:30:44.530" Score="2" Body="&lt;p&gt;Assuming you calculated $\Delta n$ correctly, the value of $x$ would  be 2, because $\Delta n= 3 - x$.  You only count the coefficients of gases in determining $\Delta n$.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-31T02:30:44.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14773" PostTypeId="1" CreationDate="2014-07-31T02:58:22.983" Score="1" ViewCount="195" Body="&lt;p&gt;One can salt CO2 out of solution by adding salt to carbonated water.  Can one do the reverse?  If one adds more and more CO2 to the water under pressure, can you drive out the salt?&lt;/p&gt;&#xA;" OwnerUserId="7290" LastActivityDate="2014-07-31T03:10:56.293" Title="Can one reduce solubility of NaCl by adding CO2 under pressure? Will the salt fall out of solution?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14774" PostTypeId="2" ParentId="14773" CreationDate="2014-07-31T03:10:56.293" Score="3" Body="&lt;p&gt;NaCl is vastly more soluble than carbon dioxide, so this method won't work.  You can get some NaCl to precipitate (assuming you have a concentrated solution) by bubbling HCl in, thanks to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Common-ion_effect&quot; rel=&quot;nofollow&quot;&gt;common-ion effect&lt;/a&gt;, so you could for example get pure NaCl from rock salt this way.  But somehow I don't think that's the situation you have in mind :-)&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-31T03:10:56.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14775" PostTypeId="2" ParentId="12259" CreationDate="2014-07-31T03:20:33.383" Score="2" Body="&lt;p&gt;The reason these salts are soluble in ammonia (including aqueous ammonia -- so why buy ammonia in EtOH?) is that they form complexes with ammonia:&#xA;$$&#xA;\ce{CuI}\text{(s)}+2\ce{NH3}\text{(aq)}\to\ce{Cu(NH3)_2^+}\text{(aq)}+\ce{I-}\text{(aq)}.$$&#xA;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-31T03:20:33.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14776" PostTypeId="2" ParentId="14764" CreationDate="2014-07-31T03:32:45.007" Score="6" Body="&lt;p&gt;Just to leave this question not unanswered. As ron already pointed out, you reasoning is correct. Based on the information given, the ethanol may perform a nucleophilic attack at the carbocation. To a very minor extent you might find the elimination product, but I guess you will have to cook it quite well. The elimination pathway is more likely to occur, when there is a stronger base than ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aqDlg.png&quot; alt=&quot;reaction scheme&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to get your terminology right. The carbocation is a reactive intermediate, not a transition state. The $\ce{S_{N}1/E1}$ reaction have always (at least) two transition states. The first one corresponds to the carbon leaving group bond break and the second one corresponds to the nucleophile carbon bond formation, proton abstraction, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LQIoT.png&quot; alt=&quot;reaction mechanisms&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-07-31T03:32:45.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14777" PostTypeId="2" ParentId="14593" CreationDate="2014-07-31T05:39:00.687" Score="4" Body="&lt;p&gt;There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal.  This notion is very sensitive to the number of electrons per atom involved in bonding.  Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4.  These all participate, in some extent, in the &quot;electron soup&quot; that holds metals together.  I haven't been able to figure out exactly to what extent, but  It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together.  (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you progress further along the transition metals the delocalization of d electrons in the metal goes down.  That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger.  As a result, the distance between neighbors (hence the metallic radius) is the same for both.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="16683" LastEditDate="2018-08-31T16:45:56.993" LastActivityDate="2018-08-31T16:45:56.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14778" PostTypeId="2" ParentId="14593" CreationDate="2014-07-31T06:10:19.313" Score="4" Body="&lt;p&gt;The series you cited belongs to so known 'metallic' radius, and it depends on crystal structure of the element, which changes through the row. In short, you cited series, that is not suited for consideration of isolated tendencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, indeed, several types of atomic radii (covalent with different valur for bonds of different order, van-der-waals radii and radius of cutoff that leaves some amount of electronic density inside the atom). When comparing atomic radii in comparable environment, two main trends are observable: growth of atomic size down the column in periodic table because more electronic shells is packed into same atom, and contraction of atoms towards the end of the row. This is a bit trickier to explain. Essentially, completed inner electronic shell isolate outer shells from nucleus, reducing effective charge of the nucleus that the outer shell 'feels'. Given that, at the beginning of the row outer electrons feels effective charge of 1 around already quite big completed shell, while at the end outer electrons feels effective charge of 8 around compacted inner shell. This is further complicated by 'mixed' status of d-electrons, that are isolated from the nucleus by inner shells much more effectively then p- and especially s- electrons, so they are valence-active in transition elements, but valence-inactive in p-elements. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-07-31T06:10:19.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14779" PostTypeId="2" ParentId="8605" CreationDate="2014-07-31T06:34:32.947" Score="1" Body="&lt;p&gt;A) Gamma &amp;lt; 1 means attractive forces dominate, and there is a decrease of vapour pressure of solute A relative to pure A. B) Gamma &gt; 1 means repulsive forces dominate, and there is an increase of vapour pressure of solute A relative to pure A.  You can think of activity coefficients as a qualifier for &quot;escaping tendency&quot; of the solute.  With case A) above, the partial molar excess Gibbs energy of mixing is negative, meaning the solvent stabilises the solute, and energy is given away as a result of the mixing event (i.e. mixing is spontaneous). With case B) above, the partial molar excess Gibbs energy of mixing is positive meaning the solvent destabilises the solute, and energy is required in order to achieve mixing (i.e. mixing is NOT spontaneous). Think of activity as escaping tendency - a small activity means the solute remains in the solvent due to attractive forces whereas a large activity indicates a large escaping tendency of the solute because repulsive forces dominate.  a = gamma x [conc.], right? So, if gamma &gt; 1, then a is going to be larger than [conc.]. It might help you to consider fugacity, which is the analogue of activity for gases (which employs similarly the fugacity coefficient). Also consider an aqueous solution of ammonium hydroxide.  You make it to be X M, right?  But you can always smell ammonia since the solution you made is losing quantities of ammonia.  Is the gamma for NH3 &gt; or &amp;lt; than 1?&lt;/p&gt;&#xA;" OwnerUserId="7291" LastActivityDate="2014-07-31T06:34:32.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14780" PostTypeId="2" ParentId="8605" CreationDate="2014-07-31T08:56:49.703" Score="1" Body="&lt;p&gt;What is meant by effective concentration is whatever number makes the equilibrium equation work.  To a first approximation it is true that, for example, [Na$^+$][Cl$^-$] is a constant in any solution saturated with NaCl at a given temperature, regardless of whatever else might be in the solution; but if you add lots of HCl, although [Na$^+$] does go down as [Cl$^-$] goes up, the product is not exactly constant. You can tabulate values for each concentration that &lt;em&gt;will&lt;/em&gt; make the product constant, and these are called activities.  As @Gordon Driver mentioned, these values can be either greater or less than the actual concentration, depending on the specific system.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-07-31T08:56:49.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14781" PostTypeId="1" CreationDate="2014-07-31T09:53:22.607" Score="1" ViewCount="16082" Body="&lt;p&gt;I understand how to calculate the change in enthalpy and entropy for a reaction in standard conditions, but is there a way to calculate these values at non-standard conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any way that a reaction with positive change in gibbs free energy will occur?&lt;/p&gt;&#xA;" OwnerUserId="7292" LastActivityDate="2018-12-06T19:50:38.567" Title="How do you calculate enthalpy for a reaction at non-standard conditions?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14782" PostTypeId="1" AcceptedAnswerId="14800" CreationDate="2014-07-31T11:56:49.447" Score="3" ViewCount="2072" Body="&lt;p&gt;I need some help naming compounds with more than one functional group based on their priority order. Here are some examples:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A 4-membered carbon chain with an aldehyde at one end and a ketone right next to the aldehyde&lt;/li&gt;&#xA;&lt;li&gt;A 4-membered carbon chain with an aldehyde at one end, and the ketone and aldehyde having 1 carbon in between.&lt;/li&gt;&#xA;&lt;li&gt;A 5-membered carbon chain with a &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; double bond at carbon number 2  and a &lt;span class=&quot;math-container&quot;&gt;$\ce{C#C}$&lt;/span&gt; triple bond at &lt;span class=&quot;math-container&quot;&gt;$\ce{C}_4$&lt;/span&gt; (maybe it is the reverse)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I think that for the one with a double bond and a triple bond, the triple bond gets higher priority and so it would be pent-4-en-2-yne but I am not sure about it.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user7951" LastEditDate="2020-09-25T09:30:31.243" LastActivityDate="2020-09-25T09:30:31.243" Title="Nomenclature of organic molecules with more than one functional group" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14783" PostTypeId="1" AcceptedAnswerId="14785" CreationDate="2014-07-31T13:37:07.093" Score="-2" ViewCount="3480" Body="&lt;p&gt;I want to know if the structure I drew is sucrose or an enantiomer of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The One I Made:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hek5x.jpg&quot; alt=&quot;The one I made&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The one on internet:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DhSVQ.png&quot; alt=&quot;The one on internet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;&lt;/h1&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-08-12T06:53:33.087" LastActivityDate="2014-08-12T06:53:33.087" Title="Is this disaccharide sucrose?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-07-17T15:48:09.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="14784" PostTypeId="1" CreationDate="2014-07-31T13:49:08.540" Score="1" ViewCount="1601" Body="&lt;p&gt;Almost all medicines I buy have storage requirements saying $T\le25^\circ C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do they really spoil at higher temperatures? What if they are stored at e.g. $30^\circ C$, which is common at summertime?&lt;/p&gt;&#xA;" OwnerUserId="2133" LastEditorUserId="4945" LastEditDate="2014-07-31T17:57:08.763" LastActivityDate="2014-07-31T17:57:08.763" Title="Why do all medicines require storage temperatures below 25 °C?" Tags="&lt;stability&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14785" PostTypeId="2" ParentId="14783" CreationDate="2014-07-31T13:53:02.553" Score="5" Body="&lt;p&gt;Those are the same molecules - the fructose part has just been rotated about the glycosidic bond (the oxygen between the two).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Enantiomer&quot; rel=&quot;noreferrer&quot;&gt;Enantiomers&lt;/a&gt; are molecules that have the same formula, but are non-superimposable - there have to be differences in &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;chirality&lt;/a&gt;. If you can superimpose the structures by rotating about a bond or by rotating the molecule, it is not an enantiomer.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-07-31T13:53:02.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14786" PostTypeId="1" CreationDate="2014-07-31T14:24:57.167" Score="1" ViewCount="562" Body="&lt;p&gt;Why is this form of p orbital occupation an excited state? I understand that there should be one up spin and one down spin in one orbital box by Pauli's exclusion rule. But in the p orbitals, it doesn't matter, whether we put up spins first or down spins first according to Hund's rule. I thought unless up and down spin are in same box, they don't affect each other. But as what I thought is wrong, I would like to know the exact reason for this. Why is it considered to be an excited state?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nTYpt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7294" LastEditorUserId="4945" LastEditDate="2014-07-31T15:17:26.980" LastActivityDate="2014-08-15T21:21:14.267" Title="Why is up spin in the px orbital and down spin in the py orbital considered an excited state?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2016-09-24T10:36:40.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="14787" PostTypeId="1" AcceptedAnswerId="18768" CreationDate="2014-07-31T15:01:25.733" Score="5" ViewCount="1396" Body="&lt;p&gt;If $\ce{C2O4}$ is very unstable no matter the configuration(that is the structure based on the bonds) and especially unstable in the form that has an oxyanion and a carbocation than how come other carbon oxides are much more stable despite the charges like $\ce{CO}$ for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 possible structures with octets for this. They are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\ce{C=O}$ with the carbon having 1 lone pair and the oxygen having 2 lone pairs. No charges here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\ce{C≡O}$ with $\ce{C}$ having 1 lone pair and $\ce{O}$ having 1 lone pair making the oxygen positive and the carbon negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is an important but very unstable structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so unstable? It is because while carbon isn't so determining as to what charge it prefers, Oxygen is and it really hates being positive because it wants to be either neutral or negative since it is more electronegative than everything else except fluorine. This is why $\ce{H3O+}$ isn't really $\ce{H3O+}$ but rather an ion in a complex of ions is because of how oxygen hates being positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why are other carbon oxides more stable than $\ce{C2O4}$ when they too have very unstable structures like $\ce{CO2}$ having as a resonance structure a single bond to 1 oxygen making it negative, a triple bond to the other making it positive, and the carbon staying neutral? And if every carbon oxide is either unstable no matter the structure or has unstable resonance structures than why do carbon oxides with just carbon and oxygen exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And would adding hydrogens to a carbon oxide make the carbon oxide even more stable? By this I mean that with the fact that carbon oxides tend to be unstable and hydrocarbons tend to be stable would a carbon oxide with hydrogens added to it be less stable than a hydrocarbon but more stable than a carbon oxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this how come glucose is so stable and 6 $\ce{CO}$s bonded together would be extremely unstable just like $\ce{C2O4}$ is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if it is too many questions but it is all about 1 thing, Carbon Oxides.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user8016" LastEditDate="2014-09-27T07:57:35.207" LastActivityDate="2016-03-11T22:25:32.843" Title="Carbon Oxides and stability" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14788" PostTypeId="2" ParentId="4211" CreationDate="2014-07-31T16:05:40.550" Score="3" Body="&lt;p&gt;There are mutants of clotting factor VIII that have longer half lives and better activity than natural Factor VIII. Factor VIII is one of the missing proteins commonly involved in hemophilia. Hemophiliacs usually use injections of Factor VIII to prevent bleeding. The mutant form should last longer in the blood, with better activity, resulting in fewer injections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;https://encrypted.google.com/patents/US7211559&quot; rel=&quot;nofollow&quot;&gt;https://encrypted.google.com/patents/US7211559&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a 2007 patent, I don't know if this product is actually being used to treat patients, drug approval takes a long time.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-31T16:05:40.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14789" PostTypeId="2" ParentId="14784" CreationDate="2014-07-31T16:22:56.123" Score="1" Body="&lt;p&gt;People do not realize that the quality requirements for a medicine is rather high. We are talking about a complex mixture of more or less reactive organic compounds that are potentially sensitive to air, moister and light. Any reaction of the original ingredient is potentially biological active and harmful, and even simple things like recrystallization or coagulation can destroy the effect of the original mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &quot;will they really spoil&quot; makes no sense in the given context: if 99.999% of the cases not spoil, but there is a handful cases when things goes bad then the public goes banana. It took only one (purposefully false) paper to convince tens of thousands of people to go anti-vaccine, and decades till it dies out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it happens with a medicine, it potentially kills the product, so easier to err (by huge margin) on the safe side. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-31T16:22:56.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14790" PostTypeId="2" ParentId="10480" CreationDate="2014-07-31T16:27:09.977" Score="0" Body="&lt;p&gt;In Protein Data Bank (RCBS), for example as molecule of the months:&#xA;&lt;a href=&quot;http://www.rcsb.org/pdb/101/motm.do?momID=23&amp;amp;evtc=Suggest&amp;amp;evta=Moleculeof%20the%20Month&amp;amp;evtl=OtherOptions&quot; rel=&quot;nofollow&quot;&gt;http://www.rcsb.org/pdb/101/motm.do?momID=23&amp;amp;evtc=Suggest&amp;amp;evta=Moleculeof%20the%20Month&amp;amp;evtl=OtherOptions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Nuclear Acid Database:&#xA;&lt;a href=&quot;http://ndbserver.rutgers.edu/&quot; rel=&quot;nofollow&quot;&gt;http://ndbserver.rutgers.edu/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-07-31T16:27:09.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14791" PostTypeId="1" AcceptedAnswerId="14802" CreationDate="2014-07-31T16:39:04.877" Score="8" ViewCount="11867" Body="&lt;p&gt;Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguments against hybridization:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{PH3}$ is less basic than $\ce{NH3}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This jibes with the supposition that $\ce{PH3}$ keeps its lone pair in what is essentially an unhybridized $\ce{s}$-orbital.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{s}$-orbitals are symmetrical and therefore do not concentrate electron density anywhere. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This lack of concentrated electron density as opposed to $\ce{NH3}$ which is $\ce{sp^3}$ hybridized makes $\ce{PH3}$ a poorer base. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, my professor writes in his book that it can be argued that $\ce{PH3}$ exhibits $\ce{sp^3}$ hybridization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another professor told me that the wave function &quot;blows up&quot; when putting an electron pair in an s-orbital. What does that mean? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-08-01T02:18:10.320" LastActivityDate="2021-01-21T05:10:09.250" Title="Hybridization in PH3" Tags="&lt;theoretical-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14792" PostTypeId="2" ParentId="14781" CreationDate="2014-07-31T17:16:41.117" Score="3" Body="&lt;p&gt;Do you know how to calculate the change in enthalpy and entropy for a substance as you change the conditions (T, P, ...) in a thermodynamically reversible fashion? Now take the reactants at &quot;standard&quot; conditions, transform them to reactant(s) at the non-standard conditions, react them, and take the product(s) back to the standard condition.  Compare with the reaction occurring at standard conditions. Figure out the enthalpy and entropy differences between reacting at 'standard' and 'non-standard' conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, ponder why some reactions do not go to completion.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-07-31T17:16:41.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14793" PostTypeId="2" ParentId="13569" CreationDate="2014-07-31T18:04:34.607" Score="5" Body="&lt;p&gt;You can determine which reaction takes place by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential&quot; rel=&quot;noreferrer&quot;&gt;standard electrode potentials&lt;/a&gt; of the half reactions. The relevant standard reduction potentials for the possible reactions (&lt;a href=&quot;http://downloads.gphysics.net/UACh/Medicina/TablasIones.pdf&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; of the values) are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^2+ +2e- -&amp;gt;~ Cu} \quad E_0=+0.342\mathrm{V}$$&#xA;$$\ce{Fe^2+ + 2e- -&amp;gt;~ Fe} \quad E_0=-0.447\mathrm{V}$$&#xA;$$\ce{Fe^3+ + e- -&amp;gt;~ Fe^{2+}} \quad E_0=+0.771\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall potential of the electrochemical cell is defined as the difference between the standard reduction potentials of reduction and oxidation reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{cell}=E_{red}-E_{ox}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spontaneous reaction to occur, $E_{cell}$ must be positive. With the potentials of copper reduction and oxidation of iron to $\ce{Fe^2+}$, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^2+/Fe}}=+0.342\mathrm{V}-(-0.447\mathrm{V})=+0.789\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious from the values that $\ce{Cu^2+}$ cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ because $E_{cell}$ is negative for this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^3+/Fe^2+}}=+0.342\mathrm{V}-(+0.771\mathrm{V})=-0.429\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced equation for the redox reaction between iron and copper(II) sulfate is therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe + CuSO4 -&amp;gt;~ FeSO4 + Cu}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-07-31T18:04:34.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14794" PostTypeId="1" CreationDate="2014-07-31T20:34:33.873" Score="1" ViewCount="1527" Body="&lt;p&gt;Sorry I understand this looks like a &quot;homework question&quot;, (I'm actually a 46yr old dentist looking to see what amount of calcium phosphate should be effective in a proprietary toothpaste!). However, if the actual question below can not be worked out can you let me know what the process of the calculation would be to determine mass of a salt to add to gain a mmol/L concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original Question:&lt;/strong&gt; &#xA;I wonder if you can help on this? I want to calculate how much calcium phosphate to add to 1g of toothpaste to have between 18mmol/L to 1.5mmol/L of calcium and phosphate on the tooth surface? Also if you are able to work this out can you let me know what assumptions you have made in this calculation?&lt;/p&gt;&#xA;" OwnerUserId="7298" LastEditorUserId="73" LastEditDate="2014-08-01T15:33:26.037" LastActivityDate="2014-08-01T16:26:28.700" Title="Calculation of mass from mmol/L" Tags="&lt;everyday-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14795" PostTypeId="2" ParentId="5352" CreationDate="2014-07-31T22:13:02.137" Score="2" Body="&lt;p&gt;If all you want is a practical answer, just try taking a constant mass of dough, dissolve in a constant amount of deionized water, and measure the pH with a standard pH meter. The pH of the measured solution should correlate with the pH of the dough, even if they are not an exact match. The key here is to perform the assay the same way every time, using slightly different amounts of dough or water can change the pH you measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also consider using a buffered solution of water instead of deionized water. I've seen &quot;pure&quot; water in our lab with pH anywhere between 6 and 7, so a dilute buffer at pH 7 should help get more consistent results. Just keep it dilute so that it doesn't buffer out the pH of the dough.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-07-31T22:13:02.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14796" PostTypeId="2" ParentId="14417" CreationDate="2014-08-01T09:42:58.833" Score="3" Body="&lt;p&gt;The strength of interaction of sigma and pi-kind developing differently with distance between two atoms. &#xA;The pi-interaction, which usually is weaker, is favoured at shorter atomic distance, because it leads to better side overlapping of orbitals. The maximum of energy for pi is by shorter atomic distance than for sigma. $\ce{N2}$ molecule has very short $\ce{N-N}$ distance, which leads to much better p-orbitals overlap and as consequence to lower energy of pi-orbital.&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:11.620" LastActivityDate="2015-02-27T17:19:11.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14797" PostTypeId="2" ParentId="5851" CreationDate="2014-08-01T10:32:25.503" Score="5" Body="&lt;p&gt;While the question has been already answered, I think it worth to mention some contraversial asspects of Paulings theoreticla work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;But first on Pauling and the ways he damaged his own credibility: he ignored MO theory to a degree that was clearly perceived by the community as blind, if not unethical. His interests (and great, great creative powers) also shifted to biological problems. And, finally, he lost&#xA;touch with the explosion in structure and reactivity, to an extent that the third edition of The Nature of the Chemical Bond is ... an embarrassment.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from an Acc. Chem. Ress. (2003) paper about VB-MO history&#xA;&lt;a href=&quot;http://www.roaldhoffmann.com/sites/all/files/493s_0.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.roaldhoffmann.com/sites/all/files/493s_0.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-01T10:32:25.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14800" PostTypeId="2" ParentId="14782" CreationDate="2014-08-01T15:09:00.753" Score="1" Body="&lt;p&gt;There are some priority rules about naming compounds with more than 1 functional group.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aldehyde has higher priority than ketone (I don't know why but I think it is because of aldehyde always take place at the end of carbon chain)&lt;/li&gt;&#xA;&lt;li&gt;Alkene has higher priority than alkyne, because -ene suffix comes before -yne&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the answer for your question is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;2-oxobutanal&lt;/li&gt;&#xA;&lt;li&gt;3-oxobutanal&lt;/li&gt;&#xA;&lt;li&gt;pent-1-en-4-yne&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="735" LastActivityDate="2014-08-01T15:09:00.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14801" PostTypeId="2" ParentId="14794" CreationDate="2014-08-01T16:26:28.700" Score="3" Body="&lt;p&gt;There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Monocalcium_phosphate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Monocalcium_phosphate&lt;/a&gt; although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known &lt;em&gt;volume&lt;/em&gt; of paste.  Instead you have a known &lt;em&gt;weight&lt;/em&gt;, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-08-01T16:26:28.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14802" PostTypeId="2" ParentId="14791" CreationDate="2014-08-01T18:04:16.213" Score="7" Body="&lt;p&gt;&lt;strong&gt;Pre:&lt;/strong&gt; My views may be clashing with some people. I express them as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To start with, hybridization is a hypothetical concept$^{*1}$ just to explain all facts and make our work easier, otherwise we would have to take into account every single interaction of electron-proton and also other phenomenons; many theories developed which explained all the major/important facts at a &lt;strong&gt;macro-level&lt;/strong&gt;. And what I have been saying to take into account was all &lt;strong&gt;micro-level&lt;/strong&gt; analysis. Attributing hybridization is purely personal choice because neither does it account all facts (for some molecules) nor it can be dropped altogether (for its $\cdots$), if then we would look for another theory$^{*2}$ to explain observed phenomenon. Parallels similarity between this and the situation of &lt;strong&gt;Ideal-Gas equation&lt;/strong&gt;, in general no perfect theory has been developed, but the most easy, practical and useful are the hybridization$^{*3}$ and &lt;strong&gt;van-der-Waals equation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we move downwards in the periodic table the hybridization concept fails at many places due to overpowering of other factors. I wonder you didn't mentioned the bond angle$^{*4}$ ($93.5^\circ$) for a $\ce{sp^3}$ hybridized (ideally $109.5^\circ$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general atom/molecule doesn't seek to follow a theory, it does what it is best comfortable with, it is us, who mould the theory to the observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, for sake of convenience, assume all molecules show hybridisation and the tendency to hybridise decreases down the group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, I would say it depends on the need of the situation and such dubious statements such as &quot;Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?&quot; should be replaced with &quot;Does hybridization explain all (actually not all) properties of $\ce{PH3}$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are weights in $\ce{sp^\alpha}$ where $\alpha$ ranges from 1 to 3(even upto 4$^{*5}$ and maybe beyond) and is in between these for $\ce{PH3}$ and some may continue to treat it has a hybridised molecule with suitable weights.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^{*1}$(as far as I know)&lt;br&gt;&#xA;$^{*2}$(probably the MOT)&lt;br&gt;&#xA;$^{*3}$(many may argue for other theories which is genuine, but in context this follows)&lt;br&gt;&#xA;$^{*4}$(One may argue as we move down the group electronegativity decreases and atomic size increases. In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)&lt;br&gt;&#xA;$^{*5}$[$\ce{O}$ in $\ce{H2O}$ which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals]&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2016-10-11T14:25:37.037" LastActivityDate="2016-10-11T14:25:37.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14803" PostTypeId="1" AcceptedAnswerId="14807" CreationDate="2014-08-01T20:13:05.607" Score="3" ViewCount="1197" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow&quot;&gt;activity&lt;/a&gt; of a species $i$ is defined as $a_i = e^\frac{\mu_i-\mu_i^\ominus}{RT}$, with $\mu_i$ the chemical potential, and $\mu_i^\ominus$ the chemical potential under standard conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very often, for a species in a non-ideal solution, the activity is written $\gamma_i x_i$, where $x_i$ is the mole fraction and $\gamma$ is a constant known as the activity coefficient for that species. This implies that $\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't have a good idea of is whether the $\gamma_i x_i$ formula is exact (under some assumptions) or an approximation. Are there situations in which the activity depends on the mole fraction in some non-linear way, or upon the concentrations of other species in addition to species $i$? Is the formula $a_i = \gamma_i x_i$ an exact one under some circumstances, or are there always (in principle) nonlinear correction terms that would make it more accurate?&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2014-08-01T22:50:57.693" Title="Is the activity coefficient always a constant?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14805" PostTypeId="2" ParentId="14764" CreationDate="2014-08-01T21:54:43.253" Score="0" Body="&lt;p&gt;The product drawn is a reduction product. You can use $\ce{LiAlH4}$ or homolytic cleavage of the $\ce{C-Br}$ bond (think radicals) to prepare it (but good luck isolating this, isopentane's b.p. is 28 $^\circ$C ). But not $\ce{EtOH}$ hehe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the point, I think a more appropriate example would be to use something like adamantyl bromide instead, as in this case you can not have competing elimination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HiXTD.png&quot; alt=&quot;adamantyl bromide&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="10185" LastEditDate="2015-06-09T16:44:11.840" LastActivityDate="2015-06-09T16:44:11.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14806" PostTypeId="2" ParentId="4085" CreationDate="2014-08-01T22:13:34.267" Score="4" Body="&lt;p&gt;I assume you mean &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrakis(acetonitrile)copper(I)_hexafluorophosphate&quot; rel=&quot;nofollow&quot;&gt;tetrakis(acetonitrile)copper(I) hexafluorophosphate&lt;/a&gt;. This compound contains a tetrahedral complex of $\ce{Cu+}$ with 4 $\ce{CH3CN}$ ligands which stabilize the copper(I) center. They are, however, easily exchanged in the presence of other potential ligands, and with water, disproportionation to $\ce{Cu^2+}$ and elemental $\ce{Cu}$ will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2[Cu(CH3CN)4]^+ + 6H2O -&amp;gt;~ [Cu(H2O)6]^2+ + Cu\downarrow + 8 CH3CN }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pale blue color of the hexaaquacopper(II) complex might not be visible at low concentrations, and the pink color that you have observed arises from fine particles of the elemental $\ce{Cu}$ precipitate.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-01T22:13:34.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14807" PostTypeId="2" ParentId="14803" CreationDate="2014-08-01T22:50:57.693" Score="2" Body="&lt;p&gt;In my copy of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1429218126&quot; rel=&quot;nofollow&quot;&gt;Atkins Physical Chemistry&lt;/a&gt;, 6th edition (it's old!) there is a chart that shows a calculation of the activity and the activity coefficient for an example system (chloroform and acetone). It shows the results using a calculation based on Henry's law (approximating ideal-dilution) and again with one based on Raoult's law (approximating a pure substance). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the activity coefficient is &lt;strong&gt;not constant&lt;/strong&gt; over the range of concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some evidence from a less-reputable but more easily accessible source, check out the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_coefficient&quot; rel=&quot;nofollow&quot;&gt;activity coefficients&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\gamma_B$ is the activity coefficient, which may itself depend on $x_B$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;($x_B$ is the mole fraction of the solute).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this means that the activity coefficient is not constant for a given substance in a given mixture at all concentrations, but that isn't the same as saying that the definition of activities is not exact - it is exact, because it is defined that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is an exact relationship, even though $\gamma_i$ is not constant.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-01T22:50:57.693" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14808" PostTypeId="2" ParentId="14771" CreationDate="2014-08-01T23:32:53.970" Score="14" Body="&lt;p&gt;I think the reaction is essentially an incomplete combustion reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a vaporizer, the idea is to get the propylene glycol, glycerine, and and &quot;flavor&quot; molecules into the vapor phase (or into aerosol droplets) without chemically degrading them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the way this is accomplished is with very high-temperature resistor coils - the idea is that if you have a small amount of heat in a confined space, and the liquid is distributed throughout an absorbent (like cotton, for example), then the heat transfer will happen very quickly and a large amount of liquid will be vaporized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that when you have a very high temperature heat source, and there is lots of oxygen in comparison to the liquid, there is a good chance that at least some of the liquid will reach the combustion temperature instead of just vaporizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that the hydrocarbons in the propylene glycol and glycerin molecules are partially oxidized - oxygen reacts with them to &quot;steal&quot; electrons. If the combustion was complete, you would wind up with just carbon dioxide and water. When combustion is incomplete, you can get any number of compounds, two of which happen to be formaldehyde and acetaldehyde. Carbon monoxide is another one I would be concerned about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this works, take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Propylene_glycol&quot; rel=&quot;nofollow noreferrer&quot;&gt;proplyene glycol&lt;/a&gt; (all pictures are from wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FKDMX.png&quot; alt=&quot;Propylene glycol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycerine&quot; rel=&quot;nofollow noreferrer&quot;&gt;glycerine&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pchv5.png&quot; alt=&quot;Glycerol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how they both have three carbon atoms in the &quot;backbone&quot; and either one or none $\ce{O-H}$ groups (called &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydroxyls&lt;/a&gt; or alcohols) attached to each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen can react with any one of those carbons and will &lt;em&gt;try&lt;/em&gt; to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{CO2}$&lt;/a&gt; and water, the most thermodynamically stable products that are possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEMzS.png&quot; alt=&quot;Carbon dioxide&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Wgu0B.png&quot; alt=&quot;Water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is plenty of oxygen, it will. However, if there is not enough, you will get some intermediate species - formaldehyde or acetaldehyde, for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Formaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;formaldehyde&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aFRRW.png&quot; alt=&quot;Formaldehyde&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, oxygen has reacted with the the other two carbons on the backbone, leaving one with a hydroxyl. The hydrogen on the oxygen goes somewhere (either water or one of the other pieces of the original molecule) and the leftover electrons end up in the oxygen-carbon bond, forming a double bond. This molecule is more thermodynamically stable than proplyene glycol or glycerol, but less stable than carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;acetaldehyde&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DPkHG.png&quot; alt=&quot;Acetaldehyde&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the similarity - it's almost the same, there is just a $\ce{CH3}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;methyl&lt;/a&gt;) group in place of one of the hydrogens. You could get this by reacting the first carbon on propylene glycol with oxygen, and then removing the hydrogen from the hydroxyl on the second carbon. This is again more stable than propylene glycol, but less stable than carbon dioxide - it is also an intermediate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does propylene glycol form more carbonylated products than glycerol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a guess - but I think it likely has to do with the reaction kinetics. The reason I think this is true is that all of the intermediate products are less thermodynamically stable than carbon dioxide and water. Therefore, if a propylene glycol (PG) gives more of a given intermediate product than glycerine does, that product is &quot;easier&quot; to make for PG. This would happen if the activation energy for forming it is &lt;em&gt;lower&lt;/em&gt;, or if the activation energy to go from the intermediate to carbon dioxide and water is &lt;em&gt;higher&lt;/em&gt; compared to the intermediates formed by glycerine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; Many people in other answers are pointing out that there are other mechanisms that could lead to these products, and that combustion is not likely at these temperatures. I think that given the mechanics of heat transfer we can't just rule it out, but I don't have any evidence for that. My answer is meant to show you how the reactant molecules contain the same structures as the products - I wouldn't put much faith in my mechanism (or anybody else's) without some real evidence.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-06-08T19:21:23.250" LastActivityDate="2015-06-08T19:21:23.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14809" PostTypeId="1" CreationDate="2014-08-02T00:10:08.693" Score="2" ViewCount="2355" Body="&lt;p&gt;I have an interesting question about the use of ethyl alcohol on polycarbonate windshields.  I have had friends asking me about the use of Pledge Multi-Surface cleaner on their windshields. Some insist it will degrade, yellow, or even crack and soften their windshields. Others have been using it for years with no apparent effect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pledge does have a small amount of ethyl alcohol, but does it have enough to degrade and ruin motorcycle windshields or does ethanol do any damage anyway?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pledges' MSDS states ethyl alcohol weight at 0.01-1.00  (&lt;a href=&quot;http://www.scjohnson.ca/pdfViewer.aspx?pkMSDSId=367&amp;amp;language=en&quot; rel=&quot;nofollow&quot;&gt;http://www.scjohnson.ca/pdfViewer.aspx?pkMSDSId=367&amp;amp;language=en&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are the effects, if any, on using this cleaner on polycarbonate windshields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7305" LastEditorUserId="305" LastEditDate="2014-08-02T00:51:43.153" LastActivityDate="2014-10-25T14:40:00.227" Title="Ethyl alcohols use on polycarbonate and acrylics" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14810" PostTypeId="1" AcceptedAnswerId="14816" CreationDate="2014-08-02T01:10:26.753" Score="2" ViewCount="1061" Body="&lt;p&gt;See this (from the Wikipedia article on dopamine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dopamine, like most amines, is an organic base. At neutral or acidic&#xA;  pH levels it is generally protonated. The protonated form is highly&#xA;  water-soluble and relatively stable, though it is capable of oxidizing&#xA;  if exposed to oxygen or other oxidants. At basic pH levels, dopamine&#xA;  becomes deprotonated. In this free base form it is less soluble and&#xA;  also highly reactive and easily oxidized&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm very curious about this since I'm curious about dopamine auto-oxidation, as it is often a cause of neurodegeneration observed in those who take amphetamines (and seems more likely in the cytosol, than the vesicles it would otherwise be in). Though I'm also curious how much auto-oxidation would happen for extracellular dopamine as well.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="7448" LastEditDate="2015-10-16T18:07:38.097" LastActivityDate="2015-10-16T18:07:38.097" Title="Why is the deprotonated form of dopamine more reactive than the protonated/neutral form of dopamine?" Tags="&lt;biochemistry&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14811" PostTypeId="1" CreationDate="2014-08-02T01:40:21.690" Score="14" ViewCount="389" Body="&lt;p&gt;I know that solid materials expand when heated, which is called thermal expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is happening to tent square or tent canvas of (&lt;em&gt;plastic?&lt;/em&gt;) camping tents? I have noticed, that during the day, when it’s hot, there is a lot of tension in the skin of a tent. But when it gets colder (when raining or during the night), the tension is less and the fabric gets sloppy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason?&lt;/p&gt;&#xA;" OwnerUserId="7306" LastActivityDate="2016-06-29T17:51:06.287" Title="Why no thermal expansion for camping tents?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14812" PostTypeId="2" ParentId="14459" CreationDate="2014-08-02T06:46:49.080" Score="1" Body="&lt;p&gt;Yes, blowing $\ce{CO2}$ over a hot object would cool it. The second law of thermodynamics says that heat cannot move from a colder material to a hotter material. So if you had a hot object and applied any material cooler than it, the hot object would cool down. The $\ce{CO2}$ offers some advantages because it's a gas. As the gas heats, it expands. By the ideal gas law, any expanding gas must cool down, assuming a constant pressure. However, this effect should apply to any gas, and compressed air would be cheaper than pure $\ce{CO2}$. You might also apply dry ice. This could be considered really compressed $\ce{CO2}$. The dry ice offers a further advantage because of the phase transition from solid to gas, not to mention its much colder initial temperature. As the dry ice sublimates its temperature remains constant throughout the phase transition, allowing to draw much more energy out of the hot object. However, water would also offer this advantage, as it has a high heat capacity, and is much cheaper than dry ice or compressed $\ce{CO2}$. It should also be safer. Suddenly heating a block of dry ice in an enclosed space could be explosive. Even if it doesn't explode, the large amount of $\ce{CO2}$ released could displace the air from the room and choke anyone in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr, use compressed air or liquid water.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="39539" LastEditDate="2017-05-29T04:38:58.897" LastActivityDate="2017-05-29T04:38:58.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14813" PostTypeId="2" ParentId="14553" CreationDate="2014-08-02T07:33:46.077" Score="7" Body="&lt;p&gt;I'll address your questions in 3 parts: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) Why amphoteric&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) How to solve the problem without making unexplained assumptions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) Why the solution in the book uses the approximate equality&#xA;    $\ce{[H3AsO4]=[HAsO4^{2-}]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PART (A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An amphoteric species (molecule or ion) is one that can either gain a proton with relative ease, or lose a proton with relative ease. &quot;Relative ease&quot; means, for example, that it can happen in aqueous solution. If your species will gain a proton in the presence of enough $\ce{H3O+}$, and if it will lose a proton in the present of enough $\ce{OH-}$, it can certainly be called amphoteric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translated into equilibrium constants, if your species has a pKa between 0 and 14, &lt;strong&gt;and&lt;/strong&gt; its conjugate acid (what you get when you protonate it) also has a (necessarily lower) $pK_a$ between 0 and 14, your species will act &lt;em&gt;in aqueous solution&lt;/em&gt; as either as an acid (losing a proton in the presence of strong base) or as a base (gaining a proton in the presence of strong acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $pK_a$ of (neutral) arsenic acid is between 2 and 3 (because the equilibrium constant for removing the first proton, $K_1$, is between $10^{-2}$ and $10^{-3}$). The $pK_a$ of the version that lost one proton, $\ce{H2AsO4-}$, is around 7 (from the value of $K_2$). So clearly $\ce{H2AsO4-}$ will gain a proton if the pH is below 1, say, and it will lose a proton if the pH is above 10. This justifies why it's amphoteric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PART (B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a monoprotic acid like HF, all the F appears either as F$^-$ or as HF (two species). You can write down 1 equilibrium equation, a mass balance for F and a charge balance, and solve all three equations simultaneously; when you combine the three you get a quadratic equation that can be solved exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with a multiprotic acid, things get complicated fast. In this problem you have &lt;em&gt;three&lt;/em&gt; equilibrium equations, plus the mass-balance and charge-balance equations.  That's 5 equations in 5 variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately it's possible to combine 4 of them easily, using only high-school algebra. They are the 3 equilibrium equations and the charge balance. This reduces the problem to two variables (the pH and the concentration of &lt;em&gt;one&lt;/em&gt; of the arsenic ions) in two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write and combine the 4 equations. There are four arsenic species with varying charges; I'll call their concentrations $A_0,A_1,A_2,A_3$ -- the index denotes the charge (which is 3 minus the number of hydrogens). So $A_0=[\ce{H3AsO4}]$, $A_1=[\ce{H2AsO4-}]$, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, $A_1[\ce{H^+}]/A_0=K_1$, or equivalently $A_0=A_1[\ce{H^+}]/K_1$. This relates two concentrations, $A_0$ and $A_1$. Similarly $A_1=A_2[\ce{H^+}]/K_2$ and $A_2=A_3[\ce{H^+}]/K_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is that all these four concentrations are &lt;em&gt;proportional&lt;/em&gt; and given one you can find all others &lt;strong&gt;if&lt;/strong&gt; you know the pH. (We don't know the pH, but I'll address that in a moment.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge balance (expressed in moles present in 1 L of solution) is simply &#xA;\begin{equation} 0\,A_0+ 1\,A_1+ 2\,A_2+ 3\,A_3+K_w/[\ce{H^+}]= [\ce{H^+}]+0.1. &#xA;\end{equation}&#xA;On the left we have the negative charges, on the right the positive ones. Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The concentration of each arsenic species is multiplied by its charge, obviously.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the left side, $K_w=10^{-14}$ is the dissociation product of water, and $K_w/[\ce{H^+}]$ represents the concentration of OH$^-$; it will turn out to be negligible, but we don't know that &lt;em&gt;a priori&lt;/em&gt;, and including it doesn't make the problem any harder.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 0.1 on the right comes from Na$^+$ ions; being an alkali metal, sodium does not exist in water solution except in ionic form, so we know [Na$^+$] exactly. This explains something that will be important later: &lt;strong&gt;The 0.10 M solution of $\ce{NaH2AsO4}$ is exactly what you obtain by dissolving 0.10 moles of NaOH in one beaker, 0.10 moles of $\ce{H3AsO4}$ in another, and mixing the two, assuming the combined volume is 1 L.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we express the left side of (1) in terms of just two&#xA;concentrations, $A_3$ and $\ce{[H^+]}$: $$&#xA;\begin{aligned}A_1+2\,A_2+3\,A_3&amp;amp;=(\ce{[H^+]}/K_1)\,A_2+2\,A_2+3\,A_3\\&#xA;&amp;amp;=(\ce{[H^+]}/K_1+2)\,A_2+3\,A_3\\&#xA;&amp;amp;=(\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)\,A_3+3\,A_3\\&#xA;&amp;amp;= \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3&#xA;. \end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last line is the key; it allows us to relate the concentration of &lt;em&gt;one&lt;/em&gt; arsenic&#xA;species to the pH.  With this (and moving the $K_w/\ce{[H+]}$ to the&#xA;right)&#xA;the charge balance becomes&#xA;$$ \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3&#xA;=\ce{[H^+]}+0.1 -K_w/\ce{[H+]}$$&#xA;or again&#xA;$$ A_3=&#xA;=\frac{\ce{[H^+]}+0.1 -K_w/\ce{[H+]}}{\bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr).}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This look complicated, but it's easy to put into a spreadsheet. The last step is this: The right [H$^+$] is the one that, with $A_3$ calculated as above, obeys the mass balance, $$ A_3 + A_2 + A_1 + A_0 =0.1.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we could declare the multiple-choice problem solved, by cheating a bit: take the various proposed values of  [H$^+$] and see which one works.  For answer (a) the value of the sum $ A_3 + A_2 + A_1 + A_0$ is 0.62; for (b) it is  0.128; for (c) it is 0.10165, and for (d) it is 0.100002.  So the best answer is (d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you didn't have multiple choice? Then you'd want to draw a graph or a table, taking (say) every integer value of pH and connecting them.  Here is such a graph: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wdUx7.jpg&quot; alt=&quot;total As concentration vs pH&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal axis is pH and the vertical one is total arsenic concentration.  The correct pH is where the vertical coordinate is exactly 0.1, and you can get a good idea of its value by eyeballing, or solve by successive approximations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this method allows you to solve with no more effort the following problem: what is the pH when you add 0.1 mol/L of $\ce{NaH2AsO4}$ (as in the original problem), &lt;strong&gt;plus&lt;/strong&gt; 0.05 mol/L of $\ce{H3AsO4}$?  The only difference is that now you'll be looking for the point on the graph where the vertical coordinate is 0.1+0.05 = 0.15, instead of 0.1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PART (C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the book say that $\ce{[H3AsO4]=[HAsO4^{2-}]}$, approximately? And is this a good approximation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equality comes from the claim that the &quot;dominant reactivity&quot; of $\ce{[H2AsO4-]}$ is $$\ce{2\,H2AsO4- \leftrightharpoons H3AsO4{}+HAsO4^{2-}}.$$ If $\ce{H2AsO4-}$ exchanges protons only with itself, and not with, say, water, then every proton lost by one $\ce{H2AsO4-}$ ion means a proton gained by another $\ce{H2AsO4-}$, and the result is that $\ce{[H3AsO4]}$ and $\ce{[HAsO4^{2-}]}$ are generated in equal amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given the claim about &quot;dominant reactivity&quot;, the equality is true, and by the way, the equality implies that the pH is the average of $pK_1$ and $pK_2$. (This follows, by taking logs, from the analysis that I'm guessing your book gives right after the part of the solution you quoted.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the claim is that it does not follow from the amphoteric nature of $\ce{H2AsO4^-}$ alone. It requires two additional conditions: You're starting from &lt;em&gt;only&lt;/em&gt; $\ce{H2AsO4-}$, and in relatively high concentrations. I'll illustrate why these conditions are necessary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) You have to start with only $\ce{H2AsO4-}$ (plus a counterion that doesn't participate further, like Na$^+$). If you started with some $\ce{H3AsO4}$ in addition, you wouldn't have the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$, even approximately. Or, to use the example you gave, if you start with H$_2$S the amphoteric character of HS$^-$ wont help you calculate the pH of the resulting mix. On the other hand if you started by dissolving NaHS alone, it would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) You have to start with a strong solution. If the original question had specified a 0.1 mM solution instead of a 0.1 M solution, the resulting H$^+$ concentration would not be $2.5\times 10^{-5}$ anymore but $3.3\times 10^{-6}$, and the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$ would be off by a factor of 60!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps answer your very interesting and perceptive questions.  Don't hesitate to ask for further clarification if necessary!&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-08-02T09:58:05.463" LastActivityDate="2014-08-02T09:58:05.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14814" PostTypeId="1" AcceptedAnswerId="14823" CreationDate="2014-08-02T09:05:42.227" Score="1" ViewCount="84" Body="&lt;p&gt;Does anyone know if the casein/whey cows milk protein can be found within processed lactose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background: our 4-month-old baby was given teething granules. She was then presented with a rash, screaming, etc. When we checked the teething granules (called teetha) we found they contain lactose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from reading it appears that lactose intolerance is extremely rare in the UK and cows milk allergy is much more common.&lt;/p&gt;&#xA;" OwnerUserId="7308" LastEditorUserId="30836" LastEditDate="2019-04-21T00:03:34.157" LastActivityDate="2019-04-21T00:03:34.157" Title="Cows milk protein in lactose" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14815" PostTypeId="1" AcceptedAnswerId="14879" CreationDate="2014-08-02T09:44:02.987" Score="5" ViewCount="1953" Body="&lt;p&gt;Suppose an initial state $A$ can transition to either state $C$ or state $D$. Suppose further that both of these two processes are rate-limited by a transition to the same intermediate state $B$, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dyVh6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\Delta G_{AB}&amp;gt; \Delta G_{BC}&amp;gt;\Delta G_{BD}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to transition state theory, rates are determined by the rate-limiting step. So the rate of each process will be the same,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;r_{A\to C}&amp;amp;=r_{A\to B} \\&#xA;r_{A\to D}&amp;amp;=r_{A\to B}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which it follows that the number of transitions made during some large $\Delta t$ will also be the same,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;N_{A\to C}&amp;amp;=r_{A\to B}\Delta t \\&#xA;N_{A\to D}&amp;amp;=r_{A\to B}\Delta t&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we were to consider the two pathways collectively then we would predict more transitions to be made to state $D$ than $B$ since $\Delta G_{BD}&amp;lt;\Delta G_{BC}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we resolve this apparent discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="7265" LastActivityDate="2016-02-16T23:04:49.107" Title="Rate-limiting in transition state theory" Tags="&lt;kinetics&gt;&lt;free-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14816" PostTypeId="2" ParentId="14810" CreationDate="2014-08-02T12:54:12.403" Score="3" Body="&lt;p&gt;In general the free amine will react as a nucleophile i.e. by &quot;donating&quot; an available lone pair of electrons. In its protonated form, this lone pair is &quot;blocked&quot; and thus it has to lose a &quot;blocking&quot; proton to react (I use the words in &quot;quotes&quot; more freely - although I'm a chemist and not biochemist but hopefully it will help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However re oxidation of dopamine, the reactive site is not the nitrogen (at least not the primary one - but will come back later) but the catechol part. Because of the electron donating effect of the 2 neighboring OH's, the aromatic ring is electron rich and more prone to oxidation to give o-quinone products. This oxidation reaction depends to an extend to the pH but mainly to the presence of (obviously enough) reactive oxygen species (ROS) and also enzymes (oxidases) or even Fe ions that can catalyse this oxidation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to the nitrogen, after oxidation occurs with creation of reactive sites on the quinone part, it can act now as a nucleophile and attack, creating indolines that will further oxidize to the more stable but toxic indole products.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-08-02T12:54:12.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14818" PostTypeId="1" AcceptedAnswerId="14822" CreationDate="2014-08-02T14:30:55.780" Score="3" ViewCount="1990" Body="&lt;p&gt;Let's take, for example: 2,3-pentadiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't this compound present geometrical isomers, like 2-butene?&lt;/p&gt;&#xA;" OwnerUserId="6998" LastActivityDate="2014-08-02T16:27:26.397" Title="Do certain cumulated dienes present geometric isomerism?" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14820" PostTypeId="1" AcceptedAnswerId="14876" CreationDate="2014-08-02T14:45:49.410" Score="1" ViewCount="72" Body="&lt;p&gt;I notice that for quite a few elements, oxygen comes before it in alphabetical order. Here is one of the rules in inorganic chemistry for molecular formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put all the elements in alphabetical order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do oxides always have what is being oxidized followed by oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes some molecular formulas like PO4 for phosphate violate the rule in inorganic chemistry that the elements are arranged in alphabetical order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess all oxides have some element(except for oxygen or fluorine) followed by oxygen because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) it clarifies that what you are talking about is an oxide of that element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the structure of an oxide is always some element with oxygen atoms around it. &lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2014-08-04T09:19:09.623" Title="Why are oxides always of the form AyOx with A representing any atom except oxygen and fluorine" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14821" PostTypeId="2" ParentId="13708" CreationDate="2014-08-02T15:12:58.950" Score="1" Body="&lt;p&gt;The sharp, intense band at 1716 $\mathrm{cm^{-1}}$ is indeed characteristic for a carbonyl group ($\ce{C=O}$ stretch, in this case aldehyde or ketone). The medium and broad bands between 2732-2957 $\mathrm{cm^{-1}}$ suggest that it is more likely an aldehyde. In this case, they would be assigned to the $\ce{=C-H}$ stretch, and the intensity of the broad band fits better to a saturated aldehyde, i.e. there is no double or triple bond in $\alpha, \beta$-position to the $\ce{CHO}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assignment of the peaks at 3624 and 3533 $\mathrm{cm^{-1}}$ to a primary amine is reasonable, since primary amines have two bands, while a secondary amine would have only one, weaker band at lower wavenumbers. One of the bands in the area of 700-940 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{N-H}$ wagging vibration mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sharp band at 3416 $\mathrm{cm^{-1}}$ could indicate a terminal alkyne ($\ce{C \equiv C-H}$ stretch), the $\ce{C-H}$ stretch band of this group at approximately 2900 $\mathrm{cm^{-1}}$ would then be included in the broad band at 2957 $\mathrm{cm^{-1}}$. The weak band in the area between 2000-2250 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{C \equiv C}$ stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/alkynesir.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/aminesir.html&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/aldehydesir.html&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-02T15:12:58.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14822" PostTypeId="2" ParentId="14818" CreationDate="2014-08-02T15:53:14.767" Score="6" Body="&lt;p&gt;Properly speaking, these sorts of allenes/cumulated dienes do not exhibit geometric isomerism in the way that alkenes can. Rather, when each of the two respective carbons bear two different substituents, the molecule is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;chiral&lt;/a&gt; and will exhibit enantiomerism. More specifically, these asymmetric allenes exhibit &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_chirality&quot; rel=&quot;noreferrer&quot;&gt;axial chirality&lt;/a&gt;. This is consequent to the two consecutive $\pi$-bonds being mutually perpendicular, which results in the molecule lacking any mirror plane of symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, an even number of successive double bonds in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cumulene&quot; rel=&quot;noreferrer&quot;&gt;cumulene&lt;/a&gt; produces a potential axis of chirality (again, assuming each carbon at the end of the double-bond system bears two different substituents), while an odd number results in the substituents at the ends of the double-bond system being coplanar, which permits geometric isomerism akin to that seen in alkenes.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-02T16:27:26.397" LastActivityDate="2014-08-02T16:27:26.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14823" PostTypeId="2" ParentId="14814" CreationDate="2014-08-02T16:00:09.007" Score="2" Body="&lt;p&gt;Yes, process lactose could still contain protein from the milk. It's equally likely that the casein could contain lactose. Whey is much more likely to contain lactose since the whey and lactose are in the water when the cheese is made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be extremely rare for a baby to be lactose intolerant since lactose is the major sugar they consume. As the child ages, the ability to digest lactose decreases. I think I heard that 5% of Caucasians adults can still digest lactose. Only a few people develop lactose intolerance.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-02T16:00:09.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14824" PostTypeId="1" AcceptedAnswerId="14827" CreationDate="2014-08-02T16:45:53.433" Score="2" ViewCount="6014" Body="&lt;p&gt;My teacher stated that allylic carbocations are comparable in stability to tertiary carbocations. $\def\SN#1{\mathrm{S_N}#1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind I am confused why question 13 of &lt;a href=&quot;http://ocw.mit.edu/courses/chemistry/5-12-organic-chemistry-i-spring-2003/exams/Practice_Exam_3_key.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this MIT practice test&lt;/a&gt; states that these two molecules are only expected to undergo $\SN 2$ reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P93nIm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P93nIm.png&quot; alt=&quot;benzylic reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u9DaJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u9DaJ.png&quot; alt=&quot;allylic reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $\SN 1$ reaction on the first molecule with the phenyl substituent would yield a primary carbocation, yes, but this primary carbocation is also benzylic, and can be stabilized through resonance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An $\SN 1$ on the second molecule similarly yields a carbocation, but this is an allylic carbocation, and can easily be stabilized through resonance and delocalization of pi-electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, would $\SN 1$ be valid pathways of reactions for both of these molecules, or is $\SN 2$ simply the only pathway for these molecules to react? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-04-08T17:55:09.810" LastActivityDate="2017-04-08T17:55:09.810" Title="Is an SN1 mechanism feasible with allylic or benzylic halides as substrates?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14825" PostTypeId="2" ParentId="14578" CreationDate="2014-08-02T17:59:54.223" Score="27" Body="&lt;h2&gt;General case&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess.&#xA;It was laid down by Courant$^{[1, 2]}$ and it states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the self-adjoint second order (partial) differential equation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \right)&#xA; u(\mathbf{x}) = 0 \end{equation}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(where $\hat{L} = L(\mathbf{\Delta}, \mathbf{x})$ is a linear,&#xA;  hermitian differential operator, $\rho(\mathbf{x})$ is positive and&#xA;  bounded, and $\lambda$ is the eigenvalue) for a domain $G$ with&#xA;  homogeneous boundary conditions, that is $u(\mathbf{x}) = 0$ on the&#xA;  boundary of the region $G$; if its eigenfunctions are  ordered&#xA;  according to increasing eigenvalues, then the nodes of the $n^{\text{th}}$ &#xA;  eigenfunction divide the domain into no more than $n$ subdomains. The&#xA;  nodal set of $u(\mathbf{x})$ is defined as the set of points&#xA;  $\mathbf{x}$ such that $u(\mathbf{x}) = 0$. No assumptions are made&#xA;  about the number of independent variables.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proof is rather involved and so I won't show it here. But if you want you can look it up in [1] or &lt;a href=&quot;http://www.civil.uwaterloo.ca/ggladwell/GW-Zhu1b.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Courant's nodal line theorem tells us, that if we order the possible energy eigenvalues of the time-independent Schroedinger equation as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then (depending on precisely how you set up the numbering) the  $n^{\text{th}}$ eigenfunction, $\Psi_{n}$ (the one with energy eigenvalue $\lambda_n$) has &lt;strong&gt;at most&lt;/strong&gt; $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$). &#xA;Unfortunately, this gives you only an upper bound for the number of nodes a wave function with a certain energy eigenvalue may possess. So, all we know is that the ground state wave function $\Psi_{1}$ cannot have any nodes within the region $G$ (in total it has one node, namely the one at $\mathbf{x} \to \infty$). Wave functions for higher $n$ may possess up to $n-1$ nodes within $G$ but may as well have less. Thus, we cannot in general say that if a wave function has more nodes than another one it will automatically correspond to a state with higher energy.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Special case: Schroedinger equation in one dimension&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is however a special case:&#xA;For the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory&quot; rel=&quot;noreferrer&quot;&gt;Sturm-Liouville eigenvalue problem&lt;/a&gt; (and thus for &lt;strong&gt;ordinary second order differential equations&lt;/strong&gt; with homogeneous boundary conditions) we can strengthen Courant's nodal line theorem such that if we order the possible eigenvalues as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then the  $n^{\text{th}}$ eigenfunction (the one with energy eigenvalue $\lambda_n$) has &lt;strong&gt;exactly&lt;/strong&gt; $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is useful since the one-dimensional time-independent Schrödinger equation is a special case of a Sturm-Liouville equation. So, in the case of the inhomogeneous radial Schrödinger equation with a local potential and node-less inhomogeneity such as the radial Schrödinger equation for the hydrogenic atom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is generally true that a wavefunction with more (radial) nodes must always correspond to a state of higher energy than a wavefunction with less radial nodes. Also, it is clear that the wavefunctions of the one-dimensional particle-in-a-box must follow this rule. But for the three-dimensional particle-in-a-box this is not true anymore, since in that case the Schroedinger equation of the system is not an ordinary second-order differential equation but a partial-differential equation for which only the general version of Courant's nodal line theorem holds.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Some concluding remarks&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For real-world system like molecules or crystals the Schroedinger equation is a partial differential equation for which the special case outlined above doesn't apply so that only Courant's nodal line theorem in its general form holds which doesn't give a strict justification for the statement that more nodes mean higher energy. Yet it is very often observed that the number of nodes indeed increases with increasing energy.&#xA;The reason for this can be motivated in the following way: The kinetic energy $E_{\mathrm{kin}}$ of a state is proportional to $\int \Psi \Delta \Psi \, d^{3} r$. Via &lt;a href=&quot;http://en.wikipedia.org/wiki/Gauss_theorem&quot; rel=&quot;noreferrer&quot;&gt;Gauss's theorem&lt;/a&gt; it can be shown that $\int \Psi \Delta \Psi \, d^{3} r \propto \int |\nabla \Psi |^{2} \, d^{3} r$ and so $E_{\mathrm{kin}} \propto \int | \nabla \Psi |^{2} d^{3} r$. Now, nodes force a wavefunction to change it's sign. This often means that the value of $\Psi$ has to increase/decrease rather rapidly thus leading to areas with high absolute values of the gradient and thus to high kinetic energy. Since the potential energies shouldn't differ too much between the different states the higher kinetic energy usually also entails a higher total energy. As an example consider the bonding and antibonding wavefunctions of a homonuclear diatomic molecule whose atoms are placed at the positions $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfYvj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding wavefunction has no nodes. Its value between the atoms doesn't have to undergo a rapid change and thus the slope is rather low. The antibonding wavefunction has one node between the atoms. Its value between the atoms must change rapidly from its positive to its negative maximum thus entailing a very high slope. The slopes of the tail regions are comparable for the bonding and antibonding wavefunctions since it can smoothly fall of to zero at infinity and is not required to go from a maximum value to zero within a very confined region of space - thus even if one wavefunction has to start of at a higher maximum value the gradient will not be much higher. It follows that the antibonding wavefunction has a higher kinetic energy than the bonding wavefunction.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;[1] R. Courant, D. Hilbert, &lt;em&gt;Methods of Mathematical Physics&lt;/em&gt;, Vol. 1, Interscience, New York, &lt;strong&gt;1953&lt;/strong&gt;, p. 451-455.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] R. Courant, &quot;Ein allgemeiner Satz zur Theorie der Eigenfunktionen Selbstadjungierter Differentialausdrücke&quot;, &lt;em&gt;Nachr. v. d. Ges. d. Wiss. zu Göttingen&lt;/em&gt; &lt;strong&gt;1923&lt;/strong&gt;, p. 81.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-26T02:23:10.773" LastActivityDate="2014-09-26T02:23:10.773" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="14826" PostTypeId="1" AcceptedAnswerId="14829" CreationDate="2014-08-02T18:02:33.590" Score="7" ViewCount="67928" Body="&lt;p&gt;I'm taking general chemistry 2 this half of the summer. We are currently going over coordination complexes, ligands and transition metal ions. In particular, this question is in regards to the ligand field stabilization energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does $\mathrm{Co}^{2+}$ have 7 valence electrons in the $3d$ orbital? Doesn't the superscript $^{2+}$ denote that it is missing 2 electrons, and therefore isoelectric with $\mathrm{Mn}$? &lt;/p&gt;&#xA;" OwnerUserId="7311" LastEditorUserId="7313" LastEditDate="2014-08-03T01:20:02.007" LastActivityDate="2018-08-15T23:58:14.217" Title="Why does Co2+ have 7 electrons in the 3d orbital, and not 5 like Mn?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14827" PostTypeId="2" ParentId="14824" CreationDate="2014-08-02T18:37:14.787" Score="4" Body="&lt;p&gt;I planned to add this as a comment but I don't have enough reputation ... A definite answer can only come by referring to the literature and/or from the lab. In the context of an undergraduate exam it is reasonable to accept both possible answers ($\mathrm{S_N1}$ and $\mathrm{S_N2}$ ) for allylic and benzylic halides. $\ce{EtOH}$ solvolysis of benzyl bromide is a known reaction and the hypothesis that it goes via $\mathrm{S_N1}$ is the most reasonable one. Is it possible that the solvolysis acually goes via $\mathrm{S_N2}$ instead? The answer lies on performing the appropriate mechanistic studies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case, having checked the actual exam question, it specifies acetone as the solvent. This polar aprotic solvent favours $\mathrm{S_N2}$ conditions, which is aprotic (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28714/solvents-in-substitution-reactions&quot;&gt;this question&lt;/a&gt; also). Incidentally, aside from the use of $\ce{KI}$ over $\ce{NaI}$, this halide exchange is basically the &lt;a href=&quot;http://en.wikipedia.org/wiki/Finkelstein_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Finkelstein reaction&lt;/a&gt;, which occurs via $\mathrm{S_N2}$. (Credit to Greg E. who wrote this in a now deleted comment.)&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-04-08T17:52:46.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14828" PostTypeId="1" CreationDate="2014-08-02T18:38:46.210" Score="1" ViewCount="3439" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An aqueous solution of a certain organic compound has a&#xA;  density of $1.063~\mathrm{g/mL}$, an osmotic pressure of $12.16~\mathrm{atm}$ at&#xA;  $25.0~\mathrm{^\circ{}C}$, and a freezing point of $-1.03~\mathrm{^\circ{}C}$. The compound is known not to dissociate in water. What is the molar mass of&#xA;  the compound?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The molarity and molality can be determined with the given information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\dfrac{\Delta T_{f}}{k_{f}}=0.554m$$ and&lt;br&gt;&#xA;$$ M=\dfrac{II}{RT}=0.497M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{mass of }\ce{H2O}=\dfrac{1000~\mathrm{g}~\ce{H2O}}{0.554~\mathrm{mol}~\text{of solute}}\cdot 0.497~\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how the last equation is used to determine the mass of $\ce{H2O}$. How are we justified in using the molarity and molality to determine the mass of water?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="5026" LastEditDate="2018-02-25T02:50:26.963" LastActivityDate="2018-02-25T02:50:26.963" Title="How to use molality and molarity to determine the molar mass of a compound?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14829" PostTypeId="2" ParentId="14826" CreationDate="2014-08-02T19:22:51.360" Score="8" Body="&lt;p&gt;Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ here and there, but at the end of the day, you just need to * memorize it*.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When d-block elements form ions, the $4s$ electrons are lost first. &lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle#The_Madelung_energy_ordering_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;$n+l$ rule&lt;/a&gt; tells you the order in which atomic orbitals are filled, and according to the rule the $4s$ orbital is occupied before the $3d$ orbital because it has lower energy.&#xA;Thus, the electron configuration of $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$.&#xA;But, the $n+l$ rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration.&#xA;For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is sometimes explained, the statement that $4s$ orbital is lower in energy than $3d$ orbital is true only when the orbitals are &lt;em&gt;unoccupied&lt;/em&gt;. But while you fill $3d$ orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the $4s$ orbital. Thus, when electrons are lost from $\ce{Co}$ atom, they are lost from the 4$s$ orbital first because it is actually higher in energy when both $3d$ and $4s$ are filled with electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="23561" LastEditDate="2018-08-15T23:58:14.217" LastActivityDate="2018-08-15T23:58:14.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14830" PostTypeId="1" AcceptedAnswerId="14832" CreationDate="2014-08-02T20:58:11.977" Score="5" ViewCount="1555" Body="&lt;p&gt;Given a 1 M concentration of $\ce{HCl}$ and $\ce{NaOH}$ here are the pHs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH of $\ce{HCl}$: 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH of $\ce{NaOH}$: 13–14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and of course water has a pH of 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a reaction that is most likely to occur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + NaOH + H2O &amp;lt;=&amp;gt; H2O + NaCl(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I think that other reactions are theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like for example theoretically sodium oxides could form from the $\ce{OH-}$ taking a proton from another hydroxide anion thus giving you $\ce{O^2-}$ which could react with the sodium to give you $\ce{Na2O}$ which is a sodium oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think that theoretically $\ce{Cl-}$ could steal electrons from the $\ce{OH-}$ since $\ce{Cl}$ can exceed an octet of electrons. This would give you neutral $\ce{OH}$ which could then react with another chloride anion to give you $\ce{OHCl-}$(not sure if formula is written right or if it would be negatively charged) or in other words chlorine hydroxide. The proton could then be taken by another hydroxide anion giving you $\ce{OCl}$ also known as chlorine monoxide which could react with another chlorine to give you $\ce{OCl2}$ which is dichlorine monoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these theoretical reactions that lead to oxyanions($\ce{O^2-}$ to be specific), chlorine oxides, chlorine hydroxides, and sodium oxides actually happen to some extent when you have $\ce{HCl}$ and $\ce{NaOH}$ in aqueous solution&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user7951" LastEditDate="2015-09-26T13:23:29.710" LastActivityDate="2015-09-26T13:23:29.710" Title="Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14831" PostTypeId="2" ParentId="6947" CreationDate="2014-08-02T21:15:23.480" Score="5" Body="&lt;p&gt;The solubility of a salt in a solvent is for a good part determined by the difference of stability in the dissolved state (solvated ions) vs. the solid state (ions in crystal lattice). The different solubility of the sodium halides in acetone can be explained by the crystal structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With iodide, the solvate $\ce{NaI \cdot 3(CH3)2CO}$ can be formed, in which each $\ce{Na+}$ is coordinated by 6 acetone ligands via oxygen lone pairs, and the $\ce{I-}$ ions fill in the gaps between these octahedral units, being surrounded by the methyl groups of the acetones. $\ce{I-}$ has a larger ionic radius compared to $\ce{Cl-}$ and $\ce{Br-}$ and thus fills the gap completely, thereby minimizing empty space in the lattice and its distance to $\ce{Na+}$, which is, however, still larger than in solvent-free sodium iodide. This larger distance between the ions decreases the lattice energy, which in turn makes the compound readily soluble in acetone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl-}$ and $\ce{Br-}$ are too small to fit in these gaps, and as the cation-anion distance becomes too large, the only way to minimize the lattice energy is the formation of a solvent-free structure $\ce{NaX}$ (X = $\ce{Cl-}$, $\ce{Br-}$). When the lattice energy is larger than the energy released by solvation of the ions, namely that of $\ce{Na+}$ by complexation with acetone, the compound becomes insoluble and precipitates from the solution (&lt;a href=&quot;http://www.ch.imperial.ac.uk/rzepa/blog/?p=439&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-08-05T17:51:15.207" LastActivityDate="2014-08-05T17:51:15.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14832" PostTypeId="2" ParentId="14830" CreationDate="2014-08-03T00:21:00.127" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a 1M concentration of $\ce{HCL}$ and $\ce{NaOH}$ here are the pHs:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;pH of $\ce{HCL}$: 1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;pH of $\ce{NaOH}$: 13-14&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, a 1 molar concentration of hydrochloric acid solution, $\ce{HCl}$, has a pH of 0.0 if we take ${-log[H_3O^+]=pH}$ to be true. (And it is mostly true for dilute solutions). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and of course water has a pH of 7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pure water at 25 degrees Celsius has a pH of 7.0. Boil the water and the water will no longer have a pH of 7.0. Cool the water and pH will similarly change. The only condition for neutrality is that the concentration of hydronium ion equals the concentration of hydroxide ion. It just so happens that water at normal lab conditions has a hydronium ion and hydroxide ion concentration of ${1.0*10^{-7}}$ M, which corresponds to a pH of 7.0. Your statement only holds at a specific temperature. At other temperatures, this reaction - the auto-ionization of water - which controls the pH of &lt;em&gt;pure&lt;/em&gt; water - will go to varying extents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H_2O &amp;lt;=&amp;gt; H_3O^+ + HO^-}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here is a reaction that is most likely to occur:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{HCl + NaOH + H2O &amp;lt;=&amp;gt; H2O + NaCl(aq)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we are talking about solutions of HCl and NaOH, your equation isn't reflective of what's actually going in on the system. HCl and NaOH both dissolve in solution (unless the solution is saturated with either). So you don't have HCl or NaOH in solution, but rather hydronium ion and chloride anions in the case of HCl and sodium ions and hydroxide ions in the case of NaOH solution. A more accurate equation would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H_3O^+ + HO^- -&amp;gt;2H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaCl(aq)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sodium chloride has a very high solubility in water and dissociates unless the solution is saturated. So it would be more accurate to write simply sodium ions and chloride ions, instead of &quot;aqueous&quot; sodium chloride (unless you are referring to the ion-pair effect). Ion pairs often occur in solution; the name describes exactly what is implied - two oppositely charged ions are attracted to each other. For sodium chloride, ion pairing is likely minimal, but in other cases, such as between a carbocation and a halide anion, the ion pair effect has noticeable effects on chemical reactions (i.e. partial racemization rather than full racemization). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Like for example theoretically sodium oxides could form from the&#xA;  $\ce{OH-}$ taking a proton from another hydroxide anion thus giving&#xA;  you $\ce{O^2-}$ which could react with the sodium to give you&#xA;  $\ce{Na2O}$ which is a sodium oxide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So hydroxide ion reacts as an acid and abstracts a proton from another hydroxide ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is at least theoretically possible, but thermodynamically unfavorable. The oxide anion is &lt;em&gt;very&lt;/em&gt; unstable in water solution to the point of being able to react with the solvent (water) itself to a large extent. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-03T16:56:41.457" LastActivityDate="2014-08-03T16:56:41.457" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="14834" PostTypeId="1" CreationDate="2014-08-03T05:27:23.993" Score="3" ViewCount="185" Body="&lt;p&gt;I do some experiments with liquid mercury and I use a special bottle that was 3d printed (ABS plastic). To prevent leaking of mercury I use 502 glue layered around the bottle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know, if it is safe to store mercury in this bottle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does mercury interact with something in 502 glue? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;502 glue: Methylene Chloride , Ethyl acetate, Toluene, Cyclohexane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a professional at chemistry.&lt;/p&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="4945" LastEditDate="2014-08-03T05:48:44.627" LastActivityDate="2014-08-03T17:07:46.917" Title="Mercury and 502 glue" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14835" PostTypeId="2" ParentId="14828" CreationDate="2014-08-03T08:28:16.200" Score="1" Body="&lt;p&gt;As:&#xA;$$m=molality\;=\;\frac{moles\; of\; solute}{mass\;of\;solvent}=\frac n{M_B}$$&#xA;$$M_B=\frac nm$$&#xA;And as :&#xA;$$M=molarity\;=\;\frac{moles\; of\; solute}{volume\;of\;solution}=\frac n{V}$$&#xA;$$n=MV$$&#xA;Now:&#xA;$$M_B=\frac {MV}m$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-08-03T12:51:52.020" LastActivityDate="2014-08-03T12:51:52.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14836" PostTypeId="1" AcceptedAnswerId="14878" CreationDate="2014-08-03T09:37:40.530" Score="12" ViewCount="2287" Body="&lt;p&gt;Why are the two molarities multiplied and not added, and why is each raised to the power of the coefficient rather than multiplied by it? What is the reasoning behind this form? Was it simply determined that the probability of reaction was proportionate to both and thus they are multiplied together? I'd like to know whether its form is purely empirical or has a theoretical backing.&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="16683" LastEditDate="2017-02-16T12:29:29.347" LastActivityDate="2017-02-16T12:35:40.773" Title="Is there a reason for the mathematical form of the equilibrium constant?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ClosedDate="2018-07-24T23:42:21.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="14837" PostTypeId="2" ParentId="14834" CreationDate="2014-08-03T10:09:06.007" Score="2" Body="&lt;p&gt;Mercury does not react with these products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider investing in a mercury spill kit if you can afford it, otherwise find some powdered sulfur and keep it in a safe place.  See &lt;a href=&quot;http://www.epa.gov/hg/spills/&quot; rel=&quot;nofollow&quot;&gt;http://www.epa.gov/hg/spills/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-08-03T10:09:06.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14838" PostTypeId="2" ParentId="14836" CreationDate="2014-08-03T10:42:14.573" Score="8" Body="&lt;p&gt;The answer is both.  The formula for the equilibrium constant was first observed experimentally (that is, it's empirical) for many many reactions.  Then it was justified theoretically in terms of what is now called the Gibbs energy. The reason the formula is multiplicative is that (very roughly) the probability of finding a system in a certain state is proportional to the inverse &lt;em&gt;exponential&lt;/em&gt; of that state's energy; and as you know when you take the exponential of a sum you get a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're curious to know more you should get a hold of Peter Rock's &#xA;&lt;em&gt;Chemical Thermodynamics&lt;/em&gt;.  It rocks! :-)&lt;br&gt;&#xA;If you like math and know calculus well you can read the derivation of the famous formula $K_{\text{equilibrium}}= \exp(-\Delta G^\circ/RT)$, where $\Delta G^\circ$ is the change in the Gibbs energy between reagents and products in their standard state, $R$ is the gas constant and $T$ is the temperature.  &lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="16683" LastEditDate="2017-02-16T12:32:42.157" LastActivityDate="2017-02-16T12:32:42.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14839" PostTypeId="2" ParentId="14836" CreationDate="2014-08-03T11:41:27.473" Score="11" Body="&lt;p&gt;Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{&amp;lt;=&amp;gt;}}}\ce{ cC + dD}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the forward and reverse processes are elementary reactions (a &lt;strong&gt;big&lt;/strong&gt; assumption), then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$&#xA;$$rate_{-1}=k_{-1} [\ce{C}]^c [\ce{D}]^d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium:&#xA;$$rate_1=rate_{-1}$$&#xA;$$k_1 [\ce{A}]^a [\ce{B}]^b=k_{-1} [\ce{C}]^c [\ce{D}]^d$$&#xA;$$\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the rate constants is the equilibrium constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-03T11:41:27.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14840" PostTypeId="1" CreationDate="2014-08-03T12:13:58.193" Score="1" ViewCount="605" Body="&lt;p&gt;What is the liquid used in instant shoe shiner products? Ive tried refilling it with WD-40 but I would like to know if there are any cheap alternatives to refill it? &lt;/p&gt;&#xA;" OwnerUserId="7318" LastActivityDate="2014-11-21T04:15:10.317" Title="What is the liquid used in instant shoe shiner?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14841" PostTypeId="2" ParentId="14834" CreationDate="2014-08-03T12:46:42.410" Score="4" Body="&lt;p&gt;Mercury does not react with this compounds. However, I strongly encourage you to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) use monolithic vial for mercury. 3d-printed pieces are very fragile in my experience. Glass vial should be fine, however I recommend double casing (keep sealed glass vial with mercury in a bigger plastic vial with sulfur dust.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) perform all operations with mercury over steel of plastic pan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) keep a piece of copper foil/plate and nitric acid at hand. Mercury drops stick to copper, freshly washed in nitric acid, which is used to collect mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) in case mercury is somewhere you don't have access to, drown it in concentrated iron (III) chloride solution. It will transform metallic mercury into mercury chloride, that does not give vapors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Note, that mercury salts may react with some organic compounds, including aromatics (toluene). While risks are low, organomercury compounds accumulates in living tissues, so caution is required.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-03T17:07:46.917" LastActivityDate="2014-08-03T17:07:46.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14842" PostTypeId="1" AcceptedAnswerId="14844" CreationDate="2014-08-03T13:29:21.440" Score="1" ViewCount="12875" Body="&lt;p&gt;According to my knowledge, I know that $\ce{AgCl}$ dissolves in dilute ammonia, $\ce{AgBr}$ dissolves in concentrated ammonia and $\ce{AgI}$ does not dissolve even with concentrated ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="4945" LastEditDate="2014-08-04T05:04:40.983" LastActivityDate="2020-05-24T15:10:28.360" Title="What is the reason for the different solubility of silver halides in ammonia?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14844" PostTypeId="2" ParentId="14842" CreationDate="2014-08-03T14:42:31.583" Score="8" Body="&lt;p&gt;Silver chloride is &lt;em&gt;very thermodynamically&lt;/em&gt; stable in ammonia solution. The &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; for the dissolution of silver chloride in ammonia solution is only &lt;span class=&quot;math-container&quot;&gt;$\pu{2.9E-3}.$&lt;/span&gt; suggesting that the product — silver ammonia complex ion — is much less thermodynamically stable than the reactant, silver chloride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{AgCl + 2 NH3 &amp;lt;=&amp;gt; [Ag(NH3)2]+ + Cl-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;K &amp;amp;= K_\mathrm{sp}(\ce{AgCl})\times K_\mathrm{f}(\ce{[Ag(NH3)2]+})\\&#xA;  &amp;amp; = \pu{1.8E-10}\times\pu{1.6E7}\\&#xA;  &amp;amp; = \pu{2.9E-3}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, silver chloride will dissolve in an excess of &lt;em&gt;concentrated&lt;/em&gt; ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt; value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good σ-electron pair donor ligand, combines with the silver ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ions themselves are not precipitates, as all bulk matter must be approximately electrically neutral, and so the silver chloride dissolves when hit with excess, concentrated ammonia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that the cyanide ion is also a good complexing agent with respect to silver. The cyanide ion has a surfeit of electrons on the carbon, and carbon is even less able to stabilize negative charge than nitrogen, making the cyanide ion reactive as a complexing agent. Hitting silver chloride and other silver salts with a source of cyanide ion will similarly cause the silver salt to dissolve into ions suspended in solution. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="41328" LastEditDate="2020-05-24T15:10:28.360" LastActivityDate="2020-05-24T15:10:28.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14845" PostTypeId="2" ParentId="14815" CreationDate="2014-08-03T14:46:24.857" Score="0" Body="&lt;p&gt;Consider the following hourglass-analogy. The radius of mouth of each sphere indicates slowness in the sense, the slowest reaction has smallest radius.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eTKmi.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Even though there is difference between the mouths at &lt;span class=&quot;math-container&quot;&gt;$B-C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B-D$&lt;/span&gt; junctions both are happening at same rate.This is what you thought.&lt;/p&gt;&#xA;&lt;p&gt;But, the energy from &lt;span class=&quot;math-container&quot;&gt;$TS_1$&lt;/span&gt;has been* released in form of heat as an exotheromic reaction.Also when there is a difference in rates of r/d step this effect is diminished.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;*Thanks to @Martin for pointing out an error&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, the rate of B-D and B-C are different and the products must be proportional to the rates.Otherways, the product must increase in which reaction is fast. Consider this:&lt;img src=&quot;https://i.stack.imgur.com/yedsg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-08-05T09:54:35.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14846" PostTypeId="1" AcceptedAnswerId="14911" CreationDate="2014-08-03T15:05:38.523" Score="10" ViewCount="421" Body="&lt;p&gt;Electrons are delocalized between the three centers in a  3c-2e bond. If there was a molecule with 3c-2e bond in each direction, e.g. a boron cluster, would it be aromatic?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5017" LastEditDate="2015-06-27T00:16:17.823" LastActivityDate="2015-06-27T19:56:46.643" Title="Can aromaticity arise from 3c-2e bonds" Tags="&lt;aromatic-compounds&gt;&lt;bent-bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14847" PostTypeId="1" AcceptedAnswerId="14856" CreationDate="2014-08-03T15:47:15.777" Score="2" ViewCount="5049" Body="&lt;p&gt;I was consulting Wade and I came across this problem and solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jA2L8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/idQpV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I agree with the solution manual, I can't help but wonder if &lt;em&gt;my&lt;/em&gt; proposed mechanism is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the water attack the hydrogens attached to the alkene. I knew that alkenes are slightly electron-withdrawing and this makes the hydrogens an appealing target for a base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the hydrogens were cleaved, I had the remaining pair of electrons attack the carbocation. Is this a valid pathway? Valid as in plausible? Would &lt;em&gt;you&lt;/em&gt; accept this mechanism on a test? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRFWV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem I see with this mechanism is that yield is exceedingly low as sp2 carbons are not especially acidic, and water is a poor base. On the other hand the original problem does specify that this product is &lt;em&gt;minor&lt;/em&gt;. But then, it appears that I chose the more minor of two minor pathways. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-03T15:53:13.273" LastActivityDate="2014-08-03T20:16:46.223" Title="Ring Formation and Alkenes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14848" PostTypeId="1" CreationDate="2014-08-03T16:11:07.393" Score="7" ViewCount="982" Body="&lt;p&gt;Why can't we find the exact position and velocity of a particle? And why is it that if the uncertainty in position is very, very large then the velocity can be determined, and vice versa? Please explain to me. &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="30836" LastEditDate="2017-11-01T04:39:20.333" LastActivityDate="2017-11-01T04:39:20.333" Title="On Heisenberg's uncertainty principle" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14849" PostTypeId="2" ParentId="7338" CreationDate="2014-08-03T16:13:28.200" Score="5" Body="&lt;p&gt;Because atomic radii of elements tend to decrease across a period (&lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/atradius.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;), and bond length increases with atomic radius, the bond lengths are expected to decrease in the following order: C (C-C bond) &gt; B, D (C-N) &gt; A (C-F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between B and D is that in D, the nitrogen is $sp^3$-hybridized, while in B, it is $sp^2$-hybridized due to conjugation of its lone pair with the aromatic ring 2. This results in partial double bond character of the C(ring 2)-N bond, similar to aniline:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gOn7c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More $s$ character leads to &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/10/13/sigma-bonds-come-in-six-varieties-pi-bonds-come-in-one/&quot; rel=&quot;noreferrer&quot;&gt;shorter bonds&lt;/a&gt;, and therefore the C-N bond B is shorter than D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the bond lengths decrease in the following order: C &gt; D &gt; B &gt; A.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-08-03T16:20:22.740" LastActivityDate="2014-08-03T16:20:22.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14850" PostTypeId="2" ParentId="14840" CreationDate="2014-08-03T16:36:09.477" Score="1" Body="&lt;p&gt;It's silicone oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a mineral oil. You can find it as baby oil and if you don't like the odor, also can be found as fleet enema products (eg. &lt;a href=&quot;http://fleetlabs.com/fleet_enema_products.php&quot; rel=&quot;nofollow&quot;&gt;http://fleetlabs.com/fleet_enema_products.php&lt;/a&gt;). &#xA;For larger and cheaper amounts, you can buy it in computer hardware store, farm supply store, or even in vet clinic.&lt;/p&gt;&#xA;" OwnerUserId="7320" LastEditorUserId="7320" LastEditDate="2014-08-03T16:42:21.377" LastActivityDate="2014-08-03T16:42:21.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14851" PostTypeId="2" ParentId="14848" CreationDate="2014-08-03T16:38:58.957" Score="2" Body="&lt;p&gt;I believe that your best course is to search the web for sites aimed at explaining the principle. &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html&quot; rel=&quot;nofollow&quot;&gt;Hyperphysics&lt;/a&gt; does a good job of explanation, but might be a little technical. I haven't read this article on &lt;a href=&quot;http://science.howstuffworks.com/innovation/science-questions/quantum-suicide2.htm&quot; rel=&quot;nofollow&quot;&gt;How Stuff Works&lt;/a&gt;, but it is aimed for the non-technical person.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-03T16:38:58.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14852" PostTypeId="2" ParentId="14848" CreationDate="2014-08-03T16:41:51.160" Score="12" Body="&lt;p&gt;If you imagine particles to be like billiard balls (the way you're taught in school) then this of course makes no sense. The only way you can understand it is to accept the quantum mechanical description of particles: that they are probability distributions that exhibit both wave-like behaviour and particle-like behaviour depending on how you measure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you accept that a particle is best imagined as a probability wave smeared out through space then the Heisenberg uncertainty principle is a rather simple mathematical result relating to &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform&quot;&gt;Fourier transforms&lt;/a&gt;. The more concentrated some function $f(x)$ is, the more spread out its Fourier transform $f(k)$ is. And since position and momentum space are Fourier transforms of each other (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Position_and_momentum_space#Relation_between_space_and_reciprocal_space&quot;&gt;here&lt;/a&gt;) then this gives rise to the uncertainty principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if you measure the position very precisely (highly peaked $\psi(x)$) then the momentum representation $\psi(k)$ will become very broad (imprecise), and vice versa. &lt;/p&gt;&#xA;" OwnerUserId="7265" LastEditorUserId="7265" LastEditDate="2014-08-03T16:52:10.813" LastActivityDate="2014-08-03T16:52:10.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14853" PostTypeId="1" AcceptedAnswerId="14855" CreationDate="2014-08-03T17:53:47.307" Score="5" ViewCount="435" Body="&lt;p&gt;Is hydrogen good to be used in cars? How much of traveling can a car do with medium sized balloon filled with hydrogen? Is it worth it?&lt;/p&gt;&#xA;" OwnerUserId="7063" LastEditorUserId="275" LastEditDate="2014-08-03T19:54:56.370" LastActivityDate="2014-08-04T22:27:33.587" Title="Hydrogen as a fuel really worth it?" Tags="&lt;hydrogen&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14854" PostTypeId="2" ParentId="14853" CreationDate="2014-08-03T18:03:05.093" Score="6" Body="&lt;p&gt;The energy density of hydrogen at atmospheric pressure (which it roughly is inside a balloon) per unit volume is a mere &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_density&quot; rel=&quot;nofollow&quot;&gt;0.01005 MJ/L&lt;/a&gt;, which is roughly a factor 3600 less then that of gasoline (at 36 MJ/L).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical car does 15 km/L so where a gasoline filled balloon (say 3 L) does 45 km for you, a hydrogen filled balloon will get you only 12.5 meters, i.e. just out of the parking lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, any realistic application of hydrogen is to store it either at high pressure (e.g. 70 bar, where the energy density is about 6 MJ/L) or in so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_hydride_fuel_cell&quot; rel=&quot;nofollow&quot;&gt;metal hydrides&lt;/a&gt; which allow much higher energy densities. Another option which to me seems to most viable, is to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fischer%E2%80%93Tropsch_process&quot; rel=&quot;nofollow&quot;&gt;Fischer-Tropsch process&lt;/a&gt; to make methane or even higher alkanes out of hydrogen and $\ce{CO}$. The big advantage of this route is that our current infrastructure of gas stations can still be used&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="1362" LastEditDate="2014-08-04T05:27:37.360" LastActivityDate="2014-08-04T05:27:37.360" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14855" PostTypeId="2" ParentId="14853" CreationDate="2014-08-03T19:23:33.427" Score="3" Body="&lt;p&gt;Hydrogen escapes from vessels more readily than just about any other gas.   All of the schemes to store hydrogen seem unwieldy.  High pressure, low temperature, or hydrides.  I've seen NH3-BH3 proposed as a way to store hydrogen.  A good way to store hydrogen might be methanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generating the hydrogen takes energy in the first place which factors in to the question of whether or not it is worth it.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-03T19:23:33.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14856" PostTypeId="2" ParentId="14847" CreationDate="2014-08-03T20:16:46.223" Score="4" Body="&lt;p&gt;Greg E.'s comments sum up the flaws with your version of the mechanism. While yours contains the same steps, it combines the nucleophilic attach and the proton transfer in a single step. Unless you have experimental evidence for a bimolecular step, the unimolecular step is much more likely. In this case, the unimolecular step is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Baldwin&amp;#39;s_rules&quot; rel=&quot;nofollow&quot;&gt;5-endo-dig&lt;/a&gt; cyclization, which is particularly fast. The alkene nucleophile is already present when the carbocation forms. The rate of the deprotonation is diffusion controlled, as we have to wait for a water molecule to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your logic relating the minority of the product to the unlikeliness of the mechanism is inherently flawed. A minor product does not occur at a lower proportion because the mechanism contains less valid steps. Instead, the distribution of products is related to the relative rates of different mechanisms and the relative energies of the products. That product is a minor product because some other product either &lt;strong&gt;forms faster&lt;/strong&gt; or is &lt;strong&gt;more stable&lt;/strong&gt; or both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this was on my exam, and I like the problem so I might use something like it, I would dock you a few points for combining steps that are not likely concerted into a concerted step. For example, if I decided this mechanism is worth 10 points, then you might receive 7 or 8. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Final note: Remember that experimental evidence trumps all else in science. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regardless of how intuitively and aesthetically unusual your mechanism appears, if experiments support it, then it becomes a more valid mechanism. As an example of some of the non-intuitive things that are in &quot;accepted&quot; mechanisms, check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Bromonium&quot; rel=&quot;nofollow&quot;&gt;halonium ions&lt;/a&gt; and sigmatropic rearrangements, especially the one in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&quot; rel=&quot;nofollow&quot;&gt;Baeyer-Villager Oxidation&lt;/a&gt;. Evidence in support of your mechanism might include a failure to observe or trap that final carbocation, a demonstration that the rate depends on the amount of water (since final deprotonation steps tend to be very fast), and others.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-03T20:16:46.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14857" PostTypeId="1" CreationDate="2014-08-03T20:29:17.403" Score="10" ViewCount="16957" Body="&lt;p&gt;I'm having difficulties understanding this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have in the problem that $K_\mathrm a$ for $\ce{HCN} = 6.2\cdot10^{-10}$ and $K_\mathrm b$ for $\ce{NH3} = 1.8\cdot10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write chemical equations to represent the dissociation of $\ce{NH4CN}$ in water, and the acid-base equilibrium equation.&lt;/li&gt;&#xA;&lt;li&gt;Will the salt react as an acid or a base?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know this is part of the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4+ + H2O &amp;lt;=&amp;gt; NH3 + H3O+}$$&#xA;$$\ce{CN- + H2O &amp;lt;=&amp;gt; HCN + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand Le Chatelier's principle and I know that the reaction will shift to the side of less energy.&lt;/p&gt;&#xA;" OwnerUserId="7321" LastEditorUserId="46306" LastEditDate="2018-02-20T10:58:55.560" LastActivityDate="2018-02-20T10:58:55.560" Title="Acid-base equilibrium of NH4CN" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14858" PostTypeId="2" ParentId="14848" CreationDate="2014-08-03T21:13:02.007" Score="1" Body="&lt;p&gt;You ask &quot;Why can't we find the exact position and velocity of a particle&quot; - this can &lt;strong&gt;not be directly answered&lt;/strong&gt; because there is no particle in this sense you use it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing &lt;strong&gt;does not have&lt;/strong&gt; an exact position and velocity - it just is &lt;strong&gt;not a particle&lt;/strong&gt; in the &quot;normal&quot; sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that things on this very small scale do not have a well defined location and velocity at the same time is just what we found out how the universe is like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, the question becomes similar to &quot;&lt;strong&gt;Why is there gravity?&lt;/strong&gt;&quot;.&lt;br&gt;&#xA;That's a very valid question too -  but a different kind of question.&lt;/p&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2014-08-03T21:13:02.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14859" PostTypeId="2" ParentId="14857" CreationDate="2014-08-03T22:23:33.613" Score="10" Body="&lt;p&gt;Fun!  weak acid versus weak base!&lt;/p&gt;&#xA;&lt;p&gt;Ammonium cyanide, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4CN}$&lt;/span&gt;, is a solid where the atoms are grouped into the same ions that are generated in solution: &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;.  Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN.  Dangerous stuff!&lt;/p&gt;&#xA;&lt;p&gt;The first part of the answer to 1. is the decomposition of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4CN}$&lt;/span&gt; into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so.  So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is &lt;strong&gt;no&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4CN}$&lt;/span&gt; as such in solution.&lt;/p&gt;&#xA;&lt;p&gt;The acid-base equilibrium equations can be the ones you wrote.  Again, you may want to add more information, like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4+ (aq) + H2O &amp;lt;=&amp;gt; NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here the equilibrium constant &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is the acid dissociation constant of the acid &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt;, the protonated form (aka conjugate acid) of ammonia.  I got its value by dividing &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm w$&lt;/span&gt;, the dissociation constant of water (&lt;span class=&quot;math-container&quot;&gt;$1.0\times10^{-14}$&lt;/span&gt;), by the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm b$&lt;/span&gt; of ammonia, given in the problem.&lt;/p&gt;&#xA;&lt;p&gt;I encourage you to add similar information for the  equilibrium involving cyanide ion.  Hint: if you keep the reaction as you wrote it, the appropriate &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is not the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm a$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; but the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm b$&lt;/span&gt; of the deprotonated form (conjugate base), &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;.&#xA;(But you can instead consider the alternate reaction &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN + H2O &amp;lt;=&amp;gt; CN- + H3O+}$&lt;/span&gt;, and then the equilibrium constant is the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm a$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;How about part 2?  The only correct answer to&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Will the salt react as an acid or a base?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;is &amp;quot;It can do either, depending on what it's reacting with&amp;quot;.  But &lt;strong&gt;maybe&lt;/strong&gt; what they mean is whether the solution of the salt is acidic or basic.  If so, one way to find out is to combine the two equilibrium equations and their &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;s.  I don't want to give problem away completely, but think of it this way:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the pH of a (say) 1 M solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt;, based on the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm a$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;(I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the pH of a 1 M solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;, based on the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm b$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; a stronger acid than &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; is a base, or is &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; a stronger base than &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; is an acid?  Based on this, which would you expect to &amp;quot;win out&amp;quot; if you mix the two in equal proportions?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;(The point here is that what you get by dissolving &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4CN}$&lt;/span&gt; is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; and ammonia.  In other words, if you mix 1 L of a 2 M solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; with 1 L of a 2 M solution of ammonia, the result is very nearly  2 L of a 1 M solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4CN}$&lt;/span&gt;.   Footnote: &amp;quot;very nearly&amp;quot; because volume is not always preserved exactly, unlike mass.)&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-10-02T11:56:39.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14860" PostTypeId="1" AcceptedAnswerId="19597" CreationDate="2014-08-03T22:27:22.890" Score="5" ViewCount="2171" Body="&lt;p&gt;Very confused by this video I stumbled across describing the cleavage of ethers with strong acids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mFEX0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can gather, the cleavage of ethers is run in &lt;em&gt;aqueous&lt;/em&gt; strong acids such as HI or HBr. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that strong acids in water &lt;em&gt;level&lt;/em&gt; to hydronium ion, why might HI or HBr be preferable to say HCl (another strong acid) in the clevage of ethers? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video from which the above picture is taken says that HI and HBr are &lt;em&gt;&quot;stronger acids&quot;&lt;/em&gt; and thus are better able to cleave the ether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds like garbage to me because if we're running this reaction in aqueous solution, there probably isn't a predominance of undissociated HI or HBr (very strong acids) reacting with the ethers. Instead, we'd only have hydronium ion and the conjugate bases of the strong acids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would the real reason HCl wouldn't work in cleaving ethers have to do with the nucleophilicity of the chloride anion in water solution (which is lower than that of either iodide or bromide ion) due to chloride anion's greater negative charge density and thus heavier solvation in water? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-03T23:30:45.247" LastActivityDate="2014-11-16T14:21:52.273" Title="Cleavage of Ethers" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14861" PostTypeId="1" AcceptedAnswerId="14867" CreationDate="2014-08-03T23:47:36.853" Score="2" ViewCount="664" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jDGf4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jDGf4.png&quot; alt=&quot;Reaction of (S)-2,2-dimethyl-3-propyloxirane with hydrochloric acid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above reaction, I am told that the nucleophilic chloride anion attacks the trisubstituted carbon because it has the most partial positive character due to its ability to stabilize positive charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that statement make any sense? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not following the train of &quot;logic,&quot; which is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The nucleophile will want to attack the location with the greatest partial positive charge. Okay, Coulomb's law. I'm good with this statement. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The carbon with the greatest partial positive charge is the one that can best stabilize positive charge. (?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alkyl groups are slightly electron-donating; the tertiary carbon will be able to better stabilize partial positive charge. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thus, the chloride anion attacks the tertiary rather than the secondary carbon. I don't see how this follows from the previous assertions. Doesn't stability correspond negatively with reactivity? If so, what is the real reason the chloride anion attacks the tertiary carbon rather than the secondary carbon? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-10-19T00:19:17.270" LastActivityDate="2018-10-19T00:19:17.270" Title="Why should epoxides be more reactive at a position where positive charge is stabilised?" Tags="&lt;organic-chemistry&gt;&lt;nucleophilic-substitution&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="14862" PostTypeId="1" AcceptedAnswerId="14864" CreationDate="2014-08-04T01:05:40.653" Score="5" ViewCount="2549" Body="&lt;p&gt;How many different possible structures (isomers) are there for the complex,&#xA;$\ce{[Mn(OH2)3(NH3)3]^{3+}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't know how to begin with this. I know the answer is 2, but i don't have the faintest idea how this is the case. Since $\ce{Mn}$ has a charge of $3+$, I know that its valence electron configuration is $3d^4$. I also know that $\ce{NH3}$ and $\ce{H2O}$ are monodentate, which means they contribute 1 lone pair. &lt;/p&gt;&#xA;" OwnerUserId="7311" LastEditorUserId="275" LastEditDate="2014-08-04T01:21:42.820" LastActivityDate="2014-08-04T16:42:30.513" Title="How many different possible structures (isomers) are there for the complex, [Mn(OH2)3(NH3)3] 3+?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14863" PostTypeId="1" CreationDate="2014-08-04T01:26:20.950" Score="0" ViewCount="682" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hFbLs.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I don't follow this reaction. This is after all an acid &lt;em&gt;catalyzed&lt;/em&gt; dehydration reaction; why then, is the final product the alkene and water? Water wasn't the original acid. Hydronium ion is the original acid. Shouldn't the final product be not water but hydronium ion (produced through deprotonation which in turn forms the alkene)?&lt;/p&gt;&#xA;&lt;p&gt;In addition, this is the book's reaction mechanism, which clearly shows the formation of hydronium ion:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZXCq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;But this isn't a one-off sort of thing with the water as a &amp;quot;product&amp;quot; of &lt;em&gt;acid-catalyzed&lt;/em&gt; dehydration.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/55cmc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, is there a reason water is listed as the product in &lt;em&gt;acid-catalyzed dehydration&lt;/em&gt;? Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-08-05T10:15:58.393" Title="Acid &quot;Catalysis&quot;" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14864" PostTypeId="2" ParentId="14862" CreationDate="2014-08-04T01:32:19.963" Score="7" Body="&lt;p&gt;Metal complexes with three each of two types of ligands $(\ce{MX3Y3})$ have two isomers: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fac%E2%80%93mer_isomerism#Cis.E2.80.93trans_isomerism_and_facial.E2.80.93meridional_isomerism&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;mer&lt;/em&gt;- and &lt;em&gt;fac&lt;/em&gt;-isomers&lt;/a&gt;. For example, for $\ce{Co(NH3)3Cl3}$ there are two isomers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Fac&lt;/em&gt; for facial, in which all three ligands of each type are clustered on a single face of the octahedron and the three other three ligands are on the opposite face.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHTK9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Mer&lt;/em&gt; for meridional, in which the three ligands of each type are arranged meridianally in a row around the metal and the other three ligands are arranged likewise on the opposite side in a perpendicular meridian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0aLUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7187" LastEditDate="2014-08-04T01:43:34.480" LastActivityDate="2014-08-04T01:43:34.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14865" PostTypeId="1" AcceptedAnswerId="14872" CreationDate="2014-08-04T02:04:07.437" Score="7" ViewCount="4024" Body="&lt;p&gt;I have a commercial lens cleaning solution. The manufacturer of the solution claims it is not only distilled water and that the formula is confidential. I suspected it was only or mostly distilled water after testing distilled water's cleaning abilities on different types of lenses with different coatings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that you can ship a sample to a lab and they will conduct a quantitative analysis of the solution using a mass spectrometer, but there are so many different kinds of mass spectrometry that I am not sure what I would want to order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of mass spectrometry (or other lab analysis service) would be best suited to discover the formula for a non-volatile, non-corrosive, liquid sample of lens cleaner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not possible, I just want to know how much of the solution is essentially water, and would like to know what kind of service would be best to determine that.&lt;/p&gt;&#xA;" OwnerUserId="7324" LastActivityDate="2018-05-26T13:44:40.023" Title="How can I reverse engineer a commercial cleaning solution's &quot;other ingredients&quot;?" Tags="&lt;analytical-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14867" PostTypeId="2" ParentId="14861" CreationDate="2014-08-04T02:41:13.530" Score="4" Body="&lt;p&gt;To the question mark in (2): this implication is simple:  if the charge were on a neighboring carbon you'd have a state of higher energy.  &quot;Can better stabilize&quot; is a synonym for &quot;gives lower overall energy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the objection in (4): the stability ranking refers to various electronic distributions in the possible transition states. In the transition state of an S&lt;sub&gt;N&lt;/sub&gt;2 reaction, and when there is a positively charged leaving group, the carbon undergoing nucleophilic substitution will bear a partial positive charge in the transition state. So, the degree to which this positive charge is stabilised settles the matter of which carbon will be attacked preferentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactivity ranking refers to the fitness of each atom in the intermediate as a site of attack by the nucleophile. The most likely target is the one which leads to the lower-energy transition state, i.e. has a lower activation energy barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no contradiction.  The carbon most able to stabilize a positive charge is also the one most likely to be the target of a nucleophilic attack (other things being equal; of course there could be factors, like sterics, changing the ranking).&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="16683" LastEditDate="2018-10-19T00:13:23.987" LastActivityDate="2018-10-19T00:13:23.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14868" PostTypeId="2" ParentId="14815" CreationDate="2014-08-04T04:37:39.897" Score="1" Body="&lt;p&gt;On remark: just because activation energy high/low it doesn't necessarily mean that reaction is slow or fast. You cannot tell this without the pre-exponential factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer: Off course, the original statement should not applied as you do it. &#xA;&quot;According to transition state theory, rates are determined by the rate-limiting step.&quot; doesn't mean automatically that all the reaction rates are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A to B conversion is slow compared to other processes, is true, but still B to C and B to D are competing processes. The original statement only means that B will not build up in bigger quantities, so during the reaction all the final products (C and D together) are approximately equal with the used A after you consider stoichimetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited for clarification (sorry, pretty new to LaTex):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should consider all the transformations as independent equations. Black magic is not involved, &quot;rate determining step&quot; and such only helps making approximations so one can solve the coupled differential equations on napkin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume a reaction system with $A \to B \to C$. The kinetics of the system can be describe by two independent equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \to B$ with a corresponding $k1$ and  $B \to C$ with a corresponding $k2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe conversions we can get different equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[A]}{dt} = -k1 [A] $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[B]}{dt} = k1 [A] - k2 [B]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[C]}{dt} = k2 [B] $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that it is already pretty messy, so assumptions like $k1 &amp;lt;&amp;lt; k2$ can help. The general way is that we say, this difference in the rate constants allows us to asume  $$ \frac{d[B]}{dt} \approx 0 $$. So we have  pipeline, and in first approximation have that much $B \to C$ conversion going as $A \to B$. I.e. the is no &quot;clogging&quot; in the middle. Not much magic here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what if we have e.g. two slow reactions as second step instead of one? Noting special:&#xA;$A\to B$ , with $k1$,&#xA;$B\to C$ , with $k2$,&#xA;$B\to D$ , with $k3$ &#xA;which translates to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[A]}{dt} = -k1 [A] $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B])$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[C]}{dt} = k2 [B] $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[D]}{dt} = k3 [B] $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the same assumption that the first step is much slower than the others than &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B]) \approx 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can try it from here and see how the rates comes out.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2014-08-04T08:14:18.633" LastActivityDate="2014-08-04T08:14:18.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14869" PostTypeId="1" AcceptedAnswerId="14871" CreationDate="2014-08-04T05:39:09.767" Score="2" ViewCount="1061" Body="&lt;p&gt;This doesn't look like a traditional inversion of configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTxia.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I dismissed the textbook author as making a typo. But then I realized that this &quot;typo&quot; was repeated throughout the solutions manual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried drawing a Sawhorse projection of the textbook's answer and my answer, and I found that both are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would my answer also be acceptable? In my answer I change the methyl from a wedge to dashes and the hydrogen from dashes to wedge and I place the SH group appropriately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4qQby.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="305" LastEditDate="2014-08-04T06:26:03.940" LastActivityDate="2014-08-04T06:26:03.940" Title="Inversion of Configuration - Opening of Epoxide" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14870" PostTypeId="2" ParentId="14848" CreationDate="2014-08-04T05:51:58.600" Score="3" Body="&lt;p&gt;It's not that you cant measure it -- it's that particles just don't have well defined positions and velocities at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I believe the uncertainty principle actually refers to the position and momentum of a wave. But I don't want to split hairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, particles, until they interact with something, appear to exist as probabilities, (or I prefer to think them as &quot;possibilities&quot;, it's more &quot;magical&quot; to me, but anyway...), and their probability functions, the areas where the particles might pop up.. for some reason or another they act like waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The think is, with a wave, how do you get the exact position? A wave can take on many forms. It can have just one peak, or it can have multiple peaks and troughs. Now, it it has just one peak, it's position is pretty well defined, at least, if you ask me. But if it has a bunch of peaks, it's position is more &quot;spread out&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if all you have is a glimpse in one instant, the more &quot;spread out&quot; the wave is, the easier to be to get an idea of where it's going. If the wave is all clumped up into one wavelet, it will be really hard to tell where it wants to go. But if it's spread out, like ripples in a pond, and you happened to get a glimpse, you can kind of tell where the wave is going, you will be able to tell that it's spreading in a certain direction, from the studying the ripples you will be able to guess it's speed, etc. But what is the exact position of the ripples? How do you define the position of .. &quot;ripples&quot;? Yeah, you can make up a suitable definition for the macroscopic world, but in the quantum case, the waves aren't actual, physical things. They are waves of &quot;probability&quot;, and the actual particle, the actual physical thing we are interested in, well it's busy chilling out somewhere in these probability waves, at least until it decides to manifest itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, the way it works out, the more spread out these probability waves are, the easier it is to guess where the particle wants to go. The more you know about where the particle wants to go, then the better you can define it's velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the probability wave is bottled up into a small spot, you have a good idea of where the particle is likely to be in that instant, but you will have no idea of where it wants to go. To know where it wants to go, the probability wave has to be smeared out. But then you lose the information of where it is, presently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's just the way it is.&lt;/p&gt;&#xA;" OwnerUserId="7326" LastActivityDate="2014-08-04T05:51:58.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14871" PostTypeId="2" ParentId="14869" CreationDate="2014-08-04T06:17:32.260" Score="1" Body="&lt;p&gt;It's slightly difficult for me to see clearly, but your drawing appears to be correct, and it corresponds to a $180^{\circ}$ rotation about the $\ce{C_1-C_2}$ bond axis relative to the textbook rendering ($\ce{C_1}$ being the carbon bearing the oxygen atom and $\ce{C_2}$ the sulfur atom, respectively). An unambiguous method of determining whether the two representations are equivalent is to assign the absolute configuration for the chiral carbon(s) in question. It may also help to use other projections (Fischer or Newman) if visualization of bond rotation and stereochemistry in the skeletal formula becomes difficult, as you did with the Sawhorse projection.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-04T06:25:03.207" LastActivityDate="2014-08-04T06:25:03.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14872" PostTypeId="2" ParentId="14865" CreationDate="2014-08-04T06:55:10.933" Score="8" Body="&lt;p&gt;Reverse engineering, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Deformulation&quot; rel=&quot;noreferrer&quot;&gt;deformulation&lt;/a&gt;, is a common industry practice. This is a very common type of request for &lt;a href=&quot;http://www.intertek.com/consumer/analysis/deformulation-product/&quot; rel=&quot;noreferrer&quot;&gt;commercial analytical laboratories&lt;/a&gt;, and is generally not cheap. You can discover just how widespread it is, and how many commercial providers there are with a quick &lt;a href=&quot;http://www.google.com#q=deformulation+services&quot; rel=&quot;noreferrer&quot;&gt;Google search&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get yourself a reasonable starting point, you should request a product MSDS from your suppplier. All suppliers should be able to provide a full MSDS, which will give an approximate ingredient listing and component concentration. Sometimes, you will find ingredients listed as &quot;Trade secret&quot; or &quot;Proprietary&quot;, but many times you will get more information. It is very likley that your lens cleaner solution is, in fact, predominantly distilled water; perhaps up to 90-95%. However it is the other minor ingredients that are the active ingredients, and may include alcohols or glycols, anti-fog and anti-static agents, anti-baceterials and preservatives. Examples of lens cleaner MSDS can be found via a &lt;a href=&quot;http://www.google.com/#q=lens+cleaner+MSDS&quot; rel=&quot;noreferrer&quot;&gt;google search&lt;/a&gt;, and this may give you a useful bit of background information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&#xA;1. all suppliers are legally bound to providing MSDSs upon request in the local idiom (official language).&#xA;2. this document is safety-based which means it will (a) provide (useful) information on possibly hazardous materials, (b) assess the hazardous component(s), their type, level and correlate to concentration.&#xA;3. websites such as state Chemical authorities (ECHA, FDA, ...) publish ratings, studies and lists (of registered/authorized/forbidden) of chemicals for use in particular applications such as food contact, medical, pharmaceutical devices/products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to determine what information you really are after, as this will determine the budget you will need to outlay. If you want to just determine whether the solution is predominantly water, you could try a few things yourself (if you are experiment-savvy). Try smelling it, as a first step; check for evaporation if you run a thin film on a glass slide, test density or melting point. These will all tell you much the same thing, which is, yes, the solution is predominantly water. A simple commercial &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_NMR&quot; rel=&quot;noreferrer&quot;&gt;1H NMR&lt;/a&gt; analysis will show you this also, and confirm water concentration up to about 99.9%. This means minor ingredients will show up, but as long as they make up about 0.1% of the ingredient composition, and they contain hydrogen (sodium bicarbonate will not show up for instance). This method is commonly used when you have a reasonable idea about the identity of the minor ingredients, but can involve extensive analysis if you really are shooting in the dark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this, if you actually want to determine the ingredients, and have no idea what they might be, you will need to head down the avenue of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_chromatography%E2%80%93mass_spectrometry&quot; rel=&quot;noreferrer&quot;&gt;GC-MS&lt;/a&gt; and/or &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_chromatography%E2%80%93mass_spectrometry&quot; rel=&quot;noreferrer&quot;&gt;LC-MS&lt;/a&gt;; a combination of chromatography to separate all of the ingredients, and then Mass Spectrometery to try to determine the mass of each component. This type of analysis is generally charged by the hour, and the more variables you are trying to evaluate, the more expensive the procedure. Generally, when analysing a competitor's products, a reasonable knowledge of the types of raw materials exists, and fingerprinting chromatography and NMR traces against raw material libraries saves a lot of time (and money). Sometimes, other techniques such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fourier_transform_infrared_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;FTIR&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;AAS&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/UV-Vis&quot; rel=&quot;noreferrer&quot;&gt;UV-Vis&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductively_coupled_plasma&quot; rel=&quot;noreferrer&quot;&gt;ICP&lt;/a&gt; may also be used, but these would really only be if the MS methods failed to give you suitable leads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial analytical labs will be able to give you a very good summary of the best techniques for your sample, but be prepared; the very first question you are likely to be asked is &quot;what is your budget?&quot;. Good luck.  &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="-1" LastEditDate="2018-05-26T13:44:40.023" LastActivityDate="2018-05-26T13:44:40.023" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14873" PostTypeId="1" CreationDate="2014-08-04T08:40:20.223" Score="1" ViewCount="66" Body="&lt;p&gt;Would I be correct in saying that the Bohr effect is ONLY related to the concentration of [H+] in the tissues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the concentration of carbon-dioxide can contribute to this concentration by producing protons (through carbonic anhydrase) but it isn't directly involved in the effect? &lt;/p&gt;&#xA;" OwnerUserId="7123" LastActivityDate="2014-08-09T14:38:28.640" Title="Bohr Effect - related to H+ only?" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14874" PostTypeId="2" ParentId="14862" CreationDate="2014-08-04T09:04:31.650" Score="2" Body="&lt;p&gt;The compound you talked about is of the type $MA_3B_3$. I will illustrate a similar example here.&#xA;There shall be 2 geometrical isomers, '&lt;strong&gt;&lt;em&gt;FACIAL&lt;/em&gt;&lt;/strong&gt;' and '&lt;strong&gt;&lt;em&gt;MERIDONAL&lt;/em&gt;&lt;/strong&gt;'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OfenJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To identify which is which, remember that in &lt;em&gt;MERIDONAL&lt;/em&gt; maximum bond&#xA;  angles between the groups is &lt;strong&gt;180&lt;/strong&gt; whereas in facial it is lesser&#xA;  than that. Also &lt;strong&gt;none&lt;/strong&gt; of them is optically active as both have a&#xA;  Plane of symmetry (&lt;em&gt;merdional&lt;/em&gt; has &lt;em&gt;diagonal POS whereas facial has&#xA;  horizontal&lt;/em&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the compound has total &lt;strong&gt;2 STRUCTURAL ISOMERS&lt;/strong&gt; (&lt;em&gt;2 GEOMETRICAL + 0 OPTICAL&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorDisplayName="user2619" LastEditDate="2014-08-04T16:42:30.513" LastActivityDate="2014-08-04T16:42:30.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14875" PostTypeId="1" CreationDate="2014-08-04T09:10:13.230" Score="5" ViewCount="5639" Body="&lt;p&gt;Specifically I live in the U.S. so I'm asking about an ordinary rock that you might find in your backyard or near a playground. Humans have discovered gun powder, petroleum, silica, iron, and many more types of useful materials by observing and experimenting with materials which lead to said discoveries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I'm walking outside and think to myself how people largely just ignore stuff outside like the rocks, dirt, grass and things like that, but I think if you knew how to separate the chemicals in these materials, then it could possibly lead to discoveries that might have otherwise been overlooked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially rocks, because they can be cracked open and looked at with a microscope. Can't these rocks be used as raw materials to engineer something? If so, shouldn't they have inherent value? If not, why? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the typical chemical elements that you will find in these types of materials, and how can you observe the chemical compositions by either isolating them or via some other physical process?&lt;/p&gt;&#xA;" OwnerUserId="7327" LastEditorUserId="8083" LastEditDate="2016-02-02T11:20:38.727" LastActivityDate="2016-02-02T11:20:38.727" Title="How can I determine the chemical composition of local rocks and minerals?" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;analytical-chemistry&gt;&lt;geochemistry&gt;&lt;rocks&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14876" PostTypeId="2" ParentId="14820" CreationDate="2014-08-04T09:19:09.623" Score="1" Body="&lt;p&gt;Generally the &lt;strong&gt;least electronegative&lt;/strong&gt; element is written &lt;strong&gt;first&lt;/strong&gt;. And while naming also, &lt;strong&gt;+ide&lt;/strong&gt; is attached as &lt;em&gt;prefix&lt;/em&gt; to the &lt;strong&gt;more&lt;/strong&gt; electronegative element.&#xA;For Ex- $ICl$ is written as &lt;strong&gt;&lt;em&gt;Iodine Chloride&lt;/em&gt;&lt;/strong&gt; not &lt;strong&gt;&lt;em&gt;Chlorine Iodide&lt;/em&gt;&lt;/strong&gt;. Similarly $OF_2$ is written as &lt;em&gt;Oxygen Difluoride&lt;/em&gt;.&#xA;Coming back to your question, Oxygen is the second most electronegative element in the periodic table after Fluorine. Therefore it is generally written after all the other elements &lt;strong&gt;(except fluorine)&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastActivityDate="2014-08-04T09:19:09.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14877" PostTypeId="2" ParentId="12427" CreationDate="2014-08-04T09:31:24.760" Score="0" Body="&lt;p&gt;The oxidation process either produces positive ions or removes negative ions from the solution at the anode (or it may change one ion to a more positive one), and the reduction process either removes positive ions or produces negative ions in the solution at the cathode.  This produces electrically charged solutions, and very quickly stops the process before a measureable number of electrons are transferred.&#xA;There must be a path for the ions to move between the two solutions in order for electrons to flow continuously through the wire.  &lt;strong&gt;This produces an &quot;ion current&quot; within the battery with cations (positively - charged ions) moving from anode to cathode, and anions (negatively - charged ions) moving from the cathode toward the anode.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8fhwd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, more the no. of salt bridges, more will be the ion current. And from the standard ohms law equation, $$V=IR$$ , V being directly proportional to I (Current), will increase if the ion current increases.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastActivityDate="2014-08-04T09:31:24.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14878" PostTypeId="2" ParentId="14836" CreationDate="2014-08-04T09:56:21.963" Score="6" Body="&lt;p&gt;After much research and work, I wrote a little explanation (not that the other answers weren't good, they just weren't well written for someone who didn't know about thermodynamics and other concepts …):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The below results were determined experimentally but this explanation gives some insight into why they are the way they are. Whilst many factors affect the probability of a reaction occurring, and thus the rate, the affect of concentration can be quite easily determined. Consider the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{a X + b Y -&amp;gt; c Z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of it occurring can be broken down into the probability of the particles reacting in a space, and the probability of them being in a space, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(reaction)=P(\ce{a X} \in \Delta V)\times P(\ce{b Y} \in \Delta V) \times P(\ce{X\bond{-&amp;gt;}Y} \in \Delta V)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wherein the probability $P$ of an atom being in certain volume $\Delta V$ is multiplied with the final probability of a collision (here displayed as $\ce{X\bond{-&amp;gt;}Y}$) happening in said volume. The probability of a molecule being in a set space can be determined from its concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=\frac nV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(\ce{1X} \in \Delta V)= C \times 6.02 \times 10^{23} \times 10^3 \frac{L}{m^3} \times \Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation assumes that $\Delta V$ is in cubic meters and concentration is in $\mathrm{mol/L}$, however this may not be the case. The main concept is that the probability of one particle being present in some area is proportional to the concentration and some scaling factor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(\ce{X} \in \Delta V)=K_S×C(\ce{X})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on basic probability, it is known that the probability of a particles being in the volume is equal to the probability of one particle being in there to the power of $a$. Thus from this, assuming the scaling constant is $K_S$, and that the probability of a reaction occurring at the given temperature in the given space if particles are present is equal to $K_R$, the probability of a reaction occurring based on concentration is as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(reaction)=K_R K_S^{a+b} [\ce{X}]^a [\ce{Y}]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a reversible reaction is considered, the overall direction can be determined by finding the ratio between the rates of the reactions in each direction. Thus the ratio of backward reaction to forward reaction for the below reaction can be identified. (Note the constants on the top and bottoms are different)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{a W + b X &amp;lt;=&amp;gt; c Y + d Z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{tot}= \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamic equilibrium is the state for a reversible reaction, in which the rate of both the forward and backward direction is equal and thus the overall change is zero. In this case, the ratio would equal one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 = \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be easier however, to simplify repetitive calculations to just the concentrations, and create a new constant from all the constants shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{K_{R\ce{B}} K_S^{c+d}}{K_{R\ce{A}} K_S^{a+b} }=\frac{[\ce{W}]^a [\ce{X}]^b}{[\ce{Y}]^c [\ce{Z}]^d }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equilibrium constant, although it does change with temperature as the average kinetic energy of particles changes. The calculations made on the concentrations come up with a value called the concentration quotient ($Q$) which is equal to the equilibrium constant when a dynamic equilibrium is reached. &lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="7475" LastEditDate="2016-12-08T21:29:18.473" LastActivityDate="2016-12-08T21:29:18.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14879" PostTypeId="2" ParentId="14815" CreationDate="2014-08-04T10:06:32.683" Score="3" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;nofollow&quot;&gt;transition state theory&lt;/a&gt; (TST, &lt;a href=&quot;http://goldbook.iupac.org/T06470.html&quot; rel=&quot;nofollow&quot;&gt;goldbook&lt;/a&gt;) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us&#xA;\begin{align}\ce{&#xA;A &amp;amp;&amp;lt;=&amp;gt; [AB]^{$\ddagger$} -&amp;gt; B\\&#xA;B &amp;amp;&amp;lt;=&amp;gt; [AB]^{$\ddagger$} -&amp;gt; A\\&#xA;A &amp;amp;&amp;lt;=&amp;gt; B,&#xA;}\end{align}&#xA;and in addition to this, also&#xA;\begin{align}\ce{&#xA;B &amp;amp;&amp;lt;=&amp;gt; [BC]^{$\ddagger$} -&amp;gt; C &amp;amp; B &amp;amp;&amp;lt;=&amp;gt; [BD]^{$\ddagger$} -&amp;gt; D\\&#xA;C &amp;amp;&amp;lt;=&amp;gt; [BC]^{$\ddagger$} -&amp;gt; B &amp;amp; D &amp;amp;&amp;lt;=&amp;gt; [BD]^{$\ddagger$} -&amp;gt; B\\&#xA;B &amp;amp;&amp;lt;=&amp;gt; C                      &amp;amp; B &amp;amp;&amp;lt;=&amp;gt; D,&#xA;}\end{align}&#xA;as well as&#xA;\begin{align}\ce{&#xA;C &amp;lt;=&amp;gt; [BC]^{$\ddagger$} -&amp;gt; &amp;amp; B &amp;lt;=&amp;gt; [BD]^{$\ddagger$} -&amp;gt; D\\&#xA;D &amp;lt;=&amp;gt; [BD]^{$\ddagger$} -&amp;gt; &amp;amp; B &amp;lt;=&amp;gt; [BC]^{$\ddagger$} -&amp;gt; C\\&#xA;C &amp;lt;=&amp;gt; &amp;amp; B &amp;lt;=&amp;gt; D,&#xA;}\end{align}&#xA;and &#xA;$$\ce{&#xA;C &amp;lt;=&amp;gt; A &amp;lt;=&amp;gt; D\\&#xA;[AB]^{$\ddagger$} &amp;lt;=&amp;gt; [BC]^{$\ddagger$} &amp;lt;=&amp;gt; [BD]^{$\ddagger$}&#xA;}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use TST to estimate rate constants.&#xA;$$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence:&#xA;$$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} =  \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="24052" LastEditDate="2016-02-16T23:04:49.107" LastActivityDate="2016-02-16T23:04:49.107" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14880" PostTypeId="2" ParentId="14875" CreationDate="2014-08-04T10:50:01.820" Score="7" Body="&lt;p&gt;Let's talk about rocks. The art and science of getting useful things from rocks is one of the oldest in human civilization. All of the example products you mention (gunpowder, petroleum, silica, and iron) were discovered or developed from experimentation on rocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ancient world, composition was determined by &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_analysis#Qualitative_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;qualitative analysis&lt;/a&gt;, or a series of chemical tests and determination of important physical properties. For example, if a lump of shiny yellow stuff was soft, melted easily, did not lose its lustre in air or water, did not dissolve in vitriol (sulfuric acid) or aqua fortis (nitric acid), but did dissolve in aqua regia (made then by dissolving sal ammoniac (ammonium chloride) in aqua fortis), then we probably have gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the modern age, the field of &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;analytical chemistry&lt;/a&gt; was developed to answer the very questions you asked: What is this stuff? What are its constituents? How can we separate them? How do we determine that we successfully separated/purified them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To continue the rock example, the modern technique of &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_fluorescence&quot; rel=&quot;nofollow noreferrer&quot;&gt;x-ray fluorescence&lt;/a&gt; is used to determine the elemental compositions of rocks (and other things, like ceramic fragments found at archaeological sites). In this technique, high frequency x-rays are used to excite the core electrons in the atoms of the sample. When these core electrons relax, they spit out x-rays of characteristic frequency/wavelength that is specific to the element. These emitted x-rays are collectively measured and converted into a spectrum like the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ONcWD.jpg&quot; alt=&quot;example x-ray fluorescent spectrum&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spectrum indicates that many elements will have more than one spectral line because there is more than one electronic transition that is possible. It also demonstrates the clear resolution of one element from another. After some maths (and better yet a calibration), you can use the heights of these peaks to determine the specific ratios of elements in your sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not everyone is ignorant of the utility of natural objects. Most of us who become scientists do so to study the natural world and learn from it. For example, organic and medicinal chemists do not design new drug molecules from scratch. Rather, they take cues from nature. A folk legend about the curative properties of the leaves of a plant that grows along the riverside might turn in to an effort to extract and analyze the phytochemicals in those leaves, including throwing those compounds at bacteria, viruses, cancer, etc. Any that look promising are studied further. &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taxol (paclitaxel)&lt;/a&gt; is a successful chemotherapeutic agent that was isolated from the bark of the pacific yew. It is now mass-produced by a &lt;a href=&quot;http://en.wikipedia.org/wiki/Taxol#Commercially_viable_paclitaxel_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;semi-synthetic pathway&lt;/a&gt; which isolates a late-stage precursor from the needles of the European yew. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are specifically interested in rocks, then you should study &lt;a href=&quot;http://en.wikipedia.org/wiki/Geoscience&quot; rel=&quot;nofollow noreferrer&quot;&gt;geoscience&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-04T10:50:01.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14881" PostTypeId="1" AcceptedAnswerId="14882" CreationDate="2014-08-04T12:23:28.883" Score="1" ViewCount="194" Body="&lt;p&gt;From 1,5-diphenylpentan-3-ol (above), how to add the two hydroxy groups in red (below)?&lt;br&gt;&#xA;Is there any proven successful synthesis route to 1,5-diphenylpentane-1,3,5-triol with higher yield (40 % or above)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Owk4H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorDisplayName="user7951" LastEditDate="2016-11-05T17:43:54.963" LastActivityDate="2016-11-05T17:43:54.963" Title="How to add two hydroxy groups to 1,5-diphenylpentan-3-ol with as high as possible yield?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14882" PostTypeId="2" ParentId="14881" CreationDate="2014-08-04T14:24:59.263" Score="6" Body="&lt;p&gt;A few possibilities for direct hydroxylation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{SeO2}$ immediately comes to mind as one potential option; it is well known to selectively oxidize allylic positions to yield alcohols and carbonyl compounds, and there appear to be references suggesting it reacts at the benzylic position as well.&lt;/li&gt;&#xA;&lt;li&gt;In &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt;, March notes that cerium(IV) triflate &quot;converts benzylic arenes to benzylic alcohols, although the major product is the ketone when &gt; 15% of water is present.&quot; Ref: Laali, K.K.; Herbert, M.; Cushnyr, B.; Bhatt, A.; Terrano, D. &lt;em&gt;J. Chem. Soc., Perkin Trans. 1&lt;/em&gt; &lt;strong&gt;2001&lt;/strong&gt;, 578.&lt;/li&gt;&#xA;&lt;li&gt;Intriguingly, March also reports a variety of oxidations accomplished enzymatically. I have no experience and very limited knowledge of these reactions, so I'll just give the relevant primary literature references:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Adam, W.; Lukacs, Z.; Saha-Moller, C.R.; Weckerle, B.; Schreier, P. Eur. &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;2000&lt;/strong&gt;, 2923.&lt;/li&gt;&#xA;&lt;li&gt;Hamada, H.; Tanaka, T.; Furuya, T.; Takahata, H.; Nemoto, H. &lt;em&gt;Tetrahedron Lett.&lt;/em&gt; &lt;strong&gt;2001&lt;/strong&gt;, 42, 909.&lt;/li&gt;&#xA;&lt;li&gt;Adam, W.; Lukacs, Z.; Harmsen, D.; Saha-Moller, C.R.; Schreier, P. &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;2000&lt;/strong&gt;, 65, 878.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If those approaches aren't appealing, there are various less direct routes. Radical bromination (with, e.g., NBS under photochemical conditions), followed either by direct substitution with hydroxide, or alternatively elimination followed by subsequent acid-catalyzed hydration, would likely work well. If you need stereoselectivity, there are are numerous epoxidation reactions (that is, after bromination and subsequent elimination) that would likely be appropriate (Sharpless, Jacobsen, Shi, and certainly numerous others). Conveniently, your molecule would already have the allylic alcohol necessary for Sharpless asymmetric epoxidation, should you want to go that route. In any case, it should be possible to open the epoxides at the less hindered carbon with a suitable hydride donor (e.g., LAH) to yield the alcohols after work-up.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-08-04T14:24:59.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14883" PostTypeId="1" CreationDate="2014-08-04T14:49:27.127" Score="4" ViewCount="3256" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r03A4.jpg&quot; alt=&quot;Nitration of Benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is mechanism of Nitration of Benzene. In the first step, Nitric Acid (pKa = -1.4) accepts a proton from sulfuric acid which is stronger acid (pKa = -9). According to Brønsted-Lowry theory we consider acids dissociate increasing proton concentration and bases dissociate increasing hydroxy concentration. On the other hand bases also accept proton from acids. Therefore should we accept nitric acid as a base in this case ? If not, why does an acid accept a proton from another &quot;stronger&quot; acid ? &lt;/p&gt;&#xA;" OwnerUserId="792" LastActivityDate="2018-07-31T03:51:43.923" Title="Why Does A Brønsted–Lowry Acid Accept Proton from Stronger Acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14884" PostTypeId="2" ParentId="14883" CreationDate="2014-08-04T14:58:52.083" Score="4" Body="&lt;p&gt;Nitric acid is definitely a strong acid, but that does not exclude the possibility of protonating it with a stronger acid. It can be effectively protonated by a stronger acid such as sulfuric acid. Simply because something has &quot;acid&quot; in its name does not mean it has to dissociate into protons and its conjugate base, or protonate something else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO_3 + H2SO4 &amp;lt;=&amp;gt;H2NO3^+ +HSO4-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that even strong acids in their pure forms - i.e. liquid nitric acid or liquid sulfuric acid - can and do auto-ionize to a fair extent. So if nitric acid can react with itself, of course it can react with a stronger acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="15410" LastEditDate="2018-07-31T03:51:43.923" LastActivityDate="2018-07-31T03:51:43.923" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="14885" PostTypeId="1" CreationDate="2014-08-04T16:19:55.703" Score="2" ViewCount="1707" Body="&lt;p&gt;Is it possible to predict a double displacement reaction between two ions where the products remain in aqueous solution? I see how you can predict precipitation reactions if you know whether the product(s) is/are soluble, but that doesn't apply to all possible double displacement reactions.&lt;/p&gt;&#xA;" OwnerUserId="7334" LastEditorUserId="11367" LastEditDate="2019-06-24T20:08:12.813" LastActivityDate="2019-06-24T20:08:12.813" Title="How can you predict a double displacement reaction with no precipitate?" Tags="&lt;aqueous-solution&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14886" PostTypeId="1" CreationDate="2014-08-04T16:20:20.513" Score="8" ViewCount="32970" Body="&lt;p&gt;How to determine order of nucleophilicity among commonly used nucleophiles? For example, in Finklestein reaction, $\ce{NaI}$ (reagent) - taken in acetone (solvent) - displaces $\ce{SH-}$ ion from alkyl group via S&lt;sub&gt;N&lt;/sub&gt;2 mechanism to form an alkyl iodide. But in a separate question we may be asked to write the product formed when $\ce{NaSH}$ reacts with an alkyl iodide in the presence of acetone. Since I do not know which of the two is a stronger nucleophile, I will displace $\ce{I-}$ and form $\ce{alkyl-SH}$. But definitely only one of the reactions shall occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is that is there an order of strength of nucleophilicity (like a metal reactivity series)? I am preparing for JEE Advanced 2015 (college entrance exam in India), so I am expecting answers within its scope.&lt;/p&gt;&#xA;" OwnerUserId="7335" LastEditorUserId="5026" LastEditDate="2018-05-06T13:50:22.087" LastActivityDate="2018-10-20T07:19:21.337" Title="How to determine order of nucleophilicity among commonly used nucleophiles?" Tags="&lt;organic-chemistry&gt;&lt;nucleophilicity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14887" PostTypeId="2" ParentId="14886" CreationDate="2014-08-04T16:25:29.610" Score="3" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; nucleophilicity is different from leaving group strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, both are possible it depends what drives the reaction. Though it usually is not possible to displace $\ce{SH^-}$ ion, but precipitation drives it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Alkyl_Halides/Properties_of_Alkyl_Halides/Haloalkanes/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Nucleophilicity_and_Solvent_Effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;UCDavis article on nucleophilicity and solvent effects&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5026" LastEditDate="2018-05-06T13:57:14.170" LastActivityDate="2018-05-06T13:57:14.170" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14888" PostTypeId="2" ParentId="14886" CreationDate="2014-08-04T16:52:08.333" Score="3" Body="&lt;p&gt;This article most probably clear all your doubts on the topic: &lt;a href=&quot;https://www.youtube.com/watch?v=pzHEtjAUVtY&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Types Of Reactions with Alkyl Halide&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a supplement :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This would clear your concept of strength of nucleophile- &lt;a href=&quot;http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;What Makes a good nucleophile&lt;/strong&gt;&#xA;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;And this will teach you how to compare nucleophilicity &lt;a href=&quot;http://www.khanacademy.org/science/organic-chemistry/substitution-elimination-reactions/nucleophilicity-basicity-sal/v/nucleophilicity--nucleophile-strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Comparison of nucleophilicity&lt;/strong&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would recommend you go through these links and clarify your doubts.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="5026" LastEditorDisplayName="user2619" LastEditDate="2018-05-06T13:56:42.513" LastActivityDate="2018-05-06T13:56:42.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14889" PostTypeId="1" CreationDate="2014-08-04T16:52:52.400" Score="3" ViewCount="5159" Body="&lt;p&gt;How is it possible for &lt;em&gt;m&lt;/em&gt;-cresol (3-methylphenol) to be synthesized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methylation of phenol would lead to the &lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-isomers instead.&lt;/p&gt;&#xA;" OwnerUserId="7030" LastEditorUserId="16683" LastEditDate="2017-04-08T18:10:47.653" LastActivityDate="2017-04-08T18:11:34.277" Title="How can meta-cresol be made, given that the hydroxyl group is ortho,para-directing?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14890" PostTypeId="2" ParentId="14883" CreationDate="2014-08-04T17:01:28.483" Score="3" Body="&lt;p&gt;Echoing Dissenter's comment, don't be hung up on the name &quot;nitric acid&quot; to mean that it can only release a proton. It's typical reactivity is as an acid, which is why it gained that name. Most compounds will change their reactivity under different reaction conditions. Take water for example. Under acidic conditions, such as when reacted with HCl, it will act as a base, becoming protonated to form hydronium. Under basic conditions, such as when reacted with butyl lithium, water will serve as an acid, being deprotonated to form hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There also seems to be some confusion about acid-base concepts. &lt;em&gt;Arrhenius&lt;/em&gt; theory considers compounds that increase hydrogen ion concentration to be acids and compounds that increase hydroxide concentration to be bases. That is a very limited definition. In the reaction between nitric acid and sulfuric acid, the Brønsted-Lowry concept is more useful. &lt;em&gt;Brønsted-Lowry&lt;/em&gt; considers any compound that donates a proton (hydrogen ion) to be an acid (sulfuric acid here) and any compound that accepts a proton (nitric acid here) to be a base.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-04T17:01:28.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14891" PostTypeId="2" ParentId="14881" CreationDate="2014-08-04T17:17:17.790" Score="3" Body="&lt;p&gt;Actually, hell with that. I recall that organolithium compounds may be carefully oxidated at very low temperatures by different things, namely halogens, sulfur and - tadam! - molecular oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BuLi}$ should selectively deprotonate each benzyl position once, and then the lithium compound may be very carefully oxidized with molecular oxygen at very low temperature followed by water hydrolysis into peroxide. Reduction of the peroxide should give the target compound, but it highly preferable to do it in basic conditions and avoid contact of target compound with acids as it looks to be able to dehydrate in acidic conditions. Another possible direction is to transform organolithium compound into boric acid and then oxidate it with hydrogen peroxide in basic conditions.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-04T18:52:31.227" LastActivityDate="2014-08-04T18:52:31.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14892" PostTypeId="2" ParentId="14889" CreationDate="2014-08-04T17:36:19.947" Score="-1" Body="&lt;p&gt;From the comments above, its clear you are asking how m-cresol is prepared.&#xA;I know one method, which involves diazotization of m-toluidine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way we should make m-toluidine first but it &lt;strong&gt;making m-toluidine is more difficult&lt;/strong&gt; than m-cresol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;m-toluidine is prepared by m-nitrotoluene&lt;/strong&gt; but unfortunately m-nitro toulene is minor isomer in nitration of toluene and this is really so difficult to separate it.&#xA;But you can prepare M-cresol thorugh this process, though the &lt;strong&gt;yield of the reaction will be small.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I will be looking for more answers on this question if someone knows another way of preparing this particular compound.&#xA;Hope this helps.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="11367" LastEditDate="2017-04-08T18:11:34.277" LastActivityDate="2017-04-08T18:11:34.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14893" PostTypeId="2" ParentId="14885" CreationDate="2014-08-04T18:52:44.503" Score="3" Body="&lt;p&gt;If there is no precipitate, then there is nothing to drive the reaction. Both possible products would be soluble, and so there is no net reaction (it's just the reactant ions in solution).&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-04T23:22:25.667" LastActivityDate="2014-08-04T23:22:25.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14894" PostTypeId="2" ParentId="9328" CreationDate="2014-08-04T20:03:23.327" Score="2" Body="&lt;p&gt;It was recently pointed out to me that there's a way to do this in a principled manner. Let's assume the lamp is $4\ \mathrm W$ strong and distributes its power over 10 plates ($1000\ \mathrm{mm^2}$). The energy of a $254\ \mathrm{nm}$ photon is $7.8 \times 10^{-19}\ \mathrm J$. If the lamp is at $4\ \mathrm W$, it's putting out $4\ \mathrm{J/s}$ of energy. This corresponds to $$ \frac{4\ \mathrm J}{\mathrm s}\times\frac{\text{photon}}{7.8 \times 10^{-19}\ \mathrm J}  = 5.11 \times 10^{18}$$ photons per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the absorbance of the initiator is 0.0001, we have that the absorbed fraction of the photons is: $$ \% Abs = 1 - \%T = 1 -  \frac{I_1}{I_0}  = 1 - 10^{-A} = 2 \times 10^{-4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have $5.11 \times 10^{18}$ photons per second are incoming, roughly $1 \times 10^{15}$ of them will be absorbed, which should be more than enough to initiate polymerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a very, very rough calculation, but I thought it was cool that you could, in theory, calculate that this experiment should work (which it did!)&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorDisplayName="user7951" LastEditDate="2016-11-05T17:48:02.133" LastActivityDate="2016-11-05T17:48:02.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14895" PostTypeId="1" AcceptedAnswerId="14901" CreationDate="2014-08-04T20:28:39.093" Score="0" ViewCount="226" Body="&lt;p&gt;I am told that chromic acid is exceedingly unstable and that it cannot be purchased as a pure reagent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't find any mentions of stability or lack thereof on Wikipedia, and hey, if I wanted to, I could purchase it here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grainger.com/product/LABCHEM-Chromic-Acid-8FV43&quot; rel=&quot;nofollow&quot;&gt;http://www.grainger.com/product/LABCHEM-Chromic-Acid-8FV43&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the deal with having to make chromic acid in situ? Does this have to do with chromic acid's stability? Or something else? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: oops I realized that the linked reagent for purchase is just chromic trioxide, which can be used to make chromic acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-02T16:37:08.390" LastActivityDate="2015-10-02T16:37:08.390" Title="Stability of Chromic Acid" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14896" PostTypeId="2" ParentId="14881" CreationDate="2014-08-04T20:29:30.557" Score="1" Body="&lt;p&gt;It's worth trying Greg's suggestion to do benzylic oxidation or bromination/hydrolysis. If it works it's cool, but a lot of times trying to change 2 functionalities of a molecule in one go has poor results with poor yields and formation of mixtures of products. Additionally you might find that you will need to protect your alcohol first (that's more possible with permeakra's enolate oxidation approach).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach could be to start from acetone or an acetone equivalent (like its silyl enol ether) and do a double condensation with benzaldehyde. I'm sure you can find similar reactions in the literature.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-08-04T20:29:30.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14897" PostTypeId="2" ParentId="14883" CreationDate="2014-08-04T21:40:56.837" Score="2" Body="&lt;p&gt;The first step of the mechanism would not, on its own, happen to a large extent - protonating nitric acid is unfavorable, so the equilibrium would lie far to the left. However, the second step of the mechanism is highly favorable (entropically), so it quickly consumes what little H&lt;sub&gt;2&lt;/sub&gt;NO&lt;sub&gt;3&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; is produced by the first step. This drives the overall reaction forward by Le Chatelier's principle&lt;/p&gt;&#xA;" OwnerUserId="7337" LastActivityDate="2014-08-04T21:40:56.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14898" PostTypeId="2" ParentId="14853" CreationDate="2014-08-04T22:27:33.587" Score="1" Body="&lt;p&gt;Hydrogen is very easy to leak. Any amount of hydrogen leak is dangerous. It's highly flammable, so if there's fire, naked light or electric spark nearby, it'll cause explosion and of course be dangerous for traffic.&#xA;By the way, hydrogen leak will eventually float up and deplete ozone layer. &#xA;Based on these facts, hydrogen isn't a promising fuel to be used in daily life.&lt;/p&gt;&#xA;" OwnerUserId="7158" LastActivityDate="2014-08-04T22:27:33.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14899" PostTypeId="1" AcceptedAnswerId="14928" CreationDate="2014-08-04T22:40:16.137" Score="3" ViewCount="996" Body="&lt;p&gt;I'd like to know whether the activation energy in Arrhenius equation is the enthalpy difference between the transition state and the reactant. If it's not enthalpy, is it Gibbs energy or something else?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastEditorUserId="9961" LastEditDate="2015-10-23T17:55:45.570" LastActivityDate="2019-08-11T04:38:44.300" Title="Is the E in Arrhenius equation the change of enthalpy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14900" PostTypeId="2" ParentId="4515" CreationDate="2014-08-04T23:27:35.357" Score="1" Body="&lt;p&gt;High concentrations of ethanol might work well if you just want to put it on skin. Ethanol should be safe, it's the main ingredient in those hand sanitizers. &lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-08-04T23:27:35.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14901" PostTypeId="2" ParentId="14895" CreationDate="2014-08-04T23:43:09.793" Score="2" Body="&lt;p&gt;&quot;chromic acid&quot; is also a nickname for chromic trioxide.  It is stable enough to be sold commercially.  Chromic acid in the strict sense ($\ce{H2CrO4}$) is not sold commercially, I think.  For most purposes what you're looking for is Cr(VI) plus acidity, so the usual practice of adding concentrated sulfuric acid to sodium or potassium dichromate (which have multiple uses and can be found in every lab) is preferable to having to buy another reagent.&lt;/p&gt;&#xA;" OwnerUserId="7187" LastActivityDate="2014-08-04T23:43:09.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14902" PostTypeId="1" CreationDate="2014-08-05T00:08:03.513" Score="1" ViewCount="49" Body="&lt;p&gt;What are the fundamental thermodynamic &amp;amp; transport properties that the discharge cycle of a rechargeable battery with an intercalating cathode (like Li-ion) should depend upon? The few that I could think of are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;transport properties of the ion through electrolye &amp;amp; cathode material.&lt;/li&gt;&#xA;&lt;li&gt;stress/strain properties of the cathode material due to intercalation.&lt;/li&gt;&#xA;&lt;li&gt;chemical interaction of the solvent &amp;amp; ions with the cathode material.&lt;/li&gt;&#xA;&lt;li&gt;electrochemical potential difference between anode &amp;amp; cathode.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is what formalism relates all these properties systematically to the experimentally observed discharge curve such as voltage as a function of capacity, repeatability of discharge curves etc.&lt;/p&gt;&#xA;" OwnerUserId="4941" LastActivityDate="2014-08-05T00:08:03.513" Title="Relating Thermodynamic &amp; Transport Properties to Electrochemical discharge curve" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14903" PostTypeId="2" ParentId="14899" CreationDate="2014-08-05T00:25:07.530" Score="3" Body="&lt;p&gt;You can understand it as a difference in the potential energies of the reactants and the transition state on the potential energy surface of the reaction. Worth of note is the fact that the energy of activation should be treated more as an experimental parameter than a &quot;real&quot; value - for example you can write the Arrhenius equation for a complex reaction mechanistically composed of a few elementary ones - the E won't be a true barrier height then, more of a combination of the barriers for all elementary reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the Eyring equation as well, that one uses the Gibbs energy change. &lt;/p&gt;&#xA;" OwnerUserId="7338" LastActivityDate="2014-08-05T00:25:07.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14906" PostTypeId="1" CreationDate="2014-08-05T03:26:54.483" Score="0" ViewCount="102" Body="&lt;p&gt;I am a chemistry super noob and I was watching a MIT video where the professor states that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Certain molecules become charged when they come in contact with water in the natural state&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give me an example of one or more such molecules ?&lt;/p&gt;&#xA;" OwnerUserId="7339" LastActivityDate="2014-08-05T14:36:59.180" Title="Molecules that get charged in Water" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14907" PostTypeId="2" ParentId="14906" CreationDate="2014-08-05T04:04:19.850" Score="1" Body="&lt;p&gt;It depends on the context how he meant it. I could imagine two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Certain materials can produce ions when they dissolved in water. A typical example is table salt, NaCl. Table salt is solid form has a lattice that contains Na+ and Cl+ ions, and when you put it in water, the water molecules can help releasing these ions into the solution, where you now can find the ions. Note, that in this case the ions are atomic entities but molecules with charge are also ions, like $\ce{MnO4-}$ or $\ce{SO}_4^{2-}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another scenario when the charged entities are formed as a reaction with water by e.g. giving an electron (oxidation) to water molecules or to other dissolved components e.g. oxygen in water. As you can imagine, it happens generally with very reactive materials (materials that are very sensitive to moisture), but essentially the rusting of iron is a similar reaction.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-05T04:04:19.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14908" PostTypeId="2" ParentId="14906" CreationDate="2014-08-05T04:13:28.090" Score="1" Body="&lt;p&gt;Terrible phrasing. Molecules by definition have no &lt;em&gt;overall&lt;/em&gt; charge. Ions have overall charge. Not molecules. Molecules becoming charged become ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get into the nitty gritty you can try looking up more information about &lt;em&gt;zwitterions&lt;/em&gt;, which are actually &lt;strong&gt;not&lt;/strong&gt; ions in the strict sense as they possess no &lt;em&gt;net&lt;/em&gt; charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the IUPAC definition of &lt;strong&gt;molecule&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;An electrically neutral entity&lt;/strong&gt; consisting of more than one atom (n &gt; 1). Rigorously, a molecule, in which  n &gt; 1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the IUPAC definition of &lt;strong&gt;ion&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An atomic or molecular particle having a &lt;strong&gt;net electric charge&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the IUPAC definition of &lt;strong&gt;zwitterion&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Neutral compounds&lt;/strong&gt; having formal unit electrical charges of opposite&#xA;  sign. Some chemists restrict the term to compounds with the charges on&#xA;  non-adjacent atoms. Sometimes referred to as inner salts, dipolar ions&#xA;  (a misnomer). E.g. H3N+CH2C(=O)O− ammonioacetate (glycine),&#xA;  (CH3)3N+–O− trimethylamine oxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/M04002.html&quot; rel=&quot;nofollow&quot;&gt;http://goldbook.iupac.org/M04002.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-05T14:36:59.180" LastActivityDate="2014-08-05T14:36:59.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14910" PostTypeId="1" AcceptedAnswerId="14968" CreationDate="2014-08-05T07:00:35.500" Score="7" ViewCount="7299" Body="&lt;p&gt;In a galvanic cell with Cu and Zn electrodes, I was taught that each electrode is immersed in a solution containing it's own positive ions. I understand why you need copper ions, so they can gain electrons, but why do you need zinc ions - the reaction is going to produce zinc ions anyway, and the negative ions from the salt bridge will migrate to neutralize them, so I don't see why you need zinc ions at the first place. Is there anything I'm missing in how the cell works?&lt;/p&gt;&#xA;" OwnerUserId="7344" LastEditorUserId="7187" LastEditDate="2014-08-06T20:04:53.633" LastActivityDate="2015-10-07T13:45:13.003" Title="Why put zinc solution in a galvanic cell?" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14911" PostTypeId="2" ParentId="14846" CreationDate="2014-08-05T07:09:42.100" Score="7" Body="&lt;p&gt;The cyclopropenyl (1) cation is believed to be aromatic as it demonstrates abnormal stability and NMR-shifts typical for aromatic compounds. Moreover, there are a good number of derived systems, where aromatic conjugation is maintained despite extra methylene insertions into the ring, such as is typical for the nornbornadienyl (2) cation. Some of these cations can be isolated in salts or observed via NMR in solution. Signs of aromaticity are found also for the cyclooctatetraene dianion and the cyclononatetraene dianion, both produced by reduction of the corresponding alkene with metallic sodium. Remember, please, that even though all systems noted above are 'aromatic' in some sense and are abnormally stable for their family of species, they are still very reactive, because they all belong to a very reactive family of species. While the neutral boron analog of the cyclopropenyl cation has not been isolated to my knowledege, a similar structure was found in &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.201402815/abstract&quot; rel=&quot;noreferrer&quot;&gt;organometalic compounds&lt;/a&gt;. Aromaticity of some ring systems may result in some weird properties of some compounds, like squaric acid (abnormally strong for ketoenol and a stable enol), high polarity and basicity of cyclononatrienol and cyclopropenone and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) - cyclopropenyl cation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D52iy.png&quot; alt=&quot;cyclopropenyl cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) nornbornadienyl cation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BFHAg.png&quot; alt=&quot;nornbornadienyl cation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="10185" LastEditDate="2015-06-26T14:36:13.447" LastActivityDate="2015-06-26T14:36:13.447" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14912" PostTypeId="2" ParentId="9085" CreationDate="2014-08-05T08:56:34.517" Score="2" Body="&lt;p&gt;Alcohols are weak acids and donate protons to the strong bases forming alkoxide. &#xA;Alkoxide ion is a strong conjugate base which shows that alcohols are weak acid but alcohols are strong lewis bases and donate electrons to hydrogen in reaction with HX.&lt;/p&gt;&#xA;" OwnerUserId="7345" LastEditorUserId="5531" LastEditDate="2014-08-05T16:10:05.493" LastActivityDate="2014-08-05T16:10:05.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14913" PostTypeId="1" AcceptedAnswerId="14919" CreationDate="2014-08-05T09:19:05.327" Score="3" ViewCount="473" Body="&lt;p&gt;I think this &lt;a href=&quot;http://en.wikipedia.org/wiki/Yttria-stabilized_zirconia&quot; rel=&quot;nofollow&quot;&gt;picture from wikipedia&lt;/a&gt; is wrong because I count:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\left.&#xA;\begin{array}{ccccr}&#xA;1&amp;amp;×&amp;amp;1 &amp;amp;= &amp;amp;\ce{1~Y} \\&#xA;6&amp;amp;×&amp;amp;\frac{1}{8} &amp;amp;= &amp;amp;\ce{\frac{3}{4}O}&#xA;\end{array}&#xA;\right\} \text{ gives }\ce{Y4O3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would be the correct elementary cell of $\ce{Y2O3}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I didn't found any correct elementary cell using google)&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="4945" LastEditDate="2014-08-05T13:38:40.623" LastActivityDate="2014-08-05T13:38:40.623" Title="Elementar cell of Y2O3" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14915" PostTypeId="2" ParentId="9085" CreationDate="2014-08-05T09:38:23.543" Score="4" Body="&lt;p&gt;Alcohols can act as both.  Many reactions of alcohols are believed to start with their protonations or deprotonation. However, isolating salts where they act as bases is much harder, while alcoholates are quite easy both to make and isolate. Protonated alcohols are very active alkylating agents and if fitting $\beta$-hydrogen is present, are prone to beta-elimination.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-05T09:38:23.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14916" PostTypeId="2" ParentId="14885" CreationDate="2014-08-05T09:39:15.433" Score="-1" Body="&lt;p&gt;Prediction, you say, is possible if you know the reactivity series like &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for an elemtary level.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-05T09:39:15.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14917" PostTypeId="2" ParentId="14863" CreationDate="2014-08-05T10:15:58.393" Score="1" Body="&lt;p&gt;Consider systematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism is an $\bf E_1$ reaction in which the substrate is a protonated alcohol.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step I&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this step, an acid–base reaction, a proton is rapidly transferred from the acid to one of the unshared electron pairs of the alcohol. In dilute sulfuric acid the acid is a hydronium ion; in concentrated sulfuric acid the initial proton donor is sulfuric acid itself. The leaving group is a molecule of water.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step II&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The carbon–oxygen bond breaks heterolytically. The bonding electrons depart with the&#xA;water molecule and leave behind a carbocation. The carbocation is, of course, highly reactive because the central carbon atom has only six electrons in its valence level, not eight.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Step III&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Finally, in step 3, a water molecule removes a proton from the $\beta$ carbon of the carbocation. The result is the formation of a hydronium ion and an&#xA;alkene.&#xA;In step 3, also an acid–base reaction, any one of the nine protons available at the three&#xA;methyl groups can be transferred to a molecule of water. The electron pair left behind when&#xA;a proton is removed becomes the second bond of the double bond of the alkene. Notice&#xA;that this step restores an octet of electrons to the central carbon atom. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Though water gives proton in step one and other molecule of water takes proton in last step, it is usually written as a produce of alkene and water until a mechanism is specifically mentioned.This is also reminding of the catalytic nature of water in this reaction.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-05T10:15:58.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14919" PostTypeId="2" ParentId="14913" CreationDate="2014-08-05T10:55:04.257" Score="3" Body="&lt;p&gt;The picture in Wikipedia doesn't give you the primitive cell. In Yttria, $\ce{Y2O3}$, 8 metal ions are in the special positions $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{4}$; the remaining 24 metals occupy the sites $u$, $0$, $\frac{1}{4}$. The 48 oxygen ions are in general positions $x$, $y$, $z$, and are arranged in distorted octahedra around the metal ions, the metal-oxygen bonding distances being unequal. So, the stoichiometry is $\ce{Y32O48}$ which gets simplified to $\ce{Y2O3}$. This structure is generally referred to as the &lt;em&gt;C&lt;/em&gt;-type metal oxide structure. All the sesquioxides of the rare earths belong to this system, as do $\ce{Ga2O3}$, $\ce{In2O3}$, $\ce{Tl2O3}$ and the mineral bixbyite, $\ce{(Fe, Mn)2O3}$. The primitive cell looks like this (oxygen ions are red, metal ions are gray)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bcWSO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tl2O3#mediaviewer/File:Tl2O3structure.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; another view which might show more easily that indeed the positions of all the 48 oxygen atoms are different in such a way that no smaller primitive cell can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M. G. Paton, E. N. Maslen, &lt;em&gt;Acta Cryst.&lt;/em&gt; &lt;strong&gt;1965&lt;/strong&gt;, &lt;em&gt;19&lt;/em&gt;, 307-310&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-05T11:11:38.337" LastActivityDate="2014-08-05T11:11:38.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14920" PostTypeId="1" CreationDate="2014-08-05T11:46:18.920" Score="2" ViewCount="489" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The equation :&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH4Cl + AgNO3 -&amp;gt; NH4NO3 + AgCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that oxidation number of &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4Cl}$&lt;/span&gt; is -3 and in &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; is +5.&lt;/p&gt;&#xA;&lt;p&gt;But what I am confused about, is the oxidation number of &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt; in the product &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4NO3}$&lt;/span&gt;. Should I consider &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; in this compound or should I get its oxidation number in &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; and in &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt; and then get the sum of them?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="16683" LastEditDate="2020-12-22T01:05:50.980" LastActivityDate="2020-12-22T01:05:50.980" Title="What is the oxidation state of nitrogen in ammonium nitrate?" Tags="&lt;physical-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14922" PostTypeId="2" ParentId="14920" CreationDate="2014-08-05T13:35:03.447" Score="6" Body="&lt;p&gt;There are 2 different nitrogen atoms in &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4NO3}$&lt;/span&gt;, one having oxidation number -3 and one with +5. They are different, having separate oxidation numbers independently of each other. It is perfectly fine to have several atoms of same element having different oxidation state in same compounds, for example in thiosulphate  &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3S^{2-}}$&lt;/span&gt; one (central) sulfur atom has oxidation number +4 and another has 0, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe3O4}$&lt;/span&gt; is actually &lt;span class=&quot;math-container&quot;&gt;$\ce{\overset{(II)}{Fe}\overset{(III)}{Fe2}O4}$&lt;/span&gt; and many more.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="95133" LastEditDate="2020-09-12T11:04:55.837" LastActivityDate="2020-09-12T11:04:55.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14923" PostTypeId="1" AcceptedAnswerId="14924" CreationDate="2014-08-05T15:14:18.200" Score="10" ViewCount="5291" Body="&lt;p&gt;I've bought $\ce{NaOH}$ or $\ce{KOH}$ (I don't remember which one!) for my electrolyzer and I've tested it. Now I want to improve its performance and I was searching for a better electrolyte when I realized I've forgotten what my current electrolyte is! It is white and has crystal shape, but which one is it, $\ce{KOH}$ or $\ce{NaOH}$? If they are both white how can I know?&lt;/p&gt;&#xA;" OwnerUserId="7310" LastEditorUserId="4231" LastEditDate="2014-10-31T19:20:58.510" LastActivityDate="2014-10-31T20:05:44.053" Title="How can I identify whether a substance is KOH or NaOH?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14924" PostTypeId="2" ParentId="14923" CreationDate="2014-08-05T15:45:11.920" Score="19" Body="&lt;p&gt;A flame test should tell you pretty quickly. Make a small wire loop (preferably steel or nichrome), heat in a bunsen burner or propane torch flame until red hot and all the contaminants are burned off, then dip it in your solution and put it in the flame. If sodium is present, you should get a bright yellow flame, and if it is potassium, the flame will be a dim purple (lilac) color. But a small amount of sodium present can mask that color.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-08-05T15:45:11.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14925" PostTypeId="1" AcceptedAnswerId="14930" CreationDate="2014-08-05T16:02:35.767" Score="3" ViewCount="5786" Body="&lt;p&gt;Does having 8 or 2 electron in the outmost shell mean its outmost shell is full and its valency is zero? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the 3rd and 4th shell can contain 18 and 32 electrons. &#xA;Then how can Argon's (2,8,8) outmost shell be full, though it does not contain the highest number of electron (18) in it's 3rd shell? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the same in the cases of other noble gases too? Why?  &lt;/p&gt;&#xA;" OwnerUserId="7347" LastEditorUserId="4945" LastEditDate="2014-08-05T17:09:07.717" LastActivityDate="2017-10-03T19:36:28.933" Title="What is meant by complete outer shell? Why do the noble gases have zero valency?" Tags="&lt;electronic-configuration&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14926" PostTypeId="1" AcceptedAnswerId="15033" CreationDate="2014-08-05T16:04:45.713" Score="1" ViewCount="2007" Body="&lt;p&gt;I have a test comming up on Chemistry in Advanced Level and I sincerly ask for a book recommendation for chemistry. I already got the book chemistry by Ramond Chang. And I am pretty near to the exam. What I really want is a book with Lesson and Questions related to that specific lesson book. &lt;/p&gt;&#xA;" OwnerUserId="7348" LastActivityDate="2018-08-28T18:26:38.437" Title="Advanced Level Chemistry Book Recommendation" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2018-08-28T22:46:42.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="14928" PostTypeId="2" ParentId="14899" CreationDate="2014-08-05T17:05:18.983" Score="5" Body="&lt;p&gt;The &quot;E&quot; is the height of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;noreferrer&quot;&gt;thermodynamic potential energy&lt;/a&gt; barrier. Which potential that is (enthalpy, Gibbs energy, Helmholtz, or internal energy) depends on the conditions under which the reaction takes place - which thermodynamic state variables are held constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a system where the number of moles, the pressure, and the temperature are held constant (called NPT), then the thermodynamic potential energy (and thus the activation energy) is measured using Gibbs free energy.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-05T17:05:18.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14929" PostTypeId="1" AcceptedAnswerId="14937" CreationDate="2014-08-05T17:22:34.617" Score="1" ViewCount="170" Body="&lt;p&gt;Could any ingredients from this product be psychoactive in some people, if the are sensitive to them?&#xA;&lt;a href=&quot;https://www.lush.co.uk/products/honey-i-washed-kids&quot; rel=&quot;nofollow&quot;&gt;https://www.lush.co.uk/products/honey-i-washed-kids&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found the product to be relaxing as advertised, but more so that I would expect from a nice scent alone.&lt;/p&gt;&#xA;" OwnerUserId="7350" LastEditorUserId="73" LastEditDate="2014-08-05T20:59:25.953" LastActivityDate="2014-08-05T20:59:25.953" Title="Psychoactive ingredients in this soap?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14930" PostTypeId="2" ParentId="14925" CreationDate="2014-08-05T17:24:08.377" Score="6" Body="&lt;p&gt;When we refer to &quot;outer shells&quot; we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow&quot;&gt;effective nuclear charge&lt;/a&gt; &quot;felt&quot; by the valence electrons changes as you move through successively higher energy configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the &lt;em&gt;next&lt;/em&gt; n fill before the &lt;em&gt;current&lt;/em&gt; n (they have lower energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Ar: \space [Ne]}3s^23p^6&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next noble gas, krypton, the condensed electron configuration is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Kr: \space [Ar]}4s^23d^{10}4p^6&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking  for the valence or &quot;outermost&quot; electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T10:28:29.293" LastActivityDate="2015-06-12T10:28:29.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14931" PostTypeId="2" ParentId="14491" CreationDate="2014-08-05T18:09:04.683" Score="4" Body="&lt;p&gt;It sounds like you're needing help with chemical graph theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert in this topic, but it is one that has interested me in the past. Hence I can recommend 'Topological Methods in Chemistry' by R. E. Merrifield and H. E. Simmons—a book which explicitly deals with chemical graph theory in chapters 3 and 4; various topological indices feature throughout.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a complement to this book, I would recommend the book on Topology by Munkres. The above book isn't rigorous enough to approach this topic perhaps, and so a mathematical treatise is also needed. I'm currently working through these topics myself for my own research.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't quite answer your question, but I hope it helps in your lofty quest!&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-05T18:09:04.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14932" PostTypeId="1" AcceptedAnswerId="14933" CreationDate="2014-08-05T18:27:18.097" Score="0" ViewCount="12378" Body="&lt;p&gt;My professor noted that bromination with light and methylene &lt;em&gt;chloride&lt;/em&gt; gives you exclusively the monobromination product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, bromination without light will give you the multi-brominated product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no found any reference to bromination in my textbook nor online. Is the second statement an accurate statement? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-02T16:36:14.527" LastActivityDate="2015-10-02T16:36:14.527" Title="Monobromination vs. Dibromination" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14933" PostTypeId="2" ParentId="14932" CreationDate="2014-08-05T18:37:37.500" Score="2" Body="&lt;p&gt;Bromination of alkanes with Br2 will not occur in a dark environment. In the presence of light, multiple halogenations can occur given enough time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, including the full mechanism, check out: &lt;a href=&quot;https://en.wikipedia.org/wiki/Free-radical_halogenation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Free-radical_halogenation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7352" LastActivityDate="2014-08-05T18:37:37.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14934" PostTypeId="2" ParentId="14932" CreationDate="2014-08-05T18:55:45.657" Score="1" Body="&lt;p&gt;The rate determining step (rds) in the free radical bromination of an alkane involves hydrogen abstraction to yield the corresponding alkyl radical.  Whether the chain initiating bromine radical is generated in the dark (thermally or by using N-bromosuccinamide for example) or light (where the Br-Br bond is homolytically cleaved using light energy), does not particularly matter; the same mechanism and (roughly the same) product distribution will result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selectivity (which hydrogen will be abstracted) will be determined by the relative strengths of the various C-H bonds in the molecule.  A tertiary C-H bond (tBu-H) has a bond dissociation energy (BDE) of around 93 kcal/mol, a secondary C-H bond (iPr-H) has a BDE around 95 kcal/mol, while a primary C-H bond (Et-H) has a BDE around 98 kcal/mol and methane has a BDE of 103 kcal/mol.  I wouldn't expect a bromine atom attached to carbon to &lt;em&gt;increase&lt;/em&gt; the BDE for a hydrogen also attached to this carbon, so multibromination at that carbon should also occur.  Further, even in a simple molecule like ethane, once bromoethane is formed, a second bromination at the non-brominated carbon should occur as readily as bromination at the initially brominated carbon.  So both 1,1 and 1,2-dibromoethane should result as the secondary products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, both reactions will proceed through the same free radical - hydrogen abstraction mechanism, and, if 1) there is not a large excess of alkane to bromine and 2) the reaction is not terminated shortly after initiation, then both reactions will produce multi-brominated products&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-05T22:59:34.430" LastActivityDate="2014-08-05T22:59:34.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14935" PostTypeId="1" CreationDate="2014-08-05T18:55:51.973" Score="0" ViewCount="628" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJXtb.jpg&quot; alt=&quot;What are the parent compounds of the following chains of carbon atoms?&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just need these two answered correctly so I have examples to work with, also if someone could label the longest continuous chain, that would be great too! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7353" LastActivityDate="2014-08-05T19:27:51.240" Title="What are the parent compounds of the following?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-08-06T01:48:43.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="14936" PostTypeId="2" ParentId="14935" CreationDate="2014-08-05T19:27:51.240" Score="3" Body="&lt;p&gt;Find the longest chain by starting at one end of the molecule and counting the number of carbon atoms along each possible path.  I've attached a figure with numbering of the longest chain.  Hopefully it will answer your question and you will understand how to find and number the longest chain.  In the future, post your thoughts, so we can better understand how to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rp6bc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-05T19:27:51.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14937" PostTypeId="2" ParentId="14929" CreationDate="2014-08-05T20:21:32.617" Score="3" Body="&lt;p&gt;None of the ingredients listed would be classified as a psychotropic substance.  However, many of the compounds, when studied in mice, do have effects on brain activity and behavior characteristics.  For example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rapeseed oil (in a mixture with other vegetable oils): &quot;the rapeseed oil group exhibited much higher locomotor activity than that expected from the alpha-linolenate/linoleate ratio. Additionally, the rapeseed oil group exhibited unusual behavior patterns, including higher ambulation and rearing activities, faster acquisition of the water maze task and slower habituation behavior as compared with the control group. Susceptibility to pentobarbital anesthesia tended to be higher in the rapeseed oil group.&quot; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/8924909&quot; rel=&quot;nofollow&quot;&gt;abstract link&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;bergamot: &quot;produces dose-dependent increases in locomotor and exploratory activity that correlate with a predominant increase in the energy in the faster frequency bands of the EEG spectrum&quot; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/19775539&quot; rel=&quot;nofollow&quot;&gt;abstract link&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;linalool: &quot;Inhaled linalool showed anxiolytic properties in the light/dark test, increased social interaction and decreased aggressive behavior&quot; &lt;a href=&quot;http://www.researchgate.net/publication/40449279_Effects_of_inhaled_Linalool_in_anxiety_social_interaction_and_aggressive_behavior_in_mice&quot; rel=&quot;nofollow&quot;&gt;abstract link&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I am not surprised to read that when used in combination they have a noticeable affect on you.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-05T20:21:32.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14938" PostTypeId="1" AcceptedAnswerId="14939" CreationDate="2014-08-05T21:49:31.700" Score="1" ViewCount="1276" Body="&lt;p&gt;Is the photoelectric effect possible for semiconductors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that this might be, if possible, a two-photon process: excitation of an electron from a filled to conductance band, and then on to ejection. Is anyone familiar with such materials, or even whether this is possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't readily think of a theoretical reason against it. Two-photon events are often less likely than one-photon, however I could imagine irradiation by a laser of bandwith covering both the band-gap wavelength and that necessary for ejection would do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does indeed happen, I'd be glad to hear a short summary of how it differs from the photoelectric effect of metals.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorUserId="5026" LastEditDate="2018-07-03T02:56:33.820" LastActivityDate="2018-07-03T02:56:33.820" Title="Is the photoelectric effect possible for semiconductors?" Tags="&lt;spectroscopy&gt;&lt;electronic-configuration&gt;&lt;materials&gt;&lt;semiconductors&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14939" PostTypeId="2" ParentId="14938" CreationDate="2014-08-05T22:24:50.113" Score="3" Body="&lt;p&gt;Given that Jaques Pankove's excellent book on 'Optical Processes in Semiconductors' devotes chapter 13 to 'Photoelectric Emission', you can rest assured that the photoelectric effect is alive and well in semiconductors.  While somewhat more complex than emission from simple metals, there is no need for two-photon processes. The 'tricky' part is that you have to look at where the electron states are, and then how much energy is needed to take it out of the semiconductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work function is the energy difference from the Fermi energy to the vacuum level.  There may not be any states at the Fermi energy though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron affinity is the energy difference from the bottom of the conduction band to the vacuum level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to take the band structure into account, since direct and indirect transitions will likely have different energies required. &#xA;Also, surface band bending may make interpretation of experimental results more difficult. Again, perusing Pankove may prove useful.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-05T22:24:50.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14940" PostTypeId="1" AcceptedAnswerId="14943" CreationDate="2014-08-06T00:16:08.463" Score="1" ViewCount="40" Body="&lt;p&gt;Could the aerobic respiration reaction occur outside the body to produce water? If it is possible how would it occur&lt;/p&gt;&#xA;" OwnerUserId="7355" LastActivityDate="2014-08-06T01:50:32.177" Title="Aerobic respiration to produce water" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-08-06T07:41:09.000" ContentLicense="CC BY-SA 3.0" />
  <row Id="14941" PostTypeId="1" CreationDate="2014-08-06T01:19:55.237" Score="1" ViewCount="510" Body="&lt;p&gt;It is to my understanding that certain amount of joules may change an element's/compound's physical state given a certain temperature. How does a certain added number of joules affect the orbital of a bonded electron ( I realize my question's ambiguity by not stating a specific type of bond).&lt;/p&gt;&#xA;" OwnerUserId="2197" LastEditorUserId="2197" LastEditDate="2014-08-06T01:28:40.290" LastActivityDate="2014-08-06T02:14:26.037" Title="In what ways does the transfer of heat affect bond stability?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14943" PostTypeId="2" ParentId="14940" CreationDate="2014-08-06T01:50:32.177" Score="3" Body="&lt;p&gt;If you look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellular_respiration#Aerobic_respiration&quot; rel=&quot;nofollow&quot;&gt;net reaction&lt;/a&gt;, it is just combustion of glucose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{C6H12O6 + 6O2 -&amp;gt; 6CO2 + 6H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could just burn a sample of glucose, and it would produce water.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-06T01:50:32.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14944" PostTypeId="2" ParentId="14941" CreationDate="2014-08-06T02:14:26.037" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a certain added number of joules affect the orbital of a&#xA;  bonded electron&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you pump enough energy into an element\molecule you can promote an electron from one orbital to the next highest.  For example, the hydrogen atom has a single 1s electron, put enough energy in and you can promote that electron to the 2s orbital.  That means that, on average, the electron would spend more time further away from the nucleus.  In molecules, there are numerous &quot;sub-levels&quot; between each (molecular) orbital.  These &quot;sub-levels&quot; relate to rotational and vibrational energy.  So as you pump energy into the system the molecule will rotate and vibrate more violently.  If you put enough energy in, the bond will break and the atoms connected by that bond will fly apart.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-06T02:14:26.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14946" PostTypeId="1" CreationDate="2014-08-06T05:39:24.207" Score="0" ViewCount="188" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the subshells $\ce{1s, 2s, 2p, 3s, 3p}$ and $\ce{3d}$, identify those&#xA;  that meet the following descriptions:&lt;br&gt;&#xA;  a) has $l=2$ &#xA;  b) Can contain two electrons with spin $m_{s}=\pm\frac{1}{2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For a)&lt;/strong&gt;: from my understanding $l=n-1$. Therefore $\ce{3s, 3p}$ and $\ce{3d}$ should be correct but according to my solutions manual, the answer is only $\ce{3d}$. Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For b)&lt;/strong&gt; The answer is $\ce{2p, 3p}$ and $\ce{3d}$. Why are these the answers?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="41556" LastEditDate="2018-02-27T14:13:20.923" LastActivityDate="2018-02-27T14:13:20.923" Title="Subtle implications of quantum numbers" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14947" PostTypeId="2" ParentId="14946" CreationDate="2014-08-06T06:15:51.020" Score="2" Body="&lt;p&gt;a) You are wrong. $l=2$ simply means $\ce{d}$-orbitals, thus, only $\ce{3d}$ is the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Assuming that the meaning of the question is to specify subshells that can contain two electrons in the same spin state, i.e. both with $m_s = +1/2$ or both with $m_s = -1/2$, $\ce{s}$-orbitals should indeed be excluded.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="5026" LastEditDate="2018-02-27T11:48:40.093" LastActivityDate="2018-02-27T11:48:40.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14948" PostTypeId="2" ParentId="14932" CreationDate="2014-08-06T07:23:35.387" Score="0" Body="&lt;p&gt;My first guess is that in light alkanes forms alkyl halides and in dark alkenes form dihalo-alkanes with bromine in non polar solvent as $CCl_4$; anyways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromine is less reactive than other halogens toward alkanes in general but bromine is&#xA;more selective in the site of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromine shows a much greater ability to discriminate among the different types of hydrogen&#xA;atoms. The reaction of 2-methylpropane and bromine, for example, gives almost exclusive&#xA;replacement of the tertiary hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\small\text{2-methylpropane}\overbrace{\longrightarrow}^{\text{Br}_2,h\nu,127^oC}\text{tertiarybutylbromide}+(99\%)+\text{sec-butylbromide(trace)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstraction of hydrogen by a bromine atom is endothermic in both cases. The transition states are more product-like and possess more radical character; therefore, the difference in radical stability is more strongly expressed, and $\Delta E_{act}$is larger. The larger $\Delta E_{act}$ is associated with greater product selectivity, since the tertiary bromide is obtained from the tertiary free radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the laboratory it is more convenient to use light, either visible or ultraviolet, as&#xA;the source of energy to initiate the reaction. Reactions that occur when light energy is&#xA;absorbed by a molecule are called photochemical reactions. Photochemical techniques&#xA;permit the reaction of alkanes to be performed at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-08-06T16:14:59.187" LastActivityDate="2014-08-06T16:14:59.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14949" PostTypeId="1" AcceptedAnswerId="14954" CreationDate="2014-08-06T09:47:22.967" Score="10" ViewCount="6571" Body="&lt;p&gt;Straight-chain octane ($\ce{CH3(CH2)6CH3}$) is also called &lt;em&gt;n&lt;/em&gt;-octane. What does the &lt;em&gt;n&lt;/em&gt; stand for, and where did the term originate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, see the &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=356&quot; rel=&quot;nofollow noreferrer&quot;&gt;PubChem entry for octane&lt;/a&gt; (beside &quot;Chemical Names:&quot;)&lt;/p&gt;&#xA;" OwnerUserId="5022" LastEditorUserId="5026" LastEditDate="2018-04-15T12:18:49.083" LastActivityDate="2018-04-15T12:18:49.083" Title="What is the meaning of n in n-octane and similar compounds?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14950" PostTypeId="1" AcceptedAnswerId="14959" CreationDate="2014-08-06T09:48:38.493" Score="4" ViewCount="78" Body="&lt;p&gt;If you dissolve two different ionic compounds (that don't make a precipitate) in water and then boil the solution so all the water evaporates, the salts will re-crystallize. Is it possible that you will get two compounds that are different from the ones you orignially dissolved? Is there any way to predict whether you will get the original compounds or new ones?&lt;/p&gt;&#xA;" OwnerUserId="7344" LastActivityDate="2014-08-06T15:27:42.073" Title="Boiling solutions with two salts" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14951" PostTypeId="1" CreationDate="2014-08-06T09:50:05.770" Score="1" ViewCount="281" Body="&lt;p&gt;Is there an efficient method to compute the two-electron integrals over the basis set represented by a product of a Gaussian function $g(r)$ and plane wave:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\psi(r)=g(r)e^{ikr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda=1/k$ is approximately one fifth of the width of gaussians. &lt;/p&gt;&#xA;" OwnerUserId="1696" LastActivityDate="2014-08-07T19:01:36.007" Title="Two-electron Integrals over Gaussian Plane Waves" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2014-08-15T05:26:14.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="14952" PostTypeId="1" CreationDate="2014-08-06T11:01:30.957" Score="1" ViewCount="8453" Body="&lt;p&gt;The correct arrange of them is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be &amp;lt; B &amp;lt; Li ... But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li will have the greatest IE&lt;sub&gt;2&lt;/sub&gt; (second ionization energy) because that will involve removing a core electron, but what I am confused about is about Be and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Be will have greater IE&lt;sub&gt;2&lt;/sub&gt; than B, because B has 2 electrons in the same orbital, so the repulsive force between them will make it easier to remove one of them, but in Be there is only one electron in the orbital and there is no repulsive force in the same orbital, therefore&lt;br&gt;&#xA;$$\ce{ B &amp;lt; Be}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="46306" LastEditDate="2018-03-17T10:49:41.630" LastActivityDate="2018-03-17T10:49:41.630" Title="Order of second iozination energies of lithium,berillium,neon,carbon and boron?" Tags="&lt;periodic-trends&gt;&lt;periodic-table&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14953" PostTypeId="1" CreationDate="2014-08-06T11:04:37.843" Score="3" ViewCount="169" Body="&lt;p&gt;I'm currently a chemistry major and now I am considering doing a minor in materials science. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before making a decision, I would like to know how the two disciplines are related and what are their differences such as techniques, approaches, emphasis, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will my knowledge in chemistry be useful in my understanding of materials science, and how useful is a knowledge of materials science to a chemist?&lt;/p&gt;&#xA;" OwnerUserId="7360" LastEditorUserId="73" LastEditDate="2014-08-06T17:52:43.280" LastActivityDate="2014-08-06T17:52:43.280" Title="How are chemistry and materials science related?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14954" PostTypeId="2" ParentId="14949" CreationDate="2014-08-06T13:15:01.160" Score="14" Body="&lt;p&gt;In this case, &lt;em&gt;n&lt;/em&gt; means &quot;normal&quot;, i.e. straight chain. This is somewhat of a holdover from the olden days, but in the case of octane in particular, it makes sense to specify &lt;em&gt;n&lt;/em&gt;-octane if that is what you have. For example, the &quot;octane&quot; used as a standard gasoline engine fuel against which gasolines are rated (by definition, 100 octane) is actually not &lt;em&gt;n&lt;/em&gt;-octane, but rather 2,2,4-trimethylpentane - a branched isomer of octane - in contrast to the straight chain version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gqltv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gqltv.png&quot; alt=&quot;Isomers of octane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="16683" LastEditDate="2016-09-24T17:32:39.930" LastActivityDate="2016-09-24T17:32:39.930" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14955" PostTypeId="2" ParentId="14952" CreationDate="2014-08-06T13:49:25.520" Score="7" Body="&lt;p&gt;To reach the second ionization potential we remove a second electron from an already ionized element &#xA;$$\ce{M^{+1} -&amp;gt; M^{+2}}$$&#xA;Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol)&#xA;\begin{aligned}&#xA;\ce{&amp;amp; Li^{+1} &amp;amp; 1s^2~~ &amp;amp;~~ 7297}\\&#xA;\ce{&amp;amp; Be^{+1} &amp;amp; 1s^2~ 2s^1~~&amp;amp;~~ 1757}\\&#xA;\ce{&amp;amp; B^{+1} &amp;amp; 1s^2~ 2s^2~~&amp;amp;~~ 2426}\\&#xA;\end{aligned}&#xA;We see that both $\ce{Li^{+1}}$ and $\ce{B^{+1}}$ have completely filled shells, the stable &quot;inert gas&quot; configuration.  We would expect it to be difficult to remove one electron from these two ions and destroy this stable configuration.  Further, we would expect removal of one electron from $\ce{Li^{+1}}$ to be more difficult than from $\ce{B^{+1}}$ because in the former case we are removing a 1s electron which is closer to the nucleus than the 2s electron in the case of $\ce{B^{+1}}$.  On the other hand, if $\ce{Be^{+1}}$ loses one electron it will have a $\ce{1s^2}$ electron configuration and achieve a stable &quot;inert gas&quot; configuration.  This should make it much easier to remove one electron from $\ce{Be^{+1}}$ than from the other two ions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-06T13:49:25.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14956" PostTypeId="1" CreationDate="2014-08-06T14:08:15.817" Score="2" ViewCount="2151" Body="&lt;p&gt;my water electrolyzer which I've built using &lt;a href=&quot;http://www.free-energy-info.com/Smack.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; pdf, produces less gas than I expected. In that file the author insists on using distilled water instead of tap water and I have used tap water. would it be the reason? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so important to use distilled water since it has high resistance for electricity?  &lt;/p&gt;&#xA;" OwnerUserId="7310" LastActivityDate="2014-08-06T15:18:00.050" Title="tap water and distilled water for electrolysis" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14957" PostTypeId="2" ParentId="14956" CreationDate="2014-08-06T15:18:00.050" Score="3" Body="&lt;p&gt;This article tells you to use distilled water because the salt in tap water will damage the booster. The author also tell you to add NaOH or KOH to the distilled water. Did you add these salts?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastActivityDate="2014-08-06T15:18:00.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14959" PostTypeId="2" ParentId="14950" CreationDate="2014-08-06T15:27:42.073" Score="3" Body="&lt;p&gt;Mixing and matching different anions and cations can easily produce different compounds.  For instance, sodium and potassium chloride, and sodium and potassium nitrate, are all soluble in water and do not react to form a precipitate. When the aqueous solution dries, all four compounds are possible.  So, mixing sodium chloride and potassium nitrate will result in 2 anions and 2 cations in solution, allowing all four pairs to possible precipitate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, will the precipitate as pure compounds or not is another question.  That requires looking at the relevant phase diagrams and seeing what is thermodynamically favored (and consideration of kinetic factors like how fast you evaporate the liquid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the phase diagram for KCl and NaCl exhibits a miscibility gap starting at 505C.  This means that below that temperature they will phase separate into a Na-rich chloride and a K-rich chloride.  At 100C the concentration of the 'impurity' phase is less than 1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrate diagram also shows a two phase region below 100C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, mixing (Na,K) (Cl,NO3) will result in all four precipitating out.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-06T15:27:42.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14960" PostTypeId="2" ParentId="14953" CreationDate="2014-08-06T17:28:15.790" Score="3" Body="&lt;p&gt;An undergraduate chemistry course here in the UK typically has a materials component to it in the form of a module or two, so you bet they're related!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's sometimes not that fruitful to view materials science and chemistry as a different subject—much like physic vs chemistry—however we can associate certain motivations to either topic as we see fit. Please note that this is quite subjective; this view does indeed vary from person to person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, materials is more interested in the solid state than chemistry in general. We might like to look at crystal structures via x-ray crystallography (see Bravais lattices, space groups etc.) as well as the rather intriguing quasi-crystals (for which Dan Shectman won the Nobel in Chemistry in 2011). The skills gained in a materials crystallography course would no doubt help with chemistry. The converse is also true since many substances in chemistry are amenable to study via crystallography, such as proteins. You therefore may also have a crystallography module in your chemistry course—looking at molecules in the solid state not typically viewed as a 'material'. If it has appreciable electron density, you can study it by x-ray crystallography!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of materials science looks at properties of materials, such as stress/strain curves etc.I've had such a lecture series specifically dealing with polymers in a chemistry course; one of the challenges of modern day chemistry is to invent and modify polymer architectures for specific industrial applications. It's clear then that chemistry and materials science complement each other here. Whilst materials looks at interesting properties of given materials, chemistry constructs those polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials science can differ from chemistry in that it focuses (or at least did in my case) on the band-structure of solids more than chemistry does. So it's possible that a materials course would give you a more nuanced appreciation of semi-conductors and the like. Again, this would help overwhelmingly if you decided to go into the field of conducting polymers—a currently very active field in R&amp;amp;D, especially if it involves replacing rare earths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise: materials science and chemistry go hand in hand in my quick overview, however some aspects of chemistry (such as organic) view the solid-state as the 'graveyard of chemistry'—you'd be hard pressed to find reactions which occur in the solid state! If you want to find out a little more before delving into materials, then finding a quick Youtube course might help to sway you either way. In my opinion there are far more ridiculous pairings in the world than chemistry and materials!&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-06T17:28:15.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14961" PostTypeId="1" AcceptedAnswerId="14997" CreationDate="2014-08-06T17:52:53.410" Score="5" ViewCount="6229" Body="&lt;p&gt;Basically the term which we use to refer them is lone pair. In Lewis structure why we represent those unbonded electron in pairs. Like here &lt;strong&gt;(structure of SO&lt;sub&gt;2&lt;/sub&gt;)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcTgZKyTJloyh_Nnp62uVQ01BwafSNNHKHOlKcU3iYKK2U_PVJ6D&quot; height=&quot;200&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here if we assume both the unbonded electrons to be existing in pairs then it's geometry will be different from what we would observe in case when both electrons exists freely (without pair).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why we basically consider those electrons to be existing in pair? Is it any kind of observation or any theory or both?&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="5021" LastEditDate="2014-11-25T19:07:02.340" LastActivityDate="2014-11-25T19:07:02.340" Title="Why do unbonded electrons exist in pairs?" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;lewis-structure&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14962" PostTypeId="2" ParentId="14961" CreationDate="2014-08-06T18:01:39.547" Score="3" Body="&lt;p&gt;A single unpaired electron is called a radical, and usually aren't stable. Each electron orbital can hold two electrons, one spin up, other spin down. So a lone pair is just a full electron orbital, and cannot accept another electron. An orbital with 1 electron will readily react to get another electron and form a pair. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this: &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Free_radical&quot; rel=&quot;nofollow&quot;&gt;Free Radical&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reaction they show is UV light being used to break $\mathrm{Cl}_2$ gas into $2\mathrm{Cl}$ radicals, each with a single lone electron, forming free radicals. These radicals could be used as free radical polymerization initiators, or for other reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the structure you show, the molecule forms a trigonal planar configuration, with the lone pair forming the third arm. If you could stably achieve two free radicals instead of the lone pair, it would have tetrahedral configuration, with the radicals forming 2 of the 4 arms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory behind it is more quantum mechanics than I can remember, but I know it involves the Pauli exclusion principle and Hund's rule.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="670" LastEditDate="2014-08-07T14:23:37.503" LastActivityDate="2014-08-07T14:23:37.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14963" PostTypeId="1" CreationDate="2014-08-06T18:04:24.023" Score="0" ViewCount="57" Body="&lt;p&gt;If you were to investigate crystal structure changes as a function of pressure, which metals would you study first? Last? Why? &lt;/p&gt;&#xA;" OwnerUserId="7362" LastActivityDate="2014-08-06T20:28:06.487" Title="Crystal structure in function of pressure" Tags="&lt;crystal-structure&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14964" PostTypeId="1" CreationDate="2014-08-06T18:25:13.543" Score="-1" ViewCount="3116" Body="&lt;p&gt;Is the question in a correct form? I can't add anything as a comment.&lt;/p&gt;&#xA;" OwnerUserId="7363" LastEditorUserId="41328" LastEditDate="2019-05-17T19:56:08.160" LastActivityDate="2019-05-17T19:57:04.800" Title="Are protons or electrons responsible for the chemistry of an atom?" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-05-17T19:56:18.480" ContentLicense="CC BY-SA 4.0" />
  <row Id="14965" PostTypeId="2" ParentId="14964" CreationDate="2014-08-06T18:41:08.227" Score="3" Body="&lt;p&gt;The electrons are responsible for the actual chemistry happening around an atom, however the number of electrons in an atom is determined by the charge of the nucleus, i.e. by the protons. The protons are not really participating in the reactions, but they are the ones who compensate the charge of the electrons, therefore they are important to determine the energy levels of the free atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an illustration, you can take helium &lt;span class=&quot;math-container&quot;&gt;$(\ce{He})$&lt;/span&gt; and a hydride ion &lt;span class=&quot;math-container&quot;&gt;$(\ce{H-}).$&lt;/span&gt; Both have 2 electrons on an 1s, however the &lt;span class=&quot;math-container&quot;&gt;$\ce{He}$&lt;/span&gt; is neutral, while the nucleus of &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; is compensating only for one charge. This result in totally different energy levels and stability, different affinity for oxidation or other chemical reactions. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="41328" LastEditDate="2019-05-17T19:57:04.800" LastActivityDate="2019-05-17T19:57:04.800" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="14966" PostTypeId="2" ParentId="14889" CreationDate="2014-08-06T18:53:30.767" Score="3" Body="&lt;p&gt;Here is a synthesis of &lt;em&gt;m&lt;/em&gt;-cresol starting from benzene.  It achieves the desired &lt;em&gt;meta&lt;/em&gt; orientation of the final substituents by introducing a &lt;em&gt;meta&lt;/em&gt;-directing substituent (the aldehyde group) early-on in the synthetic scheme. In the nitration step, the &lt;em&gt;meta&lt;/em&gt; isomer is by far the major product (o:m:p 19:72:9), but a distillation or recrystallization would be needed (or you could carry all the isomers along and separate later, but you'd probably have the widest range in boiling points at this stage with the nitro group present). The only novel step is the &lt;a href=&quot;http://www.tara.tcd.ie/bitstream/handle/2262/60702/One-step%20double%20reduction%20of%20aryl%20nitro%20and%20carbonyl%20groups%20using%20hydrazine.pdf;jsessionid=8BF41AFF57A857B4FE5D785692513EB0?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;double Wolff-Kishner&lt;/a&gt; used to reduce both the aldehyde and nitro groups in a single reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H2q7n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-06T18:53:30.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14967" PostTypeId="2" ParentId="14910" CreationDate="2014-08-06T19:19:23.417" Score="3" Body="&lt;p&gt;It's tempting to look at the whole cell here and ask: hmm what's going on? But try taking a reductionist approach and look at each electrode system one at a time, in isolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a basic model.If we immerse an electrode of $\ce{Cu}$ metal into a solution of $\ce{Cu^2+}$, the following can be thought to take place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu(s) &amp;lt;=&amp;gt; Cu^2+(aq) + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the ions in the solution are bombarding the electrode surface, and are being taken up by gaining two electrons from the metal, whilst the metal is giving up ions from its surface, keeping hold of two electrons. At equilibrium, this forward and backward bombardment occurs at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say to me: okay but we can't actually probe this system with a voltmeter to say that this is going on, because this would require a circuit to be made with an external system, and hence would perturb any effect we wish to observe. I'd then say to you: you're correct but bear with my argument, just noting what I've said. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a similar token, in our other isolated beaker we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(s) &amp;lt;=&amp;gt; Zn^2+(aq) + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the whole idea behind an electrochemical cell is to extract chemical energy by systems which are &lt;em&gt;not&lt;/em&gt; at equilibrium. If a system is at equilibrium then no energy will be available. The equation which gives the free energy available from an electrochemical cell is (derivation not given):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G=−RT\ln K_\mathrm{eq}=−nFE_\text{cell}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $R$ is the ideal gas constant, $T$ is the parameter of temperature, $F$ is the Faraday constant (the amount of charge carried by a mole of electrons) and $E_\text{cell}$ is your available electrochemical energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for these two individual single-electrode systems at equilibrium, can we extract chemical energy? Well no, because Gibbs at equilibrium is zero, and hence so is $E_\text{cell}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What then do we do? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete the circuit between the two, and if the potential between two half-reactions is sufficient to warrant the moving of electrons, a dis-equilibrium will be created until the reaction has completed (which is indeed what we get for a system of zinc/copper). Once the electron transfer between zinc and copped has been achieved i.e. equilibrium has been achieved, then we can no longer extract any useful chemical energy. Our cell is dead!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary: the dual metal electrode/same-metal ion solution is needed for your initial half-reactions, which you then perturb by completing your circuit. This perturbation of the individual half-reactions results in our being able to extract useful chemical energy by shifting the equilibrium until a new equilibrium is established. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7951" LastEditDate="2015-10-07T13:45:13.003" LastActivityDate="2015-10-07T13:45:13.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14968" PostTypeId="2" ParentId="14910" CreationDate="2014-08-06T20:04:11.433" Score="3" Body="&lt;p&gt;You're right: you don't &lt;em&gt;need&lt;/em&gt; a zinc solution to start with.  In fact the lemon battery experiment (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lemon_battery&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Lemon_battery&lt;/a&gt;) shows you don't even need the copper solution to generate a current. (The reaction at the copper end in that case is 2H$^+$ + 2e$^-$ = H$_2$ and doesn't involve reduction of copper.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the textbook description is the way it is has to do with reproducibility and simplicity.  I'll address the question of reproducibility later on, but first consider: &#xA;What's the simplest thing the zinc could be immersed in? If you put the zinc electrode in the same copper solution as the other electrode, you'll get copper-plated zinc instead of a battery.  If you put it in distilled water you won't have a circuit.  There are other things you could put it into, such as salt water, but in any case two things are true: It needs to be a solution different from the one where the copper electrode is sitting; and it &lt;em&gt;will&lt;/em&gt; have zinc ions as soon as you have a closed circuit, as the zinc electrode starts to dissolve and the copper electrode starts acquiring more copper.  So in a way the simplest answer is, stick it in a solution of zinc ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now about reproducibility.&#xA;The potential of the cell depends on (among other things) the concentration of Zn$^{2+}$ in solution near the Zn electrode, and the concentration of Cu$^{2+}$ in solution near the Cu electrode.  Other things being equal, if you increase the concentration of Zn$^{2+}$ by a factor of 10, the potential will drop by about 0.118 volts (Nernst equation).   So in a way it's actually good to have a &lt;em&gt;low&lt;/em&gt; concentration of Zn$^{2+}$; but that concentration will never be zero, at least after you close the circuit, and when the concentration is very very low it will be growing very fast, relatively speaking (the first micromole of Zn to dissolve might increase the concentration a million times, say), which means the potential will drop quickly in the initial moments.   You  might as well start with a concentration of  Zn$^{2+}$ that is not going to change so fast (in relative terms).&lt;/p&gt;&#xA;" OwnerUserId="7187" LastEditorUserId="7187" LastEditDate="2014-08-06T20:51:24.850" LastActivityDate="2014-08-06T20:51:24.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14969" PostTypeId="1" AcceptedAnswerId="14974" CreationDate="2014-08-06T20:14:17.027" Score="3" ViewCount="407" Body="&lt;p&gt;Many sulfur-containing compounds (like glutathione and cysteine groups) have high antioxidant activity. Why?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="305" LastEditDate="2014-08-06T23:59:10.777" LastActivityDate="2014-08-06T23:59:10.777" Title="What makes sulfur-containing compounds such good antioxidants?" Tags="&lt;redox&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14970" PostTypeId="2" ParentId="13971" CreationDate="2014-08-06T20:15:38.613" Score="2" Body="&lt;p&gt;In your first spectrum, the band at 3517 $\mathrm{cm^{-1}}$ indeed fits best to the $\ce{N-H}$ stretch of a primary amine (&lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/aminesir.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). However, I would not assign the shoulder of this band to a $\ce{#C-H}$ stretch. Such a band would be sharper and perhaps more intense, and as Greg E. already pointed out in his comment to your question, there is also no corresponding $\ce{C#C}$ stretch band present (&lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/alkynesir.html&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;). The absence of bands in the range between 2500-2000 $\mathrm{cm^{-1}}$ allows you to rule out some other functional groups with double or triple bonds, like nitriles, thiocyanates, isocyanates, carbodiimides or azides. The band at 1715 $\mathrm{cm^{-1}}$ is a typical $\ce{C=O}$ stretch, and while it could be a ketone, the presence of the broad band at 2932 $\mathrm{cm^{-1}}$ (which could actually be a double band) makes it likely to be an aldehyde, since the double band can then be assigned to the $\ce{=C-H}$ stretch of the CHO group (it could also include other $\ce{C-H}$ stretches due to its width).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second spectrum appears a bit more complex. The broad, intense band at 3298 $\mathrm{cm^{-1}}$ in combination with the band at approximately 1680 $\mathrm{cm^{-1}}$ could be assigned to an amide, but I would expect a stronger carbonyl band in this case. Nevertheless, the intensity and width of the 3298 $\mathrm{cm^{-1}}$ band suggest the presence of a functional group capable of H-bonding. The bands between 2900-2800 $\mathrm{cm^{-1}}$ could be assigned to alkyl $\ce{C-H}$ stretches (&lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/alkanesir.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). For the compound to be aromatic, there should be aromatic $\ce{C-H}$ stretch bands present, slightly above 3000 $\mathrm{cm^{-1}}$ (&lt;a href=&quot;http://orgchem.colorado.edu/Spectroscopy/irtutor/aromaticsir.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-06T20:15:38.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14971" PostTypeId="1" CreationDate="2014-08-06T20:19:27.670" Score="9" ViewCount="1125" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Superoxide_dismutase&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction of superoxide with non-radicals is spin forbidden. In biological systems,   this means its main reactions are with itself (dismutation) or with another biological radical such as nitric oxide (NO) or with a transition-series metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that one of the configuration bands is paired, and one is unpaired. Does the unpaired one have to bond with an unpaired one from another radical? What does this also say about reactions of the excited state of singlet oxygen (shown in the middle below) with non-radicals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y6Afg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is superoxide a doublet?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="275" LastEditDate="2014-08-07T02:11:55.127" LastActivityDate="2014-08-07T18:36:53.010" Title="What exactly does it mean for the reaction of superoxide with non-radicals to be spin-forbidden?" Tags="&lt;radicals&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14972" PostTypeId="2" ParentId="14963" CreationDate="2014-08-06T20:28:06.487" Score="1" Body="&lt;p&gt;Seems like homework, so I'll give some hints:&#xA;What do you expect to happen to a crystal under pressure?&#xA;What does the crystal structure have to do with the response to pressure?&#xA;Why would the crystal structure want to change under pressure?&#xA;What crystal structures are common for metals? (there aren't many)&#xA;Since there aren't many, what distinguishes their responses to pressure that might make the metal want to change crystal structure (a phase change).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should get you going. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-06T20:28:06.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14973" PostTypeId="1" AcceptedAnswerId="14975" CreationDate="2014-08-06T20:34:41.873" Score="2" ViewCount="1306" Body="&lt;p&gt;I was wondering if anyone knows where I can find UV-VIS spectra of the n-decanes.  Specifically tetradecane and dodecane.  I can't seem to find such information using my regular search methods, so I was wondering if there was a common database that contains such things that I am simply unaware of (likely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I really need to know is if these chemicals will absorb radiation with wavelengths between 355-375 nm.  A source to back this up will only help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Sam&lt;/p&gt;&#xA;" OwnerUserId="874" LastActivityDate="2014-08-06T21:51:02.730" Title="UV-VIS spectra of n-decanes" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14974" PostTypeId="2" ParentId="14969" CreationDate="2014-08-06T21:07:42.900" Score="6" Body="&lt;p&gt;There are several factors that combine to make sulfur compounds such effective antioxidants.  First, sulfur compounds often contain S-H bonds (mercaptans) and are excellent radical scavengers because sulfur radicals are thermodynamically more stable than the analogous oxygen radical.  That is to say that the following equilibrium lies far to the right&#xA;$$\ce{ROS^{.} + RS-H &amp;lt;=&amp;gt; ROS-H + RS^{.}}$$&#xA;where ROS represents a &quot;reactive oxygen species&quot; such as hydroxyl radical ($\ce{HO^{.}}$), superoxide radical ($\ce{O_2^{-~.}}$, etc.  Sulfur compounds (sulfides and disulfides, for example) are also very easily oxidized to their corresponding sulfoxides.&lt;br&gt;&#xA;$$\ce{R-S-R +ROS -&amp;gt; R-SO-R + deactivated ~ROS}$$&#xA;If multiple mercaptan groups exist within a molecule (for example multiple cysteine residues in an amino acid) they can be oxidized to the corresponding disulfide while reducing an oxidizer. &#xA;$$\ce{2~ RS-H + [oxidized~molecule] -&amp;gt; RS-SR + [H-reduced~molecule-H]}$$&#xA;This oxidative process provides sulfur compounds with a second pathway to trap or transform, and consequently inactivate, ROS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to being very effective ROS scavengers, sulfur compounds also bind strongly to many metal ions.  These same metal ions often play a role in the generation of ROS.  By binding to them, sulfur compounds effectively remove them from the ROS generating process thereby further lowering the concentration of ROS in the cell.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-06T21:07:42.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14975" PostTypeId="2" ParentId="14973" CreationDate="2014-08-06T21:25:42.453" Score="3" Body="&lt;p&gt;To the best of my knowledge, ordinary alkanes do not typically absorb anywhere in the UV/Vis region. The organic molecules that do typically contain electrons in $\pi$- or non-bonding orbitals, which have energy level transitions corresponding to UV/Vis wavelengths. That said, I did find one paper&lt;sup&gt;1&lt;/sup&gt; describing absorptions around the far UV ambit, where $\lambda &amp;lt; 200\mathrm{nm}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;DOI: &lt;a href=&quot;http://dx.doi.org/10.1021/jp903435c&quot; rel=&quot;nofollow&quot;&gt;10.1021/jp903435c&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-06T21:51:02.730" LastActivityDate="2014-08-06T21:51:02.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14976" PostTypeId="1" CreationDate="2014-08-07T02:41:01.373" Score="1" ViewCount="725" Body="&lt;p&gt;There were two questions that I've listed below that I've had troubles on. I got an answer in both, but both of them weren't correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first question, they list the answer to be D and then for the second they list the answer as A. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Determine the net ionic equation for the reaction between $\ce{HNO3}$ and $\ce{Mg(OH)2}$.&lt;br&gt;&#xA;   a. $\ce{2H+(aq) + (OH)2^{2-}(aq) -&amp;gt; H2(OH)2(aq) }$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b. $\ce{2HNO3(aq) + Mg(OH)2(aq) -&amp;gt; Mg(NO3)2(aq) + 2H2O(l) }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;c. $\ce{Mg2+(aq) + 2NO3 (aq) -&amp;gt; Mg(NO3)2(s) }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;d. $\ce{H+(aq) + OH(aq) -&amp;gt; H2O(l) }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;e. $\ce{Mg+(aq) + NO3 (aq) -&amp;gt; MgNO3(s) }$&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Determine the number of ions that are present in $38.8~\mathrm{mL}$ of $0.113~\mathrm{M}$ $\ce{(NH4)2CO3. }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a. $7.92\cdot 10^{21}~\text{ions }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b. $2.64\cdot 10^{21}~\text{ions }$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;c. $2.07\cdot 10^{23}~\text{ions }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;d. $6.20\cdot 10^{23}~\text{ions }$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;e. $1.32\cdot 10^{23}~\text{ions }$&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first question, the work that I did was $\ce{2 H+ + 2NO3- +Mg(OH)2 -&amp;gt; 2H2O + Mg+2 + 2NO3-}$. The problem that I have in this problem is why the $\ce{Mg(OH)2}$ splits up. I don't get it because the things you split up are the things that are aqueous or soluble. $\ce{Mg(OH)2}$ isn't soluble so I listed it as a solid and thought that you wouldn't break it down. All I need to know is why I have to break the $\ce{Mg}$ and the $\ce{OH2}$ down because it isn't soluble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second, I got an answer of $2.64\cdot 10^{21}~\text{ions}$ but the review said it was A. I set it up as $.113~\mathrm{M} = \text{moles} / 0.0388~\mathrm{L}$ and then multiplied $.113$ by $0.0388$ and then multiplied that amount by $6.022\cdot 10^{23}~\text{ions}$ which got me the answer B. &lt;/p&gt;&#xA;" OwnerUserId="7364" LastEditorUserId="305" LastEditDate="2014-08-07T05:02:46.147" LastActivityDate="2014-08-07T05:50:56.333" Title="Balancing ionic equations and determining number of ions" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14977" PostTypeId="1" AcceptedAnswerId="14982" CreationDate="2014-08-07T02:44:02.763" Score="2" ViewCount="5924" Body="&lt;p&gt;To start with its a homework problem, quite lengthy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A particle of mass equal to 208 times the mass of an electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr model of the atom is applicable to this system,&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Derive an expression for radius of $n$th Bohr orbit.&lt;/li&gt;&#xA;  &lt;li&gt;Find the value of $n$ for which the radius is equal to the radii of first orbit of hydrogen.&lt;/li&gt;&#xA;  &lt;li&gt;Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I did the first part and got the answer correct. Here's what I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the mass of the particle revolving is $M$, its speed is $v$, and $M = 208 m_{e}$. Electrostatic force is the centripetal force. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\frac{Mv^2}{r} &amp;amp;= \frac{(ke)(3e)}{r^2} \\&#xA;v^2 &amp;amp;= \frac{3ke^2}{208m_{e}r}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Bohr model,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_{e}vr=\frac{nh}{2\pi}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is Planck's constant. Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v = \frac{nh}{2\pi m_{e}r}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Squaring it,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v^2 = \frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the two equations which have $v^2$ in them,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2} = \frac{3ke^2}{208m_{e}r}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After solving for $r$, we get something like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \frac{n^2h^2}{4(\pi)^{2}3ke^{2}208m_e}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above is correct. The problem is in the second and third parts; when I put $r =\pu{0.53*10^{-10} m}$ I do NOT get the required answer. To approach the third part, I started with the standard Rydberg equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{\lambda} = \mathcal{R}Z^2 \left( \frac{1}{n_f^2}-\frac{1}{n_i^2} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged in each value, $n_i = 3, n_f = 1, Z = 3$; but again didn't get the answer correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer to the second part is 25 $(n=25)$; and to the third is 55.2 picometers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="194" LastEditorDisplayName="user2619" LastEditDate="2017-09-24T15:43:59.720" LastActivityDate="2020-09-08T15:11:11.260" Title="Finding orbit radius using the Bohr model and Rydberg equation" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14979" PostTypeId="1" AcceptedAnswerId="14984" CreationDate="2014-08-07T05:24:17.073" Score="7" ViewCount="17863" Body="&lt;p&gt;I read in the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0582446953&quot; rel=&quot;noreferrer&quot;&gt;Guidebook to Mechanism in Organic Chemistry&lt;/a&gt; that increasing the polarity of the solvent slightly reduces the rate of an $S_N2$ reaction. Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question:&#xA;We know that in $S_N2$, using a polar protic solvent would reduce the rate as the nucleophile gets solvated by hydrogen bonding, but I found in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1118133579&quot; rel=&quot;noreferrer&quot;&gt;Solomon's Organic Chemistry&lt;/a&gt; that the $S_N1$ reaction rate increases in a polar protic solvent. Why does this occur? The nucleophile still gets solvated right?&lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorUserId="73" LastEditDate="2014-08-07T21:28:09.857" LastActivityDate="2014-08-07T21:28:09.857" Title="What is the result when the polarity of the solvent is increased in nucleophilic substitution?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14980" PostTypeId="2" ParentId="14976" CreationDate="2014-08-07T05:50:56.333" Score="1" Body="&lt;p&gt;Actually $Mg(OH)_2$ has (low) solubility ($K_{sp}=1.5\cdot 10^{-11}$). Since it is asking for the net ionic equation, the only ions reacting (since $NO_3^-$ stays in solution) are $H^+$ and $OH^-$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You forgot that each molecule of ammonium carbonate produces 3 ions, so multiply your answer by 3.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-07T05:50:56.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14981" PostTypeId="1" AcceptedAnswerId="14995" CreationDate="2014-08-07T06:29:28.273" Score="47" ViewCount="73941" Body="&lt;p&gt;I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how?&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="5026" LastEditDate="2018-03-02T15:53:33.150" LastActivityDate="2019-01-21T23:29:17.747" Title="Why does bond angle decrease in the order H2O, H2S, H2Se?" Tags="&lt;bond&gt;&lt;periodic-trends&gt;&lt;hybridization&gt;&lt;molecular-structure&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="14982" PostTypeId="2" ParentId="14977" CreationDate="2014-08-07T07:21:12.053" Score="5" Body="&lt;p&gt;To answer the second part:&lt;/p&gt;&#xA;&lt;p&gt;We know &lt;span class=&quot;math-container&quot;&gt;$M=208m_e$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Z=3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\hbar=\frac{h}{2\pi}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Part one has a mistake, as it is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;&amp;amp;&amp;amp;\frac{Mv^2}{r}&amp;amp;=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\&#xA;&amp;amp;&amp;amp;Mvr&amp;amp;=n\hbar\\&#xA;\implies&amp;amp;&amp;amp; r &amp;amp;=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We also know the Bohr radius:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore we can write and cancel:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;&amp;amp;&amp;amp; r&amp;amp;=a_0\\&#xA;&amp;amp;&amp;amp; \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &amp;amp;=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\&#xA;\therefore&amp;amp;&amp;amp; Z\frac{M}{m_e}&amp;amp;=n^2\\&#xA;\therefore&amp;amp;&amp;amp; n&amp;amp;=\sqrt{Z\cdot208}\approx25&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The third part:&lt;/p&gt;&#xA;&lt;p&gt;The Rydberg Formula is given as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with the Rydberg &lt;span class=&quot;math-container&quot;&gt;$\mathcal{R}$&lt;/span&gt; constant defined for a photon emitted by an electron. We'll assume that the mass of the nucleus is 7 atomic units (three protons + four neutrons). Taking into account that &lt;span class=&quot;math-container&quot;&gt;$m_p\approx 1836m_e$&lt;/span&gt;, we arrive at&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{7\cdot1836m_e}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now the Rydberg constant has to be modified to include the mass of the particle:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$&lt;/span&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;), I got to &lt;span class=&quot;math-container&quot;&gt;$\lambda_\mathrm{vac}=55.6~\mathrm{pm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Without taking the reduced mass into account, i.e. &lt;span class=&quot;math-container&quot;&gt;$\mathcal{R}\approx\mathcal{R}_\infty$&lt;/span&gt; I arrived at &lt;span class=&quot;math-container&quot;&gt;$\lambda_\mathrm{vac}=54.8~\mathrm{pm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Both values are reasonably close to the solution given.&lt;/p&gt;&#xA;&lt;p&gt;(If the question was really about the muon, the more accurate weight ratio is 206.77 and the corresponding wavelengths 55.1 pm and 56.0 pm.)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="95509" LastEditDate="2020-09-08T15:11:11.260" LastActivityDate="2020-09-08T15:11:11.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14983" PostTypeId="2" ParentId="14828" CreationDate="2014-08-07T08:08:40.873" Score="2" Body="&lt;p&gt;You can use 3 different approaches to solve that. One I've given in the comments, another one that I will use here that hopefully sheds some light to the equation which by itself out of context here doesn't make much sense.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In say 1000g of water you have 0.554 mol of compound (definition of molality)&lt;/li&gt;&#xA;&lt;li&gt;In 1000 ml of solution you have 0.497 mol of compound (definition of molarity)&lt;/li&gt;&#xA;&lt;li&gt;Thus in 1115 ml of solution you have 0.554 mol (linearity)&lt;/li&gt;&#xA;&lt;li&gt;So you have 1000g of water and 0.554 mol of compound in 1115 ml of solution or 1115*1.063 = 1185 g of solution (using the density)&lt;/li&gt;&#xA;&lt;li&gt;Obvioulsy enough you have 1185-1000 = 185g of compound. But we know already that there are 0.554 mol of the compound. So its molecular weight is 185/0.554 = 333.8 (so it's not sugar?)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Same result gives my (preferred) method in the comments&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-08-07T08:08:40.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14984" PostTypeId="2" ParentId="14979" CreationDate="2014-08-07T08:27:28.940" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I read in a (book) that increasing the polarity of the solvent slightly reduces the rate of SN2 reaction. Why does this happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Increasing the polarity of the solvent increases the solvation of the nucleophile. There will be increased dipole-dipole interactions between a solvent and a nucleophile as polarity of the solvent increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will reduce the mobility of the nucleophile and in turn reduce the rate of reaction in an SN2 reaction. The SN2 mechanism's rate is determined by the rate at which the nucleophile and the substrate can randomly collide into each other (in the right way - i.e. you want the nucleophile to attack the backside of the leaving group). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But I found in another book that the SN1 reaction rate increases in polar protic solvent. Why does this occur? The nucleophile still gets solvated right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the nucleophile is still solvated, but in an SN1 reaction, the rate-determining step is the rate at which the leaving group leaves. That is by far the slowest step in an SN1 reaction because of the high activation energy involved in creating a carbocation, which is highly unstable. So it doesn't really matter if the nucleophile is heavily solvated or not so heavily solvated - the slowest step determines the rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, regarding why polar protic solvent increase the rate of SN1 reactions - think about the carbocation. There is a high activation energy barrier in forming this carbocation - it's extremely unstable. Protic solvents can stabilize this carbocation through hydrogen bond formation. This lowers the activation energy barrier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpIGu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNgzi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-07T18:13:00.823" LastActivityDate="2014-08-07T18:13:00.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14985" PostTypeId="1" CreationDate="2014-08-07T08:28:37.170" Score="2" ViewCount="63" Body="&lt;p&gt;Non-chemist here.  I'm curious how difficult it is to determine concentration of PAH in a sample of soil or powdered charcoal, or a sample of the two mixed together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be involved in doing so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it particularly specialized, or the sort of thing that any trained chemist could do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it particularly difficult?  Costly?  Specialized?  Or would most any chemistry lab have the tools to do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea if there are any commercial labs where samples can be shipped for analysis?  Anywhere in the world, or in the US in particular?&lt;/p&gt;&#xA;" OwnerUserId="1570" LastActivityDate="2014-08-07T08:28:37.170" Title="What sort of lab can determine PAH concentration in a soil or charcoal sample? How difficult is it to do?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;aromatic-compounds&gt;&lt;toxicity&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14986" PostTypeId="1" AcceptedAnswerId="14991" CreationDate="2014-08-07T11:06:05.230" Score="1" ViewCount="576" Body="&lt;p&gt;I understand Cu&lt;sub&gt;0.75&lt;/sub&gt;Al&lt;sub&gt;0.25&lt;/sub&gt; - that's just 75% Cu, 25% Al. I understand integer indexes - that's a normal chemical formula after all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea how to interpret Cu&lt;sub&gt;1.5&lt;/sub&gt;Al, I've never seen a fraction in a chemical formula, and if it was a composition-percentage it should be like in the 0.75-0.25 example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at between 50% and 60%, below 630°C in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;http://www1.asminternational.org/asmenterprise/apd/help/PhaseLabels.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; but still don't understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/smS0v.png&quot; alt=&quot;Cu Al phase diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5022" LastActivityDate="2017-07-31T21:32:54.663" Title="Meaning of Cu1.5Al in Cu-Al phase diagram (I understand all other indexes incl. fractions like Cu0.75Al0.25)" Tags="&lt;nomenclature&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14987" PostTypeId="1" AcceptedAnswerId="14993" CreationDate="2014-08-07T12:59:36.707" Score="1" ViewCount="1725" Body="&lt;p&gt;What are the resonance structures of $\ce{CH3COCN}$ molecule?&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="1878" LastEditDate="2014-08-07T14:25:28.823" LastActivityDate="2014-08-07T20:21:58.500" Title="How to draw resonance structures for CH3COCN" Tags="&lt;bond&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14988" PostTypeId="1" AcceptedAnswerId="14994" CreationDate="2014-08-07T13:03:52.440" Score="2" ViewCount="704" Body="&lt;p&gt;Is there any tutorial for &lt;strong&gt;how to manually write the topology file for any Ligand file(PDB for mol2)&lt;/strong&gt;? There are so many sites which provides it like ProDrug or ATB but no where the actual process is written. Can anyone explain the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am actually not sure this question belong to which SO.&lt;/p&gt;&#xA;" OwnerUserId="7120" LastActivityDate="2017-02-20T12:53:15.533" Title="Writing Topology file for Ligand for Gromacs" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14989" PostTypeId="1" AcceptedAnswerId="14990" CreationDate="2014-08-07T13:20:13.877" Score="2" ViewCount="815" Body="&lt;p&gt;There are some large tanks of ammonia in nearby iron works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0Anj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder, what do they need it for? What is the application of ammonia in steel industry - what processes of making or processing steel require this much ammonia?&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2014-08-07T13:26:47.807" Title="How is ammonia used in production or processing of steel?" Tags="&lt;inorganic-chemistry&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14990" PostTypeId="2" ParentId="14989" CreationDate="2014-08-07T13:26:47.807" Score="2" Body="&lt;p&gt;Likely, for nitriding &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitriding&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nitriding&lt;/a&gt; . Essentially, ammonia is used as nitrogen donor. Nitrogen atoms diffuse into steel, producing hardened, corrosion-resistant surface layer.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-07T13:26:47.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14991" PostTypeId="2" ParentId="14986" CreationDate="2014-08-07T13:28:42.663" Score="3" Body="&lt;p&gt;As the page you referenced mentioned, things can get a bit tricky.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will start with your example of $\ce{Cu_{0.75}Al_{0.25}}$ - that is not just $75%$ Cu and $25%$ Al - that phase has a particular crystal structure, with the prototype (basically the first identified alloy with that crystal structure) of W (tungsten). So, it has the arrangement of atoms in the unit cell in the same places as pure tungsten, but $1/4$ of them are Al, and $3/4$ of them are Cu. There is no ordering of these atoms on the W lattice sites, or else the symmetry would be different, and the prototype might just be $\ce{Cu3Al}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{Cu_{1.5}Al}$ it is similar, but perhaps even more confusing.  The 'Published phase label' is $\ce{Cu_{1.5}Al}$, so that is what it is usually called in the literature.  The 'formula' is $\ce{Cu_{1.7}Al}$, which just means when people started referring to the phase they didn't have an accurate composition measurement.  Then, the 'Prototype' is $\ce{Co_{1.75}Ge}$.  That means that the first similar crystal structure was measured in the Co-Ge system.  The non-integer subscript(s) indicate that the unit cell has vacancies in the ideal structure (without looking it up, I would speculate that the Co sites are $7/8$'s occupied). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further complicate matters, if you look at the Co-Ge phase diagram you will find that the phase label is now $\ce{Co_{1.67}Ge}$, Formula is $\ce{Co_{1.5}Ge}$, and the prototype is still called $\ce{Co_{1.75}Ge}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: crystal structures can be complicated, identification of composition has been getting better over time (folks have been looking at the Al-Cu system for a long time!), and if you really want to know what is going on you probably have to dive into the x-ray diffraction literature and look at the published atom positions in a given phase.       &lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="41328" LastEditDate="2017-07-31T21:32:54.663" LastActivityDate="2017-07-31T21:32:54.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14992" PostTypeId="2" ParentId="14979" CreationDate="2014-08-07T13:29:17.687" Score="7" Body="&lt;p&gt;The standard textbook explanation is based on a comparison of charge separation in the ground state and the transition state (TS).  If there is more charge separation in the TS than the ground state, then polar solvents stabilize the TS more than the ground state and the reaction will be accelerated. On the other hand, if there is more charge in the ground state than in the TS, then polar solvents will stabilize the ground state more than the TS and consequently the reaction will slow down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look further at the effect of charge separation and solvent on these two specific reactions and see how they are affected.  Let's start with the $\ce{S_{N}1}$ reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I found in another book that the SN1 reaction rate increases in polar&#xA;  protic solvent. Why does this occur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the $\ce{S_{N}1}$ reaction, the rate determining step (rds) is formation of the carbocation intermediate. Let's say we start with a charged nucleophile and a neutral compound.  Moving along the reaction coordinate, by the time we reach the carbocation intermediate, the starting materials (charged nucleophile plus neutral compound) have transformed into a charged carbocation plus a charged leaving group plus the charged nucleophile.  Moving from starting materials to the intermediate has created additional charged species (carbocation plus leaving group).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -&amp;gt; [R^+ +X^{-} + Nu^- ]~~ (the~ TS)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the transition state leading to these intermediate species will strongly resemble the intermediate, it will also have more charge separation then the starting materials and therefor, will be more stabilized by polar solvents then the starting materials.  This stabilization of the transition state will lower its energy and speed up the rds.  In other words a polar solvent has sped up the rate of the $\ce{S_{N}1}$ because more charge exists in the transition state than in the ground state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I read in a (book) that increasing the polarity of the solvent&#xA;  slightly reduces the rate of SN2 reaction. Why does this happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, lets perform the same analysis on the $\ce{S_{N}2}$ reaction.  Again we start out with a charged nucleophile and a neutral compound.  The rds in this case is the formation of a trigonal bipyramid species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -&amp;gt; [\delta^{-} ~~ X...R...Nu ~~\delta^{-} ]~~ (the~ TS)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ground state there was one unit of negative charge residing fully on the nucleophile. In the TS that unit of charge is now spread out over both the nuceophile and the leaving group; the charge has become much more diffuse, the charge density per unit volume has decreased in the TS compared to the ground state.  In this case, we would expect a polar solvent to stabilize the ground state (starting materials) slightly more than the still charged - but more diffusely charged - TS.  Stabilizing the starting materials slightly more than the TS would slightly increase the energy required to pass over the TS.  Hence we would expect polar solvents to slightly slow down the rate of an $\ce{S_{N}2}$ reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-07T16:15:23.423" LastActivityDate="2014-08-07T16:15:23.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14993" PostTypeId="2" ParentId="14987" CreationDate="2014-08-07T13:49:22.110" Score="6" Body="&lt;p&gt;Here are the significant resonance structures for pyruvonitrile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/to8xU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-07T20:21:58.500" LastActivityDate="2014-08-07T20:21:58.500" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14994" PostTypeId="2" ParentId="14988" CreationDate="2014-08-07T14:48:57.550" Score="5" Body="&lt;p&gt;First off, you should not be manually writing topology files whenever you can avoid it - it's far too prone to making silly errors. That's why there are so many programs out there that attempt to automate the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however, you just want to understand HOW the topology file is structured, I recommend reading the GROMACS manual (chapter 5). &lt;a href=&quot;http://www.bevanlab.biochem.vt.edu/Pages/Personal/justin/gmx-tutorials/&quot; rel=&quot;noreferrer&quot;&gt;These tutorials&lt;/a&gt; also do a good job of describing some of the key aspects of the topology files.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastEditorUserId="2008" LastEditDate="2014-08-07T15:01:34.507" LastActivityDate="2014-08-07T15:01:34.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14995" PostTypeId="2" ParentId="14981" CreationDate="2014-08-07T14:52:38.413" Score="41" Body="&lt;p&gt;Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:&#xA;\begin{array}{lcc}&#xA;\text{molecule} &amp;amp; \text{bond angle}/^\circ &amp;amp; \text{bond length}/\pu{pm}\\&#xA;\hline&#xA;\ce{H2O} &amp;amp; 104.5 &amp;amp;  96 \\&#xA;\ce{H2S} &amp;amp;  92.3 &amp;amp; 134 \\&#xA;\ce{H2Se}&amp;amp;  91.0 &amp;amp; 146 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$.  On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization.  That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively.  Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons.  If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$.  We see from the above table that we are very close to the measured values.  We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider.  This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum.  The basic concept introduced is that &quot;orbitals only hybridize in response to bonding.&quot; So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2018-06-19T11:17:24.630" LastActivityDate="2018-06-19T11:17:24.630" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="14996" PostTypeId="1" AcceptedAnswerId="14998" CreationDate="2014-08-07T16:57:27.020" Score="0" ViewCount="2085" Body="&lt;p&gt;You spend a lot of time in chemistry talking about molecular amd ionic compounds, but I'm interested in what you can do with metals only. What is the difference between the properties (physical &amp;amp; chemical) of pure metals and those of alloys? Is it possible to predict the properties of alloys before making them? Can you make alloys with mercury?&lt;/p&gt;&#xA;" OwnerUserId="7344" LastActivityDate="2014-08-07T17:45:42.183" Title="What are the properties of alloys?" Tags="&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-03-24T01:42:28.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="14997" PostTypeId="2" ParentId="14961" CreationDate="2014-08-07T17:37:47.010" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why unbonded electrons exits in pair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Unbonded, non-bonded or lone pair electrons are terms used to describe electrons that surround an atom, but don't play a direct role in bond formation.  Importantly, these electrons do not always exist as lone pairs where the electrons reside in a common orbital.  Sometimes they exist as single non-bonding electrons residing in separate orbitals.&lt;/p&gt;&#xA;&lt;p&gt;Let's compare the non-bonding electrons in water to those in molecular oxygen.  Here is an illustration of the molecular orbital arrangement for water.  As we fill the orbitals with electrons according to the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations/Aufbau_Principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aufbau Principle&lt;/a&gt; we arrive at the point where we have two electrons remaining and the next available molecular orbital is the &lt;span class=&quot;math-container&quot;&gt;$\ce{1b_1}$&lt;/span&gt; orbital.  According to &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Electronic_Configurations/Hund%27s_Rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hund's Rule&lt;/a&gt;, we must place these two electrons into this orbital with their spins paired.  The result is that we have two non-bonding electrons paired up in the same orbital - a lone pair of electrons.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBt4x.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's next examine the situation in &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;.  Here is the molecular orbital diagram for &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;.  As we again fill the orbitals with electrons according to the Aufbau Principle we again arrive at the point where we have two electrons remaining and the next available molecular orbital is the &lt;span class=&quot;math-container&quot;&gt;$\ce{\pi^{x}_{g}}$&lt;/span&gt;.  But we see that there is a second orbital at exactly the same energy (degenerate) and it is the &lt;span class=&quot;math-container&quot;&gt;$\ce{\pi^{y}_{g}}$&lt;/span&gt; orbital.  According to Hund's Rule, rather than put both electrons into one of these orbitals, we place just one electron in each of these two degenerate orbitals.  So while&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; does have two non-bonding electrons, they are not paired up - there is no lone pair, these electrons exist in separate orbitals and &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; consequently has a triplet ground state resulting from these two unpaired non-bonding electrons.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXuqa.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So sometimes a non-bonding pair of electrons will exist as a lone pair (as in the case of water), sometimes they won't exist as a lone pair as in the case of oxygen.  To solve these kinds of problems one must know (or guess) the structure of the molecule and then determine the molecular orbital arrangement.  Then, when we fill these orbitals with electrons we will see if any non-bonding electrons will exist as a lone pair in a single orbital, or if the two non-bonding electrons will exist in separate orbitals and remain unpaired.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-08-07T17:37:47.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14998" PostTypeId="2" ParentId="14996" CreationDate="2014-08-07T17:45:42.183" Score="2" Body="&lt;p&gt;What seems to be a set of simple questions leads one into the field of metallurgy, computational thermodynamics, mechanical properties, and goodness knows what else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, steel has very different mechanical properties from pure iron.  There are about a bazillion different kinds of steel, each with different impurities mixed in to give specific mechanical properties, optimizing strength, fatigue life, corrosion resistance, etc. Much of the development has been through insight gained by experience.  More recently, the CALPHAD approach is giving more fundamental insight through thermodynamic modeling, but predicting the mechanical properties $\it a\ priori$ is still difficult to impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for mercury, there are a large number of binary phases of Hg with other elements.  These are readily found in binary alloy references, or online at the ASM Alloy Phase Diagram Database (assuming your library has access).  ASM has info on &gt;60 Hg binary diagrams, and a number of ternary systems as well. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-07T17:45:42.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14999" PostTypeId="2" ParentId="14971" CreationDate="2014-08-07T18:36:53.010" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What exactly does it mean for the reaction of superoxide with&#xA;  non-radicals to be spin-forbidden?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For clarity, let's start with a simpler case.  If we take a ground state singlet molecule (singlet means all electrons are paired) and we irradiate it with photons, photon absorption will excite an electron to an excited state.  Will the excited state be a singlet or triplet immediately after it is produced?  For the transition to the excited state to occur readily it must be spin-allowed.  That means that spin must be conserved.  If we started with a singlet, then our excited state must still be a singlet.  In that excited state the excited electron will still have its spin paired with the electron that was its partner that remains in a lower energy orbital.  Creation of a triplet state, where the excited electron and its ground state partner no longer have their spins paired is spin-forbidden and is therefor much less likely to occur.  Spin must be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well rules are made to be broken.  What can we do to increase the probability of a singlet -&gt; triplet conversion that is formally spin-forbidden. &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin-orbit_coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spin-Orbit coupling&lt;/a&gt; (the coupling of the electron spin with the orbital angular momentum of non-circular orbits) is our way out. We need to find a spot where the Morse potential curves overlap for both the singlet and triplet. Then, if there is a large spin-orbit coupling interaction we can readily cross over from singlet to triplet, or back.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q5BBK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large atoms such as bromine and iodine, as well as transition metals with unpaired electrons have large spin-orbit couplings.  If such groups are attached to our molecule of interest, then spin need not be conserved and formally &quot;spin-forbidden&quot; reactions become more likely.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the unpaired one have to bond with an unpaired one from another&#xA;  radical? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, if spin is to be conserved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does this also say about reactions of the excited state&#xA;  of singlet oxygen (shown in the middle below) with non-radicals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The singlet oxygen shown in the middle of your figure ia a singlet species and will react readily with other singlet (non-radical) molecules.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is superoxide a doublet&#xA;  Yes superoxide is a doublet.  Ground state $\ce{O2}$ is a triplet - it has two unpaired electrons in energetically degenerate orbitals, if we add one electron to form superoxide radical anion, that electron will pair with one of those two electrons leaving only one unpaired electron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-07T18:36:53.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15001" PostTypeId="2" ParentId="14951" CreationDate="2014-08-07T19:01:36.007" Score="2" Body="&lt;p&gt;I'm not sure if anyone has done it, but the closest I can think of is the GPW (mixed Gaussian Plane Wave) method. I think that involves a mapping between Gaussian and PW basis. It's DFT, but just look for the Coulomb integrals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Lippert, Gerald, et al. &quot;A hybrid Gaussian and plane wave density functional scheme.&quot; Molecular Physics 92.3 (1997): 477-488.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implementation can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;VandeVondele, Joost, et al. &quot;Quickstep: Fast and accurate density functional calculations using a mixed Gaussian and plane waves approach.&quot; Computer Physics Communications 167.2 (2005): 103-128.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2014-08-07T19:01:36.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15002" PostTypeId="1" CreationDate="2014-08-07T21:10:26.077" Score="5" ViewCount="1207" Body="&lt;p&gt;I'm doing Grade 12 Chemistry and I'm unsure of this, taking the hydrogen electorde as having a potential of 0 volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use the copper-zinc cell as an example. Just let me know if I've got the full logic right:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Zinc has higher electronegativity, so due to the equilibrium $\ce{Zn &amp;lt;=&amp;gt; Zn^{2+} + 2e-}$, we have more charge building up in the Zinc electrode than in the Copper electrode. (Given both half-cells are of same concentration and in same temp)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We have a potential difference between the two electrodes. This means some charge from the Zinc electrode will move to the Copper electrode in such a way that both electrodes will momentarily have equal charges.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the charge in the Copper electrode has gone up, we have added electrons to the equilibrium reaction $\ce{Cu^{2+} + 2e- &amp;lt;=&amp;gt; Cu}$. We therefore create more copper.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this why an electrode with a higher potential will &quot;pull&quot; the electrons from the electrode with a lower potential?&lt;/p&gt;&#xA;" OwnerUserId="7371" LastEditorUserId="4945" LastEditDate="2014-08-08T02:43:41.833" LastActivityDate="2020-01-05T16:14:19.390" Title="In a galvanic cell, why does an electrode with a higher potential &quot;pull&quot; the electrons from the electrode with a lower potential?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15003" PostTypeId="1" AcceptedAnswerId="15004" CreationDate="2014-08-07T22:34:20.760" Score="4" ViewCount="5760" Body="&lt;p&gt;What can be the effects for the body to be immersed in Coca Cola for a while? A friend commented me that the coke is quite acidic (pH 2.5 if I recall properly), and we were guessing what could be the effects for the body after a time of immersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;http://www.collective-evolution.com/2013/09/22/what-happens-to-our-body-after-drinking-coca-cola/&quot; rel=&quot;nofollow&quot;&gt;20 Practical Uses for Coca Cola – Proof That Coke Does Not Belong In The Human Body&lt;/a&gt; and I see some things like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E338 – Orthophosphoric Acid. This can cause irritation of the skin and&#xA;  eyes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I am guessing: what are the problems of a long contact of coke with the body - not drinking, but being in touch with it.&lt;/p&gt;&#xA;" OwnerUserId="7374" LastEditorUserId="7374" LastEditDate="2014-08-11T09:37:38.897" LastActivityDate="2014-08-11T09:37:38.897" Title="How dangerous is taking a bath in Coca Cola?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15004" PostTypeId="2" ParentId="15003" CreationDate="2014-08-08T00:35:38.993" Score="8" Body="&lt;p&gt;Cola and soft drink products are generally in the pH=2.5-3.5 range.&#xA;This is the same pH range as vinegar, lemon juice and most fruit juices. Note apple itself is around 3.0-3.5, so it is not about some magical trick of food industry to kill us all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Cola dissolve your body traceless? No.&#xA;Do you want to soak your body in it for hours? Would you soak your body in vinegar for hours? Or even water... &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-08T00:35:38.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15005" PostTypeId="2" ParentId="15002" CreationDate="2014-08-08T02:15:40.583" Score="5" Body="&lt;p&gt;There are not many problems where electronegativity helps you find the answer.  It is popular for some reason amongst newer chem students but it shouldn't be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what happens in the Daniell cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons.  Both of these processes are endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc ions are hydrated. This is exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper ions are dehydrated. This is endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the above processes also involves an entropy change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, silver's electronegativity is 1.93 but the reduction potential for &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.8 V}$&lt;/span&gt; while &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.34 V}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="61578" LastEditDate="2020-01-05T16:14:19.390" LastActivityDate="2020-01-05T16:14:19.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15006" PostTypeId="1" CreationDate="2014-08-08T03:52:05.437" Score="2" ViewCount="137" Body="&lt;p&gt;Is it wrong way to use WATER to relieve the big disaster(propylene leak).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a better way to relieve propylene leak in a &lt;strong&gt;street about large area&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any possible choice to prevent explosion?&lt;/p&gt;&#xA;" OwnerUserId="7376" LastEditorUserId="7376" LastEditDate="2014-08-08T04:49:04.473" LastActivityDate="2014-09-04T14:32:43.353" Title="Taiwan propylene explosions one week ago" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15007" PostTypeId="1" CreationDate="2014-08-08T04:00:06.510" Score="46" ViewCount="23711" Body="&lt;p&gt;Why is the recipe for Coca-Cola still a secret?&#xA;I think that given the current state of technology, it should be proficient enough to find any of the secret ingredients in Coca Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts?&#xA;Can we make a 100% identical clone of Coca-Cola next week?&lt;/p&gt;&#xA;" OwnerUserId="7376" LastEditorUserId="5633" LastEditDate="2014-08-10T12:15:58.893" LastActivityDate="2018-07-21T00:15:18.523" Title="Why is the recipe of Coca Cola still a secret?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="8" CommentCount="8" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15009" PostTypeId="2" ParentId="15007" CreationDate="2014-08-08T06:31:02.613" Score="3" Body="&lt;p&gt;If we have high technology to find the secret of Coca-Cola then they also have high technology and money to hide it. If it would be possible then don't you think today there would hundreds of brand making the same drink?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Till now there are many brands which have tried to clone Coca-Cola but they haven't succeeded, and that means till today they are able to hide their secret, so we don't have the technology that can find the secret of Coca Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links which claim that they have cloned Coca-Cola &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.topsecretrecipes.com/Coca-Cola-Classic-Coke-Copycat-Recipe.html&quot; rel=&quot;nofollow&quot;&gt;http://www.topsecretrecipes.com/Coca-Cola-Classic-Coke-Copycat-Recipe.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.homebrewersassociation.org/forum/index.php?topic=6017.0&quot; rel=&quot;nofollow&quot;&gt;https://www.homebrewersassociation.org/forum/index.php?topic=6017.0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.instructables.com/id/Fakola-cola-clone/&quot; rel=&quot;nofollow&quot;&gt;http://www.instructables.com/id/Fakola-cola-clone/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find many more on Google, but I don't think any of them are able to find the secret of Coca-Cola, because if they had found it, then they would be billionaires and their drink would be in the market.  &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2014-08-09T09:50:29.120" LastActivityDate="2014-08-09T09:50:29.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15010" PostTypeId="2" ParentId="15007" CreationDate="2014-08-08T10:01:25.543" Score="61" Body="&lt;p&gt;There are different angles this question can be answered:&lt;/p&gt;&#xA;&lt;h1&gt;Chemical point of view:&lt;/h1&gt;&#xA;&lt;p&gt;A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution...  Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components.&lt;/p&gt;&#xA;&lt;p&gt;That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor:&lt;/p&gt;&#xA;&lt;h1&gt;Business point of view:&lt;/h1&gt;&#xA;&lt;p&gt;The recipe of Coca Cola itself worth nothing.&#xA;Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.&#xA;The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow &lt;strong&gt;DIFFERENT&lt;/strong&gt; than the other ones!&lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="-1" LastEditorDisplayName="user5764" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-03-05T02:06:40.830" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15013" PostTypeId="2" ParentId="6628" CreationDate="2014-08-08T12:03:37.593" Score="3" Body="&lt;p&gt;Sulphuryl Fluoride is chemically fairly stable, considering it's a pesticide - toxic effects are caused by chemical reactions which means they can't be inert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the Vikane documentation, it's not UV stable. UV lighting is fairly widely available, but TBH a sunny day will do far more.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastActivityDate="2014-08-08T12:03:37.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15014" PostTypeId="1" CreationDate="2014-08-08T12:04:38.507" Score="3" ViewCount="595" Body="&lt;p&gt;I have a sample of very fine graphite powder suspended in light oil. I've attempted to separate the two by placing the mix in a 2 micron filter sock. The oil will slowly drip through. If I apply pressure the graphite and oil will force their way through. No pressure, I believe the graphite will collect on the filter and eventually blind the filter preventing oil from coming through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how I can separate and re-use both items?&lt;/p&gt;&#xA;" OwnerUserId="7379" LastEditorUserId="42330" LastEditDate="2018-04-01T19:55:33.050" LastActivityDate="2018-04-01T19:55:33.050" Title="How do I separate graphite powder from light oil?" Tags="&lt;home-experiment&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15015" PostTypeId="2" ParentId="15007" CreationDate="2014-08-08T13:13:06.297" Score="24" Body="&lt;p&gt;With modern technology it would be a relatively straightforward process to determine all of the ingredients in Coca Cola.  It would require time and money, but very doable.  My guess is that companies that compete with Coca Cola, like Pepsi, know exactly all of the ingredients in Coke.  This is part of the process known as &quot;reverse engineering&quot; and is done all the time in industry.  Pharmaceutical companies, food companies, plastics companies, any industry that uses specialized formulations do this type of chemical analysis to better understand what the &quot;competition&quot; is up to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical analysis is actually the easy part.  However, Coke and all of these other manufacturers don't just take a list of chemicals or ingredients, mix them in a tub and then put it in the bottle.  The manufacturing &lt;strong&gt;process&lt;/strong&gt; is every bit as important as the ingredients and much harder to determine.  For example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in what order are the ingredients added&lt;/li&gt;&#xA;&lt;li&gt;how long do you mix the batch before adding the next ingredient&lt;/li&gt;&#xA;&lt;li&gt;at what rpm do you mix them&lt;/li&gt;&#xA;&lt;li&gt;what kind of mixer do you use&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and the list goes on and on.  There might be 100 ingredients in the Coke formula, but there are probably over 1,000 variables in the manufacturing process, and there is no straightforward analytical method you can use to specify all of the process parameters.  The ingredients is the easy part, the process is what stops someone from making a replica product - well that and legalities.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-08T13:13:06.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15016" PostTypeId="1" AcceptedAnswerId="15019" CreationDate="2014-08-08T14:16:07.707" Score="3" ViewCount="7265" Body="&lt;p&gt;I know that the standard textbook answer is that a polar protic solvent is able to better stabilize/solvate carbocations through hydrogen bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a solvent such as water or methanol is capable of hydrogen bonding, but what about the substrate? The substrate is a hydrocarbon; how is the substrate capable of &lt;em&gt;reciprocal&lt;/em&gt; hydrogen bonding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this stabilization better described as a dipole-ion interaction rather than a hydrogen bond interaction because I don't see how a hydrocarbon can form &lt;em&gt;reciprocal&lt;/em&gt; hydrogen bonding interactions. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-10-09T19:12:34.497" LastActivityDate="2016-07-24T10:49:45.330" Title="SN1 Polar Protic Solvent Stabilization of Substrate" Tags="&lt;organic-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15017" PostTypeId="1" AcceptedAnswerId="15026" CreationDate="2014-08-08T14:44:30.553" Score="7" ViewCount="10725" Body="&lt;p&gt;What is the composition of liquid nitrogen fumes? Is direct contact with these fumes harmful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l2jHL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l2jHL.jpg&quot; alt=&quot;beaker with liquid nitrogen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7120" LastEditorUserId="11367" LastEditDate="2017-01-09T19:11:53.410" LastActivityDate="2017-01-09T22:02:24.730" Title="What are the fumes that form above liquid nitrogen?" Tags="&lt;physical-chemistry&gt;&lt;safety&gt;&lt;molecules&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15019" PostTypeId="2" ParentId="15016" CreationDate="2014-08-08T17:00:59.000" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that the standard textbook answer is that a polar protic&#xA;  solvent is able to better stabilize/solvate carbocations through&#xA;  hydrogen bonding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an $\ce{S_{N}1}$ reaction, a polar protic solvent can stabilize the leaving group through both hydrogen bonding and solvent dielectric effects.  While these same solvent dielectric effects can also stabilize the carbocation intermediate, I don't think hydrogen bonding plays a role in the stabilization of the carbocation itself.  First off, any protic molecule close enough to the carbocation to hydrogen bond would simply react with it instead of forming a hydrogen bond. Second, since the carbocation is positively charged it cannot form a hydrogen bond.  Again, solvent stabilization of the carbocation occurs through dielectric (dipolar) effects of the medium.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-08T17:00:59.000" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="15020" PostTypeId="2" ParentId="13727" CreationDate="2014-08-08T17:13:05.513" Score="1" Body="&lt;p&gt;In simple terms, Full CI uses all of the possible orbitals (a &quot;full&quot; active space). Thus the number of active orbitals should equal the size of your basis set. If you have 10 basis functions, and then perform a Hartree-Fock calculation, you get 10 orbitals out (some occupied, some virtual).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete Active Space (CAS) methods, on the other hand, only use some of the orbitals and then perform a Full CI on that subset of orbitals, leaving the other orbitals (inactive space) alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full CI is CAS in the total orbital / total active space limit.&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2014-08-08T17:13:05.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15021" PostTypeId="2" ParentId="15007" CreationDate="2014-08-08T17:27:15.770" Score="15" Body="&lt;p&gt;Greg hits most of the points, but there's one other angle I'd like to add:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Practical Point of View:&lt;/strong&gt; One of the well established Coca-Cola ingredients is decocainized coca leaf extract. The DEA is, to say the least, not exactly generous in allowing this import. This effectively prevents the manufacture of a Coca-Cola equivalent anywhere in the United States. A competitor producing an exact clone of Coca-Cola but forced to import the finished product from outside the country is likely to find it difficult to significantly undercut Coca Cola's efficient distribution network.&lt;/p&gt;&#xA;" OwnerUserId="7392" LastActivityDate="2014-08-08T17:27:15.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15023" PostTypeId="1" CreationDate="2014-08-08T17:41:45.153" Score="2" ViewCount="948" Body="&lt;p&gt;I want to use FAS (fluoroalkylsilane) in my molecular dynamics study. How can I find it's real structure? So that I can find proper forcefield for simulating the molecule. It would be very nice if you mention a better method for simulating this molecule. Like making a initial configuration in a software and then optimizing the geometry. I don't know about DFT so much, but can I use that somehow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ke69w.png&quot; alt=&quot;FAS17:&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Kazem&lt;/p&gt;&#xA;" OwnerUserId="5573" LastEditorUserId="5573" LastEditDate="2014-08-08T19:09:49.230" LastActivityDate="2014-08-08T19:09:49.230" Title="Structure of fluoroalkylsilane" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15024" PostTypeId="2" ParentId="14886" CreationDate="2014-08-08T17:52:46.217" Score="4" Body="&lt;p&gt;The relative strengths of nucleophiles can be correlated with three structural features:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A negatively charged nucleophile is always a more reactive nucleophile than its conjugate&#xA;acid. Thus $\ce{HO^-}$ is a better nucleophile than $\ce{H_2O}$ and $\ce{RO^-}$ is better than $\ce{ROH}$.&lt;/li&gt;&#xA;&lt;li&gt;In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the following order of reactivity:&#xA;$$\ce{RO^- &amp;gt; HO^- &amp;gt;&amp;gt; RCO_2^- &amp;gt; ROH &amp;gt; H2O}$$&#xA;This is also their order of basicity. An alkoxide ion ($\ce{RO-}$) is a slightly stronger base than a hydroxide ion ($\ce{HO^-}$), a hydroxide ion is a much stronger base than a carboxylate ion ($\ce{RCO2^-}$), and so on.&lt;/li&gt;&#xA;&lt;li&gt;When the nucleophilic atoms are different, nucleophilicities may not parallel&#xA;basicities. For example, in protic solvents $\ce{HS, CN}$, and $\ce{I}$ are all weaker bases than $\ce{HO^-}$, yet they are stronger nucleophiles than $\ce{HO^-}$.&#xA;$$\ce{HS^- &amp;gt; CN^- &amp;gt;  I^- &amp;gt; HO^-}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nucleophilicity versus Basicity:&lt;/strong&gt; While nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by $\mathrm{p}K_\mathrm{a}$, is measured by the &lt;em&gt;position of an equilibrium&lt;/em&gt; involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by &lt;em&gt;relative rates of reaction&lt;/em&gt;, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the hydroxide ion $\ce{OH^-}$ is a stronger base than a cyanide ion $\ce{CN^-}$; at equilibrium it has the greater affinity for a proton ($\mathrm{p}K_\mathrm{a}(\ce{H_2O})= 16$, while $\mathrm{p}K_\mathrm{a}(\ce{HCN})= 10$). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5026" LastEditDate="2018-05-06T13:54:45.870" LastActivityDate="2018-05-06T13:54:45.870" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15026" PostTypeId="2" ParentId="15017" CreationDate="2014-08-08T18:25:22.577" Score="26" Body="&lt;p&gt;The visible fog that forms when liquid nitrogen is poured into an open container is almost entirely &lt;em&gt;water&lt;/em&gt; fog: the boiling nitrogen chills the air above it, causing the humidity in the air to condense into fog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some microdroplets of liquid nitrogen in the fog too, but the air, even after it's been chilled, is still much warmer than the boiling point of nitrogen, so there won't be much.  (You &lt;em&gt;can&lt;/em&gt; condense air by running a continuous stream of liquid nitrogen through a heat exchanger and blowing air over it, but mostly you get liquid &lt;em&gt;oxygen&lt;/em&gt; when you do that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid nitrogen has a relatively small heat of vaporization, as these things go: 199.2 kJ/kg according to &lt;a href=&quot;http://encyclopedia.airliquide.com/Encyclopedia.asp?GasID=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Air Liquide&lt;/a&gt;.  (Compare &lt;a href=&quot;http://www.continentalcarbonic.com/dry-ice-material-safety-data-sheet---msds.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;572.2 kJ/kg for dry ice&lt;/a&gt;.)  That plus the &lt;a href=&quot;https://en.wikipedia.org/wiki/Leidenfrost_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leidenfrost effect&lt;/a&gt; means it's actually rather difficult to freeze yourself with the stuff. There's a standard demo where you pour liquid nitrogen into an audience member's cupped hands. The Leidenfrost effect prevents it ever coming in contact with their skin, and it only leaves a mild chill behind.  Try this with dry ice and your &quot;volunteer&quot; will need emergency treatment for frostbite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaseous nitrogen is second only to the noble gases in its inertness, and it makes up 78% (by volume) of the air you're breathing as you read this, so obviously the &quot;fumes&quot; aren't harmful in themselves.  However, as was mentioned in an earlier answer (which seems to have been deleted), if you let liquid nitrogen evaporate in an enclosed space, it can &quot;displace&quot; the oxygen that makes up another 21% of the air, at which point anyone in the area may &lt;a href=&quot;https://en.wikipedia.org/wiki/Inert_gas_asphyxiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;asphyxiate&lt;/a&gt;.  The sensation of suffocation is triggered by having &lt;em&gt;too much&lt;/em&gt; CO&lt;sub&gt;2&lt;/sub&gt; in your bloodstream, &lt;em&gt;not&lt;/em&gt; by inadequate oxygen supply, so you might not even notice that anything was wrong until you collapsed.  That's the second most dangerous thing about evaporating liquid nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;most&lt;/em&gt; dangerous thing is, if you let liquid nitrogen evaporate in a &lt;em&gt;sealed container&lt;/em&gt;, &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2006/03/08/how_not_to_do_it_liquid_nitrogen_tanks&quot; rel=&quot;nofollow noreferrer&quot;&gt;it's liable to explode on you&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4921" LastEditorUserId="4921" LastEditDate="2017-01-09T22:02:24.730" LastActivityDate="2017-01-09T22:02:24.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15028" PostTypeId="1" AcceptedAnswerId="15030" CreationDate="2014-08-08T20:42:45.120" Score="3" ViewCount="2107" Body="&lt;p&gt;I wonder if sublimation can be done only on certain substances like iodine, or it can be done on all substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, can sublimation be done, for example, on water? If it cannot, then why?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="39539" LastEditDate="2017-04-12T22:35:45.140" LastActivityDate="2017-04-12T22:35:45.140" Title="Is sublimation a property of all substances?" Tags="&lt;physical-chemistry&gt;&lt;vapor-pressure&gt;&lt;sublimation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15029" PostTypeId="2" ParentId="15007" CreationDate="2014-08-08T20:56:41.533" Score="-2" Body="&lt;p&gt;Universities have HPC's that could probably do it in a week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point is, someone has to pay for it and it would be very costly and illegal.  Perhaps the reason it has not been done.&lt;/p&gt;&#xA;" OwnerUserId="7396" LastActivityDate="2014-08-08T20:56:41.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15030" PostTypeId="2" ParentId="15028" CreationDate="2014-08-08T20:59:44.643" Score="7" Body="&lt;p&gt;Look up the term “phase diagram”. At different temperatures and pressures different phases are favored. Think about water. At 0 °C and 1 atm water and ice are equally favorable. There can be freezing or melting. At other temperatures and pressures there can be sublimation and deposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sublimation of water is involved in freeze drying and freezer burn of ice cream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that iodine sublimes at room temperature and atmospheric pressure but there’s nothing special about iodine compared to other substances on a fundamental level.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T12:03:48.837" LastActivityDate="2015-04-26T12:03:48.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15031" PostTypeId="2" ParentId="15028" CreationDate="2014-08-08T21:37:30.837" Score="3" Body="&lt;p&gt;Sublimation refers to molecules passing from the solid to the gas state without going through the liquid state. This happens all the time to every solid, including water, it's just a question of at what rate it sublimes and how that rate compares to its rate of melting in the same conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Snow, for example, often noticeably sublimes instead of melting, especially when in sunlight only a few degrees above freezing temperature. It is more prone to sublime than ice is due to its lower density, which allows air to pass into it and warm more than just the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason iodine, for example, so noticeably sublimes is because at normal temperature and pressure it cannot exist for more than an infinitesimal amount of time in the liquid state. There are regions in the phase diagrams of most substances where the liquid state cannot exist in equilibrium with either the solid or gas state when temperature or pressure changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water, this is at very low pressure, below 1/100 of normal atmospheric pressure. When you heat ice at this pressure, it will go directly to gas without being in the liquid state in any appreciable amount.&lt;/p&gt;&#xA;" OwnerUserId="6907" LastActivityDate="2014-08-08T21:37:30.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15033" PostTypeId="2" ParentId="14926" CreationDate="2014-08-08T22:47:54.143" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.thestudentroom.co.uk/showthread.php?t=2017277&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; are a few recommendations for A-level chemistry, made by fellow students such as yourself. TSR is a good student resource for A-level prep since there is are masses of students talking about the same topics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1285199030&quot; rel=&quot;nofollow&quot;&gt;Zumdahl&lt;/a&gt; is a good general chemistry textbook as well, with quite the number of practice problems. However it is a full-fledged textbook you probably don't need a few days before an exam.)&lt;/p&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-08-08T22:47:54.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15034" PostTypeId="1" AcceptedAnswerId="15042" CreationDate="2014-08-09T03:30:18.107" Score="6" ViewCount="6355" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is it so that the size of $\ce{Al^{3+}}$ is lesser than $\ce{Li+}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a bit confused about this one as the aluminium(III) ion is composed of 10 electrons and 13 protons, while the lithium ion has only 2 electrons and 3 protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronic configuration of $\ce{Al^{3+}}$ is $\mathrm{1s^2 2s^2 2p^6}$ and that of $\ce{Li+}$ is $\mathrm{1s^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it seems obvious that the size of $\ce{Al^{3+}}$ should be greater than that of $\ce{Li+}$, but it is actually the other way around. &lt;strong&gt;Why is this?&lt;/strong&gt; I think that it is a kind of exception, but if anyone can explain the reason behind this, it will help me in my upcoming tests.&lt;/p&gt;&#xA;" OwnerUserId="3794" LastEditorUserId="73" LastEditDate="2014-08-23T03:13:07.847" LastActivityDate="2015-09-25T19:18:50.477" Title="Why is the size of Al3+ less than that of Li+?" Tags="&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15035" PostTypeId="1" AcceptedAnswerId="15045" CreationDate="2014-08-09T04:14:12.693" Score="33" ViewCount="10412" Body="&lt;p&gt;I have noticed that liquid $\ce{O2}$ (I clarify it as $\ce{O2}$ because oxygen exists in several other forms which have different colors in the liquid state than $\ce{O2}$) has a light blue color to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NW9kM.jpg&quot; alt=&quot;liquid oxygen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solid state it has a light blue color like it does in the liquid state and then finally a red color if enough pressure is applied. It can also be pink, black, or metallic again depending on the pressure and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it also have a light blue color in the gas phase? &lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user15489" LastEditDate="2015-04-24T20:29:48.753" LastActivityDate="2016-08-13T19:31:20.287" Title="Does O2 have a color in the gas phase" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15036" PostTypeId="1" AcceptedAnswerId="15064" CreationDate="2014-08-09T04:50:21.577" Score="2" ViewCount="37985" Body="&lt;p&gt;What is the name of this formula? Google could not provide me with an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E=q+w$$&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="4945" LastEditDate="2014-08-11T05:15:29.797" LastActivityDate="2019-02-24T12:57:42.550" Title="What is the name of the formula ΔE = q + w?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15037" PostTypeId="2" ParentId="15035" CreationDate="2014-08-09T05:01:52.993" Score="0" Body="&lt;p&gt;no, in gas phase it's colorless.&#xA;someone suggests me to be specific,&#xA;then let me show you oxygen gas's UV absorption spectrum calculated in Gaussian and visualized in GaussView. charge=0 triplet 1 atm 298K zindo&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6GSfK.png&quot; alt=&quot;charge=0 triplet 1 atm 298K zindo&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7158" LastEditorUserId="7158" LastEditDate="2014-08-10T01:29:28.480" LastActivityDate="2014-08-10T01:29:28.480" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15038" PostTypeId="2" ParentId="15036" CreationDate="2014-08-09T06:32:57.690" Score="3" Body="&lt;p&gt;The first law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-08-09T06:32:57.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15040" PostTypeId="2" ParentId="15017" CreationDate="2014-08-09T10:24:02.250" Score="1" Body="&lt;p&gt;The fog / vapor / fumes that come off a container of any liquefied gas will be a combination of condensed water vapor and whatever it is that is boiling. It is no more or less dangerous than the gas itself. In the case of nitrogen, oxygen, helium, neon, argon, sulfur hexaflouride, methane etc. you can put your nose right up to the liquid's surface and take a really deep breath, no harm will come to you unless you are stupid enough to breathe it for an extended period. Oxygen, of course, is the exception but too much O2 is bad for you as well. Don't try it with &lt;a href=&quot;https://www.youtube.com/watch?v=dIJDfNO3d5E&quot; rel=&quot;nofollow&quot;&gt;chlorine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the Leidenfrost effect and pouring liquid N2 into someone's cupped hand, lets be sure the palm is facing down, not up. Leidenfrost only works for a couple of seconds on room-temperature objects like people.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-08-09T10:24:02.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15042" PostTypeId="2" ParentId="15034" CreationDate="2014-08-09T13:45:43.497" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, it seems obvious that the size of Al3+ should be greater than that of Li+&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No it does not. It is true that electrons repulse each other, however, they all are attracted towards the nucleus.  The real 'size' of an atom or ion is a result of compromise between electron-electron repulsion and electron-nuclear attraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think that it is a kind of exception&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, it is not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is this? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Commonly, there are three consideration, used to rationalize basic trends in atomic/ionic radii.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;electron-electron repulsion (negative charges repulse each other)&lt;/li&gt;&#xA;&lt;li&gt;electron-nucleus attraction (negative charges are attracted towards positive ones)&lt;/li&gt;&#xA;&lt;li&gt;electron shielding (electrons in shells closer to nucleus 'shield' outer electrons from some charge of the nucleus.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;While electrons repulse each other, increasing nuclear charge more then compensates it; so if not for shell organisation of electrons, atoms would monotonically decrease in size toward the end of the periodic table, as it is observable in each period 1—3.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, inner electronic shells effectively shield the outer electrons from the nucleus, so they 'see' a nucleus that is a) larger in size and b) has charge equal to nuclear charge minus number of electrons in inner shells. Given that, outer electrons 'see' a nucleus with low charge density, so outer shells progressively increase in size within the same column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, electrons still repulse each other, so positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider $\ce{Li+}$ and $\ce{Al^{3+}}$. For the former, 2 outer electrons feel the charge of three protons. For the latter, 8 outer electrons feel the charge of 11 protons in the nucleus. Such great charge easily beats all other factors, so the latter cation is much smaller than the former. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="7475" LastEditDate="2015-09-25T19:18:50.477" LastActivityDate="2015-09-25T19:18:50.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15043" PostTypeId="1" AcceptedAnswerId="15051" CreationDate="2014-08-09T13:47:11.920" Score="-1" ViewCount="2096" Body="&lt;p&gt;$$\ce{FeCl3 + KI + H2SO4 -&amp;gt; ?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reaction to proceed, is it necessary to have the medium acidified, here with sulfuric acid?&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="194" LastEditDate="2017-09-23T01:42:10.173" LastActivityDate="2017-09-23T01:42:10.173" Title="Conditions needed for reaction between FeCl3 and KI" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15044" PostTypeId="2" ParentId="8638" CreationDate="2014-08-09T13:57:19.980" Score="6" Body="&lt;p&gt;As Uncle Al already pointed out, the mechanism starts with protonation of the ester carbonyl oxygen. This yields a resonance-stabilized cation whose charge can be delocalized into the furan ring, where it is subsequently captured by nucleophilic attack of methanol in $\alpha$ position to the ring oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ukj1h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cyclic acetal thus formed undergoes [1,5]-H shift which is driven by rearomatization of the furan ring. The reactive 2-methoxyfuran is protonated at the 3 position, which yields a resonance-stabilized acetal cation. During aqueous workup, this cation is captured by water, which results in ring opening. The protonated ester carbonyl group is deprotonated by the ketone enolate, and after subsequent keto-enol tautomerization, the final product is eventually obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c0ZOt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-08-10T13:05:25.310" LastActivityDate="2014-08-10T13:05:25.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15045" PostTypeId="2" ParentId="15035" CreationDate="2014-08-09T14:11:54.683" Score="26" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_oxygen&quot;&gt;solid oxygen&lt;/a&gt; with color ranging from pale blue to red to black.  In the liquid phase it has a light blue color.  This color is due to light absorption by the ground state triplet according to the following equation&#xA;$$\ce{2 O2(^3\Sigma_{g}) -&amp;gt;C[{h\nu}]\   2 O2(^1\Delta_{g})}$$&#xA;This absorption requires light from the red region of the spectrum (~630 nm).  If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller.  Hence, the probability of photon absorption in the gas phase is much reduced.  To the human eye oxygen gas will appear colorless (at normal and reduced pressures).  However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-09T16:00:58.003" LastActivityDate="2014-08-09T16:00:58.003" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15046" PostTypeId="1" AcceptedAnswerId="15298" CreationDate="2014-08-09T14:13:31.000" Score="2" ViewCount="927" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GAusK.png&quot; alt=&quot;Assume that this is agar medium and &amp;#39;Fa&amp;#39; in picture should be &amp;#39;Fe&amp;#39;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know which metals oxidize and which is reduced and changes in medium and observations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my knowledge more corrosion will take place in Zn and Fe also corrode. Then copper should reduce. So, for both reactions, will copper acts as the cathode?&#xA;I need general but complete explanation. Thank you&lt;/p&gt;&#xA;" OwnerUserId="5776" LastActivityDate="2014-08-15T16:24:16.383" Title="Corrosion of metals and observations in the agar medium?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15047" PostTypeId="2" ParentId="14926" CreationDate="2014-08-09T14:19:36.600" Score="0" Body="&lt;p&gt;Try 'Advanced Chemistry by Philip Mathews ' - a red large book.&#xA;That explain very much in chemistry for advanced level.&lt;/p&gt;&#xA;" OwnerUserId="5776" LastActivityDate="2014-08-09T14:19:36.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15048" PostTypeId="2" ParentId="5568" CreationDate="2014-08-09T14:32:51.113" Score="-4" Body="&lt;p&gt;Yes, all exergonic reactions are exothermic. Consider a reaction occurring spontaneously we know that energy would be released i.e. '$\ce{\Delta H}$ is negative' (since a reaction or process absorbing energy makes it non-spontaneous) and according to the second law of thermodynamics, entropy(or disorder) of the system must increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Negative $\ce{\Delta H}$ and increasing, positive entropy together make $\ce{\Delta G}$ negative&lt;/strong&gt; according to the equation: $\ce{\Delta G = \Delta H~ - ~T\Delta S}$ (where $\ce{\Delta }$ = change; G= Gibb's free energy; H= enthalpy; T= Thermodynamic temperature and S= entropy).&#xA;Therefore, if enthalpy change is negative and change in free energy is negative, they are both (respectively) exothermic and exergonic.&#xA;The same applies for endothermic and endergonic.&lt;/p&gt;&#xA;" OwnerUserId="7414" LastEditorUserId="4231" LastEditDate="2014-08-09T15:39:02.903" LastActivityDate="2014-08-09T15:39:02.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15050" PostTypeId="2" ParentId="15043" CreationDate="2014-08-09T15:00:29.637" Score="0" Body="&lt;p&gt;So far, the only information I found is presented &lt;a href=&quot;https://facultystaff.richmond.edu/~rdominey/301/local/iodine_titrations.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction should be able to proceed in both acidic and neutral solutions. Since iron hydroxides are slightly soluble, the solution is acidified so the hydroxides cannot form.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-09T15:00:29.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15051" PostTypeId="2" ParentId="15043" CreationDate="2014-08-09T15:12:31.490" Score="1" Body="&lt;p&gt;No, it does not, you can drop $\ce{H2SO4}$ from the equation. However, in practives it is necessary to acidify $\ce{FeCl3}$ water solutions to suppress hydrolysis. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-09T15:12:31.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15052" PostTypeId="2" ParentId="14305" CreationDate="2014-08-09T16:31:29.540" Score="4" Body="&lt;p&gt;While the combustion is a correct answer, if you are looking for &quot;retrograde electrolysis&quot; than you best check out hydrogen fuel cells where are real electrochemical reaction, opposite of water splitting, takes place and produce directly electricity:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Fuel_cell&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fuel_cell&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different types, even different fuels, but the basic idea is the same: the oxidation and reduction process is separated by a membrane and the two reactions takes place on separate electrodes. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-09T16:31:29.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15053" PostTypeId="1" AcceptedAnswerId="15054" CreationDate="2014-08-09T18:18:12.953" Score="1" ViewCount="370" Body="&lt;p&gt;By using electrolysis one can separate hydrogen and oxygen, but when burning hydrogen and oxygen together steam is released, therefore creating water. My question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how much of the original quantity of water would be lost after this&#xA;  process?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7416" LastEditorUserId="6953" LastEditDate="2014-08-09T18:47:44.423" LastActivityDate="2017-01-05T14:30:43.037" Title="How much water would one loose if one burns hydrogen and oxygen?" Tags="&lt;water&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15054" PostTypeId="2" ParentId="15053" CreationDate="2014-08-09T18:55:30.020" Score="3" Body="&lt;p&gt;Theoretically, none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, since the gases are produced in a damp environment, there would be a lot of water vapor in the gases. You could easily remove most of the water vapor, but some (maybe as much as 1%) would still be in the gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be best to elaborate on what you mean as water being lost?&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-09T18:55:30.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15055" PostTypeId="1" AcceptedAnswerId="22017" CreationDate="2014-08-09T18:59:06.933" Score="10" ViewCount="1676" Body="&lt;p&gt;This is the electrostatic potential for ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sxPH7.png&quot; alt=&quot;Electrostatic potential map of ozone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the electrostatic potential for sulfur dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xIPOh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xIPOht.png&quot; alt=&quot;SO2&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;(Click to enlarge)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They look almost exactly the same except one is all oxygen atoms and the other has sulfur in the center instead of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does ozone have an MEP (molecular electrostatic potential) similar to that of SO2?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="11367" LastEditDate="2016-04-06T15:33:39.367" LastActivityDate="2016-04-06T15:33:39.367" Title="Why does ozone have an electrostatic potential map similar to that of SO2?" Tags="&lt;inorganic-chemistry&gt;&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15056" PostTypeId="2" ParentId="15014" CreationDate="2014-08-09T20:19:21.653" Score="1" Body="&lt;p&gt;&lt;strong&gt;&quot;Any ideas on how I can seperate and re-use both items?&quot;&lt;/strong&gt;&#xA;The devil is in the details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;what is particle size and concentration?&lt;/li&gt;&#xA;&lt;li&gt;what is the light oil?&lt;/li&gt;&#xA;&lt;li&gt;what do you mean by reusing them?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Depending on particle size (pumped) filtration (paperfilter, colloid filter, celite ) should work as well as centrifugation as was recommended by @Jon Custer. Is it that urgent? Also, cannot you just try it instead of asking it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the oil?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever you do, your graphite will be oily. It does matter if this oil has high-boiling point/waxy components... On the other hand, you may able to use something with low boiling point and just remove it under vacuum / heating, maybe can skipp the whole filtration part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do you mean by reusing them?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, there will be remains of oil on the graphite. Also, most probably, you will have fine powder of graphite remains in the oil whatever you do. How do you clean those up it depends on what do you need it for. Do you need to wash the graphite and re-filter the oil? You should know/test that. Also, cost-wise does all this fuss make any sense? Neither graphite nor the oil sound something expensive/special. Also, the more unique, expensive they are, most probably their cleanup will be more demanding (as you need to shoot for higher purity). &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2014-08-20T08:00:20.537" LastActivityDate="2014-08-20T08:00:20.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15057" PostTypeId="1" AcceptedAnswerId="15071" CreationDate="2014-08-09T21:18:53.323" Score="0" ViewCount="17547" Body="&lt;p&gt;I read that melting of gold is a physical property, but I wonder if this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to melt gold we will need to add heat, and once we add heat to gold to melt it, there must be a chemical change, not a physical change, right?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="5633" LastEditDate="2014-08-10T10:46:50.330" LastActivityDate="2014-08-10T10:46:50.330" Title="Is melting of gold a physical property?" Tags="&lt;physical-chemistry&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15058" PostTypeId="1" AcceptedAnswerId="15061" CreationDate="2014-08-09T21:54:44.193" Score="7" ViewCount="14630" Body="&lt;p&gt;$\ce{O2}$ has a double bond in its normal form. That is $\ce{O=O}$. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However 1 resonance structure would be $\ce{O-O}$ (result of homolytic cleavage of double bond) where each $\ce{O}$ is a free radical (a negatively charged one at that). If you have this in hydrogen it is likely going to form hydrogen peroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also have $\ce{O-O}$ where 1 is positive and the other is negative and this is also 2 free radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally there is $\ce{O#O}$ where both oxygens are positively charged and are free radicals. Why are both positively charged? It is because 3 bonds already to oxygen means 1 lone pair and 5 electrons around oxygen is +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of these resonance structures giving 2 free radicals in $\ce{O2}$ that $\ce{O2}$ is considered a biradical?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="4945" LastEditDate="2014-08-15T02:59:29.037" LastActivityDate="2014-08-15T02:59:29.037" Title="Why is O2 a biradical?" Tags="&lt;stability&gt;&lt;lewis-structure&gt;&lt;radicals&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15059" PostTypeId="2" ParentId="15058" CreationDate="2014-08-09T22:08:06.463" Score="3" Body="&lt;p&gt;There is no way to adequately describe $\ce{O2}$ molecule using naive valence bond theory. You need to learn Molecular Orbital Theory. Relevant links&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Molecular_orbital_theory&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-09T22:08:06.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15060" PostTypeId="2" ParentId="15057" CreationDate="2014-08-09T22:51:19.167" Score="5" Body="&lt;p&gt;It is a phase change - gold exists as a solid crystal (fcc I believe off the top of my head), as a liquid, and as a vapor (gas).  First order phase transitions exhibit a volume change, and an enthalpy of the phase change.  In the case of gold, both the crystal and the liquid are metallic, with similar numbers of nearest neighbors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All elements exhibit various phases.  Since pure gold is, well, pure gold, I'm not sure how a phase change in gold could be considered a 'chemical change', particularly in the solid-to-liquid.  The type of bonding does not change, and there are no 'chemical reactions' going on. This is all fairly basic thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-09T22:51:19.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15061" PostTypeId="2" ParentId="15058" CreationDate="2014-08-09T23:13:02.477" Score="10" Body="&lt;p&gt;We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BooW.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since an oxygen atom has 6 electrons,&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons&lt;/li&gt;&#xA;&lt;li&gt;B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical&lt;/li&gt;&#xA;&lt;li&gt;C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So while structure A would indicate a biradical, we wouldn't &amp;quot;expect&amp;quot; it to count for much since the oxygens do not have octets.  This inability to clearly predict the biradical nature of &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt; illustrates one of the failings of both Lewis structures and resonance theory.&lt;/p&gt;&#xA;&lt;p&gt;In order to correctly predict the biradical nature of &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt; we must move up to molecular orbital theory.  Below is the molecular orbital diagram for &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt;.  As you can see it does predict that &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt; should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXuqa.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Edit: response to OP's comment&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When I think of triple bond I don't think of 2 3 electron bonds(which&#xA;is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond&#xA;and 2 pi bonds)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This type of triple bond would make the oxygen positive with 5&#xA;electrons around it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, the formal charge on the oxygen in structure C is&lt;/p&gt;&#xA;&lt;p&gt;Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,&lt;/p&gt;&#xA;&lt;p&gt;there is no formal charge on oxygen in the &amp;quot;triple bond&amp;quot; structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I am assuming electrons are shared equally with half around 1 atom and&#xA;half around the other(which is the basis for formal charge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, that's correct.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-08-09T23:38:09.760" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15062" PostTypeId="1" AcceptedAnswerId="15068" CreationDate="2014-08-10T00:12:41.073" Score="11" ViewCount="7508" Body="&lt;p&gt;Photovoltaic panels can have 20 or 25 year underwritten warranties with a guaranteed remaining efficiency of 80 % of the new panel. That means, that photovoltaic panels seem to degrade somehow.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why do they degrade? What exactly is reducing their efficiency?&lt;/li&gt;&#xA;&lt;li&gt;How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7306" LastEditorUserId="5633" LastEditDate="2014-08-10T10:31:26.797" LastActivityDate="2017-04-20T03:36:49.477" Title="What causes photovoltaic (solar) cells to degrade?" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15063" PostTypeId="2" ParentId="14618" CreationDate="2014-08-10T02:23:28.137" Score="7" Body="&lt;p&gt;There are several different things going on here in this rather long series. I will go through them in turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acid chlorides and acid anhydrides&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are going to be reactive because they have a good leaving group. Chloride ion is so stable that it may break away from the molecule and give $\ce{RCO+}$ and $\ce{Cl-}$ spontaneously, not to a great extent, but sufficient for this to have no limiting effect on the reaction rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly an acid anhydride may break down to $\ce{RCO+}$ and $\ce{RCOO-}$. We know that the anion is again quite stable, but not as stable as $\ce{Cl-}$. This is apparent from the fact that $\ce{RCOOH}$ is an acid, but not as strong an acid as $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H}$ and $\ce{R}$ are not stable anions, so aldehydes and ketones are much less reactive than acid anhydrides and chlorides. Of course, the $\ce{H}$ and $\ce{R}$ groups remain attached to the molecule, which is not the case with the chlorides and anhydrides. Being an addition reaction instead of a substitution reaction, there is a decrease in entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting case here is the iodoform reaction where $\ce{R-CO-CH3}$ reacts with iodine to form $\ce{R-CO-CI3}$. $\ce{CH3-}$ is a poor leaving group, but the $\ce{CI3-}$ ion is a good leaving group and breaks away from the molecule, leaving $\ce{RCO+}$, which then reacts with water to form $\ce{RCOOH}$, while the $\ce{CI3-}$ reacts with water to form $\ce{HCI3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing to note here is that there is significant thermodynamic drive here for the good leaving group to be substituted by a poorer one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aldehydes and ketones&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important factor that makes aldehydes more reactive than ketones is that there is less bulk blocking the approach of a nucleophile. There is less steric hindrance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Esters and carboxylic acids&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are now back to substitution reactions. But $\ce{RO-}$ or $\ce{OH-}$ are much poorer leaving groups than the $\ce{Cl-}$ or $\ce{RCOO-}$ that we saw with acid chlorides and anhydrides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another factor at play here. The $\ce{C=O}$ double bond is conjugated with the $\pi$ electrons on the other oxygen, which significantly stabilises the molecule. This causes a resistance to both $\ce{S_{N}1}$ reactions (where the $\ce{RO-}$ or $\ce{ROH}$ spontaneously detaches from the molecule leaving $\ce{RCO+}$) and $\ce{S_{N}2}$ reactions (where the nucleophile attacks the molecule, giving a tetrahedral intermediate before the $\ce{ROH}$ or $\ce{H2O}$ is kicked out.) Either way that $\pi$ system has to be broken down, giving a kinetic barrier to reaction (high activation energy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amides and carboxylate ions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In certain reaction mechanisms OH- may spontaneously leave a carboxylic acid. $\ce{NH2-}$ is more basic and so is very unlikely to leave of its own accord. $\ce{O^{2-}}$ is even more basic (unheard of in solution.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other reaction mechanisms require that the molecule is protonated first. The next step may be nucleophilic attack ($\ce{S_{N}2}$ type mechanism) or spontaneous cleavage of the molecule into $\ce{RCO+}$ and a leaving group ($\ce{S_{N}2}$ type mechanism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{COOH}$ is protonated, the leaving group will be $\ce{H2O}$, a neutral compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{CONH2}$ is protonated, the leaving group will be $\ce{NH3}$, a basic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{COO-}$ is protonated (to an acid) the leaving group will be $\ce{{}^{-}OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the more acidic/less basic a group is, the more stable it will be when separated from the $\ce{RCO+}$ carbocation. Here we have a neutral molecule water (a good leaving group), a weak base ammonia (poor leaving group) and a strong base (very poor leaving group) so the reactivity order is as would be expected.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="4945" LastEditDate="2014-08-11T03:13:08.917" LastActivityDate="2014-08-11T03:13:08.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15064" PostTypeId="2" ParentId="15036" CreationDate="2014-08-10T02:44:40.350" Score="10" Body="&lt;p&gt;This is a statement of the first law of thermodynamics. The first law says that the energy of an isolated system is conserved. An isolated system is one in which neither mass nor energy is exchanged with the surroundings. This equation shows the two means by which a closed system (meaning a system in which mass is conserved but energy is not) can exchange energy with its surroundings - by heat, or by work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta E=q+w&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All energy transfers can be described as either heat flow or as work done on or by a system, and therefore the total amount of heat energy that goes in or out ($q$), plus the total amount of work energy that goes in or out ($w$), must be equal to the total internal energy change of the system ($\Delta E$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is particularly important in chemical thermodynamics because we have no way of measuring the total internal energy of a real system directly - all that we can measure is the amount of heat and work that goes in or out of the system, and therefore the &lt;em&gt;change&lt;/em&gt; in internal energy.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-10T11:59:42.740" LastActivityDate="2014-08-10T11:59:42.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15065" PostTypeId="2" ParentId="15007" CreationDate="2014-08-10T02:52:05.003" Score="7" Body="&lt;p&gt;Because it makes no sense to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if someone uncovered the exact ingredient of Coca Cola or even if you made a Coca Cola recipe that tastes better than real Coca Cola, they wouldn't able to match the existing brand power of Coca Cola, nor the consumer's brand recognition and history with the general public. Coca Cola had poured billions and decades into making one of the most recognisable brand in the world, their brand is ingrained very deeply into the public's minds. A new competitor wouldn't able to replicate that overnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of those companies that do may have the brand power to match Coca Cola and enough R&amp;amp;D budget to match, say PepsiCo, they wouldn't want to be seen dead advertising that they had made an exact clone of their competitors. It works better if they make and market themselves as a better drink than as a clone of a competitor. It's very likely that PepsiCo and Coca Cola knows pretty well their competitor's recipes, but they have absolutely no incentive to advertise that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, even if you manage to uncover the exact composition and manufacturing process of the product, the public still won't believe it unless The Coca Cola company itself admitted that a given recipe is 100% truthful. A large percentage of the populace would always remain skeptical. Noone would give you research grant to reverse engineer Coca Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tasting isn't an objective experience, the way you experience a drink is affected by what's in your head just as much as what the chemistry of the drink. Even &lt;a href=&quot;http://www.youtube.com/watch?v=JdvJOF-2mm0&quot;&gt;water from the same exact garden hose tastes different&lt;/a&gt; depending on how you are presented it with. For the same reason, even if you put an actual Coca Cola into a supermarket brand Cola bottle, a lot of people would still think that that bottle of Cola tastes worse than actual Coca Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, consumers won't buy fake products when the real products cost something like \$1-\$2 a can. If you drink a can of fake cola in public, that says something about your character, probably not in a way most people want themselves to be perceived as.&lt;/p&gt;&#xA;" OwnerUserId="7420" LastActivityDate="2014-08-10T02:52:05.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15066" PostTypeId="1" CreationDate="2014-08-10T03:29:49.707" Score="4" ViewCount="53623" Body="&lt;p&gt;My question relates to dissolving $\ce{NH3}$ in water and how it produces $\ce{OH-}$ for the reaction $\ce{NH3 + H2O -&amp;gt; NH4+ + OH-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the $\mathrm{p}K_b$ of $\ce{NH3}$ is $4.8$ and $\mathrm{p}K_a$ of $\ce{NH4+}$ is $9.2$ so $\ce{NH3}$ is a stronger base than $\ce{NH4+}$ is an acid. But why the free $\ce{OH-}$? I am now taking organic chemistry and all I am reading is how strong $\ce{OH-}$ is and how it can attack anything slightly positive. So why is the above reaction favorable towards creating $\ce{OH-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually any acid-base reaction that is creating $\ce{OH-}$ as a free species?&lt;/p&gt;&#xA;" OwnerUserId="7421" LastEditorUserId="4945" LastEditDate="2014-08-13T16:12:44.853" LastActivityDate="2014-08-13T16:14:11.637" Title="Why does ammonia react with water to produce hydroxide?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15067" PostTypeId="1" CreationDate="2014-08-10T06:37:28.210" Score="3" ViewCount="73205" Body="&lt;p&gt;Observe the following parts of an electron transfer equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{2Li + S &amp;amp;-&amp;gt; Li2S}\tag{1}\\&#xA;\ce{2Al + 3Br2 &amp;amp;-&amp;gt; 2AlBr3}\tag{2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation says $\ce{Li}$ with a small subscript 2, however the second one says Al with a coefficient 2 instead of $\ce{Al2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain this discrepancy and inconsistency I'm seeing? Why is lithium in the first reaction written with a small $2$, while in the second one aluminium is written with a big $2$? Why can't both of them be written in the same way? Does it matter? Is there a difference? How do I tell which one to use?&lt;/p&gt;&#xA;" OwnerUserId="7425" LastEditorUserId="41328" LastEditDate="2017-08-06T09:23:35.700" LastActivityDate="2017-08-06T09:23:35.700" Title="What is the difference between a coefficient and a subscript in a chemical equation?" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15068" PostTypeId="2" ParentId="15062" CreationDate="2014-08-10T09:54:10.513" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do they degrade? What exactly is reducing their efficiency?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.pveducation.org/pvcdrom/modules/degradation-and-failure-modes&quot;&gt;link&lt;/a&gt; outlines several modes of solar panel degradation, and &lt;a href=&quot;http://www.nrel.gov/docs/fy12osti/51664.pdf&quot;&gt;this report&lt;/a&gt; by the National Renewable Energy lab is a very detailed review of studies on solar panel degradation worldwide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Internal resistance of the cell can increase due to infiltration of contaminants (usually water vapor) when the encapsulating material cracks due to long-term UV exposure or temperature cycling.&lt;/li&gt;&#xA;&lt;li&gt;The shunt resistance (resistance of the path to ground) can decrease when metal ions migrate (diffuse) through the cell. This happens naturally, with the rate depending on temperature and the type of cell. In general, higher temperatures lead to faster diffusion, and amorphous silicon has higher diffusion rates.&lt;/li&gt;&#xA;&lt;li&gt;The anti-reflective coating can deteriorate. Anti-reflective coatings can be made of lots of different types of materials, but in general, heat, UV exposure, and exposure to contaminants are what cause them to degrade.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are all degradation mechanisms of the cells themselves - any of the other components of the panel can also degrade. In all cases, the mechanisms are related to thermal stress or mechanical damage due to things like wind or hail, or corrosion from humidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all of the above, the main causes of solar panel degradation that can be (sort of) controlled by you are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Temperature&lt;/li&gt;&#xA;&lt;li&gt;UV exposure&lt;/li&gt;&#xA;&lt;li&gt;Mechanical damage&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anything you can do to reduce these three factors would help. However, covering a solar panel with an opaque tarp would only reduce UV exposure. The temperature (and cyclical thermal stress) would likely increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, both 1 and 2 increase with exposure to sunlight - and exposure to sunlight is the whole point of solar panels. Mechanical damage can be reduced by installing panels in places and orientations that make it unlikely that they will be stressed by winds, or have things hitting them very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it looks like the only effective option would be to completely enclose the panels in a temperature-controlled low humidity environment when they are not in use. Whether or not the benefit of doing so is worth the cost will depend on the degradation rate and cost of the particular panels.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-10T09:54:10.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15069" PostTypeId="2" ParentId="15066" CreationDate="2014-08-10T10:27:45.543" Score="3" Body="&lt;p&gt;The shortest answer is &quot;because the change in free energy is negative,&quot; but maybe you want to know how the mechanism works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_acid&quot; rel=&quot;nofollow&quot;&gt;Lewis base&lt;/a&gt;, and hydroxide is a stronger Lewis base. However, both are stronger Lewis bases than water is. Pure water always has some amount of &quot;free&quot; $\ce{H+}$ (it's actually hydronium ion) and $\ce{OH-}$ due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow&quot;&gt;self-ionization&lt;/a&gt; ($pK_W \approx 14$): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_W = \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add ammonia to the system, some of those hydronium ions (which are very strong acids) react with the ammonia to form ammonium. Since the concentration of hydronium decreases, the concentration of hydroxide must increase to maintain equilibrium. Those hydroxide ions are stronger bases than ammmonia is, and so they will &quot;win&quot; the tug-of-war over protons. However, in the meantime, some other hydronium reacts with some other ammonia, and the net result is a dynamic equilibrium in which there is more hydroxide then there would be in pure water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same mechanism is responsible for any other acid/base reaction where a base accepts protons from water and leaves hydroxide as a free species, but the net concentration of hydroxide will be different for different strengths of bases.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-12T19:23:45.627" LastActivityDate="2014-08-12T19:23:45.627" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15070" PostTypeId="1" AcceptedAnswerId="15073" CreationDate="2014-08-10T10:33:57.153" Score="16" ViewCount="5974" Body="&lt;p&gt;I opened up a bottle of champagne that was left in a wine cooler for approx. 8 years. To my astonishment there were small crystals at the underside of the cork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystals are colorless, odorless and have no specific taste at all. What is going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5fOWy.jpg&quot; alt=&quot;Colorless crystals at the bottom of a champagne cork&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kE1jE.jpg&quot; alt=&quot;Close up&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2016-09-20T07:17:58.300" Title="What are these crystals on my champagne cork?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15071" PostTypeId="2" ParentId="15057" CreationDate="2014-08-10T10:42:46.617" Score="4" Body="&lt;p&gt;Melting is a &lt;strong&gt;physical change&lt;/strong&gt;, not a &lt;strong&gt;physical property&lt;/strong&gt;. The distinction is subtle, but the way I think of it is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A physical change is one that does not involve a change in any chemical bonds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compare that to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A physical property is one that can be measured without requiring a chemical change&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Melting is the physical change, temperature and melting points are physical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply adding heat to a substance doesn't necessarily mean that chemical bonds will be broken or formed. Metallic bonds are not usually considered to be chemical bonds, although it really depends on how you look at them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later you will learn that the distinction between physical and chemical bonds is fairly arbitrary, since they all have to do with changes in electron density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, melting an ionic solid (like table salt) results in ionic bonds breaking, but this is still considered a physical change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to identify a physical vs. chemical change when you don't know the chemical details is to imagine what would happen if you reversed the operation. If the system could go back to the way it was fairly easily, the chances are good that it is a physical change. This doesn't work for all situations, but it does work for the most common ones. The way I remember it is: &quot;you can't unbake a cake,&quot; so &quot;baking&quot; is probably a chemical change. Going back to the melting example - if you cool molten metal down, you get back to where you started, and so that is probably a physical change.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-10T10:42:46.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15072" PostTypeId="1" CreationDate="2014-08-10T10:47:12.210" Score="3" ViewCount="62093" Body="&lt;p&gt;I am trying to formulate a gel hand sanitizer for a thesis project. The sanitizer that I have right now is watery and will not gel.  My teacher told me to change our formulation because it is not good. Our ingredients are ethyl alcohol, aloe vera gel, and lavender scent. What else could we add to this mixture to really &quot;gel&quot; the hand sanitizer?&lt;/p&gt;&#xA;" OwnerUserId="7426" LastEditorUserId="4945" LastEditDate="2019-01-08T16:55:31.583" LastActivityDate="2020-03-16T12:15:42.370" Title="How can I get a hand sanitizer solution to form a gel?" Tags="&lt;home-experiment&gt;&lt;colloids&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15073" PostTypeId="2" ParentId="15070" CreationDate="2014-08-10T11:02:51.273" Score="12" Body="&lt;p&gt;They are &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_bitartrate&quot;&gt;potassium bitartrate&lt;/a&gt; crystals (&lt;a href=&quot;http://www.jordanwinery.com/files/FlexibleFile/289/Wine_Tartrates_FAQ.pdf&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystals form because the potassium bitartrate is not very soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since solubility is a function of temperature, when wine is chilled the solution can become saturated, causing the precipitate to form in the bottle. Stabilizers are often &lt;a href=&quot;http://www.winegrowers.info/wine_making/Cold%20stabilisation.htm&quot;&gt;added to prevent this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the precipitate is on the cork, and so what I think has happened is that some wine soaked into the cork, perhaps while the bottle was moved. Then later, the bottle was stood upright or on its side where the wine could not reach the cork. As the wine evaporated back into the bottle, the local concentration of potassium bitartrate increased until it reached the saturation point, at which point the crystals grew. Since the crystals take a very long time to dissolve once formed, they will persist even if wine reaches the cork again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is, they are harmless (especially good news given that you tasted them - please don't make tasting strange crystals a habit!). They are made of the same substance that cream of tartar, which is a common baking ingredient, is made of.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-10T11:02:51.273" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15075" PostTypeId="2" ParentId="15067" CreationDate="2014-08-10T11:33:46.260" Score="2" Body="&lt;p&gt;The question is quickly answered when you consider the ionic forms of the elements you present. It is easy if you break down the reaction into simple oxidation and reduction reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Oxidations&lt;/strong&gt;&#xA;$$ \ce{Li -&amp;gt; Li+ + e-} $$&#xA;$$ \ce{Al -&amp;gt; Al^{3+} + 3e-} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reductions&lt;/strong&gt;&#xA;$$ \ce{S + 2e- -&amp;gt; S^{2-}} $$&#xA;$$ \ce{Br2 + 2e- -&amp;gt; 2Br-} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the resulting compounds of your reactions are neutral, all we have to do is to match up the starting compounds in the right amounts to get to the right final products. Retrosynthetically speaking:&#xA;$$ \ce{Li2S &amp;lt;- 2Li+ + S^{2-} &amp;lt;-[\mathrm{red}] 2Li+ + 2e- + S &amp;lt;-[\mathrm{ox}] 2Li + S } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you go from the start:&#xA;$$ \ce{2Al + 3Br2 -&amp;gt;[\mathrm{ox}] 2Al^{3+} + 6e- + 3Br2 -&amp;gt;[\mathrm{red}] 2Al^{3+} + 6 Br- -&amp;gt; 2AlBr3 (-&amp;gt; Al2Br6)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put the last step in parentheses because usually we give formulae of ionic compounds with the smallest integer amount possible, which would be $\ce{AlBr3}$ and not $\ce{Al2Br6}$. This should answer your question about which one to use.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. For example $\ce{AlCl3}$, a lewis acid, can dimerize to $\ce{Al2Cl6}$. Which one you mean depends on the formula you give and can be a barrier of understanding in a conversation.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-08-10T11:33:46.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15076" PostTypeId="2" ParentId="15067" CreationDate="2014-08-10T11:57:58.443" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't both of them be written in the same way? Does it matter? Is there a difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there is a difference, and it does matter! The placement of subscripts and coefficients tells you how the substances are put together on the atomic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the first equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2Li + S -&amp;gt; Li2S} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one indicates that two moles of lithium combine with one mole of sulfur to form one mole of lithium sulfide. $\ce{Li2S}$ is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_unit&quot; rel=&quot;noreferrer&quot;&gt;formula unit&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Emprical_formula&quot; rel=&quot;noreferrer&quot;&gt;empirical formula&lt;/a&gt;, and tells you the smallest whole-number ratio of atoms that make up a &quot;building block&quot; of the compound. In this case, the smallest building block of a lithium sulfide crystal would contain two lithium ions for every one sulfide ion. Here's a picture from &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_sulfide&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt; to demonstrate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0gKDX.png&quot; alt=&quot;Lithium sulfide crystal structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can learn that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Coefficients&lt;/strong&gt; tell us the number of moles, molecules, or formula units of a species&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Subscripts&lt;/strong&gt; tell us how many atoms or ions there are &lt;em&gt;within&lt;/em&gt; a compound&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if we look at the second equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2Al + 3Br2 -&amp;gt; 2AlBr3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can see that in this case, it take 2 moles of aluminum to combine with three moles of bromine to produce two moles of aluminum tribromide. If it were instead $\ce{Al2Br6}$, that would indicate a different formula unit - it would imply a different compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is Lithium in the first reaction written with a small 2, while in the second one Aluminium is written with a big 2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is because in the first equation, the product has two lithium ions for every one sulfur in the formula unit, while in the second, two moles of aluminum tribromide are produced for every two moles of aluminum, but the formula unit contains only one aluminum atom and three bromine atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your final question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I tell which one to use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You would have to know what the formula of the product is. In these examples, you can predict it by assuming that they are ionic compounds and balancing charges. You would get the correct answer, even though this assumption is not true for the second equation (it is actually a covalent compound, and you would just get the right answer by luck).&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-10T12:26:01.547" LastActivityDate="2014-08-10T12:26:01.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15077" PostTypeId="2" ParentId="15007" CreationDate="2014-08-10T12:27:39.670" Score="3" Body="&lt;p&gt;There are two very strong reasons, legal and marketing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way that Coca Cola is created is a trade secret. It is not patented, the company just keeps it secret. As long as it is secret, other companies are free to create their own recipes, even their own, identical recipes, but it is illegal to create such a recipe by breaking into some office and stealing the original recipe, or by analysing the original drink. If you reverse engineered the recipe, you could expect to be taken to court. And your engineers would have to appear as witness and have the choice of saying the truth or turning themselves into criminals by lying. Not a good business strategy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is that a company like Pepsi doesn't &lt;em&gt;want&lt;/em&gt; to create something that tastes exactly like Coca Cola. Their customers buy their drinks because they &lt;em&gt;prefer&lt;/em&gt; the different taste. If Pepsi changed their taste to be exactly like Coca Cola, they would just annoy their own customers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to make this clear: Reverse engineering in itself is possibly not illegal. Discovering a trade secret by reverse engineering and then using the reverse engineered recipe to make your own product is most likely illegal. Creating the impression that you discovered a trade secret by reverse engineering and then used it to create your own product will get you sued, which will cost you more money than it is worth. &lt;/p&gt;&#xA;" OwnerUserId="7427" LastEditorUserId="7427" LastEditDate="2014-08-10T12:33:02.613" LastActivityDate="2014-08-10T12:33:02.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15078" PostTypeId="1" CreationDate="2014-08-10T14:37:48.297" Score="6" ViewCount="510" Body="&lt;p&gt;According to my teacher, these three &lt;a href=&quot;https://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis bases&lt;/a&gt; generally decrease in strength in the following order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ammonia&lt;/li&gt;&#xA;&lt;li&gt;Water&lt;/li&gt;&#xA;&lt;li&gt;Hydroxide ion&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I agree. However, he argues that their ability to act as Lewis bases stems from Coulombic repulsions. For example, nitrogen has only &lt;em&gt;one&lt;/em&gt; lone pair. Water however has &lt;em&gt;two&lt;/em&gt; lone pairs, and when water uses one of its lone pairs in a Lewis base fashion, the other lone pair weakens the resulting Lewis adduct through exerting a repulsive force on the lone pair used in forming the Lewis adduct. Same with hydroxide, except that the repulsions are even greater in hydroxide since it has &lt;em&gt;three&lt;/em&gt; lone pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say that in addition to this effect, there is also the effect of the Lewis base being able to stabilize its valence electron density. Nitrogen is less electronegative than oxygen. Thus, any valence electron density on nitrogen will be more reactive than that on oxygen. On the other hand, oxygen in hydroxide ion has a greater valence electron density than that of oxygen in water, and hydroxide ion is a poorer Lewis base than hydroxide ion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="17768" LastEditDate="2018-05-24T13:01:57.087" LastActivityDate="2018-05-24T13:01:57.087" Title="Ligand strength - Coulombic effect and/or Electronegativities?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15079" PostTypeId="1" AcceptedAnswerId="15086" CreationDate="2014-08-10T15:27:13.497" Score="2" ViewCount="414" Body="&lt;p&gt;Why is HOOH such a good oxidizer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the oxygens in HOOH are in a negative 1 oxidation state, and that oxygen generally adopts an oxidation state of negative 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that oxidizers themselves are &lt;em&gt;reduced&lt;/em&gt;, and that reduction entails a &lt;em&gt;drop&lt;/em&gt; in oxidation state. This suggests to me that oxygen is more stable when found with a negative two oxidation state rather than a negative one oxidation state. This makes sense; oxygen is the second-most electronegative element on the periodic table, so it likely would be okay with a &lt;em&gt;more&lt;/em&gt; negative oxidation state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from organic chemistry I understand that the O-O linkage in peroxides is relatively weak; light can rather easily homolytically cleave the O-O linkage. And HOOH is generally stored in dark bottles as to block light. This says something about thermodynamic stabilities to me - that HOOH has relatively weak bonds and that it would be &lt;em&gt;enthalpically&lt;/em&gt; favorable for the oxygens to form other, stronger, bonds. Is this how HOOH is used in bombs? The O-O bond is weak; bond formation is exothermic, thus providing a way for the O-O bond to break and rearrange into stronger bonds releases vasts amount of energy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to tie together redox with thermodynamic stabilities. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-10T15:35:11.373" LastActivityDate="2014-08-10T18:31:19.283" Title="Hydrogen Peroxide as an Oxidizer" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15080" PostTypeId="1" AcceptedAnswerId="15115" CreationDate="2014-08-10T15:39:53.797" Score="2" ViewCount="1740" Body="&lt;p&gt;What is the purpose of calculating an average oxidation state for a certain element in a compound? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hence the “average oxidation state” of carbon in acetic acid is 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the oxidation state value for carbon atoms in acetic acid is, however, computed as zero. This latter value represents an average of the values for the two carbon atoms derived by the rigorous assignment rules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does this tell me? Why not just stick with individual oxidation states? Is there something the overall oxidation state is supposed to tell me about acetic acid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I can think of is that perhaps this average oxidation state of 0 for the carbon atoms tells me that acetic acid has some ionic character. Hence the oxidation state of 0 for carbon (as opposed to +4 for carbon in its elemental form). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2021-05-12T06:32:40.977" Title="Average Oxidation State - Utility?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15081" PostTypeId="2" ParentId="15080" CreationDate="2014-08-10T15:52:55.683" Score="1" Body="&lt;p&gt;It helps you when you want to balance a red-ox reaction by the mean of the oxidation number change. For example, if you want to balance the reaction of thiosulfate with iodine to tetrathionate and iodide, you realize that the average oxidation number of sulfur changed from +2 to +2.5. The oxidation half-reaction looks like this:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{4S^{+2}-&amp;gt; 4S^{2.5+} +2e-}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="98422" LastEditDate="2021-05-12T06:32:40.977" LastActivityDate="2021-05-12T06:32:40.977" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15082" PostTypeId="1" AcceptedAnswerId="15113" CreationDate="2014-08-10T16:12:58.600" Score="1" ViewCount="3035" Body="&lt;p&gt;There is a mixture of aniline and phenol. I need to obtain (just the procedure) pure aniline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the mixture should heat with Zn powder(to convert phenol to benzene) and then I thought to synthesize aniline from benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about this method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="485" LastEditDate="2014-08-10T16:40:03.190" LastActivityDate="2014-08-11T04:25:57.800" Title="How to obtain pure aniline from a mixture of phenol and aniline?" Tags="&lt;organic-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15084" PostTypeId="1" AcceptedAnswerId="15094" CreationDate="2014-08-10T18:12:37.717" Score="17" ViewCount="6261" Body="&lt;p&gt;Many transparent plastic materials, rigid or not, turn opaque (white) when subjected to stress, deformations or when folded (if not rigid) . The effect seems to be irreversible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the cause?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have guessed that some internal crystalline structure is broken, but plastic (like glasses) have no crystal structure, no?&lt;/p&gt;&#xA;" OwnerUserId="7431" LastEditorUserId="5633" LastEditDate="2014-08-11T15:34:08.173" LastActivityDate="2014-08-11T16:23:30.097" Title="Why do many transparent plastics turn opaque when deformed?" Tags="&lt;plastics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15085" PostTypeId="2" ParentId="6778" CreationDate="2014-08-10T18:25:09.827" Score="1" Body="&lt;p&gt;During copper &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper_extraction_techniques&quot; rel=&quot;nofollow&quot;&gt;electrorefining&lt;/a&gt;, impure $\ce{Cu}$ from the anode is oxidized and dissolved, while $\ce{Cu^2+}$ is reduced at the cathode, forming the deposit of refined copper. Because anodic impurities which are less noble than copper – most notably $\ce{Ni}$, $\ce{As}$ and $\ce{Sb}$ – also dissolve in the electrolyte, the cell voltage is kept in the range of 0.2-0.4 V to avoid co-deposition of these metals at the cathode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper dissolution is the main oxidation half-reaction at the anode, and the oxidation of anions from the electrolyte needs to be minimized, since it is an unwanted side reaction which reduces current efficiency of the process. In a $\ce{CuSO4}$ electrolyte, this would be &lt;a href=&quot;http://www.xtremepapers.com/revision/gcse/chemistry/electricity_in_chemistry.php&quot; rel=&quot;nofollow&quot;&gt;oxidation of $\ce{OH-}$&lt;/a&gt; (from water dissociation, so concentration will be low) to $\ce{O2}$, while in a $\ce{CuCl2}$ solution, $\ce{Cl-}$ will be oxidized to chlorine gas, which would present a greater hazard because of its toxicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric conductivity of the electrolyte is another important factor, which varies with different copper salts and their concentrations. $\ce{CuCl2}$ and $\ce{CuSO4}$ are both readily soluble in water; however, the relative molar conductivities of these solutions will be different. In &lt;a href=&quot;http://tedprints.tedankara.k12.tr/5/1/2009-Ayse%20Melis%20Aygar.pdf&quot; rel=&quot;nofollow&quot;&gt;this study&lt;/a&gt;, a copper chloride electrolyte proved to be inferior to copper sulfate and copper nitrate in terms of deposit quality and stability of cell potential.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-10T18:25:09.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15086" PostTypeId="2" ParentId="15079" CreationDate="2014-08-10T18:31:19.283" Score="2" Body="&lt;p&gt;From your argument I assume the real question you want to ask is &quot;Why HOOH is such a good oxidizer compared to molecular oxygen?&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From thermodynamical point of view, molecular oxygen is a very potent oxidizer. However when you are talking about reactivity, you should always take a look at kinetics, not only at thermodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important part of the answer to your question is that not HOOH a good oxidizer, but O2 is a very bad oxidizer from kinetic point of view. Molecular oxygen is in the triplet state, therefore its reactions are often spin-forbidden. Especially the oxygen-&gt;superoxide-&gt;peroxide steps require often level crossings between different spin surfaces. One very important reason why transition metal complexes catalyse oxidation in biology is this: as they often have electrons to flip and large spin-orbit coupling, the level crossings became easy. There is literally tons of literature on this in bioinorganic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in flames and in high temperature burning that reaction is essentially a huge variety of radical reactions, so the mechanism is completely different from the oxidation reactions in aqueous solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular oxygen has also 4 electrons to take when turned to oxides. This sometimes require complicated mechanism where every single step is thermodynamically and kinetically is feasible, and there is always a oxidizable component around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note, that HOOH itself need catalyst to be an effective catalyst, like in Fenton reactions and similar.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-10T18:31:19.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15087" PostTypeId="1" AcceptedAnswerId="77090" CreationDate="2014-08-10T18:32:22.023" Score="4" ViewCount="7111" Body="&lt;p&gt;I am thinking cheap material that can be used between two layers of carton that absorbs water. &#xA;I want to use air but I still think that there may be other materials available that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;absorb water&lt;/li&gt;&#xA;&lt;li&gt;can be easily removed when the outer carton layer is damaged&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Possible materials&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;carbon (well absorbing)&lt;/li&gt;&#xA;&lt;li&gt;air (cheap and easy)&lt;/li&gt;&#xA;&lt;li&gt;sand (should be bad)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is cheap material that absorbs water in packaging?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="292" LastActivityDate="2017-06-30T23:38:20.437" Title="What is cheap material that absorbs water in packaging?" Tags="&lt;materials&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15088" PostTypeId="2" ParentId="15087" CreationDate="2014-08-10T18:52:40.277" Score="7" Body="&lt;p&gt;What you are looking for is most probably silica gel.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Silica_gel&quot;&gt;http://en.wikipedia.org/wiki/Silica_gel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a highly water absorbent material with large surface area. It is often used in packaging as well as in laboratory in desiccators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually buy it packed or in bulk easily (let me not link explicit offers here). You can regenerate used (old) silica by heating / drying / pumping, so you don't need to buy every single time new one if the old gets less effective.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-10T18:52:40.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15089" PostTypeId="1" CreationDate="2014-08-10T19:08:05.293" Score="4" ViewCount="37360" Body="&lt;p&gt;What are lattice points exactly? For what purpose are they used in solid state chemistry?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="41556" LastEditDate="2017-08-15T16:45:45.940" LastActivityDate="2021-01-12T02:17:09.910" Title="What are Lattice Points?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15091" PostTypeId="1" AcceptedAnswerId="15095" CreationDate="2014-08-10T21:03:50.840" Score="10" ViewCount="71017" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Arrange these compounds: $\ce{CO2}$, $\ce{CH3OH}$, $\ce{RbF}$, $\ce{CH3Br}$ in order of increasing boiling points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think I should consider the forces between them, that is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CO2}$: dispersion forces&lt;/li&gt;&#xA;&lt;li&gt;$\ce{RbF}$: dispersion and  ionic forces&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CH3OH}$: Dipole-dipole interactions, Hydrogen bonding and dispersion forces&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CH3Br}$: Dipole-dipole interactions and dispersion forces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is obvious that $\ce{CO2}$ is the smallest one and because $\ce{CH3OH}$ stronger than $\ce{CH3Br}$ it will have higher boiling point &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to arrange the rest? Or how to compare ionic forces with Dipole?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorDisplayName="user10259" LastEditDate="2015-05-21T23:19:49.297" LastActivityDate="2015-05-21T23:19:49.297" Title="Arrange these compounds: CO2, CH3OH, RbF, CH3Br in order of increasing boiling points" Tags="&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15092" PostTypeId="2" ParentId="15089" CreationDate="2014-08-10T21:06:24.687" Score="-1" Body="&lt;p&gt;Lattice points are the positions (or coordinates if you wish) where you can place an atom. Therefore you can find either atoms or vacancies in the lattice points of a crystal structure.&#xA;I dont quite understand what you mean by &quot;How many Lattice Points are there is each of the Unit Cells?&quot;&lt;/p&gt;&#xA;" OwnerUserId="5086" LastActivityDate="2014-08-10T21:06:24.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15093" PostTypeId="2" ParentId="15084" CreationDate="2014-08-10T21:44:27.403" Score="10" Body="&lt;p&gt;@Philipp already mentioned microscopic cracks. This is called &lt;em&gt;crazing&lt;/em&gt;. When a plastic is stretched, the extra space is filled randomly/irregularly. The gap between two polymer chains will grow, but most chains won't break. Also, if the material is already crystallized locally, the crystals will pull apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the overall structure is non-crystalline, the cracks will have different sizes, and in particular quite a few will have sizes smaller than a micrometer - optical wavelengths, in other words. And with the variety in sizes, you don't get oen dominant color but a mixture of all colors which is of course white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason for the optical change is &lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/mats05/mats05096.htm&quot;&gt;crystals forming&lt;/a&gt;. Again, these small crystals have different sizes around the wavelengths of visible light, so there's no dominant color from these either. &lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditDate="2014-08-11T16:23:30.097" LastActivityDate="2014-08-11T16:23:30.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15094" PostTypeId="2" ParentId="15084" CreationDate="2014-08-10T22:29:19.163" Score="18" Body="&lt;p&gt;The process of forming an opaque area when you apply stress to a polymer film is known as &lt;em&gt;crazing&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polymers are long chains and, when coated, often form tangled, amorphous, randomly aligned polymer regions.  When stress is applied to the film, the stress can be relieved by the polymer chains aligning.  In other words, energy introduced into the film by stressing it can be used to align some polymer chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the alignment in the stressed region means that the density of polymer in this region of the film will be different from the density in non-stressed regions.  This difference in density will cause a difference in light reflection and scattering accounting for the different optical properties between these two regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, within the stressed area of the film, mechanical stresses are produced between areas that are more and areas that are less aligned.  These stresses can be relieved through the formation of micro-cracks between the regions.  These small cracks also result in different light reflecting and scattering properties of the film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as the polymer chains align, they are more likely to crystallize.  The formation of small crystallites within the stressed (aligned) region will also alter the light reflecting and scattering properties of the film.  Often, upon heating or standing at room temperature (for a longer time) the optical changes may reverse as the thermal energy supplied will:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;cause the polymer structure to randomize back to its amorphous structure,&lt;/li&gt;&#xA;&lt;li&gt;allow the polymer to &quot;flow&quot; and fill the micro-cracks, and\or&lt;/li&gt;&#xA;&lt;li&gt;the crystallites can redissolve.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; The following links provide more information about crazing and associated phenomena at different levels of complexity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/mats05/mats05096.htm&quot;&gt;reference 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/course/3/3.11/www/Lectures/Lecture31.pdf&quot;&gt;reference 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mate.tue.nl/mate/pdfs/1064.pdf&quot;&gt;reference 3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-11T13:52:50.993" LastActivityDate="2014-08-11T13:52:50.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15095" PostTypeId="2" ParentId="15091" CreationDate="2014-08-10T22:30:39.517" Score="11" Body="&lt;p&gt;You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol  forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (&gt; 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) interaction between the ions, so that one will be at the top. Ionic forces can be seen as extreme dipoles in a certain way, there is a grey area when electronegativity becomes large enough, that it can be seen either as a molecular structure or ionic structure. Consulting online information about the boiling points of these compounds (i.e. just check Wikipedia or some MSDS site) confirms the theory.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-08-10T22:53:52.330" LastActivityDate="2014-08-10T22:53:52.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15096" PostTypeId="2" ParentId="4274" CreationDate="2014-08-11T00:06:03.387" Score="2" Body="&lt;p&gt;There are people who need &lt;em&gt;&quot;easily removable material that sticks to skin&quot;&lt;/em&gt; virtually every working day, so take a look at what they use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Makeup artists use a special glue to stick special effects elements like Spock ears to the skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glue is based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Mastic_%28plant_resin%29&quot; rel=&quot;nofollow&quot;&gt;mastic&lt;/a&gt;, the resin of the mastic tree. So it is mainly a solution of the resin compounds in something like alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most difficult problem to solve is finding a glue that is strong, but not harmful to skin - even when using it for a full day.&lt;br&gt;&#xA;You can remove mastic glue with a &lt;em&gt;&quot;slightly strong pull&quot;&lt;/em&gt;. It will hurt the first time if there was hair.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also water-soluble skin glues, but as far as I know the mastic based is the strongest of them.&lt;br&gt;&#xA;For examples, see &lt;a href=&quot;https://us.kryolan.com/products/special-effects/adhesives-removals&quot; rel=&quot;nofollow&quot;&gt;Kryolan skin adhesives&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2014-08-11T00:06:03.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15097" PostTypeId="5" CreationDate="2014-08-11T03:07:03.200" Score="0" Body="&lt;p&gt;Molecular orbital theory (MO) is a theoretical model for bonding in molecules that may be used to deduce/rationalize molecular structure and reactivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike traditional theories of bonding, in MO theory, electrons are considered to be delocalized into ‘molecular orbitals’, formed through the linear combination of atomic orbitals (mathematically through a combination of the wavefunctions corresponding to the atomic orbitals). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Definition:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://goldbook.iupac.org/M03996.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book:&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No IUPAC definition exists for molecular orbital theory, however, the concept of a molecular orbital is defined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A one-electron wavefunction describing an electron moving in the effective field provided by the nuclei and all other electrons of a molecular entity of more than one atom. Such molecular orbitals can be transformed in prescribed ways into component functions to give 'localized molecular orbitals'. Molecular orbitals can also be described, in terms of the number of nuclei (or 'centers') encompassed, as two-centre, multi-centre, etc. molecular orbitals, and are often expressed as a linear combination of atomic orbitals. An orbital is usually depicted by sketching contours on which the wavefunction has a constant value (contour map) or by indicating schematically the envelope of the region of space in which there is an arbitrarily fixed high (say 96%) probability of finding the electron occupying the orbital, giving also the algebraic sign (+ or −) of the wavefunction in each part of that region.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Molecular orbital (MO) theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. The spatial and energetic properties of electrons within atoms are fixed by quantum mechanics to form orbitals that contain these electrons. While atomic orbitals contain electrons ascribed to a single atom, molecular orbitals, which surround a number of atoms in a molecule, contain valence electrons between atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Applicability of the &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; tag:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; should be applied to any question dealing with the underlying orbitals behind structure or reactivity, whether dealing with the qualitative use of a molecular orbital scheme, or with the mathematical derivation of the orbitals themselves. &lt;/li&gt;&#xA;&lt;li&gt;In most cases, the &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; tag is not standalone and should be applied in addition to other relevant tags. For example: &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A question dealing with the computational generation of molecular orbitals should be tagged with &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A question dealing with the molecular orbitals of a transition metal complex should be tagged with &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; and an additional tag such as &lt;a href=&quot;/questions/tagged/coordination-compounds&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;coordination-compounds&amp;#39;&quot; rel=&quot;tag&quot;&gt;coordination-compounds&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Even a question dealing &lt;em&gt;only&lt;/em&gt; with the theory itself would need tagging with &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt; and the &lt;a href=&quot;/questions/tagged/quantum-mechanics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;quantum-mechanics&amp;#39;&quot; rel=&quot;tag&quot;&gt;quantum-mechanics&lt;/a&gt; tag, as this is the underlying basis for MO theory. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related tags:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/molecular-orbital-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;molecular-orbital-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;molecular-orbital-theory&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/orbitals&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;orbitals&amp;#39;&quot; rel=&quot;tag&quot;&gt;orbitals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/bond&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;bond&amp;#39;&quot; rel=&quot;tag&quot;&gt;bond&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/quantum-mechanics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;quantum-mechanics&amp;#39;&quot; rel=&quot;tag&quot;&gt;quantum-mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;theoretical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Further reading:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Many books have been published on the topic of molecular orbital theory, some of the more readable are listed below: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Atkins', P; &lt;em&gt;Quanta, Matter and Change&lt;/em&gt;; W. H. Freeman: New York, 2009&lt;/li&gt;&#xA;&lt;li&gt;Albright,  T. A.; &lt;em&gt;Orbital Interactions in Chemistry&lt;/em&gt;; Wiley: New Jersey, 2013&lt;/li&gt;&#xA;&lt;li&gt;Fleming, I; &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions (Reference edition)&lt;/em&gt;; Wiley: Sussex, 2010&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="402" LastEditDate="2017-04-08T15:49:02.017" LastActivityDate="2017-04-08T15:49:02.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15098" PostTypeId="4" CreationDate="2014-08-11T03:07:03.200" Score="0" Body="Use of molecular orbital (MO) theory allows for an understanding of the observed properties (shape, reactivity) of molecules. The tag should be applied to all questions related to MO theory, from questions about the qualitative use of the theory to questions about the underlying quantum mechanics and applications in computational chemistry. " OwnerUserId="7448" LastEditorUserId="402" LastEditDate="2017-04-08T15:34:23.293" LastActivityDate="2017-04-08T15:34:23.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15099" PostTypeId="5" CreationDate="2014-08-11T03:09:52.267" Score="0" Body="&lt;p&gt;Means to cut (&quot;-lysis&quot;) with electricity (&quot;electro&quot;). Electrolysis is the use of a direct electrical current to drive an otherwise non-spontaneous reaction. Electrolysis is often used to extract reactive elements from their ores. It can also be used to split water into hydrogen and oxygen gas.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-16T16:54:33.513" LastActivityDate="2018-03-16T16:54:33.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15100" PostTypeId="4" CreationDate="2014-08-11T03:09:52.267" Score="0" Body="For questions about the chemical concepts of electrolysis." OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-16T16:54:28.690" LastActivityDate="2018-03-16T16:54:28.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15101" PostTypeId="5" CreationDate="2014-08-11T03:12:22.277" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-11T03:12:22.277" LastActivityDate="2014-08-11T03:12:22.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15102" PostTypeId="4" CreationDate="2014-08-11T03:12:22.277" Score="0" Body="The amount of energy released when an electron is added to a species in the gaseous phase to form a negative ion. Not to be confused with [ionization-energy]." OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-03-19T13:55:36.223" LastActivityDate="2018-03-19T13:55:36.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15103" PostTypeId="5" CreationDate="2014-08-11T03:13:47.500" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-11T03:13:47.500" LastActivityDate="2014-08-11T03:13:47.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15104" PostTypeId="4" CreationDate="2014-08-11T03:13:47.500" Score="0" Body="Enzymes are macromolecular biological catalysts that alter the rate of a reaction. Use this tag for question regarding enzyme kinetics, classification and action." OwnerUserId="5084" LastEditorUserId="33991" LastEditDate="2017-06-24T06:00:38.303" LastActivityDate="2017-06-24T06:00:38.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15105" PostTypeId="5" CreationDate="2014-08-11T03:15:27.693" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-11T03:15:27.693" LastActivityDate="2014-08-11T03:15:27.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15106" PostTypeId="4" CreationDate="2014-08-11T03:15:27.693" Score="0" Body="Hydrolysis refers to any reaction with water. For example, it may be said that ammonia hydrolyzes in water to produce ammonium ion and hydroxide ion. " OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-11T04:11:10.980" LastActivityDate="2014-08-11T04:11:10.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15107" PostTypeId="5" CreationDate="2014-08-11T03:16:48.020" Score="0" Body="&lt;p&gt;Formal charge is a bookkeeping tool in chemistry. A formal charge (FC) is the hypothetical charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, &lt;strong&gt;regardless of relative electronegativity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining the best Lewis structure (or predominant resonance structure) for a molecule, the structure is chosen such that the formal charge on each of the atoms is as close to zero as possible.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Adapted from &lt;a href=&quot;https://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-17T07:58:50.620" LastActivityDate="2018-03-17T07:58:50.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15108" PostTypeId="4" CreationDate="2014-08-11T03:16:48.020" Score="0" Body="For questions about formal charge in any molecule. Do not confuse with [oxidation-state]." OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-17T07:58:08.190" LastActivityDate="2018-03-17T07:58:08.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15109" PostTypeId="5" CreationDate="2014-08-11T03:19:02.363" Score="0" Body="&lt;h2&gt;Tag usage&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This tag should be used on questions such as (but not limited to):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The properties of stereoisomers, both enantiomers and diastereomers (see also:  &lt;a href=&quot;/questions/tagged/isomers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;isomers&amp;#39;&quot; rel=&quot;tag&quot;&gt;isomers&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/chirality&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chirality&amp;#39;&quot; rel=&quot;tag&quot;&gt;chirality&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;The nomenclature of such isomers according to Cahn-Ingold-Prelog rules or other conventions (e.g. +/-, ᴅ/ʟ) (see also:  &lt;a href=&quot;/questions/tagged/nomenclature&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;nomenclature&amp;#39;&quot; rel=&quot;tag&quot;&gt;nomenclature&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;The study of different conformations of a molecule (see also:  &lt;a href=&quot;/questions/tagged/conformers&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;conformers&amp;#39;&quot; rel=&quot;tag&quot;&gt;conformers&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Particular spatial arrangements of atoms required for a reaction to proceed (see also: &lt;a href=&quot;/questions/tagged/stereoelectronics&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereoelectronics&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereoelectronics&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;The creation of new stereoisomers in a chemical reaction (see also: &lt;a href=&quot;/questions/tagged/selectivity&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;selectivity&amp;#39;&quot; rel=&quot;tag&quot;&gt;selectivity&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/reaction-mechanism&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;reaction-mechanism&amp;#39;&quot; rel=&quot;tag&quot;&gt;reaction-mechanism&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Stereochemistry&lt;/em&gt; is the discipline of chemistry which describes the relative spatial arrangement of atoms in molecules.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Further reading&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Stereochemistry is a concept most commonly encountered in the field of organic chemistry, and aside from the usual major textbooks, a classic reference is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471016705&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eliel's &lt;em&gt;Stereochemistry of Organic Compounds&lt;/em&gt;&lt;/a&gt;. However, it is equally applicable to inorganic compounds and is well-described in all major textbooks.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2017-04-08T14:52:35.940" LastActivityDate="2017-04-08T14:52:35.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15110" PostTypeId="4" CreationDate="2014-08-11T03:19:02.363" Score="0" Body="Applicable to questions about the spatial arrangement of atoms in a molecule and how they affect its physical or chemical properties." OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-01-10T07:35:30.323" LastActivityDate="2017-01-10T07:35:30.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15112" PostTypeId="2" ParentId="14767" CreationDate="2014-08-11T03:31:27.183" Score="1" Body="&lt;p&gt;Sterics can affect basicity not only through solvation but also through &lt;em&gt;inhibition of resonance&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoin acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case of &lt;em&gt;meta-xylene&lt;/em&gt; &lt;strong&gt;with an $\ce{-NH2}$ group between its methyl groups,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/va8zi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resonance is inhibited because the methyl groups prevent the amine group from adopting a perfectly planar conformation. This inhibition of resonance decreases the electron-withdrawing capabilities of the aromatic ring. As a result, electron density is concentrated on the amine group, and &lt;em&gt;meta-xylene&lt;/em&gt; with such a configuration is unusually basic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=my3fqYtt0hwC&amp;amp;pg=SA1-PA16&amp;amp;lpg=SA1-PA16&amp;amp;dq=steric+inhibition+of+resonance+benzene&amp;amp;source=bl&amp;amp;ots=-ZFR_PcpMD&amp;amp;sig=hoE2NQxBP-KjpEgDQIlvRIT2gwQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=mDnoU7K7DovMsQSsnoHQBA&amp;amp;ved=0CGwQ6AEwCA#v=onepage&amp;amp;q=steric%20inhibition%20of%20resonance%20benzene&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://books.google.com/books?id=my3fqYtt0hwC&amp;amp;pg=SA1-PA16&amp;amp;lpg=SA1-PA16&amp;amp;dq=steric+inhibition+of+resonance+benzene&amp;amp;source=bl&amp;amp;ots=-ZFR_PcpMD&amp;amp;sig=hoE2NQxBP-KjpEgDQIlvRIT2gwQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=mDnoU7K7DovMsQSsnoHQBA&amp;amp;ved=0CGwQ6AEwCA#v=onepage&amp;amp;q=steric%20inhibition%20of%20resonance%20benzene&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-11T03:31:27.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15113" PostTypeId="2" ParentId="15082" CreationDate="2014-08-11T04:25:57.800" Score="6" Body="&lt;p&gt;First dissolve phenol and aniline in diethyl ether and add excess HCl solution .&#xA;Then aniline will react with HCl and dissolve in water layer.&#xA;Easily remove organic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the water layer and add excess NaOH to it and aniline will be precipitated. &lt;/p&gt;&#xA;" OwnerUserId="5776" LastActivityDate="2014-08-11T04:25:57.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15114" PostTypeId="2" ParentId="15036" CreationDate="2014-08-11T04:48:59.603" Score="-1" Body="&lt;p&gt;This is the first law of thermodynaics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta E=q+w&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delta E donates change of internal energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q - Amount of energy supplied&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w - work done by the gas&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="5776" LastEditDate="2014-08-11T14:33:42.400" LastActivityDate="2014-08-11T14:33:42.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15115" PostTypeId="2" ParentId="15080" CreationDate="2014-08-11T07:39:53.803" Score="1" Body="&lt;p&gt;Ionic half-reactions method is represented here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Oxidation : 2S_{2}{O}_{3}^{2-}\rightarrow S_{4}{O}_{6}^{2-} +2 e^{-}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Reduction : I_2 +2e^{-}\rightarrow 2I^-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then, as the same number of electrons is transfered in both processes, you just sum these equations to get the balanced reaction. If the number of transfered electrons wasn't the same, then you need to find the greatest common denominator and multiply each equation with a number so the number of transfered electrons gets equal to the greatest common denominator. Then you sum the equations.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="98422" LastEditDate="2021-05-12T06:32:28.780" LastActivityDate="2021-05-12T06:32:28.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15116" PostTypeId="1" CreationDate="2014-08-11T08:20:00.897" Score="2" ViewCount="41" Body="&lt;p&gt;Lets say I want to model a water system as the following set of equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O &amp;lt;=&amp;gt; OH- + H3O+}$ | $\ce{A=[OH^{-}][H3O+]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3}$ | $\ce{B=\frac{[H2CO3]}{[CO2]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2CO3 + H2O &amp;lt;=&amp;gt; HCO3- + H3O+}$ |  $\ce{C=\frac{[HCO3-][H3O+]}{[H2CO3]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCO3 + H2O &amp;lt;=&amp;gt; CO3^{2-} + H3O+}$ |  $\ce{D=\frac{[CO3^{2-}][H3O+]}{[HCO3-]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole system starts off as just $\ce{H2O}$ and $\ce{CO2}$ from the atmosphere. Now consider that calcium carbonate ($\ce{CaCO3}$) from limestone is going to react with carbonic acid in the water ($\ce{H2CO3}$) to cause the dissolution of calcium ions, increasing the water hardness (We will ignore the small affect of calcium carbonate dissolving on its own very slowly). Thus the following equilibrium also exists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaCO3 + H2CO3 &amp;lt;=&amp;gt; Ca^{2+} + 2HCO3-}$ |  $\ce{E=\frac{[Ca^{2+}][HCO3-]^2}{[H2CO3]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using basic math, I can find the relationship between hydronium and calcium concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\frac{(E[H3O+]^2)}{(C^2 B[CO2] )}=[Ca^{2+}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be linked to pH:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\frac{E}{(C^2 B[CO2 ](10^{2pH}) )}=[Ca^{2+}]}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, can I change this to take into consideration the change in pH over time caused by the change in Carbonic Acid etc. without a computer?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="7317" LastEditDate="2014-08-11T10:09:38.773" LastActivityDate="2014-08-11T10:09:38.773" Title="Analytically combining Dynamic Equilibria that have a two way affect?" Tags="&lt;equilibrium&gt;&lt;water&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15117" PostTypeId="2" ParentId="6580" CreationDate="2014-08-11T10:14:57.237" Score="11" Body="&lt;p&gt;The concept of linear combination of atomic orbitals (LCAO) to form molecular orbitals (MO) is probably best understood, while digging a little deeper into quantum chemistry. The method is an approximation that was introduced for &lt;em&gt;ab initio&lt;/em&gt; methods like Hartree Fock. I don not want to go into too much detail, but there a some points that need to be considered before understanding what LCAO actually does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time independent Schrödinger equation can was postulated as $$\mathbf{H}\Psi=E\Psi,$$&#xA; with the Hamilton operator $\mathbf{H}$, the wave function $\Psi$ and the corresponding energy eigenvalue(s) $E$. I will just note that we are working in the framework of the Born-Oppenheimer approximation and refer to many textbooks for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a set of rules, the wave function has to obey. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is a scalar, that can be real or complex, but the product of itself with its complex conjugated version is always positive and real. $$0\leq \Psi^*\Psi=|\Psi|^2$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The probability of finding all $N$ electrons in all space $\mathbb{V}$ is one, hence to function is normed. $$N=\int_\mathbb{V} |\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)|^2 \mathrm{d}(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The value of the wave function has to vanish at infinity. $$0 = \lim_{\mathbf{x}\to\infty}|\Psi(\mathbf{x})|$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The wave function has to be continuous and continuously differentiable, due to the second order differential operator for the kinetic energy $\mathbf{T}_c$ included in $\mathbf{H}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Pauli Principle has to be obeyed. $$\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N) = -\Psi(\mathbf{x}_2,\mathbf{x}_1,\dots,\mathbf{x}_N)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also the variational principle should hold for this approximation, stating that the expectation value for the energy of any trial wave function is larger that the energy eigenvalue of the true ground state. One of the most basic methods to approximately solve this problem is Hartree Fock. In it the trial wave function $\Phi$ is set up as a Slater determinant.&#xA;$$&#xA;\Phi(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_N) =&#xA;\frac{1}{\sqrt{N!}}&#xA;\left|&#xA;   \begin{matrix} &#xA;      \phi_1(\mathbf{x}_1) &amp;amp; \phi_2(\mathbf{x}_1) &amp;amp; \cdots &amp;amp; \phi_N(\mathbf{x}_1) \\&#xA;      \phi_1(\mathbf{x}_2) &amp;amp; \phi_2(\mathbf{x}_2) &amp;amp; \cdots &amp;amp; \phi_N(\mathbf{x}_2) \\&#xA;      \vdots               &amp;amp; \vdots               &amp;amp; \ddots &amp;amp; \vdots \\&#xA;      \phi_1(\mathbf{x}_N) &amp;amp; \phi_2(\mathbf{x}_N) &amp;amp; \cdots &amp;amp; \phi_N(\mathbf{x}_N)&#xA;   \end{matrix} \right| &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expectation value of the energy in the Hartree Fock formalism is set up, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation&quot;&gt;bra-ket notation&lt;/a&gt; as $$E =\langle\Phi|\mathbf{H}|\Phi\rangle.$$&#xA;Skipping through some major parts of the deviation of HF, well end up at an expression for the energy.&#xA;$$E=\sum_i^N \langle\phi_i|\mathbf{H}^c|\phi_i\rangle +\frac12 \sum_i^N\sum_j^N \langle\phi_i|\mathbf{J}_j-\mathbf{K}_j|\phi_i\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the best one-electron wave functions $\phi_i$ we introduce &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot;&gt;Lagrange multiplicators&lt;/a&gt; $\lambda$ minimising the energy with respect to our chosen conditions. These conditions include that the molecular orbitals are ortho normal. $$\langle\phi_i|\phi_j\rangle =\delta_{ij} =\left\{&#xA;\begin{matrix}  0 &amp;amp; , \text{for}~i \neq j\\ 1 &amp;amp; , \text{for}~i = j\\\end{matrix}\right.$$&#xA;I will again skip through most of it and just show you the end expression.&#xA;$$\sum_j\lambda_{ij}|\phi_j\rangle = \mathbf{F}_i|\phi_i\rangle$$&#xA;with the Fock operator set up as&#xA;$$\mathbf{F}_i = \mathbf{H}^c +\sum_j (\mathbf{J}_j − \mathbf{K}_j)$$&#xA;with $i\in1\cdots{}N$, the total number of electrons.&lt;br&gt;&#xA;We can transform these trial wavefunctions $\phi_i$ to canonical orbitals $\phi_i'$ (molecular orbitals) and obtain the pseudo eigenwertproblem $$\varepsilon_i\phi_i'=\mathbf{F}_i\phi_i'.\tag{Fock}$$ &#xA;This equation is actually only well defined for occupied orbitals and these are the orbitals that give the lowest energy. In practice this formalism can be extended to include virtual (unoccupied) molecular orbitals as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now we did not use any atomic orbitals at all. This is the next step to find an approximation to actually solve these still pretty complicated systems.&#xA;LCAO a superposition method. In this approach we map a finite set of $k$ atomic (spin) orbitals $\chi_a$ onto another finite set of $l$ molecular (spin) orbitals $\phi_i'$. They are related towards each other via the expression &#xA;\begin{align}&#xA;\phi_i'(\mathbf{x}) &amp;amp;= c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) + \cdots + c_{k,i}\chi_k(\mathbf{x})\\&#xA;\phi_i'(\mathbf{x}) &amp;amp;= \sum_a^k c_{a,i}\chi_a(\mathbf{x})\\\tag{LCAO}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $\text{(Fock)}$ you can see, that there will be $l$ possible equations depending on the chosen set of orbitals, in the form of&#xA;$$\sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\mathbf{F}_i|\chi_b\rangle = \varepsilon_i \sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\chi_b\rangle$$&#xA;or for short there are $l$ equations $$C_{ab}F_{ab}=\varepsilon_i C_{ab}S_{ab}.$$ Since $a,b \in [1,k]$ you can gather $C_{ab}$ in a $k\times k$ matrix $\mathbb{C}$. Since $i\in[1,l]$, there will only be $l$ $\varepsilon_i$ and therefore the matrix of the Fock elements $F_{ab}$ has to be a $l\times l$ matrix $\mathbb{F}$. The whole problem reduces a matrix equation $$\mathbb{F}\mathbb{C}=\mathbb{S}\mathbb{C}\epsilon\!\!\varepsilon,\tag{work}$$&#xA;from which it is obvious, that the dimension of the involved matrices have to be the same. Hence $k=l$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Too long, didn't read&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The total number of elements in the finite set of atomic orbitals is equal to the total number of elements in the finite set of molecular orbitals. Any linear combination is possible, but only the orbitals that minimise the energy will be occupied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficients $c$ from $\text{(LCAO)}$ will be chosen to minimise the energy of the wave function. This will always be constructive interference. This is also independent of the &quot;original&quot; phase of the orbital that is combined with another orbital. In other words, from $\text{(LCAO)}$&#xA;$$\phi_i'(\mathbf{x}) = c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) \equiv c_{1,i}\chi_1(\mathbf{x}) - c_{2,i}[-\chi_2(\mathbf{x})].$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-11T10:14:57.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15118" PostTypeId="1" CreationDate="2014-08-11T10:24:12.413" Score="4" ViewCount="801" Body="&lt;p&gt;I’m currently working on an underwater robot and was hoping to use the gas bubbles produced by water electrolysis for buoyancy control. The operating depths range from 200–1000 m.&lt;/p&gt;&#xA;" OwnerUserId="7442" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T18:32:20.627" LastActivityDate="2015-05-02T18:32:28.523" Title="Whats the effect of pressure on water electrolysis of sea water with increase in depth?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15119" PostTypeId="1" AcceptedAnswerId="15122" CreationDate="2014-08-11T11:51:00.760" Score="7" ViewCount="87850" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following compounds has the highest melting point: $\ce{PCl5}$, $\ce{H2O}$, $\ce{NaCl}$, $\ce{SrCl2}$, $\ce{CaF2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, ionic compounds will have higher boiling points than those of dipole-dipole interaction compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the compound above that will have the highest melting point will certainly be an ionic compound, either $\ce{NaCl}$, $\ce{CaF2}$ or $\ce{SrCl2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that I should calculate the molar weight but that won't work, because the correct answer is $\ce{CaF2}$, and $\ce{SrCl2}$ has a greater molar weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between ionic compounds, how can the one with the highest boiling point be predicted?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastEditorUserId="194" LastEditDate="2018-03-03T15:32:30.623" LastActivityDate="2020-12-16T00:22:13.373" Title="How to calculate which ionic compound has the highest melting point?" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15120" PostTypeId="2" ParentId="15119" CreationDate="2014-08-11T12:17:31.223" Score="5" Body="&lt;p&gt;You can deduce it from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb&amp;#39;s_law&quot; rel=&quot;noreferrer&quot;&gt;Coulomb law&lt;/a&gt;. As you can see, the bigger the charge of the respective ions, the electrostatic interaction between them is stronger. The charge of calcium and strontium cations is twice as that of sodium. &lt;strong&gt;r&lt;/strong&gt; is the distance between the charges, so for smaller ions the distance would be smaller i.e. the interaction would be greater. Calcium cation is smaller than strontium cation, therefore the correct answer is indeed calcium fluoride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be explained with more sophisticated equations: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kapustinskii_equation&quot; rel=&quot;noreferrer&quot;&gt;Kapustinskii equation&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Born-Land%C3%A9_equation&quot; rel=&quot;noreferrer&quot;&gt;Born-Lande equation&lt;/a&gt;. I will let you analyze them.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-11T12:17:31.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15121" PostTypeId="2" ParentId="15118" CreationDate="2014-08-11T12:35:05.443" Score="4" Body="&lt;p&gt;The higher pressure itsels doent make it much more difficult. But...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water electrolysis you need high overpotential, preferably large electrodes with small electrode gap between them. You want clear, filtered water and no fishes electrocuted, burning down the whole submarine due to an accidental short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you will need plenty of gas for buoyancy control, especially at $200{-}1000\ \mathrm{m}$ depths, therefore you need a lot of electricity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Faraday's law $$m=(Q/F)(M/z)$$ where $F = 96485\ \mathrm{C/mol}$ you can make like $4.5\ \mathrm{l}$ of hydrogen gas (1 atm, 25 °C) with a $10\ \mathrm{A}$ current in 1 hour! That is like $0.04\ \mathrm{l}$ at $1000\ \mathrm{m}$ ($100\ \mathrm{atm}$) of a potentially explosive gas.  &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T18:32:28.523" LastActivityDate="2015-05-02T18:32:28.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15122" PostTypeId="2" ParentId="15119" CreationDate="2014-08-11T13:05:52.703" Score="12" Body="&lt;p&gt;The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the larger the charge the greater the force, and the smaller the radius the smaller the distance between the ions (smaller $r$ in the equation) again leading to a stronger force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first you should calculate the charge on each ion in a molecule. As can see, the charge on strontium ($\ce{Sr^2+}$) and calcium ($\ce{Ca^2+}$) ions is more than the charge on sodium ions ($\ce{Na+}$). Also, calcium ions are smaller than strontium ions. Fluorine ions are also smaller than Chlorine ions. So $\ce{CaF2}$ will have higher bond strength which means it will have higher boiling point since stronger bonds need more energy to break.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="4945" LastEditDate="2014-08-12T07:17:25.287" LastActivityDate="2014-08-12T07:17:25.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15123" PostTypeId="1" CreationDate="2014-08-11T13:07:04.790" Score="5" ViewCount="1938" Body="&lt;p&gt;So the determining forces in the conformational analysis (what will be the more stable state and what will be the more excited state) according to my textbook are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steric interaction&lt;/li&gt;&#xA;&lt;li&gt;Electron repulsion&lt;/li&gt;&#xA;&lt;li&gt;Interaction of bonding and anti bonding molecular orbitals (for example $\sigma$ and $\sigma^{*}$ in ethane)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Electron repulsion is quite clear. Filled (molecular) orbitals repulse each other. I have no clue however what is meant with steric interaction, my textbook makes a small remark about it saying it is a repulsive force when atoms come to close to each other. It notes that in the conformation of ethane this hardly plays a role because the hydrogen atoms are to small to interact. But isn't this the same as electron repulsion? What else makes (neutral) atoms want to avoid each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the last part of bonding and anti bonding orbital interaction (overlap) stabilizing the staggered conformation of ethane is also not so clear to me. How does partially putting electrons in a higher energy molecular orbital (anti bonding) make a molecule's conformation lower in energy?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-08-12T07:21:34.507" LastActivityDate="2014-08-12T07:21:34.507" Title="Conformational analysis determining forces" Tags="&lt;stability&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15124" PostTypeId="2" ParentId="15089" CreationDate="2014-08-11T13:19:22.087" Score="2" Body="&lt;p&gt;Lattice Points are the points in a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Solids/Crystal_Lattice&quot; rel=&quot;nofollow&quot;&gt;crystal lattice&lt;/a&gt; where atoms or ions can be placed. Lattice Points &lt;strong&gt;can be&lt;/strong&gt; vancant. (See &lt;a href=&quot;http://www.britannica.com/EBchecked/topic/145211/crystal-defect&quot; rel=&quot;nofollow&quot;&gt;Crystal Defects&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many Lattice Points are there in each of the Unit Cells?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its depends on which type of Unit Cell we are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Solids/Unit_Cell&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; for full refrence of Unit Cells including &lt;strong&gt;their types&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2014-08-11T13:19:22.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15125" PostTypeId="1" CreationDate="2014-08-11T15:49:13.503" Score="3" ViewCount="1623" Body="&lt;p&gt;I was wondering how one could test the hypothesis that boric acid behaves not as a Brønsted acid but as a Lewis acid in that boric acid cleaves water molecules as to make hydrogen protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfJSk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rad&#xA;Could one run an experiment involving boric acid in heavy water ($\ce{D_2O}$)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if boric acid behaves as a Brønsted acid, the concentration of deuterium cation should stay the same, right? Boric acid might release hydrogen protons but those wouldn't affect the deuterium protons, right? Wouldn't heavy water still undergo auto-ionization and create a certain equilibrium concentration of the heavy equivalents of hydronium and hydroxide ions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus, if boric acid behaves as a Lewis acid, couldn't we somehow precipitate the $\ce{B(OH)_3(DO)^-}$, and test whether the precipitate has D rather than H? I.e. introduce sodium chloride into the system and evaporate to dryness. Wouldn't that leave behind a precipitate consisting partly of sodium ion and partly of the borate? On the other hand what would happen to the excess chloride ion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:56:19.750" LastActivityDate="2020-08-17T16:15:23.193" Title="Boric Acid - Mechanism of Acidity" Tags="&lt;acid-base&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15126" PostTypeId="1" CreationDate="2014-08-11T16:32:13.153" Score="3" ViewCount="2662" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction between hydrogen and oxygen to yield water vapor has&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ = \pu{- 484 kJ}$&lt;/span&gt;. How much &lt;span class=&quot;math-container&quot;&gt;$pV$&lt;/span&gt; work is done, and what is the value of &lt;span class=&quot;math-container&quot;&gt;$\Delta E$&lt;/span&gt; in kilojoules for the reaction of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.50 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\pu{0.25 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; at atmospheric pressure if the volume change is &lt;span class=&quot;math-container&quot;&gt;$-\pu{5.6L}?$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2(g) + O2(g) -&amp;gt; 2H2O(g)} \qquad \Delta H^\circ = \pu{- 484 kJ}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I use the formula &lt;span class=&quot;math-container&quot;&gt;$\Delta E=\Delta H - p\Delta V$&lt;/span&gt; to determine &lt;span class=&quot;math-container&quot;&gt;$\Delta E.$&lt;/span&gt; However, when determining the enthalpy, the solutions manual does this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta H = \frac{\pu{-121 kJ}}{\pu{0.50 mol}~\ce{H2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where does the &lt;span class=&quot;math-container&quot;&gt;$-121$&lt;/span&gt; come from? From my understanding, since there are two moles &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; should be &lt;span class=&quot;math-container&quot;&gt;$-\pu{242 kJ}$&lt;/span&gt;. Or do we take into account all four hydrogen atoms? That would give us &lt;span class=&quot;math-container&quot;&gt;$\pu{-484 kJ}/4 = -\pu{121 kJ}.$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="41328" LastEditDate="2020-05-26T13:06:00.457" LastActivityDate="2020-05-26T13:06:00.457" Title="Enthalpy of the reaction between hydrogen and oxygen" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15127" PostTypeId="5" CreationDate="2014-08-11T17:03:56.050" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-11T17:03:56.050" LastActivityDate="2014-08-11T17:03:56.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15128" PostTypeId="4" CreationDate="2014-08-11T17:03:56.050" Score="0" Body="Refers to ability for an atom in a covalent framework to attract electron density to itself. Do not conflate with electron affinity, which is the ability of a lone atom (or molecule) to attract an electron to itself. Both are measured in joules/mole. " OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-11T20:06:45.637" LastActivityDate="2014-08-11T20:06:45.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15129" PostTypeId="1" CreationDate="2014-08-11T18:18:23.803" Score="1" ViewCount="8559" Body="&lt;p&gt;The problem is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which of the following compounds does Cl have the most positive partial charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SCl2 ,, HCl  ,, BrCl  ,, OCl2 ,,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;before I go through how I will solve this problem .. There is a question : Can Cl have a partial &lt;strong&gt;positive&lt;/strong&gt; charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, all of them are polar, so there is a partial charge on each one of them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how to compare or specify the partial charges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think about is electronegativity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it like .. the greater the difference in electronegativity the more the partial charges ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or .. ?&lt;/p&gt;&#xA;" OwnerUserId="7278" LastActivityDate="2014-08-13T20:14:10.807" Title="In which of the following compounds does Cl have the most positive partial charge?" Tags="&lt;electrons&gt;&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15130" PostTypeId="1" AcceptedAnswerId="15132" CreationDate="2014-08-11T18:21:24.320" Score="6" ViewCount="2333" Body="&lt;p&gt;If I understand everything correctly, carboxylic acids behave like acids because oxygen is much more electronegative than hydrogen, so it takes his electron and releases H+. The problem is that alcohols have an OH group too, so why don't they behave like acids?&lt;/p&gt;&#xA;" OwnerUserId="7344" LastEditorUserId="7475" LastEditDate="2015-10-16T16:41:04.867" LastActivityDate="2015-10-17T14:17:20.523" Title="Why alcohols don't behave as acids?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15131" PostTypeId="2" ParentId="15123" CreationDate="2014-08-11T18:32:12.360" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;what is meant with steric interaction&lt;/p&gt;&#xA;&lt;p&gt;isn't this the same as electron repulsion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I agree with you Jori, the basis of steric interactions is electron repulsion.  When anyone stops to think about &amp;quot;why are adjacent bulky groups destabilizing&amp;quot; they'll soon arrive at the destabilizing interpenetration of electron clouds.  However, the use of the term &amp;quot;steric&amp;quot; to denote interactions due to size and bulk has been around so long that it has simply become common usage - the way things are said. The following figure shows trans-1,2-di-&lt;em&gt;t&lt;/em&gt;-butylethylene to be more stable than its cis counterpart.  It's easier to say and understand that the &lt;strong&gt;bulky&lt;/strong&gt; &lt;em&gt;t&lt;/em&gt;-butyl groups produce steric strain in the cis isomer, rather than to say electronic clouds interpenetrate in the cis isomer.  The word &amp;quot;steric&amp;quot; immediately creates a picture of sorts in a chemist's mind.  Tetra-&lt;em&gt;t&lt;/em&gt;-butylethylene has yet to be synthesized; attributing this to steric bulk around the double bond provides a clear image of the problem.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XhZBc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Then the last part of bonding and anti bonding orbital interaction&#xA;(overlap) stabilizing the staggered conformation of ethane is also not&#xA;so clear to me. How does partially putting electrons in a higher&#xA;energy molecular orbital (anti bonding) make a molecule's conformation&#xA;lower in energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;First off, let's recognize that the orbital interaction you are referring to is a way to understand the effect of hyperconjugation in various conformations.  Also, we are not putting electrons in a higher energy molecular orbital.&lt;/p&gt;&#xA;&lt;p&gt;Here are some diagrams that may help explain this.  The first diagram on the left doesn't relate to hyperconjugation, but since it is part of the picture, let's explain it.  The Pauli Exclusion principle says that we can't have electrons with the same quantum numbers in the same orbital. In ethane, each pair of electrons in one C-H bond have the same quantum numbers as the pair electrons in any other C-H bond.  Therefor, the closer we bring 2 C-H bonds, the closer we are to violating the Pauli Exclusion principle.  This argument tells us that the eclipsed conformation should therefor, be higher in energy than the staggered conformation.&lt;/p&gt;&#xA;&lt;p&gt;The figures in the middle and the right show the effects of hyperconjugation upon the eclipsed and staggered conformations of ethane, respectively.  According to Frontier Molecular Orbital theory we examine the interaction between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) to see if stabilization or destabilization results. In the hyperconjugated resonance structure for ethane we have a proton and a pair of electrons in the carbon &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; orbital, that &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; orbital with two electrons is our HOMO.  The only LUMO for it to interact with is the LUMO for the C-H (the &lt;span class=&quot;math-container&quot;&gt;$\sigma^{*}$&lt;/span&gt; orbital) bond on the adjacent carbon atom.  We can mix the &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sigma^{*}$&lt;/span&gt; orbitals to produce 2 new MOs, one bonding (no nodes between these 2 orbitals) and one antibonding (1 node between these two orbitals).  What is pictured below is the bonding (in-phase, no nodes) combination of these two orbitals.  We see that in the eclipsed (middle) conformation the interaction in this orbital is destabilizing, there is no constructive overlap.  However, in the staggered (right) conformation the overlap within this orbital is now constructive - there is positive overlap.  The hyperconjugation argument also favors the staggered conformation being at lower energy.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zFsIq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/juMR5.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-08-11T19:31:10.340" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15132" PostTypeId="2" ParentId="15130" CreationDate="2014-08-11T18:46:59.123" Score="9" Body="&lt;p&gt;The figure below illustrates these two ionization reactions.  The key difference between the two is that when the carboxylic ionizes the carboxylate anion is formed.  The carboxylate anion is much more stable than the alkoxide anion formed when the alcohol ionizes.  The carboxylate anion is much more stable because we can draw resonance structures to describe it.  This resonance stabilization of the carboxylate anion explains why carboxylic acids are so much more acidic than an alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6vk9m.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6vk9m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-10-17T14:17:20.523" LastActivityDate="2015-10-17T14:17:20.523" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15133" PostTypeId="2" ParentId="15129" CreationDate="2014-08-11T19:06:41.157" Score="3" Body="&lt;p&gt;The greater the difference in electronegativity the more the partial charges is true. Compare the electronegativities of the atoms bonded with chlorine and you can easily determine in which compound chlorine has the biggest partial charge. Chlorine can have a partial positive charge when it's bonded with a more electronegative element like oxygen or fluorine.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-11T19:06:41.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15134" PostTypeId="1" AcceptedAnswerId="15139" CreationDate="2014-08-11T19:25:41.130" Score="5" ViewCount="33985" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the heat that will be released in kJ per gram when aluminium (&lt;span class=&quot;math-container&quot;&gt;$\ce{Al}$&lt;/span&gt;) reacts with &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2O3}$&lt;/span&gt; as follows:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The information given is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H = 12.40 \space \rm{kJ/mol}$&lt;/span&gt; (for &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H = -822.2 \space \rm{kJ/mol}$&lt;/span&gt; (for &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2O3}$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H = -1669.8 \space \rm{kJ/mol}$&lt;/span&gt; (for &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2O3}$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If it asked me to find the heat released in kJ/mol that would be easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\Delta H^\circ = \sum \nu_{i}\Delta H^\circ_f (\rm{products}) - \sum \nu_{i}\Delta H^\circ_f (\rm{reactants})&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that there is one mole and I found the mass in grams by multiplying the molar mass of each compound by one mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I divided the &lt;span class=&quot;math-container&quot;&gt;$H^\circ_f$&lt;/span&gt;'s by those amounts to get the new values &quot;expressed in kJ per gram&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I used the previous equation and I got an answer around 10 but the correct answer is 15.24 kJ/g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is the way I used wrong, or is there something else I didn't pay attention to?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="4945" LastEditDate="2019-09-16T12:19:55.480" LastActivityDate="2019-09-16T12:19:55.480" Title="How can I find the heat that will be released in kJ per gram for the reaction of Al with Fe2O3?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15136" PostTypeId="1" AcceptedAnswerId="15138" CreationDate="2014-08-11T20:09:40.853" Score="5" ViewCount="1126" Body="&lt;p&gt;One of the primary components of Dalton's theory is to &quot;express&quot; an atom as an unbreakable spherical shape filled with whatever. The question is, why didn't he suggest any other shape, like for example, cubes?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5633" LastEditDate="2014-08-11T20:29:33.427" LastActivityDate="2018-06-01T17:24:39.137" Title="Why did Dalton assume that atoms are spherical?" Tags="&lt;atoms&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15138" PostTypeId="2" ParentId="15136" CreationDate="2014-08-11T20:27:38.247" Score="9" Body="&lt;p&gt;I don't know what Dalton was thinking exactly, but I do know that for a very long time philosophers and scientists have assumed that things are spherical in the absence of any indication that they are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this was originally related to religious beliefs, and the idea that God or (The Gods) would only create perfect objects, and the sphere (&lt;a href=&quot;http://en.wikipedia.org/wiki/Perfection#Aesthetics&quot;&gt;being perfect&lt;/a&gt;) was the obvious choice for heavenly objects and subatomic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption that &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_cow&quot;&gt;things are spherical&lt;/a&gt; is still used very often, although it is not usually directly related to religious beliefs anymore. These days, the idea is that you should always use the simplest possible explanation that works. The simplest geometric shape is a sphere - you only need one dimension to describe it, and it has some &lt;a href=&quot;http://en.wikipedia.org/wiki/Sphere&quot;&gt;other interesting properties&lt;/a&gt; like minimized surface area-to-volume ratio. Mathematics can also be greatly simplified by using spherical coordinates when things are spherical, which is another big advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when we imagine particles that we can't see, a sphere is a natural choice, and I would guess that Dalton was using one or both of these lines of reasoning when he came up with the theory. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-11T20:27:38.247" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15139" PostTypeId="2" ParentId="15134" CreationDate="2014-08-11T20:40:39.477" Score="4" Body="&lt;p&gt;When you converted the heats to kJ/g, you inadvertently &quot;broke&quot; the Hess's law equation that you plugged them into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the enthalpies of formation in that equation is multiplied by the number of moles of the matching species, and so the units of the enthalpy of formation are kJ/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume one mole of product and convert that to grams, you then would have to convert all of the other species to grams (two moles of Al, two moles of Fe, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to just do the whole calculation in moles, then divide the final answer by grams. However, you need to specify which species is the gram basis. In other words, you are asking for the heat per gram, but per gram of what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the answer is, then find the total heat output in terms of kJ/mol of that substance, then divide it by the molar mass of the substance (in g/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to avoid these kinds of problems is to always explicitly include the units in any calculation that you do. In this case, you would have quickly seen that the Hess's Law equation would not give you the correct units.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-11T20:40:39.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15140" PostTypeId="2" ParentId="15134" CreationDate="2014-08-11T20:49:58.387" Score="3" Body="&lt;p&gt;I'm guessing that you're looking for the heat released per gram of &lt;strong&gt;aluminum&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct to here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔH° = Σ(ν × ΔHf°) (products) - Σ(ν × ΔHf°) (reactants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are 2 moles of aluminum, you need to divide this ΔH° by 2, then you have heat released in kJ/mol of aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert this to kJ/gram you need to now divide by the atomic weight of aluminum (26.98 gm/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I did this I got an answer of 15.25 kJ/gm aluminum heat released.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-11T20:49:58.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15141" PostTypeId="1" AcceptedAnswerId="15142" CreationDate="2014-08-11T21:10:06.043" Score="8" ViewCount="1623" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15119/which-compound-has-the-highest-melting-point&quot;&gt;this question&lt;/a&gt;, what is the substance or compound with the highest melting point? At standard atmospheric pressure and ever-increasing temperature which substance a) eventually becomes liquid b) does so at the highest temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about glasses or related semisolids, so perhaps a reference viscosity? Glass transition temperatures are kind of like melting points, but some materials have a glass transition temperature and a melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some initial casting about on the internet finds that rhenium and tungsten have melting points at about 3400 °C, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hafnium_carbide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hafnium carbide&lt;/a&gt; claims to be 'one of the highest' at 4100 kelvin. Does it actually melt or just decompose and then sublimate? 4100 kelvin is its 'refractory temperature', but I'm not certain how that's defined.&lt;/p&gt;&#xA;" OwnerUserId="7103" LastEditorDisplayName="user7951" LastEditDate="2018-08-13T08:46:07.717" LastActivityDate="2019-03-24T12:05:30.890" Title="The highest melting point" Tags="&lt;physical-chemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15142" PostTypeId="2" ParentId="15141" CreationDate="2014-08-11T21:37:47.477" Score="8" Body="&lt;p&gt;Well, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Tantalum_hafnium_carbide&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;, it might be an alloy of tantalum carbide and hafnium carbide, specifically &lt;span class=&quot;math-container&quot;&gt;$\ce{Ta4HfC5}$&lt;/span&gt;.  A melting point of 4215 °C (4488 K) is listed.  But who really knows the answer to your question with so many new, proprietary alloys with complex compositions being fabricated these days. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2019-03-24T12:05:30.890" LastActivityDate="2019-03-24T12:05:30.890" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15145" PostTypeId="2" ParentId="15072" CreationDate="2014-08-11T23:46:31.790" Score="2" Body="&lt;p&gt;According to the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hand_sanitizer&quot; rel=&quot;nofollow&quot;&gt;hand sanitizers&lt;/a&gt;, polyacrylic acid and polyethelyne glycol are commonly used thickeners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try any hydrophilic polymer if those aren't available. Cornstarch might even work if you dissolved it in the alcohol before adding the other ingredients.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-11T23:46:31.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15146" PostTypeId="1" AcceptedAnswerId="24083" CreationDate="2014-08-12T01:14:24.110" Score="3" ViewCount="168" Body="&lt;p&gt;On this table I realized that elemental potassium could adopt oxidation states of -1, 0, and +1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that elemental potassium can function as both an oxidizer and a reducer? I understand that oxidizers are reduced (gain electrons), and that reducers are oxidized (lose electrons). How can potassium anion be formed? The potassium anion must be extremely unstable due to potassium's extremely low electronegativity, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in what compound does potassium actually have a negative oxidation state?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-12T04:34:22.360" LastActivityDate="2015-01-23T15:06:13.093" Title="Elemental Potassium as both an oxidizer and a reducer" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15147" PostTypeId="5" CreationDate="2014-08-12T01:24:18.247" Score="0" Body="&lt;p&gt;Free radicals, or commonly abbreviated as radicals, are atoms, molecules or ions that bear unpaired electrons in their valence shell, usually in their highest occupied molecular orbital (HOMO). They are therefore usually highly reactive species, that are often unstable. They react with each other or the system to form more stable diamagnetic products, or they fragment into smaller molecular entities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzoyl_peroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Benzoyl peroxide&lt;/a&gt; $\ce{Ph-CO-O-O-OC-Ph}$:&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b8/Benzoyl-peroxide.svg/200px-Benzoyl-peroxide.svg.png&quot; alt=&quot;structure&quot;&gt;&lt;br&gt;&#xA;The peroxyl bond can be split thermally or photoelectrically, resulting in a chain reaction to form carbon dioxide and the phenyl radical. $$\ce{Ph-CO-O-O-OC-Ph -&amp;gt;[h\nu/\Delta T] 2 Ph-COO. -&amp;gt;[][-2\ce{CO2}] 2Ph.}$$&#xA;It is most commonly used in radical polymerisation as a starting agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many examples of small persistent or stable molecules based on carbon nitrogen and oxygen. For example, $\ce{NO}$ and $\ce{NO2}$ have been vigorously investigated, both, experimentally and theoretically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radicals can also be stabilised with sterically demanding moieties, at least long enough to investigate these compounds. One of the earliest examples dates back to the 19th century, the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Triphenylmethyl_radical&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gomberg radical, triphenylmethyl radical, $\ce{Ph3C.}$&lt;/a&gt;. This can be considered kinetically stable since recombination with itself is prevented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, a combination of electrical effects, partial spin delocalisation and/or steric hindrance is used to make radicals more persistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radicals can be synthesised from homolytic bond cleavage, and one-electron reductions or oxidations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prominent example for molecules with more than one unpaired electrons is the dioxygen molecule, $\ce{O2}$, which has a triplet ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether a species has unpaired electron(s) can be experimentally determined through electron pair resonance spectroscopy, as Radicals are paramagnetic. Analysis of the Lewis structure and molecular orbital theory can also predict unpaired electrons. It is also possible to predict the stability of radicals with modern quantum chemical investigations.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-04T02:32:20.750" LastActivityDate="2017-05-04T02:32:20.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15148" PostTypeId="4" CreationDate="2014-08-12T01:24:18.247" Score="0" Body="Radicals are highly reactive species, that have one or more unpaired electrons. Whether a species has unpaired electron(s) can be experimentally determined through electron pair resonance spectroscopy. Radicals are paramagnetic. This tag should be applied to questions that involve free radicals as a species or radicals in reactions and their mechanisms." OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-08-12T04:30:13.137" LastActivityDate="2014-08-12T04:30:13.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15150" PostTypeId="1" CreationDate="2014-08-12T05:58:31.130" Score="2" ViewCount="418" Body="&lt;p&gt;Is forming metal coins (specifically ones with copper, nickel, and zinc) possible with these steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Carve a sheet of sodium chloride so that it is big but not too big and so that the design sticks out of the sheet of sodium chloride.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cut the sodium chloride so that you have 1 big circle with the design&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Form a mold out of clay from the sodium chloride design&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Scale it down without losing detail using a carving machine(instead of hand carving like with the sodium chloride) that carves the design on each side on a separate piece of clay(which is much easier to carve than sodium chloride).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once the clay is in the leather hard stage(usually takes about a day) put a piece of clay of uniform thickness and hardness around each side of the coin so that you can hand carve the rim(make sure the rim is not too wide) This will make there be space for the metal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once the rim is in the leather hard stage attach the 2 sides to a base, again made out of clay&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fire the clay&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Smelt the metal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;pour the metal into the clay mold&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;get the metal coin out of the clay mold&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat the metal again but don't melt it&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cool the hot metal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wash and dry the metal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;reuse the clay mold so that you don't have to start all over again from hand carving sodium chloride but rather only from step 8 which is smelting the metal&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One of my big concerns of this process is that a copper, nickel, zinc mixture might melt the clay as soon as the metal is poured in but at the same time I don't want to use a metal that you can bend with the pressure from your hand like you can with aluminum. &lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2014-08-12T08:03:48.800" LastActivityDate="2014-08-12T08:03:48.800" Title="Are forming coins starting with sodium chloride possible without the clay melting from the molten metal?" Tags="&lt;physical-chemistry&gt;&lt;melting-point&gt;&lt;metallurgy&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15151" PostTypeId="1" CreationDate="2014-08-12T06:17:09.327" Score="2" ViewCount="194" Body="&lt;p&gt;As everyone knows a diamond won't last forever. It will eventually turn into a graphite over a long period of time, as the free energy of graphite is lower than that of diamond, hence making the reaction energetically favorable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is it possible to do the reverse, by using graphite and the applying energy onto graphite so as to synthesize diamond. If so, how is it done and can it be done in a lab by a chemist? Does that mean that people who managed to do it can become rich? As a diamond costs so much more than a graphite. Shouldn't we synthesize diamonds in this way? Can I become  rich by synthesizing diamonds and selling them?&lt;/p&gt;&#xA;" OwnerUserId="7422" LastEditorUserId="4945" LastEditDate="2014-08-12T06:49:18.750" LastActivityDate="2014-08-12T07:05:41.503" Title="Synthetical formation of diamonds" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15152" PostTypeId="2" ParentId="15151" CreationDate="2014-08-12T07:05:41.503" Score="1" Body="&lt;p&gt;Natural Diamonds are formed under high pressure and temperature at depths of 140 to 190 kilometers (87 to 118 mi) in the Earth's mantle. Carbon-containing minerals provide the carbon source, and the growth occurs over periods from 1 billion to 3.3 billion years (25% to 75% of the age of the Earth). Diamonds are brought close to the Earth's surface through deep volcanic eruptions by a magma, which cools into igneous rocks known as kimberlites and lamproites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The formation of natural diamond requires very specific conditions—exposure of carbon-bearing materials to high pressure, ranging approximately between 45 and 60 kilobars (4.5 and 6 GPa), but at a comparatively low temperature range between approximately 900 and 1,300 °C (1,650 and 2,370 °F). These conditions are met in two places on Earth; in the lithospheric mantle below relatively stable continental plates, and at the site of a meteorite strike.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So yes, Diamonds can be produced synthetically if you can simulate almost similar conditions as in Earth's Mantle.&lt;/strong&gt; (See &lt;a href=&quot;http://en.wikipedia.org/wiki/HPHT_diamond&quot; rel=&quot;nofollow&quot;&gt;high-pressure high-temperature process; Synthetic Diamond&lt;/a&gt;) An alternative, and completely different growth technique is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_vapor_deposition&quot; rel=&quot;nofollow&quot;&gt;chemical vapor deposition (CVD)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2014-08-12T07:05:41.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15153" PostTypeId="1" AcceptedAnswerId="15155" CreationDate="2014-08-12T07:36:43.793" Score="0" ViewCount="2647" Body="&lt;p&gt;Diamagnetism exists in every molecule or ionic compound. It is the ability to be repelled by a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you might think from this, that liquid oxygen would either be repelled by the magnet or not affected at all, yet it is attracted to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2}$ is paramagnetic, but how can a paramagnetic material like oxygen be attracted to a ferrimagnetic material like Iron other than the fact, that neutral attracts both positive and negative charges and $\ce{O2}$ is neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just this, or is there another part to it, like charged iron atoms inducing a dipole in $\ce{O2}$ or something?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="4945" LastEditDate="2014-08-12T09:57:39.300" LastActivityDate="2014-08-12T10:13:57.740" Title="Why are paramagnetic materials attracted to ferrimagnetic materials(such as iron, nickel, and cobalt)?" Tags="&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15154" PostTypeId="1" CreationDate="2014-08-12T08:00:14.363" Score="3" ViewCount="119" Body="&lt;p&gt;By &lt;a href=&quot;http://en.wikipedia.org/wiki/Henry%27s_law&quot; rel=&quot;nofollow&quot;&gt;Henry’s Law&lt;/a&gt;, the concentration of a dissolving gas in a liquid is proportional to the partial pressure of that gas in contact with the liquid, mathematically,&#xA;$ p = k_\mathrm{H} c$,&#xA;where $p$ is the partial pressure of the dissolving gas, $c$ is the concentration of the dissolving gas in the liquid (solution), and $k_\mathrm{H}$ is a constant (Henry’s constant?) which depends on temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the value of $k_\mathrm{H}$ for common noble gases (in particular, xenon) in concentrated phosphoric acid, at, say $298~\mathrm{K}$? Can the solubility at other temperatures be calculated using an extrapolation formula?&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorDisplayName="user7951" LastEditDate="2015-06-30T19:11:18.780" LastActivityDate="2015-06-30T19:11:18.780" Title="What is the Solubility of Common Noble Gases in Concentrated Phosphoric Acid, 85% w/v?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15155" PostTypeId="2" ParentId="15153" CreationDate="2014-08-12T08:28:07.903" Score="7" Body="&lt;p&gt;First of all, in your question i feel a confusion about the nature of magnetism and paramagnetism. These magnetic interactions have nothing to do with attraction/repulsion of electric charges, and magnets do not attract or repel Coulomb charges (when we discuss the magnetic part of the interaction). In short static, magnetic and electric properties are two different stories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, iron is ferrOmagnetic and not ferrImagnetic (capitalisation is by me). Forget ferrimagnetic materials for a second, and let's just talk about ferromagnetic, paramagnetic and diamagnetic materials: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diamagnetic materials have no free electronic spin in the system, therefore they do not interact with magnetic field as a first approximation. There is a very small repulsive interaction coming from the polarisation of paired spins, but it is very small compared to any other interactions. Most organic compounds are in this category.&lt;/li&gt;&#xA;&lt;li&gt;Paramagnetic materials have free electronic spins, which can flip / rotate. When put the sample in magnetic field, this field polarises the spins which turns into an attractive interaction. This is analogous with the electric field induced dipole moments, but this is a magnetic interaction, not an electronic interaction. The alignment of spins are constantly changing, as simple thermal excitation can rotate them. Therefore paramagnetic materials has no permanent magnetic moments like magnets. Typical examples are molecular oxygen, transition metal complexes, etc. &lt;/li&gt;&#xA;&lt;li&gt;In ferromagnetic materials the free electronic spins are having a strong interaction with each other, and aligned parallel in strict order. This alignment is so stable that the magnetic moment of spins adds up, and there can be a macroscopical magnetic field around the sample (i.e. it is a magnet). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you see the attraction between paramagnetic and magnetic materials, you see the polarisation of the paramagnetic materials in the magnetic field of ferromagnetic materials, which is an attractive itneraction. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2014-08-12T10:13:57.740" LastActivityDate="2014-08-12T10:13:57.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15161" PostTypeId="1" AcceptedAnswerId="15240" CreationDate="2014-08-12T12:59:17.007" Score="4" ViewCount="220" Body="&lt;p&gt;[&lt;em&gt;Sorry if the question looks trivial but I have thought about it for days and red a number of books all with no success.&lt;/em&gt;]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In almost every thermodynamics book that has something to say about chemical reactions there is a proof for the fact that at constant pressure and temperature the process is spontaneous in the direction in which the Gibbs free energy decreases. In some of the books it is also remarked that this conclusion is for when there is no &quot;additional work&quot; (beside the work of pressure in expanding the system), if there is an additional work then except for $dG=0$ the reversible process will yield into that additional work gaining its maximum value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{The only work is expansive:}\qquad &amp;amp; dG\le 0\\&#xA;\text{There are some additional works present:}\qquad &amp;amp; dG\le \delta W_{\text{additional}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this regard I have a question:&#xA; When we have some additional works there in the problem (&lt;em&gt;like electric work in an electrochemical cell&lt;/em&gt;), so that the second inequality holds and the maximum work is assigned to $dG$ as $dG=\delta W_{Add,Max}$, why still we feel free to use the first inequality and claim $dG$ should be negative?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the case of electrochemical cells we have: $\delta W_{E,Max}=-n\,F\,E_{cell}\,d\zeta$ wherein $\zeta$ measure the advancement of the reaction. So we would obtain in a reversible (i.e. infinitely slow) reaction $\Delta_rG=\frac{dG_{p,t}}{d\zeta}=-n\,F\,E_{cell}$. But all the books then state that since the reaction is spontaneous in the direction in which $\Delta_rG\le 0$ so we would obtain $E_{cell}\ge 0$, while we know the inequality $\Delta_rG\le0$ is obtained from $dG\le0$ which doesn't hold here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question in other wording reads like this. Does the electrochemical cell problem involve an electric work or not? If &quot;yes&quot; then from where do we know the Gibbs free energy should decrease? And if &quot;no&quot; then from where do we obtain the equality $\Delta_rG=-n\,F\,E_{cell}$?&lt;/p&gt;&#xA;" OwnerUserId="6980" LastActivityDate="2014-08-13T18:36:42.903" Title="What can we say about the Gibbs free energy if it decreases or not?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15162" PostTypeId="2" ParentId="14562" CreationDate="2014-08-12T14:01:12.427" Score="0" Body="&lt;p&gt;Sucrose does melt at &lt;span class=&quot;math-container&quot;&gt;$\pu{366.8 ^\circ F}$&lt;/span&gt;. It decomposes once you get to hotter temperatures, which is why caramel darkens till it becomes burnt sugar. Thus unlike sodium chloride and most other substances sucrose and lots of other sugars from trioses upwards in numbers of carbons don't have a boiling point, just a melting point because heat breaks the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; bonds more in sugars than other organic molecules. The exact reason? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;span class=&quot;math-container&quot;&gt;$\ce{R-C-O}$&lt;/span&gt; it is easy to form carbon dioxide, and there is also an &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; attached to one of the carbons which water can form easily from.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="16683" LastEditDate="2018-12-30T19:16:08.237" LastActivityDate="2018-12-30T19:16:08.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15163" PostTypeId="1" AcceptedAnswerId="15165" CreationDate="2014-08-12T14:22:20.127" Score="-1" ViewCount="4604" Body="&lt;p&gt;I have read from several sources about sucrose and some say it is 2 glucose molecules and some others say it is 1 glucose molecule and 1 fructose molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that both of these are related disaccharides since they have the same molecular formula of $\ce{C12H22O11}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which one is actually sucrose and what is the other disaccharide. &lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user7951" LastEditDate="2016-12-23T19:55:05.720" LastActivityDate="2016-12-23T19:55:05.720" Title="Is sucrose made of 2 glucose molecules or 1 glucose and 1 fructose?" Tags="&lt;organic-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2014-08-15T01:00:32.337" ContentLicense="CC BY-SA 3.0" />
  <row Id="15165" PostTypeId="2" ParentId="15163" CreationDate="2014-08-12T15:04:44.233" Score="3" Body="&lt;p&gt;Sucrose is composed of one glucose and one fructose monosaccharide.  Two glucose molecules form maltose.&lt;/p&gt;&#xA;" OwnerUserId="5762" LastActivityDate="2014-08-12T15:04:44.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15166" PostTypeId="1" AcceptedAnswerId="15167" CreationDate="2014-08-12T15:44:13.120" Score="-3" ViewCount="670" Body="&lt;p&gt;Resonance structures are about moving electrons to form anions, cations, and new bonds. Structural isomers can be resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example take phenyl ketone. Of course you can form resonance structures by moving the pi electrons from benzene. However you can also move the electrons from the carbonyl double bond to the oxygen forming phenyl oxide. From there you can again move the electrons from the benzene ring. All of these are resonance structures and the ones with the alkoxide anion are structural isomers as well as resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way you can move the pi electrons from the $\ce{O-Cl}$ double bond to the oxygen to form $\ce{O-Cl-O}$(I don't know how to make the negative charge on the left oxygen not look like a 4 e bond(which unless you are talking about a double bond is nonexistent)) which can then form a peroxide bond($\ce{O-O}$ single bond) to make a cyclic peroxide. Resonance structures are all about moving electrons and in the process forming anions within molecules(like the alkoxide anion), cations within molecules, and new bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how are chlorine dioxide($\ce{O=Cl=O}$) and this cyclic peroxide that forms not resonance structures and just structural isomers?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="4945" LastEditDate="2014-08-16T17:02:27.973" LastActivityDate="2014-08-16T17:02:27.973" Title="O=Cl=O and O-Cl-O in a ring are not resonance structures?" Tags="&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15167" PostTypeId="2" ParentId="15166" CreationDate="2014-08-12T15:45:52.983" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Resonance structures are about moving electrons to form anions, cations, and new bonds. Structural isomers can be resonance structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, resonance structures do not form anions or cations. Resonance structures should all have the same overall formal charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also moving electrons around when drawing resonance forms does not create new bonds. These bonds always have existed; it's just that one picture is inadequate to represent them all. Hence the resonance forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structural isomers also cannot be resonance structures. Structural isomers differ fundamentally in terms of the bonding between or among their constituent atoms. Resonance structures are just things we draw on paper to better represent the true nature of bonding. For one chemical entity, there may be multiple resonance forms, but that doesn't mean that each is a discrete form that is able to be isolated. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the same way you can move the pi electrons from the O−Cl double bond to the oxygen to form O−Cl−O(I don't know how to make the negative charge on the left oxygen not look like a 4 e bond(which unless you are talking about a double bond is nonexistent)) which can then form a peroxide bond(O−O single bond) to make a cyclic peroxide. Resonance structures are all about moving electrons and in the process forming anions within molecules(like the alkoxide anion), cations within molecules, and new bonds.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So how are chlorine dioxide(O=Cl=O) and this cyclic peroxide that forms not resonance structures and just structural isomers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by this. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-12T15:58:23.670" LastActivityDate="2014-08-12T15:58:23.670" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15168" PostTypeId="1" AcceptedAnswerId="15198" CreationDate="2014-08-12T15:55:37.513" Score="0" ViewCount="487" Body="&lt;p&gt;If the question seems rather &quot;unscientific&quot;, be kind to accept my apology. My textbook is being highly unclear about the topic and I need to know the answer to this question before I go on with my studying.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2014-08-13T05:20:39.597" Title="Is it possible for a reaction to have an entropy change of zero?" Tags="&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15169" PostTypeId="1" AcceptedAnswerId="15185" CreationDate="2014-08-12T16:06:14.623" Score="3" ViewCount="839" Body="&lt;p&gt;I'm currently preparing my exam and while I did never think about this question during the course, it begins to confuse me right now...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it the structure of $\ce {ClO_{2}-}$ like $\ce {O-Cl-O}$ and not like $\ce {O-O-Cl}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to place the covalent bonds and the electrons, but I don't know how to figure out the basic structure of the molecule. Is there a relatively simple rule to solve (and understand) such relatively simple molecules?&lt;/p&gt;&#xA;" OwnerUserId="7456" LastActivityDate="2018-01-23T00:09:27.710" Title="Why is it O-Cl-O and not O-O-Cl?" Tags="&lt;molecules&gt;&lt;lewis-structure&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15170" PostTypeId="1" AcceptedAnswerId="15176" CreationDate="2014-08-12T16:40:02.970" Score="2" ViewCount="3794" Body="&lt;p&gt;Can one convert from cyclic convert to linear compound?&#xA;Like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Cyclohexane$ $+$ $?=Hexane$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in &lt;strong&gt;High&lt;/strong&gt; Temp and Pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it is very tough since we are going from a stable compound to a relatively more unstable compound.But still...???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And can we also convert $Benzene$ to some $Unsaturated Compounds$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if possible can anyone please tell the &lt;strong&gt;reaction mechanism&lt;/strong&gt; of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="7434" LastEditDate="2014-08-12T19:07:00.600" LastActivityDate="2019-03-28T20:23:21.480" Title="Converting cyclic compounds to linear compounds (possible ?)" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15171" PostTypeId="1" AcceptedAnswerId="15183" CreationDate="2014-08-12T16:40:30.310" Score="5" ViewCount="385" Body="&lt;p&gt;Let's cut to the chase. I've been researching through numerous, &lt;em&gt;reliable and unreliable&lt;/em&gt;, resources. According to the CRT official site and wiki, the first cathode ray tube (CRT) was invented by &lt;strong&gt;Carl Ferdinand Braun&lt;/strong&gt;, year 1897. Yet &lt;em&gt;undeniable&lt;/em&gt; proofs of CRTs being used in experiments in 1860s have been collected.&lt;br&gt;&#xA;So this leaves the question unanswered. Noting that one of the fundamental progresses in chemistry (discovery of electron) is based on this useful tube, I will be thankful to anyone willing to share some info.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorDisplayName="user7951" LastEditDate="2018-11-09T22:13:59.677" LastActivityDate="2018-11-09T22:13:59.677" Title="Inventor of CRT?" Tags="&lt;electrons&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15172" PostTypeId="1" AcceptedAnswerId="15181" CreationDate="2014-08-12T16:56:33.023" Score="14" ViewCount="16946" Body="&lt;p&gt;How are perchlorate or chlorate or chlorite ions and their respective acids or compounds formed. $\ce{Cl}$ can't form more than one bond but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/7/7c/Perchlorate-2D-dimensions.png/170px-Perchlorate-2D-dimensions.png&quot; alt=&quot;Perchlorate ion from Wikipedia&quot;&gt; $\rightarrow$'Perchlorate ion'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/2f/Chlorous-acid-2D.png/200px-Chlorous-acid-2D.png&quot; alt=&quot;Chlorous acid containing Chlorite ion from wikipedia&quot;&gt; $\rightarrow$'Chlorous acid containing Chlorite ion'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/10/Chlorate-2D.png/200px-Chlorate-2D.png&quot; alt=&quot;Chlorate ion from Wikipedia&quot;&gt; $\rightarrow$'Chlorate ion'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorine is making more than 1 bond... How is it possible?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.293" LastActivityDate="2015-10-20T16:28:33.997" Title="How does chlorine form more than 1 bond?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15173" PostTypeId="2" ParentId="15169" CreationDate="2014-08-12T17:42:18.813" Score="0" Body="&lt;p&gt;Try to draw the structure of $\ce{ClO2-}$. It will like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wzjbu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see the structure it is $\ce{O−Cl−O}$  and not $\ce{O−O−Cl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorine is in the center because Least electronegative atom will be at center. You can refer this links: &lt;a href=&quot;https://www.youtube.com/watch?v=GTKW7nDzVCA&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube&lt;/a&gt; and  &lt;a href=&quot;http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text7/Tx72/tx72.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemistry education&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-08-13T04:43:18.357" LastActivityDate="2014-08-13T04:43:18.357" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15174" PostTypeId="1" AcceptedAnswerId="15265" CreationDate="2014-08-12T18:09:06.233" Score="5" ViewCount="3585" Body="&lt;p&gt;i have recently got a question in an assignment, which was somewhat like this&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what would be the shape of curve obtained in a graph between vapour pressure &amp;amp; temperature of a binary solution in a closed vessel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;generally we do all the experiments regarding vapour pressure in a closed vessel, so the volume must remain constant and from the formula $PV=nRT$, the number of moles remains constant in a binary solution, as volume is remaining constant we get the equation as $Y=mX$, which is a straight line passing through the origin, &#xA;but the answer was a sigmoid curve, how did we get that, can i get an elaborate answer for this quiery.&lt;/p&gt;&#xA;" OwnerUserId="6953" LastActivityDate="2014-08-15T04:14:01.630" Title="why do we obtain a sigmoid curve in vapour pressure versus temperature graph" Tags="&lt;solutions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15175" PostTypeId="2" ParentId="15172" CreationDate="2014-08-12T18:17:48.180" Score="4" Body="&lt;p&gt;The Cl-O bond has a dissociation enthalpy of 269.1 kJ/mole in the gas phase. That's not very low. This suggests that yes, the Cl-O bond is stable. Similarly, the Cl-Cl bond has a dissociation enthalpy of 242.580 kJ/mole. Again, not very low, but not very high. For comparison, consider triply bound diatomic nitrogen - the third strongest covalent bond known. This species has a dissociation enthalpy of 945 kJ/mole - about three times higher than the aforementioned energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.webelements.com/chlorine/bond_enthalpies.html&quot; rel=&quot;nofollow&quot;&gt;http://www.webelements.com/chlorine/bond_enthalpies.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorine can have multiple bonds, as you've shown in your Lewis structures. Try looking into molecular orbital theory. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-12T18:22:58.013" LastActivityDate="2014-08-12T18:22:58.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15176" PostTypeId="2" ParentId="15170" CreationDate="2014-08-12T18:56:26.153" Score="4" Body="&lt;p&gt;Yes, this can be done.  Unsaturated cycloalkanes can be ring opened using fairly straightforward methods.  For example, heating cyclobutene will produce 1,3-butadiene in an orbital symmetry allowed reaction.  But since your example involved cyclohexane, I'm guessing that you are more interested in the ring opening of saturated cycloalkanes, so that's where I'll focus my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalysts are needed to ring-open cycloalkanes.  The catalyst complexes with a cycolalkane bond and then in a second step, hydrogen is added across this bond.  Mechanistically, it is basically the reverse of the hydrogenation reaction (e.g. catalytic hydrogenation of cyclohexene to cyclohexane).  Some of these catalysts are very indiscriminate, so the initially formed linear alkane will rebind to the catalyst and undergo subsequent isomerization and fragmentation reactions.  More recently catalysts have been developed that can open cycloalkanes to the corresponding linear alkane in fairly high yield and relatively free of other hydrogenolysis products. These reactions are run in solution at temperatures slightly above ambient.  Here is a &lt;a href=&quot;http://www.researchgate.net/publication/257861657_Hydrogenolysis_of_cycloalkanes_over_TbCl33H2O3(RO)2AlOH&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to a full text article&lt;/a&gt; on one such process that uses Lanthanide\Aluminum complexes;  Terbium seems to be the preferred lanthanide.  Yields are typically in the 65-80% range (cyclohexane, 67%; cyclopentane, 77%; cyclodecane, 77%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s77jW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-12T18:56:26.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15179" PostTypeId="2" ParentId="15066" CreationDate="2014-08-12T20:06:45.440" Score="1" Body="&lt;p&gt;Simply put, this reaction involves the formation of hydroxide, because it is relatively more stable than the alternative products: $\ce{{}^{-}NH2}$ and $\ce{H3O+}$. You're right; hydroxide is generally a reactive species, so you probably wouldn't expect it to be the product of the reaction. However, when you have two reactants, in this case $\ce{NH3}$ and $\ce{H2O}$ in an acid-base reaction, the stronger base dictates the outcome of the reaction, which of course is $\ce{NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this reaction is in equilibrium and there are more reactants than the products. Thus, this solution is hardly composed of hydroxides.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastEditorUserId="4945" LastEditDate="2014-08-13T16:14:11.637" LastActivityDate="2014-08-13T16:14:11.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15180" PostTypeId="1" AcceptedAnswerId="15182" CreationDate="2014-08-12T20:09:29.200" Score="11" ViewCount="29546" Body="&lt;p&gt;Many sources state that one of the primary rules for molecular structures is that the least electronegative element is in the center. However, none of them explain &lt;em&gt;why&lt;/em&gt; this is the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this because it is most conducive to sharing electrons and forming bonds, or is it the most stable, or is there a better reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why are many molecules symmetric. Is there a reason why symmetry in nature and in chemistry is preferred over non-symmetry?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2014-08-12T22:48:27.917" Title="Why are least electronegative elements usually in center of lewis structure" Tags="&lt;molecules&gt;&lt;molecular-orbital-theory&gt;&lt;lewis-structure&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15181" PostTypeId="2" ParentId="15172" CreationDate="2014-08-12T20:12:13.403" Score="18" Body="&lt;p&gt;The order of bonding, and so the valence state of Cl in $\ce{ClO_{x}-}, x&amp;gt;1$ compounds is very debatable. Generally, two models exist. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Cl}$ atom, just like $\ce{S, P}$ and some others has unoccupied $d$-orbitals in the valence shell. It is possible to move some electrons from $p$-orbitals to $d$-orbitals, producing half-occupied orbitals that can participate in covalent bond formation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{O}$ atom has 6 electrons at the valence shell with four orbitals. It is possible to move one electron from chlorine atom to oxygen atom to form $\ce{O^{\dot{-} }}$ ion, that can than form a bond with newly formed half-filled orbital of chlorine. However, being electronegative and having high positive charge, chlorine partially pulls back the electron pair of newly formed bond.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Currently the second option &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp906294d&quot;&gt;is in favor&lt;/a&gt; of theoreticians, both by electron population analysis and relatively high charges in the oxochloric compounds, however, in school-level chemistry people traditionally resort to the first option. On the other hand, the second approach is the only option to adequately describe nitric acid and nitrates as nitrogen does not have $d$-orbitals on the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the stability of oxochloric compounds is a completely different matter. All of them are not very stable compounds, easily freeing their oxygen. Still, the process has significant activation energy, the state in between is even less favorable then oxochloric molecule. both oxygen and chlorine have electron-rich shells that repel potential reactants. Perchlorate ion is especially stable, at its surface is completely covered by outer negative shells of oxygen, so electrons cannot reach positively charged chlorine. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-12T20:53:19.150" LastActivityDate="2014-08-12T20:53:19.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15182" PostTypeId="2" ParentId="15180" CreationDate="2014-08-12T20:14:44.220" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this because it is most conducive to sharing electrons and forming&#xA;  bonds, or is it the most stable, or is there a better reason?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your first rationalization is a good one. That is the likely reason the least electronegative element generally becomes the central element. That being said, there are exceptions. One being water. Oxygen is &lt;em&gt;much&lt;/em&gt; more electronegative than hydrogen, which suggests that hydrogen should be the central atom, but the hydrogens are clearly peripheral atoms. This stems from the limitation that hydrogen can only form one sigma bond. That being said, hydrogen has also been found in unconventional bonding arrangements (see diborane for an example). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, why are many molecules symmetric. Is there a reason why symmetry in nature and in chemistry is preferred over non-symmetry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Symmetrical conformations of molecules may or may not have lower energies than their asymmetrical counterparts. Consider the ternary interhalogen, $\ce{IBrF2}$. The symmetrical conformation is likely lower in energy than the asymmetrical conformation (draw out the Lewis structures to see). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wNh9E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, in the symmetrical conformation (left), there are only 90 degree fluorine/lone pair interactions, as opposed to the conformation on the right, in which there is a fluorine/lone pair interaction in addition to a bromine/lone pair interaction. The equatorially placed atoms have been left out of this consideration as are ~120 degrees apart from the lone pairs and thus their interactions are not as significant as the 90 degree axial/lone pair interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there are myriad exceptions. Considered the eclipsed (left) and non-eclipsed (right) conformations of 2,3-dibromobutane. The eclipsed conformation is symmetric (it is also &lt;em&gt;meso&lt;/em&gt;; all meso compounds are symmetrical but not all symmetrical compounds are meso). The non-eclipsed conformation is asymmetric. However, the non-eclipsed conformation is energetically preferred since eclipsing the bromines places them closer together, which creates more van der Waals repulsions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4Vno.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, symmetry is a complicated topic in chemistry, especially with respect to more complex molecules and entities. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-12T20:37:30.493" LastActivityDate="2014-08-12T20:37:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15183" PostTypeId="2" ParentId="15171" CreationDate="2014-08-12T20:45:41.920" Score="7" Body="&lt;h2&gt;Invention of CRT&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the year 1855, German inventor &lt;strong&gt;Heinrich Geissler&lt;/strong&gt; developed mercury pump which is later modified by &lt;strong&gt;Sir William Crookes&lt;/strong&gt; who became the first to produce cathode rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/bf/Cabinet_V_-_Geissler_tubes_1862.jpg/1024px-Cabinet_V_-_Geissler_tubes_1862.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 1:&lt;/strong&gt; Geissler's Vaccum tubes(1862)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, then vacuum tubes are being used to study cathode rays, Braun's tube was actually an improvement to the these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, all of them have used some kind of Vacuum tube which produces cathode rays.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;In 1858 &lt;strong&gt;Julius Plücker&lt;/strong&gt; showed that cathode rays bend under the influence of a magnet.&lt;/li&gt;&#xA;  &lt;li&gt;In 1865 &lt;strong&gt;H. Sprengel&lt;/strong&gt; improves the Geissler vacuum pump.&lt;/li&gt;&#xA;  &lt;li&gt;In 1869 &lt;strong&gt;J.W. Hittorf&lt;/strong&gt; establishes that &quot;rays&quot; from the cathode travel in straight lines.&lt;/li&gt;&#xA;  &lt;li&gt;In 1871 &lt;strong&gt;C.F. Varley&lt;/strong&gt; is first to publish suggestion that cathode rays are composed of particles.[Fig 2]&lt;/li&gt;&#xA;  &lt;li&gt;In 1874 &lt;strong&gt;George Johnstone Stoney&lt;/strong&gt; estimates the charge of the then unknown electron.&lt;/li&gt;&#xA;  &lt;li&gt;In 1876 &lt;strong&gt;Eugen Goldstein&lt;/strong&gt; introduces the term cathode ray to describe the light emitted.&lt;/li&gt;&#xA;  &lt;li&gt;In 1881 &lt;strong&gt;Herman Ludwig von Helmholtz&lt;/strong&gt; suggested the idea that there is a smallest unit of electricity.&lt;/li&gt;&#xA;  &lt;li&gt;In 1890 &lt;strong&gt;Arthur Schuster&lt;/strong&gt; calculates the ratio of charge to mass of the particles making up cathode rays (today known as electrons) by&#xA;  measuring the magnetic deflection of cathode rays.&lt;/li&gt;&#xA;  &lt;li&gt;In 1894 &lt;strong&gt;J.J. Thomson&lt;/strong&gt; announces that he has found that the velocity of cathode rays is much lower than that of light.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemteam.info/AtomicStructure/Disc-of-Electron-History.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemteam.info/AtomicStructure/Disc-of-Electron-History.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemteam.info/AtomicStructure/Disc-of-Electron-Intro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemteam.info/AtomicStructure/Disc-of-Electron-Intro.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QQIcN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 2:&lt;/strong&gt; &lt;a href=&quot;http://rspl.royalsocietypublishing.org/content/19/123-129/236.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;First published paper on cathode rays.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7120" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.310" LastActivityDate="2014-08-12T20:45:41.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15184" PostTypeId="2" ParentId="15180" CreationDate="2014-08-12T22:48:27.917" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are least electronegative elements usually in center of lewis&#xA;  structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there is a &quot;central atom&quot;, then there are multiple (more than 1) bonds to the central atom.  If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would.  Also, electronegative atoms tend to carry multiple lone pairs of electrons.  If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why are many molecules symmetric. Is there a reason why symmetry in&#xA;  nature and in chemistry is preferred over non-symmetry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The symmetric arrangements (linear, tetrahedral, octahedral, etc.) minimize destabilizing bonding electron - bonding electron repulsions.  Remember methane is a perfect tetrahedron - that is the lowest energy arrangement because all of the bonding electrons are as far apart from one another as possible in this geometry.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-12T22:48:27.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15185" PostTypeId="2" ParentId="15169" CreationDate="2014-08-12T22:55:32.417" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a relatively simple rule to solve (and understand) such&#xA;  relatively simple molecules?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most electropositive atom is usually at the center of a molecule because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a &quot;central atom&quot; implies that there are multiple (more than 1) bonds to the central atom. If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would. &lt;/li&gt;&#xA;&lt;li&gt;electronegative atoms tend to carry multiple lone pairs of electrons. If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-12T22:55:32.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15186" PostTypeId="1" AcceptedAnswerId="15189" CreationDate="2014-08-13T01:06:58.777" Score="5" ViewCount="4676" Body="&lt;p&gt;I understand that in general, when in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons. And when in basic media, the nucleophile attacks the less substituted side of the epoxide due to steric reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the hydride anion attacks an epoxide? I'd imagine the hydride anion to be quite small, and able to attack both the more substituted and less substituted sides of an asymmetric epoxide. What are the experimental results? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I have my doubts about hydride anion even being able to open an epoxide; I am told by Clayden that hydride anion &lt;em&gt;never&lt;/em&gt; acts as a nucleophile because its 1s orbital is too small. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2015-10-09T16:37:40.857" LastActivityDate="2015-10-09T16:37:40.857" Title="Epoxide Opening with Sodium Hydride" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15187" PostTypeId="1" AcceptedAnswerId="15370" CreationDate="2014-08-13T01:23:41.133" Score="3" ViewCount="1497" Body="&lt;p&gt;Oxygenated blood is bright red and deoxygenated blood is dark red or brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take oxygenated blood and leave it in the air it will turn dark red, then brown, then finally a bluish green from exposure to atmospheric oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it oxidize this far to get a copper oxide color to it after it has passed the phase where it is the same color as iron oxide? Why doesn't it stay at the iron oxide color?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This often happens during the last few days of my period that the bleeding is extremely light and the color of it when it first comes out is reddish brown (sort of like iron oxide) and after a few hours, it starts turning bluish green (sort of like copper oxide). However, this happens very slowly and after 24 hours of exposure to atmospheric oxygen still only a fraction of the blood is further oxidized. I don't take medications or drugs that have a lot of sulfur so I don't think it is from $\ce{H2S}$ or $\ce{SO4-}$ (which would give it a dark green color, not light bluish green) and I don't have excess copper in my blood either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it is Iron(II,III) oxide with the Iron(III) released and the autoionization of water (which happens in both the liquid and gas phase) forming Iron(II) Hydroxide which can be further oxidized to Iron(III) hydroxide which has that bluish green color along with a small amount of Copper oxide forming (Due to the fact that our bodies need some copper and like Iron there is always some copper in the bloodstream).&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="402" LastEditDate="2017-01-14T00:32:17.433" LastActivityDate="2017-01-14T00:32:17.433" Title="Why does blood get further oxidized than Iron oxide?" Tags="&lt;redox&gt;&lt;medicinal-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15188" PostTypeId="2" ParentId="15187" CreationDate="2014-08-13T01:42:09.673" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This often happens during the last few days of my period that the&#xA;  bleeding is extremely light and the color of it when it first comes&#xA;  out is reddish brown(sort of like iron oxide) and after a few hours it&#xA;  starts turning bluish green(sort of like copper oxide).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, period blood isn't necessarily just blood; it's also sloughed off tissue. That would complicate the analysis; the green color could arise not necessarily from what's in your blood but also what's in your tissues, and I wouldn't be surprised if there were copper in your tissue. Copper is after all an essential trace element for health.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Copper_in_health&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Copper_in_health&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-13T01:42:09.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15189" PostTypeId="2" ParentId="15186" CreationDate="2014-08-13T02:07:44.413" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is an $\ce{S_{N}1}$ like reaction. First we would protonate the epoxide oxygen.  If we examine the case of an unsymmetrical epoxide and draw the resonance structures for this situation (see below) we would expect the secondary carbocation structure to count more towards the actual structure than the primary carbocation structure.  Therefore, we would expect a nucleophile to react preferentially at the tertiary epoxide carbon, quite selectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZV43j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in basic media, the nucleophile attacks the less substituted side of&#xA;  the epoxide due to steric reasons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would be a traditional $\ce{S_{N}2}$ reaction.  Attack at the least substituted (sterically most accessible) carbon would be expected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What if the hydride anion attacks an epoxide? I'd imagine the hydride&#xA;  anion to be quite small, and able to attack both the more substituted&#xA;  and less substituted sides of an asymmetric epoxide. What are the&#xA;  experimental results?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The hydride anion is not free, but complexed, so it has reasonable steric bulk - enough for stereoselective reactions to occur.   Further, the reaction between LAH and epoxides is quite facile, no doubt accelerated by the release of steric strain in the 3-membered epoxide ring system. A traditional $\ce{S_{N}2}$ reaction takes place.  In the figure below, the alcohol shown is formed in &lt;em&gt;ca&lt;/em&gt;. 90% yield.  Because of the &quot;backside&quot; attack nature of the $\ce{S_{N}2}$ reaction axial attack is strongly preferred in cyclic systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Q407.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-13T02:19:49.423" LastActivityDate="2014-08-13T02:19:49.423" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15190" PostTypeId="2" ParentId="15174" CreationDate="2014-08-13T02:14:28.873" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9XS8g.gif&quot; alt=&quot;Here is a Maxwell Boltzmann distribution at several temperatures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is a graph of the Maxwell Boltzmann distribution at several temperatures.  Imagine a point on the x axis being the point at which a liquid particle has enough kinetic energy to leave the solution.  As you can see, each increase in temperature results in a larger number of particles with enough energy to vaporize.  This generates an exponential graph for VP vs T.  However, eventually you reach the point where all of the liquid has reached the gas phase.  Then you have a linear graph of P vs T.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2014-08-13T03:53:09.310" LastActivityDate="2014-08-13T03:53:09.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15191" PostTypeId="2" ParentId="15072" CreationDate="2014-08-13T02:16:22.100" Score="1" Body="&lt;p&gt;One component that has gelling effect and is added to cosmetic products is hydroxyethylcellulose. It creates clear gel and is soluble in EtOH, but possibly not in highly concentrated.  It is safe and inexpensive.&lt;/p&gt;&#xA;" OwnerUserId="7424" LastActivityDate="2014-08-13T02:16:22.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15193" PostTypeId="1" AcceptedAnswerId="15195" CreationDate="2014-08-13T02:45:06.930" Score="4" ViewCount="282" Body="&lt;p&gt;How do I figure out what is the oxidizing agent and the reducing agent in the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\overset{-3}{N}\overset{+1}{H}_3 + \overset{+1}{H}\overset{-2}{O}\overset{+1}{Cl} &amp;lt;=&amp;gt; \overset{-1}{N}\overset{+1}{H}_2\overset{-1}{Cl} + H2O}~?$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assigned known oxidation numbers for the elements, but I'm not sure how to proceed with the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; the reducing agent? Or nitrogen? Going from &lt;span class=&quot;math-container&quot;&gt;$-3$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt; &#xA;and &lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl}$&lt;/span&gt; the oxidizing agent, chlorine going from &lt;span class=&quot;math-container&quot;&gt;$+1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-1$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="41328" LastEditDate="2020-03-04T16:07:19.507" LastActivityDate="2020-03-04T16:09:44.247" Title="How to determine oxidising and reducing agents in the reaction between ammonia and hypochlorous acid?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15194" PostTypeId="1" CreationDate="2014-08-13T02:53:37.673" Score="4" ViewCount="14808" Body="&lt;p&gt;&lt;a href=&quot;http://lifehacker.com/5644007/use-charcoal-briquettes-for-cheap-and-non-toxic-odor-removal&quot; rel=&quot;nofollow&quot;&gt;I've read&lt;/a&gt; that using blocks of charcoal can remove some odors from the air by placing them in the vicinity of the odors. (Maybe not as efficiently as other forms of carbon, as pointed out in the comments of that link, but something nonetheless.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the chemical process by which charcoal can do that, and once the odors are 'gone', where do they 'go'? In the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Charcoal#Purification_and_filtration&quot; rel=&quot;nofollow&quot;&gt;'toxins in gases'&lt;/a&gt;, what typically is done to them upon 'deodorizing' by the charcoal block - are they collected, neutralised / transformed upon contact, or something else?&lt;/p&gt;&#xA;" OwnerDisplayName="user1469" LastActivityDate="2019-02-27T16:29:58.500" Title="Why does charcoal absorb smells from the air, and where do those smells 'go' once absorbed?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15195" PostTypeId="2" ParentId="15193" CreationDate="2014-08-13T03:50:40.327" Score="3" Body="&lt;p&gt;You're on the right track with assigning oxidation numbers to each element. And you've assigned them successfully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To successfully complete the task you need to define a few terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's oxidation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's reduction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the oxidizing agent? Try thinking of this in terms of what oxidation is, and then try figuring out what an oxidizing agent would do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the reducing agent? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-13T03:50:40.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15196" PostTypeId="1" AcceptedAnswerId="15199" CreationDate="2014-08-13T04:48:54.540" Score="3" ViewCount="13582" Body="&lt;p&gt;Boron has an empty $2p$ orbital.&lt;br&gt;&#xA;But is it an electroplhile?&lt;br&gt;&#xA;I know it is a Lewis acid.&lt;/p&gt;&#xA;" OwnerUserId="5776" LastEditorUserId="5591" LastEditDate="2015-11-13T19:27:25.383" LastActivityDate="2017-03-11T15:15:12.880" Title="Is BF3 an electrophile?" Tags="&lt;acid-base&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15197" PostTypeId="1" AcceptedAnswerId="15202" CreationDate="2014-08-13T05:05:07.570" Score="2" ViewCount="167" Body="&lt;p&gt;I have 2 statements , and i am not sure, whether they are true or false.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Proteins like albumin and haemoglobin are insoluble in water&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(in my view , almost all proteins are soluble- so this is a wrong  statement)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Proteins which are soluble in water have complex 3D structure and they are globular molecules. - (in my view I think this is also wrong)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please explain these statements. &lt;/p&gt;&#xA;" OwnerUserId="5776" LastActivityDate="2014-08-13T06:53:06.097" Title="Proteins and their solubility" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15198" PostTypeId="2" ParentId="15168" CreationDate="2014-08-13T05:20:39.597" Score="0" Body="&lt;p&gt;Yes, entropy change of reversible process is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gitam.edu/eresource/Engg_Phys/semester_1/THERMODYNAMICS/entr_%20rev_irrev.htm&quot; rel=&quot;nofollow&quot;&gt;This will also help&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5776" LastActivityDate="2014-08-13T05:20:39.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15199" PostTypeId="2" ParentId="15196" CreationDate="2014-08-13T05:57:58.780" Score="6" Body="&lt;p&gt;Lewis acids are by definition electrophiles. Electrophiles love electrons, or negative charge. Boron has an empty 2p orbital and there exists a strong partial positive charge on the boron due to the extremely electronegative fluorine atoms covalently bound to boron. This strong partial positive character, coupled with a vacant orbital, makes BF3 a potent Lewis acid and thus an electrophile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, note that acidity and basicity are both thermodynamic properties, while electrophilicity and nucleophilicity are both kinetic properties. That is, while you can label a Lewis acid an electrophile, it would be incorrect to say, for example, that because a certain Lewis acid is a &quot;strong&quot; Lewis acid, then it must be a strong electrophile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/14284/formation-constants-lewis-acidity&quot;&gt;Formation Constants; Lewis Acidity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-08-13T13:24:38.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15201" PostTypeId="2" ParentId="15194" CreationDate="2014-08-13T06:39:45.353" Score="5" Body="&lt;p&gt;It is a form of carbon that has very large surface. The odors and impurities are held to the surface with &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_forces&quot; rel=&quot;noreferrer&quot;&gt;Van der Waals forces&lt;/a&gt;. The capability of removing notable amounts of these impurities or odors is thanks to the extraordinarily large surface. &lt;/p&gt;&#xA;" OwnerUserId="7424" LastActivityDate="2014-08-13T06:39:45.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15202" PostTypeId="2" ParentId="15197" CreationDate="2014-08-13T06:53:06.097" Score="3" Body="&lt;p&gt;All proteins have complex structures. There are three broad classes: globular, membrane and scleroproteins. Globular proteins are generally water soluble while membrane and scleroproteins are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Albumins and haemoglobin are globular and water soluble. &lt;/p&gt;&#xA;" OwnerUserId="5220" LastActivityDate="2014-08-13T06:53:06.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15203" PostTypeId="1" AcceptedAnswerId="15204" CreationDate="2014-08-13T08:14:55.130" Score="0" ViewCount="191" Body="&lt;p&gt;I would like to explain chemical formulas is not representable in terms of the molecular structure of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good suggestions of examples?&lt;/p&gt;&#xA;" OwnerUserId="7464" LastEditorUserId="7448" LastEditDate="2015-09-18T22:50:40.403" LastActivityDate="2016-03-11T23:03:55.773" Title="Need a pair of compounds that is similar in formula but different in structure" Tags="&lt;reference-request&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15204" PostTypeId="2" ParentId="15203" CreationDate="2014-08-13T08:46:34.593" Score="2" Body="&lt;p&gt;The classical example is dimethyl ester $\ce{CH3-O-CH3}$ and etanol $\ce{CH3-CH2-OH}$ pair. They have exactly same brutto-formula, but different structure and very different properties. The first is a gas, the second is a liquid easily reacting with sodium metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon is also interesting example. It has two common forms with exactly same chemical formula $\ce{C}$, but very different properties: a very hard and transparent diamond and soft black graphite.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-13T08:46:34.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15205" PostTypeId="1" AcceptedAnswerId="15207" CreationDate="2014-08-13T09:23:20.763" Score="9" ViewCount="6337" Body="&lt;p&gt;How are the 3p orbitals of chlorine lower in energy than the 1s orbital of hydrogen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8XwpL.png&quot; alt=&quot;MO diagram of HCl&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7448" LastEditDate="2019-06-06T10:55:50.470" LastActivityDate="2019-06-06T10:55:50.470" Title="Energetic Placement of Atomic Orbitals in the HCl Molecular Orbital Diagram" Tags="&lt;theoretical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15206" PostTypeId="1" CreationDate="2014-08-13T10:30:23.813" Score="1" ViewCount="39131" Body="&lt;p&gt;If I am given a phase diagram, how can I determine which phase will have more density?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I consider to solve these kinds of questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the answer have to do with the area that each phase occupies?  Or the more area a phase occupies, the more the density?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="13184" LastEditDate="2016-02-12T18:53:04.720" LastActivityDate="2016-02-12T18:53:04.720" Title="How to figure out which phase on a phase diagram will be the most dense?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15207" PostTypeId="2" ParentId="15205" CreationDate="2014-08-13T10:39:02.190" Score="7" Body="&lt;p&gt;I'd say there are two important things to consider here: Firstly, $\ce{Cl}$ is more electronegative than $\ce{H}$. And secondly, as you proceed from the left to the right in any row of the periodic table, the s and p orbitals go down in energy. This is a consequence of the fact that the valence electrons do not screen each other effectively. Thus, the addition of one proton to the nucleus and one electron does not cancel; instead, the valence electrons &quot;feel&quot; an increased nuclear charge.&#xA;The electronegativity difference and the reduced screening among the valence electrons seems to be enough to shift the $3\ce{s}$ and even the $3\ce{p}$ orbital of $\ce{Cl}$ below the energy of the $1\ce{s}$ orbital of $\ce{H}$. The experimental state averaged ionization potentials (IP) could serve as a good guide for the placement of starting orbitals in an orbital interaction diagram. For the $\ce{H}$ $1\ce{s}$ orbital you have an IP of $-13.6 \, \mathrm{eV}$. For the $\ce{Cl}$ $3\ce{s}$ and $3\ce{p}$ orbitals you have IP's of $-25.2 \, \mathrm{eV}$ and $-13.7 \, \mathrm{eV}$, respectively.$^{[1]}$ So, this indicates that the $3\ce{p}$ orbital of $\ce{Cl}$ does indeed lie slightly below the $1\ce{s}$ orbital of $\ce{H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a table of the experimental state averaged ionization potentials, in electron volts, for the main group atoms. However, those $\ce{p}$ atomic orbitals for groups 1 and 2 and for the $\ce{s}$ and $\ce{p}$ orbitals of the sixth row are not experimentally known and hence, calculated values (indicated by an asterisk) have been used.$^{[2,3]}$ The latter include relativistic corrections. The table is taken from the 2nd Edition of the book &lt;em&gt;Orbital Interactions in Chemistry&lt;/em&gt; by T. A. Albright, J. K. Burdett, and M.-H. Whangbo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VnDfA.png&quot; alt=&quot;State averaged ionization potentials, in electron volts&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] L. C. Allen, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1989&lt;/strong&gt;, &lt;em&gt;111&lt;/em&gt;, 9003 ; L. C. Allen, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1992&lt;/strong&gt;, &lt;em&gt;114&lt;/em&gt;, 1510; L. C. Allen, &lt;em&gt;Acc. Chem. Res.&lt;/em&gt; &lt;strong&gt;1990&lt;/strong&gt;, &lt;em&gt;23&lt;/em&gt;, 175.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] A. Vela, J. L. Gazquez, &lt;em&gt;J. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;92&lt;/em&gt;, 5688.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] J. P. Desclaux, &lt;em&gt;At. Nucl. Data Tables&lt;/em&gt; &lt;strong&gt;1973&lt;/strong&gt;, &lt;em&gt;12&lt;/em&gt;, 311.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-13T11:49:56.367" LastActivityDate="2014-08-13T11:49:56.367" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15208" PostTypeId="1" AcceptedAnswerId="15213" CreationDate="2014-08-13T11:00:23.760" Score="5" ViewCount="5896" Body="&lt;p&gt;Its known that the radius of maximum probability of 2s orbitals is more than that of 2p orbitals. It means that the maximum probability of finding an electron in an 2s is further away from electron than that of 2p. This means 2p electron are mostly found nearer to atom than the 2s atom. Well! thats totally wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want to know the factors due to which 2p orbital lie further to nucleus than the 2s one.   &lt;/p&gt;&#xA;" OwnerUserId="5608" LastActivityDate="2014-08-13T14:06:28.347" Title="Difference between actual position of electron and Radial Distribution Probability" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15209" PostTypeId="2" ParentId="15208" CreationDate="2014-08-13T13:15:24.907" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This means 2p electron are mostly found nearer to atom than the 2s atom. Well! thats totally wrong.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this idea totally wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consult a radial probability graph. Here's one for hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OxdsS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph shows that the 2p electrons have a good probability of getting closer to the nucleus than 2s electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be wondering why then s-electrons are lower in energy than p-electrons of the same subshell. Notice the little bump in the pink line on the left end of the graph? The 2s electrons can &lt;em&gt;penetrate&lt;/em&gt; closer to the nucleus than the 2p electrons can. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-13T13:22:14.877" LastActivityDate="2014-08-13T13:22:14.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15210" PostTypeId="1" CreationDate="2014-08-13T13:37:02.590" Score="0" ViewCount="488" Body="&lt;p&gt;&quot;diamond structure. Every carbon atom is bonded&#xA;to four other carbon atoms forming a three-dimensional lattice of chair conformations.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SENTENCE FROM A BOOK&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean? &#xA;Where I am struck is that I knew carbons occupy zinc blende like lattice so, how can this line be true.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5424" LastEditDate="2014-08-14T09:57:43.527" LastActivityDate="2014-08-14T09:57:43.527" Title="Structure of diamond" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15211" PostTypeId="2" ParentId="15210" CreationDate="2014-08-13T13:53:20.943" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where I am struck is that I knew carbons occupy zinc sulfide like&#xA;  lattice so, how can this line be true.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Zinc sulfide can exist in two different crystalline forms, Zincblende (sphalerite) and wurtzite.  Zincblende is face-centered cubic (also known as diamond cubic), each ion is tetracoordinate and has local tetrahedral geometry - just like the carbons in diamond.  The wurtzite structure is hexagonal close-packed with interconnected 6-membered rings.  You may have been thinking of the wurtzite structure, while it is the zincblende structure that is related to diamond.  You can see pictures of the two structures &lt;a href=&quot;https://www.princeton.edu/~maelabs/mae324/glos324/zns.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-13T13:53:20.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15213" PostTypeId="2" ParentId="15208" CreationDate="2014-08-13T13:57:48.997" Score="8" Body="&lt;p&gt;Ok, let's start out by looking at the hydrogenic atom. The wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability of finding the electron in the hydrogen-like atom, with the&#xA;distance $r$ from the nucleus between $r$ and $r + \mathrm{d}r$, with angle $\theta$ between $\theta$ and $\theta + \mathrm{d}\theta$, and with the angle $\varphi$ between $\varphi$ and $\varphi + \mathrm{d}\varphi$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  | \Psi(r, \theta, \varphi) | \, \mathrm{d} \tau = \left[R_{n,\ell}(r) \right]^{2} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} r^{2} \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\varphi \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the probability $D_{n, \ell}(r) \, \mathrm{d}r$ that the electron is between $r$ and $r + \mathrm{d}r$ regardless of the direction, you integrate over the angles $\theta$ and $\varphi$ to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  D_{n, \ell}(r) \, \mathrm{d}r &amp;amp;= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \underbrace{\int_{0}^{\pi} \! \! \int_{0}^{2\pi} \left|Y_{\ell,m} (\theta, \varphi ) \right|^{2} \sin \theta  \, \mathrm{d}\theta \, \mathrm{d}\varphi}_{= \, 1} \\&#xA;  &amp;amp;= r^{2} \left[R_{n,\ell}(r) \right]^{2} \, \mathrm{d}r \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the spherical harmonics are normalized, the value of the double integral&#xA;is unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radial distribution function $D_{n, \ell}(r)$ is the probability density for the&#xA;electron being in a spherical shell with inner radius r and outer radius $r + \mathrm{d}r$.&#xA;For the $2\ce{s}$ and $2\ce{p}$ states, these functions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  D_{2, 0}(r)  &amp;amp;= \frac{1}{8} \left( \frac{Z}{a_{0}} \right)^{3} r^{2} \left( 2 - \frac{Z r}{a_{0}} \right)^{2} \mathrm{e}^{-Zr/a_{0}} \\&#xA;  D_{2, 1}(r)  &amp;amp;= \frac{1}{24} \left( \frac{Z}{a_{0}} \right)^{5} r^{4}  \mathrm{e}^{-Zr/a_{0}} \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_{0}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bohr radius&lt;/a&gt;.&#xA;The graph of those two function can be seen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6iQOB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most probable value $r_{\mathrm{mp}}$  of $r$ is found by setting the derivative of $D_{n, \ell}(r)$ with respect to $r$ equal to zero. From the graph above it can be seen that $r_{\mathrm{mp}}(2\ce{p})$ is smaller than $r_{\mathrm{mp}}(2\ce{s})$ (and if you calculate it you can check that for yourself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $r_{\mathrm{mp}}$ is not the quantity you are actually interested in. Because although $r_{\mathrm{mp}}$ gives you the most probable distance of the electron from the nucleus you want to know the average distance $\langle r \rangle$ between the electron and the nucleus as that is the distance you will get when you measure $r$ repeatedly and average over your results.&#xA;The radial distribution functions may be used to calculate the expectation values&#xA;of functions of the radial variable r. You can get them via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \langle r \rangle_{n, \ell}  &amp;amp;= \int_{0}^{\infty} r D_{n, \ell}(r) \, \mathrm{d} r \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you evaluate this integral for the $2\ce{s}$ and $2\ce{p}$ orbitals you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \langle r \rangle_{2\ce{s}} = \langle r \rangle_{2, 0}  &amp;amp;= \frac{6 a_{0}}{ Z } \\&#xA;  \langle r \rangle_{2\ce{p}} = \langle r \rangle_{2, 1}  &amp;amp;= \frac{5 a_{0}}{ Z } \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You find that the $2\ce{s}$ electrons are on average further away from the nucleus than the $2\ce{p}$ electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the hydrogenic atom your claim that $2\ce{p}$ electrons are further away from the nucleus than the $2\ce{s}$ electrons is wrong. But in real atoms you have more than one electron and those electrons will influence each other. One important effect here is, that the inner electrons (or electron density) screens nuclear charge from the outer electrons. Thus the outer electrons feel a lower effective nuclear charge $Z^{\mathrm{eff}}$ than the inner electrons. The extend of this screening is distance dependent, so $Z^{\mathrm{eff}} &#xA;= Z^{\mathrm{eff}} (r)$.&#xA;Now, if you look back at the formula for $\langle r \rangle_{n, \ell}$ things will certainly change since $D_{n, \ell}(r)$ will have a different form and the additional $r$-dependence of $Z^{\mathrm{eff}}$ will make things more difficult. Assuming that the general form of $D_{n, \ell}(r)$ will be similar in the real atom, it can be expected that the parts of $D_{n, \ell}(r)$ for small $r$ now have a higher weight in the integral of $\langle r \rangle_{n, \ell}$ than before since these parts of the electron density suffer less from the screening effect. Since $2\ce{s}$ electrons have a small portion of the electron density that lies very close to the nucleus (&quot;penetrates deep into the core region&quot;) and $2\ce{p}$ electrons don't this will lead to the observation that $2\ce{p}$ electrons are on average further away from the nucleus than the $2\ce{s}$ electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more involved explanation of the effects at work in multi-electron atoms can be found at &lt;a href=&quot;https://chemistry.stackexchange.com/a/376/189&quot;&gt;this answer of mine&lt;/a&gt;. The important bit starts at the paragraph &quot;Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?&quot;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-08-13T14:06:28.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15214" PostTypeId="2" ParentId="15206" CreationDate="2014-08-13T14:42:42.237" Score="1" Body="&lt;p&gt;Look at the lines on the phase diagram.  If you go to a line and then increase pressure you move to a new phase, the denser phase.  Edit:  Add the below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JUr0b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to the dotted green line.  Now, go straight up (increase in P).  What happened?  You went from solid and liquid being in equilibrium to just liqui.  The liquid is more dense than the solid.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2014-08-13T15:40:11.643" LastActivityDate="2014-08-13T15:40:11.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15215" PostTypeId="1" CreationDate="2014-08-13T14:55:03.570" Score="6" ViewCount="3094" Body="&lt;p&gt;Two electron of opposite spin can lie in a single orbital.. But what about the electron-electron repulsion. &#xA;Okay! I got that the nuclear charge rather the large Z-effective overcome this repulsion by pulling them together towards the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing more turned out in mind that it may be the attraction of the two unlike pole of the magnet developed due to opposite spin of the electrons of an orbital, which may be reason for the two opposite spin electron lying in an orbital..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right ? &lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="189" LastEditDate="2014-08-13T16:57:39.757" LastActivityDate="2014-09-08T20:45:27.143" Title="How can two electrons lie together in an orbital?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15216" PostTypeId="2" ParentId="9410" CreationDate="2014-08-13T15:13:15.107" Score="2" Body="&lt;p&gt;It's not that ionic bonds are weaker than hydrogen bonds, but that the energy cost of breaking the ionic bonds is well compensated by the gain in solvation energy when you put NaCl in solution.  Solvation energy drives the spontaneity of this dissolution process, because of entropic considerations.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-08-13T15:13:15.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15218" PostTypeId="2" ParentId="15206" CreationDate="2014-08-13T15:52:42.593" Score="1" Body="&lt;p&gt;The phase which has the higher density is the phase which exists under higher pressure and lower temperature. Thus, in a phase diagram showing pressure as a function of temperature, look for the phase which lies in the upper left region.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-08-13T15:52:42.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15219" PostTypeId="2" ParentId="12315" CreationDate="2014-08-13T15:56:41.310" Score="3" Body="&lt;p&gt;I would add to the &lt;a href=&quot;https://chemistry.stackexchange.com/a/12371/13409&quot;&gt;first answer&lt;/a&gt; that the formation of the hydroxyl radical can also form a small amount $\ce{H2O2}$ via&#xA;$\ce{OH⋅ + OH⋅ &amp;lt;=&amp;gt; H2O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unmatched hydrogen radicals could react forming $\ce{H2}$ and leave the solution. The newly form $\ce{H2O2}$ could attack, for example, a copper electrode contributing to anodic (or cathodic) corrosion.&#xA;Galvanic cells frequently have many non-electrochemical based side reactions occurring as well.&lt;/p&gt;&#xA;" OwnerUserId="7471" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-03-25T18:10:36.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15220" PostTypeId="1" AcceptedAnswerId="15232" CreationDate="2014-08-13T16:00:25.423" Score="0" ViewCount="8787" Body="&lt;p&gt;How exactly does sodium form an alkoxide from an alcohol? I understand the general idea - sodium has a low first ionization energy. Sodium can thus give up its electron to something that'll take it. Sodium can therefore somehow cause hydrogen gas to leave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly does this mechanism work? I'm imagining that because alcohols are weak acids, that some hydrated protons exist in solution, and that sodium can donate an electron to these hydrated protons, which then can pair up with each other and form hydrogen gas. Is this accurate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ogUIo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-13T17:04:05.910" LastActivityDate="2018-04-15T09:38:39.073" Title="Mechanism of reduction by elemental sodium" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15221" PostTypeId="1" CreationDate="2014-08-13T16:09:07.240" Score="3" ViewCount="1912" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a total chlorine concentration  &lt;span class=&quot;math-container&quot;&gt;$c(\ce{Cl})_\text{tot}= \pu{0.1 M}$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}(\ce{Cl}) = 8$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}(\ce{HOCl}) = 7.5$&lt;/span&gt;, how would you find the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{OCl-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl}$&lt;/span&gt; disinfectants?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far I think that I should start this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a} (\ce{HOCl}) = 10^{-7.5}=3.16\cdot 10^{-8}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote up the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + 2e- + OCl &amp;lt;=&amp;gt; Cl- + 2OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\ce{OCl-}$&lt;/span&gt; is the conjugate acid and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; the conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HOCl&amp;lt;=&amp;gt; OCl- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugated acid is &lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl}$&lt;/span&gt;, the conjugated base &lt;span class=&quot;math-container&quot;&gt;$\ce{OCl-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I use &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log\left(\frac{[\text{conj base}]}{[\text{conj acid}]}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and/or for weak acids &#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_\mathrm{a}= [\text{products}]/[\text{reactants}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure if this is correct and what values to use.&lt;/p&gt;&#xA;" OwnerUserId="7459" LastEditorUserId="83758" LastEditDate="2020-06-05T13:08:33.743" LastActivityDate="2020-06-05T13:08:33.743" Title="Concentration of hypochlorous acid and hypochlorite" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15223" PostTypeId="2" ParentId="15215" CreationDate="2014-08-13T16:50:49.820" Score="5" Body="&lt;p&gt;You're thinking is correct, is does cost energy to pair electron spins in an orbital (the Pauli energy or exchange energy required to pair spins).  However, the alternative is to place that next electron into a higher energy orbital - that requires energy too.  Usually it takes less energy to pair spins than to place the next electron into a higher energy orbital.  But note cases like $\ce{O2}$ where there are two orbitals at the same energy (degenerate orbitals).  Now when we add the last electron it doesn't cost any extra energy to put that electron into the next available empty orbital because it is at the same energy.  So in cases like this (or where the next orbital is very close in energy to the partially filled orbital) we can put that next electron into a different orbital and leave their spins unpaired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGW2i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-13T16:50:49.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15224" PostTypeId="2" ParentId="15215" CreationDate="2014-08-13T16:51:25.933" Score="7" Body="&lt;h2&gt;Preludium&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain &lt;strong&gt;discrete&lt;/strong&gt; energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Main answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| &amp;gt; \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look &lt;a href=&quot;https://chemistry.stackexchange.com/a/16070/189&quot;&gt;here&lt;/a&gt;: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given &lt;a href=&quot;https://chemistry.stackexchange.com/a/15207/189&quot;&gt;here&lt;/a&gt; (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does &quot;cost&quot; energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-09-08T20:45:27.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15225" PostTypeId="2" ParentId="15220" CreationDate="2014-08-13T16:54:36.803" Score="0" Body="&lt;p&gt;The alcohol splits like this:&#xA;$$ROH\rightarrow RO^-+H^+$$&#xA;since the pka of alcohol is around 15-17 and then a redox reaction happens:&#xA;$$2Na+2H^+\rightarrow 2Na^++H_2$$&#xA;Since $H_2$ is released it is entropically favourable and drives the reaction.&#xA;Better, alkoxide is represented as:&#xA;$$RO^-Na^+$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-08-13T17:00:44.133" LastActivityDate="2014-08-13T17:00:44.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15226" PostTypeId="1" AcceptedAnswerId="15234" CreationDate="2014-08-13T17:07:23.580" Score="1" ViewCount="931" Body="&lt;p&gt;I can't find any data about the Kb value of the hydride anion except that it's huge. Does anyone know what it is, or how I can characterize it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can use electrochemical means but the only reaction I find with hydride ion is this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydride#Hydride_ion&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydride#Hydride_ion&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-05-02T15:57:35.157" Title="Kb of hydride anion" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15227" PostTypeId="1" CreationDate="2014-08-13T17:08:17.947" Score="0" ViewCount="89" Body="&lt;p&gt;Can anyone explain the reaction mechanism of the following synthesis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NH4Cl + Ca(OH)2 -&amp;gt; 2NH3 + 2H2O + CaCl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(NH4)2SO4 + 2NaOH -&amp;gt; 2NH3 + 2H2O + Na2SO4}$&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="5016" LastEditDate="2015-01-11T17:24:33.130" LastActivityDate="2015-01-11T17:24:33.130" Title="Can anyone explain the reaction mechanisms of these synthesis?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15228" PostTypeId="1" AcceptedAnswerId="15229" CreationDate="2014-08-13T17:10:03.617" Score="1" ViewCount="946" Body="&lt;p&gt;Please look at the image on this website: &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=765853&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.physicsforums.com/showthread.php?t=765853&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b) and (c), what the solution manual says makes no sense to me.&#xA;You're reacting a primary alkyl halide with an unhindered base &lt;span class=&quot;math-container&quot;&gt;$\ce{EtONa}$&lt;/span&gt;. They say that the only product is the alkene. Why do they not consider the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; reaction that WILL take place in larger quantity?&#xA;I'm uploading the question too on the other website&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="61578" LastEditDate="2019-05-29T02:37:40.123" LastActivityDate="2019-05-29T02:37:40.123" Title="Reaction of alkyl halides with strong unhindered bases" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15229" PostTypeId="2" ParentId="15228" CreationDate="2014-08-13T17:13:26.510" Score="1" Body="&lt;p&gt;I agree with you, I'd expect some &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; product too. I'm not sure about the relative quantities of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{E2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; products though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't flat out say that you'd expect more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; product than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{E2}$&lt;/span&gt; product, because although the substrate is unhindered, the solvent is ethanol, which is protic. The solvent interacts with the nucleophile through hydrogen bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hinders the nucleophile. This slows down the rate of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; reaction because now the nucleophile is dragging along lots of solvent molecules, and it also increases the bulk of the nucleophile. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="61578" LastEditDate="2019-05-29T02:36:34.567" LastActivityDate="2019-05-29T02:36:34.567" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="15230" PostTypeId="2" ParentId="15226" CreationDate="2014-08-13T17:18:18.257" Score="1" Body="&lt;p&gt;Recall that $\mathrm{p}K_\text{b}$ of hydride is actually $14-\mathrm{p}K_\text{a}$ of hydrogen gas. The value I found is $\mathrm{p}K_\text{a}=35$, so $\mathrm{p}K_\text{b}=14-35=-21$.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T15:57:32.190" LastActivityDate="2015-05-02T15:57:32.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15231" PostTypeId="1" CreationDate="2014-08-13T17:21:31.773" Score="7" ViewCount="1250" Body="&lt;p&gt;I uploaded pictures on &lt;a href=&quot;http://www.physicsforums.com/showthread.php?t=765106&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.physicsforums.com/showthread.php?t=765106&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at the answers in my textbook. They seem to make incongruent choices of solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In e) and j), why choose EtOH and H2O/CH3OH? I would rather choose an aprotic solvent like DMF to better allow Sn2 to occur. Especially with j), there is a lot of competition with E2 because OH- is a strong base and because we are attacking a secondary carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see this again in k). Secondary carbon attacked: why not add DMF to prevent E2 and get best yield?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ED8nW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="5084" LastEditDate="2014-08-13T17:49:53.137" LastActivityDate="2014-08-13T20:13:58.247" Title="Synthesis via Sn2/Sn1, E2, E1: Choice of solvent" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15232" PostTypeId="2" ParentId="15220" CreationDate="2014-08-13T17:24:41.303" Score="5" Body="&lt;p&gt;When alcohol molecules adsorb on the surface of the sodium metal, an electron can be transferred from sodium to the alcohol O-H bond with simultaneous disproportionation  (homolytic cleavage) of the O-H bond thereby generating the alkoxide anion and a hydrogen radical.  The hydrogen radical (probably still on or near the metal surface) can then go on to dimerize and produce hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/842sK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Support of Mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism proposed above involves a single electron transfer (SET) from sodium.  This is based on the fact that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All dissolving (sodium) metal reactions, such as the Birch reduction of alkynes and aromatics, &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/radikale/birch_red.vlu.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;proceed by the SET mechanism&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;Other reactions involving sodium metal such as the &lt;a href=&quot;http://www.metallacycle.com/chemistry/organic/set/&quot; rel=&quot;nofollow noreferrer&quot;&gt;acyloin condensation&lt;/a&gt;, the &lt;a href=&quot;http://www.metallacycle.com/chemistry/organic/set/&quot; rel=&quot;nofollow noreferrer&quot;&gt;pinacol coupling&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bouveault%E2%80%93Blanc_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bouveault–Blanc reduction&lt;/a&gt;, etc., proceed through the SET mechanism&lt;/li&gt;&#xA;&lt;li&gt;The reaction of sodium and water still proceeds in the gas phase where ionization seems less likely, but electron transfer still viable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Searching on-line I found that water can solvate electrons just like ammonia can.  The chief difference being one of stability; in ammonia solvated electrons are stable for days, whereas in water they are stable for less than 1 msec.  Given the preceding mechanistic evidence and the fact that solvated electrons can exist in water, it seems reasonable to suggest that the mechanism of alkoxide formation, when sodium is reacted with an alcohol, also involves the SET mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-14T15:46:19.603" LastActivityDate="2014-08-14T15:46:19.603" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15233" PostTypeId="2" ParentId="15227" CreationDate="2014-08-13T17:25:17.400" Score="2" Body="&lt;p&gt;The ammonium chloride, calcium hydroxide, ammonium sulfate, and sodium hydroxide are all ionic compounds, and when dissolved in water they dissociate into $\ce{NH4+}$, $\ce{Cl-}$, $\ce{Ca^2+}$, $\ce{SO4^2-}$, and $\ce{Na+}$ ions. The $\ce{NH4+}$ ammonium ion can donate a proton to the $\ce{OH-}$ ion, creating $\ce{NH3}$ and $\ce{H2O}$. The other ions don't really do anything, so these spectator ions remain in solution but the charges are still balanced. If you removed the ammonia and water, which is easy with a vacuum, you'd be left with the $\ce{CaCl2}$ and $\ce{Na2SO4}$ salts. These reactions will have some equilibrium, with $\ce{H2O}$ donating a proton back to $\ce{NH3}$ to form $\ce{NH4+}$ and $\ce{OH-}$ again.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="5016" LastEditDate="2015-01-11T17:22:12.380" LastActivityDate="2015-01-11T17:22:12.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15234" PostTypeId="2" ParentId="15226" CreationDate="2014-08-13T17:28:01.857" Score="2" Body="&lt;p&gt;I think I got it figured out using electrochemical means, thanks to Aditya’s suggestion to consult $K_\text{a}$ values rather than $K_\text{b}$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider these two half reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2e- +H2-&amp;gt;2H-}\tag{$E^\circ=-2.25~\mathrm{V}$}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2 + 2H2O -&amp;gt; 2H3O+ + 2e-}\tag{$E^\circ=0.00~\mathrm{V}$}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coupling these two half reactions results in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2 +2H2O -&amp;gt;2H3O+ +2H- }\tag{$E^\circ=-2.25~\mathrm{V}$}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Application of the Nernst equation can help us find an equilibrium constant for this reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^\circ = -nFE^\circ = -(2)(96\,500~\mathrm{C/mol})(-2.25~\mathrm{V}) = +434\,250~\mathrm{J/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Value makes sense; we’d expect the reaction of hydrogen gas as an acid with water to be highly disfavorable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^\circ = -RT\ln K=-\left(8.31~\mathrm{J/(mol\cdot K)}\right)(298~\mathrm{K})\ln K=+434\,250~\mathrm{J/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = 6.97464\times10^{-77}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this $K$ correspond to this equilibrium: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2 +2H2O -&amp;gt;2H3O+ +2H- }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we must take the square root of the found equilibrium constant to generate a value for $K_\text{a}(\ce{H2})= 8.35\times10^{-39}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally this lines up well with Aditya’s finding that the $\mathrm{p}K_\text{a}$ of $\ce{H2}$ is 35; the −log of the above $K_\text{a}$ value I found is 38. Nice.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T15:57:35.157" LastActivityDate="2015-05-02T15:57:35.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15236" PostTypeId="1" CreationDate="2014-08-13T18:04:20.643" Score="1" ViewCount="491" Body="&lt;p&gt;Which is the stronger Lewis base, and why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that the answer is &quot;it depends.&quot; With regard to the hydrogen proton, which hydroxide ion is obviously the stronger Lewis base. However, with regard to other, more complex Lewis bases, the distinction may not be so clear, as hydroxide ion has several lone pairs (three) and these can weaken the stability of Lewis adducts through lone pair/&quot;bonding pair&quot; (the pair that forms the coordinate covalent bond) repulsions. Ammonia only has one lone pair so there isn't an issue of lone pair/&quot;bonding pair&quot; repulsion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="12860" LastEditDate="2015-11-13T19:35:53.333" LastActivityDate="2018-03-04T02:56:12.630" Title="Ammonia and hydroxide ion - Lewis basicity" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15237" PostTypeId="2" ParentId="14533" CreationDate="2014-08-13T18:05:34.697" Score="-1" Body="&lt;p&gt;Oxygen can have less than an octet. Here's why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the triple bond resonance structure of $\ce{CO}$ the $\ce{O}$ has 5 electrons around it giving it a +1 formal charge. The same thing is true for the $\ce{O}$ to the right in structure 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O}$ has the equivalent of 3 lone pairs of electrons (that is 6 electrons) to form a bond with. If it uses 1 lone pair out of the 2,2,1,1 arrangement for $\ce{O}$ to form a single bond (which it does) than the $\ce{O}$ has used 2 electrons to form a bond leaving 4 nonbonding electrons. If you separate the $\ce{O}$ and the $\ce{C}$ by breaking the bond the $\ce{O}$ is going to have its original 6. It is exceeding an octet that is invalid for period 2, not less than an octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{O}$ has less than an octet than yes it is unstable but it is still possible. This is why positive ions of $\ce{O}$ exist as well as $\ce{O^-}$. There is also a small amount of neutral atomic oxygen (which is a biradical). Carbocations exist for the same reason that oxygen ions with less than 8 electrons do. That reason is because having less than an octet is not invalid.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorDisplayName="user15489" LastEditDate="2015-05-23T09:15:10.847" LastActivityDate="2015-05-23T09:15:10.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15238" PostTypeId="5" CreationDate="2014-08-13T18:30:16.730" Score="0" Body="&lt;p&gt;Questions relating to reactions of oxides and other questions about oxides should have this tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic definitions of oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound with oxygen and any element less electronegative than oxygen(Iron(III) oxide for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any organic compound where oxygen is bonded to 2 carbon containing groups such as an ether or a special case of it known as an epioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Oxide&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Oxide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Oxides&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Oxides&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wou.edu/las/physci/ch412/oxides.html&quot; rel=&quot;nofollow&quot;&gt;https://www.wou.edu/las/physci/ch412/oxides.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2014-08-13T20:43:38.380" LastActivityDate="2014-08-13T20:43:38.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15239" PostTypeId="4" CreationDate="2014-08-13T18:30:16.730" Score="0" Body="A binary compound of oxygen with another element or group." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T11:55:18.277" LastActivityDate="2014-08-17T11:55:18.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15240" PostTypeId="2" ParentId="15161" CreationDate="2014-08-13T18:36:42.903" Score="2" Body="&lt;p&gt;I think the confusion may be because you are conflating two ideas about Gibbs free energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that Gibbs energy is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamic potential&lt;/a&gt; - this means that on a given surface in &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase-space&lt;/a&gt; (the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isothermal%E2%80%93isobaric_ensemble&quot; rel=&quot;nofollow noreferrer&quot;&gt;NPT&lt;/a&gt; surface), it is the &quot;driving&quot; force for, well, everything. Including chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second idea is that the net change in Gibbs energy when going from reactants to products indicates the direction that the reaction will (eventually) go, and that its magnitude tells you the maximum amount of work that could be done by that reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second idea follows from the first - this becomes easy to understand if you just picture the thermodynamic potential as a gravitational potential, since we all have a good intuitive understanding of gravitational potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a hill representing the thermodynamic potential (in this case Gibbs energy):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzqxR.png&quot; alt=&quot;dG vs Delta G&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rock represents a reaction coordinate - say, the concentration of a reactant. What will happen to the rock? It will go downhill. What will its &lt;strong&gt;maximum&lt;/strong&gt; energy be when it reaches the bottom? Whatever the difference in potential energies is. What is the &lt;strong&gt;minimum&lt;/strong&gt; energy it would need to get back the top? Again, whatever the net difference is in potential energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask about $dG$ being negative, you are asking about the &lt;strong&gt;differential&lt;/strong&gt; change - in other words, what is the slope? Obviously, if the reaction proceeds, the slope must be negative - &quot;things roll downhill&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you ask about $\Delta G$ being negative, you are asking whether the net change is positive or negative. The system can never do more work than the net change in $\Delta G$, because that is the maximum change in thermodynamic potential energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem, the reaction involves moving an electron against a potential - the cell potential. The reaction can provide $\Delta G$ amount of energy with which to move that electron.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-13T18:36:42.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15241" PostTypeId="2" ParentId="15236" CreationDate="2014-08-13T18:46:52.053" Score="0" Body="&lt;p&gt;$pKa$ of conjugate acid($NH_4^+$) of ammonia($NH_3$) is $9.9$ and of conjugate acid($H_2O$) of hydroxide($OH^-$) is $15.7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, hydroxide is a strong base.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-13T18:46:52.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15242" PostTypeId="2" ParentId="10650" CreationDate="2014-08-13T19:37:19.347" Score="-1" Body="&lt;p&gt;Strong acids readily give up their protons, because the resulting conjugate base becomes more stable with respect to its new electron distribution. An example of this would be the acid halides with the exception of HF. The conjugate bases of these acids achieve their octets and diffuse polarizability, rendering them electronically stable.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-08-13T19:37:19.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15243" PostTypeId="2" ParentId="15236" CreationDate="2014-08-13T19:49:01.993" Score="1" Body="&lt;p&gt;With respect to Brønsted basicity, definitely hydroxide is the stronger base. With respect to Lewis basicity, I think the same scenario holds. As you said, hydroxide has three lone pairs, whereas ammonia has only one lone pair. As far as the energetics of these situations are concerned, the magnitude of lone pair to lone pair repulsion is far greater than that of lone pair to bond repulsion. Thus, it would make sense for hydroxide to grab the proton instead of ammonia.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastEditorUserId="5026" LastEditDate="2018-03-04T02:56:12.630" LastActivityDate="2018-03-04T02:56:12.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15244" PostTypeId="2" ParentId="15231" CreationDate="2014-08-13T19:55:24.807" Score="1" Body="&lt;p&gt;In both e and j, you would choose the protic solvent over aprotic solvent, because the protic solvent will stabilize the resulting bromide anion. Sometimes, the stability of the other product should also be concerned, not just the main product.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-08-13T19:55:24.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15245" PostTypeId="2" ParentId="15231" CreationDate="2014-08-13T19:56:39.143" Score="2" Body="&lt;p&gt;For reaction (e), given the strength of the nucleophile, the primary benzylic leaving group, and the lack of any protons beta to the leaving group, SN2 is the only mechanism that can operate. The choice of ethanol as solvent likely has much more to do with getting everything into solution than influencing one pathway over another. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_cyanide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, NaCN is soluble in EtOH but only slightly soluble in DMF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reaction (j), I would guess that the water/methanol mixture was chosen to get everything into solution. There might be some concern about hydroxide deprotonating methanol, and then methoxide carrying out the displacement. My answer would probably be to use a water/THF mixture, if you were really worried about that.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-13T19:56:39.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15246" PostTypeId="2" ParentId="15231" CreationDate="2014-08-13T19:57:35.820" Score="3" Body="&lt;p&gt;Your thinking is not wrong. For (e) DMF is a better choice of solvent - this doesn't mean that MeOH won't work just that DMF is better and is typically used in these reactions. For even better results you can use DMSO. If you dare, as DMSO/cyanide is a very toxic cocktail.&#xA;(j) I don't think it is a practical synthetic reaction since it can possibly give you elimination and/or $S_N1$ (ie no inversion of stereochemistry) products.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="7284" LastEditDate="2014-08-13T20:13:58.247" LastActivityDate="2014-08-13T20:13:58.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15247" PostTypeId="2" ParentId="15129" CreationDate="2014-08-13T20:14:10.807" Score="0" Body="&lt;p&gt;$\ce{Cl}$ can have a positive charge if it is bonded to something more electronegative like oxygen for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You specify the partial charges by looking at electronegativity AND how many are bonded to 1 atom. For example take Chlorine Heptoxide($\ce{Cl2O7}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $\ce{O}$ has a partial negative with the 1 in the center having the largest of the negative charges since it is bonded to 2 chlorines whereas the other 6 have smaller negative charges since they are each bonded to 1 chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each $\ce{Cl}$ has a partial positive and they are equal since both of them are bonded to 4 oxygens.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2014-08-13T20:14:10.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15248" PostTypeId="1" AcceptedAnswerId="15249" CreationDate="2014-08-13T20:15:57.913" Score="1" ViewCount="10491" Body="&lt;p&gt;I think ionic compounds dissolve in just about any compound including nonpolar. My reasoning is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutral electric charge attracts both positive and negative (This I know from doing things with static electricity and putting it on a neutral surface).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds like strong acids and bases disassociate into individual ions, or if it is not dissolved but melted ionic dimers (like an individual Na+Cl- dimer for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do ionic compounds such as NaCl dissolve in just about every solvent from polar to nonpolar to liquid ionic?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="9961" LastEditDate="2016-08-13T00:48:44.897" LastActivityDate="2016-08-13T00:48:44.897" Title="Ionic compounds dissolve in nonpolar compounds?" Tags="&lt;solubility&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15249" PostTypeId="2" ParentId="15248" CreationDate="2014-08-13T21:04:41.650" Score="7" Body="&lt;p&gt;First, I think you have some misconceptions about electrical charge. There isn't neutral charge, it's either positive or negative, if it doesn't have either, it has no charge. Why are you seeing static charge interaction with neutral surfaces? It might be polarization, where a charged particle warps the electron cloud around a neutral atom generating partial charges around the neutral atom. Without knowing more about your experiment I can't say more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile I did an experiment of my own. I took 5 20mL scintillation vials, and added about 0.5g of NaCl to each. I then added 5mL of deionized water, methanol, 200 proof denatured ethanol, isopropanol, or acetone to each vial and tried to dissolve the salt. Only water dissolved all of the salt. I can't say there isn't ANY salt dissolved in the other solvents, but clearly the solubility is solvent dependent, and only water is polar enough to dissolve all the salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y1NzI.jpg&quot; alt=&quot;Results of the solubility experiment&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="137" LastEditDate="2014-08-14T00:42:46.577" LastActivityDate="2014-08-14T00:42:46.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15250" PostTypeId="1" CreationDate="2014-08-13T23:37:19.997" Score="5" ViewCount="6380" Body="&lt;p&gt;I’m trying to understand how breaking chemical bonds turn into kinetic energy. If AB is a chemical which breaks into A and B which zip away from each other, is it because when the bond breaks they suddenly repel each other? Another way I'm thinking of this is what happens if one single molecule of AB alone in a vacuum breaks apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the point I'm trying to make is the reaction is making the products move faster than when they started. It takes a force to make things move. So it has to be either gravity, electric, magnetic, or one of the nuclear. I'm pretty sure we can rule out nuclear and gravity.&lt;/p&gt;&#xA;" OwnerUserId="6830" LastEditorUserId="6830" LastEditDate="2014-08-15T23:13:14.620" LastActivityDate="2014-08-16T02:56:19.117" Title="How do breaking chemical bonds turn into kinetic energy" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15251" PostTypeId="2" ParentId="10415" CreationDate="2014-08-13T23:41:42.273" Score="10" Body="&lt;p&gt;(a) Correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Let me start with Linear Combination of Atomic Orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a quantum calculation method that assumes that all molecular orbitals can be derived from (basic) atomic orbitals by a method called linear combination. You might have heard of linear combination from vector geometry. It's basically just adding things up with different preceding factors. So &lt;code&gt;2x - 3y&lt;/code&gt; is a linear combination of &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about linearly combining atomic orbitals is that &lt;em&gt;you must end up with the same amount of orbitals as you started out with&lt;/em&gt; (so you combine 6 atomic orbitals to make 6 molecular orbitals). There's also additional stuff you need to take care of (the squares of all the preceding factors of one AO need to sum up to give one), but that's only of marginal importance here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of linear combination would be your average H–H sigma bond. You have two atomic orbitals (1s of each of the hydrogens) which you combine. I'll call the 1s orbital of the left hydrogen &lt;em&gt;l&lt;/em&gt; and that of the right hydrogen &lt;em&gt;r&lt;/em&gt;, for simplicity. The two simplest linear combinations are &lt;em&gt;l&lt;/em&gt; + &lt;em&gt;r&lt;/em&gt; and &lt;em&gt;l&lt;/em&gt; - &lt;em&gt;r&lt;/em&gt;. In the first one, you get positive interference, i.e. you add a positive phase to a positive phase – think of it as the positive half of a sine wave added onto another positive half of a sine wave: it'll get bigger. This type of interaction is favourable, and will thus give you a molecular orbital energy that is less than that of a separated atomic orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you also need to remember to create a second MO, the &lt;em&gt;l&lt;/em&gt; - &lt;em&gt;r&lt;/em&gt;. Here, you would be adding plus to minus, which gives zero. So you have a plane in between the two protons where the resulting wave function has zero value, is non-existant (a nodal plane). Because the nodal plane is between the atoms, the resulting molecular orbital is antibonding, or higher in energy. The basic rule is that the more nodal planes an orbital contains, the higher its energy level will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks you to find the further five linear combinations that will lead to those group orbitals capable of overlapping withthe  central carbon's orbitals. So you basically know what to look for; and yes: the CLi&lt;sub&gt;6&lt;/sub&gt; is analogous enough to an octahedral complex, that you can just copy that image. If you didn't have either the picture or the hint, you would need to use group theory and reduction formulae, which I won't cover here (for those interested: the group orbitals will transform as a&lt;sub&gt;1g&lt;/sub&gt;, e&lt;sub&gt;g&lt;/sub&gt; and t&lt;sub&gt;1u&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand your reasoning for putting the d-type orbitals lower in energy (i.e. more stable) than s- or p-type ones. Usually, it's s &amp;lt; p &amp;lt; d. But the better way to answer the question would be using nodal plane counting: The orbital that will mix with the carbon'S s orbital has no nodal plane, and will therefore be the lowest (a&lt;sub&gt;1g&lt;/sub&gt;, for those interested). The three that will mix with the carbon's p orbitals are second, because they contain one nodal plane each (t&lt;sub&gt;1u&lt;/sub&gt;). Finally, those that will mix with d orbitals are highest, because they contain two nodal planes (either parallel or perpendicular to each other. Their energy level will be the same, though, says group theory) (you guessed it, those are e&lt;sub&gt;g&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Yes, that can be regarded as the easiest answer. Carbon, being an element of the second period, only 'has' (term is incorrect, but you get the picture) 1s, 2s and 2p orbitals. 3s and 3p might be accessible by irradiation. 3d is not, as it is even higher in energy than 4s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same rationale applies to the higher order main group elements. Even though silicon, being an element of the third period, could be considered to access d orbitals to create anions like $\mathrm{SiF}_6^{2-}$, the d orbitals' energy is too far removed from that of the s and p orbitals, that it cannot be considered to take part in bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that transition metals – the elements usually associated with octahedral surroundings and d-orbitals – usually use 4s and 3d (and sometimes 4p) orbitals. 3d is a lot closer to 4s and 4p than to 3p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) Start off by thinking what the relative energy levels of the orbitals in question are. Carbon, as was pointed out, is more electronegative than lithium, so all of its orbitals will be lower in energy than the lithium orbitals (the difference is significant here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should give you a picture of carbon orbitals on one side (2s, rather far-ish down, three 2p's, still pretty far down) and the three sets of lithium group orbitals on the other side (the one that I called a&lt;sub&gt;1g&lt;/sub&gt; being the lowest, but probably higher than carbon's 2p's; the three I called t&lt;sub&gt;1u&lt;/sub&gt; next and the two e&lt;sub&gt;g&lt;/sub&gt; last; those inside a group at the same height, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you need to find out which orbitals can recombine favourably. If you attempted to combine the orbital shown in figure 1 with a carbon 2p orbital, that intereference would be non-bonding or zero (you have bonding interactions on one side cancelled out by exactly the same amount of antibonding interactions on the opposite side). Note that they helped you by saying that the figure 1 orbital (and thus also carbon's 2s) and the two lithium orbitals that would bond with d-type orbitals do not take part in favourable interactions; so only the p-type ones remain, each of them creating a bonding-antibonding pair of molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving from the edges to the centre of the diagram, you would move the carbon's 2s plus the three lithium group orbitals not taking part horizontally into the middle. Carbon's 2p and the other three lithium group orbitals are then mixed, lowering the 2p's energy while raising the lithium orbitals' energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should end up with an energy diagram of the following type:&lt;br /&gt;&#xA;    C(2s) - 3 C(2p) -- 1st Li -- 3 2nd Li - 2 3rd Li.&#xA;It's not clear (nor can it be deducted &lt;em&gt;a priori&lt;/em&gt;) whether the 2p orbitals will end up lower than the 2s (or the 2nd Lithium group higher than the third), but it is to be assumed that they shouldn't change their places too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, fill in the electrons in the way you're used to. Two go into C(2s), six into the three C(2p), and two into the first Li orbital (the one shown in figure 1). To answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;non-combining MOs can or can not be present in the MO diagram. They should, if they are part of the valence orbitals. They should not if they are core orbitals (note we always omitted the seven 1s orbitals). Not combining however means, that they do not change height from where you put them initially.&lt;/li&gt;&#xA;&lt;li&gt;Yes, that is correct, but that is more of a consequence we can think of after mixing has been done.&lt;/li&gt;&#xA;&lt;li&gt;If we include the nonparticipating orbitals, the question is no longer a question. If they weren't present, though, you would do exactly what you proposed. You would fill in three electrons spin up into three antibonding MOs and one further electron spin down. The resulting system would be degenerated. It would not be stable, and would reduce symmetry to force the three orbitals apart, creating a discrete energy level for the single paired electron.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(e) Yeah, that was a pretty good guess into the wild, that would always be correct ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the molecular orbital diagram as a reasoning, though: Consider the state before bonding. The four carbon orbitals have four electrons between them (2 in 2s, 2 degenerated in 2p). The six lithium orbitals have six electrons between them (2 in the figure 1 orbital, 4 in the three orbitals that will take part in bonding; this state is also degenerated). After the bonding, not only have we moved four electrons out of relatively unstable orbitals into a lot more stable ones (those from the lithium orbitals into the carbon-centered ones), but we have also removed two degenerated states and turned them into clearly defined ones. Both are good for the system's stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) I think your guesswork is pretty much spot on. (The orbital diagrams would be totally different ones, though; as NLi&lt;sub&gt;5&lt;/sub&gt; would be a trigonal bipyramid (symmetry group D&lt;sub&gt;3h&lt;/sub&gt;) and OLi&lt;sub&gt;4&lt;/sub&gt; a tetrahedron (symmetry group T&lt;sub&gt;d&lt;/sub&gt;) as opposed to the carbon octahedron (O&lt;sub&gt;h&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(g) I'm pretty sure this is a yes-but case. Yes, just looking at what we have for Li would suggest that H would do the same thing. But, hydrogen's atomic orbitals are lower than lithium's, as it is a lot more electronegative. That in turn means that there is significant mixing to be observed between C(2s) and the figure 1 type orbital. Therefore, the C(2s) orbital would again be stabilised, while the hydrogen group orbital would be destabilised – most likely so far that it reaches the energy level of the three higher group orbitals. While they were clearly empty in the lithium case, they would most likely be populated in the hydrogen case, destabilising the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is the relative stability of CH&lt;sub&gt;4&lt;/sub&gt; versus CLi&lt;sub&gt;4&lt;/sub&gt;. Again, the all equal phase group orbital of the lithiums would be too high to significantly mix with the C(2s) orbital in a tetrahedric case. The all equal phase group orbital of hydrogens does mix quite significantly, creating a much more stable CH&lt;sub&gt;4&lt;/sub&gt; molecule than CLi&lt;sub&gt;4&lt;/sub&gt;. Thus, CH&lt;sub&gt;6&lt;/sub&gt; would eliminate hydrogen to create CH&lt;sub&gt;4&lt;/sub&gt;, while CLi&lt;sub&gt;6&lt;/sub&gt; is stable enough (and CLi&lt;sub&gt;4&lt;/sub&gt; unstable enough) to not eliminate Li&lt;sub&gt;2&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2014-08-13T23:41:42.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15252" PostTypeId="1" AcceptedAnswerId="15264" CreationDate="2014-08-14T00:23:40.937" Score="1" ViewCount="448" Body="&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/eladd/symh2so4.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/mechanisms/eladd/symh2so4.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading my textbook and this website. I just cannot find the mechanism for what happens when we go from CH3CH2OSO3H to ethanol by the addition of water and heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the electrophillic addition part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what happens next? Is there some kind of Sn2 reaction to make HSO4- the leaving group? Then, why does the EtOH2+ molecule give out one of its acidic protons to HSO4- and not to H2O which is a more powerful base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hate when my textbook leaves out the details and I obsess on them :(&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-08-14T14:58:40.650" Title="Ethene to ethanol: electrophillic additions involving ethyl sulfates" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15253" PostTypeId="1" AcceptedAnswerId="15260" CreationDate="2014-08-14T00:27:44.420" Score="6" ViewCount="726" Body="&lt;p&gt;Regarding chemical synergies, I read: if someone ingests various products containing sodium benzoate (for example soft drinks that exist) AND if one has had some quantity of vitamin C there might be 'reactions' that produce benzene, a carcinogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people have told me you'd have to drink a lot of pop for this to happen (assuming you have had some vitamin C during the day), but how much pop is a lot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given one has had a glass of orange juice say, how much pop would one have to drink before a benzene molecule is formed?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="402" LastEditDate="2015-05-25T03:34:50.583" LastActivityDate="2015-06-18T17:52:18.683" Title="Sodium benzoate in soft drinks reacting to produce benzene?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15254" PostTypeId="2" ParentId="15250" CreationDate="2014-08-14T02:16:12.713" Score="6" Body="&lt;p&gt;You are thinking about it a little wrong.  The fact that the bond exists means it is in a state of lower energy than when A and B are separated -- therefore you would have to input energy into the system to get the bond to break.  But if you had two species A-B and C and you could form a more stable bond B-C which means it has a lower energy level, then the excess energy E(A-B) - E(B-C) for the reaction:&#xA;A-B + C -&gt; A + B-C &#xA;could be released as kinetic energy of the A and B-C species moving away from each other.  &lt;/p&gt;&#xA;" OwnerUserId="7478" LastActivityDate="2014-08-14T02:16:12.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15255" PostTypeId="2" ParentId="15250" CreationDate="2014-08-14T02:17:31.413" Score="8" Body="&lt;p&gt;Breaking chemical bonds does &lt;strong&gt;not&lt;/strong&gt; turn their potential energy into kinetic energy. All bond dissociation enthalpies are positive (for stable bonds that actually exist), which means that breaking them is an &lt;em&gt;endothermic&lt;/em&gt; or heat absorbing process, and therefore they &lt;em&gt;absorb&lt;/em&gt; kinetic energy (heat is a transfer of molecular kinetic energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you might be thinking of is exothermic reactions, where the enthalpy of the reactants is lower than the products, and so the reaction releases heat, or increases the kinetic energy of the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer to your question is &quot;because energy is conserved, it has to go somewhere, and heat and light are the only places it can go.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to think about a better way to explain it that isn't too far away from reality, but at the same time isn't too complicated. If I can, I'll edit this answer later to include it.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-14T02:17:31.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15256" PostTypeId="1" CreationDate="2014-08-14T03:59:22.650" Score="3" ViewCount="3577" Body="&lt;p&gt;I would like to know what the difference between medicinal chemistry, pharmaceutical chemistry,  pharmacy and pharmacology. All of them seem to involve the development of drugs, so I am confusd as to how to distinguish them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does specialising in medicinal chemistry require a biology background? It seems to lean more towards biology.&lt;/p&gt;&#xA;" OwnerUserId="7422" LastEditorUserId="865" LastEditDate="2015-11-09T13:52:37.743" LastActivityDate="2015-11-09T13:52:37.743" Title="How do the fields of medicinal chemistry, pharmaceutical chemistry, pharmacy, and pharmacology differ?" Tags="&lt;organic-chemistry&gt;&lt;medicinal-chemistry&gt;&lt;pharmacology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15258" PostTypeId="1" CreationDate="2014-08-14T05:48:06.840" Score="10" ViewCount="6521" Body="&lt;p&gt;Rutherfords experiment showed that most of the alpha particles passed through a thin gold foil undeflected. So why can't light pass through the foil, which is supposedly thinner than the heavy alpha particles? &lt;/p&gt;&#xA;" OwnerUserId="7480" LastEditorUserId="1499" LastEditDate="2014-08-14T12:38:56.253" LastActivityDate="2017-03-02T19:06:46.130" Title="Why can't light pass through a gold foil, but alpha particles can?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;nuclear&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15259" PostTypeId="1" CreationDate="2014-08-14T10:28:00.637" Score="5" ViewCount="273" Body="&lt;p&gt;I am looking for the names of some environmental alkylating agents (in context of DNA damages) found for example in the fume produced by industry. Most of the papers which I have found is concentrated on cigarette smoke, chemotherapy ect. I found something about methyl chloride, that is produced by plants and fungi, but also industrially. However there is no more details about that industrial part and in general the amount of methyl chloride in the air is rather low. Are there more relevant examples? Also I came across the information that alkylating agents may be found in food, but again there was no details.&lt;/p&gt;&#xA;" OwnerUserId="7482" LastActivityDate="2014-10-24T08:45:07.533" Title="Environmental Alkylating Agents" Tags="&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15260" PostTypeId="2" ParentId="15253" CreationDate="2014-08-14T10:50:46.717" Score="8" Body="&lt;p&gt;The reaction taking place is described in article &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jf00029a001&quot; rel=&quot;nofollow&quot; title=&quot;Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst&quot;&gt;Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst&lt;/a&gt; and its importance for food safety recently reviewed in &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1750-3841.2010.01536.x/full&quot; rel=&quot;nofollow&quot; title=&quot;Evaluation of Accelerated UV and Thermal Testing for Benzene Formation in Beverages Containing Benzoate and Ascorbic Acid&quot;&gt;Evaluation of Accelerated UV and Thermal Testing for Benzene Formation in Beverages Containing Benzoate and Ascorbic Acid&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the reaction CAN occur, namely at elevated temperatures (&gt;50 °C) and UV-light and in presence of transition metal. Even if the reaction takes place, the amount of benzene formed is either below detection limit, or below limit for drinking water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the end user it means, that in extremely unlucky situation, the benzene can form in quantity, which is not considered safe to drink in unlimited amount for the whole life (1 ppb). As such, I would consider the benzene formation as very weak argument against such beverages (mainly if compared with other problems, such as sugar content and associated obesity).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorDisplayName="user7951" LastEditDate="2015-06-18T17:52:18.683" LastActivityDate="2015-06-18T17:52:18.683" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15261" PostTypeId="2" ParentId="15258" CreationDate="2014-08-14T12:31:18.243" Score="9" Body="&lt;p&gt;Light &lt;strong&gt;can&lt;/strong&gt; pass through a gold foil though, it just has to be thin enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure gold is a very malleable substance and can be beaten with a hammer into foils of around 100 nm thickness. &lt;a href=&quot;http://myweb.usf.edu/~mhight/goldfoil.html&quot;&gt;Sources&lt;/a&gt; &lt;a href=&quot;http://www.chemteam.info/AtomicStructure/Rutherford-Model.html&quot;&gt;suggest&lt;/a&gt; the gold foil used in the Geiger–Marsden experiment (known more commonly as the Rutherford gold foil experiment) was about 86 nm thick. Somewhere around this region of thickness, gold starts letting light through, as well as alpha particles. It is well known that ~20 nm thick gold (or any other metal) deposited onto a surface by thermal evaporation in a vacuum is essentially invisible and can be used as a transparent conducting surface/electrode. As the foil gets thicker, the gold starts to absorb a noticeable amount of photons, especially in the violet/blue part of the electromagnetic spectrum, while scattering away the red light. &lt;a href=&quot;http://en.wikipedia.org/wiki/Gold#Characteristics&quot;&gt;The light scattered off a semi-transparent gold film acquires a reddish/yellow tinge, while the light that still does make it through is blueish&lt;/a&gt;. When the film reaches 100 nm or so, it becomes practically as opaque as the bulk metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In truth, comparing the mean free path or collision cross-sections of photons and alpha particles through a gold sheet is somewhat more complicated than just saying alpha particles should penetrate less because they're massive, so it would be conceivable for a film of appropriate thickness to block photons yet still allow alpha particles through. A very coarse, hand-waving explanation is that photons are absorbed in the electrosphere of a gold atom, while an alpha particle is only significantly pushed away if it bangs almost straight into the nucleus, a region composing approximately $10^{-15}$ of the volume of the atom.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-08-14T12:31:18.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15262" PostTypeId="2" ParentId="15256" CreationDate="2014-08-14T13:30:18.630" Score="1" Body="&lt;p&gt;The difference between the fields is significant, but as all become more interdisciplinary, some simplification are inevitable. So better look what an expert in given field does:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pharmacy - general overview of all drugs, their action and interactions&lt;/li&gt;&#xA;&lt;li&gt;pharmacology - detailed study of a given drug, how it is distributed in body, which enzymes it affects, how it is transformed and excreted, ...&lt;/li&gt;&#xA;&lt;li&gt;medicinal chemistry - detailed study of the mechanism of action at molecular level, enzyme variations which affect the efficiency, ...&lt;/li&gt;&#xA;&lt;li&gt;pharmaceutical chemistry - how to synthesize the drug, more or less specialized organic chemistry&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-08-14T13:30:18.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15263" PostTypeId="2" ParentId="15248" CreationDate="2014-08-14T14:54:07.360" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I think ionic compounds dissolve in just about any compound including nonpolar. My reasoning is this:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds don't always dissolve in just about any compound. Consider sodium chloride. This &lt;em&gt;isn't&lt;/em&gt; soluble in acetone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neutral electric charge attracts both positive and negative(This I&#xA;  know from doing things with static electricity and putting it on a&#xA;  neutral surface).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If something has no net charge (i.e. is neutral) it won't be electrically attracting anything. Only oppositely charged things attract. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ionic compounds like strong acids and bases disassociate into&#xA;  individual ions(Or if it is not dissolved but melted ionic dimers(like&#xA;  an individual Na+Cl- dimer for example).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not all strong acids or bases are ionic. Consider sulfuric acid. The S-O and O-H bonds, granted, have some ionic character, but are generally characterized as covalent. Also consider the halide series of binary acids - HCl, HBr, HI. Strength increases in that order, yet ionic character &lt;em&gt;decreases&lt;/em&gt; in that order. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus a base isn't necessarily ionic. NaOH is an ionic salt, and yes, it's a strongly basic salt. But the hydroxide ion is also a strong base. In isolation, it's not ionic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you should define what a dimer is before incorrectly applying it. The definition of a dimer according to IUPAC is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The transformation of a molecular entity $\ce{A}$ to give a molecular entity $\ce{A2}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So no, the sodium and chloride ions cannot dimerize in the fashion you suggested - with sodium ion coming together with chloride ion. That's &lt;em&gt;not&lt;/em&gt; dimerization. Now, if a chlorine radical joined with &lt;em&gt;another&lt;/em&gt; chlorine radical, that would be dimerization, since both starting molecular entities are the &lt;em&gt;same&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do ionic compounds such as NaCl dissolve in just about every solvent&#xA;  from polar to nonpolar to liquid ionic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, and this has been illustrated by &lt;em&gt;user137&lt;/em&gt;'s excellent demonstration. And consider this table of solubility products. A bunch of ionic compounds, yes. But all of them are considered &lt;em&gt;semi-soluble&lt;/em&gt; salts because of their low degree of dissociation into their constituent ions (hence the miniscule Ksp values). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQ329.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-14T17:21:02.173" LastActivityDate="2014-08-14T17:21:02.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15264" PostTypeId="2" ParentId="15252" CreationDate="2014-08-14T14:58:40.650" Score="3" Body="&lt;p&gt;Yes, I believe you are correct. Water attacks at the sulfate containing carbon by an SN2 mechanism. You are also right that sulfuric acid is a stronger acid than hydronium (or protonated ethanol), which means that in solution the bisulfate anion will predominate. Often people (including textbook authors) will draw a mechanism that goes to neutral products. Most likely water will carry out the final deprotonation (especially given that water is the solvent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, most likely this reaction is self catalyzed, as one sulfate protonates another, activating the sulfate for SN2 reaction by a weak nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/POW3H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-14T14:58:40.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15265" PostTypeId="2" ParentId="15174" CreationDate="2014-08-14T16:36:39.617" Score="3" Body="&lt;p&gt;the answer already present over here is absolutely excellent but, as an extention i would add some points to it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the equation of $PV=nRT$ is made for gases and for a solution of gas and liquid in which the volatile solvent shows some vapour pressure on its solution &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the curve sigmoid curve you get is through the Clausius Clapeyron equation which suits this condition [i.e. the situation of solutions] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wtym1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jbgUR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after the whole solution is evaporated then you get pure solvent in the vapour phase after which your equation for an ideal gas $PV=nRT$ is applicable and you will get a striaght line there after&lt;/p&gt;&#xA;" OwnerUserId="6954" LastEditorUserId="6954" LastEditDate="2014-08-15T04:14:01.630" LastActivityDate="2014-08-15T04:14:01.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15266" PostTypeId="2" ParentId="15256" CreationDate="2014-08-14T16:54:32.067" Score="7" Body="&lt;p&gt;I'm in a medicinal chemistry department, the floor below is the pharmaceutics department. We are both located in our university's college of pharmacy. The pharmacology department is under the college of medicine in same building as biochemistry, microbiology, etc. There is little interaction between departments here and that's not good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can break the differences in each discipline down into some order based on drug development. Medicinal chemistry and pharmacology come first, with the discovery of molecules that bind to receptors and other proteins and have biological effect. This used to be based almost entirely on natural products, and those still play a large role, but high throughput screening and computer modeling are becoming more important for discovery of drug candidates. While pharmacology focuses on the effects of drugs and other molecules on cells and the body, medicinal chemistry has traditionally focused on the chemistry behind them. A medicinal chemist might look at a natural product with desirable activity, and ask &quot;how can I make this better?&quot; and apply organic synthesis to produce new versions of that drug, with better activity, reduced toxicity, and longer half-life. Of course the biology cannot be ignored, and has become more important in recent decades. Our department used to be pure synthesis, but now only 2 of our 8 labs are pure synthesis, and 2 almost entirely biological.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the medicinal chemists have developed a drug candidate to the point they think it can be used in the real world, it moves on to pharmaceutical chemists. They not only have to refine the synthesis of the drug to produce it efficiently and reliably on large scale, they have to develop the formulation the drug is delivered as. If it's a tablet, they have to determine what to make the bulk of the tablet from, how to press the tablets, how to store them, etc. A large part of pharmaceutics is stability, how long can the product sit on a shelf and not degrade? How does temperature and humidity affect degradation? What does it degrade into and are those products dangerous? Pharmaceutics is also concerned with drug absorption and excretion. They often try to develop &quot;extended release&quot; formulations so that a tablet takes a longer time to break down in your gut, releasing the drug over a longer time and keeping an effective concentration of drug in your blood longer, thus requiring fewer pills and less variation in concentration. Pharmaceutics is a lot of organic chemistry, physical chemistry, analytical chemistry, and quality control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pharmacy is the end product of all of these. After a drug is developed and approved the pharmacist is responsible for distribution. While doctors prescribe drugs, they usually have very little training in how the drugs work and interact. The pharmacist needs to know more than just how to sell pills, they must have at least a working knowledge of the pharmacology behind the drug, be aware of potential interactions and side effects, and advise the patient on the best practices for using the drug. They must also be aware of the rules and regulations, and make sure the controlled substances under their care are not stolen or lost or abused. Substance abuse is actually the biggest workplace hazard for pharmacists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some knowledge on these things because I'm in a med chem department, my father is a chemist for a pharmaceutical company, and my brother just got his pharmacy license, so we see all parts of the process.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="137" LastEditDate="2014-08-14T23:40:54.037" LastActivityDate="2014-08-14T23:40:54.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15267" PostTypeId="1" AcceptedAnswerId="15271" CreationDate="2014-08-14T17:25:50.520" Score="2" ViewCount="2050" Body="&lt;p&gt;I was having a revision on periodic table I was thinking about different properties I could connect most of I could not connect metallic property and oxidation potential .&lt;/p&gt;&#xA;" OwnerUserId="7190" LastActivityDate="2014-08-14T20:24:19.563" Title="Does there exist any relation between metallic property and oxidation potential?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15268" PostTypeId="1" AcceptedAnswerId="15275" CreationDate="2014-08-14T17:26:07.563" Score="2" ViewCount="540" Body="&lt;p&gt;In medicinal chemistry, there are multiple reasons why one would want to experimentally determine the effect of a particular molecule on a receptor e.g. measuring how 'strongly' that molecule brings about a chemical response (an agonist) or how strongly it prevents a response (an antagonist). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were looking at an insulin receptor antagonist (for whatever reason), then it would be easy to quantify the strength of the antagonist by a simple blood test of the test subject (human or animal), to see how much insulin was present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when we're faced with a scenario inherently difficult to quantify? How do we measure the action of a drug then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we've found a natural neurotransmitter receptor agonist and want to compare this to a synthetic drug of a similar nature. I'd struggle to see how we'd experimentally be able to analyse the amount of neurotransmitter present in the synapses in each of the cases to compare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we then look at less direct, perhaps qualitative methods of analysis? Such as 'state of agitation' or 'how many times the subject's arm twitched in a minute'? &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-15T00:46:50.653" Title="How to Experimentally Quantify the Response of a Receptor to a Ligand?" Tags="&lt;medicinal-chemistry&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15269" PostTypeId="2" ParentId="15267" CreationDate="2014-08-14T17:46:07.173" Score="0" Body="&lt;p&gt;Oxidation is defined as an increase in oxidation state, and an increase in oxidation state can come from the loss of an electron. According to IUPAC:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;oxidation &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1) The complete, net removal of one or more electrons from a&#xA;  molecular entity (also called 'de-electronation'). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2) An increase in the&#xA;  oxidation number of any atom within any substrate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/O04362.html&quot; rel=&quot;nofollow&quot;&gt;http://goldbook.iupac.org/O04362.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it might be useful to examine first ionization energies of metals to determine their oxidation potentials. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider for example the oxidation of the alkali metals. Sodium has a highly positive oxidation potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na -&amp;gt; Na^+ + e^-}$ $E^o=2.71 V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This highly positive oxidation potential makes intuitive sense; a positive potential indicates a spontaneous reaction. The above reaction is definitely exothermic, thereby satisfying one criterion of spontaneity; solvated sodium ion is more stable than elemental sodium. The above reaction is also likely entropically favorable; solvated ions have more degrees of freedom than a solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, note that the first ionization energy of sodium is $496 ~kJ/mole$. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, consider the oxidation of potassium metal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K -&amp;gt; K^+ + e^-}$ $E^o=2.93V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first ionization energy of potassium is $418 ~kJ/mole$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As we can see, a lower first ionization energy &lt;em&gt;seems to&lt;/em&gt; corresponds with a more positive standard electrode potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes intuitive sense; the oxidation half-reactions examined above are &lt;em&gt;fundamentally&lt;/em&gt; ionization reactions. And the easier it is to strip an electron; the lower the first ionization energy, the greater the standard cell potential. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, this trend doesn't necessarily hold in all cases. Consider, for example, the oxidation of lithium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Li -&amp;gt; Li^+ + e^-}$ $E^o=3.05V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium is the smallest alkali metal on the periodic table. Accordingly, it has the highest first ionization energy of all the alkali earth metals ($520 kJ/mole$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, it still has the highest oxidation potential (greater than three volts!). So it seems that we can only use ionization energies as a rule of thumb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example in which ionization potential can work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider fluorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think that fluorine would rather lose an electron (be ionized) or gain an electron? Gain is the answer. By gaining an electron fluorine becomes isoelectronic with a noble gas. Sure, fluorine could also &lt;em&gt;lose&lt;/em&gt; all seven of its valence electrons to be isoelectronic with helium, but that's obviously a more roundabout path to stability. Given this information, you can probably deduce that fluorine would rather be &lt;em&gt;reduced&lt;/em&gt; than oxidized. And this is true; fluorine is a powerful &lt;em&gt;oxidizing agent&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-14T20:24:19.563" LastActivityDate="2014-08-14T20:24:19.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15270" PostTypeId="1" AcceptedAnswerId="15281" CreationDate="2014-08-14T18:29:12.283" Score="1" ViewCount="234" Body="&lt;p&gt;I understand that this molecule can only act as a Brønsted base because it is sterically hindered by its multiple methyl groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TNx3k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, isn't the proton bound up in solution? Doesn't that increase the effective size of the proton? How then exactly is a proton expected to reach the site of basicity if it's tied up with multiple water molecules? Or is the size of the hydrated proton still quite small? Or does the proton separate from the water molecules once the proton is sufficiently close to the site of basicity (the lone pair on the nitrogen)? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:12:42.587" LastActivityDate="2018-03-03T11:12:42.587" Title="2,2,6,6-Tetramethylpiperidine - Brønsted Basicity" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15271" PostTypeId="2" ParentId="15267" CreationDate="2014-08-14T19:00:50.760" Score="2" Body="&lt;p&gt;The short answer is yes. The reason for this is that metals exhibit metallic properties because their electrons are relatively loosely bound to the atoms. A &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bonding&quot; rel=&quot;nofollow noreferrer&quot;&gt;metallic bond&lt;/a&gt; is one in which the valence electrons are free to move throughout the lattice structure of the metal. Here's a schematic diagram from wiki commons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOQHA.png&quot; alt=&quot;Metallic Bonds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that the positively charged metallic &quot;cores&quot; (which includes the nucleus and the more tightly bound inner core electrons) are sort of suspended in a crystalline structure within a &quot;sea&quot; of mobile electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model explains most of the  metallic properties we observe, from heat and electrical conductance, to malleability and ductility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about oxidation - in an oxidation-reduction (&lt;a href=&quot;http://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow noreferrer&quot;&gt;redox&lt;/a&gt;) reaction, one element gains electrons (it is reduced) and the other loses electrons (it is oxidized). Here's an animation, also from wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wySen.gif&quot; alt=&quot;Animation of sodium oxidation by fluorine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for an element to be metallic, its valence electrons need to be relatively &quot;loose&quot; - it can't take too much energy to detach them. For an element to be oxidized, something else has to &quot;want&quot; the electrons more - the energy cost of removing the electron has to be lower then the gain from adding them to the reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result there is a very strong correlation between the metallic nature of elements and their relative ease of oxidation. There is also a very strong correlation between both of these properties and the trends in &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionization energy&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron affinity&lt;/a&gt;, and  &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronegativity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in chemistry, things are never that simple. There lots of details and it is tough to make general statements that are always true. For example, you can say that often times the more &quot;metallic&quot; an element is, the more easily it will be oxidized. However, when you compare elements using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard electrode potential table&lt;/a&gt; (which gives us a way of measuring relative ease of oxidation) you find that it is easier to oxidize calcium ($-3.8 \space \rm{V}$) than it is to oxidize potassium ($-2.9 \space \rm{V}$), even though potassium is closer to the &quot;metallic side&quot; of the periodic table, it has a lower first ionization energy, and it is less electronegative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates a limitation of thinking of elements in terms of metals and non-metals: it is really a shorthand way of describing all of the things that affect how tightly those valence electrons are bound to the atom, in comparison to how badly other elements &quot;want&quot; to remove them. With that amount of generalization, we lose too much detail to be able to make accurate predictions in every situation.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-14T19:00:50.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15272" PostTypeId="1" AcceptedAnswerId="15273" CreationDate="2014-08-14T19:37:05.697" Score="16" ViewCount="9648" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dt5oc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dt5oc.png&quot; alt=&quot;Resonance forms of tropone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at resonance structure &lt;strong&gt;1a&lt;/strong&gt;, it doesn't seem that tropone is aromatic. However, the resonance structure &lt;strong&gt;1b&lt;/strong&gt; with $\ce{C+-O-}$ satisfies Hückel's rule for aromaticity. Is it correct to say that tropone is then aromatic?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2018-01-29T18:47:38.830" LastActivityDate="2018-02-06T16:27:16.727" Title="Is tropone aromatic?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15273" PostTypeId="2" ParentId="15272" CreationDate="2014-08-14T20:26:35.237" Score="14" Body="&lt;p&gt;Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon.  If you look at the resonance structures in the drawing below you'll see that structure B &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jTitv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons.  Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is considered aromatic, or stabilized, if it contains 4n+2 pi-electrons and anti-aromatic if it contains 4n pi-electrons.  On this basis, molecules like the cyclopropenyl cation (4n+2, n=0), benzene (n=1), cycloheptatriene (tropylium) cation (n=1), cycooctatetraene dianion (n=2) would be considered aromatic systems.  Correspondingly, the cyclopropenyl anion (4n, n=1), cyclobutadiene (n=1), cyclopentadienyl cation (n=1), cyclooctatetraene (if it were a planar molecule, n=2) are considered destabilized or anti-aromatic molecules.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-14T20:43:23.380" LastActivityDate="2014-08-14T20:43:23.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15274" PostTypeId="1" CreationDate="2014-08-14T21:33:39.893" Score="5" ViewCount="170" Body="&lt;p&gt;I'm doing a book for the African market and the syllabus asks for discussions on proper disposal of chemicals and waste products, particularly in the classroom setting   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors said something along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemicals numbered 1 to 4 can be rinsed down the drain with water. Chemicals labeled 5 to 8 must be disposed of with the help of an Environmental Agent or official.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the numbers they mention? Is there some numerical system (similar to scheduling on prescription meds) which guides us on how to dispose of certain chemicals? &lt;/p&gt;&#xA;" OwnerUserId="5177" LastEditorUserId="5026" LastEditDate="2018-03-05T13:35:00.327" LastActivityDate="2018-03-05T16:02:03.467" Title="Disposing of hazardous waste indicated by numbers on chemicals" Tags="&lt;safety&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15275" PostTypeId="2" ParentId="15268" CreationDate="2014-08-15T00:46:50.653" Score="0" Body="&lt;p&gt;What you're asking about is a wide topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the basic ideas of agonist and antagonist, but there are also molecules called inverse agonists. Every receptor has some base level of activity, even without an agonist to activate them, or in the presence of an antagonist. Inverse agonists push this activity even lower. There are even &quot;partial agonists&quot; which are molecules that activate receptors, but not as well as the endogenous ligand, so act like competitive inhibitors because they occupy some of the receptors the endogenous ligand would have. In addition, the binding site is important. The endogenous ligand binds to a location called the orthosteric binding site, but other molecules may bind to other locations on the receptor. These are allosteric binding sites. Allosteric binders may activate or inhibit a receptor, but they may also be modulators. Modulators don't directly activate a receptor, but allow a stronger response when another ligand binds at the orthosteric site. Here's a schematic to help show how different types of ligand affect receptor activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DxcHc.png&quot; alt=&quot;Dose Response Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, ligands have two important properties, affinity and efficacy. Affinity is how strongly a ligand binds to a receptor, and efficacy is how the ligand affects the receptor's activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we can dig too deep into agonists or antagonists, we should think about receptor theory. Receptors, like all proteins, have complex 3 dimensional structure, or conformation. This conformation is affected by solvent, pH, temperature, and other factors, but always has some fluctuation. Classical receptor theory used a lock and key model, where an inactive receptor bound a ligand and became active. However, this did not explain the base activity and activity of inverse agonists, so an equilibrium based model was developed. Here, the population of receptors has an ensemble of conformation states, some of these are active. When an agonist binds, it drives the equilibrium of the population toward the active state by stabilizing the active conformation. This diagram shows a simplified example of how drugs affect the equilibrium. Ri is inactive receptor, Ra is active. D is drug. Ra leads to some activity, while Ra-D leads to strong activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PcsxF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Affinity is relatively straightforward to measure becuase the same idea can be applied to almost any receptor-ligand system. Usually, a radiolabled version of the drug is added to a known amount of receptors in a binding assay. The goal is to determine a dissociation constant, Kd. Kd is used to quantify how well a ligand binds to a receptor, and smaller numbers are tighter binders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at this reaction: D + R &amp;lt;--&gt; DR with forward rate constant k1, reverse rate constant k2. ( how to get the right arrows here? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, k1[D][R] = k2[DR]&#xA;and $K_d =\frac{[D][R]}{[DR]}$ = $\frac{k_2}{k_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So incubate the radioligand and receptor until some DR complex is formed, then seperate the bound and unbound ligand so you can measure D and DR. The seperation method is important because if it's too slow the DR complex will revert to D and R, creating inaccurate data. But this is commonly done using dialysis, centrifugation, and vacuum filtration. Dialysis works well when dissociation rate is slow. You put receptor in one side and drug in both sides. It's slow, but it's theoritically most accurate becuase you can measure the true steady state by taking an equal volume of solution from each side of the membrane without affecting equilibrium. But you also need a large supply of receptor and drug.&#xA;Centrifugation works when you can immobilize the receptor on a large particle, usually you do this by using whole cells. You add an excess of of drug, then spin it down and take off the supernatent. Timing is still an issue and free drug can get caught in the pellet.&#xA;Vacuum filtration is usually considered the best because it can seperate the components very quickly, by pulling the solution through a filter and trapping the receptor on the surface. However this requires specialized and expensive equipment.&#xA;Once you measure bound and free drug at several concentrations, you can do a double reciprocal plot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring efficacy is different because you need to choose an activity to measure. Idealy it will be objective and easy to quantify. If the receptor is an enzyme, you can measure its enzyme activity. If it activates a gene transcription factor, you can use cells modified to express green fluorescent protien or luciferase when that transcription factor is activated. If it's an ion channel you can measure a cell's membrane potential. If you cannot measure anything else, you can use behavioral assays. There are several assays used to measure mouse activity and try to get at certain aspects of their brain and are often used to study depression and other mental illnesses. These are subjective. Efficacy is quantified by EC50, the concentration at which response is half of maximum response. Smaller EC50s are more efficacious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This field is much much larger than my answer. For more you should read this article : &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/15063082&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/15063082&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-08-15T00:46:50.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15277" PostTypeId="1" CreationDate="2014-08-15T03:19:26.390" Score="7" ViewCount="328" Body="&lt;p&gt;How much pressure will a 16 gram cartridge of $\ce{CO2}$ (carbon dioxide) provide in a bicycle tire? Remembering my high school chemistry (forty years ago) and googling, I came up with this answer:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure the volume of the tube (a torus): area of a cross section times the circumference, for a $30\:\mathrm{mm}$ tube on a $70\:\mathrm{cm}$ wheel, I got $1,554\:\mathrm{cm^3}$, or $1.554\:\mathrm{L}$. One mole is $24.47\:\mathrm{L}$. $1.554 / 24.47 = .06353$. One mole of $\ce{CO2}$ is $44\:\mathrm{g}$ (using atomic weights of carbon, 12, and oxygen, 16). $.06353 \cdot 44 = 2.795\:\mathrm{g}$. So the volume of the tube at one atmosphere is $2.795\:\mathrm{g}$ of $\ce{CO2}$. $16 / 2.795 = 5.7\:\mathrm{atm}$. One atmosphere is $14.7\:\mathrm{psi}$. $5.7 \cdot 14.7\:\mathrm{psi} = 84\:\mathrm{psi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="7486" LastEditorUserId="7448" LastEditDate="2015-10-30T17:18:50.277" LastActivityDate="2015-10-30T17:18:50.277" Title="Bicycle tube inflation with 16g CO2 cartridge" Tags="&lt;everyday-chemistry&gt;&lt;stoichiometry&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15278" PostTypeId="2" ParentId="15186" CreationDate="2014-08-15T03:48:30.350" Score="0" Body="&lt;p&gt;There's much detail to it beyond the scope and nit necessary but the final result is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Bases (or ions with negative charge) attack the sterically unhindered positions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Acid catalyzed ring opening happens from the position best to stabalize positive charge even though it doesn't form&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="1878" LastEditDate="2014-08-15T04:00:35.277" LastActivityDate="2014-08-15T04:00:35.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15279" PostTypeId="2" ParentId="6638" CreationDate="2014-08-15T03:51:21.760" Score="-2" Body="&lt;p&gt;Structure A is more stable. Reasons are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The O has a complete octet with 2 electrons per bond around it(based off of octet rule, not formal charge or oxidation state) + 2 lone pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The O is not positive. O prefers to be negative if charged at all because of its electronegativity being so high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Carbon is not so determinant on charge. It is fine being positive, negative, or neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Each carbon that isn't positively charged has its octet from C-C-C or O-C-C + 2 H's. &lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2014-08-15T03:51:21.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15280" PostTypeId="1" CreationDate="2014-08-15T04:21:47.460" Score="2" ViewCount="1488" Body="&lt;p&gt;O likes to be negative if charged at all. Most metal oxides are of the form $\ce{A^{x+} O^{2-}_{x/2}}$ (where $x$ is the magnitude of the positive charge and $\ce{A}$ is a metal.) Here, despite the $\ce{O}$ preferring covalent bonds it most often forms ionic bonds because of how metals behave and EN difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some compounds that have $\ce{O^+}$ in them. Hydronium is an example of that. When the O is positively charged the ions form a cation complex which itself is in a complex of ions and so on and so forth. This is why $\ce{H3O+}$ isn't really $\ce{H3O+}$ as individual ions but rather a complex formed from $\ce{H3O+}$ ions spreading the positive charge across all positively charged oxygen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the oxide anion despite being really reactive more stable than the $\ce{O^+}$ ion since in the oxide anion the $\ce{O}$ already has an octet and it is negatively charged which is what $\ce{O}$ prefers?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="485" LastEditDate="2014-08-15T06:27:19.780" LastActivityDate="2014-08-15T06:27:19.780" Title="Is Oxide Anion more stable than O^+" Tags="&lt;ions&gt;&lt;stability&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15281" PostTypeId="2" ParentId="15270" CreationDate="2014-08-15T04:24:25.643" Score="2" Body="&lt;p&gt;I don't see much hinderance[note that they all have been optimised to stable configurations]&#xA;Blue: nitrogen,White: Hydrogen,Grey: Carbon&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Side View:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXPZ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Top View:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h2TyY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-15T04:24:25.643" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15283" PostTypeId="2" ParentId="14327" CreationDate="2014-08-15T10:50:31.920" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$|Q|=|(ms+C)\Delta T|=(3\times4.18+2.71)\times1.941\;\mathrm{kJ}=29.6\;\mathrm J$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta H_n=\Delta H/n=-29.6/(1/46)\;\mathrm{kJ}=-1361\;\mathrm{kJ}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2018-12-10T18:02:42.940" LastActivityDate="2018-12-10T18:02:42.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15284" PostTypeId="2" ParentId="15154" CreationDate="2014-08-15T11:07:38.097" Score="1" Body="&lt;p&gt;Value seems to be $k_H^{\ominus}=4.3\times10^{-3}\;M/atm$ for Xenon in water.For it in conc. phosporic acid seems not to be available as probably no one has carried the experiment in this solvent since water is most common and used commonly.Maybe you would?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, A simple way to describe Henry’s law as a function of temperature is:&#xA;$$\LARGE k_H=k_H^{\ominus}\times e^{\left(\frac{-\Delta_{soln}H}R\left(\frac1T-\frac1{T^{\ominus}}\right)\right)}$$&#xA;where $\Delta_{soln}H=$enthalpy of solution. Here, the temperature dependence is:&#xA;$$\frac{-d\ln k_H}{d(1/T)}=\frac{\Delta_{soln}H}R$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Useful Links:&#xA;&lt;a href=&quot;http://www.henrys-law.org/henry.pdf&quot; rel=&quot;nofollow&quot;&gt;Table of constants&lt;/a&gt;, &lt;a href=&quot;http://webbook.nist.gov/chemistry&quot; rel=&quot;nofollow&quot;&gt;The NIST Chemistry WebBook&lt;/a&gt;, &lt;a href=&quot;http://www.arsusda.gov/rsml/ppdb2.html&quot; rel=&quot;nofollow&quot;&gt;The Pesticide Properties Database (PPD)&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-15T11:07:38.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15285" PostTypeId="2" ParentId="13932" CreationDate="2014-08-15T11:16:47.780" Score="4" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ciPQo.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the first phase of chlorophyll biosynthesis, the amino acid &lt;strong&gt;glutamic acid&lt;/strong&gt; is converted to &lt;strong&gt;5-aminolevulinic acid&lt;/strong&gt; &lt;strong&gt;(ALA)&lt;/strong&gt; . This reaction is unusual in that it involves a covalent intermediate in which the &lt;strong&gt;glutamic acid&lt;/strong&gt; is attached to a transfer &lt;strong&gt;RNA&lt;/strong&gt; molecule. This is one of a very small number of examples in biochemistry in which a &lt;strong&gt;tRNA&lt;/strong&gt; is utilized in a process other than protein synthesis. Two molecules of &lt;strong&gt;ALA&lt;/strong&gt; are then condensed to form &lt;strong&gt;porphobilinogen (PBG)&lt;/strong&gt;, which ultimately form the &lt;strong&gt;pyrrole&lt;/strong&gt; rings in chlorophyll. The next phase is the assembly of a &lt;strong&gt;porphyrin&lt;/strong&gt; structure from four molecules of &lt;strong&gt;PBG&lt;/strong&gt;. This phase consists of six distinct enzymatic steps, ending with the product &lt;strong&gt;protoporphyrin IX&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;All the biosynthesis steps up to this point are the same for the synthesis of both &lt;strong&gt;chlorophyll&lt;/strong&gt; and &lt;strong&gt;heme&lt;/strong&gt;. But here the pathway branches, and the fate of the molecule depends on which metal is inserted into the center of the &lt;strong&gt;porphyrin&lt;/strong&gt;. If &lt;strong&gt;magnesium&lt;/strong&gt; is inserted by an enzyme called &lt;strong&gt;magnesium chelatase&lt;/strong&gt;, then the additional steps needed to convert the molecule into &lt;strong&gt;chlorophyll&lt;/strong&gt; take place; if &lt;strong&gt;iron&lt;/strong&gt; is inserted, the species ultimately becomes heme.&lt;/p&gt;&#xA;&lt;p&gt;The next phase of the chlorophyll biosynthetic pathway is the formation of the fifth ring (ring E) by cyclization of one of the &lt;strong&gt;propionic acid&lt;/strong&gt; side chains to form &lt;strong&gt;protochlorophyllide&lt;/strong&gt;. The pathway involves the reduction of one of the double bonds in ring D, using &lt;strong&gt;NADPH&lt;/strong&gt;. This process is driven by light in angiosperms and is carried out by an enzyme called protochlorophyllide &lt;strong&gt;oxidoreductase (POR)&lt;/strong&gt;. Non-oxygen-evolving photosynthetic bacteria carry out this reaction without light, using a completely different set of enzymes. &lt;strong&gt;Cyanobacteria&lt;/strong&gt;, &lt;strong&gt;algae&lt;/strong&gt;, lower plants, and &lt;strong&gt;gymnosperms&lt;/strong&gt; contain both the light-dependent &lt;strong&gt;POR&lt;/strong&gt; pathway and the light-independent pathway. Seedlings of &lt;strong&gt;angiosperms&lt;/strong&gt; grown in complete darkness lack &lt;strong&gt;chlorophyll&lt;/strong&gt;, because the &lt;strong&gt;POR&lt;/strong&gt; enzyme requires light. These &lt;strong&gt;etiolated&lt;/strong&gt; plants very rapidly turn green when exposed to light. The final step in the &lt;strong&gt;chlorophyll&lt;/strong&gt; biosynthetic pathway is the attachment of the phytol tail, which is catalyzed by an enzyme called &lt;strong&gt;chlorophyll synthetase&lt;/strong&gt; (&lt;em&gt;Malkin&lt;/em&gt; and &lt;em&gt;Nyogi&lt;/em&gt; &lt;strong&gt;2000&lt;/strong&gt;).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sub&gt;Source: &lt;strong&gt;&lt;a href=&quot;http://5e.plantphys.net/article.php?ch=0&amp;amp;id=76&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plant Physiology Online: Chlorophyll Biosynthesis&lt;/a&gt;&lt;/strong&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-08-15T11:16:47.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15286" PostTypeId="1" CreationDate="2014-08-15T11:28:44.717" Score="1" ViewCount="2896" Body="&lt;p&gt;From the attached picture (MO diagram of cyclopentadienyl anion) I can figure out where the nodal planes are, but why are they exactly at these positions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z7aeG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this nodal plane invalid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ao2L0.gif&quot; alt=&quot;N&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2014-08-15T20:27:54.743" LastActivityDate="2014-08-15T20:27:54.743" Title="Position of nodal planes in molecular orbitals" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15287" PostTypeId="2" ParentId="14037" CreationDate="2014-08-15T11:46:34.713" Score="1" Body="&lt;p&gt;The work would be area of triangle, which is no way related to the hypotenuse (other than $\small H^2=P_0^2+V_0^2$)$$W=\frac12P_0V_0$$&#xA;Heat is :&#xA;$$Q_{in:A-C}=(C_V+R/2)\left(3\frac{P_0V_0}{R}\right)+C_V\left(2\frac{P_0V_0}{R}\right)$$&#xA;Efficiency is :&#xA;$$\eta=\frac{W}{Q_1}=\frac{\frac12P_0V_0}{(C_V+R/2)\left(3\frac{P_0V_0}{nR}\right)+C_V\left(2\frac{P_0V_0}{R}\right)}=\frac12.\frac{R}{3C_V+3R/2+2C_V}$$&#xA;$$\eta=\frac{R}{10C_V+3R}=\frac{R}{5fR+3R}=\frac1{5f+3}$$&#xA;which for  a monoatomic,diatomic and polyatomic(linear/non-linear) are:&#xA;$$\begin{array}{|c|c|}\text{Monoatomic}&amp;amp;1/18\\&#xA;\text{Diatomic}&amp;amp;1/28\\&#xA;\text{Polyatomic-Linear}&amp;amp;1/28\\&#xA;\text{Polyatomic-Non-Linear}&amp;amp;1/33\\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-08-15T11:46:34.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15289" PostTypeId="1" CreationDate="2014-08-15T12:59:09.970" Score="1" ViewCount="12164" Body="&lt;p&gt;I have read &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2879/what-is-the-definition-of-of-compound-mixture-element-and-molecule&quot;&gt;the Chem.SE question&lt;/a&gt; about the difference between 'compound', 'mixture', 'element' and 'molecule', but I would like to know the precise definitions for more terms (see below). From the article, these are the definitions:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Molecules are a group of atoms covalently bonded to each other.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compounds are made up of two or more elements. &lt;strong&gt;Compound is often used interchangeably with the word substance.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixtures are collectives of different molecules or atoms.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How are compounds substances? Isn't 'substances' a general term for any chemical &lt;em&gt;thing&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;p&gt;These are the terms that I would like definitions for:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;chemical species&lt;/li&gt;&#xA;&lt;li&gt;substance&lt;/li&gt;&#xA;&lt;li&gt;particle&lt;/li&gt;&#xA;&lt;li&gt;atom/isotope&lt;/li&gt;&#xA;&lt;li&gt;salt&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I am rather confused about these terms, because my tutor uses particles and molecules (and elements) as if they were the same thing. As for the rest, I want to know how to distinguish between general (such as chemical species, substance) and specific terms (such as particle, atom, isotope, salt).&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-04-18T15:16:35.683" Title="What is the difference between 'particles', 'molecules' and 'chemical species'?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15290" PostTypeId="1" AcceptedAnswerId="15291" CreationDate="2014-08-15T12:59:31.333" Score="1" ViewCount="16132" Body="&lt;p&gt;I'm supposed to balance the following equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Potassium chlorate (heated)} \rightarrow \text{Potassium chloride + Oxygen}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But I don't understand what is meant by the label &amp;quot;heated&amp;quot; in front of the first element. Do I ignore it, or am I supposed to add another element to the reactants of the equation?&lt;/p&gt;&#xA;" OwnerUserId="7490" LastEditorUserId="69830" LastEditDate="2021-05-19T06:05:07.633" LastActivityDate="2021-05-19T06:05:07.633" Title="What does (heated) mean in a chemical equation?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15291" PostTypeId="2" ParentId="15290" CreationDate="2014-08-15T13:22:41.633" Score="3" Body="&lt;p&gt;In terms of balancing the equation, the fact that heat must be supplied can be ignored.  It is just providing information about the conditions necessary to run the reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-15T13:22:41.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15293" PostTypeId="2" ParentId="15286" CreationDate="2014-08-15T13:27:39.970" Score="1" Body="&lt;p&gt;&lt;strong&gt;First a descriptive explanation:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Molecule is symmetric. Thus the resulting mocular orbitals are symmetric too. If you are constructing different MOs where the nodal planes are different than in the picture above they wouldnt be symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematical Explanation:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These nodal planes result from a LCAO-MO consideration with hückel approximation. You have to form the secular determinant for such a problem at which the coulomb integrals are Alpha and the resonance is only dissimilar zero when the two considered atoms are neighbors. After solving this hückel-matrix you can determinate the coefficents of the several molecular orbitals. Doing that you will see that the nodal planes exist as you drew them above.&lt;br&gt;&#xA;(More Information about the Hückel-Method: &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7491" LastActivityDate="2014-08-15T13:27:39.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15294" PostTypeId="2" ParentId="7695" CreationDate="2014-08-15T15:00:32.227" Score="5" Body="&lt;p&gt;The reason is the same as in the case of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2795/why-does-radium-have-a-higher-first-ionisation-energy-than-barium/&quot;&gt;barium and radium&lt;/a&gt;, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-08-15T15:00:32.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15295" PostTypeId="1" CreationDate="2014-08-15T15:16:56.900" Score="0" ViewCount="712" Body="&lt;p&gt;What would cause vapour release from elemental mercury? Would it be pressure, friction... could one stop the vapour release through cooling the mercury?&lt;/p&gt;&#xA;" OwnerUserId="7493" LastEditorUserId="5084" LastEditDate="2014-08-15T15:44:29.493" LastActivityDate="2019-09-08T21:39:05.730" Title="Mercury vapor release from amalgam fillings" Tags="&lt;reaction-mechanism&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2019-09-10T12:41:21.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="15296" PostTypeId="1" CreationDate="2014-08-15T15:24:16.697" Score="1" ViewCount="43" Body="&lt;p&gt;I have conducted many searches on this subject with regards to agents which cross the blood brain barrier. I have read about lipoic acid. Is anybody on here aware of any studies or findings which show effective mercury binding and transportation from the brain?&lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2014-08-15T15:24:16.697" Title="removing mercury from the brain" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15297" PostTypeId="2" ParentId="15295" CreationDate="2014-08-15T15:42:15.867" Score="2" Body="&lt;p&gt;All solids have a certain vapor pressure, and if this vapor pressure exceeds the atmospheric pressure, then evaporation of the solid occurs. Some of the solid turns into gas. More specifically, it's the uppermost layer of the solid that evaporates. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would cause vapour release from elemental mercury? Would it be&#xA;  pressure, friction...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Friction could definitely increase the kinetic energy of the mercury particles and increase the solid's vapor pressure. An increase in pressure could also increase the temperature and therefore kinetic energy of the solid's constituent mercury atoms. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;could one stop the vapour release through cooling the mercury?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that would decrease the kinetic energy of the particles and lower the solid's vapor pressure. Here's a chart relating temperature (&lt;em&gt;in Kelvins&lt;/em&gt;) to vapor pressure. At physiological temperatures, we're still talking about an extremely low vapor pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3fCb8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.physics.rutgers.edu/~eandrei/389/NISTIR.6643.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.physics.rutgers.edu/~eandrei/389/NISTIR.6643.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-15T15:56:42.237" LastActivityDate="2014-08-15T15:56:42.237" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15298" PostTypeId="2" ParentId="15046" CreationDate="2014-08-15T16:00:41.907" Score="2" Body="&lt;p&gt;You have correctly identified zinc and iron as the anodes, since they are both less noble than copper. In the corrosion reaction, they are both oxidized and dissolved in the aqueous gel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{Zn -&amp;gt;&amp;amp; Zn^2+ + 2e-} \\&#xA;\ce{Fe -&amp;gt;&amp;amp; Fe^2+ + 2e-} \\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons flow to the copper cathode, where oxygen, which is dissolved in the liquid medium, is reduced to hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 +2H2O + 4e- -&amp;gt; 4OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, the pH of the solution around the copper cathode rises, and the phenolphthalein indicator changes its color from colorless to violet-red in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2LMZM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe^2+}$ reacts with $\ce{K3[Fe(CN6)]}$ to form a deep blue &lt;a href=&quot;http://en.wikipedia.org/wiki/Prussian_blue&quot; rel=&quot;nofollow noreferrer&quot;&gt;iron(II,III) hexacyanoferrate(II,III)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K+ + Fe^2+ + [Fe(CN6)]^3- -&amp;gt; KFe[Fe(CN6)]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the solution near the iron surface turns blue. Zinc does not react with $\ce{K3[Fe(CN6)]}$ and the solution around the zinc anode remains colorless (&lt;a href=&quot;http://www.corrosion-doctors.org/Corrosion-History/Action.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-08-15T16:24:16.383" LastActivityDate="2014-08-15T16:24:16.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15299" PostTypeId="1" CreationDate="2014-08-15T16:07:15.053" Score="2" ViewCount="8785" Body="&lt;p&gt;How do I solve for oxidation state in something like CH3-O-F? The fluorine is more electronegative than the oxygen which is more electronegative than the carbon which is more electronegative than the hydrogen. In this case would everything except the fluorine be oxidized?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="9961" LastEditDate="2016-11-29T00:47:19.843" LastActivityDate="2017-10-29T08:33:44.217" Title="How to find out oxidation state when compound has multiple electronegativities" Tags="&lt;molecules&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15300" PostTypeId="2" ParentId="15295" CreationDate="2014-08-15T16:08:04.233" Score="2" Body="&lt;p&gt;There will always be some molecules with enough kinetic energy to escape (evaporate) from the surface of a solid or liquid.  For each substance and at each temperature there is a specific &quot;vapor pressure&quot; that when reached, the solid or liquid and the vapor are said to be in equilibrium.  Again, there are always molecules escaping from a solid or liquid, but once there are enough of the these molecules in the vapor above the solid or liquid, and the equilibrium vapor pressure has been attained, then the number of molecules returning from the vapor to the solid or liquid will equal the number of molecules escaping from the solid or liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with mercury vapor in an open environment like the mouth is that the air in the mouth is continually exchanged with fresh air.  Therefor, whatever concentration of mercury vapor has built up will, to some degree, be decreased by air exchange.  Hence mercury will, more or less, continuously evaporate in such an environment. Cooling will reduce the equilibrium vapor pressure and consequently reduce the rate of mercury evaporation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-15T16:08:04.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15301" PostTypeId="1" AcceptedAnswerId="15308" CreationDate="2014-08-15T16:12:10.673" Score="0" ViewCount="2974" Body="&lt;p&gt;$\ce{HCO_2^-}$ exhibits delocalized bonding and $\ce{CO2}$ does not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am confused about is, that both of them have resonance structures. Therefore, why doesn't $\ce{CO2}$ exhibit delocalized bonding?&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="4945" LastEditDate="2014-08-15T16:30:05.190" LastActivityDate="2014-08-15T17:46:36.043" Title="Why does HCO₂⁻ exhibit delocalized bonding while CO₂ does not?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15303" PostTypeId="2" ParentId="15299" CreationDate="2014-08-15T16:23:49.143" Score="5" Body="&lt;p&gt;You can assign oxidation states with a method similar to assigning formal charges.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw the Lewis structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assign each atom all of its lone pair electrons.&#xA;For bonding electrons assign all of the electrons to the more electronegative element.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each atom perform the following calculation:&#xA;number of electrons in the atom in its elemental state - assigned electrons = oxidation state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In your example, the oxidation number of each $\ce{H}$ is +1 and of $\ce{F}$ is -1.  The oxidation number of $\ce{C}$ is -2 and of $\ce{O}$ is 0.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="46208" LastEditDate="2017-10-29T08:33:44.217" LastActivityDate="2017-10-29T08:33:44.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15304" PostTypeId="1" CreationDate="2014-08-15T16:31:43.310" Score="2" ViewCount="79" Body="&lt;p&gt;The two substances I have come across are cilantro and lipoic acid. How effective would either of them be? Are there any other agents known which produce the desired results? &lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2014-08-15T16:31:43.310" Title="Mercury binding agents which cross the blood brain barrier" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15305" PostTypeId="2" ParentId="15301" CreationDate="2014-08-15T16:46:38.927" Score="-1" Body="&lt;p&gt;$\ce{CO_2}$ does not show delocalization because the p-orbitals of the carbon atom are orthogonal. They can't interact. The hydroxymethylidene ion shows delocalization because the lone pair on the carbon atom is &quot;upgraded&quot; to a full p-orbital, just as (one) the lone pair on each oxygen because that would stabilize the negative charge. Now you have a system of three p-orbitals linked together. There is no need for $\ce{CO_2}$ to stabilize anything, so the energy required to &quot;upgrade&quot; to a full p orbital (on the oxygens) is too high of a barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I'm wrong (then I'll remove the answer). This is a very simple analysis, there could be things in MO theory that I'm not aware about.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-08-15T17:23:57.447" LastActivityDate="2014-08-15T17:23:57.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15306" PostTypeId="2" ParentId="15299" CreationDate="2014-08-15T16:47:09.833" Score="1" Body="&lt;p&gt;To successfully assign oxidation state you should define what oxidation is. Oxidation is defined as the loss of electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, along with the fact that oxidation state is represented as a charge - i.e. something may have a &quot;+2&quot; oxidation state or a &quot;-8&quot; oxidation state, it follows that oxidation state is the charge that an atom possesses &lt;em&gt;if&lt;/em&gt; all bonding were ionic. &lt;em&gt;Ionic&lt;/em&gt; interactions involve the taking of electrons; &lt;em&gt;covalent&lt;/em&gt; bonds involve the sharing of electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what &quot;takes&quot; the electrons? Clearly, the element which can best stabilize the electrons. This would be the most electronegative element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would start by drawing out the Lewis structure of $\ce{CH3OF}$. Assign electrons to element based on relative electronegativities, and then think about the charges. I.e. if an element normally has 7 electrons in its valence (such as fluorine) - but is assigned 8 electrons, what do you think the charge on the fluorine will be? It has &lt;em&gt;surfeit&lt;/em&gt; of electrons. This should suggest to you that the fluorine has a negative 1 oxidation state. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-15T16:47:09.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15307" PostTypeId="1" AcceptedAnswerId="15311" CreationDate="2014-08-15T17:26:49.887" Score="2" ViewCount="102" Body="&lt;p&gt;I am writing a small simulator for situations where water is the medium of all reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick googling confirmed that there are water models available in computer readable formats. Being a beginner in molecular chemistry, I was hoping someone can direct me to already mature libraries which I can use to read water models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following seems to be good candidates:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://msl-libraries.org/index.php/Main_Page&quot; rel=&quot;nofollow&quot;&gt;http://msl-libraries.org/index.php/Main_Page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ball-docs.bioinf.uni-sb.de&quot; rel=&quot;nofollow&quot;&gt;http://ball-docs.bioinf.uni-sb.de&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7494" LastActivityDate="2019-01-01T13:00:04.410" Title="Using water model in a computer program" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15308" PostTypeId="2" ParentId="15301" CreationDate="2014-08-15T17:46:36.043" Score="1" Body="&lt;p&gt;Here is an orbital diagram for $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/clUBV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be &quot;orthogonal&quot; to one another.  This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$).  Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.  &lt;img src=&quot;https://i.stack.imgur.com/0gboC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, we can restate the drawings from above in terms of resonance structures.  In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$.  The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hufEl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question).  Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$.  But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array.  This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon.  Just like the &quot;particle in a box&quot; problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron.  So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion.  Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-15T17:46:36.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15309" PostTypeId="1" AcceptedAnswerId="15384" CreationDate="2014-08-15T18:28:34.623" Score="7" ViewCount="1532" Body="&lt;p&gt;From &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clayden et al.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, iodine is expensive and a way round that problem is to use a&#xA;  catalytic amount of iodide. The next phosphonium salt is formed slowly&#xA;  from benzyl bromide but the addition of a small amount of LiI speeds&#xA;  up the reaction considerably.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QSxYd.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The iodide reacts as a better nucleophile than Ph3P and then as a&#xA;  better leaving group than Br–. Each iodide ion goes round and round&#xA;  many times as a nucleophilic catalyst.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H8d5z.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a private conversation with &lt;a href=&quot;https://chemistry.stackexchange.com/users/4231/ron&quot;&gt;ron&lt;/a&gt; I asked him how this catalyzing effect worked, he said that question would be more suited here and I agreed with him. It seems a bit strange at first that a strong nucleophile can also be a good leaving group (he really wants that nucleus, right, why would he leave?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the explanation I came up with (probably wrong, but it is always better to first think self): Iodine is a stronger nucleophile here than bromine because of its higher energy HOMO, which lies closer to the energy of the $\ce{C-X}$ $\sigma^{*}$ (here $\ce{X}$ = $\ce{Br}$) LUMO. So it will react more rapidly than bromine, resulting in an equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now iodine is also a good leaving group, firstly because the $\ce{C-I}$ bond is higher is energy than the $\ce{C-Br}$ bond (less stable bond). And secondly (quoting Ron from &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13609/hydrogen-halide-acidity&quot;&gt;another answer&lt;/a&gt;. Well explained, no need to rewrite it again): the larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. So $\ce{I^{-}}$ will be more stable, making it easier for iodine to leave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought was that iodine can easily knock of bromide but not remain in that position for very long (therefore the equilibrium). Now, if $\ce{PPh_3}$ happens to meet the benzyl iodide it is more eager to react with it than with the benzyl bromide (lower energy $\ce{C-X}$ $\sigma^{*}$). The resulting phosphonium salt is either to stable to be attacked by iodine again or precipitates out (or both?), so that benzyl bromide continues to be transformed into benzyl iodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-08-18T11:03:09.290" LastActivityDate="2014-08-18T22:10:25.830" Title="Catalytic iodine in synthesis of phosphonium salts" Tags="&lt;reaction-mechanism&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15310" PostTypeId="2" ParentId="12326" CreationDate="2014-08-15T19:18:37.643" Score="5" Body="&lt;p&gt;As long as your experimental setup did not contain any leakages or impurities through which air or moisture could have gotten into the mixture, the reaction should have worked fine under these conditions. The color of the hexaamminsodium electride solution is actually deep blue and intense, so at higher concentrations it can appear black at first sight (see picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gI6Zl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-15T19:18:37.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15311" PostTypeId="2" ParentId="15307" CreationDate="2014-08-15T19:24:31.133" Score="2" Body="&lt;p&gt;Libraries which support Water model Simulation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ambermd.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Assisted Model Building with Energy Refinement(AMBER)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.charmm.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry at HARvard Macromolecular Mechanics(CHARMM)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dirac.cnrs-orleans.fr/MMTK/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Molecular Modelling Toolkit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cool Site for Water Model Generation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/models.html#back1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water Models&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOO1i.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image Source: &lt;a href=&quot;http://biomodel.uah.es/en/water/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://biomodel.uah.es/en/water/index.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7120" LastEditorUserId="15751" LastEditDate="2019-01-01T13:00:04.410" LastActivityDate="2019-01-01T13:00:04.410" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15312" PostTypeId="1" AcceptedAnswerId="15318" CreationDate="2014-08-15T20:30:56.110" Score="5" ViewCount="1611" Body="&lt;p&gt;$\ce{NO}$ can't dimerize and that is explained because the free electron is in an antibonding orbital. The same is true for $\ce{NO2}$, but it can dimerize, how?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5084" LastEditDate="2014-08-17T14:28:50.777" LastActivityDate="2014-08-17T14:28:50.777" Title="Nitrogen dioxide dimerization" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15313" PostTypeId="2" ParentId="14786" CreationDate="2014-08-15T20:38:11.467" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hund&amp;#39;s_rules#Rule_1&quot; rel=&quot;nofollow&quot;&gt;Hund's rule of maximum multiplicity&lt;/a&gt; states that for a given electron configuration, the term with the lowest energy has a maximum multiplicity $2S+1$. $S$ is the total &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum_operator&quot; rel=&quot;nofollow&quot;&gt;spin angular momentum&lt;/a&gt; of all electrons in the partially-filled subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the application of the rule, we consider the 2 electrons in the $p$ subshell and add up their spin quantum numbers to obtain $S$ and the multiplicity. For the ground state configuration, with both electrons &quot;spin-up&quot;, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=+\frac{1}{2}+(+\frac{1}{2})=1$$&#xA;$$2S+1=3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your proposed configuration with one $p$ electron &quot;spin-up&quot; and the other one &quot;spin down&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S=+\frac{1}{2}+(-\frac{1}{2})=0$$&#xA;$$2S+1=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the multiplicity value is smaller, this configuration is not the one which is lowest in energy, and therefore cannot be regarded as the ground state.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-15T20:38:11.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15314" PostTypeId="2" ParentId="15312" CreationDate="2014-08-15T21:01:51.833" Score="0" Body="&lt;p&gt;Are you sure that nitric oxide can't dimerize? It has a rather high bond order - 2.5 - and we can see that from molecular orbital theory. Therefore even if its free electron is in an antibonding orbital, that wouldn't decrease the bond order to 0 or below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitric oxide dimer has actually been extensively studied too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00381a052&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ja00381a052&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-15T21:01:51.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15315" PostTypeId="1" AcceptedAnswerId="15317" CreationDate="2014-08-15T21:12:07.747" Score="1" ViewCount="7273" Body="&lt;p&gt;This is the molecular orbital diagram for nitric oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FW8T1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know that nitric oxide can dimerize. However, why does nitric oxide dimerize as to form an $\ce{N-N}$ bond? Why not an $\ce{O-O}$ bond? Wouldn't the lone electron be more likely &quot;possessed&quot; by the oxygen since oxygen is more electronegative (if we were to step away from molecular orbital theory for a second). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a Coulombic effect? Oxygen, by nature of being more electronegative, attracts more electron density to itself. This in turn would weaken any hypothetical $\ce{O-O}$ bond due to electron-electron repulsions. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-08-17T05:52:09.857" LastActivityDate="2014-08-17T13:48:14.103" Title="Nitric Oxide Dimerization" Tags="&lt;bond&gt;&lt;energy&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15316" PostTypeId="2" ParentId="14786" CreationDate="2014-08-15T21:14:20.373" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;nofollow&quot;&gt;This is a weird effect resulted from Pauli principle.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naive oversimplificated version: since no two electrons with same spin may occupy the same place, if two electrons have opposite spin, a part of their time they spend very close to each other, and this results in increased electrostatic repulsion term in energy of the state, while in case two electrons have the same spin, they are always away of each other, so the global energy is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this effect to manifest, a a significant overlapping of orbitals occupied by electrons in consideration must occur. This usually is observed on atomic or molecular level, and sometimes on &lt;a href=&quot;http://en.wikipedia.org/wiki/Ferromagnetism#Explanation&quot; rel=&quot;nofollow&quot;&gt;larger scale&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-15T21:21:14.267" LastActivityDate="2014-08-15T21:21:14.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15317" PostTypeId="2" ParentId="15315" CreationDate="2014-08-15T22:04:41.747" Score="3" Body="&lt;p&gt;Something is not correct with your MO diagram.  Here is a correct MO diagram for NO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrHQ8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in this diagram the oxygen atomic orbitals are lower in energy than the nitrogen AOs.  This makes sense since the AOs of the more electronegative element should always be at lower energy than the AOs of the more electropositive element (in your diagram the relative energies of the oxygen and nitrogen AOs are somehow reversed).  The energetic ordering of the starting AOs is important as we'll see in a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer in energy a MO is to an AO from which it was formed, the more character of that AO it shows.  So let's examine the nature of the single electron in the $\ce{\pi_{2p}^{\ast}}$ orbital since this electron will be involved in forming the new bond when $\ce{NO}$ dimerizes.  This MO is formed from both the nitrogen and oxygen p atomic orbitals, but it is closer in energy to the nitrogen p AO.  Therefore, we would expect this electron to have more nitrogen and less oxygen character (in other words it will reside more around nitrogen than oxygen).  Because this electron has greater density on nitrogen, dimerization to form an N-N bond is more likely than dimerization to form an O-O bond.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-17T13:48:14.103" LastActivityDate="2014-08-17T13:48:14.103" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="15318" PostTypeId="2" ParentId="15312" CreationDate="2014-08-15T22:25:49.167" Score="3" Body="&lt;p&gt;Having an electron in an antibonding orbital means that the electron exists in a high energy orbital.  If anything, this means the electron will be &quot;motivated&quot; to react and form a bond, particularly if bond formation lowers the energy of the electron.  An electron in an antibonding orbital can form a bonding (as opposed to antibonding) bond with positive overlap.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-15T22:25:49.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15319" PostTypeId="1" AcceptedAnswerId="15331" CreationDate="2014-08-15T22:33:12.027" Score="0" ViewCount="204" Body="&lt;p&gt;Consider the following unbalanced redox equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO_2^- +e^- -&amp;gt; NO3^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This occurs in &lt;em&gt;acidic&lt;/em&gt; media. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor would note that an negative two oxidation state oxygen &quot;disappears&quot; from left to right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxide anions are strong bases, and since this process occurs in acidic media, and since an oxide anion has &quot;disappeared,&quot; it must have been consumed by the strongest acid present - hydronium ion. Specifically, two equivalents of hydronium ion. These two equivalents of hydronium ion react with oxide anion to produce three waters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we balance the equation as such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO_2^- +2e^- +2H_3O^+ -&amp;gt; NO3^- + 3H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the &lt;em&gt;validity&lt;/em&gt; of this method, given that redox reactions likely don't involve an oxide anion intermediate? I can find no mention of any $\ce{O^2-}$ intermediates in redox literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, how can this method still yield a redox equation that is valid in terms of mass and charge? Is there some likely intermediate that is close enough in character to the oxide anion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-18T12:54:35.620" Title="Oxide Anions in Redox" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15320" PostTypeId="1" AcceptedAnswerId="15335" CreationDate="2014-08-15T22:44:57.563" Score="3" ViewCount="1091" Body="&lt;p&gt;Why are peroxide linkages generally weak? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it due to Coulombic repulsions? I can't see how placing two oxygens with high degrees of partial negative character next to each other is a good thing in terms of bond strength. I suppose a way to test this hypothesis is to compare the relative stabilities of HOOH and FOOF, as the hydrogens are electron donating while the fluorines are electron withdrawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does this have to do with promotion of a bonding electron to an antibonding orbital? I know that peroxide linkages are easy to break with application of light/heat ... this suggests to me that we can easily promote an electron to anti-bonding status and weaken the O-O link. Still, this does not answer why it might be easy to promote an electron to antibonding status. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2014-08-16T17:03:33.527" LastActivityDate="2018-02-13T05:40:54.387" Title="Peroxide Linkages" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15321" PostTypeId="2" ParentId="15250" CreationDate="2014-08-16T00:02:29.363" Score="5" Body="&lt;p&gt;As the other answers have mentioned, a simple A-B bond dissociation reaction is significantly endothermic and does not happen spontaneously. A single molecule may break apart and its pieces may be scattered if energy is introduced in the molecule (for example, a collision with a very energetic target or the absorption of a photon may trigger strong vibrational excitations which end up breaking the molecule apart). It is, however, also possible for an unstable molecule to spontaneously rearrange into a more stable isomer, and part of the released energy may cause the rearranged molecule to shake apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event it's a case of causing a vibration to stretch atoms strongly enough that the bond between them snaps and the pieces fling away, not unlike a rubber string being pulled until it tears. Perhaps &lt;a href=&quot;https://www.youtube.com/watch?v=3RqEIr8NtMI#t=542s&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; may help clarify what's going on, though it doesn't specifically show reactions occurring. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: I somehow had completely misread the question as $\ce{A +B -&amp;gt; AB}$ instead of $\ce{AB -&amp;gt; A +B}$, so I apologize for that. My previous answer isn't actually an answer, but I figure I might as well leave the information below in case anyone is interested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, most often the chemical potential energy released when two molecules react exothermically turns into vibrational potential energy in the bonds of the product molecule/entity for a small amount of time, and this vibrational energy is quickly dispersed into other molecules by collisions (timescale of a few picoseconds), turning mostly into rotational and translational kinetic energy dispersed over several molecules, as vibrational excitations are &quot;frozen out&quot; at temperatures close to ambient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two molecules meet in an absolute vacuum and can react exothermically but have nowhere to dump the energy, then the molecules actually have a large chance of not reacting at all, even if it's extremely spontaneous! The only way they can meet and stay bound is if, during the short time they are in proximity, a low-probability event may occur where the reaction transition state emits a photon carrying away most of the releasable chemical potential energy, otherwise the molecules will just fly apart again. This is a very important consideration in astrochemistry. You can read more about this specific topic in &lt;a href=&quot;http://www.crcnetbase.com/doi/abs/10.1201/9780585368115.ch3&quot; rel=&quot;nofollow&quot;&gt;&quot;The Physics of the Interstellar Medium&quot;, 2nd edition, 1997 J.E. Dyson and D.A. Williams (IoP), chapter 3&lt;/a&gt;, section 3.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I should clarify that the second paragraph is only really valid for reactions of the type $\ce{A +B-&amp;gt;C}$. A reaction of the type $\ce{A +B-&amp;gt;C +D}$ can happen more easily as one of the species being formed can also act as a dump for excess energy.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2014-08-16T02:56:19.117" LastActivityDate="2014-08-16T02:56:19.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15322" PostTypeId="1" AcceptedAnswerId="15323" CreationDate="2014-08-16T01:47:49.283" Score="4" ViewCount="9521" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBTMU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5y0eO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this mechanism is false. &#xA;The protonated ether would in my opinion give a proton to methanol rather than $\ce{H2SO4}$ because the protonated $\ce{CH3OH}$ is more stable than $\ce{H2SO4}$.&#xA;Can someone confirm my thought?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="5017" LastEditDate="2015-04-30T18:27:54.223" LastActivityDate="2015-04-30T18:27:54.223" Title="Acid catalysed mechanism of the addition of alcohols to alkenes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;erratum&gt;" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="15323" PostTypeId="2" ParentId="15322" CreationDate="2014-08-16T01:49:11.233" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The protonated ether would in my opinion give a proton to methanol&#xA;  rather than H2SO4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think sulfuric acid is gaining any protons here. You mean hydrogen sulfate ion, correct? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone confirm my thought?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mechanism is acid-&lt;em&gt;catalyzed&lt;/em&gt; dehydration, so the acid must be regenerated. Catalysts are by definition not consumed over the course of a reaction. Now, whether the acid is &lt;em&gt;actually&lt;/em&gt; sulfuric acid is debatable, and in my mind, unlikely, given the leveling effect and that the solvent is methanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leveling effect states that the strongest acid that can exist in high concentration in a particular solvent is the conjugate acid of the solvent. The conjugate acid of methanol has a pKa of ~ -2.5, while the pKa of sulfuric acid is ~ -9. Thus, sulfuric acid can't exist in high concentration in methanol. Sulfuric acid would immediately protonate the strongest base in the system. Therefore, it appears to me that the &lt;em&gt;acting&lt;/em&gt; acid is &lt;em&gt;not&lt;/em&gt; sulfuric acid. Perhaps it is protonated methanol ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps it is the case that sulfuric acid reacts directly with the alkene. Whether sulfuric acid would be regenerated in this particular solvent, however, is questionable at best, given sulfuric acid's extreme thermodynamic &lt;em&gt;instability&lt;/em&gt; in methanol. If it were regenerated, then the sulfuric acid would again protonate the strongest base in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Of course, thermodynamic considerations are important, but so are kinetic considerations.&lt;/strong&gt; Do remember that there is likely &lt;em&gt;a lot&lt;/em&gt; more solvent than there is sulfuric acid and substrate (the alkene). So it seems most likely that the &lt;em&gt;acting&lt;/em&gt; acid is protonated methanol - it is a lot more likely for sulfuric acid to bump into a methanol molecule rather than an alkene molecule. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-19T17:28:16.630" LastActivityDate="2014-08-19T17:28:16.630" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="15324" PostTypeId="1" CreationDate="2014-08-16T03:44:40.227" Score="1" ViewCount="79" Body="&lt;p&gt;I have a 420-liter (14.9 cubic ft) helium cylinder and a pressure regulator with a gauge supporting up to 160 psi. My requirement is to ensure a flow of 10 to 15 liters / min. How do I determine the pressure setting on my pressure regulator to fulfill my flow rate requirement?&lt;/p&gt;&#xA;" OwnerUserId="7498" LastEditorUserId="7498" LastEditDate="2014-08-16T03:58:21.473" LastActivityDate="2014-08-16T17:07:01.523" Title="How to determine the pressure to output a gas to achieve a particular gas flow rate?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15325" PostTypeId="1" AcceptedAnswerId="15327" CreationDate="2014-08-16T05:10:14.153" Score="2" ViewCount="1256" Body="&lt;p&gt;I need some help. From what I understand, elementary reactions are reactions which only has one transition state. Another definition that I have come across is that elementary reactions obey the rate law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that $S_N1$ are NOT elementary reactions because they proceed through more than one transition state. However I am not so sure about the other definition (the rate law one). Could someone explain to me what exactly an elementary reaction (or a one step reaction) is? And can someone explain to me if an $S_N1$ reaction is an elementary reaction or not. &lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="7448" LastEditDate="2015-10-09T15:47:30.950" LastActivityDate="2015-10-09T15:47:30.950" Title="Is an SN1 reaction an elementary reaction/one step reaction?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15326" PostTypeId="2" ParentId="15320" CreationDate="2014-08-16T05:49:56.867" Score="2" Body="&lt;p&gt;The effect is not limited to peroxide linkages. Let's look at energy of single bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|} \hline&#xA;%header&#xA;\text{Bond} &amp;amp;&#xA;\text{Energy}~/ \pu{(kJ/mol)} &amp;amp;&#xA;\text{Bond} &amp;amp;&#xA;\text{Energy}~/ \pu{(kJ/mol)} \\&#xA;\hline&#xA;\ce{C-C} &amp;amp; 347 &amp;amp; \ce{Si-Si} &amp;amp; 340 \\ \hline&#xA;\ce{N-N} &amp;amp; 160 &amp;amp; \ce{P-P} &amp;amp; 201 \\ \hline&#xA;\ce{O-O} &amp;amp; 146 &amp;amp; \ce{S-S} &amp;amp; 206 \\ \hline&#xA;\ce{F-F} &amp;amp; 154 &amp;amp; \ce{Cl-Cl} &amp;amp; 240 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can clearly see, that the effect manifests between $\ce{C/Si}$ and $\ce{N/P}$, strengthens at $\ce{O}$ and somewhat reduces at $\ce{F/Cl}$. How $\ce{C-C}$ and $\ce{N-N}$ bonds differ? The answer is simple: in $\ce{N-N}$ bonds both atoms have unbound electron pairs. So, the reason for the effect is interaction of electron pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rationalization for how the interaction affects the bond energy is more open to debates. It is well known, that in VSEPR electron pairs take more place than atoms, so it may be said to be purely steric. It is known from MO theory that antibonding orbitals unbond a bit more than bonding bond, so it may be attributed for details of MO structure. Finally, according to Pauli principle, electrons cannot be in same 'place', so electrons with same spin in two electron pairs are repelled from each other, forcing them from each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How this effect is partially compensated in $\ce{F-F}$ and $\ce{Cl-Cl}$ bonds is more interesting, I would say that it is because of their smaller orbitals, resulting in less effective &amp;pi;-interactions and stronger p-p &amp;sigma; bonding&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2018-02-13T05:40:54.387" LastActivityDate="2018-02-13T05:40:54.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15327" PostTypeId="2" ParentId="15325" CreationDate="2014-08-16T06:05:56.020" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://goldbook.iupac.org/E02035.html&quot;&gt;http://goldbook.iupac.org/E02035.html&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So you're right that elementary reactions occur in a single step/have a single transition state. I'm not sure what you mean by &quot;obey the rate law&quot;, but necessarily, a reaction that has only a single transition state is first order in each reactant. Because there is only a single step, the concentration of every molecule involved in that step effects the rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, because the probability of three molecules colliding together in a way that can cause a reaction is very low, all elementary reactions involve one or two molecules. A single molecule reaction is just a dissociation, e.g. $A_2 \rightarrow A + A$, with a first order rate in A and an overall order of one. A two molecule reaction could be something like $A+B\rightarrow AB$, with still a first order rate in each of the reactants for an overall order of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, you've identified that SN1 reactions have multiple transition states, so SN1 cannot be an elementary reaction. In fact, the 1 in SN1 means that the overall order of the reaction is one, but since it involves two reactants, the order would be two for a single step reaction. Naturally, an SN2 reaction has an overall order of two and is an elementary reaction, proceeding through a single transition state where a bond between the incoming nucleophile and the carbon centre is formed at the same time as the leaving group's bond is broken.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-08-16T06:05:56.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15328" PostTypeId="2" ParentId="5421" CreationDate="2014-08-16T06:29:48.007" Score="2" Body="&lt;p&gt;Hydrogen is flammable only when mixed with oxygen, hence this can be mixed with compressed oxygen in petrol engines (hydrogen gas &amp;amp; oxygen gas) to be used as fuel.&lt;/p&gt;&#xA;" OwnerUserId="7500" LastEditorUserId="4934" LastEditDate="2014-08-17T20:20:43.580" LastActivityDate="2014-08-17T20:20:43.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15330" PostTypeId="2" ParentId="8389" CreationDate="2014-08-16T14:23:26.207" Score="2" Body="&lt;p&gt;Baking soda is essentially &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_bicarbonate&quot; rel=&quot;nofollow&quot;&gt;sodium hydrogen carbonate&lt;/a&gt;, which should not be expected to react with oil because of its poor solubility. $\ce{NaHCO3}$ melts at 50 °C and starts to decompose slowly to &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_carbonate&quot; rel=&quot;nofollow&quot;&gt;sodium carbonate&lt;/a&gt;, carbon dioxide and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NaHCO3 -&amp;gt;~ Na2CO3 + H2O + CO2\uparrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the release of carbon dioxide, $\ce{NaHCO3}$ acts as a raising agent when used in baking. Under the conditions of deep frying (150-200 °C), the decomposition is fast as the reaction is entropically favored (generation of gases). $\ce{CO2}$ and $\ce{H2O}$ escape from the baking mixture, leaving sodium carbonate behind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ingestion of small amounts of $\ce{Na2CO3}$, which are generated from typically used amounts of baking soda, should not be a health problem. Acidic ingredients in the baking mixture will convert sodium carbonate into the sodium salts of the respective acids, carbon dioxide and water. Higher amounts of sodium carbonate, stemming from the use of too much baking soda, would cause an unpleasant metallic taste of the baked food. Sodium carbonate is actually used as a food additive (E 500) and in toothpastes as a foaming agent and abrasive.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-16T14:23:26.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15331" PostTypeId="2" ParentId="15319" CreationDate="2014-08-16T16:30:23.443" Score="2" Body="&lt;p&gt;Your equation is wrong. In acidic medium, water is the one which gives an oxygen to the nitrite. And that should be an oxidation half-reaction (as nitrite oxidizes to nitrate), but you wrote a reduction where two electrons are spent, not released.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-16T16:30:23.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15332" PostTypeId="1" AcceptedAnswerId="15334" CreationDate="2014-08-16T16:58:27.063" Score="3" ViewCount="1332" Body="&lt;p&gt;This question seems silly, but this just came in my mind while i was learning about hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Means will there be any reaction when $\ce{H2O}$ will be mixed with $\ce{D2O}$?&lt;/p&gt;&#xA;" OwnerDisplayName="user7484" LastEditorUserId="9961" LastEditDate="2015-10-26T22:34:49.203" LastActivityDate="2015-10-26T22:34:49.203" Title="Normal and heavy water mixing" Tags="&lt;water&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15333" PostTypeId="2" ParentId="15324" CreationDate="2014-08-16T17:07:01.523" Score="1" Body="&lt;p&gt;Well, since the pressure in the tank will decrease with usage, you can't ultimately do what you seem to want to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a given backing pressure, there are equations to determine the orafice size to get a flow rate - google 'choked flow'. Downstream of the regulator, your orafice should be good for quite some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an experimental setup, one would use either a mass flow controller (MFC) to actively control the flow rate, or (much cheaper) one of those little flow indicators with the ball floating in the He (air) stream. Note the second option here is basically an adjustable orafice with integrated flow indicator.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-16T17:07:01.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15334" PostTypeId="2" ParentId="15332" CreationDate="2014-08-16T17:31:18.670" Score="9" Body="&lt;p&gt;Yes, a reaction known as &quot;hydrogen exchange&quot; will take place.  When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture.  Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-16T17:57:39.947" LastActivityDate="2014-08-16T17:57:39.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15335" PostTypeId="2" ParentId="15320" CreationDate="2014-08-16T18:34:06.280" Score="4" Body="&lt;p&gt;Dissenter, I've been thinking about this question since you posted it yesterday.  My first thought was that this question has a straightforward answer based on lone pair repulsions.  However, I ran into problems when I tried to correlate this hypothesis with physical properties.  For example, look at the following Table.  As we move from hydrogen peroxide with 2 lone pairs on each of the adjacent oxygens to hydrazine with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|} \hline&#xA;%Header&#xA;\text{Molecule} &amp;amp; &#xA;\text{Bond Strength}~/ (\pu{kcal/mol}) &amp;amp; &#xA;\text{Bond~Length}~/ \pu{pm}\\ \hline&#xA;\ce{H3C-CH3} &amp;amp; 85 &amp;amp; 154\\ \hline&#xA;\ce{H2N-NH2} &amp;amp; 60 &amp;amp; 145 \\ \hline&#xA;\ce{HO-OH} &amp;amp; 51 &amp;amp; 149 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one lone pair on each of the adjacent nitrogens, the bond strength increases and the bond length decreases as one would expect if lone pair repulsions were the dominant controlling factor.  However, as we move further on to ethane with no lone pairs, the bond length increases.  This suggests that if lone pair repulsions are playing a role, then, at least with regard to bond length, some other variable is also playing a significant role and obscuring the &quot;expected&quot; trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I examined an an approach based on MO analysis. Let's examine the MO diagrams for $\ce{HO-H, H2N-H}$ and $\ce{H3C-H}$, they are pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zEztZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qqtHP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4e1Q.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that water has a (roughly) p-hybridized lone pair (the $\ce{1b_1}$ orbital) that is non-bonding.  For ammonia, the lone pair is in an (roughly) $\ce{sp^3}$ hybridized orbital (the $\ce{2a_1}$ orbital) that is lower in energy than the non-bonding atomic p-orbitals.  In the case of methane, there are no lone pairs and all MOs are bonding (lower in energy than the p AOs).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine what would happen to these MOs if we removed a hydrogen atom from each of them and  remixed the resultant MOs with themselves (e.g. dimerize $\ce{HO{.},~ H2N{.}~ and~ H3C{.}}$).  My (simple) analysis suggests the in forming hydrogen peroxide the $\ce{1b_1}$ orbital would mix and split to form 2 new MOs, one bonding and one (slightly more) antibonding, both filled with 2 electrons, for a net destabilization relative to water.  Similarly when we form hydrazine, two $\ce{2a_1}$ orbitals will mix to produce two new MOs, one lower in energy than a non-bonding p AO, and the other either slightly lower or slightly higher (depending on the degree of splitting) than a non-bonding p AO.  Energetically speaking, the hydrazine case shouldn't be as destabilized as hydrogen peroxide.  Finally, when we form ethane, all orbitals will be more stable than a non-bonding p AO.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analysis, based on the &lt;em&gt;relative energetic positioning&lt;/em&gt; of the lone pairs in these molecules, appears to be in accord with the observed bond strengths of these molecules.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2018-02-13T05:36:13.140" LastActivityDate="2018-02-13T05:36:13.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15336" PostTypeId="1" CreationDate="2014-08-16T22:14:10.553" Score="3" ViewCount="1131" Body="&lt;p&gt;If have read conflicting reports on the half-life of mercury in the brain. I would like to hear some estimates from this community. &#xA;I am also wondering how long chelation therapy would take to remove mercury from the brain. &lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2017-06-24T16:52:48.177" Title="Mercury brain half-life" Tags="&lt;biochemistry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15337" PostTypeId="1" AcceptedAnswerId="15338" CreationDate="2014-08-16T22:44:22.733" Score="4" ViewCount="626" Body="&lt;p&gt;There are oxidation states from -4 to +8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the negative oxidation states not go all the way down to -8? I mean if an atom is hypervalent meaning that it can have more than an octet than that means it can have potentially 8 bonds(counting a double bond as 2 bonds and a triple bond as 3 bonds respectively) to other atoms more electronegative than itself and thus Xenon can have a +8 oxidation state and some metals can as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't that also mean that a hypervalent atom can have 8 bonds to other atoms less electronegative than itself and thus have a -8 oxidation state?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2014-08-16T23:07:39.540" Title="Why do negative oxidation states not extend to -8?" Tags="&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15338" PostTypeId="2" ParentId="15337" CreationDate="2014-08-16T23:00:33.887" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do the negative oxidation states not go all the way down to -8?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's think about what we would need for this to occur - i.e. an oxidation state of -8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would need a central atom that has considerable electronegativity - one that is at least more electronegative than its attached atoms. This already excludes numerous molecules; most molecules' central atoms are, on a relative basis, more &lt;em&gt;electropositive&lt;/em&gt; rather than electronegative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, this atom would need to have one of the following configurations; &quot;allocated&quot; refers to the assignment of oxidation state - since oxidation states are assigned on the premise that all bonding is ionic - the more electronegative element in a bonding configuration is &quot;allocated&quot; all the bonding electrons in the calculation of oxidation states. I.e. in H2O, the oxygen would be allocated both pairs of bonding electrons in the H-O bonds, and oxygen would then &quot;have&quot; 8 electrons, and its oxidation state would be -2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 valence electron and be &quot;allocated&quot; 9 electrons &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 valence electrons and be &quot;allocated&quot; 10 electrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 valence electrons and be &quot;allocated&quot; 11 electrons,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and on the other extreme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 valence electrons and be &quot;allocated&quot; 14 electrons,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7 valence electrons and be &quot;allocated&quot; 15 electrons,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8 valence electrons and be &quot;allocated&quot; 16 electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, the first set of criteria are basically incompatible. Elements only having 1 or 2 valence electrons are going to be small (on a relative basis). These elements would also &lt;em&gt;much rather&lt;/em&gt; be oxidized than reduced. Oxidation allows these elements to easily become isoelectronic with a noble gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other set of criteria for a negative 8 oxidation state is also hard to fulfill. None of the group 6 elements except the possibility of sulfur and elements below it are hypervalent - i.e. they can accommodate more than 8 valence electrons. Even if we consider sulfur as able to accommodate &lt;em&gt;12&lt;/em&gt; valence electrons (as some depictions of sulfuric acid suggest) we must remember that sulfur is not very electronegative and unlikely to be allocated any electrons. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-16T23:07:39.540" LastActivityDate="2014-08-16T23:07:39.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15339" PostTypeId="1" AcceptedAnswerId="15405" CreationDate="2014-08-17T01:04:49.483" Score="8" ViewCount="883" Body="&lt;p&gt;This question is directed towards laboratory scientists and technicians who actually deal with ethers that can develop peroxides upon concentration (e.g. diethyl ether, THF, dioxane). I am already very familiar with the literature, the normal precautions (dispose of after 1 year, test for peroxides, etc.), how to remove peroxides (shake with ferrous salt or metabisulfate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how hazardous are peroxides in reality if stored properly. If a peroxide-able ether upon concentration is stored in a cool dark cabinet, is rarely opened, has an inhibitor (BHT or ethanol), and argon is used above the headspace, it is likely that the ether can be used for longer than a year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I need the input of an actual practitioner. Simply quoting written material will not help me, as I have consulted LOTS of written material and I believe I understand the theoretical issues. I am looking for practical experience.  I have heard informally that some ethers are available in European countries as OTC solvents, and so it leads me to think that the standard warnings may be too severe. Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-09-12T00:14:20.907" Title="How hazardous are peroxide-able ethers if stored properly?" Tags="&lt;safety&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15340" PostTypeId="1" AcceptedAnswerId="15348" CreationDate="2014-08-17T02:48:21.707" Score="4" ViewCount="4869" Body="&lt;p&gt;I have been assigned with the task of figuring out whether a reaction is $\ce{S_{N}1}$ or $\ce{S_{N}2}$. The information that I have is, that the yield is dependent on solvent polarity. More specifically, the higher the solvent polarity, the higher the yield. &#xA;The problem is that both $\ce{S_{N}1}$ or $\ce{S_{N}2}$ products are the same. Both result in charge separation (therefore both are favored by polar solvents).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I proposed to repeat the experiment. I proposed to make the site of substitution chiral. That way it is possible to check for $\ce{S_{N}2}$ inversion.&lt;br&gt;&#xA;Another proposal was to use the fact that the rates of $\ce{S_{N}2}$ and $\ce{S_{N}1}$ were different. However, I am not sure how to use this fact and perform an experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I do in this situation? The people that I am working for do not want to repeat the experiment at all. Using the rates approach seems to be the best option since there won't be a need to obtain new materials. But I have no idea on how to do this. What are your suggestions? How can I figure out whether the reaction proceeded through $\ce{S_{N}1}$ or $\ce{S_{N}2}$ by using rates approach?&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="4945" LastEditDate="2014-08-18T12:38:44.003" LastActivityDate="2014-08-18T12:54:49.720" Title="How to figure out if the reaction proceeds via SN1 or SN2 mechanism?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;experimental-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15341" PostTypeId="2" ParentId="15340" CreationDate="2014-08-17T02:51:25.677" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Another proposal was to use the fact that the rates of SN2 and SN1&#xA;  were different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;SN2, as its name suggests, is a bimolecular mechanism and its rate depends on not just substrate concentration but also nucleophile concentration. SN1, however, is solely dependent on substrate concentration, as carbocation formation has a high activation energy barrier, and overcoming this barrier takes time. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I proposed to repeat the experiment. I proposed to make the site of&#xA;  substitution chiral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would the results of this proposed experiment apply to the results of your previous experiment with the achiral substrate? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The information that I have is that the yield is dependent on solvent&#xA;  polarity. More specifically, the higher the solvent polarity, the&#xA;  higher the yield. The problem is that both SN1 or SN2 products are the&#xA;  SAME. Both result in charge separation (therefore both are favored by&#xA;  polar solvents)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, try thinking about the intermediates, and try thinking about the nucleophile. Nucleophiles can be aided or hindered by the solvent. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-17T02:51:25.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15342" PostTypeId="1" AcceptedAnswerId="15502" CreationDate="2014-08-17T03:34:29.330" Score="12" ViewCount="2014" Body="&lt;p&gt;I have a quest to explain a physical, chemical theory to kids of age 8-12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The topic is &quot;Ionization&quot;. We all understand from early experiments of Physics and Chemistry (the Millikan oil drop experiment) how a neutral atom in gas becomes a charged atom where radiation knocking the electrons out of the neutral atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's the story line I came up with. Can you guys please help me to make this a success and correct the wrong impressions, expressions and examples in the narrative?&lt;/strong&gt;&#xA;I am open for any better examples.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There's a high-class (=high voltage) shopping mall area (=radio activity area metaphorically) in the country I am living right now. One particular shopping mall called ION. If you enter that shopping mall, you will shop and get excited (=charged) and spend your pocket money (=losing electrons) by the temptation (=radiation) from fancy shoppers around. &lt;em&gt;Shopping means getting exposed to radiation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some shoppers are highly influential (=alpha rays with +2 charge) and there's a high tendency you will spend money (=lose electrons). e.g. Movie theater. You get excited even looking at it from &lt;strong&gt;far&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other shoppers are less influential (=beta rays with -1 charge). e.g. Junk food, a piece of cake you like to eat. You have to go &lt;strong&gt;near&lt;/strong&gt; to see the food and then spend your money on it. The excitement of eating the food lasts for shorter period compared to the movie you watched. So the impact from movie is higher. You will talk about it in school for a week or two. Unlike the food you had and forgotten by the next meal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last type of shoppers are least influential (=gamma with 0 charge). e.g. Manga shop. You don't find it as exciting as watching a moving or eating favourite food. But you may go there as a matter of need (for a manga copy) or peer-pressure from your friends to buy manga and spend your money. You are kinda &lt;strong&gt;forced&lt;/strong&gt; to get excited (=charged) to spend a buck on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's &lt;em&gt;quite dangerous&lt;/em&gt; for you (=neutral atom) to enter or go near a shopping area (= radiation area) where you will be losing your money and get excited for nothing...&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2014-08-19T05:17:05.663" LastActivityDate="2019-04-05T20:16:38.817" Title="Explaining Ionization to kids" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15343" PostTypeId="2" ParentId="15340" CreationDate="2014-08-17T03:44:01.767" Score="3" Body="&lt;p&gt;You should be very careful about changing the reagents or solvent, because then you are looking at a different reaction than the one proposed which may go by a different mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea of looking at rates is a good one. Since an SN2 reaction depends on the concentration of nucleophile, while SN1 does not, set up two experiments exactly the same (same concentration of electrophile, same solvent, same temperature, etc) but double the amount of nucleophile in one of the experiments. If the reaction rates differ (by a factor of two), then the mechanism is SN2. If the rates are the same, then the mechanism is SN1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method that chemists use to learn about reaction mechanisms is by isotopic labeling. Depending on the exact electrophilic substrate, you could consider labeling a proton as deuterium (or other atom as necessary), rendering the substrate chiral. Then you could take advantage of the difference in stereochemical outcomes of SN1 and SN2 mechanisms. However, whether this can be utilized depends on the exact substrates of the reaction being considered.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-17T03:44:01.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15344" PostTypeId="5" CreationDate="2014-08-17T03:54:37.230" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T03:54:37.230" LastActivityDate="2014-08-17T03:54:37.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15345" PostTypeId="4" CreationDate="2014-08-17T03:54:37.230" Score="0" Body="Ionization is the process by which an electrically neutral atom, molecule, or radical loses or gains one or more electrons and becomes an ion. Ionization can occur in gases, liquids, or solids. Physicists often call a highly ionized gas a plasma. Metals tend to form positive ions, nonmetals negative ions." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T10:10:28.243" LastActivityDate="2014-08-17T10:10:28.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15346" PostTypeId="5" CreationDate="2014-08-17T04:01:58.073" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T04:01:58.073" LastActivityDate="2014-08-17T04:01:58.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15347" PostTypeId="4" CreationDate="2014-08-17T04:01:58.073" Score="0" Body="Radiation is the emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles that cause ionization." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T10:10:34.247" LastActivityDate="2014-08-17T10:10:34.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15348" PostTypeId="2" ParentId="15340" CreationDate="2014-08-17T04:28:15.150" Score="3" Body="&lt;p&gt;As the other answers state, kinetic analysis is one way to go, especially if you want to avoid the hassle of inserting a chiral label. If you're interested at looking at the relative rates, your best bet for a practical technique is probably NMR, depending on your reactants and solvent, so long as the reaction isn't too fast. It's easy enough to get 30 s resolution for arrayed proton NMR experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ESI-MS may also be an option if your compounds ionize well and you're using a compatible solvent. A reaction in a 500 µL syringe should be good for an hour or two of measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your exact reaction, optical techniques could potentially be used, as well. If you're lucky, it may be as simple as absorbance measurements.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-08-17T04:28:15.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15349" PostTypeId="1" CreationDate="2014-08-17T08:22:34.077" Score="1" ViewCount="101" Body="&lt;p&gt;I am a freshman in college. My major is chemical engineering. I would like some recommendations for a textbook that's an introduction to polymers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering whether a major in materials science is more related to polymers than chemical engineering. &lt;/p&gt;&#xA;" OwnerUserId="7509" LastEditorUserId="73" LastEditDate="2014-09-04T17:03:54.043" LastActivityDate="2014-09-04T17:03:54.043" Title="What are some of the canonical texts for polymers? Is materials science more related to polymers than chemical engineering?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15351" PostTypeId="5" CreationDate="2014-08-17T10:19:56.587" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:19:56.587" LastActivityDate="2014-08-17T10:19:56.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15352" PostTypeId="4" CreationDate="2014-08-17T10:19:56.587" Score="0" Body="Hydroxide is a compound of a metal with the hydroxide ion OH− (as in many alkalis) or the group —OH." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T11:56:52.110" LastActivityDate="2014-08-17T11:56:52.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15353" PostTypeId="5" CreationDate="2014-08-17T10:21:31.290" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:21:31.290" LastActivityDate="2014-08-17T10:21:31.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15354" PostTypeId="4" CreationDate="2014-08-17T10:21:31.290" Score="0" Body="Dimer is a molecule or molecular complex consisting of two identical molecules linked together." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T11:57:43.207" LastActivityDate="2014-08-17T11:57:43.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15355" PostTypeId="5" CreationDate="2014-08-17T10:23:02.947" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:23:02.947" LastActivityDate="2014-08-17T10:23:02.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15356" PostTypeId="4" CreationDate="2014-08-17T10:23:02.947" Score="0" Body="In mineralogy and crystallography, a crystal structure is a unique arrangement of atoms or molecules in a crystalline liquid or solid." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T11:55:01.047" LastActivityDate="2014-08-17T11:55:01.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15357" PostTypeId="5" CreationDate="2014-08-17T10:24:53.627" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:24:53.627" LastActivityDate="2014-08-17T10:24:53.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15358" PostTypeId="4" CreationDate="2014-08-17T10:24:53.627" Score="0" Body="Catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change." OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-08-17T11:56:34.197" LastActivityDate="2014-08-17T11:56:34.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15359" PostTypeId="5" CreationDate="2014-08-17T10:27:11.240" Score="0" Body="&lt;p&gt;The molecular structure, geometry, or the structure of a molecule is the particular relative arrangement of atoms in a molecule. The term geometry is widely used in &lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;theoretical-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt; and related branches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very common format for sharing geometries are cartesian coordinates (in Ångstrom). For example, in ammonia, $\ce{NH3}$ the structure is well defined by the following code (including the calculated electronic energy at the BP86/cc-pVDZ level of theory).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    4&#xA;AMMONIA, BP86/cc-pVDZ:   -56.553652 E(H)&#xA; N     0.000000     0.000000     0.000000&#xA; H     0.000000     0.000000     1.034364&#xA; H     1.006493     0.000000    -0.238496&#xA; H    -0.300271    -0.960659    -0.238496&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A geometry can also be shared in different coordinates. The following is a z-Matrix in the Gaussian09 file format. It makes use of the internal symmetry of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; n&#xA; h                    1    hn2&#xA; h                    1    hn2      2    hnh3&#xA; h                    1    hn2      2    hnh3     3    dih4     &#xA;&#xA;  hn2                   1.0344&#xA;  hnh3                103.3308&#xA;  dih4                107.3576&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Most commonly the geometry is displayed in a picture:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/39aBe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-04-09T16:21:54.547" LastActivityDate="2017-04-09T16:21:54.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15360" PostTypeId="4" CreationDate="2014-08-17T10:27:11.240" Score="0" Body="This tag should be applied to questions that deal with the particular arrangement of atoms in a molecule, the so-called molecular structure or geometry. " OwnerUserId="1878" LastEditorUserId="402" LastEditDate="2017-04-09T16:22:04.040" LastActivityDate="2017-04-09T16:22:04.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15361" PostTypeId="5" CreationDate="2014-08-17T10:36:30.047" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:36:30.047" LastActivityDate="2014-08-17T10:36:30.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15362" PostTypeId="4" CreationDate="2014-08-17T10:36:30.047" Score="0" Body="The adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface." OwnerUserId="1878" LastEditorUserId="305" LastEditDate="2014-08-17T11:54:48.883" LastActivityDate="2014-08-17T11:54:48.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15363" PostTypeId="5" CreationDate="2014-08-17T10:37:54.430" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-08-17T10:37:54.430" LastActivityDate="2014-08-17T10:37:54.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15364" PostTypeId="4" CreationDate="2014-08-17T10:37:54.430" Score="0" Body="Each of two or more different physical forms in which an element can exist. Graphite, charcoal, and diamond are all allotropes of carbon." OwnerUserId="1878" LastEditorUserId="305" LastEditDate="2014-08-17T11:57:32.060" LastActivityDate="2014-08-17T11:57:32.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15365" PostTypeId="1" CreationDate="2014-08-17T11:30:53.050" Score="1" ViewCount="735" Body="&lt;p&gt;As the title states, what is the name of the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G=\Delta G^{\circ}+RT \ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook does not explicitly state the name and I could not find anything on google.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="16683" LastEditDate="2016-10-22T16:22:00.863" LastActivityDate="2016-10-22T16:22:00.863" Title="What is the name for the equation ΔG = ΔG° + RT ln Q?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15366" PostTypeId="1" AcceptedAnswerId="15376" CreationDate="2014-08-17T12:12:19.070" Score="15" ViewCount="10044" Body="&lt;p&gt;Black powder is composed of charcoal, sulfur and a nitrate salt. The charcoal and sulfur serve as fuel and the nitrate is the oxidizing agent.&lt;/p&gt;&#xA;&lt;p&gt;However, wouldn't it be cheaper to leave out the sulfur, thus having only charcoal as fuel?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gunpowder#Composition_and_characteristics&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says about this (emphasis added):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Black powder is a granular mixture of&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;[...]&lt;/li&gt;&#xA;&lt;li&gt;sulfur (S), which, while also serving as a fuel, &lt;strong&gt;lowers the temperature required to ignite the mixture&lt;/strong&gt;, thereby increasing the rate of combustion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I guess after this initial research my question boils down to: &lt;strong&gt;Why is sulfur required as an ingredient in gun powder, and how does it lower the ignition temperature of the mixture?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-03-29T08:32:17.687" Title="Why is there sulfur in black powder / gun powder?" Tags="&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15368" PostTypeId="1" AcceptedAnswerId="15374" CreationDate="2014-08-17T12:38:10.607" Score="6" ViewCount="6460" Body="&lt;p&gt;Why is the addition of a singlet carbene to an alkene stereospecific, but not when a triplet carbene is added?&lt;/p&gt;&#xA;" OwnerUserId="7510" LastEditorUserId="7510" LastEditDate="2016-07-06T06:14:02.367" LastActivityDate="2017-02-12T03:19:55.027" Title="Why is the addition of a singlet carbene to an alkene stereospecific?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15369" PostTypeId="1" AcceptedAnswerId="15375" CreationDate="2014-08-17T13:26:02.250" Score="2" ViewCount="3499" Body="&lt;p&gt;What is the mole fraction of $\ce{C2H6} / \ce{C3H8}$ in mixture, if both compounds reacted with  $\ce{O2}$ in stoichiometric ratio? Mass of produced $\ce{CO2}$ is $1.819$ times larger than the mass of $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H6_{(g)} + O2_{(g)} -&amp;gt; CO2_{(g)} + H2O_{(g)}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H8_{(g)} + O2_{(g)} -&amp;gt; CO2_{(g)} + H2O_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I balanced the equations and tried the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{X4M(CO_2)+(X-1)4M(CO_2)}{X8M(H_2O)+(X-1)6M(H_2O)}=1.819$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which led me to the wrong result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone kick me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="7511" LastEditorUserId="5016" LastEditDate="2014-08-17T15:19:15.973" LastActivityDate="2016-03-12T00:43:23.130" Title="What is the mole fraction in mixture before combustion?" Tags="&lt;stoichiometry&gt;&lt;combustion&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15370" PostTypeId="2" ParentId="15187" CreationDate="2014-08-17T13:54:16.910" Score="4" Body="&lt;p&gt;There are two misconception behind your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You assume you can directly compare the colors of different compounds and derive oxidation states e.g of a heme (porphyrin) comparing to iron-oxide. No, it is not true.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You assume this whole story has anything to do with copper. No, the blueish-green is nothing to do with copper, it is a (porphyrin) degradation product. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The color of transition metal compounds mostly can be derived from the crystal field (ligand field) around the central metal ion. In inorganic iron-oxo/hydroxo compounds have different structure and hence different field than heme compounds. In heme, the N coordination of porphyrin ring is generating a very strong planar field, and the oxygen coordinates as molecule on the top of it. Charge transfer to the ring can also be an important factor in heme derivatives. As you see it is a whole complicated mass, and not much direct connection with different inorganic salts you see in lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colors you see are coming from oxo-hemoglobin (which you have in your healthy blood), and most probably several steps of enzymatic degradation of heme. The porphyrin ring is oxidized and cut open by an oxygen on the the iron. Several colorful compounds are produced, including the iron complexes themselves and biliverdin (green), bilirubin (yellow). These are the same colors you see under your skin, of you hit yourself hard and have some bleeding in your tissues. First the injured area is deep red/blue, than go to greenish, finally yellowish patch. This also happens continuously in your body, when your blood cells get old, and heme is enigmatically degraded so the whole thing can be recycled:  &lt;a href=&quot;http://www.elmhurst.edu/~chm/vchembook/634bilirubin.html&quot; rel=&quot;nofollow&quot;&gt;http://www.elmhurst.edu/~chm/vchembook/634bilirubin.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-17T13:54:16.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15374" PostTypeId="2" ParentId="15368" CreationDate="2014-08-17T15:58:19.533" Score="7" Body="&lt;p&gt;Here is a diagram that might help explain things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pdOBL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbenes have two electrons that, for the moment, are not being used for bonding.  These two electrons can exist in the same orbital with their spins paired (singlet carbene) or in separate orbitals with their spins unpaired (triplet carbene).  Normal hydrocarbons exist as singlets, spins are paired in all bonds, as they must be according to the Pauli exclusion principle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look first at the pi electrons in ethylene, their spins are paired since ethylene is a singlet in the ground state.  When we add a singlet carbene to ethylene, a bond is formed by using one electron from the carbene and one from ethylene.  Those two electrons are selected such that their spins will be paired in the newly-formed bond (Pauli exclusion principle).  Either 1) a second bond is formed at the same time using the two remaining spin-paired electrons (one on ethylene and one on the carbene) to form cyclopropane in a spin-conserving, concerted addition or 2) a transient (short-lived), intermediate 1,3-biradical is formed that still contains two spin-paired electrons is formed, since the spins are paired on these two remaining electrons in the biradical and since they are held in relatively close proximity they can close quickly to form a three-membered ring without violating the Pauli exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now contrast this to the triplet carbene case.  The spins are &lt;strong&gt;unpaired&lt;/strong&gt; in the triplet carbene.  When it forms a bond with one of the ethylene pi electrons, in order to obey the Pauli exclusion principle we must select our electrons so that they are paired in the newly formed bond.  This requires that the two remaining electrons in our 1,3-biradical have their spins unpaired.  Before the biradical can form the final bond and close to cyclopropane, we must wait for one of the electron spins to flip so that we have a singlet biradical.  The triplet cannot close directly since this would violate the Pauli exclusion principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;waiting&quot; for the spins to pair in the triplet biradical means that the triplet biradical has an appreciable lifetime.  Had we run our reaction using cis-2-butene instead of ethylene.  We could see the effects of biradical lifetime by observing the stereochemistry in the resultant products.  The singlet biradical, which either reacts in one concerted step or passes through a short-lived, singlet biradical would produce only cis-dimethylcyclopropane.  In the case of the triplet carbene, the triplet biradical exists long enough (waiting for one electron spin to flip) that rotation about the various carbon-carbon bonds is now possible.  As a result, we obtain both cis- and trans-dimethylcyclopropane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2017-02-12T03:19:55.027" LastActivityDate="2017-02-12T03:19:55.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15375" PostTypeId="2" ParentId="15369" CreationDate="2014-08-17T16:56:52.790" Score="2" Body="&lt;p&gt;It's always helpful to show as much work as you can so that we can see where you went off track,  In any case it appears you made an effort, so let's start at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to write out the balanced equations&#xA;\begin{aligned}&#xA;\ce{&amp;amp;C2H6(g) + $3.5$~ O2(g) -&amp;gt; 2 CO2(g) + 3 H2O(g)} \\&#xA;\ce{&amp;amp;C3H8(g) + 5 O2(g) -&amp;gt; 3 CO2(g) + 4 H2O(g)}&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second we're given that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass ($\ce{CO2}$) = 1.819 mass ($\ce{H2O}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can converting this to a molar ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac{[mass ~(CO2)]}{[$44.009$]}~=~ $1.819$ * \frac{[mass~ (H2O)]}{[$18.015$]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we find that the molar ratio of&#xA;$$\ce{\frac{[moles~(CO2)]}{[moles~(H2O)]}~=~ $0.744604$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, just to get a feel for things, let's look at the ethane case and we see that we have a mole ratio of $\ce{CO2~/~ H2O}$ of 2/3, whereas in the propane case the ratio is 3/4.  The observed molar ratio of 0.74 indicates that the propane reaction is the major contributor. We can write a general equation&#xA;$$\ce{\frac{[2/3 + (r * 3/4)]}{[1+r]}~=~$0.744604$}$$&#xA;where &quot;r&quot; is the molar ratio of propane to ethane being burned. Solving for &quot;r&quot; yields&#xA;$$\ce{r~=~$13.59$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:43:23.130" LastActivityDate="2016-03-12T00:43:23.130" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15376" PostTypeId="2" ParentId="15366" CreationDate="2014-08-17T17:17:12.827" Score="21" Body="&lt;p&gt;Black powder is a mixture of solids. As solids, they are not particularly inclined toward fast reactions between each other. Elevated temperature generally liven things up, because it allows diffusion of solids into each other. However, both carbon (charcoal) and potassium nitrate are relatively stable compounds, so even when potassium nitrate is melted, they still do not react immediately,  further heating is required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur, on the other side, is quite easy to melt. Nitrates are not very powerful oxidizers, but potassium nitrate in particular is a low-melting compound. And at the moment both potassium nitrate and sulfur are molten, their interdiffusion becomes quite easy, so they ignite, and charcoal follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing it up: charcoal is almost impossible to melt and potassium nitrate is a relatively weak oxidizer and also hard to melt, so the mixture is hard to ignite. Sulfur is easily flammable, easy to melt, so it is much easier to react with potassium nitrate solid, and its presence makes mixture much more ignitable. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="3846" LastEditDate="2019-03-29T08:32:17.687" LastActivityDate="2019-03-29T08:32:17.687" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="15377" PostTypeId="1" CreationDate="2014-08-17T17:20:29.290" Score="1" ViewCount="223" Body="&lt;p&gt;If darker surfaces absorb/store more light/energy per square inch than light surfaces, then does dark skin, on average, store energy more efficiently than light skin? (edited).&lt;/p&gt;&#xA;" OwnerUserId="7512" LastEditorUserId="7512" LastEditDate="2014-08-17T22:10:11.657" LastActivityDate="2014-08-17T23:39:57.053" Title="If dark surfaces absorb/store more light/energy than light surfaces, then does dark skin, on average, store energy from light most efficiently?" Tags="&lt;physical-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15378" PostTypeId="1" AcceptedAnswerId="15411" CreationDate="2014-08-17T17:44:00.063" Score="3" ViewCount="1809" Body="&lt;p&gt;I recently completed making a simple water electrolysis setup to generate hydrogen which uses two graphite rods from a standard AA battery. The power source is a $24~\mathrm{V}$ $6~\mathrm{A}$ adapter. I am using salt water as electrolyte (The salt is iodized). Strangely, a minute or so after running, a black precipitate starts to form on the negative terminal.&lt;br&gt;&#xA;Also, besides the hydrogen, another gas is generated which smells sour, not very sour, but somewhat like lemonade. I have no idea what is causing this.&lt;br&gt;&#xA;I thoroughly cleaned the graphite rods before using them. Has anybody an idea about what reaction might be happening here?&lt;/p&gt;&#xA;" OwnerUserId="7513" LastEditorUserId="4945" LastEditDate="2014-08-19T01:39:05.847" LastActivityDate="2014-08-19T01:43:44.863" Title="What are possible side reactions for the generation of hydrogen via electrolysis?" Tags="&lt;electrochemistry&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15379" PostTypeId="2" ParentId="9773" CreationDate="2014-08-17T19:06:49.573" Score="2" Body="&lt;p&gt;The standard enthalpy of formation $\Delta H_\mathrm f$ refers to the formation of 1 mole of product, with the educts and product being in their respective standard states. The fourth equation is therefore correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H_\mathrm f$. The first two equations can be ruled out because they do not contain the educts nitrogen and/or oxygen in their standard states, which are $\ce{N2}$ and $\ce{O2}$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-02-25T02:58:41.903" LastActivityDate="2018-02-25T02:58:41.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15380" PostTypeId="1" CreationDate="2014-08-17T19:08:07.203" Score="3" ViewCount="4160" Body="&lt;p&gt;I'm really confused as none of the definitions of radical on the internet compare to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from my college textbook of Applied Chemistry-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radicals&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Every inorganic compound is made up of two electrically charged parts&#xA;  called &lt;em&gt;radicals&lt;/em&gt;. One part has positive charge and is called basic&#xA;  radical, while the other part has negative charge and is called acid&#xA;  radical. Hence,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;An atom or a group of atoms which forms a part of an inorganic compound is called a radical.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;For example :&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Sodium Chloride is made up of two parts:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;(i)&lt;/strong&gt; Sodium radical (Na&lt;sup&gt;+&lt;/sup&gt;)  and   &lt;strong&gt;(ii)&lt;/strong&gt; Chloride radical (Cl&lt;sup&gt;-&lt;/sup&gt;) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The two radicals combine to form sodium chloride molecule. &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Copper sulphate is made up of two parts:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;(i)&lt;/strong&gt; Cupric radical (Cu&lt;sup&gt;2+&lt;/sup&gt;)  and   &lt;strong&gt;(ii)&lt;/strong&gt; Sulphate radical (SO&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;2-&lt;/sup&gt;) &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The two radicals combine to form copper sulphate molecule. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It should be noted that a radical cannot exist independently.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This definitely doesn't mean the same as free radicals. I don't think it means &quot;functional group&quot; because &lt;strike&gt;how can sodium be a functional group?!&lt;/strike&gt;. Functional groups are in organic chemistry, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain what the term &quot;radical&quot; means here?&lt;/p&gt;&#xA;" OwnerUserId="7515" LastActivityDate="2016-11-06T12:20:53.600" Title="What does a radical mean here?" Tags="&lt;inorganic-chemistry&gt;&lt;applied-chemistry&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15381" PostTypeId="2" ParentId="15380" CreationDate="2014-08-17T19:11:26.847" Score="4" Body="&lt;p&gt;A radical, according to the IUPAC, has an unpaired electron. Radicals &lt;em&gt;may or may not&lt;/em&gt; be ions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A molecular entity such as $\ce{.CH3}$, $\ce{.SnH3}$, $\ce{Cl.}$ possessing an unpaired electron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/R05066.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://goldbook.iupac.org/R05066.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrary to your book's claim, the sodium ion is not considered a radical. It actually has all its electrons paired. It is isoelectronic with helium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your book's usage of &quot;radical:&quot; your book's terminology is outdated. Here is the IUPAC statement on your book's usage of the term:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the past, the term 'radical' was used to designate a substituent&#xA;  group bound to a molecular entity, as opposed to 'free radical', which&#xA;  nowadays is simply called radical. &lt;strong&gt;The bound entities may be called&#xA;  groups or substituents, but should no longer be called radicals.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T12:20:53.600" LastActivityDate="2016-11-06T12:20:53.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15383" PostTypeId="1" AcceptedAnswerId="15537" CreationDate="2014-08-17T20:13:30.943" Score="-1" ViewCount="537" Body="&lt;p&gt;I know that the heating of water drives the auto ionization of water. It also drives the evaporation of water. My question is where do the ions go as the water evaporates as it is heated up. Ions can't be isolated in the solid phase. Do the ions enter the gas phase? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2015-10-23T17:25:39.187" LastActivityDate="2015-10-23T17:25:39.187" Title="Auto ionization of water and evaporation of water" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15384" PostTypeId="2" ParentId="15309" CreationDate="2014-08-17T22:08:07.457" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Iodine is a stronger nucleophile here than bromine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much.  As long as iodide is a strong &lt;em&gt;enough&lt;/em&gt; nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen.  Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h&quot;&gt;unhybridized just like phosphine&lt;/a&gt; ($\ce{PH3}$).  If it is unhybridized than our lone pair on phosphorous is in an s-orbital.  Since the s-orbital has little &quot;directionality&quot;, this lone pair on phosphorous will be a weak nucleophile.  On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile.  So now matter how TPP is hybridized, we expect it to be a weak nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodide is, as you note, a much better leaving group then bromide.  In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is run with just a small (catalytic) amount of iodide.  As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP.  During this last step, iodide is regenerated and the cycle starts again.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="1887" LastEditDate="2014-08-18T22:10:25.830" LastActivityDate="2014-08-18T22:10:25.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15385" PostTypeId="2" ParentId="15377" CreationDate="2014-08-17T23:20:53.180" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Melanin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Melanin is a chemical compound&lt;/a&gt;(s) found in the skin and is responsible for skin pigmentation.  The darker the skin tone, the more melanin is present.  In light-toned-skin, melanin is produced as part of the tanning process when the skin is exposed to UV radiation.  Melanin is a very effective filter for UV radiation.  The more melanin present in the skin, the less light is available for light-catalyzed chemical processes that occur in the skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dissenter noted in his comment, light energy is not stored &lt;em&gt;per se&lt;/em&gt; in the human body.  Rather light catalyzes certain reactions in the skin and the products of these reactions can be stored (at least for a while) for use in the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vitamin_D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vitamin D&lt;/a&gt; production is an example of an extremely important light-mediated process that begins in the skin.  Vitamin D is thought to play a role in a large number of regulatory processes throughout the body ranging from bone health to cancer.  Because melanin is such an effective filter of sunlight, &lt;a href=&quot;http://www.health.harvard.edu/newsweek/vitamin-d-and-your-health.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-white populations are at an increased risk of vitamin D deficiency&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPeLK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-17T23:39:57.053" LastActivityDate="2014-08-17T23:39:57.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15386" PostTypeId="1" AcceptedAnswerId="15390" CreationDate="2014-08-18T00:59:00.843" Score="2" ViewCount="139" Body="&lt;p&gt;How were molecular formulas determined before spectroscopy and spectronomy? I know that John Dalton thought that the formula of water was &quot;HO&quot; but we now know it to be &quot;H2O.&quot; How did Dalton even know that water had hydrogen and oxygen in the first place? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-18T02:18:58.950" Title="How were the first molecular formulas determined?" Tags="&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15387" PostTypeId="1" AcceptedAnswerId="15388" CreationDate="2014-08-18T01:21:53.503" Score="5" ViewCount="33581" Body="&lt;p&gt;When my textbook talks about hydrogenation using Pt, Pd or Ni heterogeneous catalysts, it never mentions if it is anti or syn addition.&#xA;It simply jumps on to say that for alkynes, NiB2 (P-2) catalysts and Lindlar's catalysts are syn hydrogenation.&#xA;Can someone tell me if using Pt gives syn or anti for alkynes?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4945" LastEditDate="2014-08-18T02:23:40.677" LastActivityDate="2014-08-18T04:06:51.763" Title="Hydrogenation stereochemistry-Pd,Pt, Ni catalysts" Tags="&lt;organic-chemistry&gt;&lt;hydrogen&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15388" PostTypeId="2" ParentId="15387" CreationDate="2014-08-18T01:26:45.477" Score="4" Body="&lt;p&gt;Using any &lt;em&gt;transition&lt;/em&gt; metal catalyst (Pt, Pd, Ni) for hydrogenation of alkynes results in syn hydrogenation to not alkenes but alkanes. Using Lindlar's catalyst results in syn hydrogenation of an alkyne to an alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of sodium metal (&lt;em&gt;not&lt;/em&gt; a transition metal) in ammonia, hydrogenation of alkynes results in a trans alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M0dSO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ukZQv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-18T04:06:51.763" LastActivityDate="2014-08-18T04:06:51.763" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="15389" PostTypeId="2" ParentId="15387" CreationDate="2014-08-18T01:40:42.610" Score="3" Body="&lt;p&gt;All catalytic hydrogenations involving solid metal catalysts transfer hydrogen in a syn fashion to the alkyne to produce the corresponding cis-olefin.  The following figure provides a mechanism that illustrates why the   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3yAW4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;addition must be syn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, with alkynes there are problems.  First, reduction of the resultant alkene is faster than reduction of the starting alkyne, so product mixtures of alkenes and alkanes will result unless poisoned catalysts (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lindlar_catalyst&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lindlar catalyst&lt;/a&gt;) are used or the reaction is stopped early.  A second problem often observed is that the resultant alkene usually remains adsorbed on the catalyst surface.  Consequently cis-trans isomerization of the alkene can take place leading to further contamination of the initial cis-product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-18T01:40:42.610" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15390" PostTypeId="2" ParentId="15386" CreationDate="2014-08-18T02:18:58.950" Score="4" Body="&lt;p&gt;Lavoisier burned hydrogen and oxygen together and found that water was produced.  Dalton incorporated this information into his thinking that the most stable arrangement of any two atoms involved a 1:1 combination, therefore water would be &quot;OH&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In slightly more recent times, but prior to modern-day spectroscopy, chemists would chemically chop molecules into identifiable fragments.  They would also combust the molecule and get a C,H,N,O ratio.  Putting all of this together they would (often) come up with remarkably accurate molecular formulas.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-18T02:18:58.950" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15391" PostTypeId="1" AcceptedAnswerId="15490" CreationDate="2014-08-18T02:38:17.187" Score="1" ViewCount="654" Body="&lt;p&gt;Does passing chlorine gas through water form hydrochloric acid or bleach? Also, will a mixture of hydrogen and chlorine gas from electrolysis of brine do the same?&lt;/p&gt;&#xA;" OwnerUserId="7513" LastEditorUserId="1878" LastEditDate="2014-08-18T03:35:43.260" LastActivityDate="2014-08-21T19:18:22.570" Title="Chlorination of Water" Tags="&lt;water&gt;&lt;electrolysis&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15392" PostTypeId="1" AcceptedAnswerId="15404" CreationDate="2014-08-18T06:03:00.930" Score="1" ViewCount="68" Body="&lt;p&gt;I was wondering if it was possible to create Silane from Methane using the process of siliconization to replace the Carbon with Silicon. For example Silicon Carbide is produced by replacing C-C bonds with a Si-C bond.&lt;/p&gt;&#xA;" OwnerUserId="7355" LastActivityDate="2014-08-18T16:03:31.787" Title="Siliconization for Organic Compounds" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15393" PostTypeId="2" ParentId="15322" CreationDate="2014-08-18T06:28:06.823" Score="4" Body="&lt;p&gt;The short version: You are correct, the schematic drawing is not the correct mechanism and it does also not provide the most likely path. For a training exercise it is close enough though. It shows the most important steps and it conserves stoichiometry.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The depicted scheme does not show a reaction mechanism in the sense of the definition given in the &lt;a href=&quot;http://goldbook.iupac.org/M03804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;goldbook&lt;/a&gt;. A reaction mechanism always also has to include transition states, that lead to the respective products. It is rather a reaction sequence, that shows the most important steps. In this sense, the stoichiometry has to be conserved and in this sense it is as correct as it could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noteworthy is, that the catalytically active species is not &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;sulfuric acid&lt;/a&gt;, but instead a mixture of hydronium and protonated methanol, $\ce{H3+O/H2+OCH3}$. Sulfuric acid is such a strong acid, that it is capable of protonating any solvent that has a lone pair, even itself via autoprotolysis (in concentrated $98\%$ solution). $$\ce{2H2SO4 &amp;lt;=&amp;gt; [H3SO4]+ + [HSO4]- &amp;lt;=&amp;gt; H3+O + [HS2O7]- &amp;lt;=&amp;gt; H2O + H2S2O7}$$&#xA;The use of concentrated sulfuric in organic synthesis is usually not advisable, since its dehydration property can quickly decompose many organic compounds that contain hydroxyl groups. The use of diluted sulfuric acid (either in water or alcohols) is therefore much more common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only logical, that sulfuric acid will protonate methanol, which can be considered a preceding equilibrium. $$\ce{H2SO4 + HOCH3 &amp;lt;=&amp;gt; H2+OCH3 + [HSO4]- }$$&#xA;Since methanol is also the solvent and the acid is only available in catalytic amounts, it is feasible to assume, that the equilibrium lies entirely on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important, that the last step must be also in equilibrium, since it is only the protonation/deprotonation of an ether. Here again we have much more methanol than product, so that the equilibrium will most likely be entirely on the product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more descriptive mechanism would be the following:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EKblB.png&quot; alt=&quot;long mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity and sufficiently accurate bond formation and breaking is indicated by arrows, omitting the transition states. Resulting a slightly simplified picture. However, the preceding equilibrium - as you noted - should not be omitted. This formulation of the mechanism is particularly important for many simple organic reactions, as it focusses entirely on the reactants and products (and to a minor extent on the reactive intermediates).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wtPoF.png&quot; alt=&quot;short mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to more complex reactions, e.g. with asymmetric catalysts it is often much better to include the transition states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it can also be seen, that the choice of sulfuric acid is somewhat arbitrary, as most acids would do the same. However, sulfuric acid is often used, since bisulfate and sulfate are very weak nucleophiles.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Additional note&lt;/em&gt;&lt;br&gt;&#xA;In all schemes only the most likely path is shown. Any possible side reactions, like concerted [2+2] cycloaddition, have been omitted.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-18T06:28:06.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15394" PostTypeId="1" AcceptedAnswerId="15449" CreationDate="2014-08-18T07:37:58.697" Score="8" ViewCount="131311" Body="&lt;p&gt;Consider the reaction of $\ce{AgNO3}$ and $\ce{HCl}$. I read that silver chloride would be formed. But $\ce{H} &amp;gt; \ce{Ag}$ in reactivity, then how could $\ce{Ag}$ displace $\ce{H}$ from $\ce{HCl}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was a reaction, $\ce{AgCl}$ and $\ce{HNO3}$ were to form. But then these would react again to give us the original compounds back. The actual question was to write down the reaction between $\ce{AgNO3}$ and $\ce{HCl}$. I (wrongly) realized that the compounds wouldn't react because of the reasons stated.&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="5026" LastEditDate="2018-02-25T07:33:23.057" LastActivityDate="2018-06-06T08:37:41.653" Title="Reaction of silver nitrate and hydrochloric acid" Tags="&lt;inorganic-chemistry&gt;&lt;precipitation&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15395" PostTypeId="1" CreationDate="2014-08-18T08:02:02.187" Score="4" ViewCount="325" Body="&lt;p&gt;I want to make $\ce{N2O}$ in my school chem lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of making it from $\ce{Zn_{(s)}}$ in $\ce{HNO3_{(dil)}}$ :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{&#xA;Zn_{(s)} + HNO3_{(dil)} -&amp;gt; N2O + Zn(NO3)2 + H2O&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not able to get $\ce{Zn_{(s)}}$, so I would like to make it on my own in lab.&lt;/p&gt;&#xA;" OwnerUserId="7518" LastEditorUserId="305" LastEditDate="2014-08-18T09:21:52.450" LastActivityDate="2014-08-18T11:21:40.533" Title="Making Zn in school lab" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15396" PostTypeId="1" CreationDate="2014-08-18T08:31:37.563" Score="1" ViewCount="143" Body="&lt;p&gt;Why are (non-polar) hydrogen atoms generally neglected in a protein-ligand (or drug) binding energy calculation? They do have charge and mass. Doesn't that affect the free energy calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant equations are given here:&#xA;$$ \Delta G_\text{bind} = -RT\ln K_\text{d} $$&#xA;$$ K_\text{d} = \mathrm{\frac{[Receptor][Acceptor]}{[Complex]}} $$&#xA;$$ \Delta G_\text{bind} = \Delta G_\text{desolvation} + \Delta G_\text{motion} + \Delta G_\text{configuration} + \Delta G_\text{interaction} $$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A system where this is the case would be the following, note that only the polar hydrogen atoms are explicitly represented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWYdm.png&quot; alt=&quot;System with polar hydrogen atoms marked out.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7120" LastEditorUserId="3846" LastEditDate="2015-06-16T20:33:38.550" LastActivityDate="2015-06-16T20:33:38.550" Title="Why are non-polar hydrogens neglected in computational protein-drug binding studies?" Tags="&lt;biochemistry&gt;&lt;computational-chemistry&gt;&lt;free-energy&gt;&lt;structural-biology&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15397" PostTypeId="2" ParentId="15395" CreationDate="2014-08-18T08:38:41.030" Score="7" Body="&lt;p&gt;Ahem, forget it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically speaking, you can get $\ce{Zn}$ from alkaline batteries (see details &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_battery#Construction&quot;&gt;here&lt;/a&gt;), but you have no chance to get pure $\ce{N2O}$ this way and reaction may easily go out of control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attention! Detonation hazard !!!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual and easy way to produce $\ce{N2O}$ is to heat very carefully widely available $\ce{NH4NO3}$, to $180-200~^\circ\mathrm{C}$. It is an absolute imperative to keep temperature of the reaction mixture below $240~^\circ\mathrm{C}$. Decomposition of ammonia nitrate ($\ce{NH4NO3}$) is exothermic and may proceed explosively. To be on the safe side, use sand bath to heat the reaction flask and put thermometer into the sand, so its temperature never risen above safe limits. Liquid bath with appropriate liquid may be better, but I'm not sure if a school lab would have proper liquid (silicone oil with high flash point). The produced gases should be cooled in Liebig condenser to separate water, and $\ce{N2O}$ may be further dried with $\ce{H2SO4}$. Of course, all joints must be sealed using proper oil. And keep a bath of cold water to cool the reaction mixture in case if temperature raises above safe limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is advised to use lab grade, not fertilized grade ammonia nitrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible safer, but costlier route&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is reported that it is possible to produce $\ce{N2O}$ by heating mixture of   $5~\mathrm{mL}$ $6~\mathrm{M}~\ce{HNO3}$, $1~\mathrm{g}~\ce{NH4NO3}$ and $0.05~\mathrm{g}~\ce{NaCl}$ on water bath at $\approx 80~^\circ\mathrm{C}$. If this works, it is a safer route, as hot ammonia nitrate is diluted and autospeeding reaction cannot occur. &lt;a href=&quot;http://mattson.creighton.edu/N2O/&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2014-08-18T11:21:40.533" LastActivityDate="2014-08-18T11:21:40.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15399" PostTypeId="2" ParentId="14618" CreationDate="2014-08-18T11:08:04.880" Score="14" Body="&lt;p&gt;For this approach I am basically employing &lt;a href=&quot;http://en.wikipedia.org/wiki/Frontier_molecular_orbital_theory&quot; rel=&quot;noreferrer&quot;&gt;Frontier Molecular Orbital Theory (FMO)&lt;/a&gt; to predict the reactivity of carbonyl compounds towards nucleophiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack the positively polarised carbonyl carbon to form the tetrahedral reactive intermediate. The activation barrier of following step, i.e. dissociation of the leaving group (or proton for aldehydes and ketones), must be lower in energy, because otherwise the reaction will not proceed. It is therefore sufficient to look at only the initial step of the reaction, and more rigorous we can even only look at the ground state properties of the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In FMO theory there are basically three main observations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Occupied molecular orbitals of the reactants repel each other&lt;/li&gt;&#xA;&lt;li&gt;Opposite (partial) charges in the reactants attract each other&lt;/li&gt;&#xA;&lt;li&gt;Occupied orbitals of one reactant interact with unoccupied orbitals of the other reactant (and vice versa) causing attractive interaction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first point is fairly obvious and will not cause much change in the interaction of the two molecules.&lt;br&gt;&#xA;The second point is also straight forward. In all carbonyl compounds the carbon will be carrying a (partial) positive charge, while the nucleophile will be having a (partial) negative charge.&#xA;The third point is not so obvious. It can be expected, the LUMO of the carbonyl compound (electron poor carbon) and the HOMO of the nucleophile (electron rich) will interact to form the new bond. This will also be the strongest interaction. The opposite combination is also thinkable, but the energy difference between these orbitals will be much larger, as we will see. In other words, the closer the energy difference between these orbitals will be, the more attractive the interaction will be. Since we keep the nucleophile constant, we can predict a order of reactivity towards that one specific nucleophile based on the energy of the LUMO of the carbonyl compound. (This comparison will hold, even if we change the nucleophile.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following table assembles the energies of the frontier orbitals in $\mathrm{eV}$ as well as the most important partial natural atomic charges. As a model system, the ethyl substituted compounds $\ce{Et-CO-R}$ have been chosen. The calculations were performed with &lt;a href=&quot;http://www.gaussian.com/g_prod/g09.htm&quot; rel=&quot;noreferrer&quot;&gt;Gaussian09 Rev. D1&lt;/a&gt; employing &lt;a href=&quot;https://journals.aps.org/pra/abstract/10.1103/PhysRevA.38.3098&quot; rel=&quot;noreferrer&quot;&gt;Becke's exchange functional&lt;/a&gt; in conjunction with &lt;a href=&quot;https://journals.aps.org/prb/abstract/10.1103/PhysRevB.34.7406&quot; rel=&quot;noreferrer&quot;&gt;Perdew's correlation functional&lt;/a&gt;. The density fitted &lt;a href=&quot;http://dx.doi.org/10.1039/b508541a&quot; rel=&quot;noreferrer&quot;&gt;def2-SVP basis set&lt;/a&gt; was used. For charges the Natural Bond Orbital (NBO) analysis was used.&lt;br&gt;&#xA;For convenience I have not included any solvent, so these calculations assume gas phase, vacuum and absolute zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrrrr}&#xA;\ce{R =}       &amp;amp;\ce{HOMO}&amp;amp;\ce{LUMO}&amp;amp;\Delta E &amp;amp;\mathbf{q}(\ce{C})&amp;amp;\mathbf{q}(\ce{O})\\\hline&#xA;\ce{-Cl}       &amp;amp; -7.00  &amp;amp; -2.11  &amp;amp; 4.89  &amp;amp; 0.55  &amp;amp; -0.46  \\\hline&#xA;\ce{-O-OC-Et}  &amp;amp; -6.35  &amp;amp; -1.85  &amp;amp; 4.50  &amp;amp; 0.82  &amp;amp; -0.52  \\&#xA;               &amp;amp; &amp;amp; \text{(L+1)} -0.85 &amp;amp;  &amp;amp;  &amp;amp;  \\\hline&#xA;\ce{-H}        &amp;amp; -5.79  &amp;amp; -1.79  &amp;amp; 4.00  &amp;amp; 0.41  &amp;amp; -0.51  \\\hline&#xA;\ce{-Et}       &amp;amp; -5.55  &amp;amp; -1.48  &amp;amp; 4.07  &amp;amp; 0.57  &amp;amp; -0.54  \\\hline&#xA;\ce{-OH}       &amp;amp; -6.33  &amp;amp; -0.88  &amp;amp; 5.45  &amp;amp; 0.81  &amp;amp; -0.58  \\&#xA;               &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;(\ce{OH})  -0.68 \\\hline&#xA;\ce{-OEt}      &amp;amp; -6.15  &amp;amp; -0.74  &amp;amp; 5.40  &amp;amp; 0.81  &amp;amp; -0.58  \\&#xA;               &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;(\ce{OEt}) -0.56 \\\hline&#xA;\ce{-NH2}      &amp;amp; -5.53  &amp;amp; -0.24  &amp;amp; 5.29  &amp;amp; 0.67  &amp;amp; -0.60  \\&#xA;               &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;(\ce{N}) -0.83 \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kind of calculations cannot handle anions very well, so I have also omitted the carboxylate ion. For obvious reasons, I would say its tendency to react as an electrophile is negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HOMO of water is one of the lone pairs at oxygen, at $-6.35~\mathrm{eV}$. Correlating the LUMO energies from the table, the expected order of reactivity is&#xA;$$\ce{acid~chloride &amp;gt; anhydride\approx aldehyde &amp;gt; ketone &amp;gt; acid\approx ester &amp;gt; amide}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With NBO we can look at localised orbitals, which all look very similar and also give very similar numerical results. The $\pi^*$ antibonding LUMO is composed of $67\pm2\%$ carbon $\ce{p}$ AO and $33\pm2\%$ oxygen $\ce{p}$ AO for all compounds. Here are the FMO shown for the anhydride.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/o5Zes.jpg&quot; alt=&quot;nbo pi orbital&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mneZC.jpg&quot; alt=&quot;nbo pi star orbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I was currently not able to find any reaction barriers for this approach. My educated guess is, that the chosen nucleophile is too weak to coordinate to this compound in the gas phase. I believe I will have to include a solvent for transition state stabilisation, but I am not sure if I can afford the calculations at the moment.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-18T11:08:04.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15401" PostTypeId="2" ParentId="15340" CreationDate="2014-08-18T12:09:16.050" Score="6" Body="&lt;p&gt;As you have the information that the yield depends on solvent polarity. It is the key point to be considered in analysis of type of reaction. In case of a $\ce{S_{N}1}$ reaction, the polar solvent favors the reaction. $\ce{S_{N}1}$ is a two stepped reaction, in which the 1st step is carbocation formation, such that there appears the dipole dipole interaction between the carbocation and polar solvent which favors the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas in the case of $\ce{S_{N}2}$ reaction, the dipole dipole interaction develops between polar solvent and the nucleophile rather than with intermediate product. It is because the charge is dispersed in two centres in intermediate while charge is centered in single center in nucleophile. This deactivates the $\ce{S_{N}2}$ reaction and the reaction cannot proceed forward smoothly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have another specific information that the higher the solvent polarity, the higher the yield. Since higher solvent polarity favors the $\ce{S_{N}1}$ reaction, the higher will be the yield through a $\ce{S_{N}1}$ mechanism. So the reaction with tendency with higher solvent polarity and higher yield is a $\ce{S_{N}1}$ reaction.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4945" LastEditDate="2014-08-18T12:54:49.720" LastActivityDate="2014-08-18T12:54:49.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15402" PostTypeId="1" AcceptedAnswerId="15403" CreationDate="2014-08-18T13:24:46.393" Score="1" ViewCount="145" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At $800\:\mathrm{K}$ a reaction mixture contained $0.5\:\mathrm{mol}$ of $\ce{SO2}$, $0.12\:\mathrm{mol}$ of $\ce{O2}$ and $5\:\mathrm{mol}$ of $\ce{SO3}$ at equilibrium. $K_\mathrm c$ for the equilibrium $$\ce{2SO2 + O2 -&amp;gt; 2SO3}$$ is $833\:\mathrm{L/mol}$. If the volume of the container is $1\:\mathrm{L}$, calculate how much $\ce{O2}$ is to be added at this equilibrium in order to get $5.2\:\mathrm{mol}$ of $\ce{SO3}$ at the same temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer comes out to be $0.34\:\mathrm{mol}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have understood the whole question and I am getting the answer. I just have a confusion why we are taking $K_\mathrm c$ as $833$ and not $1/833$ which makes it $\mathrm{mol/L}$. I think $K_\mathrm c$ should be in $\mathrm{mol/L}$. &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="5026" LastEditDate="2018-02-23T06:26:20.837" LastActivityDate="2018-02-23T06:26:59.507" Title="Why is the equilibrium constant given in litre per mole?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15403" PostTypeId="2" ParentId="15402" CreationDate="2014-08-18T14:03:50.813" Score="6" Body="&lt;p&gt;$K_\mathrm c$ is defined as (for more information see &lt;a href=&quot;https://chemistry.stackexchange.com/a/5814/189&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; K_{c} = \prod_i [c_{i}]^{\nu_{i}} = \frac{\prod \limits_{i \in \text{products}} [c_{i}]^{|\nu_{i}|}}{\prod \limits_{j \in \text{reactants}} [c_{j}]^{|\nu_{j}|}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu_{i}$ and $c_{i}$ are the stochiometric coefficient and the concentration of the $i^{\text{th}}$ component in the reaction, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use this definition for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{2SO2 + O2 -&amp;gt; 2SO3}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; K_\mathrm c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^{2} [\ce{O2}]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the notation $[i] = c_{i}$ was used.&#xA;This equation tells you that $K_\mathrm c$ has the unit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \text{unit of }K_\mathrm c = \frac{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2}}{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2} \left(\frac{\mathrm{mol}}{\mathrm{L}} \right)} = \frac{\mathrm{L}}{\mathrm{mol}} \ .&#xA;\end{equation}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="5026" LastEditDate="2018-02-23T06:26:59.507" LastActivityDate="2018-02-23T06:26:59.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15404" PostTypeId="2" ParentId="15392" CreationDate="2014-08-18T16:03:31.787" Score="4" Body="&lt;p&gt;A key difference between carbon and silicon (and why you usually can't make an all-silicon version of an organic compound where all its carbons are replaced with silicon) is that long chains of silicon atoms are not stable. Another key difference is that a hydrogen attached directly to silicon, as in silane, is typically more reactive--say toward hydrolysis or oxidation--that one attached to a carbon atom. I do not know of any process to directly &quot;silanize&quot; a carbon compound, as stated in your question.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-08-18T16:03:31.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15405" PostTypeId="2" ParentId="15339" CreationDate="2014-08-18T16:50:32.337" Score="6" Body="&lt;p&gt;Peroxides are no fun.  There have been explosions when ether solutions were concentrated to dryness.  What you really want to do is store your ether over potassium hydroxide - hydroperoxides are slightly acidic, and the base forms hydroperoxide salts and pulls them out of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you keep solvents for more than a year you clearly aren't turning them over fast enough.  Keeping flammable material that you don't need just increases the danger in case of fire.&lt;/p&gt;&#xA;" OwnerUserId="7522" LastActivityDate="2014-08-18T16:50:32.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15407" PostTypeId="1" CreationDate="2014-08-18T18:08:05.713" Score="1" ViewCount="61" Body="&lt;p&gt;I don't understand why it varies between Group 10 metals (as with Palladium, Platinum and Darmstadtium) and Copper (Group 11). If each of Palladium, Platinum and Darmstadtium entirely filled the 3d/4d/5d shells then it would make more conceptual sense, but both Platinum and Darmstadtium still have +1/2 spin in the final 4d/5d shell.&lt;/p&gt;&#xA;" OwnerUserId="5217" LastActivityDate="2014-08-18T18:08:05.713" Title="Why do Group 10 transition metals (and Cu) lose an S electron to fill the D shell?" Tags="&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-09-27T07:58:01.433" ContentLicense="CC BY-SA 3.0" />
  <row Id="15408" PostTypeId="1" CreationDate="2014-08-18T18:14:20.340" Score="7" ViewCount="2219" Body="&lt;p&gt;The other day I saw one of my friends, who was mixing lemon juice directly into the flavour chimney (or wind guard) and water pot of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hookah&quot; rel=&quot;nofollow noreferrer&quot;&gt;hooka&lt;/a&gt;. When I asked why are you doing as such, he said that it makes inhaling hooka smoke passive.  By passive, I mean easily inhalable without any hiccups or harshness.  I have tried it, it really makes the smoke passive. But how did this happen?&lt;/p&gt;&#xA;&lt;p&gt;Here is a diagrammatic view of working principle of hooka:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pqVV4.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is a list of ingredients of hooka flavour which he used, it was of about 200 grams:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;glycerol 70 %&lt;/p&gt;&#xA;&lt;p&gt;18 % mollases&lt;/p&gt;&#xA;&lt;p&gt;5 % synthetic edible flavours&lt;/p&gt;&#xA;&lt;p&gt;5 % sodium xanthate&lt;/p&gt;&#xA;&lt;p&gt;0.05 % nicotine and 1.95 % water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;the temperature at which glycerol boils (and hence the smoke appears) is 290 °C,&lt;/p&gt;&#xA;&lt;p&gt;By adding lemon juice, we are actually adding citric acid to it, so what would be the products obtained heating that flavour at the given temperature which reduces harshness of hooka.&lt;/p&gt;&#xA;&lt;p&gt;My friend says the water present at the bottom in the hooka pot, serves for the purification of nicotine present in the flavour (which I can't believe), and he further adds that adding lemon juice to it will act as a filter for nicotine, hence reducing its harshness. [I think there shouldn't be any harshness due to nicotine as it is in a very low concentration i.e. measurable in the order of milligrams per puff, see the comments beside the answer below]&lt;/p&gt;&#xA;&lt;p&gt;So the harshness must be due to other substances present in the mixture, such as glycerol ( which wouldn't cause harshness as it is in the vapour state), mollases and other flavoured substances along with sodium xanthate, here is molecular structure of mollasses&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mhWRF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;is mollasses causing harshness, then what is the left over debris after all the flavour is exhausted, many individuals say that the left over &amp;amp; exhausted part of the flavour is nothing but the molasses itself.&lt;/p&gt;&#xA;&lt;p&gt;and coming to the other substances present in the flavour.&lt;/p&gt;&#xA;&lt;p&gt;the knowledge which i have gathered from online sources is, the reaction between glycerol and citric acid yields a polymer in a controlled and catalysed reaction, called PGC, which is generally obtained at the temperature of 150°C,  but here in this case the temperature is above 290°C i.e. boiling point of glycerol.&lt;/p&gt;&#xA;&lt;p&gt;so the fact is that, no polymer is obtained while inhaling hooka, as the glycerol is in its vapour state and citic acid being, very acidic {1M citric acid has a pH value of 2}, so it is nearly in its dissociated state {the lemon juice was mixed in water present in the water pot},&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r5pHY.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;Citric acid&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNygq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;Glycerol&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-10-30T17:27:48.253" Title="How does mixing lemon juice in hooka water affect the composition of the mixture?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2015-10-30T19:32:52.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="15409" PostTypeId="1" AcceptedAnswerId="15410" CreationDate="2014-08-18T21:29:17.647" Score="5" ViewCount="5764" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is an SN1 reaction possible in halobenzene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Will the phenyl carbocation be stabilized via resonance? Suppose the leaving group is very electronegative and the environment is protic. What then would be a suitable nucleophile?&lt;/p&gt;&#xA;" OwnerUserId="7524" LastEditorUserId="5026" LastEditDate="2018-03-13T06:58:17.747" LastActivityDate="2018-03-13T07:01:24.653" Title="Is an SN1 reaction possible in halobenzene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15410" PostTypeId="2" ParentId="15409" CreationDate="2014-08-18T22:54:50.857" Score="11" Body="&lt;p&gt;Yes, that reaction occurs, but generation of the phenyl carbocation was much more difficult than anyone might have guessed. Here is a drawing of the phenyl carbocation.&#xA;First note that the 6 p orbitals making up the aromatic pi system are all still intact and overlapping - the aromatic nature of the benzene ring has not been tampered with.  Then notice that the cationic center exists as an empty $\ce{sp^2}$ orbital where the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hIhO8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C-H}$ bond used to be. The empty (cationic) $\ce{sp^2}$ orbital is orthogonal to the aromatic pi system; as a consequence there is no resonance interaction or stabilization of any sort between the two systems.  Further, being constrained by the 6-membered ring, the cationic $\ce{sp^2}$ orbital cannot rehybridize.  This results in considerable (low energy) s-character being used to stabilize an &lt;em&gt;empty&lt;/em&gt; orbital, and is not available to stabilize other orbitals that do contain electrons! The vinylic carbocation, the ethylenic analogue of the phenyl cation, is much more stable and common (it can even be observed by nmr) because it is not constrained by a ring and can rehybridrize, thereby lowering its energy. It is both this total lack of stabilization and relatively high (higher than need be if rehybridization could occur) energy of the molecule that has made the phenyl carbocation such an elusive target.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To stack the deck in their favor, investigators have run experiments using extremely good leaving groups such as nitrogen and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Trifluoromethanesulfonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;triflate anion&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_acid#Observations_of_Stable_Carbocations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magic acid&lt;/a&gt; and its variations, a very powerful acid that stabilizes carbocations, was often employed as well.  The nitrogen leaving group route, using thermal decomposition of aryl diazonium cations has been one of the few methods that has allowed access to this transient intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few references if anyone would like to read more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://elib.uni-stuttgart.de/opus/volltexte/2009/4227/pdf/eff107.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;On Attempts at Solvolytic Generation of Aryl Cations - 1976&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/hlca.19830660610/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Attempts at Solvolytic Generation of Phenyl Cations - 2004&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.com/books?id=z3jXKOYuqQAC&amp;amp;pg=PA289&amp;amp;lpg=PA289&amp;amp;dq=phenyl%20cation%20from%20solvolysis&amp;amp;source=bl&amp;amp;ots=dM5pv9h1Sq&amp;amp;sig=kz_CGMh2-MCWrJgxWzznuWwBLLI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=6njyU8eeBIqNyASguoAY&amp;amp;ved=0CDAQ6AEwAzgK#v=onepage&amp;amp;q=phenyl%20cation%20from%20solvolysis&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;A one paragraph review - 2000&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-13T07:01:24.653" LastActivityDate="2018-03-13T07:01:24.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15411" PostTypeId="2" ParentId="15378" CreationDate="2014-08-18T22:56:53.833" Score="9" Body="&lt;p&gt;First of all, check again your electrodes to see if this 'black precipitate' is formed at the negative electrode as you state; I predict it's actually happening at the positive electrode for the following reason: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're using &lt;em&gt;two&lt;/em&gt; graphite electrodes, it isn't a black precipitate that's forming, but &lt;em&gt;is&lt;/em&gt; indeed a side reaction as you state. I suggest what you're actually making is oxygen gas at the positive terminal (as one would predict) which then reacts with the graphite electrode surface to generate $\ce{CO2}$. To verify this,take a look at your electrode—it should have started to break up, now having a non-uniform surface. Perhaps even a bit crumbly.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As LDC3 points out in his comment, you're forming $\ce{Cl2}$ as well as $\ce{O2}$ which is the characteristic smell you're noticing. To understand this we need to look at the standard electrode potential for each of these reactions. Forming $\ce{O2}$ from water at constant $\mathrm{pH}$ (compared to the standard hydrogen electrode) requires a potential difference of $-1.23~\mathrm{V}$ (this value is negative since this is the reverse of the tabulated version). Chlorine evolution on the other hand requires a p.d. of $-1.36~\mathrm{V}$. Remember that the larger (the less negative in this case) the reduction potential, the higher the tendency for reduction. So since $\ce{Cl-}$ reduction potential is more negative, it's less likely to undergo reduction and hence more likely to be oxidised (lose electrons) than water, which agrees with our chemical intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we say here is that $\ce{Cl-}$ is preferentially oxidised over water, and as a consequence $\ce{Cl2}$ is formed preferentially over $\ce{O2}$. The reason why you have any $\ce{O2}$ forming at all is due to the electrolyte being the limiting reagent. Your supporting solvent (water) is in such abundance, essentially completely surrounding the electrode, that the system simply has to oxidise some water molecules some of the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, experiment should rule the day, so revise your experiment to provide evidence for/refute my theoretical framework. Replace your positive terminal with a high purity metal electrode (copper should work and be readily available at high purity). As for the choice of electrolyte, test a couple of salts whose reduction potentials are less negative than water. Hint: perhaps look at sulphate salts and try and find a further reason why these are very useful in electrochemistry.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorUserId="4945" LastEditDate="2014-08-19T01:43:44.863" LastActivityDate="2014-08-19T01:43:44.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15412" PostTypeId="1" AcceptedAnswerId="15460" CreationDate="2014-08-19T05:32:45.413" Score="11" ViewCount="933" Body="&lt;p&gt;I am told that for 2,6-xylidine (2,6-dimethylaniline), the amino group cannot line up in such a way that its p-orbital is parallel with respect to the p-orbitals of the carbons in the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uZfdh.png&quot; alt=&quot;2,6-dimethylaniline&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked for papers and no one has actually &lt;em&gt;proved&lt;/em&gt; to me why I should believe that two methyl groups are bulky enough to interfere with the amino group. I've come across various statements regarding this molecule, but &lt;em&gt;no&lt;/em&gt; proof. For example, here's a Wiki article on steric hinderance and resonance. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoic acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But there are &lt;em&gt;no&lt;/em&gt; citations to back any of that up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I've found is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steric effects of ortho methyl groups are base strengthening. This is not due to steric inhibition of resonance since the conformation remains planar in most derivatives. Two ortho methyl groups are necessary to distort the planarity; their steric effect is more than doubled compared with one methyl group. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/poc.642/abstract&quot; rel=&quot;noreferrer&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/poc.642/abstract&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note again – there isn't really any demonstration – just a statement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I supposed to do some Euclidean geometry and use van der Waals radii and trigonometry or something? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T15:28:39.253" LastActivityDate="2016-11-06T19:49:56.810" Title="Why can the lone pair not align with the phenyl moiety in 2,6-xylidine?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15413" PostTypeId="1" AcceptedAnswerId="15416" CreationDate="2014-08-19T05:34:12.853" Score="4" ViewCount="4001" Body="&lt;p&gt;I was wondering about vodka vs. whiskey mostly. The way I understand it, a distilled beverage is created by heating a non-distilled beverage, say, beer, and then collecting and cooling the vapors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would these vapors contain anything but alcohol and some water? Certainly the impurities that color, say, whiskey do not vaporize at the low temperatures involved? Why don't all distilled beverages taste and look like vodka (i.e. water and alcohol)?&lt;/p&gt;&#xA;" OwnerUserId="7527" LastEditorUserId="73" LastEditDate="2014-08-21T08:50:31.237" LastActivityDate="2014-08-21T08:50:31.237" Title="Why do distilled alcoholic beverages taste different from each other?" Tags="&lt;everyday-chemistry&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15414" PostTypeId="1" AcceptedAnswerId="51371" CreationDate="2014-08-19T06:14:18.443" Score="5" ViewCount="1271" Body="&lt;p&gt;I have Stainless Steel welded to Titanium. I need a common etchant to etch both S.S. and $\ce{Ti}$ so that I can do further analysis (like SEM etc). Right now I'm etching Ti Side separately and S.S. side separately and doing analysis twice, once for each metal. The etchants I'm using are mentioned below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{HNO3 + 3HCl}$ - for S.S. 304L&lt;/li&gt;&#xA;&lt;li&gt;$5\%~\ce{HCl} + 6\%~\ce{HNO3} + 16\%~\ce{HF} + 74\%~\ce{H2O}$ - for $\ce{Ti}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible to combine these two etchants to form a common etchant, or are there any standard etchants available to etch both S.S. and $\ce{Ti}$?&lt;/p&gt;&#xA;" OwnerUserId="1356" LastEditorUserId="1356" LastEditDate="2015-06-16T16:14:33.400" LastActivityDate="2016-05-18T05:45:10.967" Title="Common etchant for Stainless Steel 304L and Titanium Grade 2" Tags="&lt;acid-base&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15415" PostTypeId="2" ParentId="15413" CreationDate="2014-08-19T06:22:52.880" Score="7" Body="&lt;p&gt;You compare incomparable. Vodka is distilled and immediately bottled and sold. Whiskey is stored for &gt;3 years in wooden casks, often used for wine previously, thereby getting lot of fragrances from wood and also from environment (Scotch whiskey from islands, e.g.). The fresh distilled whiskey is clear colorless liquid. As for the color, 99% of the whiskeys contain added colorants, as the customers think that darker color means more matured and therefore more tasty spirit. AFAIK, the Ardbeg distillery does not add them that much, so this is the genuine whiskey color (more like white wine): &lt;a href=&quot;http://thewhiskyphiles.files.wordpress.com/2014/07/ardbeg-blasda.jpg?w=339&quot;&gt;http://thewhiskyphiles.files.wordpress.com/2014/07/ardbeg-blasda.jpg?w=339&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aromatic compounds are nevertheless distilled together with the spirit, so you can distinguish fruit distillates, Calvados, Williams, Kirsch, etc...&#xA;These compounds (mostly esters) have boiling point somewhere between alcohol and water, thereby during batch distillation, you have to carefully watch the process and collect the smelly fraction, as it is immediately followed by acids.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-08-19T06:22:52.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15416" PostTypeId="2" ParentId="15413" CreationDate="2014-08-19T06:26:57.060" Score="8" Body="&lt;p&gt;Two different aspects come into play here. For once, most of the whiskey aroma stems from aging the fresh destillate in oak barrels previously used to store wines, e.g. sherry. Colour an aroma thus, at least in part, stem from a  slow extraction process using the fresh destillate as a solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But thinking in other beverages, particular the clear fruit spirits, there's more in it. The key to a beverage that doesn't smell and taste just like ethanol but has a cherry aroma is the sloppyness of the destillation :) In the lab, a destillation or rectification is performed to isolate a particular compound from a mixture at a &lt;strong&gt;sharp boiling point&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the production of alcoholic beverages, the destillation typically aims to obtain three fractions over different &lt;strong&gt;boiling point ranges&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At lower boiling points, methanol, acetone and ethylacetate are removed.&lt;/li&gt;&#xA;&lt;li&gt;The upper range mostly furnishes long chain alcohols.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two fractions aren't meant to be digested ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The central fraction, however, will contains all the fragrances, ethanol and water. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-08-19T06:26:57.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15417" PostTypeId="1" AcceptedAnswerId="15419" CreationDate="2014-08-19T10:19:35.363" Score="3" ViewCount="12074" Body="&lt;p&gt;Is the following correct?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gibbs free energy of activation is used in energy profiles where the stabilities of the species are expressed as changes in Gibbs energy, while the activation energy Ea is used in energy profiles where the stabilities of the species are expressed as changes in enthalpies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-08-19T11:40:03.417" LastActivityDate="2014-08-19T11:40:03.417" Title="What is the correct definition of the Gibbs free energy of activation?" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15418" PostTypeId="2" ParentId="15417" CreationDate="2014-08-19T10:45:29.037" Score="1" Body="&lt;p&gt;I think that your definitions are correct, if not perhaps slightly trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might like to look at the Eyring equation and the terms deriving from that to provide some more thought. It might help to provoke an understanding of the physical intuition of these terms in order to know whether your definitions at least make some sense. Learning how to learn, so to speak!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some further questions to ask, beyond your definition question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does an activation energy mean? What does it mean if we have a negative entropy term of activation? Not in terms of numbers but in terms of what the molecules are actually doing.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-19T10:45:29.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15419" PostTypeId="2" ParentId="15417" CreationDate="2014-08-19T11:37:28.693" Score="5" Body="&lt;p&gt;There are a lot of possible definitions and it possibly depends very much on the case and the context you are looking at. However, there are the definitions provided by the IUPAC goldbook, which can be considered the official ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first definition you suggested is correct so far and known as (&lt;a href=&quot;http://goldbook.iupac.org/S05916.html&quot; rel=&quot;noreferrer&quot;&gt;standard&lt;/a&gt;) &lt;a href=&quot;http://goldbook.iupac.org/G02631.html&quot; rel=&quot;noreferrer&quot;&gt;Gibbs energy of activation&lt;/a&gt; (standard free energy of activation) $\Delta^\ddagger{}G^\circ{}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The standard Gibbs energy difference between the transition state of a reaction (either an elementary reaction or a stepwise reaction) and the ground state of the reactants. It is calculated from the experimental rate constant  $k$ via the conventional form of the absolute rate equation:&#xA;  $$\Delta^\ddagger{}G^\circ{} = \mathcal{R} T \left[\ln\left(\frac{\mathcal{k}_\mathrm{B}}{h}\right) − \ln \left(\frac{k}{T} \right)\right]$$&#xA;  where  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant and  $h$ the Planck constant ($\frac{\mathcal{k}_\mathrm{B}}{h} = 2.08358\cdot 10^{10}~\mathrm{K^{−1}s^{−1}}$). The values of the rate constants, and hence Gibbs energies of activation, depend upon the choice of concentration units (or of the thermodynamic standard state).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your second definition refers to the definition of &lt;a href=&quot;http://goldbook.iupac.org/E02142.html&quot; rel=&quot;noreferrer&quot;&gt;enthalpy of activation&lt;/a&gt;, $\Delta^\ddagger{}H^\circ{}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The standard enthalpy of activation $\Delta^\ddagger{}H^\circ{}$ is the enthalpy change that appears in the thermodynamic form of the rate equation obtained from conventional transition state theory. This equation is only correct for a first order reaction, for which the rate constant has the dimension reciprocal time. For a second order reaction, for which the rate constant has the dimension (reciprocal time) × (reciprocal concentration), the left hand side should be read as  $k c^\circ$, where $c^\circ$ denotes the standard concentration (usually  $\mathrm{1 mol\cdot dm^{−3}}$).&#xA;  $$k = \frac{\mathcal{k}_\mathrm{B} T}{h} \exp\left\{\frac{\Delta^\ddagger{}S^\circ}{\mathcal{R}}\right\} \cdot\exp\left\{\frac{-\Delta^\ddagger{}H^\circ}{\mathcal{R}T}\right\}$$&#xA;  The quantity  $\Delta^\ddagger{}S^\circ$ is the standard entropy of activation, and care must be taken with standard states. In this equation  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant,  $T$ the absolute temperature,  $h$ the Planck constant, and $\mathcal{R}$ the gas constant. The enthalpy of activation is approximately equal to the activation energy; the conversion of one into the other depends on the molecularity. The enthalpy of activation is always the standard quantity, although the word standard and the superscript $^\circ$ on the symbol are often omitted. The symbol is frequently (but incorrectly) written  $\Delta{}H^\ddagger{}$, where the standard symbol is omitted and the  $\ddagger$ is placed after the  $H$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/A00102.html&quot; rel=&quot;noreferrer&quot;&gt;activation energy (Arrhenius activation energy)&lt;/a&gt; $E_a$ itself is defined as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$,  $$E_a = \mathcal{R} T^2 \frac{\mathrm{d}(\ln k)}{\mathrm{d} T}$$, where $\mathcal{R}$ is the gas constant and $T$ the thermodynamic temperature. The term is also used for threshold energies in electronic potential surfaces, in which case the term requires careful definition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And while we are at it, &lt;a href=&quot;http://goldbook.iupac.org/E02150.html&quot; rel=&quot;noreferrer&quot;&gt;entropy of activation&lt;/a&gt; $\Delta^\ddagger{}S^\circ$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The entropy change that appears in the thermodynamic form of the rate equation obtained from conventional transition-state theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional note and general warning&lt;/strong&gt;&lt;br&gt;&#xA;Although there are the official definitions, the term activation energy is often used as a synonym for many kinds of different barriers. One has to be very careful when reading articles, which definition is used.&lt;br&gt;&#xA;In computational chemistry articles you may also find the terms electronic energy of activation, zero-point corrected energy of activation, solvent corrected energy of activation, ..., etc., pp., and many more.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-19T11:37:28.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15420" PostTypeId="2" ParentId="15412" CreationDate="2014-08-19T11:40:23.917" Score="4" Body="&lt;p&gt;A simple space-filling model shows the plausibility of the effect being steric interaction between the methyl groups and the amino-hydrogens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This view is a non-optimised space-filling picture of the molecule generated by Chemdoodle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gQblX.png&quot; alt=&quot;amino xylene space-filling picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this crude picture shows the plausibility of significant spatial interaction between the left hand methyl group and the amino hydrogens. This is likely enough to constrain the orientation of the amino group giving the effect you query.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-08-19T11:40:23.917" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15421" PostTypeId="1" AcceptedAnswerId="15431" CreationDate="2014-08-19T14:24:08.533" Score="16" ViewCount="6758" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwBrl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a representation of magic acid. The left part is the acid; the right part is the solvent. Why would this combination make for a particularly good acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left part looks like a sulfuric acid derivative, except one of the oxygens has been replaced by an electronegative fluorine. Why can't we replace &lt;em&gt;all&lt;/em&gt; the oxygens with fluorines? Wouldn't that make for an even &lt;em&gt;stronger&lt;/em&gt; acid through inductive stabilization of negative charge after proton loss? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the solvent play a role in magic acid's acidity? Does the highly partially positive Sb central atom play a role in stabilizing negative charge on the solute after the solute has been deprotonated? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2015-11-06T19:00:27.607" LastActivityDate="2018-03-03T11:25:50.850" Title="Why is magic acid a superacid?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15422" PostTypeId="1" AcceptedAnswerId="15423" CreationDate="2014-08-19T14:42:53.670" Score="9" ViewCount="18939" Body="&lt;p&gt;Graphite has hexagonal parallel planes. In a hexagonal structure the carbon atom has three bonds. Since the valancy of carbon is 4 is the valency satisfied in graphite?&lt;/p&gt;&#xA;" OwnerUserId="7531" LastEditorUserId="7448" LastEditDate="2015-01-01T17:19:06.637" LastActivityDate="2019-04-27T11:20:32.990" Title="Is the valency of carbon satisfied in graphite?" Tags="&lt;bond&gt;&lt;structural-formula&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15423" PostTypeId="2" ParentId="15422" CreationDate="2014-08-19T14:46:08.853" Score="7" Body="&lt;p&gt;Just as much as in benzene or any other &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; carbon case.&#xA;The concept of valency doesn't mean that the given atom has to form bonds with the given number of neighbors, but refer to a simple way of counting electrons and formed bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In organic chemistry, we often find carbon forming 4 (single) bonds to neighbors. For example you can find four &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds from the central carbon in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt; or one &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt; and three &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds from each carbon in &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H6}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a benzene or ethene &lt;span class=&quot;math-container&quot;&gt;$(\ce{H2C=CH2})$&lt;/span&gt; molecule there is only three neighbors of each carbons. In ethene the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C}$&lt;/span&gt; is a double bond (unlike &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt; in ethylene) which is formed by using two-two valence electrons of the carbons. When you count valence, you count these valence electrons, therefore a double bond counts twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation of benzene is a little trickier, as it does not have a double or a single bond between the carbons. If the carbons would form a single bond, there would be a extra dangling electron remaining on each of them. In reality these electrons are delocalized (shared) over the whole ring. If you count the valence electrons of a carbon, you should count this delocalized electron, too. You can understand graphite along this line: three bonds to the three neighbors plus the remaining one valence electron delocalized over the ring system (the whole graphite sheet) are together gives back the four valence electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more on hybridization in the other answer. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="41328" LastEditDate="2019-04-27T11:20:32.990" LastActivityDate="2019-04-27T11:20:32.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15424" PostTypeId="1" CreationDate="2014-08-19T14:46:30.690" Score="4" ViewCount="2671" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which has higher rate of hydrogenation- methyl propene or trans-2-butene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that cis-2-butene has a higher rate of hydrogenation than both of these due to steric reasons, but here, both of these compounds have the same steric hindrance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when they come in contact with the hydrogen adsorbed catalyst, they both should behave in a similar manner. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="4103" LastEditorUserId="5026" LastEditDate="2018-03-12T12:51:14.583" LastActivityDate="2018-03-12T12:51:14.583" Title="Which has higher rate of hydrogenation - methyl propene or trans-2-butene?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;&lt;hydrocarbons&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15426" PostTypeId="2" ParentId="15422" CreationDate="2014-08-19T14:56:01.070" Score="18" Body="&lt;p&gt;In the hexagonal graphite structure the carbon atoms are $\ce{sp^2}$ hybridized, just like in benzene.  In graphite, this p-orbital is used for bonding just as it is in benzene, resulting in an extended pi system in the graphite structure.  Graphite is really just a large number of benzene rings annelated together to form a continuous, ring structure.  Just as carbon satisfies its valence requirement of 4 in benzene (3 sigma bonds and 1 pi bond), carbon's valency requirements are satisfied the same way in graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9uPI3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l6tB1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-19T14:56:01.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15427" PostTypeId="2" ParentId="15336" CreationDate="2014-08-19T15:03:11.817" Score="6" Body="&lt;p&gt;Since you're concerned about mercury from fillings, you may have already taken a look at the wikipedia article about &lt;a href=&quot;http://en.wikipedia.org/wiki/Dental_amalgam_toxicity&quot; rel=&quot;nofollow&quot;&gt;dental amalgam toxicity&lt;/a&gt;. It would be worth reading through some of the sources listed, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you're an adult, I don't know if you're male or female. Children are more sensitive to mercury than adults because their brains are actively developing. If you're female, you may become pregnant at some point and there would be some concern about fetal mercury exposure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to a 2003 WHO report, the amount of mercury ingested from dental amalgam is low, with exposures estimated from 1ug per day to 12.5ug per day, but the vast majority of people are exposed to less than 5ug per day. The mercury from fillings is elemental mercury, which is poorly absorbed, less than 0.01% is taken up, with smaller amounts reaching the brain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing your existing fillings probably isn't worth it, as the removal process would expose you to more mercury than just letting the filling sit there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this will make you feel better or worse, but coal fired power plants are probably a larger source of mercury exposure than your fillings. &lt;a href=&quot;http://en.wikipedia.org/wiki/Environmental_impact_of_the_coal_industry#Mercury_emissions&quot; rel=&quot;nofollow&quot;&gt;The mercury from the coal plants is methylmercury, which is much more efficiently absorbed than elemental mercury.&lt;/a&gt; This bioaccumulates in the environment, so that fish have relatively high levels of mercury in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your perceived concentration problems, we really can't say if it's caused by the mercury or not. If it concerns you, you should see a doctor. And it may be more efficient to treat the symptoms through ADD medication than trying to remove mercury. &lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="4945" LastEditDate="2014-08-20T03:29:27.193" LastActivityDate="2014-08-20T03:29:27.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15428" PostTypeId="1" AcceptedAnswerId="15434" CreationDate="2014-08-19T15:48:47.713" Score="1" ViewCount="2960" Body="&lt;p&gt;A student was contemplating a question regarding elements with &quot;no empty valence orbitals.&quot; An argument was made that elements all have infinite number of valence orbitals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every atom basically has an infinite number of shells. The thing is that almost all of those shells are empty (they don’t have electrons in them).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=1224&quot; rel=&quot;nofollow&quot;&gt;https://van.physics.illinois.edu/qa/listing.php?id=1224&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is facile to say that, say, the fluoride anion has &lt;em&gt;no&lt;/em&gt; empty valence orbitals on the basis of it having 8 valence electrons and the $2n^2$ rule (where n = 2)? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-08-19T16:37:34.980" Title="How many empty orbitals do elements have?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15429" PostTypeId="1" AcceptedAnswerId="15432" CreationDate="2014-08-19T15:56:44.600" Score="-1" ViewCount="4213" Body="&lt;p&gt;How is positive charge delocalized? I understand how negative charge can be delocalized. Electrons are mobile and they can move around. But what about positive charge? For example, consider this resonance structure and its composite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aq3rb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GNwaq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly is that positive charge delocalized? Protons aren't mobile. I can't see any protons moving through the remaining pi-system. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2020-02-02T08:01:13.740" Title="Delocalization of positive charge" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15430" PostTypeId="2" ParentId="15421" CreationDate="2014-08-19T16:05:26.290" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is magic acid a superacid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because it meets the &lt;a href=&quot;http://goldbook.iupac.org/S06135.html&quot;&gt;definition of a superacid&lt;/a&gt;: A medium having a high acidity, generally greater than that of 100 wt.% sulfuric acid&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would this combination make for a particularly good acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm simplifying a bit, but let's examine the following equilibrium (it is one of the early equilibria involved in the formation of Magic acid)&#xA;$$\ce{HSO_{3}F + SbF5 &amp;lt;=&amp;gt; H^+ + [SbF_{5}-O-SO_{2}F]^-}$$&#xA;The extreme stability of the hexacoordinate antimony anion is what drives the equilibrium to the right.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't we replace all the oxygens with fluorines?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, the oxygen(s) is used to coordinate to the $\ce{SbF_5}$ structures.  In some structures there is coordination to multiple $\ce{SbF5}$ units through multiple oxygens.  In other structures the oxygens have been removed and replaced with fluorine. Also, you need at least one oxygen so you can have a proton. The equilibria that occur in magic acid are very complex and not entirely understood (for example, &lt;a href=&quot;http://books.google.com/books?id=BH2vPZBIUXcC&amp;amp;pg=PA529&amp;amp;lpg=PA529&amp;amp;dq=Chemistry%20in%20super%20acids%20II,%20commeyras,%20olah&amp;amp;source=bl&amp;amp;ots=PJcREh-vRl&amp;amp;sig=_0x88Ff6y08ZnoGYitOfzCfGH04&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=fnHzU6SNIIKcjAKzzoCYAw&amp;amp;ved=0CFEQ6AEwBg#v=onepage&amp;amp;q=equilibria&amp;amp;f=false&quot;&gt;see p 52&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does the solvent play a role in magic acid's acidity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no solvent.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-19T16:25:50.283" LastActivityDate="2014-08-19T16:25:50.283" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15431" PostTypeId="2" ParentId="15421" CreationDate="2014-08-19T16:08:16.230" Score="21" Body="&lt;p&gt;The explanation is that sufficiently strong Lewis acids like $\ce{SbF5}$ will form adducts even with very weak, traditionally non-coordinating Lewis bases like $\ce{SO3F-}$. The resulting anionic complexes are extremely stable, with the negative charge distributed over a large, highly polarizable molecule and inductively stabilized by the fluorine atoms. The formation of this complex, in turn, promotes the autoprotolysis equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HSO3F + HSO3F &amp;lt;=&amp;gt; SO3F- + H2SO3F+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protonated species above is the most active molecule in sufficiently concentrated solutions. (In reality, the composition of these mixtures may be much more complex, with dimers of the form $\ce{Sb2F10(SO3F)-}$ and possibly more complex oligomeric structures. See, for example, this &lt;a href=&quot;http://books.google.com/books?id=2g49AAAAIAAJ&amp;amp;pg=PA344&amp;amp;lpg=PA345#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;book&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are some more recent and exotic examples of such Brønsted/Lewis acid mixtures, the idea is actually a fairly old one. Perhaps the most common example: sulfur trioxide in sulfuric acid gives a solution known as oleum, which contains a proportion of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrosulfuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;disulfuric acid&lt;/a&gt; molecules. The resulting mixture is considerably more acidic than pure sulfuric acid alone and has various uses in industrial applications as well as laboratory-scale synthetic organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why the oxygens cannot be subsituted with fluorines to yield an even stronger acid system, I can only give a speculative answer. It seems to me that a hypothetical molecule such as, e.g., $\ce{SF3OH}$ (if this is what you're imagining) would probably be very liable to decompose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SF3OH -&amp;gt; SOF2 + HF}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I can find absolutely no relevant literature, which in and of itself says something about the stability of such a molecule (or perhaps my ineptitude, as the case may be).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-03T11:25:50.850" LastActivityDate="2018-03-03T11:25:50.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15432" PostTypeId="2" ParentId="15429" CreationDate="2014-08-19T16:14:44.823" Score="5" Body="&lt;p&gt;There are no protons moving - protons cannot move in resonance structures.  In resonance structures all nuclei must remain fixed, only electrons can move.  As you progress from the top-left resonance structure to the top-middle resonance structure you have taken the two pi electrons in what I'll call the 3-4 bond (position 1 being where the bromine is attached) and moved them between carbons 2 and 3, creating a pi bond between carbons 2 and 3 and simultaneously moving the electron vacancy (the positive charge, what a physicist would call a &quot;hole&quot;) to carbon 4.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-19T16:14:44.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15433" PostTypeId="2" ParentId="15336" CreationDate="2014-08-19T16:35:11.897" Score="0" Body="&lt;p&gt;I have read reports of amalgam fillings producing as much as 50 µg per day of vapor with a 80% uptake through the lungs. This is what worries me. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0041008X13005644&quot; rel=&quot;nofollow&quot;&gt;There is evidence that the half-life of mercury in the brain is many years&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7493" LastEditorUserId="1887" LastEditDate="2014-08-19T17:25:11.940" LastActivityDate="2014-08-19T17:25:11.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15434" PostTypeId="2" ParentId="15428" CreationDate="2014-08-19T16:37:34.980" Score="6" Body="&lt;p&gt;In my view there is an important difference between empty orbitals and empty valence orbital.  There are an infinite number of orbitals, a finite number of filled orbitals, but only one valence orbital.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of a valence electron is usually something like, &quot;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/&quot;&gt;The electrons in the outermost shell are the valence electrons--the electrons on an atom that can be gained or lost in a chemical reaction&lt;/a&gt;.&quot;  Therefore, a valence orbital would be the orbital holding those electrons.  If the valence orbital is full, it is full; there are no other valence orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this understanding, it would be correct to say &quot;the fluoride anion has no empty valence orbitals.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-19T16:37:34.980" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15435" PostTypeId="1" CreationDate="2014-08-19T16:49:33.020" Score="9" ViewCount="10035" Body="&lt;p&gt;For the Electrolysis of Copper Chloride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{Cu^{2+} + 2e- &amp;lt;=&amp;gt; Cu}$&lt;br&gt;&#xA;Anode: $\ce{2Cl &amp;lt;=&amp;gt; Cl2 + 2e- }$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about the reaction taking place at the anode. Wouldn't $\ce{H2O}$ rather undergo oxidation since it has an increased reducing ability? eg. It is above $\ce{2Cl-}$ on the Table of Standard Reduction Potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Or do all halogens discharge more readily than water?&lt;/p&gt;&#xA;" OwnerUserId="5705" LastEditorUserId="4945" LastEditDate="2014-08-20T02:41:58.060" LastActivityDate="2016-05-05T13:20:38.003" Title="Why is chloride oxidised instead of water in copper chloride electrolysis?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;&lt;electrolysis&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15436" PostTypeId="1" AcceptedAnswerId="15484" CreationDate="2014-08-19T17:25:25.680" Score="3" ViewCount="93" Body="&lt;p&gt;I'm about to conduct a study of possible toxic traces from the substance M-ENDO-Agar LES in substrate after destruction with heat (I think it is 160℃) for 180 minutes. The substance is carcinogenic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find a way to measure if the substrate is still toxic in any way after the heat treatment. How could I conduct such a study?&lt;/p&gt;&#xA;" OwnerUserId="7533" LastEditorUserId="73" LastEditDate="2014-08-21T08:47:39.197" LastActivityDate="2014-08-21T17:04:56.767" Title="How can I measure carcinogenic risks of an M-ENDO-Agar LES substrate after heat destruction?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15437" PostTypeId="1" AcceptedAnswerId="15441" CreationDate="2014-08-19T17:33:29.227" Score="1" ViewCount="870" Body="&lt;p&gt;I am told by Dr. Klein that in the nitration of benzene, hydrogen sulfate anion acts as the base and deprotonates the sigma complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3xXIX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why would the hydrogen sulfate anion be doing the deprotonation? We've &lt;em&gt;just&lt;/em&gt; established that sulfuric acid is a stronger acid than nitric acid; hence sulfuric acid is able to protonate nitric acid to some extent. Now are we expected to believe that the conjugate base of a stronger acid is &lt;em&gt;stronger&lt;/em&gt; than the conjugate base of a weaker acid? Why not have nitrate anion do the deprotonation? It is a stronger base than hydrogen sulfate anion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="305" LastEditDate="2014-08-19T19:23:14.493" LastActivityDate="2014-08-19T19:23:14.493" Title="Nitration of Aromatic Compounds" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15438" PostTypeId="1" AcceptedAnswerId="15446" CreationDate="2014-08-19T17:52:53.920" Score="6" ViewCount="39776" Body="&lt;p&gt;We just had chemistry practical few days a go. The topic was strong acid-base titration, in which we had to find the normality and concentration of $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher instructed me to add 2 drops of Phenolphthalein  in 10 ml $\ce{HCl}$, and then add $\ce{NaOH}$ from buret and take a reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We needed to repeat the process 4 times and then find the constant reading, but in that I never used to take 2 drops of Phenolphthalein, instead I used to take 5-6 drops and then also used to get almost same reading as of pilot reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether it is compulsory to take 2 drops of Phenolphthalein or it was just a instruction so students don't waste it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://chemmovies.unl.edu/chemistry/smallscale/SS044.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; same instruction given. (Procedure point 2)&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="73" LastEditDate="2014-09-04T17:03:14.013" LastActivityDate="2016-06-07T22:40:21.697" Title="Amount of Phenolphthalein required" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15439" PostTypeId="1" AcceptedAnswerId="15443" CreationDate="2014-08-19T17:56:22.897" Score="0" ViewCount="1469" Body="&lt;p&gt;I am told that the sulfur trioxide molecule exhibits charge separation because of poor p-orbital overlap. Sulfur's 3p orbitals are much bigger than oxygen's 3p orbitals, and thus the $\ce{S=O}$ bond is best described as $\ce{S+ - O-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p3qvh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is the author's argument suggesting that sulfur trioxide is unhybridized, and that it uses pure s orbitals to form sigma bonds and pure p orbitals to form pi bonds? Because that's the only way I can see sulfur as having three empty p orbitals to utilize. &lt;em&gt;If&lt;/em&gt; we were to assign hybridization through counting the number of sigma pairs of electrons (3) we would arrive at sp2 hybridization, and there aren't enough p-orbitals to go around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also while we're on this topic, Wikipedia says that sulfur trioxide is best represented as having one $\ce{S=O}$ bond and two $\ce{S+ - O-}$ bonds. Now, does bond length differ between an  $\ce{S+ - O-}$ and an  $\ce{S=O}$ bond? Wikipedia suggests not, as only one bond length is given. If bond length is equivalent between the two, why? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-19T18:08:31.057" LastActivityDate="2014-08-19T18:58:51.700" Title="Sulfur Trioxide - Ionic Character" Tags="&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15440" PostTypeId="2" ParentId="15394" CreationDate="2014-08-19T18:24:32.913" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But H&gt;Ag in reactivity, then how could Ag displace H from HCl ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What do you mean that hydrogen is more reactive than silver? In what context? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there was a reaction, AgCl and HNO3 were to form. But then these&#xA;  would react again to give us the original compounds back.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you know? Why would these form? Why not silver nitrate? Why not hydrochloric acid? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-19T18:35:23.640" LastActivityDate="2014-08-19T18:35:23.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15441" PostTypeId="2" ParentId="15437" CreationDate="2014-08-19T18:25:23.947" Score="5" Body="&lt;p&gt;It's not advisable to consider these types of reaction schemes in a slavishly literal fashion. In reality, there are multiple proton transfer equilibria occurring simultaneously, and the reactions will generally be quite rapid. While it's logical and technically correct to depict the most basic of the molecules (among those present in sufficiently large concentration) performing the proton abstraction, it's also of literal practical interest or importance in many cases. I suspect the reason that $\ce{HSO4-}$ was chosen is to emphasize that the sulfuric acid is actually catalytic, so it should be regenerated in the course of the depicted reaction mechanism. Of course, water is produced as a byproduct, so the sulfuric acid will not exist primarily in its neutral molecular form.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-08-19T18:25:23.947" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15442" PostTypeId="1" CreationDate="2014-08-19T18:41:17.293" Score="17" ViewCount="6065" Body="&lt;p&gt;I know sugars have $\ce{-OH}$ groups, and that sulfuric acid is a strong acid. What I'm failing to see is how exactly sulfuric acid dehydrates sugars. Does it protonate the hydroxyl groups, making them good leaving groups? What about the case of sucrose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TMcOe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how are the oxygens &quot;taken care of&quot; by the sulfuric acid? Are they doubly protonated? Is there some established mechanism for the dehydration of sugars by sulfuric acid, because I can't find any - I just find general equations and I've looked on Google and through textbooks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some further research I learned that the gas, sulfur dioxide, is evolved. This suggests to me that this reaction is also redox in nature. So is this reaction both an acid-base reaction &lt;em&gt;and&lt;/em&gt; a redox reaction? Sulfur in sulfuric acid has an oxidation state of +6; sulfur in sulfur dioxide has an oxidation state of +4. Plus the elemental carbon formed has an oxidation state of 0 while in sucrose, carbon has various oxidation states. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4231" LastEditDate="2015-03-07T00:01:11.033" LastActivityDate="2019-01-22T08:33:05.583" Title="How does sulfuric acid dehydrate sugars?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="4" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15443" PostTypeId="2" ParentId="15439" CreationDate="2014-08-19T18:58:51.700" Score="4" Body="&lt;p&gt;Sulfur trioxide is planar with 120 degree O-S-O bond angles.  If the molecule were unhybridized and we used the 3 orthogonal p-orbitals for our sigma bonds, then we would expect the O-S-O angle to be 90 degrees.  Further, in the unhybridized scheme, there are not enough s orbitals to form 3 sigma bonds that are pure s. Just considering s and p orbital involvement, we could say that the sulfur is $\ce{sp^2}$ hybridized.  This would produce 3 $\ce{sp^2}$ orbitals for sigma bonds from sulfur to oxygen and leave one p orbital for one pi bond.  We can draw 3 resonance structures for this arrangement, so we would expect all bond lengths to be equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nkrfF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-19T18:58:51.700" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15444" PostTypeId="2" ParentId="7720" CreationDate="2014-08-19T19:24:15.783" Score="11" Body="&lt;p&gt;Knowledge of the crystal structure implies that you are also able to describe the unit cell of the compound. To determine its formula, count all sorts of atoms inside the unit cell, also considering fractional atoms which belong to more than one unit cell. For example, an atom at the corner of the unit cell counts as $\frac{1}{8}$ atom because it is shared between 8 adjacent unit cells. Likewise, an atom at the edge of the unit cell counts as $\frac{1}{4}$ as it belongs to 4 unit cells etc. When you obtain fractional values for the atom numbers, multiply the values of all atoms with a certain number so that all values are integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us take sodium chloride ($\ce{NaCl}$) as a simple example. Its unit cell, which is shown below, contains $\ce{Na+}$ (violet) and $\ce{Cl-}$ ions (green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pKEdH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Counting the number of chloride ions in the unit cell gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$12 \cdot \frac{1}{4}+1=4 \ \ce{Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because there are 12 $\ce{Cl-}$ on the edges and 1 $\ce{Cl-}$ in the center of the unit cell. For the sodium ions, which are situated at the corners and on the faces (count as $\frac{1}{2}$ because they are shared between 2 cells) of the unit cell, we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8 \cdot \frac{1}{8}+6 \cdot \frac{1}{2}=4 \ \ce{Na+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition of the unit cell is therefore $\ce{Na4Cl4}$. Since the number of formula units per unit cell is 4, the formula of the compound is $\ce{NaCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compounds in which the building blocks of the crystal structure are molecules, the procedure is essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-19T19:24:15.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15446" PostTypeId="2" ParentId="15438" CreationDate="2014-08-19T19:51:55.163" Score="9" Body="&lt;p&gt;Using &lt;em&gt;a little&lt;/em&gt; extra phenolphthalein won't matter. You just need enough so that the color is visible once the pH becomes basic enough. If you were titrating a large volume, you might need more than a couple of drops, but 2 drops is enough to be sure that the color will be visible in a typical lab-scale titration vessel. One drop would probably even work. The reason they say &quot;two drops&quot; and not &quot;at least two drops&quot; is that they don't want you to be wasteful.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-20T15:52:34.257" LastActivityDate="2014-08-20T15:52:34.257" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15447" PostTypeId="1" AcceptedAnswerId="15453" CreationDate="2014-08-19T20:03:22.720" Score="12" ViewCount="1031" Body="&lt;p&gt;It's been documented that NASA hope to capture an asteroid in 2025, and have subsequent aims to mine that asteroid. If if this is successful, we would expect other asteroids to be mined in the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consequence of this is that relative atomic masses of elements mined—those with two or more stable isotopes—will no longer be faithful to our current periodic table. Ruthenium alone, a high-value rare earth, has five stable isotopes alone, ranging from $^{98}\ce{Ru}$ to $^{102}\ce{Ru}$. The relative abundances of these isotopes are bound to differ in other places other than Earth, and so would the weighted average. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say after mining our ruthenium, we use it in an electronic device and consequently that device is thrown away. For argument's sake, also say that this ruthenium—of an isotopic distribution never before seen—leaches into the environment and later comes into contact with biological systems. Could it be possible that compounds hitherto considered non-toxic on Earth become toxic to biological systems by virtue of a newly realised isotopic discrimination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, $^{13}\ce{CO2}$  is effectively discriminated against in uptake by plants in comparison to $^{12}\ce{CO2}$  (due to it being a diffusion limited reaction). Ruthenium is purported to be carcinogenic, but the most abundant isotope on Earth of Ruthenium is $^{102}\ce{Ru}$ which could be effectively non-toxic by virtue of the fact that  compounds containing these atoms takes so long to diffuse, they don't diffuse across biological membranes at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a sample of Ruthenium from the moon found its most abundant isotope in $^{98}\ce{Ru}$, then this would be expected to diffuse much faster than its heavier counterparts. Therefore could it be possible that a compound previously thought of as non-toxic could become toxic via this effect. Does this sound plausible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: I've deliberately missed out other routes of exposure such as ingestion to only consider diffusion controlled reactions.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-20T09:26:48.787" LastActivityDate="2016-03-12T00:43:17.600" Title="Biological Consequences of Asteroid Mining—Death by Isotope?" Tags="&lt;biochemistry&gt;&lt;periodic-table&gt;&lt;isotope&gt;&lt;astrochemistry&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15448" PostTypeId="2" ParentId="15435" CreationDate="2014-08-19T20:07:55.623" Score="1" Body="&lt;p&gt;It depends upon the concentration of chloride.  If a dilute NaCl solution is electrolysed, $\ce{O2}$ is produced at first, along with $\ce{H2}$. As the volume of the solution decreases $\ce{[NaCl]}$ increases, and eventually $\ce{Cl2}$ is formed. You can use the Nernst equation to find out the chloride concentration at which $\ce{Cl2}$ is formed over $\ce{O2}$.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="4945" LastEditDate="2014-08-20T02:45:57.500" LastActivityDate="2014-08-20T02:45:57.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15449" PostTypeId="2" ParentId="15394" CreationDate="2014-08-19T20:12:29.947" Score="13" Body="&lt;p&gt;One of the most difficult parts of chemistry is learning to recognize a type of reaction based solely on its reactants. This has to be done before you can apply a reaction pattern to the problem, and so it is critical to get this step right in the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you are looking at the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{AgNO3 + HCl -&amp;gt;  AgCl + HNO3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already know the products, but are questioning how these products can be formed since the activity of H is greater than Ag, implying that Ag is easier to oxidize than H. This is true - but take a look at the oxidation numbers of each species in this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ag: +1 -&amp;gt; +1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO3: -1 -&amp;gt; -1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H: +1 -&amp;gt; +1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl: -1 -&amp;gt; -1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no overall transfer of electrons, this can't be a redox reaction, and that means activities won't matter here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means there must be another driving force for this reaction - another reaction pattern that fits better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to give you the answer directly, but I will give you some advice that should help: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the full equation - &lt;em&gt;including the phases.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{AgNO3(aq) + HCl(aq) -&amp;gt;  AgCl(s) + HNO3(aq)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if you can find another reaction pattern that fits this equation better. Later, I'll update this answer to show how you can identify the correct pattern for aqueous reactions using only the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned, one of the hardest parts of chemistry is learning to identify reaction patterns based only on the reactants. This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_metathesis_reaction&quot;&gt;double-displacement&lt;/a&gt; (or metathesis, or precipitation) reaction. It is easy to see now, given that the reactants are aqueous and at least one of the products is solid (the precipitate). Formation of the solid is the driving force for this reaction - the quick explanation is that the forces attracting silver and chloride ions together are stronger than the solvation forces between those ions and water, as well as the forces holding them to sodium and nitrate ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is good, but how do we predict that this is a precipitation reaction ahead of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is to have a good understanding of the three common types of salt/acid/base reactions that occur in aqueous solution, and to learn to recognize clues in the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, the three types that are most commonly seen are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Precipitation&lt;/li&gt;&#xA;&lt;li&gt;Acid/base (Arrhenius definition)&lt;/li&gt;&#xA;&lt;li&gt;Redox (single displacement)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can recognize each of these by the reactants if you know what to look for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Precipitation&lt;/strong&gt; - Two soluble salts (ionic compounds), or a salt with an acid or base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acid/base&lt;/strong&gt; - An Arrhenius acid &lt;em&gt;and&lt;/em&gt; base (compound containing hydroxide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Redox&lt;/strong&gt; - a salt or acid and an elemental metal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other types of reactions that occur in aqueous solution, and many variations of the acid/base and redox category, but these three cover the cases most commonly seen in a classroom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have identified the likely pattern that the reaction will follow, the next step is to predict the products using that pattern to see if they make sense. If they do, then you have likely chosen correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this algorithm for more advanced chemistry as well - in organic chemistry, for example, one of the major goals is to learn to predict reactions based on functional groups. Once you can identify functional groups and have memorized reaction patterns for them, it becomes possible to predict a huge range of reactions. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-10-02T18:10:18.003" LastActivityDate="2015-10-02T18:10:18.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15450" PostTypeId="2" ParentId="15447" CreationDate="2014-08-19T23:05:03.730" Score="11" Body="&lt;p&gt;Whatever the isotopes are for asteroidal material (and they are mostly close to those seen down here on Earth), they are contained in the 5 to 100 tons of meteoritic material that falls onto the Earth's atmosphere (and thus filters down to us on the surface) &lt;strong&gt;every day&lt;/strong&gt;. It will be a long time before the cumulative pollution from asteroid (or lunar) mining can match the meteoritic rate. &lt;/p&gt;&#xA;" OwnerUserId="7535" LastActivityDate="2014-08-19T23:05:03.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15451" PostTypeId="2" ParentId="15408" CreationDate="2014-08-19T23:07:21.693" Score="8" Body="&lt;p&gt;The acidic water is probably protonating nicotine, causing it to carry a positive charge. Once charged, the resulting nicotinium cation will not be volatile, but rather, highly soluble. Based on this, I would expect that acidifying the water causes a much lower amount of nicotine  to be pulled through the exit hose (because it is trapped in the acidic water).&lt;/p&gt;&#xA;" OwnerUserId="7211" LastActivityDate="2014-08-19T23:07:21.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15452" PostTypeId="1" AcceptedAnswerId="15454" CreationDate="2014-08-19T23:49:09.833" Score="2" ViewCount="197" Body="&lt;p&gt;I would like to use non-food grade $\ce{NaOH}$ for $\ce{pH}$ water treatment. Any problem with this mixing with water that contains iron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{NaOH}$ will be pumped into a well. As a $25\%$ solution. Then sent through an ozone filter and a water softener.  Eventually, it will be used for pet and human consumption.&lt;/p&gt;&#xA;" OwnerUserId="7536" LastEditorUserId="4945" LastEditDate="2014-08-20T01:51:30.160" LastActivityDate="2014-08-26T18:02:10.140" Title="Using non-food grade NaOH for pH control" Tags="&lt;inorganic-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15453" PostTypeId="2" ParentId="15447" CreationDate="2014-08-19T23:59:36.780" Score="13" Body="&lt;p&gt;This is an interesting question and you raise a number of points, let's step through them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A consequence of this is that relative atomic masses of elements&#xA;  mined—those with two or more stable isotopes—will no longer be&#xA;  faithful to our current periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this is already happening.  $\ce{^235U}$ constitutes 0.72% of uranium found on earth and decays to the stable isotope $\ce{^207Pb}$, which is found in a natural abundance of 22.1%.  Before we get to the asteroid the abundance of various isotopes and their weighted average mass is already shifting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;98 Ru , then this would be expected to diffuse much faster than its&#xA;  heavier counterparts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diffuse much faster?  First off, we're talking about diffusion, not chemical reaction.  Elements or compounds usually enter the body by ingestion of one sort or another (eating, breathing) rather than diffusion, and these ingestion pathways wouldn't involve isotopic discrimination.  That is, if we breath air that contains isotopes in a certain ratio, then that's the ratio that will initially appear in our lungs.  Nonetheless if diffusion were found to be an issue, my guess is that isotopic discrimination by diffusion would be a very small effect.  For a gas the maximum separation of two isotopes is given by &#xA;$$\mathrm{\sqrt{\frac{[MW~of~compound ~with ~isotope~1]}{[MW~of~compound ~with ~isotope~2]}}}$$&#xA;In the case of separating $\ce{^235UF6/^238UF6}$ this amounts to a fractionation ratio of 1.0043 after 1 pass.  Ruthenium is lighter, so the effect would be larger, about 1.02 ($\ce{^102Ru/^98Ru}$) after 1 pass, even less if it is in a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical reaction within our body of ingested isotopic compounds would also show  discrimination due to primary kinetic isotope effects.  The maximum primary kinetic isotope effect is proportional to the reduced mass as follows:&#xA;$$k~ \thicksim ~\sqrt{\frac{m_1 + m_2}{m_1 m_2}}$$&#xA;Applying this to $\ce{^102Ru}$ and $\ce{^98Ru}$ yields a primary kinetic isotope effect of 1.020, again, even less if the element is incorporated into a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, the effects look small and since shifts in natural abundance have been occurring for a long time here on earth, with no one raising a flag, my guess (and that's all it is, a guess) is that it's not something to worry about.  Still, I'd feel better if a NASA toxicologist looked at the question and said &quot;no problem&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:43:17.600" LastActivityDate="2016-03-12T00:43:17.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15454" PostTypeId="2" ParentId="15452" CreationDate="2014-08-20T00:19:23.980" Score="5" Body="&lt;p&gt;Iron hydroxide is insoluble, so it will precipitate out. If there are other metals that aren't  alkali metals or alkaline earth metals heavier than calcium, they will also precipitate out. Most of these reactions will release some heat, but if the concentration of the metal ions is low, that shouldn't matter too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plan to drink the water, using non-food grade $\ce{NaOH}$ might cause a problem, because you won't know what else is in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add more details, we can probably tell if there are any other things to be concerned about. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-20T00:19:23.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15455" PostTypeId="2" ParentId="15421" CreationDate="2014-08-20T05:28:01.603" Score="11" Body="&lt;p&gt;In addition to the answers of ron and Greg E. I just want to add another concept, that was not mentioned so far: &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-coordinating_anion&quot;&gt;non-coordinating anions&lt;/a&gt;. You can stop reading here or continue and see a short summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second remark, it is important to know, that &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorosulfuric_acid&quot;&gt;fluorosulfuric acid&lt;/a&gt; is already a &lt;a href=&quot;http://en.wikipedia.org/wiki/Superacid&quot;&gt;superacid&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Acidity is strongly dependent on the strength of the proton-conjugated-base bond. Basically stating, the weaker the bond, the higher is the acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All currently known superacids operate on this principle and &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Magic_acid&quot;&gt;magic acid&lt;/a&gt;&quot; is one of the strongest. The reason for the extraordinary acidity is the high potency of &lt;a href=&quot;http://en.wikipedia.org/wiki/Antimony_pentafluoride&quot;&gt;antimony pentafluoride&lt;/a&gt; as a lewis acid. It will coordinate to almost anything with electrons, especially itself in liquid phase, forming chain oligo-/polymers $\ce{[SbF5]_{n}}$ with a length between $\ce{n=5-10}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is something more potent than this, it will go for it. Antimony pentafluoride is a very dangerous substance. This is exactly what is happening in magic acid, forming weakly or non-coordinating anions as conjugated base.&#xA;$$\ce{SbF5 + HSO3F &amp;lt;=&amp;gt; H+ + [SbF5SO3F]^{-}}.$$&#xA;Now this is a fairly large anion with a highly delocalised (due to symmetry constraints) charge. With excess of antimony pentafluoride, even larger anions can be formed, which is the case when used as a solvent. &#xA;$$\ce{[SbF5SO3F]^{-} + SbF5 &amp;lt;=&amp;gt; [(SbF5)2SO3F]^{-}}$$&#xA;As a consequence, the bond to the proton is extremely weak. In other words, there is almost no coordination of this proton, leaving it basically naked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another superacid, using the same concept, which is even stronger: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot;&gt;Fluoroantimonic acid&lt;/a&gt;. It is in principle that, what you already suggested, substituting the oxygens by fluorines. Instead of $\ce{HSO3F}$, hydrofluoric acid $\ce{HF}$ is used, giving rise to the (to date) most powerful acid. It also makes use of the high affinity of $\ce{SbF5}$ towards fluoride anions. &#xA;$$\ce{2SbF5 + HF &amp;lt;=&amp;gt;&amp;gt; H+ + [Sb2F11]- }$$&#xA;With these naked protons it is possible to protonate aliphatic hydrocarbons to form carbocations.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-08-20T05:34:32.750" LastActivityDate="2014-08-20T05:34:32.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15456" PostTypeId="2" ParentId="9793" CreationDate="2014-08-20T08:40:06.863" Score="2" Body="&lt;p&gt;The book &lt;em&gt;Chemistry of the Elements&lt;/em&gt; by N. N. Greenwood is an absolutely huge handy tome full of information, which is organised by element or clusters of similar elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find this as a handy overall chemistry reference, but most common ores of the elements (where applicable) and geological extraction information are also given. An essential to have on any chemistry shelf. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-20T08:40:06.863" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15457" PostTypeId="5" CreationDate="2014-08-20T09:29:59.017" Score="0" Body="&lt;p&gt;Current astrochemical research has been directed at interstellar ice and dust grains; crystals of &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Volatiles&quot; rel=&quot;nofollow&quot;&gt;volatiles&lt;/a&gt;&lt;/em&gt; that can take hundreds of millions to form. This gives rise to myriad crystal types not often seen on Earth owing to the very specific conditions required for their formation (extremely low temperatures being one such criterion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further topics might include heterogeneous catalysis on these icy grain surfaces, a cause of some very unique interstellar chemistry (see &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2006/FD/b516770a#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;Laboratory investigations of the role of the grain surface in astrochemical models&lt;/a&gt; for more info). Life on other planets is also another subject, although still of a highly speculative nature. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-20T11:02:44.083" LastActivityDate="2014-08-20T11:02:44.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15458" PostTypeId="4" CreationDate="2014-08-20T09:29:59.017" Score="0" Body="Chemistry occurring on asteroids, comets, in the interstellar medium or on other planets.  " OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-20T11:02:47.623" LastActivityDate="2014-08-20T11:02:47.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15460" PostTypeId="2" ParentId="15412" CreationDate="2014-08-20T11:01:42.167" Score="15" Body="&lt;p&gt;There is no prove for this statement, as it is most likely incorrect for 2,6-xylidene. This is based on an electronic structure theory approach. However, there are non-negligible steric interactions, that affect the basicity as demonstrated in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases&quot;&gt;Ortho-effect in substituted aromatic acids and bases&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever a distance between two atoms becomes (significantly) shorter than the sum of their respective van der Waals radii, this has to be considered. Because of the strain in the xylidene molecules, this is the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a very rough estimate you can use the pen and pencil approach with tabulated values for &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;noreferrer&quot;&gt;van der Waals radii&lt;/a&gt; (&lt;a href=&quot;http://de.wikipedia.org/wiki/Van-der-Waals-Radius&quot; rel=&quot;noreferrer&quot;&gt;german&lt;/a&gt; - much better overview) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_radius&quot; rel=&quot;noreferrer&quot;&gt;covalent radii&lt;/a&gt;. Another approach is building the molecule with a molecular modeling set (&lt;a href=&quot;http://de.wikipedia.org/wiki/Molek%C3%BClbaukasten&quot; rel=&quot;noreferrer&quot;&gt;Molekülbaukasten -  german wikipedia&lt;/a&gt;), rotate and wiggle bonds, to see if there are any close contacts.&#xA;There is also some free software that lets you build molecules and check them, they usually come equipped with a set of standard bond lengths and angles. It is always a good approach to visualise a molecule.&lt;br&gt;&#xA;If you have access to quantum chemical tools you can perform calculations, but this is usually too much of an effort. For this educational purpose I ran some quick calculations, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following table is composed from values from the german wikipedia pages (values in Å):&#xA;\begin{array}{rrr}\hline&#xA;             &amp;amp; \text{van der Waals} &amp;amp; \text{covalent}\\\hline&#xA;\ce{H}       &amp;amp; 1.10  &amp;amp; 0.32 \\&#xA;\ce{C}       &amp;amp; 1.70  &amp;amp; 0.77 \\&#xA;\ce{N}       &amp;amp; 1.55  &amp;amp; 0.71 \\\hline&#xA;\sum\ce{H-H} &amp;amp; 2.20  &amp;amp; 0.64 \\&#xA;\sum\ce{C-H} &amp;amp; 2.80  &amp;amp; 1.09 \\&#xA;\sum\ce{N-H} &amp;amp; 2.65  &amp;amp; 1.03 \\&#xA;\sum\ce{C-N} &amp;amp; 3.20  &amp;amp; 1.48 \\&#xA;\sum\ce{C-C} &amp;amp; 3.40  &amp;amp; 1.54 \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can have a look at the optimised (DF-BP86/def2-TZVPP) geometry of the molecule. (Of course the pen and paper approach will result in a different geometry.)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/msRp3.jpg&quot; alt=&quot;molecular geometry&quot;&gt;&lt;br&gt;&#xA;Here you can see, that the $\ce{H\cdots{}H}$ distance is very close to the sum of the van der Waals radii. The $\ce{C\cdots{}H}$ distance with about $2.5~\mathrm{Å}$ and the $\ce{C\cdots{}N}$ distance with about $2.9~\mathrm{Å}$ are already shorter than the sum f their van der Waals radii. While these interactions are not strong, the will effect the behaviour of this molecule in solution (and gas phase). It is very important to understand, that these are only snapshot values. In reality the molecule is very flexible and bonds will rotate and vibrate. Steric interactions are basically electronic and dispersive interactions in disguise - they are most likely to influence the kinetic behaviour of a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best evidence for this is the transition state of the rotation of the amine moiety. Here we can clearly see a bonding interaction between the nitrogen and the neighbouring hydrogen (see linked post), which is evidently influencing the basicity of this molecule.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/97JJF.jpg&quot; alt=&quot;molecular geometry&quot;&gt;&lt;br&gt;&#xA;This state is only about $7~\mathrm{kcal/mol}$ higher in energy, which means it is readily accessible at room temperature. This will mean that there are many different conformations present, there is no true one. This affects the reactivity of a molecule (in this case lowered basicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, your gut feeling was correct. However, there are more effects at play and the whole picture is often not easy to see. It is usually a case by case decision/ interpretation how strong which effects affect reactions and molecules (properties and barriers). When substituents get larger the more dominant steric effects will obviously be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question sufficiently.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-08-20T11:01:42.167" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15461" PostTypeId="2" ParentId="15447" CreationDate="2014-08-20T17:33:02.100" Score="7" Body="&lt;p&gt;Also, if you follow what @ron said, for large isotope effect you generally need lighter elements. So the isotope effect for deuterium and hydrogen can be large. But even in these cases biological effects can be mostly observed only when one use high concentrations of D2O. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that diffusion control would play any role in the biology of metal ions. If we are talking about transition metals with biological role, nature is generally relies on transport enzymes and similar to carry around them. Isotope effects are most pronounced in chemical reactions with relatively high activation barrier and strong vibrational coupling with the given atom. In practice it means enzymatic reactions with active centers where there is a significant role of the given metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we even know any enzymes in which's active center Ru plays important role? You can ask this for any elements that we expect to mine, In practice, isotope effects hardly can seen even for iron or cobalt atoms, which are pretty common active centers compared to most other metals. If you study natural iron enzymes, the isotopic ratio of 56/57 iron is not anomalous. So I wouldn't expect much biological effect to anything that heavy or heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are going beyond biological function, and talking about metals and toxicity, most of the time the complex formation between e.g. proteins and the given metal ions is the important question to study. I don't think is it much influenced by isotope effects, and I definitely would not expect that a toxic heavy metal became harmless or similar drastic change.    &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-20T17:33:02.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15462" PostTypeId="1" CreationDate="2014-08-20T17:42:46.413" Score="3" ViewCount="537" Body="&lt;p&gt;I have a PolyA tailing kit for putting PolyA tails on the 3' end of RNA. The kit comes with a proprietary 5X buffer and a 25mM MnCl2 solution, and in the standard set up I use 20uL of 5X buffer and 10uL of 25mM MnCl2. When trying to force the enzyme to accept an alternative substrate I did the reaction without the MnCl2 and it worked. Further research revealed that the enzyme also works with Mg++ ions, even though the rate is faster with Mn++, and some old papers on the enzyme mention using both Mg and Mn in the buffer. This leads me to believe that the buffer contains Mg++, but I don't know for sure because the manufacturer won't tell me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I immediately thought I could use atomic absorption spectroscopy, but our department does not have one. I looked up colorimetric assays online, but the kits cost about $300, which is more than I want to spend on this idea. What I'm asking is if anyone knows a relatively simple method to detect magnesium in a solution? I have access to a UV-Vis spectrophotometer and fluorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: After digging for other phone numbers for the manufacturer I found someone who would tell me a little about the buffer, and it does contain magnesium. He couldn't me how much though. Spectro-Phone-otemotry.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="137" LastEditDate="2014-08-20T17:58:49.950" LastActivityDate="2014-08-20T21:49:25.417" Title="Detection of magnesium in a buffer?" Tags="&lt;analytical-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15463" PostTypeId="2" ParentId="15462" CreationDate="2014-08-20T18:25:54.807" Score="3" Body="&lt;p&gt;$\ce{Mg^2+}$ forms complexes with 4-(2-Pyridylazo)resorcinol that can be detected spectrophotometrically: &lt;em&gt;Simultaneous spectrophotometric determination of calcium and magnesium in water&lt;/em&gt;, described by E. Gómez, J.M. Estela, V. Cerdà in &lt;em&gt;Analytica Chim. Acta&lt;/em&gt; &lt;strong&gt;1991&lt;/strong&gt;, &lt;em&gt;249&lt;/em&gt;, 513-518.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(DOI: &lt;a href=&quot;http://dx.doi.org/10.1016/S0003-2670(00)83027-5&quot; rel=&quot;nofollow&quot;&gt;10.1016/S0003-2670(00)83027-5&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-08-20T18:25:54.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15464" PostTypeId="1" AcceptedAnswerId="15468" CreationDate="2014-08-20T19:19:56.483" Score="1" ViewCount="2265" Body="&lt;p&gt;1) When can't the rate-determining step be applied, and why? How do you recognize these cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) When can't the steady state approximation be applied, and why?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4231" LastEditDate="2014-08-21T01:54:59.153" LastActivityDate="2014-08-21T01:54:59.153" Title="Rate-determining step and steady state approximation failure" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15465" PostTypeId="1" AcceptedAnswerId="15466" CreationDate="2014-08-20T20:52:47.913" Score="3" ViewCount="1166" Body="&lt;p&gt;I was wondering about this specific scenario:&lt;br&gt;&#xA;If you have a cyclohexane, for example, and the cyclohexane has two consecutive substituent groups, for example a methyl group and an ethyl group, how do you decide where the numbering starts?&lt;br&gt;&#xA;Would it be 1-ethyl-2-methylcyclohexane or would it be 2-ethyl-1-methylcyclohexane?&#xA;Whoever answers this, could you just give some sort of general rules on what to do in this situation (e.g. does it follow alphabetical order, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="7546" LastEditorUserId="4945" LastEditDate="2014-08-21T02:28:50.230" LastActivityDate="2014-08-21T02:28:50.230" Title="What are the priorities for substituent groups on a cyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15466" PostTypeId="2" ParentId="15465" CreationDate="2014-08-20T21:11:55.027" Score="4" Body="&lt;p&gt;If the numerical sequence of locants is equivalent, the numbering should be assigned to correspond with alphabetical order. Hence, in your example, correct would be 1-ethyl-2-methylcyclohexane. Quoting the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm&quot; rel=&quot;nofollow&quot;&gt;IUPAC Blue Book&lt;/a&gt;, Rule A-2.4:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-08-20T21:11:55.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15467" PostTypeId="2" ParentId="15462" CreationDate="2014-08-20T21:49:25.417" Score="0" Body="&lt;p&gt;There is also the venerable complexometric determination of magnesium with EDTA.  Determining hardness of tap water is a popular lab exercise, your chemistry department's teaching lab might even have EDTA solution and complexometric indicator in stock.  It's worth a shot.&lt;/p&gt;&#xA;" OwnerUserId="7522" LastActivityDate="2014-08-20T21:49:25.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15468" PostTypeId="2" ParentId="15464" CreationDate="2014-08-21T01:34:44.660" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When can't the rate-determining step be applied, and why? How to recognize those mechanisms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rate determining step (often abbreviated as &quot;rds&quot;) of a reaction is the slowest step in a reaction.  Every reaction has a &quot;slowest step&quot;, therefore, &lt;strong&gt;every reaction has an rds&lt;/strong&gt;.  The slowest step will be the step where we pass over a transition state with the highest energy, e.g. the highest energy point along the reaction coordinate. Look at the figure below, the reaction coordinate shown in the top left diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8N6nH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;illustrates a reaction with a transition state, but no intermediate.  There is only one step in this reaction, passing over that transition state (marked by the double-headed arrow) is the rds for that reaction.  The drawing in middle and right side of the top row show a reaction with an intermediate, and consequently, two transition states.  In each case the double-headed arrow identifies the rds - the highest energy point along the reaction coordinate.  The 3 drawings in the bottom row illustrate the case of a reaction with 2 intermediates (and therefore 3 transition states) along the reaction coordinate - in each case the arrow identifies the transition state with the highest energy, the rds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When can't the steady state approximation be applied, and why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider the following reaction&#xA;$$\ce{A &amp;lt;=&amp;gt;C[{k_1}]\   [B] &amp;lt;=&amp;gt;C[{k_2}]\ C}$$&#xA;$\ce{A}$ is the starting material, $\ce{[B]}$ is an intermediate and $\ce{C}$ represents the product.  Either the drawing in the middle or right in the top row represent the reaction coordinate for this reaction.  The steady state assumption applies to reactions that have one or more intermediates along the reaction coordinate.  The steady state assumption states that the &lt;em&gt;rate of change&lt;/em&gt; of the concentration of the intermediate is approximately zero.  In other words, the intermediate is being created at about the same rate at which it is &#xA;$$\ce{\frac{d[B]}{dt} = 0}$$&#xA;being destroyed.  The steady state assumption is best applied to situations where the intermediate is present in low concentration (e.g. $\ce{k_2 &amp;gt;&amp;gt; k_1}$).  A general rule of thumb for the assumption to be valid is for&#xA;$$\ce{\frac{k_2}{k_1} &amp;gt; 10 }$$&#xA;Looking back at the figure, we now see that this condition is met in the top row, center drawing.  Here is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Steady_state_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on the steady state assumption, the section on &quot;Validity&quot; is nice and clear if you'd like to read further on the subject.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-21T01:53:46.793" LastActivityDate="2014-08-21T01:53:46.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15469" PostTypeId="1" CreationDate="2014-08-21T02:56:18.440" Score="13" ViewCount="2198" Body="&lt;p&gt;I'm trying to write a program to calculate fixed-point Hartree-Fock level energies of molecules (for my amusement) and everything makes sense but this. I've been agonizing over this for almost 3 hours now. I've tried pretty much every Google search I can think of (mostly returning results that were either too vague or without enough detail) and looked in almost every book/ebook I own (pretty much always too vague). Any help whatsoever would be extremely appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, an STO-NG contracted Gaussian basis function has the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{\mu}^{\textrm{CGF}}(\vec{r}) = \sum_{p}^{N_{\textrm{PGF}}} d_{p\mu} (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the contraction coefficients, $d_{k\mu}$ and exponents, $\alpha_{k\mu}$ are chosen such that $\phi_{\mu}$ provides the 'best fit' to a Slater-type orbital having a Slater exponent $\zeta$, $\vec{R_{A}}=(X_A,Y_A,Z_A)$ is a fixed reference centre (usually a nucleus) and $l_{p\mu} = i_{p\mu} + j_{p\mu} + k_{p\mu}$ defines the angular momentum of the primitive Gaussian function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_{p\mu}^{\textrm{PGF}} = (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having problems interpreting data files documenting contraction coefficients and exponents for various basis sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, please consider the following snippet from the Gaussian 94 STO-3G basis set file documenting contraction parameters for Carbon:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;****&#xA;C     0 &#xA;S   3   1.00&#xA;     71.6168370              0.15432897       &#xA;     13.0450960              0.53532814       &#xA;      3.5305122              0.44463454       &#xA;SP   3   1.00&#xA;      2.9412494             -0.09996723             0.15591627       &#xA;      0.6834831              0.39951283             0.60768372       &#xA;      0.2222899              0.70011547             0.39195739       &#xA;****&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I have read in various places that there are enough coefficients here to describe 5 contracted functions, but I see only enough for 3 maximum (if the exponents are the same, but the coefficients change between 2S and 2P). I'd be eternally grateful if someone could explain for me, precisely how one would determine the following information from such a basis set data file:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Which EXACT basis functions appear within each contracted function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How many contracted Gaussian basis functions there are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, would you be able to do the same for the following snippet, again for Carbon, Gaussian 94, but this time for the more complicated 6-31++G** basis set?:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;****&#xA;C     0 &#xA;S   6   1.00&#xA;   3047.5249000              0.0018347        &#xA;    457.3695100              0.0140373        &#xA;    103.9486900              0.0688426        &#xA;     29.2101550              0.2321844        &#xA;      9.2866630              0.4679413        &#xA;      3.1639270              0.3623120        &#xA;SP   3   1.00&#xA;      7.8682724             -0.1193324              0.0689991        &#xA;      1.8812885             -0.1608542              0.3164240        &#xA;      0.5442493              1.1434564              0.7443083        &#xA;SP   1   1.00&#xA;      0.1687144              1.0000000              1.0000000        &#xA;SP   1   1.00&#xA;      0.0438000              1.0000000              1.0000000        &#xA;D   1   1.00&#xA;      0.8000000              1.0000000        &#xA;****&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7549" LastEditorUserId="194" LastEditDate="2016-04-05T22:26:36.820" LastActivityDate="2020-06-25T11:50:01.530" Title="Coefficients and Parameters for contracted Gaussian basis sets" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;&lt;basis-set&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15470" PostTypeId="1" AcceptedAnswerId="15472" CreationDate="2014-08-21T04:53:28.393" Score="5" ViewCount="11794" Body="&lt;p&gt;My professor commented that water has a higher dielectric constant than alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, but on what basis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He didn't elaborate; he asked us which one seemed &quot;more electric&quot; and most answered water. I guess from intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why exactly is water &quot;more electric&quot;? I think it has to do with the relatively high extent of auto-ionization in water, correct? Alcohols are generally less polar than water and thus cannot stabilize ions as effectively as water can. And I've even seem some computers submerged completely in oil. I'm guessing that only works because oil doesn't ionize much and thus isn't able to conduct a lot of electricity - or enough to short out any circuits. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2020-06-02T04:53:22.270" Title="What properties does a solvent with a high dielectric constant have?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15471" PostTypeId="1" AcceptedAnswerId="15478" CreationDate="2014-08-21T05:18:58.953" Score="1" ViewCount="804" Body="&lt;p&gt;The density of $\ce{CuCl}$ is given – $x\ \mathrm{g/cm^3}$&lt;br&gt;&#xA;The crystal structure is assumed to be fcc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher is claiming the we can apply the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho=\frac{Z \cdot M}{a^3 \cdot N_\mathrm A}$$&#xA;And he took $M = 35.5\ \mathrm{g\ mol^{-1}}+63.5\ \mathrm{g\ mol^{-1}} = 99\ \mathrm{g\ mol^{-1}}$,&#xA;$Z = 4$ as it is fcc.&#xA;But my question is how can we take $M = 99\ \mathrm{g\ mol^{-1}}$? It means every particle in the fcc has mass $99\ \mathrm u$ doesn't it? &lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T15:18:03.153" LastActivityDate="2016-11-06T15:18:03.153" Title="What is the length of a unit cell of CuCl assuming that it is fcc?" Tags="&lt;homework&gt;&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15472" PostTypeId="2" ParentId="15470" CreationDate="2014-08-21T05:22:11.553" Score="5" Body="&lt;p&gt;Before I start, please note that I am not a chemist so be sure to verify what I say...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric conductivity of water depends on what is dissolved in it. For example, salt water will be far more conductive than distilled water. So if you're professor was talking about sea water, the answer could actually be different from purified water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the dissolved salt ions, sea water is a better conductor of electricity and so by definition it will be a poor dielectric, since according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Dielectric&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article: to be a good dielectric a substance must be both a good &lt;em&gt;insulator&lt;/em&gt; and able to be polarized by an electric field&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Purified water on the other hand will be a better dielectric because it does not contain any significant amount of ions and is therefor not very conductive (i.e. is a better insulator than sea water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (but don't quote me on an exam) that the reason alcohol is a poor dielectric is because it is not nearly as polar as water is, and according to the above, the second requirement for a dielectric is that it could be polarized by an electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This might also have to do with the size of the molecules, since water molecules are generally smaller than many kinds of alcohols, and can therefore re-orient themselves more easily in the presence of an electric field, but this is a guess.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Solvent Classifications&lt;/em&gt; section of &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvent#Solvent_classifications&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia page&lt;/a&gt; also mentions that the polarity of a solution can be roughly gauged by the solution's dielectric constant. So this too seems to support the idea that polarity and dielectric properties are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has any improvements or corrections, please let me know in the comments and I'll update the answer accordingly.&lt;/p&gt;&#xA;" OwnerUserId="7550" LastEditorUserId="7550" LastEditDate="2015-12-11T07:39:59.333" LastActivityDate="2015-12-11T07:39:59.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15473" PostTypeId="1" CreationDate="2014-08-21T05:44:43.823" Score="2" ViewCount="902" Body="&lt;p&gt;How good of a thermal insulator is superglue compared to other materials like ceramics, plastics, and metals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this might not seem like a chemistry question, but I was hoping someone here could provide a more meaningful answer than what I found online, which was basically along the lines of, &quot;It's not a good heat conductor so don't use it in cooling applications.&quot; I was hoping for a little more substance than that, so that's why I wanted to ask it on here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I would love for the answer to include one of the following things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An approximate numeric comparison between superglue (liquid or gel) and other materials (for example plastics, metals, and ceramics).&lt;/li&gt;&#xA;&lt;li&gt;Or a qualitative answer, like the example I gave above, but with a short explanation based on the chemical characteristics.&lt;/li&gt;&#xA;&lt;li&gt;Or perhaps a chart of thermal conductivities of various superglues relative to other common materials.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If a similar question already exists -- or there is a free resource available online that can answer it -- then a link with a short explanation would be a perfectly good answer as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it sounds like I'm asking for a very big answer, but the suggested answers above were just suggestions. Really, the most important thing (for me at least) is to get a feel for where superglue falls in relation to other common materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to Greg's comment: I would have liked to share my finding on the subject, but unfortunately I have not been able to locate any formal research on the matter. The best I was able to find so far were forums and blog articles, which were usually either vague or offered little-to-no supporting evidence of their answers. Perhaps I just don't know the right places to look for information like this, and if this really is the case then I hope you all can give me some direction as to where I can find some research on this matter.&lt;/p&gt;&#xA;" OwnerUserId="7550" LastEditorUserId="7550" LastEditDate="2014-08-21T07:10:46.287" LastActivityDate="2014-08-21T07:10:46.287" Title="Is superglue a good thermal insulator?" Tags="&lt;thermodynamics&gt;&lt;materials&gt;&lt;conductivity&gt;" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15474" PostTypeId="1" AcceptedAnswerId="15477" CreationDate="2014-08-21T08:33:22.350" Score="10" ViewCount="2936" Body="&lt;p&gt;Someone painted an original graffiti:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2D9zU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jbzd.pl/obr/206549/wandale&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what kind of reaction was depicted; what's that product substance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graffiti reads:&#xA;$\ce{C2H2 + C2H5OH -&amp;gt; C_2H_5-O-CH=CH_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have any more or less common uses?&lt;/p&gt;&#xA;" OwnerUserId="1676" LastEditorUserId="24052" LastEditDate="2016-03-11T22:24:32.313" LastActivityDate="2016-03-11T22:24:32.313" Title="What is C2H5-O-CH=CH2?" Tags="&lt;organic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15475" PostTypeId="2" ParentId="14057" CreationDate="2014-08-21T09:41:59.243" Score="12" Body="&lt;p&gt;I had some free time and resources at hand and did some highly accurate quantum chemical calculations for this system. I took it as a chance to compute some benchmarks and play a little with density functional theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally I also wanted to compute transition barriers, but based on the previous findings I had to omit that, since I currently cannot afford &lt;em&gt;ab initio&lt;/em&gt; calculations. However, I'd like to share these results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your reading convenience, here is the chart from longs post, in order to see the numeration ($\ce{X=F}$).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NJl3w.png&quot; alt=&quot;numeration scheme&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronic Energies $\Delta{}E^\circ$ are as usual in $\mathrm{kcal/mol}$. Thermodynamic data $\Delta{}E^\circ$ was estimated at the DF-BP86/def2-TZVP and DF-BP86/def2-SVP level of theory with $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$. Experimental data was taken from the cited paper. Product ratios were calculated from the electronic energies (except for BP86 where data was available) assuming that the contribution of thermodynamic corrections is neglectable, i.e. $\Delta{}E^\circ = \Delta{}G^\circ$. The level of degeneracy is for each conformer $g(\mathbf{1})=g(\mathbf{2})=g(\mathbf{3})=3$ and $g(\mathbf{4})=1$.&lt;br&gt;&#xA;(I was somehow unable to reproduce the product ratio that ron calculated with &lt;a href=&quot;https://chemistry.stackexchange.com/q/14436/4945&quot;&gt;his approach&lt;/a&gt;. I used &lt;a href=&quot;https://chemistry.stackexchange.com/q/14617/4945&quot;&gt;How to obtain product ratio from energy differences via Boltzmann statistics&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrrrr}\hline&#xA;\text{method}                &amp;amp; \mathbf{1} &amp;amp; \mathbf{2} &amp;amp; \mathbf{3} &amp;amp; \mathbf{4} \\\hline&#xA;\text{Exptl.}~\Delta G^\circ &amp;amp;  0.0   &amp;amp; -2.67  &amp;amp; -2.21  &amp;amp;  -3.41  \\&#xA;                       \%    &amp;amp;  0.4   &amp;amp; 38.0   &amp;amp; 17.5   &amp;amp;  44.2   \\\hline&#xA;\text{BP86/SVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -3.3   &amp;amp; -3.5  &amp;amp;  -2.6   \\&#xA;\phantom{\text{BP86/SVP}}~\Delta G^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -3.0   &amp;amp; -3.0  &amp;amp;  -2.6   \\&#xA;                       \%    &amp;amp;  0.3   &amp;amp; 43.5   &amp;amp; 48.0  &amp;amp;  8.2    \\\hline&#xA;\text{BP86/TZVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -3.2   &amp;amp; -3.0  &amp;amp;  -3.1   \\&#xA;\phantom{\text{BP86/TZVP}}~\Delta G^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -2.9   &amp;amp; -2.6  &amp;amp;  -3.0   \\&#xA;                       \%    &amp;amp;  0.4   &amp;amp; 48.6   &amp;amp; 30.6  &amp;amp;  20.5   \\\hline&#xA;\text{PBE0/TZVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -3.0   &amp;amp; -2.8  &amp;amp;  -3.3   \\&#xA;                       \%    &amp;amp;  0.3   &amp;amp; 44.7   &amp;amp; 30.9  &amp;amp;  24.1   \\\hline&#xA;\text{M06//BP86/TZVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -3.1   &amp;amp; -2.9  &amp;amp;  -3.7   \\&#xA;                       \%    &amp;amp;  0.2   &amp;amp; 38.5   &amp;amp; 27.2  &amp;amp;  34.1   \\\hline&#xA;\text{MP2//BP86/TZVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -1.3   &amp;amp; -1.0  &amp;amp;  -3.9   \\&#xA;                       \%    &amp;amp;  0.4   &amp;amp;  3.7   &amp;amp;  2.1  &amp;amp;  93.8   \\\hline&#xA;\text{CCSD(T)//BP86/TZVP}~\Delta E^\circ &#xA;                             &amp;amp;  0.0   &amp;amp; -1.3   &amp;amp; -1.0  &amp;amp;  -3.9   \\&#xA;                       \%    &amp;amp;  0.4   &amp;amp;  3.7   &amp;amp;  2.1  &amp;amp;  93.8   \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is astonishing that MO6 almost exactly reproduces the experimental values, while the &lt;em&gt;ab initio&lt;/em&gt; calculations CCSD(T) and MP2 conclude that there is almost only the $\mathbf{4}$ conformer. This was quite unexpected, maybe frequency calculations have to be done to obtain more fitting results.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-08-22T03:09:16.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15477" PostTypeId="2" ParentId="15474" CreationDate="2014-08-21T11:24:36.870" Score="9" Body="&lt;p&gt;The particular reaction is of acetylene ($\ce{C2H2}$) and ethanol ($\ce{C2H5OH}$) to yield ethyl vinyl ether (IUPAC name ethoxyethene). It's one example of &lt;a href=&quot;http://en.wikipedia.org/wiki/Walter_Reppe#Reppe_chemistry&quot;&gt;Reppe chemistry&lt;/a&gt;, a fairly diverse set of reactions, often operating on alkenes and alkynes (though certainly not exclusively). The specific reaction in the question is a vinylization reaction, though it was Reppe's carbonylation reactions which were probably of greater practical (read: economic) import historically, as they were famously used in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monsanto_acetic_acid_process&quot;&gt;Monsanto process&lt;/a&gt; and in production of various acrylates and other acrylic acid derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vinylization reactions could still be synthetically useful, as they provide an extremely simple alternative route to vinyl ethers without having to proceed via elimination, substitution, coupling, olefination, or substrate-specific oxidation reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any specific uses for ethyl vinyl ether, however it could certainly be used as a building block in any number of organic syntheses.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-08-21T11:24:36.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15478" PostTypeId="2" ParentId="15471" CreationDate="2014-08-21T13:47:04.510" Score="1" Body="&lt;p&gt;Well, yes, depending on the detail you want to go into. You can find a representation of the crystal structure of CuCl at, for example, &lt;a href=&quot;http://www.webelements.com/compounds/copper/copper_chloride.html&quot; rel=&quot;nofollow&quot;&gt;http://www.webelements.com/compounds/copper/copper_chloride.html&lt;/a&gt; - you will see that the Cu atoms are indeed a simple fcc lattice.  The Cl atoms are on sites that are an interpenetrating fcc lattice.  The end result is the zinc blende (ZnS) lattice prototype, very familiar to semiconductor folks as the crystal structure for GaAs an associated III-V compounds. If all sites were occupied by the same atom, this is known as diamond cubic, commonly seen in diamond, Si, and Ge (amongst others). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, yes, one way to consider the crystal (particularly for your purposes here) is an fcc crystal with a Cu-Cl basis. Just don't expect the x-ray diffraction to correspond to fcc. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-21T13:47:04.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15479" PostTypeId="1" CreationDate="2014-08-21T13:47:47.340" Score="5" ViewCount="5110" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Ammonium carbamate dissociates as shown below. &#xA;  $$\ce{NH2COONH4 (s) -&amp;gt; 2NH3 (g) + CO2 (g)}$$&#xA;  In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of $\ce{NH3}$ now equals to the original total pressure. What is the ratio of the new total pressure to the original pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried solving the problem and I know I am quite close to the answer which is $\frac{31}{27}$. &#xA;Firstly I took out $K_p$ for equilibrium assuming total pressure to be $p$.&#xA;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\mathrm{p} = \left(\frac{2p}{3}\right)^2 \times \frac{p}{3} = \frac{4p^3}{27}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this the par pressure of ammonia becomes $p$ so let's say that of $\ce{CO2}$ is $\frac{p}{3} + x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since $K_\mathrm{p}$ doesn't change so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p}{3} + x = \frac{4p^2}{27}$ (cancelling one of the $p$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But solving further and finding new pressure and dividing by the old one doesn't give the answer&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="5026" LastEditDate="2018-03-05T13:32:43.763" LastActivityDate="2018-03-05T13:32:43.763" Title="How to calculate the equilibrium composition after a change in partial pressure of one component?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15480" PostTypeId="2" ParentId="15442" CreationDate="2014-08-21T13:49:58.487" Score="-3" Body="&lt;p&gt;I think sulfuric acid is hygroscopic. When you put sugar in it, it will take out water of crystallization from it and if not, most probably it will dehydrate it by taking $\ce{-OH}$ groups from it.&lt;/p&gt;&#xA;" OwnerUserId="7555" LastEditorUserId="4231" LastEditDate="2015-03-07T00:01:08.337" LastActivityDate="2015-03-07T00:01:08.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15482" PostTypeId="2" ParentId="15479" CreationDate="2014-08-21T15:28:54.530" Score="3" Body="&lt;p&gt;As follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to simplify the math a little bit, assume pressure of $\ce{CO2_{(g)}}$ is p. Thus, pressure of $\ce{NH3_{(g)}}$ is 2p at equilibrium. Total pressure in this case is 3p. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let 3p equal new pressure of $\ce{NH3_{(g)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added pressure of $\ce{NH3_{(g)}}$ would shift the equilibrium to the left, so say we lose 2x pressure of $\ce{NH3_{(g)}}$ and x pressure of $\ce{CO2_{(g)}}$. New equilibrium pressures will be 3p-2x for $\ce{NH3_{(g)}}$ and p-x for $\ce{CO2_{(g)}}$. Thus, ratio of new to old is $\frac{4p-3x}{3p}$. You can solve for x in terms of p using $K_p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for x and find the ratio, which should be $\frac{31}{27}$&lt;/p&gt;&#xA;" OwnerUserId="7402" LastEditorUserId="7402" LastEditDate="2014-08-21T15:51:51.033" LastActivityDate="2014-08-21T15:51:51.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15483" PostTypeId="2" ParentId="15469" CreationDate="2014-08-21T16:48:05.283" Score="10" Body="&lt;p&gt;Take a look at the carbon STO-3G first. It means for each AO, you form it with three Gaussians. For carbon, you have a 1S, 2S and (three) 2P. 5 functions when finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top line says you are defining the basis for carbon; no big deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next line says you are creating an S shell, with three primitives, with a 1x scale factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you list your three rows of primitives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first column is the Gaussian exponent. The next column is the S expansion coefficient. These are the three Gaussians that make up your 1S &quot;Slater&quot; atomic orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the next section, you define SP. What this is, and I quote from the Gaussian website now: &quot;Note that 2SP requests the best least-squares fit simultaneously to S and P slater orbitals and is not equivalent to separately specifying the best S and the best P expansions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S and P (and all higher angular momenta, actually) can be related by a recursion relationship. It's an efficient way of computing integrals. I believe the Obara-Saika recursion takes advantage of this, though in Gaussian the integrals are formed using the PRISM algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first column is the exponent for your Gaussian, the second column is the S expansion coefficient, and the third column is the P expansion coefficient. They just share the same exponent. Thus you have 4 + 1 (from before) = 5 AOs total formed with three primitives each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To jump ahead to your last example, the pattern simply continues. The first section gives six primitives to form one 1S (the 6 in 6-311++G**). The second section gives three Gaussians for you valence S and P, the third section is one Gaussian for valence SP. This is the 31 in 6-31++G**. You have a split valence description, one with three primitives, one with a single primitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next section is the diffuse function, hence the really small exponent in the first column. This is the ++ part of the 6-31++G**. The small exponent gives a really wide spreading Gaussian. Helpful for describing electronic activity far from the nucleus. (It's helpful to compare with the huge exponents in the first section, which gives really tight functions that describe the core electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you have the D functions which are polarization functions. Carbon doesn't have a D valence shell, but we include them to describe polarization of electronic density in molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that depending on your program, you either have 5D AOs or 6D AOs. This is the difference between spherical and cartesian representations, and is more of a computational trick than anything else. Gaussian uses 5D by default, except in some cases.&lt;/p&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="3798" LastEditDate="2014-08-21T17:05:25.470" LastActivityDate="2014-08-21T17:05:25.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15484" PostTypeId="2" ParentId="15436" CreationDate="2014-08-21T17:04:56.767" Score="3" Body="&lt;p&gt;Most assays of carcinogenicity are designed to measure DNA damage. The 2 most common of those assays are the Ames test and the Comet Assay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ames test uses a special set of strains of bacteria that have known phenotypes. You expose the bacteria to the suspected carcinogen and then look for changes in phenotype. Different strains will be susceptible to different types of mutations, deletions, insertions, frameshifts, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comet assay uses mammalian cell lines. You have to suspend the cells in an agar and break them up. This will create little pockets of cell debris in the agar. You fluorescently stain the DNA and run an electric current through it. If the DNA is damaged and broken up, it will migrate through the gel, creating tails, hence the &quot;comet&quot;. If the DNA is intact, it will not migrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never done either assay, just heard about them in class. Wikipedia has more detail than I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ames_test&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ames_test&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Comet_assay&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Comet_assay&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-08-21T17:04:56.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15485" PostTypeId="1" AcceptedAnswerId="15767" CreationDate="2014-08-21T17:24:22.590" Score="4" ViewCount="663" Body="&lt;p&gt;There are many types of defects like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vacancy&lt;/li&gt;&#xA;&lt;li&gt;Interstitial&lt;/li&gt;&#xA;&lt;li&gt;Schottky&lt;/li&gt;&#xA;&lt;li&gt;Frenkel&lt;/li&gt;&#xA;&lt;li&gt;Non-Stoichiometric&lt;/li&gt;&#xA;&lt;li&gt;Alien Impurities&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But given a sample, how do scientists know which kind of defect is present in it [practically]?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="1878" LastEditDate="2014-09-01T04:01:27.997" LastActivityDate="2014-09-01T04:29:56.990" Title="How do people check the type of defect in a crystal?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15486" PostTypeId="1" AcceptedAnswerId="15487" CreationDate="2014-08-21T17:34:48.783" Score="1" ViewCount="71" Body="&lt;p&gt;There are many types of unit cells like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tetragonal        Monoclinic      Orthorhombic        Rhombohedral&lt;br&gt;&#xA;Hexagonal         Triclinic       Cubic    &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But how do scientists identify the type when a sample of the compound is given to them?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastActivityDate="2014-08-21T17:36:40.410" Title="How do scientists check which kind of unit cell is present in a crystal?" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15487" PostTypeId="2" ParentId="15486" CreationDate="2014-08-21T17:36:40.410" Score="5" Body="&lt;p&gt;X-ray diffraction. Usually computer will tell you automatically nowadays but in the old days you had to check whether the values you got for the miller planes were possible with the symmetry of different unit cells. They discarded the ones that didnt match until the would find the one unit cell which matched the miller planes from XRD&lt;/p&gt;&#xA;" OwnerUserId="5086" LastActivityDate="2014-08-21T17:36:40.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15488" PostTypeId="1" AcceptedAnswerId="15491" CreationDate="2014-08-21T18:10:52.450" Score="0" ViewCount="1955" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.Assume that you construct a thermometer using gallium as the fluid instead of mercury, and that you define the melting point of gallium as 0 °G and the boiling point of gallium as 1000 °G. What is the melting&#xA;  point of sodium chloride (801 °C) on the gallium scale?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First we determine the ratio of $^{\circ}\mathrm G/^{\circ}\mathrm C$ by dividing the melting points. This states that $0.4599\ ^{\circ}\mathrm G=1\ ^{\circ}\mathrm C$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1000\ ^{\circ}\mathrm G-0\ ^{\circ}\mathrm G}{2204\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C}=\dfrac{1000\ ^{\circ}\mathrm G}{2174.22\ ^{\circ}\mathrm C}=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the boiling point, this is done: $$T=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C\times\left(801\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C\right)=355\ ^{\circ}\mathrm G$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why are we subtracting by $29.78\ ^{\circ}\mathrm C$? &lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T14:56:35.037" LastActivityDate="2016-11-06T14:56:35.037" Title="Determining boiling point on a created temperature scale" Tags="&lt;homework&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15489" PostTypeId="2" ParentId="15485" CreationDate="2014-08-21T18:13:31.993" Score="1" Body="&lt;p&gt;Given amount of atoms in typical macroscopic crystal, there is no need to check if any type of defect is present - &lt;em&gt;all&lt;/em&gt; of them are presence because of sheer numbers and the law of big numbers. For estimating rough amounts different spectroscopic methods may be used. However, it is an indirect technique, which requires somehow obtained idea of the spectrum of the defect in question. Unless a very rough idea is enough, like detecting impurities using different nuclear spectroscopy methods, the only universal way is usage of quantum chemistry methods, and they are often frustratingly inaccurate when spectrum is a property of interest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nanosized structures, surface,near-surface and surface-influenced defects may be detected using hi-definition microscopy techniques, like scanning tunneling microscopy, which may provide subatomic resolution. However, in this case a different problem arises. Humanity usually does not need specific knowledge about some particular crystal, more a typical picture for some kind of systems is of interest. In such a case the only way is to make a lot of observations and somehow reduce it to typical values, and such a procedure is usually debatable, as samples for atomic-resolved spectroscopy must be prepared in very specific ways (see method descriptions for details), and so are very different from what should be present in conditions of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was this answer of any use?&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-21T18:13:31.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15490" PostTypeId="2" ParentId="15391" CreationDate="2014-08-21T18:21:17.787" Score="1" Body="&lt;p&gt;Passing &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine&quot; rel=&quot;nofollow&quot;&gt;chlorine&lt;/a&gt; gas through water yields a solution of $\ce{Cl2}$, hydrochloric acid ($\ce{HCl}$) and hypochlorous acid ($\ce{HOCl}$). The latter two are formed by reaction of chlorine with water under disproportionation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 + H2O \rightleftharpoons HCl + HOCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bleach&quot; rel=&quot;nofollow&quot;&gt;Bleaches&lt;/a&gt; are chemicals which remove color by oxidation of dyes. $\ce{Cl2}$ and $\ce{HOCl}$ themselves are already strong oxidizers, but adding alkali metal hydroxides to an aqueous chlorine solution increases the solubility of $\ce{Cl2}$, and chlorine bleach is obtained, which contains alkali metal hypochlorite as active ingredient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 + OH- \rightleftharpoons ClO- + Cl- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hypochlorite&quot; rel=&quot;nofollow&quot;&gt;Sodium hypochlorite&lt;/a&gt; is produced by this reaction and common household bleach is an aqueous 3-8 mass-% solution of the salt. The necessary chlorine gas is actually produced by electrolysis of $\ce{NaCl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While gaseous $\ce{Cl2}$ itself can act as a bleaching agent by oxidation, the use of hypochlorite solutions has the benefit of avoiding handling of the poisonous gas. $\ce{H2}$ is a much better reducing agent, and therefore unlikely to be useful for bleaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mixture of $\ce{H2}$ and $\ce{Cl2}$ will react to form gaseous &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_chloride&quot; rel=&quot;nofollow&quot;&gt;hydrogen chloride&lt;/a&gt; when exposed to UV light. Hydrochloric acid is obtained when gaseous $\ce{HCl}$ is dissolved in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + H2O \rightleftharpoons H3O+ + Cl-}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-08-21T19:18:22.570" LastActivityDate="2014-08-21T19:18:22.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15491" PostTypeId="2" ParentId="15488" CreationDate="2014-08-21T18:32:45.693" Score="1" Body="&lt;p&gt;Unlike an absolute temperature scale like degrees Kelvin, Celsius and your hypothetical °G are relative scales, where the zero points are set to the melting points of water and gallium, respectively. In your first step, you've accounted for the difference in the size of one degree Celsius and one °G, but not the difference in the zero points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of sodium chloride, think about what the melting point of gallium is on each scale. You've defined it as 0 °G, but you know it's not 0 °C. What do you have to do to correct for this to get a value in Celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is also why the conversion between Celsius and Fahrenheit is not just multiplying by a constant.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-08-21T18:32:45.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15492" PostTypeId="2" ParentId="15488" CreationDate="2014-08-21T18:32:55.523" Score="1" Body="&lt;p&gt;It zeroes things out.  It is analogous to the use of &quot;32&quot; in the equation to convert Fahrenheit to Celsius, where &quot;32&quot; represents the melting point of ice on the Fahrenheit scale.  In your G scale you must subtract the melting point of gallium in the Centigrade scale in order to get 0G at the melting point of gallium&#xA;$$\ce{^{o}C = (^{o}F-32)*(5/9)}$$&#xA;$$\ce{^{o}G = (^{o}C-29$.$78)*($0.4599$)}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-21T18:32:55.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15493" PostTypeId="2" ParentId="15424" CreationDate="2014-08-21T18:45:34.960" Score="1" Body="&lt;p&gt;The Cis-2-butene has the higher hydrogenation rate because both hydrogens on the double bond are on the same side, so you can get these close to the catalyst and transfer two more hydrogens on relatively easily. You can mimic this by building a molecular model of the cis-2-butene and use a table as your catalyst. Clearly, the hydrogen side of the double bond fits pretty well. Now build the 2-methyl-propene and trans-2-butene, and try to fit those against the surface as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be good to keep in mind that the 2-methyl-propene has a terminal alkene. I'm trying not to answer the question outright and make you do the work, but that terminal alkene is probably very important for determining which reacts more readily.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-08-21T18:45:34.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15494" PostTypeId="1" AcceptedAnswerId="15501" CreationDate="2014-08-21T18:53:22.647" Score="3" ViewCount="119" Body="&lt;p&gt;A &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/app.1979.070230319/pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; talks about the residual stress when coating polymers.&lt;br&gt;&#xA;It says that residual stress is the sum of the &quot;heat stress&quot; (due to the difference in thermal expansion coefficients) and the &quot;shrinkage stress&quot; (due to the solvent evaporation and changes in volume). It then states that heat stress usually increases with temperature while shrinkage stress decreases with temperature. Therefore the shape of the graph of residual stress vs temperature would be a parabola with the minimum changing depending on the polymer/system system.&lt;br&gt;&#xA;It then goes to find the starting point of the shrinkage stress and states that it is dependant on the glass transition temperature.&lt;br&gt;&#xA;From what I understand, that would happen when the glass transition temperature is between room temperature and the drying temperature.&lt;br&gt;&#xA;However, if the glass transition temperature is below both room temperature and drying temperature, what would happen? How this would affect the shrinkage stress? Or did I get the whole connection between glass transition temperature and shrinkage stress wrong?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="4945" LastEditDate="2014-08-22T04:38:12.623" LastActivityDate="2014-08-26T23:58:05.617" Title="Residual stress in polymer coatings" Tags="&lt;physical-chemistry&gt;&lt;polymers&gt;&lt;solutions&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15495" PostTypeId="1" CreationDate="2014-08-21T18:58:26.913" Score="5" ViewCount="340" Body="&lt;p&gt;How is the following true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \frac{∂S}{∂P} \right )_{T}=\frac{-C_{P}}{T}\left ( \frac{∂T}{∂P} \right )_{S}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{P}=T\left ( \frac{∂S}{∂T} \right )_{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means the right-hand side of the first equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left ( \frac{∂S}{∂T} \right )_{P}\left ( \frac{∂T}{∂P} \right )_{S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but how is that equal to $\left ( \frac{∂S}{∂P} \right )_{T}$ ?&lt;/p&gt;&#xA;" OwnerUserId="7562" LastEditorUserId="189" LastEditDate="2014-08-21T20:13:09.327" LastActivityDate="2014-10-25T04:45:35.000" Title="Problems with thermodynamic derivatives" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15496" PostTypeId="2" ParentId="15435" CreationDate="2014-08-21T19:15:49.853" Score="1" Body="&lt;p&gt;Yes water has a higher &lt;em&gt;reduction&lt;/em&gt; potential. But what is happening at your anode isn't reduction, it's oxidation. So a lower reduction potential for $\ce{Cl-}$means that it's more readily oxidised than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For additional information, see my &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15378/what-are-possible-side-reactions-for-the-generation-of-hydrogen-via-electrolysis/15411#15411&quot;&gt;other answe&lt;/a&gt;r on this recent question on electrolysis and $\ce{Cl2}$ production.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-08-21T19:15:49.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15497" PostTypeId="2" ParentId="15424" CreationDate="2014-08-21T19:22:21.490" Score="0" Body="&lt;p&gt;You need to consider an actual hydrogenation system rather than generics. Hydrogenation is often much more difficult than adding a substrate, finely divided metal and then throwing in hydrogen gas for good fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at Wilkinson's catalyst and its corresponding catalytic cycle (an old catalyst, but one which has plenty of literature). I expect tentatively for the transition state of the migratory insertion of the alkene to play a huge part in the relative reaction rates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a four-membered transition state at this stage, then we can possibly imagine a build up of charge on the carbon where the hydride inserts (the more hindered end, the metal on the less) which would be more stabilised if there are additional alkyl chains due to hyperconjugation.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-08-21T19:22:21.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15498" PostTypeId="2" ParentId="15495" CreationDate="2014-08-21T19:28:22.260" Score="8" Body="&lt;p&gt;It's a consequence of the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_product_rule&quot;&gt;triple product rule&lt;/a&gt;&lt;/strong&gt; for partial derivatives of thermodynamic state variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, if you have three interdependent (not &lt;em&gt;independent&lt;/em&gt;) variables, then this is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left ( \frac{\partial{X}}{\partial{Y}} \right )_Z \left ( \frac{\partial{Z}}{\partial{X}} \right )_Y = - \left ( \frac{\partial{Z}}{\partial{Y}} \right )_X&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in your variables, if we let $X = T$, $Y = P$, and $Z = S$, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;- \left ( \frac{\partial{S}}{\partial{T}} \right )_P \left ( \frac{\partial{T}}{\partial{P}} \right )_S =  \left ( \frac{\partial{S}}{\partial{P}} \right )_T&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the order of multiplication of the two partial derivatives doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;strong&gt;why&lt;/strong&gt; this rule holds for thermodynamic variables is a little bit more complicated - but I think the wikipedia article that I linked to has a pretty good derivation if you are interested.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-21T19:28:22.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15499" PostTypeId="2" ParentId="15424" CreationDate="2014-08-21T20:20:21.913" Score="3" Body="&lt;p&gt;We should be able to determine which molecule will hydrogenate faster by comparing the activation energies for hydrogenation of these two molecules.  A catalyst does not alter the reaction mechanism, it merely reduces the activation energy for the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of formation of 2-methyl-propene is &lt;em&gt;ca&lt;/em&gt;. 1.3 kcal/mol higher than that of trans-2-butene, presumably because of the adverse steric interaction between the geminal methyl groups in 2-methyl-propene.  The transition state for this reaction will start to resemble the corresponding alkane in its eclipsed conformation (due to the cis-hydrogen delivery).  If we only consider methyl-methyl steric interactions, then for 2-methyl-propene as we approach the transition state the two geminal methyl groups, initially separated by a 120 degree ($\ce{sp^2}$) angle, will displace towards a 109 degree ($\ce{sp^3}$) angular separation.  Therefore, the methyl-methyl steric interaction should increase in 2-methyl-propene as we approach the transition state.  2-methyl-propene's  transition state will be more destabilized than the ground state, consequently increasing the activation energy required to reach it.  With trans-2-butene, there are no methyl-methyl steric interactions in the ground state or transition state; its activation energy, the energy separating the ground state and transition state, should not be increased or decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this analysis, the ground state - transition state energy separation would be expected to increase (slightly) for 2-methyl-propene as compared to trans-2-butene.  Consequently, faster hydrogenation of trans-2-butene would be predicted.  However the effect is expected to be very small.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-21T21:00:05.257" LastActivityDate="2014-08-21T21:00:05.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15500" PostTypeId="2" ParentId="15349" CreationDate="2014-08-21T21:05:15.950" Score="5" Body="&lt;p&gt;I don't know of any undergraduate level texts that would be suitable for a freshman, unless you have already taken physics, statistics, and chemistry up through organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can tell you that &lt;a href=&quot;http://books.google.com/books/about/Fundamentals_of_Polymer_Science.html?id=Fb4zSQAACAAJ&quot;&gt;Fundamentals of Polymer Science&lt;/a&gt; by Painter and Coleman is a good introductory text, but I used it in the final year of my Chemical Engineering program. There could be better resources as well - that's just the only one I am personally familiar with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also depends a lot on which specific aspects of polymers you are interested in. This actually matters a lot for your second question as well. The three main branches of polymers studies are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Polymer chemistry&lt;/li&gt;&#xA;&lt;li&gt;Polymer physics&lt;/li&gt;&#xA;&lt;li&gt;Polymeric materials&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are sub-fields in each of these, and a lot of overlap, but essentially they focus on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Polymer chemistry&lt;/strong&gt; - How are polymers made? (this would include engineering and production of polymers)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Polymer physics&lt;/strong&gt; - How does the behavior on a molecular level affect the macroscopic properties?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Polymeric materials&lt;/strong&gt; - How can we use the unique material properties of polymers to make useful stuff?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At the undergraduate level, you won't get too specific on any one of these areas, no matter which degree you choose. If you majored in chemistry, you would learn a little more about polymerization reactions. If you majored in physics (and took electives related to polymers) you might learn more about the statistical mechanics of polymers. If you majored in materials science and/or engineering, you would learn more about how we use the polymers, and less about how they are made and why they act the way they do. Chemical engineers learn a little bit of everything, but they don't get too deep into any one particular branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real differences show up if you go on to graduate school or work in a particular role in industry. There, you will definitely focus more, and your degree can make more of a difference. So I would say, for your second question, major in whatever aspect interests you more - if you want to know the details about how polymers are made in a lab, then maybe chemistry is a better option. If you really like math and physics and want to understand how they relate to the behavior of polymers, then physics is good. If you are more interested in how to characterize and use polymers, then materials science/engineering is a good bet. And if you like a little bit of everything, or are interested in industrial-scale production of polymers, then chemical engineering is a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind as well: it is hard to predict now what you will really want to do in four or more years - so if you make a choice based on which aspects of polymer science you are interested in, and then later decide that polymers are actually not what you really like, you will still be studying the &quot;right&quot; stuff for you. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-21T21:05:15.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15501" PostTypeId="2" ParentId="15494" CreationDate="2014-08-21T21:17:25.223" Score="4" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the paper linked in your comment, and see that my initial guess was incorrect (see below) - for this system, it is as you say - for temperatures above the bulk pure polymer glass transition temperature, you would &lt;em&gt;not&lt;/em&gt; expect to see shrinkage stress due to evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;END EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have access to that paper - so this is just an educated guess, but usually physicists use the glass transition temperature to characterize polymer behavior as a function of temperature because things &lt;em&gt;roughly&lt;/em&gt; (very roughly) scale with $T_g$. In other words, if I know how the behavior of one linear homopolymer depends on temperature, and I know a certain response curve scales with $T_g$, then I can scale the temperature curve of another linear homopolymer using the $T_g$ and &lt;em&gt;sometimes&lt;/em&gt; get accurate results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They could be linking it to shrinkage stress because $T_g$ also characterizes the dynamic behavior of polymers (how they &quot;move&quot; in response to thermal fluctuations and external stress). This is essentially the principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Time%E2%80%93temperature_superposition&quot; rel=&quot;nofollow&quot;&gt;time-temperature superpostion&lt;/a&gt; - or the idea that as temperature increases, polymers move the same way, only faster. So heating up is like fast-forwarding time, and cooling down is like slowing time down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass transition temperature is a discontinuity in this behavior - at a certain low temperature, the polymers freeze in place (on large length scales). This phenomenon is not well understood, but we have found empirically that the temperature at which it occurs can be used to scale response curves in certain situations, and since it is fairly easy to measure, it gets used a lot in these types of models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, and again this is an educated guess, I would imagine that the glass transition temperature would be used in this model regardless of the temperature you were actually operating at - it is probably a fundamental parameter of the curve that is used to put the &quot;reality&quot; (the type of polymer) into the model.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-26T23:58:05.617" LastActivityDate="2014-08-26T23:58:05.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15502" PostTypeId="2" ParentId="15342" CreationDate="2014-08-21T21:36:45.683" Score="6" Body="&lt;p&gt;I think the key concepts that are explainable at this age level are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Atoms are made of protons, neutrons, and electrons&lt;/li&gt;&#xA;&lt;li&gt;Protons have positive charge, electrons have negative charge, neutrons have no charge&lt;/li&gt;&#xA;&lt;li&gt;Opposite charges attract&lt;/li&gt;&#xA;&lt;li&gt;Atoms can gain or lose electrons, but not protons or neutrons&lt;/li&gt;&#xA;&lt;li&gt;Atoms hold onto their electrons more or less tightly depending on which atom it is (you won't be able to really explain why, but you could just show them a periodic table and explain that non-metals (the right side) like electrons more than metals (the left side))&lt;/li&gt;&#xA;&lt;li&gt;Atoms can &quot;steal&quot; electrons from each other&lt;/li&gt;&#xA;&lt;li&gt;Electrons can be &quot;knocked off&quot; of atoms by high-energy photons (light or EM radiation)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The difficulty with explaining these things by analogy to regular-life experiences is that there is no regular-life experience that captures the whole picture. This means you are limited to activity-type demonstrations, videos/pictures, or analogies for specific pieces of the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend a mixture of all three approaches. Kids like hands-on stuff, so maybe you could explain the basic concepts with very simple analogies, then do an activity where the kids are given &quot;electrons&quot; (tennis balls, maybe) and are assigned to be metals or non-metals. Then you could have them give each other electrons and count charges as they go. If you had containers that could hold 8 electrons, you could even represent valence shells and the octet rule - metals could start with two or three electrons, non-metals with 5, 6, or 7, and the metals would need to empty their shells while the non-metals would need to fill them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would demonstrate a very important concept that is hard to teach in the abstract: Electrons don't disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are also interested in teaching about ionization by radiation, you could assign each student a number representing their ability to &quot;hold on&quot; to an electron. Here is where you might be able to use a shopping-type analogy - only instead of students giving up electrons &lt;em&gt;as money&lt;/em&gt;, they would give them up &lt;em&gt;in exchange&lt;/em&gt; for money. In this case, the money represents the energy, and you would be giving them the energy (money) needed to remove (sell) an electron. Atoms that charged a higher &quot;price&quot; (require more energy) would wind up with more money at the end of the exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many variations on these ideas that you could use, but I think anything that covered the seven points I mentioned above while simultaneously developing an intuitive understanding that energy and electrons are conserved could be called successful.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-21T21:36:45.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15503" PostTypeId="1" CreationDate="2014-08-21T23:31:11.033" Score="2" ViewCount="922" Body="&lt;p&gt;Given a nonideal substance inside a box with diathermal walls, the pressure is increased reversibly. How do you calculate the heat absorbed?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I assume we use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dE = dq + dw = dq - pdV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=&gt; dq = dE + pdV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to deal with the &quot;dE&quot; term.&lt;/p&gt;&#xA;" OwnerUserId="7564" LastActivityDate="2014-10-27T21:51:22.513" Title="How to calculate the heat absorbed by a reversible process?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15504" PostTypeId="1" AcceptedAnswerId="15508" CreationDate="2014-08-22T01:49:29.280" Score="1" ViewCount="85" Body="&lt;p&gt;Based on the comments and answers, this question wasn't very clear, so I have completely overhauled this question to make it much clearer in another post:&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15532/how-to-create-the-equivalent-of-luminol-for-a-compound&quot;&gt;How to create the equivalent of luminol for a compound?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Original Post/Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to detect trace amounts of the chemical 'monomorine I' (5-methyl-3-butyloctahydroindolizine) on a surface. (It is an ant pheromone)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to determine if the molecule is visible in IR or UV? (before I spend the money on this equipment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a strategy for causing a visible reaction? or even a reaction visible by IR or UV?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advice on where to even begin would be a huge help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I do not have a chemistry lab with a Mass Spectrometer at my disposal)&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-11-15T20:02:30.150" Title="How can I detect a specific molecule on a surface?" Tags="&lt;organic-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15505" PostTypeId="1" AcceptedAnswerId="15506" CreationDate="2014-08-22T03:22:09.663" Score="3" ViewCount="5722" Body="&lt;p&gt;It is generally said that reactants react so that they can achieve a lower energy state. Then why does a reversible reaction occur in the first place? Since turning reactants into products increases their energy - an undesirable effect from the point of view of reacting molecules.&lt;/p&gt;&#xA;" OwnerUserId="7567" LastEditorUserId="1878" LastEditDate="2014-08-22T04:06:29.963" LastActivityDate="2019-12-03T15:25:37.047" Title="Causes of reversible reaction" Tags="&lt;reaction-mechanism&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15506" PostTypeId="2" ParentId="15505" CreationDate="2014-08-22T05:25:22.333" Score="6" Body="&lt;p&gt;First, let's tackle what an irreversible reaction is. Baking a cake, for example, is commonly considered &quot;irreversible.&quot; The eggs, flour, milk, etc. which you started with are no longer present in their original forms after the cake has been baked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can you &lt;em&gt;reconstitute&lt;/em&gt; the eggs, flour, milk, etc.? Of course you could take the molecules of the cake and piece them together to make eggs, flour, and milk, but doing so would be a &lt;em&gt;very&lt;/em&gt; formidable task. A &lt;em&gt;lot&lt;/em&gt; of energy will have to be expended to reconstitute the egg, for example, because eggs are made of protein and with heat (baking) proteins are denatured (lose their original 3D conformation and possibly their linear makeup or amino acid sequence as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of an irreversible reaction is &lt;em&gt;combustion&lt;/em&gt; of hydrocarbons - i.e. burning gasoline in your car. This reaction is definitely reversible &lt;em&gt;in the long run&lt;/em&gt; - it just takes a few million years, a lot of heat, and a lot of pressure for the hydrocarbons to be formed again. But for semantic's sake we say that combustion is irreversible because it's not reversible on a really relevant time scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general, the difference between irreversible and reversible isn't exactly well defined - i.e. it might be &lt;em&gt;hard&lt;/em&gt; to reverse a reaction (turn a cake back into eggs, etc.) or it might take a &lt;em&gt;long&lt;/em&gt; time to reverse a reaction, but what's hard? What's long? Tellinging, the IUPAC (&lt;em&gt;International Union of Pure and Applied Chemistry&lt;/em&gt;) does not try to define irreversible vs. reversible reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A deeper treatment would involve discussion of equilibrium constants (symbolized by capital K) but I'll let another poster get into this if necessary. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is generally said that reactants react so that they can achieve a&#xA;  lower energy state. Then why does a reversible reaction occur in the&#xA;  first place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good question. Remember that we can always add energy to make an unfavorable reaction proceed. For example, the sodium ion, which is isoelectronic with neon, is stable with a full octet of electrons. However, we can still take away more electrons. It just takes a rather sizable application of energy. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-22T05:36:10.493" LastActivityDate="2014-08-22T05:36:10.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15507" PostTypeId="2" ParentId="15505" CreationDate="2014-08-22T05:48:49.147" Score="3" Body="&lt;p&gt;There are several reasons for the reverse reaction to be favored; one is temperature. In the following reaction:&#xA;$$\ce{2NO_2&amp;lt;=&amp;gt; N_2O_4 + energy}$$&#xA;The right is the favored product, unless the temperature is increased. With enough heat, then the reactant is favored.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2014-08-22T06:05:38.277" LastActivityDate="2014-08-22T06:05:38.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15508" PostTypeId="2" ParentId="15504" CreationDate="2014-08-22T05:58:48.350" Score="3" Body="&lt;p&gt;Simple UV spectroscopy is not an option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't find anything at all above 254 nm - probably just a a weak band in the 220 nm range. Indolizidines are pretty dead in UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derivatisation towards a fluorescent compound would require clevage of a $\ce{C-N}$ bond, formally a dealkylation of the tertiary bridgehead amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea if a reagent like 2-naphthyl chloroformate, which does form fluorescent carbamates upon heating with tertiary amines will work here and honestly, I wouldn't bother to try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in Martin's comment, GC/MS after washing/extraction probably is the way to go. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-08-22T05:58:48.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15511" PostTypeId="1" CreationDate="2014-08-22T08:47:08.140" Score="20" ViewCount="485" Body="&lt;p&gt;In reading about chemical oscillations such as those that occur in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction&quot; rel=&quot;noreferrer&quot;&gt;Belousov–Zhabotinsky reaction&lt;/a&gt; (BZ), it's often reported that these reactions were initially not taken seriously, because of a result that a homogeneous system cannot oscillate around its equilibrium point. It is usually claimed that the assumption of detailed balance guarantees this. I want to understand this result, but I can't find any specific references to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand very well why this argument doesn't apply to the BZ reaction, since the latter doesn't oscillate around its equilibrium point. However, I'm interested in understanding what this argument is, and whether it &lt;em&gt;does&lt;/em&gt; really apply to all oscillations around the equilibrium point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kondepudi and Prigogine (2002) imply that the argument is as follows: for a single reaction, one cannot move from one side of the equilibrium to the other without passing &lt;em&gt;through&lt;/em&gt; the equilibrium. Once the system reaches equilibrium it can't then keep on reacting, as this would contradict the second law, and therefore an oscillatory approach to equilibrium is impossible in such a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course makes sense for a system that only involves one reaction. However, for a system in which multiple reactions can take place it doesn't seem to be enough. For example, let's imagine that I have reactions $\ce{A &amp;lt;=&amp;gt; B}$, $\ce{B &amp;lt;=&amp;gt; C}$ and $\ce{A &amp;lt;=&amp;gt; C}$, whose kinetics are determined by some complicated set of catalytic effects. Let's also suppose that the Gibbs energies of A, B and C are the same, so that the equilibrium contains all three reagents in a 1:1:1 ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it doesn't seem impossible that I could start off with a high concentration of A and watch the following sequence of events: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;most of the A is converted into B, leaving a little bit of A unconverted; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;most of the B is converted into C, leaving a little bit of B unconverted;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;most of the C is converted into A, leaving a little bit of C unconverted;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;this cycle repeats, with the unconverted material building up over time until the system converges to its equilibrium concentrations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If I were to plot the concentration of the three reagents against one another as a 3D plot, it would look something like this (with apologies for my shaky drawing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tasaA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black dot is the equilibrium point to which the concentrations eventually converge. The path is confined to a plane, because the total concentration of A+B+C cannot change. The free energy decreases as the path approaches the equilibrium point, so the second law by itself doesn't seem to rule out this behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a system behaved this way, it would be an example of oscillations around equilibrium in a multi-reaction system. My question is, &lt;strong&gt;is there a result that says this cannot happen, if we assume the kinetics obey detailed balance at the equilibrium point?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is it called, how is it derived, and where can I read about it? If not, is there a known example of a chemical system that does approach equilibrium through oscillations in this way? (A 'toy' system will do, as long as the kinetics obey detailed balance.) Note that the BZ reaction is again not an example of this, since its oscillations do not take place around the equilibrium point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps part of the argument is that the reasoning described by Kondepudi and Prigogine must hold for &lt;em&gt;each&lt;/em&gt; reaction, independently. For the three-reagent system I described, this seems to be enough to rule out oscillations around equilibrium: the $\ce{A &amp;lt;=&amp;gt; B}$ reaction cannot pass through its equilibrium point, regardless of what happens to the concentration of C. But for a four-species system this no longer seems true: then the $\ce{A &amp;lt;=&amp;gt; B}$ reaction &lt;em&gt;can&lt;/em&gt; move away from its equilibrium, as long as it's coupled to the $\ce{C &amp;lt;=&amp;gt; D}$ reaction. If oscillations around equilibrium really are impossible in every conceivable system this seems quite non-trivial, and it would be nice to see a proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;Reference:&lt;/strong&gt; Dilip Kondepudi and Ilya Prigogine (2002) Modern Thermodynamics: From Heat Engines to Dissipative Structures. Wiley. 2nd edition.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="10185" LastEditDate="2015-09-28T13:32:41.427" LastActivityDate="2019-02-24T12:00:25.857" Title="Can a multi-species system oscillate around equilibrium?" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15512" PostTypeId="2" ParentId="15504" CreationDate="2014-08-22T09:26:35.330" Score="2" Body="&lt;p&gt;Is it possible? Sure. Could you, considering practical issues, use it here? No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is definitely one spectroscopic technique but it's fairly difficult to do and requires the use of ultrafast lasers. Such lasers require vibration-damped tables, constant (and low) relative humidity as well as a constant temperature—this is not therefore a field approach. Furthermore, the whole set up (optical parametric amplifiers etc.) costs hundreds of thousands of pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those interested anyway, &lt;em&gt;Sum Frequency Generation spectroscopy&lt;/em&gt; only works for non-centro-symmetric systems. So if you have a bulk solution, on average the system is symmetric and no spectrum is obtained. At surfaces however this symmetry is broken and molecules even along with their orientations can be observed. This works for both UV/vis and IR. Because the lasers are ultrafast, transient species can often be tracked as a given chemical transformation is occurring, whatever that may be (as long as there is a given chemical signature to lock onto).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A short review of this technique can be found &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs00216-006-0957-5&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-22T09:38:23.283" LastActivityDate="2014-08-22T09:38:23.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15514" PostTypeId="1" CreationDate="2014-08-22T10:00:46.860" Score="10" ViewCount="4177" Body="&lt;p&gt;I would like to know when working in the Chemistry lab would be hazardous to health in the long term. Because as everyone who worked in the chemistry lab knows that they would be handling many compounds that are harmful to human health. Many of those reagents for example especially organic ones are carcinogenic. While some are toxic to humans. Even with safety precaution, we would inevitably be exposed to those chemicals when working with them such as inhaling them. So actually I wonder whether it is safe to work in the chemistry lab especially with so many carcinogens. Im afraid of getting cancer due to this.&lt;br&gt;&#xA;(I'm already halfway through my undergrad chemistry course.)&lt;/p&gt;&#xA;" OwnerUserId="7422" LastEditorUserId="4957" LastEditDate="2018-02-12T13:15:25.257" LastActivityDate="2018-02-12T13:15:25.257" Title="Under what circumstances is working in the chemistry lab hazardous to health in the long term?" Tags="&lt;safety&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-08-20T13:34:37.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="15515" PostTypeId="1" AcceptedAnswerId="15521" CreationDate="2014-08-22T11:24:12.077" Score="4" ViewCount="3152" Body="&lt;p&gt;What are the most negative and the most positive values for proton chemical shifts recorded till present?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-22T20:02:40.633" Title="Most negative and most positive value for proton chemical shifts" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15517" PostTypeId="1" AcceptedAnswerId="15518" CreationDate="2014-08-22T14:58:17.900" Score="3" ViewCount="13989" Body="&lt;p&gt;Will the hydrogen atom always disassociate if the pH of the solution is higher than the pKa of the acid?&lt;/p&gt;&#xA;" OwnerUserId="7569" LastActivityDate="2016-06-10T05:32:21.550" Title="Is it true that a pH higher than the pKa of an acid make the H automatically disassociate?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15518" PostTypeId="2" ParentId="15517" CreationDate="2014-08-22T15:31:15.377" Score="1" Body="&lt;p&gt;The $\text{p}K_{\text{a}}$ of an acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ &lt;strong&gt;mainly&lt;/strong&gt; exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists mainly as undissociated $\ce{HA}$. This doesn't mean that at $\text{pH}$s higher than the $\text{p}K_{\text{a}}$ suddenly every molecule of the acid dissociates, it means that a greater portion of the initial concentration of $\ce{HA}$ will be dissociated as you go higher in $\text{pH}$. There will always be some (potentially tiny) amount of the undissociated acid left even at very high $\text{pH}$ because it is an equilibrium. For more information on $\text{p}K_{\text{a}}$ see &lt;a href=&quot;https://chemistry.stackexchange.com/a/6143/189&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-08-22T18:26:06.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15519" PostTypeId="1" CreationDate="2014-08-22T15:48:15.167" Score="0" ViewCount="4401" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the pH of an aqueous solution obtained by mixing &#xA;  20 ml of $\ce{HCOONa}$ 0,1 M and 5 ml of $\ce{HNO_3}$ 0.2 M.&#xA;  ($K_\mathrm a$ of $\ce{HCOOH}$ $= 10^{-4}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My try:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of the moles involved are: $n_{\ce{HCOONa}}=0.002~\mathrm{mol}$, $n_{\ce{HNO3}} = 0.001~\mathrm{mol}$.&#xA;We have the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCOONa + HNO3 -&amp;gt; HCOOH + NaNO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where 0.001 moles of $\ce{HNO3}$ si is consumed by reacting with 0.001 moles of $\ce{HCOONa}$ giving 0.001 moles of $\ce{HCOOH}$. Because remains $0.002\ \mathrm{mol}-0.001\ \mathrm{mol}=0.001~\mathrm{mol}$ of $\ce{HCOONa}$, we'll have a buffer solution with the acid $\ce{HCOOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It'll start the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCOOH + H2O &amp;lt;=&amp;gt; HCOO- + H3O+}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the initial concentration of $\ce{HCOO-}$ is equal to $\ce{HCOONa}$ and is $\frac{0.001~\mathrm{mol}}{(0.02+0.005)~\mathrm l} = 0.04~\mathrm{mol~l^{-1}}$. So we have, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H3O+}]=K_\mathrm a \frac{[\ce{HCOO-}]}{[\ce{HCOOH}]} \Rightarrow \mathrm{pH} = \mathrm pK_\mathrm a - \log_{10}{\frac{[\ce{HCOO-}]}{[\ce{HCOOH}]}} = \mathrm pK_\mathrm a = 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct?&lt;/p&gt;&#xA;" OwnerUserId="7570" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T14:37:28.317" LastActivityDate="2016-11-06T14:37:28.317" Title="What is the pH of a solution of HCOONa and HNO3?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15520" PostTypeId="1" AcceptedAnswerId="15522" CreationDate="2014-08-22T15:59:55.383" Score="2" ViewCount="920" Body="&lt;p&gt;Why does the standard 10gm/15ml lactulose solution turn dark when exposed to heat above 30 degrees Celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a simple question. It is funny that the warnings/important information on the drug label of lactulose state that this effect is harmless, but then go on to warn not to use if it becomes too dark. How do I know if it is &quot;too dark&quot;? This is obviously very subjective and simply want to know why it becomes dark or &quot;too dark&quot; so it is scientifically understood why it is safe (or unsafe) to use when it is dark.&lt;/p&gt;&#xA;" OwnerUserId="7571" LastEditorUserId="30836" LastEditDate="2020-06-29T02:13:31.643" LastActivityDate="2020-06-29T02:13:31.643" Title="Why does lactulose turn dark?" Tags="&lt;organic-chemistry&gt;&lt;heat&gt;&lt;carbohydrates&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15521" PostTypeId="2" ParentId="15515" CreationDate="2014-08-22T19:11:56.830" Score="7" Body="&lt;p&gt;I don't know if the following two examples present the &quot;largest&quot; upfield and downfield proton-nmr chemical shifts, but I suspect they're in the running.  The dihydropyrene dianion example has 16 pi electrons &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UNux.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;delocalized around the periphery of the pyrene frame. It fits the 4n rule with n=4, so it is antiaromatic.  The [16]-annulene dianion has 18 pi electrons and fits the 4n+2 rule with n=4, so it is aromatic.  Note how the direction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromatic_ring_current&quot; rel=&quot;nofollow noreferrer&quot;&gt;ring current&lt;/a&gt; reverses between antiaromatic (paramagnetic current) and aromatic (diamagnetic current) systems.  More intesting proton chemical shifts can be found in this &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;compilation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guilty of thinking organic.  If we're including inorganic proton shifts, then how about &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_metal_hydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;IrHCl2(PMe(t-Bu)2)2&lt;/a&gt; which has a chemical shift of -50.5!&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-22T20:02:40.633" LastActivityDate="2014-08-22T20:02:40.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15522" PostTypeId="2" ParentId="15520" CreationDate="2014-08-22T19:42:43.460" Score="3" Body="&lt;p&gt;When milk is heat-processed, some of the lactose will isomerize to lactulose, a disaccharide.  Lactulose &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LQH2p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can decompose by the same routes that most sugars decompose.  These are complicated processes with many products and pathways, so sugar decomposition is rarely, if ever, fully understood.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It can hydrolyze to its constituent monosaccharides fructose and galactose.&lt;/li&gt;&#xA;&lt;li&gt;Lactulose or its released monosaccharides equilibrate with their ring-opened isomers and the alpha-hydroxyl can air oxidize to produce an alpha-dicarbonyl.  These alpha-dicarbonyls are prone to further oxidation to generate formic acid.  Over time, as the solution becomes more acidic from the formic acid build-up, further oxidations and eliminations (dehydration) can take place to produce a mixture of colored products.&lt;/li&gt;&#xA;&lt;li&gt;In addition to degrading to smaller molecules, sugars (and the dehydration and oxidation products mentioned above) can also oligomerize to produce a variety of viscous, colored products.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Caramel#Chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Caramelization&lt;/a&gt; (see the &quot;Chemistry&quot; section in the link) is a term applied to this chemical polymerization and coloration process.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you would like to read more about lactulose and its decomposition, take a look at this &lt;a href=&quot;http://books.google.com/books?id=hz9U9nzy-rQC&amp;amp;pg=PA239&amp;amp;lpg=PA239&amp;amp;dq=degradation%20mechanism%20of%20lactulose&amp;amp;source=bl&amp;amp;ots=8ruhGrB5ez&amp;amp;sig=1ggEhqxLQ32xzOd_4fg7hohj3NA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=AoD3U9WcB4eAygStm4GYAg&amp;amp;ved=0CEIQ6AEwAw#v=onepage&amp;amp;q=degradation%20mechanism%20of%20lactulose&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;, especially section 7.2.1 starting on page 236.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-22T20:10:01.193" LastActivityDate="2014-08-22T20:10:01.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15523" PostTypeId="2" ParentId="15519" CreationDate="2014-08-22T21:00:03.787" Score="1" Body="&lt;p&gt;Yes. You could also put the molar ratio into the equation for calculating pH (then you didn't need to calculate the concentrations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those are not initial concentrations, but equilibrium concentrations.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-22T21:00:03.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15526" PostTypeId="1" AcceptedAnswerId="15656" CreationDate="2014-08-22T22:33:54.667" Score="4" ViewCount="172" Body="&lt;p&gt;When trying to develop mathematical models to show the concentration in a reaction, where can one find a database or good source of known equilibrium constants if it is not necessarily very convenient to find them empirically?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2014-08-28T08:04:57.520" Title="Where can I find Equilibrium Constants?" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15527" PostTypeId="1" AcceptedAnswerId="15534" CreationDate="2014-08-23T00:20:25.547" Score="0" ViewCount="1160" Body="&lt;p&gt;When reading about making glycerine soap, the recipes suggested to melt animal fat and mix it with $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use Sunflower oil instead? Or a olive oil for nice flavor? Is heating necessary in that case?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2019-03-27T08:58:19.643" Title="Can I use Sunflower oil to make soap?" Tags="&lt;home-experiment&gt;&lt;fats&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15528" PostTypeId="1" AcceptedAnswerId="15530" CreationDate="2014-08-23T01:29:11.477" Score="-2" ViewCount="219" Body="&lt;p&gt;I am confused about significant figures. In a teaching video, the teacher showed this math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 22.2 + 22.2 + 30.10 + 30.10 = 104.6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is, of course, the actual answer, but in using significant figure count the answer would be $105$, because:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;22.2---- 3 significant numbers&#xA;&#xA;22.2---- 3 significant numbers&#xA;&#xA;30.10--- 4 significant numbers&#xA;&#xA;30.10--- 4 significant numbers&#xA;&#xA;+______&#xA;&#xA;104.6 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first number has the fewest significant figures, so we would &quot;round&quot; to 3 significant numbers in our answer. I emailed the course helpline, but they sent a confusing email back about differences in counting the significant numbers whether you are adding and subtracting, or multiplying and dividing. I need help explaining this.&lt;/p&gt;&#xA;" OwnerUserId="7575" LastEditorUserId="5026" LastEditDate="2018-02-23T15:57:05.940" LastActivityDate="2018-02-23T15:57:05.940" Title="How to use signicant figures?" Tags="&lt;significant-figures&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15529" PostTypeId="2" ParentId="15527" CreationDate="2014-08-23T02:05:57.450" Score="3" Body="&lt;p&gt;You can use any triglyceride to make soap.  The process is called saponification.  In this process fatty acid molecules are cleaved from glycerol (glycerine).  In order for the reaction to proceed it is heated.  Regardless of the fatty acid source, the soap will probably not taste very good.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-23T02:05:57.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15530" PostTypeId="2" ParentId="15528" CreationDate="2014-08-23T02:09:16.583" Score="3" Body="&lt;p&gt;The rule you are applying is for multiplication and division.  For addition and subtraction look at the numbers in the problem.  22.2 is accurate to the tenths place so the answer can be reported to the tenths place regardless of how many sig figs are in the answer.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-23T02:09:16.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15531" PostTypeId="1" CreationDate="2014-08-23T02:20:28.267" Score="3" ViewCount="1414" Body="&lt;p&gt;Can ionic and atomic radii be compared?  For example, this is a common problem in some intro chem courses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rank the following from smallest to largest:  sodium ion, magnesium ion, neon, fluoride, oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any student who has been trained in trends should be able to get the &quot;answer&quot;, but if one looks at the actual figures, there are data that show that sodium ion is larger than neon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am of the opinion that it is comparing apples to oranges and that the problem can be rescued by removing neon from the list.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-23T12:54:16.463" Title="Can ionic and atomic radii be compared?" Tags="&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15532" PostTypeId="1" AcceptedAnswerId="15536" CreationDate="2014-08-23T04:33:41.197" Score="2" ViewCount="1312" Body="&lt;p&gt;How would you go about creating a compound, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Luminol&quot; rel=&quot;nofollow&quot;&gt;luminol&lt;/a&gt;, that would visibly react with &lt;a href=&quot;http://www.pherobase.com/database/compound/compounds-detail-monomorine%20I.php&quot; rel=&quot;nofollow&quot;&gt;monomorine I&lt;/a&gt; (5-methyl-3-butyloctahydroindolizine) on a household surface (like a kitchen floor)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is ok if the compund reacts with other similar compounds as well, as long as they aren't common to a household.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Luminol&lt;/strong&gt; is the chemical used by forensic investigators (like those on CSI) to detect trace amounts of blood at crime scenes. The luminol is sprayed over an area and when it comes into contact with blood (specifically, the iron), it exhibits a blue glow.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Monomorine I&lt;/strong&gt; is an ant trail pheromone (a trail guiding other ants to food).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Objective:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to apply a compound on a surface and see the trails left by ants.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastActivityDate="2014-08-23T08:11:44.020" Title="How to create the equivalent of luminol for a compound?" Tags="&lt;organic-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15533" PostTypeId="1" CreationDate="2014-08-23T04:55:02.840" Score="2" ViewCount="194" Body="&lt;p&gt;For example, why doesn't AgOH exist as a stable compound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that it rapidly decomposes to silver oxide and water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2AgOH -&amp;gt; Ag2O + H2O}$ $~~~~~~~~~pK_{decomposition}= 2.875$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://actachemscand.org/pdf/acta_vol_14_p0717-0725.pdf&quot; rel=&quot;nofollow&quot;&gt;http://actachemscand.org/pdf/acta_vol_14_p0717-0725.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above paper gives the pK value for the above equilibrium but doesn't bother to really delve into why silver hydroxide might be so unstable in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there repulsion between the silver cation and the proton?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-23T17:49:32.623" LastActivityDate="2014-08-23T17:49:32.623" Title="Why don't plus one oxidation state metals form stable complexes with hydroxide ion?" Tags="&lt;solubility&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15534" PostTypeId="2" ParentId="15527" CreationDate="2014-08-23T05:09:32.880" Score="3" Body="&lt;p&gt;You can use any triglyceride, but the touch, the firmness, the cleaning effectiveness of the resulting soap will be different. Soaps also can go rancid just like their corresponding fat/oil. DIY soapmakers generally mix oil or use specifically selected ones. Coconut oil, castor oil, olive oil, animal fats are generally popular.&#xA;Without the intent to advertise any specific site:&#xA;&lt;a href=&quot;http://www.lovinsoap.com/oils-chart/&quot; rel=&quot;nofollow&quot;&gt;http://www.lovinsoap.com/oils-chart/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Olive oil is actually a pretty common component, and you can buy olive oil based soaps even is shops. Flavor? If you do it well, most probably not much remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating is generally necessarily, saponification is a slow reaction and most home-made soap is not fully reacted (containing lot of NaOH). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note on safety:&#xA;Many people make soap at home, because it is &quot;more natural/organic&quot; just following some youtube videos without knowing much about lab work or chemistry. Using your own blender, doing in your kitchen or using your own kitchenware is a very bad idea. NaOH can make you blind, so do not ignore the &quot;wear googles!&quot; signs. Also just mixing everything up in a highly corrosive mixture contains or not remains of NaOH or produce all kind of unknown byproducts then putting everything on your skin may not the healthiest choice of living. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-23T05:09:32.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15535" PostTypeId="1" CreationDate="2014-08-23T07:16:11.963" Score="1" ViewCount="14205" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The successive ionization energies of an unknown element X are:&lt;br&gt;&#xA;  1st - 9534; 2nd - 11790; 3rd - 13683; 4th - 15309; 5th - 16458.&lt;br&gt;&#xA;  To what group number does X belong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There aren't any significant jumps so that means X is in groups 5-8. After that I don't know.&lt;/p&gt;&#xA;" OwnerUserId="6988" LastEditorUserId="5026" LastEditDate="2018-02-28T16:54:17.037" LastActivityDate="2018-03-01T07:28:17.083" Title="Determining the group number of an element from its successive ionization energies" Tags="&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15536" PostTypeId="2" ParentId="15532" CreationDate="2014-08-23T08:11:44.020" Score="3" Body="&lt;p&gt;Probably not easily, especially if the concentrations are really low. Luminol + $\ce{H2O2}$ only works with blood because it's a catalytic reaction, i.e. it doesn't take much iron to create a lot of chemiluminescence. I'm not an organic chemist, but I'm not aware of any sort of similar reaction specific for this class of compounds. I did see this reaction: &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-4168(19980501)21:5%3C315::AID-JHRC315%3E3.0.CO;2-V/abstract&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-4168(19980501)21:5%3C315::AID-JHRC315%3E3.0.CO;2-V/abstract&lt;/a&gt; Unfortunately, it's not open access and I can't share my institutional access, but it's basically a photo-generated ruthenium complex that reacts with tertiary amines and indoles which might be applicable here, though I'm guessing is not sufficiently sensitive. It's not terribly selective so you might run into interference problems and in that paper they use it for HPLC  detection so it may not be bright enough to see with the naked eye. (I have seen similar chemiluminescent compounds before and they're pretty dim, even with the lights out)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are ways to produce visible changes in response to small concentrations of an analyte: LSPR, ELISA, etc. but they're a lot of work to implement for new analytes and probably too expensive for looking at a large area. Chemical imaging based on mid-IR or Raman spectroscopy may be applicable, though as far as I know, they're not geared for measuring large areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that directly visualizing pheromone traces is a non-trivial task. Groups seem to do things like use image analysis (&lt;a href=&quot;http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.1002592&quot; rel=&quot;nofollow&quot;&gt;http://www.ploscompbiol.org/article/info:doi/10.1371/journal.pcbi.1002592&lt;/a&gt;) to estimate the levels of pheromones. I would like to have seen a comparison of their model with actual pheromone measurements and the way I would propose to do it is to take surface samples after the experiment. If a big sheet of paper was used as the surface, you could cut it up in a grid and extract the pheromone for analysis by LC-MS or whatever (or maybe even used directly as a paperspray emitter for MS). An ambient-ionization mass spec technique like LTP or DART would also probably be a good way to construct a surface map of pheromone concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Briefly&lt;/em&gt;: There are certainly ways of measuring trace amounts of things like pheromones, but a simple CSI-esque naked eye visualization probably isn't easily achievable.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-08-23T08:11:44.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15537" PostTypeId="2" ParentId="15383" CreationDate="2014-08-23T10:28:11.613" Score="2" Body="&lt;p&gt;The auto-ionization is an equilibrium process, means that it works in both direction. As water evaporates, the auto-ionization converts back the ions into water molecules. You can take it as an example to le chatelier's principle. You gradually remove water from the system, therefore the concentration of hydroxide and oxonioum would go up if they do not turn to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ions in gas phase are rather high energy things, because the dielectric properties of air are much less favorable in stabilizing charges than water can.&#xA;It is much more easy to evaporate water molecules. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-08-23T10:28:11.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15538" PostTypeId="2" ParentId="15535" CreationDate="2014-08-23T12:15:54.440" Score="4" Body="&lt;p&gt;There are other little jumps to look out for, like the removal of the first electron from  a filled subshell. Consider the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionization energies (in eV)&lt;/a&gt; for the third period elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline&#xA;\text{electron}&amp;amp;\ce{Ar}&amp;amp;\ce{Cl}&amp;amp;\ce{S}&amp;amp;\ce{P}&amp;amp;\ce{Si}&amp;amp;\ce{Al}&amp;amp;\ce{Mg}&amp;amp;\ce{Na}\\\hline&#xA;\ce{3p}&amp;amp;1521&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;2666&amp;amp;1251&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;3931&amp;amp;2298&amp;amp;1000&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;5771&amp;amp;3822&amp;amp;2252&amp;amp;1012&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;7238&amp;amp;5159&amp;amp;3357&amp;amp;1907&amp;amp;787&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;8781&amp;amp;6542&amp;amp;4556&amp;amp;2914&amp;amp;1577&amp;amp;578&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3s}&amp;amp;11995&amp;amp;9362&amp;amp;7004&amp;amp;4964&amp;amp;3233&amp;amp;1817&amp;amp;738&amp;amp;-\\\hline&#xA;\ce{3s}&amp;amp;13842&amp;amp;11018&amp;amp;8496&amp;amp;6374&amp;amp;4356&amp;amp;2745&amp;amp;1451&amp;amp;496\\\hline&#xA;\ce{2p}&amp;amp;40760&amp;amp;33604&amp;amp;27107&amp;amp;21267&amp;amp;16091&amp;amp;11577&amp;amp;7733&amp;amp;4562\\\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how, in addition to the big jump between $\ce{3s}$ and $\ce{2p}$, there is a small jump between $\ce{3p}$ and $\ce{3s}$ for Ar, Cl, S, P, and Si. If you have five ionization energies, and you don't see the large jump, look for the small jump. If you don't see the small jump, what group is X in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another even smaller jump to look for. Half-filled subshells are also slightly more stable than other configurations (except filled subshells). Thus, it should require a slightly higher increase in ionization energy to remove an electron form any configuration with $\ce{np^3}$ than from $\ce{np^2}$ or $\ce{np^4}$. Let's look at the step wise &lt;strong&gt;increase&lt;/strong&gt; in ionization energy (in other words, how much higher is the ionization energy for the second electron than the ionization energy for the first electron and so on?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline&#xA;\text{electron}&amp;amp;\ce{Ar}&amp;amp;\ce{Cl}&amp;amp;\ce{S}&amp;amp;\ce{P}&amp;amp;\ce{Si}&amp;amp;\ce{Al}&amp;amp;\ce{Mg}&amp;amp;\ce{Na}\\\hline&#xA;\ce{3p}&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;1145&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;1265&amp;amp;1047&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;1840&amp;amp;1524&amp;amp;1252&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;1467&amp;amp;1337&amp;amp;1105&amp;amp;895&amp;amp;-&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3p}&amp;amp;1543&amp;amp;1383&amp;amp;1199&amp;amp;1007&amp;amp;790&amp;amp;-&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3s}&amp;amp;3214&amp;amp;2820&amp;amp;2448&amp;amp;2050&amp;amp;1656&amp;amp;1239&amp;amp;-&amp;amp;-\\\hline&#xA;\ce{3s}&amp;amp;1847&amp;amp;1656&amp;amp;1492&amp;amp;1410&amp;amp;1123&amp;amp;928&amp;amp;713&amp;amp;-\\\hline&#xA;\ce{2p}&amp;amp;26918&amp;amp;22586&amp;amp;18611&amp;amp;14893&amp;amp;11735&amp;amp;8832&amp;amp;6252&amp;amp;4066\\\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the increase in ionization energy for the third electron to be removed from argon is larger than the increase for the second or fourth electrons. Your numbers don't have this little increase in them at all. Have you considered that X might be a transition metal?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-01T07:28:17.083" LastActivityDate="2018-03-01T07:28:17.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15539" PostTypeId="2" ParentId="15531" CreationDate="2014-08-23T12:54:16.463" Score="2" Body="&lt;p&gt;A periodic table coupled with simple electrostatic arguments &lt;em&gt;should&lt;/em&gt; suffice to provide at least a rough ordering of ions and atoms. For the ions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons =&gt; an excess of POSITIVE charge =&gt; the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' =&gt; $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.&lt;/li&gt;&#xA;&lt;li&gt;We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons =&gt; an excess of NEGATIVE charge =&gt; the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions =&gt; $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: I wouldn't be willing to stake my life on this, so some community validation or corrections would be much appreciated)&lt;/p&gt;&#xA;" OwnerUserId="7549" LastActivityDate="2014-08-23T12:54:16.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15541" PostTypeId="1" AcceptedAnswerId="15547" CreationDate="2014-08-23T13:30:03.910" Score="18" ViewCount="10762" Body="&lt;p&gt;I was looking at the isotope's table and noticed that one of lead's isotopes can actually turn into stable gold through the following mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{_{82}^{197}Pb -&amp;gt; _{81}^{197}Tl -&amp;gt; _{80}^{197}Hg -&amp;gt; _{79}^{197}Au}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know (or at least guess) that such a process must be awfully ineffective. Still, I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How could we get a great quantity of $\ce{_{82}^{197}Pb}$?&lt;/li&gt;&#xA;&lt;li&gt;How long would it take for it to turn into gold? Can that be accelerated?&lt;/li&gt;&#xA;&lt;li&gt;Once most of it has been turned into gold, how can we extract the gold (there are still remaining $\ce{Pb,Tl,Hg}$ molecules)?&lt;/li&gt;&#xA;&lt;li&gt;What would be the yield of the process? How much money would it approximately cost compared to how much we win in gold?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="7475" LastEditDate="2017-10-22T14:50:25.597" LastActivityDate="2018-04-29T06:15:38.417" Title="Turn lead into gold via radioactive decay" Tags="&lt;metal&gt;&lt;radioactivity&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15542" PostTypeId="1" AcceptedAnswerId="15543" CreationDate="2014-08-23T13:48:27.243" Score="5" ViewCount="9220" Body="&lt;p&gt;My understanding was that since the acceptance of &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Antoine_Lavoisier&amp;amp;oldid=619964211#Easter_memoir&quot;&gt;Lavoisier's revised Easter Memoir of 1778&lt;/a&gt;, all acids were understood to be oxidising agents.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, formic acid is often &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidizing_and_Reducing_Agents&quot;&gt;listed as a reducing agent&lt;/a&gt; (i.e. the opposite of an oxidising agent), despite being an acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please can you reconcile this apparent contradiction for me?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* &lt;em&gt;What is meant by &quot;oxidising&quot; or &quot;oxidising agent&quot; has, of course, changed since then, now referring to &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Oxidizing_agent&amp;amp;oldid=621550293&quot;&gt;electron transfer rather than necessarily involving the element oxygen&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7579" LastEditorUserId="1878" LastEditDate="2014-08-23T14:44:23.600" LastActivityDate="2016-03-07T16:35:55.583" Title="Why is formic acid considered an acid if it is a reducing agent?" Tags="&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15543" PostTypeId="2" ParentId="15542" CreationDate="2014-08-23T13:55:05.480" Score="2" Body="&lt;p&gt;Formic acid is a carboxylic acid with the formula of $\ce{CH_2O_2}$ and is easily oxidized to carbon dioxide. Other carboxylic acids are not as easy to oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;If a compound gets oxidized (as formic acid can), then another compound has to be reduced. The compound that gets oxidized is called a reducing agent. The compound that gets reduced is called an oxidizing agent. Most carboxylic acids cannot oxidize other compounds (the acid is not strong enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lavoisier was referring to mineral acids ($\ce{HNO_3, H_2SO_4, }$...) in the text. These acids can oxidized other compounds (such as changing iron into an iron ion).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-08-23T14:16:40.790" LastActivityDate="2014-08-23T14:16:40.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15544" PostTypeId="2" ParentId="15542" CreationDate="2014-08-23T15:10:35.013" Score="3" Body="&lt;p&gt;A molecule might be an oxidising agent under the electron transfer definition, and simultaneously be a reducing agent under the alternative definition: i.e. that reduction is a loss of oxygen &lt;em&gt;or&lt;/em&gt; a gain in hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific question, formic acid can be deemed to be an oxidisation agent since it loses a hydrogen during protonation of another molecule. The really silly way to think about this is following deprotonation, formic acid has a higher percentage of oxygen than it did before, i.e. it &lt;em&gt;possesses&lt;/em&gt; more oxygen than before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of formic acid being a reducing agent is due to the more modern definition of electron transfer. To see the mechanism of this for clarity, see &lt;a href=&quot;http://www.organic-chemistry.org/chemicals/reductions/formicacid.shtm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. It's clear that the formic acid is donating an electron pair (i.e. itself is being oxidised) and is making the other molecule accept those electrons (i.e. that molecule is reduced). The formic acid is therefore a reducing agent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your original issue is due to two different definitions of oxidisation and reduction which still exist to this day, and which are by no means mutually exclusive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Of course, the case of formic acid is more simple because it isn't the acid doing the reducing, it's its conjugate base i.e. &lt;em&gt;formate&lt;/em&gt;. So formic acid is an oxidising agent under the proton definition, yet formate is an reducing agent under the electron transfer definition.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-23T17:27:44.200" LastActivityDate="2014-08-23T17:27:44.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15545" PostTypeId="2" ParentId="15289" CreationDate="2014-08-23T16:19:19.747" Score="5" Body="&lt;p&gt;A &lt;a href=&quot;http://goldbook.iupac.org/CT01038.html&quot; rel=&quot;noreferrer&quot;&gt;chemical species&lt;/a&gt; consists of chemically identical atoms, molecules or ions (more generally &lt;a href=&quot;http://goldbook.iupac.org/M03986.html&quot; rel=&quot;noreferrer&quot;&gt;molecular entities&lt;/a&gt;) and is identified as a single species in a spectroscopic experiment. The latter part of the definition is important, because, for example, a compound which is identified as a single chemical species by NMR spectroscopy can still contain different enantiomers (that are separate molecular entities) which cannot be distinguished by an achiral spectroscopic method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.chemicool.com/definition/substance.html&quot; rel=&quot;noreferrer&quot;&gt;substance&lt;/a&gt; is matter which has a specific and distinguishable chemical composition and specific properties. From this follows that every pure element and chemical compound is a substance. For example, pure $\ce{Fe}$ and $\ce{CH4}$ are substances, while an aqueous solution of $\ce{NaCl}$ and gasoline are not. The latter two are a mixture of two ($\ce{NaCl}$ and $\ce{H2O}$) or more different substances and have therefore properties which depend on the composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/particle&quot; rel=&quot;noreferrer&quot;&gt;Particles&lt;/a&gt; can be defined as microscopic constituents of matter, like atoms or nuclei. The term can also refer to even smaller constituents like electrons, protons, or neutrons, which are called elementary particles. It is also used in the definition of an &lt;a href=&quot;http://goldbook.iupac.org/A00493.html&quot; rel=&quot;noreferrer&quot;&gt;atom&lt;/a&gt;, the smallest particle which is characteristic for a chemical element. Atoms with the same number of protons $Z$ (and equal number of electrons) represent the same chemical element. &lt;a href=&quot;http://goldbook.iupac.org/I03331.html&quot; rel=&quot;noreferrer&quot;&gt;Isotopes&lt;/a&gt; are atoms of the same chemical element which have different &lt;a href=&quot;http://goldbook.iupac.org/M03726.html&quot; rel=&quot;noreferrer&quot;&gt;mass numbers&lt;/a&gt;, i.e. the same number of protons $Z$, but different numbers of neutrons in the nucleus. In the three &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_hydrogen&quot; rel=&quot;noreferrer&quot;&gt;isotopes of hydrogen&lt;/a&gt;, for example, the nuclei contain different number of neutrons (0, 1 and 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://goldbook.iupac.org/S05447.html&quot; rel=&quot;noreferrer&quot;&gt;salt&lt;/a&gt; is a chemical compound which consists of cations and anions. It should be distinguished from a &lt;a href=&quot;http://goldbook.iupac.org/Z06752.html&quot; rel=&quot;noreferrer&quot;&gt;zwitterionic compound&lt;/a&gt;, which contains charged but overall neutral molecules as building blocks. An example for the latter is &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethylamine_N-oxide&quot; rel=&quot;noreferrer&quot;&gt;trimethylamine &lt;em&gt;N&lt;/em&gt;-oxide&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-23T16:19:19.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15546" PostTypeId="2" ParentId="15517" CreationDate="2014-08-23T16:23:08.337" Score="7" Body="&lt;p&gt;I propose the following  answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the following equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$&#xA;The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we rewrite this equation as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}&#xA;$$&#xA;From this equation, we can deduce that when $\ce{pH}&amp;gt;\ce{pK}_A$, $[A^-]&amp;gt;[\ce{H}A]$,  i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2016-06-10T05:32:21.550" LastActivityDate="2016-06-10T05:32:21.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15547" PostTypeId="2" ParentId="15541" CreationDate="2014-08-23T17:03:54.937" Score="20" Body="&lt;p&gt;Interesting idea, but it has already been done, and not cheaply - read on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$.  First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so you can't get a large quantity of $\ce{_{82}^{197}Pb}$'s precursor to begin with.  Second, once $\ce{_{82}^{197}Pb}$ is formed, it has a half-life of 8.1 minutes, so it transmutes quickly to $\ce{_{81}^{197}Tl}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How long would it take for it to turn into gold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{_{82}^{197}Pb}$ half-life = 8.1 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{_{81}^{197}Tl}$ half-life = 2.84 hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{_{80}^{197}Hg}$ half-life = 64.14 hours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 10 half-lives, &lt;em&gt;ca.&lt;/em&gt; 0.1% of the starting material will be left&#xA;$$\ce{(1/2)^10 = $0.0009766$}$$&#xA;The last step is the slowest by far, so after about 641.4 hours (26.73 days), you should have something around 99.9% pure gold.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can that be accelerated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unlike chemical reactions that can heated, catalyzed, etc., this type of nuclear transformation keeps a set schedule. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Once most of it has been turned into gold, how can we extract the gold&#xA;  ? (there are still remaining Pb,Tl,Hg  molecules)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As noted above, you can get whatever purity you desire, just wait.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the yield of the process?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be high for the 3 nuclear transformations you listed.  Each of the elements you listed decays directly and only to the daughter isotope you've shown.  However, as noted above, you can't start with $\ce{_{82}^{197}Pb}$, you generate it from $\ce{_{83}^{197}Bi}$, the decay of which adds some impurity along with lead. And then since the bismuth isotope is not long-lived you'd probably start with its precursor, and so on until you find something that has a long enough life that you could assemble a reasonable quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back around 1980 &lt;a href=&quot;http://en.wikipedia.org/wiki/Glenn_T._Seaborg&quot;&gt;Glenn Seaborg&lt;/a&gt; actually transmuted bismuth to gold, but only a few thousand atoms (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_transmutation&quot;&gt;this reference&lt;/a&gt; also).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much money would it approximately cost compared to how much we win&#xA;  in gold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article I referenced directly above notes, &quot;the expense far exceeds any gain.&quot;  There are other ways (fission and fusion) to produce gold, but at least with the methods available today, the cost would be astronomical.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-23T17:03:54.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15548" PostTypeId="1" AcceptedAnswerId="15552" CreationDate="2014-08-23T17:40:33.500" Score="7" ViewCount="18705" Body="&lt;p&gt;The combustion of methanol tends to generate flames that are less visible than the flames generated by the combustion of other substances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that methanol burns quickly as it has a relatively simple structure that provides some of the oxygen needed for combustion.  I am assuming that the flame comes from a mixture of fuel and air that is undergoing reaction.  If the reaction is fast enough, then there will be less flame.  However, methane and hydrogen both have visible flames, much more visible than methanol, leading me to wonder how the one oxygen in methanol makes so much of a difference.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2021-03-13T22:14:58.913" Title="Why are methanol flames less visible than other flames?" Tags="&lt;kinetics&gt;&lt;combustion&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15549" PostTypeId="1" AcceptedAnswerId="15550" CreationDate="2014-08-23T17:59:22.200" Score="2" ViewCount="1129" Body="&lt;p&gt;I have seen this organic compound named in two different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7vPCk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7vPCk.jpg&quot; alt=&quot;pentan-3-one&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some sources it is written &lt;strong&gt;pentan-3-one.&lt;/strong&gt;. Other sources call it &lt;strong&gt;3-pentanone.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is correct?&lt;/p&gt;&#xA;" OwnerDisplayName="user7484" LastEditorUserId="7475" LastEditDate="2016-09-25T23:20:54.027" LastActivityDate="2016-09-25T23:20:54.027" Title="Which is the correct way to name pentan-3-one?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15550" PostTypeId="2" ParentId="15549" CreationDate="2014-08-23T18:05:01.643" Score="5" Body="&lt;p&gt;Whether you write pentan-3-one or 3-pentanone both are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to IUPAC rule pentan-3-one this is correct format of writing the name of organic compound. So I prefer you to write pentan-3-one rather than 3-pentanone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also read &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorDisplayName="user7951" LastEditDate="2014-10-07T10:15:20.523" LastActivityDate="2014-10-07T10:15:20.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15552" PostTypeId="2" ParentId="15548" CreationDate="2014-08-23T18:40:39.640" Score="9" Body="&lt;p&gt;Here's a &lt;a href=&quot;https://www.youtube.com/watch?v=cxd_CH2NIWE&quot; rel=&quot;noreferrer&quot;&gt;link to a YouTube video&lt;/a&gt; showing a dish of methanol burning next to a dish of ethanol.  The methanol flame is more difficult to see.  Methanol was also widely used as a fuel in the auto racing circuit, until fires that resulted from crashes or fuel filling accidents resulted in &quot;&lt;a href=&quot;http://blog.chembark.com/2010/11/05/methanol-fires-are-invisible/&quot; rel=&quot;noreferrer&quot;&gt;invisible fires&lt;/a&gt;&quot; that were difficult to see during daylight, resulting in severe burns to crews and drivers (&quot;you can't fight the fire, if you can't see it&quot;, this is one reason why it was replaced by ethanol as a fuel for Indy racing).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flame color (visibility) is dependent upon the elemental composition of the material being heated and the temperature to which it is heated.  This is because heat will excite electrons in the element(s) to an excited state; when these electrons return to the ground state visible light (flame test) will often be emitted&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the carbon content of methanol and ethanol we find the following weight percents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;methanol - 38% carbon by weight&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ethanol - 52% carbon by weight&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the heats of combustion for methanol and ethanol we find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;methanol $\ce{~~}$ -726 kJ/mol (exothermic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ethanol $\ce{~~}$ -1368 kJ/mol (much more exothermic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's just not much carbon there to burn in methanol!  Consequently burning methanol doesn't get that hot and its flame will not be as colored as materials that burn more exothermically and result in higher temperatures.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-23T21:35:52.903" LastActivityDate="2014-08-23T21:35:52.903" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15553" PostTypeId="1" AcceptedAnswerId="15555" CreationDate="2014-08-23T18:50:57.047" Score="6" ViewCount="1095" Body="&lt;p&gt;I am having issues understanding the consequence(s) of a decrease in pressure on a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have understood that when we increase the pressure, the side having more number of moles wants its moles to go to the other side where it is more empty(this is how I learnt it). But I fail to understand why if we decrease the pressure by increasing volume the side having less moles goes to the side having more moles? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the case of adding inert gases why does the equilibrium shift in the direction in which larger no of moles of gas are formed?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="33991" LastEditDate="2016-11-26T08:29:35.173" LastActivityDate="2019-02-24T00:28:44.947" Title="Applying the Le Chatelier principle" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15554" PostTypeId="2" ParentId="15548" CreationDate="2014-08-23T19:08:19.350" Score="5" Body="&lt;p&gt;The yellow 'flame' is actually an aerosol of black particles heated to temperatures around 1000-1500 Celsium. Depending on temperature, they can emit deep red, orange, yellow and almost white light. So, for yellow flame to occur, black particles must be present. In case of carbon containing fuels the particles are usually carbon particles (soot). Soot is a product of incomplete combustion of carbon containing molecules. Now, let's see the difference between ethanol and methanol, specifically, at amount of oxygen required for combustion of same volume of vapors. Equations of burning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2CH3OH + 3O2 = 2CO2 + 4H2O}$&#xA;$\ce{C2H5OH + 3O2 = 2CO2 + 3H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it can be seen, same volume of vapors in case of ethanol requires twice as much of air to fully burn, so in case of ethanol incomplete combustion is more common. It is even more common in case of larger molecules, say, higher hydrocarbons, like solid paraffines used in some candles. In case the flame of organic substance has right amount of oxygen (incoming fuel-air mixture contains proper amount of air), it burns with pale flame. It is easily observable in case of hand propane torches, where user regulates amount of air in incoming mixture, depending on amount of air allowed, it is possible to have bright yellow or pale blue flame and everything in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pale blue emission has completely different nature, it is atomic emission of excited radicals and molecules. For example, it is known that $CH$ particle emits a blue light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yellow flame may also happen if another source of particles is available and the temperature is right. For example, $\ce{CuO + Al}$ mixture burns explosively, producing bright yellow flash. However, if temperature is too high, bright white flame should be observed. Indeed, magnesium burns with blinding white flame, especially if it is premixed with solid oxidizer (potassium nitrate). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, atomic/molecular emission is &lt;em&gt;usually&lt;/em&gt; low in common flames. However, even small amounts of sodium in flame makes it bright yellow even if it normally almost colorless. Some other elements may color flame into different colors, ranging from pale violet to deep red. This, however, is a different situation from what you considered.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-23T19:19:32.560" LastActivityDate="2014-08-23T19:19:32.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15555" PostTypeId="2" ParentId="15553" CreationDate="2014-08-23T19:52:25.817" Score="6" Body="&lt;p&gt;The system doesn't &quot;want&quot; anything.  This is a major misconception with Le Chatelier's principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about the following reversible reaction, everything in the gas phase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A + B &amp;lt;=&amp;gt; C}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll simplify the rate law of the forward reaction to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{rate} = K_\mathrm{fwd}[\ce{A}][\ce{B}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse rate law is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{rate} = K_\mathrm{rev}[\ce{C}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the volume is doubled, what happens to the rate of the forward reaction?  It is quartered.  What happens to the forward reaction?  It is halved.  So, the reaction &quot;shifts to reactants&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the volume is halved, what happens to the forward rate?  It is quadrupled.  The reverse rate is doubled.  The reaction &quot;shifts to products&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when an inert gas is added?  Since the inert gas appears in neither the forward nor the reverse rate law, nothing happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid you have learned some misconceptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle was developed before much of what is studied in a general chem course today was known.  It has been used to manipulate the outcome of reactions.  All of the predictions made by Le Chatelier's principle can be made by understanding how reactions work at the particle level.  Le Chatelier's Principle is from the macroscopic perspective and leads many students to generate misconceptions.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="41328" LastEditDate="2019-02-24T00:28:44.947" LastActivityDate="2019-02-24T00:28:44.947" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="15556" PostTypeId="2" ParentId="13871" CreationDate="2014-08-23T20:45:00.613" Score="1" Body="&lt;p&gt;One of the features of a solid is that there is very little movement of the molecules. This prevents most reactions from occurring due to the lack of energy to achieve the activation energy for the reaction. If the coal is already burning, then some surface molecules could absorb the energy (of the fire) to break away from the solid so the reaction with the oxygen is far easier.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-23T20:45:00.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15558" PostTypeId="1" AcceptedAnswerId="15588" CreationDate="2014-08-24T04:22:31.523" Score="9" ViewCount="3588" Body="&lt;h2&gt;Data Point 1 - &lt;em&gt;quantum chemical&lt;/em&gt; calculations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlbaE.png&quot; alt=&quot;Data Point 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Key points in paper:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is some sort of interaction between the hydrogen and the terminal oxygen in the &lt;em&gt;cis&lt;/em&gt; isomer; the author leave it up to the reader to decide whether one should consider that a hydrogen bonding interaction or a dipole-dipole interaction. &lt;strong&gt;Question&lt;/strong&gt; - the interaction seems to fit the definition of H-bonding; why raise the possibility of it being a dipole-dipole interaction? &lt;/li&gt;&#xA;&lt;li&gt;The authors discuss resonance contributors and use bond lengths along with other factors to determine the main contributors for the cis and trans isomers. Their analysis reflects ron's analysis perfectly. &lt;/li&gt;&#xA;&lt;li&gt;This is perhaps the most significant point and it is also discussed on ron's excellent analysis:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jpyzZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bent's rule is not explicitly invoked in the paper's analysis but it is implied. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Data Point 2 - Professor's claim - unsourced&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9qpAv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Data Point 3 - Another Book's Claim&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B9W8T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;My questions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1) I have three sets of bond angles for the HONO molecule. The first and the last agree with each other in a general sense: &lt;em&gt;cis&lt;/em&gt;-HONO exhibits a larger $\ce{O-N-O}$ bond angle than does the trans conformer. The middle picture - my professor's claim - contradicts the other two. &lt;em&gt;Cis&lt;/em&gt;-HONO exhibits a &lt;em&gt;smaller&lt;/em&gt; $\ce{O-N-O}$ bond angle than &lt;em&gt;trans&lt;/em&gt;-HONO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who's right?&lt;/strong&gt; From the replies I have gleaned that hydrogen bonding is likely to have a small to non-existent role in HONO given that hydrogen-bonding might actually &lt;em&gt;increase&lt;/em&gt; (pseudo) ring strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think that my professor's argument regarding free-rotation along a single bond axis is a bit simplistic at best; in isolation, yes, single bonds can rotate, but so can double bonds (you just have to expend enough energy to break the pi-bond). Plus, there are &lt;em&gt;barriers&lt;/em&gt; to free rotation of single bonds. You can't rotate single bonds willy nilly all the time. And according to the quantum mechanical calculation, the &lt;em&gt;cis&lt;/em&gt; isomer exhibits a slightly higher energy than the &lt;em&gt;trans&lt;/em&gt; isomer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a) Over what maximum distance can hydrogen bonding still exert a notable force? I'm curious. With this information I might try to do some trig using the QM data and see if the H and the O in the &lt;em&gt;cis&lt;/em&gt;-conformer even come close enough for there to be hydrogen bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Can &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; terminology be used to describe HONO? Usually these two terms are used with respect to identical atoms or functional groups. How about &lt;em&gt;anti&lt;/em&gt; and &lt;em&gt;syn&lt;/em&gt;? Or &lt;em&gt;eclipsed&lt;/em&gt; and &lt;em&gt;non-eclipsed&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Tautomerization. Why can't the proton be attached to the N? What's the barrier to tautomerization? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Ionic resonance structures - are these valid in the aqueous phase? Seems that if we have a lone proton hanging around in water it would be immediately solvated and taken away from the rest of the molecule ... this was the issue the prof raised. An obvious counter to his issue is that resonance structures don't exist discreetly; the ionic resonance structure is just a contributor - a partial representation of the true bonding nature in nitrous acid. It doesn't actually suggest that the proton is separated from the rest of the molecule. Dunno why he raised that issue ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Authors of one paper also mention that &lt;em&gt;nitric&lt;/em&gt; acid exhibits a similar widening of the &lt;em&gt;cis&lt;/em&gt; isomer's ONO bond angle ... what might the two different bond angles for nitric acid be? Hell, does nitric acid even have cis/trans isomers? I don't think so ... &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Regarding &lt;em&gt;intramolecular&lt;/em&gt; hydrogen bonding in HONO&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The below paper is based on the premise that hydrogen bonding exists within nitrous acid. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The influence of intramolecular hydrogen bonding&lt;/strong&gt; on the&#xA;  photodissociation of cis-HONO is probed via measurements on the OH&#xA;  fragment ejected by specific [...] stretching (ν2) vibrational levels&#xA;  of the Ã state. Due to hydrogen bonding between the H atom and the&#xA;  terminal oxygen, the ν2 motion is coupled to the in-plane HON bend ν3.&#xA;  Since the latter evolves into fragment rotation, the rotational energy&#xA;  and anisotropy of the OH product increases with the number of ν2&#xA;  quanta. By contrast, ν2 in trans HONO is a relatively isolated  [...]&#xA;  vibration and thus does not influence the OH photofragment's&#xA;  properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-31T23:14:58.937" LastActivityDate="2014-08-31T23:14:58.937" Title="Why does nitrous acid exhibit (unexpectedly) different bond angles?" Tags="&lt;bond&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15559" PostTypeId="2" ParentId="15541" CreationDate="2014-08-24T04:34:38.517" Score="7" Body="&lt;p&gt;One thing Ron didn't cover is the side effects of all that decay. All that short-half-life stuff will be rather radioactive, to the point where you might not survive to see the results. It's also possible that if you get enough together to be profitable that it will simply vaporize itself. The stable gold should eventually condense so as long as you contain the vapor it might work. Or you'll have a very pretty gold-lined flask.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The stable isotopes of lead are 204, 206, 207 and 208, you'd have to knock a lot of &lt;strong&gt;protons&lt;/strong&gt; and neutrons out to get down to 197.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shame! Knock a proton off of lead and it turns into thallium. But as that's the point of the exercise just bump off two more and we're done.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-08-24T04:34:38.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15560" PostTypeId="1" CreationDate="2014-08-24T06:03:02.357" Score="5" ViewCount="544" Body="&lt;p&gt;To be specific, what I mean is: is the following situation possible? At given temperature T, there are two substances, X and Y, such that pure X is a solid, pure Y is a liquid, and X and Y mix to form a single phase in all proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I could find was this phase diagram for silver and gold (two very similar metals, I suppose):&#xA;&lt;a href=&quot;http://www.crct.polymtl.ca/fact/phase_diagram.php?file=Ag-Au.jpg&amp;amp;dir=SGTE&quot;&gt;http://www.crct.polymtl.ca/fact/phase_diagram.php?file=Ag-Au.jpg&amp;amp;dir=SGTE&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it looks like there is still a narrow region in which there are two phases.&lt;/p&gt;&#xA;" OwnerUserId="5490" LastActivityDate="2014-08-24T12:28:58.673" Title="Can a solid and liquid be miscible?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15561" PostTypeId="2" ParentId="15549" CreationDate="2014-08-24T07:24:22.287" Score="2" Body="&lt;p&gt;Adding to Freddy's answer I would say that, pentan-3-one is preferred overall due to the fact that as molecules start to carry more and more functional groups, it becomes ambiguous of which functional group the number in the front is referring to without knowing the order of priority of naming functional groups.&lt;/p&gt;&#xA;" OwnerUserId="7584" LastActivityDate="2014-08-24T07:24:22.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15563" PostTypeId="1" CreationDate="2014-08-24T10:47:06.767" Score="-1" ViewCount="1082" Body="&lt;p&gt;Here is the unbalanced equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -&amp;gt; Fe(OH)3_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, one half-reaction is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(OH)2_{(s)} + O2_{(g)} -&amp;gt; Fe(OH)3_{(s)}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other half-reaction is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2_{(g)} -&amp;gt; 2H2O_{(l)}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, how are we justified in making the latter half-reaction?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="23561" LastEditDate="2018-09-02T02:50:55.120" LastActivityDate="2018-09-02T11:40:39.227" Title="Writing half-reaction for the oxidation of iron (III) hydroxide" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15564" PostTypeId="2" ParentId="15560" CreationDate="2014-08-24T10:59:38.817" Score="2" Body="&lt;p&gt;It's not necessarily a strict definition of a single intermingling of phases, however the closest thing I can think of corresponding to this in reality would be interstitial metal hydrides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These materials are being investigated as hydrogen storage materials, so essentially the hydrogen molecules interact with the metal to form hydrides, &quot;$\ce{H-}$&quot;, which recombine under different conditions to reform $\ce{H2}$ as required. A review of this technique is given &lt;a href=&quot;http://www.chem.uw.edu.pl/people/WGrochala/H_review.pdf&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be interesting, to answer your question more directly, if liquid hydrogen could be used instead of hydrogen gas (this requires temperatures of &amp;lt;33K). I imagine this process would give exactly the same result, albeit would happen more slowly than the gas phase. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-08-24T12:28:58.673" LastActivityDate="2014-08-24T12:28:58.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15565" PostTypeId="2" ParentId="9213" CreationDate="2014-08-24T11:03:12.837" Score="-3" Body="&lt;p&gt;n factor is atomic weight or molecular weight by equivalent weight. n factor for acids is its basicity, ie, number of replaceable hydrogen ion per molecule. For exampleg $\ce{HCl}$ has n factor 1 n $\ce{H2SO4}$ has 2 $\ce{H3PO3}$ has n factor 2 because when $\ce{H3PO3}$ dissociates to form H+ and HPO3^-2. Same ways for bases its n factor is its acidity or no of replaceable OH ions per molecule. Eg: for $\ce{NaOH}$ it is 1 n for $\ce{Ca(OH)2}$ its 2.&lt;/p&gt;&#xA;" OwnerUserId="7585" LastEditorUserId="42741" LastEditDate="2017-05-13T17:46:30.777" LastActivityDate="2017-05-13T17:46:30.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15566" PostTypeId="2" ParentId="15558" CreationDate="2014-08-24T11:10:25.173" Score="6" Body="&lt;p&gt;What potential stabilization might be derived from an intramolecular hydrogen bond in &lt;em&gt;cis&lt;/em&gt;-HONO, is counteracted by strain generated by the bad geometry. In order for the H and the distal O to get close enough to have a bonding interaction, &lt;em&gt;cis&lt;/em&gt;-HONO starts to suffer from van der Waals repulsion between the two oxygen atoms. Note that &lt;em&gt;cis&lt;/em&gt;-HONO is higher in energy than &lt;em&gt;trans&lt;/em&gt;-HONO on your table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hydrogen bond existed, &lt;em&gt;cis&lt;/em&gt;-HONO would be a pseudo-four-member ring. For cycloalkanes, four-member rings have almost as much &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycloalkane#Ring_strain&quot;&gt;ring strain&lt;/a&gt; (110 kJ/mol) as three-member rings (120 kJ/mol). The origin of this ring strain is much more likely van der Waals repulsion than any sort of &quot;angle strain&quot; due to &quot;improper&quot; bond angles. Meanwhile,  $\ce{O-H-O}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond#Bonding&quot;&gt;hydrogen bonds&lt;/a&gt; average at 21 kJ/mol. The ring strain that would result from hydrogen bonding in &lt;em&gt;cis&lt;/em&gt;-HONO would destabilize the molecule more than any stabilization from a hydrogen bond. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-24T11:10:25.173" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15567" PostTypeId="2" ParentId="15563" CreationDate="2014-08-24T11:18:06.330" Score="5" Body="&lt;p&gt;Oxygen should not be in both half reactions. You'll get yourself in trouble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(OH)2 -&amp;gt;Fe(OH)3}$$ &#xA;$$\ce{O2 -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are justified in making the latter half reaction because balancing the first half reaction will likely require water added to the reactant side to help balance the $\ce{H}$ and $\ce{O}$ atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an equally appropriate second half reaction might be &#xA;$$\ce{O2 -&amp;gt; OH-}$$&#xA;noting that there is an extra hydroxide in the products of your overall redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-24T11:18:06.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15568" PostTypeId="1" CreationDate="2014-08-24T11:24:24.070" Score="4" ViewCount="232" Body="&lt;p&gt;The ocean is becoming acidic. This is not a natural evolutionary change but happens through human negative effect on the environment. What element or compound could be used to create a more base ocean? How much of it would be needed? How could it be distributed? &lt;/p&gt;&#xA;" OwnerDisplayName="user1464" LastEditorUserId="402" LastEditDate="2016-03-16T17:18:59.653" LastActivityDate="2016-03-16T17:18:59.653" Title="What element or compound could be used to make the ocean more basic?" Tags="&lt;acid-base&gt;&lt;environmental-chemistry&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15569" PostTypeId="1" AcceptedAnswerId="15597" CreationDate="2014-08-24T12:10:34.163" Score="7" ViewCount="4893" Body="&lt;p&gt;Sometimes the Mulliken and NBO turn out to be so different that I can't decide which one I can trust. I've heard that Mulliken is inaccurate, but is NBO always accurate? &#xA;And should I use Gaussian or other software to calculate charge distribution? Are there methods more accurate than Mulliken or NBO?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And charge distribution doesn't represent charge density, right?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastActivityDate="2018-11-14T12:34:58.413" Title="Which one, Mulliken charge distribution and NBO, is more reliable?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15570" PostTypeId="2" ParentId="15541" CreationDate="2014-08-24T12:59:55.840" Score="3" Body="&lt;p&gt;I can’t guess at the cost, but consider this: if it were profitable, many people would do it, and then the value of gold would quickly reduce to a level consistent with the cost of producing it.&lt;/p&gt;&#xA;" OwnerUserId="7586" LastActivityDate="2014-08-24T12:59:55.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15571" PostTypeId="1" AcceptedAnswerId="15575" CreationDate="2014-08-24T13:01:52.743" Score="2" ViewCount="22050" Body="&lt;p&gt;Depending on the textbooks looked at, the energy change is described as either Q+W or the signage is changed to Q-W. Which is correct? Is it solely context dependent? Could anyone explain the background?&lt;/p&gt;&#xA;" OwnerUserId="5217" LastActivityDate="2014-08-24T15:35:49.630" Title="Why is ΔE described as Q-W and Q+W in different contexts?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15572" PostTypeId="1" CreationDate="2014-08-24T13:10:44.267" Score="4" ViewCount="629" Body="&lt;p&gt;I have a simple electrolysis setup which uses aqueous sodium chloride as electrolyte. The problem is that, the setup makes tons of chlorine gas along with hydrogen (I only want hydrogen). I didn't even realize it until I recognized that the odor smells like that of ozone. I am using a power supply of &lt;code&gt;24V @ 6A&lt;/code&gt;. Since, it's a pain to buy a new power supply, changing the supply isn't an option. Also, I have experimented with other electrolytes like aqueous sodium bicarbonate but they are far slower than the brine solution. So, the only other option is to consume the chlorine gas. I am not sure what to use here. I don't have a lot of reducing agents on hand. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7513" LastActivityDate="2014-08-24T15:57:35.277" Title="Reduce chlorine gas" Tags="&lt;electrochemistry&gt;&lt;electrons&gt;&lt;hydrogen&gt;&lt;electrolysis&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15573" PostTypeId="2" ParentId="8717" CreationDate="2014-08-24T14:34:58.003" Score="5" Body="&lt;p&gt;As we move down the group, the intervening d and f electrons shield the ns2 electrons poorly and hence the ns2 electrons are pushed inside towards the nucleus This results in the reluctance of ns electrons to participate in bond formation and this phenomenon is called Inert pair effect.&lt;/p&gt;&#xA;" OwnerUserId="7587" LastActivityDate="2014-08-24T14:34:58.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15574" PostTypeId="2" ParentId="15568" CreationDate="2014-08-24T14:56:41.447" Score="7" Body="&lt;p&gt;The ocean is becoming more acidic due to the dissolution of excess $\ce{CO2}$ from anthropogenic activities to form carbonic acid, $\ce{H2CO3}$, which even though mildly acidic is decreasing the pH of the ocean as you rightly point out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we're trying to reduce the effect of this $\ce{CO2}$ on the ocean, we might consider removing it via some sink. One of these theoretical methods could be to react a base, such as $\ce{CaCO3}$,  with this carbonic acid in a conventional neutralisation. In symbols:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2CO3 + CaCO3 -&amp;gt; Ca(HCO_3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could work, and we could derive everything we wanted to know about how much base to use from information such as the amount of residual $\ce{CO2}$ left over from emissions after photosynthesis has been taken into account (and other sinks), and then assuming that all other $\ce{CO2}$ will be taken up by the sea. From here we can calculate the amount of $\ce{H2CO3}$ present in seawater from the equilibrium constant between $\ce{CO2}$ and $\ce{H2CO3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we won't do that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notwithstanding these nonsense assumptions which could grossly overpredict/underpredict the amount of base required, neutralisation reactions don't deal with the excess carbonate effectively. What you really want is a permanent sink (a sink so stable its contents are practically constant through time) at the bottom of the sea. See the long-term carbon cycle for more information on this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question becomes: how do we get the $\ce{CO2}$ dissolved in the ocean to the deep-ocean sediments? And how do we do it &lt;em&gt;'permanently'&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1029/GB004i001p00005/abstract;jsessionid=1BE7D4AFEDADB2B1729EA9E2FF87FACE.f03t04?deniedAccessCustomisedMessage=&amp;amp;userIsAuthenticated=false&quot;&gt;interesting paper&lt;/a&gt; came out in the Journal of Global Biogeochemical Cycles in 1990 trying to address the question about why phytoplankton in the Antarctic don't use the myriad nutrients in the Southern Ocean to photosynthesise. One might expect that it could be temperature or sunlight which limit photosynthesis in the Antarctic, but this actually isn't the case. The limiting factor in the Antarctic ocean is actually a lack of iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has since be proposed to increase the amount of iron in the ocean (simply by adding it ourselves) which would promote the photosynthesis of phytoplankton in such locations. When these organisms die, they sink and take their biomass with them which contributes to the ocean sediments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both neutralisation and iron supplementation may work to offset ocean acidification, however we really don't know the full consequences of these bioremediation efforts—there are simply too many variables involved. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorUserId="4945" LastEditDate="2014-08-25T08:50:37.103" LastActivityDate="2014-08-25T08:50:37.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15575" PostTypeId="2" ParentId="15571" CreationDate="2014-08-24T15:35:49.630" Score="4" Body="&lt;p&gt;The difference in sign in the two versions of the first law of thermodynamics is to handle the two ways in which work can be defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done (assuming only pressure-volume work) can be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=P\Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the definition often used in in scenarios when we care about the fate of the work. This definition is about the &lt;strong&gt;surroundings&lt;/strong&gt;. Thus, when the system expands, it does positive work on the surroundings, which is a good thing. Think about an internal combustion engine. When the reaction occurs in a cylinder, gas is produced causing the volume to expand and move the piston, turning the drive shaft, etc. Since we care about the work after it leaves the cylinder, it is positive because the surroundings are gaining energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done by a system can also be defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=-P\Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use this definition often in chemistry and in general when we care more about how much energy is left in the system than we care about what the energy that left the system is doing. This definition is about the &lt;strong&gt;system&lt;/strong&gt;. In the case of the engine above, energy has left the system, so the work done is negative to reflect the fact that the system has less energy than it used to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign convention in the definition of the change in internal energy (and I prefer $\Delta U$ to $\Delta E$) reflects that we can define work done with respect to the system or the surroundings. The sign in front of $w$ guarantees that we always get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta U=q-P\Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With respect to the surroundings:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=P\Delta V$$&#xA;$$\Delta U = q-w=q-P\Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With respect to the system:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=-P\Delta V$$&#xA;$$\Delta U = q+w=q-P\Delta V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better question might be &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why then do we not change the sign of $q$ when we switch reference frames from system to surroundings?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-24T15:35:49.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15576" PostTypeId="2" ParentId="15572" CreationDate="2014-08-24T15:48:37.193" Score="7" Body="&lt;p&gt;The issue is that oxidation of chloride to chlorine has a lower overpotential than oxidation of water to oxygen. You will always have to have an oxidation reaction going on while reducing water to hydrogen. The one with the lowest standard potential + overpotential will win. The reason the overpotential exists is that electrolysis is pushing the system away from equilibrium, so we need to provide additional electromotive force to get that to happen. The value of the overpotential is related to the activation energy of the redox event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with a graphite electrode, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Overpotential#Activation_overpotential&quot;&gt;overpotential&lt;/a&gt; for the oxidation of water is +0.95 V, while the overpotential for the oxidation of chloride is +0.12 V. When you compare the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot;&gt;standard reduction potentials&lt;/a&gt; for these two reactions, we can see that chlorine is produced more readily (smaller overall voltage needed). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;product    Red. Pot.    Overpot.     Total Pot.&#xA;O2         +1.30 V      +0.95 V      +2.25 V&#xA;Cl2        +1.36 V      +0.12 V      +1.48 V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To avoid chloride production, you need to get ride of chloride. Switch to sodium hydroxide. You have plenty of voltage in your power supply to overcome the overpotential of water/hydroxide.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2014-08-24T15:57:35.277" LastActivityDate="2014-08-24T15:57:35.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15577" PostTypeId="1" AcceptedAnswerId="15591" CreationDate="2014-08-24T16:27:02.010" Score="9" ViewCount="5331" Body="&lt;p&gt;I've been reading about cracking and I've found various images and animations that show a process like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwcJR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I haven't been able to find out the sequence of events to get to this, and why this would occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that cracking involves thermal decomposition, where a CC bond is cut. But that's as far as I can truthfully say I understand. To get this reaction below, you need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;em&gt;Two&lt;/em&gt; cuts. One cut would be between two Cs, and the other cut between a C and a H. Why does a H have to be cut off as well? Is this example I found actually one of many random and messy ones that just happens to 'give us what we want in the end'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The free H would then have to join the alkane. I get why this happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) A double bond is made. Why?&lt;/p&gt;&#xA;" OwnerUserId="7588" LastEditorUserId="9961" LastEditDate="2015-10-26T22:16:41.680" LastActivityDate="2015-10-26T22:16:41.680" Title="In cracking, why does C form double bonds?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15578" PostTypeId="1" AcceptedAnswerId="15595" CreationDate="2014-08-24T16:57:15.593" Score="4" ViewCount="1358" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Show, by means of curly arrows and electron shifts, the resonance contributors to the structure given below. Suggest which one would not be a major contributor and state why.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VPsjg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VPsjg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is a question from my exam. I do not understand where I went wrong. My effort is placed at the end (along with a request).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Also, my TA removed 1 point out of 7 for a reason. Can anyone decipher for what? There's the writing in red but it is illegible to me. I don't understand to what he refers by 8 and 6 t or +, or whatever he wrote.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3AEUQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3AEUQ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="95133" LastEditDate="2021-05-17T09:10:42.623" LastActivityDate="2021-05-17T13:15:13.843" Title="What are the most important resonance structures of 5-aminocycloheaxa-2,4-dienone?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15579" PostTypeId="1" AcceptedAnswerId="15598" CreationDate="2014-08-24T17:19:01.447" Score="25" ViewCount="24644" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What should be the oxidation state of oxygen in $\ce{HOF}$ (hypofluorous acid)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/2d/Hypofluorous-acid-2D-dimensions.png/200px-Hypofluorous-acid-2D-dimensions.png&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources on the internet have confused me. Most state its oxidation state to be 0, while the others state that it is -2 (considering the oxidation state of fluorine to be +1, which is a bit surprising for me as it's the most electronegative element in the periodic table)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source stating the oxidation state to be -2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://doi.org/10.1021/ed049p299.2&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1972&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt; (4), 299&lt;/a&gt;:&lt;br&gt;&#xA;  The two possibilities brought forth in the Note are (1) assign hydrogen +1, oxygen -2, and fluorine +1 or (2) assign H +1, oxy- gen 0, and fluorine -1. On the grounds that $\ce{HOF}$ is a powerful oxidizing agent, the conclusion reached in the Note is that (1) should be favored.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other sources stating the oxidation state to be 0:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20090320153009AAGkgOk&quot; rel=&quot;noreferrer&quot;&gt;Yahoo Answers!&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wiki.answers.com/Q/What_is_Oxidation_state_of_oxygen_in_HOF&quot; rel=&quot;noreferrer&quot;&gt;Answers.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="670" LastEditorUserId="5026" LastEditDate="2018-05-16T13:22:00.100" LastActivityDate="2020-07-22T10:21:42.463" Title="What should be the oxidation state of oxygen in HOF (hypofluorous acid)?" Tags="&lt;oxidation-state&gt;&lt;electronegativity&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="15580" PostTypeId="2" ParentId="15578" CreationDate="2014-08-24T17:38:56.687" Score="4" Body="&lt;p&gt;i think your teacher said &quot;but all have it&quot; refering to the split charge (?) I would personally say that the ones with less contribution would be the ones with carbon with positive charges because while N+ and O- are relatively stable, C+ is certainly not. Being a bit more specific, I would say (not 100% sure) that the 4th one because is the one in which the positive charge is in an unstable carbocation and close to a Nitrogen atom which would destabalise it even more due to its electronegativity.&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="5086" LastEditDate="2014-08-24T20:30:30.920" LastActivityDate="2014-08-24T20:30:30.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15581" PostTypeId="1" CreationDate="2014-08-24T17:56:52.993" Score="1" ViewCount="548" Body="&lt;p&gt;It is known that addition of a non-volatile solute to a volatile solvent(liquid) to give a solution reduces the &lt;strong&gt;Vapour Pressure&lt;/strong&gt; of the solution (well solvent actually as only solvent is volatile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the &lt;strong&gt;Elevation of Boiling Point&lt;/strong&gt; and &lt;strong&gt;&quot;Deppresion&quot; of Freezing Point&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am clear with elevation of BP. But it is said that a liquid freezes when the vapour pressure of the liquid phase attains the vapour pressure of its solid state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how possibly does the freezing point falls? For instance let the Vapour Pressure of the solid state be at an arbitrary point &quot;&lt;strong&gt;x&lt;/strong&gt;&quot;(at a temperature &lt;strong&gt;A&lt;/strong&gt; Kelvin) and that of the liquid phase be at &quot;&lt;strong&gt;y&lt;/strong&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we add a solute to the liquid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vapour pressure of liquid(solution) = y - $\alpha $ (where $\alpha $ is the depression in the vapour pressure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the vapour pressure of the liquid(solution) reaches &quot;&lt;strong&gt;x&lt;/strong&gt;&quot; rapidly that is at a higher temperature or a temperature greater than &lt;strong&gt;A&lt;/strong&gt; which is an elevation in the freezing point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am always confused with this part.&lt;/p&gt;&#xA;" OwnerUserId="7577" LastActivityDate="2014-08-25T14:18:39.697" Title="How does the Freezing Point fall in a solution?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15582" PostTypeId="1" CreationDate="2014-08-24T18:35:24.447" Score="4" ViewCount="1966" Body="&lt;p&gt;If a mixture of water with acetone, methanol, ethanol, or other water miscible solvent, is cooled below 0 degrees, would the water freeze and aggregate out as ice, separating out of the mixture. Or would the frozen 'water particles' be dissolved in the solvent, similar to how solid organics dissolve in solvents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it really depends on the concept of solubility vs miscibility. I've been researching this for a while now and haven't been able to find a clear definite answer.&lt;/p&gt;&#xA;" OwnerUserId="7589" LastActivityDate="2014-08-24T20:38:34.357" Title="Can a mixture of water miscible solvent and water be separated by fractional freezing on a dry ice bath?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15583" PostTypeId="1" CreationDate="2014-08-24T18:49:35.423" Score="8" ViewCount="1584" Body="&lt;p&gt;I know that electrons can move from say 2s orbital to an unoccupied 2p orbital, as in Carbon atom which can form 4 bonds this way. But I want to know is it possible for an electron say in orbital 2p to switch its place with an electron in 2s? I know only 2 electrons are allowed in an orbital, but is it possible for one electron to oust another and take its place so the ousted electron would have to take the other one's place? Is this theoretically possible?&lt;/p&gt;&#xA;" OwnerUserId="1313" LastActivityDate="2018-04-09T03:22:03.183" Title="Can electrons switch orbitals within a shell?" Tags="&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15584" PostTypeId="1" AcceptedAnswerId="15587" CreationDate="2014-08-24T18:49:45.140" Score="6" ViewCount="17889" Body="&lt;p&gt;Phenyl is a functional group with an aromatic ring bonded to another group. And, phenol is a molecule that is just a phenyl bonded to a hydroxyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://www.harpercollege.edu/tm-ps/chm/100/dgodambe/thedisk/qual/9back7.htm&quot; rel=&quot;noreferrer&quot;&gt;some sources&lt;/a&gt; consider phenol itself a functional group. Are they both functional groups, or is phenol just a molecule?&lt;/p&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2014-08-24T20:28:26.490" Title="Phenyl vs phenol functional group?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15585" PostTypeId="2" ParentId="15578" CreationDate="2014-08-24T18:55:42.447" Score="3" Body="&lt;p&gt;I think those are &quot;6E&quot; and &quot;8E&quot;. &quot;6E&quot; indicates that there is a carbon atom with only 6 total electrons in that resonance structure. &quot;8E&quot; indicates that all atoms bear octets of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining importance of resonance structures, maximizing the number of atoms that follow the octet rule is the priority. It may be counterintuitive, but the structure with a positive charge on nitrogen contributes more to the electronic structure of the compound than the structures with positive charge on carbon.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-24T18:55:42.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15586" PostTypeId="1" AcceptedAnswerId="16030" CreationDate="2014-08-24T20:14:48.337" Score="18" ViewCount="48076" Body="&lt;p&gt;I recently answered &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14079/crystal-meth-in-breaking-bad/14081#14081&quot;&gt;a question about Breaking Bad's initial methamphetamine production method&lt;/a&gt; (i.e. the reduction of (pseudo)ephedrine). The reaction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOYnj.png&quot; alt=&quot;Ephedrine to methamphetamine using HI/P&quot;&gt;&#xA;(&lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Methamphetamine&amp;amp;oldid=588492419#Synthesis&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is infamously known as the $\ce{HI/P}$ reduction and was (and still is somewhat) a real plague in the United States until cough medicine containing ephedrine came under tightened control. That for an introduction, now the question: what is actually the reaction mechanism of the $\ce{HI/P}$ reduction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step seems like a simple $\ce{S_{N}2}$ substitution of $\ce{I^{-}}$, catalyzed by the protonation of the alcohol group by hydroiodic acid (making it a far better leaving group). But what is the mechanism after that? How is phosphorus involved? Does the $\ce{HI/P}$ reduction have any legal use?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="7448" LastEditDate="2015-08-30T09:36:38.273" LastActivityDate="2018-04-07T08:21:39.880" Title="HI/P reduction mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15587" PostTypeId="2" ParentId="15584" CreationDate="2014-08-24T20:28:26.490" Score="4" Body="&lt;p&gt;Phenol is both a &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol&quot; rel=&quot;nofollow&quot;&gt;molecule&lt;/a&gt; and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenols&quot; rel=&quot;nofollow&quot;&gt;functional group&lt;/a&gt; (or at least a class of molecules). &lt;a href=&quot;http://goldbook.iupac.org/P04539.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC&lt;/a&gt; has the definitive answer on the matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phenols:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g. 2-naphthol&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-24T20:28:26.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15588" PostTypeId="2" ParentId="15558" CreationDate="2014-08-24T20:32:31.200" Score="6" Body="&lt;p&gt;These are very small effects that we are discussing, so who really knows what's behind them.  The experimental measurement data was reported in, &quot;Microwave spectrum, structure, dipole moment and quadrupole coupling constants of cis and trans nitrous acids, A. Peter Cox, Alan H. Brittain and David J. Finnigan. Trans. Faraday Soc., 1971,67, 2179-2194, DOI:  10.1039/TF9716702179.&quot;  The &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1971/TF/tf9716702179#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;abstract&lt;/a&gt; states, &quot;The structural parameters are markedly different in the two isomers and have been discussed, together with the torsional barrier and relative stability of the isomers, in terms of a cis interaction.&quot;  If anyone has access to the paper, perhaps they can let us know what the authors have to say about the nature of this &quot;cis interaction.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some of the resonance structures that can be drawn for the two isomers.  First note that the cis-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JevFW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;isomer has the lone pairs on the same side of the molecule (lone pair - lone pair interactions), presumably responsible for the cis isomer being about 1/2 kcal/mol less stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we examine the (hyperconjugative) resonance structures III and IV in the cis isomer, we might expect some stabilizing, through-space electrostatic attraction of the opposite charges between the positively-charged proton and the negatively-charged, terminal oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For arguments sake, let's assume that the trans isomer is adequately described by resonance structure I, while the cis isomer is best described by resonance structures III-IV.  In this case we would expect the following structural differences between the cis and trans isomers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A longer O-H bond in the cis isomer due to the larger O-H separation in cis resonance structures III and IV, than in trans I (this is analogous to longer C-H bonds present on the carbon next to a carbocation due to hyperconjugation)&lt;/li&gt;&#xA;&lt;li&gt;A shorter internal N-O bond in the cis isomer due to resonance structure III which has a double bond in this position&lt;/li&gt;&#xA;&lt;li&gt;A longer terminal N-O bond in the cis isomer due resonance structure III which has a single bond in this position&lt;/li&gt;&#xA;&lt;li&gt;An increased H-O-N bond angle in the cis isomer because the structures cis III and IV involve roughly $\ce{sp^2}$ hybridized O-H oxygen, whereas in resonance structure trans I the O-H oxygen is roughly $\ce{sp^3}$ hybridized (actually shifted towards an even more constrained angle due to the effect of the lone pair); as we shift hybridization of the O-H oxygen from $\ce{sp^3}$ to $\ce{sp^2}$, we expect the H-O-N angle to widen.&lt;/li&gt;&#xA;&lt;li&gt;An increased O-N-O bond angle in the cis isomer due to Bent's rule (a tendency to direct hybrids of increased s-character towards more electropositive substituents); a partial negative charge resides on both oxygen atoms in cis III and IV, whereas trans-1 has no  negative charge on oxygen; therefor, the oxygens in cis III and IV are more electropositive (less electronegative) and the nitrogen will direct hybrid orbitals higher in s-character towards them as compared to trans I; higher s character in the bond means a larger O-N-O bond angle in cis III and IV compared to trans I &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So perhaps very different contributions from the possible resonance structures to the real descriptions of cis- and trans-nitrous acid can account for the observed structural differences between these two isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side Notes: 1) hydrogen bonding is unlikely due to the attendant ring strain as noted by Ben Norris, 2) the oxygen-oxygen interaction should be electrostatically attractive.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-24T20:32:31.200" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="15589" PostTypeId="2" ParentId="15582" CreationDate="2014-08-24T20:38:34.357" Score="3" Body="&lt;p&gt;The short answer is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is exactly how acetic acid is purified: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2506/acetic-acid-freezing-distilation&quot;&gt;Acetic acid freezing distilation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, acetic acid (freezing point $16\ ^\circ \text{C}$) freezes first, leaving a water-enriched fluid behind. Conceivably you could remove water from methanol or acetone in the same way (except that water would freeze first). In practice, fractional distillation can be done more economically on a large scale. Fractional distillation requires 1 step (with lots of theoretical plates), while you need to repeat fractional freezing multiple times to get to high purity. Acetic acid is purified by fractional freezing perhaps because it may azeotrope with water. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-08-24T20:38:34.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15590" PostTypeId="1" CreationDate="2014-08-24T21:25:52.213" Score="38" ViewCount="1010" Body="&lt;p&gt;With higher period the d-electrons of the metal are less strong bonded and therefore oxidative addition is easier for $\ce{Ir(I)}$ than for $\ce{Rh(I)}$ and much easier than for $\ce{Co(I)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For metal carbonyl hydrides, electrons with higher period are more stronger bound to the nucleus, so $\mathrm{p}K_\mathrm{a}$ increases in this order: $\ce{HCrCp(CO)3 &amp;lt; HMoCp(CO)3 &amp;lt; HWCp(CO)3}$&#xA;This explanation sounds logical!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wrote that for $\ce{CO}$-complexes because of the higher effective atomic number for higher period atoms (higher ionization energy) the $\pi$-backbonding is weak and therefore less stable as e.g. for $\ce{Cr(3d)}$.&#xA;This explanation sounds logical as well!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is the written tendency of the oxidative addition not a contradiction to the other tendencies?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2017-04-05T23:31:04.203" LastActivityDate="2018-07-25T10:36:26.440" Title="Relativistic effect: d-electrons in metallorganic complexes" Tags="&lt;transition-metals&gt;&lt;organometallic-compounds&gt;&lt;relativistic&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15591" PostTypeId="2" ParentId="15577" CreationDate="2014-08-24T21:39:36.163" Score="8" Body="&lt;p&gt;When we heat hydrocarbons to high temperatures (somewhat lower temperatures if we use a catalyst) they crack.  That means that carbon-carbon bonds are broken and smaller fragments are formed.  In the top part of the following drawing, I've picked n-heptane as an example.  You can see that there are 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZBMlW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;different bonds we can break and a number of different fragments will result.  Each of the fragments that result from breaking a C-C bond has an unpaired electron (there were two electrons in the bond we broke, each fragment retains one electron).  These fragments with an unpaired electron are called radicals.  The carbon with the unpaired electron does not have a stable octet configuration, therefor radicals are extremely reactive species and will react rapidly in a variety of ways to regain their stable octet configuration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've picked one of the possible radicals generated in the top drawing and show how it may react further in the bottom half of the figure.  Sometimes it will break a C-H bond and lose a hydrogen atom and form a stable compound with a double bond (an olefin), sometimes it will remove (abstract) a hydrogen from a hydrocarbon (heptane in our example) to generate a new alkane (pentane) and a new radical (perhaps a radical not formed in the initial cracking step).  Many of the radicals formed can rearrange to more stable radicals, this adds further complexity to the overall process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So starting with just heptane we can get a large variety of saturated hydrocarbons and unsaturated hydrocarbons.  The situation becomes exponentially more complex if, as usual, we do have many more hydrocarbons present in addition to heptane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that radicals are very reactive species - they want to stabilize by regaining the octet of electrons around the carbon atom that has the radical.  They can get that octet and become stable by adding a hydrogen atom (alkane formation) or breaking a C-H bond and eliminating a hydrogen atom to form an alkene.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-24T21:39:36.163" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15592" PostTypeId="1" CreationDate="2014-08-24T21:42:17.237" Score="7" ViewCount="39346" Body="&lt;p&gt;The best reducing agents are located at the bottom left of the periodic table (low electronegativity) and the best oxidizing agents are located at the top right of the periodic table (high electronegativity), excluding noble gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is electronegativity the factor that determines the strength of reducing agents and oxidizing agents? &lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="402" LastEditDate="2016-11-30T04:56:20.717" LastActivityDate="2019-06-21T13:43:58.633" Title="What determines the strength of reducing and oxidizing agents?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15595" PostTypeId="2" ParentId="15578" CreationDate="2014-08-25T05:56:23.633" Score="8" Body="&lt;p&gt;The handwriting of your TA probably stems from ancient egyptian script, I had a hard time reading it. In the end I agree with jerepierre, interpreting it as &amp;quot;6E&amp;quot; or &amp;quot;8E&amp;quot; short for 6/8 electrons. The last statement then translates to &amp;quot;but all have 8E.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Before continuing to answer your question, I want to make a statement about resonance structures in general. Resonance structures are used to express a complicated bonding situation in terms of simpler, Lewis type structures. None of these structures exist individually, only a mixture of all possible resonance structures at the same time exists. I am not a fan of declaring single resonance structure as having more or less impact on the overall structure. Of course one can go ahead and calculate the total wave function in terms of resonance structures, generally known as valence bond theory, but this is overkill for such simple molecules. Then you get a mathematical representation of the electronic structure which you can decompose into contributions of the individual resonance structures.&lt;/p&gt;&#xA;&lt;p&gt;A simpler approach is to analyse the molecule with Natural Bond Orbital (NBO) Analysis. I have done this for you, but only as an educational effort. Maybe after understanding this concept it is easier to grasp why certain structures are make a smaller contribution.&lt;/p&gt;&#xA;&lt;p&gt;Let me go ahead and explain a little NBO to you.  The electronic structure of a molecule is in general highly delocalised. All orbitals are stretched over the whole molecule. Without changing the electron density, it is possible to localise (transform) these orbitals to represent two centre interactions, that can be interpreted as bonds. This gives an approximate picture in terms of a Lewis structure. There will be electron density which cannot be assigned to a bond. The smaller this density is, the better is the fit as a Lewis structure.&lt;/p&gt;&#xA;&lt;p&gt;For this educational approach I picked the 5 resonance structures you have posted plus an additional ionic form. (I would have detracted a point for the last structure.) There are many more ionic forms, but these will not have big effects, e.g. cleavage of a carbon carbon bond.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8zJ5l.png&quot; alt=&quot;resonance structures&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The following is based on NBO analyses from DF-BP86/def2-SVP.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{crr}\hline&#xA;\text{Structure} &amp;amp; \text{% Lewis} &amp;amp; \text{% non-Lewis}\\\hline&#xA;\mathbf{1}   &amp;amp; 96.4  &amp;amp;  3.6  \\&#xA;\mathbf{2}   &amp;amp; 96.8  &amp;amp;  3.2  \\&#xA;\mathbf{3}   &amp;amp; 96.2  &amp;amp;  3.8  \\&#xA;\mathbf{4}   &amp;amp; 95.8  &amp;amp;  4.2  \\&#xA;\mathbf{5}   &amp;amp; 96.3  &amp;amp;  3.7  \\&#xA;\mathbf{6}   &amp;amp; 95.2  &amp;amp;  4.7  \\\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The first statement that should be made is that this molecule is quite well described by Lewis resonance structures. However, delocalised bonding would describe the molecule much better. All structures are reasonably close to each other, as one would expect for a very delocalised system.&lt;/p&gt;&#xA;&lt;p&gt;The ionic structure has the smallest impact, which is feasible because nitrogen-hydrogen bonds are better described as covalent. However, it proves that there can (in the right medium) be a proton shift from nitrogen to oxygen, and is therefore worth consideration.&lt;/p&gt;&#xA;&lt;p&gt;The structure with the second largest error (% non-Lewis) is &lt;span class=&quot;math-container&quot;&gt;$\mathbf{4}$&lt;/span&gt;. A positively charged carbon would be by far more electronegative than nitrogen. And many &lt;a href=&quot;http://en.wikipedia.org/wiki/Imine&quot; rel=&quot;nofollow noreferrer&quot;&gt;imine&lt;/a&gt;-type structures are known, which makes &lt;span class=&quot;math-container&quot;&gt;$\mathbf{5}$&lt;/span&gt; more important. (In this sense I would disagree with your explanation, which could also justify the detraction of a point.)&lt;/p&gt;&#xA;&lt;p&gt;In general: Carbocation structures that are not in close proximity to (much) more electronegative elements should be considered as having a lesser influence on the electronic structure of a molecule.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="69830" LastEditDate="2021-05-17T13:15:13.843" LastActivityDate="2021-05-17T13:15:13.843" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15597" PostTypeId="2" ParentId="15569" CreationDate="2014-08-25T08:41:51.233" Score="9" Body="&lt;p&gt;As I do not want to drag this out in the comments, I decided to give a brief answer. Let me first cite the very helpful comments to keep the for (relative) eternity and add my thoughts at the end.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no proper, strict and agreed definition of atom border, that's the real problem. Gaussian includes option to fit charges on atoms (assumed for the sake of the procedure as points) so they produced electrostatic field as accurately fitting once calculated from the electronic density distribution as possible, and maybe (I haven't used Gaussian for a long time, I'm not sure) dipoles too. –  permeakra&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NBO is more accurate than Mulliken in very many ways. However, NBO is not always accurate. Generally speaking, you would never want to use Mulliken charges for any sort of production level work unless its been calibrated. –  LordStryker&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Mulliken is of course the cheapest and fastest way to compute charges. However, this method tends to give qualitative results at best. The reason for this is obvious. It will divide the canonical orbitals equally amongst the participating atoms. There is no polarisation whatsoever. Another problem is, that they are very basis set dependent. And that being said, the worst part is, that the description becomes worse by increasing the basis set.&lt;/p&gt;&#xA;&lt;p&gt;NBO charges are much more reliable, since the operate on the electron density instead. Localised natural atomic orbitals can be used to describe the computed electron density. Polarisation of bonds is therefore considered. This method obviously only works very well for systems, that can actually be (quite well) described by a Lewis structure. However, they usually give a nice qualitative picture and are quite robust when it comes to increasing the basis set.&lt;/p&gt;&#xA;&lt;p&gt;There are a couple of more population analysis tools, that can be used and should be considered when it is necessary to derive some conclusions from charges.&lt;/p&gt;&#xA;&lt;p&gt;Quite common is also the Hirshfeld charges approach, which divides the electron density into spherical basins around the nucleus. Maybe interesting to read is &amp;quot;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.21901/abstract&quot; rel=&quot;noreferrer&quot;&gt;Are the Hirshfeld and Mulliken population analysis schemes consistent with chemical intuition?&lt;/a&gt;&amp;quot; (I can't tell, I do not have access.)&lt;/p&gt;&#xA;&lt;p&gt;There are also several schemes to obtain partial charges from the electrostatic potential.&lt;/p&gt;&#xA;&lt;p&gt;Probably the most rigorous approach to atomic partial charges is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;noreferrer&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt;  (QTAIM). Here the Laplacian of the electron density is used to describe the partitioning in electronic basins. The best feature of this analysis is, that it can be also used for experimentally obtained electron densities. However, as everything, this has also major drawbacks. One of the most important is the computational effort. Unfortunately the integration scheme is not always completely robust when it comes to some particular molecules. However, it is almost completely basis set independent and results could be cross referenced with experimental data.&lt;/p&gt;&#xA;&lt;p&gt;Since permeakra's initial statement holds true, especially for the &lt;em&gt;agreed&lt;/em&gt; part, the best way to deal with partial charges is to be careful and maybe check with different methods. In gaussian there are several schemes implemented, most commonly, Mulliken, Hirshfeld and NBO (3.1) and some others, see the &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_population.htm&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;population&lt;/code&gt;&lt;/a&gt; keyword for more detail.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-08-25T08:41:51.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15598" PostTypeId="2" ParentId="15579" CreationDate="2014-08-25T09:16:05.967" Score="30" Body="&lt;p&gt;Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to &lt;a href=&quot;https://chemistry.stackexchange.com/q/10944/4945&quot;&gt;Electronegativity Considerations in Assigning Oxidation States&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you apply the &lt;a href=&quot;https://web.archive.org/web/20141102154648/http://goldbook.iupac.org/O04365.html&quot; rel=&quot;noreferrer&quot;&gt;official pre-2016 definition (via the Internet Archive)&lt;/a&gt; from the IUPAC gold book, then you have to assign &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{H}) = +1$&lt;/span&gt; as it is not a compound with a metal, &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{O}) = -2$&lt;/span&gt; as it is no peroxide compound, and leaving the disturbing &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{F})= \color\red{+1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going with the on electronegativity based alternative description, you will still assign &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{H}) = +1$&lt;/span&gt; as it has the lowest electronegativity. Then you assign &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{F})= -1$&lt;/span&gt;, because of the highest electronegativity. This leaves a oxidation state of &lt;span class=&quot;math-container&quot;&gt;$\text{OS}(\ce{O}) = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assignment is matched by the 2016 definition, the summary version of which is: &lt;em&gt;The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.&lt;/em&gt; The definition refers to the use of Allen electronegativities (see &lt;a href=&quot;https://doi.org/10.1515/pac-2015-1204&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2016,&lt;/strong&gt; &lt;em&gt;88&lt;/em&gt; (8), 831–839&lt;/a&gt; for more detail). Consequently the heteronuclear approximation yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6DT9K.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6DT9K.png&quot; width=&quot;300&quot; alt=&quot;heteronuclear approximation of HOF with oxidation states&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with &lt;a href=&quot;http://sobereva.com/multiwfn/&quot; rel=&quot;noreferrer&quot;&gt;Multiwfn 3.4.1 (a newer version is available)&lt;/a&gt;. The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lrrr}\hline&#xA;\text{Method}    &amp;amp; \ce{H} &amp;amp; \ce{O} &amp;amp; \ce{F} \\\hline&#xA;\text{Hirshfeld} &amp;amp; +0.18  &amp;amp; -0.10  &amp;amp; -0.09  \\&#xA;\text{ADCH}      &amp;amp; +0.38  &amp;amp; -0.29  &amp;amp; -0.09  \\&#xA;\text{VDD}       &amp;amp; +0.18  &amp;amp; -0.10  &amp;amp; -0.09  \\&#xA;\text{Mulliken}  &amp;amp; +0.20  &amp;amp; -0.06  &amp;amp; -0.15  \\&#xA;\text{Löwdin}    &amp;amp; +0.09  &amp;amp; +0.01  &amp;amp; -0.10  \\&#xA;\text{Becke}     &amp;amp; +0.38  &amp;amp; -0.21  &amp;amp; -0.17  \\&#xA;\text{ADC Becke} &amp;amp; +0.38  &amp;amp; -0.29  &amp;amp; -0.09  \\&#xA;\text{CM5}       &amp;amp; +0.35  &amp;amp; -0.26  &amp;amp; -0.08  \\\hline&#xA;\text{QTAIM}     &amp;amp; +0.61  &amp;amp; -0.44  &amp;amp; -0.17  \\\hline&#xA;\text{NBO}       &amp;amp; +0.45  &amp;amp; -0.32  &amp;amp; -0.12  \\\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2019-06-18T13:11:28.613" LastActivityDate="2019-06-18T13:11:28.613" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="15599" PostTypeId="1" AcceptedAnswerId="15603" CreationDate="2014-08-25T13:53:42.700" Score="9" ViewCount="29181" Body="&lt;p&gt;For this question, assume all things not mentioned are perfectly controlled, all at 14.7lbs of atmospheric pressure, and water is 100% pure of anything not mentioned, with no dissolved air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a quantity of water that is 1 liter at 20°, what would the volume be when heated to 30°?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the same question, except the water being heated is fully saturated with salt.&lt;/p&gt;&#xA;" OwnerUserId="7597" LastEditorUserId="4945" LastEditDate="2014-08-26T04:09:04.817" LastActivityDate="2018-01-22T15:42:24.613" Title="How much does water expand when heated, and does salt affect that?" Tags="&lt;water&gt;&lt;solutions&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15600" PostTypeId="2" ParentId="15581" CreationDate="2014-08-25T14:18:39.697" Score="1" Body="&lt;p&gt;Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all).  For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures).  On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water &lt;em&gt;has&lt;/em&gt; to be higher than for pure water.  The presence of the solute makes it happy to stay liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water.  The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable?  Well, it only does while it has a higher concentration of salt in it than is in the solid.  It forms a classic eutectic point on the binary phase diagram (see, for example &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif&quot; rel=&quot;nofollow&quot;&gt;http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif&lt;/a&gt;).  The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt.  Given the outline of free energy above, this is likely to hold for some range of temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-08-25T14:18:39.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15601" PostTypeId="2" ParentId="15599" CreationDate="2014-08-25T14:42:00.853" Score="1" Body="&lt;p&gt;For a starting point I would suggest &lt;a href=&quot;http://web.mit.edu/seawater&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/seawater&lt;/a&gt; although this will not help much on the 'saturated' end of things since the salinity values in the tables do not reach the 20C saturation values. I leave finding the saturation value to the requester, since google brings it up quite easily.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2018-01-22T15:42:24.613" LastActivityDate="2018-01-22T15:42:24.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15602" PostTypeId="2" ParentId="6469" CreationDate="2014-08-25T15:05:58.363" Score="6" Body="&lt;p&gt;Fundamentally, although ionic radii may offer some guidance as to defect formation, the bottom line is that the silver halides have complex bonding energetics that have to be taken into consideration.  I might suggest &lt;a href=&quot;http://journals.aps.org/prb/pdf/10.1103/PhysRevB.77.064115&quot; rel=&quot;nofollow&quot;&gt;'Defect structures in the silver halides', D.J. Wilson &lt;em&gt;et al.&lt;/em&gt;, Phys. Rev. B77, 064115 (2008)&lt;/a&gt; (presuming you have APS journal access). There the authors present DFT calculations on the formation energies of both Frenkel and Schottky defects in AgCl and AgBr.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="4945" LastEditDate="2014-08-26T04:12:40.470" LastActivityDate="2014-08-26T04:12:40.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15603" PostTypeId="2" ParentId="15599" CreationDate="2014-08-25T15:12:59.203" Score="10" Body="&lt;p&gt;The volumetric expansion of a liquid is given by the following equation&#xA;$$\Delta V=V_0\beta\Delta T$$&#xA;where $\beta$ is the coefficient of thermal expansion and $\Delta T$ is the change in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;http://publishing.cdlib.org/ucpressebooks/view?docId=kt167nb66r&amp;amp;chunk.id=d3_4_ch03&amp;amp;toc.id=ch03&amp;amp;brand=eschol%20&quot; rel=&quot;nofollow noreferrer&quot;&gt;a link&lt;/a&gt; to a nice table containing coefficients of expansion for water, both pure and salinated.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; shows the maximum solubility of salt in water to be around 36% (w/w).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first link we find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for pure water at $25\ \mathrm{^\circ C}$: $\beta = 257\times10^{-6}/\mathrm K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for water with 35% $\ce{NaCl}$ dissolved at $25\ \mathrm{^\circ C}$: $\beta= 297\times10^{-6}/\mathrm K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should be able to calculate the values you are interested in.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2018-01-20T12:59:11.977" LastActivityDate="2018-01-20T12:59:11.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15604" PostTypeId="1" AcceptedAnswerId="15608" CreationDate="2014-08-25T17:24:11.203" Score="12" ViewCount="802" Body="&lt;p&gt;I hear this motif - nucleophile attacks electrophile - all the time with regard to organic chemistry. Are there any exceptions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this statement is always true &lt;em&gt;in general&lt;/em&gt; because it's a restatement of Coulomb's law. But then are there any organic reactions that cannot be characterized as nucleophile attacking electrophile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I learned: Radical termination does not involve nucleophiles attacking electrophiles. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-06-09T01:33:14.777" LastActivityDate="2018-06-09T01:33:14.777" Title="Exceptions to motif: nucleophile attacks electrophile" Tags="&lt;organic-chemistry&gt;&lt;radicals&gt;&lt;pericyclic&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="15606" PostTypeId="1" AcceptedAnswerId="15609" CreationDate="2014-08-25T19:24:24.750" Score="16" ViewCount="109942" Body="&lt;p&gt;According to the novel &lt;a href=&quot;http://en.wikipedia.org/wiki/Fight_Club_%28novel%29&quot; rel=&quot;noreferrer&quot;&gt;Fight Club&lt;/a&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Chuck_Palahniuk&quot; rel=&quot;noreferrer&quot;&gt;Chuck Palahniuk&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The three ways to make napalm: One, you can mix equal parts of&#xA;  gasoline and frozen orange juice concentrate. Two, you can mix equal&#xA;  parts of gasoline and diet cola. Three, you can dissolve crumbled cat&#xA;  litter in gasoline until the mixture is thick.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Napalm&quot; rel=&quot;noreferrer&quot;&gt;napalm&lt;/a&gt; is originally called napalm because is mixture of &lt;a href=&quot;http://en.wikipedia.org/wiki/Naphthenic_acid&quot; rel=&quot;noreferrer&quot;&gt;&lt;b&gt;na&lt;/b&gt;phthenic acid&lt;/a&gt;  and &lt;a href=&quot;http://en.wikipedia.org/wiki/Palmitic_acid&quot; rel=&quot;noreferrer&quot;&gt;&lt;b&gt;palm&lt;/b&gt;itic acid&lt;/a&gt;. This makes me very skeptic about it being created by gasoline and juice. Also I see no reason for it to be frozen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing that sounds realistic to achieve the sticky effect is the use of cat litter - but you could as well just use dry dirt, couldn't you? And dirt doesn't really burn so you're weakening the burning effect.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2019-07-11T12:15:07.820" LastActivityDate="2019-07-11T12:15:07.820" Title="Can you make napalm out of gasoline and orange juice concentrate?" Tags="&lt;chemistry-in-fiction&gt;&lt;pyrotechnics&gt;&lt;chemical-weapons&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15607" PostTypeId="1" AcceptedAnswerId="15610" CreationDate="2014-08-25T19:49:30.353" Score="14" ViewCount="9918" Body="&lt;p&gt;What makes these two seemingly identical topics separate? What does each field more focus on, like do chemical physics researchers study more the atomic/molecular interactions while the physical chememists study more macroscopic properties? I assume they have quite a bit of overlap. &lt;/p&gt;&#xA;" OwnerUserId="7600" LastEditorUserId="4945" LastEditDate="2014-08-26T03:10:36.870" LastActivityDate="2019-01-08T13:17:11.717" Title="Differences between chemical physics and physical chemistry?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15608" PostTypeId="2" ParentId="15604" CreationDate="2014-08-25T20:05:31.047" Score="18" Body="&lt;p&gt;I believe that you are asking if there are any reactions that are essentially &quot;non-polar&quot;. There are two classes that immediately come to mind: (1) radical reactions and (2) pericyclic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common radical reaction studied in introductory organic chemistry is the radical halogenation of alkanes. Instead of cationic and anionic reactants/intermediates/products, there are radical intermediates that do not fall into the nucleophile/electrophile classification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iaFcc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iaFcc.png&quot; alt=&quot;Radical halogenation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pericyclic reactions involve the flow of pi-electrons and do not require any charged reactants/intermediates/products. Reaction 1 below is a Diels-Alder reaction and reaction 2 is an electrocyclization. Depending on your definition of nucleophile and electrophile, a Diels-Alder reaction might fit the motif of &quot;nucleophile attacks electrophile.&quot; That is, the HOMO of the diene (1,3-butadiene in the example) interacts with the LUMO of the &quot;dienophile&quot; (ethylene). If you define a nucleophile as the reaction component providing the HOMO, then a diene is essentially a nucleophile in the Diels-Alder reaction (and vice-versa for the &quot;dienophile&quot;). However, since these reactions don't use the same types of nucleophiles as say SN2 reaction, they are often considered to fall outside of the usual &quot;nucleophile attacks electrophile&quot; motif. That's the way they were first introduced to me, at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5nlvE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5nlvE.png&quot; alt=&quot;Pericylic reactions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that many organometallic mediated reactions fall outside of the &quot;nucleophile attacks electrophile&quot; motif as well. My answer was confined to organic reactions.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="5026" LastEditDate="2018-06-09T01:28:34.627" LastActivityDate="2018-06-09T01:28:34.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15609" PostTypeId="2" ParentId="15606" CreationDate="2014-08-26T02:13:29.690" Score="17" Body="&lt;p&gt;Here's an excerpt from an interview with the author of Fight Club, Chuck Palhniuk.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to make Napalm with Frozen Orange Juice and Gasoline?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Well, Ed Norton changed one ingredient in every one to make them useless. So, that really pissed me off because I really research those really well. Actually its styrofoam and gasoline - it make the most incredible explosive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dvdtalk.com/interviews/chuck_palahniuk.html&quot;&gt;http://www.dvdtalk.com/interviews/chuck_palahniuk.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, Fight Club does not accurately convey any of the recipes. All were clearly modified. And most likely for safety reasons - i.e. you wouldn't want someone to walk out the theatre with a &quot;bright&quot; idea. On the other hand I would still expect a mixture of gasoline and orange juice concentrate to be rather flammable ... &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-26T07:57:59.100" LastActivityDate="2014-08-26T07:57:59.100" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15610" PostTypeId="2" ParentId="15607" CreationDate="2014-08-26T03:45:44.607" Score="17" Body="&lt;p&gt;You are of course right, there is a lot of overlap. There are also the scientific publication series &lt;a href=&quot;http://onlinelibrary.wiley.com/journal/10.1002/(ISSN)1439-7641&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemPhysChem&lt;/a&gt; and &lt;a href=&quot;http://pubs.rsc.org/en/journals/journalissues/cp#!recentarticles&amp;amp;all&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical Chemistry Chemical Physics&lt;/a&gt; amongst others, that deal mainly with subjects from the both areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle they are most likely dealing with almost the same subjects. And you often find physicists working in chemistry and vice versa. However, there is a little difference in the general approach to problems.&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical chemistry&lt;/a&gt; usually approaches the problems at hand from a chemist's point of view, focussing (initially) mainly on macroscopic properties of an ensemble and use physical laws to determine them. It is often considered the parent field for many other subjects, like physical organic chemistry, chemical kinetics, and spectroscopy, among others. Most common publication titles are &lt;a href=&quot;http://www.degruyter.com/view/j/zpch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zeitschrift für physikalische Chemie&lt;/a&gt; and &lt;a href=&quot;http://pubs.acs.org/journal/jpcafh&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Journal of Physical Chemistry A&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/journal/jpcbfk&quot; rel=&quot;nofollow noreferrer&quot;&gt;B&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/journal/jpccck&quot; rel=&quot;nofollow noreferrer&quot;&gt;C&lt;/a&gt;, and &lt;a href=&quot;http://pubs.acs.org/journal/jpclcd&quot; rel=&quot;nofollow noreferrer&quot;&gt;letters&lt;/a&gt;. And many others...&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_physics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical physics&lt;/a&gt; on the other hand studies chemical ensembles and reactions from a physicist's point of view. They focus more or less, like you assumed, on ions, atoms, clusters, surfaces, free radicals, and molecules interacting with each other. This does include, but is not limited to, the study of solvation effects and single entities like &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_dot&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum dots&lt;/a&gt;. Here are fewer specialist titles, like &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Journal of Chemical Physics&lt;/a&gt; and &lt;a href=&quot;http://www.journals.elsevier.com/chemical-physics-letters&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical Physics Letters&lt;/a&gt;. Of course publications are often widely accepted also in other journals of neighbouring areas. (Sometimes it is also treated as a subfield of physical chemistry.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modern times the various fields are not separated by a sharp line, they most likely blur into each other. You could probably ask the same question for Biochemistry and Chemical Biology, or Computational Chemistry and Quantum Chemistry, or Polymer Chemistry and Material Sciences. And there are more. Interdisciplinary cooperation is encouraged by many researchers and institutions, it is important not to care too much about the labels anymore.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2019-01-08T13:17:11.717" LastActivityDate="2019-01-08T13:17:11.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15612" PostTypeId="1" AcceptedAnswerId="15633" CreationDate="2014-08-26T10:49:42.430" Score="0" ViewCount="2283" Body="&lt;p&gt;We have always studied about metals reacting with dilute and concentrated acids, but I do not understand what will happen if the same metal reacts with an acid that is neither concentrated nor dilute?&lt;/p&gt;&#xA;" OwnerUserId="7603" LastEditorUserId="4945" LastEditDate="2014-08-27T03:28:55.950" LastActivityDate="2016-02-24T03:12:37.513" Title="What will happen when a metal reacts with an acid which is neither dilute nor concentrated?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2017-10-16T14:56:10.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="15613" PostTypeId="2" ParentId="12315" CreationDate="2014-08-26T11:16:19.363" Score="4" Body="&lt;p&gt;I'm adding another answer, representative of another mechanism, which I think is probably more likely than the one I proposed. It follows a similar line to the previous mechanism, but involves the adsorption of water molecules onto the cathode and anode. I will consider the reactions at the anode and the cathode separately, like in my previous mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is a polar molecule - the oxygen atom is more electronegative (the AOs on oxygen are lower in energy than the 1s orbital on the hydrogens, which contribute to the bonding MOs) than the hydrogen atoms, and thus the molecular orbitals will have a higher electron density on the oxygen than on the hydrogen. This gives the oxygen a small negative charge, and each hydrogen a small positive charge - not a full, ionic charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this charge liquid water has a network of hydrogen bonding, as the positively charged hydrogen attracts a lone pair from another oxygen, thus producing a hydrogen bond - an attraction between the water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode (-ve electrode) the water molecules align themselves with this negative charge on the metal cathode. The hydronium ions are reduced by the negative metal electrode. Thus we have $$\ce{H3O+--M + e- -&amp;gt; H2O + H-M}$$&#xA;This explains the production of hydroxide ions - turning the solution basic at the cathode. The hydrogen is bound to the metal , using a d-orbital. (If you're interested look up hydrogen adsorption onto metal surfaces). These hydrogen atoms are able to diffuse across the metal surface, in two dimensions, kind of like skating across an ice rink - you can't jump, but you can skate easily across it (unless you're really good!). These hydrogen atoms can then react together, forming hydrogen gas. Consider the equilibrium of water:&#xA;$$\ce{H2O + H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&#xA;Decreasing the concentration of $\ce{H3O+}$ means there is proportionally more $\ce{OH-}$ than $\ce{H3O+}$, which results in the solution near the cathode becoming basic. The hydrogen bound to the metal is not particularly stable - therefore a pair of hydrogen atoms can react, becoming $\ce{H2}$, as the hydrogens are able to diffuse across the surface of the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode (+ve), we know that $\ce{O2}$ is produced. Again the water molecules align themselves to the charge - the negative oxygen is attracted to the positive metallic anode. The positive metal will interact with the electrons involved in bonding in water. We can envisage this as the positive metal &quot;pulling&quot; electron density from the oxygen. This in turn pulls electron density from the two bonds, joining the oxygen and the two hydrogen atoms. This is the inductive effect - where a positive charge is transmitted through atoms - the electron withdrawing effect. This weakens the $\ce{H-O}$ bonds. The hydrogen atoms are then solvated, to form the hydronium ion, $\ce{H3O+}$. Reaction:&#xA;$$\ce{2H2O + H2O -&amp;gt; 2H3O+ + O + 4e-}$$&#xA;The 4 electrons are then used to reduce the hydronium ion at the cathode - remember this is a circuit. The oxygen remains bound to the metal, and then reacts with another oxygen to form $\ce{O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this mechanism is more likely than the other I proposed. I am not sure about how the oxygen is bonded to the metal - in a normal coordination complex the metal donates electrons, forming a coordinate/dative bond, instead of oxygen donating a lone pair to form a dative bond. However, in this case the metal is positively charged, with an active potential difference - energy is being put into the system. I would be interested for more experienced/knowledgable people to improve this mechanism and point out reading on this.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2014-08-26T11:16:19.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15614" PostTypeId="1" AcceptedAnswerId="15619" CreationDate="2014-08-26T11:35:40.067" Score="6" ViewCount="2727" Body="&lt;p&gt;My book defines &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thermodynamic irreversible process is defined as the process in which the changes take place rapidly, instead of occurring in successive stages with infinitesimal changes in quantities so that the system does not get a chance to attain equilibrium and there is an appreciable difference between the driving and the opposing force.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have some problems in understanding it.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which equilibrium are they speaking of? &lt;/li&gt;&#xA;&lt;li&gt;Why does rapidity of the change cannot help it to attain equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)? &lt;/li&gt;&#xA;&lt;li&gt;What does they want to mean by driving and opposing force?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="189" LastEditDate="2014-08-26T12:55:26.960" LastActivityDate="2014-08-27T00:35:26.053" Title="What is the cause for thermodynamic reversible and irreversible process?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15615" PostTypeId="2" ParentId="6798" CreationDate="2014-08-26T12:33:28.767" Score="5" Body="&lt;p&gt;The partial double bond character of the C-O bond in phenol becomes obvious from the resonance structures. Contributions from the structures with $\ce{C=O}$ double bond explain why the character of the C-O bond is in between single and double bond, and also show that oxygen lone pairs are less available for protonation to form $\ce{Ph-OH2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MSQdS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protonated phenols are therefore fairly acidic, with a $pK_a$ around -6. $\ce{HBr}$ and $\ce{HI}$ have significantly lower $pK_a$ values (-9 and -10, &lt;a href=&quot;http://www2.onu.edu/~b-myers/organic/2511_Files/pKa%20table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), so they would indeed be able to protonate phenol. However, $\ce{H2O}$ is unlikely to be eliminated from this compound due to the instability of the resulting phenyl cation. Its positive charge cannot be delocalized into the aromatic system as it resides in an empty $sp^2$ hybrid orbital which is perpendicular to the delocalized $\pi$ bonds (&lt;a href=&quot;http://isites.harvard.edu/fs/docs/icb.topic93502.files/Lectures_and_Handouts/33-Carbocations-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). Formation of the phenyl cation by elimination therefore requires much better leaving groups like $\ce{N2}$, which can be eliminated from aryl &lt;a href=&quot;http://en.wikipedia.org/wiki/Diazonium_salt&quot; rel=&quot;nofollow noreferrer&quot;&gt;diazonium cations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution of the protonated OH group by halides with an $S_N2$-like mechanism is hindered by Coulomb repulsion between the $\pi$ electron cloud of the benzene ring and the lone pairs of the nucleophile. This &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_aromatic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction&lt;/a&gt; only works with aromatic compounds in which the electron density in the $\pi$ system is decreased by multiple electron-withdrawing substituents, like nitro groups.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-26T12:33:28.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15616" PostTypeId="1" CreationDate="2014-08-26T14:26:51.900" Score="4" ViewCount="8326" Body="&lt;p&gt;I'm currently a chemistry major and I know that physical chemistry requires lots of calculus.  Which mathematics courses should I take so that I am well prepared for my physical chemistry courses? Specifically, I need to take courses in thermodynamics, quantum chemistry and statistical thermodynamics and beyond, as I want to specialize in physical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true that the more mathematics courses that one takes, the easier physical chemistry becomes? I'm currently taking a multivariate calculus course now, and I wonder which is the next level of math courses I should take. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to take as many mathematics courses as possible. I'm thinking of taking ordinary differential equations after i finish multivariate calculus. What additional courses beyond ordinary differential equations should I take? My school offers courses in complex analysis, real analysis, and partial differential equations. Will they be useful?&lt;/p&gt;&#xA;" OwnerUserId="7422" LastEditorUserId="73" LastEditDate="2014-08-27T07:48:21.823" LastActivityDate="2014-08-27T07:48:21.823" Title="What are the standard math prerequisites for physical chemistry?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-10-23T18:27:40.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="15617" PostTypeId="2" ParentId="15616" CreationDate="2014-08-26T16:48:30.060" Score="6" Body="&lt;p&gt;In my experience, the most helpful advanced math courses you can take for Physical Chemistry are Differential Equations and Linear Algebra. Multivariate Calculus is helpful, but not as helpful as Diff. Eq. would be, especially in the beginning. Linear Algebra is very helpful for quantum chemistry, Diff. Eq. is also helpful for quantum, but more so for statistical mechanics. In the comments you mentioned several different Diff. Eq. options - as @permeakra said, taking all of them is a good idea, but ODEs and PDEs will probably be the most useful.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-08-26T16:48:30.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15618" PostTypeId="2" ParentId="15452" CreationDate="2014-08-26T18:02:10.140" Score="4" Body="&lt;p&gt;In addition to the other answers, I'd like to add a comment about the risks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 25% solution of NaOH in large quantities is very very dangerous. On drop in the eye and you will be permanently blind.&#xA;It is much worse than acid. Acid causes something akin to a burn that keeps the acid away from the unburned tissue below. NaOH on the other hand dissolves the tissue and penetrates very deeply. The drop in your eye will liquefy your eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you pump it you'll have a lot of pressure, the solution might corrode the plastic and suddenly you have a leak that sprays corrosive lye all over you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you mix NaOH with water, it will heat up. Maybe enough to make it boil explosively. Hot NaOH solution is even more corrosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about well treatment, if you really need to use NaOH, use at least both a full face shield and safety goggles + raingear and have someone stand by with a water hose to wash of any spills quickly. Keep kids and pets away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you check if using a carbonate or hydrogen carbonate (e.g. sodium carbonate, sodium hydrogen carbonate) would work. These are much safer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the non-food grade, I would be careful. Some industrial stuff is just as clean, some is heavily contaminated. Maybe the people selling it have an analysis of the batch. In any case, make sure you rinse your system very well afterwards.&lt;/p&gt;&#xA;" OwnerUserId="7604" LastActivityDate="2014-08-26T18:02:10.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15619" PostTypeId="2" ParentId="15614" CreationDate="2014-08-26T18:02:14.167" Score="4" Body="&lt;p&gt;Let me start by reiterating what @Phillip said in his comment: this is &lt;strong&gt;not&lt;/strong&gt; a good definition for an irreversible processes - a better definition is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An &lt;strong&gt;irreversible process&lt;/strong&gt; is one in which the net entropy generation term is greater than zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All real processes are irreversible. The &quot;process speed&quot; comes into play when we want to make approximations for idealized systems. It turns out that one way to convert a process into a theoretically reversible one is to set the condition that it moves very slowly between thermodynamic states, so that at any given step, it is in equilibrium. The reason that this isn't a good &lt;em&gt;definition&lt;/em&gt; for a reversible process is that there are cases where a rapid change is reversible, and cases where a very slow change is irreversible. It's just that in general use, the &quot;very slow process&quot; approximation is commonly used to obtain a reversible process.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which equilibrium are they speaking of?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically they mean equilibrium between the system and surroundings, but it can also mean within the system itself. Equilibrium is another way of saying &quot;maximum entropy&quot; - the matter and energy are as &quot;spread out&quot; as they are going to get. If the entropy is maximized, no more can be generated. In &quot;real life&quot; this means the system is well-mixed and that matter and thermal energy has dissipated throughout the system - there are no temperature or concentration/density gradients. This is easier to visualize using particles instead of energy - imagine an ideal gas in one side of a partitioned chamber, where the partition has a hole in it to allow the molecules to cross into the other section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TidCb.png&quot; alt=&quot;Particles are all on one side in a low-entropy state, but are equally distributed at high entropy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of an &lt;strong&gt;internal equilibrium&lt;/strong&gt; - the particles cannot cross the system boundary, but they can reach an equilibrium &lt;em&gt;within&lt;/em&gt; the system. An example of a system in equilibrium with its surroundings would be a container filled with water at room temperature - there are no thermal gradients across the container boundary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does rapidity of the change prevent it from attaining equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whenever a system is perturbed, meaning moved away from an equilibrium state, it takes some amount of time to reach a new equilibrium. This amount of time is called the relaxation time. Different systems have different relaxation times, and if you make changes quickly enough, the system won't have time to relax, and therefore won't be in equilibrium. Using a piston in a cylinder as an example, if you compress the piston by a very small amount, then the pressure inside will be in equilibrium with the pressure outside almost immediately.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What do they mean by driving and opposing force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the piston as an example again, in this case the driving force would be the external pressure on the piston. The opposing force would be the pressure of the gas inside the cylinder. For a very slow process, at any given time the outside pressure and inside pressure would be the same - the volume would change to make sure this is true. For a very rapid change, the gas inside the cylinder would take longer to reach equilibrium, and so the pressure outside would not equal the pressure inside - the driving and opposing forces would not be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to remember that these are approximations that are intended to allow us to convert really difficult (usually unsolvable) differential equations that describe &lt;em&gt;non-equilibrium&lt;/em&gt; or &lt;strong&gt;dynamic&lt;/strong&gt; systems into something that is solvable. The trade-off is that we go from a realistic, fully dynamic system to one with constraints - we treat it like a series of tiny jumps between systems that are in equilibrium. In many cases, these approximations don't affect the accuracy of the solution. However, whether or not it will is determined by the time scale of the change relative to the relaxation time of the system.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-08-27T00:35:26.053" LastActivityDate="2014-08-27T00:35:26.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15620" PostTypeId="1" AcceptedAnswerId="15622" CreationDate="2014-08-26T18:37:14.547" Score="87" ViewCount="33207" Body="&lt;p&gt;I'm all bent out of shape trying to figure out what Bent's rule means. I have several formulations of it, and the most common formulation is also the hardest to understand. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atomic s character concentrates in orbitals directed toward electropositive substituents&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would this be true? Consider $\ce{H3CF}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the carbon and the fluorine are roughly $\ce{sp^3}$ hybridized. Given that carbon is more electropositive than fluorine, am I supposed to make the conclusion that because carbon is more electropositive than fluorine, there is a great deal of s-character in the $\ce{C-F}$ bond and most of this s-character is around the carbon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this a misunderstanding of &quot;orbitals directed toward electropositive substituents&quot;? The fluorine is $\ce{sp^3}$ hybridized and these orbitals are &quot;directed&quot; toward the carbon in that the big lobe of the hybrid orbital is pointing toward the carbon. So does electron density concentrate near the fluorine? Because that would make more sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this s-character concentrated toward the fluorine has the effect of what on the bond angle? I understand that the more s-character a bond has, the bigger the bond angle - consider $\ce{sp}$ vs $\ce{sp^2}$. But since the $\ce{C-F}$ bond now has less s-character around the carbon, the $\ce{H-C-F}$ bond angle can shrink, correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4231" LastEditDate="2015-04-30T23:55:33.207" LastActivityDate="2016-10-21T22:54:46.083" Title="What is Bent's rule?" Tags="&lt;bond&gt;&lt;orbitals&gt;&lt;hybridization&gt;&lt;stereochemistry&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="42" ContentLicense="CC BY-SA 3.0" />
  <row Id="15621" PostTypeId="2" ParentId="15620" CreationDate="2014-08-26T19:05:05.133" Score="15" Body="&lt;p&gt;Have you read the Wikipedia article to Bent's rule (especially the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent%27s_rule#Justification&quot;&gt;Justification&lt;/a&gt; paragraph). I think it explains the things rather well. In the example of $\ce{H3CF}$ the $\ce{H}$ is more electropositive than $\ce{C}$ and the $\ce{F}$ is more electronegative than $\ce{C}$. So, using the assumption that like in $\ce{CH4}$ the $\ce{C}$ atom is $\mathrm{sp}^3$ hybridized as a starting point, Bent's rule tells us that the $\ce{C}$-orbitals that are used to form bonds between $\ce{C}$ and $\ce{H}$ will not be &quot;pure&quot; $\mathrm{sp}^3$ orbitals but will contain a higher $\mathrm{s}$ character whereas the $\ce{C}$-orbital that is used to form the bond between $\ce{C}$ and $\ce{F}$ will contain a higher $\mathrm{p}$ character than a &quot;pure&quot; $\mathrm{sp}^3$ orbital. As for bond angles: the consequences of Bent's rule for the bond angles are also explained rather well in the Wikipedia article.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-26T21:09:48.267" LastActivityDate="2014-08-26T21:09:48.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15622" PostTypeId="2" ParentId="15620" CreationDate="2014-08-26T19:09:02.303" Score="58" Body="&lt;p&gt;That's a good, concise statement of Bent's rule.  Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$.  Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine how Bent's rule might be applied to your example of methyl fluoride.  In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine.  Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected.  Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is this?&lt;/strong&gt;  s orbitals are lower in energy than p orbitals.  Therefore electrons are more stable (lower energy) when they are in orbitals with more s character.  The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon.  If that's the case (and it is), why &quot;waste&quot; precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize.  Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds).  So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The physically observable result&lt;/strong&gt; of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5°  (reflective of more p character).  In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-18T23:31:17.713" LastActivityDate="2015-06-18T23:31:17.713" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15623" PostTypeId="1" AcceptedAnswerId="15627" CreationDate="2014-08-26T20:05:09.610" Score="5" ViewCount="787" Body="&lt;p&gt;I have recently started learning chemistry, and I've been reading a GCSE (UK high school level) textbook on the stuff :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of it has been fairly straightforward, though there has been a reoccurring theme that has been bothering me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've encountered many instances where the author simply presents a reaction with little explanation, as if to suggest that the reason as to why it would occur is obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give you an example I have come across just today. Ethene in bromine water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YWgJx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think to myself, 'why would this happen?'. There are no 'free electrons' that I can see in ethene. OK, bromine is highly reactive, but what will it do? Will it react with H, or C, and how? What about the bromine bond? It's all very uncertain in my mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always try to reduce it to an energy problem. We just want the most stable state, right? But, there are many possible outcomes. Am I supposed to apply some logic, follow some flowchart, or work out all possible outcomes and calculate which has the lowest energy? Do chemists just remember all simple reactions, and reduce problems to those? Or, am I just overthinking it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side problem to this is why, if we place an &lt;em&gt;alkane&lt;/em&gt; in bromine water, we don't get the same thing happening with ethane.&lt;/p&gt;&#xA;" OwnerUserId="7588" LastEditorUserId="194" LastEditDate="2017-09-23T01:49:41.910" LastActivityDate="2017-09-23T01:49:41.910" Title="How can I work out what reaction will happen?" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15624" PostTypeId="1" AcceptedAnswerId="15628" CreationDate="2014-08-26T21:05:08.233" Score="4" ViewCount="1068" Body="&lt;p&gt;I have a sample of Mercury in a plastic vial,&#xA;The vial is a BD Vacutainer (genuine), made of polyethylene terephthalate. It is used for blood samples, and the one I used was new.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the vial was new, it was almost as transparent as glass, as seen in the image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mm.claflinequip.com/catalog/product/cache/2/image/750x750/9df78eab33525d08d6e5fb8d27136e95/i/2/i20885.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;vacutainer http://mm.claflinequip.com/catalog/product/cache/2/image/750x750/9df78eab33525d08d6e5fb8d27136e95/i/2/i20885.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after about a month, here is what the vial looks like:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MOhRr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the vial has darkened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vial was sealed with a small initial layer of hot glue about 1 centimeter from the top, then the last centimeter was filled to the brim with epoxy, to ensure it won't leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been handling the vial quite a lot, but it has been stored in the padded box as seen in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is causing the plastic to darken? Once again the vial is made out of polyethylene terephthalate if that could be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5772" LastActivityDate="2014-08-26T23:11:32.890" Title="The walls of a plastic vial with Mercury inside are darkening. What is causing this?" Tags="&lt;plastics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15626" PostTypeId="2" ParentId="15623" CreationDate="2014-08-26T21:10:43.723" Score="0" Body="&lt;p&gt;You must learn some basic things in organic chemistry to understand mechanisms for such simple reactions and these are: functional groups, nucleophiles, electrophiles, leaving groups, curly arrows. You should easily find explanations of these in any organic chemistry textbook. Then you could use this link &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html#top&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html#top&lt;/a&gt; to understand what is happening in this reaction.&#xA;Here are the other basic types of reaction in organic chemistry:&#xA;&lt;a href=&quot;http://www.chemguide.co.uk/mechmenu.html#top&quot; rel=&quot;nofollow&quot;&gt;http://www.chemguide.co.uk/mechmenu.html#top&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-26T21:10:43.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15627" PostTypeId="2" ParentId="15623" CreationDate="2014-08-26T21:24:20.870" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There are no 'free electrons' that I can see in ethene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct, but...  In ethene there are two bonds between the two carbons that hold the two carbons together.  A chemist calls this a &quot;double&quot; bond.  The important thing about a double bond (that's not obvious from the drawing) is that the two bonds are not the same.  One is a strong sigma bond, just like in ethane; the other is a weaker pi bond.  While the electrons in the pi bond are not free, they are not held as tightly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gtHLH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the electrons in the sigma bond and are, therefore, more available for reaction.  This is why ethene is so much more reactive than ethane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I supposed to apply some logic... Or, am I just overthinking it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there is logic and that body of logic is what organic chemistry is all about.  In the case of bromine reacting with ethene, the reaction belongs to a class of reactions known as electrophilic additions.  Most reactions in organic chemistry involve a molecule that has electrons available to share (a nucleophile) and a molecule that would like more electrons (an electrophile).  These two molecules react (form a chemical bond) by the one donating\sharing its electrons and the other molecule accepting\sharing those electrons.  Here, the pi bond in ethene has electrons that it is willing to share (they're not free, but they're not tightly held either), while the bromine molecule is highly polarizable, much like $\ce{Br^{\delta +}-Br^{\delta -}}$, with the positively polarized bromine atom wanting electrons.  The reaction proceeds as pictured below, to yield 1,2-dibromoethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KiJpc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A side problem to this is why, if we place an alkene in bromine water,&#xA;  we don't get the same thing happening with ethane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I bet you can answer that one now.  Ethene has a pi bond with electrons loosely held and available for reaction.  Ethane, on the other hand, only has a sigma bond where the electrons are more tightly held and less available for reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-26T22:15:45.407" LastActivityDate="2014-08-26T22:15:45.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15628" PostTypeId="2" ParentId="15624" CreationDate="2014-08-26T23:11:32.890" Score="5" Body="&lt;p&gt;Mercury doesn't react with PET under normal conditions, but mercury vapor will &lt;strong&gt;diffuse&lt;/strong&gt; into plastics.  The diffusion will change the optical density of the plastic in the area where diffusion occurred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://books.google.com/books?id=HXXxw2djkcsC&amp;amp;pg=PA243&amp;amp;lpg=PA243&amp;amp;dq=mercury%20diffusion%20into%20plastics&amp;amp;source=bl&amp;amp;ots=p4UyDHh3Xm&amp;amp;sig=kbQnrnuunBIZ-CFFqwG_mJHyR3U&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=BhD9U4fSGcipyATE6ILIAw&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=mercury%20diffusion%20into%20plastics&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to a nice study where the mercury was doped with a radioactive isotope of mercury.  This doped mercury was placed in a glass chamber which also contained plastic discs held at a known distance above the mercury.  The plastic discs were removed after various times and then microtomed.  The various layers of the microtomed plastic disc were measured for radioactivity.  Analysis of the resulting data allowed the investigators to measure the rate constant for diffusion of mercury into the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final point, plastics are sometimes contaminated with small amounts of various impurities.  Sometimes these impurities were intentionally introduced, such as radical chain terminating materials (sulfur is an example) in order to produce polymers with a certain molecular weight distribution.  So while the plastic itself might not react with mercury, an impurity might.  The above link also provides analysis of plastic contaminated with sulfur and finds a much higher mercury diffusion constant in such materials.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-26T23:11:32.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15629" PostTypeId="2" ParentId="15614" CreationDate="2014-08-26T23:25:49.797" Score="1" Body="&lt;p&gt;I'll try and somehow answer this backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Driving and opposing forces: Each process is driven by some kind of 'force' (please do not understand 'force' in the traditional physical sense; it's more of 'reasoning, why something would want to do something'). For example, when having a saturated sodium chloride solution, the sodium and chloride ions that are still bound in their ion structure in the precipitate would like to dissolve, the driving force being the increase of global entropy by a greater disorder of the entire system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, every process also has an opposing force, that attempts to force the process back into its original state. In the case of the saturated sodium chloride solution, the opposing force would probably be something along the lines of releasing water molecules out of a solvation shell, and thereby increasing local entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for an (essentially) irreversible process, like the reaction of hydrogen and chlorine to form hydrogen chloride gas: This is a radical chain reaction. For each reaction step (formation of chlorine radicals, homolytic cleavage of hydrogen by a chlorine radical to form hydrogen chloride and a hydrogen radical, homolytic cleavage of chlorine by a hydrogen radical to form hydrogen chloride and a chlorine radical, and finally radical recombination in whichever combination) there are of course driving and opposing forces. However, the driving force is usually much stronger than the opposing force. For example, while it is energetically favourable to cleave a chlorine molecule with a hydrogen radical to form hydrogen chloride and a chlorine radical, a chlorine radical would never attack hydrogen chloride to liberate a hydrogen radical as HCl is far more stable than either hydrogen or chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The smaller and less rapid a change happens—in our case, dissolving one pair of sodium and chloride is a very small change—the easier it is to just go the other way. If a change is 'rapid' (I don't really like the word here, but I'm not good enough to find a better one), then obviously the driving force is stronger than the opposing force, creating a bias towards the product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I would agree that this part of the definition is especially weak and hard to understand …&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Any process can be understood as a dynamic one, with the change happening in both forward and backward direction. So even if you were to burn methane—clearly an irreversible process—you can still consider very minute quantities of carbon dioxide and water recombining to form a methane molecule or two (and oxygen). You won't, however, observe a proper equilibrium between methane/oxygen and carbon dioxide/water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while reversible processes are always at equilibrium between one side and the other side, irreversible processes are shifted so far to one side, that it doesn't really make sense to define an equilibrium for them, i.e. they 'do not get a chance to attain equilibrium'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I could help in some way, and I think you should get a better book ;)&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2014-08-26T23:25:49.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15630" PostTypeId="1" AcceptedAnswerId="15641" CreationDate="2014-08-27T02:09:09.763" Score="-1" ViewCount="6465" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PNEP1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one was a bit involved. The alkyl halide turning into an alcohol with retention of stereochemistry suggested that an SN1 mechanism was at work. The addition of NaOH and the inversion of stereochem followed by &quot;only&quot; suggested again an SN2 mechanism. No E2 products; hydroxide isn't basic enough to deprotonate an alkane to a significant extent. The transformation of the alcohol into a nitrile group with inversion of stereochemistry is SN2; attacked with a non-protic solvent for good measure. The transformation into a nitrile group with retention of stereochem is SN1; treatment with tosyl chloride turns the -OH into a good leaving group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQZZE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-27T03:31:25.233" LastActivityDate="2014-08-27T16:23:20.470" Title="Reaction Pathways - Substitution and elimination" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2014-08-27T16:32:19.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="15631" PostTypeId="1" AcceptedAnswerId="23806" CreationDate="2014-08-27T02:41:15.460" Score="11" ViewCount="18141" Body="&lt;p&gt;I have been trying to find out the bond angle of $\ce{H2O}$, but every site I visit has a different answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have found the following angles listed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chem.ufl.edu/~myers/chm2045/shapes.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Site 1&lt;/a&gt;: 104.4º&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://witcombe.sbc.edu/water/chemistrystructure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Site 2&lt;/a&gt;: 107.5º OR 104.5º, depending on where you are in the article.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/14981/trends-in-bond-angle-of-h2o-h2s-h2se-how-to-explain&quot;&gt;Site 3&lt;/a&gt;: 104.5º&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-06-06T15:42:22.230" Title="What is the bond angle of water?" Tags="&lt;bond&gt;&lt;water&gt;&lt;electrons&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15632" PostTypeId="2" ParentId="15631" CreationDate="2014-08-27T02:41:15.460" Score="6" Body="&lt;p&gt;While researching the above, I found some reliable sources that suggest the bond angle is 104.48º.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0022285279900195#&quot; rel=&quot;noreferrer&quot;&gt;Source 1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cccbdb.nist.gov/expangle2.asp?descript=aHOH&amp;amp;all=0&quot; rel=&quot;noreferrer&quot;&gt;Source 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;EDIT:&lt;/h1&gt;&#xA;&lt;p&gt;As DavePhD says, these values are for water in the gas phase. For liquid water, the bond angle is around 105.5-106º.&lt;/p&gt;&#xA;" OwnerUserId="551" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-01-16T22:38:07.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15633" PostTypeId="2" ParentId="15612" CreationDate="2014-08-27T03:06:20.500" Score="7" Body="&lt;p&gt;This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that the half-reaction with iron is always the same—iron is being oxidized:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe_{(s)}&amp;lt;=&amp;gt; Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+_{(aq)} +2e- &amp;lt;=&amp;gt; H2_{(g)}}\ E^\circ=0\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- &amp;lt;=&amp;gt; SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T08:11:25.513" LastActivityDate="2015-05-02T08:11:25.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15634" PostTypeId="2" ParentId="15630" CreationDate="2014-08-27T03:08:49.783" Score="4" Body="&lt;p&gt;In theory, one can substitute a leaving group with retention of stereochemistry by performing two successive $\mathrm{S_N2}$ reactions. The first one inverts stereochemistry, and the second one inverts it again, thus restoring the original configuration. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Finkelstein_reaction&quot; rel=&quot;nofollow&quot;&gt;Finkelstein reaction&lt;/a&gt; might be applicable: reaction of your 2-bromobutane with sodium iodide in acetone should yield 2-iodobutane via $\mathrm{S_N2}$, which can then be reacted with sodium hydroxide. Reaction of a secondary alkyl halide with $\ce{NaOH}$ will definitely yield competing elimination products, however. Note that, with secondary substrates, some competition will exist between $\mathrm{S_N1}$ and $\mathrm{S_N2}$, but the latter can be favored by use of an aprotic solvent and an excess of nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction with $\ce{H2O}$ is going to be slow and would yield a racemic mixture of $\mathrm{S_N1}$ products, as well as the competing $\mathrm{E1}$ product (with possibly some minor side-products from bimolecular pathways as well, depending on conditions). If you're satisfied with a racemic mixture, I would think elimination followed by acid-catalyzed hydration or oxymercuration-demercuration would be faster and more reliable in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyanation with inversion can be done by direct reaction of your 2-bromobutane with, e.g., sodium cyanide. Cyanide is less basic basic than hydroxide, so competing eliminations are much less of a problem. If you want retention of configuration, replace the bromine first (as described above), then perform the substitution with $\ce{NaCN}$. The same ideas apply if starting from a tosylate instead.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-27T16:23:20.470" LastActivityDate="2014-08-27T16:23:20.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15637" PostTypeId="2" ParentId="12315" CreationDate="2014-08-27T08:08:24.037" Score="6" Body="&lt;p&gt;Swedish Architect's answers already proposed quite thoughtful insight into this question. The only thing I was missing there were some solid research results on the matter to backup these reasonable claims. In course of my own research I came across a &lt;a href=&quot;http://www1.lsbu.ac.uk/water/electrolysis.html&quot;&gt;website&lt;/a&gt;, that provided me with a quite recent density functional study of the subject: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0301010405002053&quot;&gt;&quot;Electrolysis of water on (oxidized) metal surfaces&quot; by J. Rossmeisl, , A. Logadottir and J.K. Nørskov. &lt;em&gt;Chemical Physics&lt;/em&gt; &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;319&lt;/em&gt; (1–3), 178–184.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will here try to outline their reaction mechanism, which they used for their quantum chemical treatment. In their work they considered platinum and gold surfaces as electrode material. I will not go into detail in analysing their results, since this exceeds the scope of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water splitting follows the general net reaction process of $$\ce{H2O -&amp;gt; 1/2O2 + H2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode the following mechanism is proposed in acidic media:&#xA;\begin{align}\ce{&#xA;2H+ + 2e- &amp;amp;-&amp;gt; H^{$*$} + H+ + e- \\&#xA;          &amp;amp;-&amp;gt; H2&#xA;}\end{align}&#xA;The asterisk superscript denotes, that the formed Hydrogen $\ce{H}^*$ is chemically adsorbed to the active site at the metal surface. They continue that hydrogen is only weakly bound to the metal surface and therefore the hydrogen evolution is much faster than the rate of oxygen production. Hence they focus on the reaction on the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen evolution takes place at the anode and follows a more complicated mechanism. Again the asterisk denotes that the molecule is chemically adsorbed the the metal surface.&#xA;\begin{align}\ce{&#xA;2H2O &amp;amp;-&amp;gt; HO^{$*$} + H2O + H+ + e- \\&#xA;     &amp;amp;-&amp;gt; O^{$*$} + H2O + 2H+ + 2e- \\&#xA;     &amp;amp;-&amp;gt; HOO^{$*$} + 3H+ + 3e- \\&#xA;     &amp;amp;-&amp;gt; O2 + 4H+ + 4e- &#xA;}\end{align}&#xA;The key step in this reaction series is the formation of the active peroxide intermediate as they found out. It is therefore also the rate determining step. The doubly bonded  $\ce{O}^*$ and singly bonded $\ce{HO}^*$ species are relatively strongly bound to the surface. They also found out, that a coverage of up to $2/3$ of the surface with a monolayer of oxygen is necessary in order for the peroxyl species to become more easier than $\ce{O}^*$ and $\ce{HO}^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an alkaline electrolyte they propose the following mechanism, which only slightly deviates from the above. The intermediate species are the same, that's why they did not need to treat them separately. On the cathode they just summarised as $$\ce{2H2O + 2e- -&amp;gt; 2{}^{-}OH + H2}$$ as the mechanism is bound to be the same. On the anode on the other hand the following steps should be observed:&#xA;\begin{align}\ce{&#xA;4{}^{-}OH &amp;amp;-&amp;gt; {}^{$*$}OH + 3{}^{-}OH + e- \\&#xA;          &amp;amp;-&amp;gt; {}^{$*$}O + H2O + 2 {}^{-}OH + 2e- \\&#xA;          &amp;amp;-&amp;gt; {}^{$*$}OOH + H2O + {}^{-}OH + 3e- \\&#xA;          &amp;amp;-&amp;gt; O2 + 2H2O + 4e- \\&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Additional note and criticism&lt;/em&gt;&lt;br&gt;&#xA;Water is only liquid at room temperature due to a strong network of hydrogen bonds. As the included monolayers of water molecules it is sufficient to say that the system is treated extensively. However, what they failed to describe in their model is the conduction of electricity through the electrolyte. The reason why electric current passes quickly through an aqueous medium can be benefitted to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grotthuss_mechanism&quot;&gt;Grotthuss mechanism&lt;/a&gt;. This being said, I would think that the treatment of more than one monolayer should be considered, however I understand that this might not be possible due to technical limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another detailed description of the mechanism based on cyclovoltametrie can be found in &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S092058611200346X&quot;&gt;&lt;em&gt;Catalysis Today&lt;/em&gt; &lt;strong&gt;2013&lt;/strong&gt; &lt;em&gt;202&lt;/em&gt;, 105–113&lt;/a&gt;: &quot;Water dissociation on well-defined platinum surfaces: The electrochemical perspective&quot; by Maria J.T.C. van der Niet, Nuria Garcia-Araez, Javier Hernández, Juan M. Feliu and Marc T.M. Koper. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-27T08:08:24.037" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15639" PostTypeId="1" AcceptedAnswerId="15640" CreationDate="2014-08-27T14:22:52.057" Score="1" ViewCount="5770" Body="&lt;p&gt;I was experimenting with sulfur few years ago. The test tube I used for that purpose has still sulfur stuck onto it. I was searching for a way to dissolve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an idea to try warm acetone. Sounded like cheap solution, but doesn't work - or it will take ages to dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a chemist, so I don't have any strange chemicals at home.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="16683" LastEditDate="2016-04-12T19:34:22.210" LastActivityDate="2018-05-20T00:29:16.373" Title="How can I clean solid sulfur and its oxides from test tube?" Tags="&lt;equipment&gt;&lt;cleaning&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15640" PostTypeId="2" ParentId="15639" CreationDate="2014-08-27T14:49:41.743" Score="5" Body="&lt;p&gt;There are a couple of ways to do it depending on what chemicals are available to you. Since sulfur melts around $115~\mathrm{^\circ C}$, you might first melt it and let as much liquid drip out of the test tube as possible in order to make the following step easier. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Perhaps the most straightforward method involves toluene or xylene&#xA;which you can probably purchase at a hardware store.  Put some&#xA;toluene or xylene in the test tube and heat it up (note: be careful,&#xA;toluene and xylene are flammable), the sulfur should slowly dissolve.&#xA;Pour the waste liquid out while hot.  Repeat the process if&#xA;necessary.  If you can find both toluene and xylene, pick the xylene&#xA;because it has a higher boiling point - it can get hotter and&#xA;dissolve the sulfur better.&lt;/li&gt;&#xA;&lt;li&gt;Chromic acid (sulfuric acid and potassium dichromate) will remove sulfur.  Pour the chromic acid into the test tube and let it sit for a couple of days, then pour it out, and with the help of a spatula, you should be able to remove the sulfur as a solid chunk.&lt;/li&gt;&#xA;&lt;li&gt;Carbon disulfide dissolves sulfur, but $\ce{CS_2}$ is toxic and extremely flammable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-04-12T19:35:40.477" LastActivityDate="2016-04-12T19:35:40.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15641" PostTypeId="2" ParentId="15630" CreationDate="2014-08-27T15:30:51.573" Score="5" Body="&lt;p&gt;Since Greg E. focused on the first parts of the roadmap, I'll focus on the cyanation reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have shown for the sequence from (R)-2-butanol to (R)-2-cyanobutane (retention of configuration), actually gives inversion of configuration. It is probably more clear if we look at this as two discrete steps. Step one is reaction with TsCl (and pyridine typically) to give the tosylate with retention. Step 2 is reaction with NaCN to give the cyano compound (S)-2-cyanobutane with inversion. Perhaps that can all be done in one step, but I've never seen it or tried it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bp4De.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from (R)-2-butanol to (S)-2-cyanobutane (inversion of configuration) by direct reaction with NaCN will not give the target compound. Hydroxide is not a suitable leaving group for SN2 reaction. The hydroxyl must be converted into a better leaving group first, such as the tosylate above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves the problem of how to synthesize (R)-2-cyanobutane (retention of configuration). Since the SN2 reaction with NaCN is going to invert configuration, you need a method of converting -OH into a leaving group &lt;em&gt;with inversion&lt;/em&gt;. Two successive inversions give overall retention of configuration. The two most common ways taught for this transformation are &lt;a href=&quot;http://en.wikipedia.org/wiki/Thionyl_chloride#With_oxygen_species&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOCl2&lt;/a&gt; (-OH to -Cl) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorus_tribromide#Reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;PBr3&lt;/a&gt; (-OH to -Br). Both of these reagents install a halide with inversion of configuration, allowing SN2 reaction with NaCN to invert back to the original stereochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vzB3U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-08-27T16:00:09.823" LastActivityDate="2014-08-27T16:00:09.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15642" PostTypeId="1" CreationDate="2014-08-27T16:38:56.880" Score="1" ViewCount="75" Body="&lt;p&gt;There are water-soluble balls for soap baths. Then there are medical capsules that disappear in mouth. Is it possible do this kind of capsules and water-soluble containers at home? If not, what do I need to do them? What is the chemistry behind them?&lt;/p&gt;&#xA;" OwnerUserId="966" LastActivityDate="2014-08-27T16:38:56.880" Title="How to produce hot water-soluble soap balls? What is their chemical background? What about medical capsules?" Tags="&lt;water&gt;&lt;solubility&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15643" PostTypeId="2" ParentId="15639" CreationDate="2014-08-27T20:06:56.187" Score="0" Body="&lt;p&gt;I once had this problem in the lab. I put some NaOH (I guess it can be found in some bleaches or cleaning supplies for declogging pipes) and water in it. It warms in the process, but it's not a problem. After an hour or so, and after bit of scratching (watch out, the brush might decompose in such basic solution), the sulfur was gone.&lt;/p&gt;&#xA;" OwnerUserId="7610" LastEditorUserId="16683" LastEditDate="2016-04-12T19:35:58.150" LastActivityDate="2016-04-12T19:35:58.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15644" PostTypeId="1" AcceptedAnswerId="15693" CreationDate="2014-08-27T20:25:42.427" Score="4" ViewCount="9761" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{p}$&lt;/span&gt; value of a reaction is &lt;span class=&quot;math-container&quot;&gt;$\pu{10 atm}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{300K}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{4atm}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{400K}$&lt;/span&gt;. The incorrect statement about the reaction is-&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a. The reaction is exothermic&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b. The rate of forward direction is more than that of backward&#xA;      reaction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;c. The rate of backward reaction increases more than that of forward&#xA;      reaction with increase of temperature&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;d. The difference between heat of reaction at constant pressure and that at constant volume is &lt;span class=&quot;math-container&quot;&gt;$RT$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer given is option b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then how can we say that the difference between the heat of reaction at constant pressure and that at constant volume is &lt;span class=&quot;math-container&quot;&gt;$RT$&lt;/span&gt; for the above reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we go about approaching this problem? Do we need to use the van't Hoff  equation? I tried using the ideal gas equation also but didn't get this &lt;span class=&quot;math-container&quot;&gt;$RT$&lt;/span&gt;. Also I think since change in enthalpy is negative so the reaction is exothermic.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="83758" LastEditDate="2020-04-26T11:59:15.500" LastActivityDate="2020-04-26T11:59:15.500" Title="The difference between heat of reaction at constant pressure and that at constant volume" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15645" PostTypeId="1" CreationDate="2014-08-27T21:12:57.310" Score="3" ViewCount="262" Body="&lt;p&gt;I was just in my AP Chemistry class earlier today learning about hydrates and their properties. It occurred to me that a compound with so much water could aid in moisturizing the skin. Is this a valid thought? And if so, could hydrates be irritating to the skin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through research, I personally thought Natron may be an option in a solution added to a cream or lotion.&lt;/p&gt;&#xA;" OwnerUserId="7613" LastEditorUserId="73" LastEditDate="2014-08-28T11:58:34.420" LastActivityDate="2014-08-30T05:11:49.140" Title="Using Hydrates as a skin moisturizer" Tags="&lt;water&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15646" PostTypeId="1" AcceptedAnswerId="15647" CreationDate="2014-08-27T21:24:15.027" Score="1" ViewCount="222" Body="&lt;p&gt;It is taught that the orbital shapes derive from wave functions with different numbers of nodes. For example, the &quot;s&quot; orbital comes from a wave that has one node. But what are the waves modeling? A light wave for example is showing a relationship between an electric field and magnetic field. What relationship does an atomic orbital wave show? In other words, if you were to label the axes, what would the labels be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7612" LastActivityDate="2014-08-28T03:01:14.883" Title="What are the Waves Modeling when Referring to the Atomic Orbitals" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15647" PostTypeId="2" ParentId="15646" CreationDate="2014-08-27T22:27:41.360" Score="7" Body="&lt;p&gt;In a one-dimensional system (I choose this to keep things as simple as possible) the state of the system is described by the wave function $\psi = \psi(x, t)$, which depends on the coordinate $x$ of the particle at time $t$. Wave functions in general are complex functions of real variables. The symbol $\psi^{∗}(x, t)$ denotes the complex conjugate of $\psi(x, t)$. The quantity $P(x, t) = \psi^{∗}(x, t) \psi(x, t) \mathrm{d}x = | \psi^{∗}(x, t) |^{2} \mathrm{d}x$ gives the probability that at time $t$ the $x$ coordinate of the particle lies in the small interval $[x, x + \mathrm{d}x]$. The probability of the particle being in the interval $(a, b)$ on the $x$ axis is given by $\int_{a}^{b} | \psi^{∗}(x, t) |^{2} \mathrm{d}x$. The probabilistic interpretation of the wave function was proposed by Max Born. By analogy with the formula: mass = density $\times$ volume, the quantity $\psi^{∗}(x, t) \psi(x, t)$ is called the probability density that a particle at time $t$ has position $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Atomic orbitals&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_coordinate_system&quot;&gt;spherical coordinates&lt;/a&gt; instead of cartesian ones. The wave function solving the time-independent Schroedinger equation is then expressed in terms of the radial distance $r$, the polar angle $\theta$ and the azimuthal angle $\varphi$.&#xA;This wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$. Those radial wave functions are the part of the atomic wave functions $\Psi$ that determines their &quot;extent&quot; and their radial nodes, i.e. the nodes that grow with the Principal quantum number $n$: for example a $1\mathrm{s}$ orbital has no radial nodes (it is a simple sphere without any sign-change in the wavefunction) while the $2\mathrm{s}$ orbital has 1 radial node (it is a smaller sphere wrapped in a bigger spherical shell whereby the inner sphere has a different sign than the outer spherical shell such that you get one nodal plane between inner and outer part). The angular wave functions $Y_{\ell,m} (\theta, \varphi )$ are the so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Spherical_harmonics&quot;&gt;Spherical Harmonics&lt;/a&gt;. They determine the angular momentum of the atomic wave function $\Psi$ and its characteristic shape, i.e. $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist is used to comes from the usage of the so called real Spherical Harmonics, which are just linear combinations of the complex-valued Spherical Harmonics).&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-28T03:01:14.883" LastActivityDate="2014-08-28T03:01:14.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15648" PostTypeId="1" CreationDate="2014-08-27T23:07:32.963" Score="13" ViewCount="63473" Body="&lt;p&gt;I am trying to understand this mechanism for nitrobenzene reduction with $\ce{Sn/HCl}$. I do not understand why it implies that you pull off the $\ce{C-H}$ hydrogen instead of the $\ce{O-H}$ hydrogen in the methanol.  Is it actually suggesting that the resulting $\ce{.CH2-OH}$ radical formed from the methanol eliminates to form formaldehyde or do you get a rearrangement where you form $\ce{(CH3-O)-}$? If the latter is the case, why not just pluck the proton from the alcohol group instead of breaking the $\ce{C-H}$ bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, it mentions that organic solvents are preferred for this reaction, why exactly is that over water?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AlA9T.gif&quot; alt=&quot;nitrobenzene reduction to aniline with Sn/HCl&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7614" LastEditorUserId="5026" LastEditDate="2018-05-11T01:06:24.497" LastActivityDate="2018-05-11T01:06:24.497" Title="Nitrobenzene reduction with (tin) Sn catalyst: Why is C-H bond cleavage preferred over O-H bond cleavage?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nitro-compounds&gt;&lt;organic-reduction&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15650" PostTypeId="2" ParentId="15648" CreationDate="2014-08-27T23:39:29.567" Score="11" Body="&lt;p&gt;I don't buy that mechanism.  Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism.  Reductions of nitro compounds, be they catalytic hydrogenations or metal surface - acid catalyzed reductions, usually proceed through nitroso and hydroxylamine intermediates.  &lt;a href=&quot;http://www.ch.ic.ac.uk/widdowson/teach_files/nitrogen/dw1.html&quot; rel=&quot;noreferrer&quot;&gt;This link&lt;/a&gt; provides some discussion of the reaction and presents the following (more reasonable) mechanism where electrons are transferred from the metal surface to the various intermediates and the acid provides a ready source of protons.  These reactions are usually run in an organic solvent in order to bring the organic nitro compound into solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8kK8F.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-28T03:31:22.737" LastActivityDate="2014-08-28T03:31:22.737" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15651" PostTypeId="1" AcceptedAnswerId="15652" CreationDate="2014-08-28T03:03:20.993" Score="3" ViewCount="1063" Body="&lt;p&gt;I'm confused about halogenation pathways. From what I understand, there are multiple halogenation pathways. I'll try summarizing them below; let me know if I do anything incorrectly. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Halogenation with the diatomic halogen - i.e. $\ce{Br2}$. This reaction proceeds through a halonium intermediate. &lt;/li&gt;&#xA;&lt;li&gt;Halogenation with the diatomic halogen in the presence of light or heat. This reaction proceeds via radical intermediates. From what I understand, the light or heat cleaves the weak halogen-halogen bond and we get two radicals, which can initiate the radical reactions. &lt;em&gt;Also&lt;/em&gt; the cool thing about this reaction is that it doesn't require an nucleophilic pi bond, so if we were starting from an alkane, we don't even need to first form an alkene or alkyne; we can free-radical halogenate that alkane and then eliminate a beta proton as to make the alkene. All the other reactions mentioned here &lt;em&gt;require&lt;/em&gt; some sort of nucleophile to start with. &lt;/li&gt;&#xA;&lt;li&gt;Halogenation in the presence of water - i.e. $\ce{Br2}$ with $\ce{H2O}$. Again, we go through a bromonium intermediate. The $\ce{Br-Br}$ bond is instantaneously polarized and at that point we can form the bridged halonium ion and that gives water an entrance - the ring is highly strained and has partial charges all over the place - perfect for water to do something. &lt;/li&gt;&#xA;&lt;li&gt;Are these all the halogenation pathways I need to know for an elementary understanding of organic chemistry? Did I miss any? I think I &lt;em&gt;finally&lt;/em&gt; clarified all of them to myself through typing up this post ... &lt;/li&gt;&#xA;&lt;li&gt;Oops yes I missed a few. NBS for example is used for allylic (or benzylic) bromination, and this directly implies the need for a starting alkene ... so this isn't as synthetically useful as free-radical bromination in the presence of light. &lt;/li&gt;&#xA;&lt;li&gt;Addition of $\ce{HX}$ - the pi bond acts as a simple nucleophile and attacks the proton. Then the anionic halogen adds to the resultant carbocation. &lt;/li&gt;&#xA;&lt;li&gt;Addition of $\ce{HX}$ in the presence of peroxides - the pi bond again acts as a simple nucleophile but this reaction proceeds via a radical intermediate and as a result the reaction is overall &lt;em&gt;anti-Markovnikov.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="305" LastEditDate="2014-08-28T06:55:02.743" LastActivityDate="2014-08-28T06:55:02.743" Title="Bromination Pathways with alkane, alkene, and alkyne substrates" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15652" PostTypeId="2" ParentId="15651" CreationDate="2014-08-28T03:55:08.167" Score="5" Body="&lt;p&gt;I wanted to offer some commentary on the points you've enumerated:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You mentioned halogenation of alkenes in the presence of $\ce{H2O}$. This reaction is known as halohydrin formation. It's worth mentioning that this reaction can also be performed in the presence of other nucleophiles, perhaps most commonly alcohols and salts of other halogens to give mixed dihalogenated products. (Other types of nucleophiles may also successfully react, but I can't recall seeing any specific examples.) The reaction isn't always efficacious. Frequently, addition of both halogen atoms (i.e., ordinary halogen addition) will occur; the proximity of the halide leaving group to the halonium bridge will often result in a rapid reaction before any other nucleophile can intercept.&lt;/li&gt;&#xA;&lt;li&gt;Radical halogenation can be very difficult to control, even when dealing with substrates that should exhibit high regioselectivity. NBS (and its chlorinated counterpart, NCS) is actually an extremely useful and versatile reagent which mitigates some of the practical pitfalls of radical halogenation reactions. Under photocatalytic conditions, or with sufficient heating and a radical initiator, it can react by free radical pathways. It's often preferable to using the diatomic halogen directly (unless exhaustive halogenation is really the goal) because it generates lower concentrations of $\ce{Br2}$ &lt;em&gt;in situ&lt;/em&gt;, so the reactions are somewhat more controllable and side-products from direct addition to $\pi$-bonds are minimized. It can also react analogously to $\ce{Br2}$ with alkenes to give a halonium ion, which makes it preferable for halohydrin formation (since there is no $\ce{Br-}$ leaving group present to immediately attack the bromonium bridge).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Overall, my opinion is that you have a good core knowledge base re: halogenation reactions within the confines of introductory organic chemistry. Of course, there are a multitude of other halogenation reactions operating on myriad functional groups, but I presume those aren't germane to this particular question.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-28T04:32:52.207" LastActivityDate="2014-08-28T04:32:52.207" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15653" PostTypeId="1" AcceptedAnswerId="15654" CreationDate="2014-08-28T04:46:34.100" Score="17" ViewCount="4259" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ttT3W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to synthesize &quot;methamphetamine.&quot; Note that there is no configuration designated for this molecule, and methamphetamine refers to a very specific configuration. In any case, here are the steps I took: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reduction with hydrogen gas and Lindlar's catalyst. This transforms the alkyne into a &lt;em&gt;cis&lt;/em&gt; alkene. &lt;/li&gt;&#xA;&lt;li&gt;Epoxide formation with &lt;em&gt;m&lt;/em&gt;-CPBA.  &lt;/li&gt;&#xA;&lt;li&gt;Ring opening with sodium methylamide. This is a base, and it should attack the less sterically hindered side of the epoxide - the side without the phenyl substituent. This opens the ring and gives me &lt;em&gt;almost&lt;/em&gt; what I want. Now I just have to remove the hydroxyl group. &lt;/li&gt;&#xA;&lt;li&gt;Removal of hydroxyl group: I decided to use TsCl and heat. The tosylate group should leave upon heating and a secondary benzylic carbocation that'd form. Sodium hydride was added as to add to the carbocation formed when the TsCl leaves. However, a possible issue is that the hydride anion, being a good base, would give me the elimination product as well. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lY46t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alternate pathway below&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I89OBl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-03-26T11:56:31.387" LastActivityDate="2017-03-26T11:56:31.387" Title="Designing a theoretical synthesis of methamphetamine from phenylpropyne" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15654" PostTypeId="2" ParentId="15653" CreationDate="2014-08-28T06:06:45.740" Score="16" Body="&lt;p&gt;I think the synthetic route you've proposed is quite good for the most part. However, I think your use of $\ce{NaH}$ to remove the tosylate is incorrect. $\ce{NaH}$ is a very strong base, but not a good reductive hydride donor in most instances. Here, it will almost certainly just react by eliminating the tosylate to give a double bond. However, LAH (and various similar reagents) should achieve the desired effect. Alternatively, one of the more popular dehydroxylation reactions is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Barton%E2%80%93McCombie_deoxygenation&quot;&gt;Barton-McCombie deoxygenation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative route might begin with hydroboration-oxidation of the alkyne directly. Generally, when the two ends of the alkyne are equally substituted, one obtains a mixture of isomers, and electronic factors are usually more decisive than sterics. That being said, in this case, I suspect that it might be possible to control the regiochemistry by selecting a sterically hindered borane, perhaps 9-BBN or catecholborane. I'm not certain, however. Assuming one can indeed install the ketone at the end of the alkyne further from the phenyl group, then reductive amination should give the desired product. Direct reductive amination should be possible in this case (e.g., with $\ce{NH2Me}$, catalytic acid, and $\ce{NaBH3CN}$), allowing for a one-pot reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the proposed reaction scheme would be:&#xA;\begin{align}\ce{&#xA;Ph-C#C-CH3 &amp;amp;-&amp;gt;[\mathrm{1)~9-BBN}][\mathrm{2)}~\ce{H2O2,~NaOH}] Ph-CH2-C(O)CH3\\\\ &#xA;           &amp;amp;-&amp;gt;[\ce{NH2Me,~[H+]}][\ce{NaBH3CN}] Ph-CH2-CH(CH3)NHMe&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional point in favor of this approach is that it can be made highly enantioselective by use of asymmetric catalysts (q.v., &lt;a href=&quot;http://en.wikipedia.org/wiki/Noyori_asymmetric_hydrogenation&quot;&gt;Noyori asymmetric hydrogenation&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to your second reaction scheme, jerepierre's comment above is definitely correct about a competing elimination reaction in the last step being a problem. It's especially likely because the proton transfer is fast, $\ce{NaNHMe}$ is extremely basic, the elimination is highly entropically favored, and the resulting $\pi$-bond is conjugated to the ring, giving an especially stable product (the major elimination product by a large margin should be the Zaitsev one). In sum, I think both kinetic and thermodynamic factors are heavily against you here. If you really want to go the route of nucleophilic substitution to attach your amine, better approaches are probably the &lt;a href=&quot;http://en.wikipedia.org/wiki/Delepine_reaction&quot;&gt;Delépine reaction&lt;/a&gt;, any of the azide reduction methods, or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gabriel_synthesis&quot;&gt;Gabriel synthesis&lt;/a&gt; (approximately in that order, I think). All of those suffer from the fact that you would then need to methylate the resulting primary amine, and avoiding extensive polyalkylation is more or less impossible, as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also an issue before that stage, in that hydroboration-oxidation will not predominantly install the alcohol at the correct position, giving mainly the benzylic alcohol instead, unless you use a suitably bulky borane to achieve the desired regioselectivity. Since you only want to replace the alcohol with a bromine, I think $\ce{HBr}$ in the presence of peroxide (or some other radical initiator) should work, with no intermediate step and a reasonably low proportion of unwanted side-products.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-08-28T23:32:17.720" LastActivityDate="2014-08-28T23:32:17.720" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15655" PostTypeId="1" CreationDate="2014-08-28T06:59:30.237" Score="4" ViewCount="343" Body="&lt;p&gt;Okay, we've established that &lt;em&gt;intramolecular&lt;/em&gt; hydrogen bonding can exist in molecules, and doesn't always serve the purpose of closing the bond angle at least &lt;em&gt;relative&lt;/em&gt; to other stereoisomers of the same compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider &lt;em&gt;nitrous&lt;/em&gt; acid and its &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; configuration; hydrogen bonding is only a possibility in the &lt;em&gt;cis&lt;/em&gt; configuration, yet the &lt;em&gt;cis&lt;/em&gt; conformation has (rather unexpectedly) a larger bond angle than in the &lt;em&gt;trans&lt;/em&gt; conformation. The same goes for nitric acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15558/why-does-nitrous-acid-exhibit-unexpectedly-different-bond-angles&quot;&gt;Why does nitrous acid exhibit (unexpectedly) different bond angles?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there actually any examples of molecules in which internal hydrogen bonding serves the purpose of closing up the bond angle relative to other conformations of the same molecule?  &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-10-02T16:51:29.607" Title="Examples of intramolecular hydrogen bonding closing bond angles" Tags="&lt;theoretical-chemistry&gt;&lt;stereochemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15656" PostTypeId="2" ParentId="15526" CreationDate="2014-08-28T08:04:57.520" Score="5" Body="&lt;p&gt;First, a question: what kinds of reactions are we talking about? Is it a more specialized (in this context) field like biochemistry, or more general chemical reactions? In practice, finding data for the latter is probably simpler than finding data for the former without doing any experimental work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I would use the information about Gibbs free energy of formation that can be found &lt;a href=&quot;http://kinetics.nist.gov/janaf/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (among many other places) for the reactants and products, and then calculate the equilibrium constant from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K = \exp \left(\frac{-\Delta G^\circ_{\mathrm{rx}}}{RT}\right) $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta G_{\mathrm{rx}}$ is the change in Gibbs free energy for the reaction. Beware that if your reaction conditions - e.g. temperature or pressure - is different from standard conditions, this should be taken into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a bit!&lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2014-08-28T08:04:57.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15658" PostTypeId="2" ParentId="15639" CreationDate="2014-08-28T10:16:47.970" Score="11" Body="&lt;p&gt;Since you stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm not a chemist, so I don't have any strange chemicals at home.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;you will neither have facilities for their proper storage and  &lt;strong&gt;waste management&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, the only reasonable solution is: Get some new test tubes (about 0.20 Euro for a 160 x 16 mm tube), but don't risk your own health or pollute the environment by messing around with carcenogenic ($\ce{CrO3}$ in $\ce{H2SO4}$) or neurotoxic ($\ce{CS2}$) agents.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-04-12T19:34:59.163" LastActivityDate="2016-04-12T19:34:59.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15660" PostTypeId="2" ParentId="15645" CreationDate="2014-08-28T13:24:35.627" Score="1" Body="&lt;p&gt;I do not think that hydrates will serve as good moisturizing agents, because they usually do not loose their crystal water unless when being heated. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0040603181850484&quot; rel=&quot;nofollow&quot;&gt;Magnesium sulfate heptahydrate&lt;/a&gt;, for example, starts to loose water at temperatures above 50 °C, and complete dehydration of the salt is only achieved at temperatures above 310 °C. Some hydrates can even have a dehydrating effect when they react with moisture to hydrates with higher $\ce{H2O}$ content. &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_sulfate&quot; rel=&quot;nofollow&quot;&gt;Calcium sulfate&lt;/a&gt; hemihydrate, for example, reacts with water to form the dihydrate in an exothermic reaction. The stability of hydrates is the reason why many anhydrous salts which are easily hydrated are used as drying agents, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_sulfate&quot; rel=&quot;nofollow&quot;&gt;sodium sulfate&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_sulfate&quot; rel=&quot;nofollow&quot;&gt;magnesium sulfate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-28T13:24:35.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15661" PostTypeId="2" ParentId="13939" CreationDate="2014-08-28T15:47:03.657" Score="1" Body="&lt;p&gt;You have correctly identified the amount of substance &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; which were necessary to neutralize all &lt;span class=&quot;math-container&quot;&gt;$\ce{HCO3-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt; in the sample. With the given sample volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{10 cm^3}$&lt;/span&gt;, we can calculate the concentration of both ions in the sample:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}&#xA;= \pu{4.4E-4 mol//cm^3}&#xA;= [\ce{HCO3-}]+[\ce{CO3^{2-}}]&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the given ratio of &lt;span class=&quot;math-container&quot;&gt;$0.958$&lt;/span&gt;, we can express &lt;span class=&quot;math-container&quot;&gt;$[\ce{CO3^{2-}}]$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$[\ce{HCO3^{-}}]$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{gather}&#xA;[\ce{CO3^2-}] &#xA;  = 0.958\times[\ce{HCO3-}]\\&#xA;\pu{4.4E-4 mol//cm^3} &#xA;  = [\ce{HCO3-}] + [\ce{CO3^2-}] &#xA;  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}] &#xA;  = 1.958[\ce{HCO3-}]\\&#xA;[\ce{HCO3-}]&#xA;  = \pu{2.25E-4 mol//cm3}&#xA;\end{gather}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then use the calculated hydrogen carbonate concentration and the ratio to find &lt;span class=&quot;math-container&quot;&gt;$[\ce{CO3^2-}]$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="4945" LastEditDate="2018-10-25T17:59:08.390" LastActivityDate="2018-10-25T17:59:08.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15662" PostTypeId="1" AcceptedAnswerId="15664" CreationDate="2014-08-28T16:15:41.360" Score="18" ViewCount="58324" Body="&lt;p&gt;I am told that because the methyl group is electron donating in the conjugate base of acetic acid, this destabilizes the conjugate base by exacerbating the existing negative formal charge on the deprotonated oxygen, while in formic acid the electron donating methyl is absent in lieu of a hydrogen, which is neither withdrawing nor donating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this reasoning correct, and are there any other reasons why formic acid might be stronger than acetic acid?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5591" LastEditDate="2015-11-13T19:32:20.283" LastActivityDate="2019-03-12T00:30:21.117" Title="Why is formic acid a stronger acid than acetic acid?" Tags="&lt;acid-base&gt;&lt;stability&gt;&lt;hydrogen&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15663" PostTypeId="2" ParentId="15662" CreationDate="2014-08-28T16:33:46.527" Score="4" Body="&lt;p&gt;Yes the reasoning is correct . The more the charge is increased on a molecule the more it gets destabilised.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-08-28T16:33:46.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15664" PostTypeId="2" ParentId="15662" CreationDate="2014-08-28T17:22:31.683" Score="26" Body="&lt;p&gt;We are discussing the following equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pz0ga.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can make the acid a stronger acid by pushing the equilibrium to the right.  To push the equilibrium to the right we can &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;destabilize the starting acid pictured on the left side of the&#xA;equation, and \ or&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;stabilize the carboxylate anion pictured on the right side of the&#xA;equation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen.  Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon.  This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen.  Consequently, acetic acid is a weaker acid than formic acid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-28T17:22:31.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15665" PostTypeId="1" AcceptedAnswerId="15667" CreationDate="2014-08-28T19:41:41.157" Score="11" ViewCount="10727" Body="&lt;p&gt;I've always thought that the two-form resonance structure of ozone was the scientific concensus (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone#Structure&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;]:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/etSDt.png&quot; alt=&quot;Two-form resonance structure of ozone&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then today I found two other opinions. A &lt;a href=&quot;http://academic.reed.edu/chemistry/roco/resonance/hybrid.html&quot; rel=&quot;noreferrer&quot;&gt;three-form structure&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vaXDT.gif&quot; alt=&quot;Three-form resonance structure of ozone&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;And a &lt;a href=&quot;http://www.scielo.br/scielo.php?script=sci_arttext&amp;amp;pid=S0100-40422003000600017&quot; rel=&quot;noreferrer&quot;&gt;four-form structure&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/91OEH.gif&quot; alt=&quot;Four-form resonance structure of ozone&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is there a consenus? Are the two and three-form representations just simplifications of the four-form structure? Please include references.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: The second resonance structure on the four-form image above is drawn incorrectly, as pointed out by @ron. The upper oxygen atom should have a lone pair and a positive charge, just like the second structure of the two-form.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7122" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-03-26T05:57:43.160" Title="Ozone resonance structure: differing opinions" Tags="&lt;resonance&gt;&lt;lewis-structure&gt;&lt;ozone&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15666" PostTypeId="1" AcceptedAnswerId="15668" CreationDate="2014-08-28T20:14:09.317" Score="8" ViewCount="1611" Body="&lt;p&gt;I was posed this question from a friend, however I am not necessarily sure. I think: yes, because a molecule of any compound is composed of a whole-number ratio of atoms of its elements, per the law of definite proportions. However, I feel as if I am wrong. &lt;/p&gt;&#xA;" OwnerUserId="7621" LastEditorUserId="73" LastEditDate="2014-08-29T22:42:17.053" LastActivityDate="2016-10-04T21:19:50.347" Title="Does knowing the ratio of masses of the elements in a compound lead to the unique chemical identity of the compound?" Tags="&lt;analytical-chemistry&gt;&lt;identification&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15667" PostTypeId="2" ParentId="15665" CreationDate="2014-08-28T20:46:04.557" Score="12" Body="&lt;p&gt;Those are all valid resonance structures for ozone (except the second from the left in the &quot;four-form structure&quot; row, that one has two negative charges, maybe it's a typo).  In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LixO7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these structures contribute to the &quot;real&quot; structure of ozone - however, some resonance structures contribute more and some resonance structures contribute less.  Here is &lt;a href=&quot;http://www.chem.ucla.edu/harding/notes/notes_14C_resonance.pdf&quot; rel=&quot;noreferrer&quot;&gt;a link to a nice set of rules&lt;/a&gt; (see page 2, the &quot;Resonance Contributor Preference Rule&quot; section) that describes how to weight the contribution of various resonance structures.  The most important rule is that each atom in the resonance structure should have an octet of electrons.  Only the structures labeled I and II in your &quot;three-form structure&quot; row meet this requirement, so they are the two most important resonance structures for ozone - but all of the others do contribute, just to a lesser extent.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-28T20:46:04.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15668" PostTypeId="2" ParentId="15666" CreationDate="2014-08-28T20:48:03.903" Score="16" Body="&lt;p&gt;No, knowing the mass ratios is not sufficient by itself. In the absence of additional information (for example, molar mass) that would only be enough information to determine the &lt;a href=&quot;http://en.wikipedia.org/wiki/Empirical_formula&quot;&gt;empirical formula&lt;/a&gt;, which is the formula that contains the smallest integral ratios of atoms. (E.g., the empirical formula of $\ce{C2H6}$ would be $\ce{CH3}$.) There are (theoretically) infinitely many molecules with the same atomic and mass ratios, yet differing in the actual molecular formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, even given the additional information of molar mass (so that the molecular formula can be determined), the existence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Isomer&quot;&gt;isomers&lt;/a&gt; precludes identification of a singular, specific compound without more sophisticated analysis. For example, $\ce{C3H6}$ could be either cyclopropane or propene. $\ce{C3H6O}$ could be as innocuous as acetone, or it could be highly carcinogenic epoxypropane. With more atoms, the number of possible isomers grows very rapidly. There's a nice table giving the number of isomers of alkanes of different lengths in &lt;a href=&quot;http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf&quot;&gt;this article&lt;/a&gt;. By the time you get to $\ce{C30H62}$, the number is $4111846763$. (Note, this is only considering structural isomers; accounting for stereoisomerism, the number would be even more astronomical.)&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-08-28T20:48:03.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15670" PostTypeId="2" ParentId="4461" CreationDate="2014-08-28T21:15:22.467" Score="2" Body="&lt;p&gt;The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-28T21:15:22.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15671" PostTypeId="1" AcceptedAnswerId="15674" CreationDate="2014-08-28T22:17:01.253" Score="15" ViewCount="4619" Body="&lt;p&gt;Does Bent's rule have any utility? I got a vehement earful for &quot;NO.&quot; Points raised by my professor: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule. &lt;/li&gt;&#xA;&lt;li&gt;Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)&lt;/li&gt;&#xA;&lt;li&gt;The Coulombic effect is also real as opposed to orbitals. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Consider carbon tetrachloride and chloromethane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, carbon tetrachloride has the &lt;em&gt;stronger&lt;/em&gt; and &lt;em&gt;shorter&lt;/em&gt; carbon-halide bond because in chloromethane, the methyl group is slightly electron donating, and this repels the terminal chlorine, which lengthens the carbon-halide bond. No application of Bent's rule is necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, you can justify this difference with Bent's rule; s-character is concentrated in orbitals &lt;em&gt;directed&lt;/em&gt; toward more electropositive substituents; the carbon is more electropositive than chlorine; thus, the bond has more s-character around the chlorine and more p-character around the carbon in chloromethane. This greater p-character around the carbon in chloromethane weakens the carbon-halide bond relative to the bond in carbon tetrachloride. And greater p-character in a bond is also correlated with a compression in bond angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, bond angles are &lt;em&gt;also&lt;/em&gt; explained nicely through Coulombic effects. The C-H bonds in chloromethane are obviously shorter due to the electron-withdrawing nature of chlorine and the carbon-chlorine bond is obviously expanded due to methyl group/chlorine lone pair repulsion. This shortening of the C-H bond lengths explains the H-C-H bond angle expansion in chloromethane, and the lengthening of the C-X bond explains the X-C-H bond angle expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any case in which other elementary considerations (Coulombic, VSEPR, van der Waals repulsions) fall short in the realm of molecular geometry and Bent's rule can step in to pick up the slack? Anyone got some sort of &lt;strong&gt;&lt;em&gt;vindication&lt;/em&gt;&lt;/strong&gt; for Bent's rule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWBHS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-08-29T04:25:13.397" LastActivityDate="2014-12-29T23:33:42.933" Title="Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?" Tags="&lt;theoretical-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15674" PostTypeId="2" ParentId="15671" CreationDate="2014-08-29T08:17:55.597" Score="19" Body="&lt;p&gt;There is a reason for everything.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does Bent's rule have any utility?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;YES!&lt;/em&gt;&lt;/strong&gt; It wouldn't be there if there was not. But I will get back to this at the end of this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me go through the points raised by your &lt;a href=&quot;http://www.chem.ufl.edu/research/facultypage.shtml?photo=horvath&quot;&gt;teacher&lt;/a&gt; first:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This might be true to some extent, but that only means that coulombic considerations give the same result as a complete orbital analysis. Boltzmann, Planck, Einstein, Schrödinger, Pauling, etc. worked the field for a reason - classical mechanics cannot be applied to molecular systems. Any theory that simplifies it beyond quantum mechanics, is a model to understand or comprehend - not a model to predict.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Occam&amp;#39;s_razor&quot;&gt;Occam's razor&lt;/a&gt; would suggest picking molecular orbital or valence bond theory, as it states that from competing hypotheses the one with the fewest assumptions should be selected. Coulombic consideration is purely based on assumptions, it is a rather crude approximation to the correct picture.&lt;br&gt;&#xA;The statement, that orbitals are not real is clearly wrong. There are many books that explain how one can comprehend this. Teaching that orbitals are not real is, with all due respect, just plain wrong.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Coulombic effect is also real as opposed to orbitals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The coulombic effect is not more real than orbitals. Orbitals should be used to explain the coulombic effect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But let's talk about &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule&quot;&gt;Bent's rule&lt;/a&gt;. The &lt;a href=&quot;http://goldbook.iupac.org/BT07000.html&quot;&gt;IUPAC goldbook&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is important to know, that Bent's rule is, like Lewis rule of two (or eight, known as octet rule), a rule (not a law) based on observation. Hence its character is more or less one of a guideline. The key paper was published in &lt;a href=&quot;http://dx.doi.org/10.1021/cr60211a005&quot;&gt;Henry A. Bent, &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;1961&lt;/strong&gt;, &lt;em&gt;61&lt;/em&gt; (3), 275–311&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bent's rule originates from &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot;&gt;valence bond theory&lt;/a&gt; (VB). Early forms of VB theory proposed that a molecule can entirely be described by a set (linear combination) of resonance Lewis structures, i.e. every bonding orbital is a directed localised orbital, which will be occupied by two electrons and the same applies to lone pairs. The advantage of this approach is that it is easily comprehensible. The disadvantage is, that it is quite inflexible. The reason for this was the use of either pure $\ce{s}$ and $\ce{p}$ functions or fixed hybrid orbitals $\ce{sp^3,~sp^2,~sp}$, which only allowed a subset of angles. A direct quote from Bent's original paper states this as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For atoms that satisfy the octet rule, and it is with such atoms (e.g., carbon, nitrogen, oxygen, fluorine) that this review is primarily concerned, the four valence-shell orbitals of an atom may be thought of as compounded from one spherically symmetrical $\ce{2s}$ atomic orbital and three mutually perpendicular dumbbell-shaped $\ce{2p}$ orbitals, designated $\ce{2p_$x$, 2p_$y$, 2p_$z$}$, ($x, y, z$ indicate mutually orthogonal axes). The $\ce{2s}$ orbital is lower in energy than the $\ce{2p}$ orbital; hence the ground state of a carbon atom, for example, is written $\ce{ls^2~2s^2~2p^2}$ (not $\ce{ls^2, 2s~2p^3}$. However, in a molecule such as methane, which is known to have the shape of a regular tetrahedron, the $\ce{2s}$ orbital of carbon may be considered to mix (or “hybridize”) with the three $\ce{2p}$ orbitals to form four equivalent orbitals that point to the corners of a regular tetrahedron and make angles with each other of $109^\circ28’$. These are called hybrid orbitals. Each contains one-quarter $\ce{s}$ character and three-quarters $\ce{p}$ character, abbreviated $\ce{s^{$1/4$}p^{$3/4$}}$ or $\ce{sp^3}$, or $\ce{te}$ (for tetrahedral) (223). In ethylene, on the other hand, the $\ce{2s}$ orbital of carbon may be considered to mix with but two of the three $\ce{2p}$ orbitals to form three equivalent hybrid orbitals that lie in a plane and make angles with each other of $120^\circ$; these three hybrids, which contain 33 per cent $\ce{s}$ character, are abbreviated $\ce{s^{$l/2$}p^{$2/3$}}$ or $\ce{sp^2}$, or $\ce{tr}$ (for trigonal); perpendicular to the plane of these three orbitals is the axis of cylindrical symmetry of the remaining pure $\ce{2p}$ orbital. The geometry of acetylene is often explained by assuming that the carbon $\ce{2s}$ orbital mixes with but one of the three $\ce{2p}$ orbitals, forming two $\ce{di}$ (for digonal) hybrids that point in diametrically opposed directions; these are abbreviated $\ce{s^{$1/2$}p^{$1/2$}}$ or $\ce{sp}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The original approach allowed already good results. However, Bent already draw the conclusion, that hybridisation is connected to molecular geometry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Molecular geometry provides then a direct clue to interorbital angles and&#xA;  (table 1) the distribution of $\ce{s}$ character. For example, from the bond angles in ammonia ($106^\circ46'$) (284) and water ($104^\circ27'$) (134, 284), it is inferred that the nitrogen atom of $\ce{NH}$, devotes slightly more $\ce{s}$ character to its bonding orbitals than does the oxygen atom of $\ce{H2O}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In conclusion to this (unfortenuately I ran out of time, so this may be extended soon) he suggested to use fractional linear combinations of hybrid orbitals to add more flexibility to the VB approach. Therefore Bent's rule is an extension to VB theory, that refers to the linear combination of atomic orbitals, that eventually form hybrid orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulated rule was later merely a result of the approach to flexibilize this theory. It is therefore a very important part of theoretical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are now many more modifications to VB theory, and it is nowadays considered complementary to molecular orbital theory, but this would certainly exceed the scope of this question.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-08-29T08:17:55.597" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15675" PostTypeId="2" ParentId="11148" CreationDate="2014-08-29T08:51:15.153" Score="4" Body="&lt;p&gt;The ideal crystal is terminated by planes, which are described by &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Miller_index&quot; rel=&quot;nofollow&quot;&gt;Miller Index&lt;/a&gt;. Some of these planes are more favorable than others, and as a result, the crystals do get macroscopic shape which is different from the shape of unit cell. For some discussion on reasons for different stability, I have found this article: &lt;a href=&quot;http://iopscience.iop.org/0022-3719/12/22/036/pdf/jcv12i22p4977.pdf&quot; rel=&quot;nofollow&quot;&gt;iopscience.iop.org/0022-3719/12/22/036/pdf/jcv12i22p4977.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditDate="2014-08-29T08:57:26.303" LastActivityDate="2014-08-29T08:57:26.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15676" PostTypeId="1" CreationDate="2014-08-29T09:23:32.620" Score="6" ViewCount="7582" Body="&lt;p&gt;What I have been told is that the &lt;strong&gt;acidity&lt;/strong&gt; of the four oxyacids of Chlorine increase in the order &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl} &amp;lt; \ce{HClO2} &amp;lt; \ce{HClO3} &amp;lt; \ce{HClO4}$&lt;/span&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also been told that the oxidising strength goes in the reverse order. In my knowledge acidic strength should increase with oxidising strength. Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your reference, the Ka values of the acids are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl} = 2.9 \times 10^{-8}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO2} = 1.1 \times 10^{-2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO3} = 5.0 \times 10^2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO4} = 1 \times 10^3$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7624" LastEditorUserId="68148" LastEditDate="2019-02-23T10:17:44.533" LastActivityDate="2019-02-23T10:17:44.533" Title="What's the order of oxidising strength of the oxyacids of Chlorine?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15678" PostTypeId="2" ParentId="15676" CreationDate="2014-08-29T13:06:11.143" Score="4" Body="&lt;p&gt;For many acids, the proton $\ce{H+}$ is the oxidizing agent. For example, in the oxidation of zinc with $\ce{HCl}$, the oxidizing agent is $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + 2H+ -&amp;gt; Zn^{2+} + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for some acids, the conjugate base (anion) is also an oxidant (and usually a more powerful oxidant than $\ce{H+}$. The oxyacids of chlorine are all oxidizing agents through their anions, as is nitric acid and some others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + 2H+ + 2OCl- -&amp;gt; Zn^{2+} + H2O + 2Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example, hypochlorite is the oxidizing agent. For the oxyacids of chlorine and others, the strength of the oxidizing agent is independent of the strength of the acid.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-08-29T13:06:11.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15679" PostTypeId="1" AcceptedAnswerId="15683" CreationDate="2014-08-29T14:41:35.327" Score="2" ViewCount="2618" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic&#xA;solution:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As a general rule:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Strong acid + Strong base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; Neutral solution&lt;/p&gt;&#xA;&lt;p&gt;Strong acid + weak base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; acidic solution&lt;/p&gt;&#xA;&lt;p&gt;Weak acid + Strong base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; Basic solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to my solutions manual.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; is the conjugate base of a weak acid, therefore  the solution is basic.&lt;/p&gt;&#xA;&lt;p&gt;b) &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; is the anion of strong acid, therefore the solution is neutral.&lt;/p&gt;&#xA;&lt;p&gt;c) &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; is the conjugate acid of a weak base, therefore the solution is acidic&lt;/p&gt;&#xA;&lt;p&gt;d) &lt;span class=&quot;math-container&quot;&gt;$\ce{[K(H2O)6]+}$&lt;/span&gt; is the conjugate base of a weak acid, therefore the soltuion is basic.&lt;/p&gt;&#xA;&lt;p&gt;e) &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3^2-}$&lt;/span&gt; is the conjugate base of a weak acid, therefore the solution is basic&lt;/p&gt;&#xA;&lt;p&gt;f) &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cr(H2O)6]^3+}$&lt;/span&gt; is an acidic solution, therefore the solution is acidic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, all these answers imply that water is acting as a strong base. How is this possible when water pH is &lt;span class=&quot;math-container&quot;&gt;$\sim 7$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="95133" LastEditDate="2021-05-12T06:27:03.440" LastActivityDate="2021-05-12T06:28:19.043" Title="When to regard water as a strong base" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;water&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15680" PostTypeId="1" CreationDate="2014-08-29T14:42:13.917" Score="3" ViewCount="3606" Body="&lt;p&gt;I was asked the following question in my introductory chemistry class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The diameter of metal wire is often referred to by its American wire gauge number. A 16-gauge wire has a diameter of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.05082 in}$&lt;/span&gt;. What length of wire, in meters, is there in a &lt;span class=&quot;math-container&quot;&gt;$\pu{1.00 lb}$&lt;/span&gt; spool of 16-gauge copper wire? The density of copper is &lt;span class=&quot;math-container&quot;&gt;$\pu{8.92 g/cm^3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First I converted the imperial measurements to metric. &lt;span class=&quot;math-container&quot;&gt;$\pu{0.05082 in} = \pu{0.12908 cm}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{1 lb} = \pu{0.45359 kg}$&lt;/span&gt;, according to Google. So I can use &lt;span class=&quot;math-container&quot;&gt;$\pi r^2 h$&lt;/span&gt; to calculate the volume (and thus, using density, calculate the mass) of one meter of wire, I figure the radius to be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.06454 cm}$&lt;/span&gt;. The equation becomes &lt;span class=&quot;math-container&quot;&gt;$V = \pi \times (\pu{0.06454 cm})^2 \cdot \pu{100 cm}$&lt;/span&gt;, which equals &lt;span class=&quot;math-container&quot;&gt;$\pu{0.41654 cm^3}$&lt;/span&gt;. So that's the volume of one meter of wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the density of copper &lt;span class=&quot;math-container&quot;&gt;$(\pu{8.92 g/cm^3})$&lt;/span&gt;, I figure the mass of one meter of wire to be &lt;span class=&quot;math-container&quot;&gt;$\pu{3.72 g}$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$\pu{1  lb} = \pu{453.59 g}$&lt;/span&gt;, then converting &lt;span class=&quot;math-container&quot;&gt;$\pu{3.72 g}$&lt;/span&gt; to lbs gives me &lt;span class=&quot;math-container&quot;&gt;$\pu{3.72 g} \cdot \frac{\pu{1  lb}}{\pu{453.59 g}} = \pu{0.01 lbs}$&lt;/span&gt;. So assuming I did everything correct, one meter of wire is a hundredth of a pound, so in one pound of wire there is &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; meters.&lt;/p&gt;&#xA;" OwnerUserId="7626" LastEditorUserId="4945" LastEditDate="2020-05-26T17:33:42.653" LastActivityDate="2020-05-26T17:33:42.653" Title="Length of copper wire in 1 lb spool" Tags="&lt;materials&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15683" PostTypeId="2" ParentId="15679" CreationDate="2014-08-29T15:59:22.810" Score="2" Body="&lt;p&gt;You have to take into consideration the pKa and pKb, not the pH, when making judgments about the acidity or basicity, respectively, of a molecule. In the examples you pulled out, for water to act as a base, it must have a higher propensity to grab a proton or donate electrons than the competing molecule. Thus, it must have a higher pKa or lower pKb than the other molecule to act as a base. &lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-08-29T15:59:22.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15684" PostTypeId="2" ParentId="15679" CreationDate="2014-08-29T16:02:13.293" Score="4" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic&#xA;solution:&lt;/p&gt;&#xA;&lt;p&gt;As a general rule:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Strong acid + Strong base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; Neutral solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Strong acid + weak base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; acidic solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Weak acid + Strong base &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; Basic solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That's not true. The only time in which these rules are true is when you have an equimolar concentration of the acid and base in question, and the acids and bases &lt;em&gt;must&lt;/em&gt; both be monoprotic and monobasic, or diprotic and dibasic, etc. Otherwise there are &lt;em&gt;too&lt;/em&gt; many variables to consider for these guidelines to always ring true. Consider mixing 1.0 molar &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; with 0.10 molar &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; ... I guarantee you that the pH of that solution is not &amp;quot;neutral.&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;d) &lt;span class=&quot;math-container&quot;&gt;$\ce{K(H2O)6+}$&lt;/span&gt; is the conjugae base of a weak acid, therefore the soltuion is basic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I didn't know that potassium hydrated to a significant extent in water; even so, what the solution manual has written down is &lt;em&gt;not&lt;/em&gt; the conjugate base of hydrated potassium cation - the weak acid.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;However, all these answers imply that water is acting as a strong&#xA;base. How is this possible when water pH is &lt;span class=&quot;math-container&quot;&gt;$\sim 7$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How do these answers imply that water is acting as a strong base?&lt;/p&gt;&#xA;&lt;p&gt;Also regarding how water can be a strong base in general - it just depends on your solvent system.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="95133" LastEditDate="2021-05-12T06:28:19.043" LastActivityDate="2021-05-12T06:28:19.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15685" PostTypeId="1" CreationDate="2014-08-29T16:15:45.343" Score="1" ViewCount="364" Body="&lt;p&gt;In our electrolysis of brine, $\ce{Cl2}$ and $\ce{H2}$ are produced as gaseous products along with aqueous $\ce{NaOH}$. In this scenario I require what amount of maximum moisture can $\ce{Cl2}$ and $\ce{H2}$ hold up at specific temperature &amp;amp; pressure (precisely 85 celsius &amp;amp; 1.2bar absolute). Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched hard but without proper psychrometric data am unable to even estimate preliminary values. Link to data or other alternatives are highly welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards, &lt;/p&gt;&#xA;" OwnerUserId="7583" LastEditorUserId="5456" LastEditDate="2014-08-29T18:00:14.457" LastActivityDate="2014-08-29T18:00:14.457" Title="Moist Chlorine &amp; Hydrogen Psychrometric or Mollier Chart" Tags="&lt;thermodynamics&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15687" PostTypeId="2" ParentId="15685" CreationDate="2014-08-29T17:59:59.687" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have searched hard but without proper psychrometric data am unable to even estimate preliminary values. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to google for 'vapor pressure' of water and ideal gas model. In cases ideal gas model is close enough to true, the total pressure is divided into partial pressures, attributed to different gases, and the partial pressures are proportional to amount of molecules in the gas mixture. So, in the case the equilibrium partial pressure of water vapors above the liquid can be found in any table of vapor pressures, the rest of the pressure would be other gases. In the case of vapor above saline water, the equilibrium pressure is reduced by the fact that the salt is present. The exact reduction is guarded by Raoult's Law and is linearly proportional to the concentration of the salt. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two coolers (shell/tube exchanger with chilled water on shell side) are planned to dehumidify these two gases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To dehumidify these gases effectively, if this is really required, consider bubbling them through layer of concentrated $\ce{H2SO4}$, it is a quite common and easy way to dehumidify gases.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-08-29T17:59:59.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15688" PostTypeId="1" CreationDate="2014-08-29T18:47:19.227" Score="3" ViewCount="1301" Body="&lt;p&gt;I am currently working on a mechanism for forming paracetamol (or N-(4-hydroxyphenyl)-ethanamide). The typical reaction to nitrate the phenol (starting product) is $\ce{NaNo3}$ with a catalyst of sulfuric acid. The sulfuric acid is required as a catalyst to form the nitronium cation - an electrophile - to attack the aromatic phenol. Unlike the nitration of toluene (for TNT) or benzene, it is possible to use $\ce{NaNO3}$ because the oxygen in the hydroxy group is activating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proposed mechanism for the formation of the nitronium is this:&#xA;$$\ce{NaNO3 + H2SO4 + 2H2O &amp;lt;=&amp;gt; Na+ +NO3- + SO4- + 2H3O+}$$&#xA;I think this is a likely equilibrium between the starting ions. Sulfuric acid is a strong acid, and therefore to use a mechanism with it undissociated I think unlikely - the reaction would be slower, due to lower concentrations of the undissociated sulfuric acid. I know that $\ce{NaNO3}$ is fairly soluble, so would dissociate into ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest attack of the $\ce{NO3-}$ by $\ce{H3O+}$, protonating one of the oxygens:&#xA;$$\ce{NO3- + H3O+ -&amp;gt; HNO3 + H2O}$$ &#xA;$\ce{HNO3}$ has a Ka of ${24 * 10}$ - 24 moles of $\ce{HNO3-}$ present for every $\ce{NO3}$ we can attribute this to the stability of the $\ce{NO3-}$ ion, but it also is fairly low compared to the Ka of sulfuric acid - ${1*10^3}$. By adding sulfuric acid, a strong acid, it increases the concentration of $\ce{H3O+}$, therefore shifting the equilibrium to the right. This favors the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\ce{HNO3}$ molecule the oxygen-hydrogen bonds are all different lengths - $\ce{N-O}$ is 119.9pm, or 121.1pm, and the $\ce{N-O-H}$ bond is 140.6pm. The $\ce{N-O-H}$ bond is clearly weaker than the other two bonds. This is attributed to the resonance effect - a valence electron from each oxygen interacts with a lone pair on nitrogen favorably, shortening and strengthening the bonds. Protonating the oxygen in $\ce{NO3-}$ thus weakens the bond. Protonating the oxygen a second time:&#xA;$$\ce{HNO3 + H3O+ &amp;lt;=&amp;gt; NO2+ + H2O + H2O}$$&#xA;Would probably break that bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydronium protonates the oxygen already protonated - resulting in an unstable intermediate. The positive charge pulls electron density from the $\ce{N-O}$ sigma bond onto the oxygen (inductive effect). This breaks the bond, and stabilizes the positive charge on the oxygen, and it becomes $\ce{H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the hydronium may protonate one of the other two oxygens. Why is the one already protonated more basic? The other two oxygens have a higher electron density, so would be more nucleophilic - better for the electrophilic $\ce{H3O+}$ to attack. A fraction would surely react in the way I have described, but I would think a much higher proportion would react on other oxygens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this mechanism likely/correct? Could you explain how the second protonation occurs?&lt;/p&gt;&#xA;" OwnerUserId="4222" LastEditorUserId="73" LastEditDate="2014-08-30T11:05:41.643" LastActivityDate="2014-08-30T11:05:41.643" Title="Is this the correct mechanism of the formation of the nitronium (NO2+) ion from sodium nitrate and sulfuric acid?" Tags="&lt;reaction-mechanism&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15690" PostTypeId="1" AcceptedAnswerId="15773" CreationDate="2014-08-29T19:11:49.753" Score="27" ViewCount="9485" Body="&lt;p&gt;In the book &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0198503466&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Chemistry&lt;/a&gt; by  J. Clayden, N. Greeves, S. Warren, and P. Wothers I found the following reasoning:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may have wondered why it is that, while methyl chloride (chloromethane) is a reactive electrophile that takes part readily in substitution reactions, dichloromethane (DCM) is so unreactive that it can be used as a solvent in which substitution reactions of,other alkyl halides take place. You may think that this is a steric effect: Indeed, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; is bigger than &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;.&#xA;  But &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2Cl2}$&lt;/span&gt; is much less reactive as an electrophile than ethyl chloride or propyl chloride: there must be more to its unreactivity. And there is: Dichloromethane benefits from a sort of 'permanent anomeric effect.' One lone pair of each chlorine is always anti-periplanar to the other &lt;span class=&quot;math-container&quot;&gt;$\ce{C–Cl}$&lt;/span&gt; bond so&#xA;  that there is always stabilization from this effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in MO-terms the situation would look something like this.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ALRXb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning looks plausible to me. The interaction between the free electron pair on &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; and the &lt;span class=&quot;math-container&quot;&gt;$\sigma^{*}$&lt;/span&gt; orbital of the neighboring &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bond, which would be the LUMO of DCM, lowers the energy of the free-electron-pair-orbital, thus stabilizing the compound and it raises the energy of the LUMO, thus making DCM less reactive towards nucleophiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But how important is this anomeric effect actually for explaining the unreactiveness of DCM toward nucleophiles, especially compared to the steric effect? And how important is the steric effect actually: Is the steric hindrance exerted by the second &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2Cl2}$&lt;/span&gt; really that much larger than the steric hindrance exerted by the methyl group in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2Cl}$&lt;/span&gt; (which is a good electrophile for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; reactions)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little skeptical because if this anomeric effect was very important I would have expected that the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bond length in DCM would be slightly higher than in methyl chloride because there should be some transfer of electron density from the free electron pair into the antibonding &lt;span class=&quot;math-container&quot;&gt;$\sigma^{*}$&lt;/span&gt; orbital, which should weaken the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bond. But the actual bond lengths don't show this. They show rather the contrary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{c|c}&#xA;\hline&#xA;\text{Species} &amp;amp; \text{Average }\ce{C-Cl}\text{ bond length / Å} \\&#xA;\hline&#xA;\ce{CH3Cl} &amp;amp; 1.783 \\&#xA;\ce{CH2Cl2} &amp;amp; 1.772 \\&#xA;\ce{CHCl3} &amp;amp; 1.767 \\&#xA;\ce{CCl4} &amp;amp; 1.766 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&#xA;(source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bent%27s_rule#Bond_lengths&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know that the stronger polarization of the &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; atomic orbitals in di-, tri-, and tetrachlorinated methane as compared to methyl chloride (due to the electronegativity of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt;) should lead to an overall strengthening of the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Cl}$&lt;/span&gt; bonds in those compounds. But I would have expected a trend that would show only a slight decrease (or maybe even an increase) of bond length when going from methyl chloride to DCM and then a more pronounced decrease when going from DCM to chloroform followed by a similar decrease when going from chloroform to tetrachloromethane. But instead the polarization effect seems to only slightly increase on adding more chlorine atoms.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="61578" LastEditDate="2019-04-07T17:40:46.437" LastActivityDate="2019-04-07T17:40:46.437" Title="Why are DCM and chloroform so resistant towards nucleophilic substitution?" Tags="&lt;organic-chemistry&gt;&lt;stereoelectronics&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="15691" PostTypeId="2" ParentId="13916" CreationDate="2014-08-29T20:38:54.257" Score="1" Body="&lt;p&gt;You need 15 g HCl and 6 g NaI per 100 g of the solution. Treating 31.45% (I don't know much about practice) as exact, there is 15 g HCl in 15 g / 31.45% = 47.69 g of your muriatic acid. So you need 6 g of sodium iodide, 47.69 g of your muriatic acid and the rest, i.e. 46.31 g, of pure water per 100 g of the solution.&lt;/p&gt;&#xA;" OwnerUserId="7630" LastEditorUserId="7630" LastEditDate="2014-08-29T20:59:45.583" LastActivityDate="2014-08-29T20:59:45.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15692" PostTypeId="2" ParentId="15688" CreationDate="2014-08-29T20:41:53.817" Score="2" Body="&lt;p&gt;The electrons on the two other oxygens are delocalized making them much less basic.&lt;img src=&quot;https://i.stack.imgur.com/oyHQ6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-29T20:41:53.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15693" PostTypeId="2" ParentId="15644" CreationDate="2014-08-29T21:07:15.807" Score="6" Body="&lt;p&gt;So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First Part: heat of reaction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From the first law of thermodynamics we know&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume.&#xA;So, for a reaction at constant volume you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mathrm{d}U = \mathrm{d} Q \ .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a reaction at constant pressure you can invoke the relationship&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d}H &amp;amp;= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\&#xA;              &amp;amp;= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\&#xA;              &amp;amp;= \delta Q + V \mathrm{d}p \\&#xA;              &amp;amp;\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H$  is the enthalpy.&#xA;This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure.&#xA;The so called heat of reaction is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\xi$ is the extent of reaction.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second Part: temperature dependence&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives&quot; rel=&quot;nofollow&quot;&gt;equality of mixed partials&lt;/a&gt; and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &amp;amp;= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\&#xA;  &amp;amp;= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &amp;amp;= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\&#xA;  &amp;amp;= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where  $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you integrate the equations for the temperature dependences of the reaction heats you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \text{constant volume:  } &amp;amp;\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\&#xA;  \text{constant pressure:  } &amp;amp;\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the difference of the reaction heats at constant volume and constant pressure would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Third Part: ideal gases&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Relations_between_heat_capacities#Ideal_gas&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  c_{p} - c_{V} = R \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the integral simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &amp;amp;= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\&#xA;  &amp;amp;= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\&#xA;  &amp;amp;= R \sum_{j} \nu_{j} \int \mathrm{d} T \\&#xA;  &amp;amp;= R T \sum_{j} \nu_{j}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction.&#xA;This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ .&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-30T16:26:55.153" LastActivityDate="2014-08-30T16:26:55.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15694" PostTypeId="1" AcceptedAnswerId="15695" CreationDate="2014-08-29T21:32:10.173" Score="7" ViewCount="5443" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Acetone_peroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acetone peroxide&lt;/a&gt; is a very dangerous explosive, easily detonated by mild heating, friction or shock. It appears in the form of small white crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBoLa.png&quot; alt=&quot;acetone peroxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This compound forms from the mixture of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_peroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen peroxide&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetone&quot; rel=&quot;nofollow noreferrer&quot;&gt;acetone&lt;/a&gt; and hydrochloric or sulphuric (and maybe any other) acid. The acid, however, is only a &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalyst&lt;/a&gt; in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3lcQ.png&quot; alt=&quot;formation of trimmer acetone peroxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious why the acid is required for the reaction to happen. Is it actually necessary? Does the acid change during the reaction?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4231" LastEditDate="2015-06-19T22:18:46.403" LastActivityDate="2016-05-31T11:48:00.840" Title="Why do we need acid when making acetone peroxide?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;catalysis&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15695" PostTypeId="2" ParentId="15694" CreationDate="2014-08-29T22:14:43.143" Score="11" Body="&lt;p&gt;The role of the acid in this reaction (really a polymerization) is two-fold.  First, it protonates the acetone carbonyl oxygen making the acetone carbonyl carbon more electrophilic.  This speeds up the rate of attack by the nucleophilic hydrogen peroxide (or other hydroperoxides formed later in the reaction) on this carbon.  Second, the acid also protonates hydroxyl groups in intermediates formed later in the reaction.  This makes them better leaving groups and allows nucleophiles like hydrogen peroxide or other hydroperoxides formed in the reaction to displace them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acid is a catalyst, the more acid the faster the reaction, but the reaction will proceed, albeit more slowly, without the acid.  Any proton acid can be used, but as expected, the stronger the acid, the faster the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are links to two articles on the subject (&lt;a href=&quot;http://www.ukessays.com/essays/biology/synthetic-and-spectroscopic-studies-of-triacetone-triperoxide-biology-essay.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 1&lt;/a&gt;, &lt;a href=&quot;http://www.academia.edu/7629854/Factors_Influencing_Triacetone_Triperoxide_TATP_and_Diacetone_Diperoxide_DADP_Formation_Part_2&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 2&lt;/a&gt;) along with a diagram of the proposed reaction mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M9FZu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-29T22:14:43.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15696" PostTypeId="2" ParentId="5299" CreationDate="2014-08-29T22:45:47.437" Score="11" Body="&lt;p&gt;You're right - water vapor can exist at temperatures such as $\pu{50^\circ C}$. This is a phenomenon known as evaporative cooling, where molecules of water with higher kinetic energies tend to &quot;release&quot; themselves from the system, and as a result, less and less water molecules are held within that system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is a sort of proxy for kinetic energy, and vice versa. Reference, also, the Maxwell-Boltzmann distribution, which reveals that at $\pu{100^\circ C}$, for example, not &lt;em&gt;all&lt;/em&gt; molecules in the system possess the kinetic energy that they would at $\pu{100^\circ C}$, but rather the average exists at that temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, never in a real scenario.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="4945" LastEditDate="2017-11-01T15:46:25.517" LastActivityDate="2017-11-01T15:46:25.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15697" PostTypeId="1" CreationDate="2014-08-29T22:48:41.860" Score="5" ViewCount="107" Body="&lt;p&gt;I've noticed that some groups of metals with similar properties tend to belong to the same cubic structure group, such as $\text{FCC, BCC, or HCP}$. What is a good explanation as to why these metals have similar cubic structures, and how can one determine a metal's expected cubic structure just from knowing its identity (if there is a way)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, copper, silver, and gold all have face centered cubic structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trend seems to break down at some point, I think, because $\frac{2}{3}^{rd}$ of the platinum group have $\text{FCC}$ structures and $\ce{Ru}$ and $\ce{Os}$ have simple cubic structures.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="73" LastEditDate="2014-08-30T11:08:17.257" LastActivityDate="2014-10-02T14:21:15.737" Title="Is there a systematic way to determine the expected cubic structure of metals?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15698" PostTypeId="2" ParentId="15645" CreationDate="2014-08-29T23:05:36.967" Score="0" Body="&lt;p&gt;This isn't necessarily true (that hydrates will hydrate your skin). For example, $\ce{MgSO4*7H2O}$ is Epsom salt. Do salts sound particularly hydrating? No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, dehydrated hydrates have a high attraction for water, as seen in $\ce{CaSO4}$. When hydrated, it forms plaster of Paris, which can be dried to make drywall.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="418" LastEditDate="2014-08-30T05:11:49.140" LastActivityDate="2014-08-30T05:11:49.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15699" PostTypeId="1" AcceptedAnswerId="41933" CreationDate="2014-08-29T23:40:45.447" Score="3" ViewCount="82" Body="&lt;p&gt;I'm curious about some of the wondrous materials available today are able to repel liquid water but permit water vapor to pass through. I've heard that housewrap-type plastics accomplish this by being perforated with microscopic holes that are slightly too large for liquid water, but large enough for the vaporous form to pass through. I have no idea how &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicate_mineral_paint&quot; rel=&quot;nofollow&quot;&gt;silicate mineral paints&lt;/a&gt; do it, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Housewraps and paints are thin films, though. I'd like to learn about how such properties could be imparted to a masonry unit like a compressed earth block. Could you cast such a block with a mixture of soil and some additive that would grant this kind of water repellency but allow water vapor to pass though?&lt;/p&gt;&#xA;" OwnerUserId="7633" LastEditorUserId="7633" LastEditDate="2014-08-30T03:58:41.660" LastActivityDate="2015-12-09T14:19:59.593" Title="What additives could make a masonry unit water-repellent but vapor-permeable?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15700" PostTypeId="2" ParentId="15653" CreationDate="2014-08-30T00:21:56.187" Score="11" Body="&lt;p&gt;There has been explosive growth in the use of &lt;strong&gt;hydroamination&lt;/strong&gt; reactions to synthesize organic, nitrogen-containing compounds over the past 10-15 years.  It seems that regioselective addition to produce either the Markovnikov or anti-Markovnikov product can be attained by proper selection of catalyst and/or reaction conditions.  Both terminal and internal alkynes can react, aliphatic or aromatic amines can react, and the reaction can be run in an inter- or intra-molecular (e.g. the amine is already part of the molecule containing the alkyne, so a ring is formed in the process) fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two model reactions that suggest this methodology can be used to produce your synthetic target in a one pot, two step reaction process.  The catalyst used in these reactions is bis(indenyl)dimethyltitanium(IV).  The first step in the reaction involves addition of the amine to the triple bond, and the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f36dD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second step involves reduction of the resultant imine to the corresponding amine.  It appears that if you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vVbXH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;were to run this reaction using your alkyne and methylamine you could expect overall high yield and regioselectivity to produce methamphetamine in a one pot process.  Here is a &lt;a href=&quot;http://books.google.com/books?id=xCGGAwAAQBAJ&amp;amp;pg=PA291&amp;amp;lpg=PA291&amp;amp;dq=hydroamination%20of%20internal%20alkynes%20with%20methylamine&amp;amp;source=bl&amp;amp;ots=UF1ag0BZyQ&amp;amp;sig=kPEzaZJ314SJREBzShDP3n3VuPY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=QgoBVOeABZecygSHuoHQBQ&amp;amp;ved=0CEIQ6AEwBA#v=onepage&amp;amp;q=hydroamination%20of%20internal%20alkynes%20with%20methylamine&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;link to a review&lt;/a&gt; (see page 291-2 for more information on these specific examples) that includes more references to this reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-08-30T18:22:45.833" LastActivityDate="2014-08-30T18:22:45.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15701" PostTypeId="2" ParentId="15688" CreationDate="2014-08-30T04:36:26.457" Score="2" Body="&lt;p&gt;As you described, the reaction depends on formation of the nitronium intermediate. I think you are right that the (partially) negatively charged oxygens are more basic. However, nitronium formation is much much faster by using water as a leaving group rather than hydroxide, which is what would be required otherwise.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-30T04:36:26.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15702" PostTypeId="1" CreationDate="2014-08-30T06:15:48.200" Score="6" ViewCount="4588" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In $\ce{SO3}$, 2 $p\pi-d\pi$ bonds are present. But which $d$ orbitals of&#xA;  sulfur take part in these $\pi$ bonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer says $d_{xy}$ and $d_{yz}$. Someone also told me that crystal field theory supports it as while splitting $d_{x^2-y^2}$ and $d_{x^2}$ are raised in energy compared to the other 3, but that theory has nothing to do here. &lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="402" LastEditDate="2017-01-14T20:32:13.443" LastActivityDate="2017-01-14T20:32:13.443" Title="Which d orbitals of sulfur take part in the pi bonds of SO3?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15703" PostTypeId="2" ParentId="15274" CreationDate="2014-08-30T06:31:09.837" Score="3" Body="&lt;p&gt;The numbering clearly is &lt;strong&gt;not&lt;/strong&gt; a part of the international &lt;strong&gt;GHS&lt;/strong&gt; (Globally Harmonized System of Classification and Labelling of Chemicals) labeling scheme.&#xA;&lt;hr&gt;&#xA;On a side note, GHS requires &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hazard pictograms, &lt;/li&gt;&#xA;&lt;li&gt;coded hazard statements (&lt;em&gt;Hnnn&lt;/em&gt; and &lt;em&gt;EUHnnn&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;li&gt;coded precautionary statements (&lt;em&gt;Pnnn&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but typically does not give any information on the how/where to dispose chemicals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For classroom settings, the questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many different waste containers are needed?&lt;/li&gt;&#xA;&lt;li&gt;How to label them?&lt;/li&gt;&#xA;&lt;li&gt;Which chemicals go where?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, a global or EC-wide (European Community) scheme does not exist for this and I'd be astonished to find that the African Union has agreed upon a common scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, this is typically regulated on a nation-wise basis only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an example: In Germany, the classification scheme is provided by the &lt;em&gt;Deutsche Gesetzliche Unfallversicherung&lt;/em&gt; (mandatory accident insurance) and you may find their &lt;em&gt;suggestions for waste disposal in schools&lt;/em&gt; &lt;a href=&quot;http://publikationen.dguv.de/dguv/pdf/10002/213-098.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (PDF in German, &lt;em&gt;Stoffliste zur Regel&#xA;„Unterricht in Schulen&#xA;mit gefährlichen Stoffen“&lt;/em&gt;, 2017).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, without knowing the (national) origin of the source you are citing in your question, it is very difficult to say which number refers to which compound class.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="24052" LastEditDate="2018-03-05T16:02:03.467" LastActivityDate="2018-03-05T16:02:03.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15704" PostTypeId="1" CreationDate="2014-08-30T11:05:21.813" Score="0" ViewCount="4936" Body="&lt;p&gt;It is well known fact that Barium Hydroxide is a base. How does the Brønsted-Lowry concept explain $\ce{Ba(OH)2}$  as a base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that $\ce{Ba(OH)2}$ first accepts a proton then leaves behind $\ce{H2O}$?&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="73" LastEditDate="2014-08-30T12:56:13.700" LastActivityDate="2016-10-29T16:59:51.130" Title="Explaining Ba(OH)2 as a base?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15705" PostTypeId="2" ParentId="15704" CreationDate="2014-08-30T13:15:11.830" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Barium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;Barium hydroxide&lt;/a&gt; – or more exactly, the hydroxide ion – is a Brønsted-Lowry base, because in aqueous solution, it is completely dissociated into $\ce{Ba^2+}$ and $\ce{OH-}$ ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ba(OH)2 -&amp;gt; Ba^2+ + 2OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{OH-}$ is a proton acceptor, forming $\ce{H2O}$. In aqueous solution, the following reaction takes place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{OH- + H3O+ -&amp;gt;~ 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barium hydroxide thus reacts with acids in a neutralization reaction, with water and the barium salt of the respective acid as the products. For example, with hydrochloric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ba(OH)2 + 2HCl -&amp;gt;~ BaCl2 + 2H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2016-10-29T16:59:51.130" LastActivityDate="2016-10-29T16:59:51.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15706" PostTypeId="1" AcceptedAnswerId="15708" CreationDate="2014-08-30T15:20:09.550" Score="2" ViewCount="488" Body="&lt;p&gt;To complete homework I am working on, I need help finding the density of a penny. I'm not given the mass or the volume. All I know about this penny is it's zinc coated in copper, and I am given the percentage of copper in the penny: 3.0% (by mass). We are to use these values for the density of copper and zinc:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper = 8.93, Zinc = 7.14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friends did this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{8.98 \times 0.03 + 7.14 \times 0.97 = average~density~of~a~penny}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I told them &quot;I think you're wrong, you can only calculate the average density of something if you use the volume&quot;, but I couldn't explain it any better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $M$ be the mass of this penny in grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.03M~\mathrm{g~mL^{-1}}}{8.98~\mathrm{g}} = \frac{0.03}{8.98}M~\mathrm{ml}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.97M~\mathrm{g~mL^{-1}}}{7.14~\mathrm{g}} = \frac{0.973}{7.14}M~\mathrm{ml}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding them together we get $7.1952M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what this number represents but we can use it to find the ratios of volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\frac{0.03}{8.98 M}}{7.1952 M} = 0.0374$$ $$\frac{\frac{0.97}{7.14 M}}{7.1952 M} = 0.963$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these numbers, we do what my friends did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$8.98 \times 0.0374 + 7.14 \times 0.963 = \mathrm{average~density~of~a~penny}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am, is there a formula or any other simpler way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first question on this site, so let me know if I did anything wrong.&lt;/p&gt;&#xA;" OwnerUserId="7636" LastEditorUserId="69830" LastEditDate="2021-05-08T13:02:02.047" LastActivityDate="2021-05-08T13:02:02.047" Title="Density of an object" Tags="&lt;metal&gt;&lt;density&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15708" PostTypeId="2" ParentId="15706" CreationDate="2014-08-30T16:02:28.790" Score="1" Body="&lt;p&gt;You have the right idea, but somewhere your math is incorrect.&lt;/p&gt;&#xA;&lt;p&gt;Let the mass of the penny be &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The mass of the copper is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.03M}$&lt;/span&gt; and the mass of the zinc is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.97M}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The volume of copper is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.03M}/(\pu{8.93g/cm^3})$&lt;/span&gt; and the volume of the zinc is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.97M}/(\pu{7.14g/cm^3})$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The total volume is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The density of the penny is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {M}{\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Factoring out M gives&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {1}{0.03/(\pu{8.93g/cm^3}) + 0.97/(\pu{7.14g/cm^3})}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So the density of a mixture works out to be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="69830" LastEditDate="2021-05-08T13:01:48.603" LastActivityDate="2021-05-08T13:01:48.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15709" PostTypeId="1" AcceptedAnswerId="15721" CreationDate="2014-08-30T16:12:17.880" Score="4" ViewCount="2298" Body="&lt;p&gt;I'm just blindly experimenting. I tried to dissolve some &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_permanganate&quot; rel=&quot;nofollow&quot;&gt;potassium permanganate&lt;/a&gt; in water before performing electrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't see much difference. How is $\ce{KMnO4}$ dissolved in water? How does it react during electrolysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added salt and the electrolysis was so violent that the copper started to burn (probably reacting with something).&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2014-08-31T03:47:59.607" LastActivityDate="2019-10-06T13:49:23.340" Title="How does KMnO₄ affect electrolysis?" Tags="&lt;aqueous-solution&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15710" PostTypeId="1" CreationDate="2014-08-30T16:58:08.700" Score="0" ViewCount="247" Body="&lt;p&gt;Is this reaction possible: $\ce{C6H8O7 + 12 H3O -&amp;gt; C6H6 + 19 H2O}$?&#xA;Considering that the $\ce{H+}$ can be induced by ionization of the citric acid solution the benzene can be produced in this way?&lt;/p&gt;&#xA;" OwnerUserId="7638" LastEditorUserId="4945" LastEditDate="2014-09-04T07:42:36.877" LastActivityDate="2014-09-04T07:42:36.877" Title="Can citric acid protonation produce benzene and water" Tags="&lt;experimental-chemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15711" PostTypeId="2" ParentId="15710" CreationDate="2014-08-30T17:05:22.510" Score="4" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since citric acid is a 5 carbon carbon chain, you would need to break a C-C bond and reattach the carbon on the end of the chain. This does not happen in acidic media.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-30T17:05:22.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15712" PostTypeId="2" ParentId="14376" CreationDate="2014-08-30T17:22:43.683" Score="4" Body="&lt;p&gt;A lot of metal oxides decompose under liberation of $\ce{O2}$ when subjected to sufficiently high temperatures. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury(II)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{HgO}$&lt;/a&gt; decomposes at temperatures above $\pu{500^\circ C}$ into elemental mercury and oxygen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HgO -&amp;gt;~ 2Hg + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, disproportionation is another possible reaction pathway. An example is the thermal decomposition of iron(II) oxide to elemental iron and iron(II, III) oxide at temperatures below $\pu{575^\circ C}$ ($\ce{FeO}$ is more stable above that temperature; sources: &lt;a href=&quot;http://books.google.de/books?id=i9nyvTYBQtAC&amp;amp;pg=PA305&amp;amp;lpg=PA305&amp;amp;dq=thermal%20decomposition%20of%20metal%20oxides%20%2B%20disproportion&amp;amp;source=bl&amp;amp;ots=Dr9JNqyiXN&amp;amp;sig=FXfQ2mF_zhLgraSXF27OscDZMYs&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=b_0BVKPAMMiB4gTUsoDYBw&amp;amp;ved=0CC4Q6AEwAQ#v=onepage&amp;amp;q=thermal%20decomposition%20of%20metal%20oxides%20%2B%20disproportion&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(II)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4FeO \rightleftharpoons Fe + Fe3O4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the decompostion of $\ce{SnO}$ at temperatures between $700\mathrm -\pu{1000K}$, which proceeds via an intermediate $\ce{Sn3O4}$ mixed oxide. Tin disproportionates from the oxidation state +II in $\ce{SnO}$ to $0$ in $\ce{Sn}$ and +IV in $\ce{SnO2}$ (&lt;a href=&quot;http://books.google.de/books?id=i9nyvTYBQtAC&amp;amp;pg=PA305&amp;amp;lpg=PA305&amp;amp;dq=thermal%20decomposition%20of%20metal%20oxides%20%2B%20disproportion&amp;amp;source=bl&amp;amp;ots=Dr9JNqyiXN&amp;amp;sig=FXfQ2mF_zhLgraSXF27OscDZMYs&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=b_0BVKPAMMiB4gTUsoDYBw&amp;amp;ved=0CC4Q6AEwAQ#v=onepage&amp;amp;q=thermal%20decomposition%20of%20metal%20oxides%20%2B%20disproportion&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin(II)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{4SnO -&amp;gt;~&amp;amp; Sn3O4 + Sn} \\&#xA;\ce{Sn3O4 -&amp;gt;~&amp;amp; 2SnO2 + Sn}&#xA;\end{aligned}&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="11367" LastEditDate="2017-04-08T16:29:35.770" LastActivityDate="2017-04-08T16:29:35.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15713" PostTypeId="2" ParentId="15710" CreationDate="2014-08-30T17:24:11.563" Score="5" Body="&lt;p&gt;Here's a drawing of citric acid and benzene.  You would have to spend a lot of energy in order to break a large number of bonds (carbon-carbon bonds, carbon-oxygen bonds and carbon-hydrogen bonds) to rearrange citric acid into benzene.  This would be unlikely to occur in a natural process and would require many steps in the laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5SzIz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-08-30T17:24:11.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15714" PostTypeId="1" AcceptedAnswerId="15716" CreationDate="2014-08-30T18:05:41.293" Score="1" ViewCount="2123" Body="&lt;p&gt;How would one prepare a one molar sodium hydroxide solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see a few issues with this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{NaOH}$ absorbs carbon dioxide. So old NaOH samples might have a significant portion of $\ce{NaHCO3}$ or sodium bicarbonate. How would get get rid of the sodium bicarbonate? &lt;/li&gt;&#xA;&lt;li&gt;I am told that $\ce{NaOH}$ is hygroscopic. So it absorbs water, apparently. What is the mechanism for this? I can see the hydroxide attacking carbon dioxide in the above bullet point, that's easy, but I can't really see $\ce{NaOH}$ attacking water. Also couldn't we circumvent the water absorption problem by cooking the $\ce{NaOH}$? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="73" LastEditDate="2014-08-31T12:21:59.197" LastActivityDate="2014-08-31T13:23:18.887" Title="Preparation of NaOH Solution" Tags="&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15716" PostTypeId="2" ParentId="15714" CreationDate="2014-08-30T18:29:05.230" Score="6" Body="&lt;p&gt;There is a difference in preparing a solution and determining its concentration. As you said, NaOH solutions absorb carbon dioxide over time. It is possible to remove the absorbed carbon dioxide by boiling, but it is just as easy to use the solution as is and re-determine its concentration when you use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of sodium hydroxide solutions can be determined by titrating a known amount of dried potassium hydrogen phthalate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need an exact concentration of the solution, make it slightly more concentrated, standardize it and then dilute it to get the concentration desired. It would only be accurate for a few hours.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am told that NaOH is hygroscopic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The powder and pellets of sodium hydroxide are hygroscopic, but the solutions of sodium hydroxide already have water, so it is unlikely more will be absorbed.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-08-30T18:29:05.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15717" PostTypeId="2" ParentId="4130" CreationDate="2014-08-30T18:36:23.830" Score="4" Body="&lt;p&gt;I'll just explain the effect of changing temperature, as this isn't covered so well as the others in Brian's answer.&#xA;Gibbs energy and the equilibrium constant of a reaction is given by this identity:&#xA;$$\ce{\Delta G = -RT * ln(K)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gibbs energy is also given by:&#xA;$$\ce{\Delta G = \Delta H - T\Delta S}$$&#xA;Equating the two RHS:&#xA;$$\ce{-RT*ln(K) = \Delta H - T\Delta S}$$&#xA;Divide through by $\ce{-RT}$:&#xA;$$\ce{ln(K) = \Delta H/-RT} + \Delta S/R$$&#xA;$$\ln(K) = \frac{-\Delta H}{R}*\frac{1}{T} + \frac{\Delta S}{R}$$&#xA;If $\Delta H$ is positive - an endothermic reaction, then $-\Delta H$ is negative. Increasing the temperature makes $\Delta H$ less negative, and therefore makes $ln(K)$ less negative, and thus $K$ bigger- it therefore shifts the equilibrium towards the side of the products:&#xA;$$K = \frac{Products}{Reagents}$$&#xA;Meaning more products.&#xA;If $\Delta H$ s negative - and exothermic reaction - then $-\Delta H$ is positive. Increasing the temperature makes $-\Delta H$ less positive, and therefore makes $ln(K)$ smaller, and $K$ is therefore smaller, shifting the equilibrium to the side with the reagents on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H$ and $\Delta S$ also depend on temperature, but over a small temperature range this can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2014-08-30T18:36:23.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15720" PostTypeId="2" ParentId="9001" CreationDate="2014-08-30T20:36:51.750" Score="14" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Permanganate&quot;&gt;Permanganate&lt;/a&gt; as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in $\ce{MnO4-}$ to +II in $\ce{Mn^2+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{8H+ + MnO4- + 5e- -&amp;gt;~ Mn^2+ + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is the largest, and the least amount of $\ce{KMnO4}$ is needed for the reaction with a certain amount of alcohol. This makes oxidation in acidic medium the best choice for economic reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation in neutral medium is less efficient, as manganese(VII) is only reduced to manganese(IV), yielding insoluble $\ce{MnO2}$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product (carboxylic acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O + MnO4- + 3e- -&amp;gt;~ MnO2\downarrow + 4OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation in strongly alkaline medium is least efficient in terms of transferred electrons per mole permanganate, as the latter is only reduced to manganate(VI).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4- +e- -&amp;gt;~ MnO4^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_of_primary_alcohols_to_carboxylic_acids&quot;&gt;reaction&lt;/a&gt; is done under these conditions, a higher reaction speed and selectivity can be achieved, presumably because the alcohol will react more readily with $\ce{MnO4-}$ when deprotonated. Nevertheless, oxidation in acidic medium can be preferred for reasons stated above and substrates which are sensitive to strong bases.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-30T20:36:51.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15721" PostTypeId="2" ParentId="15709" CreationDate="2014-08-30T22:04:57.763" Score="8" Body="&lt;p&gt;A lot of questions have been posed on electrolysis and electrochemistry on this forum recently, with many recurring themes. I will explain electrolysis in depth in order for a true understanding to come out of this; if you're in a hurry then skip to the last three paragraphs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most important things about learning electrochemistry is the abstract property of voltage. Voltage can be considered a driving force, which denotes which direction charges are going to flow in a system. A high positive voltage means that charges most definitely wish to flow in that direction, and a high negative voltage means that the charges wish to flow strongly in the &lt;em&gt;opposite&lt;/em&gt; direction. If we really want charges to flow in the direction of negative voltage, we must supply energy to the system in order for that to happen—in other words, we're supplying energy to move a charged particle against an electric field (work must therefore be done to do this, as with mass in a gravitational field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it is with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;! The reaction corresponding to our notion of electrolysis is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2O -&amp;gt; O2 + 2H2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least two of the intermediate species in electrolysis are going to be &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, and so the energy requirement for this process takes into account the energy required to move an electron from &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; to the anode, and also to move an electron from the cathode to &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;.The voltage requirement for this process is  &lt;span class=&quot;math-container&quot;&gt;$\pu{-1.23V}$&lt;/span&gt;. In reality the voltage requirement is closer to  &lt;span class=&quot;math-container&quot;&gt;$\pu{-2V}$&lt;/span&gt; taking into account overpotential (the electrochemical equivalent to activation energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this negative voltage tells us that this process will not happen spontaneously, and we will need to supply a voltage of  &lt;span class=&quot;math-container&quot;&gt;$\pu{2V}$&lt;/span&gt; in order for this to happen. So far so good: we never observe electrolysis in a standard beaker of water just sitting in the lab!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question asked what happened to the &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; during electrolysis. &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt; is quite happy in a &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^7+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^4+}$&lt;/span&gt; oxidation state. Therefore in solution, the following reaction can happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Mn^7+ + 3e- -&amp;gt; Mn^4+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above is a simplification, &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^7+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^4+}$&lt;/span&gt; don't occur simply as isolated ions in solution. In our case, these oxidation states of manganese are tied up in molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{MnO4^- -&amp;gt; MnO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the above isn't balanced, so we ask ourselves what other likely molecules are going to be produced in order to balance our oxygens. It's common for these redox systems to produce &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;, so we add &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; on the right to balance the oxygens, and add &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; on the left so we have enough hydrogens. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{MnO4^- + 4 H+ -&amp;gt; MnO2 + 2 H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage requirement for this reaction is &lt;span class=&quot;math-container&quot;&gt;$\pu{+1.68 V}$&lt;/span&gt;, so this means that the electron transfer happens spontaneously. In other words, this reaction is more likely to occur than the electrolysis of water at &lt;em&gt;a less than neutral pH&lt;/em&gt;. If you start your electrolysis of water using distilled water (neutral pH implied) then the reduction of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^7+}$&lt;/span&gt; doesn't happen initially. It's only when the electrolysis starts and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; is generated that the reduction of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^7+}$&lt;/span&gt; can happen; this is then favoured over electrolysis on thermodynamic grounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last point, it's unlikely your copper electrode was burning, but these can get &lt;em&gt;very&lt;/em&gt; hot if the voltage is high enough and hence melt. Turn the voltage down as only &lt;span class=&quot;math-container&quot;&gt;$\pu{2V}$&lt;/span&gt; is needed for water electrolysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental hint: try using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; in combination and see if there is any change in your observations. Then try increasing the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; relative to &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; to see if you observe any changes which would suggest my theory. Note that other &lt;a href=&quot;https://web.archive.org/web/20151010161326/https://www.epa.gov/safewater/mdbp/pdf/alter/chapt_5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;manganese reactions (pdf via The Internet Archive)&lt;/a&gt; can still happen at lower voltages, which does complicate things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question directly, nothing additional is observed using &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; as it reacts instead of behaving as a charge carrier for electrolysis. A quality of a good electrolyte is that it does not interfere with your reaction of interest.&lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorUserId="4945" LastEditorDisplayName="user7232" LastEditDate="2019-10-06T13:49:23.340" LastActivityDate="2019-10-06T13:49:23.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15722" PostTypeId="1" AcceptedAnswerId="15732" CreationDate="2014-08-31T00:21:49.790" Score="4" ViewCount="687" Body="&lt;p&gt;When calculating bond energy or molecular binding energy, which value should I use, enthalpy or total energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, A+B=C  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be C's enthalpy of formation minus (A's+B's)enthalpies of formation, &#xA;or C's total energy minus (A's+B's) total energies?&lt;/p&gt;&#xA;" OwnerUserId="7158" LastActivityDate="2016-03-11T23:51:14.797" Title="When it comes to bond energy or molecular binding energy, do people mean enthalpy?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15723" PostTypeId="1" AcceptedAnswerId="15726" CreationDate="2014-08-31T00:30:44.667" Score="5" ViewCount="3371" Body="&lt;p&gt;Why is the oxidation of certain metals such as iron favorable from a thermodynamic standpoint? What is it about these metal oxides that makes them inherently more thermodynamically stable than their starting, elemental forms? The only reason I can imagine is that iron, being a d-block element, has a low electronegativity and cannot stabilize valence electron density as well as oxygen, the second most electronegative element on the table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is oxidation always thermodynamically favorable? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2020-10-19T09:08:53.540" Title="Why are metal oxides generally more stable than their corresponding elemental forms?" Tags="&lt;thermodynamics&gt;&lt;redox&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15724" PostTypeId="1" AcceptedAnswerId="15725" CreationDate="2014-08-31T01:56:36.610" Score="1" ViewCount="1111" Body="&lt;p&gt;If the points are non-polar, I don't see how making the eluent more polar will make them travel faster. Anybody care to explain?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-08-31T03:01:13.450" Title="TLC: Why is a more polar eluent more effective at displacing the points?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15725" PostTypeId="2" ParentId="15724" CreationDate="2014-08-31T03:01:13.450" Score="3" Body="&lt;p&gt;Most likely you are using &quot;normal phase&quot; TLC plates. The stationary phase is silica gel, which has free hydroxyl groups making it a polar surface. In TLC, there is a competition for the plate between the analytes and the eluent. As you make the eluent more polar, the eluent will interact more with the plate leaving the analytes to be carried with the mobile phase.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-08-31T03:01:13.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15726" PostTypeId="2" ParentId="15723" CreationDate="2014-08-31T04:25:29.787" Score="4" Body="&lt;p&gt;The oxidation of gold is not favorable.  For many other metals the metallic bonding is weaker than the bonding with oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the steps in oxidation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking metallic bonds (endothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionizing the metal atoms (endothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron affinity of oxygen (exo for the first, endo for the second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lattice energy of the oxide (exothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lattice energy is so high that the process is favorable.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-31T04:25:29.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15727" PostTypeId="1" AcceptedAnswerId="15730" CreationDate="2014-08-31T05:19:22.137" Score="-3" ViewCount="354" Body="&lt;p&gt;From very elementary chemistry I understand that atoms are mostly empty space, with a few tiny little objects whizzing around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must then be a vacuum in between these particles. I always thought that nature abhorred a vacuum, and rushed to fill it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't everything clump together in one little ball and get rid of the vacuum? Or are my premises wrong?&lt;/p&gt;&#xA;" OwnerDisplayName="user7642" LastEditorUserId="5456" LastEditDate="2014-08-31T06:14:01.357" LastActivityDate="2019-12-29T20:53:17.327" Title="How does the vacuum in atoms persist?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15728" PostTypeId="1" AcceptedAnswerId="15737" CreationDate="2014-08-31T05:36:32.497" Score="36" ViewCount="39472" Body="&lt;p&gt;What is the basic difference between aqueous and alcoholic $\ce{KOH}$? Why does alcoholic $\ce{KOH}$ prefer elimination whereas aqueous $\ce{KOH}$ prefers substitution?&lt;/p&gt;&#xA;" OwnerUserId="7510" LastEditorUserId="5026" LastEditDate="2018-03-13T16:37:36.133" LastActivityDate="2019-11-05T16:32:12.887" Title="Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;&lt;nucleophilic-substitution&gt;&lt;elimination&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="15730" PostTypeId="2" ParentId="15727" CreationDate="2014-08-31T05:54:10.573" Score="2" Body="&lt;p&gt;I think what you are asking is why electron do not fall in nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this you should have general idea of quantum mechanics. Because in the microscopic world nature follows quantum mechanics equations and not classical mechanics equations. Quantum mechanics equations include electromagnetic fields, and their solutions are stable and allow for the existence of atoms, which is what we experimentally observed to start with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is general idea of answer given to question physics stack exchange. You can see the detailed answer over in &lt;a href=&quot;https://physics.stackexchange.com/a/9418/46239&quot;&gt;Physics S.E.&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem1.com/acad/webtut/atomic/WhyTheElectron.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; Link will  explain diffrent theories which failed to explain why electron do not fall in nucleus.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-08-31T09:17:54.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15731" PostTypeId="2" ParentId="15728" CreationDate="2014-08-31T05:57:39.373" Score="8" Body="&lt;p&gt;$\ce{OH-}$ acts as a nucleophile. Reactions carried out in alcohol tend to be elimination reactions, and reactions carried out in water (aqueous) tend to be substitution reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If water were used as a solvent in an elimination reaction involving $\ce{KOH}$, the equilibrium would be shifted towards the reactants (water reacting with product), so substitution is favored.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="5026" LastEditDate="2018-03-13T16:41:39.967" LastActivityDate="2018-03-13T16:41:39.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15732" PostTypeId="2" ParentId="15722" CreationDate="2014-08-31T06:28:00.000" Score="4" Body="&lt;p&gt;When in doubt, consult &lt;a href=&quot;http://goldbook.iupac.org/index.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC Gold Book&lt;/a&gt; for definitions (italics below is mine).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/B00699.html&quot; rel=&quot;nofollow&quot;&gt;bond-dissociation energy&lt;/a&gt;, $D$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;enthalpy&lt;/em&gt; (per mole) required to break a given bond of some specific molecular entity by homolysis, e.g. for $\ce{CH4 -&amp;gt; CH3^{.} + H^{.}}$, symbolized as  $D(\ce{CH3−H})$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/B00701.html&quot; rel=&quot;nofollow&quot;&gt;bond energy (mean bond energy)&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average value of the gas-phase bond dissociation energies (usually at a temperature of  298 K) for all bonds of the same type within the same chemical species. The mean bond energy for methane, for example, is one-fourth the &lt;em&gt;enthalpy&lt;/em&gt; of reaction for: $$ \ce{CH4_{(g)} -&amp;gt; C_{(g)} + 4H_{(g)}} \, .$$ Tabulated bond energies are generally values of bond energies averaged over a number of selected typical chemical species containing that type of bond.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, yes, &lt;em&gt;conceptually&lt;/em&gt; you should use &lt;em&gt;enthalpy&lt;/em&gt;. Note, however, that the enthalpy of a molecular specie in gas phase at, say 298K, is calculated as follows &#xA;$$H_{298} = E_{\mathrm{e}} + E_\mathrm{ZPE} + H_{\mathrm{trans}} + H_{\mathrm{rot}} + H_{\mathrm{vib}} + RT \, ,$$&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E_{\mathrm{e}}$ is the electronic energy;&lt;/li&gt;&#xA;&lt;li&gt;$E_\mathrm{ZPE}$ is the zero point energy;&lt;/li&gt;&#xA;&lt;li&gt;$H_{\mathrm{trans}}$, $H_{\mathrm{rot}}$ and $H_{\mathrm{vib}}$ are the temperature-dependent contributions to enthalpy from translation, rotational and vibrational motion, respectively.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the problem (as Martin mentioned in comments) is that basically only the first term out of this five can be calculated accurately enough when the ultimate goal is to calculate energies (bond energies in our case) with the so-called &quot;chemical accuracy&quot; which is about $\pm 1$ kcal/mol.&#xA;For relatively large molecules harmonic frequencies and consequently $E_\mathrm{ZPE}$, as well as a rigid-rotor harmonic-oscillator approximation used to calculate thermal corrections, become inaccurate, and thus, the change of electronic energy should always be used as a reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="24052" LastEditDate="2016-03-11T23:51:14.797" LastActivityDate="2016-03-11T23:51:14.797" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15733" PostTypeId="2" ParentId="15723" CreationDate="2014-08-31T08:07:35.113" Score="1" Body="&lt;p&gt;When you think about it, metals tend to want to lose electrons to finish with full outer shells.  When they bond with oxygen, they do lose those electrons, ending in a more stable electron configuration.  These ions don't go off and violently react with other things because they're ionically attracted to the oxygens they've just given their electrons to.  So all their electrons are happily occupied, meaning they have no need to react and share them off.  Also, because of oxygen's high electronegativity, it is unlikely to give those electrons up to another element, causing a reaction.&lt;/p&gt;&#xA;" OwnerUserId="7643" LastActivityDate="2014-08-31T08:07:35.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15734" PostTypeId="1" CreationDate="2014-08-31T09:51:09.380" Score="1" ViewCount="593" Body="&lt;p&gt;We have a school experiment, where I must report on the following question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does salinity affect the solubility of $\ce{CO2}$ in saltwater? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we selected four different concentrations of salt (in the range of 1-4 %) in 100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely. We then used titration to figure out the amount of $\ce{CO2}$ (reacting with water to give weak carbonic acid) that is dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Will the amount of $\ce{CO2}$ that reacts with water depend upon the salt concentration?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will the amount of titrant used to figure out the amount of $\ce{CO2}$ dissolved also change depending on the salt concentrations?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7644" LastEditorUserId="3846" LastEditDate="2014-08-31T23:01:42.330" LastActivityDate="2016-01-02T16:55:12.660" Title="Does CO2 dissolution in different salt concentrations affects the amount of carbonic acid formed?" Tags="&lt;water&gt;&lt;solubility&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15735" PostTypeId="2" ParentId="6728" CreationDate="2014-08-31T12:08:35.210" Score="3" Body="&lt;p&gt;For the p &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;nofollow&quot;&gt;subshell&lt;/a&gt;, the &lt;a href=&quot;http://n.wikipedia.org/wiki/Azimuthal_quantum_number&quot; rel=&quot;nofollow&quot;&gt;azimuthal quantum number&lt;/a&gt;, which describes the angular momentum and shape of the orbitals, is $l=1$. For the associated &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_quantum_number&quot; rel=&quot;nofollow&quot;&gt;magnetic quantum number&lt;/a&gt;, discrete values in the range $-l, -l+1, ... 0 ..., l-1, l$ are allowed, so the number of different $m_l$ values is $2l+1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $l=1$, it follows that there are $2 \cdot 1+1=3$ p orbitals with the $m_l$ values of -1, 0 and +1, respectively. Each orbital can hold 2 electrons, so the p subshell can contain a maximum of 6 electrons. This also explains why there are 6 elements in the p block of each period. With increasing atomic number $Z$, one p electron is added per element in this block.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-08-31T12:08:35.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15736" PostTypeId="2" ParentId="15714" CreationDate="2014-08-31T12:42:03.663" Score="2" Body="&lt;p&gt;1)In short, you can't prepare exactly $1M$ solution of $\ce{NaOH}$. It is common to prepare roughly $1M$ solution and then titrate it to get an exact value of its molarity. For preparation of standards people usually avoid usage of $\ce{NaOH}$ and similar compounds that adsorb something from air. For exapmple, $\ce{NaHCO3}$ is a nice option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)$\ce{NaOH}$ granules adsorb water from air, forming a solution of the hydroxide in water. The driving force behind it is energy of solvation of $\ce{Na^+}$ and $\ce{OH^-}$ ions (as far as these words are applicable to concentrated solutions)&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-08-31T13:23:18.887" LastActivityDate="2014-08-31T13:23:18.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15737" PostTypeId="2" ParentId="15728" CreationDate="2014-08-31T13:00:41.070" Score="33" Body="&lt;p&gt;In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations. &lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="5026" LastEditDate="2018-03-13T16:41:00.233" LastActivityDate="2018-03-13T16:41:00.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15738" PostTypeId="2" ParentId="15690" CreationDate="2014-08-31T14:24:25.513" Score="8" Body="&lt;p&gt;Thinking about it some more made me realize how the problem with the bond lenghts might be resolved. Let's look at the orbital picture again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TcLE1.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still think, that the in-phase combination of the $\sigma^{∗}(\ce{C-Cl})$ orbital and the free-electron-pair-orbital $n(\ce{e-})$ will have to some extent - depending on the energetic difference between the interacting orbitals and their orbital overlap - an $\ce{C-Cl}$-antibonding character, thus weakening the blue $\ce{C-Cl}$ bond (a simplified view would be, that on interaction electron density flows from the free electron pair into antibonding $\sigma^{∗}(\ce{C-Cl})$ orbital and more electron density in an antibonding orbital mean a weakening of the corresponding bond). But my mistake was not to also consider the $\pi$-bonding interaction between the red $\ce{Cl}$ and $\ce{C}$ in the in-phase combination. This $\pi$-bonding interaction will of course strengthen the red $\ce{C-Cl}$ bond. So resulting from this one interaction pictured above there should be a bond-shortening of the red $\ce{C-Cl}$ bond and a bond-lengthening of the blue $\ce{C-Cl}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the anomeric effect works not only in one direction in DCM (or Chloroform). The blue $\ce{Cl}$ atom has free electron pairs, too. And there is also a $\sigma^{∗}(\ce{C-Cl})$ orbital for the red bond that can interact with one of those free electron pairs in exactly the same way described before. Now, this interaction will strengthen the blue $\ce{C-Cl}$ bond while weakening the red $\ce{C-Cl}$ bond. So, it counteracts the effects of the interaction presented above and the bond-shortening and bond-lengthening effects should cancel each other out to some extent. But the raising of the LUMO, which makes DCM so unreactive remains unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only thing that needs some thinking is: Do the $\pi$-bonding and the $\sigma$-antibonding in the in-phase combination have a comparable magnitude? This question I can't answer in any quantitative way. So I will only list some factors of influence here. The $\ce{C-Cl}$ bond is rather long. This means the $\pi$-overlap won't be optimal. And the  $n(\ce{e-})$ orbital will lie somewhat lower in energy than the $\sigma^{∗}(\ce{C-Cl})$ orbital thus weakening the covalency of the $\pi$-interaction. On the other hand, the same factors will also lower the amount of $\sigma^{∗}(\ce{C-Cl})$-character contained in the in-phase combination MO. All this leaves me with the gut feeling that the $\pi$-bonding might indeed be comparable to or even a little stronger than the $\sigma$-antibonding effect. And this would mean that the observed bond lengths are in accord with the orbital situation and the anomeric effect might very well be quite important for explaining the unreactiveness of DCM towards nucleophiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But the main question remains: Is the anomeric effect the main cause for the unreactiveness of DCM towards nucleophiles or is it the steric hinderance exerted by the second $\ce{Cl}$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe one could go about it by employing the van der Waals radii of a chlorine atom and a methyl group. According to Wikipedia, the radius of a chlorine atom is 175 pm while the radius of a carbon atom (without any hydrogen atoms around) is 170 pm (so it should be safe to say that the radius of a methyl group is higher than that). Here is a picture of the van der Waals surface of methyl chloride created with &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FLAeA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to support the view that the spacial requirements of a chlorine atom and a methyl group are rather similar (but I don't know how reliable the van der Waals surface calculations by Avogadro actually are).&#xA;If those lengths mirror the real spacial requirements of the groups in the context of a nucleophilic substitution reaction then a chlorine atom wouldn't exert a larger steric hinderance than a methyl group and the anomeric effect would clearly be the determining factor when explaining the unreactiveness of DCM. But I'm unsure whether resorting to van der Waals radii is justified.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-31T22:58:40.060" LastActivityDate="2014-08-31T22:58:40.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15739" PostTypeId="1" CreationDate="2014-08-31T14:44:23.870" Score="1" ViewCount="68" Body="&lt;p&gt;This is quote from OxyChem document on Hypochlorites:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steel tanks in caustic soda service become passivated with a&#xA;  &lt;strong&gt;gray-black film&lt;/strong&gt; forming on the metal. Thereafter, caustic soda at less&#xA;  than 130° F will dissolve very little iron from the tank. The&#xA;  protective film should be preserved whenever practical. Therefore,&#xA;  when sediment is rinsed from a tank, only the bottom should be washed.&#xA;  Caustic Soda solution should then be put into the tank as soon as the&#xA;  washing is finished. Otherwise a &lt;strong&gt;soft rust&lt;/strong&gt; will form and contaminate&#xA;  any future storage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot understand what can be this gray-black film chemical formula. Plus is he referring soft rust to common Iron (II) Oxide when it comes to O2 rich atmosphere. May be thats why he instructs to refill cleansed container as soon as possible?&lt;/p&gt;&#xA;" OwnerUserId="7583" LastActivityDate="2014-08-31T14:44:23.870" Title="Identifying probable compound using its colour &amp; reactants" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15742" PostTypeId="1" CreationDate="2014-08-31T16:01:17.660" Score="0" ViewCount="2800" Body="&lt;p&gt;At home I make experiments on salt (as it is easily available at home). &#xA;One day I came to know that on mixing salt and water I will get Hydrochloric Acid and Sodium Hydroxide. So I became really excited, but my excitement broke into small pieces because I don't know how to separate $\ce{HCl}$ and $\ce{NaOH}$ from the solution of $\ce{NaCl}$ and water.&lt;/p&gt;&#xA;" OwnerUserId="7646" LastEditorUserId="4945" LastEditDate="2014-08-31T17:20:02.473" LastActivityDate="2018-10-01T20:26:52.207" Title="How to separate HCl and NaOH from an aqueous solution of NaCl?" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15743" PostTypeId="2" ParentId="15690" CreationDate="2014-08-31T16:43:48.147" Score="1" Body="&lt;p&gt;The anomeric effect where σ(C−C) orbital overlaps with the σ∗(C−Cl) should be considerably less than the overlap of n(e−) and σ∗(C−Cl). Therefore by comparing the relative rates of substitution of DCM, and a primary alkyl chloride, whose alkyl group is similar in size as the chlorine atom, against MeCl should give us the answer. I would be thankful if someone could provide the data.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-08-31T16:43:48.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15744" PostTypeId="1" CreationDate="2014-08-31T16:52:55.807" Score="2" ViewCount="713" Body="&lt;p&gt;When we add tollens reagent (diamminesilver(I) cation, $\ce{[Ag(NH3)2]+}$) to glucose it forms a silvery mirror on the walls of the container. Since the sides are coated with actual Silver, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;is there any possible reagent/method to take the silver out from the sides of the test tube.?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;will it be in its pure form or will there be organic/inorganic  impurities?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7647" LastEditorUserId="4945" LastEditDate="2014-08-31T17:16:58.950" LastActivityDate="2014-09-01T02:44:52.500" Title="Extracting silver from Tollens test" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;silver&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15745" PostTypeId="1" AcceptedAnswerId="15748" CreationDate="2014-08-31T16:53:35.087" Score="0" ViewCount="1548" Body="&lt;p&gt;Normally phenols have the $\ce{OH}$ group at the 1 or 4 position. Can they be made at 2 or 3 positions, where nitro groups and $\ce{-CN}$ can attach? If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="5217" LastEditorUserId="189" LastEditDate="2014-08-31T18:05:27.963" LastActivityDate="2014-08-31T19:04:09.313" Title="Is it possible to make meta-phenol?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15746" PostTypeId="2" ParentId="15742" CreationDate="2014-08-31T18:15:16.527" Score="5" Body="&lt;p&gt;The solution can be electrolyzed to produce chlorine and sodium hydroxide solution.  The chlorine can be added to water to produce HCl and HOCl.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-08-31T18:15:16.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15748" PostTypeId="2" ParentId="15745" CreationDate="2014-08-31T18:39:20.837" Score="6" Body="&lt;p&gt;The hydroxyl groups in phenols are &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-directing. So, if you want to introduce a group in the &lt;em&gt;meta&lt;/em&gt; position with respect to a hydroxyl group you will have to &quot;overwrite&quot; the &lt;em&gt;ortho&lt;/em&gt;-/&lt;em&gt;para&lt;/em&gt;-preference. This can be done by starting out with &lt;em&gt;meta&lt;/em&gt;-directing group which can later be transformed into $\ce{OH}$ in the position where the hydroxyl group shall be. Then you only need to introduce the group you want to be in the &lt;em&gt;meta&lt;/em&gt;-position by a simple electrophilic aromatic substitution reaction and subsequently convert the &lt;em&gt;meta&lt;/em&gt;-directing group into $\ce{OH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example for this procedure would be to start out with a carbonyl group at the position of the $\ce{OH}$ group, then nitrate the aromatic ring in the &lt;em&gt;meta&lt;/em&gt; position and employ the Baeyer-Villiger oxidation to introduce a $\ce{OH}$ group:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q9ZCW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you start out with a nitro group at the position of the $\ce{OH}$ group, then introduce chlorine in the &lt;em&gt;meta&lt;/em&gt; position of the aromatic ring, reduce the nitro group to an amine, which is then coverted into an diazonium salt which in turn is then transformed into a $\ce{OH}$ group using a Schiemann-type reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3bwO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-08-31T19:04:09.313" LastActivityDate="2014-08-31T19:04:09.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15750" PostTypeId="1" AcceptedAnswerId="15753" CreationDate="2014-08-31T20:40:19.873" Score="1" ViewCount="190" Body="&lt;p&gt;I believe that electrolysis of salt water creates lye ($\ce{NaOH}$ - sodium hydroxide). In salt water, there's $\ce{Na+}$, $\ce{Cl-}$, $\ce{H+}$ and $\ce{OH-}$ as of ionts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at cathode, you get $\ce{Na2}$ and that should react with water into $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to test, if this is true - how do I do it with household items? The solution didn't react with aluminium.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-08-31T21:25:38.107" LastActivityDate="2014-08-31T22:44:11.727" Title="Home-made way to detect lye in solution" Tags="&lt;home-experiment&gt;&lt;analytical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15751" PostTypeId="1" CreationDate="2014-08-31T20:57:45.847" Score="1" ViewCount="133" Body="&lt;p&gt;There are two containers connected to each other but separated by a cork. If &lt;strong&gt;container(1)&lt;/strong&gt; has a gas $G_1$ at a volume $V_1$,  pressure $P_1$ and temperature $T_1$ and similarly in &lt;strong&gt;container(2)&lt;/strong&gt; there is a gas $G_2$ at volume $V_2$, pressure $P_2$ and temperature $T_2$. If the cork between the containers is removed and the temperature changed to $T_3$ what is the pressure of the mixture inside the system of containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook gives it as:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}=\frac{P_3V_3}{T_3}$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;where $V_3$ = $V_1$+$V_2$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;$P_3=\frac{T_3}{V_3}\left(\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}\right)$&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I don't understand how that comes out. Can we just add the constants(of combined gas law) like that? &lt;/p&gt;&#xA;" OwnerUserId="7577" LastActivityDate="2014-08-31T21:12:18.953" Title="Addition of combined gas law constant" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15752" PostTypeId="2" ParentId="15751" CreationDate="2014-08-31T21:12:18.953" Score="2" Body="&lt;p&gt;&lt;strong&gt;Caution:&lt;/strong&gt; This answer is exclusively from my perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moments after uploading this question I realized the answer myself. For those who strike upon this question. I 'think' the answer goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combined gas law states that : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{PV}{T}=nR$ ----------  [or $PV=nRT$ ideal gas equation] (where 'n' is the number of moles of the gas in consideration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}$ = $n_1R+n_2R$ = $R\left(n_1+n_2\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_1+n_2=n_3$ (NOTE: $n_3$ is not the number of moles of a single gas but the gases $G_1$ and $G_2$ combined)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be said as the volume gets added up. Therefore further referring to Avogadro's Law (of equal volume) we the latter equation yields the equation in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLEASE DO CORRECT THIS ANSWER IF IT IS FAULTY.&lt;/p&gt;&#xA;" OwnerUserId="7577" LastActivityDate="2014-08-31T21:12:18.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15753" PostTypeId="2" ParentId="15750" CreationDate="2014-08-31T22:44:11.727" Score="1" Body="&lt;p&gt;The easiest way to measure the pH of a solution is to use an acid-base indicator. Limited to household supplies only, I would suggest a solution made from &lt;a href=&quot;http://en.wikipedia.org/wiki/Red_cabbage&quot; rel=&quot;nofollow&quot;&gt;red cabbage&lt;/a&gt;, which should turn blue-greenish in the alkaline region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ions you actually get from your reaction is a higher concentration of $\ce{Na+}$ and $\ce{OH-}$, due to the fact that you are removing $\ce{H2}$ and $\ce{Cl2}$ (from $\ce{H+}$ and $\ce{Cl-}$ ions respectively) via the performed electrolysis. Under no circumstances would you generate a $\ce{Na2}$ species, you would rather write it as $\ce{2Na}$. Also, in aqueous solutions the $\ce{NaOH}$ most certainly dissociates. So speaking of the formation of solid $\ce{NaOH}$ from $\ce{Na}$ in water is quite a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: &lt;strong&gt;Since the salt is already dissolved when you start the electrolysis, all you get is a change in concentration of the ions involved. The thereby induced change in pH can be &lt;s&gt;measured&lt;/s&gt; observed using an acid-base indicator such as the juice of the red cabbage.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-08-31T22:44:11.727" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15754" PostTypeId="1" AcceptedAnswerId="15757" CreationDate="2014-08-31T23:11:49.010" Score="14" ViewCount="6334" Body="&lt;p&gt;Why does hydrogen peroxide exhibit a dihedral angle of &lt;span class=&quot;math-container&quot;&gt;$111.5^\circ$&lt;/span&gt; in the gaseous state? And a dihedral angle of &lt;span class=&quot;math-container&quot;&gt;$90.2^\circ$&lt;/span&gt; in the crystalline phase?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VquYQ.png&quot; alt=&quot;Dihedral angle of gaseous and crystalline HOOH&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I know that in general, there is likely to be more hydrogen bonding in the crystalline phase than in the vapor phase.&lt;/p&gt;&#xA;&lt;p&gt;But what's up with 1) the bond angles themselves in the first place and 2) the bond angle expansion in the gas phase? How can one rationalize them?&lt;/p&gt;&#xA;&lt;p&gt;I know that there exist lone pairs on the oxygens and lone-pair/lone-pair repulsion should probably be minimized, but what it is about these conformers that minimizes energy? I'm guessing that we'd also want to minimize eclipsed and &lt;em&gt;gauche&lt;/em&gt; interactions ... Newman projection time?&lt;/p&gt;&#xA;&lt;p&gt;It seems that there is only one Newman projection possible for &lt;span class=&quot;math-container&quot;&gt;$\ce{HOOH}$&lt;/span&gt;; &lt;em&gt;i.e.&lt;/em&gt; we'll have the inevitable lone-pair/lone-pair interaction, and the lone-pair/hydrogen interaction. I guess all we &lt;em&gt;can&lt;/em&gt; do is minimize these interactions; &lt;em&gt;i.e.&lt;/em&gt; the energy-minimized conformer is probably not the one with them all eclipsed. However, why does it seem that the energy-minimized conformer for vapor &lt;span class=&quot;math-container&quot;&gt;$\ce{HOOH}$&lt;/span&gt; is one that puts the lone-pairs so close to each other in terms of dihedral angles?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4AoH.jpg&quot; alt=&quot;My explanation&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="61578" LastEditDate="2020-06-11T06:34:36.233" LastActivityDate="2020-06-11T06:45:32.787" Title="Dihedral angle of gaseous and crystalline HOOH" Tags="&lt;inorganic-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;crystal-structure&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15756" PostTypeId="2" ParentId="15680" CreationDate="2014-08-31T23:46:35.330" Score="2" Body="&lt;h2&gt;Find a function of length depending on mass&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Conversion factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2.54 ~\mathrm{\frac{cm}{in}}$&lt;/li&gt;&#xA;&lt;li&gt;$454 ~\mathrm{\frac{g}{lb}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Constants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wire diameter $d = 0.129~\mathrm{cm}$&lt;/li&gt;&#xA;&lt;li&gt;Density of copper $\rho = 8.92~\mathrm{g\, cm^{-3}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, our function $l(m)$ for the wire length from the wire mass in pounds reads:&#xA;$$ l(m) = m \cdot 454~\mathrm{\frac{g}{lb}} \cdot \frac{1}{\pi d^2 \rho}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Evaluate the function&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since I'm too lazy to evaluate something like this by hand (and since I somehow had to record this thought process and did not want to waste the processed remnants of a dead tree) there is a small python script available to do this for you, which you can find &lt;a href=&quot;https://gist.github.com/tschoppi/f2f2e372f080478d3da4&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and which, if you run it using the command &lt;code&gt;python3 wires.py&lt;/code&gt;, will give you the following output:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wire length of 1.00 lb of 16-gauge wire: 9.71 m&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Bonus: You can run the command as &lt;code&gt;python3 -i wires.py&lt;/code&gt; and will then be dropped into an interactive python shell, where you can call the &lt;code&gt;wirelength&lt;/code&gt; function as many times as you wish using different masses for the wire.)&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-08-31T23:46:35.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15757" PostTypeId="2" ParentId="15754" CreationDate="2014-09-01T01:01:38.307" Score="14" Body="&lt;p&gt;You've already answered your own question - hydrogen bonding.&lt;/p&gt;&#xA;&lt;p&gt;First, don't think of it as to why is the gas state different than solid-state.  Rather, think about why is the solid-state different than the gas state. In the gas state, there is very little intermolecular interaction (very, very few species do not follow the ideal gas law to many significant digits). The gas state geometry is close to what is expected for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; tetrahedral geometry and may be considered normal. The link below is a paper discussing the gas phase microwave spectroscopy of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; in far greater detail than I can hope to here. The description of the torsional rotation energies was not trivial.&lt;/p&gt;&#xA;&lt;p&gt;Internal‐Rotation in Hydrogen Peroxide: The Far‐Infrared Spectrum and the Determination of the Hindering Potential by&#xA;Robert H. Hunt, Robert A. Leacock, C. Wilbur Peters, Karl T. Hecht, &lt;em&gt;The Journal of Chemical Physics&lt;/em&gt; &lt;strong&gt;1965&lt;/strong&gt;, &lt;em&gt;42&lt;/em&gt;, 1931-1946, &lt;a href=&quot;https://doi.org/10.1063/1.1696228&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1063/1.1696228&lt;/a&gt;,&#xA;&lt;a href=&quot;http://deepblue.lib.umich.edu/bitstream/handle/2027.42/71115/JCPSA6-42-6-1931-1.pdf?sequence=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;mirror&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;When condensed to the solid-state, molecules get close enough to allow intermolecular forces (in this case hydrogen bonding to a much greater extent than van der Waals). Additionally, the energies of the crystal lattice must be considered.  In this case, th&lt;/p&gt;&#xA;" OwnerUserId="7651" LastEditorUserId="61578" LastEditDate="2020-06-11T06:45:32.787" LastActivityDate="2020-06-11T06:45:32.787" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15758" PostTypeId="1" AcceptedAnswerId="15762" CreationDate="2014-09-01T01:36:41.283" Score="4" ViewCount="13551" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;For the following reaction&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{W(s) + 4 Br(g) &amp;lt;=&amp;gt; WBr4(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;at &lt;span class=&quot;math-container&quot;&gt;$\pu{900 K},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$K_p = 100$&lt;/span&gt; and at &lt;span class=&quot;math-container&quot;&gt;$\pu{2800 K}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{p} = 5.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Determine if the reaction is endothermic or exothermic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At higher temperatures, the equilibrium has shifted towards the reactants since &lt;span class=&quot;math-container&quot;&gt;$K_p$&lt;/span&gt; is significantly smaller. However, how do we determine if a reaction is endothermic or exothermic from this information?&lt;/p&gt;&#xA;&lt;p&gt;If the reaction was endothermic, an increase in temperature would push the equilibrium to products to counteract the increase in heat. However, the equilibrium has been pushed to the reactants thus signifying the reaction is exothermic. But this assertion is on the assumption that the forward reaction is exothermic.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-04-10T07:34:20.810" Title="Determining if reaction is endothermic or exothermic using equilbrium constant" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15759" PostTypeId="2" ParentId="15744" CreationDate="2014-09-01T01:48:13.530" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Silver's pretty soft—it's possible to just scrape it off with a metal spatula or something. Also, not all of the silver deposits on the glass. You might be able to adjust the conditions of the reaction to make more silver precipitate into the liquid where it's easier to get at. You can, of course, redissolve the silver in nitric acid or the like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the byproducts of the reaction are soluble, it should be pretty pure, depending on how pure the starting materials are. If other easily reducible metal cations are present, you might get some impurities from that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a side-note, this isn't a particularly efficient way of getting elemental silver. The typical films produced in a test tube probably contain a couple hundred milligrams at most. Even a whole undergrad lab's worth of tubes is probably not worth recovering.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-01T01:48:13.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15760" PostTypeId="1" CreationDate="2014-09-01T01:49:49.113" Score="-3" ViewCount="332" Body="&lt;p&gt;I found that reaction: HNO3 + 4.1633363423443E-17 H3O = NO2{+} + OH{-}&#xA;Please help me understand &quot; 4.1633363423443E-17&quot; term.&lt;/p&gt;&#xA;" OwnerUserId="7654" LastActivityDate="2014-09-01T02:04:52.113" Title="Protonation of nitric acid" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-09-01T16:08:35.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="15761" PostTypeId="2" ParentId="15760" CreationDate="2014-09-01T01:57:27.370" Score="7" Body="&lt;p&gt;It seems to me like you were asked to balance a chemical equation and you used some computer program to do it for you instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.1633363423443E-17 is the result of the limitations of floating point arithmetic (ie that computers only have finite precision). The real answer should be 0 (which you could have figure out by balancing it yourself). &lt;/p&gt;&#xA;" OwnerUserId="5220" LastEditorUserId="5220" LastEditDate="2014-09-01T02:04:52.113" LastActivityDate="2014-09-01T02:04:52.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15762" PostTypeId="2" ParentId="15758" CreationDate="2014-09-01T02:02:14.143" Score="2" Body="&lt;p&gt;Think about it this way.  In an exothermic reaction that is at equilibrium the forward reaction has a lower activation energy.  So, if you increase T the forward reaction rate increases but the reverse rate increases even more.  That's because the rates of reactions with higher activation energy are more sensitive to temperature.  With the reverse rate higher more reactants will be made until the concentration reactants reaches a point where the rate of the forward reaction equals the reverse rate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an endothermic reaction increasing T increases the forward rate more than the reverse, producing more product until the increased product concentration causes the reverse rate to equal the forward rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant can be thought of as a ratio of products to reactants needed to equalize the rates of two reactions with different activation energies.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-01T02:02:14.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15763" PostTypeId="2" ParentId="15758" CreationDate="2014-09-01T02:04:14.247" Score="1" Body="&lt;p&gt;I think you're looking for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;nofollow&quot;&gt;van't Hoff equation&lt;/a&gt;. Since you have the equilibrium constants for two temperatures you can integrate the equation and get $\Delta H^{0}$ from it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d} \ln K}{\mathrm{d} T} &amp;amp;= \frac{\Delta H^{0}}{R T^{2}} \\&#xA;  \int\limits_{\ln K(T_{0})}^{\ln K(T_{1})}\mathrm{d} \ln K &amp;amp;= \frac{\Delta H^{0}}{R} \int\limits_{T_{0}}^{T_{1}} \frac{1}{T^{2}} \mathrm{d} T \\&#xA;  \ln \frac{K(T_{1})}{K(T_{0})} &amp;amp;= \frac{\Delta H^{0}}{R} \left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right) \\&#xA;  \Delta H^{0} &amp;amp;= \frac{R \ln \frac{K(T_{1})}{K(T_{0})}}{\left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which will enable you to decide whether the reaction is exothermic or endothermic.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-01T02:04:14.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15765" PostTypeId="2" ParentId="15744" CreationDate="2014-09-01T02:44:52.500" Score="-1" Body="&lt;p&gt;1) Silver is pretty easily dissolved in concentrated nitric acid (attention, nitrogen oxides fumes). The resulting solution may be gently evaporated to get rid of excess of nitric acid and silver nitrate may be then reduced with zinc and then washed from excess zinc with diluted sulfuric acid. Alternatively, silver nitrate may be used as is for similar tests in future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)As Michael noted, it is pretty pure.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-01T02:44:52.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15766" PostTypeId="1" CreationDate="2014-09-01T03:36:31.620" Score="3" ViewCount="30" Body="&lt;p&gt;So I was just exploring some ideas on my own and am having trouble analytically solving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $D(n,m,T)$ be the time it takes to dissolve an n x n x n cube of concrete in m molar HcL at a temperature of T.  Find $D(n,m,T)$. Over the temperature range of $32 ^o &amp;lt; T &amp;lt; 273^o$ in fahrenheit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically how does mechanism of dissolving work? Intuitively I understand the exchange of $\ce{H+}$ and $\ce{OH-}$ but still don't understand how to use this to generate said function.&lt;/p&gt;&#xA;" OwnerUserId="7657" LastEditorUserId="1878" LastEditDate="2014-09-01T03:41:26.720" LastActivityDate="2014-09-01T03:41:26.720" Title="Dissolving Time Function for Concrete" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15767" PostTypeId="2" ParentId="15485" CreationDate="2014-09-01T04:29:56.990" Score="3" Body="&lt;p&gt;As you said there are many different kind of defects in crystals. There are basically two different methods to find the Crystallographic defect.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mathematical classification methods&lt;/li&gt;&#xA;&lt;li&gt;Computer simulation methods&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mathematical classification methods:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A successful mathematical classification method for physical lattice defects, which works not only with the theory of dislocations and other defects in crystals but also, e.g., for disclinations in liquid crystals and for excitations in superfluid $\ce{^{3}He}$, is the topological homotopy theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Computer simulation methods:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density-functional theory, classical molecular dynamics and kinetic Monte Carlo simulations are widely used to study the properties of defects in solids with computer simulations. Simulating jamming of hard spheres of different sizes and/or in containers with non-commeasurable sizes using the Lubachevsky-Stillinger algorithm can be an effective technique for demonstrating some types of crystallographic defects.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystallographic_defect&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-01T04:29:56.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15768" PostTypeId="1" CreationDate="2014-09-01T05:01:59.027" Score="0" ViewCount="99" Body="&lt;p&gt;I want to know other than SEM is there any other technique or instrument to detect the surface area of particle esp eletrochemical way.&lt;/p&gt;&#xA;" OwnerUserId="7660" LastActivityDate="2014-09-01T05:38:54.397" Title="Is there way to detect surface area of nanoparticle by electrochemical technique?" Tags="&lt;electrochemistry&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15769" PostTypeId="2" ParentId="15768" CreationDate="2014-09-01T05:14:43.997" Score="2" Body="&lt;p&gt;Dynamic light scattering is commonly used to measure nanoparticle size: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_light_scattering&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dynamic_light_scattering&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you can measure this sort of thing electrochemically depends a lot on the type of nanoparticle. Conductive nanoparticles, you might be able to estimate the surface area by measuring adsorptive behaviours of various things, like CO for Pt and Pd: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1388248113000933&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/S1388248113000933&lt;/a&gt; Non-conductive particles might be measured by how they impede diffusion to a surface. Electrochemical techniques could suffer from problems related to the shapes of the nanoparticles, as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most straightforward ways of examining the surface features of nanoparticles would be electron microscopy or AFM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further details of your application, it's difficult to recommend a specific technique. If the nanoparticles are being used as an electrode, the apparent electrochemical surface area may be all you care about, even though this is not necessarily the same as the geometric surface area.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-09-01T05:38:54.397" LastActivityDate="2014-09-01T05:38:54.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15770" PostTypeId="1" AcceptedAnswerId="15771" CreationDate="2014-09-01T06:02:57.700" Score="2" ViewCount="145" Body="&lt;p&gt;The standard way of representing molecules with letters like $\ce{C}$ for Carbon, $\ce{O}$ for Oxygen etc.; and a lot of lines and double lines like a graph from Graph theory where each letter is a 'vertex', the lines like edges. I forget what this chemical notation system is called. Yet it is 2-dimensional. The formula for '$\ce{H2O}$' ,water doesn't indicate the 'angle' of 'separation' between the two Hydrogen atoms that are bonded with the Oxygen atom. Similarly a formula for some molecule like a protein might not indicate all the 'involved angles' in 3-dimensional space that 'allow' it to be folded in a precise way. So does the 2-dimensional notation system for molecules miss details of how a molecule is twisted in 3-D space?  Also can two distinct molecules 'look' the same in 2-D notation?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="5456" LastEditDate="2014-09-01T06:10:56.857" LastActivityDate="2014-09-01T06:15:29.157" Title="About the way chemicals and molecules are represented" Tags="&lt;molecules&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15771" PostTypeId="2" ParentId="15770" CreationDate="2014-09-01T06:15:29.157" Score="3" Body="&lt;p&gt;Below is 2 types of sugars; glucose and mannose.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nvGfN.png&quot; alt=&quot;Glucose an mannose&quot;&gt;&lt;br&gt;&#xA;If you know how to interpret the lines, then you can determine which one is which.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-01T06:15:29.157" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15772" PostTypeId="1" AcceptedAnswerId="17096" CreationDate="2014-09-01T08:22:27.470" Score="11" ViewCount="40578" Body="&lt;p&gt;I'm just learning about significant figures (sig figs) in my chemistry class, and I'm confused about the rules. An example to my problem: if I had the following expression:&#xA;$$\left ( \frac{2.378 - 1.2}{1.03} \right )$$&#xA;I've thought of 2 ways to approach this. Which one is correct (or is neither correct)?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Approach 1:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Round each intermediate operation as you go. This is what my textbook does for its examples, but it itself suggests &quot;Approach 2&quot; below. I'm just listing this approach here as an option since addition/subtraction and multiplication/division have their own rules for rounding (hence, maybe do intermediate-operation rounding).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;1) Round the subtraction part.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \frac{1.2}{1.03} \right )$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2) Do the division and round.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$1.2\,\,\, (2\ \mathrm{sig\ figs})$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Approach 2:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In actual practice, the book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a series of calculations, carry the extra digits through to the final result, THEN round.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it means this:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;1) Do the subtraction, but keep the *exact* number while noting that the rounded number was supposed to have 2 sig figs.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\left ( \frac{1.178}{1.03} \right )$$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2) Do the division with the exact number and the denominator, and since this step produces the final result, NOW you round, rounding to 2 sig figs per the note in Step 1.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$1.1\,\,\,(\text{2 sig figs})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that these two approaches produce 2 different results. So, to repeat, which one is correct (or is it neither)? The textbook doesn't provide an example like this. I found two different websites, but they too conflict with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm&lt;/a&gt;&#xA;(From the solution to problem #3, I should use &quot;Approach 1&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fabice.com/misc/significant_figures.html&quot; rel=&quot;noreferrer&quot;&gt;http://fabice.com/misc/significant_figures.html&lt;/a&gt;&#xA;(It quotes, &quot;In a composed operation, intermediate results are not rounded.&quot; So that means I should use &quot;Approach 2?&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the other questions on this forum, this question seems silly, but I find the concept of sig figs important if I want to understand chemistry at all.&lt;/p&gt;&#xA;" OwnerUserId="7661" LastEditorUserId="16683" LastEditDate="2017-09-06T10:54:49.407" LastActivityDate="2017-09-06T10:54:49.407" Title="How do I round a calculation involving both addition/subtraction and multiplication/division to the correct number of significant figures?" Tags="&lt;numerical-analysis&gt;&lt;significant-figures&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15773" PostTypeId="2" ParentId="15690" CreationDate="2014-09-01T10:39:41.947" Score="22" Body="&lt;h1&gt;Introduction (and Abstract, TLDR)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Analysis of Molecular Orbitals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lr}\hline&#xA;\text{Compound} &amp;amp; \mathbf{d}(\ce{C-Cl})\\\hline&#xA;\ce{ClCH3}      &amp;amp;  1.797\\&#xA;\ce{Cl2CH2}     &amp;amp;  1.786\\&#xA;\ce{Cl3CH}      &amp;amp;  1.783\\\hline&#xA;\ce{ClCH2CH3}   &amp;amp;  1.797\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X9micm.jpg&quot; alt=&quot;lumo chloromethane&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfKZ9m.jpg&quot; alt=&quot;lumo dichloromethane&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tkuG9m.jpg&quot; alt=&quot;lumo chloroform&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pxpOQm.jpg&quot; alt=&quot;lumo chloroethane&quot;&gt;&lt;br&gt;&#xA;In the  mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted.&lt;br&gt;&#xA;The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond. &#xA;\begin{array}{lrr}\hline&#xA;\text{Compound} &amp;amp;\sigma-\ce{C-Cl} &amp;amp; \sigma^*-\ce{C-Cl}\\\hline&#xA;\ce{ClCH3}      &amp;amp;  45\ce{C}(21s79p) 55\ce{Cl}(14s85p)  &#xA;                &amp;amp;  55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\&#xA;\ce{Cl2CH2}     &amp;amp;  46\ce{C}(22s77p) 54\ce{Cl}(14s85p)  &#xA;                &amp;amp;  54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\&#xA;\ce{Cl3CH}      &amp;amp;  48\ce{C}(24s76p) 52\ce{Cl}(14s86p)  &#xA;                &amp;amp;  52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline&#xA;\ce{ClCH2CH3}   &amp;amp;  44\ce{C}(19s81p) 56\ce{Cl}(14s85p)  &#xA;                &amp;amp;  56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline&#xA;\end{array}&#xA;As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well.&lt;br&gt;&#xA;In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the &quot;backside&quot; of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon.&lt;br&gt;&#xA;In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2ITVDm.jpg&quot; alt=&quot;localised orbitals chloromethane&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0Qf4m.jpg&quot; alt=&quot;localised orbitals dichloromethane&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmgBZm.jpg&quot; alt=&quot;localised orbitals chloroform&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GlHxGm.jpg&quot; alt=&quot;localised orbitals chloroethane&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;(Note that in chloroform two lone pair orbitals are shown.)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule#Bond_lengths&quot; rel=&quot;noreferrer&quot;&gt;Bent's Rule&lt;/a&gt; (compare also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/15671/4945&quot;&gt;Utility of Bent&amp;#39;s Rule - What can Bent&amp;#39;s rule explain that other qualitative considerations cannot?&lt;/a&gt;). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-09-01T10:39:41.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15775" PostTypeId="1" CreationDate="2014-09-01T11:15:57.940" Score="1" ViewCount="299" Body="&lt;p&gt;Is it possible to make organic acids concentrated like inorganic acids like concentrated sulfuric acid and similar mineral acids?&lt;/p&gt;&#xA;" OwnerUserId="7603" LastEditorUserId="41328" LastEditDate="2019-08-04T07:07:24.627" LastActivityDate="2019-08-04T07:07:24.627" Title="Can organic acids be concentrated and used like concentrated sulfuric acid and similar mineral acids?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15776" PostTypeId="1" CreationDate="2014-09-01T11:19:25.200" Score="1" ViewCount="183" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_aluminate&quot; rel=&quot;nofollow&quot;&gt;sodium aluminate&lt;/a&gt; is formed when aluminium or aluminium hydroxide reacts with water solution of lye. After drying the water you'll end up with white crystals - sodium aluminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it's possible to get pure Na out of this compound - by heating or a redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-09-01T11:19:25.200" Title="How to get sodium from sodium aluminate?" Tags="&lt;redox&gt;&lt;home-experiment&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15778" PostTypeId="1" AcceptedAnswerId="15788" CreationDate="2014-09-01T11:53:11.300" Score="4" ViewCount="3180" Body="&lt;p&gt;When reading up on electrophilic aromatic substitution, most resources refer to electrophiles as positively-charged species, &quot;in general&quot; or &quot;mostly&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are non-positively charged electrophiles? Are they simply neutral atoms/molecules more willing to gain electrons due to having less grasp on their own electrons? Are there any, as strange as it sounds, negatively charged electrophiles?&lt;/p&gt;&#xA;" OwnerUserId="5217" LastActivityDate="2015-01-11T01:37:21.457" Title="Are there any negatively charged electrophiles, or simply neutral and positive?" Tags="&lt;organic-chemistry&gt;&lt;electrons&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15779" PostTypeId="1" CreationDate="2014-09-01T12:22:58.580" Score="6" ViewCount="23649" Body="&lt;p&gt;I used normal paper from my notebook to perform filtration. The filtration is 100% perfect, but takes incredibly long to happen, because the paper is too dense.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sgzGY.jpg&quot; alt=&quot;filtration isung notebook paper&quot;&gt;&lt;br&gt;&#xA;I'm not planning to buy filtration papers. They're cheap, but I don't filter something every day, so they'd be laying around useless. I would prefer to be able to use some household item as a filtration paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea I had was to apply some substance on normal writing paper to make it softer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tissues which are very soft, but these are too soft and will leave paper chunks in my filtrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any idea what should I use for filter paper, or how could I alter normal paper to suit my needs?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2021-05-26T09:18:35.930" Title="Homemade filter paper" Tags="&lt;home-experiment&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15781" PostTypeId="2" ParentId="15775" CreationDate="2014-09-01T12:33:23.190" Score="6" Body="&lt;p&gt;They can. Vinegar is typically 4-18% acetic acid by mass, but of course concentrated acetic acid also exists with a concentration &gt;90% by mass. It is corrosive and flammable, but it surely exists.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="73" LastEditDate="2014-09-05T20:17:41.937" LastActivityDate="2014-09-05T20:17:41.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15783" PostTypeId="1" AcceptedAnswerId="15786" CreationDate="2014-09-01T13:27:10.240" Score="6" ViewCount="1628" Body="&lt;p&gt;How does benzene react with $\ce{NaNO2/HCl}$? Will $\ce{PhNO}$ be obtained? I &lt;a href=&quot;http://www.nature.com/srep/2013/131211/srep03481/full/srep03481.html?WT.ec_id=SREP-639%2C638-20140102&quot; rel=&quot;nofollow&quot;&gt;read&lt;/a&gt; that pyridine can be produced under special conditions, but what will happen in some less exotic conditions?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-09-04T07:37:18.930" LastActivityDate="2014-09-04T07:37:18.930" Title="Nitrosation of benzene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15784" PostTypeId="1" CreationDate="2014-09-01T13:39:57.207" Score="3" ViewCount="2098" Body="&lt;p&gt;I know basic cell representations of galvanic cells and can identify the anode/cathode. But here's a weird representation that I encountered and couldn't understand it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Pt | H2(g)}(\pu{1.0 atm})\ce{ | HX || KCl(1M) | Hg2Cl2(s) | Hg | Pt}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this actually translate into?&lt;/p&gt;&#xA;" OwnerUserId="4995" LastEditorUserId="72973" LastEditDate="2020-04-14T13:31:47.297" LastActivityDate="2020-04-14T13:32:45.837" Title="Cell Representation of a galvanic cell:" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15786" PostTypeId="2" ParentId="15783" CreationDate="2014-09-01T14:12:59.830" Score="9" Body="&lt;p&gt;As you noted, the link you provided was to a reaction run under rather exotic conditions (low-temperature plasma). Under more everyday laboratory conditions, for example, in solution, activated aromatic rings can react with $\ce{NO^+}$ as shown below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S7QwN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if there is a ring-activating amino substituent present, then the $\ce{NO^+}$ will react with the nitrogen to produce the synthetically important diazonium ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wfakm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-01T14:18:43.780" LastActivityDate="2014-09-01T14:18:43.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15787" PostTypeId="1" AcceptedAnswerId="16117" CreationDate="2014-09-01T15:18:47.697" Score="8" ViewCount="461" Body="&lt;p&gt;I'm continuing with my electrolysis experiments. Over a long time, I seek a &lt;a href=&quot;https://chemistry.stackexchange.com/q/13881/1677&quot;&gt;perfect electrodes&lt;/a&gt;. It seems, that when you pick the right anode and cathode, it might be a lot cheaper than universal super-electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cathodes are usually OK, since (+) ions do not hurt metals. However, $\ce{O-}$ and $\ce{Cl-}$ are both very agresive and will, sooner or later, destroy any anode I found so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My latest experiment was Zinc. Zinc is often used as non-corrosive layer added on iron via electro-plating. So I used it as an anode in $\ce{NaCL}$ aqueous solution. The anode I used was a piece of iron plated with zinc layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scary jelly is forming on it. I also do not see any bubbles at all coming from the electrode - which mean that both chloride and oxygen ions are reacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mt7y7.jpg&quot; alt=&quot;jelly on zinc anode&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://i.stack.imgur.com/4Eg8e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;anode is consumed&lt;/a&gt; during the process - but what is produced? The jelly does not burn. When heated intensively, it bubbles and smells like chlorine. Does it decompose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tQ4Dt.jpg&quot; alt=&quot;jelly formed on zinc anode now collected together&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orange colour is caused by the iron centre of the anode being oxidized. I tried other pieces of zinc I could find - they all produced this jelly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, that solution does contain following additional compounds formed from salt and water:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{NaOH}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HCl}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cl2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-09T18:52:47.053" Title="Scary jelly forming on zinc anodes" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15788" PostTypeId="2" ParentId="15778" CreationDate="2014-09-01T15:30:49.980" Score="6" Body="&lt;p&gt;Yes, there are many neutral electrophiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One large class are the compounds with leaving groups. Some that you have likely encountered are: acids (HCl, HBr, sulfuric acid, etc) which are electrophilic at the hydrogen; unhindered alkyl halides/sulfonates (e.g., methyl iodide for SN2 reaction); dihalides (e.g., Br2 for electrophilic addition to an alkene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KE200.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second large class are compounds with pi bonds to an electronegative atom. Most importantly this class includes the carbonyls (aldehydes, ketones, esters, etc). Michael acceptors fall into this class as well, which have an extended pi system containing an electronegative atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VstDT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are covering electrophilic aromatic substitution, soon you may learn about nucleophilic aromatic substitution, where a highly electron poor aromatic ring (e.g., having multiple nitro groups) bearing a leaving group can be attacked by a strong nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/baaQh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all of these cases, although the compounds are neutral, there is a partial positive charge at the electrophilic site explained by a polarized bond (acids, alkyl halides), resonance (carbonyl), or spontaneously (dihalides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutral electrophiles are less reactive than their positively charged counterparts, which means that the nucleophiles are typically stronger for a reaction to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for negatively charged electrophiles, I can think of only a couple of cases that could be considered that way. For example, in the reduction of a carboxylic acid with lithium aluminum hydride, the first step is deprotonation of the acid to give a  carboxylate. LiAlH4 is very reactive and adds a hydride to the carboxylate. That could be seen as nucleophilic attack on a negatively charged electrophile, and many people will draw it that way. In reality though, there is something stabilizing the negative charge, either the lithium or aluminum. So it's not really a negatively charged carboxylate that's being attacked. I don't think that there are really any negatively charged electrophiles, but someone may have an example I haven't seen or thought of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEYK2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-09-01T15:50:11.807" LastActivityDate="2014-09-01T15:50:11.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15789" PostTypeId="1" AcceptedAnswerId="15790" CreationDate="2014-09-01T16:00:17.260" Score="3" ViewCount="7617" Body="&lt;p&gt;Here is the excerpt of my lab manual:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Simple distillation is more useful for purifying a liquid that contains either a non-volatile impurity or small amounts of higher or lower boiling impurities. Fractional distillation allows for several condensation-condensation cycles in a single operation. It can be used to separate liquids with comparable volatilities and to purify liquids that contain relatively large amounts of volatile impurities.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does simple distillation put emphasis on small amounts and fractional distillation, on large volumes?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-09-01T19:37:02.557" Title="Distillation: Simple vs Fractional" Tags="&lt;organic-chemistry&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15790" PostTypeId="2" ParentId="15789" CreationDate="2014-09-01T16:19:51.220" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, why does simple distillation put emphasis on small amounts and&#xA;  fractional distillation, on large volumes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not small and large volumes of material being distilled, but rather small and large volumes of impurities present in the material to be distilled.  Both simple and fractional distillations can be used to distill large quantities of material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency of a distillation column is measured in &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Theoretical_plate&quot; rel=&quot;nofollow&quot;&gt;height-equivalent theoretical plates&lt;/a&gt;&quot; (HETP), the more plates the more efficient the column, the more efficient the column the more likely that it will be able to separate two closely boiling materials.  If you have a small amount of a lower boiling impurity in your solution and you want to separate it from a large amount of the desired material, then you can probably get away with using a simple distillation column and discarding some of the forerun (the first drops of liquid that distill over).  On the other hand, if there is a large amount of impurity and it boils close to the desired product, then you will need a column with a higher efficiency (fractional distillation column) to effect a good separation between the two fractions (or &quot;cuts&quot; as they're sometimes called) and get pure product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, perhaps the original question would be better written as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, why does simple distillation put emphasis on small amounts and&#xA;  fractional distillation, on large amounts of impurities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the answer would be that even an inefficient (fewer HETP) simple distillation column has enough power (enough theoretical plates) to separate a small amount of impurity from a large amount of product, especially if you are willing to discard the initial fore-run from the distillation.  If there is a larger amount of impurity, then we may need a more powerful (higher efficiency column) to produce the desired separation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher the efficiency of a distillation column, the greater its ability to separate two things.  If we have to separate two materials that are both present in large quantity, or distill at boiling points close together, then we need a more efficient column than if they boiled further apart or one was present in only a small amount and we were willing to throw away the forerun (e.g. throw away some of the desired product in that early fraction that contained the impurity).  When we call a column a fractionation column, we're really saying that it has greater separating efficiency than a simple distillation column.  The more efficient the column, the better the job it will do on difficult separations.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-01T19:37:02.557" LastActivityDate="2014-09-01T19:37:02.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15791" PostTypeId="5" CreationDate="2014-09-01T16:38:36.813" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T16:38:36.813" LastActivityDate="2014-09-01T16:38:36.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15792" PostTypeId="4" CreationDate="2014-09-01T16:38:36.813" Score="0" Body="Questions asking for identification of compounds in photos, by description, or given by a formula." OwnerUserId="1677" LastEditorUserId="402" LastEditDate="2017-04-06T23:03:11.457" LastActivityDate="2017-04-06T23:03:11.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15793" PostTypeId="5" CreationDate="2014-09-01T16:39:28.900" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T16:39:28.900" LastActivityDate="2014-09-01T16:39:28.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15794" PostTypeId="4" CreationDate="2014-09-01T16:39:28.900" Score="0" Body="Questions discussing explosive nature of chemicals. Also, consider using the safety tag." OwnerUserId="1677" LastEditorUserId="402" LastEditDate="2017-05-03T00:12:09.337" LastActivityDate="2017-05-03T00:12:09.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15795" PostTypeId="5" CreationDate="2014-09-01T16:40:40.850" Score="0" Body="&lt;p&gt;For questions about cleaning agents and methods used to remove dirt, stains, residues, and bad smells, including proper waste disposal. For particular questions about surfactants, use the &lt;a href=&quot;/questions/tagged/surfactants&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;surfactants&amp;#39;&quot; rel=&quot;tag&quot;&gt;surfactants&lt;/a&gt; tag.&lt;/p&gt;&#xA;" OwnerUserId="15751" LastEditorUserId="15751" LastEditDate="2020-01-12T19:00:24.213" LastActivityDate="2020-01-12T19:00:24.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15796" PostTypeId="4" CreationDate="2014-09-01T16:40:40.850" Score="0" Body="For questions about cleaning agents and methods used to remove dirt, stains, residues, and bad smells, including proper waste disposal. For particular questions about surfactants, use the [surfactants] tag." OwnerUserId="1677" LastEditorUserId="15751" LastEditDate="2020-01-12T19:00:13.777" LastActivityDate="2020-01-12T19:00:13.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15797" PostTypeId="5" CreationDate="2014-09-01T16:43:23.167" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T16:43:23.167" LastActivityDate="2014-09-01T16:43:23.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15798" PostTypeId="4" CreationDate="2014-09-01T16:43:23.167" Score="0" Body="Glassy chemical equipment - bakers, flasks and other." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T16:48:46.070" LastActivityDate="2014-09-01T16:48:46.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15799" PostTypeId="2" ParentId="15784" CreationDate="2014-09-01T16:45:29.507" Score="2" Body="&lt;p&gt;You will probably know that the notation for galvanic cell reads from left to right as &lt;em&gt;anode&lt;/em&gt; || &lt;em&gt;cathode&lt;/em&gt;, where &lt;strong&gt;||&lt;/strong&gt; typically denotes a salt bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within each half cell, different phases are separated by &lt;strong&gt;|&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the anode is a platinum rod in an acidic solution under a hydrogen atmosphere. This is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_hydrogen_electrode&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard hydrogen electrode (SHE)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cathode consists of a platinum rod which is contact with a layer of mercury and a layer of solid mercury(I) chloride. It is an aqueous cell with 1M KCl solution as the medium. This electrode is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Saturated_calomel_electrode&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturated calomel electrode (SCE)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both electrodes are used as reference electrodes.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="72973" LastEditDate="2020-04-14T13:32:45.837" LastActivityDate="2020-04-14T13:32:45.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15800" PostTypeId="5" CreationDate="2014-09-01T16:45:54.467" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T16:45:54.467" LastActivityDate="2014-09-01T16:45:54.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15801" PostTypeId="4" CreationDate="2014-09-01T16:45:54.467" Score="0" Body="Fats are triesters of glycerol and fatty acids. They are generally soluble in organic solvents." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:57:06.160" LastActivityDate="2014-09-01T17:57:06.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15802" PostTypeId="5" CreationDate="2014-09-01T17:01:44.507" Score="0" Body="&lt;p&gt;Isomers are mostly common amongst hydrocarbons, where the amount of hydrogen and carbon can be the same, yet the structure is different.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Allene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Allene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Propyne&quot; rel=&quot;nofollow noreferrer&quot;&gt;propyne&lt;/a&gt; have both molecular formula $\ce{C3H4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YD6lP.png&quot; alt=&quot;Allene&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fbQ5V.png&quot; alt=&quot;Propyne&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      Allene                 Propyne&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:57:44.193" LastActivityDate="2014-09-01T17:57:44.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15803" PostTypeId="4" CreationDate="2014-09-01T17:01:44.507" Score="0" Body="Compounds that have identical molecular formula but different chemical structure." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:57:17.933" LastActivityDate="2014-09-01T17:57:17.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15804" PostTypeId="5" CreationDate="2014-09-01T17:06:11.413" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:06:11.413" LastActivityDate="2014-09-01T17:06:11.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15805" PostTypeId="4" CreationDate="2014-09-01T17:06:11.413" Score="0" Body="Process where atom or a molecule gains negative or positive charge by acquiring or losing electrons, respectively." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:56:51.783" LastActivityDate="2014-09-01T17:56:51.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15806" PostTypeId="5" CreationDate="2014-09-01T17:14:29.707" Score="0" Body="" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-01T17:14:29.707" LastActivityDate="2014-09-01T17:14:29.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15807" PostTypeId="4" CreationDate="2014-09-01T17:14:29.707" Score="0" Body="Photochemistry is a branch of chemistry concerned with the chemical effects of ultraviolet, visible, or infrared radiation." OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-01T17:14:29.707" LastActivityDate="2014-09-01T17:14:29.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15808" PostTypeId="5" CreationDate="2014-09-01T17:23:31.870" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:23:31.870" LastActivityDate="2014-09-01T17:23:31.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15809" PostTypeId="4" CreationDate="2014-09-01T17:23:31.870" Score="0" Body="A laboratory method to determine the concentration of an analyte in solution. The solution volume plays a key role in the analysis." OwnerUserId="1677" LastEditorUserId="402" LastEditDate="2017-04-09T16:22:08.910" LastActivityDate="2017-04-09T16:22:08.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15810" PostTypeId="5" CreationDate="2014-09-01T17:25:36.753" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:25:36.753" LastActivityDate="2014-09-01T17:25:36.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15811" PostTypeId="4" CreationDate="2014-09-01T17:25:36.753" Score="0" Body="The reagents are gaining electrons and therefore reducing their oxidation state." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:57:13.677" LastActivityDate="2014-09-01T17:57:13.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15812" PostTypeId="5" CreationDate="2014-09-01T17:27:13.353" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:27:13.353" LastActivityDate="2014-09-01T17:27:13.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15813" PostTypeId="4" CreationDate="2014-09-01T17:27:13.353" Score="0" Body="Questions inspired by Breaking Bad TV series." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:57:08.517" LastActivityDate="2014-09-01T17:57:08.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15814" PostTypeId="5" CreationDate="2014-09-01T17:28:19.303" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:28:19.303" LastActivityDate="2014-09-01T17:28:19.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15815" PostTypeId="4" CreationDate="2014-09-01T17:28:19.303" Score="0" Body="Questions where color (wavelengths associated with the visible region of the electromagnetic spectrum of about 400-800nm), or a color change is discussed." OwnerUserId="1677" LastEditorUserId="14" LastEditDate="2014-09-24T20:21:24.410" LastActivityDate="2014-09-24T20:21:24.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15816" PostTypeId="5" CreationDate="2014-09-01T17:29:22.297" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:29:22.297" LastActivityDate="2014-09-01T17:29:22.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15817" PostTypeId="4" CreationDate="2014-09-01T17:29:22.297" Score="0" Body="A material's ability to conduct electric current or thermal energy by any means." OwnerUserId="1677" LastEditorUserId="23561" LastEditDate="2018-08-11T20:00:41.560" LastActivityDate="2018-08-11T20:00:41.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15818" PostTypeId="5" CreationDate="2014-09-01T17:31:16.630" Score="0" Body="&lt;p&gt;In thermodynamics, entropy is a measure of the number of specific ways in which a thermodynamic system may be arranged, commonly understood as a measure of disorder.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-17T11:08:32.143" LastActivityDate="2018-03-17T11:08:32.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15819" PostTypeId="4" CreationDate="2014-09-01T17:31:16.630" Score="0" Body="For questions about the thermodynamical state function entropy, commonly understood as a measure of disorder." OwnerUserId="1677" LastEditorUserId="5026" LastEditDate="2018-03-17T15:50:43.867" LastActivityDate="2018-03-17T15:50:43.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15820" PostTypeId="5" CreationDate="2014-09-01T17:32:41.767" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-01T17:32:41.767" LastActivityDate="2014-09-01T17:32:41.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15821" PostTypeId="4" CreationDate="2014-09-01T17:32:41.767" Score="0" Body="Filtering is a process in which chemicals are being separated using their different state of matter." OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-01T17:56:39.077" LastActivityDate="2014-09-01T17:56:39.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15822" PostTypeId="1" AcceptedAnswerId="15833" CreationDate="2014-09-01T18:01:55.990" Score="5" ViewCount="541" Body="&lt;p&gt;I was making a solution of spermidine, and noticed that Sigma Aldrich warns that spermidine solutions deaminate over time. I tried to think of how that would happen, and made a diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VX8Vm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming some of the spermidine has the secondary amine in an uncharged state, the lone pair electrons can attack either charged terminal amine group. I don't know which is more likely, I have drawn the 4 membered ring, but the other amine would make a 5 membered ring, which would probably be more stable. Either way, this would kick off an ammonia, which would be stable enough in water. Then a water molecule would attack a carbon on the ring and open it, making an OH2+ charged group, which would give up a proton. Of course the protons would be balanced because the ammonia would pick up a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make sense? It reminds me of nitrogen mustard chemistry, but must not be as reactive because spermidine is used to condense DNA, and nitrogen mustards are used to damage DNA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, spermidine seems hard to work with. I tried to make 10mL of a 10mM solution, and calculated what volume to use to get the right mass, because spermidine melts at 20-25C. However, when I used that volume, I had about 2.5 times as much mass as I wanted. I assume this is because the spermidine was not totally thawed out, and the density was higher than I assumed (925mg/mL at 25C). Has anyone worked with this compound and has suggestions for how to handle it better?&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="865" LastEditDate="2015-10-16T18:07:56.423" LastActivityDate="2015-10-16T18:07:56.423" Title="Deamination of Spermidine?" Tags="&lt;reaction-mechanism&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15823" PostTypeId="2" ParentId="15690" CreationDate="2014-09-01T18:12:08.770" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But the main question remains: Is the anomeric effect the main cause&#xA;  for the unreactiveness of DCM towards nucleophiles or is it the steric&#xA;  hinderance exerted by the second Cl ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assessment of Steric Effects&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonds, like humans, have a length and a girth.  In some reactions, the reaction is most likely to occur when the attacking group strikes an atom or substituent &quot;head on&quot;.  In other reactions, like the $\ce{S_{N}2}$, the attacking group approaches from the back side and slips by the various substituents in a side-ways manner as it approaches the carbon under attack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice way to measure this side-ways (girth of a bond, if you will) interference in organic systems.  Axial substituents in a cyclohexane ring are subject to what is termed &quot;A-strain&quot; produced by the interactions depicted in the figure below.&#xA;These interactions are very analogous to the gauche interactions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LrzQr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in butane and arise from side-ways interaction of the substituent and the two diaxial hydrogens located two carbons away.  Equatorial substituents avoid these repulsive interactions.  The smaller the substituent, the less repulsive the 1,3 interaction will be and, at equilibrium, more of the substituent should exist in the axial position.  Conversely, the larger the substituent, more of it should reside in the equatorial position.  By measuring the axial/equatorial ratio for various substituents we can meaningfully assess the side-ways steric bulk of the substituent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For chlorocyclohexane the axial/equatorial ratio is &lt;em&gt;ca&lt;/em&gt;. 31/69 ($\ce{\Delta G = 0$.$48 kcal/mol}$) at room temperature.  For methylcyclohexane the axial/equatorial ratio is &lt;em&gt;ca&lt;/em&gt;. 5/95 ($\ce{\Delta G = 0$.$1.7 kcal/mol}$) at room temperature.  This analysis clearly suggests that the chloro substituent is &quot;smaller&quot; than a methyl substituent when we use this side-ways steric probe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the interested reader, here is a Table of axial/equatorial $\ce{\Delta G's}$ (in kcal/mol) for other substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ajm9c.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this analysis suggests that a chlorine substituent is smaller than a methyl substituent, in terms of the side-ways interaction expected in a back side $\ce{S_{N}2}$ attack, it would be difficult to use a steric argument to explain the reduced reactivity in the polyhalomethanes towards  $\ce{S_{N}2}$ reaction.  This leaves the anomeric effect as a reasonable explanation for the reduced reactivity of the polyhalomethanes. But...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Thoughts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How unreactive are these compounds to $\ce{S_{N}2}$ reaction?  Dichloromethane, chloroform and carbon tetrachloride will all undergo the $\ce{S_{N}2}$ reaction with various nucleophiles, under various conditions.  Before I sign up for the anomeric effect explanation, it would be nice to know exactly what reaction Clayden, &lt;em&gt;et. al.&lt;/em&gt; were discussing.  Perhaps some other factor (hydrogen bonding, who knows what) suppressed the nucleophile's reactivity in Clayden's series of compounds.&lt;/li&gt;&#xA;&lt;li&gt;Bond strengths:  $\ce{C-Cl}$ bond strengths decrease from chloromethane (80 kcal/mol) to carbon tetrachloride (70 kcal/mol).  This would cause us to suspect the chloromethane might react the slowest and carbon tetrachloride the fastest.  That this is not the case is another argument in support of the anomeric effect playing a key role.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-01T18:12:08.770" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15824" PostTypeId="1" AcceptedAnswerId="15827" CreationDate="2014-09-01T18:21:25.377" Score="2" ViewCount="2439" Body="&lt;p&gt;If I take the electrodes as nickel and silver, I've according to the nernst equation that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ni(s) + 2Ag+(aq) -&amp;gt; Ni^2+(aq) + Ag(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $E=E_{0} - \frac{RT}{2F} \ln\left(\frac{[\ce{Ni^{2+}}]}{[\ce{Ag+}]^2} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I double the stoichiometric coefficients, the value inside the logarithm would square up. So if I write the same reaction but with different coefficients, I'll get different values of potential of the cell?&lt;/p&gt;&#xA;" OwnerUserId="6912" LastEditorUserId="275" LastEditDate="2014-12-02T14:35:30.743" LastActivityDate="2014-12-02T14:35:30.743" Title="Does changing the stoichiometric coefficients affect the Nernst equation?" Tags="&lt;electrochemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15825" PostTypeId="1" CreationDate="2014-09-01T18:41:36.790" Score="5" ViewCount="3483" Body="&lt;p&gt;In my exam, my teacher asked me if the following is true:&#xA;&quot;The chair cyclohexane has more axial hydrogens&quot;&#xA;I did not have time to think, so I just put false.&#xA;And apparently I was right.&#xA;But, at home, I thought to myself: Does this question even make sense? Axial hydrogens are by definition only seen on cyclohexanes according to what I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/e8bhPvE.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorDisplayName="user7951" LastEditDate="2016-08-31T22:22:20.000" LastActivityDate="2016-08-31T22:22:20.000" Title="Boat conformation: axial hydrogens" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;cyclohexane&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15826" PostTypeId="2" ParentId="15825" CreationDate="2014-09-01T18:53:20.710" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In my exam, my teacher asked me if the following is true: &quot;The chair cyclohexane has more axial hydrogens&quot; I did not have time to think, so I just put false. And apparently I was right. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the complete question? The &lt;em&gt;chair&lt;/em&gt; form has more axial hydrogens than what other form? If the test were referring to the &lt;em&gt;other&lt;/em&gt; chair conformation, then you are right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexane has 12 hydrogens, and in its two chair conformations, six substituents are axial while the other are equatorial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BuN1N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Basically, the question is does boat have more axial H's than&#xA;  cyclohexane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The boat conformation doesn't have any true axial positions. I'm not sure what your professor considers to be axial or not, but if we take the &lt;em&gt;flagpole&lt;/em&gt; hydrogens to be &quot;axial,&quot; then both the chair and the boat should still have the same number of axial substituents. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;TRUE OR FALSE: Boat cyclohexane is a slightly higher energy conformer&#xA;  than chair cyclohexane because: (a).... (a sub question) T OR F&#xA;  (b).... (a sub question) T OR F (c)chair cyclohexane has more axial&#xA;  hydrogens. T OR F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Regarding the question now &lt;em&gt;in context&lt;/em&gt; the instability of the boat conformation likely has more to do with flagpole-eclipsing interactions rather than the number of hydrogens. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-01T18:58:50.167" LastActivityDate="2014-09-01T18:58:50.167" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15827" PostTypeId="2" ParentId="15824" CreationDate="2014-09-01T19:05:47.267" Score="4" Body="&lt;p&gt;No, you won't get a different value. The number of electrons transfered in that case is 4, so it will be $E=E_{0} - \frac{RT}{4F} \ln\left(\frac{[\ce{Ni^{2+}}]^{2}}{[\ce{Ag+}]^{4}} \right)$.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="189" LastEditDate="2014-09-01T19:25:09.027" LastActivityDate="2014-09-01T19:25:09.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15828" PostTypeId="2" ParentId="15825" CreationDate="2014-09-01T19:08:12.463" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It says fully: TRUE OR FALSE: Boat cyclohexane is a slightly higher&#xA;  energy conformer than chair cyclohexane because: (a).... (a sub&#xA;  question) T OR F (b).... (a sub question) T OR F (c)chair cyclohexane&#xA;  has more axial hydrogens. T OR F&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to part (c) of that question is &quot;False.&quot;  It turns out that the chair and boat conformations of cyclohexane both have 6 axial hydrogens - so the chair does not have more axial hydrogens.  Here's a picture of the chair (on the left) and boat (in the middle), you should be able to see the 6 axial hydrogens in both the chair and the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDFuo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-01T19:08:12.463" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15830" PostTypeId="2" ParentId="15825" CreationDate="2014-09-01T20:55:48.510" Score="6" Body="&lt;p&gt;If you want to be very precise (perhaps pedantic), then I would conclude that the question itself is a little dubious without additional clarification. Consider the IUPAC recommendation on &lt;a href=&quot;http://www.iupac.org/publications/pac/1996/pdf/6812x2193.pdf&quot;&gt;Basic Terminology of Stereochemistry&lt;/a&gt;. It contains the following definition of &lt;em&gt;axial&lt;/em&gt; with respect to ring conformations (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In the chair form of cyclohexane&lt;/strong&gt; ring bonds to ring atoms (and molecular entities attached to such bonds) are termed axial or equatorial according to whether the bonds make a relatively large or small angle, respectively, with the plane containing or passing closest to a majority of the ring atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point here is that the definition is contextually limited to &lt;strong&gt;chair&lt;/strong&gt; conformations. Granted, this is only a recommendation, but I think it's a logical and authoritative one. Seen in that light, the question itself is malformed inasmuch as it admits either answer in the absence of an agreement on exact terms. To wit, &quot;true&quot; might be a valid answer, since by the strictest definition a boat conformation has &lt;strong&gt;zero&lt;/strong&gt; axial hydrogens; on the other hand, if a similar definition (based on bond angles relative to a plane containing a majority of ring atoms) is applied to the boat conformation, one can identify &lt;strong&gt;six&lt;/strong&gt; axial hydrogens roughly analogous to those in the chair conformation and the correct answer would be &quot;false.&quot; Really, the question is murky unless you (and more importantly, your professor) are clear on the formal definition being used.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-09-01T20:55:48.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15831" PostTypeId="1" AcceptedAnswerId="15832" CreationDate="2014-09-01T21:48:25.393" Score="4" ViewCount="2774" Body="&lt;p&gt;I glued a sheet of paper to a cardboard package using superglue (labelled with ethyl 2-cyanoacrylate).&#xA;To my surprise the paper became very warm.&#xA;What kind of reaction is it?&lt;/p&gt;&#xA;" OwnerUserId="7666" LastEditorUserId="41328" LastEditDate="2020-01-12T06:21:00.080" LastActivityDate="2020-01-12T06:21:00.080" Title="What is the exothermic reaction of superglue with cardboard?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15832" PostTypeId="2" ParentId="15831" CreationDate="2014-09-01T22:22:47.560" Score="10" Body="&lt;p&gt;I believe this is due to the fact that cyanoacrylates polymerize by anionic (or sufficiently strong neutral nucleophile) addition. In particular, they react with hydroxyl ($\ce{-OH}$) groups; paper, being primarily &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot;&gt;cellulose&lt;/a&gt;, is abundant in hydroxyls. The bond formation involved in the polymerization reaction is inherently exothermic (generating a $\ce{C-O}$ bond on initiation and many subsequent $\ce{C-C}$ bonds as the polymerization proceeds), and with a large surface area and many reactive groups, I could certainly imagine the reaction being surprisingly violent. The Wikipedia page on cyanoacrylates notes its &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyanoacrylate#Reaction_with_cotton&quot;&gt;reactivity with cotton fibers&lt;/a&gt; (again, being chiefly composed of cellulose), and the MSDSs for various similar compounds warn of exothermic reaction with paper and similar materials (q.v., &lt;a href=&quot;http://www.tedpella.com/msds_html/10035%20msds%20mfg.pdf&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.adhesive.com/resources/MSDS/i5-Instant-Adhesive-MSDS.pdf&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://cyberbond1com.blob.core.windows.net/multimedia/pdfs/5008%20MSDS.pdf&quot;&gt;3&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-09-01T22:58:12.503" LastActivityDate="2014-09-01T22:58:12.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15833" PostTypeId="2" ParentId="15822" CreationDate="2014-09-01T22:42:01.993" Score="5" Body="&lt;p&gt;Here are some links to the mechanism of the degradation of spermidine and some analogues in living systems.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.plantphysiology.org/content/132/4/1973.full&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ornithine_decarboxylase&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tec-digital.itcr.ac.cr/file/3295531/regulacion%20genica%20luis%20sabino%20y%20arturo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 3&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.biocyc.org/META/NEW-IMAGE?type=PATHWAY&amp;amp;object=PWY-6117&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on these reactions I would guess that an oxidative pathway is followed here as well, once the sample has been exposed to air.  Instead of a nice enzyme catalyzed oxidation, here we would have a free radical oxidation. Here is a general outline of such a process.  The first step would involve generation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNxyn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of the more stable secondary nitrogen radical (as opposed to the primary terminal nitrogens) and formation of the nitrogen hydroperoxide.  Elimination of hydrogen peroxide could produce the two possible imines which could be further oxidized by oxygen to produce both possible diamines and both possible amino-aldehydes.  You could test your material for the presence of hydrogen peroxide, but the smell of putrescine ought to be a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Putrescine&quot; rel=&quot;nofollow noreferrer&quot;&gt;dead&lt;/a&gt;&quot; giveaway.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-01T22:47:22.183" LastActivityDate="2014-09-01T22:47:22.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15834" PostTypeId="1" AcceptedAnswerId="15843" CreationDate="2014-09-01T23:25:56.943" Score="10" ViewCount="1350" Body="&lt;p&gt;Most &quot;invisible inks&quot; turn brown when heated and they take a while to transform.  I would like to make (or obtain) an &quot;ink&quot; that responds to the heat of an iron and permanently turns black or a similarly dark color - instantly.  Thermal receipt paper does this where the paper is white until it is heated with almost instant response times AND the &quot;ink&quot; has a permanence of sorts (e.g. sticking a receipt in a freezer won't revert the reaction).  What chemical combinations in what amounts will do an equivalent transformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus question:  What chemical combinations would create similar effects for standard palette colors?  That is, transparent/hidden until heat from an iron is applied, whereupon it instantly appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to use such an ink with paintbrushes and fountain pens on canvas.  Black is probably sufficient, but knowing how to make standard colors would dramatically expand my options.  The ink ideally shouldn't be toxic to touch after drying nor emit toxic fumes when heating with an iron at a &quot;Cotton&quot; setting.  It can be semi-toxic while wet but nothing that will melt nitrile gloves, fingers, or human lungs.  I will, of course, wear sufficient protection at all times as I don't really need any chemicals leeching through my skin.  Having a drawing/painting appear instantly in front of an audience from a seemingly blank canvas would be very dramatic and painting with invisible ink will be an interesting adventure.&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2014-09-02T06:48:02.697" Title="What chemicals are used in receipt paper?" Tags="&lt;heat&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15835" PostTypeId="1" AcceptedAnswerId="15836" CreationDate="2014-09-02T02:31:09.443" Score="5" ViewCount="3852" Body="&lt;p&gt;I have a piece of copper mesh from which I would like to remove all coatings/oils. The copper comes from a pot scrubber like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B002YLT5A4&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. In order to remove any residue, I held a standard bic lighter up to it. It gave off black smoke. I moved the lighter around the mesh, waiting for it to stop giving off any such smoke, but seemed to do so indefinitely. Even when I just held the lighter in one place, it would not stop giving off a steady stream of black smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then did the same thing with a torch lighter on a fresh piece of copper from the same source. This time it was hard to tell if there was any smoke at all-- and if there was, it stopped being produced very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With both the normal lighter and the torch lighter, the surface of the copper turned gray-black. I assume part of this is from the formation of copper oxides, but part of it may also come from the deposition of incompletely combusted carbon on the surface. In both cases I was able to restore the shiny surface by putting the mesh in $\ce{HCl}$-based toilet cleaner for a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the normal lighter producing black smoke where the torch lighter is not? What does this smoke consist of?&lt;/p&gt;&#xA;" OwnerUserId="289" LastEditorUserId="4945" LastEditDate="2014-09-04T07:34:34.113" LastActivityDate="2020-05-22T11:11:29.133" Title="Black Smoke Produced from Regular Lighter but Not Torch Lighter Applied to Copper" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15836" PostTypeId="2" ParentId="15835" CreationDate="2014-09-02T03:36:01.653" Score="5" Body="&lt;p&gt;Common copper wire, especially used in electronic components like inductors and transformers, is usually covered with insulating varnish. This is the common source of black smoke, sometimes it may even burn on its own. I doubt, however, that kitchenware is coated with such varnish, there is no point in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of devices with flames: ones where the fuel is premixed with air, that usually produce blueish flame, and ones where fuel is not pre-mixed with air, but burns and mixes with it in same time, that may produce bright yellow flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many organic compounds, especially ones that require large amount of oxygen per molecule to fully burn, &lt;strong&gt;in case the fuel is not premixed with enough air&lt;/strong&gt;, the flame usually contains oxygen-deficient region where tiny particles of carbon and molecular hydrogen are formed. These carbon particles, since they are heated, give thermal light emission and are responsible for bright non-transparent orange-yellow organic flames. A typical example of such flame is a flame of a candle. It is a flame of heavy hydrocarbons or fatty acids (c20+), that require tens of molecules of oxygen per molecule of the fuel to fully burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case gases from this yellow part of the flame are cooled &lt;strong&gt;without mixing with more air&lt;/strong&gt;, this carbon particles may be obtained in form of deposits or black smoke. For this, a cold effective conductor of heat (i.e. metal, especially copper and silver) or a solid surface with high heat capacity introduced into yellow part of the flame may be used. In some cases, when too many of such particles are produced and the flame is not particularly hot, there is no need even to cool the flame. This is the usual case of hydrocarbons with big amount of carbon, like aromatics, and chlorinated hydrocarbons, like PVC (as long as they burn at all, which often is not the case even if there is no much chlorine in the compound). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propane torches usually have a mean to regulate amount of air premixed with fuel. Torch lighters are small propane torches, usually without such means. When propane in such torch is premixed with excess of air, a very hot blueish short flame is formed. However, when air is not premixed with propane, the flame is organized in layers with more oxidized outer layers and more reduced inner layers. In inner layers, the dominating reactions are thermal decomposition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C3H8 = 3C + 4H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and incomplete combustion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C3H8 + 2O2= 3C + 4H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(this is an oversimplification, as numerous other intermediates are formed, but that's advanced stuff)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In outer layers, propane combusts completely, just as if it was premixed with air&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C3H8 + 5O_2= 3CO2 + 4H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and carbon particles (and other reduced products) from inner layers also combust here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to get carbon particles from the flame in form of smoke the key is to cool them before they mix with more air. An ideal solution would be to suck them with tiny pipe and cool with water. However, it seems, that skein of copper mesh, at least until it is heated enough, works good for this purpose, both cooling the inner layers and hindering their mixing with air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since flames of fuels that are not pre-mixed with air have very efficient emitter of visible and and infrared light in it (many carbon particles), they tend to be cooler, especially the inner part. The flames with fuel pre-mixed with air (like in torch lighters with blueish flame like &lt;a href=&quot;http://www.southsmoke.com/pd-shark-series-torch-lighter.cfm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;), the flame is thin, quite short and much hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper wire, when heated in air, also oxidizes. If done properly, it is a good way to produce pure black copper (II) oxide. However, it usually stay on the surface of the wire and indeed, can be washed with hydrochloric acid. If there are black particles that are not dissolved in hydrochloric acid, they are carbon particles.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-03T07:17:33.770" LastActivityDate="2014-09-03T07:17:33.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15837" PostTypeId="1" AcceptedAnswerId="15838" CreationDate="2014-09-02T04:15:08.353" Score="0" ViewCount="138" Body="&lt;p&gt;The kinetic energy of an electron is $1.67 \times 10^{-17}\mathrm{J}$. Calculate the wavelength ($\lambda$) of the electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the formula $\lambda= \frac{h}{mv}$; where h is Planck's constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find $\lambda$ without $v$ (velocity) and $m$ (mass)?&lt;/p&gt;&#xA;" OwnerDisplayName="user7484" LastEditorUserId="186" LastEditDate="2015-10-02T15:37:48.350" LastActivityDate="2017-02-05T16:22:09.823" Title="What is the wavelength of an electron?" Tags="&lt;homework&gt;&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15838" PostTypeId="2" ParentId="15837" CreationDate="2014-09-02T04:23:36.030" Score="3" Body="&lt;p&gt;The mass of electron is never given in question you should remember it. Mass of electron is $9.109 \times 10^{-31} \mathrm{kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now read the question Kinetic Energy is given, so from that you can easily find velocity of electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K.E.=\frac{1}{2}mv^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so You will get $v= 6 \times10^6 \mathrm{m/s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can substitute value of m and v in formula $\lambda = \frac{h}{mv}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get wave length of electron as $0.12 \times 10^{-9} \mathrm{m}= 0.12 \mathrm{nm}$&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorDisplayName="user7484" LastEditDate="2014-10-14T18:17:26.480" LastActivityDate="2014-10-14T18:17:26.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15840" PostTypeId="1" AcceptedAnswerId="15842" CreationDate="2014-09-02T04:42:33.453" Score="1" ViewCount="22600" Body="&lt;p&gt;I was wondering about the amazing conductive properties of graphene, lets assume a large copper bar that is 10 kg, current of 1kA and probably more can flow in it, what about graphene? It certainly can since its a better conductor, but it would be much lighter? In the milligrams possibly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can graphene replace the heavy busbars that carry MW of power?&#xA;Making it much lighter? &lt;/p&gt;&#xA;" OwnerUserId="7669" LastEditorUserId="7448" LastEditDate="2015-04-02T18:27:10.510" LastActivityDate="2020-04-06T02:29:59.377" Title="Graphene Vs Copper?" Tags="&lt;conductivity&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15842" PostTypeId="2" ParentId="15840" CreationDate="2014-09-02T06:29:36.370" Score="11" Body="&lt;p&gt;The trick with graphene is that a lot of its amazing properties only work when you have continuous perfect sheets of it, and making graphene like this is currently beyond us, for large scales anyways. It is true that graphene &lt;a href=&quot;http://journals.aps.org/prb/abstract/10.1103/PhysRevB.77.115449&quot;&gt;has very high electron mobility $\approx10^{5}~\mathrm{cm^2/Vs}$ at room temperature&lt;/a&gt;, which works out to on the order of $10~\mathrm{n\Omega\cdot m}$ in bulk (which assumes you can make a perfect multilayer structure that maintains the properties of a single sheet). An impressive figure, but only $\approx40\%$ better than copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, copper is about 6 times more dense than graphene, so if mass is the main concern, graphene would be a pretty good improvement. Still, we're not talking about replacing $10~\mathrm{kg}$ of copper with a few milligrams. Given how much cheaper and easier it is to make metal wires, we're not at the point of using graphene for this kind of bulk application. There are also some more mundane problems to sort out, e.g. graphene is brittle like a ceramic, which might cause mechanical issues.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="4945" LastEditDate="2014-09-02T07:11:46.090" LastActivityDate="2014-09-02T07:11:46.090" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15843" PostTypeId="2" ParentId="15834" CreationDate="2014-09-02T06:48:02.697" Score="6" Body="&lt;p&gt;Have a look at wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_paper&quot;&gt;Thermal Paper&lt;/a&gt;. The whole trick is in the three components, which are mixed as solid substances and applied as coating to the paper. When you heat the paper, the solid substances melt and the coloring reaction occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this you need fine pulverized chemicals, stirred in a form of suspension in some volatile solvent and apply them on the canvas. It should not be a big problem, see e.g. &lt;a href=&quot;http://www.google.com/patents/US3953659&quot;&gt;patent on Thermal Paper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all this effort is for an artistic purpose, I would suggest as a first try to approach the companies manufacturing the thermal paper (there should be lot of them) and ask them for a bottle of the prepared suspension, telling them your plans. The success rate would not be high, but hopefully non-zero and it saves you 99% of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the painting itself could be quite fun, I would expect the components to be UV-active, so the black light could help you.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-09-02T06:48:02.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15844" PostTypeId="1" CreationDate="2014-09-02T08:20:38.853" Score="3" ViewCount="94" Body="&lt;p&gt;I am working on Enzyme Assay.I assayed the enzyme in different buffers from $pH\ 1-12.5$ However,the enzyme has good activities starting from $pH\ 1-10.5$ Is that possible to have enzyme activity in such broad range ?&lt;/p&gt;&#xA;" OwnerUserId="7492" LastEditorUserId="5456" LastEditDate="2014-09-02T10:31:33.953" LastActivityDate="2014-09-02T10:31:33.953" Title="Enzyme Assay - pectinase" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;spectrophotometry&gt;&lt;enzymes&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15845" PostTypeId="1" AcceptedAnswerId="54157" CreationDate="2014-09-02T09:52:40.703" Score="1" ViewCount="4396" Body="&lt;p&gt;I'm trying to get as pure citric acid as possible from a mandarin. It's not because I desperately need citric acid, but because I want to achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NFZpr.jpg&quot; alt=&quot;citric acid filtration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started by grating the mandarine and filtering the juice. After filtering overnight, I was left with slightly yellow liquid. I heated it while avoiding boiling and I reduced the volume to 50%. When volume was reduced, the solution started to become saturated and crystals appeared. At this point, I stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm looking for a way to remove sugar from the solution, as this is the most annoying part. It's sticky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When crystals appeared while I was boiling the solution, I came to think, that perhaps that could allow to separate sugar. &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Citric acid&lt;/a&gt; has solubility in water 1477.6 g/L (20 °C), while &lt;a href=&quot;http://en.wikipedia.org/wiki/Sucrose#cite_ref-ICSC_1-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;sugar&lt;/a&gt; has 2000 g/L (25 °C). Wikipedia unfortunately doesn't provide values for the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above it seems that I can have more sugar dissolved than citric acid. I think that means that citric acid will crystallise first. Does it work like this? Are the crystals I've produced the citric acid crystals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another way to separate the compounds? Does something react with sugar to produce precipitate that can be filtered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sbnIz.jpg&quot; alt=&quot;filtrated and concentrated citric acid solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-02T10:12:24.910" LastActivityDate="2016-06-24T17:52:31.707" Title="Separating citric acid from sugar" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;&lt;filtering&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15847" PostTypeId="1" AcceptedAnswerId="15989" CreationDate="2014-09-02T13:15:48.430" Score="4" ViewCount="6826" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Knowing that at 25 °C the following galvanic cell:&#xA;  $$\ce{Pb~|~Pb(NO_3)_2~1M~||~PbS~saturated~|~Pb}$$ shows an $\mathrm{EMF} =0.413~\mathrm{V}$, find the $K_\mathrm{sp}$ of $\ce{PbS}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Approach&lt;/strong&gt;&lt;br&gt;&#xA;This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the cathode and the right one the anode.&lt;br&gt;&#xA;So we have the following semireactions:&#xA;\begin{align}&#xA; \ce{Pb^2+ + 2e- &amp;amp;-&amp;gt; Pb}  &amp;amp;&amp;amp; \text{(cathode)} \\&#xA; \ce{Pb  &amp;amp;-&amp;gt; Pb^2+ +2e- } &amp;amp;&amp;amp; \text{(anode)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the anode we also have $$\ce{PbS &amp;lt;=&amp;gt; Pb^2+ + S^2- }$$ where $\ce{[Pb^2+]} = \ce{[S^2- ]}= \sqrt{K_\mathrm{sp}}$.&#xA;So the semicell potentials are:&#xA;\begin{align}&#xA;E_\text{cathode} &amp;amp;= E^\circ\\&#xA;E_\text{anode}   &amp;amp;= E^\circ - \frac{0.059}{2} \log_{10}{[\ce{Pb^2+}]}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$0.413~\mathrm{V} = E_\text{cathode} - E_\text{anode} = \frac{0.059}{2} \log_{10}\ce{[Pb^2+]} \Rightarrow \ce{[Pb^2+]} = 10^{14} $$&#xA;And: $$K_\mathrm{sp} = \ce{[Pb^2+]}^2 = 10^{28}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that I'm wrong for a sign but I don't understand where is the error. &lt;/p&gt;&#xA;" OwnerUserId="7570" LastEditorUserId="41328" LastEditDate="2019-11-18T08:12:31.340" LastActivityDate="2019-11-18T08:18:36.703" Title="How to calculate the solubility product for PbS from galvanic potential?" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15848" PostTypeId="2" ParentId="15697" CreationDate="2014-09-02T13:35:20.507" Score="1" Body="&lt;p&gt;There is no systematic way (besides brute force - that is a system, right?) to predict the crystal structure of the metals.  Even sticking just to the transition metals, as you have noted, there are quite a variety.  The primary ones are as you have noted $fcc$, $bcc$, and $hcp$ (ignoring the rhombohedral Hg).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several issues arise here. First, there is little difference between $fcc$ and $hcp$ - they are both close packed structures with the only the stacking order changing (ABCABC for $fcc$ and ABABAB for $hcp$). An estimate of the free energy difference between these two crystal structures can be obtained 'SGTE Data for Pure Elements', A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991), the compilation of elemental thermodynamic data. Here one finds that for Ag ($fcc$), the $hcp$ phase is only 300J/mol higher free energy (at T=0K).  Similarly, for Co ($hcp$), the $fcc$ phase is only 427 J/mol higher. So, the free energy differences aren't necessarily that much, making an $a\ priori$ differtiation impossible. (I'll note here that Ru and Os are $hcp$ crystals, not simple cubic as you stated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second factor that is difficult to determine is the impact of magnetic free energy on the equilibrium crystal structure (as included in the above reference).  Iron is the most obvious candidate to consider this, and it does not disappoint.  As observed, the equilibrium crystal structure at room temperature is $bcc$.  Heating leads to a change to $fcc$ at $910C$, then back to $bcc$ at $1400C$.  If the magnetic contributions to the free energy are ignored, the observed crystal structures would instead be $hcp$ at low temperatures, a transition to $fcc$ at around $200C$, and a brief appearance of $bcc$ right before melting. I see no way that one could predict these particular changes in phases with temperature.   &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-09-02T13:35:20.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15849" PostTypeId="2" ParentId="15847" CreationDate="2014-09-02T13:41:21.303" Score="1" Body="&lt;p&gt;You swapped the cathode an anode. In a galvanic cell diagram, on the left salt of the bridge is the anode, and on the right side is the cathode. &#xA;Also, you should have written&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E(\text{anode})= -E^\circ - 0.059/2 \ldots$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{EMF} = E(\text{cathode}) + E(\text{anode}),$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E(\text{anode}) = E^\circ + 0.059/2\ldots$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{EMF} = E(\text{cathode}) - E(\text{anode}).$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways of writing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nernst equation&lt;/a&gt; and you are mixing them, so you get something wrong. One way is by using the reaction coefficient &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; and the second by using &lt;span class=&quot;math-container&quot;&gt;$[\text{Ox}]/[\text{Red}].$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure that you first understand these two ways of writing the Nernst equation, then try to derive the &lt;span class=&quot;math-container&quot;&gt;$\text{EMF}$&lt;/span&gt; by each method. You need to make these things clear and then return to the problem again. The red-ox potential of the cathode must be higher than the one of the anode.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="41328" LastEditDate="2019-11-18T08:12:06.197" LastActivityDate="2019-11-18T08:12:06.197" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="15850" PostTypeId="1" AcceptedAnswerId="15858" CreationDate="2014-09-02T16:50:41.073" Score="6" ViewCount="3460" Body="&lt;p&gt;The mechanism for the reaction:&#xA;\begin{align}&#xA;\ce{NO_{(g)} + O_{2(g)}  &amp;amp;&amp;lt;=&amp;gt;[k_{-1}][k_{1}] NO_{3(g)}} &amp;amp;\text{(fast)}\\&#xA;\ce{NO_{3(g)} + NO_{(g)} &amp;amp;&amp;lt;=&amp;gt;[K_{2}] 2\,NO_{2(g)}}      &amp;amp;\text{(slow)}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental rate law:&#xA;$$Rate=-\dfrac{\Delta\ce{[NO]}}{\Delta t}=k\ce{[NO]}^2\ce{[O_{2}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, the forward and reverse rates are equal to each other. We then have&#xA;$$k_{1}\ce{[NO]}\ce{[O_{2}]}=k_{-1}\ce{[NO_{3}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be re-arranged to&#xA;$$\ce{[NO_{3}]}=\dfrac{K_{1}}{K_{-1}}\ce{[NO]}\ce{[O_{2}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $2\,\ce{NO}$ disappear in the overall reaction for every NO that reacts in the second step, we multiply the overall reaction rate by two and thus have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2k_{2}\ce{[NO_{3}]}\ce{[NO]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand everything up to this point. What I do not understand is how we get to this equation from the previous one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Rate=2k_{2}\dfrac{k_{1}}{k_{-1}}\ce{[NO]}^2\ce{[O_{2}]}$$&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="21296" LastEditDate="2017-10-15T12:14:49.047" LastActivityDate="2017-10-15T12:14:49.047" Title="Show proposed mechanism is consistent with the experimental rate law" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15851" PostTypeId="1" CreationDate="2014-09-02T17:49:15.587" Score="2" ViewCount="607" Body="&lt;p&gt;It's fun to consider bond angles of molecules in isolation. For example, what is the number of different bond angles exhibited by the most polar form of $\ce{SIBrClF2}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, can bond angles really be considered in isolation? The method I would employ to determine the number of different bond angles only considers the molecule in isolation; I just consider van der Waals repulsions between the component atoms. Molecules generally do not exist in isolation. We can study molecules in the gas phase as to minimize intermolecular forces, but still, intermolecular forces exist, as it is hard to isolate exactly one molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{SIBrClF2}$, I would expect a number of rather strong dipole-dipole interactions among the $\ce{SIBrClF2}$ molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore my question is: are there any cases in which considering solely &lt;em&gt;intramolecular&lt;/em&gt; forces leads one astray when determining the bond angles of a molecule? I.e. one might look at a molecule and expect a certain bond angle to be &amp;lt; 180 degrees but find out the angle is in fact &gt; 180 degrees due to &lt;em&gt;intermolecular&lt;/em&gt; forces prying the bond angle open? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5456" LastEditDate="2014-09-02T18:23:58.933" LastActivityDate="2014-09-02T22:07:17.793" Title="Bond Angles and intermolecular interactions" Tags="&lt;bond&gt;&lt;intermolecular-forces&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15852" PostTypeId="1" AcceptedAnswerId="15857" CreationDate="2014-09-02T18:10:58.780" Score="15" ViewCount="125062" Body="&lt;p&gt;I was working on something in school and came across the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does the temperature not change much during a phase change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm really not sure why this happens in matter and I couldn't find an answer in my school resources. Does anyone here know?&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="5026" LastEditDate="2018-02-19T10:47:46.237" LastActivityDate="2018-02-19T10:47:46.237" Title="During phase change in matter, why doesn't the temperature change?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15853" PostTypeId="2" ParentId="15852" CreationDate="2014-09-02T18:34:15.713" Score="8" Body="&lt;p&gt;For a first-order phase transition, you need to add the enthalpy of the phase transition.  As an example, starting with ice below the melting point, you pump heat in, and raise the temperature.  When you hit the melting temperature, the heat you put in goes towards the enthalpy of melting, and starts converting ice (sold) to water (liquid).  Additional heat continues to melt more of the ice.  Once all of the ice is converted (still at the melt temperature), than more heat starts increasing the temperature of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second-order phase transition does not have an enthalpy associated with it.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-09-02T18:34:15.713" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15854" PostTypeId="1" AcceptedAnswerId="15871" CreationDate="2014-09-02T19:52:55.143" Score="8" ViewCount="425" Body="&lt;p&gt;Is there a group that would attached to a benzene ring, donate electrons by inductive effect and withdraw them by conjugation?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="73" LastEditDate="2014-09-05T20:13:04.903" LastActivityDate="2014-09-05T20:13:04.903" Title="Donation by inductive effect and withdrawal by conjugation" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15855" PostTypeId="2" ParentId="15851" CreationDate="2014-09-02T20:13:51.410" Score="4" Body="&lt;p&gt;When people talk about bond angles in molecules, they are usually talking about the equilibrium bond angle in isolation. This can be derived from the energy minimized structure of the molecule in isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you point out, the angles in a real, non-isolated molecule will fluctuate due to interactions with neighboring molecules. However, the actual angle formed by three successive atoms in a molecule will naturally fluctuate even if the molecule is isolated and has some kinetic energy. Everything's always jiggling around.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastEditorUserId="2008" LastEditDate="2014-09-02T22:07:17.793" LastActivityDate="2014-09-02T22:07:17.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15856" PostTypeId="1" AcceptedAnswerId="15942" CreationDate="2014-09-02T20:25:48.240" Score="3" ViewCount="3223" Body="&lt;p&gt;Here is the experimental data: Note all concentrations are the initial amounts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{cccc}&#xA;\text{Experiment} &amp;amp; \ce{[C2H4Br2]} &amp;amp;\ce{[I-]} &amp;amp; \text{Initial rate of formation of }\ce{I3-}\\\hline&#xA;1 &amp;amp; 0.127 &amp;amp; 0.102 &amp;amp; 6.45\times 10^{-5}\\&#xA;2 &amp;amp; 0.343 &amp;amp; 0.102 &amp;amp; 1.74\times 10^{-4}\\&#xA;3 &amp;amp; 0.203 &amp;amp; 0.125 &amp;amp; 1.26\times 10^{-4}\\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Rate}=k\ce{[C2H4Br2]}^m\ce{[I-]}^n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can determine the exponent $m$ easily because experiment one and two have the same initial $\ce{[I-]}$. However, when determining $n$, there is no data that has the same initial concentration. What the solution manual does is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=&#xA;\dfrac&#xA;{\ln \left(\dfrac{\mathrm{Rate}_{3}\cdot [\ce{C2H4Br2}]_2}&#xA;{\mathrm{Rate}_{2} \cdot [\ce{C2H4Br2}]_3}\right)}&#xA;{\ln \left(\dfrac{[\ce{I-}]_3}{[\ce{I-}]_2}\right)}=\dfrac{\ln \dfrac{(1.26\times 10^{-4})(0.343)}{(1.74\times 10^{-4})(0.203)}}{\ln \left(\dfrac{0.125}{0.102}\right)}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why are they multiplying the rate of the third experiment with the initial concentration from the second experiment and then multiply the rate from the second experiment with the initial concentration of the third experiment. I don't see any correlation with these numbers and how they can be used to determine the exponent.  Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="5026" LastEditDate="2018-03-06T14:23:29.877" LastActivityDate="2018-03-06T14:25:14.197" Title="Determining exponents in a rate law with different data" Tags="&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15857" PostTypeId="2" ParentId="15852" CreationDate="2014-09-02T20:49:52.347" Score="14" Body="&lt;p&gt;From &lt;a href=&quot;http://zonalandeducation.com/mstm/physics/mechanics/energy/heatAndTemperature/changesOfPhase/changeOfState.html&quot;&gt;Changes of Phase (or State)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So, how could there be a change in heat during a state change without a change in temperature?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&quot;During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same.&quot;&lt;/p&gt;&#xA;" OwnerUserId="7677" LastActivityDate="2014-09-02T20:49:52.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15858" PostTypeId="2" ParentId="15850" CreationDate="2014-09-02T21:04:16.440" Score="7" Body="&lt;p&gt;The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into the second to last equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2k_2\ce{[NO_3][NO]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.&lt;/p&gt;&#xA;" OwnerUserId="7653" LastActivityDate="2014-09-02T21:04:16.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15859" PostTypeId="2" ParentId="13989" CreationDate="2014-09-02T21:09:03.300" Score="1" Body="&lt;p&gt;The solutions which you have already found look good so far. For solid &lt;strong&gt;A&lt;/strong&gt;, we need to find a compound which thermally decomposes into $\ce{N2}$ and an element which is only soluble in oxidizing acids like $\ce{HNO3}$, and whose chloride is insoluble in water or acids. &lt;a href=&quot;http://en.wikipedia.org/wiki/Silver_azide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silver azide&lt;/a&gt; could possibly be a good fit, as it can be cleaved into $\ce{Ag}$ and $\ce{N2}$ by heating, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Silver_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;silver chloride&lt;/a&gt; has a very low solubility in water. Care would have to be taken with the temperature when actually performing the decomposition of $\ce{AgN3}$, as it is explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2AgN3 -&amp;gt; 3N2\uparrow + 2Ag}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolving silver (solid &lt;strong&gt;B&lt;/strong&gt;) in nitric acid gives a &lt;a href=&quot;http://en.wikipedia.org/wiki/Silver_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;silver nitrate&lt;/a&gt; solution &lt;strong&gt;D&lt;/strong&gt; and $\ce{NO}$ as colorless gas &lt;strong&gt;E&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag + 4HNO3 -&amp;gt; 3AgNO3 + 2H2O + NO\uparrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO}$ turns red on contact with air when it is oxidized to &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;nitrogen dioxide&lt;/a&gt; by atmospheric oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NO + O2 &amp;lt;=&amp;gt; 2NO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From solution &lt;strong&gt;D&lt;/strong&gt;, insoluble $\ce{AgCl}$ (&lt;strong&gt;F&lt;/strong&gt;) precipitates when $\ce{HCl}$ is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag+ + Cl- -&amp;gt; AgCl\downarrow}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-07-24T01:51:16.590" LastActivityDate="2018-07-24T01:51:16.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15860" PostTypeId="1" AcceptedAnswerId="15861" CreationDate="2014-09-02T21:42:35.807" Score="6" ViewCount="2821" Body="&lt;p&gt;Of 4-nitrobenzaldehyde and 4-aminobenzaldehyde, which has the higher $\ce{K_{eq}}$ of hydration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dLcZf.png&quot; height=&quot;150&quot;&gt; $\qquad \qquad$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PchZS.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steric factors can likely be ignored in this problem since we are dealing with rather structurally similar compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the problem comes down to recognizing that a more polar compound is more favorably solvated by water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the nitro group is &lt;em&gt;always&lt;/em&gt; electron-withdrawing, and I drew a few resonance structures as well. The resonance structures for both, however, seem to suggest that both are rather polarized (given the distribution of formal charges). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Resonance forms for 4-nitrobenzaldehyde&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I expect the ones with the oxygen possessing the incomplete octet are very minor resonance contributors. &lt;/li&gt;&#xA;&lt;li&gt;Therefore it seems that the main thing about this molecule is that the nitro group is withdrawing electron density from the ring and the aldehyde group. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SK3Zt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Resonance forms for 4-aminobenzaldehyde&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The resonance structures here are all somewhat reasonable. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/up8if.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2015-08-13T09:13:53.447" LastActivityDate="2016-07-08T13:22:43.530" Title="Is 4-nitrobenzaldehyde or 4-aminobenzaldehyde more likely to be hydrated?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;nitro-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15861" PostTypeId="2" ParentId="15860" CreationDate="2014-09-02T23:28:45.737" Score="7" Body="&lt;p&gt;I think you're asking about the following equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kczf9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equilibrium will shift to the left as the aldehyde becomes more stabilized.  So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most?  A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing.  Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-02T23:28:45.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15862" PostTypeId="1" AcceptedAnswerId="15867" CreationDate="2014-09-03T00:56:10.967" Score="5" ViewCount="1666" Body="&lt;p&gt;The question reads, &quot; express each of the following quantities using significant figures to imply the stated error&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)$2.3 \pm 0.001$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)$1.989 \pm 0.0003$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the first one is 2.300 because the zero in the thousands place implies an error of $\pm 0.001$ but how do I use sig figs to imply an error of $\pm 0.0003$?&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="73" LastEditDate="2014-09-09T20:02:53.287" LastActivityDate="2014-09-09T20:02:53.287" Title="Significant figures used to imply stated error?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15865" PostTypeId="2" ParentId="10478" CreationDate="2014-09-03T05:29:55.793" Score="2" Body="&lt;p&gt;Also the actinides and lanthanides have very similar chemical behaviour- I think nearly all have a possible +3 oxidation state and similar metallurgical behaviour. For a long time, lanthanides was added as mishmetal to steel alloys.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-09-03T05:29:55.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15866" PostTypeId="1" AcceptedAnswerId="15892" CreationDate="2014-09-03T05:49:20.240" Score="9" ViewCount="10247" Body="&lt;p&gt;Many molecular biology protocols advice against sterilizing HEPES (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) solutions by autoclave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, HEPES is heat labile, and the high temperatures used for sterilization somehow impair the function of the buffer, or perhaps generate unwanted chemical species (this is speculation on my part; I've often seen the advice to not autoclave HEPES, but never seen anything about &lt;em&gt;why not&lt;/em&gt;). Typically autoclaves heat solutions to 121C and increase pressure to slightly above 2 atm, which is about a degree short of the boiling point of water. This environment is maintained for 20 minutes in order to destroy biological contaminants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen this claim often, but had trouble locating an actual source or reasoning for why HEPES should not be autoclaved. Has there been a publication that actually tried to heat a HEPES buffer to autoclave temperatures (121C) and see what happens? What exactly is the consequence, especially with regard to the biological uses of HEPES (ie. pH buffer)?&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2021-01-21T02:32:58.817" Title="What happens if you autoclave HEPES?" Tags="&lt;biochemistry&gt;&lt;heat&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15867" PostTypeId="2" ParentId="15862" CreationDate="2014-09-03T06:49:30.440" Score="7" Body="&lt;p&gt;The use of significant figures always implies some error, that is not explicitly stated. It is therefore irrelevant how big this error is. I merely states that the last digit is rounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case the quantity is given as $2.3\pm0.001$, which gives you the interval $(2.301;2.999)$, which is expressed as $2.300$ in significant figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to the second quantity, which is given as $1.989\pm0.0003$, which in turn gives you the interval $(1.9893;1.9887)$, which is expressed as $1.9890$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-09-03T06:49:30.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15868" PostTypeId="1" CreationDate="2014-09-03T06:58:58.643" Score="1" ViewCount="433" Body="&lt;p&gt;On one hand, the larger the &lt;strong&gt;reduction potential&lt;/strong&gt; of a chemical reaction and the smaller its &lt;strong&gt;activation energy&lt;/strong&gt;, the larger the &lt;em&gt;reaction reaction&lt;/em&gt;. on the other hand the rate of a reaction depends on &lt;strong&gt;thermodynamic conditions&lt;/strong&gt;, such as temperature and pressure, and the concentration of the species involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now given that both the reduction potential and the activation energy of the reaction depend on thermodynamic conditions, can we deduce that the whole thermodynamic dependency of the rate of the reaction lies inside the reduction potential and the activation energy?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the sake of clarification, that I said the larger is the reduction potential and the smaller is the activation energy the greater will be the rate of the reaction is a conclusion based on the analogy of a redox reaction and an electric circuit, the reduction potential being in place of the voltage difference, the activation energy as the energy consumer being in place of the electric resistance and the electron transferred during the reaction being in place of the electric current. It is clear that the greater the electric current, the greater the rate of reaction should be.&lt;/p&gt;&#xA;" OwnerUserId="6980" LastEditorUserId="194" LastEditDate="2017-04-01T01:44:41.657" LastActivityDate="2017-04-01T01:44:41.657" Title="Implicit dependence of reaction rate through reduction potential and activation energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15869" PostTypeId="1" AcceptedAnswerId="15874" CreationDate="2014-09-03T07:21:41.010" Score="3" ViewCount="75" Body="&lt;p&gt;Let we have a reaction that starts from some reactants to produce some products. When the activation energy is provided to the reactants they will be atomized and produce a gas of individual atoms. The gas contain the whole atoms that can either form again the reactants, or form the products, or instead form the elemental molecules of themselves from which the reactants or products can be produced in other reactions. Now that the &lt;strong&gt;gas of all individual atoms&lt;/strong&gt; is common for all these different reactions my question is what conditions decide which reaction to happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I say the reaction happens in the direction at which the Gibbs free enthalpy decreases more rapidly? Or do thermodynamic conditions like temperature, pressure and especially concentration of the species also play a role, independent of the variation in the Gibbs function?&lt;/p&gt;&#xA;" OwnerUserId="6980" LastEditorUserId="194" LastEditDate="2017-09-23T02:04:28.980" LastActivityDate="2017-09-23T02:04:28.980" Title="What reaction is expectable from the atomized reactants?" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;ions&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15870" PostTypeId="2" ParentId="15852" CreationDate="2014-09-03T10:41:11.727" Score="1" Body="&lt;p&gt;Mix water and ice in a container. Keep stirring the mix while the ice is melting. Then the following three statements are true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The temperature of the water is greater than or equal to 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The temperature of the ice is less than or equal to 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The water and the ice have roughly the same temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It follows that the temperature of the mix is roughly 0.&lt;/p&gt;&#xA;" OwnerUserId="7684" LastActivityDate="2014-09-03T10:41:11.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15871" PostTypeId="2" ParentId="15854" CreationDate="2014-09-03T13:11:35.633" Score="9" Body="&lt;p&gt;I guess a $\ce{-BH2}$ group attached to a benzene ring would have this effect. It has an empty $\ce{p}$ orbital that would withdraw electrons via conjugation, but $\ce{B}$ is less electronegative than $\ce{C}$ and so it would have a positive inductive effect.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-03T13:11:35.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15872" PostTypeId="1" CreationDate="2014-09-03T13:45:49.653" Score="7" ViewCount="16024" Body="&lt;p&gt;Why would a hydrogen atom &quot;donate&quot; to fluorine in an ionic bond but not in $\ce{HCl}$?  Why would $\ce{H}$ and chlorine share instead of $\ce{Cl}$ just stripping it away like $\ce{F}$ does?&lt;/p&gt;&#xA;" OwnerUserId="7687" LastEditorUserId="9961" LastEditDate="2017-01-19T20:23:13.160" LastActivityDate="2017-01-19T20:23:13.160" Title="Is the bond in HF ionic while it is covalent in HCl?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15873" PostTypeId="1" AcceptedAnswerId="15878" CreationDate="2014-09-03T15:05:25.787" Score="5" ViewCount="13411" Body="&lt;p&gt;My lecturer said that in a $\ce{C-S}$ bond, sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$, although they have (almost) the same electronegativity. What is the cause of this phenomena if it is not electronegativity? &lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="189" LastEditDate="2014-09-03T15:35:15.480" LastActivityDate="2019-06-28T01:22:12.933" Title="Electron distribution in a C-S bond" Tags="&lt;bond&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15874" PostTypeId="2" ParentId="15869" CreationDate="2014-09-03T15:21:20.117" Score="5" Body="&lt;p&gt;The rate for a reaction is dependent upon the rate constant and, depending upon what type of reaction it is, the concentrations of the reactants.  The rate constant can be further broken down into an entropic factor (how likely are the reactants to collide in an orientation that will produce a reaction) and an energy factor (the activation energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming for the moment that the reactant concentrations are such that they will not control the direction of reaction, and that the entropic factors don't differ dramatically for one pathway over another (that's usually the case for similar reactions), then the first reaction to happen will be the one involving the reaction with the &lt;strong&gt;lowest activation energy&lt;/strong&gt;.  This will lead to the &lt;strong&gt;kinetically controlled product.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the back (reverse) reaction from the kinetically controlled product also has a low enough activation energy such that it happens rapidly under our reaction conditions, then essentially we are setting up an equilibrium.  In this case, thermodynamics will eventually control the product distribution and we will eventually wind up with the product that has the &lt;strong&gt;lowest free energy&lt;/strong&gt;, the &lt;strong&gt;thermodynamically controlled product&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many reactions first produce the kinetically controlled product, but with time, the thermodynamic product becomes the major product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-03T15:21:20.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15876" PostTypeId="1" AcceptedAnswerId="16422" CreationDate="2014-09-03T15:57:54.837" Score="7" ViewCount="4797" Body="&lt;p&gt;It is some kind of everyday lab science and with a feel of shame I cannot answer this observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed a pH drift (~0.5 units) of my 10X phosphate buffer (PBS), when I adjust the pH with $\ce{HCl}$ and dilute the 10X buffer afterwards. &#xA;When I prepare another buffer by mixing $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$ to achieve the desired pH value, there is virtually no drift after dilution. &#xA;If using carbonate buffer, $\ce{CO2}$ is created if I set the pH with $\ce{HCl}$, but none with the also acidic $\ce{NaH2CO3}$ is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know how the phosphate buffer system is working and why there is a pH drift after dilution if I use $\ce{HCl}$ to set the pH. This might answer the general question, why adjusting the pH with $\ce{HCl}$ in PBS is not suggested. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel there could be something wrong with the counter ion maybe?!&lt;/p&gt;&#xA;" OwnerUserId="7082" LastEditorUserId="7082" LastEditDate="2014-09-04T17:19:35.677" LastActivityDate="2017-07-12T14:57:23.070" Title="Adjusting PBS with HCl: pH drift after dilution of phosphate buffer" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15877" PostTypeId="1" AcceptedAnswerId="15879" CreationDate="2014-09-03T16:49:45.660" Score="7" ViewCount="16956" Body="&lt;p&gt;I know that water is 2 parts hydrogen and 1 part oxygen. Hydrogen and oxygen are quite light gasses, they don't weigh very much compared to water. Water is very heavy and it made me wonder exactly how much oxygen I'd need to make a cup of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please answer in terms of cubic meters of air, assuming the air is 21 % oxygen, and that my small cup contains only 100 ml. I suppose the exact question is: &lt;strong&gt;How many cubic meters of air contains the same amount of oxygen as found in 100 ml of water?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="974" LastEditorDisplayName="user7951" LastEditDate="2015-11-09T12:10:44.133" LastActivityDate="2015-11-09T12:10:44.133" Title="How much oxygen is in a cup of water?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15878" PostTypeId="2" ParentId="15873" CreationDate="2014-09-03T16:54:57.517" Score="9" Body="&lt;p&gt;I think your lecturer would have been more correct had he said that the carbon-sulfur bond &lt;strong&gt;reacts&lt;/strong&gt; as if the sulfur is slightly $\delta^{-}$ and the carbon slightly $\delta^{+}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulphur is a larger atom so it has more, loosely held electrons than carbon.  This means that a sulfur atom is more polarizable than a carbon atom.  Although carbon and sulfur have similar electronegativities, &lt;strong&gt;the effect of this increased polarizability for sulfur is that sulfur behaves or reacts as if the bond were polar&lt;/strong&gt;.  Take the case of nucleophilic attack on a $\ce{C-S}$ bond.  As the nucleophile approaches, the $\ce{C-S}$ bond will tend to polarize because of sulfur's polarizability.  And in which direction will the bond polarize?  Electron density will prefer to shift such that it resides on the larger atom (charge to size effect), sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same argument used in the case of a carbon-iodine bond.  Again the two atoms have similar electronegativities (carbon = 2.55, iodine = 2.66), so although the bond is not polar based on electronegativity considerations, the polarizability of the iodine allows the bond to react as if it were polar when a nucleophile approaches.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-03T16:54:57.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15879" PostTypeId="2" ParentId="15877" CreationDate="2014-09-03T17:09:15.810" Score="10" Body="&lt;p&gt;Since the molecular weight of water is $18\ \mathrm{g/mol}$, $100\ \mathrm{ml}$ (density = $1\ \mathrm{g/ml}$, so $100\ \mathrm{ml} \overset{\wedge}{=} 100\ \mathrm{g}$ of water) of water contains &#xA;$$\frac{100}{18} = 5.56$$&#xA;moles of water, or 5.56 moles of atomic oxygen ($\ce{O}$), or 5.56/2 = 2.78 moles of molecular oxygen ($\ce{O2}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At standard temperature and pressure, 1 mole of a gas occupies 22.4 liters.  Therefore, 2.78 moles of molecular oxygen will occupy 62.22 liters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If air is 21% molecular oxygen, then we would need 62.22/0.21, or 296.3 liters of air to provide that much molecular oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 1 cubic meter equals 1000 liters, this volume converts to 0.3 cubic meters of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, 0.3 cubic meters of air contains the same amount of oxygen as is present in 100 ml of water.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-11-09T12:06:44.110" LastActivityDate="2015-11-09T12:06:44.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15881" PostTypeId="2" ParentId="14412" CreationDate="2014-09-03T18:12:22.517" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Microbial_fuel_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;microbial fuel cell&lt;/a&gt; (MFC) is an electrochemical cell that converts chemical into electrical energy, and in which one redox half-reaction is catalyzed by enzymes of microorganisms. The fuel cell consists of anode and cathode compartments which are separated by a membrane that is only permeable for specific cations. At the anode, bioorganic fuel is oxidized to $\ce{CO2}$ and $\ce{H3O+}$. The gained electrons flow from the anode through an external electrical circuit to the cathode, where they are consumed together with protons by the reduction of $\ce{O2}$ to $\ce{H2O}$. The $\ce{H3O+}$ ions are transferred from the anode to the cathode compartment through the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fuel is a medium rich in nutrients which can be oxidatively degraded by the microbes, like, for example, sugars (in the following formula, &lt;a href=&quot;http://en.wikipedia.org/wiki/C12H22O11&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{C12H22O11}$&lt;/a&gt; stands for a variety of disaccharides):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C12H22O11 + 13H2O -&amp;gt; 12CO2 + 48H+ + 48e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons from this half-reaction are either transferred directly to the anode (mediator-free MFC) or via a mediator (mediator MFC), which integrates into the electron transport chain of the cell and transfers electrons to the anode by passing the cell membrane. The mediator gains and loses electrons in every transport cycle, and switches between its reduced form (laden with electrons from the cell) and oxidized form (has given up electrons at the anode).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scheme below (&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:SoilMFC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;: MFCGuy2010/Wikimedia Commons) shows a MFC which uses soil as biofuel. Nutrients from the soil, like acetic acid, are oxidized by microbes in the anodic biofilm to $\ce{CO2}$ and $\ce{H3O+}$. Protons are then consumed in the reduction of supplied oxygen at the cathode, and electrons flow through the external circuit from anode to cathode. In this case, the soil acts simultaneously as nutrient medium and proton exchange membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YyHYz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-03T18:12:22.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15882" PostTypeId="2" ParentId="10959" CreationDate="2014-09-03T18:39:27.180" Score="6" Body="&lt;p&gt;If this is in regard to mechanisms with organometallic complexes or coordination compounds, there are a set of specific mechanisms that are well defined. These are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ligand dissociation / association&lt;/li&gt;&#xA;&lt;li&gt;Reductive elimination / oxidative addition&lt;/li&gt;&#xA;&lt;li&gt;Intramolecular insertion (e.g. CO migratory insertion) / elimination (e.g. &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;–hydride elimination)&lt;/li&gt;&#xA;&lt;li&gt;Metallacycle processes (e.g. &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;-bond metathesis)&lt;/li&gt;&#xA;&lt;li&gt;Nucleophilic / electrophilic addition and abstraction are also possible, similar to organic chemistry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These mechanisms should be in inorganic textbooks but can also be viewed on the first page &lt;a href=&quot;http://www.scs.illinois.edu/white/lectures/week2.2012.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The later slides in this pdf cover more of the details of each specific mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many reaction mechanisms with organometallic complexes, it is a process of a number of these smaller mechanisms occurring, leading to the overall chemical transformation that is stated in a reaction equation. These overall mechanisms can be complicated and need to be evaluated for each case. Knowing the basic underlying mechanisms will certainly help clarify processes that seem quite complex at first glance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are subtleties in all of these mechanisms to consider beyond the basic types above. Significantly more detail can be found in these textbooks that I have previously utilized which you may also find informative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collman, J.; Hegedus, L.; Norton, J.; Finke, R. &lt;em&gt;Principles and Applications of Organotransition Metal Chemistry&lt;/em&gt;. 1987. University Science Books. Mill Valley, California.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spessard, G.; Miessler, G. &lt;em&gt;Organometallic Chemistry&lt;/em&gt;. 1997. Prentice-Hall. Upper Saddle River, New Jersey.&lt;/p&gt;&#xA;" OwnerUserId="7653" LastEditorUserId="5017" LastEditDate="2019-12-29T16:47:35.380" LastActivityDate="2019-12-29T16:47:35.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15883" PostTypeId="1" CreationDate="2014-09-03T18:45:03.947" Score="4" ViewCount="670" Body="&lt;p&gt;I asked my general chem instructor this and he didn't know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mendeleev was able to group elements and discovered a bunch of them and put them in the correct order in the mid-1800s.  What equipment did he use to do this?  Was he just weighing stuff and guessing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, how did he figure out atomic weight?&lt;/p&gt;&#xA;" OwnerUserId="1101" LastEditorUserId="73" LastEditDate="2014-09-03T21:53:56.480" LastActivityDate="2014-09-04T03:11:49.143" Title="How was Mendeleev able to develop his table?" Tags="&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15884" PostTypeId="2" ParentId="15883" CreationDate="2014-09-03T19:25:52.683" Score="4" Body="&lt;p&gt;This quote from the &lt;a href=&quot;http://en.wikipedia.org/wiki/History_of_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article on History of Chemistry&lt;/a&gt; might provide some insight into the determination of atomic mass:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemists in different countries developed several different incompatible atomistic systems. A paper that suggested a way out of this difficult situation was published as early as 1811 by the Italian physicist Amedeo Avogadro (1776-1856), who hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, from which it followed that relative molecular weights of any two gases are the same as the ratio of the densities of the two gases under the same conditions of temperature and pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this implies that you could take equal volumes of different gases at the same temperature and pressure, then weigh them and take the ratios. This would allow you to get relative masses for $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$. Other gases could be used, but they wouldn't be as useful here. You could then determine that $\ce{H2}$ is the lightest gas, and that $\ce{O2}$ is 16 times heavier, $\ce{N2}$ is 14 times heavier, and so on. You don't even need to know that these are diatomic gases in order to make these measurements. Once you had these determined you could move on to gases like $\ce{CO2}$, or $\ce{H2S}$. Once you knew the mass of the gas molecule and what elements were in it, you could calculate how much of each element must be there, and by subtracting the mass of the known element, the remaining mass must be the atomic mass for the other element. As you accumulated more data you could approach substances that don't easily form gases, such as $\ce{NaCl}$. If you know the mass for $\ce{Cl}$ you could determine the mass for $\ce{Na}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I suggest might not be the way atomic masses were determined, but it would probably work. Looking at this image of Mendeleev's original table shows they already had pretty good estimates of atomic mass by 1869.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p6KNE.png&quot; alt=&quot;Mendeleev&amp;#39;s Periodic Table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="4945" LastEditDate="2014-09-04T03:11:49.143" LastActivityDate="2014-09-04T03:11:49.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15885" PostTypeId="1" AcceptedAnswerId="15887" CreationDate="2014-09-03T19:27:30.543" Score="9" ViewCount="12642" Body="&lt;p&gt;I noticed, and perhaps many others have too, that the application of hand sanitizer (mainly ethanol), leaves one's hands feeling rather chilly after application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is responsible for this phenomena? Is it the high heat of vaporization of hand sanitizer? However, explanation doesn't hold water; water has a standard heat of vaporization of 40.65 kJ/mol while ethanol has a heat of vaporization of 38.56 kJ/mol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be the low boiling point of ethanol? Hand sanitizer disappears (vaporizes) within seconds upon rubbing the hands together. Water, however, does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, how does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-23T17:03:11.723" LastActivityDate="2018-10-24T18:19:48.483" Title="Why does hand sanitizer leave your skin feeling cool?" Tags="&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15887" PostTypeId="2" ParentId="15885" CreationDate="2014-09-03T20:12:38.300" Score="7" Body="&lt;p&gt;As user137 noted, this is from evaporation of the alcohol. Evaporation of a liquid takes the enthalpy of vaporization away with it, making you cooler.  Just like how sweating helps keep you cool in the summer, and how swamp coolers (evaporative coolers) keep houses cool in the southwest.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-09-03T20:12:38.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15889" PostTypeId="2" ParentId="15872" CreationDate="2014-09-03T22:04:07.363" Score="18" Body="&lt;p&gt;Bonds can be completely covalent as in $\ce{Cl-Cl}$, $\ce{H-H}$, etc. In these cases the electron density is shared equally between the two atoms.  Bonds can also be ionic as in $\ce{Na^{+} Cl^{-}}$ where much (~80%) of an electron has been transferred from the sodium atom to the chlorine atom.  Between these two extreme cases exist a continuum of bonds that are called polar, covalent bonds.  In such cases the electron density between the two atoms is not shared equally, but neither is the electron transferred extensively from one atom to the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as we can draw resonance structures for acetone and discuss how the real molecule is a weighted superposition of these two structures, we can draw resonance structures for inorganic compounds and, again, discuss how much the various structures count towards the &quot;real&quot; description of the molecule.  In the diagram below, two resonance structures are shown for sodium chloride.  The structure on the left represents a covalent bond and the structure on the right represents an ionic bond. Just as acetone is a mix of the two resonance structures, so is sodium chloride.  When the weighted mix of the resonance structures lies far to the left, we can say the compound or bond is covalent; when it lies far to the right we can say the compound or bond is ionic.  Everything in between represents a polar, covalent bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RUAZH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can crudely estimate how polar a diatomic covalent bond is by using the following methodology.  The dipole moment of a molecule is given by&#xA;$$\ce{\mu~=~Q*d}$$&#xA;where the dipole moment is in Debye (D), Q is the partial charge on an atom measured in coulombs and d is the distance between the two atoms (the bond length) in meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{HF}$, the dipole moment is 1.86 D and the bond length is 91.7 x 10^(-12) m.  Noting that 1 D = 3.3356 × 10^(−30) C·m, solving for Q we find a partial charge on one of the atoms of &#xA;$$\mathrm{Q~=~\frac{[1.86 ~x~3.3356 ~x~ 10^{-30} ]}{[91.7~x~10^{-12}]}~=~6.77~x~10^{-20}~C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since one electron equals 1.6022 × 10^(−19) C, by taking the ratio&#xA;$$\mathrm{\frac{[6.77~x~10^{-20}]}{[1.6022 × 10^{−19}]}~=~ 0.42}$$&#xA;we estimate that roughly 42% of an electron has been transferred from the hydrogen atom to the fluorine atom.  This gives us a rough idea of how ionic the polar, covalent bond is in $\ce{HF}$.  Repeating this calculation for $\ce{HCl}$ suggests that roughly only 18% of an electron has been transferred in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the polar, covalent bond in $\ce{HF}$ has more ionic character than the analogous polar, covalent bond in $\ce{HCl}$.  The reason why the $\ce{HF}$ bond has more ionic character is the greater electronegativity of fluorine (3.98) compared to chlorine (3.16).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:50:06.380" LastActivityDate="2016-03-12T00:50:06.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15890" PostTypeId="1" AcceptedAnswerId="15893" CreationDate="2014-09-03T22:04:31.827" Score="3" ViewCount="470" Body="&lt;p&gt;I need to compute the root mean square deviation (RMSD) between two protein structures: the X-ray determined protein structure and the MOPAC optimized structure. Which software do you suggest?&lt;/p&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="4945" LastEditDate="2014-09-04T03:04:44.613" LastActivityDate="2014-09-12T18:11:25.483" Title="Which software is suitable to calculate the root mean square deviation between two protein structures?" Tags="&lt;biochemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15891" PostTypeId="2" ParentId="15890" CreationDate="2014-09-03T23:35:07.210" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;nofollow&quot;&gt;VMD&lt;/a&gt;.  The built-in scripting language is insanely powerful.&lt;/p&gt;&#xA;" OwnerUserId="7522" LastEditorUserId="73" LastEditDate="2014-09-04T09:06:47.097" LastActivityDate="2014-09-04T09:06:47.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15892" PostTypeId="2" ParentId="15866" CreationDate="2014-09-04T01:27:46.857" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;https://www.promega.com/resources/product-guides-and-selectors/protocols-and-applications-guide/buffers-for-biochemical-reactions/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some buffers (e.g., MOPS and HEPES) cannot be autoclaved because they degrade upon heating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and from &lt;a href=&quot;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma/Product_Information_Sheet/2/h3784pis.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[HEPES] Solutions may be autoclaved under standard conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess nobody knows the answer.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-04T01:27:46.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15893" PostTypeId="2" ParentId="15890" CreationDate="2014-09-04T03:17:56.693" Score="6" Body="&lt;p&gt;All of these are freely available packages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Visual Molecular Dynamics&lt;/strong&gt;&lt;/a&gt; - good visualization tool that can also do quick-and-dirty calculations from a graphical interface. The scripting interface has a learning curve, but yes, it can be quite powerful.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Avogadro&lt;/strong&gt;&lt;/a&gt; - not as powerful as VMD, but can be useful for certain things.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rasmol.org/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Rasmol&lt;/strong&gt;&lt;/a&gt; - Similar to VMD, although I've never used it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://lammps.sandia.gov/&quot; rel=&quot;nofollow&quot;&gt;LAMMPS&lt;/a&gt;, &lt;a href=&quot;http://www.gromacs.org/&quot; rel=&quot;nofollow&quot;&gt;Gromacs&lt;/a&gt;, or &lt;a href=&quot;http://www.ks.uiuc.edu/Research/namd/&quot; rel=&quot;nofollow&quot;&gt;NAMD&lt;/a&gt;&lt;/strong&gt; - If one of these is installed on a computer you have access to, each of them can do this sort of computation and will allow you to do more advanced things if needed (they are actually molecular dynamics simulators). Each of these packages requires some familiarity with running code from the command-line and will have a learning curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could always write your own code. RMSD is a pretty straightforward computation, and it's always nice to have more control over your analysis!&lt;/p&gt;&#xA;" OwnerUserId="7693" LastEditorUserId="73" LastEditDate="2014-09-04T09:04:23.610" LastActivityDate="2014-09-04T09:04:23.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15894" PostTypeId="1" AcceptedAnswerId="15896" CreationDate="2014-09-04T04:47:50.173" Score="3" ViewCount="2602" Body="&lt;p&gt;Question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prior to 1961, the atomic mass unit was defined as 1/16 the&#xA;  mass of the atomic mass of oxygen; that is, the atomic mass&#xA;  of oxygen was defined as exactly &lt;span class=&quot;math-container&quot;&gt;$\pu{16 amu}$&lt;/span&gt;. What was the&#xA;  mass of a &lt;span class=&quot;math-container&quot;&gt;$\ce{^{12}C}$&lt;/span&gt; atom prior to 1961 if the atomic mass of oxygen on today’s scale is &lt;span class=&quot;math-container&quot;&gt;$\pu{15.9994 amu}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't fully understand the logic behind the question. Here's my thinking. Atomic mass of all substances were related to &lt;span class=&quot;math-container&quot;&gt;$\pu{16.0000 amu}$&lt;/span&gt; prior to 1961. Therefore, we can make the following expression where &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; is mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{^{12}C}$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{X}{\pu{16.0000amu}}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, all atomic mass is related to &lt;span class=&quot;math-container&quot;&gt;$\pu{12.0000 amu}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{^{12}C}$&lt;/span&gt;. We can model the relationship as &lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the solution manual does is make the two equations equal to each other and solves for &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{X}{\pu{16.0000amu}}=\dfrac{\pu{12.0000amu}}{\pu{15.9994amu}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are we justified in doing this?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="83758" LastEditDate="2020-05-01T09:05:16.013" LastActivityDate="2020-05-01T09:05:16.013" Title="Atomic Mass of ¹²C before 1961" Tags="&lt;units&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15895" PostTypeId="2" ParentId="15894" CreationDate="2014-09-04T04:59:54.860" Score="1" Body="&lt;p&gt;Think about what happened to the scale. For $\ce{^{16}O}$, it decreased from $16.0000\,\mathrm{amu}$ to $15.9994\,\mathrm{amu}$. So the scale was stretched out (since 0 needs to stay constant). That would mean that $\ce{^{12}C}$ would need to decrease from some number greater than $12.0000\,\mathrm{amu}$. With $0$ being constant, there is no addition factor (like when converting from Fahrenheit to Centigrade); it is just a ratio from one scale to the other.&#xA;$$\frac {\text{unknown}}{16.0000\,\mathrm{amu}}=\frac {12.0000\,\mathrm{amu}}{15.9994\,\mathrm{amu}}$$&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2014-09-04T07:31:21.000" LastActivityDate="2014-09-04T07:31:21.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15896" PostTypeId="2" ParentId="15894" CreationDate="2014-09-04T05:05:51.723" Score="5" Body="&lt;p&gt;Here is another equivalent way of thinking about the problem. The mass of oxygen on the old scale is 16 amu. The mass of oxygen on the new scale is 15.994 amu. When you divide them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} = 1.0000375 \frac{\textrm{old}}{\textrm{new}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, taking a ratio of any value in the old scale to the equivalent value in the new scale (or vice versa) gives you a conversion factor. So for every new amu, there is 1.0000375 old amus. The question is asking you to convert 12 new amus to old amus. Just by looking at the units, it should be obvious what to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{16\ \textrm{old}}{15.9994\ \textrm{new}} * 12\ \textrm{new}=12.00045\ \textrm{old}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you explicitly label old and new, it's easy to see that the new units cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Incidentally, if you keep track of your significant figures, you can see why this redefinition of the amu didn't cause too many problems for chemists—even to five sig figs, the change is negligible.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-04T05:05:51.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15897" PostTypeId="1" AcceptedAnswerId="15902" CreationDate="2014-09-04T06:10:01.917" Score="0" ViewCount="2008" Body="&lt;p&gt;Although I am now taking a pretty fundamental chemistry course, I am really looking to gain an extremely comprehensive knowledge of the subject, specifically, in inorganic and organic chemistry. I am looking for two texts, keeping in mind that the most important thing is &lt;strong&gt;comprehensiveness&lt;/strong&gt; and the most information about the most topics, not ease of readability. I have the time and dedication to do it, but I really don't want to start with a book that doesn't cover everything and I have to fill in the holes as I go or individually seek out topics it misses. &lt;/p&gt;&#xA;" OwnerUserId="7694" LastEditorUserId="4945" LastEditDate="2014-09-04T08:07:43.503" LastActivityDate="2014-09-04T08:31:06.177" Title="Most Comprehensive Chemistry Book(s)?" Tags="&lt;books&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ClosedDate="2014-11-04T05:30:55.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="15898" PostTypeId="2" ParentId="15742" CreationDate="2014-09-04T06:54:58.577" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One day I came to know that on mixing salt and water I will get Hydrochloric Acid and Sodium Hydroxide [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you won't!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, electrolysis of a sodium chloride solution and photolysis of the combined gases formed (hydrogen, chlorine) will provide you with hydrogen chloride, as I have outlined &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8948/how-to-separate-hcl-and-naoh-from-salt-and-water-solution/8950#8950&quot;&gt;here&lt;/a&gt;.  In reality, you might want to avoid this - unless you're planning to jump out of the genetic pool ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-09-04T06:54:58.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15899" PostTypeId="1" CreationDate="2014-09-04T08:08:08.977" Score="5" ViewCount="946" Body="&lt;p&gt;If I raise the temperature of a solution of a gas in a liquid solvent, why won't the solubility increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the pressure will increase the solubility, but increasing the pressure doesn't mean increasing the temperature?  What is wrong with this argument?&lt;/p&gt;&#xA;" OwnerUserId="7695" LastEditorUserId="73" LastEditDate="2014-09-04T17:00:20.020" LastActivityDate="2014-09-04T17:00:20.020" Title="Why won't increasing the temperature increase the solubility of a gas in a liquid solvent?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15900" PostTypeId="2" ParentId="15648" CreationDate="2014-09-04T08:10:21.583" Score="4" Body="&lt;p&gt;In the mechanism you are considering, there are two options for reduction of R-$\ce{N^{.}H}$ (and similar for other neutral radical species)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CH3OH + R-N^{.}H -&amp;gt; RNH2 + C^{.}H_2OH }$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CH3OH + R-N^{.}H -&amp;gt; RNH2 + CH3O^{.}} $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the first option (1) nitrogen oxidizes carbon - rare, but possible, as carbon has lower electronegativity. In option (2) nitrogen oxidizes oxygen - an element that has higher electronegativity, and this is unlikely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, in option (1) free radical is stabilized by forming 3-electron 2-center bond $\pi$-bond C...O, while in option (2) such stabilization cannot occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;em&gt;proton&lt;/em&gt; (i.e. hydrogen cation) from $\ce{OH}$ is not an option for reduction, and $\ce{R-N^{.}H}$ is a radical, and expects hydrogen &lt;em&gt;atom&lt;/em&gt;, not ion. Taking proton will produce $\ce{R-N^{+.}H2}$ and $\ce{CH3O-}$ and the first particle is not much better than the original $\ce{R-N^{.}H}$ while heterolitic bond breaking is generally less favorable, than radical one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proton may be accepted by negatively charged radical anions, like $\ce{R-N(O)O^{-.}}$ but after that they will seek a hydrogen &lt;em&gt;atom&lt;/em&gt; for reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it IS theoretically possible, that the radicals are actually reduced by tin and all the reaction happens on metal surface, and this is likely the case if the reaction happens in water, but in case it happens in organic solvent, especially reducing one like methanol, it is theoretically possible that it does participate in reduction. Until proper proofs, like trapping of the intermediate species, are obtained, any mechanism is no more then speculation anyway.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="7614" LastEditDate="2014-09-10T18:12:53.223" LastActivityDate="2014-09-10T18:12:53.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15901" PostTypeId="2" ParentId="15897" CreationDate="2014-09-04T08:21:20.317" Score="0" Body="&lt;p&gt;Here is the list some books which i fill it is good and detailed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.amazon.in/Chemistry-Course-Book-2014-Programme/dp/0198392125?tag=googinhydr18418-21&quot; rel=&quot;nofollow&quot;&gt;IB Chemistry Course Book 2014 edition: Oxford IB Diploma Programme&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486656225&quot; rel=&quot;nofollow&quot;&gt;General Chemistry (Dover Books on Chemistry)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.amazon.in/Comprehensive-Chemistry-Dr-N-K-Verma/dp/938038694X/ref=sr_1_2?ie=UTF8&amp;amp;qid=1409818775&amp;amp;sr=8-2&amp;amp;keywords=comprehensive+chemistry&quot; rel=&quot;nofollow&quot;&gt;Comprehensive Chemistry by Dr. N.K. Verma, Prof. S.K. Khanna and Dr. B. Kapila &lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-04T08:21:20.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15902" PostTypeId="2" ParentId="15897" CreationDate="2014-09-04T08:31:06.177" Score="4" Body="&lt;p&gt;For the inorganic chemistry I recommend Holleman-Wiberg's text, also exists in english. It is reasonably readable and very thick and contains a lot. &#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0123526515&quot; rel=&quot;nofollow&quot;&gt;Holleman-Wiberg's Inorganic Chemistry&lt;/a&gt;. If you prefer, the original is in German and updated very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the organic chemistry, besides being thick also very nicely readable,  &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199270295&quot; rel=&quot;nofollow&quot;&gt;Organic Chemistry by Clayden, Greeves and Warren &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides that, you would probably need some nice general chemistry book, but I have no good personal experience in this area.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-09-04T08:31:06.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15903" PostTypeId="1" AcceptedAnswerId="16136" CreationDate="2014-09-04T08:43:26.487" Score="1" ViewCount="159" Body="&lt;p&gt;I'm doing homework for Inorganic Chemistry, and I am a little confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Question is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the electronic configuration of $\ce{Ru^{2+}}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I were to remove from the highest energy level it would be $\ce{[Kr]~5s^1~4d^5}$&#xA;leaving $6$ unpaired electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to take into account electron pairing it would be $\ce{[Kr]~4d^6}$ with $4$ unpaired electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the correct approach?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="865" LastEditDate="2021-05-22T20:52:20.507" LastActivityDate="2021-05-22T20:52:20.507" Title="What would the Electronic Configuration of Ru²⁺ be?" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15904" PostTypeId="2" ParentId="15899" CreationDate="2014-09-04T10:12:43.913" Score="5" Body="&lt;p&gt;A gas has bigger entropy when it's free, than in a solution. By increasing the temperature the molecules have bigger energy and they tend to be in a state with more freedom. Therefore, the solubility would decrease.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-04T10:12:43.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15905" PostTypeId="2" ParentId="15903" CreationDate="2014-09-04T11:07:05.557" Score="3" Body="&lt;p&gt;You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to know know is the &lt;strong&gt;order of energies&lt;/strong&gt; for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aufbau principle&lt;/a&gt; (and possible pitfalls/exceptions) and another &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7959/electron-configuration-of-tellurium&quot;&gt;thread&lt;/a&gt; on this site. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-09-04T11:07:05.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15907" PostTypeId="1" AcceptedAnswerId="15909" CreationDate="2014-09-04T13:23:40.580" Score="19" ViewCount="6113" Body="&lt;p&gt;A hybrid sp3 orbital is drawn with one lobe smaller than its other half, the latter which is of equal size when drawing the p orbital. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="14" LastEditDate="2014-09-04T15:25:26.007" LastActivityDate="2017-04-29T10:37:04.777" Title="Why is one lobe of an sp3 hybridized orbital smaller than its other half?" Tags="&lt;orbitals&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15908" PostTypeId="2" ParentId="2644" CreationDate="2014-09-04T13:50:48.483" Score="3" Body="&lt;p&gt;Electrons cannot survive in aqueous state. Being a charged subatomic particle, the electron&#xA;has to stay close to protons which are located at the center of the atom. Hence, the electron&#xA;can move from one atom to another which are closely-packed, what we have in a solid.&lt;/p&gt;&#xA;" OwnerUserId="7698" LastActivityDate="2014-09-04T13:50:48.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15909" PostTypeId="2" ParentId="15907" CreationDate="2014-09-04T14:21:49.250" Score="26" Body="&lt;p&gt;When combined at a given atomic center, any two atomic orbitals add in a vectorial&#xA;manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbital addition can be pictured like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nwsvP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the two cases $c_1 = c_2 &amp;gt; 0$ and $c_l = -c_2 &amp;gt; 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mixing of atomic orbitals with different angular momentum quantum number is&#xA;also controlled by a vectorial addition and leads to various types of hybrid orbitals&#xA;some examples of which are shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dW1J8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the  $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger. &lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-04T14:21:49.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15910" PostTypeId="2" ParentId="15899" CreationDate="2014-09-04T14:22:38.430" Score="2" Body="&lt;p&gt;When a gas dissolves in a liquid, bonds are formed between the gas particles and the liquid particles. The formation of these bonds liberates energy. Increasing the temperature of the solution will shift  the equilbrium backwards. Hence less gas can dissolve and solubility&#xA;decreases.&lt;/p&gt;&#xA;" OwnerUserId="7698" LastActivityDate="2014-09-04T14:22:38.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15911" PostTypeId="2" ParentId="15006" CreationDate="2014-09-04T14:32:43.353" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propene&quot; rel=&quot;nofollow&quot;&gt;Propene&lt;/a&gt; gas is flammable and has a higher density ($1.81 \ \mathrm{kg/m^3}$) than air. It therefore easily forms explosive mixtures with air when it accumulates near ground level or, like in this case, inside an underground channel without much ventilation. While propene itself is color- and odourless, it is described in &lt;a href=&quot;http://en.wikipedia.org/wiki/2014_Kaohsiung_gas_explosions&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; that &quot;eyewitnesses reported a smell of gas (...) coming out of manholes&quot;. This smell comes from volatile compounds which are added to the gas in small amounts to make the detection of leaking gas easier. However, by the time that the smell was noticed, enough propene had already accumulated underground to exceed explosion limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detection of underground gas leaks by odour alone is therefore not a reliable means to prevent possible explosions, namely when the gas is heavier than air. It is also not possible to exactly localize the leak when the gas escapes from multiple manholes which are spread over a large area. Gas sensors in the underground pipeline network could provide earlier warnings about leaks, and gas accumulation could be prevented in this case by shutting down the affected pipeline. However, prevention of leaks in the first place (by regular inspections) should still have higher priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.gasinnovations.com/literature/MSDS-Sheets/MSDS-PROPYLENE.pdf&quot; rel=&quot;nofollow&quot;&gt;MSDS data sheet&lt;/a&gt; for propene states that shutting down gas supply and letting the fire burn out while cooling the surroundings with water is the recommended way of extinguishing burning propene. So using water to cool down the roads with the aim to reduce ground temperature was not completely off, but trying to fight the fire directly with water is unadvisable.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-04T14:32:43.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15912" PostTypeId="1" AcceptedAnswerId="15914" CreationDate="2014-09-04T15:19:02.613" Score="5" ViewCount="1909" Body="&lt;p&gt;A kid asked me this question just the other day, and I said that the positive part of $\ce{H2O}$ attracts the negative $\ce{Cl}$ ion and the negative part of the $\ce{H2O}$ attracts the positive Na ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he asked me several questions which stumped me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why doesn't $\ce{NaCl}$ dissolve water instead (break water into $\ce{H}$ and $\ce{O}$)?&lt;/li&gt;&#xA;&lt;li&gt;If $\ce{H2O}$ is electrically neural, why would it attract any ions?&lt;/li&gt;&#xA;&lt;li&gt;Why doesn't $\ce{NaCl}$  stick together, how can one know if one molecule will attract the ions hard enough that the ions breaks away?&lt;/li&gt;&#xA;&lt;li&gt;Why is Na positive and $\ce{Cl}$ negative?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hopefully someone can help me answer the above questions as I have not done chemistry in years!&lt;/p&gt;&#xA;" OwnerUserId="7699" LastEditorUserId="14" LastEditDate="2014-09-04T15:38:23.697" LastActivityDate="2014-09-04T18:35:28.363" Title="Why does water break salt really" Tags="&lt;molecules&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15913" PostTypeId="2" ParentId="15899" CreationDate="2014-09-04T15:19:10.207" Score="1" Body="&lt;p&gt;In reference to Henry's Law,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;At a constant temperature, the amount of a given gas that dissolves&#xA;  in a given type and volume of liquid is directly proportional to the&#xA;  partial pressure of that gas in equilibrium with that liquid.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, increasing partial pressure of gas above the liquid increases solubility of gas, which does NOT mean that temperature of the solution is increasing. There is a distinction to be made between the dynamics of the solution and the gas above the solution.&lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-09-04T15:19:10.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15914" PostTypeId="2" ParentId="15912" CreationDate="2014-09-04T15:30:40.917" Score="5" Body="&lt;p&gt;1.) The covalent OH bond strength in water is slightly stronger than the NaCl ionic bond strength.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Water might be neutral but each atom exhibits a partial charge.  Oxygen sucks up most of the electron density away from the hydrogen atoms leaving O with a partial negative charge and hydrogen with a partial positive charge.  This gives rise to the famous hydrogen bonding behavior of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) You would have to analyze the energies of each of the interacting components to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) Cl is more electronegative than Na due to the larger number of protons in the nucleus.  This means that Cl will suck electron density away from Na as opposed to the other way around.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-09-04T15:30:40.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15915" PostTypeId="1" AcceptedAnswerId="15917" CreationDate="2014-09-04T15:51:29.913" Score="2" ViewCount="22200" Body="&lt;p&gt;Is the amine group a donating or withdrawing group? I think it depends on resonance. If the amine group can participate in resonance, the amine group is inductively donating. Resonance depictions below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tjcWp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-02T16:49:39.820" LastActivityDate="2015-10-02T16:49:39.820" Title="Amine group - electron donating or withdrawing group?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15917" PostTypeId="2" ParentId="15915" CreationDate="2014-09-04T16:21:57.320" Score="6" Body="&lt;p&gt;When the atome of nitrogen participates through its free electron pair in resonance, the amine group plays the role of a donating group by conjugation. When it cannot participate in conjugation through resonance, the amine groupe plays the role of inductive electron-withdrawer group, due to the higher electronegativity of nitrogen atom compared to carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the amine group plays the role of electron donating group through conjugation, its inductive withdrawing effect (still existing) is weaker and masked by the first effect.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2014-09-04T17:29:29.597" LastActivityDate="2014-09-04T17:29:29.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15918" PostTypeId="1" AcceptedAnswerId="15919" CreationDate="2014-09-04T16:26:18.237" Score="3" ViewCount="4602" Body="&lt;p&gt;I am told that this method yields ONE equivalent of water, but I seem to get two equivalents of water. Also does this mechanism actually work? I've never seen anything like it, but it seems to make sense. Would another Lewis acid such as boron trifluoride also work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l0o7p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/s1Ciq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="194" LastEditDate="2017-09-23T01:51:48.417" LastActivityDate="2017-09-23T01:51:48.417" Title="Dehydration of alcohol with Lewis acid aluminum trichloride" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15919" PostTypeId="2" ParentId="15918" CreationDate="2014-09-04T16:46:56.743" Score="8" Body="&lt;p&gt;In the reaction $\ce{{}^{$i$}Pr-OH + H2O -&amp;gt;[\ce{AlCl3}] C3H6 + 2H2O}$ the $\ce{H2O}$ on the reactant side and one of the $\ce{H2O}$s on the product side cancel, so you get $\ce{{}^{$i$}Pr-OH -&amp;gt;[\ce{AlCl3}] C3H6 + H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the reaction would also work with other Lewis acids. But it will rather proceed through the $\mathrm{E}1$ mechanism in any case. And it is likely that you'll get $\mathrm{S}_\mathrm{N}1$ side reactions.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-04T16:55:00.620" LastActivityDate="2014-09-04T16:55:00.620" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15920" PostTypeId="1" AcceptedAnswerId="15922" CreationDate="2014-09-04T16:47:00.360" Score="1" ViewCount="228" Body="&lt;p&gt;I am told that for these two molecules, one of them is not as resonance stabilized as the other. Apparently it's the chlorine one, and it's because of the mismatch in the size of chlorine and carbon. This makes sense &lt;em&gt;prima facie&lt;/em&gt;, but can someone confirm this is true? I can see how carbon and nitrogen have similarly sized p orbitals and thus have good overlap, and how chlorine and carbon might have poor p orbital overlap, like a BB pellet with a basketball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1mGzt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-05T04:23:47.550" LastActivityDate="2014-09-05T04:23:47.550" Title="Resonance stabilization and size of ligand atoms" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15921" PostTypeId="2" ParentId="15918" CreationDate="2014-09-04T16:50:24.427" Score="3" Body="&lt;p&gt;$\ce{BF3*OEt2}$ in $\ce{CH2Cl2}$ can be used too, at least for the dehydration of tertiary alcohols [&lt;a href=&quot;http://dx.doi.org/10.1016/0040-4039(91)80200-P&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the moles of water yielded, please see Philipp's answer and use your common sense on whether the formation of a second molecule of water &lt;em&gt;out of nothing&lt;/em&gt; is likely.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-04T16:50:24.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15922" PostTypeId="2" ParentId="15920" CreationDate="2014-09-04T17:11:24.383" Score="1" Body="&lt;p&gt;Yes, your analysis is accurate. The nitrogen is donating electrons through a 2p orbital which is similar in size to the carbon and oxygen 2p orbitals. The chlorine is donating electrons through a 3p orbital, which is significantly larger than the second row elements' 2p orbitals.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-04T17:11:24.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15923" PostTypeId="2" ParentId="15920" CreationDate="2014-09-04T17:11:54.643" Score="3" Body="&lt;p&gt;Yes, this is true. The $\pi$ overlap between a valence $\ce{p}$ orbital of a second row element and one of a higher row element is much smaller than the $\pi$ overlap between two second row elements. One reason is that the bond length is generally higher when higher row elements are involved (due to their size). Another reason is, that the $3 \ce{p}$ or $4 \ce{p}$ orbitals are much more diffuse than the $2 \ce{p}$ orbitals. If you want to read deeper into the topic have a look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] W. Kutzelnigg, &lt;em&gt;Angew. Chem. Int. Ed. Engl.&lt;/em&gt;, &lt;strong&gt;23&lt;/strong&gt;, 272 (1984).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] N. C. Norman, &lt;em&gt;Polyhedron&lt;/em&gt;, &lt;strong&gt;12&lt;/strong&gt;, 2431 (1993).&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-04T17:11:54.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15924" PostTypeId="1" AcceptedAnswerId="15925" CreationDate="2014-09-04T17:45:07.643" Score="0" ViewCount="50" Body="&lt;p&gt;I'm a very beginner in chemistry. The book I'm studying says that an ion is an atom, or a collection of atoms, that has an electric charge, then it writes about the difference between molecules and ions. What I'm wondering is, if I consider a unit of, say, sodium chloride (by unit I mean one atom of sodium plus one atom of chlorine), should I say &quot;a molecule of s. chloride&quot; or &quot;an ion of s. chloride&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7665" LastActivityDate="2014-09-04T17:58:45.830" Title="Chemical nomenclature of ions" Tags="&lt;molecules&gt;&lt;ions&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15925" PostTypeId="2" ParentId="15924" CreationDate="2014-09-04T17:58:45.830" Score="0" Body="&lt;p&gt;It is said: &lt;a href=&quot;http://en.wikipedia.org/wiki/Formula_unit&quot; rel=&quot;nofollow&quot;&gt;one formula unit of sodium chloride&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-04T17:58:45.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15926" PostTypeId="1" CreationDate="2014-09-04T18:31:36.683" Score="3" ViewCount="4653" Body="&lt;p&gt;I was wondering why the elements in the periodic table were disposed the way they are. I understand, of course, that they are put in increasing atomic number fashion, but I'd like to know more about the topic. Could you link some reference that examine in detail the structure of the periodic table?&lt;/p&gt;&#xA;" OwnerUserId="7665" LastEditorUserId="7475" LastEditDate="2016-11-02T16:01:15.350" LastActivityDate="2017-12-06T03:39:36.723" Title="According to what criteria was the periodic table of elements ordered?" Tags="&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15927" PostTypeId="2" ParentId="15912" CreationDate="2014-09-04T18:35:28.363" Score="1" Body="&lt;p&gt;The main reason about why water breaks salt is related to the relative permittivity $\epsilon_r$ of  water compared to its value in air. In fact,$\epsilon_r= 80$ in water, while its value in air is $\epsilon_r= 1$.&#xA;Now, if we look at the formula of the Coulomb force, which is responsible of the electrostatic attraction between ${\rm Na}^+$ and ${\rm Cl}^-$ :&#xA;$$f = \frac{q.q'}{4\pi \epsilon_0\epsilon_r r^2}$$&#xA;(Where q and q' are the charges of  ${\rm Na}^+$ and ${\rm Cl}^-$, and r is the distance between the two punctual charges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We notice that, this force is 80 times bigger in water than it is in air.&#xA;So, the attraction between the two charges is  80 times weaker in water than in air.C&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-04T18:35:28.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15928" PostTypeId="1" AcceptedAnswerId="15929" CreationDate="2014-09-04T21:16:11.463" Score="1" ViewCount="113" Body="&lt;p&gt;I was wondering if this actually happens in a chemical reaction&#xA;$$\ce{AgCl_{(s)} + 2NH_4OH_{(aq)} \rightleftharpoons  Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)} + 2H_2O_{(l)}}$$&#xA;My teacher wrote something like this but on Internet I saw that it yields to $\ce{Ag(NH_3)_2Cl}$ and water, so can someone explain it to me.&lt;/p&gt;&#xA;" OwnerUserId="7113" LastEditorDisplayName="user15489" LastEditDate="2015-06-25T10:19:27.860" LastActivityDate="2015-06-25T10:23:14.413" Title="Silver identification of the group I of cations" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2015-10-31T16:47:53.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="15929" PostTypeId="2" ParentId="15928" CreationDate="2014-09-04T21:49:45.643" Score="2" Body="&lt;p&gt;Both formulas for diamminesilver(I) chloride essentially mean the same. Writing it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag(NH_3)_{2(aq)}^+ + Cl^{-}_{(aq)}}$$ emphasizes that the compound is dissolved in water as discrete ions (the cationic silver complex and chloride anions). One can also write $\ce{Ag(NH_3)_2Cl}$ instead because it is known that the compound exists in the aforementioned way in aqueous solution. In the latter case, it would still be more precise to use brackets to indicate the complex, i.e. $[\ce{Ag(NH3)2}]\ce{Cl}$, so that it becomes clear that the two $\ce{NH3}$ are the ligands of silver.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorDisplayName="user15489" LastEditDate="2015-06-25T10:23:14.413" LastActivityDate="2015-06-25T10:23:14.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15930" PostTypeId="2" ParentId="10792" CreationDate="2014-09-04T22:34:54.507" Score="13" Body="&lt;p&gt;The fundamental misunderstandings here are related to two concepts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Conservation (atoms are conserved, molecules are not)&lt;/li&gt;&#xA;&lt;li&gt;A mole is just a number (like the chemist's dozen)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would start with conservation, then move on to moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explaining why atoms are conserved, but not moles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For second-graders, hands-on activities are probably best, so molecular models or something similar would be a good approach. I like your example using two molecules - if you can get some models, you could try having groups of students start with two (or more - 4+ would make it easier to see) molecules of $\ce{SO3}$. Have them count the number of molecules and the number of each atom. Write the results down where everyone can see them. Then, have them do the &quot;reaction&quot; - let them take apart the molecules and put them back together to make$\ce{O2}$ and $\ce{SO2}$. Again, have them count the molecules and the atoms. Show them that the number of atoms stays the same (for each element and in total), but the number of molecules can change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once your students are comfortable with this idea, introduce the concept of moles as a way of counting very large numbers. This is the more difficult part, and in my experience is the &quot;make or break&quot; concept for most chemistry students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A mole is just a very large number&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Goal: The number of atoms in macroscopic samples is huge&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can motivate the discussion by calculating the number of molecules/atoms in very small samples. In my classes (college level) I use the number of molecules in one mL of hydrogen gas at STP, and show that if you counted one molecule per second, it would take longer than the age of the universe (a lot longer) to count them all. Sometimes I use the number of silicon atoms in a bag of sand and show them that it is larger (much larger) than the number of stars in the universe. Other examples would work as well - the idea is to get them thinking about the largest quantity that they can possibly imagine and show them that the number of atoms in even a small amount of matter is much larger than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Goal: A mole is just a number&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you want to link this to moles. I start with the concept of a dozen. I ask things like &quot;how many donuts are in a dozen?&quot; &quot;Ok, how many cookies?&quot; &quot;How many cats?&quot; &quot;How many people&quot; until someone points out that it's always 12. Then I ask them why we say a dozen instead of 12. If they don't figure it out quickly on their own, I tell them that for bakers and their customers, it is a convenient way to count stuff. Then I tell them chemists also have a convenient way to count atoms and molecules, which is good because there are a lot of them! The number they use is called a mole, and similarly to how a dozen = 12 of anything, a mole = 6.022 x 10^23 of anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Goal: A mole is a HUGE number&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how good the students are with scientific notation, you might want to actually write out that number of zeroes - it's an impressive demonstration! When there is time, I also like to demonstrate the scale by comparing it to different things as the number of zeroes grows. For example, I say (or have them say) &quot;thousand,&quot; &quot;million,&quot; &quot;billion,&quot; &quot;trillion,&quot; for every three zeroes. At around 11 zeroes I start comparing it to unimaginably large things. As examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number of stars in the Milky Way galaxy is about 100 billion ($1 \times 10^{11}$). &lt;/li&gt;&#xA;&lt;li&gt;The number of sheets of paper that would fit in a stack between Earth and the moon is about 4 trillion ($4 \times 10^{12}$). &lt;/li&gt;&#xA;&lt;li&gt;The number that would fit between Earth and Mars (at their average distance) is about $2 \times 10^{15}$. &lt;/li&gt;&#xA;&lt;li&gt;Age of the universe in seconds is $4 \times 10^{17}$.&lt;/li&gt;&#xA;&lt;li&gt;There are $10^{22}$ stars in the universe. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think that this is effective because it connects the concept of moles being huge and the number of atoms being huge to a bunch of astonishing facts - people tend to remember weird little surprising facts, and because of the way memories form, the moles stuff gets stored along with it, leading to an artificially created &quot;intuitive&quot; understanding of the size scale of moles and their relationship to counting atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to tie it all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Goal: Moles are used to count atoms and molecules&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I connect the concept of a mole being a very large &quot;convenience&quot; number used for counting to the stoichiometry and conservation concepts with a simple exercise. I write out a chemical equation, which in your case would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2SO3 &amp;lt;=&amp;gt; 2SO2 + O2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Underneath, I write the number of molecules of each, starting with the smallest. So, I would write...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|}\hline&#xA;\ce{2SO3} &amp;amp; \ce{&amp;lt;=&amp;gt;} &amp;amp; \ce{2SO2} &amp;amp; \ce{O2} \\ \hline&#xA;2 \space \rm{molecules} &amp;amp;  &amp;amp;  2 \space \rm{molecules} &amp;amp; 1 \space \rm{molecules} \\ \hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and say &quot;If we have two molecules of $\ce{SO3}$, we will end up with two molecules of $\ce{SO2}$ and one of $\ce{O2}$.&quot; Then I ask them: &quot;What if I start with 10 molecules $\ce{SO3}$?&quot; and let them figure it out, writing on the board as I go. Next, I ask &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;What if I start with 12?&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;What about a dozen?&quot; &lt;/li&gt;&#xA;&lt;li&gt;&quot;How about 100?&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;A billion?&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;$6.022 \times 10^{23}$?&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;A mole?&quot; - at this point, most of them will have the idea&lt;/li&gt;&#xA;&lt;li&gt;&quot;10 moles?&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the end, the board looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|}\hline&#xA;\ce{2SO3} &amp;amp; \ce{&amp;lt;=&amp;gt;} &amp;amp; \ce{2SO2} &amp;amp; \ce{O2} \\ \hline&#xA;2 \space \rm{molecules} &amp;amp;  &amp;amp;  2 \space \rm{molecules} &amp;amp; 1 \space \rm{molecules} \\ \hline&#xA;10\space \rm{molecules} &amp;amp;  &amp;amp;  10\space \rm{molecules} &amp;amp; 5 \space \rm{molecules} \\ \hline&#xA;12\space \rm{molecules} &amp;amp;  &amp;amp;  12\space \rm{molecules} &amp;amp; 6 \space \rm{molecules} \\ \hline&#xA;1\space \rm{dozen} &amp;amp;  &amp;amp;  1\space \rm{dozen} &amp;amp; 0.5 \space \rm{dozen} \\ \hline&#xA;100 \space \rm{molecules} &amp;amp;  &amp;amp;  100 \space \rm{molecules} &amp;amp; 50 \space \rm{molecules} \\ \hline&#xA;1 \times 10^9 \space \rm{molecules} &amp;amp;  &amp;amp;  1 \times 10^9\space \rm{molecules} &amp;amp; 0.5 \times 10^9 \space \rm{molecules} \\ \hline&#xA;6.022 \times 10^{23} \space \rm{molecules} &amp;amp;  &amp;amp;  6.022 \times 10^{23}\space \rm{molecules} &amp;amp; 3.011 \times 10^{23} \space \rm{molecules} \\ \hline&#xA;1 \space \rm{mole} &amp;amp;  &amp;amp;  1 \space \rm{mole} &amp;amp; 0.5 \space \rm{mole} \\ \hline&#xA;10 \space \rm{mole} &amp;amp;  &amp;amp;  10 \space \rm{mole} &amp;amp; 5 \space \rm{mole} \\ \hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point most everyone will &quot;get it&quot; - and the advantage of this method is that they sort of figure it out for themselves along the way, which means they will remember it more clearly. If anyone still struggles with the conservation part, you can have them calculate the number of each atom and each molecule for all or some of the examples above, in the same manner as at the beginning of the lesson.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="43942" LastEditDate="2017-05-08T02:28:49.797" LastActivityDate="2017-05-08T02:28:49.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15932" PostTypeId="1" AcceptedAnswerId="15936" CreationDate="2014-09-05T01:22:55.907" Score="8" ViewCount="2065" Body="&lt;p&gt;Can anyone explain why my textbook only gives the following resonance structures?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the important resonance structures for each of the following:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sEeML.png&quot; alt=&quot;question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Ftd9.png&quot; alt=&quot;answer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also put two resonance case where the double bond electrons are moved on the individual carbon atoms.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorDisplayName="user10259" LastEditDate="2015-05-21T17:17:43.757" LastActivityDate="2015-05-21T17:17:43.757" Title="What are the correct resonance structures of bromoethene?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15933" PostTypeId="1" AcceptedAnswerId="15934" CreationDate="2014-09-05T01:55:42.587" Score="8" ViewCount="23512" Body="&lt;p&gt;The definition of atomic radius is half the average internuclear distance of an atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But many times I've been told that it is not possible to handle a single atom. Thus how can we measure the internuclear distance of an atom?&lt;/p&gt;&#xA;" OwnerDisplayName="user7484" LastEditorUserId="4945" LastEditDate="2016-02-26T05:48:41.490" LastActivityDate="2017-09-30T03:46:52.743" Title="How can we measure the atomic radius of an atom?" Tags="&lt;atoms&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15934" PostTypeId="2" ParentId="15933" CreationDate="2014-09-05T02:03:58.927" Score="7" Body="&lt;p&gt;As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pY6D0.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This internuclear distance can be determined by two methods:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;X-rays method&lt;/li&gt;&#xA;&lt;li&gt;Spectroscopy method&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Atomic radius is not a set value for a given atom. Eg. $\ce{Fe^{+2}}$ ion have different atomic radius than  $\ce{Fe+}$. Moreover,  $\ce{Fe^{+2}}$&amp;lt;  $\ce{Fe+}$&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-10-30T04:38:12.370" LastActivityDate="2014-10-30T04:38:12.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15935" PostTypeId="2" ParentId="15866" CreationDate="2014-09-05T02:22:28.730" Score="4" Body="&lt;p&gt;Today I tested a set of competent &lt;em&gt;E. coli&lt;/em&gt; cells that were prepared using a HEPES-based calcium chloride buffer solution. The solution was autoclaved as I described in the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I obtained a fair transformation efficiency from these cells, and viability was also not perturbed noticeably. There are several confounds in the preparation of my solution that could affect the transformation efficiency, but in conclusion, autoclaving the HEPES has not abrogated its ability to make cells chemically competent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will speculate further and say that autoclaving HEPES may degrade some of the HEPES molecules or produce reactions and generate unexpected subspecies - but both of these will be generated in small, insignificant amounts which should not affect most biological applications.&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2014-09-05T02:22:28.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15936" PostTypeId="2" ParentId="15932" CreationDate="2014-09-05T02:42:54.947" Score="4" Body="&lt;p&gt;Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible. This is simply a matter of electrostatics - seperating charges and holding them apart from one another costs energy. A second reason has to do with the &quot;electronic origin&quot; of this resonance charge: It describes a conjugation between a free electron pair of $\ce{Br}$ and the $\pi$ system of the double bond - more accurately put the filled free-electron-pair-orbital can interact with the $\pi^{*}$ orbital of the $\ce{C=C}$ bond to form two new orbitals: an in-phase combination that is lower in energy and will contain the two electrons from the free electron pair and an out-of-phase combination that is higher in energy and will remain unoccupied. This stabilizes the system. The stronger the overlap between the free-electron-pair-orbital and the $\pi^{*}$ orbital of the $\ce{C=C}$ bond and the closer they are in energy the better the stabilization of the system through this conjugation is and the more important is the second resonance structure. But in the case at hand the $\pi$-overlap between the bromine orbital and the $\ce{C=C}$ bond is very weak (the reason for this you can find &lt;a href=&quot;https://chemistry.stackexchange.com/a/15923/189&quot;&gt;here&lt;/a&gt;) and so the conjugation is very weak, too. Thus, this resonance structure is not very important.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-09-05T03:23:40.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15937" PostTypeId="2" ParentId="14131" CreationDate="2014-09-05T03:45:05.247" Score="2" Body="&lt;p&gt;I used to teach an introductory chemistry course and found that the understanding of most chapters would have been improved by having already knowing other chapters, but you have to start somewhere with something.  Different texts have things ordered differently and different instructors present material in different orders even using the same text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introductory chemistry by nature covers a very wide variety of topics.  As an example, I had to cover the basics of organic nomenclature as part of an introductory chemistry class and I had about 5 weeks to cover the organic section, maybe less if I needed more time to cover the material I taught earlier in the semester.  Obviously, even the nomenclature part of organic chemistry will be more rigorous in a two semester Organic sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introductory chemistry is a challenging class to teach - if you don't plan well enough how much time to devote to a specific topic you are at risk of either running out of time to cover something or most of your class not understanding something important.&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2014-09-05T03:45:05.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15939" PostTypeId="1" AcceptedAnswerId="15948" CreationDate="2014-09-05T04:08:05.013" Score="6" ViewCount="1423" Body="&lt;p&gt;$\ce{CO2}$ dissolves in water and goes through chemical reactions and exists at different concentrations of, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;dissolved $\ce{CO2}$, &lt;/li&gt;&#xA;&lt;li&gt;carbonic acid, &lt;/li&gt;&#xA;&lt;li&gt;bicarbonate, and &lt;/li&gt;&#xA;&lt;li&gt;carbonate. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Bjerrum_plot&quot; rel=&quot;noreferrer&quot;&gt;Bjerrum plot&lt;/a&gt; says that a high concentration of carbonate results in an increased pH (i.e., basic). But many diagrams in the Internet about ocean acidification show that as carbonate is produced, it becomes more acidic (although they also shows as carbonate is produced $\ce{H+}$ is also produced). &#xA;&lt;img src=&quot;https://i.stack.imgur.com/3lqJ3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;a href=&quot;http://theotherco2problem.files.wordpress.com/2009/11/ocean-chemistry.gif&quot; rel=&quot;noreferrer&quot;&gt;&#xA;Source&lt;/a&gt;&lt;br&gt;&#xA;I don't know how to reconcile this kind of figure and Bjerrum plot as I feel they are not consistent.&lt;/p&gt;&#xA;" OwnerUserId="7707" LastEditorUserId="4945" LastEditDate="2014-09-05T04:19:38.170" LastActivityDate="2014-09-05T07:40:58.957" Title="Is the Ocean acidification inconsistent with the Bjerrum plot?" Tags="&lt;acid-base&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15940" PostTypeId="1" CreationDate="2014-09-05T04:16:16.493" Score="7" ViewCount="1304" Body="&lt;p&gt;For the following double displacement half reaction, the half reactions can be determined easily using the oxidation numbers.&#xA;$$\ce{MnO4- {(aq)} + IO3- {(aq)} -&amp;gt; MnO2 {(s)} + IO4- {(aq)}}$$&#xA;We know this is the reduction half reaction because the oxidation number of $\ce{Mn}$ changes from +7 to +2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4- {(aq)} -&amp;gt; MnO2 {(s)}}$$&#xA;and we know this is the oxidation half reaction because the oxidation number of $\ce{I}$ changes from +5 to +7:&#xA;$$\ce{IO3- {(aq)} -&amp;gt; IO4- {(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the following reactions, this strategy cannot be used. So how do we determine the half reaction equations for the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{Mn(OH)2 {(s)} + H2O2 {(aq)} -&amp;gt;T[Base] Mn(OH)3 {(s)}}$     &lt;/li&gt;&#xA;  &lt;li&gt;$\phantom{iiiiiiiiiiiiiiiii}\ce{MnO4^2- {(aq)} -&amp;gt;[Acid] MnO2 {(s)} + MnO4- {(aq)}}$   &lt;/li&gt;&#xA;  &lt;li&gt;$\phantom{iiiiiiiii}\ce{IO3- {(aq)} + I- {(aq)} -&amp;gt;T[Acid] I3 {(aq)}}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\phantom{iiiiiiiiii}\ce{P {(s)} + PO3^3- {(aq)} -&amp;gt;T[Base] HPO3^2- {(aq)}}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="5026" LastEditDate="2018-06-10T03:16:49.193" LastActivityDate="2018-06-10T03:20:58.590" Title="Determining the redox half-reactions of single displacement reactions" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15941" PostTypeId="1" AcceptedAnswerId="15952" CreationDate="2014-09-05T05:07:40.147" Score="1" ViewCount="1553" Body="&lt;p&gt;I am to turn 1-octyne to 1-hexene. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Method 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQHw2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Reduction using sodium metal in pure ammonia to an alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Halohydrin formation; the -OH installs at the more substituted carbon due to its greater ability to stabilize positive charge (this always confused me; the more substituted carbon is both more positive in character and has more electron-donating methyl groups?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Conversion of OH to a good leaving group (-OTs) using a sulfonate ester and a weak base, pyridine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) E2 elimination using NaOH and heat. Because of resonance and sheer bulk and negative induction the OTs(-) should make for a better leaving group than Br-. So the internal alkene is formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Two-part one step ozonolysis; creation of aldehyde. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Reduction of carbonyl with hydrogen and transition metal to terminal alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Conversion of terminal alcohol to a better leaving group using tosyl chloride and pyridine again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) Use of LDA to deprotonate and create the terminal alkene. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Method 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Also we can take another route with the brominated alkene; we can use OsO4 and basic hydrogen peroxide to create a 1,2 (vic) diol, which can be cleaved by periodic acid, and this creates a primary alcohol, which we convert to a good leaving group using tosyl chloride and pyridine, and we can eliminate. Same number of steps, but just something different that I thought of. Well, perhaps slightly simpler since there isn't a need for a two-step process (ozonolysis). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPdPp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-05T15:45:39.570" Title="Carbon decrease from 1-octyne to 1-hexene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15942" PostTypeId="2" ParentId="15856" CreationDate="2014-09-05T05:28:11.620" Score="3" Body="&lt;p&gt;If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this equation, one obtains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $n$, one finds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the final equation from your textbook's solutions manual.&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="5026" LastEditDate="2018-03-06T14:25:14.197" LastActivityDate="2018-03-06T14:25:14.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15943" PostTypeId="2" ParentId="15856" CreationDate="2014-09-05T05:48:40.547" Score="2" Body="&lt;p&gt;I think that there is a simple method to solve this question. I did this question with these 6 steps and my answer is right (although I had to do some approximation in calculation)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Take values from experiment 1 and put them in rate equation. You will obatain an equation with $m$, $n$ and $k$ as varibales. Lets assume it equation 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now take values from experiment 2 and put them in rate equation. You obtain another eqaution like equation 1. Lets assume it equation 2.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Divide equation 1 by 2 or 2 by 1 (whatever you like.) Solve it and you will easily get $m$ (equals to 1).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now put the value of $m$ in equation 2. You will obtain an equation with $n$ and $k$ as variables. Lets assume it equation 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now take values from experiment 3 and put them in rate equation along with value of $m$. You will get another equation like equation 3. Lets assume it equation 4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Divide equation 3 by 4 or 4 by 3 (whatever you like.) Solve it and you will easily get n (equals to 1).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS, this might not be a 'by-the-book' method but I will solve it this way when I don't have too much time to solve a question and also because I am lazy enough to solve natural logarithm and stuff.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2014-09-05T05:57:43.763" LastActivityDate="2014-09-05T05:57:43.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15944" PostTypeId="1" CreationDate="2014-09-05T06:15:13.820" Score="3" ViewCount="4317" Body="&lt;p&gt;I know it expands when hot and grows compact when cool (that's about all I know), but I suppose many other substances do, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, and excuses for my ignorance, what the term used to describe the function of mercury in a thermometer? I mean, would you call it an &quot;indicator&quot; (insofar as it indicates the temperature), or is there some technical term? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7708" LastActivityDate="2016-08-31T08:27:02.247" Title="Why is mercury used in thermometers?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15946" PostTypeId="2" ParentId="15940" CreationDate="2014-09-05T06:34:40.547" Score="7" Body="&lt;p&gt;Let's take the first equation in the question by example:&#xA;$$\ce{Mn(OH)2 (s) + H2O2 (aq) -&amp;gt; Mn(OH)3- (s)}$$&#xA;In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2):&#xA;$$&#xA;\begin{align}&#xA;\ce{Mn^{2+} &amp;amp;-&amp;gt; Mn^{3+} +e^- \tag 1}\\&#xA;\ce{H_2O2 (aq) + 2e^- &amp;amp;-&amp;gt; 2OH^- \tag 2}&#xA;\end{align}&#xA;$$ &#xA;If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$: &#xA;$$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -&amp;gt; 2 Mn^{3+} + 2OH^- \tag 3}$$ &#xA;By adding four ions hydroxide to the two sides of equation $(3)$: &#xA;$$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -&amp;gt; Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$ &#xA;By rearranging the equation $(3)$, we find:&#xA;$$\ce{2Mn(OH)2 (s) + H2O2 (aq) -&amp;gt; 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$&#xA;We do the same for the other equations.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-06-10T03:19:25.590" LastActivityDate="2018-06-10T03:19:25.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15947" PostTypeId="2" ParentId="15944" CreationDate="2014-09-05T07:25:15.503" Score="9" Body="&lt;p&gt;In addition to the answer and the comment given, the use of mercury in thermometers had another advantage frequently used in the lab:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury, as a metal, has a high electrical conductivity. Imagine two thin wires inside the capillary of your thermometer. One of them is fixed. &#xA;The other wire is can move along a threaded rod. Typically, this was realized via magnetical coupling to a freely rotating knob on top of the (hermetically sealed) thermometer. Upon heating, the mercury expands and finally reaches the upper wire previously adjusted to a particular working temperature. At this point, an electrical current may flow through the thermometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EhqmY.jpg&quot; alt=&quot;contact thermometer&quot;&gt;&#xA;Source: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/4/48/Kontth.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/4/48/Kontth.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These &lt;strong&gt;contact thermometers&lt;/strong&gt;, nowadays replaced by mercury-free electronic thermoswitches, once were the working horses in the lab when it came to controlled heating of oil baths or monitoring cooling circuits for uv lamps.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-05T07:25:15.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15948" PostTypeId="2" ParentId="15939" CreationDate="2014-09-05T07:40:58.957" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fsUut.png&quot; alt=&quot;Bjerrum Plot&quot;&gt; &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Carbonate_system_of_seawater.svg&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I see the confusion. You are right that directly adding carbonate, perhaps in the form of $\ce{Na2CO3}$, would increase the pH. However, ocean acidification, as shown in your figure, is from atmospheric $\ce{CO2}$ dissolving in the ocean. The relevant equilibrium: $$\ce{CO2_{(aq)} + H2O &amp;lt;=&amp;gt; H2CO3_{(aq)} &amp;lt;=&amp;gt; HCO3^{-}_{(aq)} +H+_{(aq)}&amp;lt;=&amp;gt;CO_{3(aq)}^{2-} +H+_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bjerrum plot is simply a graphical representation of this equilibrium. As you add $\ce{CO2}$, you push the equilibrium towards the right, forming $\ce{H+}$, lowering the pH. Conversely, when you add $\ce{CO3^{2-}}$, the equilibrium move towards the left, consuming some $\ce{H+}$ and raising the pH. Although adding $\ce{CO2}$ leads to an increase in the absolute concentration of carbonate, the ratio of concentrations goes the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a change in pH, you can move along the x-axis to predict how the concentrations of the carbonate species change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you change the ratios of the species, e.g. by adding $\ce{CO2}$, you can predict how the pH will change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-05T07:40:58.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15949" PostTypeId="2" ParentId="15941" CreationDate="2014-09-05T07:47:47.650" Score="5" Body="&lt;p&gt;In principle, this can be realized in two steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Oxidative cleavage using catalytic amounts of $\ce{RuO2}$ and excess of oxone converts 1-octyne to heptanoic acid. (&lt;a href=&quot;http://dx.doi.org/10.1021/jo0357925&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heating a mixture of acetic anhydride and a carboxylic acid in the presence of $\ce{Pd}$ catalyst turns the acid to a terminal alkene lacking one carbon atom (&lt;a href=&quot;http://dx.doi.org/10.1021/jo00053a008&quot;&gt;DOI&lt;/a&gt;). Admittedly, the high temperatures needed might be a problem here and may force you to perform the reaction in a sealed tube in an autoclave.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-05T07:47:47.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15951" PostTypeId="2" ParentId="15941" CreationDate="2014-09-05T11:20:35.840" Score="5" Body="&lt;p&gt;Your second step will not work as written. Reaction of an alkene with aqueous HBr will produce the bromide (not the bromohydrin) with Markovnikov regioselectivity and the possibility for carbocation rearrangement. To get the bromohydrin, se $\ce{Br2/H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your fourth step will be challenging also. There are two leaving groups, and LDA is a very strong base. There are regioselectivity and chemoselectivity problems, and unless the reaction is controlled well, you will also get 1-octyne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better to install just one functional group in step 2 (oxymercuriation, while old-fashioned, gives you just 2-octanol with little to no rearrangement), and then your elimination in step 4 will behave better.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-09-05T11:20:35.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15952" PostTypeId="2" ParentId="15941" CreationDate="2014-09-05T13:39:57.310" Score="2" Body="&lt;p&gt;This might be a shorter route that employs common synthetic techniques you're probably already familiar with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bkei4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The route begins by forming the acetylide anion, followed by alkylation with pentyl bromide.  Oxidation of the resultant acetylene will produce the two possible carboxylic acids.  These two acids should be readily separable by distillation (C6 b.p. 206C; C7 b.p. 223C).  Further reaction (reduction, tosylation, elimination) of the hexanoic acid fraction would produce 1-hexene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Non-Subverting Route&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m2s0Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you've studied radical allylic oxidation (maybe it was radical allylic bromination that you posted about?  Well, same idea.).  In any case, it is a well-known reaction &lt;a href=&quot;http://books.google.com/books?id=4PdYwo25VP8C&amp;amp;pg=PA513&amp;amp;lpg=PA513&amp;amp;dq=allylic%20oxidation%20of%20alkynes&amp;amp;source=bl&amp;amp;ots=uWPC5hldNh&amp;amp;sig=K24Lk61o5l921w-P0Ax5SzQlMC4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=As8JVOrRLcKTyATWu4CoBg&amp;amp;ved=0CFcQ6AEwCQ#v=onepage&amp;amp;q=allylic%20oxidation%20of%20alkynes&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;(here are some examples)&lt;/a&gt;.  You might get a mixture of the propargylic alcohol and ketone - no matter, carry them both along to the next oxidation step. In this second step we oxidize the triple bond and any remaining propargylic alcohol to the alpha-ketoacid, which will decarboxylate (see the first few sentences in this &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CCMQFjAA&amp;amp;url=http%3A%2F%2Fwww.researchgate.net%2Fpublication%2F221676840_Peroxy_Acid_Oxidations_A_Kinetic_and_Mechanistic_Study_of_Oxidative_Decarboxylation_of_-Keto_Acids_by_Peroxomonophosphoric_Acid%2Flinks%2F0c960524867b864988000000&amp;amp;ei=6dkJVKzpGY2MyATEo4LwCw&amp;amp;usg=AFQjCNG4FBhp1LYGfFfWnvmdwt4ZldVN8A&amp;amp;sig2=67ExqhX-OQogkbcrwNYlCQ&amp;amp;bvm=bv.74649129,d.aWw&amp;amp;cad=rja&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;) under the reaction conditions.  This yields hexanoic acid which can then be converted to the desired olefin.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-05T15:45:39.570" LastActivityDate="2014-09-05T15:45:39.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15953" PostTypeId="1" CreationDate="2014-09-05T14:24:06.587" Score="3" ViewCount="183" Body="&lt;p&gt;I'm trying to write some software that I can use to determine, roughly, what the physical properties of a pure substance are. I know I could just use a database of the known properties of each element, but that doesn't do much for me when I go to create molecules and combine substances. A database that lists how each element will react to another is likely impossible, so I got to thinking... there has to be a mathematical way of determining these properties!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I understand that there are trends on the periodic table that show that the stronger the force between the molecules is, the higher the melting point will be. I also understand that we can calculate how strong that force is based on what we know about the atom. So why can't I find any information relating the two in some mathematical way? The same goes for color, hardness, malleability, etc. Do we really only know these values through testing? We know the values but we don't know why they are what they are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is deeply complex topic and may not even be computational possible. In those cases, I'm trying to do as much as I can to approximate.&lt;/p&gt;&#xA;" OwnerUserId="7711" LastEditorUserId="7711" LastEditDate="2019-12-11T01:30:04.700" LastActivityDate="2019-12-11T01:30:04.700" Title="Can you calculate the properties of a substance based solely on its atomic properties?" Tags="&lt;molecules&gt;&lt;computational-chemistry&gt;&lt;atoms&gt;&lt;periodic-trends&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="15954" PostTypeId="1" AcceptedAnswerId="15958" CreationDate="2014-09-05T15:06:22.147" Score="0" ViewCount="2564" Body="&lt;p&gt;I am told that a carboxylic ester would have a higher carbonyl stretching wave number than a carboxylic amide due to there being more double bond character in the carbonyl group of the ester. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense prima facie; the resonance structures suggest that there is somewhat more double bond character in the carbonyl of the ester. And nitrogen and oxygen are nearly identical in mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this actually true and applicable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t5NkY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-05T16:21:22.177" LastActivityDate="2014-09-05T16:38:06.433" Title="Resonance contributors and IR stretching frequency" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15955" PostTypeId="2" ParentId="15953" CreationDate="2014-09-05T15:24:21.920" Score="3" Body="&lt;p&gt;For &lt;strong&gt;atomic and molecular-level properties&lt;/strong&gt; (bond, angle, torsion energies, etc), quantum mechanics provides the equations for solving for these properties. However, there are a number of approximations that are employed, especially for larger, more complex molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;bulk properties&lt;/strong&gt; of a substance, thermodynamics and statistical mechanics provide equations for solving for a variety of material properties (critical properties, phase equilibria, various types of enthalpy, and so on). Read up on &lt;strong&gt;equations of state&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the option of &lt;strong&gt;molecular simulation&lt;/strong&gt;. In this case you are solving relatively simple, well-established equations (eg, Newton's equations of motion in molecular dynamics simulations) repeatedly in order to compute the (usually) equilibrium properties of a substance of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, the accuracy varies, and validating results by comparing various methods is considered best practice. While there are &quot;purists&quot; out there who believe that all parameters in a model should be derived from first principles alone, in practice models are often fit to experimental data (or even data from quantum mechanical calculations) to improve accuracy.&lt;/p&gt;&#xA;" OwnerUserId="7693" LastActivityDate="2014-09-05T15:24:21.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15956" PostTypeId="1" AcceptedAnswerId="15959" CreationDate="2014-09-05T16:03:18.933" Score="0" ViewCount="50" Body="&lt;p&gt;I am told that an intramolecular proton abstraction can occur. For some reason this seems at least sterically unlikely. Is it possible sterically? Or is the proton abstraction best described as a equilibrium process between the nitrogen's hydrogens and water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NCkgc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LkWs1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-02T16:49:03.647" LastActivityDate="2015-10-02T16:49:03.647" Title="Ammonia to tri substituted nitrogen" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15957" PostTypeId="1" CreationDate="2014-09-05T16:10:20.240" Score="4" ViewCount="325" Body="&lt;p&gt;I asked this in biology, but I think they don't want to deal with the chemistry behind UV-Vis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question has to do with the Poly A tails on mRNA. I've been doing in vitro transcription and tailing for a while now and have noticed that the untailed RNA has a 260/280 of about 2.0, like you'd expect, but successful tailing results in ratios from 2.15 to 2.45, with higher ratios correlating to better band shifts on agarose gels, implying a longer tail. I suspect the reason for this is because the adenine has a much higher 260/280 than the other bases, so if you add a lot of adenines to the 3' end of the RNA you add much more 260 absorbance than 280 absorbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After seeing this I tried to come up with a mathematical method to estimate tail length based on 260/280, since gels are annoying, while I can get the ratio directly from the nanodrop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started by using the 260/280 for each nucleotide, A: 4.50, G: 1.15, C: 1.51, U: 4.00, T: 1.47 source here: &lt;a href=&quot;http://www.bio.davidson.edu/projects/gcat/protocols/NanoDrop_tip.pdf&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;. I then took the sequence for the Luciferase gene I'm using for mRNA and counted the nucleotides, 449 A, 566 C, 470 G, 374 U. I tried to get the weighted average of the 260/280 for my RNA. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A: 449 x 4.50 = 2020.50&#xA;C: 566 x 1.51 =  854.60&#xA;G: 470 x 1.15 =  540.50&#xA;U: 374 x 4.00 = 1496.00&#xA;Sum:            4911.60&#xA;&#xA;Nucleotides: 449 + 566 + 470 + 374 = 1859&#xA;Sum / Nucleotides : 4911.60 / 1859 = 2.64&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;2.64 is clearly not 2.0. Even accounting for the potential effect of pH and ionic strength on RNA 260/280 ( acidic solutions have lower ratios ) I can't explain why the calculated ratio is so high. Even if we assumed an RNA with 25% of each base, the weighted average is still (1.15 + 4.50 + 1.51 + 4.00)/4 = 2.79, even higher than for my real sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the same calculation for DNA, (1.15 + 4.50 + 1.51 + 1.47)/4 = 2.16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why don't my calculated ratios for RNA and DNA 260/280 match the &quot;ideal&quot; ratios of 2.0 and 1.8? It probably has something to do with base stacking interactions, but I don't enough to compensate for it.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="4945" LastEditDate="2014-09-09T08:29:52.757" LastActivityDate="2019-01-03T14:17:33.983" Title="Predicting 260/280 of DNA or RNA from sequence?" Tags="&lt;biochemistry&gt;&lt;spectrophotometry&gt;&lt;dna-rna&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15958" PostTypeId="2" ParentId="15954" CreationDate="2014-09-05T16:38:06.433" Score="3" Body="&lt;p&gt;Yes, it is true and applicable. In fact, the resonance structure you've drawn with the double bond between the carbonyl carbon and the amide nitrogen is so significant that one can actually observe restricted rotation about this bond on the nmr timescale.  That is to say, if a methyl group was attached to the amide nitrogen, you can see distinct resonances for the syn and anti isomers.  The rotational barrier about the C-N bond is around 20 kcal/mol, depending upon the substituents, etc.  In the case of the ester, for the reason you provided, the barrier to rotation about the carbonyl carbon - ester oxygen bond is much lower, typically in the range of 10-12 kcal/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case of the amide, the second resonance structure is more important than it is for the ester.  Consequently, the amide carbonyl bond is weaker than the carbonyl bond is for the ester.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-05T16:38:06.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15959" PostTypeId="2" ParentId="15956" CreationDate="2014-09-05T16:53:30.077" Score="1" Body="&lt;p&gt;Until there's water present (I'm assuming that it's not the solvent), the proton can't come from water. The proposed intramolecular proton transfer would go through a 5-membered transition state, which is reasonable. If you don't like that mechanism, then two of the intermediates could transfer protons to each other. Depending on the concentration, one or the other or both could be occuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact nature of these proton transfers are difficult to prove and typically inconsequential, so you will often see &quot;shortcut&quot; mechanisms drawn which imply intramolecular proton transfer even when it is more unlikely than the one you show here.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-05T16:53:30.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15960" PostTypeId="2" ParentId="15957" CreationDate="2014-09-05T17:20:11.143" Score="3" Body="&lt;p&gt;The problem with doing this, especially with RNA, is that secondary structure can have a large impact on the absorption properties of the molecule. Interactions between bases will change the molar absorptivities in ways that may not be easily predictable. The &quot;ideal&quot; ratios are only guidelines and I wouldn't worry too much that a long string of a single base produces a deviation from them. If very accurate measurements are required, hydrolysis into nucleotides before measurement is usually done to avoid these interactions. For your specific case, it may be possible to construct a calibration curve of sorts to get an estimate if you can measure the tail length by some other means. So long as the tail is sort relative to the rest of the molecule, I would expect to see some predictable change in ratios.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-05T17:20:11.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15961" PostTypeId="1" CreationDate="2014-09-05T17:23:42.957" Score="5" ViewCount="10299" Body="&lt;p&gt;I am asking this question for my son.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(&lt;a href=&quot;https://chemistry.stackexchange.com/questions/4593/buffers-and-ph-question-for-exam-review&quot;&gt;This question&lt;/a&gt; is related but we still  have our confusions.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have one liter buffer solution of 4.74 pH which contains 0.1 mole $\ce{CH_3COOH}$ and &#xA;0.1 mole $\ce{CH_3COONa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the resulting pH when 0.02 mole $\ce{NaOH}$ is added to this solution? (Given $\ce{Ka} = 1.8 \times 10^{-5}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the definition of a buffer solution&lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt; [A&lt;/a&gt;, &lt;a href=&quot;http://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;B]&lt;/a&gt; goes, pH value should not change when small quantity of acid or base is added to it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, one way of answering the question  will be to simply comment that since&#xA;a small quantity of base has been added, the pH value will not change.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;But, we find another solution (presented and accepted in some &#xA;moderately acceptable study-sheet) in the following line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will have a reaction like,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH_3COOH} + \ce{NaOH} = \ce{CH_3COONa} + \ce{H_2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after the reaction, we have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH_3COONa}$: $(0.1 + 0.02)\ \text{mole} = 0.12\ \text{mole}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH_3COOH}$: $(0.1 - 0.02)\ \text{mole} = 0.08\ \text{mole}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $pH = pKa + \log{\text{[Salt]}\over\text{[Acid]}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &#xA;$pH = pKa + \log{{{n_\text{Salt}}\over V}&#xA;\over&#xA;{{n_\text{Acid}}\over V}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which after plugging in the values gives, &#xA;$pH = 4.92$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We are  confused about which of the above solutions is&#xA;correct.&lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-09-06T05:13:13.163" Title="What happens when 0.02 mole NaOH is added to a buffer solution?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15962" PostTypeId="1" CreationDate="2014-09-05T18:08:44.580" Score="2" ViewCount="2263" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given $400.0\ \mathrm{g}$ of hot tea at $80.0 °\mathrm{C}$, what mass of ice at $0 °\mathrm{C}$ must&#xA;  be added to obtain iced tea at $10.0 °\mathrm{C}$? The specific heat of the&#xA;  tea is $4.18\ \mathrm{J}/(\mathrm{g}\ \cdot°\mathrm{C})$ , and $\Delta H_{\mathrm{fusion}}$ for ice is $+6.01\ \mathrm{kJ\ mol}^{-1}$ .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$q$ is the amount of heat transferred. Since the solution is decreasing heat, then $q_{\mathrm{ice\ tea}}=-q_{\mathrm{ice}}$ We can then solve q using the information given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_{\mathrm{ice}\,\mathrm{tea}}=\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\cdot ^{\circ}\mathrm{C}}\right)(400\,\mathrm{g})(-70\,^{\circ}\mathrm{C})=-1.17\times 10^5\,\mathrm{J}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next part cofuses me, this is what the solution manual does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q_{\mathrm{ice}}=1.17\times 10^{5}\,\mathrm{J}=(6.01\,\mathrm{kJ\ mol}^&#xA;{-1})\left(\dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\right)\biggl(m_{\mathrm{ice}}\times\dfrac{1\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}{18.02\,\mathrm{g}\,\mathrm{H}_{2}\mathrm{O}}\biggr)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\ \cdot\ ^{\circ}\mathrm{C}}\right)(m_{\mathrm{ice}})(10.0\ ^{\circ}\mathrm{C})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not too sure where this equation is derived from or of its purpose. The top part is determining the the enthalpy of ice but why is it being added to the specific heat of ice?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="5026" LastEditDate="2018-02-25T03:20:36.057" LastActivityDate="2018-02-25T03:21:15.250" Title="Calculating amount of ice required for heat loss" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15963" PostTypeId="1" AcceptedAnswerId="15965" CreationDate="2014-09-05T19:43:59.560" Score="6" ViewCount="3645" Body="&lt;p&gt;Why are imines more susceptible to reduction than carbonyl compounds? Imines react with $\ce{NaCNBH3}$ while the carbonyls don't.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="73" LastEditDate="2014-09-05T21:59:01.293" LastActivityDate="2017-04-29T12:14:04.793" Title="Why are imines more susceptible to reduction than carbonyl compounds?" Tags="&lt;redox&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15965" PostTypeId="2" ParentId="15963" CreationDate="2014-09-05T20:19:48.160" Score="10" Body="&lt;p&gt;Generally speaking, I think it is the reverse - carbonyls are reduced faster than imines with various sources of $\ce{H^-}$.  The reason usually cited (&lt;a href=&quot;http://chemweb.bham.ac.uk/~coxlr/Teaching/1st_Year/Carbonyl_Group/Carbonyl_pdfs/lecture%207%20(student%20HO).pdf&quot; rel=&quot;noreferrer&quot;&gt;see pages 2-4 of this reference&lt;/a&gt;) for this behavior is that the imine double bond is less polarized than the carbonyl double bond.  In the figure below resonance structure II contributes less when X=N as compared to the case X=O, due to the decreased electronegativity of nitrogen compared to oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pRQSi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the reaction is run under mildly acidic conditions, since the imine nitrogen is more basic than the carbonyl oxygen, more of the imine nitrogen will be protonated than the carbonyl oxygen.  This will make the protonated imine carbon much more electrophilic (positively polarized).  Consequently, under acidic conditions, an imine may be more reactive than a carbonyl to nucleophilic reduction (in addition to the above reference, also see &lt;a href=&quot;http://books.google.com/books?id=vrITBZ-_jwEC&amp;amp;pg=PA26&amp;amp;lpg=PA26&amp;amp;dq=borohydride%20reduction%20of%20imines%20vs.%20carbonyls&amp;amp;source=bl&amp;amp;ots=_SNHDXjJ2z&amp;amp;sig=FAcWPrs2AsZDpgy2K9cTrOOxpC0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=EBQKVLuJL4WuyATF3oH4Bg&amp;amp;ved=0CGcQ6AEwBw#v=onepage&amp;amp;q=borohydride%20reduction%20of%20imines%20vs.%20carbonyls&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;this reference&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-05T20:19:48.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15966" PostTypeId="2" ParentId="15856" CreationDate="2014-09-05T22:33:23.273" Score="1" Body="&lt;p&gt;From the equation : $\text{rate}=k[\ce{C2H4Br2}]^m[\ce{I-}]^n$&#xA;you can get by applying logarithms : $\ln\text{rate}=\ln k + m\ln[\ce{C2H4Br2}]+n\ln[\ce{I-}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That has the form : $\mathrm Y= w_0+w_1x_1+w_2x_2 $&#xA;where $y=\ln\text{rate}$,  $x_1=\ln[\ce{C2H4Br2}]$, $x_2=\ln[\ce{I-}]$ and $w_1=m$, $w_2=n$ and you can consider $x_0=1$ and $w_0=\ln k$&#xA;That becomes a &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_regression&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear regression&lt;/a&gt; problem, you can apply least squares or for better visualising it, you can draw the experimental values on logarithmic paper and fit a line. The result will be: $m=0.99884$ and $n = 0.98919$&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T18:00:24.947" LastActivityDate="2016-11-06T18:00:24.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15967" PostTypeId="1" AcceptedAnswerId="15968" CreationDate="2014-09-06T02:41:30.357" Score="9" ViewCount="4082" Body="&lt;p&gt;Due to Hund's rule, electrons start filling up the orbitals without pairing up. When this is happening, do the electrons all fill up the 'up' spin? Could they fill in the 'down' spin? Why do they prioritize the 'up' spin if they do?&lt;/p&gt;&#xA;" OwnerUserId="7716" LastEditorUserId="189" LastEditDate="2014-09-06T03:02:44.777" LastActivityDate="2018-01-01T22:32:54.800" Title="Do electrons only fill 'spin up' first? Or could it start filling 'down spins' first?" Tags="&lt;electrons&gt;&lt;orbitals&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-08-22T20:41:15.703" ContentLicense="CC BY-SA 3.0" />
  <row Id="15968" PostTypeId="2" ParentId="15967" CreationDate="2014-09-06T02:51:28.403" Score="7" Body="&lt;p&gt;No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this for all the orbitals you want to fill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: The convention is that you start with 'spin up' as this usually is taken to represent the $\frac{1}{2} \! \hbar$ eigenvalue of the $z$ component of the spin angular momentum operator and you should adapt to that in order to avoid confusion.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-06T03:22:39.733" LastActivityDate="2014-09-06T03:22:39.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15969" PostTypeId="1" AcceptedAnswerId="15984" CreationDate="2014-09-06T02:54:24.837" Score="1" ViewCount="110" Body="&lt;p&gt;I noticed Heinz Ketchup has a salt substitute version that uses potassium chloride instead of the usual salt. If this is safe and tastes like salt could it be a better additive than sodium chloride? Since it has potassium wouldn't this be a benefit for Heart health or against stroke if not used too much?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastActivityDate="2014-09-06T15:06:58.547" Title="About possibly healthy salt substitutes" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15970" PostTypeId="2" ParentId="15961" CreationDate="2014-09-06T05:13:13.163" Score="1" Body="&lt;p&gt;The second solution using the Henderson-Hasselbach equation is correct. While buffer solutions change pH very little near the pka of their constituent weak acid, it would not be correct to say that they do not change at all. Also, an addition of 20 mmol of NaOH is not insignificant to a 100 mM buffer—a 20% change in concentration. The Henderson-Hasselback equation in the form you've shown is an acceptable estimation of the buffer pH in a moderately concentrated buffer with an acid/base ratio close to 1. Deviations from it in that form can occur for very dilute or concentrated solutions or if it's well away from the pka. &lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-06T05:13:13.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15971" PostTypeId="1" CreationDate="2014-09-06T08:06:37.250" Score="1" ViewCount="534" Body="&lt;p&gt;Can someone please check my table to make sure I have the correct information. I am especially unsure of the parts highlighted in yellow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iv1tI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4906" LastActivityDate="2014-09-06T08:06:37.250" Title="Fuel Cell Chart - Anode, Cathode, Electrolyte" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;oxidation-state&gt;&lt;fuel&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15972" PostTypeId="1" CreationDate="2014-09-06T08:46:05.237" Score="2" ViewCount="3985" Body="&lt;p&gt;We did an experiment involving different $\ce{NaCl}$ concentration(ranging, 1-4%) on 100 mL of water. Our aim of the experiment was to find the dissolution rate of $\ce{CO2}$in the brine solvent. We did some research and found that $\ce{NaCl}$ increases the viscosity of the solvent, which in-turn decreases the dissolution rate. We want to explain how the entire process works, in a microscopic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers with good source would be good.&lt;/p&gt;&#xA;" OwnerUserId="7644" LastEditorUserId="5456" LastEditDate="2014-09-06T09:19:30.270" LastActivityDate="2014-11-18T14:56:16.407" Title="Effect of water viscosity (increased due to NaCl concentration) on dissolution rate of CO2" Tags="&lt;water&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15973" PostTypeId="1" CreationDate="2014-09-06T09:20:04.480" Score="12" ViewCount="314" Body="&lt;p&gt;Are phenylboranes susceptible to ipso or meta electrophilic aromatic substitution, i.e. what's the regioselectivity?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="189" LastEditDate="2014-09-13T20:39:12.087" LastActivityDate="2018-03-05T16:13:23.713" Title="Are phenylboranes susceptible to ipso or meta substitution?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15974" PostTypeId="1" CreationDate="2014-09-06T10:21:16.223" Score="20" ViewCount="940" Body="&lt;p&gt;I found a sentence in book which states the reverse step of this reaction ( I forgot what was it!)  has the faster step as rate determining .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate-determining_step&quot;&gt;Rate determining step-Wikipedia&lt;/a&gt;  states:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemical kinetics, the rate (or velocity) of a reaction mechanism with several steps is &quot;OFTEN&quot;  determined by the slowest step .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I think  'OFTEN' means not 'everytime' so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if faster step of the reaction is rate determining. Then why it is? &lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5424" LastEditDate="2014-09-06T18:04:59.057" LastActivityDate="2014-09-21T17:23:08.173" Title="Can in any case the faster step of the reaction be rate determining?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15975" PostTypeId="2" ParentId="15962" CreationDate="2014-09-06T10:29:36.010" Score="1" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pu{4.18 J/g K}$ is the specific heat of water. Also question assumes that due to addition of tea leaves the specific heat doesn't change much and remains equal to that of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you mix ice and tea and want the final temperature at $\pu{10^\circ C}$ then what do you expect of ice. Shouldn't it must melt and reach to $\pu{10^\circ C}$ and that will require a further energy to raise it temperature from $\pu{0^\circ C}$ to $\pu{10^\circ C}$. Also the mass of ice will remain same even if it get converted to water&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5026" LastEditDate="2018-02-25T03:21:15.250" LastActivityDate="2018-02-25T03:21:15.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15976" PostTypeId="2" ParentId="15972" CreationDate="2014-09-06T10:32:53.373" Score="1" Body="&lt;p&gt;By dissolving NaCl you increase the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_strength&quot; rel=&quot;nofollow&quot;&gt;ionic strength&lt;/a&gt; of the solution, i.e. there is more electrostatic attraction between the layers of water increasing the viscosity. Carbon dioxide is non-polar, so by increasing the polarity of the solvent, the carbon dioxide tend to form less interactions with the solvent.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-06T10:32:53.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15977" PostTypeId="1" AcceptedAnswerId="15985" CreationDate="2014-09-06T11:54:07.387" Score="4" ViewCount="216" Body="&lt;p&gt;One of the oldest recipes to create lye is this one:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take well burnt ashes and put them in a barrel&lt;/li&gt;&#xA;&lt;li&gt;Fill the rest of the barrel with water&lt;/li&gt;&#xA;&lt;li&gt;Leave the suspension for a week or two&lt;/li&gt;&#xA;&lt;li&gt;You can now try the lye concentration with egg - if 2/3 remain above the water, the concentration is sufficient for domestic uses&lt;/li&gt;&#xA;&lt;li&gt;Separate the lye solution from ashes by carefully draining the water. You can also use filtration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now I know there are traces of various metals in us, as well in flowers and trees. But how can ashes accumulate such a big amount of sodium? And how is the sodium bound into it. How does it turn into hydroxide?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2018-02-28T01:01:07.823" Title="How does lye appear in ashes?" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15978" PostTypeId="1" AcceptedAnswerId="15981" CreationDate="2014-09-06T12:11:17.143" Score="4" ViewCount="10987" Body="&lt;p&gt;Why does Birch reduction on benzene derivatives with an electron donating group happen at ortho and meta positions? When the substitute is electron withdrawing, ipso and para reduction take place because the electron withdrawing group stabilizes the anion in these positions. What's the explanation for electron donating group directivity?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-09-09T08:23:55.813" LastActivityDate="2017-09-05T13:47:29.983" Title="Regioselectivity of Birch reduction (electron donating groups)" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15981" PostTypeId="2" ParentId="15978" CreationDate="2014-09-06T13:09:01.110" Score="8" Body="&lt;p&gt;The explanation lies in the distribution of electron density in the intermediate radical anions that appear during the Birch reduction.&#xA;I use the following reactions as guiding examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1N25g.png&quot; height=&quot;75&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v8Kxc.png&quot; height=&quot;75&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron-withdrawing groups stabilize electron density at the &lt;em&gt;ipso&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions through conjugation and so the negative charge will mainly be found in these positions; subsequent protonation occurs &lt;em&gt;para&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VTF2t.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron-donating groups destabilize a negative charge (when compared to simple benzene) through conjugation and this effect would be strongest at the &lt;em&gt;ipso&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfkgJ.png&quot; height=&quot;125&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the negative charge will mainly be found at the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;meta&lt;/em&gt; position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FC4ll.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-06T14:32:41.423" LastActivityDate="2014-09-06T14:32:41.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15982" PostTypeId="1" AcceptedAnswerId="16095" CreationDate="2014-09-06T13:21:51.660" Score="1" ViewCount="3990" Body="&lt;p&gt;I have water contaminated with dissolved chlorine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + Cl2 &amp;lt;=&amp;gt;HCl + HOCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During electrolysis, this is an issue. These are my products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl + H2O -&amp;gt;T[electrolysis] NaOH + Cl2 + O2 + H2 }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\ce{NaOH}$ will be neutralised by $\ce{HCl}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaOH + HCl-&amp;gt; NaCl + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this can be considered hydrochloric acid. I want to remove the chlorine ions from water and prevent future dissolving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does increasing temperature help to remove the chlorine? Does boiling make the chlorine leave the mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + HCl + HOCl -&amp;gt;T[???] H2O + Cl2(gas)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;PS.:&lt;/strong&gt; I understand that there are means of separating the electrolysis columns. But this question is really about removing $\ce{Cl2}$ or $\ce{HCl}$ from aqueous solution.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2014-09-09T12:43:47.220" LastActivityDate="2014-09-09T12:43:47.220" Title="Removing HCl from water" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15983" PostTypeId="1" AcceptedAnswerId="15993" CreationDate="2014-09-06T13:44:07.343" Score="1" ViewCount="695" Body="&lt;p&gt;I just watched &lt;a href=&quot;https://www.youtube.com/watch?v=z08VMebsyNE&quot; rel=&quot;nofollow&quot;&gt;this YouTube video for a DIY key duplication system&lt;/a&gt; and was trying to figure out what all the material components could be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The &quot;mold solution&quot; (parts 1 and 2)&lt;/li&gt;&#xA;&lt;li&gt;The &quot;hardener&quot;&lt;/li&gt;&#xA;&lt;li&gt;The &quot;low melting point ingots&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The mold gel has to be resilient enough to withstand molten metal (even if the melting point is low), but I can't figure out what materials would fit the bill. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7724" LastActivityDate="2017-05-25T14:21:19.793" Title="Solution, hardener and metal required to cast key duplicates" Tags="&lt;experimental-chemistry&gt;&lt;metal&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15984" PostTypeId="2" ParentId="15969" CreationDate="2014-09-06T13:55:38.050" Score="1" Body="&lt;p&gt;The taste of potassium chloride is not close enough to sodium chloride so the salt substitutes usually contain a mix of potassium and sodium chloride.  This is healthier than pure sodium chloride.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-06T13:55:38.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15985" PostTypeId="2" ParentId="15977" CreationDate="2014-09-06T13:59:33.583" Score="3" Body="&lt;p&gt;There is sodium and potassium in wood.  When the wood burns some of the sodium and potassium become bound to oxygen.  The oxides become hydroxides when exposed to water.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-06T13:59:33.583" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15986" PostTypeId="1" AcceptedAnswerId="15988" CreationDate="2014-09-06T14:59:24.813" Score="4" ViewCount="4737" Body="&lt;p&gt;Can someone explain what reaction is going on here?&#xA;I looked in my list of mechanisms learnt: I find nothing related!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://imgur.com/poqRBxo&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/poqRBxo.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2019-11-19T10:10:08.810" Title="2-methylpropene to tert-butyl alcohol Reaction Mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15987" PostTypeId="2" ParentId="15969" CreationDate="2014-09-06T15:06:58.547" Score="0" Body="&lt;p&gt;The taste of $\ce{KCl}$ is a bitter and salty, but is not as salty as that of $\ce{NaCl}$, so the idea seems questionable. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Special Remarks on other Toxic Effects on Humans: Actute Potential&#xA;  Health Effects: Skin: May cause skin irritation Eye: Dust may cause&#xA;  eye irritation. Inhalation: Dust may cause respiratory tract&#xA;  irritation. Low hazard for usual industrial handling Ingestion: May&#xA;  affect behavior (coma, change in motor activity, listlessness,&#xA;  vertigo, mental confusion, paresthesias, general weakness, flaccid&#xA;  paralysis), metabolism, blood (change in clotting factor, electrolytic&#xA;  imbalance), cardiovascular (hypotension, circulatory disturbances,&#xA;  cardiac arrhythmias, heart block), and respiratory, gastrointestinal&#xA;  (irritation of GI tract, nausea, vomiting, diarrhea, abdominal&#xA;  discomfort, purging), and urinary(impairment of renal function)&#xA;  systems. Acute potassium intoxication by mouth is rare because large&#xA;  single doses usually induce vomiting, and because in the absence of&#xA;  pre-existing kidney damage potassium is rapidly excreted. Maximal&#xA;  nontoxic oral dose of KCl in man varies from 0.2g to 1 g of&#xA;  potassium/kg/day depending upon efficiency of individual excretory&#xA;  mechanism; lower doses sometimes cause impairment of renal function as&#xA;  shown by reduced inulin, and urea clearance. Chronic Potential Health&#xA;  Effects: May affect blood and cardiovascular system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(c) $\ce{KCl}$ MSDS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, If you are into this, I recommend to consult your doctor. $\ce{K,Na,Mg,Ca}$ ions are vital for cardiovascular and nervous system and affect kidneys. They are not highly toxic, and our body can deal with reasonable doses of each, but in case you have chronic condition, it may cause individual vulnerability.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-06T15:06:58.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15988" PostTypeId="2" ParentId="15986" CreationDate="2014-09-06T15:11:02.093" Score="6" Body="&lt;p&gt;It is an acid-catalyzed hydration of the double bond. First, the double bond is protonated to yield the &lt;em&gt;tert&lt;/em&gt;-butyl cation which is most stabilized by hyperconjugation (protonation at C2 would form a less stable primary cation). The cation is captured by nucleophilic attack of water, which after deprotonation yields &lt;em&gt;tert&lt;/em&gt;-butanol and the regenerated proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{(H3C)2C=CH2 + H3O+ &amp;lt;=&amp;gt; (H3C)3C+ + H2O} \tag 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{(H3C)3C+ + H2O &amp;lt;=&amp;gt; (H3C)3C-OH2^{+}} \tag 2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{(H3C)3C-OH2^{+} + H2O -&amp;gt;~ (H3C)3C-OH + H3O+} \tag 3$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="24052" LastEditDate="2019-11-19T10:10:08.810" LastActivityDate="2019-11-19T10:10:08.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15989" PostTypeId="2" ParentId="15847" CreationDate="2014-09-06T15:17:16.237" Score="4" Body="&lt;p&gt;The galvanic cell must be written like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in order to have a positive  &lt;span class=&quot;math-container&quot;&gt;$\mathrm{EMF} = \pu{0.413 V}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your approach is right with a slight error.&#xA;This is a concentration cell based on &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb^2+}$&lt;/span&gt;. Since &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb(NO3)2}$&lt;/span&gt; dissociates completely, while &lt;span class=&quot;math-container&quot;&gt;$\ce{PbS}$&lt;/span&gt; is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.&lt;br&gt;&#xA;So, we have the following semireactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA; \ce{Pb^2+ + 2 e^- &amp;amp;-&amp;gt; Pb} &amp;amp;\quad &amp;amp;\text{(cathode)} \\&#xA; \ce{Pb &amp;amp;-&amp;gt; Pb^2+ + 2 e-} &amp;amp;\quad &amp;amp;\text{(anode)}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the anode we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{PbS &amp;lt;=&amp;gt; Pb^2+ + S^2-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$[\ce{Pb^2+}] = [\ce{S^2-}] = \sqrt{K_\mathrm{sp}}$&lt;/span&gt;.&#xA;So the semicell potentials are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;E_\text{cathode} &amp;amp;= E^\circ\\&#xA;E_\text{anode}   &amp;amp;= E^\circ + \frac{0.059}{2}\log_{10}[\ce{Pb^2+}]\\&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="41328" LastEditDate="2019-11-18T08:18:36.703" LastActivityDate="2019-11-18T08:18:36.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15991" PostTypeId="1" CreationDate="2014-09-06T16:40:56.487" Score="5" ViewCount="557" Body="&lt;p&gt;I was looking at substances that produce the highest flame temperatures upon combustion. I note that the equations for combustion are producing carbon monoxide rather than carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CN)2 + O2 -&amp;gt; 2CO + N2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't carbon dioxide produced? Why doesn't the production of carbon dioxide give a hotter flame? It seems that a higher flame temperature would be generated by complete combustion. My thinking is that the heat of formation of carbon dioxide is around −400 kJ/mol while the heat of formation of carbon monoxide is around −100 kJ/mol so the additional heat released by producing carbon dioxide should account for carbon dioxide's greater heat capacity.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorDisplayName="user7951" LastEditDate="2015-04-29T21:27:44.087" LastActivityDate="2015-04-29T21:27:44.087" Title="Why is carbon monoxide the product in reactions that produce high flame temperatures?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15992" PostTypeId="2" ParentId="83631" CreationDate="2014-09-06T16:48:41.273" Score="8" Body="&lt;p&gt;You need to enforce the boundary conditions of your problem to evaluate the integral. I think that might be the material point here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's start at the beginning: You have the rate equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    - \frac{\mathrm{d}c_{\mathrm{A}}}{\mathrm{d}t} = k c_{\mathrm{A}} c_{\mathrm{B}} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, introduce the conversion $x = c_{\mathrm{A}0} - c_{\mathrm{A}} = c_{\mathrm{B}0} - c_{\mathrm{B}}$. This means that an infinitesimal change in the conversion can be written as $\mathrm{d} x = -\mathrm{d} c_{\mathrm{A}} = -\mathrm{d} c_{\mathrm{A}}$. Substituting this into the rate equation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \frac{\mathrm{d}x}{\mathrm{d}t} = k ( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x ) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this differential equation it is sufficient to separate the variables and integrate both sides of the equation. And this is the point where the boundary conditions enter the game. To evaluate the integrals fully, i.e. without leaving a constant of integration $C$, you need to use some limits for your integrals for which you know the values of you variables beforehand. So the question is: What is known beforehand? You know that at the start of the reaction, and we will define this point in time as $t=0$, none of you reactants have reacted with one another. So, $c_{\mathrm{A}}(t\!=\!0) = c_{\mathrm{A}0}$ or in terms of the conversion $x(t\!=\!0) = 0$. This will set the lower limit of the integrals. For the upper limit you can use the value that you actually want to calculate, i.e. the concentration $c_{\mathrm{A}}(t)$ or conversion $x(t)$ at some time $t$. Using all this leads to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \int \limits_{x(t=0) = 0}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x = \int \limits_{t=0}^{t} k \, \mathrm{d}t \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the integral on the left-hand side you can use the hint already given to (although calculating it yourself isn't that difficult if you use partial-fraction decomposition) but you can leave out the constant $C$ as this will be fixed by setting explicit integration limits. The integral on the right-hand side is no problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \int \limits_{x(t=0)}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x &amp;amp;= \int \limits_{0}^{t} k \, \mathrm{d}t \\&#xA;    \left[ \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{B}0} - x }{c_{\mathrm{A}0} - x} \right) \right]^{x(t)}_{0} &amp;amp;= \Bigl[ k t \Bigr]^{t}_{0} \\&#xA;    \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} \right) = k t&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exponentiating this gives nearly the result that you are after:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} = \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the only work that is left is to make the back-substitution $x(t) = c_{\mathrm{A}0} - c_{\mathrm{A}}(t)$ and solve the resulting equation for $c_{\mathrm{A}}(t)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) }{c_{\mathrm{B}0} c_{\mathrm{A}}(t)} &amp;amp;= \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\&#xA;    c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) &amp;amp;= c_{\mathrm{B}0} c_{\mathrm{A}}(t) \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\&#xA;    c_{\mathrm{A}0} c_{\mathrm{B}0} - c_{\mathrm{A}0}^{2} &amp;amp;= c_{\mathrm{A}}(t) \left( c_{\mathrm{B}0} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - c_{\mathrm{A}0} \right) \\ &#xA;    c_{\mathrm{B}0} - c_{\mathrm{A}0} &amp;amp;= c_{\mathrm{A}}(t) \left( \frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1 \right) \\ &#xA;   \Rightarrow \qquad  c_{\mathrm{A}}(t) &amp;amp;= \frac{c_{\mathrm{B}0} - c_{\mathrm{A}0}}{\frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-08T20:27:25.530" LastActivityDate="2014-09-08T20:27:25.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15993" PostTypeId="2" ParentId="15983" CreationDate="2014-09-06T17:05:37.613" Score="1" Body="&lt;p&gt;From the video, the material looks like some kind of silicone. There aren't many other materials that have that translucent elastomeric quality. If that's the case, the larger of the two mould solutions will be mostly some kind of silane precursor such as SiOH. The second part will contain the catalyst for the polymerization. Depending on the exact chemistry used it could be perhaps tin-based. My guess for the &quot;hardener&quot; is that it's simply more catalyst. Two part silicones with longer working lives are more common than very rapid cure ones so adding more/stronger catalyst as the third part may be just to make it cure quickly. Many silicones can withstand temperatures of 200+ °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the low melting point alloy, there are a number of things it could be. Given that the application is keymaking and people touch keys with their hands, I hope it's safe to assume that the alloy doesn't contain lead or cadmium. The video says its melting point is around 90 °C. The two closest common alloys without toxic metals are 58% Bi, 42% Sn @ 132 °C and 66.3% In and 33.7% Bi @ 72 °C, from the wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Fusible_alloy&quot; rel=&quot;nofollow&quot;&gt;fusible alloys&lt;/a&gt;, so it's likely some alloy with similar composition to one of these. Bismuth and tin are far cheaper than indium, so it wouldn't surprise me if the alloy was some mixture of those two elements.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-06T17:05:37.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15994" PostTypeId="2" ParentId="15991" CreationDate="2014-09-06T17:14:54.300" Score="3" Body="&lt;p&gt;This is just a guess... But I wonder if this is due to entropic considerations. In the equation you have written, there are two molecules of gas getting converted to three molecules of gas. $\Delta S$ is positive, so it decreases $\Delta G$ as the reaction temperature increases. However, in the reaction $\ce{(CN)2 + 2 O2 = 2 CO2 + N2}$ you have an equal number of gas molecules, so the entropic component of $\Delta G$ will be small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure why that would matter. At very high temps, the first reaction will presumably have a very large, negative free energy. Of course, the $\Delta G$ for the $\ce{CO2}$ reaction is most likely negative, so you might have to control the feed stoichiometry carefully to ensure that you are getting only the first reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the $\Delta G$ values from the &lt;a href=&quot;http://kinetics.nist.gov/janaf/&quot; rel=&quot;nofollow&quot;&gt;NIST-JANAF&lt;/a&gt; website, and it doesn't look like the free energy for these reactions is much different at 4000 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_f G_{(CN)_2}^{0} = 136.036\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_f G_{CO}^{0} = -446.485\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_f G_{CO_2}^{0} = -393.183\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G_{rxn1}^{0} = -1029.005\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G_{rxn2}^{0} = -922.397\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I just read one of the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01098a517&quot; rel=&quot;nofollow&quot;&gt;original papers&lt;/a&gt; on this type of flame, and it sounds like the thermal stability of the products is an issue as well. I take it that at these very high temperatures, some of the energy can be transferred to bond-breaking processes, which would lower the total flame temperature. $\ce{CO}$ and $\ce{N2}$ have very strong covalent bonds, so perhaps they do not readily dissociate at these temps.  &lt;/p&gt;&#xA;" OwnerUserId="2109" LastEditorUserId="2109" LastEditDate="2014-09-06T19:58:27.757" LastActivityDate="2014-09-06T19:58:27.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15995" PostTypeId="1" AcceptedAnswerId="15998" CreationDate="2014-09-05T12:29:31.367" Score="15" ViewCount="904" Body="&lt;p&gt;Let's say a room is filled with butane, I then throw a cigarette into the room. What happens to the atoms/molecules of the butane when they are in contact with the heat from the cigarette?&lt;/p&gt;&#xA;" OwnerUserId="7731" OwnerDisplayName="wtoh" LastEditorUserId="30836" LastEditDate="2019-04-14T11:43:01.250" LastActivityDate="2019-04-14T11:43:01.250" Title="What really happens atomically in an explosion?" Tags="&lt;energy&gt;&lt;combustion&gt;&lt;explosives&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15996" PostTypeId="2" ParentId="15995" CreationDate="2014-09-05T12:44:02.650" Score="7" Body="&lt;p&gt;The butane and oxygen molecules' electrons gain enough kinetic energy to overcome the potential barrier that prevented them from disassociating from their respective molecules making the hydrogen and carbon atoms bond with oxygen – producing $\ce{CO2}$ and $\ce{H2O}$ and extra heat. Which would go to increase more electron's kinetic energy as well as increase the pressure of the air (since that what happens when gasses heat). This sudden increase in pressure is an explosion.&lt;/p&gt;&#xA;" OwnerUserId="7726" OwnerDisplayName="Mr.WorshipMe" LastEditorDisplayName="user7951" LastEditDate="2016-11-06T16:44:10.667" LastActivityDate="2016-11-06T16:44:10.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15997" PostTypeId="2" ParentId="15995" CreationDate="2014-09-05T13:47:47.453" Score="12" Body="&lt;p&gt;The butane and oxygen in the room are in a metastable state.  An example energy diagram of a metastable state is shown in the image below.  Before throwing in the cigarette, the butane and oxygen molecules are in the state labeled as 1.  This state is metastable because there is a lower energy state (which makes it statistically more probable) available if they were to combine to form $CO_2$ and $H_2O$.  This lower energy state is represented by the 3 in the image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that there is an energy barrier which must be overcome for them to get to this more stable state.  This is where the cigarette comes in; it adds some thermal energy which pushes the system over the energy barrier (state 2 in the image) and allows it to fall into its more stable state.  In the process the excess energy gained from moving from state 1 to state 3 is given off as heat.  This excess heat causes more molecules in the area to combine, and you end up with a runaway reaction known as an explosion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8BuT6.png&quot; alt=&quot;Image from Wikimedia Commons&quot;&gt;&#xA;*Image taken from the &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Meta-stability.svg&quot; rel=&quot;noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4188" OwnerDisplayName="Chris Mueller" LastActivityDate="2014-09-05T13:47:47.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15998" PostTypeId="2" ParentId="15995" CreationDate="2014-09-05T15:37:04.637" Score="28" Body="&lt;p&gt;The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a &lt;a href=&quot;https://www.iwr.uni-heidelberg.de/groups/reaflow/Diss/Heghes.pdf&quot;&gt;dissertation that presents a nice summary of the process here&lt;/a&gt; (this is a 1MB PDF).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Radical_(chemistry)&quot;&gt;free radicals&lt;/a&gt;. These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction.  You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust.&lt;/p&gt;&#xA;" OwnerUserId="1470" OwnerDisplayName="John Rennie" LastActivityDate="2014-09-05T15:37:04.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15999" PostTypeId="2" ParentId="7173" CreationDate="2014-09-06T17:50:49.037" Score="1" Body="&lt;p&gt;Your first statement is correct. The basicity increases with $pK_a$, so a base with $pK_a &amp;gt; 16$ will be able to deprotonate your alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second statement is also in principle correct; however, the value of the sum $pK_a+pK_b$ is solvent-dependent, and 14 is the value for water at standard conditions. $pK_a$ and $pK_b$ values of acids and bases depend on the solvent as well. For example, sodium amide is a strong base in liquid ammonia, but not so in water due to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Leveling_effect&quot; rel=&quot;nofollow&quot;&gt;leveling effect&lt;/a&gt; of the solvent. Dissolving sodium amide in water will give a solution of &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;sodium hydroxide&lt;/a&gt;, which is not basic enough ($pK_a=13$) to deprotonate the alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the amount of deprotonated alcohol will depend on the difference between $pK_a$ values of alcohol and base, since the acid-base reaction is an equilibrium. For example, a base with $pK_a=33$ will deprotonate your alcohol to a greater extent than one with a $pK_a$ of 16.5 .&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-06T17:50:49.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16000" PostTypeId="1" AcceptedAnswerId="16002" CreationDate="2014-09-06T18:43:32.783" Score="3" ViewCount="1150" Body="&lt;p&gt;I am told that because of the poor overlap between a chlorine atom and a carbon atom and chlorine's relatively high electronegativity, there is a strong partial positive character on the central carbon in an acyl chloride molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense. I am then told that because of this high partial positive character created through the inductive effect of chlorine and oxygen and the poor overlap between chlorine and oxygen, the oxygen's lone pair(s) actually participate in a sort of resonance with the central carbon, thereby giving the C=O bond in an acyl chloride partial triple bond character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Also what exactly is this called? Backbonding? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-16T16:54:56.073" LastActivityDate="2015-10-16T16:54:56.073" Title="Triple Bond Character in Acyl Chlorides" Tags="&lt;orbitals&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16001" PostTypeId="1" CreationDate="2014-09-06T19:37:25.553" Score="1" ViewCount="556" Body="&lt;p&gt;I am trying to determine the number of nanorods that I would need to measure out to get a 1 mM solution.  For the gold nanorods that we ordered, it was mentioned that the single nanorods dimensions are 40 nm length by 10 nm in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product's concentration is $6.3\times10^{11}$ nanorods per ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I measure out the solution to obtain a 1 mM gold nanorod solution?&lt;/p&gt;&#xA;" OwnerUserId="7727" LastEditorDisplayName="user7951" LastEditDate="2015-10-21T20:01:58.760" LastActivityDate="2019-09-07T20:41:27.580" Title="How should I measure out a solution given in nanorods per ml to make a solution of a set number of mols?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16002" PostTypeId="2" ParentId="16000" CreationDate="2014-09-06T20:53:55.347" Score="5" Body="&lt;p&gt;The various resonance structures that one can draw for an acid chloride are shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EDnSE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure IV, is the &quot;triple bond&quot; structure that you mentioned.  The acylium ion ($\ce{R-C#O^{+}}$) is very stable, so structure IV does contribute to the overall description of an acid chloride.  Note that the $\ce{C#O}$ bond in IV is not a real triple bond as the carbon remains $\ce{sp^2}$ hybridized, with the attendant 120 degree (non-linear) bond angles.  I've also drawn a fifth resonance structure involving hypeconjugation.  This is probably a reasonable structure as well since it places negative charge on oxygen and positive charge on the hydrogen.  It also helps explain why hydrogens on the carbon alpha to a carbonyl have increased acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the term &quot;backbonding&quot; is used when electron density is transferred from one atom to another through the use of antibonding orbitals.  See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14034/difference-between-backbonding-and-hyperconjugation-and-conjugation/14043#14043&quot;&gt;here&lt;/a&gt; for a better description.  In resonance structure IV, the lone pair from oxygen is being shared with an empty carbon $\ce{sp^2}$ orbital, the one that had been used to bond to the chlorine.  I've never heard that referred to as backbonding.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-09-06T20:53:55.347" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16003" PostTypeId="1" AcceptedAnswerId="16070" CreationDate="2014-09-06T20:58:06.877" Score="6" ViewCount="1108" Body="&lt;p&gt;What is a ballpark figure for the difference in energy for an atom that follows Hund's rule vs one that has two electrons with opposite spins?  I'd be interested to know carbon and nitrogen.  Is there a bigger difference when there are two electrons with one spin and one electron with the opposite spin vs two electrons with opposite spins? &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="189" LastEditDate="2014-09-09T02:02:06.187" LastActivityDate="2014-09-09T02:02:06.187" Title="How much energy does it cost to have electron configurations that are not in accordance with Hund's rules?" Tags="&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16004" PostTypeId="1" AcceptedAnswerId="16006" CreationDate="2014-09-06T21:32:40.527" Score="-1" ViewCount="631" Body="&lt;p&gt;Can ionic character also be construed as a poor sharing of electrons rather than simply as the taking of electrons? For example, NaCl is ionic, and that's because the chlorine takes sodium's one valence electron. On the other hand, C-Br and C-I are sometimes referred to as ionic (at least informally). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it proper to refer to C-Br and C-I as ionic because of the poor overlap between carbon and the two respective halides rather than any electron-withdrawing effect per se; the EN difference between C and I is less than the EN difference in the &quot;non-polar&quot; C-H bond. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-02T16:53:52.663" LastActivityDate="2015-10-02T16:53:52.663" Title="Ionic Character" Tags="&lt;bond&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16005" PostTypeId="2" ParentId="6130" CreationDate="2014-09-06T21:50:11.797" Score="2" Body="&lt;p&gt;In P element, the energy of 3d is lower than 4s. Look at the link provided below. Only in K and Ca, 4s has lower energy than 3d. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Chemistry%201/Slide/lect6/6_6.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Chemistry%201/Slide/lect6/6_6.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason could be because the symmetry of hybridized orbital with d orbital will be more preferred in bonding with that of s orbital, and the molecule obtain more energy compensation with sp3d over sp3s.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2014-09-06T21:50:11.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16006" PostTypeId="2" ParentId="16004" CreationDate="2014-09-06T22:28:04.507" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it proper to refer to C-Br and C-I as ionic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've never thought of those bonds as ionic.  For example, consider electrophilic aromatic substitution in bromobenzene. The sigma bond between the bromine and benzene carbon allows for inductive withdrawal of electron density from the benzene ring, hence bromobenzene typically reacts slower than benzene.  However, to explain the  ortho-para directing effect of the halogen you also need to invoke resonance interaction between the halogen and the benzene carbon p orbital. So there is both sigma and pi interaction between a halogen substituent and an aromatic carbon, that doesn't sound ionic to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, even in the case of sodium chloride the electron is not completely transferred from sodium to chlorine.  Crude estimates using &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15872/why-is-the-bond-in-hf-ionic-while-it-is-covalent-in-hcl/15889#15889&quot;&gt;this methodology&lt;/a&gt; put it around 80%.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-06T22:28:04.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16007" PostTypeId="1" CreationDate="2014-09-06T22:33:52.900" Score="4" ViewCount="1996" Body="&lt;p&gt;Let's consider a typical agarose gel electrophoresis experiment. An appropriate quantity (not so much that it is overloaded, but not so little that it cannot be visualized) of linear DNA pieces with identical sequence and chemical structure is loaded onto the gel, and constant voltage is applied for a given time. I suspect that there is a function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x = f(V, t, l, c, k)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the variables are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x$, cm, distance from well&lt;/li&gt;&#xA;&lt;li&gt;$V$, Volt, potential difference across the gel&lt;/li&gt;&#xA;&lt;li&gt;$t$, minutes, time over which the voltage is applied&lt;/li&gt;&#xA;&lt;li&gt;$l$, bp, length of DNA fragments&lt;/li&gt;&#xA;&lt;li&gt;$c$, gr/100 ml, density of the gel given in grams of agarose per 100 ml of buffer&lt;/li&gt;&#xA;&lt;li&gt;$k$, a constant which is empirically determined for a given running buffer and gel tank&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically, I want a formula into which I can plug in the size of my DNA and a few other parameters, and figure out how long I have to run it to get decent separation (the output variable is x because I intend to plot $x$ over $t$ for a given planned electrophoresis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a formula possible? What is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have previously worked with gel documentation software which fits linear, polynomial and exponential functions to the DNA (I couldn't figure out exactly what kind of fitting the software was doing). However, after training a best-fit model on my ladder (with ~12 bands) and applying it to my samples of known size, I found all of these fits to be very inaccurate (5-10% error). Therefore, I suspect the equation governing the migration rate of DNA is in fact not a simple exponential function (as is often taught in college) but more complicated. Unfortunately, I lack the background in physical chemistry to derive the equation myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a good argument for why the relation should be a simple exponent, eg. $x = e ^ {\alpha \cdot l}$, that is also an acceptable answer.&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2020-03-28T19:02:50.407" Title="What is the formula for migration rate of DNA in agarose gel electrophoresis?" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16009" PostTypeId="1" CreationDate="2014-09-07T00:04:25.403" Score="3" ViewCount="118" Body="&lt;p&gt;I'm starting to prepare for my research project on photoelectrochemical (PEC) splitting of water into H and O using (F doped) titanium dioxide as the photoanode. I've been unsuccessfully searching in the literature for any particular reason(s) which could make PEC better than other methods of photocatalytic water splitting. What is, if anything, special about PEC water splitting compared to other approaches? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2014-09-07T00:04:25.403" Title="Photoelectrochemical (PEC) water splitting" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;photochemistry&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16010" PostTypeId="1" AcceptedAnswerId="31482" CreationDate="2014-09-07T00:06:10.987" Score="3" ViewCount="699" Body="&lt;p&gt;For a solution such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luria-Bertani (LB) Broth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 g tryptone&lt;br&gt;&#xA;5 g yeast extract&lt;br&gt;&#xA;10 g of NaCl ($M = 58.44\ \mathrm{g/mol}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q.s. to 1 l, pH to 7.2, autoclave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to make a bulk amount of premixed powder, i.e. combine 1 kg tryptone, 500 g yeast extract and 1 kg NaCl in a big bottle, shake it and keep it on shelf. Whenever I need to make LB now I just weight 25 g of this premix and add to 1 l of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m concerned about a couple things thought. First how thoroughly it gets mixed, and second any settling of different chemicals over time. Either would cause many different batches to be messed up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m wondering if anyone has experience with this and ways to avoid those two problems. I’m interested in doing this for other solutions as well. Do manufacturers who make these premixed powders re pulverize them or process them in a certain way to make it work?&lt;/p&gt;&#xA;" OwnerUserId="7729" LastEditorDisplayName="user7951" LastEditDate="2015-05-15T17:32:22.643" LastActivityDate="2019-11-27T11:29:38.440" Title="Mixing of powders to make stock solutions" Tags="&lt;solutions&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16012" PostTypeId="2" ParentId="16010" CreationDate="2014-09-07T01:10:04.073" Score="2" Body="&lt;p&gt;Once the powders are mixed, it is unlikely that they will separate unless the container is subjected to vibrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for mixing, companies that need to mix batches of material use a mixer to make their batches. The second photo in the &lt;a href=&quot;http://www.pharmpro.com/articles/2011/08/continuous-solid-oral-dosage-processing&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; is similar to what I have seen. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-07T01:10:04.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16013" PostTypeId="1" AcceptedAnswerId="18500" CreationDate="2014-09-07T01:45:32.513" Score="1" ViewCount="720" Body="&lt;p&gt;I am trying to find visual ways to demonstrate water contaminants. For example, chlorine reacts to N,N-diethyl-p-phenyldiamine (DPD), forming a magenta color. The chemical doesn't need to be accurate, as this is more of pass-fail test of the water. &lt;strong&gt;What reacts with fluoride to provide similar results?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7728" LastEditorUserId="7920" LastEditDate="2014-10-16T06:20:27.127" LastActivityDate="2018-06-01T13:24:00.863" Title="Fluoride Color Reaction" Tags="&lt;water&gt;&lt;analytical-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16014" PostTypeId="1" AcceptedAnswerId="16022" CreationDate="2014-09-07T02:00:48.717" Score="3" ViewCount="106" Body="&lt;p&gt;There are 4 different timescales in material characterization (as classified in the Atkins textbook):&lt;/p&gt;&#xA;&lt;p&gt;a. Time during which a quantum of radiation or a particle can interact with a molecule.&lt;/p&gt;&#xA;&lt;p&gt;b. Lifetime of excited state of a molecule.&lt;/p&gt;&#xA;&lt;p&gt;c. Minimum lifetime species must have to be seen as a distinct species.&lt;/p&gt;&#xA;&lt;p&gt;d. Total time of experiment in which the species is observed.&lt;/p&gt;&#xA;&lt;p&gt;Now the question is how to determine each of these? Say I have a certain compound and want to know what is the time of interaction of a photon with the molecule (part a). Is there an experimental method or instrument to determine these timescales?&lt;/p&gt;&#xA;" OwnerUserId="6985" LastEditorUserId="95133" LastEditDate="2021-05-04T06:44:01.680" LastActivityDate="2021-05-04T06:44:01.680" Title="Determination of timescales using characterization methods" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;&lt;crystal-structure&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16015" PostTypeId="1" AcceptedAnswerId="16016" CreationDate="2014-09-07T02:35:15.230" Score="0" ViewCount="348" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Primary Cells - pH changes during discharge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;Alkaline Cell:&lt;/code&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Manganese Dioxide Zinc Alkaline Cell&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anode (-)&lt;/strong&gt; &#xA;$\ce {Zn(s) + 2OH^{−}(aq) → ZnO(s) + H2O(l) + 2e^{−}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cathode (+)&lt;/strong&gt;&#xA;$\ce {2MnO2(s) + H2O(l) + 2e^{−} → Mn2O3(s) + 2OH^{−}(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overall&lt;/strong&gt;&#xA;$\ce {Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce {OH-}$ ions are consumed at the anode&lt;/li&gt;&#xA;&lt;li&gt;$\ce {OH-}$ ions are formed at the cathode&lt;/li&gt;&#xA;&lt;li&gt;$\ce {same[OH^{−}]}$&lt;/li&gt;&#xA;&lt;li&gt;pH - no change&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Do pH changes occur during the discharge process?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="4945" LastEditDate="2014-09-09T08:23:17.313" LastActivityDate="2014-09-09T08:23:17.313" Title="pH Changes in primary cells" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;ph&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16016" PostTypeId="2" ParentId="16015" CreationDate="2014-09-07T02:45:44.967" Score="1" Body="&lt;p&gt;No, the reaction does not produce hydroxide ions. But the solution is already basic (probably between pH 12 and 14).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-07T02:45:44.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16017" PostTypeId="1" AcceptedAnswerId="16023" CreationDate="2014-09-07T02:54:02.043" Score="2" ViewCount="40" Body="&lt;p&gt;Do attributes such as heat of formation have to be found through experimentation or can they be calculated based on the attributes of the bonds and the atoms involved?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2014-09-07T07:32:27.260" Title="Can the thermodynamic information of a compound be calculated or must it be found emperically?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16018" PostTypeId="1" CreationDate="2014-09-07T03:07:42.743" Score="1" ViewCount="4032" Body="&lt;p&gt;The equation to measure work is $$w=F\cdot D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the equation for force is  $$F=(-P \cdot A)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the work equation we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=-P\cdot A\cdot D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why is pressure negative in force?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastActivityDate="2019-01-17T20:43:50.983" Title="Why is pressure negative in force and work" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16019" PostTypeId="2" ParentId="16018" CreationDate="2014-09-07T03:47:15.747" Score="2" Body="&lt;p&gt;In this context we are talking about the work done by a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than representing it as $w = -PAD$, think of the formula as representing an area (that the gas is pressing against) moving a distance $D$. Since this a change in volume, we can set $w = -P\Delta V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are putting work into the system (say pressing a piston into a cylinder containing a gas at a given pressure), the volume is becoming smaller, so $\Delta V$ is negative.  The negative of the pressure times a negative $\Delta V$ is a positive value, which matches with the sign convention for putting work in.  Vice versa, an expansion of a compressed volume at a given pressure will push the piston back out of the cylinder and allow the gas to do work, so $w$ is negative which follows the sign convention for work output.&lt;/p&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-09-07T03:47:15.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16022" PostTypeId="2" ParentId="16014" CreationDate="2014-09-07T06:57:42.317" Score="4" Body="&lt;p&gt;The absorption of a photon en route to an electronically excited Franck-Condon state typically typically proceeds in the femtosecond range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lifetime of an electronically excited singlet state mostly is in the range of several nanoseconds. In some cases, however, fluorescence lifetimes in the range of several tenth of nanoseconds have been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cases where the initial excited singlet state undergoes intersystem crossing to a triplet state, much longer phosphorescence lifetimes (seconds range) are observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to detect a transient, its lifetime must be longer than the laser pulse needed to generate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've populated a state, you may either use &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;time-resolved absorption spectroscopy&lt;/strong&gt; to record it's spectrum and the decay of the absorption at a particular wavelength over time or&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;use &lt;strong&gt;time-resolved emission spectroscopy&lt;/strong&gt; to measure the decrease of fluorescence (or phosphorescence emission) intensity at a particular wavelength over time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-22T08:39:18.913" LastActivityDate="2014-09-22T08:39:18.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16023" PostTypeId="2" ParentId="16017" CreationDate="2014-09-07T07:32:27.260" Score="2" Body="&lt;p&gt;For gaseous compounds heat of formation can be calculated using quantum chemistry packages as a difference between energies, assigned by the package, though inaccuracies are possible. Entropy of formation is a bit trickier, it requires calculation of vibrational and rotational states, but still pretty straightforward. For condensed phase (liquids, solids) the situation is harder, as knowledge of structure of solid state must be known beforehand, or full exhaustive search of possible crystal structures must be performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, both heat and entropy of formation are found experimentally. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-07T07:32:27.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16024" PostTypeId="2" ParentId="15991" CreationDate="2014-09-07T08:27:13.033" Score="-1" Body="&lt;p&gt;mainly because in high flame areas, oxygen is used up by the fire more rapidly,thus there is less oxygen available,and thus CO carbon monoxide forms.&lt;/p&gt;&#xA;" OwnerUserId="7732" LastActivityDate="2014-09-07T08:27:13.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16025" PostTypeId="1" CreationDate="2014-09-07T10:30:46.240" Score="-1" ViewCount="67" Body="&lt;p&gt;If you have read this question and you thought that what is this stupid question then sorry to ask this question to you. Hey reader if you want to increase one student's knowledge then please answer this. Hey reader please answer this in an easy way. I am only in 10th grade\class.&lt;/p&gt;&#xA;" OwnerUserId="7646" LastEditorUserId="418" LastEditDate="2014-09-07T17:43:10.470" LastActivityDate="2014-09-07T17:52:27.583" Title="How to photolyze anything?" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-09-08T04:25:50.617" ContentLicense="CC BY-SA 3.0" />
  <row Id="16026" PostTypeId="1" AcceptedAnswerId="16041" CreationDate="2014-09-07T11:00:14.343" Score="5" ViewCount="4006" Body="&lt;p&gt;&quot;I heard&quot; that gray lines forming on porcelain indicate usage of aluminium in the coating. I would primarily like to verify this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bowl #1 and bowl #2 have been used both for several months and were bought in different shops. The same observation goes for other dishes. Some don't feature lines, other dishes feature those after days already. The dishes who tend to display those patterns have been bought all at the same place (a large chain of stores specializing on furniture). The lines are most likely caused by cutlery as in cups the same lines are found forming ellipses at the bottom from the stirring. Again some cups don't feature those patterns others do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bowl #1 with no lines despite usage for a while:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDGDc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bowl #2 with lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0A01c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bowl #2 with lines - picture taken HDR mode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/72Bsf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditDate="2014-09-12T09:10:09.123" LastActivityDate="2017-10-22T17:49:11.050" Title="Do dark lines on porcelain dishes indicate usage of aluminium for the color?" Tags="&lt;materials&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16027" PostTypeId="1" CreationDate="2014-09-07T11:06:39.100" Score="5" ViewCount="270" Body="&lt;p&gt;What amount of oxidization is required to refer to some coal as oxidized coal, and how long does it take for coal to become oxidized in temperatures of roughly 25-40&amp;deg;C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found a research paper from 1940 that I don't have access to &lt;a href=&quot;https://dx.doi.org/10.1021/ie50362a021&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt; But other than that, I have had no luck. Does anyone have any ideas or references I can refer to.&lt;/p&gt;&#xA;" OwnerUserId="7733" LastEditorUserId="7475" LastEditDate="2017-09-23T04:33:34.107" LastActivityDate="2017-09-23T04:33:34.107" Title="When is coal considered to be oxidized?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16028" PostTypeId="1" AcceptedAnswerId="16038" CreationDate="2014-09-07T12:19:55.490" Score="9" ViewCount="2370" Body="&lt;p&gt;Imagine the police (or army or whoever's job is that) found out that some maniac placed a lot of primary explosives under a building. They have evacuated the building and deactivated the trigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What now? You have a barrel full of Acetone peroxide (terrorist's favourite explosive). You move it - it might blow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it doesn't - can you react acetone peroxide slowly, to avoid explosion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do factories do with this compound formed as a side waste?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you answer the same for nitroglycerine?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2015-11-14T18:48:55.297" Title="How do we get rid of dangerous explosive acetone peroxide and nitroglycerine?" Tags="&lt;safety&gt;&lt;cleaning&gt;&lt;explosives&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16029" PostTypeId="1" AcceptedAnswerId="16051" CreationDate="2014-09-07T13:10:51.073" Score="15" ViewCount="2113" Body="&lt;p&gt;All the elements with an atomic number more than 100 are given temporary names by IUPAC according to nomenclature rules. For example, element 101 was temporarily  named &quot;Unnilunium&quot; until they give it a permanent name (101 is now known as Mendelevium). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the IUPAC give a temporary name to elements?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2017-06-19T08:51:42.400" LastActivityDate="2017-06-19T08:51:42.400" Title="Why is a temporary name given to an element with an atomic number above 100?" Tags="&lt;nomenclature&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16030" PostTypeId="2" ParentId="15586" CreationDate="2014-09-07T13:28:10.300" Score="21" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.beilstein-journals.org/bjoc/single/articleFullText.htm?publicId=1860-5397-8-36&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FDxDP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.beilstein-journals.org/bjoc/single/articleFullText.htm?publicId=1860-5397-8-36&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-07T13:28:10.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16031" PostTypeId="2" ParentId="16028" CreationDate="2014-09-07T14:14:54.277" Score="1" Body="&lt;p&gt;I can think of one way: Dilute it in ethanol (or other flammable solvent) and burn the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, It would be very unlikely that either of these would be a generated in a side reaction.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-07T14:14:54.277" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16033" PostTypeId="1" AcceptedAnswerId="16036" CreationDate="2014-09-07T14:43:56.857" Score="2" ViewCount="19852" Body="&lt;p&gt;I know that in an oxide oxygen's oxidation number is $-2$, in a peroxide oxygen's oxidation number is $-1$ and in a superoxide oxygen's oxidation number is $-\frac12$. So, how will I be able to detect that any compound is oxide or peroxide or superoxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to get the oxidation number of $\ce{Mn}$ in $\ce{KMnO4}$ how will I determine whether $\ce{KMnO4}$ is an oxide, superoxide or peroxide? Is there any way?  &lt;/p&gt;&#xA;" OwnerUserId="7347" LastEditorUserId="4945" LastEditDate="2014-09-08T05:21:53.183" LastActivityDate="2018-11-10T13:55:59.127" Title="Is there any way to determine if the compound is oxide or peroxide or superoxide?" Tags="&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16034" PostTypeId="1" CreationDate="2014-09-07T14:45:32.493" Score="0" ViewCount="1879" Body="&lt;p&gt;Well, according to the proton-electron ratio $\ce{O-}$ should be bigger than $\ce{F-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the charge/electron density in $\ce{F}$? Will it not affect the size of the atom of $\ce{F-}$? &lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="73" LastEditDate="2014-09-08T04:45:42.897" LastActivityDate="2014-09-08T05:10:37.057" Title="Which is the bigger ion, F- or O-?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;quantum-chemistry&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16036" PostTypeId="2" ParentId="16033" CreationDate="2014-09-07T16:30:03.490" Score="6" Body="&lt;p&gt;generally the most universal way to prove this is X-ray crystallography. It allows to obtain crystal structure of target compound directly. It has, however, an inherent flaw: it can't reliably detect position of hydrogen atoms, so in some cases some uncertainty may remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case obtaining crystal is impossible, the second option is NMR, that allows to get various info about atom's neighborhood, but interpreting NMR data is really hard. In case of superoxide, EPR may prove direct evidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally there is no way to detect such things from brutto-formula. For example, compound $\ce{CH2O2}$ may be either dioxirane (methylene peroxide) or formic acid. In practice, however, peroxides are really unstable, so in most cases if compound is not overloaded with oxygen (like, $\ce{H2SO5}$) and is reasonably stable towards thermal decomposition, it is most likely oxide and not peroxide. Peroxides are usually formed either from another peroxide or electrochemically, or (in rare cases) via cold oxidation with molecular oxygen. And there are exactly three common superoxides. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-08T09:16:44.463" LastActivityDate="2014-09-08T09:16:44.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16037" PostTypeId="2" ParentId="16034" CreationDate="2014-09-07T16:32:16.400" Score="4" Body="&lt;p&gt;definitely $\ce{O-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-07T16:32:16.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16038" PostTypeId="2" ParentId="16028" CreationDate="2014-09-07T17:23:56.657" Score="4" Body="&lt;p&gt;I don't know if it's practical to dissolve or otherwise stabilize acetone peroxide, but I could see absorbing nitroglycerine with sawdust or trapping it in gelatine, rendering it more stable, which is how dynamite is made. Certainly if it can be done safely, detonating in place is be the preferred solution for dealing with unstable explosives. It's done frequently with unexploded munitions if they're deemed too dangerous to move, e.g. &lt;a href=&quot;http://www.theguardian.com/world/video/2012/aug/29/munich-war-bomb-video&quot; rel=&quot;nofollow&quot;&gt;WWII bomb in Munich&lt;/a&gt;. I also visited a potash mine once. Old dynamite tends to &quot;sweat&quot; nitroglycerine, but the explosives lockers are designed such that the contents can be detonated in place without causing damage, if this happens as it's safer than attempting to remove them.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-07T17:23:56.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16039" PostTypeId="2" ParentId="16025" CreationDate="2014-09-07T17:52:27.583" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Find a suitable solvent.&lt;/li&gt;&#xA;&lt;li&gt;Record a UV-VIS spectrum to determine at which wavelength you need to irradiate.&lt;/li&gt;&#xA;&lt;li&gt;Chose the right glassware. For $\lambda$ = 254 nm, quartz is mandatory. Irradiations at $\lambda$ = 300 nm and above may be performed in pyrex.&lt;/li&gt;&#xA;&lt;li&gt;Chose the right light source. For $\lambda$ = 254 nm, use a low-pressure mercury vapour lamp. For irradiations at $\lambda$ around 300 or 360 nm, use low-pressure mercury vapour lamps with specific coatings. Medium- or high pressure lamps typically show much broader emission spectra. For experiments at $\lambda$ = 589 nm, use a sodium vapour lamp.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-07T17:52:27.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16040" PostTypeId="1" AcceptedAnswerId="20512" CreationDate="2014-09-07T17:59:24.717" Score="2" ViewCount="103" Body="&lt;p&gt;I’m an undergraduate Chemistry student and I’m currently writing my dissertation about semi-empirical methods, particularly PM7, applied to protein structure refinement from X-ray structure. Could you recommend me any book/article about it?&lt;/p&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="9961" LastEditDate="2016-09-24T12:51:25.657" LastActivityDate="2016-09-24T12:51:25.657" Title="Book recommendations for protein structure refinement" Tags="&lt;biochemistry&gt;&lt;computational-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16041" PostTypeId="2" ParentId="16026" CreationDate="2014-09-07T18:22:30.180" Score="7" Body="&lt;p&gt;If the two bowls look different although treated similar, the material must differ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your bowls are almost fully covered with a glaze - except for the ring on the bottom on which the ceramic was standing while fired in an oven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, water can enter the ceramic core. If the ceramic material of one bowl is more porous, water can be taken up easier. Upon heating, as in a dishwasher, the water absorbed expands. The glaze can't withstand the tension build up and breaks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in the cracks that you observe.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-07T18:22:30.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16042" PostTypeId="1" CreationDate="2014-09-07T18:47:53.253" Score="3" ViewCount="511" Body="&lt;p&gt;I don't know much about analytical chemistry, but I need to determine the chemical structure of a water soluble polysaccharide which is not pure and may have D-glucose, mannose and fructose. How do I actually know its chemical structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's from a plant and I know it has fructan and a random polysaccharide which consists of glucose and mannose as the main part and it also has less than 20% impurities such as ash, fat, protein or other sugars. I need to know the arrangement between monosaccharides in the random polymer and the percentage of each polysaccharide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that it's amorph and linear but I'd rather to check if it is branched. The compound can be consumed with the testing.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastEditorDisplayName="user7951" LastEditDate="2016-12-23T17:16:18.447" LastActivityDate="2016-12-23T17:16:18.447" Title="Chemical structure of water-soluble polysaccharide" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16043" PostTypeId="1" AcceptedAnswerId="16067" CreationDate="2014-09-07T21:18:29.193" Score="1" ViewCount="892" Body="&lt;p&gt;This is my synthetic pathway. For one, is it plausible? For another, is it efficient? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Reduction of 1-octyne with hydrogen gas and Lindlar's catalyst to 1-octene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Oxymercuration to install a hydroxyl group at the most substituted carbon. This creates 2-octanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Reaction with tosyl chloride and pyridine to create a sulfonate ester - a much better leaving group than an alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) E2 elimination with sodium hydroxide and strong heating to create the internal, Zaitsev alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Ozonolysis to create two aldehydes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Reduction of longer aldehyde with hydrogen gas over transition metal to primary alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Dehydration of primary alcohol to alkene using a Lewis acid (aluminium trichloride in this case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZNLIg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-09T00:15:55.263" Title="Synthesis of 1-hexene from 1-octyne" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16044" PostTypeId="1" CreationDate="2014-09-07T22:05:03.387" Score="2" ViewCount="566" Body="&lt;p&gt;A slightly bruised apple will rot extensively in about 3.5 days at room temperature (22.0 degrees Celsius). If it is kept in the refrigerator at 0.5 degrees Celsius, the same extent of rotting takes about 12 days. What is the activation energy for the rotting reaction?&lt;/p&gt;&#xA;" OwnerUserId="7738" LastActivityDate="2014-09-07T22:05:03.387" Title="Activation Energy of a rotting apple" Tags="&lt;energy&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-09-08T02:57:30.220" ContentLicense="CC BY-SA 3.0" />
  <row Id="16045" PostTypeId="1" AcceptedAnswerId="16053" CreationDate="2014-09-08T02:49:36.190" Score="2" ViewCount="2745" Body="&lt;p&gt;To be IR active, the vibration of a bond must result in a substantial change in dipole moment. Since trans-2-butene is symmetrical, will the C=C stretch show up on IR? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something tells me no because if the C=C bond vibrates, there will be no net change in the bond dipole moment, unlike in cis-2-butene. Any polarization one way in trans-2-butene is cancelled out by a polarization in the opposite direction in trans-2-butene. Cis-2-butene, however, has a permanent dipole and any vibration only exacerbates this dipole. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-08T08:06:41.943" Title="Is trans-2-butene IR active with regard to its C=C bond?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16046" PostTypeId="1" AcceptedAnswerId="16048" CreationDate="2014-09-08T03:05:12.420" Score="2" ViewCount="834" Body="&lt;p&gt;In para-Toluenesulfonylchloride, there are 5 pi bonds and 1 ring. This suggests a IHD of 6. However, the actual IHD is 4. How come? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because the formula IHD = rings + pi bonds does not apply to compounds containing sulfur? If not, what formula do I use? I have this formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$IHD = \frac{2C + N - X - H + 2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this formula yields a IHD of 4. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2015-10-16T17:26:22.120" LastActivityDate="2015-10-16T17:26:22.120" Title="Index of Hydrogen Deficiency for para-Toluenesulfonylchloride" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16047" PostTypeId="2" ParentId="16046" CreationDate="2014-09-08T03:33:32.060" Score="1" Body="&lt;p&gt;There are 5 pi bonds: three C-C in the ring and two S-O pi bonds. Add the ring, and the IHD is 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of it's ability to exist as higher oxidation states than the second row elements, I don't believe that there is any way to include sulfur in the IHD formula unambiguously. Consider the following two compounds, toluenesulfonyl chloride and an isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d4NIh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas are the same, but the IHD of the left structure is 6 but the right structure is 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the best way to handle third row (and beyond) is to treat the large atom as if it were it's second row analog (P-&gt;N, S-&gt;O, etc.), and use the result as the lower limit on the IHD.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-09-08T04:24:29.133" LastActivityDate="2014-09-08T04:24:29.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16048" PostTypeId="2" ParentId="16046" CreationDate="2014-09-08T04:53:46.520" Score="3" Body="&lt;p&gt;You have to start from the general formula for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Degree_of_unsaturation&quot; rel=&quot;nofollow&quot;&gt;degree of unsaturation&lt;/a&gt;&#xA;$$DU = 1 + \frac{1}{2} \sum n_i(v_i-2),$$&#xA;where $n_i$ is the number of atoms with a valency of $v_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For para-toluenesulfunylchloride ($\ce{C7H7SO2Cl}$) you will therefore arrive at&#xA;\begin{align}DU &amp;amp;= 1 + \frac{1}{2}\Big[n_\ce{C}(4-2) + n_\ce{H}(1-2) + n_\ce{S}(6-2) + n_\ce{O}(2-2) + n_\ce{Cl}(1-2)\Big]\\&#xA;&amp;amp;=1+\frac12\Big[7(4-2) + 7(1-2) + 1(6-2) + 2(2-2) +1(1-2)\\&#xA;&amp;amp;=1 + \frac12\Big[14 -7 +4 +0 -1\Big]\\&#xA;&amp;amp;=6,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because sulfur uses all six valence electrons in this case.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-09-08T04:53:46.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16050" PostTypeId="2" ParentId="16033" CreationDate="2014-09-08T05:12:47.460" Score="1" Body="&lt;p&gt;If you want to distinguish oxide, superoxide and peroxide ions, generally you can use their X-ray structure well. Oxide is a single oxygen ion, superoxide and peroxide is $O-O$ with approximate bond distance of 1.33 and 1.49 A respectively. Superoxides also have an unpaired spin which can be often detected (even if it is coupling with some other spins). &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-09-08T05:12:47.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16051" PostTypeId="2" ParentId="16029" CreationDate="2014-09-08T06:47:27.553" Score="15" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Systematic_element_name&quot;&gt;systematic element names&lt;/a&gt; refer to a temporary name and symbol for hypothetical and newly synthesized elements. Because of naming controversies in the past, these names were introduced by the IUPAC to discuss the elements without ambiguity. After the discovery is confirmed, they receive a permanent name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC states in the &lt;a href=&quot;http://www.iupac.org/nc/home/publications/iupac-books/books-db/book-details.html?tx_wfqbe_pi1[bookid]=5&quot;&gt;red book&lt;/a&gt; this controversy as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the past, some elements were given two names because two groups claimed to have discovered them. To avoid such confusion it was decided in 1947 that after the existence of a new element had been proved beyond reasonable doubt, discoverers had the right to suggest a name to IUPAC, but that only the Commission on Nomenclature of Inorganic Chemistry (CNIC) could make a recommendation to the IUPAC Council to make the final decision. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As this text suggests, this is a highly political process, that can be dragged out quite easily. The systematic nomenclature and symbols for new elements are stated in IR-3.1.1 as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Newly discovered elements may be referred to in the scientific literature but until they have received permanent names and symbols from IUPAC, temporary designators are required. Such elements may be referred to by their atomic numbers, as in ‘element 120’ for example, but IUPAC has approved a systematic nomenclature and series of three-letter symbols (see Table II).2 The name is derived directly from the atomic number of the element using the following numerical roots:&lt;br&gt;&#xA;  0 = nil&lt;br&gt;&#xA;  1 = un&lt;br&gt;&#xA;  2 = bi&lt;br&gt;&#xA;  3 = tri&lt;br&gt;&#xA;  4 = quad&lt;br&gt;&#xA;  5 = pent&lt;br&gt;&#xA;  6 = hex&lt;br&gt;&#xA;  7 = sept&#xA;  8 = oct&lt;br&gt;&#xA;  9 = enn&lt;br&gt;&#xA;  The roots are put together in the order of the digits which make up the atomic number and terminated by ‘ium’ to spell out the name. The final ‘n’ of ‘enn’ is elided when it occurs before ‘nil’, and the final ‘i’ of ‘bi’ and of ‘tri’ when it occurs before ‘ium’.&lt;br&gt;&#xA;  The symbol for the element is composed of the initial letters of the numerical roots which make up the name.&lt;br&gt;&#xA;  &lt;em&gt;Example:&lt;/em&gt;&lt;br&gt;&#xA;  1. element 113 = ununtrium, symbol Uut&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the transfermium elements, their names have been finalised only in 1994 (&lt;a href=&quot;http://dx.doi.org/10.1351/pac199466122419&quot;&gt;&lt;em&gt;Pure Appl. Chem.&lt;/em&gt;, &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;Vol. 66&lt;/em&gt;, No. 12, pp. 2419-2421&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for Mendelevium the first publication already suggested the name and a symbol. It was in use before officially recommended by the IUPAC (&lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.98.1518&quot;&gt;A. Ghiorso, B. G. Harvey, G. R. Choppin, S. G. Thompson, and G. T. Seaborg, &lt;em&gt;Phys. Rev.&lt;/em&gt; &lt;strong&gt;1955&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt;, 1518&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-09-08T06:47:27.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16053" PostTypeId="2" ParentId="16045" CreationDate="2014-09-08T08:06:41.943" Score="6" Body="&lt;p&gt;The $\ce{C=C}$ stretch will not show in IR. The frequency is $\nu_{\ce{C=C}}=1700~\mathrm{cm^{-1}}$ and has A gerade symmetry (DF-BP86/def2-SVP). There is no change in the dipole moment.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yrJZa.gif&quot; alt=&quot;c=c stretch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some bending frequencies involving the $\ce{C=C}$ double bond will show up. For example $\nu_{\ce{HC=CH}}=963~\mathrm{cm^{-1}}$ with a medium intensity.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SlBIQ.gif&quot; alt=&quot;c=c bend&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-09-08T08:06:41.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16054" PostTypeId="1" CreationDate="2014-09-08T08:28:45.867" Score="20" ViewCount="24275" Body="&lt;p&gt;I just went camping and someone brought a game from Australia. There was a wooden box with the top made of a thin, soft sheet of lead with a hole in it. The participants then throw coins at the box, and getting the coin inside gives some points, while landing a coin on top gives a little less. I've been looking online, and I think the game is called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Toad_in_the_hole_%28game%29&quot; rel=&quot;noreferrer&quot;&gt;Toad in the hole&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that lead is toxic, but I don't know in what way. Is it toxic to touch this lead sheet? And to eat afterwards without washing hands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe lead is only toxic in some form?&lt;/p&gt;&#xA;" OwnerUserId="2423" LastEditorUserId="16683" LastEditDate="2017-06-02T17:35:31.277" LastActivityDate="2017-06-02T17:35:31.277" Title="Toxicity of a sheet of lead?" Tags="&lt;toxicity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16055" PostTypeId="1" AcceptedAnswerId="16059" CreationDate="2014-09-08T08:45:36.897" Score="39" ViewCount="66672" Body="&lt;p&gt;While I was looking at the periodic table today, I realised that there were gases that were much lighter than helium such as hydrogen. If hydrogen is lighter than helium, why do we insist on using helium in balloons?&lt;/p&gt;&#xA;" OwnerUserId="7741" LastEditorUserId="23561" LastEditDate="2018-09-30T22:14:24.847" LastActivityDate="2020-05-16T10:37:48.583" Title="Why do we use helium in balloons?" Tags="&lt;elements&gt;&lt;hydrogen&gt;&lt;density&gt;&lt;noble-gases&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16056" PostTypeId="2" ParentId="16055" CreationDate="2014-09-08T08:51:51.247" Score="9" Body="&lt;p&gt;Yes, hydrogen is lighter than helium but helium, on the other hand, is an &lt;strong&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Inert_gases&quot; rel=&quot;noreferrer&quot;&gt;inert gas&lt;/a&gt;&lt;/strong&gt; (very less reactive). Also, hydrogen is highly flammable so that would make it unsafe to play with balloons.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T09:49:31.453" LastActivityDate="2015-06-26T09:49:31.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16057" PostTypeId="2" ParentId="16055" CreationDate="2014-09-08T08:52:29.137" Score="49" Body="&lt;p&gt;Actually, hydrogen is the only gas that is lighter than helium. However, it has a very big disadvantage: It is highly flammable. On the other hand, helium is almost completely inert - this is why it is very much safer to use the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might happen when you use hydrogen instead of helium was impressively proven by history  when &lt;a href=&quot;http://en.wikipedia.org/wiki/Hindenburg_disaster&quot;&gt;the &quot;Hindenburg&quot; zeppelin exploded on 6 May 1937&lt;/a&gt;. There is video footage, that can be seen on &lt;a href=&quot;http://youtu.be/IjSawn7jKfQ&quot;&gt;youtube&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some of the comments it was mentioned, that hydrogen alone might not be the cause of the Hindenburg disaster, there were other contributing factors. However, using hydrogen remains dangerous, as this &lt;a href=&quot;http://youtu.be/cQvpK9cl0No&quot;&gt;weather balloon experiment&lt;/a&gt; shows. In a more &lt;a href=&quot;http://youtu.be/s66fVHQ4-68&quot;&gt;scientific setup&lt;/a&gt; the burning of a hydrogen balloon is compared to oxygen and a mixture of oxygen and hydrogen. Unfortunately a video of a helium filled balloon is not available, but it basically only ruptures and pops because of the different pressures on the in- and outside.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-09-10T02:59:30.807" LastActivityDate="2014-09-10T02:59:30.807" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16058" PostTypeId="2" ParentId="14385" CreationDate="2014-09-08T11:53:21.027" Score="4" Body="&lt;p&gt;There are a lot of questions in this one post which is probably why there has not yet been an attempt to answer this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.) What is the Schrödinger wave equation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave equation (i.e. wave function) is a function which contains all the information about a chemical system.  With an appropriate operator, one can extract this information and gain insight into the system.  For example, the Hamiltonian operator will extract the energy of a system from the wave function.  The wave function is an eigenfunction while the energy (in this example) is the eigenvalue of the eigenfunction.  For example, in the Schrödinger equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathcal{H}\Psi = E\Psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal{H}$ is the Hamiltonian operator, $\Psi$ is the wave function (eigenfunction) and $E$ is the energy (eigenvalue).  Notice that by operating on $\Psi$ with $\mathcal{H}$ returns $E$ AND $\Psi$, which by definition, makes this an eigenfunction-eigenvalue equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.) What are radial probability functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are probability distributions for the probability of an electron residing some distance from the nucleus of an atom such as the hydrogen atom.  Consider this plot of various radial probability distribution functions of the hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNelW.gif&quot; alt=&quot;Radial probability distribution of an electron in the hydrogen atom.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we see that the probability of an electron (in the hydrogen atom) in a 1s orbital is zero at the nucleus (at 0,0) and a maximum a hair shy of $2 a_0$.  Interestingly an electron in a 2p orbital has a small probability of residing close to the nucleus ($&amp;lt; 1a_0$) but is more likely to reside further away from the nucleus than an electron in a 1s orbital.  This makes sense given that a 2p orbital is higher in energy because 'it' is further away from the nucleus than a 1s orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.) What are nodes in radial and angular functions?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A node is where the function is zero at a particular point.  This simply means in a physical sense that at this point (at the node), the probability of an electron existing at that point is ZERO.  Nodes are a result of the separation of variables technique when solving the appropriate equation(s). When any one component (or variable) is zero then the resulting function will be zero at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.) What is electron cloud?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron 'cloud' is just a nicer away of saying electron density.  Electrons are not strictly particles.  Electrons do not just 'orbit' a nucleus.  Electrons also exhibit wavelike behavior.  (see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;nofollow noreferrer&quot;&gt;double-slit experiment&lt;/a&gt; - wave-like behavior and J. J. Thompsons &lt;a href=&quot;http://en.wikipedia.org/wiki/J._J._Thomson#Discovery_of_the_electron&quot; rel=&quot;nofollow noreferrer&quot;&gt;cathode ray tube experiments&lt;/a&gt; - particle-like behavior).  The 'cloud' is obtained simply by 'squaring' the wave function.  More technically, its taking the product of the wave function and its complex conjugate.  The 'cloud' is just a probability distribution of where an electron is likely to be (usually given in 3-dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.) What is the Hamiltonian operator?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hamiltonian operator is the energy operator to the Schrodinger equation.  Because it is Hermitian, it will return a 'real' result (as opposed to a complex result).  There are different versions of the Hamiltonian depending upon what approximations you want to make.  The full Hamiltonian operator can be written as (in terms of atomic units)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{H} = -\sum\limits_{i=1}^{N}\dfrac{1}{2}\nabla_i^2&#xA;              -\sum\limits_{A=1}^{M}\dfrac{1}{2M_A}\nabla_A^2&#xA;              -\sum\limits_{i=1}^{N}\sum\limits_{A=1}^{M}\dfrac{Z_A}{r_{iA}}&#xA;              +\sum\limits_{i=1}^{N}\sum\limits_{j&amp;gt;i}^{N}\dfrac{1}{r_{ij}}&#xA;              +\sum\limits_{A=1}^{M}\sum\limits_{B&amp;gt;A}^{M}\dfrac{Z_A Z_B}{R_{AB}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each term on the right hand side of the equation (going from left to right) is defined as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the kinetic energy of the electrons&lt;/li&gt;&#xA;&lt;li&gt;the kinetic energy of the nuclei&lt;/li&gt;&#xA;&lt;li&gt;the potential energy of the electron-nuclear attractions&lt;/li&gt;&#xA;&lt;li&gt;the potential energy of the electron-electron repulsions&lt;/li&gt;&#xA;&lt;li&gt;the potential energy of the nuclear-nuclear repulsions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can write the Hamiltonian using a bit more compact notation such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathcal{H} = \hat{T}_{e} + \hat{T}_{N} + \hat{V}_{eN} + \hat{V}_{ee} + \hat{V}_{NN}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where negative signs are assumed for each $\hat{T}$ operator.  The terms denoted here correspond to the respective terms in the equation before this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take into account the Born-Oppenheimer approximation (the B.O. approximation assumes that the electron motion is orders of magnitude larger than the nuclei they belong to... so much faster that the nuclei are, relatively, standing still with respect to the electrons), we can make two changes to the full Hamiltonian giving us the electronic Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathcal{H} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee} +\ \mathrm{constant}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nuclei are treated as frozen in space when the B.O. approximation is applied for reasons explained above.  Therefore the kinetic energy of the nuclei goes to zero.  The nuclear-nuclear repulsion is now a constant because the nuclei are fixed.  The repulsion will never change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.) What is a Laplace operator?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Laplace operater (or Laplacian) is simply a differential operator of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\nabla^2 = \dfrac{\partial^2}{\partial x^2} +  \dfrac{\partial^2}{\partial y^2} +  \dfrac{\partial^2}{\partial z^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you are working with Cartesian coordinates. This is simply a set of second-order derivative operators within the kinetic energy operator.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While I may have addressed each question, there is a lot more depth to these answers that lie well beyond the scope of this post.  I encourage you to find more detailed and insightful answers in well-established texts.  I highly recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486691861&quot; rel=&quot;nofollow noreferrer&quot;&gt;Szabo &amp;amp; Ostlund&lt;/a&gt; for a solid introduction to quantum mechanics.  You can snag a paperback version of this excellent work for about $10.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-08T12:57:52.800" LastActivityDate="2014-09-08T12:57:52.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16059" PostTypeId="2" ParentId="16055" CreationDate="2014-09-08T11:59:36.557" Score="60" Body="&lt;p&gt;As other answers have noted, the only gas lighter than helium is hydrogen, which has some &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen#Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;flammability issues&lt;/a&gt; that make it more difficult to handle safely than helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in practice, hydrogen is not significantly &quot;lighter&quot; than helium.  While the molecular mass (and thus, per the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas law&lt;/a&gt;, the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the &lt;em&gt;difference&lt;/em&gt; between the density of the gas inside the balloon and the air outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_of_air&quot; rel=&quot;nofollow noreferrer&quot;&gt;density of air&lt;/a&gt; at &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;STP&lt;/a&gt; is about &lt;span class=&quot;math-container&quot;&gt;$\rho_{\ce{air}}=\pu{1.2754 kg m-3}$&lt;/span&gt; , while the densities of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium&quot; rel=&quot;nofollow noreferrer&quot;&gt;helium&lt;/a&gt; gas are &lt;span class=&quot;math-container&quot;&gt;$\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rho_{\ce{He}}=\pu{0.1786 kg m-3}$&lt;/span&gt;  respectively.  The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences  &lt;span class=&quot;math-container&quot;&gt;$\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$&lt;/span&gt;.  Thus, helium is only about &lt;span class=&quot;math-container&quot;&gt;$7.5\%$&lt;/span&gt; less buoyant in air than hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant.  For example, if you wished to go ballooning on &lt;a href=&quot;//en.wikipedia.org/wiki/Jupiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jupiter&lt;/a&gt;, which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply &lt;em&gt;sink&lt;/em&gt;, and even a pure hydrogen balloon (at ambient temperature) would not lift much weight.  Of course, you could always just fill the balloon with ambient Jovian air and &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot_air_balloon&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat it up&lt;/a&gt; to produce a &lt;a href=&quot;https://en.wikipedia.org/wiki/A_Meeting_with_Medusa&quot; rel=&quot;nofollow noreferrer&quot;&gt;hot hydrogen balloon&lt;/a&gt; (not to be confused with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rozi%C3%A8re_balloon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rozière balloon&lt;/a&gt;, which are used on Earth and have separate chambers for hot air and hydrogen / helium).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ps.&lt;/strong&gt; A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about &lt;span class=&quot;math-container&quot;&gt;$\pu{2 Da}$&lt;/span&gt;, while a helium atom has two protons and two neutrons, for a total mass of about &lt;span class=&quot;math-container&quot;&gt;$\pu{4 Da}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about  &lt;span class=&quot;math-container&quot;&gt;$\pu{32 Da}$&lt;/span&gt; (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to  &lt;span class=&quot;math-container&quot;&gt;$\pu{28 Da}$&lt;/span&gt; (one proton and one neutron per atom less than oxygen).  Thus, the average molecular mass of air should be between &lt;span class=&quot;math-container&quot;&gt;$28$&lt;/span&gt; and  &lt;span class=&quot;math-container&quot;&gt;$\pu{32 Da}$&lt;/span&gt;; in fact, since air is about three quarters nitrogen, it's about  &lt;span class=&quot;math-container&quot;&gt;$\pu{29 Da}$&lt;/span&gt;, and so the buoyancies of hydrogen and helium in air are proportional to &lt;span class=&quot;math-container&quot;&gt;$29 - 2 = 27$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$29 - 4 = 25$&lt;/span&gt; respectively.  Thus, hydrogen should be about &lt;span class=&quot;math-container&quot;&gt;$\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$&lt;/span&gt; more buoyant than helium, or, in other words, helium should be about &lt;span class=&quot;math-container&quot;&gt;$\frac{2}{27} \approx 7.5\%$&lt;/span&gt; less buoyant than hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pps.&lt;/strong&gt; To summarize some of the comments below, there are other possible &lt;a href=&quot;https://en.wikipedia.org/wiki/Lifting_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;lifting gases&lt;/a&gt; as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Methane&quot; rel=&quot;nofollow noreferrer&quot;&gt;methane&lt;/a&gt; (molecular mass  &lt;span class=&quot;math-container&quot;&gt;$\approx \pu{16 Da}$&lt;/span&gt;) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas.  However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's &lt;em&gt;more&lt;/em&gt; dangerous by others (total energy content per volume).  In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much less flammable choice would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Water&quot; rel=&quot;nofollow noreferrer&quot;&gt;water&lt;/a&gt; vapor which, with a molecular mass of  &lt;span class=&quot;math-container&quot;&gt;$\approx \pu{18 Da}$&lt;/span&gt;, is only slightly less buoyant than methane at the same temperature and pressure.  The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all.  This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster &amp;mdash; whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite these drawbacks, &lt;a href=&quot;http://www.flyingkettle.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;hot steam balloons&lt;/a&gt; have &lt;a href=&quot;https://www.halfbakery.com/idea/Steam_20balloon&quot; rel=&quot;nofollow noreferrer&quot;&gt;certainly&lt;/a&gt; been &lt;a href=&quot;https://www.flyingkettle.com/priorart.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;suggested&lt;/a&gt;, &lt;a href=&quot;https://www.flyingkettle.com/jbfa.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;studied&lt;/a&gt; and &lt;a href=&quot;https://www.flyingkettle.com/experims.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;tried&lt;/a&gt; in the past &amp;mdash; alas, &lt;a href=&quot;https://www.youtube.com/watch?v=BX37syDt-sw&quot; rel=&quot;nofollow noreferrer&quot;&gt;not always particularly successfully&lt;/a&gt; (although, apparently, there have been &lt;a href=&quot;http://www.aeroix.de/fileadmin/user_upload/aeroix/redakteure/Dokumente_HeiDAS/HeiDAS_DGLRKongress2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;much more successful attempts&lt;/a&gt; as well).  There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope.  So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other potential lifting gases, with molecular mass similar to methane and water, include &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonia&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonia&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Neon&quot; rel=&quot;nofollow noreferrer&quot;&gt;neon&lt;/a&gt;.  Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia, on the other hand, while &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonia#Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;much less flammable&lt;/a&gt; than methane, is rather toxic and corrosive (not to mention &lt;em&gt;really&lt;/em&gt; stinky, which, given its other properties, is probably a &lt;em&gt;good&lt;/em&gt; thing).  I don't think I'd like to &lt;a href=&quot;https://web.archive.org/web/20150131045358/http://www.balloonlife.com/publications/balloon_life/9703/freeflt.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;fly in an ammonia balloon&lt;/a&gt;, but &lt;a href=&quot;https://www.alifeintheair.com/?_escaped_fragment_=flying-on-fertilizer/c1b2p#!flying-on-fertilizer/c1b2p&quot; rel=&quot;nofollow noreferrer&quot;&gt;apparently&lt;/a&gt;, &lt;a href=&quot;https://web.archive.org/web/20081007204640/http://www.balloonlife.com/publications/balloon_life/9511/ammonia.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;some people do!&lt;/a&gt;  It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second.  If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical.  Then again, so would hydrogen &amp;mdash; its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.⠀⠀⠀⠀⠀⠀&lt;/p&gt;&#xA;" OwnerUserId="283" LastEditorUserId="283" LastEditDate="2020-05-16T10:37:48.583" LastActivityDate="2020-05-16T10:37:48.583" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="16061" PostTypeId="2" ParentId="16054" CreationDate="2014-09-08T14:50:25.333" Score="16" Body="&lt;p&gt;No, it is not toxic to touch solid lead.  Lead poisoning results from&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;ingestion:&lt;/strong&gt; paints used to contain lead-based materials, and kids would sometimes eat the peeling paint leading to health problems.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;inhalation:&lt;/strong&gt; if you were grinding or polishing a block of lead and fine particles were generated, you might inhale them if you weren't using the proper protective equipment and health problems could develop.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;dermal contact:&lt;/strong&gt; if you were working with &lt;strong&gt;organic lead compounds&lt;/strong&gt; that were easily absorbed through the skin, serious health problems would result if you weren't using the proper protective equipment.  Gasoline used to contain lead compounds, but they were removed from gasoline in order to prevent problems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Being around or touching solid lead would not lead to ingestion, inhalation or dermal contact issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link to &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead#Biochemistry_of_poisoning&quot;&gt;Wikipedia article&lt;/a&gt; on lead, see the section on &quot;Health effects&quot; if you'd like to read more or read about the biochemical mechanism of action.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-08T14:50:25.333" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16063" PostTypeId="1" AcceptedAnswerId="16072" CreationDate="2014-09-08T17:30:35.987" Score="2" ViewCount="3000" Body="&lt;p&gt;I am learning about zwitterions and am confused about the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;From what I understand, when amino acids are in their solid form, they exist as zwitterions which is why they have such a high melting point and are so soluble in polar solvents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then when placed in a solution with pH lower than their isoelectric point, they act as a base and form positive ions, and when placed in a solution with pH above their isoelectric point, they act as acids and form a negatively charged ion and a water molecule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the above is correct, what I am confused about is what happens at the isoelectric point. In my textbook it says that the isoelectric point is 'the pH value at which the amino acid exists as a zwitterion', however I thought it also exists as a zwitterion when not in solution. Are they refering just to the case of the amino acid being in solution and ignoring the solid state?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If this is so, I still do not understand how it remains as a zwitterion at the isoelectric point. Why does it not act as both a base and an acid and accept a H+ ion onto the O- atom in the carbonyl group and donate the extra H+ ion from its amine group? Does this have something to do with the fact that then it would no longer have any charges on it so it would not be soluble in the polar solvent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="7448" LastEditDate="2015-10-16T18:06:15.497" LastActivityDate="2016-07-25T21:20:29.710" Title="What happens to the amino acid when it is in a solution with the pH of its isoelectric point? (zwitterions)" Tags="&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16065" PostTypeId="2" ParentId="7259" CreationDate="2014-09-08T18:12:21.153" Score="1" Body="&lt;p&gt;You are right, you can use Le Chatelier's principle to predict the shift of the equilibrium when temperature or pressure is in- or decreased. It is stated that the reaction is exothermic, and this tells you on which side of the reaction equation the heat goes. And as both products and educts include gases, it would also be easy to predict how the equilibrium shifts when pressure is varied. I presume that the question, in its current form, is asking for the shifts that happen in all possible combinations of in- or decreased pressure or temperature.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-09-08T18:24:15.687" LastActivityDate="2014-09-08T18:24:15.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16066" PostTypeId="1" AcceptedAnswerId="16069" CreationDate="2014-09-08T18:31:21.673" Score="1" ViewCount="4890" Body="&lt;p&gt;In the Ncert Chemistry Tb, it is given that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\Delta H$ is negative for exothermic reactions which evolve heat during the reaction and $\Delta H$ is positive for endothermic reactions which absorb heat from the surroundings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In enthalpy change video lesson which I got access to online, it is given that &#xA;for $\ce{NH4NO3(s) -&amp;gt; NH4NO3(aq)}$ &lt;strong&gt;since the temperature decreases during the reaction, it is positive enthalpy of solution&lt;/strong&gt;.                            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it positive? I guess both $\Delta H$ and enthalpy change of a solution is almost change. Since the temp decreases here, it is endothermic, so it should be positive right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are both of the above processes different from each other and are the sign conventions vice versa for each other?&lt;/p&gt;&#xA;" OwnerUserId="7760" LastEditorUserId="194" LastEditDate="2017-05-12T03:08:28.680" LastActivityDate="2017-05-12T03:08:28.680" Title="Is sign convention for delta H and enthapy change of a solution different?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16067" PostTypeId="2" ParentId="16043" CreationDate="2014-09-08T19:10:43.320" Score="4" Body="&lt;p&gt;On the plus side, you've clearly thought about this problem and have worked out a detailed synthetic route.  I think most of the steps would work as desired.  The two steps that concern me the most are 4 and 7.  I suspect your step 4 will give a mixture of products.  Perhaps with further experimentation you could find reaction conditions that would give you the desired product without  other products from undesired side reaction (elimination in the other direction and perhaps some rearrangement products too).  I see your step 7 as the most problematical.  Aluminum chloride, heat and the alcohol will produce a carbocation and it will undergo significant rearrangement creating a variety of elimination and addition (polymer formation) products.  So while your reaction scheme is plausible, in my opinion it is not efficient.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-08T19:10:43.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16068" PostTypeId="1" AcceptedAnswerId="16075" CreationDate="2014-09-08T19:12:18.757" Score="2" ViewCount="5782" Body="&lt;p&gt;I have been learning about zwitterions and the isoelectric point, and have been trying to apply this to hydrolysis of peptides. From the diagram in my textbook, I can see that when acid hydrolysis occurs, the amino acids formed from the hydrolysis act as bases because the pH is below their isoelectric point, so they accept the $\ce{H+}$ ions forming positively charged ions.&lt;br&gt;&#xA;Now when it comes to alkaline hydrolysis, I don't understand why the peptide would not just be hydrolysed, and then act as an acid and donate the extra $\ce{H+}$ ion on its amine group to the $\ce{OH-}$ ions to form negative ions and a $\ce{H2O}$ molecule. Instead, in my text book it shows something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{-NH-CH(CH2SH)-C(=O)-NH-CH(CH3)-C(=O)- + NaOH \\-&amp;gt; -NH-CH(CH2SH)-C(=O)-O^{-}Na^{+} + H2N-CH(CH3)-C(=O)-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really sorry if this makes no sense whatsoever...&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="4945" LastEditDate="2014-09-09T06:46:08.057" LastActivityDate="2014-09-09T06:46:08.057" Title="How does alkaline hydrolysis of peptides occur?" Tags="&lt;acid-base&gt;&lt;hydrolysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16069" PostTypeId="2" ParentId="16066" CreationDate="2014-09-08T19:52:34.540" Score="3" Body="&lt;p&gt;The reaction is endothermic, meaning that it takes energy from the surrounding. The &quot;surrounding&quot; in your example is the solvent, so in the dissolution process, energy is taken from the solvent,  thereby decreasing its temperature.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-08T19:52:34.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16070" PostTypeId="2" ParentId="16003" CreationDate="2014-09-08T20:03:25.877" Score="6" Body="&lt;p&gt;Each energy level of a given electron configuration is described by a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Term_symbol&quot; rel=&quot;nofollow noreferrer&quot;&gt;Russell-Saunders term symbol&lt;/a&gt;, assuming LS coupling. The ground state term symbol is predicted by Hund's rules. Tables of atomic energy levels identified by their term symbols have been compiled by &lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/levels_form.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST&lt;/a&gt;. So, the only thing you have to do, is to compare the energies of the terms describing the electron configurations you are looking for with their respective ground states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following pictures the notation $m_{S} = + \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \uparrow$ and $m_{S} = - \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \downarrow$ will be used for the spin quantum number $m_{S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data for carbon can be found &lt;a href=&quot;http://physics.nist.gov/PhysRefData/Handbook/Tables/carbontable5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The ground state for carbon is ${}^{3}\mathrm{P}_{0}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{3}\mathrm{P}_{2}$ or ${}^{1}\mathrm{S}_{0}$ therefore I will give the values for both of them (for comparison: the thermal energy at room temperature is about $200 \, \mathrm{cm}^{-1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wbmvK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data for nitrogen can be found &lt;a href=&quot;http://physics.nist.gov/PhysRefData/Handbook/Tables/nitrogentable5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The ground state for nitrogen is ${}^{4}\mathrm{S}_{3/2}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{2}\mathrm{D}_{5/2}$ or ${}^{2}\mathrm{P}_{3/2}$ therefore I will give the values for both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bSCFE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-08T20:12:02.340" LastActivityDate="2014-09-08T20:12:02.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16071" PostTypeId="2" ParentId="16068" CreationDate="2014-09-08T20:03:56.797" Score="3" Body="&lt;p&gt;The unprotonated amino group ($\ce{R-NH2}$) does not protonate $\ce{OH-}$ because it is an extremely weak acid. Primary and secondary amines have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid_dissociation_constant&quot; rel=&quot;nofollow&quot;&gt;$pK_a$&lt;/a&gt; of around 35, while the $pK_a$ of hydroxide is about 13-14 in aqueous solution. Deprotonating $\ce{NH2}$ groups requires much stronger bases in non-aqueous medium. Therefore, the carboxylic acid group of the amino acid is deprotonated in aqueous alkaline solution, as shown in the scheme from your textbook.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-09-08T20:30:35.043" LastActivityDate="2014-09-08T20:30:35.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16072" PostTypeId="2" ParentId="16063" CreationDate="2014-09-08T20:07:21.227" Score="1" Body="&lt;p&gt;The accepted definition by IUPAC gold book is: &lt;em&gt;The pH value at which the net electric charge of an elementary entity is zero. pI is a commonly used symbol for this kind-of-quantity. It should be replaced by pH(I) because it is a pH determined under that particular condition.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of an amino acid is the smallest at its isoelectric point and it easily precipitates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 4 you need to be familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamic_equilibrium&quot; rel=&quot;nofollow&quot;&gt;dynamic equilibirum&lt;/a&gt;. The zwitterion accepts and loses protons all the time, but at the isoelectric point, its concentration is the biggest among all the forms of the amino acid, that is, the species that has a protonated carboxylic group, would easier lose that proton than the zwitterion's carboxylic group would accept a proton.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2014-09-08T20:26:47.090" LastActivityDate="2014-09-08T20:26:47.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16075" PostTypeId="2" ParentId="16068" CreationDate="2014-09-08T22:26:06.670" Score="3" Body="&lt;p&gt;In the drawing shown below, the first 3 lines show a small peptide and how it will exist under&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;acidic conditions: the basic amino nitrogen is protonated&lt;/li&gt;&#xA;&lt;li&gt;neutral conditions&lt;/li&gt;&#xA;&lt;li&gt;basic conditions: the carboxylic acid group is deprotonated&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A peptide contains multiple amide bonds ($\ce{-RCO-NHR'-}$).  When a peptide is hydrolysed under basic conditions, the amide bond is broken.  Note that, as mentioned above, under basic conditions any acid groups in the peptide will exist as very stable (due to resonance delocalization) carboxylate groups.  Line 4 in the drawing shows this in more detail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual mechanism for the hydrolysis of the amide group under basic conditions is pictured in the bottom line of the drawing.  The reaction occurs by a nucleophilic bimolecular substitution reaction ($\ce{S_{N}2}$).  The nucleophilic hydroxide anion attacks the relatively electrophilic amide carbonyl carbon to form a tetrahedral intermediate.  This intermediate can either return back to the amide or go forward to the carboxylic acid (hydrolysed amide).  This same $\ce{S_{N}2}$ reaction can be repeated until all of the amide bonds in the peptide are hydrolysed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0vZDM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-08T22:26:06.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16076" PostTypeId="1" AcceptedAnswerId="16077" CreationDate="2014-09-08T23:07:36.653" Score="8" ViewCount="4185" Body="&lt;p&gt;Why is hydrated carbon dioxide - the predominant form of acid that one gets upon dissolution of carbon dioxide in solution - so unstable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the below rationale valid? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Carbon in carbon dioxide has two (empty? no, but still vulnerable to attack) p-orbitals and bears a strong partial positive charge. &lt;/li&gt;&#xA;&lt;li&gt;Oxygen's lone pair can attack an empty p-orbital and form a formal charge-separated complex with the carbon dioxide. &lt;/li&gt;&#xA;&lt;li&gt;This is &quot;hydrated&quot; carbon dioxide or &quot;carbonic acid.&quot; This form is extremely unstable and subject to disproportionation due to an unfavorable charge and the unfavorable nature of charge separation itself. The change in entropy also favors the products of disproportionation. &lt;/li&gt;&#xA;&lt;li&gt;However, there exists a pathway to stability - that is - protonation of the oxygen with the negative formal charge by the oxygen bearing the positive formal charge. &lt;/li&gt;&#xA;&lt;li&gt;This, however, is akin to a forbidden fruit; the ephemeral three-membered ring that would have to be formed exhibits &quot;ring strain&quot; (&lt;em&gt;if you object to this term, can you please elaborate on your objection&lt;/em&gt;), and as a result, disproportionation is overwhelmingly favored, especially from an entropic standpoint (reconstitution of carbon dioxide &lt;em&gt;gas&lt;/em&gt; is highly entropically favorable). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P9ewQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-08T23:13:47.760" LastActivityDate="2015-04-26T16:20:28.733" Title="The instability of hydrated carbon dioxide or &quot;carbonic acid&quot;" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16077" PostTypeId="2" ParentId="16076" CreationDate="2014-09-08T23:19:22.900" Score="10" Body="&lt;p&gt;The situation is described by the following equilibrium&#xA;$$\ce{H2O + CO2 &amp;lt;=&amp;gt; HO-CO-OH}$$&#xA;In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.).  However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all is said and done, water and carbon dioxide are two extremely stable molecules.  Since&#xA;$$\Delta G = -RT\ln K$$&#xA;Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T16:20:28.733" LastActivityDate="2015-04-26T16:20:28.733" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16078" PostTypeId="1" CreationDate="2014-09-08T23:53:49.600" Score="0" ViewCount="51245" Body="&lt;p&gt;What is the difference between having a volatile versus a non-volatile solute at the boiling point of the solution?&lt;/p&gt;&#xA;&lt;p&gt;I have been getting mixed answers online concerning the differences outlined above. This is what I have gathered so far:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A non-volatile solute does not produce vapour at the boiling point of the solution.&lt;/p&gt;&#xA;&lt;p&gt;A volatile solute does produce vapour at the boiling point of the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can someone confirm these definitions?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-11-07T19:17:52.437" Title="Volatile and non-volatile solutes in solution" Tags="&lt;solutions&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16079" PostTypeId="1" AcceptedAnswerId="16081" CreationDate="2014-09-09T00:12:16.040" Score="2" ViewCount="1249" Body="&lt;p&gt;I was introduced to this thing today, and saw a graph showing the detection of ions with different relative abundances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as an example here, $\ce{Br2}$ is shown a detection chart. And at $160~\mathrm{m/z}$ the ion has the highest relative abundance, but why? Since the $\mathrm{m/z}$ ratio is just the mass, for $\mathrm{z}$ is always $+1$, the one which is supposed to have the highest peak should be at $162$?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4325" LastEditDate="2014-11-13T18:51:54.263" LastActivityDate="2014-11-13T18:51:54.263" Title="What is the highest relative abundance for diatomic bromine?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16080" PostTypeId="2" ParentId="16043" CreationDate="2014-09-09T00:15:55.263" Score="4" Body="&lt;p&gt;I guess this is just a paper exercise, since trying to make 1-hexene from octyne means you forget efficiency. A very useful and interesting exercise though. I agree with Ron about potential issues with step 4. I think sulfuric acid is worth trying for step 7 though and if it fails you can try stronger dehydrating agents that are less/non acidic or an indirect method like your tosylate elimination or even selenium chemistry. For step 6 I would use the standard $\ce{NaBH4}$ or $\ce{LiAlH4}$ for reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, after step 1, you can apply your ozonolysis-aldehyde reduction - elimination cycle twice to get to the product. Generally you sacrifice 2 of your 8 carbons ie 25% &quot;wasted&quot; from your starting material.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-09-09T00:15:55.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16081" PostTypeId="2" ParentId="16079" CreationDate="2014-09-09T00:25:37.813" Score="8" Body="&lt;p&gt;In $\ce{Br2}$, the possible combinations of isotopes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4945" LastEditDate="2014-09-09T06:26:20.817" LastActivityDate="2014-09-09T06:26:20.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16082" PostTypeId="1" AcceptedAnswerId="26412" CreationDate="2014-09-09T00:26:24.000" Score="5" ViewCount="674" Body="&lt;p&gt;I asked this question today and my professor was unable to provide any rationale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that sodium hydroxide is hygroscopic, but why not magnesium hydroxide (at least to the same extent as $\ce{NaOH}$)? Could it be that $\ce{Mg(OH)2}$ has a higher lattice energy; the hydroxide is more closely associated with the magnesium ion and less able to react with carbon dioxide? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4231" LastEditDate="2015-02-20T16:25:16.203" LastActivityDate="2015-02-20T16:25:16.203" Title="Why is magnesium hydroxide not hygroscopic?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16083" PostTypeId="2" ParentId="16079" CreationDate="2014-09-09T01:05:56.227" Score="4" Body="&lt;p&gt;jerepierre's answer is correct, you should see a statistical distribution of 3 different peaks, but also keep in mind that the m/z is not always the same as the mass. Depending on the molecule you're looking at and the ionization technique, the charge can be much larger than 1. Electrospray ionization, commonly used for ionizing biomolecules, routinely produces multiply charged ions. +20 charge is not uncommon. Additionally, many types of ionization cause fragmentation of molecules. For electron impact ionization, it's often the case that a peak for the full molecule is not observed at all.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-09T01:05:56.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16084" PostTypeId="1" AcceptedAnswerId="16085" CreationDate="2014-09-09T02:10:14.270" Score="0" ViewCount="3193" Body="&lt;p&gt;In the ICE table pure solids and liquids have a value of one in the equilibrium expression but when we create an ICE table, why is the $\ce{H_{2}O}$ column empty? Should it not have a value of one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://lh5.googleusercontent.com/Anz-rz_EkR_cVKjZwc9KfbII4Q5ZQ2V19x3qi7hHAsvWGTmvIRU2n_KH-ZhDEalhTg3ZkD-6mnYq8xD_zk3K-vhKJz7CzsvyP4NphJB44Wnl6Yqqdig&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="4945" LastEditDate="2014-09-09T06:00:49.380" LastActivityDate="2014-09-09T06:00:49.380" Title="Why is the water column empty in ICE table" Tags="&lt;homework&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16085" PostTypeId="2" ParentId="16084" CreationDate="2014-09-09T02:20:58.860" Score="5" Body="&lt;p&gt;Presumably water is the solvent for this reaction, which means that it will be in extreme excess and the amount will not change to a measurable extent.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-09T02:20:58.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16086" PostTypeId="1" CreationDate="2014-09-09T02:46:52.707" Score="5" ViewCount="1197" Body="&lt;p&gt;How does it damage proteins/lipids/DNA? There seems to be surprisingly little literature on its toxicity.&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2018-04-02T16:00:23.100" Title="What is the molecular mechanism of ammonia's toxicity?" Tags="&lt;toxicity&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16087" PostTypeId="1" CreationDate="2014-09-09T02:49:49.707" Score="3" ViewCount="198" Body="&lt;p&gt;It has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Catechol&quot; rel=&quot;nofollow&quot;&gt;catechol&lt;/a&gt; group, and it's the oxidation of the catechol group that makes dopamine produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactive_oxygen_species&quot; rel=&quot;nofollow&quot;&gt;reactive oxygen species (ROS)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxytyrosol&quot; rel=&quot;nofollow&quot;&gt;Hydroxytyrosol&lt;/a&gt; isn't that different from &lt;a href=&quot;http://en.wikipedia.org/wiki/Dopamine&quot; rel=&quot;nofollow&quot;&gt;dopamine&lt;/a&gt; (the distal N being replaced with an OH). I'm not sure if that really affects anything. Yet - it's often sold as an antioxidant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't it have the same ROS potential that dopamine and other catecholamines have?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="4945" LastEditDate="2014-09-09T05:56:25.890" LastActivityDate="2014-09-09T06:06:31.103" Title="Does hydroxytyrosol have the same neurotoxicity potential as dopamine?" Tags="&lt;toxicity&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16088" PostTypeId="1" CreationDate="2014-09-09T02:51:51.227" Score="2" ViewCount="1258" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Butylated_hydroxytoluene&quot; rel=&quot;nofollow&quot;&gt;Butyl Hydroxy Toluene (BHT)&lt;/a&gt; has two tert-butyl groups that can stabilize a free radical. &lt;a href=&quot;http://en.wikipedia.org/wiki/Butylated_hydroxyanisole&quot; rel=&quot;nofollow&quot;&gt;Butyl Hydroxy Anisol (BHA)&lt;/a&gt; doesn't, though it has the 4-substituted methoxy group (I'm not sure what that does).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the stronger antioxidant?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="194" LastEditDate="2017-09-23T01:55:36.200" LastActivityDate="2017-09-23T01:55:36.200" Title="What has stronger antioxidant potential: BHT or BHA?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16089" PostTypeId="2" ParentId="16078" CreationDate="2014-09-09T04:22:02.040" Score="2" Body="&lt;p&gt;A substance with higher vapour pressure vaporizes more readily than a substance with a lower vapour pressure. In case of a volatile solute this has a high vapour pressure and hence produces vapour. While in the case of a non-volatile solute due to its lower vapour pressure it does not produce vapour. The vapor pressure of a substance is the pressure at which its gas phase is in equilibrium with its condensed phases (liquid or solid).&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="7475" LastEditDate="2015-12-14T02:37:44.510" LastActivityDate="2015-12-14T02:37:44.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16091" PostTypeId="1" AcceptedAnswerId="16102" CreationDate="2014-09-09T04:53:41.003" Score="12" ViewCount="28077" Body="&lt;p&gt;I am unable to understand why &lt;strong&gt;d-d&lt;/strong&gt; electronic transitions are forbidden? Why are they weakly absorbing and apart from that, why do they occur at all?&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="4231" LastEditDate="2014-09-10T20:49:39.343" LastActivityDate="2019-03-29T18:49:39.403" Title="Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all?" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16092" PostTypeId="2" ParentId="16087" CreationDate="2014-09-09T06:06:31.103" Score="3" Body="&lt;p&gt;while hydroxytyrosol, ( lets call it HT because I can't keep spelling that ) is a catechol, it lacks the amine of dopamine. Dopamine gets metabolized to an aldehyde, this aldehyde can form schiff bases with amines on proteins. The catechol can also react with thiols on proteins. So in addition to dopamine's oxidative damage potential, it can crosslink proteins. HT doesn't have the amine, so it probably doesn't get metabolized in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/?term=j%20doorn%20dopamine&quot; rel=&quot;nofollow&quot;&gt;Here's a paper&lt;/a&gt; with more information on dopamine metabolism and toxicity. Look up other papers by that group for more.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-09-09T06:06:31.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16093" PostTypeId="1" AcceptedAnswerId="16096" CreationDate="2014-09-09T08:40:30.390" Score="4" ViewCount="86" Body="&lt;p&gt;Carbon ($\ce{^{14}C}$) is used to determine how long a particular species has lived or how old it is. Can we use the same isotope to derive how long a species may survive? (My understanding is still quite at infant level on half-life concept)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it prohibited or is it impossible? If not possible, why?&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2014-09-09T11:52:10.830" LastActivityDate="2014-11-18T17:26:40.800" Title="Can carbon dating tell how long a species would survive?" Tags="&lt;isotope&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16095" PostTypeId="2" ParentId="15982" CreationDate="2014-09-09T08:48:01.423" Score="4" Body="&lt;p&gt;1) Do NOT heat the solution, if you want chlorine out. Electrolysis of hot $\ce{NaCl}$ solution produces $\ce{NaClO3}$, then $\ce{NaClO4}$, not chlorine. You apparantly don't want this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) traditionally, in such setups some barrier between cathode and anode spaces is used. The easiest one would be &lt;strong&gt;unglazed&lt;/strong&gt; ceramic pot (with no holes and no direct contact of anode and cathode solutions), though I'm not sure if it will work with chlorine ions. It is recommended to perform electrolisys of $\ce{MgSO4}$ solution first. In this case chlorine ions would migrate from cathode space. With barrier introduced you may expect chlorine to evolve from anode space and sodium hydroxide to accumulate in cathode space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) do NOT blow air atop sodium hydroxide solution. It reacts with $\ce{CO2}$ from air.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-09T09:02:56.687" LastActivityDate="2014-09-09T09:02:56.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16096" PostTypeId="2" ParentId="16093" CreationDate="2014-09-09T08:50:12.033" Score="3" Body="&lt;p&gt;No, it can't be used for that. Read the mechanism of &lt;a href=&quot;http://en.wikipedia.org/wiki/Radiocarbon_dating&quot; rel=&quot;nofollow&quot;&gt;radiocarbon dating&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radiocarbon, often abbreviated as $\ce{^{14} C}$, is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting radiocarbon combines with atmospheric oxygen to form radioactive carbon dioxide. This is then incorporated into plants by photosynthesis, and animals acquire $\ce{^{14}C}$ by eating the plants. When the animal or plant dies, it stops exchanging carbon with its environment, and from that point the amount of $\ce{^{14} C}$ it contains begins to reduce as the $\ce{^{14}C}$ undergoes radioactive decay. Measuring the amount of $\ce{^{14}C}$ in a sample from a dead plant or animal such as piece of old wood or a fragment of bone provides information that can be used to calculate when the animal or plant died. The oldest dates that can be reliably measured by radiocarbon dating are around 50,000 years ago, though special preparation methods occasionally permit dating of older samples.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-09-09T10:16:23.303" LastActivityDate="2014-09-09T10:16:23.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16099" PostTypeId="1" CreationDate="2014-09-09T13:28:20.923" Score="3" ViewCount="3612" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following statements is true about the trend &lt;strong&gt;down&lt;/strong&gt; group 7 (halogens) with increase in atomic number?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The reactivity increases.  &lt;/li&gt;&#xA;  &lt;li&gt;The atoms get smaller  &lt;/li&gt;&#xA;  &lt;li&gt;The boiling points decrease  &lt;/li&gt;&#xA;  &lt;li&gt;The melting points increase  &lt;/li&gt;&#xA;  &lt;li&gt;The reactivity decreases  &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For me D and E must be right but only one can be right.&lt;/p&gt;&#xA;" OwnerUserId="7747" LastEditorUserId="5026" LastEditDate="2018-07-16T02:37:04.380" LastActivityDate="2018-07-16T02:37:04.380" Title="Which of the following statements is true about the trend down group 7?" Tags="&lt;inorganic-chemistry&gt;&lt;halides&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16101" PostTypeId="1" CreationDate="2014-09-09T14:03:01.403" Score="0" ViewCount="60" Body="&lt;p&gt;I'm trying to find a (free is possible) code to perform Dynamic Monte Carlo calculations over chemistry systems. Which one is the most intuitive one?&lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2014-09-10T04:32:10.257" Title="Which is the most intuitive to perform Dynamic Monte Carlo calculations?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16102" PostTypeId="2" ParentId="16091" CreationDate="2014-09-09T14:23:55.853" Score="12" Body="&lt;p&gt;In order for an electronic transition to be allowed (occur with strong intensity), certain &quot;selection rules&quot; must be obeyed.  You may already be familiar with the rule that the electron spin quantum number cannot change during a transition. It is this rule that forbids singlet-to-triplet absorption or emission (emission follows the same rules as absorption) and consequently makes phosphorescent emission so much weaker than fluorescent emission.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{\Delta S = 0}$ the Spin rule&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another rule, often referred to as the &quot;LaPorte&quot; rule or the &quot;Orbital&quot; rule, states that in molecules with a center of symmetry (centrosymmetric molecules), transitions within a subshell are forbidden when $\ce{\Delta l =0}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{\Delta l = \pm 1}$ the LaPorte rule&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The orbital angular momentum quantum number &quot;l&quot; describes the type of orbital (s, p, d, f) within the subshell.  In centrosymmetric molecules, the various s, p, d, f orbitals cannot mix within a subshell.  For example, the 4d orbital cannot mix with the 4p orbital in a centrosymmetric molecule, such mixing being symmetry forbidden.  Octahedral complexes can be centrosymmetric.  In such a case 3p-&gt;3p, 3d-&gt;3d transitions would be forbidden by LaPorte's rule.  Tetrahedral molecules do not have a center of symmetry and p-d orbital mixing is allowed, so in the case of tetrahedral molecules 3p-&gt;3p and 4d-&gt;4d transitions may appear stronger because a small amount of another orbital may be mixed into the p or d orbital thereby removing the violation of LaPorte's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we can do something to remove the centrosymmetric nature of an octahedral complex, then we &quot;relax&quot; LaPortes rule and weak transitions can be observed.  For example, if a molecular vibration removes the molecular center of symmetry, then a d-&gt;d transition can occur if light is absorbed at that instant.  Things such as spin-orbit coupling, vibronic coupling, or ligands that can mix with the metal's d orbitals would all allow relaxation of LaPorte's rule and thereby permit weak transitions to be observed.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-09T22:27:42.277" LastActivityDate="2014-09-09T22:27:42.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16103" PostTypeId="1" AcceptedAnswerId="16105" CreationDate="2014-09-09T15:23:57.203" Score="2" ViewCount="3516" Body="&lt;p&gt;One suggestion I received is that I prepare a solution of a known activity and probe the solution with the pH meter. The problem I see with this is that any error in solution preparation can lead us to conclude that the pH meter is faulty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other way to test the accuracy of a pH meter? We need to make extremely accurate measurements (we are studying the activities of ionic solutes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what solution has an activity of &lt;em&gt;exactly&lt;/em&gt; 1.0? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-09T15:33:33.607" LastActivityDate="2018-01-20T20:19:06.520" Title="How to verify the accuracy of a pH meter" Tags="&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16104" PostTypeId="2" ParentId="16103" CreationDate="2014-09-09T15:47:11.940" Score="5" Body="&lt;p&gt;There are usually standard buffer solutions available at pH 4 and 10 used for calibrating pH meters. I assume the manufacturer takes great care to verify the accuracy of these solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ph-meter.info/pH-electrode-calibration-buffers&quot; rel=&quot;noreferrer&quot;&gt;Here is a list of standard solutions and their pH&lt;/a&gt;. If you trust your scale more than your pH meter, you could make some of these buffers from dry reagents and see how your meter reads them.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-09-09T15:47:11.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16105" PostTypeId="2" ParentId="16103" CreationDate="2014-09-09T15:50:06.633" Score="8" Body="&lt;p&gt;Your school should have a purchased buffer solutions that have been calibrated against a NIST standard.  If they don't, have them purchase several in order to span the entire pH range, here's one from &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/fluka/82598?lang=en&amp;amp;region=US&quot;&gt;Sigma-Aldrich&lt;/a&gt;.  For your calibration, pick a standard solution that is close to the range you plan to work in.  Here's a couple of links with some more &quot;tips&quot; on how best to actually perform the calibration.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wikihow.com/Calibrate-and-Use-a-pH-Meter&quot;&gt;tips #1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/PH_meter#Calibration_and_use&quot;&gt;tips #2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-09T15:50:06.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16106" PostTypeId="1" AcceptedAnswerId="16109" CreationDate="2014-09-09T16:26:53.947" Score="1" ViewCount="8720" Body="&lt;p&gt;I was watching an episode of Man vs Wild where steel wool was rubbed with phone battery. To my surprise, the wool caught fire. What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="7510" LastEditorUserId="73" LastEditDate="2014-09-09T20:46:15.937" LastActivityDate="2019-07-25T07:03:16.570" Title="Why does steel wool catch fire when rubbed with a battery?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16107" PostTypeId="1" AcceptedAnswerId="16108" CreationDate="2014-09-09T16:30:04.193" Score="0" ViewCount="904" Body="&lt;p&gt;The negative oxygen is the site of protonation. Size and EN are the same for both oxygens. We are mainly looking at resonance (and possibly inductive) effects here. We want the least resonance withdrawal of electron density from the oxygen. And if possible, the least inductive withdrawal of electron density from the oxygen (although this is not a major issue; resonance trumps induction). Recall that the nitro group cannot withdraw electron density via resonance effects at the meta position. Therefore meta-nitrophenoxide should be more basic than para¬-nitrophenoxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2019-01-02T19:41:30.810" Title="Basicities of meta-nitrophenoxide and para-nitrophenoxide" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16108" PostTypeId="2" ParentId="16107" CreationDate="2014-09-09T16:42:31.303" Score="2" Body="&lt;p&gt;Yes, it is. These resonance structures describe basicity in para-nitrophenoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYQLE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="15751" LastEditDate="2019-01-02T19:41:30.810" LastActivityDate="2019-01-02T19:41:30.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16109" PostTypeId="2" ParentId="16106" CreationDate="2014-09-09T16:43:44.647" Score="3" Body="&lt;p&gt;Steel wool is a mat of very fine steel fibres. Because of their small size, it's easy to heat them up quickly to the point of burning and their relatively large surface area to volume ratio lets them oxidize quickly. If you try it with a flame, you'll find that it's quite easy to burn steel wool. When you short the terminals a battery across steel wool, a lot of current passes producing heat (think about incandescent light bulbs), and especially if you touch the battery such that only a few fibres are touching, enough heat is produced to cause them to burn.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-09T16:43:44.647" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16110" PostTypeId="1" AcceptedAnswerId="16118" CreationDate="2014-09-09T16:45:22.353" Score="4" ViewCount="2007" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At high temperatures the Van Der Waals equation has the form &#xA;  $$P\approx\frac{RT}{V_\mathrm{M}-b}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What would happen with the equation if in addition to high T, molar volume of the gas ($V_\mathrm{m}$) will get very large too? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only sense I was able to make out of this was that the pressure would be reduced, but if $V_\mathrm{m}$ stands for the volume occupied by a mole of a substance, than wouldn't a larger $V_\mathrm{M}$ mean less available void space, less volume, and thus a &lt;em&gt;greater&lt;/em&gt; pressure?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="5026" LastEditDate="2018-04-21T15:40:15.440" LastActivityDate="2018-04-21T15:42:05.280" Title="Van der Waals equation at high temperature and high molar volume" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16111" PostTypeId="1" AcceptedAnswerId="16120" CreationDate="2014-09-09T16:51:17.810" Score="3" ViewCount="5974" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ugtzI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that the first of the two acids is the &lt;em&gt;stronger&lt;/em&gt; acid. According to this person,deprotonation for both occur at the nitrogen.The resulting conjugate bases include,respectively a neutral amine group,and a negatively charged amine group.Because negatively charged nitrogens are highly unstable from an acid/base standpoint (consider lithium diisopropyl amide)... the first must be the stronger acid (because of its more stable conjugate base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have with this line of reasoning is that it's based on a faulty premise - that the nitrogen is the site of deprotonation.I see a &lt;em&gt;much&lt;/em&gt; better site of deprotonation for both - and that is the carboxylic acid site. A negative charge here will not only be resonance stabilized but also inductively stabilized by oxygen.Plus, the Ka of ammonium ion is 5.6*10^-10, while the Ka of acetic acid is 1.8*10^-5.In other words - orders of magnitude of difference. I can't see why something that is structurally analogous to ammonium ion (much less ammonia) would be deprotonated preferentially over a carboxylic acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-16T18:06:53.220" LastActivityDate="2015-10-16T18:06:53.220" Title="Acidities of amines and carboxylic acids" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16112" PostTypeId="2" ParentId="16111" CreationDate="2014-09-09T17:21:08.293" Score="6" Body="&lt;p&gt;Compound 1's conjugate base IS compound 2. The positive NH3 loses a proton to become the neutral NH2, but deprotonation of the NH2 group is extremely unlikely. Compound 2 will lose a proton at the carboxylic acid group first, so your friend is wrong to say deprotonation will occur at the nitrogen in both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will likely form a zwitterion, where the nitrogen is NH3+ and the carboxylic acid is COO-. This compound will have 2 pKas, one for the carboxylic acid, 1 for the amine. The carboxylic acid will be the lower of the 2 pKas. So if we start at very low pH and add base, the compound will start like compound 1. As we approach the pKa for the carboxylic acid, it will deprotonate to form the zwitterion. Then, as we approach the pKa of the NH3+, it will also deprotonate, forming a compound with an NH2 and COO-.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-09-09T17:21:08.293" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16113" PostTypeId="1" AcceptedAnswerId="16121" CreationDate="2014-09-09T18:22:11.513" Score="2" ViewCount="889" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9OVnv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the oxygen in an ester participate in resonance donation? I think so, although the donation is slight. The hybridization is correct. However, resonance withdrawal by the ring via the oxygen doesn't seem possible as my picture depicts. Therefore, because of resonance donation from the oxygen in the conjugate base of the ester, would that make benzoic acid the stronger acid? It seems that the resonance donation would destabilize the conjugate base of the ester. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-11-13T19:45:51.860" LastActivityDate="2015-11-13T19:45:51.860" Title="Acidity of ester and benzoic acid" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16115" PostTypeId="2" ParentId="16101" CreationDate="2014-09-09T18:33:00.303" Score="1" Body="&lt;p&gt;If you want to perform Kinetic Monte Carlo for a chemical reaction system, you can use &lt;a href=&quot;http://www.e-lucid.com/i/software/material_modelling/Zacros.html&quot; rel=&quot;nofollow&quot;&gt;ZACROS&lt;/a&gt;. I am not sure what your system is? You should follow some good text book like &quot;&lt;a href=&quot;http://www.sciencedirect.com/science/book/9780122673511&quot; rel=&quot;nofollow&quot;&gt;Understanding Molecular Simulation&lt;/a&gt;&quot; to identify the type of your problem and appropriate software for that. There are plenty of freely available code on the internet but they all are for some specific purpose, So basically it is your responsibility to choose the best available software that will suit your need.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="4945" LastEditDate="2014-09-10T04:32:10.257" LastActivityDate="2014-09-10T04:32:10.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16116" PostTypeId="1" AcceptedAnswerId="16184" CreationDate="2014-09-09T18:33:28.167" Score="5" ViewCount="913" Body="&lt;p&gt;I have some oil-pastel sketches on newsprint I would like to treat for framing (and longevity). My plan is to rinse them in a deacidifying bath, then use wheat paste to wet-mount them to heavyweight pH-neutral 100% rag paper. And finally, to frame them with UV-protective glass recommended for art. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Is there any reason not to use baking soda for the deacidifying bath? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have experimented with it, and it &lt;em&gt;seems&lt;/em&gt; to work great. I used a garden-soil pH test solution before and after, and sure enough, the treated paper shows no sign of acidity. The wheat paste bond seems just as good with the treated newsprint as with the untreated newsprint. The oil pastel colors seem unaffected (not surprising, oil pastel being very water-resistant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything about sodium bicarbonate that might, over time, cause a visible effect on the paper?&lt;/p&gt;&#xA;" OwnerUserId="7778" LastEditorUserId="73" LastEditDate="2014-09-10T00:31:12.420" LastActivityDate="2014-09-11T16:31:28.447" Title="Baking soda bath to deacidify newsprint?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16117" PostTypeId="2" ParentId="15787" CreationDate="2014-09-09T18:52:47.053" Score="4" Body="&lt;p&gt;Judging from the color, this clearly looks like more or less hydrated &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(III)_oxide&quot; rel=&quot;nofollow&quot;&gt;$\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$&lt;/a&gt;, and it could also contain dissolved &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron(III)_chloride&quot; rel=&quot;nofollow&quot;&gt;$\ce{FeCl3}$&lt;/a&gt; formed by the reaction of iron with $\ce{Cl2}$ evolved at the anode. The fact that it smells like chlorine when you heat the gelatinous precipitate would support the latter assertion, but the smell could also arise from residual $\ce{Cl2}$ which was still dissolved in the water of the jelly. Heating $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$ intensively above 200 °C will give you $\ce{Fe2O3}$ by thermal decomposition (dehydration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned}&#xA;\ce{2Fe(OH)3 -&amp;gt;&amp;amp; Fe2O3 + 3H2O} \\&#xA;\ce{2FeO(OH) -&amp;gt;&amp;amp; Fe2O3 + H2O} \\&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During electrolysis, the zinc layer on the electrode is degraded by &lt;a href=&quot;http://en.wikipedia.org/wiki/Pitting_corrosion&quot; rel=&quot;nofollow&quot;&gt;pitting corrosion&lt;/a&gt; of $\ce{Cl-}$ and oxidation in the presence of oxygen. When the coating is completely removed by the corrosion in some places, the exposed iron reacts with oxygen and chlorine to the products mentioned above. On anodes made of pure zinc, &lt;a href=&quot;http://en.wikipedia.org/wiki/Zinc_hydroxide&quot; rel=&quot;nofollow&quot;&gt;$\ce{Zn(OH)2}$&lt;/a&gt; will likely be the primary product, which will form a transparent to white, jelly-like precipitate. Zinc hydroxide formed by oxidative degradation of the $\ce{Zn}$ layer would also be a constituent of the precipitate on a zinc-coated iron anode.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-09T18:52:47.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16118" PostTypeId="2" ParentId="16110" CreationDate="2014-09-09T20:13:17.550" Score="7" Body="&lt;p&gt;First, let's look at this from a purely mathematical standpoint. This isn't going to be rigorous at all, but I'm happy to clarify anything if asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume we have 1 mole, and we say that $V_\mathrm{m}$ and $T$ become very large, then the question becomes what happens to $P$ (our only remaining variable)? Well as $T$ and $V_\mathrm{m}$ become very large with respect to (w.r.t) our excluded volume parameter, $b$, (also written as $T$, $V_\mathrm{m} &amp;gt;&amp;gt; b$), the -b term in the denominator becomes negligible and we approach the conventional ideal gas law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P \approx \frac{R T}{V_{\mathrm{m}} - b} \implies P \approx \frac{R T}{V_\mathrm{m}} = \frac{n R T}{V}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not clear, try plugging in some numbers to the equation and calculating what $P$ would be for small $P$ and $T$ (same order of magnitude as $b$) vs. large $P$ and $T$ (100 times larger than $b$). I will let you determine if this leads to greater or less pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's switch gears to a chemistry point of view. What Johannes van der Waals did, is add this little $b$ parameter in the denominator of the ideal gas law. This parameter is a correction to account for the actual volume that each mole of particles can occupy. You can think of it equivalently as the actual volume each particle can occupy, since a mole is just a really big number of particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $b$ correction is taking into account the fact atoms can repel each other at a relatively large distance. If we approximate these atoms to be hard sphere bouncing off of each other, the radii of these hard sphere is what we call the van der Waals radii - the distance at which atoms &quot;bounce&quot; off of each other if they come into contact. So the  parameter $b$ in the denominator is derived purely from the van der Waals radii of the atoms in question. So in the case of large $T$ and $V_\mathrm{m}$, we find that this correction is no longer necessary to make a reasonable approximation, because its presence makes such a small difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend looking at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_equation#Derivation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="4267" LastEditorUserId="5026" LastEditDate="2018-04-21T15:42:05.280" LastActivityDate="2018-04-21T15:42:05.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16119" PostTypeId="1" CreationDate="2014-09-09T20:34:26.647" Score="9" ViewCount="470" Body="&lt;p&gt;&lt;strong&gt;Short question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How (quantitatively speaking) does temperature influences rate of decay of proteins? I am looking for some general number/function, the influence of temperature on an average protein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For computer modeling purposes, I am looking for some referenced quantitative measurements of the effect(s) of temperature on the dynamic of biochemical reactions. In particular, my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does temperature influences rate of protein decay?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am looking for a function (eventually expressing the michaelis-menten constant as a function of the temperature) that I can plug in my algorithms to incorporate the influence of various temperatures on the process I want to simulate. The model under simulation is an eukaryote such as Saccharomyces telluris (a yeast) for example living at some common range of temperature [5-35 °C]. I am not thinking to any particular protein rather I am looking for some approximative effect of temperature on the decay rate on an &quot;average protein&quot;. I welcome any kind of expression of how temperature influences the rate of decay. Some values using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eyring_equation&quot; rel=&quot;nofollow&quot;&gt;Erying equation&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;nofollow&quot;&gt;Van't Hoff equation&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Q10_(temperature_coefficient)&quot; rel=&quot;nofollow&quot;&gt;$Q10$&lt;/a&gt; (measured in some reasonable range [5°C - 35°C]) for examples would also fit my needs. I would be very surprise that there is not enough literature on the subject in order to get some average impact of temperature on protein decay rate but as a biologist (and not a biochemist) I can't wrap my head around this problem and can't find any satisfying article. Thanks for your precious help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not looking for...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am NOT looking for a theoretical explanation of how temperature influences these reactions (I have basic understanding of the importance of the activation energy as displayed on a Maxwell-Boltzman distribution and of the michaelis-menten equation). I am just looking for some empirical observations of how temperature influences proteins decay rate. I am not looking for any accurate value but only for some estimation that I can plug into my algorithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the comments I realize that my question is too broad in the sense that there might have too much variance in how different proteins respond to temperature in order to draw in general tendency. If this is so, someone willing to answer may reduce the question to transcription factor, to &lt;a href=&quot;http://en.wikipedia.org/wiki/General_transcription_factor&quot; rel=&quot;nofollow&quot;&gt;General Transcription Factor (GTF)&lt;/a&gt; or even to &lt;a href=&quot;http://en.wikipedia.org/wiki/Transcription_factor_II_A&quot; rel=&quot;nofollow&quot;&gt;TFIIA&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about &lt;a href=&quot;http://www.ploscompbiol.org/article/info%3Adoi%2F10.1371%2Fjournal.pcbi.1003689&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for example? I am not really able to extract from it the info I am looking for but it might be in here.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="5691" LastEditDate="2014-09-17T01:36:23.953" LastActivityDate="2014-09-17T01:36:23.953" Title="Influence of Temperature on Biochemical Reactions" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;enzymes&gt;&lt;dna-rna&gt;" AnswerCount="0" CommentCount="15" ClosedDate="2015-04-16T22:46:00.827" ContentLicense="CC BY-SA 3.0" />
  <row Id="16120" PostTypeId="2" ParentId="16111" CreationDate="2014-09-09T21:05:35.810" Score="3" Body="&lt;p&gt;The first compound is more acidic. The ammonium functionality is inductively electron withdrawing, stabilizing the conjugate base (carboxylic acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer is assuming that we are talking about the drawn structures exactly. User137 is correct that it's not possible to have the second compound, because the amine is a strong enough base to deprotonate the carboxylic acid, and it exists as the zwitterion. However, assuming that we could hold the amine as neutral, it would be less inductively electron withdrawing than the cationic ammonium.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-09T21:05:35.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16121" PostTypeId="2" ParentId="16113" CreationDate="2014-09-09T21:22:33.123" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the oxygen in an ester participate in resonance donation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sure, structure III in the following diagram is a valid resonance structure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, resonance withdrawal by the ring via the oxygen doesn't seem&#xA;  possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that's possible too.  It might not be a major contributor, but structure V is a valid resonance structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L09NZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would that make benzoic acid the stronger acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you're asking which is the stronger acid, benzoic acid or the half-ester of carbonic acid. Resonance structures like III and VII will stabilize both the acid and the corresponding carboxylate anions.  One molecule doesn't appear to have a more resonance stabilized starting acid or a more resonance stabilized carboxylate anion than the other.  So resonance arguments don't seem to suggest that one molecule should be more acidic than the other.  This conclusion, that resonance factors do not control the acidity in these two compounds, can also be arrived at by noting that the starting acids and particularly the carboxylate anions are analogous to the allyl MO system which has a node on the central carbon atom in the HOMO ($\ce{\Psi 2}$). The node effectively  disconnects the substituent (-OR or phenyl) from the MO, meaning that the substituent cannot interact &lt;em&gt;via&lt;/em&gt; resonance with the MO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductively one can argue that the electron withdrawing oxygen in the half-ester would destabilize the already positively polarized carbonyl carbon in the starting acid and then also provide some inductive stabilization of the electron-rich carboxylate anion, more so than a phenyl group would.  So perhaps, based mainly on inductive arguments, one could argue that the carbonic acid half-ester would be more acidic than benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any pKa values for carbonate half-esters, but we can compare the pKa of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;carbonic acid 3.6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;benzoic acid 4.2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The values are close, as expected - even in the right direction, but that's probably just luck.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-09T21:22:33.123" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16122" PostTypeId="1" CreationDate="2014-09-09T23:09:56.233" Score="9" ViewCount="11527" Body="&lt;p&gt;As an end of high school treat, my class is trying to convince our chemistry teacher to make an explosion as a demonstration. What sort of reactions are beautifully explosive but safe enough to do in a school lab environment, when performed by a teacher who knows what they're doing?&lt;/p&gt;&#xA;" OwnerUserId="7784" LastEditorUserId="4945" LastEditDate="2014-09-10T04:49:23.807" LastActivityDate="2016-02-25T21:25:12.880" Title="Safe explosion demonstrations" Tags="&lt;teaching-lab&gt;&lt;explosives&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16123" PostTypeId="2" ParentId="16055" CreationDate="2014-09-09T23:33:01.120" Score="8" Body="&lt;p&gt;One counter-argument: Helium is essentially a &quot;fossil gas&quot;, and there's a limited supply of easy-to-get helium (until we get practical fusion reactors running, at least). Hydrogen, on the other hand, is universally available in $\ce{H2O}$ and needs only a bit of electricity to break it out. Since helium has important industrial uses &lt;em&gt;other&lt;/em&gt; than balloons, I expect that we will eventually find it becomes too expensive to throw away on toys.&lt;/p&gt;&#xA;" OwnerUserId="4022" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T09:50:04.263" LastActivityDate="2015-06-26T09:50:04.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16124" PostTypeId="2" ParentId="16122" CreationDate="2014-09-10T01:01:59.333" Score="13" Body="&lt;p&gt;From our friends at Periodic Videos, the following come to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.periodicvideos.com/videos/mv_guncotton.htm&quot;&gt;Gun cotton&lt;/a&gt;, more precisely known as nitrocellulose, which is a very simple nitration that you could show prior to the pyrotechnics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.periodicvideos.com/videos/011.htm&quot;&gt;Sodium&lt;/a&gt; and &lt;a href=&quot;http://www.periodicvideos.com/videos/019.htm&quot;&gt;Potassium&lt;/a&gt; are also good, though you may not always get an explosion, depending upon the available hydrogen above the metal.  Which suggests...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.periodicvideos.com/videos/001.htm&quot;&gt;Hydrogen&lt;/a&gt;! This is the classic explosion in oxygen. In fact, on a &lt;em&gt;small&lt;/em&gt; scale, the electrolysis of water into hydrogen and oxygen can be used to show that the mixture of these gases is very explosive. But keep it small!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these experiments avoid outright explosives, which seems prudent in a high school setting.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-09-10T01:01:59.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16126" PostTypeId="1" CreationDate="2014-09-10T02:23:28.937" Score="3" ViewCount="10791" Body="&lt;p&gt;I understand that talcum powder is insoluble to water, would alcohol or something be able to dissolve it?&lt;/p&gt;&#xA;" OwnerUserId="7787" LastActivityDate="2018-02-19T19:50:26.693" Title="Are there any solvents that dissolve talcum powder?" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16127" PostTypeId="1" AcceptedAnswerId="16351" CreationDate="2014-09-10T02:57:48.303" Score="4" ViewCount="1354" Body="&lt;p&gt;While reading a chapter on transition metals, I came to know that d-Block elements have colour compounds due to d-d transitions.  A question came to my mind as to whether there is any d-d kind of transitions in plants as they also show characteristics colours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the green colours of leaves is due to presence of chlorophyll and also there are coloured pigments in plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be wrong in my thought, but I really want to clarify my doubt.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="418" LastEditDate="2014-09-16T07:53:30.923" LastActivityDate="2014-09-16T08:39:23.337" Title="Is the colour of leaves due to d-d transitions in chlorophyll?" Tags="&lt;everyday-chemistry&gt;&lt;transition-metals&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16128" PostTypeId="1" CreationDate="2014-09-10T03:28:01.467" Score="5" ViewCount="603" Body="&lt;p&gt;I understand this is a basic question, but I'm having such a hard time wrapping my head around it. I'm trying to avoid thinking about it as an actual &quot;particle&quot; but as a wave, but that confuses me too.  Wouldn't a wave be the entire area the electron travels ? Or does the associated wave represent the path and not the electron? If so, where is the electron? &#xA;Here's a picture to refer to in your explanation if it helps.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X5KOg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7788" LastEditorUserId="189" LastEditDate="2014-09-10T03:30:17.660" LastActivityDate="2017-10-01T12:47:00.667" Title="How do electrons travel through nodes" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16129" PostTypeId="2" ParentId="16122" CreationDate="2014-09-10T03:44:06.620" Score="1" Body="&lt;p&gt;I am giving you the link where you can find many interesting explosion experiments.but let me tell you these are little dangerous too.so be careful while performing these.&#xA;&lt;a href=&quot;http://youtu.be/sfW7KztrA8I&quot; rel=&quot;nofollow&quot;&gt;http://youtu.be/sfW7KztrA8I&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2014-09-10T03:44:06.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16130" PostTypeId="1" CreationDate="2014-09-10T03:48:53.193" Score="1" ViewCount="163" Body="&lt;p&gt;&lt;a href=&quot;http://spacecollege.org/isee3/isee-3-post-lunar-flyby-status-and-modification-of-mission-goals.html&quot; rel=&quot;nofollow&quot;&gt;For context I'm talking about ISEE-3&lt;/a&gt; and how long it would take to dump the now useless and hazardous fuel it still carries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a heated tank of hydrazine (somewhere near $0~^\circ\mathrm{C}$) connected by way of &lt;a href=&quot;http://images.spaceref.com/news/spacecollege/isee3/propulsion.l.jpg&quot; rel=&quot;nofollow&quot;&gt;a long slender unheated tube&lt;/a&gt; to a infinite high vacuum, how fast will the tank vent to ~zero pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that rather quickly, the boiling action will produce a plug of frozen hydrazine in the tube. Once that happens, the sublimation on the outer face will cause condensation and/or solidification on the other face and the plug will walk back towards the tank until it either reaches it or the solid portion of the plug get to thin in which case it will burst and vent like in the beginning and start over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how fast will this happen?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: on further inspection, it looks like the &quot;tube&quot; may be heated after all.&lt;/p&gt;&#xA;" OwnerUserId="7400" LastEditorUserId="4945" LastEditDate="2014-09-10T04:42:33.200" LastActivityDate="2014-09-10T04:42:33.200" Title="How fast will hydrazine boil out of a tube into a vacuum?" Tags="&lt;thermodynamics&gt;&lt;sublimation&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16131" PostTypeId="2" ParentId="16122" CreationDate="2014-09-10T04:07:04.380" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_triiodide&quot; rel=&quot;nofollow&quot;&gt;Nitrogen triiodide&lt;/a&gt; is pretty spectacular and not dangerous so long as you keep the batch small. I remember my high school chem teacher had a batch left in the back of the classroom to go off randomly to scare us all. It's basically a matter of mixing aqueous ammonia and iodine crystals, leaving them to react a couple hours, then gravity filter. The result is pretty stable so long as it's damp, but explodes with a cloud of purple smoke if even slightly disturbed when dry. Even a feather or something touching is likely to set it off. As I mentioned above, keep the batch small. It doesn't take much to give a good show.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-10T04:07:04.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16132" PostTypeId="2" ParentId="16126" CreationDate="2014-09-10T04:21:19.450" Score="3" Body="&lt;p&gt;Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-10T04:21:19.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16133" PostTypeId="1" AcceptedAnswerId="16155" CreationDate="2014-09-10T04:26:35.113" Score="5" ViewCount="42936" Body="&lt;p&gt;When you turn a balanced equation into a net ionic equation, you ignore all the ions that are spectator ions (aqueous on one side as well as the other). You only write the solid and the aqueous ions that form that solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what happens when you try to write a net ionic equation if &lt;strong&gt;ALL&lt;/strong&gt; the ions are aqueous? Would you write down every ion in the reactants/products, or write nothing?&lt;/p&gt;&#xA;" OwnerUserId="7789" LastActivityDate="2014-09-10T21:55:05.817" Title="Net ionic equation with only aqueous ions" Tags="&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16134" PostTypeId="2" ParentId="16099" CreationDate="2014-09-10T04:29:15.790" Score="6" Body="&lt;p&gt;The answer is D, the melting points increase. This is absolutely true (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_17%3A_The_Halogens&quot; rel=&quot;nofollow noreferrer&quot;&gt;source for values&lt;/a&gt;):&#xA;\begin{array}{lrr}&#xA;\text{Halogen} &amp;amp; \text{Melting point}/^\circ\mathrm{C}&amp;amp; \text{Boiling point}/^\circ\mathrm{C}\\\hline&#xA;\text{fluorine}   &amp;amp; -220  &amp;amp; -188  \\&#xA;\text{chlorine}  &amp;amp; -101  &amp;amp; -35   \\&#xA;\text{bromine}  &amp;amp; -7.2  &amp;amp; 58.8  \\&#xA;\text{iodine}   &amp;amp; 114   &amp;amp; 184   \\&#xA;\text{astatine}  &amp;amp; 302   &amp;amp; 337   \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the boiling point increases, too, so answer C is definitely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another obviously wrong answer is B, the atoms get smaller. With increasing main quantum number the atoms obviously have to become bigger since the main electron density moves further away from the nucleus. In numbers:&#xA;\begin{array}{lrr}&#xA;\text{Halogen} &amp;amp; \text{Covalent radius}/\mathrm{pm}&amp;amp; \text{Ionic radius }\ce{(X^{-})}/\mathrm{pm}\\\hline&#xA;\text{fluorine} &amp;amp;  71  &amp;amp; 133  \\&#xA;\text{chlorine}   &amp;amp;  99  &amp;amp; 181  \\&#xA;\text{bromine}  &amp;amp; 114  &amp;amp; 196  \\&#xA;\text{iodine}   &amp;amp; 133  &amp;amp; 220  \\&#xA;\text{astatine}  &amp;amp; 150  &amp;amp;      \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For answers A and E it cannot be unambiguously answered, because reactivity is not a well defined concept, see &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/14172/4945&quot;&gt;What is reactivity really, and can it be quantified?&lt;/a&gt;&quot;  for example. The IUPAC goldbook states, that &lt;a href=&quot;http://goldbook.iupac.org/R05180.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reactivity&lt;/a&gt; is a kinetic property. It goes on to describe, that it can only be absolutely used in a given context or in reference to another system. However, it is also often used in expressing general trends, which is the case when looking at this question.&lt;br&gt;&#xA;Given this context, you can find the following portion on UC Davis ChemWiki:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Reactivity of Elements: decreases down the group&lt;/strong&gt;&lt;br&gt;&#xA;  The reactivities of the halogens decrease down the group ( At &amp;lt; I &amp;lt; Br &amp;lt; Cl &amp;lt; F). This is due to the fact that atomic radius increases in size with an increase of electronic energy levels. This lessens the attraction for valence electrons of other atoms, decreasing reactivity. This decrease also occurs because electronegativity decreases down a group; therefore, there is less electron &quot;pulling.&quot; In addition, there is a decrease in oxidizing ability down the group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this context, answer A is wrong. However, I would not consider E to be strictly true either. Have a look at the electron affinities, and you will find, that fluorine behaves anomalous. This was also discussed in &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/6901/4945&quot;&gt;Why does chlorine have a higher electron affinity than fluorine?&lt;/a&gt;&quot; &#xA;However the general statement will usually be given as: &quot;Electron affinity decreases down the group.&quot;&#xA;\begin{array}{lc}&#xA;\text{Halogen} &amp;amp; \text{Electron Affinity}/\mathrm{kJ\cdot mol^{-1}}\\\hline&#xA;\text{fluorine}  &amp;amp;  -328.0  \\&#xA;\text{chlorine} &amp;amp;  -349.0  \\&#xA;\text{bromine}   &amp;amp;  -324.6  \\&#xA;\text{iodine}    &amp;amp;  -295.2  \\&#xA;\text{astatine}  &amp;amp;  -270.1  \\\hline&#xA;\end{array}&#xA;The same trend can be observed with bond enthalpies of the $\ce{X2}$ series (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_17%3A_The_Halogens/Properties_of_the_Halogens/Physical_Properties_of_the_Group_17_Elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OHsQw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHsQw.png&quot; alt=&quot;bond enthalpies graph halogens&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;This trend is broken for the hydrogen halides $\ce{HX {(g)}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your assumption is therefore correct, maybe you have already done the same analysis. However, we can only definitely state and proof, that answer D must be correct, while answer E might be correct depending on the reference system.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-07-16T02:19:01.737" LastActivityDate="2018-07-16T02:19:01.737" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16135" PostTypeId="1" AcceptedAnswerId="27980" CreationDate="2014-09-10T05:13:03.563" Score="21" ViewCount="267157" Body="&lt;p&gt;I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, &lt;a href=&quot;http://www.madsci.org/posts/archives/1997-12/878572346.Ch.r.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;structural isomers of hexane&lt;/a&gt; is 5, while the number structural isomers of decane is 75.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the possible number of structural isomers of a given organic compound?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="402" LastEditDate="2019-05-17T06:39:31.977" LastActivityDate="2019-05-17T06:40:20.430" Title="How to determine number of structural isomers?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="5" CommentCount="11" FavoriteCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="16136" PostTypeId="2" ParentId="15903" CreationDate="2014-09-10T06:33:21.407" Score="3" Body="&lt;p&gt;I don't know if it's kosher to provide the answer like this but it turns out (after i got my homework back) that the correct answer was;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Ru^{2+}: [Kr]$ $4d^6$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reasoning provided that was behind it is &quot;When the d or f orbitals start to fill up, the energy of these orbitals is lowered such that they have lower energy than the corresponding s orbitals.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-10T06:33:21.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16137" PostTypeId="2" ParentId="15903" CreationDate="2014-09-10T07:57:49.857" Score="1" Body="&lt;p&gt;There is another simple version of Aufbau principle. n+l rule, where n is principle quantum number and l is azimuthal quantum number. For 4d n+l= 4+2=6 and for 5s n+l=5+0=5, so electron will enter 4d orbital first and then it will fill 5s if 4d is filled or half filled. Note: This principle is for filling the orbital only. But for removing electron it must be removed from the outer shell first, in this case 5s. So, electron will be removed from 5s and then 4d and so on. &#xA;NB: Order of energy should only be applied for filling the orbital and for removing electron it must be removed from the outer shell and so on.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2014-09-10T07:57:49.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16138" PostTypeId="2" ParentId="16126" CreationDate="2014-09-10T07:59:29.540" Score="-1" Body="&lt;p&gt;I consider that there is no solvent that can dissolve talcum powder, indeed if you consider &quot;solving&quot; as having your material into solution but maintaining the crystalline structure, only some inorganic compounds (most of them transition metal complexes) are soluble even in water or in other organic solvents (like alcohol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting wikipedia, we find: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Talc is not soluble in water, but it is slightly soluble in dilute mineral acids. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any case, solving it will keep intact its structure. This is naturally true for the case of silicates, which are characteristics by their insolubility. &lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2014-09-10T07:59:29.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16139" PostTypeId="2" ParentId="16122" CreationDate="2014-09-10T08:08:13.873" Score="1" Body="&lt;p&gt;Explosions usually happen too fast to be overly fascinating. Explosions with true detonation are quite dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is quite a set  of flash-powders that produce a cloud of flame and smoke. They are not true explosives as they do not detonate. The easiest one is $\ce{1Zn +1S}$ equimolar mix, it produces bright green-blue burst. I also found funny $\ce{3CuO +2Al}$ mix - it burns with a report, producing yellowish flame and black smoke, and covering nearby surfaces with yellow layer of copper, and $\ce{1Bi(OH)2NO3 +5Mg}$ - in small doses (and I mean really small) it burns with a report, bright (may be even blinding) yellow flash and mushroom cloud of bright yellow smoke. Bismuth oxynitrate is quite easy to prepare via hydrolysis of commercially available bismuth trinitrate. $\ce{2Fe +KNO3}$ mix burns without explosion, producing fountain of yellow sparks.  Generally, however, mixes with metal powders must be handled with great care as the mixes may be sensitive, burn without oxygen and produce toxic fumes. Metal powder are available via specialized pyro- or chemical stores and aluminium powder is available also via painting store. However, in reciepes above the finest powder is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are very few explosives that are easy to make. If you have time, you can go with synthesis of barium styphnate (lead is much more toxic, and explosion products will contain lead oxide smoke. Barium is toxic too, but to much, much lesser extent, so it is reasonable safe). It IS sensitive, and will explode if left on heating plate, but is quite easy to produce (styphnic acid is produced by nitration of resorcinol in two stage process with $\ce{NaNO2 + HCl}$ first and $\ce{HNO3}$ second, barium styphate is easily produced from barium chloride, styphic acid and a base, though don't do it without finding proper procedure), so no need to store it. I tested it and found... funny, if not very spectacular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a funny video for you &lt;a href=&quot;http://youtu.be/7gnC0mXE0Vk&quot; rel=&quot;nofollow&quot;&gt;http://youtu.be/7gnC0mXE0Vk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-10T08:20:02.500" LastActivityDate="2014-09-10T08:20:02.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16140" PostTypeId="1" CreationDate="2014-09-10T09:27:34.777" Score="3" ViewCount="4475" Body="&lt;p&gt;Silica found in &quot;do-not-eat&quot; packets is a desiccant (I'm not sure about this); and so is activated charcoal. Would consuming it have the same effect if one were to have a stomach ache? &lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T18:34:04.613" LastActivityDate="2017-09-07T16:01:13.827" Title="Would silica have the same effect as activated charcoal?" Tags="&lt;everyday-chemistry&gt;&lt;absorption&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16141" PostTypeId="2" ParentId="16128" CreationDate="2014-09-10T10:34:13.033" Score="10" Body="&lt;p&gt;From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;There is no concept in QM for &quot;path of a particle&quot;:&lt;/strong&gt; position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$.&#xA;$$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$&#xA;So it's meaningless in QM to talk about trajectories.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Meaning of the wave function:&lt;/strong&gt; Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$&#xA;is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$&#xA;, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx&#xA;, gives you the probability for finding the particle between $x$&#xA;and $x+dx$&#xA;$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$&#xA;Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$&#xA;$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Stationary state:&lt;/strong&gt; If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like&#xA;$$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$&#xA;In this case it's easy to prove that the probability density is time independent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;The quantization:&lt;/strong&gt; For this simple problem, you must solve the eigenvalue equation for the energy&#xA;$$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$&#xA;By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$&#xA;. Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$&#xA;. The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$&#xA;. They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$&#xA;is zero the probability of finding the particle is of course zero (for example in 0&#xA;and 3&#xA;, like we requested before).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;So where is the electron?&lt;/strong&gt; Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by&#xA;$$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$&#xA;$$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="7692" LastEditDate="2014-09-12T17:29:09.337" LastActivityDate="2014-09-12T17:29:09.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16142" PostTypeId="1" CreationDate="2014-09-10T10:59:55.067" Score="5" ViewCount="472" Body="&lt;p&gt;Is it possible that blowing on water (hot/cold) can cause a reaction between the $\ce{CO2}$ from expiration and the water, to form $\ce{H2CO3}$? Is $\ce{H2CO3}$ dangerous?&lt;/p&gt;&#xA;" OwnerUserId="7793" LastEditorUserId="9961" LastEditDate="2015-11-13T19:48:10.573" LastActivityDate="2015-11-13T19:48:55.987" Title="Formation of carbonic acid from breath?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16143" PostTypeId="2" ParentId="16142" CreationDate="2014-09-10T11:24:14.193" Score="5" Body="&lt;p&gt;It is possible that &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonic_acid&quot; rel=&quot;nofollow&quot;&gt;carbonic acid&lt;/a&gt; will form; however, only in small amounts because the equilibrium of the hydration of $\ce{CO2}$ to $\ce{H2CO3}$ lies mainly on the educt side. As more $\ce{CO2}$ is soluble in cold water, more carbonic acid will be present in the solution at lower temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 + H2O \rightleftharpoons H2CO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $25~\mathrm{^\circ C}$ and in pure water, the quotient of the concentrations is $[\ce{H2CO3}]/[\ce{CO2}] \approx 1.7 \times 10^{-3}$, so most carbon dioxide remains physically dissolved in water. This makes carbonic acid - despite a $\mathrm{p}K_\mathrm{a}$ of $3.6$ for the first dissociation step - a fairly weak acid. It is not a dangerous compound, otherwise it would not be used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonated_water&quot; rel=&quot;nofollow&quot;&gt;carbonated drinks&lt;/a&gt;. It is also a natural constituent of blood, where it acts as a pH buffer. In red blood cells, it is produced as an intermediate by the enzyme &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonic_anhydrase&quot; rel=&quot;nofollow&quot;&gt;carbonic anhydrase&lt;/a&gt;, which catalyzes the hydration of carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="16683" LastEditDate="2015-11-13T19:48:55.987" LastActivityDate="2015-11-13T19:48:55.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16144" PostTypeId="2" ParentId="16140" CreationDate="2014-09-10T13:34:21.387" Score="2" Body="&lt;p&gt;The silica gel granules are granular silicon dioxide (SiO2), the same matrial from what is the normal sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;pure&lt;/strong&gt; SiO2 has the same effect to health as when you eat sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But because the granules are larger, may cause appendix problems (im not an doc). Anyway, this is not chemical related problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can cause dehydratation or other problems, what is caused by it absorbtion capabality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger problem is, than the standard silica gel granules &lt;strong&gt;aren't pure&lt;/strong&gt; SiO2.  The common additivas are what causes color change when it is moistured,  like cobalt chloride or others and thise compounds can be toxic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, don't eat it!&lt;/p&gt;&#xA;" OwnerUserId="7792" LastActivityDate="2014-09-10T13:34:21.387" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16145" PostTypeId="1" CreationDate="2014-09-10T15:24:24.633" Score="9" ViewCount="2248" Body="&lt;p&gt;The molecular orbital theory dictates that when two atomic orbitals form molecular orbitals, then two molecular orbitals must form (i.e number of atomic orbitals = number of molecular orbitals). For this to occur there must be bonding molecular orbitals and anti-bonding molecular orbitals. The bonding MO's are caused when the orbitals constructively interfere and the anti-bonding orbitals form as the orbitals destructively interfere. I understand the concept of interference but I do not understand how two atomic orbitals (one from each bonding atom) can simultaneously interfere constructively and destructively. Moreover, I am under the impression that the anti-bonding orbitals exist even if they're not occupied. As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can two orbitals interfere constructively and destructively?&lt;/li&gt;&#xA;&lt;li&gt;Do anti-bonding orbitals exist when not occupied?&lt;/li&gt;&#xA;&lt;li&gt;Is it the orbitals themselves that are interfering with each other like waves?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2016-01-12T15:03:52.167" LastActivityDate="2016-01-15T20:43:54.430" Title="How can two orbitals constructively and destructively interfere simultaneously?" Tags="&lt;bond&gt;&lt;atoms&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16146" PostTypeId="2" ParentId="16135" CreationDate="2014-09-10T16:21:18.710" Score="0" Body="&lt;p&gt;This is a start towards a partial answer, but I tried to count the isomers of methane through nonane, and only by moving methyl groups around, each &quot;branch&quot; is only 1 CH3. I noticed a pattern in the number of 1 branch, 2 branch, 3 branch, and 4 branch isomers. Don't know how accurate these numbers are or how well this pattern holds up.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N   Straight    1 Branch    2 Branch    3 Branch    4 Branch&#xA;1          1           0           0           0           0&#xA;2          1           0           0           0           0&#xA;3          1           0           0           0           0&#xA;4          1           1           0           0           0&#xA;5          1           1           1           0           0&#xA;6          1           2           2           0           0&#xA;7          1           2           4           0           0&#xA;8          1           3           6           1           1&#xA;9          1           3           9           4           3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-09-10T16:21:18.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16148" PostTypeId="2" ParentId="16145" CreationDate="2014-09-10T19:11:51.643" Score="8" Body="&lt;p&gt;They do not &lt;em&gt;simultaneously interfere constructively and destructively&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain. Since the time-independent &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow&quot;&gt;Schrödinger equation&lt;/a&gt; is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Variational_method_(quantum_mechanics)&quot; rel=&quot;nofollow&quot;&gt;variational method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals:&#xA;$$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function:&#xA;$$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Of course the anti-bonding orbital exists when not occupied.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that the previous procedure can be generalized for all kind of atomic orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to clarify something about the meaning of the wave function since you're asking: &lt;em&gt;As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$,&#xA;is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$&#xA;$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$.&#xA;Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$&#xA;$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.&lt;/p&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="470" LastEditDate="2016-01-15T20:43:54.430" LastActivityDate="2016-01-15T20:43:54.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16149" PostTypeId="1" AcceptedAnswerId="16154" CreationDate="2014-09-10T20:35:10.900" Score="6" ViewCount="13932" Body="&lt;p&gt;I don't know where to begin with this question.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A function &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt; can be expressed as a special power series called the Taylor expansion. The common expansion is &lt;span class=&quot;math-container&quot;&gt;$(1-x)^{-1}=1+x+x^2$&lt;/span&gt; Note that the higher terms become neglibible small and usually are not included in expansion. Find the correlation between the virial coefficients and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; in the corresponding van der Waals equation of state. Start with the equation shown below and compare it with the virial equation of state.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p=\frac{RT\left(1-\frac{b}{V_{\mathrm m}}\right)^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The virial equation of state is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p=\frac{nRT}{V}\left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+\ldots\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I get that there are &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;… coefficients, and that the &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; coefficient should be 1.&lt;/p&gt;&#xA;&lt;p&gt;The best I have is that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;A &amp;amp;= 1 \\&#xA;B &amp;amp;= -\left(\frac{b}{V_{\mathrm m}}\right)^{-1} \\&#xA;C &amp;amp;= -\frac{a}{V_{\mathrm m}^2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I were to relate that to the terms in the Taylor expansion, then &lt;span class=&quot;math-container&quot;&gt;$p=(1-x)^{-1}$&lt;/span&gt; and then &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;… are the terms on the right side of the equation.&lt;/p&gt;&#xA;&lt;p&gt;Is this the correct approach? Can the virial coefficients have a negative value?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-05-09T13:14:03.560" Title="Relation between the virial coefficients and van der Waals constants" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16150" PostTypeId="1" CreationDate="2014-09-10T20:58:40.620" Score="7" ViewCount="9929" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would equal masses of gold and copper result in the same volume change if added to water?&lt;br&gt;&#xA;  Density of gold: $19.3~\mathrm{g/mL}$&lt;br&gt;&#xA;  Density of copper: $8.96~\mathrm{g/mL}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was thinking the answer would be no since the density is different between gold and copper but was unsure if this was a trick question since it maybe has to do with the mass of the substance that is placed into the water. &lt;/p&gt;&#xA;" OwnerUserId="7795" LastEditorUserId="5026" LastEditDate="2018-02-22T11:19:48.877" LastActivityDate="2018-02-22T11:19:48.877" Title="Would equal masses of gold and copper result in the same volume change if added to water?" Tags="&lt;physical-chemistry&gt;&lt;density&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16151" PostTypeId="1" AcceptedAnswerId="16153" CreationDate="2014-09-10T21:06:35.450" Score="5" ViewCount="153" Body="&lt;p&gt;What's the name of this ion: $\ce{CN2^2-}$? I think that it has this structure:&#xA;$\ce{{}^{-}N=C=N-}$.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorDisplayName="user7951" LastEditDate="2016-01-13T18:19:46.177" LastActivityDate="2016-01-13T18:19:46.177" Title="Unknown anion (one carbon and two nitrogen atoms, charge 2−)" Tags="&lt;nomenclature&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16152" PostTypeId="2" ParentId="16150" CreationDate="2014-09-10T21:22:46.503" Score="10" Body="&lt;p&gt;You're reasoning and your answer (&quot;no&quot;) are both correct; this is not a trick question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we had 8.96 gm of gold and the same amount of copper and dropped them both into separate graduated cylinders that were filled to the 10 ml mark with water.  Submerged items displace a volume equal to their own volume. Therefore, this amount of copper should displace&#xA;$$\mathrm{\frac{[8.96]}{[density~ of~ Cu]} = \frac{[8.96]}{[8.96]} =1 ml~displaced}$$&#xA;whereas 8.96 gm of gold would displace&#xA;$$\mathrm{\frac{[8.96]}{[density~ of~ Au]} =\frac{[8.96]}{[19.3]}= 0.46 ml~displaced}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:50:10.473" LastActivityDate="2016-03-12T00:50:10.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16153" PostTypeId="2" ParentId="16151" CreationDate="2014-09-10T21:37:45.707" Score="7" Body="&lt;p&gt;The parent molecule is cyanamide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5cAo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ion is usually referred to as either the cyanamide dianion or the carbimide dianion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MGjcn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-10T21:43:08.153" LastActivityDate="2014-09-10T21:43:08.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16154" PostTypeId="2" ParentId="16149" CreationDate="2014-09-10T21:46:55.180" Score="9" Body="&lt;p&gt;You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you wrote: The virial equation of state is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;p &amp;amp;=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\&#xA;  &amp;amp;= \frac{RT}{V_{\mathrm{m}}} \left(1+ B\frac{1}{V_{\mathrm{m}}}+ C\frac{1}{V_{\mathrm{m}}^2}+ \ldots \right)&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the van der Waals equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA; p &amp;amp;=\frac{RT(1-\frac{b}{V_{\mathrm m}})^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2} \ .&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have to bring the van der Waals equation into the same form as the virial equation by factoring out &lt;span class=&quot;math-container&quot;&gt;$\frac{RT}{V_{\mathrm{m}}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA; p &amp;amp;= \frac{RT}{V_{\mathrm{m}}}\left( \left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} - \frac{a}{R T V_{\mathrm{m}}} \right) \ ,&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the Taylor expansion &lt;span class=&quot;math-container&quot;&gt;$\left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} = 1 + \frac{b}{V_{\mathrm{m}}} + \frac{b^2}{V_{\mathrm{m}}^2} + \ldots$&lt;/span&gt; and ordering the terms according to their power in &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{V_{\mathrm{m}}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA; p &amp;amp;= \frac{RT}{V_{\mathrm{m}}}\left( 1 + \left(b - \frac{a}{R T} \right) \frac{1}{V_{\mathrm{m}}} + b^{2} \frac{1}{V_{\mathrm{m}}^{2}} + \ldots \right) \ ,&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that you can compare the coefficients in both equations to get the result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA; B &amp;amp;= b - \frac{a}{R T} \\&#xA; C &amp;amp;= b^2 \ .&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2019-05-09T13:14:03.560" LastActivityDate="2019-05-09T13:14:03.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16155" PostTypeId="2" ParentId="16133" CreationDate="2014-09-10T21:55:05.817" Score="5" Body="&lt;p&gt;If you have a balanced equation like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCl_{(aq)} + KBr_{(aq)} -&amp;gt; NaBr_{(aq)} + KCl_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solubility products for all the species are big enough that there's essentially no solid, there's really no distinction between the left and right sides of the equation. It's a balanced equation, but nothing has changed. We started with $\ce{Na+_{(aq)} + K+_{(aq)} + Br^{-}_{(aq)} + Cl^{-}_{(aq)}}$ in solution and we still have that on the right side. So there's really not a net ionic equation because there's no net change.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-10T21:55:05.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16156" PostTypeId="1" AcceptedAnswerId="16157" CreationDate="2014-09-11T03:28:04.727" Score="0" ViewCount="20123" Body="&lt;p&gt;If I dissolve 1  g of sodium bicarbonate powder ($\ce{NaHCO3}$) into 400 ml of distilled water, what will the pH of the resulting solution be?  &lt;/p&gt;&#xA;" OwnerUserId="7778" LastEditorUserId="39539" LastEditDate="2017-01-31T22:29:16.780" LastActivityDate="2017-01-31T22:29:16.780" Title="pH of baking soda in water?" Tags="&lt;home-experiment&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-09-11T09:34:19.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="16157" PostTypeId="2" ParentId="16156" CreationDate="2014-09-11T05:22:18.260" Score="4" Body="&lt;p&gt;1 g of sodium bicarbonate in 400 mL of water will give an approximately 30 mM solution with a pH of ~8.4. The interesting thing about the pH of solutions of amphiprotic compounds is that so long as the concentration of the salt is significantly larger than its Ka and Kb, the solution pH only depends on the dissociation constants. Basically any amount of bicarbonate above 10 mM or so will have the same pH in solution, so pH is not a good indicator of how much acid you can neutralize with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure what good this information will do you. How much this solution will be able to affect the paper is going to depend on how much paper you use for a given volume of solution and how acidic the paper is. I don't know much about paper—it could be that it doesn't really matter if you go overboard on the bicarbonate. It might be as simple as making a concentrated solution, testing periodically that your bath hasn't become acidic, similar to photographic stop bath, which has a pH indicator that changes colour when the bath is not acidic anymore. If residual bicarbonate is a problem, rinsing with water afterwards is probably advisable.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-11T05:22:18.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16158" PostTypeId="2" ParentId="16116" CreationDate="2014-09-11T05:26:24.790" Score="1" Body="&lt;p&gt;It seems that adding bicarbonate is how acid-free paper is made, though more modern paper is neutralized by the use of chalk as filler. The wiki page says that magnesium or calcium bicarbonate are commonly used, but that may be just due to cost rather than any specific issues with using sodium bicarbonate. &lt;a href=&quot;http://en.wikipedia.org/wiki/Acid-free_paper&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-11T05:26:24.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16159" PostTypeId="1" AcceptedAnswerId="16198" CreationDate="2014-09-11T05:56:11.883" Score="3" ViewCount="2628" Body="&lt;p&gt;Until now to execute &lt;a href=&quot;http://www.phytotechlab.com/TechInfo/B141-Info.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; recipe, each time i had to count those milligrams/Liter. An enough difficult procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am posting here this question to help me understand the procedure of creating stock solutions. So the next time i will create that plant substrate, to take a specific volume of that stock solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="7800" LastEditorUserId="7717" LastEditDate="2014-09-13T14:36:21.593" LastActivityDate="2014-09-13T14:36:21.593" Title="What is the procedure to create stock solutions?" Tags="&lt;aqueous-solution&gt;&lt;analytical-chemistry&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16160" PostTypeId="1" AcceptedAnswerId="18707" CreationDate="2014-09-11T07:19:05.303" Score="10" ViewCount="445" Body="&lt;p&gt;I am currently working on a mechanistic study that deals with stereoselective $\beta$-lactam formation. I am well aware that there is a class of &lt;a href=&quot;http://en.wikipedia.org/wiki/%CE%92-Lactam_antibiotic&quot;&gt;antibiotics&lt;/a&gt;, that involve this particular functional group. However, the &lt;a href=&quot;http://en.wikipedia.org/wiki/%CE%92-Lactam&quot;&gt;wikipedia article&lt;/a&gt; of this compound class does not offer very much to continue, apart from the antibiotics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other uses of the $\beta$-lactam moiety in general, apart from drug design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to view my research in a broader scope. Since my part is more focussed on the reaction mechanism of the synthesis, I have not been too much concerned about the main purpose of this particular functional group, yet. This is of course apart from the obvious drug design process. For me this is just a question of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am resurrecting this question, I am in particular looking for some interesting publications about the above mentioned question. However, I am willing to recognise, that the most common use is in drug design. If someone from this area of expertise could provide a source as a starting point for more research in this area. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-10-14T09:21:35.843" LastActivityDate="2014-10-22T00:04:32.560" Title="What general uses do beta-lactam functional groups have?" Tags="&lt;reference-request&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16161" PostTypeId="2" ParentId="16150" CreationDate="2014-09-11T07:51:20.833" Score="13" Body="&lt;p&gt;&quot;Added to water&quot; is ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the metal objects are shaped and placed such that they float&lt;/strong&gt; (eg: as a metal boat does, or a light hollow sphere), then they will each displace the &lt;strong&gt;same&lt;/strong&gt; volume of water (the volume of water equivalent to their equal common masses).  So long as they float, they will displace the same amount of water no matter their shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the masses objects are shaped and placed such that they sink&lt;/strong&gt; (eg: consider a solid sphere or cube), then they will &lt;em&gt;in general&lt;/em&gt; displace &lt;strong&gt;different&lt;/strong&gt; volumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus.&lt;/strong&gt;  The second part is slightly more complex than it seems when it comes to calculating how much water they would displace.  If they have no dry internal voids, their respective displaced volumes are simply the mass of each divided by its density.  Others have given the simple calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they were shaped such that they had dry voids yet were small enough to sink (think of a heavy hollow sphere with overall density greater than water so it sinks, or an overturned metal boat with small air pockets), then the displaced water will be somewhere between the above two calculations (fully floating calc and no-voids calc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it would be possible to construct the two objects such that they sank and yet displaced the same amount of water, eg: two hollow spheres with the same outer diameter but different internal diameters, both heavier than the same volume of water.  But that would be a very trick question indeed!&lt;/p&gt;&#xA;" OwnerUserId="7802" LastEditorUserId="7802" LastEditDate="2014-09-11T08:02:01.057" LastActivityDate="2014-09-11T08:02:01.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16162" PostTypeId="1" AcceptedAnswerId="16170" CreationDate="2014-09-11T08:09:40.790" Score="2" ViewCount="61" Body="&lt;p&gt;My friend works at a headphone company at the support service and today she came across someone who claimed that he threw his broken headphone away because he had heard dangerous chemicals could be released when the cable burned through (because they would be exposed to the outside world possibly) which was according to him affirmed at a local waste disposal service. As I am unable to find any confirmation for this claim online, I decided to go ahead and post the question here. I was a bit unsure whether to post on chemistry.SE, biology.SE or engineering.SE, but went ahead with this one in the end, hope it's not off topic, but if it is then my apologies.&lt;/p&gt;&#xA;" OwnerUserId="7803" LastActivityDate="2014-09-11T11:09:13.653" Title="&quot;Chemicals&quot; released in headphone with burned cable" Tags="&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16163" PostTypeId="1" AcceptedAnswerId="16165" CreationDate="2014-09-11T08:22:52.723" Score="1" ViewCount="1051" Body="&lt;p&gt;I am currently looking through a revision sheet, however it has caught me on a question 'Explain why a potassium atom can act as a reducing agent yet a potassium ion cannot'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done brief searches, however none have given the answer I'm looking for. So please, why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the explanation doesn't already include it; is this only the case with Potassium, and how is this to be determined? &lt;em&gt;(How is it known whether it is referring to a cation or anion?)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="7804" LastEditorUserId="7804" LastEditDate="2014-09-11T08:59:06.863" LastActivityDate="2014-09-11T09:40:37.713" Title="Potassium Atom can act as reducing agent, yet ion cannot?" Tags="&lt;homework&gt;&lt;redox&gt;&lt;ions&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16165" PostTypeId="2" ParentId="16163" CreationDate="2014-09-11T09:40:37.713" Score="5" Body="&lt;p&gt;The ability of K to act as a reducing agent is mainly due to its electronic structure. &#xA;$K = [Ar] 4s^1$ . In this case, such atom tends to donate his $4s^1$ electron (acting as a reducing agent). This is due to the gain in energy that produces to have its octet complete and stay in the cationic form $K^{+}=[Ar]$ which is quite more stable that the pure element one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is common to all elements in the first column of the periodic table, just because all of them have a quite similar electronic structure, just with more electrons on their inner orbitals.&lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2014-09-11T09:40:37.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16166" PostTypeId="2" ParentId="16159" CreationDate="2014-09-11T09:58:18.257" Score="0" Body="&lt;p&gt;The best way of creating a stock solution (even more importante if it is quite diluted) is to start from a highly concentrated solution, and thus dilute it in subsequent steps taking aliquots from the previous one. &#xA;It is very important that you check the final dilution to be sure of its exact concentration. That is the use of the &quot;analysis standard&quot; sets that permits you to guarantee the concentration of your stock solution.&lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2014-09-11T09:58:18.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16167" PostTypeId="1" AcceptedAnswerId="16169" CreationDate="2014-09-11T10:17:15.037" Score="5" ViewCount="868" Body="&lt;p&gt;&lt;strong&gt;Ethanethiol&lt;/strong&gt;, commonly known as &lt;strong&gt;ethyl mercaptan&lt;/strong&gt;, is a colorless gas added to &lt;strong&gt;LPG&lt;/strong&gt; in order to detect its leakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from ethyl mercaptan, is there any gas or liquid that may be used as a replacement to this?&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="30836" LastEditDate="2019-04-13T17:42:02.320" LastActivityDate="2021-02-16T11:14:28.960" Title="Is there any replacement for ethyl mercaptan in LPG?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;smell&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16168" PostTypeId="2" ParentId="16162" CreationDate="2014-09-11T10:38:36.203" Score="1" Body="&lt;p&gt;A headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.&#xA;The degradation of the plastic could give some hazardous materials for the environment. We cannot be sure of which ones just because each plastic is different. But I do not think it can be dangerous for human health.&lt;/p&gt;&#xA;" OwnerUserId="7774" LastActivityDate="2014-09-11T10:38:36.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16169" PostTypeId="2" ParentId="16167" CreationDate="2014-09-11T10:49:12.113" Score="5" Body="&lt;p&gt;Why would one want to replace it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of ethanethiol is a reasonable compromise between its toxicity and the extremely low detection limit around 1 ppb (1mg/1000 kg).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-11T10:49:12.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16170" PostTypeId="2" ParentId="16162" CreationDate="2014-09-11T11:09:13.653" Score="4" Body="&lt;p&gt;Burning or Incineration of Plastic cause carcinogenic chemical products if without suitable catalysts... To pyrolyze plastics such as polyethylene(PE) and polypropylene(PP) of olefin series, the catalysts with high cracking property are&#xA;needed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dioxin- carcinogenic and antibiotic effect to the environment.&lt;/li&gt;&#xA;&lt;li&gt;Flyash- may produce respiratory illnesses.&lt;/li&gt;&#xA;&lt;li&gt;Sulfur dioxide&lt;/li&gt;&#xA;&lt;li&gt;Nitrogen oxide- green plastics.&lt;/li&gt;&#xA;&lt;li&gt;Toluene- aromatic compound that is related in the burning of cigarettes.&lt;/li&gt;&#xA;&lt;li&gt;Carbon monoxide- main product of anerobic/ lack of ventilation combustion.&lt;/li&gt;&#xA;&lt;li&gt;carbonaceous char, oils and gases.&lt;/li&gt;&#xA;&lt;li&gt;paraffins, olefins, naphthenes and aromatics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Plastic and Diesel&#xA;&lt;a href=&quot;http://www.chemtec.org/proddetail.php?pr&quot; rel=&quot;nofollow&quot;&gt;http://www.chemtec.org/proddetail.php?pr&lt;/a&gt;...&#xA;&lt;a href=&quot;http://www.p-fuel.com/index_files/faq.ht&quot; rel=&quot;nofollow&quot;&gt;http://www.p-fuel.com/index_files/faq.ht&lt;/a&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermofuel&#xA;&lt;a href=&quot;http://www.cynarplc.com/thermo-fuel.asp&quot; rel=&quot;nofollow&quot;&gt;http://www.cynarplc.com/thermo-fuel.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mael is right that a  headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.so even i agree it can not be dangerous for human health to a great extent.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2014-09-11T11:09:13.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16171" PostTypeId="1" AcceptedAnswerId="16176" CreationDate="2014-09-11T11:34:11.483" Score="1" ViewCount="425" Body="&lt;p&gt;What is the advance of OFET's (Organic Field effect transistors) over the traditional CMOS Transistors?&lt;/p&gt;&#xA;" OwnerUserId="7774" LastEditorUserId="4231" LastEditDate="2014-09-11T13:54:24.420" LastActivityDate="2014-09-11T13:54:24.420" Title="OFET vs Traditional Transistors" Tags="&lt;physical-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16172" PostTypeId="2" ParentId="15514" CreationDate="2014-09-11T11:56:33.467" Score="2" Body="&lt;p&gt;May I start with a disclaimer saying that whatever long term studies can be done, this will always be a subjective question. All I can do here is offer a few comparisons and examples to show that you shouldn't (in my opinion) be worried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There answer to the question &quot;Am I more likely to come into contact with carcinogenic/hazardous substances than a regular person?&quot; is quite likely yes. At undergraduate level you will be using a range of strong, harmful, poisonous etc. substances that you could not get outside of a lab environment or at school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer to the question &quot;Am I more likely to suffer poor health as a result of working in a lab?&quot; is more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly state in your question, there are always many safety procedures in place when you undertake experiments which have been carefully thought, require the scientists to have some minimum level of training in the techniques (perhaps learning with less dangerous substances) out and always err on the side of safe rather than sorry. I myself have been working with a new nanotube based material recently and though unlikely to be carcinogenic, all experiments were to be carried out using gloves in a fume cupboard with extraction as the material hasn't been fully classified or tested yet (again: &quot;safe rather than sorry&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are unfortunate cases (for example &lt;a href=&quot;http://en.wikipedia.org/wiki/Marie_Curie#Death&quot; rel=&quot;nofollow&quot;&gt;Marie Curie&lt;/a&gt;), but in this day and age the risk assessments are so extensive and always erring on the side of caution that the chance of something going terribly wrong are no worse, I feel, than something catastrophic happening in many other walks of life (e.g. mountaineering accident, plane crash, shark attack).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One perhaps reassuring and rather amusing case is that you are far less likely to succumb to terminal cancer if you work in the nuclear industry by virtue of the extensive risk assessments, precautions and health checks. There is so much safety in place at a reactor, that exposure is minimal (indeed, the background radiation from radon in &lt;a href=&quot;http://www.bloomberg.com/news/2011-04-01/hong-kong-radiation-exceeds-tokyo-even-after-japan-crisis.html&quot; rel=&quot;nofollow&quot;&gt;Cornwall&lt;/a&gt; is higher than that next to a nuclear power plant!!) and the industry provides so many health checks for workers that and cancer (radiation induced or natural) is often picked up at a very early, curable stage. Earlier than if you simply went to a doctor when you began to feel ill as a regular person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore depending on the policy of your employer, you could in fact be at a lower risk of dying from cancer, and I see no need for you to worry for your health in the lab!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned when I started, this is a subjective issue, but I really see no need for worry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jack&lt;/p&gt;&#xA;" OwnerUserId="7805" LastActivityDate="2014-09-11T11:56:33.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16173" PostTypeId="2" ParentId="16159" CreationDate="2014-09-11T12:00:45.950" Score="2" Body="&lt;p&gt;For the easy reference,you prepare a concentrated stock solution and you dilute it according to your requirement by the below given procedure.You prepare a solution by dissolving a known mass of solute (often a solid) into a specific amount of a solvent. One of the most common ways to express the concentration of the solution is M or molarity, which is moles of solute per liter of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prepare 1 liter of 1.00 M NaCl solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First calculate the molar mass of NaCl which is the mass of a mole of Na plus the mass of a mole of Cl or 22.99 + 35.45 = 58.44 g/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Weigh out 58.44 g NaCl.&#xA;   2) Place the NaCl in a 1 liter volumetric flask.&#xA;   3) Add a small volume of distilled, deionized water to dissolve the salt.&#xA;   4) Fill the flask to the 1 L line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a different molarity is required, then multiply that number times the molar mass of NaCl. For example, if you wanted a 0.5 M solution, you would use 0.5 x 58.44 g/mol of NaCl in 1 L of solution or 29.22 g of NaCl. &#xA;once your stock solution is prepared you can dilute it to a particular volume to make a new, more dilute solution.  The key equation here is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M1V1 = M2V2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that number of moles you are working with is constant whether in the  stock solution or in the more dilute new solution.  Since molarity times volume equals moles, that it what the equation above is talking about.  M1V1 is the molarity (moles per liter) times the volume of the stock solution and M2V2 is the molarity times the volume of the new solution.  So you are usually given three of the variables in the problem and you simply plug them in and solve for the missing fourth variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of actual laboratory practice to make a new solution from a stock solution, first you calculate how many milliliters of the stock solution that you need to make the new solution.  Then measure out the required number of mL's of the stock solution in a graduated cylinder and pour into into a new flask.  Then add fresh water (or whatever solvent you are working with) to the flask until the level reaches the required number of mL's of the new solution.  Then stir the solution to mix evenly.  Now you have a new, more dilute solution of a know concentration.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2014-09-11T12:00:45.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16174" PostTypeId="2" ParentId="16150" CreationDate="2014-09-11T12:01:29.950" Score="5" Body="&lt;p&gt;Historical anecdote time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;David Richerby made an offhand comment about Archimedes of Syracuse. Archimedes had to figure out whether the king's crown was made of actual gold, but obviously he couldn't damage the crown because that would make the king angry. He had to determine the density, but you need the volume for that, and there wasn't really a way to determine the volume of an irregular object back then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He was wondering how to do this, and at one point, he had to take a bath. Now, his servant filled the bath to the brim, and when Archimedes got into the water, some of it spilled over the edge. The legend after this goes that Archimedes saw this, jumped out of the water, and still naked, ran out onto the streets yelling &quot;Eureka&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He had the crown weighed and asked for gold equal in mass to the weight of the crown. Then he submerged the crown and the gold into water, he saw that the gold displaced a different amount of water, despite weighing as much. This means that the volume, and thus the density of the object, was not the same, and thus that the crown was not made of solid gold. (note)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your answer is absolutely correct: the displaced amount of water is dependent on the volume, not the weight.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(note) This is a legend. People who know more about this than me have stated that Archimedes didn't use the volume measurement, since this required measurement accuracy beyond contemporary means. Instead, the more likely method was that he tied them to a balanced scale, then submerged them, subjecting them to a buoyant force equal to the weight of the water displaced. The denser object would encounter less force and thus sink.&lt;/p&gt;&#xA;" OwnerUserId="7809" LastActivityDate="2014-09-11T12:01:29.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16175" PostTypeId="1" AcceptedAnswerId="16211" CreationDate="2014-09-11T12:13:06.347" Score="12" ViewCount="2732" Body="&lt;p&gt;Let us for simplicity discuss RHF formalism. For $2n$-electron system we have $n$ Hartree-Fock equations written for $n$ spatial orbitals $\{ \phi_{k} \}_{k=1}^{n}$&#xA;$$&#xA;\newcommand{\mat}[1]{\boldsymbol{\mathbf{#1}}}&#xA;$$&#xA;\begin{equation}&#xA;    \hat{F}(1) \phi_{k}(1) = \varepsilon_{k} \phi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .&#xA;\end{equation}&#xA;Once we introduce finite basis $\{ \chi_{q} \}_{q=1}^{m}$ and express spatial orbitals as a linear combination of basis functions $\chi_{q}$&#xA;\begin{equation}&#xA;    \phi_{k}(1) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(1) \, , \quad k = 1, 2, \dotsc, n \, .&#xA;\end{equation}&#xA;we end up with $n$ Roothaan–Hall equations&#xA;\begin{equation}&#xA;    \sum\limits_{q=1}^{m} F_{pq} c_{qk} = &#xA;    \varepsilon_{k} \sum\limits_{q=1}^{m} S_{pq} c_{qk} \, , &#xA;    \quad k = 1, 2, \dotsc, n \, ,&#xA;\end{equation}&#xA;which can be rewritten in the following matrix form&#xA;\begin{equation}&#xA;    \mat{F} \mat{c}_{k} = \varepsilon_{k} \mat{S} \mat{c}_{k} \quad k = 1, 2, \dotsc, n \, .&#xA;\end{equation}&#xA;The Fock matrix $\mat{F}$ and the overlap matrix $\mat{S}$ are both $m \times m$ square matrices, $\mat{c}_{k}$ is a column $m \times 1$ matrix, $\varepsilon_{k}$ is just a scalar value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uofqt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then collecl all $n$ $\mat{c}_{k}$ column $m \times 1$ matrices into one $m \times n$ matrix $\mat{C}$ and all $n$ values $\varepsilon_{k}$ into $n \times n$ square matrix $\mat{\varepsilon}$&#xA;\begin{equation}&#xA;    \mat{F} \mat{C} = \mat{S} \mat{C} \mat{\varepsilon} \, .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Dvga.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, we extend both $\mat{C}$ and $\mat{\varepsilon}$ to $m \times m$ matrices from $m \times n$ and $n \times n$ respectively, which results in having $m-n$ virtual (unoccupied) orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account that virtual orbitals are even more unphysical than their occupied counterparts the question is what is the point of such extension of $\mat{C}$ and $\mat{\varepsilon}$? Why do not we just leave them of $m \times n$ and $n \times n$ sizes respectively? &lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-09-11T12:20:36.353" LastActivityDate="2021-05-21T19:55:40.863" Title="What is the point of introducing virtual orbitals in Hartree-Fock calculations?" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16176" PostTypeId="2" ParentId="16171" CreationDate="2014-09-11T13:42:33.563" Score="2" Body="&lt;p&gt;The advantages of OFET's over traditional CMOS transistors all derive from how OFET's can be manufactured.  Traditional transistors are manufactured using &lt;strong&gt;vacuum deposition&lt;/strong&gt; techniques.  These are slow, time consuming, batch processes, hence the higher cost for electronic components produced by such methods.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OFET's can be manufactured using standard &lt;strong&gt;coating&lt;/strong&gt; technologies that have been widely used in industry for a long time, and are well understood.  Therefore, the manufacturing equipment needed is readily available. With this coating equipment uniform layers ranging from sub-micron to tens of microns (you don't need thicker layers for an OFET, but the coating technology would allow you to coat much thicker layers if you wanted to) can be coated at high speed on thin, flexible, web supports such as aluminized mylar. Multiple layers can be coated by making multiple passes down the coating line, into a drying oven and then on to the next coater.  Thousands of OFETS can be obtained from a small, coated, mylar spool. Since the process is continuous and high speed the manufacturing cost of the device is much reduced compared to similar CMOS devices.  Each layer contains an active organic compound dissolved in a low-cost plastic (for example, polycarbonate) binder, so material costs are reduced as well.  Another key advantage is that since organic materials, plastic binders and flexible web materials are used, the final product is flexible and can be easily cut to any desired shape.  This allows the manufacturer to use the product in new applications where flexibility (newspapers) and shape (electronic clocks) might preclude the use of rigid CMOS devices.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-11T13:42:33.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16177" PostTypeId="2" ParentId="15944" CreationDate="2014-09-11T13:51:55.553" Score="3" Body="&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It expands linearly (i.e a 1 kelvin temperature increase almost always produces the same change in volume over a reasonable range of temperatures). You might think that this is the norm - but not always. For example, water does not expand linearly. It has a maximum density at $4~\mathrm{^\circ C}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It can measure over a reasonable range of temperatures, from $-39~\mathrm{^\circ C}$ up to its boiling point. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is quite dense in the first place. This is so that it can be used in configurations such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Six&amp;#39;s_thermometer&quot; rel=&quot;nofollow&quot;&gt;Six's Thermometer&lt;/a&gt; — a type of thermometer that records the maximum and minimum temperatures of an area (I won't go into schematics now, although it is quite simple)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="7475" LastEditDate="2016-08-31T08:27:02.247" LastActivityDate="2016-08-31T08:27:02.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16178" PostTypeId="2" ParentId="16135" CreationDate="2014-09-11T13:58:56.143" Score="1" Body="&lt;p&gt;As far as I am aware, there is no straight up formula. Obviously there are very general trends, but not precise enough to tell you the number of isomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to decide how many isomers something has, I suggest you either do it manually or use a programme like this: &lt;a href=&quot;http://www-jmg.ch.cam.ac.uk/tools/isomercount/&quot; rel=&quot;nofollow&quot;&gt;http://www-jmg.ch.cam.ac.uk/tools/isomercount/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7813" LastActivityDate="2014-09-11T13:58:56.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16180" PostTypeId="1" AcceptedAnswerId="16182" CreationDate="2014-09-11T14:12:10.547" Score="2" ViewCount="432" Body="&lt;p&gt;I cannot find much guidance in the literature. This is the reaction between free chlorine in water and triclosan. &#xA;I understand ether cleavage and electrophilic aromatic substitution (maybe) involved. Any ideas as to what the mechanism of this reaction might be?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/p69F8.jpg&quot; alt=&quot;Degradation pathway of triclosan&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7815" LastActivityDate="2014-09-11T15:27:55.520" Title="Formation of 2,4-Dichlorophenol by free chlorine mediated oxidation of Triclosan" Tags="&lt;organic-chemistry&gt;&lt;kinetics&gt;&lt;aromatic-compounds&gt;&lt;environmental-chemistry&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16181" PostTypeId="2" ParentId="16180" CreationDate="2014-09-11T15:07:54.723" Score="2" Body="&lt;p&gt;I'm not sure if this qualifies as an answer since it's more of a guess. Catechols (1,2-dihydroxybenzene derivatives) can be oxidized to ortho-quinones. Applying that transformation here (chlorine in water is certainly oxidizing conditions) gives the oxonium of the ortho-quinone, which can be hydrolyzed to 2,4-DCP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8HQpu.jpg&quot; alt=&quot;proposed mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-11T15:07:54.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16182" PostTypeId="2" ParentId="16180" CreationDate="2014-09-11T15:27:55.520" Score="2" Body="&lt;p&gt;Just as ethers can be formed when alcohols are protonated, the reverse reaction, formation of alcohols from ethers can also occur in the presence of acid.  In the case at hand, when chlorine is mixed with water hypochlorous acid is one of the compounds formed, and it can serve as the proton source for the following reaction.  The nucleophile can be $\ce{OCl^-}$, or any other nucleophile present.  In any case this mechanism explains the formation of 2,4-dichlorophenol from TCS, chlorine and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cJIui.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to the a study entitled, &quot;&lt;a href=&quot;http://scholar.google.com/citations?view_op=view_citation&amp;amp;hl=en&amp;amp;user=NADpe3wAAAAJ&amp;amp;citation_for_view=NADpe3wAAAAJ:u-x6o8ySG0sC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formation of chloroform and chlorinated organics by free-chlorine-mediated oxidation of triclosan&lt;/a&gt;&quot;.  If you click on the &quot;PDF&quot; link in the upper-right corner, you'll get the full text pdf of the article.  The article discusses the reaction pathways to the various products and also comments on the formation of 2,4-dichlorophenol from TCS as well as other reaction products, as you've illustrated.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-11T15:27:55.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16184" PostTypeId="2" ParentId="16116" CreationDate="2014-09-11T16:16:10.623" Score="0" Body="&lt;p&gt;As the sodium bicarbonate solution dries, it gives off carbon dioxide and water, leaving a residue of sodium carbonate. Sodium carbonate is hygroscopic, meaning it will attract moisture from the air. Moisture can cause paper to buckle and mildew, and that is good a reason not to use a baking soda bath to deacidify newsprint for archival purposes.&lt;/p&gt;&#xA;" OwnerUserId="7778" LastEditorUserId="7778" LastEditDate="2014-09-11T16:31:28.447" LastActivityDate="2014-09-11T16:31:28.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16185" PostTypeId="1" AcceptedAnswerId="16194" CreationDate="2014-09-11T16:40:42.163" Score="3" ViewCount="361" Body="&lt;p&gt;Is there a chemical input format that is-&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Supported by Open Babel (and hence supports visualization in avogadro??)&lt;/li&gt;&#xA;&lt;li&gt;Can represent a bond connecting an atom to its partner in a neighbouring unit cell, for crystals/systems with periodic boundary conditions&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h1&gt;Follow up (added later):&lt;/h1&gt;&#xA;&lt;p&gt;I could not understand how &lt;code&gt;cif&lt;/code&gt; does it, or &lt;code&gt;cml&lt;/code&gt;. But it would be nice if I could get rid of the symmetry folds in the markup given below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.crystallography.net/cif/1/00/00/1000022.cif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is the &lt;code&gt;cif&lt;/code&gt; file. The &lt;code&gt;cml&lt;/code&gt; output using &lt;code&gt;babel&lt;/code&gt; is --&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;?xml version=&amp;quot;1.0&amp;quot;?&amp;gt;&#xA;&amp;lt;molecule id=&amp;quot;Calcium titanate&amp;quot; xmlns=&amp;quot;http://www.xml-cml.org/schema&amp;quot;&amp;gt;&#xA; &amp;lt;crystal&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;a&amp;quot; units=&amp;quot;units:angstrom&amp;quot;&amp;gt;5.380000&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;b&amp;quot; units=&amp;quot;units:angstrom&amp;quot;&amp;gt;5.440000&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;c&amp;quot; units=&amp;quot;units:angstrom&amp;quot;&amp;gt;7.639000&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;alpha&amp;quot; units=&amp;quot;units:degree&amp;quot;&amp;gt;90.000003&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;beta&amp;quot; units=&amp;quot;units:degree&amp;quot;&amp;gt;90.000003&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;scalar title=&amp;quot;gamma&amp;quot; units=&amp;quot;units:degree&amp;quot;&amp;gt;90.000003&amp;lt;/scalar&amp;gt;&#xA;  &amp;lt;symmetry spaceGroup=&amp;quot;-P 2c 2ab&amp;quot;&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;1 0 0 0.5 0 -1 0 0.5 0 0 -1 0 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;-1 0 0 0 0 -1 0 0 0 0 1 0.5 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;-1 0 0 0.5 0 1 0 0.5 0 0 -1 0.5 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;-1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;-1 0 0 0.5 0 1 0 0.5 0 0 1 0 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;1 0 0 0 0 1 0 0 0 0 -1 0.5 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;   &amp;lt;transform3&amp;gt;1 0 0 0.5 0 -1 0 0.5 0 0 1 0.5 0 0 0 1&amp;lt;/transform3&amp;gt;&#xA;  &amp;lt;/symmetry&amp;gt;&#xA; &amp;lt;/crystal&amp;gt;&#xA; &amp;lt;atomArray&amp;gt;&#xA;  &amp;lt;atom id=&amp;quot;a1&amp;quot; elementType=&amp;quot;Ti&amp;quot; formalCharge=&amp;quot;4&amp;quot; xFract=&amp;quot;0.000000&amp;quot; yFract=&amp;quot;0.500000&amp;quot; zFract=&amp;quot;0.000000&amp;quot;/&amp;gt;&#xA;  &amp;lt;atom id=&amp;quot;a2&amp;quot; elementType=&amp;quot;Ca&amp;quot; formalCharge=&amp;quot;2&amp;quot; xFract=&amp;quot;0.006480&amp;quot; yFract=&amp;quot;0.035600&amp;quot; zFract=&amp;quot;0.250000&amp;quot;/&amp;gt;&#xA;  &amp;lt;atom id=&amp;quot;a3&amp;quot; elementType=&amp;quot;O&amp;quot; formalCharge=&amp;quot;-2&amp;quot; xFract=&amp;quot;0.571100&amp;quot; yFract=&amp;quot;-0.016100&amp;quot; zFract=&amp;quot;0.250000&amp;quot;/&amp;gt;&#xA;  &amp;lt;atom id=&amp;quot;a4&amp;quot; elementType=&amp;quot;O&amp;quot; formalCharge=&amp;quot;-2&amp;quot; xFract=&amp;quot;0.289700&amp;quot; yFract=&amp;quot;0.288800&amp;quot; zFract=&amp;quot;0.037300&amp;quot;/&amp;gt;&#xA; &amp;lt;/atomArray&amp;gt;&#xA; &amp;lt;bondArray&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&amp;quot;a1 a4&amp;quot; order=&amp;quot;1&amp;quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&amp;quot;a4 a2&amp;quot; order=&amp;quot;1&amp;quot;/&amp;gt;&#xA; &amp;lt;/bondArray&amp;gt;&#xA;&amp;lt;/molecule&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-09-12T14:55:57.127" Title="Represent bonds spanning a unit cell boundary plane" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16186" PostTypeId="1" CreationDate="2014-09-11T17:01:45.397" Score="3" ViewCount="3189" Body="&lt;p&gt;If you look at the trend in orbital energies as you go across a period the pattern is clear (orbital energy decreases with increasing effective nuclear charge) and, to my knowledge, it has no anomalies like the trend in ionization energies does when a 2p orbital is filled for the first time with two electrons (when this happens the ionization energy decreases). I expected that the graphs for ionization energy would be mirror images in the x-axis given that orbital energies are negative and measured relative to an ionized electron (which equals 0) albeit with different y-scales because ionization energy is the energy to ionize a mole of a substance in a gaseous state. Is my understanding that orbital energy = -(energy required to remove one electron from that shell) incorrect? Fundamentally, the question is, why doesn't the trend in orbital energies mirror the trend in ionization energies? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-09-12T15:59:42.247" Title="Whats the difference between ionization energy and orbital energy?" Tags="&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16187" PostTypeId="1" AcceptedAnswerId="16202" CreationDate="2014-09-11T20:35:51.003" Score="5" ViewCount="3130" Body="&lt;p&gt;I really like electrolysis. Today I bought 30% sulphuric acid to electroplate something with pieces of copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xv6li.png&quot; alt=&quot;copper electroplating&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've poured half water and half acid in a glass. I connected some cuprextit-copper plate to the anode and outdated aluminium coin to the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After starting the process, I could see that nothing is appearing on the cathode. I scratched the aluminium of the oxides. After repeating the electrolysis, aluminium turned black wherever scratched. &lt;strong&gt;Then I found out I swapped the wires.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once I had the wires correct, this happened to my cathode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3Eec.jpg&quot; alt=&quot;copper dust on cathode&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dust can be easily washed off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the image from wikipedia, any metal should be &quot;electroplateable&quot;. I actually thought that any conductor that doesn't react with the solution can be electroplated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the reality? Was I just impatient? Is electroplating only limited to some conductors? What about non metal ones? Can I electroplate carbon electrode?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1677" LastEditDate="2014-09-11T21:37:03.633" LastActivityDate="2014-09-12T01:38:21.133" Title="Can every conductor be electroplated using copper and sulphuric acid?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16188" PostTypeId="1" AcceptedAnswerId="16223" CreationDate="2014-09-11T21:09:11.977" Score="6" ViewCount="5484" Body="&lt;p&gt;I have seen in textbooks and videos that an electron must absorb energy (become excited) to enter a farther-away orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of energy that must be gained is equal to the difference in energy between the orbitals. When the electron relaxes and returns to its ground state, it emits that energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am not sure I understand why &quot;higher&quot; orbitals are of greater energy, for a couple of reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate &lt;em&gt;less&lt;/em&gt; energy as distance increases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also know that the electron is attracted to the nucleus due to opposing charge. By this logic, an electron closer to the nucleus would be more stable (have a lower energy state) than an electron farther from the nucleus (which would thus have a higher energy state). This line of thinking is somewhat similar to how a ball that's still high in the air is in a higher energy state than a ball that's closer to the ground.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So #1 is inconsistent with what we are taught about electrons and orbitals, while #2 would support it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attempting to clarify the misunderstanding I must have somewhere (possibly not making a proper analogy) - and better understand why higher orbitals are of greater energy.&lt;/p&gt;&#xA;" OwnerUserId="7820" LastActivityDate="2014-09-12T15:44:00.973" Title="Why do higher orbitals have more energy?" Tags="&lt;orbitals&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16189" PostTypeId="1" CreationDate="2014-09-11T21:43:14.727" Score="1" ViewCount="1072" Body="&lt;p&gt;I am interested in writing and revealing a message in invisible ink for theatrical purposes. I found a list of potential compounds &lt;a href=&quot;http://chemistry.about.com/od/chemistryhowtoguide/a/invisibleinks.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; but I am having a hard time prioritizing which to experiment/rehearse with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The message will be written on a lab coat and revealed in the dim lighting of a night-time bonfire so the ink must be dark enough to see easily. Because I will be writing on cloth, I will need an ink that will not run, so that when the message is revealed, it will be legible and not distorted. &#xA;Heat-activated inks are preferable because I will have the bonfire on hand for the reveal but there is the danger of revealing the writing too soon.&#xA;Chemically-activated inks are less desirable because of the less-theatrical reveal and because many of the revealing-solutions look like they will stain the coat making the message difficult to read but I will keep them as candidates because I have not tested my staining hypothesis and because of the added control they offer.&#xA;Blacklight-activated inks are not suitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have enough lab coats to test all the compounds listed so my question(s) are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which of the inks listed are least likely to run when applied to cloth?&lt;/li&gt;&#xA;&lt;li&gt;Of the remaining inks, which will appear darkest when revealed?&lt;/li&gt;&#xA;&lt;li&gt;Of the remaining heat-activated inks, which are easiest/hardest to reveal, in terms of temperature/time required?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="5026" LastEditDate="2018-07-24T01:42:30.527" LastActivityDate="2018-07-24T01:42:30.527" Title="What is the best invisible ink for clothing?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;applied-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16190" PostTypeId="1" AcceptedAnswerId="16193" CreationDate="2014-09-11T21:48:32.240" Score="4" ViewCount="1029" Body="&lt;p&gt;Why is a phosphorus atom with three different substituents chiral, but nitrogen isn't? Nitrogen inverts fast, while phosphorus obviously much slower. How is that explained?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-11T22:14:33.277" Title="Chirality of heteroatoms" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2016-09-08T07:52:10.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="16191" PostTypeId="1" AcceptedAnswerId="16199" CreationDate="2014-09-11T21:49:05.680" Score="2" ViewCount="1450" Body="&lt;p&gt;There are no other electrons to collide, repel and &lt;em&gt;kick&lt;/em&gt; Hydrogen's single electron to a distant nucleus. And that a single electron is tightly attracted to the nucleus by the electrostatic energy between them. So it seems to me, that Hydrogen does not require ionization energy at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I checked on the sources - the Hydrogen's ionization energy is relatively high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even Hydrogen makes $NaH$, which just doesn't stay for long and has extreme need for becoming $Na$ metal and Hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case why can't we say that Hydrogen does not require ionization energy? &lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2014-09-12T16:27:47.710" LastActivityDate="2014-09-12T16:27:47.710" Title="Why is the ionization energy for Hydrogen non-zero?" Tags="&lt;hydrogen&gt;&lt;covalent-compounds&gt;&lt;ionization-energy&gt;&lt;electrostatic-energy&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16192" PostTypeId="2" ParentId="16190" CreationDate="2014-09-11T22:02:58.000" Score="4" Body="&lt;p&gt;It's exactly as you say. Some heteroatoms can be chiral if they have 4 different substituents (or 3 different substituents and a lone pair).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with nitrogen is that it rapidly interconverts at normal temperatures. The inversion barrier is ~24 kJ/mol and quantum effects also play a role in the conversion. The barrier for phosphorous is much, much higher. It's a larger atom so the bonds are usually longer, and the lone pair occupies a larger volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's actually a nice Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;nofollow&quot;&gt;nitrogen inversion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, molecules like &lt;a href=&quot;http://en.wikipedia.org/wiki/Helicene&quot; rel=&quot;nofollow&quot;&gt;helicenes&lt;/a&gt; are chiral (i.e., with a right-handed or left-handed helicity), but like nitrogen can interconvert between forms through inversion.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-11T22:02:58.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16193" PostTypeId="2" ParentId="16190" CreationDate="2014-09-11T22:09:16.387" Score="7" Body="&lt;p&gt;Typical amines ($\ce{NR3}$) are roughly $\ce{sp^3}$ hybridized.  A chiral amine must undergo nitrogen inversion in order to racemize. The transition state for nitrogen inversion is a planar $\ce{sp^2}$ structure.  So in order to invert the ligands attached to nitrogen we must move from an $\ce{sp^3}$ ground state with &lt;em&gt;ca&lt;/em&gt;. 109 degree bond angles to an $\ce{sp^2}$ transition state with 120 degree bond angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand typical $\ce{PR3}$ compounds are roughly unhybridized with 90 degree bond angles in the ground state.  In order to racemize they must also invert at phosphorous which again requires an $\ce{sp^2}$ transition state with 120 degree bond angles.  Changing the geometry from 90 to 120 degrees in the case of phosphorous requires more energy that a distortion from 109 to 120 degrees in the case of nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Put a chiral nitrogen in a 3-membered ring and chirality can be preserved because you've now made it so much more difficult to expand the 60 degree bond angle in the ring to 120 degrees.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-11T22:14:33.277" LastActivityDate="2014-09-11T22:14:33.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16194" PostTypeId="2" ParentId="16185" CreationDate="2014-09-11T22:11:15.893" Score="6" Body="&lt;p&gt;As the developer of both Open Babel and Avogadro I can say the answer is &quot;not yet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main thing isn't the file format. It's that Avogadro at the moment doesn't have support at the moment to show bonds across a unit cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there was a patch for that, but it was too slow for typical use:&#xA;&lt;a href=&quot;https://github.com/dlonie/avogadro/commits/ENH_intercell_bonds&quot;&gt;https://github.com/dlonie/avogadro/commits/ENH_intercell_bonds&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-11T22:11:15.893" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16195" PostTypeId="2" ParentId="5173" CreationDate="2014-09-11T22:31:54.170" Score="3" Body="&lt;p&gt;Actually, it's not too hard. One of the display types in Avogadro is &quot;van der Waals&quot; which will just show atoms - no bonds. So you'd enable the vdW display type and then turn off the &quot;balls and sticks&quot; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you could write code which removes bonds, but why?&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-11T22:31:54.170" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16196" PostTypeId="2" ParentId="15339" CreationDate="2014-09-12T00:06:52.747" Score="4" Body="&lt;p&gt;Having accepted Abel Friedman's very sound answer, I wanted to append a coda of sorts. After posting this question, I tested all my opened peroxide-able ethers (diethyl, THF), and there were no measurable peroxides despite being well past the 1 year mark. I double-checked that the test was valid with hydrogen peroxide, which produced a strong positive for peroxides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I conclude that the standard recommendations for peroxides are similar to the warnings on MSDS sheets, in that they err on the side of extreme caution. As Abel Friedman says, peroxides are no fun! So testing and careful storage (full airtight containers, argon in the headspace, dark cool location, inhibitors) are essential. But if you do these things, the &quot;store-by&quot; dates do seem to be excessive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the valid point that I might not be turning over my solvents fast enough, I realized that the converse could be true in a working lab. That is, solvents are used frequently, and thus are left out in the light and opened frequently. If these are not used quickly enough, there is a real risk of peroxides. But it would seem that dark airtight storage (and inhibitors!) does retard peroxide formation well beyond the recommended limits.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2014-09-12T00:14:20.907" LastActivityDate="2014-09-12T00:14:20.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16197" PostTypeId="2" ParentId="5173" CreationDate="2014-09-12T00:38:39.070" Score="2" Body="&lt;p&gt;This is a minor comment, but I don't have the reps for it. (Any mods prepared to change this into a comment?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility is to convert the structure file to &lt;code&gt;.xyz&lt;/code&gt; using &lt;code&gt;babel&lt;/code&gt; and re-loading it in Avogadro, if the bond information is to be removed deeply. Otherwise adjusting the bond and atom radii should be able to make the bonds unobtrusive!&lt;/p&gt;&#xA;" OwnerUserId="1650" LastEditorUserId="1650" LastEditDate="2014-09-12T00:43:46.680" LastActivityDate="2014-09-12T00:43:46.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16198" PostTypeId="2" ParentId="16159" CreationDate="2014-09-12T00:50:08.210" Score="3" Body="&lt;p&gt;Since it contains agar, you wouldn't be able to make a stock solution with the agar in it; but you can make a stock solution of the other items and add the agar when you need the working solution. Also, since there is a fair amount of sugar in the recipe, it would be best to add that to the working solution as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to make a 100 times concentrate (leaving out the sugar and agar) and then sterilize it in an autoclave. When you need to make a working solution, weigh 5 g of agar and 20 g of sugar, add 10 mL of stock solution, and about 900 mL of water. Boil the solution to dissolve the agar, cool (to about 50°C) and add water (before it sets) to get 1000 mL of solution.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-12T00:50:08.210" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16199" PostTypeId="2" ParentId="16191" CreationDate="2014-09-12T01:03:12.047" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With ionization energy, an electron is not &quot;kicked out&quot; by other electrons, but rather it is &quot;the energy required for the electron to 'climb out' and leave the atom.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the electron &quot;is drawn inwards by positive electrostatic potential,&quot; it would make sense to infer that the more &quot;tightly drawn&quot; the electron is to the nucleus, the more energy it would require to &quot;climb out.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More of this can be found on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy#Electrostatic_explanation&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-09-12T02:11:21.000" LastActivityDate="2014-09-12T02:11:21.000" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16202" PostTypeId="2" ParentId="16187" CreationDate="2014-09-12T01:38:21.133" Score="8" Body="&lt;p&gt;While you can use basically any conductive material as an electrode to reduce the metal ions in solution, to actually plate the electrode with the metal, the metal has to be able to attach to the surface. If you have contaminants like surfactants, it can prevent deposition. The trouble with using aluminum as a cathode is that it readily forms an oxide layer that can be difficult to stick things to. I understand that it's possible to electroplate copper onto aluminum, but it's usually done with copper cyanide baths, so you may find it easier to plate copper onto steel or nickel. The other issue is that reduction of the copper ion is not the only possible reaction at the cathode. If your voltage is too negative, you may get significant hydrogen evolution and the bubbles will prevent a solid film from adhering properly. (which looks like it may be the problem in your photo) What you want is to deposit very slowly by keeping your voltage down so that you minimize gas evolution, though you may still have trouble with an aluminum cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the sorts of things that can be electroplated, many metals have similar behaviour where it's possible to produce quite high quality films of a second metal over top. Other metals like mercury and bismuth tend to form alloys or intermetallic compounds. It is certainly possible to deposit thin metal films on materials like indium tin oxide, graphite or glassy carbon, thought adherence can be a problem.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-12T01:38:21.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16203" PostTypeId="1" AcceptedAnswerId="16285" CreationDate="2014-09-12T06:12:33.330" Score="2" ViewCount="755" Body="&lt;p&gt;Background information:&#xA;I'm not sure if this question belongs here; but i'll risk the down-votes to ask it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently doing an undergraduate program, and it's in chemistry. I am torn between doing chemistry and biology and ideally the nicest compromise would be biochemistry. However I have a horrible ineptitude for biochemistry. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question is, what subfields of chemistry work very closely with biology? &lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorUserId="11367" LastEditDate="2017-01-14T14:47:40.317" LastActivityDate="2020-01-08T02:49:18.500" Title="What are the subfields of chemistry that work very closely with biology?" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16204" PostTypeId="2" ParentId="16191" CreationDate="2014-09-12T06:21:59.877" Score="3" Body="&lt;p&gt;The ionization energy of an atome is the minimum energy required to remove an elelectron from the atome in the gaseous state.&#xA;In spite of the fact that there are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. That  single electron is tightly attracted to the nucleus by the electrostatic energy between them. So the electron would require energy to get rid of the electrostatic attractive force of the nucleus and escape from the atome. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-12T06:21:59.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16205" PostTypeId="1" CreationDate="2014-09-12T09:37:50.853" Score="1" ViewCount="778" Body="&lt;p&gt;Since we are able to determine the enthapy of formation experimentally, why it is impossible to determine the lattice enthalpy directly by experiment?&lt;/p&gt;&#xA;" OwnerUserId="7760" LastEditorUserId="73" LastEditDate="2014-09-13T08:49:20.767" LastActivityDate="2014-09-14T07:40:59.433" Title="Can lattice enthalpy only be determined by the Born-Haber cycle?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16207" PostTypeId="1" CreationDate="2014-09-12T11:01:28.933" Score="18" ViewCount="10353" Body="&lt;p&gt;How do we know that there are different types of orbitals? For example, what evidence is there for the existence of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals instead of there being multiple &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbitals (for example, why isn't the electronic configuration of sodium &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1s^1, 2s^2, 2s^2, 2s^2, 2s^2, 3s^2}$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1s^2 2s^2 2p^6 3p^1}$&lt;/span&gt;)? &lt;/p&gt;&#xA;" OwnerUserId="7825" LastEditorUserId="-1" LastEditDate="2019-11-10T11:59:15.693" LastActivityDate="2021-02-21T12:35:14.473" Title="Evidence of orbitals?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16208" PostTypeId="1" AcceptedAnswerId="16210" CreationDate="2014-09-12T11:27:35.880" Score="1" ViewCount="2435" Body="&lt;p&gt;$\ce{Mg}$ powder is stirred into $\ce{CuSO4}$, what is the balanced molecular equation for this reaction ? I have $\ce{Mg + CuSO4 -&amp;gt; Mg2SO4 + Cu}$. Also, is the ionic equation the same as the net ionic equation? I believe it is not, because Copper would act as a spectator ion, however, I am not too sure. &lt;/p&gt;&#xA;" OwnerUserId="7621" LastEditorUserId="865" LastEditDate="2021-05-22T19:05:10.040" LastActivityDate="2021-05-22T19:05:10.040" Title="Magnesium powder mixed with CuSO4 solution" Tags="&lt;aqueous-solution&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16209" PostTypeId="2" ParentId="16207" CreationDate="2014-09-12T11:33:25.817" Score="11" Body="&lt;p&gt;The answer lies with experimental chemistry, specifically successive ionisation energies (i.e how much energy is required to remove the first electron, the second electron, the third electron and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4QOvp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QOvp.gif&quot; alt=&quot;Plot of first ionistaion energy vs. atomic number&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each point on the graph corresponds to an element. The first one is hydrogen, the second is helium.&#xA;The height of each point shows how much energy is required to remove the first electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can note that generally, the energy increases over a period. This is because In each successive element there is one more proton, and this stronger nuclear charge 'hold the outer electrons' more tightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to explain your question. Observe that within a period (e.g the 3rd dot to the 10th), it is not a constant increase. You can see that between the 4th-5th there is a slight drop, likewise between the 7th-8th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation for this is the sub orbitals. Some knowledge you need to know is that electron sub shells are only stable when empty, full, or half full (if you need an explanation for this comment on it later). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us examine the fourth dot, which represents Beryllium. It has an electronic config of 1s2 2s2. All of its sub orbitals are full, meaning that it is quite stable. &#xA;Compare it to the 5th dot, boron. Boron has a config of 1s2 2s2 3p1. Now the P orbital has room for 6 electrons, but this only has 1! It is not happy. It is not full or half full. Because of this it is trying to 'get rid' of the electron to be more stable. That's why it doesn't require so much energy to remove the outer electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decrease between the 6th and 7th is explained by the fact that the p orbital is stable when empty, full or half full. The 7th dot (nitrogen) has 3 electrons in its p orbital (half full). Contrasted with oxygen, which has 4/6. This is not stable, so less energy is required to remove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL-DR:  By analysing the ionisation energy graphs, we can see patterns that can be explained by sub orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a more basic/complex explanation, comment.&lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="7813" LastEditDate="2019-05-30T21:03:37.693" LastActivityDate="2019-05-30T21:03:37.693" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16210" PostTypeId="2" ParentId="16208" CreationDate="2014-09-12T11:45:59.437" Score="6" Body="&lt;p&gt;Your equation is very close, but you it is not $\ce{Mg2SO4}$. $\ce{Mg^{2+}}$ has a charge of $2+$, and $\ce{SO4^{2-}}$ has a charge of $2-$. They combine in a ratio of 1:1, so the result is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + CuSO4 -&amp;gt; MgSO4 + Cu} ~ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ionic equation does not include the $\ce{SO4^{2-}}$, and in the case of this equation the net ionic equation is the same as the ionic equation. Ionic equations look at what happens to electrons.&#xA;In this case, electrons are being transferred from the magnesium to the copper (since the copper is bonded to $\ce{SO4^{2-}}$ it is in the form $\ce{Cu^2+}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mg + Cu^{2+} -&amp;gt; Mg^{2+} + Cu} ~ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the charge on the $\ce{SO4^{2-}}$ doesn't change, it's just a spectator ion. &lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="4945" LastEditDate="2014-09-13T11:44:04.850" LastActivityDate="2014-09-13T11:44:04.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16211" PostTypeId="2" ParentId="16175" CreationDate="2014-09-12T12:48:57.927" Score="8" Body="&lt;p&gt;Virtual orbitals are automatically 'given' by the solved eigenvalue equations and are necessary for solving said equations. These orbitals are important for some post-HF methods such as CI and perturbation theory and coupled-cluster theory.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-12T19:53:33.547" LastActivityDate="2014-09-12T19:53:33.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16212" PostTypeId="1" CreationDate="2014-09-12T13:13:05.667" Score="1" ViewCount="252" Body="&lt;p&gt;Which type of citric acid crystalline powder do they sell in grocery stores?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the concentration of citric acid in the crystals 100%?Is it anhydrous or a monohydrate?&lt;/p&gt;&#xA;" OwnerUserId="7827" LastEditorUserId="73" LastEditDate="2014-09-13T08:47:39.473" LastActivityDate="2014-09-16T07:13:21.393" Title="Grocery store crystalline citric acid" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16213" PostTypeId="2" ParentId="16207" CreationDate="2014-09-12T13:45:22.843" Score="10" Body="&lt;p&gt;Oh, I suspect someone could come up with a &quot;s-orbital only&quot; view of chemistry, much like people came up with complicated models for an Earth-centered view of the universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the simple fact that solving quantum mechanics for a hydrogen atom gives you solutions for s, p, d, f, g.. orbitals, and even the degeneracy (i.e., that there is one s-type orbital, three p-type, five d-type, etc.) based on the angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As described above, solutions for the many-electron equations (albeit approximate) match up very nicely with experimental observations for ionization energies, electron affinities, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an elegant theory and experiment agree to a remarkable degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, we know that there must be non-spherical orbitals because we see molecular shapes with bond angles. I cannot come up with any way to describe a tetrahedral methane (let alone anything else) without some sort of non-spherical orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, when we look at reactivities of molecules, we see reactions occurring where we predict lone pairs or radical spin-density, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While orbitals are truly a mathematical construct, we find they are incredibly predictive of a wide range of chemistry. So my response would be &quot;how can we think there are &lt;strong&gt;not&lt;/strong&gt; different kinds of orbitals?&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-12T13:45:22.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16214" PostTypeId="2" ParentId="7765" CreationDate="2014-09-12T14:38:07.340" Score="9" Body="&lt;p&gt;A few more links to those listed above:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1186/1758-2946-1-8&quot;&gt;Peter Ertl's synthetic accessibility score (SAscore)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.molecular-networks.com/products/sylvia&quot;&gt;SYLVIA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.simbiosys.com/caesa/&quot;&gt;CAESA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.simbiosys.ca/archem/&quot;&gt;ArChem&lt;/a&gt; - also called Route Designer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There was also a lot of hype a few years ago about an effort called &lt;a href=&quot;http://en.wikipedia.org/wiki/Chematica&quot;&gt;Chematica&lt;/a&gt;, but I have not seen an actual shipping product (commercial or otherwise).&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-12T14:38:07.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16215" PostTypeId="1" AcceptedAnswerId="16219" CreationDate="2014-09-12T14:49:23.090" Score="31" ViewCount="32588" Body="&lt;p&gt;In my textbook it is written that the order of basic strength of pnictogen hydrides is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH3 &amp;gt; PH3 &amp;gt; AsH3 &amp;gt; SbH3 &amp;gt; BiH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried but could not find any explanation as to why this happens. What is the explanation?&lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorUserId="5026" LastEditDate="2018-04-22T10:40:34.737" LastActivityDate="2019-06-18T11:09:26.713" Title="Why does basicity of group 15 hydrides decrease down the group?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;pnictogen&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16216" PostTypeId="2" ParentId="16185" CreationDate="2014-09-12T14:55:57.127" Score="2" Body="&lt;p&gt;As pointed out by Geoff Hutchinson, this isn't currently possible by Avogadro and OpenBabel - and he should know best ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you want to visualize a feature of crystalline systems, you might want to have a look at specialized programmes that predominantly/only operate on &lt;strong&gt;CIF&lt;/strong&gt; files. We have used &lt;a href=&quot;http://www.krist.uni-freiburg.de/ki/Mitarbeiter/Keller/schakal.html&quot; rel=&quot;nofollow&quot;&gt;SCHAKAL&lt;/a&gt;, written by Egbert Keller, for this purpose in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of my mind, I don't remember how it was done. But it can be done, just like almost anything else you can imagine, such as centroid-centroid bonds between rings, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be warned: &lt;strong&gt;SCHAKAL&lt;/strong&gt;'s GUI violates any design rules (even if you apply Klingon standards). Once you have it compiled and running, switch the GUI off and operate it through chains of unituitive one-key shortcut commands. If you have overcome the horrors, you'll be rewarded with a fantastic postscript output in publication quality.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-12T14:55:57.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16217" PostTypeId="2" ParentId="16207" CreationDate="2014-09-12T15:14:38.840" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do we know that there are different types of orbitals?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the early days of atomic spectroscopy scientists were able to explain the spectral properties of elements like hydrogen and sodium.  Atoms of these elements had a single electron in the outermost shell and produced spectra that fit with (what we now view as) the relatively simplistic &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow&quot;&gt;Bohr theory&lt;/a&gt; of the atom.  When more complex atoms were examined spectroscopically, fine structure was observed that the Bohr theory could not explain.  These spectral lines were often referred to as &lt;strong&gt;s&lt;/strong&gt;harp, &lt;strong&gt;p&lt;/strong&gt;rincipal, &lt;strong&gt;d&lt;/strong&gt;iffuse and &lt;strong&gt;f&lt;/strong&gt;undamental.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As quantum mechanics emerged, the concept of the 4 quantum numbers being needed to describe an electron surfaced.  It was found that the spectral fine structure could now be explained when these various quantum states were taken into account.  To honor the work done by the early spectroscopists (that led to the development of a better theory), scientists used their &quot;&lt;strong&gt;s, p, d, f&lt;/strong&gt;&quot; notation to describe the various values for the angular momentum quantum number $\ell$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So although the observations made by the early spectroscopists couldn't be explained at the time, it turns out that they were actually observing spectral transitions involving other orbitals (p, d, f) in addition to those involving the s orbital.  &lt;strong&gt;The observation of this spectral fine structure requires the presence of orbitals that are different from s orbitals.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-09-24T03:35:21.203" LastActivityDate="2016-09-24T03:35:21.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16218" PostTypeId="1" CreationDate="2014-09-12T15:19:23.670" Score="0" ViewCount="1362" Body="&lt;p&gt;Copper is a metallic element, so it is likely to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. Poor conductor of electricity       B. A substance with a low melting point      C. Malleable     D. Poor conductor of heat     E. An element that forms negative ions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, C should be the only right. But are metallic elements really likely to be malleable?&lt;/p&gt;&#xA;" OwnerUserId="7747" LastEditorUserId="7717" LastEditDate="2014-09-12T17:00:43.583" LastActivityDate="2014-09-12T17:00:43.583" Title="Copper is a metallic element, so it is likely to be?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16219" PostTypeId="2" ParentId="16215" CreationDate="2014-09-12T15:25:11.343" Score="32" Body="&lt;p&gt;Each of these molecules has a pair of electrons in an orbital - this is termed a &quot;lone pair&quot; of electrons.  It is the lone pair of electrons that makes these molecules nucleophilic or basic.  As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell).  As the &lt;strong&gt;electron density of the lone pair&lt;/strong&gt; is spread over a greater volume and is consequently &lt;strong&gt;more diffuse&lt;/strong&gt;, the lone pair of electrons becomes less nucleophilic, &lt;strong&gt;less basic&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-04-16T19:30:41.907" LastActivityDate="2018-04-16T19:30:41.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16221" PostTypeId="2" ParentId="16218" CreationDate="2014-09-12T15:27:05.520" Score="1" Body="&lt;p&gt;Yes, It is C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Generally good conductors of electricity &#xA;-Substances with high melting points since they are mostly in liquid/solid states. Only liquid metal I can recall offhand is probably mercury, at room temperature and pressure of course.&#xA;-They form positive ions because they usually lose electrons to gain octet configuration.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;well, Wikipedia says &quot;Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples of malleable metals are Gold, Silver and Iron.&lt;/p&gt;&#xA;" OwnerUserId="7787" LastActivityDate="2014-09-12T15:27:05.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16222" PostTypeId="2" ParentId="16218" CreationDate="2014-09-12T15:35:19.547" Score="0" Body="&lt;p&gt;yes your answer is right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;well for your second question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Malleability, a similar property, is a material's ability to deform under compressive stress; this is often characterized by the material's ability to form a thin sheet by hammering or rolling&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so basically you have to deal with some physics of elasticity here. Yes generally metals are ductile though there are exceptions like sodium, potasium etc.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-09-12T15:35:19.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16223" PostTypeId="2" ParentId="16188" CreationDate="2014-09-12T15:44:00.973" Score="2" Body="&lt;p&gt;potential energy is negative between infinity to nucleus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It varies with .Thus, U∝(−1)1/r so it means the father you go from nucleus the value become less negative that is higher in energy than equilibrium energy.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-09-12T15:44:00.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16224" PostTypeId="2" ParentId="16215" CreationDate="2014-09-12T15:45:00.873" Score="6" Body="&lt;p&gt;The hydrides of nitrogen family have one lone pair of electrons on their central atom. Therefore,they act as Lewis bases.As we go done the group, the basic character of these hydrides decreases. Nitrogen atom has the smallest size among the hydrides.Therefore the lone pair is concentrated on a small region and electron density on it is the maximum.Consequently, its electron releasing tendency is maximum.As the size of the central atom increases down the family, the electron density also decreases. As a result, the electron donating capacity or the basic strength decreases down the group.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="7475" LastEditDate="2016-11-08T16:54:01.723" LastActivityDate="2016-11-08T16:54:01.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16225" PostTypeId="2" ParentId="16186" CreationDate="2014-09-12T15:59:42.247" Score="5" Body="&lt;p&gt;First off, the idea that the orbital energy is the negative of the ionization energy is actually &lt;a href=&quot;http://en.wikipedia.org/wiki/Koopmans&amp;#39;_theorem&quot; rel=&quot;noreferrer&quot;&gt;Koopmans' theorem&lt;/a&gt;. It works qualitatively because of cancelation of errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I digress...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitals are single-electron constructions. That is, in Hartree-Fock theory, an orbital is a single-electron wavefunction assuming the average electrostatic potential of the remaining electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionization energy trends are experimental observables and you see effects of multi-electron interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that orbital energies are &lt;em&gt;not&lt;/em&gt; physical quantities (i.e., being single-electron mathematical results) and ionization energies are based on multi-electron interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the simplest way to understand the trends in ionization energies vs. orbital energies is to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater&amp;#39;s_rules&quot; rel=&quot;noreferrer&quot;&gt;Slater's rules&lt;/a&gt; and also remember &lt;a href=&quot;http://en.wikipedia.org/wiki/Hund&amp;#39;s_rules&quot; rel=&quot;noreferrer&quot;&gt;Hund's rules&lt;/a&gt;. That is, when we add electrons, we are partially shielding the nuclear charge, but we also have to consider the stabilization due to exchange of electrons and destabilization when we pair electrons in an orbital. These are multi-electron effects, and only some of that can be recovered in a simple one-electron picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $2p$ orbitals, as you mention, there's a decrease in the ionization potential when you put two electrons in the same orbital. This happens because this new electron does not benefit from the exchange effect (Hund's rules) and has an electrostatic repulsion with the electron in the same orbital. Thus it's easier to remove it than might be expected from a one-electron orbital picture.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-12T15:59:42.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16226" PostTypeId="2" ParentId="16218" CreationDate="2014-09-12T16:21:32.577" Score="1" Body="&lt;p&gt;Metals generally have properties like good conductor of electricity,good Thermal conductivity,shiny,sonorous,lustrous,malleable(bitten into thin sheets under stress),ductile(drawn into wires),brittle,have high density,high melting and boiling point,generally solid in nature,forms positive ions etc...         &quot;exceptions are there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;well your question is regarding choosing the right option.so,you are right in your thought.correct answer is option c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer regarding the second part of your,i think is &lt;strong&gt;no&lt;/strong&gt; as exceptions are always there in periodic table.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastActivityDate="2014-09-12T16:21:32.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16228" PostTypeId="2" ParentId="16207" CreationDate="2014-09-12T16:35:36.840" Score="21" Body="&lt;p&gt;Let me approach this another way than the others: orbitals are NOT physical objects! They do  not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn&amp;ndash;Sham / etc. equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitals are not unique: given linear combinations are equivalent with each other, and there is no &quot;correct orbitals&quot;, one can choose whichever they like. Canonical orbitals, natural orbitals etc are all good to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the evidence they exist? They do not exist, they are just mathematical solutions for given equations, and it is a purely mathematical question if they are good solutions for those equations or not. The theories themselves are consistent with experimental data, e.g. spectroscopic properties, geometries, reactivity. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="15751" LastEditDate="2019-05-29T16:04:47.043" LastActivityDate="2019-05-29T16:04:47.043" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="16229" PostTypeId="5" CreationDate="2014-09-12T16:41:57.910" Score="0" Body="&lt;p&gt;The ionization energy or ionization potential is the energy necessary to remove an electron from the neutral atom. It is a minimum for the alkali metals which have a single electron outside a closed shell. It generally increases across a row on the periodic maximum for the noble gases which have closed shells. For example, sodium requires only $\pu{496 kJ/mol}$ or $\pu{5.14 eV/atom}$ to ionize it while neon, the noble gas immediately preceding it in the periodic table, requires $\pu{2081 kJ/mol}$ or $\pu{21.56 eV/atom}$. The ionization energy is one of the primary energy considerations used in quantifying chemical bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See also:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/ionize.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperphysics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="5026" LastEditDate="2018-03-19T13:17:34.597" LastActivityDate="2018-03-19T13:17:34.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16230" PostTypeId="4" CreationDate="2014-09-12T16:41:57.910" Score="0" Body="The ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state. Do not confuse with [electron-affinity]." OwnerUserId="7450" LastEditorUserId="5026" LastEditDate="2018-03-19T13:02:15.070" LastActivityDate="2018-03-19T13:02:15.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16231" PostTypeId="1" CreationDate="2014-09-12T16:42:27.520" Score="2" ViewCount="372" Body="&lt;p&gt;What are the delta G values for the mammalian (human) biosynthesis of alanine from pyruvate, and for the conversion of oxaloacetate to asparagine? (assuming no particular cellular restriction or other systemic pressures like catalysis or anabolism)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that oxaloacetate must be converted to aspartate and then to asparagine.  Based on the fact there are two steps to that conversion, I assume that cumulatively the delta G value is higher for the oxaloacetate to finally get to asparagine. So I want to know how much of a difference there is in the delta G values of these two conversions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is NOT a homework question, it is to better my own understanding.  I would welcome peer-reviewed literature, or other resources if you deem them applicable.  &lt;/p&gt;&#xA;" OwnerUserId="7369" LastActivityDate="2014-09-12T16:42:27.520" Title="Delta G values for Pyruvate--&gt; Alanine vs. Oxaloacetate -&gt; Aspartate -&gt;Asparagine" Tags="&lt;biochemistry&gt;&lt;energy&gt;&lt;free-energy&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16232" PostTypeId="2" ParentId="16111" CreationDate="2014-09-12T16:46:31.317" Score="1" Body="&lt;p&gt;These are two possible forms of &lt;strong&gt;Alanine&lt;/strong&gt;(Ala||A), an amino acid with hydrophobic side chain. Here are the exact $pK_A$ values:&#xA;$$-COOH:2.33\\-NH_3^+:9.71$$&#xA;Compund two is stronger as you anticipated.Cheers!&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-09-12T16:46:31.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16233" PostTypeId="5" CreationDate="2014-09-12T16:47:00.617" Score="0" Body="&lt;p&gt;Electrostatic potential energy is created based on the law of electrostatic attraction and repulsion, or Coulomb's Law. It describes the force exerted on a charged object(e.g.electron) due to the presence of another charged object (e.g.nucleus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;aka electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system. An object may have electric potential energy by virtue of two key elements: its own electric charge and its relative position to other electrically charged objects. &quot;electrostatic potential energy&quot; is used to describe the potential energy in systems with time-invariant electric fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_potential_energy&quot; rel=&quot;nofollow&quot;&gt;1 ; description&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://guweb2.gonzaga.edu/faculty/cronk/CHEM198pub/L02-index.cfm?L02resource=energy&quot; rel=&quot;nofollow&quot;&gt;2 ; excerpt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2014-09-12T17:00:08.620" LastActivityDate="2014-09-12T17:00:08.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16234" PostTypeId="4" CreationDate="2014-09-12T16:47:00.617" Score="0" Body="Electrostatic potential energy is created based on the law of electrostatic attraction and repulsion, or Coulomb's Law. It describes the force exerted on a charged object(e.g.electron) due to the presence of another charged object (e.g.nucleus)." OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2014-09-12T17:00:01.230" LastActivityDate="2014-09-12T17:00:01.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16235" PostTypeId="1" AcceptedAnswerId="16250" CreationDate="2014-09-12T16:52:26.800" Score="2" ViewCount="822" Body="&lt;p&gt;Is this a valid mechanism? I think it is. It is just a simple acid base mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I increase yield though by using a stronger base than heavy water? Perhaps heavy sodium hydroxide? Or is chloroform sufficiently acidic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yb5xG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-12T21:14:37.237" LastActivityDate="2014-09-12T21:46:31.107" Title="Hydrogen and deuterium exchange" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16236" PostTypeId="5" CreationDate="2014-09-12T16:54:47.907" Score="0" Body="&lt;p&gt;A covalent bond is formed when atoms share one or more pairs of electrons. A molecule is the smallest unit of a substance that keeps all of the properties of that substance, and is formed of atoms covalently bonded to one another. Examples: $\ce{O2}$ and $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A covalent compound is a chemical substance made up of atoms of more than one different elements held together by chemical bonds. So $\ce{H2O}$ is a compound, but $\ce{O2}$ is not because it does not have more than one different element in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See also:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/add_edexcel/covalent_compounds/compoundsrev1.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;BBC&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://education-portal.com/academy/lesson/covalent-compounds-properties-naming-formation.html#lesson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Education portal&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="5026" LastEditDate="2018-03-19T17:13:01.267" LastActivityDate="2018-03-19T17:13:01.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16237" PostTypeId="4" CreationDate="2014-09-12T16:54:47.907" Score="0" Body="For questions about covalent compounds - which are compounds in which all intermolecular bonds between atoms are considered mostly covalent - have a stronger covalent than ionic or metallic component. This tag is not to be confused with [bond]." OwnerUserId="7450" LastEditorUserId="5026" LastEditDate="2018-03-19T17:13:15.750" LastActivityDate="2018-03-19T17:13:15.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16238" PostTypeId="2" ParentId="15890" CreationDate="2014-09-12T18:11:25.483" Score="2" Body="&lt;p&gt;There are a number of command-line utilities that can do this, but I'd suggest &lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;nofollow&quot;&gt;Open Babel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you can align the molecules and append or print the RMSD as described in the documentation &lt;a href=&quot;https://open-babel.readthedocs.org/en/latest/Command-line_tools/babel.html#aligning-molecules-or-substructures&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll point out that calculating an RMSD can be difficult if the frame of reference has changed or the order of atoms is different in the calculation. The Open Babel code will ignore such differences.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-12T18:11:25.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16242" PostTypeId="1" CreationDate="2014-09-12T19:42:00.950" Score="1" ViewCount="84" Body="&lt;p&gt;If all life relies on water, and water freezes at 32F and boils at 212F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why isn't the nominal temperature for life 122F?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;122F will literally kill some animals, including us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do our biochemical reactions need &quot;cold&quot; (most people prefer ~72F) for some reason?&lt;/p&gt;&#xA;" OwnerUserId="7073" LastActivityDate="2014-09-12T20:44:02.023" Title="How does life choose it's nominal temperature preferences?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16243" PostTypeId="1" AcceptedAnswerId="16249" CreationDate="2014-09-12T20:23:47.380" Score="3" ViewCount="2499" Body="&lt;p&gt;Apparently, water or &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; combines in some way with nitrogen (from the air or elsewhere?) to form ammonia, or  &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;. Where/how in the body is this &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; (or whatever nitrate) produced before it is excreted with urine?&lt;/p&gt;&#xA;" OwnerUserId="4957" LastEditorUserId="30836" LastEditDate="2019-04-14T14:09:24.173" LastActivityDate="2019-06-22T09:21:10.530" Title="Why does urine smell like ammonia?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16244" PostTypeId="1" AcceptedAnswerId="16256" CreationDate="2014-09-12T20:30:40.097" Score="9" ViewCount="344" Body="&lt;p&gt;It is (mostly) used as an &quot;oxidizer&quot; in motors and engines, I understand that much. The oxygen &quot;oxidizes&quot; with other chemicals, while the nitrogen returns to the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\ce{N2O}$ also produces &quot;euphoria&quot; in human beings, works as a mild anesthetic, and causes people to laugh. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="4957" LastEditorUserId="73" LastEditDate="2014-09-13T08:46:44.463" LastActivityDate="2016-04-29T08:27:26.727" Title="Why N2O is a &quot;laughing gas&quot;?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16245" PostTypeId="2" ParentId="16242" CreationDate="2014-09-12T20:44:02.023" Score="2" Body="&lt;p&gt;Here's a link to a brief &lt;a href=&quot;http://www.scientificamerican.com/podcast/episode/986-trades-metabolic-cost-for-funga-10-12-27/&quot; rel=&quot;nofollow&quot;&gt;Scientific American article&lt;/a&gt; that presents the results from a study of this question.  Basically, for every $\ce{1^{o}C}$ increase in body temperature, 6% fewer fungal species can survive - so the hotter, the better (the longer you're likely to survive).  The downside of &quot;hot&quot; is the increased metabolic rate required, which translates into more time spent consuming food.  If you were to plot the effect of these two competing effects on longevity &lt;em&gt;versus&lt;/em&gt; body temperature you would get something like an inverted (upside-down) parabola.  The study found a maximum longevity at 98.1 F, not far off from the observed average body temperature of 98.6 F.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-12T20:44:02.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16248" PostTypeId="1" CreationDate="2014-09-12T21:09:37.107" Score="2" ViewCount="9859" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A vinegar sample is found to have a density of $1.006 \mathrm{\frac{g}{mL}}$ and to contain $8.7~\%$ acetic acid by mass. How many grams of acetic acid are present in $\pu{2.80L}$ of this vinegar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I multiplied: $$\left(\frac{1.006~\mathrm{g}}{\mathrm{ml}}\right)\cdot \left(\frac{1000~\mathrm{ml}}{1~\mathrm{L}}\right)\cdot 2.80~\mathrm{L} = 2816.8~\mathrm{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then converted $8.7~\%$ into $0.087$ and multiplied it into $2816.8$, to get the answer $245~\mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="7475" LastEditDate="2018-02-21T04:50:17.173" LastActivityDate="2018-02-21T04:50:17.173" Title="How many grams of acetic acid are present in 2.80 L of this vinegar?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16249" PostTypeId="2" ParentId="16243" CreationDate="2014-09-12T21:12:43.753" Score="7" Body="&lt;p&gt;All animals (including humans) consume significant amount of proteins. Proteins consist of aminoacids, which include amine nitrogen. This nitrogen traces back to mineral fertilizers humans use in agriculture. As proteins in our bodies are destroyed, the waste is utilized, formine urea (carbamide), so ammonia didn't cause too much trouble. However, urea is slowly hydrolized in water back into ammonia, so old urine stinks of (among other things) ammonia.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="30836" LastEditDate="2019-06-22T09:21:10.530" LastActivityDate="2019-06-22T09:21:10.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16250" PostTypeId="2" ParentId="16235" CreationDate="2014-09-12T21:46:31.107" Score="3" Body="&lt;p&gt;If you are interested in making deutero-chloroform, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloroform#Deuterochloroform&quot; rel=&quot;nofollow&quot;&gt;better ways&lt;/a&gt; to do it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a purely mechanistic standpoint, chloroform isn't a very strong acid ($\ce{pKa \sim 15}$) so $\ce{D2O}$ is unlikely to deprotonate it at a useful rate.  Further, chloroform isn't very soluble in water making your desired reaction even less likely (but you could engineer your way around this problem by using a phase transfer catalyst).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your suggestion to use a stronger base is reasonable, but the problem you run into here is that a different reaction is brought into play.  Specifically, once a stronger base like potassium &lt;em&gt;t&lt;/em&gt;-butoxide (in &lt;em&gt;t&lt;/em&gt;-butanol) is used, dichlorocarbene formation occurs.  Once the trichloromethylcarbanion is formed, it quickly ejects $\ce{Cl^-}$ to form dichlorocarbene.  Dichlorocarbene is an extremely energetic species and would quickly react with whatever else is around - perhaps the t-butanol to produce dichloromethyl-t-butyl ether.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-12T21:46:31.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16251" PostTypeId="2" ParentId="9555" CreationDate="2014-09-12T22:28:19.000" Score="1" Body="&lt;p&gt;No, it's not an error. Take a look at the Tanabe Sugano diagram for $d^2$ (from Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/98/D2_Tanabe-Sugano_diagrams.png&quot; alt=&quot;Tanabe Sugano diagram for d2 from Wikipedia&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that again, the $^1D$ state is below the $^3P$ state. Why? If we were considering a $p^2$ configuration, the $^3P$ is indeed lower in energy than $^1D$. And again anyone &lt;strong&gt;might&lt;/strong&gt; predict that $^3P$ is lower for $d^2$ too, for the exact reasons you give.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know this result from spectroscopy. The difference is because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Racah_parameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Racah parameters&lt;/a&gt; for the different terms (i.e., the electrostatic repulsion between electrons in the same orbital. The Wikipedia article had a &lt;a href=&quot;http://www.cond-mat.de/events/correl12/manuscripts/eder.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice link&lt;/a&gt; to how you'd compute the difference in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now of course a $^3F \rightarrow ^1D$ transition is not an allowed transition, which is shown in the textbook on the next page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral of the story is that Hunds' rules and the basics will always get you the lowest energy term symbols for multi-electron atoms, but the details can be tricky for higher-level states.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.343" LastActivityDate="2014-09-12T22:28:19.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16252" PostTypeId="2" ParentId="16248" CreationDate="2014-09-13T01:05:25.767" Score="2" Body="&lt;p&gt;Your math and stoichiometry are correct.&#xA;$2.80L*(\frac{1000 ml}{1 L})*(\frac{1.006 g}{ ml})*8.7$%$=245g \checkmark$&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-13T01:05:25.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16256" PostTypeId="2" ParentId="16244" CreationDate="2014-09-13T06:39:12.477" Score="13" Body="&lt;p&gt;In contrast to $\ce{O2}$ (or $\ce{CO}$), $\ce{N2O}$ does not bind to hemoglobin and it is not metabolized in the human body. It is taken up through the lungs, gets in the blood and leaves the body the same way being exhaled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a 2004 study by &lt;em&gt;P. Nagele&lt;/em&gt; in &lt;em&gt;PNAS&lt;/em&gt; (&lt;a href=&quot;http://dx.doi.org/10.1073/pnas.0402825101&quot;&gt;DOI&lt;/a&gt;), $\ce{N2O}$ acts as an antagonist for the  $N$-methyl-$D$-aspartate (&lt;strong&gt;NMDA&lt;/strong&gt;) receptor. Other infamous antagonists are drugs like &lt;em&gt;ketamine&lt;/em&gt; or &lt;em&gt;phencyclidine&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information on NDMA receptor antagonists can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/NMDA_receptor_antagonist&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-13T06:39:12.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16258" PostTypeId="2" ParentId="16205" CreationDate="2014-09-13T09:43:11.957" Score="1" Body="&lt;p&gt;Because you can't take an ionic solid and convert it into a gas of ions. That's not a physically realistic process. You have to infer it using a Born-Haber cycle and other quantities that can be determined experimentally.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-13T09:43:11.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16259" PostTypeId="1" AcceptedAnswerId="16286" CreationDate="2014-09-13T12:02:43.127" Score="-4" ViewCount="812" Body="&lt;p&gt;How does $\ce{K2Cr2O7}$ become $\ce{Cr2O7^2-}$ when dissolved in water?&lt;/p&gt;&#xA;" OwnerUserId="6845" LastEditorUserId="16683" LastEditDate="2018-08-17T18:18:13.623" LastActivityDate="2018-08-27T11:13:28.473" Title="How is the dichromate anion generated from potassium dichromate?" Tags="&lt;aqueous-solution&gt;&lt;ions&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ClosedDate="2018-08-27T16:04:58.377" ContentLicense="CC BY-SA 4.0" />
  <row Id="16260" PostTypeId="1" AcceptedAnswerId="19689" CreationDate="2014-09-13T13:09:25.463" Score="3" ViewCount="28392" Body="&lt;p&gt;I have a formula for finding the final temperature in an irreversible adiabatic expansion:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_2=\left[\frac{C_V+\frac{P_2}{P_1}}{C_p}\right]T_1,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; is the initial temperature.&#xA;&lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_p$&lt;/span&gt; are the molar heat capacities at constant volume and constant pressure, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in how this formula is derived.&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="41328" LastEditDate="2020-04-29T13:32:46.437" LastActivityDate="2020-04-29T13:32:46.437" Title="Derivation of the relation between temperature and pressure for an irreversible adiabatic expansion" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16261" PostTypeId="1" AcceptedAnswerId="16274" CreationDate="2014-09-13T13:38:12.943" Score="16" ViewCount="24397" Body="&lt;p&gt;I would like to ask if anyone has a list or knows which covalent compounds have metals in them. For example, beryllium and aluminium are both metals but they bond covalently with chlorine to form beryllium chloride and aluminium chloride respectively. Other than these 2 compounds, I do not know any other and thus would like to ask this question.&lt;/p&gt;&#xA;" OwnerUserId="7835" LastEditorUserId="4945" LastEditDate="2014-09-13T15:56:02.713" LastActivityDate="2017-12-01T14:58:37.910" Title="Metal Compounds that bond covalently" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="4" ClosedDate="2017-02-26T23:33:54.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="16262" PostTypeId="1" AcceptedAnswerId="16266" CreationDate="2014-09-13T14:59:31.377" Score="4" ViewCount="152" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At $1000~\mathrm{K}$, $K_\mathrm p = 1.85$ for the reaction. &#xA;  $$\ce{SO2 + \frac12O2 &amp;lt;=&amp;gt; SO3}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the value of $K_p$ for the reaction&lt;br&gt;&#xA;  $$\ce{SO3 &amp;lt;=&amp;gt; SO2 + \frac12O2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, since the task gives us $K_\mathrm p$, it seems strange to ask for it again with the equilibrium put the other way around. Does this make sense? Is it a typo? Either that or my understanding is flawed. It would make more sense to ask for $K_\mathrm c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chapter has this equation: $K_\mathrm p = K_\mathrm c(RT)^{\Delta n}$, I think this is relevant given the temperature is available to solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; task 15.21 in the 11th edition of Chemistry: The Central Science, and the answer is suppose to be $0.541$.&lt;/p&gt;&#xA;" OwnerUserId="1448" LastEditorUserId="5026" LastEditDate="2018-02-25T02:35:25.413" LastActivityDate="2018-02-25T02:35:25.413" Title="Finding the pressure equilibrium constant of the reverse reaction" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16265" PostTypeId="2" ParentId="16261" CreationDate="2014-09-13T15:44:17.550" Score="0" Body="&lt;p&gt;Covalent bonding can be either two atoms of the same element becoming a molecule (single bond) or atoms from different elements bonding (covalent compound). Examples of atoms becoming a molecule are basically diatomic molecules such as H2, Cl2, I2, etc. Examples of more complex covalent compounds are methane, water, ammonia and carbon dioxide. There are also solid covalent compounds which are mainly allotropes of carbon such as diamond and graphite.&lt;/p&gt;&#xA;" OwnerUserId="7741" LastActivityDate="2014-09-13T15:44:17.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16266" PostTypeId="2" ParentId="16262" CreationDate="2014-09-13T15:53:36.400" Score="3" Body="&lt;p&gt;They are actually just looking for the equilibrium constant for the reverse reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p(-\xi) = K_p^{-1}(\xi) = \frac{1}{1.85} = 0.541$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-09-13T15:53:36.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16267" PostTypeId="1" AcceptedAnswerId="16268" CreationDate="2014-09-13T16:30:54.117" Score="55" ViewCount="9740" Body="&lt;p&gt;We've learnt that the electropositive element is written first. Then why is ammonia written as $\ce{NH3}$ ?&lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorUserId="16683" LastEditDate="2016-01-08T05:52:07.970" LastActivityDate="2016-02-11T17:47:44.493" Title="Why do we write NH3?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16268" PostTypeId="2" ParentId="16267" CreationDate="2014-09-13T16:45:14.707" Score="45" Body="&lt;p&gt;According to current nomenclature rules, $\ce{H3N}$ would be correct and acceptable.  However some chemical formulas, like $\ce{NH3}$ for ammonia, that were in use long before the rules came out, are still accepted today.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-13T16:45:14.707" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16273" PostTypeId="1" AcceptedAnswerId="16277" CreationDate="2014-09-13T20:38:46.230" Score="2" ViewCount="6776" Body="&lt;p&gt;I am told that para-dichlorobenzene exhibits &lt;em&gt;two&lt;/em&gt; proton NMR &lt;em&gt;signals&lt;/em&gt;. This doesn't jibe with me; all the protons on the molecule are equivalent. Here is a book claiming the same thing. Am I wrong, or is the person who told me that para-dichlorobenzene has two signals on its proton NMR wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/evfKh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-11-18T20:58:39.290" Title="para-dichlorobenzene - number of proton NMR signals" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16274" PostTypeId="2" ParentId="16261" CreationDate="2014-09-13T20:38:54.573" Score="16" Body="&lt;p&gt;There is no sharp line between ionic, metallic or covalent bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of transition metal oxides and sulfates have strong covalent characteristics. For example, the whole reason of using ligand field theory instead of crystal field theory is that the ionic description (CFT) breaks down, and the covalent effects are important in all but the simplest cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to take even more covalent molecules, look for organometallic compounds, metal carbonyls, metallocenes, etc.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5354" LastEditDate="2015-09-12T17:03:15.607" LastActivityDate="2015-09-12T17:03:15.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16276" PostTypeId="1" AcceptedAnswerId="16278" CreationDate="2014-09-13T21:16:21.607" Score="22" ViewCount="1938" Body="&lt;p&gt;Can someone explain to me &lt;em&gt;charge shift&lt;/em&gt; bonding? I recognize that it is proposed as another domain of bonding - one different from ionic and covalent bonding. I am also told that $\ce{C-F}$ exhibits charge shift bonding. But why—and firstly, what exactly &lt;em&gt;is&lt;/em&gt; charge shift bonding? All the literature I have consulted are at too high of a level for me, for example &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00046a035&quot; rel=&quot;noreferrer&quot;&gt;this 1992 article&lt;/a&gt; in the Journal of the American Chemical Society. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="11367" LastEditDate="2017-02-28T16:39:01.913" LastActivityDate="2017-08-22T08:26:51.890" Title="What is charge shift bonding?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16277" PostTypeId="2" ParentId="16273" CreationDate="2014-09-13T21:22:58.097" Score="7" Body="&lt;p&gt;Be very careful with how you interpret this molecule. Although the 1H NMR spectrum appears as a single line, this is actually an AA'A''A''' system, meaning that, although chemically equivalent,  all four protons are magnetically non-equivalent, but have coincidental chemical shift. The coupling between all protons is different; J&lt;sub&gt;A-A'''&lt;/sub&gt; is different to J&lt;sub&gt;A-A'''&lt;/sub&gt; is different to J&lt;sub&gt;A'-A''&lt;/sub&gt; is different to J&lt;sub&gt;A'-A'''&lt;/sub&gt; and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the spectrum is as you thought. p-dichlorobenzene IS a single peak in the 1H NMR spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a set of simulations for a generic para disubstituted benzene system, which is an AA'BB' system. A is chemically equivalent to A', but magnetically non-equivalent. A is chemically AND magnetically non-equivalent to B. Your system is just a special case example of these types of systems, with all protons being chemically equivalent, but magnetically non-equivalent. These are second order systems, which have a lot of complex additional splitting structure in their spectrum. Like all spin systems, as the chemical shift difference becomes less (the peaks move closer together), the tilting, (or tenting or roofing, depending on which text book you read) becomes more pronounced, and the wings on the main peaks become smaller and smaller. Eventually, they collapse into a single peak at very small chemical shift difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fine text which you can't read (sorry about that), gives the coupling constants for the system, which are bog-standard ortho/meta/para couplings: Jortho=7Hz, Jmeta=3Hz; Jpara=1Hz. The only variable in the simulations is the chemical shift of the downfield peak, which slowly moves towards its upfield partner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/D98Gz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-09-13T22:02:59.443" LastActivityDate="2014-09-13T22:02:59.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16278" PostTypeId="2" ParentId="16276" CreationDate="2014-09-13T21:45:04.547" Score="17" Body="&lt;p&gt;OK, this was a new one to me too, but I found a few resources which might be useful:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://comporgchem.com/blog/?p=406&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Computational Organic Chemistry blog post&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nasw.org/users/mslong/2009/2009_10/ChargeShift.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Scott Long blog post&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nchem/journal/v1/n6/full/nchem.327.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nature Chemistry review&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The basic idea is this. Most bonds are ionic (i.e., both atoms are charged, but electrostatically attracted) or covalent (i.e., the electron pair is shared).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is largely cribbed from the Comp. Org. Chem. blog post. Thanks to Steven Bachrach)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were writing a multi-configurational wavefunction to describe the bond, we might have something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi(\mathrm{VB}) = c_1\psi_\mathrm{covalent} + c_2\psi_{\ce{A+B-}} + c_3\psi_{\ce{A-B+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So normally, we'd look at which is the dominant configuration - maybe it's the covalent one ($c_1$) or one of the ionic ones ($c_2$ or $c_3$). Instead, sometimes the covalent form isn't even stabilizing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my perspective, the classic example given for this charge-shift bond is $\ce{F2}$ which of course has many $\pi^*$ anti-bonding electrons, so the classical bond order is 1, but ignores that there's a lot of antibonding character. But you can't really consider $\ce{F2}$ an ionic compound!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the big question is the future - can this new type of bonding show general predictability and utility. Clearly most chemists have a general intuition (if sometimes wrong) of covalent and ionic bonding. Besides the authors, I'm not sure if anyone else has an intuition of charge-shift systems.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="16683" LastEditDate="2017-08-22T08:21:54.317" LastActivityDate="2017-08-22T08:21:54.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16280" PostTypeId="1" AcceptedAnswerId="16283" CreationDate="2014-09-14T04:55:01.763" Score="-2" ViewCount="4013" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Battery acid has a density of $\pu{1.285 g cm^-3}$ and contains $38\%$ by weight sulfuric acid. How many grams of pure sulfuric acid are contained in a liter of battery acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I just took $1.285 \times 0.38$ and got how many grams of sulfuric acid are in $\pu{1 cm^3}$ of battery acid. So now I just converted it into grams per liter and got $\pu{488.3 gL^-1}$ of sulfuric acid in battery acid. I'm not sure if I did it right though.&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="4945" LastEditDate="2018-02-21T10:49:31.497" LastActivityDate="2018-02-21T10:49:31.497" Title="How to calculate the mass of a component in battery acid based on the density and mass fraction?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16283" PostTypeId="2" ParentId="16280" CreationDate="2014-09-14T06:44:18.663" Score="3" Body="&lt;p&gt;Yes, you did do it correctly. However, I would refer you to the homework policy of this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textrm{density} = \frac{\textrm{mass}}{\textrm{volume}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textrm{mass}=\textrm{density} \times  \textrm{volume} = 1.285 \times 1000=1285\ \textrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you just multiply by 0.38 (which is the same as 38%) and you get the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a note, next time you ought to phrase your question by asking about the specific topic of the question (e.g How do I calculate the mass given the percentage...... etcetc), rather than the question itself. &lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="14" LastEditDate="2014-09-15T12:58:37.707" LastActivityDate="2014-09-15T12:58:37.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16284" PostTypeId="2" ParentId="16205" CreationDate="2014-09-14T07:40:59.433" Score="0" Body="&lt;p&gt;What about theoretically calculating it from principles in electrostatics ex: as briefly noted on U Waterloo's &lt;a href=&quot;http://www.science.uwaterloo.ca/%7Ecchieh/cact/applychem/lattice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;webpage here&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;p&gt;For instance, for crystalline ionic compounds: &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Land%C3%A9_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Born-Lande&lt;/a&gt; and/or &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Mayer_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Born-Mayer&lt;/a&gt; equations can provide results close to experimentally determined values. The two equations both utilize the Madelung constant, which simplifies the electrostatic calculations for organized crystals. Of course these equations bring up new constants :)&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Kapustinskii_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kapustinskii&lt;/a&gt; took a slightly different approach to determining this.)&lt;/p&gt;&#xA;&lt;p&gt;EDIT: In retrospect, perhaps this doesn't answer the question since it asks for experimental solutions.&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-09-14T07:40:59.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16285" PostTypeId="2" ParentId="16203" CreationDate="2014-09-14T07:56:18.190" Score="4" Body="&lt;p&gt;In addition to jonsca's link of chemistry subfields, &lt;a href=&quot;http://en.wikipedia.org/wiki/Outline_of_biochemistry#Branches_of_biochemistry&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; also has a list of biochem subfields. &lt;strong&gt;Find what research interests you&lt;/strong&gt;, perhaps not based on a course you've taken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look up what professors are doing in their respective fields of research by visiting faculty research group sites online. Check out their publications, current projects, probably would be good to join a lab too. &lt;strong&gt;(I'm blindly assuming you're interested in research and/or graduate studies)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list on Wikipedia first lists (for whatever reason these are &quot;main branches&quot; but I'm surprised biotech is categorized under &quot;other branches&quot;...):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Animal Biochemistry&lt;/li&gt;&#xA;&lt;li&gt;Plant Biochemistry&lt;/li&gt;&#xA;&lt;li&gt;Molecular Biology&lt;/li&gt;&#xA;&lt;li&gt;Cell Biology&lt;/li&gt;&#xA;&lt;li&gt;Metabolism&lt;/li&gt;&#xA;&lt;li&gt;Immunology&lt;/li&gt;&#xA;&lt;li&gt;Genetics&lt;/li&gt;&#xA;&lt;li&gt;Enzymology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Getting into more or less random scopes of research topics which may or may not be within the above fields:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Biotechnology,&lt;/li&gt;&#xA;&lt;li&gt;Bioluminescence,&lt;/li&gt;&#xA;&lt;li&gt;Molecular chemistry,&lt;/li&gt;&#xA;&lt;li&gt;Enzymatic Chemistry, &lt;/li&gt;&#xA;&lt;li&gt;Genetic engineering, &lt;/li&gt;&#xA;&lt;li&gt;Pharmaceuticals, &lt;/li&gt;&#xA;&lt;li&gt;Endocrinology, &lt;/li&gt;&#xA;&lt;li&gt;Hematology, &lt;/li&gt;&#xA;&lt;li&gt;Nutrition, &lt;/li&gt;&#xA;&lt;li&gt;Photosynthesis, &lt;/li&gt;&#xA;&lt;li&gt;Environmental Chem,&lt;/li&gt;&#xA;&lt;li&gt;Toxicology&lt;/li&gt;&#xA;&lt;li&gt;(Probably many, many others...)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3704" LastActivityDate="2014-09-14T07:56:18.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16286" PostTypeId="2" ParentId="16259" CreationDate="2014-09-14T08:13:39.580" Score="7" Body="&lt;p&gt;You don't seem to be overly familiar with the concepts of salts (made of cations and anions), ionic bonds and solvation of salts in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets's walk through some examples in which a solid material made from cations and anions is dissolved in water in such a way that the anions and cations are each surrounded and stabilized by water molecules (i.e. they are in an &lt;em&gt;aqueous&lt;/em&gt; state, denoted as &lt;em&gt;aq&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{NaCl (s) &amp;amp;-&amp;gt;[H2O] Na+ (aq) + Cl- (aq)} \\&#xA;\ce{MgSO4 (s) &amp;amp;-&amp;gt;[H2O] Mg^2+ (aq) + SO4^2- (aq)} \\&#xA;\ce{Na2SO4 (s) &amp;amp;-&amp;gt;[H2O] 2Na+ (aq) + SO4^2- (aq)} \\&#xA;\ce{K3PO4 (s) &amp;amp;-&amp;gt;[H2O] 3K+ (aq) + PO4^3- (aq)} \\&#xA;\ce{Mg3(PO4)2 (s) &amp;amp;-&amp;gt;[H2O] 3Mg^2+ (aq) + 2PO4^3- (aq)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule for the solvation of the salts, we can derive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{M_mX_n (s) -&amp;gt;[H2O] mM^n+ (aq) + nX^m- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this to potassium dichromate, we recognise that the cation is $\ce{K+}$ and the anion is $\ce{Cr2O7^2-}$. Therefore, it is the &lt;em&gt;dissolution&lt;/em&gt; of potassium dichromate in water which generates the dichromate anion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K2Cr2O7 (s) -&amp;gt;[H2O] 2K+ (aq) + Cr2O7^2- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the physical reason behind why the salts dissolve, you may find a more in-depth discussion here: &lt;a href=&quot;https://chemistry.stackexchange.com/q/5915/16683&quot;&gt;Why do salts such as NaCl dissolve?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2018-08-27T11:13:28.473" LastActivityDate="2018-08-27T11:13:28.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16287" PostTypeId="1" CreationDate="2014-09-14T08:57:19.573" Score="1" ViewCount="1360" Body="&lt;p&gt;What happens when $\ce{KCl}$, $\ce{K2Cr2O7}$, and concentrated $\ce{H2SO4}$ reacts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to approach these questions. I'd love if anybody can help me see the logic behind this.&lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:22.680" LastActivityDate="2015-02-27T17:19:22.680" Title="What happens when KCl, K₂Cr₂O₇, and concentrated H₂SO₄ reacts?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16290" PostTypeId="2" ParentId="16260" CreationDate="2014-09-14T10:05:31.380" Score="2" Body="&lt;p&gt;$$\color{purple}{X-axis: Pressure},\color{green}{Y-axis: Volume}$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HPEC0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Use conditions for reversible adiabatic process:&lt;/strong&gt;&#xA;$$P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma|\quad P_1V_1^\gamma=P_2V_2^\gamma|\quad T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}|\quad\gamma=\frac {C_p}{C_v}$$&#xA;&lt;strong&gt;and the combined gas law:&lt;/strong&gt;&#xA;$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}=\frac{P_2V_3}{T_3}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-09-14T10:15:26.967" LastActivityDate="2014-09-14T10:15:26.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16291" PostTypeId="2" ParentId="16287" CreationDate="2014-09-14T11:07:29.593" Score="5" Body="&lt;p&gt;Have a look at the properties of the compounds involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dichromate ($\ce{Cr(VI), Cr2O7^{2-}}$) is a strong oxidant and is reduced to $\ce{Cr(III)}$ if it finds a suitable partner (which then is oxidized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, a potential partner in a redox reaction is chloride: $\ce{2 Cl- + 2 e^{-} -&amp;gt; Cl2}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-14T11:07:29.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16293" PostTypeId="1" AcceptedAnswerId="16295" CreationDate="2014-09-14T13:15:15.633" Score="0" ViewCount="272" Body="&lt;p&gt;The definition of spontaneous process states that &quot;A spontaneous process is an irreversible process and maybe reversed by some external agency&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we say that all spontaneous reactions are irreversible reactions under the absence of some external force? &lt;/p&gt;&#xA;" OwnerUserId="7760" LastEditorUserId="73" LastEditDate="2014-09-14T14:13:18.870" LastActivityDate="2014-09-14T15:15:09.857" Title="Nature of spontaneous reactions" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16294" PostTypeId="2" ParentId="16261" CreationDate="2014-09-14T13:40:10.310" Score="6" Body="&lt;p&gt;There is no clear separation between ionic and covalent bonds.&#xA;Electronegativity values are useful in determining if a bond is to be classified as nonpolar covalent, polar covalent or ionic. &#xA;The rule is that when the electronegativity difference is greater than 2.0 (Electronegativity according to Pauling), the bond is considered ionic.&#xA;If the electronegativity difference  is less than 0.5, then the bond is nonpolar covalent.&#xA;If the electronegativity difference is between 0.5 and 1.6, the bond is considered polar covalent.&#xA;For this reason $\ce{CaO}$ is ionic, while $\ce{Be3N2}$  is covalent.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-14T13:40:10.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16295" PostTypeId="2" ParentId="16293" CreationDate="2014-09-14T15:15:09.857" Score="1" Body="&lt;p&gt;A spontaneous reaction, as contrasted with a non-spontaneous reaction, is a reaction quantified by a negative free energy change. A reaction is said to be spontaneous if it occurs without being driven by an outside force. So, we can say:  In any spontaenous process, the path between the reactants and products is irreversible&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-14T15:15:09.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16296" PostTypeId="1" AcceptedAnswerId="16297" CreationDate="2014-09-14T15:26:49.707" Score="3" ViewCount="511" Body="&lt;p&gt;Why does methane have two different ionization potentials? How does this work? I understand that MO theory predicts C-H bonds of differing strength, while hybridization predicts C-H bonds of varying strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=gY-Sxijk_tMC&amp;amp;pg=PA827&amp;amp;dq=methane+ionization&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=VrMVVN3JLI-vogTv8IGABw&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q=methane%20ionization&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;http://books.google.com/books?id=gY-Sxijk_tMC&amp;amp;pg=PA827&amp;amp;dq=methane+ionization&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=VrMVVN3JLI-vogTv8IGABw&amp;amp;ved=0CCgQ6AEwAA#v=onepage&amp;amp;q=methane%20ionization&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is methane ionized successively and is are there two different ionization energies found? Doesn't this make intuitive sense; by ionizing methane we end up with a fundamentally different molecule which would more than likely have different ionization energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how are these ionizations performed? The reference given above makes no mention of whether the bonds are broken homolytically or heterolytically (although &lt;em&gt;ionization&lt;/em&gt; implies heterolytic bond cleavage). Even then, there are two ways bonds can be cleaved &lt;em&gt;heterolytically&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="189" LastEditDate="2014-09-14T16:29:10.723" LastActivityDate="2014-09-14T16:58:38.807" Title="Unequal ionization energies of methane" Tags="&lt;theoretical-chemistry&gt;&lt;spectroscopy&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16297" PostTypeId="2" ParentId="16296" CreationDate="2014-09-14T16:58:38.807" Score="4" Body="&lt;p&gt;In the photoelectron spectroscopy (PES) experiment a sample is irradiated with photons of differing energy and the kinetic energy of any ejected electrons is measured.  By knowing the energy of the incoming photon and the kinetic energy of any ejected electrons, we can calculate the energy required to ionize the electron as the difference between these two recorded values. An electron will only be ejected if the photon energy is equal to or greater than the energy required to ionize an electron in the molecule.  See the section entitled &quot;Photoelectron Spectroscopy&quot; in &lt;a href=&quot;http://www.lasalle.edu/~prushan/MO-PES%20Handout.pdf&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; if you would like to read more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of methane, all of the bonds are equivalent as required by symmetry.  There is only one $\ce{C-H}$ bond strength for methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we run the PES experiment with methane we see 3 signals (as pictured in your link; two of the signals are due to photoionization of electrons from the $\ce{C-H}$ bond, and the third signal, at much higher photon energy, is due to photoionization of the carbon 1s electrons).  We know that the 4 bonds in methane can be described by 8 orbitals, 4 bonding, 4 antibonding.  We can transform these 8 localized orbitals into 8 molecular orbitals; the 4 bonding MO's are pictured in the link you provided.  Notice from the picture in your link that 3 of the 4 MO's are degenerate, but there is also a different 4th MO.  All of these bonding MO's are filled and ionization can occur from either one of the 2 types - hence 2 signals from the occupied $\ce{C-H}$ MO's (in the PES experiment, these are not sequential ionizations, e.g. these are not first and second ionizations; rather they are independent ionizations that will occur only when our incoming photon energy matches the energy of the orbital containing electrons).  During the PES experiment with methane, no bonds are broken, rather an electron from an orbital is ejected and the methyl radical cation is formed.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-14T16:58:38.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16298" PostTypeId="1" CreationDate="2014-09-14T17:22:01.893" Score="2" ViewCount="301" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In a &lt;span class=&quot;math-container&quot;&gt;$\pu{45.9 g}$&lt;/span&gt; sample of borane, a compound containing only boron and hydrogen, that's &lt;span class=&quot;math-container&quot;&gt;$85.63\%$&lt;/span&gt; boron, what is the mass of hydrogen in this sample?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I was just going to take &lt;span class=&quot;math-container&quot;&gt;$\pu{45.9 g}$&lt;/span&gt; and multiply by &lt;span class=&quot;math-container&quot;&gt;$0.1437$&lt;/span&gt; to get the mass of hydrogen, but I don't think it works that way because they don't equally contribute to the mass of the compound. Then I thought, maybe I should consider the molar masses of both atoms, hydrogen being &lt;span class=&quot;math-container&quot;&gt;$\pu{1.008 g mol-1}$&lt;/span&gt; and boron being &lt;span class=&quot;math-container&quot;&gt;$\pu{10.811 g mol-1}$&lt;/span&gt;. I am stuck at this point. How to solve the question.&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="4945" LastEditDate="2020-09-03T09:07:47.317" LastActivityDate="2020-09-03T09:24:11.687" Title="How to calculate the mass of hydrogen in a borane sample?" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16299" PostTypeId="1" CreationDate="2014-09-14T17:44:47.003" Score="1" ViewCount="56" Body="&lt;p&gt;The question I was doing tells you that the compound consist of 49.0% carbon, 2.7% of hydrogen ,and 48.2% of chlorine. And ask you to find the empirical formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the usual procedure of dividing the percentages by each elements atomic mass. Because the result for carbon was 4.0796, I thought I should rounding to four moles of carbon. And for hydrogen and chlorine the numbers of mole for each are 3 and 1.  I want to check the formula is correct, so I make it 8 moles of carbon, but then the number of mole for hydrogen turns out to be 5 and chlorine 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is a new empirical formula. Which one is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PC: when I divided the percentages by atomic mass, I rounded the results to ten thousandth place.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iTZcJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="73" LastEditDate="2014-09-15T03:49:13.530" LastActivityDate="2014-09-15T03:49:13.530" Title="Why does rounding off the mol values when finding an empirical formula lead to a different result?" Tags="&lt;stoichiometry&gt;" AnswerCount="0" CommentCount="10" ClosedDate="2014-09-15T22:05:32.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="16300" PostTypeId="1" CreationDate="2014-09-14T18:09:37.673" Score="4" ViewCount="928" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A glass vessel weighs &lt;span class=&quot;math-container&quot;&gt;$\pu{20.2367 g}$&lt;/span&gt; when empty and &lt;span class=&quot;math-container&quot;&gt;$\pu{20.3102 g}$&lt;/span&gt; when filled to the etched mark with water at &lt;span class=&quot;math-container&quot;&gt;$\pu{4 °C}$&lt;/span&gt;. The same vessel was then dried and filled to the same mark with a solution at &lt;span class=&quot;math-container&quot;&gt;$\pu{4 °C}$&lt;/span&gt;, the vessel was now found to weigh &lt;span class=&quot;math-container&quot;&gt;$\pu{20.3300 g}$&lt;/span&gt;. What is the density of the solution? (Assume the density of water is &lt;span class=&quot;math-container&quot;&gt;$\pu{1.00 g/ml}$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here’s what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the water is just &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0735 g}$&lt;/span&gt; when you subtract the weight of (water + vessel) and (empty vessel). Then I found how many &lt;span class=&quot;math-container&quot;&gt;$\pu{ml}$&lt;/span&gt; of water there was in the vessel by realizing that the density (&lt;span class=&quot;math-container&quot;&gt;$m/V$&lt;/span&gt;) must be equal to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0735 g}$&lt;/span&gt; of water must be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0735 ml}$&lt;/span&gt; of water too. So with that, I figured out how many grams of the unknown solution were in the vessel (&lt;span class=&quot;math-container&quot;&gt;$\pu{0.0993 g}$&lt;/span&gt;), and knowing that it was filled to the same “etched mark,” we can say that the density of the unknown solution is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0933 g}/\pu{0.0735 ml} = \pu{1.27 g/ml}$&lt;/span&gt;. I went through all of that because I am not sure if my thoughts are on track with logic and I am pretty confused.&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="61578" LastEditDate="2019-12-01T01:01:47.020" LastActivityDate="2019-12-01T01:10:04.313" Title="Density of unknown solution?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16301" PostTypeId="1" CreationDate="2014-09-14T19:24:07.607" Score="1" ViewCount="2273" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Gold is present in seawater to the extent of 0.15 mg/ton. Assume the density of seawater is 1.03 g/ml and determine how many gold atoms could be extracted from 0.250 L of seawater. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;($\pu{1 ton}=\pu{2000lbs}, \pu{1kg}=\pu{2.205lbs}, \pu{1 mole}=\pu{6.022E23 atoms}, \text{1 mole of gold}=\pu{197g}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just multiplied everything together using dimensional analysis and got $5.2\times10^{14}$ atoms of Au per liter of seawater, but they want how many atoms of Au in $\pu{0.250L}$ of seawater so I just multiplied $5.2\times10^{14}$ by $0.250$ to get $1.30\times10^{14}$ atoms of Au per $\pu{0.250L}$ of seawater. Can you simply multiply everything together like this? When can you do this and when can't you?&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorDisplayName="user7951" LastEditDate="2018-09-02T08:13:24.630" LastActivityDate="2018-09-02T08:13:24.630" Title="How many Au atoms can be extracted from 0.250 L of seawater?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16302" PostTypeId="2" ParentId="16301" CreationDate="2014-09-14T20:11:30.817" Score="3" Body="&lt;p&gt;You also need to convert the grams of gold into moles in order to use Avogadro's number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic weight of gold = $\pu{196.967 g/mol}$&#xA;$$&#xA;\begin{align}&#xA;\frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&amp;amp;= \pu{7.615E-07 mol/ton}\\&#xA;\pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &amp;amp;= \pu{4.586E+17 atoms/ton}\\&#xA;\frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&amp;amp;=\pu{2.293E{+14} atoms/lb}\\&#xA;\frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&amp;amp;=\pu{5.056E{+11} atoms/g}\\&#xA;\pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&amp;amp;=\pu{5.208E{+11} atoms/ml}\\&#xA;\pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&amp;amp;= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\&#xA;\end{align}&#xA;$$&#xA;There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2018-02-28T20:20:00.687" LastActivityDate="2018-02-28T20:20:00.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16303" PostTypeId="1" AcceptedAnswerId="16310" CreationDate="2014-09-14T21:02:03.747" Score="-1" ViewCount="230" Body="&lt;p&gt;The reaction of carbon dioxide with water yields carbonic acid to some extent (0.4 %) depending on pressure (and temperature). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process, according to @ron, goes through a four-membered transition stage in which a proton transfer occurs, thereby relieving the unstable &quot;carbonic acid&quot; or hydrated carbon dioxide of its charge separation (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16076/the-instability-of-hydrated-carbon-dioxide-or-carbonic-acid&quot;&gt;The instability of hydrated carbon dioxide or &amp;quot;carbonic acid&amp;quot;&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/en0bx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the resulting species is a di-enol; we have hydroxyl groups attached to a carbon that bears a pi bond. This species is likely &lt;em&gt;highly&lt;/em&gt; unstable due to the highly positive central carbon atom. On the other hand, neither hydroxyl group makes for a good leaving group, so I can't think of any &quot;forward&quot; (rather than &lt;em&gt;reverse&lt;/em&gt;) pathways to more stable products. Also, I can't see the di-enol tautomerizing into a keto. Or can it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I was arguing that this species is analogous to &lt;em&gt;cis&lt;/em&gt;-nitrous acid, in that both species involve a &lt;em&gt;four-membered ring&lt;/em&gt; (or at least a &lt;em&gt;psuedo&lt;/em&gt; four-memebred ring in the case of &lt;em&gt;cis&lt;/em&gt;-HONO).&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15558/why-does-nitrous-acid-exhibit-unexpectedly-different-bond-angles&quot;&gt;Why does nitrous acid exhibit (unexpectedly) different bond angles?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My points were:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Proton transfer within the charge-separated form of carbonic acid mirrors the electrostatic attraction between the hydrogen and the terminal oxygen of &lt;em&gt;cis&lt;/em&gt;-HONO. &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Because&lt;/em&gt; the extent of this proton transfer is 0.4%, this suggests that forming a four membered ring is unfavorable (including the case of &lt;em&gt;cis&lt;/em&gt;-HONO). Well, of course. &lt;/li&gt;&#xA;&lt;li&gt;This &lt;em&gt;more importantly&lt;/em&gt; suggests that any &lt;em&gt;interaction&lt;/em&gt; between any terminal oxygen and hydrogen in a four-membered ring is &lt;em&gt;unfavorable&lt;/em&gt;); just look at the extent of proton transfer between an acidic oxygen bearing a &lt;em&gt;positive&lt;/em&gt; formal charge of and an oxygen bearing a &lt;em&gt;negative&lt;/em&gt; formal charge. In isolation, we would expect the below to be &lt;em&gt;highly&lt;/em&gt; favorable (at least thermodynamically):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{RO^+H + RO^- -&amp;gt;RO + ROH}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My professor, however, countered that the resulting product, while not charge separated, is a &lt;em&gt;&quot;di-enol&quot;&lt;/em&gt;. This is highly unstable; the carbon has a strong partial positive charge. So this distorts the comparison I'm trying to draw between &lt;em&gt;cis&lt;/em&gt;-HONO and carbonic acid; the sheer instability of the resulting product from the &lt;em&gt;intramolecular&lt;/em&gt; reaction may be hiding any thermodynamic gains to be had from the proton transfer, and thus does not rule out the possibility that the &lt;em&gt;net&lt;/em&gt; result of hydrogen bonding in &lt;em&gt;cis&lt;/em&gt;-HONO is unfavorable, and that H-bonding is &lt;em&gt;cis&lt;/em&gt;-HONO is unable to close up the ONO bond angle. &lt;/li&gt;&#xA;&lt;li&gt;The &quot;di-enol&quot; also reminds me of acid/acyl chlorides, which are stable enough to be isolated, but not exceedingly stable to be completely unreactive. Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge. &lt;/li&gt;&#xA;&lt;li&gt;In addition, the charge-separated form is prone to disproportionation to carbon dioxide &lt;em&gt;gas&lt;/em&gt; and water, which is highly favorable from an entropic standpoint. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is carbonic acid a &quot;di-enol&quot; (if that even is a thing)?&lt;/li&gt;&#xA;&lt;li&gt;Can &lt;em&gt;true&lt;/em&gt; (non-charge separated) carbonic acid tautomerize into more stable forms? &lt;/li&gt;&#xA;&lt;li&gt;Does this system provide a good parallel with regard to the case of &lt;em&gt;cis&lt;/em&gt;-HONO? &lt;/li&gt;&#xA;&lt;li&gt;Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge. &lt;/li&gt;&#xA;&lt;li&gt;Is true carbonic acid stable? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2019-05-18T15:26:09.247" LastActivityDate="2019-05-18T15:26:09.247" Title="The stability of a &quot;Di-onol&quot; - carbonic acid - and analogies to cis-nitrous acid" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="16304" PostTypeId="1" AcceptedAnswerId="16309" CreationDate="2014-09-14T21:16:48.000" Score="5" ViewCount="5524" Body="&lt;p&gt;Mercury can form amalgams with metals, such as gold, however how exactly does the &quot;reaction&quot; if you can call it that take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of metal bonds is that the atoms of a metal are surrounded by a &quot;sea&quot; of electrons, however I don't understand how mercury can &quot;penetrate&quot; metals and weaken their structure or &quot;dissolve&quot; them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it do this? Also does this apply to Gallium?&lt;/p&gt;&#xA;" OwnerUserId="5772" LastEditorUserId="5772" LastEditDate="2016-02-27T19:39:40.220" LastActivityDate="2019-04-15T21:12:07.477" Title="What is the chemistry behind amalgamation?" Tags="&lt;metal&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16305" PostTypeId="1" AcceptedAnswerId="16306" CreationDate="2014-09-14T21:34:55.893" Score="16" ViewCount="19213" Body="&lt;p&gt;What are the symmetry labels for the p and d orbitals of platinum in $\ce{[PtCl4]^2-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to find the point group of a molecule, but am not sure how to use the character table to find the irreducible representations corresponding to the orbitals.&lt;/p&gt;&#xA;" OwnerUserId="6985" LastEditorUserId="16683" LastEditDate="2018-09-06T07:53:45.903" LastActivityDate="2018-09-06T07:53:45.903" Title="How can I find the symmetry labels of atomic orbitals in a molecule?" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="16306" PostTypeId="2" ParentId="16305" CreationDate="2014-09-14T23:14:34.080" Score="22" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found &lt;a href=&quot;http://symmetry.otterbein.edu/common/images/flowchart.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). The point group of the complex $\ce{[PtCl4]^2-}$, for example, is $D_\mathrm{4h}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now you consult the character table of the point group you determined in the previous step. The character tables for many common point groups can be found &lt;a href=&quot;http://symmetry.jacobs-university.de/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. The character table of the point group $D_\mathrm{4h}$ is shown below and can also be found &lt;a href=&quot;http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=604&amp;amp;option=4&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|cccccccccc|cc} \hline&#xA;D_\mathrm{4h} &amp;amp; E &amp;amp; 2C_4 &amp;amp; C_2 &amp;amp; 2C_2' &amp;amp; 2C_2'' &amp;amp; i &amp;amp; 2S_4 &amp;amp; \sigma_\mathrm{h} &amp;amp; 2\sigma_\mathrm{v} &amp;amp; 2\sigma_\mathrm{d} &amp;amp; &amp;amp; \\ \hline&#xA;\mathrm{A_{1g}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; &amp;amp; x^2+y^2,z^2 \\&#xA;\mathrm{A_{2g}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; R_z &amp;amp; \\&#xA;\mathrm{B_{1g}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; &amp;amp; x^2-y^2 \\&#xA;\mathrm{B_{2g}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; &amp;amp; xy \\&#xA;\mathrm{E_g} &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; (R_x,R_y) &amp;amp; (xz,yz) \\&#xA;\mathrm{A_{1u}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; &amp;amp; \\&#xA;\mathrm{A_{2u}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; z &amp;amp; \\&#xA;\mathrm{B_{1u}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; &amp;amp; \\&#xA;\mathrm{B_{2u}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; &amp;amp; \\&#xA;\mathrm{E_u} &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0 &amp;amp; (x,y) &amp;amp; \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Finally you need to determine the symmetry labels of the p- and d-orbitals within the point group by identifying the irreducible representations (the labels on the far left of the character table) which match the transformational properties of the basis vectors ($\color{red}{x}$, $\color{red}{y}$, $\color{red}{z}$) and some of their quadratic combinations ($\color{blue}{xy}$, $\color{blue}{xz}$, $\color{blue}{yz}$, $\color{blue}{z^2}$, $\color{blue}{x^2 - y^2}$). These can be found in the two columns on the right columns. Again this will be shown for the example compound $\ce{[PtCl4]^2-}$ where the relevant functions are highlighted.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\require{bbox} \begin{array}{c|cccccccccc|cc} \hline&#xA;D_\mathrm{4h} &amp;amp; E &amp;amp; 2C_4 &amp;amp; C_2 &amp;amp; 2C_2' &amp;amp; 2C_2'' &amp;amp; i &amp;amp; 2S_4 &amp;amp; \sigma_\mathrm{h} &amp;amp; 2\sigma_\mathrm{v} &amp;amp; 2\sigma_\mathrm{d} &amp;amp; &amp;amp; \\ \hline&#xA;\mathrm{A_{1g}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; &amp;amp; x^2+y^2,\bbox[cyan]{z^2} \\&#xA;\mathrm{A_{2g}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; R_z &amp;amp; \\&#xA;\mathrm{B_{1g}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; &amp;amp; \bbox[cyan]{x^2-y^2} \\&#xA;\mathrm{B_{2g}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; &amp;amp; \bbox[cyan]{xy} \\&#xA;\mathrm{E_g} &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; (R_x,R_y) &amp;amp; \bbox[cyan]{(xz,yz)} \\&#xA;\mathrm{A_{1u}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; &amp;amp; \\&#xA;\mathrm{A_{2u}} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; \bbox[pink]{z} &amp;amp; \\&#xA;\mathrm{B_{1u}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; &amp;amp; \\&#xA;\mathrm{B_{2u}} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; &amp;amp; \\&#xA;\mathrm{E_u} &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0 &amp;amp; \bbox[pink]{(x,y)} &amp;amp; \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the three p-orbitals point directly along the $x$, $y$, and $z$ axes, they show the same transformational properties as the basis vectors, and consequently get the same symmetry labels, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\left.\begin{array}{c}&#xA;\mathrm{p}_x \\ \mathrm{p}_y \end{array}\right\} &amp;amp;\rightarrow \mathrm{E_u} \\&#xA;\mathrm{p}_z &amp;amp;\rightarrow \mathrm{A_{2u}} \ .&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\mathrm{d}_{xy} &amp;amp;\rightarrow \mathrm{B_{2g}} \\&#xA;\left.\begin{array}{c}&#xA;\mathrm{d}_{xz} \\ \mathrm{d}_{yz} \end{array}\right\} &amp;amp;\rightarrow \mathrm{E_g} \\&#xA;\mathrm{d}_{z^2} &amp;amp;\rightarrow \mathrm{A_{1g}} \\&#xA;\mathrm{d}_{x^2 - y^2} &amp;amp;\rightarrow \mathrm{B_{1g}} \ .&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another worked example can be found in &lt;a href=&quot;https://chemistry.stackexchange.com/a/5642/189&quot;&gt;this answer of mine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-09-06T07:52:58.007" LastActivityDate="2018-09-06T07:52:58.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16307" PostTypeId="1" AcceptedAnswerId="16329" CreationDate="2014-09-14T23:14:40.923" Score="6" ViewCount="1659" Body="&lt;p&gt;I'm basically thinking whether I could potentionally poison myself while neutralising strong acid/base by some weak base/acid when I accidentally spill it, touch it or when I'm cleaning reaction product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combinations of chemicals that are non-toxic in my opinion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{HCl + NaHCO3}$  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2SO4 + NaHCO3}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NaOH + CH3COOH}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;None of these are poisonous. I just thought these up, for example the first one produces table salt, $\ce{NaCl}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But could there be a combination of non-toxic acid and base that would produce some nasty salt when neutralised? Something worse than what I would be getting rid of?&lt;br&gt;&#xA;Is there a combination where I should better just wash myself with water?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2014-09-28T07:13:36.667" LastActivityDate="2014-09-28T07:13:36.667" Title="Could neutralisation of safe chemicals produce dangerous ones?" Tags="&lt;acid-base&gt;&lt;safety&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16309" PostTypeId="2" ParentId="16304" CreationDate="2014-09-14T23:54:07.953" Score="5" Body="&lt;p&gt;An &lt;a href=&quot;http://en.wikipedia.org/wiki/Amalgam_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;amalgam&lt;/a&gt; is simply an alloy of some metal with mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alloy&quot; rel=&quot;noreferrer&quot;&gt;Alloys&lt;/a&gt; are simply solid solutions with metals. Since the mixture of multiple elements disrupts the metal bonding, the properties are often quite different than the components. (In some cases, alloys form so-called intermetallic compounds with defined stoichiometry, but more often, the ratios are very flexible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't describe the mixture as a reaction, in much the same way that I wouldn't really call dissolving sodium chloride in water as a reaction. You're forming a mixture. Water disrupts the ionic bonding in NaCl, and mercury and other metals can also accept other components or impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gallium does alloy with a lot of metals, and indium-gallium &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;noreferrer&quot;&gt;eutectic&lt;/a&gt; alloy is a liquid at room temperature, as is &lt;a href=&quot;http://en.wikipedia.org/wiki/Galinstan&quot; rel=&quot;noreferrer&quot;&gt;&quot;galinstan&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question about mercury dissolving the metal atoms, consider that mercury is &lt;em&gt;also&lt;/em&gt; a metal, so it also participates in similar metal bonding.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-14T23:54:07.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16310" PostTypeId="2" ParentId="16303" CreationDate="2014-09-15T00:14:48.720" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.Is carbonic acid a &quot;di-enol&quot; (if that even is a thing)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think most folks would understand what you mean, di-onal (alkene vs. ketone), since it is a carbonyl double bond, not a carbon-carbon double bond, that is involved, might be preferred.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.Can true (non-charge separated) carbonic acid tautomerize into more stable forms? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can undergo keto-onol tautomerization to regenerate itself, but now with a different oxygen involved in the carbonyl and a different oxygen in one of the hydroxyls.  I can also imagine a tautomer with a 2-oxygen, 3-membered ring, but I doubt this tautomer is more stable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3.Does this system provide a good parallel with regard to the case of cis-HONO?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first reaction is that it is not a close analogy.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The carbonic acid situation involves proton &lt;strong&gt;transfer&lt;/strong&gt;.  In this case with water around, the proton transfer is likely &lt;strong&gt;intermolecular&lt;/strong&gt;, involving nearby water molecules.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The nitrous oxide problem involved a &lt;strong&gt;static&lt;/strong&gt;, &lt;strong&gt;intramolecular&lt;/strong&gt; hydrogen bond&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first reaction here was - no way.  But the more I think about it, the more I see your point.  They both react via similar mechanisms, you can draw that extra resonance structure with hydroxide or chloride ionized for both of them, you have to protect them both in order to store them.  Guess I'd say there are a number of analogies between the two compounds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5.Is true carbonic acid stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, when stored under the appropriate conditions (just like an acid chloride). To compare it's kinetic stability to some other compound, we'd need to find out what the activation energy for decomposition is.  To compare it's thermodynamic stability, we'd need to calculate it's heat of formation and compare that to the heat of formation of water and carbon dioxide.  P.S What is untrue carbonic acid?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-15T01:31:45.717" LastActivityDate="2014-09-15T01:31:45.717" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16311" PostTypeId="1" AcceptedAnswerId="16320" CreationDate="2014-09-15T02:40:31.337" Score="2" ViewCount="4729" Body="&lt;p&gt;I had a question in my textbook to convert propane to methane. The basic idea was chlorination in presence of sunlight. However, two products are obtained in this case - namely 2-chloropropane and 1-chloropropane. So it didn't work out at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with another idea i.e. dehydrogenation, but it has not been taught to us at all. In addition the dehydrogenation reactions I searched on google were too complex. So, I hope there's another method for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me to sort this out?&lt;/p&gt;&#xA;" OwnerUserId="4306" LastEditorUserId="5026" LastEditDate="2018-04-10T07:03:33.233" LastActivityDate="2020-04-29T06:18:37.727" Title="Conversion of propane to methane" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;hydrocarbons&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="1" ClosedDate="2018-01-01T20:11:45.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="16313" PostTypeId="2" ParentId="16311" CreationDate="2014-09-15T03:43:20.623" Score="-2" Body="&lt;p&gt;If you are trying to do a dehydrogenation, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehydrohalogenation&quot; rel=&quot;nofollow&quot;&gt;dehydrohalogenation&lt;/a&gt; of both your products would get you to propene, even though it takes two steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck getting to methane. Of course, if you were in industry you would just &lt;a href=&quot;http://en.wikipedia.org/wiki/Cracking_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;heat the crap out of it&lt;/a&gt; and watch the money roll in... Well, maybe not with methane as the product...&lt;/p&gt;&#xA;" OwnerUserId="2109" LastActivityDate="2014-09-15T03:43:20.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16314" PostTypeId="1" CreationDate="2014-09-15T03:45:05.893" Score="3" ViewCount="282" Body="&lt;p&gt;I was wondering if the length of the carbon backbone affects hydrophobicity of amino acids (i.e. longer carbon backbone of side group = less polar = more hydrophobic).&lt;/p&gt;&#xA;" OwnerUserId="4465" LastActivityDate="2014-11-22T18:27:55.480" Title="What trends affect hydrophobicity in amino acids?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16315" PostTypeId="2" ParentId="16311" CreationDate="2014-09-15T06:18:02.970" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s8Ya8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use:&#xA;$$\begin{array}{}&#xA;\ce{Br_2}/h\nu&amp;amp; \text{Bromination}\\&#xA;\ce{^{\large \circleddash} OEt/EtOH}&amp;amp; \text{Elimination: Dehydrohalogenation}\\&#xA;\ce{O_3/H_2O}&amp;amp; \text{Ozonolysis}\\&#xA;\ce{NaOH/CaO}/\Delta&amp;amp; \text{Decarboxylation}\\&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-09-15T08:05:16.620" LastActivityDate="2014-09-15T08:05:16.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16317" PostTypeId="2" ParentId="21626" CreationDate="2014-09-15T07:23:59.017" Score="3" Body="&lt;p&gt;Did you react aniline (or its hydrochloride) with sodium nitrite in the presence of 2-naphthol?&lt;/p&gt;&#xA;&lt;p&gt;I'm afraid there will not be any table with &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{f}$&lt;/span&gt; values.&#xA;These values are not only dependent on the compounds and the mobile phases, but also on the stationary phase: e.g. silica, aluminium oxide, etc.&#xA;Your best option is to use reference samples and measure the &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{f}$&lt;/span&gt; value yourself for the conditions you want to apply or transfer from TLC to column chromatography.&lt;/p&gt;&#xA;&lt;p&gt;My suggestion is to collect the crude reaction product (Büchner funnel, etc.) dry it, note the yield and &lt;strong&gt;measure the melting point&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For purification, recrystallize from ethanol. You should receive red needles with a melting point of &lt;span class=&quot;math-container&quot;&gt;$\pu{134 ^\circ C}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Wear proper protection when handling the azo dye, which is a cancer suspect.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;References:&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ioc-praktikum.de/iocp/versuche/7_2_5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;IOC-Praktikum&lt;/a&gt; (PDF in German)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://gestis-en.itrust.de/nxt/gateway.dll?f=templates%7B0eab7102-d09f-4a9a-86d2-da09178535d2%7Dvid=gestiseng:sdbeng&quot; rel=&quot;nofollow noreferrer&quot;&gt;GESTIS-database on hazardous substances&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2020-09-08T13:46:18.953" LastActivityDate="2020-09-08T13:46:18.953" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16318" PostTypeId="1" CreationDate="2014-09-15T08:41:18.733" Score="2" ViewCount="739" Body="&lt;p&gt;Will Sodium metal donate its electrons to bromine in 1,3-dibromopropane or to the hydroxyl group in ethanol in a polar protic solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question comes from a problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'What is the major product formed when 1,3-dibromopropane is added to &lt;strong&gt;Na&lt;/strong&gt;(sodium)/&lt;strong&gt;EtOH&lt;/strong&gt;(ethanol)?&#xA;(a)cyclopropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)3-bromoprop-1-ene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c)propyne&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d)prop-1,2-diene'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer was (a)cyclopropane but I was not convinced as then it would have to follow Wurtz reaction- where sodium donates its electron to bromine forming a alkyl radical. But, Wurtz reaction takes place in an ether solvent and here ethanol is used as solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ethanol is shown to readily react with Na to form a sodium ethoxide and liberate hydrogen. The ethoxide ion would react with 1,3-dibromopropane in either Sn1 or E2 mechanism and so it would not form cyclopropane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If sodium donates electron to bromine however, then it would follow Wurtz reaction mechanism and form (a)cyclopropane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know whether an alkyl halide or alcohol reacts more readily with sodium metal and what your reasoning is to assert your statement. &lt;/p&gt;&#xA;" OwnerUserId="7858" LastEditorUserId="7858" LastEditDate="2014-09-15T11:49:42.833" LastActivityDate="2014-09-15T11:49:42.833" Title="Is Sodium metal more reactive towards an alkyl halide or an alcohol in a protic polar solvent?" Tags="&lt;organic-chemistry&gt;&lt;electrons&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16319" PostTypeId="2" ParentId="16304" CreationDate="2014-09-15T08:41:58.443" Score="0" Body="&lt;p&gt;To my knowledge aluminum treated with gallium will happily reduce water to hydrogen the way aluminum amalgam does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to reduce a nitroalkane you have to consider hydrogen overpotential of the gallium-aluminium compared to aluminium amalgam. That is what dictates the selectivity (substrate reduction vs. hydrogen evolution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same idea as in electrolytic reductions where you have to use electrodes made of suitable materials (mercury and lead are traditional choices) to be able to actually reduce your compound instead of making hydrogen gas.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-15T08:41:58.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16320" PostTypeId="2" ParentId="16311" CreationDate="2014-09-15T08:45:06.087" Score="5" Body="&lt;p&gt;This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C3H8 +O2 +H2O -&amp;gt; CO + H2 + CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO + 3 H2 -&amp;gt; CH4 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-15T12:58:31.080" LastActivityDate="2014-09-15T12:58:31.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16321" PostTypeId="1" AcceptedAnswerId="16322" CreationDate="2014-09-15T12:41:10.177" Score="3" ViewCount="7107" Body="&lt;p&gt;Regarding solvents, the only thing I remember from school is my teacher saying: &quot;The same dissolves the same.&quot; regarding the polar/non-polar solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this I understand that water is &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvent#Solvent_classifications&quot; rel=&quot;nofollow&quot;&gt;polar solvent&lt;/a&gt;, because it has &lt;a href=&quot;http://en.wikipedia.org/wiki/Polar_bond&quot; rel=&quot;nofollow&quot;&gt;polar bonds&lt;/a&gt; and therefore it dissolves &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chloride&quot; rel=&quot;nofollow&quot;&gt;table salt&lt;/a&gt;, which also has a polar bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do &lt;em&gt;not&lt;/em&gt; understand for example is, why does water perfectly mix with &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetone&quot; rel=&quot;nofollow&quot;&gt;acetone&lt;/a&gt;? &lt;sup&gt;&lt;strong&gt;X&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing is, that non-polar organic solvents seem to have some very complicated rules. For example, acetone dissolves most paints and all ABS plastics. &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethanol&quot; rel=&quot;nofollow&quot;&gt;Ethanol&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrachloroethylene&quot; rel=&quot;nofollow&quot;&gt;tetrachloroethylene&lt;/a&gt; do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does ethanol dissolve &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetone_peroxide&quot; rel=&quot;nofollow&quot;&gt;acetone-peroxide&lt;/a&gt; while acetone doesn't? &lt;sup&gt;&lt;strong&gt;X&lt;/strong&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; Are there some general rules to this, or is this big science that requires years of study?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;strong&gt;X&lt;/strong&gt;: these are just rhetoric questions. You can answer in comments but what I'm looking for are rules.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="33991" LastEditDate="2016-11-29T08:33:01.360" LastActivityDate="2016-11-29T08:33:01.360" Title="How can I predict what dissolves what?" Tags="&lt;solubility&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16322" PostTypeId="2" ParentId="16321" CreationDate="2014-09-15T13:41:29.167" Score="9" Body="&lt;p&gt;As you note, there are some general rules about solubility. For example, acetone is a highly polar organic molecule, so it's not surprising it has high solubility/mixibility in water. The &quot;like dissolves like&quot; rule is generally based on polarity and size/shape considerations, and it's a fairly good rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But solubility is complex because you have two big sources of driving force:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;enthalpy&lt;/em&gt;: The molecular and intermolecular interactions between the solvent and solute are favorable. Consider methanol dissolving in water: it forms hydrogen bonds with the water, so the enthalpy is favorable.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;entropy&lt;/em&gt;: This is often the larger driving force. You create a &lt;strong&gt;far&lt;/strong&gt; more disordered result with a mixture than with two pure components.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While most chemists develop reasonably accurate intuitions of &lt;em&gt;enthalpy&lt;/em&gt;, the entropic effects are harder to predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many efforts have been made to predict solubility, both qualitatively and quantitatively. The comments above give some good links. Consider that this problem is important to chemical synthesis broadly, but also to drug delivery (i.e., can a drug be given orally and dissolve in the stomach, can a drug go through the blood-brain barrier?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most successful is the &lt;a href=&quot;http://onschallenge.wikispaces.com/Benzoic+acid+-+Abraham&quot;&gt;Abraham general solubility model&lt;/a&gt; which attempts to predict solubility of organic compounds in 84 different solvents. &lt;a href=&quot;http://dx.doi.org/10.1002/jps.21922&quot;&gt;Here’s the article.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it breaks down solubility into multiple components (grabbed from Andrew Lang and Jean-Claude Bradley):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt; is the solute excess molar refractivity in units of (cubic cm per mol)/10. It represents the solute's polarizability and gives a measure of the ability of a solute to interact with a solvent through n- and $\pi$- electron pairs.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;S&lt;/strong&gt; is the solute dipolarity/polarizability. It gives a measure of the solute's ability to stabilize a charge or dipole&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A&lt;/strong&gt; is the overall (summation) hydrogen bond acidity. The hydrogen bond acidity descriptor measures the extent of hydrogen bonding by the solute in a basic solvent. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;B&lt;/strong&gt; is the overall (summation) hydrogen bond basicity. The hydrogen bond basicity descriptor measures of the extent of hydrogen bonding by the solute in an acidic solvent.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;V&lt;/strong&gt; is the McGowan characteristic volume in units of (cubic cm per mol)/100.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The nice thing about this model is that it makes sense -- you want a solute that matches lone pairs, hydrogen bond donating or accepting ability, polarizability, and volume for the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch is that surprisingly little data is available. The late Jean-Claude Bradley encouraged his students to &lt;em&gt;make&lt;/em&gt; accurate solubility measurements, but also to publish them on the web. In this way, they gathered a large spreadsheet of &quot;compound X dissolves Y amount in Z solvent.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-15T13:41:29.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16323" PostTypeId="1" AcceptedAnswerId="16333" CreationDate="2014-09-15T13:43:45.043" Score="10" ViewCount="2484" Body="&lt;p&gt;I know that in piperidine, the lone pair on the nitrogen can occupy either an axial or a equatorial position in the two chair forms. I also read that the axial position for the lone pair is preferred by a small margin &lt;em&gt;in the gas phase&lt;/em&gt; (so intermolecular forces are minimized). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WRcIv.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this was presented as an argument against the idea that lone pairs are &quot;sterically demanding&quot; or &quot;bulky&quot; or repel other entities of negative charge more strongly than do bonding pairs, etc. The conclusion was that a lone pair must occupy less space than a hydrogen atom since it prefers the axial position in piperidine! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This argument seems correct at first, but could there be other factors affecting why the axial lone pair conformation might be more stable than the equatorial lone pair conformation? Also, are there any other examples in which the lone pair behaves &quot;unexpectedly,&quot; at least with respect to elementary theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: On the other hand one could make the argument that this isn't a apples to apples comparison; we are comparing a lone pair of electrons with a bonding pair of electrons plus a hydrogen atom. So it may be that this is a steric rather than an electronic (i.e. repulsive) effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or could there be some sort of hyperconjugative effect going on? Carbon withdraws electron density from hydrogen; hydrogen becomes slightly partially positive, is attracted to lone pair? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-15T23:00:34.930" LastActivityDate="2017-02-14T13:28:14.220" Title="Steric effects of a lone pair and piperidine" Tags="&lt;organic-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16324" PostTypeId="2" ParentId="16287" CreationDate="2014-09-15T13:51:17.820" Score="1" Body="&lt;p&gt;I like the question above, since it gives a general strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll point out that mixing dichromates and sulfuric acid makes &quot;dichromix&quot; and there are a complicated mixture of (chromic acid)[&lt;a href=&quot;http://en.wikipedia.org/wiki/Chromic_acid]&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chromic_acid]&lt;/a&gt; and related compounds. Depending on pH and other factors, you can have chromates, dichromates, and even sulfur-chromate mixed ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one point chromic acid was used to clean glassware, since they are very strong oxidizing agents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SAFETY NOTE&lt;/strong&gt; The use of chromic acid (aka &quot;chromerge&quot;) is highly unsafe and &lt;a href=&quot;http://chemlab.truman.edu/Miscellaneous_files/Cleaning.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;not&lt;/strong&gt; recommended!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-15T13:51:17.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16325" PostTypeId="2" ParentId="16307" CreationDate="2014-09-15T15:36:19.407" Score="3" Body="&lt;p&gt;While not an acid-base reaction, an example of mixing two relatively &quot;safe&quot; chemicals to form dangerous ones is bleach (sodium hypochlorite) plus ammonia - both common household chemicals. This reaction yields chloramine, dichloramine, and nitrogen trichloride, all more toxic and hazardous than bleach or ammonia.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-09-15T15:36:19.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16326" PostTypeId="2" ParentId="16261" CreationDate="2014-09-15T16:04:32.557" Score="15" Body="&lt;p&gt;There are examples of metal compounds that are regarded by a majority of chemists as covalent in organometallic chemistry. The problem is that not everyone regards some of these compounds as covalent, because the border between ionic and covalent isn't so strict and it doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkyllithium compounds and Grignard reagents ($\ce{R-Li}, \ce{RMgX}$) are considered to be covalent, and they are represented like this $\ce{CH_3CH_2-Li, CH_3-Mg^+I^-}$ (no +,- signs above the carbon and lithium and only the magnesium-halogen bond in Grignard reagents is ionic), while alkyl-sodium and potassium compounds are considered to be ionic (like sodium-acetylide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Organoaluminium example of a covalent compound is $\ce{Al(C_2H_5)_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organocopper exhibits covalent bonds between copper and carbon, like in $\ce{(CH_3)_2CuLi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more examples in organometallic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The d-block metals have significant polarizability and thus the bonds they participate in can have a significant covalent character like in $\ce{RuO_4}$, $\ce{Hg(CH_3COO)_2}$, $\ce{Pb(CH_3COO)_4}$, $\ce{CuI}$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury(I) exists in solution as $\ce{Hg_2^{2+}}$ where the two mercury ions are covalently bonded.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="17175" LastEditDate="2017-12-01T14:58:37.910" LastActivityDate="2017-12-01T14:58:37.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16327" PostTypeId="1" AcceptedAnswerId="27362" CreationDate="2014-09-15T16:19:38.797" Score="-1" ViewCount="93" Body="&lt;p&gt;I'm trying to find a reason for an experimental observation, I noticed that when the absorbance of $\ce{Ca}$ is measured with FAS(flame atomic spectroscopy) that it decreases when metals such as $\ce{Al}$ and $\ce{K}$ is present(the ions). I used a $\ce{Ca}$ hallow cathode lamp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea as to why? &lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorUserId="5456" LastEditDate="2014-09-15T18:39:22.137" LastActivityDate="2020-03-30T19:16:44.917" Title="Why does the absorbance of Ca decrease in the presence of certain metals?" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16328" PostTypeId="2" ParentId="16307" CreationDate="2014-09-15T16:24:24.177" Score="1" Body="&lt;p&gt;I can't think of any acid base examples where you will make something more toxic than what you are trying to neutralize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said I think you should always be washing with copious water. There's no reliable way to know how much material is on your skin, which means you won't know how much neutralizing agent to use. You are better off trying to dilute the acid or base dramatically to avoid its adverse effects. Furthermore, this gives you an immediate procedure to follow if a spill should occur. Instead of going through the process of considering what the nature of the contamination is, finding the appropriate counter-reagent, and approximating the amount needed for neutralization, just immediately go to the sink. The time saved is critical for minimizing injury.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-15T16:24:24.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16329" PostTypeId="2" ParentId="16307" CreationDate="2014-09-15T16:25:45.440" Score="2" Body="&lt;p&gt;For &quot;everyday chemistry&quot;, I cannot think of an example besides &lt;a href=&quot;https://chemistry.stackexchange.com/a/16325/1677&quot;&gt;that with bleach&lt;/a&gt; above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, mixing hydrogen peroxide and sulfuric acid to make &lt;a href=&quot;http://en.wikipedia.org/wiki/Piranha_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;piranha solution&lt;/a&gt; is another unsafe mixture. &lt;a href=&quot;http://en.wikipedia.org/wiki/Aqua_regia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aqua regia&lt;/a&gt;, mixing nitric and $HCl$ would be another problem. Both reagents should be avoided. That's not what you're asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that typical &quot;strong&quot; acids and bases are highly ionic compounds. That is, the definition of a strong acid or base is one that dissociates completely in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So typical neutralization with weak acids and weak bases will still produce soluble salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are caveats. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Neutralization often is exothermic, and with common acids ($HCl$, $H_2SO_4$ and $HNO_3$ the heating effect can drive release of the gases, which are corrosive. So if you have concentrated acids, you should dilute &lt;em&gt;significantly&lt;/em&gt; by pouring the acid into water before neutralization with weak base.&lt;/li&gt;&#xA;&lt;li&gt;Nitric acid. The various nitrous oxides are dangerous, and care must be used to prevent them from being released. Worse, nitric acid is a strong oxidizer and shouldn't be used with organic reagents, especially organic bases.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the usual rules apply. Dilute, then neutralize carefully.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-15T16:39:48.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16330" PostTypeId="1" AcceptedAnswerId="16369" CreationDate="2014-09-15T18:29:15.807" Score="4" ViewCount="373" Body="&lt;p&gt;I would like to address the compound below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCcJf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This compound on elaborate elongation would result in this structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hicyi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is the below given numbering correct? I am just wondering that what if the same compound is given numbering, starting from that carbon which is assigned 7th number in the first elaborate diagram (second diagram), by doing so, triple bonded substituent will also get the least number, as shown in the below given diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/brVkF.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have a similar problem with this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oz8M3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if you just reverse the numbering? By reversal I mean, the carbon assigned 1st number should be given the last number, am I violating any rule? If I do that, by numbering the compound in reverse order, we get least number for the methyl chain substituent, so is it necessary that methyl group substituted in the side chain should get the highest possible number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I mean, is the below given numbering valid or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWqdq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6953" LastEditorUserId="5026" LastEditDate="2018-07-17T03:06:53.120" LastActivityDate="2018-07-17T03:06:53.120" Title="How to define the atom numbers in unsaturated hydrocarbons correctly according to IUPAC?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="16331" PostTypeId="1" CreationDate="2014-09-15T21:40:04.670" Score="2" ViewCount="924" Body="&lt;p&gt;I poured some water in a balloon, inflated, and tied it up.  Over a period of couple weeks, all water in the balloon is gone.  But the balloon is still at about 2/3 of original inflated size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the the air molecules are much smaller than water vapor.  Why the water is all vaporized?&lt;/p&gt;&#xA;" OwnerUserId="7861" LastActivityDate="2014-12-14T23:18:22.060" Title="water vapor vs air (in a balloon)" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16332" PostTypeId="2" ParentId="16331" CreationDate="2014-09-15T22:09:29.310" Score="0" Body="&lt;p&gt;Guessing, I would say that the water diffused through the plastic. Many plastics absorb water quite well. However I'm only guessing.&lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2014-09-15T22:09:29.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16333" PostTypeId="2" ParentId="16323" CreationDate="2014-09-15T22:57:40.600" Score="9" Body="&lt;p&gt;The energy difference between the two conformations in your example is extremely small.  This makes it difficult to say that &quot;this&quot; is the reason why things are the way they are. Nonetheless, it is a very interesting question. Perhaps, like in the &quot;how many angels can you draw on the head of a pin&quot; question, the discussion around the question might be as significant as the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we need is more data.  The phosphorus analogue of piperidine, phosphacyclohexane, has also been examined and the hydrogen resides in the axial position (no evidence for the equatorial isomer was observed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lVP4j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are important differences between these nitrogen and phosphorus analogues.  To start with, phosphorus-carbon and phosphorus-hydrogen bond lengths (180-85 pm and 145 pm respectively) are typically longer than the analogous nitrogen-carbon and nitrogen-hydrogen bonds lengths (150 pm and 102 pm respectively).  More important is the fact that the phosphorus is basically unhybridized, so the lone pair of electrons around phosphorous exists in a non-directional s-orbital.  This makes conformational analysis of phosphacyclohexane easier because we don't have competing (hydrogen and lone pair) substituent effects to consider and dissect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation that the hydrogen in phosphacyclohexane exists in the axial position suggests that there is an attractive 1-3 synaxial H-H interaction and/or a sterically repulsive 1-2 H-H gauche interaction.  Given the long length of bonds to phosphorous, repulsive interactions seem unlikely and the attractive 1-3 synaxial H-H explanation was favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another data point comes with N-methyl piperidine where the lone pair strongly prefers the axial position.  This was attributed to a repulsive 1-3 synaxial Me-H interaction.  The fact that the methyl group adopts the equatorial position also shows that the repulsive 1-2 Me-H gauche interaction must be of little consequence and any repulsion in the corresponding H-H case would be even smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piperidine itself appears to be somewhere in the middle, in between N-methyl piperidine and phosphacyclohexane.  Certainly the repulsive 1-3 synaxial Me-H interaction observed in N-methyl piperidine has been reduced, but not eliminated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether just the repulsive 1-3 synaxial H-H interaction has been reduced or whether there is now also a contribution from the attractive 1-3 synaxial H-H interaction seen in phosphacyclohexane cannot be determined. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-15T22:57:40.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16334" PostTypeId="1" AcceptedAnswerId="16335" CreationDate="2014-09-15T23:10:24.807" Score="1" ViewCount="1421" Body="&lt;p&gt;In discussing how best to deuterate chloroform, it occurred to me that we &lt;em&gt;could&lt;/em&gt; use sodium metal to reduce hydrogen into hydrogen gas. If this &lt;em&gt;could&lt;/em&gt; actually happen, this would be a great thing - we wouldn't have much if any residual hydrogen in the liquid phase of the system after treatment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? I had a possible mechanism in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcrbi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems I foresee are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Won't reduction of chlorine to chlorine gas be a competitive, side reaction? From this page (&lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&lt;/a&gt;) I see that the reduction potential for chloride ion to to chlorine gas is -1.36 V (unfavorable) and reduction of hydride ion to hydrogen gas is a +2.23 V (favorable). Am I interpreting and applying the data correctly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What happens to the $\ce{CCl_3}$ radicals after hydrogen is reduced? Is dimerization going to occur to a large extent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How then do we introduce deuterium and effectively incorporate it with $\ce{CCl_3}$ radicals? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-16T00:32:46.720" LastActivityDate="2014-09-16T00:32:46.720" Title="Sodium metal reduction of chloroform" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16335" PostTypeId="2" ParentId="16334" CreationDate="2014-09-15T23:51:32.840" Score="4" Body="&lt;p&gt;Sodium + Chloroform:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=du3ksrfnCcg&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=du3ksrfnCcg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prepare deuterochloroform I'd recommend the haloform reaction.&lt;/p&gt;&#xA;" OwnerUserId="7862" LastActivityDate="2014-09-15T23:51:32.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16338" PostTypeId="1" AcceptedAnswerId="16349" CreationDate="2014-09-16T00:39:26.930" Score="0" ViewCount="1375" Body="&lt;p&gt;Can someone at least give me a hint to push me in the right direction of how to estimate the magnitude of the absolute and percentage errors when using a smaller amount in mL of a substance for a lab measurement?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="10185" LastEditDate="2015-10-02T15:50:12.397" LastActivityDate="2015-10-02T15:50:12.397" Title="Estimating magnitude of percentage and absolute error" Tags="&lt;homework&gt;&lt;analytical-chemistry&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16339" PostTypeId="1" AcceptedAnswerId="16341" CreationDate="2014-09-16T01:24:16.710" Score="11" ViewCount="7943" Body="&lt;p&gt;The monomer in polyethene seems to drop the $\ce{C=C}$ bond and end up with a $\ce{C-C}$ bond. The polymer chain then looks like $\ce{[CH2-CH2]_{n}}$. If I were to look at that polymer chain I would probably call it polyethane not polythene (or I could even call it polymethane) because it doesn't have a double bond, so I wouldn't classify it as an alkene. Am I doing something wrong with my analysis here? Or is there some sort of rule about nomenclature that change when things become polymers and drop a functional bond (in this case the $\ce{C=C}$ bond)?&lt;/p&gt;&#xA;" OwnerUserId="7864" LastEditorUserId="4945" LastEditDate="2015-01-16T02:05:00.257" LastActivityDate="2020-03-26T23:15:21.070" Title="What is the difference between polyethene and polyethane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;polymers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16340" PostTypeId="2" ParentId="16330" CreationDate="2014-09-16T01:35:24.033" Score="2" Body="&lt;p&gt;Your first compound should be (3E)-5-ethynylhepta-1,3,6-triene. The 2 conjugated double bonds at the end determine the first carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second compound should be (4-methylpentyl)cyclobutane. The aliphatic group has less precedence than the cyclobutane group.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-16T01:35:24.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16341" PostTypeId="2" ParentId="16339" CreationDate="2014-09-16T01:43:20.593" Score="9" Body="&lt;p&gt;The reason it is called polyethene (or polyethylene) is because it was produced from the monomer of ethylene. To produce polyethylene, a radical initiator is added to ethylene, which attacks one carbon and opens up the double bond leaving a radical at the end. This radical attacks another ethylene molecule, etc. The chain stops forming when it encounters another radical. So, from ethylene, we get polyethylene (even though it looks like polymethylene).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="17952" LastEditDate="2020-03-26T23:15:21.070" LastActivityDate="2020-03-26T23:15:21.070" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16342" PostTypeId="1" AcceptedAnswerId="16374" CreationDate="2014-09-16T02:48:02.403" Score="17" ViewCount="5833" Body="&lt;p&gt;I need someone to back me up on this before I go confront my teacher:&#xA;I was doing some analysis of the dipole moment of cis-2-butene.&#xA;Let's say that the alkyl groups are both on top.&#xA;Would the dipole moment be a vector that has its tail on the top and its head on the bottom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one calculate that?&#xA;I made some calculations assuming all C-H bonds have the same dipole moment vector and came to the conclusion that the three H's attached to the alkyl groups actually end up overcoming the dipole moments of the C-H bonds of the two unsubstituted H's on the double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="14" LastEditDate="2014-09-17T13:20:38.393" LastActivityDate="2014-10-02T12:48:40.263" Title="Dipole moment of cis-2-butene" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16343" PostTypeId="2" ParentId="16342" CreationDate="2014-09-16T03:17:33.063" Score="12" Body="&lt;p&gt;This was actually an interesting problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I ran a quick calculation using &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt; and GAMESS, although other packages would work. This is a CCSD/aug-cc-pVTZ calculation, pretty much the gold standard in quantum chemistry. (CCSD(T) and a larger basis set might be better, but are unlikely to differ much here and I can't run them on my laptop.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall dipole moment is 0.305 Debye for cis-2-butene, so it's weakly polar. Here's the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tCR0d.png&quot; alt=&quot;dipole diagram for cis-2-butene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you probably know, the standard in chemistry is that the dipole moment is drawn from + pointing to - (e.g. +--&gt;) so the dipole arrow is as drawn in the figure, with the &quot;tail&quot; of the arrow from the methyl groups to the double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it's very similar to your thought process. Also, there are some comments below with links very similar to your discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the interesting part. The differences in partial charges are &lt;em&gt;very&lt;/em&gt; small. My initial calculations used the PM6 and PM7 &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-empirical_quantum_chemistry_method&quot; rel=&quot;noreferrer&quot;&gt;semiempirical methods&lt;/a&gt;. I also tried B3LYP, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;noreferrer&quot;&gt;density functional&lt;/a&gt; method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These get the size of the dipole moment right, but the direction opposite. So who knows, maybe your teacher did the same thing I did at first.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-16T13:05:32.300" LastActivityDate="2014-09-16T13:05:32.300" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="16345" PostTypeId="1" AcceptedAnswerId="16348" CreationDate="2014-09-16T04:56:21.043" Score="14" ViewCount="26398" Body="&lt;p&gt;As asked in the question and also, a bonus question: which property of a chemical would make it 'permanent' in terms of markers?&lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorUserId="7717" LastEditDate="2014-09-16T09:47:57.863" LastActivityDate="2017-11-20T13:44:13.523" Title="Which chemical is responsible for making marker ink permanent?" Tags="&lt;everyday-chemistry&gt;&lt;identification&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16346" PostTypeId="2" ParentId="16345" CreationDate="2014-09-16T05:36:53.523" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Basically, ink is a liquid or paste made up of pigments or dyes that is used to color a surface to create printed words or pictures. Modern inks are often made of complex combinations of various chemicals, including solvents, pigments, dyes, resins and other materials. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(From &lt;a href=&quot;http://wonderopolis.org/wonder/what-makes-ink-permanent/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The chemicals found in permanent markers--&lt;strong&gt;xylene, toluene and urethane resin&lt;/strong&gt;--are what give these markers their characteristic abilities to handle unusual surfaces and leave long-lasting messages. While substantial or prolonged exposure to these chemicals is harmful and toxic, the amount present in a permanent marker is minimal for the casual user.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(From &lt;a href=&quot;http://www.ehow.com/list_6193137_chemicals-permanent-markers.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I got from google search. The above mentioned chemicals are most oftenly used, which make ink permanent. However If you have notice you would realise that the ink actually is not permanent, it fade away after certain period of time!&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-09-16T08:41:42.807" LastActivityDate="2014-09-16T08:41:42.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16348" PostTypeId="2" ParentId="16345" CreationDate="2014-09-16T06:15:49.973" Score="14" Body="&lt;p&gt;The &quot;permanence&quot; is just a reference to it being waterproof. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, the ink comprises a main carrier solvent, a glyceride, a pyrrolidone, a resin and a colorant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is mostly the result of solubility. There are other markers sold on the market with labels such as &quot;buff proof&quot; (slang for removal proof), that are really just pens that use a different carrier solvent. With those, typically organic solvents would be the correct solvent for removal.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-16T06:15:49.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16349" PostTypeId="2" ParentId="16338" CreationDate="2014-09-16T06:30:58.500" Score="3" Body="&lt;p&gt;When you take a small amount of a liquid in mL, you you usually use a pipette. On every pipette, at the top of it, the absolute error is written. As for the relative error, it is the ratio of the absolute error to the measured value. The percent error is the relative error multiplied by 100.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-16T06:30:58.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16350" PostTypeId="2" ParentId="16212" CreationDate="2014-09-16T07:13:21.393" Score="4" Body="&lt;p&gt;The box in my kitchen is labelled &quot;reine Zitronensäure&quot; (= pure citric acid) and &quot;lebensmittelsauber&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the manufacturer, the latter means that citric acid in this quality can be used to descale (decalcify) kitchen items, but it is &lt;strong&gt;not&lt;/strong&gt; intended to be used as a food supplement and should &lt;strong&gt;not&lt;/strong&gt; be digested! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may have real technical reasons, i.e. the product may in fact contain impurities resulting from production, processing and packaging. The manufacturer however does not specify the kind or amount of impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, this kind of labelling may just reflect different legal treatment (certification, monitoring, etc.) of products intended as food supplement and those only used to clean kitchen equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning waterfree vs monohydrate, neither the package in my kitchen, nor the website of the manufacturer provide any information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know a chemist, this question is easily solved: Let him measure the melting point of a sample. Waterfree citric acid melts at 153 °C, the monohydrate at 100 °C. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-16T07:13:21.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16351" PostTypeId="2" ParentId="16127" CreationDate="2014-09-16T07:52:31.883" Score="2" Body="&lt;p&gt;Let's rephrase your hypothesis and test it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The green colour of leaves results from a d-d transition of the central atom of metal complexes known as chlorophylls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Green leaves contain two types of chlorophyll, &lt;em&gt;Chl a&lt;/em&gt; and &lt;em&gt;Chl b&lt;/em&gt;, both have $\ce{Mg}$ as the central atom and show two distinct bands in the UV-VIS absorption spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stronger ($\epsilon \approx 100000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) band in the 400 nm (blue) region is known as the &lt;strong&gt;Soret band&lt;/strong&gt;, the weaker  $\epsilon \approx 30000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) in the 660 nm (red) region is known as the &lt;strong&gt;Q band&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that &lt;em&gt;Chl a&lt;/em&gt; is treated with weak $\ce{HCl}$. In the resulting product, known as &lt;em&gt;Pheophytin a&lt;/em&gt;, the central atom ($\ce{Mg^{2+}}$) is replaced with two protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UV-VIS absorption spectrum, however, we would still observe the two bands. They are slightly shifted and show somewhat different extinction coefficients, but the pattern is fundamentally the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may conclude that our initial hypothesis is wrong. If it is not the absorption of the central atom in our complex, the ligand must be responsible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our new working hypothesis therefore is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The colour of green leaves results from the absorption of the polycyclic pheophytin ligand and its extended $\pi$ system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's simplify the &lt;em&gt;Pheophytin a&lt;/em&gt; molecule: Removal of the &lt;em&gt;E ring&lt;/em&gt; and replacing all the fancy substituents with ethyl groups leads to &lt;em&gt;octaethylporphyrin&lt;/em&gt; as the core structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zYxmM.png&quot; alt=&quot;structure of octaethylporphyrin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UV-VIS spectrum, we would still be able to see the strong &lt;strong&gt;Soret band&lt;/strong&gt;, which corresponds to a $\pi,\pi^*$ transition from the ground state to the second excited singlet state ($S_0 \rightarrow S_2$).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-16T08:39:23.337" LastActivityDate="2014-09-16T08:39:23.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16352" PostTypeId="1" AcceptedAnswerId="16425" CreationDate="2014-09-16T08:10:08.577" Score="0" ViewCount="7170" Body="&lt;p&gt;I've got a homework question i'm stumped on.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The viscosity of the ethylene gas was measured at $99.8\times10^{-7}\ \mathrm{N\ s\ m^{-2}}$ at $288\ \mathrm K$. Calculate the collision diameter in $\overset{\circ}{\mathrm{A}}$ for this gas molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta=\frac{m\overline v}{3\sqrt2\pi d^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then i said&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline v=\sqrt{\frac{8RT}{\pi M}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar mass $M=28.05\ \mathrm{g\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my equation looks like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta=\frac{m}{3\sqrt2\pi d^2}\sqrt{\frac{8RT}{\pi M}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but $m$ is still a variable. What should I do from here? Is there another substitution I can do?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So apparently my confusion about solving this problem, came from a lack of understanding what the $m$ variable was supposed to represent. I had thought that it was the total mass of the system, but it is actually the mass of one molecule.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorDisplayName="user7951" LastEditDate="2016-11-07T16:24:29.720" LastActivityDate="2016-11-07T16:24:29.720" Title="Calculating collision diameter with viscosity, but not having the mass given" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16353" PostTypeId="2" ParentId="16345" CreationDate="2014-09-16T08:18:10.977" Score="2" Body="&lt;p&gt;In my experience, ink of a permanent marker can be removed by drawing over it with a non-permanent marker and wiping it off (success depends on surface). The same goes for non-permanent writing that can't be removed because it is too old. Expo (apparently a company that sells markers and related stuff) confirms this in their &lt;a href=&quot;http://www.expomarkers.com/faq&quot; rel=&quot;nofollow&quot;&gt;FAQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the big question:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;how do I get permanent marker off a whiteboard?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;No worries if those pesky permanent markers make their way into your whitespace. Just draw over it with an Expo Dry Erase Marker and poof! It's gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this I would conclude that non-permanent markers contain some solvent that permanent markers lack. But obviously, permanent markers need some solvent, too. Otherwise, you wouldn't be able to write in the first place. So it must be some solvent that evaporates quickly while the solvent in whiteboard markers doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search for ingredients of whiteboard markers, the organic solvents &lt;a href=&quot;http://en.wikipedia.org/wiki/Xylene&quot; rel=&quot;nofollow&quot;&gt;Xylene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Toluene&quot; rel=&quot;nofollow&quot;&gt;Toluene&lt;/a&gt; keep popping up, so that's obviously not the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whiteboard cleaners contain all kinds of &lt;a href=&quot;http://www.answerbag.com/q_view/1904278&quot; rel=&quot;nofollow&quot;&gt;organic&lt;/a&gt; &lt;a href=&quot;http://old.akronschools.com/dotAsset/70957.pdf&quot; rel=&quot;nofollow&quot;&gt;solvents&lt;/a&gt; that evaporate much faster:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the main ingredient is isopropyl alcohol, which is sometimes called &quot;rubbing alcohol.&quot; Other ingredients may include 2-butoxy ethanol and (di)chloroacetic acid.&lt;br&gt;&#xA;  ~~~&lt;br&gt;&#xA;  ISOPROPANOL (ISOPROPYL ALCOHOL), 2-PROPANOL, DIMETHYL CARBINOL&lt;br&gt;&#xA;  2-BUTOXYETHANOL (ETHYLENEGLYCOL MONOBUTYL ETHER), BUTYL CELLOSOLVE, BUTYL GLYCOL, GLYCOL ETHER EB&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the wide range of possibilities, the actual chemicals used probably depend on the marker's brand.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Isopropyl_alcohol&quot; rel=&quot;nofollow&quot;&gt;Isopropyl alcohol&lt;/a&gt; seems to be a good bet, though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To answer your more generic &quot;bonus&quot; question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which property of a chemical would make it 'permanent' in terms of markers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ink needs to waterproof, i.e. not water soluble&lt;/li&gt;&#xA;&lt;li&gt;The solvent needs to vaporize quickly&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7868" LastEditorUserId="7868" LastEditDate="2014-09-16T13:35:43.790" LastActivityDate="2014-09-16T13:35:43.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16354" PostTypeId="1" CreationDate="2014-09-16T09:08:46.130" Score="0" ViewCount="95" Body="&lt;p&gt;I need to determine the amount of chlorine in a swimming pool. How do I get started? What kind of experiment would I need to carry out?&#xA;What equipment would I need for this experiment?&lt;/p&gt;&#xA;" OwnerUserId="7869" LastEditorUserId="402" LastEditDate="2016-11-20T22:21:08.087" LastActivityDate="2016-11-20T22:21:08.087" Title="Determining the amount of chlorine in a swimming pool" Tags="&lt;water&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16355" PostTypeId="2" ParentId="16354" CreationDate="2014-09-16T09:31:24.617" Score="4" Body="&lt;p&gt;If you need a no-frills solution, get some quick-test paper, such as &lt;a href=&quot;http://www.mn-net.com/Testpapers/SpezielleTestpapiere/ChlorTest/tabid/10490/language/en-US/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly more complex solution would be a test kit like &lt;a href=&quot;http://www.merckmillipore.com/INTL/en/product/chemicals/chlorine-test,MDA_CHEM-114434&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, which makes use of the formation of &lt;em&gt;Bandrowski's base&lt;/em&gt; from 1,4-phenylenediamine in the presence of oxidants, such as chlorine.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-16T09:31:24.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16358" PostTypeId="1" AcceptedAnswerId="16839" CreationDate="2014-09-16T11:26:29.880" Score="0" ViewCount="214" Body="&lt;p&gt;&lt;strong&gt;Pemoline&lt;/strong&gt; is a psychostimulant that is no longer used in most countries. Similarly to the psychostimulants amfetamine and methylphenidate I can clearly see a &lt;strong&gt;phenethylamine backbone&lt;/strong&gt; to this compound. I know it is also an &lt;strong&gt;oxazoline&lt;/strong&gt;, but I am writing an article on my Wiki about substituted phenethylamines and their pharmacologic and structural properties so I would like to know what its chemical name would be in terms of its &lt;strong&gt;phenethylamine backbone&lt;/strong&gt;. If you want an example of what I mean, see the current revision of my Wiki article on said drug [&lt;a href=&quot;http://health-and-medicine.wikia.com/wiki/Substituted_phenethylamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] and look at the chemical names I have written. For example for methylphenidate the chemical name is &lt;em&gt;N&lt;/em&gt;,α-butyl-β-methoxycarbonylphenethylamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry knowledge is only first-year university chemistry level, hence why I need your help as I don't know all the adjective forms of functional groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the 2D structure of pemoline if you would like it in order to deduce its name in terms of phenethylamine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pa8eB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2014-09-28T07:08:10.443" LastActivityDate="2014-09-28T17:05:23.567" Title="How would one express pemoline in terms of phenethylamine?" Tags="&lt;nomenclature&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16359" PostTypeId="2" ParentId="16358" CreationDate="2014-09-16T12:26:32.343" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to know what its chemical name would be in terms of its phenethylamine backbone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Artificial numbering and naming schemes that only exist in ones own cloud cuckoo land are pointless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IUPAC nomenclature exists for a reason.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the IUPAC nomenclature, other naming schemes - based on established (trivial) names exists and are widely used - for good reason and with acceptance by the scientific community. Typical examples are found in the field of natural products chemistry (steroids, alkaloids, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however crucial to understand that in these cases, prefixes and/or suffixes denote a &lt;strong&gt;variation&lt;/strong&gt; of the underlying core structure, such as &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;additional substituents&lt;/li&gt;&#xA;&lt;li&gt;lack of methyl groups&lt;/li&gt;&#xA;&lt;li&gt;an elongated side chain, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the current case, the central structure is a &lt;strong&gt;completely different heterocyclus&lt;/strong&gt;, so &lt;strong&gt;markedly different&lt;/strong&gt; from the original phenethylamine that it would not make any sense to apply a phenethylamine-based name here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither would one try to apply such a naming scheme to the &lt;strong&gt;morphine&lt;/strong&gt; skeleton - despite the fact that one might spot a six-membered ring with $\ce{CH2-CH2-NCH3}$ side chain here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean that a member of the phenethylamine family is not among the ancestors of morphine in the biosynthesis! In fact, reticuline, a morphine precursor, most likely is a grandchild of dopamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, at a certain point structures have become so markedly different that a valid nomenclature (= valid for a particular compound class) no longer is applicable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to describe similarities between structures is not wrong, but trying to apply an inappropriate naming scheme is not helpful here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarities between molecular entities should be expressed otherwise.&#xA;For sequences (DNA, RNA, etc.), Hamming and Levenshtein distances can be used. For the title molecule and the whole dopamine/amphetamine class, I'd consider the comparison of their fingerprints (Tanimoto coefficients).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-17T15:05:05.950" LastActivityDate="2014-09-17T15:05:05.950" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16360" PostTypeId="2" ParentId="16261" CreationDate="2014-09-16T13:33:37.913" Score="10" Body="&lt;p&gt;There is no sharp border between ionic and covalent bond. Generally, single-atom anions are not stable enough to justify full transfer of an electron from cation, so simple binary compounds usually have significant covalent character, even if their crystal structure suggests otherwise. For example, $\ce{CsCl}$ said to have 25% covalent character. On the opposite end, heteroatomic bonds usually have some ionic character. Given that, we don't have clear groups of ionic and covalent metal compounds, but rather a spectrum of them.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CsClO4}$ would be an almost pure ionic compound&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CsCl}$ would be mostly ionic&lt;/li&gt;&#xA;&lt;li&gt;$\ce{AlCl3}$ would be highly polar, but significantly covalent&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CuS}$ would be almost purely covalent&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Li2}$ in gas phase is purely covalent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To get a structure with purely covalent bond we have to consider homoatomic bonds with atoms in identical neighborhood, otherwise some amount of polarity and consequently ionic bonding will arise. Such molecules with metal-metal bonds do exist. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let's start with the most clear (and, admittedly, least chemically significant) example: group one metal diatomic molecules. Such molecules do exist in gas phase and can be detected spectroscopically, though the bond energy is quite low because of low density and relatively insignificant overlapping of the orbitals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For group two there is a &lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2007/November/08110702.asp&quot; rel=&quot;nofollow&quot;&gt;report&lt;/a&gt; of a rather complex compound with a Mg-Mg bond&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the third row there are reports of various compounds with the most simple being $\ce{((((CH3)3Si)2CH)2Al)2}$ with a direct Al-Al bond. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are reports of $\ce{Sn2H6}$, i.e. $\ce{(SnH3)2}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, it must be noted that with atoms of the lower-left corner of periodic table homoatomic bonds are very unstable, to the point of impossibility to obtain stable compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many more examples are known for d-elements. For starters, $\ce{(Mn(CO)5)2}$ is a relatively stable compound with a singular Mn-Mn bond. However, d-elements favor complex structures with unusual delocalized, but clearly covalent bonding, similar to boranes. Such structures are especially common for late transition metals, say, from the Cr-W-Pt-Ni rectangle. An example would be $\ce{Rh6(CO)16}$, adopting structure with $\ce{Rh}$ atoms forming an (almost) perfect octahedron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking from another side, we can look for molecules with metal as a central atom. They are much more common in regular chemistry, though the nature of bonding may be sometimes disputed. For example, copper acetate adopts an unusual paddle-wheel &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper(II)_acetate&quot; rel=&quot;nofollow&quot;&gt;structure&lt;/a&gt;, but the copper-oxygen bond may be argued to be ionic. On the other hand, molecules with high d-element oxidation state usually have an effective charge of the central atom far below its formal oxidation state, so such compounds may be considered highly polar, but still covalent in nature. This said, many higher chlorides, oxochlorides and fluorides may be considered covalent, such as $\ce{TiCl4}$, $\ce{VCl4}$, $\ce{CrO2Cl2}$, $\ce{OsO4}$, $\ce{WF6}$, $\ce{AuF5}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking from third side we can argue to be mostly covalent compounds where metal is bound to radical, that clearly cannot stabilize a negative charge, such as $\ce{CO}$ molecule or $\ce{CH3}$ and some other organic fragments. Such compounds are known for most metals in the periodic table, say, $\ce{LiCH3}$, $\ce{CH3MgCl}$ (the nature of these compounds is more complex than is usually written), $\ce{Al2(CH3)6}$, $\ce{PbEt4}$ (tetraethyllead), and many d-element compounds, such as $\ce{(CH3)2CuLi}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I mostly avoided a huge can of worms named 'coordination compounds', which includes (mostly) interactions of d-element atoms with different electron pair donors. Some of such compounds may be argued to be mostly covalent, such as most metal carbonyles, like $\ce{Cr(CO)6}$, but some may be argued to be mostly ionic, such as $\ce{[FeF6]^{3-}}$, or ion-dipole, such as $\ce{[Co(NH3)6]^{2+}}$&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5354" LastEditDate="2015-09-12T16:41:12.130" LastActivityDate="2015-09-12T16:41:12.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16361" PostTypeId="1" AcceptedAnswerId="16366" CreationDate="2014-09-16T15:26:03.327" Score="2" ViewCount="1941" Body="&lt;p&gt;According to the Molecular Orbital Theory, the method to check whether a given compound is paramagnetic or diamagnetic is to draw the molecular orbitals and then to see whether we have all the electrons paired or is there some unpaired electron. However, I want to know, is there a quick way to find the magnetic nature without drawing the orbital diagrams?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastActivityDate="2014-09-16T16:37:40.553" Title="Quick way to find magnetic nature of compounds or ions" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16362" PostTypeId="1" AcceptedAnswerId="16364" CreationDate="2014-09-16T15:50:38.853" Score="7" ViewCount="5906" Body="&lt;p&gt;I keep seeing the term cropping up, but cannot seem to find a definition for it. My understanding is that it is an environment in which, when the isomers interact with the environment, the two enantiomers can be distinguished. The only example I can think of is with the human body, where one type of enantiomer can form an enzyme-substrate complex and the other cannot. I can't think of any other examples though. Can you create a chiral environment in a lab?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="5456" LastEditDate="2014-09-16T16:02:52.757" LastActivityDate="2014-09-17T00:13:19.797" Title="What are chiral environments?" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16363" PostTypeId="1" AcceptedAnswerId="16365" CreationDate="2014-09-16T15:51:10.327" Score="1" ViewCount="863" Body="&lt;p&gt;For the reaction of 1,3-butadiene with DBr (D = heavy hydrogen), I found that several possible products could be formed. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Kinetic products&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A pair of enantiomers is formed. One of the pi bonds attacks the heavy proton and the leaving group, bromide anion, comes back around to take care of the allylic carbocation. Because the bromide can attack from both above and below the plane of the carbocation, both enantiomers are formed. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Thermodynamic products&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;2) Now, because there is resonance, the carbocation possesses partial primary allylic carbocation character. As a result, bromide anion &lt;em&gt;may&lt;/em&gt; also attack at a different position. This results in an internal alkene - the more stable alkene (the Zaitsev alkene). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diastereomers are formed as well because single bonds can rotate, and the partial, terminal double bond can be &quot;flipped&quot; upward in the resonance depiction (I know that resonance depictions do not exist independently; hence &quot;partial&quot;). This allows the Z alkene to be formed in addition to the E alkene, as illustrated by my mechanism. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1) Are the products all valid and correctly categorized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is the E and Z isomer each produced, and in the way I think both are produced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BwmB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-16T16:49:01.877" LastActivityDate="2014-09-16T16:49:01.877" Title="Free rotation of sigma bonds; diastereomer formation; thermodynamic and kinetic products" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16364" PostTypeId="2" ParentId="16362" CreationDate="2014-09-16T16:23:27.903" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My understanding is that it is an environment in which, when the&#xA;  isomers interact with the environment, the two enantiomers can be&#xA;  distinguished.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's exactly right.  Some examples of chiral environments that could be created in a lab would include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Running a reaction in a chiral solvent&lt;/li&gt;&#xA;&lt;li&gt;Running a reaction in a chiral cavity&lt;/li&gt;&#xA;&lt;li&gt;Running a reaction with a chiral reagent&lt;/li&gt;&#xA;&lt;li&gt;Running a photochemical reaction with a polarized light source&lt;/li&gt;&#xA;&lt;li&gt;Carrying out a reaction on a surface to which chiral molecules have been attached (certain clays have this property)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these items are like a glove, and each enantiomer will fit differently.  This different &quot;fit&quot; leads to transition states of different energy as each enantiomer moves along the reaction coordinate towards the product(s).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-17T00:13:19.797" LastActivityDate="2014-09-17T00:13:19.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16365" PostTypeId="2" ParentId="16363" CreationDate="2014-09-16T16:33:12.697" Score="3" Body="&lt;p&gt;You have only considered reaction from the s-trans conformation of 1,3-butadiene, where the 2 double bonds are in a cis relationship about the central single bond.  The s-trans conformation is in equilibrium with the s-cis conformation.  The s-cis conformation can react also react to the terminal olefin, but also  (by the mechanism you described) it can react to produce the z 2-butene as one of its kinetic products.  Hence, all 3 butenes (1-butene and e and x 2-butene) can be formed as the kinetic products. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-16T16:33:12.697" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16366" PostTypeId="2" ParentId="16361" CreationDate="2014-09-16T16:37:40.553" Score="3" Body="&lt;p&gt;Well, for some compounds it's easy, because there are an odd number of electrons - thus there's no way to pair them all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, there is no quick-fire way without doing the MO diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider $O_2$ which has an even number of electrons, but is a ground-state triplet with two unpaired electrons. The only way to know is to look at the MO diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you practice, you'll realize certain patterns and develop more of an intuition.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-16T16:37:40.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16367" PostTypeId="1" AcceptedAnswerId="16368" CreationDate="2014-09-16T16:52:15.140" Score="9" ViewCount="908" Body="&lt;p&gt;What's wrong with ionic resonance structures? I asked one professor about them once and all he commented was &quot;I've seen them too, and I think they're wrong.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another professor rejected an ionic resonance structure showing the &quot;H+&quot; ion and asked me what phase the experiment was run in, in an attempt to figure out why the authors reported a resonance contributor with a H+ proton. He specifically asked if the experiment was run in the gas phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this guy forgot what resonance structures are because I know that he emphasizes that bare protons do not exist in solution, and he was probably thinking about how there could be a bare proton shown in a resonance structure if the molecule is in solution. The resonance structure I showed him was this (which I could see him picturing in solution, given that it is &lt;em&gt;nitrous acid&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVkuN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, resonance structures are &lt;em&gt;not&lt;/em&gt; discrete forms of the molecule in question - they're just contributors and show that there is &lt;em&gt;partial&lt;/em&gt; ionic character; that there is a highly electronegative atom that is withdrawing electron density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Are ionic resonance structures valid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If so, why might they be valid, or what counterarguments might you present to someone who argues they are invalid?  &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-16T17:09:15.750" LastActivityDate="2015-04-21T13:18:40.607" Title="Validity of ionic resonance structures" Tags="&lt;resonance&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16368" PostTypeId="2" ParentId="16367" CreationDate="2014-09-16T19:23:05.817" Score="7" Body="&lt;p&gt;I can't imagine anyone arguing that ionic resonance structures are invalid.  The often contribute significantly to the structure of a molecule and help explain the various physical aspects of the molecule.  The following ionic resonance structure contributes significantly to the description of the carbonyl compound.  $\ce{NaCl}$ is not totally ionic and is described by a blend of both the ionic (major) and covalent (minor) resonance structures shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDJ0g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structures with a positive hydrogen (proton) are important in explaining hyperconjugative effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hdwdn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-16T19:23:05.817" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16369" PostTypeId="2" ParentId="16330" CreationDate="2014-09-16T19:29:35.577" Score="2" Body="&lt;p&gt;For your first structure, I think your numbering is correct. Your choice is between including the 6-alkene or 6-alkyne in the main chain. The rule is &quot;tie goes to the alkene&quot;.  &lt;a href=&quot;http://www.cpp.edu/~psbeauchamp/pdf/314_supp_8_nomen_2.pdf&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; may provide useful examples starting around p. 9.  Also see  page 5, item 7, &quot;middle&quot; example in the above link for another relevant example. Note how in this example (6-ethyl-6-methylocta-trans-2-cis-4,7-triene) they do not start numbering from the side with the terminal double bond, but rather from the side with the lowest overall double bond numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your second compound, your first numbering scheme has the lower substituent numbering (1,4 vs. 2,5).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-12-30T15:29:10.783" LastActivityDate="2015-12-30T15:29:10.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16370" PostTypeId="1" CreationDate="2014-09-16T20:04:57.843" Score="6" ViewCount="1893" Body="&lt;p&gt;Electrolysis of water, means that the water is separated into hydrogen and oxygen. As beer also includes water, I initially thought electrolysis of beer would cause the water to separate, and the hydrogen and oxygen given off gas, leaving behind a higher concentrate alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However then realise that ethanol is also in beer (and is what makes you drunk it seems), and like water has an OH bond, so would the electrolysis also cause that to break?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also does the rest of the molecule affect the strength of the OH bond, either stronger or weaker?)&lt;/p&gt;&#xA;" OwnerUserId="17" LastActivityDate="2014-09-16T22:33:46.077" Title="Electrolysis of Beer" Tags="&lt;water&gt;&lt;molecules&gt;&lt;electrolysis&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16371" PostTypeId="1" AcceptedAnswerId="16373" CreationDate="2014-09-16T20:27:46.430" Score="2" ViewCount="1843" Body="&lt;p&gt;Is the carbon in a carbonyl functional group &lt;em&gt;electropositive&lt;/em&gt;? I had a professor tell me so, but I immediately refuted him, because he defined electronegativity earlier as the ability to stabilize negative charge ... and taking electronegativity to be a converse property, the strong partial positive character on the carbon probably isn't able to stabilize any further positive charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, I know that lone pairs are defined as having an electronegativity of 0 because electrons cannot stabilize other electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guy agreed with me. So, to confirm, am I correct, and, as an extension, can we describe the electron deficient carbon as &lt;em&gt;electronegative&lt;/em&gt; instead? I could very well see a electron-deficient carbon stabilizing negative charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we stick with the IUPAC definition of electronegativity - which is the ability to &lt;em&gt;attract&lt;/em&gt; electrons, the carbonyl carbon still seems to lose to oxygen. A charge-separated resonance structure can be drawn for any carbonyl group with a + on carbon and a - on oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that leads me to another question - is &lt;em&gt;electronegativity&lt;/em&gt; the ability to &lt;em&gt;stabilize&lt;/em&gt; negative charge or the ability to &lt;em&gt;attract&lt;/em&gt; electrons within a covalent framework? I guess the two go hand in hand - i.e. you wouldn't &lt;em&gt;attract&lt;/em&gt; electrons if you weren't able to &lt;em&gt;stabilize&lt;/em&gt; them ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question: can we calculate EN for ions? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-16T20:42:15.313" LastActivityDate="2018-05-08T17:10:01.643" Title="Carbonyl carbon electropositivity?" Tags="&lt;organic-chemistry&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16372" PostTypeId="1" AcceptedAnswerId="16375" CreationDate="2014-09-16T21:03:09.157" Score="3" ViewCount="1096" Body="&lt;p&gt;My professor emphasizes the concept of the carboxylate &quot;cage&quot; or the idea that electrons participate in resonance within a carboxylate group such as in the benzoate anion but the electrons in the carboxylate group &lt;em&gt;cannot&lt;/em&gt; leave this &quot;cage&quot; - i.e. participate in resonance with the phenyl substituent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this idea valid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already poked a hole into it by pointing out that electron donation to the &quot;cage&quot; is possible - i.e. consider carbonic acid with one hydroxyl group replaced by oxygen connected to a phenyl group. Valid resonance structures can be drawn in which electron density from the oxygen in the ester linkage participates in resonance with the electrons &quot;in&quot; the carboxylate group. So donation into the &quot;cage&quot; is possible. Now, this begs the question - is withdrawal from the cage possible as well? He insists no. Now, he might be hiding some edge cases for pedagogical purposes, but I don't care - if there is an exception - I want to know about it (because chemistry always gets interesting at the exceptions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4eo3I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-17T00:29:59.970" Title="Carboxylate &quot;cage&quot; concept" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16373" PostTypeId="2" ParentId="16371" CreationDate="2014-09-16T21:12:45.893" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the carbon in a carbonyl functional group electropositive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Electronegativity and electropositivity are usually used in a relative sense, X is more electronegative than Y.  The carbonyl carbon in acetone is more electropositive than the central carbon atom in isobutane, and less electropositive than the methyl carbocation carbon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;because he defined electronegativity earlier as the ability to&#xA;  stabilize negative charge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Weak definition, wouldn't you say; that's why you won that argument.  Atoms can be electronegative in the abstract, charge doesn't need to be involved.  The statement, &quot;fluorine is more electronegative than carbon according to the Pauling Electronegativty scale&quot; is accurate and understandable, yet no one is describing a situation where charge stabilization is involved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;because electrons cannot stabilize other electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand what you mean, but you should know by now that you can't make such sweeping statements.  Don't the electrons in one of the double bonds in 1,3-butadiene interact with and stabilize the electrons in the other double bond?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we describe the electron deficient carbon as electronegative&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, I understand what you mean, but... Yes, it is more electronegative than the carbon in an alkane, but not as electronegative as $\ce{F+}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is electronegativity the ability to stabilize negative charge or the&#xA;  ability to attract electrons within a covalent framework? I guess the&#xA;  two go hand in hand - i.e. you wouldn't attract electrons if you&#xA;  weren't able to stabilize them&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the two go hand in hand.  More electronegative atoms are better able to stabilize negative charge than less electronegative atoms.  It's all about lowering the overall system energy. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a side question: can we calculate EN for ions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure there is a way.  For example, using studying the $\ce{^{13}C}$-nmr spectrum of various carbocations leads to a relationship of (somewhere around) 180 ppm shift/ electron.  So you can estimate the fractional positive charge on any carbon atom, the carbonyl carbon in acetone for example.  Once you've estimated the amount of charge on an atom, I suspect there is a way to relate it to EN.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="40947" LastEditDate="2018-05-08T17:10:01.643" LastActivityDate="2018-05-08T17:10:01.643" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16374" PostTypeId="2" ParentId="16342" CreationDate="2014-09-16T21:25:34.107" Score="27" Body="&lt;h1&gt;&lt;strong&gt;Mini Research Project Time&lt;/strong&gt;&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Updates&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Preamble&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This response is in no way meant to be contrary to what Geoff has already posted.  I happen to enjoy these types of questions and I like to tackle them as little research projects for fun.  Therefore, my answer will be brutally detailed and in-depth.  That said, this should be a good introduction to how I would approach this problem if I were going to do this at the 'production-level' of research but this definitely goes far beyond the purpose of any reasonable response for an SE answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Its All About the Dipoles Baby&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Okay so I wouldn't use THAT heading in a paper but maybe a talk depending on who my audience was...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily determine the dipole moment of cis-2-butene via electronic structure theory.  I have modeled two conformers of cis-2-butene as shown below.  I will refer to the geometry on the left as &lt;strong&gt;StructA&lt;/strong&gt; and the geometry on the right as &lt;strong&gt;StructB&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8m2RS.png&quot; alt=&quot;Two conformers of cis-2-butene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of things to note since I don't include any captions to tables.  Energies are always reported in $\mathrm{kcal\ mol}^{-1}$ and dipole moments ($\mu$) are given in Debye.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Computational Methods&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with second-order Moller-Plesset perturbation (MP2) theory and a variety of density functional theory (DFT) methods using the &lt;code&gt;Gaussian 09&lt;/code&gt; software package for the conformers of cis-2-butene.  Both conformers were characterized in $C_{2v}$ symmetry.  The DFT methods implemented include B3LYP, B3LYP-GD3(BJ), M06-2X, MN12SX, N12SX, and APFD.  The B3LYP-GD3(BJ) method employs Grimme's 3rd generation dispersion correction as well as the Becke-Johnson damping function. All DFT computations employed a pruned numerical integration grid having 90 radial shells and 590 angular points per shell.  The heavy-aug-cc-pVTZ basis set was employed for these computations where the heavy (non-hydrogen) atoms were augmented with diffuse functions (e.g. cc-pVTZ for H and aug-cc-pVTZ for carbon).  This basis set is abbreviated as haTZ.  The CCSD(T) (i.e. coupled-cluster method that includes all single and double substitutions as well as a perturbative treatment of the connected triple excitations) was similarly employed using the &lt;code&gt;CFOUR&lt;/code&gt; software package.  The magnitudes of the components of the residual Cartesian gradients of the optimized geometries were less than $6.8\times 10^{-6} E_h\ a_0^{-1}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single point energies were computed with the explicitly correlated MP2-F12 [specifically MP2-F12 3C(FIX)] and CCSD(T)-F12 [specifically CCSD(T)-F12b with unscaled triples contributions] methods in conjunction with the haTZ.  These computations were performed with the &lt;code&gt;Molpro 2010.1&lt;/code&gt; software package using the default density fitting (DF) and resolution of the identity (RI) basis sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural bond orbital (NBO) analyses were performed for the MP2 optimized structures using the haTZ basis set and the SCF density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All computations employed the frozen core approximation (i.e. 1s$^2$ electrons frozen in carbon).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Computational Methods in &lt;em&gt;English&lt;/em&gt;&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Two different conformations of cis-2-butene were characterized with a variety of cheap (but usually okay) approximations (i.e. DFT methods) as well as reliable (but generally more expensive) wave function methods (i.e. MP2 and CCSD(T)).  The wave function methods are necessary to validate the DFT results.  CCSD(T) is the gold-standard and gives very good results for single-reference closed-shell well-behaved systems so we will use this as our 'best estimate'.  We use a variety of methods in order to look for agreement in the results.  If we see good agreement across the board, we can be confident in our results.  If we see massive discrepancies then we will have to be careful when we analyze the results.  Geometries have been converged to a tight threshold (i.e. we have good molecules) and our DFT computations use a relatively dense integration grid (which leads to more accurate results).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I employ a heavy-aug-cc-pVTZ (haTZ) basis set.  Why leave the diffuse functions off hydrogen?  The purpose of diffuse functions is to describe electron density far away from the nucleus of an atom.  Therefore we slap these functions onto carbon which are relatively large atoms compared to hydrogen.  Hydrogen, on the other hand, has only one electron and therefore has a small electron density when isolated.  In cis-2-butene, hydrogen is bonded to carbon via a rather small bond distance.  The electron density around the hydrogen is even more reduced than an isolated hydrogen atom in the gas phase.  Therefore, it would be impractical to include diffuse functions on hydrogen.  Doing so may even lead to erroneous results since we will be trying to describe electron density far away from the hydrogen nucleus when in reality there is virtually none to be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we perform single point energies using explicitly correlated methods.  Because the CCSD(T) opts and freqs will likely not be done in time for this posting, we can gauge how the resulting geometry from each optimization procedure will vary from another geometry given by a different method.  If all of the energies are similar (within a few tenths of a $\mathrm{kcal\ mol}^{-1}$), then we can be confident that our geometries are not only very similar but that small deviations in the geometry will have little effect on the corresponding energies at least in this region of the potential energy surface (PES). Large deviations will usually mean that the method which produced the 'outlier' is not a good approximation for the system (I do not expect cis-2-butene to be problematic at all).  Explicitly correlated methods accelerate convergence to the CBS limit.  These methods have been shown to give results that a large basis set and a canonical method would provide but with a much smaller basis set.  For example,  the result that I get with a regular CCSD(T)/aug-cc-pV5Z basis set could be obtained using CCSD(T)-F12/aug-cc-pVTZ.  This makes the computations less intensive and much more feasible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Results&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The number of imaginary frequencies ($n_i$), relative MP2-F12 and CCSD(T)-F12 energetic  ($\Delta E^{\mathrm{MP2-F12}}$ and $\Delta E^{\mathrm{CC-F12}}$, respectively, in $\mathrm{kcal\ mol^{-1}}$) and dipole moment ($\mu_z$ in Debye) are given in the following table for both conformers of cis-2-butene for a variety of a methods.&#xA;The relative energies were determined by taking the difference of the respective geometry and the reference CCSD(T) geometry [e.g. E(CCSD(T))-E(MP2)].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/EDKKe6t.png&quot; alt=&quot;Number of imaginary frequencies, relative MP2-F12 and CCSD(T)-F12 energetics and dipole moments for the two conformers of cis-2-butene with a variety of methods and the haTZ basis  set.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;StructB&lt;/strong&gt; is a second-order saddle point ($n_i = 2$) on every single PES considered and therefore is not a minimum energy structure.  &lt;strong&gt;StructA&lt;/strong&gt; is, however, a minimum ($n_i = 0$) on every PES considered.  The characterization of the nature of the stationary point is consistent between CCSD(T), our best estimate, MP2, and DFT methods.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single point energies reveal negligible differences in the optimized geometries.  The energies associated with the MP2 optimized structures is the reference point for all other relative energies.  Deviations grow no larger than 0.27 $\mathrm{kcal\ mol}^{-1}$ for the MP2-F12 and CCSD(T)-F12 relative energies.  In addition, there is good agreement between the MP2-F12 and CCSD(T)-F12 relative energies, suggesting that higher-order correlation effects are small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dipoles for &lt;strong&gt;StructA&lt;/strong&gt; and &lt;strong&gt;StructB&lt;/strong&gt; are very similar with very small magnitudes, on the order of a couple tenths of a Debye.  To put these quantities into perspective, the dipole moment of water is 1.85 D.  We all know that water has a pretty large dipole moment so by comparison, cis-2-butene has a very WEAK dipole moment.  You can compare to the dipole moments of other molecules by referring to this &lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS-10.pdf&quot; rel=&quot;noreferrer&quot;&gt;NIST reference&lt;/a&gt;.  Clearly, the rotation of the methyl groups have a very small effect on the dipole moments of each conformer.  The MP2 and DFT dipole moments deviate from the best estimate by no more than 0.03 D, but remains in qualitative agreement for both StructA and StructB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below shows the directionality of the dipole.  The head of the (unscaled) arrow points toward the negative pole while the tail of the (unscaled) arrow is oriented to the positive pole.  The numbers on the atoms represent '&lt;a href=&quot;http://www.chem.wisc.edu/~nbo5/tutorial.html&quot; rel=&quot;noreferrer&quot;&gt;natural charges&lt;/a&gt;' from an Natural Bond Orbital (NBO) analysis of the MP2 optimized &lt;strong&gt;Struct A&lt;/strong&gt;.  Clearly the carbon atoms have a small negative charge to them as these atoms are sucking (great scientific term here) electron density away from the neighboring hydrogen atoms.  This is because the nuclear charge (i.e. the number of protons) on carbon is much greater than that of hydrogen (6 vs. 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YW7tp.jpg&quot; alt=&quot;Unscaled dipole moment of MP2/haTZ optimized cis-2-butene **StructA**.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was suggested that I add some data highlighting the energy difference between the two different conformers.  The following table presents the energy difference (where $\Delta E_{\mathrm{A-B}}$ is equivalent to E(A)-E(B)) of each optimized geometry using the MP2-F12 and CCSD(T)-F12 single point energies (in $\mathrm{kcal\ mol}^{-1}$).  We can see that &lt;strong&gt;StructA&lt;/strong&gt; is about 1.5 $\mathrm{kcal\ mol}^{-1}$ lower in energy than &lt;strong&gt;StructB&lt;/strong&gt;.  This makes sense because &lt;strong&gt;StructB&lt;/strong&gt; is a higher order saddle point and &lt;strong&gt;StructA&lt;/strong&gt; is a minimum on the PES. (I'm actually quite surprised that the energy difference is this large for a couple of methyl rotations...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aP1qp.png&quot; alt=&quot;Energy difference of **StructA** and **StructB**&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Conclusions&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The dipole moment for two conformers of cis-2-butene has been examined using seven different computational approaches and a triple-$\zeta$ quality basis set.  The performance of these methods have been tested by evaluating the energies of each geometry with the 'gold standard' CCSD(T) method (the explicitly correlated variant).  Good agreement is seen across the board with respect to the Hessian indices, energies, and dipole moments. Only &lt;strong&gt;StructA&lt;/strong&gt; was a minimum on each PES.  Both conformers of cis-2-butene have a very weak dipole moment on the order of 0.2 -- 0.3 D.  The positive pole is in the vicinity of the methyl groups whereas the negative pole is centered around the sp2 hybridized carbons.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;FAQ&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So you may be asking yourself (or rather, &lt;em&gt;should be&lt;/em&gt; asking yourself) questions such as these listed below.  I will tackle them one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.) Why did we look at two conformers of cis-2-butene and why does it matter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;StructA&lt;/strong&gt; is a minimum which means that if you were to characterize this guy in the gas phase, you'd find &lt;strong&gt;StructA&lt;/strong&gt; rather than &lt;strong&gt;StructB&lt;/strong&gt;.  This is important because if we were to report these results to other scientists, they will want to know what they can expect to find without wondering.  Therefore, &lt;strong&gt;StructA&lt;/strong&gt; is going to be our conformer of interest rather than &lt;strong&gt;StructB&lt;/strong&gt;.  The latter still will provide insightful results but that is about it for the purposes of this examination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.)  Why did we use a variety of methods to characterize these molecular systems?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational methods are simply approximations.  They are not guaranteed to give the 'correct' answer.  So we try to address this by using a &lt;em&gt;variety&lt;/em&gt; of methods (i.e. approximations) and we analyze the results accordingly.  If the results are in agreement, then we can feel confident that the results are correct since we tested them against a list of methods.  You can &lt;strong&gt;never&lt;/strong&gt; rely on just one method unless it is rigorous, well-tested and has been shown repeatedly to perform well in the literature.  When we say 'perform well', this usually means that the computational results are in reasonable agreement with experimental results.  This is important because experimental results is the LAW (for all intents and purposes).  If the computations disagree with experiment, 99.9% of the time this means that your computational approach sucked and that your approximation was either flawed or misapplied.  By using a host of methods, we can put a little more faith into the results we observe because the chances of massive disagreement between the results of the methods is very unlikely in a normal situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.) What is the point of doing a bunch of energy points?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, because we used a slew of methods to characterize the geometries of cis-2-butene, we end up with non-identical geometries each time we use a new approximation.  For instance, the methyl C-H bond lengths from B3LYP are going to be a little bit different from those obtained with MP2.  So then that begs the question, &quot;How do these small differences effect the property of the system that we are interested in?&quot;  Generally, minute differences will have little effect on the resulting energies of each molecule under consideration.  Energy is a very important property that chemists love to look at.  So, if we take each geometry (and each one is unique from the other) and we evaluate the energy of the molecule at the &lt;strong&gt;same level of theory&lt;/strong&gt; (in this case, MP2-F12 and CCSD(T)-F12), then we can quickly see how 'resolved' each geometry was.  There should be very good agreement between the relative energies of each geometry (probably to within a few tenths of a $\mathrm{kcal\ mol}^{-1}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.) Okay, so why MP2-F12 AND CCSD(T)-F12?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use MP2-F12 AND CCSD(T)-F12 to test for 'higher-order correlation effects'.  MP2 methods are much cheaper than CCSD(T) methods but MP2 is not as rigorous and can be error prone in a host of molecular systems.  Therefore, we test the performance of MP2 by busting out our 'gold-standard' which is the CCSD(T) method.  If MP2 agrees well with CCSD(T), then we can feel confident in our MP2 results and never have to revisit the difficult, time-consuming CCSD(T) computations ever again.  ALso, CCSD(T) will also tell us how DFT performed as well.  DFT methods &lt;strong&gt;must always&lt;/strong&gt; be calibrated against something more rigorous since DFT is known for 'getting the right answer for the wrong reasons' and it isn't always right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'F12' bit just means that these are 'explicitly-correlated' methods.  Rather than give an introduction to what this means, you should understand why we use it instead.  You may have noticed that whenever we do a computational job, we specify a method AND a basis set (e.g. heavy-aug-cc-pVTZ).  These basis sets can be measured by how many atomic orbitals (or functions) are given in the set.  The more that are given, the better the 'basis-set approximation'.  Think of this in terms of Riemann sums where you try to approximate the area under a curve using a set of rectangles.  Each rectangle is a basis function and the number of rectangles you use forms a basis set.  The more rectangles you use, the better your approximation of the area under that curve will be.  Basis sets in computational chemistry behave the same way.  When you approach an infinite set of rectangles, you approach the exact answer.  When you approach an infinite number of basis functions, you approach what is called the CBS (complete basis set) limit.  At the CBS limit, you have an exact answer.  We cannot implement an infinite basis set in chemistry (for obvious reasons), and very large basis sets are cost prohibitive.  Therefore, people have devised these F12 approximations that are constructed in such a way to give results that are comparable to those that you'd get with a large basis set, but you can get them by using a relatively small basis set instead!  This is a powerful approach to convergent quantum chemistry that saves you a lot of time while maintaining a set of very good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.) Why didn't you provide more pretty pictures?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just the nature of the beast.  Computational chemistry is usually short on graphics but very dense on spreadsheets.  Plus... I'm no artist.  I actually spent a good couple hours trying to get some electrostatic potentials posted but the new version of G09 hates the molecular viewer programs I currently use so I ditched that idea.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-10-02T12:48:40.263" LastActivityDate="2014-10-02T12:48:40.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16375" PostTypeId="2" ParentId="16372" CreationDate="2014-09-16T21:45:12.810" Score="2" Body="&lt;p&gt;If I understand correctly, then I think your professor is correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;within a carboxylate group such as in the benzoate anion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it is phenyl carboxylate that we are discussing, then I think your professor is saying the same thing I wrote &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16113/acidity-of-ester-and-benzoic-acid/16121?noredirect=1#comment25297_16121&quot;&gt;here&lt;/a&gt;.  Namely, that the carboxylate anion (O-C-O with a negative charge) is analogous to the allyl system, and with 4 pi electrons $\ce{\Psi_2}$ is the HOMO.  In the allyl system (and to a first approximation in this dioxallyl analogue) $\ce{\Psi_2}$ has a node (a &quot;0&quot; coefficient) at $\ce{C_2}$ (the central carbon atom).  If there truly is a zero or very small coefficient at $\ce{C_2}$, then the substituent connected to $\ce{C_2}$ (your benzene ring) is effectively disconnected from the system and cannot interact with the MO &lt;em&gt;via&lt;/em&gt; resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your drawing up above, you have not drawn a carboxylate anion, rather you've drawn an un-ionized acid.  In this molecule there is no carboxylate anion, no dioxallyl system and no $\ce{\Psi_2}$ with a node at $\ce{C_2}$.  In this molecule, resonance interaction between the phenyl substituent and the carbonyl carbon is possible.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-09-17T00:29:59.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16376" PostTypeId="2" ParentId="16370" CreationDate="2014-09-16T22:33:46.077" Score="8" Body="&lt;p&gt;It's hard to say what exactly you would end up with, but it probably wouldn't much resemble beer. The trick with electrochemistry is that a lot of things are electroactive and what reactions actually occur in a given reaction are not always easy to predict and depend on electrodes, solution composition, pH, applied potential, etc. Depending on your electrodes, ethanol may be significantly more easily oxidized than water. A quick literature search shows that platinum and gold are both decent catalysts for ethanol oxidation (platinum is commonly used in fuel cells for this reason), but even with other materials, I can't imagine any conditions in which one could easily enrich an ethanol/water solution in this way. Many organic compounds, probably including some that impart flavour, are likely to be oxidized as well, even if the ethanol wasn't, so it would certainly affect the taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the electrolysis of water requires a large amount of energy, far more than distillation, so even if it would work, it would be an expensive way of doing it.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-16T22:33:46.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16377" PostTypeId="1" CreationDate="2014-09-16T23:51:15.987" Score="1" ViewCount="45" Body="&lt;p&gt;Is there a way to detect the concentration of chlorpromazine in a small sample ; ex: human saliva, using electro-chemical analysis? Does the sensor necessarily have to be a bulky piece of lab equipment or are there smaller sensors available for that kind of test?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="7872" LastActivityDate="2014-09-17T04:54:31.610" Title="Detecting the concentration of chlorpromazine" Tags="&lt;electrochemistry&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16378" PostTypeId="2" ParentId="16372" CreationDate="2014-09-17T00:19:59.397" Score="6" Body="&lt;p&gt;Let's start by deciding what compounds we're talking about here. You reference an example by your instructor of benzoate, which is the conjugate base of benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPudV.jpg&quot; alt=&quot;benzoic acid and benzoate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your instructor is suggesting is correct for benzoate. The most important resonance structures for benzoate are the two equivalent structures that place negative charge on the oxygen atoms. It is also possible to draw a resonance structure where electrons are removed from the ring and both oxygens carry a negative charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YZ0rZ.jpg&quot; alt=&quot;predominant resonance structures of benzoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;carboxylate cage&quot; essentially means that there is no way to delocalize the negative charge of the carboxylate into the aromatic ring. The only way to move electrons into the ring is from the C-O pi bond, which leads to (what I would consider) an unreasonable resonance structure, as it renders an oxygen both positively charged AND non-octet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fz1JR.jpg&quot; alt=&quot;unreasonable resonance structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbonic acids are either a subset of carboxylic acids or a closely related functional group to carboxylic acids. Depending on what property you are interested in, they may be useful to compare or the difference might be too great. In this case, starting from the conjugate base of phenyl carbonic acid (as drawn in the question), there will be no way to delocalize the charge into the aromatic ring.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-17T00:19:59.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16379" PostTypeId="1" CreationDate="2014-09-17T00:43:19.593" Score="12" ViewCount="11027" Body="&lt;p&gt;There are many reduction methods of going from an alcohol to an alkane: &lt;a href=&quot;http://www.transtutors.com/chemistry-homework-help/hydrocarbons/reduction-of-alcohols.aspx&quot;&gt;http://www.transtutors.com/chemistry-homework-help/hydrocarbons/reduction-of-alcohols.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are catalytic methods of adding hydrogen  gas to aldehydes and keystones to convert them into alcohols: &lt;a href=&quot;http://jmcct.com/products-services/product_p404.html&quot;&gt;http://jmcct.com/products-services/product_p404.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have not found a catalytic method of converting an alcohol into an alkane using a catalyst and possibly hydrogen. The fewer non-catalytic entities the better. &lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2020-11-04T04:47:12.447" Title="Is there a catalyst that will reduce an alcohol to an alkane?" Tags="&lt;catalysis&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16381" PostTypeId="1" AcceptedAnswerId="16385" CreationDate="2014-09-17T02:12:09.140" Score="8" ViewCount="2079" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWH3m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to the above professor's claim, I disagree. Bent's rule, which states that &lt;em&gt;atomic p-character is concentrated in orbitals directed toward more electronegative substituents&lt;/em&gt;, implies that the $\ce{Br-C-Br}$ bond angle is going to be &lt;em&gt;smaller&lt;/em&gt; than 109.5 degrees since the &quot;$\ce{sp^3}$&quot; hybrid orbitals involved in C-Br bonding have more p-character than 75%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the $\ce{H-C-H}$ bond angle should be bigger than 109.5 degrees. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;1) Does Bent's rule work here (i.e. am &lt;em&gt;I&lt;/em&gt; correct?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does Bent's rule always work? &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Revision&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;HCH bond angle is ~110 degrees. &lt;/li&gt;&#xA;&lt;li&gt;HCBr bond angle is ~108 degrees. &lt;/li&gt;&#xA;&lt;li&gt;Br-C-Br bond angle is 112 degrees. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Rationales&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;vdW repulsions play a role in Br-C-Br bond angle expansion, although the amount of p-character in the C-Br bonds might be &gt;75%. &lt;/li&gt;&#xA;&lt;li&gt;HCH angle is expanded because of Bent's rule; s-character in the H-C bond exceeds 25%. &lt;/li&gt;&#xA;&lt;li&gt;The expansion of the Br-C-Br bond angle and the expansion of the H-C-H bond angle squeezes together the remaining H-C-Br bond angle. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-17T05:42:43.863" LastActivityDate="2014-09-17T13:06:46.890" Title="Dibromomethane bond angles" Tags="&lt;inorganic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16382" PostTypeId="1" AcceptedAnswerId="16384" CreationDate="2014-09-17T02:33:00.487" Score="17" ViewCount="4602" Body="&lt;p&gt;The 1993 IUPAC Blue Book claims that &quot;methane&quot; is a semisystematic name:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Semisystematic names also exist, such as &quot;methane&quot;, &quot;propanol&quot;, and &quot;benzoic acid&quot;, which are so familiar that few chemists realize that they are not fully systematic. They are retained, and indeed, in some cases there are no better systematic alternatives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Propan-1-ol and phenylmethanoic acid are obvious systematic versions of propanol and benzoic acid. (Though I disagree that they are &quot;better&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &quot;fully systematic&quot; name for methane—or is this a case where there is no &quot;better systematic alternative&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7821" LastEditorUserId="4945" LastEditDate="2014-09-17T06:22:58.627" LastActivityDate="2017-02-08T13:34:41.217" Title="What is the IUPAC systematic name for methane?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16383" PostTypeId="2" ParentId="16381" CreationDate="2014-09-17T03:03:49.823" Score="3" Body="&lt;p&gt;Hybridization occurs in response to a bonding interaction.  Bent's rule adds that the hybridization will occur in a direction that will lower the energy of the electrons in the system.  I can't imagine a case were a molecule will hybridize in a fashion to raise the energy of (all) the electrons in the system, so Bent's rule should always apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water molecule has an H-O-H angle around 104.5 degrees; in the direction Bent's rule would suggest.  Bromine is much less electronegative than oxygen (Br = 2.96, O = 3.44, C = 2.55), so for dibromomethane we might expect a Br-C-Br bond angle larger than 104.5 and less than 109.5.  Then we would need to also factor in the steric repulsion between the two bulky bromines which would open the bond angle even more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that experimental values for the H-C-H and Br-C-Br bond angles are all fairly close to the tetrahedral angle with the Br-C-Br angle around 112 degrees. (&lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/carbon-compounds&quot; rel=&quot;nofollow&quot;&gt;ref. 1&lt;/a&gt;, &lt;a href=&quot;http://cccbdb.nist.gov/exp2.asp?casno=74953&quot; rel=&quot;nofollow&quot;&gt;ref. 2&lt;/a&gt;).  It's hard to separate out the effects due to sterics and hybridization, but I could believe that the experimental values reflect a slightly sub-tetrahedral angle (due to Bent's rule) being opened up by sterics to the experimentally observed value.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-17T03:03:49.823" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16384" PostTypeId="2" ParentId="16382" CreationDate="2014-09-17T03:03:59.987" Score="20" Body="&lt;p&gt;The latest edition of the Blue Book, the 2013 Revision, clarifies this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Methane is a retained name (see P-12.3) that is preferred to the systematic name ‘carbane’, a name never recommended to replace methane, but used to derive the names ‘carbene’ and carbyne for the radicals $\ce{H2C^2-}$ and $\ce{HC^3-}$, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems to be the case that the systematic IUPAC name is only there for consistency with the ions, and is not recommended (and not used) to describe methane. Similarly, &quot;ethane&quot;, &quot;propane&quot;, and &quot;butane&quot; are preferred to their systematic equivalents &quot;dicarbane&quot;, &quot;tricarbane&quot;, and &quot;tetracarbane&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7821" LastActivityDate="2014-09-17T03:03:59.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16385" PostTypeId="2" ParentId="16381" CreationDate="2014-09-17T03:19:36.340" Score="8" Body="&lt;p&gt;Here's the B3LYP/6-31G* geometry. On something like this, I'd suspect this is fairly reliable, although I'll try a few other methods in a second. It gives the Br-C-Br angle as a healthy 113.4 degrees, and H-C-H as 112.5, both in agreement with experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ET36O.png&quot; alt=&quot;image of CH2Br2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see how the Br-C-Br angle is ever going to be &lt;em&gt;smaller&lt;/em&gt; than 109.5. It's the Br-C-H angles that shrink (107.7 degrees), since breaking the symmetry forces the molecule into a &quot;squashed&quot; tetrahedron (point group $D_{2d}$ for the squashed tetrahedron, $C_{2v}$ for $CH_2Br_2$ in particular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Does &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bent's rule&lt;/a&gt; work here (i.e. am I correct?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Does &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bent's rule&lt;/a&gt; always work?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I typically consider VSEPR and Bent's rule to be very useful heuristics. Indeed, it's used in one of my favorite force field methods, &lt;a href=&quot;http://en.wikipedia.org/wiki/Universal_force_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;UFF&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm holding back on &quot;always&quot; because chemistry is incredibly varied. A quick Google search of &quot;Bent's rule fail&quot; turns up &lt;a href=&quot;http://books.google.com/books?id=vQA3V_AK4CoC&amp;amp;pg=PT90&amp;amp;lpg=PT90&amp;amp;dq=bent&amp;#39;s+rule+fails&amp;amp;source=bl&amp;amp;ots=MsOl8mcxwM&amp;amp;sig=4WnTlE3al9pZa5tREjAoDcJZos8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=yPsYVIqrJK2asQSAwoGAAg&amp;amp;ved=0CD0Q6AEwAw#v=onepage&amp;amp;q=bent&amp;#39;s%20rule%20fails&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this interesting nugget from Weinhold's book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SY03G.png&quot; alt=&quot;Section from Discovering Chemistry with Natural Bond Orbitals&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-17T13:06:46.890" LastActivityDate="2014-09-17T13:06:46.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16387" PostTypeId="1" CreationDate="2014-09-17T04:53:30.017" Score="5" ViewCount="9007" Body="&lt;p&gt;I've heard the commercials for this stuff for years, but what exactly does it do?  Does it do what the commercials say it does (clean out deposits in your tank)?  How well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is this just an advertising campaign making a huge deal out of nothing, or is there actually an advantage to Chevron fuel?&lt;/p&gt;&#xA;" OwnerUserId="7876" LastEditorUserId="7475" LastEditDate="2015-10-30T16:53:42.233" LastActivityDate="2015-10-30T16:53:42.233" Title="What is Chevron with Techron fuel?" Tags="&lt;everyday-chemistry&gt;&lt;fuel&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16388" PostTypeId="2" ParentId="16377" CreationDate="2014-09-17T04:54:31.610" Score="3" Body="&lt;p&gt;A cursory online search furnished  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Graphene paste electrode for detection of chlorpromazine&lt;/em&gt; by &lt;em&gt;Mohammad Hadi Parvin&lt;/em&gt; in &lt;em&gt;Electrochemistry Communications&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;13&lt;/em&gt;, 366-369 (&lt;a href=&quot;http://dx.doi.org/10.1016/j.elecom.2011.01.027&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;A multipurpose Voltammetric Sensor for the Determination of &#xA;Chlorpromazine in Presence of Acetaminophen, Uric Acid, &#xA;Dopamine and Ascorbic Acid&lt;/em&gt; by &lt;em&gt;Binesh Unnikrishnan, Pei-Chi Hsu, and Shen-Ming Chen&lt;/em&gt; in &lt;em&gt;Int. J. Electrochem. Sci.&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;7&lt;/em&gt;, 11414-11425 (&lt;a href=&quot;http://www.electrochemsci.org/papers/vol7/71111414.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-17T04:54:31.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16389" PostTypeId="2" ParentId="16387" CreationDate="2014-09-17T05:05:21.823" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;clean out deposits in your tank&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully, there won't be any deposits in your tank (unless you refueled during a sandstorm). As for deposits in your engine, yes it is likely that techron will clean out deposits. If you always use it, there will probably be very few deposits in the engine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is there actually an advantage to Chevron fuel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compared to what? Nothing, yes. To other fuels? It's had to say since other companies also add cleaners to their fuel.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-17T05:05:21.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16390" PostTypeId="5" CreationDate="2014-09-17T06:28:37.260" Score="0" Body="" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-09-17T06:28:37.260" LastActivityDate="2014-09-17T06:28:37.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16391" PostTypeId="4" CreationDate="2014-09-17T06:28:37.260" Score="0" Body="This tag should be applied to questions about the organisation IUPAC. For questions about nomenclature, the appropriate tag should be applied." OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-02T21:30:23.350" LastActivityDate="2017-05-02T21:30:23.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16392" PostTypeId="1" AcceptedAnswerId="16394" CreationDate="2014-09-17T07:59:06.847" Score="11" ViewCount="2035" Body="&lt;p&gt;What is the significance of the quantity &lt;strong&gt;4π&lt;/strong&gt; in the Heisenberg uncertainty principle?&#xA;$Δp \times Δx = \frac{h}{4π}$ &#xA;The focus of the question is to learn why the ratio of a circle's circumference to its diameter, universally known as $π$, is used in the denominator in the term $\frac{h}{4π}$. To put it another way, why is this geometric number relevant in Heisenberg's formula?&lt;/p&gt;&#xA;" OwnerUserId="5367" LastEditorUserId="16683" LastEditDate="2016-09-27T09:28:54.117" LastActivityDate="2016-09-27T14:31:00.167" Title="Significance of 4π in the uncertainty principle" Tags="&lt;quantum-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16393" PostTypeId="1" CreationDate="2014-09-17T08:08:02.690" Score="1" ViewCount="58" Body="&lt;p&gt;What kind of salts may I add in my GPC Eluent?&lt;/p&gt;&#xA;" OwnerUserId="7879" LastEditorUserId="7879" LastEditDate="2014-09-17T11:23:13.533" LastActivityDate="2017-07-15T00:15:07.740" Title="What kind of salts may I add in my GPC Eluent?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2017-08-06T04:56:49.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="16394" PostTypeId="2" ParentId="16392" CreationDate="2014-09-17T08:51:41.323" Score="13" Body="&lt;p&gt;The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of &quot;uncertainty&quot; used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)&#xA;$$ \Delta x \Delta p \sim \hbar \, .$$&#xA;To make the statement more precise, one have to define what is actually meant by &quot;uncertainty&quot; and usually uncertainties are defined as the &lt;em&gt;standard deviations&lt;/em&gt;.&#xA;$$&#xA;\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\&#xA;\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\&#xA;$$&#xA;where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that&#xA;$$&#xA;\Delta x \Delta p \geq \hbar / 2 \, .&#xA;$$&#xA;Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity&#xA;$$&#xA;\sigma_x \sigma_p \geq \hbar / 2 \, .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact, you can prove a more general &lt;em&gt;Robertson uncertainty relation&lt;/em&gt; that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ &#xA;$$&#xA;    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.&#xA;$$&#xA;Now, for $x$ and $p$ it is postulated by so-called &lt;em&gt;canonical commutation relation&lt;/em&gt; that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to &#xA;$$&#xA;    \sigma_x \sigma_p \geq \hbar / 2 \, .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 &lt;a href=&quot;https://chemistry.stackexchange.com/a/16396/186&quot;&gt;noted&lt;/a&gt;, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation&#xA;$$&#xA;    p = \hbar k \, ,&#xA;$$&#xA;where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows&#xA;$$&#xA;    \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, ,&#xA;$$&#xA;and its derivative with respect to $x$ is&#xA;$$&#xA;    \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, .&#xA;$$&#xA;If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get &#xA;$$&#xA;    \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, ,&#xA;$$&#xA;or&#xA;$$&#xA;    p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, ,&#xA;$$&#xA;which suggest that &#xA;$$&#xA;    \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, .&#xA;$$&#xA;$\hat{x}$ is simple multiplicative operator in this picture &#xA;$$&#xA;    \hat{x} \psi(x) = x \psi(x) \, ,&#xA;$$&#xA;so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-09-18T17:16:36.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16395" PostTypeId="1" AcceptedAnswerId="16401" CreationDate="2014-09-17T10:17:18.223" Score="12" ViewCount="39594" Body="&lt;p&gt;I have a very basic question. Energy is absorbed to break bonds. Bond-breaking is an endothermic process. Energy is released when new bonds form. Bond-making is an exothermic process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ATP breaks down to ADP and releases energy right? Is there anything wrong in my understanding?&lt;/p&gt;&#xA;" OwnerUserId="7847" LastActivityDate="2020-12-30T14:05:04.227" Title="ATP break down and energy release?" Tags="&lt;bond&gt;&lt;biochemistry&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16396" PostTypeId="2" ParentId="16392" CreationDate="2014-09-17T10:19:21.703" Score="9" Body="&lt;p&gt;It seems that you are looking for a more mathematical answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually Heisenberg's inequality naturally comes from using Fourier Analysis. $2\pi$ comes from deriving a Fourier transform and the $2$ that remains comes from playing around with the terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a detailed proof of the Heisenberg inequality that explains every term, you can check GB Folland's Lectures on Partial Differential Equations (pdf) &lt;a href=&quot;http://www.math.tifr.res.in/~publ/ln/tifr70.pdf&quot;&gt;http://www.math.tifr.res.in/~publ/ln/tifr70.pdf&lt;/a&gt; and go to page 11. &lt;/p&gt;&#xA;" OwnerUserId="7880" LastActivityDate="2014-09-17T10:19:21.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16397" PostTypeId="1" AcceptedAnswerId="16398" CreationDate="2014-09-17T10:43:41.670" Score="0" ViewCount="281" Body="&lt;p&gt;There are two stable isotopes of iridium, $\ce{^191Ir}$ and $\ce{^193Ir}$. If you randomly pick an iridium atom from a large collection of iridium atoms, which isotope are you more likely to select?&lt;/p&gt;&#xA;" OwnerUserId="7881" LastEditorUserId="15410" LastEditDate="2015-05-12T01:13:51.400" LastActivityDate="2015-05-12T01:13:51.400" Title="Relative abundance of the isotopes of iridium" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16398" PostTypeId="2" ParentId="16397" CreationDate="2014-09-17T11:16:33.197" Score="3" Body="&lt;p&gt;An urn contains nothing but&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;63 white balls and&lt;/li&gt;&#xA;&lt;li&gt;37 black balls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You grab into the urn, pick one item, look at it and put it back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the probability that you drew:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a ball&lt;/li&gt;&#xA;&lt;li&gt;a white ball&lt;/li&gt;&#xA;&lt;li&gt;a black ball&lt;/li&gt;&#xA;&lt;li&gt;a red ball?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-17T11:16:33.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16399" PostTypeId="1" CreationDate="2014-09-17T12:18:56.467" Score="5" ViewCount="166" Body="&lt;p&gt;Short: How to get &lt;strong&gt;proton affinities&lt;/strong&gt; for a large list of chemicals? A database? A programming API?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ca. 2000 chemicals (downloaded from &lt;a href=&quot;http://mcm.leeds.ac.uk/MCM/&quot; rel=&quot;nofollow&quot;&gt;MCM&lt;/a&gt;, mostly organics that occur in the atmosphere and their reaction products). I would like to try to get proton affinities for as many of the 2000 chemicals as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to look up the proton affinity for a chemical in the NIST Chemistry WebBook (&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&amp;amp;Units=SI&amp;amp;Mask=20#Ion-Energetics&quot; rel=&quot;nofollow&quot;&gt;for example isoprene&lt;/a&gt;). But I would prefer to get a machine readable database, or a Python API (or any programming language), to get these nicely in a program code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What options do I have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am not going to get 2000 values by hand, so my last option would be to write a small web scraper to make these 2000 web queries)&lt;/p&gt;&#xA;" OwnerUserId="1253" LastEditorUserId="31313" LastEditDate="2017-03-31T21:47:17.343" LastActivityDate="2017-03-31T21:47:17.343" Title="Proton affinities in machine readable format?" Tags="&lt;organic-chemistry&gt;&lt;ions&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16400" PostTypeId="1" AcceptedAnswerId="16409" CreationDate="2014-09-17T12:44:07.043" Score="1" ViewCount="1304" Body="&lt;p&gt;I am currently learning about the iodoform reaction, i.e. when CH&lt;sub&gt;3&lt;/sub&gt;COR compounds react with iodine and sodium hydroxide to form iodoform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My course book and &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/carbonyls/iodoform.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemguide&lt;/a&gt; both says that the reaction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this picture from &lt;a href=&quot;https://en.wikipedia.org/wiki/Iodoform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; suggests that hydrogen iodide is formed instead of water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IwwK6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="7882" LastEditorUserId="5026" LastEditDate="2018-03-16T05:56:47.453" LastActivityDate="2018-03-16T05:56:47.453" Title="Does the iodoform reaction form hydrogen iodide or water?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16401" PostTypeId="2" ParentId="16395" CreationDate="2014-09-17T13:10:45.290" Score="13" Body="&lt;p&gt;A bond is formed between the oxygen of water and the phosphorus of the gamma-phosphate.  &lt;a href=&quot;http://employees.csbsju.edu/hjakubowski/classes/ch331/oxphos/olcouplingoxphos.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a good link.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g6cG5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g6cG5.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Bonds are both broken and made in chemical reactions but many biology teachers and textbooks state that &amp;quot;Breaking ATP bonds releases energy.&amp;quot;  In reactions bonds are broken and made.  If the strength of the bonds formed exceeds the strength of the bonds broken the reaction is exothermic. Under physiological conditions (with respect to pH and magnesium concentration), the reaction is exergonic as well (i.e. negative Gibbs free energy of reaction).&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="72973" LastEditDate="2020-12-30T14:05:04.227" LastActivityDate="2020-12-30T14:05:04.227" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16402" PostTypeId="1" AcceptedAnswerId="31318" CreationDate="2014-09-17T13:14:41.873" Score="3" ViewCount="153" Body="&lt;p&gt;I have a couple of big barrels with a $50\,\mathrm{L}$ demi-water solution of the following nutrients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{rl}&#xA;200~\mathrm{\mu M} &amp;amp; \ce{NH4NO3} &amp;amp; +\\&#xA;30~\mathrm{\mu M} &amp;amp; \ce{KCl} &amp;amp; +\\&#xA;10~\mathrm{\mu M} &amp;amp; \ce{CaCl2} &amp;amp; +\\&#xA;10~\mathrm{\mu M} &amp;amp; \ce{KH2PO4} &amp;amp; +\\&#xA;\\&#xA;10~\mathrm{\mu M} &amp;amp; \ce{Fe-EDTA} &amp;amp; +\\&#xA;\\&#xA;0.7~\mathrm{\mu M} &amp;amp; \ce{ZnSO4} &amp;amp; +\\&#xA;0.8~\mathrm{\mu M} &amp;amp; \ce{MnCl2} &amp;amp; +\\&#xA;0.2~\mathrm{\mu M} &amp;amp; \ce{CuSO4} &amp;amp; +\\&#xA;0.8~\mathrm{\mu M} &amp;amp; \ce{H3BO4} &amp;amp; +\\&#xA;0.008~\mathrm{\mu M} &amp;amp; \ce{(NH4)6Mo7O24}\\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After mixing those nutrients, I adjust the $\mathrm{pH}$ of the barrels to $5.5$ by carefully adding $\ce{HCl}$ and continuing mixing until the $\mathrm{pH}$ stabilizes at $5.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when I measure the $\mathrm{pH}$in the barrels again after four days, I notice that it has risen to about $5.8$-$5.9$. This is contrary to my expectation. I'd have expected $\ce{CO2}$ from the air to be dissolved in the solution, thus lowering the $\mathrm{pH}$ rather than raising it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain to me what could be happening here?&lt;/p&gt;&#xA;" OwnerUserId="7883" LastEditorUserId="7883" LastEditDate="2015-05-14T19:12:54.563" LastActivityDate="2015-05-14T19:12:54.563" Title="Why does the pH of this solution rise instead of fall when exposed to the atmosphere?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16403" PostTypeId="2" ParentId="16399" CreationDate="2014-09-17T13:39:28.150" Score="4" Body="&lt;p&gt;To my knowledge, a (Python) API for the NIST chemistry WebBook does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than filling out the &lt;a href=&quot;http://webbook.nist.gov/chemistry/cas-ser.html&quot; rel=&quot;nofollow&quot;&gt;Search for Species Data by CAS Registry Number&lt;/a&gt; by hand, you might want to use &lt;code&gt;mechanize&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since you're only interested in &lt;strong&gt;Gas phase ion energetics data&lt;/strong&gt; anyway, you might want to construct the relevant URLs directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isoprene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&amp;amp;Units=SI&amp;amp;Mask=20#Ion-Energetics&quot; rel=&quot;nofollow&quot;&gt;http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&amp;amp;Units=SI&amp;amp;Mask=20#Ion-Energetics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Styrene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C100425&amp;amp;Units=SI&amp;amp;Mask=20#Ion-Energetics&quot; rel=&quot;nofollow&quot;&gt;http://webbook.nist.gov/cgi/cbook.cgi?ID=C100425&amp;amp;Units=SI&amp;amp;Mask=20#Ion-Energetics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameter pattern is obvious: ID is the CAS number of the compound minus the dashes, prefixed by &lt;strong&gt;C&lt;/strong&gt;. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-17T13:39:28.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16406" PostTypeId="2" ParentId="16402" CreationDate="2014-09-17T14:14:50.863" Score="1" Body="&lt;p&gt;Unless the water to make the solution was freshly boiled then it was probably saturated with carbon dioxide to begin with.  It's likely that ammonia is being lost to the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-17T14:14:50.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16409" PostTypeId="2" ParentId="16400" CreationDate="2014-09-17T14:38:11.927" Score="3" Body="&lt;p&gt;Your balanced equation is correct, both water and iodide (and iodide in water will be in equilibrium with hydrogen iodide) are formed in the iodoform reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article that you mention, could have been written a bit clearer.  Specifically, there are two key steps in the iodoform reaction.  First, there is formation of the tri-iodo ketone and water is generated in this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Id8DT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a subsequent step, the carbonyl undergoes an SN2 reaction to generate iodoform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wb41J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia drawing in your post seems to focus on the mechanism of the second step (&lt;strong&gt;3&lt;/strong&gt; to &lt;strong&gt;4b&lt;/strong&gt;).  The first step (&lt;strong&gt;2&lt;/strong&gt; to &lt;strong&gt;3&lt;/strong&gt;) is just illustrated in a cursory fashion and the formation of water is not included in the figure.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-17T14:38:11.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16410" PostTypeId="1" CreationDate="2014-09-17T14:40:33.133" Score="0" ViewCount="1146" Body="&lt;p&gt;Here are four observations of the reactions of four metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal F and G react slowly with water&#xA;Metal H displaces metal G from its chloride salt solution&#xA;Metal E does not react with dilute acids&#xA;Metal H does NOT displace metal F from its sulphate salt solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is their order of reactivity, starting from the most reactive metal to the least reactive metal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F&gt;H&gt;G&gt;E;  H&gt;G&gt;F&gt;E;  H&gt;E&gt;F&gt;G;  G&gt;H&gt;F&gt;E;  None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really unsure how to order them     I would say F&gt;H&gt;G&gt;E but thats half guessing.&#xA;Can someone give me a clue to arrange them right?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="10185" LastEditDate="2015-03-18T17:52:35.373" LastActivityDate="2015-03-18T17:52:35.373" Title="Here are four observations of the reactions of four metals" Tags="&lt;inorganic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16411" PostTypeId="2" ParentId="14557" CreationDate="2014-09-17T14:42:43.637" Score="2" Body="&lt;p&gt;The Cambridge Cluster Database can give you a start with theoretical binding energies, but only one isomer for each size is shown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-wales.ch.cam.ac.uk/~wales/CCD/TIP4P-water.html&quot; rel=&quot;nofollow&quot;&gt;http://www-wales.ch.cam.ac.uk/~wales/CCD/TIP4P-water.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7886" LastActivityDate="2014-09-17T14:42:43.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16414" PostTypeId="2" ParentId="16387" CreationDate="2014-09-17T15:43:07.103" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Techron&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on Techron&lt;/a&gt; suggests the additive is polyetheramine. I suspect there's probably more, but this is confirmed by the &lt;a href=&quot;http://www.techron.com/what-is-techron/proven-science.aspx&quot; rel=&quot;nofollow&quot;&gt;Techron website.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are probably other things, but yes, it's in other additives as well:&lt;br&gt;&#xA;&lt;a href=&quot;http://householdproducts.nlm.nih.gov/cgi-bin/household/brands?tbl=chem&amp;amp;id=947&amp;amp;query=polyether+amine&amp;amp;searchas=TblChemicals&quot; rel=&quot;nofollow&quot;&gt;NIH Household Products Search&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-17T15:43:07.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16415" PostTypeId="1" AcceptedAnswerId="16424" CreationDate="2014-09-17T16:19:22.877" Score="3" ViewCount="191" Body="&lt;p&gt;If a lattice contains a solvent, lets say $\ce{NaCl}$ and water of crystallisation, what is the condition for a solvent molecule to change it's location?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my current point of view I assume that such a water molecule is bound to a lattice ion by ion-dipole forces. It vibrates at that location until, randomly, the particle energy exceeds the binding energy. Then it goes to a new location, in the direction that minimizes Gibbs free energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the water molecule energy distributed? Currently I use Maxwell-Boltzmann distribution. However, Maxwell-Boltzmann is for non-interacting particles and therefore not quite a good fit for molecules that interact with a ion. Further, my chemistry book suggest that ion-dipole bonds have a binding energy of &gt;$50~\mathrm{kJ~mol^{-1}}$, which will in my calculations rarely allow a jump at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask this question because I'm working on a particle-drying simulation. I intend to use a Random Walk simulation to obtain diffusion coefficients under special circumstances: e.g. if the lattice salvation behaviour is somehow weird. The model must not be physically correct in all details. Even bigger &quot;model&quot; error will most likely be small in the simulation results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I asked this question already &lt;a href=&quot;https://physics.stackexchange.com/questions/135600/approach-to-model-solvent-in-a-lattice-e-g-h2o-in-nacl&quot; title=&quot;here&quot;&gt;here&lt;/a&gt; but have not received an answer. As suggested there, I open it in Chem.SE.&lt;/p&gt;&#xA;" OwnerUserId="5325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-05-12T01:03:11.453" Title="Approach to model solvent in a lattice: e.g. H2O in NaCl" Tags="&lt;water&gt;&lt;crystal-structure&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16421" PostTypeId="1" AcceptedAnswerId="16423" CreationDate="2014-09-17T19:39:41.557" Score="4" ViewCount="713" Body="&lt;p&gt;I want to do an experiment to investigate factors affecting the synthesis of chiral molecules. The only problem is, my institution doesn't have much money, and it seems like the chemicals required for any experiment involving chiral molecules are very expensive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial research found an experiment based on the Bürgi-Dunitz trajectory (&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/sterslct.htm&quot; rel=&quot;nofollow&quot;&gt;see Addition to Carbonyl Double Bonds&lt;/a&gt;), but with the use of L-Selectride it isn't the cheapest. I then researched the Diels-Alder reaction, but the conjugated dienes are also relatively expensive, or are only available in large quantities!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am only an amateur in the field of chemistry, and much of the interesting material is very hard to find unless one knows the specific terms to use. Now I come to you for help - are there any cheap experiments involving the synthesis of enantiomers? Other than cost, the constraints are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Equipment: standard laboratory equipment is available, but nothing too fancy&lt;/li&gt;&#xA;&lt;li&gt;Aim: the experiment needs to investigate something, for example factors affecting the synthesis of L and D enantiomers. Two different chiral auxilliaries or catalysts would do well here.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7892" LastEditorUserId="5017" LastEditDate="2014-09-17T20:37:08.457" LastActivityDate="2014-09-17T20:37:08.457" Title="Simple experiments involving enantiomer synthesis" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16422" PostTypeId="2" ParentId="15876" CreationDate="2014-09-17T19:56:27.447" Score="3" Body="&lt;p&gt;The differences between PBS and the buffer obtained by mixing $\ce{Na2HPO4}$ and&#xA; $\ce{NaH2PO4}$ are in ionic strength ( because of the presence of $\ce{KCl}$ d $\ce{NaCl}$ in PBS) and in the ratio of [acid]/[conjugated base]. We have to signal that the buffer capacity (buffer effectiveness) is maximum when $\ce{pH=pKa}$.&#xA;The best way to dilute 10x PBS into 1PBS (let's assume we want to prepare 1L of 1xPBS from 10xPBs) is to begin by &lt;strong&gt;diluting&lt;/strong&gt; 100 mL of 10xPBS solution with 800mL water, then &lt;strong&gt;adjusting&lt;/strong&gt; the pH and finally &lt;strong&gt;completig&lt;/strong&gt; the volume to 1000mL. In doing so, there will be no drift in pH.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-17T19:56:27.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16423" PostTypeId="2" ParentId="16421" CreationDate="2014-09-17T19:59:03.937" Score="5" Body="&lt;p&gt;If you have some way of measuring the enantiomeric excess of the product mixture, I suggest trying proline catalysis of an aldol reaction. The materials are cheap / readily available, and the reaction setup is straightforward. Proline is probably the cheapest chiral catalyst available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a procedure for an organic chemistry lab run at Rutgers that might get you started:&#xA;&lt;a href=&quot;http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf&quot;&gt;http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-17T19:59:03.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16424" PostTypeId="2" ParentId="16415" CreationDate="2014-09-17T20:16:51.657" Score="2" Body="&lt;p&gt;This is a bit outside my realm of expertise - you'd want to look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_dynamics&quot; rel=&quot;nofollow&quot;&gt;molecular dynamics (MD)&lt;/a&gt; simulations for more details. Let me take your question in a few parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A water molecule in a lattice is likely bound by electrostatic (and hydrogen-bond) interactions. So your general assumption is a good one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) MD simulations usually have a &quot;thermostat&quot; depending on the type of ensemble used. Your thermal energy will be distributed into different degrees of freedom, such as rotations, translations, and vibrations. So I'd work out the amount of translational free energy from statistical mechanics, and then use the electrostatic interactions to determine the path for jumps. (Your main question here seems like the activation barrier, since each lattice site should have the same energy, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If the moves are rare, you might want to look into &lt;a href=&quot;http://en.wikipedia.org/wiki/Rare_Event_Sampling&quot; rel=&quot;nofollow&quot;&gt;&quot;rare event sampling&quot;&lt;/a&gt; methods, which preserve the correct statistics, but allow you to more efficiently run your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these techniques are already implemented in simulation codes. You might want to look at something like &lt;a href=&quot;http://lammps.sandia.gov&quot; rel=&quot;nofollow&quot;&gt;LAMMPS&lt;/a&gt; which can handle different levels of particles (e.g., atoms, molecules, crystal domains, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps - I can only talk in broad generalities here.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-17T20:16:51.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16425" PostTypeId="2" ParentId="16352" CreationDate="2014-09-17T20:17:53.603" Score="0" Body="&lt;p&gt;Answering my own question here (just received the answer to the problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution came out looking like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-17T20:17:53.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16427" PostTypeId="1" AcceptedAnswerId="16776" CreationDate="2014-09-17T20:50:27.233" Score="3" ViewCount="275" Body="&lt;p&gt;On wikipedia, this &lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_candy#Additives&quot; rel=&quot;nofollow&quot;&gt;statement exists about rocket propellant aditives&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal oxides have been found to increase the burn rate of sugar propellants. [...] Most often used are iron oxides. Red iron oxide is used most often as it is somewhat easier to obtain than the yellow, brown, or black versions. &lt;strong&gt;Brown iron oxide exhibits unusual burn rate acceleration properties under pressure.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea how would already oxidized compound increase burn rate. I have failed to research which compound is brown iron oxide. All I could find was &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28III%29_oxide&quot; rel=&quot;nofollow&quot;&gt;red iron oxide&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_oxide&quot; rel=&quot;nofollow&quot;&gt;iron monoxide&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II,III%29_oxide&quot; rel=&quot;nofollow&quot;&gt;magnetite&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do oxides help combustion generally? And what makes brown iron oxide so excellent for that purpose?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;I would appreciate help with adding proper tags to this question. Couldn't think of any.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="4945" LastEditDate="2014-09-28T07:14:18.357" LastActivityDate="2014-09-28T07:14:18.357" Title="Electron Transfer in Organic Reactions: Why Does Brown Iron Oxide Improve Burn Rate of Rocket Candy?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16428" PostTypeId="1" CreationDate="2014-09-17T20:52:16.760" Score="43" ViewCount="8041" Body="&lt;p&gt;It is well known (the simplest textbook example) that a diamond has a well-defined arrangement of &lt;code&gt;sp3&lt;/code&gt; carbon atoms, as each atom is connected to four others in a tetrahedral structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the last carbon atoms at the edge? For each of these, three bonds are missing; it has a single bond with one carbon atom inside the structure. What is the hybrid and atomic (electron) structure of the carbon atoms at the edge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some references explaining this in detail (books or research articles).&lt;/p&gt;&#xA;" OwnerUserId="7730" LastEditorUserId="4021" LastEditDate="2014-09-18T20:20:50.090" LastActivityDate="2019-04-22T14:43:50.000" Title="Carbon atoms at the edge of a diamond" Tags="&lt;crystal-structure&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16429" PostTypeId="1" AcceptedAnswerId="16430" CreationDate="2014-09-17T21:03:10.087" Score="6" ViewCount="769" Body="&lt;p&gt;I was thinking about making a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_candy&quot; rel=&quot;nofollow&quot;&gt;rocket candy&lt;/a&gt;&quot;. It's relatively inexpensive rocket based on sugar and some oxidiser, mostly $\ce{KNO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fuel is usually formed by heating the mix of both compounds so that the sugar sticks onto the potassium nitrate. They can also stick together by being dissolved and dried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what makes sugar the best fuel? Paraffin or resin can be melted more easily and the oxidizer could be poured on them while they are hot. To me, that seems like a much better idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there were many people before me and they have surely tried many things. So, what is the reason the sugar is more advised than paraffin or resin?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="5137" LastEditDate="2014-09-18T18:19:34.240" LastActivityDate="2014-09-18T18:19:34.240" Title="Why is sugar used in low-cost homemade rockets?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16430" PostTypeId="2" ParentId="16429" CreationDate="2014-09-17T21:24:55.687" Score="10" Body="&lt;p&gt;the best fuel is actually sorbitol. It is a bit costlier than sugar, but still cheap. It melts easily without decomposition and can dissolve some amount of potassium nitrate. Common sugar decomposes before melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paraffin has one bad habit: it quite easily vaporizes. It is bad, because it makes mixture much harder to ignite and it becomes prone to self-extinguishing because of it. Many polymers also easily decompose producing much gases. Paraffin is used in some specific formulations with more active oxidizers or as a non-fuel additive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black powder is actually quite common, but it has very high combustion rate and is prone to explosive combustion, so it is not recommended for beginners. It also requires ball mill to make, and constructing one is time consuming.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: The main advantages of rocket candy are readily available components, simplicity of preparation (no special equipment required), stable burn and safety in comparison with many other compositions. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-18T05:55:46.913" LastActivityDate="2014-09-18T05:55:46.913" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16431" PostTypeId="1" AcceptedAnswerId="16438" CreationDate="2014-09-17T21:29:13.310" Score="5" ViewCount="4814" Body="&lt;p&gt;Just a quick question; I was looking up the &lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9925286&quot; rel=&quot;nofollow&quot;&gt;MSDS for o-Toluic Acid&lt;/a&gt; for an experiment when I came across this in the &quot;Hazards Identification&quot; section.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Very hazardous in case of skin contact (irritant), ....&#xA;...&#xA;Slightly hazardous in case of skin contact (permeator).&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How should this warning be interpreted?&lt;/p&gt;&#xA;" OwnerUserId="4969" LastEditorUserId="5017" LastEditDate="2014-09-18T23:01:29.693" LastActivityDate="2014-09-18T23:01:29.693" Title="How to interpret the MSDS warning?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16432" PostTypeId="2" ParentId="16428" CreationDate="2014-09-17T21:33:31.407" Score="18" Body="&lt;p&gt;You asked a question, belonging to surface chemistry. It is a relatively new area of research, as it relies heavily on atomic-resolved microscopy and computational methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the answer depends on prehistory of the surface and its environment. In case you crack a diamond, making new surface, two processes happens.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;so-known reconstruction, or reorganization of bonds on the surfaces. I'm not aware of articles on reconstruction of diamond, but silicon has diamond-like structure, so we can speculate from &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_reconstruction#Example_1:_Silicon&quot;&gt;there&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;reaction of the surface with environment. In case of air, oxygen and water may insert into strained bonds, and react from there. It may result in all kinds of surface species up to carboxy-groups, ketone fragments, alkyl fragments and so on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-17T21:33:31.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16433" PostTypeId="2" ParentId="16428" CreationDate="2014-09-17T21:36:11.297" Score="36" Body="&lt;p&gt;Atoms at the edge of a crystal that have an unsatisfied valence are said to have &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dangling_bond&quot;&gt;dangling bonds&lt;/a&gt;.&quot;  Many elements, in addition to carbon, can have dangling bonds.  Dangling bonds is a subject of current interest because of the impact these structures can have on semiconductor properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These dangling bonds are very similar to free radicals, except since they are immobilized in a solid, they are somewhat less reactive than free radicals in solution. Nonetheless, they can react with whatever materials they are exposed to, such as hydrogen, water vapor, oxygen, etc. In addition, if there is a neighboring dangling bond then they can both react with one another to form a bond and satisfy there valence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When carbon or silicon surfaces are prepared under clean room conditions, the dangling bonds can persist.  In the semiconductor industry this clean room preparation technique is followed by bringing in a doping gas in order to purposefully alter the electronic band structure of the substrate material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since unpaired electrons have magnetic properties, in carbon (or any other element) nanostructures where there is a lot more surface area to volume, the concentration of dangling bonds is much higher.  Consequently, the unpaired electrons in the dangling bonds confer magnetic properties on these materials that are large enough to be easily  detectable and to manipulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since dangling bonds represent a non-equilibrium situation, surfaces containing dangling bonds undergo a relaxation or reshaping that is referred to as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_reconstruction&quot;&gt;surface reconstruction&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a full-text article that should help you get started:&#xA;1) &lt;a href=&quot;http://homepages.uni-paderborn.de/wgs/Dpubl/PRB53_13725_96.pdf&quot;&gt;Structure of the diamond 111 surface: Single-dangling-bond versus triple-dangling-bond face&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-17T21:36:11.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16434" PostTypeId="1" AcceptedAnswerId="19214" CreationDate="2014-09-17T22:21:31.517" Score="6" ViewCount="25702" Body="&lt;p&gt;We know that the electric flux density &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; can be related to electric field intensity &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D = \epsilon(\omega) E$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\epsilon(\omega) = \epsilon_0 \epsilon_r(\omega)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;span class=&quot;math-container&quot;&gt;$\epsilon_0$&lt;/span&gt; is the permittivity of vacuum and &lt;span class=&quot;math-container&quot;&gt;$\epsilon_r(\omega)$&lt;/span&gt; is the relative permittivity of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the low-frequency limit, I would like to know how the value of &lt;span class=&quot;math-container&quot;&gt;$\epsilon_r (\omega)$&lt;/span&gt; of water changes with temperature and salt (&lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;) concentration. Please explain,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the rationale of the variation (from quantum mechanics or statistical mechanics etc)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;observed trend holds good for any liquid or only for highly polar solvents&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;any references that you know of which talks in detail about these variations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4941" LastEditorUserId="69977" LastEditDate="2020-04-30T08:35:50.967" LastActivityDate="2020-04-30T08:35:50.967" Title="Salt concentration and electrical permittivity of water" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;electrolysis&gt;&lt;solvents&gt;&lt;electrical-properties&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16435" PostTypeId="2" ParentId="16431" CreationDate="2014-09-17T22:21:45.233" Score="1" Body="&lt;p&gt;As canadianer said, wear gloves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irritant means just as it sounds, it will irritate your skin (ie: itching, rashes, strange sensations, ect.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permeator means that it can pass through the skin.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-17T22:21:45.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16438" PostTypeId="2" ParentId="16431" CreationDate="2014-09-18T01:52:11.643" Score="10" Body="&lt;p&gt;An irritant is a chemical, which is not corrosive, but which causes a reversible inflammatory effect on living tissue by chemical action at the site of contact. There are specific lab tests to determine the ranking of an irritant, but something which is rated as very hazardous is likely to cause a very severe reaction upon exposure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permeator is a chemical which can pass through the outer protective dermal layers and into the body, and can either expose the body to toxic effects of that chemical OR act as a carrier for other toxic/hazardous chemicals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other common classes of hazardous chemicals include corrosive and sensitizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;o-Toluic acid is a strong irritant, but moderate permeator. Depending on your level of experience in handling chemicals, simply getting the advice to 'wear gloves' may leave you a little underprepared. You should take this to mean that you should not allow any skin contact; full length sleeves (lab coat), eye protection and THE CORRECT TYPE of gloves. Using the incorrect type of gloves is worse than using no gloves at all, as it gives you a false sense of security. Select the correct type of gloves using a suitable selection tool or guide, &lt;a href=&quot;http://www.ansellpro.com/product-catalog/chemical-liquid-protection/assessment-process.aspx&quot;&gt;such as this one&lt;/a&gt; from Ansell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a strong irritant, you should also avoid breathing any vapours or fumes, so it means handling this chemical in a suitably ventilated area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AS with any risk assessment, it also means that &lt;em&gt;before&lt;/em&gt; you handle the chemical you have an understanding of the response procedures in the event of things going wrong. That is, from your MSDS, you need to know Section 4:First Aid Measures prior to any procedure you undertake. It is no good having to find and read the MSDS &lt;em&gt;after&lt;/em&gt; you have splashed yourself with this stuff.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-09-18T01:52:11.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16439" PostTypeId="1" AcceptedAnswerId="64269" CreationDate="2014-09-18T04:42:10.037" Score="29" ViewCount="17107" Body="&lt;p&gt;Which is the densest gas known to exist under normal &quot;shirt sleeve&quot; conditions: room temperature, one atmosphere, won't explode on contact with air, won't kill everyone in the building in trace amounts, etc.&lt;/p&gt;&#xA;" OwnerUserId="7900" LastEditorUserId="30836" LastEditDate="2019-04-07T05:28:40.557" LastActivityDate="2021-05-28T14:53:35.813" Title="Which is the densest gas known?" Tags="&lt;materials&gt;&lt;density&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16440" PostTypeId="2" ParentId="16439" CreationDate="2014-09-18T05:46:39.193" Score="21" Body="&lt;p&gt;I would guess radon is the densest gas ($9.73\; \text{kg/m}^3$) that is not directly lethal. It is radioactive however, emitting alpha radiation, so you don't want to breath it in. Given that it is much heavier than air, as long as you don't hug the ground it think it would be possible to be in the same room without killing you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fully non-lethal stuff, the first that comes to mind is sulfur hexafluoride, $\ce{SF6}$ ($6.17~\mathrm{kg/m^3}$). It is the stuff that people use to artificially make their voice lower (similar to helium making it higher) and it has a 0000 NFPA rating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle you can dream up your own densest gas, because the density of a gas at given temperature and pressure is directly related to its molecular mass. So the higher the molecular mass of the gas, the denser it is. &lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="16683" LastEditDate="2017-02-04T16:11:49.390" LastActivityDate="2017-02-04T16:11:49.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16441" PostTypeId="1" AcceptedAnswerId="16891" CreationDate="2014-09-18T05:53:46.333" Score="2" ViewCount="2467" Body="&lt;p&gt;I have an air mass which consists of 78.1% $\ce{N_2}$, 20.9% $\ce{O_2}$, 0.964% $\ce{Ar}$, and 0.0378% $\ce{CO_2}$ in a container which volume $\mathrm{V=40~L}$ at a temperature $\mathrm {T = 298.15~K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total pressure is $\mathrm{p_{total} = 1.5~atm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am removing all the $\ce{O_2}$ in the container. What will the partial pressure of $\ce{N_2}$ be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the formula $\mathrm{pV = nRT}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I remove all the $\ce{O_2}$ I can calculate the amount of substance $$\mathrm{{n_{total} = \frac{pV}{RT} = \frac{1.5 \cdot 40}{0.082 \cdot 298.15}~mol = 2.45~mol}}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of $\ce{N_2}$ and $\ce{O_2}$ is $\ce{M(N_2)} = \mathrm{28~g\ mol^{-1}}$ and $\ce{M(O_2)} = \mathrm{32~g\ mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remove all the $O_2$ the new mass/volume percentage of $\ce{N_2}$ will be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{m_\%(N_2) = \frac{78.1}{78.1+0.964+0.0378} = 98.7 \%}$$. &#xA;So I thought I could find the amount of substance of $\ce{N_2}$  by saying &#xA;$$\ce{n(N_2)} = \mathrm{\frac{98.7~g}{28~g/mol} = 3.525~mol}$$ but this will exceed the total amount of substance before I removed $\ce{O_2}$ so this makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I have found the amount of substance of $\ce{N_2}$ I think I can just replace $\ce{n}$ in $\mathrm{p = \frac{nRT}{V}}$ and get the partial pressure of $\ce{N_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I have the right percentage of $\ce{N_2}$ but I am calculating the amount of substance wrong when it's a gas? Could the correct answer be $\mathrm{98.7 \% \cdot n_{total} = 2.422~mol}$ and then inserting this in $\mathrm{p = \frac{nRT}{V}}$ which gives me around $\ce{p(N_2)} = \mathrm{1.5~atm}$?&lt;/p&gt;&#xA;" OwnerUserId="5193" LastEditorUserId="4356" LastEditDate="2014-09-29T07:38:39.927" LastActivityDate="2014-09-29T17:26:36.727" Title="Find the partial pressure in a container" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16444" PostTypeId="1" CreationDate="2014-09-18T08:06:44.117" Score="0" ViewCount="1054" Body="&lt;p&gt;In organic detergents, when I add Hydrogen peroxide to increase the stain removing feature a gas is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I prevent the gas production process?&lt;/p&gt;&#xA;" OwnerUserId="7902" LastActivityDate="2014-09-18T14:40:55.273" Title="Preventing the gas when Hydrogen peroxide is added to a detergent" Tags="&lt;hydrogen&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16445" PostTypeId="1" AcceptedAnswerId="18962" CreationDate="2014-09-18T08:25:46.670" Score="3" ViewCount="644" Body="&lt;p&gt;I am having some trouble with qualitative phase identification using x-ray diffraction (XRD). Here follows some basic information about the experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Instrument&lt;/code&gt;: PANalytical X'Pert$^3$ Powder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Software&lt;/code&gt;: PANalytical Highscore Plus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Database&lt;/code&gt;: PAN ICSD &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Sample&lt;/code&gt;: respirable fraction aerosol samples from ferro- and silica manganese plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The goal&lt;/code&gt;: which manganese phases are present, and which dominate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grasping the software is not what I am concerned about. I lack an analytical strategy. I know how to determine the background, but I am not sure when the background is appropriately determined. Too high? Too low? Does this strongly affect the phase identification? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I do a peak search, and 50 peaks are detected. Should I focus just on the major peaks? Are the minor ones important too? When I do a search-match analysis, I get 400 suggested phases, where 20 phases share the highest score. Then another 20 phases share the highest score minus 1. I lack criteria needed to accept a phase as present, and discard other phases as not present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to find good literature on this subject. I did find an article called &quot;Hints on Phase Identification Using Powder X-ray Diffraction&quot;, which certainly raised a lot of potential problems and procedures, but which did not necessarily solve these problems or explain why the procedures would help. Does anyone know of any books or other literature that could help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem boils down to lack of &quot;strategy&quot; or &quot;analytical procedure&quot;, and maybe someone here could offer some advice. I hope my question is &quot;answerable&quot;, and not too broad; I know there is a &quot;specific-and-precise-question&quot; policy in here.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastActivityDate="2014-10-29T22:57:51.060" Title="X-Ray Diffraction: Phase Identification Strategy" Tags="&lt;analytical-chemistry&gt;&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16446" PostTypeId="1" AcceptedAnswerId="16449" CreationDate="2014-09-18T09:53:38.867" Score="9" ViewCount="938" Body="&lt;p&gt;In the lecture our professor told us that nuclei with $S&amp;gt;1/2$ have a quadrupole and are therefore detected with NQR instead of NMR.&#xA;The majority of all elements have nuclei with $S&amp;gt;1/2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered why he used $S$ (electron spin) instead of $I$ (nuclear spin) in the slides?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7475" LastEditDate="2016-12-20T00:30:16.643" LastActivityDate="2016-12-20T00:30:16.643" Title="Difference between S and I in the context of magnetic resonance experiments" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16447" PostTypeId="1" AcceptedAnswerId="16490" CreationDate="2014-09-18T10:03:01.697" Score="0" ViewCount="927" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a sample of silver alloy and would like to know how much silver is in it. Dissolve the silver sample in nitric acid. Filter to remove any insoluble material. Add an excess of $\ce{KCl}$ solution, which causes the formation of silver chloride precipitate. Filter, dry, and weigh the precipitate. What do you estimate the percentage of silver in the original sample is if its mass was $5.2~\mathrm{g}$, and the mass of silver chloride was $6.4~\mathrm{g}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The atomic mass of silver is $M(\ce{Ag})= 107.87~\mathrm{\frac{g}{mol}}$. That of chlorine is $M(\ce{Cl}) = 35.45~\mathrm{\frac{g}{mol}}$.&lt;br&gt;&#xA;My answer is $81\%$, because $5.2/6.4 \cdot 100 = 81$. However, I am quite unsure.&lt;/p&gt;&#xA;" OwnerUserId="7621" LastEditorUserId="4945" LastEditDate="2014-09-19T04:49:59.043" LastActivityDate="2014-09-19T04:53:25.113" Title="How to find the percentage contribution of a compound via gravimetric analysis?" Tags="&lt;homework&gt;&lt;analytical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16449" PostTypeId="2" ParentId="16446" CreationDate="2014-09-18T10:21:20.873" Score="7" Body="&lt;p&gt;The nomenclature you mean has nothing to do with the electron spin S, it is something completely different. In heteronuclear experiments with two spins it is common to call them $I$ and $S$, with $I$ being the more sensitive nucleus. They're still both spin 1/2 nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I$ is one spin, and $S$ is a different spin in this nomenclature, with a scalar coupling between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This nomenclature is usually introduced with the INEPT transfer and experiments like HSQC or HMQC.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-09-18T10:21:20.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16450" PostTypeId="2" ParentId="10385" CreationDate="2014-09-18T11:03:15.740" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;S = 1, L = 1, so (S+1)(L+1) = 6&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are using here not just alternative physics, but also alternative math. For, even if the formula were right, you should get $2 \cdot 2 = 4$ and not $6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the right formula for the degeneracy is $(2S + 1)(2L + 1)$ which being used for the ground $^1D$ state of oxygen atom (i.e. for $S = 0, L=2$) indeed gives $1 \cdot 5 = 5$.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-09-18T11:03:15.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16451" PostTypeId="1" CreationDate="2014-09-18T11:25:08.523" Score="2" ViewCount="431" Body="&lt;p&gt;I have created a solution of $\ce{KNO3}$ &lt;del&gt;and table sugar&lt;/del&gt; in order to create fuse from hemp rope. I wet the rope in the solution and let it dry - I repeat this process several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works and produces very slowly burning rope which is easy to set off. The problem is that after 24 hours, my solution started to stink. Something between garlic, onion and vinegar. I have a bad feeling about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does my solution degrade, or is this stink caused by some impurities (the solution isn't very pure)?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="13184" LastEditDate="2015-05-08T20:04:38.103" LastActivityDate="2016-03-11T23:25:13.983" Title="Does potassium nitrate solution degrade?" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16452" PostTypeId="1" AcceptedAnswerId="16457" CreationDate="2014-09-18T11:36:48.447" Score="0" ViewCount="335" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A chemist mixes 24.0 g $\ce{H2}$ with 8.0 g $\ce{N2}$. Assuming the reaction goes to completion, which reactant should she use to calculate the yield?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3H2 + N2 -&amp;gt; 2NH3}$$&#xA;I believe it should be nitrogen. With 24 g of $\ce{H2}$, we have 11.9 mol of $\ce{H2}$. With 8 g of $\ce{N2}$, we have 0.3 mol of $\ce{N2}$. So, when I express the moles of $\ce{N2}$ in terms of $\ce{H2}$, I see that I only need 0.9 moles of $\ce{H2}$ – which means that $\ce{N2}$ is the limiting reacting and then used to calculate the theoretical yield, right?&lt;/p&gt;&#xA;" OwnerUserId="7621" LastEditorDisplayName="user7951" LastEditDate="2016-11-07T15:49:50.220" LastActivityDate="2016-11-07T15:49:50.220" Title="Which reactant to use to calculate yield?" Tags="&lt;homework&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16453" PostTypeId="2" ParentId="16428" CreationDate="2014-09-18T11:42:06.213" Score="10" Body="&lt;p&gt;To add to the current answers about the individual electrons, there is also a global effect at the lattice level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have a monocrystal with no (or low density of) defects, the lattice will be regular until around 100 atoms from the surface. The reason is obvious: the atoms at the surface have neighbours only at one side, so they have to be displaced to balance the forces, which in turn forces a smaller displacement on the second layer, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this has powerful implications in the study of nanostructures, like nanotechnology, as well as suspensions and colloids; as they are structures that are all close to the boundary, with no proper bulk. Also,they are fundamental in crystal growth, and, in Chemistry in catalystic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further reading:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: they are excelent classical books, but they are old, and there has been quite a lot of research in surfaces in the late year. Unfortunately, I am not aware of any modern reference.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://books.google.se/books/about/Solid_State_Physics.html?id=1C9HAQAAIAAJ&amp;amp;redir_esc=y&quot;&gt;Solid State Physics, Ashcroft and Mermin.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://archive.org/details/IntroductionToSolidStatePhysics&quot;&gt;Introduction to Solid State Physics, Charles Kittel&lt;/a&gt; (I'd avoid the parts of introduction to SS, as I find them extremely boring, but some other people like it).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7903" LastActivityDate="2014-09-18T11:42:06.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16457" PostTypeId="2" ParentId="16452" CreationDate="2014-09-18T12:41:42.380" Score="1" Body="&lt;p&gt;You have the right thought process - you need to find which is the limiting reagent. &#xA;Your calculation is correct, $N_2$ is limiting, so it can be used to calculate yield. Looks like we weren't really needed!&lt;/p&gt;&#xA;" OwnerUserId="7813" LastActivityDate="2014-09-18T12:41:42.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16458" PostTypeId="1" AcceptedAnswerId="16542" CreationDate="2014-09-18T14:04:50.067" Score="7" ViewCount="3776" Body="&lt;p&gt;Why is it $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-1}n\mathrm{s}^2$.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only exception I thought to this formula was Thorium $\ce{5f^{0} 6d^{2} 7s^{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I feel it should be $(n-2)\mathrm{f}^{1-14}(n-1)\mathrm{d}^{0-2}n\mathrm{s}^2$.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="46306" LastEditDate="2018-03-01T04:01:14.027" LastActivityDate="2018-03-01T04:01:14.027" Title="Outermost electronic configuration of f block elements" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16459" PostTypeId="2" ParentId="16444" CreationDate="2014-09-18T14:40:55.273" Score="1" Body="&lt;p&gt;While hydrogen peroxide is fairly stable if pure in water solution, it can quickly break down to water plus oxygen in the presence of a catalyst such as transition metal ions--probably what is causing the gas evolution you observe. One approach you might try is to test other types of detergent - perhaps you can find one that does not catalyze breakdown of peroxide. Liquid facial soap or glycerine soap--or even dishwashing liquid--might be worth a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach would be to mix the peroxide in right before use rather than try to store the mixture--the more dilute, the better.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-09-18T14:40:55.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16460" PostTypeId="1" AcceptedAnswerId="16462" CreationDate="2014-09-18T15:28:48.517" Score="1" ViewCount="54" Body="&lt;p&gt;Suppose I would like to create a one molar solution of NaCl in water. The molar concentration is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_i=\frac{n_i}{V} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_i=1$ in this case. I need to take 58.5 grams of NaCl (atomic weight of Na=23 &amp;amp; Cl=35.5) and dissolve it in water until the total solvent volume is 1 litre. The total volume of the water added is less than 1 litre. But what exactly is that amount of water and how can I determine it apriori from thermodynamics?&lt;/p&gt;&#xA;" OwnerUserId="4941" LastActivityDate="2014-09-18T16:12:22.737" Title="Can we determine apriori the volume of solvent in an electrolyte" Tags="&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16461" PostTypeId="2" ParentId="16446" CreationDate="2014-09-18T15:54:07.793" Score="7" Body="&lt;p&gt;Mad Scientist has provided a nice answer to your question about S and I.  I'd like to comment on another aspect of your post.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;our professor told us that nuclei with S&gt;1/2 have a quadrupole and&#xA;  are therefore detected with NQR instead of NMR&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Instead of nmr&quot; might be too strong.  Most nuclei with a nuclear spin &gt;1/2 can also be observed by nmr.  Some of these nuclei may have a low relatively sensitivity to the nmr experiment, or due to nuclear quadrupole relaxation might produce signals where information has been lost, in which case NQR might be preferred.  While there are many cases where the NQR experiment would be more informative than the NMR experiment, there are also many cases where NMR would work just fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting example related to proton-nmr compares the following 3 compounds, pentadeuterioacetone, N-methylaniline and chloroform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bV6sF.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deuterium, nitrogen and chlorine are quadrupolar nuclei. Deuterium and nitrogen-14 (the major isotope of nitrogen) both have a nuclear spin of 1, while chlorine-35 and -37 (the 2 predominant isotopes of chlorine) have a nuclear spin of 3/2. If one examines the proton-nmr of pentadeuterioacetone, one can observe the coupling between the deuterium and hydrogen nuclei.  By the same token, if one examined the deuterium-nmr of this compound, one would observe the same H-D coupling.  If one examines the proton-nmr of N-methylaniline, the coupling between nitrogen and the various protons is washed out, broadened signals where the coupling information has been lost are observed.  In the case of chloroform, the proton-nmr shows only a sharp singlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the change in coupling to a proton as we change the coupled nucleus from D to N to Cl?  The situation is very similar to physical exchange phenomena.  If a proton is attached to a nucleus, then coupling can be observed between that proton and adjacent protons.  If the proton is exchanging with the environment at a rate similar to the timescale of the nmr experiment, then the coupling will broaden.  If the proton is exchanging very rapidly compared to the timescale of the nmr experiment, then the proton is effectively decoupled from the system and a sharp singlet will be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the 3 compounds discussed above, instead of physical exchange, the proton is being relaxed (decoupled) at varying rates by the attached quadrupolar nuclei.  The rate of quadrupolar relaxation caused by a coupled deuterium atom is slow on the nmr timescale, so all coupling is preserved and observed.  The quadrupolar relaxation rate of nitrogen-14 is comparable to the timescale of the nmr experiment and so the coupling is beginning to wash out.  The chlorine nuclei undergo rapid quadrupolar relaxation on the nmr timescale and are effectively decoupled from other nuclei.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-12-19T16:36:52.443" LastActivityDate="2016-12-19T16:36:52.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16462" PostTypeId="2" ParentId="16460" CreationDate="2014-09-18T16:12:22.737" Score="3" Body="&lt;p&gt;You can calculate it using &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_molar_property&quot; rel=&quot;nofollow&quot;&gt;partial molar volumes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial molar volume is defined as the differential change in volume with respect to change in mole fraction, and is determined experimentally. From the definition of partial molar properties, you can calculate the amount of one component needed to achieve a certain volume, if you know all of the partial molar volumes and pure component molar volumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some complications to that for this system because you are talking about mixing a solid-phase ionic compound with water, instead of two liquids directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it is much easier to just use a volumetric flask so that you know the total volume is 1 L. If you measure the amount of water you actually used you will find that it is very close to 1 L.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-09-18T16:12:22.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16463" PostTypeId="1" AcceptedAnswerId="16466" CreationDate="2014-09-18T16:25:33.680" Score="1" ViewCount="1986" Body="&lt;p&gt;So &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulbutiamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;sulbutiamine&lt;/a&gt; is said to be a thiamine substitute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SKuLP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SKuLP.png&quot; alt=&quot;Structures of sulbutiamine and thiamine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet - it doesn't contain the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_14%3A_Reactions_with_stabilized_carbanion_intermediates_II/Section_14.5%3A_Thiamine_diphosphate-dependent_reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;thiazole ring&lt;/a&gt; that's so important for many of thiamine's functions (highlighted in red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it function as a thiamine substitute?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="16683" LastEditDate="2017-03-28T13:12:09.727" LastActivityDate="2017-03-28T13:12:09.727" Title="How does sulbutiamine work in the same way as thiamine, even if it doesn't have a thiazole ring?" Tags="&lt;biochemistry&gt;&lt;aromatic-compounds&gt;&lt;heterocyclic-compounds&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16464" PostTypeId="2" ParentId="16439" CreationDate="2014-09-18T16:39:00.900" Score="25" Body="&lt;p&gt;Actually the densest gas which has an actual use is Tungsten hexafluoride, clocking in at an astounding 13 grams per liter. Tungsten hexafluoride isn't as well known as its lighter cousin, sulfur hexafluoride, because it has a narrow range of uses. It is useful in the field of electronics manufacturing as it can be used to coat circuit boards with Tungsten. Also, as fun it may sound, don't try to dip your hand in it or try to swallow it, as it's quite corrosive. Besides being corrosive itself, it also produces hydrogen fluoride, a nasty chemical, when it comes in contact with any moisture. Although it's a pretty unsafe chemical, it's cousin, sulfur hexafluoride, is safe enough to inhale in reasonable amounts.&lt;/p&gt;&#xA;" OwnerUserId="7513" LastEditorUserId="7513" LastEditDate="2014-09-18T16:58:10.077" LastActivityDate="2014-09-18T16:58:10.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16465" PostTypeId="1" CreationDate="2014-09-18T17:18:25.973" Score="4" ViewCount="68" Body="&lt;p&gt;Is the corrosive nature of acid relative to some substance quantifiable like acid dissociation constant ($K_\mathrm{a}$)? If yes, what can be used to describe the corrosive nature of an acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is known that citric acid is far less corrosive than, say, sulfuric acid. But, can it be quantified? Also, if corrosiveness can be quantified, then can the caustic nature of bases be quantified too?&lt;/p&gt;&#xA;" OwnerUserId="7513" LastEditorUserId="5026" LastEditDate="2018-04-10T06:59:10.987" LastActivityDate="2018-04-10T06:59:46.050" Title="Quantitative relationship between corrosiveness of acid and other properties" Tags="&lt;acid-base&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16466" PostTypeId="2" ParentId="16463" CreationDate="2014-09-18T17:39:45.153" Score="3" Body="&lt;p&gt;To get started, here are the structures of some compounds that are relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thiamine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eSNMS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulbutiamine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v8ugX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allithiamine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abm5O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at these structures we see that you are of course correct, sulbutiamine does not contain a 5-membered thiazole ring like thiamine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thiamine (vitamin B1) was an early focus of interest because sailors at sea for long periods of time developed &lt;a href=&quot;http://en.wikipedia.org/wiki/Beriberi&quot; rel=&quot;nofollow noreferrer&quot;&gt;beriberi&lt;/a&gt; due to a nutritional thiamine deficiency.  Since those early days, thiamine substitiutes were of medicinal interest.  One of the early substitutes was &lt;a href=&quot;http://en.wikipedia.org/wiki/Allithiamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;allithiamine&lt;/a&gt;.  It is a natural product that can be extracted from garlic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the structural similarities between allithiamine and sulbutiamine.  Sulbutiamine has the advantage that both sides of the disulfide (S-S) bond contain the same active ingredient, so one would expect increased pharmacological properties from sulbutiamine.  The advantage that both allithiamine and sulbutiamine possess over thiamine is that they are more soluble in lipids (the positively charged thiazole ring in thiamine makes it very soluble in water, but not so soluble in lipids).  This enhanced solubility in lipids means that they can cross the blood-brain barrier much more effectively than thiamine, this improves their bioavalabilty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that background we can now answer your question, &quot;how does sulbutiamine work in the same way as thiamine.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once in the brain, sulbutiamine can be easily metabolized to thiamine.  Look at the structure of sulbutiamine and here is what happens:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reductively break the disulfide bond and form two identical mercaptans&#xA;$$\ce{RS-SR -&amp;gt;C[{[H]}]\   2 ~RS-H}$$&lt;/li&gt;&#xA;&lt;li&gt;protonate the acetyl (that's on the nitrogen that will become part of our thiazole ring) oxygen to make it more attractive to nucleophilic substitution&lt;/li&gt;&#xA;&lt;li&gt;Let the mercaptan sulfur nucleophilically attack the acetyl carbonyl carbon&lt;/li&gt;&#xA;&lt;li&gt;eliminate water and form the 5-membered thiazole ring&#xA;and we now have thiamine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But a much larger dosage in the brain than had we simply ingested thiamine.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-18T17:48:57.873" LastActivityDate="2014-09-18T17:48:57.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16467" PostTypeId="1" AcceptedAnswerId="16483" CreationDate="2014-09-18T17:54:36.727" Score="10" ViewCount="3381" Body="&lt;p&gt;I am aware that the square of the Wavefunction gives the probability density of finding an electron at a particular point in space. I have also heard that it's a complex number but since it's a function I am unsure as to how that could be the case (perhaps someone could please clear that up for me as a sub-question). Moreover, I have seen it described as the amplitude of something but I literally have no idea what is meant by that. However, the crux of the question is what is the Wavefunction itself (i.e not what the square of it is). I have read in the book &quot;Why Chemical Reactions Happen&quot; that it is essentially synonymous with the word/concept: orbital. I have also seen in various places w(x,y,x) or w(r,theta,thi) [where &quot;w&quot; represents the sign for a wavefunction] so, from what I can gather it is a function in three dimensions that represents the shape of a particular orbital where the function of (x,y,z) or (r,theta,thi) would, of course, be different for the s,p,d and f orbitals. However, in the book &quot;Why Chemical Reactions Happen&quot; it goes onto say that wavefunctions of different atoms &lt;strong&gt;interfere&lt;/strong&gt; with each other to form molecular orbitals and the atomic orbitals somehow have a phase?! For my idea of a wavefunction to be correct, the idea of interference of these wavefunctions must be an idea that was thought of to try and rationalize the process of the formation of molecular bonding and anti-bonding orbitals from atomic ones to make it easier to understand and conceptualize. If I am right about the wavefunction being a purely mathematical representation of the space where an electron is likely to reside then the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from (i.e nobody really knows &lt;strong&gt;why&lt;/strong&gt; they are that shape but when atoms or molecules are arranged in a particular way, orbitals &quot;just are&quot; that shape). Also the idea of phase seemed to bring my argument down but then I thought that perhaps phase is simply defined as being opposite relative to another part of an orbital if it is the other side of an angular node (I suspect this is wrong but could someone please shed some light on this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to when I said - &quot;the molecular wavefunctions are just as curious in terms of their shape and suchlike as the shapes of the atomic orbitals they came from&quot; - I thought this because I found it bizarre that empty orbitals could &quot;interfere&quot; with an orbital from another atom. For example one p orbital is empty yet all three interact with the four hydrogen s orbitals. Unless, the electrons can jump between the the degenerate p orbitals at such a frequency that all can be considered occupied I find it incomprehensible that just a portion of space can interfere with an orbital. As a result, I came to the conclusion that I quoted at the start of the paragraph. Basically meaning that when atoms come together to form molecules the electrons find the lowest possible energy spaces to reside; or in other words, a molecuar orbital / molecular wavefunction forms which is equally as curious as the atomic wavefunctions with their strange shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with this concept would be greatly appreciated - I am off to Oxford university this October and Molecular Orbital Theory is a small part of the suggested reading list before arrival into the first year. Also, on that note, please refrain from heavily mathematical answers because I simply won't understand as I have only done A-level Maths and A-level Chemistry. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2017-07-14T10:38:59.197" Title="What actually is the Wavefunction?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;quantum-chemistry&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16470" PostTypeId="2" ParentId="10833" CreationDate="2014-09-18T19:32:56.030" Score="23" Body="&lt;p&gt;In the pinacol rearrangement, a 1,2-diol is treated with acid and rearranges to a carbonyl compound. Here is a reaction scheme showing a mechanism for the rearrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ACnxd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the molecule is symmetric and methyl migration is the only reaction pathway available.  Methyl migration can 1) help stabilize the developing carbocation center and 2) once fully migrated, produces a resonance stabilized hydroxyl-carbocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one or more of the methyl groups is replaced with some other substituent (for example, alkyl, hydrogen, aryl), then our reaction possibilities become more complex; which hydroxyl group will protonate, which group will migrate?  Generally, both hydroxyls will protonate and lead to separate products.  In terms of which group will migrate, the only thing agreed upon is that a tertiary carbon will migrate in preference to a secondary carbon which will migrate in preference to a primary carbon.  &lt;strong&gt;Clearly, the ability of the migrating group to stabilize positive charge plays a role&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is reversible and product distribution can change as reaction conditions are changed (thermodynamic control &lt;em&gt;vs.&lt;/em&gt; kinetic control).  All of these complexities make it difficult to sort things out in detail, thereby making it difficult to determine the true relative migratory preference of other groups like H and phenyl.  This is why the migratory aptitudes of these groups (H and phenyl) can change from textbook to textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See section 2. Pinacol Rearrangement in &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/rearrang.htm&quot; rel=&quot;noreferrer&quot;&gt;this reference&lt;/a&gt; for a very nice, straightforward discussion of this reaction mechanism, and all its subtleties.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-18T19:38:24.053" LastActivityDate="2014-09-18T19:38:24.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16472" PostTypeId="1" AcceptedAnswerId="16474" CreationDate="2014-09-18T19:42:09.503" Score="3" ViewCount="908" Body="&lt;p&gt;I am told that $\ce{MnO_2}$ oxidizes allylic and benzylic alcohols. However, what is the mechanism of this oxidation? All the professor had to offer me was that the metal ion, manganese 2+, chelates the substrate; there is some sort of interaction between a pi bond, the metal ion, and the electrophilic carbon atom bearing the hydroxyl group. No further explanation was given, and a Google search hasn't unearthed anything of use. What, therefore, is the mechanism of this oxidation? I would like to know more about the mechanism to better understand why $\ce{MnO_2}$ only touches allylic and benzylic alcohols rather than any vanilla alcohol. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2020-03-03T18:19:52.000" Title="Method of allylic and benzylic oxidation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16473" PostTypeId="1" AcceptedAnswerId="16476" CreationDate="2014-09-18T19:52:38.697" Score="4" ViewCount="18185" Body="&lt;p&gt;Why does $\ce{C=O}$ stretching frequency increase with ring strain? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does conjugation decrease $\ce{C=O}$ ring strain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told that cyclopropanone exhibits a higher $\ce{C=O}$ stretching frequency than does cyclohexanone. I know that higher stretching frequencies are correlated with a shorter, stiffer, and stronger bond. What I can't see is why would the $\ce{C=O}$ bond have more of these characteristics in a strained ring system than in a less strained ring system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there hyperconjugative effects going on? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other rationalizations that we can make? I.e. would smaller $\ce{C-C-C}$ angles suggest higher $p$-character in the $\ce{C-C}$ bonds and therefore higher $s$-character in the $\ce{C=O}$ bond? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="15410" LastEditDate="2015-05-20T03:07:03.740" LastActivityDate="2017-05-05T02:14:00.570" Title="Ring Strain and C=O Stretching Frequency" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16474" PostTypeId="2" ParentId="16472" CreationDate="2014-09-18T20:09:21.517" Score="4" Body="&lt;p&gt;This is one of those heterogenous reactions occurring on a surface which are  mechanistically messy. As expected for a reaction that occurs on the surface of a reagent, factors such as surface area and reagent preparation (freshness, surface contamination) are significant variables.  A free radical mechanism is generally accepted and a reasonable mechanism is posted below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Ni6T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Ni6T.png&quot; alt=&quot;reaction mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQLBumtq_7DseSIOlZuuX8Wn1PV7isYjyHuUGf9-7kb3-mfL0hq&amp;amp;s&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvymZ.png&quot; alt=&quot;rxn example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQLBumtq_7DseSIOlZuuX8Wn1PV7isYjyHuUGf9-7kb3-mfL0hq&amp;amp;s&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that we are not really dealing with a free, long-lived radical intermediate, but rather a free radical held in close association with the manganese ion.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-03-03T18:19:52.000" LastActivityDate="2020-03-03T18:19:52.000" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16475" PostTypeId="1" CreationDate="2014-09-18T20:27:24.977" Score="2" ViewCount="57" Body="&lt;p&gt;When oxygen is adsorbed on iron metallic surface, does it create $\ce{Fe-O}$ bonds like the one in iron oxide? Can it be considered as a monolayer of iron oxide on the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how is the atomic bond of adsorbed oxygen and iron? How much is it strong? and what is its difference with normal chemical bond in iron oxide molecule?&lt;/p&gt;&#xA;" OwnerUserId="7730" LastEditorUserId="14" LastEditDate="2015-06-17T15:54:45.427" LastActivityDate="2015-06-17T15:54:45.427" Title="Chemical bond of adsorbed atoms" Tags="&lt;bond&gt;&lt;adsorption&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16476" PostTypeId="2" ParentId="16473" CreationDate="2014-09-18T20:34:35.800" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does C=O stretching frequency increase with ring strain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The carbonyl carbon prefers to be (more or less) $\ce{sp^2}$ hybridized.  Imagine taking the two other bonds attached to the carbonyl carbon and bending them towards each other so that the R-C-R angle is forced from 120 to 90 degrees (the same thing as incorporating these bonds into a smaller and smaller ring).  As we do this, the hybridization of the orbitals the carbonyl carbon uses to form these bonds changes from $\ce{sp^2}$ to $\ce{p}$.  As these bonds use more p-character, more s-character remains for use in the C-O sigma bond.  This would strengthen the CO bond and raise the energy required to stretch it (increase its ir stretching frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate explanation can be made based on resonance structures.  Look back at the acylium ion resonance structure used in our earlier &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16000/triple-bond-character-in-acyl-chlorides/16002#16002&quot;&gt;acid chloride discussion&lt;/a&gt; (resonance structure IV).  As we strain the carbonyl by placing it in a small ring, the bonds between the carbonyl carbon and the adjacent carbons become weaker.  This means that resonance structure IV, with a strong $\ce{C#O}$ triple bond, becomes more important.  Here, in the present case, we would say that resonance structure IV involves carbon-carbon hyperconjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xq1MJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does conjugation decrease C=O ring strain?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you mean, &quot;why does conjugation to the carbonyl decrease the ir stretch frequency&quot;?  Look at the resonance structure on the right side below for the conjugated ketone, cyclohexenone.  It's inclusion leads to a weakening of the carbonyl pi bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KTolg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-09-19T15:20:33.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16477" PostTypeId="1" CreationDate="2014-09-18T20:36:57.970" Score="4" ViewCount="10487" Body="&lt;p&gt;It is well known that single atom of oxygen is not stable, and it forms $\ce{O2}$ molecule. But elements like carbon form a network of repeated bonds. As answered in another &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16428/carbon-atoms-at-the-edge-of-a-diamond&quot;&gt;question&lt;/a&gt;, last atoms in such network are unstable and need to create some bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have a (free-standing) single atom of carbon? Stable in the form of a stable powder like graphite or diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the minimum size of a carbon particle (diamond or graphite network)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Further Clarification:&lt;/em&gt; I meant normal chemical stability to have a powder material at room temperature. I do not mean highly reactive carbon, quickly turning into a compound.&lt;/p&gt;&#xA;" OwnerUserId="7730" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-04-14T09:07:55.983" Title="Is a single carbon atom stable?" Tags="&lt;atoms&gt;&lt;stability&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16478" PostTypeId="2" ParentId="16465" CreationDate="2014-09-18T21:10:12.657" Score="3" Body="&lt;p&gt;In addition to the acid dissociation constant, some other factors that add to the corrosiveness of an acid are its ability to act as an oxidizing agent and its affinity for water (hence, ability to dehydrate material). Concentrated sulfuric acid has both these abilities in addition to simply being a strong acid. I do not know if these factors have been quantified, but they could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing chromic oxide ($\ce{CrO3}$) with sulfuric acid makes it an even better oxidizing agent, especially when hot - so-called - &quot;cleaning solution&quot; is great in the lab for removing residues of tar from your glassware, but it is rather dangerously corrosive and has to be handled very carefully.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5026" LastEditDate="2018-04-10T06:59:46.050" LastActivityDate="2018-04-10T06:59:46.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16479" PostTypeId="1" AcceptedAnswerId="16489" CreationDate="2014-09-18T21:11:01.070" Score="9" ViewCount="36666" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the molecular formula of a compound containing only carbon and hydrogen if combustion of $1.05~\mathrm{g}$ of the compound produces $3.30~\mathrm{g}~\ce{CO2}$ and $1.35~\mathrm{g}~\ce{H2O}$ and its molar mass is about $70~\mathrm{g}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{cccccc}&#xA;                  &amp;amp; \ce{C_{a}H_{b}} &amp;amp; \ce{-&amp;gt;} &amp;amp; \ce{CO2} &amp;amp; + &amp;amp; \ce{H2O} \\&#xA;\text{masses (g)} &amp;amp; 1.05            &amp;amp;         &amp;amp; 3.30     &amp;amp;   &amp;amp; 1.35&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\ce{CO2} &amp;amp;\rightarrow \ce{C} \\&#xA;44~\mathrm{g} &amp;amp;\rightarrow 12~\mathrm{g} \\&#xA;3.30~\mathrm{g} &amp;amp;\rightarrow x&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = 0.9~\mathrm{g},~\text{moles of C} = \frac{0.9}{12} = 0.075&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\ce{H2O} &amp;amp;\rightarrow \ce{2H} \\&#xA;18~\mathrm{g} &amp;amp;\rightarrow 2~\mathrm{g} \\&#xA;1.35~\mathrm{g} &amp;amp;\rightarrow y&#xA;\end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = 0.15~\mathrm{g},~\text{moles of H} = \frac{0.15}{1} = 0.15&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{empirical formula}~\ce{C_{0.075/0.075}H_{0.15/0.075} -&amp;gt; CH2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{70}{14} = 5&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{molecular formula is}~\ce{C5H10}&#xA;$$&#xA;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="7881" LastEditorUserId="194" LastEditDate="2018-02-26T02:50:07.447" LastActivityDate="2019-06-14T12:46:02.630" Title="How to find the molecular formula for a hydrocarbon compound from known combustion products?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16480" PostTypeId="2" ParentId="16477" CreationDate="2014-09-18T21:19:12.773" Score="4" Body="&lt;p&gt;A single atom of helium is stable in nearly all circumstances.  Stability isn't a binary thing, though; whether an atom will react to form a molecule depends on what other atoms are available to react with (for instance, xenon is generally non-reactive, but will form molecules with fluorine), as well as temperature (as Wildcat mentions) and pressure.&lt;/p&gt;&#xA;" OwnerUserId="7912" LastEditorUserId="7912" LastEditDate="2014-09-18T21:30:59.333" LastActivityDate="2014-09-18T21:30:59.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16481" PostTypeId="2" ParentId="16477" CreationDate="2014-09-18T21:20:45.577" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to have a (free-standing) single atom of carbon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is just the matter of having appropriate physical conditions. So, for instance, atomic carbon exist in the gas phase above 3,642 °C (according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_carbon&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-09-18T21:20:45.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16482" PostTypeId="2" ParentId="16477" CreationDate="2014-09-18T21:39:14.317" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to have a (free-standing) single atom of carbon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's definitely possible, but such an atom would be highly unstable and likely would have a short lifetime. An example of this happening is if you blasted some carbon material with a laser. Here's a really cool paper where researchers blasted carbon nanotubes and graphene with lasers and observed single-atom chains (not an individual atom in this case, but the same principle applies) of carbon exhibiting some very interesting electronic properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pa.msu.edu/cmp/csc/eprint/DT086.pdf&quot;&gt;http://www.pa.msu.edu/cmp/csc/eprint/DT086.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If not, what is the minimum size of a carbon particle (diamond or&#xA;  graphite network)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Buckyballs (60 carbon atoms) come to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tangentially related....&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few years ago, a team of Australian scientists fabricated a single-atom transistor (the basic building block of integrated circuits). The atom was Phosphorus. Granted, the atom was not free-standing, but still very impressive. Such a device is extremely difficult to fabricate, and requires cryogenic temperatures, so it's doubtful that this would ever become a viable technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nnano/journal/v7/n4/full/nnano.2012.21.html&quot;&gt;http://www.nature.com/nnano/journal/v7/n4/full/nnano.2012.21.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7693" LastActivityDate="2014-09-18T21:39:14.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16483" PostTypeId="2" ParentId="16467" CreationDate="2014-09-18T21:48:43.523" Score="14" Body="&lt;p&gt;Let me see if I can get at some of your questions. As mentioned above, it's much easier when you ask individual specific questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem with books on introductory quantum mechanics is that, put simply, the language of quantum mechanics is math. Specifically, most people use the Schrödinger equation which involves second derivatives and differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I teach quantum chemistry to undergraduates, there are typically problems due to the mathematical nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple &quot;interpretations&quot; of quantum mechanics. That gets into philosophy. &lt;a href=&quot;http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics&quot;&gt;Wikipedia&lt;/a&gt; has some nicely-written descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mention, the most widely used interpretation (the Copenhagen interpretation) of the wave function centers on the square of the wave function, or rather $|\psi^*\psi|$ as a probability density. As you mention, the wave function could be imaginary or complex, so this notation indicates a mathematical way of getting a real number for a probability density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask about the shape and why atomic orbitals have particular shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that's amazing about quantum mechanics is that we put in some relatively simple equations to describe the motion of an electron in a hydrogen atom, and the solutions are &lt;strong&gt;exactly&lt;/strong&gt; these atomic orbitals (s, p, d, f, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I disagree that &quot;no one knows&quot; why atomic orbitals have particular shapes. The answer is that they are the mathematical solution to the Schrödinger equation. That is, we want to know the energies of the system and there are only certain &lt;strong&gt;quantized&lt;/strong&gt; solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, the whole notion of chemistry is that electrons (and hence orbitals) on different atoms interact. Otherwise, there is no bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave you with this. Without the math, it can be exceptionally hard to understand quantum mechanics. When you do understand the math, I think the results are quite beautiful. The shapes of orbitals aren't arbitrary. This is actually just a reflection of the mathematical nature of quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-18T21:48:43.523" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16484" PostTypeId="1" CreationDate="2014-09-19T00:25:37.800" Score="0" ViewCount="665" Body="&lt;p&gt;I noticed while looking through a couple of books (specifically, Wade, Carey, and Vollhardt/Schore) that the mechanism of concerted hydride and alkyl shifts is always shown with alcohol leaving groups.  I understand that these are difficult, equilibrium-driven processes.  But, I would assume they can happen with other leaving groups, such as a halogen, right?  Is it just the matter of fact that dehydrations are usually done at elevated temperatures while SN2 is not?  If that's all it is, though, we would expect to see rearranged primary E2 products, which Carey mentions does happen, but the other books remain silent on.  &lt;/p&gt;&#xA;" OwnerUserId="7914" LastEditorUserId="4945" LastEditDate="2014-09-19T03:37:21.863" LastActivityDate="2014-09-19T03:37:21.863" Title="Concerted hydride and alkyl shifts" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16486" PostTypeId="2" ParentId="16484" CreationDate="2014-09-19T01:31:03.353" Score="2" Body="&lt;p&gt;The SN2 reaction involves an attacking group and a leaving group moving in concert. A pentacoordinate transition state is involved.  Typically, there is no rearrangement in an SN2 reaction.  Similarly for an E2, as a base removes a proton. a leaving group is ejected in concert to create an olefin, no rearrangement products are observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the SN1 and E1 reactions that proceed stepwise through distinct carbocation intermediates.  As the carbocation is being formed in these reactions, a hydride or alkyl group can migrate to produce a more stable cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leaving group can be things other than a hydroxyl group, for example chloro, bromo, or iodo groups make good leaving groups.  Back in the early days of physical-organic chemistry many alcohols were used in reactions simply because they were readily available as natural products.  Alcohols could also be readily converted into acetates,  &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-1.html&quot; rel=&quot;nofollow&quot;&gt;tosylates&lt;/a&gt;, &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/Anchimeric.htm&quot; rel=&quot;nofollow&quot;&gt;brosylates&lt;/a&gt;, or other sulfonate esters, which were even better leaving groups and allowed for reactions to proceed under even milder conditions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-19T01:31:03.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16487" PostTypeId="1" CreationDate="2014-09-19T01:36:13.807" Score="0" ViewCount="287" Body="&lt;p&gt;In class, my professor went over a method for finding the net ionic equation &lt;em&gt;without&lt;/em&gt; determining the balanced equation (e.g., if finding the balanced equation was too difficult for us students). It involved writing the ions of each of the solutions being mixed and determining which ions would be spectator ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, finding the net ionic equation for the following mixture of solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaC_2H_3O_3 + BaCl_2 +NH_4NO_3 + K_2CrO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't quite grasp the process. How does one determine which ions and compounds will remain in the net ionic equation?&lt;/p&gt;&#xA;" OwnerUserId="7918" LastEditorUserId="13339" LastEditDate="2015-06-26T18:40:24.130" LastActivityDate="2015-06-26T18:40:24.130" Title="How do you solve a net ionic equation without a balanced equation?" Tags="&lt;ions&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16489" PostTypeId="2" ParentId="16479" CreationDate="2014-09-19T03:41:01.453" Score="7" Body="&lt;p&gt;$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -&amp;gt; $a$CO2 + $\frac{b}{2}$ H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$:&#xA;$$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$&#xA;So the empirical formula is $\ce{CH2}$&#xA;Now for actual formula $\frac{70}{14}=5$&#xA;Yes the formula is $\ce{C5H10}$.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-09-19T04:39:08.637" LastActivityDate="2014-09-19T04:39:08.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16490" PostTypeId="2" ParentId="16447" CreationDate="2014-09-19T03:51:15.850" Score="2" Body="&lt;p&gt;\begin{align}\ce{&#xA;3Ag + 4HNO3 &amp;amp;-&amp;gt; 3AgNO3 + NO + 2H2O\\&#xA;AgNO3 + KCl &amp;amp;-&amp;gt; AgCl + KNO3\\&#xA;}\end{align}&#xA;Moles of $\ce{AgCl} = 6.4/143.5 = 0.0446~\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of $\ce{Ag} = 0.0446 \cdot 108 = 4.816~\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Percentage of $\ce{Ag} = 4.816/5.2 \cdot 100=92.62\%$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-09-19T04:53:25.113" LastActivityDate="2014-09-19T04:53:25.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16491" PostTypeId="1" AcceptedAnswerId="16493" CreationDate="2014-09-19T05:11:03.260" Score="4" ViewCount="4030" Body="&lt;p&gt;How is Bent's rule applied on $\ce{SO3}$ and $\ce{SO2}$?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond length of $\ce{SO3}$ &amp;lt; $\ce{SO2}$. Is this a direct consequence of applying the Bent's rule?&lt;/p&gt;&#xA;" OwnerUserId="2586" LastEditorUserId="5456" LastEditDate="2014-09-19T05:21:30.580" LastActivityDate="2018-03-04T07:39:28.087" Title="Bent's rule for Sulfur trioxide and Sulfur dioxide" Tags="&lt;bond&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16492" PostTypeId="1" AcceptedAnswerId="16496" CreationDate="2014-09-19T07:52:27.253" Score="2" ViewCount="331" Body="&lt;p&gt;I need a simple explanation of dative bonding. I am a student that has learnt simple covalent bonding and ionic bonding, but I need help to learn dative bonding. This if for an urgent assignment, so please help. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7836" LastEditorUserId="7836" LastEditDate="2014-09-19T08:04:41.607" LastActivityDate="2014-09-19T10:30:27.033" Title="Simple explanation of dative bonding" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2014-09-19T11:52:51.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="16493" PostTypeId="2" ParentId="16491" CreationDate="2014-09-19T07:57:20.103" Score="6" Body="&lt;p&gt;Here are structures of dominant resonance structure (only one of 3 and 2 differencing only by rotation) for sulfur tri- and dioxide. Despite common misconception, d-orbitals apaprantely play very insignificant role in chemistry of p-block elements, so the structures include formal charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tc5eb.png&quot; alt=&quot;sulfur tri- and dioxide structures without d-orbitals involved using formal charges&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen, the average order of sulfur-oxigen bond lowers with move from di- to trioxide from $1.5$ to $1.(3)$. However, average formal charges grows from $+1$ to $+2$ on sulfur and $-0.5$ to $-0.(6)$,  on oxygen, thus greatly increasing ionic bonding (roughly from $1*0.5 = 0.5$ to $2*0.(6)=1.(3)$) Generally, ionic bonding greatly outperforms covalent bonding, and here charges grown considerably so bond shortening may be attributed to that. The point, however, is, that in this particular structure more than one effect is in play, and their effects outperform whatever small effects are considered when Bent's rule is applied.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-19T07:57:20.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16494" PostTypeId="2" ParentId="16467" CreationDate="2014-09-19T08:59:11.767" Score="5" Body="&lt;p&gt;I know, I would better write a comment, but suddenly comments are limitted in size, so I will write my thoughts (rather than actual answer) here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, on that note, please refrain from heavily mathematical answers&#xA;  because I simply won't understand as I have only done A-level Maths&#xA;  and A-level Chemistry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is the root of your problems. But do not be afraid, you are not alone! As I pointed out in comment to Geoff Hutchison answer, many chemistry students (and let us be honest here, not only students) are strongly reluctant to the following two simple very well established and no longer questioned facts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In order to understand &lt;em&gt;chemistry&lt;/em&gt; you have to understand &lt;em&gt;quantum chemistry&lt;/em&gt;. In other words, at the very fundamental level, there is no any other chemistry except for quantum chemistry.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order to understand &lt;em&gt;quantum chemistry&lt;/em&gt; you have to understand &lt;em&gt;quantum mechanics&lt;/em&gt; which requires a lot of &lt;em&gt;math&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, the right way to proceed is to learn some required math first (at least calculus), then physics (at least classical and quantum mechanics in the form of wave mechanics), and only then quantum chemistry. Thus, my advice is: do not put the cart before the horse! If you have some free time, learn some math. &lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-09-19T08:59:11.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16496" PostTypeId="2" ParentId="16492" CreationDate="2014-09-19T10:30:27.033" Score="2" Body="&lt;p&gt;Alright, I will keep this extremely brief. You know the following things I presume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ionic bonding&lt;/strong&gt; - formed when electrons are transferred from one atom to another, forming a cation and anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Covalent bonding&lt;/strong&gt; - formed when two atoms share an electron pair&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dative covalent bonding&lt;/strong&gt; - formed when two atoms share an electron pair, but this electron pair comes entirely from one of the atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a Hydrogen to Hydrogen covalent bond: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;H-H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They share a pair of electrons, one electron coming from each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,in the carbon monoxide molecule, there are two normal covalent bonds and one dative bond. In the dative bond, both electrons are initially from the oxygen, and are then shared between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5tN9x.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="7813" LastActivityDate="2014-09-19T10:30:27.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16497" PostTypeId="1" CreationDate="2014-09-19T11:18:29.383" Score="19" ViewCount="88912" Body="&lt;p&gt;If zinc is less reactive than sodium, then why does it still react with sodium hydroxide?&lt;/p&gt;&#xA;" OwnerUserId="4602" LastEditorUserId="5026" LastEditDate="2018-04-10T06:52:52.193" LastActivityDate="2018-08-25T15:22:03.423" Title="Why does zinc react with sodium hydroxide?" Tags="&lt;redox&gt;&lt;metal&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16498" PostTypeId="2" ParentId="16497" CreationDate="2014-09-19T11:46:11.320" Score="30" Body="&lt;p&gt;Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction &lt;em&gt;won't&lt;/em&gt; take place&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, however, that $\ce{Zn}$ is &lt;em&gt;amphoteric&lt;/em&gt;, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + 2 H2O + 2 NaOH -&amp;gt; Na2Zn(OH)4 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I forget a lot of good ol' chemistry, and I have just realized that usually we do not call metals themselves amphoteric, but rather use this term for their oxides/hydroxides. For metals we just say that some of them (such as $\ce{Zn}$) dissolve both in acids and in bases. The point is that $\ce{Zn}$ does not react as an acid in the reaction above since the reaction is not an acid-base one, but rather oxidation-reduction. So, the answer then is that $\ce{Zn}$ reacts with aqueous solution of sodium hydroxide because it can be oxidized in these condition in accordance with the reaction mentioned.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="5026" LastEditDate="2018-04-10T06:56:48.763" LastActivityDate="2018-04-10T06:56:48.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16499" PostTypeId="1" CreationDate="2014-09-19T11:49:34.507" Score="7" ViewCount="1189" Body="&lt;p&gt;Of course I know that both 1s orbitals will combine and that the electrons will occupy the bonding MO and that the two atoms will reach a lower energy state – but how do we really &lt;em&gt;know&lt;/em&gt; that is the case?&lt;/p&gt;&#xA;" OwnerUserId="7924" LastEditorUserId="865" LastEditDate="2015-10-30T17:41:43.217" LastActivityDate="2015-10-30T17:41:43.217" Title="How do we know that hydrogen gas consists of two hydrogen atoms?" Tags="&lt;thermodynamics&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16501" PostTypeId="2" ParentId="16467" CreationDate="2014-09-19T12:09:18.800" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://tvtropes.org/pmwiki/pmwiki.php/Main/MathematiciansAnswer&quot; rel=&quot;nofollow&quot;&gt;Mathematician's answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavefunction is quantum mechanical conterpart to trajectory in classical mechanics. Unfortunately, since we interact with the world in a way exposing to us mostly classical newtonian physics, there is no way to find pleasant, easily understandable everyday analogy. However, you may consider atomic orbitals as building blocks for 'trajectory' of electrons. They don't have to be filled to be used as building blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To truly understand the role wavefunction plays in quantum mechanics, at least rudimentary understanding of quantum mechanics is required.  In includes understanding of what is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;coordinate space&lt;/li&gt;&#xA;&lt;li&gt;matrix operator&lt;/li&gt;&#xA;&lt;li&gt;eigenvalues and eigenvectors&lt;/li&gt;&#xA;&lt;li&gt;Knowledge of at least one example of basis in some function space, say, fourier series.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To understand MO theory, a very rudimentary understanding of Hartree-Fock formalism may help greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, wikipedia provides basic knowledge on the topics I mentioned. It is OK to have very vague understanding of concepts I mentioned until 3rd year in university. There is no need to have deep understanding of quantum mechanics to successfully apply MO theory.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-19T12:09:18.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16502" PostTypeId="1" CreationDate="2014-09-19T12:14:54.070" Score="3" ViewCount="1230" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Explain, using a chemical equation to aid your answer, why a mixture of $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ ammonia solution and $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ hydrochloric acid cannot be used as a buffer solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So writing the equation down: $$\ce{NH4OH + HCl -&amp;gt; NH4Cl + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are $0.005~\mathrm{mol}$ of each reactant. From my notes, I think there has to be double the amount of the weak base, but I don't really understand why. Even after googling explanations, I can't find one that answers my question. Why do you need double the amount (if this is correct) of the weak base? An answer to the question above would also be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="5026" LastEditDate="2018-02-25T03:21:48.927" LastActivityDate="2018-02-25T03:21:48.927" Title="Why can't this mixture be used as a buffer solution?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16503" PostTypeId="2" ParentId="16499" CreationDate="2014-09-19T13:40:23.800" Score="11" Body="&lt;p&gt;Assuming ideal gas behavior we can determine the molar mass $M$ of the hydrogen gas from its density $\rho$ (which can be measured) by the following formula&#xA;$$&#xA;    M = {{RT\rho}\over{p}}\ \, .&#xA;$$&#xA;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot;&gt;STP&lt;/a&gt; $\rho \approx 0.09 \, \mathrm{g} \, \mathrm{L}^{−1}$ so the following back-of-the-envelope calculation &#xA;$$&#xA;    M = {{8.31 \, \mathrm{L} \, \mathrm{kPa} \, \mathrm{K}^{−1} \, \mathrm{mol}^{−1} \cdot 273.15 \, \mathrm{K} \cdot 0.09 \, \mathrm{g} \, \mathrm{L}^{-1}}\over{101.325 \, \mathrm{kPa}}}\ = 2.02 \, \mathrm{g} \, \mathrm{mol}^{−1} \, .&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shows that $M$ is indeed close to 2 g/mol which means that hydrogen gas consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-09-19T14:02:16.017" LastActivityDate="2014-09-19T14:02:16.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16506" PostTypeId="2" ParentId="16499" CreationDate="2014-09-19T14:59:10.360" Score="8" Body="&lt;p&gt;I think you're asking a historical question too. This largely comes from a similar question of how we get the modern periodic table and atomic masses. The &lt;a href=&quot;http://www.chemheritage.org/discover/online-resources/chemistry-in-history/themes/the-path-to-the-periodic-table/index.aspx&quot;&gt;Chemical Heritage Foundation&lt;/a&gt; has a nice outline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the big breakthrough here was due to &lt;a href=&quot;http://www.chemheritage.org/discover/online-resources/chemistry-in-history/themes/the-path-to-the-periodic-table/dalton.aspx&quot;&gt;John Dalton&lt;/a&gt; (the &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Dalton&quot;&gt;Wikipedia page&lt;/a&gt; is also good).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dalton really pushed the atomic theory to molecular formulas. At the time, they knew from various analyses the various compositions - there is 8 times the mass of oxygen in water than hydrogen. So Dalton reasoned if everything is made of atoms, they must combine to form different compounds (molecules) in whole number ratios. For example, if an atom can't be divided, we can't have $H_{1.5}O$ -- that violates the idea of indivisible atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he guessed and reasoned through the different compounds and compositions known and published a table of atomic masses. As mentioned above, he was wrong about some things, assuming $\ce{HO}$ instead of $\ce{H2O}$, etc. But you can make some fairly good reasoning of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_multiple_proportions&quot;&gt;law of multiple proportions&lt;/a&gt; to determine stoichiometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Amedeo_Avogadro&quot;&gt;Avogadro&lt;/a&gt; was the next step. He reasoned that equal volumes of gases should have the same number of molecules. So if you weigh equal volumes, you can get the (molar) mass. Unfortunately, his work was largely ignored. (Interesting note, &lt;a href=&quot;http://en.wikipedia.org/wiki/Johann_Josef_Loschmidt&quot;&gt;Loschmidt&lt;/a&gt; is actually the person who worked out $N_A$, not Avogadro.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other people, notably &lt;a href=&quot;http://en.wikipedia.org/wiki/Stanislao_Cannizzaro&quot;&gt;Cannizzaro&lt;/a&gt; helped to clarify and expand Avogadro's work at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Karlsruhe_Congress&quot;&gt;Karlsruhe Congress&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro believed that most gases were binary, i.e., $\ce{H2, O2, N2}$ not $\ce{H, O, N}$ as assumed by Dalton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you carefully measure the reaction of hydrogen gas and oxygen gas to get water, you realize it's $\ce{H2O}$ based on the proportions, and then the atomic masses fall into place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We get further proof of $\ce{H2}$ from other compounds. Sodium bicarbonate clearly has a small amount of hydrogen and you can work out $\ce{NaHCO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, this all required many people and a lot of careful measurements and deduction. But the law of multiple proportions and Avogadro's theory will eventually give you $\ce{H2}$.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-19T14:59:10.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16507" PostTypeId="1" CreationDate="2014-09-19T15:27:19.637" Score="10" ViewCount="54684" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;5) Which one of the following metals does not react with hydrochloric acid?       &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. zinc&lt;br&gt;&#xA;  B. magnesium&lt;br&gt;&#xA;  C. iron&lt;br&gt;&#xA;  D. copper&lt;br&gt;&#xA;  E. aluminium      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The right answer must be copper, but why doesn't copper react with hydrochloric acid while the other metals do?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="46306" LastEditDate="2018-03-01T02:15:18.723" LastActivityDate="2021-04-13T00:29:08.130" Title="Why doesn't copper react with hydrochloric acid while the other metals do?" Tags="&lt;inorganic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16508" PostTypeId="1" CreationDate="2014-09-19T15:34:15.413" Score="0" ViewCount="1460" Body="&lt;p&gt;An element burns in oxygen to form a solid oxide which dissolves in water. The pH of the resulting solution is greater than seven. Which one of the following could the element be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. sulphur&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. phosphorus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C. magnesium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D. sodium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right answer is sodium. But I don't really understand why it can't be magnesium as well.&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="9344" LastEditDate="2016-09-27T00:00:41.233" LastActivityDate="2016-09-27T00:00:41.233" Title="Which of the following elements forms a water-soluble, alkaline oxide?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16509" PostTypeId="2" ParentId="16508" CreationDate="2014-09-19T15:41:03.553" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;solid oxide which dissolves in water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Magnesium oxide $\ce{MgO}$ is (practically) insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be interested to know that burning of magnesium was the classical flash light in the early days of photography:  &lt;a href=&quot;http://en.wikipedia.org/wiki/Flash_(photography)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-19T15:49:21.013" LastActivityDate="2014-09-19T15:49:21.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16511" PostTypeId="2" ParentId="16507" CreationDate="2014-09-19T16:21:43.830" Score="14" Body="&lt;p&gt;You might want to look up some terms, such as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_metal&quot;&gt;noble and less noble metals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reduction_potential&quot;&gt;reduction potential&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_series&quot;&gt;galvanic series&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, reaction means that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hydrogen gas is formed&lt;/li&gt;&#xA;&lt;li&gt;the metal is dissolved&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H+ + 2 e- -&amp;gt; H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal serves as an electron donor and is oxidized, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn -&amp;gt; Zn^{2+} + 2 e- }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more noble a metal is, the more reluctant it is to lose electrons. This is the case for copper, which is therefore not oxidized under these conditions.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-19T16:21:43.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16513" PostTypeId="2" ParentId="16446" CreationDate="2014-09-19T18:59:00.187" Score="1" Body="&lt;p&gt;I'd like to reply to &quot;Mad Scientists&quot; post:&#xA;Now, because of your answers, I understand the topic better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) An electron can has a spin of s=1/2 (lower case s).&#xA;The total spin quantum number S=1/2 (upper case S) is the sum of the single spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Although protons are nuclei, they are elementar particles (which have usually a spin of s=1/2). Protons have only a spin s=1/2 and therefor S=1/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) In contrast to the mentioned two cases of elementar particles examples an atom (or ion with at least one electron; a proton is also an ion) have a spin quantum number s and a angular momentum quantum number l. So the two quantum numbers sum up to the nuclear spin which is usually determined by I= sum over all s + sum over all l. Very often we find as well instead of I the letter J = L + S in spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &quot;long&quot; wrote, I belief also that the lecturer wrote S as an abreviation of Spin withouth distinguishing I und S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any thinking mistake don't hesitate to directly edit this post.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastActivityDate="2014-09-19T18:59:00.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16514" PostTypeId="2" ParentId="16508" CreationDate="2014-09-19T19:00:18.347" Score="1" Body="&lt;p&gt;The answer is sodium. In fact, burning sodium in oxygen atmosphere will produce sodium oxide. When water is added to sodium oxide $\ce{NaOH}$ is produced. The $pH$ of the resulting solution is greater than seven. It can not be $\ce{Mg}$, because $\ce{MgO}$ is insoluble in water. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4231" LastEditDate="2014-09-19T19:10:38.433" LastActivityDate="2014-09-19T19:10:38.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16515" PostTypeId="2" ParentId="16487" CreationDate="2014-09-19T21:40:53.527" Score="2" Body="&lt;p&gt;This sort of question is really all about solubility. What you're really asking when you're trying to get a net ionic equation is &lt;em&gt;What combinations of the ions are insoluble?&lt;/em&gt; Net ionic equations involve only ions that end up as precipitate on the right side of the equation. If you do enough of these problems, you get a feel for the general solubility trends of common ions, but there should be a table in your textbook or have a look &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this specific example, salts with alkali metal and ammonium cations are soluble. Nitrate and acetate anion salts are soluble. Therefore, all of those end up as spectator ions and we're left with just $\ce{Ba^2+}$ and $\ce{CrO_4^2-}$. Chromate salts other than those of alkali metal cations or ammonia are insoluble (rule 8 in the link), so the net ionic equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ba^2+_{(aq)} + CrO_4^{2-}_{(aq)} -&amp;gt; BaCrO_{4(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it's already balanced, but that may not necessarily be true.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-19T21:40:53.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16516" PostTypeId="2" ParentId="15973" CreationDate="2014-09-19T21:45:14.580" Score="7" Body="&lt;p&gt;A SciFinder search turns up very little on this type of transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qui and coworkers run halogenations of phenyl boronate esters using N-halosuccinimide using gold(III) catalysis. The regioselectivity was essentially 1:1:1 for ortho:meta:para. When the reaction was carried out with other substituents as well, the other substituent's directing group preference dominated the regioselectivity.&#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ol102350v&quot; rel=&quot;noreferrer&quot;&gt;Gold(III)-Catalyzed Halogenation of Aromatic Boronates with N-Halosuccinimides&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one other reaction in SciFinder that is relevant, however I don't have access to the full text (and it's in Russian, so I wouldn't be able to read it anyway). The substrate is a benzene ring substituted with a borane cluster. The reported regioselectivity is 80:20 for meta:para.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrophilic substitution in the benzene ring of 9-phenyl-o- and -m-carborane. Zakharkin, L. I. et al &lt;em&gt;Zhurnal Organicheskoi Khimii&lt;/em&gt;, 23(8), 1691-5; &lt;strong&gt;1987&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-19T21:45:14.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16517" PostTypeId="1" CreationDate="2014-09-19T23:59:23.330" Score="0" ViewCount="1113" Body="&lt;p&gt;which level corresponds to n1 and which level corresponds to n2 in the Rydberg equation?&lt;/p&gt;&#xA;" OwnerUserId="7931" LastActivityDate="2014-09-20T06:36:30.653" Title="If the electron of a hydrogen atom transitions from energy level n=5 to n=3, which level corresponds to n1 and which level corresponds to n2" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-09-20T06:52:28.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="16518" PostTypeId="1" CreationDate="2014-09-20T00:53:25.873" Score="0" ViewCount="2199" Body="&lt;p&gt;0.845g of the unknown dissolved in 4.250 grams of camphor had a freezing point of 1.36C. The freezing point depression constant for camphor is 40C*kg/Mol and the melting point of pure camphor is 129C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the freezing temperature depression, T? Tf=Kf*m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calulate the molality? Which i know is mol/kg solvent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the number of moles present of the unknown compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the molar mass of the unknown compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance and please lists steps if possible. I really want to understand and to get this!! :)&lt;/p&gt;&#xA;" OwnerUserId="7934" LastActivityDate="2014-09-20T06:31:43.607" Title="Freezing point depression to find molecular weight" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-09-20T06:52:03.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="16519" PostTypeId="2" ParentId="16518" CreationDate="2014-09-20T06:31:43.607" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mass of solute $W_b=0.845\times10^{-3}\;kgm$&lt;br&gt;&#xA;  Mass of solvent $W_a=4.250\times10^{-3}\;kgm$&lt;br&gt;&#xA;  Depression in freezing point $\Delta T_f=1.36\;K$&lt;br&gt;&#xA;  Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$&lt;br&gt;&#xA;  Freezing point of Camphor $T_f=129\;^\circ C$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part I:&lt;/strong&gt; $\Delta T_f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part II:&lt;/strong&gt; Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part III:&lt;/strong&gt; Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part IV:&lt;/strong&gt; Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-09-20T06:31:43.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16520" PostTypeId="2" ParentId="16517" CreationDate="2014-09-20T06:36:30.653" Score="0" Body="&lt;p&gt;$\bar\nu=\bar R(1/3^2-1/5^2){}{}{}{}{}{}{}{}$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-09-20T06:36:30.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16522" PostTypeId="1" AcceptedAnswerId="16525" CreationDate="2014-09-20T13:01:12.167" Score="7" ViewCount="9426" Body="&lt;p&gt;Why does the difference of energy between the 2s and 2p orbitals of the second period elements increase with increasing atomic number? Does this difference increases by moving down a group, e.g. is the difference between 3s and 3p bigger than 2p and 2p and why?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Atomic Orbital Ionization Energies, eV&lt;/strong&gt;&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{lccccc}&#xA;\hline&#xA;\text{Atom} &amp;amp; \mathrm{1s} &amp;amp; \mathrm{2s} &amp;amp; \mathrm{2p} &amp;amp; \mathrm{3s} &amp;amp; \mathrm{3p} \\&#xA;\hline&#xA;\ce{H} &amp;amp; 13.6 &amp;amp; &amp;amp; &amp;amp; &amp;amp; \\&#xA;\ce{He} &amp;amp; 24.5 &amp;amp; &amp;amp; &amp;amp; &amp;amp; \\&#xA;\ce{Li} &amp;amp; &amp;amp; 5.45 &amp;amp; &amp;amp; &amp;amp; \\&#xA;\ce{Be} &amp;amp; &amp;amp; 9.30 &amp;amp; &amp;amp; &amp;amp; \\&#xA;\ce{B} &amp;amp; &amp;amp; 14.0 &amp;amp; 8.30 &amp;amp; &amp;amp; \\&#xA;\ce{C} &amp;amp; &amp;amp; 19.5 &amp;amp; 10.7 &amp;amp; &amp;amp; \\&#xA;\ce{N} &amp;amp; &amp;amp; 25.5 &amp;amp; 13.1 &amp;amp; &amp;amp; \\&#xA;\ce{O} &amp;amp; &amp;amp; 32.3 &amp;amp; 15.9 &amp;amp; &amp;amp; \\&#xA;\ce{F} &amp;amp; &amp;amp; 46.4 &amp;amp; 18.7 &amp;amp; &amp;amp; \\&#xA;\ce{Ne} &amp;amp; &amp;amp; 48.5 &amp;amp; 21.5 &amp;amp; &amp;amp; \\&#xA;\ce{Na} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 5.21 &amp;amp; \\&#xA;\ce{Mg} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 7.68 &amp;amp; \\&#xA;\ce{Al} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 11.3 &amp;amp; 5.95 \\&#xA;\ce{Si} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 15.0 &amp;amp; 7.81 \\&#xA;\ce{P} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 18.7 &amp;amp; 10.2 \\&#xA;\ce{S} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 20.7 &amp;amp; 11.7 \\&#xA;\ce{Cl} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 25.3 &amp;amp; 13.8 \\&#xA;\ce{Ar} &amp;amp; &amp;amp; &amp;amp; &amp;amp; 29.2 &amp;amp; 15.9 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Data source: &lt;a href=&quot;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.colby.edu&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="41328" LastEditDate="2019-07-09T12:05:56.227" LastActivityDate="2019-07-09T12:05:56.227" Title="Periodic trend in difference of energy between the s and p orbitals" Tags="&lt;orbitals&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16524" PostTypeId="2" ParentId="16522" CreationDate="2014-09-20T14:10:48.523" Score="9" Body="&lt;p&gt;In naive terms I think this could be explained as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$2s$ orbital is lower in energy than $2p$ orbitals because it is closer to the nucleus, and thus, experiences a bigger attractive force to the positive nucleus. The so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;effective nuclear charge&lt;/em&gt;&lt;/a&gt; is bigger for an electron on $2s$ orbital than on one of $2p$ orbitals since in the later case there will be an additional shielding from electrons on $2s$ orbital.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the nuclear charge increases, both $2s$ orbital and $2p$ orbitals become lower in energy due to the increased attractive force to the nucleus, but the $2s$ orbital is affected to a larger degree for the same reason: it is closer to the nucleus and experiences a bigger gain in attractive force to it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, the energy gap between $2s$ orbital and $2p$ orbitals becomes larger as the atomic number increases.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That is all what they can tell you in school. But, of course, the situation is more complicated, and i don't think it can be addressed without doing electronic structure calculations. Yes, &lt;em&gt;effective nuclear charge&lt;/em&gt; experienced by an electron is often expressed as $Z_{\mathrm{eff}}= Z - s$, where $Z$ is the real nuclear charge and $s$ is the so-called &lt;em&gt;screening constant&lt;/em&gt;; $s$ is different for electrons at orbitals corresponding to different values of the principal quantum number $n$ and there are some empirical &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;noreferrer&quot;&gt;Slater's rules&lt;/a&gt; which provide numerical values of $s$ for different values of $n$. But these picture is &quot;not quite right&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, as I already pointed out and as one indeed can see from the Wikipedia page linked above, $s_{2s} = s_{2p}$ (same value of $n=2$). But surely if one calculate these screening constant &lt;a href=&quot;http://dx.doi.org/10.1063%2F1.1733573&quot; rel=&quot;noreferrer&quot;&gt;by solving the corresponding Schrödinger equation&lt;/a&gt; he/she will indeed found that $s_{2s} \neq s_{2p}$. But even without solving the Schrödinger equation the naive argument given above in (1) favour this inequality. So, $s$ is not just a function of $n$, but also of $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, screening constants in Slater's rules does not change their values as atomic number increases which is also wrong. Loosely speaking, the more nuclear charge electron can screen, the more will it screen. Below is the related graph from the article with screening constant calculated at the Hartree-Fock level linked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ptrG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $s$ is also a function of $Z$, and besides for different $n$ and $l$ it depends on $Z$ in different ways. The &quot;real&quot; picture (in quotes since orbitals aren't real anyway) as we can see is much more complicated, and, honestly speaking, I do not even believe that there is a trend you are looking an explanation for.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-09-20T15:35:44.100" LastActivityDate="2014-09-20T15:35:44.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16525" PostTypeId="2" ParentId="16522" CreationDate="2014-09-20T15:24:01.207" Score="3" Body="&lt;p&gt;Let's start with neon, the attractive force of 10 protons, screened to varying degrees by the 9 other electrons, must be overcome in order to remove the outermost p electron, this requires 21.5 eV.  There are only 9 protons holding the outermost electron in fluorine in place, again screening plays a role, but with only 9 protons it is easier to remove fluorine's outermost electron (18.7 eV).  This trend of decreasing IP continues for the remaining elements with 2p electrons as we move to the left across this row of elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding 2s IP's are much higher.  For example, the neon 2s IP=48.5 eV.  To measure this IP, we have already removed neon's 6 outermost 2p electrons.  The 2s electron is still held in place by 10 protons, but now only 3 other electrons remain to screen these protons.  Consequently it is much more difficult to remove one of neon's inner 2s electrons.  As we move to the left across this row of elements looking at the 2s IP's we see a trend similar to what we saw with the 2p IP's, but 2 factors are now involved. It get's easier to remove the 2s electron as the number of protons decreases (reduced nuclear attraction) &lt;strong&gt;and&lt;/strong&gt; as the number of already removed outer 2p electrons becomes smaller (more screening).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same trends are observed in the n=3 series, but the IP's are lower to start with and differences between successive ionizations are compressed because the number of protons, and therefore the number of screening electrons, has increased by 8.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everytime we add a proton to the nucleus, we also add an electron to keep the charge balanced.  In the case of neon, that 10th electron does not feel the full attraction of the 10 protons - the other electrons screen this last electron from the nucleus.  The bigger and bigger we make our nucleus, the more electrons there are between the last electron we added and the nucleus, therefore that last electron feels less and less of a full proton's attraction, as the number of electrons between it and the nucleus increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good example of compression.  The first IP of neon is 21.5 ev, the IP to pull out a fifth electron (after the first 4 electrons have already been removed) is 126.2 eV - about 6 times more energy is required.  Now consider radon, its first IP is only 10.7 eV.  That electron in radon has so many more electrons between it and the nucleus (76 more to be exact) that screen that last electron from the nucleus, that it is much easier to remove that outermost electron in radon.  The fifth IP for radon is 55 eV - only about 5 times more energy (less than 6) is required here, that is compression.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a nice, thorough discussion on ionization energies and their trends that you may find of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally here is a pictorial representation of what we have been discussing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/exR9e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the spacing between the lines decreases as we move to higher atomic number; also note that the slope of the lines tends to increase as the atomic numbers increases.  Both of these observations illustrate what we have been discussing.  As the atomic number increases there are more and more electrons between the outermost electron and the nucleus making it easier to remove that last electron &lt;strong&gt;and&lt;/strong&gt; making the both the absolute and relative differences between successive ionizations (e.g. 1st IP and 2nd IP, 1st IP and 5th IP, etc.) smaller.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-20T18:45:25.510" LastActivityDate="2014-09-20T18:45:25.510" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16526" PostTypeId="1" AcceptedAnswerId="16608" CreationDate="2014-09-20T15:25:12.177" Score="0" ViewCount="1986" Body="&lt;p&gt;I have a question with $\ce{NaOH}$ initially present with given concentration and conductivity, then an equal volume of given concentration of HCl is added and the final conductivity is given, again a same volume of HCl of same concentration is added, the conductivity of which is again given. Now I need to find molar conductivity of $\ce{HCl ,NaOH, NaCl, H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar conductivity($\lambda_m$), conductivity($\kappa$), Molarity($M$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula $\lambda_m=\kappa/M$,firstly I found the molar conductivity of NaOH, afterwards when HCl was added, there will be only NaCl remaining, the rest being used up, so I again found molarity of NaCl and put in the formula and found molar conductivity, but in the last part I am confused, there is only HCl and NaCl remaining and I think one of these must be true:&#xA;$$\kappa_{solution}=\kappa_{\ce{NaCl}}+\kappa_{\ce{HCl}}\\\lambda_{solution}=\lambda_{\ce{NaCl}}+\lambda_{\ce{HCl}}$$&#xA;Probably the second one due to Kolrauch's Law? Also I am not aware of what is going on here because I'm just using a pretty formula. Please explain in detail the physical situation and the processes involved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The conductivity of $0.1$ M $\ce{NaOH}$ is $0.0221\;\Omega^{-1}cm^{-1}$. When an equal volume of $0.1$ M $\ce{HCl}$ solution is added, the conductivity decreases to $0.0056\;\Omega^{-1}cm^{-1}$. A further addition of $\ce{HCl}$ solution, the volume of which is equal to that of first portion added, the conductivity increases to $0.017\;\Omega^{-1}cm^{-1}$. Calculate $\Lambda_m(\ce{NaOH}),\Lambda_m(\ce{NaCl}),\Lambda_m(\ce{HCl}),\Lambda_m(\ce{H^+,OH^-})$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T08:33:26.777" LastActivityDate="2015-06-20T08:33:26.777" Title="Conductance and Conductivity of acid base titration?" Tags="&lt;acid-base&gt;&lt;titration&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16529" PostTypeId="2" ParentId="15974" CreationDate="2014-09-20T18:13:43.660" Score="8" Body="&lt;p&gt;In the past I was also explicitly told that the slowest reaction is &quot;often&quot; the rate determining step, and like you I figured some day I would find a reaction where the fastest step determines the rate. Thinking over it a bit more now, however, I believe we may have misinterpreted what was meant. Rather than looking for an opposite situation, it is likely we were indirectly told that in some cases the rate-determining-step picture simply isn't applicable in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, reactions can't be modelled though such simple kinetic theory; in truth, when a bunch of reactants are brought together, &lt;em&gt;all&lt;/em&gt; steps in &lt;em&gt;every&lt;/em&gt; possible reaction route matter. It just happens that for several simple but relevant chemical systems there are few possible steps and few accessible reaction routes, and they have such wildly different rate constants that we can approximate by looking only at the slowest step of the fastest route. A more complex reaction with several steps, several side-reactions and similar rate constants will likely be poorly described by a RDS framework, so one can say that the reaction rate is not determined by the slowest step because &lt;em&gt;the reaction rate won't be well determined by&lt;/em&gt; &lt;strong&gt;any&lt;/strong&gt; &lt;em&gt;single step&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2014-09-20T18:13:43.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16530" PostTypeId="1" AcceptedAnswerId="16537" CreationDate="2014-09-20T18:44:36.063" Score="6" ViewCount="865" Body="&lt;p&gt;I'm still learning about MO theory – and I thought that I would do some calculations with GAMESS to become more familiar with the concept. Even though I look forward to delving into the realm of quantum math in the not-so-distant future – it's very difficult to get a grip on all of the terminology. I'm using the GAMESS Avogadro extension to write some input files, but I don't know which options to use. Which options should I use if I want to study the orbitals of, say, shikimic acid? RHF, MP2, AM1 / MINI 6-31G(d)? Computer specs: 3.6GHz i5 with 16GB of RAM&lt;/p&gt;&#xA;" OwnerUserId="7942" LastActivityDate="2014-09-21T03:51:52.530" Title="GAMESS crash course?" Tags="&lt;computational-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16531" PostTypeId="1" AcceptedAnswerId="16532" CreationDate="2014-09-20T20:56:10.747" Score="1" ViewCount="139" Body="&lt;p&gt;Can anyone help me with a quick question regarding MestReNova 9.0. I have been fiddling with this thing all day and before you ask yes, I have consulted the users guide to no avail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do three things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Set a threshold so that only peaks above a certain intensity will automatically label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated! &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2014-09-20T22:43:01.647" Title="Using MestReNova: An Introduction" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16532" PostTypeId="2" ParentId="16531" CreationDate="2014-09-20T22:43:01.647" Score="2" Body="&lt;p&gt;Testing this on version 9 will have to wait until tomorrow, but on version 8 (which I have in front of me, but which has limited differences to version 9 by memory)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a manual zoom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt;View &gt;Zoom &gt;Manual Zoom (or just type M, or click the manual zoom button)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type in the values you want&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Set a threshold so that only peaks above a certain intensity will automatically label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do Pick Picking via a Manual Threshold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt;Analysis &gt;Peak Picking &gt;Manual Threshold (or just type K)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left click and hold to define a box within which you will have your peaks picked&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to resize the text to a smaller font size&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First select the parameters box. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Left click the parameters (which should have the green selected boxes shown) and choose properties. Make sure under the geometry tab, the Auto box is selected for size.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then from the text toolbar (by default is usually at the bottom of the screen, but if not shown go to &gt;View&gt;Toolbars&gt;Text) change the font size to something small enough to fit on the screen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Try that for your version - if no luck I'll check on one of my other processing stations tomorrow which has version 9 installed. Let me know how you go.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-09-20T22:43:01.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16533" PostTypeId="1" AcceptedAnswerId="16534" CreationDate="2014-09-20T22:55:07.553" Score="0" ViewCount="1613" Body="&lt;p&gt;&lt;em&gt;This is a part of my online chemistry homework (mastering chemistry)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do it but when I enter $\pu{1.36E-5}$ into the program, it says that it's incorrect and that the correct answer does not depend on $\times10$. I'm not sure what it wants from me. The answer should be given in moles with $3$ sig figs.&lt;/p&gt;&#xA;" OwnerUserId="7945" LastEditorUserId="5026" LastEditDate="2018-02-16T12:31:43.553" LastActivityDate="2018-02-16T12:31:43.553" Title="How many moles of aspartame are present in 4.00mg of aspartame? (3 sig figs)" Tags="&lt;mole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-09-21T08:51:25.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="16534" PostTypeId="2" ParentId="16533" CreationDate="2014-09-20T23:07:02.540" Score="2" Body="&lt;p&gt;Try putting in 0.0000136. Remember that zeroes between the decimal point and the last numbers do not contribute to significant figures. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-09-20T23:07:02.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16537" PostTypeId="2" ParentId="16530" CreationDate="2014-09-21T03:51:52.530" Score="8" Body="&lt;p&gt;First off, welcome to Chemistry.SE! For formatting help visit the &lt;a href=&quot;/help&quot;&gt;help&lt;/a&gt; center and for more information about this page, take the &lt;a href=&quot;/tour&quot;&gt;tour&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I do think you should learn some theory, it's definitely possible to learn some practical computational chemistry through experimentation. And yes, there are a &lt;em&gt;lot&lt;/em&gt; of abbreviations to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you should track while you explore is how long different calculations take. In general, more accurate quantum chemical methods are also more computationally costly. This means that they take longer, but also as systems get bigger, the cost rises faster. You might hear some people talk about &quot;linear scaling&quot; (i.e., double the size of the system, twice as much time) or $N^2$, $N^3$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good resource, IMHO, is &lt;em&gt;Molecular Modeling Basics&lt;/em&gt; by Jensen. It's available through &lt;a href=&quot;http://www.crcnetbase.com/isbn/9781420075274&quot; rel=&quot;noreferrer&quot;&gt;CRC Press&lt;/a&gt; but Jan also &lt;a href=&quot;http://molecularmodelingbasics.blogspot.com&quot; rel=&quot;noreferrer&quot;&gt;blogs extensively&lt;/a&gt;. He uses GAMESS and Avogadro but also Jmol and other programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the Avogadro project is working on some more detailed tutorials, I'd suggest starting some calculations with HF/6-31G* or B3LYP/6-31G* and see how things go for different molecules. I &lt;strong&gt;strongly&lt;/strong&gt; suggest running some calculations and then consider how the results compare to experiment. Benchmarking is important. Computational models all have systematic and other errors, and for real work it's critical to run a set of molecules and carefully choose methods that work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few key terms to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Method&lt;/strong&gt;: There are several types of methods.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Semiempirical methods&lt;/strong&gt;: These include AM1, PM3, RM1, (and PM6, PM7, and others that I don't think are in GAMESS-US). Such methods are fast, since they approximate some integrals and use empirical parameters. They also have an implied basis set which is usually very small. (Avogadro doesn't support viewing semiempirical calculations with GAMESS yet.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Wavefunction methods&lt;/strong&gt;: These include HF and relatives. These are similar to typical quantum chemistry methods taught in school.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;MP2, MP3, MP4&lt;/strong&gt;: These are perturbation methods that improve on Hartree Fock and introduce some level of electron correlation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;CCSD, CCSD(T), etc.&lt;/strong&gt;: These are &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupled_cluster&quot; rel=&quot;noreferrer&quot;&gt;&quot;coupled cluster&quot;&lt;/a&gt; methods that are highly accurate, but often slow, and do not handle large molecules.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Density Functional&lt;/strong&gt;: This includes another large set of methods, including BLYP, B3LYP, and newer methods like M06-X. Rather than approach things through the Schrödinger equation, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Basis&lt;/strong&gt;: Most methods (except AM1, RM1, PM3, etc.) also require a &quot;basis set,&quot; a set of functions to describe the atomic orbitals. &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; There are a few families, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pople Type&lt;/strong&gt;: 3-21G(d), 6-31G(d), 6-311+G(d), etc. These are older, more traditional basis sets. Generally 6-31G(d) or 6-31G* is considered minimally acceptable today. The &quot;*&quot; or (p,d,f) included at the end indicate that higher &lt;em&gt;polarization&lt;/em&gt; is included on some or all atoms. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Correlation Consistent&lt;/strong&gt;: cc-pVDZ, aug-cc-pVTZ, etc. These are newer basis sets, designed to function better with methods that treat electron correlation. The &quot;aug-&quot; includes various diffuse functions, which are useful particularly for anions and electron density that is particularly, well, diffuse. (In Pople basis sets, these are indicated with &quot;+&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-21T03:51:52.530" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16538" PostTypeId="1" AcceptedAnswerId="16543" CreationDate="2014-09-21T09:31:10.070" Score="12" ViewCount="689" Body="&lt;p&gt;Why is disilyne $\ce{Si2H2}$ bent when the steric number of silicon is 2?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qTRtn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5456" LastEditDate="2014-09-21T09:44:44.970" LastActivityDate="2014-09-21T14:15:39.450" Title="Why is disilyne bent?" Tags="&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16539" PostTypeId="2" ParentId="16502" CreationDate="2014-09-21T09:35:10.727" Score="2" Body="&lt;p&gt;You have an example of an alkaline buffer solutions here. They are commonly made from a &lt;em&gt;weak base&lt;/em&gt; ($\ce{NH3}$ aqueous solution) and one of &lt;em&gt;its salts&lt;/em&gt; ($\ce{NH4Cl}$) and work in a little simplified picture as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia is a weak base, so the position of the equilibrium for the following reaction is well to the left&#xA;$$&#xA;    \ce{NH3_{(aq)} + H2O_{(aq)} &amp;lt;&amp;lt;=&amp;gt; NH4+_{(aq)} + OH^{-}_{(aq)}} \, ,&#xA;$$&#xA;and once you add some $\ce{NH4Cl}$ it pushes the equilibrium of the reaction above even more to the left (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier's Principle&lt;/a&gt;). As the result you have both lots of ammonia (due to the equilibrium of the above reaction being well to the left) &lt;em&gt;and&lt;/em&gt; of ammonium ions (coming from $\ce{NH4Cl}$). And you will still have some hydroxide ions due to the reaction above, so that solution will be alkaline. The actual $pH$ depends on in which molar proportions the ingredients were mixed. &lt;em&gt;Usually&lt;/em&gt;, $\ce{NH3_{(aq)}}$ and $\ce{NH4Cl}$ mixed in &lt;em&gt;equal molar proportions&lt;/em&gt;, the solution would have a pH of 9.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When some acid is added to such buffer it will mostly be removed by the following process&#xA;$$&#xA;    \ce{NH3_{(aq)} + H+_{(aq)} &amp;lt;=&amp;gt;&amp;gt; NH4+_{(aq)}} \, ,&#xA;$$&#xA;and when some base is added it will mostly be consumed by the following reaction with ammonium ions&#xA;$$&#xA;    \ce{NH4+_{(aq)} + OH^{-}_{(aq)} &amp;lt;=&amp;gt;&amp;gt; NH3_{(aq)} + H2O_{(aq)}} \, ,&#xA;$$&#xA;where in both cases the equilibrium is well to the right.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, as you see, &lt;em&gt;both the weak base and its salt should present in the solution&lt;/em&gt;. Thus, there should be some excess of weak base when you mix it with a strong acid, otherwise there will be no weak base at the end of neutralization reaction. The idea that one should have double amount of a weak base comes from the fact that, as I mentioned, &lt;em&gt;usually&lt;/em&gt; a weak base and its salt are mixed in equal molar proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="24052" LastEditDate="2016-03-11T23:50:39.687" LastActivityDate="2016-03-11T23:50:39.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16540" PostTypeId="1" AcceptedAnswerId="16599" CreationDate="2014-09-21T10:29:32.340" Score="5" ViewCount="569" Body="&lt;p&gt;I'm wondering is there any branches of Chemistry that make use of manifolds, Riemann geometry, tensors etc. So far I have not come across anything but hoped that the wider community would know of something? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive me if this is too vague a question but I'm just looking for some topics I can research myself.&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7475" LastEditDate="2015-11-09T15:06:51.630" LastActivityDate="2015-11-09T15:06:51.630" Title="Are higher branches of mathematics used in sub-fields of chemistry?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16541" PostTypeId="2" ParentId="15974" CreationDate="2014-09-21T11:33:45.683" Score="10" Body="&lt;p&gt;I can provide you an example.&#xA;The oxidation of formate ion by peroxydisulfate in water solution:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{HCOO^{-} +  S_2O_8^{2-} -&amp;gt; CO_2 + 2SO_4^{2-} + H^+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;has the following mechanism:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OdWpn.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: &lt;span class=&quot;math-container&quot;&gt;$r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]$&lt;/span&gt;. The  first  reaction  is  very  slow,  so  most  of  peroxydisulfate  is consumed  in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.&lt;/p&gt;&#xA;&lt;p&gt;(This isn't the true mechanism but it is good for showing the point. A  more complex  mechanism  includes  the  formation  of  OH  radicals  and  several  chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-09-21T17:23:08.173" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16542" PostTypeId="2" ParentId="16458" CreationDate="2014-09-21T13:55:00.777" Score="6" Body="&lt;p&gt;As you said, if you look at the electron configurations of &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/f-Block_Elements/The_Lanthanides#Electron_Configuration&quot; rel=&quot;noreferrer&quot;&gt;Lanthanides&lt;/a&gt; and &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/f-Block_Elements/The_Actinides#Actinide_Properties_Table&quot; rel=&quot;noreferrer&quot;&gt;Actinides&lt;/a&gt;, Thorium seems to be the only exception from this empirical rule. You could expand the rule to also include the case of Thorium but I would say it is more useful to have the rule describe the majority of cases and remember $\ce{Th}$ as an exception.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-21T13:55:00.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16543" PostTypeId="2" ParentId="16538" CreationDate="2014-09-21T14:15:39.450" Score="11" Body="&lt;p&gt;The following disilyne has been prepared and found to be stable to ~100 C.  An X-ray crystal structure found that the two silicon atoms in the triple bond, Si(1) and Si(1'), along with the two attached silicon atoms, Si(2) and Si(2') are coplanar and the Si(1)-Si(1')-Si(2') angle is 137.44 degrees (&quot;trans-bent&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDSNE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full text of the article describing the preparation, characterization and reactivity of the compound can be found &lt;a href=&quot;http://www.mycrandall.ca/faculty/MSchriver/coursewebs/BCSJ2006SiSi.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  See figure 4 in the paper for a MO explanation as to why the trans-bent geometry occurs. Basically, the authors suggest that, &quot;the bending is thought to be the result of the mixing of an in-plane $\ce{\pi}$-orbital with a $\ce{\sigma^{\ast}}$ orbital whose energies are close enough to cause the interaction.&quot;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-21T14:15:39.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16544" PostTypeId="1" CreationDate="2014-09-21T14:29:06.780" Score="5" ViewCount="1464" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the pH of a mixture of $75\:\mathrm{cm^3}$ of methanoic acid solution of concentration $0.10\:\mathrm{mol\:dm^{-3}}$ and $75\:\mathrm{cm^3}$ of sodium hydroxide solution of concentration $0.050\:\mathrm{mol\:dm^{-3}}$. The $\mathrm{p}K_\mathrm{a}$ of methanoic acid is $3.75$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got $0.00375\:\mathrm{mol}$ of $\ce{HCOONa}$, the salt, made. The concentration of that is $\dfrac{0.00375}{150/1000}=0.025\:\mathrm{mol\:dm^{-3}}$(?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Henderson–Hasselbalch: $\mathrm{pH}=3.75+\log\dfrac{0.025}{0.10}=3.15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer says $3.75$, is that a typo?&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="5026" LastEditDate="2018-02-28T15:13:33.840" LastActivityDate="2018-11-28T15:42:17.003" Title="pH of a buffer solution calculation" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16545" PostTypeId="1" AcceptedAnswerId="16546" CreationDate="2014-09-21T15:04:41.740" Score="2" ViewCount="125" Body="&lt;p&gt;Can someone draw what the missing structure is? I can't find it.&#xA;&lt;a href=&quot;https://imgur.com/8FaYuXk&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/8FaYuXk.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-09-21T17:50:46.457" Title="Resonance of 4-isopropylbenzenethiol" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16546" PostTypeId="2" ParentId="16545" CreationDate="2014-09-21T15:16:06.067" Score="6" Body="&lt;p&gt;The missing resonance structure is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3TPj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replace &quot;SH&quot; with &quot;OH&quot; in the 4 structures you've drawn (your 5th structure is the same as the one pictured on the exam) and the one I've drawn, you'll have all of the analogous resonance structures for phenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my view, it doesn't really matter how you order the structures, but your suggestion (between the 3rd and 4th structures from the left) seems logical.  I've redrawn the figure to show the electron movement.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-21T17:50:46.457" LastActivityDate="2014-09-21T17:50:46.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16548" PostTypeId="1" CreationDate="2014-09-21T18:12:40.200" Score="3" ViewCount="5577" Body="&lt;p&gt;&lt;a href=&quot;https://biology.stackexchange.com/q/21514/4466&quot;&gt;To exterminate a coackroach infestation&lt;/a&gt;, I bought solid boric acid (in powder form). Rather than awaiting the cockroaches' encounter with the boric acid, I thought about spraying boric acid onto affected areas or cockroaches. I only took high school chemistry, but remembered solubility so &lt;a href=&quot;http://chemicals.etacude.com/b/more/h3bo3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;checked its solubility in water&lt;/a&gt;. I then compared it with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table#B&quot; rel=&quot;nofollow noreferrer&quot;&gt;solubility of other compounds&lt;/a&gt;, to see if this were an effective idea:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table#B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boric acid&lt;/a&gt;: At 100 Celsius, 27.53 g/100 g of water.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table#S&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sodium chloride&lt;/a&gt;: At 100 Celsius, 38.99 g/100 g of water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This looks fine, so I poured boiling water into a bottle with the boric acid powder. I shook it and saw the boric acid dissolve. Yet two weeks later, the boric acid seems to have reversed dissolution. Now, it's separated from the water and sits at the bottom of the water.&lt;/p&gt;&#xA;&lt;p&gt;Can substances do this? What should I do now? I'd need a pot to boil the mixture over the stove. But then, in the interest of health and safety, I couldn't use this pot anymore, so don't want to sacrifice one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; With thanks to and in view of the instructive answers, would it help to refill the bottle, to benefit from the low solubility of boric acid at room temperature? There's only around 10% of water left, with the boric acid sitting at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="7952" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-11-17T15:53:45.693" Title="What to do about a un-dissolved substance?" Tags="&lt;solubility&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16549" PostTypeId="1" AcceptedAnswerId="16556" CreationDate="2014-09-21T18:31:28.957" Score="2" ViewCount="1930" Body="&lt;p&gt;We make many assumptions in sciences which are not always true but as they make little difference we ignore them for simplicity. But today I have decided that I can no longer be lured into using expression of $K_{a}$ and $K_{w}$ by assumptions which I deem to make a significant difference.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We say that $[H_{2}O]$ is constant. Why I think it is not justified?&lt;br&gt;&#xA;$$\ce{H2O + HCL &amp;lt;=&amp;gt; H3O+ + Cl-}$$&#xA;$[H_{2}O]$ is almost $ 55.6 \:\mathrm{mol\:dm^-3}{}$.&lt;br&gt;&#xA;At around $20^{o}C$ $20\:\mathrm{mol}$ of $HCl_{aq}$ can be dissolved. Let's say we dissolve $20 \:\mathrm{mol}$ of $HCl$ and try to find it's &lt;strong&gt;WE ASSUME&lt;/strong&gt; that % disassociation of water is $100 \%$. So we must take $[HCl]\approx0$ that means that $20\:\mathrm{mol}$ of $H_{2}O$ has reacted and $[H_{2}O]$ is no longer a constant but has rather changed to $35.6\mathrm{mol}$.&lt;br&gt;&#xA;So this expression can no longer be used. $$K_{a}=\frac{[H_{3}O^{+}].[Cl^{-}]} {[HCl]}=K_{c}.[H_{2}O]$$&#xA;&lt;strong&gt;&lt;em&gt;Note:&lt;/strong&gt; you might point out that $K_{a}$ is not used for strong acids but I just used $HCl$ as an example but make my point easier to explain.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Assumption made about concentrations when calculating $Ka$ of buffer solutions.&lt;br&gt;&#xA;$$CH_{3}COOH_{aq} \rightleftharpoons CH_{3}COO^{-}_{aq} + H^{+}_{aq}$$&#xA;We have added $CH_{3}COONa$  to make a buffer solution.&lt;br&gt;&#xA;$$CH_{3}COONa_{aq} \rightarrow CH_{3}COO^{-}_{aq} + Na^{+}_{aq}$$&#xA;So,&lt;br&gt;&#xA;$$[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa} + [CH_{3}COO^{-}]_{CH_{3}COOH}$$&#xA;$$[CH_{3}COO^{-}]_{CH_{3}COOH} = [H^{+}_{aq}]$$&#xA;&lt;strong&gt;BUT WE ASSUME&lt;/strong&gt; that $[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa}$ because we say $[CH_{3}COO^{-}]_{CH_{3}COOH}$ is negligible yet we include [$H_{aq}^{+}]$ in the expression which has same value as $[CH_{3}COO^{-}]_{CH_{3}COOH}$. Same value is considered negligible at one place and not negligible at another in the same expression. How is that justified?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6791" LastEditorUserId="22" LastEditDate="2014-09-21T21:21:40.243" LastActivityDate="2014-09-22T09:40:14.680" Title="Can assumptions made during calculation of ka of buffer solutions be justified?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16550" PostTypeId="2" ParentId="16548" CreationDate="2014-09-21T18:35:05.600" Score="2" Body="&lt;p&gt;Looking at your link, the solubility of boric acid at room temperature is around 5g / 100g of water. You can either heat it to boiling point to redissolve or you can filter it and this way you will get a 5g / 100g solution of boric acid that you can still use and also recover solid boric acid. You may need to experiment with different DIY filtering devices depending on the particle size of your precipitated boric acid but I guess you made some nice crystals that shouldn't be very hard to filter.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-09-21T18:35:05.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16551" PostTypeId="2" ParentId="16548" CreationDate="2014-09-21T18:46:52.043" Score="10" Body="&lt;p&gt;I think what you forgot is that the solubility of boric acid (or sodium chloride.. or most solids) is lower at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you dissolved the boric acid in boiling water, there's no problem at first. As it cooled, you will have a &lt;a href=&quot;http://en.wikipedia.org/wiki/Supersaturation&quot; rel=&quot;nofollow&quot;&gt;supersaturated&lt;/a&gt; solution until eventually the boric acid will recrystallize or crash out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd create a new mixture using &lt;strong&gt;room temperature&lt;/strong&gt; water and add only a small amount of boric acid (as mentioned by @K_P, about 5g / 100 mL of water). Let it dissolve slowly while you mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you spray, you shouldn't leave large amounts of boric acid around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Updated Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, if there's only ~10% water, you will want to add more (room temperature) water. This will dissolve more boric acid. Stir/shake to dissolve as much as possible, wait a bit and then you can spray again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, you don't want to add water to a hot acid. In this case, adding more room temperature water to a room temperature acid will probably produce heat, but add slowly and carefully. &lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-22T14:49:44.293" LastActivityDate="2014-09-22T14:49:44.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16553" PostTypeId="1" AcceptedAnswerId="16558" CreationDate="2014-09-21T20:42:53.350" Score="3" ViewCount="9014" Body="&lt;p&gt;Can the volume of a solid (specifically a granular crystal or powder) be accurately determined using the displacement method with a liquid in which the solid is soluble? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally I thought by dissolving in the liquid there would (or could) somehow be a change to the density of the mixture which would affect the displacement. As such, it would be a good idea to ensure the substance is not soluble in the liquid being used to determine the solid's volume. But if the dissolution of the solid is only a physical process (i.e., dissolving sugar or salt in water), would this at all affect one's ability to measure the volume of the amount of substance added? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess when dealing with an unknown substance, we might not be able to easily determine if the dissolution took place as a physical or chemical process without further testing, so it would make more sense to conduct the measurement with a liquid in which the substance was insoluble.&lt;/p&gt;&#xA;" OwnerUserId="7953" LastEditorUserId="69977" LastEditDate="2020-04-29T11:40:24.890" LastActivityDate="2020-04-29T11:40:24.890" Title="Does solubility matter when measuring density with displacement method?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16554" PostTypeId="1" CreationDate="2014-09-21T20:59:20.860" Score="1" ViewCount="500" Body="&lt;p&gt;Consider calcium carbonate &lt;em&gt;saturated&lt;/em&gt; in $1.0\:\mathrm{M}$ ammonium chloride solution. I have calculated the $\mathrm{pH}$ of this solution analytically (with mass and charge balance) to be $6.92$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how would the &lt;em&gt;actual&lt;/em&gt; solution $\mathrm{pH}$ compare to the theoretical, calculated $\mathrm{pH}$ - in a general sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would one expect the &lt;em&gt;actual&lt;/em&gt; solution $\mathrm{pH}$ to be &lt;em&gt;higher&lt;/em&gt; or &lt;em&gt;lower&lt;/em&gt; than $6.92$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are going to be a host of solute-solute interactions in this system that are more than likely to cause deviations from ideal solute behavior. Here's a list of species in the system:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{Ca^2+}$ and its respective hydrates &lt;/li&gt;&#xA;&lt;li&gt;$\ce{CO_3^2-}$ and its respective protonation products (i.e. bicarb and carbonic acid)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H_4N+}$ and its respective deprotonation products (i.e. ammonia)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cl-}$ and perhaps (unlikely) a bit of its protonation product - hydrochloric acid. &lt;/li&gt;&#xA;&lt;li&gt;$\ce{H_3O+}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HO-}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, the question I'm fundamentally getting at is how would all these solutes interact with each other. I understand that the calcium ion and the carbonate ion with their +2 charge magnitudes is likely going to cause this solution to deviate significantly from ideal behavior. A quick leafing through of an analytical text lead me to this hypothesis. However, is it possible to actually predict the direction, so to speak, of $\mathrm{pH}$ deviation? Can we model this system with a computer program? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I plan on actually taking $\mathrm{pH}$ measurements on this system. Here is a general outline of the things I plan to do; is there anything else that I need to take into account? &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use of triple distilled water. Our laboratory's &quot;deionized&quot; water runs strongly basic. Water will be distilled with a high partial pressure of nitrogen gas as to prevent carbon dioxide from dissolving in water.  &lt;/li&gt;&#xA;&lt;li&gt;Should we measure $\mathrm{pH}$ in a sealed vessel? I don't want any carbon dioxide to be dissolving into the system, again. Also, I don't want any ammonia gassing off from the system (since ammonia is a possible product upon saturating ammonium chloride solution with calcium carbonate). How would measuring $\mathrm{pH}$ in a sealed vessel work? &lt;/li&gt;&#xA;&lt;li&gt;Any other lab practical tips/suggestions? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="418" LastEditDate="2016-10-31T07:03:25.267" LastActivityDate="2016-10-31T07:03:25.267" Title="Reasonings on the pH of calcium carbonate in ammonium chloride solution" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16555" PostTypeId="1" AcceptedAnswerId="16577" CreationDate="2014-09-21T21:33:05.527" Score="5" ViewCount="1529" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_nitrate&quot; rel=&quot;noreferrer&quot;&gt;Potassium nitrate&lt;/a&gt; is very popular compound in pyrotechnics, used as an oxidiser. It can be mixed with sugar to make a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rocket_candy&quot; rel=&quot;noreferrer&quot;&gt;rocket candy&lt;/a&gt;&quot;, with coal and sulphur to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Gunpowder&quot; rel=&quot;noreferrer&quot;&gt;black powder&lt;/a&gt;. You can stick it on rope to create &lt;a href=&quot;http://www.instructables.com/id/Homemade-Waterproof-Fuse,-For-Homemade-Fireworks/&quot; rel=&quot;noreferrer&quot;&gt;functional fuse&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The saltpeter releases oxygen in all these reactions. But what are the results of reaction? I learned that &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_nitrite&quot; rel=&quot;noreferrer&quot;&gt;$\ce{KNO2}$&lt;/a&gt; is also oxidiser. So does the nitrate turn tu nitrite and then reacts again? Or does it form something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that some strange bubbles come from my homemade fuse (photo will follow). Since potassium nitrate has melting point at 334°C (&lt;a href=&quot;https://www.google.cz/search?q=potassium%20nitrate%20melting%20point&quot; rel=&quot;noreferrer&quot;&gt;according to Google&lt;/a&gt;) it's likely the saltpeter melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about fumes? Does potassium nitrate or nitrite evaporate to turn the air poisonous or irritating?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2020-02-21T02:58:54.287" Title="How does potassium nitrate work as an oxidiser?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16556" PostTypeId="2" ParentId="16549" CreationDate="2014-09-21T21:35:11.567" Score="3" Body="&lt;p&gt;Answering your first question, $[\ce {H2O}]$ is constant because the square brackets indicate a concentration. Concentration is the number of moles of the solute on the volume of the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except in this case, water &lt;em&gt;is&lt;/em&gt; the solvent. So the concentration will always be constant, since the volume is proportional to the number of moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, while dealing with the law of mass action, for liquids we don't take &lt;em&gt;concentration&lt;/em&gt;, we take &lt;em&gt;activity&lt;/em&gt;. Which is approximately one for water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, we basically have the following product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{([CH3COO- ]_{CH3COOH} + [CH3COO- ]_{CH3COONa})[H+ ]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be rewritten as $(a + b)b$, where $b &amp;lt;&amp;lt; a$. Let's say $a$ was $1$ and $b$ was $0.0001$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual value of the expression is $1.0001\times 10^{-4}$. Neglecting only the first $b$ will give $1\times 10^{-4}$, which is not too much different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neglecting only the second $b$ will give $1.0001$, and neglecting both will give $1$. Both of these answers are very wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a small number is added to a large one, the small number can be ignored. But not when it is multiplied.&lt;/p&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2014-09-22T09:40:14.680" LastActivityDate="2014-09-22T09:40:14.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16558" PostTypeId="2" ParentId="16553" CreationDate="2014-09-21T21:48:17.207" Score="3" Body="&lt;p&gt;Volume is not a conserved extensive property of mixtures in the same way that other quantities like mass, net electric charge, number of particles, etc. are (within closed systems). That is to say, in reality, the volumes of components physically mixed together are not simply additive, with the total volume not necessarily equal to the sum of the individual volumes. I.e., $\sum{V_i} \not= V_{mix}$ generally. This occurs chiefly because intermolecular forces of attraction between particles will vary with the composition of the mixture. Mixtures of ethanol and water, for example, are always reduced in volume by comparison to the summed volumes of ethanol and water, respectively, in isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the total volume can be calculated given the partial molar volumes of the individual components &lt;em&gt;for a specific composition&lt;/em&gt; under given conditions. Specifically, $V_{mix} = \sum{n_i \bar{V_i}}$, where $\bar{V_i}$ and $n_i$ are the partial molar volumes and moles, respectively, of the separate constituents of the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few good references I found for additional study: &lt;a href=&quot;http://www2.stetson.edu/~wgrubbs/datadriven/fchen/bartender/partialmolarvolumechen.html&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://infohost.nmt.edu/~jaltig/PartVol.pdf&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;. Any decent physical chemistry textbook should also cover partial molar properties.&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-09-21T21:48:17.207" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16559" PostTypeId="5" CreationDate="2014-09-21T21:51:58.587" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-21T21:51:58.587" LastActivityDate="2014-09-21T21:51:58.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16560" PostTypeId="4" CreationDate="2014-09-21T21:51:58.587" Score="0" Body="Activities are a reflection of the ability of a solute to behave independently in solution. " OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-22T15:14:53.283" LastActivityDate="2014-09-22T15:14:53.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16561" PostTypeId="2" ParentId="16548" CreationDate="2014-09-21T22:09:06.223" Score="5" Body="&lt;p&gt;Usually, when you want to create saturated solution the approach is following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You heat the water (does not need to boil though)&lt;/li&gt;&#xA;&lt;li&gt;You dissolve as much solid substance as possible&lt;/li&gt;&#xA;&lt;li&gt;Let the water cool down&lt;/li&gt;&#xA;&lt;li&gt;Pour the water through filter to filter out any recrystallised solid compound&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The amount you can dissolve is the amount that can be ever dissolved. If the temperature decreases, some of the dissolved compound will recrystallize.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-09-21T22:09:06.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16562" PostTypeId="1" AcceptedAnswerId="16567" CreationDate="2014-09-22T00:17:03.090" Score="33" ViewCount="15523" Body="&lt;p&gt;Pyrene doesn't seem to be aromatic. However, sources claim that it is aromatic.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sXrAG.png&quot; width=&quot;200&quot; alt=&quot;Lewis structure of pyrene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considerations: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pyrene is cyclic. ✓&lt;/li&gt;&#xA;&lt;li&gt;Pyrene is flat (planar). ✓&lt;/li&gt;&#xA;&lt;li&gt;Pyrene has &lt;strong&gt;16&lt;/strong&gt; π electrons. &lt;/li&gt;&#xA;&lt;li&gt;Every atom in the ring structure of pyrene is $\ce{sp^2}$ hybridized. ✓&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The problem is that pyrene fails the $4n+2$ rule. $4n + 2 \neq 16$ where $n$ is an integer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something leads me to suspect that Hückel's rule is an oversimplification here. Or is pyrene indeed non-aromatic? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="23561" LastEditDate="2018-11-07T19:32:48.513" LastActivityDate="2019-01-18T12:58:58.790" Title="Is pyrene aromatic despite failing Hückel's rule?" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;&lt;aromaticity&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="16563" PostTypeId="1" CreationDate="2014-09-20T16:34:23.707" Score="14" ViewCount="27839" Body="&lt;p&gt;When I dissolve sugar in my cup of tea/coffee, does the sugar go from being a solid to being a liquid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJgVJ.jpg&quot; alt=&quot;Figure 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea&quot;&gt;https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7588" OwnerDisplayName="User 17670" LastEditorUserId="9961" LastEditDate="2017-09-29T16:29:40.447" LastActivityDate="2018-05-28T15:23:56.527" Title="When I dissolve sugar in my cup of tea/coffee, does it become a liquid?" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16564" PostTypeId="2" ParentId="16563" CreationDate="2014-09-20T23:09:26.207" Score="10" Body="&lt;p&gt;When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best fit is a &lt;strong&gt;liquid&lt;/strong&gt; since the molecules are free to move independently and will take on the shape of the container.&lt;/p&gt;&#xA;" OwnerUserId="4934" OwnerDisplayName="LDC3" LastActivityDate="2014-09-20T23:09:26.207" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16565" PostTypeId="1" AcceptedAnswerId="16570" CreationDate="2014-09-22T00:24:43.253" Score="5" ViewCount="7942" Body="&lt;p&gt;I saw an interesting spectrum in my biology class the other day. On the left end was methane - the least oxidized form of carbon - and the most energetic form of carbon (at least relative to what else was on the slide). On the right end was carbon dioxide - the most oxidized form of carbon - and the form with the lowest energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hCGMJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I do agree with the slide - we can combust methane but we can't combust carbon dioxide - at least I don't think we can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does increasing oxidation state correlate with lower potential energy for all elements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it seem that oxygen forms rather strong bonds with other elements in many cases? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-22T13:46:16.333" LastActivityDate="2017-10-03T01:48:25.233" Title="Why does potential chemical energy decrease with increasing oxidation of carbon?" Tags="&lt;redox&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16566" PostTypeId="2" ParentId="16562" CreationDate="2014-09-22T00:27:53.247" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel&amp;#39;s_rule&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic, even though they fail the 4n+2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-22T00:27:53.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16567" PostTypeId="2" ParentId="16562" CreationDate="2014-09-22T00:29:20.137" Score="29" Body="&lt;p&gt;Pyrene is aromatic. The Hückel $4n+2$ rule works best with monocyclic ring systems. If you look at the following resonance structure for pyrene with a central double bond, the monocyclic periphery has 14 π electrons (ignoring the greyed-out central double bond), but that is a rationalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pbnEv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pbnEv.png&quot; alt=&quot;Periphery of pyrene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, pyrene undergoes reactions characteristic of aromatic systems and has ring currents expected from aromatic systems.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-01-06T18:03:19.450" LastActivityDate="2018-01-06T18:03:19.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16568" PostTypeId="2" ParentId="16562" CreationDate="2014-09-22T00:49:54.380" Score="10" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1nYK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;These are both aromatic, but only one is flat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;The molecule on the right is not flat because of the 2 hydrogens overlapping.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-22T00:49:54.380" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16569" PostTypeId="2" ParentId="16562" CreationDate="2014-09-22T00:50:18.747" Score="31" Body="&lt;p&gt;To clarify @ron's point, the general &quot;cheminformatics&quot; rule for deducing aromatic fused-ring systems is whether there is &lt;em&gt;one&lt;/em&gt; path (usually the peripheral one) that satisfies $4n+2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for naphthalene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DSI13.png&quot; alt=&quot;naphthalene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;bold&lt;/strong&gt; bonds give you a ring path of $\mathrm{sp^2}$ carbons with 10 π electrons, so it's aromatic. There are some arguments about whether the central bond is aromatic or not. (I say baloney, but it's fair to say aromaticity generates lots of arguments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JmVqk.png&quot; alt=&quot;pyrene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the &lt;strong&gt;bold&lt;/strong&gt; path gives you 14 π electrons for a $4n+2$ path. Again, I've heard people say the central two carbons aren't aromatic. (IIRC, the $\ce{^13C}$ NMR disagrees.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote the code for Open Babel and aside from some tricky nitrogen-heteroaromatic rings, this rule generally matches chemical reality. (Setting aside the question of the interior carbons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think another fairly safe rule is &quot;if a fused aromatic system is made up of individually aromatic rings, aromaticity is retained.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="16683" LastEditDate="2018-01-06T18:04:41.607" LastActivityDate="2018-01-06T18:04:41.607" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16570" PostTypeId="2" ParentId="16565" CreationDate="2014-09-22T00:54:46.640" Score="3" Body="&lt;p&gt;From a very basic explanation from general chemistry:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increasing your oxidation state increases the charge. So that means that you move to the left on the periodic table increasing your atom radius and decreasing your ionization energy and electronegativity. &lt;/li&gt;&#xA;&lt;li&gt;If your IE decreases and you become more electronegative, your atom is more attractive for chemical reactions so it has a lower potential energy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a less handwaving explanation from inorganic chemistry, but that is  a little harder to understand and I don't want to formulate a MO diagram. &lt;/p&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2014-09-22T00:54:46.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16571" PostTypeId="2" ParentId="16544" CreationDate="2014-09-22T01:28:13.890" Score="2" Body="&lt;p&gt;One will need the Henderson Hasselbach equation which is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where HA and A- are the concentrations of the &lt;strong&gt;&lt;em&gt;conjugate&lt;/em&gt;&lt;/strong&gt; acid and &lt;strong&gt;&lt;em&gt;conjugate&lt;/em&gt;&lt;/strong&gt; base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we find our moles of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;same thing for base: &lt;span class=&quot;math-container&quot;&gt;$\mathrm {mol_{base} = 375} $&lt;/span&gt; mmol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{750 mmol -375 mmol} = \pu{375 mmol}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{375 mmol - 375 mmol} = \pu{0}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;: --------------------&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{A-}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{0 + 375 mmol} = \pu{375 mmol}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}=3.75$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7957" LastEditorUserId="7957" LastEditDate="2018-11-28T15:42:17.003" LastActivityDate="2018-11-28T15:42:17.003" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="16572" PostTypeId="1" AcceptedAnswerId="16574" CreationDate="2014-09-22T02:23:32.560" Score="3" ViewCount="12410" Body="&lt;p&gt;Like for group I metals, it is understood that all the elements in the group would have an oxidation number of +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, oxidation number is how much electron the element loses/gains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence my question is, why does an element like Iron, Manganese or even oxygen have more than one oxidation state unlike elements like Magnesium or Potassium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If I have stated any terms incorrectly, please let me know because I'd want to know where I went wrong. &lt;/p&gt;&#xA;" OwnerUserId="7787" LastActivityDate="2021-02-09T18:13:50.957" Title="Why does a chemical have more than one oxidation number?" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16573" PostTypeId="2" ParentId="16565" CreationDate="2014-09-22T03:01:59.253" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is there an intuitive explanation of why increasing oxidation state&#xA;  correlates with lower potential energy, at least for carbon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass).  As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does increasing oxidation state correlate with lower potential energy&#xA;  for all elements?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't checked extensively, but the trend seems to hold up.  I'd expect it to hold up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BeH_2}$, $\ce{BeF_2}$  127.2, -796.7 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaO}$, $\ce{CaF_2~~}$      39.1, -789.1 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaH}$, $\ce{NaF~~}$      126.2, -288.0 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does it seem that oxygen forms rather strong bonds with other&#xA;  elements in many cases?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean by &quot;strong bonds&quot;.  The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane.  Perhaps you're alluding to molecular stability which takes us back to the top, &lt;em&gt;e.g&lt;/em&gt;. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-22T03:01:59.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16574" PostTypeId="2" ParentId="16572" CreationDate="2014-09-22T04:33:56.773" Score="3" Body="&lt;p&gt;As you told oxidation number(oxidation state) is number assigned to atom, ion or compound which represent electron gain or lost by atom, ion or compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some element have more than one oxidation number, because they act differently when they combine with other element. Eg. Oxygen (act differently when combined with peroxide or superoxide.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some element have more than one oxidation number, because they can gain as well as loss electrons.So they have negative as well as positive oxidation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all the elements have more than one oxidation state including Magnesium and potassium. You can see the list of oxidation state of all the elements &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-09-22T04:55:00.683" LastActivityDate="2014-09-22T04:55:00.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16575" PostTypeId="1" AcceptedAnswerId="18765" CreationDate="2014-09-22T07:03:14.443" Score="13" ViewCount="10683" Body="&lt;p&gt;From searching online I found this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The apparatus consists of two sets of bulbs. The first set of three bulbs is filled with solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like $\displaystyle P_2O_5$, conc. $\displaystyle H_2SO_4$ etc. The bulbs of solution and pure solvent are kept in a thermostat maintained at a constant temperature.          &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution, i.e. $\displaystyle p_0 – p_s$. The two sets of bulbs are weighed again. The guard tubes are also weighed.          &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Loss in mass in the solution bulbs $\displaystyle \propto p_s $         &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Loss in mass in the solvent bulbs $\displaystyle \propto (p_0 – p_s) $        &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Total loss in both sets of bulbs $\displaystyle \propto [p_s +(p_0 – p_s)] \propto p_0 $         &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.          &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Thus, $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass in solvent bulbs}}{\text{Total loss in mass in both sets of bulbs}}&#xA;= \frac{\text{Loss in mass in solvents bulbs}}{\text{Gain in mass of guard tubes}}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Further we know from Raoult’s law          &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A} + \frac{w_B}{m_B}}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The above relationship is used for calculation of molecular masses of non-volatile solutes. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For very dilute solutions, the following relationship can be applied. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass of solvent bulbs}}{\text{Gain in mass of guard tubes}}= \frac{w_Am_B}{w_Bm_A}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why &quot;The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution&quot;?&lt;/li&gt;&#xA;&lt;li&gt;How can we add $\propto p_s$ and $\propto p_0-p_s$ when we don't know what the proportionality constants are or even if they are equal?&lt;/li&gt;&#xA;&lt;li&gt;It would be better if you could explain in detail the process because I'm not getting it properly from here.For your information the text after &quot;From Raoult's Law&quot; and about &quot;the setup&quot; is purely clear to me.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-09-28T07:16:39.653" LastActivityDate="2016-10-29T12:23:15.483" Title="How does Ostwald-Walker method work?" Tags="&lt;experimental-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16576" PostTypeId="2" ParentId="16549" CreationDate="2014-09-22T07:35:47.657" Score="0" Body="&lt;p&gt;For the first part I was forgetting this simple this:&#xA;$$[H_{2}O]=\frac{n(H_{2}O)}{Volume_{H_{2}O}} mol.dm^{-3}$$&#xA;So as the $n(H_{2}O)$ will decrease the $Volume_{H_{2}O}$ will decrease too making $[H_{2}O]$ a constant.&lt;br&gt;&#xA;The second part have been answered by ManishEarth. &lt;/p&gt;&#xA;" OwnerUserId="6791" LastActivityDate="2014-09-22T07:35:47.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16577" PostTypeId="2" ParentId="16555" CreationDate="2014-09-22T08:00:45.373" Score="5" Body="&lt;p&gt;If we assume potassium carbonate ($\ce{K2CO3}$) as the only solid reaction product, the &lt;strong&gt;overall&lt;/strong&gt; reaction can be described as:&#xA;$$\ce{2 KNO3 + C12H22O11 + 11 O2 -&amp;gt; N2 + 11 CO2 + 11 H2O + K2CO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when using &lt;em&gt;sucrose&lt;/em&gt; for candy. With &lt;em&gt;sorbitol&lt;/em&gt;, an equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 KNO3 + C6H14O6 + O2 -&amp;gt; N2 + 5 CO2 + 7 H2O + K2CO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seems fitting. In both cases, it is assumed that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the nitrate is completely reduced to nitrogen&lt;/li&gt;&#xA;&lt;li&gt;additional oxygen is needed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The latter is plausible, the rocket candy is supposed to burn layer-wise at the surface in contact with air. This allows for propulsion rather than one instant explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether the nitrite is formed as an intermediate (or which organic intermediates are formed) is beyond my knowledge. However, once this partial reduction (nitrate to nitrite) has happened, I rather doubt that the reaction goes back again (to regenerate the nitrate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the bubbles are concerned, is it conceivable that the fuse wasn't completely dried and thus water was driven out?&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-22T11:53:41.243" LastActivityDate="2014-09-22T11:53:41.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16578" PostTypeId="2" ParentId="16572" CreationDate="2014-09-22T08:11:05.610" Score="5" Body="&lt;p&gt;Because a lot of what you have been taught in school is a lie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, you've probably been taught about the octet rule, a rule which makes it seem as if any given element should have exactly two oxidation states; 'n' and '8-n', where n is a simple function of it's location in the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth, as always, is a little more complicated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the (somewhat) well-behaved non-metals in the upper right corner of the periodic table, most metals have a &lt;em&gt;lot&lt;/em&gt; for electrons. At 26, iron is a lightweight, but already there you have 4 (common!) oxidation states, the two most common ones being iron(II) and iron(III).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason for this is that while iron atoms have a &quot;&lt;em&gt;the&lt;/em&gt; lowest energy state&quot;, it's difficult to get at without first coming across several &quot;&lt;em&gt;a&lt;/em&gt; (locally) lowest energy state&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its four oxidation states represent local minima, and moving between them requires an activation energy, making each of the states relatively stable, but possible to dislodge, i.e. metastable.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="5638" LastEditDate="2014-09-22T08:17:57.290" LastActivityDate="2014-09-22T08:17:57.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16579" PostTypeId="1" AcceptedAnswerId="16588" CreationDate="2014-09-22T09:36:29.813" Score="1" ViewCount="6271" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balance the following redox reaction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{FeS2 + O2 -&amp;gt; Fe2O3 + SO2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Efforts:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I tried balancing with oxidation number method.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;First of all i determined the oxidation state as follows:&#xA;$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -&amp;gt; Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, all the element are either oxidized or reduced, so how to move ahead? &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I tried balancing with Half reaction method.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{(2)} -&amp;gt; Fe2^{(3)}} \text Oxidation$$ &#xA;$$\ce{S2^{(-1)} -&amp;gt; 2S^{(4)} } \text Oxidation$$&lt;br&gt;&#xA;$$\ce{O2^{(0)} -&amp;gt; 2O^{(-2)}} \text Reduction$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same here, how to move ahead?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. Oxidation state are mentioned in the brackets. Correct me if i am wrong anywhere.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="865" LastEditDate="2021-05-22T19:02:23.733" LastActivityDate="2021-05-22T19:02:23.733" Title="Balancing the redox reaction" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16580" PostTypeId="2" ParentId="16563" CreationDate="2014-09-22T09:51:20.183" Score="2" Body="&lt;p&gt;When we dissolve sugar in tea or coffee,the sugar particle get dissociated into smaller particle. As you might be knowing there is inter molecular distance in case of solid,liquid and gas as well. In case of liquid the particles are loosely held compared to solid and also inter molecular distance is more as to solid. So finally particles get packed in those interstitial spaces.so the sugar particles do not get transformed to liquid or gas rather it get surrounded by water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7717" LastEditorUserId="7717" LastEditDate="2014-09-23T14:54:52.710" LastActivityDate="2014-09-23T14:54:52.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16581" PostTypeId="5" CreationDate="2014-09-22T09:55:08.167" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-22T09:55:08.167" LastActivityDate="2014-09-22T09:55:08.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16582" PostTypeId="4" CreationDate="2014-09-22T09:55:08.167" Score="0" Body="Question related to balancing chemical reaction.  " OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-09-22T15:15:03.123" LastActivityDate="2014-09-22T15:15:03.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16583" PostTypeId="1" CreationDate="2014-09-22T11:08:33.373" Score="4" ViewCount="1205" Body="&lt;p&gt;I'm looking for a Chemical database that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is up to date (via a firm, experts or crowdsourcing)&lt;/li&gt;&#xA;&lt;li&gt;is open and free to use (best via an API)&lt;/li&gt;&#xA;&lt;li&gt;contains the CLP/GHS classifications for chemicals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have not found any database that has these three points (e.g. ChemSpider is actually not really open and does not contain CLP/GHS classifications)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was really hoping any of you might know of one? &lt;/p&gt;&#xA;" OwnerUserId="1538" LastEditorUserId="4945" LastEditDate="2017-01-14T07:54:45.257" LastActivityDate="2017-01-14T07:54:45.257" Title="Is there an open Chemical database that includes GHS/CLP classifications?" Tags="&lt;reference-request&gt;&lt;databases&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16584" PostTypeId="2" ParentId="16583" CreationDate="2014-09-22T11:17:34.860" Score="2" Body="&lt;p&gt;Since you're a Dutch neighbour, the &lt;a href=&quot;http://echa.europa.eu/information-on-chemicals&quot; rel=&quot;nofollow&quot;&gt;European Chemicals Agency&lt;/a&gt; might be an option.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-09-22T11:17:34.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16586" PostTypeId="2" ParentId="16379" CreationDate="2014-09-22T13:43:59.957" Score="6" Body="&lt;p&gt;The following not catalytic and does not use hydrogen but tin hydrides:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Convert the alcohol to a xanthate&lt;/li&gt;&#xA;&lt;li&gt;Treat with $\ce{Bu3SnH}$ and AIBN&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Yes, it's the &lt;a href=&quot;http://dx.doi.org/10.1039/P19750001574&quot; rel=&quot;noreferrer&quot;&gt;Barton-McCombie&lt;/a&gt; reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't mind a bit of photochemistry, use the method described by &lt;strong&gt;Isao Saito&lt;/strong&gt; (&lt;a href=&quot;http://dx.doi.org/10.1021/ja00271a057&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Convert the alcohol to a &lt;em&gt;m&lt;/em&gt;-trifluoromethylbenzoate&lt;/li&gt;&#xA;&lt;li&gt;Photolyze (high pressure $\ce{Hg}$ lamp) in THF/water in the presence of &lt;em&gt;N&lt;/em&gt;-methylcarbazole as an electron donor&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-09-22T13:58:35.843" LastActivityDate="2014-09-22T13:58:35.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16587" PostTypeId="2" ParentId="16379" CreationDate="2014-09-22T13:46:48.833" Score="11" Body="&lt;p&gt;Here is a method for converting secondary and tertiary alcohols into the corresponding alkane using $\ce{InCl3}$ as a catalyst.  Chlorodiphenylsilane is the only reagent used in addition to the catalyst. The method works in the presence of common functional groups leaving them untouched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uI9rd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since primary alcohols are unreactive, this reaction can be used to selectively reduce secondary and tertiary alcohols in their presence.  An abstract showing a number of examples of the reaction can be found &lt;a href=&quot;http://www.organic-chemistry.org/abstracts/literature/431.shtm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-22T13:46:48.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16588" PostTypeId="2" ParentId="16579" CreationDate="2014-09-22T13:58:36.563" Score="3" Body="&lt;p&gt;FeS&lt;sub&gt;2&lt;/sub&gt; has an odd structure; the iron atom has a +2 oxidation number and each of the sulfurs has a -1 oxidation number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be balanced by inspection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 FeS&lt;sub&gt;2&lt;/sub&gt; + 11 O&lt;sub&gt;2&lt;/sub&gt; --&gt; 2 Fe&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; + 8 SO&lt;sub&gt;2&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to check, using oxidation numbers we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur goes from -1 to +4.  (Total change from sulfur = +40)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron goes from +2 to +3.    (Total change from iron = +4)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure where you got 4 for the oxidation number of iron in FeS&lt;sub&gt;2&lt;/sub&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen goes from 0 to -2.   (Total change from oxygen = -44)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA:&#xA;Half reactions are messy here, since both iron and sulfur are oxidized.  Similarly, using oxidation numbers is problematic (except to check the solution) because there are three substances changing oxidation states.  You could write a system of equations to describe it, but that's a lot more trouble than it's worth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know from the structure of iron(II) disulfide that there are twice as many sulfur atoms as iron atoms.  That means that the number of sulfur dioxide molecules must be four times the number of iron(III) oxide molecules.  With that relationship in mind, the smallest ratio of molecules that fits the pattern is the one I wrote above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to use oxidation numbers, here's what I've come up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let a equal the number of iron atoms, b equal the number of sulfur atoms, and c equal the number of oxygen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron increases its oxidation state by 1, sulfur by 5, and oxygen decreases by 2. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a + 5b - 2c = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that iron and sulfur are in a 1:2 ratio because they come from pyrite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 a = b &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a + 5(2a) - 2c = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11a - 2c = 0      Iron and oxygen are in a 2:11 ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen must be even since it comes as O&lt;sub&gt;2&lt;/sub&gt;, so its smallest possible value is 22 atoms.  Using the ratios listed above, we get 22 O: 4 Fe: 8 S. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;ETA #2: Half Reactions&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe&lt;sup&gt;+2&lt;/sup&gt; --&gt; Fe&lt;sup&gt;+3&lt;/sup&gt; + e&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S&lt;sup&gt;-1&lt;/sup&gt; --&gt; S&lt;sup&gt;+4&lt;/sup&gt; + 5 e&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;O&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;0&lt;/sup&gt; + 4 e&lt;sup&gt;-&lt;/sup&gt; --&gt; 2 O&lt;sup&gt;-2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from FeS&lt;sub&gt;2&lt;/sub&gt; that there must be twice as many sulfur atoms as iron, so the second equation has to be multiplied by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 S&lt;sup&gt;-1&lt;/sup&gt; --&gt; 2 S&lt;sup&gt;+4&lt;/sup&gt; + 10 e&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add together all of the species that are being oxidized (Fe and S) then we get an oxidation half reaction of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe&lt;sup&gt;+2&lt;/sup&gt; + 2 S&lt;sup&gt;-1&lt;/sup&gt; --&gt; Fe&lt;sup&gt;+3&lt;/sup&gt; + 2 S&lt;sup&gt;+4&lt;/sup&gt; + 11 e&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This must be multiplied by 4 and the reduction equation by 11 to balance the number of electrons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 Fe&lt;sup&gt;+2&lt;/sup&gt; + 8 S&lt;sup&gt;-1&lt;/sup&gt; --&gt; 4 Fe&lt;sup&gt;+3&lt;/sup&gt; + 8 S&lt;sup&gt;+4&lt;/sup&gt; + 44 e&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;11 O&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;0&lt;/sup&gt; + 44 e&lt;sup&gt;-&lt;/sup&gt; --&gt; 22 O&lt;sup&gt;-2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding this mess together gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 Fe&lt;sup&gt;+2&lt;/sup&gt; + 8 S&lt;sup&gt;-1&lt;/sup&gt; + 11 O&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;0&lt;/sup&gt; --&gt; 4 Fe&lt;sup&gt;+3&lt;/sup&gt; + 8 S&lt;sup&gt;+4&lt;/sup&gt; + 22 O&lt;sup&gt;-2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know the structure of the molecules, putting it all back together is fairly straightforward, and it gives the same result as listed above.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-09-22T19:22:48.090" LastActivityDate="2014-09-22T19:22:48.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16589" PostTypeId="1" AcceptedAnswerId="16600" CreationDate="2014-09-22T14:10:17.307" Score="4" ViewCount="1945" Body="&lt;p&gt;I know that in acyl chlorides, there is partial triple bond character in the $\ce{C=O}$ bond &lt;em&gt;not&lt;/em&gt; because the chlorine is electronically donating (due to size mismatch with carbon) but because the chlorine is highly electron withdrawing and this stimulates the oxygen into partially donating one of its lone pairs to the partially positive carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supported by IR stretch data for the $\ce{C=O}$ bond in acid chlorides; we're looking at frequencies from 1810-1775 reciprocal centimeters - somewhat higher than a vanilla $\ce{C=O}$ bond stretch frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in acid fluorides, in which the fluorine is of comparable size to the central carbon, it seems that resonance donation is &lt;em&gt;at least&lt;/em&gt; a possibility. However, resonance must compete with the inductive withdrawal effect. So, which one wins out in acid fluorides, and does the $\ce{C=O}$ bond exhibit more single bond character or triple bond character? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried looking up acid fluorides, and I can't find any mention of them. So I guess a good starting point would be: are they stable enough to even exist? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="15410" LastEditDate="2015-05-20T02:57:23.687" LastActivityDate="2015-05-20T02:57:23.687" Title="Do acyl fluorides exhibit resonance donation or inductive withdrawal?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16590" PostTypeId="1" CreationDate="2014-09-22T14:32:01.420" Score="21" ViewCount="7728" Body="&lt;p&gt;I'm doing a research project where we analyzing the structure of sand after a certain process. To get a better idea of the internal structure of the sand, we use different colors to visualize the movement of the particles. However we can't actually observe that structure since sand is not a solid substance. Therefore we need a way to solidify the sand without changing the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We figured we would need some sort of liquid which we can pour into the sand container (the container is 2x1x0.2 meters) so that it will solidify. In other words, how can we solidify sand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a transparent container is not really an option since we are curious about the structure in the center which we cannot see... unless we use transparent sand ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: we are using 'regular' beach sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;I have read through all the answers and they are great! I can't accept any answers just yet since we need to figure out which method we will use, but it looks promising! We started with 0 and now we have some great ideas going with your help!&lt;/p&gt;&#xA;" OwnerUserId="7961" LastEditorUserId="7448" LastEditDate="2015-09-19T21:34:16.817" LastActivityDate="2015-09-19T21:34:16.817" Title="How to solidify sand" Tags="&lt;inorganic-chemistry&gt;&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16591" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T14:38:36.077" Score="1" Body="&lt;p&gt;Sand particles are held together by covalent bonds.  The only way to get 2 separate sand particles to join is to heat them to a temperature at which covalent bonds form between particles.  This is the process used in making glass.  So, to do what you are proposing requires you to make glass from the sand.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-22T14:38:36.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16592" PostTypeId="1" AcceptedAnswerId="16593" CreationDate="2014-09-22T14:44:57.167" Score="19" ViewCount="2546" Body="&lt;p&gt;Why is proton transfer always kinetically favored? In other words, why are Brønsted acid-base reactions so quick?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because protons are generally unhindered, sterically? This seems plausible; consider an alcohol. The proton isn't blocked by any bulky groups. It's on the end of a oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else could contribute to the kinetic favorability of a proton transfer? Reasoning backwards - given that protons transfers are very fast - this implies that the activation energy of a proton transfer reaction is small. And as outlined above - the activation energy is small in part because most of the time, protons are unhindered. What else might contribute to the small activation energy of a proton transfer/kinetic favorability of a proton transfer? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:26:52.553" LastActivityDate="2018-03-03T11:26:52.553" Title="Why is proton transfer so fast?" Tags="&lt;acid-base&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16593" PostTypeId="2" ParentId="16592" CreationDate="2014-09-22T15:15:25.383" Score="19" Body="&lt;p&gt;Due to their low mass, protons can tunnel through barriers, they don't always have to go over the barrier in an activated process.  Proton tunneling becomes particularly important at low temperatures where the available thermal energy may be inadequate for an activated process (&lt;em&gt;e.g.&lt;/em&gt; over the barrier) to occur.  A nice, brief and not overly-mathematical introduction to tunneling can be found &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_States_of_Atoms_and_Molecules/6._Vibrational_States/Quantum_Mechanical_Tunneling&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dashed line - thermally activated process&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solid line - tunneling process&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjGIu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second reason why proton transfer is so facile relates to hydrogen bonding.  If hydrogen bonding is present, then the proton only needs to elongate one bond (the initial bond) while simultaneously shortening the hydrogen bond in order to effect transfer from one molecule to the next.  This energetically compensating process reduces the activation energy required for proton transfer.  In solvents with hydrogen bonded clusters or chains (water would be one such example), this process can be repeated within the cluster thus allowing the proton to cover large distances (from one side of the cluster to the other) on a very short timescale.  In other words, only minor mass movement is required; by simply shortening some bonds a bit, and slightly elongating others, a proton at the left side of a group of molecules (top row in the following figure) can &quot;reappear&quot; on the right side of the chain (bottom row) extremely rapidly. This process is referred to as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Grotthuss_mechanism&quot; rel=&quot;noreferrer&quot;&gt;Grotthuss mechanism&lt;/a&gt; after its discoverer.  Some of the finer points of the mechanism still constitute an active areas of research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ivK9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-24T23:56:58.977" LastActivityDate="2014-09-24T23:56:58.977" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16594" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T15:16:49.940" Score="17" Body="&lt;p&gt;Based on your comments, it sounds like you have some sort of colored sand and you want to immobilize it. My suggestion would be some sort of polymer matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple possibilities, but one of the easiest is &lt;a href=&quot;http://en.wikipedia.org/wiki/Polydimethylsiloxane&quot;&gt;polydimethylsiloxane PDMS&lt;/a&gt; which is optically clear and pretty easy to work with (i.e., safe). There's the &quot;base&quot; and a curing agent which creates cross linking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the easiest how-to comes from &lt;a href=&quot;http://education.mrsec.wisc.edu/background/PDMS/&quot;&gt;U. Wisconsin&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If done right (i.e., no air bubbles) you should be able to pour in the PDMS, cure, and have a polymerized matrix around the sand. Then you could cut it up or do whatever needed to see different regions.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-22T15:16:49.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16595" PostTypeId="1" AcceptedAnswerId="16605" CreationDate="2014-09-22T15:23:28.050" Score="0" ViewCount="3944" Body="&lt;p&gt;What is the net ionic equation for the reaction between phosphoric acid and sodium hydroxide? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag1 \ce{H+ + H2PO4- + OH- &amp;amp;-&amp;gt; PO4^3- + H2O},\\&#xA;\tag2 \ce{H+ + OH- &amp;amp;-&amp;gt; H2O}.&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it, since phosphoric acid is a triprotic acid, whose ionization is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H3PO4 &amp;lt;=&amp;gt;3H+ + PO4^3-},$$&lt;/span&gt; &#xA;therefore the net ionic equation will be &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{3H+ + 3OH- -&amp;gt; 3H2O}?$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7881" LastEditorUserId="4945" LastEditDate="2018-09-27T12:46:25.667" LastActivityDate="2018-09-27T12:50:09.270" Title="What is the correct net ionic equation for the neutralization of phosphoric acid?" Tags="&lt;acid-base&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-09-22T18:04:39.990" ContentLicense="CC BY-SA 4.0" />
  <row Id="16596" PostTypeId="1" CreationDate="2014-09-22T15:28:39.287" Score="5" ViewCount="6200" Body="&lt;p&gt;Since the pressure of a real gas is less than that of the ideal gas  and its volume is more than that of ideal gas, I am assuming that the real gas is difficult to compress in comparison to an ideal gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my logic right? Also I'm looking for a clearer explanation!&lt;/p&gt;&#xA;" OwnerUserId="4486" LastActivityDate="2014-09-22T17:40:03.237" Title="Which gas is easier to compress, the ideal gas or a real gas?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16597" PostTypeId="1" CreationDate="2014-09-22T15:39:30.657" Score="2" ViewCount="386" Body="&lt;p&gt;I understand that the square of the wave function is the probability density of finding an electron. However, what is the wave function itself? I am trying to understand molecular orbital theory but this concept is getting in the way of me completely understanding it. Any help would be great. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-09-22T17:20:48.950" Title="What is the wave function?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-09-22T18:03:42.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="16598" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T15:41:18.760" Score="12" Body="&lt;p&gt;Geologists prepare soil profiles for a living.  The task is easy in principle - dig a pit, apply glue to the wall, then transfer the glued layer to a sheet of cloth.  The practice is really difficult, the glue recipe and transfer technique are trade secrets.  See here (warning, German language): &lt;a href=&quot;http://www.sand-abc.de/sandkorn/lackprofil/lackprofil.htm&quot;&gt;http://www.sand-abc.de/sandkorn/lackprofil/lackprofil.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not remove layer by layer and then take photos of the cross-sections?  That would be easier than messing about with casting resin and vacuum chambers.  Alternatively, enquire in your friendly Department of Geology.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-09-22T15:41:18.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16599" PostTypeId="2" ParentId="16540" CreationDate="2014-09-22T15:44:00.363" Score="7" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We use tensors in my research lab to analyze solid state NMR. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In my research laboratory we use manifolds all the time for stimulating annealing calculations. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are several research groups in physical chemistry which utilize what you are describing, but it will never be an undergraduate topic. One particular group that comes to mind that uses this sort of stuff a lot is the Hirata group at the University of Illinois. They actually made a piece of software called the Tensor Contraction Engine. &lt;a href=&quot;http://www.scs.illinois.edu/~sohirata/software.html&quot;&gt;http://www.scs.illinois.edu/~sohirata/software.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2014-09-22T15:44:00.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16600" PostTypeId="2" ParentId="16589" CreationDate="2014-09-22T15:50:30.743" Score="3" Body="&lt;p&gt;Yes, acid fluorides exist.  You can purchase acetyl fluoride from Sigma-Aldrich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look back at these two questions for background on how the increased contribution from the triple-bonded carbonyl resonance structure (see below) will lead to an increase in the carbonyl ir stretching frequency (e.g. the triple-bonded carbonyl resonance structure will lead to a stiffening of the carbonyl bond and consequently shift it absorption to higher energy).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16473/ring-strain-and-c-o-stretching-frequency/16476#16476&quot;&gt;Ring Strain and C=O Stretching Frequency&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16000/triple-bond-character-in-acyl-chlorides/16002#16002&quot;&gt;Triple Bond Character in Acyl Chlorides&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are the resonance structures we considered in the case of an acid chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zh4xE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structure IV is what we're calling the &quot;triple-bonded&quot; carbonyl structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbonyl stretch for a acetyl chloride (I, R=Methyl) occurs at 1818 ${cm^{-1}}$.  In the case of acetyl fluoride (replace Cl with F in the above structures) this stretch occurs at 1869 ${cm^{-1}}$, significantly higher.  This suggests that resonance structure IV plays a larger role in the case of acid fluorides, despite resonance structure III presumably also playing a larger role compared to the acid chloride case.  The stability of $\ce{F-}$ vs. $\ce{Cl-}$ seems to be a very significant factor.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-22T15:50:30.743" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16601" PostTypeId="2" ParentId="14376" CreationDate="2014-09-22T16:26:04.313" Score="4" Body="&lt;p&gt;In addition to the examples given by Jannis Andreska, metal peroxides, such as barium peroxide and lithium peroxide also release oxygen upon heating:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2 BaO2 -&amp;gt;[\text{T = 700 °C}] 2 BaO + O2} \\&#xA;\ce{2 Li2O2 -&amp;gt;[\text{T = 195 °C}] 2 Li2O + O2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was once used to produce pure oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, some, if not most, metal oxides are extremely heat resistant. For example, once sintered at temperatures between $1700$ and $\pu{2000°C}$, magnesium oxide $(\ce{MgO})$ can be heated up to its melting point ($\sim\pu{2800°C}$) without decomposition and can thus be used as a lining for heat sensors. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="11367" LastEditDate="2017-04-08T16:34:55.227" LastActivityDate="2017-04-08T16:34:55.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16602" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T17:04:59.840" Score="7" Body="&lt;p&gt;I don't know what kind of resolution you're looking for here.  If you want a true 3d structure you might be able to create one by taking slices off of the sand bed using adhesive paper.  Adding a glue or epoxy to the sand bed will also disturb the sand while holding it in place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do whatever process you've got to do to the sand to get it to its final state, then take a picture of the surface.  Apply an adhesive sheet to the top layer and remove.  Take another picture.  Apply a sheet, remove, picture, repeat.  It's going to take a lot of sheets to empty the bed, but if you want a really fine structure of the sand, it ought to work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be a container with walls and bottom that can move independently.  Image, slide the sides down by 1mm, use a wiper to remove the protruding sand, repeat.  It might prove difficult to avoid disturbing the lower layers as the upper sand is wiped, but the consumable cost is virtually nothing with this method.  (This idea is basically the reverse of powder bed 3d printers, if it's not clear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: A cheaper version of my first idea - Prepare your sand bed.  Image, spray a heavy layer of spray glue on top.  Apply a sheet of printer paper.  Wait a minute, remove.  Repeat.  I think you would pick up more sand with each pass this way and it shouldn't cost too much.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-09-22T17:04:59.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16603" PostTypeId="2" ParentId="13688" CreationDate="2014-09-22T17:08:12.233" Score="3" Body="&lt;p&gt;This is a great question and I'm sorry it wasn't answered sooner. I'll give the caveat that I am not a card-carrying electrochemist, and I'm &lt;em&gt;definitely&lt;/em&gt; not a battery researcher, even though I have worked with them in the past. I'll give my best effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, in principal, the capacity of a battery is tracked with units of mA$\cdot$hr. In theory, this should be constant no matter the current draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean the capacity is uniform across a particular size/shape designation. The &quot;AA&quot; type specifies a &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_battery_sizes&quot; rel=&quot;nofollow&quot;&gt;particular shape/volume dimension&lt;/a&gt;, but different chemistry will give different electrochemical capacities. For example, I have plastic shells that you put a AA battery to fit &quot;C&quot; or &quot;D&quot; types - the downside is less electrochemical capacity (~2500 mA$\cdot$hr for a typical AA vs. ~8000 mA$\cdot$hr for a C cell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One part of the electrochemical capacity is how much of the volume is filled with electro-active materials, rather than electrolyte, fillers designed to dissipate heat, or for other safety margins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, as seen in the showdown, nothing ever matches the &lt;em&gt;theoretical&lt;/em&gt; capacity, and different batteries and different chemistries yield different &lt;em&gt;practical&lt;/em&gt; capacities depending on the current draw.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.batteryshowdown.com/results-lo.html&quot; rel=&quot;nofollow&quot;&gt;Results @ 200 mA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.batteryshowdown.com/results-high.html&quot; rel=&quot;nofollow&quot;&gt;Results @ 1000 mA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We see that the lithium cells get ~3000 mA$\cdot$hr under either current. The alkaline cells get ~2500 mA$\cdot$hr at low current, and &lt;em&gt;much&lt;/em&gt; less under high current ~1500 mA$\cdot$hr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, as we mentioned, not every battery has the same capacity, even of the same battery type (size/shape/volume). Assume for the sake of argument that we have two 3000 mA$\cdot$hr batteries: one lithium (of some sort) and one alkaline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above a certain current threshold, lower currents will let batteries get closer to their theoretical capacities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is actually simple. Besides electrons, in every battery &lt;em&gt;ions&lt;/em&gt; move through an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolyte#Electrochemistry&quot; rel=&quot;nofollow&quot;&gt;electrolyte&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alkaline_battery&quot; rel=&quot;nofollow&quot;&gt;Alkaline cells&lt;/a&gt; use mobile $\ce{OH-}$ ions.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium-ion_battery&quot; rel=&quot;nofollow&quot;&gt;Lithium cells&lt;/a&gt; use mobile $\ce{Li+}$ ions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, discharge requires ionic conductivity to match the electrical current. So which is likely to have a higher ionic conductivity? Well, $\ce{Li+}$ is &lt;em&gt;smaller&lt;/em&gt; and usually has higher ionic conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principal, at low current, an alkaline cell with the same capacity rating as a lithium cell, say 3000 mA$\cdot$hr should have about the same capacity. I'm not sure I've ever seen an alkaline AA battery with such high capacity -- they usually come around 2500-2700 mA$\cdot$hr theoretical capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you ask if manufacturers &quot;tune&quot; the batteries. Put simply, yes. Besides safety considerations, different cells add compounds to minimize self-discharge, corrosion, shelf-life, etc. These can affect the amount of electrochemical capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized I should clarify the very low current comment. If the current is low enough, then the rate of ionic mobility will be high enough for the &lt;em&gt;chemical&lt;/em&gt; part of a battery to keep up with the electrical draw. So decreasing current draw will improve the capacity, up to this point - and then the real capacity should remain the same no matter how low the current draw.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-22T18:35:41.450" LastActivityDate="2014-09-22T18:35:41.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16604" PostTypeId="2" ParentId="16597" CreationDate="2014-09-22T17:20:48.950" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It is often asked &quot;what is waving when a particle is described as a wave?&quot; and the correct response is &quot;existence itself is waving.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The wavefunction is most elegantly described as a description of reality for your system. You then act on the wavefuntion with a hamiltonian to yeild an eigenvalue to quantify the energy. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2014-09-22T17:20:48.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16605" PostTypeId="2" ParentId="16595" CreationDate="2014-09-22T17:23:11.197" Score="2" Body="&lt;p&gt;Phosphoric acid dissociates into hydrogen ions and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4^-}$&lt;/span&gt; ions in water.  This then can dissociate further, but &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4^-}$&lt;/span&gt; are the primary ionic species you'd find in a phosphoric acid solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; dissociates completely into &lt;span class=&quot;math-container&quot;&gt;$\ce{Na^+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH^-}$&lt;/span&gt; ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction would then be between &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH^-}$&lt;/span&gt;, your option 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty arises because adding more &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; to the reaction pulls off the second then the third hydrogen from the phosphoric acid.  In all cases it is an &lt;span class=&quot;math-container&quot;&gt;$\ce{H^+}$&lt;/span&gt; reacting with an &lt;span class=&quot;math-container&quot;&gt;$\ce{OH^-}$&lt;/span&gt; ion with the phosphate ions as spectators, but that really doesn't tell the story of the reaction well.  This is a case where the net ionic equation really isn't your best option.&lt;em&gt;emphasized text&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="4945" LastEditDate="2018-09-27T12:50:09.270" LastActivityDate="2018-09-27T12:50:09.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16606" PostTypeId="2" ParentId="16596" CreationDate="2014-09-22T17:40:03.237" Score="3" Body="&lt;p&gt;&lt;strong&gt;Let's use the van Der Waals equation for this. (Real gasses also described elegantly by the Viral equation)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h2&gt;&lt;img src=&quot;https://i.stack.imgur.com/y1lFM.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;You were false that &amp;quot;pressure of a real gas is less than that of the ideal gas&amp;quot;. It is greater. Also the volume is smaller.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Thus if it has a higher pressure and smaller volume, it typically is harder to compress (i.e. the compression factor increases) but it also depends on temperature. Here is a nice plot of compression factors vs temp for some common gasses from the cambridge chem guide.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/physical/kt/realgases.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemguide.co.uk/physical/kt/realgases.html&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cx6ob.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7957" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-09-22T17:40:03.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16607" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T17:41:00.223" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the container is 2x1x0.2 meters&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess you want something cheap in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;option 1: water. Wet fine sand keeps shape. Not strongly, mind you, but still.&#xA;option 2: drop it, push glass tube into sand (reasonably wide with thin walls)&#xA;option 3: use diluted silicate glue. Would work, but you will have troubles drying it without oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything else is too costly for fun personal project.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-22T17:41:00.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16608" PostTypeId="2" ParentId="16526" CreationDate="2014-09-22T17:48:09.017" Score="3" Body="&lt;p&gt;As the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are equal, that concentration can be regarded as the &quot;concentration&quot; of the solution. If you divide both sides of your first equation (which includes $κ$) with that concentration you would get your second equation (which includes $λ$). So both equations are valid. The problem would arise if the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are different, you couldn't define the concentration of the solution, and you couldn't calculate it's molar conductivity, therefore only the first equation would be usable. Note that the Kolrauch's Law can also be expressed through $κ$.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T08:30:42.197" LastActivityDate="2015-06-20T08:30:42.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16609" PostTypeId="1" CreationDate="2014-09-22T18:24:00.790" Score="2" ViewCount="1675" Body="&lt;p&gt;How the Gibbs free energy of a chemical compound is theoretically calculated from scratch? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When saying that the Gibbs free energy of NaCl is -384.1 kJ/mol, this value is related to the chemical bonds of Na-Cl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should be able to calculate this value from the sum of energies in chemical bonds of one mole of NaCl? Right? How?&lt;/p&gt;&#xA;" OwnerUserId="7730" LastActivityDate="2014-09-23T13:28:53.763" Title="Gibbs free energy of chemical compounds" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16610" PostTypeId="1" CreationDate="2014-09-22T18:54:16.447" Score="0" ViewCount="612" Body="&lt;p&gt;So I have an unsaturated air parcel in the atmosphere that ascends adiabatically from the $1000\:\mathrm{mbar}$ level to the $700\:\mathrm{mbar}$ level with its initial temperature being $10\:°\mathrm{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to find the initial specific volume in $\mathrm{m^3}\:\mathrm{kg^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the specific volume formula is $v=1/\rho$ and that the equation of the state of an ideal gas can be written as $p=\rho RT$ where $R$ is $8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have $1000\:\mathrm{mbar}=\rho(8.3143\:\mathrm{J\:mol^{-1}\:kg^{-1}})(283.15\:\mathrm{K})$ and solve then for $\rho$ and then get $0.42477$ and plug that for $v=1/0.42477$ and get $2.354\:\mathrm{m^3}\:\mathrm{kg^{-1}}$ as the initial specific volume. Am I doing this right? &lt;/p&gt;&#xA;" OwnerUserId="7963" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T10:25:24.327" LastActivityDate="2015-05-20T10:25:24.327" Title="Specific volume of an air parcel" Tags="&lt;homework&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16611" PostTypeId="2" ParentId="7306" CreationDate="2014-09-22T18:57:53.503" Score="4" Body="&lt;p&gt;You are acting under two false assumptions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrode potentials are measured relative to real zero. They are not, there is no practical procedure to measure absolute electron potential. All standard potentials are measured relative to standard hydrogen electrode (well, actually some more pragmatic standard electrode, but recalculated to be against standard hydrogen electrode)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That values got for gas has anything to do with values for condensed matter. Hydration energy can and does shift electrode potentials greatly, as well as energy of crystal lattice. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-22T18:57:53.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16612" PostTypeId="2" ParentId="16590" CreationDate="2014-09-22T20:15:12.087" Score="0" Body="&lt;p&gt;How about hairspray ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While dumping the sand (slowly) using a flat and wide board slowly pour the sand while two others generously spray the hairspray onto the falling sand. This way when the sand lands it's sticky, and you capture almost all the sand particles. They will stick together where they land so you don't disturb the original state of the experiment too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dry, you should have a rock-like blob inside a mold.&lt;/p&gt;&#xA;" OwnerUserId="7964" LastActivityDate="2014-09-22T20:15:12.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16613" PostTypeId="1" AcceptedAnswerId="17281" CreationDate="2014-09-22T20:57:21.377" Score="11" ViewCount="217" Body="&lt;p&gt;I'm working on a project to create a large open repository of quantum calculations, largely for teaching purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get thousands of common organic compounds easily from sources like PubChem or drug databases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for sources for teaching or other uses with VSEPR examples or other types of inorganic and organometallic &lt;strong&gt;&lt;em&gt;molecules&lt;/em&gt;&lt;/strong&gt;. Ideally I'd want 100s or 1000s with good coverage of the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, there are lots of databases of solids. I know most of those, although feel free to suggest some for archival. I really want isolated complex ions, molecules, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there are many, since I haven't turned anything up besides the Cambridge database, which is restrictive about reuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-22T22:15:28.857" LastActivityDate="2017-08-25T23:39:59.227" Title="Where to find common inorganic/organometallic molecules?" Tags="&lt;inorganic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16614" PostTypeId="1" CreationDate="2014-09-22T21:08:22.930" Score="3" ViewCount="577" Body="&lt;p&gt;I'm a sculptor and I have a chemistry/plastic question. I thought here would be a good place to ask.&#xA;I'm looking for a substance (probably a plastic) that has these characteristics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Doesn't melt at the annealing temperature of Copper (doesn't melt at 700 degrees Fahrenheit)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can be dissolved/made non-solid/melted by a process that will not affect copper (a specific acid?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is available in a wire form&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is relatively flexible&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I want to wind long coils of tiny copper wire around a flexible core (the plastic I'm looking for), form that coil into complex shapes, then get rid of the core so as to be left only with the copper wire showing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand these requirements are perhaps impossible to meet...Just wondering if someone with good knowledge of plastics maybe has an idea.&lt;/p&gt;&#xA;" OwnerUserId="7965" LastEditorUserId="4945" LastEditDate="2014-10-23T04:12:14.747" LastActivityDate="2015-01-21T15:56:57.040" Title="Material that doesn't melt at 800 deg F, but can be dissolved?" Tags="&lt;acid-base&gt;&lt;plastics&gt;&lt;melting-point&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16615" PostTypeId="1" CreationDate="2014-09-22T21:25:14.237" Score="7" ViewCount="232" Body="&lt;p&gt;Is it possible to work-harden non-ferrous metals without direct contact?&#xA;I know induction hardening is used on ferrous metals, but would like to know if a technique exists for non-ferrous metals.&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2015-05-31T13:03:26.807" Title="Contactless non-ferrous metals hardening?" Tags="&lt;metallurgy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16616" PostTypeId="2" ParentId="16613" CreationDate="2014-09-22T21:28:24.043" Score="0" Body="&lt;p&gt;This might be exactly what you're looking for and it has a nice API to help do whatever you want with it:&#xA;&lt;a href=&quot;https://www.materialsproject.org/&quot; rel=&quot;nofollow&quot;&gt;https://www.materialsproject.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2008" LastActivityDate="2014-09-22T21:28:24.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16617" PostTypeId="1" CreationDate="2014-09-22T21:53:25.227" Score="1" ViewCount="2991" Body="&lt;p&gt;Can anyone help me with this question I have tried everything.&#xA;I know hydrogen bonding is with $\ce{F}$, $\ce{O}$, or $\ce{N}$, but every time I select those is says it's wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uIk1A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7966" LastEditorUserId="10185" LastEditDate="2015-10-02T17:12:34.540" LastActivityDate="2016-09-05T18:36:30.580" Title="Which atoms in a given amino acid are able to form hydrogen bonds with water?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16619" PostTypeId="2" ParentId="16379" CreationDate="2014-09-22T23:01:41.710" Score="8" Body="&lt;p&gt;This is a very limited situation, but benzylic alcohols can be reduced to the corresponding alkyl group with simple hydrogenolysis conditions, hydrogen gas with Pd/C. This is similar to the cleavage of benzyl ethers, a common protecting group. Looking at it this way, the reaction deprotects water. An example of this reaction that I ran long ago is depicted below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nuki6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Developing a more general method for this transformation is an ongoing field of research, with many methods published. That often means that there isn't one single best method (yet). A 2013 publication from Li and coworkers used an iridium complex with hydrazine as the terminal oxidant. Benzylic, allylic, and notably primary alcohols were successfully reduced, although the benzylic/allylic substrates worked much better. The introduction offers a brief description on the state of the field: &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ejoc.201301293/abstract&quot; rel=&quot;noreferrer&quot;&gt;Iridium-Catalyzed Direct Dehydroxylation of Alcohols&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-09-22T23:01:41.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16620" PostTypeId="2" ParentId="16617" CreationDate="2014-09-23T00:02:29.167" Score="5" Body="&lt;p&gt;The positive end of water can hydrogen bond with N, O, or F, as you said, but only if A) the geometry of the atom allows the water and N/O/F to get close to one another and B) the N/O/F has a lone pair of electrons.  This happens at each of the green circled nitrogen and oxygen atoms.  The red circled nitrogen is a decoy, as it is neither geometrically available nor does it have the appropriate configuration of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative end of water can hydrogen bond with hydrogen atoms that are attached to the N/O/F atoms, including the ones that are attached to the N atom that was not correct for the first part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPs2K.png&quot; alt=&quot;Solution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-09-23T00:02:29.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16621" PostTypeId="2" ParentId="16614" CreationDate="2014-09-23T00:22:56.610" Score="1" Body="&lt;p&gt;Alright, seems like aluminium wire and hydrochloric acid is the way to go!&#xA;Thank you so much, it would have taken me days of googling to figure this one out!&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2014-09-23T00:22:56.610" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16623" PostTypeId="1" AcceptedAnswerId="16626" CreationDate="2014-09-23T00:49:49.873" Score="28" ViewCount="1233" Body="&lt;p&gt;If one wants to calculate a moderate size Alkane (with say 10-15 Carbons , assuming 100 electrons , with Restricted Hartree Fock based methods) we can simply say that electron-electron part will be $100^4 = 100$ million integrals. If we remove the non-unique ones it will become $\frac{100M}{8}= 12.5 M$ Integrals. If we do a pre-screening it may become around 6 million.  Even if we go for DFT it will be around $\frac{100^3}{8} =125,000$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated a small system like Methane with just 10 electrons (around 400 unique/pre-screened integrals) with an efficient C++ code  (based on Rys Integration which is among the most efficient methods) and it took around 1 hour. When 400 cycles takes 1 hour I can assume that 6 million cycles may take years! But software packages do that just in seconds!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if I just want to do a simple math like 2+2 for 6 million times it takes way longer than what these packages do with such complicated integrals!!&#xA;Anyone knows what their trick is? Do they have pre-calculated database or some approximation methods?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&#xA;there are many good hints in this question and thanks to all for that. A combination of symmetry, finding negligible integrals and using optimized methods work fine. The only unanswered part is an algorithm for finding the symmetric situations.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="7448" LastEditDate="2015-04-24T14:37:32.150" LastActivityDate="2015-04-24T14:43:05.913" Title="How do Quantum Software Packages Work?" Tags="&lt;quantum-chemistry&gt;&lt;software&gt;&lt;density-functional-theory&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16624" PostTypeId="1" AcceptedAnswerId="16625" CreationDate="2014-09-23T01:44:08.373" Score="1" ViewCount="27671" Body="&lt;p&gt;I've read online in multiple sites that the least electronegative atom goes in the middle with the exception of H, which always goes on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the molecule NaCN, C is in the middle. Could anyone explain to me what is going on with this molecule?&lt;/p&gt;&#xA;" OwnerUserId="7971" LastActivityDate="2014-09-24T19:09:38.817" Title="Which atom goes in the middle of a lewis dot structure?" Tags="&lt;electrons&gt;&lt;lewis-structure&gt;&lt;electronegativity&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16625" PostTypeId="2" ParentId="16624" CreationDate="2014-09-23T02:10:41.440" Score="6" Body="&lt;p&gt;&lt;b&gt;ETA:&lt;/b&gt; I'm unclear about your question.  Are you asking why nitrogen isn't the central atom or why sodium isn't?  If it's the former, the explanation is below.  If it is the latter, sodium, then it's because this is an ionic compound, and we build the lewis structures of the cation and anion separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First and foremost, carbon (EN=2.5) is less electronegative than nitrogen (EN=3.1), so by your rule it should be in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete explanation of why that is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your molecule is sodium cyanide, an ionic compound.  The cyanide group, CN&lt;sup&gt;-&lt;/sup&gt;, carries a negative charge.  If we consider where that charge lies in the ion, it is a bit confusing.  The basic structure of the cyanide is  ${\ce :C:::N:^-}$  .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance we might think that the nitrogen would be the more negative end of the ion, since it is the more electronegative atom.  However, if we consider the formal charges of the atoms in the ion, we see that nitrogen has a formal charge of 0 and carbon a formal charge of -1.  Formal charge is not the same as actual charge, but it does tell us that the carbon will be the negative end of the ion, and that is what we see in its bonding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the sodium ion, Na&lt;sup&gt;+&lt;/sup&gt;, bonds with the carbon in the cyanide ion.  Even if it weren't a sodium ion, but something like hydrogen cyanide (with covalent bonding) the bond would be formed at the carbon end.  This is because that structure would allow carbon to minimize its electron excess and reach a lower energy state than if the hydrogen stuck to the nitrogen.  Again, formal charges predict the HCN structure over the HNC structure.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-09-24T19:09:38.817" LastActivityDate="2014-09-24T19:09:38.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16626" PostTypeId="2" ParentId="16623" CreationDate="2014-09-23T02:30:03.657" Score="20" Body="&lt;p&gt;There are many tricks. Since I'm not a developer, I'm not privy to all of them, but there are some simple things to start.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Some integrals are predictably close to zero&lt;/em&gt;: Even in fully first-principals calculations, there are parts of the code that prune out integrals that will clearly be negligible. This cuts out a lot of work. Consider, for example, overlap or Coulomb integrals between atoms that are very far apart.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Gaussian Orbitals&lt;/em&gt;: Using &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_orbital&quot;&gt;Gaussian-type orbitals&lt;/a&gt; are incredibly fast, because most quantum integrals can be derived as simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Recurrence_relation&quot;&gt;recurrence relationships&lt;/a&gt;. That is, you almost never need to do the integration. See, for example, &lt;a href=&quot;http://dx.doi.org/10.1063%2F1.450106&quot;&gt;Obara and Saika&lt;/a&gt; and the recent PRISM algorithm from &lt;a href=&quot;http://dx.doi.org/10.1002%2Fqua.560400605&quot;&gt;Gill and Pople&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I believe there are some actual quantum developers on here and perhaps they'd be willing to share other tricks. But these are big - eliminate any integral you don't need, and don't actually take the integrals in the first place. ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there are numeric tricks (using BLAS, LAPACK, etc.) and use of multi-core processing. But these are actually minor optimizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that I forgot to talk about convergence of the self-consistent field which is also optimized. There are various convergence strategies, most notably &lt;a href=&quot;http://en.wikipedia.org/wiki/DIIS&quot;&gt;DIIS&lt;/a&gt;, which helps to extrapolate the final orbital coefficients and electron density.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-09-23T03:14:33.540" LastActivityDate="2014-09-23T03:14:33.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16628" PostTypeId="1" CreationDate="2014-09-23T03:47:20.317" Score="1" ViewCount="71" Body="&lt;p&gt;i made a self-assembled monolayer of dodecanethiol on copper, before that i let the copper in a solution of $\ce{NaOH}$ for 5 min, and i need to know what this step is for, anyone can help me?&lt;/p&gt;&#xA;" OwnerUserId="7973" LastEditorUserId="5017" LastEditDate="2014-09-23T13:36:31.293" LastActivityDate="2014-09-23T14:08:19.827" Title="Letting Copper sit in NaOH for 5 min (for monolayer assembly)" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16630" PostTypeId="1" AcceptedAnswerId="16653" CreationDate="2014-09-23T07:18:09.977" Score="1" ViewCount="75" Body="&lt;p&gt;I'm looking for some guidance in selecting the right material for dance shoes heels and women's self-defense accessories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a material that &lt;strong&gt;1) has a superb strength-to-weight ratio&lt;/strong&gt;, can withstand punishing and frequent impact, and very light (like plastic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part for the shoe will be very, very slender (stiletto heel), but will have to be able to endure hours of jumping, sliding, stomping, spinning, and skidding dancers… is compression strength the right term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The material must also be &lt;strong&gt;safe&lt;/strong&gt; to use as jewelry, so nothing toxic. Also, I am looking for something that tends to &lt;strong&gt;snap, rather than shatter&lt;/strong&gt;, when it fails under pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Lastly, it would be great if this material can be 3D printed, although that is not a must.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inexpensive material would be ideal, of course, but I am open to learning about a fuller range of materials, including the more expensive ones. I am willing to consider a higher priced material if it will offer substantial value to the accessories in terms of strength, quality, and safety.&lt;/p&gt;&#xA;" OwnerUserId="7977" LastEditorUserId="5456" LastEditDate="2014-09-23T08:23:32.993" LastActivityDate="2014-09-23T23:20:18.443" Title="Which material to use for athletic accessories?" Tags="&lt;materials&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16631" PostTypeId="2" ParentId="16590" CreationDate="2014-09-23T09:29:27.023" Score="10" Body="&lt;p&gt;I upvoted the water idea, after all it is possible to build sandcastles over 0.2 m high with just water as a binder. I also think coring as mentioned in that answer is a good idea. I also upvoted the comment about the freezer, although expansion could be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m going to suggest a hot &lt;strong&gt;gelatine solution&lt;/strong&gt; (or even dessert jelly just to try it out.) There are various other food gelling agents (pectin, etc.) If it’s strong enough (and you don’t want it too strong, because you want to cut it open) it would be a good option. Gelatine is cheap, and you ought to be able to recover it by heating your used sand and filtering if cost is really an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m thinking you may have a serious problem with air trapped at the bottom, if you are planning to distribute the sand using air then add the liquid afterwards. Chromatography columns have a porous frit at the bottom, which initially lets the air out (you see the colour darken as you add the first solvent) and then obviously the liquid, too. A wire or plastic mesh over a platform grating would probably work in your case. Pressure is usually applied above a chromatography column to push the solvent through, but in your case you may find it easier to suck from below, if a pressure difference turns out to be required. A bottom connection could also be useful for preheating the bed with steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to think about how to distribute the liquid without disturbing the sand. A matrix of pipes with sprinkler holes, and/or a plastic mesh on top of the sand is probably the best way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1: some more ideas, which while more expensive, will give you a sharp, predictable freezing point. &lt;strong&gt;“Glauber’s salt” (sodium sulphate decahydrate)&lt;/strong&gt; melts at 32 °C, and &lt;strong&gt;sodium thiosulphate pentahydrate&lt;/strong&gt; melts at 48 °C. &lt;strong&gt;Sodium acetate trihydrate&lt;/strong&gt; melts at 58 °C. These compounds can all be supercooled below their melting point, then made to suddenly freeze by adding a single crystal for nucleation (though I expect with sand about they will nucleate immediately.) Videos of this can be found on Youtube. The compounds have been tried for medium term storage of low grade solar heat for space heating, and/or are used in personal heat packs, as all the heat is released at the same temperature (the melting/freezing point) when they cool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sodium carbonate decahydrate “washing soda”&lt;/strong&gt; apparently melts at 32 °C. I’ve never seen any fun experiments with supercooling melted washing soda, so maybe it does not supercool. The situation is confused by the existence of a heptahydrate too. However it is certainly cheap and readily available. It’s strongly alkaline, so would tend to pick up carbon dioxide from the air to become the bicarbonate, but in the volumes required I doubt that would occur to the extent required to produce any change in characteristics. Alkalies are harder on the skin than acids, because they turn your natural oils to soap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought about candle wax, but with the quantities involved &lt;em&gt;extreme&lt;/em&gt; caution would be required to keep it between its melting point and flash point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any contact between molten salts and metals should be checked carefully for corrosion. In particular alkali + $\ce{H2O}$ dissolves the protective oxide layer on aluminium, leading it to corrode extremely rapidly with evolution of hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: even though the OP has now indicated that the height is 1 m not 0.2 m, I still like the gelatine idea, though it’s probably not strong enough to support a metre column. It would probably mean tipping the tank on its side and removing the front (which will have become the top.) It’s a question of balancing the strength of the matrix against the ease of cutting (which will be extremely arduous for some of the higher strength suggestions mentioned here.)&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorDisplayName="user7951" LastEditDate="2015-06-21T08:29:41.563" LastActivityDate="2015-06-21T08:29:41.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16632" PostTypeId="2" ParentId="16579" CreationDate="2014-09-23T10:39:13.073" Score="0" Body="&lt;h2&gt;Balancing With Half Reaction method&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt; Determine oxidation number of each element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -&amp;gt; Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$ &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Step 2:&lt;/strong&gt; Determine total increase and decrease in oxidation number. Also maintain $\ce{Fe}$ and $\ce{S}$ ratio(1:2) on both side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{(2)} -&amp;gt; Fe2^{(3)}} \text Oxidation -----------(A)$$ &#xA;$$\ce{S2^{(-1)} -&amp;gt; 2S^{(4)} } \text Oxidation -----------(A)$$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2^{(0)} -&amp;gt; 2O^{(-2)}} \text Reduction -----------(B)$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Step 3:&lt;/strong&gt; Balance Oxidation number of of reaction (A) and (B)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are total 11 electron transfer take place in (A) and 4 electron in (B)&#xA;$$4\ce{[Fe, S2] -&amp;gt;4[Fe, 2S] }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{11[O2] -&amp;gt; 11[2O]}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Step 5:&lt;/strong&gt; Finally combine above two reactions &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4FeS2 + 11O2 -&amp;gt; 2Fe2O3 + 8SO2}$$&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-23T10:39:13.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16633" PostTypeId="1" AcceptedAnswerId="16636" CreationDate="2014-09-23T12:00:46.487" Score="227" ViewCount="31784" Body="&lt;p&gt;Bulk gold has a very characteristic warm yellow shine to it, whereas almost all other metals have a grey or silvery color. Where does this come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that this property arises from relativistic effects, and I assume that it has to do with some distinct electronic transition energies in the gold atoms. But what changes with the &quot;introduction&quot; of relativistic effects, that then changes the energy of the frontier orbitals in such a drastic manner?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="7475" LastEditDate="2017-12-16T15:14:22.527" LastActivityDate="2020-06-05T14:45:50.180" Title="Why is gold golden?" Tags="&lt;inorganic-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;color&gt;&lt;relativistic&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="87" ContentLicense="CC BY-SA 3.0" />
  <row Id="16634" PostTypeId="2" ParentId="16633" CreationDate="2014-09-23T13:28:50.487" Score="58" Body="&lt;p&gt;This effect comes from the band structure of the metal, not atoms.  In simplest terms (see, &lt;em&gt;e.g.&lt;/em&gt;, Ashcroft &amp;amp; Mermin, Solid State Physics, chapter 1) the electrons in the conduction band have a plasma frequency &lt;span class=&quot;math-container&quot;&gt;$\omega_{p}^{2} = 4\pi n e^{2}/m$&lt;/span&gt; - this is a collective excitation of the metal. For &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; (reddish), you start exciting the plasmons around &lt;span class=&quot;math-container&quot;&gt;$\pu{2eV}$&lt;/span&gt; (red), while for &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; you don't hit it until &lt;span class=&quot;math-container&quot;&gt;$\pu{4eV}$&lt;/span&gt; (near UV).  &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt; is a bit higher than &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; but still in the visible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an excellent mirror across the visible, you need to use &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag},$&lt;/span&gt; not &lt;span class=&quot;math-container&quot;&gt;$\ce{Au-Au}$&lt;/span&gt; will absorb the blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT - additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming from a solid state background, I'll add some more on band theory. Again, following Ashcroft and Mermin's discussion of noble metals (Chapter 15): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the case of copper (and the other noble metals) at least six bands are required (and six turn out to be enough) to accommodate the eleven additional electrons.  ...&#xA;  For almost all waver vectors &lt;strong&gt;k&lt;/strong&gt; the six bands can be seen to separate into five lying in a relatively narrow range of energies from about &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{5eV}$&lt;/span&gt; below E_F, and a sixth, with an energy anywhere from about &lt;span class=&quot;math-container&quot;&gt;$\pu{7eV}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{9eV}$&lt;/span&gt; below E_F. ... It is conventional to refer to the set of five narrow bands as the &lt;em&gt;d&lt;/em&gt;-bands, and the remaining set of levels as the &lt;em&gt;s&lt;/em&gt;-band.  However, these designations must be used cautiously, since at some values of &lt;strong&gt;k&lt;/strong&gt; all size levels are close together and the distinction between &lt;em&gt;d&lt;/em&gt;-band and &lt;em&gt;s&lt;/em&gt;-band levels is not meaningful. The nomenclature reflects the fact that at wave vectors where the levels do clearly group into sets of five and one, the five are derived from the five orbital atomic &lt;em&gt;d&lt;/em&gt;-levels, in the sens of tight binding...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it is tempting to jump on the &lt;em&gt;d&lt;/em&gt;-band desciption and say, hey, since there are relativistic effects in &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt;, than that must carry over into the band structure.  I'm really not convinced.  Yes, while the levels may well derive from &lt;em&gt;d&lt;/em&gt;-levels of the atom, that does not mean that the resulting band structure is or is not different from a non-relativistically impacted band structure. As an example, look at the Fermi surfaces of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag},$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt; as plotted in Ashcroft and Mermin (Fig. 15-5) - they are all free-electron like, except that contact is made along &amp;lt;111&gt; directions.  It is very difficult to differentiate the Fermi surfaces of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt;, while &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; is different (actually more free-electron like). This shows up in the table of Belly-to-Neck areas of the Fermi surface (from De Haas-van Alphen oscillations), where &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}=27,$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}=29,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}=51$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you overlap the orbitals to make bands, all that matters is their relative placement with respect to the Fermi energy. In the big picture, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag},$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt; are all very similar - it is just that our eyes are sensitive to just the small energy region where their plasma frequencies differ, so we see a big difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to ascribe &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt;'s golden color to relativistic effects on the orbitals, than I'm going to have to ask how &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; has such a similar band structure without the relativistic effects.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="61578" LastEditDate="2020-06-05T14:45:50.180" LastActivityDate="2020-06-05T14:45:50.180" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="16635" PostTypeId="2" ParentId="16609" CreationDate="2014-09-23T13:28:53.763" Score="3" Body="&lt;p&gt;The value you have reported is the &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_standard_Gibbs_free_energies_of_formation&quot; rel=&quot;nofollow&quot;&gt;standard Gibbs free energy change of formation&lt;/a&gt;, which has a specialized definition. A standard free energy change of formation is the free energy change that occurs when 1 mole of substance is formed by reaction of its constituent elements in their standard states at 1 atmosphere of pressure and 25 degrees Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for sodium chloride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na(s) +\frac{1}{2} Cl2(g) -&amp;gt; NaCl(s)}\ \ \ \Delta_f G^\circ =-384.1 \text{ kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all we need to be able to do is measure $\Delta G$ for this reaction at standard temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, we will separately detemrine $\Delta_f H^\circ$ and $\Delta_f S^\circ$ and plug them into $\Delta_f G^\circ = \Delta_f H^\circ -T\Delta_f S^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant pressure (which we are by definition operating at),  $\Delta H$ is equal to the heat transfer  $\Delta H = q = -C_p\Delta T$, which is equal to the heat capacity of the surrounding (in practice the calorimeter, which is reasonably isolated from the universe) multiplied by the change in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S$ is still tricky to measure. At equilibrium, for a reversible process at constant temperature $\Delta S = \frac{q}{T} = \frac{\Delta H}{T}$. If the process is not reversible or you are not at equilibrium, then... then it is challenging.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-09-23T13:28:53.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16636" PostTypeId="2" ParentId="16633" CreationDate="2014-09-23T13:29:24.227" Score="178" Body="&lt;p&gt;Yes, this is a beautiful question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, in lower rows of the periodic table, there are relativistic effects for the electrons. That is, for core electrons in gold, the electrons are traveling at a significant fraction of the speed of light (&lt;em&gt;e.g.&lt;/em&gt;, ~58% for &lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1s}$&lt;/span&gt; electrons). This contracts the Bohr radius of the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1s}$&lt;/span&gt; electrons by ~22%. &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_quantum_chemistry&quot; rel=&quot;noreferrer&quot;&gt;Source: Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also contracts the size of other orbitals, including the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6s}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption you see is a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d \rightarrow 6s}$&lt;/span&gt; transition. For the silver &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4d \rightarrow 5s}$&lt;/span&gt; transition, the absorption is in the UV region, but the contraction gives gold a blue absorption (&lt;em&gt;i.e.&lt;/em&gt; less blue is reflected). Our eyes thus see a yellow color reflected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a very readable article by &lt;a href=&quot;https://pubs.acs.org/doi/abs/10.1021/ar50140a002&quot; rel=&quot;noreferrer&quot;&gt;Pekka Pyykkö and Jean Paul Desclaux&lt;/a&gt; that goes into more detail (if you subscribe to ACS &lt;em&gt;Acc. Chem.Res.&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="61578" LastEditDate="2020-06-05T13:35:09.977" LastActivityDate="2020-06-05T13:35:09.977" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="16637" PostTypeId="2" ParentId="16633" CreationDate="2014-09-23T13:56:13.947" Score="124" Body="&lt;p&gt;&lt;strong&gt;See the footnotes I've included if you would like to see more of the detail behind a specific statement.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below compares the reflectance spectrum for silver and gold (let's forget about aluminum, it's not relevant to this discussion; also keep in mind that where reflectance is low, absorbance is high and &lt;em&gt;vice-versa&lt;/em&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pLrd3.png&quot; alt=&quot;Reflectance of gold vs silver&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption (reduced reflectance) seen at lower wavelengths by silver and gold is caused by an electron transition (caused by the absorption of light) between their &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4d \rightarrow 5s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d \rightarrow 6s}$&lt;/span&gt; orbitals respectively. Silver absorbs (reflects less) in the ultraviolet, no visible light is appreciably absorbed. Therefore, if we shine white light on silver, none of the visible wavelengths are absorbed, they are all reflected and silver appears as a shiny, white metallic substance. In fact, jewelers say that silver is the &lt;a href=&quot;http://www.keytometals.com/page.aspx?ID=CheckArticle&amp;amp;site=ktn&amp;amp;NM=239&quot; rel=&quot;noreferrer&quot;&gt;whitest of the metals&lt;/a&gt;. On the other hand, gold absorbs (reflects less) blue light in the visible region.  When white light shines on gold, the blue light is absorbed and yellow light is reflected (see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Complementary_colors&quot; rel=&quot;noreferrer&quot;&gt;article on complementary colors&lt;/a&gt; for more detail on how removing certain colors from white light produces different colors).  So &lt;strong&gt;our question becomes, why is gold's &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d \rightarrow 6s}$&lt;/span&gt; electronic transition shifted into the visible region?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-Electrons are strongly attracted to the nucleus because their probability density is higher near the nucleus than other types of orbitals (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p, d, f,}$&lt;/span&gt; etc.). Simple Bohr theory tells us that these strongly attracted &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; electrons in atoms with a large number of protons (heavy nuclei) would have to travel at a considerable fraction of the speed of light to avoid &quot;falling into&quot; the nucleus (1). Special relativity takes things a step further and explains that, for these heavy nuclei with their &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; electrons traveling over half the speed of light, the electron mass will increase and as mass increases (2), orbital radius decreases (3).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, as the orbital radius decreases and the electron is closer to the nucleus it becomes more stabilized and the energy of the orbital decreases. As the various &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; electrons (and to a lesser degree &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; electrons, but &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{f}$&lt;/span&gt; electrons don't have enough electron density close to the nucleus so they are not directly affected by these relativistic effects and their orbits are not contracted) move closer to the nucleus, the other non-&lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;) electrons are better screened from the nucleus. So although the various &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbitals are contracted and placed at lower energy, the various &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{f}$&lt;/span&gt; orbital radii shift further out and occur at higher energy due to this increased screening effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, &lt;strong&gt;in the gold atom&lt;/strong&gt;, we would expect &lt;strong&gt;the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d}$&lt;/span&gt; orbital to be raised in energy&lt;/strong&gt; (screening effect) and &lt;strong&gt;the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6s}$&lt;/span&gt; orbital to be lowered in energy&lt;/strong&gt; (relativistic orbital contraction effect). The &lt;strong&gt;net effect is a smaller energy separation&lt;/strong&gt; between these two orbitals and a &lt;strong&gt;red-shift in the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{5d \rightarrow 6s}$&lt;/span&gt; absorption to the lower energy&lt;/strong&gt; (compared to the absorption of silver which is not subject to the same magnitude of relativistic effects) blue region of the spectrum, &lt;strong&gt;resulting in a yellowish color for light reflected from gold&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(1) The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_\mathrm{rel} = \frac{Z}{137}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; is the atomic number of the element under consideration and &lt;span class=&quot;math-container&quot;&gt;$137$&lt;/span&gt; is &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_units#Related_physical_constants&quot; rel=&quot;noreferrer&quot;&gt;the speed of light in atomic units&lt;/a&gt;, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant&quot; rel=&quot;noreferrer&quot;&gt;the fine structure constant&lt;/a&gt;.  Consequently, a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1s}$&lt;/span&gt; electron in silver &lt;span class=&quot;math-container&quot;&gt;$(Z=47)$&lt;/span&gt; will travel around &lt;span class=&quot;math-container&quot;&gt;$\frac{47}{137} = 34\%$&lt;/span&gt; the speed of light, while a similar electron in gold &lt;span class=&quot;math-container&quot;&gt;$(Z=79)$&lt;/span&gt; will travel at &lt;span class=&quot;math-container&quot;&gt;$58\%$&lt;/span&gt; the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The relativistic mass of an electron is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m_\mathrm{rel} = \frac{m_\mathrm{e}}{\sqrt{1-\left(\frac{v_\mathrm{rel}}{c}\right)^2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$m_\mathrm{e}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_\mathrm{rel}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; are the electron rest mass, the velocity of the electron, and the vacuum speed of light, respectively. The following figure provides a graphical representation of the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zfGNz.gif&quot; alt=&quot;mass ratio vs speed of light&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The following equation relates the ratio of the relativistic radius of the first Bohr orbit &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{rel}$&lt;/span&gt; to the normal radius &lt;span class=&quot;math-container&quot;&gt;$R_\circ$&lt;/span&gt;, to the relativistic velocity of the electron:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{R_\mathrm{rel}}{R_\circ} = \sqrt{1-\left(\frac{v_\mathrm{rel}}{c}\right)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~18%.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2020-06-05T14:21:24.177" LastActivityDate="2020-06-05T14:21:24.177" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16639" PostTypeId="2" ParentId="16628" CreationDate="2014-09-23T14:08:19.827" Score="1" Body="&lt;p&gt;Such steps are usually cleaning steps to remove deposited organic materials. $\ce{NaOH}$ is also known as lye and will dissolve basically all organics. In this case, it should leave the $\ce{Cu}$ and copper oxides alone, leaving a clean surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have as much experience with copper monolayers, but similar cleaning steps are used in my lab, where you might treat with ozone or some other step to remove any deposited organic &quot;gunk&quot; from the metal surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I believe you &lt;em&gt;don't&lt;/em&gt; want to use an acid, because the thiols react with the copper oxide surface, not strictly the copper itself.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-23T14:08:19.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16640" PostTypeId="1" AcceptedAnswerId="16643" CreationDate="2014-09-23T14:16:59.307" Score="102" ViewCount="79124" Body="&lt;p&gt;At the Renaissance fair a few years back I was watching a smith forge metal into shapes. During this time a very odd question came to me. I was wondering what the furnace was made of. My logic stated that whatever the furnace was made of must have a higher melting point than the materials he was melting. This quickly turned into an elemental arms race resulting in an odd question of how do we melt stuff like refractory metals (more specifically the one with the highest melting point) so we can melt other things inside of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that (for some odd reason I don't understand) rapid cooling can manipulate the strength of an item. Is there a similar property to manipulate the melting point? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: My current best guess (like can be done to make weapons harder) is that we take two elements, melt them, and the resulting compound has a higher melting point.&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="7149" LastEditDate="2014-09-23T20:15:47.477" LastActivityDate="2021-04-19T17:04:11.307" Title="How do you melt metals with super high melting points?" Tags="&lt;melting-point&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="16643" PostTypeId="2" ParentId="16640" CreationDate="2014-09-23T17:32:53.733" Score="86" Body="&lt;p&gt;Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon (3550 °C) among the elements. This is why tungsten is used in rocket nozzles and reactor linings.  There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum carbide at 3800 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon cannot be used to hold molten tungsten because they will react to form tungsten carbide.  Sometimes ladles and crucibles used to prepare or transport high melting point materials like tungsten are lined with the various higher melting ceramics or alloys.  More typically tungsten and other refractory materials are fabricated in a non-molten state.  A process known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Powder_metallurgy&quot; rel=&quot;noreferrer&quot;&gt;powder metallurgy&lt;/a&gt; is used.  This process uses 4 basic steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;powder manufacture - a variety of techniques are available to generate small particles of the material being worked&lt;/li&gt;&#xA;&lt;li&gt;powder blending - routine procedures are used to blend the constituent particles into a uniform mixture&lt;/li&gt;&#xA;&lt;li&gt;compacting - the blended powder is placed in a mold and subjected to high pressure&lt;/li&gt;&#xA;&lt;li&gt;sintering - the compacted material is subjected to high temperature and some level of bonding occurs between particles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2017-03-04T22:17:18.960" LastActivityDate="2017-03-04T22:17:18.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16645" PostTypeId="2" ParentId="16640" CreationDate="2014-09-23T19:00:31.127" Score="50" Body="&lt;p&gt;Sorry, can't comment here, but I wanted to more directly answer your question.&lt;/p&gt;&#xA;&lt;p&gt;Blacksmiths avoid melting their forges because the &amp;quot;heat&amp;quot; that can melt or oxidize iron and steel is actually contained in a ball in the center of the coal. In fact, maintaining coal &amp;quot;structure&amp;quot; is an important skill in blacksmithing.&lt;/p&gt;&#xA;&lt;p&gt;To clarify better, imagine a hollow in the center of a pile of coal. This is where the temperatures rise past 2000 °F, since the heat reflects back in on itself due to the coal molding into a refractory ball of sorts.&lt;/p&gt;&#xA;&lt;p&gt;And yes, sometimes your ball does fall apart, or you've structured it poorly - and then you notice that the cast iron drain cover that protects your airflow intake has melted through.&lt;/p&gt;&#xA;" OwnerUserId="7989" LastEditorUserId="41328" LastEditDate="2021-04-19T17:03:33.690" LastActivityDate="2021-04-19T17:03:33.690" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16647" PostTypeId="2" ParentId="10988" CreationDate="2014-09-23T20:23:59.380" Score="3" Body="&lt;p&gt;What is the range of the components? This is very important because this will determine if titration or photometry is used (or a combination of both). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also interfering components should be known. For eaxample. Barium is known to react with a lot of other components (&lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table&quot; rel=&quot;nofollow&quot;&gt;solubility table&lt;/a&gt;). Also EDTA reacts with oter components (Also depending on the pH (&lt;a href=&quot;http://www2.chemistry.msu.edu/courses/cem262/EDTATable.html&quot; rel=&quot;nofollow&quot;&gt;EDTA complexation constants&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Titration:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper and sulfate can be determined in one analysis using and copper ISE. This is done by adding an NH3 buffer (pH of solution = 10) and adding an exxess of Barium nitrate. Barium will react with SO4 and the free barium is titrated. Titrate the solution with EDTA. This first inflection point will be copper, the second one the free barium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that EDTA reacts with a lot of other components. Also if the concentration difference (molar) between copper and barium is large, high titrant usages are to be expected&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photometry:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper can be determined using an variaty of chemical substances (e.g. cuprizone, Bathocuproine). The choise depends on range, interfering components ect..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfate is possible. However, titration is preferred.&lt;/p&gt;&#xA;" OwnerUserId="7993" LastEditorUserId="7993" LastEditDate="2014-09-25T09:49:35.017" LastActivityDate="2014-09-25T09:49:35.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16648" PostTypeId="2" ParentId="10777" CreationDate="2014-09-23T20:28:52.163" Score="1" Body="&lt;p&gt;Chemicals are probably not the good way to go. What is the temperature of the solution? Increasing the temperature speeds up the reaction. Btw: take the proper safety precautions when etching pcb's!&lt;/p&gt;&#xA;" OwnerUserId="7993" LastActivityDate="2014-09-23T20:28:52.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16649" PostTypeId="1" AcceptedAnswerId="16674" CreationDate="2014-09-23T21:04:04.687" Score="6" ViewCount="2424" Body="&lt;p&gt;I was always wondering why iron only rusts when it gets wet. I know that for example aluminium oxidises on its own on regular basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also heard many complains that salt used to remove snow from roads increases corrosion rate of car undercarriage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, I was researching how to produce iron oxide to improve rocket candy fuel - and among with oxidising substances, especially hydrogen peroxide, salt water was suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do these things have to do with aerial oxygen reacting with iron? I have also noticed that copper roofs turn green. Copper stuff kept away from rain doesn't do that so fast.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="1362" LastEditDate="2014-11-10T06:40:45.350" LastActivityDate="2014-11-10T06:40:45.350" Title="How does salt and water promote corrosion?" Tags="&lt;redox&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16651" PostTypeId="1" CreationDate="2014-09-23T22:29:49.103" Score="10" ViewCount="2745" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;TL:DR conditions below&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Highschool to first year undergraduate level&lt;/li&gt;&#xA;&lt;li&gt;Theory based(Experimentation not wanted)&lt;/li&gt;&#xA;&lt;li&gt;General - Giving a good introduction to all or most areas of Chemistry&lt;/li&gt;&#xA;&lt;li&gt;Rigorous - Professional formatting - Not a casual textbook&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am looking for a textbook for someone(me) who has done no chemistry, but isn't too basic as it progresses. What are some of the better textbooks for starting to learn chemistry? I don't know any specific fields, so I would like to get some exposure to a number of different topics. I am a third year Mathematics student, so I am not worried about it being too concise, I am willing to reread areas one hundred times if need be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One note: I am not interested in doing experimentation at all, and would just like to learn theory, as I have (sometimes) embarrassingly bad knowledge of the field as I didn't take chemistry in highschool(which I greatly regret).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highschool through to the end of first year undergraduate level preferably, or alternatively multiple textbooks by the same author of successive difficulty.&lt;/p&gt;&#xA;" OwnerUserId="7996" LastEditorUserId="7996" LastEditDate="2014-09-24T00:49:40.450" LastActivityDate="2018-05-24T21:51:23.183" Title="Introduction to Chemistry Textbook - Theory based" Tags="&lt;reference-request&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16652" PostTypeId="1" AcceptedAnswerId="16708" CreationDate="2014-09-23T22:50:00.743" Score="4" ViewCount="949" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Determine the packing geometry of the following compounds by using the radius ratio rules. For your convenience assume a coordination number of six for both cation and anion.&lt;/p&gt;&#xA;&lt;p&gt;Be specific in the your answer, &amp;quot;body-centered cubic with the anion occupying the corners of the cell [1 atom] and the cation occupying the central hole [1 atom].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Below is the table we are given, and we're are choosing between geometries of cubic, body-centered, or face-centered.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tSZfw.png&quot; alt=&quot;the table&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The compound I'm on is &lt;span class=&quot;math-container&quot;&gt;$\ce{LiBr}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I did &lt;span class=&quot;math-container&quot;&gt;$\frac{r^+}{r^-}=\frac{79}{182}=0.43$&lt;/span&gt; for the Radius-Ratio value.&lt;/p&gt;&#xA;&lt;p&gt;But from here would I choose square planar because it's closer to 0.414 or do I choose octahedral because we're assuming coordination number 6? And how would I determine the locations of the cation/anion?&lt;/p&gt;&#xA;&lt;p&gt;My best guess is cubic geometry because the cation is too large to fit tetrahedral holes and octahedral holes, with &lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; taking occupying complementary corners.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I'm making the assumption that the coordination number 6 is used as a reference to the atomic radii.&lt;/p&gt;&#xA;&lt;p&gt;I've currently adjusted my answer based off of other examples I could find to&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;FCC with OH holes occupied by &lt;span class=&quot;math-container&quot;&gt;$\ce{Li^+}$&lt;/span&gt; [4 atoms] and the &lt;span class=&quot;math-container&quot;&gt;$\ce{Br^-}$&lt;/span&gt; forms the FCC Lattice [4 atoms (by stoichiometry}] for a 100% occupancy rate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;But I still have an issue as to how to conceptually approach the problem.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-07-24T01:49:10.730" Title="How to determine the packing geometry of compounds with contradicting coordination number and radius-ratio values?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16653" PostTypeId="2" ParentId="16630" CreationDate="2014-09-23T23:20:18.443" Score="1" Body="&lt;p&gt;For plastics, my first instinct for the first two properties would be a polycarbonate plastic. Polycarbonates are known to be very strong plastics (tensile strength up to ~70 MPa) and deform without breaking under many conditions, which is why they're commonly used in things like safety glasses and bullet-resistant glass. It also looks very interesting as it's quite optically clear. PC is 3D printable, I've seen a few people using it in hobby printers (&lt;a href=&quot;http://www.protoparadigm.com/blog/2011/12/printing-polycarbonate/&quot; rel=&quot;nofollow&quot;&gt;http://www.protoparadigm.com/blog/2011/12/printing-polycarbonate/&lt;/a&gt;) and in very high end StrataSys printers, but it's not a super common material so it might take some experimenting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is a nylon. There are a number of different types available and many have similar properties to polycarbonates (minus the optical clarity). Nylons seem to be a bit more popular for 3D printing and I've heard good things about the &lt;a href=&quot;http://taulman3d.com&quot; rel=&quot;nofollow&quot;&gt;Taulman&lt;/a&gt; filaments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you're not limited to a single material and there are many things that could potentially be used in this kind of application. Often, the combination of multiple materials makes sense (e.g. a metal insert in a plastic part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like carbon fibre-reinforced epoxy construction might be appropriate. It's not 3D printable, but can be constructed in layers around a lightweight (expanded polystyrene or the like) form. This approach is often used for parts that need to be very light while having high tensile strength (think bicycle frames, ski poles, etc.) I could see a carbon fibre-wrapped plastic insert making a very strong and light shoe heel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, these things are balancing games between strength, mass, fabrication cost, aesthetics, etc. but these are a few potential materials that come to mind.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-23T23:20:18.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16655" PostTypeId="1" AcceptedAnswerId="16660" CreationDate="2014-09-24T00:41:44.327" Score="2" ViewCount="47" Body="&lt;p&gt;If an unsaturated air parcel ascends adiabatically in the atmosphere with an initial temperature, would the final temperature of the air parcel, when it reaches it's destination, be the same at the initial temperature since it does not exchange heat with it's surroundings?&lt;/p&gt;&#xA;" OwnerUserId="7963" LastEditorUserId="189" LastEditDate="2014-09-24T03:11:17.330" LastActivityDate="2014-09-24T17:46:55.420" Title="Dry adiabatic lifting" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16656" PostTypeId="1" AcceptedAnswerId="16657" CreationDate="2014-09-24T01:07:14.087" Score="4" ViewCount="2291" Body="&lt;p&gt;Whys is the standard electrode potential (-) for electrolytic cells but positive for (+)  galvanic cells.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If E°cell &gt; 0, then the process is spontaneous (galvanic cell)&lt;/li&gt;&#xA;  &lt;li&gt;If E°cell &amp;lt; 0, then the process is non-spontaneous (electrolytic cell)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Thus in order to have a spontaneous reaction, E°cell must be positive,&#xA;  where:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;E°cell = E°cathode − E°anode&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;^ above taken form wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also can someone please explain the follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cells is the electron flow from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;anode -&gt; cathode ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electrolytic cells isn't the electron flow from the battery (-) anode to cathode (-) of the cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we then show this by electron flow as cathode (-) ------&gt; anode (+) [in the actual cell itself]??&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="4945" LastEditDate="2014-09-28T07:14:07.697" LastActivityDate="2014-09-28T07:14:07.697" Title="Standard electrode potential for electrolytic vs galvanic cells" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16657" PostTypeId="2" ParentId="16656" CreationDate="2014-09-24T01:27:47.203" Score="3" Body="&lt;p&gt;Potential refers to the amount of energy produced or consumed by the reaction. If the sign for potential is (+) than the reaction is spontaneous (moves on its own). If the sign is (-) then the reaction requires energy to proceed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of electrolytic cells, the cell is used to decompose a compound. this can require energy depending on the reaction. Think of water and how much energy it takes to produce hydrogen gas and oxygen, that's an example of electrolytic cell doing work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of an anode is an electrode in which a positive charge flows into an electrical device, this is the direction of the electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both types of cells direction of electron flow would remain the same, but with electrolytic cells, there would be an additional object, the electrolyte, in the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a decent picture i found that also does some explanation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxUFS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-24T01:27:47.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16658" PostTypeId="1" AcceptedAnswerId="16659" CreationDate="2014-09-24T01:44:24.623" Score="3" ViewCount="1343" Body="&lt;p&gt;How is the &quot;correct&quot; answer &lt;em&gt;not&lt;/em&gt; a redox reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gn712.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the &quot;correct&quot; choice a textbook example of an organic redox reaction? After all, we are breaking a C-C bond and replacing that with a C-O bond. If that's not redox I don't know what is redox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I missing something fundamental here? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-24T02:06:52.300" Title="Redox reactions - alkene to alcohol" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16659" PostTypeId="2" ParentId="16658" CreationDate="2014-09-24T02:06:52.300" Score="5" Body="&lt;p&gt;The reaction in question involves the addition of water across a double bond. The proton from the water is added to one carbon in the double bond - this carbon is reduced.  The hydroxyl group is added to the other end of the double bond - this carbon is oxidized. Even though the oxidation number of one olefinic carbon (the one that winds up with 2 hydrogens attached) changes from-1 to -2, and the other from -1 to 0, &lt;strong&gt;overall&lt;/strong&gt;, for the entire molecule, there is no &lt;strong&gt;net&lt;/strong&gt; oxidation or reduction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-24T02:06:52.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16660" PostTypeId="2" ParentId="16655" CreationDate="2014-09-24T03:11:05.750" Score="4" Body="&lt;p&gt;No. When the gas - let's assume it is an ideal gas - rises up it will expand because the pressure $p$ sinks the higher it comes. If it is allowed to expand quasi-statically under adiabatic conditions ($\delta Q = 0$) then it does work $W$ (pressure-volume work $W = - p \mathrm{d}V$ to be more precise) on its environment, and, hence, its internal energy $U$ is reduced, and its temperature $T$ changes because according to the first law of thermodynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d} U &amp;amp;= \delta Q + \delta W \\&#xA;  &amp;amp;\overset{\delta Q = 0}{=} \mathrm{d} W = - p \mathrm{d}V &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also for an ideal gas you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d} U &amp;amp;= C_{V} \mathrm{d}T &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_{V}$ is the heat capacity. Hence the temperature changes according to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d}T = - \frac{p}{C_{V}} \mathrm{d}V \ .&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-24T17:46:55.420" LastActivityDate="2014-09-24T17:46:55.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16661" PostTypeId="2" ParentId="16651" CreationDate="2014-09-24T04:44:58.623" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Chemistry by Zumdahl is very good. I teach General Chemistry I and this is the book we use. I still reference it frequently. You will want the instructor solutions to accompany it. I let students check this out for classroom use, but you should have it for yourself at least.&lt;/strong&gt; (I saw for the first time this semester a version that is loose leaf; this is great since it is sort of big. We have a custom version that takes out parts not covered. There is also a nice ebook.) This book is lovely, except for electron affinity. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;For introductory chemistry, Zumdahl and DeCoste wrote Introductory Chemistry: A Foundation. We use this for Chem 101.&lt;/strong&gt; Also very good, but a little too elementary for my taste and there is not as many topics discussed. But this may be what you want. You are probably somewhere between these two. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The next two are probably more rigorous than you would like, but worth mentioning. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Principles of Instrumental Analysis by Douglas A. Skoog is outstanding, but probabbly more advanced than you would like.&lt;/li&gt;&#xA;&lt;li&gt;Physical Chemistry - Atkins. This book is very dense and is definitly rigorous. This will take an undergrad 2-3 semesters to get through. This may be a little overkill. You will actually see many general chemistry textbooks take exercises from this book. Really need the solutions bad for this book, it is that hard. A great place to find exam questions. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2014-09-24T04:44:58.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16664" PostTypeId="1" AcceptedAnswerId="16668" CreationDate="2014-09-24T07:51:57.533" Score="8" ViewCount="11899" Body="&lt;p&gt;To have a fire, you need 1) oxygen, 2) fuel and 3) heat. You can have different fuels that burn at different temperatures, but you always need some amount of heat to have a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can use any fuel, what is the lowest temperature at which you can have a fire with an open flame? Is it possible to have a fire that's so cold that you can hold your hand in it without burning yourself?&lt;/p&gt;&#xA;" OwnerUserId="7809" LastEditorUserId="7809" LastEditDate="2014-09-24T09:42:08.327" LastActivityDate="2019-08-14T21:58:04.657" Title="What is the lower bound to the temperature at which a fire can burn?" Tags="&lt;thermodynamics&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16665" PostTypeId="1" AcceptedAnswerId="16666" CreationDate="2014-09-24T09:22:07.930" Score="0" ViewCount="1122" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many atoms of carbon are needed to produce $\pu{0.45 mol} ~\ce{Al}$?&#xA;  $$\ce{3C + 2Al2O3 -&amp;gt; 4Al + 3CO2}$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure where to start, whether or not I convert moles straight to atoms or convert moles to grams of $\ce{Al}$, then to moles of $\ce{C}$ and then to atoms? I tried the latter of my suggestions and ended up with $\pu{9.8E24}$, but  I don't know if this is correct.&lt;/p&gt;&#xA;" OwnerUserId="7621" LastEditorUserId="5026" LastEditDate="2018-02-27T08:16:16.650" LastActivityDate="2018-02-27T08:16:16.650" Title="How many atoms of carbon are needed to produce 0.45 mol aluminium?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16666" PostTypeId="2" ParentId="16665" CreationDate="2014-09-24T09:30:26.410" Score="1" Body="&lt;p&gt;The key to this is by examining the mole ratio. Your balanced equation shows that 4 mols aluminium: 3 mols carbon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with ratios from maths, you can therefore find the specific number of moles of carbon in this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4\ce{Al}: 3\ce{C}\implies0.45\ce{Al} :~?$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you should be able to get the answer as 0.3375 moles of carbon (again, it's just ratios)! From here, you use moles = mass / mr to get your answer. &lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="5026" LastEditDate="2018-02-27T08:13:54.167" LastActivityDate="2018-02-27T08:13:54.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16667" PostTypeId="2" ParentId="16664" CreationDate="2014-09-24T09:38:30.627" Score="2" Body="&lt;p&gt;It depends on your definition of fire.  You can have hand warmers based on iron filings, the iron oxidises slowly and gives off heat as it does.  So you have a redox reaction and heat is given off, but its not enough to burn a human hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hometrainingtools.com/a/homemade-hand-warmers&quot; rel=&quot;nofollow&quot;&gt;http://www.hometrainingtools.com/a/homemade-hand-warmers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a &quot;fire&quot; has to involve actual flames or smoke, then of course that's different.&lt;/p&gt;&#xA;" OwnerUserId="5410" LastActivityDate="2014-09-24T09:38:30.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16668" PostTypeId="2" ParentId="16664" CreationDate="2014-09-24T11:21:15.147" Score="14" Body="&lt;p&gt;Fires which burn below about 400°C are known as cool flames, and this is a well known and widely researched phenomenon. Most hydrocarbons and many alcohols can produce cool flames, and the conditions under which they burn depends on the oxygen content available. The most common encounter that most people have with the cool flame effect is through incomplete combustion of a petrol engine, leading to engine knocking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest recorded cool flame temperatures are between 200 and 300°C; the &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Cool_flame&quot;&gt;Wikipedia&lt;/a&gt; page references n-butyl acetate as 225°C. You can read a lot more about cool flames on that page. One thing to note is that cool flames are very hard to visibly detect at lower temperatures - both heat and light being two of the by products of the combustion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasa.gov/mission_pages/station/research/news/cool_flame.html#.VCKopyNdaSN&quot;&gt;NASA&lt;/a&gt; have conducted a number of experiments in space on low temperature combustion of nano droplets of heptane, and this is an interesting area of science still being investigated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting point to make relates to your question of whether you could hold your hand in the flame and not burn yourself. Clearly the answer is no, but for some cool flames it may take noticeably longer to feel the pain of heat. Paper starts to discolour at about 150°C, but may not burn as ignition temperature is between 220-250°C. &lt;a href=&quot;http://www.tcforensic.com.au/docs/article10.html&quot;&gt;For reference&lt;/a&gt;, candle flames are about 600-1400°C. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="189" LastEditDate="2014-09-24T18:11:34.783" LastActivityDate="2014-09-24T18:11:34.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16670" PostTypeId="1" CreationDate="2014-09-24T12:47:58.270" Score="2" ViewCount="71" Body="&lt;p&gt;I've never understood how to systematically write out IUPAC names. For example, how do I write out 6-aminohexanoic acid ? I tried searching for IUPAC related terms, but I didn't get the gist of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me please? &lt;/p&gt;&#xA;" OwnerUserId="8014" LastEditorUserId="4945" LastEditDate="2014-09-28T07:23:33.437" LastActivityDate="2014-09-28T07:23:33.437" Title="Something I've never understood in chemistry" Tags="&lt;nomenclature&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ClosedDate="2014-09-24T14:30:41.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="16671" PostTypeId="1" CreationDate="2014-09-24T12:50:02.540" Score="28" ViewCount="2361" Body="&lt;p&gt;A single molecule can't be solid, liquid or gas. It's just a molecule. A mole of something can be any of the three. So, how many molecules does it take for phases to be meaningful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that this may depend on the molecules in question. The minimum number that can form a solid might depend on the configuration that molecules adopt when solid. The smallest crystal of a lattice for example. How about a gas though? Can two molecules be called a gas if they exist under conditions where a larger quantity of the same moiety would be in a gaseous state?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, can a minimum number of molecules necessary to adopt a particular phase be defined?&lt;/p&gt;&#xA;" OwnerUserId="4710" LastEditorUserId="4710" LastEditDate="2014-09-24T21:26:52.047" LastActivityDate="2021-01-22T14:17:51.590" Title="How many molecules does it take to have a phase?" Tags="&lt;theoretical-chemistry&gt;&lt;phase&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16672" PostTypeId="1" AcceptedAnswerId="16706" CreationDate="2014-09-24T13:18:27.947" Score="2" ViewCount="49282" Body="&lt;p&gt;I don't get how does this reaction of hydrochloric acid and potassium permanganate proceeds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + KMnO4 -&amp;gt; KCl + MnCl2 + Cl2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{HCl}$ dissociates into $\ce{H+}$ and $\ce{Cl-}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^{2-}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I don't get: How $\ce{MnCl2}$ and $\ce{Cl2}$ can appear there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone, please, guide me through this reaction, step by step?&lt;/p&gt;&#xA;" OwnerUserId="8015" LastEditorUserId="5026" LastEditDate="2018-07-12T02:39:54.973" LastActivityDate="2018-07-12T02:42:37.877" Title="How does the reaction of HCl and KMnO4 proceed?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16673" PostTypeId="1" CreationDate="2014-09-24T13:30:57.640" Score="6" ViewCount="1601" Body="&lt;p&gt;Making a &quot;carbon snake&quot; requires the mixing of strong sulphuric acid with sugar, as explained &lt;a href=&quot;http://chemistry.about.com/od/chemistrydemonstrations/a/acidsugardemo.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this about.chemistry webpage&lt;/a&gt; - the reaction causes rapid dehydration of the sugar and produces steam in the process, which expands and forms the 'carbon snake', following the incomplete chemical reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C12H22O11 + H2SO4 → 12C + 11H2O}$  + mixture of excess unreacted water, sulfur dioxide and acid (and presumably a bundle of other chemicals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what physical/chemical processes cause this rapid dehydration and exothermic reaction?&lt;/p&gt;&#xA;" OwnerDisplayName="user8016" LastEditorUserId="402" LastEditorDisplayName="user8016" LastEditDate="2017-04-01T01:17:00.850" LastActivityDate="2017-04-01T01:18:49.400" Title="&quot;Carbon snakes&quot; - what physico-chemical process causes the exothermic response?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16674" PostTypeId="2" ParentId="16649" CreationDate="2014-09-24T13:32:04.730" Score="6" Body="&lt;p&gt;I think a tantalising clue can be found in the principle of &lt;em&gt;cathodic protection&lt;/em&gt; used in industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathodic protection is used in the shipping industry to prevent iron/steel hulls from rusting. Essentially you make your hull into a cathode by connecting a (typically) zinc anode to the hull. How do we know the ship's hull is the cathode and zinc the anode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And zinc in fact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets go back to the standard reduction potential tables. The following reaction has a reduction potential of $\ce{-0.7618V}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {Zn^2+(aq)} + 2{e}  \ce{-&amp;gt;} \ce{Zn(s)}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the following has a reduction potential of $\ce{-0.44V}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {Fe^2+(aq)} + 2{e}  \ce{-&amp;gt;} \ce{Fe(s)}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the rusting process works in the opposite direction (and thus so do the above equations) which just means we reverse the voltage to get the oxidation value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A higher reversed voltage means that that process has a greater tendency to occur, which means that zinc here is preferentially oxidised over iron. Hence if we connect an iron hull to a zinc rod, the zinc acts as the anode whilst the hull is the cathode i.e any $\ce{Fe^2+}$ directly formed by reaction with sea-water is pushed back to iron metal by the production of $\ce{Zn^2+}$, a process which liberates two electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should suggest that some electrolytic process contributes to salt water affecting the oxidation of iron. The most basic explanation is that the sea water acts as an electrolyte, which facilitates this process much more easily than, say, distilled water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean however when we don't have an electrochemical cell as above, and only have a single electrode i.e. the ship's hull? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might like to look up the standard reduction potentials for both chloride and iron, however we quickly see that this isn't the case; both chloride and iron metal have to lose electrons in our case, which isn't going to form an electrochemical reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an interesting case that bleach (alkali salts of chlorite, $\ce{ClO-}$)undergoes the following reaction with iron:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {Fe(s)} + \ce{ClO^-(aq)} +\ce{H2O}  \ce{-&amp;gt;} \ce{Fe^2+(aq)} + \ce{HClO2(g)} +\ce{2H+(aq)}  $$ which has a favourable overall voltage of &lt;em&gt;around&lt;/em&gt; $\ce{2.11V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've seen that chloride doesn't directly react with iron, but chlorite does. This is a possible route for corrosion, although it's not obvious by which natural process chlorite comes about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to note is that seawater just doesn't contain sodium chloride, but myriad other ions which can contribute to this corrosion process,especially sulphate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {SO4^2-(aq)} + \ce{4H^+(aq)} + \ce{Fe(s)}  \ce{-&amp;gt;} \ce{Fe^2+(aq)} + \ce{SO_2(g)} + \ce{2H2O(g)} $$ which occurs with a total voltage of $\ce{0.61V}$, which tells us that the chlorite process is much more favourable thermodynamically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, there are many different reactions that can utilise seawater to oxidise iron—it's not incredibly clear which ones do contribute, and which ones don't, but the above provides the means for some educated guesswork, so to speak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final point, iron &lt;em&gt;can&lt;/em&gt; oxidise in air without water, it just takes a long time. The presence of an electrolyte will speed up the process however. Corrosion also needn't be a reaction with oxygen either, but any process which makes iron ions from iron metal. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-09-24T20:57:59.903" LastActivityDate="2014-09-24T20:57:59.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16675" PostTypeId="2" ParentId="16671" CreationDate="2014-09-24T13:38:27.100" Score="11" Body="&lt;h1&gt;無&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;How many grains of sand does it take to make a pile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is nonsensical, because a phase of matter is an emergent property of a collection of atoms or molecules, just like being a pile is an emergent property of a bunch of sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to that the notion that &quot;molecule&quot; is kind of vague in any case, and you've got a world-class mess on your hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, it's quite clear what the unit of C&lt;sub&gt;2&lt;/sub&gt;H&lt;sub&gt;5&lt;/sub&gt;OH is, and that it is a molecule, but for NaCl, what constitutes the unit? What constitutes the &quot;molecule&quot;? Is there one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about polymers, or diamonds? Is a diamond with its multitudes of covalent bonds a single molecule? Is it many? Is (C&lt;sub&gt;3&lt;/sub&gt;H&lt;sub&gt;6&lt;/sub&gt;)&lt;sub&gt;n&lt;/sub&gt; n molecules, or one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid the question leads you down a philosophical rabbit hole that goes far beyond the science of chemistry.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="9961" LastEditDate="2016-08-12T20:29:25.747" LastActivityDate="2016-08-12T20:29:25.747" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16676" PostTypeId="1" CreationDate="2014-09-24T14:06:43.543" Score="12" ViewCount="4229" Body="&lt;p&gt;Is mathematics important to learn concepts in inorganic chemistry, such as point groups, point symmetry, because doing these topics requires one to determine the symmetry of the elements and to visualize 3D (such as determining the symmetry of a water molecule?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As these concepts are derived from mathematics, I wonder whether taking maths courses would be helpful for me to understand these concepts better and if so which maths course would be most useful? Such as linear algebra, abstract algebra?&lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="186" LastEditDate="2014-09-24T19:35:24.173" LastActivityDate="2016-02-09T18:13:41.093" Title="Mathematics in inorganic and quantum chemistry" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16677" PostTypeId="2" ParentId="16672" CreationDate="2014-09-24T14:08:05.907" Score="0" Body="&lt;p&gt;I'm assuming that you were expecting a double replacement.  Clearly that doesn't happen here, and it's because of the permanganate ion.  $\ce{KMnO4}$ is a powerful oxidizing agent.  That means that it likes to steal electrons from other species, ultimately producing $\ce{Mn^2+}$ ions and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the species in solution with it, $\ce{K^{+}}$, $\ce{H^{+}}$, and $\ce{Cl^{-}}$, the only one that has electrons to get rid of is the chloride.  If chloride loses its surplus electrons it will turn into chlorine, leaving behind manganese(II) ions and some oxygen ions that need a home.  Those stick to hydrogen ions from the acid, making water. A portion of the chloride ions are not oxidized and remain in solution to balance the charges of the potassium ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've simplified the mechanism, of course.  For example, the permanganate ion doesn't just fly apart into $\ce{Mn^7+}$ and $\ce{O^2-}$ ions once the reaction starts, but the final effect is as if it did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation-reduction reactions often have products that are difficult to predict without a significant amount of chemistry experience.  Balancing these reactions often requires some new techniques (beyond inspection) such as balancing by oxidation number and balancing by half reactions.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="5026" LastEditDate="2018-07-12T02:41:05.673" LastActivityDate="2018-07-12T02:41:05.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16678" PostTypeId="2" ParentId="16676" CreationDate="2014-09-24T14:27:22.010" Score="7" Body="&lt;p&gt;My personal advice would be to do as many maths courses as you possibly can, they never hurt and even if they are not directly relevant to your chemistry career then they sharpen the mind. There's no such thing as too much maths, although my personal opinion aside there are two school of thought on this—that of the chemist, and that of the mathematician.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In perhaps crude terms, chemists often use very simplified versions of mathematics which would probably make a professional mathematician cry. Linear algebra is very important for quantum mechanics. This topic doesn't really differ too much (what I mean by that is a mathematician inverts a matrix just the same as a chemist does). Linear algebra is therefore useful regardless of where you learn it from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A professional mathematician would tell you to learn group theory from a mathematician's point of view, but this can often be highly inaccessible to beginners (new notation to learn, much stricter logic which unintuitively seems less obvious to the untrained). The second point is that it's also not necessary for your chemistry career to learn all of group theory if all you need to do is follow the literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the more maths you learn, the more open your options are. Taking abstract algebra for example means that you are exposed to ideas which may be useful to your chemistry career, but such ideas haven't been realised in the chemistry community yet. Abstract algebra lays the foundations for topological chemistry and other higher study, of which the benefits to chemistry are only starting to be realised. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-09-24T14:27:22.010" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16679" PostTypeId="2" ParentId="16676" CreationDate="2014-09-24T14:28:58.040" Score="14" Body="&lt;p&gt;For better understanding of molecular symmetry you indeed need some mathematics, namely, some basics of &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Group_theory&quot;&gt;group theory&lt;/a&gt;&lt;/em&gt;, but aside from that I doubt that you need some mathematics apart form &lt;em&gt;arithmetic&lt;/em&gt; and &lt;em&gt;calculus&lt;/em&gt; for general, inorganic, organic, and even physical chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;quantum mechanics&lt;/em&gt; you absolutely need to familiarize yourself with the concept of &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Complex_number&quot;&gt;complex numbers&lt;/a&gt;&lt;/em&gt; and you have to learn some basic approaches for solving &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_equation&quot;&gt;differential equations&lt;/a&gt;&lt;/em&gt;. For an advanced insight into &lt;em&gt;quantum chemistry&lt;/em&gt; you need some &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_algebra&quot;&gt;linear algebra&lt;/a&gt;&lt;/em&gt;, by which I mean the theory of vector spaces &lt;em&gt;both finite- and infinite-dimensional&lt;/em&gt;, not just finite-dimensional part of it, which, in a sense, reduces to matrix algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. &lt;em&gt;Abstract algebra&lt;/em&gt; is nothing but a common name for different areas of mathematics which studies different so-called &lt;em&gt;algebraic structures&lt;/em&gt;. Say, group theory studies groups, linear algebra studies vector spaces, etc.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-09-24T16:51:23.717" LastActivityDate="2014-09-24T16:51:23.717" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16681" PostTypeId="2" ParentId="16671" CreationDate="2014-09-24T15:22:43.043" Score="21" Body="&lt;p&gt;This is actually an active area of research for &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;water clusters&lt;/a&gt;. In principle, for &lt;span class=&quot;math-container&quot;&gt;$\ce{(H2O)_{n}}$&lt;/span&gt; there should be a &amp;quot;melting&amp;quot; phase transition, much like for ice to liquid water.&lt;/p&gt;&#xA;&lt;p&gt;So, in principle, if you had an accurate enough theoretical method, you could do molecular dynamics and see when the melting point of the cluster matches bulk water.&lt;/p&gt;&#xA;&lt;p&gt;Defining a line between liquid and gas seems tricky to me, but I don't do statistical mechanics, and perhaps there are molecular dynamics techniques for defining a boiling point.&lt;/p&gt;&#xA;&lt;p&gt;Defining the solid-liquid phase transition is much easier, because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Radial_distribution_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;radial distribution function&lt;/a&gt; of the system can be monitored to see the transition (i.e., the density changes).&lt;/p&gt;&#xA;&lt;p&gt;At the moment, I don't think the best dynamics methods get the melting transitions accurately yet, but I think n~20 is about right. I'll revise this when I get a better answer.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: &lt;a href=&quot;https://www.chemistryworld.com/news/how-many-water-molecules-make-a-droplet/4013080.article&quot; rel=&quot;nofollow noreferrer&quot;&gt;It turns out that 21 seems to be the number for water drops.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;How many water molecules are needed to solvate one?&amp;quot;&#xA;&lt;a href=&quot;https://doi.org/10.1039/D0SC05785A&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Chem. Sci.&lt;/em&gt; &lt;strong&gt;2021&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The minimal structure eventually responsible for proper solvation is made of a total of 21 water molecules and includes two complete solvation shells, of which the whole first one is tetrahedrally coordinated to the central molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2021-01-22T14:17:51.590" LastActivityDate="2021-01-22T14:17:51.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16682" PostTypeId="2" ParentId="16672" CreationDate="2014-09-24T15:23:17.203" Score="0" Body="&lt;p&gt;If you are interested in the mechanism it should look something like this ($\ce{HBr}$ was used instead of $\ce{HCl}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{2H+ + Br- + MnO4- &amp;amp;&amp;lt;-&amp;gt; H2MnO4Br} \tag{$K$ fast}\\&#xA;\ce{H2MnO4Br + H+ + Br- &amp;amp;-&amp;gt; H3MnO4 + Br2} \tag{$k$ limiting}\\&#xA;\ce{H3MnO4 &amp;amp;-&amp;gt; \text{products}} \tag{fast}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that the first two steps can be divided into more steps. &lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5026" LastEditDate="2018-07-12T02:42:37.877" LastActivityDate="2018-07-12T02:42:37.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16683" PostTypeId="1" AcceptedAnswerId="16684" CreationDate="2014-09-24T15:48:36.007" Score="1" ViewCount="1019" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An oxide of chromium used in chrome plating has a formula mass of&#xA;  100.0 u and contains four atoms per formula unit. Establish the formula of this compound, with a minimum of calculation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Express your answer as a chemical formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the question tells me that the mass of the Chromium Oxide is 100.0 amu and has 4 atoms for very unit. Thus, the question wants to know a formula that has 4 atoms? Right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if $\ce{Cr}$'s (Chromium) molecular mass is 51.9961 and since the question is stating that we are dealing with an oxide, which would mean that it is a ionic compound (Chromium gives two electrons to Oxygen [15.999 amu]), wouldn't the answer be $\ce{Cr2O2}$; but that would equal 135.9902 grams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if I am supposed to view a formula unit as an empirical formula, wouldn't that require me to write the simplest formula for a compound? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, $\ce{CrO}$ would give me (51.9961 + 15.999) 67.9951 g/mol and that is not correct, or is it? &lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="865" LastEditorDisplayName="user15489" LastEditDate="2021-05-22T21:30:27.837" LastActivityDate="2021-05-22T21:30:27.837" Title="Establish the formula of an oxide of chromium" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16684" PostTypeId="2" ParentId="16683" CreationDate="2014-09-24T16:02:07.670" Score="4" Body="&lt;p&gt;Rounding judiciously, the mass of chromium is 52 amu and oxygen is 16 amu.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our compound is Cr&lt;sub&gt;x&lt;/sub&gt;O&lt;sub&gt;y&lt;/sub&gt; and has 4 atoms.  The sum of those 4 atoms' mass is 100 amu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x + y = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;52x + 16y = 100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve the system of equations to find the answer.  Alternatively, you can just guess and check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any species with 2 chromium atoms must have a mass greater than 100 amu, since Cr&lt;sub&gt;2&lt;/sub&gt; = 104 amu and we haven't included any oxygens.  Thus we know for certain that there is only 1 chromium atom in the compound.  The remaining three must be oxygen.  I leave it to you to check that the answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your confusion stems from assuming that something called &quot;chromium oxide&quot; must have 1 chromium atom for every 1 oxygen atom, and that isn't the case.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-09-24T16:02:07.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16685" PostTypeId="2" ParentId="16665" CreationDate="2014-09-24T16:02:42.167" Score="0" Body="&lt;p&gt;You can't reduce aluminium oxide to aluminium using carbon; aluminium, like many metals, forms a stable carbide.  The equation looks like this: $\ce{2 Al2O3 + 9 C \rightarrow Al4C3 + 6 CO}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why in real life aluminium is prepared through the Hall–Héroult process.  Before the electrochemical process was discovered the reducing agent used was sodium.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-09-24T16:02:42.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16686" PostTypeId="1" AcceptedAnswerId="16689" CreationDate="2014-09-24T16:28:04.277" Score="5" ViewCount="906" Body="&lt;p&gt;Just wondering if anyone can tell me why bubbling nitrogen through a solution of $\ce{HCl}$/diethyl ether catalyst would remove said mix from my solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current thought is that the $\ce{N2}$ will form ammonia and deprotonate the $\ce{HCl}$. But what of the Diethyl ether? I can't think of how it would be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only final explanation is that it is the $\ce{HCl}$ which is the important species to remove and the diethyl ether is just solvent ... but I would really like a second opinion!   &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7475" LastEditDate="2015-10-09T16:14:37.237" LastActivityDate="2015-10-09T16:14:37.237" Title="HCl/diethyl ether + Nitrogen?" Tags="&lt;solutions&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16687" PostTypeId="1" AcceptedAnswerId="16780" CreationDate="2014-09-24T16:46:12.287" Score="7" ViewCount="1282" Body="&lt;p&gt;My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure $P_i$ to $P_f$. Suppose,the process is done with innumerable &lt;strong&gt;infinitesimal steps&lt;/strong&gt;. Then, say at each step, $dp$ is changed in the pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the step, gas and the piston were exerting same pressure $P_i$ . Now, when the piston is raised, gas does expand to equalize the pressure,which is $P_i - dp$ . So, at the end of the step, the gas &amp;amp; the piston will exert same pressure $P_i - dp$ and hence will be in mechanical equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the step,the gas and the piston was not in equilibrium;it was at the end of the step  the equilibrium is achieved. So, &lt;strong&gt;during each step of the process the gas would be at disequilibrium&lt;/strong&gt; . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by $dp$ . So, why does the definition tell that at each step,there is equilibrium?       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have three questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is maximum work only obtained from reversible processes? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please help. &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="4356" LastEditDate="2014-12-03T00:11:28.313" LastActivityDate="2014-12-03T15:28:57.037" Title="Thermodynamically Reversible Processes: How does an infinitesimal step make a process reversible and how is maximum work derived from the process?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16688" PostTypeId="1" CreationDate="2014-09-24T16:58:32.080" Score="2" ViewCount="1156" Body="&lt;p&gt;I'm a higher secondary school studet. See this following image, I can't understand how water can rises through the bed of permutit, my textbook says - Hard water introduced from the top of tank and then reaches at the bottom and then rises through the bed of permutit.!!, please tell me how it is possible.&lt;img src=&quot;https://i.stack.imgur.com/dgomw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8024" LastEditorUserId="21964" LastEditDate="2016-08-17T17:38:54.013" LastActivityDate="2017-02-27T03:46:25.873" Title="Zeolite process" Tags="&lt;water&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16689" PostTypeId="2" ParentId="16686" CreationDate="2014-09-24T17:11:39.237" Score="7" Body="&lt;p&gt;HCl is a gas at room temperature and diethyl ether is a low boiler.  Both will have significant vapor pressures and, therefore, exist in the vapor phase above your  solution.  The HCl and diethyl ether in solution will continuously maintain an equilibrium with the vapor phase. As you bubble nitrogen through the system it will continuously sweep the vapor out, removing HCl and diethyl ether.  But since the HCl and diethyl ether in solution continuously maintain an equilibrium in the vapor phase, they will be continuously swept out until all of the HCl and diethyl ether in solution have been removed.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7475" LastEditDate="2015-10-09T16:13:58.457" LastActivityDate="2015-10-09T16:13:58.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16690" PostTypeId="1" AcceptedAnswerId="16713" CreationDate="2014-09-24T18:54:51.443" Score="7" ViewCount="1198" Body="&lt;p&gt;Consider the series reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A -&amp;gt;[k_1] B -&amp;gt;[k_2] C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d}{\mathrm d t} [\ce{B}] = k_1 [\ce{A}] - k_2 [\ce{B}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We already know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{A}]=[\ce{A}]_{0} \exp(-k_1 t)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far, I have used the method integrating factors, with $u=\exp(k_2t)$, to arrive at the equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} &#xA;[\ce{B}]= \frac{k_1}{k_2-k_1} \left( \exp(-k_1 t) + C \exp(-k_2 t) \right) [\ce{A}]_{0}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the constant of integration be in this situation?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="30836" LastEditDate="2017-10-15T13:10:33.713" LastActivityDate="2017-10-17T11:37:33.447" Title="What is the constant of integration in a series reaction?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16691" PostTypeId="1" AcceptedAnswerId="16693" CreationDate="2014-09-24T18:55:04.357" Score="5" ViewCount="4141" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  If all of the silver in a $7.563~\mathrm{g}$ sample of a mineral, which was dissolved by using nitric acid, is precipitated as silver thiocyanate ($\ce{AgSCN}$) by using $82.46~\mathrm{mL}$ of a $0.2500~\mathrm{M}$ potassium thiocyanate solution, what is the percent silver (by mass) in the sample?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;&lt;br&gt;&#xA;I'm pretty sure the relevant chemical equation here is $\ce{Ag^+ + SCN- -&amp;gt; AgSCN}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molarity is mols solute over L solution, so the number of mols of $\ce{KSCN}$ used is $0.25~\mathrm{M} \times 0.08246~\mathrm{L} = 0.020615~\mathrm{mol\ KSCN}$. According to the chemical equation above, the molar ratio is 1 to 1 so there is $0.020615~\mathrm{mol\ Ag}$  involved. That accounts for all of the silver in the original mineral sample, I believe. Therefore, there are $2.22370~\mathrm{g\ Ag}$, so the percent silver by mass is $2.22370~\mathrm{g\ Ag} \div 7.563 ~\mathrm{g} = 0.294$. Put into percent, that's $29.4\%~\mathrm{Ag}$ by mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure about this because  I totally ignored the nitric acid.&lt;/p&gt;&#xA;" OwnerUserId="7626" LastEditorUserId="46306" LastEditDate="2018-03-05T02:42:44.687" LastActivityDate="2018-03-05T02:42:44.687" Title="Percent of silver by mass in an unknown sample" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16692" PostTypeId="2" ParentId="16691" CreationDate="2014-09-24T19:12:19.083" Score="1" Body="&lt;p&gt;It's correct. You need to assume that the mineral will be completely dissolved in nitric acid, as in the opposite case, you wouldn't have enough data to calculate the mass of silver.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-24T19:12:19.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16693" PostTypeId="2" ParentId="16691" CreationDate="2014-09-24T19:22:39.657" Score="3" Body="&lt;p&gt;Noah--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastActivityDate="2014-09-24T19:22:39.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16694" PostTypeId="2" ParentId="16688" CreationDate="2014-09-24T19:34:20.610" Score="2" Body="&lt;p&gt;Hard water has minerals (mostly calcium and magnesium salts). The hard water is pumped into the coarse sand layer at the bottom. As the coarse sand settles, the water comes into contact with the Permutit zeolite. As the water rises through zeolite, it binds the calcium and magnesium and replaces them with sodium.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastActivityDate="2014-09-24T19:34:20.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16695" PostTypeId="5" CreationDate="2014-09-24T19:43:11.420" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-24T19:43:11.420" LastActivityDate="2014-09-24T19:43:11.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16696" PostTypeId="4" CreationDate="2014-09-24T19:43:11.420" Score="0" Body="A set of (basis) functions make up a basis set.  These functions (generally) represent atomic orbitals whose linear combinations are used to make molecular orbitals (LCAO-MO) in computational chemistry." OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-24T21:32:17.777" LastActivityDate="2014-09-24T21:32:17.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16697" PostTypeId="5" CreationDate="2014-09-24T19:52:06.177" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-09-24T19:52:06.177" LastActivityDate="2014-09-24T19:52:06.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16698" PostTypeId="4" CreationDate="2014-09-24T19:52:06.177" Score="0" Body="Mass spectrometry (MS) is an analytical technique which takes advantage of the mass-to-charge ratio of gas-phase ions in order to elucidate the composition of a sample." OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-24T19:59:28.053" LastActivityDate="2014-09-24T19:59:28.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16699" PostTypeId="5" CreationDate="2014-09-24T19:53:59.073" Score="0" Body="&lt;p&gt;A Lewis structure (usually referred to as a Lewis dot structure) is a simple model or approximation used to determine the bonding behavior of atoms within molecules.  Dots represent valence electrons for each atom while solid lines represent shared pair of electrons between atoms (i.e. covalent bonds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this is the Lewis structure of the water molecule $\ce{H2O}$ (See also &lt;a href=&quot;https://simple.wikipedia.org/wiki/Lewis_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/91vkR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/91vkR.png&quot; alt=&quot;Lewis structure of water&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the two lone pairs (= 4 electrons) on the oxygen atom shown by dots and the two bond pairs (O-H) shown by solid lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis structures may also be drawn for charged species. Here is the Lewis structure for the ammonium ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3jJp.png&quot; alt=&quot;Lewis structure of the ammonium ion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that a solid line with an arrow indicates a &lt;em&gt;coordinate&lt;/em&gt; covalent bond, with the arrow pointing from the donor atom to the receiver atom.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="4945" LastEditDate="2018-03-19T04:50:02.223" LastActivityDate="2018-03-19T04:50:02.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16700" PostTypeId="4" CreationDate="2014-09-24T19:53:59.073" Score="0" Body="For questions about determining Lewis structures (or their validity) for different chemical species. Do not use this tag if your question is about determining [oxidation-state] or [formal-charges], use the other respective tags instead." OwnerUserId="14" LastEditorUserId="5026" LastEditDate="2018-03-17T11:10:21.803" LastActivityDate="2018-03-17T11:10:21.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16701" PostTypeId="5" CreationDate="2014-09-24T19:55:01.873" Score="0" Body="&lt;p&gt;A dipole is the result of an uneven distribution of electron density within a molecule.  Perfectly symmetric molecules (such as benzene) will have no dipole moment while a non-symmetric one like hydrogen fluoride (HF) will have a very large dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-19T17:12:43.397" LastActivityDate="2018-03-19T17:12:43.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16702" PostTypeId="4" CreationDate="2014-09-24T19:55:01.873" Score="0" Body="For questions about the dipole moment of a bond or a molecule; its calculation, comparison with other bonds or molecules, affect on molecular properties, etc." OwnerUserId="14" LastEditorUserId="5026" LastEditDate="2018-03-19T17:12:37.430" LastActivityDate="2018-03-19T17:12:37.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16703" PostTypeId="1" AcceptedAnswerId="16714" CreationDate="2014-09-24T20:45:19.663" Score="3" ViewCount="204" Body="&lt;p&gt;I'd like to be able to find the crystal structures of drugs, for example at the moment I want to find the crystal structure of (&lt;em&gt;R&lt;/em&gt;)-fluoxetine so that I can use Jmol to create an accurate 3D model of it. This is so I can then upload it to Wikimedia Commons so it may be used alongside Benjah-bmm27's structure of (&lt;em&gt;S&lt;/em&gt;)-fluoxetine on the Wikipedia page for fluoxetine. I have tried this database [&lt;a href=&quot;http://www.crystallography.net/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;] but when I searched for the SMILES of (&lt;em&gt;R&lt;/em&gt;)-fluoxetine, CNCCC@HOC2=CC=C(C=C2)C(F)(F)F (from ChemSpider [&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.1265981.html&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]), I obtained no search results. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2014-10-25T08:15:49.277" Title="What's the most efficient way to find crystal structures of drugs?" Tags="&lt;crystal-structure&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16704" PostTypeId="2" ParentId="16687" CreationDate="2014-09-24T21:28:26.200" Score="5" Body="&lt;p&gt;I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My mathematical reasoning:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use.  Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being &lt;em&gt;reversible&lt;/em&gt;. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My analogy:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has &lt;em&gt;still&lt;/em&gt; ended up in the same place as the irreversible case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: How to Achieve Equilibrium &lt;em&gt;and&lt;/em&gt; still have Movement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the &lt;em&gt;oscillation&lt;/em&gt; of position. Even if you've physically pushed your piston a distance &lt;em&gt;ds&lt;/em&gt;, it can still be in equilibrium by the very fact that it can still move forwards &lt;em&gt;and&lt;/em&gt; backwards in its next spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in fact the periodicity of position is maintained at each step, which is what we mean by the &lt;em&gt;equilibrium is maintained&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic &lt;em&gt;equilibrium&lt;/em&gt;. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastEditorDisplayName="user7232" LastEditDate="2014-09-25T11:29:29.687" LastActivityDate="2014-09-25T11:29:29.687" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16705" PostTypeId="1" CreationDate="2014-09-24T21:50:34.323" Score="8" ViewCount="2152" Body="&lt;p&gt;I know we all think of things like uranium as being radioactive in the nasty radiation poisoning sort of way. But do elements like silver have similar effects in their radioactive isotopes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an author doing some research for a book I'm working on. Would there be a particular quantity or exposure time that would be necessary to make it toxic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we did know the amount to make it toxic. I know regular silver can cause Argyria which is mostly a cosmetic illness, but would the side effects of the radioactive silver be similar to that of other radioactive materials?&lt;/p&gt;&#xA;" OwnerUserId="8032" LastEditorUserId="10185" LastEditDate="2016-07-16T16:18:09.047" LastActivityDate="2016-07-16T16:18:09.047" Title="Are radioactive isotopes of silver dangerous?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;isotope&gt;&lt;radioactivity&gt;&lt;silver&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16706" PostTypeId="2" ParentId="16672" CreationDate="2014-09-24T22:07:26.567" Score="2" Body="&lt;p&gt;As others have pointed out, you don't have a &quot;normal&quot; reaction, you have a oxidation-reduction reaction (if you want to sound especially geeky, calls them RedOx (pronounced &quot;ree-dox&quot;) reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons (which we don't normally balance in a equation) feature prominently because there are elements changing oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, split your reaction into its half reactions. Bring the &quot;spectator ions&quot; along for the ride and make sure they balance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: 2 $\ce{Cl-}$ gets oxidized to $\ce{Cl2}$ and gives up 2 electrons. $\ce{H+}$ is your spectator ion. $$\ce{2 H+ + 2 Cl- -&amp;gt; Cl2 + 2e- + 2H+}$$&#xA;Reduction: $\ce{Mn^7+}$ needs 5 electrons to reduce it to $\ce{Mn^2+}$. $\ce{K+}$ and $\ce{O^2-}$ are your spectator ions. $$\ce{K+ + MnO4- + 5e- -&amp;gt; Mn^2+ + K+ + 4O^2-}$$&#xA;Now, just add water. Oxidation reactions get $\ce{OH-}$; reduction reactions get $\ce{H+}$:&#xA;$$\begin{align}&#xA;\ce{2OH- + 2 H+ + 2 Cl- &amp;amp;-&amp;gt; Cl2 + 2e- + 2H2O}\\&#xA;\ce{8H+ + K+ + MnO4- + 5e- &amp;amp;-&amp;gt; Mn^2+ + 4 H2O + K+}&#xA;\end{align}$$&#xA;Now we balance the electrons (remember least-common-multiple from math class? You thought you'd never use that, didn't you?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LCM for 2 and 5 is 10, so multiply the Cl equation by 5 and the Mn reaction by 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{10OH- + 10 H+ + 10 Cl- &amp;amp;-&amp;gt; 5Cl2 + 10e- + 10H2O}\\&#xA;\ce{16H+ + 2K+ + 2MnO4- + 10e- &amp;amp;-&amp;gt; 2Mn^2+ + 8 H2O + 2K+}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add everybody back together, make more water, and cancel out the electrons:&#xA;$$\ce{10H2O + 10 HCl + 6H+ + 2KMnO4 -&amp;gt; 5Cl2 + 10H2O + 2Mn^2+ + 8 H2O + 2K+}$$&#xA;Take out the excess water:&#xA;$$\ce{10 HCl + 6H+ + 2KMnO4 -&amp;gt; 5Cl2 + 2Mn^2+ + 8 H2O + 2K+}$$&#xA;Looks like we're short something to balance out the charges. We can add more chloride spectator ions to both sides of the equation. They're not part of the redox since they are $\ce{Cl-}$ on both sides of the equation:&#xA;$$\ce{10 HCl + 6HCl + 2KMnO4 -&amp;gt; 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$&#xA;Collect terms one last time and we're done:&#xA;$$\ce{16HCl + 2KMnO4 -&amp;gt; 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$&lt;/p&gt;&#xA;" OwnerUserId="8030" LastEditorUserId="5026" LastEditDate="2018-07-12T02:40:31.347" LastActivityDate="2018-07-12T02:40:31.347" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16707" PostTypeId="2" ParentId="16705" CreationDate="2014-09-24T22:26:54.270" Score="7" Body="&lt;p&gt;A glance through the table of the isotopes in the venerable CRC handbook reveals that the longest lived radioactive isotope of Ag is $Ag^{108m}$, made by neutron capture by $Ag^{107}$.  this has a listed half life of &quot;&gt; 5y&quot;, but the capture cross section is only $35\pm5$ barns, pretty small for thermal neutrons.  So, it would be really hard to get a lot of radioactive silver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, all isotopes listed decay by $\beta, \beta^{+}$, EC, or IT (electron capture of internal transitions for the metastable nuclei), with decay energies ranging from 93 keV ($Ag^{107m} \rightarrow Ag^{107}$) to 6MeV ($Ag^{116}$).  Those are ionizing radiation, which would increase your probability of getting cancer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'nasty radiation poisoning' things are actually nasty primarily because of being heavy metals, the radiation comes along for the ride, so to speak.  The normal Uranium isotopes are very long lived, so emit very little radiation per unit time.  Polonium, the scary poison of choice in stories and perhaps by the KGB, in contrast, has a short lifetime (138 days for $Po^{210}$), and spews lots of ~5MeV $\alpha$ particles, heavily damaging tissue (much more so than gammas). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the difficulty in making the stuff, assuming you would try $Ag^{108m}$, (small cross sections, can't chemically separate from the stable $Ag^{107}$, etc., your best bet would just be acute heavy metal poisoning from Ag salts (which may or may not be highly toxic). Or, using a silver candlestick to bludgeon the poor victim to death.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-09-24T22:26:54.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16708" PostTypeId="2" ParentId="16652" CreationDate="2014-09-24T23:06:03.300" Score="5" Body="&lt;p&gt;You've calculated your radius ratio value and you know you're between 0.414 and 0.732, so cubic is off the table, based on your chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're looking at either square planar or octahedral. Your chart also says that ionic compounds do not use square planar packing and the coordination number would have to be 4. LiBr is an ionic compound (like NaCl) with a coordination number of 6, therefore you're looking at octahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can determine your &quot;specific&quot; answer.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastEditorUserId="5026" LastEditDate="2018-07-24T01:49:10.730" LastActivityDate="2018-07-24T01:49:10.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16709" PostTypeId="1" CreationDate="2014-09-25T01:33:23.100" Score="1" ViewCount="5670" Body="&lt;p&gt;From my lab book it says it is black? I don't understand how we know the color of it is black. Can anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="7795" LastEditorUserId="865" LastEditDate="2021-05-22T21:05:57.857" LastActivityDate="2021-05-22T21:05:57.857" Title="What is the color of elemental silver?" Tags="&lt;color&gt;&lt;elements&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16710" PostTypeId="1" CreationDate="2014-09-25T02:23:07.503" Score="2" ViewCount="73" Body="&lt;p&gt;My Google-fu hasn't got me anywhere. All results have been quite fuzzy about it, nothing explicit.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="4592" LastEditDate="2014-09-26T00:25:00.177" LastActivityDate="2014-09-26T00:25:00.177" Title="Is there a compound where one element has a CCP structure with all voids occupied by others?" Tags="&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16711" PostTypeId="1" AcceptedAnswerId="16712" CreationDate="2014-09-25T02:27:53.687" Score="3" ViewCount="36399" Body="&lt;p&gt;I am asked to determine what is the O.S. of oxygen in &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2O2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was under the impression that oxygen had a charge of -2 and I multiplied that charge by 2 giving me a total of -4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the computer program I am using, states that I am wrong. Can someone clarify why? &lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="83758" LastEditDate="2020-03-20T20:36:47.433" LastActivityDate="2020-03-20T20:36:47.433" Title="What is the oxidation state of oxygen in Na2O2?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16712" PostTypeId="2" ParentId="16711" CreationDate="2014-09-25T02:37:46.690" Score="11" Body="&lt;p&gt;Oxygen can take multiple oxidation states. This particular compound is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_peroxide&quot; rel=&quot;noreferrer&quot;&gt;sodium peroxide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that usually oxygen has a charge of -2, but in this case, there's no way that &lt;em&gt;each&lt;/em&gt; $\ce{Na}$ can have an oxidation state of +2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you then work backwards, deciding if it's $\ce{Na+}$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxide&quot; rel=&quot;noreferrer&quot;&gt;peroxides&lt;/a&gt; ($\ce{O2^{2-}}$) there are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Superoxide&quot; rel=&quot;noreferrer&quot;&gt;superoxides&lt;/a&gt; ($\ce{O2^{-}}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not surprising that these compounds are usually very reactive.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-11-17T00:22:53.753" LastActivityDate="2015-11-17T00:22:53.753" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16713" PostTypeId="2" ParentId="16690" CreationDate="2014-09-25T02:50:29.093" Score="9" Body="&lt;p&gt;There might or there might not be a problem with your final equation. Either you have made a mistake while using the integrating factor to solve your differential equation or you made some further substitutions so that your constant $C$ contains something more than just the &quot;standard stuff&quot; I would have expected from the method. Thus it's best I go through the derivation from the beginning so that it'll become clear what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of solving differential equation using an integrating factor is explained well &lt;a href=&quot;http://en.wikipedia.org/wiki/Integrating_factor&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. You have used the correct integrating factor $u(t)$. For the differential equation at hand, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &amp;amp;= k_1[\ce{A}] - k_2[\ce{B}] \\&#xA;  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} + k_2[\ce{B}] &amp;amp;= k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is indeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  u(t) = \exp\biggl( \int_{0}^{t} k_2 \, \mathrm{d} t \biggr) = \mathrm{e}^{k_2 t} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the integrating factor you can get $[\ce{B}]$ according to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  u(t) [\ce{B}] &amp;amp;= \int_{0}^{t} u(t) k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \, \mathrm{d} t + \tilde{C} \\&#xA;  [\ce{B}] \mathrm{e}^{k_2 t} &amp;amp;= \int_{0}^{t} k_1 [\ce{A}]_{0} \mathrm{e}^{(k_2 - k_1) t} \, \mathrm{d} t + \tilde{C} \\&#xA;  &amp;amp;= k_1 [\ce{A}]_{0} \biggl[\frac{1}{k_2 - k_1} \mathrm{e}^{(k_2 - k_1) t} \biggr]_{0}^{t} + \tilde{C} \\&#xA;  &amp;amp;= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{(k_2 - k_1) t} - 1 \bigr) + \tilde{C}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging for $[\ce{B}]$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  [\ce{B}] &amp;amp;= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \underbrace{\frac{ \mathrm{e}^{(k_2 - k_1) t} - 1 }{\mathrm{e}^{k_2 t}}}_{= \, \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t}}  + \frac{\tilde{C}}{\mathrm{e}^{k_2 t}} \\&#xA; &amp;amp;= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) + \tilde{C} \mathrm{e}^{-k_2 t} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation differs from yours a little if your integration constant $C$ would be the same as my $\tilde{C}$. But if you used some substitutions so that your integration constant is $C = \frac{k_2 - k_1}{k_1 [\ce{A}]_{0}}  \tilde{C} - 1$ then your solution and mine would be equivalent. If you indeed made a mistake then I hope my derivation cleared things up a bit and if you had intended the equation to look that way by introducing the substitution then at least you know what integration constant I mean when I am going to explain the method for determining its value in the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration constant can now be determined by enforcing boundary conditions. In this particular case this means that at the start of your reaction ($t=0$) you require that $[\ce{B}]$ is equal to the initial concentration of your reactant $\ce{B}$, i.e. $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$. So, plugging $t=0$ and $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$ into the equation for $[\ce{B}]$ you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  [\ce{B}]_0 &amp;amp;= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{0} - \mathrm{e}^{0} \bigr) + \tilde{C} \mathrm{e}^{0} \\&#xA;  [\ce{B}]_0 &amp;amp;= \tilde{C} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your reaction $\ce{A -&amp;gt; B -&amp;gt; C}$ the initial concentration of $[\ce{B}]$ will usually be $0$ as $\ce{B}$ is only an intermediate product, so $\tilde{C} = [\ce{B}]_0 = 0$ and the final equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  [\ce{B}] &amp;amp;= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) \ .&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-25T11:47:20.617" LastActivityDate="2014-09-25T11:47:20.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16714" PostTypeId="2" ParentId="16703" CreationDate="2014-09-25T03:07:10.707" Score="2" Body="&lt;p&gt;I think my first question is why you need the crystal structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For organic molecules, the definitive resource is the proprietary (and fairly expensive) &lt;a href=&quot;http://www.ccdc.cam.ac.uk/pages/Home.aspx&quot; rel=&quot;nofollow&quot;&gt;Cambridge Structure Database&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://crystallography.net/&quot; rel=&quot;nofollow&quot;&gt;Crystallographic Open Database&lt;/a&gt; which you linked is fairly new and expanding, but does not have the many, many molecular crystals in the CSD. For example, American Chemical Society journals required submission to the CSD for publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to start with the &lt;a href=&quot;http://cactus.nci.nih.gov/chemical/structure&quot; rel=&quot;nofollow&quot;&gt;NIH Chemical Resolver&lt;/a&gt; or &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/&quot; rel=&quot;nofollow&quot;&gt;PubChem&lt;/a&gt;. The latter has a 3D computed structure of &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=1548970#itabs-3d&quot; rel=&quot;nofollow&quot;&gt;(&lt;em&gt;R&lt;/em&gt;)-fluoxetine&lt;/a&gt;, which I'm sure is pretty decent. Jmol should be able to load these directly, or you could save SDF files and open them in Jmol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll point out that &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow&quot;&gt;Avogadro&lt;/a&gt; can also do these tasks, but since I'm the creator, I'm biased.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-25T03:07:10.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16715" PostTypeId="1" AcceptedAnswerId="16721" CreationDate="2014-09-25T03:20:00.997" Score="11" ViewCount="787" Body="&lt;p&gt;I understand the difference between empirical and chemical formula. But I always thought that compounds and molecules took their specific shape because it's the lowest energy configuration. How then, can fructose ($\ce{C6H12O6}$) and formaldehyde ($\ce{CH2O}$) both exist? &#xA;If for example formaldehyde had overall the smaller energy, why then wouldn't fructose just decompose in 6 $\ce{CH2O}$ molecules? What makes both of those molecules so stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is a very general question of what makes certain elements combine the way they do to form compounds. &lt;/p&gt;&#xA;" OwnerUserId="8038" LastEditorUserId="5456" LastEditDate="2014-09-25T04:03:03.037" LastActivityDate="2014-09-25T15:37:10.147" Title="By what mechanisms can molecules with the same empirical formula be so different?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16716" PostTypeId="1" CreationDate="2014-09-25T03:22:00.577" Score="2" ViewCount="1447" Body="&lt;p&gt;Traces of nitrogen make diamonds yellow while traces of boron make them blue.  What accounts for the differences in color for a given element?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="189" LastEditDate="2014-09-26T23:53:26.460" LastActivityDate="2014-09-26T23:53:26.460" Title="Why do different impurities create different colors in diamonds?" Tags="&lt;electronic-configuration&gt;&lt;materials&gt;&lt;solid-state-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16717" PostTypeId="1" AcceptedAnswerId="16743" CreationDate="2014-09-25T03:26:07.257" Score="1" ViewCount="631" Body="&lt;p&gt;AESA (Actively Electronically Scanned Array) radars use thousands of individual transmit/receive elements to create a radar.  All in service AESAs use gallium arsenide as the material in the t/r elements.  It is said that gallium nitride allows a tenfold increase in the power emitted by these elements?  What allows GaN to emit so much more radio frequency energy than GaAs?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="189" LastEditDate="2014-09-25T04:01:48.110" LastActivityDate="2014-09-25T17:17:23.613" Title="Why does gallium nitride allow an AESA radar to be more powerful than a radar with elements made of gallium arsenide?" Tags="&lt;electronic-configuration&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16718" PostTypeId="2" ParentId="16709" CreationDate="2014-09-25T03:26:57.750" Score="6" Body="&lt;p&gt;Here is the reflectance spectrum for silver&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r0R4J.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It reflects the entire visible spectrum, absorbing light only in the ultra-violet region.  An ingot of elemental silver will have a silvery-white color.  In fact silver is said to be the &quot;whitest&quot; of all metals.  If you grind the silver up into smaller and smaller particles, the silver will appear progressively grayer or darker.  This is because smaller particles become less effective at reflecting light and produce more scattering.  Eventually when the particles are very small, even less light is reflected to the eye and the material appears black.  So depending upon the particle size, the color of silver can change from white to black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, silver can react with chemicals present in air and tarnish.  It is the reaction of silver with sulfides in the air to produce silver sulfide that turns silver black, but this is through a chemical reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-25T03:26:57.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16719" PostTypeId="2" ParentId="16716" CreationDate="2014-09-25T03:50:42.080" Score="5" Body="&lt;p&gt;There are two mechanisms by which diamonds can become colored.  The first mechanism involves impurities.  As the diamond crystal is growing, materials in the atmosphere or soil can become incorporated into the diamond structure.  The bonds that form between carbon atoms adjacent to the impurity and the impurity can absorb photons and result in electronic transitions.  If these transitions happen to occur in the visible region of the spectrum, then any reflected light will not contain the absorbed wavelength and the diamond will no longer appear white (as it does when all wavelengths are reflected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second mechanism that can impart color to diamonds involves structural deformation of the diamond crystal.  As the diamond crystal is growing, tectonic forces or radiation can cause the diamond crystal to grow in an irregular manner.  There are no impurities as described above, but a carbon atom may be missing here and there.  These structural anomalies also result in a shift in the frequency of light absorption by the diamond.  Again, if the light absorption shift is into the visible region, the diamond will appear colored.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-25T03:50:42.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16720" PostTypeId="2" ParentId="16716" CreationDate="2014-09-25T03:58:33.087" Score="9" Body="&lt;p&gt;If you had googled a little bit you would have found &lt;a href=&quot;http://www.webexhibits.org/causesofcolor/11A0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this wonderful resource&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a yellow diamond, a few carbon atoms per million have been replaced&#xA;  by nitrogen atoms, each containing five valence electrons. The&#xA;  structure of the diamond crystal does not change significantly, but&#xA;  the extra electrons occupy a donor level.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The nitrogen donor level energy in a diamond is large, peaking at&#xA;  about 4 eV. With a concentration of a few nitrogen atoms per million,&#xA;  instead of a clean &quot;spike&quot; donor level energy, the nitrogen donor&#xA;  level energy broadens into a band because of a number of complex&#xA;  factors, including thermal vibrations. This broadened donor energy&#xA;  band results in the difference between donor and conduction bands&#xA;  being as low as 2.2 eV. The most likely transition allows incident&#xA;  light with energy of 4 eV per photon to excite electrons from the&#xA;  donor level to the conduction band. However, it is possible for&#xA;  electrons to be excited to the conduction band with energy of 2.2 eV&#xA;  and upwards. This means that blue and violet light are absorbed from&#xA;  the full spectrum of light normally transmitted, and the resulting&#xA;  color is yellow.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYQf9l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unlike blue boron-doped diamonds, which conduct electricity,&#xA;  nitrogen-doped diamonds remain insulators. This is because nitrogen is&#xA;  a deep impurity: a relatively high energy (2.2 eV, as compared with&#xA;  0.4 eV for boron) is required to excite electrons from the donor energy level to the conduction band, and only a fraction of the&#xA;  available electrons are freed to carry a charge.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An extremely rare green color can result from a higher nitrogen&#xA;  content of about 1 atom per 1000 atoms of carbon. At even higher&#xA;  nitrogen concentrations, the donor level broadens so that all visible&#xA;  light can be absorbed, resulting in a black color.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Synthetic blue diamonds are created by adding boron as an impurity.&#xA;  Boron is trivalent, having three valence electrons; where a boron atom&#xA;  replaces a carbon atom in the diamond structure there is one fewer&#xA;  electron than usual. This missing electron, or hole, creates an&#xA;  acceptor energy level above the valence band. The boron acceptor&#xA;  energy is only 0.4 eV, so light of any energy can be absorbed during&#xA;  excitation. The boron acceptor band is broadened, and the absorption&#xA;  tapers off throughout the visible light energies, resulting in&#xA;  stronger absorption at the red end of the spectrum. At a level of one&#xA;  or a few boron atoms for every million carbon atoms, an attractive&#xA;  blue color results. Natural diamonds of this color, such as the Hope&#xA;  Diamond, are rare and highly priced.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/piKEh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-25T03:58:33.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16721" PostTypeId="2" ParentId="16715" CreationDate="2014-09-25T04:32:54.330" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What makes both of those molecules so stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a fundamental question. Let's start by looking at the following diagram; a chemist would call it a &quot;potential energy diagram.&quot; The y-axis shows the energy of a molecule as it travels along the reaction coordinate (x-axis; transforms, reacts) from the starting material on the left to the product on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/44DBX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that there are 3 &quot;potential energy wells&quot; (valleys) in the diagram, these reflect molecules that are stable enough to have a finite lifetime.  Notice too that to move from one well to another, an energy barrier must be surmounted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to return to your question, a molecule would be &quot;kinetically stable&quot; if it was protected on all sides by high energy barriers.  A molecule would be &quot;thermodynamically stable&quot; if its potential energy well was located at an energy that had a very low value on the y-axis.  A molecule that is very stable in a thermodynamic sense, could still be kinetically unstable (and therefore very reactive) if it was surrounded by low energy barriers.  By the same token, a molecule that is very unstable in a thermodynamic sense (exists at a high energy value on the y-axis) could be kinetically stable if it were surrounded by high energy barriers to reduce the likelihood of further reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Millions of molecules are stable at room temperature, but if you supply enough energy to the system (heat them for example) they can all transform into other compounds.  At room temperature there are high enough energy barriers separating fructose and formaldehyde and they will both be stable in a jar, but under more energetic conditions such transformations might be feasible.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-25T15:37:10.147" LastActivityDate="2014-09-25T15:37:10.147" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16723" PostTypeId="2" ParentId="16705" CreationDate="2014-09-25T09:29:10.547" Score="5" Body="&lt;p&gt;The most important radioactive silver nuclide is Ag-110m (half-life: 249.9 d), which is generated in nuclear reactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective dose coefficient of Ag-110m for ingestion by adult members of the public is 2.8E−09 Sv/Bq. However, this committed dose is evaluated over 50 years. In order to estimate the deterministic short-term effects of radiation sickness, only the dose that is actually accumulated during a few days is taken into account. For a period of seven days (time after intake), the effective dose coefficient of Ag-110m for ingestion by adults is 1.8E−09 Sv/Bq.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first symptoms of acute radiation syndrome may occur at a dose of about 1 Sv. Thus, you would need to eat at least 5.6E+08 Bq (or 560 MBq) of Ag-110m in order to find such symptoms within seven days after acute intake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding mass of pure Ag-110m amounts to 3.2 µg. However, any real sample of Ag-110m would be “diluted” with a larger amount of stable silver.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2014-09-25T15:23:50.250" LastActivityDate="2014-09-25T15:23:50.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16724" PostTypeId="2" ParentId="16715" CreationDate="2014-09-25T09:48:03.277" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess this is a very general question of what makes certain elements&#xA;  combine the way they do to form compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the fundamental level, I think, a good answer to this question would be the &lt;em&gt;principle of minimum potential energy&lt;/em&gt;, which basically states that a system has a &lt;em&gt;tendency&lt;/em&gt; to be in a configuration that &lt;em&gt;minimizes&lt;/em&gt; its total &lt;em&gt;potential energy&lt;/em&gt;. (Please, note the word tendency!)&#xA;For the case of molecules it means that if forming a molecule out of atoms lowers the potential energy comparing to the situation when these atoms are infinitely far apart of each other, than there is a natural tendency for this molecule to be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a one important detail that complicates matters: given some set of atoms, there is no guarantee that there exist only one possible arrangement of them which minimizes the potential energy. In fact, the situation is usually the opposite: there are &lt;em&gt;many different minima&lt;/em&gt; in the potential energy corresponding to different possible arrangements of the same set atoms. Often we can clearly identify one &lt;em&gt;global minimum&lt;/em&gt; (the lowest one) and call all other minima &lt;em&gt;local minima&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any local minima in potential energy (directly on indirectly, i.e. through other local minima) is connected with the global minimum in a sense that there exist some way for a system to transit from a local local minima to the global one. But on it is way to the global minimum from a local one potential energy first rises and only then falls. It falls to the lower value, but first the system need to overcome a barrier. Thus, we say that a structure that corresponds to a global minimum is &lt;em&gt;stable&lt;/em&gt;, while structures that correspond to local minima are &lt;em&gt;metastable&lt;/em&gt; in a sense that a system (a molecule) actually may reside in these local minima for quite a long time if the barriers are relatively high. Relatively to what? Basically, to the amount of kinetic energy a molecule posses so that it can be converted to the potential energy to overcome the barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good analogy may be drawn with a ball resting in a hollow on a slope (image below courtesy of Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iScBe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ball being in a hollow (1) does not have enough kinetic energy to be converted into potential with the aim to overcame the barrier (2) and move on to the lower energy minimum (3) it will stay at (1) for ages. But if the ball has enough kinetic energy it will roll to the lower level. Here (3) might be just another local minimum or the global minimum, it does not matter. If (3) is not the global minimum then this process is just one of the steps on a way to the global minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to molecules, think about (1) as of 6 molecules of formaldehyde being far apart of each other and about (3) as of 1 molecule of fructose. Both these different combinations of atoms correspond to minima in potential energy, and here I assumed that fructose is lower in potential energy. Note that I could not guarantee that, as well as, I could not guarantee that of of these minima is the global minimum for a system composed of 6 atoms of $\ce{C}$, 12 atoms of $\ce{H}$, and 6 atoms of $\ce{O}$. &lt;em&gt;Apriori&lt;/em&gt; we do not know what form is lower in potential energy, so we have to do some calculations, say. But here let me just assume that $\ce{C6H12O6}$ is lower in potential energy than $\ce{6 CH2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now recall that (as I noted at the very beginning) the principle of minimum potential energy only tells us that there is a &lt;em&gt;tendency&lt;/em&gt; for a system to favour a configuration with the lowest potential energy (i.e. $\ce{C6H12O6}$ out of two choices we discuss). But starting from $\ce{6 CH2O}$ the natural tendency to form $\ce{C6H12O6}$ would not necessarily be realized in reality if the barrier between these configurations is relatively high in a sense already mentioned. So, it might be the case that at, say, room temperature, most of the $\ce{CH2O}$ molecules do not posses enough kinetic energy to be converted to potential energy to overcame the barrier. But you can increase the amount if kinetic energy of $\ce{CH2O}$ molecules by rising the temperature, so that the system will roll to overcome the barrier. Chemical reaction will take place, although, I'm not sure that these two ($\ce{6 CH2O}$ and $\ce{C6H12O6}$ are directly connected). But conceptually, the situation is similar to just pushing the ball from the picture above. :D&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-09-25T10:14:19.023" LastActivityDate="2014-09-25T10:14:19.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16725" PostTypeId="1" AcceptedAnswerId="22076" CreationDate="2014-09-25T10:47:24.390" Score="7" ViewCount="6749" Body="&lt;p&gt;I'm a Mechanical Engineering student and I have been working on a school project but I want to go a bit beyond that and it has led me here. I'm designing an engine cylinder, and I figured I should be able to calculate the heat of combustion of the working fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please remember that I have no chemistry textbook and all I have been doing up to now is using the internet and some vague memory from high school.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wrote the balanced equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C8H18 + 12.5\left(O2 + 3.76 N2\right) -&amp;gt; 8 CO2 + 9 H2O + 23.5 N2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using that I was able to compute the air-to-fuel ratio, individual masses for the $1.8\ \mathrm L$ volume, etc. I settled on a compression ratio of 8 which after some thermodynamic computation (if neccessary I'll present them) I got to the following state,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_2 = 650.9 \; \text{K} \\&#xA;P_2 = 1779 \; \text{kPa} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically those are the conditions prior to the spark. I'm trying to compute the heat of combustion using those properties. Using Mathematica I found out that for Octane the heat of combustion is somewhere along,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta H=5474\;\frac{\text{kJ}}{\text{mol}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I assume this is for STP, my interest is to calculate for my conditions at state 2. After some bit of research I noticed one has to use the specific heats of the participants, but the website stated 'find these at the end of the book' – which I don't have one. To be honest, it will probably take me days to figure out this problem on my own using the internet. So does anyone know of a shortcut such as an online database or some public tables that display specific heats (maybe as a function of temp) for octane. Or how would you go about solving my problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that I don't have much knowledge regarding advanced chemistry, and most of my interest lies in the field of engineering.&lt;/p&gt;&#xA;" OwnerUserId="8039" LastEditorDisplayName="user7951" LastEditDate="2018-05-22T06:53:55.033" LastActivityDate="2018-05-22T06:53:55.033" Title="Heat of combustion for octane" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16726" PostTypeId="1" AcceptedAnswerId="16732" CreationDate="2014-09-25T11:54:50.557" Score="6" ViewCount="133" Body="&lt;p&gt;I am writing a lesson plan for high school students who are studying group theory,&#xA;and I am having difficulty in relating the subject matter to the real world.&#xA;I have quite a few examples using the Rubik's cube, but I would like to use&#xA;some statement that might provide some intuition to students of how group theory&#xA;is useful for chemists.&#xA;I understand that group theory is used with spectroscopy, but I am not sure how to &#xA;relate this back to the theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen some good reasonable non rigorous attempts to relate group theory to &#xA;the insolubility of quintic polynomials, but I think it would great to have a second &#xA;more concrete example, such as chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this may be a big thing to ask, and would be happy with just a broad brush strokes answer without being too technical or trivial.&lt;/p&gt;&#xA;" OwnerUserId="8040" LastEditorUserId="5017" LastEditDate="2014-09-25T13:27:05.953" LastActivityDate="2014-09-25T13:53:53.253" Title="Accessible statement on the intuition and usefulness of group theory to chemistry" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16727" PostTypeId="2" ParentId="14809" CreationDate="2014-09-25T12:44:29.950" Score="1" Body="&lt;p&gt;Not able to comment.... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main problem is propyl alcohol (not present in this cleaner). Propyl alcohol (propanol) is present in a lot of cleaning solutions and known to damage polycarbonate and acrylics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this chemical resistance sheet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.enista.com/nl/assets/Uploads/Chemische-resistentie-type-M.pdf&quot; rel=&quot;nofollow&quot;&gt;Sheet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;30% solution containing ethanol (ethyl alcohol) lightly attacks the plastic. However, a 10% (propanol) propyl alcohol already does the same damage. That is a factor 3!&lt;/p&gt;&#xA;" OwnerUserId="7993" LastEditorUserId="4934" LastEditDate="2014-10-25T14:40:00.227" LastActivityDate="2014-10-25T14:40:00.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16728" PostTypeId="1" CreationDate="2014-09-25T12:56:33.457" Score="0" ViewCount="125" Body="&lt;p&gt;is there a general word equation so I can learn how to write out the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aEXR2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With chlorine, it seems to be: Hydrocarbon + Chlorine ---&gt; ____ + HCL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Hexane, it seems to be: Hydrocarbon + Br ---&gt; ____ + HBr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to derive the first part of the final reaction (____)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that methane has lost one hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that hexane has lost one hydrogen atom too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean, if it was say Pentane with Chlorine, the first part of the final reaction would be CH4CL (not ch5cl) ? Likewise, would Pentane with Bromine be C6H12BR (and not c6h13br)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It feels like i'm rote-learning a skill, wonder where my knowledge is missing hmm..&lt;/p&gt;&#xA;" OwnerUserId="8014" LastActivityDate="2014-11-18T11:54:35.600" Title="Seeking word equations" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16729" PostTypeId="2" ParentId="16711" CreationDate="2014-09-25T13:05:21.350" Score="5" Body="&lt;p&gt;There are &quot;special rules&quot; for determining the formal charge in some situations.  Peroxides fall into one of these special cases, &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/redox/oxidnstates.html&quot; rel=&quot;noreferrer&quot;&gt;see here&lt;/a&gt; (see the section entitled &quot;Oxygen in peroxides&quot;).  Each oxygen in a peroxide is assigned a formal charge of -1.  Since sodium peroxide is neutral overall, each sodium atom is assigned an oxidation number of +1.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-25T13:05:21.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16730" PostTypeId="2" ParentId="16726" CreationDate="2014-09-25T13:33:46.373" Score="1" Body="&lt;p&gt;I realize my answer renders this off-topic for Chemistry, but what about general relativity? The fact that there is no &quot;special&quot; point in the universe is made concrete by saying that general relativity is invariant under the group of rigid motions (rotations and translations). That's an interesting real world group on its own.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-09-25T13:33:46.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16731" PostTypeId="2" ParentId="16728" CreationDate="2014-09-25T13:47:08.833" Score="1" Body="&lt;p&gt;I don't understand your question completely, but If you want to understand what's happening in an organic reaction you need to know which type of reaction is in question and you have to look at the mechanism. This is a &lt;a href=&quot;http://www.chemguide.co.uk/mechanisms/freerad/ch4andcl2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;radical reaction&lt;/a&gt; with the following mechanism (excluding the termination step):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9dy8F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-09-25T13:47:08.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16732" PostTypeId="2" ParentId="16726" CreationDate="2014-09-25T13:53:53.253" Score="6" Body="&lt;p&gt;Individual molecules rely on &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_symmetry&quot; rel=&quot;nofollow noreferrer&quot;&gt;point group symmetry&lt;/a&gt; - that is, the symmetry about a point. Crystal structures rely on &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;space group symmetry&lt;/a&gt; including the translational symmetry in a crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many molecules, particularly important ones, are highly symmetric. The different symmetry operations (reflections, rotations, etc.) form a finite group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of interesting group theory results, but a simple statement comes down to two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Understanding Orbitals, Bonding, Vibrations..&lt;/strong&gt;: Using point groups, any function of the molecule corresponds to a linear combination of irreducible representations. This means I can quickly understand vibrations of a molecule, spectroscopy (as you mentioned) and orbital diagrams and bonding using just pencil and paper.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Faster Calculations&lt;/strong&gt;: Using symmetry, you can greatly reduce the amount of time required for quantum calculations. For one, you only need to consider the symmetry-unique atoms and bonds. For another, you can show that certain types of integrals and calculations will always be zero and can be completely ignored. This leads to immense speedups.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, chemists label orbitals with symmetry labels based on the point group. Using the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16547/what-is-the-product-of-two-irreducible-representations-on-the-character-table&quot;&gt;direct product&lt;/a&gt; we can realize that certain interactions will never occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, vibrations have symmetry labels, and we can determine what types of spectroscopy will interact with those vibrations, or whether they are &quot;silent.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-09-25T13:53:53.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16733" PostTypeId="2" ParentId="16711" CreationDate="2014-09-25T13:55:38.107" Score="4" Body="&lt;p&gt;The structure of peroxides - $\ce{ROOR}$ - implies that the oxygens will generally exhibit -1 oxidation states. Oxidation states are assigned as if all bonding is ionic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that oxygen is the second most electronegative element on the periodic table, we would expect that all the electrons in the $\ce{R-O}$ bond to be &quot;taken&quot; by oxygen (unless $\ce{R=F}$ as in the case of $\ce{FOOF}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\ce{O-O}$ bond, however, the electrons are split 50/50 between the two oxygens (since both are of equal atomic electronegativities). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore each oxygen has an oxidation state of -1 in peroxides (usually): each oxygen has 3 bonding electrons and 4 lone pair electrons for a total of 7 electrons, and oxygen by itself only has 6 electrons. 6 minus 7 is -1. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-25T15:47:47.613" LastActivityDate="2014-09-25T15:47:47.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16735" PostTypeId="1" AcceptedAnswerId="16736" CreationDate="2014-09-25T14:02:42.257" Score="6" ViewCount="7340" Body="&lt;p&gt;I'm trying to understand what halogens do in electrophilic aromatic substitution. Consider the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromobenzene is introduced to a carbocation. Bromobenzene will likely react, but at a slower rate than unsubstituted benzene, because the bromine is inductively withdrawing. The bromine is also &lt;em&gt;not&lt;/em&gt; resonance donating/activating, but can stabilize the intermediate product through resonance donation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bromine is an ortho/para director, &lt;strong&gt;but&lt;/strong&gt; not because it donates electron density to these positions. Rather, when bromobenzene attacks a carbocation, the meta positions are formally positive, and through resonance donation, bromine &lt;em&gt;can&lt;/em&gt; complete an octet in one of the resonance depictions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/16Xgp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an accurate depiction of what a halogen does to benzene? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-04-10T06:51:02.200" LastActivityDate="2018-04-10T06:51:37.897" Title="The role of halogens in electrophilic aromatic substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;halides&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16736" PostTypeId="2" ParentId="16735" CreationDate="2014-09-25T14:45:53.427" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The bromine is also not resonance donating/activating&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't have it both ways.  If the bromine can resonance interact with the intermediate, then it can also resonance interact in the starting bromobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's back up to the starting bromobenzene. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bromobenzene will likely react, but at a slower rate than plain&#xA;  benzene, because the bromine is inductively withdrawing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct, but let's compare further.  The oxygen in phenol is electron withdrawing (oxygen's electronegativity is even greater than bromine's), yet phenol is a very reactive aromatic compound towards electrophilic aromatic substitution (it will decolorize a solution of bromine, it doesn't even need a Lewis acid catalyst).  The difference between bromobenzene and phenol is that the phenolic oxygen can interact strongly &lt;em&gt;via&lt;/em&gt; resonance and donate electrons into the aromatic ring, while the bromine in bromobenzene also resonance interacts with the aromatic ring, but much less efficiently (the larger size of bromine's $\ce{4p}$ orbitals makes overlap much less efficient with carbon's 2p orbitals).  Although the interaction is much less efficient, there still is some resonance interaction between the bromine substituent and the ring.  Look at the analogous resonance structures for bromobenzene that you drew for the intermediate, in the contest between inductive deactivation and weak resonance activation, the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions are the &lt;strong&gt;least deactivated&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZY05d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When bromobenzene undergoes electrophilic aromatic substitution, the whole ring is deactivated because (unlike phenol, where the resonance effect overwhelms the inductive effect) the weak resonance effect doesn't offset the strong inductive effect.  However, since the bromine substituent does interact with the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions by resonance, these two positions are less deactivated that the &lt;em&gt;meta&lt;/em&gt; position which cannot interact &lt;em&gt;via&lt;/em&gt; resonance with the bromine substituent.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-04-10T06:51:37.897" LastActivityDate="2018-04-10T06:51:37.897" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16742" PostTypeId="1" CreationDate="2014-09-25T17:11:22.717" Score="0" ViewCount="143" Body="&lt;p&gt;For up to period 3, elements' electron configurations are relatively easy to work out until the transition metals begin. Then some elements may have $\ce{4s}$ orbitals which are preferred by electrons over $\ce{3d}$ orbitals. These small rules don't make sense all the time and only cover some elements. Is their any &lt;em&gt;master&lt;/em&gt; method for working out the electron configurations of &lt;em&gt;any&lt;/em&gt; given element?&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="14" LastEditDate="2014-09-25T17:31:53.367" LastActivityDate="2014-09-26T14:55:09.580" Title="Is there a general method for working out electron configurations?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16743" PostTypeId="2" ParentId="16717" CreationDate="2014-09-25T17:17:23.613" Score="3" Body="&lt;p&gt;The short answer is that GaN transistors have a much higher breakdown voltage and saturation velocity than GaAs, allowing them to operate at higher voltages at high frequency, handling more power. The large band gap (~2x GaAs) means that GaN still behaves like a semiconductor, even at relatively high temperatures, allowing for higher operating powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GaN has really been a breakthrough in semiconductor technology. It's what allows us to have 405 nm violet laser diodes and could potentially replace magnetrons in microwave ovens. NXP has a little document &lt;a href=&quot;http://www.nxp.com/documents/external_articles/NXP_goes_with_GaN_July_2011.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; comparing GaN with a few older technologies that's a quick read.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-25T17:17:23.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16747" PostTypeId="1" CreationDate="2014-09-25T18:44:25.013" Score="2" ViewCount="1032" Body="&lt;p&gt;$\ce{W(CH3CN)3(CO)3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking: &#xA;$\ce{CO}$ is neutral so $\ce{O}$ is -2 and $\ce{C}$ +2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CN}$ is negative one so $\ce{C}$ is +2 and $\ce{N}$ -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3}$ is positive one so $\ce{H}$ is +1 and $\ce{C}$ is -2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{W}$ is neutral so $\ce{W}$=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this right?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="4231" LastEditDate="2014-09-25T22:10:33.470" LastActivityDate="2015-02-14T08:18:36.827" Title="What is the oxidation state of all the atoms in this compound, W(CH3CN)3(CO)3?" Tags="&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16748" PostTypeId="1" AcceptedAnswerId="16750" CreationDate="2014-09-25T19:25:29.563" Score="1" ViewCount="949" Body="&lt;p&gt;I was doing some problems from the textbook on converting between units using dimensional analysis and I came across this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains&#xA;455 grams of oxygen atoms. How many&#xA;molecules of ($\ce{H2O2}$) are in the vat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if I didn`t format the question correctly, I am a still a newbie.&#xA;But how would I set this up? I would not like the answer to this question but I would like some input on how would I get how many molecules of &lt;em&gt;hydrogen peroxide&lt;/em&gt; from grams of &lt;em&gt;oxygen atoms&lt;/em&gt;? Any help would be much appreciated. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5409" LastEditorUserId="4231" LastEditDate="2014-09-25T19:39:21.497" LastActivityDate="2014-09-25T19:39:21.497" Title="Converting from Grams to Molecules" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16750" PostTypeId="2" ParentId="16748" CreationDate="2014-09-25T19:37:52.437" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains 455 grams of oxygen atoms.&#xA;  How many molecules of ($\ce{H2O2}$) are in the vat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many moles of oxygen atoms are in 455 gms of oxygen atoms? Let's call the answer $x$.&lt;/li&gt;&#xA;&lt;li&gt;How many moles of $\ce{O2}$ are in $x$?  Let's call the answer $y$.&lt;/li&gt;&#xA;&lt;li&gt;How many moles of $\ce{H2O2}$ could you make from $y$ moles of $\ce{O2}$?  Let's call the answer $z$.&lt;/li&gt;&#xA;&lt;li&gt;1 mole of any substance contains Avogadro's number of molecules.  How many molecules are in $z$ moles?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-25T19:37:52.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16751" PostTypeId="2" ParentId="16664" CreationDate="2014-09-25T21:33:31.050" Score="3" Body="&lt;p&gt;Neither of the previous 2 answers mention a phosphorus flame, which behaves visibly like other flames we are familiar with, but burns at a temperature so cool that you can hold your finger in the flame without getting burned. This only works with white (not red or black) phosphorus. It is (with proper precautions!!) placed into the chamber of an alcohol burner or similar device with the wick removed from the vent tube and warmed gently on a hot plate. The phosphorus (P4) vapor will, if ignited (maybe spontaneously), burn with a very cool flame. The same holds true for phosphine (PH3)and related low molecular weight phosphorus hydrides.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-09-25T21:33:31.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16752" PostTypeId="2" ParentId="16705" CreationDate="2014-09-25T21:42:21.477" Score="2" Body="&lt;p&gt;The energy of radioactive emission does depend on nucleus decaying. This, however, does not depend only on number of protons in the nucleus, so the fact that silver is, well, silver, does not make its radioactive isotopes less or more dangerous if they somehow enter biological system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it must be noted, that if a radioactive nucleus belongs to element/compound that can deposit/accumulate in human system, its damage potential increases as it stays there longer. This applies to elements/compounds that can deposite in bones (calcium, stroncium, phosphorous) or accumulate by different means (mercury, some other heavy metals, some hydrocarbons. And yes, silver also can accumulate in animals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another particulary nasty habit is capability to produce radioactive gases, usually radon. Unlike most common radioactive elements, radon can be inhaled, effectively bypassing skin, that is some (even if little) defense against radioactivity. This is why uranium and torium mines must have especially effective ventilation systems.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-25T21:42:21.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16753" PostTypeId="1" CreationDate="2014-09-26T01:33:02.773" Score="2" ViewCount="7823" Body="&lt;p&gt;I'm confused, in the reaction:$$\ce{K_2CrO_{4(aq)} + Ba(NO_3)_{2(aq)}\rightleftharpoons BaCrO_{4(s)} + 2KNO_{3(aq)}}$$&#xA;I got a white precipitate in an opaque yellow solution and I was told that the white precipitate was $\ce{BaCrO_4}$ but while searching on Internet I found that its color is yellow, and that the $\ce{KNO_{3(aq)}}$ is white, so my question is what is true? or what causes the color of the final solution?&lt;/p&gt;&#xA;" OwnerUserId="7113" LastEditorUserId="186" LastEditDate="2014-09-29T09:19:06.243" LastActivityDate="2014-11-21T19:15:41.463" Title="reaction with Barium Chromate, Potassium nitrate" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16754" PostTypeId="1" CreationDate="2014-09-26T01:37:35.220" Score="2" ViewCount="1634" Body="&lt;p&gt;I was given $\ce{K2CrO4}$ and was asked to name the compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Determining the molecular compounds oxidation state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does one need to memorize the polyatomic ions and that is the only way to know what is what? &lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorDisplayName="user8016" LastEditDate="2014-09-26T08:01:06.240" LastActivityDate="2014-09-26T09:35:29.503" Title="Why is K2CrO4 Potassium chromate?" Tags="&lt;everyday-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16755" PostTypeId="2" ParentId="16613" CreationDate="2014-09-26T02:14:18.343" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gmelin_database&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gmelin&lt;/a&gt; is supposed to be the largest database of inorganic/organometallic chemical information (as Beilstein is for organics). Both are now part of Elsevier's &lt;a href=&quot;https://www.elsevier.com/solutions/reaxys&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reaxys&lt;/a&gt;, which is as far from open source as you can get. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PubChem does list &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pccompound?term=%22BORON%20TRICHLORIDE%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;inorganic molecules&lt;/a&gt; and &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pccompound/?term=%22Wilkinson%27s%20catalyst%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;coordination compounds&lt;/a&gt;, but I am not sure how easy it is to systematically mine them. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="41328" LastEditDate="2017-08-25T23:39:59.227" LastActivityDate="2017-08-25T23:39:59.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16756" PostTypeId="2" ParentId="16754" CreationDate="2014-09-26T05:17:06.683" Score="3" Body="&lt;p&gt;Definitely you need to remember it. Potassium Chromate  is IUPAC name of $\ce{K2CrO4}$. There are other names such as Chromic acid and dipotassium salts. But mostly IUPAC names are accepted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you should try to mug up all 24 names instead you should keep getting familiar with them when they come in reactions or somewhere else. After seeing them at few different places you will be remember them all. They will come very often while you are learning chemistry.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-09-26T05:45:13.467" LastActivityDate="2014-09-26T05:45:13.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16757" PostTypeId="1" AcceptedAnswerId="16759" CreationDate="2014-09-26T05:40:06.430" Score="17" ViewCount="9312" Body="&lt;p&gt;Why is aniline aromatic? Doesn't it have 8 π electrons including the lone pair on nitrogen, thereby violating Hückel's rule? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zcj0W.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zcj0W.png&quot; alt=&quot;Structure of aniline&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see it, there are 6 π electrons from the benzene ring, and an additional 2 from the $\mathrm{sp^2}$hybridized nitrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am suspecting that I am mis-applying Hückel's rule. I asked an instructor today and the only rationalization he had to offer was that &quot;we just don't count those electrons on the nitrogen&quot; ... which is deeply unsatisfying. He still insisted that aniline is aromatic. Can someone offer an alternate, more through explanation? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-01-07T02:50:06.983" LastActivityDate="2018-01-07T02:55:12.210" Title="Do lone pairs on substituents (e.g. in aniline) count towards Hückel's rule?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;aromaticity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16758" PostTypeId="1" AcceptedAnswerId="16766" CreationDate="2014-09-26T05:51:05.273" Score="3" ViewCount="4826" Body="&lt;p&gt;Given the reaction between $\ce{CH_{3}COOH {(l)}}$ and $\ce{NaHCO_3 {(s)}}$, which happens at the same pressure but at different temperatures, is there any way to mathematically compare the rate of reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;rate equations of order 2&lt;/a&gt;, but I use the same concentration of materials. Suppose I could have found the connection between temperature of he material and its concentration, I would be able to find my required rate-indicators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to find the connection between temperature and concentration, or how to describe the rate of the reaction more exactly?&lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="5026" LastEditDate="2018-04-10T06:48:08.233" LastActivityDate="2018-04-10T06:48:55.840" Title="Comparing rate of reaction of acetic acid and sodium bicarbonate at different temperatures" Tags="&lt;acid-base&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16759" PostTypeId="2" ParentId="16757" CreationDate="2014-09-26T06:01:04.767" Score="17" Body="&lt;p&gt;It is only the electrons from atoms &lt;strong&gt;in&lt;/strong&gt; the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for only using the ring electrons traces back to the origin of aromaticity. Take a look at the π-orbital energy levels of benzene (figure taken from p 36 of Fleming, &lt;em&gt;Molecular Orbitals and Organic Chemical Reactions, Reference Edition&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LeTAz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LeTAz.png&quot; alt=&quot;Frost circle for benzene pi-type MOs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\alpha$ is the energy of an isolated p-orbital and $\beta$ is the stabilization energy of the π-bond in ethylene. You can see that for benzene you get the best stabilization if you have 6 π electrons ($4n + 2$ for $n=1$) in the system because then you fill the stabilized MOs $\psi_1$, $\psi_2$ and $\psi_3$ which are lower in energy than an isolated p-orbital and gain the maximum stabilization energy possible. If you'd have more electrons in the system you would have to fill the destabilized MOs $\psi^{*}_{4}$, $\psi^{*}_{5}$ and $\psi^{*}_{6}$ which would cost a lot of energy because of the large energy gap between the bonding and antibonding MOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens if you add a cross-conjugating substituent to the ring. This can be seen in the (Huckel-)MO-diagram of styrene (Fleming, p 71).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z5UVA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z5UVA.png&quot; alt=&quot;Pi-type MOs of styrene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main effects of the substituent are that the degeneracies of some orbitals are lifted and that the orbital energies shift a bit, especially those of HOMO and LUMO are of interest. In the case of styrene the HOMO lies higher in energy and the LUMO lies lower in energy compared to benzene. Now there are 4 stabilized energy levels instead of 3 - any electron that occupies one of those levels will lead to a stabilization of the molecule. But that is nice and well as you get 2 electrons from the substituent which will want to occupy one of the 4 levels and that leaves you with 3 levels that have to be occupied by ring electrons. Thus, your system will be most stable if the π-system of the ring contains 6 electrons, exactly like in benzene. Any more than 6 electrons will lead to destabilization as there is still quite a big HOMO–LUMO gap (even though it is smaller than in benzene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument I made for styrene will remain valid for other molecules with different substituents (like aniline). The MO scheme of aniline will look a little different but the essential features should be the same.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-01-07T02:49:55.040" LastActivityDate="2018-01-07T02:49:55.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16760" PostTypeId="1" CreationDate="2014-09-26T07:30:14.140" Score="4" ViewCount="3954" Body="&lt;p&gt;So any data sheet will tell you; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Group 1: methane boils at: -164 deg C&#xA;         methanol boils at: 65 deg C &#xA;&#xA;&#xA;Group 2: decane boils at: 174 deg C&#xA;         1-decanol boils at 229 deg C &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;that being said; why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well is there anything else other than difference in non-polarity and IMF's that contribute to differences between molecules in individual groups (i.e. between methane and methanol; between decane and 1-decanol) &lt;/p&gt;&#xA;" OwnerUserId="8057" LastEditorUserId="5456" LastEditDate="2014-09-26T09:35:46.523" LastActivityDate="2014-09-26T10:57:27.520" Title="Why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane?" Tags="&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16762" PostTypeId="2" ParentId="16754" CreationDate="2014-09-26T09:35:29.503" Score="1" Body="&lt;p&gt;2 - Yes and no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is recommended to known common ions. They are, however, derived in uniform fashion from name of the element in latin. For example, for $Fe$ the name is Ferrum so anions are Ferrates. The 'fancy' part of the anion may consist either of naming the oxidation state (and optionally the number of oxygens and other atoms), i.e. (tetraoxo)ferrate(VI) or known suffixes and prefixes&#xA;(hypochlorite-chlorite-chlorate-perchlorate, also hydro-, pyro-, orto- and meta- prefixes). Probably the easiest way to remember the second nomenclature would be to learn several (tens of) common ions and assume that isoelectronic ions have same prefixes and suffixed, but are derived from names of respective elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, even though the first nomenclature is normative and is understood universally, the second is traditional and more common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some oxoions, however, have special names, like dithionite-ion.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-26T09:35:29.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16763" PostTypeId="2" ParentId="16760" CreationDate="2014-09-26T10:57:27.520" Score="3" Body="&lt;p&gt;The difference in boiling point, between methanol and methane on one hand and between 1-decanol and decane on the other hand, is primary due to the presence of hydrogen bonds (HB) in methanol and 1-decanol and not in methane and decane. The density of these intermolecular forces (HB) is higher in methanol than in 1-decanol (In 1-decanol, for every ten atoms of carbon there's one possibility to have HB. While in methanol for every atom of carbon there's one  possibility to have HB.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-26T10:57:27.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16764" PostTypeId="2" ParentId="16757" CreationDate="2014-09-26T11:15:26.720" Score="5" Body="&lt;p&gt;I agree with you that in aniline there are 6 π electrons from the benzene ring, and an additional 2 from the lone pair of the amine group. But aniline is still aromatic, because there are only 6 π electrons resonating, i.e. the 6 π electrons in the benzene ring are conjugated in a plane perpendicular to the ring. While the additional lone pair of electron on the nitrogen atom cannot resonate with them, as it is not in that plane perpendicular to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to notice that the nitrogen atom in aniline is not $\mathrm{sp^2}$ hybridized but $\mathrm{sp^3}$.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16683" LastEditDate="2018-01-07T02:55:12.210" LastActivityDate="2018-01-07T02:55:12.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16765" PostTypeId="2" ParentId="16757" CreationDate="2014-09-26T13:03:14.070" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel&amp;#39;s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hückel's rule&lt;/a&gt; (or the $4n+2$ rule) suggests that systems with $4n+2$ π electrons will be stabilized (aromatic), while systems with $4n$ π electrons will be destabilized (antiaromatic).  It was written with planar, monocyclic, π-systems in mind.  Applied without care, it would suggest that &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclooctatetraene&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclooctatetraene&lt;/a&gt; with 8 π electrons is antiaromatic, yet the molecule is a stable, ethylenic like, cyclic, polyolefin - because it is non-planar (it exists in a tub shape conformation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n2ovb.png&quot; alt=&quot;Conformation of cyclooctatetraene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hückel rule also has trouble with polycyclic π systems, as we discussed in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16562/is-pyrene-aromatic/16567#16567&quot;&gt;your question about pyrene&lt;/a&gt;.  The rule was not intended to take substituents into account at all.  We can say that furan or pyrrole has aromatic properties because the lone pair on oxygen or nitrogen contributes to the π-system.  But note that in these cases the oxygen or nitrogen is part of the monocyclic ring, not a substituent appended to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of aniline, the molecule is aromatic. The benzene ring is, in agreement with Hückel's rule, aromatic. The lone pair of electrons on the amino substituent clearly interact with the π system of the benzene ring as evidenced by the reduced basicity of aniline in comparison to ammonia. But the amino group is a substituent attached to the ring and the lone pair on nitrogen is not counted as part of the $4n+2$ system. The same concept applies to phenol, halobenzenes, etc.: these substituents interact with the benzene ring, but their lone pairs are not part of the electron count when Hückel's rule is applied. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-01-07T02:53:16.820" LastActivityDate="2018-01-07T02:53:16.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16766" PostTypeId="2" ParentId="16758" CreationDate="2014-09-26T13:28:30.737" Score="6" Body="&lt;p&gt;Yes, there is, it has been done by others.  Typical acid-base reactions that occur in solution are &lt;a href=&quot;http://en.wikipedia.org/wiki/Diffusion-controlled_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;diffusion controlled reactions&lt;/a&gt;. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. The reaction rate is limited only by how fast the acid and base can diffuse and encounter each other. It is usually easy to identify diffusion controlled reactions because they are extremely dependent upon the rate of stirring or mixing.  They are also very sensitive to the solution viscosity. Rate constants for these reactions are typically $&amp;gt;\pu{10^11 sec^-1}$.  Here is a &lt;a href=&quot;http://www.vanderbilt.edu/langlab/Publications/EPines-etal(1997).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to a full paper&lt;/a&gt; that provides considerable experimental detail on one way to measure the rate of diffusion controlled acid-base reactions. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-04-10T06:48:55.840" LastActivityDate="2018-04-10T06:48:55.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16767" PostTypeId="1" AcceptedAnswerId="16771" CreationDate="2014-09-26T13:38:48.423" Score="11" ViewCount="2458" Body="&lt;p&gt;Is the move from pi bonds to sigma bonds &lt;em&gt;always&lt;/em&gt; favorable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor claims so, except in the case of conjugated pi bonds. I can see how pi bonds going to sigma bonds might be favorable; I know that double bonds are actually not twice as strong as single bonds because of poorer overlap ... but are there any exceptions to this guideline that the reduction of pi bonds to sigma bonds is favorable except in the case of conjugated systems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: what about in the case of organic chemistry? It certainly seems that &lt;strong&gt;hydrogenation&lt;/strong&gt; (a way of transforming pi bonds to sigma bonds) is generally a favorable move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BmVb5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="14" LastEditDate="2014-09-26T14:39:45.750" LastActivityDate="2014-09-26T21:33:25.287" Title="Is the transformation of pi bonds to sigma bonds always energetically favorable?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;energy&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16768" PostTypeId="2" ParentId="16767" CreationDate="2014-09-26T13:54:59.417" Score="5" Body="&lt;p&gt;This is certainly not generally true. Have a look &lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for example: There you find that the bond energy for an $\ce{O-O}$ single bond is 142 kJ/mol while that of an $\ce{O=O}$ double bond is more than thrice as large with 494 kJ/mol. So, a reduction of $\ce{O=O}$ that would produce two $\ce{O-O}$ bonds would not be favorable. The same is true for $\ce{N-N}$ (167 kJ/mol) and $\ce{N=N}$ (418 kJ/mol) bonds. For nitrogen you even have the rare case that the $\ce{N#N}$ triple bond with 942 kJ/mol packs more bond strength than two isolated $\ce{N=N}$ double bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of organic chemistry you often have the situation that a reaction destroys a $\pi$ bond while in the process creating a new $\sigma$ bond. A double bond will always be stronger than a single bond, so going from double to single costs energy in any case. But usually when you reduce a double or triple bond you create new bonds at the expense of them. Only when those new bonds + the reduced bond are stronger than the old double/triple bond then the reaction is favorable. There are reactions where the opposite is the case. Phosphorous is infamous for its affinity to oxygen and very prone to form $\ce{P=O}$ double bonds. In the Wittig reaction (see for example &lt;a href=&quot;https://chemistry.stackexchange.com/q/5313/189&quot;&gt;here&lt;/a&gt;) you have an oxaphosphetane intermediate with a strained 4-membered ring with 4 $\sigma$ bonds that springs open to form one $\ce{C=C}$ and one $\ce{P=O}$ double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another case comes to mind. Hemiacetals (or 1,1-geminal diols), having two $\ce{C-O}$ $\sigma$ bonds, are usually unstable and prefer to destroy one of the $\ce{C-O}$ bonds by expelling the $\ce{R-O}$ group (or one of the $\ce{O-H}$ groups) thus forming a $\ce{C=O}$ bond. So, in this case one $\ce{C=O}$ bond beats two $\ce{C-O}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to read more on the matter of the relation between single and multiple bonds concerning the strengths of the $\sigma$ and the $\pi$ interactions have a look at &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.198402721/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper by Kutzelnigg.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-26T15:43:24.770" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16769" PostTypeId="1" AcceptedAnswerId="16773" CreationDate="2014-09-26T13:55:31.980" Score="11" ViewCount="3841" Body="&lt;p&gt;I came across a Frost diagram for cyclic compounds in my book, and all my book had to offer was that it geometrically reproduces the solutions of the wave equation, and can therefore determine the relative energies of each pi molecular orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How, exactly, does a Frost diagram work? My book was of no use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWcH1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6y1EA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-01-07T14:28:48.360" LastActivityDate="2018-01-07T14:31:04.600" Title="How does a Frost diagram reproduce the solutions to the wave equation?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16770" PostTypeId="2" ParentId="16767" CreationDate="2014-09-26T13:58:43.853" Score="7" Body="&lt;p&gt;When thinking about questions along these lines, I always try to remember the bent-bond description of ethylene.  In all aspects related to physical reality (structure, electron distribution, etc.) the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJHl0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;nice&quot; thing about the bent bond model is that it casts ethylene as a highly-strained, two-membered ring.  When wouldn't releasing this strain and forming sigma bonds be an energy lowering process.  Aside from conjugated examples, I think that transformation of a strained, ethylenic system to a saturated, sigma-bonded system would always be energy lowering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Further Thoughts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of analyzing hydrogenations or other reactions where the starting material and product have a different number of atoms, let's consider the following isomerization reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6qCcj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically, cyclohexane is the most stable $\ce{C6H12}$ isomer in this series by a significant margin.  All we've done is convert a C-C pi bond into a C-C sigma bond and convert 3 $\ce{C(sp^2)-H}$ bonds into 3 $\ce{C(sp^3)-H}$ bonds.  Which (at least as I read it) was the OP's premise.  So I would say that, in carbon-hydrogen systems, pi bonds generally prefer to transform into sigma bonds, as pi bonds are inherently destablilized due to poor overlap (pi model) or strain (bent bond model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I retract what I wrote above that carbon-carbon pi to sigma transformations would &quot;always&quot; be energy lowering.  Consider the photolysis of benzene to produce dewar benzene, prismane and benzvalene.  These conversions are examples where one or more pi bonds have been converted to one or more sigma bonds and the reaction is not energetically (thermodynamically) favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HqKky.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The products are much less stable (due to the strain in the small rings that have been created) than the starting benzene.  Because of these relative stabilities, the forward reaction does not occur thermally (in fact, thermally, each of those 3 products will revert to benzene), but by pumping in excess energy photochemically, the transformation can be brought about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, I would maintain that generally, in model, carbon-hydrogen systems, ethylenic linkages are destabilized (due to poor overlap in the pi model, or strain in the bent bond model) and will thermodynamically prefer to exist as thermodynamically more-stable saturated linkages.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-26T21:33:25.287" LastActivityDate="2014-09-26T21:33:25.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16771" PostTypeId="2" ParentId="16767" CreationDate="2014-09-26T14:01:45.217" Score="5" Body="&lt;p&gt;The answer to your question is no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of organic chemistry, consider the two hydrocarbons, $\ce{H_3C-CH_3}$ and $\ce{HC\equiv CH}$.  The $\ce{C-C}$ bond dissociation energy is 368 kj/mol while the $\ce{C\equiv C}$ triple bond has a bond dissociation energy of 682 kj/mol, nearly twice that of the single bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation of this triple to single bond may be thermodynamically favorable via hydrogenation but may not &lt;em&gt;always&lt;/em&gt; be the case.  How does temperature affect hydrogenation and can $\Delta H$ be positive under other conditions? Furthermore, if we want to talk about the 'strength' of the CC bond, then clearly the triple bond is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phillip brought up a good point in the comments that merits attention.  Hydrogenation contains an appreciable barrier for this process to occur.  When we take into account the formation of the additional single bonds by hydrogen then the energy is lower.  However, if we consider &lt;strong&gt;strictly&lt;/strong&gt; a pi to sigma transformation (between two carbons in this example) &lt;strong&gt;and not the formation of other bonds&lt;/strong&gt;, based on the energetics of the bond we are transforming, &lt;strong&gt;the energy will increase.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference for bond energies: &lt;a href=&quot;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2014-09-26T14:44:06.200" LastActivityDate="2014-09-26T14:44:06.200" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16773" PostTypeId="2" ParentId="16769" CreationDate="2014-09-26T14:29:44.733" Score="10" Body="&lt;p&gt;Frost developed this mnemonic patterning as an extension of the Hückel ($4n+2$) rule. A Frost diagram is usually applied to all-carbon, monocyclic, π systems.  It allows one to find the number of molecular orbitals in the molecule's π system and their energetic positions. To construct a Frost diagram, proceed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n3wf4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n3wf4.png&quot; alt=&quot;Example Frost diagrams&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Draw a circle and inscribe a regular polygon with a vertex located at the bottom of the circle. The polygon has the same shape as the ring you are interested in. For example, if you are interested in benzene, draw a hexagon; for the tropylium ion, draw a heptagon.&lt;/li&gt;&#xA;&lt;li&gt;Energy-wise, the top and bottom of the circle are defined as $\alpha+2\beta$ and $\alpha-2\beta$ respectively (so the circle has radius $2\beta$); the center of the circle is located at $\alpha$; other points can be interpolated accordingly; the bottom of the circle is at lower energy than the top of the circle.&lt;/li&gt;&#xA;&lt;li&gt;Wherever a vertex of the polygon touches the circle, that is the energetic location of a molecular orbital.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using benzene as an example, the lowest MO has energy $\alpha-2\beta$; the HOMO is degenerate (2 MO's) and located at $\alpha-\beta$; the LUMO is also degenerate and located at $\alpha+\beta$.  Any orbital below the center of the circle is bonding, any orbital at the center is non-bonding and any orbital in the top-half of the circle is antibonding.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-01-07T14:27:17.853" LastActivityDate="2018-01-07T14:27:17.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16774" PostTypeId="2" ParentId="16742" CreationDate="2014-09-26T14:55:09.580" Score="2" Body="&lt;p&gt;There is a semi-empirical method to fill the atomic orbitals: atomic orbitals are filled  in the order of increasing &lt;em&gt;n&lt;/em&gt;+&lt;em&gt;l&lt;/em&gt;. If the sum &lt;em&gt;n&lt;/em&gt;+&lt;em&gt;l&lt;/em&gt; is equal for two atomic orbitals, they are filled in order of increasing n.&#xA;This gives the following order for filling the orbitals:&#xA;1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, 5g, 6f, 7d, 8p, and 9s&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-09-26T14:55:09.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16775" PostTypeId="2" ParentId="16769" CreationDate="2014-09-26T14:59:41.067" Score="12" Body="&lt;p&gt;Algebraically, the levels of cyclic polyenes may be derived using simple&#xA;&lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=H%C3%BCckel_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hückel theory&lt;/a&gt; (see also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/76809/16683&quot;&gt;Pi molecular orbitals of polyenes&lt;/a&gt;). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ orbital before interaction (Coulomb integral), $\beta$ is the interaction energy between two adjacent $\mathrm p_{\pi}$ orbitals (the resonance integral) and $j= 0, \pm 1, \pm 2,\ldots, \pm \frac{N - 1}{2}, +\frac{N}{2}$ for even $N$, and $j= 0, \pm 1, \pm 2, \ldots, \pm \frac{N - 1}{2}$ for odd $N$. The very simple form of this equation leads to a useful mnemonic for remembering the energy levels of these molecules. Draw a circle of radius $2\beta$ and inscribe an $N$-vertex polygon such that one vertex lies at the bottom position. The points at which the two figures touch define the Hückel energy levels. And that is what is called a Frost diagram.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-01-07T14:31:04.600" LastActivityDate="2018-01-07T14:31:04.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16776" PostTypeId="2" ParentId="16427" CreationDate="2014-09-26T15:06:31.180" Score="3" Body="&lt;p&gt;First of all, you're thinking of one of the two definitions of oxidation—breaking bonds in molecular oxygen and rejoining them in a different way in another molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the electron transfer definition of oxidation however. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, to see what is happening we need to look at how many electrons are being transferred. In glucose the oxidation state of carbon is $\ce{0}$, whilst in carbon dioxide it is $\ce{+4}$. So during our rocket burning process, each carbon gives up four electrons. But what is it giving them to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use molecular oxygen as our accepting species—our oxidisation agent—we have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {6O_2} + \ce {C6H12O6}  \ce{-&amp;gt;} \ce{6CO2} + \ce{6H2O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With oxygen here going from an oxidation state of $\ce{0}$ in $\ce{O_2}$ to $\ce{-2}$ in both $\ce{H2O}$ and $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if each $\ce{O2}$ molecule  is picking off two electrons and this is all it's doing, then why can't we replace this with two lots of $\ce{Fe^3+}$ to form $\ce{Fe^2+}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I write this in equation form, it's worth noting that organic electrochemical equations approximate &lt;em&gt;organic matter&lt;/em&gt; (OM) with formaldehyde, $\ce{CH2O}$ since OM is biologically equivalent to a polymer (albeit complex ones!) of $\ce{CH2O}$. Note that glucose is therefore equivalent to six lots of this monomer. This simplifies writing balanced chemical equations a lot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{2Fe_2O_3} + \ce {CH2O}  \ce{-&amp;gt;} 4\ce{FeO} + \ce{H2O} + \ce{CO2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now molecular oxygen is by far a better oxidisation agent than $\ce{Fe_2O_3}$ however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm warranting that red iron oxide ($\ce{Fe_2O_3}$) can be used as well as brown iron oxide (I &lt;em&gt;think&lt;/em&gt; this means $\ce{FeO}$). The difference here is that in the former reaction each $\ce{Fe^3+}$ only captures one electron from your OM to form $\ce{Fe^2+}$, whilst in the latter case you're moving from $\ce{Fe^2+}$ to $\ce{Fe}$ metal which captures two electrons—so you need less. Note that formation of iron from $\ce{Fe^2+}$ will need really high temperatures or pressures, which is what you note in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a super question in terms of getting people thinking about redox not just as a process which happens with ions, but with entire organic species. Such organic reactions can be characterised by a reaction voltage, just like the other types of electrochemical reaction you've been looking at. This is actually what happens in the field of biogeochemistry, where scientists look at which types of oxidisation agent are responsible for degrading OM in different environments. More can be found about this in this paper &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/bk-2011-1071.ch007&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user7232" LastActivityDate="2014-09-26T15:06:31.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16777" PostTypeId="1" AcceptedAnswerId="16778" CreationDate="2014-09-26T16:20:45.557" Score="1" ViewCount="566" Body="&lt;p&gt;Does the bonding pair in $\ce{FeCl4^-}$ attack a proton or does a lone pair on the chlorine attack the proton? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLmU9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My prof has an argument for the first one; the first one involves fewer steps, so it is probably more favorable. My proposed mechanism has three steps of electron pushing versus two, so okay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can see an argument for the second mechanism - the first one involves a bonding pair acting as a nucleophile rather than a lone pair on chlorine. The bonding pair is stabilized by the chlorine and the iron atom, while the lone pair is stabilized by only the chlorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which one wins? I'm guessing both mechanisms are operative here. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-27T02:55:13.320" LastActivityDate="2014-09-27T02:55:13.320" Title="Nucleophilicity of bonding vs lone pairs" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16778" PostTypeId="2" ParentId="16777" CreationDate="2014-09-26T16:48:10.927" Score="2" Body="&lt;p&gt;I don't know the answer to your question, but we can speculate.  A lone pair is more sterically exposed than a bonding pair of electrons and therefor more accessible.  Also a lone pair is generally more nucleophilic because the electrons are concentrated in a smaller region since they are not being shared.  These two arguments would favor the second mechanism.  Finally, when we consider the SN2 reaction where, for example, water is the attacking nucleophile; I've never seen the mechanism drawn where the attacking electrons are from the O-H bond, but rather they always seem to show involvement of the lone pair electrons.  All of these arguments favor the second mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-26T16:48:10.927" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16780" PostTypeId="2" ParentId="16687" CreationDate="2014-09-26T22:26:36.563" Score="3" Body="&lt;p&gt;Reversibility is a statement that means, the system and the surroundings are perfectly balanced (in equilibrium) at ANY AND ALL instants in time. This allows us to define $T_{\mathrm{sys}} = T_{\mathrm{surr}}$ and $p_{\mathrm{sys}} = p_{\mathrm{ext}}$ ALWAYS. The latter statement on pressure allows then that we can calculate work in terms of system pressure rather than external pressure. The proof from this point is the mathematics to show, irreversible systems always do less work than reversible systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the approach you take with your analysis of pressure in steps is the same one that allowed the tortoise to beat Achilles in a race. It is called Zeno's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In specific answers to your numbered questions ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Accepting the above definitions allows the mathematics to prove that reversible work is always greater than irreversible work. As a physical answer, we can define reversible work as $W_{\mathrm{rev}}$. Any irreversible process must have a point in time where the system and surroundings are not at equilibrium with each other. Let's use this to get work $W_{\mathrm{extra}}$, and call $W_{\mathrm{irr}} = W_{\mathrm{rev}} + W_{\mathrm{extra}}$. The final proof that $|W_{\mathrm{irr}}| &amp;lt; |W_{\mathrm{rev}}|$ is that $W_{\mathrm{extra}}$ is always work done ON the system over a full cycle (where the system returns to its starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You mistake that a process taking (very small but) finite steps as a valid replacement for a process that takes infinitesimal steps. We would never go anywhere in life (literally!) when your approach would be valid. See Zeno's paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Infinitesimal steps are not part of the definition of reversibility. They are part of the way to solve a reversible process properly. Reversibility is a definition about the state of system and surroundings during any point in time.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastEditorUserId="189" LastEditDate="2014-09-26T22:34:18.740" LastActivityDate="2014-09-26T22:34:18.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16781" PostTypeId="2" ParentId="15503" CreationDate="2014-09-26T22:35:16.640" Score="2" Body="&lt;p&gt;I assume you mean the first law as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU = \delta q + \delta w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $dU$ is the (state) change in internal energy, $\delta q$ is the (path dependent) work, and $\delta w$ is the (path dependent) work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying the walls are diathermal says something about temperature (as you should confirm). Then, you should find a relationship between temperature change and internal energy change for the substance in your system.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastEditorUserId="8067" LastEditDate="2014-09-27T21:39:03.947" LastActivityDate="2014-09-27T21:39:03.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16782" PostTypeId="1" CreationDate="2014-09-26T22:38:42.613" Score="5" ViewCount="158" Body="&lt;p&gt;&lt;a href=&quot;http://books.google.com/books?id=W9-ggnMUd4QC&amp;amp;pg=PA28&amp;amp;lpg=PA28&amp;amp;dq=polyphenol+dna+intercalate&amp;amp;source=bl&amp;amp;ots=8TWeVCnjIN&amp;amp;sig=KhYbighEgXU46j0e2p1rqZ3KjVM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=fOclVKeGGoruoAT2moGABw&amp;amp;ved=0CDkQ6AEwAg#v=onepage&amp;amp;q=polyphenol%20dna%20intercalate&amp;amp;f=false&quot;&gt;This&lt;/a&gt; shows that polyphenols (like quercetin) have the potential to intercalate DNA, but at potency 100x less than that of ethidium bromide. How does one analytically measure/quantify &quot;100x weaker&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested in how this could apply for the compound &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0300483X09003370#&quot;&gt;9-fluorenol&lt;/a&gt; (hydrafinil) - and of the relative potency of 9-fluorenol and quercetin in intercalating DNA.&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2014-11-05T22:22:02.607" Title="How to analytically quantify a chemical's potential to intercalate DNA?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16783" PostTypeId="2" ParentId="16676" CreationDate="2014-09-26T22:39:37.680" Score="2" Body="&lt;p&gt;For inorganic chemistry and group theory, learn linear algebra and matrix mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantum chemistry, learn how to solve partial differential equations. Depending here on how far and what direction you want to go, you may need either a science approach or a numerical methods approach to the higher-level math methods. At an undergraduate level, a solid course on differential equations should be a good start.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2014-09-26T22:39:37.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16784" PostTypeId="2" ParentId="16676" CreationDate="2014-09-27T01:33:46.613" Score="3" Body="&lt;p&gt;for inorganic chemistry you need arithmetic and stereometry. Group theory may be fancy, but more or less ignorable as long as you don't draw molecular orbital diagrams in complicated hi-symmetry cases (understanding ones that are already drawn for you does not require this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for physical chemistry (including understanding of equilibrium) you also need differential and integral calculus and linear differential equations are required. In case of statistical physics approach familiarity with series is of use. complex numbers may be of use in differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for quantum chemistry you also need partial differential equations, linear algebra and function approximation by series (generalized Fourier series). It is highly recommended to be familiar with complex numbers, but technically speaking it is possible to get basic understanding of quantum chemistry with very basic understanding of complex numbers. For advanced applications calculus of variations and basic understanding of numerical methods, including ones for function approximations, integration, and iterative solution of eigenvalue problems are of use. Group theory is optional, it is touched in simplification of cases with symmetry, but other than that is of no use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also recommend to consider memory excercises. You'll have to remember a huge amount of assorted, badly systematized facts and names.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-27T21:33:45.463" LastActivityDate="2014-09-27T21:33:45.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16785" PostTypeId="1" AcceptedAnswerId="16788" CreationDate="2014-09-27T02:37:55.700" Score="53" ViewCount="545055" Body="&lt;p&gt;In a galvanic (voltaic) cell, the anode is considered negative and the cathode is considered positive. This seems reasonable as the anode is the source of electrons and cathode is where the electrons flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in an electrolytic cell, the anode is taken to be positive while the cathode is now negative. However, the reaction is still similar, whereby electrons from the anode flow to the positive terminal of the battery, and electrons from the battery flow to the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does the sign of the cathode and anode switch when considering an electrolytic cell? &lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="402" LastEditDate="2017-01-22T09:55:52.670" LastActivityDate="2020-07-01T11:28:41.233" Title="Positive or Negative Anode/Cathode in Electrolytic/Galvanic Cell" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="36" ContentLicense="CC BY-SA 3.0" />
  <row Id="16786" PostTypeId="2" ParentId="15583" CreationDate="2014-09-27T03:29:56.117" Score="2" Body="&lt;p&gt;Firstly, you seem to have a misconception about atomic orbitals and molecular orbitals. When carbon takes part in bonding within a molecule the electrons don't move from its 2s orbital into an unoccupied 2p orbital. I think what you are referring to is hybridization: You take the 2s and 2p orbitals and build linear combinations from them to get a set of e.g. four energetically equivalent $\ce{sp^3}$ orbitals and then you use these orbitals to form bonds between carbon and, say, the 1s orbitals of four hydrogen atoms thus getting methane $\ce{CH4}$. But the problem is that the carbon $\ce{sp^3}$ or hydrogen 1s orbitals are not actually present in a physical sense in the $\ce{CH4}$ molecule. Those orbitals are merely a mathematical tool, a basis, used to describe the actual molecular orbitals of the methane molecule. A more in-depth explanation of that can be found &lt;a href=&quot;https://chemistry.stackexchange.com/a/16148/189&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so there is no moving electrons from one orbital to another when forming bonds. But what about your question concerning the switching of electrons? Well, electrons are indistinguishable particles. That means if you have two electrons, one on the left and one on the right, you could exchange them without ever noting a difference in the whole system: it will still have the same energy, same momentum, same wavefunction, etc. Let's take the example of lithium: lithium has two 1s electrons and one 2s electron, so three electrons in total. If you'd label those electrons with the names Peter, Paul, and Mary you couldn't say whether Peter and Paul are hanging around in the 1s orbital while Mary has a good time in the 2s orbital or whether Peter and Mary have fun in the 1s orbital while Paul sits lonely in the 2s orbital because electrons are indistinguishable. Both situations would look absolutely the same to you. So, even if two electrons would switch their orbital &quot;positions&quot;, you really couldn't tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point that might raise confusion is that there is still the Pauli exclusion principle, saying that two electrons having the same spin can't occupy the same orbital. But that is a completely different matter that has nothing to do with Peter, Paul and Mary - to stay in the picture. You can't say Peter and Paul are both spin-up and Mary is spin-down, so Peter and Paul can never be together in the 1s orbital and one of them is always paired up with Mary. The only thing you can say is that among Peter, Paul, and Mary you must have two with spin-up and one with spin-down (two spin-down and one spin-up would be equivalent). But they can't all be spin-up (at least not in the ground state of $\ce{Li}$) because then you would have two electrons with the same spin in the 1s orbital which is forbidden by the Pauli exclusion principle.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-09-27T04:46:23.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16787" PostTypeId="2" ParentId="16785" CreationDate="2014-09-27T04:02:38.313" Score="1" Body="&lt;p&gt;The anode is the electrode at which the oxidation half-reaction takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a galvanic cell, the reaction is spontaneous, there is no external potential applied, and when the anode material is oxidized that makes the anode the negative electrode.  In an electrolytic cell, it is the external potential that drives the reaction, the anode is the electrode where the oxidation reaction happens, consequently this time it is the electrode with the positive potential.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="402" LastEditDate="2017-01-22T10:06:55.013" LastActivityDate="2017-01-22T10:06:55.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16788" PostTypeId="2" ParentId="16785" CreationDate="2014-09-27T04:02:47.953" Score="44" Body="&lt;p&gt;The anode is the electrode where the oxidation reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \ce{Red -&amp;gt; Ox + e-}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;takes place while the cathode is the electrode where the reduction reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; \ce{Ox + e- -&amp;gt; Red}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;takes place. That's how cathode and anode are defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Galvanic cell&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in a galvanic cell the reaction proceeds without an external potential helping it along. Since at the anode you have the oxidation reaction which produces electrons you get a build-up of negative charge in the course of the reaction until electrochemical equilibrium is reached. Thus the anode is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode, on the other hand, you have the reduction reaction which consumes electrons (leaving behind positive (metal) ions at the electrode) and thus leads to a build-up of positive charge in the course of the reaction until electrochemical equilibrium is reached. Thus the cathode is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrolytic cell&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electrolytic cell, you apply an external potential to enforce the reaction to go in the opposite direction. Now the reasoning is reversed. At the negative electrode where you have produced a high electron potential via an external voltage source electrons are &quot;pushed out&quot; of the electrode, thereby reducing the oxidized species $\ce{Ox}$, because the electron energy level inside the electrode (Fermi Level) is higher than the energy level of the LUMO of $\ce{Ox}$ and the electrons can lower their energy by occupying this orbital - you have very reactive electrons so to speak. So the negative electrode will be the one where the reduction reaction will take place and thus it's the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the positive electrode where you have produced a low electron potential via an external voltage source electrons are &quot;sucked into&quot; the electrode leaving behind the the reduced species $\ce{Red}$ because the electron energy level inside the electrode (Fermi Level) is lower than the energy level of the HOMO of $\ce{Red}$. So the positive electrode will be the one where the oxidation reaction will take place and thus it's the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A tale of electrons and waterfalls&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is some confusion concerning the principles on which an electrolysis works, I'll try a metaphor to explain it. Electrons flow from a region of high potential to a region of low potential much like water falls down a waterfall or flows down an inclined plane. The reason is the same: water and electrons can lower their energy this way. Now the external voltage source acts like two big rivers connected to waterfalls: one at a high altitude that leads towards a waterfall - that would be the minus pole - and one at a low altitude that leads away from a waterfall - that would be the plus pole. The electrodes would be like the points of the river shortly before or after the waterfalls in this picture: the cathode is like the edge of a waterfall where the water drops down and the anode is like the point where the water drops into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, what happens at the electrolysis reaction? At the cathode, you have the high altitude situation. So the electrons flow to the &quot;edge of their waterfall&quot;. They want to &quot;fall down&quot; because behind them the river is pushing towards the edge exerting some kind of &quot;pressure&quot;. But where can they fall down to? The other electrode is separated from them by the solution and usually a diaphragm. But there are $\ce{Ox}$ molecules that have empty states that lie energetically below that of the electrode. Those empty states are like small ponds lying at a lower altitude where a little bit of the water from the river can fall into. So every time such an $\ce{Ox}$ molecule comes near the electrode an electron takes the opportunity to jump to it and reduce it to $\ce{Red}$. But that does not mean that the electrode is suddenly missing an electron because the river is replacing the &quot;pushed out&quot; electron immediately. And the voltage source (the source of the river) can't run dry of electrons because it gets its electrons from the power socket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the anode: At the anode, you have the low altitude situation. So here the river lies lower than everything else. Now you can imagine the HOMO-states of the $\ce{Red}$ molecules as small barrier lakes lying at a higher altitude than our river. When a $\ce{Red}$ molecule comes close to the electrode it is like someone opening the floodgates of the barrier lake's dam. The electrons flow from the HOMO into the electrode thus creating an $\ce{Ox}$ molecule. But the electrons don't stay in the electrode, so to speak, they are carried away by the river. And since the river is such a vast entity (lots of water) and usually flows into an ocean, the little &quot;water&quot; that is added to it doesn't change the river much. It stays the same, unaltered so that everytime a floodgate gets opened the water from the barrier lake will drop the same distance.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="402" LastEditDate="2017-01-22T10:00:39.720" LastActivityDate="2017-01-22T10:00:39.720" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16789" PostTypeId="2" ParentId="16785" CreationDate="2014-09-27T04:19:16.173" Score="14" Body="&lt;p&gt;The electrode at which oxidation takes place is known as the anode, while the electrode at which reduction take place is called the cathode.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Reduction -&amp;gt; cathode  &#xA;Oxidation -&amp;gt; anode&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you see galvanic cell reduction take place at the left electrode, so the left one is the cathode. Oxidation takes place at the right electrode, so the right one is the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in electrolytic cell reduction takes place at the right electrode, so right one is the cathode. Oxidation takes place at the left electrode, so the left one is the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p1mke.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="402" LastEditDate="2017-01-22T10:02:51.663" LastActivityDate="2017-01-22T10:02:51.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16790" PostTypeId="2" ParentId="15583" CreationDate="2014-09-27T05:59:31.523" Score="3" Body="&lt;p&gt;The short answer to this is: according to our current understanding of quantum mechanics, we can't tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now for the long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, if you haven't already seen Richard Feynman's 60-second primer to the scientific method, &lt;a href=&quot;https://www.youtube.com/watch?v=viaDa43WiLc&quot; rel=&quot;nofollow noreferrer&quot;&gt;you should definitely watch it now.&lt;/a&gt; It's short. I promise I won't go anywhere while you're watching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright. So the way science works is to test ideas by experiment. This means that in order to distinguish between which of two ideas is right, we'll need to come up with some difference in their behavior. This can sometimes be problematic in science because the difference in behavior between two models can be incredibly small. To take just one example, it wasn't until the 20th century that an apparent error in the orbit of Mercury was confirmed, which was eventually explained by relativity. Before that, if you had given someone the theory of relativity and the Newtonian theory of gravity and asked them to tell you which one was right, they would have shrugged their shoulders. There simply wasn't enough high-quality evidence to decide between one or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, what if there's a nightmare situation in which we simply can't tell the difference no matter how good our instruments are? Imagine that I give you a small cube which lights up when you shake it, but I've used magic so that no matter what experiments you try to use to figure out what's inside of the cube, you can't get any information. You can try to use x-rays to image the inside, or weigh it, or spin it around to see if the insides are unevenly distributed, or look at it with an IR camera...but every single time, the magic renders your results useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose I asked you how the box worked. Without being able to get any info, you're stuck! No matter how clever you are in making your guess or how many guesses you come up with, you can't really say anything about any of them, because you're being magically prevented from getting information about what's in the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important idea, so I want to make sure you really understand it. Without some sort of actual behavioral difference that can be looked at, science has no way of distinguishing between alternate ideas.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, quantum mechanics is sometimes a lot like our magic box (to pretty much anyone who tries to study it, it is a magic box). Our current understanding of QM says that we have this &lt;strong&gt;wavefunction&lt;/strong&gt; that tells us what the system is doing. We can't actually look at the wavefunction. The best we can do is to use some &lt;strong&gt;operators&lt;/strong&gt; on it, which will give us the values of certain properties (don't worry if this all sounds very mysterious, it takes years of study to understand some of this stuff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BvN35.png&quot; alt=&quot;a wavefunction equation with some operators&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can we see if the wavefunction matches what we think it is? Well, we can look at what we should get from the operators, and what we actually do get from measuring it. For instance, by applying something called the Hamiltonian operator to the wavefunction, we get the total energy of the system. We can then measure the total energy of the system and see if theory and experiment agree. We can do that a whole bunch of times, measuring the momentum, and kinetic energy, and the average position of the system. If we do it enough times, we can say &quot;eh, it's probably correct&quot; and then move on to the next problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to get back to your original question: can electrons ever switch places? Well, our current understanding of quantum mechanics is that, when electrons switch places, the wavefunction changes sign.&lt;sup&gt;2&lt;/sup&gt; &quot;Aha!&quot; I hear you say. &quot;So we can watch the wavefunction and see if it changes sign, and if we ever see it flip, we can say that the electrons switched places!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well...not really. Because we can't look at the wavefunction directly, we can only measure things through the operators. And here's where the nasty trick comes into play: for all operators that we can actually observe, there won't be any change if the wavefunction flips its sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're stuck! Just like our magic box, we can't get any meaningful information about what's going on inside, and so we can't really tell if the electrons switched places. Even if they did, it would look exactly the same to us (no matter how good our instruments are, we cannot see a difference, because there literally isn't one). It's an undecidable problem!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you're anything like me, this is a really crummy and disappointing answer. It took me a long time to accept the fact that there are some things in science that we just can't decide. So let's get a little creative: what would have to change in science for us to be able to tell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, and most directly, we would need an alternate theory to QM that allows us to distinguish between the two cases. Seeing how incredibly complex current theories are, I doubt that will happen anytime soon (though I could be pleasantly surprised). The second is that we would need some way to measure the wavefunction directly, instead of looking at it through operators. I've heard whispers of groups making progress on that front, though I honestly don't know if there's any truth to that or not. One way or the other, I don't expect we'll be able to answer this question tomorrow, or anytime relatively soon.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Some time back, I heard about a very clever paper that had to deal with this problem. There was an effect that could either depend on the temperature or density of water. Of course, you can't control the temperature and density of water independently (the stuff expands when you heat it up), and so everyone had just assumed that temperature was responsible. This group realized that, if you supercool water, it takes on densities similar to those found above 0C. By carefully supercooling the water and measuring this effect, they found that at similar densities of water, the effect was the exact same, even when the temperature differed by over 50C. It was density that was responsible all along!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is one of the most mysterious things about all of physics. Even after two years of studying quantum mechanics, I have yet to find anyone who can offer a satisfying explanation for why fermions (half-integer spin particles) should flip their wavefunction upon exchange of (ostensibly identical) particles. If you know of a good read on the subject, please put it in the comments and I'll add it here!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="5026" LastEditDate="2018-04-09T03:20:52.933" LastActivityDate="2018-04-09T03:20:52.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16791" PostTypeId="2" ParentId="15583" CreationDate="2014-09-27T08:00:54.717" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this theoretically possible?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Theoretically your question from the title is meaningless so science would not even answer it and that is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons are &lt;a href=&quot;http://en.wikipedia.org/wiki/Identical_particles&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;indistinguishable, or identical, particles&lt;/em&gt;&lt;/a&gt; in a sense that they all have exactly the same physical properties so they cannot be distinguished from one another, even in principle. Thus, there is no observable difference between the two situations you are trying to distinguish:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Electron-one is at $\ce{2s}$ orbital while electron-two is at $\ce{2p}$ orbital.&lt;/li&gt;&#xA;&lt;li&gt;Electron-two is at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can &lt;em&gt;not&lt;/em&gt; tell these two different configuration apart. Imagine that we play the following simple game: I show to you (how do I do so is irrelevant) an atom with one electron at $\ce{2s}$ orbital and another one at $\ce{2p}$ orbital. I then hide it from you and either do nothing, or exchange the electrons. Finally, I show you the atom once again and ask you, did I exchange electrons or not? Would you be able to answer that question? Surely, &lt;em&gt;no&lt;/em&gt;! And this is essentially what makes your question from the title meaningless.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that electrons can move from say $\ce{2s}$ orbital to an unoccupied $\ce{2p}$ orbital, as in carbon atom which can form 4 bonds this way.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of course, here you &lt;em&gt;can&lt;/em&gt; clearly tell the difference between the following two situations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two electrons at $\ce{2s}$ orbital and two electrons at $\ce{2p}$ orbital.&lt;/li&gt;&#xA;&lt;li&gt;One electron at $\ce{2s}$ orbital while electron-one is at $\ce{2p}$ orbital.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;by just counting electrons on diferent orbitals (again, how do you do so is irrelevant).&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="5026" LastEditDate="2018-04-09T03:22:03.183" LastActivityDate="2018-04-09T03:22:03.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16792" PostTypeId="1" CreationDate="2014-09-27T08:24:34.270" Score="11" ViewCount="912" Body="&lt;p&gt;I'm doing a module on my Biochemistry course looking at protein folding, and in a discussion of [folding] transition states I was a little confused at the thought of a zero-lifetime transition state - especially as for folding &quot;&lt;em&gt;the TS is probably &lt;strong&gt;an ensemble&lt;/strong&gt; of states (a population) with a lifetime of 0&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In searching for anything with mention of protein folding TS with lifetime of 0 I found &lt;a href=&quot;http://www.bio.net/mm/btk-mca/2000-June/002413.html&quot; rel=&quot;noreferrer&quot;&gt;this old discussion&lt;/a&gt; ported to Google Groups, in which this is described as an 'inherent contradiction' to TS theory.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Transition_State_Theory/Eyring_equation&quot; rel=&quot;noreferrer&quot;&gt;&quot;capital K, double dagger&quot; is defined as a *thermodynamic equilibrium&#xA;  constant*&lt;/a&gt;.  So, for example, a Transition State Theory applied to a&#xA;  bimolecular reaction involves a term corresponding to the&#xA;  *thermodynamic equilibrium constant* for the transition state [AB]**.   At the same time, the transition state itself is presumed to have&#xA;  *zero lifetime*.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In other words, the Transition State Theory is wonderfully&#xA;  self-contradictory in its very essence (it contains an *equilibrium*&#xA;  constant for something that has zero lifetime). &lt;a href=&quot;https://groups.google.com/forum/#!topic/bionet.metabolic-reg/yqqb2HOfGwA&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;(via)&lt;/code&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm just wondering if anyone can enlighten me / agree / disagree with this, and maybe suggest some of the original literature I might go to for an understanding of the meaning to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my chemistry textbooks has a small section on Hammond's postulate but without explanation to why transient is defined as zero lifetime:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dbQu7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, it's possible I've brought myself to the wrong equations but if I understand correctly the mass balance equation (described from the source above as the reason TS is only a 'quasi-steady state') says here the &lt;code&gt;dm/dt&lt;/code&gt; &lt;em&gt;does not tend to zero&lt;/em&gt;, i.e. zero lifetime is overextrapolation of the theory :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;one can assume steady state for a species as long as its production&#xA;  rate and its lifetime t have both remained approximately constant for&#xA;  a time period much longer than t. When the production rate and t both&#xA;  vary but on time scales longer than t, the steady-state assumption is&#xA;  still applicable even though the concentration of the species keeps&#xA;  changing; such a situation is called quasi steady state or dynamic&#xA;  equilibrium. The way to understand steady state in this situation is&#xA;  to appreciate that the loss rate of the species is limited by its&#xA;  production rate, so that production and loss rates remain roughly&#xA;  equal at all times. Even though dm/dt never tends to zero, it is&#xA;  always small relative to the production and loss rates. &lt;a href=&quot;http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap3.html#22210&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;(via)&lt;/code&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please let me know in comments if I can make this question more clear, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; This is a question about protein folding transition states - please can you not give me answers or comments directing me to resources covering the basics of transition state theory.&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="2119" LastEditDate="2014-09-30T20:54:09.057" LastActivityDate="2015-03-10T02:30:39.420" Title="How can a protein folding transition state have zero lifetime?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="11" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16793" PostTypeId="1" AcceptedAnswerId="16795" CreationDate="2014-09-27T08:33:11.940" Score="13" ViewCount="35469" Body="&lt;p&gt;In the field of Chemical Kinetics, terms like &lt;strong&gt;Molecularity&lt;/strong&gt; and &lt;strong&gt;Order Of Reaction&lt;/strong&gt; are often used. I am a bit confused between these two terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain to me about these two terms and their origins?&lt;/p&gt;&#xA;" OwnerUserId="670" LastEditorUserId="5026" LastEditorDisplayName="user8016" LastEditDate="2019-04-23T15:28:11.473" LastActivityDate="2021-01-27T17:33:12.183" Title="What is the difference between Molecularity and Order Of Reaction?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16794" PostTypeId="1" CreationDate="2014-09-27T09:11:53.613" Score="7" ViewCount="11055" Body="&lt;p&gt;In the kinetic theory of gases, we have &lt;strong&gt;rms&lt;/strong&gt; (root mean square), &lt;strong&gt;mean&lt;/strong&gt;, and &lt;strong&gt;mp&lt;/strong&gt; (most probable) velocities. I understand the concept well. But my question is why do we have &lt;strong&gt;three&lt;/strong&gt; different kinds of velocities? Why don't we stick onto any one of them? What is the significance of these velocities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it would be helpful if someone could explain how: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{v} = \sqrt{\frac{8RT}{\pi M}}$$&lt;/p&gt;&#xA;" OwnerUserId="7577" LastEditorUserId="16683" LastEditDate="2016-09-30T10:09:16.713" LastActivityDate="2016-09-30T10:09:16.713" Title="Why do we need the rms, mean, and most probable velocities?" Tags="&lt;physical-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16795" PostTypeId="2" ParentId="16793" CreationDate="2014-09-27T11:58:56.993" Score="14" Body="&lt;p&gt;In fact, molecularity and order of reaction are identical if the reaction is &lt;strong&gt;elementary&lt;/strong&gt;. They may become different when the reaction is complex and has a mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;molecularity&lt;/strong&gt; of a reaction is defined &lt;em&gt;as the number of molecules or ions that participate in the rate determining step&lt;/em&gt;. A mechanism in which two reacting species combine in the transition state of the rate-determining step is called bimolecular. If a single species makes up the transition state, the reaction would be called unimolecular. The relatively improbable case of three independent species coming together in the transition state would be called termolecular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;order of a reaction&lt;/strong&gt; is an experimentally derived number. Let's consider the general reaction: $\ce{A + B-&amp;gt; C + D}$&#xA; The reaction rate$=\ce{k[A]^n.[B]^m}$ where the rate constant $k$ is a proportionality constant that reflects the nature of the reaction, $\ce{[A]}$ is the concentration of reactant $\ce{A}$, $\ce{[B]}$ is the concentration of reactant $\ce{B}$, and $n$ &amp;amp; $m$ are the exponents used to fit the rate equation to the experimental data. Chemists refer to the sum $n + m $ as the kinetic order of a reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a simple bimolecular reaction $n$ and $m$ would both be $1$, and the reaction would be termed second order, supporting a mechanism in which a molecule of reactant $\ce{A}$ and one of $\ce{B}$ are incorporated in the transition state of the rate-determining step. A bimolecular reaction in which two molecules of reactant $\ce{A}$ (and no $\ce{B}$) are present in the transition state would be expected to give a kinetic equation in which $n=2$ and $m=0$ (also second order).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="51553" LastEditDate="2018-03-12T16:35:43.373" LastActivityDate="2018-03-12T16:35:43.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16797" PostTypeId="1" AcceptedAnswerId="16799" CreationDate="2014-09-27T15:50:25.807" Score="1" ViewCount="216" Body="&lt;p&gt;The answer makes sense to me. However, what if $\ce{OsO4}$ were to attack from the other side of the double bond? Why does the answer key not show that possibility? Would we not get an other molecule?&#xA;&lt;a href=&quot;https://imgur.com/Ua3Voex&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/Ua3Voex.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://imgur.com/c1KrbfE&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/c1KrbfE.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4231" LastEditDate="2014-10-31T19:05:25.007" LastActivityDate="2014-10-31T19:05:25.007" Title="Syn Dihydroxylation of Alkenes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16798" PostTypeId="1" AcceptedAnswerId="16802" CreationDate="2014-09-27T15:55:13.680" Score="2" ViewCount="9022" Body="&lt;p&gt;I've always wondered about this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can look up the exact chemical structure of (say) glucose, which is &lt;em&gt;a&lt;/em&gt; sugar. That's easy enough. But what I cannot figure out is this: if I proffer some random molecule at you, how do you determine whether it can be considered a &quot;sugar&quot; or not? I can't seem to locate anywhere a definition of what it means to &lt;em&gt;be&lt;/em&gt; a sugar; I can only find lists of specific molecules which &lt;em&gt;are&lt;/em&gt; sugars.&lt;/p&gt;&#xA;" OwnerUserId="8072" LastEditorUserId="16683" LastEditDate="2015-10-09T15:53:00.710" LastActivityDate="2015-10-09T15:53:00.710" Title="What is a sugar?" Tags="&lt;organic-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16799" PostTypeId="2" ParentId="16797" CreationDate="2014-09-27T15:58:40.413" Score="6" Body="&lt;p&gt;If you look carefully you'll see that your compound has stereocenters, yet is achiral.  Both stereocenters have the opposite configuration.  This is the &lt;em&gt;meso&lt;/em&gt;-form, these are achiral, and the hydroxy groups are prochiral.  Look carefully for the rotoinversion center!&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-09-27T15:58:40.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16800" PostTypeId="2" ParentId="16792" CreationDate="2014-09-27T16:21:37.370" Score="6" Body="&lt;p&gt;I don’t know where the person from the Google Groups you quote got the information that a transition state is &quot;presumed to have zero lifetime&quot;, but a far as I can tell it is wrong. On the contrary, the &lt;em&gt;existence&lt;/em&gt; of transition states is postulated in &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot;&gt;transition state theory&lt;/a&gt; (TST), which evidently implies that &lt;em&gt;within the TST framework&lt;/em&gt; transition states have &lt;em&gt;non-zero lifetime&lt;/em&gt;. Existence clearly contradicts zero lifetime: if some object has zero lifetime, what do you mean then by its existence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, while the lifetime of a transition state in TST is postulated to be not zero, it is also proposed to be quite &lt;em&gt;small&lt;/em&gt;, on the order of the time of a single molecular vibration ($10^{-13}$ seconds). Why do we have to make such a proposition? Well, transition state can be defined as the state corresponding to the highest potential energy between the reactants and the products (i.e. along a reaction coordinate) such that the probabilities of forming the reactants or products are equal. On potential energy surface (PES) then transition states correspond to the so-called &lt;em&gt;saddle points&lt;/em&gt; (first-order saddle points, to be more precise), which, loosely speaking, can be thought of as a &lt;em&gt;maximum&lt;/em&gt; in one direction (along the reaction coordinate) and minimum in all other directions. And structures that correspond to maxima (even in one direction) on PES) are &lt;em&gt;unstable&lt;/em&gt;: once a system is at a maximum on PES it will &lt;em&gt;almost instantly&lt;/em&gt; fall from it to one of the two neighbouring minima (either to reactants or to the products side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to small assumed lifetime of a transition state, it is often said that &quot;no physical or spectroscopic method is available to directly observe the structure of the transition state&quot; (quote from &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot;&gt;Wikipedia article on TST&lt;/a&gt;). Although, with &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1999/press.html&quot;&gt;femtosecond IR spectroscopy&lt;/a&gt; one can actually observe molecular structures extremely close to the transition state structures. The practical problem is that quite often along the reaction coordinate in addition to a maximum corresponding to transition state between the reactants and the products there exist a minimum (or few) corresponding to the so-called &lt;em&gt;reactive intermediates&lt;/em&gt;, and it is quite difficult to distinguish between transition states and reactive intermediates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note here that reactive intermediates correspond to minimum in all directions on PES, and are thus stable, or, to be more precise, &lt;em&gt;metastable&lt;/em&gt;, since they are high-energy molecules which are quickly converted into more stable molecules (either another reactive intermediate or eventually the products). Reactive intermediates are also short-lived but with lifetimes appreciably longer than a molecular vibration, so they cannot usually be isolated, but they can be observed with the already mentioned femtosecond IR spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;One should &lt;em&gt;not&lt;/em&gt; in the first place think about transition states being real (and consequently detectable). A transition state is a part of a &lt;em&gt;model description&lt;/em&gt; of a chemical reaction introduced by TST, where we just postulate their existence, and say no word on are they really exist or not. Irregardles of their real existence, transition states exist in our model description, and their existence implies &lt;em&gt;non-zero lifetime&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Besides, with femtosecond IR spectroscopy it is possible to directly observe a transition states for some chemical reactions and it was actually done. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-09-27T16:21:37.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16802" PostTypeId="2" ParentId="16798" CreationDate="2014-09-27T16:38:25.497" Score="3" Body="&lt;p&gt;There is no standard &quot;rule&quot; for identifying sugars because in chemistry there are always exceptions to any rule.  Nonetheless, the following guidelines will help you correctly identify a simple sugar (a monosaccharide) in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple sugars have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar#Chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular formula $\ce{C_{n}(H_2O)_{n}}$, where n is at least 3&lt;/a&gt;.  They also have to be capable of forming an aldehyde or ketone carbonyl group.  For example, in glucose the cyclic forms predominate, but there is an isomer with a carbonyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mcvGW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas compounds in the inositol family fit the molecular formula requirement, but are not considered to be sugars because they are incapable of forming a carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f7psz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-27T21:07:31.100" LastActivityDate="2014-09-27T21:07:31.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16803" PostTypeId="2" ParentId="16798" CreationDate="2014-09-27T17:13:24.050" Score="0" Body="&lt;p&gt;Sugars are a family of compound that includes monosaccharides and (poly-/oligo)-saccharides. The latter are derivide from the former by means of inter-esterification. Monosaccharides include mostly linear-chain molecules with three to seven carbon atoms that consist mostly of $\ce{-CH(OH){-}}$ and one $\ce{-C(O){-}}$ units terminated with hydrogen on chain ends. Some C8- and C9-molecules are also considered to be monosaccharides&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monosaccharides often exhibit linear-ring tautomery, and polysaccharides are usually built from ring forms of monosaccharides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, some molecules including $\ce{-CH2{-}}$ block, like deoxyribose, are also considered monosaccharides, though it may be debatable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While brutto-formula of sugars usually is close to $\ce{C_{n}(H2O)_{m}}$, it is not s sufficient criterion to find one. For example, dimethylcarbonate has formuls $\ce{CH3-OC(O)O-CH3}$, i.e. $\ce{C3(H2O)3}$, but is not a sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant wiki article &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sugar&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sugar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some common monosaccharides&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Deoxyribose&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Deoxyribose&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Glucose&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sucrose&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Sucrose&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-27T17:34:26.147" LastActivityDate="2014-09-27T17:34:26.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16804" PostTypeId="1" CreationDate="2014-09-27T17:23:26.257" Score="9" ViewCount="1064" Body="&lt;p&gt;What is the word to describe the number of acidic hydrogens an acid has? I am currently thinking something like &quot;proticity&quot;, but I am not sure  The use case world be for a table of things like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Acid    | Proticity&#xA;--------------------&#xA;CH₃COOH |  1&#xA;H₂SO₄   |  2&#xA;H₃BO₃   |  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or in a set of statements like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The volume of gas produced by reaction A was 4.6 L. The proticity of Acid B was 3. The tare mass of object C was 4.76 g. The density of substance D was 7.8 g/mL.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a specific word used to refer to this quantity? What word should I use to refer to it? &lt;/p&gt;&#xA;" OwnerUserId="8074" LastEditorUserId="73" LastEditDate="2014-09-27T20:19:49.340" LastActivityDate="2016-12-02T08:22:46.577" Title="What is the word for how polyprotic an acid is?" Tags="&lt;acid-base&gt;&lt;terminology&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16805" PostTypeId="2" ParentId="16804" CreationDate="2014-09-27T17:53:14.530" Score="12" Body="&lt;p&gt;The term that I learned was (ironically) &quot;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20081224234255AAJxP4y&quot; rel=&quot;noreferrer&quot;&gt;basicity&lt;/a&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that's the standard term, or even if there is a standard term, but this refers to how many bases you need to neutralize the acid completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thanks for the correction from LDC3: the &quot;basicity&quot; of an acid refers to how many bases you've already added to it. Thus, $\ce{NaH2PO4}$ would be &quot;sodium phosphate, monobasic,&quot; $\ce{Na2HPO4}$ would be &quot;sodium phosphate, dibasic,&quot; and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, not sure if this is standard terminology.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="2087" LastEditDate="2014-09-28T23:25:04.653" LastActivityDate="2014-09-28T23:25:04.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16806" PostTypeId="1" AcceptedAnswerId="16807" CreationDate="2014-09-27T18:17:02.640" Score="4" ViewCount="349" Body="&lt;p&gt;For (a) (iii), how do you get the second product listed in the answers manual? I just can't visualize it.&#xA;&lt;a href=&quot;https://imgur.com/gilOF53&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/gilOF53.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt; &lt;a href=&quot;https://imgur.com/0ZkLlvo&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/0ZkLlvo.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about (c) (i), why are there three resonance structures? If a hydrogen is taken away from a specific carbon of the initial molecule to have a radical, there is no way you can have three resonance structures. &#xA;&lt;a href=&quot;https://imgur.com/v56jefh&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/v56jefh.png&quot; title=&quot;Hosted by imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="7256" LastEditDate="2014-09-27T20:09:50.643" LastActivityDate="2014-09-27T21:21:34.453" Title="Allylic Radicals Bromination Mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16807" PostTypeId="2" ParentId="16806" CreationDate="2014-09-27T18:39:44.077" Score="3" Body="&lt;p&gt;The barrier to rotation in the allyl radical is low (~&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00405a032http://&quot; rel=&quot;nofollow noreferrer&quot;&gt;15-16 kcal/m&lt;/a&gt;) and it can undergo cis-trans isomerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQmWz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to question (c)(i), you're right, there are only two resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T8kKo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you show above with three resonance structures is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethylenemethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;trimethylenemethane diradical&lt;/a&gt;, isobutene with 2 allylic hydrogens removed. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-27T21:21:34.453" LastActivityDate="2014-09-27T21:21:34.453" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16808" PostTypeId="1" AcceptedAnswerId="16817" CreationDate="2014-09-27T19:47:45.657" Score="3" ViewCount="609" Body="&lt;p&gt;I have a question regarding a reversible elementary reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have an equilibrium where $$\ce{A&amp;lt;=&amp;gt;[k_1][k_{-1}]B}$$ , then $$-d[A]/dt= k_1[A]-k_{-1}[B]$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take $$[B]=[B]_0 + ([A]_0 - [A])$$, the equation above becomes $$d[A]/dt= -k_1[A]+k_{-1}([B]_0+([A]_0 -[A])$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I've taken the integrating factor to be $$u=\mathrm{e}^{(k_1+k_{-1})t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and emerged with the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[A]=\frac{k_{-1}}{k_1+k_{-1}}([B]_0+[A]_0)(1-\mathrm{e}^{-t(k_1+k_{-1})}) + C \mathrm{e}^{-t(k_1+k_{-1})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to put $$[A](t=0) = [A]_0$$ and the solution appears to be $$C=[A]_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there further approximations required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lecturer has produced the final equation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[A]=\frac{k_{-1}+k_1\mathrm{e}^{-t(k_1+k_{-1})}}{k_1+k_{-1}}([B]_0+[A]_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we take $$t-&amp;gt;\infty$$ our equations match, however.&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="7721" LastEditDate="2014-09-27T22:05:35.070" LastActivityDate="2014-09-28T17:39:55.453" Title="Reversible elementary reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16809" PostTypeId="1" CreationDate="2014-09-27T21:03:17.257" Score="0" ViewCount="214" Body="&lt;p&gt;Is it possible for vegetable oil to occur naturally, without being extracted? Along the same lines, it it possible for Hexane to occur naturally? &lt;/p&gt;&#xA;" OwnerUserId="8077" LastEditorUserId="7448" LastEditDate="2015-10-09T17:09:11.970" LastActivityDate="2015-10-09T17:09:11.970" Title="Naturally Occurring Vegetable Oil?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16810" PostTypeId="1" AcceptedAnswerId="16848" CreationDate="2014-09-27T22:23:34.163" Score="5" ViewCount="1607" Body="&lt;p&gt;I understand that a polar solvent will help dissociate the &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; into ions.&lt;/p&gt;&#xA;&lt;p&gt;But why a protic solvent?&lt;/p&gt;&#xA;&lt;p&gt;Why would we want a solvent that would encage the &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; ions?&lt;/p&gt;&#xA;&lt;p&gt;Why is the aprotic solvent not favored? My textbook does not discuss this and I searched online, with no result.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="95133" LastEditorDisplayName="user15489" LastEditDate="2020-08-19T07:20:13.620" LastActivityDate="2020-08-19T07:20:13.620" Title="Why would a polar protic solvent favor the addition of HBr to an alkene?" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16812" PostTypeId="1" CreationDate="2014-09-28T00:30:06.317" Score="3" ViewCount="1529" Body="&lt;p&gt;I just need an explanation of why it is a strong acid in its first ionization and a weak acid its second ionization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It becomes $\ce{HSO_4-}$. From what I have learned about oxoacids, the bond strength determines the strength of the acid. So why does it become harder for the molecule to ionize the second time after it loses a hydrogen atom. Just can't think of how it will affect the strength of the bond connecting oxygen and hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="9961" LastEditorDisplayName="user8016" LastEditDate="2015-11-13T19:25:52.510" LastActivityDate="2015-11-13T19:25:52.510" Title="Interesting ionization of sulfuric acid" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;bond&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16813" PostTypeId="1" AcceptedAnswerId="16819" CreationDate="2014-09-28T00:36:32.710" Score="2" ViewCount="1283" Body="&lt;p&gt;I took the kids to the Science Museum in Canberra (&lt;a href=&quot;https://www.questacon.edu.au/&quot; rel=&quot;nofollow&quot;&gt;Questacon&lt;/a&gt;) and one of the toys we brought back was a &lt;a href=&quot;http://www.csiroshop.com/high-bounce-ball-kit/&quot; rel=&quot;nofollow&quot;&gt;Home-Made High Bouncing Ball&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I did it with the kids and it was amazing. You pour the powder in the mould, hold it underwater for a minute. Dry it in the mould for two minutes and open the mould and let it dry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result was like the high-bouncer balls we used to play with as kids. (Lots of fun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What is the chemical reaction in a home-made high-bouncer ball?&lt;/strong&gt; (I'd also love a picture I can show my kids - but that is not essential). &lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2018-10-23T17:26:07.680" Title="What is the chemical reaction in a home-made high-bouncer ball?" Tags="&lt;reaction-mechanism&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16815" PostTypeId="1" CreationDate="2014-09-28T01:18:00.170" Score="1" ViewCount="826" Body="&lt;p&gt;Why is the dissociation of strong electrolytes complete in very dilute solutions of ionic substances but less likely to be so in more concentrated solutions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before seeing this, I thought a strong acid will always dissociate no matter the situation.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="189" LastEditDate="2014-09-28T01:34:51.130" LastActivityDate="2014-09-28T01:34:51.130" Title="Degree of dissociation of strong electrolytes in concentrated solutions of ionic substances" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16816" PostTypeId="1" CreationDate="2014-09-28T02:12:57.583" Score="5" ViewCount="1080" Body="&lt;p&gt;I'm having trouble determining which of these protons is more acidic. I was thinking the solution had something to do with the double bond in the aromatic ring, but I'm still not sure. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/fcNZQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8078" LastEditorUserId="83758" LastEditDate="2020-02-13T18:15:29.520" LastActivityDate="2020-02-13T18:15:29.520" Title="Which proton is more acidic in this compound?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16817" PostTypeId="2" ParentId="16808" CreationDate="2014-09-28T03:05:02.543" Score="2" Body="&lt;p&gt;I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t =  0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-28T17:39:55.453" LastActivityDate="2014-09-28T17:39:55.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16818" PostTypeId="1" AcceptedAnswerId="17180" CreationDate="2014-09-28T03:39:23.423" Score="13" ViewCount="6433" Body="&lt;p&gt;How do I convert $\ce{CH3CH2OH}$ (ethanol) to $\ce{CH3CH2C#CH}$ (but-1-yne) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not too good with reaction mechanisms and conversions are beyond me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched through a couple of textbooks and notes and I don't see any direct way to do the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can use thionyl chloride to to convert from alcohol to chloroalkane: &#xA;$$\ce{CH3CH2OH + SOCl2 -&amp;gt; CH3CH2Cl + HCl + SO_2\uparrow}$$&#xA;The $\ce{SO2}$ escapes but I'm not sure how I can get rid of the $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that I do get rid of it somehow, I can take the resulting chloroalkane and add an alkynyl group to it hence forming the required higher alkane.&#xA;$$\ce{CH3CH2Cl + C2HNa + NaCl -&amp;gt; CH3CH2C#CH + \dots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the mechanism of these reactions elude me. I've only figured out the above from randomly stated properties of reagents. A detailed method of the conversion would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="2325" LastEditorUserId="16683" LastEditDate="2017-09-01T07:56:24.193" LastActivityDate="2017-09-01T07:56:24.193" Title="Synthetic route for the conversion of ethanol to but-1-yne" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16819" PostTypeId="2" ParentId="16813" CreationDate="2014-09-28T05:30:48.577" Score="0" Body="&lt;p&gt;I have google it and got the link which explains the process but don't show the organic reactions(might be because they are complex)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This activity demonstrates an interesting chemical reaction, primarily between the borax and the glue. The borax acts as a “cross-linker” to the polymer molecules in the glue – basically it creates chains of molecules that stay together when you pick them up. The cornstarch helps to bind the molecules together so that they hold their shape better. - See more at: &lt;a href=&quot;http://www.sciencebob.com/blog/?p=558&quot; rel=&quot;nofollow&quot;&gt;Sciencebob&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-28T05:30:48.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16820" PostTypeId="1" CreationDate="2014-09-28T05:48:15.563" Score="2" ViewCount="592" Body="&lt;p&gt;If for some stage play/trick we would like to create some surprising effect in which something with a pungent odor has been released, but at the same time need to ensure that thing be absolutely safe for participants and audience. Is there any organic compound that can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the compound should be volatile as possible to be effective which reminds me of benzene; however it is detrimental to health. Any recommendation for safer liquids?&lt;/p&gt;&#xA;" OwnerUserId="8080" LastEditorUserId="73" LastEditDate="2014-09-28T06:53:03.230" LastActivityDate="2014-09-28T17:39:10.080" Title="Harmless &amp; pungent volatile liquid for use in a stage show" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16821" PostTypeId="2" ParentId="16820" CreationDate="2014-09-28T06:24:46.233" Score="1" Body="&lt;p&gt;Acetic acid, it's the compound in vinegar, so it is not harmful when diluted. Unfortunately, to get it to spread quickly, it would be best to spray the concentrated acetic acid in the air.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-09-28T06:24:46.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16822" PostTypeId="2" ParentId="7375" CreationDate="2014-09-28T06:48:42.837" Score="3" Body="&lt;p&gt;The salt lakes that occur in Western Australia are touted by the article &lt;a href=&quot;http://docs.lib.purdue.edu/cgi/viewcontent.cgi?article=1005&amp;amp;context=easpubs&quot; rel=&quot;nofollow&quot;&gt;Geochemical characteristics of naturally acid and alkaline saline lakes&#xA;in southern Western Australia&lt;/a&gt; (Bowen and Benison, 2009) are unique environments, in that they possess&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an extreme and rare&#xA;  range of natural hydrogeochemical conditions. Many WA lake&#xA;  waters and groundwaters are hypersaline and acidic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article discusses observations of effects due to temperature, especially cycles of dry and wet.  The high ambient temperatures would cause increased evaporation, and hence increase local evapoconcentrations of ions, resulting in a lowering of the local pH, storm/inundation events have the opposite effect.  This leads to further observations by the authors that found that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Field observations of lake pH during storm events&#xA;  illustrated that fluids that have been freshened due to meteoric input&#xA;  are able to regain acidity rapidly. Part of the acidity may be locally&#xA;  sustained by ongoing $\ce{Fe}$ cycling and redox reactions that&#xA;  generate $\ce{H+}$ ions. The amount of $\ce{Fe}$ increases significantly in the&#xA;  more acid waters, and authigenic $\ce{Fe} mineralization (oxides&#xA;  and sulfates, e.g., hematite and jarosite) is abundant throughout&#xA;  the acid lake sediments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user8016" LastActivityDate="2014-09-28T06:48:42.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16823" PostTypeId="2" ParentId="16820" CreationDate="2014-09-28T07:17:16.703" Score="0" Body="&lt;p&gt;Mercaptan gas can used. This is a gas mixed with natural gas to detect leakage because natural gas is colourless, odorless and tasteless. I don't think here will be any adverse effect on health because it mixed in a gas used in kitchen. But should read &lt;a href=&quot;http://www.environmentalhealth.ca/summer01gas.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes but of course it might happen that everybody start running here and there.  &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-28T07:17:16.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16824" PostTypeId="1" AcceptedAnswerId="16883" CreationDate="2014-09-28T07:38:08.630" Score="4" ViewCount="11443" Body="&lt;p&gt;Why must 3 equivalents of sodium amide be used here to form the alkyne? The part I specifically have issue with is in red. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, we should just use 2 equivalents of sodium amide to get rid of the bromines and to form the alkyne. Another equivalent of sodium amide will form the acetylide anion. Removing two protons effectively forms the triple bond and eliminates the two halogens. Also, take a look at example 3 here .. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pDkbt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then does the book's solution manual want me to use 3 equivalents of sodium amide to get neutral acetylene, and another equivalent to get to the acetylide ion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/liK9i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, my book's solution manual contradicts another book's discussion of obtaining alkynes from vicinal dihalides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dqqif.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w3zde.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-29T15:16:57.777" LastActivityDate="2017-02-01T21:48:52.237" Title="Alkyne formation from vicinal dihalide" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16826" PostTypeId="1" AcceptedAnswerId="16828" CreationDate="2014-09-28T08:03:29.383" Score="1" ViewCount="5182" Body="&lt;p&gt;My book asks me to consider reacting 1-butene with NBS and light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that NBS and light leads to radical formation, and this is an allylic/benzylic bromination problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my book only provides one product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't both products expected? Actually, I would expect one product in excess (the book actually listed my minor product as the sole product). There are at least two resonance contributors, and the more major one has the more substituted radical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tm1pa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-09-28T08:17:32.027" LastActivityDate="2014-09-29T00:30:13.657" Title="Alllylic bromination of 1-butene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16827" PostTypeId="2" ParentId="16623" CreationDate="2014-09-28T08:40:49.043" Score="10" Body="&lt;p&gt;To further reduce storage requirement, integrals may be calculated on the fly, emplpoying different caching strategies. For some methods storing all the integrals is plainly impossible. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-28T08:40:49.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16828" PostTypeId="2" ParentId="16826" CreationDate="2014-09-28T11:12:13.130" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Your textbook is wrong. Both products are possible. You are oversimplifying your analysis into which product is the major product.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While a secondary radical may usually be more stable than a primary radical due to inductive stabilization from alkyl groups, this is an allyl radical and the pi bond is important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember &lt;a href=&quot;http://en.wikipedia.org/wiki/Zaitsev%27s_rule#Thermodynamic_considerations&quot; rel=&quot;nofollow&quot;&gt;Zaitsev's Rule&lt;/a&gt;. Pi bonds are also stabilized by alkyl groups. So now when we compare the two resonance contributors to see which one is &quot;more important&quot;, we need to compare two competing factors:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                           Radical        Alkene&#xA;your &quot;minor&quot; radical       primary        internal&#xA;your &quot;major&quot; radical       secondary      terminal&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, as our computational chemists will tell us, this sort of analysis is goofy because the two radicals we are comparing are &lt;strong&gt;the same radical&lt;/strong&gt;. Thus a more complete picture would require some computational calculations to tell us where the &quot;radical density&quot; is or how the distribution of the SOMO (singly-occupied molecular orbital) compares to the van der Waals surface (in other words, which position is more sterically accessible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet you forgot sterics! I do not have the capability right now to plug the butenyl radical into a modeling program and generate the convincing pictures, but I can predict what we would learn:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The molecular orbitals and electron density distribution would confirm that the pi bond has more density between carbon atoms 2 and 3 than between carbon atoms 1 and 2 (in other words, the pi bond's position is more important than the radical's position).&lt;/li&gt;&#xA;&lt;li&gt;The &quot;radical density&quot; and the SOMO will be more concentrated at carbon atom 1 than at carbon atom 3.&lt;/li&gt;&#xA;&lt;li&gt;The radical density and the SOMO will extend outside of the van der Waals surface at carbon atom 1 and they will remain inside the van der Waals surface at carbon atom 3 (in other words, carbon atom 1 is more sterically accessible). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would love to see some MO pictures or electron/radical density maps compared to van der Waals surfaces to confirm this. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-09-28T11:12:13.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16831" PostTypeId="2" ParentId="4989" CreationDate="2014-09-28T13:00:26.513" Score="3" Body="&lt;p&gt;Working in the domain of the kinetic theory of gases, it was Einstein who gave the first estimates of the size of of atoms and molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.damtp.cam.ac.uk/user/gold/pdfs/teaching/einstein_brownian05.pdf&quot; rel=&quot;nofollow&quot;&gt;INVESTIGATIONS ON THE THEORY &#xA;OF THE BROWNIAN MOVEMENT&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this paper it will be shown that according&#xA;  to the molecular-kinetic theory of heat, bodies&#xA;  of microscopically-visible size in a&#xA;  liquid will perform movements of such magnitude&#xA;  that they can be easily observed in a microscope,&#xA;  on account of the molecular motions of heat.&#xA;  It is possible that the movements to be discussed&#xA;  here are identical with the so-called Brownian&#xA;  molecular motion however, the information&#xA;  available to me regarding the latter is so lacking&#xA;  in precision, that I can form no judgment in the&#xA;  matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Jean Baptiste Perrin was later awarded the Nobel Prize for his confirmations. I quote from &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/1926/press.html&quot; rel=&quot;nofollow&quot;&gt;the award ceremony speech&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It may perhaps be said that in the work which we have just summarized Perrin has offered indirect evidence for the existence of molecules. Here, follows a direct evidence. Microscopic particles in a liquid are never at rest. They are in perpetual movement, even under conditions of perfect external equilibrium, constant temperature, etc. The only irrefutable explanation for this phenomenon ascribes the movements of the particles to shocks produced on them by the molecules of the liquid themselves. A mathematical theory of this phenomenon has been given by Einstein. The first experimental proof of this theory was given by a German physicist, Seddig. After him, the problem was taken up by two scientists simultaneously. One of them was Perrin; the other Svedberg. I have to speak of Perrin only. His measurements on the Brownian movement showed that Einstein's theory was in perfect agreement with reality. Through these measurements a new determination of Avogadro's number was obtained.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5615" LastEditorUserId="5615" LastEditDate="2014-09-28T13:36:39.940" LastActivityDate="2014-09-28T13:36:39.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16833" PostTypeId="1" AcceptedAnswerId="16841" CreationDate="2014-09-28T13:49:44.740" Score="3" ViewCount="1892" Body="&lt;p&gt;Excess $\ce{NaHCO3 (s)}$ is added to 510 mL of $\ce{Cu(NO3)2 (aq)}$ $\pu{0.240M}$ for the reaction&#xA;$\ce{Cu(NO3)2 (aq) + 2NaHCO3 (s) -&amp;gt; CuCO3 (s) + 2NaNO3 (aq) + H2O (l) + CO2 (g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SAxQB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="5026" LastEditDate="2018-03-06T14:26:18.683" LastActivityDate="2018-03-06T14:26:18.683" Title="How many grams of solid sodium bicarbonate will be consumed?" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16834" PostTypeId="2" ParentId="16826" CreationDate="2014-09-28T14:43:20.037" Score="5" Body="&lt;p&gt;Yes you are correct, at least two products are formed (&lt;a href=&quot;http://books.google.com/books?id=w9lTxIaPCQIC&amp;amp;pg=PA518&amp;amp;lpg=PA518&amp;amp;dq=nbs%20bromination%20of%201-butene&amp;amp;source=bl&amp;amp;ots=Ky83wJzzS4&amp;amp;sig=fdBlv0uplF6btojnyIxHbJe8c8M&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=9RcoVMLdHoyYyATM6YCwBA&amp;amp;ved=0CCsQ6AEwBA#v=onepage&amp;amp;q=nbs%20bromination%20of%201-butene&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref. 1&lt;/a&gt;, &lt;a href=&quot;http://www.biologie.uni-hamburg.de/b-online/library/newton/Chy251_253/Lectures/DelocalizedBonding/DelocalizedFS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref. 2&lt;/a&gt;).  Unfortunately, neither of these references provide the actual product distribution between the isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say that &quot;at least two products are formed&quot; because depending on your reaction conditions 3 products might be expected.  NBS brominations can be carried out over a range of temperatures, from cold to somewhat above room temperature.  The activation energy required to rotate about an allylic bond is only ~15-16 kcal/m (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16806/allylic-radicals-bromination-mechanism/16807#16807&quot;&gt;see here&lt;/a&gt;).  Therefore, your 1-bromo isomer could exist as a cis-trans mixture, unless the reaction is performed at very cold conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the analogous case of &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NBS bromination of 1-octene&lt;/a&gt;, a mixture of 3-bromo-1-octene (~18%) and 1-bromo-2-octene (~82%) - both cis and trans are produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{RCH2CH=CH2  +  NBS \rightarrow RCHBrCH=CH2  +  RCH=CHCH2Br}$ (cis &amp;amp; trans)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-09-29T00:30:13.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16835" PostTypeId="1" CreationDate="2014-09-28T15:31:14.983" Score="3" ViewCount="202" Body="&lt;p&gt;All things in nature tells us that high entropy is easier to happen.&#xA;But then why does it takes energy to break bonds, in order to achieve high entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thing in gaseous state has higher potential energy than it is in solid state. But that creates less stability.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-09-03T10:04:24.440" Title="A contradiction, probably from confusion" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16836" PostTypeId="2" ParentId="16835" CreationDate="2014-09-28T16:25:36.713" Score="6" Body="&lt;p&gt;Processes in nature that lead to an increase in the entropy of the system are in general favourable, but entropy alone is not the deciding factor! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to follow a full proof we would have to consider Carnot engines and such but for now the result that we are interested in is called the Clausius statement. It is the Clausius statement that really defines the second law and the direction of spontaneity. The Clausius statement leads us to define two parameters depending on the experimental conditions: either the &lt;a href=&quot;https://en.wikipedia.org/wiki/Helmholtz_free_energy&quot; rel=&quot;nofollow&quot;&gt;Helmholtz free energy&lt;/a&gt; $A$ (when working at constant volume) or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;nofollow&quot;&gt;Gibbs free energy&lt;/a&gt; $G$ (under conditions of constant pressure). In practice it is much easier to work with $G$ since we can expose our reaction mixture to the atmosphere and call it constant pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is these &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow&quot;&gt;thermodynamic potentials&lt;/a&gt; $G$ and $A$ that determine the overall thermodynamic picture of spontaneity of a given process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we work with the Gibbs potential, after some mathematics we can easily express by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;G=H-TS&#xA;\end{equation}&#xA;Where $H$ is the enthalpy of the system, $T$ is the temperature and $S$ is the entropy. For a given process it is the Gibbs energy that determines the spontaneity of the reaction. Crudely put, the starting material will have a certain $G$ value and the products will have a certain $G$ value, if the change in $G$ over the course of the reaction is negative then we say it is spontaneous. If it is positive it is not spontaneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can clearly see that the condition for spontaneity is not solely dependant on the entropy alone rather it is also a function of the enthalpy too. Further the entropic contribution to the $G$ is dependent on the temperature indicating that there will be a certain threshold temperature for an endergonic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An endergonic process is characterised by having a positive enthalpy change, requiring energy to occur. If the &quot;amount&quot; of energy the process requires is very large then the change in enthalpy will also be large and as we can see this makes the change in $G$ for the process become less favourable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, we can now see that a given process will have a certain energy barrier... an activation energy if you like. Whether a process is thermodynamically spontaneous or not one has to consider the Kinetics too! A simple yet ubiquitous model for Kinetics is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow&quot;&gt;Arrhenius equation&lt;/a&gt; which in general says the rate of reaction decreases for processes with a higher energy of activation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning the later part of your question. The most stable phase (state of matter) is dependent on the conditions of the reaction. The most stable phase for the given combination of temperature and pressure will prevail in the reaction mixture. In practice this leads to the state with the lowest value of $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solids actually liberate energy when they come together, especially ionic solids! This energy is called the Lattice energy and it is relative to all the reactants in their gaseous state (as ions). (look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Born%E2%80%93Land%C3%A9_equation&quot; rel=&quot;nofollow&quot;&gt;Born-Lande equation&lt;/a&gt; for more info)! So although gaseous species will have a higher entropy due to all the rotational vibrational freedom they have ... the state of matter is dictated by $G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the full mathematical argument behind what I have said I feel my answer will be incomplete, but I hope it helps you :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2015-09-03T10:04:24.440" LastActivityDate="2015-09-03T10:04:24.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16837" PostTypeId="2" ParentId="14012" CreationDate="2014-09-28T16:45:13.397" Score="2" Body="&lt;p&gt;Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should  contain this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The short answer to your question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Somewhat longer answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$)  spontaneously only till it reaches the equilibrium condition. In this direction  $\sum_i \mu_i dN_i \le 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$  transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-09-28T16:45:13.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16838" PostTypeId="2" ParentId="16833" CreationDate="2014-09-28T16:51:48.640" Score="2" Body="&lt;p&gt;Since $\ce{NaHCO3}$ is added in excess the limiting reagent for this reaction is $\ce{Cu(NO3)2}$.&#xA;Now we calculate the weight of $\ce{Cu(NO3)2}$ which will provide grams of $\ce{NaHCO3}$ consumed by it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Weight in g} = \frac{\text{Molarity * Mol. Wt * Volume in ml}}{1000}\\ \ \ = \frac{240 \times 188 \times 510 }{1000}\ = 23.0112 \text{ g}\ce{Cu(NO3)2}$$&#xA;The given equation is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -&amp;gt; CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this balanced equation, it is clear that 1 mole of $\ce{Cu(NO3)2_{(aq)}} $ can consume 2 moles of $\ce{NaHCO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can write,&#xA;$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\188  \text{g } \ce{Cu(NO3)2} \iff 2 \times 84 \text{ g } \ce{NaHCO3}\\23.0112 \text { g} \ce{Cu(NO3)2}  \iff \frac{2 \times 84 \times 23.0112}{188}\ =20.5632 \text{g} \ce{NaHCO3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, we can conclude that 20.5632 g $\ce{NaHCO3}$ will be consumed.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4656" LastEditDate="2014-09-29T08:01:09.603" LastActivityDate="2014-09-29T08:01:09.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16839" PostTypeId="2" ParentId="16358" CreationDate="2014-09-28T17:05:23.567" Score="1" Body="&lt;p&gt;What you are really looking for is the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pharmacophore&quot; rel=&quot;nofollow&quot;&gt;pharmacophore&lt;/a&gt;, that's structural and electronic features in a molecule that allow it to bind to a certain receptor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you squint at pemoline &lt;em&gt;just right&lt;/em&gt; you can see an aromatic group, and two or three carbnos away you see a hydrogen bond acceptor, just as in phenethylamine, so you could argue that the pharmacophores are the same.  At least both compounds act in some way on the dopaminergic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important concept to remember is &lt;em&gt;polypharmacology&lt;/em&gt;.  That means that one compound can act on different targets. A medical chemist would call these &quot;off-target effects&quot;, and a physician would say &quot;side effects&quot;.  It's also that any biological system has many components - there are not only dopamine &lt;em&gt;receptors&lt;/em&gt;, there are also dopamine &lt;em&gt;transporters&lt;/em&gt; that regulate agonist levels in the synaptic cleft and allow for agonist recycling, and there is dopamine &lt;em&gt;metabolism&lt;/em&gt;.  Any compound that acts on any system might affect these components differently, consequently &lt;em&gt;target identification&lt;/em&gt; is an important step in medicinal chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is perhaps ill-posed (well, you are a first year student), but it probes really deep levels in med. chem.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-09-28T17:05:23.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16840" PostTypeId="2" ParentId="16820" CreationDate="2014-09-28T17:10:31.137" Score="2" Body="&lt;p&gt;I would be very, very &lt;strong&gt;very&lt;/strong&gt; cautious in choosing the odorant here. And speed of odour spreading would be quite low unless specific adjustments are employed to fasten air circulation. And even then, it is quite easy to screw because of minor differences in air circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice I would drop any unpleasant odours like mercaptans, sticking to powerful, but generally not unpleasant odours. An example would be benzaldehyde. It has quite strong odour, but autocatalitically oxidizes on air, so the odour should vanish reasonably fast. Some food odourant or parfume would be even better if possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible option would be acetic aldehyde. It has strong apple-like odour, and can be easily produced by pouring ethyl alcohol into cromium (VI) oxide, immidiately producing strong odour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I must say, most odoured compound are slightly toxic, so it is not impossible to overdo with troublesome consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, I strongly recommend to move from odour to light and/or smoke effect, preferably using premanufactured and commercially available devices. It is too easy to screw up with it.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-28T17:39:10.080" LastActivityDate="2014-09-28T17:39:10.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16841" PostTypeId="2" ParentId="16833" CreationDate="2014-09-28T17:38:02.210" Score="3" Body="&lt;p&gt;$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -&amp;gt; CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$&#xA;The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here,&#xA;$$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510  \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\&#xA;0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$&lt;/p&gt;&#xA;" OwnerUserId="2325" LastActivityDate="2014-09-28T17:38:02.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16842" PostTypeId="1" AcceptedAnswerId="19078" CreationDate="2014-09-28T19:47:44.107" Score="2" ViewCount="832" Body="&lt;p&gt;I am tasked by my textbook to synthesize 2-hexyne from 1-propanol. All carbon atoms must come from the starting material, 1-propanol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solutions manual (an older version) does not have this additional problem, so I am forced to ask here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that in this problem there is a double in carbon count in the product, so we are expected to somehow &quot;fuse&quot; the carbons from the starting material. I also understand that this means making an electrophilic carbon and a nucleophilic carbon from the starting materials. The electrophilic carbons that I know about are alkyl halides, alkyl tosylates, epoxides, and acid chlorides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll choose to make an alkyl halide from the starting material through reaction of the primary alcohol (1-propanol) with thionyl bromide and pyridine as a base. Now I have my electrophilic carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My nucleophilic carbon will be the acetylide anion. I could use a cyanide ion but that only gives me a one carbon increase. Plus that's not an allowed carbon source. So I dehydrate my primary alcohol (is this possible?) and creates 1-propene. I can brominate this in an inert solvent (i.e. methylene chloride). From here with the vicinal dihalide I can form the alkyne and then methyl acetylide. I can then react this with the alkyl halide I made previously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this look? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RR5F0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-02T20:16:01.043" Title="Restricted synthesis; formation of nucleophile and electrophile from one carbon source" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16843" PostTypeId="2" ParentId="16810" CreationDate="2014-09-28T19:51:22.903" Score="2" Body="&lt;p&gt;A polar protic solvent will help stabilize the carbocation that is formed in the process of electrophilic addition. When the nucleophilic $\pi$ bond grabs the hydrogen from the $\ce{HBr}$, one of the carbons that was part of the $\pi$ bond is left deficient of an octet. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T22:42:38.750" LastActivityDate="2015-06-27T22:42:38.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16844" PostTypeId="1" AcceptedAnswerId="16851" CreationDate="2014-09-28T19:57:16.207" Score="2" ViewCount="1086" Body="&lt;p&gt;I am trying to find something I can get the ingredients for and make that starts out a liquid and sets into a solid. My end goal is to create a simple 8bit art figure so I want to pour it into a mold and have it harden. Also I would really like something fully or partly transparent (I was looking at aerogel because of the nice blue translucency but it seems too brittle to work with). So my question is what would be the best compound to try to create this figure out of?&lt;/p&gt;&#xA;" OwnerUserId="8084" LastEditorUserId="4934" LastEditDate="2014-09-28T20:21:36.883" LastActivityDate="2015-12-13T18:47:52.507" Title="A semi-transparent material that is a solid" Tags="&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16845" PostTypeId="1" CreationDate="2014-09-28T20:02:34.707" Score="4" ViewCount="5009" Body="&lt;p&gt;Why will $\ce{K}$ have a greater electron affinity than $\ce{Ca}$? I know that $\ce{Ca}$ has a greater first ionization energy, and I also know that $\ce{K}$ has a greater second ionization energy, but I am confused on how to explain this.&lt;/p&gt;&#xA;" OwnerUserId="8085" LastEditorUserId="15410" LastEditDate="2015-05-19T06:32:23.547" LastActivityDate="2015-05-19T06:32:23.547" Title="Why does K have a greater electron affinity than Ca?" Tags="&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16846" PostTypeId="2" ParentId="16810" CreationDate="2014-09-28T20:11:49.850" Score="2" Body="&lt;p&gt;High-dielectric solvent assist dissociation because they shield charge.  Since the proton's electric field is so high, it always finds a base to bind to, which will in your case be $\ce{Br-}$.  So you also have to consider the solvent's basicity.  Read &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100139a045&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assertion that you need a protic solvent is wrong, you need a sufficiently basic solvent.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T22:41:58.643" LastActivityDate="2015-06-27T22:41:58.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16847" PostTypeId="1" AcceptedAnswerId="16849" CreationDate="2014-09-28T20:17:08.037" Score="7" ViewCount="15613" Body="&lt;p&gt;If an electron is removed from an atom of $\ce{K}$, why does it then have 0 valence electrons as it states in my General Chemistry textbook? I would think that if this happened, it would have the same exact electron configuration as $\ce{Ar}$, which has 8 valence electrons, correct? What's going on here?&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="15410" LastEditDate="2015-05-19T06:24:48.207" LastActivityDate="2015-05-19T06:24:48.207" Title="Why does K⁺ have 0 valence electrons?" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16848" PostTypeId="2" ParentId="16810" CreationDate="2014-09-28T20:17:10.143" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;why a protic solvent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Protic solvents tend to have higher dielectric constants than typical, &lt;strong&gt;unreactive&lt;/strong&gt;, non-protic solvents.  Look at the following table.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{|c|c|c|}\hline&#xA;\text{Solvent} &amp;amp; \text{Dielectric constant}    \\ \hline&#xA;\text{methanol} &amp;amp; 32.6    \\ \hline&#xA;\text{ethanol} &amp;amp; 24.6   \\ \hline&#xA;\text{methylene chloride} &amp;amp; 9.08 \\ \hline&#xA;\text{chlorobenzene} &amp;amp; 5.69\\ \hline&#xA;\text{chloroform} &amp;amp; 4.81    \\ \hline&#xA;\text{diethyl ether} &amp;amp; 4.27    \\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The higher dielectric constant stabilizes ions which means that 1) &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; will tend to be more dissociated and 2) (as Dissenter noted) the resultant carbocation intermediate will be more stabilized thereby lowering the energy of the transition state leading to it.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="81509" LastEditorDisplayName="user15489" LastEditDate="2020-08-19T07:16:19.563" LastActivityDate="2020-08-19T07:16:19.563" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16849" PostTypeId="2" ParentId="16847" CreationDate="2014-09-28T20:35:35.277" Score="15" Body="&lt;p&gt;What you say is correct.  The  [Ar] configuration we are left with does have 8 valence electrons.  But I think it is just semantics.  Elemental potassium has an [Ar] 4s1 electron configuration.  One would say it has one valence electron.  If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because &quot;1 - 1 = 0&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-28T20:35:35.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16851" PostTypeId="2" ParentId="16844" CreationDate="2014-09-28T21:44:26.860" Score="5" Body="&lt;p&gt;There are other plastics that have better mechanical properties, but probably nothing is easier to find than &lt;a href=&quot;http://www.dickblick.com/products/castincraft-clear-polyester-casting-resin/&quot; rel=&quot;nofollow&quot;&gt;polyester casting resin&lt;/a&gt;. It works really well in silicone moulds and there are various pigments you can buy to change its colour and opacity. It isn't as optically clear as some polyurethane or epoxy resins (you won't be making lenses out of polyester), but it's not bad and you can find it in a lot of art supply stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for something with aerogel's smokey translucency, I don't know. Aerogel's translucency (its colour, as well) is due to scattering off its nanoporous structure. It may be possible to mimic the look by blending silica particles or something into a plastic to cause scattering, but I can't recommend anything specific.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="5084" LastEditDate="2015-12-13T18:47:52.507" LastActivityDate="2015-12-13T18:47:52.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16852" PostTypeId="1" CreationDate="2014-09-28T21:56:22.993" Score="0" ViewCount="2421" Body="&lt;p&gt;I'm wanting to make  a Pine-Sol scented candle.  However, in candle making you have to add a fragrance oil.  People make them all the time by adding vanilla, etc and canola oil, vegetable oil, etc.  I'm wondering what proportions I need to do this with Pine-Sol since it has chemicals in it??  I'm using the Original Pine-Sol - it contains 8.7% Pine oil in it if that helps.  That's all I got from the label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&#xA;Mandy&lt;/p&gt;&#xA;" OwnerUserId="8088" LastActivityDate="2016-02-25T07:56:59.507" Title="How can I turn Pine-Sol into an oil?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16853" PostTypeId="1" AcceptedAnswerId="16855" CreationDate="2014-09-28T22:55:48.277" Score="8" ViewCount="723" Body="&lt;p&gt;Sulfur hexafluoride ($\ce{SF6}$) is, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfur_hexafluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;, a hypervalent substance.  However, according to the page in regards to $\ce{SF6}$ reactions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is virtually no reaction chemistry for $\ce{SF6}$. The main contribution to the inertness of $\ce{SF6}$ is the steric hindrance of the sulfur atom, whereas its heavier group 16 counterparts, such as $\ce{SeF6}$ are more reactive than $\ce{SF6}$ as a result of less steric hindrance &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linked &lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_effects#Steric_hindrance&quot; rel=&quot;nofollow noreferrer&quot;&gt;steric hindrance&lt;/a&gt; provides many examples of how that chemical process works, but specifically, what steric hindrance (or any other) mechanism results in $\ce{SF6}$ possessing virtually no reaction chemistry?&lt;/p&gt;&#xA;" OwnerDisplayName="user8016" LastEditorUserId="402" LastEditDate="2017-01-14T20:33:36.240" LastActivityDate="2017-01-14T20:33:36.240" Title="How does steric hindrance cause sulfur hexafluoride to have almost no reaction chemistry?" Tags="&lt;inorganic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16854" PostTypeId="2" ParentId="16852" CreationDate="2014-09-28T23:10:09.567" Score="1" Body="&lt;p&gt;I can answer the chemistry part of your question... the candle-making part is better suited to a candle making forum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no easy way of converting the other 90% of (older) pine-sol into pine oil and newer pine-sol has 0% pine oil. Best to purchase pure pine oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pine oil is cyclic: &lt;img src=&quot;https://i.stack.imgur.com/23KkZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other pine-sol ingredients are esters and other linear compounds. So using pine-sol to make pine oil would be ridiculous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not certain how much pine oil is needed for a candle to be scented though. It may just be enough. For that you will have to just experiment... or ask on a candle forum. &lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2014-09-28T23:10:09.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16855" PostTypeId="2" ParentId="16853" CreationDate="2014-09-28T23:19:13.080" Score="12" Body="&lt;p&gt;As you can see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJjMs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA; Sulfur (16) is significantly smaller than Se (34). That size difference&#xA; means that there is more room between the fluorine atoms on Se than on &#xA;Sulfur. That is what steric hindrance means. Think of steric hindrance&#xA; as limited access. The smaller the atom the less access there is. &#xA;Oxygen is so small that less than 6 fluorine atoms can crowd around it &#xA;because of steric hindrance... probably more like four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So because there is more room between the fluorine atoms on Se there is &#xA;more opportunity for other atoms to attack the central atom Se, but &#xA;sulfur is completely shielded by inert fluorine. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="16683" LastEditDate="2016-03-26T13:51:09.577" LastActivityDate="2016-03-26T13:51:09.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16856" PostTypeId="2" ParentId="16804" CreationDate="2014-09-28T23:33:40.707" Score="3" Body="&lt;p&gt;Monoprotic, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diprotic_acid&quot; rel=&quot;nofollow&quot;&gt;Diprotic&lt;/a&gt;, Trioprotic, tetraprotic, pentaprotic, hexaprotic, septaprotic, octaprotic....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{edit}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then suggest &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalent_concentration&quot; rel=&quot;nofollow&quot;&gt;equivalents factor&lt;/a&gt;,&quot; which is directly related to the concept of normality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Normality isn't the same like Molarity, but you can say N = M * equivalents. 1 M HCl = 1 N HCl, but 1 M H2SO4 = 2 N H2SO4...&quot; -&lt;a href=&quot;http://www.sciencemadness.org/talk/viewthread.php?tid=3798&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2014-10-22T20:57:39.607" LastActivityDate="2014-10-22T20:57:39.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16857" PostTypeId="2" ParentId="16847" CreationDate="2014-09-28T23:35:06.593" Score="5" Body="&lt;p&gt;You are correct in that it has 8 valence electrons, not 0. If we look at the electron configuration of potassium (K), we see that it has one electron: [Ar]4s&lt;sup&gt;1&lt;/sup&gt;. Obviously, removing that electron gives us [Ar] (same configuration as K&lt;sup&gt;1+&lt;/sup&gt;), which is a noble gas and has 8 electrons. &lt;/p&gt;&#xA;" OwnerUserId="8090" LastEditorUserId="8090" LastEditDate="2014-09-29T19:54:40.000" LastActivityDate="2014-09-29T19:54:40.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16858" PostTypeId="2" ParentId="16845" CreationDate="2014-09-28T23:59:38.127" Score="-1" Body="&lt;p&gt;the &lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;nofollow&quot;&gt;octet rule&lt;/a&gt;. All elements want 8 electrons in their outer shell. K normally has only one electron in its outer shell and removing it exposes the full octet shell underneath which is favorable and spontaneously occurs, which is why K exists as K+ in solution. So K actually wants to ionize to +1 but no further. However Ca has two electrons in its outer electron shell. Therefore it wants to loose two and expose the next full shell of eight electrons, which is comparable to the electron configuration of Argon (Ar).&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2014-09-28T23:59:38.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16859" PostTypeId="1" AcceptedAnswerId="16865" CreationDate="2014-09-29T00:37:47.197" Score="14" ViewCount="49556" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which isomer would have the largest heat of combustion?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;A) Propylcyclopropane&lt;/strong&gt;&lt;br /&gt;&#xA;B) Ethylcyclobutane&lt;br /&gt;&#xA;C) Methylcyclopentane&lt;br /&gt;&#xA;D) Cyclohexane&lt;br /&gt;&#xA;E) Since they are all isomers, all would have the same heat of combustion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can anyone explain why?&lt;/p&gt;&#xA;&lt;p&gt;If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion &lt;em&gt;per bond&lt;/em&gt; than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.&lt;/p&gt;&#xA;&lt;p&gt;It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;p&gt;Let's say a cyclopropane C−C bond is 10 J.&lt;br /&gt;&#xA;A cyclobutane C−C bond is 9.5 J.&lt;br /&gt;&#xA;A regular alkane bond is 8 J.&lt;/p&gt;&#xA;&lt;p&gt;For propylcyclopropane:&lt;br /&gt;&#xA;Total energy is 54 J&lt;br /&gt;&#xA;For ethylcyclobutane:&lt;br /&gt;&#xA;Total energy is 54 J&lt;/p&gt;&#xA;&lt;p&gt;The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.&lt;/p&gt;&#xA;&lt;p&gt;How do you analyze this situation? Please help.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-10-29T13:10:15.097" Title="Heats of combustion and stability of rings" Tags="&lt;organic-chemistry&gt;&lt;heat&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="16860" PostTypeId="2" ParentId="16859" CreationDate="2014-09-29T00:43:27.887" Score="3" Body="&lt;p&gt;The products of combustion are the same so the compound with the highest potential energy (least stable) will have the highest enthalpy of combustion.  All have the same number of carbons.  The cyclopropane has the most ring strain and therefore the highest potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use metacognition you can see what's being asked.  They can only write questions asking &quot;Which has the most ring strain?&quot; so many times.  This question is supposed to involve more thinking but can be distilled down to &quot;Which has the most ring strain?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good strategy in answering multiple choice questions is to determine what is the same and what is different about the answer choices.  In this case the number of carbons is the same so the products are the same.  What is different is the number of carbons in the rings and therefore the angle strain is different.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-29T00:43:27.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16861" PostTypeId="2" ParentId="16845" CreationDate="2014-09-29T00:49:42.193" Score="7" Body="&lt;p&gt;Neither K nor Ca &quot;wants&quot; to ionize.  Both ionization energies are endothermic.  This question refers to eelctron affinity, which is gaining an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K has a more exothermic electron affinity because the electron gained fills the 4s orbital.  In Ca the electron goes to 3d.  The energy gap between 4s and 3d is enough to make the process barely exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the numbers given in tables are positive but that the process is exothermic.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-09-29T00:49:42.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16863" PostTypeId="1" CreationDate="2014-09-29T01:01:21.243" Score="1" ViewCount="1697" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The three naturally occurring isotopes of potassium are $\ce{^39K}$ ($38.963707\:\mathrm{amu}$); $\ce{^40K}$ ($39.963999\:\mathrm{amu}$); and $\ce{^41K}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The percent natural abundances of $\ce{^39K}$ and $\ce{^41K}$ are $93.2581\%$ and $6.7302\%$ respectively.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) What is the natural abundance of $\ce{^40K}$? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) Determine the isotopic mass of $\ce{^41K}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I started to utilize this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{atomic\:mass = (fractional\:abundance\:isotope\:1 &#xA;\cdot mass) + (fractional\:abundance\:isotope\:2 \cdot mass)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I realized that I did not have some values. $\ce{^39K}$ is the only isotope which has two values. $\ce{^40K}$ only has the mass weight, however, if I were to multiply $\ce{^40K}$'s mass with ($1-0.932581$) would that help me arrive at an answer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, would that answer be accurate since it only takes into account that there are only to values that add up to 100%.&lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="865" LastEditDate="2015-10-30T16:55:30.087" LastActivityDate="2015-10-30T16:55:30.087" Title="What is the Natural abundance of Potassium" Tags="&lt;everyday-chemistry&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16865" PostTypeId="2" ParentId="16859" CreationDate="2014-09-29T01:41:10.247" Score="18" Body="&lt;p&gt;Compounds A-D all have the same molecular formula, $\ce{C6H12}$.  We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12 + 9O2 -&amp;gt; 6CO2 + 6H2O + heat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any &lt;strong&gt;differences&lt;/strong&gt; in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; provides the heats of combustion for some useful model compounds.  Look at the last column (&quot;Total Strain&quot;) in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained.  The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is &quot;A&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="55119" LastEditDate="2017-12-07T02:00:12.820" LastActivityDate="2017-12-07T02:00:12.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16867" PostTypeId="1" AcceptedAnswerId="16884" CreationDate="2014-09-29T02:21:41.403" Score="3" ViewCount="587" Body="&lt;p&gt;Can sugars like fructose-glucose or high-fructose corn syrup that are found in many types of processed food contribute to shortening a products shelf life?  Don't bacteria love things like sugar so if a product has a lot of fructose-glucose added to it, it will need even more preservative chemicals to compensate for any increase of sugar?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastActivityDate="2014-09-30T17:01:53.783" Title="Regarding preservatives in food" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16868" PostTypeId="1" AcceptedAnswerId="16873" CreationDate="2014-09-29T02:44:10.900" Score="3" ViewCount="4460" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WGyS8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of these two molecules, which one would be expected to have the most deshielded signal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I think the wording of this question is &lt;strong&gt;wack&lt;/strong&gt; because neither of these molecules has only one NMR signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose E, because I was looking at the benzene's signals. This one has an alkyl group, which is a &lt;em&gt;slight&lt;/em&gt; inductive donor, thereby shielding the benzene's protons (slightly). This one also has a benzylic carbonyl group, which, according to my teacher, is &lt;em&gt;only&lt;/em&gt; a resonance withdrawer and &lt;em&gt;not&lt;/em&gt; in any form a resonance donator. Based on this, I chose E as the more de-shielded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other group similarly has a C=O resonance withdrawing group, but a resonance donating -OH group. Given that the resonance effect usually trumps the inductive effect, I said that E is still the more deshielded of the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize I might have now erred in limiting my analysis of shielding to only the hydrogens on the benzene. In E, the methyl's hydrogens are fairly well-shielded still, being connected to a carbon (albeit an sp2 hybridized carbon). On the other hand molecule F has a long linear chain; the hydrogens toward the end of that linear chain will have well-shielded signals! F's linear chain is also longer than E's linear chain ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly I think this is a horrible question but in any case the answer is apparently &quot;F&quot; ... do you agree? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-09-29T06:43:56.773" Title="NMR: Most deshielded protons (electron withdrawal)" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16869" PostTypeId="1" AcceptedAnswerId="19012" CreationDate="2014-09-29T03:08:43.783" Score="3" ViewCount="1135" Body="&lt;p&gt;I propose the following synthesis.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Oxidation of aldehyde to a carboxylic acid. This sets me up for FC acylation once I ...&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Attack the -OH with thionyl chloride ad form the acid chloride.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reaction with vanilla benzene in the presence of Lewis acid catalyst. Benzene won't go after acid chlorides unless there is some catalyst in the system.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now that the acid chloride is installed, I can go through the motions of getting rid of the pesky C=O group.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given that alkyl groups are slightly activating and are ortho/para directing, I can sulfonate with fuming sulfuric acid and expect a mixture of products (including the one desired by the synthesis). Sterics might favor the para product over the ortho product given the sheer size of the SO3 group.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;General comments:&lt;/p&gt;&#xA;&lt;p&gt;I did NOT sulfonate first, as that is an electron-withdrawing group akin to the nitro group. It deactivates the ring and is also a meta-director. I want a para-substituted product. Not a meta-substituted product.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, how reasonable is my synthesis? Is it correct?&lt;/p&gt;&#xA;&lt;p&gt;Also would sulfonation before getting rid of the C=O work? Something tells me no; the C=O is able to withdraw electrons from the ring via resonance, and that would make it a meta-director.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VVnCx.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="30836" LastEditDate="2020-06-29T07:49:30.527" LastActivityDate="2020-06-29T07:49:30.527" Title="Synthesis of synthetic detergent with Friedel Crafts Acylation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;acylation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16871" PostTypeId="2" ParentId="16863" CreationDate="2014-09-29T06:14:54.623" Score="1" Body="&lt;p&gt;A). If you know the percentages of the abundance of $\frac{2}{3}$s of the isotopes then finding the percentage of the $3^{rd}$ is as simple as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 100\% - (\%_{\ce{^39K}}+\%_{\ce{^41K}}) = \%_{\ce{^40K}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B). Finding the mass of the isotope should be pretty simple. It would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\mathrm{Mass\ of\ proton} \cdot \mathrm{{\#\ of protons}} )+(\mathrm{Mass\ of\ a\ neutron} \cdot \mathrm{{\#\ of neutrons}} ) \approx \mathrm{Mass\ of\ isotope}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You neglect the weight of electrons because the mass is small relative to the others.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-09-29T06:14:54.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16872" PostTypeId="1" AcceptedAnswerId="16895" CreationDate="2014-09-29T06:22:51.787" Score="4" ViewCount="6619" Body="&lt;p&gt;Really fine metal powders are good for two things: making paints and pyrotechnics. For this purposes, I can see people usually use thing called &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_mill&quot; rel=&quot;nofollow noreferrer&quot;&gt;ball mill&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2rvgj.jpg&quot; alt=&quot;ball mill&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically a barell stuffed with heavy balls and the thing to be powdered. You leave it for a while, rotating all the time, and the balls slowly crush your substance into smaller and smaller pieces.&lt;br&gt;&#xA;There are many disadvantages to this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'd need to construct it&lt;/li&gt;&#xA;&lt;li&gt;It would be running and making loud mess all the time&lt;/li&gt;&#xA;&lt;li&gt;soft metals, like aluminium and copper don't even crush into pieces&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I have figured out a different way. More chemical one I'd say. I can dissolve metals in acid by forming salts out of them and then electrolyse those salts. Not so long ago, I complained that &lt;a href=&quot;https://chemistry.stackexchange.com/q/16187/1677&quot;&gt;instead of copper plating I got copper powder&lt;/a&gt;. That's &lt;em&gt;just what I want&lt;/em&gt; now. I'm really sorry that powder I created the other day is already flushed down the drain. It could make beautiful green flames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3Eec.jpg&quot; alt=&quot;copper powder&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now while the general idea is easy, I'm not so sure about choosing the right electrolytes. My theoretical knowledge of electro-chemistry unfortunately sucks. And I'm afraid that every metal will need different electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example I think using any acid for Aluminium could cause something nasty since it reacts with them little bit too quickly. For Iron, I tried hydrochloric acid and then I found out thet $\ce{HCl}$ dissolves iron to form iron chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there general rules to figure this out? If not, what sould I consider to find electrolyte individually for these metals:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aluminium&lt;/li&gt;&#xA;&lt;li&gt;Iron&lt;/li&gt;&#xA;&lt;li&gt;Copper&lt;/li&gt;&#xA;&lt;li&gt;Magnesium&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Powdered metals can be bought of course, but I don't want to waste good money for something, that can be easily made from trash (old nails, aluminium foils from chocolate...).&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-09-30T05:39:43.167" Title="Making metal powders using electrolysis - chosing the right electrolyte" Tags="&lt;metal&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16873" PostTypeId="2" ParentId="16868" CreationDate="2014-09-29T06:43:56.773" Score="5" Body="&lt;p&gt;I don't believe there is anything really wrong with this question. Yes, both molecules have multiple signals, but one of them contains a signal which corresponds to it being the most deshielded. In fact, one of these, molecule F, contains the most deshielded signal AND the most shielded signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ...&quot; not necessarily so. The correlation between electron withdrawing groups and shielding/deshielding is a strong rule of thumb, but can become complicated with aromatic systems and competing electronegativities of substituents. In any case, you need to be looking at &lt;em&gt;parts&lt;/em&gt; of the molecule; shielding is quite a localised effect, and as I said above, molecule F actually has the most shielded AND deshielded protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if you consider the electronegativity of oxygen; a proton attached to this will be strongly deshielded. Stick them on an aromatic ring para to a carbonyl group, and you will generate a resonance system like below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4FkHo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the phenolic proton is by far the most deshielded of all protons in the molecules shown; I'd expect a chemical shift of over 10ppm. In response to some of the comments about -OH signals being over the place; it is important not to confuse raw chemical shift and averaged chemical shift resulting from chemical exchange with other labile protons, such as the (often observed) exchange of phenol protons with water. In these cases, the resulting chemical shift depends on the relative concentrations of the exchanging species, the exchange rate, and the original chemical shifts of the non-exchanging species.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-09-29T06:43:56.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16874" PostTypeId="2" ParentId="16847" CreationDate="2014-09-29T07:55:11.357" Score="0" Body="&lt;p&gt;I think you need to consider that if you take last electron from the orbital it's there, but it's empty. Orbital itself is a bit abstract thing and if you accept that it can be empty then you can have 0 valence electrons - that's why your book says what it says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not right it will be a lesson for me too :)&lt;/p&gt;&#xA;" OwnerUserId="8095" LastActivityDate="2014-09-29T07:55:11.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16875" PostTypeId="1" CreationDate="2014-09-29T12:04:24.623" Score="3" ViewCount="724" Body="&lt;p&gt;I am currently doing the questions:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Write equations for the reaction of chlorine with propene and 2-butene using structural formulas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So I know that propene is &lt;span class=&quot;math-container&quot;&gt;$\ce{H3C-CH=CH2}$&lt;/span&gt; and chlorine is &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jow9S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jow9S.png&quot; alt=&quot;Chlorination&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From this I observe that naturally, the double bond is removed in the presence of an addition reaction and that the chlorine atom bonds to the &lt;span class=&quot;math-container&quot;&gt;$\ce{CH}$&lt;/span&gt; pair and the other &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; atom does not bond to any of them but attaches itself at the end.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why did the second &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; pair not attach itself to &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2}$&lt;/span&gt;? Does this mean that atoms can only bond to &lt;span class=&quot;math-container&quot;&gt;$\ce{CH}$&lt;/span&gt; pair and nothing more than that, e.g. not &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt;, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Second question for the reaction of 2-butene with Chlorine. I know that 2-butene is &lt;span class=&quot;math-container&quot;&gt;$\ce{H3C-CH=CH-CH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bsZuE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bsZuE.png&quot; alt=&quot;Chlorination&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I noticed, like what happened above, the chlorine atom bonds to only the &lt;span class=&quot;math-container&quot;&gt;$\ce{CH}$&lt;/span&gt; pair. Here however there are two &lt;span class=&quot;math-container&quot;&gt;$\ce{CH}$&lt;/span&gt; pairs so the Chlorine doesn't need to attach itself at the end of the chain, unlike the first picture.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Are my observations correct? My textbook has no mention of this.&lt;/p&gt;&#xA;" OwnerUserId="8099" LastEditorUserId="4945" LastEditDate="2021-05-13T16:58:31.573" LastActivityDate="2021-05-13T16:58:31.573" Title="Preference of certain positions during chlorination of an alkene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16876" PostTypeId="1" CreationDate="2014-09-29T12:12:17.353" Score="0" ViewCount="93" Body="&lt;p&gt;Calculate the pH of a solution which contains 100mL of 0.1M $\ce{HCl}$ and 9.9mL of 1.0M $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While calculating the concentration of $\ce{H+}$ ions, why shouldn't we consider the H+ ion concentration from water?&lt;/p&gt;&#xA;" OwnerUserId="8100" LastEditorUserId="12860" LastEditDate="2015-11-13T20:27:25.160" LastActivityDate="2015-11-13T20:27:25.160" Title="Ionic Equilibrium" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16877" PostTypeId="1" AcceptedAnswerId="16881" CreationDate="2014-09-29T12:25:54.617" Score="5" ViewCount="298" Body="&lt;p&gt;I have this issue with a group of people intentionally damaging a bench. So far, they would paint curses over the bench with permanent marker. However after I removed the curses with ethanol, they damaged the bench physically and threw the trash all around (we used to have a trash can there, with sign &quot;Please do not throw trash around&quot; - they threw this trash can away too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to keep this question non-personal though, so my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I protect a nitro-based paint layer from other paints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The professional solutions are way too expensive, while the thing I want to do is simple: I'm looking for non-nitro based material and solvent that will form transparent layer over the original paint. In case of damage by nitro-based paints, I could wash the layer under them and then apply it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is following: you apply the protective layer over your paint. When somebody applies any paint over it, you've got 3 layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FLZEO.png&quot; alt=&quot;anti-graffiti layer theory&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter what the top layer is, as long as it does not dissolve our protection. After this, we apply solvent for the protective layer. Assuming there are microscopic holes in the top paint, the protective layer will dissolve and the top layer will go off with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mH7w.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The layer must not be obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was considering these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ethanol-based transparent paint (too expensive, suprisingly)&lt;/li&gt;&#xA;&lt;li&gt;Wax (parafin) based layer (I don't know how to dissolve wax, it's crystalic so not transparent)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1677" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:09:51.097" LastActivityDate="2015-10-30T17:09:51.097" Title="Simple and cheap anti-graffiti layer based non-paint solvent" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16878" PostTypeId="2" ParentId="16847" CreationDate="2014-09-29T12:29:16.503" Score="2" Body="&lt;p&gt;Valence electrons are generally regarded as being 'the outermost electrons' for a given atom.  Therefore, with neutral potassium, there is one valence electron.  If we take away the outermost electron, we now have a new set of outermost electrons being the 8 electrons in the 3s and 3p orbitals.  Ron is correct in his assessment of semantics.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-09-29T12:29:16.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16880" PostTypeId="2" ParentId="16875" CreationDate="2014-09-29T13:47:00.067" Score="6" Body="&lt;p&gt;In your first example, propene plus chlorine, one of the Cl atoms bonded to a &quot;CH&quot; and the other, that you described as bonding to the end, bonded to a &quot;CH&lt;sub&gt;2&lt;/sub&gt;,&quot; so it's not just CH groups getting all of the chlorines.  Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think part of your misunderstanding lies in the way that you're writing/imagining the alkene.  The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions.  The double bonded section is more or less flat, and the CH&lt;sub&gt;3&lt;/sub&gt; is spread out in space.  The chlorine atoms attach to the ends of the double bond, wherever that may be.  The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information &lt;a href=&quot;http://www.masterorganicchemistry.com/2013/03/15/bromination-of-alkenes-the-mechanism/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i6MrI.gif&quot; alt=&quot;Propene molecule sketches&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If my drawing isn't clear, you can try building a model of the molecule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH&lt;sub&gt;3&lt;/sub&gt; carbon, and that's unlikely.  As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen.  A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks.  With easy physical access and a lower energy barrier, the reaction goes forward.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-09-29T13:47:00.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16881" PostTypeId="2" ParentId="16877" CreationDate="2014-09-29T14:02:56.417" Score="4" Body="&lt;p&gt;The &quot;professional&quot; solutions here in the US are often &lt;a href=&quot;http://en.wikipedia.org/wiki/Anti-graffiti_coating&quot; rel=&quot;nofollow&quot;&gt;fluorinated paints&lt;/a&gt;. My understanding is that the graffiti literally slides off and people give up after a few tries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the costs, my suggestion would probably be a thick layer of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyurethane&quot; rel=&quot;nofollow&quot;&gt;polyurethane&lt;/a&gt;. These are often applied to seal wood floors and can be transparent or stained. If you apply multiple layers (say 2-3 on a floor) the surface is very smooth and slick. It's hard for paint or other stains to stick.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-29T14:02:56.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16882" PostTypeId="2" ParentId="16876" CreationDate="2014-09-29T14:03:25.817" Score="1" Body="&lt;p&gt;You &lt;strong&gt;can&lt;/strong&gt; consider the hydrogen ion concentration from the ionization of water. However, that quantity is a million fold less than the amount of hydrogen ion coming from $\ce{HCl}$ and can be considered insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, in this problem you could ask the same question about hydroxide from the ionization of water, and you'd find that the added hydroxide is ten-million fold greater than from the ionization of water.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4231" LastEditDate="2014-11-01T15:52:40.470" LastActivityDate="2014-11-01T15:52:40.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16883" PostTypeId="2" ParentId="16824" CreationDate="2014-09-29T14:12:04.997" Score="3" Body="&lt;p&gt;I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-09-29T14:12:04.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16884" PostTypeId="2" ParentId="16867" CreationDate="2014-09-29T14:19:46.093" Score="4" Body="&lt;p&gt;Two factors may be at play. (1) While bacteria and molds do indeed grow on sugar, a very high concentration of sugar inhibits growth. So fruit preserves, syrup, honey, molasses etc. do not need to be refrigerated and generally do not require other preservatives to be added. Soft drinks contain less sugar and are more susceptible to spoilage, hence the use of preservatives. I would definitely refrigerate a soft drink after opening it, if not consumed immediately. Acidity also helps inhibit growth (e.g. sweet pickles). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Another factor is that bacteria and molds require water for growth, so foods such as dried fruits, that contain a moderate level of sugar but only a small amount of water, may have a long shelf life.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5577" LastEditDate="2014-09-30T17:01:53.783" LastActivityDate="2014-09-30T17:01:53.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16886" PostTypeId="1" AcceptedAnswerId="16888" CreationDate="2014-09-29T15:51:53.623" Score="2" ViewCount="668" Body="&lt;p&gt;So I created a DC power supply of 15 V (special transformator connected to the net) and I have a graphite anode and stainless steel cathode. When I put them in a beaker next to each other in water, the electrolysis takes place as I see gas bubbles. However, when i put them in another container that consists of two beakers connected with a tube to allow liquid to flow through but to separate the gases, the reaction doesn't run while it is the same setup more or less. How come? It is true that the tube connecting the beakers is relatively thin and the distance is also quite large (10 cm or sth), so does this lead to a higher required overpotential? I really don't know why it doesn't work in the larger container. Any help is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3743" LastActivityDate="2014-09-30T03:26:00.480" Title="Electrolysis not running, maybe dependant on distance of electrodes?" Tags="&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16887" PostTypeId="1" AcceptedAnswerId="16889" CreationDate="2014-09-29T16:09:43.240" Score="6" ViewCount="5679" Body="&lt;p&gt;Would his be a valid synthesis? I'm just unsure about the oxidative clevage step. I know that there is a benzylic hydrogen. So should oxidative clevage work here? What happens to the fragment that is lopped off? Is it also oxidized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YlSQa.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YlSQa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="41328" LastEditDate="2017-08-05T19:30:57.513" LastActivityDate="2018-04-07T00:14:44.927" Title="Synthesis of benzophenone" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16888" PostTypeId="2" ParentId="16886" CreationDate="2014-09-29T16:34:21.503" Score="3" Body="&lt;p&gt;Solution resistance in a two electrode cell can make a big difference as it is totally uncompensated. You may be applying 15 V between the electrodes, but the redox reaction only occurs at a very close proximity to the surface (&amp;lt;10 nm), so it's the interfacial potential we're concerned with, which can be significantly different than the applied voltage if the solution is highly resistive. A tight constriction in the cell can certainly increase the resistance and you should make sure you're using an electrolyte as plain water is very resistive.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-29T16:34:21.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16889" PostTypeId="2" ParentId="16887" CreationDate="2014-09-29T16:39:26.983" Score="5" Body="&lt;p&gt;Your basic strategy to involve a &lt;strong&gt;Friedel-Crafts acylation&lt;/strong&gt; is correct, but your route is a bit circuitous.  You could acylate benzene directly to produce benzaldehyde in one step using carbon monoxide and hydrochloric acid to generate formyl chloride &lt;em&gt;in situ&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HTp05.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is &quot;just another&quot; Friedel-Crafts acylation, it has it's own name, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gattermann_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gatterman-Koch reaction&lt;/a&gt;, after its discoverers.  Once in hand, you could oxidize benzaldehyde to benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate approach would be to 1) brominate benzene to form bromobenzene, 2) form the Grignard reagent with magnesium and 3) react it with carbon dioxide to produce benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, with benzoic acid in hand, you can procede as you've indicated, on to benzophenone.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-04-05T17:35:28.343" LastActivityDate="2018-04-05T17:35:28.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16890" PostTypeId="2" ParentId="16872" CreationDate="2014-09-29T17:15:00.860" Score="0" Body="&lt;p&gt;Regarding the suggestion in the comments above: Concentrated sulfuric and nitric acids produce $\ce{SO2}$ and $\ce{NO2}$ gases, respectively, as they oxidize metals. This is a common way of dissolving many metals, but you need to do it somewhere where you won't breathe them in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're actually electroplating, you'll want to dilute them with a different electrolyte. Potassium nitrate or citrate or phosphate buffers are common choices. Anything that doesn't complex or form an insoluble salt with the metal will do, though adding things like surfactants may change the morphology of the deposited metal structure. Depending on the metal, you might get precipitation if you raise the pH too high, but you don't want to electroplate in concentrated acid as you'll just redissolve your deposits.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-09-29T17:15:00.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16891" PostTypeId="2" ParentId="16441" CreationDate="2014-09-29T17:26:36.727" Score="0" Body="&lt;p&gt;Since we're assuming these are all ideal gases, we can use Dalton's Law of Partial Pressures to solve this fairly easily; you never need to find moles of anything.  Unless something is strange, the percentages given are by volume or by pressure rather than by mass.  If they are intended to be by mass, what follows would need to be modified.  I suspect that the use of the words &quot;air mass&quot; in the problem are what caused the confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total pressure, 1.5atm, is equal to the sum of the partial pressures, each of which is proportional to its mole fraction in the container.  Because these are gases that have the same volume, 40L, their mole fractions are their percent of the total amount of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrogen, which is 78.1% of the total, has a partial pressure of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ 1.5 \cdot 0.781 = 1.1715\ atm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's true whether the oxygen is in the container or not.  To see that this is the case, we can remove the oxygen and do the same calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we remove 20.9% of the gas (in the form of oxygen) we are left with 81.1% of the original 1.5 atm remaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.5\cdot 0.811 = 1.1865\ atm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can determine the mole fraction of nitrogen in this new mixture of gases by adding up the remaining percentages and finding nitrogen's contribution to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$78.1 + 0.964 + 0.0378 =  79.1018\ units\ of\ gas\ total$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{78.1}{79.1018} \cdot 100\% = 98.73\%\ N_2\ after\ oxygen\ is\ removed$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, Dalton's Law of Partial Pressures tells us that the pressure of nitrogen will be its mole fraction times the total pressure, or 98.73% of 1.1865 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.1865 \cdot 0.9873 = 1.1714\ atm$&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-09-29T17:26:36.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16892" PostTypeId="2" ParentId="16623" CreationDate="2014-09-29T17:49:54.730" Score="10" Body="&lt;p&gt;First things first, I would check your code: there is no reason 400 unique integrals should take an hour, even with the most naive implementation. You are computing at a rate of an integral every ten seconds. While I agree with all the comments pointing to increasing parallelization, I think efficient scalar code is more important. Why scale up inefficient code? Considering GPUs is a whole 'nother beast, and from the problems my coworkers have expressed trying to program them I'd recommend avoiding them for the time being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing you can do is to exploit molecular symmetry. Heads up though: this won't fix everything, because many interesting molecules lack symmetry. But for methane, like you tried, this should help a ton. Of course, this isn't as simple as just looking up some algorithm, because depending on your program you need to rotate the nuclear orientations to be symmetric, or develop routines to detect point group symmetry, etc. That being said, there are --- as far as I know --- two primary ways of exploiting point group symmetry for integrals. The first is to create symmetry-adapted basis functions, so your basis set is symmetric, which means your wave function/integrals/Fock matrix/whatever will be symmetric as well. &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/62/2/10.1063/1.430484&quot;&gt;Ernie Davidson did this in the seventies&lt;/a&gt;. A slightly more modern reference by Helgaker and Taylor is given &lt;a href=&quot;http://folk.uio.no/helgaker/reprints/ModElectStructTheory_GaussianBasisIntegrals_1995.pdf&quot;&gt;here&lt;/a&gt;. Daniel Crawford's &lt;a href=&quot;http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project9&quot;&gt;electronic structure tutorial&lt;/a&gt; explores this as well (it's really well written --- a great introduction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way you can do it, and I believe this is more popular, is to create symmetry unique integrals and then build a partial Fock matrix only to take care of symmetrization later. If you see something like &quot;generating petite list&quot; in the output of your favorite electronic structure program, this is probably what it's doing. This method allows for integral screening more efficiently, I believe. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.560110408/abstract&quot;&gt;Dupuis and King&lt;/a&gt; did this in the late seventies, based on earlier work by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009261470802445&quot;&gt;Dacre&lt;/a&gt; and &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.560070109/abstract&quot;&gt;Elder&lt;/a&gt;. These are the standard references you'll find. There may be some more recent work that I'm unaware of. The problem of point group symmetry had been effectively solved by 1980.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, many interesting molecules don't have symmetry to take advantage of, so symmetry won't fix all your problems. The Schwarz equality is well addressed in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1118531477&quot;&gt;Molecular Electronic Structure Theory&lt;/a&gt; (actually, &lt;em&gt;most&lt;/em&gt; electronic structure theory is well addressed in that lovely pink book). If you haven't checked that out, please do. I think it will answer many of your questions about how modern codes work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to repeat the other answers (Geoff gave you the papers I'd have suggested as well), so I'll finish with two lesser-considered references. First, there is an awesome (but tiny!) book called &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1420051644&quot;&gt;Parallel Computing in Quantum Chemistry&lt;/a&gt; that lays out the basics for efficient code including integral evaluation. It's not too hard to read, and I found the work on local MP2 to be pretty interesting. I'm often surprised more people haven't heard of it. The other place I would check out is Ed Valeev's &lt;a href=&quot;http://sourceforge.net/p/libint/home/Libint%20--%20efficient%20computation%20of%20quantum%20mechanical%20matrix%20elements%20over%20Gaussian%20basis%20sets/&quot;&gt;LibInt package&lt;/a&gt;. He's a pretty talented programmer, and it's used in many modern packages like Psi and MPQC and ORCA, not to mention free for anyone to download and use. Some of my coworkers have played around with it for their own &quot;homemade&quot; electronic structure code, and speak highly of it. The user's manual is actually quite helpful understanding integral evaluation. He has also written several &lt;a href=&quot;http://www.valeevgroup.chem.vt.edu/teaching.html&quot;&gt;exercises and tutorials&lt;/a&gt;, which you may find useful.&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2014-09-29T17:49:54.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16893" PostTypeId="1" AcceptedAnswerId="16900" CreationDate="2014-09-29T19:02:49.890" Score="1" ViewCount="2228" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Complete the following reactions and write the spectators for each if any&lt;/p&gt;&#xA;&lt;p&gt;A. cobalt(II) chloride (aq) and cesium phosphate (aq)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am not quite sure if I have the correct answer to this equation, but I have &lt;span class=&quot;math-container&quot;&gt;$\ce{ CoCl2(aq) + Cs(PO4)(aq)}$&lt;/span&gt; (for the equation setup) and for spectator ions &lt;span class=&quot;math-container&quot;&gt;$\ce{Co^2+ (aq) + 2 Cl- (aq) + Cs^+ (aq) + (PO4) (aq) -&amp;gt; CsCl(aq) + Co(PO4)2(s)}$&lt;/span&gt; and so for the spectators I have &lt;span class=&quot;math-container&quot;&gt;$\ce{Cs^+(aq)\ and\ 2Cl- (aq)}$&lt;/span&gt;, Is this correct?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;B. sodium phosphate (aq) and calcium chloride (aq)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For this equation I have &lt;span class=&quot;math-container&quot;&gt;$\ce{6Na^+\ and\ 6Cl-}$&lt;/span&gt; as spectator ions, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca3(PO4)2(s)}$&lt;/span&gt; (as a solid)&lt;/p&gt;&#xA;" OwnerUserId="8103" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-11T03:08:31.790" Title="Completing inorganic reactions and writing spectator ions" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16894" PostTypeId="2" ParentId="16886" CreationDate="2014-09-29T19:09:14.080" Score="3" Body="&lt;p&gt;Resistance is proportional to length and reverse proportional to сross section of the conductor. When you increase distance by 20 cm from 2-3 cm (10 times resistance) AND indtroduce ion bridge about 1 or less cm in diameter (probably also 10+ times resistance, likely more), you effectively increase the resistance by several orders of magnitude, proportionally reducing the current and the speed of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industrial-grade applications when speed of reaction is of essence, like eletrolitic cutting, the distance between electrodes may be as small as several millimeters. Strong electrolite flow is also used.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-30T03:26:00.480" LastActivityDate="2014-09-30T03:26:00.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16895" PostTypeId="2" ParentId="16872" CreationDate="2014-09-29T19:21:47.433" Score="1" Body="&lt;p&gt;Fine copper powder can be prepared using thermal decomposition of copper citrate (can be prepared by mixing dirt cheap citric acid, three equivalents of a base like potassium hydroxide and copper sulphate). I once made pyrophoric (!) copper powder using this technique. There are mentions of preparation of iron powder same way. Anyway, it is much more practical than electrolysis for big batches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly speaking, fine aluminium flakes are widely employed in various paints and are dirt cheap. They should be available in shops for artists. There is no reason to make them at home (and aluminium reacts with water once cleaned of oxide film)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium is probably the worst. It is quite costly and active and making fine powder is dangerous. It also reacts with water and alcohols, so making it employing electrolysis is out of question. There are mentions of home smelting of aluminium and magnesium, producing magnalium. Magnalium is brittle and can be easily milled, and it is active enough so fine milling is usually not needed. While fancy, proper pyrotechnics can be made without magnesium with easy, aluminium is usually much more preferable. Magnesium often produce huge volumes of white smoke, a habit aluminium doesn't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of above, I strongly encourage to be cautions when working with metal powders, especially magnesium. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="1677" LastEditDate="2014-09-30T05:39:43.167" LastActivityDate="2014-09-30T05:39:43.167" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16897" PostTypeId="1" AcceptedAnswerId="16902" CreationDate="2014-09-29T19:49:40.640" Score="2" ViewCount="1528" Body="&lt;p&gt;Why would the following mechanism be valid for the conversion of t-butylbenzene to benzene in highly acidic medium followed by an aqueous workup? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem I have with this mechanism is the formation of the secondary doubly allylic carbocation. &lt;em&gt;Why&lt;/em&gt; not go for the tertiary, inductively and doubly allylic stabilized carbocation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I guess the problem with going down this path it becomes hard to see how the t-butyl group might leave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V1zyn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the book's mechanism is entirely dependent on having the secondary allylic carbocation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pl77a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-16T16:51:35.203" LastActivityDate="2020-03-02T01:15:08.363" Title="t-butylbenzene in highly acidic medium" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16899" PostTypeId="1" AcceptedAnswerId="16906" CreationDate="2014-09-29T19:56:38.740" Score="7" ViewCount="121" Body="&lt;p&gt;There is ongoing research where efforts were launched to find out the original constituents of a perfume for the female pharaoh &lt;a href=&quot;http://en.wikipedia.org/wiki/Hatshepsut&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hatshepsut&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPi3Y.jpg&quot; alt=&quot;Statue of Pharaoh Hatshepsut&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Hatshepsut#mediaviewer/File:Hatshepsut.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a perfume flask that was found in her tomb, there are apparently some remnants of the (assumed once-liquid) perfume that she used to wear. The perfume flask was sealed, so there was probably a very low amount of gas exchange taking place during the millennia that the flask took to wait for being discovered with its treasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there should still be a lot of reactions that can happen in this timescale, and since most of the reactions that just happen by themselves tend to lower energies, at some point different organic molecules would lead to the same product (e.g. acetic acid, or similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me, that then the only reasonable approach is to make a list of possible constituents and then start to play with the amounts thereof and model the decay reaction pathways. Then you could check if your theory predicts the experimental results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this really how you find out what an ancient perfume was made of?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;&lt;a href=&quot;http://www.livescience.com/3436-egyptian-pharaoh-smell.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;, &lt;a href=&quot;http://www.haaretz.com/news/latest-trend-in-scents-3-400-year-old-pharaoh-s-perfume-1.273167&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;, &lt;a href=&quot;http://www.zmescience.com/research/scientists-to-recreate-the-perfume-worn-by-pharaoh-hatshepsut/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; and &lt;a href=&quot;http://www.touregypt.net/egypt-info/magazine-mag03012001-mag4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;that&lt;/a&gt; (talking about Tutankhamun, but the same principles would apply).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-09-29T21:29:06.200" Title="How can we recreate old pharaos' perfumes?" Tags="&lt;organic-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16900" PostTypeId="2" ParentId="16893" CreationDate="2014-09-29T20:00:36.257" Score="0" Body="&lt;p&gt;For B, you have correctly identified &lt;a href=&quot;http://en.wikipedia.org/wiki/Tricalcium_phosphate&quot; rel=&quot;nofollow&quot;&gt;calcium phosphate&lt;/a&gt; as the precipitate, as it is the least soluble of all possible compounds. In the case of A, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobalt_phosphate&quot; rel=&quot;nofollow&quot;&gt;cobalt phosphate&lt;/a&gt; precipitates from the solution containing cobalt(II) and phosphate salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ca^{2+} + 2PO4^{3-} -&amp;gt;~ Ca3(PO4)2\downarrow}$$&#xA;$$\ce{3Co^{2+} + 2PO4^{3-} -&amp;gt;~ Co3(PO4)2\downarrow}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-09-29T20:00:36.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16901" PostTypeId="2" ParentId="16897" CreationDate="2014-09-29T20:06:19.123" Score="5" Body="&lt;p&gt;I think the problem you are having stems from locally optimizing the energy, but disregarding the global energy well that you get in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism your book gives is valid because it explains how this reaction could take place along a minimal-energy pathway. Since the reaction is observed, it somehow has to proceed from tert-butylbenzene to the products benzene and tert-butanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this is a slight tongue-in-cheeky kind of argument, but in the end it is a valid one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you protonate elsewhere (and we haven't discussed the protonation kinetics involved here&lt;sup&gt;1&lt;/sup&gt; but they might be important) you will not end up cleaving the right $\ce{C-C}$ bond (or any bond at all). The tert-butyl cation is quite stable, and having stable reaction products means that the reaction is more happy to proceed. Since the protonation is likely to be a fast equilibrium reaction, it will happen at many sites on the aryl group.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Frankly I have no clue about those, so maybe someone can help me out here...&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-09-29T20:06:19.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16902" PostTypeId="2" ParentId="16897" CreationDate="2014-09-29T20:35:09.057" Score="7" Body="&lt;p&gt;All 6 positions on the benzene ring are being protonated (the frequency of protonation at the different sites is related to the stability of the carbocation generated) in acidic media, so all of the ions you suggested are being formed.  Protonation at 5 of these positions can only regenerate starting material when one of the two protons at the initial site of protonation is removed (an identity reaction, we could make this pathway visible by running the reaction in deuterio-acid + $\ce{D2O}$).  Protonation of the benzene ring at the carbon bearing the substituent provides a choice.  Either the proton can again be removed or the stable t-butyl cation can be ejected.  All of these things are happening in the reaction, but only removal of the t-butyl group is visible.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-29T22:05:04.630" LastActivityDate="2014-09-29T22:05:04.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16904" PostTypeId="2" ParentId="14781" CreationDate="2014-09-29T21:05:12.190" Score="1" Body="&lt;p&gt;1) Yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of entalpy of one mole with cnange of temperature is known as $C_p$, a heat capacity at constant pressure. It is, unfortunately, a function of temperature, so one has to perform integration of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H= \int\limits^{T_{final}}_{T_{standard}}{C_p(T)}dT$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, $C_p$ is usually approximated as cubic polynom $C_p=aT^3+bT^2 +cT +d $, where $a,b,c,d$ are found in reference literature. However, the situation is further complicated by phase change. In case one calculates enthalpy of water at $120^oC$, he will have to perform integration both for liquid and gas phases in their respective regions of stability and then add heat of evaporation of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change of entropy is calculated in similiar manner, using the fact that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S= \int \frac {dQ} T$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;during heating to different temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S= \int\limits^{T_{final}}_{T_{standard}}\frac{C_p(T)}TdT $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After obtaining entalpies and entropies of reagents and products at different temperature, one can obtain $\Delta G$ of reaction at same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see it requires a lot of math, so such excercises are almost universally hated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Technically speaking, most reactions move to point of equilibrium, where concentraions of reagents and products are guareded by constant of equilibrium, like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K=\frac{[A]^a[B]^b...[E]^e}{[W]^w[X]^x...[Z]^z}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{w W + x X + ... z Z = a A + b B + ... + e E}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, if one start from products, i.e. when &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[W]=[X]=...=[Z]=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a sligh amount of reagents will appear as equilibrium is achieved.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-09-29T21:05:12.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16905" PostTypeId="1" CreationDate="2014-09-29T21:26:34.420" Score="6" ViewCount="1979" Body="&lt;p&gt;I need some help in developing a method by which to calculate how many tons of &quot;evil&quot; $\ce{CO2}$ are routinely being released into the atmosphere SOLELY due to our beverage addictions, notably beer and soft drinks.  I realize that not all manufacturers and products are consistent in their carbonation levels, and that an average probably has to be applied industry-wide (unless individual companies are required to record and publish such data) per unit measure or monthly/annual volume output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the proper forum to ask such a question?  If not, suggestions on where else I go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a previous post I got part of the answer, relating to an individual 12 oz can, or 1 liter, but I'm wondering if there is an easier way in which I can go about plugging weekly, monthly or annual volumes of beverages into formulas which will yield the results I'm looking for, notably a discussable unit of measurement in tons or ?  I'm willing to do all the numbers work once I'm pointed in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if anybody's ever thought about this before, because I've never heard it discussed.  But if, according to some environmentalists, we should all watch our carbon footprint (and I'm not debating that here), then it seems like the most logical place to start would be with our hundreds of millions of daily beverage choices?  Isn't it?&lt;/p&gt;&#xA;" OwnerUserId="8107" LastEditorUserId="16088" LastEditDate="2016-09-22T12:02:58.503" LastActivityDate="2020-03-06T22:49:00.947" Title="Quantifying CO2 generated/released by the soft drink industry and breweries" Tags="&lt;food-chemistry&gt;&lt;green-chemistry&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16906" PostTypeId="2" ParentId="16899" CreationDate="2014-09-29T21:29:06.200" Score="5" Body="&lt;p&gt;The analysis won't be done before lunch, but neither will it be an extremely difficult process.  Certainly some of the researchers will have a fragrance industry background  and access to necessary equipment like  a tandem gas chromatograph - mass spectrometer.  Also, Pharoah's perfumes were undoubtedly less complex than todays, and all oils used in perfumes are available as &quot;known&quot; compounds for comparison.  Additionally, all fragrances have been &quot;fingerprinted&quot; making identification of the compound in a sample fairly routine and straightforward.  By &quot;fingerprinted&quot; I mean that if a certain pattern of compounds can be detected than they will suggest the presence of a certain oil.  For example, look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Lavender_oil&quot;&gt;lavender oil&lt;/a&gt;, there are about 30 chemical components in it - just like a fingerprint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the base solvent has evaporated through the wall of the container and oxygen has flowed into the container.  Certainly some of the ingredients will have degraded, but the container won't be full of acetic acid.  Some amount of an alcohol may have oxidized to the corresponding ketone, aldehyde or carboxylic acid, perhaps some amount of a conjugated diene may have oxidized to the corresponding dienol; but that shouldn't make the process that much more difficult.   &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-29T21:29:06.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16907" PostTypeId="1" AcceptedAnswerId="16908" CreationDate="2014-09-29T21:49:47.683" Score="10" ViewCount="1610" Body="&lt;p&gt;I want to run classical molecular dynamics simulations of a periodically replicated surface (rutile $\ce{TiO2}$ with grooves). In order to do so, I first need to solve for the partial charges residing on each atom of the surface. What electronic structure methods are available for determining the partial charges on a surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am vaguely familiar with Mulliken population analysis, but my understanding is that this method is not the most accurate. What other methods are commonly used, and how accurate are each of these methods?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, what are some good (and free) packages for performing such calculations? I've used SIESTA in the past, is that well-suited for determining partial charges of a surface?&lt;/p&gt;&#xA;" OwnerUserId="7693" LastEditorUserId="189" LastEditDate="2014-09-30T21:33:04.800" LastActivityDate="2014-09-30T21:33:04.800" Title="Methods for Determining Partial Charges" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;&lt;surface-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16908" PostTypeId="2" ParentId="16907" CreationDate="2014-09-30T01:18:01.373" Score="14" Body="&lt;p&gt;There are &lt;em&gt;lots&lt;/em&gt;, and I mean &lt;strong&gt;&lt;em&gt;lots&lt;/em&gt;&lt;/strong&gt; of methods and software programs to produce partial charges. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_charge&quot;&gt;Wikipedia&lt;/a&gt; for a small (incomplete list)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with the basics. The idea of a partial atomic charge, while useful for concept, can not be defined uniquely. Quantum chemical methods (whether wavefunction or DFT) produce some sort of electron density. Different schemes divide up that electron density in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like Cramer and Truhlar's categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Class I&lt;/strong&gt;: Some sort of intuitive or empirical approach (i.e., non-QM). These include methods based on experimental dipole moments and atomic electronegativities, including Gasteiger-Marsili partial charges and other &quot;electronegativity equalization methods&quot; (EEM).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Class II&lt;/strong&gt;: Partitioning using wave functions / orbital schemes (including &lt;a href=&quot;http://en.wikipedia.org/wiki/Mulliken_population_analysis&quot;&gt;Mulliken charges&lt;/a&gt; while very efficient to compute, are basis set dependent.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Class III&lt;/strong&gt;: Partitioning the electron density or electrostatic potential (e.g., Hirshfeld, density-fitting schemes, etc.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Class IV&lt;/strong&gt;: Semiempirical mapping from Class II or Class III (preferred) to match experimental data like dipole moments.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Almost any modern quantum package will include several schemes to fit partial charges. My personal preference is for something like Hirshfeld charges (i.e., from the electron density, so not basis-set dependent), or electrostatic potential fitting schemes like Merz-Kollman or CHelpG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the latter (electrostatic potential fitting) will be better generally for molecular dynamics, since you're attempting to produce point charges that will best represent the quantum mechanical electrostatic potential.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-09-30T01:18:01.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16909" PostTypeId="2" ParentId="16907" CreationDate="2014-09-30T01:50:17.980" Score="4" Body="&lt;p&gt;Kinda obvious, but still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you have to decide, what face you are working with. $\ce{TiO2}$ may exhibit several different faces, you'll have to consult literature to find out which are the most frequent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you have to consult literature to find which surface species exist on the surface in your conditions. Generally, oxides tend to adsorb water forming hydroxide groups and some other small molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that you have to build a model slab of the surface. Generally, two-three layers of atoms close to surface are 'relaxed' and two-three 'deeper' ones are fixed in positions, modelling unperturbed crystal structure. However, ideally you should perform several computations, gradually increasing amount of layers in slab until the characteristic you considering stabilize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally recomended to employ periodic approach using plane-wave sets, possibly in DFT method, as it greately reduces memory requirements and amount of atoms (and electrons) considered.  Given you care to use the charges to model electrostatic potential, Mulliken and Natural charges go to garbage. Ideally you should use full-scale electrostatic potential. However, if it is not an option, analog to Gaussian Pop=CHelpG should be used, that fit atomic charges to produce  electrostatic potential closest to one produced for given electronic distribution. Not aware of other programs implementing similar schemes, so you have to find one on your own.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-09-30T03:18:09.380" LastActivityDate="2014-09-30T03:18:09.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16910" PostTypeId="1" AcceptedAnswerId="16911" CreationDate="2014-09-30T04:18:22.280" Score="2" ViewCount="3589" Body="&lt;p&gt;Q)The &lt;strong&gt;derived&lt;/strong&gt; name of (CH&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;4&lt;/sub&gt;C is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A)Tetramethylmethane &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B)2,2-Dimethylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C)Neopentane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D)None of these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer was A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a derived name?Never heard of it in any book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="73" LastEditDate="2014-09-30T04:22:01.987" LastActivityDate="2014-09-30T05:49:55.307" Title="What are derived names?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-09-30T13:27:06.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="16911" PostTypeId="2" ParentId="16910" CreationDate="2014-09-30T05:26:33.830" Score="5" Body="&lt;p&gt;There are (were) three main nomenclature systems used for naming organic compounds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Common Name (or trivial name)&lt;/strong&gt;. Under this convention this molecule is neopentane. For alkanes, this naming system is based on the positioning of methyl substituents, and gives rise to the iso-, sec-, neo- namiong prefixes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Derived Name&lt;/strong&gt;. Who uses this method these days???? This system identifies the central carbon atom which has the most carbon groups attached, and then names the groups as a prefix, and ends the name with 'methane'. The formula are often written with the central carbon unit in square brackets. You may still see derived names on MSDS/SDS under the 'Other Names' section. Under this system this molecule is tetramethylmethane. I can think of more important things to teach you in class if this is what you are learning. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;IUPAC Name&lt;/strong&gt;&#xA;This is the accepted naming convention in use today, which allows for retained names, which usually have their origins in the trivial naming system. Its naming system is based on a systematic framework and is used universally. In older texts (pre 1970s), it was known as the Geneva system or congress. Under this convention, this molecule is 2,2-dimethylpropane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't even find any decent references for you at this point, but will update if I find some.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-09-30T05:49:55.307" LastActivityDate="2014-09-30T05:49:55.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16912" PostTypeId="2" ParentId="16747" CreationDate="2014-09-30T05:27:38.170" Score="3" Body="&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound is neutral, So the sum of oxidation state of all the element should be ZERO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what you have got, is sum of oxidation state of all the element is zero? This way you can check whether you are right or wrong. &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-09-30T05:27:38.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16913" PostTypeId="1" CreationDate="2014-09-30T05:45:38.383" Score="2" ViewCount="89" Body="&lt;p&gt;Even one letter can make a huge difference in Chemistry Nomenclature. So I am looking for credible resources that is specifically published for explaining Nomenclature rules, examples in Spanish. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="4945" LastEditDate="2014-09-30T11:12:28.647" LastActivityDate="2014-09-30T11:12:28.647" Title="Nomenclature in Other languages" Tags="&lt;nomenclature&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16914" PostTypeId="1" AcceptedAnswerId="16915" CreationDate="2014-09-30T08:25:50.767" Score="5" ViewCount="20358" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/08mJh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/08mJh.png&quot; alt=&quot;D-tartaric acid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Why is the isomer of tartaric acid above labelled as ᴅ and (−)?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;it rotates plane-polarized light clockwise&lt;/li&gt;&#xA;&lt;li&gt;it rotates plane-polarized light anticlockwise&lt;/li&gt;&#xA;&lt;li&gt;it is an equimolar mixture of the enantiomorphs&lt;/li&gt;&#xA;&lt;li&gt;its configuration is related to that of ᴅ-glyceraldehyde&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If I am not wrong, the ᴅ in ᴅ-glyceraldehyde refers to (+)-glyceraldehyde, and likewise ʟ-glyceraldehyde is (−)-glyceraldehyde. The (+) and (−) labels indicate that these isomers are dextrorotatory and levorotatory, respectively.&lt;/p&gt;&#xA;&lt;p&gt;So in ᴅ-(−)-tartaric acid, what does the ᴅ mean, if not &lt;em&gt;dextro&lt;/em&gt;, and what does the (−) mean, if not &lt;em&gt;levo&lt;/em&gt; (the minus sign indicating anticlockwise rotation of plane-polarized light)? How can one molecule be both?&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-09-05T13:57:33.203" Title="What do the stereochemical labels D and − mean in tartaric acid?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16915" PostTypeId="2" ParentId="16914" CreationDate="2014-09-30T09:15:45.863" Score="7" Body="&lt;p&gt;As far as I remember, small &lt;em&gt;d&lt;/em&gt; always means the same as (+), not small cap ᴅ. So, a dextrorotatory compound is prefixed with either (+) or &lt;em&gt;d&lt;/em&gt; while a levorotatory compound is prefixed with either (−) or &lt;em&gt;l&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small caps ᴅ and ʟ prefixes have different meaning: if an enantiomer can in principle be synthesized from naturally occurring (+)-glyceraldehyde without affecting its configuration, it is considered the ᴅ-form. Otherwise, it is considered the the ʟ-form. And, say, the ᴅ-form of a molecule is not necessarily a dextrorotatory one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answers to your question are options 2 (the (−) sign indicates that it rotates plane-polarized light anticlockwise), &lt;em&gt;and&lt;/em&gt; option 4 (the ᴅ label indicates that it is related to ᴅ-glyceraldehyde).&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="16683" LastEditDate="2017-09-05T13:51:23.843" LastActivityDate="2017-09-05T13:51:23.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16916" PostTypeId="1" CreationDate="2014-09-30T13:33:09.183" Score="2" ViewCount="662" Body="&lt;p&gt;Is there a relation (or formula) between the sign of the magnetic susceptibility and the gyromagnetic ratio of a nucleus in NMR?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="30836" LastEditDate="2020-04-30T08:27:42.800" LastActivityDate="2020-04-30T08:27:42.800" Title="Relation in the sign (+/-) for gyromagnetic ratio and magnetic susceptibility" Tags="&lt;nmr-spectroscopy&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16917" PostTypeId="1" CreationDate="2014-09-30T14:02:29.870" Score="4" ViewCount="1662" Body="&lt;p&gt;I know it seems to be a weird question. But for long I have been thinking whether there is any relation between thermodynamic reversible process and reversible reaction. Do they have any connection and if so how? &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="5456" LastEditDate="2014-09-30T14:07:03.297" LastActivityDate="2019-12-03T15:04:35.000" Title="Relation between thermodynamic reversible process and reversible reaction" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16918" PostTypeId="1" AcceptedAnswerId="16928" CreationDate="2014-09-30T16:37:49.630" Score="5" ViewCount="215" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9JiSu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;This is a question which I come across in an examination. I know the answer and also know the solution. Well when I first saw this problem I didn't know how to approach this problem and I think I will still not be able to predict the correct answer of a question which will be based on this question's type . I am not asking that this is &quot;this&quot; reaction and it must proceed like&quot; this &quot; ,but what I want is how will you approach this problem . What will you think on first look of the problem. What all things you will observe from question and also do write what you think the answer must be . Also BEWARE THIS HAS MULTIPLE ANSWERS CORRECT. I am not telling you the answer because it would be easier to explain that answer . Also, please those who have seen this problem earlier. Please resist yourself from posting answers.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="4231" LastEditDate="2015-03-25T19:07:51.730" LastActivityDate="2015-03-25T19:07:51.730" Title="Effects Guiding Electrophilic Aromatic Substitution" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16919" PostTypeId="2" ParentId="16623" CreationDate="2014-09-30T16:44:29.377" Score="6" Body="&lt;p&gt;There are two things here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Quantum chemistry problem  &lt;/li&gt;&#xA;&lt;li&gt;Optimization problem&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Quantum chemistry or the Schrodinger equation is simplified using DFT and some other techniques:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) ECP or effective core potentials and pseudopotentials: Instead of solving for all electrons just the outer shell electrons are solved and core electrons are considered using pre calculated integrals&lt;br&gt;&#xA;b) ri approximation: Resolution of identity, it treats two electron term as a product of single particle orbital function thus greatly reduces computational cost  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as mentioned above there are some other good techniques are used to deal with quantum chemistry part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Optimization problem:&lt;/strong&gt; In optimization usually second derivative (Hessian matrix) is used to find minima or maxima but in computational chemistry code only ist derivative or jacobian is used, thus it is highly sensitive to initial guess and computational setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating Hessian is the most costly operation of computational chemistry code, either approximate hessian is used or no hessian is used in this types of optimization problem. If you want to make sure that you got the right geometry you can calculate Hessian with your optimized geometry and for geometry optimization all the frequency should be positive and for transition state only one frequency should be negative but you will find that to meet this requirement is highly demanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover to reduce computational cost we can apply periodic boundary condition. It enables us to solve the problem for minimum non repititive unit cell ( Irreducible Kpoints). Finding the symmetry is also an crucial part of this types of problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut off energy is used to reduce the number of basis sets considered. And also optimization techniques also has great effect on computational cost, you may use fast but aggressive algorithm but that may not be able to find the minima or maxima.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read VASP or TURBOMOLE manual (available freely on online), there you will find some good techniques for reducing the computational cost. And above all as it is mentioned above efficient implementation of parallelization can help to reduce computational cost dramatically. One another potential method is to use GPU, but it is still in it's primary stage.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7448" LastEditDate="2015-04-24T14:43:05.913" LastActivityDate="2015-04-24T14:43:05.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16920" PostTypeId="2" ParentId="16917" CreationDate="2014-09-30T16:46:24.287" Score="3" Body="&lt;p&gt;In thermodynamics, a reversible process -- or reversible cycle if the process is cyclic -- is a process that can be &quot;reversed&quot; by means of infinitesimal changes in some property of the system without entropy production (i.e. dissipation of energy).[1] Due to these infinitesimal changes, &lt;strong&gt;**the system is in &lt;em&gt;thermodynamic equilibrium&lt;/em&gt; throughout the entire process.**&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamic equilibrium is an axiomatic concept of classical thermodynamics. It is an internal state of a single thermodynamic system, or a relation between several thermodynamic systems connected by permeable walls. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dO YOU OBSERVE SUCH KIND OF equilibriums IN A REACTION. SO THEY ARE NOT RELATED AT ALL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACCORDING TO WIKIPEDIA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as&#xA; aA + bB ---&gt; cC + dD&#xA;A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is &lt;strong&gt;&lt;em&gt;distinct&lt;/em&gt; from reversible process in thermodynamics&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-09-30T16:46:24.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16921" PostTypeId="1" AcceptedAnswerId="28246" CreationDate="2014-09-30T17:01:38.173" Score="1" ViewCount="115" Body="&lt;p&gt;Can anyone explain why at a conical intersection the wavefunction changes sign? My understanding is that it is a test to see if the crossing IS indeed a conical intersection or just a coincidence of potential energy surfaces, which do not exhibit such phase changes upon periodic cycling. If this is correct ... why? Further what can we gain from knowing that?&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="189" LastEditDate="2014-09-30T20:50:52.467" LastActivityDate="2015-04-05T14:27:02.320" Title="Conical intersections - phase" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16922" PostTypeId="1" AcceptedAnswerId="16930" CreationDate="2014-09-30T18:14:47.883" Score="28" ViewCount="34848" Body="&lt;p&gt;Okay, so I know that this is about filling the orbitals of the atom, and I understand that. What I don't understand is why? For example, an Oxygen atom has 8 protons and 8 electrons spinning around it. Oxygen wants to have 8 valence electrons so it takes 2 electrons from other elements when bonding. But, since Oxygen has 8 protons and now has 10 electrons around it, it has a charge of -2. I don't understand how having a charge is more stable than having no charge at all, shouldn't the most stable version of an element be its neutral state? Why do the atoms want to look like the next noble gas?&lt;/p&gt;&#xA;" OwnerUserId="5076" LastEditorUserId="189" LastEditDate="2014-09-30T19:04:39.403" LastActivityDate="2019-11-11T03:18:02.787" Title="Why do atoms &quot;want&quot; to have a full outer shell?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16923" PostTypeId="1" AcceptedAnswerId="16926" CreationDate="2014-09-30T18:31:10.737" Score="4" ViewCount="1600" Body="&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/GpBK0ia.png&quot; alt=&quot;butanone 13C&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the correct $ ^{13}C $ NMR spectrum of butanone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $ ^{13}C $ NMR spectrum of butanone, I figure that the peak locations of the first and third carbons should be switched (27.3, 35.2). Shouldn't the third carbon be shifted upfield by the terminal methyl group and therefore have a &quot;smaller ppm&quot; than the first carbon?&lt;/p&gt;&#xA;" OwnerUserId="8113" LastActivityDate="2014-09-30T21:29:59.387" Title="13C NMR peak location in butanone" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16925" PostTypeId="1" CreationDate="2014-09-30T19:59:12.160" Score="5" ViewCount="814" Body="&lt;p&gt;Basically I need to draw the molecular orbital for $\ce{F2}$ and then answer a bunch of questions about it. I have drawn it correctly, as far as I know, but I don't know how to use it to show that the vibrational frequency of $\ce{F2-}$ is lower than $\ce{F2}$. &#xA;I worked out the bond order for $\ce{F2-}= 0.5$  and for $\ce{F2}= 1$. Is this what you use to find it out, or is it something to do with the HOMO-LUMO?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="46306" LastEditDate="2018-03-04T02:38:27.977" LastActivityDate="2018-03-04T02:38:27.977" Title="Show the vibrational frequency of fluorine molecule anion is much lower than that of fluorine molecule" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16926" PostTypeId="2" ParentId="16923" CreationDate="2014-09-30T20:04:31.713" Score="4" Body="&lt;p&gt;No, substitution by an alkyl group produces a small &lt;strong&gt;downfield&lt;/strong&gt; shift in both proton and carbon nmr. Therefore, in a typical hydrocarbon a secondary carbon will be downfield of a primary carbon.  The assignments in your figure are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible explanation for the downfield shift of a carbon atom when a proton is replaced with another carbon atom is as follows.  Hydrogen is more electropositive than carbon, therefore, in a C-H bond a significant amount of the electron density in the bond resides around the carbon.  When we replace the electropositive hydrogen with a relatively more electronegative carbon, then there is less electron density around the central carbon atom.  This reduction in electron density deshields the carbon nucleus and shifts it downfield.  Every time another hydrogen is replaced by carbon a further downfield shift occurs.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-09-30T21:29:59.387" LastActivityDate="2014-09-30T21:29:59.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16928" PostTypeId="2" ParentId="16918" CreationDate="2014-09-30T20:23:47.437" Score="5" Body="&lt;p&gt;The first thing I notice is the hydroxyl group, a very powerful activating group and an ortho-para director.  The hydroxyl group is such a strong activator that if you take a solution of phenol and drip bromine into it, the phenol will react with the bromine so rapidly that the bromine color will be immediately removed from the solution as 2,4,6-tribromophenol is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing I notice is the very bulky t-butyl group, also an activator, but much milder than hydroxyl and also a mild ortho-para director.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I notice that the two substituents are arranged meta to one another.  There is one position in between the two groups that will be sterically blocked (double-ortho effect); the position on the other side of the t-butyl group will also be sterically blocked, but not as bad.  Also the directing effect of the hydroxyl group will overwhelm the directing effect of the t-butyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, in the halogen series size increases from chlorine to iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the reaction with iodine, the position ortho to the hydroxyl group and para to the t-butyl group is the only position that is electronically activated and sterically accessible to the bulky iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the reaction with smaller bromine, the position ortho to the hydroxyl and para to the t-butyl and the position para to the hydroxyl and ortho to the t-butyl are electronically activated by both groups and sterically accessible.  The &quot;double-ortho&quot; position is still sterically blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the smaller chlorine atom, all 3 positions - even the &quot;double ortho&quot; - position are sterically accessible to the smaller chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this reasoning, I would select answers A, B, and C.  I would probably also include D because the results are (weakly) reinforced by the electronic effect of the t-butyl group.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-09-30T20:23:47.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16929" PostTypeId="1" CreationDate="2014-09-30T20:43:33.810" Score="3" ViewCount="133" Body="&lt;p&gt;I was carrying out an experiment into the effect of temperature changes on the standard potential of a cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(s) +2Fe^{3+}-&amp;gt;Zn^{2+} +2Fe^{2+}}$$&#xA;I used equi-molar concentrations of $\ce{Fe(CN)_6^{4-}}$and $\ce{Fe(CN)_6^{3-}}$. Thus the Nernst equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{E^{\theta}_{cell}=E_{cell}-{\frac{RT}{nF}}*ln1}$$ reduces to $$\mathrm{E^{\theta}_{cell}=E_{cell}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We plotted our $\mathrm{E^{\theta}_{cell}}$ values versus &lt;strong&gt;T&lt;/strong&gt; and noticed that our $\mathrm{E^{\theta}_{cell}}$ values were actually &lt;em&gt;increasing&lt;/em&gt; with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have expected that these values go down as the concentration of the reactant falls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked my demonstrator and sourced information from different web resources, but can't access a definitive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints even? &lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="4356" LastEditDate="2014-09-30T22:50:02.550" LastActivityDate="2014-10-02T17:33:20.623" Title="Electrochemical potentials" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16930" PostTypeId="2" ParentId="16922" CreationDate="2014-09-30T20:46:46.967" Score="18" Body="&lt;p&gt;You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a compound in very broad strokes. Often these broad strokes are accurate enough to make some meaningful statements about molecular properties but it does not accurately describe the true electron or charge distribution in a compound. Take water for example. As you say, the 8-electron rule would predict that the hydrogen atoms each transfer one electron to the oxygen molecule which would then carry a charge of -2. But experiments show a different result: &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/112/21/10.1063/1.481541&quot;&gt;here&lt;/a&gt; you can find a paper that determined &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;that a charge of approximately $0.5 e$ is transferred from each&#xA;  hydrogen atom to the oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which would amount to a partial charge of $-1 e$ on oxygen and &lt;a href=&quot;http://www1.lsbu.ac.uk/water/molecule.html&quot;&gt;here&lt;/a&gt; you can find a similar claim:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about&#xA;  $-0.7e$ on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And there not being a $-2e$ charge on oxygen is quite understandable if you look at the electron affinities of oxygen (e.g. &lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/eas.html&quot;&gt;here&lt;/a&gt;): You see that the first electron affinity is $-142 \, \mathrm{kJ}/\mathrm{mol}^{-1}$ and thus energy is gained by the process of transferring a free electron to a neutral oxygen atom. But the second electron affinity is $+844 \, \mathrm{kJ}/\mathrm{mol}^{-1}$, so another electron transfer is very costly. The quite extreme situation of a -2-charged oxygen described by the 8-electron rule is very rarely found in real compounds. Even in ionic compounds like $\ce{NaCl}$ the electron is not 100% completely transfered from the electropositive to the electronegative element.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-09-30T20:46:46.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16931" PostTypeId="2" ParentId="16925" CreationDate="2014-09-30T21:06:23.000" Score="8" Body="&lt;p&gt;You should definitely post the MO diagram of $\ce{F2}$ that you worked out. When trying to solve such a problem you should try to analyse the question as thoroughly as possible to make sure that you collect all the information available to you beforehand. Especially important here is that the question already tells you that the vibrational frequency of $\ce{F2^{-}}$ is lower than that of $\ce{F2}$. So the first question you should ask yourself is: Under what circumstances is a vibrational frequency lowered and which relation do I know that links the vibrational frequency to some variable that is connected to the MO diagram of $\ce{F2}$? I give you a hint: You will need the equation of a harmonic oscillator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \omega = \sqrt{\frac{k}{m}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega = 2 \pi f$ is the angular frequency, $f$ is the frequency, $k$ is the force constant (in your case of the bond and thus its a measure of the bond strength) and $m$ is the reduced mass of the oscillating system. (The harmonic oscillator is one of those simple model systems that can help you quite often when you want a starting point for the investigation of much more complicated ones.) Now ask yourself: Which of the variables $k$ or $m$ might have a relation to the MO diagram and how will they be related?&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-09-30T21:19:03.067" LastActivityDate="2014-09-30T21:19:03.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16932" PostTypeId="1" CreationDate="2014-09-30T23:13:21.687" Score="1" ViewCount="798" Body="&lt;p&gt;I have some old pieces of wire. Since I want to use copper in homemade fireworks, I'm trying to gather it from the wires. I could buy it, however harvesting is much more fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulphuric acid doesn't dissolve copper in normal conditions, even when it's concentrated. However adding $\ce{H2O2}$ causes vigorous reaction during which copper sulphide is created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B8W38.jpg&quot; alt=&quot;copper sulphide solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the solution surface, remains of impurities are floating. I'm afraid, however, that most of impurities were dissolved. Were they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need my $\ce{CuSO4}$ chemically clean, but too much impurities could increase the risk of toxic gases during electrolysis.&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-09-30T23:13:21.687" Title="Can I use piranha solution to clean copper out of impurities?" Tags="&lt;safety&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16933" PostTypeId="1" CreationDate="2014-09-30T23:27:09.543" Score="2" ViewCount="6287" Body="&lt;p&gt;My mother works in a ceramics workshop. Besides giving lessons she creates various ceramic dishes, cups and plates. She often uses the effect of colored glass melting in the product and then leaving beautiful patterns at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it always cracks a lot as cooling down. Sometimes, it looks good when it's cracked, but other times she's trying to make it look like a gem - and cracked gem is not very gemmy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was curious if there is something &lt;em&gt;non toxic&lt;/em&gt;, insoluble in water and alcohol, that would melt at the similar temperature as glass and would prevent it from cracking in the cool down phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass doesn't have to be temperature resistant in the final products. We just want to avoid the cracks&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="23561" LastEditDate="2018-07-03T00:10:30.760" LastActivityDate="2021-02-07T02:16:44.293" Title="Prevent molten glass from cracking when cooling down" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16934" PostTypeId="1" AcceptedAnswerId="16937" CreationDate="2014-10-01T01:36:27.850" Score="5" ViewCount="229" Body="&lt;p&gt;Just started to study organic compounds and I am curious about functional groups. I want to know if functional groups can exist as molecules themselves.&lt;/p&gt;&#xA;" OwnerUserId="8121" LastEditorUserId="7475" LastEditDate="2015-10-09T16:07:35.620" LastActivityDate="2015-10-09T16:07:35.620" Title="Can functional groups exist by themselves" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16935" PostTypeId="2" ParentId="16929" CreationDate="2014-10-01T01:41:54.587" Score="5" Body="&lt;p&gt;A few thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$&#xA;Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-10-02T17:33:20.623" LastActivityDate="2014-10-02T17:33:20.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16936" PostTypeId="2" ParentId="16905" CreationDate="2014-10-01T02:49:17.333" Score="6" Body="&lt;p&gt;The xkcd What If? blog, by Randall Munroe, addressed a similar question.  He states the result better than I can.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For most of the history of civilization, there were about 270 parts&#xA;  per million of carbon dioxide in the atmosphere. In the last hundred&#xA;  years, industrial activity has pushed that number up to 400 parts per&#xA;  million. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One &quot;part per million&quot; of CO&lt;sub&gt;2&lt;/sub&gt; weighs about 7.8 billion tons.&#xA;  A can of soda contains in the neighborhood of 2.2 grams of CO&lt;sub&gt;2&lt;/sub&gt;, so you&#xA;  would need about 450 quadrillion cans of soda. That's enough to cover&#xA;  the Earth's land with ten layers of cans.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Randall Munroe - xkcd What If? #88:Soda Sequestration. &lt;a href=&quot;http://what-if.xkcd.com/88/&quot; rel=&quot;noreferrer&quot;&gt;http://what-if.xkcd.com/88/&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For further perspective, each breath of a healthy adult male involves about 500mL of gas exchange.  The air is, on average, 0.04% CO&lt;sub&gt;2&lt;/sub&gt; prior to inhaling.  Exhaled air contains between 4 and 5 percent carbon dioxide, I'll call it 4.04% for a nice, round 100-fold addition.  The mass of 500mL of air is about 0.6g, and though it will be slightly more dense with the added carbon dioxide, I'll assume that it remains at that mass for simplicity's sake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 4% of the 0.6g is CO&lt;sub&gt;2&lt;/sub&gt; that was added to the atmosphere by a single adult male breath.  That's 0.024g.  About 90 breaths add as much CO&lt;sub&gt;2&lt;/sub&gt; to the atmosphere as the carbonation in a can of Coke.  Where does the additional carbon in those breaths come from?  The Coca Cola itself, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sugar in the can of Coke (39g) contains about 16g of carbon atoms.  After being metabolized, that turns into 57.2g of carbon dioxide.  Not considering any of the energy required to harvest, process, or transport the sugar, that's still 26 times a much carbon dioxide going into the atmosphere from the calories in a can of Coke as opposed to its carbonation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it more succinctly, the CO&lt;sub&gt;2&lt;/sub&gt; output from the carbonation in Coke could be completely offset by reducing overall caloric intake by 6 calories per can.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-01T02:49:17.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16937" PostTypeId="2" ParentId="16934" CreationDate="2014-10-01T05:52:35.037" Score="6" Body="&lt;p&gt;A functional group on its own essentially has at least one non-bonded electron, thus making it a free radical, this makes them highly reactive molecules, making them react with other substances (including themselves) and can thus only exist in very low concentrations in inert medium/vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are molecules such as $\ce{NO}$, $\ce{NO2}$ which do have an odd electron in their molecular state, can exist in reasonably large quantities, and can be functional groups in organic chemistry, these are in some sense functional groups that exist as molecules.&lt;/p&gt;&#xA;" OwnerUserId="7584" LastEditorDisplayName="user7951" LastEditDate="2015-07-01T19:09:55.467" LastActivityDate="2015-07-01T19:09:55.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16938" PostTypeId="2" ParentId="16816" CreationDate="2014-10-01T06:16:43.747" Score="2" Body="&lt;p&gt;You could start by drawing the resonance structure of the naphthalene ring, this gives a compound in which the conditions on the two nitrogen atoms are interchanged, thus both nitrogen atoms are equivalent. As &lt;span class=&quot;math-container&quot;&gt;$\ce{N-H}$&lt;/span&gt; bonded protons are more acidic than &lt;span class=&quot;math-container&quot;&gt;$\ce{C=C-H}$&lt;/span&gt; protons and &lt;span class=&quot;math-container&quot;&gt;$\ce{N-C-H}$&lt;/span&gt; protons, the answer must be the hydrogen atom bonded to the nitrogen atom, as both nitrogen atoms are equivalent, either of the hydrogen atoms is a valid answer (as they are equivalent).&lt;/p&gt;&#xA;" OwnerUserId="7584" LastEditorUserId="83758" LastEditDate="2020-02-13T18:15:20.627" LastActivityDate="2020-02-13T18:15:20.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16939" PostTypeId="2" ParentId="16922" CreationDate="2014-10-01T09:59:59.543" Score="3" Body="&lt;p&gt;Its not a problem about being charged, its about the symmetry. The quantum states of those valence orbitals are stable, in that there is little interaction with the other electrons. Since there are already filled orbitals, electrons will want to bind to complete the shell because as it happens extra binding energy is available. With a full shell maximum binding occurs.&lt;/p&gt;&#xA;" OwnerUserId="8123" LastActivityDate="2014-10-01T09:59:59.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16940" PostTypeId="1" CreationDate="2014-10-01T10:41:48.080" Score="13" ViewCount="10747" Body="&lt;p&gt;Which of the following will be more basic, hydrazine $\ce{H2N-NH2}$ or ammonia $\ce{NH3}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry teacher said that $\ce{NH3}$ is more basic since after giving $\ce{H+}$ to hydrazine results in $\ce{H3N+-NH2}$, and accommodation of the lone pair of $\ce{-NH2}$ is not possible by $\ce{-N+H3}$, since it doesn't have the space. So instead, $\ce{-NH2}$ will pull electrons from it making it unstable. Ammonia has no such problem so it must be more basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not so pleased with this argument. In $\ce{H3N+-NH2}$, although the lone pair cannot be accommodated, but the positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable. And also, not to forget, hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a bit of clarity on this.&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="4945" LastEditDate="2014-10-01T11:18:08.200" LastActivityDate="2020-06-30T09:28:50.823" Title="Which is more basic, hydrazine or ammonia?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16941" PostTypeId="1" AcceptedAnswerId="16942" CreationDate="2014-10-01T11:39:56.637" Score="2" ViewCount="3931" Body="&lt;p&gt;From what I understand, to make an amine from an alkyl halide you need to heat the alkyl halide with a mixture of ammonia and ethanol. I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do both ammonia and ethanol need to be concentrated? Because I think I read somewhere that you use a 'concentrated mixture of ethanol and ammonia', although I'm not sure whether the 'concentrated' part refers to the ethanol or the ammonia. I'm guessing it refers to the ammonia.&lt;/li&gt;&#xA;&lt;li&gt;What is the role of the ethanol? Does it act a a catalyst, or just a solvent, or both?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="5026" LastEditDate="2018-04-11T03:01:30.290" LastActivityDate="2018-04-11T03:01:30.290" Title="What are the conditions for making amines from alkyl halides?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16942" PostTypeId="2" ParentId="16941" CreationDate="2014-10-01T14:09:14.687" Score="4" Body="&lt;p&gt;Ethanol is just a solvent often used for this reaction because 1) the reactants and products are soluble in it and 2) the reaction proceeds at a reasonable rate in boiling ethanol.  If your trying to prepare the corresponding primary amine, then a large excess of ammonia should be used.  Otherwise, if the concentration of the initial product (the primary amine) becomes comparable to the concentration of ammonia, then the primary amine could react with the starting alkyl halide and a complicated mixture of products could ensue (see &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/haloalkanes/nh3.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; for a nice description of all the possible products that could be formed if the ammonia concentration is not high enough).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is typically run at reflux (boiling ethanol).  Since ammonia is a gas at room temperature it would quickly escape from the reaction and we wouldn't have a high concentration of ammonia for long.  Therefore, the ammonia is condensed into a solution of the ethanol and alkyl halide (usually an alkyl bromide) in a glass tube at low temperature.  When the desired concentration of ammonia is obtained, the tube is sealed and then heated to allow the reaction to take place.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-01T15:17:48.047" LastActivityDate="2014-10-01T15:17:48.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16943" PostTypeId="2" ParentId="7800" CreationDate="2014-10-01T14:15:47.040" Score="-1" Body="&lt;p&gt;fluorine is high electronegativity so when it combine with other halogen which is less electronegativity then fluorine can take electron from other atom due to this compound become unstable that's why  fluorine can't be central atom in interhalogen compound.&lt;/p&gt;&#xA;" OwnerUserId="8126" LastEditorUserId="5456" LastEditDate="2014-10-01T16:30:57.717" LastActivityDate="2014-10-01T16:30:57.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16944" PostTypeId="1" CreationDate="2014-10-01T14:38:56.550" Score="11" ViewCount="16207" Body="&lt;p&gt;I'm interested in &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How many  &lt;strong&gt;unique&lt;/strong&gt; molecules (of all types: organic and inorganic, etc), exist  in the world accessible by man (let's exclude black holes and the centers of stars etc)?&lt;/li&gt;&#xA;&lt;li&gt;What subset of those distinct types are only in existence because of synthesis by man?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To make it simpler, ignore variants such as molecules with different atom isotopes and different bond configurations (e.g. cis, trans). E.g. standard water and &quot;heavy water&quot; are both $\ce{H2O}$ and count as one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a theoretical estimate, so orders of magnitude only are required. I'm not looking for a count of records in some database catalogue of chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain methods for determining your estimates. &lt;a href=&quot;http://www.answers.com/Q/How_many_different_types_of_molecules_are_there&quot;&gt;This isn't really helpful&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8127" LastEditorUserId="4325" LastEditDate="2016-04-08T06:45:30.880" LastActivityDate="2017-11-22T14:15:02.543" Title="How many unique molecules exist? And how many exist only by synthesis?" Tags="&lt;cheminformatics&gt;" AnswerCount="6" CommentCount="15" FavoriteCount="1" ClosedDate="2017-11-22T22:10:14.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="16945" PostTypeId="1" AcceptedAnswerId="17003" CreationDate="2014-10-01T15:28:50.273" Score="1" ViewCount="1224" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EbAcQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading that perovskite presents a sort of pseudocubic crystalline system, however it can be stretched and suffer modifications to become orthorhombic etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, if I consider it as cubic, it's sort of a combination between FCC and BCC, because it's an FCC with an atom in the center. I can't find a name for this sort of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me out?&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="5456" LastEditDate="2014-10-02T16:11:12.563" LastActivityDate="2014-10-02T16:11:12.563" Title="Determining perovskite crystal structure" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16946" PostTypeId="2" ParentId="16944" CreationDate="2014-10-01T16:02:28.427" Score="5" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.cas.org/content/chemical-substances&quot; rel=&quot;nofollow&quot;&gt;Chemical Abstracts Service&lt;/a&gt;, CAS REGISTRY (SM) contains more than 90 million unique organic and inorganic chemical substances, such as alloys, coordination compounds, minerals, mixtures, polymers and salts, and more than 65 million sequences—more than any other database of its kind. It includes substances reported in the literature back to the early 1800s, and is updated daily with about 15,000 substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CAS registry does not include substances yet to be thought of or discovered. It does indeed include your five examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about as close to an answer as I can think of.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5456" LastEditDate="2014-10-02T08:37:20.210" LastActivityDate="2014-10-02T08:37:20.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16947" PostTypeId="2" ParentId="16944" CreationDate="2014-10-01T17:14:50.677" Score="25" Body="&lt;p&gt;Polymers make this question more or less unanswerable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider human chromosome 1, which contains about 249,000,000 base pairs.  There is nothing that says we couldn't order those pairs in any way we like, so for DNA alone, in a quantity that exists in every person on Earth, there exists the possibility for 4&lt;sup&gt;249000000&lt;/sup&gt; different molecules.  Include things like alternative nucleotides (there are almost surely at least thousands of these that are possible) and longer chains and the number gets even more enormous, and that's only including DNA (and close relatives.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no real way to include all of them, since there have to be an enormous number of ways to put these molecules together that we haven't found yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many exist?  An awful lot.&#xA;How many only exist in our minds?  Almost all of them (the universe only contains about 10&lt;sup&gt;80&lt;/sup&gt; particles, after all.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-01T17:14:50.677" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16949" PostTypeId="1" AcceptedAnswerId="17138" CreationDate="2014-10-01T19:38:26.303" Score="0" ViewCount="867" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is meant by the lowest photometric error?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;why it is important to work in the $0.2\ \mathrm{to}\ 0.7$ absorbance range for a single beam instrument? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why $0.4\ \mathrm{to}\  1.4$ for double-beam instrument?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8133" LastEditorUserId="5456" LastEditDate="2014-10-06T15:19:32.637" LastActivityDate="2014-10-06T15:19:32.637" Title="Spectroscopy and photometric error" Tags="&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16950" PostTypeId="1" AcceptedAnswerId="52167" CreationDate="2014-10-01T19:46:08.853" Score="5" ViewCount="1954" Body="&lt;p&gt;I have put some old nails and screws in the 30% sulfuric acid to prepare an electrolyte for plating. After some bubbling and horribly irritating smell (probably &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{SO2}$&lt;/a&gt;), the reaction slowed down. Overnight, beautiful green/blue crystals formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bhfWZ.jpg&quot; alt=&quot;iron in sulfuric acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just love them. After another day, the reaction almost stopped. I suppose the acid was all bound to iron. To continue the process, I poured some concentrated sulfuric acid into the solution. That was a big mistake - the crystals reacted with concentrated acid and formed ugly white dust:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nlQa7.jpg&quot; alt=&quot;iron in concentrated sulfuric acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I create? What are the crystals and what's that dust? How to reverse the reaction? Does the dust dissolve to be electrolysed?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2016-06-03T19:48:26.297" Title="How to get back my beautiful crystals made from iron and sulphuric acid" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16951" PostTypeId="1" CreationDate="2014-10-01T21:05:58.247" Score="0" ViewCount="3320" Body="&lt;p&gt;In my lab notebook it asks us to think about the spectroscope further than the reading goes on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It asks if we use a diffraction grating that contains 600 lines/mm in the spectroscope what do we expect would happen to the lines of the line spectrum in terms of sharpness and resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the answer (assuming 600 is more than normal) that the lines would be both more sharper and have a higher resolution? My first thought was that the lines would be skinnier but more clear.&lt;/p&gt;&#xA;" OwnerUserId="7795" LastActivityDate="2014-11-05T15:45:09.350" Title="How does different diffraction grating effect resolution and sharpness of lines in a line spectrum?" Tags="&lt;homework&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16952" PostTypeId="2" ParentId="16944" CreationDate="2014-10-01T22:16:06.540" Score="3" Body="&lt;p&gt;In a database of currently available synthetic compounds that i built, there are 9.8 milion synthetic and natural compounds available to purchase, and an additional 27 milion compounds that the vendors think they can produce if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explicitly includes only pure compounds that can be described by drawing a structure, so no mixtures, and no stuff like DNA where you can only guess at the exact structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I included only around 30 trustworthy vendors, and this set is specifically not the same as the CAS register, mind you, as that one is very heavily polluted with gazillions of duplicates and variations in notation. Also, as a nice to know: nearly all vendors lie about the number of compounds they have.&lt;/p&gt;&#xA;" OwnerUserId="8135" LastActivityDate="2014-10-01T22:16:06.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16953" PostTypeId="1" AcceptedAnswerId="16954" CreationDate="2014-10-01T22:20:19.373" Score="6" ViewCount="91" Body="&lt;p&gt;$\binom{16}{8}$O (n,$\alpha$)$\binom{13}{6}$C  and $\binom{9}{4}$Be ($\alpha$,n)$\binom{12}{6}$C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had initially thought the products &amp;amp; reactants may be O-16 $\to$ C-13 + He-2 + p$^+$ from alpha decay, but I'm leaning towards this form O-16 + C-13 $\to$ ... now. Any help clarifying &amp;amp; determining products would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8136" LastEditorUserId="4231" LastEditDate="2015-04-02T15:36:43.293" LastActivityDate="2018-08-22T15:52:51.350" Title="Interpret Radiodecay Symbolism" Tags="&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16954" PostTypeId="2" ParentId="16953" CreationDate="2014-10-01T22:29:50.490" Score="10" Body="&lt;p&gt;The given expression&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{^{16}_{8}O}\,(\mathrm{n},\alpha)\,\ce{^{13}_{6}C}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;translates as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{16}_{8}O -&amp;gt;[+\ \mathrm{n}][-\ \alpha] ^{13}_{6}C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{16}O + n -&amp;gt; ^{13}C + ^4He}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the expression&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{^{9}_{4}Be}\,(\alpha,\mathrm{n})\,\ce{^{12}_{6}C}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;translates as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{9}_{4}Be -&amp;gt;[+\ \alpha][-\ \mathrm{n}] ^{12}_{6}C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^9Be + ^4He -&amp;gt; ^{12}C + n}$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="22102" LastEditDate="2018-08-22T15:52:51.350" LastActivityDate="2018-08-22T15:52:51.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16955" PostTypeId="2" ParentId="16944" CreationDate="2014-10-01T23:15:42.363" Score="13" Body="&lt;p&gt;Without some constraint on the number of atoms/molecular weight, the number of possible theoretical compounds is infinite, for the reason given by Jason Patterson: you can always extend a polymer by one more unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been estimated that there are 10^60 possible &lt;strong&gt;organic&lt;/strong&gt; compounds with molecular weight less than 500 &lt;a href=&quot;http://dx.doi.org/10.1002/(SICI)1098-1128(199601)16:1%3C3::AID-MED1%3E3.0.CO;2-6&quot;&gt;The art and practice of structure-based drug design: a molecular modelling perspective.&lt;/a&gt; CAS may have duplicates, but it is still the best database of all known compounds in the literature, and yet is not comprehensive. Almost all of the CAS database is wholly synthetic. A search for entries in CAS with molecular weight &amp;lt;500 gives 61,026,438 hits, about half of the entire database. The reference above suggests that life can exist with a few thousand small molecules (i.e. compounds with molecular weight &amp;lt;500). Nature produces far more compounds than the absolute minimum, but I expect that we have only scratched the surface of what is out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should give you an idea of the orders of magnitude that we are talking about. Keep in mind that the numbers will grow exponentially as the constraint of MW &amp;lt; 500 is relaxed. In short, chemists have synthesized literally millions of different compounds, but essentially none relative to the possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you research the topic of &quot;chemical space&quot; if you want to know more about this area.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-01T23:15:42.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16956" PostTypeId="2" ParentId="16950" CreationDate="2014-10-01T23:16:30.013" Score="2" Body="&lt;p&gt;A quick look at the Wiki suggests that you first formed iron(II) sulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe + H2SO4 → FeSO4 + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is blue-green. It is then possible that you obtained iron(III) sulfate, which is pale. But you need to oxidize the iron(II) to iron(III), which usually requires oxygen, and sulfuric acid is not an oxidizing acid. Mere exposure to oxygen works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{12 FeSO4 + 3 O2 → 4 Fe2(SO4)3 + 2 Fe2O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I would assume not that quickly. Still, this looks like the best explanation, and perhaps some impurities managed to produce the oxidation. Just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how to get back the iron(II), you would need to reduce the iron(III), and given that the oxidation direction is basically spontaneous in air, this might take some work. So why not just start again with more iron and dilute acid?&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-10-01T23:16:30.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16957" PostTypeId="1" AcceptedAnswerId="16960" CreationDate="2014-10-02T00:06:22.517" Score="1" ViewCount="1181" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yV8Mk.jpg&quot; alt=&quot;redox reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am understanding that the Hydrogen ions oxidize the calcium metal, neutralizing itself and creating a calcium cation. My question is why does the Hydrogen oxidize the calcium? Doesn't it prefer to have an empty outer shell? Does it want more to be neutral or to have an empty outer shell? &lt;/p&gt;&#xA;" OwnerUserId="8138" LastActivityDate="2014-10-02T00:58:26.387" Title="In this simple redox reaction, why does the Hydrogen oxidize the calcium metal?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16958" PostTypeId="2" ParentId="16957" CreationDate="2014-10-02T00:54:00.670" Score="1" Body="&lt;p&gt;This is a simple example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow&quot;&gt;reactivity series&lt;/a&gt;. More reactive metals will displace less reactive metal ions, and they switch places, with the more reactive metal becoming the ion and and the less reactive ion becoming a metal. This can be very cool in many situations: for example, copper metal can displace a silver salt to plate silver metal and produce a blue copper salt. See &lt;a href=&quot;http://www.periodicvideos.com/videos/047.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; video! In any case, hydrogen is consider a reference &quot;metal&quot; in the reactivity series, and it is less reactive than calcium, so the displacement reaction you cite occurs.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-10-02T00:54:00.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16959" PostTypeId="1" CreationDate="2014-10-02T00:54:17.117" Score="3" ViewCount="1591" Body="&lt;p&gt;I once saw a problem, iron ions(II) titrated with potassium dichromate to produce Fe(II) and Cr(III). If I am to write the net ionic equation of this reaction, I got to make potassium dichromate on the right into dichromate ion since potassium is a spectator.&lt;br&gt;&#xA;I got very confused over here. Why should you see potassium as a spectator if K(II) will be balanced as 2K on the right.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4945" LastEditDate="2014-10-02T04:08:24.540" LastActivityDate="2014-10-03T01:59:07.907" Title="What exactly is net ionic equation?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16960" PostTypeId="2" ParentId="16957" CreationDate="2014-10-02T00:58:26.387" Score="0" Body="&lt;p&gt;In one simple sentence, a more reactive metal is more likely to donate its electrons than a less reactive metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes it more reaction, for example francium, it has more electrons shells thus the nucleus Is less electronegative to the valence electrons.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-10-02T00:58:26.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16961" PostTypeId="5" CreationDate="2014-10-02T01:08:07.397" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T01:08:07.397" LastActivityDate="2014-10-02T01:08:07.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16962" PostTypeId="4" CreationDate="2014-10-02T01:08:07.397" Score="0" Body="The process of a substance transitioning from solid to gaseous phases, and vice versa." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T02:06:32.757" LastActivityDate="2014-10-02T02:06:32.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16963" PostTypeId="5" CreationDate="2014-10-02T01:09:15.837" Score="0" Body="&lt;p&gt;The &lt;strong&gt;thermodynamic temperature&lt;/strong&gt; (quantity symbol: &lt;em&gt;T&lt;/em&gt;) is one of the base quantities in the International System of Quantities (ISQ), on which the International System of Units (SI) is based. Thermodynamic temperature can be measured with a primary thermometer, examples of which are constant volume gas thermometers, acoustic thermometers, or total radiation thermometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit of thermodynamic temperature is the &lt;strong&gt;kelvin&lt;/strong&gt; (unit symbol: K), which is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Celsius temperature&lt;/strong&gt; (quantity symbol: &lt;em&gt;t&lt;/em&gt; or &lt;em&gt;ϑ&lt;/em&gt;) is defined as &lt;em&gt;t&lt;/em&gt; = &lt;em&gt;T&lt;/em&gt; − &lt;em&gt;T&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;, where &lt;em&gt;T&lt;/em&gt; is thermodynamic temperature (see above) and &lt;em&gt;T&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; := 273.15 K. The thermodynamic temperature &lt;em&gt;T&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; is exactly 0.01 K below the thermodynamic temperature of the triple point of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit name &lt;strong&gt;degree Celsius&lt;/strong&gt; (unit symbol: °C) is a special name for the kelvin for use in stating values of Celsius temperature: 1 °C := 1 K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units of thermodynamic and Celsius temperature difference or change are identical. Such differences or changes may be expressed either in kelvin (unit symbol K) or in degree Celsius (unit symbol °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit &lt;strong&gt;degree Rankine&lt;/strong&gt; (unit symbol °R) is a non-SI unit of thermodynamic temperature &lt;em&gt;T&lt;/em&gt;:&lt;br&gt;&#xA;1 °R := 5/9 K.&lt;br&gt;&#xA;The use of this unit is deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit &lt;strong&gt;degree Fahrenheit&lt;/strong&gt; (unit symbol °F) is a non-SI unit of Fahrenheit temperature &lt;em&gt;t&lt;/em&gt;&lt;sub&gt;F&lt;/sub&gt;:&lt;br&gt;&#xA;&lt;em&gt;t&lt;/em&gt;&lt;sub&gt;F&lt;/sub&gt;/°F := (9/5) (&lt;em&gt;t&lt;/em&gt;/°C) + 32 = (9/5) (&lt;em&gt;T&lt;/em&gt;/K) − 459.67&lt;br&gt;&#xA;The use of this unit is deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the symbol K (for kelvin), the symbol °C (for degree Celsius), the symbol °R (for degree Rankine), and the symbol °F (for degree Fahrenheit) shall be preceded by a space when one expresses values of temperature.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-07-01T15:45:19.413" LastActivityDate="2015-07-01T15:45:19.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16964" PostTypeId="4" CreationDate="2014-10-02T01:09:15.837" Score="0" Body="A measure of the average kinetic energy of the particles in a sample of matter; may be expressed either in kelvin (symbol K) or in degree Celsius (symbol °C)." OwnerDisplayName="user8016" LastEditorDisplayName="user7951" LastEditDate="2015-07-01T15:43:09.860" LastActivityDate="2015-07-01T15:43:09.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16966" PostTypeId="5" CreationDate="2014-10-02T01:27:00.073" Score="0" Body="&lt;p&gt;&lt;strong&gt;Further reading:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atmospheric Chemistry (Wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="39539" LastEditorUserId="39539" LastEditDate="2017-05-16T06:30:23.853" LastActivityDate="2017-05-16T06:30:23.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16967" PostTypeId="4" CreationDate="2014-10-02T01:27:00.073" Score="0" Body="Chemical reactions that occur in the atmosphere of Earth and other planets.  Relevant reactions include both those that take place only in the gas-phase, in the solid or liquid phases of aerosols or any reactions between the phases as well as the thermodynamics of the phase changes that take place in the atmosphere." OwnerDisplayName="user8016" LastEditorUserId="402" LastEditorDisplayName="user8016" LastEditDate="2017-05-03T01:47:10.093" LastActivityDate="2017-05-03T01:47:10.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16968" PostTypeId="2" ParentId="16940" CreationDate="2014-10-02T01:30:39.563" Score="3" Body="&lt;p&gt;Obligatory &lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to Evans' pK&lt;sub&gt;a&lt;/sub&gt; table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia is more basic than hydrazine, by about one order of magnitude.  This is expected, because the -NH&lt;sub&gt;2&lt;/sub&gt; group is more electronegative than -H or -CH&lt;sub&gt;3&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second lone pair is not involved in the acid-base reaction, it does not point towards the -NH&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; group.  You can, however, force two lone pairs into close proximity.  This destabilizes the unprotonated form.  The keyword is &quot;proton sponge&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1,8-Bis%28dimethylamino%29naphthalene&quot; rel=&quot;nofollow&quot;&gt;1,8-Bis(dimethylamino)naphthalene&lt;/a&gt; has a pK&lt;sub&gt;a&lt;/sub&gt; of 12.3, it's one of the strongest known amine bases.  Compare that to the pK&lt;sub&gt;a&lt;/sub&gt; of aniline, which is something like 4.5.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-02T01:51:55.430" LastActivityDate="2014-10-02T01:51:55.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16969" PostTypeId="5" CreationDate="2014-10-02T01:33:20.880" Score="0" Body="&lt;p&gt;The &quot;flow&quot; or current of electrons through a conductive medium across a potential, usually metals.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="23561" LastEditDate="2018-08-11T18:50:36.323" LastActivityDate="2018-08-11T18:50:36.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16970" PostTypeId="4" CreationDate="2014-10-02T01:33:20.880" Score="0" Body="For questions involving electricity in the scope of chemistry. Note that questions about electrical circuits, semi-conductor diodes, etc. are off topic and more suited for EE.SE or Phy.SE." OwnerDisplayName="user8016" LastEditorUserId="5026" LastEditorDisplayName="user8016" LastEditDate="2018-03-15T07:31:05.263" LastActivityDate="2018-03-15T07:31:05.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16971" PostTypeId="5" CreationDate="2014-10-02T01:39:00.377" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T01:39:00.377" LastActivityDate="2014-10-02T01:39:00.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16972" PostTypeId="4" CreationDate="2014-10-02T01:39:00.377" Score="0" Body="For questions relating to the dynamics and processes that occur within and interactions between the nuclei of atoms." OwnerDisplayName="user8016" LastEditorUserId="402" LastEditorDisplayName="user8016" LastEditDate="2017-05-04T00:31:37.143" LastActivityDate="2017-05-04T00:31:37.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16973" PostTypeId="1" CreationDate="2014-10-02T02:04:54.370" Score="0" ViewCount="13825" Body="&lt;p&gt;What is the charge of Silicon dioxide ions? I can't find it anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is there anywhere I can find these on my own?&lt;/p&gt;&#xA;" OwnerUserId="7996" LastEditorUserId="5456" LastEditDate="2014-10-02T04:09:24.453" LastActivityDate="2014-10-02T05:16:48.927" Title="What is the charge of Silicon dioxide?" Tags="&lt;bond&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2016-05-06T18:45:33.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="16974" PostTypeId="5" CreationDate="2014-10-02T02:20:45.373" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T02:20:45.373" LastActivityDate="2014-10-02T02:20:45.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16975" PostTypeId="4" CreationDate="2014-10-02T02:20:45.373" Score="0" Body="The chemical processes involved with the use of heat in the preparation of food." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T02:46:58.953" LastActivityDate="2014-10-02T02:46:58.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16976" PostTypeId="5" CreationDate="2014-10-02T02:24:32.783" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T02:24:32.783" LastActivityDate="2014-10-02T02:24:32.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16977" PostTypeId="4" CreationDate="2014-10-02T02:24:32.783" Score="0" Body="Chemical processes and dynamics associated with geological processes, including hydro-thermal veins and weathering on Earth. For questions about other celestial bodies, please use [astrochemistry]." OwnerDisplayName="user8016" LastEditorUserId="23561" LastEditorDisplayName="user8016" LastEditDate="2019-04-22T07:48:20.787" LastActivityDate="2019-04-22T07:48:20.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16978" PostTypeId="1" AcceptedAnswerId="16982" CreationDate="2014-10-02T02:46:07.107" Score="7" ViewCount="8247" Body="&lt;p&gt;For E2 reactions, why is a strong base like $\ce{NaOH}$ or $\ce{RONa}$ needed? Whereas for E1, even a weak base like $\ce{H2O}$ could be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Elimination_reaction&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E2 typically uses a strong base, it needs a chemical strong enough to pull off a weakly acidic hydrogen. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However it does not explain why. The lone pair of the base directly attacks the hydrogen, regardless of E1 or E2.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="7475" LastEditDate="2015-10-09T16:08:44.777" LastActivityDate="2015-10-09T16:34:11.417" Title="Why is a strong base needed for an E2 reaction and not a an E1 reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16979" PostTypeId="2" ParentId="16973" CreationDate="2014-10-02T02:51:29.403" Score="2" Body="&lt;p&gt;Like what @GeoffHutchison said,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SiO2}$ is not an ion. It is a network solid. Its net charge is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From charge balance, since oxygen holds a -2 formal charge, then silicon must hold a +4 formal charge in order to balance out. $(+4) + (+2 \times -2) = 0$&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="5456" LastEditDate="2014-10-02T05:16:48.927" LastActivityDate="2014-10-02T05:16:48.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16980" PostTypeId="5" CreationDate="2014-10-02T03:14:42.067" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T03:14:42.067" LastActivityDate="2014-10-02T03:14:42.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16981" PostTypeId="4" CreationDate="2014-10-02T03:14:42.067" Score="0" Body="The study of drugs/medicines on living organisms.  The study of drug action." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T05:05:52.447" LastActivityDate="2014-10-02T05:05:52.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16982" PostTypeId="2" ParentId="16978" CreationDate="2014-10-02T03:23:33.607" Score="11" Body="&lt;p&gt;E2 is a concerted mechanism. The alpha proton in an E2 reaction substrate is also only weakly acidic because the bond between the alpha proton and the alpha carbon is relatively strong; there is only some inductive withdrawal of electron density from the C-H bond by the halogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we need a strong base to &lt;em&gt;rip&lt;/em&gt; the proton away from the alpha carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3XFP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E1 however involves two steps. In E1 the leaving group leaves, and then the base attacks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the leaving group leaves, a carbon lacking an octet is formed. This carbon is highly electronegative and will withdraw electron density. This makes any alpha proton more acidic as hydrogen is less electronegative than plain old carbon - much less a carbon deficient of an octet in a covalent framework. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, electron withdrawal by the carbocation weakens the bond holding on to the alpha proton, and a weaker base suffices for taking away the proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I don't particularly like the below picture because it doesn't show anything abstracting the proton. Still the general idea is correct; electron density from the alpha C-H bond flows toward the hypovalent carbon. Heck, we can draw valid, hyperconjugative resonance structures for this baby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MBMf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-10-03T13:00:35.553" LastActivityDate="2014-10-03T13:00:35.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16983" PostTypeId="1" CreationDate="2014-10-02T04:44:28.913" Score="1" ViewCount="87" Body="&lt;p&gt;Is catechol (o-dihydroxybenzene) more prone to auto-oxidation than p-dihydroxybenzene or m-dihydroxybenzene? If so, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, both dopamine and &lt;a href=&quot;http://en.wikipedia.org/wiki/Caffeic_acid#Chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;caffeic acid&lt;/a&gt; are prone to auto-oxidation. So I wonder if it's a property of o-dihydroxybenzenes in general.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="402" LastEditDate="2017-05-16T02:05:14.677" LastActivityDate="2017-05-16T02:05:14.677" Title="Is catechol (o-dihydroxybenzene) more prone to auto-oxidation than p-dihydroxybenzene or m-dihydroxybenzene?" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16984" PostTypeId="5" CreationDate="2014-10-02T04:45:53.253" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T04:45:53.253" LastActivityDate="2014-10-02T04:45:53.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16985" PostTypeId="4" CreationDate="2014-10-02T04:45:53.253" Score="0" Body="Viscosity refers to the 'thickness' of a fluid as well as its resistance to shear and tensile stresses." OwnerDisplayName="user8016" LastEditorUserId="402" LastEditorDisplayName="user8016" LastEditDate="2017-05-02T22:37:30.737" LastActivityDate="2017-05-02T22:37:30.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16986" PostTypeId="5" CreationDate="2014-10-02T04:48:11.320" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T04:48:11.320" LastActivityDate="2014-10-02T04:48:11.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16987" PostTypeId="4" CreationDate="2014-10-02T04:48:11.320" Score="0" Body="Chemical processes and theories being used in practical and experimental applications." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T12:36:14.863" LastActivityDate="2014-10-02T12:36:14.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16988" PostTypeId="5" CreationDate="2014-10-02T04:55:17.237" Score="0" Body="&lt;p&gt;&lt;strong&gt;As defined in IUPAC &lt;a href=&quot;https://goldbook.iupac.org/html/C/C01022.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gold Book&lt;/a&gt;&lt;/strong&gt; (&quot;element&quot; is equivalent to &quot;chemical equivalent&quot;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A species of atoms; all atoms with the same number of protons in the atomic nucleus&lt;/li&gt;&#xA;  &lt;li&gt;A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="42352" LastEditorUserId="5026" LastEditDate="2018-03-16T15:51:57.793" LastActivityDate="2018-03-16T15:51:57.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16989" PostTypeId="4" CreationDate="2014-10-02T04:55:17.237" Score="0" Body="A pure chemical substance consisting of a unique type of atom with a distinguished by its atomic mass." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T12:32:22.870" LastActivityDate="2014-10-02T12:32:22.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16990" PostTypeId="5" CreationDate="2014-10-02T04:58:48.130" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T04:58:48.130" LastActivityDate="2014-10-02T04:58:48.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16991" PostTypeId="4" CreationDate="2014-10-02T04:58:48.130" Score="0" Body="Chemical compounds that contain halogens (group 17 in the Periodic Table) as part of their chemical structure. This tag should only be applied where the halide group is one of the main focuses of the question." OwnerDisplayName="user8016" LastEditorUserId="5026" LastEditorDisplayName="user8016" LastEditDate="2018-03-14T13:42:09.293" LastActivityDate="2018-03-14T13:42:09.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16992" PostTypeId="1" AcceptedAnswerId="17000" CreationDate="2014-10-02T07:11:32.943" Score="4" ViewCount="1611" Body="&lt;p&gt;I teach an MCAT course in chemistry. I like to explain VSEPR by saying.. first, imagine arranging electron pairs around the central atom so they are maximally distant from each other, and uniformly arranged. If there are two pairs, they'll just be across from each other. Three? You get a triangle, and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rhetorical device fails with five pairs. Five is strange because the electron pairs are most definitely &lt;em&gt;not&lt;/em&gt; equidistant from each other, with the equatorial pairs being 120 degrees from each other, but 90 degrees from the axial electron pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is this a simplifying geometry for high school / college classes, and the geometry is actually more equally arranged in 3D space among the pairs? Alternatively, if 'trigonal bipyramidal' is actually the geometry, &lt;em&gt;why&lt;/em&gt;? What induces the asymmetry? And does that imply that molecules with the same formula and connectivity, but with an atom equatorially versus axially located correspond to distinct stereoisomers?&lt;/p&gt;&#xA;" OwnerUserId="8142" LastActivityDate="2018-10-09T04:53:04.337" Title="Asymmetry in trigonal bipyramidal geometry" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16993" PostTypeId="1" AcceptedAnswerId="16994" CreationDate="2014-10-02T09:01:26.310" Score="3" ViewCount="913" Body="&lt;p&gt;For the electrophilic addition of $\ce{X2}$ to alkene in $\ce{CCl4}$ solvent, a dihaloalkane is formed. This is a halogen addition reaction, with the intermediate being a halonium ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P8cTy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if instead of $\ce{CCl4}$, $\ce{H2O}$ is used instead, a halohydrin would be formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tn6aT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the second case (halohydrin formation reaction), the $\ce{X-}$ ion is still present in the reaction. Furthermore, $\ce{X-}$ is a stronger nucleophile than $\ce{H2O}$. Why is it then that it is $\ce{H2O}$ that reacts with the halonium rather than $\ce{X-}$, the stronger nucleophile?&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="7475" LastEditDate="2015-10-16T16:52:49.957" LastActivityDate="2015-10-16T16:52:49.957" Title="Formation of halohydrin vs alkyl dihalide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16994" PostTypeId="2" ParentId="16993" CreationDate="2014-10-02T09:43:03.280" Score="3" Body="&lt;p&gt;The halohydrin is usually formed by means of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvolysis&quot; rel=&quot;nofollow&quot;&gt;solvolysis&lt;/a&gt; reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by &quot;encasing&quot; it in a large hydration shell.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-10-02T09:43:03.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16995" PostTypeId="5" CreationDate="2014-10-02T10:30:32.073" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T10:30:32.073" LastActivityDate="2014-10-02T10:30:32.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16996" PostTypeId="4" CreationDate="2014-10-02T10:30:32.073" Score="0" Body="A method to measure how much a chemical substance absorbs light by measuring the intensity of light as a beam of light passes through sample solution." OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-02T20:27:38.377" LastActivityDate="2014-10-02T20:27:38.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16997" PostTypeId="1" AcceptedAnswerId="17013" CreationDate="2014-10-02T11:05:25.277" Score="2" ViewCount="118" Body="&lt;p&gt;Is it possible to copy my chemdoodle displayed structure into mestranovea? I have tried simple dragging (mac) and all the various copy/paste options... any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="5017" LastEditDate="2014-10-02T12:31:49.800" LastActivityDate="2014-10-02T19:28:14.117" Title="Chemdoodle to mestranova" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16998" PostTypeId="1" CreationDate="2014-10-02T12:23:16.967" Score="0" ViewCount="1290" Body="&lt;p&gt;Does same amount of heat absorbed cause the same changes of temperature in the same amount of different liquids? Explain.&lt;/p&gt;&#xA;" OwnerUserId="8146" LastActivityDate="2014-10-02T12:27:44.190" Title="Heat Absorption of Liquids" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-19T14:48:41.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="16999" PostTypeId="2" ParentId="16998" CreationDate="2014-10-02T12:27:44.190" Score="1" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question we will assume certain things here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume  $C_p = C_v$ for most liquids, where $C_p$ is the heat capacity at constant pressure, and $C_v$ is the heat capacity at constant volume.&lt;/li&gt;&#xA;&lt;li&gt;Assume that $C$ is constant with respect to temperature, which is not the case in most real fluids. Here, in order to leave out the integration, we assume that it is a constant. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = C\Delta{}T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As different liquids have different heat capacities, the same amount of heat supplied to the liquid would result in a different $\Delta{}T$, the temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher the heat capacity, the more heat is required to raise a unit mass(or mole) of substance  by a unit temperature. Likewise, more heat has to be removed from the unit mass of substance in order to decrease the temperature by a unit temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the higher the heat capacity, the more &quot;reluctant&quot; it is to change its temperature, for the same amount of heat supplied.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-02T12:27:44.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17000" PostTypeId="2" ParentId="16992" CreationDate="2014-10-02T12:46:14.940" Score="4" Body="&lt;p&gt;Yes, 5-coordinate compounds are always a bit confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way I've answered this question about &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;VSEPR&lt;/a&gt; for students is to do a little demo with balloons. Blow up balloons (i.e., electron pairs) and tie them around a central knot. It will &lt;strong&gt;definitely&lt;/strong&gt; give a trigonal bipyramidal shape - it has nothing to do with chemistry, and everything to do with &quot;fitting 5 things around a central point.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree with @Martin that there's no theoretical basis to the shapes. The theory is purely geometric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple possible shapes, including square pyramidal, but these are higher in energy because you have more ~90-degree angles. As discussed in the comments, if all 5 atoms are in a plane, the angles are 72 degrees, which is clearly bad. You could in principal have the top of an octahedron, in which case all angles are 90 degrees, but that's also clearly worse. (In practice, square pyramidal structures have the central atom slightly above the plane, increasing some angles slightly beyond 90.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I do the demo with the balloons, talk about square pyramidal and how this alternative is clearly slightly higher in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Berry pseudorotation, as mentioned above, does convert between the shapes, and it's known that at room temperature the axial and equatorial positions scramble quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to do some quick calculations using &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt; and &lt;a href=&quot;http://openmopac.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MOPAC&lt;/a&gt; on $\ce{PF5}$ in square planar versus trigonal bipyramidal using the PM7 semiempirical method. This took longer to write than to run the calculations (seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the optimized square pyramidal geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8n2vq.png&quot; alt=&quot;square pyramidal PF5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phosphorous isn't in the plane, it's a bit above, to increase four of the bond angles and minimize electron-pair repulsion. Note that the largest angle is ~103 degrees, but the F-P-F angles in the basal plane are all ~87 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the trigonal bipyramidal geometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mpkyi.png&quot; alt=&quot;trigonal bipyramidal PF5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the angles are exactly what we expect from VSEPR. We see 90 degree angles between the equatorial plane and the axial F atoms. And the F-P-F in the plane are all 120 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just from &lt;em&gt;the angles&lt;/em&gt; we can guess that trigonal bipyramidal should be lower in energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, the difference in energy between these two geometries using PM7 is ~2.76 kcal/mol. So there's not a &lt;em&gt;huge&lt;/em&gt; difference in energy, but it's there.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-02T18:20:27.297" LastActivityDate="2014-10-02T18:20:27.297" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17001" PostTypeId="5" CreationDate="2014-10-02T14:27:27.317" Score="0" Body="&lt;p&gt;Valence shell electron pair repulsion (VSEPR) theory provides simple explanations for the general shapes of molecules based on the repulsion of bonding pair and lone pair electron repulsions, and we developed by Sidgwick and Powell and later Gillespie and Nyholm.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-15T06:22:39.050" LastActivityDate="2018-03-15T06:22:39.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17002" PostTypeId="4" CreationDate="2014-10-02T14:27:27.317" Score="0" Body="For questions about the valence shell electron pair repulsion (VSEPR) theory, in topics of: estimation of bond length, bond angle, bond energy, etc.; determining hybridization of atoms, or shape and geometry of molecules; and other related topics." OwnerUserId="5017" LastEditorUserId="5026" LastEditDate="2018-03-15T06:22:30.917" LastActivityDate="2018-03-15T06:22:30.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17003" PostTypeId="2" ParentId="16945" CreationDate="2014-10-02T14:51:10.573" Score="2" Body="&lt;p&gt;At Geoff's request I'll make comments into an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is a multi-component system you need to think of a crystal structure with a basis. Not surprisingly, the basis here is $\ce{ABO3}$. In your picture from Wikipedia, start with just the blue atoms - they form the simple cubic. Then each blue atom shares 6 red atoms with other blues, so pick three for a given blue. Then place the green atom at the cube center to go with the blue atom at the lower right front blue atom. This is now your basis, and placing that basis unit at each corner of the cube (repeating to infinity) gives you the crystal structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain further: each red has two blue neighbors, so each red is 'half-owned' by a blue. For a basis, you want only whole atoms to keep your mind a bit saner, so pick three to be 'wholly-owned' by a blue, and leave the others to be owned by another blue. Taking the left lower center blue atom, I'd pick the red atoms in the +x, +y, and +z directions, then the green atom in the (x,y,z) direction to make a nice basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also add some on deviations from cubic symmetry.  If the green atom is at (1/2,1/2,1/2) than it all remains simple cubic.  If the greens are big fat atoms compared to the blues, they may want a little more room, which leads to distortions.  Simply stretching along one dimension is pretty easy, since it leaves 2/3's of the blue-red pairs at the same interatomic distance, stretching the other 1/3.  This creates an orthorhombic crystal, still with the same $\ce{ABO3}$ unit as a basis, but now slightly elongated along one direction (commonly taken as z).&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5017" LastEditDate="2014-10-02T15:54:00.360" LastActivityDate="2014-10-02T15:54:00.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17004" PostTypeId="5" CreationDate="2014-10-02T14:51:40.853" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-02T14:51:40.853" LastActivityDate="2014-10-02T14:51:40.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17005" PostTypeId="4" CreationDate="2014-10-02T14:51:40.853" Score="0" Body="Cheminformatics involves the production, processing, filtering, and searching of chemical information. This includes chemical databases, computational screening, searching, etc." OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-02T23:24:33.547" LastActivityDate="2014-10-02T23:24:33.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17006" PostTypeId="2" ParentId="16959" CreationDate="2014-10-02T17:22:25.633" Score="4" Body="&lt;p&gt;The molecular equation for this titration is $$\ce{K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2SO4 -&amp;gt;&#xA;3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complete ionic equation is $$\ce{2K+ + Cr2O7^2- + 6Fe^2+ + 12NH4+ + 19SO4^2- + 14H+ -&amp;gt;&#xA;6Fe3+ + 14SO4^2- + 2Cr^3+  + 2K+  + 12NH4^+  + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you cross out everybody who shows up on both sides in the same form, you get the net ionic equation for this titration: $$\ce{Cr2O7^2- + 6Fe^2+ + 14H+ -&amp;gt; 6Fe^3+ + 2Cr^3+ + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the hydrogens in $\ce{H2SO4}$ become $\ce{2H+}$ so the potassiums in  $\ce{K2Cr2O7}$ become $\ce{2K+}$ and they don't change on the other side of the equation where they form $\ce{K2SO4}$.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastActivityDate="2014-10-02T17:22:25.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17007" PostTypeId="2" ParentId="16959" CreationDate="2014-10-02T17:25:37.713" Score="4" Body="&lt;p&gt;Well first of all instead of  $\ce{K(II)}$ it is always $\ce{K(I)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now the use of net ionic equation are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) It saves your time in writing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) It highlights the species that are undergoing change and bringing more clarity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) spectator ions have little significance in reaction but are necessary for that ion to ion to be stable , also spectator ion are to show that to make that reaction happen you do not need only those spectator ions which are given in your reaction but any other ion of similar properties could do. So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) you will see better pattern in remembering reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think instead of remembering these two reaction you would prefer to remember this one reaction only &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuSO4 + Zn -&amp;gt; ZnSo4 + Cu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu(NO3)2 + Zn -&amp;gt; Cu + Zn(NO3)2 }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would like to remember &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^{2+} + Zn -&amp;gt; Zn^{2+} + Cu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also you could even see from here that the two reactions are clearly the same only spectator ion is changed&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="4945" LastEditDate="2014-10-03T01:59:07.907" LastActivityDate="2014-10-03T01:59:07.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17008" PostTypeId="1" CreationDate="2014-10-02T17:33:01.740" Score="6" ViewCount="598" Body="&lt;p&gt;Sugammadex is a modified cyclodextrin: 8 member sugar ring with carboxyl thioether extensions. Rocuronium is an aminosteroid, 4 steroid rings and one protonated amine group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sugammadex binds rocuronium really tightly. The reason for this is straightforward - lipophilic interactions with the inner cavity of the cyclodextrin, which is the perfect size because of the thioether extensions. The negatively charged carboxylate groups lock into place with the protonated amine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question asks how the rocuronium is initially drawn into the cavity, which makes less sense. I've only found one reference to this in the literature, in a figure caption that says the carboxylate groups interact with the four steroid rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how? The rings are not aromatic, so there is no anion-pi interaction. One would think the positively charged amine would interact with the carboxylate groups far more often, leading to the wrong orientation. If anything, the hydrophobic rings should be repulsed by the negative charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xMxCH.jpg&quot; alt=&quot;sugammadex binding rocuronium&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hlRdP.jpg&quot; alt=&quot;encapsulation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 6: Encapsulation process: Sugammadex carboxyl groups interact with steroidal rings A, B, C, and D, drawing the aminosteroidal NMBA molecule into the cavity where additional non-covalent attractions hold the molecule securely in place.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;:&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2761174/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NCBI&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="83758" LastEditDate="2020-02-23T08:53:03.413" LastActivityDate="2020-02-23T08:53:03.413" Title="How does sugammadex (a modified cyclodextrin) draw rocuronium into its lipophilic cavity?" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;pharmacology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17009" PostTypeId="1" CreationDate="2014-10-02T17:38:50.203" Score="1" ViewCount="83" Body="&lt;p&gt;I need to precisely measure the internal volume of an open steel fixture.  I was planning to do this via displacement, but I'm wondering what the best fluid would be.  Ideally it should have no surface tension so that it readily penetrates every facet of the fixture.  It would also be convenient if it was non-wetting which (if I understand that term correctly) means that it would not adhere to the steel when removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One suggestion I heard was to use water-soluble cutting fluid.  Are there any other readily-available fluids with these characteristics, or well suited to this application?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2015-01-30T19:18:10.753" Title="Non-wetting fluids for precise volume measurements" Tags="&lt;home-experiment&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17010" PostTypeId="1" AcceptedAnswerId="17012" CreationDate="2014-10-02T17:59:36.627" Score="1" ViewCount="213" Body="&lt;p&gt;I am an applied mathematician interested in the dynamics of potential systems - i.e., systems with multiple unique energy minima. One of the best examples of such systems are protein folding potentials. From sources like &lt;a href=&quot;https://www.mat.univie.ac.at/~neum/ms/protein.pdf&quot; rel=&quot;nofollow&quot;&gt;This one&lt;/a&gt; I know that the potential, &#xA;$$&#xA;V(x)&#xA;$$ exists and is a function of the positions of each residue in space and their interactions with each other. Since such a function exists, there should (mathematically) be a set of ordinary differential equations that capture the dynamics described by the potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, although I have seen definitions of $V(x)$ in the literature, for the life of me I can't figure out &lt;em&gt;exactly&lt;/em&gt; what the set of differential equations associated with this system is - i.e., the system is usually written out in a generalized form that applies to all proteins, but the exact information required to build such a potential for one such protein is unclear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does anyone have an example of of a particular $V(x)$ (for a particular, short, set of amino acids, say) where all the parameters are known and the system can be solved numerically as a set of ordinary differential equations?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8148" LastActivityDate="2014-10-02T18:38:38.607" Title="Explicit differential equation based model of protein folding" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17011" PostTypeId="2" ParentId="17009" CreationDate="2014-10-02T18:00:59.793" Score="1" Body="&lt;p&gt;For a very precise volume measurement, you might want to consider using mercury because of its high density. (You would actually measure the weight of the fluid.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-02T18:00:59.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17012" PostTypeId="2" ParentId="17010" CreationDate="2014-10-02T18:38:38.607" Score="4" Body="&lt;p&gt;What you are looking for is a &lt;strong&gt;force field&lt;/strong&gt;: &lt;a href=&quot;http://en.wikipedia.org/wiki/Force_field_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Force_field_(chemistry)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard force fields for proteins include CHARRM and AMBER. These have relatively simple, well-defined expressions for defining the various types of interactions between bonded and non-bonded atoms. Parameters are available for all standard atom types, and are fit to experimental and/or electronic structure calculations. These force fields are considered &quot;classical&quot; in that they do not explicitly treat electronic interactions, but rather try to capture these interactions in a coarse-grained way. Typically, there are expressions for describing the following interactions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bonded interactions (bond stretching)&lt;/li&gt;&#xA;&lt;li&gt;angle interactions (bond bending)&lt;/li&gt;&#xA;&lt;li&gt;torsion interactions (dihedral bending)&lt;/li&gt;&#xA;&lt;li&gt;12-6 Lennard-Jones interactions (non-bonded, dispersion forces)&lt;/li&gt;&#xA;&lt;li&gt;Short-range charge-charge interactions (Coulombic interactions between charged or partially charged atoms)&lt;/li&gt;&#xA;&lt;li&gt;Long-range charge-charge interactions (see Ewald summation method for more details)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that in these classical models, partial charges are generally statically assigned to each atom. The functional form of these different terms varies depending on the force field and sometimes implementation. These interactions are typically pairwise additive, so to get the total potential energy of the system you simply add together all the various interactions between the appropriate pairs of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly more rigorous models that attempt to capture other electronic structure affects, however for large molecules such as proteins, relatively simple force fields are generally applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that finding the global minimum in potential energy of a protein is not a trivial problem. People have been working on this for years, it's extremely difficult to develop algorithms that do not get trapped in local minima.&lt;/p&gt;&#xA;" OwnerUserId="7693" LastActivityDate="2014-10-02T18:38:38.607" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17013" PostTypeId="2" ParentId="16997" CreationDate="2014-10-02T19:28:14.117" Score="4" Body="&lt;p&gt;ChemDoodle contains many options for communicating chemical data to other applications. The most common way is to use the Edit&gt;Copy as... function and choose a chemical type to be put on the system clipboard as a string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use the copy function in ChemDoodle, a wide range of common data types, from chemical to image, are placed on the system clipboard for a 3rd party application to paste. You can control what is placed on the clipboard in ChemDoodle's advanced preferences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ChemDoodle also supports round trip editing with applications that will allow it. However, if none of this works with the application you are pasting into, you will have to contact that specific vendor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information on working with ChemDoodle data in 3rd party applications can be found in Chapter 11 of the ChemDoodle User Guide: &lt;a href=&quot;http://www.chemdoodle.com/downloads/UserGuide.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.chemdoodle.com/downloads/UserGuide.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further help, please open a ticket here: &lt;a href=&quot;https://support.ichemlabs.com&quot; rel=&quot;nofollow&quot;&gt;https://support.ichemlabs.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8150" LastActivityDate="2014-10-02T19:28:14.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17017" PostTypeId="2" ParentId="16978" CreationDate="2014-10-02T21:13:24.643" Score="1" Body="&lt;p&gt;Also, if you tried using a weak base with, for example, a primary substrate then you would see $\rm S_N2$ and $E_2$ (however this is not practical due to the slowness of the reaction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may recall rate of $$\rm E_2=k[Substrate][Base]$$ while $E_1$ does not have a rate dependence on the base.&lt;/p&gt;&#xA;" OwnerUserId="8111" LastEditorUserId="7448" LastEditDate="2015-10-09T16:34:11.417" LastActivityDate="2015-10-09T16:34:11.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17018" PostTypeId="1" CreationDate="2014-10-02T21:17:30.683" Score="6" ViewCount="23010" Body="&lt;p&gt;The company I'm working with uses a fuel mixture of methanol and water. We use a hydrometer to measure specific gravity, in order to test the composition of fuel received from our suppliers. To determine the weight percent of methanol we use &lt;a href=&quot;http://www.methanol.org/Technical-Information/Resources/Technical-Information/Densities-of-Methanol-Water-Mixes.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;a chart provided by The Methanol Institute&lt;/a&gt; that correlated specific gravity to weight percent at different temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.methanol.org/Technical-Information/Resources/Technical-Information/Densities-of-Methanol-Water-Mixes.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7br5tm.jpg&quot; alt=&quot;Densities of methanol-water solutions&quot;&gt;&lt;/a&gt; &lt;br/&gt;&#xA;&lt;em&gt;&lt;sup&gt;Click image for full-size PDF&lt;/sup&gt;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to build a spreadsheet for operators, suppliers, and customers to measure the specific gravity at an arbitrary temperature, enter the temperature and specific gravity measured, and get the weight percent without having to interpret a chart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do this, I need some sort of formula to calculate the weight percent for any input of specific gravity and temperature  (within a reasonable range anyways). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the approaches I've tried so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fitting a line to the plot above and creating an empirical model. There isn't sufficient resolution in the plot to do this with any useful level of accuracy. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finding the source information used to create this plot. I called The Methanol Institute and they have no idea where the plot came from, saying it's been used for years and they don't know a source or have any other related data. I tried calling Methanex, a major supplier of methanol which has published similar tables and charts, but I haven't gotten through to anybody so far.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finding other sources for water/methanol mixes. One of my coworkers found a (printed) table with weight percent methanol in a water/methanol mixture for a wide range of temperatures and covering the weight percent range we're interested in. I scanned and OCR'd the table and created a plot. The table appears to lack sufficient resolution to develop an empirical model as well; with only 3 significant figures the lines appear wavy when plotted:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7fUn9.png&quot; alt=&quot;Weight percent methanol in a methanol/water mixture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate using the specific gravity of pure methanol and pure water at a given temperature, then calculate the specific gravity of the mixture using an adjustment for the $\Delta V_{mix}$. This approach seems to work fairly well, but I could only find data for the change in volume due to mixing for 25 &amp;deg;C (&lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanex-TISH-Guide.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 12 of this PDF from Methanex&lt;/a&gt;). This table is &quot;calculated from density and specific volume data&quot; but no additional information is given. A reference table at the end of the document does have the specific volume of methanol at various temperatures, but nothing about change of volume when mixed with water specifically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched at length for methanol/water excess molar volume data. So far I've found data from various sources, but only at 20 and 25 &amp;deg;C, not over a range of temperatures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Derivation from thermodynamics. I've pulled out my thermo textbook and am reading about equations of state and thermodynamics of mixing. I think given enough work I could probably derive a model from the thermodynamics approach, but I would prefer something based on actual measurements, since theory doesn't always match reality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The actual question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there an existing resource that would accomplish what I need?&lt;/li&gt;&#xA;&lt;li&gt;If not, is there a source of &lt;em&gt;good&lt;/em&gt; data on water/methanol mixtures at different temperatures that I might be overlooking?&lt;/li&gt;&#xA;&lt;li&gt;Otherwise, is there a good source of data for excess molar volume over a range of temperatures (at least 18 &amp;ndash; 30 &amp;deg;C, preferably wider) that I might be overlooking?&lt;/li&gt;&#xA;&lt;li&gt;Is there a different approach I should consider?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24" LastEditorUserId="24" LastEditDate="2014-10-03T17:02:41.707" LastActivityDate="2018-01-25T12:59:48.243" Title="Determining weight-percent methanol in water from specific gravity and temperature" Tags="&lt;water&gt;&lt;temperature&gt;&lt;mixtures&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17019" PostTypeId="1" CreationDate="2014-10-02T22:23:07.337" Score="2" ViewCount="2391" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The molar extinction coefficient of pure antibody is $1.4$ absorbance units per milligram antibody at $\pu{280 nm}$. You have a concentrated antibody solution, so you make a $5$-fold dilution and measure at $\pu{280 nm}$, getting an absorbance of $0.845$. What is the concentration of that solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using Beer's Law this is my work: &#xA;$$0.845 = (1.4)\cdot (1)\cdot (c) \\&#xA;\therefore c = \pu{0.604 mol L^-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I think we need to factor in the dilution, therefore $c = (5) \cdot (0.604) = \pu{3.02 mol L^-1} $. Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="8155" LastEditorUserId="4945" LastEditDate="2018-02-27T12:12:53.153" LastActivityDate="2018-02-27T12:12:53.153" Title="How to determine the concentration after a dilution with Beer's law?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17021" PostTypeId="2" ParentId="17019" CreationDate="2014-10-02T23:19:15.170" Score="1" Body="&lt;p&gt;Beers law is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E= \epsilon_{\lambda}\ \cdot c\ \cdot d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $c$ is the concentration of the solution. Then, you've found your answer once you've found $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jerepierre points out, if you were to account for units in the answer, the question is actually missing some information. The pathlength $d$ is unaccounted for (but we're assuming its $1$), however if it was say $\mathrm{1\ cm}$, then the units for $\epsilon$ would need to be $\mathrm{AU\cdot mg^{-1} \cdot cm^{-1}}$. Which would mean that the units for your concentration would be in $\mathrm{mg}$. But, for concentration we would need to know antibody per some amount of &lt;em&gt;volume&lt;/em&gt;. So taking units into account shows that something is missing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're adding at the end is a step you would take to find the original concentration of the undiluted solution. Given that they diluted the concentration 5 times, but we don't know to what factor so we would have trouble solving for that.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4945" LastEditDate="2014-12-02T10:00:59.980" LastActivityDate="2014-12-02T10:00:59.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17022" PostTypeId="1" CreationDate="2014-10-03T00:23:22.770" Score="2" ViewCount="117" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWx3c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is used to measure force or something like that. I broke it and have to buy a new one.&lt;/p&gt;&#xA;" OwnerUserId="8157" LastEditorUserId="4945" LastEditDate="2015-10-23T16:57:39.173" LastActivityDate="2015-10-23T16:57:39.173" Title="Does someone know the name of this equipment?" Tags="&lt;physical-chemistry&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17023" PostTypeId="1" CreationDate="2014-10-03T01:33:02.810" Score="2" ViewCount="177" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Sarin&quot; rel=&quot;nofollow&quot;&gt;sarin&lt;/a&gt; to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tricresyl_phosphate&quot; rel=&quot;nofollow&quot;&gt;tricresyl phosphate&lt;/a&gt;, there are organophosphates that are able to deactivate acetylcholinesterase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's so special about the simple organophosphate group that makes them able to do it?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="15410" LastEditDate="2016-05-28T05:43:40.143" LastActivityDate="2016-05-28T21:09:27.173" Title="Why are so many different organophosphates able to deactivate acetylcholinesterase?" Tags="&lt;biochemistry&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17024" PostTypeId="2" ParentId="17022" CreationDate="2014-10-03T01:53:25.087" Score="3" Body="&lt;p&gt;There are different equipment used to measure force like pivot balance, strain gauge, spring balance, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By seeing the image, we can easily figure out it is &lt;strong&gt;spring balance&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tuUXw.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-10-03T01:53:25.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17026" PostTypeId="2" ParentId="17018" CreationDate="2014-10-03T02:58:08.710" Score="4" Body="&lt;p&gt;You are looking for the &lt;em&gt;excess molar volumes&lt;/em&gt; of water and methanol.  A quick Google search shows that these quantities have been determined and re-determined dozens of times.  The art is to find which paper has the best values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CRC Handbook of Thermophysical and Thermochemical Data and Landolt-Börnstein may have done the sanity check already. &lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-03T02:58:08.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17027" PostTypeId="1" CreationDate="2014-10-03T03:52:49.563" Score="2" ViewCount="545" Body="&lt;p&gt;A question in my book asks...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a gas X that is found in quantities of $261~\mathrm{PPTV}$ (parts per trillion by volume). Compute the molar concentration at $283.15~\mathrm{K}$ and $101325~\mathrm{Pa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you just find the volume using $\frac{n\mathcal{R}T}{P}$ for $1~\mathrm{mol}$ of gas, then multiply this answer by $261/10^{12}$?&lt;/p&gt;&#xA;" OwnerUserId="8159" LastEditorUserId="4945" LastEditDate="2014-10-03T07:21:39.307" LastActivityDate="2014-10-03T07:55:37.563" Title="How can I compute molar concentration if I am given parts per trillion?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17029" PostTypeId="2" ParentId="17027" CreationDate="2014-10-03T07:55:37.563" Score="2" Body="&lt;p&gt;I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.&lt;br&gt;&#xA;First of all the &lt;a href=&quot;http://en.wikipedia.org/wiki/Parts-per_notation&quot; rel=&quot;nofollow&quot;&gt;parts-per-notation&lt;/a&gt; should be avoided as it is not compliant with SI and highly ambiguous.&lt;br&gt;&#xA;Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you can probably make your best guesses, determining how many moles of compound do you have:&#xA;\begin{align}&#xA;pV &amp;amp;=n\mathcal{R}T\\&#xA;n  &amp;amp;= \frac{pV}{\mathcal{R}T}\\&#xA;n  &amp;amp;= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\&#xA;n  &amp;amp;= 1.12\cdot10^{-8}~\mathrm{mol} &amp;amp;\implies c&amp;amp;=  1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-10-03T07:55:37.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17030" PostTypeId="1" CreationDate="2014-10-03T08:00:04.867" Score="2" ViewCount="77" Body="&lt;p&gt;I would like to use the &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1367299/&quot; rel=&quot;nofollow noreferrer&quot;&gt;charmm CMAP&lt;/a&gt; correction in another program (PyRosetta). I have looked at parmfile.doc and io.doc and &lt;a href=&quot;http://www.charmm.org/ubbthreads/ubbthreads.php?ubb=showflat&amp;amp;Number=33451&amp;amp;Searchpage=2&amp;amp;Main=8205&amp;amp;Words=CMAP%20&amp;amp;Search=true#Post33451&quot; rel=&quot;nofollow noreferrer&quot;&gt;this discussion&lt;/a&gt;, but nowhere is it distinctly written how the CMAP is laid out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PHI1 PHI2 PHI3 PHI4 PSI1 PSI2 PSI3 PSI4 24&#xA;!-180&#xA;1 2 3 …&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is this read as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PHI PSI E&#xA;-180 -180 1&#xA;-180 -165 2&#xA;-180 -150 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Or as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PHI PSI E&#xA;-180 -180 1&#xA;-165 -180 2&#xA;-150 -180 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also out of curiosity, must the CMAP be defined between two consecutive torsion angles, or can the angles be far apart (topologically)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is crossposted from &lt;a href=&quot;http://www.charmm.org/ubbthreads/ubbthreads.php?ubb=showflat&amp;amp;Number=34437#Post34437&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. And I must say using UBB forums is a pain after getting used to the wonderful SE interface:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: Would this Q find a better home at &lt;a href=&quot;https://scicomp.stackexchange.com/&quot;&gt;https://scicomp.stackexchange.com/&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="309" LastEditorUserId="-1" LastEditDate="2017-04-13T12:53:54.353" LastActivityDate="2014-10-06T20:26:45.073" Title="CMAP layout of data" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17031" PostTypeId="1" AcceptedAnswerId="17033" CreationDate="2014-10-03T09:39:50.817" Score="4" ViewCount="8639" Body="&lt;p&gt;Does hydrogen bonding only occur between water molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the dipole-dipole intermolecular interaction know as hydrogen bonding because it is between water molecules, or is called that because of the hydrogen molecules present in the bond?&lt;/p&gt;&#xA;" OwnerUserId="8163" LastEditorUserId="9961" LastEditDate="2015-10-02T16:56:07.367" LastActivityDate="2017-05-27T12:34:43.160" Title="Does hydrogen bonding only occur between water molecules?" Tags="&lt;water&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17032" PostTypeId="1" CreationDate="2014-10-03T09:49:03.130" Score="3" ViewCount="1722" Body="&lt;p&gt;How can I identify the liquid in Mathmos lava lamps?&#xA;I have two of them with broken caps, I need to get rid of them. Can I flush the stuff down the toilet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It smells funny.&lt;/p&gt;&#xA;" OwnerUserId="8164" LastEditorUserId="186" LastEditDate="2014-10-03T12:38:37.697" LastActivityDate="2014-10-03T12:38:37.697" Title="What is the liquid in Mathmos lava lamps?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17033" PostTypeId="2" ParentId="17031" CreationDate="2014-10-03T09:56:26.607" Score="8" Body="&lt;p&gt;&lt;strong&gt;NO&lt;/strong&gt;, Water is not only the example for hydrogen bonding. It's most common example for basics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonds occur in DNA, proteins, polymers, etc. But for introductory level it is easy to explain with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DNA:&lt;/strong&gt; Hydrogen bonding between guanine and cytosine, one of two types of base pairs in DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2t21M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Polymer:&lt;/strong&gt; Para-aramid structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWquK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Images: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="9961" LastEditDate="2017-05-27T12:34:43.160" LastActivityDate="2017-05-27T12:34:43.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17035" PostTypeId="2" ParentId="17032" CreationDate="2014-10-03T12:27:55.543" Score="4" Body="&lt;p&gt;The company is not very forthcoming with the ingredients in their &lt;a href=&quot;http://www.mathmos.com.au/FAQ.html&quot; rel=&quot;nofollow&quot;&gt;FAQ&lt;/a&gt;; however, they do state in regards for disposal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Q. How should I dispose of my lava lamp bottle when it reaches the end of its life?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. Please make a hole in the metal cap with a sharp point. &lt;strong&gt;Then pour&#xA;  away the liquid.&lt;/strong&gt; The bottle may then be placed in the glass&#xA;  recycling.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, personally, I go by a general rule that if I do not know what substance is, I do not pour it down the sink.  The authors of the HowStuffWorks webpage &lt;a href=&quot;http://home.howstuffworks.com/green-living/dispose-lava-lamp.htm&quot; rel=&quot;nofollow&quot;&gt;How to Dispose of Lava Lamps&lt;/a&gt; state the actual ingredients are a trade secret, but the first patent for lava lamps states the ingredients as being:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;solidified globule of mineral oil, paraffin and a dye as well as paraffin wax or petroleum jell, preferably Ondina 17 with a light paraffin, carbontetrachoride, a dye and the paraffin wax or petroleum jelly.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as for disposal, the author of that website made contact with a company, who advised that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They told me that I could wrap the lava lamp in newspaper and throw it in the trash. I was told that it was non-toxic. &lt;strong&gt;They warned me not to pour the liquid down the sink because it contained wax&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wax could make a mess of the pipes from your sink if it were to accumulate and solidify, and there is more than enough rubbish and pollutants in the water ways as it is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best to consult your local area's rules and regulations as to disposal.&lt;/p&gt;&#xA;" OwnerDisplayName="user8016" LastActivityDate="2014-10-03T12:27:55.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17036" PostTypeId="1" CreationDate="2014-10-03T17:14:24.023" Score="0" ViewCount="1602" Body="&lt;p&gt;What is the meaning of &lt;em&gt;ordinary chemical means&lt;/em&gt; in the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An element may be defined as a pure substance that can not be&#xA;  decomposed by ordinary chemical means.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8166" LastActivityDate="2014-11-02T18:03:26.047" Title="Ordinary chemical means" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17037" PostTypeId="2" ParentId="17036" CreationDate="2014-10-03T17:22:00.533" Score="1" Body="&lt;p&gt;Essentially it means that you exclude the possibility of splitting an element even further via nuclear means.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-03T17:22:00.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17038" PostTypeId="1" AcceptedAnswerId="17040" CreationDate="2014-10-03T17:46:45.883" Score="22" ViewCount="20979" Body="&lt;p&gt;When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$. It stays in this form until the water evaporates and then the ions go back to normal $\ce{NaCl}$. So why does water with salt in it still taste like salt? I am asking because if the molecule $\ce{NaCl}$ is broken down into $\ce{Na+}$ and $\ce{Cl-}$ then how does it have the characteristics of $\ce{NaCl}$?&lt;/p&gt;&#xA;" OwnerUserId="8167" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T13:04:44.113" LastActivityDate="2020-04-04T21:44:26.103" Title="Why does salt dissolved in water taste like salt? If it's just Na+ and Cl- ions" Tags="&lt;ions&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17039" PostTypeId="2" ParentId="17038" CreationDate="2014-10-03T18:08:59.607" Score="4" Body="&lt;p&gt;I'm not a biology expert, so I'm not sure how the $\ce{Na+}$ and $\ce{Cl-}$ ions actually activate certain taste receptors in your tongue such that you can  taste the $\ce{Na+}$ and $\ce{Cl-}$ ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the chemistry part, do remember that even when you are putting solid $\ce{NaCl}$ into your mouth, the saliva in your mouth will dissolve it, resulting in the breaking of the ionic bond between $\ce{Na}$ and $\ce{Cl}$. Sodium chloride is so soluble that even a tiny drop of water could dissolve a significant amount of salt - probably more than  what you would put into your food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it doesn't really matter if you taste solid sodium chloride or sodium chloride solution - they will all taste the same, but perhaps varying in intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, if you are able to find a sodium chloride crystal large enough, and ensure that your tongue is dry enough, I believe(again I'm not a biology expert) that you would not be able to taste the sodium chloride, as solid crystals cannot diffuse as easily into your tongue and activate the sodium channels.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="4945" LastEditDate="2014-10-04T08:45:22.867" LastActivityDate="2014-10-04T08:45:22.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17040" PostTypeId="2" ParentId="17038" CreationDate="2014-10-03T18:10:09.820" Score="37" Body="&lt;p&gt;When you taste salt, you're not pushing crystalline $\ce{NaCl}$ into your taste buds. It dissolves in your saliva and dissociates. When one tastes salt, the saltiness taste receptors respond specifically to the sodium cation. That type of taste receptor is a cation channel. This is why lithium and potassium cations also taste salty (though they also stimulate other receptors which make them taste somewhat different). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seem to be at least two types of receptors that respond to saltiness. One responds almost specifically to sodium at low concentrations, but at higher concentrations, the other type responds to many cations. See &lt;a href=&quot;http://www.nih.gov/researchmatters/february2010/02082010taste.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T13:05:16.323" LastActivityDate="2015-06-27T13:05:16.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17041" PostTypeId="2" ParentId="6929" CreationDate="2014-10-03T19:15:59.800" Score="2" Body="&lt;p&gt;The concept to understand here is that most theories in physics and chemistry are obtained by &lt;em&gt;observing&lt;/em&gt; then theorizing, this results in the theories and models being made in such a way that they can explain most (if not all) observable phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when Bohr made his model of the atom, it had to explain the the phenomenon of &quot;X-ray emission&quot; otherwise it wouldn't be accepted, because, X-ray emission is an observed phenomenon of atoms. Following on, it was discovered that the Bohr-model couldn't explain all observable phenomenon as equipment got better and more things could be seen (notably the splitting of spectral lines in magnetic fields). So any theory that followed on to explain this splitting of spectral lines had to explain X-ray emission as well, without disturbing the results of Bohr's theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A suitable analogy would be the discovery of special relativity by Einstein, whatever discovered had to explain why Newton's Law had stuck around for hundred's of years without people realizing it was wrong. (And it did, the correction at velocities we know are less than one part in a million and that's why we didn't think Newton was wrong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the deeper question is, even a one part in a million error is still an error and so we should know that the theory is wrong even early on... this is not the case as most instrumental errors on their own are much larger in this regard (at least in Newton's time), and the same applies here in that older instruments weren't usually sensitive enough to notice the change brought on by quantum mechanics on X-ray emission to that of Moseley's rough equation. (Though now it is usually possible to detect differences to a good degree.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to the final part of your question, there is a horrendous equation waiting if you want to plug in exact values to find the exact wavenumber, but most like the instrument barely notices a difference, so in a sense, it doesn't matter which theory we use. (It's very well possible a hundred years down the road, another genius comes up with an even smaller correction that can be observed only with futuristic devices)&lt;/p&gt;&#xA;" OwnerUserId="7584" LastActivityDate="2014-10-03T19:15:59.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17042" PostTypeId="1" CreationDate="2014-10-03T19:28:12.423" Score="12" ViewCount="15151" Body="&lt;p&gt;Why doesn’t diamond have $\Delta H_\mathrm{f}^\circ=0$, when graphite does? Is it something to do with the definition – diamonds can’t really form at STP, even though it is naturally occurring?&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorDisplayName="user7951" LastEditDate="2015-06-15T08:55:47.210" LastActivityDate="2015-06-15T08:55:47.210" Title="Why is the standard enthalpy of formation of diamond not zero?" Tags="&lt;enthalpy&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17043" PostTypeId="2" ParentId="17042" CreationDate="2014-10-03T19:38:12.470" Score="11" Body="&lt;p&gt;You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTGrm.png&quot; alt=&quot;Gibbs energy of phases of C relative to graphite&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bm0jd.png&quot; alt=&quot;P-T phase diagram for C&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In the reference above, both absolute Gibbs free energies, as well as free energies with respect to the stable phase at STP, are given for most elements.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-10-03T19:38:12.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17044" PostTypeId="2" ParentId="17042" CreationDate="2014-10-03T19:39:46.477" Score="5" Body="&lt;p&gt;Carbon naturally exists as two allotropes, graphite and diamond. By definition, the most stable allotrope at STP (the one with the lowest heat of formation at STP) is assigned a heat of formation of zero.  Graphite has the lower heat of formation and is assigned a heat of formation of zero, while diamond being slightly less stable has a heat of formation of 2.4kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-03T19:39:46.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17046" PostTypeId="1" CreationDate="2014-10-03T21:26:34.327" Score="5" ViewCount="435" Body="&lt;p&gt;Is there a standard manner of representing chemical formulas as linear strings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $H_2O$ can be represented with the string &lt;code&gt;H2O&lt;/code&gt;. Isotopes are usually specified with a number before the element. But this must be preceded by a space, or it can be get mixed with the number right before: &lt;code&gt;O4 13CH&lt;/code&gt;, which is equivalent to $O_4{^{13}C}H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just wondering if there is any set of accepted conventions when representing chemical formulas as linear strings.&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="5017" LastEditDate="2014-10-04T01:00:07.973" LastActivityDate="2014-10-04T01:00:07.973" Title="Representation of chemical formulas as strings?" Tags="&lt;computational-chemistry&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17047" PostTypeId="2" ParentId="17046" CreationDate="2014-10-03T21:40:38.343" Score="7" Body="&lt;p&gt;If I understand your question correctly, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Simplified_molecular-input_line-entry_system&quot; rel=&quot;noreferrer&quot;&gt;simplified molecular-input line-entry system (SMILES)&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Chemical_Identifier&quot; rel=&quot;noreferrer&quot;&gt;IUPAC International Chemical Identifier (InChI)&lt;/a&gt; are two examples of &quot;human-readable&quot; string notations for chemical formulas your are looking for.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-10-03T21:40:38.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17048" PostTypeId="2" ParentId="11212" CreationDate="2014-10-03T23:09:22.847" Score="1" Body="&lt;p&gt;For the sake of correct formalism, here the answer as it should be given in the textbook:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass flow rate of the leaking substance, under the assumption of linear scaling, is&#xA;$$\dot{m}_{\ce{CHF2Cl}} = 12 ~\mathrm{kg\, month^{-1}} = 144~\mathrm{kg\, a^{-1}}\; .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass flow rate of chlorine, as contained in the leaking substance, is then calculated as follows:&#xA;$$\dot{m}_{\ce{Cl}} = \dot{m}_{\ce{CHF2Cl}} \cdot \frac{M_{\ce{Cl}}}{M_{\ce{CHF2Cl}}} = 144~\mathrm{kg\, a^{-1}} \cdot \frac{35.453~\mathrm{g\,mol^{-1}}}{86.47~\mathrm{g\,mol^{-1}}} = 59.04~\mathrm{kg\, a^{-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the end result to the question &quot;how many kilograms of chlorine will be emitted into the atmosphere each year?&quot;, please note that the mass flow rate still has a dependency on time&#xA;$$ \dot{m} = \frac{\mathrm{d} m}{\mathrm{d} t}\; , $$&#xA;which means that we have to do an integration so we can calculate the mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m(\tau) = \int_0^\tau \dot{m}\,\mathrm{d}t = \dot{m}\,\tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one year, we get&#xA;$$ m_{\ce{Cl}}(1~\mathrm{a}) = \dot{m}_{\ce{Cl}}\cdot1~\mathrm{a} = 59.04~\mathrm{kg}\; , $$&#xA;which means that your attempt at answering your own question were very good, on spot and totally valid.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-03T23:09:22.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17049" PostTypeId="2" ParentId="6913" CreationDate="2014-10-04T00:15:07.870" Score="3" Body="&lt;p&gt;You can measure non-stoichiometry by determining how many atoms are present.  There are a number of ways to do this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can dissolve your unknown material in an acid and use a reagent to precipitate away one element but not another.  By keeping careful track of the number of moles of reagent you used, you can know how many atoms were precipitated and thus determine the stoichiometry of the original sample.  Once upon a time, this approach was used to demonstrate the fact that some chemicals ARE stoichiometric!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, another commonly used method is to count the number of X-rays emitted by a sample in an electron microscope.  Each atom emits X-rays with different amounts of energy so you can determine not only what atoms are present, but how many relative to other atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you can measure iron sulfide which stoichiometrically should have one Fe atom for each S.  Fe emits X-rays with about 2.5 times the energy of S, so they are easy to tell apart.  You then place a known stoichiometric iron sulfide (for example, using chemistry as above) in the microscope and measure how many Fe and S X-rays you get.  Let's say, hypothetically, you get one X-ray from Fe for each X-ray from S (just to keep the math simple).  Then you place a non-stoichiometric iron sulfide in and suddenly you get about 0.9 Fe X-rays for each S.  Presto, you know you have about 9 Fe atoms for each 10 S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: For simplicity sake, I left out all the intricacies of the experimental method which would have to be considered in a real measurement.  If you are interested you can google: Energy Dispersive Spectroscopy (EDS), wavelength dispersive spectroscopy (WDS), and wet chemical analysis.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2014-10-04T00:15:07.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17050" PostTypeId="2" ParentId="17038" CreationDate="2014-10-04T00:37:27.117" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When $\ce{NaCl}$ is dissolved into water it breaks down into $\ce{Na+}$ and $\ce{Cl-}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, not quite. Dissolving something in water does not break the chemical bonds of the molecule*. You still have $\ce{NaCl}$ molecules floating about. That would leave you with elemental sodium and elemental chlorine, and those two fellows are a bit like drunk bikers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing sodium, chlorine and water will produce all sorts of things like sodium hypochorite ($\ce{NaClO}$, best known as bleach), sodium hydroxide ($\ce{NaOH}$ - caustic soda) and some lesser-known chlorine compounds. Many are rather dangerous, and I don't think we would have survived as a lifeform if a bit of salt on our food turned into bleach (which destroys DNA) after a Perrier chaser.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;exception: some molecules will react with water rather than simply dissolve in in it. Depends on a number of factors.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5692" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T13:07:23.193" LastActivityDate="2015-06-27T13:07:23.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17052" PostTypeId="1" AcceptedAnswerId="17133" CreationDate="2014-10-04T05:40:37.757" Score="7" ViewCount="499" Body="&lt;p&gt;I am currently reading a textbook on combustion. They assume that all reactions go to completion, such that post combustion only unburned fuel/air and actual products remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that combustion is lean or perfect (no fuel left unburned) and not neglecting dissociation effects, can reverse reaction actually occur to reform fuel at equilibrium? In other words, is the equilibrium constant infinity in real life?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="5456" LastEditDate="2014-10-04T06:01:37.510" LastActivityDate="2014-10-06T14:52:04.740" Title="Do reverse reactions occur in combustion?" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17053" PostTypeId="2" ParentId="17023" CreationDate="2014-10-04T05:50:26.433" Score="2" Body="&lt;p&gt;Acetylcholine's normal function is to cleave acetylcholine's ester bond to convert it into acetic acid and choline, but certain compounds can interfere. The short answer is that these nerve agents all share similar structural components allowing them to be bound by acetylcholinesterase, where they then either react to alter the enzyme's binding site, preventing activity, or are just very slow to hydrolyze, temporarily inhibiting the enzyme until it can remove the compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all of these related acetylcholinesterase inhibitors work in exactly the same way, but if you examine their structures and compare them with acetylcholine, the normal substrate of the enzyme, you'll notice some similarities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetylcholine: &lt;img src=&quot;https://i.stack.imgur.com/1tNaR.png&quot; alt=&quot;acetylcholine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sarin: &lt;img src=&quot;https://i.stack.imgur.com/2ukcD.png&quot; alt=&quot;sarin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tricresyl phosphate: &lt;img src=&quot;https://i.stack.imgur.com/yFpZU.png&quot; alt=&quot;tricresyl phosphate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that they all have this central double-bonded oxygen (a few have sulfur instead, but are converted to oxygen metabolically), which is important for fitting into the enzyme. The ester bond holding the choline to the carbonyl in acetylcholine is relatively easily hydrolized, but phosphoesters are much slower to hydrolyze. From what I can tell, this seems to be one mechanism behind trecresyl phosphate's toxicity and probably why it's not as acutely toxic as something like sarin or VX. Sarin, on the other hand, has a pretty good fluoride leaving group. This lets it form a strong covalent bond with a serine's sidechain in ACE's binding pocket. This reaction is effectively irreversible, making sarin very dangerous as it totally disables any ACE it reacts with, meaning it takes very little to cause a significant effect.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-04T05:50:26.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17054" PostTypeId="1" AcceptedAnswerId="17057" CreationDate="2014-10-04T06:01:34.377" Score="1" ViewCount="47" Body="&lt;p&gt;This may sound weird but could one 'zap' some liquid like milk with electricity of a certain voltage for a 'few' seconds hoping to kill any bacteria in it? I know some people suggested some type of radiation on meat ( I think) but could a small charge of electricity on liquid products kill any bacteria in them?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastActivityDate="2014-10-04T07:57:09.867" Title="About electricity and liquid products" Tags="&lt;safety&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17056" PostTypeId="2" ParentId="16651" CreationDate="2014-10-04T06:16:23.780" Score="3" Body="&lt;p&gt;Silberberg or Zumdahl &lt;em&gt;Chemistry&lt;/em&gt; are great introductory reference books in my opinion. They cover the topics in just the right amount of depth and explain concepts extremely well, I highly recommend it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Inorganic Chemistry&lt;/em&gt; and &lt;em&gt;Physical Chemistry&lt;/em&gt; by Peter Atkins also seem to be decent (though very dense) (I haven't used it much though). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also suggest Clayden et al.'s &lt;em&gt;Organic Chemistry&lt;/em&gt; - it's a great textbook for organic chemistry though a bit intimidating at first- you should probably go through the organic chemistry section of Silberberg or Zumdahl first (nomenclature, functional groups and naming) before they use this. &lt;/p&gt;&#xA;" OwnerUserId="8143" LastActivityDate="2014-10-04T06:16:23.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17057" PostTypeId="2" ParentId="17054" CreationDate="2014-10-04T07:57:09.867" Score="4" Body="&lt;p&gt;Yes, this is a known phenomenon. It's a process called electroporation. &lt;a href=&quot;http://ieeexplore.ieee.org/xpl/login.jsp?tp=&amp;amp;arnumber=887601&amp;amp;url=http%3A%2F%2Fieeexplore.ieee.org%2Fiel5%2F57%2F19183%2F00887601.pdf%3Farnumber%3D887601&quot; rel=&quot;nofollow&quot;&gt;Here's&lt;/a&gt; a paper talking about using it for sterilization. Basically, when you apply an electric field to a cell membrane, you can render it porous. At low field strength, the pores are small and can be closed by the cell, but if you apply a larger potential across the cell membrane, cell lysis is the result, killing the cell. It's actually used for killing cancerous tumour cells in humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the trick with doing it with something like milk is that, to get a large enough potential across the cells in the liquid, an enormous potential (up to 1 kV/mm) needs to be applied to electrodes immersed in the liquid, even though the cell membrane is only seeing a volt or two. For this to be manageable, you would probably need a thin flow cell or something. It could be a fair bit more expensive than pasteurization. Even though only a short pulse is used, electrolysis will occur, which could affect the taste. How well it works will also be affected by the electrical properties of the solution, which you can't necessarily play with too much in foods without changing the taste, so it might not be the best technique for foods, but there are certainly other areas where it's useful.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-04T07:57:09.867" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17059" PostTypeId="1" CreationDate="2014-10-04T11:17:38.343" Score="1" ViewCount="3345" Body="&lt;p&gt;I've been taught at school that in an ionic compound like $\ce{NaCl}$, electrons are transferred from the $\ce{Na}$ to the $\ce{Cl}$ to form $\ce{Na+}$ and $\ce{Cl-}$ which both have full outer shells. But what causes these electrons to be transferred?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="4231" LastEditDate="2014-11-04T00:18:53.040" LastActivityDate="2014-11-04T00:18:53.040" Title="Why are electrons transferred in an ionic bond?" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17061" PostTypeId="2" ParentId="17059" CreationDate="2014-10-04T13:36:55.223" Score="3" Body="&lt;p&gt;The modern definition of electronegativity is due to Linus Pauling. It is the power of an atom in a molecule to attract electrons to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the electronegativity difference between the two atoms  is greater than 2.0, the bond is considered ionic; i.e. there is complete transfer (between the two atoms) of the electrons in the bond.&#xA;The electronegativity (according to Pauling scale) of chlorine is 3.0, while the electronegativity  of sodium is 0.9. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-04T13:36:55.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17062" PostTypeId="1" CreationDate="2014-10-04T14:15:17.213" Score="4" ViewCount="60" Body="&lt;p&gt;I've been looking at the MO:s of different molecules in Avogadro, fascinated by their magnificent geometries. But how is one supposed to interpret them, so as to understand their reactivity? What are the first interesting things one should note when looking at the MO:s of different molecules? Interesting patterns to recognize? There is a lot of technical information on MO theory but not much info on how to interpret them intuitively. Any other sources or books I should read?&lt;/p&gt;&#xA;" OwnerUserId="8180" LastEditorUserId="865" LastEditDate="2015-10-30T17:46:40.407" LastActivityDate="2015-10-30T17:46:40.407" Title="Interpreting the MOs of a molecule" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2017-03-25T12:29:40.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="17063" PostTypeId="1" CreationDate="2014-10-04T14:20:15.693" Score="-3" ViewCount="2157" Body="&lt;p&gt;The Nitronium ion, which is a positively charged ligand, in which the donor atom is Nitrogen. The question is, that Nitrogen has a total of 5 electrons, it shares 2 with oxygen and donates 2 electrons to another oxygen and also looses one electron to form a cationic  $\ce{NO2+}$. It does not have a lone pair, not even a single electron. How is it a ligand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Dissenter: You explain nitronium for the nitration of benzene, in which a covalent bond is formed. I would like to know this for a coordination bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have any example, but I see the nitronium ion as a example in a book of PRADEEP publication which is a TRUSTED publisher.&lt;/p&gt;&#xA;" OwnerUserId="8181" LastEditorUserId="4945" LastEditDate="2014-10-06T10:22:10.467" LastActivityDate="2014-10-06T10:22:10.467" Title="Are there ligands without lone pairs?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="10" ClosedDate="2014-10-05T07:54:28.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="17064" PostTypeId="1" AcceptedAnswerId="17072" CreationDate="2014-10-04T15:52:25.853" Score="18" ViewCount="16536" Body="&lt;p&gt;I saw somewhere (can't recall where) that KF is the most ionic compound.  I expected CsF.  Does the greater polarizability of Cs allow it to more easily form covalent bonds compared to K?  Does this overcome the fact that K being in n = 4 should bond better with F in n = 2?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="9961" LastEditDate="2018-03-05T19:00:08.770" LastActivityDate="2019-10-24T15:29:02.253" Title="Is KF the most ionic compound?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17065" PostTypeId="2" ParentId="16673" CreationDate="2014-10-04T16:11:46.147" Score="2" Body="&lt;p&gt;$\ce{C_mH_{2n}O_n -&amp;gt; m C + n H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydration reaction of sulfuric acid acts as the driving force because it is highly exothermic (95.33 kJ/mol H&lt;sub&gt;2&lt;/sub&gt;SO&lt;sub&gt;4&lt;/sub&gt;).&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="16683" LastEditDate="2017-04-01T01:18:49.400" LastActivityDate="2017-04-01T01:18:49.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17066" PostTypeId="1" AcceptedAnswerId="17077" CreationDate="2014-10-04T16:47:54.780" Score="9" ViewCount="3527" Body="&lt;p&gt;Attached is a phase diagram for carbon dioxide. At &lt;span class=&quot;math-container&quot;&gt;$p = 0$&lt;/span&gt;, it appears that the substance will exist as a gas regardless of temperature.&lt;/p&gt;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;strong&gt;any&lt;/strong&gt; liquid at a fixed temperature, is it possible to vaporize it completely by simply reducing its pressure isothermally?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;strong&gt;any&lt;/strong&gt; solid at a fixed temperature, is it possible to sublime it completely by simply reducing its pressure isothermally?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What if the solid is at &lt;span class=&quot;math-container&quot;&gt;$T \approx0\ \mathrm K$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HPtp0.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-22T00:54:57.973" Title="Does everything exist as a gas at p = 0?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17067" PostTypeId="1" AcceptedAnswerId="17082" CreationDate="2014-10-04T17:28:43.493" Score="10" ViewCount="3750" Body="&lt;p&gt;Here are the 2 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{3Cu + 2NO3- + 8H+ &amp;amp;&amp;lt;=&amp;gt; 3Cu^2+ + 2NO + 4H2O} &amp;amp; E &amp;amp;= 0.62~\mathrm{V}\\&#xA;\ce{Cu + 2NO3- + 4H+ &amp;amp;&amp;lt;=&amp;gt; Cu^2+ + 2NO2 + 2H2O}&amp;amp; E &amp;amp;= 0.45~\mathrm{V}\\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nernst equation tells us that as we increase the concentration of nitric acid the first equation becomes less spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how? What is the mechanism of the two reactions that decides between the first and second reactions depending upon how much nitric acid is available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, any qualitative explanation for what the Nernst equation shows would be appreciated. Why does having more protons available end up in fewer protons being consumed by the reaction?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It was pointed out that the second equation above is improperly balanced.  Here's the correct equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Cu + 4H+ + 2NO3- &amp;lt;=&amp;gt; Cu^2+ + 2NO2 + 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as pointed out in the comment by Neto, the more concentrated acid &lt;em&gt;does&lt;/em&gt; consume more protons per reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any insight on the mechanism of the reaction and how it changes with nitric acid concentration?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="17175" LastEditDate="2020-02-28T01:58:50.720" LastActivityDate="2020-02-28T02:15:23.843" Title="How does copper reduce dilute nitric acid to nitric oxide and concentrated nitric acid to nitrogen dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17068" PostTypeId="2" ParentId="17063" CreationDate="2014-10-04T17:40:06.240" Score="1" Body="&lt;p&gt;A ligand is ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an inorganic coordination entity, the atoms or groups joined to the central atom.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In biochemistry: if it is possible or convenient to regard part of a polyatomic molecular entity as central, then the atoms, groups or molecules bound to that part are called ligands. Biochemical usage is thus wider, in that the central entity can be polyatomic. Thus H+ may be a ligand for proteins ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/L03518.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://goldbook.iupac.org/L03518.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, yes, there can be ligands which do not possess lone pairs. One being the hydrogen proton. It doesn't have any electrons at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the nitronium ion - yes - the central nitrogen doesn't have a lone pair - but that doesn't mean it can't participate in coordination chemistry. If the N had a lone pair and could donate it to something, then that N would be a Lewis base. An example of a Lewis base is ammonia. Since the N in nitronium ion has no lone pairs, but it can form complexes with other molecules - the N in nitronium ion is instead a Lewis acid. What happens usually is that some Lewis base attacks the N in nitronium ion and one of the N=O bonding pairs becomes an oxygen lone pair. This free the central N up for bonding. Consider the bottom left part in this graphic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y59ZG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-10-04T17:45:10.887" LastActivityDate="2014-10-04T17:45:10.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17069" PostTypeId="1" CreationDate="2014-10-04T18:57:13.470" Score="-1" ViewCount="850" Body="&lt;p&gt;Can we compare the electron shells with the orbits of the planets around the Sun, is this a good comparison?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true for all the atoms that they can have only 2 electrons on the first shell, but why is it so, why 2? &lt;/p&gt;&#xA;" OwnerUserId="8185" LastEditorUserId="4945" LastEditDate="2015-10-23T17:39:14.297" LastActivityDate="2015-10-23T17:39:14.297" Title="Can we compare the electron shells with the orbits of the planets around the sun?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17070" PostTypeId="1" AcceptedAnswerId="17071" CreationDate="2014-10-04T19:29:17.797" Score="1" ViewCount="7852" Body="&lt;p&gt;Does pyridine or furan have the more shielded proton NMR signals? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that both are aromatic, so both have rather deshielded signals due to ring current combining constructively with the applied magnetic field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, which would have the more shielded signal? I know that shielded protons have more electron density around them. I guess then the question boils down to whether nitrogen or oxygen is more electron-releasing, as both nitrogen and oxygen have at a lone pair participating in resonance.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-03-02T20:23:55.673" LastActivityDate="2017-03-02T20:23:55.673" Title="Proton NMR signals and rings" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17071" PostTypeId="2" ParentId="17070" CreationDate="2014-10-04T20:40:13.293" Score="10" Body="&lt;p&gt;Here is the proton nmr spectrum for pyridine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kj9cw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the proton nmr spectrum for furan&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wd0NX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, benzene has a single proton nmr signal at 7.27 ppm.  In both pyridine and furan there are different types of protons, so multiple signals are seen in both cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the resonance structures for pyridine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RqYGh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and furan.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PND0R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pyridine removes electron density from the alpha and gamma positions by resonance while furan adds electron density to all four ring positions by resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to resonance effects there are also inductive effects and ring currents that will modify the proton chemical shifts.  Inductively, the alpha positions in both molecules will be deshielded.  Ring currents will also deshield the protons as you've noted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the signals are in the general vicinity of benzene.  The alpha-protons in both molecules are the most deshielded due to inductive effects, with furan's alpha signal upfield from pyridine's due to the electron donating resonance effect of the furan oxygen.  &lt;strong&gt;The beta proton in furan is the most shielded&lt;/strong&gt; since it is not inductively deshielded and, again, the oxygen in furan is pushing electron density into the ring &lt;em&gt;via&lt;/em&gt; resonance which will cause shielding.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-04T20:40:13.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17072" PostTypeId="2" ParentId="17064" CreationDate="2014-10-04T20:44:39.130" Score="19" Body="&lt;p&gt;Yes, this is a subtle thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pauling electronegativities&lt;/a&gt;, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be &quot;more ionic.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, an ionic bond requires separating charge, so &lt;span class=&quot;math-container&quot;&gt;$\ce{Cs+F-}$&lt;/span&gt;. The problem is that &lt;span class=&quot;math-container&quot;&gt;$\ce{Cs+}$&lt;/span&gt; is much larger than &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and so the dipole moment for a fully ionic &lt;span class=&quot;math-container&quot;&gt;$\ce{Cs+F-}$&lt;/span&gt; would be much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find that the charge isn't fully separated, likely because of the need to stabilize the large dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some people plot a curve of &quot;% ionic character&quot; as determined by the &lt;strong&gt;&lt;em&gt;actual&lt;/em&gt;&lt;/strong&gt; dipole moment vs. the expected dipole moment (i.e, full charge separation). We find that this curve reaches a limit ~75-85% ionic character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC, I think on that basis LiF is more Ionic than KF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xAl9g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more, I really like Bruce Robinson's lecture notes:&#xA;&lt;a href=&quot;https://web.archive.org/web/20160406074454/http://courses.washington.edu/bhrchem/c152/Lec23.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://courses.washington.edu/bhrchem/c152/Lec23.pdf&lt;/a&gt; via &lt;a href=&quot;https://web.archive.org/web/20160406074454/http://courses.washington.edu/bhrchem/c152/Lec23.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Internet Archive&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="4945" LastEditDate="2019-10-24T15:29:02.253" LastActivityDate="2019-10-24T15:29:02.253" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17074" PostTypeId="2" ParentId="7800" CreationDate="2014-10-04T22:51:45.543" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't fluorine be the central atom in inter-halogen compounds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, fluorine can be at the &quot;center of things.&quot;  Examples would include the strongly hydrogen bonded hydrofluoric acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ar5Gh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the &lt;strong&gt;very relevant example&lt;/strong&gt; of the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja001613e&quot; rel=&quot;noreferrer&quot;&gt;trifluoride anion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[F-F-F]^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent).  Of course, in this last example we've &quot;tricked&quot; the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to answer this question, two general cases need to be considered,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The anionic situation $\ce{[X-F-X]^-}$ &lt;em&gt;vs.&lt;/em&gt;  $\ce{[X-X-F]^-}$&lt;/li&gt;&#xA;&lt;li&gt;The cationic situation $\ce{[X-F-X]^+}$ &lt;em&gt;vs.&lt;/em&gt; $\ce{[X-X-F]^+}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where X is Cl, Br or I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Mrp1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, computational analysis of the electron distribution in the &lt;a href=&quot;http://classes.uleth.ca/200303/chem3810a/part7.pdf&quot; rel=&quot;noreferrer&quot;&gt;triiodide anion&lt;/a&gt; (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom).  Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the cation, similar arguments come into play.  Again, &lt;a href=&quot;http://www.academia.edu/847283/Electronic_Structure_and_Properties_of_Trihalogen_Cations_X3_and_XY2_X_Y_F_Cl_Br_I_&quot; rel=&quot;noreferrer&quot;&gt;calculations for a variety of trihalide cations&lt;/a&gt; (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguments that &lt;strong&gt;don't&lt;/strong&gt; explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;the lack of &quot;d-orbitals&quot;&lt;/strong&gt;; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;inefficient 2p-3p overlap&lt;/strong&gt;; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-25T20:23:19.960" LastActivityDate="2015-02-25T20:23:19.960" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17075" PostTypeId="2" ParentId="17038" CreationDate="2014-10-04T23:45:46.450" Score="9" Body="&lt;p&gt;Have you ever tasted salt that was not dissolved in water?  Think about this.  Have you ever tasted ANYTHING that was not dissolved in, or in contact with, water?  The answer is &quot;No&quot;, unless you have the worst case of chronic dry mouth ever to exist in the history of humanity, and if you do have that terrible case of dry mouth, you no longer taste anything.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because everything you taste is sampled by your taste buds after having samples of its molecules transported through pores on your tongue by water.  No water, no taste.  And saliva, of course, is mostly water.  If you place a dry grain of salt on your tongue, saliva immediately dissolves a bit of it, and transports that salt through taste pores, where receptors in your taste buds report its flavor.&lt;/p&gt;&#xA;" OwnerUserId="8190" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T13:05:48.690" LastActivityDate="2015-06-27T13:05:48.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17076" PostTypeId="2" ParentId="17069" CreationDate="2014-10-04T23:48:37.490" Score="4" Body="&lt;p&gt;The comparison with planets is problematic. In fact, it's so problematic that it's one of the reasons that the development of quantum mechanics was necessary and this classical picture of the atom was entirely discarded. If the electrons were like planets, they would act like rapidly spinning charges. That would cause them to radiate out a lot of energy. As they lost energy they would crash into the nucleus - and so destroy the atom in a flash of radiation almost instantly. In short, atoms simply could not exist if they acted like planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the shells &quot;exist&quot; in the first. The shells are the only place an electron can go. The first shell is the lowest energy an electron can take, which stops it crashing into the nucleus as it would if it were circling it like a planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second half of the question, it is true that the first shell contains, at most, two electrons. You really need to learn a bit more about quantum mechanics and quantum numbers to really get to grips with &lt;em&gt;why&lt;/em&gt;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the shortest answer is that these &quot;shells&quot; you're talking about are not a complete picture of the structure of electrons in an atom. Each shell is divided into smaller parts called &lt;strong&gt;orbitals&lt;/strong&gt; and each orbital can contain two electrons. It happens that this first shell contains only one orbital - so it takes two electrons. The second shell has four orbitals (one that is similar to the orbital in the first shell and three others) - so that adds up to eight electrons in the second shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this implied on some of these &quot;shell&quot; diagrams where electrons are grouped in pairs, representing how each orbital has two electrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.educationquizzes.com/library/GCSE/Chemistry/AtomicStructure3/Atomic-Structure-3-03.jpg&quot; alt=&quot;Atomic structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons behind &lt;em&gt;all of that&lt;/em&gt; would make for a far longer answer, though. That would range from &quot;because that's what the quantum numbers say&quot; to &quot;because those are the solutions to Laplace's equation say&quot; and more. Which I think is a bit beyond the scope of what you're asking right now, but I think it is more or less thoroughly covered &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8598/maximum-number-of-electrons-each-shell?rq=1&quot;&gt;here&lt;/a&gt; if you need it.&lt;/p&gt;&#xA;" OwnerUserId="8189" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.360" LastActivityDate="2014-10-04T23:48:37.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17077" PostTypeId="2" ParentId="17066" CreationDate="2014-10-05T01:29:40.117" Score="12" Body="&lt;p&gt;Regardless of whether a substance is in a vacuum, it will vaporize if some fraction of its particles have enough energy (due to their temperature) to exceed their binding energy.  &lt;a href=&quot;http://www.es.ucsc.edu/~fnimmo/website/binding_energies.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Binding energy&lt;/a&gt; is the energy contained in the intermolecular bonds holding a substance together.  It is definitely possible for the particles of a substance to have a binding energy that is substantially higher than thermal energy will readily provide.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No.  While most liquids have fairly low binding energies (and thus, relatively high vapor pressures) there are liquids that will not evaporate in a vacuum at an appreciable rate, one example being &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_liquid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic liquids&lt;/a&gt;.  As I wrote above, the attraction between the ions of these liquids is strong enough that they do not measurably evaporate over time.&lt;br&gt;&lt;br&gt;&#xA;If you know the Periodic Videos YouTube channel, they've had a couple of videos about these fluids over the past couple of years.  (&lt;a href=&quot;https://www.youtube.com/watch?v=EF44__0GJEY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a link to one&lt;/a&gt;, if you're interested.)  One of them featured the behavior of liquids in a vacuum, but I can't seem to find the link; it may have been on a sister channel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, consider all of the asteroids, space ships, satellites, and such floating around in space.  For all intents and purposes they will float around until the end of time without evaporating.  Again, the reasoning is as I listed above.  We also don't see the vast majority of solid objects here on Earth slowly evaporating away, which we would expect if sublimation were a concern in a vacuum.  The reasoning is the same as above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the solid is very, very cold, it is less likely to experience sublimation since its particles' average energies would be very low.  Even dry ice would not vaporize if its temperature were very close to 0 K.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is a phase diagram of both water and carbon dioxide that might help illustrate the situation.  You can see that as the temperature of the substance decreases the vapor pressure of the substance also decreases.  At a sufficiently low temperature the vapor pressure is effectively zero (below the x-axis of this chart.)  There is always the possibility of a single atom or molecule gaining enough energy to vaporize or doing something weird like tunneling off of the surface of the substance, but the process is so slow as to be unmeasurable.  Again, asteroids in the solar system have been hanging around for several billion years without subliming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wt00v.jpg&quot; alt=&quot;Phase Diagram of CO2 and H2O&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture from &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/4/40/Comparison_carbon_dioxide_water_phase_diagrams.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Be careful if you look up phase diagrams of common substances.  They often involve only a few points that are actually referenced and the rest is drawn with quite a bit of artistic license.  Compare the phase diagrams of water &lt;a href=&quot;http://stevengoddard.files.wordpress.com/2010/09/water-phase-diagram1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.phy.duke.edu/~hsg/363/table-images/water-phase-diagram.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; to see what I mean.  They are often drawn without scales (or with minimal scales) and are sloppy at best.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2020-02-21T20:50:03.803" LastActivityDate="2020-02-21T20:50:03.803" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="17080" PostTypeId="1" AcceptedAnswerId="17305" CreationDate="2014-10-05T04:08:10.430" Score="2" ViewCount="859" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;State the hybridisation of asterisked carbon in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3-CH=\overset{\ast}{C}=CH2}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am not sure between &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}$&lt;/span&gt; hybridization. Because of the presence of double bonds I think it should be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2},$&lt;/span&gt; but because of the two π-bonds I think &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="41328" LastEditDate="2021-05-23T12:06:47.663" LastActivityDate="2021-05-23T14:35:12.577" Title="What is the hybridization of the central carbon in buta‐1,2‐diene?" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;&lt;hybridization&gt;&lt;allenes&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17082" PostTypeId="2" ParentId="17067" CreationDate="2014-10-05T04:24:01.007" Score="16" Body="&lt;p&gt;If we examine the Nernst equation: $E = E^\circ-\frac{RT}{zF}\ln Q$, the logarithmic term is what changes with concentration. The reaction quotient $Q$, gets smaller as the ratio of reactants to products increases, meaning that the log term will decrease (and become negative below $Q = 1$), so both equations will become more favourable as the ratio of reactants to products increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correctly balanced second reaction is $$\ce{Cu + 2NO3- + 4H+ &amp;lt;=&amp;gt; Cu^2+ + 2NO2 + 2H2O}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at just the logarithmic term for each reaction, assuming the products part of the reaction quotient is unity and that the proton activity is equal to the nitrate activity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{gather}&#xA;\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{10}}\tag{cyan}\\&#xA;\frac{25.7~\mathrm{mV}}{2}\ln\frac{1}{[\ce{NO3-}]^6}=\frac{25.7~\mathrm{mV}}{6}\ln\frac{1}{[\ce{NO3-}]^{18}}\tag{red}&#xA;\end{gather}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y axis: log term (mV), x axis: $[\ce{NO3-}]$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2sN2J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As expected, when multiplied to normalize the number of electrons transferred, the reaction that consumes more reagents becomes more favourable more quickly as the reagent concentration is increased and less favourable more quickly as the concentration is decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually, the reason for this is that the reactions travel down a slope towards the equilibrium point (see figure) and the reaction that consumes more of the reagents to produce a given amount of the products involved in the equilibrium reaches equilibrium with fewer reactions. Because ∆G is simply the partial derivative of the Gibbs energy with respect to reaction extent, the reaction that reaches equilibrium with less reaction extent has a steeper slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZgJE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="4945" LastEditDate="2016-02-08T03:29:13.403" LastActivityDate="2016-02-08T03:29:13.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17084" PostTypeId="5" CreationDate="2014-10-05T04:40:10.020" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-10-05T04:40:10.020" LastActivityDate="2014-10-05T04:40:10.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17085" PostTypeId="4" CreationDate="2014-10-05T04:40:10.020" Score="0" Body="collective term for a set of laboratory techniques for the separation of mixtures" OwnerDisplayName="user8016" LastEditorDisplayName="user8016" LastEditDate="2014-10-05T18:17:15.723" LastActivityDate="2014-10-05T18:17:15.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17086" PostTypeId="2" ParentId="9801" CreationDate="2014-10-05T04:55:45.733" Score="1" Body="&lt;p&gt;You are actually correct, the formatting of the vertical axis in the solution graph has resulted in the vertical (Peak Height) values written below the actually gridline it is referring to - you can see the same for peak height values of 5, 10 and 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the linear equation formula $y=mx+c$ for the 3 known values yields a relationship of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = 0.25C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Where $P$ is Peak height and $C$ is concentration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the known peak height for the 4th value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$7=0.25C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C=28$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is exactly where it is shown on the graph in the answer.&lt;/p&gt;&#xA;" OwnerDisplayName="user8016" LastActivityDate="2014-10-05T04:55:45.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17088" PostTypeId="1" CreationDate="2014-10-05T05:58:03.713" Score="1" ViewCount="27372" Body="&lt;p&gt;I need to find the temperature two things equilibrate to in a constant pressure calorimeter given only their specific heats, masses, and starting temps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I know for specific heat is that. $c = \frac{J}{G\times \Delta T}$, and how to work that around to find if one is missing, and well as $\frac{q}{\Delta T}$ for doing things with calorimeters, though a different kind. That's the only equation I know for this stuff. In this question I have neither the energy nor the change in temp. I know how to solver it if the two substances have equal specific heats, then you just find the mid point between temperatures. But two different substances, I have no idea. I tried to do it the hard way of just slowly taking of/adding one *C of each and matching up the energies required but that got me nowhere.&lt;/p&gt;&#xA;" OwnerUserId="8196" LastEditorUserId="73" LastEditDate="2014-10-05T06:32:25.527" LastActivityDate="2017-01-16T23:31:40.217" Title="Finding the equilibrium temperature between two substances in a constant pressure calorimeter" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17089" PostTypeId="1" AcceptedAnswerId="17102" CreationDate="2014-10-05T06:44:57.477" Score="13" ViewCount="97806" Body="&lt;p&gt;I studied the solubility of compounds in water. I have found that &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_carbonate&quot;&gt;calcium carbonate&lt;/a&gt; doesn't dissolve in water. The teacher stated that the ionic compounds dissolve in water except some carbonates. What would be a clear elaboration of this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="8197" LastEditorUserId="470" LastEditDate="2014-10-05T10:10:38.110" LastActivityDate="2016-03-10T05:41:26.550" Title="Why doesn't calcium carbonate dissolve in water even though it is an ionic compound?" Tags="&lt;solubility&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17090" PostTypeId="2" ParentId="17088" CreationDate="2014-10-05T07:00:48.770" Score="2" Body="&lt;p&gt;Assuming the heat absorption of the calorimeter is negligible (so if it's made of an insulating material, the only changes in heat will be of the contents themselves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substance 1 (the one at the higher initial temperature) has a heat = $q_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substance 2 has a heat of $q_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since substance 1 will lose heat until it comes to equilibrium with substance 2, the heat lost by 1 will be absorbed by 2, so $q_1 = -q_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $q_1 = m_1c_{p1}(T_{{1}_{initial}} - T_{{1}_{final}})$ and $q_2 = m_2c_{p2}(T_{{2}_{initial}} - T_{{2}_{final}})$ and $T_{1_{final}} = T_{2_{final}} = T_f$ you can solve for $T_f$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heat absorption of the calorimeter was not negligible, then $q_1 + q_2 + q_{cal} = 0$&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2014-10-05T07:17:02.000" LastActivityDate="2014-10-05T07:17:02.000" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17091" PostTypeId="2" ParentId="17089" CreationDate="2014-10-05T07:03:50.067" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Teacher stated that the ionic compounds dissolve in water except some carbonates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, at best, an oversimplification. Other ionic compounds such as silver sulfide are sparingly soluble in water. Note that this isn't a carbonate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And sodium hydrogen carbonate - $\ce{NaHCO3}$ - is soluble in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium carbonate ($\ce{Na2CO3}$) is also soluble in water. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want a clear elaboration of this phenomenon can somebody help me??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solubility depends on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The strength of the intermolecular attractions within the solute and the strength of the solvent-solvent interactions. How strongly is the calcium ion and the carbonate ion bound together? You might &lt;em&gt;think&lt;/em&gt; that the force of attraction between a +2 and a -2 ion is strong, but remember, covalent bonds are generally stronger than ionic interactions. How strong are the solvent-solvent molecule interactions - which must be broken before something can be solvated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Solvation. How well can the solvent lower the energy of the dissociated solute ions? How well can calcium ion and carbonate ion be solvated (by water in this case?) How strong are the solvent-solute ion interactions? What about the entropy of solvation? (Solvation shells are highly ordered). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to predict the solubility of some compound, try using hard soft acid/base theory. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-01-03T14:40:08.187" LastActivityDate="2015-01-03T14:40:08.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17092" PostTypeId="2" ParentId="17089" CreationDate="2014-10-05T07:05:22.107" Score="15" Body="&lt;p&gt;For an ionic substance to dissolve in water, there are two competing factors that determine the enthalpy of solution $\Delta{H}_\mathrm{sol}$ which is the enthalpy (energy) change  when a solute is dissolved in a solvent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The lattice energy (LE), the energy of formation of the crystal between infinitely separated ions. As LE is proportional to the charges of its ions, for calcium carbonate, its LE would be roughly 4&amp;times; the LE of sodium chloride. $$\ce{CaCO3 (s) -&amp;gt; Ca^2+ (g) + CO3^2- (g)}$$ This value is always positive, as energy is required to separate the ions.&lt;/li&gt;&#xA;&lt;li&gt;The hydration energy of gaseous ions, which  is the enthalpy change when gaseous ions dissolve in sufficient water to give an infinitely dilute solution. $$\ce{Ca^2+ (g) + CO3^2- (g) -&amp;gt; Ca^2+ (aq) + CO3^2- (aq)}$$ These values are always negative, as energy is released upon hydration of ions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As the LE of calcium carbonate is so large, a great amount of free energy would be required to break the strongly attracted ions apart, and this energy has to come from the enthalpy of hydration. In the case of calcium carbonate,  the enthalpy of hydration is not large enough to overcome the large lattice energy, hence it exists as a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do note that solubility of ionic compounds are not absolute. In fact, solubility in water is often given as a value in g/L of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, generally, we define solubility as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Soluble substances can form a 0.1&amp;nbsp;M solution at 25&amp;nbsp;°C.&lt;/li&gt;&#xA;&lt;li&gt;Insoluble substances cannot form a 0.1&amp;nbsp;M solution at 25&amp;nbsp;°C.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="4945" LastEditDate="2016-03-10T05:41:26.550" LastActivityDate="2016-03-10T05:41:26.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17096" PostTypeId="2" ParentId="15772" CreationDate="2014-10-05T10:43:56.067" Score="12" Body="&lt;p&gt;Approach 2 is better than approach 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach corresponds with methodology employed in both web links you provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic 'rule' is to do your calculation keeping all digits as you go right until to the very end (without rounding parts as you suggest in approach 1). This ensures your calculation retains maximum accuracy. You leave the 'rounding' til the very end, that is when you report your answer, only 'so many' digits are 'significant'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use your example, when calculating the subtraction part: 2.378-1.2 = 1.178, in approach 1, by 'rounding' this down to 2 significant digits (1.2), you are reducing the accuracy of your calculation before you finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct (most accurate) way to calculate the result is to keep all significant digits in your calculation until the very end, that is, 1.178 / 1.03 = 1.14368932 and then report only the 'significant' digits. In this case, since the number 1.2 only has 2 significant digits, your final answer can only have 2 significant digits. That is, the final answer is only as precise as the least precise measurement (in your case, 1.2), so in reporting the final answer for your calculation, you can only rely on 2 significant digits, so you report 1.1 as your calculated answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you re-examine the answer provided in your link (&lt;a href=&quot;http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm&quot;&gt;http://www.occc.edu/kmbailey/chem1115tutorials/Sig_Fig_and_Math_Answers.htm&lt;/a&gt;) in which you state problem 3 follows approach 1, you will find that the solution provided in that link actually follows approach 2, similar to what I described in the above paragraph. That is, all decimal places are retained in the calculation until the very, when the final answer is reported to the number of significant figures corresponding to the least significant measurement.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-10-05T10:43:56.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17097" PostTypeId="1" AcceptedAnswerId="17098" CreationDate="2014-10-05T11:38:14.917" Score="1" ViewCount="697" Body="&lt;p&gt;In the book &quot;Foundations of Materials Science and Engineering&quot; (Smith and Hashemi) the following is written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For cases of non-steady state diffusion in which diffusivity is independent of time, Fick's second law of diffusion applies, which is&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what cases are materials' diffusivity dependent on time, and in what cases are they not?&lt;/p&gt;&#xA;" OwnerUserId="8200" LastEditorDisplayName="user7951" LastEditDate="2015-09-28T23:08:43.887" LastActivityDate="2018-02-21T13:44:49.283" Title="Fick's Second Law of Diffusion" Tags="&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17098" PostTypeId="2" ParentId="17097" CreationDate="2014-10-05T13:47:12.513" Score="1" Body="&lt;p&gt;Diffusivity is not dependent on time. It is dependent on temperature (and material). Fick's second law holds because the diffusivity, $D$, is constant with respect to depth, $x$, and time, $t$, so it is treated as a constant in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right) = D\left(\frac{\partial^2c_x}{\partial x^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fick's second law of diffusion applies to non steady state diffusion, which &lt;strong&gt;concentration at a depth $x$, $c_x$&lt;/strong&gt; is a function of time.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorDisplayName="user7951" LastEditDate="2015-09-28T23:14:19.473" LastActivityDate="2015-09-28T23:14:19.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17099" PostTypeId="1" AcceptedAnswerId="17100" CreationDate="2014-10-05T14:27:51.673" Score="0" ViewCount="4262" Body="&lt;p&gt;The problem is stated as follows:&#xA;&lt;em&gt;The surface of a steel gear made of 1022 steel (0.22 wt% C) is to be gas-carburized at $927$ °C. Calculate the time necessary to increase the carbon content to 0.30 wt% at 0.030 in below the surface of the gear. Assume the carbon content of the surface to be 1.20 wt% and&lt;/em&gt; $D_{927^{\circ}C}=1.28 \times 10^{-11}\frac{m^{2}}{s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is $\frac{C_{s}-C_{x}}{C_{s}-C_{0}}=ERF(\frac{x}{2\sqrt{Dt}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution manual is setting $C_{x}=0.30$, but should it not be $C_{x}=0.30+0.22=0.52$, since the initial carbon content in the steel is $C_{0}=0.22$?&lt;/p&gt;&#xA;" OwnerUserId="8200" LastEditorUserId="5456" LastEditDate="2014-10-05T16:06:21.487" LastActivityDate="2014-10-05T16:06:21.487" Title="Carburizing Calculation" Tags="&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17100" PostTypeId="2" ParentId="17099" CreationDate="2014-10-05T14:49:22.520" Score="0" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equation $C_x$ simply means the concentration at depth x. It does not mean the &lt;em&gt;change&lt;/em&gt; of concentration. There is no need to add its initial concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, concentration cannot be linearly added. (Does adding a 1M solution to a 2M solution give 3M?)&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-05T14:49:22.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17101" PostTypeId="2" ParentId="17089" CreationDate="2014-10-05T17:26:21.713" Score="5" Body="&lt;p&gt;Solubility of salts is not a prediction that I like to make.  It all comes down to $\Delta{G}_{sol} &amp;lt; 0$.  That condition is fulfilled when $\Delta{H}_{sol} &amp;lt; 0$ and $T \Delta{S}_{sol} &amp;gt; 0$ or one factor outweighs the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's two problems.  Lattice enthalpies are large, as are solvation enthalpies, and the common observation is that $\Delta{H}_{sol}$ in most cases is small.  We are calculating differences of large numbers, and the one fact we can confidently assert is that the difference is impossible to predict and generally meaningless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing for $\Delta{S}_{sol}$.  Lattice entropies are small, but entropies of solution are lower than one would think thanks to the short-range order the solvated cation imposes on the  solvent.  You are gaining translational entropy from dissolving the lattice and losing translational entropy from the solvent through the inner coordination shell.  You are calculating differences of numbers of equal magnitude again, and just by staring at reagents it's impossible to tell which way the pendulum is going to swing and if the result is positive or negative.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-05T17:36:09.523" LastActivityDate="2014-10-05T17:36:09.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17102" PostTypeId="2" ParentId="17089" CreationDate="2014-10-05T18:44:00.917" Score="16" Body="&lt;p&gt;As someone said here, this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The teacher stated that the ionic compounds dissolve in water except&#xA;  some carbonates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is indeed an oversimplification. First of all, the distinction between an &quot;ionic compound&quot; to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FMZlp.jpg&quot; alt=&quot;picture of a cave&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This forms &lt;em&gt;because&lt;/em&gt; calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\ce{CO2}$ as $$\ce{H2O + CO2 &amp;lt;=&amp;gt; H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\ce{CaCO3 + H2CO3 &amp;lt;=&amp;gt; Ca^2+ + 2HCO3-}.$$&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="4945" LastEditDate="2016-03-10T05:36:59.730" LastActivityDate="2016-03-10T05:36:59.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17103" PostTypeId="1" AcceptedAnswerId="17183" CreationDate="2014-10-05T18:53:36.240" Score="12" ViewCount="3285" Body="&lt;p&gt;Our professor only told us that A is favored, and B is not getting formed. I thought about the resonance structures and saw a contradiction to that what the prof told us. To be honest I think my argument with the resonance structure is not strong enough, because it does not explain the prefered product. Why is structure B disfavored?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sS81h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@jerepierre: I checked again. You're right! I'm sorry for the wrong re-drawing. Although the above discussed reaction with deprotonation was also dictactical important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct starting material is in fact a alpha,beta-unsaturated ketone which is reduced by Li/NH3!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correct reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EnmSx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="46306" LastEditDate="2018-02-20T07:24:14.257" LastActivityDate="2018-03-06T08:38:26.497" Title="Regioselective enolate formation" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;resonance&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17104" PostTypeId="2" ParentId="17080" CreationDate="2014-10-05T19:05:07.200" Score="1" Body="&lt;p&gt;As the asterisked carbon has two neighbors, the hybridization of this asterisked carbon is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp},$&lt;/span&gt; i.e. the angle &lt;span class=&quot;math-container&quot;&gt;$\ce{−CH=\overset{\ast}{C}=CH2}$&lt;/span&gt; is 180°.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="41328" LastEditDate="2021-05-23T12:04:41.610" LastActivityDate="2021-05-23T12:04:41.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17106" PostTypeId="1" AcceptedAnswerId="20350" CreationDate="2014-10-05T19:47:21.540" Score="5" ViewCount="2191" Body="&lt;p&gt;Different internet sources say different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, can someone explain why it is (or isn't) chemically reactive?&lt;/p&gt;&#xA;" OwnerUserId="7163" LastEditorUserId="194" LastEditDate="2017-05-23T17:14:05.717" LastActivityDate="2017-05-23T17:14:05.717" Title="Is graphene chemically reactive?" Tags="&lt;materials&gt;&lt;reactivity&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17108" PostTypeId="1" AcceptedAnswerId="17110" CreationDate="2014-10-05T20:17:33.357" Score="0" ViewCount="187" Body="&lt;p&gt;A Millimeter of Mercury(mmHg) is by definition the pressure exerted on a column of mercury 1 mm  in height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this definition confusing because it doesn't say how the diameter of the container will affect the pressure. &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="5137" LastEditDate="2014-10-05T22:09:47.883" LastActivityDate="2014-10-05T22:09:47.883" Title="Understanding Millimeter of Mercury" Tags="&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17109" PostTypeId="1" AcceptedAnswerId="17111" CreationDate="2014-10-05T20:31:11.877" Score="16" ViewCount="161966" Body="&lt;p&gt;Wikipedia states that waxes can be dissolved by non-polar organic solvents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I failed to dissolve candle wax in both acetone and ethanol, which are the only organic solvents I have at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some clothes with wax stuck on them. How can I dissolve the wax and drive it away without damaging my clothing?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="402" LastEditDate="2015-11-12T21:15:21.700" LastActivityDate="2015-11-12T21:15:21.700" Title="How to dissolve candle wax (paraffin)?" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;&lt;fats&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17110" PostTypeId="2" ParentId="17108" CreationDate="2014-10-05T20:38:10.580" Score="3" Body="&lt;p&gt;Pressure is defined as force exerted &lt;strong&gt;per unit area.&lt;/strong&gt; Therefore, the diameter of the mercury column is taken into account when we calculate the area of the column.  A mercury column with a large diameter will exert a greater force than a column of the same height, but with a smaller diameter - but the force divided by the area will be the same.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-05T20:38:10.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17111" PostTypeId="2" ParentId="17109" CreationDate="2014-10-05T20:56:26.763" Score="18" Body="&lt;p&gt;Ethanol and acetone are not non-polar organic solvents. Each one has  a slight dipole moment; due to the difference of electronegativity between $\ce{H}$ and  $\ce{O}$ in ethanol and between $\ce{C}$ and  $\ce{O}$ in acetone.  Wax is composed of heavy, long-chain alkanes. And as &quot;Like dissolves like&quot; try to dissolve your wax in toluene or in xylene.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-05T20:56:26.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17112" PostTypeId="2" ParentId="17103" CreationDate="2014-10-05T21:03:13.597" Score="10" Body="&lt;p&gt;Unsymmetrical ketones can produce two different enolates, as shown in your reaction scheme.  The enolate pictured in the top row of your drawing is referred to as the &lt;strong&gt;thermodynamic enolate&lt;/strong&gt;, while the enolate shown in the bottom row of your drawing is the &lt;strong&gt;kinetic enolate&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More highly substituted double bonds are thermodynamically more stable than less substituted double bonds. Therefore, the thermodynamically favored enolate is the one that gives rise to the most highly substituted double bond in the enolate intermediate. The thermodynamic enolate will be formed when reaction conditions are chosen (see below) to allow equilibration of all enolate intermediates; product formation is then thermodynamically controlled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetically favored enolate is the fastest formed enolate, the one formed by abstracting the most acidic proton (proton acidity follows primary &gt; secondary &gt; tertiary).  In your reaction, removal of the proton on the secondary carbon would be kinetically favored (more acidic) over removal of the proton on the tertiary carbon atom (less acidic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction conditions can be selected to favor kinetic (aprotic solvents, strong bases, low temperature) or thermodynamic (protic solvents, weaker bases, higher temperature) enolate formation, giving the chemist control over selecting which product will be formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your reaction were run under conditions favoring kinetic enolate formation, then we would expect to observe product formed from following the bottom reaction line.  If the reaction were run under conditions favoring thermodynamic enolate formation, then we would expect to obtain products from following the top line of your reaction scheme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your professor said that pathway &quot;A&quot;, the top line in your drawing, is favored, then the reaction is being run under conditions favoring formation of the thermodynamic enolate.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-05T23:42:07.233" LastActivityDate="2014-10-05T23:42:07.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17115" PostTypeId="1" AcceptedAnswerId="17117" CreationDate="2014-10-05T22:38:18.160" Score="2" ViewCount="1733" Body="&lt;p&gt;I have a conceptual physical chemistry question here. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A thermos bottle containing milk is shaken vigorously. Consider the milk as the system. Will the temperature rise as a result of the shaking?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer to this is yes, but I'm having trouble finding an equation that explains this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best i have is the molecular kinetic energy formula that &lt;span class=&quot;math-container&quot;&gt;$$\mathrm{ke_{avg}= \frac{1}{2}mv^2=\frac{3}{2}kT}$$&lt;/span&gt;&#xA;So a change in kinetic energy is &lt;span class=&quot;math-container&quot;&gt;$$ \mathrm{\Delta ke = \frac{3}{2} k \Delta T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the right approach?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2019-11-15T22:23:06.883" LastActivityDate="2021-04-28T03:40:11.713" Title="A thermos bottle containing milk is shaken vigorously" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17117" PostTypeId="2" ParentId="17115" CreationDate="2014-10-06T00:06:17.217" Score="7" Body="&lt;p&gt;A conceptual approach might work better, since the shaking system is not going to convert mechanical energy into thermal energy as neatly as something like &lt;a href=&quot;http://www.eoht.info/page/Paddle+wheel+experiment&quot; rel=&quot;noreferrer&quot;&gt;Joule's Paddle Wheel&lt;/a&gt;.  As the bottle is shaken, the liquid in it is swirled around into more or less randomly oriented vortices.  These cancel one another out and slow to rest via friction with their container, and the energy they contained has to go somewhere.  As you wrote, it winds up increasing the average velocity, and thereby temperature, of the particles in the milk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic mechanical to thermal energy conversion is the source of the conversion factor from joules to calories.  Joules are used to measure mechanical energy and calories were used to measure thermal energy.  It was the paddle wheel experiment I linked above that first related the two to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge presented by trying to relate the work done on the bottle of milk to the thermal energy that winds up going into the milk is less mathematical than practical.  On any given shake we would first put energy into the bottle (move it upward, let's say) then take energy back out (slow it to a stop) before putting energy back into the bottle (moving back downward) and taking energy out again (slowing to a stop a second time.)  This repeats until we get tired of shaking, and if we measured the difference in all of the energy inputs and outputs that ought to equal the thermal energy added to the milk, but actually doing that for real would be a pain.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-10-06T00:18:58.560" LastActivityDate="2014-10-06T00:18:58.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17118" PostTypeId="1" AcceptedAnswerId="17121" CreationDate="2014-10-06T00:55:45.980" Score="4" ViewCount="88916" Body="&lt;p&gt;My teacher isn't very good at helping me solve this problem. I keep getting the answer, but I don't know why I am getting it. Could you please help me understand this problem and why you should do this specific step?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{1.00 g}$&lt;/span&gt; sample of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3.10H2O}$&lt;/span&gt; was dissolved in &lt;span class=&quot;math-container&quot;&gt;$\pu{20.0 mL}$&lt;/span&gt; of distilled water. Additional water was added so as to give &lt;span class=&quot;math-container&quot;&gt;$\pu{250 mL}$&lt;/span&gt; of solution. What is the molar concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I did and got the answer and don't understand why I need to do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \frac{\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} &amp;amp;= \pu{0.003447 mol}\\&#xA;\frac{\pu{0.003447 mol}}{\pu{0.020 L}} &amp;amp;= \pu{0.174825 M}\\&#xA;(\pu{0.174825 M})(\pu{0.020 L}) &amp;amp;= (\pu{0.250 L})(x)\\&#xA;x &amp;amp;= \pu{0.013986 M}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.013986 M}$&lt;/span&gt; is the correct answer, but I don't know why and don't understand the question.&#xA;If you were to do the problem, could you explain why you did what you did to get to the answer.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="4945" LastEditDate="2020-02-22T12:31:12.110" LastActivityDate="2020-02-22T12:43:34.083" Title="How do I calculate the molarity of the Na2CO3 solution?" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17119" PostTypeId="1" CreationDate="2014-10-06T01:26:33.737" Score="0" ViewCount="2284" Body="&lt;p&gt;I am given $20 \mathrm {mL}$ of the unknown mixture of Barium Nitrate and Sodium Acetate dissolved in water. I am given the total mass of the two solids before they were put into the mixture.&#xA;I am given $0.200 \mathrm{M}$ of $\ce{NaOH}$, $0.200 \mathrm{M}$ of $\ce{Na2SO4}$, $0.200 \mathrm{M}$ of $\ce{NaCl}$ and ethanol to separate a precipitate. &#xA;I am learning about selective precipitation.&lt;/p&gt;&#xA;" OwnerUserId="8209" LastEditorUserId="5456" LastEditDate="2014-10-06T04:26:20.853" LastActivityDate="2014-10-06T17:38:55.517" Title="How do I determine the mass percent of Barium nitrate and Sodium acetate in an original unknown solution" Tags="&lt;homework&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17120" PostTypeId="2" ParentId="17118" CreationDate="2014-10-06T02:53:53.303" Score="4" Body="&lt;p&gt;The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original &lt;span class=&quot;math-container&quot;&gt;$1.00\ \mathrm  g$&lt;/span&gt; sample:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is determining the concentration of the first &lt;span class=&quot;math-container&quot;&gt;$20.0\ \mathrm{mL}$&lt;/span&gt; solution, i.e. what concentration results from dissolving the &lt;span class=&quot;math-container&quot;&gt;$0.00345\ \mathrm{mol}$&lt;/span&gt; of sodium carbonate calculated above.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third part is simply a dilution – what concentration results when we dilute the first &lt;span class=&quot;math-container&quot;&gt;$20.0\ \mathrm{mL}$&lt;/span&gt; solution to &lt;span class=&quot;math-container&quot;&gt;$250\ \mathrm{mL}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_2 = \frac{c_1V_1}{V_2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user7951" LastEditDate="2020-02-22T12:43:34.083" LastActivityDate="2020-02-22T12:43:34.083" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17121" PostTypeId="2" ParentId="17118" CreationDate="2014-10-06T02:56:35.687" Score="4" Body="&lt;p&gt;What the question is asking is the molar concentration of the given substance after dilution with the help of values given. The first step of the answer is converting the given weight of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3.10H2O}$&lt;/span&gt; into amount of substance. For that we have a formula as &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\text{amount of substance} &amp;amp;= \frac{\text{mass of substance}}{\text{molecular mass of the substance}}\\&#xA;\text{amount of substance} &amp;amp;= \frac{&#xA;\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}}&#xA;= \pu{0.003447 mol}.\\&#xA;\end{align}&lt;/span&gt;&#xA;We used this first step to determine the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3.10H2O}$&lt;/span&gt; present in the solution. It is necessary because even after dilution of the solution the amount of substance will remain constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is converting the amount of substance into molarity. As,&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \text{molarity} &amp;amp;= \frac{\text{amount of substance}}{\text{volume in L}}\\&#xA;  \text{molarity} &amp;amp;= \frac{\pu{0.003447 mol}}{\pu{0.020 L}} = \pu{0.174825 M}\\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is done to find out the concentration of the substance in terms of molarity since the question has asked us to give the answer in molarity after dilution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that the given solution is diluted up to &lt;span class=&quot;math-container&quot;&gt;$\pu{250 mL}$&lt;/span&gt;, so we have to use dilution formula &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;  \text{initial concentration}\times \text{initial volume of solution} =\\&#xA;  \text{final concentration}\times \text{final volume of solution}.&#xA;\end{multline}$$&lt;/span&gt;&#xA;So we suppose that the final concentration of the solution is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and we have values for others entities.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;(\pu{0.174825 M})(\pu{0.020 L}) &amp;amp;= (\pu{0.250 L})(x)\\&#xA;x &amp;amp;= \pu{0.013986 M}\\&#xA;\end{align}&lt;/span&gt;&#xA;This last step is carried out because after dilution the molarity of the solution changes but not the amount of substance as I stated earlier. So we equate the equation as &lt;span class=&quot;math-container&quot;&gt;$$\text{initial amount of substance} = \text{final amount of substance}.$$&lt;/span&gt;&#xA;And the final answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.013986 M}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4945" LastEditDate="2020-02-22T12:39:13.683" LastActivityDate="2020-02-22T12:39:13.683" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17122" PostTypeId="1" AcceptedAnswerId="37806" CreationDate="2014-10-06T05:11:05.110" Score="2" ViewCount="67" Body="&lt;p&gt;I read somewhere that EXAFS (Extended X-ray absorption fine structure) gives us the distances between central atom and its neighboring atoms. Is that true? Why can't we get the distance between two atoms, which none of them are central? How is EXAFS method distinguishing between central atom and other ones? So for example for CF4, I think we get one EXAFS peak for C...F distance and not F...F distance, why?&lt;/p&gt;&#xA;" OwnerUserId="6985" LastActivityDate="2015-09-23T23:33:34.193" Title="Central Atom in EXAFS" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17123" PostTypeId="2" ParentId="17109" CreationDate="2014-10-06T05:15:20.637" Score="4" Body="&lt;p&gt;A readily available mixture of various non-polar alkanes  would be petrol (gasoline).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-10-06T05:15:20.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17124" PostTypeId="1" AcceptedAnswerId="17135" CreationDate="2014-10-06T05:34:13.243" Score="30" ViewCount="23613" Body="&lt;p&gt;According to Wikipedia, $\ce{HeH+}$ and fluoroantimonic acid are the strongest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;em&gt;Nature&lt;/em&gt;, Carborane acid is the strongest, but Wikipedia says fluoroantimonic acid is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Links:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Helium_hydride_ion&quot; rel=&quot;noreferrer&quot;&gt;$\ce{HeH+}$ / Wikipedia&lt;/a&gt; - stating HeH+ the second most acidic&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Compounds_of_fluorine&quot; rel=&quot;noreferrer&quot;&gt;$\ce{HeH+}$ / Wikipedia 2nd source&lt;/a&gt; - stating $\ce{HeH+}$ the most acidic (see in 'Notes' section)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot; rel=&quot;noreferrer&quot;&gt;Fluoroantimonic Acid / Wikipedia&lt;/a&gt; - stating it is the strongest acid (super_acid)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/news/2004/041115/full/news041115-5.html&quot; rel=&quot;noreferrer&quot;&gt;Carborane Acid / nature.com&lt;/a&gt; - According to &lt;em&gt;Nature&lt;/em&gt; and even Google itself, it is the strongest acid created but according to wikipedia Flouroantimonic acid is stronger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, which is the strongest ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="5026" LastEditDate="2018-03-29T01:34:06.430" LastActivityDate="2018-03-29T01:34:06.430" Title="What is the strongest acid?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17125" PostTypeId="2" ParentId="17124" CreationDate="2014-10-06T05:49:41.677" Score="15" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Superacid&quot;&gt;Wikipedia&lt;/a&gt;&lt;br&gt;&#xA;It depends on the medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For plasma: &lt;a href=&quot;http://en.wikipedia.org/wiki/Helium_hydride_ion&quot;&gt;helium hydride ion&lt;/a&gt;&lt;br&gt;&#xA;For liquids: &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot;&gt;fluoroantimonic acid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot;&gt;Wikipedia&lt;/a&gt;&lt;br&gt;&#xA;Fluoroantimonic acid - $H_0$ Value = −31.3&lt;br&gt;&#xA;Carborane superacid - $H_0$ Value = −18.0&lt;br&gt;&#xA;100% Sulfuric Acid - $H_0$ Value = -12.0&lt;br&gt;&#xA;I didn't find any information about $\ce{HeH^+}$, I guess it's because it only exist in a plasma and difficult to study.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-10-06T06:04:38.007" LastActivityDate="2014-10-06T06:04:38.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17126" PostTypeId="2" ParentId="17124" CreationDate="2014-10-06T06:04:38.510" Score="11" Body="&lt;p&gt;Carborane Acid  is the strongest acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is technically the world’s strongest acid is because it’s the strongest acid as a &lt;strong&gt;single molecule&lt;/strong&gt;. Much stronger acids- the acids that melt through tables- can be created by mixing certain antimony compounds with &lt;strong&gt;hydrofluoric acid ($\ce{HF}$)&lt;/strong&gt;. For example when you mix $\ce{SbF5}$ with $\ce{HF}$, you get an acid with a pH of &lt;strong&gt;-31&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://chemistrytwig.com/2011/05/09/what-is-the-strongest-acid/&quot; rel=&quot;noreferrer&quot;&gt;Chemistrytwig&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T12:57:24.980" LastActivityDate="2015-06-27T12:57:24.980" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17127" PostTypeId="1" AcceptedAnswerId="17128" CreationDate="2014-10-06T08:41:06.897" Score="1" ViewCount="1749" Body="&lt;p&gt;Generally, for the addition of dry HX to an alkene, it follows Markovnikov's rule. The hydrogen atom adds to the carbon with the most hydrogen atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the case of an alkene which does not have any hydrogen atom attached to its carbon, such as 2,3-dimethyl2-butene, would the electrophilic addition still take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jfkyM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="7475" LastEditDate="2015-10-09T16:51:34.027" LastActivityDate="2015-10-09T16:51:34.027" Title="Electrophilic addition of HX to 2,3-dimethylbut-2-ene" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17128" PostTypeId="2" ParentId="17127" CreationDate="2014-10-06T10:42:19.680" Score="6" Body="&lt;p&gt;Electrophilic addition would likely still occur. Remember that Markovnikov's rule is an observational heuristic for &quot;in electrophilic addition reactions of alkenes in which carbocations form, the more/most stable carbocation forms preferentially, and thus the product originating from that cation is the major product&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a carbocation can form. Nothing about the tetrasubstituted pattern prevents addition. The molecule is still planar and can be aproached by the electrophile. The methyl groups are electron donating by &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_effect&quot;&gt;induction&lt;/a&gt;. The methyl groups 1) increase the electron density of the alkene and make it more nucleophilic, and 2) stabilize the carbocation intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 2,3-dimethyl-2-butene is likely more reactive than other alkenes toward electrophiles. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)2C=C(CH3)2 + HBr -&amp;gt; (CH3)2BrC-CH(CH3)2}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-10-06T10:42:19.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17129" PostTypeId="1" AcceptedAnswerId="17170" CreationDate="2014-10-06T14:01:37.607" Score="6" ViewCount="402" Body="&lt;p&gt;I want to use copper(II) sulfate as rocket propellant additive in hope that it will alter the flame colour. Because water is not welcome in rocket fuel, I decided to dehydrate the crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I tried stainless steel mess tin, but that produced some smoke warning me that it's reacting. So I switched to aluminium mess tin, that one didn't smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still some fumes, presumably steam was produced. I continued roasting. Almost at the end, I noticed that my nose hurts in the inside a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose copper sulfate has got in there. Do you think I risked badly, or is it just weak irritation? How can I avoid it next time, apart of using fumehood?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="7475" LastEditDate="2015-12-15T00:47:51.960" LastActivityDate="2015-12-15T00:47:51.960" Title="Why did roasting of copper(II) sulfate pentahydrate hurt my nose?" Tags="&lt;safety&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17133" PostTypeId="2" ParentId="17052" CreationDate="2014-10-06T14:52:04.740" Score="1" Body="&lt;p&gt;An equilibrium constant is never $\infty$ in theory. In practice, large values of $K_{eq}$ (and slow reverse reaction rates) lead to the equivalent of complete combustion.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2014-10-06T14:52:04.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17134" PostTypeId="1" AcceptedAnswerId="17136" CreationDate="2014-10-06T14:55:50.323" Score="3" ViewCount="17233" Body="&lt;p&gt;I was solving practice problems for electron configuration and periodic table, and I got stuck through a question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An atom of silicon in its ground state has how many electrons with quantum number $l = 1$?&lt;br&gt;&#xA;   &lt;strong&gt;a&lt;/strong&gt;) 14&lt;br&gt;&#xA;   &lt;strong&gt;b&lt;/strong&gt;) 2&lt;br&gt;&#xA;   &lt;strong&gt;c&lt;/strong&gt;) 8&lt;br&gt;&#xA;   &lt;strong&gt;d&lt;/strong&gt;) 6&lt;br&gt;&#xA;   &lt;strong&gt;c&lt;/strong&gt;) 28&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the solution the answer is &lt;strong&gt;c&lt;/strong&gt;: 8 electrons, but that wasn't my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My analysis:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Silicon is a chemical element with symbol $\ce{Si}$ and atomic number 14, this is its electron configuration:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{1s^22s^22p^63s^23p^2}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The quantum number $l = 1$ corresponds to $\mathrm{p}$ levels only. So we have:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{2p^63p^2}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The total number of orbitals is: $6 + 2 = 8$ orbitals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since each orbital has 2 electrons, the total number of electrons is $8 \times 2 = 16$ electrons. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my answer? Maybe the question has a wrong answer?&lt;/p&gt;&#xA;" OwnerUserId="8214" LastEditorUserId="41328" LastEditDate="2017-08-06T09:11:48.657" LastActivityDate="2017-08-06T09:11:48.657" Title="How many electrons with l = 1 does Si in its ground state have?" Tags="&lt;atoms&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17135" PostTypeId="2" ParentId="17124" CreationDate="2014-10-06T14:57:41.887" Score="31" Body="&lt;p&gt;The problem with this question is that the exact answer is &quot;&lt;strong&gt;it depends...&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, it depends on your definition of acidity and how you measure it. Everyone seems to be using Brønstead acids (i.e. $\ce{H+}$ donors). I see two different measures in other answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_affinity&quot; rel=&quot;noreferrer&quot;&gt;Proton affinity&lt;/a&gt;&lt;/strong&gt;: This is a gas-phase measurement of $\ce{A^{−} + H^{+} -&amp;gt; HA}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_acidity_function&quot; rel=&quot;noreferrer&quot;&gt;Hammett acidity ($H_0$)&lt;/a&gt;&lt;/strong&gt;: This is a solution measurement, given by $\mathrm pK_{\ce{BH^+}} - \log\left(\frac{[\ce{BH^+}]}{[\ce{B}]}\right)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Secondly, it depends on medium as mentioned by LDC3.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Leveling effect&lt;/strong&gt;: The &lt;a href=&quot;http://en.wikipedia.org/wiki/Leveling_effect&quot; rel=&quot;noreferrer&quot;&gt;solvent leveling effect&lt;/a&gt; reflects the lowest possible $\mathrm pK_\mathrm a$ in a particular solvent, based on the basicity of the conjugate base. So you need $\ce{HF}$ or fluorosulfuric acid to reach low $\mathrm pK_\mathrm a$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the problem in my mind is that carboranes and fluoroantimonic acid are solution measurements, but $\ce{HeH+}$ is a gas-phase measurement. It does have the highest gas-phase proton affinity. But I'd put my money on things I can use in lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, the Reed group prepared the fluorinated carborane acid this year: &lt;a href=&quot;https://doi.org/10.1002/anie.201308586&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Angew. Chem. Int. Ed.&lt;/em&gt; &lt;strong&gt;2014,&lt;/strong&gt; &lt;em&gt;53&lt;/em&gt; (4), 1131–1134&lt;/a&gt;. So that compound, $\ce{H(CHB11F11)}$ wins the crown for strongest solution-phase Brønstead acid (for the moment, at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also a nice review article &quot;Myths about the Proton. The Nature of H&lt;sup&gt;+&lt;/sup&gt; in Condensed Media&quot; by the same group in &lt;a href=&quot;https://doi.org/10.1021/ar400064q&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Acc. Chem. Res.&lt;/em&gt; &lt;strong&gt;2013,&lt;/strong&gt; &lt;em&gt;46&lt;/em&gt; (11), 2567–2575&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-03-28T22:21:25.960" LastActivityDate="2018-03-28T22:21:25.960" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17136" PostTypeId="2" ParentId="17134" CreationDate="2014-10-06T15:02:40.137" Score="4" Body="&lt;p&gt;You're on the right track, but those superscripts in the electronic configuration don't refer to orbitals. They refer to the number of electrons in that type of orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, you might want to look at this question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8598/maximum-number-of-electrons-each-shell?rq=1&quot;&gt;What are the maximum number of electrons in each shell?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-10-06T15:02:40.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17137" PostTypeId="2" ParentId="16951" CreationDate="2014-10-06T15:06:46.720" Score="1" Body="&lt;p&gt;Sharpness is colloquial for how narrow a peak is. Resolution is how well we can see two features as they move closer together. You might do well to re-read your book to learn these definitions better. As already noted, &lt;em&gt;skinnier&lt;/em&gt; and &lt;em&gt;more clear&lt;/em&gt; have no meaning in spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search on terms such as diffraction grating equation will give you further reading to support or invalidate the guess (hypothesis) that you propose.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2014-10-06T15:06:46.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17138" PostTypeId="2" ParentId="16949" CreationDate="2014-10-06T15:16:30.950" Score="0" Body="&lt;p&gt;Start with a reading about the types of errors that arise in spectroscopy. Clearly photometric error $\delta I_{pe}$ is a device uncertainty. The suggestion is, its magnitude depends on the number of &quot;beams&quot;. What changes as the number of beams increases? Path length? Number of gratings or optical components? What else? How do each of those factors contribute to an analytical expression for photometric uncertainty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end, a reasonable result might suggest that, as you double the number of beams, you double the relative uncertainty in signal $\delta I_{pe} / I$, where $I$ is the signal intensity. To obtain the same absolute confidence, you therefore have to double the measurement intensity $I$.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2014-10-06T15:16:30.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17139" PostTypeId="2" ParentId="16933" CreationDate="2014-10-06T15:25:08.327" Score="2" Body="&lt;p&gt;The surface cracks due to a mis-match in the temperatures inside and outside the ceramic. Things contract as they cool. So, the cooler outer surface is contracting against the hotter inner surface, which is still not contracting. The outer surface wants to be closer together than the inner material, but the inner material forces the outer material to stay apart. When the ceramic on the outside transitions to a solid (at what is called its &quot;glass temperature&quot;), it can no longer flow like a liquid. The only way it can continue to stay apart due to the forces from the inside is to &quot;fall apart&quot; ... i.e. to crack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mother may be able to coat the pottery with a different glaze that forms a glass at a lower temperature than the one she has now. Also, the suggestions to cool VERY SLOWLY are directly on target. The cooling should happen in a place that has no air flow, ideally in a closed box. Absent an annealing oven, a different option may be to put the pots in an conventional NON-CONVECTION oven that is held at a temperature just below the glass temperature of the glaze being used. When the entire pot reaches that temperature, it can be removed to cool further in a box at room temperature, again with no air flow.&lt;/p&gt;&#xA;" OwnerUserId="8067" LastActivityDate="2014-10-06T15:25:08.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17140" PostTypeId="1" AcceptedAnswerId="17238" CreationDate="2014-10-06T15:36:14.850" Score="3" ViewCount="528" Body="&lt;p&gt;Which series of quantum numbers describes the highest (energy) occupied orbital in a ground state of &lt;strong&gt;At&lt;/strong&gt; atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) n = 6, l = 0 &lt;br /&gt;&#xA;b) n = 6, l = 2 &lt;br /&gt;&#xA;c) n = 5, l = 2 &lt;br /&gt;&#xA;d) n = 4, l = 3 &lt;br /&gt;&#xA;&lt;strong&gt;e) n = 6, l = 1 &lt;br /&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my teacher, the answer is e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My analysis: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The electron configuration of At atom: &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}1s^22s^22p^63s^23p^6\ce{3d^10}4s^24p^64\ce{d^10}5s^25p^64\ce{f^14}\ce{5d^10}6s^26p^5&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing for levels 0,1,2,3 correspond to s,p,d,f respectively we can say that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $$\begin{aligned}6s^2\end{aligned}$$&#xA;b) $$\begin{aligned}Nothing \end{aligned}$$&#xA;c) $$\begin{aligned}\ce{5d^10}\end{aligned}$$&#xA;d) $$\begin{aligned}\ce{4f^14}\end{aligned}$$&#xA;e) $$\begin{aligned}\ce{6p^5}\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought here first is that d is the answer since 4f&lt;sup&gt;14&lt;/sup&gt; has the largest number of electrons, and thus will occupy the highest energy around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I saw on another question that the orbitals are filled so that the ones of lowest energy are filled first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1s &amp;lt; 2s &amp;lt; 2p &amp;lt; 3s &amp;lt; 3p &amp;lt; 4s &amp;lt; 3d &amp;lt; 4p &amp;lt; 5s &amp;lt; 4d &amp;lt; 5p &amp;lt; 6s &amp;lt; 4f &amp;lt; 5d &amp;lt; 6p &amp;lt; 7s ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, e is the answer since 6p&lt;sup&gt;5&lt;/sup&gt; has the highest energy regardless of the number of electrons it is holding?&lt;/p&gt;&#xA;" OwnerUserId="8214" LastEditorUserId="1362" LastEditDate="2014-11-04T08:15:25.780" LastActivityDate="2014-11-04T08:15:25.780" Title="Is my analysis of quantum numbers describing the ground state of At correct?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17141" PostTypeId="1" AcceptedAnswerId="17143" CreationDate="2014-10-06T15:48:24.670" Score="0" ViewCount="256" Body="&lt;p&gt;Normally, the electron configuration of &lt;strong&gt;Te&lt;/strong&gt; is known as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^4 \end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, one day I was asked in a exam if this can be written also as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^3 6s^1\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I answered that it couldn't. But my answer turned out to be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, why is it allowed to be written like this? And what general case can I learn when being asked such tricky questions?&lt;/p&gt;&#xA;" OwnerUserId="8214" LastActivityDate="2014-10-06T16:13:22.153" Title="Can the electron configuration of Te be written that way?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17143" PostTypeId="2" ParentId="17141" CreationDate="2014-10-06T16:13:22.153" Score="2" Body="&lt;p&gt;The second configuration allows Te to take advantage of the relative stability of half-full shells. Having the 5p and 6s orbitals both half-populated is more stable than having 4 electrons in the 5p orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar behavior is what allows carbon to form 4 bonds. The $\ce{s^1p^3}$ configuration is more stable than the $\ce{s^2p^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tellurium shows the behavior while oxygen does not as a result of the overlap between the 5p and 6s orbitals in the larger atoms. The same degree of overlap does not occur with the 2p and 3s orbitals.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastActivityDate="2014-10-06T16:13:22.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17144" PostTypeId="2" ParentId="17119" CreationDate="2014-10-06T16:40:47.763" Score="0" Body="&lt;p&gt;Start with your chemical equations: What's going to react and how; what will precipitate out of what sort of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When get one component fully precipitated (as Jason Patterson points out) you can use the stoichiometry of the exchange reaction to determine the initial mass of one of the components.&lt;/p&gt;&#xA;" OwnerUserId="8030" LastActivityDate="2014-10-06T16:40:47.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17145" PostTypeId="2" ParentId="17119" CreationDate="2014-10-06T17:38:55.517" Score="1" Body="&lt;p&gt;It occurred to me that you might have been asking how to complete this separation experimentally rather than conceptually, which is what the comments and previous answer all were pointing toward.  If that is the case, then once you've precipitated out the barium sulfate, you have to isolate it from all other substances in order to find its true mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precipitate is a solid, so you could separate it from most of its contaminants by pouring the solution through a pre-weighed filter paper, making sure to wash the walls and bottom of the reaction vessel with distilled water and pouring this wash through the filter as well in order to collect all of the precipitate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The filtrate would contain the bulk of the sodium acetate and sodium sulfate in the reaction mixture.  What remains in the filter paper would be barium sulfate and a small amount of reaction mixture, including the sodium salts.  To eliminate the sodium salts, add additional distilled water to the filter to wash the precipitate (and filter paper) clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a filter paper with clean barium sulfate, dry it thoroughly and weigh.  I don't believe that barium sulfate forms a hydrate, so the mass of the precipitate should all be barium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Barium salts are quite toxic, particularly highly soluble ones like barium nitrate.  Be careful using them and dispose of the product appropriately.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-06T17:38:55.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17148" PostTypeId="1" CreationDate="2014-10-06T19:04:14.647" Score="3" ViewCount="388" Body="&lt;p&gt;Why is LDL more easily oxidized than HDL is? By LDL and HDL, I mean the lipoproteins that carry cholesterol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation of LDL ultimately contributes to heart disease.&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="402" LastEditDate="2017-05-16T02:04:36.977" LastActivityDate="2017-05-16T02:04:53.383" Title="Why is LDL more easily oxidized/damaged than HDL?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17149" PostTypeId="1" CreationDate="2014-10-06T19:09:19.170" Score="5" ViewCount="526" Body="&lt;p&gt;Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine? &lt;a href=&quot;http://www.nap.edu/openbook.php?record_id=11919&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; the source for the claim (the half-life is 340 years at 298 K, which is much smaller than that of adenosine and guanosine). &lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="402" LastEditDate="2017-05-16T02:04:07.870" LastActivityDate="2017-05-22T18:32:56.040" Title="Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17150" PostTypeId="1" CreationDate="2014-10-06T19:13:50.517" Score="0" ViewCount="168" Body="&lt;p&gt;How do we obtain the expressions for integrated rate laws ?!? What are the limits we integrate it under and with respect to what do we integrate it?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2014-10-06T19:56:30.840" Title="Rate law and rate constant" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-10-07T01:18:48.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="17151" PostTypeId="2" ParentId="17150" CreationDate="2014-10-06T19:56:30.840" Score="4" Body="&lt;p&gt;Remember first that the integrated rate laws that we are familiar with for zeroth, first, and second order reactions are based on systems involving a single reactant in their rate determining step (or those that have been manipulated to give very large excesses of one reactant such that its concentration remains essentially fixed during the reaction process.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, there are only two variables involved, the concentration of our reactant of interest, $[A]$, and time, $t$.  We can't directly integrate the differential rate law; it's a differential equation and must be solved as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, consider the first order differential rate law:&#xA;$rate = k[A]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meaning of  $rate$ here is the rate of change of the concentration of reactant A with respect to time.  Since we want the rate to be positive, we use the opposite of that quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-\frac{d[A]}{dt} = k[A]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a differential equation, specifically one that is separable (and thus fairly easy to solve.)  Multiply both sides by $-dt$ and we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d[A]}{[A]} = -k\ dt$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int{\frac{1}{[A]}d[A]} = \int{-k\ dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating both sides and applying initial condition to find the constant gives the integrated first order rate law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ln[A]=-kt+ln[A_0]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other integrated rate laws are found in similar ways.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-06T19:56:30.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17152" PostTypeId="1" CreationDate="2014-10-06T19:56:56.017" Score="3" ViewCount="759" Body="&lt;p&gt;What is the physical meaning when we have an empirical formula with fractional subscripts such as $\ce{LiMn_{1-x}Fe_{x}PO4}$ $(0 \le x \le 1)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This an example of the paper the notation above is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bai, J.; Hong, J.; Chen, H.; Graetz, J.; Wang, F. &lt;em&gt;J. Phys. Chem. C&lt;/em&gt; &lt;strong&gt;2015&lt;/strong&gt;, 119 (5), 2266–2276. DOI &lt;a href=&quot;http://dx.doi.org/10.1021/jp508600u&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1021/jp508600u&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8219" LastEditorUserId="41328" LastEditDate="2017-08-09T11:00:40.843" LastActivityDate="2017-08-09T11:00:40.843" Title="Fractional Subscripts?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;stoichiometry&gt;&lt;solid-state-chemistry&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17153" PostTypeId="2" ParentId="17152" CreationDate="2014-10-06T20:07:01.227" Score="4" Body="&lt;p&gt;These indicate &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Doping_(semiconductor)&quot; rel=&quot;nofollow noreferrer&quot;&gt;doping&lt;/a&gt;&quot; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-stoichiometric_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-stoichiometric compounds&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, there's some fraction of Fe, and some fraction of Mn in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_iron_phosphate_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium iron phosphate battery&lt;/a&gt;. If I remember correctly, you need to dope a different ion to increase the conductivity of $\ce{LiFePO4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also here: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6913/how-are-non-stoichiometric-compounds-determined&quot;&gt;How are non-Stoichiometric compounds determined?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-10-06T20:07:01.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17154" PostTypeId="2" ParentId="17030" CreationDate="2014-10-06T20:26:45.073" Score="2" Body="&lt;p&gt;In the latest version of the &lt;a href=&quot;http://mackerell.umaryland.edu/charmm_ff.shtml&quot; rel=&quot;nofollow&quot;&gt;CHARMM forcefield&lt;/a&gt; in &lt;code&gt;par_all36_prot.prm&lt;/code&gt; the comments are a bit more informative:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CMAP&#xA;! alanine map&#xA;C    NH1  CT1  C    NH1  CT1  C    NH1   24&#xA;! phi = -180.0&#xA;     0.126790      0.768700      0.971260      1.250970      2.121010&#xA;     2.720430      2.089440      1.789790 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So from that it can be inferred that my example is read as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PHI PSI E&#xA;-180 -180 1&#xA;-180 -165 2&#xA;-180 -150 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="309" LastActivityDate="2014-10-06T20:26:45.073" CommentCount="0" CommunityOwnedDate="2014-10-06T20:26:45.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="17155" PostTypeId="1" CreationDate="2014-10-06T21:40:14.293" Score="9" ViewCount="501" Body="&lt;p&gt;The reaction is this: &lt;img src=&quot;https://i.gyazo.com/6d226e59cd826dd033fb62fe712dd265.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only idea I have thus far is if we have a chloride acid that can form an ester, then we might be able to find a suitable mechanism. This would require a chloride reagent. &lt;/p&gt;&#xA;" OwnerUserId="8220" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.377" LastActivityDate="2016-08-28T00:55:49.430" Title="Mechanism of hydrolysis of polycyclic oxetan" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17156" PostTypeId="1" CreationDate="2014-10-06T22:48:50.963" Score="0" ViewCount="7488" Body="&lt;p&gt;I'm thinking it will be linear with a negative charge on the middle bromine. Someone told me it is trigonal bipyramidal geometry with 3 lone pairs around the central bromine and the two other bromines above and below making it linear. Is this correct? Also is it just covalent bonding in the molecule?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="41328" LastEditDate="2019-02-06T20:50:01.197" LastActivityDate="2019-02-06T20:50:01.197" Title="Describe the bonding in a tribromide anion, what is its structure?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-structure&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17157" PostTypeId="2" ParentId="17129" CreationDate="2014-10-06T22:51:11.987" Score="6" Body="&lt;p&gt;For me, tschoppi really nails the right response to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you handle any chemical you are unfamiliar with, you should first read AND UNDERSTAND all facets of the MSDS (SDS) for that chemical. You need to understand the hazard potential, risk mitigation steps, emergency response strategies and disposal management of all chemicals you are handling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the capacity to legally obtain chemicals such as copper sulfate pentahydrate, you should also be able to obtain the MSDS (SDS). Google searching will also provide you with this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with regards to $\ce{CuSO4.5H2O}$, it has a GHS signal word DANGER, with the following key safety statements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard Statements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Toxic if swallowed.&lt;/li&gt;&#xA;&lt;li&gt;Causes skin irritation.&lt;/li&gt;&#xA;&lt;li&gt;Causes serious eye irritation.&lt;/li&gt;&#xA;&lt;li&gt;Very toxic to aquatic life.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Precautionary Statements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wash skin thoroughly after handling.&lt;/li&gt;&#xA;&lt;li&gt;Do not eat, drink or smoke when using this product.&lt;/li&gt;&#xA;&lt;li&gt;Avoid release to the environment.&lt;/li&gt;&#xA;&lt;li&gt;Wear protective gloves/ eye protection/ face protection.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response Statements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;IF SWALLOWED: Immediately call a POISON CENTER or doctor/physician.&lt;/li&gt;&#xA;&lt;li&gt;IF ON SKIN: Wash with plenty of soap and water.&lt;/li&gt;&#xA;&lt;li&gt;IF IN EYES: Rinse cautiously with water for several minutes. Remove contact lenses, if present and easy to do. Continue rinsing.&lt;/li&gt;&#xA;&lt;li&gt;If skin irritation occurs: Get medical advice/ attention.&lt;/li&gt;&#xA;&lt;li&gt;If eye irritation persists: Get medical advice/ attention.&lt;/li&gt;&#xA;&lt;li&gt;Take off contaminated clothing and wash before reuse.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additionally, it advises to wear correct respiratory protection (P2 or N99) and avoid breathing vapours mist or gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this information, it is unsurprising that you experienced an irritation-type reaction to breathing the vapours. So, how to avoid this next time? Read and understand the MSDS prior to use. Employ the correct protective equipment. Carry out the procedure in a suitably ventilated area, with active ventilation. This could be outside with a strong fan pushing vapours away.&#xA;If you are unsure about how to handle this chemical correctly, forgo the pretty colours on your rocket. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="5017" LastEditDate="2014-10-07T01:16:15.900" LastActivityDate="2014-10-07T01:16:15.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17159" PostTypeId="1" CreationDate="2014-10-07T00:27:19.970" Score="4" ViewCount="1459" Body="&lt;p&gt;I receive $120~\mathrm{mL}$ bags of medicine in sterile water. The medicine is refrigerated and then packed with a significant amount of ice packs in an insulated bag. Generally three times the volume of all the medicine. When I remove the ice packs, they are still solid two hours after being packed. So, they are very cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there enough energy to freeze the water directly in contact with the ice packs when packed and then melt immediately? The air temperature is around $21~\mathrm{^\circ C}$ at the time.&lt;/p&gt;&#xA;" OwnerUserId="8221" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T09:04:08.907" LastActivityDate="2015-05-11T09:04:08.907" Title="Can Ice Packs Freeze Water at 21 °C?" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17161" PostTypeId="2" ParentId="17159" CreationDate="2014-10-07T01:17:24.977" Score="8" Body="&lt;p&gt;It really depends on the initial temperature of the ice packs and their composition. If they start at a temperature far below the freezing point of water, they could conceivably freeze the water as their container comes to thermal equilibrium. That said, they’re going to have to be quite cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cool by $1°\mathrm{C}$, liquid water requires the removal of roughly double the amount of energy that ice would give up in moving up $1°\mathrm{C}$ (per gram.) If your bags of medicine started at $5°\mathrm{C}$, (stored in a refrigerator) then the ice will warm by about $3°\mathrm{C}$ as it brings the medicine down to its temperature because it has three times the mass and gives up half as much heat per degree gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the water has reached $0°\mathrm{C}$, it will begin freezing. For every gram of frozen water, about $160~\mathrm{g}$ of ice will warm by $1°\mathrm{C}$. The entire mass of ice will be enough to freeze ~$2.3~\mathrm{g}$ of water for each degree it is below freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the ice packs came from a normal freezer, their starting temperature is about $−20°\mathrm{C}$. They would warm to about $−17°\mathrm{C}$ in bringing the water to its freezing point. From there the ice could then freeze about $40~\mathrm{g}$ of the medicine before it all came to equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There are significant assumptions (notably, that the heat capacity of the ice pack is equal to that of ice and that its melting point is equal to $0°\mathrm{C}$) and some healthy rounding here. If the ice packs melted at a temperature significantly lower than $0°\mathrm{C}$, their heat of fusion could be sufficient to freeze the water, given time.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="15410" LastEditDate="2015-05-11T07:13:25.600" LastActivityDate="2015-05-11T07:13:25.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17162" PostTypeId="1" CreationDate="2014-10-07T01:29:25.977" Score="-2" ViewCount="629" Body="&lt;p&gt;What is a dipole? Also, how do you know if a molecule is symmetrical or asymmetrical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, are the following symmetrical or asymmetrical? Can you tell just from the molecular formula? Which ones have dipoles?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{O2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CO2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are these concepts related?&lt;/p&gt;&#xA;" OwnerUserId="8223" LastEditorUserId="194" LastEditDate="2018-01-29T04:46:18.640" LastActivityDate="2018-01-29T04:46:18.640" Title="Dipoles and determining molecular (a)symmetry" Tags="&lt;molecular-structure&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-10-07T04:33:27.710" ContentLicense="CC BY-SA 3.0" />
  <row Id="17163" PostTypeId="2" ParentId="17162" CreationDate="2014-10-07T02:08:05.270" Score="1" Body="&lt;p&gt;You should easily find the answers by searching the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dipole is the separation of a positive charge and a negative charge in a molecule. Any 2 different atoms bonded together will have a dipole since their electronegativities are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A symmetrical molecule will either have a mirror plane or a rotation that will superimpose the atoms. Benzene has both a mirror plane and a rotation of 60°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An asymmetrical molecule does not have a mirror plane or rotation. Ethanol is an example.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-07T02:08:05.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17165" PostTypeId="1" AcceptedAnswerId="17166" CreationDate="2014-10-07T04:18:59.093" Score="4" ViewCount="2208" Body="&lt;p&gt;Brønsted Lowry definition of an acid says that an acid gives away protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is what makes dissolved protons grant the property of acidity? Like, why would dissolved protons eat up materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question applies to bases too, I suppose. And I guess you could include Lewis-acids and bases in your argumentation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a question that has been bothering me for long. By the way I am not chemist(nor do I study chemistry).&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="5026" LastEditDate="2018-03-03T11:22:33.437" LastActivityDate="2019-12-10T10:51:28.553" Title="What makes protons give the property of acidity?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17166" PostTypeId="2" ParentId="17165" CreationDate="2014-10-07T04:48:09.557" Score="1" Body="&lt;p&gt;Whether something is a Brønsted acid is defined by its pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pH = -log_{10}[H^+]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the better a substance is at releasing protons, the higher the concentration of protons [H+], and correspondingly the lower its pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas for Lewis acid, it is the ability for it to strip electrons from anything it attacks (electrophilicity). There is no direct measurement for the strength of Lewis acidity, as many factors have to be taken into consideration, such as enthalpy change, steric factors. In fact, many substances can act as Lewis acids can also act as Lewis bases, such as water, ammonia, and even some alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, what makes acids and bases so corrosive is its ability act as an oxidant or reductant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: Lose electrons/Gain hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: Gain electrons/Lose hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="5026" LastEditDate="2018-03-03T11:22:38.440" LastActivityDate="2018-03-03T11:22:38.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17167" PostTypeId="2" ParentId="12341" CreationDate="2014-10-07T05:10:51.357" Score="1" Body="&lt;p&gt;From this &lt;a href=&quot;http://www.chemteam.info/Thermochem/Specific-Heat.html&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every substance has its own specific heat and each phase has its own distinct value. In fact, the specific heat value of a substance changes from degree to degree, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would mean that a solute in water would change the specific heat from plain water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that if the density is greater than water, then the specific heat would be greater (because you are heating the water and the solute). To be certain, you would need to determine it for yourself.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-07T05:10:51.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17168" PostTypeId="2" ParentId="14000" CreationDate="2014-10-07T06:10:08.403" Score="5" Body="&lt;p&gt;It has been done, not with bond &lt;em&gt;formation&lt;/em&gt; but rather with bond &lt;em&gt;breaking&lt;/em&gt;, and is called the Szilard–Chalmers reaction (credit to Loong for pointing out the name).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a biphasic mixture of ethyl iodide (with stable $\ce{^{127}I}$) and water is irradiated with neutrons, all the radioactivity from $\ce{^{128}I}$ is found as iodide in the aqueous phase; there is none in the organic phase.  This is because $\ce{^{128}I}$ is initially formed in an excited state that emits a γ-photon, and the recoil from that is more than enough to break the C–I bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of a γ-photon is &gt; 10 keV; in comparison, the strength of a typical chemical bond is ~ 1 to 10 eV. The C–I bond strength is roughly 2.5 eV.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="16683" LastEditDate="2017-09-07T17:21:34.050" LastActivityDate="2017-09-07T17:21:34.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17169" PostTypeId="2" ParentId="17129" CreationDate="2014-10-07T09:22:45.300" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to use copper(II) sulfate as rocket propellant additive in hope that it will alter the flame colour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It likely won't. AFAIK, common colored flame mixture using copper is copper benzoate/potassium perchlorate mix (pale blue trace), but using it involves known risks of explosive combustion in case of improper handling. In case you are interested in colored light compositions, you need specifically designed compositions, and most of them are not health-friendly. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I suppose copper sulfate has got in there. Do you think I risked badly, or is it just weak irritation? How can I avoid it next time, apart of using fumehood?**&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no much risks here, though next time stick to ceramics. Copper salts are prone to hydrolysis, so I believe you smelled some sulfuric acid vapors. Not pleasant, but not deadly. In comparison, copper chloride would hydrolyze completely, producing much hydrochloric acid vapors, and that would be much worse.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-10-07T09:50:40.280" LastActivityDate="2014-10-07T09:50:40.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17170" PostTypeId="2" ParentId="17129" CreationDate="2014-10-07T09:28:58.710" Score="17" Body="&lt;p&gt;Roasting of copper(II) sulfate at elevated temperatures generates sulfur trioxide, which would hurt your nose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{$\underset{\text{(blue)}}{\ce{CuSO4.5H2O}}$ &amp;amp;-&amp;gt;[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][100\ \mathrm{^\circ C}] CuSO4.H2O + 4H2O}\\[6pt]&#xA;\ce{CuSO4.H2O &amp;amp;-&amp;gt;[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][\gt\ 200\ \mathrm{^\circ C}] $\underset{\text{(white)}}{\ce{CuSO4}}$ + H2O}\\[6pt]&#xA;\ce{CuSO4 &amp;amp;-&amp;gt;[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][] CuO + SO3}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-12-14T14:34:10.627" LastActivityDate="2015-12-14T14:34:10.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17173" PostTypeId="1" AcceptedAnswerId="17191" CreationDate="2014-10-07T12:08:03.220" Score="2" ViewCount="4798" Body="&lt;p&gt;In order to draw chemical structures I use ChemBioDraw Ultra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an exercise I've to draw the most stable conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTxAx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect a twist-boat like structure (instead of usual chair conformation for substituted dioxanes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a freeware or an easy computer online tool to check if my propose (right side picture) is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I go to &gt;Edit &gt; Get 3D Model the program gives me a small picture with a 3D-Model:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PgflT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By double-clicking this small image it opens ChemBio3DUltra and I can rotate this molecule:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cHlW6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Martin: Is this correct (I've expected a twisted cyclic ring)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Martin I draw the structure at right with i-Bu and H exchanged in Chemdraw and do &gt;Edit &gt; Get 3D Model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pNujg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I first thought it should be twist boat like was that one C-O is pointing up and the other pointing down. &lt;strong&gt;May be someone knows why people gives this stereoinformation inside the cyclic ring?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uITmA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As I've drawn it as a twist boat I get exactly the same with ChemBioDraw. Is there a way to calculate by how many kcal/mol the first one is more stable then the second one with ChemBioDraw?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's interesting: I searched for a structure in which the twist-boat conformation is the most favorable one and found one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, when there are two large substituents in the cis-1,4 arrangement on &#xA;  a cyclohexane ring, neither of which will go easily into an axial position, then &#xA;  it appears that the twist-boat conformation (Section 12-3A) is most favorable &#xA;  [&lt;a href=&quot;http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf&lt;/a&gt;, page 458]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The A-value of a tert-Butyl is in the range of 5 kcal/mol which is around the value of the energy difference between cyclohexane in twist-boat and chair conformation.&#xA;&lt;strong&gt;I've drawn cis-1,4-di-tert-butyl-cyclohexane with ChemDraw and used &gt;Edit&gt;Get 3D Model and ChemBioDraw gave me the structure which is NOT the energetical most suitable one! Why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TpVKC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7201" LastEditDate="2014-10-07T19:05:24.750" LastActivityDate="2014-10-07T19:34:10.410" Title="Freeware that converts structure in a conformational model?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17174" PostTypeId="1" AcceptedAnswerId="19424" CreationDate="2014-10-07T13:58:49.407" Score="10" ViewCount="401" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1nX2j.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: &#xA;$r=k\ce{[HCOO^{−}]}^{1/2} \ce{[S2O8^{2−}]}$. &#xA;The first reaction is very slow, so most of the peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WHAT does this mean?&#xA;So, why is the rate constant &lt;strong&gt;NOT&lt;/strong&gt; according to the slowest step? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it is not just $k\ce{[S2O8^{2-}]}$?&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="4945" LastEditDate="2014-11-18T05:55:14.747" LastActivityDate="2014-11-18T06:03:17.070" Title="Oxidation of formate ion by peroxydisulfate in water solution: Why are there &quot;strange&quot; kinetics?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17175" PostTypeId="2" ParentId="17156" CreationDate="2014-10-07T14:11:49.350" Score="1" Body="&lt;p&gt;In fact, tribromide ion has the generic formula $\ce{AX2E3}$ according to Lewis diagram. Where  $\ce{A}$ is the central atom, $\ce{X}$ an atom bonded to $\ce{A}$ and $\ce{E}$ a lone pair on $\ce{A}$.  Using Valence Shell Electron Pair Repulsion (VSEPR) theory,  the geometry  of this ion will be&#xA; :&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5aRdd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the molecular shape is linear, while the electron pair geometry is trigonal  bipyramid.&#xA; And the bonds in the ion are covalent.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-07T14:11:49.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17176" PostTypeId="1" CreationDate="2014-10-07T14:16:54.113" Score="4" ViewCount="711" Body="&lt;p&gt;I want to calculate the dipole moment for the first excited state with a Q.C. program (TURBOMOLE: TD-DFT). The calculation lists the following output:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ground state total dipole moment:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-1.129096&#xA;-2.295685&#xA; 0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think that's $\langle\Psi_0|\vec{\mu}|\Psi_0\rangle$ with $\Psi_0$ the wavefunction for the ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st excited state electric transition dipole moment (length representation):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-2.413321&#xA; 0.367463&#xA; 0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Should be $\langle\Psi_1|\vec{\mu}|\Psi_0\rangle$ with $\Psi_1$ the wavefunction for the 1st excited state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd excited state electric transition dipole moment (length representation):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0.841417&#xA;-0.329533&#xA; 0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Should be  $\langle\Psi_2|\vec{\mu}|\Psi_0\rangle$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a summary:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             nuc           gs,e       -&amp;gt;  gs,tot          excit     -&amp;gt;  total&#xA; ------------------------------------------------------------------------------&#xA;   x     -29.158609      28.029513      -1.129096       1.370060       0.240963&#xA;   y      48.811850     -51.107535      -2.295685       0.086529      -2.209156&#xA;   z       0.000000       0.000000       0.000000       0.000000       0.000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know how to get the values denoted by &lt;code&gt;excit&lt;/code&gt; in the summary and they are not present anywhere else in the output. There's also no hint which excited state they apply to, so how do you calculate the total dipole moment for the 1st excited state e.g.?&lt;/p&gt;&#xA;" OwnerUserId="1023" LastEditorUserId="186" LastEditDate="2015-05-12T09:14:25.700" LastActivityDate="2015-05-12T09:14:25.700" Title="How to Locate the Excited State Dipole Moment in Turbomole Output (TD-DFT)" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;&lt;td-dft&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17177" PostTypeId="2" ParentId="7800" CreationDate="2014-10-07T14:56:21.640" Score="-1" Body="&lt;p&gt;Fluorine can't be the central atom in inter-halogen compounds because it is an element from the period 2 in the periodic table. It can not have more than 8 valence electrons. And as it has 7 valence electrons, it can only form one bond. This is not the case of the other halogens $\ce{Cl, Br, I,...}$ where these elements are below period 2 in the periodic table. That means they can have an expanded octet. They can have more than 8 valence electrons. So they can be central atoms in inter-halogen compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an orbital viewpoint, we can attribute this to the fact that in the case of $\ce{Cl, Br,  I,...}$  the orbitals $\ce{3d}$ , $\ce{4d}$ and $\ce{5d}$  respectively are available for hybridization and can interact with the valence orbitals $\ce{ns, np}$ of these atoms. So these halogen atoms have the possibility to form more than one bond with other atoms. In the case of fluorine, the valence orbitals $\ce{2s, 2p}$ can not interact with $\ce{3d}$ due to the big difference of energy.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2014-10-07T15:10:39.320" LastActivityDate="2014-10-07T15:10:39.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17178" PostTypeId="1" CreationDate="2014-10-07T15:19:36.500" Score="1" ViewCount="3097" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A) Calculate the mass of a mercury atom in kg.&lt;br&gt;&#xA;  B) How many atoms of mercury does a drop of mercury contain, knowing that the volume of a drop of mercury is $\pu{0.05mL}$ and that the density of mercury is $13.6 \times 10^3 \mathrm{kg/m^3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that $\mathrm{1 mL = 1 cm^3}=1.0\times10^{-6} \mathrm{m^3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the mass of a drop of mercury is $\mathrm{m=v}\times \mathrm{density}&#xA;= 5.0\times10^{-18} \times 13.6\times{10^3}&#xA;=6.8\times10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is how to proceed to find how many atoms mercury are present in its one drop? Do I simply divide the mass of a drop of mercury by the mass of a mercury atom, or do I use the Avogadro's number?  I don't want a straight-off answer, I only wish to know how to calculate it. &lt;/p&gt;&#xA;" OwnerUserId="8227" LastEditorUserId="5026" LastEditDate="2018-02-21T11:43:41.143" LastActivityDate="2018-02-21T11:43:41.143" Title="Number of mercury atoms in a drop of mercury" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17179" PostTypeId="2" ParentId="17178" CreationDate="2014-10-07T15:30:27.137" Score="3" Body="&lt;p&gt;You can use dimensional analysis. It is a very useful tool to solve equations such as these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Mass of mercury:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{}&#xA;\frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this is dimensionally equivalent. Therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Atoms of mercury per drop:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{atoms}{drop}=&#xA;\frac{N_A(atoms)}{(mol^{})}\cdot{}&#xA;\frac{(mol)}{Mr(g)}\cdot{}&#xA;\frac{1000(g)}{1(kg)}\cdot{}&#xA;\frac{\rho{(kg)}}{(m^3)}\cdot{}&#xA;\frac{(m^3)}{10^6(ml)}\cdot{}&#xA;\frac{V(ml)}{drop}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this is dimensionally equivalent. Therefore, &#xA;$$\frac{atoms}{drop}=&#xA;\frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-08T02:24:00.737" LastActivityDate="2014-10-08T02:24:00.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17180" PostTypeId="2" ParentId="16818" CreationDate="2014-10-07T15:31:51.483" Score="11" Body="&lt;p&gt;The first reaction is the conversion of the alcohol to an alkyl chloride using thionyl chloride. The &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/12/03/reagent-friday-thionyl-chloride-socl2/&quot; rel=&quot;noreferrer&quot;&gt;mechanism of the reaction&lt;/a&gt; starts with nucleophilic attack of the alcohol oxygen on sulphur, followed by displacement of $\ce{Cl-}$. Chloride then substitutes the alkyl chlorosulfite group in an $\ce{S_{N}2}$ reaction, and the latter decomposes into $\ce{SO2}$ and $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uPbWe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(&lt;a href=&quot;http://www.masterorganicchemistry.com/2011/12/03/reagent-friday-thionyl-chloride-socl2/&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second reaction, chloroethane is alkylated with sodium hydrogen acetylide to yield the desired product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H5-Cl + HC#C^{-} -&amp;gt;~ C2H5-C#CH + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple $\ce{S_{N}2}$ reaction, in which $\ce{Cl-}$ is substituted by $\ce{HC#C^{-}}$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetylide&quot; rel=&quot;noreferrer&quot;&gt;Acetylide salts&lt;/a&gt; of alkali metals can be prepared by deprotonation of ethyne with a strong base, for example, with butyllithium in THF or &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_amide&quot; rel=&quot;noreferrer&quot;&gt;sodium amide&lt;/a&gt;, which can be generated in situ by dissolving sodium metal in liquid ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fgaC4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-10-07T16:57:37.067" LastActivityDate="2014-10-07T16:57:37.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17181" PostTypeId="2" ParentId="17173" CreationDate="2014-10-07T15:33:41.270" Score="3" Body="&lt;p&gt;It's tough to see in your Chem3D picture (try turning off display of H's), but the dioxane is in a chair conformation. Truthfully, I don't think you need a 3D package to determine this, although using the energy minimization can confirm your hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach is to assume that the dioxane is going to be in a chair conformation. I'm much better at converting six membered rings to the chair if I start with the ring completely in the plane. In this case, that means redrawing the original so that the oxygens are in the plane and the isobutyl and methyl groups are viewed as substituents off the dioxane (left side of figure). Now it becomes clear that the substituents should be in a trans-1,3 relationship. (Note that you have an error in your ChemDraw pics that these groups are cis. The R/S designation even changes at the isobutyl group.) Next, I draw a chair and place the oxygens. I like to number the atoms to avoid mistakes, although many people are better at this than me and can skip this step. In my first chair drawing, I see that the isobutyl group is placed axially, and interacts with the fused cyclohexane. Going through a ring flip puts the isobutyl equatorial and the methyl axial. There's no way to avoid 1,3-diaxial interactions completely, so I would predict that the right chair structure is the most favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8m6dX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the 3D packages can really help is to see how much the chairs are distorted toward a twist conformation, but for a first guess, I think the hand drawn structures are useful.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-07T15:33:41.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17182" PostTypeId="1" CreationDate="2014-10-07T15:57:16.603" Score="6" ViewCount="1235" Body="&lt;p&gt;Let's say we have a stable solution of &#xA;$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$&#xA;and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$&#xA;, with their respective counter-ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The associated redox reactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}&amp;lt;\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$&#xA;, since no reaction between $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$&#xA;and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$&#xA;is occurring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Provided that neither $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$&#xA;nor $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$&#xA;are present in the solution, how can we determine, form the Nernst equation, the $\mathcal{E}$ needed to start the electrolysis process &#xA;$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;In other words, how can the Nernst equation deal with activities that are equal to zero?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="7692" LastEditDate="2014-10-07T16:13:04.463" LastActivityDate="2014-10-09T12:09:33.413" Title="Nernst equation and electrolysis" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17183" PostTypeId="2" ParentId="17103" CreationDate="2014-10-07T16:03:56.003" Score="7" Body="&lt;p&gt;Given that the reaction is starting from an α,β-unsaturated ketone, the dissolving metal reduction is a very useful way of generating a particular enolate regioselectively. The mechanism dictates that only one enolate can be produced. Dissolving lithium metal in ammonia releases an electron that can add to the unsaturated system giving a radical anion. The radical anion is protonated by ammonia at the beta-position, leaving a resonance stabilized radical (resonance structure not shown). Filling in another electron gives the enolate. The enolate is typically trapped by alkylation, such as by methyl iodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rDk8N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rDk8N.png&quot; alt=&quot;Mechanism of dissolving metal reduction to give enolate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method avoids the issues discussed by Ron when there is a real selectivity issue when generating an enolate by deprotonation of an asymmetrical ketone. Of course, it means that the starting material must be an enone, which may or may not be practical for a given synthesis.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="31775" LastEditDate="2018-03-06T08:38:26.497" LastActivityDate="2018-03-06T08:38:26.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17184" PostTypeId="1" AcceptedAnswerId="17185" CreationDate="2014-10-07T17:39:47.900" Score="0" ViewCount="72" Body="&lt;p&gt;For the interaction,  n$^0$ + $^{59}$Co $\to$ $^{60}$Co&lt;/p&gt;&#xA;" OwnerUserId="8136" LastActivityDate="2014-10-07T17:51:15.523" Title="How long must a sample be irradiated before all 59-Co atoms are converted to 60-Co?" Tags="&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-10-07T18:14:11.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="17185" PostTypeId="2" ParentId="17184" CreationDate="2014-10-07T17:51:15.523" Score="1" Body="&lt;p&gt;At a constant neutron flux, this is a first-order reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ - \frac{{{\text{d}}\left[ {{}^{59}{\text{Co}}} \right]}}{{{\text{d}}t}} = k \cdot \left[ {{}^{59}{\text{Co}}} \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\left[ {{}^{59}{\text{Co}}} \right] = {\left[ {{}^{59}{\text{Co}}} \right]_0} \cdot {\operatorname{e} ^{ - k \cdot t}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, $\left[ {{}^{59}{\text{Co}}} \right]$ will never become zero.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-07T17:51:15.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17186" PostTypeId="1" CreationDate="2014-10-07T18:00:20.830" Score="2" ViewCount="1843" Body="&lt;p&gt;What is the colour of $\ce{C2N2NiS2}$ (Nickel thiocyanate)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the web sites it only molecular weight,formula and other specifications but its colour is missing every where.&lt;/p&gt;&#xA;" OwnerUserId="8228" LastEditorUserId="61578" LastEditDate="2020-09-02T08:33:34.133" LastActivityDate="2020-09-02T08:33:34.133" Title="Colour of Nickel thiocyanate" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-structure&gt;&lt;transition-metals&gt;&lt;molecules&gt;" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17187" PostTypeId="1" CreationDate="2014-10-07T18:02:52.960" Score="11" ViewCount="5037" Body="&lt;p&gt;As far as I know, in order to synthesize a hydrocarbon $\ce{R-R'}$, Grignard reagent or Gilman reagent could be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{RMgX + R'X -&amp;gt; RR' + MgX2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R2CuLi + R'X -&amp;gt; RR' + LiX + RCu} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the pros and cons of using Gilman reagent over Grignard reagent? Gilman reagent is significantly harder to prepare and involves a reactive organolithium compound. I'm not looking for the pros and cons in terms of cost and yield, but rather specificity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any $\ce{RR'}$ groups that must be synthesized via Gilman reagent and not Grignard reagent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any $\ce{RR'}$  groups that must be synthesized via Grignard reagent and not Gilman reagent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Someone (outside of SE) pointed out to me how pointless it is for chemists to synthesize hydrocarbons using these methods (as they are readily available to purchase). While I agree with him, I'm still interested to find out &quot;theoretically&quot; if Gilman/Grignard reagent is specific to certain R/R' groups.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="5026" LastEditDate="2018-04-11T02:57:14.830" LastActivityDate="2018-04-11T02:59:15.707" Title="What are the pros and cons of using Gilman reagent over Grignard reagent to synthesize a hydrocarbon?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;hydrocarbons&gt;&lt;grignard-reagent&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17188" PostTypeId="2" ParentId="17186" CreationDate="2014-10-07T18:14:32.900" Score="2" Body="&lt;p&gt;After bit research i got finally one link where they mentioned the colour. The other name of(Synonym) of Nickel thiocyanate is &quot;Nickel Thiocyanate Greenpowder&quot; which suggest it is &lt;strong&gt;Green&lt;/strong&gt; in colour. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB2725417.htm&quot; rel=&quot;nofollow&quot;&gt;ChemicalBook&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-10-07T18:14:32.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17189" PostTypeId="2" ParentId="17186" CreationDate="2014-10-07T18:30:29.550" Score="0" Body="&lt;p&gt;Most hydrated Ni(II) salts are green.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For transition metals, its salts are often colored because of transitions of electrons between d-orbitals of different energy, and such transitions release protons of a specific wavelength (which happen to be within the visible range of light for most d block elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for a transition metal with the same oxidation state, its hydrated salts often exhibit a characteristic color.&#xA;In fact for most transition metal salts, the color can be predicted simply by knowing the cation of the salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Color_of_chemicals#Ions_in_aqueous_solution&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Color_of_chemicals#Ions_in_aqueous_solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Cu(II) salts are blue, Ni(II) salts are green. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As thiocyanate has the charge of -1, the nickel in $\ce{Ni(SCN)2}$ has the charge of +2, and $\ce{Ni(SCN)2}$ is therefore green, as confirmed in the answer given by @hey. &lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-07T19:35:28.917" LastActivityDate="2014-10-07T19:35:28.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17190" PostTypeId="2" ParentId="17186" CreationDate="2014-10-07T19:22:38.653" Score="1" Body="&lt;p&gt;The proper answer 'it depends'&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;PROPERTIES:&#xA;  Formula weight 174.88. Dark chocolate-colored powder. On&#xA;  addition of water, becomes first yellow and then dissolves with a&#xA;  green color.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(c) Brauer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you should be extremely cautious guessing colors of $d$-metal salts when the metal ion is not hydrated. The color may range widely. The only reasonably safe assumption is that $d^0$-salts are usually colorless, but then agains $\ce{MnO4-}$ is violet, so the colors must be memorized separately for each case.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-10-07T19:22:38.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17191" PostTypeId="2" ParentId="17173" CreationDate="2014-10-07T19:34:10.410" Score="5" Body="&lt;p&gt;Short answer: yes, the dioxane has a minimum when it's in a twist-boat form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use &lt;a href=&quot;http://avogadro.cc&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt; which offers a &quot;auto-optimize&quot; tool. You can grab atoms and throw them around to force rings into different conformations while the force field is minimizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what I did.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;File -&gt; Import -&gt; Fetch by chemical name: (2R,4R)-2-isobutyl-4-methyl-1,5-dioxaspiro[5.5]undecane. This grabs the 3D geometry from the NIH Resolver (i.e., PubChem)&lt;/li&gt;&#xA;&lt;li&gt;Auto-Optimize Tool: Force Field MMFF94&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then I started minimizing and looking at the energies. I pulled various atoms around to try to minimize steric clashes and flip rings from chair to boat, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the figures that follow, I've removed the hydrogens for clarity, but they exist as far as the energies and minimization. I just removed them to save the graphic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chair/Chair:&lt;/strong&gt; 24 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJOu7.png&quot; alt=&quot;chair-chair form&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Twist-boat/Chair:&lt;/strong&gt; 14 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHsGx.png&quot; alt=&quot;twist-chair&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, there's a decent stabilization in the twisted form of the twist-boat. You have to get the hydrogens right, or you'll end up with a bunch of eclipsed interactions between the dioxane oxygens and hydrogens on the cyclohexane.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-07T19:34:10.410" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17192" PostTypeId="1" CreationDate="2014-10-07T19:46:35.903" Score="2" ViewCount="29557" Body="&lt;p&gt;For molecular orbital theory:&#xA;How do you determine how many electrons occupy the anti bonding orbitals?&#xA;Also, why does hydrogen gas have no anti bonding orbitals- if each atom's electrons form a molecular orbital, then their energy is reduced so where does the extra energy go ? &lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="5017" LastEditDate="2014-10-08T01:04:02.717" LastActivityDate="2019-12-27T12:16:35.453" Title="How do you determine how many electrons occupy the anti bonding orbitals?" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17193" PostTypeId="2" ParentId="17192" CreationDate="2014-10-07T20:03:47.597" Score="3" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FSPke.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the MO diagram for Hydrogen. As you can see, for hydrogen atom, there is only 1 electron in the 1s orbital. Two hydrogens could come together to form a sigma bond, resulting in lowered energy. This doesn't mean that there is no sigma antiorbital, but rather due to the fact that hydrogen only contributes one electron each leading to a sum of two electrons, there are not enough electrons to fill the antiorbital, hence it is unfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine if it was helium instead. Helium has two 1s electrons, therefore, if two He atoms form a bond, 4 electrons has to be placed into the molecular orbitals. Since every orbital can hold a maximum of 2 electrons, two electrons would be in the sigma orbital, and the other two would be in the sigma antiorbital. An orbital and antiorbital would effectively cancel out, therefore, you see Helium exist in the monoatomic state.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-08T03:05:39.530" LastActivityDate="2014-10-08T03:05:39.530" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17194" PostTypeId="2" ParentId="17182" CreationDate="2014-10-07T20:26:33.947" Score="1" Body="&lt;p&gt;The Nernst equation only technically applies when the system is in electrochemical equilibrium—when there's no net current flow, but there's actually an exchange current. A system in which the reaction can only happen in one direction because there are no products wouldn't apply. Source:    Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm not sure about is the validity of ∆G values calculated from $∆G=∆G°+RT\ln Q$ at the  limits where the system is all products or all reactants. ±∞ don't seem to be meaningful values, but the system should quickly move away from these boundary values, even if only by a small amount. These chemical thermodynamics methods are well-behaved when near equilibrium, but well away from equilibrium, they can produce invalid results.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-07T20:26:33.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17195" PostTypeId="2" ParentId="16812" CreationDate="2014-10-07T20:33:30.873" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From what I have learned about oxoacids, the bond strength determines the strength of the acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. But that is not the whole picture. You also have to consider what else is floating around in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of the species (I assume aqueous solution) is given below:&#xA;$$ \ce{H2SO4 &amp;lt;=&amp;gt; HSO4- + H+ &amp;lt;=&amp;gt; SO4^{2-} + 2H+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can clearly see, the concentration of the protons is raised when going from left to right. There is a natural equilibrium that pushes backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you correctly point out, the bond stability (or &quot;strength&quot;) directly correlates to the stability of the acid species. It might well be (and I'd even bet on it) that the bond strength of $\ce{H-SO4-}$ is higher than $\ce{H-HSO4}$ (note that the bonds indicated here are not the real bonds as they occur but merely a representation) but this requires an in-depth (possibly computational) analysis, as a quick search of the internet did not yield any useful results.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I did some small, very basic geometry optimization calculations with turbomole (on default settings, so def-SV(P) basis set for all atoms) to find out the bond length of the hydrogen to its oxygen. Turns out my hunch was right, the bond is shorter in $\ce{HSO4-}$ (95.07 pm) than in $\ce{H2SO4}$ (95.72 pm), indicating stronger bonding once the first proton went flying away.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-10-09T22:27:21.887" LastActivityDate="2014-10-09T22:27:21.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17196" PostTypeId="1" AcceptedAnswerId="17198" CreationDate="2014-10-07T20:54:29.577" Score="2" ViewCount="2000" Body="&lt;p&gt;How does molecular orbital or valence bond theory explain the delocalised structure of benzene? I.e. when the carbon forms 3 bonds with other carbons/hydrogen, what happens to the remaining hybridised orbital? &lt;/p&gt;&#xA;" OwnerUserId="8176" LastActivityDate="2015-03-19T05:08:19.090" Title="Structure of benzene" Tags="&lt;molecular-orbital-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17197" PostTypeId="1" AcceptedAnswerId="17275" CreationDate="2014-10-07T21:33:14.293" Score="-1" ViewCount="259" Body="&lt;p&gt;What determines which type of hybridisation (sp3/sp2/sp) a molecule will take? Methane/ethylene/acetylene all have the same electron configuration but undergo different different types of hybridisation- why?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastActivityDate="2016-09-20T17:30:43.097" Title="Hybridised orbitals" Tags="&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17198" PostTypeId="2" ParentId="17196" CreationDate="2014-10-07T21:49:28.073" Score="3" Body="&lt;p&gt;The short answer is that it's not hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In benzene, the carbon atoms are sp2 hybridized. That means that each carbon's set of atomic orbitals is three sp2 orbitals and one p orbital. The sp2 orbitals are utilized to make sigma bonds to the adjacent carbons and hydrogens. The unhybridized p orbital contributes to the pi system.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-07T21:49:28.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17199" PostTypeId="1" CreationDate="2014-10-07T22:29:49.347" Score="1" ViewCount="172" Body="&lt;p&gt;Olive oil has &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3704564/table/T2/&quot; rel=&quot;nofollow&quot;&gt;significant levels&lt;/a&gt; of Carboxymethyllysine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does it form from the heating of olive oil, chemically speaking? Olive oil barely has any protein in it to begin with (though it still has some), so where would all the lysine come from?&lt;/p&gt;&#xA;" OwnerUserId="217" LastActivityDate="2014-10-07T23:10:35.500" Title="How is Carboxymethyllysine formed from the heating of olive oil?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17200" PostTypeId="2" ParentId="16812" CreationDate="2014-10-07T23:00:43.730" Score="1" Body="&lt;p&gt;The answer goes back to Pauling. pK&lt;sub&gt;a&lt;/sub&gt; of oxoacies of general formula XO&lt;sub&gt;m&lt;/sub&gt;(OH)&lt;sub&gt;n&lt;/sub&gt; is independent of the central atom and the number of hydroxy groups.  It only depends on the number of oxo groups, because that's the atoms the negative charge can be distributed over.  The difference between successive ionizations is about 5 pK&lt;sub&gt;a&lt;/sub&gt; units.  That's because you are removing a proton from a charged species, and work has to be expended to do that.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-07T23:00:43.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17201" PostTypeId="2" ParentId="17199" CreationDate="2014-10-07T23:10:35.500" Score="5" Body="&lt;p&gt;Given that the levels are expressed in arbitrary units (based on the results of an immunoassay), it's difficult to say how much CML there actually is and accordingly, how much lysine is needed to produce it. It may be that only a small quantity of lysine is required) In fact, a &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs00726-008-0071-4&quot;&gt;different study&lt;/a&gt; found no CML in raw olive oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CML is produced during cooking from glycation of the side chain amine, starting with the Maillard reaction between the amine and a reducing sugar's carbonyl at high temperature. However, in the paper you cited, there was already significant CML in raw olive oil that was only somewhat increased by cooking. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0963996913000781&quot;&gt;This paper&lt;/a&gt; suggests that CML formation is related to the oxidation of fatty acids and found that oils with increased concentrations of unsaturated fatty acids (olive oil is in the middle of the 5 oils they tested) were more prone to forming CML, but they deliberately added lysine to test it. It could be that this oxidation is something that happens in storage and the study you linked was using old oil or something. Also, just as with any other plant derivative, there could be significant variation in lysine content from different olives. &lt;a href=&quot;http://www.oliveoilsource.com/page/what-olive-oil&quot;&gt;This site&lt;/a&gt; indicates a lysine content of 3 mg/8.3 g whole olive, so there is some there. How much actually makes it into oil, I don't know.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-07T23:10:35.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17202" PostTypeId="2" ParentId="17155" CreationDate="2014-10-07T23:26:43.540" Score="5" Body="&lt;p&gt;I have drawn a mechanism but I haven't checked for any references as I am at home now, so it's more like paper chemistry. The chlorine atoms should play some role and possibly disfavour formation of certain carbocations that you would expect in a standard Wagner - Meerwein. It is possible to draw a concerted mechanism (not shown here) but seems very unlikely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**EDIT: I have changed my initial mechanism so it explains better the methyl ester formation (see the comments below). The stereochemistry at C* has also been corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general comment you cannot propose a non trivial mechanism without drawing one, it is handwaving. So I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kPczu.jpg&quot; alt=&quot;Mech2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="7284" LastEditDate="2014-10-10T19:24:08.893" LastActivityDate="2014-10-10T19:24:08.893" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17203" PostTypeId="1" AcceptedAnswerId="17207" CreationDate="2014-10-08T00:27:47.360" Score="4" ViewCount="2358" Body="&lt;p&gt;I understand activation energy to be the relative difference in energy between that of the activated complex/transition state (at some temperature) and the average energy of the reactants like shown here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/soCK0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/soCK0.png&quot; alt=&quot;Activation energies Ea(X→Y) or &amp;#39;Ea(Y→X)&amp;#39; need to be supplied to initiate the reactions X→Y or Y→X, respectively. When the catalyst is added, the activation energy decreases. However the Enthalpy change, ΔH (Thermodynamic factor) between X and Y remain constant.&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Activation_energy.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia&lt;/a&gt;, User &lt;a href=&quot;https://commons.wikimedia.org/wiki/User:Bkell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bkell&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct, I have questions about graph like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QBXVp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QBXVp.jpg&quot; alt=&quot;Maxwell-Boltzmann distribution indicating the activation energy&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(Source: webchem.net (not available anymore, &lt;a href=&quot;https://web.archive.org/web/20130720090743/http://webchem.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Internet Archive&lt;/a&gt;), retrieved via the &lt;a href=&quot;https://web.archive.org/web/20130617094803/http://www.webchem.net/notes/how_far/kinetics/maxwel2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Internet Archive&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the location marked activation energy supposed to be the energy level of the transition state provided the energy axis is absolute?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can the Maxwell-Boltzmann distribution be done for heterogeneous systems (i.e., two chemical reactants)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unrelated to the graph, why does negative activation energy result in a decrease in reaction rate (per Arrhenius equation)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="4945" LastEditDate="2018-02-19T05:33:04.353" LastActivityDate="2018-02-19T05:33:04.353" Title="How to interpret the Maxwell-Boltzmann distribution to find the activation energy?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17205" PostTypeId="1" AcceptedAnswerId="17208" CreationDate="2014-10-08T01:47:55.067" Score="1" ViewCount="5932" Body="&lt;p&gt;The following instructions were used to prepare magnesium sulfate crystals, $\ce{MgSO4 . 7H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Measure $50~ \mathrm{cm^3}$ of dilute sulfuric acid into a beaker and warm the solution.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using a spatula, add some magnesium oxide and stir the mixture. Continue &#xA;adding the magnesium oxide until excess is present.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Separate the excess magnesium oxide from the solution of magnesium &#xA;sulfate.  &lt;/li&gt;&#xA;&lt;li&gt;Heat the solution until crystals form. Obtain the crystals and dry them.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question I have an issue with is this: How would you know when excess magnesium oxide is present in step 2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer key says that I would know when no more solid (that is, no more magnesium oxide dissolves). However, since we are talking about dilute sulfuric acid, there should be water, and I read on the internet that $\ce{MgO}$ is soluble in water. &lt;/p&gt;&#xA;" OwnerUserId="8230" LastEditorUserId="5026" LastEditDate="2018-03-03T10:51:10.057" LastActivityDate="2018-03-03T10:51:10.057" Title="How can I know when I have added an excess of magnesium oxide?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17206" PostTypeId="2" ParentId="7477" CreationDate="2014-10-08T01:49:26.030" Score="10" Body="&lt;p&gt;I'm less familiar with safety regulations elsewhere in the world, but in the US, &quot;chemical manufacturers&quot; (which includes soap, sugar, salt, and just about everything else you can imagine) have to produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Material_safety_data_sheet&quot;&gt;Materials Safety Data Sheets (MSDS)&lt;/a&gt; and make them publicly available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of curiosity, I did some digging for soaps. &lt;a href=&quot;http://www.pg.com/&quot;&gt;Proctor and Gamble&lt;/a&gt; is a huge consumer producer, including &lt;a href=&quot;https://www.pg.com/productsafety/msds/professional_line/professional_line/Dawn_Professional_Dish_Detergent.pdf&quot;&gt;dish detergent&lt;/a&gt;. So there are &lt;a href=&quot;https://www.pg.com/productsafety/msds/professional_line/professional_line/Dawn_Professional_Dish_Detergent.pdf&quot;&gt;MSDS sheets for soap&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MSDS gives oral &lt;a href=&quot;http://en.wikipedia.org/wiki/Median_lethal_dose&quot;&gt;LD50 values&lt;/a&gt;. That is, the lethal dose required to kill half the population.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LD50 for soaps seem to vary by animal, but judging from a few MSDS I pulled up, they're ~2 g/kg or so. (That value was for rats.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I had to guess, given, say a ~70 kg adult, we'd be talking about 140 g of soap, &lt;strong&gt;&lt;em&gt;minimum&lt;/em&gt;&lt;/strong&gt; to kill a person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now.. that's &lt;em&gt;acute&lt;/em&gt; doses. I have no ideal if anyone has established long-term toxicity of soaps, and I can't find anything obvious online.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-08T01:49:26.030" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17207" PostTypeId="2" ParentId="17203" CreationDate="2014-10-08T01:52:04.800" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The marked location corresponds to a level of kinetic energy in the reactants sufficient to result in a successful collision (energy wise, it says nothing about orientation).  The energy required for a successful collision is the gap in potential energy between the reactants and transition state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A heterogeneous system is a system where the reactants are in different phases.  A common example is a gaseous reactant that collides with a solid catalyst.  In that case the Maxwell Boltzmann distribution can be applied to the gaseous reactant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you clarify this?  What reactions have negative activation energies?  The reactions you'll commonly encounter have positive activation energies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-10-08T01:52:04.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17208" PostTypeId="2" ParentId="17205" CreationDate="2014-10-08T02:00:00.473" Score="1" Body="&lt;p&gt;$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction that is taking place is an acid-base reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MgO + H2SO4 -&amp;gt; MgSO4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your question, &lt;em&gt;how would you know when excess $\ce{MgO}$ is present&lt;/em&gt;? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of your reaction,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MgO_{excess} + H2SO4 -&amp;gt; MgSO4 + H2O + MgO_{remaining}}$$&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-08T02:05:47.440" LastActivityDate="2014-10-08T02:05:47.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17209" PostTypeId="1" CreationDate="2014-10-08T02:10:13.910" Score="1" ViewCount="186" Body="&lt;p&gt;How long can a metal oxide such as magnesium oxide remain in solution before being converted to the metal hydroxide?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="73" LastEditDate="2014-10-08T04:09:49.190" LastActivityDate="2014-10-08T04:09:49.190" Title="Kinetics of the conversion from metal oxide to metal hydroxide in water" Tags="&lt;ions&gt;&lt;solutions&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17210" PostTypeId="1" AcceptedAnswerId="17213" CreationDate="2014-10-08T03:40:38.933" Score="0" ViewCount="110" Body="&lt;p&gt;What is actually meant by that concentration of reactants and products remain constant in equilibrium?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastActivityDate="2014-10-08T03:55:11.157" Title="Actual meaning of equilibrium" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17211" PostTypeId="1" AcceptedAnswerId="17306" CreationDate="2014-10-08T03:47:50.833" Score="2" ViewCount="168" Body="&lt;p&gt;In other words, I want to know why some reactions attain equilibrium early in the reaction while some reactions obtain equilibrium at the end of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="73" LastEditDate="2014-10-10T05:18:05.447" LastActivityDate="2014-10-10T17:18:13.853" Title="Why is equilibrium achieved at different stages of a reaction?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17212" PostTypeId="2" ParentId="17210" CreationDate="2014-10-08T03:48:48.013" Score="2" Body="&lt;p&gt;For an equation&#xA;$\ce{Reactant &amp;lt;=&amp;gt; Product}$, at equilibrium, the rate of reactants forming products is the same as the rate of products converting back to reactants. Therefore, the concentration of products and reactants remain constant.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-08T03:48:48.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17213" PostTypeId="2" ParentId="17210" CreationDate="2014-10-08T03:55:11.157" Score="2" Body="&lt;p&gt;t.c is right, but to expand on it, equilibrium means that a reaction is still occurring, i.e. reactants are converting to products and products are converting to reactants, but they're doing so at the same rate so there's no net change in the concentrations of any of the components. Contrast this with a metastable system, in which concentrations of components may be remaining the same, but the two aren't interconverting because of a large kinetic barrier.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-08T03:55:11.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17214" PostTypeId="2" ParentId="17211" CreationDate="2014-10-08T04:07:35.807" Score="0" Body="&lt;p&gt;The rate of reaction is dependent on the rate constant, as well as the order of reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate constant, which is approximated by the Arrhenius equation, is dependent on the Activation energy, Ea, and temperature, T. Activation energy can be lowered by using a catalyst (if any).&#xA; $$k = A e^{-E_a/(R T)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, you can see that the higher the temperature and the lower the Ea, the faster the rate. &lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-08T04:21:00.680" LastActivityDate="2014-10-08T04:21:00.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17215" PostTypeId="1" CreationDate="2014-10-08T04:18:24.757" Score="4" ViewCount="75" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pimagedine&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; says that aminoguanidine interferes with the formation of advanced glycation endproducts (AGEs) through interacting with 3-deoxyglucosone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does it interact with 3-deoxyglucosone that prevents AGE formation?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="217" LastEditDate="2014-10-08T20:41:10.053" LastActivityDate="2014-10-09T17:06:35.223" Title="How does aminoguanidine interfere with the formation of advanced glycation endproducts?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17216" PostTypeId="1" AcceptedAnswerId="17217" CreationDate="2014-10-08T04:33:18.860" Score="1" ViewCount="3375" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following isn't an elementary substance?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ozone&lt;/li&gt;&#xA;&lt;li&gt;Sulfur&lt;/li&gt;&#xA;&lt;li&gt;diamond&lt;/li&gt;&#xA;&lt;li&gt;quartz&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't get what &lt;strong&gt;elementary substance&lt;/strong&gt; means.&lt;/p&gt;&#xA;&lt;p&gt;By seeing the options and the word elementary, I guess it means something like pure or obtained in its original form.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="69830" LastEditDate="2021-05-22T07:46:44.453" LastActivityDate="2021-05-22T07:46:44.453" Title="What does being an &quot;elementary substance&quot; mean?" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17217" PostTypeId="2" ParentId="17216" CreationDate="2014-10-08T04:37:24.487" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;http://golatintos.blogspot.com/2013/03/a-term-in-chemistry-elementary-substance.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;LATINTOS&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An elementary substance is a pure chemical substance that consists of atoms belonging to a single chemical element. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you should be able to solve your question.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2017-06-13T01:18:31.760" LastActivityDate="2017-06-13T01:18:31.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17218" PostTypeId="1" AcceptedAnswerId="17220" CreationDate="2014-10-08T06:10:47.657" Score="6" ViewCount="1980" Body="&lt;p&gt;According to wikipedia $\ce{H^+3}$ and $\ce{H3}$ are possible and have been synthesized. It says that $\ce{H^+3}$ is stable too. Can anyone please tell me how is the bonding possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r9F3H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen has a valency of 1 but it still bonds like Ozone. How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Trihydrogen_cation&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{H^+3}$ - Trihydrogen Cation&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triatomic_hydrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{H3}$ - Triatomic Hydrogen&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="23561" LastEditDate="2018-09-06T13:42:47.477" LastActivityDate="2020-10-20T21:09:07.837" Title="How is bonding in the trihydrogen cation and triatomic hydrogen possible?" Tags="&lt;bond&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17220" PostTypeId="2" ParentId="17218" CreationDate="2014-10-08T09:39:22.417" Score="12" Body="&lt;p&gt;Here are the vital parts from the links about $\ce{H3}$ in the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The molecule can only exist in an excited state [...] the electronic state for a trihydrogen cation with an electron delocalized around it is a Rydberg state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and from second-level link&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rydberg states of the neutral triatomic hydrogen molecule can be viewed as&#xA;  an electron attached to a tightly bound $\ce{H3+}$ core&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed. Such bonding is actually pretty common, even if (almost never) mentioned in school. As long as orbitals of several atoms can overlap effectively and have fitting symmetry, they can form a single highly symmetrical bonding orbital, that can accept two electrons. In this case the bonding order between atoms is fractional, so in $D_\mathrm{3h}$ (triangular) $\ce{H3+}$, each of the three $\ce{H-H}$ bonds has bond order $1/3$. This kind of bonding plays a critical role in the chemistry of boranes (and Group 3 elements in general, maybe except thallium) and some families of compounds of d-block elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain more intimate and correct understanding you should invest in &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular orbital theory&lt;/a&gt;. Your link for $\ce{H3+}$ contains a comprehensive description of bonding in terms of said theory.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="41556" LastEditDate="2018-09-06T14:39:43.137" LastActivityDate="2018-09-06T14:39:43.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17221" PostTypeId="1" AcceptedAnswerId="17233" CreationDate="2014-10-08T10:19:43.630" Score="1" ViewCount="138" Body="&lt;p&gt;Well I've just started on kinetics and there are a few things I am absolutely confused about!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First of all, is a neutron actually made up of a proton and electron? If yes, then how is this possible?&lt;/li&gt;&#xA;&lt;li&gt;In K-electron capture where the electron of the K-Shell is captured by the nucleus, doesn't the species become positively charged or ionized as it is losing an electron?&lt;/li&gt;&#xA;&lt;li&gt;When metastable technetium disintegrates into normal technetium and a gamma particle, what is the change brought about? Isn't the technetium on both sides of the equation basically the same?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="10185" LastEditDate="2016-05-21T15:40:06.780" LastActivityDate="2016-05-21T15:40:06.780" Title="Radioactivity and Kinetics of disintegration" Tags="&lt;kinetics&gt;&lt;nuclear&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17222" PostTypeId="2" ParentId="17221" CreationDate="2014-10-08T10:29:13.113" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is neutron actually made up of a proton and electron?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is &lt;strong&gt;no.&lt;/strong&gt; Neutron is neutral(no charge) doesn't mean it is made up Electron and Proton. The neutron is made of two Down and one Up &lt;strong&gt;quark&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-10-08T10:29:13.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17223" PostTypeId="1" AcceptedAnswerId="17318" CreationDate="2014-10-08T11:33:04.543" Score="8" ViewCount="1968" Body="&lt;p&gt;In the quantum chemistry course that I currently attend, it was said several times that one of the key quantities derived from molecules by means of computation is the total electronic energy $E_\text{el}$, because from this many derived quantities can be calculated, such as bond length or $J$-coupling values for NMR experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've never gotten a good explanation as to how to proceed when I've calculated a value for $E_\text{el}$ and want to find out the bond lengths in the molecule in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the formulae and/or computation steps necessary for this?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-13T12:04:31.563" Title="How is the bond length calculated from the total electronic energy?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17224" PostTypeId="2" ParentId="17187" CreationDate="2014-10-08T11:37:06.420" Score="2" Body="&lt;p&gt;There is a flaw in your first reaction. $\ce{RMgX + R'X}$ is in equilibrium with $\ce{R'MgX + RX}$. This is a type of halogen metal exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction is in equilibrium you will get 3 products instead of one $\ce{RR', RR, R'R'}$ and therefore sheer wastage of Grignard reagent. On the other hand, Gilmann reagent is pretty good at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only symmetrical alkane could be prepared by Grignard reagent (Here I am talking only about synthesis of alkane). So, as you say, Gilmann reagent is harder to care so its better you use Grignard reagent there, but I am not sure of that. Though I am pretty much sure that preparation of unsymmetrical alkane leads to complete wastage if Grignard reagent is used.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5026" LastEditDate="2018-04-11T02:59:15.707" LastActivityDate="2018-04-11T02:59:15.707" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17225" PostTypeId="2" ParentId="17223" CreationDate="2014-10-08T11:51:32.997" Score="7" Body="&lt;p&gt;There is no formula. In short, for an electronic state of interest you do what is known as the &lt;em&gt;geometry optimisation&lt;/em&gt;, i.e. for the chosen electronic state you solve the electronic Schrödinger equation for a number of fixed (or clamped) nuclear configuration chosen in a certain way by the employed minimisation procedure with the aim to find the minimum of the electronic energy. And the nuclear configuration that corresponds to &lt;em&gt;minimum of the electronic energy&lt;/em&gt; for the chosen electronic state describes the &lt;em&gt;equilibrium geometry&lt;/em&gt; of a molecule being in the chosen electronic state. Then from such equilibrium nuclear configuration you can easily extract structural parameters, such as, for instance, bond lengths.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2014-10-08T11:51:32.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17226" PostTypeId="2" ParentId="17223" CreationDate="2014-10-08T12:35:28.677" Score="8" Body="&lt;p&gt;The geometry of a system is what is known (or given) and a corresponding energy is computed.  &lt;strong&gt;If you have an energy then you should have the geometry (or in your case, bond length).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wave function, $\Psi$, in the Schrodiner equation is simply a function of all the positions of the particles within the system (nuclei and electrons).  Typically chemists invoke the Born-Oppenheimer approximation which assumes that the nuclei are stationary (frozen in space) with respect to the very 'fast-moving' electrons.  Therefore, we freeze the nuclei in space and compute the energy of this &lt;strong&gt;stationary state&lt;/strong&gt;.  Move the nuclei in your molecule a little bit and you will get a new energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Wildcat was explaining, optimization procedures essentially moves the nuclei in your molecule around in space, computing an energy for each configuration. The goal of this type of routine is to find an energy that is lower than all the rest.  Take, for example, the $\ce{H2}$ molecule.  It contains one bond length and therefore the only coordinate we need to consider in an 'optimization procedure' is the internuclear separation of the two $\ce{H}$ atoms.  It is the chemist which will choose the initial positions of these atoms.  Say we initially set these atoms to be 200 pm apart.  We compute an energy.  Now, can we do better?  (i.e. can we perturb this geometry in such a way that gives rise to a lower energy?)  Well we can certainly try other bond lengths so we now set our bond length to 150 pm and compute a new energy.  Lo and behold we see that this new energy is lower than the first!  We keep trying different bond lengths until we reach a point where the energy &lt;em&gt;increases&lt;/em&gt; no matter what we do with the separation (whether we increase it or decrease it).  Once we are at this point, we have found the lowest energy configuration of $\ce{H2}$.  Of course optimization routines are smarter than doing a brute-force approach to sampling all the possible quantities each geometrical parameter can take on.  Furthermore, second-derivatives are essential for characterizing the nature of the stationary state that was determined (i.e. is the stationary state a true minimum, a transition state or some higher order saddle point?).&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-10-08T12:35:28.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17227" PostTypeId="2" ParentId="17165" CreationDate="2014-10-08T13:03:55.120" Score="2" Body="&lt;p&gt;While it is true that oxidizing strong acids like nitric and sulfuric are more corrosive than a non-oxidizing strong acid like hydrochloric, the latter still is able to dissolve materials (e.g, metals, minerals, proteins) and cause burns to the skin. A strong acid like HCl can (in the case of human skin or other biological materials) catalyze the hydrolysis of proteins, fats, and carbohydrates down to their water-soluble components. This is what happens in your stomach, where there is an abundance of hydrochloric acid, as well as enzymes to speed up the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most widely accepted measure of how strong an acid is, its its acid dissociation constant, not the pH of a solution of it.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-08T13:03:55.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17228" PostTypeId="1" AcceptedAnswerId="17234" CreationDate="2014-10-08T13:30:04.747" Score="3" ViewCount="1290" Body="&lt;p&gt;What are the intermolecular forces that can be found in Carbon Nanotubes?&lt;/p&gt;&#xA;" OwnerUserId="8233" LastActivityDate="2014-10-08T21:14:31.753" Title="Intermolecular forces in Carbon Nanotubes?" Tags="&lt;intermolecular-forces&gt;&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17229" PostTypeId="2" ParentId="17221" CreationDate="2014-10-08T13:50:21.843" Score="3" Body="&lt;p&gt;For 3., this is nuclear physics (or chemistry - Glenn Seaborg was a chemist). You clearly understand electron structure of an atom - there are various energy levels that can be occupied, and the ground state has all the electrons at the lowest states.  An excited (neutral) atom has an electron occupying a higher electron state, and it can decay back to the ground state by emitting a photon.  Well, the nucleus has various energy states for the protons and neutrons to occupy (the forces and math aren't the same, but the general idea is similar).  A metastable nuclear configuration is not in the ground state.  However, it can drop into the ground state, and has to release the energy somehow, and just like the electron it does so by emitting a photon, just a high energy one. Gammas are from nuclear transitions, x-rays from electronic transitions.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2014-10-08T18:56:12.143" LastActivityDate="2014-10-08T18:56:12.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17230" PostTypeId="2" ParentId="17228" CreationDate="2014-10-08T13:52:25.977" Score="3" Body="&lt;p&gt;Carbon nanotubes are not polar or ionic.  Therefore I do not expect any ion or dipole intermolecular interactions.  Since carbon has electrons, that only leaves us with dispersion forces (induced-dipole).  Therefore carbon nanotubes exhibit &lt;strong&gt;dispersion&lt;/strong&gt; as the only intermolecular force.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2014-10-08T13:52:25.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17231" PostTypeId="1" AcceptedAnswerId="17235" CreationDate="2014-10-08T14:27:30.790" Score="1" ViewCount="894" Body="&lt;p&gt;In a quiz, I was asked to find all of the combinations of organic molecule of $\ce{C4H6O2}$. The combination excludes -COOH, -OH, double bonds and the alkoxy group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem, as far as I am concerned, is the definition of the term &quot;alkoxy group&quot;. According to this &lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_group#Groups_containing_oxygen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;, a Peroxy or Methelenedioxy should be one of the possible combinations allowed. However, the names do have the &quot;-oxy&quot; suffix in them. So I am uncertain whether to include Peroxy's and Methelenedioxy's in the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Below is a picture of the Methelenedioxy group and the peroxy group, respectively,  from the already mentioned Wikipedia article. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MTjeN.png&quot; alt=&quot;Methelenedioxy&quot;&gt;       &lt;img src=&quot;https://i.stack.imgur.com/q83a2.png&quot; alt=&quot;Peroxy&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8234" LastEditorUserId="7475" LastEditDate="2015-10-09T16:53:17.313" LastActivityDate="2015-10-09T16:53:17.313" Title="What functional groups are subsets of the Alkoxy group?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17232" PostTypeId="2" ParentId="17231" CreationDate="2014-10-08T14:39:22.523" Score="3" Body="&lt;p&gt;Alkoxy groups are ethers, so I would consider methylenedioxy as a subset of alkoxy groups (although the reactivity is pretty different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peroxy is not an ether, so I would consider it distinct from an alkoxy group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given those functional group restrictions, it seems like the quiz was trying to force you into a particular line of thinking. Without alcohols and ethers, I think the only thing left would be peroxides, but perhaps I'm missing something.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-08T14:39:22.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17233" PostTypeId="2" ParentId="17221" CreationDate="2014-10-08T14:44:13.160" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;b&gt;No&lt;/b&gt;, as hey answered, neutrons are made of two down quarks and an up quark.  Protons are made of two up quarks and a down quark.  It is the conversion of a down quark into an up quark that results in the neutron --&gt; proton conversion.  Read &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html#c4&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information about this reaction.&lt;br&gt;&lt;br&gt;&#xA;The confusion with this probably comes from the fact that net reaction  in a neutron to proton conversion is $n^{0} \rightarrow p^{+} + e^{-} + \overline{\nu_e}$, so it looks like a neutron is falling apart into three sub-particles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;b&gt;No&lt;/b&gt;, the atom will not become ionized from this event itself.  Consider what is happening in this reaction:&#xA;$p^{+} + e^{-} \rightarrow n^{0} + \nu_e$&lt;br&gt;&lt;br&gt;&#xA;Yes, an electron is removed from the atom, but at the same time a proton is as well.  We should know that the atoms remains neutral because of conservation of charge. If the atom has no net charge at the beginning of the reaction, and no charged particles are ejected, it must be neutral at the end of the reaction as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Jon Custer's response is spot on here.  Gamma ray emission represents the movement of the nucleus from a metastable state to its ground state via the release of electromagnetic radiation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-08T14:44:13.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17234" PostTypeId="2" ParentId="17228" CreationDate="2014-10-08T14:46:32.143" Score="6" Body="&lt;p&gt;There is known, but poorly understood &lt;a href=&quot;http://en.wikipedia.org/wiki/Stacking_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;$\pi-\pi$ stacking&lt;/a&gt;. While weak compared to covalent, dipole or ionic interactions, it still beats usual van-der-vaals interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon nanotubes are, essentially, polycondensed aromatical systems, so they can and would interact with each other as aromatic systems via $\pi-\pi$ stacking&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-10-08T21:14:31.753" LastActivityDate="2014-10-08T21:14:31.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17235" PostTypeId="2" ParentId="17231" CreationDate="2014-10-08T15:39:13.487" Score="5" Body="&lt;p&gt;There is another Wkipedia article entitled &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alkoxy_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alkoxy group&lt;/a&gt;&quot;.  In it they show the following molecules as containing alkoxy groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mvbpj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They include methylenedioxy, but not peroxide in their list.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-08T15:39:13.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17236" PostTypeId="1" AcceptedAnswerId="17241" CreationDate="2014-10-08T15:41:24.777" Score="2" ViewCount="221" Body="&lt;p&gt;Can the compound butane-2,2,3,3-tetraol, $\ce{C4H10O4}$, be oxidized by $\ce{KMnO4}$? If so, in what conditions? As I have learnt in class, normally the hydroxy group would oxidize to form a ketone, but what about this particular case?&lt;/p&gt;&#xA;" OwnerUserId="8234" LastEditorUserId="7448" LastEditDate="2015-10-16T17:05:13.330" LastActivityDate="2015-10-16T17:05:13.330" Title="Can butane-2,2,3,3-tetraol be oxidized by potassium permanganate?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17237" PostTypeId="2" ParentId="14169" CreationDate="2014-10-08T16:22:14.320" Score="1" Body="&lt;p&gt;Since supercritical steam is not an ideal gas, you might want to use standard reference data (e.g. provided by NIST).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initial state&lt;/strong&gt;&lt;br&gt;&#xA;mass &lt;em&gt;m&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 10 kg&lt;br&gt;&#xA;pressure &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 500 bar = 50.0 MPa&lt;br&gt;&#xA;volume &lt;em&gt;V&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 0.01 m&lt;sup&gt;3&lt;/sup&gt;&lt;br&gt;&#xA;density &lt;em&gt;ρ&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 1000 kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;br&gt;&#xA;temperature &lt;em&gt;T&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 340.53 K&lt;br&gt;&#xA;specific enthalpy &lt;em&gt;h&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt; = 323.06 kJ/kg  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final state&lt;/strong&gt;&lt;br&gt;&#xA;mass &lt;em&gt;m&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 10 kg&lt;br&gt;&#xA;pressure &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 500 bar = 50.0 MPa&lt;br&gt;&#xA;volume &lt;em&gt;V&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 0.07 m&lt;sup&gt;3&lt;/sup&gt;&lt;br&gt;&#xA;density &lt;em&gt;ρ&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 142.86 kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;br&gt;&#xA;temperature &lt;em&gt;T&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 925.86 K&lt;br&gt;&#xA;specific enthalpy &lt;em&gt;h&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; = 3453.11 kJ/kg  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Heat transfer&lt;/strong&gt;&lt;br&gt;&#xA;heat &lt;em&gt;Q&lt;/em&gt; = Δ&lt;em&gt;H&lt;/em&gt; = (&lt;em&gt;h&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt; – &lt;em&gt;h&lt;/em&gt;&lt;sub&gt;0&lt;/sub&gt;) • &lt;em&gt;m&lt;/em&gt; = 31.3 MJ&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-08T16:22:14.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17238" PostTypeId="2" ParentId="17140" CreationDate="2014-10-08T17:41:48.927" Score="3" Body="&lt;p&gt;Yes , there is another name for it &lt;strong&gt;(n+l) rule&lt;/strong&gt; or &lt;em&gt;Aufbau principle&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Aufbau_principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though it is not correct to check this rule at such higher values of Z where the trend sometimes changes due to very less differnce between higher values of shell and orbitals .But if you are a beginner you should not heed to this statement.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-10-08T17:41:48.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17240" PostTypeId="2" ParentId="10760" CreationDate="2014-10-08T17:53:35.373" Score="1" Body="&lt;p&gt;The specified maximum carbon content in stainless steel 202 is 0.15 %.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-08T17:53:35.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17241" PostTypeId="2" ParentId="17236" CreationDate="2014-10-08T18:22:28.250" Score="5" Body="&lt;p&gt;You are talking about the dihydrate of &lt;a href=&quot;http://en.wikipedia.org/wiki/Biacetyl&quot; rel=&quot;noreferrer&quot;&gt;biacetyl&lt;/a&gt;.  &lt;em&gt;gem&lt;/em&gt;-diols are generally not stable, the equilibrium is on the side of the aldehyde/ketone, except when the keto group is next to an electronegative atom.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Glyoxal&quot; rel=&quot;noreferrer&quot;&gt;Glyoxal&lt;/a&gt; in aqueous solution is all hydrate and oligomers, as is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloral&quot; rel=&quot;noreferrer&quot;&gt;chloral&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biacetyl is more hindered than glyoxal, but in aqueous solution one would think they there is still some hydrate in the equilibrium.  What you'd think would happen is that on reaction with permanganate you'd get the cyclic diol and under sufficiently forcing conditions (read: acidic medium) you'd get cleavage to acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biacetyl is awful.  Run away!&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-08T19:38:29.940" LastActivityDate="2014-10-08T19:38:29.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17242" PostTypeId="1" CreationDate="2014-10-08T18:41:59.547" Score="0" ViewCount="3022" Body="&lt;p&gt;Consider a 1:1 mixture of propane and nitrogen at $60°\mathrm{F}$ and $75\:\mathrm{psi}$. Will the propane liquefy if pressure is increased to $200\:\mathrm{psi}$? I am experimenting in the oilfield with a shallow heavy oil lease.&lt;/p&gt;&#xA;" OwnerUserId="8238" LastEditorUserId="15410" LastEditDate="2015-05-19T07:03:49.267" LastActivityDate="2015-05-19T07:03:49.267" Title="Will propane liquefy under these conditions?" Tags="&lt;everyday-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17243" PostTypeId="1" AcceptedAnswerId="17245" CreationDate="2014-10-08T19:43:27.920" Score="1" ViewCount="323" Body="&lt;p&gt;A question asks whether C4 in 3-butyne-2-one can act as a nucleophile, electrophile, or acid (it can be more than one). The answer is that it can be both electrophilic and acidic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acidity makes sense, it's a terminal alkyne, sp orbitals stabilize the negative charge on the conjugate base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrophilicity I don't quite understand. I can write a resonance structure that puts some positive charge on C4, analogous to an alpha-beta unsaturated carbonyl compound, but it looks bizarre. I know ketenes are a real thing, but is  this resonance structure a real contributor? Is there another explanation for the electrophilicity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google searches for alkynone, ynones, etc. yielded mostly peer-reviewed literature that's above my understanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2gYhe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rov2w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="865" LastEditDate="2021-05-22T18:59:58.727" LastActivityDate="2021-05-22T18:59:58.727" Title="Why is C4 of butynone electrophilic?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17244" PostTypeId="2" ParentId="7225" CreationDate="2014-10-08T20:02:02.653" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;h2&gt;Safety Warning&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Caffeine is a toxic substance. When following the advice below, always keep in mind the factors that will help you kill yourself if you are not careful! For more details you can start out on &lt;a href=&quot;http://en.wikipedia.org/wiki/Caffeine#Toxicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Supersaturation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your idea of preparing a supersaturated solution is not that bad, but it might be that you have some nucleation sites in the bottle where you want to store it. Furthermore, it is probably not that easy to transfer the supersaturated solution from the container where you prepared it to the vial.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Acidic Medium&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've come across &lt;a href=&quot;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/c0750pis.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PDF&lt;/a&gt; that mentions something interesting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solubility in water is increased by adding dilute acid (e.g. HCl or citric acid).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be very easy to prepare an acidic solution (I'd aim at a pH of about 3 or so, in the general vicinity of &lt;a href=&quot;https://www.patientconnect365.com/dentalhealthtopics/article/energy_drinks_and_your_teeth_should_you_worry&quot; rel=&quot;nofollow noreferrer&quot;&gt;energy drinks&lt;/a&gt; and &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC162805/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coca Cola&lt;/a&gt;, beverages known to contain large amounts of caffeine) from some citric acid¹ and dissolve caffeine in there. Maybe you could even conduct a study about the solubility of caffeine at different pH values!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Water/Ethanol Mixture&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is one dissertation &lt;a href=&quot;http://arizona.openrepository.com/arizona/bitstream/10150/193915/1/azu_etd_1659_sip1_m.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(PDF)&lt;/a&gt; that touches on the solubility of caffeine in a mixture of water and ethanol. I wouldn't trust it too much, since it was written using MS Word and contains several formatting errors (it says $K_{ow}$ for example, instead of $K_\text{ow}$). The graphic containing the data can be found on page 52, but for your convenience it is reproduced below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/To6UT.png&quot; alt=&quot;Solubility of caffeine (diamonds) against the volume fraction of ethanol in the solvent mixture.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Seriously, the figure caption is useless, and the plot is so bad I'm tempted to redo it just for this answer.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never mind the linear relationship that is indicated by the line, the diamonds are the important part here. As you can see, there seems to be a solubility maximum at around $x=0.5$, where $x$ is the volume fraction of ethanol in the water-ethanol mixture. So that's a whopping 50 vol.-% of ethanol, about equivalent to a cask-strength whisky!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is probably not a good idea to use this knowledge to its fullest, because that will be quite a lot of ethanol, and as already mentioned in the comments it is probably not very wise to combine the consumption of caffeine with ethanol.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Does it have to be dissolved?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are also caffeine tablets and pills that you can take. They come with the additional guarantee that some FDA-like agency has looked into the production of these products, so you can assume they are safe to ingest. Also, they come with instructions and more than enough warning labels so you can educate yourself further.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you really want a caffeine solution, it is probably your safest and best bet to create an acidic solution where you can dissolve above-normal (i.e. pH=7) amounts of caffeine.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;¹ Though please do not ask here how to calculate the amount you need, that would be considered a homework question and get deleted pretty quickly...&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-08T20:02:02.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17245" PostTypeId="2" ParentId="17243" CreationDate="2014-10-08T20:04:42.397" Score="2" Body="&lt;p&gt;Yes, that second resonance structure, the one on the right, is a significant contributor to the overall description of the molecule.  That resonance looks a little different because the p orbitals on the alpha and beta carbons are on $\ce{sp}$ hybridized carbons, rather than the $\ce{sp^2}$ hybridized carbon more commonly seen in enones.   As this second resonance structure explains, alpha-beta unsaturated enones and ynones are both subject to attack by nucleophiles at the electrophilic beta carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structures, like the one pictured below, explain the reactivity seen in enamines and yneamines (the triple-bond analogue); however, now the positive charge is on the nitrogen and a negative charge on the beta carbon.  Therefore these nitrogen compounds are subject to attack by an electrophile at the nucleophilic beta carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ubETJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-08T21:39:47.067" LastActivityDate="2014-10-08T21:39:47.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17246" PostTypeId="1" CreationDate="2014-10-08T20:58:39.923" Score="6" ViewCount="2198" Body="&lt;p&gt;In lecture, the trans-effect was described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ligand $L^t$ with a higher trans-effect as $L$ (cis to $L^t$) leads to a faster dissociation of ligand $L^d$ (trans to $L^t$). I would expect that the attacking ligand $B$ is needed to make a stronger bond as the leaving ligand $L^d$ did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Huc9M.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to see if these assumptions are always valid in planar quadratic systems I've set up following problem according to the given trans-effect order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7vm96.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO}$ has a stronger trans-effect as $\ce{CH3-}$ and therefore the ligand trans to $\ce{CO}$ will be exchanged faster as trans to the $\ce{CH3-}$. But the leaving group $\ce{I-}$ would also have a higher trans-effect as $\ce{Cl-}$ so I wouldn't expect a substitution in b). Are these statements correct?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="5026" LastEditDate="2018-06-25T03:07:33.383" LastActivityDate="2018-06-25T03:12:00.977" Title="Dissociation rates: trans-effect rule in square planar complexes" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17247" PostTypeId="2" ParentId="7477" CreationDate="2014-10-08T21:52:22.767" Score="2" Body="&lt;p&gt;I am new here, however I have an Opinion. &#xA;Soap from the dish water would not be the only concern for me. Food particles, oils,bacteria, etc., would be what would come to my mind.&#xA;So, &quot;for me&quot; to leave soap ( or anything else on the dishes) negates the act of washing them in the first place.&#xA;Just my opinion but I believe a thorough analysis of the rinsed and air dried dish as opposed to the un-rinsed and Toweled dish would show a significant difference in how &quot;Clean&quot; they were. &#xA;In terms of Ingesting soap ( dish soap in this instance )being a health hazard I personally would consider ingesting ANY soap to be a health risk. &#xA;I would think ( without being a Doctor) that such ingestion could Quite easily lead to Diarrhea, Irritable Bowels, the killing of essential enzymes and probiotics in the digestive tract just to mention a few things off the top of my head. &#xA;There are probably many others but this Is what I thought of first.&#xA;&lt;a href=&quot;http://www.netwellness.org/question.cfm/19193.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.netwellness.org/question.cfm/19193.htm&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.merckmanuals.com/pethealth/special_subjects/poisoning/household_hazards.html&quot; rel=&quot;nofollow&quot;&gt;http://www.merckmanuals.com/pethealth/special_subjects/poisoning/household_hazards.html&lt;/a&gt;&#xA;&lt;strong&gt;&lt;em&gt;Detergents, Soaps, and Shampoos&lt;/em&gt;&lt;/strong&gt;&#xA;Exposures to products containing anionic and nonionic detergents generally cause mild gastrointestinal irritation that responds well to symptomatic care. These products include human and pet shampoos, liquid hand dishwashing soaps, bar bath soaps (except homemade soaps, which may contain lye), many laundry detergents, and many household all-purpose cleaners. Some of these, such as electric dishwasher detergents, are also alkaline corrosives (see Poisoning: Corrosives). All animals are susceptible.&#xA;These agents are not well absorbed by the body, and toxicity is limited to irritation of the eye, mouth, or gastrointestinal tract, which is usually mild and resolves on its own. Vomiting and diarrhea are the most common signs. Dehydration and metabolic abnormalities can develop in rare cases after prolonged vomiting or diarrhea.&#xA;Dilution with milk or water may reduce the risk of spontaneous vomiting. Vomiting usually resolves on its own after short periods of food and water restriction. In severe cases or in animals with sensitive stomachs, medications that prevent vomiting may be required. Rarely, fluid treatment is needed. Eyes should be flushed with water or saline for 5 minutes.&#xA;&lt;strong&gt;&lt;em&gt;www-ehow-com/facts_5802351_harmful-chemicals-dish-soap-html&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8240" LastEditorUserId="8240" LastEditDate="2014-10-08T22:15:49.090" LastActivityDate="2014-10-08T22:15:49.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17248" PostTypeId="2" ParentId="17187" CreationDate="2014-10-08T22:05:27.227" Score="4" Body="&lt;p&gt;I am not sure if Gilman reagents are harder to make than Grignard but they are softer nucleophiles and so they are used to perform 1,4-(conjugate) additions on $\ce\alpha,\beta-$unsaturated electrophiles. Grignard reagents will generally go for 1,2-attack instead. See this &lt;a href=&quot;http://en.wikipedia.org/wiki/Reactions_of_organocopper_reagents&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article for more info:&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="5026" LastEditDate="2018-04-11T02:57:34.537" LastActivityDate="2018-04-11T02:57:34.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17250" PostTypeId="2" ParentId="12404" CreationDate="2014-10-08T23:46:28.487" Score="18" Body="&lt;p&gt;Ron's answer is great, but I'd just like to touch on the mechanisms behind thermal conductivity so we can rationalize the differences between the behaviour of diamond, graphite, and metals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways in which heat is transmitted through solids: phonons and electronic conductivity. The latter occurs in electrically conductive solids, where conduction band electrons are free to move throughout the structure, carrying thermal energy along with them. This is a significant component to the thermal conductivity of metals and explains some of the in-plane conductivity of graphite. Electrically conductive solids tend to also be good thermal conductors for this reason, but this mode of conduction is not accessible to diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phonons are very complicated, but conceptually, phonons are waves of vibrations in the lattice of a solid. (quantized vibrational modes, actually, but not important for this) Vibrational energy in one part of a solid will be transmitted to other parts of the solid as phonons. What limits how quickly heat is transferred by this mechanism is phonon scattering. Phonons can scatter on many different things: crystal defects, impurities, crystal boundaries, even other phonons. This is why crystalline substances (like diamond) tend to have higher thermal conductivity than amorphous ones (like glass or polymers) as the phonons scatter more when they don't have a nice ordered crystal structure to travel through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why diamond, in particular, is an especially good thermal conductor even compared to other well-ordered crystals boils down to two factors: the mass of the carbon atoms and the strength of the bonds connecting them. I don't want to get too much into the details, but briefly, much as a harmonic oscillator has a higher frequency with a stronger spring (the strength of the C-C bonds) and a lighter mass (the C atoms), diamond can support higher energy phonons. This means that at a given temperature, fewer phonons near the material's limit will be present in diamond than a material with a smaller limit. The details aren't important, but when phonons near the limit interact, they &lt;a href=&quot;http://en.wikipedia.org/wiki/Umklapp_scattering&quot;&gt;scatter in such a way&lt;/a&gt; that the phonons don't travel through the structure. Because diamond's limit is so high, the probability of this type of scattering is lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phonons and electrons travel very well along graphite's graphene sheets, but poorly between them, due to weak inter-layer interactions and the large distance between layers, explaining the anisotropy of its thermal and electronic conductivities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals exhibit both modes of thermal conduction in varying degrees of importance and conduct isotropically since they lack the layered structure of graphite.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-08T23:46:28.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17251" PostTypeId="2" ParentId="10760" CreationDate="2014-10-09T00:06:42.497" Score="3" Body="&lt;p&gt;All  standardised grades of alloys have definite limits of composition. &#xA;If its common, then a quick search will find it. 0.15% is the specified limit for SS202. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=8209&quot; rel=&quot;nofollow&quot;&gt;http://www.azom.com/article.aspx?ArticleID=8209&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether its brittle or not depends on what you mean by brittle and its thermal processing. Its a martensitic stainless steel so note the details in azom and that it will be more brittle when hardened as described. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2014-10-09T00:06:42.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17253" PostTypeId="2" ParentId="17155" CreationDate="2014-10-09T00:46:55.090" Score="0" Body="&lt;p&gt;I haven't looked at the literature either, but it seems to go like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;protonation at the oxetane&lt;/li&gt;&#xA;&lt;li&gt;ring-opening of the oxetane. This gives a tertiary carbenium ion and the keto group (elimination of HCl from the $\alpha$-chloroalcohol).&lt;/li&gt;&#xA;&lt;li&gt;skeletal rearrangement to yield the dioxocarbenium ion; the resonance stabilization is what provides the driving force&lt;/li&gt;&#xA;&lt;li&gt;nucleophilic attack of solvent, to lose a methyl group and form the methyl ester&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-09T00:46:55.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17254" PostTypeId="1" AcceptedAnswerId="17257" CreationDate="2014-10-09T01:46:16.047" Score="1" ViewCount="435" Body="&lt;p&gt;I want to know what is the officially used file extension for files containing SMILES and SMARTS strings. Somewhere I see .SMI/.SMA and somewhere its the full .SMILES/.SMARTS.&lt;/p&gt;&#xA;" OwnerUserId="6834" LastEditorUserId="5017" LastEditDate="2014-10-09T02:05:37.630" LastActivityDate="2014-10-09T02:23:18.937" Title="What is the official file extension for SMILES and SMARTS files" Tags="&lt;computational-chemistry&gt;&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17255" PostTypeId="2" ParentId="17254" CreationDate="2014-10-09T02:05:09.563" Score="0" Body="&lt;p&gt;Slightly beside the point, but here is what &lt;a href=&quot;http://www.daylight.com/meetings/summerschool98/course/dave/smiles-relatives.html&quot; rel=&quot;nofollow&quot;&gt;Daylight says:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &quot;chemical&quot; MIME type, championed by H. Rzepa of Imperial College,&#xA;  London, is (hopefully) nearing ratification. Should this happen,&#xA;  SMILES will almost certainly appear as MIME subtype, i.e., the MIME&#xA;  type: chemical/smiles. Until that blessed day, the IETF recommendation&#xA;  is to use something like application/x-smiles or x-chemical/x-smiles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-09T02:05:09.563" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17256" PostTypeId="2" ParentId="13592" CreationDate="2014-10-09T02:11:42.827" Score="5" Body="&lt;p&gt;On the Mac, there's &lt;a href=&quot;http://chemspotlight.openmolecules.net&quot; rel=&quot;noreferrer&quot;&gt;ChemSpotlight&lt;/a&gt;. The most recent version includes a QuickLook component that will allow previews of most molecule files, including interactive 3D depictions using HTML/JavaScript.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also batch generate previews of MDL, SDF, etc. using &lt;a href=&quot;https://open-babel.readthedocs.org/en/latest/FileFormats/SVG_2D_depiction.html#svg-2d-depiction&quot; rel=&quot;noreferrer&quot;&gt;Open Babel&lt;/a&gt; to export previews to SVG or PNG formats.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-09T02:11:42.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17257" PostTypeId="2" ParentId="17254" CreationDate="2014-10-09T02:23:18.937" Score="3" Body="&lt;p&gt;I don't think I've ever seen anything proposed as &quot;official,&quot; though I admit I don't use Daylight's official toolkit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;nofollow&quot;&gt;Open Babel&lt;/a&gt; accepts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;.smi&lt;/li&gt;&#xA;&lt;li&gt;.smiles&lt;/li&gt;&#xA;&lt;li&gt;.can (ideally canonical SMILES, but that matters more for writing, not reading)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I haven't seen many SMARTS files, but I would suspect both .sma and .smarts are out there &quot;in the wild.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is probably something to bring up with the &lt;a href=&quot;http://www.opensmiles.org/opensmiles.html&quot; rel=&quot;nofollow&quot;&gt;OpenSMILES&lt;/a&gt; effort. I know there's an &lt;a href=&quot;https://github.com/timvdm/OpenSMARTS&quot; rel=&quot;nofollow&quot;&gt;OpenSMARTS&lt;/a&gt; specification started as well.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-09T02:23:18.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17258" PostTypeId="1" CreationDate="2014-10-09T02:43:51.003" Score="1" ViewCount="1259" Body="&lt;p&gt;I had a question about the carbon atom specifically. Carbon has 4 valence electrons, with electronegativity 2.55 (approx). If I was to fire electrons at the carbon atom, each carbon will occupy up to 4 electrons (8 in total in outer shell) to complete the octet. Now, let's assume that this happens slowly such that we have time to observe each carbon atom as it changes its charge, from $\ce{C-, C^2-, C^3-, C^4-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how would the electronegativity of the carbon ions will vary from each other? Will it increase as electrons are added and becomes zero after obtaining full octet? What would happen upon taking away electrons from carbons through help of highly electronegative elements?&lt;/p&gt;&#xA;" OwnerUserId="8245" LastEditorUserId="5026" LastEditDate="2018-07-10T09:38:50.737" LastActivityDate="2018-07-10T09:38:50.737" Title="How does an atom's electronegativity change as it becomes more negatively charged?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17259" PostTypeId="1" CreationDate="2014-10-09T03:26:38.260" Score="-2" ViewCount="324" Body="&lt;p&gt;How can one rationalize Avogadro's hypothesis using the equation of kinetic molecular theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$pV = \frac{1}{3} \cdot N \cdot m\overline{u^2} = \frac{2}{3} \cdot N \cdot \frac{1}{2}m\overline{u^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is the number of molecules and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is the average speed.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4945" LastEditDate="2019-01-15T18:30:57.627" LastActivityDate="2019-01-15T18:38:11.957" Title="How can one rationalize Avogadro's hypothesis using the equation of kinetic molecular theory?" Tags="&lt;gas-laws&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17260" PostTypeId="2" ParentId="17258" CreationDate="2014-10-09T03:30:42.190" Score="1" Body="&lt;p&gt;Electronegativity is a measure of the ability of an atom to attract electrons to itself (usually in a bonding framework; outside of a bonding framework the ability to attract electrons is electron affinity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from simple Coulombic considerations, we can surmise that as your carbon becomes more and more negatively charged, the less it will be able to attract electrons to itself. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-09T03:30:42.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17261" PostTypeId="1" CreationDate="2014-10-09T04:11:37.557" Score="2" ViewCount="57" Body="&lt;p&gt;I have a master degree in physical chemistry with close to 10 years experience as a chromatographer in pharma.  About two years ago I made a career change into industrial hygiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I am prepping to give a senior seminar talk about my professional experiences as an industrial hygienist at the college where I earned my BS in chemistry.  The talk will be half lecture and half lab - I'll be bringing in some equipment to demonstrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would chemistry students be particularly interested in hearing about from someone in industry?&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2014-10-09T09:32:02.753" Title="What would chemistry seniors like to hear about industrial hygiene?" Tags="&lt;safety&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17262" PostTypeId="1" AcceptedAnswerId="17280" CreationDate="2014-10-09T04:27:35.710" Score="5" ViewCount="1128" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wScQu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first of all, what's up with this &quot;plane&quot; I see two planes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis of my question is that to determine miller indices the plane cannot lie on your origin, so you have to make a translation. If I move my origin to the right then I get the miller indices that are indicated in the figure. But if I move backwards then I don't get the miller indices in the figure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question then is: what determines the appropriate translation?&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="5456" LastEditDate="2014-10-09T04:30:54.677" LastActivityDate="2017-12-12T08:14:26.410" Title="Miller indices. What determines the appropriate axis?" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17263" PostTypeId="1" CreationDate="2014-10-09T07:39:04.923" Score="5" ViewCount="2379" Body="&lt;p&gt;When phenol reacts with nitrous acid ($\ce{NaNO2 + \mathrm{conc}\ H2SO4}$), 4-nitrosophenol and not 2-nitrosophenol is formed. I cannot understand why the &lt;em&gt;para&lt;/em&gt;-isomer should be preferred over the &lt;em&gt;ortho&lt;/em&gt;-isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the reaction mechanism proceeds by electrophilic substitution by the $\ce{NO+}$ cation, and that the $\ce{OH}$ group activates the ring, which leads to &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; substitution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the &lt;em&gt;para&lt;/em&gt;-isomer should be preferred over &lt;em&gt;ortho&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="8246" LastEditorUserId="46306" LastEditDate="2018-06-15T01:59:57.050" LastActivityDate="2020-07-01T19:07:41.337" Title="Why is the para product major in the nitrosation of phenol?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17264" PostTypeId="2" ParentId="17263" CreationDate="2014-10-09T07:56:39.680" Score="-2" Body="&lt;p&gt;Due to steric hinderance and also there is some electronegativity of alcohol group due to induction it takes electron from sigma bond though not dominating therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the 2 reasons for more p- substitution&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-10-09T07:56:39.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17265" PostTypeId="2" ParentId="8827" CreationDate="2014-10-09T07:59:00.273" Score="0" Body="&lt;p&gt;While there are many great answers here, just to be sure let's point out some implicit restrictions. While in general being a Chemist may not need a licence, you are still responsible for what you may obtain as the result of the experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you are still responsible for your own and others safety, and property, in case of an accident,&lt;/li&gt;&#xA;&lt;li&gt;you still cannot prepare drugs and other illegal substances, and you can be punished even if you didn't use or sell it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So e.g. while obtaining NaOH is legal, you still can badly hurt yourself or others if you make soap based on Youtube videos, and you will be responsible for it. Similarly, preparing explosives or making fireworks can land you in jail depending on the country.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-10-09T07:59:00.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17266" PostTypeId="1" AcceptedAnswerId="17268" CreationDate="2014-10-09T09:04:35.870" Score="2" ViewCount="4979" Body="&lt;p&gt;If I dissolve sucrose sugar ($\ce{C_12H_22O_11}$) in water, would the water readily hydrolyse the sucrose sugar to form glucose and fructose sugars?&lt;/p&gt;&#xA;" OwnerUserId="8234" LastEditorUserId="13184" LastEditDate="2015-02-19T11:16:47.140" LastActivityDate="2015-02-19T11:16:47.140" Title="In water, will sucrose be hydrolysed by water to form glucose and fructose molecules?" Tags="&lt;biochemistry&gt;&lt;solutions&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2020-10-05T14:12:10.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="17267" PostTypeId="2" ParentId="17261" CreationDate="2014-10-09T09:32:02.753" Score="2" Body="&lt;p&gt;To keep this answer as far from opinionated as possible, I'd suggest some of the W's:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What&lt;/strong&gt; is industrial hygiene? (I've never heard of the term before this post and actually had to look it up myself.) &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How&lt;/strong&gt; do you do industrial hygiene? Procedures, risk assessment...&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Why&lt;/strong&gt; is industrial hygiene important? You might give examples where disaster was averted thanks to IH, and an example where it would have helped if done right.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Where&lt;/strong&gt; is industrial hygiene applied? (I guess the office building will have less rigorous guidelines than the production facility.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;When&lt;/strong&gt; did industrial hygiene start to be a thing? A small historical overview could be given, you could even give a &lt;strong&gt;who&lt;/strong&gt; to this one if it was some distinct person who came up with the idea.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Always good is a personal anecdote, something you witnessed or someone (whom you trust as a source) told you, highlighting the bullet points above.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-09T09:32:02.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17268" PostTypeId="2" ParentId="17266" CreationDate="2014-10-09T09:36:21.207" Score="6" Body="&lt;p&gt;Not without a catalyst present. The process is usually performed using &lt;a href=&quot;http://en.wikipedia.org/wiki/Invertase&quot; rel=&quot;noreferrer&quot;&gt;invertase&lt;/a&gt;, but some other enzymes with similar activity exist and addition of acid also can catalyse the reaction.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4934" LastEditDate="2014-10-10T01:11:49.367" LastActivityDate="2014-10-10T01:11:49.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17269" PostTypeId="2" ParentId="16610" CreationDate="2014-10-09T09:47:26.693" Score="2" Body="&lt;p&gt;The ideal gas law states&#xA;$$ pV = nRT \Leftrightarrow \frac{n}{V} = \frac{p}{RT}\ .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the density $\rho = m/V$ and the mass $m = M\cdot n$ with the molar mass $M$, we get a new formula for the density:&#xA;$$ \rho = M\frac{n}{V} = M\frac{p}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the definition of the specific volume we get&#xA;$$ v = \frac{RT}{Mp}\ . $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unless you know the molar mass of your parcel of air, you will not get out a numerical value for the specific volume.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-09T09:47:26.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17270" PostTypeId="1" AcceptedAnswerId="17277" CreationDate="2014-10-09T10:43:18.337" Score="9" ViewCount="7737" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Below is a picture of a titration between a weak acid, $\ce{HA}$, and a 0.150 M $\ce{NaOH}$ solution. Find the initial volume of the weak acid in milliliters in this titration. &#xA;  &lt;img src=&quot;https://i.stack.imgur.com/wCwww.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My work so far:&lt;/strong&gt;&#xA;From the graph, I think that the two most important points are (0,3.0) and (20.00,8.0) - I don't know the reason, but it looks important. &lt;br&gt; &#xA;At (0,3.0), there is no $\ce{NaOH}$. This equilibrium takes place: &#xA;$$ \ce{HA_{(aq)} + H2O_{(l)} &amp;lt;=&amp;gt; A_{(aq)}- + H3O+_{(aq)} } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant $K_\text{a}$ can be found from:&#xA;$$&#xA; K_\text{a} = \frac{[ \ce{H_3O^+}][\ce{A^-}]}{[\ce{HA}]}&#xA;$$&#xA;with $[\ce{H_3O^+}] = 10^{-3}$ since $\text{pH}=-\log_{10}[\ce{H_3O^+}]=3$ &lt;br&gt;&#xA;This is a dead end for me. &lt;br&gt;&lt;br&gt;&#xA;At (20.00, 8.0), $[\ce{H_3O^+}] =10^{-8}$, this number has decreased because some $\ce{H_3O+}$ have reacted with the newly dissolved $\ce{OH^-}$, although how much I can't tell. I have reached another dead end&lt;br&gt;&lt;br&gt;&#xA;Other facts I do know:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.150 M of $\ce{NaOH}$ means for every 1 liter of the solution, there will be exactly 0.150 moles of $\ce{NaOH}$.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NaOH}$ dissolves completely as a strong base. So, for every 1 mole of $\ce{NaOH}$ there will be 1 mole of $\ce{OH^-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8234" LastEditorUserId="5026" LastEditDate="2018-02-26T05:24:26.927" LastActivityDate="2018-02-26T05:27:38.080" Title="Find the initial volume of the weak acid given a titration graph" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17271" PostTypeId="1" AcceptedAnswerId="18683" CreationDate="2014-10-09T11:32:44.543" Score="0" ViewCount="1070" Body="&lt;p&gt;Yes it's a white ppt. which is soluble in dil. &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, but I'm not sure which of the following happens:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{BaSO3 +$dil.$HCl-&amp;gt;Ba(HSO)3 +Cl-}\tag1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{BaSO3 +2H+-&amp;gt;Ba^2+ +SO2 ^ +H2O}\tag2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which of the above actually happens? I have these two from different sources.&lt;/li&gt;&#xA;&lt;li&gt;Does it becomes &lt;span class=&quot;math-container&quot;&gt;$\ce{BaSO4}$&lt;/span&gt; on standing? Is it afterwards soluble in dil. &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, does this change require an oxidising agent, if yes then which one is more suitable - bromine water or &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="17368" LastEditDate="2020-06-21T14:36:18.463" LastActivityDate="2020-06-21T14:43:28.967" Title="How to recognize Barium sulphite?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17272" PostTypeId="1" AcceptedAnswerId="17276" CreationDate="2014-10-09T11:34:39.363" Score="0" ViewCount="185" Body="&lt;p&gt;I have been using several chemical drawing softwares including MarvinSketch, Jmol, Accelrys DS Visualizer, Avogadro, &lt;em&gt;etc.&lt;/em&gt;, all of which I have set to CPK colouring. Despite this supposedly common setting I have noticed that fluorine atoms are coloured rather varyingly from software to software. For example, MarvinSketch uses orange, Accelrys uses cyan as does Avogadro and Jmol uses algae green for fluorine atoms. I have tried Googling for the answer but I have only found incomplete tables, aside from on the Wikipedia website but I am dubious as to using Wikipedia for this information due to its reputation for inaccuracy. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2014-10-09T12:55:13.477" Title="Where can I find a complete and accurate table of CPK colours?" Tags="&lt;software&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17274" PostTypeId="2" ParentId="17182" CreationDate="2014-10-09T12:09:33.413" Score="0" Body="&lt;p&gt;$$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$$&#xA;If you don't have $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$ you must be having atleast  $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$, the reactants to produce the products when there is no product the case is:&#xA;$$\mathcal{E}=\mathcal{E}^\circ-\frac{\mathcal{R}T}{\mathcal{\nu F}}\ln{\frac{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}]}{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ 2+}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}]}}$$&#xA;when the $\ln$ term is zero, the $\mathcal{E}\to\infty$ implying an instantaneous reaction which requires no external force, also implies by the Le'Chatelirer's principe but note that when even a slightest amount of product is created the tendency to procees forward decreases until it reaches the equilibrium where the reaction attains the equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-10-09T12:09:33.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17275" PostTypeId="2" ParentId="17197" CreationDate="2014-10-09T12:23:05.863" Score="4" Body="&lt;p&gt;It's a pretty long topic and you would like to have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. In short I can present this from the same link, also some explanation at the last:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cGqBG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane $\ce{CH4}$, Ethylene $\ce{H2C=CH2}$ and acetylene $\ce{HC#CH}$ all have same electronic configuration but the number of  unhybridised orbitals( and of course the hybridised orbitals) are different in each case due to formation of $\pi$ bonds.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="29678" LastEditDate="2016-09-20T17:30:43.097" LastActivityDate="2016-09-20T17:30:43.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17276" PostTypeId="2" ParentId="17272" CreationDate="2014-10-09T12:55:13.477" Score="1" Body="&lt;p&gt;Indeed, this is annoying. Because of this problem, several years ago, one of the Jmol developers and I sat down and worked out a color scheme. It was similar to Accelrys, although we tried to make certain &quot;known colors&quot; match (e.g., rust-like for Fe, golden for Au, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly, there's been an effort to keep color consistency between &lt;a href=&quot;http://jmol.sourceforge.net&quot; rel=&quot;nofollow&quot;&gt;Jmol&lt;/a&gt; and &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow&quot;&gt;Avogadro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These colors are open source and available, e.g., through &lt;a href=&quot;https://github.com/openbabel/openbabel/blob/master/data/element.txt&quot; rel=&quot;nofollow&quot;&gt;Open Babel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are differences between Jmol and Avogadro, I'm sure both the Jmol developers and the Avogadro developers (including myself) would be happy to reconcile them.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-09T12:55:13.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17277" PostTypeId="2" ParentId="17270" CreationDate="2014-10-09T12:56:51.970" Score="7" Body="&lt;h2&gt;Outline&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are three main points of interest in the titration graphic, and two you have found yourself. The three points are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The start at $V_\text{titr}=0$&lt;/li&gt;&#xA;&lt;li&gt;Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$&lt;/li&gt;&#xA;&lt;li&gt;The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, armed with this information, here is how to proceed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find out total amount of acid&lt;/li&gt;&#xA;&lt;li&gt;Find out total concentration of acid&lt;/li&gt;&#xA;&lt;li&gt;Calculate the initial volume from the data gathered in steps 1 and 2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;1  Total amount of acid&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The total amount of acid can be gleaned from information point 3, the equivalence point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2  Total concentration of acid&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning:&#xA;$$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species:&#xA;$$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total initial concentration of all acid is then trivial:&#xA;$$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we here actually calculated the &lt;em&gt;activities&lt;/em&gt; and not &lt;em&gt;concentrations&lt;/em&gt;. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration:&#xA;$$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3  Calculation of the initial volume&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The initial volume is now calculated following the simple concentration formula:&#xA;$$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values:&#xA;$$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="5026" LastEditDate="2018-02-26T05:25:26.553" LastActivityDate="2018-02-26T05:25:26.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17279" PostTypeId="2" ParentId="15221" CreationDate="2014-10-09T13:33:22.777" Score="2" Body="&lt;p&gt;You really overcomplicate things (as I did initially, because I thought you were talking about the dissociation of $\ce{Cl2}$ gas in water and the subsequent reactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the basic pieces of information needed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reaction equation&lt;/li&gt;&#xA;&lt;li&gt;Equilibrium equation&lt;/li&gt;&#xA;&lt;li&gt;Mass balance equation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Reaction Equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The reaction is a simple acid-base equilibrium reaction:&#xA;$$ \ce{HOCl &amp;lt;=&amp;gt;[K_\text{a}] H+ + OCl-} $$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Equilibrium Equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The equilibrium equation follows easily from the reaction equation:&#xA;$$ K_\text{a} = \frac{[\ce{H+}][\ce{OCl-}]}{[\ce{HOCl}]} $$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Mass Balance Equation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The total amount of chlorine is calculated as follows:&#xA;$$ [\ce{Cl}]_\text{tot} = [\ce{HOCl}] + [\ce{OCl-}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the reaction equation also follows&#xA;$$ [\ce{OCl-}] = [\ce{H+}] $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since the concentration of protons is known, we can easily calculate the concentration of the hypochlorite anion:&#xA;$$ [\ce{OCl-}] = [\ce{H+}] = 10^{-\text{pH}} = 10^{-8} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know the total concentration, simply reforming the mass balance gives us the concentration (strictly speaking, the &lt;em&gt;activity&lt;/em&gt;) of the hypochlorous acid:&#xA;$$ [\ce{HOCl}] = [\ce{Cl}]_\text{tot} - [\ce{OCl-}] = 0.1 - 10^{-8} = 0.09999999 \approx 0.1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slap on some units and we're good to go!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{align}&#xA;c(\ce{HOCl}) &amp;amp;\approx 0.1~\mathrm{mol\, L^{-1}} \\&#xA;c(\ce{OCl-}) &amp;amp;= 10^{-8} ~\mathrm{mol\, L^{-1}}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-09T13:33:22.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17280" PostTypeId="2" ParentId="17262" CreationDate="2014-10-09T14:33:44.827" Score="5" Body="&lt;p&gt;There is no &quot;&lt;em&gt;appropriate&lt;/em&gt;&quot; translation. You must remember that everything includes translational symmetry, including the Miller planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_structure#Miller_indices&quot; rel=&quot;noreferrer&quot;&gt;crystal structures&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Miller_index&quot; rel=&quot;noreferrer&quot;&gt;Miller indices&lt;/a&gt; go into much more detail, but some key points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The unit cells possess translational symmetry, so there are a &lt;strong&gt;set&lt;/strong&gt; of Miller planes that also differ by translational symmetry.&lt;/li&gt;&#xA;&lt;li&gt;The Miller indices (hkl) (which yes should be expressed in smallest possible integers) define the normal of the Miller plane.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That is, (hkℓ) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in your case, you have to remember if you &quot;move the origin&quot;, you have to also consider the symmetrically identical unit cells &lt;em&gt;around&lt;/em&gt; the one in your picture. As I said above, there are an infinite set of identical Miller planes that simply differ by the translational symmetry of the lattice.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-09T14:33:44.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17281" PostTypeId="2" ParentId="16613" CreationDate="2014-10-09T15:08:41.020" Score="5" Body="&lt;p&gt;I appreciate the suggestions but some digging found a few possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As I mentioned in the question, there's the &lt;a href=&quot;http://www.ccdc.cam.ac.uk/Solutions/CSDSystem/Pages/CSD.aspx&quot; rel=&quot;noreferrer&quot;&gt;Cambridge Structural Database&lt;/a&gt;, which includes over 700,000 compounds (both organic and inorganic/organometallic). It's decidedly &lt;strong&gt;not&lt;/strong&gt; free, but available at many universities.&lt;/li&gt;&#xA;&lt;li&gt;There is the &lt;a href=&quot;http://webcsd.ccdc.cam.ac.uk/teaching_database_demo.php?productname=CSD%20Teaching%20Database&quot; rel=&quot;noreferrer&quot;&gt;&quot;Teaching Subset&quot; of 733 compounds&lt;/a&gt; from CCDC which offers a Java web view of interesting compounds, including plenty of VSEPR examples and inorganic and organometallic species. This is free to access, but as far as I can tell, not free to redistribute.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.stolaf.edu/depts/chemistry/mo/struc/explore.htm&quot; rel=&quot;noreferrer&quot;&gt;Cool Molecules&lt;/a&gt; from St. Olaf college, including 900+ structures from experimental data, and lots of inorganic and organometallic species (most elements are reflected), including lots of &quot;cool&quot; or unusual shapes (e.g., hexagonal bipyramidal).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.crystallography.net/browse.html&quot; rel=&quot;noreferrer&quot;&gt;Crystallography Open Database&lt;/a&gt; as mentioned by @permeakra. I was aware of this for solid-state materials, but it seems as if they've merged in data from &lt;a href=&quot;http://wwmm.ch.cam.ac.uk/crystaleye/&quot; rel=&quot;noreferrer&quot;&gt;CrystalEye&lt;/a&gt;, a resource which grabbed crystallography data from journal articles, including &lt;strong&gt;&lt;em&gt;many&lt;/em&gt;&lt;/strong&gt; molecular species now, and can be browsed by journal (e.g., &lt;a href=&quot;http://www.crystallography.net/result.php?journal=Organometallics&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Organometallics&lt;/em&gt;&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last two resources seem the best, since the data can be freely distributed. They also don't require using Java to view the structures.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-09T15:18:05.660" LastActivityDate="2014-10-09T15:18:05.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17282" PostTypeId="1" AcceptedAnswerId="17283" CreationDate="2014-10-09T15:36:10.277" Score="13" ViewCount="5156" Body="&lt;p&gt;Why is the &lt;span class=&quot;math-container&quot;&gt;$\ce{F-F}$&lt;/span&gt; bond relatively weak? And why does bond strength decrease from &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-Cl}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{I-I}$&lt;/span&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X47fQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have anything to do with &quot;lone-pair&quot; repulsion on the rather small &lt;span class=&quot;math-container&quot;&gt;$\ce{F-F}$&lt;/span&gt; molecule? I haven't asked my professor but I'm suspecting this is going to the first thing on his mind. I'm sure that many of you here will have rebuttals to this argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ur0UT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand does &lt;span class=&quot;math-container&quot;&gt;$\ce{F2}$&lt;/span&gt; even have lone pairs? There are no non-bonding orbitals populated on the molecular orbital diagram of fluorine. I wouldn't expect there to be any non-bonding orbitals populated either, as this is a simple diatomic we're dealing with and the atoms involved are identical ... so each atom's orbital should be able to interact with the other atom's orbital effectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zg18A.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="61578" LastEditDate="2019-06-16T23:02:01.110" LastActivityDate="2020-11-18T08:02:08.433" Title="Diatomic halogens - bond strength" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17283" PostTypeId="2" ParentId="17282" CreationDate="2014-10-09T15:56:36.753" Score="11" Body="&lt;p&gt;I see as I write the answer, that many comments are spot-on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are &lt;strong&gt;&lt;em&gt;higher&lt;/em&gt;&lt;/strong&gt; in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a really nice article by Peter Politzer in JACS that I like to use when discussing this &quot;anomaly&quot;: &lt;a href=&quot;http://dx.doi.org/10.1021/ja01051a006&quot; rel=&quot;noreferrer&quot;&gt;&quot;Anomalous properties of fluorine&quot; &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1969&lt;/strong&gt;, &lt;em&gt;91&lt;/em&gt; (23), pp 6235–6237&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-10T02:39:05.340" LastActivityDate="2014-10-10T02:39:05.340" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17284" PostTypeId="1" AcceptedAnswerId="17290" CreationDate="2014-10-09T16:37:30.153" Score="23" ViewCount="76919" Body="&lt;p&gt;In the crystal field theory (CFT), when the splitting of the d-orbital occurs, it gets divided into two parts.&#xA;The upper part with higher energy is the  &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e_g}$&lt;/span&gt; and the lower part with lower energy is called the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; as in:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fYDHW.png&quot; alt=&quot;t-2g and e-g&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So can anyone please explain me what is this &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_2}$&lt;/span&gt; or the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e}$&lt;/span&gt; or the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{_g}$&lt;/span&gt; below.&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="95340" LastEditDate="2021-02-11T15:57:08.800" LastActivityDate="2021-02-11T15:57:08.800" Title="What are t2g and eg in CFT?" Tags="&lt;coordination-compounds&gt;&lt;molecular-orbital-theory&gt;&lt;notation&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="17285" PostTypeId="2" ParentId="1321" CreationDate="2014-10-09T17:02:59.287" Score="0" Body="&lt;p&gt;Sodium hypochlorite is unstable in substance, but calcium hypochlorite, $\ce{Ca(ClO)_2}$, isn't, any chemical supply store won't hesitate to sell it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's other chlorine equivalents: chloroamine T, trichlorocyanuric acid, chlorosuccinimide are but a few.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-09T17:02:59.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17286" PostTypeId="2" ParentId="17215" CreationDate="2014-10-09T17:06:35.223" Score="2" Body="&lt;p&gt;I would expect the N-amino group of aminoguanidine to react with 3-deoxyglucosone at either of the adjacent, reactive carbonyl groups to form a stable guanidinylimine derivative (analogous to a hydrazone). This derivative, unlike 3-deoxyglucosone itself, would not be able to react with free amino groups on proteins and continue on the pathway to AGEs, but rather would be excreted as is.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-09T17:06:35.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17289" PostTypeId="1" CreationDate="2014-10-09T19:12:07.167" Score="2" ViewCount="2521" Body="&lt;p&gt;I'm reading a combustion text and would like help interpreting the following passage (diagram attached). It is regarding the relationship between equivalence ratio and adiabatic flame temperature:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For equivalence ratios between  $\phi$ = 1 and $\phi$ ($T_{max}$), the heat capacity decrease more rapidly with $\phi$ than $\Delta H$, while beyond $\phi$ ($T_{max}$), $\Delta H$ falls more rapidly than does the heat capacity. The decrease in heat capacity is dominated by the decrease in the number of product moles formed per mole of fuel burned, with the decrease in the mean specific heat being less significant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tOya6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m assuming that we are keeping the total mass of reactants constant (just changing the proportion between them). I get that total heat generated goes down since less fuel is reacting. But I don’t understand why heat capacity (I think this is extensive) decreases. Octane (&lt;a href=&quot;http://www.ohio.edu/mechanical/thermo/property_tables/gas/idealGas.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;at least at 300K&lt;/a&gt;&#xA;) has a higher specific heat than $\ce{CO2}$ and $\ce{H2O}$, so doesn’t having more fuel vapor increase the heat capacity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last sentence, I don’t understand why a decrease in heat capacity is more significant than the decrease in mean specific heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase “decrease in the number of product moles formed per mole of fuel burned,” does “mole of fuel burned” include fuel that technically didn’t react, since we have a fuel rich mixture?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="194" LastEditDate="2017-04-01T02:58:01.193" LastActivityDate="2017-04-01T02:58:01.193" Title="Relationship between Equivalence Ratio and Adiabatic Flame Temperature for Fuel Rich Combustion" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17290" PostTypeId="2" ParentId="17284" CreationDate="2014-10-09T19:35:08.107" Score="31" Body="&lt;p&gt;As stated in both the links Geoff and Philipp have kindly commented (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/8072/splitting-of-d-orbitals-when-ligands-approach-central-metal-ion?rq=1&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/q/16305&quot;&gt;2&lt;/a&gt;) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are &quot;allowed&quot; to interact together in reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_2}$&lt;/span&gt; groups three of the metal atom's d-orbitals into a certain class while two of the orbitals belong to the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e}$&lt;/span&gt; class. The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t}$&lt;/span&gt; means triply degenerate while the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e}$&lt;/span&gt; means doubly degenerate (degenerate means have the same energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{g}$&lt;/span&gt; is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital. It instead relates to how the orbitals behave if we hypothetically were to put a line bisecting the orbitals or &quot;invert&quot; them around a centre point (crudely put &quot;going from one corner to the other&quot;. If the phase of the wavefunction (or perhaps more pragmatic) the &quot;colour&quot; of the orbital lobes changes while doing this we label the orbital with a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{u}$&lt;/span&gt; if we have no phase change we label it with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{g}$&lt;/span&gt;. They indicate the orbitals response to an inversion about the centre of symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do lots more of these operations to investigate the symmetry of the orbitals which together lead us to assigning the molecule to a symmetry group. Knowing this is very important for a plethora of reasons.  &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="41328" LastEditDate="2020-01-02T08:42:46.140" LastActivityDate="2020-01-02T08:42:46.140" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17292" PostTypeId="1" AcceptedAnswerId="17298" CreationDate="2014-10-09T23:19:52.543" Score="8" ViewCount="166" Body="&lt;p&gt;Is there such a thing as a list of &quot;standard&quot; odors?  In everyday context, there are various descriptions to odors such as &quot;fruity&quot; and &quot;plastic-like&quot;.  But are there more accurate and normalized names of odors?&lt;/p&gt;&#xA;" OwnerUserId="8257" LastEditorUserId="12860" LastEditDate="2015-10-30T17:26:19.243" LastActivityDate="2015-10-30T17:26:19.243" Title="Standard list of odors?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17293" PostTypeId="1" AcceptedAnswerId="17313" CreationDate="2014-10-10T01:40:20.140" Score="0" ViewCount="339" Body="&lt;p&gt;Is there a way to prove that $\frac{1}{2} mv^2 $$= \frac{3}{2}RT$&#xA;The first one is kinetic energy and the second is the average kinetic energy.&#xA;Can this be proved by component of velocity.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="189" LastEditDate="2014-10-10T05:23:08.010" LastActivityDate="2014-10-10T18:07:56.837" Title="Proof of kinetic energy equation" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17294" PostTypeId="1" AcceptedAnswerId="17295" CreationDate="2014-10-10T04:17:40.133" Score="5" ViewCount="28149" Body="&lt;p&gt;Is removing the waters of hydration from &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4 * 5H2O}$&lt;/span&gt; a chemical or physical change? Sorry for the trivial question. I get varying answers across the net.&lt;/p&gt;&#xA;" OwnerUserId="8205" LastEditorUserId="41328" LastEditDate="2019-04-15T13:31:00.127" LastActivityDate="2020-04-05T15:10:36.583" Title="Is dehydrating copper(II) sulphate pentahydrate a chemical or physical change?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17295" PostTypeId="2" ParentId="17294" CreationDate="2014-10-10T04:49:49.663" Score="9" Body="&lt;p&gt;The traditional answer is that it's a chemical change when bond-making and bond-breaking is involved.  The material certainly changes appearance, and heating copper sulphate somehow releases bound water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, bonding in metal-aquo complexes runs the gamut from very weak to exceedingly strong - exchange rates span 15 orders of magnitude (Fig. 1 in Jan Reedijk, Metal-Ligand Exchange Kinetics in Platinum and Ruthenium Complexes. &lt;em&gt;Platinum Metals Rev.&lt;/em&gt; &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;52&lt;/em&gt;, 2–11. DOI: &lt;a href=&quot;https://doi.org/10.1595/147106708X255987&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1595/147106708X255987&lt;/a&gt;), and where do you draw the line between a &quot;physical&quot; electrostatic interaction (in, say, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na(OH_2)_6^+}$&lt;/span&gt;) and a chemical bond (as seen in &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(OH_2)_6^3+}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="4945" LastEditDate="2019-04-15T13:19:39.970" LastActivityDate="2019-04-15T13:19:39.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17296" PostTypeId="2" ParentId="17294" CreationDate="2014-10-10T05:15:45.627" Score="2" Body="&lt;p&gt;According to one text,&lt;sup&gt;[1]&lt;/sup&gt; a change in color, release of a gas (and some other changes), are indications of a chemical reaction. Since water vapor is released and the solid changes color, this should be considered as a chemical reaction, not a physical change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://en.wikibooks.org/wiki/General_Chemistry/Properties_of_Matter/Changes_in_Matter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikibooks, General Chemistry/Properties of Matter/Changes in Matter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4945" LastEditDate="2019-04-15T13:29:13.690" LastActivityDate="2019-04-15T13:29:13.690" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17297" PostTypeId="2" ParentId="17292" CreationDate="2014-10-10T05:33:59.677" Score="5" Body="&lt;p&gt;The problem with odors or taste is that it is a form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Qualia&quot;&gt;Qualia&lt;/a&gt;. Whether cyanide smells like bitter almonds or is odorless is subjective. Similarly, whether &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenylthiocarbamide&quot;&gt;Phenylthiocarbamide&lt;/a&gt; taste extremely bitter or completely tasteless is also subjective (depending on genetics). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in everyday life, coffee/wine/spices taste different to different people. Even for fruits, &lt;a href=&quot;http://en.wikipedia.org/wiki/Durian#Flavour_and_odour&quot;&gt;the Durian fruit can taste and smell fragrant to certain people but utterly disgusting to others.&lt;/a&gt; This is probably why the perfume industry is making billions - if everyone agree that a certain smell is &quot;nice&quot;, everyone would be using the same perfume. &lt;a href=&quot;http://www.livescience.com/17626-eau-de-dna-genes-determine-perfume-preference.html&quot;&gt;(Do Genes determine our perfume preferences?)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe it is difficult to have a standard list of odors. Even in the event that the list is created, it will not be universally agreed upon. There are bound to be disputes. Colours on the other hand, (which is also a form of Qualia), are less subjective (everyone can agree that chlorine is green and bromine is brown).&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-10T05:40:07.803" LastActivityDate="2014-10-10T05:40:07.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17298" PostTypeId="2" ParentId="17292" CreationDate="2014-10-10T05:55:03.737" Score="5" Body="&lt;p&gt;It's a long time since I worked in that field, but there are standard odours, and perfumery has a vocabulary to describe odours.  Common English is fairly weak in that department.  According to &lt;a href=&quot;http://www.fragrancesoftheworld.com/downloads/odor%20maps.pdf&quot;&gt;this paper&lt;/a&gt; consensus amongst professionals is high, although there are pathological cases (coriander/cilantro).&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-10T05:55:03.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17299" PostTypeId="1" AcceptedAnswerId="17303" CreationDate="2014-10-10T06:20:31.757" Score="5" ViewCount="1750" Body="&lt;p&gt;One child has claimed to have find a solution to all physical problems. On asking for details, he said that all periodic elements has common components, i.e. electrons, protons, neutrons.&#xA;The child has suggested a solution: convert atom to another by adding electron. This way one can get substance like &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O, Au, He}$&lt;/span&gt; in abundance.&#xA;How can it be done?&lt;/p&gt;&#xA;" OwnerUserId="720" LastEditorUserId="720" LastEditDate="2019-10-04T11:06:28.493" LastActivityDate="2020-02-07T06:32:06.967" Title="Conversion atom to another" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17300" PostTypeId="1" CreationDate="2014-10-10T06:36:19.507" Score="0" ViewCount="2375" Body="&lt;p&gt;Is there a non-toxic compound that binds with $\ce{NaCl}$ in a water medium?&lt;/p&gt;&#xA;" OwnerUserId="8259" LastEditorUserId="73" LastEditDate="2014-10-10T07:17:19.873" LastActivityDate="2014-10-10T10:05:21.903" Title="Is there a compound that binds with salt (NaCl) in solution?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-10-10T22:18:31.587" ContentLicense="CC BY-SA 3.0" />
  <row Id="17301" PostTypeId="1" CreationDate="2014-10-10T07:42:34.663" Score="3" ViewCount="3321" Body="&lt;p&gt;So, I'm confused on the notation my professor uses in a certain question; could someone explain. &#xA;He says: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the structure: (R)-(Z)-chloro-4,6,-dimethyl-3-octene &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get everything after '(z)'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just unsure what he means by '(R)-(Z)'... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input is appreciated. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="8260" LastEditorUserId="1362" LastEditDate="2014-11-04T07:58:49.960" LastActivityDate="2014-11-04T07:58:49.960" Title="What do (R) and (Z) mean in chemical names?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17302" PostTypeId="2" ParentId="17299" CreationDate="2014-10-10T08:35:21.817" Score="6" Body="&lt;p&gt;I think you are looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_transmutation&quot;&gt;nuclear transmutation&lt;/a&gt; which occurs via nuclear reactions or radioactive decay, as explained in the referenced wikipedia page. As you can imagine then, implementation has some important impediments.&lt;/p&gt;&#xA;" OwnerUserId="7770" LastActivityDate="2014-10-10T08:35:21.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17303" PostTypeId="2" ParentId="17299" CreationDate="2014-10-10T09:30:00.897" Score="2" Body="&lt;p&gt;While what you suggest may sound nice on paper, it has some serious problems.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Getting the elements. Let's say that we're talking about purifying water to remove toxic elements such as Hg or Cd. Extracting the elements out of the water is a feat by itself, for example using reverse osmosis methods. This is a method used for seawater desalination - to turn them from salt water filled with all kinds of elements into drinkable water. This process is very expensive, and a very polluting one as well. Desalination just the amount of water you need for drinking water is complicated, so desalinating an entire reservoir is simply not going to happen.&lt;/li&gt;&#xA;&lt;li&gt;Let's say you did somehow manage to extract the element in question. Now you need nuclear reactions to transmute one element to another. Not all are possible. For some elements, like Tc or Am this is the only way you can produce them. However, you are going to end up with radioactive nuclear waste.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To sum it up, even if it was possible, you would need so much power and to do it and you will produce some much pollution that it's simply not worth it. Just going and mining the gold will be orders of magnitude cheaper (and probably cleaner) than producing it using nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do manage to somehow extract the the polluting elements, you usually do something else with them (aka recycling) and you do not attempt being an alchemist. Another example is soils contaminated with lead. The solution is to just dig it up, put it somewhere where it is not hazardous to anyone and replace it with clean soil.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-10-10T09:30:00.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17304" PostTypeId="2" ParentId="17300" CreationDate="2014-10-10T09:38:03.530" Score="5" Body="&lt;p&gt;$\ce{NaCl}$ itself does not exist as a compound in water. The dissolution of salt in water means that both the sodium cation and the chloride anion are separated from each other and are each surrounded by water molecules. Therefore, no compound can bind to NaCl in water because $\ce{NaCl}$ does not exist in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, however, possible that a compound &quot;binds&quot; with either sodium or chlorine. For example, by adding copper to the solution and performing a bit of more electrochemical magic (out of the scope of this answer), you can get the solid compound copper chloride ($\ce{CuCl}$).&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="5456" LastEditDate="2014-10-10T10:05:21.903" LastActivityDate="2014-10-10T10:05:21.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17305" PostTypeId="2" ParentId="17080" CreationDate="2014-10-10T12:49:25.923" Score="2" Body="&lt;p&gt;You must be clear that the hybridization is decided by the number of σ-electrons, so for &lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt; the number of σ-electrons is four and so its hybridization is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In case of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3-CH=\overset{\ast}{C}=CH2}$&lt;/span&gt; the &lt;span class=&quot;math-container&quot;&gt;$\ce{\overset{\ast}{C}}$&lt;/span&gt; is having two σ- and two π-bonds, so its hybridization is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}.$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5175" LastEditorUserId="41328" LastEditDate="2021-05-23T12:03:54.980" LastActivityDate="2021-05-23T12:03:54.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17306" PostTypeId="2" ParentId="17211" CreationDate="2014-10-10T14:51:42.007" Score="1" Body="&lt;p&gt;If you define early or end of reaction by the how much the concentration of reactants change from initial reaction to once equilibrium is reached, it is because the equilibrium constant itself is essentially a ratio of the forward and reverse rate constants (can be approximated by the Arrhenius equation posted by t.c.). A high ratio means &quot;more&quot; products (at least a higher concentration) are present at equilibrium than reactants (so in a sense near the &quot;end of the reaction&quot;), since the forward rate is much higher than the reverse, a lower concentration of reactants compared to products is needed to maintain equilibrium. A low ratio means the opposite. &lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="8117" LastEditDate="2014-10-10T17:18:13.853" LastActivityDate="2014-10-10T17:18:13.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17307" PostTypeId="2" ParentId="17294" CreationDate="2014-10-10T14:59:33.353" Score="4" Body="&lt;p&gt;You heated copper(II) sulfate until it turned white.&#xA;This is a quite simple chemical reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{\underset{(blue)}{CuSO4.5H2O} &amp;amp;-&amp;gt;[][100 ^\circ C] CuSO4.H2O + 4 H2O}\\&#xA;\ce{CuSO4.H2O &amp;amp;-&amp;gt;[][&amp;gt; 200 ^\circ C] \underset{(white)}{CuSO4}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="4945" LastEditDate="2019-04-15T13:25:08.263" LastActivityDate="2019-04-15T13:25:08.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17308" PostTypeId="2" ParentId="17271" CreationDate="2014-10-10T15:26:36.637" Score="0" Body="&lt;p&gt;According to what I've found in russian segment of internet:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rm BaSO_3 + {dil.} HCl\rightarrow Ba(HSO)_3 +Cl^-\tag{1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rm BaSO_3 +2H^+\rightarrow Ba^{2+} +SO_2+H_2O\tag{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rm 2BaSO_3+O_2 \xrightarrow{\Delta} 2BaSO_4\tag{3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rm BaSO_3+H_2O_2 \rightarrow BaSO_4 + H_2O\tag{4}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I believe that it is stable at room temperature on standing.&lt;/p&gt;&#xA;&lt;p&gt;Reaction &lt;span class=&quot;math-container&quot;&gt;$\pu{(1)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{(2)}$&lt;/span&gt; are both possible depending on concentration from my point of view.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="17368" LastEditDate="2020-06-21T14:43:28.967" LastActivityDate="2020-06-21T14:43:28.967" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="17309" PostTypeId="1" AcceptedAnswerId="17311" CreationDate="2014-10-10T16:02:35.160" Score="6" ViewCount="2400" Body="&lt;p&gt;I have a question about &lt;a href=&quot;http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Equilibrium/equil.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this proof&lt;/a&gt; that shows the relationship between the rate constant and the equilibrium constant. In this proof, the assumption is that the reaction is $(a+b)$th order in the forward direction and $(c+d)$th order in the reverse direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had always thought reaction orders are empirically determined (and not necessarily whole numbers); can one simply add the stoichiometric coefficients like this? If not, how does one show that equilibrium constant is a ratio of the rate constants since the equilibrium constant requires the concentrations raised to the order of the species' stoichiometric coefficient?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="194" LastEditDate="2017-04-01T01:27:47.510" LastActivityDate="2017-04-01T01:27:47.510" Title="Proof that Equilibrium Constant is a Ratio of Rate Constants" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17311" PostTypeId="2" ParentId="17309" CreationDate="2014-10-10T17:42:27.810" Score="4" Body="&lt;p&gt;For a &lt;em&gt;reaction step&lt;/em&gt;, the assumption that the rate is proportional to the product of the concentration of reactants raised to their respective powers is valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, generally, reactions are made up of many reaction steps (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/SN1_reaction&quot; rel=&quot;nofollow&quot;&gt;SN1 reaction&lt;/a&gt;). For such reactions, you cannot use the assumption above. You have to understand the reaction mechanisms in order to determine the rate equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For reactions that involve decimal powers, usually the &lt;a href=&quot;http://en.wikipedia.org/wiki/Steady_state_(chemistry)#Steady_state_approximation_in_chemical_kinetics&quot; rel=&quot;nofollow&quot;&gt;steady state approximation&lt;/a&gt; is applied, and some mathematics is done along with empirical findings in order to determine the rate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;: Yes. A reaction step means an elementary reaction. From Wikipedia: &lt;em&gt;&quot;An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.&quot;&lt;/em&gt;. Therefore, there is only one reaction step.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, for the equilibrium constant,$K$, in your ratio derivation $K_c = k_f/k_r$, the chemical species must also be from that particular reaction step, and not the global reaction step. It makes sense because the equilibrium constant, K, is related to the free energy change, $\Delta{G}$ of a reaction step. In a two step reaction (e.g. SN1), there are two $\Delta{G}$ and therefore two equilibrium constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in practice, people do use the overall equilibrium constant, $\beta{}$, which is the product of the equilibrium constants at each reaction step. $\beta{} = K_1K_2K_3...$ (E.g.If the first step gives you 70% yield, and the second step gives you 70% yield, the global step would be 49% yield).(&lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, for your rate constant,$k$ and your equilibrium constant $K$, in $K_c = k_f/k_r$, they must be of that of a single reaction step.  &lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-11T05:50:51.613" LastActivityDate="2014-10-11T05:50:51.613" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="17312" PostTypeId="2" ParentId="16140" CreationDate="2014-10-10T17:53:30.373" Score="1" Body="&lt;p&gt;Silica and charcoal are both adsorbents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, silica is polar thanks to its plentiful OH groups, it preferentially adsorbs water, and that's why it is encountered in desiccant packages.  In the chemical laboratory one filters stuff through silica pads to separate polar materials out of solutions in non-polar solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activated charcoal is non-polar; it consists of aromatic rings, and it preferentially adsorbs non-polar molecules.  It is used in deodorizing packages.  In the lab, decolorizing charcoal is used in polar solvents to get rid of coloured impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, they are not equivalent.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-10T17:53:30.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17313" PostTypeId="2" ParentId="17293" CreationDate="2014-10-10T18:07:56.837" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt&quot;&gt;https://physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/molke.html#c1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/molke.html#c1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_theory#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Kinetic_theory#Properties&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These contain proof at various level . Choose according to your mathematical level&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-10-10T18:07:56.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17314" PostTypeId="2" ParentId="17242" CreationDate="2014-10-10T18:58:40.983" Score="2" Body="&lt;p&gt;First of all, you might want to consider using proper SI units.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If you know the saturation temperature (boiling point) of propane at a given pressure:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At a pressure of &lt;em&gt;p&lt;/em&gt; = 75 psi (0.52 MPa), the saturation temperature of pure propane is 37.13 °F (2.85 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a pressure of &lt;em&gt;p&lt;/em&gt; = 200 psi (1.38 MPa), the saturation temperature of pure propane is 104.54 °F (40.30 °C); thus, at a temperature of 60 °F (16 °C), propane would be liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is &lt;em&gt;w&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 0.39.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:&lt;br&gt;&#xA;&lt;em&gt;p&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; • &lt;em&gt;p&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the partial pressure of propane is approximately &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 78 psi (0.54 MPa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a pressure of &lt;em&gt;p&lt;/em&gt; = 78 psi (1.38 MPa), the saturation temperature of pure propane is 39 °F (4 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If you know the vapour pressure of propane at a given temperature:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At a temperature of 60 °F (16 °C), the vapour pressure of propane is 107.71 psi (0.74 MPa).&#xA;Thus, at a pressure of &lt;em&gt;p&lt;/em&gt; = 75 psi (0.52 MPa), pure propane would be gaseous. At a pressure of &lt;em&gt;p&lt;/em&gt; = 200 psi (1.38 MPa), it would be liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once again, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is &lt;em&gt;w&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 0.39.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:&lt;br&gt;&#xA;&lt;em&gt;p&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = &lt;em&gt;x&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; • &lt;em&gt;p&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the partial pressure of propane is approximately &lt;em&gt;p&lt;/em&gt;&lt;sub&gt;propane&lt;/sub&gt; = 78 psi (0.54 MPa). Since this partial pressure is below the vapour pressure of 107.71 psi (0.74 MPa) at a temperature of 60 °F (16 °C), propane would be gaseous.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-10T18:58:40.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17315" PostTypeId="1" CreationDate="2014-10-10T20:10:48.477" Score="4" ViewCount="143" Body="&lt;p&gt;Can the empirical formula be defined for berthollides? Is it equal to the molecular formula, or it can't be defined?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-10-14T09:40:52.483" LastActivityDate="2014-10-14T09:40:52.483" Title="Empirical formula of Berthollides" Tags="&lt;crystal-structure&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17316" PostTypeId="1" CreationDate="2014-10-10T20:16:15.227" Score="1" ViewCount="102" Body="&lt;p&gt;I want to make resin jewelry, but I'm having problems!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to create that effect, but I do not know what materials to use&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wXTbo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a mix of resin and other such things&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ketf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8263" LastEditorUserId="73" LastEditDate="2014-10-10T22:30:38.480" LastActivityDate="2014-10-10T22:30:52.280" Title="Resin for making Jewelry" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17317" PostTypeId="2" ParentId="17315" CreationDate="2014-10-10T21:25:42.947" Score="4" Body="&lt;p&gt;&quot;Berthollide&quot; seems to be a word for &quot;non-stoichiometric compound&quot; that has fallen out of use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to look at the structure, as a rule things happen for a reason.  The formula can be derived from the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the series of alums, which is  KAl(SO&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;·12H&lt;sub&gt;2&lt;/sub&gt;O, or more generic:  X&lt;sup&gt;+&lt;/sup&gt;Y&lt;sup&gt;3+&lt;/sup&gt;(SO&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;·12H&lt;sub&gt;2&lt;/sub&gt;O.  The generic structure can accomodate a monovalent cation, provided it's sufficiently large (lithium is out of the question, sodium is an edge case), and a tervalent cation (Al, Cr, Fe, pretty much anything), and the proportion can be freely chosen within the constraints of the structure.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-10T21:25:42.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17318" PostTypeId="2" ParentId="17223" CreationDate="2014-10-10T21:32:00.813" Score="7" Body="&lt;p&gt;The traditional way of getting the right geometry for your system is by taking a fixed atomic structure, perform an electronic structure minimization, calculate the forces acting on the atoms from that, use them to alter the atomic structure according to the forces and then use the altered structure to start over again. This is repeated until you converge to the equilibrium geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is an alternative (quite elegant) way for &lt;a href=&quot;http://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;DFT&lt;/a&gt; calculations using pseudopotentials or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Projector_augmented_wave_method&quot; rel=&quot;nofollow&quot;&gt;PAW method&lt;/a&gt; with plane-wave basis sets: the &lt;a href=&quot;http://en.wikipedia.org/wiki/Car%E2%80%93Parrinello_method&quot; rel=&quot;nofollow&quot;&gt;ab-initio molecular dynamics method introduced by Roberto Car and Michele Parrinello&lt;/a&gt;. The &quot;killer feature&quot; of the Car-Parrinello method is that it allows the simultaneous solution of the electronic structure problem and the equations of motion for the nuclei which allows for the relaxation of the nuclei to find stable structures, i.e. the equilibrium geometry, as well as for thermal simulations of solids and liquids - so this method does offer much more than a simple geometry optimization. How does it achieve that? In the Car-Parrinello approach, the total &lt;a href=&quot;http://en.wikipedia.org/wiki/Kohn%E2%80%93Sham_equations&quot; rel=&quot;nofollow&quot;&gt;Kohn-Sham (KS)&lt;/a&gt; energy is the potential energy as a function of the positions of the nuclei. The forces from this energy then determine the molecular dynamics (MD) for the nuclei. The special feature of the Car-Parrinello algorithm is that it also solves the quantum electronic problem using MD. This is accomplished by deriving the equations of motion from a fictitious &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrangian&quot; rel=&quot;nofollow&quot;&gt;Lagrangian&lt;/a&gt; which contains a fictitious kinetic energy of the electronic (KS) wave functions and some constaints to ensure that the wave functions remain orthonormal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathcal{L} &amp;amp;= \sum_{n} f_{n} m_{\psi} \langle \dot{\psi}_{n} | \dot{\psi}_{n} \rangle + \frac{1}{2} \sum_{i} M_{i} \dot{\vec{R}}_{i} \! {}^{2} + E_{\mathrm{DFT}}[\psi_{n}, \vec{R}_{i}] - \sum_{m, n} \Lambda_{n,m} \left( \langle \psi_{m} | \psi_{n} \rangle - \delta_{n,m} \right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_{n}$ is the occupation of the $n^{\mathrm{th}}$ Kohn-Sham eigenstate $\psi_{n}$, $M_{i}$ and $\vec{R}_{i}$ are the mass and the position of the $i^{\mathrm{th}}$ nucleus, $\delta_{n,m} = \begin{cases}1 &amp;amp; n=m \\ 0 &amp;amp; n \neq m \end{cases}$ is the Kronecker delta, $\dot{\psi}_{n} = \frac{\mathrm{d} \psi_{n}}{\mathrm{d}t}$ and $\dot{\vec{R}}_{i} = \frac{\mathrm{d} \vec{R}_{i}}{\mathrm{d}t}$ are the first derivatives with respect to time and $\Lambda_{n,m}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lagrange_multiplier&quot; rel=&quot;nofollow&quot;&gt;Lagrange multiplier&lt;/a&gt; ensuring the constraint of wave function orthonormality.&#xA;The first term in this Lagrangian represents the fictitious kinetic energy of the wave functions. There is no physical correspondence for this term. Ideally one would choose the fictitious mass $m_{\psi}$ of the wave functions equal to zero. This introduction of this unphysical quantity also lead to naming the Lagrangian fictitious.&#xA;The second term describes the classical kinetic energy of the nuclei.&#xA;The third term $E_{\mathrm{DFT}}[\psi_{n}, \vec{R}_{i}]$ is the density functional total energy, which is a functional of the electronic wave functions and the atomic positions.&#xA;The last term is the constraint of orthonormal wave functions $| \psi_{n} \rangle$.&#xA;Applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Principle_of_least_action&quot; rel=&quot;nofollow&quot;&gt;principle of least action&lt;/a&gt; to this Lagrangian leads to the following Euler-Lagrange equations (equations of motion for the ab initio MD simulation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  M_{i} \ddot{\vec{R}}_{i} &amp;amp;= \underbrace{- \nabla_{\vec{R}_{i}} E_{\mathrm{DFT}}}_{= \, \vec{F}_{i}} \\&#xA;  m_{\psi} | \ddot{\psi}_{n} \rangle &amp;amp;= - \underbrace{ \frac{1}{f_{n}} \frac{\delta E_{\mathrm{DFT}}}{\delta \langle \psi_{n} |} }_{= \, \hat{H} | \psi_{n} \rangle } + \underbrace{ \sum_{m} | \psi_{m} \rangle \frac{\Lambda_{n,m}}{f_{n}} }_{\tilde \, | \psi_{n} \rangle \epsilon_{n}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ddot{\psi}_{n} = \frac{\mathrm{d}^{2} \psi_{n}}{\mathrm{d}t^{2}}$ and $\ddot{\vec{R}}_{i} = \frac{\mathrm{d}^{2} \vec{R}_{i}}{\mathrm{d}t^{2}}$ are the second derivatives with respect to time, $\hat{H}$ is the Kohn-Sham hamiltonian, $\epsilon_{n}$ is $n^{\mathrm{th}}$ Kohn-Sham eigenvalue and $\vec{F}_{i}$ is the force acting on the $i^{\mathrm{th}}$ nucleus.&#xA;The first set of equations is just Newton's equations of motion for the nuclei moving under the forces derived from $E_{\mathrm{DFT}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stationary solution of the second set of equations is equivalent to the Kohn-Sham equations, since for a steady state all time derivatives vanish so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;   \hat{H} | \psi_{n} \rangle &amp;amp;=  \sum_{m} | \psi_{m} \rangle \frac{\Lambda_{n,m}}{f_{n}} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructing a matrix $\mathbf{\Lambda}$ from the Lagrangian multipliers $\Lambda_{n,m}$ shows that $\mathbf{\Lambda}$ is proportional to the transpose of the matrix representation of $\hat{H}$, i.e. $\Lambda_{n,m} \propto H_{m,n}$. Diagonalizing $\mathbf{\Lambda}$ leads to the eigenvalues $\epsilon_{n}$ of the Kohn-Sham equations.&#xA;So, this Lagrangian does not lead to the time-dependent Schroedinger equation for the electronic wave functions. Rather it creates a dynamic that (in the ideal case) keeps the electrons always in their ground state.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-13T12:04:31.563" LastActivityDate="2014-10-13T12:04:31.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17319" PostTypeId="2" ParentId="71" CreationDate="2014-10-10T22:36:46.727" Score="3" Body="&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1224932/&quot; rel=&quot;nofollow&quot;&gt;Permeation of protons, potassium ions, and small polar molecules through phospholipid bilayers as a function of membrane thickness&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reported values for proton permeability coefficients vary over&#xA;  a&#xA;  wide&#xA;  range,&#xA;  from&#xA;  $10^{-2} \, \mathrm{cm/s}$ to $10^{-7} \, \mathrm{cm/s}$&#xA;  (Nichols&#xA;  and&#xA;  Deamer,&#xA;  1980;&#xA;  Nichols&#xA;  et&#xA;  al.,&#xA;  1980;&#xA;  Biegel&#xA;  and&#xA;  Gould,&#xA;  1981;&#xA;  Elamrani&#xA;  and&#xA;  Blume,&#xA;  1983;&#xA;  Grzesiek&#xA;  and&#xA;  Dencher,&#xA;  1986;&#xA;  Perkins&#xA;  and&#xA;  Cafiso,&#xA;  1986;&#xA;  Norris&#xA;  and&#xA;  Powell,&#xA;  1990).&#xA;  This&#xA;  variation&#xA;  relates&#xA;  to&#xA;  the&#xA;  use&#xA;  of&#xA;  different&#xA;  techniques,&#xA;  lipids,&#xA;  and&#xA;  bilayer&#xA;  systems&#xA;  (liposomes&#xA;  and&#xA;  planar&#xA;  bilayers)&#xA;  or&#xA;  different&#xA;  pH&#xA;  ranges&#xA;  (Perkins&#xA;  and&#xA;  Cafiso,&#xA;  1986).&#xA;  The&#xA;  proton&#xA;  data&#xA;  of&#xA;  this&#xA;  study&#xA;  agree&#xA;  with&#xA;  the&#xA;  higher&#xA;  reported&#xA;  values.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As revealed by Fig. 4, proton permeability coefficients strongly&#xA;  depend on membrane thickness, starting at a maximum value of $1.3 \cdot 10^{-2} \, \mathrm{cm/s}$ for the shortest lipid and steadily decreasing to a value of $4.9 \cdot 10^{-5} \mathrm{cm/s}$ for the longest lipid. It follows that proton permeability decreases by a factor of approximately 250 as the thickness of the&#xA;  hydrophobic region is increased from $20 \, \mathring{\mathrm{A}}$ to $37 \, \mathring{\mathrm{A}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="189" LastEditDate="2014-10-16T21:50:16.173" LastActivityDate="2014-10-16T21:50:16.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17320" PostTypeId="1" CreationDate="2014-10-11T02:35:29.293" Score="3" ViewCount="4083" Body="&lt;p&gt;Can non-ferrous metals/alloys (copper, brass, bronze) be tempered (made harder) by temperature control alone, like ferrous metals? I understand it's simple to anneal copper, brass, etc., just by making them hot enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm Wondering if it's possible to temper non-ferrous metals without having to either work-harden it or media-blast it (essentially work hardening).&lt;/p&gt;&#xA;" OwnerUserId="7965" LastEditorUserId="73" LastEditDate="2014-10-12T07:59:07.023" LastActivityDate="2020-12-28T23:31:10.797" Title="How to temper a non-ferrous metal" Tags="&lt;metal&gt;&lt;materials&gt;&lt;metallurgy&gt;" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17321" PostTypeId="1" AcceptedAnswerId="17324" CreationDate="2014-10-11T02:45:01.393" Score="12" ViewCount="9381" Body="&lt;p&gt;I'm wanting to do a follow up to this Q&amp;amp;A on the cooking stack regarding &lt;a href=&quot;https://cooking.stackexchange.com/questions/46494/browning-avocados-whats-happening-what-helps&quot;&gt;browning avocados&lt;/a&gt;. I specifically want to expand my experiment to include the options in &lt;a href=&quot;https://cooking.stackexchange.com/questions/46494/browning-avocados-whats-happening-what-helps/46499#46499&quot;&gt;Wayfaring Stranger's Answer&lt;/a&gt;. Unfortunately, I managed to get the wrong stuff, and shipping rules make exchanging it problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the proud owner of $500~\mathrm{g}$ of sodium bisulfate ($\ce{NaHSO4}$), instead of sodium bisulfite ($\ce{NaHSO3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dammit Jim! I'm a cook not a chemist!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have learned is that that both compounds are used as food additives towards the same end, and that my stuff has a slightly higher LD50. That's it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone advise as to substituting my compound for the other? I'm not interested in tasty guacamole at this point, I'm really only looking at browning. If I'm trying to more or less duplicate the more successful experiments of the University of Florida, should I start with the same ratios? Or more or less of the additive? I can get sodium metabisulfite (also an anti-oxidant food additive) locally, if that would be a better option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; So far, answers have all been consistent in that I have a half-kilo of white powder that is useless to me. I will make a trip to the wine-making shop and acquire some sodium metabisulfite. That still leaves me with the question of how much? The University of Florida got the results I'd like to duplicate with $30~\mathrm{mg}$ of sodium bisulfite per $100~\mathrm{g}$ of avocado. Should I start with the same ratio of sodium metabisulfite?&lt;/p&gt;&#xA;" OwnerUserId="8265" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:30.103" LastActivityDate="2017-02-22T04:02:51.397" Title="Sodium Bisulfate vs Sodium Bisulfite" Tags="&lt;food-chemistry&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17323" PostTypeId="2" ParentId="17321" CreationDate="2014-10-11T03:40:42.190" Score="5" Body="&lt;p&gt;Sulfate will not provide the antioxidant effect you seek.  Metabisulfite has antioxidant effect.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-10-11T03:40:42.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17324" PostTypeId="2" ParentId="17321" CreationDate="2014-10-11T03:41:30.880" Score="16" Body="&lt;ul&gt;&#xA;&lt;li&gt;Sodium bisulphate (NaHSO&lt;sub&gt;4&lt;/sub&gt;): an acid&lt;/li&gt;&#xA;&lt;li&gt;Sodium bisulphiTe (NaHSO&lt;sub&gt;3&lt;/sub&gt;): an antioxidant&lt;/li&gt;&#xA;&lt;li&gt;Sodium metabisulphiTe (Na&lt;sub&gt;2&lt;/sub&gt;S&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;): another antioxidant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, bisulphate and bisulphite are not interchangeable, but bisulphite and metabisulphite are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium metabisulphite is prepared from sodium bisulphite via dehydration, as in this equation:&#xA;$$\ce{2 NaHSO3 -&amp;gt; Na2S2O5 + H2O }$$&#xA;It's reversible in aqueous solution. The assumption that 30 mg of metabisulphite contain the same amount of sulphite as 30 mg of bisulphite is OK, the error is small, about 10 %.  The correct amount would be 27 mg of metabisulphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd ask yourself if pasteurization (75 deg. C, 30 seconds) would work to deactivate the polyphenoloxidase without affecting flavour too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also suggest searching &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/&quot;&gt;Pubmed&lt;/a&gt; (it's a decent alternative to commercial citation databases) for such keywords as &quot;polyphenol oxidase&quot; (that's the enzyme that causes browning), &quot;inhibition&quot; and &quot;sulfite&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want hands-on answers I'd suggest going to the library and get (via interlibrary loan) a copy of this paper: Plant Foods Hum Nutr. 1995, 47, 245-56, &quot;The control of polyphenol oxidase activity in fruits and vegetables. A study of the interactions between the chemical compounds used and heat treatment.&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the abstract: ... Interactions between the use of ascorbic acid,&#xA;  citric acid, EDTA, sodium metabisulphite and heat treatment (70&#xA;  degrees C for 2 min) in the control of PPO activity were studied in&#xA;  avocado ..., banana ..., apple ..., pear ..., peach ..., potato ...,&#xA;  eggplant ..., mushroom ... and&#xA;  hearts-of-palm ... . The results demonstrated that&#xA;  PPO of avocado and eggplant was most resistant to inhibition by the&#xA;  methods used. ... The results indicated that,&#xA;  with the exception of PPO from avocado, the most adequate alternative&#xA;  method to substitute for the use of SO2 in the control of PPO was a&#xA;  combination of ascorbic acid, citric acid and heat treatment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This shows that how to get on top of browning while still having a presentable product is a tough problem and an active area of research.  Also, avocado seems a specially hard case.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-13T21:57:19.633" LastActivityDate="2014-10-13T21:57:19.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17325" PostTypeId="2" ParentId="17320" CreationDate="2014-10-11T04:38:52.257" Score="2" Body="&lt;p&gt;Most non-ferrous alloys are heated in order to form a solution.&lt;/p&gt;&#xA;&lt;h2&gt;Techniques:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;Annealing:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Most non-ferrous alloys that are heat-treatable are also annealed to relieve the hardness of cold working. These may be slowly cooled to allow full precipitation of the constituents and produce a refined microstructure.&lt;/p&gt;&#xA;&lt;p&gt;Different annealing techniques are subjected with non-ferrous alloys like  recrystallization annealing, partial annealing, full annealing, and final annealing.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Quenching&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Quenching is a process of cooling a metal at a rapid rate. This is most often done to produce a martensite transformation. In ferrous alloys, this will often produce a harder metal, while non-ferrous alloys will usually become softer than normal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_treating#Techniques&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-10-12T09:16:10.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17326" PostTypeId="1" AcceptedAnswerId="17336" CreationDate="2014-10-11T07:05:18.477" Score="8" ViewCount="119519" Body="&lt;p&gt;Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ &lt;strong&gt;has multiple molecules in neutral state and not dissociate&lt;/strong&gt;, thus it forms precipitates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $\mathrm{p}K_\mathrm{a}$ mean that most of the molecules of the substance $\ce{A}$ &lt;strong&gt;are in nondissociated state&lt;/strong&gt;?&#xA; Does lowering $\mathrm{pH}$ in this case causes substance $\ce{A}$ &lt;strong&gt;to dissociate&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are bold parts right? Please explain in plain English without the use of external links.&lt;/p&gt;&#xA;" OwnerUserId="4548" LastEditorUserId="15410" LastEditDate="2015-05-12T23:53:41.457" LastActivityDate="2020-12-21T14:26:29.850" Title="How do you explain pKa to non-professional?" Tags="&lt;acid-base&gt;&lt;solubility&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17327" PostTypeId="2" ParentId="17326" CreationDate="2014-10-11T07:49:25.047" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's say we have substance A which was mixed with substance B to improve shelf-life because pKa of the substance A is 7.9 and in mix the pH is 5.2.&#xA;  Does this mean that in the solution with pH near 8 the substance A has multiple molecules in neutral state and not dissociate, thus it forms precipitates?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At all pH, the substance A will exist as a mixture of HA and A-. HA may or may not be soluble, so it may or may not precipitate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does pKa mean that most of the molecules of the substance A are in nondissociated state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The higher the pKa, the lower the Ka, the lower the ratio of A- to HA, and thus the higher fraction of molecules of substance A will be in the non dissociated state&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does lowering pH in this case causes substance A to dissociate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No. Increasing pH causes substance A to dissociate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some additional information: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;From the Henderson–Hasselbalch equation,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\textrm{pH} = \textrm{pK}_{a}+ \log_{10} \left (&#xA;&amp;gt; \frac{[\textrm{A}^-]}{[\textrm{HA}]} \right )$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If pH = pKa, the concentration ratio of [A-] to [HA] is 1.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If your pH is higher than pKa, then [A-] must be greater than [HA].&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you lower your pH such that it is lower than pKa, then [HA] must be&#xA;  greater than [A-]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Therefore, by &lt;em&gt;increasing&lt;/em&gt; the pH, you are dissociating HA into H+ and&#xA;  A-. Which makes sense according to the Le Chatelier principle, where&#xA;  H+ is released to counteract the increase in pH (or the decrease in&#xA;  [H+])&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="8140" LastEditDate="2014-10-11T08:51:14.370" LastActivityDate="2014-10-11T08:51:14.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17328" PostTypeId="2" ParentId="17321" CreationDate="2014-10-11T08:11:47.543" Score="6" Body="&lt;p&gt;Generally, sodium hydrogen sulfite and sodium hydrogen sulfate are not interchangeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sodium hydrogen sulfite&lt;/strong&gt; or “sodium bisulfite” (NaHSO&lt;sub&gt;3&lt;/sub&gt;) is used in food processing as sanitising agent for food containers and fermentation equipment, preservative to reduce or prevent microbial spoilage, selective inhibitor of undesirable microorganisms&#xA;in the fermentation industries, and as an antioxidant and inhibitor of enzyme-catalysed&#xA;oxidative discoloration and non-enzymic browning. The food additive code (the E number, which is commonly found on food labels) of sodium hydrogen sulfite is E 222.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sodium hydrogen sulfate&lt;/strong&gt; or “sodium bisulfate” (NaHSO&lt;sub&gt;4&lt;/sub&gt;) is used as acidity regulator. The food additive code of sodium hydrogen sulfate is E 514.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-11T08:11:47.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17329" PostTypeId="1" CreationDate="2014-10-11T10:56:18.267" Score="4" ViewCount="289" Body="&lt;p&gt;In the following ASCI structure (single bond under &quot;$\ce{CH}$&quot; containing $\ce{CH2,CH2, CH3}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\ce{CH3-CH2-CH2-&amp;amp;CH-CH3}\\&#xA;&amp;amp;|\\&#xA;\ce{&amp;amp;CH2-CH2-CH3}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct way to name it is &quot;4-methlyheptane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I do it the way I have been taught I get the following &quot;4-propylpentane&quot;.&#xA;Here are my systematic steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I count the longest carbon chain. In this case its 5 carbons together so suffix -pentane. Then I look to see if there are any branched chains and number it from left to right. In this case the branched chain is on number 4. The branched chain contains $\ce{CH2-CH2-CH3}$, resp. $\ce{C3H7}$, aka propyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ergo, 4-propylpentane, however, the correct answer as mentioned above is 4-methylheptane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it 'heptane' ? The longest chain has FIVE carbons. I got the '4-' part correct, but how is $\ce{CH2-CH2-CH3}$ Methyl? Is this a mistake in my book?&lt;/p&gt;&#xA;" OwnerUserId="8268" LastEditorUserId="4945" LastEditDate="2014-10-14T07:11:33.057" LastActivityDate="2014-10-14T07:11:33.057" Title="Why doesn't this structure represent a substituted pentane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17330" PostTypeId="2" ParentId="17329" CreationDate="2014-10-11T11:14:15.320" Score="12" Body="&lt;p&gt;You need to find the &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Union_of_Pure_and_Applied_Chemistry#Organic_nomenclature&quot; rel=&quot;noreferrer&quot;&gt;longest continuous carbon chain&lt;/a&gt; of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;\ce{CH3-CH2-CH2-&amp;amp;CH-CH3}\\&#xA;&amp;amp;|\\&#xA;\ce{&amp;amp;CH2-CH2-CH3}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stretch that out to $\ce{CH3CH2CH2CH(CH3)CH2CH2CH3}$ (with the $\ce{-CH3}$ coming off of the $\ce{CH}$) and you'll see that the book is correct.  The &quot;side chain&quot; of $\ce{-CH2CH2CH3}$ is actually part of the main carbon chain.  As far as the numbering, counting from either side give the 4- position due to the symmetry of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1AM0n.png&quot; alt=&quot;4-methylheptane&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="4934" LastEditDate="2014-10-11T23:29:14.843" LastActivityDate="2014-10-11T23:29:14.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17331" PostTypeId="1" AcceptedAnswerId="17332" CreationDate="2014-10-11T11:25:57.183" Score="1" ViewCount="74" Body="&lt;p&gt;Assuming that electrons behave as both waves and particles, and that waves(like in double slit experiment) can interfere with themselves:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do electrons actually move in any direction, or do they just 'teleport'(similar to how when making a Quantum Leap, they instantaneously appear at a different shell, rather than actually move continuously). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do electrons 'subtract' with other electrons for the probability function of an electron being present in that location. i.e. if there is a covalent bond, will the electrons interfere with the other probably function aside from repelling it?.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do electrons only have positive probabilities. Is it more like how gravity never repels, only attracts, or more similar to how waves of light have peaks and troughs. Do electrons ever cancel out with others? If so, how?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8269" LastEditorUserId="5456" LastEditDate="2014-10-11T13:09:08.420" LastActivityDate="2014-10-11T16:12:46.423" Title="Electron interference" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17332" PostTypeId="2" ParentId="17331" CreationDate="2014-10-11T11:58:50.413" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Traditional understanding of trajectory is of no use for electrons. Granted, on macroscale you may detect something close to movement and trajectory, (electron beams and electron cannons take their names from this) but it still far from what we have in classical mechanics. Ideally you should abandon any attempts to relate quantum and classical mechanics and view quantum mechanics as an abstract mathematical construct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Incorrect question. If you have a system of two or more particles, you should consider the wavefunction of combined system, not separate wavefunctions for separate particles. Still, for electrons being fermions, there is a strict rule that wavefunction must be zero for positions in phase-space where two electrons of same spin are at same location and since electrons are of same charge, probability of having two electrons of different sping close to each other is usually lower than to have them at some distance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, only positive probabilities. Though, this does no apply to wavefunction itself, it may be negative at some point (actually, its value can be a complex number).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5456" LastEditDate="2014-10-11T16:12:46.423" LastActivityDate="2014-10-11T16:12:46.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17335" PostTypeId="2" ParentId="17301" CreationDate="2014-10-11T17:14:16.587" Score="6" Body="&lt;p&gt;The name (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;Z&lt;/em&gt;)-chloro-4,6-dimethyloct-3-ene is ambiguous. This could be (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;Z&lt;/em&gt;)-1-chloro-4,6-dimethyloct-3-ene, (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;Z&lt;/em&gt;)-3-chloro-4,6-dimethyloct-3-ene, (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;Z&lt;/em&gt;)-6-chloro-4,6-dimethyloct-3-ene, or (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;Z&lt;/em&gt;)-8-chloro-4,6-dimethyloct-3-ene. (2-chloro-4,6-dimethyloct-3-ene, 5-chloro-4,6-dimethyloct-3-ene, and 7-chloro-4,6-dimethyloct-3-ene have one additional stereocenter.) For example:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MYZrw.png&quot; alt=&quot;(R,Z)-1-chloro-4,6-dimethyloct-3-ene&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-11T17:14:16.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17336" PostTypeId="2" ParentId="17326" CreationDate="2014-10-11T17:18:14.323" Score="13" Body="&lt;p&gt;pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its &quot;conjugate acid&quot;) is more stable for long term storage than A itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT BASED ON NEW INFORMATION:&#xA;As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-10-11T20:14:59.880" LastActivityDate="2014-10-11T20:14:59.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17337" PostTypeId="1" CreationDate="2014-10-11T17:28:27.387" Score="3" ViewCount="115" Body="&lt;p&gt;Someone mistakenly poured a solution containing $15\mathrm{g}$ of gold into a nearly full waste drum that is around $30$% $\ce{HCl}$ with a little nitric acid in there as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a safe way of separating the acid from the gold via distillation? &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="5456" LastEditDate="2014-10-12T05:59:58.193" LastActivityDate="2014-10-12T05:59:58.193" Title="How to recover 15g gold from 50 gallons of 30% Acid?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17338" PostTypeId="2" ParentId="5516" CreationDate="2014-10-11T18:22:00.617" Score="2" Body="&lt;p&gt;Only half as many Ca&lt;sup&gt;2+&lt;/sup&gt; are needed as Na&lt;sup&gt;+&lt;/sup&gt; to neutralize the charge of the ion exchanger (here: clay).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The affinity of clay for bivalent ions is larger than that for monovalent ions:&lt;br&gt;&#xA;Ca&lt;sup&gt;2+&lt;/sup&gt; &gt; Na&lt;sup&gt;+&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most clays, the affinity for equally charged ions increases with the nonhydrated radius of the ions:&lt;br&gt;&#xA;Cs&lt;sup&gt;+&lt;/sup&gt; &gt; K&lt;sup&gt;+&lt;/sup&gt; &gt; Na&lt;sup&gt;+&lt;/sup&gt; &gt; Li&lt;sup&gt;+&lt;/sup&gt;&lt;br&gt;&#xA;Ba&lt;sup&gt;2+&lt;/sup&gt; &gt; Sr&lt;sup&gt;2+&lt;/sup&gt; &gt; Ca&lt;sup&gt;2+&lt;/sup&gt; &gt; Mg&lt;sup&gt;2+&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-11T18:22:00.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17341" PostTypeId="2" ParentId="17326" CreationDate="2014-10-11T19:10:04.237" Score="-1" Body="&lt;p&gt;By specifying that A is lidocaine I could say that by adding HCl to it you'd be forming a diferent compound, lidocaine hydrocloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4QMi.png&quot; alt=&quot;Which is more or less like this in the solution.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8274" LastActivityDate="2014-10-11T19:10:04.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17342" PostTypeId="1" AcceptedAnswerId="17343" CreationDate="2014-10-11T19:23:17.377" Score="15" ViewCount="9997" Body="&lt;p&gt;What are weak acids? My chemistry and biology classes skimmed over them briefly but nothing in depth. I understand that an acid is weak if it partially dissociates and breaks down into ionic compounds, but they never stated why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is why do weak acids break down in water? Do strong acids dissociate at all? What if they were in a different solution (or do they just stay themselves all the time?)&lt;/p&gt;&#xA;" OwnerUserId="8167" LastEditorUserId="402" LastEditDate="2015-11-07T06:29:03.110" LastActivityDate="2018-03-03T11:28:28.737" Title="Why are weak acids weak?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17343" PostTypeId="2" ParentId="17342" CreationDate="2014-10-11T19:30:42.073" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are weak acids? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples of weak acids in &lt;em&gt;water solution&lt;/em&gt; would be acetic acid (main component of vinegar) and ammonium ion (conjugate acid of ammonia). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that an acid is weak if it partial dissociates and breaks down into ionic compounds but they never stated why. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An acid is weak if not all of the acid molecules &lt;em&gt;ionize&lt;/em&gt; into hydrogen protons and its conjugate base in a particular solvent system. Alternately, if we were to use the broader, Brønsted definition, an acid is weak if it does not completely or nearly completely donate its proton to some base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid is a weak acid in water. Put acetic acid into pure liquid ammonia and now acetic acid will fully dissociate. The proton will be lost from acetic acid to ammonia and this is a favorable reaction. Makes sense, because the only thing that's changed is the solvent that's grabbing the proton - and ammonia is &lt;em&gt;only&lt;/em&gt; a few billion times stronger as a base than water. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That and do the strong acids dissociate at all. Like if they were in a different solution or do they just stay themselves all the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strong acids by definition ionize to an extent of 100% or nearly 100%. Strong acids will give up their protons completely to the strongest base in the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the above text I have used the word &quot;ionize&quot; to describe molecular acids because by definition when these acids react, they &lt;em&gt;ionize.&lt;/em&gt; Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to &quot;dissociate,&quot; because ionic compounds are made of &lt;em&gt;ions,&lt;/em&gt; and it wouldn't make much sense to say that ions &lt;em&gt;ionize&lt;/em&gt; or become ions.  &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:28:28.737" LastActivityDate="2018-03-03T11:28:28.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17344" PostTypeId="2" ParentId="17342" CreationDate="2014-10-11T19:40:21.233" Score="2" Body="&lt;p&gt;There is nothing magic about weak acids.  The dissociation reaction of an acid in water looks like this: $\ce{HA + H2O &amp;lt;=&amp;gt; A- + H3O+}$.  There's an equilibrium constant &lt;em&gt;K&lt;/em&gt; that's related to the free enthalpy $\Delta G$ in the usual manner:&#xA;$$\Delta G=-RT \cdot \ln K$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weak acid is simply an acid where the equilibrium lies far to the left.  A strong acid, on the contrary is an acid where the equilibrium lies far to the right, i.e. it is more acidic than the $\ce{H3O+}$ ion.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T08:42:07.827" LastActivityDate="2015-04-26T08:42:07.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17345" PostTypeId="2" ParentId="17342" CreationDate="2014-10-11T19:42:01.677" Score="1" Body="&lt;p&gt;I think you meant that weak acids don't break down in water... you see, and acid is usually considered strong if a lot of it ionizes and donate it's protons (H+) to the water in which it's dissolved. While a base is considered strong depending on it's capacity to protonate (take the H+ from the water leaving OH- on the solution).&#xA;This sort of behavior is defined by the affinity between the ions (that are in the acid/base) and the H+.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more complete definitions for acids and bases that don't necessarily require to consider it in a solution with water, but if your teacher skimmed over more basic definitions, then it's not likely that you will study it unless you get General or Inorganic Chemistry classes in college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in this sort of stuff you can always get a chemistry book on the library. I used &quot;Chemistry The Central Science&quot; by Brown during my General 'n Inorganic Chemistry classes on my first semester on college and I think it is a book that would be easily understood by someone who's still in high school given you put some effort in it.&lt;/p&gt;&#xA;" OwnerUserId="8274" LastActivityDate="2014-10-11T19:42:01.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18338" PostTypeId="1" CreationDate="2014-10-11T20:27:39.023" Score="0" ViewCount="2490" Body="&lt;p&gt;I know chemistry has been around for a long time, but I was curious who the first chemist in recorded history. I am aware that &lt;a href=&quot;http://en.wikipedia.org/wiki/Antoine_Lavoisier&quot; rel=&quot;nofollow&quot;&gt;Antoine Lavoisier&lt;/a&gt; was one of the first modern chemistry and is considered to be the &quot;Father of Modern Chemistry&quot;, but was he is the first chemist? &lt;/p&gt;&#xA;" OwnerDisplayName="user9273" LastEditorDisplayName="user9273" LastEditDate="2014-10-11T21:52:48.537" LastActivityDate="2014-10-12T02:35:18.473" Title="Who Was the First Chemist?" Tags="&lt;history-of-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2014-10-12T16:04:40.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="18339" PostTypeId="2" ParentId="17329" CreationDate="2014-10-11T21:47:29.443" Score="2" Body="&lt;p&gt;The structure is misleadingly drawn.  The longest carbon chain is &lt;em&gt;not&lt;/em&gt; this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*} &#xA;\color{red}{\ce{CH3-CH2-CH2 -}}&amp;amp;\color{red}{\ce{CH-CH3}}\\&#xA;&amp;amp;|\\&#xA;\ce{&amp;amp;CH2-CH2-CH3}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*} &#xA;\color{green}{\ce{CH3-CH2-CH2 -}}&amp;amp;\color{green}{\ce{CH}}\ce{-CH3}\\&#xA;&amp;amp;\color{green}{|}\\&#xA;&amp;amp;\color{green}{\ce{CH2-CH2-CH3}}&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 1, 2, 3, 4, 5, 6... 7 carbons.  Thus, this is a substituted heptane.  The single remaining carbon forms a methyl group, which is attached to the fourth carbon (from either end) in the heptane chain.  Thus, 4-methylheptane.&lt;/p&gt;&#xA;" OwnerUserId="283" LastEditorUserId="4934" LastEditDate="2014-10-11T23:39:03.253" LastActivityDate="2014-10-11T23:39:03.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18340" PostTypeId="1" CreationDate="2014-10-11T22:09:28.077" Score="1" ViewCount="298" Body="&lt;p&gt;I'm given $\ce{CoCl2*6H2O}$ as well as $\ce{Na3PO4*12 H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical formula for Cobalt Violet is $\ce{Co3(PO4)2}$. I haven't done stoichiometry in a long time but I think that's creatable, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some website says that I need $\ce{NaH2PO4·10H2O}$, instead of the Sodium phosphate dodecahydrate which is what I have.&lt;/p&gt;&#xA;" OwnerUserId="9274" LastEditorUserId="7499" LastEditDate="2014-10-11T22:45:06.960" LastActivityDate="2014-10-11T23:18:10.633" Title="Can I make Cobalt Violet with these materials?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18341" PostTypeId="2" ParentId="18340" CreationDate="2014-10-11T22:51:43.353" Score="1" Body="&lt;p&gt;The dihydrogen phosphate anion is actually a weak acid and will dissociate to make both $\ce{HPO4^{2-}}$ and $\ce{PO4^{3-}}$ when dissolved in solution. If it works with the dihydrogen phosphate salt, it should work with the hydrogen phosphate and phosphate salts, too. (After all, it's the phosphate anion that precipitates out with the cobalt, if your formula is correct.)&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-11T22:51:43.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18342" PostTypeId="2" ParentId="18340" CreationDate="2014-10-11T23:18:10.633" Score="1" Body="&lt;p&gt;My best guess that pH control is important - at basic pH (and $\ce{Na3PO4}$ is pretty basic) you will precipitate cobalt hydroxide as well.  To prevent that, the recipe calls for $\ce{NaH2PO4}$ to provides a lower pH.  Cobalt phosphate is so insoluble that it will precipitate even in somewhat acidic medium.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-11T23:18:10.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18343" PostTypeId="1" CreationDate="2014-10-11T23:39:58.267" Score="1" ViewCount="102" Body="&lt;p&gt;In my chemistry lab, we had 18 different reactions involving many different substances. One of the experiments involved placing a piece of copper metal in hydrochloric acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While observing it for a minute, my partner and I did not notice any reaction taking place. However, a minute was all we were allowed, and we could not keep it longer to observe it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was the only experiment that looked like it had no reaction, so I felt a little uneasy putting it down. (I didn't do anything wrong, but I guess I was unsure with myself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen some sources say that no reaction is the current result. However, I have also seen some sources saying that it does react, via $\ce{Cu + 2HCl -&amp;gt; CuCl2 + H2}$, yielding cupric chloride and dihydrogen gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct result? If no reaction is correct, what would be the correct chemical reaction for this case?&lt;/p&gt;&#xA;" OwnerUserId="9276" LastEditorUserId="5026" LastEditDate="2018-04-11T02:48:08.637" LastActivityDate="2018-04-11T02:48:08.637" Title="Does copper react with hydrochloric acid?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-10-12T16:05:00.827" ContentLicense="CC BY-SA 3.0" />
  <row Id="18344" PostTypeId="2" ParentId="18338" CreationDate="2014-10-11T23:58:09.263" Score="3" Body="&lt;p&gt;That depends how you define &quot;chemist.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today we think of chemists as working at the molecular level. However Lavoisier didn't have any concrete evidence of an atom's existence. He worked with Sulfates (SO4-2), metal oxides, and other molecular groups in an empirical an algebraic fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout much of the 1800s there was enormous debate about the very existence of atoms. Much of the problem arose from confusion over molecular weights and stoichiometry. Physicists were the most stubborn opponents of atoms, which we still can't fully explain even today. Eventually Advodagro's number and accepted stoichiometry combined to form irrefutable evidence of molecules. Many scientists were completely convinced that atoms existed long before they were proven though. Dalton was the first to introduce stoichiometry so he would be a good candidate as well. &lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2014-10-11T23:58:09.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18346" PostTypeId="2" ParentId="14312" CreationDate="2014-10-12T00:08:29.730" Score="4" Body="&lt;p&gt;For main group elements (groups 1, 2, 13-18), the highest possible oxidation state $O_\text{max}$ is given by the group number $G$ (for groups 1 and 2) or by the group number - 10 (for the rest):&#xA;$$ O_\text{max} = \begin{cases} G &amp;amp;\text{if } G \in \{1,2\} \\&#xA;G - 10 &amp;amp;\text{if } 13 \leq G \leq 18 \end{cases} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for your compound $\ce{\stackrel{VI}{X}O3}$, so with the oxidation state VI, the corresponding group would be group 16.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-12T00:08:29.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18347" PostTypeId="2" ParentId="46" CreationDate="2014-10-12T01:00:30.430" Score="1" Body="&lt;p&gt;The textbook example are crystals isotropized over geologic timescales through the effect of ionizing radiation.  Heat them up, to overcome the activation barrier, and they release the accumulated energy and turn crystalline again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant keyword is &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Wigner_effect&quot; rel=&quot;nofollow&quot;&gt;Wigner effect&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-12T01:00:30.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18348" PostTypeId="1" AcceptedAnswerId="18351" CreationDate="2014-10-12T02:30:14.760" Score="3" ViewCount="164" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Toothpaste#Alteration_of_taste_perception&quot; rel=&quot;nofollow&quot;&gt;Wikipedia states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After using toothpaste, orange juice and other juices have an unpleasant taste. This effect is attributed to products of the chemical reaction between stannous fluoride in toothpaste and the acetic acid in the juices. Sodium lauryl sulfate alters taste perception. It can break down phospholipids that inhibit taste receptors for sweetness, giving food a bitter taste. In contrast, apples are known to taste more pleasant after using toothpaste.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The primary reference for this appears to be &lt;a href=&quot;http://www.newton.dep.anl.gov/askasci/chem00/chem00216.htm&quot; rel=&quot;nofollow&quot;&gt;a comment on the US Department of Energy website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://www.sbs.com.au/news/article/2014/10/10/why-does-toothpaste-make-orange-juice-taste-so-weird&quot; rel=&quot;nofollow&quot;&gt;several&lt;/a&gt; &lt;a href=&quot;http://www.sciencemadesimple.co.uk/activities/oranges_and_toothpaste&quot; rel=&quot;nofollow&quot;&gt;references&lt;/a&gt; to this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be an untested hypothesis. My question is &lt;strong&gt;Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2014-10-12T02:58:20.050" Title="Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?" Tags="&lt;taste&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18349" PostTypeId="2" ParentId="18338" CreationDate="2014-10-12T02:35:18.473" Score="1" Body="&lt;p&gt;The first chemist? The first person who learned that if you mix crushed plant root with the runoff from fermented vegetation (alcohol) you got something you could use to make your robe a bit more colourful? If he experimented with various plants and liquid ratios he would meet the modern definition of &quot;chemist&quot;. I recall doing similar things in grade 9 Chemistry class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That individual is long lost in the mists of time. The main problem with history is we only know what people bothered to write down, or left behind in a durable form. In the days before patents or research for it's own sake all this would have been closely guarded secrets recorded (if at all) in shorthand on small pieces of whatever passed as paper in those days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first chemistry research paper was possibly found in the Great Library of Alexandria, which unfortunately burned down on more than one occasion.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-10-12T02:35:18.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18350" PostTypeId="2" ParentId="17337" CreationDate="2014-10-12T02:50:51.227" Score="2" Body="&lt;p&gt;You could distill the waste easily enough, but the distillation gases will be rather nasty. You will also need a condenser to cool everything down again or the environmental penalties (and corrosion damage to the exhaust system) will be a lot more than the gold value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the waste boiled down (literally) into a small flask you could use any of the known gold-extraction methods to get it out in elemental form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At current market prices you have about US$580 of gold in the drum. No idea how much you will spend in fuel and cooling water to recover it but it will be a substantial percentage. Why don't you just tell &quot;Someone&quot; that they can either pay for the gold they dumped or mind the still themselves over the weekend (and pay for the propane). This will also encourage others to pay more attention to what is in the beaker.&lt;/p&gt;&#xA;" OwnerUserId="5692" LastActivityDate="2014-10-12T02:50:51.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18351" PostTypeId="2" ParentId="18348" CreationDate="2014-10-12T02:58:20.050" Score="1" Body="&lt;p&gt;There may be more, but &lt;a href=&quot;http://chemse.oxfordjournals.org/content/5/4/317.abstract&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is one paper examining the effects of membrane-modifying compounds on taste receptors. It suggests that substances like sodium lauryl sulfate and gymnemic acid affect taste response by altering taste receptors' lipid membranes. Though many anionic surfactants have sulfate functional groups, not all compounds with sulfate groups act as surfactants (nor does the sulfate anion on its own). It sounds like this is something that any compound that can perturb the structure of a taste receptor's membrane can do, not something that necessarily applies to all sulfates.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-12T02:58:20.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18352" PostTypeId="1" AcceptedAnswerId="18354" CreationDate="2014-10-12T03:32:05.633" Score="5" ViewCount="573" Body="&lt;p&gt;How do you measure dissolved oxygen reliably in a culture of cells? This page has info on one sensor: &lt;a href=&quot;http://water.me.vccs.edu/concepts/domeasure.html&quot; rel=&quot;nofollow&quot;&gt;http://water.me.vccs.edu/concepts/domeasure.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how would you know that this sensor is working - meaning that only dissolved oxygen triggers the electrical change in the sensor, and not other atoms? and how would you calibrate the sensor to translate changes in electric signal to the change in number of dissolved oxygen atoms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?   &lt;/p&gt;&#xA;" OwnerDisplayName="user9277" LastEditorUserId="7499" LastEditDate="2014-10-12T05:15:20.053" LastActivityDate="2014-10-12T05:15:20.053" Title="how to measure dissolved oxygen?" Tags="&lt;electrochemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18353" PostTypeId="1" CreationDate="2014-10-12T04:28:43.620" Score="3" ViewCount="307" Body="&lt;p&gt;A match box lighter strip typically consists of powdered glass and red phosphorous. When this part is burned, it's residue is like a sticky dust that, when applying friction, will release smoke or vapor. You can do it with you fingers and it looks really cool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the chemistry behind this?&lt;/p&gt;&#xA;" OwnerUserId="9279" LastEditorUserId="10185" LastEditDate="2015-10-02T16:51:13.147" LastActivityDate="2015-10-02T16:51:13.147" Title="What are the product(s) of red phosphorous when burned?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18354" PostTypeId="2" ParentId="18352" CreationDate="2014-10-12T05:14:40.793" Score="4" Body="&lt;p&gt;To start, the figure and explanation in the link are almost comically misleading. Potassium chloride solution does not &quot;attract&quot; oxygen. The oxygen flowing through the wire is pretty silly and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where the probe joins the wire, oxygen mingles with the electricity. &#xA;  Oxygen is not very ionized, meaning that it does not have a negative&#xA;  charge as electricity does, so the oxygen dilutes the current at the&#xA;  electrode beyond the probe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is just plain ridiculous. &lt;a href=&quot;http://nett21.gec.jp/CTT_DATA/WMON/CHAP_4/html/Wmon-099.html&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; one is better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way this type of probe works is by amperometric reduction of $\ce{O2}$, the first example being the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clark_electrode&quot; rel=&quot;noreferrer&quot;&gt;Clark electrode&lt;/a&gt;. Essentially, an electrolyte is separated from the measurement solution by a porous film of poly(tetrafluoroethylene) (Teflon). These pores are small enough that only small substances like oxygen can diffuse through. On the electrolyte side, there are two electrodes in one of two possible configurations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the original Clark-type electrode, the working electrode is platinum, which is chemically-inert but catalyzes the reduction of oxygen. The counter electrode is a AgCl-coated Ag electrode that acts as a reference for the measurement. (Its potential is supposed to remain constant) To reduce the oxygen, an external potential must be applied across the two electrodes (normally using a potentiostat).&lt;/li&gt;&#xA;&lt;li&gt;The galvanic cell–type electrode has one electrode made of silver (where the oxygen reduction occurs) and one made of an easily oxidized metal like lead. It works the same way as the Clark electrode, but instead of applying an external potential, the current required to reduce the oxygen is supplied by the oxidation of the lead electrode.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any oxygen that reaches the working electrode is reduced and produces a current that is measured to produce the analytical signal. The amount of current produced depends on how much oxygen diffuses through the membrane, which depends on the concentration of dissolved oxygen in the test solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, for your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how would you know that this sensor is working - meaning that only&#xA;  dissolved oxygen triggers the electrical change in the sensor, and not&#xA;  other atoms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is a bit tricky. You're right that oxygen isn't the only thing that could produce a signal—amperometry isn't a very selective technique on its own. You can control the redox reactions that occur at the working electrode, to a limited degree, by changing the potential you apply, i.e. different substances require different potentials to be reduced and won't be reduced if an insufficient potential is applied. In the case of this specific type of electrode, selectivity comes mainly from the PTFE membrane. Since it limits diffusion to only a few very small molecules, that greatly increases the selectivity of the electrode, though depending on the solution you're trying to measure, there still may be interferences.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and how would you calibrate the sensor to translate changes in&#xA;  electric signal to the change in number of dissolved oxygen atoms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In principle, you could calculate the number of reduction reactions happening at the surface from &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis&quot; rel=&quot;noreferrer&quot;&gt;Faraday's laws&lt;/a&gt;, but to turn this into a measure of the exact concentration in solution requires a lot of thought of how the transport of oxygen to the electrode occurs, from diffusion from the electrolyte to the electrode surface and transport from the test solution into the electrolyte. In practice, the assumption that the measured current is directly proportional to the concentration in solution is made (and is valid under many conditions), and a calibration is made using a standard solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;also, how would one get pure dissolved oxygen in varying amounts as 'positive control' for the sensor?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calibration solutions &lt;a href=&quot;http://www.swrcb.ca.gov/water_issues/programs/swamp/docs/cwt/guidance/311.pdf&quot; rel=&quot;noreferrer&quot;&gt;seem&lt;/a&gt; to be commonly made using tables of oxygen saturation at different temperatures. It's also possible to measure dissolved oxygen through other means, such as titration. Solutions can be made up, titrated, then the value used to calibrate an electrode.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, for the measurement specifically in cell cultures, a membrane electrode may not be the best bet, at least not without sample pre-treatment. The PTFE membrane is quite susceptible to adsorption of proteins and the like, which will probably affect the electrode's response. It may be necessary to somehow remove these contaminants (without affecting the dissolved oxygen concentration). Other options that might be viable are colourimetric measurements or titration, though there are problems with these, as well.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-12T05:14:40.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18355" PostTypeId="1" AcceptedAnswerId="18356" CreationDate="2014-10-12T05:55:41.083" Score="4" ViewCount="8780" Body="&lt;p&gt;Maybe I am wrong, but I thought acids were covalently bonded since hydrogen doesn't form ionic compounds. For example, wouldn't $\ce{HCl}$ look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/evtWN.png&quot; alt=&quot;structure of HCl&quot;&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the electrons are shared, why does it break into $\ce{H+}$ and $\ce{Cl-}$ in solution?&lt;/p&gt;&#xA;" OwnerUserId="9281" LastEditorUserId="5456" LastEditDate="2014-10-12T06:12:02.447" LastActivityDate="2015-09-07T20:44:47.743" Title="Why do acids ionize in solution if they are covalently bonded?" Tags="&lt;acid-base&gt;&lt;bond&gt;&lt;aqueous-solution&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18356" PostTypeId="2" ParentId="18355" CreationDate="2014-10-12T06:24:23.853" Score="6" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kpcXu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it is energetically favourable ($\Delta{G}&amp;lt;0$) for hydrogen chloride to react with water to form hydronium( $\ce{ H3O+}$) and chloride ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that $\ce{ H+}$ does not exist as $\ce{ H+}$ in water, but rather as $\ce{ H3O+}$.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-12T06:24:23.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18357" PostTypeId="1" AcceptedAnswerId="18359" CreationDate="2014-10-12T06:50:55.330" Score="7" ViewCount="419" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following compound doesn't contain peroxy oxygen?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{CrO5}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{Cr2O3}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{Na2O2}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{BaO2}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think peroxy oxygen is peroxide, am I right?. I have read in &lt;a href=&quot;https://chemistry.stackexchange.com/a/16036/5456&quot;&gt;this&lt;/a&gt; answer that there is no direct way to recognise peroxide oxygen, and I think that's the reason no one has taught me that in the school. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with many peroxides like $\ce{Na2O2}$, $\ce{BaO2}$, etc. But I not able to recognise between 1 and 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way for this type of question to recognise peroxide oxygen?  And if there is no way to determine then really I need to remember all, because we can't use X-ray crystallography in exam or while studying high school chemistry!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Answer given is 2. $\ce{Cr2O3}$&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="46306" LastEditDate="2018-02-27T10:33:07.117" LastActivityDate="2018-02-27T10:33:07.117" Title="How to determine peroxy oxygen?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18358" PostTypeId="2" ParentId="18355" CreationDate="2014-10-12T06:56:45.993" Score="1" Body="&lt;p&gt;Your are right. But you have to remember that the bond between hydrogen and chlorine (for example) is covalent polar one. When you introduce this molecule into water which is a polar solvent with a dipole moment of 1.85 D, the polarization of the molecule $\ce{HCl}$ increases. We  have then two solvated ions  $\ce{H_3O^+}$  and and  $\ce{Cl^-}$. And  as water has also a high dielectric constant  $\epsilon_r=80$, the electrostatic force between  $\ce{H_3O^+}$ and $\ce{Cl^-}$ will decrease by a factor of 80. So, the the bond between $\ce{H_3O^+}$ and $\ce{Cl^-}$ will weaken  and we have ionization of the acid.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-09-07T20:44:47.743" LastActivityDate="2015-09-07T20:44:47.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18359" PostTypeId="2" ParentId="18357" CreationDate="2014-10-12T07:01:18.537" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have become familiar with many peroxide like Na2O2, BaO2, etc. But not able to recognise between 1 and 2. Is there any way for this type of question to recognise peroxide oxygen? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. In the exam, if you do not know whether (1) or (2) contains peroxy, simply look at the charges of Chromium if all the oxygen is  in the -2 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{CrO5}$, if all the oxygen have a charge of -2, then Cr would have to have a charge of +10, which is impossible, therefore you know that some of the oxygen must have a charge of -1(peroxy). On the other hand, for $\ce{Cr2O3}$, Cr has a charge of +3, which is possible (Cr(III) salts are plenty)&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-12T07:01:18.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18360" PostTypeId="1" CreationDate="2014-10-12T09:30:59.020" Score="3" ViewCount="113" Body="&lt;p&gt;How is the thermal stability of nitric acid related to its planar structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In textbooks it is written that nitric acid is unstable at room temperature and decomposes where $\ce{H2SO4}$ and $\ce{H3PO4}$ are stable and decompose at much higher temperature.$\ce{H2SO4}$ and $\ce{H3PO4}$ have tetrahedral structures.&lt;/p&gt;&#xA;" OwnerUserId="9282" LastEditorUserId="73" LastEditDate="2014-10-12T10:26:26.693" LastActivityDate="2014-10-12T10:26:26.693" Title="How is the thermal stability of nitric acid related to its structure?" Tags="&lt;stability&gt;&lt;vsepr-theory&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18361" PostTypeId="1" AcceptedAnswerId="18366" CreationDate="2014-10-12T10:28:34.550" Score="1" ViewCount="959" Body="&lt;p&gt;For a product to be called &quot;vinegar&quot;, it must contain at least $5.0$% acetic acid by mass. A 10.0 g sample of a &quot;raspberry vinegar&quot; is titrated with $0.125 \mathrm{M}$ barium hydroxide and required $37.50 \mathrm{mL}$ for complete neutralization. Can the product be called &quot;vinegar&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor said that the product is vinegar, but I did not understand how he solved it. Could you please tell me what to do? What formulas should I use to solve this kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="7921" LastEditorUserId="5456" LastEditDate="2014-10-12T12:07:25.193" LastActivityDate="2014-10-12T22:58:36.110" Title="Can the product be called &quot;vinegar&quot;?" Tags="&lt;everyday-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18362" PostTypeId="1" AcceptedAnswerId="18364" CreationDate="2014-10-12T12:30:34.933" Score="4" ViewCount="459" Body="&lt;p&gt;Can Grignards be chiral? I've seen them drawn as ionic structures in class. I've also seen little asterisks on certain positions of Grignards (a common designation of chirality). So this begs the question - are Grignards chiral? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand Grignards are best described as polar covalent rather than ionic, but that still leaves one more question: can ionic compounds be chiral? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-12T14:35:24.750" Title="Are Grignards Chiral?" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18363" PostTypeId="1" CreationDate="2014-10-12T14:01:34.493" Score="14" ViewCount="24327" Body="&lt;p&gt;Why is sand that insoluble?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sand is a polar molecule but does not dissolve in water.&#xA;Is it because of the big structure that sand makes?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="4934" LastEditDate="2014-10-12T14:06:24.403" LastActivityDate="2019-03-14T19:17:15.300" Title="Why is sand that insoluble?" Tags="&lt;solubility&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18364" PostTypeId="2" ParentId="18362" CreationDate="2014-10-12T14:35:24.750" Score="3" Body="&lt;p&gt;I read the question as asking whether a stereocenter bearing a C-Mg bond will keep its configuration. The answer is yes, but not indefinitely. Unfunctionalized (alkane) Grignards of this type racemize/epimerize with a half life on the order of hours. Specific compounds will have more or less stability. By analogy to organolithium compounds, I would expect having Lewis basic atoms in the proximity to stabilize the configuration. I'm not sure if this is still an active field of research, but it certainly was for a long time. Try searching &quot;configurational stability of Grignard reagents&quot; or similar search terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an additional complication with these chiral Grignard reagents: they can't be prepared by the usual method of oxidative addition of Mg(0) to an alkyl halide. The mechanism of that reaction goes by a single electron transfer, through a radical intermediate, destroying any stereocenter. Chiral Grignards have to be synthesized using less direct methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: Side Reactions in Organic Synthesis: A Guide to Successful Synthesis Design, pp. 200-202.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-12T14:35:24.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18365" PostTypeId="1" CreationDate="2014-10-12T14:49:13.887" Score="-2" ViewCount="796" Body="&lt;p&gt;Why are $\ce{Fe}$ and $\ce{Ni}$ ferromagnetic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe}$ has metallic bonds, which is the electrostatic attraction between the positive metal ions and delocalized electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it due to these charges, as well as with ions and the &quot;sea of electrons&quot; that makes the metal ferromagnetic?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="15410" LastEditDate="2015-05-19T06:56:26.210" LastActivityDate="2015-05-19T06:56:26.210" Title="Why are some metals ferromagnetic?" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18366" PostTypeId="2" ParentId="18361" CreationDate="2014-10-12T14:51:40.813" Score="2" Body="&lt;p&gt;0.125M of $\ce {Ba(OH)2}$ means 0.125 moles of $\ce {Ba(OH)2}$ per volume unit (so far I've only seen liters been used as volume unit) . &#xA;Assuming that all the $\ce {Ba(OH)2}$ you used was consume on the reaction with the acetic acid, and knowing the ratio of $\ce {Ba(OH)2}$ and acetic acid on the reaction is 1/2 you`d have this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce {Ba(OH)2 + 2CH3CO-OH -&amp;gt; Ba(CH3COO)2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to do is calculate the amount of acetic acid consumed. Knowing the amount of acetic acid consumed based on the amount of moles of $\ce {Ba(OH)2}$ that are present on 37.5mL. Then you need to get the molar mass of acetic acid  and calculate the amount of mass of acetic acid you had. Then divide the total mass of vinegar by the mass of acetic acid  you found, multiply it by 100 and you'll have the mass percentage of acetic acid. &lt;/p&gt;&#xA;" OwnerUserId="8274" LastEditorUserId="73" LastEditDate="2014-10-12T22:58:36.110" LastActivityDate="2014-10-12T22:58:36.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18367" PostTypeId="2" ParentId="18365" CreationDate="2014-10-12T15:20:42.740" Score="4" Body="&lt;p&gt;Now this is a really complex subject... first, magnetism is actually divided on a bunch of &quot;classes&quot;, some elements are diamagnetic and will be repelled by a magnetic field, other are paramagnetic and are attracted by the field, and there are the ferromagnetic elements like iron, nickel and others, that become permanent magnetics upon being exposed to a mag. field, other &quot;classes&quot; also exist but NVM them.&#xA;This behaviors are defined by subatomic interactions on electron fields, and charges in the nucleus, not so subatomic interactions between the molecules and the crystals they form and this sort of stuff. If you want an explanation on these topics I'd suggest you to take 6 minutes to watch the &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=hFAOXdXZ5TM&quot; rel=&quot;nofollow&quot;&gt;Magnets: how do they work?&quot;&lt;/a&gt; from minute physics, it's a pretty cool short video, and this channel has a lot of other cool stuff on this stuff.&lt;/p&gt;&#xA;" OwnerUserId="8274" LastActivityDate="2014-10-12T15:20:42.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18368" PostTypeId="2" ParentId="18363" CreationDate="2014-10-12T15:40:16.477" Score="2" Body="&lt;p&gt;By sand I'd guess you mean $\ce {SiO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see, sand or $\ce {SiO2}$ for what it matters, don't appear in nature as single molecules, they're crystals and the crystal arrangement defines if it will be quartz, sand, silica-gel and other stuff, and this crystalline structure also defines it's solubility in water.&#xA;So even if a single molecule of $\ce {SiO2}$ could be polar due to the electronegativity difference between Si and O, the sand won't be exactly polar because they only hang together as crystals, and so the polarity will vary as a function of it's structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I never heard of single $\ce {SiO2}$ molecules in labs neither.&lt;/p&gt;&#xA;" OwnerUserId="8274" LastEditorUserId="4934" LastEditDate="2014-10-12T15:42:28.490" LastActivityDate="2014-10-12T15:42:28.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18369" PostTypeId="1" CreationDate="2014-10-12T15:47:15.050" Score="1" ViewCount="86" Body="&lt;p&gt;Several month ago, I soundproofed the door of my room with two sheets of lead. Can it be dangerous to live with them for many years? Can lead &quot;decay&quot; in some way and be inhaled? (and what happen when it oxidizes?)&lt;/p&gt;&#xA;" OwnerUserId="9284" LastEditorUserId="5017" LastEditDate="2014-10-12T18:40:33.263" LastActivityDate="2014-10-12T18:40:33.263" Title="Is it dangerous to live near lead?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;materials&gt;&lt;toxicity&gt;&lt;metallurgy&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2014-10-13T13:55:38.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="18370" PostTypeId="1" CreationDate="2014-10-12T16:36:02.077" Score="6" ViewCount="81286" Body="&lt;p&gt;I used to use the explanation that $s$ orbitals penetrate better than $p$ orbitals, however, could the explanation be that $3s$ is shielding the $3p$ in aluminum?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="9961" LastEditDate="2015-10-23T16:41:19.117" LastActivityDate="2017-07-06T08:36:39.373" Title="Why does aluminum have a lower first ionization energy than magnesium?" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;&lt;ionization-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18371" PostTypeId="2" ParentId="18363" CreationDate="2014-10-12T16:58:58.993" Score="6" Body="&lt;p&gt;There are many reasons, due to which it is insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;High lattice energy:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to polar interaction with water, it get suspended but remain undissolved due to high lattice energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;size of sand particle:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of sand particle is bigger than $0.3 \mathrm{nm}$(Again as commented it depend on type of sand) which is difficult to dissolve. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some things which are bigger than 0.3 nm can squeeze inside and cause unfavourable entropic effects but significantly more favourable than being at the surface or slightly more than being in normal solution. &#xA;  However, sand particles are significantly larger and entropically it prefers being suspended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20130701235540AAc2lG2&quot;&gt;Yahoo&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Because of bond:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sand is not able to dissolve in water, because water can not break bonds between silica (sand / $\ce{SiO2}$ / Silicon dioxide)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sand is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(&lt;a href=&quot;http://www.chemguide.co.uk/atoms/structures/giantcov.html&quot;&gt;Chemguide&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. sand is a non-polar molecule.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-10-12T17:57:15.313" LastActivityDate="2014-10-12T17:57:15.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18372" PostTypeId="2" ParentId="18363" CreationDate="2014-10-12T17:06:13.143" Score="14" Body="&lt;h1&gt;What is Sand?&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Scope of this Answer&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In order to give a good answer, we first have to decide on the type of sand we want to talk about, because its &lt;a href=&quot;http://en.wikipedia.org/wiki/Sand#Composition&quot; rel=&quot;noreferrer&quot;&gt;composition&lt;/a&gt; is highly dependent on that. Is it beach sand? If so, where from? All these questions should have been addressed in the question. I will limit the scope of this answer to pure silicium dioxide $\ce{SiO2}$ sand.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Is Sand a Molecule?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Sand is, strictly speaking, not a molecule. It's a crystal. As such it has properties such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lattice_energy&quot; rel=&quot;noreferrer&quot;&gt;lattice energy&lt;/a&gt;. Maybe have a look at that link, it might clarify it all for you already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;So now we have talked about what we want to talk about and clarified the basics, so let's dive right in, shall we?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Why does $\ce{SiO2}$ not Dissolve in Water?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Basically this is an energy calculation. On one hand you have the crystal, with its lattice energy, and on the other hand you have the solvated $\ce{SiO2}$ units. A tentative reaction equation would look like this:&#xA;$$ \ce{Si_{n} O_{2n} &amp;lt;&amp;lt;=&amp;gt;[E_\text{lat}] SiO2 &amp;lt;=&amp;gt;[E_\text{solv}] SiO2(aq)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energies I've indicated are the lattice energy $E_\text{lat} = 12535~\mathrm{kJ\,mol^{-1}}$ (&lt;a href=&quot;http://database.iem.ac.ru/mincryst/s_carta.php?QUARTZ%203895&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;) and the solvation energy $E_\text{solv}$ (see figure below, where I've also included sodium chloride as a substance to compare with). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BUPOI.png&quot; alt=&quot;Solvation Energy of Quartz and NaCl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data from &lt;a href=&quot;https://www.physics.gatech.edu/files/u26/papers/0256.pdf&quot; rel=&quot;noreferrer&quot;&gt;here (PDF)&lt;/a&gt; ($\ce{SiO2}$) and &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/100/5/10.1063/1.466363&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; ($\ce{NaCl}$). It is obvious that the solvation of $\ce{SiO2}$ monomers is highly unfavored (if I got all the sign conventions right, that is...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, for ease of understanding, imagine that the $\ce{SiO2}$ unit in the middle has to decide on which side it wants to go. Either it forms a crystal with other units (to the left) or it gets solvated by (surrounded by molecules of) water. Which way will it go? We already know that it will take the route to the left. This means that the whole system can release more energy this way. That is why $\ce{SiO2}$ crystals are stable in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or somewhat. Some &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp983239z&quot; rel=&quot;noreferrer&quot;&gt;theoretical investigations&lt;/a&gt; point towards the possibility that the surface of quartz crystals undergo chemisorption of water, thereby generating $\ce{SiOH}$ terminal groups, which would then generate more favorable interactions between the crystal surface and the surrounding water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to (a part of) your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it because of the big structure that sand makes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are kind of on the right track here. &lt;strong&gt;It is because of the lower energetical state as a crystal that these big structures (crystals) are formed, vs. the units just being dissolved (solvated by water).&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-12T17:06:13.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18373" PostTypeId="1" CreationDate="2014-10-12T18:11:25.413" Score="0" ViewCount="477" Body="&lt;p&gt;The structure of an atom is well known to me as something like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CdfdD.png&quot; alt=&quot;Structure of atom&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;But what is the accurate structure of the &lt;strong&gt;Nucleus&lt;/strong&gt; ? What is the arrangement of the &lt;em&gt;Protons&lt;/em&gt; and the &lt;em&gt;Neutrons&lt;/em&gt; or any of the other nucleons? I will be glad if anybody give an image of its structure.&lt;/p&gt;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-10-12T18:44:14.410" Title="What is the structure of the Nucleus?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="0" ClosedDate="2014-10-12T21:08:54.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="18374" PostTypeId="2" ParentId="16950" CreationDate="2014-10-12T18:12:22.450" Score="0" Body="&lt;p&gt;I think you merely grew some Iron Sulfate crystals and then dissolved them by adding more solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how much sulfuric acid you added the second time you could add a proportional amount of the same mixture of nails and screws to reform the crystals  except larger. Alternately you could evaporate the acid you added by putting it on a hot plate or something. or as trb said you could just start over. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2014-10-12T18:20:32.383" LastActivityDate="2014-10-12T18:20:32.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18375" PostTypeId="2" ParentId="18373" CreationDate="2014-10-12T18:16:22.680" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The nucleus is the very dense region consisting of protons and neutrons at the center of an atom.&lt;/p&gt;&#xA;&lt;p&gt;The diameter of the nucleus is in the range of 1.75 fm (&lt;span class=&quot;math-container&quot;&gt;$1.75×10^{−15}$&lt;/span&gt; m) for hydrogen (the diameter of a single proton)&lt;a href=&quot;https://i.stack.imgur.com/SslzL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; to about 15 fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about 23,000 (uranium) to about 145,000 (hydrogen)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x7VPm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SslzL.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source:&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_nucleus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-10-12T18:44:14.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18376" PostTypeId="2" ParentId="8663" CreationDate="2014-10-12T18:21:00.083" Score="1" Body="&lt;p&gt;They would advertise $\ce{NaBPh4}$ as a reagent to determine $\ce{K+}$ gravimetrically.  Old-timers may remember the feared Kalignost.  $\ce{XBPh4-}$, for X=$\ce{Rb+, Cs+, NH4+,Tl+}$ is similarly insoluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanksfully, now there is methods like AAS, and we plague the kids with that in Instrumental Analysis/Analytical Chemistry classes instead.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-12T18:27:15.917" LastActivityDate="2014-10-12T18:27:15.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18377" PostTypeId="1" CreationDate="2014-10-12T18:33:44.187" Score="1" ViewCount="141" Body="&lt;p&gt;Static would be where a set amount of both products and reactants is formed. Dynamic would be where both the forward and backward reactions occur at equal rates, meaning no change in the overall product and reactant concentrations.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2014-10-12T21:22:22.910" Title="Are there actually any reactions during which STATIC equilibrium is reached, as opposed to dynamic?" Tags="&lt;equilibrium&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18378" PostTypeId="1" AcceptedAnswerId="18395" CreationDate="2014-10-12T18:45:29.193" Score="4" ViewCount="3269" Body="&lt;p&gt;For example the reaction of ethanoic acid with ethanol is reversible, however is this true for &lt;strong&gt;all&lt;/strong&gt; esters? I'm not quite whether it is as easy for a reversible reaction to occur both ways. Is there a limit to the complexity an ester can be? What would be the general mechanism for these reactions? I assume there is no limit on the complexity (carbon chain length) of esters, other than the frequency of longer-chain carboxylic acids and alcohols needed to form them. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorDisplayName="user7951" LastEditDate="2015-08-15T20:49:41.723" LastActivityDate="2017-06-09T01:59:11.227" Title="Are all esterification reactions reversible? What is their general mechanism?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18380" PostTypeId="1" CreationDate="2014-10-12T19:07:04.733" Score="9" ViewCount="7592" Body="&lt;p&gt;I thought that anything not in a rate equation was automatically zeroth order and therefore did not affect the reaction. However, I have heard that catalysts can be involved in a reaction while not being in a rate equation. How can this be possible?&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="30836" LastEditDate="2017-10-15T14:23:53.127" LastActivityDate="2020-05-24T04:54:31.940" Title="How can a catalyst not be included in a rate equation if, by definition, it speeds up a chemical reaction?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18382" PostTypeId="1" AcceptedAnswerId="18384" CreationDate="2014-10-12T20:35:46.247" Score="1" ViewCount="794" Body="&lt;p&gt;When discussing the solvent uptake of a metal-organic framework would the correct term be &quot;solvent molecules were enclathrated into the pores&quot;? Despite searching for a definition online I have drawn a blank and its not a term I am familiar with. Having read many papers in polymer chemistry where this term is also used I wonder if I have misunderstood its usage.&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7920" LastEditDate="2014-10-12T20:46:21.777" LastActivityDate="2014-10-12T20:48:06.663" Title="Enclathrated - definition?" Tags="&lt;inorganic-chemistry&gt;&lt;materials&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18383" PostTypeId="1" AcceptedAnswerId="18392" CreationDate="2014-10-12T20:44:45.047" Score="2" ViewCount="1048" Body="&lt;p&gt;When I was milling NaOH for some purpose, I noticed that the powder is incredibly hygroscopic. Because I have problems with drying certain chemicals safely (without extensive heating), I thought this might be a safer solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mcOew.png&quot; alt=&quot;drying machine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not so sure about that safety. Can NaOH exit into the air and pollute the compound in the second beaker? If it can, which compounds are safe and strongly hygroscopic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use this technique to increase sulfuric acid concentration? What about acetic acid?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="402" LastEditDate="2017-01-14T08:40:20.323" LastActivityDate="2017-01-14T08:40:42.680" Title="Safe drying agents for drying machine" Tags="&lt;safety&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18384" PostTypeId="2" ParentId="18382" CreationDate="2014-10-12T20:48:06.663" Score="1" Body="&lt;p&gt;The Oxford Dictionary does not have an entry for enclathrate, but does for clathrate: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A compound in which molecules of one component are physically trapped within the crystal structure of another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using the word enclathrate should be redundant.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-12T20:48:06.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18385" PostTypeId="2" ParentId="18383" CreationDate="2014-10-12T20:57:19.660" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can NaOH exit into the air and pollute the compound in the second beaker? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you leave the NaOH as pellets, it is not likely to transfer. The powder might.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could I use this technique to increase sulfuric acid concentration? What about acetic acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No and no. Glacial acetic acid is already 100%. If you could remove water from sulfuric acid, you would get fuming sulfuric acid. The excess $\ce {SO3}$ would dissipate quickly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which compounds are safe and strongly hygroscopic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more hygroscopic the chemical, the more hazardous it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need to dry a solvent, it would be better to follow the information &lt;a href=&quot;http://delloyd.50megs.com/moreinfo/drying.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="402" LastEditDate="2017-01-14T08:40:42.680" LastActivityDate="2017-01-14T08:40:42.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18386" PostTypeId="2" ParentId="18383" CreationDate="2014-10-12T20:58:37.303" Score="0" Body="&lt;p&gt;Sodium hydroxide (potassium hydroxide is even better) is indeed used as a drying agent in the chemical laboratory, as are many other things: sulfuric acid, phosphorus pentoxide, molecular sieves are but a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important consideration is compatibility with the substance to be dried.  You wouldn't even &lt;em&gt;think&lt;/em&gt; of, say drying a stream of chlorine gas with potassium hydroxide, or a propylene stream with sulfuric acid.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-12T20:58:37.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18388" PostTypeId="2" ParentId="18383" CreationDate="2014-10-12T21:52:27.930" Score="0" Body="&lt;p&gt;You want &lt;a href=&quot;http://en.wikipedia.org/wiki/Gypsum&quot; rel=&quot;nofollow&quot;&gt;gypsum&lt;/a&gt; sold here to labs as &lt;a href=&quot;http://en.wikipedia.org/wiki/Drierite&quot; rel=&quot;nofollow&quot;&gt;&quot;Drierite&quot;&lt;/a&gt; in the anhydrous $\ce{CaSO4}$ form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drierite, or anhydrous calcium sulfate, is readily used in labs for drying gases, etc. It's fairly safe (I wouldn't eat it, but it's used as plaster), hygroscopic, and easily re-usable. Generally you mix the plain white form with a little bit of indicator, and then bake at 210 °C to drive off the water before re-use.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-12T21:52:27.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18392" PostTypeId="2" ParentId="18383" CreationDate="2014-10-12T23:09:47.337" Score="9" Body="&lt;p&gt;The first part of your schematic basically describes a &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Air_dryer#Desiccant_dryer&quot;&gt;dessicant dryer&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Air_dryer#Deliquescent_dryer&quot;&gt;deliquescent dryer&lt;/a&gt;. These are common industrial air drying processes that produce very low dew point properties. Generally air (or other inert gas) is passed through a column of desiccant, under increased pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Desiccant dryers use &lt;a href=&quot;http://en.wikipedia.org/wiki/Desiccant&quot;&gt;common desiccants&lt;/a&gt; that include activated alumina, molecular sieves and silica gel. In all cases, there is a maximum amount of water that can be adsorbed/absorbed before the desiccant needs to be regenerated; typically regeneration is done by heating the desiccant column, although other methods, such as purging can be used. Industrial desiccant dryers have the heater element installed as part of the construction, although any bench chemist can probably tell you about heating red silica gel beads from desiccators in the oven until they turn blue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deliquescent dryers are similar, but involve a highly hygroscopic agent that slowly dissolves, or deliquesces, as it absorbs more water. Usually the agent cannot be regenerated, and the waste liquid is drained and discarded. Fresh drying agent then needs to be added. This is most likely the scenario for NaOH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this would need to be a major consideration for your proposed system; how to safely and effectively dry your NaOH. This is a very caustic material to have to try to continually regenerate or replace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I would be concerned that you would transfer NaOH from the first chamber. To check this, you could easily test the gas outlet with some litmus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you use dry air to remove excess water from other materials (such as acids)? In theory, this may be possible and will depend on the relative hygroscopicity of the drying agent and the material to be dried; you need to be sure that the air you are supplying your wet material with is dry enough to receive water and not simply supply more. Purge drying (passing dry air over a desiccant) is another form of regenerating industrial desiccants, but is not as effective as heating. Sulfuric acid is itself used as a desiccating agent as it is very hygroscopic. I don't know the relative hygroscopicity of NaOH and H2SO4/CH3COOH, but there is no way I would ever consider passing air from one container directly into the other chamber. &lt;strong&gt;This is a very hazardous setup, and is fraught with much danger.&lt;/strong&gt; Heating an acid solution in a suitably ventilated environment under controlled conditions is still, to my knowledge, the most effective way to increase the concentration of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also consider the energy efficiencies of trying to 'dry' a sulfuric acid solution by passing dry air over or through the solution versus heating it. The amount of time and materials required to produce sufficient dry air to effectively concentrate a sulfuric acid solution is likely to be far in excess than the time/material investment required to simply heating it. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-10-12T23:55:30.113" LastActivityDate="2014-10-12T23:55:30.113" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18394" PostTypeId="2" ParentId="18370" CreationDate="2014-10-12T23:33:09.030" Score="11" Body="&lt;p&gt;The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for &lt;em&gt;all&lt;/em&gt; the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what their orbital pairings look like: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/e9aSc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/i8l4C.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sources:&#xA;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html&quot; rel=&quot;noreferrer&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_pair&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Electron_pair&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-10-13T00:12:08.053" LastActivityDate="2014-10-13T00:12:08.053" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18395" PostTypeId="2" ParentId="18378" CreationDate="2014-10-12T23:53:28.933" Score="3" Body="&lt;p&gt;I'll start this answer off with saying that both of these answers were the first result when searching for Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/alcohols/esterification.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The esterification reaction is both slow and reversible&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is the mechanism from an image search&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y0oum.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-10-12T23:53:28.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18396" PostTypeId="2" ParentId="18370" CreationDate="2014-10-13T00:22:10.727" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;could the explanation be that 3s is shielding the 3p in aluminum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, one could say that for 2p, 3p and 4p orbitals; it becomes more confusing for 5p and 6p. One can also say that a half filled p orbital (3 electrons) shields the nucleus against the 4th electron.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-13T00:22:10.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18397" PostTypeId="1" CreationDate="2014-10-13T00:22:29.677" Score="6" ViewCount="251" Body="&lt;p&gt;It seems like all the new advances in molecular imaging/microscopy are in the visible light range of the EM spectrum. An example would be the most recent &lt;a href=&quot;http://io9.com/the-2014-nobel-prize-for-chemistry-has-been-awarded-for-1643908437&quot; rel=&quot;nofollow&quot;&gt;Nobel Prize for Chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some reason we are only looking for things in the visible part of the EM spectrum? I'd love to know why, and also what techniques look outside of this range? &lt;/p&gt;&#xA;" OwnerUserId="9291" LastEditorUserId="73" LastEditDate="2014-10-13T11:08:48.053" LastActivityDate="2014-10-14T02:51:16.867" Title="Nanoscopy Nobel Prize - why are all the tools in the visible light range?" Tags="&lt;spectroscopy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18398" PostTypeId="2" ParentId="18397" CreationDate="2014-10-13T00:30:32.013" Score="5" Body="&lt;p&gt;As far as fluorophores go, if the excitation wavelength is too short (&amp;lt; 350 nm), the incident light will get stuck in the microscope.  Near-infrared fluorophores, on the other hand, suffer from several drawbacks, especially low quantum yields from competing non-radiative processes.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-13T01:10:32.423" LastActivityDate="2014-10-13T01:10:32.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18399" PostTypeId="2" ParentId="18383" CreationDate="2014-10-13T08:47:11.583" Score="0" Body="&lt;p&gt;Common drying  chemicals for air dryiers I'm aware of are (anhydrous) copper sulfate and calcium chloride, the second awailable as granules and dirt cheap. Sodium hydroxide reacts with carbon dyoxide, so it's a questionable choice. However in case one need to dry an inert solvent, like benzene or diethyly ester, sodium hydroxide is a valid choice for the first run (with sodium metal for the second run). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the extreeme corner phosphorous pentoxide and sodium metal and metal hydrides lie. They are effective, but dangerous and should not be used as first-stage dryers, as they may react too lively. In addition, some solvents/gases may react with them, so something more inert must be used. This includes silica, molecular sieves and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of drying gases it is common to bubble the gas throw layer of concentrated sulfuric acid, using two safeguard vials before and after the vial with sulfuric acid.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-10-13T08:47:11.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18400" PostTypeId="2" ParentId="18363" CreationDate="2014-10-13T08:54:28.253" Score="2" Body="&lt;p&gt;Surely the correct answer is to apply a variation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthropic_principle&quot; rel=&quot;nofollow&quot;&gt;anthropic principle&lt;/a&gt;: sand can be made of a variety of different crystals, some of which are water soluble and some of which are not. The reason you only see the insoluble crystals when you go to the beach is that the soluble ones were dissolved into the ocean millions of years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at different sand compositions, you'll see that desert sands have a higher proportion of water soluble components. For example, some parts of the Sahara have significant amounts of gypsum. &lt;a href=&quot;http://en.wikipedia.org/wiki/White_Sands_National_Monument&quot; rel=&quot;nofollow&quot;&gt;White Sands&lt;/a&gt; is almost entirely gypsum. But gypsum never makes up more than a trace of beach sand because it dissolves in water. (Indeed, it dissolves better in cold water like the ocean than in warm water.)&lt;/p&gt;&#xA;" OwnerUserId="9293" LastActivityDate="2014-10-13T08:54:28.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18401" PostTypeId="1" CreationDate="2014-10-13T10:25:13.443" Score="5" ViewCount="15249" Body="&lt;p&gt;Carbon electrodes are very popular, because unlike metal electrodes, carbon is quite inert during electrolysis (I'm sure that has some explanation based on electronegativity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However they too are not eternally lasting. Especially the pencil graphite bars will slowly release black chunks into the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what's happening them. The black chunks seem like carbon, so it's probably not reacting. What's damaging the structure?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2014-12-12T02:53:22.430" Title="What's happening to carbon electrodes?" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18402" PostTypeId="1" CreationDate="2014-10-13T11:04:39.453" Score="9" ViewCount="1243" Body="&lt;p&gt;As the universe reaches the maximum entropy, presumably only the most stable chemicals would exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that all the elements haven't turn into iron yet, what kind of chemicals would dominate at that point?&lt;/p&gt;&#xA;" OwnerUserId="9294" LastActivityDate="2015-07-08T19:21:45.343" Title="What kind of chemicals would exist at the heat death of the universe?" Tags="&lt;enthalpy&gt;&lt;stability&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18403" PostTypeId="2" ParentId="18402" CreationDate="2014-10-13T11:52:13.190" Score="10" Body="&lt;p&gt;There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are &lt;strike&gt;expected&lt;/strike&gt; predicted to decay through at least one of several mechanisms, with a half-life &lt;a href=&quot;http://arxiv.org/abs/astro-ph/9701131&quot; rel=&quot;nofollow&quot;&gt;somewhere between the range of $10^{35}$-$10^{200}$ yr&lt;/a&gt; (far shorter than the timescale for nuclear reactions to convert all matter into iron, &lt;a href=&quot;http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.51.447&quot; rel=&quot;nofollow&quot;&gt;~$10^{1500}$ yr&lt;/a&gt;). Therefore, at some point the statistically expected number of atoms inside our observable bubble could potentially dip below 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should proton decay happen relatively quickly, perhaps the last elements in the Universe will be present near the end of the black hole era (~$10^{100}$ yr), when the last supermassive black holes decay in a flash of subatomic particles, possibly creating small amounts of new protons/neutrons. These will decay in a comparative blink of an eye at this timescale, however, and the Universe will soon become a barren place once more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last structures with any semblance to matter in the Universe will likely be &lt;a href=&quot;http://en.wikipedia.org/wiki/Positronium&quot; rel=&quot;nofollow&quot;&gt;positronium &quot;atoms&quot;&lt;/a&gt;, where an electron and a positron (anti-electron) orbit each other in a pseudo-hydrogen atom. Though ground-state positronium decays in a matter of nanoseconds or less, it will be very long-lived in the future due to the extreme sparsity of subatomic particles in the hugely expanded observable bubble, with a mean distance between a bound positron and electron measured in &lt;strike&gt;trillions of trillions of trillions of trillions of...&lt;/strike&gt; billions of lightyears or more. Even in these conditions, positronium too shall eventually decay, in a timescale of anywhere between ~$10^{140}$ yr and ~$10^{800}$ yr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before all of these nuclear processes take place, there may also be a significant amount of unusual chemical reactions happening solely due to tunnelling effects, even at arbitrarily low temperatures, since the kinetic barriers for chemical reactions are far smaller than that for nuclear reactions. For example, most molecules containing nitrogen atoms are thermodynamically unstable with respect to release of $\ce{N2}$, and so might spontaneously break up even at $\mathrm{0\ K}$ given enough time. I don't know what the timescale for these processes is, however.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-07-08T19:21:45.343" LastActivityDate="2015-07-08T19:21:45.343" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18404" PostTypeId="1" AcceptedAnswerId="18411" CreationDate="2014-10-13T14:04:41.137" Score="12" ViewCount="31551" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;My textbook says that a $\ce{C^{4+}}$ cation cannot be formed because it requires a lot of energy to remove 4 electrons. Formation of  ionic bonds involve &quot;removing&quot; electrons and there seems to be enough energy there. So what's different for carbon?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The textbook also mentions that a $\ce{C^{4-}}$ anion cannot be formed because 6 protons cannot hold on to 10 electrons. Elements like chlorine form ionic bonds and end up with 18 electrons and 17 protons. I know there's just one more electron. So is $\ce{NaCl}$ possible because it's not very hard to hold on to 1 extra electron? In that case at what exact point does it become hard to hold onto the extra electrons?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="4923" LastEditDate="2014-10-13T20:52:35.807" LastActivityDate="2019-10-30T15:56:23.360" Title="Why can't carbon form an ionic bond?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18405" PostTypeId="2" ParentId="9743" CreationDate="2014-10-13T14:25:44.787" Score="26" Body="&lt;p&gt;I will try to describe what happens when two hydrogen atoms approach each other from infinity. At infinite separation the hydrogen atoms don't feel their mutual presence and each atom has one electron localized in its atomic 1s orbital. In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and you will basically have an ensemble of systems with an equal amount of triplet (parallel spins, $S=1$) and singlet (antiparallel spins, $S=0$) states, since $E_{\uparrow \uparrow} = E_{\uparrow \downarrow}$. But when hydrogen atoms are brought closer together the situation changes: For one thing each atom has a magnetic moment associated with its electronic spin and there will be a dipolar interaction between those magnetic moments. However this dipolar interaction will be very weak (in the order of $10^{-4} \, \mathrm{eV}$ at close interatomic distance) and won't have much influence. Another factor that comes into the game is spin-orbit coupling. Yet again, since $\ce{H}$ is such a light element spin-orbit coupling will be very, very weak and so it can be ignored too. But the predominant factor to be taken into acount is the exchange energy (its effect is in the order of $0.1$ to $1 \, \mathrm{eV}$). &lt;a href=&quot;http://phycomp.technion.ac.il/~riki/H2_molecule.html&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; you can find a very nice treatment of the exchange energy for the hydrogen molecule which comes to the conclusion that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the hydrogen molecule the exchange energy is negative [...] so the state when the resultant spin is $S=0$ has the lower energy than the state when $S=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This source also provides the potential energy curves for the hydrogen singlet and triplet states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gOSVQ.jpg&quot; height=&quot;500&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These curves amount to the following electron configuration diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BBZNj.png&quot; height=&quot;350&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hhp10.png&quot; height=&quot;350&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the situation for the approaching hydrogen atoms is like this: If both atoms start out with electrons of opposite spin (i.e. the system is in a singlet state) you have the common case that the hydrogen atoms will bond together thus lowering the total energy of the system. But if both atoms start out with parallel spins (i.e. the system is in a triplet state) the total energy will rise when they get closer to one another and the atoms will not bond together but keep their distance. This can be understood from the electron configuration diagram of the triplet hydrogen molecule: The antibonding $\sigma^{*}$ MO is more destabilized than the bonding $\sigma$ orbital is stabilized by the interaction (this is generally true for 2-orbital interactions) and the energy splitting becomes larger the closer the hydrogen atoms get. But because of the Pauli principle the two electrons of like spin in the system can't both occupy the $\sigma$ MO but one is forced to occupy the $\sigma^{*}$ MO and thus the system's total energy rises when the hydrogen atoms approach each other. And because there is virtually no spin-orbit coupling for hydrogen the total angular momentum (represented by $L$) and the total spin (represented by $S$) of the electrons are essentially conserved quantities and there is no mechanism to facilitate intersystem crossing, i.e. the process of changing from the triplet to the singlet state has such a low probability that it virtually doesn't happen. The situation changes of course if you add an external bias to the system, e.g. irradiation with high-energetic light. Then a triplet-singlet transition might be possible but as I read it that was not your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the way MO-diagrams are often pictured with electrons of like spin pairing up in a bonding MO does not reflect the &quot;real&quot; situation of two atoms approaching each other from an infinite distance. But it is not meant that way although it is often pictured as such.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, in summary I established that if you have two hydrogen atoms in free space without spin-orbit coupling and relativistic effects they are stuck with the spin state they initially start out with and can't change it during their time evolution, i.e. they have to stay either on the triplet or the singlet Born-Oppenheimer surface and no switching between those is allowed. But I want to delve a little deeper into the reasons for it. If the hydrogen atoms wanted to change from triplet to singlet state, the atoms would have to transfer $1 \hbar$ of spin angular momentum somewhere else in order to comply with the law of conservation of momentum. In the absence of any particles to collide with the only possible mechanism by which this can be achieved is to emit a photon whose energy is equal to the energy difference between the triplet and the singlet state, i.e. $\Delta E = \hbar \omega = E_{\uparrow \uparrow} - E_{\uparrow \downarrow}$, and which would &quot;carry away&quot; the $1 \hbar$ of spin. But in order to emit a photon there must be a coupling between the electrons and the electromagnetic field. The magnitude of this coupling or in other words the probability for a transition from an initial to a final state under the emission of a photon is described by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_dipole_moment&quot; rel=&quot;noreferrer&quot;&gt;transition dipole moment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \Psi_{\mathrm{i}} \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \Psi_{\mathrm{f}} \big\rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q_{n}$ and $\mathbf{\hat{r}}_{n}$ are the charge and the position operator of the $n^{\mathrm{th}}$ particle in the system respectively, and the subscripts $\mathrm{i}$ and $\mathrm{f}$ signify the initial and final states of the transition. Under the conditions of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation&quot; rel=&quot;noreferrer&quot;&gt;Born-Oppenheimer approximation&lt;/a&gt; the contribution of the nuclei to $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be zero for the situation at hand, so that we only need to be concerned with the electrons in the system. If there is a sizable spin-orbit coupling then the spin degrees of freedom and the spatial degrees of freedom are coupled via the total angular momentum operator $\mathbf{\hat{J}} = \mathbf{\hat{L}} + \mathbf{\hat{S}}$ because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2} + \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\mathbf{\hat{L}} = \mathbf{\hat{r}} \times \mathbf{\hat{p}}$ where $\mathbf{\hat{p}}$ is the momentum operator. However, if there is no spin-orbit coupling then $\mathbf{\hat{L}} \cdot \mathbf{\hat{S}} = 0$ and $\mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2}$, i.e.&#xA;the spin degrees of freedom and the spatial degrees of freedom are decoupled. That means you can make a product ansatz for the wave function where you seperate the spin-dependent part $| \psi(S) \rangle$ and the spatial part $| \psi(\mathbf{r}) \rangle$ such that $| \Psi \rangle = | \psi(\mathbf{r}) \rangle | \psi(S) \rangle$. So, for the transition dipole moment you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \psi_{\mathrm{i}} (\mathbf{r}) \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \psi_{\mathrm{f}}(\mathbf{r}) \big\rangle \underbrace{\big\langle \psi_{\mathrm{i}} (S) \big| \psi_{\mathrm{f}}(S) \big\rangle}_{= \, 0} = 0&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\langle \psi_{\mathrm{i}} (S) | \psi_{\mathrm{f}}(S) \rangle = 0$ because the initial state would be the triplet state ($S =1$) and the final state would be the singlet state ($S=0$) and spin wave functions for different $S$ are orthogonal.&#xA;So, no spin-orbit coupling means that the probability for a transition from a triplet to a singlet state under the emission of a photon is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real atoms and molecules there is of course some spin-orbit coupling, so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be larger than zero. But the spin-orbit coupling in hydrogen is tiny and so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be tiny too and thus the probability of a triplet-singlet transition will be tiny as well and can be neglected for all practical purposes.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="41556" LastEditDate="2017-07-27T19:01:32.717" LastActivityDate="2017-07-27T19:01:32.717" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18406" PostTypeId="1" CreationDate="2014-10-13T14:32:20.683" Score="3" ViewCount="4327" Body="&lt;p&gt;I'm trying to reclaim the metals on circuit boards and the vast majority is underneath the plastic coating. It is a brittle plastic that chips off in small flakes when etched but that process will lose to much of the precious metal.&lt;/p&gt;&#xA;" OwnerUserId="9299" LastEditorUserId="19616" LastEditDate="2015-10-30T17:34:29.637" LastActivityDate="2015-10-30T17:34:29.637" Title="What chemical will dissolve the plastic coating on circuit boards" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18407" PostTypeId="2" ParentId="2509" CreationDate="2014-10-13T15:48:20.173" Score="5" Body="&lt;p&gt;The answer mentioned above is correct but there is also a shortcut built in which is math autocorrect. It is much like LaTeX. By default it's inactive but you can activate it and is really helpful if you want to write big equations. For example, if you want to type $\ce{H2}$H then you just have to type H_2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many more options are available. For example, if you want a superscript character, the caret (^) sign will be converted as superscript. Many such shortcuts are covered in &lt;a href=&quot;https://www.youtube.com/playlist?list=PLbTE-xLDPxtBP-TE2fS1MysSqFCkHh1N3&quot; rel=&quot;nofollow noreferrer&quot;&gt;these videos&lt;/a&gt; or you can simply search for How to insert mathematical equation (like LaTeX) in Ms-Office: Tips and tricks on YouTube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These methods will be especially helpful if you are a fast at typing. Moreover, it will save your time which is lost while switching between keyboard and mouse and searching proper option in Word.&lt;/p&gt;&#xA;" OwnerUserId="9302" LastEditorUserId="402" LastEditDate="2017-04-02T20:20:48.950" LastActivityDate="2017-04-02T20:20:48.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18408" PostTypeId="2" ParentId="18406" CreationDate="2014-10-13T15:57:36.763" Score="2" Body="&lt;p&gt;This seems to be a popular question judging from some Google queries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Printed_circuit_board#Materials&quot; rel=&quot;nofollow&quot;&gt;Printed circuit boards&lt;/a&gt; are made of a variety of materials, and from what I understand, the specifics may vary somewhat based on the specific needs of the product and the manufacturer. For example, in some cases, I know the PCBs are coated with special anti-static coatings to prevent delicate RAM and CPU chips from being damaged (e.g., &lt;a href=&quot;http://www2.evaluationengineering.com/features/1102esd.htm&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; found via Google among others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I doubt there's a definite solution - more likely you'll need to try several different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there's a recent article that looks very interesting about separating parts by dissolving and swelling the plastic resins with DMSO: &lt;a href=&quot;http://dx.doi.org/10.1021/es303264c&quot; rel=&quot;nofollow&quot;&gt;&quot;Dissolution of Brominated Epoxy Resins by Dimethyl Sulfoxide To Separate Waste Printed Circuit Boards&quot; &lt;em&gt;Environ. Sci. Technol.&lt;/em&gt;, &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;47&lt;/em&gt; (6), pp 2654–2660.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly DMSO is considered a near-universal solvent and will swell many polymers. So I'd probably try that first.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-13T15:57:36.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18409" PostTypeId="2" ParentId="17246" CreationDate="2014-10-13T16:32:13.080" Score="2" Body="&lt;p&gt;First off, there are two nice articles on the trans effect on &lt;a href=&quot;http://en.wikipedia.org/wiki/Trans_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;https://chem.libretexts.org/?title=Core/Inorganic_Chemistry/Organometallic_Chemistry/Fundamentals/The_trans//cis_Effects_%26_Influences&quot; rel=&quot;nofollow noreferrer&quot;&gt;LibreTexts&lt;/a&gt; that are worth reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to remember that there are two factors leading to the &lt;em&gt;trans&lt;/em&gt; effect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weakening the $\ce{Pt-X}$ bond &lt;em&gt;trans&lt;/em&gt; to another group (i.e., a structural effect)&lt;/li&gt;&#xA;&lt;li&gt;Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ &quot;wins&quot; the &lt;em&gt;trans&lt;/em&gt; effect &quot;competition&quot; and so the &lt;em&gt;trans&lt;/em&gt; $\ce{CH3-}$ will be replaced by the $\ce{I-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3UVd2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater &lt;em&gt;trans&lt;/em&gt; effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the prototypical example for the &lt;em&gt;trans&lt;/em&gt; effect is synthesis of &lt;em&gt;trans&lt;/em&gt;-platin or &lt;em&gt;cis&lt;/em&gt;-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller &lt;em&gt;trans&lt;/em&gt; effect, you can synthesize &lt;em&gt;cis&lt;/em&gt;-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5026" LastEditDate="2018-06-25T03:12:00.977" LastActivityDate="2018-06-25T03:12:00.977" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="18410" PostTypeId="1" AcceptedAnswerId="18434" CreationDate="2014-10-13T16:45:26.800" Score="6" ViewCount="88964" Body="&lt;p&gt;I picked up some stones from a hiking trip. What can I use to thoroughly clean the mud and other hard deposits on the surface of the stones? I have only washed them with plain water.&lt;/p&gt;&#xA;" OwnerUserId="9303" LastEditorUserId="8083" LastEditDate="2016-02-02T11:48:19.937" LastActivityDate="2020-04-16T16:08:37.000" Title="Chemicals to clean natural stones" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;&lt;rocks&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18411" PostTypeId="2" ParentId="18404" CreationDate="2014-10-13T17:44:11.670" Score="22" Body="&lt;p&gt;$\ce{C^{4+}}$ ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single ions are really only observed in the gas phase.  There is absolutely nothing that prevents a C&lt;sup&gt;4+&lt;/sup&gt; ion from being generated in the gas phase, given sufficient energy.  Each successive electron removal requires additional energy though, so by the time you get to 4 electrons removed, you're looking at quite a lot of energy required.  Note that this does not mean that $\ce{C^{4+}}$ ions are seen in compounds, just that they can be made in the gas phase.  Any cation you can imagine can be produced in gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty arises when we want to do something with that carbon cation.  If I want to make a compound with it, it can bond with other substances and make new bonds.  Once that is done we can evaluate those compounds to tell whether the bond is truly ionic or not.  In condensed phases or in solution, we look at how the parts of a compound interact with the particles surrounding them to decide whether to treat them as ions or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best chance at making a C&lt;sup&gt;4+&lt;/sup&gt; ion would be to bond it to fluorine, which is an extremely electron-hungry element.  When I actually try this, I find that the C-F bond is very polar, but the $\ce{CF_4}$ molecule that forms behaves as a covalent compound normally would.  For example, it does not dissociate into ions when dissolved or melted and it doesn't form an ionic crystal in its solid form.  As much as fluorine wants electrons, it doesn't want them enough to completely steal 4 of them from carbon, because the energy required to remove each electron becomes greater than the energy required to remove the one before it.  It never gets to the point of $\ce{C^{4+}}$ and $\ce{F^{-}}$ ions being bonded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead can form a +4 ion, $\ce{Pb^{4+}}$, in compounds such as lead fluoride.  The reason that this can happen but carbon cations cannot is that the outermost electrons in a lead atom are much farther from the nucleus than those of a carbon atom and are easier to remove as a result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C^{4-}}$ ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In gas form this does not happen.  Carbon's first electron affinity is positive, so we would expect to see it form $\ce{C-}$ ions in the gas phase if given free electrons to attract.  Adding a second electron to this substance would require it to attract an electron against its already negative overall charge, and that is not energetically favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make the $\ce{C^{4-}}$ion in a compound, a species would have to donate 4 electrons to carbon.  The attraction of carbon's nucleus to other atoms' electrons is just too low for that to happen.  The best candidates for a cation in an ionic compound of this type would be either lithium or cesium, however, when we try to actually make this compound we get another class of ions, the carbides, which feature $\ce{C2^{2-}}$ ions.  In short, it doesn't happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phillip's comment regarding the carbides is a good one.  There are a few metal carbides that feature carbon atoms bonding to a metal in the ratios we would expect if it were purely ionic, like $\ce{Mg2C}$.  In these compounds the carbon does have a formal charge of -4, but the properties of the substance are such that it behaves like a covalent network compound rather than an ionic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the limit of this process?  Anions with -3 charge, like nitride or phosphide, do exist, though they require a very weakly electronegative species, like an alkali metal or alkaline earth metal, to form an ionic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon atoms can carry a charge as part of an ion, but only when they are part of a larger species (again, like $\ce{C2^{2-}}$) and they do not carry 4 additional electrons per atom.  Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyanide&quot;&gt;cyanide&lt;/a&gt; as another good example of this type of polyatomic ion.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-10-14T11:39:59.657" LastActivityDate="2014-10-14T11:39:59.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18413" PostTypeId="1" CreationDate="2014-10-13T19:26:37.527" Score="2" ViewCount="2136" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Find the root-mean-square speed of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ne}$&lt;/span&gt; atoms at the temperature at which their kinetic energy is &lt;span class=&quot;math-container&quot;&gt;$\pu{6.24 kJ mol-1}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I tried using the kinetic energy formula&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{KE} = \frac{mv^2}{2},$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;but I don't really understand how to achieve the necessary values.&lt;/p&gt;&#xA;&lt;p&gt;I tried to do it by converting mass of one atom to mass of a mole since the given energy is per mole, but I'm still not getting the answer. I understand that I need to do &lt;span class=&quot;math-container&quot;&gt;$\displaystyle\sqrt{\frac{3RT}{M}},$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$R= \pu{8.3145 J mol^-1 K^-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M = 20,$&lt;/span&gt; but how would I get the temperature?&lt;/p&gt;&#xA;" OwnerUserId="9304" LastEditorUserId="41328" LastEditDate="2020-07-05T17:01:01.817" LastActivityDate="2020-07-05T17:04:24.063" Title="Finding root-mean-square speed of neon using kinetic energy" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;energy&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18414" PostTypeId="2" ParentId="5677" CreationDate="2014-10-13T19:53:30.327" Score="6" Body="&lt;p&gt;I doubt that this is a chemistry problem. Food safe plastics which can hold wine and coffee are so abundant and cheep, there is no reason that they went to the trouble of finding one which is not suitable for these, but suitable for water, and made the bladder out of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All beverages beside water are a good growing medium for one microorganism or another. And the bladder is hard to get clean, the sucking straw even more so. So, it looks to me that they are warning you to not put in anything beside water because you'd risk starting your own bacterial or fungal colony. Beer is as bad as other drinks - don't store it there. &lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2014-10-13T19:53:30.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18415" PostTypeId="1" AcceptedAnswerId="18418" CreationDate="2014-10-13T20:14:08.533" Score="-1" ViewCount="911" Body="&lt;p&gt;In the following equation,&#xA;$$\text{speed of sound}=c=\sqrt{\frac{\gamma RT}{M}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure term is missing or I guess is integrated in the adiabatic constant $\gamma$. But I don't know how to even derive the adiabatic constant for a certain gas.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="16622" LastEditDate="2015-09-03T10:26:22.713" LastActivityDate="2015-09-03T10:26:22.713" Title="How is the speed of sound dependent on pressure change in this formula?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-10-13T21:43:02.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="18416" PostTypeId="2" ParentId="18397" CreationDate="2014-10-13T20:33:47.680" Score="4" Body="&lt;p&gt;The reason is not in the technology, the reason is in the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are already electron microscopes which have better resolution for much smaller structures. But, for them, you have to kill and fix your specimen. But the hottest area of application of STED is biology. And there, the coolest thing is to get towards imaging in a live cell, so you can observe the biological processes as they happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you could use wavelengths outside of visible light, they disrupt the processes in the living cell a lot. In fact, the intensity of light needed for these super resolution microscopies is so high that phototoxicity becomes quite a concern in everyday research. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you also need fluorescent tissue for these methods. We don't have that many ways to produce fluorescent tissue yet, and the ones we have (mainly GFP) have their peaks in the visible spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stuff you can do with this technology is amazing. I recently attended a lecture given by Hell (one of the Nobel winners) and he showed a video of recording the electric impulses in the dendritic cells of an anesthesized mouse. It was haunting - beautiful, but also slightly unnerving, to see inside a live brain. This is why we need the light microscopy - our understanding of life itself depends on it. &lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2014-10-13T20:33:47.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18417" PostTypeId="1" CreationDate="2014-10-13T20:37:34.310" Score="5" ViewCount="1372" Body="&lt;p&gt;I am trying to find the structure of an unknown molecule. What is the splitting pattern of these sets of lines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R613Jm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R613Jm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my unknown molecule is butanol and the hydrogens this splitting pattern is referring to are the middle ones, i.e. $\ce{H3C-CH2-C\color{\red}{\mathbf{H}}_2-CH2-OH}$. &#xA;They would couple with the hydrogens on the right as a triplet and the ones on the left as a triplet, but would they also maybe do a W-coupling and couple with the methyl on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what would the coupling pattern be?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="5026" LastEditDate="2018-07-01T04:01:15.067" LastActivityDate="2018-07-01T04:01:15.067" Title="What splitting pattern is this?" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18418" PostTypeId="2" ParentId="18415" CreationDate="2014-10-13T20:45:15.033" Score="6" Body="&lt;p&gt;The best way to understand an equation in my view is to derive it from scratch ... along the way you pick up the details as to why you have certain conditions or restrictions that limit the usage of the equation. With this in mind I will outline an easy way to derive the equation in your book. Along the way you may think &quot;how on earth would you first think about doing this derivation&quot;! The answer is that it is a longer yet much more basic way of arriving at the same result as opposed to the &quot;proper way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish me to describe it the proper way I will add it in at the end, its much shorter but possibly a little harder maths/concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longitudinal sound waves cause small oscillations in the distribution of the gas molecules (hard spheres) which propagate in the direction of the oscillations. These compressions (regions of higher pressure than usual) and rarefactions (regions of lower pressure than usual) effects the local density and pressure of the gas molecules in a given region of space with respect to their equilibrium values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we imagine that a cylinder of gas, which has a cross sectional area of $A$ and a length of $vdt$ where $v$ is the velocity of the gas molecules and $dt$ is the time they spend in the cylinder, has a sound wave propagating through it length ways. Where we have a compression of the molecules: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) their pressure increases from equilibrium $P_0$ to $P$ by $dP$, where $P=P_0+dP$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the local density increases from equilibrium $\rho _0$ to $\rho$ by $d\rho$, where $\rho=\rho _0 +d\rho$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) due to all the extra collisions of lots of molecules in a small space the average velocity of each molecule $v$ is less than it is at equilibrium $v_o$ by $dv$ where $dv$ represents a deceleration (so is negative).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the opposite of the rarefied regions which experience an acceleration, and occupy a localised region of lower $P$ and $\rho$ than equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra force exerted on the molecules in the compressed zone is given by $F=dPA$. The volume of the cylinder $V$ is its length multiplied by its cross sectional area giving us $V=Avdt$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass is just volume times density (equilibrium density), $m=\rho _0 Avdt$, notice that this part is independent of local changes since rarefactions/compressions manifest themselves by making $v$ larger/smaller and correspondingly $dt$ smaller/larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now substitute that into Newton's second law $F=ma$ where $a$ is the acceleration of the gas molecules. We can go one of two ways here either compression or rarefaction ... lets pick a compression. This means that we will have a deceleration in the velocity of the gas molecules. Therefore the acceleration is now given by:&#xA;\begin{equation}&#xA;a=-\frac{dv}{dt}&#xA;\end{equation}&#xA;Further we can substitute the mass into Newton's second law: &#xA;\begin{equation}&#xA;F=-\rho _0Avdt\frac{dv}{dt}&#xA;\end{equation}&#xA;But remember from the start we said that the force is also given by $F=AdP$. &#xA;\begin{equation}&#xA;AdP=-\rho _0Avdt\frac{dv}{dt}&#xA;\end{equation}&#xA;Cancel the $A$s and $dt$s and re-arrange:&#xA;\begin{equation}&#xA;-\frac{dP}{dv}=\rho _0v&#xA;\end{equation}&#xA;Multiply both sides by $v$,&#xA;\begin{equation}&#xA;-v\frac{dP}{dv}=\rho _0v^2&#xA;\end{equation}&#xA;Right lets put this result to one side and think about the change in volume. In fact we need to consider the fractional change in volume since volume is an extensive property. Therefore we will be interested not in $dV$ but in the quantity $dV/V_0$. We already know that the equilibrium volume is given by $V_0=Avdt$, the change in this volume under a compression is given by $dV=Advdt$. Therefore,&#xA;\begin{equation}&#xA;\frac{dV}{V_0}=\frac{Advdt}{Avdt}&#xA;=\frac{dv}{v}&#xA;\end{equation}&#xA;Substituting this result into the above equation:&#xA;\begin{equation}&#xA;-\frac{V_0}{dV}dP=-\frac{dP}{\bigg(\frac{dV}{V_0}\bigg)}=\rho_0v^2&#xA;\end{equation}&#xA;Now define the quantity in the middle as $the$ &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bulk_modulus&quot; rel=&quot;nofollow&quot;&gt;bulk modulus&lt;/a&gt;&lt;/em&gt;, $B$. It is the ratio of an infinitesimal pressure change to the relative change in volume, it gives an indication of the compressibility of a material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now rearrange this for the velocity:&#xA;\begin{equation}&#xA;v= \sqrt{\frac{B}{\rho _0}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you have your equation in the most general form, the &lt;a href=&quot;http://www.thermaxxjackets.com/newton-laplace-equation-sound-velocity/&quot; rel=&quot;nofollow&quot;&gt;Newton-Laplace equation&lt;/a&gt;. From here you can pick what equation of state you wish to use and tailor this equation as needed. You can clearly see how the pressure and density information is tied up in this expression. But that being said its not so great for the experimentalists. Think about how you would take these readings and so on... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we are not quite done!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use &lt;a href=&quot;https://en.wikipedia.org/wiki/Boyle%27s_law&quot; rel=&quot;nofollow&quot;&gt;Boyle's law&lt;/a&gt; $P_1V_1=P_2V_2$ as Newton did then you will find that although close to experimental values your always slightly off. Therefore we will use a different equation of state... one that inherently assumes constant entropy (some people say that adiabatic means that &quot;no heat is transferred&quot;, this is the easy way to think about things .... what they really mean is that the process occurs at constant entropy .... an adiabatic process means constant entropy... (maybe if you study statistical mechanics and phase space this will be more apparent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation we should use is the adiabatic version of Boyle's law. The reason that a sound wave travelling through a medium is considered adiabatic is because &quot;no heat is exchanged between rarefactions and compressions&quot; i.e the equilibrium temperature is maintained because rarefaction velocity change cancels out compressions velocity change on average meaning that the kinetic energy is pretty much constant and hence so is $T$ from $E_k=\frac{3}{2} NKT$. So instead of Boyle's law we use $PV^{\gamma}=P_0V_0^{\gamma}$. But for a given temperature the entire right hand side is just a constant so lets instead use $PV^{\gamma}=K$ where $K$ is the constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;P=KV^{-\gamma}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets differentiate $P$ with respect to $V$.... lets find the rate of change oof pressure with respect to the change in volume...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{dP}{dV}=-\gamma KV^{-\gamma-1}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides by $-V$ and you will notice we have an expression for $B$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;-V\frac{dP}{dV}=KV\gamma V^{-\gamma-1}=VK\gamma V^{-\gamma}V^{-1}=\frac{K\gamma}{V^{\gamma}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V=V_0$ the equilibrium volume which we are free to do. Now show the equilibrium volume in terms of the equilibrium density $\rho_0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\rho_0=\frac{nM}{V_0}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now go back and put these in $v=\sqrt{\frac{B}{\rho_0}}$ and simplify then we have the equation stated in your book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;v=\sqrt{\frac{B}{\rho_0}}=\sqrt{\frac{K\gamma V_0}{V_0^{\gamma}nM}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $K/V_0^{\gamma}=P_0$ and then the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt; $P_0V_0=nRT$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;v=\sqrt{\frac{\gamma P_0V_0}{nM}}=\sqrt{\frac{\gamma RT}{M}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see now how pressure is tied up in the entire expression, along with the density. The assumption we had to assume was that the sound wave travelling through is adiabatic and there we have the speed of the wave in a gas of non-interacting (except for collisions) hard spheres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like I can derive the original formula in a much more professional way from elementary fluid mechanics ... but although quicker and more sophisticated it is a little harder. (if I've made any errors feel free to edit away) :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2015-09-03T09:52:35.800" LastActivityDate="2015-09-03T09:52:35.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18420" PostTypeId="2" ParentId="18417" CreationDate="2014-10-13T21:27:01.537" Score="4" Body="&lt;p&gt;This is either a quartet of triplets or quartet of doublets of doublets, with both doublets of equal magnitude. It is most definitely not butanol. The pattern you see is from a double bond, and shows 3J coupling to a methyl group. I suspect you have another set of signals near this at about 5.5, and probably a methyl signal near 2ppm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, W-coupling (or M-coupling) is only observed in rigid systems, and is typically a 4J phenomenon, although can occur over longer bond orders if the bond overlap allows.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-13T21:27:01.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18421" PostTypeId="2" ParentId="18417" CreationDate="2014-10-13T21:36:00.843" Score="7" Body="&lt;p&gt;This is the a peak from one of the vinyl protons on an allyl group. This splitting pattern is an ABX pattern from the bold hydrogen below. It is technically a triplet of doublets of doublets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-CH2-C}\textbf{H}\ce{=CH2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the following simulated NMR spectrum of 3-chloro-1-propene (allyl chloride) from &lt;a href=&quot;http://www.nmrdb.org&quot; rel=&quot;noreferrer&quot;&gt;nmrdb.org&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/I9SNp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now image if the coupling constants are a little different and the structure collapses into what you have.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-10-13T21:36:00.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18423" PostTypeId="2" ParentId="18380" CreationDate="2014-10-14T01:07:05.710" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Enzyme_kinetics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enzyme kinetics&lt;/a&gt; is one area of catalysis that has been studied extensively. The Michaelis-Menton equation being one of the central pillars and explicitly states the roles of concentration for both the substrate and enzymes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to answer your question: the rate equation is at times more interested in establishing equilibrium conditions than obtaining the time it takes to get there. The time to reach equilibrium is slightly neglected as the question &quot;are we there yet&quot; seems trivial when distance = rate times time. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="71148" LastEditDate="2020-05-24T00:38:56.963" LastActivityDate="2020-05-24T00:38:56.963" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18424" PostTypeId="2" ParentId="18397" CreationDate="2014-10-14T02:11:54.943" Score="3" Body="&lt;p&gt;You can do microscopy with other light sources, too. Actually there are X-ray, Raman and IR microscopes, but there are some issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Few or No lenses: it is difficult(not impossible) to focus IR, UV or X-ray&lt;/li&gt;&#xA;&lt;li&gt;Fewer &quot;light source&quot;: in visible region, we have already developed a lot of lasers etc, for UV or X-ray we have fewer options. Part of the reason is that no one needed them, so less effort were put to develop them. Also, high intensity, coherent X-ray sources tend to be synchrotrons or free electron lasers which are marvelous things, but very expensive and big. &lt;/li&gt;&#xA;&lt;li&gt;Less spectroscopic info: Most characteristic optical properties of organic / bioinorganic materials are in the UV-Vis region. In X-ray region generally we have either high energy electric excitation kicking off valence electrons (XPS-UPS) or excitation of core-electrons. Off course, it is not always the case, for example IR or Raman microscopy can be very informative from chemical point of view. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2014-10-14T02:51:16.867" LastActivityDate="2014-10-14T02:51:16.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18425" PostTypeId="1" CreationDate="2014-10-14T04:00:30.390" Score="0" ViewCount="107" Body="&lt;p&gt;Why does the %s character increase on the orbital with electropositive atom? &#xA;I do not seem to properly grasp this concept. Is this rule based on observation or is there any proof to it? If there is a proof can you guys explain it to me in simple terms? I currently have knowledge of basic quantum mechanics and chemical bonding.&lt;/p&gt;&#xA;" OwnerUserId="9310" LastActivityDate="2014-10-14T04:00:30.390" Title="Explanation of Bent's Rule" Tags="&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2014-10-14T11:41:11.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="18426" PostTypeId="2" ParentId="5569" CreationDate="2014-10-14T06:02:27.970" Score="1" Body="&lt;p&gt;To add to the previous answers, for a real-world example look no further than your average rocket engine.  Especially hydrogen/oxygen engines are run fuel-rich; at stoichiometric rato the chamber temperature would be too high and the combustion gases too reactive for the engine to survive.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-14T06:02:27.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18427" PostTypeId="1" AcceptedAnswerId="18544" CreationDate="2014-10-14T09:21:39.107" Score="64" ViewCount="8686" Body="&lt;p&gt;Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$? I realize that it would be more stable there than at equatorial bond, but what is the reason of its stability? Similarly in $\ce{AB4}$ type of molecules with $\ce{sp^3d}$ hybridization (4 bond pairs and 1 lone pair) why is the geometry of the molecule that of a see-saw, where the lone pair is at equatorial position rather than at axial one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book states the reason to be &quot;due to Bent's rule&quot; but I have difficulty linking these two.&lt;br&gt;&#xA;Also Bent's rule states that the hybridized orbitals of equivalent energy of central atom tend to give more %s character to the electropositive atom attached to it rather than the electronegative one. (e.g. - in $\ce{CH3F}$) Am I right? What else is more to this rule?&lt;/p&gt;&#xA;" OwnerUserId="9310" LastEditorUserId="16683" LastEditDate="2017-07-03T15:37:17.397" LastActivityDate="2020-10-16T05:51:56.040" Title="Why does F replace the axial bond in PCl5?" Tags="&lt;molecular-structure&gt;&lt;halides&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="36" ContentLicense="CC BY-SA 3.0" />
  <row Id="18428" PostTypeId="2" ParentId="18357" CreationDate="2014-10-14T09:38:59.333" Score="4" Body="&lt;p&gt;I'd like to offer an alternative method here, which is generic for all binary compounds (but may also work for ternery, quarternery, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the highest oxidation state of the metal. Then divide the negative by the number of oxygens. This is how you find the average oxidation state of oxygen. If this is greater than -2, then there has to be a peroxyl oxygen involved. If it is -2, or lower, then there is no peroxyl involved. Additional remark: If the average oxidation state of oxygen is lower than -2, then the metal is not in its highest oxidation state. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-10-14T09:38:59.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18430" PostTypeId="1" AcceptedAnswerId="18550" CreationDate="2014-10-14T14:43:05.917" Score="7" ViewCount="838" Body="&lt;p&gt;I think I am getting slightly confused about the names we give to functional groups. I am aware that there are different types of functional groups such as alkyl, aryl and acyl groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that there exist functional groups such as those of esters, ketones and amides which come under the acyl group. However, thinking about this recently I also realised that there existed a carbonyl group, which is also present in a ketone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also a bit confused as to whether the acyl group's general formula is RCO or RCOX where X could be a variety of groups. Also, is R a part of the functional group or is it there to show what the functional group is attached to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please make this hierarchy of functional groups a bit clearer?&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="73" LastEditDate="2014-10-15T10:16:58.393" LastActivityDate="2014-10-17T22:52:51.197" Title="Is there a hierarchy to functional groups?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18431" PostTypeId="2" ParentId="2879" CreationDate="2014-10-14T15:17:13.937" Score="3" Body="&lt;p&gt;I was also confused when I learned this all stuff first time. When you get any element, molecule, compound or mixture don't think about complex stuff but think on the basic level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have given the most basic definition (properties) of an element, molecule, compound or mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Element:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A substance which cannot be broken down chemically is called elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All substances listed in Periodic Table are elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Molecule:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two or more atoms or elements combined together are called a molecule.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Compound:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Two or more elements combine together to form a compound. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The smallest unit of a compound is a molecule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compounds possess different characteristics compared to constituent elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Example: $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H}$ is highly combustible&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{O}$ helps in combustion&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H2O}$ is used in fire extinguishers (different characteristics than $\ce{H}$ and $\ce{O}$  )&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mixture:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When two or more matter mix without changing their original property this is called a mixture.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixtures can be separated by a simple process, like filtration. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="66352" LastEditDate="2018-07-26T08:53:09.033" LastActivityDate="2018-07-26T08:53:09.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18433" PostTypeId="1" CreationDate="2014-10-14T16:40:20.367" Score="3" ViewCount="11086" Body="&lt;p&gt;I prepared a battery with $\ce{Zn}$ screws and a $\ce{Cu}$ wire, with water as an electrolyte. The $\ce{Zn}$ screw acted as the cathode, and the $\ce{Cu}$ wire acted as the anode. I got nearly $0.6\:\mathrm{V}$ as the emf. I placed numerous amounts of such cells in an ice cube tray. Each ice cube acts as a cell and i got nearly $6\:\mathrm{V}$ of emf. What is the reaction happening here? What is the reaction involved with the electrodes and the electrolytes?&lt;/p&gt;&#xA;" OwnerUserId="9316" LastEditorUserId="15410" LastEditDate="2015-05-19T06:45:39.663" LastActivityDate="2018-08-14T21:58:11.497" Title="What is the reaction between a copper anode and a zinc cathode?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18434" PostTypeId="2" ParentId="18410" CreationDate="2014-10-14T17:13:58.410" Score="4" Body="&lt;p&gt;Removing all of the debris from the stones without damaging them can be difficult, the following is what I've done in the past to clean geodes with debris and deposits.  Your mileage may vary, especially since we don't know exactly what is on your stones.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Soak the stones in soapy water for a day or so, then scrub them thoroughly with a stiff brush to clean dirt and most debris from them.  Be cautious if the stone has delicate structure that you're trying to preserve.  If some debris remains that appears to be washable, you can try scrubbing with a toothbrush and toothpaste to get rid of it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the stone is acid safe, soak it in vinegar for a couple of days to dissolve carbonate deposits.  You can check to see if it will react with vinegar by adding a few drops of vinegar to the stone's surface and waiting.  If it hasn't started bubbling within 10 minutes or so, it should be safe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the stone has rust deposits and is acid safe, you may be able to remove them with something like CLR Rust Stain remover.  If you need to go this route, skip the vinegar step, as this will also clean carbonate deposits.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, it's possible that your crystal has something like granite attached.  If that is the case, nothing practical, short of cutting it off, is likely to clean the crystal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you've got the crystal to the cleanliness you're happy with, you can polish it with powdered polish and a cloth, given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of nice, angular crystals you're just trying to produce something like a smooth, rounded stone, after cleaning with water you can use a series of sandpapers (start at 60-80 grit and work up to ~400 grit) and polish to gradually smooth, then polish the stone by hand.  This will clear away any deposits, but it will also change the surface of the stone itself.  That might be acceptable to you if you've got something like a quartz river rock, and it will make a very nice end product.  Take care to keep the stone wet while working, you don't want to breathe in rock dust if you don't have to.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-14T17:13:58.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18435" PostTypeId="1" AcceptedAnswerId="18440" CreationDate="2014-10-14T17:30:02.640" Score="4" ViewCount="11443" Body="&lt;p&gt;I used a spectrometer to find the absorbency of 5 solutions with different iron concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{rrrrr}&#xA;  \ce{Fe (III)} / \pu{mol L-1} &amp;amp; &#xA;  \text{Absorbance} &amp;amp; &#xA;  V(\ce{HCl}) &amp;amp;&#xA;  V(\ce{FeCl3}) &amp;amp;&#xA;  V(\ce{KSCN}) \\&#xA;  \hline&#xA;  0.00005 &amp;amp; 0.194 &amp;amp; \pu{9 mL} &amp;amp; \pu{1 mL} &amp;amp; \pu{10 mL} \\&#xA;  0.0001  &amp;amp; 0.424 &amp;amp; \pu{8 mL} &amp;amp; \pu{2 mL} &amp;amp; \pu{10 mL} \\&#xA;  0.00015 &amp;amp; 0.674 &amp;amp; \pu{7 mL} &amp;amp; \pu{3 mL} &amp;amp; \pu{10 mL} \\&#xA;  0.0002  &amp;amp; 0.89  &amp;amp; \pu{6 mL} &amp;amp; \pu{4 mL} &amp;amp; \pu{10 mL} \\&#xA;  0.00025 &amp;amp; 1.113 &amp;amp; \pu{5 mL} &amp;amp; \pu{5 mL} &amp;amp; \pu{10 mL} &#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this data I've constructed a graph and a line of best fit.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;y &amp;amp;= 4608x - 0.0322 &amp;amp;&#xA;R^2 &amp;amp;= 0.9994&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took &lt;span class=&quot;math-container&quot;&gt;$\pu{5.10g}$&lt;/span&gt; of spinach leafs and burned them, then added &lt;span class=&quot;math-container&quot;&gt;$\pu{10.00 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{KSCN}$&lt;/span&gt; then filtered the contents (so ashes weren't in the beaker).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorbency of this spinach extract was &lt;span class=&quot;math-container&quot;&gt;$0.015$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: How do I calculate how much iron is in this sample? I need to determine the mass of iron per &lt;span class=&quot;math-container&quot;&gt;$\pu{100 g}$&lt;/span&gt; of spinach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the line of best fit question, solve for &lt;span class=&quot;math-container&quot;&gt;$V(\ce{Fe(III)}) = x$&lt;/span&gt; concentration:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;0.015 &amp;amp;= 4608x - 0.0322\\&#xA;x &amp;amp;= 0.0000102431&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convert &lt;span class=&quot;math-container&quot;&gt;$\pu{mol/L}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{g/L}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{mg/L}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;0.0000102431 \times 55.845 &amp;amp;= 0.00057202591\\&#xA;\pu{0.00057202591 g} &amp;amp;= \pu{0.572 mg}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is clearly wrong because it's not taking account the amount of spinach I started with (&lt;span class=&quot;math-container&quot;&gt;$\pu{5.10 g}$&lt;/span&gt;). Anytime I try to include the amount, either by multiplying the density my answer is way off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it reasonable that I'm getting a number above &lt;span class=&quot;math-container&quot;&gt;$\pu{10 mg}$&lt;/span&gt;? The iron amount reported by the USDA says that there is &lt;span class=&quot;math-container&quot;&gt;$\pu{2.7 mg}$&lt;/span&gt; of iron per &lt;span class=&quot;math-container&quot;&gt;$\pu{100 g}$&lt;/span&gt; of spinach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone guide me in the correct direction?&lt;/p&gt;&#xA;" OwnerUserId="2356" LastEditorUserId="4945" LastEditDate="2020-05-25T16:07:11.423" LastActivityDate="2020-05-25T16:07:11.423" Title="How to determine the iron content in spinach with a spectrometer?" Tags="&lt;concentration&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18436" PostTypeId="2" ParentId="18401" CreationDate="2014-10-14T18:07:03.930" Score="5" Body="&lt;p&gt;Pencil leads aren't pure graphite, they're generally graphite powder with various polymers to hold them together and make sure they have the correct hardness (after all, you wouldn't be able to use them to write if they didn't leave graphite on the paper or if they crumbled under slight pressure. This is why the much more rigid glassy carbon is a popular material for carbon electrodes. My guess is that there's no reaction with the electrode happening but gas being produced is causing mechanical damage to the electrode. If you work at less extreme potentials, you might be able to reduce this effect. Otherwise, pure graphite rods should be more durable. Of course, if you're using pencil leads, they're obviously cheap to replace so it might be easier to just use them until they fall apart, then switch to a new one.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-14T18:07:03.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18437" PostTypeId="1" CreationDate="2014-10-14T18:08:39.917" Score="1" ViewCount="346" Body="&lt;p&gt;does reaction quotient After equilibrium has been achieved make any sense?&#xA;At equilibrium point rates of forward and backward reaction are same and the system is in dynamic equilibrium.The graph of concentration vs time shows a straight line parallel to x axis after equilibrium point.&#xA;Reaction quotient refers to ratio of concentration at any stage.This means that after equilibrium point reaction Quotient value will same as equilibrium constant as no more products are formed.So how can it have a value more than that of K ?!&#xA;I have been thinking about this for hours but I get more confused with time.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2015-08-22T10:08:23.470" Title="Equilibrium constant and reaction Quotient" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18438" PostTypeId="1" CreationDate="2014-10-14T21:10:33.787" Score="0" ViewCount="5136" Body="&lt;p&gt;Why has lithium bromide a slight solubility in organic solvents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretical, LiBr is an ionic compounds, so it should be only soluble in polar solvent like water. But why is it soluble in others?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="5017" LastEditDate="2014-10-23T19:58:02.817" LastActivityDate="2015-11-29T06:22:25.403" Title="Why has lithium bromide a slight solubility in organic solvents?" Tags="&lt;solubility&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18439" PostTypeId="2" ParentId="18438" CreationDate="2014-10-14T21:31:18.557" Score="6" Body="&lt;p&gt;Lithium bromide is a salt of small cation and large anion. They cannot interact effectively, so the crystal lattice is quite easy to break. On the other hand, lithium cation is small, so it can interact effectively with small negatively charged atoms, like oxygen. Indeed, lithium cation has highest hydration energy of all alkaly metal cations. Altogether, it means, that in solvents with oxygen atoms (alcohol, esters, acetone) lithium cation effectively bounds to solvent molecules, leaving crystall lattice, and bromide has to follow, resulting in some solubility of lithium bromide in solvents with negatively charged oxigen and to lesser extent, nitrogen (for example pyridine). Lithium bromide is still insoluble in organic solvents that cannot interact with lithium cation, like benzene. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-10-15T12:03:13.567" LastActivityDate="2014-10-15T12:03:13.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18440" PostTypeId="2" ParentId="18435" CreationDate="2014-10-14T21:49:09.490" Score="6" Body="&lt;p&gt;Here's how I approached the problem.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Plot the Data&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j8A4v.png&quot; alt=&quot;Plot of the data given by OP with a linear regression through (0,0)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Calculate the Concentration of Fe Ions Present&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You have measured an absorbance of 0.015 with a corresponding concentration of $c_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of iron in the solution is given by&#xA;$$ m_\ce{Fe} = c_\text{meas} \cdot M_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$&#xA;with the volume $V=0.02~\mathrm{L}$ and molar mass $M_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Calculate Mass Percentage&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The mass of iron in one gram of spinach is easily calculated:&#xA;$$ w_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(I could just calculate away the units, but I was too lazy to do it here.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass in $100~\mathrm{g}$ of spinach follows accordingly:&#xA;$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-10-14T22:08:31.350" LastActivityDate="2014-10-14T22:08:31.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18441" PostTypeId="1" AcceptedAnswerId="18445" CreationDate="2014-10-14T22:36:52.597" Score="4" ViewCount="1406" Body="&lt;p&gt;I am a chemistry student, and my current are of study is mass spectrometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ionization process, atoms are bombarded with electrons, but how does this work? I assume the flow of electrons attract the electrons of the atom, thereby removing them, but where do the atom's electrons go?&lt;/p&gt;&#xA;" OwnerUserId="9319" LastEditorUserId="15410" LastEditDate="2015-05-18T04:14:37.503" LastActivityDate="2015-05-18T04:14:37.503" Title="How does bombarding a molecule with electrons cause ionization?" Tags="&lt;electrons&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18442" PostTypeId="1" AcceptedAnswerId="18454" CreationDate="2014-10-15T00:05:15.843" Score="3" ViewCount="3577" Body="&lt;p&gt;I am considering the hydrogenation of 2-butanone with respect to its Re and Si faces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we look at 2-butanone as I've drawn below from &lt;em&gt;above&lt;/em&gt; we are looking at its rectus face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from below we are looking at its sinister face. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, time to consider hydrogenation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll keep the wedge and dash groups as they are and the carbon and oxygen both on lines. In other words we keep two points as is in the originally trigonal planar 2-butanone and when the hydrogen inserts itself, the other two points (a hydrogen and an alcohol group) pop up out the plane to form the classic tetrahedral shape of an $\ce{sp3}$ carbon atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it appears that hydrogenation from the sinister face adds an hydrogen to the central carbon and forms an alcohol, and the alcohol is above the added hydrogen spatially, and this gives us the R isomer. Hydrogenation from the rectal face gives the opposite S isomer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How's the reasoning? Am I right? I am fairly sure. But not completely sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lqmEe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-10-15T03:51:38.933" LastActivityDate="2014-10-15T04:31:18.417" Title="Si and Re facial descriptors" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18443" PostTypeId="1" CreationDate="2014-10-15T00:17:24.770" Score="1" ViewCount="75" Body="&lt;p&gt;I'm trying to see if such a thing exists in paper form that upon activation,  changes colour or &quot;indicates&quot; after a set configurable amount of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words a litmus or pH paper that can act as an egg timer. I'm by no means a chemist so please excuse my lack of protocol.&lt;/p&gt;&#xA;" OwnerUserId="9320" LastEditorUserId="5026" LastEditDate="2018-04-12T03:51:45.027" LastActivityDate="2018-04-12T03:51:45.027" Title="Does a timed reaction &quot;ready indicator&quot; paper exist?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18444" PostTypeId="1" AcceptedAnswerId="18448" CreationDate="2014-10-15T00:22:16.803" Score="3" ViewCount="802" Body="&lt;p&gt;I have some questions about the auto-ignition temperature for fuels. Per &lt;a href=&quot;http://www.google.ca/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;frm=1&amp;amp;source=web&amp;amp;cd=4&amp;amp;sqi=2&amp;amp;ved=0CC4QFjAD&amp;amp;url=http%3A%2F%2Fxa.yimg.com%2Fkq%2Fgroups%2F9534928%2F993743717%2Fname%2FASTM%2BE659%2B-%2B78%2B(Reapproved%2B2005).pdf&amp;amp;ei=Poc9VKrcGMGyyASPmYAo&amp;amp;usg=AFQjCNHU3HPYYiWr7X2RL79rGgMnyx7Kzg&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM E659&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;AIT (auto-ignition temperature) is the lowest temperature at which the substance will produce hot-flame ignition in air at atmospheric pressure without the aid of an external energy source such as spark or flame. It is the lowest temperature to which a combustible mixture must be raised so that the rate of heat evolved by the exothermic oxidation reaction will overbalance the rate at which heat is lost to the surroundings and cause ignition.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The last sentence seems to say that AIT is the temperature where enough of the particles are above the activation energy such that enough heat released goes to perpetuate the remaining fuel to combust (cause ignition). Is this correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Per ASTM E659, ignition is subjectively defined as the appearance of a flame with a sharp rise in temperature. Why it necessary that we observe the appearance of a flame (in other words, can we have combustion without an observed flame)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an assumption of thermal equilibrium between fuel and oxidizer when defining AIT? The testing method above says air at predetermined temperature will be used, but I'm not sure how they plan to control air temperature when heating begins. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="402" LastEditDate="2017-04-01T01:19:38.627" LastActivityDate="2018-08-24T12:08:59.403" Title="Mechanics behind auto-ignition temperature" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18445" PostTypeId="2" ParentId="18441" CreationDate="2014-10-15T00:25:20.683" Score="5" Body="&lt;p&gt;This type of ionization is known as electron ionization (there are many types of ionization methods used for MS). One possible reaction for EI is:&#xA;$$\ce{M +e- -&amp;gt;M^{•+} +2e-}$$ &lt;a href=&quot;http://goldbook.iupac.org/E01999.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the incoming electron has the correct energy to interact and passes close enough to the molecule, it's electric field can impart enough energy to one of the molecule's electrons for it to be ejected. What happens to the electrons after that isn't particularly important as far as ionization is concerned, but in most EI sources, electrons are emitted perpendicular to the axis of the sample source and are collected on the other side by an electrode, preventing charging of the source chamber and focusing the emitted electrons into a beam.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-15T00:25:20.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18446" PostTypeId="2" ParentId="18443" CreationDate="2014-10-15T00:33:51.720" Score="2" Body="&lt;p&gt;I'm not aware of any commercially available examples, but I know that the speed of a fluid wicking through a paper microfluidic device has been &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac1005537&quot; rel=&quot;nofollow&quot;&gt;used&lt;/a&gt; for timing. I'm also reminded of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pencil_detonator&quot; rel=&quot;nofollow&quot;&gt;pencil detonators&lt;/a&gt;, dating back to WWII, where the time taken to dissolve a lead wire is used for timing. Since electronic timing is now so cheap, easy, accurate, and reliable, I'm not surprised that chemical timers have pretty much died out. There are a number of chemical phenomena that could be exploited for timing, but unless you need a huge number of cheap, one time use timers with relatively poor accuracy, there are probably better choices.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-15T00:33:51.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18447" PostTypeId="2" ParentId="2879" CreationDate="2014-10-15T01:18:15.553" Score="0" Body="&lt;p&gt;Compound: a pure substance formed from the bonding of two or more different elements together. It can be decomposed into its constituent elements by chemical means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Element: a pure substance consisting of one type of atom only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixture: two or more different substances are mixed but they are not chemically bonded together and may be separated by a physical process such as filtration, distillation and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule: I'm actually not sure about this - I always read that it's a group of atoms (same or different) bonded together and is the smallest particle of an element/compound capable of independent existence; however, we treat monoatomic gases as a molecule, e.g. Neon. &lt;/p&gt;&#xA;" OwnerUserId="8143" LastActivityDate="2014-10-15T01:18:15.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18448" PostTypeId="2" ParentId="18444" CreationDate="2014-10-15T01:23:38.510" Score="5" Body="&lt;p&gt;To be clear, I'm no expert on the ASTM standards writers' thinking, and all I have to go on is the standard you've posted.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, that's exactly right.  Some of the reaction's heat will be lost to the surroundings, but enough of it will go into the reaction mixture to raise its temperature and promote further, more rapid, combustion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are reactions that produce lots of heat but no flame, hydrogen peroxide decompostion, for example.  The standards writers had to come up with a clear definition for ignition, and as you note, they did: &quot;the appearance of a flame accompanied by a sharp rise in the temperature of the gas mixture.&quot;  They even say that ignition is a subjective term, so they are careful to define it clearly.  There might be some chemical reaction that produces a large amount of heat in air with no visible flame (even in the dark, as the standard requires) and it might even qualify as an ignition event to you or I or any other sensible person watching, but it would not fit the standard.  They have to set some limit, after all, and that was what they set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My original thinking on this was flawed; I misunderstood the air temperature reference.  The reason that they mention the air at all is that they are being specific that the reaction mixture should be an air/fuel mixture, rather than something else (oxygen/fuel, for instance.)  The predetermined temperature aspect of that point has to do with the AIT determination procedure itself.  They are setting the temperature of the system to some value T, then adding the fuel, then waiting for ignition.  So yes, the air in the flask is more or less in thermal equilibrium with the substance being tested.  It can, however, leak out and exchange with the air outside the flask.&lt;br&gt;&lt;br&gt; If the substance does not autoignite, the flask is purged and the temperature increased (including the air in the flask.)  I'm not sure whether they want a completely new sample of the substance for each temperature measurement.  It seems likely, since we're dealing with small amounts of generally volatile compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2018-08-24T12:08:59.403" LastActivityDate="2018-08-24T12:08:59.403" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="18449" PostTypeId="1" AcceptedAnswerId="18458" CreationDate="2014-10-15T01:48:36.127" Score="6" ViewCount="15390" Body="&lt;p&gt;Where's the lone pair on the cyanide anion in this MO diagram? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LUmXo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the traditional Lewis picture there are two lone pairs on the cyanide ion. In the MO diagram though there appear to be no non-bonding orbitals that are populated ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also how come there appear to be two sigma bonding orbitals populated when in $\ce{NC^-}$ there is only one sigma bond? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-15T08:42:56.997" Title="Cyanide ion non-bonding/lone pair?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18450" PostTypeId="1" AcceptedAnswerId="18451" CreationDate="2014-10-15T01:57:50.413" Score="7" ViewCount="26216" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Describe how zinc metal can be obtained from zinc sulfate solution by electrolysis. &#xA;  A labelled diagram is acceptable. Include all the products of this electrolysis. The electrolysis is similar to that of copper (II) sulfate solution with inert electrodes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer key says that zinc would be formed at the cathode. However, zinc is more reactive than hydrogen, so logically hydrogen should be given off at the cathode, not zinc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the ions involved in this electrolysis are $\ce{Zn^2+}$, $\ce{H+}$, $\ce{OH-}$ and $\ce{SO4^2-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why can possible cathodes in this reaction be carbon, platinum OR zinc while the anode can only be carbon or platinum? What allows the cathode to be zinc?&lt;/p&gt;&#xA;" OwnerUserId="8230" LastEditorUserId="5026" LastEditDate="2018-06-25T03:13:45.790" LastActivityDate="2018-10-15T12:18:15.847" Title="Why would hydrogen not be generated at the cathode when electrolysing zinc sulfate?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18451" PostTypeId="2" ParentId="18450" CreationDate="2014-10-15T02:18:00.380" Score="3" Body="&lt;p&gt;In an electrolytic cell we are using the electrical energy from an outside power source to drive a chemical system against the direction it would spontaneously move.  In this case, we can take a solution of zinc ions and force them to become zinc metal.  As long as we put the zinc ions in an environment where they can be reduced, they will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the half reaction that occurs when zinc metal is produced:&#xA;$\ce{Zn^{2+}(aq) + 2e- -&amp;gt; Zn(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a reduction reaction, and reductions always occur at the cathode, whether the cell is galvanic or electrolytic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other half reaction involves hydroxide being converted into water and oxygen.&#xA;$\ce{4 OH- -&amp;gt; 2 H2O + O2 + 4e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an oxidation reaction, and oxidations always occur at the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Platinum and carbon are nonreactive in this system, so they can be used for either electrode.  Zinc can be used as a cathode because that's where it is being deposited, so we're just adding additional metallic zinc onto our already existing zinc electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the idea that oxidation occurs at the anode in mind, what would happen if we used a zinc anode?&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-15T02:18:00.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18452" PostTypeId="1" CreationDate="2014-10-15T03:14:19.577" Score="1" ViewCount="240" Body="&lt;p&gt;What kind of zinc naturally occurs in plants? Zinc oxide, zinc sulphide, or something else?&lt;/p&gt;&#xA;" OwnerUserId="9321" LastEditorUserId="73" LastEditDate="2014-10-15T10:18:32.790" LastActivityDate="2015-04-27T13:47:04.237" Title="What form of zinc's found naturally in vegetables &amp; meat?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18453" PostTypeId="2" ParentId="18452" CreationDate="2014-10-15T03:21:01.507" Score="1" Body="&lt;p&gt;The most likely form is $\ce {Zn^{2+}}$ bound inside a protein, or maybe chelated to some amino acids. It is not very likely that there is zinc oxide, or zinc sulfide, or zinc sulfate in organic matter.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="73" LastEditDate="2014-10-15T10:20:52.447" LastActivityDate="2014-10-15T10:20:52.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18454" PostTypeId="2" ParentId="18442" CreationDate="2014-10-15T04:31:18.417" Score="4" Body="&lt;p&gt;When looking at the corresponding Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Prochirality&quot; rel=&quot;nofollow noreferrer&quot;&gt;prochirality&lt;/a&gt; you will find out, that your reasoning is correct for the hydrogenation. Of course if you have a different nucleophile, then the situation would change, as the Cahn-Ingold-Prelog identifiers would change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pcg0E.jpg&quot; alt=&quot;re si faces and products of hydrogenation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://goldbook.iupac.org/P04859.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;prochirality in the IUPAC goldbook&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-10-15T04:31:18.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18455" PostTypeId="1" CreationDate="2014-10-15T04:43:22.040" Score="0" ViewCount="130" Body="&lt;p&gt;I had some followup questions regarding a previous post I made &lt;a href=&quot;https://chemistry.stackexchange.com/q/18444&quot;&gt;here&lt;/a&gt; regarding the auto-ignition temperature and &lt;a href=&quot;http://xa.yimg.com/kq/groups/9534928/993743717/name/ASTM+E659+-+78+(Reapproved+2005).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM E659&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For fuel temperature below AIT, we should still have finite reactants above activation energy, reacting, and heating the remaining mixture so additional reactants are above the activation energy. Theoretically as $t \rightarrow \infty$, could the fuel burn itself out like this, regardless of what temperature it is at (obviously the rate will be very different)? If so, is there an implicit rate requirement in defining the AIT that the above chain reaction process has to happen within a short duration?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given that ignition is defined as when a flash and temperature rise is seen, does this mean that the actual quantity of fuel burnt is unimportant and it is assumed that a significant enough amount is used?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the reason for using an open flask in ASTM E659 likely just for simplicity, as opposed to a piston-cylinder arrangement that offers more control over air-fuel mixture?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-04-01T02:41:04.873" Title="Additional Questions Regarding the Auto-Ignition Temperature" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18457" PostTypeId="2" ParentId="18433" CreationDate="2014-10-15T05:46:00.957" Score="3" Body="&lt;p&gt;Well the standard galvanic cell with $\ce{Zn}$ and $\ce{Cu}$ would usually look something like this, assuming there's a salt bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sMTam.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the reaction could be written as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{Cu^{2+}_(aq) +2e^{-} &amp;amp;&amp;lt;=&amp;gt; Cu^0} &amp;amp; E^\circ &amp;amp;=0.34\\&#xA;\ce{Zn^0 (s) &amp;amp;&amp;lt;=&amp;gt; Zn^{2+} (aq) +2e^-}&amp;amp; E^\circ &amp;amp;=+0.76\\&#xA;\ce{Cu^{2+}(aq) + Zn^0 (s) &amp;amp;&amp;lt;=&amp;gt; Cu^0 (s) + Zn^{2+}(aq)} &amp;amp; E^\circ &amp;amp;= +1.10 &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your measured emf, that can be calculated using the Nernst Equation.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4945" LastEditDate="2014-10-15T06:22:10.873" LastActivityDate="2014-10-15T06:22:10.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18458" PostTypeId="2" ParentId="18449" CreationDate="2014-10-15T08:42:56.997" Score="8" Body="&lt;p&gt;As Geoff already pointed out in the comments, the connection between a molecular orbital theory (MO) point of view and a Lewis picture (LT) is tenuous at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the best example for this is the mentioned $\ce{{}^{-}CN}$ and obviously the isoelectronic $\ce{CO}$. But there are many more molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the MO point of view the electron pair which can be donated to an acceptor orbital is of course the highest occupied molecular orbital (HOMO). Often this comes with the more trivial notation of &quot;lone pair&quot;, based on the coefficients of a certain element in that orbital (percentage contribution). In the case of $\ce{CO}$ this becomes fairly obvious, but since this is not the question at hand I will only focus on the isoelectronic counterpart.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6SdOn.png&quot; alt=&quot;MO Diagramm&quot;&gt;&lt;br&gt;&#xA;Please consider the above MO scheme only as a sketch, it is not to scale and contributions were estimated only be a quick browse of the output file. I also did not compensate the calculation for the negative charge, so even the bonding orbitals are antibonding with respect to fragmentation. But it shall still provide the point.&lt;br&gt;&#xA;We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. But this is, however, not a real lone pair since it is $\sigma$ bonding and also has quite a big coefficient at the nitrogen end. (In $\ce{CO}$ the difference is much bigger.) &#xA;Below that we see the two $\pi$ orbitals, followed again by two $\sigma$ orbitals. All five orbitals make up the bonding of the molecules and determine also its reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in the Lewis theory picture this is quite different. We traditionally describe it as a carbon oxygen $\sigma$ bond, two carbon oxygen $\pi$ bonds, a oxygen lone pair and a carbon lone pair. All made up from some kind of hybrid orbitals. In computational chemistry we can analyse the wave function and obtain localised orbitals. One of these methods is natural bond orbital theory (NBO). I have carried that out and sketched the common bonding picture along with it:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IhywE.png&quot; alt=&quot;lewis picture&quot;&gt;&lt;br&gt;&#xA;While this picture is also consistent with a valence bond theory (VB) approach, it explains bonding in a different way than MO theory. However, the conclusions we will ultimately get are essentially the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shall be noted, that MO theory and VB theory are complementary theories, but the different explanations should not be mixed, even when the conclusions are essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-10-15T08:42:56.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18459" PostTypeId="2" ParentId="18452" CreationDate="2014-10-15T10:38:09.750" Score="-1" Body="&lt;p&gt;Different vegetables contain different types of Zinc compounds, so we can't say specifically which type of zinc compound is there in vegetables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc Oxide is the most used supplement in the U.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc carbonate is also present, which our body cannot absorb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also they can contain  zinc picolinate, zinc gluconate or zinc citrate (zinc picolinate is better than zinc gluconate.), zinc glycinate, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://en.wikipedia.org/wiki/Zinc#Dietary_intake&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2015-04-27T13:47:04.237" LastActivityDate="2015-04-27T13:47:04.237" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18460" PostTypeId="1" CreationDate="2014-10-15T10:52:01.490" Score="2" ViewCount="124" Body="&lt;p&gt;In the formation of dichlorocarbene, sodium trochloroacetate (in tetrachloroethene) is refluxed at 130 deg C resulting in decarboxylation which forms a trichloromethyl anion. (Does this step happen because of the increase in entropy? What is the driving force for this step?) Alpha elimination of Cl- then occurs to give dichlorocarbene but as carbenes are so unstable surely this final step would be energetically unfavourable? Why does it still happen? Is it just because the reaction is being heated? Does this provide enough energy for this step to happen? Is this step in equilibrium?&lt;/p&gt;&#xA;" OwnerUserId="9326" LastActivityDate="2014-10-15T10:52:01.490" Title="Why does Cl- get eliminated from a trichloromethyl anion to give dichlorocarbene?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2014-10-15T14:09:28.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="18461" PostTypeId="1" CreationDate="2014-10-15T11:23:24.220" Score="2" ViewCount="168" Body="&lt;p&gt;Is it possible to create a 100% identical copy of a chemical? For example, a 100% identical copy of vitamin A would have the same shape, the same formula, and the same structure.&lt;/p&gt;&#xA;" OwnerUserId="9327" LastEditorUserId="15410" LastEditDate="2015-05-18T04:09:29.713" LastActivityDate="2015-05-18T04:09:29.713" Title="Is it possible to create a 100% identical copy of a chemical?" Tags="&lt;synthesis&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18462" PostTypeId="1" CreationDate="2014-10-15T11:27:06.047" Score="4" ViewCount="4967" Body="&lt;p&gt;I am looking for the SI units of isothermal incompressibility and specific heat capacity at constant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I found was that isothermal compression was measured in &quot; $\mathrm{m^2}\ \mathrm{N^{-1}}$&quot; so I'm guessing isothermal incompressibility would be the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found that specific heat capacity was &quot; $\mathrm{J}\ \mathrm{K^{-1}}$&quot;so is this the same units for constant volume also?&lt;/p&gt;&#xA;" OwnerUserId="9328" LastEditorUserId="16088" LastEditDate="2016-09-25T01:36:20.890" LastActivityDate="2017-11-13T23:08:22.640" Title="SI Units for Isothermal Incompressibility and Specific heat capacity at constant volume" Tags="&lt;thermodynamics&gt;&lt;units&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18463" PostTypeId="1" CreationDate="2014-10-15T13:17:30.557" Score="2" ViewCount="804" Body="&lt;p&gt;What will benezendiazonium ion become if it is not maintained at a temperature below $5\ °\mathrm{C}$ during diazotization?&lt;/p&gt;&#xA;" OwnerUserId="9329" LastEditorDisplayName="user7951" LastEditDate="2015-05-18T18:20:54.687" LastActivityDate="2019-10-05T20:28:26.007" Title="Effect of temperature on benzenediazonium during diazotization" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18464" PostTypeId="1" AcceptedAnswerId="18486" CreationDate="2014-10-15T13:33:17.483" Score="5" ViewCount="635" Body="&lt;p&gt;Is it possible to derive the nuclear spin I=3/2 for $\ce{^23Na}$ from a term scheme or from something else from spectroscopy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the nucleus spin is empirical (and cannot be calculated from J and F values) or does it?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="5026" LastEditDate="2018-07-03T08:45:47.817" LastActivityDate="2018-07-03T08:45:47.817" Title="How to derive the nuclear spin of 23Na?" Tags="&lt;nuclear&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18465" PostTypeId="2" ParentId="18463" CreationDate="2014-10-15T13:35:45.283" Score="1" Body="&lt;p&gt;Diazonium salts are prone to ejection of molecular nitrogen. The process is fairly slow for aromatic diazonium salts, so the salts are usually relatively stable around 0 celsius (alkyldiazonium salsts, though, are a completely different matter). In case of benzenediazonium resulting phenyl-cation is extremely active and will react with first nucleophile present, in many cases it is water, so the product would be phenol. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-10-15T15:13:33.773" LastActivityDate="2014-10-15T15:13:33.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18466" PostTypeId="1" AcceptedAnswerId="18514" CreationDate="2014-10-15T15:16:47.407" Score="35" ViewCount="258795" Body="&lt;p&gt;What are the definitions of these three things and how are they related? I've tried looking online but there is no concrete answer online for this question.&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorUserId="5017" LastEditDate="2014-10-15T21:06:01.313" LastActivityDate="2015-09-05T07:58:16.607" Title="Difference between shells, subshells and orbitals" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="27" ContentLicense="CC BY-SA 3.0" />
  <row Id="18467" PostTypeId="1" AcceptedAnswerId="18475" CreationDate="2014-10-15T15:16:49.090" Score="5" ViewCount="28258" Body="&lt;p&gt;Why is alkaline $\ce{KMnO4}$ used in the oxidation of toluene to benzoic acid? Can acidified or neutral $\ce{KMnO4}$ be used in this conversion?&lt;/p&gt;&#xA;" OwnerUserId="9330" LastEditorUserId="16683" LastEditDate="2018-03-25T17:02:05.787" LastActivityDate="2018-03-25T18:21:36.603" Title="Can acidified or neutral KMnO4 oxidise toluene to benzoic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;organic-oxidation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18468" PostTypeId="1" CreationDate="2014-10-15T15:23:24.793" Score="7" ViewCount="15296" Body="&lt;p&gt;I've found out about something called resonance but I'm getting the impression that this indicates the kekule structure is possibly correct as it uses double bonds (I thought that the delocalised model was the only correct one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little confused on this so could someone clarify? &lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="5456" LastEditDate="2014-10-15T17:24:53.867" LastActivityDate="2019-08-10T11:51:27.053" Title="Is the Kekule structure of benzene completely wrong?" Tags="&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18469" PostTypeId="1" CreationDate="2014-10-15T15:39:44.240" Score="1" ViewCount="1600" Body="&lt;p&gt;Dimethyl dichlorosilane will undergo hydrolysis to form siloxane polymers. But why would its carbon analog dimethyl dichlorocarbon not be able to undergo a similar hydrolysis reaction to form polymers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that $\ce{Si-Cl}$ bond being much weaker than $\ce{C-Cl}$ bond would be easier to break and undergo hydrolysis, and that the geminal diol is unstable, hence it is less likely to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure of my answer though. Could anyone please explain and elaborate?&lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="5026" LastEditDate="2018-07-11T03:26:38.583" LastActivityDate="2018-07-11T03:26:38.583" Title="Why does dimethyl dichlorosilane undergo hydrolysis to form siloxane polymers, but its carbon analog does not?" Tags="&lt;organic-chemistry&gt;&lt;hydrolysis&gt;&lt;carbon-family&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18470" PostTypeId="2" ParentId="18468" CreationDate="2014-10-15T15:53:53.163" Score="11" Body="&lt;p&gt;In benzene, all of the carbon-carbon bond lengths are equal.  Therefore, &lt;strong&gt;the Kekule structure shown below is an incorrect representation of benzene.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uv8bh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is incorrect because it suggests that there are two different types of carbon-carbon bonds in benzene, a carbon-carbon double bond and a carbon-carbon single bond. Typically, the bond length in a carbon-carbon double bond is around 135 pm, a carbon-carbon single bond is around 147 pm long, while the carbon-carbon bond length in benzene is roughly in between at 140 pm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene is often drawn as follows to indicate that all of the carbon-carbon bonds have the same length and the pi electron system is continuous and uniformly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tEvDA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, when one is discussing the resonance properties of benzene, two Kekule structures are drawn with the double headed resonance arrow between them as shown below.  This is meant to indicate that these two Kekule structures are resonance structures of benzene, and that taken together (&lt;em&gt;e.g&lt;/em&gt;. in one Kekule structure a carbon-carbon bond is a single bond, in the other structure the same bond is a double bond; so &quot;real&quot; benzene is a simultaneous, superposition of these two structures where each carbon-carbon bond is in between a single and double bond), they represent the true electronic and physical structure of benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9tig3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-15T20:22:29.720" LastActivityDate="2014-10-15T20:22:29.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18471" PostTypeId="2" ParentId="18469" CreationDate="2014-10-15T16:09:34.367" Score="-1" Body="&lt;p&gt;Your guess is correct. The geminal diol is unstable because carbon atom can't hold more than one $\ce{-OH}$ group. Where as Si atom can hold three $\ce{-OH}$  groups. It is this property of silicon that makes the formation of organosilicon polymers possible.&lt;/p&gt;&#xA;" OwnerUserId="9330" LastEditorUserId="5456" LastEditDate="2014-10-15T17:23:46.927" LastActivityDate="2014-10-15T17:23:46.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18472" PostTypeId="2" ParentId="18469" CreationDate="2014-10-15T16:09:57.460" Score="3" Body="&lt;p&gt;Actually a $\ce{Si-Cl}$ bond is much stronger than a $\ce{C-Cl}$ bond.  It takes about 90 kcal/mole to break a $\ce{Si-Cl}$ bond, but only around 81 kcal/mole to break a $\ce{C-Cl}$ bond.  The strengths of the corresponding $\ce{Si-O}$ (110 kcal/mole) and $\ce{C-O}$ (85 kcal/mole) bonds also need to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the silicon reaction we need 90 kcal/mole to break the $\ce{Si-Cl}$ bond, but we get 110 kcal/mole back when we form the $\ce{Si-O}$; the reaction is exothermic by roughly 20 kcal/mole - a large amount.  In the analogous carbon case, we need 81 kcal/mole to break a $\ce{C-Cl}$ bond, and we only get 85 kcal/mole back when we form the $\ce{C-O}$ bond; the reaction is exothermic by only 4 kcal/mole - not much.  There is quite a difference in the overall driving force for the two reactions with the silicon case being energetically much more favorable than the carbon case!&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-15T16:09:57.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18473" PostTypeId="1" AcceptedAnswerId="18478" CreationDate="2014-10-15T16:13:50.400" Score="1" ViewCount="1554" Body="&lt;p&gt;What chemical substance/component would be obtained of dissolving $\ce{C60}$ in olive oil? Would buckminsterfullerene (bucky-ball) change its spherical fullerene structure (soccer ball-like) into something else ($\ce{C60OO}$) or will it stay the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIWzv.png&quot; alt=&quot;Buckminsterfullerene 3D ball&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fullerene#Safety_and_toxicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fullerene - Safety and toxicity&lt;/a&gt; at Wikipedia&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.owndoc.com/pdf/C60-Fullerene.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fathi Moussa's experiment study on rats in 2012 which doubled their lifespan&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://c60antiaging.com/c60-faq/the-color-of-the-carbon-60-in-olive-oil/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The color of the Carbon-60 in olive oil&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://c60antiaging.com/c60-oil-production/buckybal-olive-oil-production-process/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buckminsterfullerene-olive oil production process&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="478" LastEditorUserId="15410" LastEditDate="2015-05-18T04:03:37.467" LastActivityDate="2017-10-09T04:42:34.123" Title="What chemical substance is obtained by dissolving C60 in olive oil?" Tags="&lt;solubility&gt;&lt;solutions&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18474" PostTypeId="2" ParentId="18473" CreationDate="2014-10-15T16:30:41.300" Score="5" Body="&lt;p&gt;The typical reactions of C&lt;sub&gt;60&lt;/sub&gt; happen at the double bond: [4+2] cycloaddition, [3+2] cycloaddition, oxidation.  Just dissolving it in olive oil (will it even go up, I know it does go up in aromatic solvent) won't do anything.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-15T16:30:41.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18475" PostTypeId="2" ParentId="18467" CreationDate="2014-10-15T16:39:37.893" Score="5" Body="&lt;p&gt;Here are the three equations describing the reduction of manganese (and concurrent oxidation of whatever substrate may be present) under basic, neutral and acidic conditions respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Mn^{+7}O4- +e- -&amp;gt;~ Mn^{+6}O4^2-~~~~[basic]}$$&#xA;$$\ce{2H2O + Mn^{+7}O4- + 3e- -&amp;gt;~ Mn^{+4}O2 + 4OH-~~~~[neutral]}$$&#xA;$$\ce{8H+ + Mn^{+7}O4^{-} + 5e- -&amp;gt;~ Mn^{+2} + 4H2O~~~~[acidic]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acidic conditions are the most economical as 5 electrons are transferred per mole of manganese, but oxidation can be achieved under all 3 conditions. A problem with neutral conditions is the precipitation of insoluble $\ce{MnO2}$ and the need for its subsequent separation. A problem with all 3 conditions is that oxidation will only occur if the material to be oxidized is somewhat soluble in the reaction medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The milder basic conditions are generally preferred for the oxidation of aromatic alkyl side chains because with the milder basic conditions an organic co-solvent that is soluble in water can be employed to facilitate solubility of the organic substrate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cFYVa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-03-25T18:21:36.603" LastActivityDate="2018-03-25T18:21:36.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18476" PostTypeId="1" CreationDate="2014-10-15T16:47:37.377" Score="6" ViewCount="239" Body="&lt;p&gt;I am confused about the energies we are talking about when speaking in regards to the Maxwell Boltzmann Distribution and activation energies. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there an relationship between kinetic energy and free energy for ideal gases? I am wondering if it’s possible to plot the x-axis of a Maxwell Boltzmann Distribution in terms of free energy as opposed to kinetic energy or velocity.&lt;/li&gt;&#xA;&lt;li&gt;Is the activation energy the difference in the free energy of the activated complex and the average free energy of the gas (or any reactant)?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/17203/a-few-basic-questions-about-activation-energy-and-the-maxwell-boltzmann-distribu&quot;&gt;From the answer here&lt;/a&gt;, the activation energy mark on Maxwell Boltzmann Distributions &lt;a href=&quot;http://www.webchem.net/notes/how_far/kinetics/maxwel2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one&lt;/a&gt; refers to a minimum kinetic energy level. Since the activated complex/transition state is a state, I do not understand why having a minimum kinetic energy (and right collision orientation) means reactants will reach the transition state. How come there is no free energy requirement?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="45696" LastEditDate="2017-06-16T08:33:18.140" LastActivityDate="2017-06-16T08:33:18.140" Title="Free energy and kinetic energy for reacting gases" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18477" PostTypeId="2" ParentId="16651" CreationDate="2014-10-15T17:05:50.183" Score="2" Body="&lt;p&gt;I'm surprised that no one has mentioned Huheey yet.  This text has less emphasis on descriptive chemistry but explains the underlying concepts &lt;em&gt;extremely&lt;/em&gt; well.  As an undergraduate reading it (it wasn't an assigned text) was quite an enlightenment.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-15T17:05:50.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18478" PostTypeId="2" ParentId="18473" CreationDate="2014-10-15T17:07:34.780" Score="8" Body="&lt;p&gt;There is no chemical reaction between fullerene and olive oil, that is to say that the two molecules do not break or make chemical bonds to produce a new chemical species.  However, fullerene is soluble in olive oil and so there are typical solvent-solute interactions.  A homogenous solution of fullerene in olive oil produces a light purple color typical of many fullerene solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lii9Es.jpg&quot; alt=&quot;light purple color of fullerene solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/22498298&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moussa's 2012 experiment&lt;/a&gt; where he fed fullerene in olive oil to rats and found that their life span roughly doubled, many companies have been marketing a &lt;a href=&quot;http://c60antiaging.com/c60-faq/the-color-of-the-carbon-60-in-olive-oil/&quot; rel=&quot;nofollow noreferrer&quot;&gt;dark-red liquid&lt;/a&gt; comprised of olive oil and fullerene.  It is likely that in many of these highly colored liquids, the fullerene is not completely dissolved and light scattering by the fine particulates are producing the observed color.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="478" LastEditDate="2015-01-16T17:05:01.783" LastActivityDate="2015-01-16T17:05:01.783" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18479" PostTypeId="1" CreationDate="2014-10-15T17:55:15.490" Score="2" ViewCount="158" Body="&lt;p&gt;First of all i'm no chemist, this is not homework and if you need further clarification please ask. I'll be happy to provide!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;have 1 kg of any given chemical X&lt;/li&gt;&#xA;&lt;li&gt;A car battery&lt;/li&gt;&#xA;&lt;li&gt;A semi-controlled combustion chamber (chamber temp is initially 20ºC, pressure is 1atm, the chamber is either vacuum or filled with atmosphere &quot;air&quot;...)&lt;/li&gt;&#xA;&lt;li&gt;As much atmosphere &quot;air&quot; as you want, initially at 20ºC...&lt;/li&gt;&#xA;&lt;li&gt;A not so potent air-pressurizer... you might be able to get 3atm out of it...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The goal here is to get the most energy out of reaction as possible. So the question is: What chemical is X?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X doesn't have to be consumed all at once as in a explosion reaction... Damage/Destruction is not the goal&lt;/li&gt;&#xA;&lt;li&gt;If you can combine X with something in the air, and then use it to fuel the reaction, increasing the energy output, great!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9331" LastEditorUserId="1389" LastEditDate="2014-10-15T20:52:50.240" LastActivityDate="2014-10-15T21:04:14.653" Title="Greatest Energy Output Reaction to Atmosphere Air" Tags="&lt;energy&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18480" PostTypeId="2" ParentId="18462" CreationDate="2014-10-15T18:54:19.093" Score="0" Body="&lt;p&gt;By specific heat capacity at constant volume, do you mean $c_v$? If so the units should be J/kgK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If by isothermal incompressibility you mean the coefficient of isothermal compressibility found &lt;a href=&quot;http://petrowiki.org/Isothermal_compressibility_of_gases&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it looks like the units you have is correct (1/Pa).&lt;/p&gt;&#xA;" OwnerUserId="8117" LastActivityDate="2014-10-15T18:54:19.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18481" PostTypeId="1" CreationDate="2014-10-15T20:18:32.737" Score="10" ViewCount="58440" Body="&lt;p&gt;Why does the Brady's test, the reaction of 2,4-dinitrophenylhydrazine with aldehydes and ketones, fail with carboxylic acids?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="4231" LastEditDate="2014-10-16T17:34:12.553" LastActivityDate="2018-06-04T16:55:03.730" Title="Why doesn't the Brady's test (2,4-DNPH) work on carboxylic acids?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18482" PostTypeId="2" ParentId="18481" CreationDate="2014-10-15T20:42:48.500" Score="10" Body="&lt;p&gt;2,4-Dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) reacts readily with aldehydes and ketones &lt;em&gt;via&lt;/em&gt; a condensation reaction (the lone pair of electrons on the terminal amino group in 2,4-DNPH makes it a strong nucleophile and the condensation starts by the nucleophilic 2,4-DNPH attacking the electrophilic carbonyl carbon) to produce the corresponding hydrazone.  The hydrazine is usually a brightly colored yellow, orange or red compound, so the reaction is often used to test for the presences of an aldehyde or ketone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gEdwg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2,4-DNPH does not react with amides, esters or carboxylic acids.  As shown below for the case of an ester, an extra resonance structure can be drawn for these 3 types of compounds as compared to a ketone.  This extra resonance structure delocalizes some of the positive charge away from the carbonyl carbon onto the adjacent hetero-atom (oxygen in the case of the ester or carboxylic acid, nitrogen in the case of an amide).  This makes the carbonyl carbon less electrophilic in these compounds, and consequently attack by the nucleophilic 2,4-DNPH is less favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pZ0yV.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-16T00:18:01.117" LastActivityDate="2014-10-16T00:18:01.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18483" PostTypeId="2" ParentId="18466" CreationDate="2014-10-15T20:59:23.780" Score="10" Body="&lt;p&gt;Have a look &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html#shells&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Orbitals that have the same value of the principal quantum number $n$ form a shell. Orbitals within a shell are divided into subshells that have the same value of the angular quantum number $l$. Chemists describe the shell and subshell in which an orbital belongs with a two-character code such as 2p or 4f. The first character indicates the shell (n = 2 or n = 4). The second character identifies the subshell. By convention, the following lowercase letters are used to indicate different subshells.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;s:         l = 0&lt;/li&gt;&#xA;  &lt;li&gt;p:         l = 1&lt;/li&gt;&#xA;  &lt;li&gt;d:         l = 2&lt;/li&gt;&#xA;  &lt;li&gt;f:         l = 3&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is called an orbital might differ according to the context.&#xA;With orbitals in the context of shells and subshells one usually means atomic orbitals, i.e. two-electron eigenstates of an atom's Hamilton operator which are characterized by the three quantum numbers: the principal quantum number $n$, the angular quantum number $l$ and the magnetic quantum number $m$.&#xA;But often the word orbital is also used for spin-orbitals, i.e. one-electron eigenstates of the system's one-electron Hamilton operator which are characterized not only by $n$, $l$ and $m$ but also by the spin quantum number $m_{\mathrm{s}}$ which can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-15T22:31:42.430" LastActivityDate="2014-10-15T22:31:42.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18484" PostTypeId="2" ParentId="18479" CreationDate="2014-10-15T21:04:14.653" Score="2" Body="&lt;p&gt;this is a way too generalized question. you certainly need experimental data or extensive screening to make sure you are on the top of the volcano plot. have you heard the term adiabatic flame temperature? The ratio of inert to gasoline/fuel is a determining factor to get high combustion energy. For example the adiabatic flame temperature for methane is higher for a mixture of methane and air with 5-15 mol% methane. 5-15 mol% methane can ignite without any external source of fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the attached link to compare fuels according to their energy density:  &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy_density&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Energy_density&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for your question you can't have two variables to optimized with one equation, either you have to specify the chemical X or you have to specify what is the output energy of this process. If you know the chemical X with it's thermodynamic property than using basic heat balance equation you can calculate the output energy. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2014-10-15T21:04:14.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18486" PostTypeId="2" ParentId="18464" CreationDate="2014-10-15T22:04:58.760" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_shell_model&quot; rel=&quot;noreferrer&quot;&gt;nuclear shell model&lt;/a&gt; is a useful &quot;first approach&quot; to determining nuclear spin.  It doesn't always work, but it is a relatively simple way to make a first attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a nuclear shell energy diagram.  As you can see it is somewhat analogous to an electron orbital energy diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9KOIP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of shells are filled by neutrons and a separate set is filled by protons.  The protons will pair when possible, same applies to the neutrons.  Notice that shells with J=1/2 hold 2 nucleons (a nucleon can be either a proton or a neutron), shells with J=3/2 can hold 4 nucleons, shells with J=5/2 can hold 6 nucleons, and so on.  This is because the shell with (for example) J=5/2 has 3 sub-shells: J=5/2, J=3/2 and J=1/2, each capable of holding 2 nucleons.  Higher J sub-shells are &quot;usually&quot; filled first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply this model to a few examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^2H}$ (deuterium): 1 neutron, 1 proton; the neutron goes into the neutron $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; the proton goes into the proton $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; total nuclear spin = 1/2 + 1/2 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{17}O}$: 9 neutrons, 8 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the 9th neutron goes into the neutron $\ce{1d_{5/2}}$ shell and is unpaired, spin = 5/2; all 8 protons are paired in the 1s and two 1p shells, spin = 0; total nuclear spin = 5/2 + 0 = 5/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{23}Na}$: 12 neutrons, 11 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the last 4 neutrons go into the neutron $\ce{1d_{5/2}}$ shell and each set of two is paired, spin = 0; the first 8 protons fill the proton 1s and two 1p shells and are all paired, the next 2 protons go into the $\ce{1d_{5/2}}$ J=5/2 sub-shell and are paired, spin = 0, the last proton goes into the $\ce{1d_{5/2}}$ J=3/2 sub-shell and has spin 3/2; total nuclear spin = 0 + 3/2 = 3/2&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-15T22:04:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18487" PostTypeId="2" ParentId="16331" CreationDate="2014-10-15T23:01:59.390" Score="1" Body="&lt;p&gt;The air in the balloon will be saturated (with water vapour) and the air outside not. As a result there will be a vapour density (or partial pressure) gradient between the inside and outside of the balloon that will drive the mass flow of water vapour through the balloon and to the outside.&lt;/p&gt;&#xA;" OwnerUserId="8117" LastActivityDate="2014-10-15T23:01:59.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18488" PostTypeId="1" AcceptedAnswerId="18490" CreationDate="2014-10-16T00:17:50.020" Score="3" ViewCount="17318" Body="&lt;p&gt;I am having trouble figuring out how to calculate the detection limit. Here is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You and your friend working in a different lab have been given an unknown water sample that contains potassium.  You are both asked to measure the potassium concentration in the solution (10 replicate measurements).  The methods that you and your friend are using are different.  The results that were obtained are listed below.  Determine: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) the concentration, standard deviation and relative standard deviation for the unknown as measured using the two methods (check for outliers!), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) calculate the detection limit (3sigma) for each method, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) compare the standard deviations and evaluate whether the two averages are significantly different (or not) at the 95% confidence level.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RESULTS:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Standard    Method 1    Method 2&#xA;Concentration   Intensity   Intensity&#xA;(mg/L)  (nA)    (mV)&#xA;0.000   0.624        1.955 &#xA;0.000   0.488        2.490 &#xA;0.000   0.522        2.166 &#xA;0.000   0.355        1.500 &#xA;5.000   9.245       15.644 &#xA;10.000  17.069      31.220 &#xA;15.000  26.200      44.266 &#xA;20.000  33.881      62.394 &#xA;25.000  43.826      75.611 &#xA;&#xA;Replicate           &#xA;1       26.544      46.977 &#xA;2       25.449      49.120 &#xA;3       21.053      50.998 &#xA;4       24.353      46.615 &#xA;5       23.899      49.326 &#xA;6       24.010      46.666 &#xA;7       25.554      45.291 &#xA;8       23.549      42.995 &#xA;9       26.008      43.678 &#xA;10      24.404      49.012 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So my problem is with part (b). I know the detection limit is calculated with the formula &lt;code&gt;DL = 3sigma/slope&lt;/code&gt; and that I can find the slope by calculating &lt;code&gt;Sxy/Sxx&lt;/code&gt; but I'm not sure what points I'm supposed to use? What points do I use for the standard deviation (sigma) and for Sxy/Sxx? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried looking at examples in the textbook but nothing was really similar. I also tried searching online, but I couldn't find anything that helped me.&lt;/p&gt;&#xA;" OwnerUserId="9335" LastActivityDate="2014-10-16T00:37:15.053" Title="How to calculate the detection limit?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18489" PostTypeId="1" AcceptedAnswerId="18491" CreationDate="2014-10-16T00:28:22.243" Score="18" ViewCount="957" Body="&lt;p&gt;My professor has an anti-fetish for carbocations. In any mechanism that we draw with a carbocation, the maximum earnable credit is automatically &lt;em&gt;halved.&lt;/em&gt; In any case, he draws all mechanisms involving carbocations as concerted reactions; he never wants us to draw an carbocation by itself. If a carbocation &lt;em&gt;could&lt;/em&gt; be drawn (as in the case of the formation of an acetal; specifically the step in which water leaves) then we must have electrons flowing from somewhere else to give the carbon an octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is this valid? Are all mechanisms involving carbocations &lt;em&gt;concerted&lt;/em&gt;? Excluding the obvious case of SN1; I'm pretty sure he'd accept that carbocation (and he hasn't touched on SN1 ever). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-09-30T22:53:53.310" LastActivityDate="2015-09-30T22:53:53.310" Title="Are carbocations that bad?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18490" PostTypeId="2" ParentId="18488" CreationDate="2014-10-16T00:37:15.053" Score="3" Body="&lt;p&gt;Have a look at this: &lt;a href=&quot;http://www.chem.utoronto.ca/coursenotes/analsci/stats/LimDetect.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.utoronto.ca/coursenotes/analsci/stats/LimDetect.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to calculate the limit of detection from the standard error of the regression: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{\mathrm{LOD}} = \frac{3s_{y/x}}{b}$$ These values are calculated from the regression of &lt;em&gt;all&lt;/em&gt; the data points, not the standard deviation of any subset of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case we have replicate blank measurements (four measurements at C=0), so we can also calculate the limit of detection as simply 3 times the standard deviation of the blank measurements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{\mathrm{LOD}} = y_{\mathrm{blank}}+3s_{\mathrm{blank}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in instrument response units—convert to concentration from a known standard or from the regression equation)&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-16T00:37:15.053" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18491" PostTypeId="2" ParentId="18489" CreationDate="2014-10-16T01:03:41.717" Score="18" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are all mechanisms involving carbocations concerted?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many carbocations are stable, observable entities.  Many large, delocalized carbocations like trityl perchlorate form very stable salts (&lt;a href=&quot;https://books.google.com/books?id=lXfdBwAAQBAJ&amp;amp;pg=PA196&amp;amp;lpg=PA196&amp;amp;dq=trityl%20perchlorate%20structure&amp;amp;source=bl&amp;amp;ots=wBHrFBFFh0&amp;amp;sig=oTRH7tdvrtbF4aclRhv4WdMeb9Y&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=rGw5VditCobQtQX_34HADw&amp;amp;ved=0CF8Q6AEwDQ#v=onepage&amp;amp;q=trityl%20perchlorate%20structure&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt; see p. 196), other carbocations are easily observed by nmr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9S1EL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If free carbocations didn't exist as discrete intermediates how could you explain the stereochemical results from an SN1 reaction where a chiral starting material generates a planar, long-lived carbocation that can be attacked from both sides to produce an achiral product?  Any type of concerted process would be expected to proceed with inversion of configuration (SN2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XAmvq.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(&lt;a href=&quot;http://en.wikipedia.org/wiki/SN1_reaction&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-23T22:33:16.533" LastActivityDate="2015-04-23T22:33:16.533" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18492" PostTypeId="1" CreationDate="2014-10-16T02:32:43.577" Score="15" ViewCount="27273" Body="&lt;p&gt;Redox reaction is a type of chemical reaction, and is the result of electron transfer between chemical species. But, all chemical reactions somehow involve electron transfer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there chemical reactions &lt;em&gt;without&lt;/em&gt; electron transfer?&lt;/p&gt;&#xA;" OwnerUserId="7730" LastEditorUserId="5026" LastEditDate="2018-04-11T03:18:00.937" LastActivityDate="2020-01-23T21:39:00.727" Title="Do non-redox reactions exist?" Tags="&lt;redox&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18493" PostTypeId="2" ParentId="18492" CreationDate="2014-10-16T02:38:17.100" Score="7" Body="&lt;p&gt;Yes, there are precipitation reactions (which don't have electron transfer), e.g. silver cation and chloride anion combining to make silver chloride, a slightly soluble compound. Also barium sulfate will precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrolysis of polysaccharides and proteins: with the assistance of enzymes, water is use to break up polysaccharides into simple sugars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are others.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-16T02:38:17.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18494" PostTypeId="2" ParentId="18492" CreationDate="2014-10-16T03:05:21.810" Score="26" Body="&lt;p&gt;There are many, here are some examples&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Ion interchange reactions of various types, such as precipitation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl + AgNO3 -&amp;gt; AgCl \downarrow + NaNO3 }$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PCl5 + 4 H2O -&amp;gt;  H3PO4 + 5 HCl}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reorganization of bonds between atoms of same type, such as catalytic benzene synthesis from acetylene &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3 C2H2 -&amp;gt; C6H6}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ligand exchange&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ [Fe(H2O)6]^{3+} + 6CN- -&amp;gt; [Fe(CN)6]^{3-} + 6 H2O}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;polar oligomerization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ H2O + n CH2O -&amp;gt; HO-(-CH2-O-)_{n}-H}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="485" LastEditorUserId="41328" LastEditDate="2018-01-09T11:33:53.900" LastActivityDate="2018-01-09T11:33:53.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18495" PostTypeId="1" CreationDate="2014-10-16T03:39:57.657" Score="-5" ViewCount="168" Body="&lt;p&gt;I am trying to find a SAFE way to change the colour of my urine, from what I have read online, Methylene Blue is relatively safe, and will change urine to blue or green. Is it safe? Can I put this in a food or drink? If so what are safe amounts? If not does anyone know a way to change the colour of urine safely?&lt;/p&gt;&#xA;" OwnerUserId="9338" LastEditorUserId="2462" LastEditDate="2014-10-16T04:42:27.553" LastActivityDate="2014-10-16T04:42:27.553" Title="How to change color of urine safely?" Tags="&lt;safety&gt;&lt;color&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-10-16T06:00:37.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="18496" PostTypeId="1" CreationDate="2014-10-16T04:21:25.110" Score="5" ViewCount="151" Body="&lt;p&gt;I have some major confusion on irreducible representations. An example of a problem I have is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the character table for the point group of each of the follwing molecules, which irreducible representation would be used to classify the Cu $\mathrm{d}_{xz}$ orbital in $\ce{[CuCl5]^3-}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since this complex probably has a trigonal bipyramidal geometry, I'm starting with the $D_\mathrm{3h}$ character table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|cccccc|cc} \hline&#xA;D_\mathrm{3h} &amp;amp; E &amp;amp; 2C_3 &amp;amp; 3C_2 &amp;amp; \sigma_\mathrm{h} &amp;amp; 2S_3 &amp;amp; 3\sigma_\mathrm{v} &amp;amp; &amp;amp; \\ \hline&#xA;\mathrm{A_1'} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; &amp;amp; x^2+y^2,z^2 \\&#xA;\mathrm{A_2'} &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; R_z &amp;amp; \\&#xA;\mathrm{E'} &amp;amp; 2 &amp;amp; -1 &amp;amp; 0 &amp;amp; 2 &amp;amp; -1 &amp;amp; 0 &amp;amp; (x,y) &amp;amp; (x^2-y^2,xy) \\&#xA;\mathrm{A_1''} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; &amp;amp; \\&#xA;\mathrm{A_2''} &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; z &amp;amp; \\&#xA;\mathrm{E''} &amp;amp; 2 &amp;amp; -1 &amp;amp; 0 &amp;amp; -2 &amp;amp; 1 &amp;amp; 0 &amp;amp; (R_x,R_y) &amp;amp; (xz,yz) \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i don't know what exactly I need to &lt;em&gt;do&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the irreducible representation just the row label, i.e. the Mulliken term? Or is it one of the operations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the column with quadratic functions, $xz$ is paired with $\mathrm{E''}$, so that's my current best guess.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="16683" LastEditDate="2018-09-06T07:54:25.530" LastActivityDate="2018-09-06T07:54:25.530" Title="Finding the irreducible representation which an orbital transforms under" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2014-10-17T04:56:59.383" ContentLicense="CC BY-SA 4.0" />
  <row Id="18498" PostTypeId="2" ParentId="18495" CreationDate="2014-10-16T04:23:12.327" Score="4" Body="&lt;p&gt;Lots of common foods can colour urine; beets, berries, carrots, asparagus (as well as making it smell) and rhubarb to name a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google search will give you plenty of references, but to get you started, try &lt;a href=&quot;http://www.urinecolors.com/&quot; rel=&quot;nofollow&quot;&gt;the urine colors website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methylene Blue (CAS 61-73-4) in concentrated form is harmful if swallowed and is a known irritant. Methylene Blue solution (~1%) is also harmful if swallowed, and has reported warnings which include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;May cause gastrointestinal irritation, discoloration of oral mucosa, irritation of lips,&#xA;mouth and throat, paleness of complexion, lack of coordination or drowsiness&lt;/li&gt;&#xA;&lt;li&gt;Effects may be delayed. Laboratory experiments with methylene blue have resulted in mutagenic effects. Repeated contact may cause corneal damage. May cause adverse eproductive effects.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Eat lots of carrots or liquorice instead.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-10-16T04:37:57.560" LastActivityDate="2014-10-16T04:37:57.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18499" PostTypeId="2" ParentId="13519" CreationDate="2014-10-16T04:40:36.323" Score="1" Body="&lt;p&gt;You can watch &lt;a href=&quot;https://www.youtube.com/watch?v=yDarDR4vi1A&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube&lt;/a&gt; video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aJinO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Materials:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;2 1l Plastic Containers&lt;/li&gt;&#xA;  &lt;li&gt;Cotton rope&lt;/li&gt;&#xA;  &lt;li&gt;Aluminum mesh&lt;/li&gt;&#xA;  &lt;li&gt;Paper clips&lt;/li&gt;&#xA;  &lt;li&gt;Copper wire&lt;/li&gt;&#xA;  &lt;li&gt;Alligator clips&lt;/li&gt;&#xA;  &lt;li&gt;Electrical tape&lt;/li&gt;&#xA;  &lt;li&gt;Glue gun&lt;/li&gt;&#xA;  &lt;li&gt;Glue sticks&lt;/li&gt;&#xA;  &lt;li&gt;Aquarium air pump&lt;/li&gt;&#xA;  &lt;li&gt;Duct tape&lt;/li&gt;&#xA;  &lt;li&gt;Sludge&lt;/li&gt;&#xA;  &lt;li&gt;Hand spade&lt;/li&gt;&#xA;  &lt;li&gt;Bucket&lt;/li&gt;&#xA;  &lt;li&gt;Water&lt;/li&gt;&#xA;  &lt;li&gt;Salt&lt;/li&gt;&#xA;  &lt;li&gt;Small Pot&lt;/li&gt;&#xA;  &lt;li&gt;Spoon&lt;/li&gt;&#xA;  &lt;li&gt;Voltmeter&lt;/li&gt;&#xA;  &lt;li&gt;Drill gun&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Procedure&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Collect sludge from the bottom of a still creek or pond into a bucket. Such a source will most likely have plenty of anaerobic bacteria.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Drill one hole for copper wire on lids of containers. On one of the two lids, drill one hole for the air pump tube and one small hole for ventilation (this will not be sealed). Drill one hole on one side to both containers for salt bridge.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Prepare the electrodes. Fold aluminum mesh a few times over and bind with large paper clips. Strip ends off of copper wire and attach to both electrodes.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Insert copper wire and air pump tube into drilled holes on lids. Seal with hot glue or caulk.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Prepare the salt bridge. Heat water over stovetop and dissolve in as much salt as possible. Twist a long rope around itself to create a thicker rope. If necessary, cut the   rope to approx. 15 cm. Soak the rope in the salt water. Once damp, wrap the rope in one layer of electrical tape and one of duct tape, but keep the ends exposed.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Insert each end of the salt bridge into the drilled holes on the sides of the containers. Seal with hot glue or caulk and extra tape (as needed).&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Fill one container almost to the rim with sludge and the other with water.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Submerge the electrodes into the sludge and water. Close the lids of the containers, and make sure the one over the sludge is airtight.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Anaerobic bacteria should be exposed to as little oxygen as possible.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Attach alligator clips to each loose end of copper wire. Clamp the alligator clips onto the voltmeter probes. Make sure the red probe is attached electrically to the electrode in the water. The black probe should connect to the electrode submerged in the sludge.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Turn on the aquarium air pump.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Turn the dial on the voltmeter to 20 VDC (can also be marked by the symbol,      ) to measure the force of electricity moving through the circuit in terms of volts. Turn the dial to 2000m to measure it in terms of millivolts, a thousandth of a volt, to observe a more accurate reading.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Turn the dial to 10A. This number is the flow of current the microbial fuel cell is generating, measured in amps.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;source:&lt;a href=&quot;http://www.instructables.com/id/DIY-Microbial-Fuel-Cell-easy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-10-16T04:40:36.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18500" PostTypeId="2" ParentId="16013" CreationDate="2014-10-16T05:43:37.827" Score="1" Body="&lt;p&gt;Fluoride has a high affinity for $Fe^{3+}$, and $Fe^{3+}$ forms deeply coloured complexes with phenols.  The decomposition of the latter by fluoride ion is the basis for colorimetric determination of fluoride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ligands used in this kind of assay include 5-phenylsalicylaldehyde and resacetophenone.  For occasional testing of well water I'd recommend any analytical lab.  I'd be very surprised if Merck didn't offer any fluoride test strips, at their price of course, although whether they go down to the 5 ppm range is a different question.  For those that really like to spend money, there's fluoride-sensitive electrodes.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-16T06:11:27.790" LastActivityDate="2014-10-16T06:11:27.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18502" PostTypeId="2" ParentId="18492" CreationDate="2014-10-16T08:29:26.557" Score="5" Body="&lt;p&gt;Of course there are non-redox reactions, for example, acid-base reactions. They do not involve any electron transfer only protons move around (so no oxidation numbers change). Here are some basic examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\ce{HCl + H2O &amp;amp;&amp;lt;=&amp;gt; Cl- + H3O+}\\&#xA;\ce{HNO3 + H2O &amp;amp;&amp;lt;=&amp;gt; NO3- + H3O+}\\&#xA;\ce{CH3-COOH + H2O &amp;amp;&amp;lt;=&amp;gt; CH3-COO+ + H3O+}\\&#xA;\ce{HCOOH + H2O &amp;amp;&amp;lt;=&amp;gt; HCOO- + H3O+}\\&#xA;\ce{HCl + NaOH &amp;amp;&amp;lt;=&amp;gt; NaCl + H2O}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="9344" LastEditorUserId="41328" LastEditDate="2018-01-09T11:35:35.830" LastActivityDate="2018-01-09T11:35:35.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18503" PostTypeId="1" CreationDate="2014-10-16T08:33:40.090" Score="0" ViewCount="726" Body="&lt;p&gt;When we say that the electrode potential for $\ce{Co}^2$$^+$/$\ce{Co}$ is -0.28 V and $\ce{Cu}^2$$^+$/$\ce{Cu}$ +0.337 V, what is the physical meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you conceptually explain when two electrons react with $\ce{Cu}^2$ or $\ce{Co}^2$to form the metal, what happens and what is the source of these different potentials?&lt;/p&gt;&#xA;" OwnerUserId="7730" LastEditorUserId="5456" LastEditDate="2014-10-16T08:48:18.647" LastActivityDate="2014-10-16T16:34:14.393" Title="What is the meaning of electrode potential?" Tags="&lt;electrochemistry&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18507" PostTypeId="1" CreationDate="2014-10-16T15:56:23.207" Score="2" ViewCount="94" Body="&lt;p&gt;I am trying to determine the reaction rate of pepsin at different pH values by performing an experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using egg-white suspension as the substrate and am adding differently concentrated $\ce{HCl}$ to the substrate to vary the pH of the reaction. Eventually, I add the enzyme. However, for the reaction rate I require the time and the substrate disappeared and product formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the volume of the egg-white suspension different before and after the enzymatic reaction so that I could use it to calculate the reaction rate or does the protein's volume remain the same after the reaction? &lt;/p&gt;&#xA;" OwnerUserId="9350" LastEditorUserId="95133" LastEditDate="2021-04-13T07:37:49.787" LastActivityDate="2021-04-13T07:37:49.787" Title="Is the volume of a protein substrate equal before and after an enzymatic reaction? (pepsin degrading egg white suspension)" Tags="&lt;experimental-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18508" PostTypeId="1" AcceptedAnswerId="18512" CreationDate="2014-10-16T16:04:20.807" Score="8" ViewCount="38171" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I know atoms are held together by Ionic and Covalent bonds and i understand the ionically bonded atoms are held together by electrostatic forces. What about covalent bonds?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are molecules of bonded atoms held together?Are ionic compounds just aggregates of oppositely charged ions and hence don't have any definitive &quot;molecules&quot;. What about molecules of covalently bonded atoms?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="15410" LastEditDate="2016-05-19T09:21:51.973" LastActivityDate="2016-05-19T09:21:51.973" Title="What holds atoms together?" Tags="&lt;bond&gt;&lt;atoms&gt;&lt;molecules&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ClosedDate="2017-08-26T21:55:38.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="18509" PostTypeId="2" ParentId="18507" CreationDate="2014-10-16T16:32:04.493" Score="2" Body="&lt;p&gt;The volume of the protein will probably change (although not by much) because it will be denatured as it is degraded and as the pH decreases. As it denatures, the hydrophilic surface area and hydrophobic interior volume will change, leading to a change in the local water liquid structure and density. Denatured, short polypeptides also pack into smaller volumes than native proteins do, and so that would also cause the volume to decrease in most cases. The amount of water in the system will affect the magnitude of the change. I would predict that the volume would decrease slightly, probably on the order of what happens with a few degrees of temperature increase of water (around 0.1%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the short answer is &quot;Yes, the volume will change, but probably not consistently or by enough to be used to reliably measure the extent of the reaction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not my area of expertise, so I don't know what the current standard protocols are for determining reaction rates for pepsin reactions. However, I was able to find a couple of research papers that talk about using &lt;a href=&quot;http://en.wikipedia.org/wiki/Ninhydrin&quot; rel=&quot;nofollow&quot;&gt;ninhydrin&lt;/a&gt; to measure reaction extent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bip.cnrs-mrs.fr/bip10/CBKnowles69.pdf&quot; rel=&quot;nofollow&quot;&gt;The pH-Dependence of Pepsin-Catalyzed Reactions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.agriculturejournals.cz/publicFiles/28541.pdf&quot; rel=&quot;nofollow&quot;&gt;Kinetics of Hydrolysis of Egg White Protein by Pepsin&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second study, they quenched the reaction at a specific time by adding NaOH to bring the pH up, then centrifuged the products before measuring the reaction extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not a professional scientific study (meaning it's a school project, or something like that) then you might look at the method used in this report:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.scribd.com/doc/94539412/The-Effect-of-pH-on-the-Reaction-between-Pepsin-and-Egg-White-REPORT&quot; rel=&quot;nofollow&quot;&gt;The Effect of pH on the Reaction between Pepsin and Egg – White&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, they just added pepsin and watched to see how long it took for the solution to become &quot;clear&quot;. This won't give you very good accuracy or precision, but it might work for a science project or school report.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-10-16T16:32:04.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18510" PostTypeId="1" AcceptedAnswerId="18522" CreationDate="2014-10-16T16:34:03.537" Score="3" ViewCount="470" Body="&lt;p&gt;For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the orbital approximation (ie the solutions obtained after ignoring electronic repulsions in the Hamiltonian) of the many electron atom( say helium) how does one go about finding total orbital angular momentum? &lt;/p&gt;&#xA;" OwnerUserId="9349" LastEditorUserId="189" LastEditDate="2014-10-16T20:39:03.550" LastActivityDate="2014-10-16T20:58:53.417" Title="Orbital angular momentum" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18511" PostTypeId="2" ParentId="18503" CreationDate="2014-10-16T16:34:14.393" Score="1" Body="&lt;p&gt;The potential &lt;em&gt;difference&lt;/em&gt; between two electrodes represents the voltage that is generated between the two electrodes that will allow the transfer of current/electrons. Physically it's basically a measure of how strongly one electrode attract electrons over the other which leads to the flow of current.&lt;/p&gt;&#xA;" OwnerUserId="8117" LastActivityDate="2014-10-16T16:34:14.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18512" PostTypeId="2" ParentId="18508" CreationDate="2014-10-16T17:04:04.517" Score="14" Body="&lt;p&gt;In non-nuclear chemistry, &lt;em&gt;everything&lt;/em&gt; is electrostatic interactions. This is why you can learn and predict so much just by &quot;following the electrons&quot;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Covalent bonds are also formed because of electrostatic interactions - they are just more complicated conceptually than ionic (actually, ionic bonds are more accurately described by wavefunctions, we just try to keep things simple in the beginning). Since electrons exist as waves, when you confine them they start to do weird stuff (or what seems weird to us). For example, the shapes of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomic orbitals&lt;/a&gt; are all the way they are because the electrons act like standing, 3-dimensional waves trapped between a nucleus with a positive charge and a &quot;zero-point&quot; at infinite distance. When you put two atoms close together, the electrons on different atoms interact with each other, and the wavefunction becomes much more complicated. The result is that in some pairs of atoms, the wavefunctions combine to form bonding orbitals.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the short answer to your first question is: &quot;Molecular orbitals hold atoms together in covalent bonds, and those are a result of electrostatic interactions and the quantum nature of electrons.&quot;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Yes, ionic compounds are large collections of ions, and you can't really define &quot;molecules&quot; for them - instead we talk about &quot;formula units&quot; which are the lowest possible whole-number ratio of elements that represent the compound. Groups of covalently bonded atoms are also held together by electrostatic interactions, but since the covalent bonds are so much stronger, a molecular compound can exist &quot;on its own&quot; as a single molecule. Collectively, the forces that hold collections of molecules together are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;van der Waals forces&lt;/a&gt; if they don't involve ions. In any atom or molecule, there is never a completely uniform charge density on the surface. For some molecules, this is extreme (water is a good example) and we say it is very &lt;strong&gt;polar&lt;/strong&gt;, or that it has a large &lt;strong&gt;dipole moment.&lt;/strong&gt; This is just another way of saying that one part has a negative charge and the other has a positive charge. In water it looks like this (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_polarity&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3lqA4.png&quot; alt=&quot;Electrostatic potential map of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture, red means &quot;more electrons&quot; and blue means &quot;less electrons.&quot; Water can form &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen bonds&lt;/a&gt;&lt;/strong&gt;, which are very strong electrostatic interactions. Some atoms and molecules have an &lt;em&gt;almost&lt;/em&gt; uniform charge density on the surface. We call these &quot;non-polar&quot; molecules - noble gases are good examples. However, even noble gases have what is called an &lt;a href=&quot;https://www.chem.purdue.edu/gchelp/liquids/inddip.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;induced dipole&lt;/a&gt; due to statistically correlated fluctuations in electron density when the atoms are near each other. As a result, even noble gases can be cooled to the point where they become liquid - the very, very weak electrostatic interactions will hold them together at low temperature, when they are not moving very fast. These forces are called &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;London Dispersion Forces&lt;/a&gt; - after the guy who first described them. London dispersion forces are important, because they are found in &lt;em&gt;all&lt;/em&gt; molecules - polar or not. In fact, this is what makes most plastics solid. Polyethylene, for example, is made of very long chains of essentially non-polar molecules (&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;from wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3jPCa.png&quot; alt=&quot;Polyethylene space-filling model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each chain is attracted to the others through weak London dispersion forces, however, since each chain has tens of thousands of atoms, those tiny forces quickly add up to make large forces holding the polymer together. This is why polyethylene is solid at room temperature, and can be used to make things like shopping bags!&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-10-16T17:04:04.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18513" PostTypeId="1" AcceptedAnswerId="18518" CreationDate="2014-10-16T17:08:31.947" Score="13" ViewCount="48119" Body="&lt;p&gt;I have just put together on my mind these two facts: caffeine is an alkaloid and brewed coffee is slightly acidic (pH = 5). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Biology teacher and my Chemistry teacher could not elaborate satisfying answers, but my Physics teacher said it could be because of substances to the industrial coffee powder (hillbillies in Brazil, for instance, do not display gastritis attained from too much coffee).&lt;/p&gt;&#xA;" OwnerUserId="9351" LastEditorUserId="30836" LastEditDate="2020-04-04T21:10:02.700" LastActivityDate="2020-07-03T20:54:16.533" Title="If caffeine is an alkaloid, why is coffee acidic?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;food-chemistry&gt;&lt;alkaloids&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18514" PostTypeId="2" ParentId="18466" CreationDate="2014-10-16T17:31:14.920" Score="42" Body="&lt;p&gt;Here's a graphic I use to explain the difference in my general chemistry courses:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i8V5L.png&quot; alt=&quot;Shells, sub-shells, and orbitals&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All electrons that have the same value for &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;&lt;/strong&gt; (the principle quantum number) are in the same &lt;strong&gt;shell&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Within a shell (same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;), all electrons that share the same &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;&lt;/strong&gt; (the angular momentum quantum number, or orbital shape) are in the same &lt;strong&gt;sub-shell&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;When electrons share the same &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;&lt;/strong&gt;, &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;&lt;/strong&gt;, and &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$m_l$&lt;/span&gt;&lt;/strong&gt;, we say they are in the same &lt;strong&gt;orbital&lt;/strong&gt; (they have the same energy level, shape, and orientation)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So to summarize:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; - shell&lt;/li&gt;&#xA;&lt;li&gt;same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; - sub-shell&lt;/li&gt;&#xA;&lt;li&gt;same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$m_l$&lt;/span&gt; - orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by &amp;quot;shells&amp;quot; and &amp;quot;sub-shells.&amp;quot; For you, for now, orbital means &amp;quot;place where up to two electrons can exist,&amp;quot; and they will both share the same &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$m_l$&lt;/span&gt; values, but have opposite spins (&lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-10-16T17:31:14.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18515" PostTypeId="2" ParentId="16809" CreationDate="2014-10-16T17:42:34.117" Score="1" Body="&lt;p&gt;Vegetable oil is extracted from plant. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Vegetable_oil&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the term &quot;vegetable oil&quot; can be narrowly defined as referring only to plant oils that are liquid at room temperature,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, answer is &lt;strong&gt;no.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it occurs naturally than we will not call it vegetable oil.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-10-16T17:51:26.680" LastActivityDate="2014-10-16T17:51:26.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18516" PostTypeId="2" ParentId="18513" CreationDate="2014-10-16T17:55:08.870" Score="9" Body="&lt;p&gt;Alkaloids are low molecular weight nitrogen-containing compounds, mostly from plants. The nitrogen is usually in the form of amines and this often results in a basic compound, but there are many alkaloids where this is not the case. In fact, caffeine is a pretty weak base. I calculated the pKa of the imine nitrogen with MarvinSketch as -.92, so caffeine is not going to affect the pH of coffee much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, caffeine isn't even a major component of brewed coffee, only accounting for ~1% of the dry mass. If you look &lt;a href=&quot;http://monographs.iarc.fr/ENG/Monographs/vol51/mono51-6B.pdf&quot;&gt;here&lt;/a&gt; you'll see that brewed and green coffee naturally contains a fair bit of soluble acids, including things like nicotinic acid (vitamin B3), which probably accounts for the pH of brewed coffee, rather than the caffeine or any processing.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-16T17:55:08.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18517" PostTypeId="2" ParentId="18481" CreationDate="2014-10-16T18:01:13.810" Score="5" Body="&lt;p&gt;It's for the same reason that you can't make amides from a carboxylic acid and an amine: the amine deprotonates the acid.  The carboxylate anion is resonance-stabilized and the carbonyl carbon is insufficiently electrophilic.  You couldn't form a tetrahedral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same here: the hydrazone is formed from a ketone through nucleophilic attack on the carbon, the intermediate being a hemiacetal.  You won't get a hydrazide derivative here, although from an ester with sufficiently forcing conditions you might.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-16T18:01:13.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18518" PostTypeId="2" ParentId="18513" CreationDate="2014-10-16T18:07:35.940" Score="24" Body="&lt;p&gt;Coffee contains hundreds, if not thousands, of other compounds in addition to caffeine.  Included among these other compounds are many acids.  Many small, organic acids such as  citric, malic, lactic, pyruvic and acetic acid are present, but both quinic acid and chlorogenic acid (and their derivatives) are usually present in even higher concentration.  Phosphoric acid, an inorganic acid, is also present. The exact concentration of these various acids depends upon processing variables such as roast conditions and grind size.  Here is a &lt;a href=&quot;http://www.coffeeresearch.org/science/sourmain.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to an interesting, one-page discussion of the subject.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quinic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;quinic acid&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mocoe.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorogenic_acid#:%7E:text=Structurally%2C%20chlorogenic%20acid%20is%20the,hydroxyl%20of%20L%2Dquinic%20acid.&quot; rel=&quot;nofollow noreferrer&quot;&gt;chlorogenic acid&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sCPY2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sCPY2.png&quot; alt=&quot;chlorogenic acid&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; In Season 3, episode 6 of &amp;quot;Breaking Bad&amp;quot;, we meet Walt's new lab assistant Gale, who is brewing coffee using a lab glassware set-up.  Walt says that it is the best coffee he has ever tasted and Gale responds by mentioning &amp;quot;&lt;strong&gt;quinic acid&lt;/strong&gt;&amp;quot; as something to be considered in brewing great coffee.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-07-03T20:54:16.533" LastActivityDate="2020-07-03T20:54:16.533" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18519" PostTypeId="2" ParentId="18489" CreationDate="2014-10-16T18:15:40.653" Score="12" Body="&lt;p&gt;The mention of acetal chemistry leads me to have some sympathy for your instructor, although this person is taking it to the extreme with the grading policy. Below are two mechanisms for a trans-acetalization reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k4OVR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that mechanism A is one that your instructor would give half credit (since it goes through a carbocation) and mechanism B is one that your instructor would give full credit. I prefer mechanism B, and I think it's fine for your instructor to agree. However, &lt;em&gt;these are identical mechanisms&lt;/em&gt;. How is that? Because the third structure in each mechanism are the same thing, just different resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if they are identical mechanisms, why do I prefer mechanism B? Because that resonance structure explains something about the reaction, particularly in relation to a reaction that you have learned before, the SN1 reaction. In the typical SN1 reaction, formation of a carbocation is slow. However in the trans-acetalization reaction here (really just a variant of the SN1), that carbocation formation is much faster. Why? Because the carbocation formed is stabilized by the ability of the adjacent oxygen to donate electrons, giving an intermediate that has a resonance structure where all atoms obey the octet rule. I think that's an exceptionally important consideration, and one that helped me understand and recognize reactions of carbonyls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think that it would be unfair to give different levels of credit for mechanisms A and B. However, I do appreciate what your instructor is trying to get across, although s/he may not be doing a good job at it.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2015-04-23T22:56:44.553" LastActivityDate="2015-04-23T22:56:44.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18520" PostTypeId="2" ParentId="18476" CreationDate="2014-10-16T18:22:05.527" Score="3" Body="&lt;p&gt;My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In nature there are three main distributions that &quot;things&quot; follow: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Boltzmann&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Bose-Einstein&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Fermi-Dirac&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the &quot;way&quot; energy is distributed among the energy levels of a molecule/atom/generic particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular &quot;state&quot;. The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to&#xA;\begin{equation}&#xA;e^{-\frac{E}{KT}}&#xA;\end{equation}&#xA;Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have already come across this factor in kinetic theory. &lt;strong&gt;!!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$. &#xA;\begin{equation}&#xA;f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}}&#xA;\end{equation} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Maxwell velocity distribution is just a special case of a Boltzmann distribution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any questions I will try to answer them when I can! :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2014-10-17T08:54:07.543" LastActivityDate="2014-10-17T08:54:07.543" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18521" PostTypeId="1" CreationDate="2014-10-16T19:51:23.683" Score="8" ViewCount="1800" Body="&lt;p&gt;With something like a reaction or phase change, all sources use the criterion that $dG &amp;lt; 0$ for the reaction to be spontaneous and then substitute an appropriate expression for $dG$ to the specific application. For phase change my book stats $(\mu'-\mu'')dn' &amp;lt; 0$ where $'$ is phase A and $''$ is phase B, $\mu$ is the chemical potential and $dn' &amp;gt; 0$ means gaining of molecules in phase A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, reactions require activation energy and phase change require nucleation (I think this is kind of like an activation energy); both of these processes require an increase in $G$ before a larger decrease. Why do we not use $\Delta G$ (between initial and final equilibrium states) instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused by the use of dG (a rate differential) as opposed to $\Delta G$ (a net change between end states) because I'm not sure how one would integrate this for say, a phase change, since it must pass through a potential barrier where G must increase. Is the integral just $G_2-G_1$ between the end states and thermodynamically the process ignores the barrier (since thermodynamics deals with equilibrium only)? A typical thermodynamic process I'm imagining is quasistatic compression, where we can get work from the integral of PdV and P is defined as it goes through an infinite number of equilibrium states; I'm not sure if it's possible to draw a parallel here for free energy and phase change/reacting systems.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/e/ee/Plot_of_classical_nucleation_theory_prediction_for_the_free_energy_of_a_nucleus_as_a_function_of_radius.png&quot; rel=&quot;noreferrer&quot;&gt;this graph&lt;/a&gt; it looks like G increases then decreases to stable equilibrium after phase change. If we find dG/dr from this graph and integrate dG(r), should we get the same $\Delta G$ as from thermodynamics (before and after phase equilibrium)? If so, this seems to me like finding work of a rapid piston-cylinder compression of an insulated system; PdV is undefined since we are not passing through a set of equilibrium states but we can still find work from $\Delta U$. Difference is that in this case, our initial state is technically not in equilibrium/stable since the system favours a phase change, so we are beginning from a state that technically does not exist on the phase diagram (like if we have a supercooled/superheated liquid).&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="8117" LastEditDate="2014-10-17T23:45:38.453" LastActivityDate="2015-10-25T00:10:14.347" Title="Why do we use dG &lt; 0 to describe a spontaneous process?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18522" PostTypeId="2" ParentId="18510" CreationDate="2014-10-16T20:27:31.623" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, you can't. The different components of the angular momentum operator $\hat{\mathbf{L}}$, i.e. $\hat{L}_{x}$, $\hat{L}_{y}$, and $\hat{L}_{z}$, do not commute, so you cannot simultaneously determine all of them. But the squared angular momentum operator, which gives you the squared length of the angular momentum vector commutes with the components of $\hat{\mathbf{L}}$, so you can simultaneously determine the length of the angular momentum and one of its components. Which component you choose is immaterial, usually $\hat{L}_{z}$ is chosen. What you get out of those two operators can be pictured like this: you know the length of the angular momentum vector and its z-component; the other two components are undetermined but because you know the length they can only take on values that lie on a circle centered around the ${L}_{z}$ axis, whose offset is determined by the value of ${L}_{z}$ and whose radius is determined by $| \sqrt{\mathbf{L}^{2}}|$. For more details have a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum_operator&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; or any general physical chemistry or quantum mechanics textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7V1jK.png&quot; height=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, in the orbital approximation (ie the solutions obtained after&#xA;  ignoring electronic repulsions in the Hamiltonian) of the many&#xA;  electron atom( say helium) how does one go about finding total orbital&#xA;  angular momentum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ignoring relativistic effects the Hamilton operator of a system consisting of $M$ nuclei and $N$ electrons, which are considered to be point charges takes on the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{H} &amp;amp;= \underbrace{ \sum_{j=1}^{M} - \frac{ \hbar^{2} }{ 2 M_{j} } \mathbf{\nabla}_{ \mathbf{R}_{j} }^{2} }_{ = \, \hat{T}_{\mathrm{K}} } + \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{M} \frac{ e^{2} Z_{i} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{R}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{K--K}} }  \\ &#xA;  &amp;amp; \quad \underbrace{- \sum_{i=1}^{N} \sum_{j=1}^{M} \frac{ e^{2} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{e--K}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{N} \frac{ e^{2} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{r}_{j} | } }_{ = \, \hat{V}_{\text{e--e}} }&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ \mathbf{R}_{j} $, $Z_{j}$, and $M_{j}$ are the coordinate, the charge number, and the mass of the $j^{\mathrm{th}}$ nucleus, respectively and $ \mathbf{r}_{i} $ is the coordinate of the $i^{\mathrm{th}}$ electron and $m_{\mathrm{e}}$ is the electron mass.&#xA;Furthermore $e$ is the elementary charge und $\epsilon_{0}$ is the vacuum permittivity.&#xA;The different terms of this Hamilton operator can be explained as follows: $\hat{T}_{\mathrm{e}}$ and $\hat{T}_{\mathrm{K}}$ are the terms representing the kinetic energy of the electrons and the nuclei, respectively, $\hat{V}_{\text{K--K}}$ describes the coulomb interaction between the nuclei, $\hat{V}_{\text{e--e}}$ is the interelectronic repulsion, and $\hat{V}_{\text{e--K}}$ represents the coulomb interaction between the electrons and the nuclei.&#xA;Using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Born-Oppenheimer approximation&lt;/a&gt; you can seperate the electronic degrees of freedom from the ones of the nuclei and treat the electronic problem independently and for the many-electron atom you also only have one nucleus to consider so the terms $\hat{T}_{\mathrm{K}}$ and $\hat{V}_{\text{K--K}}$ drop out of the Hamiltonian. If you now also ignore the electron-electron interaction you are basically left with a Hamilton operator describing two independent electrons moving in the electrostatic field of the nucleus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{H} &amp;amp;= \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \sum_{i=1}^{N}  \frac{- e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} | } }_{ = \, \hat{V}_{\text{e--K}} }&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this you can easily seperate the $N$-electron problem into $N$ one-electron problems&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{H} &amp;amp;= \sum_{i=1}^{N} \hat{h}_{i} \qquad \Rightarrow \qquad \Psi(\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) = \psi(\mathbf{r}_{1}) \cdots \psi(\mathbf{r}_{N})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each one-electron Hamiltonian $\hat{h}_{i} = - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} - \frac{e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} |}$ and one-electron eigenfunction $\psi_{n}(\mathbf{r}_{i})$ basically satisfy a Schroedinger equation for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen-like_atom&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen-like atom&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{h}_{i} \psi_{n}(\mathbf{r}_{i}) = \varepsilon_{n} \psi_{n}(\mathbf{r}_{i}) \qquad \text{with} \qquad \varepsilon_{n} = -\frac{m_{\mathrm{e}} e^{4} Z^{2} }{ 32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}} \frac{1}{n^{2}} \qquad n = 1, 2, ...&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using spherical coordinates, $x, y, z \to r, \theta, \phi$, you can rewrite the kinetic energy operator $- \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2}$ in terms of $\hat{\mathbf{L}}^{2}$ so that the Hamilton operator becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{h}_{i} = \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} r^{2}} \frac{ \partial }{ \partial r } \biggl( r^{2} \frac{ \partial }{ \partial r } \biggr) + \frac{ 1 }{ 2 m_{\mathrm{e}} r^{2} } \hat{\mathbf{L}}^{2} - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r }&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r$ is the distance between nucleus and electron. The operator $\hat{\mathbf{L}}^{2}$ commutes with the Hamilton operator $\hat{h}_{i}$ because $\hat{\mathbf{L}}^{2}$ commutes with itself and does not involve the variable&#xA;$r$. Likewise, the operator $\hat{L}_{z}$ commutes with $\hat{h}_{i}$ because it&#xA;commutes with $\hat{\mathbf{L}}^{2}$ and also does not involve the variable $r$.&#xA;Thus the operators $\hat{\mathbf{L}}^{2}$, $\hat{L}_{z}$, and $\hat{h}_{i}$ have simultaneous eigenfunctions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \hat{h}_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &amp;amp;= \varepsilon_{n} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \\&#xA;  \hat{\mathbf{L}}^{2} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &amp;amp;= \hbar^{2} \ell_{i} (\ell_{i} + 1) \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad \ell = 0, 1, 2,..., n-1 \\&#xA;  \hat{L}_{z} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &amp;amp;= \hbar m_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad m_{i} = -\ell_{i}, -\ell_{i} + 1, \ldots , \ell_{i} - 1, \ell_{i} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you have a well defined angular momentum for each electron in the system.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-16T20:58:53.417" LastActivityDate="2014-10-16T20:58:53.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18523" PostTypeId="2" ParentId="18521" CreationDate="2014-10-16T20:35:06.660" Score="11" Body="&lt;p&gt;Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs potential is a function of $P$, $T$ and $N$. That is to say:&#xA;\begin{equation}&#xA;G=G(T,P,N)&#xA;\end{equation}&#xA;Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles.&#xA;\begin{equation}&#xA;dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i&#xA;\end{equation}&#xA;Use some Maxwell relations (Look these up):&#xA;\begin{equation}&#xA;dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i&#xA;\end{equation}&#xA;By &lt;strong&gt;definition&lt;/strong&gt; we state the following formula at constant temperature and pressure:&#xA;\begin{equation}&#xA;dG=\sum _{i=1}^{I}\mu _idN_i&#xA;\end{equation}&#xA;Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$. &#xA;(As an aside), with a little product ruling we get:&#xA;\begin{equation}&#xA;dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big)&#xA;\end{equation}&#xA;This leaves us with the &lt;strong&gt;Gibbs-Duhem relation&lt;/strong&gt;:&#xA;\begin{equation}&#xA;\sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP&#xA;\end{equation} &#xA;Which is a handy result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits &quot;clump&quot; together) the Gibbs energy of the particle will rise ... this is due to two competing effects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the interface interaction (increasing energy) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the volume free energy (decreasing energy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch. &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2014-10-16T20:51:21.143" LastActivityDate="2014-10-16T20:51:21.143" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18524" PostTypeId="1" CreationDate="2014-10-16T20:51:38.080" Score="3" ViewCount="1917" Body="&lt;p&gt;In an ionic bond, are electrons actually transferred from one atom to the other to form a positive and a negative ion OR are ionic bonds just extremely polar covalent bonds? Sorry if this seems like a simple question but I've heard this somewhere so just wanted to make sure. &lt;/p&gt;&#xA;" OwnerUserId="8176" LastActivityDate="2015-06-30T13:15:51.543" Title="Ionic bonds- are electrons transferred?" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18527" PostTypeId="1" AcceptedAnswerId="18528" CreationDate="2014-10-16T23:18:56.630" Score="1" ViewCount="250" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/18476/confusion-about-the-relationship-between-free-energy-and-kinetic-energy-for-reac&quot;&gt;Continued from my previous question here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;With regards to the Boltzmann factor, I am confused on what type of terms we can substitute for E. I know kinetic energy is one, free energy as well, but can we also do internal energy, enthalpy etc.?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;From this link&lt;/a&gt; it looks like internal energy and molecular kinetic energy are related and for real gases the internal energy term contains other factors. Since U is a part of G, is it correct to say that kinetic energy in Maxwell Boltzmann Distributions make up a part of free energy?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am still unclear as to whether I can interpret activation energy as a difference in free energy between average reactant energy and activated complex like diagram 1?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it correct to say that, reaction cannot occur unless the reactant have free energy above that of the activated complex, but even if this was satisfied other factors such as minimum collision kinetic energy, orientation etc. must also be met for reaction to occur?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Addendum for Q3 and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18476/confusion-about-the-relationship-between-free-energy-and-kinetic-energy-for-reac&quot;&gt;Previous Post&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is with regards to the relationship between Maxwell Boltzmann Distributions like diagrams 1 and 2, if there even is one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The M-B distribution implies to me that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;activation energy $E_a$ (which I assume to be that in Arrhenius equation) is absolute given it is marked on an absolute energy scale &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;activation energy refers to a minimum absolute kinetic energy &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;activation energy is constant with temperature. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My confusion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;My impression has always been that activation energy (in the Arrhenius equation) refers to the difference in free energy between the transition state and reactant and that $E_a = \Delta G^{++}$ as noted in the latter two graphs, which appear to contradict the M-B Distribution. I'm not sure if there's a formal definition for activation energy that defines what &quot;energy&quot; it is.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The relationship between KE and activation energy (if any)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The effect temperature has on reactant and transition state free/kinetic energies. I believe that higher temperature lowers activation energy if it is the relative energy between the two. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Diagram 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMlUi.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pbhuv.png&quot; width=&quot;300&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="5026" LastEditDate="2018-03-08T12:17:20.183" LastActivityDate="2018-03-08T12:17:20.183" Title="Regarding the free energy requirements for reacting gases" Tags="&lt;physical-chemistry&gt;&lt;free-energy&gt;&lt;reaction-coordinate&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18528" PostTypeId="2" ParentId="18527" CreationDate="2014-10-17T00:33:39.070" Score="4" Body="&lt;p&gt;I think what you are fundamentally saying is you are trying to relate classical thermodynamics to the molecular properties. It is best to have some firm familiarity with classical thermodynamics before attempting this. Classical thermodynamics does not require a molecular interpretation at all! (It helps a lot but is by no means necessary)!. I personally feel the answer I am about to write will help you more than trying to wade through terms and ideas/principles.... I hope so at least. If you don't like it then maybe someone else can provide a more quantitative solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The whole point of statistical thermodynamics is to provide a molecular interpretation. To do this we have to divide the system up into smaller sub-systems as I described earlier. The key to it is that there can be many &quot;configurations&quot; that give rise to the same macroscopic properties. The aim is to find the most probable one ... which is where your distributions come in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the energy of the microstates that is what you put in so that you can calculate the various macroscopic properties you quote above. &lt;strong&gt;This is the best way to think about it&lt;/strong&gt;. When you get to advanced theories this definition may change slightly. For now just accept that it is the energy of the micro state you are putting in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my honest opinion I don't know why courses teach the Maxwell-Boltzmann velocity distribution in a statistical thermodynamics course. I feel while a nice introduction to dealing with ensembles ... it leads to confusion (when I was learning this I had very similar questions with regard to what can be related to what, it only comes with time and exposure to the subject. So I know exactly where you are coming from!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Whenever you see kinetic energy you should be thinking temperature rather than trying to get the individual molecular speeds... what does knowing the instantaneous position/velocity of a mole of gas particles at one instant in time really lead to? (not much)... I haven't really seen anyone combine the thermodynamic potentials with something like velocity. If anyone reading this has then I would love to hear what you have to say!! When I was first learning thermodynamics I clung on to what I was familiar with ... potential / kinetic energies ... now I don't really think about them when you say thermodynamics at all ... I don't think about $G$ as being a &quot;part&quot; of $U$ ... I just treat them as if they were separate (hmm how to phrase this) &quot;properties&quot; of a given system under certain conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I would go further to say that I view them as mathematical tools ... functions of variables of the system that I can manipulate to express molecular properties in terms of measurable quantities .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Yes. You can indeed &lt;strong&gt;As is done by most theoretical chemists&lt;/strong&gt; use Gibbs energies to calculate the thermodynamic activation barrier in the starting material and transition state.... The problem, which admittedly I may have confused you earlier about, is that this is done via the Arrhenius equation &lt;strong&gt;A kinetics equation&lt;/strong&gt;. When in early undergraduate years you don't mix kinetics with thermodynamics. Later on it becomes (bar a factor of $2\pi$ or so ) acceptable to do so ... but you should not attempt this yet .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crucial point is that the activation energy graph's x-axis is the reaction coordinate. The reaction coordinate is a generalised coordinate which can be used (whatever it is) to describe the progression of the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally we deal very nicely in cartesian $x,y,z$ coordinates ... the idea of a generalised coordinate is that it can be used to reduce the dimensionality of the mathematical problem with some nifty rethinking of how we choose to describe the system .... for instance when talking about a simple harmonic diatomic molecule .. instead of choosing the location of both atoms in space (requiring 6 position coordinates) to fully describe the system as a function of time (as it oscillates back and forth ) i could use the relative separation (the internuclear distance) relate to the equilibrium bond length.... in this way it becomes much easier to describe the system ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was just one example of a generalised coordinate.... something that makes the description of the problem al little easier.... (especially if the system is subject to constraints, the generalised coordinates remove the constraint equation... anyway where were we.)... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: ** there is a big difference between reaction coordinate and the Maxwell-boltzmann graph you linked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) This question in itself has hit on a very big topic. This is the realm of classical thermodynamics. What you need to understand is the difference between the standard Gibbs energy and the Gibbs energy .... For the following just imagine that $A$→$B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very brief overview it all boils down to this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs free energy is a mathematical function... don't try to pin down WHAT it is... its just another energy we can describe the system in .... It is a function of $P$, $T$ and $N$. Lets make it easier and say that the number of particles does not change. i.e $dN=0$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Gibbs energy is a function of $T$ and $P$ if we change either of these variables ... $G$ will change too .... The starting materials will have a certain $G$ and the products will have a certain $G$ ... The difference between these under standard conditions is fixed by definition that we are controlling the variables. This is like the thermodynamic drive for the process if it were to occur at our specified standard conditions .... Over the course of the reaction &lt;strong&gt;this value does not change&lt;/strong&gt; since as I said it is the difference between the Starting Material and the Products under given conditions. .... This is also for pure values of the starting material and the product... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your reaction proceeds the composition varies i.e $dN\neq 0$. As I pointed out in another answer from the definition of the Gibbs energy this means that the value Gibbs function now changes... if you want a molecular picture it is due to the entropy of mixing adding in another factor to the &quot;disorder&quot; that has so far not been accounted for...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an illustration please follow this link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set5/lecture_set5.html&quot; rel=&quot;nofollow&quot;&gt;http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set5/lecture_set5.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are therefore two measures of the Gibbs energy. The one under standard conditions... unchanging for a given reaction... it is this one that has the $\Delta G^{\small o}$.. it is this one you get your electrode potentials off... it is this one that leads to the the equilibrium constant....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other one is ever changing as the reaction proceeds since the composition of the reaction vessel is ever changing as $A$ goes to $B$. At equilibrium the net rate of change of A to B and B to A is now zero right... this in fact defines an equilibrium point.... shifting it slightly in one way or the other will cause the system to respond to re-establish equilibrium .... Most importantly we define equilibrium as the point where $dN=0$ .... for both types (in this case one would go up as the other went down ... ) but at equilibrium $dN=0$ .... This means that $dG=0$ too from the mathematics I showed you before .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore equilibrium is the point where the change in the Gibbs energies of two substances in a mixture is zero.... The reaction has no more drive if you like .... You can equivalently say their chemical potentials are equal too ... but leave that for now..... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So thats it ... Im sure if you ask question 4 to someone else they will give you a more complete answer.... but since its getting on for 2:00 am here I think I'm done. Hope that helps and again if I have made any errors ... please edit away :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are describing on the end of question 4 is know as the steric factor... you basically compare the frequency of successful collisions in the transition state geometry &quot;linear&quot; for example with respect to a full sphere of reactivity .... this gives you your factor :)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;So the problem is your definition of the energy. I dot know if you have seen this before but perhaps this will help,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes you can indeed use statistical mechanics to obtain a value of $G$ ... The best way to do it is to get an expression for the Helmholtz free energy and then just Legendre transform it to $G$ ... $G=A+pv$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst yes at a simple level the internal energy does indeed &quot;contain&quot; the kinetic energy ... when you learn a bit more about quantum mechanics .. we never talk about kinetic energy ... I think it would help to actually leave the classical results behind and focus on the actual version of events... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reference to the uncertainty relation .. how do you think kinetic energy is something we can talk about at a molecular level? The answer is you don't. Once QM comes in it is not useful to talk about velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said how do we describe motion ? after all the Schrödinger equation is just an equation of motion. All the Schrödinger equation describes is the time evolution of a vector state in a certain vector space - Hilbert space, Which is the space in which we describe quantum mechanics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind ... all motion of the system is quantised ... can only have discrete values .... non-continuous values ... etc etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translational motion is quantised... so is vibrational ... so is rotational ..... each degree of freedom has its own quantised set of discrete energy levels .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equipartition theorem uses classical ideas and distributions to distribute all the available energy across these states ... roughly it says that each has energy of $\frac 1 2 KT$....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say $U$ includes kinetic energy is a primitive way of saying the internal energy is a function that describes the energy of all the modes in the system... (you would need a distribution to see how they are populated mind) ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equipartition theorem doesn't get it quite right when you consider the low temperature theoretical and experimental heat capacity of a substance..... Even at low temperature you are still expected to populate all modes with $\frac 12 KT$ ... but this doesn't happen since $KT$ does not have enough energy to populate said modes ... i.e it doesn't reach the quantised energy level requirement .... therefore fewer modes are available at low temperature...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is: just try and separate the different preconceived competing ideas you have when it comes to statistical mechanics .... have you had a lecture course on this yet? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is certainly not about having &quot;enough Gibbs energy to react&quot;... The Gibbs energy between a reactant and transition state $\Delta G ^{\ddagger}$ is not a variable thing as you are saying ... (obviously for a given set of chemicals),.... If it changed all the time then why would we write them down to compare them? The Gibbs energy in your picture is not the same Gibbs energy that goes to zero at equilibrium ... &#xA;\begin{equation}&#xA;\bigg(\frac{dG}{dq}\bigg)_{T,P}=\Delta _r G^{\small o}+RT\ln Q&#xA;\end{equation}&#xA;at equilibrium the left hand side .... the gradient of the graph in that link i sent you before ... is zero ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of change of the Gibbs energy with respect to the extent of the reaction is by definition zero at eqm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When equilibrium is reached $Q=K$ that is to say the reaction quotient becomes the equilibrium constant ....&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2014-10-17T16:44:37.960" LastActivityDate="2014-10-17T16:44:37.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18530" PostTypeId="1" AcceptedAnswerId="18531" CreationDate="2014-10-17T07:05:01.977" Score="3" ViewCount="296" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A metal &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt; forms a carbonate and a nitrate. The molecular formula of the carbonate of &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{QCO3}$&lt;/span&gt;.  What is the molecular formula of the nitrate of &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer says it is &lt;span class=&quot;math-container&quot;&gt;$\ce{Q(NO2)3}$&lt;/span&gt; which I think is incorrect; I think the answer should be &lt;span class=&quot;math-container&quot;&gt;$\ce{Q(NO3)2}$&lt;/span&gt;. Am I right? &lt;/p&gt;&#xA;" OwnerUserId="9357" LastEditorUserId="83758" LastEditDate="2020-03-04T12:10:00.840" LastActivityDate="2020-03-04T12:10:00.840" Title="What is the molecular formula of the nitrate of a metal Q that forms a carbonate of the form QCO3?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18531" PostTypeId="2" ParentId="18530" CreationDate="2014-10-17T07:22:02.633" Score="5" Body="&lt;p&gt;The carbonate anion is $\ce{CO3^{2-}}$. This tells us that the metal cation is doubly positively charged:&#xA;$$ \ce{Q^{2+} + CO3^{2-} -&amp;gt; QCO3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two anions mentioned in your question are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the nitr&lt;strong&gt;i&lt;/strong&gt;te anion $\ce{NO2^-}$&lt;/li&gt;&#xA;&lt;li&gt;the nitr&lt;strong&gt;a&lt;/strong&gt;te anion $\ce{NO3^-}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the question you're concerned with asks for the formula of the nitrate of $\ce{Q}$, the correct answer would be $\ce{Q(NO3)2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the compound $\ce{Q(NO2)2}$ is, at least from a charge perspective, also correct. But that is the nitr&lt;strong&gt;i&lt;/strong&gt;te of $\ce{Q}$.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-17T07:22:02.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18533" PostTypeId="2" ParentId="2815" CreationDate="2014-10-17T10:36:50.717" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the energy of an X-ray with particular wavelength is known&#xA;  (for aluminum Kα X-rays, Ephoton = 1486.7 eV), and because the emitted&#xA;  electrons' kinetic energies are measured, the electron binding energy&#xA;  of each of the emitted electrons can be determined by using an&#xA;  equation that is based on the work of Ernest Rutherford (1914):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{equation}&#xA;     E_\text{binding} = E_\text{photon} - \left(E_\text{kinetic} + \phi\right)   \end{equation}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where $E_\text{binding}$ is the binding energy (BE) of the electron, $E_\text{photon}$ is&#xA;  the energy of the X-ray photons being used, $E_\text{kinetic}$ is the kinetic&#xA;  energy of the electron as measured by the instrument and $\phi$ is the&#xA;  work function dependent on both the spectrometer &amp;amp; the material. This&#xA;  equation is essentially a conservation of energy equation. &lt;strong&gt;The work&#xA;  function term $\phi$ is an adjustable instrumental correction factor&#xA;  that accounts for the few eV of kinetic energy given up by the&#xA;  photoelectron as it becomes absorbed by the instrument's detector.&lt;/strong&gt; It&#xA;  is a constant that rarely needs to be adjusted in practice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, it is the absorbtion of the photoelectron by the detector of the spectrometer that causes the work function term. The work function of the measured sample should be part of $E_\text{binding}$ as this is the ionisation energy of the electron. However, I think the equation you quote might have it slightly wrong because if $B.E._F$ is the energy difference between the energy level of the electron and the Fermi energy then there is indeed some portion of the work function of the sample missing since $B.E._F$ is not equal to the ionisation energy.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-17T18:16:25.323" LastActivityDate="2014-10-17T18:16:25.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18534" PostTypeId="1" CreationDate="2014-10-17T10:46:23.940" Score="2" ViewCount="189" Body="&lt;p&gt;I am having a problem wrapping my mind around the idea that in $230\:\mathrm{mL}$ of $0.615\:\mathrm{M}$ $\ce{KCl}$ there is $0.141\:\mathrm{mol}$ of $\ce{Cl^-}$ ions. I guess I still do not understand moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this calculation saying that there $0.141\:\mathrm{mol}$ of $\ce{K}$ and $\ce{Cl}$ in one mole of $\ce{KCl}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this may be a silly question, but I just want to make sure I am following the correct way of thinking when it comes to this concept.   &lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="15410" LastEditDate="2015-05-21T13:18:38.297" LastActivityDate="2015-05-21T13:18:38.297" Title="Understanding a concentration term" Tags="&lt;concentration&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18535" PostTypeId="2" ParentId="5015" CreationDate="2014-10-17T11:11:01.080" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1959/df/df9592700007#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by Förster himself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This coupling is strongest if the corresponding optical transitions in&#xA;  both molecules are allowed for electric dipole radiation. Then these&#xA;  transitions are coupled not only to the radiation field but also to&#xA;  each other. Naturally, the interaction energy is of a dipole-dipole&#xA;  nature, depending on an inverse proportionality to the third power of&#xA;  the molecular distance. The probability of energy transfer is then&#xA;  proportional to the square of this interaction energy and decreases,&#xA;  therefore, with the sixth power of the distance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This paper does not give a detailed derivation though. But in his &lt;a href=&quot;http://zfbb.thulb.uni-jena.de/servlets/MCRFileNodeServlet/jportal_derivate_00156309/19484370105_ftp.pdf&quot; rel=&quot;nofollow&quot;&gt;seminal paper from 1948&lt;/a&gt; Förster goes through the derivation in detail (unfortunately it is not in English but in German and the condition of the text is not very good).&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-17T11:31:36.300" LastActivityDate="2014-10-17T11:31:36.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18536" PostTypeId="2" ParentId="18534" CreationDate="2014-10-17T11:36:36.570" Score="7" Body="&lt;p&gt;Your concentration is 0.615 M. The &lt;strong&gt;M&lt;/strong&gt; stands for &lt;em&gt;molar concentration&lt;/em&gt;, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution.&lt;/p&gt;&#xA;&lt;p&gt;A mole is just like any other unit for amount, except that it is built on a very large number. Since, for example, there at &lt;span class=&quot;math-container&quot;&gt;$6.02\times 10^{23}$&lt;/span&gt; atoms of chlorine in 35.45 grams of chlorine, we generally do not discuss numbers of atoms (or molecules) as a useful measurable amount. How much would a dozen atoms of chlorine weigh? Much less than any of our balances or scales could measure!&lt;/p&gt;&#xA;&lt;p&gt;One mole of stuff is an amount of stuff equal to the number of carbon atoms in 12 grams of &lt;span class=&quot;math-container&quot;&gt;$\ce{^{12}C}$&lt;/span&gt;, most common isotope of carbon. This number happens to be &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;noreferrer&quot;&gt;Avogadro's Number&lt;/a&gt; - &lt;span class=&quot;math-container&quot;&gt;$6.02\times 10^{23}$&lt;/span&gt;, which is really large! This is also the number of atoms or molecules present in a number of grams equal to one formula weight of that substance, so it is a convenient number. We know that when we weigh out 12.01 grams of carbon, 1.008 grams of hydrogen, 15.99 grams of oxygen, etc., we know that we always have the same number of atoms in each sample, because each is one mole. We tend not to worry too much about the large number hiding in the definition of one mole.&lt;/p&gt;&#xA;&lt;p&gt;Now, if you have one mole of a compound, like &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt;, then you know that there is one mole of the &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; formula unit in that sample. Since each &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; formula unit has one &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and one &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;, then each mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; has one mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; ions and one mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; ions.&lt;/p&gt;&#xA;&lt;p&gt;Now, you have your 0.615 M solution, and you only have 230 mL of it. How much &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; is in that sample? Well, you would have 0.615 moles of each in 1 liter, but you do not have one liter. You have 230 mL. Since there are 1000 mL in a liter, we can do a little maths:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$230\text{ mL}\times \dfrac{1 \text{ L}}{1000\text{ mL}}\times\dfrac{0.615\text{ mol KCl}}{1\text{ L}}=0.141 \text{ mol KCl}\equiv 0.141 \text{ mol}\,\ce{K+}\equiv 0.141 \text{ mol}\,\ce{Cl-}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-10-17T11:36:36.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18540" PostTypeId="1" CreationDate="2014-10-17T13:47:01.163" Score="0" ViewCount="687" Body="&lt;p&gt;I have understood generally for a while there is a significant preference in &quot;life&quot; for so-called &quot;left-handed&quot; (chiral) molecules, though I'm reading things now that say &quot;life&quot; sugars are right-handed, while &quot;life&quot; amino-acids are left-handed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering: how simple do these molecules get (particularly the molecules that are in intelligent life, if there happen to be any that are exclusive to intelligent life)? &lt;/p&gt;&#xA;" OwnerUserId="9362" LastActivityDate="2014-10-17T15:05:53.160" Title="What are some of the simplest chiral molecules in biological systems" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18541" PostTypeId="2" ParentId="18540" CreationDate="2014-10-17T15:05:53.160" Score="2" Body="&lt;p&gt;Don't get hung up on the &quot;left-handed&quot; and &quot;right-handed&quot; designations for amino acids and sugars. Those are just conventions for distinguishing one mirror image (enantiomer) of a chiral molecule from another. It just so happens that based on a particular way of drawing (Fisher projection) amino acids and sugars that it appears that the configurations are &quot;left-handed&quot; or &quot;right-handed.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how simple chiral biological molecules get, the simplest ones that I can think of are &lt;a href=&quot;http://en.wikipedia.org/wiki/Alanine&quot; rel=&quot;nofollow&quot;&gt;alanine&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Lactic_acid&quot; rel=&quot;nofollow&quot;&gt;lactic acid&lt;/a&gt;. I'm not sure what you consider &quot;intelligent&quot;, but these are likely to be found in all life (as we know it).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-17T15:05:53.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18544" PostTypeId="2" ParentId="18427" CreationDate="2014-10-17T16:54:20.767" Score="71" Body="&lt;p&gt;Recently, there has been a lot of discussion of Bent's rule (see for example &amp;quot;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622&quot;&gt;What is Bent's rule?&lt;/a&gt;&amp;quot;) here in SE Chem. Simply stated, the rule suggests that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-character tends to concentrate in orbitals directed at electronegative elements.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why does &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; replace an axial bond in &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In order to answer this question, we need to start by understanding the bonding in &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt; and its fluorinated isomers. In introductory courses and texts, it is usually stated that &lt;span class=&quot;math-container&quot;&gt;$\ce{AX5}$&lt;/span&gt; type molecules adopt a trigonal bipyramid geometry and are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^{3}d}$&lt;/span&gt; hybridized. As the comments by Martin and permeakra point out, and as is learned in more advanced classes, this hybridization scheme is likely incorrect. There are several reasons that argue against &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^{3}d}$&lt;/span&gt; hybridization including: 1) the fact that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; orbitals are relatively high in energy compared to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals and therefore it is energetically quite costly to involve them in bonding; and 2) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt; orbitals in non-metals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak.&lt;/p&gt;&#xA;&lt;p&gt;A reasonable hybridization alternative involves what is termed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypercoordinated bonding&lt;/a&gt;; where 3 center, 4 electron bonds are involved. Applying this concept to &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt; we would say that the central phosphorus atom is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized.  Thus, there would be 3 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbitals (these will be used to create the equatorial bonds) and a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital (this will be used to create our axial bonds) emanating from the central phosphorus atom.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/exDHY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/exDHY.png&quot; alt=&quot;Hybridisation of phosphorus in PCl5&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals for bonding (but it could be any type of orbital, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding M.O., so the bond order in this bond is reduced.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CkSXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CkSXL.png&quot; alt=&quot;MO diagram of Cl–P–Cl 3c4e bonding&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This reduced bond order in the bond using the phosphorus &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital explains why the axial bonds in &lt;span class=&quot;math-container&quot;&gt;$\ce{AX5}$&lt;/span&gt; type molecules are longer than the equatorial bonds.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UtR02.png&quot; alt=&quot;P-Cl bond lengths&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now that we understand the bonding in &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt; we can consider the case of &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl4F}$&lt;/span&gt; and how Bent's rule applies to the situation.  First, note that fluorine is more electronegative than chlorine.  As stated above, Bent's rule suggests that more electronegative ligands prefer to form bonds with orbitals that are high in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-character. Why is this? &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-Orbitals are lower in energy than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character. The two electrons in the &lt;span class=&quot;math-container&quot;&gt;$\ce{P-F}$&lt;/span&gt; bond will spend more time around the electronegative fluorine and less time around phosphorus. If that's the case (and it is), why &amp;quot;waste&amp;quot; precious, low-energy, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital character in an orbital that doesn't have much electron density to stabilize. Instead, save that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-character for use in phosphorus hybrid orbitals that do have more electron density around phosphorus (like the &lt;span class=&quot;math-container&quot;&gt;$\ce{P-Cl}$&lt;/span&gt; bonds). So, &lt;strong&gt;as a consequence of Bent's rule, we would expect phosphorus to use the orbital with lowest &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character, the axial &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbital, to form the &lt;span class=&quot;math-container&quot;&gt;$\ce{P-F}$&lt;/span&gt; bond; and the orbitals with more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character, the equatorial &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbitals, to form &lt;span class=&quot;math-container&quot;&gt;$\ce{P-Cl}$&lt;/span&gt; bonds.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2020-10-16T05:51:56.040" LastActivityDate="2020-10-16T05:51:56.040" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="18546" PostTypeId="2" ParentId="18534" CreationDate="2014-10-17T19:42:41.093" Score="1" Body="&lt;p&gt;One thing that may help is to keep in mind that a &quot;mole&quot; is simply a name for a number, like &quot;dozen&quot; (12) or &quot;gross&quot; (144) or &quot;score&quot; (20).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that a &quot;mole&quot; is a very large number (approximately 6.02x10&lt;sup&gt;23&lt;/sup&gt;) - one that is typically only reached when you're dealing with things the size of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 0.141 moles of KCl breaks into 0.141 moles of K&lt;sup&gt;+&lt;/sup&gt; and 0.141 moles of Cl&lt;sup&gt;-&lt;/sup&gt;, just as 2 dozen eggs can be broken down into 2 dozen yolks and 2 dozen shells (and two dozen egg whites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Molar&quot; simply means &quot;moles per liter&quot;, and can be treated similarly to eggs per carton. If there are two dozen eggs per carton and you have three quarters of a carton left, you have 2 dozen eggs/carton * 0.75 carton = 1.5 dozen eggs left. You also have 1.5 dozen yolks and 1.5 dozen eggshells, because each egg has both a yolk and a shell. Likewise, having 750 mL of a 2 Molar solution of KCl means that you have 2 moles KCl/liter * 0.75 liters = 1.5 moles KCl. You also have 1.5 moles of K&lt;sup&gt;+&lt;/sup&gt; and 1.5 moles of Cl&lt;sup&gt;-&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that just like &quot;dozen&quot; and &quot;score&quot;, you have to keep track of what you're measuring. A dozen cars and a dozen wheels are two different things. Two boxes with 1.5 dozen toy cars per box have 3 dozen cars total but 12 dozen wheels, because there are 4 dozen wheels per 1 dozen cars (which is entirely equivalent to saying there are 4 wheels per 1 car). Likewise, for every mole of CaCl&lt;sub&gt;2&lt;/sub&gt; there are 2 moles of Cl&lt;sup&gt;-&lt;/sup&gt; (which is equivalent to saying there are two chlorines for each &quot;molecule&quot; of CaCl&lt;sub&gt;2&lt;/sub&gt;), so you'll have to work that factor of two into your calcuations if you're going from quantities of CaCl&lt;sub&gt;2&lt;/sub&gt; to quantities of chloride ions.&lt;/p&gt;&#xA;" OwnerUserId="9367" LastActivityDate="2014-10-17T19:42:41.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18547" PostTypeId="1" AcceptedAnswerId="18583" CreationDate="2014-10-17T19:48:33.063" Score="4" ViewCount="695" Body="&lt;p&gt;I know that electrolysis of a saturated solution of sodium chloride with an inert anode (eg., graphite) is a handy way to obtain pure chlorine. I have a sample of saturated salt solution from the Dead Sea which is reported to have &lt;a href=&quot;http://en.wikipedia.org/wiki/Bromine#Occurrence&quot; rel=&quot;nofollow&quot;&gt;up to 5% bromine concentration&lt;/a&gt;, the domination anion being $\ce{Cl}^{-}$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will I get bromine on the anode by electrolysis of Dead Sea water?&#xA;In general, if I have a mixture of chlorides and bromides, what determines the ratio of electrolytically produced pure $\ce{Cl}_2$ and $\ce{Br}_2$?&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="1613" LastEditDate="2014-10-17T20:26:49.830" LastActivityDate="2014-10-18T19:58:24.960" Title="Will I get bromine by electrolysis of the brine from the Dead Sea?" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18548" PostTypeId="2" ParentId="18461" CreationDate="2014-10-17T21:04:05.843" Score="5" Body="&lt;p&gt;If you are asking if a compound, such as vitamin A, can be resynthesized to give the same chemical structure, the answer is yes. That is routine and carried out naturally inside organisms and by chemists in the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect that you suggest in the comment is more likely related to a different phenomenon. Most likely the &lt;strong&gt;sample&lt;/strong&gt; is different if there is a different effect. It would be similar to eating a carrot versus taking a vitamin A pill. One could ingesting the same amount of vitamin A from either source, but the body might process them differently based on the other &quot;stuff&quot; in the sample. In the carrot, there's the rest of the plant. In the pill, there are excipients that are used to bind the material of the pill, make it shelf stable, and convenient to take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When drugs are synthesized, they have to be made (and analyzed) following rigorous regulations to ensure that every sample of drug that goes out meets the specifications of the previous batch. Even though the active ingredient may be structurally identical to the previous batch, impurities must be within established specifications. The sample must have the same crystalline form so that it dissolves identically to prior batches. That's just one example of the lengths that chemists to through to replicate the composition of a sample, which in many respects is more difficult than replicating a chemical structure.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-17T21:04:05.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18549" PostTypeId="2" ParentId="18461" CreationDate="2014-10-17T21:55:21.137" Score="8" Body="&lt;p&gt;I agree with most of what the commenters and @jerepierre have written, with one or two exceptions.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to create a 100% identical copy of a chemical?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your question were asked in an introductory chemistry class, I believe most instructors would answer with a simple &quot;yes&quot;, for the reasons outlined above.  On the other hand, if your question were asked in an advanced chemistry class, or perhaps in a court of law, then I believe the answer would likely be &quot;no&quot;.  A frustrating dichotomy, but such is life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain the &quot;no&quot; answer.  To begin with, realize that simple molecules carry deuterium and $\ce{^13C}$ at levels found in natural abundance. Next, note that the way an organism would synthesize a vitamin, hormone, protein, etc., is likely different than the way a chemist might synthesize it in the laboratory. The differences would lie in the raw materials used in the process as well as the actual synthetic pathway.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each step in a synthesis that requires breaking (or making) a carbon-hydrogen bond or a carbon-carbon bond will result in isotopic fractionation.  Simply put, $\ce{C-H}$ bonds and $\ce{^12C-^12C}$ bonds are easier to break than the corresponding $\ce{C-D}$ bonds and $\ce{^13C-^12C}$ bonds.  Therefore, if we repeatedly break some number of (they can be different or the same) C-H bonds and/or C-C bonds, then our final product will be enriched in molecules with lower deuterium and $\ce{^13C}$ content, the degree of enrichment being dependent upon the difference in the number of carbon-hydrogen and carbon-carbon bond breaking and bond making steps between the synthetic process as carried out within the organism and as carried out on the lab bench.  There may be further fractionation if the starting material used by the body has required C-H and/or C-C bond making and breaking and has, therefore, also suffered isotopic fractionation.  Such fractionation can easily be monitored through comparison of the mass spectra of natural and synthetic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this isotopic differentiation between synthetic and natural products, there will also exist a difference in reaction by-products and contaminants created in these two different synthetic processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both methods (isotopic fractionation and by-product/impurity differences) could be used to determine if, for example, the testosterone in an athlete's body was produced naturally within the body or synthetically outside the body.  Also, we are probably all aware of various reports were a synthetic version of a drug (often illicit, but not always) naturally produced by a plant or metabolite has produced death upon administration due to toxic by-products produced in the lab synthetic route - clearly a different side-effect profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to your question, &lt;strong&gt;no, it is not possible to synthesize a natural product in the lab that will be 100% identical to the same material produced in a living organism.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-17T23:36:13.723" LastActivityDate="2014-10-17T23:36:13.723" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18550" PostTypeId="2" ParentId="18430" CreationDate="2014-10-17T22:44:13.753" Score="4" Body="&lt;p&gt;What you may be confusing is the difference between a functional group and a moiety. A moiety is a part of the molecule that could be either the whole functional group or just a part of the structure of a functional group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Functional_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; does a decent job of explaining this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5cvU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this illustration there is an ester functional group and an acetyl moeity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Acyl&quot; rel=&quot;nofollow noreferrer&quot;&gt;acyl group&lt;/a&gt;, my understanding is that the formula is RCO- and that the X is not necessary to define the acyl group. The R represents an alkyl group.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/e8VXr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-10-17T22:52:51.197" LastActivityDate="2014-10-17T22:52:51.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18551" PostTypeId="1" CreationDate="2014-10-18T00:27:43.330" Score="4" ViewCount="236" Body="&lt;p&gt;I'm trying to figure out the following reaction mechanism &lt;img src=&quot;https://i.imgur.com/nSrosd5.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a reference paper by Yanagi and Akiyoshi(1959) showing the hydroxyethylation of imides resulting in the following &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/R5OmT6B.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Why would the hydroxylethylation occur in the 1 position instead of the imide at the 3 position? I would imagine that this reaction occurs for the more acidic amine, but I have no idea why there doesn't seem to be any literature about that reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Why would this reaction produce a HydroxyEthyl and not a urethane, as shown here : &lt;img src=&quot;https://i.imgur.com/id8kAkl.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights would be appreciated, thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yanagi and Akiyoshi : &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo01090a601&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jo01090a601&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction in red (from a patent application): &lt;a href=&quot;http://www.huntsman.com/performance_products/Media%20Library/a_MC348531CFA3EA9A2E040EBCD2B6B7B06/Products_MC348531D0B9FA9A2E040EBCD2B6B7B06/Carbonates_MC348531D1109A9A2E040EBCD2B6B7B06/files/jeffsol_alkylene_carbonates_synthesis_of_hydroxy_alkyl_urethanes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.huntsman.com/performance_products/Media%20Library/a_MC348531CFA3EA9A2E040EBCD2B6B7B06/Products_MC348531D0B9FA9A2E040EBCD2B6B7B06/Carbonates_MC348531D1109A9A2E040EBCD2B6B7B06/files/jeffsol_alkylene_carbonates_synthesis_of_hydroxy_alkyl_urethanes.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9368" LastEditorUserId="865" LastEditDate="2015-10-16T18:09:18.580" LastActivityDate="2015-10-16T18:09:18.580" Title="Mechanism for hydroxyethylation on nucleobases (e.g. Thymine)" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18552" PostTypeId="2" ParentId="18551" CreationDate="2014-10-18T00:41:51.993" Score="3" Body="&lt;p&gt;The second part of the question is easy to answer: the imide attacks at C&lt;sub&gt;3&lt;/sub&gt; of propylene carbonate instead of C&lt;sub&gt;1&lt;/sub&gt; (the carbonyl carbon) because carbonates are stabilized (double ester-resonance), whereas carboxylate is a so-so leaving group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial product from nuclephilic attack at C&lt;sub&gt;3&lt;/sub&gt; is a carbonate half-ester, which subsequently loses CO&lt;sub&gt;2&lt;/sub&gt; to form the observed &lt;em&gt;N&lt;/em&gt;-hydroxyethyl imide.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-18T15:34:49.037" LastActivityDate="2014-10-18T15:34:49.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18553" PostTypeId="1" CreationDate="2014-10-18T01:25:13.540" Score="3" ViewCount="624" Body="&lt;p&gt;I have a molecule&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rZbEt.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;and i need to assign the hnmr. I have a printed spectrum&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Et4Bu.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So far my answer is:&lt;/p&gt;&#xA;&lt;p&gt;chemical shift_______splitting pattern_______coupling constant&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt; 2.32                quartet            6.6 Hz &amp;amp; 13.2 Hz&#xA;&#xA; 2.66                singlet                   -&#xA;&#xA; 3.66                triplet                6.5 Hz&#xA;&#xA; 5.11         triplet of doublets         1.5 Hz &amp;amp; 16.8 Hz&#xA;&#xA; 5.81   doublet of doublet of triplets   6.9 Hz &amp;amp; 10.2 Hz &amp;amp; 13.8 Hz &amp;amp; 17.2 Hz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;I am having trouble assigning them to the hydrogens. I think the H on the OH is a singlet and the H on the right hand side of the double bond is a ddt as it has cis (13.8 Hz) and trans (17.2 Hz). Then on the left of the double bond, the cis H would be the quartet as it has cis coupling (13.2) and the trans H would be the td as it has trans coupling (16.8 Hz).&lt;/p&gt;&#xA;&lt;p&gt;But if these are correct then does that mean the four hydrogens on carbons 2 and three are the triplet group?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-10-18T21:33:22.420" Title="HNMR Assignment" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18554" PostTypeId="2" ParentId="18553" CreationDate="2014-10-18T02:37:52.337" Score="7" Body="&lt;p&gt;You haven't quite rationailsed the spectrum correctly; If you look at your molecule you will see that you have 6 distinct &lt;sup&gt;1&lt;/sup&gt;H species, but you have only described 5. Take particular note of the signals that you have described as a &lt;em&gt;td&lt;/em&gt; at 5.11ppm. This should be symmetrical about the centre peak if it was a true &lt;em&gt;td&lt;/em&gt;, and this signal clearly is not. So, it actually comes from two overlapping signals; one with a large coupling (~17Hz) and the other a medium coupling (~10Hz). Also, a quartet will only have one coupling constant - this will be the separation of any of two adjacent peaks in the quartet, and so is 6.6Hz, as you have described. So perhaps this is not a true quartet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When assigning your peaks, you should first look at your chemical shifts;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vinylic protons typically come in the range 3.5-8.5, but commonly 5-7ppm&lt;/li&gt;&#xA;&lt;li&gt;alkyl protons alpha to a heteroatom 3-4.5ppm&lt;/li&gt;&#xA;&lt;li&gt;alkyl protons alpha to vinyl group 1.5-2.5ppm&lt;/li&gt;&#xA;&lt;li&gt;hydroxyl protons can come just about anywhere, but true alkyl-OH 1H will typically come 1-5 ppm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So from this information, you should be able to broadly assign the 6 protons in your molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then need to look at coupling constants to further refine your assignments. For vinylic groups, you can rely on the following general coupling constants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9nDP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;trans coupling J&lt;sub&gt;bc&lt;/sub&gt; 14-19Hz&lt;/li&gt;&#xA;&lt;li&gt;cis coupling J&lt;sub&gt;ac&lt;/sub&gt; 4-12 Hz&lt;/li&gt;&#xA;&lt;li&gt;geminal coupling J&lt;sub&gt;ab&lt;/sub&gt; -4-4 Hz&lt;/li&gt;&#xA;&lt;li&gt;vicinal coupling J&lt;sub&gt;cd&lt;/sub&gt; 4-10 Hz&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This should allow you to complete the assignment of this spectrum. See how you go from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful approach to these type of assignment problems, especially if you know the structure, is to look at the molecule and try to predict the spectrum you'd expect, rather than looking at the spectrum and working out what's what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyOwm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your molecule, I'd expect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ha to come in the vinylic region, and appear as a &lt;em&gt;dd&lt;/em&gt; (cis and geminal coupling to Hb and Hc)&lt;/li&gt;&#xA;&lt;li&gt;Hb to come in the vinylic region, and appear as a &lt;em&gt;dd&lt;/em&gt; (trans and geminal coupling to Ha and Hc)&lt;/li&gt;&#xA;&lt;li&gt;Hc to come in the vinylic region, and appear as a &lt;em&gt;ddt&lt;/em&gt; (cis, trans and vicinal coupling to Ha, Hb and Hd)&lt;/li&gt;&#xA;&lt;li&gt;Hd to come in the alkyl region alpha to a double bond, and appear as a &lt;em&gt;dt&lt;/em&gt; (coupling to Hc and He)&lt;/li&gt;&#xA;&lt;li&gt;He to come in the alkyl region alpha to a heteroatom, and appear as a &lt;em&gt;t&lt;/em&gt; (coupling to Hd)&lt;/li&gt;&#xA;&lt;li&gt;Hf to come somewhere and appear most likely as an exchange broadened singlet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;==============&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edits to answer questions raised in comment to the original answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are two splitting diagrams to help you rationalise the splitting for the peaks centred around 2.3 ppm and 5.10 ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BEBu6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first look at your upfield peaks (2.10ppm), which is the splitting diagram on the left. This comes from signal H&lt;sub&gt;d&lt;/sub&gt;, and should show coupling to H&lt;sub&gt;c&lt;/sub&gt; (which will give rise to a doublet) and also H&lt;sub&gt;e&lt;/sub&gt; (giving a triplet). So this signal is actually a doublet of triplets. Normally if the doublet coupling and triplet coupling are quite different, then you observe two clear sets of triplet peaks. However, in this case, J&lt;sub&gt;dc&lt;/sub&gt; is approximately the same as J&lt;sub&gt;de&lt;/sub&gt; (within the natural linewisth of the peaks), and so you get overlap of some of peaks, creating what looks like a quartet. It is still a dt, with 6 lines, but you can't resolve them all. So, you still have two coupling constants, but both couplings are about 6.6Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second diagram (on the right) is for your sets of signals around 5.10 ppm. This is actually the overlap of two independent signals. H&lt;sub&gt;a&lt;/sub&gt; will occur as a dd, with cis coupling to H&lt;sub&gt;c&lt;/sub&gt; (~14Hz), and geminal coupling to H&lt;sub&gt;b&lt;/sub&gt; (~2Hz, so very small, and realy only broadens the peak in this case.) This is the signal on the right.&#xA;H&lt;sub&gt;b&lt;/sub&gt; will also occur as a dd, with trans coupling to H&lt;sub&gt;c&lt;/sub&gt; (~17Hz), and again, geminal coupling to H&lt;sub&gt;a&lt;/sub&gt;. This is the signal on the left. It just so happens (life is like that sometimes), that one set of lines for each signal overlap, to create what looks like an aysmmetrical triplet. So, here are your two signals, H&lt;sub&gt;a&lt;/sub&gt; and H&lt;sub&gt;b&lt;/sub&gt;. Integrating the area under the peaks serves as a valuable tool to tell you how many signals each peak represents, and in this case you would have clearly seen that the signals at 5.10ppm come from 2 protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, geminal couplings in vinylic systems is usually quite small, and often is of similar magnitude to allylic coupling (4J couplings, such as J&lt;sub&gt;ad&lt;/sub&gt;, and this is why the peaks for protons a and b actually look a little more complicated than just doublets of doublets. The main take home message about double bond couplings at an introductory level really comes from the cis and trans couplings.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2014-10-18T21:33:22.420" LastActivityDate="2014-10-18T21:33:22.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18555" PostTypeId="2" ParentId="18551" CreationDate="2014-10-18T02:54:42.547" Score="2" Body="&lt;p&gt;I would expect that the two N-H groups would be similar acidity. The conjugate base from deprotonation at N-1 is resonance stabilized through both carbonyls, through the carbonyl at C-2 directly and through the carbonyl at C-4 via the C-5/6 double bond.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-18T02:54:42.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18556" PostTypeId="2" ParentId="6774" CreationDate="2014-10-18T04:58:51.547" Score="1" Body="&lt;p&gt;It is simple. Graphite anode is used because when carbon react with oxygen liberated at anode it forms &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; gas which is a exothermic reaction and maintain the temperature of process and not allow molten aluminum to solidify. Because the density of aluminum in solid form is less dense than cryolite and hence it can't be able to extract from the bottom.&lt;/p&gt;&#xA;" OwnerUserId="9371" LastEditorUserId="41328" LastEditDate="2021-02-14T13:46:52.560" LastActivityDate="2021-02-14T13:46:52.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18557" PostTypeId="1" AcceptedAnswerId="18566" CreationDate="2014-10-18T07:20:40.907" Score="23" ViewCount="18957" Body="&lt;p&gt;Cyclobutadiene is very unstable. But, some sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some, cyclobutadiene is simply non-aromatic (as opposed to antiaromatic) because it doesn't even have a fully conjugated pi system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is/are the real reason(s) for its instability? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-01-07T13:17:21.447" LastActivityDate="2018-10-17T15:54:48.683" Title="Is cyclobutadiene antiaromatic?" Tags="&lt;organic-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18559" PostTypeId="1" CreationDate="2014-10-18T11:52:08.853" Score="1" ViewCount="834" Body="&lt;p&gt;I'm trying to combine precise amounts of citric acid and lye to get sodium citrate with minimal amount of reacted acid/base. I'm not so sure how stochiometry works, but I could sure learn that from some web article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the real problem is, that I don't know which of three &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_citrate&quot; rel=&quot;nofollow&quot;&gt;sodium citrates&lt;/a&gt; will form. Depending on the number of $\ce{Na}$ molecules, the amounts of $\ce{NaOH}$ greatly differs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which forms and why?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastEditorUserId="5601" LastEditDate="2014-10-18T13:59:57.360" LastActivityDate="2014-12-17T16:52:23.270" Title="When mixing lye and citric acid, which citrate will form?" Tags="&lt;acid-base&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18560" PostTypeId="1" CreationDate="2014-10-18T12:05:50.207" Score="3" ViewCount="514" Body="&lt;p&gt;The Fischer Tropsch reaction between $\ce{CO}$ and $\ce{H2}$ has the potential to give a wide range of products. However by judicious choice of catalytic metal it is possible to produce $\ce{CH4}$, $\ce{CH3OH}$, and other alkanes selectively. $\ce{Fe}$ and $\ce{Co}$ are most commonly used catalysts where $\ce{Fe}$ produces less $\ce{CH4}$, more oxygenated products and more alkenes and $\ce{Co}$ produces more $\ce{CH4}$, less oxygenated products and more alkanes. There are a number of other catalysts that can be used, for example $\ce{Rh}$ which can produce $\ce{C2}$ and higher oxygenates and hydrocarbons. This includes the production of $\ce{CH3OH}$. However, it is not selectively produced but is produced in a mixture of other products and in small amounts. My question is why can $\ce{CH3OH}$ not be produced selectively?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="10185" LastEditDate="2015-05-19T12:17:27.657" LastActivityDate="2016-03-11T22:24:10.530" Title="Why it is not yet possible to produce ethanol selectively in the Fischer Tropsch reaction?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18561" PostTypeId="1" AcceptedAnswerId="18564" CreationDate="2014-10-18T13:31:29.323" Score="2" ViewCount="163" Body="&lt;p&gt;I've got a question regarding 'notation' for redox reactions. Suppose you are asked to identify the oxidizing and reducing agents for a simple reaction like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2(aq)}+\ce{2KI(aq)} \longrightarrow \ce{2KCl(aq)}+\ce{I2(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now are you supposed to say $\ce{Cl}$ is the oxidizing agent or $\ce{Cl2}$? Also, is $\ce{KI}$ the reducing agent or $\ce{I}$ or $\ce{I^-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, if you were instead asked to identify which species got oxidized, would you again say that $\ce{KI}$ is oxidized or $\ce{I}$ or $\ce{I^-}$?&lt;/p&gt;&#xA;" OwnerUserId="9376" LastEditorUserId="1362" LastEditDate="2014-11-02T14:26:56.057" LastActivityDate="2014-11-02T14:26:56.057" Title="Terminology of redox reactions: which species are the oxidizing and reducing agents?" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18562" PostTypeId="2" ParentId="18560" CreationDate="2014-10-18T13:44:03.757" Score="3" Body="&lt;p&gt;in Fischer-Tropsch synthesis-like reaction there are (roughly) following processes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO \rightarrow  Surf-CO}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H2 \rightarrow  2 Surf-H}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO + Surf-(CH2)_{n}-H \rightarrow  Surf-CO-(CH2)_{n}-H}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3'. $\ce{CO + Surf-H \rightarrow  Surf-CO-H}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Surf-CO-(CH2)_{n}-H + 4 Surf-H \rightarrow&#xA;    Surf-(CH2)_{n\,+1}-H + H2O }$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Surf-CO-(CH2)_{n}-H + 3 Surf-H \rightarrow &#xA;    HO-(CH2)_{n\,+1}-H }$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Surf-CH2-CH2-(CH2)_{n}-H \rightarrow Surf-H  + &#xA;    CH2=CH-(CH2)_{n}-H}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Surf-(CH2)_{n}-H + Surf-H \rightarrow H-(CH2)_{n}-H}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All processes happen simultaneously, and tuning catalyst and reaction conditions it is possible to hinder some processes. However, as you can see, as long as you have chain growth at all (that is necessary to have $\ce{C2}$ products, like ethanol), you cannot stop $\ce{C2}$ surface particles to participate in further chain growth, producing $\ce{C_{3+}}$ products. So, if chain growth happens at all, a mixture of products with different chain length is always obtained, and there is no way around it. On the other hand, on proper catalyst in proper conditions it is possible to prevent chain growth, but still allow reduction of carbon monoxide, gaining either methane or methanol depending on catalyst. So, in practice it is possible to obtain one of the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Methane&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixture of terminal alkenes and alkanes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Methanol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mixture of linear terminal alcohols (OK, this case is a bit more complicated, as other oxygenates may be produced as well, giving rather complex mixture, but that's beyond the point)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is still possible to obtain pure ethanol indirectly (using hydration of isolated ethylene), or using other processes. It is also possible to obtain product mixture where some products, say C2-C4 alcohols, will be dominating.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="24052" LastEditDate="2016-03-11T22:24:10.530" LastActivityDate="2016-03-11T22:24:10.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18563" PostTypeId="2" ParentId="18559" CreationDate="2014-10-18T13:59:34.737" Score="1" Body="&lt;p&gt;What is the pH?  You can find the pKa's of the three acidic protons in citric acid and use the Hendersson Hasselbalch equation to find what percent of each proton is ionized at that pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemteam.info/AcidBase/HH-Equation.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemteam.info/AcidBase/HH-Equation.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-10-18T13:59:34.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18564" PostTypeId="2" ParentId="18561" CreationDate="2014-10-18T14:11:07.040" Score="2" Body="&lt;p&gt;Ignoring spectator ions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2 +2 I- -&amp;gt;2Cl- +I2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\ce{Cl2}$ is the oxidant and $\ce{I-}$ is the reductant.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-18T14:11:07.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18565" PostTypeId="1" CreationDate="2014-10-18T14:15:37.283" Score="1" ViewCount="15998" Body="&lt;p&gt;Why is exothermic reactions are favoured by low temperature while endothermic reactions are favoured by high temperature?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastActivityDate="2016-04-19T05:09:10.970" Title="Effect of temperature on equilibrium" Tags="&lt;equilibrium&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18566" PostTypeId="2" ParentId="18557" CreationDate="2014-10-18T14:17:33.800" Score="18" Body="&lt;p&gt;If we go back to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16769&quot;&gt;your earlier question on Frost diagrams&lt;/a&gt; (I've reproduced the key figure below), we see why simple molecular orbital theory or the &quot;&lt;span class=&quot;math-container&quot;&gt;$4n+2$&lt;/span&gt;&quot; rule suggests that benzene is aromatic while cyclobutadiene is antiaromatic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JWcH1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWcH1.png&quot; alt=&quot;Frost circles for cyclobutadiene, cyclopentadienyl, and benzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forming a planar, conjugated, 6-membered ring and placing 6 π-electrons in it creates a π-system that is energetically more stable than 3 conjugated C=C bonds (e.g. in hexa-1,3,5-triene). On the other hand, forming a planar, conjugated, 4-membered ring and placing 4 π-electrons in it creates a π-system that is energetically less stable than 2 conjugated C=C bonds (e.g. in butadiene).  For these reasons we say that the first system, benzene, is &quot;aromatic&quot;, while the second system, cyclobutadiene, is &quot;antiaromatic&quot;. Other measurements and physical phenomenon such as reactivity, bond lengths, ring currents, etc. support these conclusions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free cyclobutadiene has been observed as a transient intermediate.  Further theoretical analysis suggests that due to its lack of aromaticity it will distort from a square structure to a rectangular one with alternating single and double bonds (Jahn–Teller effect).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It turns out that other systems involving the cyclobutadiene skeleton have been prepared and studied.  Derivatives of both the cyclobutadiene dication&lt;sup&gt;[1]&lt;/sup&gt; (&lt;span class=&quot;math-container&quot;&gt;$4n+2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$n=0$&lt;/span&gt;) and the cyclobutadiene dianion&lt;sup&gt;[2]&lt;/sup&gt; (&lt;span class=&quot;math-container&quot;&gt;$4n+2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt;) have been prepared and studied.  Their stability is much less than that of benzene, but much more than that of cyclobutadiene. Placing two charges in these small rings results in extremely high coulombic repulsions in both the dianion and dication, and this may be a significant part of the explanation as to  why they are less stable than say, benzene.  The fact that they exist as planar structures, with appropriate ring currents and are more stable than cyclobutadiene suggests that the instability observed with cyclobutadiene is not due to ring strain and angle strain alone.  Rather the aromatic stabilization - antiaromatic destabilization suggested by simple MO theory seems to be involved. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Olah, G. A.; Staral, J. S. Novel aromatic systems. 4. cyclobutadiene dications. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1976,&lt;/strong&gt; &lt;em&gt;98&lt;/em&gt; (20), 6290–6304. &lt;a href=&quot;https://doi.org/10.1021/ja00436a037&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja00436a037&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Takanashi, K.; Inatomi, A.; Lee, V. Y.; Nakamoto, M.; Ichinohe, M.; Sekiguchi, A. Tetrakis(trimethylsilyl)cyclobutadiene dianion alkaline earth metal salts: new members of the 6π-electron aromatics family. &lt;em&gt;Eur. J. Inorg. Chem.&lt;/em&gt; &lt;strong&gt;2008,&lt;/strong&gt; &lt;em&gt;2008&lt;/em&gt; (11), 1752–1755. &lt;a href=&quot;https://doi.org/10.1002/ejic.200800066&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1002/ejic.200800066&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-10-17T15:54:48.683" LastActivityDate="2018-10-17T15:54:48.683" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="18567" PostTypeId="1" AcceptedAnswerId="18717" CreationDate="2014-10-18T14:24:09.240" Score="31" ViewCount="150106" Body="&lt;p&gt;Why does a dissociation reaction shift to the right with the addition of an inert gas?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="865" LastEditDate="2021-04-22T15:17:08.917" LastActivityDate="2021-04-22T15:17:08.917" Title="What would be the effect of the addition of an inert gas to a reaction at equilibrium?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="18568" PostTypeId="1" AcceptedAnswerId="18584" CreationDate="2014-10-18T14:35:46.017" Score="3" ViewCount="452" Body="&lt;p&gt;Today I studied Slater's rules for calculating the effective nuclear charge. One particular line, &amp;quot;The shielding power of the d and f orbital is lesser in comparison to the p and s orbitals&amp;quot;, seems to appear in every single book and website I have studied from.&lt;/p&gt;&#xA;&lt;p&gt;Slater's rules does not seem to take into account the orbitals for calculation of &lt;span class=&quot;math-container&quot;&gt;$Z_\mathrm{effective}$&lt;/span&gt;. Nevertheless, I don't disagree with the logic behind the statement.&lt;/p&gt;&#xA;&lt;p&gt;Thus my question is: Do Slater's rules actually account for this logic? If they don't, is it faulty? And, how do Slater's rules account for the shielding effect of the individual orbitals?&lt;/p&gt;&#xA;" OwnerUserId="8217" LastEditorUserId="69830" LastEditDate="2021-05-11T05:18:44.690" LastActivityDate="2021-05-11T05:21:24.340" Title="Are the rules proposed by Slater for the calculation of effective Z right?" Tags="&lt;inorganic-chemistry&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18569" PostTypeId="2" ParentId="18565" CreationDate="2014-10-18T14:55:57.520" Score="4" Body="&lt;p&gt;I'll try and simplify the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NImkl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the image above, imagine the products sitting above the letter D and the reactants sitting below the letter B.  Now, imagine the particles involved in the reaction moving since they have kinetic energy.  It is much easier for the reactants to climb up the top of the hill labeled C then it is for the products to climb up the hill.  Once a reactant climbs the hill and slides down to products it will not be able to go back to reactants by climbing up the hill unless we give them more energy by raising temperature.  Raising temperature allows the products to climb up the hill and go to reactants.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward reaction in the diagram is exothermic and the reverse is endothermic.  So, the exo is favored at low T and the endo at high T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look at the equation delta G = delta H - Tdelta S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make up some numbers for delta H and delta S.  You will see that endo reactions becomes more spontaneous as you raise T and exo reactions become more spontaneous as you lower T regardless of delta S.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-10-18T14:55:57.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18570" PostTypeId="1" AcceptedAnswerId="18576" CreationDate="2014-10-18T15:03:30.333" Score="1" ViewCount="282" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Will a reaction occur in the following case? If so, write a net ionic equation for it:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{(NH4)2SO4 (aq) + ZnCl2 (aq) -&amp;gt; ?}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The book says that all possible combinations of positive and negative ions lead to water soluble compounds, all of the ions remain in solution. No reactions occur.&lt;/p&gt;&#xA;&lt;p&gt;Now I am a little confused. Why didn't &lt;span class=&quot;math-container&quot;&gt;$\ce{(NH4)2}$&lt;/span&gt; join with &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2}$&lt;/span&gt; and why didn't &lt;span class=&quot;math-container&quot;&gt;$\ce{SO4}$&lt;/span&gt; join with &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt;? Or did they?&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-11T03:14:28.527" Title="Using solubility guidelines to predict a reaction of two ionic, water-soluble salts" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18571" PostTypeId="2" ParentId="9007" CreationDate="2014-10-18T15:46:12.777" Score="1" Body="&lt;p&gt;As mentioned in the comments, &lt;em&gt;J. Chem. Ed&lt;/em&gt; is a good resource for NMR kinetics experiments. I have had success with the experiment published by J. Orvis in &lt;em&gt;J. Chem. Ed&lt;/em&gt; 80(7), &lt;strong&gt;2003&lt;/strong&gt;, 803 &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed080p803&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.  This publication describes the synthesis of trans-dichlorotetraamminecobalt(III) and its subsequent reaction with water to produce cis-aquachlorotetramminecobalt(III).  There are several advantages to this experiment as a model study:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The synthesis is straightforward and does not require materials unlikely to be already in the laboratory, although it is critical that the directions are followed precisely.&lt;/li&gt;&#xA;&lt;li&gt;The aquation reaction occurs within about one hour - slow enough to easily monitor via NMR but not so slow as to take up precious instrument time&lt;/li&gt;&#xA;&lt;li&gt;The reaction rate is temperature sensitive, which makes it a good model for extracting thermodynamic properties from the reaction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My undergraduate students are able to complete the synthesis about 50% of the time, and the yields using the published procedure are about 400 mg, so one successful synthesis is suitable for a number of NMR experiments.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-10-18T15:46:12.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18572" PostTypeId="1" AcceptedAnswerId="18575" CreationDate="2014-10-18T16:04:23.693" Score="0" ViewCount="12837" Body="&lt;p&gt;(a) 0.008 M $\ce{Ba(OH)2}$&lt;br&gt;&#xA;(b) 0.010 M $\ce{KI}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;a&quot; but I do not understand why. The book says that &quot;a&quot; is a strong electrolyte with a total ion concentration of 0.024. How did they calculate the concentration from the information that they gave? And why is it a strong electrolyte based on that calculation. I thought strong electrolytes were (solutes) ions that dissolved completely in a solvent/solution? &lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="13339" LastEditDate="2015-06-19T19:57:07.893" LastActivityDate="2015-06-19T19:57:07.893" Title="Which solution has the highest total molarity of ions and why?" Tags="&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18573" PostTypeId="2" ParentId="18572" CreationDate="2014-10-18T16:29:04.437" Score="2" Body="&lt;p&gt;You are right in thinking that a strong electrolyte is something that dissociates completely in solution.$\ce{Ba(OH)2}$ and $\ce{KI}$ then, being a strong electrolytes, will effectively dissociate into $\ce{Ba^2+}$+ $\ce{2OH-}$ and $\ce{K+}$+ $\ce{I-}$, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going into your example, it's a just a matter of multiplication. We can assume that the volume of the solution is 1L for simplicity. Since you begin with 0.008 M $\ce{Ba(OH)2}$, we can say that you begin with 0.008 mol of $\ce{Ba(OH)2}$. In solution this becomes 0.008 mol of $\ce{Ba+}$ and 2(0.008) mol of $\ce{OH-}$, or 0.024 mol of ions in total. Likewise for KI, you are left with 0.010 mol of $\ce{K+}$ and 0.010 mol of $\ce{I-}$, or 0.020 mol of ions in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting these values back to molarity (by dividing by 1), we get that the concetration of ions in the $\ce{Ba(OH)2}$ solution is 0.024 M, whereas it is only 0.020 M in the $\ce{KI}$ solution. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="5234" LastEditDate="2014-10-18T16:39:48.693" LastActivityDate="2014-10-18T16:39:48.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18574" PostTypeId="1" AcceptedAnswerId="18577" CreationDate="2014-10-18T16:35:33.363" Score="2" ViewCount="3002" Body="&lt;p&gt;I want to create a 3D molecular file using coordinates taken from:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hrynchuk, RJ; Barton, RJ; Robertson, BE (1983). &quot;&lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v83-085&quot; rel=&quot;nofollow noreferrer&quot;&gt;The crystal structure of free base cocaine, C17H21NO4&lt;/a&gt;&quot; (PDF). &lt;em&gt;Canadian Journal of Chemistry&lt;/em&gt; &lt;strong&gt;61&lt;/strong&gt; (3): 481–487. doi:&lt;a href=&quot;http://dx.doi.org/10.1139/v83-085&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1139/v83-085&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are the coordinates:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              X             Y            Z&#xA;C(1)          0.5419        0.1631       0.7095&#xA;C(2)          0.4579        0.2047       0.5345&#xA;C(3)          0.3080        0.1713       0.5168&#xA;C(4)          0.2903        0.0286       0.5727&#xA;C(5)          0.3950        0.0004       0.7409&#xA;C(6)          0.3870        0.1080       0.8678&#xA;C(7)          0.4858        0.2192       0.8450&#xA;C(8)          0.1028        0.2512       0.3219&#xA;C(9)          0.0323        0.2713       0.1411&#xA;C(10)         0.0857        0.2213       0.0238&#xA;C(11)         0.0189        0.2435       -0.1419&#xA;C(12)        -0.0995        0.3166        -0.1875&#xA;C(13)        -0.1528        0.3671      -0.0719&#xA;C(14)        -0.0880        0.3462      0.0951&#xA;C(15)         0.5172        0.1331      0.4127&#xA;C(16)         0.7261        0.1009      0.3465&#xA;C(17)         0.6418        -0.0439     0.8543&#xA;&#xA;N(1)          0.5324        0.0158      0.7182&#xA;&#xA;O(1)          0.2257        0.1933      0.3470&#xA;O(2)          0.0560        0.2836      0.4306&#xA;O(3)          0.4616        0.0531      0.3097&#xA;O(4)          0.6481        0.1713      0.4393&#xA;&#xA;H(1)          0.6210        0.1840      0.7230&#xA;H(2)          0.4640        0.2980      0.5170&#xA;H(3)          0.2820        0.2340      0.5830&#xA;H(4)          0.1930        0.0110      0.5730&#xA;H(5)          0.3050        -0.0240     0.4930&#xA;H(6)          0.3850        -0.0960     0.7770&#xA;H(7)          0.4090        0.0810      0.9760&#xA;H(8)          0.2870        0.1410      0.8610&#xA;H(9)          0.4440        0.3140      0.8150&#xA;H(10)         0.5560        0.2330      0.9400&#xA;H(11)         0.1580        0.1680      0.0530&#xA;H(12)         0.0640        0.2050      -0.2240&#xA;H(13)         -0.1470        0.3330      -0.3060&#xA;H(14)         -0.2320        0.4140      -0.103&#xA;H(15)         -0.1280        0.3810      0.186&#xA;H(16)         0.7330        0.0190      0.3810&#xA;H(17)         0.8130        0.1470      0.3710&#xA;H(18)         0.6990        0.1060      0.2370&#xA;H(19)         0.7250        -0.0340      0.836&#xA;H(20)         0.6470        -0.0070      0.961&#xA;H(21)         0.6230        -0.1410      0.861&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Converting this to XYZ coordinates, namely to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;43&#xA;Cocaine 1983&#xA;C          0.5419        0.1631       0.7095&#xA;C          0.4579        0.2047       0.5345&#xA;C          0.3080        0.1713       0.5168&#xA;C          0.2903        0.0286       0.5727&#xA;C          0.3950        0.0004       0.7409&#xA;C          0.3870        0.1080       0.8678&#xA;C          0.4858        0.2192       0.8450&#xA;C          0.1028        0.2512       0.3219&#xA;C          0.0323        0.2713       0.1411&#xA;C          0.0857        0.2213       0.0238&#xA;C          0.0189        0.2435       -0.1419&#xA;C         -0.0995        0.3166        -0.1875&#xA;C         -0.1528        0.3671      -0.0719&#xA;C         -0.0880        0.3462      0.0951&#xA;C          0.5172        0.1331      0.4127&#xA;C          0.7261        0.1009      0.3465&#xA;C          0.6418        -0.0439     0.8543&#xA;N          0.5324        0.0158      0.7182&#xA;O          0.2257        0.1933      0.3470&#xA;O          0.0560        0.2836      0.4306&#xA;O          0.4616        0.0531      0.3097&#xA;O          0.6481        0.1713      0.4393&#xA;H          0.6210        0.1840      0.7230&#xA;H          0.4640        0.2980      0.5170&#xA;H          0.2820        0.2340      0.5830&#xA;H          0.1930        0.0110      0.5730&#xA;H          0.3050        -0.0240     0.4930&#xA;H          0.3850        -0.0960     0.7770&#xA;H          0.4090        0.0810      0.9760&#xA;H          0.2870        0.1410      0.8610&#xA;H          0.4440        0.3140      0.8150&#xA;H          0.5560        0.2330      0.9400&#xA;H          0.1580        0.1680      0.0530&#xA;H          0.0640        0.2050      -0.2240&#xA;H          -0.1470        0.3330      -0.3060&#xA;H          -0.2320        0.4140      -0.103&#xA;H          -0.1280        0.3810      0.186&#xA;H          0.7330        0.0190      0.3810&#xA;H          0.8130        0.1470      0.3710&#xA;H          0.6990        0.1060      0.2370&#xA;H          0.7250        -0.0340      0.836&#xA;H          0.6470        -0.0070      0.961&#xA;H          0.6230        -0.1410      0.861&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Opening this with Accelrys gives this structure. &lt;a href=&quot;https://i.imgur.com/7paFRgE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://i.imgur.com/7paFRgE.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="5017" LastEditDate="2014-10-18T20:37:10.027" LastActivityDate="2014-10-18T20:37:10.027" Title="How do I write a valid CIF, PDB or XYZ file from the coordinates listed below?" Tags="&lt;crystal-structure&gt;&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18575" PostTypeId="2" ParentId="18572" CreationDate="2014-10-18T16:36:48.400" Score="2" Body="&lt;p&gt;Barium hydroxide is a slightly soluble salt. It dissociates in water and gives 3 ions (one ion of barium and two ions of hydroxide) with the concentration of $0.008 \ce{ M}$ for each one. This makes the total concentration of ions to be $0.024 \ce{ M}$. While the total concentration of $\ce{KI}$ is $0.02 \ce{ M}$. So the solution of barium hydroxide is a stronger electrolyte as it has a higher concentration of dissociated ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your book, they treat barium hydroxide as a strong electrolyte, i.e. it dissolved completely in water and dissociates to its constituent ions. This is of course an approximation. While $\ce{KI}$ is a true strong electrolyte; it dissolved completely in water and dissociates to its constituent ions.&#xA;So, I agree with you with the definition of strong electrolyte, and I hope I clarify the ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-18T16:36:48.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18576" PostTypeId="2" ParentId="18570" CreationDate="2014-10-18T16:47:17.903" Score="5" Body="&lt;p&gt;Of course $\ce{NH4+}$ can join $\ce{Cl-}$ and $\ce{Zn^2+}$ can join $\ce{SO4^2-}$, but the formed salts are completely soluble in water, and they completely dissociate in water to gives the corresponding ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the starting salts:  $\ce{(NH4)2SO4}$ and  $\ce{ZnCl2}$ are completely soluble in water, and they completely dissociate in water to gives the corresponding ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, all of the ions remain in solution, as if no reaction occurs. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-04-11T03:14:28.527" LastActivityDate="2018-04-11T03:14:28.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18577" PostTypeId="2" ParentId="18574" CreationDate="2014-10-18T16:51:29.417" Score="4" Body="&lt;p&gt;Generally speaking, &lt;code&gt;babel&lt;/code&gt; from the &lt;a href=&quot;http://openbabel.org&quot; rel=&quot;nofollow&quot;&gt;Openbabel suite&lt;/a&gt; will convert chemical file formats.  The openbabel wiki has a &lt;a href=&quot;http://openbabel.org/wiki/Free_Form_Fractional&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; with examples on input in free-form fractional coordinate format, and the abstract of the paper has the unit cell dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single CIF files can be requested from the &lt;a href=&quot;http://www.ccdc.cam.ac.uk/pages/Home.aspx&quot; rel=&quot;nofollow&quot;&gt;Cambridge Crystallographic Data Centre&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-18T19:50:33.663" LastActivityDate="2014-10-18T19:50:33.663" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18578" PostTypeId="1" CreationDate="2014-10-18T17:07:21.867" Score="1" ViewCount="2483" Body="&lt;p&gt;The book I am using is &quot;General Chemistry: Principles and Modern Applications&#xA;by Ralph H. Petrucci. On page 158 the Solubility Guidelines for Common Ionic Solids is given. Rule 3 states that, &quot;Salts of silver, lead, and mercury(I) are insoluble, however, my book states that $\ce{AgClO4}$ is soluble. Can someone explain? &lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="865" LastEditDate="2015-10-30T16:56:03.407" LastActivityDate="2015-10-30T16:56:03.407" Title="Applying the Solubility Guidelines is AgClO4 soluble or insoluble?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18579" PostTypeId="2" ParentId="18565" CreationDate="2014-10-18T17:35:58.360" Score="0" Body="&lt;p&gt;You have to know that when you have a chemical equilibrium that is subjected to a change in concentration, temperature or pressure,then  this equilibrium will readjust itself to &lt;strong&gt;counteract&lt;/strong&gt; the effect of the applied change and a new equilibrium is established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the reaction is exothermic. This means it produces heat. So, if we decrease the temperature (low temperatures), the equilibrium will tend to counteract this effect and favor the reaction that produces heat to increase the temperaure of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the reaction is endothermic. This means it consumes heat. So, if we increase the temperature (high temperatures), the equilibrium will tend to counteract this effect and favor the reaction that consumes heat to decrease the temperaure of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum, exothermic reactions are favoured by low temperature while endothermic reactions are favoured by high temperature&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-18T17:35:58.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18580" PostTypeId="2" ParentId="18578" CreationDate="2014-10-18T17:54:11.317" Score="3" Body="&lt;p&gt;Different guidelines will say slightly different things to convey the same idea. Many I've seen don't explicitly mention silver, but still produce the same result for most compounds. I don't have that book but if it's anything like &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;, there's probably another line that says salts of anions like nitrate and chlorate are soluble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/solrules.htm&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; table is worded a bit better to make the possibility of exceptions more explicit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Salts containing Group I elements are soluble ($\ce{Li+, Na+, K+, Cs+, Rb+}$). Exceptions to this rule are rare. Salts containing the ammonium&#xA;  ion ($\ce{NH4+}$) are also soluble.&lt;/li&gt;&#xA;  &lt;li&gt;Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.&lt;/li&gt;&#xA;  &lt;li&gt;Salts containing $\ce{Cl^-, Br^-, I^-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag^+, Pb^{2+}}$, and $\ce{(Hg2)^{2+}}$.&#xA;  Thus, $\ce{AgCl, PbBr2}$, and $\ce{Hg2Cl2}$ are all insoluble.&lt;/li&gt;&#xA;  &lt;li&gt;Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually anything else is insoluble.&lt;/li&gt;&#xA;  &lt;li&gt;Most sulfate salts are soluble. Important exceptions to this rule include $\ce{BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$ .&lt;/li&gt;&#xA;  &lt;li&gt;Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements&#xA;  ($\ce{Ca, Sr,Ba}$) are slightly soluble. Hydroxide salts of transition&#xA;  metals and $\ce{Al^{3+}}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not&#xA;  soluble.&lt;/li&gt;&#xA;  &lt;li&gt;Most sulfides of transition metals are highly insoluble. Thus, $\ce{CdS, FeS, ZnS, Ag2S}$ are all insoluble. Arsenic, antimony, bismuth, and lead&#xA;  sulfides are also insoluble.&lt;/li&gt;&#xA;  &lt;li&gt;Carbonates are frequently insoluble. Group II carbonates ($\ce{Ca, Sr, Ba}$) are insoluble. Some other insoluble carbonates include $\ce{FeCO3}$&#xA;  and $\ce{PbCO3}$.&lt;/li&gt;&#xA;  &lt;li&gt;Chromates are frequently insoluble. Examples: $\ce{PbCrO4, BaCrO4}$&lt;/li&gt;&#xA;  &lt;li&gt;Phosphates are frequently insoluble. Examples: $\ce{Ca3(PO4)2, Ag3PO4}$&lt;/li&gt;&#xA;  &lt;li&gt;Fluorides are frequently insoluble. Examples: $\ce{BaF2, MgF2 PbF2}$.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="73" LastEditDate="2014-10-18T20:29:07.937" LastActivityDate="2014-10-18T20:29:07.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18582" PostTypeId="1" CreationDate="2014-10-18T19:54:59.873" Score="1" ViewCount="189" Body="&lt;p&gt;$$\ce{NH3 + O2 -&amp;gt; N_xO_y + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there is $\pu{1 m^3}$ of both the nitrogen oxide and the ammonia and the volume of oxygen on the left side is $\pu{1.25 m^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that I should use the ideal gas equation. It is also obvious that there is only one atom of nitrogen in the oxide - I guess that there is also only one oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone give me any hints?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="5026" LastEditDate="2018-02-26T13:16:40.277" LastActivityDate="2018-02-26T13:16:40.277" Title="Given volume of reagents find the chemical formula of nitrogen oxide" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18583" PostTypeId="2" ParentId="18547" CreationDate="2014-10-18T19:58:24.960" Score="4" Body="&lt;p&gt;The short answer is probably a bit, depending on the potential you apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least a few factors involved:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How easily are $\ce{Cl-}$ and $\ce{Br-}$ oxidized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}\ce{2Cl^{-}_{(aq)} -&amp;gt; Cl_{2(g)} +2e-}\; &amp;amp;E°_{\mathrm{ox}}=-1.36\ V\\\ce{2Br^{-}_{(aq)} -&amp;gt; Br_{2(aq)} +2e-}\; &amp;amp;E°_{\mathrm{ox}}=-1.09\ V\end{align*}$$&#xA;At standard activities, the two ions are pretty similar in oxidation potentials, so there is probably a range of potentials where both reactions occur to a similar degree or it might be possible to achieve some partial selectivity for bromine formation. The Nernst equation can be used to calculate equilibrium concentrations for a given potential, but so long as the potential applied is well over the standard oxidation potential, the exact values don't really matter.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What difference do the concentrations make?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, what determines how much of either reaction occurs is mass transport. For a redox reaction to occur, the reacting molecule must be very close to the electrode surface (&amp;lt;10 nm). For a planar electrode at a single potential in a standing solution, mass transport occurs by diffusion and the current that flows should be directly proportional to concentration in the bulk. In principle, the Cl to Br ratio should be the same as their activities in solution, though in some situations, other factors like the speed of the electron transfer can be limiting, so this may not be the case, especially since there is so much chloride compared to bromide.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What about the electrodes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As electrolysis is a surface phenomenon, the electrode and its interactions with the reactants can be important. Assuming you pick an electrode which doesn't itself react with the chloride or bromide, the exact material can affect the behaviour of the cell. Even though the Nernst equation may predict a certain oxidation potential, interactions of the reactants and the electrode may result in varying amounts of additional potential (overpotential) needed to cause the reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What happens to the bromine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A certain amount of bromine can dissolve in water, but depending on the pH, it can rapidly disproportionate back to bromate and hypobromate: $\ce{Br2 + 2OH- → Br- + OBr- + H2O}$ Also, the chlorine gas can react with the bromide to make bromine and bromine is also known to react with metals (use carbon electrodes?). In short, you can probably create an electrolytic cell that produces some bromine, but it's difficult to say in what ratio with chlorine and what its exact fate would be.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-18T19:58:24.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18584" PostTypeId="2" ParentId="18568" CreationDate="2014-10-18T20:46:26.483" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slater's rules&lt;/a&gt; are pretty good, but they're not perfect. Consider that he published his work in 1930: &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRev.36.57&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Phys Rev&lt;/em&gt; &lt;strong&gt;1930&lt;/strong&gt; v. 36, 57&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are a number of people who have modified Slater's scheme to better reflect shielding effects and correct for experimental ionization energies and electron affinities (and similar properties) in later rows of the periodic table.&lt;/p&gt;&#xA;&lt;p&gt;If I remember correctly, one such modification is that d and f orbitals shouldn't have as much shielding as suggested by Slater, but the correction is small. There's another scheme that uses an exponential correction instead.&lt;/p&gt;&#xA;&lt;p&gt;For example, Wikipedia gives &lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;effective nuclear charges&lt;/a&gt; from Clementi and Raimondi, who did full self-consistent quantum calculations and you can compare these to Slater's.&lt;/p&gt;&#xA;&lt;p&gt;The main take-home from Slater's rules is that they work pretty well for a wide range of elements, including 1st, 2nd, and 3rd row. I think the corrections generally tweak things when we get into the 4d and 5d transition metals and some of the lanthanides and actinides. Not bad for an 84-year old paper!&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="69830" LastEditDate="2021-05-11T05:21:24.340" LastActivityDate="2021-05-11T05:21:24.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18586" PostTypeId="1" CreationDate="2014-10-18T22:46:22.190" Score="2" ViewCount="96" Body="&lt;p&gt;While making dango with a friend we decided to try to make the dough a bit sweeter.  We already had a large firm lump of dough (rice flour, glutinous rice flour and water) so we sprinkled a bit of sugar on top.  While kneading the dough in my hand I started to feel it run through my fingers.  The entire lump had turned into a goopy drippy mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our best guess was that the sugar soaked up all the water, turning it into a bunch of flour particles in free flowing sugar water, but I cannot find any sources on the internet to explain what exactly happened.  Can anyone explain the reaction that causes this? &lt;/p&gt;&#xA;" OwnerUserId="9379" LastActivityDate="2014-10-18T23:27:00.173" Title="Why does sugar make dough turn to liquid" Tags="&lt;food-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18588" PostTypeId="1" AcceptedAnswerId="18591" CreationDate="2014-10-18T23:30:32.903" Score="4" ViewCount="11373" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TxwRJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TxwRJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to 1,5-dimethyl-6,8-dioxabicyclo[3.2.1]octane when it is hydrolyzed under acidic conditions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that under acidic conditions the first thing to likely happen is protonation. Does it matter which oxygen is protonated first? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once that oxygen has been protonated it is going to want to return to neutrality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have our carbonyl restored (sort of). However, we still want to return to neutrality. Water can attack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we deprotonated the attached water and regenerate the acid-catalyst and we have our original starting compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct mechanism and is the product correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2018-02-01T18:16:26.690" LastActivityDate="2018-02-01T19:05:51.793" Title="Hydrolysis of cyclic acetal" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18589" PostTypeId="1" AcceptedAnswerId="18594" CreationDate="2014-10-18T23:36:35.117" Score="2" ViewCount="24020" Body="&lt;p&gt;It's a question on my quiz which I guessed right and couldn't do the reasoning:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Based on the intermolecular forces, predict the boiling points of the following substances and list them in the order of increasing boiling point. State your reasons accordingly.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dimethyl ether &lt;span class=&quot;math-container&quot;&gt;$(\ce{CH3OCH3})$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Ethanol &lt;span class=&quot;math-container&quot;&gt;$(\ce{CH3CH2OH})$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Propane &lt;span class=&quot;math-container&quot;&gt;$(\ce{CH3CH2CH3})$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Right away, I see the hydrogen bond in ethanol, so it must have the highest boiling point.&lt;/p&gt;&#xA;&lt;p&gt;I couldn't decide on the other two, so I drew lewis structures (kind of) for both&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2IwzF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I ignored the hydrogens because they don't affect the polarity (correct me if my lewis strctures are wrong).&lt;/p&gt;&#xA;&lt;p&gt;This is what I guessed, and I was right. When my teacher takes this question up, he said that propane is non polar, which means it only interacts via london forces, which means it has the lowest boiling points out of these three.&lt;/p&gt;&#xA;&lt;p&gt;But how is propane non polar though, when I asked, he said &amp;quot;I don't have an answer for this, life is hard.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;I mean, if I draw the structure in the bad way, it does look symmetrical and balanced out:&lt;br /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZPCUY.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;But if I draw it in this way, it doesn't help me when I'm trying to figure out the polarity, because everything will look non polar. I can even say water is in a linear shape because the H's and the O line up perfectly in a line, which is not the correct picture of reality.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;My biggest question is, how do I figure out if something is polar or not.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5138" LastEditorUserId="69830" LastEditDate="2021-05-05T05:07:20.347" LastActivityDate="2021-05-05T05:07:20.347" Title="How do I figure out the relative polarity of organic compounds?" Tags="&lt;intermolecular-forces&gt;&lt;lewis-structure&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18590" PostTypeId="1" AcceptedAnswerId="18595" CreationDate="2014-10-18T23:41:35.607" Score="0" ViewCount="4911" Body="&lt;p&gt;I would like to know how do I begin to solve a problem like this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is finding the formula equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that it would be &#xA;$$ \ce{NH3 + CH3CH2COOH -&amp;gt; NH3 + CH3CH2COOH + H2O} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really confused as to how I can balance this equation. Do I give the ammonia a hydrogen? Do I break up the acid into smaller units? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there I would like to know the best way for me to switch the equation into an ionic equation and then finally a net ionic equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way the net ionic equation is &#xA;$$ \ce{CaCO3 (s) + 2HC2H3O2+ (aq) -&amp;gt; CO2 (g) + H2O (l) + Ca^2+ (aq) + 2C2H3O2^- (aq) } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just clueless as to how they got there.  &lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="73" LastEditDate="2014-10-19T01:22:32.177" LastActivityDate="2014-10-19T01:22:32.177" Title="Write a Net Ionic Equation to Represent the Reaction of Aqueous Ammonia with Propionic acid" Tags="&lt;homework&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18591" PostTypeId="2" ParentId="18588" CreationDate="2014-10-18T23:44:47.427" Score="2" Body="&lt;p&gt;Your product is correct, but the mechanism isn't quite right. In the first step, one of the oxygens will be protonated (either is possible). The other oxygen &quot;kicks out&quot; the one activated in the first step. So far so good. However, in the next step, water will attack the &quot;carbonyl&quot; carbon to give a hemi-acetal and a hydroxyl group. Repeating a similar set of steps reveals the ketone and second hydroxyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xA8Kl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xA8Kl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key difference here is that by adding to the pi system, water reacts with the C-O pi-antibonding orbital, which is the LUMO of the system&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="55803" LastEditDate="2018-02-01T19:05:51.793" LastActivityDate="2018-02-01T19:05:51.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18592" PostTypeId="2" ParentId="18589" CreationDate="2014-10-18T23:59:48.183" Score="2" Body="&lt;p&gt;Look for atoms connected to other atoms with different electronegativities.  The larger the difference in electronegativity, the greater the polarity of a bond.  In propane, we just have &lt;strong&gt;carbon (EN=2.55) connected to carbon&lt;/strong&gt; and hydrogen (EN=2.20).  Propane is weakly polar, it has a very small dipole moment, around 0.08 D. Dimethyl ether, on the other hand has &lt;strong&gt;carbon connected to oxygen&lt;/strong&gt; (EN=3.44) and hydrogen.  The electronegativity difference between carbon and oxygen is 0.89 D, pretty large.  Dimethyl ether has a dipole moment of 1.3 D.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-18T23:59:48.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18593" PostTypeId="2" ParentId="18582" CreationDate="2014-10-19T00:03:27.023" Score="2" Body="&lt;p&gt;As a hint, you could write balanced equations for oxidation to N&lt;sub&gt;2&lt;/sub&gt;O, NO, NO&lt;sub&gt;2&lt;/sub&gt; and N&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;.  (Did I forget any nitrogen oxide?)  Then you can compare moles (or volumes, since we are talking gases) of starting materials and products.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-19T00:03:27.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18594" PostTypeId="2" ParentId="18589" CreationDate="2014-10-19T00:16:22.647" Score="6" Body="&lt;p&gt;General steps to figure out the polarity (or at least relative polarity) and subsequently the relative boiling point of compounds:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Acquire the correct geometry&lt;/li&gt;&#xA;&lt;li&gt;Estimate dipole moments&lt;/li&gt;&#xA;&lt;li&gt;Sum up estimated dipole moments&lt;/li&gt;&#xA;&lt;li&gt;Check for other intermolecular interactions&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Acquire Correct Geometry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This step requires some knowledge about the geometry of hydrocarbon compounds. But it seems that you already know about the tetrahedral bonding centers of saturated hydrocarbons. This is then important later on for the dipole forces. Never try to linearize the carbon backbone just to try to make sense of this, it will do more damage than good.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Estimate Dipole Moments&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The dipole moments that you have to estimate will be those along bonds. There is a larger dipole moment in the ether than the hydrocarbon without heteroatoms, because the oxygen atom is strongly electronegative. As such, there will be a dipole along the $\ce{C-O}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sum Up Estimated Dipole Moments&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These dipoles can be understood as vectors along bonds, pointing from positive to negative partial charges, and can be summed up for the total dipole moment of a molecule. The larger the absolute value of this dipole vector is, the larger the dipole-dipole interactions between molecules: Subsequently, the higher the forces between the molecules of this substance.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Check for Other Intermolecular Interactions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One other main intermolecular action is, as you already pointed out in your question, hydrogen bonding. This one is quite strong, but isn't found for every compound out there. Basically, it is when a highly polarized $\ce{H-X}$ bond can be formed, where $\ce{X}$ can be oxygen, nitrogen, chlorine or some other electronegative atom (with respect to carbon). The hydrogen that is sandwiched in the hydrogen bond has to be (formally) bonded to such an electronegative atom as well, otherwise it wouldn't be as free to create those bonds (as you also have realized, because the hydrocarbon hydrogens are really quite boring from an intermolecular interaction perspective).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Result&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Based on the steps above, we can generate a table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ayH0.png&quot; alt=&quot;Table for the substances in the question.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more &quot;Yes&quot; a substance has, the higher the intermolecular forces. Based on these data, assumed by us, we can generate a list (from highest boiling point to lowest):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ethanol&lt;/li&gt;&#xA;&lt;li&gt;Dimethyl ether&lt;/li&gt;&#xA;&lt;li&gt;Propane&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-19T00:16:22.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18595" PostTypeId="2" ParentId="18590" CreationDate="2014-10-19T00:22:54.470" Score="3" Body="&lt;p&gt;We know that ammonia is $\ce{NH3}$ and propanoic acid is $\ce{CH3CH2COOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ammonia is a basic substance, and propanoic acid is... an acidic substance. More specifically, ammonia fits the definition of a Bronsted-Lowry base, so it will act as an $\ce{H+}$ accepter in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arrhenius acids are expected to increase the concentration of $\ce{H+}$ in solution, and propanoic acid is no exception. In water as propanoic acid dissociates...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3CH2COOH -&amp;gt;[H2O] CH3CH2COO- + H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the $\ce{H+}$ comes (and is later bonded to the ammonia via)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH3 + H+ -&amp;gt; NH4+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After those two steps occur, the products $\ce{NH4+ + CH3CH2COO-}$ are formed. Your ionic equation is as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH3 + CH3CH2COOH -&amp;gt; NH4+ + CH3CH2COO-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appear to be no spectator ions (ions that appear on both sides of the equation), so there is nothing to cancel out. As such, this is ALSO the net ionic equation. They both appear to be balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, you had to do both. You were supposed to add an $\ce{H+}$ to the ammonia, but that happens if you dissociate the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the equation with the calcium, carbonate ions sometimes create carbon dioxide and water (via carbonic acid readily decomposing, like in sodas).&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2014-10-19T00:22:54.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18596" PostTypeId="2" ParentId="18590" CreationDate="2014-10-19T00:28:24.220" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am just clueless as to how they got there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, you're not the only one. I really don't see how you get to equation 2 from equation 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out how to get the acid-base reaction equation, you first need to know which compounds react as bases (i.e. take up a proton) and acids (i.e. deliver a proton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, ammonia is the base (because the nitrogen has a lone pair that can be accessed by a proton) and propionic acid is the acid (&lt;em&gt;nomen est omen&lt;/em&gt;). Then you know that the final equation has to be&#xA;$$ \ce{NH3 + CH3CH2COOH &amp;lt;=&amp;gt;[][\ce{H2O}] NH4+ + CH3CH2COO-} $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, how often does one need to balance these types of reactions or do they usually end up balanced or working themselves out as balanced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Acids (and bases) can be multiprotic, which means that they can give off (or accept) more than just one proton. An example for a diprotic acid would be carbonic acid $\ce{H2CO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this, it is possible to write several (equilibrium) reactions. For example:&#xA;$$ \ce{2NH3 + H2CO3 &amp;lt;=&amp;gt; NH3 + NH4+ + HCO3- &amp;lt;=&amp;gt; 2NH4+ + CO3^{2-}} $$&#xA;Where this equilibrium then lies depends on the equilibrium constants for each of these reactions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second equation can be expanded some more to show &quot;intermediate&quot; steps:&#xA;$$ \ce{CaCO3 + 2 CH3COOH -&amp;gt; Ca^{2+} + H2CO3 + 2CH3COO- \\ &amp;lt;=&amp;gt; Ca^{2+} + 2CH3COO- + H2O + CO2 ^} $$ &#xA;Here carbonic acid decomposes to give carbon dioxide which bubbles out of the solution, thereby making the reaction stay out of the equilibrium zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this clarified it somewhat for you.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-10-19T00:54:13.877" LastActivityDate="2014-10-19T00:54:13.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18597" PostTypeId="1" AcceptedAnswerId="18599" CreationDate="2014-10-19T01:20:39.343" Score="5" ViewCount="6912" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0NyT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what it looks like, Gilman reagents are good nucleophiles, and will attack at the less-substituted side of an asymmetric alkane because Gilman reagents don't necessarily need a substantial amount of partial positive character at the site of attack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, don't strong nucleophiles open epoxides in an SN2 like manner? Where's the inversion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="186" LastEditDate="2015-10-02T16:28:10.223" LastActivityDate="2017-12-12T12:51:11.900" Title="How do Gilman reagents open epoxides?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18598" PostTypeId="1" AcceptedAnswerId="18600" CreationDate="2014-10-19T01:27:13.090" Score="0" ViewCount="98" Body="&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How many liters of $H_2S$ would be required to react with excess oxygen to produce $12.0$ $L$ of $SO_2$? The reaction yield is $88.5$%. Assume constant temperature and pressure throughout the reaction.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used dimensional analysis to convert from liters of $SO_2$ to $moles$ of $H_2S$ and then used Avogadro`s Law ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$) to find the volume of $H_2S$ in liters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dimensional Analysis:&lt;/strong&gt;&#xA;$\frac{1}{.885} * \frac{12.0L}{1} * \frac{1 g}{1 mL} * \frac{1 mol SO_2}{64g} * \frac{2 mol SO_2}{2 mol H_2S} * \frac{1 mol H_2S}{34 g} = 6.23$ $mol$ $of$ $H_2S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Avogadro`s Law:&lt;/strong&gt;&#xA;$\frac{12.0L}{166 mol} * \frac{ L }{6.23 mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;End Result = $.028 L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;^ The end result does not seem right and any help in remedying my approach or shedding light on another approach would be much appreciated. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="5409" LastActivityDate="2014-10-19T01:55:04.400" Title="Stoichiometry of Gaseous Reactions" Tags="&lt;homework&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18599" PostTypeId="2" ParentId="18597" CreationDate="2014-10-19T01:30:34.413" Score="5" Body="&lt;p&gt;In the first reaction, the reaction doesn't take place at a stereocenter, so there's no way to observe the inversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second reaction, the stereochemistry in the product is R, R. The Gilman reagent approaches opposite the C-O bond. After bond rotation, we see the product in the indicated conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5VWf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-10-21T20:41:21.693" LastActivityDate="2014-10-21T20:41:21.693" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18600" PostTypeId="2" ParentId="18598" CreationDate="2014-10-19T01:55:04.400" Score="0" Body="&lt;p&gt;I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your balanced equation is as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2S + 3O2 -&amp;gt; 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2014-10-19T01:55:04.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18602" PostTypeId="1" CreationDate="2014-10-19T02:34:31.853" Score="1" ViewCount="189" Body="&lt;p&gt;i have the mass spectrum for the molecule but-3-en-1-ol. The Mr of this is 72&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four peaks that are labelled which I have to name the species responsible for forming that peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peaks are:&lt;br&gt;&#xA;m/z = 66&lt;br&gt;&#xA;m/z = 67&lt;br&gt;&#xA;m/z = 71&lt;br&gt;&#xA;m/z = 73 - this peak is labelled as (M + H)^+&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer so far...&lt;br&gt;&#xA;The peak at m/z=73 is the but-3-en-1-ol with one added hydrogen, somewhere to the double bond?&lt;br&gt;&#xA;For the other peaks am I just removing hydrogens as the mass only decreases by 8 in total so i couldn't remove a carbon or the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the peaks are:&lt;br&gt;&#xA;m/z=71 is the removal of 1H&lt;br&gt;&#xA;m/z=67 is the removal of another 4H (5H in total)&lt;br&gt;&#xA;m/z=66 is the removal of another 1H (6H in total)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this is correct? Which hydrogens are removed first and what does each species look like for each peak, and are there any charges on the species produced. Also is there a special name for this sort of fragmentation, e.g alpha cleavage?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastActivityDate="2014-10-20T19:09:07.660" Title="Mass Spectroscopy peak assignments in but-3-en-1-ol" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18603" PostTypeId="2" ParentId="17069" CreationDate="2014-10-19T03:12:12.843" Score="0" Body="&lt;p&gt;We can compare the orbitals of electrons to the orbitals of planets, and although this is not a perfect model, it can be useful(such as looking at electron velocities in a similar fashion to planetary velocities, subject to relativistic effects &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden&quot;&gt;Why is gold golden?&lt;/a&gt;).&#xA;One use is to find the escape velocity of the electron, treating it like a planet trying to escape orbit. So yes, we can usefully make such a comparison. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the answer to your question is multi-layed. In maths, 5+5+5+5+5 is equivalent to 5x5 could be seen as 5 rows of 5, in total 25 objects. Alternatively, it could be represented as 5&lt;sup&gt;2&lt;/sup&gt;, and later we know that when multiplying by the same base, you add the indices: 5&lt;sup&gt;1&lt;/sup&gt; x 5&lt;sup&gt;1&lt;/sup&gt; = 5&lt;sup&gt;1+1&lt;/sup&gt; = 5&lt;sup&gt;2&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar fashion, shallowly its given that there can only be 2n&lt;sup&gt;2&lt;/sup&gt; electrons per shell, where &lt;em&gt;n&lt;/em&gt; is the shell number. For the first shell, 1&lt;sup&gt;1&lt;/sup&gt; = 1 so 2n&lt;sup&gt;2&lt;/sup&gt; = 2x1 = 2 &#xA;This also gives the second shell to have capacity for 8(2x2x2) and the third as 18(2x3x3). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a deeper level, in Quantum Mechanics, electrons have 4 quantum numbers(&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Quantum_number&lt;/a&gt;) which are the coordinates (n, l, m&lt;sub&gt;l&lt;/sub&gt;, m&lt;sub&gt;s&lt;/sub&gt;). Any electron in an atom can be represented as combinations of these 4 quantum numbers. The first, n, is the shell number, just like before. The middle 2 numbers, L and m&lt;sub&gt;L&lt;/sub&gt; are complex, but eventually cause the n&lt;sup&gt;2&lt;/sup&gt; part of the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last number, m&lt;sub&gt;s&lt;/sub&gt; is its spin. An electron can either be in spin 'up' or 'down', only 2 states are possible. This is where the 2 comes from in 2xn&lt;sup&gt;2&lt;/sup&gt;, as for every combination of (n, l, m&lt;sub&gt;l&lt;/sub&gt;) there is two possibilities, one where the fourth is 0(meaning spin is 'down') and another when it's 1(where spin is 'up). Due to the Pauli exclusion principle(&lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&lt;/a&gt;), no two electrons can have identical Quantum Numbers. This means atleast one of the 4 must be different. Due to this, once all of the combinations of the first middle two have been filled in a shell, and once an equal number with the opposite spin(but same middle two quantum numbers) have been filled, the first number, n, must change to accommodate more electrons. This establishes the limit for electrons per shell as 2n&lt;sup&gt;2&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind the 2n&lt;sup&gt;2&lt;/sup&gt; limit applies only to shells other than the valence shell(outer most shell). The valence shell is ruled by the octet rule that places the limit at 8 for the outermost shell. This is also ruled by the orbitals and the energy to fill the shells, but that requires a deeper understanding. &lt;/p&gt;&#xA;" OwnerUserId="8269" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2014-10-19T03:12:12.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18605" PostTypeId="1" CreationDate="2014-10-19T13:17:16.353" Score="6" ViewCount="3545" Body="&lt;p&gt;The vibrational modes of $\ce{XeF2}$ are at $555$, $515$, and $213~\mathrm{cm^{-1}}$ but only two are IR active. &#xA;Explain why this is consistent with $\ce{XeF2}$ having a linear structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, since it has $3$ atoms it should have $3n - 5$ vibrational modes, thus $3(3) - 5 = 4$ modes. Yet the problem tells me it only has $3$, which would suggest a non-linear shape. How to explain this?&lt;/p&gt;&#xA;" OwnerUserId="6893" LastEditorUserId="16683" LastEditDate="2016-09-29T12:12:50.910" LastActivityDate="2016-09-29T13:12:49.137" Title="How can I deduce the linearity of XeF2 from the IR spectrum?" Tags="&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18607" PostTypeId="2" ParentId="9380" CreationDate="2014-10-19T14:34:46.773" Score="1" Body="&lt;p&gt;Heating most viscous solution will decrease their viscosity. Most agar gels will liquefy when heated. Also, some polysaccharide solutions don't become viscous again when they cool down. I think alginate will become viscous, but you will have to test that out.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-19T14:34:46.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18608" PostTypeId="2" ParentId="18605" CreationDate="2014-10-19T14:47:56.530" Score="8" Body="&lt;p&gt;Let's start with the &lt;strong&gt;rule of mutual exclusion&lt;/strong&gt;, it will come in handy.  It states that,  for centrosymmetric molecules (molecules with a center of symmetry), vibrations that are IR active are Raman inactive, and &lt;em&gt;vice versa&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear $\ce{AB2}$ molecule will have $(3n-5) = 4$ vibrational modes, but because it is linear 2 of the modes are degenerate.  $\ce{CO2}$ is an example of a linear $\ce{AB2}$ molecule, look at the following diagram showing the 4 vibrations for $\ce{CO2}$ and notice that the bending vibration ($v_2$) is degenerate.  That is, there is no difference between those 2 bending modes, they are identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rBi5z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice too that $v_1$ is the only vibrational mode that does not change the dipole moment of the molecule, therefore it is the Raman active mode. Since the linear molecule is centrosymmetric, the Raman band does not appear in the IR, nor do any of the IR bands appear in the Raman spectrum. Hence, for a linear $\ce{AB2}$ molecule we would expect 2 IR active bands and 1 Raman band. &lt;strong&gt;This is consistent with the spectral data you reported.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $\ce{AB2}$ molecule is bent rather than linear (water for example), then there are $(3n-6) = 3$ vibrational modes as pictured below.  All of these vibrations result in a change in the dipole moment.  Therefore, for a non-linear $\ce{AB2}$ molecule, all 3 of the bands will be IR active. Since the molecule is not centrosymmetric all 3 bands may also appear in the Raman spectrum (note: while a change in dipole moment is required for an IR-active band to appear, a change in polarizability is required for a Raman band to appear; looking at the water vibrations pictured, they should alter both the dipole moment and the polarizability of the molecule). Hence, for a non-linear $\ce{AB2}$ molecule we would expect 3 bands in the IR and 3 in the Raman spectrum. &lt;strong&gt;This is not consistent with the spectral data you reported above.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kXGjH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary, your spectral data is consistent with $\ce{XeF2}$ having a linear geometry.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Thanks to @porphyrin for pointing out that bent $\ce{AB2}$ molecules are indeed not centrosymmetric.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7475" LastEditDate="2016-09-29T13:08:48.660" LastActivityDate="2016-09-29T13:08:48.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18609" PostTypeId="1" AcceptedAnswerId="18619" CreationDate="2014-10-19T14:52:00.097" Score="1" ViewCount="1087" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does silicon not react with dioxygen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm doing an organic lab experiment. I'm heating up silicon, carbon, and sulfur individually then placing them in pure $\ce{O2}$. I don't really know about carbon chemistry but I thought the reaction might be related to the electronegativity of silicon being lower than that of sulfur/carbon causing the bond to be more polar or something?&lt;/p&gt;&#xA;" OwnerUserId="9384" LastEditorUserId="5026" LastEditDate="2018-04-12T03:49:45.053" LastActivityDate="2018-04-12T03:50:52.923" Title="Why does silicon not react with dioxygen?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18610" PostTypeId="2" ParentId="18609" CreationDate="2014-10-19T15:13:38.147" Score="0" Body="&lt;p&gt;In order for a reaction to occur, bonds in the reactants have to break so that they can form new bonds that comprise the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you use &lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;amorphous carbon&lt;/a&gt;? Was it &lt;a href=&quot;http://en.wikipedia.org/wiki/Monocrystalline_silicon&quot; rel=&quot;nofollow noreferrer&quot;&gt;monocrystalline silicon&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article on amorphous carbon says that it is stabilized by dangling pi bonds with hydrogen, suggesting a bond that is easier to break and start an exothermic reaction that will provide further energy to break other, stronger bonds.  The silicon lacks this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you try graphite?  My guess is that it would not react either, nor would diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sulfur probably reacted due to the weaker $\ce{S-S}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, electronegativity isn't a very good explanation for much other than bond polarity. &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5026" LastEditDate="2018-04-12T03:50:52.923" LastActivityDate="2018-04-12T03:50:52.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18612" PostTypeId="1" AcceptedAnswerId="18613" CreationDate="2014-10-19T16:20:44.673" Score="16" ViewCount="31979" Body="&lt;p&gt;I am watching the Stoke vs Swansea game, &amp;amp; to be honest, it's pretty boring. So much so, that I've started to wonder why it is that my beer has turned into a slushy after being taken out of the freezer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought it this morning, it wasn't very cold, so just before the game I put it in the freezer to achieve optimal freshness. Started off pretty well, but rather than getting warmer as I drink, it's actually continued to freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in France.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The windows are open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's 22 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached a pic of the beer in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/slJ45.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I discovered the holy grail of beer chilling?&lt;/p&gt;&#xA;" OwnerUserId="9386" LastEditorUserId="7475" LastEditDate="2016-06-22T18:16:15.727" LastActivityDate="2019-09-18T19:32:15.157" Title="Beer freezing after being opened" Tags="&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18613" PostTypeId="2" ParentId="18612" CreationDate="2014-10-19T16:26:18.720" Score="13" Body="&lt;p&gt;When you opened the beer, carbon dioxide was lost to the atmosphere, increasing the mole fraction of water in the beer and raising its freezing point to a point above the temperature of the beer.  Not all of the water froze, leaving it a slush.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  There's also the possibility that the loss of CO2 from the solution is not the issue.  It could just be that when you opened the beer you disturbed a supercooled solution enough to start it freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd edit:  Instead of continuing the comments I'll add here.  The freezing on the outside of the can shows that the temperature of the beer is below the freezing point of water.  Is the temperature of the beer lower than the freezing point of the beer?  If it were, the beer should freeze instantly upon being disturbed.  If the beer is above its freezing point, but the loss of CO2 raises the freezing point it should freeze gradually as it sits out.  &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2014-10-19T16:51:34.227" LastActivityDate="2014-10-19T16:51:34.227" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18614" PostTypeId="1" CreationDate="2014-10-19T16:39:58.367" Score="1" ViewCount="56" Body="&lt;p&gt;I am not sure if the equivalents of $\ce{H2}$ in the equation &#xA;$$\ce{Mg (s) + 2HCl (aq) -&amp;gt; MgCl2 (aq) + H2}$$&lt;br&gt;&#xA;is that of dry or wet hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $\ce{H2}$ refers to wet hydrogen in the equation, how should I find the volume of dry hydrogen given the pressure of wet and dry hydrogen and the volume of wet hydrogen?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4945" LastEditDate="2017-11-27T12:24:56.610" LastActivityDate="2017-11-27T12:24:56.610" Title="How to calculate the volume of dry hydrogen in magnesium/ HCl reaction" Tags="&lt;gas-laws&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2014-10-19T19:44:57.840" ContentLicense="CC BY-SA 3.0" />
  <row Id="18615" PostTypeId="1" CreationDate="2014-10-19T16:59:03.350" Score="5" ViewCount="41339" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Volume of wet hydrogen collected at room temperature: &lt;span class=&quot;math-container&quot;&gt;$\pu{40.48 mL}$&lt;/span&gt;, ambient pressure: &lt;span class=&quot;math-container&quot;&gt;$\pu{29.7 inHg}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{754 torr}$&lt;/span&gt;), ambient temperature: &lt;span class=&quot;math-container&quot;&gt;$\pu{295 K}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{22^\circ C}$&lt;/span&gt;), vapor pressure of water at &lt;span class=&quot;math-container&quot;&gt;$\pu{22^\circ C}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\pu{19.83 torr}$&lt;/span&gt;. Partial pressure of dry hydrogen: &lt;span class=&quot;math-container&quot;&gt;$\pu{9.87 inHg}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Find the volume of hydrogen under standard conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The equation of combined gas law is &lt;span class=&quot;math-container&quot;&gt;$p_1V_1/T_1 = p_2V_2/T_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Say if the volume of wet hydrogen gas is &lt;span class=&quot;math-container&quot;&gt;$\pu{40.48 mL}$&lt;/span&gt;, does this equation still work if part of that volume is water vapor? Or, in other words, does the same volume of water vapor and hydrogen gas exert the same pressure?&lt;/p&gt;&#xA;&lt;p&gt;One of the correct way to do this problem is find the moles of dry hydrogen gas using the room temperature and volume of wet hydrogen gas, gas constant and room temp. Right here, I am kinda confused why the volume of &lt;em&gt;wet&lt;/em&gt; hydrogen if we are trying to find the moles of &lt;em&gt;dry&lt;/em&gt; hydrogen gas. Then rest of this solution is plugging the number of moles into the equation along with STP to find the volume of dry hydrogen gas.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-11-05T07:55:12.693" Title="Using combined gas law to calculate the volume of dry hydrogen gas" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="18616" PostTypeId="2" ParentId="18612" CreationDate="2014-10-19T17:11:58.087" Score="11" Body="&lt;p&gt;As you can see in the &lt;a href=&quot;http://stevengoddard.wordpress.com/2010/09/02/the-ideal-world-phase-diagrams-part-deux/&quot;&gt;phase diagram&lt;/a&gt; for water, higher pressures decrease the freezing point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Phase_Transitions/Phase_Diagrams&quot;&gt;Chemwiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;one is able to melt ice simply by applying pressure and not by adding heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you open the beer, you release the pressure in the bottle, raising the freezing point of the beer, so it starts to freeze. Since the bottle is colder than the freezing point of water, any water condensing on the outside will freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the release of $\ce {CO2}$ from the beer will actually raise the freezing point, not lower it.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-10-19T17:18:09.097" LastActivityDate="2014-10-19T17:18:09.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18617" PostTypeId="2" ParentId="18615" CreationDate="2014-10-19T17:52:02.277" Score="2" Body="&lt;p&gt;Consider the composition of the gas in your eudiometer.  Most of it is hydrogen, but some of it is water vapour.  In other words, not all the pressure is exerted by hydrogen, part of it is due to water.  You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should have been given vapour pressure of water in a table.  You should also have taken down value of air pressure and ambient temperature.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-19T17:52:02.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18619" PostTypeId="2" ParentId="18609" CreationDate="2014-10-19T18:22:21.193" Score="5" Body="&lt;p&gt;Elemental sulfur commonly occurs in form of $\ce{S8}$, with relatively low melting point and, more importantly, relatively low bond energies. So, it easily breaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon commonly occurs as graphite or structures composed of graphite-like fragments, but usually something we call carbon contains up to 10% of other elements, meaning it can produce some combustible gases on heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, more importantly, both elements mentioned above &lt;strong&gt;produce gaseous oxides on oxidation&lt;/strong&gt;, that leave the surface of the reacting compound, exposing it to further reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon, on the other hand, produces solid oxide on heating, that prevents further reaction. It is still possible to make a very fine dust of silicon to react with oxygen, but it is not an easy feat. For reaction to occur to bulk silicon, the products must be removed from the surface. Given that, bulk silicon will not give silicon oxide, but it can react with chlorine in reasonable conditions, giving volatile silicon tetrachloride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why silicon forms solid oxide with high melting point is another matter, with answer considerably more obscure. The simplest approach to rationalize it would be to consider atomic radius of carbon, sulfur and silicon and consider bonding in alternative structures for this compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon dioxide is an oxide with small central atom, that can form $\pi$-bonds effectively with oxide, so it does not form polymeric structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur dioxide is an oxide of larger atom, but it has a lone pair on central atom, so polymeric form would be considerably hindered. (on contrary, sulfur trioxide forms polymeric forms, but still has gaseous form  which is only marginally less stable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon dioxide in monomeric form has two $\pi$-bonds, and this is uncharacteristic for third row elements, and silicon is significantly larger than carbon, having enough space for four oxygens around it. So, it easily and effectively forms 3-d network of $\ce{SiO4}$ units with shared oxygens, that is quite hard to melt and vaporize, and, once formed on the surface of silicon, it prevents further reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, formation of silicon dioxide films on surface of bulk monocrystalline silicon is essential for microelectronics.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="8083" LastEditDate="2014-10-20T05:15:35.393" LastActivityDate="2014-10-20T05:15:35.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18621" PostTypeId="1" AcceptedAnswerId="18623" CreationDate="2014-10-19T19:36:42.093" Score="1" ViewCount="770" Body="&lt;p&gt;Since the $\ce{sp^2}$ hybridized orbitals are lower in energy than the p orbital shouldn't the $\ce{sp^2}$ orbitals fill first? Why is this not the case - two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are singly occupied? If you look at Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these orbitals. However, the higher energy p orbital is empty - as would be predicted is Hund's rules are followed. Is it something to do with the conjugation in amides? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2014-10-19T20:00:47.707" LastActivityDate="2014-10-19T20:05:52.727" Title="Doesn't the fact that the lone pair on the nitrogen in amides occupy a p orbital contradict Hund's rules?" Tags="&lt;electronic-configuration&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18622" PostTypeId="1" CreationDate="2014-10-19T19:38:17.647" Score="1" ViewCount="1698" Body="&lt;p&gt;How do we experimentally measure the strengths of covalent versus ionic bonds? Also, for a given molecule, how can we know if two pairs of atoms is covalently or ionically bound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question: Textbooks say that in solution of water, two neighboring water molecules (call them A and B) will be 'attracted' because the O in molecule A will attract the H of the 'nearby' molecule B. It is said that H will alternate states between A and B. How do we know this is true experimentally and how can it be measured?  &lt;/p&gt;&#xA;" OwnerUserId="9388" LastEditorUserId="73" LastEditDate="2014-10-19T19:40:01.870" LastActivityDate="2014-10-29T07:08:32.593" Title="How do we experimentally know the strength of bonds?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18623" PostTypeId="2" ParentId="18621" CreationDate="2014-10-19T19:59:33.063" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are&#xA;  singly occupied&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really, hybridization occurs in response to bonding, so we can only talk about an atom's hybridization when it is in a molecule.  In amides the nitrogen $\ce{sp^2}$ orbitals form molecular orbitals (MO's) with whatever else they are bound to.  In the case of a simple amide with the nitrogen bound to hydrogen, the MO would involve a nitrogen $\ce{sp^2}$ orbital and a hydrogen 1s orbital and 2 electrons would occupy this orbital (one from the nitrogen $\ce{sp^2}$ orbital and one from the hydrogen 1s orbital)  - so what was the $\ce{sp^2}$ orbital will be filled with 2 shared electrons when a bond is formed.  Therefore the 3 $\ce{sp^2}$ orbitals, now MO's, and the p orbital on nitrogen will each be filled with 2 electrons in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these&#xA;  orbitals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Carbenes can exist in either a singlet or triplet state, as shown below.  There are 2 possible triplets depending upon how the carbene is hybridized, a bent $\ce{sp^2}$ hybridized carbene, or a linear $\ce{sp}$ hybridized carbene. In the case of methylene ($\ce{:CH2}$), and actually for most carbenes, the triplet states are lower in energy due to Hund's rule (don't pair electrons unless you must); and the triplet has one electron in the p orbital and one in the $\ce{sp^2}$ orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wHsp6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-19T20:05:52.727" LastActivityDate="2014-10-19T20:05:52.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18625" PostTypeId="1" AcceptedAnswerId="18626" CreationDate="2014-10-19T20:02:51.863" Score="3" ViewCount="306" Body="&lt;p&gt;Nickel's electronic ground state is $\mathrm{3d^8 4s^2}$ (there's a dispute on this, but let's not worry about it for now). Oxygen's is $\mathrm{2s^2 2p^4}$. In a Rare-Earth based Nickelate with chemical formula $\mathrm{RNiO_3}$, if we assume simple ionic bonds, then what should the electronic state be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature indicate that the &quot;classical&quot; case is $\mathrm{Ni^{3+} O^{2-}}$ corresponding to $\mathrm{3d^7, 2p^6}$. But this means that Oxygen has gained 2*3=6 electrons. But how can this be possible if Nickel has donated only 3 electrons?&lt;/p&gt;&#xA;" OwnerUserId="9390" LastEditorUserId="8083" LastEditDate="2014-10-20T18:35:55.023" LastActivityDate="2014-10-20T18:35:55.023" Title="What's the electronic configuration of Nickel and Oxygen in RNiO3 (R=Rare-Earth)?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18626" PostTypeId="2" ParentId="18625" CreationDate="2014-10-19T20:35:23.880" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What's the electronic configuration of Nickel and Oxygen in RNiO3&#xA;  (R=Rare-Earth)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As an example, let's look at $\ce{LaNiO3}$.  Each oxygen has an oxidation state of -2, nickel has an oxidation state of +3 and lanthanum has an oxidation state of +3.  Overall the molecule is neutral.&#xA;$$\ce{La(+3)~ plus~ Ni(+3)~plus~ 3~oxygens (-6)~=~0}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-19T20:35:23.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18627" PostTypeId="1" CreationDate="2014-10-19T20:37:21.040" Score="6" ViewCount="29605" Body="&lt;p&gt;I need to assign the H-NMR of aspirin, below is aspirin with the H's labelled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mUkOY.png&quot; alt=&quot;structure of aspirin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peaks I have are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2.30ppm   (I this is a singlet and would be F)    &lt;/li&gt;&#xA;&lt;li&gt;7.07ppm   (I think this is E)   &lt;/li&gt;&#xA;&lt;li&gt;7.29ppm   (I think this is D)   &lt;/li&gt;&#xA;&lt;li&gt;7.53ppm   (I think this is C)  &lt;/li&gt;&#xA;&lt;li&gt;8.05ppm   (I think this is B)  &lt;/li&gt;&#xA;&lt;li&gt;11.44ppm (this is a singlet and would be A)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For B,C,D,E I need to say what kind of splitting pattern there would be and how many coupling constants are present and there approximate value. I think I know the assignments of them but I don't know the splitting pattern or coupling constants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would E and B be doublet of doublets because they couple with D and C so they would have ortho and meta coupling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would D and C be coupling with each other and B and E so would they be doublet of doublets of doublets, with two ortho and one meta coupling?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="4231" LastEditDate="2014-10-28T22:53:04.537" LastActivityDate="2014-10-28T22:53:04.537" Title="How to assign the signals in the H-NMR of Aspirin?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18628" PostTypeId="2" ParentId="18627" CreationDate="2014-10-19T21:06:02.307" Score="5" Body="&lt;p&gt;Here is a link to the actual assigned proton nmr spectrum for &lt;a href=&quot;https://www.chem.wisc.edu/~cic/nmr/NMRdatab/common_cmpd/aspirin/aspirin_H1.pdf&quot; rel=&quot;noreferrer&quot;&gt;acetylsalicylic acid&lt;/a&gt;.  Your peak assignments are correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would E and B be doublet of doublets because they couple with D and C&#xA;  so they would have ortho and meta coupling?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that is correct. Here is another &lt;a href=&quot;http://www.joshstaiger.org/amychan/Literary/Lab%20Three%20Report%20Aspirin.pdf&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; with more detail on the coupling constants (see p. 13 for example)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would D and C be coupling with each other and B and E so would they be&#xA;  doublet of doublets of doublets, with two ortho and one meta coupling?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Conceptually you are correct, however in the case where two coupling constants are very similar, the expected quartet collapses to a triplet. Since $\ce{J_{DC}}$ is very close in magnitude to $\ce{J_{BC}}$ the expected quartet appears as a triplet.  Therefore, proton C appears as a triplet from these two protons and then each of these lines is further split into a tight doublet by the smaller coupling to proton E.  The overall pattern appears as a triplet of doublets.  The same reasoning applies to proton D, and again a triplet of doublets results.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-19T21:19:40.833" LastActivityDate="2014-10-19T21:19:40.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18629" PostTypeId="1" AcceptedAnswerId="20254" CreationDate="2014-10-19T23:07:49.097" Score="3" ViewCount="2323" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem has a flame emitting a wavelength of $589\ \mathrm{nm}$, and asks for the mass of one photon of that wavelength. It tells me that $1\ \mathrm J = 1\ \frac{\mathrm{kg\cdot m^2}}{\mathrm s^2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I converted to $\mathrm J$ using $E=\frac{h\cdot c}{\lambda}$. I'm not quite sure what to do past that. I have $3.36\cdot 10^{-19}\mathrm J$. I know the answer is supposed to be $3.75\cdot 10^{-34}$.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="41328" LastEditDate="2019-07-22T02:07:47.860" LastActivityDate="2019-07-22T02:07:47.860" Title="How do I find mass equivalence of a photon?" Tags="&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18634" PostTypeId="2" ParentId="18627" CreationDate="2014-10-20T00:31:32.430" Score="5" Body="&lt;p&gt;Your chemical shift analysis is correct. There are two schools of thought concerning how to describe splitting patterns, and your analysis for splitting patterns is perfectly acceptable, and correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first school of thought is to label a splitting pattern based on the appearance of the peaks, which is why Ron describes your aromatic peaks as doublets of triplets, and even quartet. I'm not saying this is wrong; this is just one method used. This method can cause problems for the inexperienced spectroscopist, as the appearance of a spectrum is very much dependent on the strength of the magnetic field, choice of solvent, temperature, sample preparation, quality of shimming etc. This is why experimental conditions should always be reported when publishing NMR data - and something students should be entitled to ask of their teachers when given a spectrum to decipher. What one person reports as a triplet on a 60MHz spectrometer in water might be undeniably a doublet of doublets on a 800MHz in acetone. Second order splittings in aromatic systems are a great example of this. Although common on low field systems, second order splittings are far less common on high field instruments, and many AB quartets collapse to AX spin systems at high field and couplings can be analysed as first order systems. Of course, this method makes very good sense for a number of systems. For instance, labile amine or alcohols are almost always reported as broad singlets, even though they are likely to be possess 3J couplings that are not observed through exchange decoupling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second school of thought (which your original descriptions of ddd adhere to) is to look at the molecule and predict the splitting that would be expected (two ortho couplings and a meta couplings for example) and describe that. You can then rationalise quite easily why a ddd looks like a triplet (here, the two ortho couplings are almost identical and fall within the natural linewidth, and so cannot be resolved and &lt;em&gt;appear&lt;/em&gt; as a triplet). So your original suggestions are perfectly acceptable. Without seeing the spectrum you have been given, it is hard to say, but a well shimmed sample of aspirin in methanol even at medium fields clearly shows the two ortho couplings aren't quite identical, and the peak for (what you label) spin E does not look like a true triplet, with the centre line not much taller than the outside lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendations for reporting splitting patterns and, especially, if you are reporting coupling constants, is to always report descriptions of splittings in accordance with the reported couplings. A doublet of triplets arises from an AMX&lt;sub&gt;2&lt;/sub&gt; system, and cannot have three coupling constants for spin A; only J&lt;sub&gt;AM&lt;/sub&gt; and J&lt;sub&gt;AX&lt;/sub&gt;. A doublet of doublets of doublets arises from an AHPX system, and will have three distinct couplings for spin A; J&lt;sub&gt;AH&lt;/sub&gt;, J&lt;sub&gt;AP&lt;/sub&gt;, J&lt;sub&gt;AX&lt;/sub&gt;. It is perfectly acceptable for any of these to be equal, such as J&lt;sub&gt;AP&lt;/sub&gt;=J&lt;sub&gt;AX&lt;/sub&gt;. So, go with what you have originally suggested, and the only matter for you to consider is what typical values are for ortho (~7-8Hz) and meta (2-3Hz) couplings (and possibly para couplings - which you don't observe in aspirin, but can be ~1Hz)&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-20T00:31:32.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18635" PostTypeId="1" AcceptedAnswerId="19149" CreationDate="2014-10-20T01:31:26.693" Score="6" ViewCount="8400" Body="&lt;p&gt;In my organic chemistry textbook, lithium is used to create a free radical of the alkynes, to then allow the anti addition of hydrogen to get an alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Dissolving Metal Reduction of Alkynes&quot; is a reaction that then uses $\ce{NH4Cl}$. Why put $\ce{NH4Cl}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://imgur.com/H87ovV7&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/H87ovV7.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://imgur.com/zdI3lxP&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/zdI3lxP.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="5026" LastEditDate="2018-04-12T03:48:02.517" LastActivityDate="2018-04-12T03:48:32.607" Title="What is the purpose of ammonium chloride in anti hydrogenation of alkynes?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrocarbons&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18637" PostTypeId="1" CreationDate="2014-10-20T02:13:54.323" Score="1" ViewCount="5685" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wu8Z9.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found This Online. Something to do with bipolar disorder?&lt;/p&gt;&#xA;" OwnerUserId="9396" LastEditorUserId="7920" LastEditDate="2014-10-20T05:14:50.227" LastActivityDate="2014-10-20T05:14:50.227" Title="Interpreting a Formula Tattoo (Bipolar?)" Tags="&lt;organic-chemistry&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18638" PostTypeId="1" CreationDate="2014-10-20T02:15:35.983" Score="5" ViewCount="2920" Body="&lt;p&gt;Experimentally it appears that increase in pressure, fuel concentration, and chain length causes AIT to decrease until a minimum where even further increase in these parameters increases the AIT. Additionally, it seems that AIT is lowered by increasing the vessel volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.yumpu.com/en/document/view/18157279/autoignition-temperatures-for-mixtures-of-flammable-liquids-with-/5&quot; rel=&quot;nofollow noreferrer&quot;&gt;From the last page of this paper:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Semenoff's equation seems to indicate that it is possible to lower the AIT to arbitrarily low values by increasing the pressure. But the higher the total pressure of the fuel/air mixture, the higher is also the partial pressure of the fuel required to reach the most critical fuel concentration. At some temperature, the required partial pressure will become equal to the fuel vapor pressure. At even higher pressures and lower temperatures, it is impossible to reach the critical fuel concentration, and the depression of the AIT is expected to stop. A similar mechanism has been employed to explain the reincrease of the standard AIT with chain length for very long chain molecules /5/.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit: Article title is Autoignition Temperature for Mixtures of Flammable Liquids at Elevated Pressures by Brandes, Hirsh, and Stolz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) From this, it seems that pressure and fuel concentration effects are related. Is the critical fuel concentration related to the equivalence ratio and the ability to reach this equivalence ratio based on the saturation pressure of the fuel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I do not get the relation between the above explanation and the last sentence regarding re-increase of AIT for longer chain molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) I'm not sure why larger volumes leads to lower AIT as stated in &lt;a href=&quot;https://www.astm.org/Standards/E659.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM E659&lt;/a&gt; Appendix X2. &lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="69830" LastEditDate="2020-05-25T02:59:40.860" LastActivityDate="2020-05-25T02:59:40.860" Title="Relationship between auto-ignition temperature with pressure and fuel" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18639" PostTypeId="2" ParentId="18637" CreationDate="2014-10-20T02:39:03.043" Score="7" Body="&lt;p&gt;This is &lt;a href=&quot;http://en.wikipedia.org/wiki/Lamotrigine&quot;&gt;lamotrigine&lt;/a&gt;, a sodium channel blocker prescribed against epilepsy.  Apparently it's also prescribed for bipolar disorder.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-20T02:39:03.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18641" PostTypeId="1" AcceptedAnswerId="18650" CreationDate="2014-10-20T03:26:02.883" Score="4" ViewCount="7064" Body="&lt;p&gt;The entropy of mixing of ideal gas is given by this equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S_{mix} =  -nR(x_1\ln x_1 + x_2\ln x_2)$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this equation works only when the initial conditions of both compartments &lt;strong&gt;are at the same pressure&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If ideal gas 1 at initial conditions $n_1,P_1,V_1,T$ and ideal gas 2 at $n_2,P_2,V_2,T$ are mixed, which equation should I use to obtain $\Delta S_{mix}$ given that $P_1 ≠P_2$? Assume $V_1 + V_2 = V_{tot}$, and $T_1 = T_2 = T$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQQvv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8140" LastEditorUserId="189" LastEditDate="2014-10-20T15:22:19.617" LastActivityDate="2021-02-25T02:39:45.107" Title="What is the entropy of mixing of two ideal gases starting out with different pressures?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;mixtures&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18644" PostTypeId="1" AcceptedAnswerId="18645" CreationDate="2014-10-20T08:42:19.510" Score="2" ViewCount="285" Body="&lt;p&gt;Is it possible by any means or by using some reagents to convert &#xA;hexane to 1,6-dibromohexane? The yield should also be good.&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="5026" LastEditDate="2018-04-12T03:53:04.323" LastActivityDate="2018-04-12T03:53:04.323" Title="Convert hexane to 1,6-dibromohexane" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18645" PostTypeId="2" ParentId="18644" CreationDate="2014-10-20T10:14:40.463" Score="7" Body="&lt;p&gt;In principle, alkanes may be converted to alkyl bromides by photoinitiated bromination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, taking statistics (2 $\ce{CH3}$ vs 4 $\ce{CH2}$ groups) and stabilities of the radicals resulting from H abstraction by a bromine radical from your starting material into account, this approach is pretty worthless and will most likely yield 1,6-dibromohexane as a minor product in a complex mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, your reasonable options are a) a different starting material (and a different synthesis) or b) simply buy 1,6-dibromohexane (around 20 Euro for 100 g). &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-10-20T10:14:40.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18650" PostTypeId="2" ParentId="18641" CreationDate="2014-10-20T15:19:55.833" Score="4" Body="&lt;p&gt;A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \big|_{p, \{ n_{i} \}}$ later): for isothermal and isobaric conditions it is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.&#xA;So, now we need an expression for the chemical potential of an ideal gas.&#xA;It can be shown that this is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{p_{i}}{P^{0}}\biggr) \hspace{18em} (2)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $P^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).&#xA;For ideal gases we can apply Raoult's law stating that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  p_{i} = x_{i} P_{\mathrm{tot}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P_{\mathrm{tot}}$ is the total pressure of the system.&#xA;Using this with equation (2) we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{x_{i} P_{\mathrm{tot}}}{P^{0}}\biggr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \hspace{6em} (3)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  P_{\mathrm{tot}} = \sum_{i} p_{i}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.&#xA;At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.&#xA;Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the pressure $P_{\mathrm{tot}} = P_{i}$ in the system (whereby $P_{i}$ is the ambient pressure in chamber $i$), since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).&#xA;According to equation (1) the Gibbs free energy of the $i$th chamber is then given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{i} &amp;amp;= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\&#xA;        &amp;amp;= n_{i} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Gibbs free energy of the total system at the initial state is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{\mathrm{ini}} &amp;amp;= \sum_{i} G_{i} \\&#xA;                 &amp;amp;= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \biggr) \\&#xA;                 &amp;amp;= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\&#xA;                 &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \ .  \hspace{7em} (4)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a multicomponent system with more than two components I would simply use this equation. But for a system with exactly two components we can introduce some substitutions that will simplify the final equations a little. For that purpose we want to establish a relationship between the initial pressures $P_{1}$ and $P_{2}$ and the final pressure $P_{\mathrm{fin}}$. In order to get there we use the ideal gas law and the additivity relations for the volumes and amounts: From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  n_{\mathrm{tot}} = n_{1} + n_{2} \ , \qquad V_{\mathrm{tot}} = V_{1} + V_{2} \\&#xA;  P_{1} V_{1} = n_{1} R T \ , \qquad P_{2} V_{2} = n_{2} R T \ , \qquad P_{\mathrm{fin}} V_{\mathrm{tot}} = n_{\mathrm{tot}} R T&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  P_{1} V_{1} + P_{2} V_{2} = P_{\mathrm{fin}} V_{\mathrm{tot}} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can express $P_{2}$ in terms of $P_{\mathrm{fin}}$ by introducing the deviation factor $\alpha$ which is defined by $P_{2} = (1 - \alpha) P_{\mathrm{fin}}$ and use this in the previous equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  P_{1} V_{1} + P_{2} V_{2} &amp;amp;= P_{\mathrm{fin}} V_{\mathrm{tot}} \\&#xA;  P_{1} V_{1} + (1 - \alpha) P_{\mathrm{fin}} V_{2} &amp;amp;= P_{\mathrm{fin}} V_{\mathrm{tot}} \\&#xA;  P_{1} V_{1} &amp;amp;= P_{\mathrm{fin}} \bigl( \underbrace{V_{\mathrm{tot}} - V_{2}}_{= \, V_{1}} + \alpha V_{2} \bigr) \\&#xA;  P_{1} &amp;amp;= \biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) P_{\mathrm{fin}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives us the relation between $P_{1}$ and $P_{\mathrm{fin}}$. Introducing these equations into equation (4) yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{\mathrm{ini}} &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl(\frac{P_{1}}{P^{0}}\biggr) + x_{2} R T \ln\biggl(\frac{P_{2}}{P^{0}}\biggr) \biggr) \\&#xA;  &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\frac{\bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \bigr) P_{\mathrm{fin}}}{P^{0}} + x_{2} R T \ln\frac{(1 - \alpha) P_{\mathrm{fin}}}{P^{0}} \biggr) \\&#xA;  &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + \underbrace{(x_{1} + x_{2})}_{= \, 1} R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\&#xA; &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.&#xA;Also, the ambient pressure in the system changes to $P_{\mathrm{tot}} = P_{\mathrm{fin}}$.&#xA;Now, the complete system of connected chambers must be treated via equation (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  G_{\mathrm{final}} &amp;amp;= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr)}_{= \, \mathrm{constant}} \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\&#xA;                   &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Delta_{\mathrm{mix}}G &amp;amp;= G_{\mathrm{final}} - G_{\mathrm{ini}} \\&#xA;                       &amp;amp;= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\&#xA;                       &amp;amp;\quad - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\&#xA;                       &amp;amp;= n_{\mathrm{tot}} R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) - n_{\mathrm{tot}}  R T \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr) \qquad (5)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you see why we went through those substitutions for $P_{1}$ and $P_{2}$. This way the $n_{\mathrm{tot}} R T \ln\Bigl(\frac{P_{\mathrm{fin}}}{P^{0}}\Bigr)$ terms cancelled in equation (5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating equation (5) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Delta_{\mathrm{mix}} S &amp;amp;= - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} \\&#xA;   &amp;amp;= - n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + n_{\mathrm{tot}}  R \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term $- n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right)$ is just the usual entropy of mixing. The second term $n_{\mathrm{tot}}  R \Bigl( x_{1} \ln\Bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \Bigr) + x_{2} \ln(1 - \alpha) \Bigr)$ describes the contribution of the pressure equilibration to the entropy.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-10-20T15:19:55.833" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18651" PostTypeId="1" AcceptedAnswerId="18655" CreationDate="2014-10-20T15:25:40.470" Score="2" ViewCount="12084" Body="&lt;p&gt;Apparently carboxylic acid derivatives such as cyanohydrins can be taken back to &quot;daddy&quot; - a carboxylic acid - under the right conditions. This involves acid, water, and &quot;high heat.&quot; What, however, is the mechanism of a cyanohydrin going back to daddy? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2016-07-09T21:17:21.823" Title="Acidic conditions, high heat, and return to carboxylic acids" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18652" PostTypeId="1" AcceptedAnswerId="18659" CreationDate="2014-10-20T15:40:52.740" Score="5" ViewCount="29223" Body="&lt;p&gt;I didn't understand the fact that chiral objects are non-superimposable on their mirror images. I mean, if you put your right hand (apparently an achiral object) in front of a mirror, the mirror image will be like your left hand, but, when you make your hand touch the mirror, isn't it superimposable on its mirror image (and hence achiral)? Doesn't that mean that everything is achiral?&lt;/p&gt;&#xA;" OwnerUserId="9407" LastEditorUserId="2325" LastEditDate="2014-10-22T09:43:41.957" LastActivityDate="2019-04-30T09:29:09.693" Title="Concept of &quot;non-superimposable mirror image&quot; in chirality" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18653" PostTypeId="2" ParentId="18652" CreationDate="2014-10-20T15:46:37.547" Score="9" Body="&lt;p&gt;Your hands are chiral, that is why you need two different leather gloves, one that only fits your right hand, and one that only fits your left hand.  If your hands were superimposable, then you would only need one kind of glove and it would fit both hands.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-20T15:46:37.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18654" PostTypeId="1" CreationDate="2014-10-20T15:50:52.270" Score="4" ViewCount="568" Body="&lt;p&gt;Is there a chemical(s) that provides good discrimination between absorbance curves for different pHs in the UV light spectrum (100-400nm?)&lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2016-04-19T15:37:41.530" Title="Ultraviolet acid base indicator" Tags="&lt;ph&gt;&lt;color&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18655" PostTypeId="2" ParentId="18651" CreationDate="2014-10-20T16:08:27.667" Score="6" Body="&lt;p&gt;In the cyanohydrin reaction aldehydes or ketones can react with HCN to form a cyanohydrin.  Inorganic cynanides and acid can be used in place of HCN, to generate the required cyanide anion.   The reaction proceeds by typical addition of the nucleophilic cyanide anion to the carbonyl carbon to form the tetrahedral intermediate, which can ultimately be protonated to yield the corresponding cyanohydrin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eacoG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once formed, the nitrile group of the cyanohydrin can undergo acidic (or basic) hydrolysis by the same mechanism as any other common nitrile.  The only difference in the case of a cyanohydrin is that the &quot;R&quot; group contains a hydroxyl substituent on its alpha carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XmeLC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful variant of cyanohydrin formation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Strecker_amino-acid_synthesis&quot; rel=&quot;noreferrer&quot;&gt;Strecker amino acid synthesis&lt;/a&gt;.  Instead of forming a cyanohydrin, an aminonitrile is formed and the nitrile is hydrolyzed to the corresponding alpha-amino acid.  A detailed mechanism can be found in the above link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KGggX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-20T16:08:27.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18656" PostTypeId="1" AcceptedAnswerId="18658" CreationDate="2014-10-20T16:50:45.260" Score="25" ViewCount="5336" Body="&lt;p&gt;I know the history of the discovery of the rare earth elements, that they were discovered near the village Ytterby in Sweden in the 19th century. Four elements are directly named after that village: erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are they named like this? It makes sense to name one element after the locality, but why name the rest as misspellings of that village? It's not like we have ermanium, germium and ermium in addition to germanium (&lt;em&gt;although we do have holmium&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.livescience.com/38389-erbium.html&quot;&gt;This site&lt;/a&gt; mentions this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1842, Swedish chemist Carl Gustaf Mosander separated the &quot;yttria&quot; component found in the mineral gadolinite into three segments that he called yttria, erbia and terbia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the any record as to why Mosander chose these names? Because...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As might be expected considering the similarities between their names and properties, scientists soon confused erbia and terbia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-31T23:30:47.730" Title="Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?" Tags="&lt;nomenclature&gt;&lt;history-of-chemistry&gt;&lt;rare-earth-elements&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18657" PostTypeId="2" ParentId="18656" CreationDate="2014-10-20T17:24:27.393" Score="10" Body="&lt;p&gt;Found a source &lt;a href=&quot;http://periodic.lanl.gov/39.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; with some of the history and explanation,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Named after Ytterby, a village in Sweden near Vauxholm. Yttria – earth containing yttrium – was discovered by Gadolin in 1794. Ytterby is the site of a quarry which yielded many unusual minerals containing rare earths and other elements. This small town, near Stockholm, bears the honor of giving names to erbium, terbium, and ytterbium as well as yttrium.&lt;/p&gt;&#xA;&lt;p&gt;In 1843 Mosander showed that yttria could be resolved into the oxides (or earths) of three elements. The name yttria was reserved for the most basic one; the others were named erbia and terbia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Another &lt;a href=&quot;http://www.periodni.com/history_of_rare_earth_elements.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; describes that the two names were derived from the village name, but i don't quite see the relation.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Given that he separated the two new earths from yttria, Mosander derived the names of terbia and erbia from the name of the village Ytterby&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-12-31T23:30:47.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18658" PostTypeId="2" ParentId="18656" CreationDate="2014-10-20T17:24:36.760" Score="20" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The links provided below are to the website &quot;&lt;a href=&quot;http://www.vanderkrogt.net/elements/index.php&quot;&gt;Elementymology &amp;amp; Elements&#xA;  Multidict&lt;/a&gt;&quot;.  It is a fascinating site with information on the&#xA;  discovery, naming and various other facts for each element. As the author (Peter van der Krogt) says, &quot;I am not a chemist, but a (map) historian much interested in the origin of names&quot;. Enjoy!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A nice piece on the complex etymology for Erbium, Terbium and Yttrium can be found &lt;a href=&quot;http://www.vanderkrogt.net/elements/element.php?sym=Tb&quot;&gt;here&lt;/a&gt;.  In brief the article says, &quot;All three names were derived from the Ytterby quarry where the gadolinite was originally found in 1787. It is also said that since the original earth was divided into three new earths, Mosander &lt;strong&gt;split the name of Ytterby in three parts: ytt, terb, and erb&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to ytterbium's etymology see &lt;a href=&quot;http://www.vanderkrogt.net/elements/element.php?sym=Yb&quot;&gt;here&lt;/a&gt;.   Again the history is complex with the name changing as recently as 1925. Another chemist isolated this element and again, wanted to pay homage to the geographic point of discovery, the Ytterby quarry.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-21T01:02:37.043" LastActivityDate="2014-10-21T01:02:37.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18659" PostTypeId="2" ParentId="18652" CreationDate="2014-10-20T17:36:05.180" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;if you put your right hand (apparently an achiral object) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, your right and left hand are both chiral objects, as we will see below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in front of&#xA;  a mirror, the mirror image will be like your left hand, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but, when you&#xA;  make your hand touch the mirror, isn't it superimposable on its mirror&#xA;  image&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, that mirror image is not superimposable.  Just because the bottoms of your right hand and left hand can overlay one another &lt;strong&gt;does not make them superimposable&lt;/strong&gt;. If you could take that 3-dimensional  mirror image out of the mirror and place it next to your hands and compare it to your right and left hand, it would look like your left hand.  And, just like your real left hand, you couldn't pick the mirror image up and place it next to your right hand, look at them both and say, &quot;they are the same&quot; because, for one thing, the thumbs are on opposite sides.  Therefore, they are non-superimposable mirror images.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-20T17:36:05.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18661" PostTypeId="5" CreationDate="2014-10-20T18:12:53.840" Score="0" Body="&lt;p&gt;According to the IUPAC, the rare earth elements are a group that comprises the fifteen lanthanoides (&lt;a href=&quot;http://www.webelements.com/lanthanum/index.html&quot; rel=&quot;nofollow&quot;&gt;La&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/cerium/index.html&quot; rel=&quot;nofollow&quot;&gt;Ce&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/praseodymium/index.html&quot; rel=&quot;nofollow&quot;&gt;Pr&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/neodymium/index.html&quot; rel=&quot;nofollow&quot;&gt;Nd&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/promethium/index.html&quot; rel=&quot;nofollow&quot;&gt;Pm&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/samarium/index.html&quot; rel=&quot;nofollow&quot;&gt;Sm&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/europium/index.html&quot; rel=&quot;nofollow&quot;&gt;Eu&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/gadolinium/index.html&quot; rel=&quot;nofollow&quot;&gt;Gd&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/terbium/index.html&quot; rel=&quot;nofollow&quot;&gt;Tb&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/dysprosium/index.html&quot; rel=&quot;nofollow&quot;&gt;Dy&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/holmium/index.html&quot; rel=&quot;nofollow&quot;&gt;Ho&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/erbium/index.html&quot; rel=&quot;nofollow&quot;&gt;Er&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/thulium/index.html&quot; rel=&quot;nofollow&quot;&gt;Tm&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/ytterbium/index.html&quot; rel=&quot;nofollow&quot;&gt;Yb&lt;/a&gt;, &lt;a href=&quot;http://www.webelements.com/lutetium/index.html&quot; rel=&quot;nofollow&quot;&gt;Lu&lt;/a&gt;), &lt;a href=&quot;http://www.webelements.com/yttrium/index.html&quot; rel=&quot;nofollow&quot;&gt;Y&lt;/a&gt; and  &lt;a href=&quot;http://www.webelements.com/scandium/index.html&quot; rel=&quot;nofollow&quot;&gt;Sc&lt;/a&gt;. &#xA;Compare also the &lt;a href=&quot;http://media.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow&quot;&gt;IUPAC Red Book (2005)&lt;/a&gt; Rule IR-3.5. It further states the following regarding the naming scheme:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although lanthanoid means ‘like lanthanum’ and so should not include lanthanum,&#xA;  lanthanum has become included by common usage. Similarly, actinoid. The ending ‘ide’&#xA;  normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to&#xA;  lanthanide and actinide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These elements are commonly grouped together due to the similarity in their chemical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\scriptstyle\Tiny&#xA;\begin{array}{l|cccccccccccccccccc}&#xA;\text{Gr.} &amp;amp;1 &amp;amp;2 &amp;amp;3 &amp;amp;4 &amp;amp;5 &amp;amp;6 &amp;amp;7 &amp;amp;8 &amp;amp;9 &amp;amp;10 &amp;amp;11 &amp;amp;12 &amp;amp;13 &amp;amp;14 &amp;amp;15 &amp;amp;16 &amp;amp;17 &amp;amp;18\\&#xA;\text{Per.}\\\hline&#xA;1&amp;amp; \ce{^1H}	&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^2He}\\&#xA;2&amp;amp; \ce{^3Li}&amp;amp; \ce{^4Be}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^5B}&amp;amp; \ce{^6C}&amp;amp; \ce{^7N}&amp;amp; \ce{^8O}&amp;amp; \ce{^9F}&amp;amp; \ce{^{10}Ne}\\&#xA;3&amp;amp; \ce{^{11}Na}&amp;amp; \ce{^{12}Mg}&amp;amp; &amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;&amp;amp;\ce{^{13}Al}&amp;amp; \ce{^{14}Si}&amp;amp; \ce{^{15}P}&amp;amp; \ce{^{16}S}&amp;amp; \ce{^{17}Cl}&amp;amp; \ce{^{18}Ar}\\&#xA;4&amp;amp; \ce{^{19}K}&amp;amp; \ce{^{20}Ca}&amp;amp; \color{red}{\ce{^{21}Sc}}&amp;amp; \ce{^{22}Ti}&amp;amp; \ce{^{23}V}&amp;amp; \ce{^{24}Cr}&amp;amp; \ce{^{25}Mn}&amp;amp; \ce{^{26}Fe}&amp;amp; \ce{^{27}Co}&amp;amp; \ce{^{28}Ni}&amp;amp; \ce{^{29}Cu}&amp;amp; \ce{^{30}Zn}&amp;amp; \ce{^{31}Ga}&amp;amp; \ce{^{32}Ge}&amp;amp; \ce{^{33}As}&amp;amp; \ce{^{34}Se}&amp;amp; \ce{^{35}Br}&amp;amp; \ce{^{36}Kr}\\&#xA;5&amp;amp; \ce{^{37}Rb}&amp;amp; \ce{^{38}Sr}&amp;amp; \color{red}{\ce{^{39}Y}}&amp;amp; \ce{^{40}Zr}&amp;amp; \ce{^{41}Nb}&amp;amp; \ce{^{42}Mo}&amp;amp; \ce{^{43}Tc}&amp;amp; \ce{^{44}Ru}&amp;amp; \ce{^{45}Rh}&amp;amp; \ce{^{46}Pd}&amp;amp; \ce{^{47}Ag}&amp;amp; \ce{^{48}Cd}&amp;amp; \ce{^{49}In}&amp;amp; \ce{^{50}Sn}&amp;amp; \ce{^{51}Sb}&amp;amp; \ce{^{52}Te}&amp;amp; \ce{^{53}I}&amp;amp; \ce{^{54}Xe}\\&#xA;6&amp;amp;\ce{^{55}Cs}&amp;amp; \ce{^{56}Ba}&amp;amp; \color{red}{\text{*}}&amp;amp; \ce{^{72}Hf}&amp;amp; \ce{^{73}Ta}&amp;amp; \ce{^{74}W}&amp;amp; \ce{^{75}Re}&amp;amp; \ce{^{76}Os}&amp;amp; \ce{^{77}Ir}&amp;amp; \ce{^{78}Pt}&amp;amp; \ce{^{79}Au}&amp;amp; \ce{^{80}Hg}&amp;amp; \ce{^{81}Tl}&amp;amp; \ce{^{82}Pb}&amp;amp; \ce{^{83}Bi}&amp;amp; \ce{^{84}Po}&amp;amp; \ce{^{85}At}&amp;amp; \ce{^{86}Rn}\\&#xA;7&amp;amp; \ce{^{87}Fr}&amp;amp; \ce{^{88}Ra}&amp;amp; \text{**}&amp;amp; \ce{^{104}Rf}&amp;amp; \ce{^{105}Db}&amp;amp; \ce{^{106}Sg}&amp;amp; \ce{^{107}Bh}&amp;amp; \ce{^{108}Hs}&amp;amp; \ce{^{109}Mt}&amp;amp; \ce{^{110}Ds}&amp;amp; \ce{^{111}Rg}&amp;amp; \ce{^{112}Cn}&amp;amp; \ce{^{113}Uut}&amp;amp; \ce{^{114}Fl}&amp;amp; \ce{^{115}Uup}&amp;amp; \ce{^{116}Lv}&amp;amp; \ce{^{117}Uus}&amp;amp; \ce{^{118}Uuo}\\\hline&#xA;\end{array}\\&#xA;\scriptstyle\Tiny&#xA;\begin{array}{lc} &#xA;\color{red}{\text{* Lanthanoide}}&amp;amp; \ce{^{57}La}&amp;amp; \ce{^{58}Ce}&amp;amp; \ce{^{59}Pr}&amp;amp; \ce{^{60}Nd}&amp;amp; \ce{^{61}Pm}&amp;amp; \ce{^{62}Sm}&amp;amp; \ce{^{63}Eu}&amp;amp; \ce{^{64}Gd}&amp;amp; \ce{^{65}Tb}&amp;amp; \ce{^{66}Dy}&amp;amp; \ce{^{67}Ho}&amp;amp; \ce{^{68}Er}&amp;amp; \ce{^{69}Tm}&amp;amp; \ce{^{70}Yb}&amp;amp; \ce{^{71}Lu}\\&#xA;\text{** Actinoide}&amp;amp; \ce{^{89}Ac}&amp;amp; \ce{^{90}Th}&amp;amp; \ce{^{91}Pa}&amp;amp; \ce{^{92}U}&amp;amp; \ce{^{93}Np}&amp;amp; \ce{^{94}Pu}&amp;amp; \ce{^{95}Am}&amp;amp; \ce{^{96}Cm}&amp;amp; \ce{^{97}Bk}&amp;amp; \ce{^{98}Cf}&amp;amp; \ce{^{99}Es}&amp;amp; \ce{^{100}Fm}&amp;amp; \ce{^{101}Md}&amp;amp; \ce{^{102}No}&amp;amp; \ce{^{103}Lr}	\\&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rare_earth_element&quot; rel=&quot;nofollow&quot;&gt;Rare earth element (wikipedia)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="4945" LastEditDate="2014-11-18T12:17:14.157" LastActivityDate="2014-11-18T12:17:14.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18662" PostTypeId="4" CreationDate="2014-10-20T18:12:53.840" Score="0" Body="According to the IUPAC, the rare earth elements are a group that comprises the fifteen lanthanoides (La to Lu), Y and  Sc. These elements are commonly grouped together due to the similarity in their chemical properties." OwnerUserId="8083" LastEditorUserId="4945" LastEditDate="2014-11-18T10:44:38.977" LastActivityDate="2014-11-18T10:44:38.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18663" PostTypeId="1" AcceptedAnswerId="18680" CreationDate="2014-10-20T18:43:38.873" Score="0" ViewCount="1294" Body="&lt;p&gt;I asked how to draw glucose as a Fischer projection from its given chair conformation but all I got was an unsatisfactory ad hoc explanation involving rotation of my neck muscles to somehow see where the hydroxyl groups were. How does one actually draw a Fischer projection of a sugar molecule from its chair conformation? &lt;img src=&quot;https://i.stack.imgur.com/Z27l9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="19616" LastEditDate="2015-10-16T17:58:18.390" LastActivityDate="2015-10-16T17:58:18.390" Title="How to draw glucose using a Fischer projection?" Tags="&lt;organic-chemistry&gt;&lt;molecular-structure&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18664" PostTypeId="2" ParentId="18602" CreationDate="2014-10-20T19:09:07.660" Score="4" Body="&lt;p&gt;As you have pointed out, the peak at M = 73 is M+H, protonation is however more likely to occur at the oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QA2vO.png&quot; alt=&quot;MS fragmentation of butenol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, peaks at M = 66 and 67, i.e. with a rather small mass difference to $\mathrm{M^+}$, are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common sense and the link in permeakra's comment suggest that the major fragmentation of but-3-en-1-ol occurs along the red dotted line, yielding an allylic and a hydroxymethyl fragment.  &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-10-20T19:09:07.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18665" PostTypeId="2" ParentId="18652" CreationDate="2014-10-20T19:32:01.203" Score="5" Body="&lt;p&gt;I have to say that when they taught Chiral molecules to me in Chemistry, they always used confusing molecules to demonstrate. The easiest to visualise the mirror symmetry is with spirenes (spiral molecules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a spiral staircase, it can spiral clockwise, or it can spiral anticlockwise, but if you tried to put one of each in the same space (superimpose them), nobody could climb the stairs ! &lt;/p&gt;&#xA;" OwnerUserId="9410" LastActivityDate="2014-10-20T19:32:01.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18666" PostTypeId="1" AcceptedAnswerId="18690" CreationDate="2014-10-20T19:48:24.443" Score="8" ViewCount="423" Body="&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; Chapter 4 of Brunlow's 1979 Geochemistry text.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the solubility of fluorite in river water with the average composition given in table 4-1.  Assume that no fluoride complexes form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Table 4-1 lists ppm, mEq/L and moles/L values for constituents, anions and cations found in the water sample.  There is no value for $\ce{F}^-$ but there is for $\ce{Ca}^{2+}$. The molarity of $\ce{Ca^{2+}}$ is given to be $\pu{0.375E-3M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaF2 -&amp;gt; Ca^{2+} + 2Fl^{-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\mathrm {sp}$ of $\ce{CaF2}$= $10^{-10.5}$ (from a table in the book)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{F_{\text{added}}}]^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{F_{\text{added}}}] = [2\ce{Ca_{\text{added}}}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][2\ce{Ca_{\text{added}}}]^2 = 4[\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{Ca_{\text{added}}}]^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track?  I end up with a polynomial to solve for.  The answer in the back of the book is $10^{-4.64}$ which I can't arrive at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other method I've tried is to calculate the ppm of fluorite dissolved in pure water ($\pu{15.6ppm}$) and then subtract the ppm value for the river water $\ce{Ca^{2+}}$ given in the table ($\pu{15.0ppm}$) and then convert the result ($\pu{0.6ppm}$) to $\pu{7.69e-6 moles/L}$ which doesn't equal the answer in the back of the book ($10^{-4.64}$), so I believe that I'm missing something conceptually.&lt;/p&gt;&#xA;" OwnerUserId="1803" LastEditorUserId="5026" LastEditDate="2018-02-26T05:09:56.717" LastActivityDate="2018-02-26T05:10:38.160" Title="Caclulating the Solubility of Fluorite in River Water" Tags="&lt;solubility&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18667" PostTypeId="1" CreationDate="2014-10-20T20:31:17.457" Score="1" ViewCount="175" Body="&lt;p&gt;I've read that bromothymol blue can be mixed in with the electrolyte in the demonstration of electrolysis using a Hoffman apparatus. The indicator clearly shows the local differences in pH for the anode and cathode. And the blue color provides good visual contrast between the fluid and the gases that evolve. But I've found this indicator tends to break down quickly into a colorless solution while operating the apparatus, and after awhile you lose the indicating effect as well. Does anyone know of a better choice of indicator to use that will last longer?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2014-10-20T21:26:03.157" Title="pH Indicators for Hoffman apparatus" Tags="&lt;electrochemistry&gt;&lt;ph&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18668" PostTypeId="2" ParentId="18667" CreationDate="2014-10-20T21:26:03.157" Score="2" Body="&lt;p&gt;I've never used it for that particular experiment, but bromocresol purple has a similar transition range and a very obvious colour change. The trouble with using indicators here is that many are organic compounds and are subject to redox reactions near the potentials at which electrolysis of water is occurring, so they break down or polymerize if they reach the electrodes. Bromothymol blue is &lt;a href=&quot;http://www.electrochemsci.org/papers/vol4/4030471.pdf&quot; rel=&quot;nofollow&quot;&gt;known&lt;/a&gt; to oxidize at below 1 V and deposit on the electrode. Bromocresol purple tends to polymerize, but it &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/elan.201000260/full&quot; rel=&quot;nofollow&quot;&gt;seems&lt;/a&gt; to be at a relatively high potential (&gt;1.5 V), so it may fare better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be possible to limit the diffusion of the indicator to the electrode surfaces. Perhaps using an indicator trapped in polymer beads or a block of agarose gel.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-20T21:26:03.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18669" PostTypeId="1" CreationDate="2014-10-20T21:32:06.110" Score="3" ViewCount="309" Body="&lt;p&gt;I have a question concerning metal ligand complexes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every chemistry text book I have read which treats metal ligand complexes, following situation is discussed:&#xA;When 6 ligands (in octahedral coordination) approaches a metal ion with d-orbitals, 2 of the three orbitals are energetically raised because these orbitals have their lobs in direction of the ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is this always true? &#xA;Why does always point the lobes of the z² and x²-y² orbital towards the ligands coming from &quot;above&quot; and &quot;below&quot;? Isn't it possible, that the z-axis of orbital system of the metal is a little bit inclined with respect to the octahedral ligand system surrounding the middle metal ion?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of, is that this is because of energetic reasons. That in total, it's energetically more favourable when 3 of the 5 d-orbitals are arranged, so that electron occupying these orbitals have the lowest possible energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But up to now, I haven't found anything discussing this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance,&#xA;Jürgen &lt;/p&gt;&#xA;" OwnerUserId="9413" LastActivityDate="2014-10-20T23:34:03.340" Title="Metal-Ligand-Complex and bond directions: What defines coordinate system of the complex ion?" Tags="&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2014-10-21T07:34:15.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="18670" PostTypeId="2" ParentId="18669" CreationDate="2014-10-20T23:34:03.340" Score="1" Body="&lt;p&gt;The answer lies in symmetry.  When placed in an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are directed at the surrounding ligands because they have $e_g$ symmetry.  The remaining three orbitals have $t_{2g}$ symmetry and are directed between the ligands.  This explanation can be found in an undergraduate inorganic chemistry textbook such as that by Miessler, Fischer and Tarr.  (In the 5th edition, that's section 10.2.1 - Crystal Field Theory).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-10-20T23:34:03.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18672" PostTypeId="1" AcceptedAnswerId="18674" CreationDate="2014-10-21T00:42:19.300" Score="3" ViewCount="186" Body="&lt;p&gt;Does it change the compound if the element letters are switched? ex. $\ce{HO}$ or $\ce{OH}$. Are those two different compounds or are they one in the same?&lt;/p&gt;&#xA;" OwnerUserId="9416" LastEditorUserId="5026" LastEditDate="2018-07-25T14:21:14.450" LastActivityDate="2018-07-26T10:38:00.897" Title="Does it change the compound if the element letters in its molecular formula are switched?" Tags="&lt;physical-chemistry&gt;&lt;elements&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18673" PostTypeId="1" AcceptedAnswerId="18676" CreationDate="2014-10-21T01:03:45.980" Score="1" ViewCount="304" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An experiment is done to determine the rate of the following reaction:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{2 Al (s) + 6 HCl (aq) -&amp;gt; 3 H2 (g) + 2 AlCl3 (aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The following data is collected: &#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{array}{rr} &#xA;\text{time}/\pu{s} &amp;amp; (\text{Mass of beaker + contents})/\pu{g}\\\hline&#xA;   0.0 &amp;amp; 270.230\\&#xA;  30.0 &amp;amp; 270.200\\&#xA;  60.0 &amp;amp; 270.170 \end{array}&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Calculate the rate of consumption of &lt;span class=&quot;math-container&quot;&gt;$\ce{Al}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\pu{g/min}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.534 g/min}$&lt;/span&gt;, but I don't know how to get there.&lt;/p&gt;&#xA;" OwnerUserId="9417" LastEditorUserId="4945" LastEditDate="2018-10-12T14:25:26.843" LastActivityDate="2018-10-12T14:36:45.783" Title="How to calculate the rate of consumption of aluminium with hydrochloric acid?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-10-21T04:00:11.137" ContentLicense="CC BY-SA 4.0" />
  <row Id="18674" PostTypeId="2" ParentId="18672" CreationDate="2014-10-21T01:04:30.440" Score="2" Body="&lt;p&gt;In some cases it can matter. $\ce{COH}$ usually refers to an alcohol. $\ce{CHO}$ refers to an aldehyde. But since acetic acid is a carboxylic acid, it is written as $\ce{CH3COOH}$&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="5026" LastEditDate="2018-07-25T14:29:12.223" LastActivityDate="2018-07-25T14:29:12.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18675" PostTypeId="1" CreationDate="2014-10-21T01:57:47.150" Score="3" ViewCount="2093" Body="&lt;p&gt;I'd like to know the temperatures that various hydrocarbons such as methane, butane, and propane, etc. will dissociate into hydrogen and carbon.  Is there a reference where I can look up these temperatures?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does pressure or partial vacuum effect these temperatures?&lt;/p&gt;&#xA;" OwnerUserId="9418" LastActivityDate="2021-05-18T09:45:43.417" Title="Bond Dissociation Temperatures for Hydrocarbons" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18676" PostTypeId="2" ParentId="18673" CreationDate="2014-10-21T02:24:58.307" Score="3" Body="&lt;p&gt;Since one of the products is a gas &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;, the decrease in system mass must reflect the escape of hydrogen. Thus, after thirty seconds, the system has lost &lt;span class=&quot;math-container&quot;&gt;$\pu{0.030 g}$&lt;/span&gt; of hydrogen, which is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.01488 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the rate &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; of hydrogen production is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;r (\ce{H2})&#xA;= \frac{\Delta n(\ce{H2})}{\Delta t}&#xA;= \frac{\pu{0.01488 mol}}{\pu{30 s}}&#xA;= \pu{4.96E-4 mol//s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how can you use this rate and the stoichiometry of the reaction to determine the rate of aluminum consumption?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{2 Al (s) + 6 HCl (aq) -&amp;gt; 3 H2 (g) + 2 AlCl3 (aq)}$&lt;/span&gt;&lt;br&gt;&#xA; For every three moles of hydrogen produced two moles of aluminium are consumed, therefore &lt;span class=&quot;math-container&quot;&gt;$r(\ce{Al}) = -\frac{2}{3}r(\ce{H2}) = \pu{3.31E-4 mol//s}$&lt;/span&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditDate="2018-10-12T14:36:45.783" LastActivityDate="2018-10-12T14:36:45.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18677" PostTypeId="1" AcceptedAnswerId="18687" CreationDate="2014-10-21T02:32:39.207" Score="1" ViewCount="261" Body="&lt;p&gt;I am asked to consider the reduction of acetaldehyde from the re and si faces. I am also asked whether the added hydrogen (labeled H* in my picture) is pro-R or pro-S in both cases. In my attached picture I replaced the added hydrogen (H*) with D to determine the absolute configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did this correctly, addition of hydrogen from the si face results in a pro-S product with respect to the added hydrogen (H*). And addition of hydrogen from the re face results in a pro-R product with respect to the added hydrogen. Am I correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oe04i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="2462" LastEditDate="2014-10-21T05:53:08.403" LastActivityDate="2014-10-21T12:57:45.790" Title="Reduction of acetaldehyde from rectus and sinister faces" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18679" PostTypeId="2" ParentId="589" CreationDate="2014-10-21T04:51:03.993" Score="3" Body="&lt;p&gt;Oxygen solubility is 10 mg/L for water, and 14.5 mg/L for ice. Even after the ice melts to room temperature, the oxygen level in the melted ice water remains higher than the water that kept a constant room temperature(~12 mg/L vs 10 mg/L). The extra oxygen present is most likely the cause for the difference in taste.&lt;/p&gt;&#xA;" OwnerUserId="9420" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T15:37:05.520" LastActivityDate="2015-04-24T15:37:05.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18680" PostTypeId="2" ParentId="18663" CreationDate="2014-10-21T06:12:16.893" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://biochemjm.wordpress.com/2013/02/13/fischer-and-haworth-projections-anomers/&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a web page that shows how to interpret the position from the structure.&lt;br&gt;&#xA;The UP groups go to the left side of the backbone and the DOWN groups go to the right side.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-21T06:12:16.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18681" PostTypeId="1" AcceptedAnswerId="18685" CreationDate="2014-10-21T08:41:43.453" Score="2" ViewCount="159" Body="&lt;p&gt;I'm a bit confused having heard two different terms in different lecture parts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;J-coupling&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is the J-coupling (= indirect dipole dipole coupling = scalar coupling) which arises from a hyperfine interaction between the nuclei and the interaction. The J-coupling is a coupling between the two nuclear spins of two nuclei (e.g. a C-H bond). Because of the Pauli principle the two binding electrons are anti-parallel located so the nuclear spins $I_1$ and $I_2$ also have an anti-parallel nuclear spin.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;j,j-coupling&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is as well the jj-coupling (= spin orbit coupling) which describes the coupling of the different $j_i$. This coupling scheme is valid for heavier atoms. (For leight atoms the ls-coupling is valid). In atoms with bigger nuclear charges, spin-orbit interactions are frequently as large as or larger than spin-spin interactions or orbit-orbit interactions. &#xA;$\mathbf J = \sum_i \mathbf j_i = \sum_i (\boldsymbol{\ell}_i + \mathbf{s}_i).$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum_coupling&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any commonalities. Why is the term &quot;J-coupling&quot; similar to the term &quot;j,j-coupling&quot;?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="4945" LastEditDate="2014-10-23T05:13:00.913" LastActivityDate="2014-10-23T05:13:00.913" Title="Are there some similarities between J-coupling and j,j-coupling?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18682" PostTypeId="1" AcceptedAnswerId="18696" CreationDate="2014-10-21T09:27:02.630" Score="0" ViewCount="881" Body="&lt;p&gt;I know that the salt bridge produces ions to balance out the positive and negative ions in each solution, but why won't it work without a salt bride? Would it even work at the start since there's few electrons gained and lost and few ions produced?&lt;/p&gt;&#xA;" OwnerUserId="9421" LastEditorUserId="4945" LastEditDate="2014-10-21T10:21:53.703" LastActivityDate="2014-10-21T17:33:02.237" Title="Why would electrode reaction in galvanic cell stop when there's no neutrality in each cell?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18683" PostTypeId="2" ParentId="17271" CreationDate="2014-10-21T09:31:59.410" Score="1" Body="&lt;p&gt;&lt;strong&gt;Sulphite&lt;span class=&quot;math-container&quot;&gt;$(\ce{SO3^2-})$&lt;/span&gt; and Barium Choride&lt;span class=&quot;math-container&quot;&gt;$(\ce{ BaCl2})$&lt;/span&gt; or Strontium Chloride&lt;span class=&quot;math-container&quot;&gt;$(\ce{ SrCl2})$&lt;/span&gt;:&lt;/strong&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SO3^2- +Ba^2+ -&amp;gt;BaSO3 v}$$&lt;/span&gt;&#xA;the precipate disoolves in dilute hydrochloric acid, when sulphur dioxide evolves:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{BaSO3 v +2H+ -&amp;gt;Ba^2+ +SO2 ^ +H2O}$$&lt;/span&gt;&#xA;On standing, the precipitate is slowly oxidises to the sulphate and is then insoluble in dilute mineral acids; this change is rapidly effected by warming with bromine water or a little concentrated nitric acid or with hydrogen peroxide:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2BaSO3 v +O2 -&amp;gt;2BaSO4 v }\\&#xA;\ce{2BaSO3 v +Br2 + H2O-&amp;gt;BaSO4 v +2Br- +2H+}\\&#xA;\ce{2BaSO3 v +2HNO3 -&amp;gt;3BaSO4 v +2NO ^ +H2O}\\&#xA;\ce{2BaSO3 v +H2O2 -&amp;gt;BaSO4 v +H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solubilites at &lt;span class=&quot;math-container&quot;&gt;$\pu{18 ^\circ C}$&lt;/span&gt; of the sulphites of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca,Sr,Ba}$&lt;/span&gt; are respectively &lt;span class=&quot;math-container&quot;&gt;$1.25, 0.033, 0.022$&lt;/span&gt; all in &lt;span class=&quot;math-container&quot;&gt;$\pu{g/l}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Vogel’s Qualitative Inorganic Analysis for the JEE&lt;/em&gt;, by Svehla/Sivasankar,  7/e Paperback, &lt;strong&gt;1 January 2013&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="17368" LastEditDate="2020-06-21T14:40:05.167" LastActivityDate="2020-06-21T14:40:05.167" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18685" PostTypeId="2" ParentId="18681" CreationDate="2014-10-21T12:07:18.547" Score="3" Body="&lt;p&gt;I don't believe there are any commonalities, and the terms have been derived independently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &lt;em&gt;J&lt;/em&gt; coupling in NMR was first proposed in 1951 by Erwin Hahn in his &lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.84.1246&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;  describing the spin echo experiment, in which two frequencies are observed; one at the chemical shift (which they called delta, as was convention, even back then) as expected, and a second frequency which they labelled as frequency &lt;em&gt;J&lt;/em&gt;. There is no explanation for why the choice of J is given. &lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.84.589.2&quot; rel=&quot;nofollow&quot;&gt;Gutowsky and McCall&lt;/a&gt;  who observed the same phenomenon independently did not give it a label, and the term &lt;em&gt;J&lt;/em&gt; splitting stuck. &lt;em&gt;J&lt;/em&gt; coupling describes the relationship between nuclear spin, the intrinsic angular momentum of atomic nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term jj coupling relates orbital angular momentum of electrons. I'm not going to pretend to know anything about the origins of the term, nor anything about its application in physics and astronomy. However, the term J is sometimes used to represent total angular momentum, which is made up of individual components j&lt;sub&gt;i&lt;/sub&gt;. It is the interaction between these individual j terms that give rise to jj coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than being types of interactions between entities of angular momentum, the two couplings are quite unrelated, as are the terms used to describe them. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-21T12:07:18.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18686" PostTypeId="1" AcceptedAnswerId="18694" CreationDate="2014-10-21T12:41:00.453" Score="0" ViewCount="451" Body="&lt;p&gt;Is there a specification for MDL Connection Table format somewhere. At some places it is mentioned that the format writes the line length as a character then the actual line characters. If I save a structure using ChemDraw, I see no such thing and the lines are separated using CarriageReturn and LineFeed characters. Can someone please point me to the correct format? Is it that the format is different while SAVING TO FILE and while COPYING TO CLIPBOARD?&lt;/p&gt;&#xA;" OwnerUserId="6834" LastActivityDate="2016-11-08T17:25:51.833" Title="MDLCT (Connection Table) Format specification" Tags="&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18687" PostTypeId="2" ParentId="18677" CreationDate="2014-10-21T12:57:45.790" Score="2" Body="&lt;p&gt;Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vs71f.jpg&quot; alt=&quot;Reduction of deuterated acetaldehyde from both the re and si face.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we add from the &lt;em&gt;si&lt;/em&gt; face, we get an (&lt;em&gt;R&lt;/em&gt;) center, and if we add from the &lt;em&gt;re&lt;/em&gt; face, it's the other way around (as it should be!) and we have (&lt;em&gt;S&lt;/em&gt;) configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we reduce &quot;normal&quot; acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-&lt;em&gt;R&lt;/em&gt; or pro-&lt;em&gt;S&lt;/em&gt;. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-21T12:57:45.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18688" PostTypeId="1" CreationDate="2014-10-21T13:17:48.097" Score="3" ViewCount="411" Body="&lt;p&gt;I'm trying to understand the following data being presented in &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/01904169709365252?journalCode=lpla20#preview&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper measuring the mineral content of the &lt;em&gt;Cannabis sativa&lt;/em&gt; plant (leaves and inflorescences):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2GAO7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to interpret these data readings in terms of simple percentages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the first two examples - sodium and potassium - translate to in simple percentage / ppm etc.?&lt;/p&gt;&#xA;" OwnerDisplayName="user1469" LastEditorDisplayName="user1469" LastEditDate="2014-10-21T13:24:13.177" LastActivityDate="2017-04-07T23:44:17.543" Title="What do chemical constituent measurement units 'g%g' and 'µg g⁻¹' mean?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18689" PostTypeId="2" ParentId="18675" CreationDate="2014-10-21T13:35:08.520" Score="5" Body="&lt;p&gt;When hydrocarbons undergo thermal decomposition carbon-hydrogen and carbon-carbon bonds are broken.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R-CH3 -&amp;gt; R-CH2. + ~H.}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R-R' -&amp;gt; R. + ~R'.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bond-dissociation_energy#Tabulated_data&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link to a table&lt;/a&gt; that shows the energies required to break these bonds (the bond dissociation energy). These reactions are unimolecular decompositions and are governed by first order kinetics.  The rate of the reaction is given as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{reaction rate} = k[\ce{A}]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where k is the rate constant and &lt;span class=&quot;math-container&quot;&gt;$[\ce{A}]$&lt;/span&gt; represents the concentration of the hydrocarbon.  The rate constant can be described using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arrhenius equation&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$k= A e^{-E_a/(R T)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the pre-exponential factor, T the absolute temperature in kelvin, &lt;span class=&quot;math-container&quot;&gt;${E_a}$&lt;/span&gt; is the activation energy, and R is the universal gas constant.  In the thermal unimolecular decomposition of a hydrocarbon, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; will have a value around &lt;span class=&quot;math-container&quot;&gt;$\pu{10^{12}sec^-1}$&lt;/span&gt; and the bond dissociation energy can be used for the activation energy.&lt;/p&gt;&#xA;&lt;p&gt;The Arrhenius equation tells us that &lt;strong&gt;this decomposition will be occurring at all temperatures above absolute zero&lt;/strong&gt;.  As we increase or decrease the reaction temperature the rate constant will increase or decrease respectively, but the reaction is always occurring as long as we are above absolute zero.&lt;/p&gt;&#xA;&lt;p&gt;As to pressure, yes it will affect the reaction rate.  Pressure is directly related to concentration, the &lt;span class=&quot;math-container&quot;&gt;$\ce{[A]}$&lt;/span&gt; term in our rate equation.  As we increase or decrease the pressure, we are increasing or decreasing the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;, and thereby increasing or decreasing the rate of the reaction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="69830" LastEditDate="2021-05-18T09:45:43.417" LastActivityDate="2021-05-18T09:45:43.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18690" PostTypeId="2" ParentId="18666" CreationDate="2014-10-21T14:15:30.840" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;the ppm of fluorite in pure water (15.6ppm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the ppm of fluorite in pure water is zero!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and  $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;cubic equation&lt;/a&gt;, which can be solved exactly or by approximate methods.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-02-26T05:10:38.160" LastActivityDate="2018-02-26T05:10:38.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18692" PostTypeId="2" ParentId="374" CreationDate="2014-10-21T15:22:21.933" Score="2" Body="&lt;p&gt;In the brown ring complex, $\ce{[Fe(H2O)5(NO)]^{2+}}$ five water molecules are present, they are harder to stabilise the $+3$ oxidation state of $\ce{Fe}$ and hence here $\ce{NO}$ present as $-1$.&lt;br&gt;&#xA;If we consider that $\ce{NO}$ is $+1$, then $\ce{Fe}$ is $+1$, which is very unstable in the co-ligand region like $\ce{H2O}$.&lt;br&gt;&#xA;i.e, in brown ring complex oxidation state of Iron is $+3$.&lt;/p&gt;&#xA;" OwnerUserId="9429" LastEditorUserId="4945" LastEditDate="2015-01-05T03:05:19.293" LastActivityDate="2015-01-05T03:05:19.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18693" PostTypeId="1" CreationDate="2014-10-21T16:41:01.703" Score="1" ViewCount="458" Body="&lt;p&gt;Law of equivalence states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One equivalent of an element combines with one equivalent of the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Equivalent of $\ce{Na}$ is $22.99$ and that of $\ce{O}$ is $8$. By the law of equivalence, $1$ equivalent of $\ce{Na}$ combines with $1$ equivalent of $\ce{O}$. But in $\ce{Na2O2}$, $$ 23 \times 2 ~\text{parts by weight of sodium combines with } 16 \times 2~\text{parts by weight of oxygen} \\\implies 23~\text{parts by weight of sodium combines with } 16~\text{parts by weight of oxygen}$$ which means &lt;strong&gt;one equivalent of sodium combines with two equivalents of oxygen&lt;/strong&gt; which is contradiction to law of equivalence. So why does this compound not follow the law of equivalence?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:05:26.953" LastActivityDate="2015-10-23T17:05:26.953" Title="Confusion with Law of Equivalence" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18694" PostTypeId="2" ParentId="18686" CreationDate="2014-10-21T16:47:09.290" Score="1" Body="&lt;p&gt;MDL &quot;CT&quot; is a synonym for part of a MDL Molfile or SD file. ChemDraw has a &quot;Connection Table (.ct)&quot; format, which is different and deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specification is online:&#xA;&lt;a href=&quot;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&quot; rel=&quot;nofollow&quot;&gt;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in any format, something saved to the clipboard should be the same as saved to a file.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-21T19:30:19.140" LastActivityDate="2014-10-21T19:30:19.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18695" PostTypeId="2" ParentId="18693" CreationDate="2014-10-21T17:15:27.947" Score="1" Body="&lt;p&gt;The idea of a chemical equivalent relies on a compound's properties with respect to the usual reactions that occur with a few select species that tend to behave in very predictable ways.  In this case, the number of grams of sodium that would react with 8 grams of elemental oxygen, 23g Na, are one equivalent of sodium for oxidation-reduction purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na_2O_2}$ is sodium peroxide.  It contains oxygen (normally a very well behaved element, chemically speaking) behaving in an unusual way, bonding not only with sodium but also with another oxygen atom.  The oxygen can be thought of as part of a peroxide ion, $\ce{O2^2-}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know about oxidation and reduction, then you'll remember that oxygen usually has an oxidation number of -2, but when it is in a peroxide it has an oxidation number of -1.  Because this particular compound involves the -1 state, it takes twice as much oxygen to react with a given amount of sodium (which in both compounds has an oxidation state of +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, and perhaps more important in answering the root source of your trouble here, a scientific law is a thing that describes the behavior of the universe in specific conditions.  Outside of those conditions the law does not apply.  Even inside those conditions laws are often incorrect to some degree (though still useful approximations.)  It's not unexpected that we would find some exceptions to this law.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-21T17:15:27.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18696" PostTypeId="2" ParentId="18682" CreationDate="2014-10-21T17:33:02.237" Score="1" Body="&lt;p&gt;It would work vanishingly briefly. The separation of charge that occurs when you connect two separated half cells without a salt bridge produces an electrostatic potential difference that opposes the electrochemical potential difference. The cell reaction would proceed until the two voltages balance, and it doesn't take much charge separation to hit the 1–2 volt electrochemical potential of typical cells, so this process is very fast.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-21T17:33:02.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18697" PostTypeId="1" CreationDate="2014-10-21T18:33:11.743" Score="13" ViewCount="11321" Body="&lt;p&gt;How do bond angles vary in molecules with a lone pair and central atom of different electronegativity, but in the same period so that electronegativity matters more than orbital size? Let's assume that the molecules I am comparing have the same substituents and one lone pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the molecule with the more electronegative atom in the middle have the lone pair electrons closer to the nucleus, so they take up more bonding space and the angles for the other substituents will be smaller?&lt;/p&gt;&#xA;" OwnerUserId="9430" LastEditorUserId="16683" LastEditDate="2016-11-20T19:01:27.390" LastActivityDate="2017-03-06T17:31:41.503" Title="How do electronegativity and lone pairs affect bond angles?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18700" PostTypeId="1" AcceptedAnswerId="18704" CreationDate="2014-10-21T20:51:04.247" Score="4" ViewCount="52646" Body="&lt;p&gt;For example, the rate of a chemical reaction can be expressed in $\mathrm{mol}/\mathrm{L}^{-1}/\mathrm{sec}^{-1}$. Why is it ‘−1’ and not, say, ‘−2’? Does it change the meaning if the minus is removed and we simply express the rate in $\mathrm{mol}/\mathrm{L}/\mathrm{sec}$?&lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorUserId="7475" LastEditDate="2017-11-18T03:07:05.283" LastActivityDate="2018-09-15T09:00:41.727" Title="What does the ‘−1’ superscript mean in units?" Tags="&lt;notation&gt;&lt;units&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18701" PostTypeId="2" ParentId="18688" CreationDate="2014-10-21T21:21:59.157" Score="5" Body="&lt;p&gt;$\pu{g\%g}$ literally means grams per cent grams or grams of mineral per $\pu{100g}$ of sample. So, in population 1 there is $\pu{2.33g}$ of $\ce{K}$ per $\pu{100g}$ of leaves. Not bad when you consider cooked spinach only has about$\pu{0.05\%}$ $\ce{K}$! Eat your greens, kids. About $\pu{2\%}$ $\ce{K}$ in plants is average across a broad spectrum of plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pu{ug g-1}$ is micrograms of mineral per gram of sample. Population 1 has 23 micrograms of $\ce{Na}$ per gram of leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to convert these to percents or ppm, as needed, and check in the text as to whether this refers to dry weight of plant or not. I'm guessing with this multivariate array of mineral analyses, the origin of particular cannabis crops could be determined.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="39539" LastEditDate="2017-04-07T23:44:17.543" LastActivityDate="2017-04-07T23:44:17.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18702" PostTypeId="2" ParentId="18700" CreationDate="2014-10-21T21:23:26.587" Score="12" Body="&lt;p&gt;The $^{-1}$ superscript can be thought of as saying &quot;per&quot; or as being the denominator of the fraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your example $\mathrm{mol \cdot L^{-1} sec^{-1}}$ can be thought of as saying moles per liter per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easier than writing $\mathrm{\frac{mol}{(L \cdot sec)}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the super script from $1$ to $2$ or $3$ would change the meaning of the value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1 \mathrm{cm^{3}\ is\ 1mL}$$ &#xA;So, $\mathrm{cm}^{-1}$ is per centimeter, which would be a measurement of something per distance, but $\mathrm{cm^{-3}}$ would be talking about something in a given volume.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-10-21T21:23:26.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18703" PostTypeId="2" ParentId="18700" CreationDate="2014-10-21T21:24:29.833" Score="8" Body="&lt;p&gt;It may have its roots even earlier than that, but this was mainly due to people using typewriters to write scientific papers, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the ability to format things like $\mathrm{\frac{mol}{L}}$, both on-screen and in print, but adjusting the carriage and line feed knob every time you had to type a complicated formula was tedious, so it was easier to type &quot;mol-L-1&quot; instead.  Even when the -1s became superscripts, as John points out in his answer, it was still used in typesetting to keep formulas, etc. all on the same line in books.&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="41556" LastEditDate="2017-11-18T18:29:40.360" LastActivityDate="2017-11-18T18:29:40.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18704" PostTypeId="2" ParentId="18700" CreationDate="2014-10-21T21:33:05.030" Score="15" Body="&lt;p&gt;The -1 means &quot;per&quot; unit. So your first example mol/L&lt;sup&gt;-1&lt;/sup&gt;/s&lt;sup&gt;-1&lt;/sup&gt; is not correct - it would actually be written as mol L&lt;sup&gt;-1&lt;/sup&gt; s&lt;sup&gt;-1&lt;/sup&gt;, OR mol/(L s). It is also sometimes written as mol/L/s, but the double division is ambiguous and should be avoided unless parentheses are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were mol L&lt;sup&gt;-1&lt;/sup&gt; s&lt;sup&gt;-2&lt;/sup&gt;, this would mean moles per litre per second per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really just a question of notation, and is not chemistry-specific at all. Yes, all the minus/plus signs and the value of numbers are important. Good examples of units can include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;area, measured in m&lt;sup&gt;2&lt;/sup&gt;, or metres squared&lt;/li&gt;&#xA;&lt;li&gt;volume, measured in m&lt;sup&gt;3&lt;/sup&gt;, or metres cubed&lt;/li&gt;&#xA;&lt;li&gt;pressure, measured in N m&lt;sup&gt;-2&lt;/sup&gt;, or Newtons per metre squared&lt;/li&gt;&#xA;&lt;li&gt;velocity, measured in m s&lt;sup&gt;-1&lt;/sup&gt;, or metres per second&lt;/li&gt;&#xA;&lt;li&gt;acceleration, measured in m s&lt;sup&gt;-2&lt;/sup&gt;, or metres per second per second&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="16683" LastEditDate="2018-09-15T09:00:41.727" LastActivityDate="2018-09-15T09:00:41.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18705" PostTypeId="1" AcceptedAnswerId="18706" CreationDate="2014-10-21T22:08:49.037" Score="3" ViewCount="1706" Body="&lt;p&gt;Would an acceptable way to protect a phenolic alcohol be through a Williamson ether synthesis? As in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Deprotonating phenol with sodium hydroxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Addition of methyl iodide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Deprotection of ether with strong, hot, concentrated acid such as HI. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are more sophisticated ways of protecting alcohols, but what's wrong with this method (if any)? It certainly seems relatively simple and effective. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="10185" LastEditDate="2016-05-24T22:08:33.117" LastActivityDate="2016-05-24T22:08:33.117" Title="Phenol protection" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18706" PostTypeId="2" ParentId="18705" CreationDate="2014-10-21T22:30:51.567" Score="6" Body="&lt;p&gt;Protecting a phenol by using the Williamson ether synthesis to make the methyl ether is an acceptable method.  However your deprotection step using strong, hot acid is undesirable since the molecule might contain acid or thermally sensitive groups.  A milder way to deprotect a phenolic methyl ether involves the use of &lt;a href=&quot;http://www.commonorganicchemistry.com/Rxn_Pages/O_Demethylation/O-Demethylation_BBr3_Mech.htm&quot;&gt;boron tribromide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a nice &lt;a href=&quot;http://www.synarchive.com/protecting-group/Phenol_Methyl_ether&quot;&gt;link&lt;/a&gt; to a page listing a variety of methods to protect and deprotect phenols, with examples for each route.  Depending on your actual molecule, the dimethyl sulfate protection route might be preferred.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-21T22:30:51.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18707" PostTypeId="2" ParentId="16160" CreationDate="2014-10-22T00:04:32.560" Score="8" Body="&lt;p&gt;&amp;beta;-lactams, also known as 2-azetidinones, have experienced a resurgence of interest in the last decade or so, with a number of new applications being exploited. Of course, antibiotic activity is well-known, but I hope you don't limit your acceptance of drug design just to this class of chemical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;beta;-Lactams have found use in a number of fields of medical application, inlcuding gene activation, enzyme inhibition and cholesterol absorption inhibition, and recently there has been growing interest in the use of these molecules as synthetic templates for accessing other target moelcules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of sources of references if you go digging, but some key reviews that may interest you include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/journal/00404020/56/31&quot;&gt;TETRAHEDRON Volume 56, Issue 31, Pages 5553-5742 (July 2000)&lt;/a&gt;. The entire issue is concerned with developments in &amp;beta;-lactam chemistry; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0223523410006835&quot;&gt;EUROPEAN JOURNAL OF MEDICINAL CHEMISTRY  Volume 45, Issue 12, Pages 5541-5560   (DEC 2010)&lt;/a&gt; &quot;2-Azetidinone - A new profile of various pharmacological activities&quot;  gives some very good insights into new medical applications of this class of compound, including anticancer, antiparkinson, antidiabetic, antiinflammatory activities. I think they can also do the washing up and buy your groceries.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0040402003010998&quot;&gt;TETRAHEDRON  Volume 59, Issue 39, Pages 7631-7649 (Sep 2003)&lt;/a&gt; &quot;Recent progress in the synthesis and chemistry of azetidinones&quot; gives a pretty good summary of synthetic approaches to these molecules.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alcaide and Almendros do a lot of work using 2-azetidones as reactive synthons in catalytic cyclisations. A recent review by them is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ar4002558&quot;&gt;ACCOUNTS OF CHEMICAL RESEARCH  Volume 47, Issue 3, Pages 939-952 (Mar 2014)&lt;/a&gt; &quot;Gold-Catalyzed Cyclization Reactions of Allenol and Alkynol Derivatives&quot;, but they have a heap of other papers also. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example, probably somewhat similar, of their use as a reactive synthon is &#xA;BULLETIN OF THE KOREAN CHEMICAL SOCIETY  Volume 28, Issue 1, Pages 17-28 (2007) &quot;Indium and gallium-mediated addition reactions&quot; - I have to admit I only read the abstract of this paper, but they appear to use azetidones as synthons in metal-mediated syntheses of a range of compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, the August edition of Chemical Reviews Vol 114, Issue 16 looks at small heterocycles in synthesis and has some interesing papers, inparticular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr4005549&quot;&gt;CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7930-7953&lt;/a&gt; &quot;Chemical Synthesis of beta-Lactams: Asymmetric Catalysis and Other Recent Advances&quot;. This is a very good paper for you, I think.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr500127b&quot;&gt;CHEMICAL REVIEWS Volume 114, Issue 16, Pages 8257-8322&lt;/a&gt; &quot;Four-Membered Ring-Containing Spirocycles: Synthetic Strategies and Opportunities&quot;. An interesting application of lactams in spiro molecule formation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr400582d&quot;&gt;CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7954-8015&lt;/a&gt; &quot;Synthesis and Reactivity of 2-(Carboxymethyl)aziridine Derivatives&quot;. This looks at lactams being used as synthons for a range of molecules, here focussing on aziridines.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope some of these point you in the direction of what you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-22T00:04:32.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18708" PostTypeId="1" CreationDate="2014-10-22T01:24:16.653" Score="16" ViewCount="12755" Body="&lt;p&gt;I'm disagreeing with my chemistry teacher over this enthalpy of formation problem, so can someone tell me if I am right or why I am wrong?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The enthalpy of formation of gaseous sulfur trioxide is $-396~\mathrm{kJ/mol}$. What is the enthalpy of the reaction represented by the following balanced equation? &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce {2 SO_3 (g) -&amp;gt; 2 S (s) + 3 O2 (g) }$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A) $-396~\mathrm{kJ}$ &amp;emsp; (B) $396~\mathrm{kJ}$ &amp;emsp; (C) $792~\mathrm{kJ}$ &amp;emsp; (D) $-792~\mathrm{kJ}$ &amp;emsp; (E) $198~\mathrm{kJ}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got (C) by flipping the sign of the enthalpy of formation given (because the equation given proceeds in the opposite direction from the formation of sulfur trioxide) and multiplying it by two (because there are 2 moles of sulfur trioxide).&lt;br&gt;&#xA;My teacher and most of my other classmates said that it was (B) instead. They said that I shouldn't multiply by 2 because the equation given is at it's lowest realistic coefficient values.&lt;br&gt;&#xA;I don't understand this for a few reasons:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the enthalpy of formation is defined as the enthalpy change per mole of substance &lt;/li&gt;&#xA;&lt;li&gt;in previous problems, we would change the chemical equation to make the product only have one mole &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For example, here's one problem:   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which of the following reactions is the heat of reaction equal to the heat of formation of the product? &#xA;  \begin{align}&#xA;\text{(A)} &amp;amp;&amp;amp;\ce{N_2 (g) + 3 H_2 (g) &amp;amp;-&amp;gt; 2 NH_3 (g)}\\&#xA;\text{(B)} &amp;amp;&amp;amp;\ce{1/2 N_2 (g) + O_2 (g) &amp;amp;-&amp;gt; NO_2 (g) }\\&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We narrowed it already down to (A) and (B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher said the answer was (B), because while both equations had the elements in their elemental form, the first reaction produces 2 moles of product while the second produces 1 mole of product. Thus the fraction coefficients were necessary.&lt;/p&gt;&#xA;" OwnerUserId="9437" LastEditorUserId="24052" LastEditDate="2018-04-12T05:59:31.933" LastActivityDate="2018-04-12T12:42:36.720" Title="Enthalpy of formation and reaction equations" Tags="&lt;enthalpy&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18709" PostTypeId="2" ParentId="18708" CreationDate="2014-10-22T02:16:33.090" Score="11" Body="&lt;p&gt;As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By &quot;enthalpy of the reaction&quot; it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, but even if it did, the most common convention for balancing equations is to use the smallest whole numbers for stoichiometric coefficients, which is what is written there and would give the answer you got. You can calculate an enthalpy change of reaction for any reaction conditions you like and the value you get will change depending on the quantities of the substances involved, but if the question is implicitly assuming moles for coefficients, you still get the same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, there's no reason you couldn't use some other definition of &quot;standard&quot; in which you set the coefficients such that the smallest ones are equal to 1, though I can't remember seeing anyone do it this way. IUPAC convention is that whenever a quantity referring to some standard state is used, the conditions of the standard should be specified (though chemists like to ignore IUPAC).&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-10-22T02:23:27.270" LastActivityDate="2014-10-22T02:23:27.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18710" PostTypeId="2" ParentId="18708" CreationDate="2014-10-22T02:43:16.483" Score="7" Body="&lt;p&gt;I found an &lt;a href=&quot;http://mrsmcnicholspage.weebly.com/uploads/1/5/0/3/15034344/sch_4u_enthalpy_problem_set_2_solutions.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;enthalpy problem set PDF file&lt;/a&gt;:&#xA;$$\ce {2 S (s) + 3 O2 (g) -&amp;gt; 2 SO3 (g)}\qquad \Delta H = -794.45 \, \mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the best answer is C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading all the answers and comments, I have noticed that the question asks for the enthalpy of the reaction. Since the reverse reaction has a value of $\pu{-396 kJ/mol}$, and the current reaction has 2 moles of $\ce{SO3}$, then the answer is $\pu{792 kJ}$ (the energy for the total reaction). The answer depends on the number of moles in the reaction&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="5026" LastEditDate="2018-04-12T12:42:36.720" LastActivityDate="2018-04-12T12:42:36.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18714" PostTypeId="1" CreationDate="2014-10-22T06:10:59.730" Score="1" ViewCount="37" Body="&lt;p&gt;i am working with a multicomponent alloy system for soft magnetic application,it is iron based and have composition of Fe Mn Si B Mo,how can i calculate the net magnetic moment or saturation magnetization of the system,consider proper mixing of alloys in melt takes place.&#xA;kindly tell me the theoritical way of calculating,and individual contribution&lt;/p&gt;&#xA;" OwnerUserId="9442" LastActivityDate="2014-10-22T06:10:59.730" Title="magnetism in metallic alloys" Tags="&lt;magnetism&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18715" PostTypeId="1" AcceptedAnswerId="18716" CreationDate="2014-10-22T09:22:30.187" Score="3" ViewCount="8145" Body="&lt;p&gt;I'm a highschool student and as part of an investigatory project, I have decided to analyse which metals are present in our everyday coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose this because in my previous academic year, I had learnt how to analyze the cations present in inorganic salts and I employed the same systematic analysis method that I had learnt to test the presence of metal cations in some coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, I cannot expect the coin analysis to go exactly like the salt analysis. I would have to ignore certain tests like the 'flame test'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please tell me what precaustions I should take for this experiment? I wish to write up a clear and systematic process by which any student / coin collector can easily do in a laboratory with only common chemicals and salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All advice, resources and tips will be most appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2325" LastEditorUserId="5017" LastEditDate="2014-10-22T12:54:27.910" LastActivityDate="2014-11-13T14:33:17.887" Title="Precautions and tips for Qualitative Inorganic Analysis of Coins" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18716" PostTypeId="2" ParentId="18715" CreationDate="2014-10-22T09:48:44.673" Score="5" Body="&lt;p&gt;1) The coins must be dissolved (at least partially). There is a good amount of metals that won't dissolve in common  acids (hydrochloric acids), some will require nitric acid, some will require aqua regia and some will require even worse, though that's unlikely. So, most likely you will have to deal with concentrated nitric acid, that produce nitric oxide (fox's tail) when react with metals. Given that &lt;strong&gt;you must have fume hood&lt;/strong&gt;. In case something undissolved remains, you can try aqua regia, and if even that fail, you have something rare, like nioubium, with no common method of wet analisys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Most metals used in coins have toxic salts. So, be cautios, and avoid skin contact. Latex gloves and protective goggles are not vital, but certainly recommended unless you believe you can be accurate enough. Usual safety requirements for work with conc. acids applies as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In case of non-destructive analisys, X-ray fluorescent spectrometry may be employed. It will give accurate analysis in almost any realistic case, but requires quite costly equipment. It is not extremely costly per analysis, though.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-10-22T13:41:10.550" LastActivityDate="2014-10-22T13:41:10.550" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18717" PostTypeId="2" ParentId="18567" CreationDate="2014-10-22T10:09:25.153" Score="52" Body="&lt;p&gt;Dissociation obviously increases the number of moles.&lt;br&gt;&#xA;The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two cases on which equilibrium depends. These are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Addition of an inert gas at constant volume&lt;/strong&gt;:&lt;br&gt;&#xA;When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Addition of an inert gas at constant pressure&lt;/strong&gt;:&lt;br&gt;&#xA;When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Consider the following reaction in equilibrium:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$&lt;/span&gt;&#xA;The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please read about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier&amp;#39;s_principle#Effect_of_adding_an_inert_gas&quot; rel=&quot;noreferrer&quot;&gt;Effect of adding an Inert Gas&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2325" LastEditorUserId="68148" LastEditDate="2020-03-30T00:14:28.380" LastActivityDate="2020-03-30T00:14:28.380" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="18720" PostTypeId="2" ParentId="18567" CreationDate="2014-10-22T10:55:13.753" Score="42" Body="&lt;p&gt;Nick's answer is good. Let's add a little maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an example dissociation reaction $\ce{A&amp;lt;=&amp;gt;B + C}$ for which $K_p=1$. Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p = \frac{P_B P_C}{P_A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one equilibrium scenario is $P_A =P_B = P_C = 1 \text{ atm}$. The total pressure $P_T = 3 \text{ atm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add an inert gas &lt;strong&gt;at constant pressure&lt;/strong&gt;, the total pressure cannot change. Thus, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ must decrease. If we add some inert gas $\ce{D}$ so that $P_A=P_B=P_C=P_D=\frac{3}{4} \text{ atm}$, the total pressure is still $P_T=3\text{ atm}$. The reaction quotient $Q_p$ is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_p = \frac{P_B P_C}{P_A}=\frac{(\frac{3}{4}\text{ atm})(\frac{3}{4}\text{ atm})}{\frac{3}{4}\text{ atm}}=\frac{3}{4}\text{ atm}&amp;lt;K_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Q_p&amp;lt;K_p$, the equilibrium will shift toward products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &lt;strong&gt;constant volume&lt;/strong&gt;, the partial pressures of the three components $\ce{A}$, $\ce{B}$, and $\ce{C}$ can remain constant and equilibrium will be unchanged. Adding inert gas $\ce{D}$ will increase the pressure of the system.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="41556" LastEditDate="2017-11-06T14:28:02.660" LastActivityDate="2017-11-06T14:28:02.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18723" PostTypeId="1" AcceptedAnswerId="18725" CreationDate="2014-10-22T13:01:14.287" Score="9" ViewCount="2678" Body="&lt;p&gt;Why is atomic size mismatch a sign of poor bond strength? Wouldn't orbital overlap be proportional whether the two atoms are small or the two atoms are big? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example I heard someone comment that an argument that the C-I bond is weaker than the C-F bond based on orbital overlap made no sense to them, because why would the size of the individual orbitals matter in overlap? Small atom, big atom and small atom, small atom - shouldn't their overlaps be proportional? Or something along the lines of that. I'll have to pull up the post again ... &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-10-22T13:59:58.070" LastActivityDate="2014-10-22T14:57:55.260" Title="Orbital overlap and size &quot;mismatch&quot;" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18724" PostTypeId="1" AcceptedAnswerId="18734" CreationDate="2014-10-22T14:07:31.180" Score="7" ViewCount="2275" Body="&lt;p&gt;Why is the C-S bond in $\ce{CH_2SH^-}$ longer than in $\ce{CH_3SH}$? Wouldn't there be resonance in the former, forcing the C-S bond to contract? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lPkEZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LtXPC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also while we're at it are there any non &lt;em&gt;ab initio&lt;/em&gt; calculations/empirical studies on this interesting thiol ion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-10-22T18:39:01.263" LastActivityDate="2014-10-23T15:22:52.257" Title="Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18725" PostTypeId="2" ParentId="18723" CreationDate="2014-10-22T14:57:55.260" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Small atom, big atom and small atom, small atom - shouldn't their&#xA;  overlaps be proportional&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  Look at the figure below and let's consider s-orbital overlap, just to keep it simple.  Remember that these are spheres with the electron density distributed throughout the entire volume.  In the &quot;small-small&quot; case there is effective overlap of the electron density in both 1s orbitals; we would expect the resultant bond to be strong.  In the &quot;small-big&quot; case, the small 1s orbital involves a significant fraction of its electron density in overlap, but not much of the electron density in the large s orbital is involved in overlap.  In the &quot;small-big&quot; case the overlap of electron clouds is ineffective and any resultant bond would be weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qi5x1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-22T14:57:55.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18726" PostTypeId="2" ParentId="8140" CreationDate="2014-10-22T15:19:08.413" Score="0" Body="&lt;p&gt;Maybe, Molpro has one of the easiest ways to do it&#xA;&lt;a href=&quot;http://www.molpro.net/info/2012.1/doc/manual/node194.html&quot; rel=&quot;nofollow&quot;&gt;http://www.molpro.net/info/2012.1/doc/manual/node194.html&lt;/a&gt;&#xA;though, not free&lt;/p&gt;&#xA;" OwnerUserId="9449" LastActivityDate="2014-10-22T15:19:08.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18727" PostTypeId="1" AcceptedAnswerId="18729" CreationDate="2014-10-22T16:17:58.890" Score="3" ViewCount="3971" Body="&lt;p&gt;I am told that all anomers are diastereomers. I think this is BS. If there is a carbohydrate with only one chiral center, then changing the configuration of the anomeric carbon gives us enantiomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it impossible to have carbohydrates with only one chiral center? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-22T20:36:03.063" Title="Are all anomers diastereomers?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18728" PostTypeId="1" AcceptedAnswerId="18730" CreationDate="2014-10-22T16:46:23.857" Score="5" ViewCount="14700" Body="&lt;p&gt;Why are D and L-glucose enantiomers???! I thought D/L refers to the &quot;configuration&quot; of the last chiral carbon on the open chain Fischer projection. In the D form, the higher priority group is on the right in the Fischer projection; in the L form the higher priority group is on the left. &lt;strong&gt;So&lt;/strong&gt; why are D and L glucose &lt;strong&gt;enantiomers&lt;/strong&gt;? Shouldn't they be &lt;strong&gt;diastereomers??!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Af7VY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-22T17:41:27.097" Title="D/L Configuration" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18729" PostTypeId="2" ParentId="18727" CreationDate="2014-10-22T17:13:10.080" Score="7" Body="&lt;p&gt;The only saccharides I know of that have only a single chiral centre are the trioses L- and D-glyceraldehyde, but, at least according to the Gold Book &lt;a href=&quot;http://goldbook.iupac.org/A00373.html&quot;&gt;definition&lt;/a&gt;, anomers are specifically diastereomers of cyclic forms of sugars. Trioses are too short for a cyclic form to normally arise and therefore can't anomerize.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-22T17:13:10.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18730" PostTypeId="2" ParentId="18728" CreationDate="2014-10-22T17:31:53.720" Score="5" Body="&lt;p&gt;Simply by inspecting the two figures of the molecules, you can see that they are in fact enantiomers—non-superimposable mirror images and were they diastereomers instead of enantomers, they wouldn't both be named &quot;Glucose&quot;. The D-/L- system is simply an antiquated naming system for enantiomers based upon the relation of the structures to glyceraldehyde. D-glucose refers to the enantiomer of glucose in which the C5 stereocentre has the same configuration as (R)-glyceraldehyde and L-glucose that of (S)-glyceraldehyde. Because this notation refers specifically to enantiomers, it is assumed that the configuration at all stereocentres is inverted and the configuration at the last stereocentre was chosen to assign unambiguous names.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-22T17:31:53.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18731" PostTypeId="2" ParentId="18728" CreationDate="2014-10-22T17:41:27.097" Score="4" Body="&lt;p&gt;For D-glucose, it has the configuration (2R,3S,4R,5R). For L-glucose, it has the configuration (2S,3R,4S,5S).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, for an enantiomer, all S will change to R and vice versa at each respective carbon centres. All the other configurations are diastereomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, (R,R) and (S,S) are enantiomers, (R,S) and (S,R) are enantiomers, but (R,R) and (R,S) are diastereomers.&lt;/p&gt;&#xA;" OwnerUserId="8140" LastActivityDate="2014-10-22T17:41:27.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18732" PostTypeId="2" ParentId="18724" CreationDate="2014-10-22T18:26:39.073" Score="3" Body="&lt;p&gt;First, the difference we are talking about is only 0.074 A, it's always difficult to be sure what causes such small changes.  That said...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of negative charge on the carbon atom in $\ce{CH3-SH}$ (-0.639) surprises me; it's actually more than in the corresponding anion $\ce{CH2^{-}-SH}$ (-0.620). Note too the significantly greater positive charge on the carbon hydrogens in $\ce{CH3-SH}$.  Perhaps this suggests that hyperconjugated resonance structures like I are important in $\ce{CH3-SH}$.  This would result in the $\ce{C-S}$ bond being shortened to begin with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Td8Mh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in the case of the anion, resonance structures II and III suggest that there is appreciable negative charge on both carbon and sulfur, consistent with the high $\ce{q_{b}}$ and $\ce{q_{c}}$ values reported in Table I (-0.620, -0.328, respectively).  The appreciable negative charge on both carbon and sulphur, in the anion, results in coulombic repulsion and a slight elongation of the carbon-sulphur bond, from the somewhat shortened bond length observed in the neutral compound.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-22T18:26:39.073" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18733" PostTypeId="2" ParentId="16753" CreationDate="2014-10-22T18:58:20.653" Score="1" Body="&lt;p&gt;If you used distilled water to prepare the solutions of potassium chromate and barium nitrate, it is more likely that you got a yellow precipitate of $\ce{BaCrO4}$ in a yellow solution of excess $\ce{K2CrO4}$. The yellow color originates from $\ce{CrO4^{2-}}$, and a small amount of finely divided precipitate may look white at first sight (namely when the yellow color of the solution is more intense), when it is actually pale-yellow (it can look like &lt;a href=&quot;http://www.public.asu.edu/~jpbirk/qual/qualanal/barium.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-10-22T18:58:20.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18734" PostTypeId="2" ParentId="18724" CreationDate="2014-10-22T20:21:24.327" Score="5" Body="&lt;h2&gt;Quick Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431&quot; rel=&quot;nofollow noreferrer&quot;&gt;linked paper (from 1975)&lt;/a&gt; is probably &lt;strong&gt;wrong&lt;/strong&gt;. A &lt;em&gt;slightly&lt;/em&gt; more recent paper (&lt;a href=&quot;http://dx.doi.org/10.1021/ja00203a007&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. Am. Chem. Soc., 1989, 111 (21), pp 8112–8115&lt;/a&gt;) suggests the bond lengths are about equal. Some quick calculations (below) using multiple modern methods suggest that, as expected, the C-S bond length in $\ce{CH2SH-}$ is &lt;em&gt;shorter&lt;/em&gt; than in $\ce{CH3SH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both older papers, the computational results (using basic Hartree Fock) would be considered highly suspect under modern peer review, since Hartree Fock is known to have problems with anions. Certainly the &lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431&quot; rel=&quot;nofollow noreferrer&quot;&gt;1975 paper&lt;/a&gt; doesn't even mention the &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;basis set&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen below, it's clear that the results are method dependent, but the best current methods, with reasonable basis sets indicate the bond length is at least slightly shorter due to some partial double-bond character.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;More Details&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first thing to consider in this is that there are two forms of $\ce{CH2SH-}$, &lt;em&gt;syn&lt;/em&gt; and &lt;em&gt;anti&lt;/em&gt;, based on the orientation of the formally C-bound anion and the SH group:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;em&gt;syn&lt;/em&gt; form&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w7PuI.png&quot; alt=&quot;syn-CH2SH- anion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;em&gt;anti&lt;/em&gt; form&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3s4Wp.png&quot; alt=&quot;anti-CH2SH- anion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These will have different bond lengths, since the &lt;em&gt;anti&lt;/em&gt; form will have contribution of resonance structures &lt;strong&gt;II&lt;/strong&gt; and &lt;strong&gt;III&lt;/strong&gt; from Ron's answer. This gives some double-bond character and should likely have a shorter bond, although electrostatic repulsion could also be invoked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;syn&lt;/em&gt; form won't have a clear effect, since the C anion lone pair and the S lone pairs are on the opposite side of the bond. My guess is that the bond length is about the same as $\ce{CH3SH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the clear message is that we can &lt;em&gt;explain&lt;/em&gt; a lot of things, but it's better to know if our results are accurate before trying to explain them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I see with more modern computational methods is that the &lt;em&gt;syn&lt;/em&gt; form has similar C-S bond length to $\ce{CH3SH}$ and the &lt;em&gt;anti&lt;/em&gt; form is slightly shorter. I suspect the previous result is simply &lt;strong&gt;&lt;em&gt;wrong&lt;/em&gt;&lt;/strong&gt;. Calculations involving anions are tricky, since the extra electron is usually highly diffuse. I expect the basis set was simply insufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in energy between the two forms is ~1 kcal/mol, so experimentally, these will interconvert readily, and an average bond length would be observed. As mentioned in the comments, this species would probably be hard to observe, even in the gas phase, since $\ce{CH3S-}$ is more stable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Computed C-S bond lengths (in Å):&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For all methods, full geometry optimization was performed using the methods indicated. I would consider the CCSD calculations to be high quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SoZ7G.png&quot; alt=&quot;bond length table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-23T15:22:52.257" LastActivityDate="2014-10-23T15:22:52.257" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18735" PostTypeId="2" ParentId="18727" CreationDate="2014-10-22T20:36:03.063" Score="4" Body="&lt;p&gt;Adding a bit to Michael's answer. Although it would be highly strained, let's look at what happens if glyceraldehyde cyclizes. Since glyceraldehyde has one stereocenter, the anomeric position becomes a second, giving diastereomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/877kt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order for the anomers to be enantiomers, the compound actually must have zero stereocenters in the open form. If you consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycolaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;glycoaldehyde&lt;/a&gt; a carbohydrate and you consider the three-membered ring cyclic hemiacetal a reasonable structure, then you have found a set of anomers that are enantiomers.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-22T20:36:03.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18736" PostTypeId="2" ParentId="18708" CreationDate="2014-10-22T22:25:06.597" Score="0" Body="&lt;p&gt;While ambiguously worded, if by the &quot;enthalpy of reaction&quot; the teacher is implying the standard enthalpy of reaction, this being the energy to form one mole of product, then by the most common usage of the words used in the problem, &quot;B&quot; is a better solution than &quot;C&quot;, as it utilizes the more precise definition of energy per mole of formation.  However as the word &quot;standard&quot; was never used you certainly have a valid argument as to why &quot;C&quot; should also be considered correct.&lt;/p&gt;&#xA;" OwnerUserId="9453" LastActivityDate="2014-10-22T22:25:06.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18737" PostTypeId="1" CreationDate="2014-10-23T00:07:53.460" Score="2" ViewCount="809" Body="&lt;p&gt;I have a molecule and I need to assign the carbon NMR to each of the carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My molecule is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kgIun.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peaks I have for the carbon NMR are shown below. I have also used the DEPT to work out if it's a $\ce{CH3}$, $\ce{CH2}$, $\ce{CH}$ or quaternary C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)&lt;br&gt;&#xA;19.5ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)&lt;br&gt;&#xA;25.5ppm $\ce{CH2}$&lt;br&gt;&#xA;25.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)&lt;br&gt;&#xA;32.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;36.7ppm $\ce{CH2}$&lt;br&gt;&#xA;39.7ppm $\ce{CH2}$&lt;br&gt;&#xA;66.0ppm $\ce{CH2}$&lt;br&gt;&#xA;122.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;124.4ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;128.2ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;128.3ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;128.6ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;131.5ppm quaternary carbon&lt;br&gt;&#xA;136.2ppm quaternary carbon&lt;br&gt;&#xA;149.0ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)&lt;br&gt;&#xA;166.4ppm quaternary carbon  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I can assign the quaternary carbon with the $\ce{C=O}$ in the ester as the 166.4ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the three methyl groups, would the lowest ppm of 17.7ppm be the one at the top of the molecule by itself and then would the other two at the bottom of the molecule on the alkene be 19.5ppm and 25.7ppm? I don't know which one is which, I know it is to do with cis/trans but I'm not sure which way around it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that on the benzene ring, only three $\ce{CH}$ peaks show up as the two ortho and the two meta carbons are equivalent. &lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorDisplayName="user10259" LastEditDate="2015-05-16T05:18:51.927" LastActivityDate="2015-05-16T05:18:51.927" Title="Assigning Carbon NMR" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18738" PostTypeId="1" CreationDate="2014-10-23T00:17:00.383" Score="0" ViewCount="759" Body="&lt;p&gt;I have a nutty neighbor that said someone told them to put liquid lye into their above ground pool.  I said it would eat away the lining, their answer was to only use a little bit.&lt;/p&gt;&#xA;" OwnerUserId="9454" LastEditorUserId="4356" LastEditDate="2014-10-23T04:48:23.640" LastActivityDate="2014-12-22T06:55:30.197" Title="Will lye damage vinyl pool liners?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18739" PostTypeId="2" ParentId="18737" CreationDate="2014-10-23T01:17:50.080" Score="2" Body="&lt;p&gt;Have you tried a &lt;a href=&quot;http://www.nmrdb.org/13c/&quot; rel=&quot;nofollow&quot;&gt;C13 NMR predictor&lt;/a&gt;?&lt;br&gt;&#xA;Highlights according to the predictor:&lt;br&gt;&#xA;166.4ppm quaternary carbon - carbonyl&lt;br&gt;&#xA;136.2ppm quaternary carbon - benzene ring&lt;br&gt;&#xA;131.5ppm quaternary carbon - aliphatic carbon at far end&lt;br&gt;&#xA;149.0ppm CH carbon beta to carbonyl&lt;br&gt;&#xA;66.0ppm CH2 between ring and ester&lt;br&gt;&#xA;The chemical shifts are slightly different (as much as 5 ppm), but I think the order will be correct.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-23T01:17:50.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18740" PostTypeId="1" AcceptedAnswerId="18942" CreationDate="2014-10-23T01:36:41.563" Score="2" ViewCount="341" Body="&lt;p&gt;I'm solving problems related to X-ray diffraction, and I get these results which apparently are correct. My question may be dumb, but I just can't see how the interplanar distance might be shorter than the atoms radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's up with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it might make sense that its shorter than the cells parameter, but it's not immediately obvious to me, so if someone could prove it to me, i'd be quite glad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic radius of chromium is 124pm. THe distance between (310) planes is .0912nm. &#xA;See &lt;a href=&quot;http://uqu.edu.sa/files2/tiny_mce/plugins/filemanager/files/4220193/material_science_and_material_1/solved_problems/01-solved_problem.pdf&quot; rel=&quot;nofollow&quot;&gt;problem 6&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="5456" LastEditDate="2014-10-29T07:10:44.603" LastActivityDate="2014-10-29T07:10:44.603" Title="XRD-Shorter distance between planes than actual radius of atoms/or cell parameter" Tags="&lt;crystal-structure&gt;&lt;atomic-radius&gt;&lt;x-ray-diffraction&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18741" PostTypeId="1" CreationDate="2014-10-23T02:04:31.313" Score="1" ViewCount="8554" Body="&lt;p&gt;If you have a simple alkene for example 1-methyl-But-2-ene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you will have two methyl groups on one side of the double bond, which one of these will have the highest. Will the one on the same side as the rest of the chain of butene have the higher ppm or will it be the one in trans with the butene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, on my spectra, these two methyl groups are not equivalent. I need to know if the E or Z will have a higher chemical shift and why, thanks.&lt;/p&gt;&#xA;" OwnerUserId="5181" LastActivityDate="2014-10-23T03:13:14.847" Title="Which has the higher chemical shift E/Z alkene?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18742" PostTypeId="2" ParentId="18741" CreationDate="2014-10-23T02:21:00.013" Score="1" Body="&lt;p&gt;cis-butene: $\delta$(H2): 5.37 ppm &lt;a href=&quot;http://sdbs.db.aist.go.jp/sdbs/cgi-bin/img_disp.cgi?disptype=disp3&amp;amp;imgdir=hpm&amp;amp;fname=HPM02471&amp;amp;sdbsno=8087&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;trans-butene: $\delta$(H2): 5.55 ppm &lt;a href=&quot;http://sdbs.db.aist.go.jp/sdbs/cgi-bin/img_disp.cgi?disptype=disp3&amp;amp;imgdir=hpm&amp;amp;fname=HPM02472&amp;amp;sdbsno=8088&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lists of increments (Schoolery, Tobey-Simon...) out there to predict chemical shifts in systems such as there, but for methyl groups in vinylic these are very similar.  In those cases it really doesn't bear to assign isomers through chemical shifts when you can get the correct assignment through coupling constants.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-23T02:21:00.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18743" PostTypeId="2" ParentId="18737" CreationDate="2014-10-23T02:37:55.837" Score="6" Body="&lt;p&gt;If this is a real sample, and not a provided coursework spectrum, then the best recommendation is to run a HSQC and/or HMBC. This will provide complete data to allow this assignment fully, and in much the same time as it takes to run a DEPT135. I don't understand the urge to run DEPTS any more - I haven't run one for years. A phase edited HSQC will give you all of the information that a DEPT can provide, as well as connectivities between proton and carbons. Anyway, off my soapbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a simulation of this molecule, which has reasonable alignment with your peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/25Qji.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A methyl centre trans to a alkyl chain will be downfield shifted to an equivalent methyl group trans to a proton. So 19 will be downfield shifted from 20, and be fairly similar in chemical shift to 16. Simulations have their limitations, and carbon shifts are solvent dependent to +/- a few ppm at least. To avoid any uncertainty, run HSQC and HMBC. These should be part of your normal procedure for assigning structures.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-23T02:37:55.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18744" PostTypeId="2" ParentId="18741" CreationDate="2014-10-23T03:13:14.847" Score="2" Body="&lt;p&gt;Are you talking about 1H NMR? In the general case, it is not so easy to say - it depends on all substituents, with geminal substitution usually contributing most to change in chemical shift. &lt;em&gt;trans&lt;/em&gt; and &lt;em&gt;cis&lt;/em&gt; substitution will affect chemical shifts differently, but will depend on the electronegativities of all groups concerned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your example, there will be very little difference between chemical shift for these two methyl groups, but the methyl trans to the alkyl chain is probably slightly downfield shifted. Coupling constants for the two methyl groups would be slightly different. In acyclic systems, 4J couplings from a methyl to a &lt;em&gt;cis&lt;/em&gt; or &lt;em&gt;trans&lt;/em&gt; proton can range from about -3 to +2.5 (but usually 0.8 to 2), and usually |J&lt;sub&gt;&lt;em&gt;trans&lt;/em&gt;&lt;/sub&gt;|&gt;|J&lt;sub&gt;&lt;em&gt;cis&lt;/em&gt;&lt;/sub&gt;|. Of course, nOe information would also be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are talking about 13C NMR, a methyl peak trans to a carbon chain will be downfield shifted from a methyl group trans to a proton.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-23T03:13:14.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18745" PostTypeId="1" CreationDate="2014-10-23T03:18:36.827" Score="3" ViewCount="5793" Body="&lt;p&gt;Consider the equilibrium below. If $K = 4$ and initial pressures of $\ce{CO}$ and $\ce{H2O}$ are equal, giving a total pressure of $1.5~\mathrm{bar}$ at equilibrium. What was the initial pressure of $\ce{CO}$ or $\ce{H2O}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO(g) + H2O (g) &amp;lt;=&amp;gt; CO2 (g) + H2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to get the initial pressures in this situation. Do I have to set up an ICE chart? I'm not sure how to start, any help would be appreciated. The answer is apparently $0.75~\mathrm{bar}$.&lt;/p&gt;&#xA;" OwnerUserId="6842" LastEditorUserId="69830" LastEditDate="2021-05-09T11:22:11.100" LastActivityDate="2021-05-09T11:41:01.743" Title="Finding initial pressure in a chemical equilibrium" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;ideal-gas&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18746" PostTypeId="1" CreationDate="2014-10-23T03:44:40.483" Score="2" ViewCount="915" Body="&lt;p&gt;I've read that powdered elemental sulfur will react to the presence of mercury vapor in the air creating red mercury sulfide, and that an open container of sulfur can be used as a means to verify leaks in a container of mercury that's being kept in a storage cabinet. But is the sulfur reaction/color change sensitive enough relative to the permissible levels of mercury vapor that might exist?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2014-10-23T04:37:25.383" Title="How effective is sulfur in detecting mercury vapors" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18747" PostTypeId="2" ParentId="18746" CreationDate="2014-10-23T04:37:25.383" Score="1" Body="&lt;p&gt;The vapour pressure of mercury is appreciable, but the evaporation kinetics are slow.  In a well-ventilated lab there shouldn't be concern about mercury poisoning from broken thermometers, Schlenk lines &amp;amp;c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there is &lt;a href=&quot;http://en.wikipedia.org/wiki/Alfred_Stock&quot; rel=&quot;nofollow&quot;&gt;Alfred Stock's&lt;/a&gt; famous review (Angewandte Chemie, 1926) on mercury poisoning.  He suffered from a chronic case of mercury poisoning, which IIRC really took off after the war, when they had to save on ventilation.  In old laboratory buildings the quantities found under the wood flooring during renovation are at times staggering.  Our safety guys liked to show pictures of &lt;em&gt;puddles&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To detect mercury there are &lt;a href=&quot;http://link.springer.com/content/pdf/10.1007%2FBF03215020.pdf&quot; rel=&quot;nofollow&quot;&gt;commercially available detectors&lt;/a&gt;, usually build around the resistance of a vacuum-deposited gold film (a few 100 atoms thick).  A &lt;a href=&quot;http://www.youtube.com/watch?v=lpZF88fqrl8&quot; rel=&quot;nofollow&quot;&gt;low-pressure mercury lamp&lt;/a&gt; might also work.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-23T04:37:25.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18748" PostTypeId="2" ParentId="18745" CreationDate="2014-10-23T05:45:16.070" Score="1" Body="&lt;p&gt;If the initial pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H_2O}$&lt;/span&gt; are equal, then let &lt;span class=&quot;math-container&quot;&gt;$p_{\ce{H_2O}}=p_{\ce{CO}}=x$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Since their total is 1.5 bar, i.e. &lt;span class=&quot;math-container&quot;&gt;$\ce{p_{H_2O} + p_{CO}}=1.5$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Can you solve this equation?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.&lt;/p&gt;&#xA;&lt;p&gt;It is given that &lt;span class=&quot;math-container&quot;&gt;$K=4$&lt;/span&gt;. From the equation for the equilibrium constant, &lt;span class=&quot;math-container&quot;&gt;$K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;It is also given that initially, &lt;span class=&quot;math-container&quot;&gt;$\ce{p_{H2O} = p_{\ce{CO}}}$&lt;/span&gt;; from the ideal gas law,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$PV=nRT \implies P=\frac{nRT}{V}=kC  \tag{1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is some constant and C is concentration.&lt;/p&gt;&#xA;&lt;p&gt;So initially, &lt;span class=&quot;math-container&quot;&gt;$[\ce{H2O}]=[\ce{CO}]=y$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.&lt;/p&gt;&#xA;&lt;p&gt;At equilibrium, &lt;span class=&quot;math-container&quot;&gt;$\frac{\ce{[CO2][H_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Solving this equation gives &lt;span class=&quot;math-container&quot;&gt;$x=\frac{2y}{3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Now, from &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: &lt;span class=&quot;math-container&quot;&gt;$[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$&lt;/span&gt;, therefore &lt;span class=&quot;math-container&quot;&gt;$[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So then &lt;span class=&quot;math-container&quot;&gt;$p_{\ce{H2}}= p_{\ce{CO2}} = 2p_{\ce{CO}} = 2p_{\ce{H2O}}$&lt;/span&gt; at equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;So from substitution into Martin's equation, &lt;span class=&quot;math-container&quot;&gt;$\ce{p_{H_2O} + p_{CO} + p_{\ce{CO2}} + p_{\ce{H2}}}=1.5$&lt;/span&gt;, we find that &lt;span class=&quot;math-container&quot;&gt;$p_{\ce{H2O}} = p_{\ce{CO}} = \pu{0.25 bar}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.75 bar}$&lt;/span&gt; each.&lt;/p&gt;&#xA;" OwnerUserId="8143" LastEditorUserId="69830" LastEditDate="2021-05-09T11:21:31.687" LastActivityDate="2021-05-09T11:21:31.687" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18749" PostTypeId="2" ParentId="18738" CreationDate="2014-10-23T05:56:01.870" Score="1" Body="&lt;p&gt;Lye (Sodium Hydroxide) is sometimes added to pool water to bring about increase in pH, without making significant increases to total alkalinity. It will bring about quite rapid changes in pH. I assume the intention is to bring the overall pH back within a sensible range of 7.4-7.6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding high concentrations of sodium hydroxide (liquid lye is about 50% in water) might degrade the vinyl over long periods, but so too will high levels of acid, bleach, exposure to sun, and just about every other chemical used in pools. Vinyl liner can be damaged by low pH (below 7.0), but there will be other symptoms of high pH (above 7.8) (cloudiness, calcification, precipitation) long before damage to the vinyl liner kicks in, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the nutty neighbour may be not too far from the mark to say &quot;only use a little bit&quot;. All things in moderation.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-23T05:56:01.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18750" PostTypeId="2" ParentId="18745" CreationDate="2014-10-23T07:03:55.970" Score="0" Body="&lt;p&gt;Here's what I found using an ICE table. We let the initial partial pressures of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; be &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, and assign an arbitrary value &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; to the change.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{array}{|c|c|c|c|c|}&#xA;\hline&#xA; &amp;amp;\ce{CO}  &amp;amp; \ce{H2O}  &amp;amp; \ce{CO2}  &amp;amp; \ce{H2}   \\&#xA;\hline&#xA; p_0      &amp;amp; x  &amp;amp; x   &amp;amp; 0    &amp;amp; 0    \\&#xA;\hline&#xA; \text{change}  &amp;amp; -y &amp;amp; -y  &amp;amp; +y   &amp;amp; +y  \\&#xA;\hline&#xA; p_{\text{eq}}   &amp;amp; x-y &amp;amp; x-y  &amp;amp; y   &amp;amp; y   \\&#xA;\hline&#xA;\end{array}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So we find that &lt;span class=&quot;math-container&quot;&gt;$2(x-y+y)=1.5 \Rightarrow x = 0.75$&lt;/span&gt; bar, as required.&lt;/p&gt;&#xA;" OwnerDisplayName="user9457" LastEditorUserId="69830" LastEditDate="2021-05-09T11:41:01.743" LastActivityDate="2021-05-09T11:41:01.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="18752" PostTypeId="1" CreationDate="2014-10-23T08:23:48.163" Score="7" ViewCount="4818" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A piece of electronic equipment originally contained $\pu{1.250 g}$ of a valuable metal. Corrosion over time has given a compound of formula $\ce{M(OH)3}$, which is isolated but not weighed. This compound is treated with sulfuric acid to give a compound that has beautiful red crystals and is found to have a molecular formula of $\ce{M2(SO4)2}$ and a mass of $\pu{1.860 g}$. What is this valuable metal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt;&lt;br&gt;&#xA;$$ 1.860 ~\mathrm{g} - 1.250~\mathrm{g}= 0.61~\mathrm{g}\, \ce{SO4} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the $\pu{0.61 g}$ and converted them to moles of $\ce{SO4}$, then I converted 2 moles $\ce{SO4}$ to 2 moles $\ce{M}$ and got $0.0064~\mathrm{M}$. &#xA;Then I divided that by $\pu{1.250 g}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got 195.31, so is platinum the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="9458" LastEditorUserId="5026" LastEditDate="2018-02-22T11:17:10.277" LastActivityDate="2018-02-22T11:17:10.277" Title="Determine what the unknown element is" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18753" PostTypeId="2" ParentId="18752" CreationDate="2014-10-23T09:34:07.263" Score="10" Body="&lt;p&gt;The mass of the sulfate is $m_{\ce{SO4}} = 0.610~\mathrm{g}$, the mass of the metal is $m_\text{M} = 1.250~\mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the molar mass of the sulfate anion $M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}}$ we can calculate the amount of sulfate present in the final compound:&#xA;$$ n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sulfate has the formula $\ce{M2(SO4)2}$ we know that we have the same amount of metal as sulfate: $n_\text{M} = n_{\ce{SO4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now calculate the molar mass of the metal using the information above:&#xA;$$ M_\text{M} = \frac{m_\text{M}}{n_\text{M}} = 196.85~\mathrm{g\, mol^{-1}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This number resembles most closely gold with a molar mass of $M_\text{Au} = 196.97~\mathrm{g\, mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: It is important that you don't round your numbers while doing the calculations, but round them at the end. So you should keep as much precision as possible in between to minimize rounding errors along the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.35\ mmol}$ you will get $\mathrm{196.85\ g/mol}$ (gold).&lt;br&gt;&#xA;If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.4\ mmol}$ you will get $\mathrm{195.31 g/mol}$ (platinum).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="7475" LastEditDate="2015-12-13T21:48:50.507" LastActivityDate="2015-12-13T21:48:50.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18754" PostTypeId="1" AcceptedAnswerId="18787" CreationDate="2014-10-23T10:53:47.960" Score="3" ViewCount="483" Body="&lt;p&gt;On this website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_copernicium&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Isotopes_of_copernicium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a sentence &quot;This isomer is unconfirmed&quot; as a footnote explaining two isotopes.&#xA;What do they mean by 'isomer' in this context?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that an element may have different spatial distribution of neutrons and protons that could have an impact on its stability?&lt;/p&gt;&#xA;" OwnerUserId="9459" LastActivityDate="2014-10-24T09:02:56.540" Title="Mass number of isotopes of copernicium denoted with 'm'" Tags="&lt;elements&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18756" PostTypeId="1" CreationDate="2014-10-23T13:13:58.080" Score="0" ViewCount="4796" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A simple mixture of $\ce{NaCl}$ and $\ce{NaBr}$ weighing $0.180~ \mathrm{g}$ is treated with $\ce{AgNO3}$ solution to give $0.3715~ \mathrm{g}$ of precipitate. Calculate the content of $\ce{NaCl}$ and $\ce{NaBr}$ in the mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer given in book are $\ce{NaCl}$ = $0.0682~ \mathrm{g}$ and $\ce{NaBr}$ = $0.1118~ \mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers I got are $0.1519~ \mathrm{g}$ $\ce{NaCl}$ and $0.0281~ \mathrm{g}$ $\ce{NaBr}$. Please tell me where I made mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Working:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCl + AgNO3 -&amp;gt; NaNO3 + AgCl}$$&#xA;$58.5~ \mathrm{g}$ $\ce{NaCl}$ give $143.5~ \mathrm{g}$ $\ce{AgCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.3715~ \mathrm{g}$ $\ce{AgCl}$ give $58.5/143.5 \cdot 0.3715~ \mathrm{g}$ $\ce{NaCl}$ = $0.1519~ \mathrm{g}$ $\ce{NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaBr}$ = $0.180 ~ \mathrm{g}$ - $0.1519~ \mathrm{g}$ = $0.0281~ \mathrm{g}$ $\ce{NaBr}$&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="5026" LastEditDate="2018-03-03T10:45:52.967" LastActivityDate="2018-03-03T10:47:00.350" Title="Given the mass of precipitate, how can I calculate the relative amount of the original dissolved compounds?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18758" PostTypeId="2" ParentId="18756" CreationDate="2014-10-23T15:05:51.187" Score="1" Body="&lt;p&gt;x = number of moles of chloride&#xA;y = number of moles of bromide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="5026" LastEditDate="2018-03-03T10:47:00.350" LastActivityDate="2018-03-03T10:47:00.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18759" PostTypeId="2" ParentId="16314" CreationDate="2014-10-23T17:14:23.807" Score="3" Body="&lt;p&gt;Your assumption that hydrophobicity of amino acids increases with length of the carbon chain is correct. Also, a branched chain can have a lower hydrophobicity than a linear chain with the same number of carbons, because it has a decreased exposed surface area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When polar acidic/basic functional groups are introduced, the side chain becomes more hydrophilic. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenylalanine&quot; rel=&quot;nofollow noreferrer&quot;&gt;phenylalanine&lt;/a&gt; is classified as non-polar because of the hydrophobic benzyl side chain. Attaching a hydroxy group in &lt;em&gt;para&lt;/em&gt;-position in the phenyl ring leads to &lt;a href=&quot;http://en.wikipedia.org/wiki/Tyrosine&quot; rel=&quot;nofollow noreferrer&quot;&gt;tyrosine&lt;/a&gt;, which is more polar. This is due to its acidic phenol group which is deprotonated at physiological pH. Note that in both cases, the number and arrangement of carbon atoms in the side chain is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9HX3T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-10-23T17:30:31.527" LastActivityDate="2014-10-23T17:30:31.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18760" PostTypeId="1" CreationDate="2014-10-23T17:20:49.033" Score="3" ViewCount="2032" Body="&lt;p&gt;Why can't unimolecular reactions be second order? For example, doubling the concentration of a single reactant would result in four times the original rate of reaction. &lt;/p&gt;&#xA;" OwnerUserId="9462" LastEditorUserId="25316" LastEditDate="2016-05-06T09:05:40.493" LastActivityDate="2016-05-06T09:05:40.493" Title="Why can't unimolecular reactions be 2nd order?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18761" PostTypeId="2" ParentId="18760" CreationDate="2014-10-23T17:48:53.290" Score="5" Body="&lt;p&gt;A first order reaction involves the behavior of individual particles, usually their decomposition into smaller, more reactive species.  The rate of such a reaction relies solely on the number of particles present, so if we double the number of particles then the rate doubles simply because there are twice as many particles available to decompose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second order reaction involves the interaction of two particles with one another. (Even if both particles are the same it is still a bimolecular reaction.)  If we double the concentration of either species we double the likelihood of a collision between those species.  If both species are the same (ex: $\ce{2A -&amp;gt; A2}$) then the result of doubling the concentration is a quadrupling of the rate because we simultaneously double the number of particles that can interact and the number of interactions that can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An metaphorical example involving people:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unimolecular&lt;/strong&gt;:&#xA;A couple can spontaneously break up into two separate people because of its internal properties.  (Assume there are no intercouple issues involved here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C -&amp;gt; M + W}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we double the number of couples we would expect breakups to occur twice as often because there are twice as many couples to be broken up.  However, since there are no intercouple issues in this example, adding additional couples to our consideration does not increase the likelihood of existing couples breaking up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bimolecular with Two Different Species&lt;/strong&gt;:&#xA;A man and a woman can join together to make a couple if they collide in the right orientation with enough energy. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{M + W -&amp;gt; C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I increase the number of either men or women, it is more likely that I will form a couple simply because there are more interactions between men and women occurring.  Each species' increase is first order though, since they require a partner of the other gender to couple with.  i.e. Doubling the number of women only doubles the rate at which couples are formed because they still have to connect with a man, and the number of men has not changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bimolecular with a Single Species&lt;/strong&gt;:&#xA;Homosexual men can join together to form a couple.&#xA;$\ce{M + M -&amp;gt; C'}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I double the number of men, then each additional man can interact with every other man in the room, greatly increasing the likelihood of a couple forming.  This means that not only are there more men available to make couples (the first doubling of the rate), but that the existing men are also more likely to be involved in a connection that results in a new couple being formed (the second doubling of the rate.)&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-10-23T19:11:08.500" LastActivityDate="2014-10-23T19:11:08.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18762" PostTypeId="1" CreationDate="2014-10-23T18:53:26.317" Score="0" ViewCount="800" Body="&lt;p&gt;I am trying to find the structures of thiosulfate, Dithionate &amp;amp; Tetrathionate Ions but I am not getting an exact answer. Please help.&lt;/p&gt;&#xA;" OwnerUserId="9464" LastActivityDate="2014-10-23T19:51:04.293" Title="Structures of Thiosulfate, Dithionate &amp; Tetrathionate Ions" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2014-10-24T01:10:41.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="18763" PostTypeId="2" ParentId="18762" CreationDate="2014-10-23T19:30:01.280" Score="3" Body="&lt;p&gt;Here you go...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thiosulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thiosulfate&lt;/a&gt; $\ce{[S2O3]^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8jZ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There ia a negative charge on the apical sulfur and another negative charge distributed across the 3 oxygens by resonance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dithionate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dithionate&lt;/a&gt; $\ce{[S2O6]^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOuzh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one negative charge distributed by resonance over each group of 3 oxygens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dithionite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dithionite&lt;/a&gt; $\ce{[S2O4]^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/veO4Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one negative charge distributed by resonance over each group of 2 oxygens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrathionate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tetrathionate&lt;/a&gt; $\ce{[S4O6]^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QFSuf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one negative charge distributed by resonance over each group of 3 oxygens.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-23T19:51:04.293" LastActivityDate="2014-10-23T19:51:04.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18764" PostTypeId="1" CreationDate="2014-10-23T20:27:05.820" Score="2" ViewCount="10962" Body="&lt;p&gt;Is it two isomers with at least one stereogenic center each but the corresponding center on the other isomer will rotate the plane of polarized light in the opposite direction. Is this applicable to meso compounds which aren't chiral overall but have stereogenic centers?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-10-23T22:36:05.543" Title="What is a dl pair in stereochemistry?" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18765" PostTypeId="2" ParentId="16575" CreationDate="2014-10-23T21:34:54.583" Score="15" Body="&lt;h1&gt;How are the vapors transported?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The additional question from the comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why [is] the amt. of vapours taken up proportional to vapour press. Why directly proportional? Why not nothing else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First things first, let's get a definition (&lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, italics added):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Vapor pressure&lt;/strong&gt; [...] is defined as the pressure exerted by a vapor in &lt;em&gt;thermodynamic equilibrium&lt;/em&gt; with its condensed phases (solid or liquid) at a given temperature in a &lt;em&gt;closed system&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So basically, before you start the experiment (see below) the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor &quot;layer&quot; (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do &lt;strong&gt;not&lt;/strong&gt; have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also why it is directly proportional to the vapor pressure: There is no other mechanism that moves solvent molecules from the condensed phase to the vapor phase. Of course you could add a supersonic vibrating membrane and force more solvent molecules into a vapor phase, but that is not what you want to do, because you want to quantify the relative lowering of the vapor pressure.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Is relative vapor pressure lowering a thing?&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why &quot;The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I guess the question you really want to ask is: &lt;strong&gt;Why is the vapor pressure of a solution lower than the vapor pressure of a pure substance?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is not simple, but the explanation of the &lt;a href=&quot;http://chem.wisc.edu/deptfiles/genchem/sstutorial/Text11/Tx117/tx117.html&quot; rel=&quot;noreferrer&quot;&gt;vapor pressure lowering&lt;/a&gt; effect can be given by assuming a simple model: Imagine that in the solution, there are two types of molecules; the solvent and the solute. In the pure solvent, there is only one kind. Statistically speaking, that means that there is also a smaller amount of solvent molecules at the boundary between the bulk phase and the vapor. Since solute molecules don't go over into the vapor, there is a smaller area of solvent molecules that can go over into the vapor, which means that the vapor pressure of the solution is smaller when compared to the pure solvent.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What about the proportionality constant?&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can we add $\propto p_s$ and $\propto p_0−p_s$ when we don't know what the proportionality constants are or even if they are equal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While we don't know the magnitude of the proportionality constants, we know that they are equal. The proportionality constant is most likely dependent on &lt;strong&gt;ambient pressure&lt;/strong&gt;, &lt;strong&gt;temperature&lt;/strong&gt; and &lt;strong&gt;flow rate&lt;/strong&gt; of the gas. These three variables are being controlled during the experiment and kept constant for all glass bulbs. (Well, I don't know about keeping ambient pressure constant. But if you perform the experiment fast enough, that shouldn't pose much of a problem.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;The Ostwald-Walker Process&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It would be better if you could explain in detail the process because I'm not getting it properly from here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Why do we do this?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, we have to remember: Why are we doing this process; what do we want to find out? The answer is: &lt;strong&gt;We want to quantify the relative lowering of the vapor pressure.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How do we do this?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here's what happens during the process:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dry gas arrives at the first set of bulbs, filled with a solution.&lt;/li&gt;&#xA;&lt;li&gt;The dry gas gets saturated with solvent vapor.&lt;/li&gt;&#xA;&lt;li&gt;The solvent-vapor-saturated gas arrives at the second set of bulbs which contain the pure solvent.&lt;/li&gt;&#xA;&lt;li&gt;Because of the vapor pressure lowering effect (here now applied in reverse) the vapor pressure of the pure solvent is higher. As such, the gas that enters is not saturated. When it exits the second set of bulbs, it is saturated again.&lt;/li&gt;&#xA;&lt;li&gt;The pure-solvent-saturated gas enters the desiccator and loses all solvent vapor to it.&lt;/li&gt;&#xA;&lt;li&gt;Dry gas exits the apparatus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This might still be hard to understand, so here a graphic that shows you what happens to the solvent vapor partial pressure in the gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UJSn.png&quot; alt=&quot;Ostwald-Walker&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time 1 the gas enters the first bulb. It exits the first cluster of bulbs at about time 4. Then it travels to the next cluster of bulbs (containing the pure solvent) and loads up there, up until time 8. At time 9 the gas enters the desiccator where it loses all the solvent vapor and exits dry at time 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we weigh all components, we know exactly how much mass we lost from solution and solvent, and how much mass the desiccant gained (both those numbers [the total of the mass loss of course] should be equal):&#xA;$$ \Delta m_\text{solution} + \Delta m_\text{solvent} = \Delta m_\text{desicc.} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following proportionalities hold:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ \Delta m_\text{solution} \propto p_s $&lt;/li&gt;&#xA;&lt;li&gt;$ \Delta m_\text{solvent} \propto (p - p_s) $&lt;/li&gt;&#xA;&lt;li&gt;$ \Delta m_\text{desicc.} \propto p $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;relative lowering of the vapor pressure&lt;/strong&gt; can now be calculated, it is simply&#xA;$$ \frac{p-p_s}{p} = 1 - \frac{p_s}{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the vapor pressure of the pure solvent, you can simply plug it in the equation, get the proportionality constant and from that calculate the vapor pressure of the solution.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In conclusion, I can only say that I hope to have clarified things for you. I myself spent quite some time researching this method, as I have never heard of this question before. Thank you for the opportunity!&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="33991" LastEditDate="2016-10-29T12:23:15.483" LastActivityDate="2016-10-29T12:23:15.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18766" PostTypeId="2" ParentId="18764" CreationDate="2014-10-23T21:43:09.260" Score="2" Body="&lt;p&gt;The &lt;em&gt;d&lt;/em&gt;- and &lt;em&gt;l&lt;/em&gt;- nomenclature is &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29#By_optical_activity:_.28.2B.29-_and_.28.E2.88.92.29-_or_d-_and_l-&quot; rel=&quot;nofollow&quot;&gt;exactly the same&lt;/a&gt; as the (+)- and (-)- nomenclature. It is used to indicate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation&quot; rel=&quot;nofollow&quot;&gt;optical rotation&lt;/a&gt; of enantiomers. As such it is not applicable to &lt;em&gt;meso&lt;/em&gt;- compounds (or diastereomers in general), because it is not unambiguous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diastereomers still have chiral centers, and as such they exhibit optical rotation, but it is absolutely useless to specify the direction of it in the name.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that the &lt;em&gt;d&lt;/em&gt;- and &lt;em&gt;l&lt;/em&gt;- nomenclature is discouraged by IUPAC. (+)- and (-)- prefixes should be used, so as to minimize confusion with the &lt;em&gt;D&lt;/em&gt;- and &lt;em&gt;L&lt;/em&gt;- nomenclature.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-23T21:43:09.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18768" PostTypeId="2" ParentId="14787" CreationDate="2014-10-23T22:25:13.880" Score="4" Body="&lt;h1&gt;Why is $\ce{C2O4}$ unstable?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Have you tried drawing it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get four different kinds of structures for it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{(O=)O^{.}C-CO^{.}(=O)}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{(O^{.})2C-C(O^{.})2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{O#C-O-CO^{.}(=O)}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{O#C-O-O-C#O}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;They are unstable for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Biradical&lt;/li&gt;&#xA;&lt;li&gt;Tetraradical (This won't really exist anywhere near anything else.)&lt;/li&gt;&#xA;&lt;li&gt;Radical&lt;/li&gt;&#xA;&lt;li&gt;Peroxo bond, very easy to break homolytically.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these compounds can easily decompose to give (the thermodynamical Mariana trench) carbon dioxide:&#xA;$$\ce{C2O4 -&amp;gt; 2CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why it is so unstable.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Reduce 'em!&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And would adding hydrogens to a carbon oxide make the carbon oxide even more stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. If you start reducing the compounds above, you will end up with ketones and water (because geminal hydroxy groups tend to give water and ketones), or even alcohols (if you continue reducing).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="24052" LastEditDate="2016-03-11T22:25:32.843" LastActivityDate="2016-03-11T22:25:32.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18769" PostTypeId="2" ParentId="18740" CreationDate="2014-10-23T22:29:10.157" Score="2" Body="&lt;p&gt;First of all, diffraction is best thought of in reciprocal space rather than from planes of atoms, but let me answer your question directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may not understand the meaning of a reflection index. If one imagines a simple cubic lattice made up of atoms of diameter a, then the lattice dimension of the cube is also a. The spacing between (100) planes is also a. The spacing between (200) planes is a/2, as these planes have double the density of (100) planes. As we keep increasing h,k or l the planes get closer. &lt;/p&gt;&#xA;" OwnerUserId="9466" LastActivityDate="2014-10-23T22:29:10.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18770" PostTypeId="2" ParentId="18764" CreationDate="2014-10-23T22:36:05.543" Score="3" Body="&lt;p&gt;&lt;em&gt;d-&lt;/em&gt; &lt;em&gt;l-&lt;/em&gt; Nomenclature as well as (+), (-) refer to a physical property of the molecule, its optical rotation. Optical rotation is a result of chirality, a topological property of the molecule. So &lt;em&gt;d-&lt;/em&gt;, &lt;em&gt;l-&lt;/em&gt; is something that you can generally measure with a polarimeter. If you have a 1:1 mixture of the 2 enantiomers then you will see zero rotation and this mixture is called racemic and named with the dl- prefix. A &lt;em&gt;meso&lt;/em&gt; compound also has zero optical rotation since it is achiral but as it is not a (racemic) mixture the dl- notation doesn't make sense but you might see it used sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now is if you are given a pair of enantiomeric molecules that are mirror images how do you name them so you can tell them apart every time? This is why the D-, L- and R-, S- notations were introduced. D- L- is the older and more naive approach that was later improved by the R-, S- (CIP) notation. Unfortunately as chemistry is a science with long history that is also used by other non chemistry-trained professionals, several times older naming conventions have survived. So you have to know all of them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So very simply, &lt;em&gt;d-, l-&lt;/em&gt; reflect a real physical property of the molecule and D-, L- (and R-, S-) is a way to distinguish between molecules that contain stereogenic centres (and might be or not be chiral). Actually even the R-, S- system has been criticized for not being rigorous enough. I saw a paper where a new naming system was suggested but it is not possible to read this without knowledge of advanced maths (group theory, algebraic topology)   &lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-10-23T22:36:05.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18771" PostTypeId="1" AcceptedAnswerId="18783" CreationDate="2014-10-23T23:55:06.390" Score="2" ViewCount="1565" Body="&lt;p&gt;Our chemistry club has, for the last few years, been showing off some reactions to prospective students. One of the most popular tricks we have is a bit of an explosive one: we mix some potassium permanganate with ethylene glycol in a small bottle. The permanganate oxidizes the glycol, and we usually end up with an intense, three foot flame that shoots out of the bottle, followed by some popping, and finally, the whole thing burning to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently, we've been using the exact same mixtures with the same ratios and getting wildly differing results. This entire week, all we've gotten is the glycol bubbling a little and maybe spitting out some of the permanganate, and not much more. The ratios we use are almost the same, so I'm not sure what could be causing this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this reaction sensitive to age of the reactants? Ambient temperature? Humidity? What could cause the reaction to not work normally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. In spite of the home-experiment tag, the chemicals used here are lab-grade (from Sigma-Aldrich), though I'd be interested in how the purity might affect the reaction as well.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2014-10-24T04:37:24.027" Title="What factors might affect the reaction between potassium permanganate and ethylene glycol?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18774" PostTypeId="1" AcceptedAnswerId="18789" CreationDate="2014-10-24T00:36:53.133" Score="1" ViewCount="12203" Body="&lt;p&gt;Given the question &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sample of $\pu{1.50 g}$ of lead(II) nitrate is mixed with $\pu{125 mL}$ of $\pu{0.100 M}$ sodium sulfate solution. What is the limiting reactant in the reaction? Calculate the concentrations of all ions that remain in the solution after reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I first found the chemical equation for this problem to be:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Pb(NO3)2 (aq) + Na2SO4 (aq) -&amp;gt; PbSO4 (s) + 2NaNO3 (aq)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the limiting reactant to be $\ce{Pb(NO3)2}$ because there are fewer moles of that substance than the $\ce{Na2SO4}$, with there being  $0.00453 \,\mathrm{mol}\, \ce{Pb(NO3)2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then used the limiting reactant to find the concentration of each ion. I multiplied 0.00453 mol by the number of moles of each ion in the equation, and put that over 0.125 L to get the molarity. However, the correct answers are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;no $\ce{Pb^{2+}}$ ions are left in solution&lt;/li&gt;&#xA;  &lt;li&gt;$[\ce{SO4^{2-}}] = 0.0638 \,\mathrm{M}$&lt;/li&gt;&#xA;  &lt;li&gt;$[\ce{Na^+}] = 0.200 \,\mathrm{M}$&lt;/li&gt;&#xA;  &lt;li&gt;$[\ce{NO3^-}] = 0.0725 \,\mathrm{M}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I only got the concentration of $\ce{NO3^-}$ correct. What have I done wrong? I think my mistake has something to do with using the limiting reactant in the equation. Also, I understand why there is no $\ce{Pb^{2+}}$ in the solution, because it is in solid form, but doesn't that mean the $\ce{SO4^{2-}}$ should also produce no ions since it is bonded to the $\ce{Pb^{2+}}$?&lt;/p&gt;&#xA;" OwnerUserId="9281" LastEditorUserId="9281" LastEditDate="2018-04-16T02:13:10.460" LastActivityDate="2018-04-16T02:13:10.460" Title="How does a limiting reactant relate to the concentration of ions in solution?" Tags="&lt;ions&gt;&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18775" PostTypeId="2" ParentId="18760" CreationDate="2014-10-24T01:32:01.890" Score="0" Body="&lt;p&gt;This question is provocative, but there appears to be a contradiction in terms. If a reaction is unimolecular, its rate must not depend on the concentration of any other molecule including molecules of the same identity. Certainly in dilute solution, a reacting molecule in a unimolecular reaction cannot be influenced by its neighbors. At the level of stochastics there appears to be no information in the electronic (or nuclear) structure of the reacting molecule that could impart some &quot;knowledge&quot; of the presence of more or fewer neighboring molecules of the same kind. The rate law for a single molecule becomes just a probability; the only reason the macroscopic rate law is proportional to concentration is that to get the overall flux you have to apply that probability to all the molecules present. For example, the classic unimolecular, first order process is radioactive decay. Here the reaction is entirely nuclear (unless we are talking chain reactions - but those are bi-&quot;molecular&quot;), completely insulated from its neighbors by electrons, and its observed rate is first order over many orders of magnitude in initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could consider extremely crowded molecular environments such as some investigators think exists in the cell cytosol, but I think we could go all the way to solid state chemistry and not really address the question being asked because if you get the neighboring molecules close enough to influence the reaction rate in any way, you have, by definition, a bimolecular reaction. If we read the question as &quot;What prevents the reaction kinetics of a unimolecular reaction from depending on the square of the concentration?&quot;, I think the answer must be that there is no physical force that could influence the local electronic chemical interactions based on the macroscopic concentration of reacting molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One situation that almost meets the requirements of the question is the case where the unimolecular reaction is enzyme-catalyzed. In this case it's possible that the molecule itself is an allosteric activator and the kinetic law would have a term in S^2. A typical kinetic law for a reaction where S is the substrate and A is the activator is Flux = (Vm&lt;em&gt;S/Ks + b&lt;/em&gt;Vm&lt;em&gt;A&lt;/em&gt;S/(Ka&lt;em&gt;Ks)) / (1 + S/Ks + (A&lt;/em&gt;S)/(KaKs)). You can see that if A=S then the rate law has both numerator and denominator terms in S^2.&lt;/p&gt;&#xA;" OwnerUserId="9468" LastActivityDate="2014-10-24T01:32:01.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18777" PostTypeId="1" AcceptedAnswerId="18782" CreationDate="2014-10-24T02:13:34.553" Score="4" ViewCount="204" Body="&lt;p&gt;I'd imagine the answer to the question posed in the title is no - bond angles are not constant. Covalent bonds are not infinitely rigid; hence IR as a spectroscopic technique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, how can we accurately determine bond angles? What margin of error might we expect from an experimentally measured bond angle? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-24T04:05:49.073" Title="Are bond angles constant?" Tags="&lt;bond&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18778" PostTypeId="1" CreationDate="2014-10-24T02:24:29.800" Score="3" ViewCount="7174" Body="&lt;p&gt;Could someone explain in detail as to why exactly recrystallization gives a more pure solid sample? As well as why is would be better 'technique' to allow for the crystal to form slowly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that it would be because being a purification technique it essentially breaks the lattice and allows the structure to reform again. As for why it needs to be done slowly, well, because rapid changes in heat would cause the structure to reform into the impure solid it was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a more detailed, better expressed, answer to this.&lt;/p&gt;&#xA;" OwnerUserId="8260" LastEditorUserId="41328" LastEditDate="2017-08-15T00:33:00.493" LastActivityDate="2020-12-14T04:54:11.527" Title="Recrystallization and purification techniques for getting a pure sample" Tags="&lt;materials&gt;&lt;purification&gt;&lt;recrystallization&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18779" PostTypeId="1" CreationDate="2014-10-24T03:12:11.007" Score="0" ViewCount="26650" Body="&lt;p&gt;The compound in question is &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetanilide&quot; rel=&quot;nofollow&quot;&gt;acetanilide&lt;/a&gt;, and&lt;br&gt;&#xA;the solvents in question are: water, ethanol, and ethyl acetate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking all three would be suitable since upon experimentation, acetanilide dissolved in all three solvents when heated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to rank the suitability? &#xA;As, well could someone provide an explanation as to why one is more suitable than the other ? &lt;/p&gt;&#xA;" OwnerUserId="8260" LastEditorUserId="5017" LastEditDate="2014-10-24T03:42:56.000" LastActivityDate="2018-01-03T12:56:49.173" Title="What would be the most suitable solvent for a single-solvent recrystallization?" Tags="&lt;purification&gt;&lt;recrystallization&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18780" PostTypeId="2" ParentId="18779" CreationDate="2014-10-24T03:42:38.677" Score="1" Body="&lt;p&gt;It's not enough to be soluble. You want a solvent to have high solubility at high temperature, plus you want to have low solubility of the solute at low temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, in an ideal world, you'd like the solubility of any impurities to remain higher than the target compound (i.e., they'll stay dissolved while your target crystallizes out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it should be possible to rank the suitability, but that requires:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;knowing the solubility at high temperature (i.e., very high)&lt;/li&gt;&#xA;&lt;li&gt;knowing the solubility at low temperature (i.e., very low)&lt;/li&gt;&#xA;&lt;li&gt;knowing potential impurities&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two are absolutely necessary to rank different solvents. The third is useful, but sometimes hard to know completely.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-24T03:42:38.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18781" PostTypeId="2" ParentId="18779" CreationDate="2014-10-24T03:45:36.923" Score="2" Body="&lt;p&gt;The ideal properties for a recrystallization solvent are that the compound of interest is soluble at high temperature but insoluble at low temperature. You have addressed the first part, so you just have to assess the second part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that acetanilide is a non-polar compound, I would expect that it would have the lowest solubility in the most polar solvent. For this reason, water is probably the best, then ethanol. I would expect acetanilide to be readily soluble in ethyl acetate at room temperature, making it unsuitable for this recrystallization.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-10-24T04:41:39.150" LastActivityDate="2014-10-24T04:41:39.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18782" PostTypeId="2" ParentId="18777" CreationDate="2014-10-24T03:53:48.330" Score="11" Body="&lt;p&gt;No, bond angles are not constant. For example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water#Vibrational_spectrum&quot;&gt;low-energy IR absorption of water&lt;/a&gt; (at $1595 \:\mathrm{cm^{-1}}$) is from the &lt;a href=&quot;http://www.chemtube3d.com/vibrationsH2O.htm&quot;&gt;H-O-H bend&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few techniques to measure bond angles. For polar compounds, &lt;a href=&quot;http://en.wikipedia.org/wiki/Rotational_spectroscopy&quot;&gt;microwave rotational spectroscopy&lt;/a&gt; can often be used on small molecules in the gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess though, that most bond angles are measured from crystal structures using &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot;&gt;X-ray diffraction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that spectroscopic techniques always have a timescale associated with them. So any experimental measurement is time-averaged. In the case of bond angles, while the angle will change, the average value is the equilibrium - at some points, the vibration will create a larger bond angle, and at some points in time, the bond angle will be smaller.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-24T04:05:49.073" LastActivityDate="2014-10-24T04:05:49.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18783" PostTypeId="2" ParentId="18771" CreationDate="2014-10-24T04:37:24.027" Score="3" Body="&lt;p&gt;Capsules of potassium permanganate injected with ethylene glycol are used as aerial ignition devices in bush and forest firefighting. They are considered a reasonably safe option because of the delayed reponse prior to ignition. They are sometimes known as dragon eggs or fireballs. The rate of chemical reaction is very much dependent upon &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the particle size of the potassium permanganate. Powdered permanganate (greater surface area) works much better than larger crystals.&lt;/li&gt;&#xA;&lt;li&gt;the concentration of the ethylene glycol. Dilute glycol may not lead to ignition at all. Perhaps your glycol solution is very wet. &lt;/li&gt;&#xA;&lt;li&gt;ambient temperature. Unless you are trying this in the middle of winter outside in the snow, this is unlikely to be the cause of disappointment.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-24T04:37:24.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18784" PostTypeId="2" ParentId="18778" CreationDate="2014-10-24T04:40:59.633" Score="0" Body="&lt;p&gt;Crystallization is an equilibrium process where we count on the pure compound interacting with itself to be a lower energy state than interacting with impurities. If impurities become trapped in the growing precipitate, we want the system to have enough time and energy to redissolve the surface of the crystal and trapped impurities in the bulk solution. Said another way, effective crystallization achieves the lowest energy state, so we should use thermodynamic conditions: long time and high temperature (i.e., without cooling if possible).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-24T04:40:59.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18785" PostTypeId="2" ParentId="15259" CreationDate="2014-10-24T08:45:07.533" Score="1" Body="&lt;p&gt;A much more significant example of this is methyl bromide, which has historically been used as a soil fumigant. Its use has been slowed phased out due to its toxicity and ozone depletion abilities.&lt;/p&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2014-10-24T08:45:07.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18786" PostTypeId="1" CreationDate="2014-10-24T08:53:21.863" Score="5" ViewCount="6834" Body="&lt;p&gt;In the scope of the analysis of organic compounds in column-switching liquid chromatography, what does the following statement mean?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sample preconcentration techniques with high enrichment factor(EFs)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word  preconcentration is not understandable for me.&#xA;Is it a technique for extracting the components?&lt;/p&gt;&#xA;" OwnerUserId="9472" LastEditorUserId="4945" LastEditDate="2014-10-24T09:47:37.510" LastActivityDate="2014-11-02T23:08:13.717" Title="What does 'preconcentration techniques' mean?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18787" PostTypeId="2" ParentId="18754" CreationDate="2014-10-24T09:02:56.540" Score="3" Body="&lt;p&gt;The 'm' does indeed stand for metastable. The word 'isomer' in this sense refers to nuclear isomers, in which at least one of the nucleons is in an excited state. Think of it as somewhat analogous to an electron in an excited state, but occurring within the nucleus. Just like electrons release photons when an excited state decays to the ground state, when nuclear isomers decay to the ground state (which is different than 'normal' nuclear decay like $\alpha$ or $\beta$ decay), a gamma ray is release. &lt;/p&gt;&#xA;" OwnerUserId="9471" LastActivityDate="2014-10-24T09:02:56.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18788" PostTypeId="1" CreationDate="2014-10-24T09:12:28.187" Score="1" ViewCount="43" Body="&lt;p&gt;Does anyone know of groups working on water sorbtion testing and/or the use of inverse gas chromatography for characterizing Perovskites (as a class of materials)? Especially with regard to applications eg solar PV&lt;/p&gt;&#xA;" OwnerUserId="9473" LastActivityDate="2014-10-24T09:12:28.187" Title="Perovskite characterization" Tags="&lt;chromatography&gt;&lt;adsorption&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18789" PostTypeId="2" ParentId="18774" CreationDate="2014-10-24T09:29:53.353" Score="3" Body="&lt;p&gt;Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the  main &lt;strong&gt;reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$&lt;/strong&gt;.&#xA;The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of mols  of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$.&#xA;The concentration of ion nitrate is  $ 0.00906/0.125=0.0725 \ce{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go back to the main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate  $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate  $0.0125 \ce{mol}$.&#xA;So, after the reaction, there  are no  $\ce{Pb^2+}$ ions left in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the remaining $\ce{SO4^2-}$; its concentration is:  $(0.0125-0.00453)/0.125=0.06376 \ce{M}$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-24T09:29:53.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18790" PostTypeId="1" CreationDate="2014-10-24T13:09:37.233" Score="5" ViewCount="7701" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many liters of $\ce{H2(g)}$ at STP is produced per gram of $\ce{Al(s)}$ consumed in the following reaction?  $$\ce{2Al(s) + 6HCl(aq) -&amp;gt; 2AlCl3(aq) + 3H2(g)}$$&#xA;  Express your answer to four significant figures and include the appropriate units. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempted solution is included below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1.00~\mathrm{g~Al} \left( \dfrac{1~\mathrm{mol~Al}}{26.9815385} \right)&#xA;\left(\dfrac{3~\mathrm{mol~H_2}}{2~\mathrm{mol~Al}} \right)&#xA;\left( \dfrac{22.4~\mathrm{L}}{1~\mathrm{mol~H_2}}\right) = 1.245~\mathrm{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is being marked as the wrong answer.  Where is my error?&lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="5026" LastEditDate="2018-03-02T04:57:12.613" LastActivityDate="2019-03-29T23:39:16.527" Title="How many liters of hydrogen gas is produced per gram of aluminium?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18791" PostTypeId="1" AcceptedAnswerId="18796" CreationDate="2014-10-24T13:59:42.217" Score="3" ViewCount="886" Body="&lt;p&gt;As the title says. What's the point of learning everything in 2D and trying to visualize and interpret stereochemistry, newman structures, isomers etc, with hundreds of hours wasted, when a 3d visualization suite could be used to teach everything instead, without any ambiguities as to what the structure looks like? The 2D format seems completely irrelevant besides the communication of chemical structures on paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely a better system would teach everything in 3D first to gain a true visual and theoretical understanding of the subject, and ONLY THEN taught in 2D on how to simplify things for proper communication?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastActivityDate="2014-10-24T16:07:20.170" Title="Why isn't organic chemistry taught in 3D?" Tags="&lt;organic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ClosedDate="2014-10-24T21:30:58.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="18792" PostTypeId="2" ParentId="18790" CreationDate="2014-10-24T13:59:48.480" Score="6" Body="&lt;p&gt;Your work is correct.  I have to assume that this is being graded by a computer, so trivial differences between your answer and the answer that it is expecting matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible source of error 1:&lt;/strong&gt; Ideal Gas Law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using $22.4\ \mathrm{L/mol}$ they may have been expecting you to use the ideal gas law.&#xA;$$PV = nRT$$&#xA;$$101.3\ \mathrm{kPa}\times V = 0.05559\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 273.15\ \mathrm K$$&#xA;$$V = 1.246\ \mathrm L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible source of error 2&lt;/strong&gt;: Units&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computer may be looking for units in liters per gram of aluminum, something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm L}{\mathrm g} \ \text{or}\ \frac{\mathrm L}{\mathrm g\ \ce{Al}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that this is the case, but it's possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible source of error 3:&lt;/strong&gt; Typo&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either you or the person who set up the quiz made a mistake and typed in the answer incorrectly.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-02T04:39:01.057" LastActivityDate="2018-03-02T04:39:01.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18793" PostTypeId="2" ParentId="18791" CreationDate="2014-10-24T14:15:13.700" Score="12" Body="&lt;p&gt;Organic chemistry is often taught in 3D.  There have long been 3D molecular models used in classrooms, as well as Fischer projections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes 2D is enough for a given situation.  Maybe curved 4-dimensional spacetime is needed for some situations, vicinity of neutron stars or black holes lets say.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the simplist description that includes all the relavent concepts.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-10-24T16:05:40.180" LastActivityDate="2014-10-24T16:05:40.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18794" PostTypeId="1" AcceptedAnswerId="18802" CreationDate="2014-10-24T14:40:41.087" Score="0" ViewCount="511" Body="&lt;p&gt;Considering a general amphiprotic substance NaHA. I know that the maximum concentration of the specie $HA^-$ is found at $pH = \dfrac{pKa1+pKa2}{2}$. Why? How i can confirm this statment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{HA^- + H_2O &amp;lt;=&amp;gt; A^{2-} + H_3O^{+}} \quad Ka2$$&#xA;and&#xA;$$ \ce{HA^- + H_2O &amp;lt;=&amp;gt; H_2A + OH^{-}} \quad Kb=\dfrac{Kw}{Ka1}$$&lt;/p&gt;&#xA;" OwnerUserId="9475" LastEditorUserId="7920" LastEditDate="2014-10-25T17:49:59.013" LastActivityDate="2014-10-25T17:49:59.013" Title="find the pH where is maximum the concentration of a amphiprotic substance" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18795" PostTypeId="2" ParentId="18791" CreationDate="2014-10-24T14:46:48.300" Score="8" Body="&lt;p&gt;The problem it boils down to is &lt;em&gt;ease of communication&lt;/em&gt;. When you develop a new synthetic pathway, or you find the newest cancer drug, you want to tell the whole scientific community about it, and the medium for that is scientific articles, published in scientific journals after rigorous peer-review.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those articles were traditionally printed on dead trees (i.e. paper) and are now mostly just read on-screen in PDF format. With both of these cases, it is not easily feasible to include 3D visualisation more than what is already done in class on a blackboard. Wedges, bold lines and parallel perspective are used all over the world and they mean the same thing. We already have a system that works well enough, why change?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I too sometimes run into the problem that I cannot easily visualize the 3D structure of some molecule. Then I build it from a model by hand, or on the computer. Those few cases where I had to do that simply do not justify the overhead if you want to do this for every molecule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely a better system would teach everything in 3D first to gain a true visual and theoretical understanding of the subject, and ONLY THEN taught in 2D on how to simplify things for proper communication?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are absolutely free to teach organic chemistry the way you want to. But you will then see, that this generates a bunch of work for the set-up of that system (figuring out which software to use, writing guides for the students to use the software) and basically everyone needs a screen in the lecture hall. A blackboard is readily available, easily erasable and as such the best option.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-24T14:46:48.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18796" PostTypeId="2" ParentId="18791" CreationDate="2014-10-24T16:07:20.170" Score="4" Body="&lt;p&gt;The problem is that 3d is not enough either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most organic molecules can have multiple conformations, and drawing them all is impossible. However, influence of groups quickly reduces with distance growth, so only local 3d neighborhood is required most of the time, and it can be adequately described in 2d using shaded and unshaded triangles, even if needed. On the other hand, 2d sketches can be done quickly, while adequate 3d structures are not that easy to come with, and in some cases the exact 3d structure is debatable at best. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, visualisation and 3d-models for some key molecules and transition states are useful.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-10-24T16:07:20.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18797" PostTypeId="2" ParentId="18794" CreationDate="2014-10-24T16:19:15.477" Score="0" Body="&lt;p&gt;The usual answer for this kind of problem is to write down all species ($HA, A^-, H_2A^+, H_3O^+, OH^-$) and all equations for the various equilibria (don't forget autoionization!) and then find the maximum concentration of $HA$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it!  Try also various approximations.  The exercise will increase your understanding of that kind of equilibrium problems, and of buffers, titration curves, acid-base reactions in general. &lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-24T16:19:15.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18798" PostTypeId="1" CreationDate="2014-10-24T17:01:53.233" Score="17" ViewCount="35082" Body="&lt;p&gt;$\ce{N2}$ is a gas&lt;br&gt;&#xA;$\ce{P4}$ is a solid&lt;br&gt;&#xA;$\ce{N}$ and $\ce{P}$ are in the same group in the periodic table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond enthalpies are:&lt;br&gt;&#xA;$\ce{N#N}$ triple bond is $949~ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{P#P}$ triple bond is $490~ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{P-P}$ single bond is $200 ~ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N-N}$ single bond is $159-296 ~ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does $\ce{P}$ go to $\ce{P4}$ not $\ce{P2}$ and why does $\ce{N}$ go to $\ce{N2}$ and not $\ce{N4}$?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="4231" LastEditDate="2015-02-26T20:26:18.333" LastActivityDate="2018-08-04T21:23:56.490" Title="Why does Nitrogen form N₂ (and not N₄) while but Phosphorus forms P₄ (and not P₂)?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18800" PostTypeId="2" ParentId="18790" CreationDate="2014-10-24T18:12:37.297" Score="3" Body="&lt;p&gt;The answer is 1.263 One has to use 1 mol gas = 22.414 (at 0 degrees Celsius, 1 atm) = 22.711 (at STP) in the calculation. &lt;/p&gt;&#xA;" OwnerUserId="7938" LastActivityDate="2014-10-24T18:12:37.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18801" PostTypeId="2" ParentId="6193" CreationDate="2014-10-24T20:23:59.167" Score="0" Body="&lt;p&gt;In this particular case it means that the product (a peroxide, explosive!) has been phlegmatized with 25 % water.  The product is a solid, of which 25 % are water.  The dry solid has been analyzed for peroxide content, and 97 % of the dry solid is benzoyl peroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would &lt;em&gt;think&lt;/em&gt; it came in a plastic container because glass is basic and may leach ions that could catalyze decomposition of the product over time.  But that is just a hunch. &lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-24T20:23:59.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18802" PostTypeId="2" ParentId="18794" CreationDate="2014-10-24T20:45:48.180" Score="1" Body="&lt;p&gt;Let's take an aqueous solution of a salt $\ce{NaHA}$ with the initial concentration $C$ when added to water. It will completely dissociate  according to the eaquation: $\ce{NaHA(s) \rightarrow Na^+ +HA^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA^-}$ will participate in three equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2HA^- \leftrightarrows H2A +A^{2-}\quad \quad \quad  }$ ${K_1^0=K_{A2}/K_{A1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA^- +H2O\leftrightarrows H3O^+ +A^{2-}\quad }$ $K_2^0=K_{A2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA^- +H2O\leftrightarrows OH^- +H_2A^\quad }$ $K_3^0=K_{B1}={K_w}/{K_{A1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, $K_1^0$ is far bigger than $K_2^0$ and $K_3^0$. So, the first equilibrium is &lt;strong&gt;&lt;em&gt;the preponderant reaction&lt;/em&gt;&lt;/strong&gt;, and this reaction will impose the pH of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's now calculate the product $K_{A2}\times K_{A1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{A2}\times K_{A1}= \frac{[\ce{}A^{2-}].[\ce{H3O+}]}{\ce{[HA^-]}}.\frac{[\ce{}HA^{-}].[\ce{H3O+}]}{\ce{[H_2A]}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the stoichiometry  of the preponderant reaction, we have $\ce{[A^{2-}]=[H_2A^]}$. So the product&#xA;$K_{A2}\times K_{A1}= \ce{[H3O+}]^2}$ .&#xA;i.e. $\ce{pH}=0.5(\ce{p}K_{A2} +\ce{p} K_{A1})$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2014-10-24T20:50:57.607" LastActivityDate="2014-10-24T20:50:57.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18803" PostTypeId="1" AcceptedAnswerId="18808" CreationDate="2014-10-24T22:21:42.887" Score="1" ViewCount="62" Body="&lt;p&gt;Not sure if my question is better suited to a software related group, but here goes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a software program that lets you add differing amounts of elements, minerals, etc., then tells you what the items added might make?  Specifically, I am looking for information about what boron, silicon, calcium and magnesium together (no idea about amounts for each) would make.&lt;/p&gt;&#xA;" OwnerUserId="9481" LastEditorUserId="5017" LastEditDate="2014-10-25T03:26:41.257" LastActivityDate="2014-10-25T03:26:41.257" Title="Finding compounds with B, Si, Ca, and Mg in differing amounts" Tags="&lt;everyday-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18804" PostTypeId="1" AcceptedAnswerId="18807" CreationDate="2014-10-25T00:23:19.687" Score="1" ViewCount="140" Body="&lt;p&gt;Can Grignards react with diatomic halogens as to perform an substitution reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd imagine that the first part of making the reaction work - polarizing the halogen as to prepare to pull it apart - would be easy for a Grignard. Hell, even alkene pi bonds can polarize halogens sufficiently; why can't a Grignard? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in essence, can phenylmagnesiumbromide react with $\ce{Cl2}$ to make chlorobenzene? The other product of this reaction would be $\ce{Cl-}$ ... which I could see combining with the $\ce{MgBr+}$ in some way. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-09T17:00:33.557" LastActivityDate="2015-10-09T17:00:33.557" Title="Grignards and diatomic halogens" Tags="&lt;reactivity&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18805" PostTypeId="1" AcceptedAnswerId="18806" CreationDate="2014-10-25T01:18:30.960" Score="2" ViewCount="910" Body="&lt;p&gt;So, whenever there is abstraction of H&lt;sup&gt;+&lt;/sup&gt; by an acid-base reaction, I see that the proton on the sp3-hybridized carbon is taken. Why is this if you take into account that vinylic hydrogens are more acidic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, on a sidenote, what is the hybridization state of a carbon in a C=C double bond where the vinylic proton was removed by a base?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="7920" LastEditDate="2014-10-25T17:21:24.250" LastActivityDate="2014-10-25T17:21:24.250" Title="Abstraction of a proton in cycloheptatriene" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18806" PostTypeId="2" ParentId="18805" CreationDate="2014-10-25T01:45:53.843" Score="2" Body="&lt;p&gt;The pKa for a vinylic proton is ~ 43, the pKa for propene is ~ 40, and the pKa for 1,4-pentadiene is ~ 35.  Consequently it's not the vinylic protons that are the most acidic, it's the allylic ones.  Now the question whether tropylium anion is aromatic/antiaromatic is &lt;em&gt;&lt;a href=&quot;http://web.utk.edu/~rmagid/360antiaromatic.pdf&quot; rel=&quot;nofollow&quot;&gt;complicated&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-10-25T02:00:11.267" LastActivityDate="2014-10-25T02:00:11.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18807" PostTypeId="2" ParentId="18804" CreationDate="2014-10-25T02:51:12.017" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can Grignards react with diatomic halogens as to perform an&#xA;  substitution reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but it looks like a mixture of products results, including something explosive.  Here is a &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022328X00836220&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to an abstract of some early work that reports,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Chlorination of Grignard and Iotsich reagents, RMgX, (where X = Br, I) gives considerable quantities of RI and RBr together with RCl. Bromination of RMgI results in RI and RBr. Bromination of RMgCl leads to RBr, and iodination of RMgX (X =Cl, Br), to RI.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &quot;Iotsich&quot; reagent is an acetylenic Grignard ($\ce{RC#CMgBr}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by the fact that 1) the abstract suggests that product mixtures of the various possible alkyl\aryl halides are often formed and 2) no further reports claiming synthetic utlity have emerged, I suspect that the reaction is of limited synthetic value.  Nonetheless, your initial suspicion that reaction should occur seems supported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One further note, many MSDS's and &quot;&lt;a href=&quot;http://books.google.com/books?id=bLV55kunjA4C&amp;amp;pg=PA137&amp;amp;lpg=PA137&amp;amp;dq=phenylmagnesium%20bromide%20reacts%20with%20chlorine,%20explosive&amp;amp;source=bl&amp;amp;ots=Lq7PdjpWab&amp;amp;sig=qp2G1W1M1DPeFLfQHZTbkdNWbBo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=5AdLVLroFcL_yQTMo4DQAg&amp;amp;ved=0CDIQ6AEwAw#v=onepage&amp;amp;q=phenylmagnesium%20bromide%20reacts%20with%20chlorine%2C%20explosive&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Hazardous Laboratory Chemicals Disposal Guide, Third Edition&lt;/a&gt;&quot; report that when phenylmagnesium bromide is reacted with chlorine, a solid is formed that explodes when shaken.  The reference provided is the same J. Organomet. Chem. paper cited above.  Given the lack of detail provided on the solid, my guess is that it wasn't characterized.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-25T02:51:12.017" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18808" PostTypeId="2" ParentId="18803" CreationDate="2014-10-25T03:25:35.633" Score="4" Body="&lt;p&gt;This would be an exceedingly difficult problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion might be to use a database like the &lt;a href=&quot;http://www.crystallography.net&quot; rel=&quot;nofollow&quot;&gt;Crystallography Open Database&lt;/a&gt; and perform searches with the elements you indicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.crystallography.net/search.html&quot; rel=&quot;nofollow&quot;&gt;the search interface&lt;/a&gt; ensures B, Ca, Mg, and Si are all in the compound. I get 141 matches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be incredibly hard to predict compounds with arbitrary stoichiometry. There are several genetic algorithm search packages that let one attempt to find low-energy crystal structures given a formula, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://xtalopt.openmolecules.net/wiki/index.fcgi/&quot; rel=&quot;nofollow&quot;&gt;XtalOpt&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://theory.mse.cornell.edu/software/gasp.html&quot; rel=&quot;nofollow&quot;&gt;GASP&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://uspex.stonybrook.edu&quot; rel=&quot;nofollow&quot;&gt;USPEX&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is, however, an incredibly active area of research. I'm not sure if any of these packages can handle arbitrary amounts of different elements, or how they would handle missing elements. (For example, the elements you describe could be found in various borosilicate minerals &lt;em&gt;given&lt;/em&gt; that oxygen is also allowed.)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-25T03:25:35.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18810" PostTypeId="2" ParentId="18778" CreationDate="2014-10-25T04:50:29.363" Score="2" Body="&lt;p&gt;Recrystallization is a purification method because it is a &lt;em&gt;slow&lt;/em&gt; selective generation of a crystal framework, (mostly) free of trapped impurities. It is only effective as a purification method if done properly. Poor recrystallization techniques; rapid cooling, dramatic changes in solubility with temperature, bad choice of solvent, over evaporation of solvent, can all produce crystals of low purity which are not much better than the starting crop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impurities in crystals can occur via three mechanisms;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;inclusion: impurities occupy a lattice site within the crystal framework; that is, they become part of a defective crystal unit themselves&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;occlusion: impurities become physically trapped within the crystal framework. This can be a common occurence for some solvent molecules&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;adsorption: impurities form a weakly bound interaction to the crystal surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The faster the crystallization takes place, the greater the chances these mechanisms have of occurring.  Crystals are specific odd-shaped units that need to pack in nicely to their neighbours to form a good crystal. This takes time. A new molecule may attach to the growing lattice, but if it is not the right fit (such as because it is an impurity), it will leave in favour of a better suited fit attaching to the lattice. I may be showing my age here, but we used to liken crystal growth to a game of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetris&quot; rel=&quot;nofollow&quot;&gt;Tetris&lt;/a&gt;. Heck you can still &lt;a href=&quot;https://www.freetetris.org/game.php&quot; rel=&quot;nofollow&quot;&gt;play it!&lt;/a&gt; Done slowly, there is time to rotate all the pieces and get the perfect fit. As you speed things up, it becomes harder to get the right fit, holes are left, and before long the whole crystal goes to the dogs. That is essentially what is happening with rapid crystal growth. It's why you get smaller crystals and greater levels of impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there other methods for purifying solids which are effective also, if used correctly. &lt;a href=&quot;http://en.wikipedia.org/wiki/Precipitation_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;Precipitation&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sublimation_%28phase_transition%29&quot; rel=&quot;nofollow&quot;&gt;sublimation&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_crystallization_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;fractional crystallization&lt;/a&gt; are all methods for purifying solids, which can give high levels of purity, matching purities obtained be recrystallization.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-25T04:50:29.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18811" PostTypeId="2" ParentId="16703" CreationDate="2014-10-25T07:33:46.720" Score="1" Body="&lt;p&gt;Consider &lt;a href=&quot;http://www.rcsb.org/pdb/home/home.do&quot; rel=&quot;nofollow&quot;&gt;http://www.rcsb.org/pdb/home/home.do&lt;/a&gt; , that contains structures for proteing-drug complexes and &lt;a href=&quot;http://www.drugbank.ca/&quot; rel=&quot;nofollow&quot;&gt;http://www.drugbank.ca/&lt;/a&gt; . Also, don't drop plain google searches&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-10-25T07:33:46.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18812" PostTypeId="2" ParentId="16703" CreationDate="2014-10-25T08:15:49.277" Score="0" Body="&lt;p&gt;Are there any crystal structures of the free base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of the hydrochloride was published by D. W. Robertson and co-workers in  &lt;em&gt;J. Med. Chem.&lt;/em&gt;, &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;31&lt;/em&gt;, 185–189 (&lt;a href=&quot;http://dx.doi.org/10.1021/jm00396a030&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of 1:1 co-crystals of the hydrochloride with different acids was published by S. L. Childs and co-workers in &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;126&lt;/em&gt;, 13335–13342 (&lt;a href=&quot;http://dx.doi.org/10.1021/ja048114o&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://pubs.acs.org/doi/suppl/10.1021/ja048114o/suppl_file/ja048114osi20040613_103412.cif&quot; rel=&quot;nofollow&quot;&gt;CIF&lt;/a&gt; of the 1:1 co-crystal of the hydrochloride with benzoic acid is freely available.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-10-25T08:15:49.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18813" PostTypeId="1" AcceptedAnswerId="18879" CreationDate="2014-10-25T10:10:55.353" Score="16" ViewCount="5156" Body="&lt;p&gt;For the first time I'm doing &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_td.htm&quot; rel=&quot;noreferrer&quot;&gt;TD-DFT&lt;/a&gt; calculations (wB97XD functional) in Gaussian 09 for an &lt;em&gt;open-shell&lt;/em&gt; system and the results look like the hell of a mess for me. The molecule is rather big, so I started with just 10 excited states to see how it would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the ground electronic state is $^2A_2$, orbital 453 is HOMO and 454 is SOMO. Now to the excited states. In the output I found, for example, the following piece that looks strange to my eyes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;Excited State   3:  2.139-B2     0.9431 eV 1314.65 nm  f=0.0816 &amp;lt;S**2&amp;gt;=0.894    &#xA;453B -&amp;gt; 454B       0.98895  &#xA;453B &amp;lt;- 454B       0.17190&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first line is fine: it says that the third excited state is $^2B_2$ (taking spin contamination into account). &lt;/li&gt;&#xA;&lt;li&gt;The second line is also fine: it tells that the determinant corresponding to excitation from HOMO to SOMO has the largest coefficients in the CI expansion.&lt;/li&gt;&#xA;&lt;li&gt;But what is the meaning of the third line?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-12T09:09:41.550" LastActivityDate="2019-08-17T08:14:35.573" Title="Interpretation of TD-DFT results" Tags="&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;&lt;density-functional-theory&gt;&lt;td-dft&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18814" PostTypeId="1" AcceptedAnswerId="18816" CreationDate="2014-10-25T10:34:56.700" Score="10" ViewCount="19580" Body="&lt;p&gt;Can you help me understand the difference between absolute and relative configuration?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="46306" LastEditDate="2018-02-20T10:59:08.330" LastActivityDate="2021-03-31T11:32:36.123" Title="What is the difference between absolute and relative configuration in stereochemistry?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18815" PostTypeId="1" CreationDate="2014-10-25T14:29:21.307" Score="3" ViewCount="1938" Body="&lt;p&gt;When you boil tomato soup it very easily &quot;explodes&quot; when reaching the boiling point. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question has been answered here before: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4103/why-do-some-solutions-eventually-stop-boiling-and-start-popping-as-they-get-mo&quot;&gt;Why do some solutions eventually stop boiling and start &amp;quot;popping&amp;quot; as they get more concentrated?&lt;/a&gt;&#xA;However, the answer to that question does not fully answer the current question since it does not address what properties in a solution causes superheating/bumping. That answer describes what we see but does not describe WHY it happens (except for the words &quot;heated too rapidly&quot;, but many solutions can be heated with the same speed as tomato soup without &quot;exploding&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to my knowledge tomato soup only suffers the risk of &quot;exploding&quot; THE FIRST TIME it reaches the boiling point. Therefore I have thought the reason for the explosion could be that it contains gases that leave the solution violently. (I have read this is the reason that milk also may &quot;explode&quot; at the boiling point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friends if mine suggest that the reason for the explosive boiling is the viscosity of tomato soup. But this answer would not explain why tomato soup only risks &quot;exploding&quot; the FIRST time it boils. (Without extensive experience I would also say that NOT ALL viscous solutions boil explosively.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tha tomatoes are naturally peeled so their elastic peel/membrane does not come into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous question also relates to &lt;em&gt;gradual concentration&lt;/em&gt; which is not the issue of the current question. The current question relates to reaching the boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this question is not answered. To rephrase it slightly:&#xA;What properties in tomato soup causes it to boil &quot;explosively&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9485" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-09-08T21:12:12.020" Title="Why does tomato soup &quot;explode&quot; when boiled?" Tags="&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18816" PostTypeId="2" ParentId="18814" CreationDate="2014-10-25T14:53:52.300" Score="14" Body="&lt;ol&gt;&#xA;&lt;li&gt;Let's say that we have synthesized a new compound, C(1)(2)(3)(4), where (1)-(4) represent 4 different groups attached to the central carbon, so the molecule is chiral.&lt;/li&gt;&#xA;&lt;li&gt;Let's say that we find a method to separate the two enantiomers present in the new compound and that we placed the enantiomers into separate test tubes.&lt;/li&gt;&#xA;&lt;li&gt;Next we might take each enantiomer by itself and place it in a polarimeter.  We would find that the material in one test tube rotates light in the (+) direction and the material in the other test tube rotates light in an equal, but opposite direction (-).&lt;/li&gt;&#xA;&lt;li&gt;At this point, all we know is that we have two enantiomers that, as expected, rotate light in equal and opposite directions.  We don't know which test tube contains the &quot;R&quot; enantiomer or which contains the &quot;S&quot; enantiomer.  We just know their &lt;strong&gt;relative&lt;/strong&gt; configuration with respect to each other, they are mirror images of each other.&lt;/li&gt;&#xA;&lt;li&gt;To find out which is &quot;R&quot; and which is &quot;S&quot; we would either need to 1) put a crystal of one of enantiomers in an X-ray diffractometer and get a three-dimensional picture of the structure of the molecule from which we could see whether it is the &quot;R&quot; or &quot;S&quot; enantiomer or 2) carry out a series of chemical transformations where the stereochemical consequences of each step are known (e.g. does each step proceed with inversion or retention of configuration at the chiral center) and convert it to a compound of known absolute configuration&lt;/li&gt;&#xA;&lt;li&gt;Let's say we pick one of our test tubes and are able to convert that enantiomer to a chiral compound C(A)(B)(C)(D) where we know that R-C(A)(B)(C)(D) rotates light in the (+) direction.&lt;/li&gt;&#xA;&lt;li&gt;We find that the C(A)(B)(C)(D) we created from our C(1)(2)(3)(4) enantiomer rotates light in the (-) direction; therefore we have produced the &quot;S&quot; enantiomer of C(A)(B)(C)(D).&lt;/li&gt;&#xA;&lt;li&gt;For simplicity, let's say that all of the reactions we used in our conversion of C(1)(2)(3)(4) into C(A)(B)(C)(D) are known to proceed with retention of configuration.  Then we know that the test tube of C(1)(2)(3)(4) used in our reaction contains the S-C(1)(2)(3)(4) enantiomer. Now we know the &lt;strong&gt;absolute&lt;/strong&gt; configuration of our sample.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: d, l and D,L&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower case &quot;d&quot; and &quot;l&quot; can be used interchangeably with &quot;+&quot; and &quot;-&quot; respectively.  The letters stand for dextrorotatory and levorotatory.  They simply indicate how the plane of polarized light will be rotated when a sample of the material is placed in a polarimeter.  So l-2-butanol is the same as (-)-2-butanol.  Use of &quot;d&quot; and &quot;l&quot; is discouraged because it can create confusion with &quot;D&quot; and &quot;L&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper case &quot;D&quot; and &quot;L&quot; are somewhat archaic.  Way back when, the two enantiomers of glyceraldehyde were separated, purified and placed in a polarimeter.  One enantiomer rotated the light in a dextrorotatory manner; this enantiomer was called D-glyceraldehyde.  The other enantiomer rotated light in the levorotatory direction and was referred to as L-glyceraldehyde.  If you take what ever chiral molecule you are studying and convert the chiral carbon to glyceraldehyde, and it rotates light in the &quot;+&quot; direction, then you've made D-glyceraldehyde, and if all of your reaction steps involved retention of configuration (see above) then you would have the &quot;D&quot; enantiomer of your material. The terms &quot;D&quot; and &quot;L&quot; relate whatever chiral center you are studying to the chiral carbon in glyceraldehyde.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-25T18:53:41.093" LastActivityDate="2014-10-25T18:53:41.093" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18818" PostTypeId="2" ParentId="14121" CreationDate="2014-10-25T17:33:54.427" Score="2" Body="&lt;p&gt;You'd think that the main source of hydrogen is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_anode&quot; rel=&quot;nofollow&quot;&gt;sacrificial anode&lt;/a&gt; in your water heater.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-25T17:33:54.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18820" PostTypeId="1" CreationDate="2014-10-25T19:43:18.680" Score="3" ViewCount="1346" Body="&lt;p&gt;When we mix 1-bromopropane and $\ce{NaNH2}$, what happens? Which reactions wins out to make the major product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook says that primary alkyl halides give mainly SN2 products unless the base is hindered ($\ce{t-BuO-}$) (where E2 would win out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it contradicts itself when you have for example 1,2-dichlorobutane and you add three molar equivalents of $\ce{NaNH2}$ and then $\ce{NH4Cl}$ to have an alkyne. Why is there no SN2 going on in that case?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4231" LastEditDate="2014-10-26T14:48:23.697" LastActivityDate="2014-10-26T14:57:44.383" Title="NH2- and Primary Alkyl Halides: Sn2 vs E2" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18822" PostTypeId="1" CreationDate="2014-10-25T20:17:12.240" Score="2" ViewCount="351" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/gNgk4T8&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/gNgk4T8.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher took off marks because he said the reaction was SN1. Before I go confront him, please tell me if my reasoning is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, I thought: we have two possibilities:&#xA; 1. SN1: a 2°benzylic has the same stability of carbocation as a tertiary carbocation so, SN1 could work.&#xA; 2. SN2: we have a secondary carbocation and a weak base cyanide: SN2 could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I just randomly put the rate law for SN2 since the problem did not specify SN1 or SN2 and there is no way I can know which mechanism is prominent. &lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-10-27T15:41:33.770" Title="Rate Expression for Reactions Involving 2° Benzylic Alkyl Halides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18823" PostTypeId="1" CreationDate="2014-10-25T21:32:07.013" Score="5" ViewCount="1963" Body="&lt;p&gt;I don't understand why they would be so different, when the only structural difference is the orientation of hydroxyl groups.&lt;/p&gt;&#xA;" OwnerUserId="9486" LastActivityDate="2014-10-27T15:03:47.437" Title="Why do glucose, galactose and mannose differ in solubility in water?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18824" PostTypeId="1" AcceptedAnswerId="18875" CreationDate="2014-10-25T22:18:20.420" Score="5" ViewCount="1428" Body="&lt;p&gt;The following image is from 'Organic Chemistry Second Edition' by Clayden, Greeves, and Warren.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P7ZWY.jpg&quot; alt=&quot;Oxymercuration-Reduction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final step is the demercuration, but I'm unable to find a definitive mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page lists a mechanism sourced from the Journal of the American Chemical Society:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Oxymercuration_reaction#Oxymercuration.E2.80.93reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia Page for Oxymercuration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've found videos on YouTube which outline an SN2 displacement of the acetoxymercuri group by hydride. e.g.: &lt;a href=&quot;https://www.youtube.com/watch?v=arbynGlF-yk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanism of Oxymercuration Demercuration (new)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to know which mechanism is correct.&lt;/p&gt;&#xA;" OwnerUserId="7185" LastEditorUserId="7185" LastEditDate="2014-10-26T10:14:31.633" LastActivityDate="2014-10-27T12:04:43.917" Title="What is the mechanism for demercuration?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-01-06T20:12:46.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="18825" PostTypeId="1" CreationDate="2014-10-25T23:16:55.837" Score="3" ViewCount="1840" Body="&lt;p&gt;I have a the proton NMR of 2⁠&lt;em&gt;H&lt;/em&gt;-1-benzopyran-2-one (2⁠&lt;em&gt;H&lt;/em&gt;-chromen-2-one):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3wdJ.png&quot; alt=&quot;2H-1-benzopyran-2-one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 individual hydrogens and none of them are equivalent. In the spectrum I have been given there are only four peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to assign two double peaks at chemical shifts 6.43 ppm and 7.73 ppm to the two hydrogens on the alkene in the ring on the left. This leaves the four hydrogens on the benzene ring which should all be non-equivalent, but this spectrum shows them as two 2H peaks. Does this mean they have overlapped, and if so please help me figure out how. Should the multiplicities have been two ddd for the two H's in the middle and then the two H's on the ends should be a dd or ddd? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the expansion of those two 2H peaks in my spectra shown below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HAAU7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorDisplayName="user7951" LastEditDate="2016-11-09T12:57:50.123" LastActivityDate="2016-11-09T12:57:50.123" Title="Assigning 1H NMR spectrum and predicting peak coupling patterns for 2H-1-benzopyran-2-one" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18826" PostTypeId="2" ParentId="18825" CreationDate="2014-10-26T01:35:11.803" Score="5" Body="&lt;p&gt;Don't you have an integral for this spectrum? This would show you clearly that you have two sets of overlapping signals, just as you have surmised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the numbering for your molecule described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WdmGT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct with the assignment for the vinyl protons. Proton E will be strongly downfield shifted, and F only moderately shifted from the base double bond chemical shift of 5.25. These values could be determined using additive tables found in lots of NMR textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton shifts for the aromatic ring can also be roughly calculated using additive tables, but what you have are indeed overlapping signals; Protons A and C are at ~7.3 and protons B and D are at ~7.55 ppm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To predict the splitting patterns you anticipate for the aromatic system, consider first the general couplings you'd expect:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ortho coupling ~7-8 Hz&lt;/li&gt;&#xA;&lt;li&gt;meta coupling ~2-3 Hz&lt;/li&gt;&#xA;&lt;li&gt;para coupling ~1Hz or less&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's ignore the para couplings for simplicity - they are not often seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Proton A will have one ortho coupling and one meta coupling, so a will be a &lt;em&gt;dd&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Proton B will have two ortho couplings and one meta coupling, so will be a &lt;em&gt;ddd&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Proton C will have two ortho couplings and one meta coupling, so will be a &lt;em&gt;ddd&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Proton D will have one ortho couplings and one meta coupling, so will be a &lt;em&gt;dd&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Let's now build in double bond :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Protons E and F have a cis coupling (~10 Hz). &lt;/li&gt;&#xA;&lt;li&gt;Proton F will appear as a doublet, but proton E is likely to have a homoallylic type coupling to proton D (4J), splitting both of these signals to an additional doublet with a coupling of about 1-2 Hz. This means that proton D may be a &lt;em&gt;ddd&lt;/em&gt;, and proton E a &lt;em&gt;dd&lt;/em&gt; if this coupling is observable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So that should be the expected splitting pattern from this molecule. Now to justify the appearance of the peaks. Overlapping signals can often make things difficult to interpret. Without knowing what frequency this spectrum was recorded at, you also need to consider whether there are second order effects starting to manifest, certainly with the meta couplings. There will definitely be tenting of the peaks giving rise to distorted peak intensities from what would normally be expected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I wouldn't be devoting too much time to the interpretation of overlapping signals like these. One thing I will say is that if you label the peaks as (say) a &lt;em&gt;ddd&lt;/em&gt;, then you should be able to report all couplings for each &lt;em&gt;d&lt;/em&gt;. Otherwise, if you cannot report couplings (and this means couplings, not splittings) report them as a multiplet (or in this case overlapping multiplets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complex splitting patterns of second order and overlapping signals can be analysed and simulated, but it takes a lot of work.You can try drawing out a splitting diagram for each signal using the approximate couplings for ortho and meta couplings, and then start overlapping through constructive addition to see what you get. It gets very complicated very quickly. &lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-26T01:35:11.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18827" PostTypeId="1" CreationDate="2014-10-26T01:42:22.963" Score="-1" ViewCount="1856" Body="&lt;p&gt;It's impossible to produce an infinite amount products if the k constant was infinity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;no right?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastActivityDate="2014-10-26T02:05:39.167" Title="In a reaction if the equilibrium constant K was infinity, could you produce an infinite number of products at equilibrium?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18829" PostTypeId="2" ParentId="18827" CreationDate="2014-10-26T02:05:39.167" Score="5" Body="&lt;p&gt;It's impossible, law of conservation of mass and energy immediately comes to mind. All an infinite K value means is that the forward reaction will go to completion. &lt;/p&gt;&#xA;" OwnerUserId="8143" LastActivityDate="2014-10-26T02:05:39.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18832" PostTypeId="1" AcceptedAnswerId="18892" CreationDate="2014-10-26T05:36:56.953" Score="1" ViewCount="197" Body="&lt;p&gt;I know it is convention to take liquid/solid concentrations as constants and lump it with the equilibrium constant. However, if we were trying to compare the equilibrium constants of two different reactions and if one of the reactions had a solid reactant, doesn't lumping the solid reactant concentration with the equilibrium constant make one reaction seem more product favoured than it actually is?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="8117" LastEditDate="2014-10-27T15:30:19.830" LastActivityDate="2014-10-27T17:20:21.507" Title="Does Combining Solid/Liquid Concentration with $K_c$ Matter when Comparing $K_c$ of Different Reactions?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18833" PostTypeId="1" CreationDate="2014-10-26T07:15:59.717" Score="1" ViewCount="4392" Body="&lt;p&gt;Is benzene acidic? It has $sp^2$ hybridization so doen it mean that it is acidic in nature? Are alkenes acidic?                               &lt;/p&gt;&#xA;" OwnerUserId="9491" LastEditorUserId="7448" LastEditDate="2015-10-16T16:45:50.057" LastActivityDate="2015-10-16T16:45:50.057" Title="Acidic nature of benzene and other alkenes" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18834" PostTypeId="2" ParentId="6367" CreationDate="2014-10-26T08:30:14.667" Score="-1" Body="&lt;p&gt;the correct definition is &#xA;&quot;a tertiary carbon atom has 3 bonds to ANY number of carbon atoms&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the carbon atoms in acetylene are tertiary, not primary (this came as a shock to me too the first time i heard it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reason for this is, you can further improve the classification by adding the hybridisation of the C atom, so in acetylene, the carbon atoms are both SP tertiary, while in isobutane, the central carbon atom is sp3 Tertiary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if we would follow the flawed definition &quot;a tertiary carbon is bound to 3 other carbon atoms&quot; , then adding the hibridisation would be useless, as it would not alter the classification, which is why this definition is flawed.&lt;/p&gt;&#xA;" OwnerUserId="9492" LastActivityDate="2014-10-26T08:30:14.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18836" PostTypeId="1" AcceptedAnswerId="18844" CreationDate="2014-10-26T09:34:14.220" Score="8" ViewCount="89012" Body="&lt;p&gt;I understand conformation to be any spacial arrangement of atoms that can be achieved through rotation of a single bond. I understand configuration to be the disposition of atoms/groups in space. However, would a conformational isomer also be a configurational isomer since the groups are arranged differently in space?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2015-04-12T10:55:14.400" LastActivityDate="2018-11-19T03:18:36.370" Title="What is the difference between configuration and conformation in stereochemistry?" Tags="&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18838" PostTypeId="2" ParentId="8642" CreationDate="2014-10-26T10:57:55.920" Score="-3" Body="&lt;p&gt;I'm a plasma magrav physicist.  and something to try would be to take a copper tube fill it with neodymium magnets that that look like a donut. seal both ends of the copper tubing.  1 and should be permanently sealed.  nano coat the outside by using the vaporization method.  this will take about two weeks to accomplish.  once that has been done unseal the end that is not sealed perminently and remove the magnets. solder a wire to the inside of the nano coated tube however long the lead you may need. make copper oxide gas at nano state and fill the tube 30% of the way with this material, top it off with distilled water.  take a zinc wire or nickel wire and put that into the tube to make however long a lead you may need. now add a copper wire to the centerand keeping the copper wire and the nickel or zinc wire away from the walls of the copper tubeand from touching each other and seal the top with wax or epoxy or some other method to your choosing.  what this will do is it will give you the ability to pull an AC and DC voltage.  with this placed in the water with the leads coming out even partially in the water you will be able to measure the frequency of that water at that volume and then determine the properties necessary for your electrolysis to make it the most efficient.&lt;/p&gt;&#xA;" OwnerUserId="9495" LastActivityDate="2014-10-26T10:57:55.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18839" PostTypeId="1" CreationDate="2014-10-26T11:02:09.443" Score="4" ViewCount="673" Body="&lt;p&gt;It's easy to apply CIP rules to chiral carbons but what about in chiral molecules without these carbon cneters like biphenyls, sulphoxides and allenes? Are they applied at all?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-10-27T10:46:43.393" Title="How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes" Tags="&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18840" PostTypeId="2" ParentId="18815" CreationDate="2014-10-26T12:57:01.537" Score="2" Body="&lt;p&gt;I think it's probably a combination of things you mention. First, tomato soup is a suspension of solids (tomato tissue) in water. So it's the &lt;em&gt;water&lt;/em&gt; that's boiling. That water is turning into steam, which bubbles out of the water. The solid suspension likely traps some of that steam until the pressure builds and you get a big explosive bubble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I happen to cook, and make a fair bit of tomato sauce, which is very similar to tomato soup. I disagree with your statement that tomato soup explodes only once upon boiling. I see bubbles popping quite continuously, and you have to stir often to prevent this. Which happens to also be the way you prevent bumping in laboratory conditions. Cooking is probably the most comparable chemistry application most people are exposed to.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-10-26T12:57:01.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18841" PostTypeId="2" ParentId="18833" CreationDate="2014-10-26T14:30:36.057" Score="5" Body="&lt;p&gt;A sp&lt;sup&gt;2&lt;/sup&gt;-hybridized carbon is more electronegative than an sp&lt;sup&gt;3&lt;/sup&gt;-hybridized carbon, and you would expect it to be more acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments support this prediction.  Benzene has a pK&lt;sub&gt;a&lt;/sub&gt; of ~ 43, compared to alkanes, which have a pK&lt;sub&gt;a&lt;/sub&gt; of ~ 50.  You can prepare lithioaromatics from bromo/iodoaromatics and butyllithium via metal-halogen exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation with alkenes is slightly more complicated; the pK&lt;sub&gt;a&lt;/sub&gt; of vinylic protons is in the same ballpark as of aromatic protons, but allylic protons are slighty more acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consult &lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;noreferrer&quot;&gt;any&lt;/a&gt; of the &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/pkatable/&quot; rel=&quot;noreferrer&quot;&gt;usual&lt;/a&gt; pK&lt;sub&gt;a&lt;/sub&gt; tables on the web.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-26T14:30:36.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18842" PostTypeId="2" ParentId="18820" CreationDate="2014-10-26T14:57:44.383" Score="1" Body="&lt;p&gt;IIRC, Sodamide is generally insoluble, except in liquid ammonia.  And why do people like to prepare amines with ammonia as a nucleophile?  The basicity complicates things, as does the fact that the primary amine initially formed will react further to give secondary and tertiary amines, even quaternary ammonium compounds.  It's not a reaction useful in preparative chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the balance of nucleophilicity and basicity is also influenced by the solvent.  It's well known that KO&lt;em&gt;t&lt;/em&gt;-Bu acts mostly as a base in THF and mostly as a nucleophile in DMSO, there's plenty of examples in the literature.  Hydrogen-bonding solvents tip the balance towards elimination.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-26T14:57:44.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18843" PostTypeId="1" AcceptedAnswerId="18847" CreationDate="2014-10-26T15:01:18.273" Score="13" ViewCount="3485" Body="&lt;p&gt;My teacher told me that carbon's tetravalency and high catenation ability makes it special and is the reason why there are millions of compounds of carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Silicon is tetravalent too but doesn't create as many compounds as carbon.Why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The textbook says that no other element exhibits the property of catenation to the extent of carbon. Why is this so? Does it just happen magically?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="7448" LastEditDate="2015-10-02T15:50:48.983" LastActivityDate="2015-10-02T15:50:48.983" Title="What makes carbon special and versatile?" Tags="&lt;organic-chemistry&gt;&lt;elements&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18844" PostTypeId="2" ParentId="18836" CreationDate="2014-10-26T15:46:07.467" Score="8" Body="&lt;p&gt;I recommend you this webpage: &lt;a href=&quot;http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html&quot; rel=&quot;noreferrer&quot;&gt;http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html&lt;/a&gt;.&#xA;Configuration and conformation are two concepts are often used interchangeably, but represent very different things. The conformation regards the relative spatial orientation of a portion of a molecule relative to another. Thus, it is an aspect that is not directly related to the covalent bonds that are established within the molecule, but with their possible rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UC4uz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about rotation around covalent bonds, we are only referring to the single bonds, as they are the only ones that can suffer rotation. Basically this concept is easily understood if we think that the bounds work as an axis...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that when we speak of different conformations, it does not necessarily involve all the covalent bonds of a molecule, it can account only for one or few of them.&#xA; Taken together, it is possible to convert one conformation to another without &lt;strong&gt;cleaving or forming chemical bonds&lt;/strong&gt;, simply by rotating some simple covalent bonds. &lt;strong&gt;&lt;em&gt;Thermal energy at room temperature is sufficient to rotate some simple covalent bonds&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The configuration is a concept that is related to the order by which different substituents linked to the same central atom establish covalent bonds. That means, in this case it is clearly an aspect that is a direct consequence of the covalent backbone of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Od9gd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change the configuration, you &lt;strong&gt;must always cleave and form new covalent bonds.&lt;/strong&gt;..&lt;strong&gt;&lt;em&gt;This costs a lot of energy!!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the concept conformation encompasses portions of a molecule which are not directly linked to the same atom and do not involve the covalent backbone of the molecules, while the configuration comprehends parts of the molecule which are bound to the same atom, which means that there is a direct involvement of the covalent bounds of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-10-26T15:46:07.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18845" PostTypeId="1" AcceptedAnswerId="18853" CreationDate="2014-10-26T16:31:32.057" Score="3" ViewCount="47984" Body="&lt;p&gt;The answer i know is delocalized electrons. But how?&lt;a href=&quot;http://www.chemguide.co.uk/atoms/structures/giantcov.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; tells me that:&lt;img src=&quot;https://i.stack.imgur.com/1oE4i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook  tells me that one carbon atom is bonded to three others and two of these bonds are single while the third is a double bond.So all the electrons are bonded and there are no free moving electrons right? Where do these delocalized elctrons come from?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="4945" LastEditDate="2015-01-06T06:59:16.910" LastActivityDate="2015-01-06T06:59:16.910" Title="Why does graphite conduct electricity?" Tags="&lt;organic-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-11-02T14:40:43.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="18846" PostTypeId="1" AcceptedAnswerId="18856" CreationDate="2014-10-26T16:57:25.827" Score="1" ViewCount="1212" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Grignards don't add to alkyl halides to create linear carbon extensions because Grignards do transmetallation rather than SN2 displacement of halide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought transmetallation a) involved two metals and b) involved ligand exchange between the two metals … correct me if I'm wrong, but if I have a Grignard and an alkyl halide, I only have one metal.  &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-16T17:13:45.133" LastActivityDate="2015-10-16T17:13:45.133" Title="Grignards do transmetallation?" Tags="&lt;reactivity&gt;&lt;halides&gt;&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18847" PostTypeId="2" ParentId="18843" CreationDate="2014-10-26T17:02:19.350" Score="16" Body="&lt;p&gt;There are two key factors that account for the ubiquity of carbon compounds.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bond strengths:&lt;/strong&gt; Look at the following table of bond strengths and notice how the strength of both carbon-carbon single and double bonds is much greater than the bond strengths found in other molecules.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;\begin{array}\hline&#xA;Typical~Single ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline&#xA;\end{array}&#xA;\begin{array}{|c|c|c|}\hline&#xA;C-C &amp;amp; 83    \\ \hline&#xA;N-N &amp;amp; 38    \\ \hline&#xA;O-O &amp;amp; 35    \\ \hline&#xA;Si-Si &amp;amp; 52    \\ \hline&#xA;P-P &amp;amp; 50   \\ \hline&#xA;S-S &amp;amp; 54    \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}\hline&#xA;Typical~Double ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline&#xA;\end{array}&#xA;\begin{array}{|c|c|c|}\hline&#xA;C=C &amp;amp; 146    \\ \hline&#xA;N=N &amp;amp; 109    \\ \hline&#xA;O=O &amp;amp; 119    \\ \hline&#xA;P=P &amp;amp; 84   \\ \hline&#xA;\end{array}&#xA;&lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot;&gt;link to reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers tell us that carbon-carbon bonds are significantly &lt;strong&gt;more stable&lt;/strong&gt; than other bonds and that &lt;strong&gt;the driving force for their formation will be more exothermic&lt;/strong&gt;.  Therefore, large collections of carbon atoms can be bonded together to produce very stable systems.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hybridization:&lt;/strong&gt; Saying that carbon is tetravalent doesn't fully convey the versatility carbon can display in forming bonds. Hybridization involves a re-mixing of s- and p-orbitals in response to bond formation and this leads to the formation of more stable bonds. Carbon is capable of extensive variation in the hybridization of the bonds it forms, much more so than most other atoms.  This wide range in hybridization that is possible for carbon allows it to form much more stable bonds in a wide variety of different structural situations.&lt;br&gt;&#xA;\begin{array}\hline&#xA;Typical~Carbon-Carbon ~Bond~ Strengths~ (kcal/mole)\\ \hline&#xA;\end{array}&#xA;\begin{array}{|c|c|c|}\hline&#xA;alkane &amp;amp; sp^3 &amp;amp; 84    \\ \hline&#xA;alkene &amp;amp; sp^2 &amp;amp; 150    \\ \hline&#xA;alkyne &amp;amp; sp &amp;amp; 200    \\ \hline&#xA;\end{array}&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-26T17:02:19.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18848" PostTypeId="2" ParentId="18846" CreationDate="2014-10-26T17:18:34.887" Score="2" Body="&lt;p&gt;You'd think that reaction of an alkyl grignard with a (primary) alkyl bromide/iodide gave all sorts of grignards via metal/halogen exchange.  I guess that's what your professor means by transmetallation.  You'd also get some elimination products.  Certainly that's what would happen with alkyllithiums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why for carbon-carbon bond formation reactions you'd use cuprates.  Reduced basicity, increased nucleophilicity, all that.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-26T17:18:34.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18850" PostTypeId="2" ParentId="10938" CreationDate="2014-10-26T19:15:45.387" Score="1" Body="&lt;p&gt;You are looking for the Clausius-Clapeyron equation.  It comes in various forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact version is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm dp}{\mathrm dT}=\frac{\Delta H}{\Delta V T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, since the volume of the gas phase is so much larger than the volume of the condensed phase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm dp}{\mathrm dT}=\frac{p \Delta H}{R T^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or even more succinctly, assuming $\Delta H$ remains constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d \ln p}{\mathrm d \frac{1}{T}}=-\frac{\Delta H}{R}$$&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorDisplayName="user7951" LastEditDate="2017-01-16T11:28:18.670" LastActivityDate="2017-01-16T11:28:18.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18851" PostTypeId="2" ParentId="14467" CreationDate="2014-10-26T20:00:44.637" Score="0" Body="&lt;p&gt;We had what we would call &quot;filtersticks&quot;, they were a cannula with a metal cylinder with a bore at one end, around which we would tie glass filter paper.  Larger quantities we transfer through polyethylene tubing to a filtration column and filter through a pad of Celite.  Both methods worked well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your product doesn't sound air-sensitive, you might consider quenching with wet THF.  Have an ice bath and a fire extinguisher ready.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-26T20:00:44.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18853" PostTypeId="2" ParentId="18845" CreationDate="2014-10-26T21:45:45.337" Score="7" Body="&lt;p&gt;Carbon has four spare electrons. In graphite structure, two of them make two single (sigma) bonds while the other pair form a double bond (consists of one sigma and one pi bond). Electron that forms pi bond is freer to travel between atoms compared to the others and is theoretically associated with all adjacent atoms (imagine the 3d structure of graphite) rather than a single atom. Because of that, it is called delocalized electron. In other words, electrons that form pi bonds continously replace each other causing temporary, local crystallizations within the structure. According to Wikipedia &quot;It is increasingly appreciated that electrons in sigma bonding levels are also delocalized&quot;. But it is not the subject we are discussing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric current is basically the flow of charged particles (ions and electrons, but electric currents in everyday life can simply be defined as flow of electrons). If graphite is exposed to an electric current in a circuit, &quot;the pi electrons&quot; which are already traveling between carbon atoms that forms the graphite structure begin to flow through the circuit and leave the structure from one point of connection between graphite and the circuit. At the same time, other electrons come via the other connection point to replace them as new delocalized electrons. So, the new delocalized electrons simply come from the current.&lt;/p&gt;&#xA;" OwnerUserId="9474" LastActivityDate="2014-10-26T21:45:45.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18854" PostTypeId="1" CreationDate="2014-10-26T22:29:16.690" Score="6" ViewCount="765" Body="&lt;p&gt;I have several old coffee stains on a carpet. Googling &quot;coffee stain removal&quot; reveals a variety of suggested cleaning substances-- vinegar, baking soda, laundry detergent, etc. I have used all of these on stains in the past to varying success, but always have a hard time totally eliminating the stain. Typically the instructions suggest iterative applications of the substance, and I usually end up with a huge pile of wet, wadded-up paper towels, each tinged slightly brown from the tiny amount of coffee absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the most effective way to remove old coffee stains from carpet using common household chemicals? Chemically, why is this the most effective means?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="289" LastActivityDate="2018-09-07T04:33:43.243" Title="Most effective means of removing coffee stains from carpet" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18855" PostTypeId="1" CreationDate="2014-10-26T22:35:53.983" Score="2" ViewCount="2490" Body="&lt;p&gt;One description of the concept is that heat of reaction (exothermic) is the quantity of heat that would given off to the surrounding by the product of a reaction if we allow the product to cool down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this description is valid for all exothermic reaction in a isolated system, would it also be valid for all endothermic reactions??&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="5456" LastEditDate="2014-10-28T06:24:22.147" LastActivityDate="2014-10-28T06:24:22.147" Title="Concept of heat of reaction in a isolated system" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18856" PostTypeId="2" ParentId="18846" CreationDate="2014-10-26T22:47:34.110" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Prof told me Grignards don't add to alkyl halides&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Grignards can react with alkyl halides to form the expected coupling product.  However yields are usually very low except under special conditions.  Improved yields (but yields that would still be considered poor for a synthetic route) result when the halide involved is a good leaving group (I ~ Br &gt; Cl), or when the Grignard is made from a stabilized &quot;R&quot; group such as allyl or benzyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an interesting &lt;a href=&quot;http://www.mdpi.org/ecsoc/ecsoc-7/papers/a019/a019.htm&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt; where the reaction of an aryl Grignard reagent with an alkyl halide was examined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When a THF solution of p-tolylmagnesium bromide 1a and an equimolar amount of 1-iodoheptane under nitrogen was stirred for 48 h at room temperature, only a very small amount (7% by GC) of the expected coupling product, 1-p-tolylheptane, was formed.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in principle, Grignard reagents are capable of reactions that look like $\ce{S_{N}2}$ reactions (who knows what the mechanism is when a metal is involved), the yields are just very low.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-26T22:47:34.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18857" PostTypeId="1" CreationDate="2014-10-27T01:02:34.610" Score="1" ViewCount="214" Body="&lt;p&gt;Concerning the liberation of energy when an atom that is close to the configuration of a noble gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the energy dissipated from an atom upon receiving an electron come from?&lt;/p&gt;&#xA;" OwnerUserId="9501" LastEditorUserId="5017" LastEditDate="2014-10-27T01:21:52.530" LastActivityDate="2015-01-25T16:34:08.990" Title="Electron affinity" Tags="&lt;electrons&gt;&lt;energy&gt;&lt;noble-gases&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18860" PostTypeId="1" AcceptedAnswerId="18864" CreationDate="2014-10-27T02:35:45.420" Score="0" ViewCount="59" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYkpi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the last step in there is just something to throw us off. From what I understand, the first step will give us a Grignard, which will attack the carbonyl in formaldehyde to give us a &lt;em&gt;primary&lt;/em&gt; alcohol, and the workup is just to neutralize any remaining Grignard and of course protonate the oxide ion that was just formed. I don't see any possibility of an intramolecular attack either ...... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The H2/Pt step looks like a &lt;em&gt;serious&lt;/em&gt; red herring from here; I know it doesn't touch the phenyl substituent. And there are no other pi bonds in this system for me to even consider. So is the last step just something to throw me off? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-27T03:34:50.817" Title="Synthesis; effect of hydrogen gas and platinium" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18861" PostTypeId="1" AcceptedAnswerId="18865" CreationDate="2014-10-27T02:53:26.563" Score="0" ViewCount="1167" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/auZ8b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the LAH will reduce the nitrile into a primary amine. The addition of acid will likely destroy the acetal mask and leave us with ethylene glycol and a molecule with both a ketone and an amine functional group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what does the sodium cyanoborohydride do here with the acid and heat? I'm again thinking this is a red herring; $\ce{NaBH3CN}$ only reduces iminium ions as far as we are concerned... and we have an amine. Amines don't exist in equilibrium with imines. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-16T17:49:49.983" LastActivityDate="2015-10-16T17:49:49.983" Title="Sodium cyanoborohydride and iminium reductions" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18864" PostTypeId="2" ParentId="18860" CreationDate="2014-10-27T03:34:50.817" Score="1" Body="&lt;p&gt;The selective debenzylation of benzyl ethers can be carried out under very mild conditions, leaving other functional groups intact. Commonly carried out under H2 using a heterogeneous catalyst such as Pd/C or Raney Ni, Pt can also be used. Solvent choice is critical, and reactions times are longer, but it is still effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like red herrings.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-27T03:34:50.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18865" PostTypeId="2" ParentId="18861" CreationDate="2014-10-27T03:50:22.583" Score="2" Body="&lt;p&gt;I believe the reaction effected in the third step is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reductive_amination&quot; rel=&quot;nofollow noreferrer&quot;&gt;reductive amination&lt;/a&gt;. In this case, the ketone and amine moieties are both present within the same molecule, so the reaction proceeds intramolecularly. The $\ce{TsOH}$ is an acid, presumably serving to protonate the carbonyl oxygen and activate the carbon toward nucleophilic attack by the amine. As the iminium group forms (with concomitant loss of water), it is rapidly reduced to an amine by the $\ce{NaBH3CN}$. After workup, the final product is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ltsp9.png&quot; alt=&quot;Bicyclic amine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="305" LastActivityDate="2014-10-27T03:50:22.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18866" PostTypeId="2" ParentId="18798" CreationDate="2014-10-27T04:04:47.930" Score="8" Body="&lt;p&gt;Due to larger atomic size $\ce{P}$ is unable to form pi bonds and so it is tetra-atomic in which each $\ce{P}$ atom is linked with 3 other $\ce{P}$ atoms by 3 sigma bond. But, due to smaller atomic size $\ce{N}$ forms 1 sigma and 2 pi bonds i.e. triple bonds with other $\ce{N}$ atom and exists as diatomic molecule.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mYD1G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CZsC9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q9kM3.png&quot; alt=&quot;Thats why Phosphorus exists as $P_4$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;a href=&quot;http://philmintz.tripod.com/Chemistry/page4.html&quot; rel=&quot;noreferrer&quot;&gt;http://philmintz.tripod.com/Chemistry/page4.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="73" LastEditDate="2015-02-27T00:35:16.267" LastActivityDate="2015-02-27T00:35:16.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18867" PostTypeId="2" ParentId="18861" CreationDate="2014-10-27T04:07:48.333" Score="1" Body="&lt;p&gt;Cyanoborohydride has a number of functions as a selective reducing agent, which are very much pH dependant:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reduction of aldehydes and ketones. &lt;em&gt;At low pH (3-4)&lt;/em&gt;, it effectively reduces aldehydes and ketones to the parent hydrocarbon.&lt;/li&gt;&#xA;&lt;li&gt;Reductive amination of aldehydes and ketones. &lt;em&gt;At pH 6-8&lt;/em&gt;, aldehydes and ketones undergo reductive amination&lt;/li&gt;&#xA;&lt;li&gt;Reduction of oximes, at low pH&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have a close look at the pH of your system.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-10-27T04:07:48.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18869" PostTypeId="2" ParentId="18353" CreationDate="2014-10-27T04:29:16.127" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The major product formed is -  Tetraphosphorus decaoxide ($\ce{P4O10}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Infact you could have googled it. There are many interesting videos on youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this out -  &lt;a href=&quot;https://www.youtube.com/watch?v=CsvSUAwq0uM&quot; rel=&quot;nofollow&quot;&gt;Burning of Red Phosphorus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="5456" LastEditDate="2014-10-28T06:02:58.903" LastActivityDate="2014-10-28T06:02:58.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18870" PostTypeId="1" AcceptedAnswerId="18873" CreationDate="2014-10-27T06:51:44.703" Score="19" ViewCount="23405" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;I am asking this question on behalf of my seven year old daughter, so if possible please take that into consideration in the answer!&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemist, but I do try to teach my daughter the basics of diverse subjects. I explained to her that the right-most elements on Mendeleev's chart do not generally react with other elements as they have all the electrons that they need. Those to the left of them need a single electron to be satisfied, those another column to the left need &lt;em&gt;two&lt;/em&gt; electrons, and so on. Additionally, those elements on the left of the chart &lt;strong&gt;have an extra electron that they want to be rid of&lt;/strong&gt;, and those one column to the right of them have &lt;em&gt;two&lt;/em&gt; electrons to get rid of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this basis she understands why water has a single oxygen atom, but &lt;em&gt;two&lt;/em&gt; hydrogen atoms. Then she asks my why hydrogen peroxide has only two hydrogen atoms. What satisfies the second oxygen atom? I don't know myself, I'm not a chemist!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does hydrogen peroxide use a different type of bond? Is the electron-swapping somehow satisfied for hydrogen peroxide in a different way?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling around I discovered &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-why-cant-it-exist-with-atoms-other-than-f&quot;&gt;hydrogen bonding&lt;/a&gt; but this doesn't seem to work with Oxygen. What else should I be reading? I've googled and found many &quot;Chemistry for Kids&quot; websites, but they seem to be either (fun) experiments, or explanations that are no more simple than those in wikipedia and obviously not aimed at children.&lt;/p&gt;&#xA;" OwnerUserId="6987" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-10-10T13:48:58.773" Title="How does H2O2 bond" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18872" PostTypeId="1" CreationDate="2014-10-27T09:43:18.837" Score="1" ViewCount="50" Body="&lt;p&gt;Thermally evaportated $\ce{Al}$ under high vacuum is free of oxidized species and highly active. But is it active enough to even substitute the alkali cationic metal in alkali salts, like $\ce{LiOH, NaOH, Cs2CO3}$.&#xA;Here $\ce{Al}$ is directly depositely onto the salt substrate.&lt;/p&gt;&#xA;" OwnerUserId="9507" LastEditorUserId="4231" LastEditDate="2014-10-27T13:10:21.747" LastActivityDate="2014-10-27T14:56:38.100" Title="Is atom-clean aluminum active enough to react with alkali salts?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18873" PostTypeId="2" ParentId="18870" CreationDate="2014-10-27T09:47:35.407" Score="22" Body="&lt;p&gt;Time's up for you to explain &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;Covalent Bond&lt;/a&gt; to your daughter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting rid of, or capturing an electron (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_bonding&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic bond&lt;/a&gt;) is not the only way that atoms can get their outermost electron shells filled. If it were the only way, we would not have H&lt;sub&gt;2&lt;/sub&gt; or O&lt;sub&gt;2&lt;/sub&gt;. Atoms can share an electron (or more) with another atom, if that fills the shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, H&lt;sub&gt;2&lt;/sub&gt; does not bond by giving one Hydrogen atom's electron to another. When two Hydrogen atoms get together, they share the two electrons between them (little dots denote electrons).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Before bonding                   After bonding&#xA;H ·      · H                          H : H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, each Hydrogen atom has two electrons to enjoy, and their shells are (sort of) filled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, H&lt;sub&gt;2&lt;/sub&gt;O isn't made by Oxygen capturing Hydrogen's electrons. The H--O bond is a covalent bond as well (hope you can see the electron pairs &lt;code&gt;¨&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Before bonding                  After bonding&#xA;         ¨                                ¨&#xA;H ·    · O ·      · H                 H : O : H &#xA;         ¨                                ¨&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;H&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;2&lt;/sub&gt; has a covalent bond between the two Oxygen atoms&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;           ¨   ¨&#xA;       H : O : O : H &#xA;           ¨   ¨&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;See the structure of Hydrogen Peroxide from &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_peroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YxIr1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br/&gt;&#xA;&lt;sup&gt;source: &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Hydrogen-peroxide-3D-balls.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/File:Hydrogen-peroxide-3D-balls.png&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9506" LastEditorUserId="9506" LastEditDate="2017-10-10T13:48:58.773" LastActivityDate="2017-10-10T13:48:58.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18874" PostTypeId="2" ParentId="18839" CreationDate="2014-10-27T10:46:43.393" Score="3" Body="&lt;p&gt;Structures with axial chirality (e.g. an allene or a 2,2′,6,6′-tetrasubstituted biphenyl) with two pairs of substituents (ab and cd) are regarded as elongated tetrahedra and viewed along the chiral axis. The substituents (a, b, c, and d) are arranged in the elongated tetrahedral system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each pair (ab as well as cd), one higher-ranking substituent (a &gt; b and c &gt; d) is chosen using the sequence rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the path going from a to b to c (while looking toward d) is clockwise, the chirality is described by the symbol &lt;em&gt;R&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;. The symbol is &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; if this path is anticlockwise.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-27T10:46:43.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18875" PostTypeId="2" ParentId="18824" CreationDate="2014-10-27T12:04:43.917" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.google.ru/books?id=mbAh-X4bQgYC&amp;amp;lpg=PA1&amp;amp;ots=v7X37hwKnk&amp;amp;dq=NaBH4%20%20demercuration%20mechanism&amp;amp;lr&amp;amp;pg=PA36#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Apparently, the current assumption implies formation of $\ce{R-Hg-H}$ intermediate, followed by radical chain decomposition&lt;/a&gt; This, however, should be taken with a grain of salt, i.e. very skeptically. Proving mechanism is an incredible amount of work, and considering relative unimportance of organomercury chemistry in modern organic synthesis, it is unlikely any considerable resources will be allocated to the task. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-10-27T12:04:43.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18878" PostTypeId="1" CreationDate="2014-10-27T13:57:09.540" Score="2" ViewCount="136" Body="&lt;p&gt;I want to estimate rates of substitution reaction of esters with nucleophiles in water.&#xA;How can I estimate them with a computational chemistry software (preferably GAMESS(US))?&#xA;Or is it practically impossible?&#xA;I know that this is possible in gas phase, but I don't know how to do it in solution.&lt;/p&gt;&#xA;" OwnerDisplayName="user7013" LastActivityDate="2014-10-27T14:36:06.327" Title="How can I estimate reaction rates in water with computational chemistry?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18879" PostTypeId="2" ParentId="18813" CreationDate="2014-10-27T14:04:05.313" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Time-dependent_density_functional_theory&quot;&gt;Time-dependent DFT&lt;/a&gt; can be used to predict excitation energies through a linear-response formulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_td.htm&quot;&gt;Gaussian result&lt;/a&gt;, beyond the first line, you are looking at the largest coefficients in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Configuration_interaction&quot;&gt;configuration-interaction (CI)&lt;/a&gt; style expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It's not strictly CI, but the implementation of time-dependent HF or &lt;a href=&quot;http://en.wikipedia.org/wiki/Random_phase_approximation&quot;&gt;RPA&lt;/a&gt; is essentially the same for TDDFT or Tamm-Dancoff Approximation TDDFT.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, since you're performing a calculation on an open-shell system, the formulation for the state will include all one-electron excitations &lt;strong&gt;&lt;em&gt;AND&lt;/em&gt;&lt;/strong&gt; all one-electron de-excitations. That state has a de-excitation from 454B into 453B (i.e., the SOMO will fill the hole left by the excitation). Now you may think &quot;but that doesn't accomplish anything.&quot; Remember that Gaussian only prints the dominant coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, Rich Martin created the &lt;a href=&quot;http://dx.doi.org/10.1063/1.1558471&quot;&gt;Natural Transition Orbitals method&lt;/a&gt; as a nice way to express the excitation, rather than a set of coefficients and the ground-state orbitals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oftentimes, there is no dominant configuration in the list of excitation amplitudes, thereby making a straightforward interpretation of the excited state difficult. This is particularly unsatisfactory when attempting to determine the qualitative nature of an excited state. An additional complication in the DFT case is that in principle all orbitals in DFT but the HOMO are devoid of physical significance. However, chemical intuition is built on the orbital construct, and a simple orbital interpretation of &quot;what got excited to where&quot; is important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is available in Gaussian from the &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_population.htm&quot;&gt;Population&lt;/a&gt; keyword and in other codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, there's a nice review of TDDFT in: &lt;a href=&quot;http://dx.doi.org/10.1146/annurev-physchem-032511-143803&quot;&gt;&quot;Progress in Time-Dependent Density-Functional Theory&quot; &lt;em&gt;Annual Review of Physical Chemistry&lt;/em&gt;&#xA;Vol. 63: 287-323.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-27T14:04:05.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18881" PostTypeId="2" ParentId="18857" CreationDate="2014-10-27T14:14:44.427" Score="2" Body="&lt;p&gt;Well everything in nature wants to or tries to attain a configuration where it is at its lowest energy.&#xA;When an atom receives an electron and completes its octet it becomes more stable or in other words we can say that it has acquired a stable configuration.&#xA;The energy released was always present in the species before bonding took place. &lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2014-10-27T14:14:44.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18882" PostTypeId="2" ParentId="18878" CreationDate="2014-10-27T14:36:06.327" Score="1" Body="&lt;p&gt;Finding solvent  effect in computational chemistry is still a crucial part and subject to deep research effort. Now there are two types of models that are being used in computational chemistry.&lt;br&gt;&#xA;One is explicit solvation model: Here you will use explicit solvent molecule with your reactant or products and then using first principles you will find the equilibrium configuration but the problem is, it is very expensive and the configuration space is so much high that it is practically impossible. For that you have to use molecular dynamics or monte carlo techniques. In our group we use 1000 to 5000 explicit water molecule but we can't do it for large reaction.&lt;br&gt;&#xA;Another method is implicit solvation model: Here electrostatic potential is applied to account the solvent effect but this procedure is so expensive that now that potential is replaced by the solvent induced reaction field energy.&lt;br&gt;&#xA;In our group we are using a hybrid approach that involves communication of several software at different level of theory. In my quantum chemistry class i used spartan and qchem and they have some implicit solvation but i didn't use GAMESS.&lt;br&gt;&#xA;From GAMESS manual i found that they have solvation module. They are using QM/MM approach. That's actually a very good approach. Please see page 9 of this file:  &lt;a href=&quot;http://www.sdsc.edu/~kimb/gamesstut.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sdsc.edu/~kimb/gamesstut.pdf&lt;/a&gt;&lt;br&gt;&#xA;In this following file at page 3 &amp;amp; 4 you will also find the list of keywords that are needed to start a solvation model run&lt;br&gt;&#xA;&lt;a href=&quot;http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2014-10-27T14:36:06.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18883" PostTypeId="2" ParentId="18872" CreationDate="2014-10-27T14:56:38.100" Score="1" Body="&lt;p&gt;Even normal metallic aluminum (with its spontaneously-formed corrosion-resistant oxide coating) reacts with concentrated sodium or potassium hydroxide to give the aluminate (e.g. NaAl(OH)4), plus hydrogen gas. Lithium hydroxide should also react (lithium is likely functioning as only a spectator ion) whereas alkali metal carbonates may not be sufficiently reactive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is not what you refer to in your question as &quot;substitution&quot; reactions, where the alkali metal ion would be reduced to metallic form. Note that aluminum chloride is reduced by sodium or potassium metal to give aluminum metal, so the reverse reaction is unlikely to occur, but the energetics may be different for the hydroxides or carbonates in a non-aqueous environment.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-27T14:56:38.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18884" PostTypeId="2" ParentId="18823" CreationDate="2014-10-27T15:03:47.437" Score="4" Body="&lt;p&gt;The solubility of these sugars will depend on the relative energies of the crystalline form versus the dissolved form of the sugar. Without knowing the specifics, one could surmise that the crystal structures of these three sugars may differ substantially from each other - one may be more tightly or efficiently packed into a crystal lattice than another due to the specific shape of the molecule, resulting in a lower energy state than that of the other sugar's crystal form, and hence giving a lower solubility in water.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-27T15:03:47.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18885" PostTypeId="1" AcceptedAnswerId="18888" CreationDate="2014-10-27T15:41:15.733" Score="1" ViewCount="323" Body="&lt;p&gt;In lecture we discussed the nature of the nucleophile for the attack of the carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LiAlH4 acts as a better nucleophile (hydride) than NaBH4. I know there is no hydroboration! So I would have expected that LiAlH4 gives more Felkin Anh product as the NaBH4 gives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However LiAlH4 gives a Felkin Anh : Anti Felkin Anh ratio of 3:1 and&#xA;NaBH4 gives a ratio of 5:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/xv7mlhooiu46x9d/2014-10-27%2016.21.56.jpg?dl=0&quot; rel=&quot;nofollow&quot;&gt;detailed reaction&lt;/a&gt;&#xA;Source: Carreira, E.M.; Classics in Stereoselective Synthesis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good rational?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastActivityDate="2014-10-27T16:38:07.610" Title="Why gives NaBH4 more Felkin-Anh product than LiAlH4 gives?" Tags="&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18886" PostTypeId="2" ParentId="18822" CreationDate="2014-10-27T15:41:33.770" Score="1" Body="&lt;p&gt;My inclination is that you are correct and the reaction proceeds by an SN2 reaction. Secondary alkyl halides can react by either an SN1 or SN2 mechanism, so the other reaction conditions are important to make a decision of the predominant pathway. Since cyanide is a strong nucleophile (and often utilized as the prototypical SN2 nucleophile), that suggests that the reaction should go through an SN2 pathway. One other piece of information that we don't have is the solvent. If this is run in a protic solvent, that could tip the pathway toward SN1, while polar aprotic will favor SN2. I also would not be surprised if both pathways are operating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the aromatic ring will stabilize a carbocation at the benzylic position, but it will also stabilize the transition state of the SN2 reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief literature search has revealed less insight than I expected. The closest result I have been able to find is that an optically active benzylic bromide treated with trimethylsilyl cyanide in acetonitrile (polar aprotic solvent) does give an optically active benzylic cyanide, suggesting SN2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOI: &lt;a href=&quot;http://dx.doi.org/10.1055/s-0030-1260187&quot; rel=&quot;nofollow&quot;&gt;10.1055/s-0030-1260187&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-27T15:41:33.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18887" PostTypeId="1" AcceptedAnswerId="18890" CreationDate="2014-10-27T16:04:07.243" Score="1" ViewCount="396" Body="&lt;p&gt;Why is the pi star orbital of the C=C bond higher in energy than the antibonding pi orbital of the C=O bond? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it was the other way around; it's relatively easy to add something to a C=C alkene bond (i.e. halogenation) but halogenation with diatomic halogens doesn't occur with C=O bonds. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-27T16:52:51.480" Title="C=O and C=C antibonding orbitals" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18888" PostTypeId="2" ParentId="18885" CreationDate="2014-10-27T16:09:56.760" Score="2" Body="&lt;p&gt;Lithium aluminum hydride is such a reactive nucleophile that it reacts readily with any conformation of the ketone. The reduced nucleophilicity of sodium borohydride allows for some discrimination of the conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the difference in selectivity between aluminum hydride and borohydride small. A large boost in stereoselectivity can be achieved by increasing steric bulk of the reducing agent (L-selectride gives a 10-fold increase in selectivity).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-10-27T16:38:07.610" LastActivityDate="2014-10-27T16:38:07.610" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18889" PostTypeId="1" CreationDate="2014-10-27T16:43:20.247" Score="0" ViewCount="37627" Body="&lt;p&gt;For a metal cup as opposed to a styrofoam cup calorimeter, do you expect the enthalpy of the reaction to be higher, lower, or the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the enthalpy of the reaction would be the same in the sytrofoam cup as the metal cup because enthalpy is a state function? So if the same products are yielded at the end along with the same reactants being used in the beginning, they'll have the same enthalpy because it doesn't matter HOW you got that enthalpy, just the final result? That's my reasoning but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does it matter that metal cup will conduct heat better because if they're using the same reactants it'll give you the same change as a styrofoam cup?&lt;/p&gt;&#xA;" OwnerUserId="9512" LastEditorUserId="7960" LastEditDate="2014-10-27T17:22:40.727" LastActivityDate="2017-11-09T03:09:44.727" Title="Metal vs. Styrofoam cup calorimeter?" Tags="&lt;homework&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18890" PostTypeId="2" ParentId="18887" CreationDate="2014-10-27T16:52:51.480" Score="5" Body="&lt;p&gt;It's relatively easy to add an electrophilic species, such as X-X, to an alkene. As @Philipp points out, the alkene is operating through the (filled) pi orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's relatively easy to add a nucleophilic species to a carbonyl. In that case, the carbonyl operates through its (unfilled) pi* orbital.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-10-27T16:52:51.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18892" PostTypeId="2" ParentId="18832" CreationDate="2014-10-27T17:12:18.430" Score="1" Body="&lt;p&gt;The equilibrium constant of a reaction containing solids or liquids doesn't contain hidden values for their concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that we use the concentration (or pressure) of a substance in the equilibrium expression because it is a shortcut.  What we are actually putting into the expression is the activity of that substance, a more complicated idea.  It very often reduces to the ratio of the concentration of a solution to the concentration at standard state.  Since the concentration of a solution at standard state is 1M, there's no point in dividing through by 1 for each term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For solids and liquids, the concentration at standard state is its molar density. ($C = \frac{mol}{L}$)  This value doesn't depend on the quantity of substance available, so the ratio between the density at standard state and the density at any point in a reaction is 1.  The values ARE in the equilibrium expression; they're just all 1's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your concern about comparing equilibrium constants, we can only do that directly for reactions with the same number of reactants and products.  Adding (or removing) any term to either side interferes with the comparison.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare:&lt;br&gt;&#xA;$\ce{A(aq) +B(aq)-&amp;gt;AB(aq)}$&lt;br&gt;&#xA;$\ce{2A(aq) +B(aq)-&amp;gt;A2B(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just one of the limitations of the equilibrium constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the wiki on &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow&quot;&gt;Thermodynamic Activity&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-10-27T17:20:21.507" LastActivityDate="2014-10-27T17:20:21.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18895" PostTypeId="2" ParentId="18889" CreationDate="2014-10-27T17:37:14.927" Score="1" Body="&lt;p&gt;You're right that IF the metal calorimeter and the styrofoam calorimeter were both perfect and behaved in the way that we usually pretend calorimeters do (they don't absorb or lose heat) then they should produce identical results, for the enthalpy reason you provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since we're asking about specific materials, metal vs styrofoam, we're considering real calorimeters instead of ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few things that calorimeters can do that will affect how large a temperature change you measure.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat loss - As you note, metal conducts heat much more effectively than styrofoam.  This means that the metal calorimeter will carry the heat of the reaction away from the reaction vessel to the outside world faster than styrofoam will.  If that is the case, which reaction vessel will get hotter, styrofoam or metal?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat absorption - The calorimeter (and thermometer) gets heated up along with the reaction mixture.  A calorimeter that absorbs large amounts of energy will have a relatively lower final temperature, since more of the reaction's heat went into heating up the calorimeter than went into heating up the reaction mixture.  This quantity is known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Calorimeter_constant&quot; rel=&quot;nofollow&quot;&gt;calorimeter constant&lt;/a&gt; and varies depending on the materials and mass of the calorimeter.  Styrofoam absorbs more heat per gram than steel, but the steel calorimeter is probably heavier.  With a fast reaction there is the additional fact that the styrofoam calorimeter probably never comes up to the full temperature of the reaction before you quit recording temperatures.  This winds up leading to a small source of lost heat for the metal calorimeter as compared to the styrofoam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The quality of the equipment -  Metal calorimeters are often made specifically for the purpose of calorimetry.  Styrofoam calorimeters are often a couple of coffee cups jammed together.  If the calorimeter allows a significant amount of heat loss to the atmosphere due to its construction, then it will result in a cooler reaction mixture and a lower calculated enthalpy for the reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You've got to take all three of these into consideration in order to determine the particulars for your experiment.  &lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-27T17:37:14.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18896" PostTypeId="1" CreationDate="2014-10-27T19:29:43.573" Score="1" ViewCount="135" Body="&lt;p&gt;How can the definition of the pH value $\mathrm{pH}=-\log\left[\ce{H+}\right]$ yield a unitless result? Since $\left[\ce{H+}\right]$ is a concentration, wouldn't it have units $\mathrm{mol/dm^3}$ and therefore following the product rule you would get: $-\log\left(x\ \mathrm{mol/dm^3}\right)=-\log\left(x\right)-\log\left(\mathrm{mol/dm^3}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can these units in a mathematical sense suddenly disappear?&lt;/p&gt;&#xA;" OwnerUserId="9517" LastEditorDisplayName="user7951" LastEditDate="2015-09-20T00:35:25.303" LastActivityDate="2015-09-20T00:35:25.303" Title="How can the pH value be unitless?" Tags="&lt;ph&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2014-10-27T20:15:54.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="18897" PostTypeId="1" AcceptedAnswerId="18908" CreationDate="2014-10-27T23:26:39.980" Score="9" ViewCount="26887" Body="&lt;p&gt;Why cathode rays are produced if pressure of the gas discharge tube is lowered to about $10^{-4}$ atm?&#xA;How bombardment of glass of gas discharge tube from cathode ray result in a faint glow?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="4231" LastEditDate="2014-10-28T13:11:38.407" LastActivityDate="2019-01-04T14:55:56.633" Title="How are cathode rays produced?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18898" PostTypeId="1" CreationDate="2014-10-27T23:38:05.830" Score="6" ViewCount="7935" Body="&lt;p&gt;Why is the standard enthalpy of formation of $\ce{O2}$ and other reference forms of elements equal to zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, from what I know, bonding releases energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does anybody know a source for finding the reference form of an element at standard conditions?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="402" LastEditDate="2017-04-01T01:24:37.547" LastActivityDate="2019-06-10T05:58:57.087" Title="Why is the standard enthalpy of formation of elements in their native forms zero?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18899" PostTypeId="2" ParentId="18855" CreationDate="2014-10-27T23:40:48.583" Score="1" Body="&lt;p&gt;When chemical bonds are broken potential energy increases.  When they are made potential energy decreases. A loss in PE raises kinetic energy and a gain in PE lowers KE.  Since all reactions involve bond breaking and bond making the reaction will involve an increase or decrease in PE and a decrease or increase in KE.  When KE of the system increases, a system in contact with the surroundings will lose energy as heat (exothermic).  When KE of the system decreases a system in contact with the surroundings will gain energy as heat (endothermic).  In an isolated system &quot;exothermic&quot; reactions just raise temperature and &quot;endothermic&quot; reactions just lower temperature.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-10-27T23:40:48.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18900" PostTypeId="2" ParentId="18898" CreationDate="2014-10-28T00:13:41.197" Score="11" Body="&lt;p&gt;It is purely a matter of definition. A standard enthalpy of formation describes the change in enthalpy during the formation of 1 mol of a target compound by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature. Now take the case of oxygen as an example: At standard conditions &lt;span class=&quot;math-container&quot;&gt;$\ce{O2(g)}$&lt;/span&gt; is the most stable modification of oxygen. That means at standard conditions the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2O &amp;lt;=&amp;gt;&amp;gt; O2(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is defined in such a way that each of the elemental &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;s are defined to be exactly the same state as an oxygen atom that is part of an &lt;span class=&quot;math-container&quot;&gt;$\ce{O2(g)}$&lt;/span&gt;, i.e. their enthalpy &lt;span class=&quot;math-container&quot;&gt;$2H^{\circ}(\ce{O})$&lt;/span&gt; matches exactly that of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2(g)}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$H^{\circ}(\ce{O}) = \frac{1}{2}H^{\circ}(\ce{O2(g)})$&lt;/span&gt;. This gives you a reference point for &lt;span class=&quot;math-container&quot;&gt;$H^{\circ}(\ce{O})$&lt;/span&gt; which is otherwise nearly impossible to measure much as the Standard Hydrogen Electrode is used to set up a reference system for the electrode potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A table showing some standard enthalpies of formation can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. There it is also indicated which phases are used as reference form.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="24052" LastEditDate="2019-06-10T05:58:57.087" LastActivityDate="2019-06-10T05:58:57.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18904" PostTypeId="1" AcceptedAnswerId="18917" CreationDate="2014-10-28T02:50:57.467" Score="16" ViewCount="60497" Body="&lt;p&gt;If I understand correctly, when we go from a keto to an enol form, we simply abstract an alpha proton and thereby create a double bond to a carbon-bearing the hydroxyl functional group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When evaluating the stabilities of the two enol forms, it appears that the enol derivative of 2,4-pentadione is more stable because of intramolecular hydrogen bonding. Both have long-range conjugation, so that cannot be a factor. The only difference really is the hydrogen bonding ... and of course ring strain (a 5-membered ring with &lt;em&gt;two&lt;/em&gt; double bonds can't be very stable). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BE3vL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BE3vL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2019-08-13T04:04:33.430" LastActivityDate="2019-08-13T04:06:11.880" Title="Which is the more stable enol form?" Tags="&lt;organic-chemistry&gt;&lt;tautomer&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="18906" PostTypeId="1" CreationDate="2014-10-28T05:05:29.847" Score="2" ViewCount="1317" Body="&lt;p&gt;I'm reading something about nickel affinity chromatography and it talks about using imidazole to affect pH. In which direction does it change the pH?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="16683" LastEditDate="2017-04-21T22:28:36.263" LastActivityDate="2017-04-21T22:28:36.263" Title="In which direction does imidazole affect the pH?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18907" PostTypeId="2" ParentId="18906" CreationDate="2014-10-28T05:46:19.327" Score="5" Body="&lt;p&gt;Imidazole is an amphoteric molecule and can therefore act as an acid and a base.&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b8/Imidazole_2D_numbered.svg/100px-Imidazole_2D_numbered.svg.png&quot; alt=&quot;imidazole&quot;&gt; (courtesy of wikipedia)&lt;br&gt;&#xA;It's acidity is given with $\mathrm{p}K_a=14.5$ for the proton as $\ce{N~1}$. The basic site is the lone pair at $\ce{N~3}$. Its basicity is given through the following equilibrium: $$\ce{Im + H+ &amp;lt;=&amp;gt; IMH+}$$ with a $\mathrm{p}K_a=7.05$.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EExHx.png&quot; alt=&quot;DOI: 10.1021/jo01066a039&quot;&gt;(&lt;a href=&quot;http://dx.doi.org/10.1021/jo01066a039&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;You can obtain more information about this nice little molecule on its &lt;a href=&quot;http://en.wikipedia.org/wiki/Imidazole#Amphoterism&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, it very much depends on the solution with which you are working, which effect of imidazole takes over.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.387" LastActivityDate="2014-10-28T05:46:19.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18908" PostTypeId="2" ParentId="18897" CreationDate="2014-10-28T06:09:17.173" Score="12" Body="&lt;p&gt;&quot;Cathode rays&quot; are simply beams of electrons. The name exists because cathode ray devices predate the discovery of the electron. The sharp shadows produced by Crookes tubes meant that something was travelling in a straight line from the cathode down the tube, so they were named cathode rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum doesn't have much to do with the production of electrons, per se, but without high vacuum, the electrons collide with gas particles before they travel very far. If there's enough gas, you get a plasma rather than a beam of electrons. The electron beam is produced by thermionic emission in hot cathodes—electrons are liberated from a hot filament when the thermal energy is enough for electrons to exceed the filament's work function. In cold cathodes, electrons are released by field emission, where a large applied electric field allows electrons to tunnel away from a metal electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glow of the glass envelope is luminescence—either fluorescence or phosphorescence (or both). Basically, the electrons striking the glass excite components of the glass to an elevated electronic state which releases light upon returning to the ground state. Normal glass isn't particularly luminescent so tubes designed to produce light (like a CRT) are coated with phosphors.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-28T06:09:17.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18911" PostTypeId="2" ParentId="18904" CreationDate="2014-10-28T13:54:10.973" Score="7" Body="&lt;p&gt;The 2,4-pentanedione enol form shown is the more stable enol, since the lone pair on the enol oxygen can delocalize across the five atoms to the electronegative carbonyl oxygen. Hydrogen bonding is also a minor factor in its favor. The enol form of cyclopent-2,3-en-1-one is indeed (also) a conjugated system, but the lone pair on the enol oxygen cannot delocalize to an electronegative atom, rather only to a carbon atom, and the delocalization occurs only over three atoms instead of five. In thinking about this problem, it may help to draw out and look at the various &quot;resonance structures&quot; of these enols, but from a molecular orbital standpoint, the governing principle is that a more delocalized orbital is lower in energy (when it contains electrons, that is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is more pronounced in comparing the enolate anions of the two species.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="402" LastEditDate="2019-08-13T04:06:11.880" LastActivityDate="2019-08-13T04:06:11.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18913" PostTypeId="1" AcceptedAnswerId="18919" CreationDate="2014-10-28T15:12:59.077" Score="1" ViewCount="202" Body="&lt;p&gt;I understand that to create an enol, you simply eliminate a proton alpha to the relevant $\ce{C=O}$ and then tack the proton onto the $\ce{O}$ and there you have it — an alkene and an alcohol — or &lt;em&gt;enol.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, between pyruvate and lactate, which is &lt;em&gt;more&lt;/em&gt; enolizable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that pyruvate doesn't even have an enol tautomer … there are no protons alpha to a carbonyl group. Sure, there is the carboxylate group … but that doesn't have anything to do with keto-enol tautomerism (does it?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jer0i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-09T16:36:21.257" LastActivityDate="2015-10-09T16:36:21.257" Title="Enolizablity of pyruvate and lactate" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18914" PostTypeId="1" AcceptedAnswerId="18918" CreationDate="2014-10-28T15:20:29.740" Score="2" ViewCount="1408" Body="&lt;p&gt;I am told that &lt;em&gt;all&lt;/em&gt; three sugars below &lt;em&gt;can&lt;/em&gt; be oxidized to their corresponding carboxylic acids using potassium permanganate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This strikes me as bloody wrong because only hemiacetal sugars interconvert in solution with their open chain forms, and we need that open chain form to access the aldehyde/ketone for oxidation. Potassium permanganate as far as we know creates C-O bonds while not breaking any C-C bonds ... hence ... the general rule in our class is that ketones cannot be further oxidized (further oxidation would rupture C-C bonds). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEija.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2015-10-16T16:58:23.363" LastActivityDate="2015-10-16T16:58:23.363" Title="Potassium permanganate oxidation of sugars" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18915" PostTypeId="2" ParentId="16086" CreationDate="2014-10-28T16:20:04.047" Score="0" Body="&lt;p&gt;Ammonium salts are routinely used in protein purification.  It does not damage proteins.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-10-28T16:20:04.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18917" PostTypeId="2" ParentId="18904" CreationDate="2014-10-28T17:22:34.453" Score="29" Body="&lt;p&gt;Keto-enol equilibria reflect a &lt;strong&gt;delicate thermodynamic balance&lt;/strong&gt; between the two forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rROhq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding differences between the keto and enol structures shown above are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;keto: C=O double bond, C-C single bond, C-H bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;enol: C=C double bond, C-O single bond, O-H bond&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look up the bond energies for these bonds we &lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles/Bond_Energies&quot; rel=&quot;noreferrer&quot;&gt;find&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;keto: 745 + 347 + 413 = 1505 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;enol: 614 + 358 + 467 = 1439 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rough calculation tells us that, generally, the keto form will predominate.  However, the energy difference between the two forms is only 66 kJ/mol, a relatively small number; so it is likely that small differences can cause a dramatic shift in the relative concentrations of the two species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some general guidelines for predicting shifts in keto-enol equilibria&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nnkxx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromaticity and conjugation - The keto-enol equilibrium for phenol (I) lies entirely on the enol side due to thermodynamic stabilization provided by aromaticity. Compound III would be expected to have a higher enol content than compound II due to the extended conjugation present in the enol of III&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution - Replacing hydrogen with alkyl groups on a double bond stabilizes the double bond. Therefore, we would expect compound V to have a higher enol content than compound IV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dipolar repulsion - The carbonyl dipoles in butane-2,3-dione (VI) can reduce their electrostatic repulsion by adopting the geometry pictured with the carbonyl groups oriented away from one another.  Due to the constraints of the 5-membered ring, cyclopentane-1,2-dione (VII) cannot adopt a similar geometry.  In this case, the dipolar repulsion is lessened by increasing the enol content in the equilibrium to a point where the enol predominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen Bonding - Hydrogen bonding can stabilize the enol form.  If the hydrogen bond is strong enough, and particularly if other factors also stabilize the enol form, the enol form can predominate.  Your molecule, 2,4-pentanedione (IX), is a good example.  There are two possible enol forms, VIII and X. X is by far the predominant enol as it has 1) extended conjugation and 2) substitution on the carbon-carbon double bond.  In addition, X can form a very stable hydrogen bond involving a 6-membered structure between the two oxygen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvent Effects - Especially in cases where hydrogen bonding is involved in stabilizing the enol, the solvent can have a dramatic effect. In benzene, where intramolecular hydrogen bonding predominates, the IX:X ratio is approximately 5:95. In water, where the intramolecular hydrogen bond is replaced by hydrogen bonding to the solvent, the ratio is roughly reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to your question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclopent-2-enone can form two different enols, XI and XIII. Neither enol has any features that would be expected to provide significant enol stabilization.  There is no more conjugation than what is found in the ketone, there is no special hydrogen bonding, and no substitution of the carbon-carbon double bonds.  Indeed, the &lt;a href=&quot;http://www.sigmaaldrich.com/spectra/fnmr/FNMR009636.PDF&quot; rel=&quot;noreferrer&quot;&gt;proton nmr of cyclopentenone&lt;/a&gt; is just as expected, no evidence for a significant amount of enol.  On the other hand, 2,4-pentanedione exists primarily as the enol form, at least in non-polar solvents, for the 3 reasons (conjugation, hydrogen bond, double bond substitution) discussed above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, you mentioned that a 5-membered ring with two double bonds (a cyclopentadiene) might not be very stable; actually, they are quite stable and quite common, 2 double bonds in a 5-membered ring do not create a lot of strain.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2019-08-13T04:05:26.420" LastActivityDate="2019-08-13T04:05:26.420" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="18918" PostTypeId="2" ParentId="18914" CreationDate="2014-10-28T17:31:43.913" Score="4" Body="&lt;p&gt;Each of your molecules also contains at least one primary alcohol appendage.  Primary alcohols are readily oxidized by potassium permanganate to the corresponding carboxylic acid (see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_of_primary_alcohols_to_carboxylic_acids&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; for more information).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/58nxU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-28T17:31:43.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18919" PostTypeId="2" ParentId="18913" CreationDate="2014-10-28T17:51:49.893" Score="3" Body="&lt;p&gt;You're right and wrong.  Your right that pyruvate anion has a higher enol content than lactate anion, at least that would be my guess (pyruvate anion has 3 enolizable hydrogens, while lactate anion has one and needs to destroy the carboxylate resonance in order to create the carbonyl required for enolization...read on).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think that lactic acid could enolize, after all esters can enolize.  I suspect that the enol content in the lactate carboxylate anion would be non-measurable - you would need to disrupt the carboxylate anion resonance in order to generate the carbonyl required for enolization.  If we use 36 kcal/mol as an estimate of the carboxylate resonance energy and say that for every 1.4 kcal/mol we can shift an equilibrium by a power of 10, then we might expect 1 enol molecule in every $10^{26}$ lactate carboxylate molecules.  In the lactic acid itself, we have no carboxylate resonance to disrupt, the enol content would be higher.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-28T17:51:49.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18920" PostTypeId="2" ParentId="2499" CreationDate="2014-10-28T18:09:23.210" Score="-2" Body="&lt;p&gt;High voltage and low current is more efficient but must be oscillated at specific frequencies to be effective low cost electrolysis. it is the oscillation that breaks the bond of hydrogen and oxygen not the current or voltage. Current can be below .5 amp at millions of volts and still be non life threatening like a stun gun. But you need to experiment with the frequency of oscillation to find the most efficient electrolysis.&lt;/p&gt;&#xA;" OwnerUserId="9528" LastActivityDate="2014-10-28T18:09:23.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18921" PostTypeId="1" AcceptedAnswerId="18930" CreationDate="2014-10-28T18:54:23.760" Score="3" ViewCount="1191" Body="&lt;p&gt;In diatomic nitrogen&lt;br&gt;&#xA;There are two electron configurations I'd like to compare:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1s2 2s2(lone pair) 2p3(the bonding electrons sigma,pi,pi)&lt;/li&gt;&#xA;&lt;li&gt;1s2 2sp(lone pair) 2sp(sigma bonding) 2p2(pi,pi bonding)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why is option 2 favored over option 1?&lt;br&gt;&#xA;I thought having the lone pair in a 2s orbital would be favored over 2sp.&lt;/p&gt;&#xA;" OwnerUserId="9324" LastEditorUserId="9324" LastEditDate="2014-10-28T19:08:14.383" LastActivityDate="2014-10-28T22:12:15.123" Title="Why hybridization occurs in diatomic molecule" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18922" PostTypeId="1" CreationDate="2014-10-28T19:26:36.643" Score="1" ViewCount="1018" Body="&lt;p&gt;I understand that imines can be hydrolyzed in acidic medium. But what about in basic medium? Can imines still be hydrolyzed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a plausible mechanism ... nitrogen's withdrawal of electron density makes the alpha proton relatively acidic; base abstracts the proton in an E2 reaction; nitrogen then takes back a proton to regenerate the base catalyst. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Does this actually happen? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-10-28T19:26:36.643" Title="Imine stability in acids and bases" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18923" PostTypeId="1" AcceptedAnswerId="19428" CreationDate="2014-10-28T19:54:19.220" Score="4" ViewCount="129" Body="&lt;p&gt;Can these be used to induce acetal ring closures? For example, when dealing with a molecule that has both a C=O functional group and an alkene as part of its structure and there is the possibility of creating a 5 or 6 membered ring through intramolecular -OH attack on the C=O ... can we &lt;strong&gt;just&lt;/strong&gt; use the $\ce{OsO4}$ to create the diol and expect it to provide us sufficient acidity to catalyze ring closure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same question goes for meta-chloroperoxybenzoic acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-12T15:35:05.953" Title="Osmium tetroxide and meta-chloroperoxybenzoic acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18924" PostTypeId="1" AcceptedAnswerId="18941" CreationDate="2014-10-28T20:02:03.027" Score="9" ViewCount="2443" Body="&lt;p&gt;Are there any reactions with &lt;em&gt;no&lt;/em&gt; activation energies? Our professor just told us there are no chemical cliffs. Is this true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-04-21T05:43:56.147" LastActivityDate="2018-04-21T05:43:56.147" Title="Are there any reactions with no activation energies?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="6" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18925" PostTypeId="2" ParentId="18924" CreationDate="2014-10-28T20:25:52.007" Score="3" Body="&lt;p&gt;So of course it is hard to tell the difference between the case where there is no activation energy and the case where there is a very small activation energy.  Given that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many reactions, such as simple acid-base reactions are diffusion controlled.  This means that reaction occurs at every encounter, and that this is not an activated process.  Another example involves benzene.  Say we generated the molecule 1,3,5-cyclohexatriene, it would immediately isomerize to benzene (unequal bond lengths, localized pi electrons -&gt; equal bond lengths, delocalized pi electrons).  If we put this on a reaction coordinate it might look something like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vfpe3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always thought of this as a process with a &lt;strong&gt;negative&lt;/strong&gt; activation energy, but maybe zero would be a more accurate description (can you have a negative barrier?).  I suspect it &lt;strong&gt;applies to many reactions where there is a large thermodynamic difference in stability between product and reactant favoring product and where there is minimal structural reorganization&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-28T21:42:40.240" LastActivityDate="2014-10-28T21:42:40.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18927" PostTypeId="2" ParentId="18924" CreationDate="2014-10-28T20:47:54.220" Score="0" Body="&lt;p&gt;I seem to remember (haven't got the reference on hand) that singlet methylene (|CH&lt;sub&gt;2&lt;/sub&gt;) inserts into the C-H bond with zero activation enthalpy.  It's certainly an enormously reactive and high-energy species.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-10-28T20:47:54.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18929" PostTypeId="1" CreationDate="2014-10-28T22:00:27.267" Score="6" ViewCount="102" Body="&lt;p&gt;Is there a heat capacity of an object when it is melting? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m questioning it since the temperature doesn’t change during phase transitions and the unit of heat capacity is joule per kelvin.&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorDisplayName="user7951" LastEditDate="2015-06-14T14:59:08.473" LastActivityDate="2015-06-14T14:59:08.473" Title="Is there a heat capacity of an object when it is melting?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18930" PostTypeId="2" ParentId="18921" CreationDate="2014-10-28T22:12:15.123" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought having the lone pair in a 2s orbital would be favored over&#xA;  2sp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's true, placing the lone pair in a 2s orbital would be put the lone pair electrons at lower energy then if they were in a 2sp orbital.  However you're only looking at a part of the puzzle.  As you note, in this case the sigma bonding electrons would be in a pure p orbital and they would then be higher in energy then if they were in an sp orbital. You need to optimize the total energy of the molecule, so you need to consider all electrons and find the hybridization that gives the lowest possible energy to both the bonding and non-bonding electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which hybridization scheme a molecule adopts is the result of a delicate balance between electron stabilization and other factors such as bond strengths, steric and electrostatic effects.  The balance of these factors can shift and that is why some molecules hybridize one way and other molecules hybridize differently.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-10-28T22:12:15.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18934" PostTypeId="2" ParentId="18929" CreationDate="2014-10-28T22:50:01.963" Score="6" Body="&lt;p&gt;Yes there is a heat capacity even at the melting point.  The reason the temperature doesn't change is that you need to add the enthalpy of melting to convert solid to liquid in equilibrium.  However, under the right conditions you can supercool the liquid under the melting point, and the temperature will behave as you would expect, needing to pull a certain amount of energy out to supercool by a degree.  But this merely delays having to deal with the enthalpy of melting until solid uncleared - it is a kinetics issue, not a thermodynamics issue.  So, it is clear that the heat capacity of a phase is continuous through the phase transition (and beyond).&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-10-28T22:50:01.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18935" PostTypeId="1" AcceptedAnswerId="18936" CreationDate="2014-10-29T00:06:00.567" Score="2" ViewCount="2382" Body="&lt;p&gt;In experiments to detect the photoelectric effect, a clean metal was irradiated by monochromatic light and electrons were emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why was monochromatic light used in the experiment, and why does the frequency of the light have to be above a threshold frequency?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="16683" LastEditDate="2017-07-29T10:52:25.957" LastActivityDate="2017-07-29T10:52:25.957" Title="Why does the photoelectric effect depend on the frequency of the light?" Tags="&lt;electrons&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18936" PostTypeId="2" ParentId="18935" CreationDate="2014-10-29T01:02:17.440" Score="7" Body="&lt;p&gt;The photoelectric effect is described by the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\mathrm{max} = h\nu - \mathrm{WF_M}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_\mathrm{max}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\mathrm{WF_M}$ is the workfunction for the particular metal. The kinetic energies of all electrons emitted are distributed from $0$ to $E_\mathrm{max}$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why was monochromatic light used in the experiment? Why is a proper&#xA;  frequency used and not any other frequency?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The experiment is typically performed by scanning through a continuous range of monochromatic wavelengths from lower to higher energy.  At some specific wavelength, the observer will notice that electrons start to be emitted (the threshold).  As the scan continues to wavelengths with even more energy, the emitted electrons will increase in kinetic energy.  From the threshold energy and the above equation, the experimenter can determine the workfunction of the metal.  If the experiment had been run with light containing many different wavelengths (non-monochromatic light), electrons would still be ejected, but you wouldn't know what the threshold wavelength was and you wouldn't be able to determine the workfunction of the metal being studied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like more information on the photoelectric effect, &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_States_of_Atoms_and_Molecules/2._Foundations_of_Quantum_Mechanics/Photoelectric_Effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's a good, concise reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-07-29T10:50:28.100" LastActivityDate="2017-07-29T10:50:28.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18938" PostTypeId="2" ParentId="18935" CreationDate="2014-10-29T02:30:24.837" Score="2" Body="&lt;p&gt;The physical background of the experiment is that inside the metal the electrons can occupy states up till a point. These states are more or less continuous, however the electrons fill these states till an energy, called Fermi level. This energy is lower than the energy of electron in vacuum, so the electron needs an extra energy to jump. If the Fermi energy was not lower than vacuum, the electrons would just leave the metal, and fly away. When PE happens and a photon is absorbed, and it gives enough energy, and electron can jump out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point here is that a single photon gives all the energy to a single electron. In a classical theory, light is just wave, it has no reason to excite a specific electron. Also, if it is continuous wave, electrons could collect enough energy just &quot;waiting longer&quot; if the frequency of light is lower. But PE works this way, because light is quantized: one photon is not the same as two photons with half energy. To jump, the electron needs enough energy from a single photon, because the chance that it is hit by two photons is practically zero, and there may not be empty states along the way and so cannot just add up energy piece by piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Monochromatic light&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need monochromatic light. If you use monochromatic light, than all the photons correspond to a given energy, therefore easier to interpret the data, but it is just a practical requirement. If you use normal light, all kind of photos will come, some bellow threshold, some above, so all you see is the mixture of zillion energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Threshold energy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, this is the whole point. The electron cannot just absorb 3 and a half photons to get enough energy. If you have enough energy from a photon, you see PE, if not, there is no PE. The threshold energy which tells you if the energy is enough. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-10-29T02:30:24.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18939" PostTypeId="1" AcceptedAnswerId="18948" CreationDate="2014-10-29T02:38:20.650" Score="5" ViewCount="8913" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  If a metallic oxide has $40\,\%$ oxygen, find the equivalent weight of&#xA;  the metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This amounts to finding the atomic weight of the metal and the charge on the cation (in effect identifying the metal itself). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't know how to do this, so I looked up the solution. It goes as follows. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;em&gt;Assume the oxide has the formula $\ce{MO}$&lt;/em&gt;. Let the atomic weight of $\ce{M}$ be $x$&#xA;  Then &#xA;  $$\frac{16}{x + 16} = \frac{40}{100}$$&#xA;  Simplifying and solving for $x$ yields&#xA;  $$x = 24$$&#xA;  Thus the metal is magnesium and the equivalent weight is $12$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My problem is, how can we assume that the metallic oxide has the formula $\ce{MO}$? For all we know, the metallic oxide may be $\ce{M2O}$ or $\ce{M2O3}$. In fact, if we assume the latter formula we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{48}{2x + 48} = \frac{40}{100}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = 72$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which I believe is Hafnium.  &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="46306" LastEditDate="2018-03-05T01:07:26.427" LastActivityDate="2018-03-05T01:07:26.427" Title="What does &quot;a metallic oxide has 40 % oxygen&quot; mean?" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18941" PostTypeId="2" ParentId="18924" CreationDate="2014-10-29T03:42:25.013" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Marcus_theory&quot; rel=&quot;noreferrer&quot;&gt;Marcus-Hush theory&lt;/a&gt; describes electron/charge transfer rates using conventional &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;noreferrer&quot;&gt;transition state theory&lt;/a&gt;, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{A+ + B -&amp;gt; A + B+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one might naively assume that there's no activation barrier in a simple electron transfer, but that's not true. Since the geometry of $\ce{A+}$ and $\ce{A}$ as well as $\ce{B}$ and  $\ce{B+}$ are not identical, there is a &lt;strong&gt;&lt;em&gt;reorganization energy&lt;/em&gt;&lt;/strong&gt; of both the molecular species and the solvent or environment. This reorganization energy ($\lambda$) serves as an activation barrier for the charge transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the convenient thing about charge transfer reactions is that you can tune the $\Delta G$ of the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marcus predicted that the maximum rate would occur when the $\Delta G^0 = -\lambda$ and thus the $\Delta G^\ddagger = 0$. If the reaction becomes more thermodynamically favorable, you enter the &lt;strong&gt;inverted&lt;/strong&gt; region and the activation barrier goes up and the reaction rate goes down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a long-winded answer to say there are &lt;strong&gt;&lt;em&gt;some&lt;/em&gt;&lt;/strong&gt; chemical reactions with no activation barrier, in addition to the other cases here. Usually the key phrase describing these in papers is &quot;barrier-less reactions&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-10-29T13:16:05.317" LastActivityDate="2014-10-29T13:16:05.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18942" PostTypeId="2" ParentId="18740" CreationDate="2014-10-29T05:24:57.243" Score="3" Body="&lt;p&gt;Does this help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4IgUz.jpg&quot; alt=&quot;X-Ray diffraction from a cubic crystal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between planes will be less than the distance between atoms in many cases because the atoms don't have to stack directly on top of one another in each plane.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2014-10-29T05:24:57.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18943" PostTypeId="2" ParentId="18622" CreationDate="2014-10-29T06:04:05.047" Score="1" Body="&lt;p&gt;You measure the energy of any bond by carrying out reactions and seeing how much energy (usually in the form of heat) has to be invested.  For example, if you react one mole of $\ce{C}$ with one mole of $\ce{O2}$, you get $\ce{CO2}$ and heat.  Measure the energy with a calorimeter and divide by how many atoms and you get the energy you invested to produce one bond.  Not coincidentally, bonds are generally measured in units of energy/mol (e.g. kJ/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether a bond is ionic or covalent has to do with whether the bonding electron resides on one atom or both.  A perfectly covalent bond finds the electron half on one atom and half on the other.  A perfectly ionic bond has the electron 100% on one atom and not at all on the other.  Real systems are a mix of both.  When we say a bond is ionic (such as $\ce{NaCl}$), we are really referring to the fact that we have measured a circumstance where the electron truly favors one atom.  There are various ways to measure this, one of the most common is to melt the solid and see if it conducts electricity.  In the case of $\ce{NaCl}$, the bond is so ionic that a small voltage will cause the ions to wander from their neighbor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a covalent bond such as you find in $\ce{SiO2}$, atoms can't wander without breaking the bond, so a small voltage doesn't cause a current.  A large voltage may break the covalent bonds and cause a current, but it also changes the substance (in this case releasing oxygen gas and producing silicon metal).&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="5456" LastEditDate="2014-10-29T07:08:32.593" LastActivityDate="2014-10-29T07:08:32.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18945" PostTypeId="1" CreationDate="2014-10-29T11:18:56.590" Score="7" ViewCount="12238" Body="&lt;p&gt;In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number'l' (0,1...n-1 or s,p,d..). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2020-04-30T14:06:30.537" Title="Why are all the orbitals that have the same principal number in Hydrogen degenerate?" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18947" PostTypeId="2" ParentId="18945" CreationDate="2014-10-29T13:21:57.217" Score="5" Body="&lt;p&gt;The answer is right there in your question. It's only the interaction of &lt;strong&gt;&lt;em&gt;multiple&lt;/em&gt;&lt;/strong&gt; electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this.. For the one electron system, why should a $p$ or $d$ orbital differ in energy from an $s$. What makes them differ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the multi-electron case, the $p$ orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the &quot;shape&quot; of the $p$ and $d$ and $f$ orbitals to matter to the other electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the $s$, $p$, and $d$ orbitals.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-10-29T13:21:57.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18948" PostTypeId="2" ParentId="18939" CreationDate="2014-10-29T16:21:38.970" Score="5" Body="&lt;p&gt;Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily.  However, you can find the equivalent weight of the metal without &lt;em&gt;any&lt;/em&gt; assumptions about its formula.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen.  We know that the compound contains 40% oxygen, so it must contain 60% metal.  Suppose we had 100g of this compound; that would be 40g O and 60g M.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)&lt;br&gt;&#xA;$8 g\ \ce{O}:12g\ \ce{M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equivalent weight of the metal must be 12g.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given an equivalent weight for M of 12g:&lt;br&gt;&#xA;$\ce{M2O-&amp;gt; 12g/mol, carbon}$ (not a metal)&lt;br&gt;&#xA;$\ce{MO-&amp;gt; 24g/mol, magnesium}$&lt;br&gt;&#xA;$\ce{M2O3-&amp;gt; 36g/mol, chlorine}$ (not a metal)&lt;br&gt;&#xA;$\ce{MO2-&amp;gt;48g/mol, titanium}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-29T16:21:38.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18949" PostTypeId="2" ParentId="13540" CreationDate="2014-10-29T16:30:41.467" Score="2" Body="&lt;p&gt;Yes, because though the liquid phase initially contains no water, water molecules may enter the liquid phase from the gas phase.  The more hygroscopic the liquid is, the more the evaporation rate will be affected by humidity.  If the liquid is very hygroscopic and has a low vapor pressure, the initial evaporation rate &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2013/nj/c3nj00213f#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;may even be negative&lt;/a&gt; in humid conditions, as water enters the liquid phase at a rate greater than liquid enters the gas phase.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-10-29T16:30:41.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18953" PostTypeId="1" AcceptedAnswerId="18960" CreationDate="2014-10-29T19:19:03.773" Score="1" ViewCount="707" Body="&lt;p&gt;What happens during this reductive amination? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the first product; the lithium aluminium hydride will reduce the nitrile to an amine. But what about the reductive animation step? As far as I can see the C=O will be protonated by the strong tosic acid, and then there will be a competition between the methanol and the internal amine to attack C=O. Actually, the amine is able to attack the C=O without the C=O being protonated, but still, this intramolecular seem &lt;em&gt;too&lt;/em&gt; favorable given that we are forming a 4-membered ring. So will the methanol outcompete the amine? In that case what's up with the reductive animation step? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SpR5z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I &lt;em&gt;think&lt;/em&gt; happens after the second step ... I am going through the intramolecular route ... since that's the only way we get an iminium ion, on which the sodium cyanoborohydride can act. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xqPmD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="865" LastEditDate="2015-10-16T17:50:14.797" LastActivityDate="2015-10-16T17:50:14.797" Title="Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18955" PostTypeId="1" CreationDate="2014-10-29T20:11:50.663" Score="3" ViewCount="472" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JITx5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph comes from a so-called &quot;differential thermal analysis&quot;. It is studying the effects of temperature on glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what the difference between &quot;glass transition&quot; and &quot;devitrification&quot; is, and why one would be endothermic and the other would be exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble analysing these types of graphs. Does anyone know of a simple source online that I could check out?&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="23561" LastEditDate="2019-04-15T18:19:31.933" LastActivityDate="2019-04-15T18:19:31.933" Title="Devitrification vs glass transition temperature" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;equipment&gt;&lt;glass&gt;&lt;calorimetry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18956" PostTypeId="2" ParentId="18953" CreationDate="2014-10-29T20:34:59.823" Score="2" Body="&lt;p&gt;I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-10-29T20:34:59.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18957" PostTypeId="1" AcceptedAnswerId="18972" CreationDate="2014-10-29T20:35:45.947" Score="3" ViewCount="13242" Body="&lt;p&gt;I know that technically two or more gases in one container should have different volumes. But when you imagine the gas molecules being mixed together, it look sorta like the two gases have the exact same volume, that is the volume of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in one of the questions I did on gas laws, wet hydrogen and hydrogen are shown to have the same volume in a container. Given the barometric pressure, I subs tract the vapor pressure at the certain temperature from it. Then I sub the pressure of dry hydrogen gas into the ideal gas law along with the volume of the container to find the number of mole of hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaBj4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;This is the part in my textbook that I am confused about. It shows how partial pressure fraction is equal to the corresponding mole fraction. But this is all under the assumption that the volume and other things are constant. I am lost right here, because Avogadro's theory states the volume and mole number of a gas are directly proportional. &#xA;( wait...wouldn't this mean that pressure, mole and volume are all directly proportional to one another???)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give a answer, along with a proof/ a source?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-11-02T17:44:46.043" LastActivityDate="2014-11-02T17:44:46.043" Title="Volume of different gas in a container" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18958" PostTypeId="1" CreationDate="2014-10-29T21:13:49.713" Score="2" ViewCount="1389" Body="&lt;p&gt;The questions states: Ammonia has the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;. Household ammonia is a dilute aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;. Aqueous ammonia is a base that can neutralize acids.&lt;/p&gt;&#xA;&lt;p&gt;If 18.2 mL of 0.800 M &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?&#xA;Express your answer with the appropriate units.&lt;/p&gt;&#xA;&lt;p&gt;I used the values I was given; converted the &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; to moles; divided the moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; by the liters of ammonia solution. However, I do not understand why this is correct.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; is an acid and &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; is a base. How does finding the molar mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; help one calculate the molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="1782" LastEditorDisplayName="user9974" LastEditDate="2021-05-23T20:42:36.197" LastActivityDate="2021-05-23T20:42:36.197" Title="Do acids and bases have the same coefficients if they neutralize one another?" Tags="&lt;physical-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18959" PostTypeId="1" CreationDate="2014-10-29T21:40:19.393" Score="0" ViewCount="1115" Body="&lt;p&gt;Taking Glycine and aspartic acid. As both the amino acids are fully protonated at pH 0 ad pH 1, how does the peptide bond form? Is it a two step mechanism where the OH from an alpha carboxyl group bonds with a hydrogen from the alpha amino group of the other amino acid to form water then the H+ from the amino group being released to form a peptide bond?&lt;/p&gt;&#xA;" OwnerUserId="9552" LastEditorUserId="4325" LastEditDate="2014-10-29T23:23:20.877" LastActivityDate="2014-10-30T16:38:58.813" Title="Net charge of Peptides at pH1 and pH0" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18960" PostTypeId="2" ParentId="18953" CreationDate="2014-10-29T21:45:21.367" Score="3" Body="&lt;p&gt;Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8YVgb.png&quot; height=&quot;60&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with &lt;a href=&quot;http://en.wikipedia.org/wiki/Baldwin%27s_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baldwin's rules&lt;/a&gt; and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uVxpb.png&quot; height=&quot;160&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;tether&quot; by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Yp3k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RWtq.png&quot; height=&quot;100&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-10-30T00:15:40.293" LastActivityDate="2014-10-30T00:15:40.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18962" PostTypeId="2" ParentId="16445" CreationDate="2014-10-29T22:57:51.060" Score="3" Body="&lt;p&gt;I don't have much information on your particular system: are they nanocrystals?  Are you expecting just one or two phases?  Dozens of phases? Do you have a lot of amorphous material present?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one publication you can look at which addresses a rather complicated sample with amorphous, nanocrystalline, and crystalline objects.  It is (hopefully) overkill for your sample, but it will give you ideas on how to partition the problem (and some of the references may help too):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/maps.12148/abstract&quot; rel=&quot;nofollow&quot;&gt;Gainsforth, Z., Brenker, F. E., Simionovici, A. S., Schmitz, S., Burghammer, M., Butterworth, A. L., et al. (2013). Stardust Interstellar Preliminary Examination VIII: Identification of crystalline material in two interstellar candidates. Meteoritics and Planetary Science, 49(9), 1645–1665. doi:10.1111/maps.12148&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also find it helpful to play with simple fitting softwares such as CrystalMaker/CrystalDiffract where you can try out different mixtures and fits.  Sometimes the peak-finding softwares are a little simplistic with a large number of false positives.  If you try manually combining the possible phases, you can probably rule many of them out and simplify your list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few tips:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Look for sets of peaks with FWHM that are similar.  This works with nanocrystalline materials.  If you have a nanocrystal with a larger crystal, then the larger crystal will have sharp peaks and the nanocrystal will have broad peaks.  Then you can select which peaks apply to Phase A and Phase B and simplify your search space drastically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If Peak A is present in your experimental data, and you get a &quot;hit&quot; for phase X, but Phase X also predicts Peak B but you don't see it in your experimental data, then you can rule Phase X out.  You may need to understand the characteristics of the phase in a little detail because you can have situations where one atom substitutes for another and you get a change in peak intensities or positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Do you know the chemical composition of your sample?  If one of your &quot;found&quot; phases is a Yttrium based superconductor, and you're pretty sure you don't have Yttrium, then you can drop that phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Try looking at it with an electron microscope (SEM or TEM).  This will provide basic constraints as to whether you have many phases, or just a couple, and how big the crystals are.  This is a lot of work, but maybe less work than spending the next month chasing false positives.  TEM can also give you diffraction on any phase, no matter how small, and will be pretty unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) You can also provide constraints using techniques such as XANES, NMR, FTIR, etc., depending on what kind of sample you have.  Which of these are useful is also dependent on the resources you have present at your university/company/facility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2014-10-29T22:57:51.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18963" PostTypeId="1" CreationDate="2014-10-29T22:58:04.330" Score="0" ViewCount="49" Body="&lt;p&gt;I recognize that the 3d orbital decreases in energy to lower than the 4s once it becomes occupied (even if I don't completely understand why?!). However, how is the difference in electron configuration of Ca and V^3+ explained by the radial distribution functions? This is a question from a problem sheet at my University and I can't seem to find an answer relating to RDF's anywhere.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-10-29T22:58:04.330" Title="Using knowledge of radial distribution functions, how is it possible to explain the different electron configurations of V3+ and Calcium" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2014-10-31T16:16:56.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="18964" PostTypeId="1" CreationDate="2014-10-29T23:35:59.193" Score="0" ViewCount="5469" Body="&lt;p&gt;What is the smallest whole number coefficent of $\ce {Cr^{2+}}$ when the equation is balanced?&lt;br&gt;&#xA;$$\ce {Cr^{2+}_{(aq)} + Al_{(s)} -&amp;gt; Cr_{(s)} + Al^{3+}_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand, how can this be balanced? And why would 3 be the smallest coefficient? &lt;/p&gt;&#xA;" OwnerUserId="9537" LastEditorUserId="865" LastEditDate="2015-10-02T16:05:58.880" LastActivityDate="2015-10-02T16:05:58.880" Title="Smallest whole number coefficient?" Tags="&lt;homework&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18965" PostTypeId="2" ParentId="18964" CreationDate="2014-10-30T00:57:18.680" Score="2" Body="&lt;p&gt;You need to factor in the electrons that are being moved around&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromium is doing this:&lt;br&gt;&#xA;$\ce {Cr^{2+}_{(aq)} + 2 e- -&amp;gt;  Cr_{(s)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum is doing the reverse of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Al^{3+}_{(aq)} + 3 e- -&amp;gt; Al_{(s)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, like a math equation flip around the aluminum reaction.&lt;br&gt;&#xA;And balance out the electrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {[Cr^{2+} + 2 e- -&amp;gt;   Cr_{(s)}] \times 3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {[Al_{(s)} -&amp;gt; Al^{3+} + 3 e- ] \times 2}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$\ce {3Cr^{2+} + 6 e- + 2 Al_{(s)} -&amp;gt; 3Cr_{(s)} + 2 Al^{3+} + 6 e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps &lt;/p&gt;&#xA;" OwnerUserId="9554" LastEditorUserId="4934" LastEditDate="2014-10-30T02:51:43.323" LastActivityDate="2014-10-30T02:51:43.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18966" PostTypeId="2" ParentId="18958" CreationDate="2014-10-30T01:20:18.560" Score="2" Body="&lt;p&gt;Look at the chemical equation. One mole of $\ce{HCl}$ will neutralize one mole of $\ce{NH3}$. $\ce{HCl}$ is an acid so it loses a hydrogen ion and ammonia is a base so it gains a hydrogen ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + NH3 -&amp;gt; NH4Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you found the number of moles $\ce{HCl}$ added, and this equals the number of moles in the 5.00 mL sample of $\ce{NH3}$. Divide that by the volume and you get concentration, which is moles per liter. You don't need to find the molar mass of $\ce{HCl}$ and you didn't find it as described in your calculations above. &lt;/p&gt;&#xA;" OwnerUserId="4696" LastEditorUserId="5017" LastEditDate="2014-10-31T19:41:54.283" LastActivityDate="2014-10-31T19:41:54.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18967" PostTypeId="1" CreationDate="2014-10-30T01:42:50.423" Score="4" ViewCount="6609" Body="&lt;p&gt;I would like to know under what circumstances is Trouton's rule obeyed by liquids and why certain systems i.e. substances may deviate from Trouton's rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way I know what Trouton's rule is and I have a vague idea that sometimes Trouton's rule is not obeyed due to the presence of hydrogen bonding in liquids such as water but I would like to know more about other reasons why certain situations the rule is invalid.&lt;/p&gt;&#xA;" OwnerUserId="9464" LastEditorUserId="5456" LastEditDate="2014-10-30T07:07:21.843" LastActivityDate="2016-10-06T15:50:06.960" Title="Validity of Trouton's Rule" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18968" PostTypeId="1" AcceptedAnswerId="18983" CreationDate="2014-10-30T03:22:48.057" Score="9" ViewCount="867" Body="&lt;p&gt;When I use a torch to split a glass tube, there's always a hazy white band which appears. Does anyone know what the cause of this is? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a hydrogen-oxygen torch. The glass is fused quartz glass, which should be pure silica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0jm5.jpg&quot; alt=&quot;white band on glass tube apparent after being heated with a torch&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9557" LastEditorUserId="5078" LastEditDate="2016-01-27T18:53:55.587" LastActivityDate="2016-01-27T18:53:55.587" Title="Why does a glass tube show white bands after being heated with a torch?" Tags="&lt;heat&gt;&lt;color&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18969" PostTypeId="1" AcceptedAnswerId="18974" CreationDate="2014-10-30T04:12:28.797" Score="1" ViewCount="1281" Body="&lt;p&gt;When liquid(water/butter) is added to starch lumps of starch are formed. Which are completely dry from inside. Adding more solvent doesn't seem to help therefore it is not a case of saturated solution. So what does cause the lump formation ? and how can it be prevented ?&lt;/p&gt;&#xA;" OwnerUserId="9558" LastActivityDate="2014-10-30T05:43:29.837" Title="What causes Starch's lump creation when solvent (butter/water/oil) is added and how to prevent them" Tags="&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18970" PostTypeId="2" ParentId="18969" CreationDate="2014-10-30T04:15:56.107" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Put the starch in a jar with a lid.&lt;/li&gt;&#xA;&lt;li&gt;Add a little cold water.&lt;/li&gt;&#xA;&lt;li&gt;Put on lid and shake well.&lt;/li&gt;&#xA;&lt;li&gt;Use starch mixture as desired.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-10-30T04:15:56.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18972" PostTypeId="2" ParentId="18957" CreationDate="2014-10-30T04:24:07.040" Score="5" Body="&lt;p&gt;Your second idea, that they have the same volume, is correct.  Each gas in a mixture of gases stored in the same container has the volume of that container.  Remember that gases are mostly empty space, so it is not a problem that their volumes overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As is implied by your comment about subtracting the vapor pressure of water from the pressure of wet hydrogen, the gases in a mixture can have different partial pressures. Perhaps that is your point of confusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only case where they would have different volumes would be in a more accurate gas model than ideal gases, something like the Van der Waals model, where gas volume is corrected by a small amount to account for the portion of space that is taken up by the gas particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, each component of a mixture of gases stored in a given container shares the same volume and temperature and may (or may not) have different partial pressures and number of moles.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-10-30T04:24:07.040" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18974" PostTypeId="2" ParentId="18969" CreationDate="2014-10-30T05:43:29.837" Score="2" Body="&lt;p&gt;The trick here is that you're not actually dissolving the starch. Starch is made up of small semi-crystalline granules of amylose and amylopectin which are actually pretty much insoluble at room temperature. Some water can get in as amylose breaks up the crystallinity, but it doesn't really dissolve. When you drop it into water, it hydrates a bit on the outside, holding it all together, but water doesn't make it into the inside. If you stir or shake it, you can break the clumps up into smaller pieces, eventually getting at most individual hydrated granules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not until you heat the mixture that the granules really break apart and allow water in, in a process called gelatinization. This is what makes starch useful in cooking as the starch can form a more diffuse network, thickening the liquid.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-30T05:43:29.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18976" PostTypeId="2" ParentId="8964" CreationDate="2014-10-30T07:11:46.260" Score="2" Body="&lt;p&gt;For the reaction $\ce{2H3O+ + Mg -&amp;gt; H2 + Mg^{-2}}$; the actual rate law is &#xA;$r=k \times [\ce{H3O+}]^2 \times (\ce{Mg})^1.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious error is that HCl is a strong acid, thus not in equilibrium. Cl- is merely a spectator ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt; more accurate reaction step method is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3O+ + Mg -&amp;gt; H2O + MgH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3O+ + MgH -&amp;gt; H2 + Mg^{-2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface area of magnesium directly affects the k value, but magnesium is still a valid part of the reaction. The magnesium term may appear zeroth order if the magnesium is in far excess, so significantly lowering the initial input of magnesium will probably result in the correct rate law. &lt;/p&gt;&#xA;" OwnerUserId="9564" LastEditorUserId="5456" LastEditDate="2014-10-30T07:26:58.657" LastActivityDate="2014-10-30T07:26:58.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18977" PostTypeId="1" CreationDate="2014-10-30T09:14:51.037" Score="1" ViewCount="268" Body="&lt;p&gt;I have performed an experiment where I had some test tubes with the reagents and different amounts of catalyst. I started the reactions and measured the time they took.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I calculated the rate of reaction and plotted it in function of the amount of catalyst (in mL) in each test tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I can't decide between a logarithmic fit line and a linear one (though the logarithmic one has a higher $R^{2}$). The data points are below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, the reaction we used was the one of potassium permanganate with oxalic acid. We also added sulfuric acid and the catalyst used was manganese sulphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would make more sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bj6um.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0   4e-05&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.2 7.22e-05&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.4 0.000136&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.6 0.000159&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.8    0.000183&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1    0.0002&lt;/p&gt;&#xA;" OwnerUserId="9565" LastEditorUserId="9565" LastEditDate="2014-10-30T11:14:25.597" LastActivityDate="2014-10-30T11:14:25.597" Title="What is the type of dependence of the rate of reaction on the amount of catalyst?" Tags="&lt;catalysis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18978" PostTypeId="1" CreationDate="2014-10-30T10:23:02.210" Score="0" ViewCount="5128" Body="&lt;p&gt;Now I know it's something related to the fact that $\ce{CCl4}$ does not have a net dipole moment but I can't connect the dots. &lt;/p&gt;&#xA;" OwnerUserId="9566" LastEditorUserId="5456" LastEditDate="2014-10-30T13:59:12.033" LastActivityDate="2015-09-19T16:30:53.870" Title="Why is NaCl insoluble in CCl4?" Tags="&lt;everyday-chemistry&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18980" PostTypeId="1" CreationDate="2014-10-30T12:42:35.223" Score="0" ViewCount="440" Body="&lt;p&gt;A $\ce{[M(H2O)6]^2+}$ complex typically absorbs at around $600\ \mathrm{nm}$. It is allowed to form a new complex $\ce{[M(NH3)6]^2+}$ that should have absorbtion at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorbtion is because of transition of $\mathrm{e_2g}$ electron to $\mathrm{t_{2g}}$ and vice versa which depend on crystal field stabilization energy as:&#xA;$$\Delta = \frac{hc}{\lambda=600\ \mathrm{nm}}\approx2.0\ \mathrm{eV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know how to convert it in terms of $\Delta_\text o$ so that I can know the actual configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{H2O}$ there will be no pairing of electrons as it is weak and reverse with $\ce{NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks useful but isn't without the metal configuration:&#xA;$$\frac{\Delta_{\text o,\ce{H2O}}}{\Delta_{\text o,\ce{NH3}}}=\frac{\lambda_{\ce{NH3}}}{\lambda_{\ce{H2O}}}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-05-16T18:56:16.123" LastActivityDate="2015-05-16T18:56:16.123" Title="Absorbtion of light by a coordination compound" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18981" PostTypeId="1" AcceptedAnswerId="33328" CreationDate="2014-10-30T12:43:35.270" Score="7" ViewCount="635" Body="&lt;p&gt;The mass balance for a generic amphiprotic substance $[\ce{AH^-}]$ this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\ce{HA^-}]=C \cdot \dfrac{K_{a1} [\ce{H_3O^+}]}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.1&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &#xA; C = initial concentration of the amphitrofic substance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Eq.1 derivative  is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} = -\dfrac{C \cdot [\ce{H_3O^+}] \cdot K_{a1} \cdot (2 [\ce{H_3O^+}] + K_{a1}}{(K_{a1}\cdot [\ce{H_3O^+}] + [H_3O^+]^2 + K_{a1} \cdot K_{a2})^2} + \dfrac{C \cdot \cdot K_{a1}}{K_{a1}\cdot [\ce{H_3O^+}] + [\ce{H_3O^+}]^2 + K_{a1} \cdot K_{a2}} \quad Eq.2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero of this derivative is found at $[\ce{H_3O^+}] = \sqrt{K_{a1} \cdot K_{a2}}$, where is the maximum of $[\ce{AH^-}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buffer power (BP) can be defined $BP = \dfrac{dC_B}{dp \ce{H}} = -\dfrac{dC_A}{dp \ce{H}} = - 2.3 \cdot [\ce{H_3O^+}] \cdot \dfrac{d[\ce{HA^-}]}{d[\ce{H_3O^+}]} \quad Eq.3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K_{a1} = 6.2  \times 10^{-8}$ and $K_{a2} = 4.8 \times 10^{-13}$ and C=0.1, the maximum oncentration of the specie $\ce{AH^-}$ is found at $[\ce{H_3O^+}] = \sqrt{6.2  \times 10^{-8} \cdot 4.8 \times 10^{-13}}  = 1.725 \times 10^{-10} \quad pH=9.76$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve that represents the buffer power, plotting the Eq.3, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBI4f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way I can say that at maximum concentration of the amphiprotic substance the buffer power is zero. Is this statement true?&lt;/p&gt;&#xA;" OwnerUserId="9475" LastEditorUserId="19383" LastEditorDisplayName="user15489" LastEditDate="2015-11-06T18:58:26.600" LastActivityDate="2015-11-06T18:58:26.600" Title="Buffer Power of the amphiprotic substance" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18982" PostTypeId="2" ParentId="18978" CreationDate="2014-10-30T12:59:36.853" Score="5" Body="&lt;p&gt;In order for a substance to dissolve, the free energy change for the dissolution must be negative.  For the enthalpy change to be negative the heat of solvation (the solvent solvating the solute) must be greater in magnitude than the heat needed to break up the forces holding the solid solute together (in the case of NaCl lattice energy).  The entropy change of the solution process is accounted for by the gain in entropy from breaking up the solid and the loss of entropy from organizing solvent around the solute.  In forming most solutions, the entropy change is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the ability of carbon tet to solvate sodium and chloride ions is too low to make up for the lattice energy.  Carbon tet is nonpolar.  A good solvent for NaCl would be polar and the negative end of the solvent would surround sodium ions and the positive end would surround chloride ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2015-09-19T16:30:53.870" LastActivityDate="2015-09-19T16:30:53.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18983" PostTypeId="2" ParentId="18968" CreationDate="2014-10-30T14:26:17.737" Score="5" Body="&lt;p&gt;It is sublimed silica according to the following reference at page 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.public.asu.edu/~aomdw/GLASS/book.pdf&quot;&gt;http://www.public.asu.edu/~aomdw/GLASS/book.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-10-30T14:26:17.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18985" PostTypeId="2" ParentId="18959" CreationDate="2014-10-30T16:38:58.813" Score="2" Body="&lt;p&gt;First, note that the protonated amino group of an amino acid like glycine or aspartic acid has a pKa of about 9.6 - the number for glycine. At pH 1 it is indeed &quot;fully protonated&quot;, but more correctly about 1 molecule in 10^(8.6) will have a free amino group at any given point in time (always in flux due to the rapidity of proton transfer). This amino group can react with a carboxyl group (from another amino acid) at the carbonyl carbon to give a tetrahedral intermediate, which in turn can eliminate water to give the peptide bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under equilibrium at pH 0 or 1, the energetics typically favor the reverse of this reaction - hydrolysis of the peptide bonds to give the amino acids - this is what goes on in the stomach when you digest a protein - but it is possible that these conditions could produce a small amount of material from the reverse reaction - the amide-bond forming process.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-30T16:38:58.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18987" PostTypeId="1" AcceptedAnswerId="18989" CreationDate="2014-10-30T17:05:34.010" Score="11" ViewCount="21957" Body="&lt;p&gt;Two organic chemistry textbooks I've read (by Bruice and by Tandon) say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Esters are more acidic than ketones, because the resonance between the two oxygen atoms gives less opportunity for the delocalization of the electron pair on the alpha carbon in esters contrary to what happens in ketones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, in esters, when the resonance happens between the two oxygen atoms, the oxygen atom gains a positive charge which is highly unstable, given that oxygen is highly electronegative. So, shouldn’t the order of acidity be the other way round?&lt;/p&gt;&#xA;" OwnerUserId="8217" LastEditorUserId="16683" LastEditDate="2017-12-30T21:32:25.113" LastActivityDate="2017-12-30T21:32:25.113" Title="Is an ester or a ketone more acidic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18988" PostTypeId="1" AcceptedAnswerId="19263" CreationDate="2014-10-30T17:24:24.363" Score="8" ViewCount="935" Body="&lt;p&gt;When Vanadium is ionised it loses the 4s electron first, meaning that it's 3+ ion has a different electron configuration to Calcium despite it being isoelectronic. Can it be explained in terms of radial distribution functions? Maybe the maxima of the 3d and 4s become further apart as the effective nuclear charge increases?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5017" LastEditDate="2014-11-08T16:48:14.530" LastActivityDate="2014-11-08T16:48:14.530" Title="How does the radial distribution function of Vanadium differ from that of Calcium and how does this affect the ionic electron configurations?" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18989" PostTypeId="2" ParentId="18987" CreationDate="2014-10-30T17:47:26.807" Score="19" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So, shouldn’t the order be the other way round?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is the other way around.  Ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25).  The figure below shows the resonance structures for the enolate anion of a ketone and an ester.  Resonance structures II and IV stabilize the enolate anion in the ketone and ester respectively.  In the ester, there is an additional resonance structure, V. Through resonance structure V, the alkoxy group in the ester donates electrons to the carbonyl.  This movement of electron density towards the carbonyl will tend to destabilize the formation of any negative charge on the carbon attached on the other side of the carbonyl (electrostatic repulsion).  This reduces the likelihood of the formation of an enolate anion on that carbon. Therefore, in the ester, to whatever extent structure V contributes, structure IV will contribute somewhat less.  Hence, the ester enolate anion resonance structure IV plays a smaller role and the enolate anion is less stabilized in the ester compared to the ketone.  Consequently, the ester is less acidic than the ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjL35.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-10-30T18:07:52.833" LastActivityDate="2014-10-30T18:07:52.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18991" PostTypeId="1" AcceptedAnswerId="18993" CreationDate="2014-10-30T20:39:47.573" Score="14" ViewCount="5906" Body="&lt;p&gt;From what I understand, bleach works to weaken the ability of a substance to absorb light. Thus this seems to be &quot;whitening&quot; i.e. reflecting all light. In contrast, colorlessness (according to Wikipedia) implies that light can pass through the object without scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is '&lt;strong&gt;colorless&lt;/strong&gt;' a misnomer when used to describe bleach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, to double check when you bleach something, is it still there? Only that it is now white?)&lt;/p&gt;&#xA;" OwnerUserId="9573" LastEditorUserId="-1" LastEditDate="2014-10-31T03:01:02.567" LastActivityDate="2016-05-25T21:39:07.300" Title="Does bleach turn a substance white or colorless?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18992" PostTypeId="1" CreationDate="2014-10-30T20:44:17.230" Score="4" ViewCount="90" Body="&lt;p&gt;When a truck comes into the shop with extreme ice that is rock hard, is there a way to remove it fast? Any tricks of the trade?&lt;/p&gt;&#xA;" OwnerUserId="9574" LastEditorUserId="7475" LastEditDate="2015-10-30T17:15:50.750" LastActivityDate="2015-10-30T17:15:50.750" Title="Melting ice off of trucks fast" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18993" PostTypeId="2" ParentId="18991" CreationDate="2014-10-30T21:37:49.600" Score="20" Body="&lt;p&gt;First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene: &lt;img src=&quot;https://i.stack.imgur.com/1R8aL.png&quot; alt=&quot;beta carotene&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The details aren't dreadfully important, but you'll notice that the molecule has a lot of conjugated double bonds. This is what allows the molecule to absorb visible light and this conjugation is what produces the colour in many things that might stain clothing or the like. The size of this conjugated system affects how it absorbs light. In general, the more conjugation, the more absorption occurs and at longer wavelengths (towards red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substances used as bleaches tend to be oxidizing agents. Again, the details aren't too important, but one thing oxidation can do to these conjugated molecules is break them apart or convert double bonds to other functionalities. By disrupting the conjugated system, one changes the absorption properties of the molecule. Breaking the molecule apart or removing double bonds shrinks the extent of conjugation, reducing absorption and shifting it to shorter wavelengths. If the wavelength is short enough, the absorption is the in the UV range, which is invisible to the human eye, making it colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to your questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From what I understand, bleach works to weaken the ability of a substance to absorb light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is as described above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thus this seems to be &quot;whitening&quot; i.e. reflecting all light.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reflective/scattering properties are distinct from absorption. e.g. if I add bleach to an aqueous solution of a dye, it may go from a coloured clear solution to a colourless clear solution, but if I bleach a shirt stained with a dye, the shirt doesn't become transparent—the reflective properties are determined by the fabric, not the dye. Bleach simply changes the structure and thus, absorption of the dye, it doesn't render them reflective.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So is 'colorless' a misnomer when used to describe bleach?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Colourless is the correct term to describe the effects of bleach—it makes coloured substances colourless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, to double check when you bleach something, is it still there? Only that it is now white??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The coloured molecules undergo a chemical change so they aren't the same substances with which you started. Whether the products of the reaction are still there depends on what they are. It may very well be that the oxidation products of a dye are more soluble and easier to remove when you wash a piece of clothing. The bleaching itself implies a chemical reaction that alters the absorption of coloured substances, not their removal.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-10-30T21:37:49.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18995" PostTypeId="1" AcceptedAnswerId="18996" CreationDate="2014-10-31T00:10:55.580" Score="2" ViewCount="430" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A $4~\mathrm{g}$ mixture of sucrose ($\ce{C12H22O11}$) and zinc nitrate ($\ce{Zn(NO3)2}$) is dissolved in $150~\mathrm{g}$ of water. If the resulting solution freezes at $-0.768~^\circ\mathrm{C}$, what is the mass percent of sucrose in the mixture? Assume that zinc nitrate dissociates completely in water, $K_\mathrm f = 1.86$. (Answer: $2.8\%$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhgs5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9576" LastEditorUserId="5026" LastEditDate="2018-02-28T14:42:24.257" LastActivityDate="2018-02-28T14:42:24.257" Title="How to determine the composition for a sucrose/ zinc nitrate solution from its freezing point?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18996" PostTypeId="2" ParentId="18995" CreationDate="2014-10-31T01:19:53.380" Score="2" Body="&lt;p&gt;First find the molality of solute particles.  Assume i = 1; we're just counting how many things are dissolved and it doesn't matter whether they are ions or molecules at this point.  I'm using m for mass and b for molality throughout to avoid confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T = ibk_f\implies\frac{\Delta T}{k}=b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b=\frac{0.768}{1.86}=\pu{0.413 molal}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next find how many moles of particles were dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{\text{solute}}=b\times m_{\text{solvent}}=0.413\times 0.150 = 0.0619~\text{moles of particles}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now determine the mass of each species based on this total number of moles.&lt;br&gt;&#xA;$n_{\text{particles}}=3n_{\ce{Zn(NO3)2}}+n_{\ce{C12H22O11}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_{\text{particles}}=3\frac{m_{\ce{Zn(NO3)2}}}{M_{\ce{Zn(NO3)2}}}+\frac{m_{\ce{C12H22O11}}}{M_{\ce{C12H22O11}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the total number of moles of particles, 0.0619, and you know the molar masses of both zinc nitrate and sucrose, but you don't know either the actual mass of zinc nitrate or of sucrose.  However, you do have one other piece of information that allows you to set up a system of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{\text{solute}}=m_{\ce{Zn(NO3)2}}+m_{\ce{C12H22O11}}=\pu{4g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system of equations should give you the masses of both solutes.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="5026" LastEditDate="2018-02-28T14:41:26.967" LastActivityDate="2018-02-28T14:41:26.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18997" PostTypeId="1" AcceptedAnswerId="19493" CreationDate="2014-10-31T06:46:28.860" Score="3" ViewCount="803" Body="&lt;p&gt;Till now I have heard $\ce{CO2}$ is the main reason of global warming. I have studied the remedies to control $\ce{CO2}$ many times through books, awarness programs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But today I learned about Chlorofluorocarbon ($\ce{CFC}$). It have &lt;strong&gt;38 million&lt;/strong&gt; times more &quot;Global Warming Potential&quot; than $\ce{CO2}$    according to UN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my book they have mention the sources of $\ce{CFC}$ like aerosols, refrigerators, air-conditioners, foams in fire extinguisher, etc.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it does not tell what effort we should make or scientist are doing in order to reduce harmful of $\ce{CFC}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that $\ce{CFC}$ is replaced by &lt;a href=&quot;http://en.wikipedia.org/wiki/1,1,2,2-Tetrachloroethane&quot; rel=&quot;nofollow&quot;&gt;1,1,2,2-Tetrachloroethane&lt;/a&gt;, nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know what else we are trying to reduce harmful effect of  $\ce{CFC}$?   &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-04-19T10:07:59.027" Title="Efforts made to reduce the amount of CFC" Tags="&lt;environmental-chemistry&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18998" PostTypeId="1" AcceptedAnswerId="19007" CreationDate="2014-10-31T09:15:33.080" Score="1" ViewCount="3807" Body="&lt;p&gt;We know that in an atom charge of an electron at infinity is zero. As it approaches the nucleus it become more and more negatively charged.&lt;br /&gt;&#xA;We also know that E.P.E is positive when charges are like and negative when charges are opposite.&#xA;Well does it mean that as an electron approaches nucleus there is repulsion which causes the approaching electron to go away. As the electron go away it is getting closer to to infinity point because of which electron become less and less negatively charged. So the E.P.E is positive .Nucleus of atom is like earth and the electron are objects.The farther the object move from earth it has more Potential energy.&lt;br /&gt;&lt;/p&gt;&#xA;" OwnerUserId="8231" LastActivityDate="2014-10-31T17:25:16.153" Title="Why is Electrostatic Potential Energy positive when the charges are like?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18999" PostTypeId="2" ParentId="18995" CreationDate="2014-10-31T09:16:52.623" Score="0" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&#xA;$$i\neq3\text{, but }\underbrace{\frac x{342}}_{\text{sucrose}}+\underbrace{\frac{4-x}{189}\times3}_{\text{zinc-nitrate}}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5026" LastEditDate="2018-02-28T14:42:02.593" LastActivityDate="2018-02-28T14:42:02.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19001" PostTypeId="1" AcceptedAnswerId="32370" CreationDate="2014-10-31T11:05:00.080" Score="13" ViewCount="25647" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Active mass&lt;/strong&gt; is defined as the molar concentration ie. number of Gram-moles per litre.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My book then wrote&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Active mass of pure solid/liquid is always 1 .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book reasoned that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Molar concentration is directly proportional to density. Since density of solid or liquid always remains constant, the active mass is taken as 1 .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Really? Doesn't the density of the solid reactant decrease when the forward reaction is about to reach the equilibrium??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a word, I am greatly confused. What is the reason for taking the active mass 1 ? Please help.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="9961" LastEditDate="2015-10-23T16:58:17.850" LastActivityDate="2021-01-08T06:03:44.640" Title="Why is active mass of a pure solid or liquid always taken as unity?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19002" PostTypeId="1" AcceptedAnswerId="19005" CreationDate="2014-10-31T14:08:44.463" Score="4" ViewCount="5147" Body="&lt;p&gt;From the &lt;a href=&quot;http://sodeliciousdairyfree.com/products/coconut-milk-creamers/creamer-french-vanilla&quot; rel=&quot;nofollow&quot;&gt;manufacturers website&lt;/a&gt; of a coconut french vanilla coffee creamer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ingredients: ORGANIC COCONUT MILK (WATER, ORGANIC COCONUT CREAM), DRIED CANE SYRUP, NATURAL FLAVORS, &lt;strong&gt;COLORED WITH TITANIUM DIOXIDE&lt;/strong&gt;, DIPOTASSIUM PHOSPHATE, CARRAGEENAN, GUAR GUM.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My guess, from looking at the abstract of the report (which is behind a paywall), &quot;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3288463/&quot; rel=&quot;nofollow&quot;&gt;Titanium Dioxide Nanoparticles in Food and Personal Care Products&lt;/a&gt;&quot; is that the the metal is simply used for the (white) coloring. Is this all it is used for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the paper abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... human exposure analysis to $\ce{TiO2}$ through foods identified children as having the highest exposures because $\ce{TiO2}$ content of sweets is higher than other food products, and that a typical exposure for a US adult may be on the order of 1 mg Ti per kilogram body weight per day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there any potential risk (toxicity) to consumption of $\ce{TiO2}$ at these levels?&lt;/p&gt;&#xA;" OwnerUserId="9582" LastEditorUserId="4356" LastEditDate="2014-10-31T16:36:36.193" LastActivityDate="2017-09-23T16:54:45.803" Title="Why does my coffee creamer have titanium dioxide in it and is it safe?" Tags="&lt;food-chemistry&gt;&lt;toxicity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19003" PostTypeId="1" AcceptedAnswerId="20256" CreationDate="2014-10-31T14:39:17.140" Score="15" ViewCount="4801" Body="&lt;p&gt;I'm trying to grow oyster mushrooms that like high temperatures. They look great in an indoor terrarium, and can eventually yield several tasty meals :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However oyster mushrooms are hard to grow indoors, because they need low $\ce{CO2}$ levels (&amp;lt;1000 ppm according to Paul Stamets), AND high humidity (&gt;95%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warm climate oysters tend to grow fast, and can generate quite a lot of $\ce{CO2}$ fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I vent the air in tent about 10 times a day, humidity will suffer (and I can't do that all the time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I close the tent, mushrooms will be stunted and club-like, and yields will be next to nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about putting a tray with $\ce{NaOH}$ crytals on it, probably with a fan to slightly move the humid air around a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other method would be to use an aquarium pump to bubble air through NaOH solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does this look like something feasible? Can some $\ce{NaOH}$ remove enough $\ce{CO2}$ from the air to keep it under 1000ppm?&lt;/li&gt;&#xA;&lt;li&gt;Which of the above methods seem to be the better one given the aquarium pump can bubble about 100-200 litres of air in every hour and the tent is about 500 L ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9583" LastEditorUserId="7475" LastEditDate="2015-10-30T12:31:33.947" LastActivityDate="2019-11-07T05:25:22.483" Title="CO2 removal from mushroom growing tent" Tags="&lt;everyday-chemistry&gt;&lt;extraction&gt;" AnswerCount="3" CommentCount="14" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19004" PostTypeId="2" ParentId="18992" CreationDate="2014-10-31T14:54:37.520" Score="6" Body="&lt;p&gt;Water has a high heat capacity, so I agree with JoeHobbit that a spray of hot water via hose would probably work best. A power washer that is capable of heating the water might be worth getting. A secondary advantage of doing this is that the water spray would remove accumulated road salt, which accelerates corrosion. Short of that, a space heater that blows hot air on the truck would be a less effective but also less messy solution. A moving stream of hot water or air should be far more effective than static immersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small amounts of ice deposits can be removed with an aerosol deicer spray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, there is actually a lot of chemistry going on here.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-10-31T14:54:37.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19005" PostTypeId="2" ParentId="19002" CreationDate="2014-10-31T16:18:38.467" Score="9" Body="&lt;p&gt;$\ce{TiO2}$ is is the most widely used white pigment out there, and it is considered to be non toxic. You'll find it used as food coloring commonly in candies, but in this instance  it's in your coconut sweetener. Aside from food, it's used as a primary pigment in paint (the more $\ce{TiO2}$ generally the higher the quality of paint, because it results in higher opacity), solar cells, toothpaste, and as you've mentioned, sunscreen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only health concerns I found were from inhalation &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Titanium dioxide dust, when inhaled, has been classified by the International Agency for Research on Cancer (IARC) as an IARC Group 2B carcinogen, meaning it is possibly carcinogenic to humans. The findings of the IARC are based on the discovery that high concentrations of pigment-grade (powdered) and ultrafine titanium dioxide dust caused respiratory tract cancer in rats exposed by inhalation and intratracheal instillation. The series of biological events or steps that produce the rat lung cancers have also been seen in people working in dusty environments. Therefore, the observations of cancer in animals were considered, by IARC, as relevant to people doing jobs with exposures to titanium dioxide dust.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-10-31T16:18:38.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19007" PostTypeId="2" ParentId="18998" CreationDate="2014-10-31T17:19:09.407" Score="4" Body="&lt;p&gt;There is no way to know the actual potential energy of a system, we can only determine it relative to some predefined zero value.  In the case of electrostatic potential, zero potential energy is defined as the energy between charges when their separation is infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge of the electron doesn't change in this scenario; it is always $1.6\times 10^{-19}\ C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When oppositely charged particles near one another, they lose potential energy (their energy becomes more negative.)  When particles with the same charge near one another, they gain potential energy.  The attractive force between oppositely charged particles does work as their separation decreases, and the repulsive force between similarly charged particles requires work to be done on them in order to move closer to one another (so potential energy goes up.)&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-10-31T17:25:16.153" LastActivityDate="2014-10-31T17:25:16.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19009" PostTypeId="2" ParentId="14923" CreationDate="2014-10-31T20:05:44.053" Score="15" Body="&lt;p&gt;An alternative, non-destructive, physical method:&lt;br&gt;&#xA;Potassium hydroxide (KOH) is radioactive due to the presence of naturally occurring &lt;sup&gt;40&lt;/sup&gt;K. With sufficient counting time, even a simple radiation detector can be used to distinguish the radiation of a nice big sample of potassium hydroxide from the natural background radiation in the room.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-10-31T20:05:44.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19010" PostTypeId="1" AcceptedAnswerId="19011" CreationDate="2014-10-31T20:39:50.107" Score="1" ViewCount="185" Body="&lt;p&gt;A follow up question from this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18832/does-combining-solid-liquid-concentration-with-k-c-matter-when-comparing-k-c/18892#18892&quot;&gt;answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compare:&lt;br&gt;&#xA;  $\ce{A(aq) +B(aq)-&amp;gt;AB(aq)}$&lt;br&gt;&#xA;  $\ce{2A(aq) +B(aq)-&amp;gt;A2B(aq)}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a way to &quot;normalize&quot; the $K_c$ of reactions like that shown above, such that a fair comparison can be made, regarding whether one reaction is more product favoured than the other?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-11-26T05:17:37.107" Title="Is it possible to normalize the concentration equilibrium constant between different reactions for comparison?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19011" PostTypeId="2" ParentId="19010" CreationDate="2014-10-31T21:06:47.973" Score="2" Body="&lt;p&gt;Since the equilibrium constant depends on the concentration of the species involved, there is no way to normalize it in the way I think you imagine it. A fair comparison of equilibrium constants is already given. The equilibrium constant is, as its name gives away already, &lt;em&gt;constant&lt;/em&gt;. As such there is no normalization to be made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the quote is slightly off the point where it says that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the equilibrium constant might actually be the same (except for the unit, of course). The way you calculate the constant changes, but inferring from that that the value changes drastically is simply overreaching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $K &amp;gt; 1$ then the equilibrium lies on the side of the products, if $K&amp;lt;1$ then the equilibrium lies on the side of the reactants. &lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-31T21:06:47.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19012" PostTypeId="2" ParentId="16869" CreationDate="2014-10-31T21:08:55.363" Score="3" Body="&lt;p&gt;Your synthesis looks reasonable, and it is correct to perform sulfonation as the very last step, because the alkanoyl group would be &lt;em&gt;meta&lt;/em&gt;-directing for the reason you have provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can simplify your synthesis by reducing the carbonyl group away in one step with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clemmensen_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clemmensen reduction&lt;/a&gt;, which works well for aryl-alkyl ketones produced by Friedel-Crafts acylation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vAiHE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wolff%E2%80%93Kishner_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolff-Kishner&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Mozingo_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mozingo&lt;/a&gt; reaction could be used. Reducing the number of steps in your synthesis would in practice have the benefit of increased overall yield for the final product, as the yield of each single step is usually &amp;lt; 100 %.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-10-31T21:08:55.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19013" PostTypeId="1" AcceptedAnswerId="19015" CreationDate="2014-10-31T21:35:45.990" Score="3" ViewCount="2320" Body="&lt;p&gt;Is a molecule with no symmetry necessarily polar? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a polar molecule still have some elements of symmetry (i.e. mirror image is the same as itself)? I think so ... because we can have achiral tetrahedral carbon molecules that are polar - i.e. methylene chloride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it proper to say that a molecule that lacks symmetry is necessarily polar? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or can a more nuanced statement based on symmetry be made? Is there a &quot;minimum&quot; level of symmetry needed to achieve non-polarity? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-01T01:19:01.853" Title="&quot;Asymmetric molecules are necessarily polar&quot;" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19014" PostTypeId="2" ParentId="19013" CreationDate="2014-10-31T21:43:45.393" Score="2" Body="&lt;p&gt;Consider the following (absolutely imaginary) molecule: $\ce{CX^1X^2X^3X^4}$ where the $\ce{X}$ atoms are all of a different kind but have the same electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This molecule is asymmetrical but has no dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While an imaginary example is nice and well, I think that in reality, electronegativities will always differ, if even ever so slightly. As a result, there will be a total dipole moment on a molecule with no symmetry. The question you really have to ask now is: &lt;b&gt;What is the threshold for a molecular dipole moment to consider the molecule polar?&lt;/b&gt;&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-10-31T21:43:45.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19015" PostTypeId="2" ParentId="19013" CreationDate="2014-10-31T23:20:24.807" Score="5" Body="&lt;p&gt;Let's start by &lt;strong&gt;defining&lt;/strong&gt; a polar molecule as a molecule with a dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is a molecule with no symmetry necessarily polar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, a molecule with no symmetry belongs to point group $\ce{C_1}$ (e.g. the only symmetry element it contains is a $\ce{C_1}$ axis, rotation of the molecule 360/1 degrees converts the molecule back into itself).  The molecule is asymmetric and polar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a polar molecule still have some elements of symmetry (i.e. mirror&#xA;  image is the same as itself)? I think so ... because we can have&#xA;  achiral tetrahedral carbon molecules that are polar - i.e. methylene&#xA;  chloride. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;, methylene chloride, vinyl chloride, methanol, chloroform, etc. all contain planes of symmetry, but are still polar.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, is it proper to say that a molecule that lacks symmetry is&#xA;  necessarily polar? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, this is a repeat of your first question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or can a more nuanced statement based on symmetry be made? Is there a&#xA;  &quot;minimum&quot; level of symmetry needed to achieve non-polarity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All asymmetric molecules are polar.  Such molecules belong to point group $\ce{C_1}$&lt;/li&gt;&#xA;&lt;li&gt;Molecules belonging to point group $\ce{C_{n}}$ with n&gt;1 are polar. 1,3-dimethylallene is an example (it has $\ce{C_2}$ symmetry, rotate it by 360/2 degrees about its $\ce{C_{2}}$ axis and it converts back to its original representation)&lt;/li&gt;&#xA;&lt;li&gt;Molecules belonging to point group $\ce{C_{s}}$ are polar.  A sigma plane either contains or bisects molecules in this point group, and this plane is the only symmetry element.  Methanol, bromocyclopropane, vinyl chloride, etc. belong to this point group and are polar.&lt;/li&gt;&#xA;&lt;li&gt;Molecules belonging to point group ${C_{nv}}$ are polar.  Formaldehyde and methylene chloride (both ${C_{2v}}$), chloroform (${C_{3v}}$) and hydrogen chloride (${C_{\infty v}}$) are examples.  This point group contains a ${C_{n}}$ axis and n sigma planes containing the axis.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the molecule is not in one of these point groups, then it is not polar and does not have a dipole moment&lt;/strong&gt; (or said differently, molecules in all other point groups [${D_n, S_n, C_{nh}, D_{nd}, D_{nh}, T_d, O_h}$ and higher] are not polar and do not have dipole moments).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-01T01:19:01.853" LastActivityDate="2014-11-01T01:19:01.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19016" PostTypeId="1" CreationDate="2014-11-01T03:53:50.260" Score="1" ViewCount="613" Body="&lt;p&gt;I'm working out the best process to chemically strip silver brazing from tungsten carbide (WC) bits. My main objective is to recover the silver as a relatively pure metal. When I use $\ce{HNO3}$ at anywhere from 20-70%, it dissolves the solder and I end up with a red solution.  When I try to use cementation (putting copper metal in the solution) to precipitate the $\ce{Ag}$ as sponge, it isn't working. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any one help me figure out what's going on here?  The brazing being used should be 50-60% $\ce{Ag}$ by weight. I'm not sure yet if there is $\ce{Pd}$ or not.&lt;/p&gt;&#xA;" OwnerUserId="9587" LastEditorUserId="5026" LastEditDate="2018-05-05T01:35:17.720" LastActivityDate="2019-01-01T15:01:05.233" Title="Dissolving silver braze from tungsten carbide" Tags="&lt;metallurgy&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="19017" PostTypeId="1" AcceptedAnswerId="19051" CreationDate="2014-11-01T04:40:39.520" Score="0" ViewCount="299" Body="&lt;p&gt;Can imines be catalytically hydrogenated? According to the DEA this appears to be possible. However I cannot find any other references to catalytic hydrogenation of imines anywhere else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mIuZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thedea.org/synthesis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://thedea.org/synthesis.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4356" LastEditDate="2014-12-03T04:10:01.250" LastActivityDate="2014-12-03T04:10:01.250" Title="Can imines be catalytically hydrogenated?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19018" PostTypeId="1" AcceptedAnswerId="19019" CreationDate="2014-11-01T06:52:48.840" Score="0" ViewCount="148" Body="&lt;p&gt;This seems stupid question, but from many years I am studying that burning/combustion fossil fuel cause pollution because it releases $\ce{CO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have recently read about $\ce{CFC}$ &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18997/efforts-made-to-reduce-the-amount-of-cfc&quot;&gt;here&lt;/a&gt;, which is more dangerous than $\ce{CO2}$ and seems it is not release by burning of fossil fuel then why are scientist more concerned about fossil fuel instead of $\ce{CFC}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want proper explanation Why fossil fuel is too dangerous?&lt;/p&gt;&#xA;" OwnerDisplayName="user7484" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-11-24T06:34:20.890" Title="Why fossil fuel is too dangerous?" Tags="&lt;experimental-chemistry&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19019" PostTypeId="2" ParentId="19018" CreationDate="2014-11-01T07:33:26.313" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why fossil fuel is too dangerous?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Combustion of fossil fuel does not only release $\ce{CO2}$ but many other harmful substances(pollutants) like oxides of sulphur ($\ce{SO_{x}}$), oxides of nitrogen($\ce{NO_{x}}$), Carbon Monoxide($\ce{CO}$), etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;oxides of sulphur ($\ce{SO_{x}}$):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When sulphur containing fossil fuel is burnt, then oxides of sulphur is produced. Sulphur Dioxide($\ce{SO2}$) is generally produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low concentration of $\ce{SO2}$ causes diseases like asthma, swelling and irritation of respiratory tract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High concentration $\ce{SO2}$ leads to stiffness of flower buds, which eventually fall off! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Oxides of Nitrogen($\ce{NO_{x}}$):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in automobile engine, fossil fuel is burnt at high temperature significant amount of nitric oxide($\ce{NO}$) and nitrogen dioxide($\ce{NO2}$) is released. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO2}$ is lung irritant. it is highly toxic to living tissue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO2}$ decrease the rate of photosynthesis, cause respiratory diseases in children, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carbon monoxide($\ce{CO}$):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO}$ is produced due to incomplete combustion of carbon. It is major constituent in smoke of vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO}$ binds with haemoglobin and form carboxyhemoglobin, which is 300 times more stable than usual oxygen-haemoglobin! If concentration of carboxyhemoglobin reaches 3 to 4% then oxygen carrying capacity of blood reduces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Few more!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO2}$ with smoke and fog is responsible for classical smog(&lt;a href=&quot;http://en.wikipedia.org/wiki/Great_Smog&quot; rel=&quot;nofollow&quot;&gt;London smog&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NO2}$ and $\ce{NO}$ is responsible foe photochemical smog(&lt;a href=&quot;http://en.wikipedia.org/wiki/User:Natalie_cali222/Los_Angeles_Smog&quot; rel=&quot;nofollow&quot;&gt;Los-Angeles smog&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading more about environmental chemistry and i discovered that $\ce{NO_{x}}$ is responsible for 40% depletion  of concentration of Ozone layer!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why are scientist more concerned about fossil fuel instead of CFC?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As commented by LDC3 The production of the more hazardous CFC's has been banned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 95 derivatives of chloro and bromo fluorocarbon which are listed as ODS(Ozone Depleting Substances).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On 16th September, 1987 93 countries sign Montreal Protocol and accept not to use ODS. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://gujarat-education.gov.in/textbook/Images/11sem2/chemistry-11eng/chap7.pdf&quot; rel=&quot;nofollow&quot;&gt;GSEB textbook&lt;/a&gt; &lt;sup&gt;pdf&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2014-11-24T06:34:20.890" LastActivityDate="2014-11-24T06:34:20.890" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19020" PostTypeId="1" CreationDate="2014-11-01T08:26:49.670" Score="0" ViewCount="847" Body="&lt;p&gt;My question is about the condenser used in a Soxhlet extractions. It's unusual looking compared to the rest of condensers, used in distillation for example, I've noticed it has a thin and short tube on top and my question is, do one leave it open when performing a Soxhlet extraction or should it be stopped with a stopper or something? &lt;/p&gt;&#xA;" OwnerUserId="9590" LastEditorUserId="7475" LastEditDate="2015-10-09T17:02:50.833" LastActivityDate="2015-10-09T17:02:50.833" Title="Is the condenser for a Soxhlet extraction left open on top" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19022" PostTypeId="2" ParentId="19020" CreationDate="2014-11-01T10:07:11.017" Score="5" Body="&lt;p&gt;First thing, you should NEVER perform any heated reaction with a stoppered condenser. You shouldn't even be allowed to work independently in the lab if this rule is not coming automatic to you. I am not sure what you mean by a special condenser being used in Soxhlet extractions. You can use any condenser that fits. Since you don't do a distillation you can use a shorter condenser so your whole Soxhlet set-up has a reasonable height.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-11-01T10:07:11.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19023" PostTypeId="1" AcceptedAnswerId="19025" CreationDate="2014-11-01T13:43:46.210" Score="12" ViewCount="8510" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/kspgjhV&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/kspgjhV.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct mechanism here? Would S&lt;sub&gt;N&lt;/sub&gt;1 occur here? I chose S&lt;sub&gt;N&lt;/sub&gt;2 because we have a secondary alkyl halide with a weak base $\ce{CN-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution says S&lt;sub&gt;N&lt;/sub&gt;1 is major. Can you explain why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My friend got full credit for this:&lt;br&gt;&#xA;&lt;a href=&quot;https://imgur.com/IXVgRH7&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/IXVgRH7.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-04-03T15:59:07.270" LastActivityDate="2018-04-03T15:59:07.270" Title="Will this beta-branched secondary alkyl halide undergo SN1 or SN2?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19024" PostTypeId="2" ParentId="19017" CreationDate="2014-11-01T14:14:29.953" Score="0" Body="&lt;p&gt;Several methods on the reduction of imines are outlined on &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogenation_of_carbon%E2%80%93nitrogen_double_bonds&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;. Another method using catalytic hydrogenation is given &lt;a href=&quot;http://en.wikipedia.org/wiki/Imine#Imine_reductions&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-01T14:14:29.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19025" PostTypeId="2" ParentId="19023" CreationDate="2014-11-01T14:36:35.690" Score="13" Body="&lt;p&gt;Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full mechanism:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/19nO3.jpg&quot; alt=&quot;full mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.&lt;br&gt;&#xA;I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T22:22:35.683" LastActivityDate="2015-06-27T22:22:35.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19026" PostTypeId="1" CreationDate="2014-11-01T15:10:19.723" Score="2" ViewCount="85" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/WsT42zQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/WsT42zQ.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could find two products:&#xA;The first one where is $\ce{Cl}$ is replaced by $\ce{CN}$.&#xA;AND the second one where $\ce{CH3CH2O}$ replaces $\ce{Cl}$.&#xA;Did I make a mistake or miss anyone?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4231" LastEditDate="2014-11-01T15:45:01.810" LastActivityDate="2014-11-04T22:14:52.260" Title="Predict the Product(s) of a reaction of tertiary benzylic alkyl halide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19027" PostTypeId="1" AcceptedAnswerId="19032" CreationDate="2014-11-01T18:18:23.103" Score="15" ViewCount="20200" Body="&lt;p&gt;Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon - i.e. methyl groups donate negative charge for example. Inductively we can argue that carbon is of greater electronegativity than hydrogen, and so carbon withdraws negative charge from hydrogen, and thus negative charge is &quot;donated&quot; to whatever that the methyl is attached to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a hyperconjugative effect we can argue that there is some interaction between the C-H orbitals and some antibonding orbital ... how is this supposed to be stabilizing if it involves an antibonding orbital? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-03-09T22:14:30.013" LastActivityDate="2017-03-09T22:14:30.013" Title="Inductive effect and hyperconjugation - one elephant, different parts?" Tags="&lt;electronegativity&gt;&lt;hyperconjugation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19028" PostTypeId="1" AcceptedAnswerId="19030" CreationDate="2014-11-01T18:27:09.687" Score="2" ViewCount="545" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/DzoY7AG&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/DzoY7AG.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would one get 1-(1-chloroethyl)-1-methylcyclobutane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if the driving force identification is correct? Is my going from a 2° to a 3° carbocation correct? Does anyone have any tips for knowing when to look out for rearrangements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://imgur.com/oCmOP0a&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/oCmOP0a.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-11-01T18:52:17.567" Title="Carbocation Rearrangements in SN1 reactions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19029" PostTypeId="1" AcceptedAnswerId="19031" CreationDate="2014-11-01T18:40:15.837" Score="2" ViewCount="953" Body="&lt;p&gt;Does resonance always stabilize a molecule or can resonance effects destabilize molecules? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only example I can think of in which resonance (more accurately, conjugation) destabilizes molecules is in the cases of anti-aromatic compounds. Is this a valid example, and are there any more such examples? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-01T18:59:02.853" Title="Is resonance always stabilizing" Tags="&lt;resonance&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19030" PostTypeId="2" ParentId="19028" CreationDate="2014-11-01T18:52:17.567" Score="3" Body="&lt;p&gt;I think the solvolysis would involve rearrangement to the cyclopentyl carbocation as indicated in the figure below.  The driving force for the rearrangement is two-fold&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;without rearrangement, the cyclobutyl compound would generate a secondary carbocation; with rearrangement a more stable tertiary carbocation is generated instead&lt;/li&gt;&#xA;&lt;li&gt;release of strain as the 4-membered ring opens to a 5-membered ring&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/deoW8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tertiary cyclopentyl carbocation ultimately captures a solvent molecule as indicated.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-01T18:52:17.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19031" PostTypeId="2" ParentId="19029" CreationDate="2014-11-01T18:59:02.853" Score="3" Body="&lt;p&gt;Resonance only occurs when the result is beneficial, that is when a more-stabilized molecule will result. If resonance (delocalization) occurred in a compound such as cyclobutadiene to produce a square molecule with equivalent bond lengths, an antiaromatic molecule would be the result.  Therefore, resonance does not occur and a square molecule with equivalent bond lengths is not produced.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-01T18:59:02.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19032" PostTypeId="2" ParentId="19027" CreationDate="2014-11-01T19:25:57.133" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are the inductive effect and hyperconjugation both different ways of&#xA;  looking at the same phenomenon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around.  I tend to separate the two effects as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;inductive effects are associated with electron movement through sigma bonds due to electronegativity differences&lt;/li&gt;&#xA;&lt;li&gt;resonance effects are associated with electron movement through p orbitals due to electronegativity differences&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, we often say that a methyl group stabilizes a double bond (carbon-carbon or carbonyl) or a carbocation.  The methyl group is roughly $\ce{sp^3}$ hybridized while the double bond carbon or carbocation carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow from $\ce{sp^3}$ to $\ce{sp^2}$ orbitals.  The direction of this electron flow also indicates  that $\ce{sp^2}$ orbitals are more electronegative than $\ce{sp^3}$ orbitals.  How can the $\ce{sp^3}$ methyl group shift electrons to the lower energy, more electronegative $\ce{sp^2}$ carbon?  It can do it inductively (through sigma bonds) or through resonance (through p orbitals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added a picture showing resonance structures involving hyperconjugation, just to be clear as to what hyperconjugation involves.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VK1PH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-01T19:25:57.133" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19034" PostTypeId="1" CreationDate="2014-11-01T21:02:34.080" Score="4" ViewCount="1061" Body="&lt;p&gt;I have researched three books: Brown's, Solomons' and Wade's. I could not find enough information about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My prof. says that secondary alkly halides undergo SN1 only under acidic environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had this quiz. I got a terrible mark on one question.&#xA;&lt;a href=&quot;https://imgur.com/lNvg3rS&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/lNvg3rS.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think these are the missing structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://imgur.com/kscJzgC&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/kscJzgC.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am wondering, did Sn2 occur here, if yes, at what rate? Was it more important than Sn1? What was the major pathway?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the major pathway for all secondary alkyl halides with acidic environment and with NONacidic environments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone also just check my missing structures?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-11-03T03:09:20.297" Title="SN2 vs SN1 in Acidic Environments for Secondary Alkyl Halides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19035" PostTypeId="1" AcceptedAnswerId="19038" CreationDate="2014-11-01T21:00:07.843" Score="1" ViewCount="5553" Body="&lt;p&gt;I'd like to understand the chemical reaction in a typical alkaline battery a little better and how that produces a flow of electrons.  I read wiki, how stuff works, and some more advanced articles.  They all seemed to either stop short of explaining the nitty gritty details, or went straight to chemical formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a chemical reaction occurs between the cathode and the anode and that the electrolyte helps make that happen.  I don't really understand the flow of electrons though, and how the anode/cathode are consumed.  I read a bit about ions but not sure I grasped how that was moving electrons about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does this reaction start right away or only after a circuit is completed and electrons can flow?  Or has the reaction already started, and there is an excess of electrons on one plate that only start flowing after the circuit is completed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What forces the potential to stop at 1.5V?&lt;/p&gt;&#xA;" OwnerUserId="9597" OwnerDisplayName="confused" LastEditorUserId="865" LastEditDate="2015-11-08T19:00:34.150" LastActivityDate="2015-11-08T19:00:34.150" Title="How do chemical reactions taking place in an alkaline battery produce electrons?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19036" PostTypeId="1" AcceptedAnswerId="19064" CreationDate="2014-11-01T21:23:16.383" Score="2" ViewCount="133" Body="&lt;p&gt;I did this quiz and I forgot the hydride shift.&#xA;&lt;a href=&quot;https://imgur.com/uRLkE3U&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/uRLkE3U.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the hydride shift, would these be the right answer?&#xA;&lt;a href=&quot;https://imgur.com/GUoURrc&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/GUoURrc.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4356" LastEditDate="2014-12-03T00:11:10.037" LastActivityDate="2014-12-03T00:11:10.037" Title="Am I correctly showing a hydride shift on the alkyl halide?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19038" PostTypeId="2" ParentId="19035" CreationDate="2014-11-02T01:10:54.243" Score="3" Body="&lt;p&gt;An alkaline battery is a specific type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_cell&quot; rel=&quot;nofollow&quot;&gt;galvanic cell&lt;/a&gt;.  As I am sure you found in your research, it involves an oxidation-reduction reaction between zinc($\ce{Zn}$), and manganese(IV) oxide ($\ce{MnO2}$.)  They are called &lt;em&gt;alkaline&lt;/em&gt; batteries because these chemicals are surrounded by an alkaline (basic) solution of potassium hydroxide, $\ce{KOH}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is really one that is more about galvanic cells in general.  When an oxidation-reduction (a.k.a. redox) reaction occurs, electrons are transferred between species.  This sort of reaction is exceedingly common.  That's all that happens in a battery, it's just that people have figured out how to catch the electrons on their way from species A to species B and use their energy to do work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I took a piece of zinc and placed it in a slurry of $\ce{MnO2}$ and $\ce{KOH}$, it would react according to the following balanced chemical equation and produce a lot of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + 2MnO2 &amp;lt;-&amp;gt; ZnO + Mn2O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that I separated the two pieces of the reaction into their own jars.  A piece of zinc placed in a solution of $\ce{KOH}$ in one jar and some $\ce{MnO2}$ mixed with $\ce{KOH}$ in another.  The reactants are all present, but they have no way of interacting, so nothing would happen - there's no way for electrons to move from the zinc to the $\ce{MnO2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I attach a piece of wire to the zinc on one end and into the $\ce{MnO2 - KOH}$ slurry at the other. Now electrons can move from one jar to the other.  If I have a voltmeter that can measure a very brief signal, I can actually see a quick pulse of electrons flowing through the wire, from the zinc to the $\ce{MnO2}$.  The pulse ends almost immediately though, because the movement of electrons away from the zinc and toward the $\ce{MnO2}$ creates a charge separation.  The former becomes positively charged and the latter negatively charged.  Continuing to move electrons against this charge separation is progressively more difficult, and it doesn't take long for the electrostatic force of the charge separation to balance out the electrochemical potential of the reaction and the entire thing just stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this problem, I need a way for the charges to balance back out without allowing the reactants to mix directly.  This is done with some sort of &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_bridge&quot; rel=&quot;nofollow&quot;&gt;salt bridge&lt;/a&gt;.  A salt bridge allows positive and negative ions to move between the jars, but it still keeps the jars separate from one another.  This could be as fancy as an actual salt bridge (a glass tube filled with a gel containing some sort of salt) or as simple as a piece of paper soaked in the $\ce{KOH}$ solution.  Now when I connect the wire, electrons flow from the zinc to the $\ce{MnO2}$, negatively charged $\ce{OH-}$ ions move through the salt bridge toward the zinc, and positively charged $\ce{K+}$ ions move through the salt bridge toward the $\ce{MnO2}$.  The charge stays balanced in both jars, so there is no electrostatic force preventing electrons from moving, and they flow freely through the wire.  If I put a motor in the middle of the wire, it turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Specifics of Alkaline Batteries&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all was regarding galvanic cells in general.  A battery has additional requirements.  It needs to be compact and safe, in particular.  The construction of an alkaline battery involves a zinc core with a paste of manganese(IV) oxide and potassium hydroxide wrapped around it.  The paste and zinc are separated by a paper sheet (the salt bridge in this galvanic cell.)  Because all of the reactants and products are solid, there is no concern that they will diffuse through the cardboard; it just keeps the reactants in their places.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction does not start until the battery is put into a device and the device is turned on.  Flipping that switch gives the electrons from the zinc a pathway to reach the manganese(IV) oxide.  Until the circuit is complete, there is no buildup of electrons at one end of the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical potential (voltage) is the ratio of the energy carried by an object to that object's charge.  One electron in this system has a potential of 1.43V, two electrons carry twice as much energy, but they also have twice the charge, so they also have a potential of 1.43V.  Electrical potential is sometimes called electromotive force, and that idea, that we're measuring how hard the electrons are being pushed through the wire, may help you understand the battery's behavior better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why 1.43V though?  It's because of the properties of the chemical reactions the battery is based on.  You probably know that if you burn some amount of propane it will produce a given amount of heat, and if you burn an equal amount of acetylene, it will produce a different amount of heat.  The same thing is true here.  The energy associated with the &lt;em&gt;half reactions&lt;/em&gt; in each part of the battery depends on the chemicals involved.  Rather than list a total amount of energy available, batteries list the energy per charge, voltage.  This battery's half reactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + 2OH- -&amp;gt; ZnO + H2O + 2e-}$  E°=1.28V&lt;br&gt;&#xA;and&lt;br&gt;&#xA;$\ce{2MnO2 + H2O + 2e- -&amp;gt; Mn2O3 + 2OH-}$ E°=0.15V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each half reaction gives a specific amount of energy to the electrons involved in it.  Since the energy:charge ratio (E° above, called a reduction or oxidation potential) is constant, we can use that to find the energy:charge ratio of the entire battery just by adding them together.  Why doesn't it go above 1.43V?  For the same reason that a fire doesn't burn hotter.  The chemicals involved only release so much energy per reaction.  Note: The potential can be changed in some systems by playing around with the concentrations of the chemicals in the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the battery is used up, the zinc slowly but surely turns into zinc oxide and the manganese(IV) oxide turns into manganese(III) oxide.  As their amounts fall to zero, the battery's potential falls and it quits working.  &lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-11-02T02:03:07.560" LastActivityDate="2014-11-02T02:03:07.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19042" PostTypeId="1" AcceptedAnswerId="19046" CreationDate="2014-11-02T02:42:28.803" Score="2" ViewCount="4581" Body="&lt;p&gt;I have a question that asks me to describe why adding distilled water to my vinegar sample would not affect my calculated Molarity (determining amount of acid in the vinegar be titration with a base). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first thought, it wouldn't, because the same amount of acid is still there, we'll still have the same amount of base reacting. So you'll still get the same amount by measuring how much base was needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that got me thinking. There is a lot of diluting acids with water, but isn't water amphiprotic? Why doesn't it react with the acid, or react with it enough to matter at least? What information am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="1362" LastEditDate="2014-11-04T07:57:32.277" LastActivityDate="2014-11-04T07:57:32.277" Title="Does the amount of acid change when I add water to vinegar?" Tags="&lt;acid-base&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19043" PostTypeId="1" CreationDate="2014-11-02T02:42:57.550" Score="0" ViewCount="5685" Body="&lt;p&gt;Find the heat of combustion for 1 mol of $\ce {C21H44}$ in the reaction where some wax is oxidized to elemental C (soot) and water vapor. The Heat of Formation of $\ce {C21H44}$ is -476 kJ/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it says some, does that necessarily change the molar ratio?&#xA;I thought the balanced equation would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C21H44 + 11O2 -&amp;gt; 21C (graphite) + 22H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {O2}$ and elemental carbon are zero for heat of formation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22(-241.818 kJ/mol) - (-476 kJ/mol) = -4843.996 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="9598" LastEditorUserId="4934" LastEditDate="2014-11-02T02:53:52.617" LastActivityDate="2015-01-01T04:06:54.047" Title="Combustion of paraffin" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19044" PostTypeId="2" ParentId="19043" CreationDate="2014-11-02T03:30:45.973" Score="2" Body="&lt;p&gt;The question is ambiguous, unfortunately.  There are two possibilities that I can see.  Either the question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;is asking about a reaction in which some quantity of wax undergoes the reaction you give, in which case you are 100% correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO&lt;sub&gt;2&lt;/sub&gt;, the usual reaction that occurs.  Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information.  The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -&amp;gt; 21 CO2 + 22H2O}$.  &lt;br&gt;&lt;br&gt;(I guess that would be an upper limit and a lower, numerically, since the values are both negative, but this reaction would release far more heat.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-11-02T03:30:45.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19045" PostTypeId="1" CreationDate="2014-11-02T04:10:06.447" Score="-1" ViewCount="64" Body="&lt;p&gt;If someone has an average of 1 litre of Coke or Pepsi a day, would this form too much benzene oxide in the person's body or stomach and eventually effect part of their genetics?&lt;/p&gt;&#xA;" OwnerUserId="5440" LastEditorUserId="41328" LastEditDate="2019-05-21T13:14:15.280" LastActivityDate="2019-05-21T13:14:15.280" Title="Regarding common preservatives and their effect" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19046" PostTypeId="2" ParentId="19042" CreationDate="2014-11-02T04:16:20.387" Score="3" Body="&lt;p&gt;First off, your thinking about the titration volume of a dilute solution vs the concentrate from which it was made is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we don't need to worry about water acting as an acid or a base in this context:  Consider the reaction that occurs between a weak acid, HA, and water (a very weak base.)&#xA;$\ce{HA + H2O &amp;lt;-&amp;gt; A- + H3O+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The products of the reaction are a weak base and a strong acid (hydronium, the strongest acid that can exist in aqueous solution.)  If I add a base to this mixture, it doesn't matter whether that base reacts with an undissociated HA molecule or with a hydronium ion, in either case it has removed a particle of one acidic species from the solution, leaving behind A&lt;sup&gt;-&lt;/sup&gt;, the conjugate base of HA.  The total number of such events that has to occur before the titration is done is a fixed value, regardless of how many of the events involved hydronium or HA.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-11-02T04:16:20.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19047" PostTypeId="2" ParentId="19045" CreationDate="2014-11-02T04:37:32.783" Score="3" Body="&lt;p&gt;Benzene oxide isn't present in Coke or Pepsi.  Some drinks contain sodium benzoate though.  If that is what you're concerned about, there's not really any cause.  Sodium benzoate is not remotely harmful in the concentrations that humans ingest.  &lt;a href=&quot;http://www.fda.gov/Food/IngredientsPackagingLabeling/GRAS/SCOGS/ucm260074.htm&quot; rel=&quot;nofollow&quot;&gt;Here is the FDA's statement&lt;/a&gt; on benzoates and benzoic acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were claims that a mixture of ascorbic acid and benzoates would react to form benzene.  That has been shown to not be an issue.  There were ~100 drinks containing benzoates &lt;a href=&quot;http://www.fda.gov/Food/FoodborneIllnessContaminants/ChemicalContaminants/ucm055815.htm#table1&quot; rel=&quot;nofollow&quot;&gt;tested by the FDA&lt;/a&gt;, and only 5 showed benzene levels higher than allowed in drinking water.  These were reformulated and now show less than 5ppb benzene.  I would suspect that their benzene came from another source rather than this unlikely reaction, but I don't have any evidence for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, neither Coke nor Pepsi even contain benzoate, benzoic acid, or any other compound that is chemically close to benzene.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-11-02T04:37:32.783" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19048" PostTypeId="2" ParentId="19023" CreationDate="2014-11-02T07:40:41.250" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;It is in $2^\circ$ position, so both, $\ce{S_{N}1}$ and $\ce{S_{N}2}$ can happen.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Br}$ is a good leaving group probably the third best in the groups very common(depends on opinion, but it is a good one anyways though):&#xA;$\ce{OTs&amp;gt;I&amp;gt;Br&amp;gt;Cl&amp;gt;...}$ but LG accelerates both reactions&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CN-}$ is a very good nucleophile in the list of common ones(depends on opinion, but it is a good one anyways though): $\ce{PhS- &amp;gt; CN- &amp;gt; I- &amp;gt; ...}$, the second best so $\ce{S_{N}2}$ is favourable.&lt;/li&gt;&#xA;&lt;li&gt;The solvent is water, which is polar protic which highly favours $\ce{S_{N}1}$&lt;/li&gt;&#xA;&lt;li&gt;The temperature is normal so there's no elimination&lt;/li&gt;&#xA;&lt;li&gt;Finally  fourth point overthrows thirds and moreover water reduces nucleophilicity of $\ce{CN^-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-14T06:19:19.953" LastActivityDate="2014-11-14T06:19:19.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19049" PostTypeId="2" ParentId="19026" CreationDate="2014-11-02T07:43:03.500" Score="0" Body="&lt;p&gt;nucleophilicity $CN-&amp;gt;I-&amp;gt;EtO-$ which implies cyanide attacks, no heating means no elimination generally&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-02T07:43:03.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19051" PostTypeId="2" ParentId="19017" CreationDate="2014-11-02T07:54:57.560" Score="0" Body="&lt;p&gt;You can use $\ce{NaBH_3CN ,LiAlH_4,H_2/Ni}$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="73" LastEditDate="2014-11-02T10:22:20.937" LastActivityDate="2014-11-02T10:22:20.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19052" PostTypeId="2" ParentId="19001" CreationDate="2014-11-02T07:56:27.420" Score="1" Body="&lt;p&gt;Density is an intensive property that is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, refractive index, &lt;strong&gt;density&lt;/strong&gt;, and hardness of an object. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably that answers your question, for more detail &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Intensive_and_extensive_properties&quot; rel=&quot;nofollow&quot;&gt;see this&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-20T10:59:16.460" LastActivityDate="2014-11-20T10:59:16.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19055" PostTypeId="1" CreationDate="2014-11-02T08:41:20.383" Score="1" ViewCount="415" Body="&lt;p&gt;Why does Be have a higher ionization energy than B? I know that it has a filled 2s orbital, but what's the more accurate explanation?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-04-19T10:08:19.600" Title="Be and B ionization energies" Tags="&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19058" PostTypeId="1" CreationDate="2014-11-02T11:16:47.220" Score="4" ViewCount="1005" Body="&lt;p&gt;How does a redox reaction between $\ce{NaClO}$ and $\ce{KBr}$ take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will bromide be oxidised to bromine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Br- + CIO- +  H2O  -&amp;gt;  Cl- + 2 OH- +  Br2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or will bromide be oxidised to $\ce{BrO-}$?      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br- + CIO-   -&amp;gt;  BrO- + Cl-}$&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="5456" LastEditDate="2014-11-02T11:58:16.530" LastActivityDate="2014-12-03T17:04:10.573" Title="How does a redox reaction between ClO- and Br- take place?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19059" PostTypeId="1" CreationDate="2014-11-02T11:25:03.570" Score="12" ViewCount="1005" Body="&lt;p&gt;I had the opportunity to work with a NMR spectrometer and decided to learn some of the theory behind it. After finishing &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clayden's&lt;/a&gt; chapter on &lt;span class=&quot;math-container&quot;&gt;$^{1}\textrm{H}$&lt;/span&gt; NMR (Ch. 11) I'm still left with some questions, mostly concerning coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proton coupling is described as the interaction between neighboring (2-4 bonds away) protons magnetic field. This can split the signals by a constant amount (the coupling constant), which depend on the environment of the protons but not on the strength of the magnetic field. Ideally creating a binomial distribution when the neighboring protons have the same shift.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?&lt;/li&gt;&#xA;&lt;li&gt;Why do identical protons not influence each other? Or more generally: why is it that the more identical the influencing protons are with the influenced proton (i.e. &lt;span class=&quot;math-container&quot;&gt;$\Delta\delta$&lt;/span&gt; getting equal to/lower than &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;), the more their &quot;outer&quot; signals vanish? Especially since &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="7448" LastEditDate="2019-09-23T10:38:17.023" LastActivityDate="2019-09-23T10:39:02.190" Title="Why is proton coupling through bonds and how does this affect identical protons?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19061" PostTypeId="1" CreationDate="2014-11-02T14:00:10.207" Score="1" ViewCount="196" Body="&lt;p&gt;for example mercury oxide has neurotoxic properties but isn't organic so, it isn't soluble in lipids and so doesn't cross the blood brain barrier when ingested, however, methylmercury chloride is organic and soluble in lipids thus it can cross the blood brain barrier&lt;/p&gt;&#xA;" OwnerUserId="9603" LastEditorUserId="4231" LastEditDate="2014-11-02T19:32:46.813" LastActivityDate="2014-11-03T14:11:56.193" Title="is there a lead compound that crosses the blood-brain barrier when ingested?" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19062" PostTypeId="2" ParentId="19034" CreationDate="2014-11-02T14:11:08.323" Score="4" Body="&lt;p&gt;Note that the starting material is drawn as a chiral molecule.  Solvolysis of this starting material will generate a symmetric, &lt;strong&gt;achiral&lt;/strong&gt;, allylic carbocation.  Since a stabilized allylic carbocation is generated, the mechanism likely proceeds through an $\ce{S_{N}1}$ mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ends (there is positive charge delocalized at both ends of an allylic carbocation - draw the resonance structures and convince yourself)  &lt;strong&gt;and&lt;/strong&gt; both faces (top and bottom) of the allylic carbocation can capture the nucleophilic solvent molecule to restore neutrality as shown below. This results in the formation of two diastereomers of the 3,5-disubstituted cyclohexene - the &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;tran&lt;/em&gt; isomers. For each diastereomer, both enantiomers are also produced, as they must be, having passed through an achiral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1lLH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-02T14:56:16.067" LastActivityDate="2014-11-02T14:56:16.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19064" PostTypeId="2" ParentId="19036" CreationDate="2014-11-02T14:49:53.350" Score="3" Body="&lt;p&gt;When you solvolyze the starting bromide a secondary carbocation would be generated.  Since a secondary carbocation is not particularly stable, the molecule will explore other pathways such as hydrogen shift, alkyl shift, &lt;em&gt;etc&lt;/em&gt;., in order to &lt;strong&gt;produce a more stable carbocation&lt;/strong&gt;.  The figure below, explores these options for your starting bromide.  Realize that although I've drawn a discreet secondary carbocation that then undergoes rearrangement, in reality cation generation and alkyl or hydrogen shift may occur in a more or less concerted fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/14XDf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case there are 3 options (actually there are 4 options, but the fourth one would produce a primary carbocation, so I left it out - can you find it?), a, b and c. Pathways a and c produce secondary carbocations; however pathway b, involving hydrogen shift, produces a more stable tertiary carbocation.  Capture of the tertiary carbocation from pathway c by solvent should produce the expected product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-02T14:49:53.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19065" PostTypeId="1" AcceptedAnswerId="19075" CreationDate="2014-11-02T14:58:28.373" Score="1" ViewCount="2041" Body="&lt;p&gt;Why are &lt;span class=&quot;math-container&quot;&gt;$\ce{F2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; not considered isoelectronic? Both have 18 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{align}&#xA;\ce {H}\!:&amp;amp;\  \ce{1 e-}\\&#xA;\ce{Cl}\!:&amp;amp;\  \ce{17 e-}\\&#xA;\ce{F}\!:&amp;amp;\  \ce{9 e-}\\&#xA;\end{align}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{F2} \implies 2 \times \ce{9 e- = 18\ e-}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl \implies 1 e- + 17 e- = 18 e-}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="6915" LastEditorUserId="16683" LastEditDate="2018-10-05T17:44:58.267" LastActivityDate="2018-10-05T17:44:58.267" Title="Are difluorine and hydrogen chloride considered isoelectronic?" Tags="&lt;electronic-configuration&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19066" PostTypeId="1" CreationDate="2014-11-02T15:34:37.090" Score="2" ViewCount="1594" Body="&lt;p&gt;use these data to suggest the possible structure for the compound PF3(Ph)(NMe2):   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;19F Spectrum, two signals:&lt;br&gt;&#xA;a doublet of doublets, J=56 and 819Hz, -39.5ppm, relative intensity 2&lt;br&gt;&#xA;a doublet of triplets J=959 and 56Hz, -68.3ppm, relative intensity 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;31P Spectrum, one signal:&lt;br&gt;&#xA;a doublet of triplets, J=959 and 819Hz, -53.0ppm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part b,&lt;br&gt;&#xA;    suggest an experiment or measurement that might be helpful to discern possible isomers of the compound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer so far:&lt;br&gt;&#xA;There are two types of structure that a Phosphorus with five bonds could be, trigonal bipyramidal and square based pyramid. I think that because there are two equivalent fluorines that this must be the trigonal bipyramidal structure as it had the two axial positions that would be equivalent. &lt;/p&gt;&#xA;" OwnerUserId="5181" LastActivityDate="2014-11-02T19:31:18.240" Title="Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19067" PostTypeId="1" AcceptedAnswerId="19071" CreationDate="2014-11-02T15:38:35.770" Score="-1" ViewCount="2500" Body="&lt;p&gt;How do I predict the products and balance the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li(s)+N2(g)---___&lt;/p&gt;&#xA;" OwnerUserId="8256" LastActivityDate="2014-11-02T16:33:09.620" Title="Balancing Chemical Equations + predicting products" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-11-02T19:43:40.613" ContentLicense="CC BY-SA 3.0" />
  <row Id="19068" PostTypeId="2" ParentId="11105" CreationDate="2014-11-02T16:02:01.190" Score="8" Body="&lt;p&gt;To add to the other answers that only address the mathematical behavior of the first order rate equation close to $t=\infty$, let me address what actually happens physically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Before you reach $[A]_t=0$, the first order rate equation actually breaks down.&lt;/strong&gt; It isn't valid anymore, because one of the assumptions on which it is based, namely that the reaction rate is deterministic, breaks down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to really low quantities of (one of) your reactants, reaction events, i.e. two molecules meeting, are blind luck. Actually, for high quantities this is also the case, but then the behavior is a physical version of the law of large numbers: a probabilistic system behaves deterministic on average. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can simulate this basically by flipping coins whether a reaction occurs or not. If you do this for a simple $A$ to $B$ reaction you get graphs like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGEJQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the chance of reaction per timestep for each molecule $A$ is 10%. You can see that the curves at low number of molecules $N$, are jumpy (and actually would look different everytime I rerun my calculation), while for $N=1000$ you already see the fairly smooth and familiar first order behavior. Just note that close to t=10 also the $N=1000$ case is behaving jumpy again as you can see from the zoom below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6VI4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-11-02T16:02:01.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19069" PostTypeId="1" CreationDate="2014-11-02T16:25:49.500" Score="1" ViewCount="49" Body="&lt;p&gt;I have this reaction and need to state what the products would be:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(CF3)2PH + [W(CO)5(THF)] -&amp;gt; }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THF stands for Tetrahydrofuran.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this just be a simple ligand exchange, and if so which ones would be exchanged? How would you know? &lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="4231" LastEditDate="2014-11-04T00:17:10.400" LastActivityDate="2014-11-04T00:17:10.400" Title="Which ligands would be exchanged in this reaction?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19071" PostTypeId="2" ParentId="19067" CreationDate="2014-11-02T16:33:09.620" Score="1" Body="&lt;p&gt;If $\ce{Li_{(s)} + N2_{(g)}}$ react (I am not sure if they will or not), the product will be $\ce{LiN}$ and the balanced reaction will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Li_{(s)} + N2_{(g)} -&amp;gt; 2 LiN}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/reactions/faq/predicting-products-nicl2-nh42s.shtml&quot; rel=&quot;nofollow&quot;&gt;this tutorial&lt;/a&gt;, it contains all the steps you should follow to predict the products when formula of reactants are given.&lt;/p&gt;&#xA;" OwnerUserId="1878" LastActivityDate="2014-11-02T16:33:09.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19072" PostTypeId="2" ParentId="19066" CreationDate="2014-11-02T16:41:14.767" Score="2" Body="&lt;p&gt;I agree that the trigonal bipyramid structure is what we are dealing with, so let me focus on that structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nmr data shows that 2 of the fluorines are equivalent and one is different. This results in 3 possible isomers as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f0Kqb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $\ce{^31P}$ and $\ce{^19F}$ are spin 1/2 nuclei, so each of these isomers would produce phosphorous and fluorine nmr spectra with splitting patterns and integration ratios consistent with the reported data. That is, the phosphorous will split each different fluorine into a doublet, the two equivalent fluorines will split the singular  fluorine's doublet lines further into triplets.  The one non-equivalent fluorine will split the other two equivalent fluorine doublet lines into doublets.  Similar analysis of the phosphorous signal leads to the expectation of a doublet (from the singular fluorine) of triplets (from the two equivalent fluorines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two additional pieces of information:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fluorine prefers the axial position in pentacoordinate phosphorous&#xA;(see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;here&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Because the orbitals involved in axial and equatorial bonding in pentacoordinate phosphorous compounds are different, the axial and equatorial coupling constants will be different.  Generally, $\ce{J_{PF(eq)}}$ &gt; $\ce{J_{PF(ax)}}$ because there is increased s-orbital density in the equatorial bond (see &lt;a href=&quot;http://books.google.com/books?id=4KfDNuJfBFAC&amp;amp;pg=PA76&amp;amp;lpg=PA76&amp;amp;dq=pf5,%20p-f%20coupling%20constants&amp;amp;source=bl&amp;amp;ots=32wZmFGpXG&amp;amp;sig=jCbBELiDvteXziAsK4i2u76vCGY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=SVRWVPDSNJGcygSEyYCABw&amp;amp;ved=0CCYQ6AEwAg#v=onepage&amp;amp;q=pf5%2C%20p-f%20coupling%20constants&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, for example)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The larger reported P-F coupling constant (959 Hz) is associated with the intensity = 1 fluorine signal.  This suggests that structure I, with one equatorial fluorine (with the larger $\ce{P-F}$ coupling constant) and 2 axial fluorines (with the smaller coupling constant and as preferred electronically - point 1 above) is the structure being observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part B Experiments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trigonal bipyramid structure could be confirmed by heating the sample while in the nmr spectrometer.  As &lt;a href=&quot;http://en.wikipedia.org/wiki/Berry_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Berry pseudorotation&lt;/a&gt; commences the 2 fluorine signals should coalesce and determination of a barrier height around 35-40 kcal/m would be expected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, compounds II and III must have similar energies and both would be expected to be present at equilibrium.  If we heated our sample so that pseudorotation took place, and then upon cooling we saw an additional set of signals, that would be consistent with II or III being our starting compound.  No new signals upon cooling would support structure I.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-11-02T19:31:18.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19073" PostTypeId="1" AcceptedAnswerId="19126" CreationDate="2014-11-02T18:00:22.643" Score="0" ViewCount="3112" Body="&lt;p&gt;Can anyone describe the process through which we determine the heat capacity of calorimeter, for me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is &#xA;$q_{calorimeter} = C_{calorimeter} \times \Delta T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of bomb calorimeter is shown to be  the same as  negative heat of reaction. But without mass, their unit would be different, no?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="7960" LastEditDate="2014-11-04T03:03:59.167" LastActivityDate="2014-11-04T03:14:24.340" Title="Bomb calorimeter and heat capacity of calorimeter" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19074" PostTypeId="1" AcceptedAnswerId="19080" CreationDate="2014-11-02T18:34:50.520" Score="3" ViewCount="101" Body="&lt;p&gt;This morning I was boiling carrot dices. And saw something quite interesting happening to the dices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is not boiling. But before it boils, the dices started gathering at the center  of the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any one has any answer as to why this happened?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-11-04T03:15:39.990" LastActivityDate="2014-11-04T03:15:39.990" Title="Real life chemistry :boiling carrot dices" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-30T19:35:49.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="19075" PostTypeId="2" ParentId="19065" CreationDate="2014-11-02T18:50:08.767" Score="4" Body="&lt;p&gt;If we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have the same number of electrons, they are actually isoelectronic. Please see &quot;Ultraviolet Spectroscopy And Uv Lasers&quot;&#xA; edited by Prabhakar Misra, Mark A. Dubinskii, page 35 (available in google books)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have a similar electron configuration and the same structure, they are not.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-02T18:50:08.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19078" PostTypeId="2" ParentId="16842" CreationDate="2014-11-02T20:16:01.043" Score="2" Body="&lt;p&gt;Your proposed synthesis looks reasonable. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Synthesis_of_Alkenes/Alkenes_by_Dehydration_of_Alcohols&quot; rel=&quot;nofollow&quot;&gt;Dehydration&lt;/a&gt; of the primary alcohol is indeed possible, it proceeds via the E2 mechanism (although primary alcohols are less reactive than secondary or tertiary ones). Chlorinated solvents, like methylene chloride and chloroform, are commonly used for the &lt;a href=&quot;http://www.masterorganicchemistry.com/2013/03/06/bromination-of-alkenes-how-does-it-work/&quot; rel=&quot;nofollow&quot;&gt;bromination of alkenes&lt;/a&gt;, for the reason you have provided (inertness).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-11-02T20:16:01.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19079" PostTypeId="1" AcceptedAnswerId="26348" CreationDate="2014-11-02T20:31:32.450" Score="6" ViewCount="24476" Body="&lt;p&gt;$$H=U+pV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is the formula for enthalpy or the sum of the internal energy and pressure volume product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-p\Delta V$ is work done by or on the system. But $pV$, yet very similar to work, I think, calculates the whole “cylinder”. Instead of just calculating the height difference of the cylinder before and after the air is condensed, it calculate the height of whole cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But anyway this is just my guess, what is it really?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorDisplayName="user7951" LastEditDate="2015-05-06T09:25:17.950" LastActivityDate="2015-05-06T09:25:17.950" Title="Enthalpy (H) formula: what is the pressure volume product" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19080" PostTypeId="2" ParentId="19074" CreationDate="2014-11-02T20:42:23.050" Score="7" Body="&lt;p&gt;My guess would be natural convection. Hot water is less dense than cold water so when the heat source is at the bottom, the freshly heated water rises and the cooler water at the top sinks. The cooler water can't sink through the rising hot water, so the two flows run along side each other. Depending on the fluid properties, this can result in circular &quot;rolls&quot; of flowing water, known as convection cells. If you use a wide pot and small particles, you may be able to see many, but in a relatively narrow pot, you're more likely to observe a single recirculating flow around the centre, as shown in &lt;a href=&quot;https://www.youtube.com/watch?v=7gm-iVyGQko&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; simulation. The liquid loses more heat around the sides of the pot, so the descending flow rides down the sides and the ascending flow is in the middle where it can stay the hottest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that your dice is too heavy to ride the ascending flow, but the flow from the edge to the centre pushes all the carrots towards the middle.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-11-03T03:59:07.227" LastActivityDate="2014-11-03T03:59:07.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19082" PostTypeId="1" AcceptedAnswerId="19091" CreationDate="2014-11-02T23:07:22.167" Score="14" ViewCount="16121" Body="&lt;p&gt;I’m teaching chemistry to a home-ed 15 year old.  Occasionally we do some practical work, since I believe very strongly that physical science learning should have some real lab work to bring it alive, and also to teach things like observational skills and attention to detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My background is a PhD in organic chemistry, so I’m familiar with hazard assessment, and all the lab work I do with my student is something I’ve done myself already.  Any time she touches the apparatus it’s under close supervision by myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the practical involved taking hydrated copper sulphate, heating it to drive off the water, weighing before and after, and thus calculating the number of water molecules of crystallisation, based on the respective molecular weights of the anhydrous salt and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a porcelain evaporating dish to contain the hydrate during heating, electronic scales that will weigh to 300 g in increments of 10 mg, and commercially bought copper sulphate that is allegedly 99.5 % pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started with the hydrate, heated it in the evap dish over a flame (using a gas cooker ring) until it had all turned greyish white, then allowed it to cool and weighed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Starting wt hydrate: 13.03 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wt after heating: 8.22 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hence water lost: 4.81 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that’s 4.81/18 = &lt;strong&gt;0.267 mol water&lt;/strong&gt;, and &#xA;8.22/(63.5 + 32 + 64) = &lt;strong&gt;0.0515 mol anhydrous copper sulphate&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence 0.267/0.0515 = 5.18 mol water per mol copper sulphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my question: why not 5.0?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic scales (quite cheap) were an obvious first candidate for a source of error.  I’ve compared its readings at weights of about 2, about 20, and about 50 g, with another similar model, and they agree to within 10 mg (one digit in the final decimal place) at all weights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could perhaps have not heated the hydrate enough, and not driven off all the water – but then I’d have less than 5, not more, waters per mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the hydrate, as supplied, could be a little wet.  I haven’t investigated that yet, but I plan to weigh a sample before and after leaving in a warm airing cupboard overnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you’ve read this far, I’m already grateful, but I’d be even more so, if anyone could suggest where the difference between 5 and 5.18 is coming from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did it a second time, and got 5.16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between 5.00 and 5.16 is a weight error of about 200 mg, and I was really careful, so this seems unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight after cooling of the evap dish is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (2)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There was no sign of $\ce{CuO}$ after the heating.  No trace of black, and indeed we dissolved the anhydrous salt in water (for growing some seed crystals later), and there was no insoluble residue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I’ve now heated the dish over a hot blue natural gas flame for ten minutes.  Its weight before cooling was 0.02 g (on my scales) less than its empty weight cold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit (3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I weighed out some of the hydrate into another dry dish, flattened it out as much as possible within the confines of the dish to expose as much surface area as possible, and left it in the airing cupboard for 24 h at 26 degrees Celsius.  Then I re-weighed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pre drying wt: 42.68 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Post drying wt: 41.98 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumed water content: &lt;strong&gt;42.68 − 41.98 = 0.7 g&lt;/strong&gt; (!!!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming that 26 degrees is not warm enough to cause partial loss of water of crystallisation, that suggests that at best my hydrate is only 98.36 % pure, not 99.5 as it says on the bottle (typo corrected above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So repeating my calculation above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wt before&lt;/strong&gt;: 13.03 × 0.9836 = &lt;strong&gt;12.82 g&lt;/strong&gt; hydrate&lt;br&gt;&#xA;&lt;strong&gt;Wt after: 8.22 g&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Water lost: 4.60 g&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles water: 4.60/18 = 0.256&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles $\ce{CuSO4}$: 8.22/159/5 = 0.052&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ratio: 4.96&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m much happier with that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorDisplayName="user7951" LastEditDate="2015-07-01T20:15:43.147" LastActivityDate="2015-07-01T20:15:43.147" Title="Where's my experimental error coming from?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19083" PostTypeId="2" ParentId="18786" CreationDate="2014-11-02T23:08:13.717" Score="4" Body="&lt;p&gt;Preconcentration means to increase the concentration of a sample prior to analysis or detection. For example you can do preconcentration for an non-volatile organic sample with evaporation its solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An operation (process) as the result of which microcomponents are transferred from the sample of larger mass into the sample of smaller mass, so that the concentration of the microcomponents is increased. Examples include the decrease in solvent volume during distillation or evaporation, and the transfer of microcomponents from an aqueous solution into a smaller volume of organic solvent by extraction.&lt;a href=&quot;http://goldbook.iupac.org/A00027.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(IUPAC)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XuXqT.jpg&quot; alt=&quot;In a simple word, its means: decreasing the volume of solvent.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of methods for preconcentration: solvent evaporation, using a different phase (ex: a solid or liquid that could absorbs sample from the solvent) and ...&lt;/p&gt;&#xA;" OwnerUserId="9476" LastActivityDate="2014-11-02T23:08:13.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19085" PostTypeId="1" CreationDate="2014-11-02T23:57:59.990" Score="7" ViewCount="1872" Body="&lt;p&gt;I am trying to predict the products of both the hydroboration/oxidation and acid-catalyzed hydration of 1-phenylpropene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4LKKU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LKKU.png&quot; alt=&quot;Hydration of 1-phenylpropene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both C1 and C2 of the double bond are secondary carbons. However, I have the feeling that there are slight differences in the electron density on each carbon, which will affect the addition of the boron electrophile and the stability of the carbocation for the two reactions respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which regioisomers are predominantly formed from both reactions?&lt;/p&gt;&#xA;" OwnerUserId="9611" LastEditorUserId="16683" LastEditDate="2018-10-01T15:27:53.237" LastActivityDate="2018-12-07T00:02:40.527" Title="Hydroboration/oxidation vs acid-catalyzed hydration of 1-phenylpropene" Tags="&lt;organic-chemistry&gt;&lt;regioselectivity&gt;&lt;c-c-addition&gt;&lt;organoboron-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19086" PostTypeId="2" ParentId="19082" CreationDate="2014-11-03T02:56:28.717" Score="2" Body="&lt;p&gt;Did you heat the dish before weighing it to find the tare?  Water adheres to the dish when the dish is at room temperature.  This is probably a small effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger effect is decomposing copper sulfate to copper oxide and sulfur trioxide.  You can check for this by looking for the telltale brown/black color of copper oxide.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-11-03T02:56:28.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19087" PostTypeId="1" AcceptedAnswerId="19093" CreationDate="2014-11-03T03:54:57.620" Score="0" ViewCount="475" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following pairs are isomers?&lt;/p&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$\ce{C5H10}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C10H20}$&lt;/span&gt;&lt;br /&gt;&#xA;b) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3(CH2)4CH3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3(CH2)3CH3}$&lt;/span&gt;&lt;br /&gt;&#xA;c) &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH(CH3)(CH2)2CH3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3(CH2)2CH(CH3)2}$&lt;/span&gt;&lt;br /&gt;&#xA;d) &lt;span class=&quot;math-container&quot;&gt;$\ce{(CH3)3CH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My textbook says the answer is C). I beg to differ. I think it is D). C) is just a different way of writing 2-methylpentane.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="-1" LastEditorDisplayName="user10259" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-14T05:13:12.563" Title="Isomer Identification Using Condensed Structural Formulae" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19088" PostTypeId="2" ParentId="19085" CreationDate="2014-11-03T04:03:48.513" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YBYej.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: $H_3O^+$ looks like  $H_3O^-$ due to thin width of vertical bar in $+$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During HBO, the carbon nearer to the phenyl ring is more sterically hindered than the other one next to it, otherwise these two are similar in many respects exscluding some, this can be understood by imagining the intermediate $R_3B$ by taking R as branched through both carbons one by one, maybe then you'll appreciate the fact of difference in steric hindrance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of acid catalyzed hydration, the carbon nearer to it has more carbocation stabilization than the one next to it, because of the conjugation with the benzene/phenyl ring. Even if the carbocation is formed on the $\beta$ wrt phenyl, it'll rearrange to $\alpha$/allylic position.Thus the product.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Herbert C. Brown (1912–) won the Nobel Prize in 1979 for his invention and development of hydroboration, mostly carried out at Purdue University, which made this new chemistry familiar to the practising organic chemist. The prize was shared with Georg Wittig, showing how important organic chemistry with main group elements had become. He vigorously opposed ‘non-classical’ carbocation theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;most substrates&lt;/strong&gt;, the addition in hydroboration is stereospecific and syn,&#xA;with attack taking place from the less-hindered side. &#xA;&lt;sup&gt;Brown, H.C.; Zweifel, G. J. Am. Chem. Soc. 1961, 83, 2544; Bergbreiter, D.E.; Rainville, D.P.&#xA;J. Org. Chem. 1976, 41, 3031; Kabalka, G.W.; Newton, Jr., R.J.; Jacobus, J. J. Org. Chem. 1978,&#xA;43, 1567.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only &lt;strong&gt;special case&lt;/strong&gt; known (atleast to me) is the treatment of iPrCHCHMe with borane which&#xA;gave 57% of product with boron on the methyl-bearing carbon and 43% of the&#xA;other&lt;sup&gt;Brown, H.C.; Zweifel, G. J. Am. Chem. Soc. 1961, 83, 1241.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In all cases&lt;/strong&gt;, the boron goes to the side of the double bond that has more hydrogens,&#xA;whether the substituents are &lt;strong&gt;aryl&lt;/strong&gt; or alkyl&lt;sup&gt;For a thorough discussion of the regioselectivity with various types of substrate and hydroborating&#xA;agents, see Cragg, G.M.L.Organoboranes in Organic Synthesis Marcel Dekker, NY, 1973, pp.63–84,&#xA;137–197. See also, Brown, H.C.; Vara Prasad, J.V.N.; Zee, S. J. Org. Chem. 1986, 51, 439.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-03T17:30:01.687" LastActivityDate="2014-11-03T17:30:01.687" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19089" PostTypeId="2" ParentId="19079" CreationDate="2014-11-03T04:54:07.123" Score="2" Body="&lt;p&gt;The best thing I found was on &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy#Physical_Interpretation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, I couldn't explain it in a better way:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The U term can be interpreted as the energy required to create the system, and the pV term as the energy that would be required to &quot;make room&quot; for the system if the pressure of the environment remained constant. When a system, for example, n moles of a gas of volume V at pressure p and temperature T, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy U plus pV, where pV is the work done in pushing against the ambient (atmospheric) pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-20T10:57:36.497" LastActivityDate="2014-11-20T10:57:36.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19090" PostTypeId="1" AcceptedAnswerId="19162" CreationDate="2014-11-03T07:01:42.123" Score="43" ViewCount="19565" Body="&lt;p&gt;Can someone explain this to me by drawing resonance structures for the cyclopropylmethyl carbocation please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also one more question, is the tricyclopropylmethyl carbocation more stable than tropylium ion?&lt;/p&gt;&#xA;" OwnerUserId="9310" LastEditorUserId="16683" LastEditDate="2016-09-24T06:40:12.533" LastActivityDate="2020-12-31T17:18:25.823" Title="What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="19091" PostTypeId="2" ParentId="19082" CreationDate="2014-11-03T09:01:42.023" Score="10" Body="&lt;p&gt;I would go with a full fledged error propagation analysis on this one, because without systematic approach you might be guessing forever. So if I look at the calculation process you have the following equation to get from known quantities to the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X=\left(\frac{W_0}{W_e}-1\right)\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right) \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W_0$ is the start weight, $W_e$ the end weight and $M_i$ is the respective molecular mass. $X$ is the desired answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the expected error in the answer, $\Delta X$, assuming no error in the molecular masses, we calculate:&#xA;$$\Delta X=\sqrt{\left(\frac{\partial X}{\partial W_0}\Delta W_0\right)^2+\left(\frac{\partial X}{\partial W_e}\Delta W_e\right)^2} \tag{2}$$&#xA;which works out to:&#xA;$$\Delta X=\frac{\sqrt{(\Delta W_0)^2 W_e^2+(\Delta W_e)^2 W_0^2}}{W_e^2}\left(\frac{M_{CuSO_4}}{M_{H_2O}}\right)  \tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now the final thing left to do is estimate how big $\Delta W_0$ and $\Delta W_e$ are. For $\Delta W_0$ I think it is safe to say that this is only the inaccuracy of the scale, so perhaps $0.01$ g. For $\Delta W_e$ I would go for the worst case scenario: that all of the 0.6% impurity in the original copper sulfate has evaporated, meaning that the weight will be off by 0.6% times $W_0$ i.e. $0.08$ g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now plugging in the numbers in Eq.$(3)$ I find $\Delta X=0.14$ mol water per mol copper sulphate. Which is very close to the actual error you find. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This error is almost completely determined by the error in $W_e$. If, like @brinnb suggests, an additional reduction in $W_e$ could be expected due to adhered water say that the error in $W_e$ is $0.1$ g you already get to $\Delta X=0.17$ mol water per mol copper sulphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error you find falls within the worst-case expected error due to impurity and potentially adhered water.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-11-03T09:01:42.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19093" PostTypeId="2" ParentId="19087" CreationDate="2014-11-03T09:17:06.913" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;isomer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source:&lt;br /&gt;&#xA;&lt;em&gt;Pure and Applied Chemistry&lt;/em&gt;, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994))&lt;br /&gt;&#xA;&lt;em&gt;Pure and Applied Chemistry&lt;/em&gt;, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996))&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt;&lt;br /&gt;&#xA;C&lt;sub&gt;5&lt;/sub&gt;H&lt;sub&gt;10&lt;/sub&gt;&lt;br /&gt;&#xA;and&lt;br /&gt;&#xA;C&lt;sub&gt;10&lt;/sub&gt;H&lt;sub&gt;20&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;different chemical formula → not isomeric&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt;&lt;br /&gt;&#xA;CH&lt;sub&gt;3&lt;/sub&gt;(CH&lt;sub&gt;2&lt;/sub&gt;)&lt;sub&gt;4&lt;/sub&gt;CH&lt;sub&gt;3&lt;/sub&gt; = hexane (C&lt;sub&gt;6&lt;/sub&gt;H&lt;sub&gt;14&lt;/sub&gt;)&lt;br /&gt;&#xA;and&lt;br /&gt;&#xA;CH&lt;sub&gt;3&lt;/sub&gt;(CH&lt;sub&gt;2&lt;/sub&gt;)&lt;sub&gt;3&lt;/sub&gt;CH&lt;sub&gt;3&lt;/sub&gt; = pentane (C&lt;sub&gt;5&lt;/sub&gt;H&lt;sub&gt;12&lt;/sub&gt;)&lt;/p&gt;&#xA;&lt;p&gt;different chemical formula → not isomeric&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;c)&lt;/strong&gt;&lt;br /&gt;&#xA;CH&lt;sub&gt;3&lt;/sub&gt;CH(CH&lt;sub&gt;3&lt;/sub&gt;)(CH&lt;sub&gt;2&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;CH&lt;sub&gt;3&lt;/sub&gt; = 2-methylpentane (C&lt;sub&gt;6&lt;/sub&gt;H&lt;sub&gt;14&lt;/sub&gt;)&lt;br /&gt;&#xA;and&lt;br /&gt;&#xA;CH&lt;sub&gt;3&lt;/sub&gt;(CH&lt;sub&gt;2&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt;CH(CH&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;2&lt;/sub&gt; = 2-methylpentane (C&lt;sub&gt;6&lt;/sub&gt;H&lt;sub&gt;14&lt;/sub&gt;)&lt;/p&gt;&#xA;&lt;p&gt;same chemical formula and same structural formula → identical, not isomeric&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;d)&lt;/strong&gt;&lt;br /&gt;&#xA;(CH&lt;sub&gt;3&lt;/sub&gt;)&lt;sub&gt;3&lt;/sub&gt;CH = isobutane (C&lt;sub&gt;4&lt;/sub&gt;H&lt;sub&gt;10&lt;/sub&gt;)&lt;br /&gt;&#xA;and&lt;br /&gt;&#xA;CH&lt;sub&gt;3&lt;/sub&gt;CH&lt;sub&gt;2&lt;/sub&gt;CH&lt;sub&gt;2&lt;/sub&gt;CH&lt;sub&gt;3&lt;/sub&gt; = butane (C&lt;sub&gt;4&lt;/sub&gt;H&lt;sub&gt;10&lt;/sub&gt;)&lt;/p&gt;&#xA;&lt;p&gt;same chemical formula and different structural formula → isomeric&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-11-03T09:17:06.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19096" PostTypeId="1" AcceptedAnswerId="19107" CreationDate="2014-11-03T13:41:16.527" Score="16" ViewCount="4958" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WMhlb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WMhlb.png&quot; alt=&quot;1,1',1''-(1,3-dichloropropane-1,2,3-triyl)tribenzene&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How can I determine the number of possible pairs of diastereomers here?&lt;/p&gt;&#xA;&lt;p&gt;My first guess was two: RR with RS and SS with RS. But what about the potential chirality center middle carbon? There are two identical branches attached to it; however, those two branches can change R/S. So, is it a chirality center?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorDisplayName="user7951" LastEditDate="2020-08-08T18:54:37.837" LastActivityDate="2020-08-09T07:13:30.993" Title="Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="19097" PostTypeId="2" ParentId="19096" CreationDate="2014-11-03T14:05:42.110" Score="-1" Body="&lt;p&gt;I don't think it is hindered enough to prevent rotation around the central carbon, so I think it only has two stereogenic centres, leading to four possible stereocentre combinations as you have identified: RR, SS, SR, RS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in more than two pairs of diastereomers though:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) RR with SR&lt;br&gt;&#xA;2) RR with RS&lt;br&gt;&#xA;3) SS with SR&lt;br&gt;&#xA;4) SS with RS&lt;/p&gt;&#xA;" OwnerUserId="5075" LastActivityDate="2014-11-03T14:05:42.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19098" PostTypeId="2" ParentId="19061" CreationDate="2014-11-03T14:11:56.193" Score="1" Body="&lt;p&gt;Tetraethyl lead (bp 180 °C) and tetramethyl lead (bp 110 °C) are easily resorbed through the lungs and the intact (!) skin. They pass the blood-brain barrier without any problems. The half-life for the elimination from the brain is about 500 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Günter Fred Fuhrmann, Toxikologie Für Naturwissenschaftler: Einführung in Die Theoretische und Spezielle Toxikologie, Springer-Verlag, 2006, p 209)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These compounds are neurotoxic and the literature is full of reported cases of intoxications and - sometimes - successful treatment, such as &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/20306169&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-03T14:11:56.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19099" PostTypeId="2" ParentId="19096" CreationDate="2014-11-03T14:33:46.580" Score="13" Body="&lt;p&gt;Let's number the 3 aliphatic carbons to make our discussion clearer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3ADQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using the definition of a chiral carbon as a carbon that has 4 different substituents, carbons 1, 2 and 3 are all capable of being chiral. Carbons 1 and 3 are relatively straightforward chiral carbons, it's easy to see the 4 different substituents and assign &amp;quot;&lt;em&gt;R&lt;/em&gt;&amp;quot; or &amp;quot;&lt;em&gt;S&lt;/em&gt;&amp;quot; configurations.  However carbon 2 is problematic.&lt;/p&gt;&#xA;&lt;p&gt;If we assign &lt;em&gt;R&lt;/em&gt;- and &lt;em&gt;S&lt;/em&gt;- configurations to carbons 1 and 3 respectively, then there are 4 different groups attached to carbon 2, but the difference in 2 of the groups is just &amp;quot;&lt;em&gt;R&lt;/em&gt;&amp;quot; or &amp;quot;&lt;em&gt;S&lt;/em&gt;&amp;quot;.  In fact, in this situation there is actually a plane of symmetry bisecting the molecule and the molecule is achiral - it is a &lt;em&gt;meso&lt;/em&gt;-isomer.  The plane of symmetry contains carbon 2 and the hydrogen and phenyl group attached to carbon 2, and, as expected, the &lt;em&gt;R&lt;/em&gt;- and &lt;em&gt;S&lt;/em&gt;-substituents are enantiomeric.  In such cases it is correct to say that carbon 2 is &lt;a href=&quot;https://goldbook.iupac.org/html/P/P04921.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;pseudo&lt;/em&gt;-asymmetric&lt;/a&gt; (thanks to @GaurangTandon for pointing this out!). The Cahn–Ingold–Prelog priority rules can still be used to describe carbon 2, the &lt;em&gt;R&lt;/em&gt; substituent is given priority over the &lt;em&gt;S&lt;/em&gt; substituent and lowercase &amp;quot;&lt;em&gt;r&lt;/em&gt;&amp;quot; and &amp;quot;&lt;em&gt;s&lt;/em&gt;&amp;quot; are used to describe carbon 2, the &lt;em&gt;pseudo&lt;/em&gt;-asymmetric carbon.&lt;/p&gt;&#xA;&lt;p&gt;In the case where we assign &lt;em&gt;R&lt;/em&gt;- and &lt;em&gt;R&lt;/em&gt;- (or &lt;em&gt;S&lt;/em&gt;- and &lt;em&gt;S&lt;/em&gt;-) configurations to carbons 1 and 3 respectively, then there is no longer a plane of symmetry bisecting the molecule.  Carbon 2 is now &lt;a href=&quot;http://en.wikipedia.org/wiki/Prochirality&quot; rel=&quot;nofollow noreferrer&quot;&gt;prochiral&lt;/a&gt; (this means changing one substituent is all that is required to make carbon 2 chiral), but the molecule as a whole is now chiral.&lt;/p&gt;&#xA;&lt;p&gt;Going in the order carbon 1-2-3 here are the various possibilities:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1&lt;em&gt;R&lt;/em&gt;,3&lt;em&gt;R&lt;/em&gt; is chiral and it has an enantiomer 1&lt;em&gt;S&lt;/em&gt;,3&lt;em&gt;S&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;1&lt;em&gt;R&lt;/em&gt;,2&lt;em&gt;r&lt;/em&gt;,3&lt;em&gt;S&lt;/em&gt; is an achiral (there is a plane of symmetry bisecting the molecule), &lt;em&gt;meso&lt;/em&gt;-compound and a diastereomer of 1&lt;em&gt;R&lt;/em&gt;,3&lt;em&gt;R&lt;/em&gt;;&lt;/li&gt;&#xA;&lt;li&gt;1&lt;em&gt;R&lt;/em&gt;,2&lt;em&gt;s&lt;/em&gt;,3&lt;em&gt;S&lt;/em&gt; is a second achiral, &lt;em&gt;meso&lt;/em&gt;-compound.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;So there are 3 diastereomers; one is chiral and exists as a &lt;em&gt;d,l&lt;/em&gt; pair and the two other  diastereomers are achiral, &lt;em&gt;meso&lt;/em&gt;-compounds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2020-08-09T07:13:30.993" LastActivityDate="2020-08-09T07:13:30.993" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="19100" PostTypeId="1" AcceptedAnswerId="19103" CreationDate="2014-11-03T15:25:21.280" Score="6" ViewCount="5350" Body="&lt;p&gt;I am a PhD student in materials engineering. What is D3O? It is not a chemical, but rather a &quot;polyurethane energy-absorbing material&quot; (from &lt;a href=&quot;https://en.wikipedia.org/wiki/D3o&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). I would wish to know and study the chemical composition of this impact shock absorbing materials.&lt;/p&gt;&#xA;" OwnerUserId="9615" LastEditorUserId="5026" LastEditDate="2018-04-22T08:38:55.153" LastActivityDate="2018-04-22T13:32:32.120" Title="What new material is D3O?" Tags="&lt;materials&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19102" PostTypeId="2" ParentId="19069" CreationDate="2014-11-03T15:27:03.117" Score="2" Body="&lt;p&gt;Complexes like $\ce{M(CO)5(solv)}$ are typically generated by photolysis ($\lambda$ = 350 nm) of $\ce{M(CO)6}$ in the presence of a non-inert solvent. This proceeds through a highly reactive &quot;naked&quot; $\ce{M(CO)5}$ intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{W(CO)6 + solv -&amp;gt;[h\nu] W(CO)5(solv) + CO}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When offered another ligand $\mathrm{L}$, the weakest bond  ligand in the $\ce{W(CO)5(solv)}$ complex is replaced. This is THF.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking the lone pair on the phosphorous atom into account, is is conceivable that $\ce{(CF3)2PH}$ will do the job:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{W(CO)5(thf) + (CF3)2PH -&amp;gt; W(CO)5((CF3)2PH) + THF}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-03T15:27:03.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19103" PostTypeId="2" ParentId="19100" CreationDate="2014-11-03T15:41:39.287" Score="5" Body="&lt;p&gt;The inventor, Richard M. Palmer, has filed at number of patents on &quot;energy absorbing materials&quot;, one of which is US patent (&lt;a href=&quot;http://www.google.com/patents/US7381460&quot; rel=&quot;noreferrer&quot;&gt;7,381,460&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, there does not seem to be any peer-reviewed publication by Mr Palmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's always the option to ask the manufacturer D30 for further information and/or material samples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible downside of such a cooperation is that you might not be allowed to report on the progress of your experimental work on conferences or in publications due to an NDA previously signed!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-11-03T18:58:11.423" LastActivityDate="2014-11-03T18:58:11.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19104" PostTypeId="2" ParentId="19058" CreationDate="2014-11-03T16:11:18.847" Score="2" Body="&lt;p&gt;In the pH range between 10 and 14, chloride and hypobromide are formed (&lt;a href=&quot;http://dx.doi.org/10.1021/ja01174a025&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{OCl- +Br- -&amp;gt;  Cl- + OBr-}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-03T16:11:18.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19105" PostTypeId="1" CreationDate="2014-11-03T16:14:02.703" Score="0" ViewCount="817" Body="&lt;p&gt;(3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the Fischer Projection of the compound above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kgYsV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kgYsV.png&quot; alt=&quot;Answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mistake at the upper stereocenter? I think so.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="5591" LastEditDate="2015-10-09T16:58:04.413" LastActivityDate="2015-10-09T16:58:04.413" Title="Is this Fischer Projection of (3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene wrong?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;stereochemistry&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19106" PostTypeId="2" ParentId="19105" CreationDate="2014-11-03T17:29:12.427" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a mistake at the upper stereocenter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes you are thinking correct, the R configuration will be the diastereomeric pair of this compund, the second chiral carbon is in correct configuration. Don't you want to ask why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iaJsi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-04T03:55:27.230" LastActivityDate="2014-11-04T03:55:27.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19107" PostTypeId="2" ParentId="19096" CreationDate="2014-11-03T17:31:29.327" Score="18" Body="&lt;p&gt;I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is &lt;em&gt;&lt;a href=&quot;http://goldbook.iupac.org/S05980.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-stereogenic&lt;/a&gt;&lt;/em&gt; but it may be &lt;em&gt;&lt;a href=&quot;http://goldbook.iupac.org/C01065.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chirotopic&lt;/a&gt;&lt;/em&gt; depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a &lt;a href=&quot;http://goldbook.iupac.org/P04921.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pseudo-asymmetric carbon atom&lt;/a&gt;. Carbon-2 is not a stereocenter, because it does not have four different &quot;things&quot; attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RVRNX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: (1​&lt;em&gt;R&lt;/em&gt;,2​&lt;em&gt;s&lt;/em&gt;,3​&lt;em&gt;S&lt;/em&gt;)-1,3-dichloro-1,2,3-triphenylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: (1​&lt;em&gt;S&lt;/em&gt;,3​&lt;em&gt;S&lt;/em&gt;)-1,3-dichloro-1,2,3-triphenylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: (1​&lt;em&gt;R&lt;/em&gt;,2​&lt;em&gt;r&lt;/em&gt;,3​&lt;em&gt;S&lt;/em&gt;)-1,3-dichloro-1,2,3-triphenylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F: (1​&lt;em&gt;R&lt;/em&gt;,3​&lt;em&gt;R&lt;/em&gt;)-1,3-dichloro-1,2,3-triphenylpropane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small &lt;em&gt;r&lt;/em&gt; and &lt;em&gt;s&lt;/em&gt; are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the &lt;em&gt;r&lt;/em&gt;/&lt;em&gt;s&lt;/em&gt; designations, but I believe that the &lt;em&gt;R&lt;/em&gt; configured substituent receives higher priority than the &lt;em&gt;S&lt;/em&gt; configured substituent, and Cahn–Ingold–Prelog system is followed from there.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2017-03-27T16:46:41.260" LastActivityDate="2017-03-27T16:46:41.260" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19110" PostTypeId="1" CreationDate="2014-11-03T19:25:22.280" Score="3" ViewCount="1147" Body="&lt;p&gt;Just going through my thermodynamics chapter I came across 2 contrasting statements:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Change in enthalpy=enthalpy of (products-reactants) at constant pressure and temperature.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Change in enthalpy=Energy required to break bonds in molecules of reactants -energy&#xA;released in the formation of bonds in molecules of products.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, if I consider these statements individually then these sound correct to me but they are different in the sense that in the first one it is (products-reactants) and in the second one (reactant -products)?&lt;/p&gt;&#xA;&lt;p&gt;Is the book wrong or am I mistaken in my fundamentals  ?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-30T13:29:03.907" Title="Are these statements about change in enthalpy contradictory?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19111" PostTypeId="2" ParentId="19110" CreationDate="2014-11-03T20:26:23.113" Score="1" Body="&lt;p&gt;They are basically same statements. We can use Hess law to elaborate these two statements. For example, we can use the following reaction,&lt;br&gt;&#xA;                                 CH4=C+2H2&lt;br&gt;&#xA;Here, Delta reaction = Enthalpy of CH4 - Enthalpy of C + 2(Enthalpy of H2).......(1)&lt;br&gt;&#xA;We can break the reaction events as,&lt;br&gt;&#xA;1) Breaking of 4 C-H bonds &#xA;2) Formation of 2 H-H bonds&lt;br&gt;&#xA;So, delta H = Energy required to break 4 C-H bonds + Energy required to forming two H-H bonds.......(2)&lt;br&gt;&#xA;Now we always take the reactor to be the system and that's why heat input to the reactor is positive and heat going out of the reactor is negative. In the formation of bonds, the energy are released, so the formation energy of H-H is negative with respect to our thermodynamic system.From 2&lt;br&gt;&#xA;So,  delta H = Energy required to break 4 C-H bonds - Energy released in the formation of two H-H bonds.......(3)&lt;br&gt;&#xA;Now, energy required to break 4 C-H bond= Energy of C + Enrgy of 4 H - energy of CH4&lt;br&gt;&#xA;                                        =-Energy of reactants...(4)&lt;br&gt;&#xA;Similarly,  energy required to form 2 H-H bond= Energy of 2 H2 - energy of 4 H&lt;br&gt;&#xA;                                              =Energy of products (As C is in reference state).........(5)&lt;br&gt;&#xA;Adding 4 and 5 yields, equation (1), which proves that both statements are same.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2014-11-03T20:26:23.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19112" PostTypeId="2" ParentId="19027" CreationDate="2014-11-03T21:12:59.470" Score="2" Body="&lt;p&gt;Inductive effect and hyperconjugation are two distinct phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive effects are purely through sigma bonds. For example, in an organofluorine compound, the fluorine withdraws electrons from carbon by not sharing the electrons of the sigma bond equally. We explain this using the rationale that fluorine is more electronegative than carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in a carbocation, the electron-deficient carbon is more electronegative than any substituent carbons. Thus, the carbocation will withdraw electrons from substituents &lt;em&gt;through the sigma bonds&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperconjugation is electron donation of a filled bonding orbital to a nearby low-lying, unfilled orbital. Considering t-butyl cation as an example, a filled C-H bonding orbital can donate electrons to the empty p-orbital on the electron-deficient carbon. From a molecular orbital view, this essentially serves to lower the energy of the C-H sigma orbital and raise the energy of the p-orbital. With only two electrons in this system, the effect is stabilizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g3uNU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this means there are two effects that explain why greater substitution increases carbocation stability: inductive effect and hyperconjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperconjugation can also be invoked for the substitution of alkenes. In this case, the filled C-H sigma orbital overlaps with the empty C-C pi-antibonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n9rto.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good rule of thumb is that a filled orbital overlapping with an unfilled orbital is a stabilizing interaction.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-03T21:12:59.470" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19113" PostTypeId="1" CreationDate="2014-11-03T23:45:09.767" Score="1" ViewCount="1367" Body="&lt;p&gt;In terms of IR spectrum, why does an increase in reduced mass lead to a decrease in frequency?&lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2014-11-04T00:13:35.833" Title="Why does an increase in reduced mass lead to a decrease in frequency?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-11-04T06:36:44.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="19114" PostTypeId="2" ParentId="19113" CreationDate="2014-11-04T00:13:35.833" Score="3" Body="&lt;p&gt;Consider a simple molecule &lt;span class=&quot;math-container&quot;&gt;$\ce{A-B}$&lt;/span&gt; composed of atoms &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and let the bond connecting them be described as a spring.  Then the stretching and contracting of the &lt;span class=&quot;math-container&quot;&gt;$\ce{A-B}$&lt;/span&gt; bond (spring) can be considered as a simple harmonic oscillation and is described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's law&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;${\nu}$&lt;/span&gt; is the frequency of the stretch,  &lt;span class=&quot;math-container&quot;&gt;${k}$&lt;/span&gt; is the force constant describing the specific spring and &lt;span class=&quot;math-container&quot;&gt;${m}$&lt;/span&gt; is the mass of the system.  In our &lt;span class=&quot;math-container&quot;&gt;$\ce{A-B}$&lt;/span&gt; example, the mass of the system is described by the reduced mass &lt;span class=&quot;math-container&quot;&gt;${\mu}$&lt;/span&gt;, where&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\mu = \frac{m_A m_B}{m_A + m_B}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So our equation becomes&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;From this equation we can see that as the reduced mass increase's the stretching frequency will become smaller.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-11-04T00:13:35.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19116" PostTypeId="1" CreationDate="2014-11-04T00:32:56.330" Score="5" ViewCount="9490" Body="&lt;p&gt;I'm trying to create a reasonable reaction mechanism for the reaction of nitrosobenzene and N-hydroxyaniline, which happens as part of the reduction of nitrobenzene to azobenzene using magnesium as reductor and methanol as solvent. &lt;a href=&quot;https://chemistry.stackexchange.com/users/4231/ron&quot;&gt;Ron&lt;/a&gt; was very helpful to give me a general outline of this entire mechanism:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key step is formation of nitrosobenzene (this is just a reasonable proposal).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGEDG.png&quot; alt=&quot;Reduction of nitrobenzene to yield azobenzene&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some of the nitrosobenzene can be further reduced to aniline and&#xA;  N-phenylhydroxylamine, both of which are nucleophiles and can react&#xA;  with the nitroso N=O bond (analogous to a carbonyl). Elimination of&#xA;  water in both reactions produces azobenzene and azoxybenzene&#xA;  respectively. The azoxybenzene can be further reduced to azobenzene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I went on to try and create a reasonable mechanism for the reactions of N-hydroxyaniline with nitrosobenzene and aniline with nitrosobenzene (both in the presence of magensium with methanol as solvent):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9MgaG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder whether this is a correct reasonable reaction mechanism. I tried to move radicals as much as possible from positively charged/highly electronegative atoms (e.g. the nitrogen in the nitro functional group / oxygen) to the less electronegative atoms (e.g. carbon). My reasoning was as follows (I have not yet learned about in radical chemistry, but I give it a try): radicals can be removed in two ways: 1) the radical electron leaves the molecule, which results in a positive charge. This is less likely to happen with a more electronegative atom like oxygen in comparison to for example carbon, so that the radical is more stable on oxygen. 2) The radical reacts with an other molecule and obtains an extra electron forming a new covalent bond (in this case the electron is donated by methanols C-H bond), resulting in a neutral molecule with no radicals. Since oxygen is more electronegative than carbon it is more &quot;greedy&quot; to fill the singly occupied orbital, making the radical more stable on carbon.&#xA;Furthermore the second way to lose the radical is much more likely than the first, since it leaves a neutral molecule instead of a charged one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At last, I was told that reactions involving metal surfaces (like magnesium in this case) are often not well understood. Why is that? Because there are too many pathways?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2014-11-17T03:49:17.923" LastActivityDate="2014-11-17T03:49:17.923" Title="Formation of azobenzene from nitrosobenzene and N-hydroxyaniline" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19117" PostTypeId="1" AcceptedAnswerId="19124" CreationDate="2014-11-04T00:40:14.033" Score="2" ViewCount="3471" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f8O9u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow I am told that my product is incorrect. Why is my product incorrect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see this is a Wittig reaction resulting in a two carbon increase with removal of the oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing that strikes me as odd is that the Wittig is conjugated to a bromide salt. The only way this can happen is if the phosphorous part has an overall positive formal charge. Does this mean that the n-butyllithium will abstract a proton and leave the carbon backbone of the Wittig as $\ce{CH3CH}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even so this doesn't seem to affect my final product ... &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2016-08-01T20:50:42.737" Title="Wittig reaction with benzaldehyde" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19118" PostTypeId="2" ParentId="19110" CreationDate="2014-11-04T00:52:46.037" Score="1" Body="&lt;p&gt;Bond breaking is endothermic, bond making is exothermic. To use bond strengths, sum the strengths of the broken bonds and sum the made bonds. Since the breaking is endothermic and the making is exothermic you can add the positive to the negative and get the enthalpy change for the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With heats of formation, it’s as if the above process has already been done with reference to the constituent elements of the substances involved. So, subtracting the heats of formation of the reactants from the heats of formation of the products is like doing the above process but with the actual values of the bonds in the substances, which can vary depending upon molecular structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at water as an example. The strength of a $\ce{O-H}$ bond is about 110 kcal/mol. $\ce{O=O}$ is about 120 and $\ce{H-H}$ is about 105. To make 2 mol of water you have to raise the potential energy of the system by 330 kcal by breaking the oxygen and hydrogen molecules into atoms. You then make 4 mol of $\ce{O-H}$ bonds in making 2 mol of water, lowering the potential energy by 440. This is a net change of −110 kcal for the process. −110 kcal is about 460 kJ. I rounded a lot and there’s always a difference in the bond strengths in a table and those in a molecule. A heat of formation table will tell you that a mol of water is about 240 kJ/mol lower in potential energy than the oxygen and hydrogen it was made from. You can take the 2(−240 kJ/mol) and subtract 1(0) for the oxygen and 2(0) for the hydrogen and end up with −480 kJ/mol, about what we got from the bond strength calculation.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T13:27:41.327" LastActivityDate="2015-04-30T13:27:41.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19119" PostTypeId="1" AcceptedAnswerId="19121" CreationDate="2014-11-04T00:53:09.080" Score="11" ViewCount="64331" Body="&lt;p&gt;Wouldn't CO2 be IR inactive because of its non-polar bonds?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2020-02-11T16:14:44.553" LastActivityDate="2020-02-11T16:14:44.553" Title="Is carbon dioxide IR inactive?" Tags="&lt;spectroscopy&gt;&lt;dipole&gt;&lt;ir-spectroscopy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19120" PostTypeId="2" ParentId="19117" CreationDate="2014-11-04T01:00:17.243" Score="1" Body="&lt;p&gt;I suspect they  want you to specify the stereochemistry too. The expected product should be the &lt;em&gt;cis&lt;/em&gt; (predominantly at least). That's the textbook answer, reality is not that ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: If I remember correctly the Z- isomer is favoured in higher temperatures and in lower temperatures you will get worse selectivity, or even predominantly &lt;em&gt;trans&lt;/em&gt;. In any case your answer is not &quot;wrong&quot; maybe incomplete.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="7284" LastEditDate="2014-11-04T01:06:23.933" LastActivityDate="2014-11-04T01:06:23.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19121" PostTypeId="2" ParentId="19119" CreationDate="2014-11-04T01:14:45.010" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am told that carbon dioxide is IR inactive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're right, &lt;strong&gt;that's not true&lt;/strong&gt;.  Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x8d35.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The symmetric stretch does not result in a change (of the initially&#xA;zero dipole moment), so it is ir-inactive.&lt;/li&gt;&#xA;&lt;li&gt;The asymmetric stretch does result in a change in dipole moment so it is ir-active.&lt;/li&gt;&#xA;&lt;li&gt;The bend also results in a change in dipole moment so it too is ir-active.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We expected 4 vibrations and I've only listed 3. This is because the &quot;bend&quot; (let's start by placing the molecule along the x-axis) can occur in the y direction and the z direction.  But these two motions are the same, just deforming in different directions, the bend is said to be degenerate, accounting for the &quot;fourth&quot; vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, carbon dioxide has 2 ir-active vibrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; - response to example added (question d) by OP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question d is incorrect.  Either the author 1) inadvertently switched the column headings (IR active, IR inactive) or 2) meant to use some molecule other than carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first 3 rules you learn for interpreting IR and Raman spectra are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number of molecular vibrational modes equals 3n-6 (3n-5 for linear molecules), where n is the number of atoms.&lt;/li&gt;&#xA;&lt;li&gt;An ir active band will be observed if a vibration results in a &lt;strong&gt;change&lt;/strong&gt; of the dipole moment. The initial dipole moment in the molecule's equilibrium geometry can be zero; all you need is a &lt;strong&gt;change&lt;/strong&gt;.  The terms &quot;polar&quot; and &quot;non-polar&quot; can be confusing, they often mean different things to different people.  Leave &quot;polar&quot; out of the criteria for ir activity and stick with dipole moment, it is a much better understood term. &lt;/li&gt;&#xA;&lt;li&gt;the rule of &lt;strong&gt;mutual exclusion&lt;/strong&gt;, it states that, for centrosymmetric molecules (molecules with a center of symmetry, like carbon dioxide), vibrations that are IR active are Raman inactive, and vice versa.  So for carbon dioxide there is 1 Raman band and two IR bands.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's a link to a recent SE Chem question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18605/how-can-i-deduce-the-linearity-of-xef2-from-the-ir-spectrum/18608#18608&quot;&gt;How can I deduce the linearity of XeF2 from the IR spectrum?&lt;/a&gt; where these rules were used to determine the structure of a molecule.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="23561" LastEditDate="2018-09-08T05:35:09.350" LastActivityDate="2018-09-08T05:35:09.350" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19123" PostTypeId="1" CreationDate="2014-11-04T01:32:02.477" Score="5" ViewCount="268" Body="&lt;p&gt;(First question on stackexchange, apologies for any faux pas)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ebola workers get drenched in sweat while working inside their personal protective equipment.  There's a discussion happening over on OpenIdeo about whether the desiccant packs used in shipping containers might help reduce humidity:&#xA;&lt;a href=&quot;https://openideo.com/challenge/fighting-ebola/ideas/desiccant-embedded-ppe-suit-to-keep-ppe-dry-and-bearable&quot; rel=&quot;noreferrer&quot;&gt;https://openideo.com/challenge/fighting-ebola/ideas/desiccant-embedded-ppe-suit-to-keep-ppe-dry-and-bearable&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions for the chemists:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the reaction products of a calcium chloride + starch desiccant pack with humid air?  In particular, would it generate any gases?  (That'd obviously be a problem inside an impermeable suit)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The worker sweats to carry away heat evaporatively.  Presumably most of that body heat would need to pass through the desiccant pack as the sweat gets absorbed, and would need to be transmitted outside the suit in order to be worthwhile.  It looks like the reaction is also exothermic.  How much total heat would need to be removed from the pack, and is this going to be too much to dissipate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any other risks to be aware of with this idea?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9621" LastActivityDate="2020-08-22T07:23:29.957" Title="Calcium chloride + starch desiccant packs inside PPE for ebola workers?" Tags="&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19124" PostTypeId="2" ParentId="19117" CreationDate="2014-11-04T01:39:33.117" Score="5" Body="&lt;p&gt;Below is the preparation of the Wittig reagent you are given. Yes, the intermediate is a phosphonium salt, with a formal positive charge on phosphorus. Treatment with base (BuLi) deprotonates at the adjacent carbon giving a phosphorus ylide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AbVSe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the Wittig reaction itself, one postulated mechanism (depicted in Organic Chemistry by Maitland Jones) is shown below. Nucleophilic attack of the ylide onto the aldehyde gives an open betaine intermediate, which closes to form an oxaphosphatane. The strained 4-membered ring decomposes readily to give an alkene and triphenylphosphine oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6IjHd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stereochemistry of the Wittig is related to the nature of the ylide used. Unstabilized ylides, such as the one here, predominantly give the Z-configured alkene, while stabilized ylides (usually through conjugation to a carbonyl) give an E-configured alkene as the major product. With substrates similar to the ones here, I obtained a 4:1 ratio of Z:E products, so the selectivity is not perfect, as K_P points out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stereochemical outcome is intimately related to the mechanism, and according to my copy of Jerry March's Advanced Organic Chemistry (4th ed), there is no evidence for the betaine intermediate. Instead, the oxaphosphatane is formed directly through a [2+2] cycloaddition mechanism. This mechanism is attractive because the stereochemical demands of the cycloaddition account for the preference of the (less stable) Z-product. However, this does not explain why different ylides give different selectivities.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-04T01:39:33.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19125" PostTypeId="1" AcceptedAnswerId="19128" CreationDate="2014-11-04T02:49:21.937" Score="5" ViewCount="432" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mixture of three gases N&lt;sub&gt;2&lt;/sub&gt;, CO&lt;sub&gt;2&lt;/sub&gt; and O&lt;sub&gt;2&lt;/sub&gt; has a total pressure of 106000 Pa. The mixture is cooled to -100°C and the resulting pressure is found to have decreased to 82500 Pa. What gas is condensed and what is its partial pressure?&lt;br&gt;&#xA;  (BP N&lt;sub&gt;2&lt;/sub&gt; = -196°C, BP O&lt;sub&gt;2&lt;/sub&gt; = -183°C and BP CO&lt;sub&gt;2&lt;/sub&gt; = -78°C)    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've had a look at all the formulas and laws that have been covered in my Chemistry textbook like Dalton's law of partial pressure and the combined gas equation but it appears that there is not enough information to solve it. I mean, if it asked to find the initial temperature, I could do it, but I'm not sure how to find the partial pressure of the gas(es) from the information provided.&lt;/p&gt;&#xA;" OwnerUserId="9623" LastEditorUserId="7448" LastEditDate="2018-03-07T10:14:45.690" LastActivityDate="2018-03-07T10:14:45.690" Title="How to use gas laws to calculate the partial pressure of a gas that has partly condensed from a mixture?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19126" PostTypeId="2" ParentId="19073" CreationDate="2014-11-04T03:14:24.340" Score="1" Body="&lt;p&gt;The heat capacity of a calorimeter, $C_{calorimeter}$ is given in units of kJ/°C, with no units of mass included.  This is because the calorimeter's mass is a constant and each calorimeter's heat capacity (and mass) is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general procedure that is used to determine the heat capacity of a calorimeter is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place a carefully measured quantity of a compound whose combustion energy is well known and that is available in a pure form in the bomb, usually benzoic acid.  &lt;/li&gt;&#xA;&lt;li&gt;Seal the bomb and add oxygen so that the sample will burn completely.&lt;/li&gt;&#xA;&lt;li&gt;Submerge the bomb and ignite the sample.&lt;/li&gt;&#xA;&lt;li&gt;Measure the temperature change of the water bath in which the bomb is submerged.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since you know the energy of the reaction and you know the heat energy in the water, you can find the heat energy of the calorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E = q_{water} + q_{calorimeter}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of the calorimeter is the difference between the energy we know we put into the system through the combustion of our standard and the heat that was measured in the water bath.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-11-04T03:14:24.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19127" PostTypeId="1" CreationDate="2014-11-04T03:22:48.803" Score="0" ViewCount="1510" Body="&lt;p&gt;What is the difference between the absolute thermodynamic temperature scale and the practical temperature scale: ITS-90?&lt;/p&gt;&#xA;" OwnerUserId="9624" LastActivityDate="2014-12-04T05:13:07.953" Title="Difference between the absolute thermodynamic scale and ITS-90" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19128" PostTypeId="2" ParentId="19125" CreationDate="2014-11-04T03:24:10.517" Score="2" Body="&lt;p&gt;You would need the initial temperature to be able to solve this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use the combined gas law (or Gay-Lussac's Law) to determine the total pressure of the gas at -100°C, assuming none had condensed.  This ideal gas pressure would differ from the 82.5kPa by the partial pressure of the condensed gas at -100°C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question doesn't specify at which conditions it wants to know the partial pressure of the missing gas.  If it's -100°C, you'd be done at that point.  If it's at the original temperature, you would have to apply the same gas law in reverse to get back to your original conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiling point information you added should allow you to determine which of the three gases condensed.  Only one would be a liquid or solid at -100°C.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2014-11-04T03:31:53.193" LastActivityDate="2014-11-04T03:31:53.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19129" PostTypeId="1" CreationDate="2014-11-04T03:35:13.440" Score="2" ViewCount="578" Body="&lt;p&gt;If it's OD of 600nm I know that's the wavelength used, but what are the other numbers in something like  &quot;OD 600nm of 0.6-0.7&quot;. The 0.6-0.7?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="418" LastEditDate="2015-04-18T05:31:07.867" LastActivityDate="2015-04-18T05:31:07.867" Title="Optical Density numbers" Tags="&lt;biochemistry&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19130" PostTypeId="2" ParentId="19129" CreationDate="2014-11-04T03:40:20.647" Score="3" Body="&lt;p&gt;That is the optical density (&lt;a href=&quot;http://en.wikipedia.org/wiki/Absorbance&quot; rel=&quot;nofollow&quot;&gt;absorbance&lt;/a&gt;). It's just a range of between 0.6 and 0.7. Without the context, it's hard to say why they don't have a precise value, but in something like a protocol, it's common to see that kind of thing when there is some natural variation or the exact number isn't important.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-04T03:40:20.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19131" PostTypeId="2" ParentId="19110" CreationDate="2014-11-04T04:17:03.507" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) change in enthalpy = enthalpy of (products-reactants) at constant pressure and temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H_{p,T} = H_\text{products}-H_\text{reactants}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) change in enthalpy = energy required to break bonds in molecules of reactants − energy released in the formation of bonds in molecules of products&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H_{p,T} = (-H_\text{reactants})-(-H_\text{products})= H_\text{products}-H_\text{reactants}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because the energy required is negative of enthalphy of formation/ enthalpy of substance.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T13:29:03.907" LastActivityDate="2015-04-30T13:29:03.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19133" PostTypeId="1" AcceptedAnswerId="19157" CreationDate="2014-11-04T04:20:33.477" Score="1" ViewCount="406" Body="&lt;p&gt;I know that Grignards react with carbon dioxide to give carboxylate anions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But will Gilmans do the same? Will Gilmans react with carbon dioxide in a productive fashion? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-16T17:09:57.577" LastActivityDate="2015-10-16T17:09:57.577" Title="Gilmans and carbon dioxide" Tags="&lt;reactivity&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19134" PostTypeId="1" AcceptedAnswerId="19455" CreationDate="2014-11-04T04:49:10.090" Score="4" ViewCount="618" Body="&lt;p&gt;What are the factors to consider when predicting the EPR spectrum of $\ce{Cu(Ox)P(CH3)3}$ where $\ce{OX}$ is the oxalate ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that there are several factors to consider for EPR (Electron Paramagnetic Resonance) when we have transition metal complexes like: coupling, anisotropy, high spin/low spin, Jahn Teller distortion, Zero Field Splitting, or Kramer's rule. My question is which of these are important for this special copper complex and why?&lt;/p&gt;&#xA;" OwnerUserId="6985" LastEditorUserId="69830" LastEditDate="2021-05-04T09:25:10.967" LastActivityDate="2021-05-04T09:25:10.967" Title="Factors to consider to predict EPR" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;epr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19135" PostTypeId="1" CreationDate="2014-11-04T05:03:10.730" Score="9" ViewCount="623" Body="&lt;p&gt;Intrinsic defects occur when a pure material crystal shows either vacancy or interstitialcy. When an ionic compound has vacancy, the imbalance in charge has to be compensated with another defect. When a vacancy is balanced with another vacancy it is called Schottky defect , when it is compensated by an interstitial defect, it is called Frenkel defect. When species are exchanged on the sites, the defect is called anti-site defect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing all of that, how do we predict the type of intrinsic defects in a given material such as $\ce{HgSe}$ or $\ce{Ba3N2}$ ?&lt;/p&gt;&#xA;" OwnerUserId="6985" LastEditorUserId="6985" LastEditDate="2017-07-05T16:10:40.810" LastActivityDate="2017-07-05T16:10:40.810" Title="How to predict the type of intrinsic defect?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2018-03-19T16:15:11.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="19136" PostTypeId="2" ParentId="19127" CreationDate="2014-11-04T05:12:53.557" Score="1" Body="&lt;p&gt;The absolute thermodynamic temperature scale is a scale fixed by absolute zero and (at least for the kelvin, I'm not sure about the Rankine scale) the triple point of water, where absolute zero is defined as 0 K and the triple point of water is defined as 273.16 K. The rest of the scale is just a straight line from zero through 273.16 K off to infinity. However, this scale relies on only two defined points (only one of which can actually be measured, absolute zero being established from thermodynamic relations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practical use, for the purposes of calibrating thermometres and the like, this is not a great scale as the two calibration points may be far away from the temperature region one is interested in actually measuring. The usual way to deal with this is to use other reference points outside the actual definition. So that different people do this as consistently as possible, ITS-90 was created as a practical implementation of the absolute scale, setting additional &lt;a href=&quot;http://en.wikipedia.org/wiki/International_Temperature_Scale_of_1990#Defining_points&quot; rel=&quot;nofollow&quot;&gt;reference points&lt;/a&gt; (all phase transitions) to defined values, interpolating between them to get other values. It's technically not a scale in itself, just a standard for calibrating temperatures that are as close to the absolute scale as possible. By making sure everyone uses the same reference points, measurements can be more consistent than if reference points are chosen arbitrarily and possibly assigned different values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one wanted to make a thermometre for general lab use, one might pick the triple point of mercury, the triple point of water, and the melting point of gallium from the ITS-90 standard for reference points, knowing that other thermometres in this range conforming to the standard would also be calibrated this way, rather than choosing say the melting and boiling points of water, increasing the likelihood of consistent measurements with other thermometres.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-04T05:12:53.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19137" PostTypeId="2" ParentId="19129" CreationDate="2014-11-04T07:10:46.577" Score="1" Body="&lt;p&gt;It's a (sloppy) spectroscopists' way to denote a concentration range &lt;em&gt;without&lt;/em&gt; actually knowing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It happens when you don't start from a properly prepared stock solution and use an UV-VIS spectrometer instead of a balance. It typically goes like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Add some compound (tip of a spatulum) to a cuvette and dissolve it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Measure the absorption. Typically it's much too high! ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Remove a part of the solution from the cuvette and add some solvent&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Record the absorption spectrum again.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Repeat 3. and 4. until your absorption is in a medium range: far from saturation but still high enough to observe smaller signals (as the result of a possible reaction).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-04T07:10:46.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19138" PostTypeId="1" AcceptedAnswerId="19147" CreationDate="2014-11-04T08:13:56.933" Score="1" ViewCount="318" Body="&lt;p&gt;I'm searching for a chemical daily life experience that can be modelled by superior mathematics so that the origin of the equations involved can be easily explained to a beginner in chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By superior mathematics I mean things like partial and ordinary differential equation, Integral Calculus, group theory, differential geometry. But I want to understand how are the equations derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in physics, I could model the trajectory of the rear wheel of a bicycle (a daily life experience) and I fully understand all the equations involved, without being a physicist. I want something similar to find in chemistry. &lt;/p&gt;&#xA;" OwnerUserId="9628" LastEditorUserId="4945" LastEditDate="2015-12-14T09:42:31.590" LastActivityDate="2015-12-14T09:42:31.590" Title="What is a chemical daily life experience that can be modelled by superior mathematics?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2017-04-16T18:40:53.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="19140" PostTypeId="1" CreationDate="2014-11-04T11:46:44.203" Score="-1" ViewCount="131" Body="&lt;p&gt;I want to know if:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A -&amp;gt; B &amp;lt;- C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a multi-step reaction OR are multiple reactions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A -&amp;gt; B, C -&amp;gt; B&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9630" LastActivityDate="2014-11-04T13:10:23.120" Title="Is (A-&gt;B&lt;-C) a multi-step reaction?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19141" PostTypeId="2" ParentId="19140" CreationDate="2014-11-04T13:10:23.120" Score="5" Body="&lt;p&gt;A reaction is generally something along the lines of&#xA;$$\ce{A-&amp;gt;Z}$$&#xA;where A is your starting material and Z the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction can also proceed stepwise, here shown via an intermediate Y:&#xA;$$\ce{A-&amp;gt;Y-&amp;gt;Z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also add intermediate &quot;steps&quot; in the reaction, where you introduce more reactants (here B) to give the product:&#xA;$$\ce{A-&amp;gt;Y -&amp;gt;[\ce{B}] Z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One commonality that all these reactions share is that the arrow is always pointing one way (and usually to the right). This allows us to reorder your reaction as follows:&#xA;$$\ce{A-&amp;gt;B&amp;lt;-C} \begin{cases} \ce{A-&amp;gt;B}\\ \ce{C-&amp;gt;B} \end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to figure it out from here.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-11-04T13:10:23.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19143" PostTypeId="1" AcceptedAnswerId="19148" CreationDate="2014-11-04T16:23:15.113" Score="7" ViewCount="187" Body="&lt;p&gt;What methods other than linear combination of atomic orbitals does MO theory use? Which one is the most accurate and comprehensive?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-11-04T17:48:07.013" Title="MO theory methods other than LCAO" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19144" PostTypeId="1" AcceptedAnswerId="19146" CreationDate="2014-11-04T16:36:11.300" Score="6" ViewCount="7356" Body="&lt;p&gt;Suppose I have a phenyl Grignard and methyl benzoate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I react the two, can I expect benzophenone and a triphenylmethoxide ion as my products? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because I know that Grignards react with ketones and aldehydes. So as the benzophenone is generated, couldn't more Grignard attack the benzophenone as well as the original reactant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected products and mechanism shown below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QS6Ki.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-04-05T13:02:58.480" LastActivityDate="2017-04-05T13:02:58.480" Title="Reaction of Grignard reagents with esters" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;organometallic-compounds&gt;&lt;esters&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19145" PostTypeId="2" ParentId="19144" CreationDate="2014-11-04T16:49:01.670" Score="4" Body="&lt;p&gt;Yes, that is exactly what happens. And since ketones are typically more reactive than esters, you should expect that the tertiary alcohol (after workup) will be the predominant product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to access the ketone, add the Grignard to a nitrile, which will give an imine (after protonation) that can be hydrolyzed to a ketone. Alternatively, add the Grignard to an aldehyde and oxidize the resulting alcohol to the ketone.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-04T16:49:01.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19146" PostTypeId="2" ParentId="19144" CreationDate="2014-11-04T16:53:23.053" Score="5" Body="&lt;p&gt;Reaction of a Grignard reagent with an ester is a standard method for producing tertiary alcohols where at least two of the substituents (the &quot;$\small\ce{R_2}$&quot; group attached to the Grignard) are the same.  Judicious choice of the starting ester allows for the preparation of a tertiary alcohol where all 3 substituents are the same. Alternately, you could react 3 equivalents of the Grignard with a cheap dialkyl carbonate (a &quot;double&quot; ester, if you will) to obtain a tertiary alcohol where all 3 groups are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzHkC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically a slight excess of Grignard reagent is used and the tertiary alcohol is produced after acid work-up.  If an excess of Grignard is not used then a mixture (ugh!) of ester, intermediate ketone and alcohol may result.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-04T18:44:53.683" LastActivityDate="2014-11-04T18:44:53.683" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19147" PostTypeId="2" ParentId="19138" CreationDate="2014-11-04T17:23:49.200" Score="5" Body="&lt;p&gt;Diffusion is one example from the daily life. For example, the absorption of ethanol from the small intestine to blood can be modeled using differential equations. The extraction of caffeine from grounded coffee is also a diffusion phenomenon.&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2014-11-04T17:23:49.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19148" PostTypeId="2" ParentId="19143" CreationDate="2014-11-04T17:48:07.013" Score="5" Body="&lt;p&gt;Well, this gets into the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;basis sets&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_combination_of_atomic_orbitals&quot; rel=&quot;noreferrer&quot;&gt;LCAO&lt;/a&gt; implies that the number of molecular orbitals will be the same as the number of atomic orbitals involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we usually use &lt;strong&gt;&lt;em&gt;more&lt;/em&gt;&lt;/strong&gt; basis functions than just the simple number of atomic orbitals. We might add diffuse or polarization functions to better describe electron density away from the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, rather than the physically correct &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater_orbitals&quot; rel=&quot;noreferrer&quot;&gt;Slater-type orbitals&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_l(r) = A(l,\alpha) r^l e^{-\alpha r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually use &lt;a href=&quot;http://en.wikipedia.org/wiki/Gaussian_orbital&quot; rel=&quot;noreferrer&quot;&gt;Gaussian-type orbitals&lt;/a&gt; to express the radial component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_l(r) = B(l,\alpha) r^l e^{-\alpha r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now GTOs don't express the radial part of the wave function well -- they decay too quickly as you get far from the nucleus. So we use a linear combination of multiple Gaussian orbitals to approximate the physically-correct Slater terms. Still, it's much faster to use a set of multiple GTOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond STOs and GTOs, people do use other types of orbitals. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://arxiv.org/abs/1110.4853&quot; rel=&quot;noreferrer&quot;&gt;Wavelets&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)#Plane-wave_basis_sets&quot; rel=&quot;noreferrer&quot;&gt;Plane waves&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Muffin-tin_approximation&quot; rel=&quot;noreferrer&quot;&gt;Muffin-tin&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-04T17:48:07.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19149" PostTypeId="2" ParentId="18635" CreationDate="2014-11-04T18:10:36.733" Score="5" Body="&lt;p&gt;Products from reaction step (1) are the &lt;em&gt;trans&lt;/em&gt;-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{LiNHEt + NH4Cl -&amp;gt;~ EtNH2 + NH3 + LiCl}$$&#xA;$$\ce{Li + H2O -&amp;gt;~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the &lt;em&gt;trans&lt;/em&gt;-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-04-12T03:48:32.607" LastActivityDate="2018-04-12T03:48:32.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19150" PostTypeId="1" AcceptedAnswerId="19158" CreationDate="2014-11-04T18:45:13.673" Score="6" ViewCount="32448" Body="&lt;p&gt;So the two conditions for a reaction to be spontaneous are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;tendency to achieve minimum energy&lt;/li&gt;&#xA;&lt;li&gt;tendency to achieve maximum randomness.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both these things are contrasting and the only way I see them working are when they work independent of each other. Do these work simultaneously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="275" LastEditDate="2014-11-04T19:03:34.087" LastActivityDate="2015-03-01T14:44:56.460" Title="Conditions for spontaneity" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19154" PostTypeId="1" AcceptedAnswerId="19160" CreationDate="2014-11-04T19:26:10.957" Score="12" ViewCount="5406" Body="&lt;p&gt;The Rehm-Weller equation, which defines the energy change for photoinduced electron transfer according to Lakowicz &amp;quot;Principles of Fluorescence Spectroscopy&amp;quot; [1, p. 337] should look like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=E^\mathrm{red}(D^+/D)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \frac{e^2}{\epsilon d} \tag{1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So you can see the first two components are reduction potentials of donor &lt;span class=&quot;math-container&quot;&gt;$D^+$&lt;/span&gt; and acceptor &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. But sometimes you can find another definition:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=E^\mathrm{ox}(D/D^+)-E^\mathrm{red}(A/A^-)-\Delta G_{00} - \dfrac{e^2}{\epsilon d} \tag{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which one is correct? I'm totally confused.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lakowicz, J. R. Principles of Fluorescence Spectroscopy, 3rd ed.; &lt;em&gt;Springer: New York&lt;/em&gt;, &lt;strong&gt;2006&lt;/strong&gt;. ISBN 978-0-387-31278-1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-09-19T05:10:22.570" Title="What should the Rehm-Weller equation look like?" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;photochemistry&gt;&lt;fluorescence&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19157" PostTypeId="2" ParentId="19133" CreationDate="2014-11-04T21:03:26.627" Score="3" Body="&lt;p&gt;Yes, cuprates do react with carbon dioxide to give access to carboxylates / carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running a SciFinder search, I do not see any examples of traditional Gilman reagents (alkyl cuprates) reacting directly with carbon dioxide. This shouldn't be surprising though, since the alkyl cuprates are generated from the corresponding alkyl lithium compound. If you have the alkyl lithium compound, then you can just react that with carbon dioxide directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more synthetically useful situation is when the alkyl cuprate is reacted with an alkyne, resulting in a syn-addition of the alkyl group and copper across the alkyne. The alkenyl (vinyl) cuprate can then be trapped with carbon dioxide to give a carboxylate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FmYCZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple examples from the literature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0040-4039(00)98174-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Z,​Z-​Dienes via acetylene carbocupration:  synthesis of navel orangeworm pheromone&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/P19810002527&quot; rel=&quot;nofollow noreferrer&quot;&gt;The silylcupration of acetylenes: a synthesis of vinylsilanes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check the write up in &lt;em&gt;Organic Reactions&lt;/em&gt; by Lipshutz (the world's expert in organocopper chemistry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complicated (non-Gilman) organocopper compounds have been reacted with carbon dioxide to give copper carboxylates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/0022-328X(88)80403-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbon-​carbon coupling of activated alkanes with carbon dioxide by [(phen)​(Ph3P)​Cu(O2COH)​]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/om034368r&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthesis, Structure, and CO2 Reactivity of a Two-​Coordinate (Carbene)​copper(I) Methyl Complex&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-04T21:03:26.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19158" PostTypeId="2" ParentId="19150" CreationDate="2014-11-04T21:11:46.577" Score="11" Body="&lt;p&gt;Luckily for us, a really smart guy named &lt;a href=&quot;http://en.wikipedia.org/wiki/Josiah_Willard_Gibbs&quot; rel=&quot;noreferrer&quot;&gt;Josiah Willard Gibbs&lt;/a&gt; figured out how these two tendencies work together back in the late 1800's. As a result, in chemical thermodynamics we now have a &lt;a href=&quot;http://en.wikipedia.org/wiki/State_function&quot; rel=&quot;noreferrer&quot;&gt;state function&lt;/a&gt; called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Gibbs Free Energy&lt;/a&gt; that describes what is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;noreferrer&quot;&gt;thermodynamic potential&lt;/a&gt; of a system at constant pressure, temperature, and number of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the systems we work with under lab conditions are at constant pressure and temperature and are typically closed to mass transfer (abbreviated as NPT), Gibbs free energy is a very convenient way of determining what will happen to a given chemical system under normal lab conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it works is that it is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy&quot; rel=&quot;noreferrer&quot;&gt;potential energy&lt;/a&gt; function - it describes how a system's potential energy will change as different state variables change, and therefore lets us predict where the system will &quot;try&quot; to go. An analogy that works pretty well is to imagine it like the surface of a planet, where gravity is the potential energy. If the derivative or slope is negative in a given direction, then things will tend to &quot;roll downhill&quot; at that point. On the other hand, to make things go uphill, you need to add energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing we can get from this analogy is that given the opportunity, things will eventually move from a state of higher potential energy to a state of lower potential energy. So we can compare the energy at two different states (the reactants and products, for example), and decide based on the difference in potential energy whether the reaction would be spontaneous under those conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for Gibbs free energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = H - TS$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $H$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy&quot; rel=&quot;noreferrer&quot;&gt;enthalpy&lt;/a&gt; (a.k.a heat of reaction), $T$ is absolute temperature (usually measured in Kelvin) and $S$ is &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;noreferrer&quot;&gt;entropy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the derivative at constant temperature and pressure, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = \Delta H - T\Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is for large changes. for small changes replace $\Delta$ with $\delta$). What this tells us is that change in Gibbs Free Energy is a function of change in enthalpy, change in entropy, and the absolute temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate how this works, take a look at the following diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fry9V.png&quot; alt=&quot;Delta G&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the line represents the Gibbs Free energy &quot;surface&quot; - the thermodynamic potential as a function of some variable (a reaction coordinate, for example). The ball represents the system at some point along that coordinate axis. If this were regular potential energy, it would be easy to see what happens - the ball will roll downhill if it gets the chance, and it will stop at the lower energy state. To get back up the hill, someone would have to put in that amount of energy. For thermodynamic potentials, it works the same way - the system will move along the coordinate axis in the direction of decreasing thermodynamic potential energy ($G$ in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that under constant NPT conditions, any process that involves an overall &lt;em&gt;decrease&lt;/em&gt; in Gibbs Free energy will be spontaneous. It will also tend to move in the &quot;direction&quot; that has the most negative slope in G at any given time. In mathematical terms,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G &amp;lt; 0$ - overall process is spontaneous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dG &amp;lt; 0$ - process will be moving in that direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dG = 0$ - process is at equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does this all fit in with what you described about systems trying to reach minimum energy, and maximum entropy? The answer is: those are both thermodynamic potentials for different types of systems. For a system with constant entropy, volume, and number of particles, (NSV) the &lt;em&gt;internal energy&lt;/em&gt; (or total energy) is the thermodynamic potential. For a system at constant entropy and pressure (NSP), &lt;em&gt;enthalpy&lt;/em&gt; is the thermodynamic potential. And for a system at constant volume and internal energy (NVE), the &lt;em&gt;negative of entropy&lt;/em&gt; is the thermodynamic potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, all of these thermodynamic variables are interconnected, and a change in one affects all of the others. You can hold three constant at any given time and still allow the system to &quot;move&quot; through &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_space&quot; rel=&quot;noreferrer&quot;&gt;phase-space&lt;/a&gt;. Which thermodynamic potential you need to describe how the system will move depends on which variables you choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the equation for $\Delta G$ and see how changes in enthalpy and entropy affect it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\downarrow \Delta G = \space \downarrow \Delta H - T \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\Delta H$ decreases, it will make $\Delta G$ decrease as well. This makes sense, since we know that exothermic processes tend to be spontaneous, because they are releasing energy and therefore the final system energy is lower than the initial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\downarrow \Delta G = \Delta H - T \space \uparrow \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\Delta G$ tends to decrease as $\Delta S$ &lt;strong&gt;increases&lt;/strong&gt; - this is because the change in entropy is &lt;strong&gt;subtracted&lt;/strong&gt; in the equation. This also matches up with what you know - an increase in entropy indicates a spontaneous process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your last question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it can! In fact, under these conditions, the process is &lt;strong&gt;guaranteed&lt;/strong&gt; to be spontaneous - if $\Delta H$ is negative, and $\Delta S$ is positive, then $\Delta G$ &lt;em&gt;has to be&lt;/em&gt; negative - the reaction would be spontaneous at any temperature under these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the other possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H &amp;gt; 0; \Delta S &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, since $\Delta H$ is always positive, and we are &lt;em&gt;subtracting a negative&lt;/em&gt; $\Delta S$, the reaction can &lt;strong&gt;never be spontaneous&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H &amp;gt; 0; \Delta S &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H &amp;lt; 0; \Delta S &amp;lt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these two cases, the reaction could be spontaneous or it could be non-spontaneous - it depends on the relative &lt;em&gt;magnitudes&lt;/em&gt; of the enthalpy and energy terms as well as the temperature at which it occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize - all thermodynamic variables are related to each other in fairly complicated ways. When you hold three of them constant (two if you don't count number of molecules), you can derive thermodynamic potential energy functions that describe the behavior of the system in terms of a single quantity. For most cases in chemistry, Gibbs Free Energy is the thermodynamic potential that we use. It gives us the relationship between enthalpy, entropy and temperature under constant pressure conditions. Since it's the thermodynamic potential, it also lets us predict how the system will behave - how it will move through phase space. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-11-04T21:11:46.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19159" PostTypeId="2" ParentId="19116" CreationDate="2014-11-04T21:27:10.667" Score="3" Body="&lt;p&gt;Below is my proposed mechanism. You'll see that we go through some of the same key intermediates, so I think you are on the right track. In your second step, it is much more likely that the proton is removed by acid base reaction instead of through the electron transfer process you've shown. Since the initial intermediate (my second structure) is zwitterionic, we can just say that it will neutralize itself. If the reaction conditions are not acidic (from what I see, it's just methanol and magnesium), then there will not be an appreciable amount of the protonated hydroxylamine. The mechanism doesn't require it, as nitrogen is strong enough to &quot;kick out&quot; hydroxide. The result is essentially an azoxybenzene, so we just need to figure out how to lose the other hydroxyl. As the cationic form, it will want to accept electrons, such as from the magnesium as a free electron. Back to a neutral (radical) compound, the other hydroxyl can be kicked out, giving a radical cation on nitrogen. Filling in a second electron from magnesium results in the azobenzene product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ltPSd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-04T21:27:10.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19160" PostTypeId="2" ParentId="19154" CreationDate="2014-11-04T22:14:32.383" Score="6" Body="&lt;p&gt;Both expressions are identical. It's just juggling of the signs, since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^{\mathrm{red}}(D^{\cdot+}/D) = - E^{\mathrm{ox}}(D/D^{\cdot+})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant primary references would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Rehm, D.; Weller, A. H. &lt;em&gt;Isr. J. Chem.&lt;/em&gt; &lt;strong&gt;1970&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 259-271.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rehm, D.; Weller, A. H. &lt;em&gt;Ber. Bunsen-Ges. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1969&lt;/strong&gt;, &lt;em&gt;73&lt;/em&gt;,&#xA;834-839.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Weller, A. H. &lt;em&gt;Z. Phys. Chem. NF&lt;/em&gt; &lt;strong&gt;1982&lt;/strong&gt;, &lt;em&gt;133&lt;/em&gt;, 93-98.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A normative glossary of terms in photochemistry is available from the &lt;a href=&quot;http://www.iupac.org/publications/pac/2007/pdf/7903x0293.pdf&quot; rel=&quot;noreferrer&quot;&gt;IUPAC&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-11-04T22:22:24.947" LastActivityDate="2014-11-04T22:22:24.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19161" PostTypeId="2" ParentId="19026" CreationDate="2014-11-04T22:14:52.260" Score="1" Body="&lt;p&gt;$\ce{CN-}$ is a significantly stronger nucleophile than $\ce{EtOH}$ and will therefore primarily displace the chloride. The facts that substitution takes place at a tertiary benzylic carbon, and that a polar protic solvent (ethanol) is used indicate that the reaction follows the $\ce{S_{N}1}$ mechanism (sources: &lt;a href=&quot;https://www.google.de/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=5&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CEAQFjAE&amp;amp;url=http%3A%2F%2Fhighered.mheducation.com%2Fsites%2Fdl%2Ffree%2F0073402745%2F990527%2Fsample_chapter.pdf&amp;amp;ei=FUJZVJLUOIzXarqcgegI&amp;amp;usg=AFQjCNGWAs8cJBU0l2Xy52elFQrzImeYqA&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://pages.towson.edu/ladon/orgrxs/reagent/bases.htm&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.wikipremed.com/03_organicmechanisms.php?mch_code=030204_030&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;). Due to the achiral intermediate (carbocation), the nitrile will be obtained as a racemic mixture, i.e. 2 products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx3.htm&quot; rel=&quot;nofollow&quot;&gt;Elimination&lt;/a&gt; is also a possible reaction pathway which needs to be considered, since cyanide is a good nucleophile but a weak base, which favors an &lt;a href=&quot;http://www.masterorganicchemistry.com/2012/11/30/deciding-sn1sn2e1e2-2-the-nucleophilebase/&quot; rel=&quot;nofollow&quot;&gt;E2 reaction&lt;/a&gt; unless an excess of $\ce{CN-}$ is present. In this case, the product with the more substituted double bond should be favored.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-11-04T22:14:52.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19162" PostTypeId="2" ParentId="19090" CreationDate="2014-11-04T23:01:52.297" Score="34" Body="&lt;p&gt;It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the $\pi^*_{\ce{C=C}}$ orbital while the carbene $\mathrm{sp^2}$ orbital interacts with the $\pi_{\ce{C=C}}$ orbital, so this 'virtual' double bond behaves somewhat like a normal double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the structure of the cyclopropylmethyl cation is downright weird. It is well known that both cyclopropylmethyl and cyclobutyl derivatives give very similar product mixture under $\mathrm{S_N1}$ hydrolysis conditions, resulting in both cyclopropylmethyl and cyclobutyl derivatives (see, for example, &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01150a029&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1951,&lt;/strong&gt; &lt;em&gt;73&lt;/em&gt; (6), 2509–2520&lt;/a&gt;). This is commonly described by the conjugation in following manner (here, the 1-cyclopropylethyl cation is depicted):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lbmbD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lbmbD.png&quot; alt=&quot;Cyclopropylethyl cation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here bonding of $\ce{C-4}$ with $\ce{C-1}$ and $\ce{C-2}$ can be roughly described as an interaction of the vacant $\mathrm{p}$-orbital of $\ce{C-4}$ with filled orbital of $\pi$-bond between $\ce{C-1}$ and $\ce{C-2}$. It does not matter much, where the original leaving group was - at  $\ce{C-2}$ or $\ce{C-1}$. Since the positive charge is more or less symmetrically distributed between three atoms and the small ring is somewhat relieved of its geometrical strain (both cyclopropane and cyclobutane are very strained molecules, not only due to angle strain, but also considerable steric interactions between hydrogens), the cation has remarkable stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar effects are common in the chemistry of small bicyclic systems, with norbornane derivitives being the chosen test subjects for decades with the &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Norbornyl_cation&quot; rel=&quot;noreferrer&quot;&gt;2-norbornyl cation&lt;/a&gt; being probably the most well-known example. March's &lt;em&gt;Advanced Organic Chemistry&lt;/em&gt;, 7th ed., Section 10.C.i discusses such nonclassical carbocations in great detail, with the cyclopropylmethyl system being described on pp 404–406.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With further addition of multiple cyclopropyl groups, however, the full conjugation becomes sterically hindered, so extra groups beyond the first have less of an effect. Of course, the stability of these cations is far below that of the tropylium cation, which has very little strain and also possesses aromatic character, distributing the positive charge over seven(!) carbon atoms.  In fact, the stability of the tropylium system is so high, that even the &lt;a href=&quot;http://en.wikipedia.org/wiki/Homoaromaticity&quot; rel=&quot;noreferrer&quot;&gt;cyclooctatrienyl cation (also known as the homotropylium cation)&lt;/a&gt; adopts a similar structure. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2016-09-24T10:42:30.903" LastActivityDate="2016-09-24T10:42:30.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19163" PostTypeId="1" AcceptedAnswerId="19166" CreationDate="2014-11-05T00:10:24.667" Score="2" ViewCount="505" Body="&lt;p&gt;I'm messing around with building a solar distillation apparatus, and one of the parts I think I need looks kind of like the following, except I have no idea what it is called (and thus can't look it up to see if that site even sells them).&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pxTRi.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pxTRi.gif&quot; alt=&quot;this&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can anyone tell me what this thing is commonly called?&lt;/p&gt;&#xA;" OwnerUserId="9638" LastEditorDisplayName="user7951" LastEditDate="2018-05-14T09:42:23.857" LastActivityDate="2018-05-14T09:42:23.857" Title="What is this glassware for a distillation apparatus called?" Tags="&lt;equipment&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19164" PostTypeId="1" AcceptedAnswerId="19169" CreationDate="2014-11-05T04:46:07.493" Score="4" ViewCount="351" Body="&lt;p&gt;Can't we perform Fischer esterification using methanol to make a methyl ester from a carboxylic acid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, why might we use diazomethane? Would the release of nitrogen gas confer an entropic advantage to the reaction? As I understand, the Fischer esterification process is an equilibrium process and at equilibrium there is often significant amounts of both product and reactant. So we might have to add excess alcohol to drive the reaction to the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because my book just presents the use of diazomethane as a way of making methyl esters without any discussion of why might someone do this rather than just use a few gallons of methanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand I know from working in a research lab that methanol is poisonous and protocol calls for its use in a fume hood ... could that be another reason? On the other hand diazomethane is also toxic (and explosive to boot) ... so I would definitely be inclined to use that in a fume hood (perhaps more so than when using methanol). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user7951" LastEditDate="2015-08-13T12:54:19.030" LastActivityDate="2015-08-13T12:54:19.030" Title="Formation of methyl esters" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19165" PostTypeId="1" AcceptedAnswerId="19172" CreationDate="2014-11-05T05:30:04.993" Score="1" ViewCount="192" Body="&lt;p&gt;When HBr reacts with ethene it proceeds by the attack of proton on Ethene to give Ethyl carbocation and then attack of nucleophilic bromide ion to give Bromoethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does the proton react with Ethene earlier than the bromide ion? Is a proton more reactive than bromide ion or is there some other reason?&lt;/p&gt;&#xA;" OwnerUserId="9310" LastEditorUserId="186" LastEditDate="2015-10-02T16:18:27.400" LastActivityDate="2015-10-02T16:18:27.400" Title="Electrophilic Addition Reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19166" PostTypeId="2" ParentId="19163" CreationDate="2014-11-05T06:21:59.323" Score="2" Body="&lt;p&gt;The &quot;A&quot; on the glassware gives it away as an Aldrich glassware component;&#xA;and Ron is correct calling it a Kugelrohr bulb. It is actually a &lt;em&gt;single bulb ball tube&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This specific item can be found in the &lt;a href=&quot;http://www.sigmaaldrich.com/labware/labware-products.html?TablePage=17215996&quot; rel=&quot;nofollow&quot;&gt;catalogue at Aldrich's website&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2014-11-05T06:21:59.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19169" PostTypeId="2" ParentId="19164" CreationDate="2014-11-05T08:42:38.740" Score="5" Body="&lt;p&gt;Both methods are used in practice (and a a lot more exist). Diazomethane is a last resort method when everything else fails. For example you might have a molecule that is acid sensitive or heat sensitive and then $\ce {MeOH/H^+}$ is out of the question. Although diazomethane is an excellent reagent to do the job, as it works selectively with almost every substrate, to give good yields under mild conditions, generally no extensive work-up or purification are required, it has some major disadvantages: It is not commercially available, it has to be prepared fresh before the reaction using a (somewhat) specialised equipment and a toxic starting material (&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/d28000&quot;&gt;Diazald&lt;/a&gt;). It is explosive and very toxic and obviously can not be stored. An alternative to using diazomethane is TMS diazomethane that is stable enough to be stored as a solution and it is commercially available however it is not innocent at all as &lt;a href=&quot;http://pipeline.corante.com/archives/2009/05/14/tms_diazomethane_update_on_a_fatality.php&quot;&gt;this article &lt;/a&gt; shows.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-11-05T08:42:38.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19170" PostTypeId="1" CreationDate="2014-11-05T09:31:17.193" Score="0" ViewCount="1089" Body="&lt;p&gt;Is 95% ethanol plus 5% water consider as food grade ethanol?&lt;/p&gt;&#xA;" OwnerUserId="9642" LastActivityDate="2015-08-24T16:02:46.397" Title="what is the acceptable percentage for food grade ethanol?Is 95% ethanol plus 5% water consider as food grade ethanol?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19171" PostTypeId="2" ParentId="19170" CreationDate="2014-11-05T10:17:31.843" Score="2" Body="&lt;p&gt;&lt;strong&gt;The grade of ethanol does not depend on the water and ethanol percentages, but rather on &lt;em&gt;what else&lt;/em&gt; is in there.&lt;/strong&gt; If your mixture is truly 95-5 ethanol-water, with nothing else in it, than that would be safe to consume (accounting for the health hazards of consuming too much ethanol of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there might be traces of e.g. methanol or benzene in the ethanol, which are both highly toxic to humans and therefore cannot be present in food grade ethanol mixtures. According to &lt;a href=&quot;http://www.grainprocessing.com/alcohol/alcohol-grades.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt; the FCC grade and above are food grade, in particular due to some heavy metal specifications, so apparently there might also be trace heavy metals in the ethanol.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-11-05T10:17:31.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19172" PostTypeId="2" ParentId="19165" CreationDate="2014-11-05T11:18:08.703" Score="3" Body="&lt;p&gt;The mechanism of the addition of hydrogen bromide to ethene is called &lt;em&gt;electrophilic addition.&lt;/em&gt; The region of highest electron density in ethene is the double bond, where the π electrons are. Hydrogen bromide acts as an &lt;em&gt;electrophile&lt;/em&gt; toward the π electrons of the double bond. The proton of hydrogen bromide is positively polarized and thus &lt;em&gt;electrophilic&lt;/em&gt;. (Obviously, the negatively charged bromide ion is not electrophilic.) Hence, the first step is an acid-base reaction in which hydrogen bromide donates a proton to ethene, forming a carbocation.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-11-05T11:18:08.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19177" PostTypeId="1" CreationDate="2014-11-05T16:50:03.747" Score="0" ViewCount="4153" Body="&lt;p&gt;Got a paper due tomorrow in chemistry, and my friends and I are totally stuck on this:&#xA;Basically we mixed some regular salt (NaCl) into a glass of water and stuck a zinc nail and a copper nail into the solution. We connected the nails and measured a voltage of .6V. Why does this happen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand galvanic cells of other types and electrolysis to a pretty OK extent, but I can't for the life of me understand what's happening at a chemical level in this reaction. I also need chemical formulas and the likes. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="9646" LastActivityDate="2017-10-05T17:31:12.790" Title="Galvanic cell using NaCl, Cu and Zn, why does it work?" Tags="&lt;homework&gt;&lt;electrolysis&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19179" PostTypeId="1" AcceptedAnswerId="19180" CreationDate="2014-11-05T17:22:42.970" Score="6" ViewCount="3542" Body="&lt;p&gt;I ran across a problem in which I was asked to predict the result of a reaction between $\ce{Rb}$ and $\ce{N2}$. I was frustrated until I found a note in my textbook which stated that out of all the alkali metals, only $\ce{Li}$ reacts with $\ce{N2}$. I am curious to know the reason for this. &lt;/p&gt;&#xA;" OwnerUserId="9639" LastActivityDate="2014-11-05T17:47:12.910" Title="Why doesn't rubidium react with nitrogen gas?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19180" PostTypeId="2" ParentId="19179" CreationDate="2014-11-05T17:47:12.910" Score="4" Body="&lt;p&gt;Lithium is the only group 1 element that can form a stable alkali metal nitride (at standard conditions), the other group 1 elements (such as sodium, potassium, rubidium,...) will react (at different reaction conditions) to form their respective nitrides (in the case of rubidium, $\ce{Rb3N}$ (rubidium nitride)), but the compound will be unstable and decompose quickly.  The reactions of the alkali metals (other than lithium) with nitrogen would not release enough lattice energy and would thus be endothermic, so they do not form nitrides at standard conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more here : &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkali_metal#Nitrides_and_pnictides&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Alkali_metal#Nitrides_and_pnictides&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user2117" LastActivityDate="2014-11-05T17:47:12.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19182" PostTypeId="2" ParentId="19177" CreationDate="2014-11-05T19:12:40.760" Score="3" Body="&lt;p&gt;I want you to try and work out the details yourself, but there's one aspect of this kind of cell that's not straightforward. At one electrode, you have a reduction reaction and at the other, an oxidation, just like any other galvanic cell. In the oxidation reaction, one of the metal electrodes is oxidized to put cations in solution. (think about which metal is more easily oxidized) In the type of galvanic cell that you may be more familiar with, a metal cation in solution is reduced onto the other electrode, but in this case, the only metal cation is $\ce{Na+}$, which is extremely difficult to reduce. Therefore, some other reduction reaction must be happening. The only other reducible substance present is water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two reactions occurring can be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; table, and using the potentials in the table, you should be able to work out which reactions are occurring and at which electrodes.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-05T19:12:40.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19184" PostTypeId="1" AcceptedAnswerId="19185" CreationDate="2014-11-05T20:19:55.180" Score="1" ViewCount="89" Body="&lt;p&gt;Chemical elements are said to be immutable at &lt;em&gt;normal&lt;/em&gt; energies, but not at the energies that are orders of magnitude higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that nucleus at &lt;em&gt;normal&lt;/em&gt; energies gives us the Periodic Table and at &lt;em&gt;higher&lt;/em&gt; energies becomes unstable, can we say that (with fusion) we will be able to transform an arbitrary chemical element into some other chemical element (if we have enough energy)?&lt;/p&gt;&#xA;" OwnerUserId="7251" LastEditorUserId="5017" LastEditDate="2014-11-05T20:58:11.270" LastActivityDate="2014-11-05T21:04:56.407" Title="&quot;Orthogonality&quot; of chemical elements" Tags="&lt;energy&gt;&lt;elements&gt;&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19185" PostTypeId="2" ParentId="19184" CreationDate="2014-11-05T21:04:56.407" Score="3" Body="&lt;p&gt;Certainly &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fusion&quot; rel=&quot;nofollow&quot;&gt;nuclear fusion&lt;/a&gt; and nuclear &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_fission&quot; rel=&quot;nofollow&quot;&gt;fission&lt;/a&gt; can convert elements into other elements. Indeed, this is how &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleosynthesis&quot; rel=&quot;nofollow&quot;&gt;different elements form&lt;/a&gt; in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one issue I have is with your &quot;arbitrary chemical element.&quot; You can use fusion to put together lighter nuclei or elements into heavier ones - up to iron, or use fission to divide more massive nuclei into lighter ones.. up to about iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit here is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy&quot; rel=&quot;nofollow&quot;&gt;binding energy of the nucleus&lt;/a&gt; which maximizes with $\ce{^{56}Fe}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, though, your point is a good one. Given enough energy (e.g., in a supernova), you can create lots of different elements.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-05T21:04:56.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19187" PostTypeId="2" ParentId="16782" CreationDate="2014-11-05T22:22:02.607" Score="2" Body="&lt;p&gt;Our lab uses a thiazole orange exclusion assay to measure how well our &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/22786534&quot; rel=&quot;nofollow&quot;&gt;polyintercalating peptides&lt;/a&gt;* bind DNA. The assay works by mixing thiazole orange with DNA so it becomes fluorescent. Then we measure fluorescence while titrating in peptide and the loss of flourescence indicates binding. This is because thiazole orange is only fluorescent when intercalated into DNA, and as the competing compound displaces it, the free thiazole no longer fluoresces. Note that this assay is not foolproof, polycationic compounds can condense DNA without intercalating, but this conformation change still forces the thiazole out and reduces the fluorescence, so loss of fluorescence implies binding, not necessarily intercalation. This method works for RNA too, but you'll need more thiazole to see signal because there are usually fewer double helices to intercalate into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to measure viscosity, because &lt;a href=&quot;http://en.wikipedia.org/wiki/Intercalation_%28biochemistry%29&quot; rel=&quot;nofollow&quot;&gt;intercalators&lt;/a&gt; make the DNA longer and the solution becomes more viscous, but you either need a very sensitive method to measure the change or a very concentrated DNA solution, and the viscosity can probably also be changed by condensing the DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible method MIGHT be &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorescence_anisotropy&quot; rel=&quot;nofollow&quot;&gt;fluorescent anisotropy&lt;/a&gt;, if your intercalator is fluorescent. It would spin rapidly outside the DNA, but slow down once it binds the double helix. However, conformational changes in the DNA can change its rotation speed and confound the results.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;That paper doesn't really mention the assay, it's just to show off the peptides so you understand what I'm talking about a little more. Those peptides also bind by electrostatic interaction and condense DNA, so it's not a pure measurement of intercalation.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="137" LastActivityDate="2014-11-05T22:22:02.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19189" PostTypeId="1" CreationDate="2014-11-05T23:16:38.267" Score="1" ViewCount="623" Body="&lt;p&gt;Why is it that when you consume something that is sugary-sticky (like candy, chocolate, fruit juice, etc.) and get it on your skin, when you try to clean it off with saliva (particularly, licking/sucking it off) it doesn't work?  What is it about the makeup of saliva that doesn't dissolve the sugar?&lt;/p&gt;&#xA;" OwnerUserId="9652" LastActivityDate="2014-11-07T00:25:24.927" Title="Why can't saliva clean &quot;stickiness&quot;?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19190" PostTypeId="1" CreationDate="2014-11-05T23:22:35.173" Score="10" ViewCount="25356" Body="&lt;p&gt;Why do krypton and xenon have high electronegativity? Noble gases are supposed to be &quot;happy&quot; with the amount of electrons they have, because they have 8 valence electrons (thus, most noble gases have no electronegativity). So why do krypton and xenon have electronegativity? Why do they tend to &quot;want&quot; one more electron? It makes no sense... their outer shells are already full!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to do a research paper on this question, so I would really like some good resources that could help me answer this question for myself. Know any? &lt;/p&gt;&#xA;" OwnerUserId="9653" LastEditorUserId="9961" LastEditDate="2016-07-24T22:33:51.190" LastActivityDate="2016-07-24T22:33:51.190" Title="Electronegativity in krypton and xenon?" Tags="&lt;electronegativity&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19191" PostTypeId="1" AcceptedAnswerId="19197" CreationDate="2014-11-05T23:27:10.617" Score="0" ViewCount="2676" Body="&lt;p&gt;It's a well-known fact that the dissociation of $\text{NaCl}$ can be represented by the series of equations $$ \begin{eqnarray*} \text{Na} &amp;amp;\rightarrow&amp;amp; \text{Na}^+ + \text{e}^- \qquad \Delta H = 495 \, \text{kJ} \\ \text{e}^- + \text{Cl} &amp;amp;\rightarrow&amp;amp; \text{Cl}^- \qquad \qquad \Delta H = -348 \, \text{kJ} \end {eqnarray*} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding gives $ \text{Na} + \text {Cl} \rightarrow \text{Na}^+ + \text{Cl}^- $ with $ \Delta H = 147 \, \text{kJ} $, which is positive so it is not spontaneous. What explains this? &lt;/p&gt;&#xA;" OwnerUserId="3842" LastActivityDate="2016-08-14T12:25:53.810" Title="NaCl dissociation not spontaneous?" Tags="&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19192" PostTypeId="1" CreationDate="2014-11-05T23:29:18.823" Score="2" ViewCount="1212" Body="&lt;p&gt;How does a redox reaction between $\ce{NaClO}$ and $\ce{KBr}$ take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will bromide be oxidised to bromine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Br^{−} + ClO^{−} + H2O ⟶ Cl^{−} + 2 OH^{−} + Br2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or will bromide be oxidised to $\ce{BrO^{−}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br^{−} + ClO^{−} ⟶ BrO^{−} + Cl^{−}}$&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="4231" LastEditDate="2014-11-05T23:44:45.063" LastActivityDate="2015-04-19T10:05:27.020" Title="How does a redox reaction between NaClO and KBr take place? will bromide be oxidised to bromine?" Tags="&lt;redox&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19193" PostTypeId="1" CreationDate="2014-11-05T23:52:16.303" Score="1" ViewCount="178" Body="&lt;p&gt;I have to decide which one has a higher oxidation strength $\ce{ClO-}$ or $\ce{Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO- + H2O + 2e- -&amp;gt; Cl- + 2OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2 + 2e- -&amp;gt; 2Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this reaction occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO- + H2O + 2Cl- -&amp;gt; Cl- + 2OH- + Cl2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion not, because otherwise the $\ce{ClO-}$ is reduced to $\ce{Cl-}$. If the reaction works, the $\ce{Cl-}$ could react as reduction agent again. Therefore if I would add $\ce{Br2}$ to $\ce{ClO-}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 ClO- + 2 H2O + 4Br- -&amp;gt; \mathbf{2 Cl- }  + 4OH- + 2Br2 }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ClO- + H2O + 2Br- + \mathbf{2 Cl- } -&amp;gt; \mathbf{Cl- } + 2OH- + Br2  + \mathbf{Cl2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So therefore chlorine has a higher oxidation strength than $\ce{ClO-}$?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T12:19:20.300" LastActivityDate="2016-07-04T23:08:19.667" Title="Which substance has the higher oxidation strength?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19195" PostTypeId="2" ParentId="19190" CreationDate="2014-11-06T03:43:41.090" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do Krypton and Xenon have high electronegativity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family.  They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron.  However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules.  In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Xe + 3F2 → XeF6}~~(1)$$&#xA;$$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common definition of electronegativity is based on Pauling's work and is given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot;&gt;equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$.  From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, &lt;strong&gt;for any element that can react and form a molecule&lt;/strong&gt;, we can calculate its electronegativity.  Since Kr and Xe do react to form molecules, &lt;strong&gt;we can use the above methodology to calculate its electronegativity&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity.  Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2).  We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user15489" LastEditDate="2015-06-25T01:49:24.833" LastActivityDate="2015-06-25T01:49:24.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19196" PostTypeId="1" CreationDate="2014-11-06T04:29:13.783" Score="2" ViewCount="468" Body="&lt;p&gt;Are there any molecules that are chiral that are also gases at room temperature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, what chiral molecules have a high vapor pressure? Propylene oxide has a boiling point of only 34 C, so that seems like a good one. Are there others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I appreciate all of the great suggestions! I should have mentioned before, but I am looking for something that can be purchased in both R and S enantiomers.&lt;/p&gt;&#xA;" OwnerUserId="4037" LastEditorUserId="4037" LastEditDate="2014-11-06T16:10:48.763" LastActivityDate="2014-11-06T18:17:03.010" Title="Gas phase chiral molecules" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;chirality&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19197" PostTypeId="2" ParentId="19191" CreationDate="2014-11-06T05:12:36.997" Score="2" Body="&lt;p&gt;You might be confusing $\ce{\Delta G}$ and $\ce{\Delta H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\Delta G = \Delta H - T \Delta S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\Delta G}$, free energy, measures the &quot;usefulness&quot; of a system. If $\ce{\Delta G \gt 0}$, then the reaction is not thermodynamically favored - it's NOT SPONTANEOUS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{\Delta H}$ being positive means that the reaction is endothermic, or makes the surroundings cooler. While this would CONTRIBUTE to a positive value of $\ce{\Delta G}$, it DOES NOT GUARANTEE it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your example above, I would assume the $\ce{T \Delta S}$ factor would make $\ce{\Delta G \lt 0}$ if that reaction really is SPONTANEOUS (I don't have an entropy table with me).&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2014-11-06T05:12:36.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19198" PostTypeId="2" ParentId="19196" CreationDate="2014-11-06T05:18:22.477" Score="5" Body="&lt;p&gt;The molecule has to be small. &lt;em&gt;trans&lt;/em&gt;-1,2-difluorocyclopropane is certainly chiral, and its predicted boiling point is ~ 5 deg C.  &lt;em&gt;trans&lt;/em&gt;-1,1,2,3-tetrafluorocyclopropane is another likely candidate.  This one will be easier to prepare (from perfluoropropylene oxide and &lt;em&gt;trans&lt;/em&gt;-1,2-difluoroethylene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separation of enantiomers is the hard part.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-11-06T18:17:03.010" LastActivityDate="2014-11-06T18:17:03.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19199" PostTypeId="2" ParentId="19189" CreationDate="2014-11-06T06:30:55.603" Score="6" Body="&lt;p&gt;So I could have sworn I've cleaned stuff off with saliva before, but I wanted to make sure, so I ran a quick impromptu experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I slathered the underside of my left arm with some maple syrup I had lying around. I waited about 10 minutes for it to dry, then I mentally sectioned off my arm into three parts and cleaned each part off in different ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section closest to my hand was cleaned by putting my other hand under running water and wiping the syrup off (basically, getting my right hand wet and making wiping motions across the arm, then returning the hand to the water, until I felt like I was no longer making any progress).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle section was left untouched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section closest to my elbow was cleaned by licking my hand, then using it to wipe that section down (essentially the same as the water, except I used my mouth as the source of liquid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the cleaning, I let everything dry again for 3 minutes. Then I placed some toilet paper over my entire arm and attempted to remove it. Here are the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry for the horrid image quality) &lt;img src=&quot;https://i.stack.imgur.com/7jaGn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I didn't have any skin marking tools on me at the time (I use pretty much only water-soluble inks), so I had to add the lines afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see, the section of my arm that was cleaned with water and with saliva are pretty much TP free, aside from a small section near my elbow that I think was just me missing some spots. The section cleaned with water was similarly TP-free. The middle section is pretty heavily TP-ed. What you're seeing is what got left behind when I attempted to rip the TP free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like saliva actually &lt;em&gt;can&lt;/em&gt; clean off stickiness pretty well! This is backed up by the comment that Michiel made, &lt;a href=&quot;https://en.wikipedia.org/wiki/Saliva&quot; rel=&quot;nofollow noreferrer&quot;&gt;which points out that saliva is 99.5% water.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, running this experiment gave me a possible explanation for why it might seem like it doesn't: it takes a LOT of saliva to do this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the water, I was easily able to wash off all of the syrup off of my hand and get a new coating of water on my fingers. Unfortunately, I can't generate that much saliva at once, and so I couldn't wash off the sugar that was already on my fingers that effectively. That made the saliva cleaning take a LOT longer than washing with water, both in terms of how many wipes it took to clean off most of the stuff and how much time it took overall (sadly, I cannot drool at the same rate a bathroom sink outputs water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another possible explanation that springs to mind: when you have syrupy stuff stuck to you, you probably just ate some of the syrupy stuff. That means your mouth probably still has a good bit of sugar in it, which would also make it harder to clean off additional sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, based off of this experiment, I feel fairly confident in saying that saliva cleans off sugar stickiness almost as well as regular old water, as long as you have enough of it lying around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I initially tried this with a flavored whiskey, hoping the alcohol would help it dry faster. Terrible mistake! Not only was there not enough sugar in the whiskey to get any sort of real stickiness, most of it basically ran off of my arm right away, so I ended up losing a lot of it. Save your flavored sprits for drinking, use a syrup!&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="2087" LastEditDate="2014-11-07T00:25:24.927" LastActivityDate="2014-11-07T00:25:24.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19202" PostTypeId="1" CreationDate="2014-11-06T07:45:48.567" Score="9" ViewCount="5287" Body="&lt;p&gt;There exist many home made pH scales like these&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8XTe.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/avn1a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WqLjj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do any of you know if there is an &quot;official&quot; scale that defines if a neutral detergent is 7 sharp or within a range _______?&lt;/p&gt;&#xA;" OwnerUserId="9659" LastActivityDate="2019-05-05T03:27:06.433" Title="Is there a pH range for &quot;neutral&quot; detergents?" Tags="&lt;ph&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19204" PostTypeId="1" CreationDate="2014-11-06T10:09:29.403" Score="3" ViewCount="15575" Body="&lt;p&gt;What does oxidizing strength mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean the strength of a oxidation agent that is needed to oxidize the reduction agent, or does it just mean the tendency of a substance to lose electrons (oxidate)?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="7159" LastEditDate="2014-11-06T16:37:16.477" LastActivityDate="2018-11-08T23:50:41.943" Title="What does oxidizing strength mean?" Tags="&lt;redox&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19205" PostTypeId="1" AcceptedAnswerId="19230" CreationDate="2014-11-06T13:04:46.823" Score="34" ViewCount="51980" Body="&lt;p&gt;Phenol has a $\mathrm pK_\mathrm a$ approximately equal to $9.9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one studies the acidity of chlorophenols, one notices the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ot2Lp.png&quot; alt=&quot;2-, 3-, and 4-chlorophenol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, chlorophenols are more acidic than phenol, due the negative inductive effect (−I) of chlorine, that reduces the negative charge, located on the oxygen of the phenolate anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the different chlorophenols, the observed trend is explained by the fact that the stabilizing −I effect of chlorine decreases with the distance from oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now focusing on another class of substituted phenols, such as methylphenols (= cresols), the $\mathrm pK_\mathrm a$ varies as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRmt2.png&quot; alt=&quot;2-, 3-, and 4-methylphenol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the methyl group yields a positive inductive effect (+I), thus increasing the negative charge on phenolate oxygen. That's why cresols are less acidic than phenol. One can further realize that the meta compound is more acidic, and this is due to the resonance structures which show a negative charge in ortho and para, but not in meta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate to learn whether such a reasoning, for these two type of substituted phenols, is plausible or whether further more solid arguments should be invoked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I would appreciate to understand why the difference in acidity between 2- and 4-chlorophenols is much more pronounced than the same difference between &lt;em&gt;o&lt;/em&gt;- and &lt;em&gt;p&lt;/em&gt;-cresols.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorUserId="5026" LastEditDate="2018-02-23T02:45:37.313" LastActivityDate="2020-03-01T05:59:31.337" Title="Acidity of substituted phenols" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;quantum-chemistry&gt;&lt;resonance&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19206" PostTypeId="2" ParentId="19196" CreationDate="2014-11-06T14:06:05.993" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1-Chloro-1,2,2,2-tetrafluoroethane&quot; rel=&quot;nofollow&quot;&gt;1-Chloro-1,2,2,2-tetrafluoroethane&lt;/a&gt; has bp -12 Celsium. For comparison, &lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_US_CB7446771.aspx&quot; rel=&quot;nofollow&quot;&gt;1-chloro-1-fluoroethane&lt;/a&gt; has bp +16 Celsium. In most cases, replacing good share of hydrogen in an organic molecule with fluorine leads to significant decrease of bp. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexafluoropropylene_oxide&quot; rel=&quot;nofollow&quot;&gt;perfluoropropylene oxide&lt;/a&gt;  has bp -27 Celsium, comparing to +34 of vanilla propylene oxide you noted.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-06T14:06:05.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19207" PostTypeId="2" ParentId="19196" CreationDate="2014-11-06T14:28:18.057" Score="2" Body="&lt;p&gt;If you use deuterium and tritium, you can make stereocenters from methyl groups. For example, 1-deutero-1-tritioethane is be chiral.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-06T14:28:18.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19209" PostTypeId="2" ParentId="19196" CreationDate="2014-11-06T14:50:57.920" Score="4" Body="&lt;p&gt;&lt;strong&gt;Considering molecules that have been prepared and resolved:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1,3-dimethylallene&lt;/strong&gt; (2,3-pentadiene) boils a bit above room temperature at 48.5 C (see  page 11 of this &lt;a href=&quot;https://scholarspace.manoa.hawaii.edu/bitstream/handle/10125/11395/uhm_phd_6713714_r.pdf?sequence=1&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; for the experimental preparation), but likely has a reasonable vapor pressure since it has a lower molecular weight than pentane.  The advantage to this molecule is that it has already been (at least) partially resolved (see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01026a032?journalCode=jacsat&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0040403900898305&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) in a one-step hydroboration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simple hydrocarbon to consider is &lt;strong&gt;&lt;em&gt;trans&lt;/em&gt;-cyclooctene&lt;/strong&gt;.  It has a higher boiling point (143 C) but a significant vapor pressure, it flashes at 25 C.  The molecule is chiral and has also been resolved.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-06T15:05:01.237" LastActivityDate="2014-11-06T15:05:01.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19211" PostTypeId="1" AcceptedAnswerId="19364" CreationDate="2014-11-06T15:14:10.157" Score="3" ViewCount="814" Body="&lt;p&gt;Why does $\ce{Ca}$ have a negative electron affinity, i.e. energy is released when it gets an electron (ca. $\pu{-2 kJ/mol}$), when $\ce{Be}$ and $\ce{Mg}$ have positive values? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the electron will be settled in the $\ce{3d}$ orbital which has a lower principal quantum number than $\ce{4s}$, but still, why would energy be released?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="41328" LastEditDate="2017-08-06T04:26:21.523" LastActivityDate="2017-08-06T04:26:21.523" Title="Ca energy release when gaining an electron" Tags="&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19212" PostTypeId="1" AcceptedAnswerId="19232" CreationDate="2014-11-06T15:23:14.920" Score="5" ViewCount="3550" Body="&lt;p&gt;I know that for atoms with big values of Z, relativistic effects have to be applied. Why does electron velocity increase with the increase of atomic number?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5017" LastEditDate="2015-05-15T15:41:30.973" LastActivityDate="2015-05-15T15:41:30.973" Title="Why does electron velocity increase with the increase of atomic number?" Tags="&lt;electrons&gt;&lt;relativistic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19213" PostTypeId="1" AcceptedAnswerId="19215" CreationDate="2014-11-06T15:40:52.240" Score="1" ViewCount="2957" Body="&lt;p&gt;When 50.00 mL of an unknown 0.2461 M weak acid are titrated with 0.1968 M NaOH, it was found after the addition of exactly 16.00 mL of base that the pH of the solution was 3.79. From this information, calculate the Ka of the acid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have an idea of how to do this question, but looking in my book, I seem to be arriving at a wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$50 \text{mL of }0.2461\text{M weak acid} = 0.2461 \cdot 0.05 = 0.012305\text{ mol weak acid}$$&#xA;$$16 \text{mL of }0.1968\text{M NaOH} = 0.2461 \cdot 0.013 = 0.0031488\text{ mol NaOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|cccc|}\hline&#xA;   &amp;amp; \ce{HA} &amp;amp; \ce{NaOH} &amp;amp; \ce{A^{-}} &amp;amp; \ce{H_{2}O} \\ \hline&#xA;  \mathrm I &amp;amp; 0.0123 &amp;amp; 0.0032 &amp;amp; 0 &amp;amp; -\\&#xA;  \mathrm C &amp;amp; -0.0032 &amp;amp; -0.0032 &amp;amp; +0.0032 &amp;amp; - \\&#xA;  \mathrm E &amp;amp; 0.0091 &amp;amp; 0 &amp;amp; 0.0032 &amp;amp; -\\ \hline&#xA; \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{molarity } \ce{A^-} = \frac{0.0032~\textrm{mol}}{0.066~\textrm L} = 0.0485~\textrm M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from here, I work on the assumption that the weak conjugate base will associate in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm p\ce{OH} = 14 - \mathrm p\ce H = 14 - 3.79 = 10.21$$&#xA;$$[\ce{OH^-}] = 10^{-\mathrm p\ce{OH}} = 10^{-10.21} = 6.17\times 10^{-11}~\mathrm M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{|c|cccc|}\hline&#xA;   &amp;amp; \ce{A^-} &amp;amp; \ce{H_2O} &amp;amp; \ce{HA} &amp;amp; \ce{OH^-} \\ \hline&#xA;  \mathrm I &amp;amp; 0.0485 &amp;amp; - &amp;amp; 0 &amp;amp; 0\\&#xA;  \mathrm C &amp;amp; -6.17\times 10^{-11} &amp;amp; - &amp;amp; +6.17\times 10^{-11} &amp;amp; +6.17\times 10^{-11} \\&#xA;  \mathrm E &amp;amp; 0.0485 \text{ (negligible change)} &amp;amp; - &amp;amp; 6.17\times 10^{-11} &amp;amp; 6.17\times 10^{-11}\\&#xA; \hline\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}K_\mathrm {b} &amp;amp;= \frac{(6.17\times 10^{-11})^{2}}{0.0485} = 7.85\times 10^{-20}\\ &amp;amp; = \frac{K_\mathrm W}{K_\mathrm b}\\ &amp;amp; = \frac{1.0\times 10^{-14}}{7.85\times 10^{-20}} = 127388.54 \leftarrow\text{This is obviously wrong}\;.\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I go wrong? Any help is much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="9604" LastEditorDisplayName="user5764" LastEditDate="2016-06-26T19:40:15.957" LastActivityDate="2017-03-07T20:16:28.793" Title="Finding the Ka of a weak acid after addition of a strong base" Tags="&lt;homework&gt;&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19214" PostTypeId="2" ParentId="16434" CreationDate="2014-11-06T15:52:15.303" Score="5" Body="&lt;p&gt;The relative electric permittivity of water at $20^oC$ under the effect of a static or low frequency is $\epsilon_r\approx80$. Hasted (1948) conducted an experimental study on the dielectric properties of salt-water solutions and observed a dielectric decrement with salt concentration. That is, the addition of sodium chloride to water results in a drop in electrical permittivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E60ir.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Source: Hasted, Ritson and Collie (1948).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be understood by examining the following diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KQcSH.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sodium and chloride ions dissociate in solution, resulting in an an electric field between them (shown in bold line). The water molecules, which are polarised, get oriented under the effect of the electric field so that their oxygen atoms (carrying a partial negative charge) face toward the sodium ion and their hydrogens (carrying partial positive charges) face toward the chloride ion. This orientation of the polar water molecules causes its own electric field (dashed lines), which cancels out most of the electric field that would exist if the ions were in a vacuum. The end result is that the sodium and chloride ions are effectively 'sheilded' by a 'hydration shell' of polarized water molecules which effectively lowers their to the applied external field and hence lowering the dielectric constant (electric permittivity) as the ionic concentration increases ().&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dilute solution (less than 2.0M) the dielectric decrement is linear. At higher salt concentrations, the dielectric decrement is observed to saturate at about $\epsilon_r=45$. The effect of sodium chloride concentration on electric permittivity is shown in the graph below at various temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g277H.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Gavish and Promislow (2012).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above graph indicates that for a given salt concentration below about 3M, the electric permittivity decreases with temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;REFERENCES:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hasted, J.B., Ritson, D.M., Collie, C.H., &quot;Dielectric Properties of Aqueous Ionic Solutions. Parts I and II.&quot; Journal of Chemical Physics 16, 1 (1948). &lt;a href=&quot;http://dx.doi.org/10.1063/1.1746645&quot; rel=&quot;noreferrer&quot;&gt;http://dx.doi.org/10.1063/1.1746645&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gavish, N., Promislow, K, &quot;Dependence of the dielectric constant of electrolyte solutions on ionic concentration&quot;, &lt;a href=&quot;http://arxiv.org/pdf/1208.5169.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://arxiv.org/pdf/1208.5169.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-11-06T15:52:15.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19215" PostTypeId="2" ParentId="19213" CreationDate="2014-11-06T16:05:01.597" Score="1" Body="&lt;p&gt;When you used the ICE method the second time, you forgot that HA does not have an initial concentration of 0. Instead, the initial concentration of HA is 0.0091 as you calculated previously. For simplicity's sake, you can just use Henderson-Hasselbalch equation after your first ICE method to determine the answer. &lt;/p&gt;&#xA;" OwnerUserId="7402" LastActivityDate="2014-11-06T16:05:01.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19216" PostTypeId="2" ParentId="19212" CreationDate="2014-11-06T16:45:32.170" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does electron velocity increase with the increase of atomic&#xA;  number?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Newton's second law tells us that for an electron moving in a stable circular orbit around a nucleus the centripetal force pushing out must equal the electrostatic attraction pulling in, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\frac{m_e~ v^2}{r}} = {\frac{Z~k_e ~e^2}{r^2}}~~(1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_e$ is the electron's mass, e is the electron charge, $k_e$ is Coulomb's constant and $Z$ is the atom's atomic number.  Rearranging equation (1) yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{ Z~k_e~ e^2 \over m_\mathrm{e} r}~~(2) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From equation (2) we can see that the electron velocity increases as the atomic number, $Z$, increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am expecting some quantum mechanical explanation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the electron velocity given by equation (2) becomes comparable to the speed of light, the mass of the electron increases due to relativistic effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_{rel}=\frac{m_{e}}{\sqrt{1-(v_e/c)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the electron mass increases, the s (and somewhat for p) orbital radius contracts.  So relativistic effects affect the electron mass and the orbital radius.  In terms of electron velocity, these 2 relativistic effects offset one another and the electron velocity is not affected; equation (2) is still valid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-06T19:08:43.547" LastActivityDate="2014-11-06T19:08:43.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19217" PostTypeId="1" AcceptedAnswerId="19223" CreationDate="2014-11-06T18:16:33.513" Score="4" ViewCount="1148" Body="&lt;p&gt;My main purpose is to simulate excited state calculations on a molecule with Time Dependent DFT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for free software that makes input files for GAMESS. I have tried Avogadro and it works fine (in Extensions&gt; GAMESS &gt; Input Generator &gt; Advanced tab) but it can not make &lt;strong&gt;TD-DFT&lt;/strong&gt; input codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also checked the recommended software by the Gordon Group (&lt;a href=&quot;http://www.msg.ameslab.gov/graphics/other.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) and tried installing some of them but couldn't install Gabedit at all (I was unable to compile the binary, maybe because I am using Mac OS X Yosemite).  I also tried WebMO but it is available just on iTunes for iOS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a software that makes the input code interactively to avoid possible errors.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="73" LastEditDate="2014-11-06T23:32:53.210" LastActivityDate="2015-06-18T02:28:44.417" Title="Is there any free GAMESS input file maker for Mac OS?" Tags="&lt;quantum-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19218" PostTypeId="1" AcceptedAnswerId="24283" CreationDate="2014-11-06T19:00:53.543" Score="3" ViewCount="152" Body="&lt;p&gt;I have some trouble visualising scattering of x-rays in 3D, since all textbooks and examples present it in 2D. Under I will share some reflections (no pun intended) about the elastic scattering of x rays, and the diffraction effects observed, which is the basis of XRD analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2nIB.jpg&quot; alt=&quot;Common figure used to derive Bragg&amp;#39;s law of constructive interference.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you see a figure of two crystal planes, separated by the distance d. Two parallel and in-phase x-rays hit an atom in the planes. The scattering cones are drawn. The darker area represents those scattering directions going into the crystal itself, and thus not detectable and uninteresting for XRD analysis. The light area are those scattered rays exiting the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in three dimensions is different from thinking in one dimension. To achieve constructive interference, the rays must travel in the same direction and also be superimposed on each other. Even though the scattered rays ' angle is, say, 45 degrees, they may not travel in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if Bragg's law is valid for 45 degrees, all rays of 45 degrees scattering angle will interfere constructively. So we would get a diffraction cone of increased intensity. However, we cannot detect the whole diffraction cone, partly because some of the rays travel into the crystal, and partly because the detector surface area only covers parts of the diffraction cone; we would measure a curved segment of the diffraction cone. This is called the Debye-Scherrer ring, yes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though we get constructive interference in all directions of 45 degrees scattering angle, the scattering angle itself is random. This means there are scattered rays which do not undergo constructive interference in all directions, regardless of angle (though smaller angles are more probable). So, in an ocean of random scattering, we have diffraction cones of much greater intensity, since Bragg's law is fulfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a correct assessment of the XRD situation?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2014-11-17T03:57:34.940" LastActivityDate="2015-01-28T04:14:57.617" Title="Elastic scattering of x rays in 3D: the principle for XRD analysis" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19219" PostTypeId="2" ParentId="19204" CreationDate="2014-11-06T19:08:06.907" Score="0" Body="&lt;p&gt;Oxidizing strength is a &lt;em&gt;relative&lt;/em&gt; term used to denote how certain elements oxidize other elements with respect to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19204/what-does-oxidizing-strength-mean#comment29396_19204&quot;&gt;t.c&lt;/a&gt; said in the comments, if $\text{X}$ has a higher oxidizing strength than $\text{Y}$, then $\text{X}$ can a higher likelihood oxidize $\text{Y}$ faster than $\text{Y}$ has to oxidize $\text{X}$. The same definition could apply for reducing strength, and you could say that if $\text{X}$ has a higher oxidizing strength than $\text{Y}$, then $\text{Y}$ has a higher reducing strength than $\text{X}$. &lt;/p&gt;&#xA;" OwnerUserId="3842" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2016-04-05T21:48:00.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19220" PostTypeId="2" ParentId="19059" CreationDate="2014-11-06T19:29:07.983" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Through space coupling is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnetic_dipole%E2%80%93dipole_interaction#Dipolar_coupling_and_NMR_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;dipolar coupling&lt;/a&gt; and is actually much greater than through bond coupling (&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-03-jcoupl.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;J-coupling&lt;/a&gt;).  However, in low viscosity liquids the chemical shift effects of dipolar coupling are averaged to essentially a single value by molecular rotation.  Relaxation effects of dipolar coupling are still observed.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do identical protons not influence each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For two protons there will be four states &lt;span class=&quot;math-container&quot;&gt;$\alpha \alpha , \alpha \beta , \beta \alpha$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\beta \beta$&lt;/span&gt;.   In the general case, each state may have a different energy and there can be four possible one quantum transitions (&lt;span class=&quot;math-container&quot;&gt;$\alpha \alpha$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\alpha \beta$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\alpha \alpha$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\beta \alpha$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\alpha \beta$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\beta \beta$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\beta \alpha$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\beta \beta$&lt;/span&gt;), each transition having a unique energy and hence 4 peaks (two doublets). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the two protons are equivalent, &lt;span class=&quot;math-container&quot;&gt;$\beta \alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\alpha \beta$&lt;/span&gt; will be the same energy, &lt;span class=&quot;math-container&quot;&gt;$E_2$&lt;/span&gt; let's say.  So, letting the energy of  &lt;span class=&quot;math-container&quot;&gt;$\alpha \alpha$&lt;/span&gt; be &lt;span class=&quot;math-container&quot;&gt;$E_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta \beta$&lt;/span&gt; be &lt;span class=&quot;math-container&quot;&gt;$E_3$&lt;/span&gt;, and given a one quantum transition, the possible energies for transtions are only &lt;span class=&quot;math-container&quot;&gt;$E_3$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$E_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E_2$&lt;/span&gt;- &lt;span class=&quot;math-container&quot;&gt;$E_1$&lt;/span&gt;.  But &lt;span class=&quot;math-container&quot;&gt;$E_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E_1$&lt;/span&gt; are equally spaced above and below &lt;span class=&quot;math-container&quot;&gt;$E_2$&lt;/span&gt;, so there is only one energy of transition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more-rigorous explanation see: &lt;a href=&quot;http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7448" LastEditDate="2019-09-23T10:39:02.190" LastActivityDate="2019-09-23T10:39:02.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19222" PostTypeId="2" ParentId="19217" CreationDate="2014-11-06T21:15:21.963" Score="3" Body="&lt;p&gt;If you want to use Gabedit, it's in &lt;a href=&quot;http://brew.sh&quot; rel=&quot;nofollow&quot;&gt;Homebrew&lt;/a&gt;. Just &lt;code&gt;brew install gabedit&lt;/code&gt; after you install Homebrew. I just installed it without issue on Yosemite and it was pretty fast. It seems to have a lot of options for GAMESS, but it's been a long time since I've done computational chem, so I can't tell you whether it has all the TD-DFT options. I didn't get any weird errors or anything, so it seems to run fine.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-06T21:15:21.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19223" PostTypeId="2" ParentId="19217" CreationDate="2014-11-06T21:17:18.830" Score="5" Body="&lt;p&gt;If you can give me the requisite keywords for TD-DFT in GAMESS, it's not hard to add that to &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never done TDDFT in GAMESS because I use other programs, but given a link to the docs, it's easy enough to enable more drop-down menus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you can interactively edit the text in the Avogadro input for &lt;em&gt;any&lt;/em&gt; quantum program (not just GAMESS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DGuGj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to use Avogadro, the &quot;native&quot; program for GAMESS is &lt;a href=&quot;http://www.scl.ameslab.gov/MacMolPlt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MacMolPlt&lt;/a&gt; which was developed in the Gordon group itself. I haven't checked, but I'd assume it supports TDDFT options, since it enables pretty much everything else in GAMESS.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-06T21:17:18.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19224" PostTypeId="1" CreationDate="2014-11-06T21:29:52.767" Score="2" ViewCount="288" Body="&lt;p&gt;What is the pOH of a 0.0100 F solution of dipotassium phtalate? pKa1 = 2.950 and pKa2 = 5.408 for phtalic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did it for phthalic acid by calculating pH and then just doing pOH = 14.00 - pH, which gave me pOH = 11.60. Not sure if I did it correctly, I'm only just (re)learning this. I just need a clear and basic explanation, if possible. The two pKa's are what confuse me... which one do I use? I've been searching on google and through my textbook for 2 days now but I'm just confusing myself even more at this point. I know it's probably really easy stuff, but I haven't done chem in 4 years... trying to refresh my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="9661" LastActivityDate="2015-05-14T07:25:31.507" Title="What is the pOH of a 0.0100 F solution of dipotassium phtalate?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19225" PostTypeId="1" CreationDate="2014-11-06T22:29:27.260" Score="1" ViewCount="53" Body="&lt;p&gt;I'm working on upgrading my current skin adhesive emulsion. It is a roll-on skin adhesive used to keep clothing attached to the skin. Ingredients are; Alcohol Denae,Ethyl Ester of PM/MA Copolymer,Water, Butylene Glycol, and Aminomethyl Proponal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What ingredient(s) could be used to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Improve formula so that it is more pliable(as far as the skin moving with fabric). &lt;/li&gt;&#xA;&lt;li&gt;Still maintain a strong bond/hold(possibly even stronger then current formula).&lt;/li&gt;&#xA;&lt;li&gt;Have the overall formula be slightly less wet/runny (as it sometimes absorbs through clothing).&lt;/li&gt;&#xA;&lt;li&gt;And when the formula sets, remains fairly matte/dry to the touch.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9662" LastEditorUserId="73" LastEditDate="2014-11-06T23:34:32.077" LastActivityDate="2014-11-06T23:34:32.077" Title="Skin Adhesive Emulsion" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19226" PostTypeId="1" AcceptedAnswerId="27792" CreationDate="2014-11-06T23:25:38.973" Score="2" ViewCount="637" Body="&lt;p&gt;I have a shaver that runs off a rechargeable battery that is dying. Would it be feasible to make a silver zinc battery and use it to replace the existing battery? Cost is not an object, anything less than $2500 I would consider doable. I have a machine shop and a small chemistry laboratory with the standard equipment and glassware, including high vacuum capability, a centrifuge and simple glass blowing capability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a book on silver-oxide zinc chemistry and battery design, but it is $500 and I don't want to spend that if the information can be obtained just as easily elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to find commercial options, but had no luck. Most silver zinc batteries seem to be just for large (multi-million dollar) military or satellite applications. Sony makes a line of silver oxide primary cells for hearing aids, but these are not rechargeable. There is a company called &lt;a href=&quot;http://ultralifecorporation.com/&quot; rel=&quot;nofollow&quot;&gt;Ultralife&lt;/a&gt; that makes medical and military batteries and might have something viable, but before I call them I wanted to check out the opinion of the experts here.&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorUserId="5017" LastEditDate="2014-11-07T23:08:15.913" LastActivityDate="2019-11-03T14:10:28.397" Title="Make a silver zinc battery" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19227" PostTypeId="2" ParentId="19059" CreationDate="2014-11-06T23:54:28.813" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does a proton create a magnetic field that influences the neighboring protons (mainly) trough bonding and not trough space?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A proton doesn't interact with another proton by producing a magnetic field &lt;em&gt;per se&lt;/em&gt;, but it interacts via the bonding electrons &lt;em&gt;via&lt;/em&gt; the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_contact_interaction&quot; rel=&quot;noreferrer&quot;&gt;Fermi contact interaction&lt;/a&gt;. And because the Wikipedia article is kind of dry on this matter, here a figure that very well illustrates what is going on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/R3j3T.png&quot; alt=&quot;Fermi Contact Interaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This interaction is not the only source for indirect coupling of spins. If we only consider protons, then we're fine. But if you'd like to consider fluorine-fluorine couplings, you also have to take into account the dipolar interaction between the nuclear and electron spins. This dipolar interaction also provides a source for anisotropic $J$ interactions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do identical protons not influence each other? Or more general: why is it that how more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their &quot;outer&quot; signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also here, a picture is worth more than a thousand words (I'll spare you the calculation of this spectrum):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K8JYI.png&quot; alt=&quot;Spectrum for an interacting two-spin system&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k$ is defined as follows:&#xA;$$ k = \left| \frac{\Omega_2 - \Omega_1}{2\pi J} \right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can clearly see, the outer peaks start getting smaller the smaller the difference in chemical shift ($\Omega$) of the two spins tends towards zero. This is sometimes called &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/NMR%3A_Interpretation#High-order_splitting_pattern&quot; rel=&quot;noreferrer&quot;&gt;&quot;roof effect&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity of the left- and rightmost lines can be calculated (assuming $J&amp;gt;0$) with the following formula:&#xA;$$ I_{13} = I_{34} = \frac{1 - \sin(2\alpha)}{4} $$&#xA;where&#xA;$$ 2\alpha = \tan^{-1}\left(\frac{2\pi J}{\Omega_1 - \Omega_2} \right) $$&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-11-06T23:54:28.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19228" PostTypeId="1" AcceptedAnswerId="19233" CreationDate="2014-11-07T00:32:09.930" Score="3" ViewCount="588" Body="&lt;p&gt;There are various compounds which have common names that end in &quot;ylene&quot;, like acetylene and ethylene. However, how come acetylene is an alkyne whereas ethylene is an alkene. Both of these have the same &quot;ylene&quot; ending though. What is the cause for this discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="4231" LastEditDate="2014-11-07T01:17:28.783" LastActivityDate="2014-11-07T04:29:25.040" Title="Naming system -&quot;ylene&quot;" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19229" PostTypeId="1" AcceptedAnswerId="19240" CreationDate="2014-11-07T00:49:18.220" Score="22" ViewCount="5851" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/49FkW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you name the above compound according to IUPAC rules? I see some sources say it is 2-phenyloctane, but ChemDraw and a few other sources say it is 2-octanylbenzene.&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="16683" LastEditDate="2017-04-11T00:19:36.757" LastActivityDate="2020-12-08T03:20:27.850" Title="Is the ring or the chain chosen as the parent structure in octan-2-ylbenzene?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;aromatic-compounds&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19230" PostTypeId="2" ParentId="19205" CreationDate="2014-11-07T01:15:04.680" Score="18" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would appreciate learning whether such reasoning, for these two&#xA;  type of substituted phenols is plausible or whether furthermore&#xA;  solid arguments should be invoked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your reasoning seems generally well thought out.  You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series.  Structural effects don't play a role with these compounds, but let me mention them for completeness.  If we have a compound such as I with a carbonyl ortho to the hydroxyl substituent, then in addition to inductive (stabilizes the anion) and resonance (stabilizes the anion) effects, we should also consider hydrogen bonding; which, in this case, would stabilize the phenol making it less acidic.  On the other hand, in compound 2, the bulky t-butyl group would force the ester carbonyl to twist out of the plane of the benzene ring.  In II, the inductive effect of the carbonyl would still operate, but the resonance and hydrogen bonding effects on acidity would be diminished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ZJrM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I would appreciate learning why the difference in acidity between&#xA;  2- and 4-chlorophenols is much more pronounced than the same&#xA;  difference between o- and p-cresols&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; discusses electrophilic aromatic substitution, where again both inductive and resonance effects need to be considered.  About two-thirds of the way down the page we see that chlorobenzene undergoes electrophilic nitration about 30 times &lt;strong&gt;slower&lt;/strong&gt; than benzene, but is still &lt;em&gt;ortho-para&lt;/em&gt; directing.  We also see that toluene reacts about 25 times &lt;strong&gt;faster&lt;/strong&gt; than benzene and is also &lt;em&gt;ortho-para&lt;/em&gt; directing.  From these and other examples, we learn that chlorine donates electrons weakly by resonance and withdraws electrons strongly through its inductive effect; and that a methyl group is a mild electron donor through resonance (hyperconjugation), while its inductive effect is also electron-donating, but is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply these concepts from electrophilic aromatic substitution to your two phenol series.  In the chlorophenols, we would expect the chloro substituent to influence the series primarily through a strong electron-withdrawing inductive effect.  Indeed, as the chlorine moves closer to the hydroxyl substituent, the inductive electron withdrawal increases in power and the &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; increases dramatically.  On the other hand, in the methyl series we expect mild resonance (hyperconjugation) effects to dominate (even though the inductive effect is in the same direction), therefore in the &lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-cresols we expect the anion to be weakly destabilized and consequently a bit less acidic than phenol itself; while the &lt;em&gt;meta&lt;/em&gt; isomer, with minimal resonance and inductive effects affecting the anion, should have an acidity close to phenol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2020-03-01T05:59:31.337" LastActivityDate="2020-03-01T05:59:31.337" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19232" PostTypeId="2" ParentId="19212" CreationDate="2014-11-07T02:26:46.457" Score="1" Body="&lt;p&gt;You can easily understand that based on the particle in the box model. Large Z corresponds to large electrostatic field around the nucleus which act on the electrons. This leads to a smaller &quot;box&quot;, ie. the core electrons of heavy elements are moving on orbitals with radius orbitals than electrons of light elements.&#xA;Since the energy of the electron is (if we use the simple 1D version) $$E = \frac{n^2h^2}{8mL^2}$$&#xA;where $L$ corresponds to the box size and the energy ($E$) is purely kinetic energy. You can see that smaller box corresponds to larger kinetic energies.  &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-11-07T02:26:46.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19233" PostTypeId="2" ParentId="19228" CreationDate="2014-11-07T04:29:25.040" Score="5" Body="&lt;p&gt;These are common names, and predate our knowledge about bonds. My guess is that the &quot;ylene&quot; comes from reduction -- reducing ethyl alcohol gives ethylene, reducing acetic acid  gives acetylene.&lt;/p&gt;&#xA;" OwnerUserId="22" LastActivityDate="2014-11-07T04:29:25.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19234" PostTypeId="1" AcceptedAnswerId="60796" CreationDate="2014-11-07T05:38:02.793" Score="7" ViewCount="245" Body="&lt;p&gt;Starting from the equations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_mass_action&quot; rel=&quot;nofollow&quot;&gt;mass action kinetics&lt;/a&gt;, one may show that (for suitably chosen numbers $A_i$), the function $\sum_i x_i(A_i + \log x_i)$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lyapunov_function&quot; rel=&quot;nofollow&quot;&gt;Lyapunov function&lt;/a&gt;, i.e. it decreases over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from an irrelevant factor of $RT$, this is the expression for the free energy of an ideal solution (or ideal gas, etc). So it seems that mass action kinetics carries with it an assumption that the reactions are taking place in an ideal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in how to think about kinetics in &lt;em&gt;non&lt;/em&gt;-ideal solutions, but I haven't been able to find any information about it. Is there a standard set of kinetic equations that remain valid when the chemical potential is given not by $\mu_i^0 + RT\log x_i$ but by $\mu_i^0 + RT\log \gamma_i x_i$, for activity coefficients $\gamma_i$ that depend on the concentrations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several alternatives to mass action kinetics that I know about, such as Michaelis Menton kinetics for enzymes - but these are actually &lt;em&gt;derived&lt;/em&gt; from mass action kinetics applied to elementary steps, so they don't seem to be able to escape from the assumption of ideality.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="148" LastEditDate="2016-10-08T14:35:30.367" LastActivityDate="2016-10-15T11:18:38.323" Title="What is the non-ideal equivalent of mass action kinetics?" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19235" PostTypeId="1" AcceptedAnswerId="19246" CreationDate="2014-11-07T06:14:05.487" Score="2" ViewCount="8652" Body="&lt;p&gt;I have just started learning thermodynamics. I have learned that The system which is not able to exchange energy and matter on its own with surrounding is called isolated system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to know that practically is it possible to make 100% isolated system? &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2021-01-25T04:58:38.780" Title="Thermodynamics: Is Isolated system possible?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19236" PostTypeId="5" CreationDate="2014-11-07T06:15:49.190" Score="0" Body="&lt;p&gt;Green chemistry, also known as sustainable chemistry, is a philosophy intended to reduce the environmental impact of chemical industry and research. This includes most commonly synthetic products and practices, but also other aspects of chemistry such as analytical techniques. Common goals are to reduce the production of hazardous materials and to use chemicals efficiently. The US EPA provides an introduction to green chemistry including a set of guiding principles &lt;a href=&quot;http://www2.epa.gov/green-chemistry/basics-green-chemistry#twelve&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tag usage:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Green chemistry should not be confused with environmental chemistry, though there is a certain amount of overlap. A question about environmental interactions of a hazardous material released by an industrial process falls more under the &lt;a href=&quot;/questions/tagged/environmental-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;environmental-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;environmental-chemistry&lt;/a&gt; tag, while a discussion of possible modifications of the process to reduce the production of the hazardous material should be tagged as green-chemistry. Of course, questions may deal with both green chemistry as well as the environment itself and could have both tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usage of this tag includes sustainability-related issues such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pollution reduction&lt;/li&gt;&#xA;&lt;li&gt;CO2 capture and recycling (CCR)&lt;/li&gt;&#xA;&lt;li&gt;better fuels&lt;/li&gt;&#xA;&lt;li&gt;the chemistry of photovoltaics&lt;/li&gt;&#xA;&lt;li&gt;photochemical splitting of water&lt;/li&gt;&#xA;&lt;li&gt;catalysts for hydrolysis of water and other routes to hydrogen generation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For questions that are primarily about sustainability and less about chemistry you can try &lt;a href=&quot;http://sustainability.stackexchange.com&quot;&gt;Sustainability SE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.epa.gov/greenchemistry&quot; rel=&quot;nofollow&quot;&gt;EPA Green Chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rsc.org/journals-books-databases/about-journals/green-chemistry/&quot; rel=&quot;nofollow&quot;&gt;Journal of Green Chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/journal/10.1002/(ISSN)1864-564X&quot; rel=&quot;nofollow&quot;&gt;Wiley ChemSusChem Journal&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/journal/ascecg&quot; rel=&quot;nofollow&quot;&gt;ACS Journal of Sustainable Chemistry &amp;amp; Engineering&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="24093" LastEditDate="2016-07-21T09:52:59.087" LastActivityDate="2016-07-21T09:52:59.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19237" PostTypeId="4" CreationDate="2014-11-07T06:15:49.190" Score="0" Body="Green chemistry, also known as sustainable chemistry, is a philosophy intended to reduce the environmental impact of chemical industry and research. A discussion of possible modifications of the process to reduce the production of the hazardous material should be tagged as green-chemistry. A question concerning the environmental interactions of hazardous chemicals should rather be tagged with `environmental-chemistry`." OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2016-07-25T07:41:42.657" LastActivityDate="2016-07-25T07:41:42.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19238" PostTypeId="2" ParentId="19235" CreationDate="2014-11-07T06:31:05.683" Score="1" Body="&lt;p&gt;It is an idealized situation, an extreme, just as well all the other thermodynamic models.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-11-07T06:31:05.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19239" PostTypeId="1" CreationDate="2014-11-07T10:34:08.263" Score="1" ViewCount="7890" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$5.00~\mathrm{g}$ of nitrogen is completely converted into an oxide of nitrogen. The mass of the oxide formed is $19.3~\mathrm{g}$.&lt;br&gt;&#xA;  The empirical formula of the oxide would be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My working:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2 + O_2 -&amp;gt;  N$_x$O$_y$}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$m(\ce{N2}) = 5~\mathrm{g}$&lt;/li&gt;&#xA;&lt;li&gt;$n(\ce{N2}) = 5/28~\mathrm{mol}$&lt;/li&gt;&#xA;&lt;li&gt;$n(\ce{N$_x$O$_y$}) = 19.3~\mathrm{g}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do I approach this? &lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="4945" LastEditDate="2014-12-02T05:11:06.763" LastActivityDate="2014-12-13T00:22:24.360" Title="How to find the empirical formula for an nitrogen oxygen compound from the given mass of nitrogen and the product?" Tags="&lt;homework&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;&lt;elements&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19240" PostTypeId="2" ParentId="19229" CreationDate="2014-11-07T10:35:10.537" Score="15" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.1.2.2&lt;/strong&gt; Systems composed of rings and chains (exclusive of linear phanes)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two methods are recognized to name systems composed of rings and chains&#xA;  (exclusive of linear phanes).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…) For selection of a preferred IUPAC name, see P-52.2.8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-52.2.8&lt;/strong&gt; Selection between a ring and a chain as parent hydride&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method (1): “octan-2-ylbenzene” (preferred IUPAC name; ring preferred to chain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method (2): “2-phenyloctane” (chain has greater number of skeletal atoms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function &lt;em&gt;Convert Structure to Name&lt;/em&gt; of &lt;em&gt;ChemDraw&lt;/em&gt; correctly yields “octan-2-ylbenzene”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFKc3.png&quot; alt=&quot;octan-2-ylbenzene&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-28T13:49:44.113" LastActivityDate="2015-03-28T13:49:44.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19241" PostTypeId="2" ParentId="19211" CreationDate="2014-11-07T11:00:07.727" Score="-1" Body="&lt;p&gt;Please verify your data.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Be}$ has negative electron affinity (precisely -2.4 eV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also calcium has positive electron affinity (precisely 0.0245 eV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check the Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_affinity_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electron Affinity Values&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you got confused between electron affinity and gain enthalpy. Electron gain enthalpy is negative of electron affinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1: DAVE PhD, you misunderstood me. Electron affinity has absolute definite values. Only one convention is used. Electron affinity can assume both positive and negative values. But beryllium has negative electron affinity, by any convention you can't assume it to be positive. You can even check the Wikipedia page.&lt;/p&gt;&#xA;" OwnerDisplayName="user2619" LastEditorUserId="39539" LastEditorDisplayName="user2619" LastEditDate="2017-02-05T21:02:18.533" LastActivityDate="2017-02-05T21:02:18.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19242" PostTypeId="2" ParentId="19235" CreationDate="2014-11-07T13:25:07.533" Score="0" Body="&lt;p&gt;An isolated system does not exchange energy or matter with its surroundings. For example, if soup is poured into an insulated container and closed, there is no exchange of heat or matter. The fact that, in reality, a thermos is not perfect in keeping things warm/cold illustrates the difficulty in creating an truly isolated system. In fact, there are a few, if any, systems that exist in this world that are completely isolated systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The universe of course is an another approximation of an isolated system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,  practically  it is impossible to make 100% isolated system.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-07T13:25:07.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19243" PostTypeId="2" ParentId="19239" CreationDate="2014-11-07T14:22:59.403" Score="2" Body="&lt;p&gt;$\ce{x/2 N2 +y/2 O2 \rightarrow N$_x$O$_y$}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two unknowns $x$ and $y$. So, to determine them we need two independent  equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - The conservation of nitrogen mass: $5=14x$. So,  $x= 0.35714$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- According to the stoichiometry of the above balanced chemical equation: The number of moles of  $\ce{ N$_x$O$_y$}$ equals $n= \frac{5}{28}\times\frac{2}{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the number of moles of  $\ce{ N$_x$O$_y$}$ equals   $n= \frac{19.3}{14x+16y}=\frac{10}{28x} $.&#xA;By substituting $x= 0.35714 $ into the last equation, we find $y=0.89375$. &#xA;We find $\ce{ N_{0.35714}O_{0.89375}}$. We divide by $ 0.35714$, We obtain  $\ce{ NO_{2.5}}$.i.e. the formula of nitrogen oxide is $\ce{ N_2O_5}$. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="14" LastEditDate="2014-11-10T14:30:28.803" LastActivityDate="2014-11-10T14:30:28.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19244" PostTypeId="2" ParentId="19239" CreationDate="2014-11-07T14:25:08.543" Score="1" Body="&lt;p&gt;If the question specifies that only one N&lt;sub&gt;x&lt;/sub&gt;O&lt;sub&gt;y&lt;/sub&gt; is formed (so not also N&lt;sub&gt;p&lt;/sub&gt;O&lt;sub&gt;q&lt;/sub&gt; in some ratio) , you need to calculate how many atoms of Oxygen you will have added. Then, you can deduce the number of Oxygen atoms per Nitrogen atom and find x and y. If you find decimal values, adjust the stoichiometry accordingly.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-11-07T14:25:08.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19245" PostTypeId="2" ParentId="15926" CreationDate="2014-11-07T15:02:40.917" Score="2" Body="&lt;p&gt;I do not address the question of history, of Mendeleyev's investigations, that is answered in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15883/how-was-mendeleev-able-to-develop-his-table&quot;&gt;How was Mendeleev able to develop his table?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern understanding of basic principles follows. Let us have a piece of periodic table&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  ... A0 B0 ...&#xA;  ... A1 B1 ...&#xA;  .............&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Element &quot;B0&quot; must have the next atomic number after &quot;A0&quot;, &lt;em&gt;i.e.&lt;/em&gt; one more electron. It is different chemically, although may have (albeit not necessarily) similar physical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Element &quot;A1&quot; must be &lt;em&gt;chemically&lt;/em&gt; similar to &quot;A0&quot;, although physical properties may differ significantly (as a general rule, &quot;.1&quot; elements are denser than &quot;.0&quot;). It happens that for each element there exist its &quot;analog&quot; in the next period, that has similar chemical properties (based on similar outer-shell electron configuration). This pattern, once noticed, permitted for good predictions (search for Mendeleyev's &quot;eka&quot; elements in Internet for more information).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Periods have different lengths. Analog's atomic number may be 8, 18, or 32 ahead off the prototype's atomic number (the number of electrons expended to &quot;raise&quot; all the configuration one shell up). This difference is due to quantummechanical effects.&lt;/p&gt;&#xA;" OwnerUserId="9666" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-11-07T15:02:40.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19246" PostTypeId="2" ParentId="19235" CreationDate="2014-11-07T15:03:03.257" Score="2" Body="&lt;p&gt;There are no 100% isolated systems - even the universe itself may not be truly isolated (depending on how you define &lt;em&gt;universe&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in practice, many systems are &quot;isolated enough&quot; that we can treat them as conserving mass and energy, especially over short time frames or small distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we often treat a well-insulated and closed reaction vessel as isolated when doing thermodynamic calculations, and this approximation works very well for fast reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example where this works is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_process&quot; rel=&quot;nofollow&quot;&gt;adiabatic process&lt;/a&gt;. Many processes can be successfully modeled as adiabatic even though they happen in poorly-insulated containers - for example, the rapid expansion or compression of a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the use of small sub-domains of a non-isolated system, which over short time scales behave as isolated systems themselves. For example, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Finite_element_method&quot; rel=&quot;nofollow&quot;&gt;finite element methods&lt;/a&gt; (FEM), a system is broken up into very small sub-domains. Each one acts as a small system in which mass and energy can transfer across the boundaries. Locally, an element is an open system, and over long time scales the entire system is open, but on short time scales, the sets of domains that share boundaries are treated as isolated. If energy leaves one domain, it enters the neighboring domain - it doesn't completely leave the system. In effect, you can model an open system as a large collection of isolated systems, with the outermost having open boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, even though the assumption of an isolated system is not &lt;em&gt;really&lt;/em&gt; true, you can find many, many places where it is &quot;true enough&quot; that we can use it practically. In fact, it would be very hard to do anything in thermodynamics without it!&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-11-07T15:03:03.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19247" PostTypeId="2" ParentId="19229" CreationDate="2014-11-07T15:21:29.060" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Well basically there are &lt;strong&gt;8&lt;/strong&gt; &lt;strong&gt;carbons&lt;/strong&gt; in the &lt;em&gt;long chain&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;Phenyl Group&lt;/strong&gt; is actually considered to be kind of a &lt;em&gt;Functional Group&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hence the name should be:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;octan-2-ylbenzene&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although some sources say that the name should be &lt;em&gt;2-phenyloctane&lt;/em&gt; , They do it on the basis that :the unsaturated acyclic structure is recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure, which is kind of not used and considered to be &lt;em&gt;wrong&lt;/em&gt; by many according to IUPAC rules.&lt;/p&gt;&#xA;" OwnerUserId="7434" LastActivityDate="2014-11-07T15:21:29.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19248" PostTypeId="2" ParentId="10855" CreationDate="2014-11-07T15:55:36.547" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In many cases, they weren't - the ancient... maybe we can't call them scientists. Let's call them &quot;natural philosophers.&quot; The ancient natural philosophers (up until about 1700) &lt;a href=&quot;http://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries&quot;&gt;only knew about a handful of elements&lt;/a&gt;. This is part of the reason that it took so long to put the periodic table together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time, an element was something that was not reducible to smaller components. The ancient philosophers tended to conflate properties of things with this idea of fundamental components, and as a result, things like earth, air, fire, and water were thought of as elements. When we say they &quot;knew about&quot; modern elements, what we mean is that they knew of the substances - they did not necessarily know that they were chemical elements in the sense that we define them today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That modern concept of an element defined by the number of protons in the nucleus wasn't developed until around 1913. The wikipedia article on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_element#Evolving_definitions&quot;&gt;history of chemical element definitions&lt;/a&gt; gives a good overview of how the definition has evolved over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time period you are probably most interested in is that between the &quot;ancient&quot; era (pre-history up until the mid-1660's or so) to the beginnings of the modern era (say early 1900's). During this period, chemistry was evolving from alchemy into a modern science. The systematic investigation and communication of results was occurring, and a large body of known but poorly understood chemical &quot;recipes&quot; was available to chemists. Using a range of reactions and isolation techniques, they basically just randomly tried things (maybe not completely randomly) until they reached a point where it seemed like something could not be decomposed further, or until they identified a new substance. In many cases, they would discover the new element, but not be aware that it was an element - in other words, they could isolate and identify a substance, but because our collective understanding of chemistry was still very limited, they had no way of knowing that it was an &lt;em&gt;elemental&lt;/em&gt; substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries&quot;&gt;list of elemental discoveries&lt;/a&gt; gives details about the discovery of each element - although you should keep in mind that &quot;discovery&quot; in this case means &quot;identified as a new substance,&quot; not necessarily &quot;identified as a chemical element.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-11-07T15:55:36.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19249" PostTypeId="1" CreationDate="2014-11-07T17:25:26.103" Score="3" ViewCount="1530" Body="&lt;p&gt;I am a student from a school and recently I came across the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Na2O2 + 2CO2 -&amp;gt; 2Na2CO3 + O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't we use this reaction to reduce $\ce{CO2}$ in the atmosphere by placing some $\ce{Na2O2}$ filters in places where there is high level $\ce{CO2}$ emission? I then realized that the filter would have to be loaded with lot of $\ce{Na2O2}$ to achieve this. So why not adsorb as much $\ce{CO2}$ as much as possible using a good adsorbent for $\ce{CO2}$. I am not sure if there are other methods better than these but I would like to know if there is any?&lt;/p&gt;&#xA;" OwnerUserId="9668" LastEditorUserId="5633" LastEditDate="2014-11-10T22:07:48.097" LastActivityDate="2014-11-10T22:07:48.097" Title="Using sodium peroxide to scrub carbon dioxide emissions" Tags="&lt;environmental-chemistry&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19250" PostTypeId="1" CreationDate="2014-11-07T17:25:37.503" Score="3" ViewCount="2317" Body="&lt;p&gt;The synthesis of Pseudopelletierine starts from glutardialdehyde, which reacts with methylamine and acetonedicarboxylic acid in a two-fold Mannich reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S13hO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to write the mechanism for this reaction. It needs to be under basic conditions also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MY ANSWER SO FAR:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a Mannich reaction followed by a second intramolecular Mannich reaction in order to form a ring by joining al three reagents together. Finally there will be a decarboxylation to remove the carboxylic acid groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the mechanism I have done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/StXdB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ePTec.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? Any help would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5181" LastActivityDate="2014-11-07T19:58:08.583" Title="Synthesis of Pseudopelletierine, Mannich reaction base catalysed" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19251" PostTypeId="2" ParentId="3949" CreationDate="2014-11-07T17:47:30.770" Score="0" Body="&lt;p&gt;There's an interesting documentary on how to dissolve organic tissue on YouTube: &lt;a href=&quot;https://www.youtube.com/watch?v=ENrZM5ORsPo&quot; rel=&quot;nofollow&quot;&gt;How to Commit the Perfect Murder, a BBC Horizon Documentary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They demonstrate that a washing powder is the best for dissolving organic tissue provided you can heat it up a bit.&lt;/p&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="73" LastEditDate="2014-11-07T18:37:55.647" LastActivityDate="2014-11-07T18:37:55.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19253" PostTypeId="1" AcceptedAnswerId="19272" CreationDate="2014-11-07T19:15:04.660" Score="0" ViewCount="74" Body="&lt;p&gt;So there has been a lot of noise over whether or not certain molecules activate SIRT1. Resveratrol was thought to be a specific SIRT1 activator, but many have argued against this claim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, this has also &lt;a href=&quot;http://en.wikipedia.org/wiki/SRT1720&quot; rel=&quot;nofollow noreferrer&quot;&gt;happened&lt;/a&gt; to SRT1720 too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is - why is it so hard to quantify whether or not a molecule is an SIRT1 activator? How would one go about doing this?&lt;/p&gt;&#xA;" OwnerUserId="217" LastEditorUserId="402" LastEditDate="2017-05-16T02:02:54.953" LastActivityDate="2017-05-16T02:03:09.207" Title="How does one analytically quantify whether or not a compound is a SIRT1 activator?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19254" PostTypeId="2" ParentId="19250" CreationDate="2014-11-07T19:58:08.583" Score="3" Body="&lt;p&gt;You are very close. The only issue I see is that the Mannich reaction takes place between an enol(ate) and an imine (iminium). Your mechanism goes to the hemiaminal, and then the enolate must displace a hydroxide leaving group. Essentially that's an SN2 with a hydroxide as a leaving group, which does not occur. Instead, from the hemiaminal, the nitrogen kicks out the hydroxide to form an iminium, which is a much more reactive electrophile. The enolate attacks readily. Repeat that process, decarboxylate &amp;amp; tautomerize twice, and you're there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IdatB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-07T19:58:08.583" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19255" PostTypeId="1" AcceptedAnswerId="19256" CreationDate="2014-11-07T20:03:18.887" Score="1" ViewCount="11174" Body="&lt;p&gt;I am given $\ce{H2SO4}$ in a reaction vessel of about $50~\mathrm{mL}$. I need to solve for the molarity of $\ce{H2SO4}$. A $10~\mathrm{mL}$ sample of $\ce{H2SO4}$ is removed and then titrated with $33.26~\mathrm{mL}$ of standard $0.2643\ \mathrm{M}\ \ce{NaOH}$ solution to reach the endpoint. How do I solve for titration of the $50~\mathrm{mL}$ sample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I can use the formula: &#xA;$$M_i \times V_i = M_f \times V_f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which brings me to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_i \times 10~\mathrm{mL} = 0.2643~\mathrm{M} \times 33.26~\mathrm{mL}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_i = (0.2643~\mathrm{M} \times 33.26~\mathrm{ml}) / (10~\mathrm{mL})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will find the molarity of the $10~\mathrm{mL}$ sample of $\ce{H2SO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how do I find the molarity of the $50~\mathrm{mL}$ sample of $\ce{H2SO4}$ from this? Would I just do five times the $10~\mathrm{mL}$ sample's molarity? &lt;/p&gt;&#xA;" OwnerUserId="9673" LastEditorUserId="7475" LastEditDate="2016-10-09T19:04:21.570" LastActivityDate="2019-12-07T04:08:21.620" Title="Titration of H2SO4 w NaOH: Solving for the molarity of H2SO4?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19256" PostTypeId="2" ParentId="19255" CreationDate="2014-11-07T20:23:50.003" Score="3" Body="&lt;p&gt;The molarity would be the same whether you have $5~\mathrm{mL}$ of $\ce{H2SO4}$ or a swimming pool full of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you wanted to solve for moles of $\ce{H2SO4}$ in $50~\mathrm{mL}$, you would have to multiply the number of moles in the $10~\mathrm{mL}$ sample by $5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Make sure you're working with molarity and not moles. I'm in analytical chem right now and often we're multiplying the number of moles in our sample by the total volume of the volumetric flask from which the sample was drawn, so we're doing calculations similar to this.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="34388" LastEditDate="2017-01-28T05:36:34.860" LastActivityDate="2017-01-28T05:36:34.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19257" PostTypeId="2" ParentId="19255" CreationDate="2014-11-07T20:27:19.607" Score="1" Body="&lt;p&gt;A couple things to think about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How many protons can one molecule of sulfuric acid give? Will this affect the amount of NaOH it takes to neutralize a given amount of sulfuric acid?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Molarity is the number of moles in a Litre of solution. If I double the volume, it doubles the number of moles. Does this change the ratio of moles to litres?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-07T20:27:19.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19258" PostTypeId="1" AcceptedAnswerId="19426" CreationDate="2014-11-07T20:32:20.717" Score="9" ViewCount="799" Body="&lt;p&gt;It is known, that decomposition of Lead(II) nitrate is one of the ways of generating $\ce{NO2}$ for lab use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently did this in order to acquire $\ce{NO2}$ in liquid form (ambient temperature was ~0 °C), and got quite strange results. First, liquid $\ce{NO2}$ had 2 visible layers, usual brown one at the top and transparent at the bottom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLM0I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, during condensation liquid initially had green color (I know that Graham condenser is not safe in this configuration - I wouldn't see this color change in conventional condenser setup).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G94lm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone clarify why I got such variety of products instead of pure $\ce{NO2}$?&lt;/p&gt;&#xA;" OwnerUserId="9675" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T15:51:25.427" LastActivityDate="2021-05-28T05:21:03.533" Title="Products of Lead(II) nitrate decomposition" Tags="&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19259" PostTypeId="1" AcceptedAnswerId="19260" CreationDate="2014-11-07T20:53:25.280" Score="0" ViewCount="354" Body="&lt;p&gt;Would I be correct if I state: Thin layer chromatography (TLC) is a separation method for non-volatile mixtures. Because distillate samples are pure, this would not be appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure and would like some insight into this. &lt;/p&gt;&#xA;" OwnerUserId="8260" LastEditorUserId="4945" LastEditDate="2014-12-05T03:25:37.987" LastActivityDate="2014-12-05T03:25:37.987" Title="Is thin layer chromatography suitable to characterize distillate samples?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19260" PostTypeId="2" ParentId="19259" CreationDate="2014-11-07T21:39:03.020" Score="3" Body="&lt;p&gt;Your statement is unnecessarily complicated and not easy to follow. You can use TLC as a quick way to assess the purity of a sample. It doesn't matter if your sample is a distillate or not. It only has to be non volatile as you said and soluble in some solvent or a mixture of solvents that can (the solvent/s) make it move on the TLC plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you want to use distillation to separate a mixture and TLC in order to assess the purity of your fractions. Obviously TLC becomes an unsuitable method for low b.p. distillates&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-11-07T21:39:03.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19261" PostTypeId="1" CreationDate="2014-11-07T22:32:16.687" Score="4" ViewCount="1304" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Write the mechanism for this reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EnPbQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EnPbQ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am trying to do a dehydration of an alcohol using the reagents sulfuric acid and acetic acid. I don't know how I am supposed to use the acetic acid, instead I've been using water. This is my proposed mechanism:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OTD9n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OTD9n.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;But it doesn't include the acetic acid, and it doesn't say in the question whether it is done in solution with water or not but I only know how to do it with &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;. My main issue is that I need to put the acetic acid into the mechanism.&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-06-25T05:51:29.983" Title="Dehydration of secondary alcohol via E1 mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19262" PostTypeId="1" CreationDate="2014-11-07T22:36:38.420" Score="5" ViewCount="1590" Body="&lt;p&gt;I have the ability to use many kinds of reactions, and the use of most common acids/bases.&lt;/p&gt;&#xA;" OwnerUserId="9678" LastActivityDate="2018-04-12T04:12:39.853" Title="How do I isolate calcium from calcium carbonate?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19263" PostTypeId="2" ParentId="18988" CreationDate="2014-11-07T23:05:51.597" Score="4" Body="&lt;p&gt;I think @NicolauSakerNeto mentioned a number of excellent points in his &lt;a href=&quot;https://chemistry.stackexchange.com/a/8426/5017&quot;&gt;linked answer&lt;/a&gt;. Frankly, I think he deserves a bounty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several effects occurring, which give different electron configurations to the transition metals. A very nice review article is available: &lt;a href=&quot;http://dx.doi.org/10.1021/ed8001286&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;The Full Story of the Electron Configurations of the Transition Elements&quot; &lt;em&gt;J. Chem. Educ.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;87&lt;/em&gt; (4), pp 444–448&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;strong&gt;d orbital collapse&lt;/strong&gt;&quot;: Because of the difference in angular-momentum &quot;centrifugal force&quot; in quantum mechanics, incomplete shielding of core electrons, and nuclear attraction, the $Z_{eff}$ - effective nuclear charge - for 3d and 4s electrons are different. As a rule of thumb, electrons are typically lost from the 4s first, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;electrostatic repulsion in 4s vs. 3d&lt;/strong&gt;. It's interesting you mention V, since $\ce{V^0}$ is $[3d^3 4s^2]$ and $\ce{V+}$ is $[3d^4]$. The usual explanation is that while 3d is closer to the nucleus and energetically favorable, once you start filling the 3d electrons, the electrostatic repulsion is high, and for the neutral atoms, it's favorable to have one or two 4s electrons, which are more diffuse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are other things going on, particularly later in the periodic table, where relativistic and spin-orbit coupling effects occur. Also, many times the electron configuration can change in a free atom or ion compared to a complex.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2014-11-08T16:47:41.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19264" PostTypeId="1" CreationDate="2014-11-07T23:36:25.620" Score="12" ViewCount="3396" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Write the mechanism for the following reaction (bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/utHGj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/utHGj.png&quot; alt=&quot;que&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;My answer so far:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KYKNt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYKNt.png&quot; alt=&quot;my answer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have only shown one of the reactions, I presume it is the same for adding the other &lt;span class=&quot;math-container&quot;&gt;$\ce{Me2NH}$&lt;/span&gt; on the other side.&lt;/p&gt;&#xA;&lt;p&gt;How do I get rid of the second bromine, or does just one of the Br's add? If it is only one, why, because you usually use &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; for that?&lt;/p&gt;&#xA;" OwnerUserId="5181" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-18T11:46:22.553" Title="Bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19267" PostTypeId="2" ParentId="19264" CreationDate="2014-11-08T04:21:44.597" Score="2" Body="&lt;p&gt;Two possible mechanisms exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either the molecule, undergoed 1,6 bromine addition or 1,2 addition, in which the first case easily gives the product and in case of second, the final addition can be 1,6. It was 1,6 addition in the first or second step is hard to say, as I'm concerned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pgN9V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't be confused if the final structure is looking as if it had no rotation, it had michael addition.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-12-12T08:52:08.377" LastActivityDate="2014-12-12T08:52:08.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19268" PostTypeId="2" ParentId="19239" CreationDate="2014-11-08T04:54:37.973" Score="0" Body="&lt;p&gt;$$\ce{N2 +O2 -&amp;gt; N_{$x$}O_{$y$}}$$&#xA;Now using POAC (Principle of Atomic conservation),&#xA;$$\frac5{28}\cdot2=\frac{19.3}{14x+16y}\cdot x\implies \frac{y}{x}\approx2.5025\sim2.5\implies y'=5,x'=2\implies \ce{N2O5}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-12-02T05:05:26.183" LastActivityDate="2014-12-02T05:05:26.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19271" PostTypeId="1" CreationDate="2014-11-08T06:57:06.817" Score="1" ViewCount="96" Body="&lt;p&gt;I have 100% acetic acid (glacial) with a pH of 2.4 and I want to know how much DM water is required to dilute the acetic acid and obtain a solution with a pH of 4.3.&lt;/p&gt;&#xA;&lt;p&gt;As per observation 0.0001 M is equal to pH 4.3. So how can I implement this calculation using the Nernst equation?&lt;/p&gt;&#xA;" OwnerUserId="9681" LastEditorUserId="69830" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T07:28:59.857" LastActivityDate="2021-05-22T07:28:59.857" Title="How much water is required to dilute glacial acetic acid to given pH?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19272" PostTypeId="2" ParentId="19253" CreationDate="2014-11-08T07:21:45.387" Score="1" Body="&lt;p&gt;It's important to distinguish SIRT1, which is a gene, from the protein it encodes, sirtuin 1 (some papers use SIRT1 for the protein too, but to avoid confusion, I'll use Sirt1). When looking at compounds that affect gene expression, one method of measuring the effects is measuring the change in mRNA levels for that gene when the compound is administered, usually with northern blotting or qPCR. In this case, because the gene encodes a protein, quantifying the protein levels also tells us about the expression of the gene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SIRT1 gene doesn't have any regulatory role itself, so in this case, people are more interested in the Sirt1 protein. While we can measure protein concentrations, Sirt1 is an enzyme that deacetylates proteins and resveratrol has been proposed to directly alter the activity of the enzyme (it's also known to increase SIRT1 expression, but this isn't the controversial aspect), so it's important to examine the effects of resveratrol on Sirt1's ability to deacetylate certain proteins. The way this is done is to expose various proteins/peptides to Sirt1 and look at whether they deacetylate. One method of measuring this is to use a special peptide as the substrate to which a fluorophore can be attached if the acetyl group is cleaved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason resveratrol's effects have been debated is that the &lt;a href=&quot;http://www.nature.com/nature/journal/v425/n6954/full/nature01960.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;early studies&lt;/a&gt; that showed large activation effects were &lt;a href=&quot;http://www.jbc.org/content/280/17/17187.long&quot; rel=&quot;nofollow noreferrer&quot;&gt;later found&lt;/a&gt; to only occur because of the fluorophore on the substrate and not resveratrol alone. Some work since then using other detection methods have shown activation effects: &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3629287/#!po=30.0000&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; tried many different substrates on a microarray using fluorescently labelled antibodies as well as kinetic measurements by mass spectrometry and found that resveratrol increased Sirt1's activity for some substrates, decreased it for others, and left many the same, though they didn't offer much in the way of structural insights that might be responsible. I haven't done a thorough literature search, but it seems like the debate is less contentious now that people are using other assays to examine activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to summarize your questions, it's not hard to quantify whether a molecule increases the expression of SIRT1, but because Sirt1 is an enzyme, any kind of label you add to measure activity could have its own effects on that activity. Any detection method that can measure the deacetylation of substrates could potentially be used for studying Sirt1 activity, and there are many, but label-free techniques like mass spectrometry are especially useful because they reduce complications in simulating biological conditions.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="402" LastEditDate="2017-05-16T02:03:09.207" LastActivityDate="2017-05-16T02:03:09.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19273" PostTypeId="1" CreationDate="2014-11-08T11:52:06.383" Score="0" ViewCount="1105" Body="&lt;p&gt;Recently I did a distillation. At a certain point an azeotropic mixture of alcohol, water and toluene evaporated and was caught in a beaker with anhydrous potassium carbonate. I had to shake it and then pour it back into the reaction mixture (while filtering out the solid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did I have to catch it in a $\ce{K2CO3}$-mixture? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The product was diethyladipate)&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="5591" LastEditorDisplayName="user15489" LastEditDate="2015-10-09T17:29:44.223" LastActivityDate="2015-10-09T17:29:44.223" Title="Potassium carbonate and water/alcohol/toluene?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19274" PostTypeId="1" CreationDate="2014-11-08T13:12:53.487" Score="1" ViewCount="110" Body="&lt;p&gt;There are a lot of tutorials on the net on how to make colloidal silver. Many of them basically go like this - Hook up two 99.99% silver coins or wires to three 9V batteries, and put the silver in saline, an inch apart, for a few minutes. Does this make real colloidal silver or silver ionic silver or silver chloride?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example:&#xA;&lt;a href=&quot;http://www.instructables.com/id/How-to-make-COLLOIDAL-SILVER-Easy/&quot; rel=&quot;nofollow&quot;&gt;http://www.instructables.com/id/How-to-make-COLLOIDAL-SILVER-Easy/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9684" LastActivityDate="2014-11-08T13:12:53.487" Title="Colloidal Silver?" Tags="&lt;home-experiment&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19275" PostTypeId="1" AcceptedAnswerId="19280" CreationDate="2014-11-08T13:53:53.197" Score="0" ViewCount="2255" Body="&lt;p&gt;I think $\ce{HF}$ can be hydrolysis and get $\ce{F-}$ as a conjugate base, so this solution should be a buffer solution. But the correct answer is that this solution is not a buffer. Please explain it to me. I'm very confused.&lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="4231" LastEditDate="2014-11-08T19:15:18.037" LastActivityDate="2014-11-09T00:52:56.167" Title="Why HF plus H2O is not a buffer solution?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19276" PostTypeId="1" CreationDate="2014-11-08T14:00:57.273" Score="3" ViewCount="1640" Body="&lt;p&gt;Is there a specific chemical reaction that enables dimethyltryptamine to cross the capillary edothelial cells?&lt;/p&gt;&#xA;" OwnerUserId="9685" LastEditorUserId="186" LastEditDate="2015-10-02T16:25:07.753" LastActivityDate="2015-10-02T16:25:07.753" Title="How does dimethyltryptamine cross the blood-brain barrier?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19277" PostTypeId="1" CreationDate="2014-11-08T14:48:39.497" Score="6" ViewCount="535" Body="&lt;p&gt;Suppose water under neutral conditions is confined in a virtual spherical nanocontainer with a radius of 25 nm. To calculate the number of hydronium ions, one uses water dissociation constant which is derived from the law of mass action. In this case, considering the temperature at 25°C I come up with a number which is smaller than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytcAs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A logical suggestion here is probability. Is there any limitation to the law of mass action, i.e. does it break down at the nanoscale level and quantum mechanics and thus probability kick in? If yes, is there any boundary to specify things with respect to the size of the system? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2015-02-28T17:33:32.890" Title="Dissociation of water into H+ and OH-: Does the law of mass action hold at nanoscale?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;ph&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19278" PostTypeId="1" CreationDate="2014-11-08T14:52:37.860" Score="-1" ViewCount="6460" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Each atom is at the corner of a cube and 8 cubes meet at each corner. Therefore each atom is shared by 8 cubes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I can't imagine it in space, so please tell how I visualize it. &lt;/p&gt;&#xA;" OwnerUserId="9687" LastEditorUserId="41328" LastEditDate="2019-02-27T11:52:39.230" LastActivityDate="2019-02-27T14:46:11.320" Title="Cubic lattice: How corner atom of a cube is shared by 8 atoms?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;crystallography&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ClosedDate="2019-02-28T01:11:25.097" ContentLicense="CC BY-SA 4.0" />
  <row Id="19279" PostTypeId="1" AcceptedAnswerId="19294" CreationDate="2014-11-08T14:59:26.903" Score="1" ViewCount="56" Body="&lt;p&gt;Good day,&#xA;i had organic chemistry labs. And we had this reaction:&#xA;$$\ce{(CH3)3COH + HCl -&amp;gt; (CH3)3CCl + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;how do i explain my yield of reaction is only 40%? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory is:&#xA;reaction is reversible so the K isn't so high and that lead to yield 40%. &#xA;If my theory is right please explain me how to calculate K?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had these informations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V$\ce{((CH3)3COH)}$ = 10 ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V$\ce{(HCl)}$ = 20 ml&#xA;HCl was concentrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank for answers.&lt;/p&gt;&#xA;" OwnerUserId="9688" LastEditorUserId="4231" LastEditDate="2014-11-08T17:06:20.077" LastActivityDate="2014-11-08T19:48:07.020" Title="can i calculate equilibrium constant of this reaction?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19280" PostTypeId="2" ParentId="19275" CreationDate="2014-11-08T15:17:54.907" Score="1" Body="&lt;p&gt;What are you talking about?&#xA;$\ce{HF}$ in water solution is acting like moderately strong acid. It's because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HF + H2O \rightleftharpoons H3O+ + F-}$$&#xA;$$\ce{H3O+ + F- \rightleftharpoons [H3O]F}$$&#xA;which is undissociated.Then:&#xA;$$\ce{F- + HF \rightleftharpoons HF2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the large affinity of $\ce{F^-}$ to $\ce{H^+}$ won't allow it to dissociate.&#xA;The $K_a(HF)(6.6x10^{-4})$ is strongly favoring reactants.&#xA;So the $\ce{HF}$ in $\ce{H2O}$ isn't buffer because there isn't almost any $\ce{F-}$ to be considered as conjugated acid. So if you add some acid there will not be enough $\ce{F-}$ for it's neutralisation.&#xA;So if you add some acid or base it will react with water and equilibrium constant will be changed and that will lead to drastic change of pH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the buffer is solution which consist of: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;salt of strong acid and weak base and the weak base (for example $\ce{NH_{4}Cl}$ and $\ce{HN3}$ ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;salt of weak acid and strong base and the weak acit (for example  $\ce{CH3COONa}$ and $\ce{CH3COOH}$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope i helped you. If you have any other questions please ask.&lt;/p&gt;&#xA;" OwnerUserId="9688" LastEditorUserId="9688" LastEditDate="2014-11-09T00:52:56.167" LastActivityDate="2014-11-09T00:52:56.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19281" PostTypeId="1" AcceptedAnswerId="19282" CreationDate="2014-11-08T15:46:21.423" Score="0" ViewCount="3118" Body="&lt;p&gt;Adding $\ce{HCl}$ to silver nitrate solution causes silver chloride to precipitate. Adding more $\ce{HCl}$ causes the silver to re-dissolve. What is happening to the silver chloride when it re-dissolves?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4231" LastEditDate="2014-11-08T17:02:02.560" LastActivityDate="2014-11-08T17:02:02.560" Title="Why does HCl both precipitate and dissolve Silver?" Tags="&lt;silver&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19282" PostTypeId="2" ParentId="19281" CreationDate="2014-11-08T15:49:48.003" Score="3" Body="&lt;p&gt;The precipitation reaction is:&#xA;$$\ce {Ag+ + Cl- -&amp;gt; AgCl_{(s)}}$$&#xA;The next reaction is:&#xA;$$\ce {AgCl_{(s)} +Cl- -&amp;gt; AgCl2- }$$&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-08T15:49:48.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19283" PostTypeId="2" ParentId="19278" CreationDate="2014-11-08T16:06:12.263" Score="1" Body="&lt;p&gt;Surprisingly, this simple question does not seem to have been asked before. &lt;a href=&quot;http://en.wikipedia.org/wiki/Bravais_lattice&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Bravais_lattice&lt;/a&gt; shows all the two dimensional Bravais lattices in extended form (but only the unit cell for the three dimensional ones.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the 2-dimensional image from wikipedia. As you can see in the simple square lattice, each sphere is a the corner of a square and 4 squares meet at each corner. Therefore each sphere is shared by 4 squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VfPeK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple cubic lattice is the same, except that the atoms are found at the corner of the cubic unit cell. Therefore each atom is shared by 8 unit cells. There are eight eighths (equal to one whole atom) in each unit cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any elements that crystallise in the simple cubic lattice. It is not energetically favoured because it is not spatially efficient. The more complex face-centred and body-centred lattices are preferred. Caesium Chloride, Bromide and Iodide form interpenetrating simple cubic lattices with the halide ion at the centre and the caesium ion at the corners of the unit cell (or vice versa depending on your viewpoint.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the relevant wikipedia links&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Caesium_chloride&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_bromide&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Caesium_bromide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_iodide&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Caesium_iodide&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2014-11-08T16:06:12.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19284" PostTypeId="1" AcceptedAnswerId="26250" CreationDate="2014-11-08T16:10:04.433" Score="3" ViewCount="20394" Body="&lt;p&gt;I can't seem to find what the difference between these two is. In my text book, it says that 'arenes are aromatic hydrocarbons containing one or more benzene rings' which to me suggests that arenes are a subgroup of aromatic hydrocarbons and aromatic hydrocarbons need not contain one or more benzene rings. However later on in my book it says that 'any compound with the benzene ring is classified as an aromatic compound', which to me is the exact same definition as that given for an arene. Is there a difference between arenes and aromatic compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance :)&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="418" LastEditDate="2015-02-17T09:23:25.007" LastActivityDate="2018-08-27T16:37:19.947" Title="Difference between any arene and an aromatic compound?" Tags="&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19285" PostTypeId="1" CreationDate="2014-11-08T16:18:50.720" Score="1" ViewCount="346" Body="&lt;p&gt;What should be a suitable mechanism for the reaction between aluminium and iodine, catalysed by water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my initial approaches, I had first thought of repeated attacks of the lone pairs of oxygen in water on aluminium, then formation of $\ce{Al(OH)3}$ by losing $\ce{H+}$, then $\ce{H+}$ attacking $\ce{I2}$, and formation of $\ce{HI}$, and finally formation of $\ce{AlI3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But actually $\ce{Al2I6}$ is formed, so how can I accommodate its formation in that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anything wrong in the above suggested mechanism? If the actual mechanism is something else, please write it in your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Lewis dot structure of the mechanism would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="34388" LastEditDate="2017-01-24T15:17:33.490" LastActivityDate="2019-10-01T18:58:53.227" Title="Mechanism for reaction between aluminium and iodine" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2019-10-07T00:05:55.163" ContentLicense="CC BY-SA 3.0" />
  <row Id="19287" PostTypeId="1" AcceptedAnswerId="19289" CreationDate="2014-11-08T17:02:24.683" Score="6" ViewCount="4165" Body="&lt;p&gt;I have been looking at the symmetry of $\ce{B2Cl4}$ and was cannot understand how it has two perpendicular $C_2'$ axes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand it has a $C_2(z)$, $S_4(z)$ and two dihedral planes bisecting each of the $\ce{Cl-B-Cl}$ bonds. However my book says it also has an additional two $C_2'$ axes in the $x$ and $y$ planes centred at the middle of the $\ce{B-B}$ bond. I just can't see it. Any help? ( I understand it will be difficult online but I would very much appreciate your assistance. &lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="4356" LastEditDate="2014-11-15T00:15:33.757" LastActivityDate="2018-02-23T00:25:14.257" Title="How does B2Cl4 have two perpendicular c2' axes?" Tags="&lt;symmetry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19288" PostTypeId="2" ParentId="14417" CreationDate="2014-11-08T17:08:49.300" Score="8" Body="&lt;p&gt;This phenomenon is explained by s-p mixing. All the elements in the second period before oxygen have the difference in energy between the 2s and 2p orbital small enough, so that s-p mixing (combination) can occur lowering the energy of the σ(2s) and σ*(2s) and increasing the energy of the σ(2p) and σ*(2p) molecular orbitals. By moving towards right in a period, the s orbital gets more stabilized than the p orbital and the difference in their energies increases, making the s-p mixing for oxygen much smaller.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-03-19T18:55:40.457" LastActivityDate="2015-03-19T18:55:40.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19289" PostTypeId="2" ParentId="19287" CreationDate="2014-11-08T17:25:51.620" Score="8" Body="&lt;p&gt;Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, a picture tells more than I thousand words I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iaCuG.png&quot; alt=&quot;Allene C2 axis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two $\ce{C_2^{'}}$ axis are turned by 45° in comparison to the C-H bonds. It therefore also has two dihedral planes, thus belonging to the $\ce{D_{nd}}$ point group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have difficulties picturing this in your head that is not strange. Many people make mistakes in assigning point groups. You get a better visualization by &lt;a href=&quot;http://symmetry.otterbein.edu/challenge/index.html&quot; rel=&quot;noreferrer&quot;&gt;practicing online&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-08T18:07:41.070" LastActivityDate="2014-11-08T18:07:41.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19290" PostTypeId="2" ParentId="19264" CreationDate="2014-11-08T17:56:12.743" Score="10" Body="&lt;p&gt;This is a quote from Solomons and Frylhe Organic Chemistry 10th Ed, Chapter 13, p.614&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to cycloocta-1,3,5-triene.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kUGb1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kUGb1.png&quot; alt=&quot;Bromination of cycloocta-1,3,5-triene&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A plausible mechanism for the bromination step would be the initial formation of a bromonium ion which is attacked distally by bromide:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U6hMU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U6hMU.png&quot; alt=&quot;Plausible bromination mechanism&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Sequential nucleophilic substitution with dimethylamine then leads to the desired product.&lt;/p&gt;&#xA;&lt;p&gt;Echoing &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19264/bromination-of-1-3-5-cyclooctatriene-followed-by-nucleophilic-substitution-by-ex/19267#comment175501_95408&quot;&gt;Martin's comment&lt;/a&gt; on &lt;a href=&quot;https://chemistry.stackexchange.com/a/95408/16683&quot;&gt;Siddharth Porwal's answer&lt;/a&gt;, there is a distillation step between the bromination and the substitution with dimethylamine. It is not possible for a putative bromonium ion to survive. Also, the isolation of the 1,6-dibromide also disproves the possibility of 1,2-addition mentioned in &lt;a href=&quot;https://chemistry.stackexchange.com/a/19267/16683&quot;&gt;RE60K's answer&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The reaction above was originally reported by Cope &lt;em&gt;et al.&lt;/em&gt;:&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Bromine (3 g) in 20 ml of chloroform was added to a stirred solution of 2 g of 1,3,5-cyclooctatriene in 20 ml of chloroform at −15 to −25° in a nitrogen atmosphere in a period of one hour. The product was distilled rapidly under reduced pressure; b.p. 95° (0.3 mm), &lt;em&gt;n&lt;/em&gt;&lt;sup&gt;25&lt;/sup&gt;ᴅ 1.5954. [...] Previous work had shown that dibromocyclooctadiene prepared by the two routes gave the same bis-dimethylaminocyclooctadiene by reaction with dimethylamine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cope, A. C.; Stevens, C. L.; Hochstein, F. A. Cyclic polyolefins. v. preparation of bromocyclooctadienes and 1,3,5-cyclooctatriene from 1,5-cyclooctadiene. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1950,&lt;/strong&gt; &lt;em&gt;72&lt;/em&gt; (6), 2510–2514. &lt;a href=&quot;https://doi.org/10.1021/ja01162a047&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja01162a047&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-04-18T11:46:22.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19291" PostTypeId="1" CreationDate="2014-11-08T18:59:08.273" Score="0" ViewCount="383" Body="&lt;p&gt;Is it plausible that free chlorine could be formed in concentrated hydrochloric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background to this is that I have bought some 30% HCl to remove calcium carbonate scale from a reverse osmosis water purifier. The latest batch of HCl seems to have a very slight yellowish tint which I had not noticed on previous occasions. Having used some of this acid I find that the performance of the membrane seems to have deteriorated (these membranes are notably sensitive to hypochlorite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could some factor (perhaps exposure to direct sunlight) be responsible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way that I could detect the presence of dissolved chlorine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest an alternative explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Chris.&lt;/p&gt;&#xA;" OwnerUserId="9692" LastActivityDate="2020-11-04T09:15:46.880" Title="chlorine formed in hydrochloric acid" Tags="&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19293" PostTypeId="2" ParentId="19291" CreationDate="2014-11-08T19:45:25.957" Score="1" Body="&lt;p&gt;You could buy the chlorine test strips from a pool supply store to test for hypochlorite. You may need to neutralize the acid before testing, so ask if the test is accurate in a strong acidic solution.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-08T19:45:25.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19294" PostTypeId="2" ParentId="19279" CreationDate="2014-11-08T19:48:07.020" Score="1" Body="&lt;p&gt;$\ce{H2O}$ is a very poor nucleophile and we are dealing with a tertiary substituted carbon, so we can minimize the scope of possible reaction mechanisms to the $\ce{E 1}$ and $\ce{S_{n}1}$ type mechanisms. $\ce{E 1}$ will compete here with $\ce{S_{n}1}$, forming tert-Butyl chloride and isobutylene respectively. That is probably where most of your loss is coming from. Other reasons can be indeed that a bit is left in the equilibrium, but I don't think that would be much since $\ce{H3O+}$ is a much better leaving group than $\ce{Cl-}$. Note that is chemistry papers the yield obtained is almost never a 100% of theory.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-11-08T19:48:07.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19295" PostTypeId="1" CreationDate="2014-11-08T20:17:11.393" Score="0" ViewCount="270" Body="&lt;p&gt;I am interested in disaccarides of D-glucose linked 1-4 style. Anomerization is the interconversion of alpha-D-glucose into beta-D-glucose. I realize that this could only be possible on one of the glucose sub-units: The other glucose is obviously unable to inter-convert due to the glycosidic bond preventing ring-opening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIi1V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not see what would prevent it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gyi7y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4945" LastEditDate="2014-11-17T04:00:41.157" LastActivityDate="2017-08-31T05:03:58.013" Title="Do disaccharides undergo anomerization?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19296" PostTypeId="2" ParentId="19295" CreationDate="2014-11-08T20:25:50.573" Score="1" Body="&lt;p&gt;There is no reason to think that anomerization on the terminal glucose unit does not occur. Although there may be a slight difference in the equilibrium constant between a monosaccharide and a disaccharide due to the change in environment.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-08T20:25:50.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19297" PostTypeId="1" CreationDate="2014-11-08T20:52:16.200" Score="2" ViewCount="206" Body="&lt;p&gt;When HF meets glass, Hexafluorosilic acid is formed (assuming some water is present). It appears that Hexafluorosilic acid and further attack glass? How is this so, considering that all F atoms are &quot;used up&quot; (already bonded to Si atoms)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cdc.gov/niosh/ipcsneng/neng1233.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cdc.gov/niosh/ipcsneng/neng1233.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1211" LastEditorUserId="23561" LastEditDate="2018-07-03T11:52:17.657" LastActivityDate="2018-07-03T11:52:17.657" Title="Hexafluorosilicid acid attacks glass?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;glass&gt;" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19300" PostTypeId="1" AcceptedAnswerId="70309" CreationDate="2014-11-08T22:22:30.970" Score="4" ViewCount="181" Body="&lt;p&gt;What can we say about a molecule from the position and number of bands in FTIR spectrum. I know that the more symmetric a molecule is, the less number of bands it has. What else can we say, for example about the position of the band? Does it show the frequency of vibrations and therefore the strength of the bonds? &#xA;For example, the spectrum of MoF4 has one band at 674 cm -1. XeF4 has 3 bands at 586, 291, and 161. How can we compare the structure of these materials from the above information?&lt;/p&gt;&#xA;" OwnerUserId="6985" LastEditorUserId="6985" LastEditDate="2017-03-12T01:09:57.533" LastActivityDate="2017-03-12T11:41:33.397" Title="General methods of Analysis for FTIR Results" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19302" PostTypeId="1" CreationDate="2014-11-09T00:44:09.260" Score="0" ViewCount="5169" Body="&lt;p&gt;What is difference between organosilyl, alkoxysilane and silanes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these groups make a surface hydrophobic?&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="4945" LastEditDate="2014-11-17T03:38:15.173" LastActivityDate="2014-11-18T04:45:04.640" Title="What is the difference between organosilyl, alkoxysilane and silane compounds?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19303" PostTypeId="1" AcceptedAnswerId="19304" CreationDate="2014-11-09T02:05:01.977" Score="0" ViewCount="19033" Body="&lt;p&gt;I just bought a bottle of sulfuric acid in order to simulate acid rain for an experiment. I was wondering what $0.12\:\mathrm{N}$ on the bottle means. I need a simplified version of the explanation as I'm currently a freshman in high school taking biology, I have never taken a chemistry course before. I've read that the $\mathrm{N}$ means normal, but the explanation included a ton of jargon that I couldn't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ebay.com/itm/400700051478&quot; rel=&quot;nofollow&quot;&gt;The time in question&lt;/a&gt;: does the $0.12\:\mathrm{N}$ mean concentration, as in $12\%$ concentration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you guys suggest making simulated acid rain with this? My guess is that I would keep titrating the sulfuric acid into the water until the $\mathrm{pH}$ gets to my preferred $\mathrm{pH}$.&lt;/p&gt;&#xA;" OwnerUserId="9697" LastEditorUserId="15410" LastEditDate="2015-05-16T15:56:08.463" LastActivityDate="2018-05-23T03:15:57.887" Title="What does 0.12 N mean on a bottle of sulfuric acid?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19304" PostTypeId="2" ParentId="19303" CreationDate="2014-11-09T02:17:29.843" Score="6" Body="&lt;p&gt;0.12N is indeed 0.12 normal. In short, there are 0.12 moles of hydrogen ions per liter. Since this is sulfuric acid, it is 0.06 moles of sulfuric acid per liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.12 moles of hydrogen ions per liter would give a pH just less than 1. To obtain a pH near 3, you would need to dilute 1 mL of this solution to 100 mL. For a pH of 4, dilute 1 mL to 1000 mL.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-09T02:17:29.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19305" PostTypeId="1" AcceptedAnswerId="19307" CreationDate="2014-11-09T02:32:38.203" Score="8" ViewCount="1141" Body="&lt;p&gt;I'm writing a report on the spectrophotometric determination of iron. In my introduction I want to briefly explain the importance of iron in the human body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states that &quot;Iron is essential to life due to its unusual flexibility to serve as both an electron donor and acceptor&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. But there are so many elements that can bind with iron! What is special about iron?&lt;/p&gt;&#xA;" OwnerUserId="8205" LastActivityDate="2019-07-16T11:14:19.237" Title="Why is iron used in the body?" Tags="&lt;biochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19306" PostTypeId="2" ParentId="16794" CreationDate="2014-11-09T02:53:50.993" Score="6" Body="&lt;p&gt;Alright, here's my best attempt at it.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Root Mean Squared $v_\mathrm{rms}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by angle brackets $\langle\rangle$), and finally take the square root of the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;v_{rms} &amp;amp;= \langle v^2\rangle^{1/2}&#xA;\\&amp;amp;= \left(\frac{v_1^2+v_2^2+...+v_N^2}{N}\right)^{1/2}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mean Speed $\bar{v}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The r.m.s speed might at first sight seem to be rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass $m$ travelling at a speed $v$ is $E_\mathrm{k} = \frac{1}{2}mv^2$, which implies that the mean kinetic energy of a collection of molecules, $\langle E_\mathrm{k} \rangle$, is the average of this quantity, or $\frac{1}{2}mv_{rms}^2$. It follows from the relation $\frac{1}{2}mv_\mathrm{rms}^2= \langle E_\mathrm{k}\rangle$, that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_\mathrm{rms} = \left(\frac{2\langle E_\mathrm{k}\rangle }{m}\right)^{1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, wherever $v_\mathrm{rms}$ appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s speed is quite close in value to another and more readily visualized measure of molecular speed, the &lt;strong&gt;mean speed&lt;/strong&gt;, $\bar{v}$, of the molecules: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{v} = \frac{v_1+v_2+...+v_N}{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For samples &lt;strong&gt;consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed&lt;/strong&gt;. The precise relation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{v} = \left(\frac{8}{3 \pi}\right)^{1/2}v_\mathrm{rms} = 0.921v_\mathrm{rms}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For elementary purposes, and for qualitative arguements, we don't need to distinguish between the two measures of average speed, but &lt;strong&gt;for precise work the distinction is important.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Most Probable $v_\mathrm{mp}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$v_\mathrm{mp}$ is the most probable speed of the molecules, and is the maximum value for the velocity distribution curve. When putting all of of these equations together we see the relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_\mathrm{rms} &amp;gt; \bar{v} &amp;gt; v_\mathrm{mp}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $v_\mathrm{mp}$ is the smallest of the three speeds is due to the asymmetry of the curve.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Atkins, P.; de Paula, J. &lt;em&gt;Elements of Physical Chemistry&lt;/em&gt; 6th ed. W.H. Freeman and Company. New York, NY. &lt;strong&gt;2013&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chang, R. &lt;em&gt;Physical Chemistry for the Chemical and Biological Sciences&lt;/em&gt;. University Science Books. Sausalito, CA. &lt;strong&gt;2000&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="16683" LastEditDate="2016-09-30T10:08:21.243" LastActivityDate="2016-09-30T10:08:21.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19307" PostTypeId="2" ParentId="19305" CreationDate="2014-11-09T05:11:11.587" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, iron is not the only metal that will fit inside the porphyrin structure. But its ability to be an electron acceptor and donor makes it very useful for oxygen and carbon dioxide to bind to the hemoglobin. Since iron is also very abundant, it was easy to incorporate it and utilize its properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other macrocyclic structures include chlorophyll with magnesium in the center and vitamin B12 (cobalamin) with cobalt in the center.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. ... What is special about iron?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Iron is abundant. The iron in the hemoglobin can weakly bind both oxygen and carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hemoglobin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="7399" LastEditDate="2019-07-16T11:14:19.237" LastActivityDate="2019-07-16T11:14:19.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19308" PostTypeId="2" ParentId="8862" CreationDate="2014-11-09T07:30:26.777" Score="0" Body="&lt;p&gt;It can be, since the $\ce{HNO2}$ is very unstable and decompose very easily (disproportioning): &#xA;$$\ce{3HNO2 \rightarrow HNO3 + 2NO +H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="9688" LastEditorUserId="7448" LastEditDate="2015-09-29T19:27:35.440" LastActivityDate="2015-09-29T19:27:35.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19310" PostTypeId="1" AcceptedAnswerId="19312" CreationDate="2014-11-09T08:43:23.550" Score="1" ViewCount="771" Body="&lt;p&gt;How to know that reaction that occur at the anode electrode is $\ce{2H2O -&amp;gt; 4H+ + O2}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the reaction is not the oxidation of  $\ce{SO4^{2-}}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem does not give the E0 of $\ce{SO4^{2-}}$ too. &lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="5456" LastEditDate="2014-11-09T12:24:06.653" LastActivityDate="2014-11-10T14:24:18.367" Title="Reactions that occur at the anode electrode of electrolysis dilute H2SO4" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19312" PostTypeId="2" ParentId="19310" CreationDate="2014-11-09T11:17:24.593" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How to know that reaction that occur at the anode electrode ... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two kinds of anions: oxidizable (ones that contain central element in lower oxidation states), like $\ce{ClO3-}$, $\ce{Cl-}$, $\ce{NO2-}$, and unoxidaziable, where central element is in hightest oxidation state ($\ce{ClO4-}$, $\ce{NO3-}$. Fluoride obviously is unoxidizable as well.  &lt;strong&gt;As a rule of thumb, it is usually assumed that unoxidizable anions do not react in electrolysis.&lt;/strong&gt; A rationalization for it is that in such cases an oxygen should be oxidized, and oxygen in water reacts faster. For the sake of basic chemistry course it is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is not entirely true: electrolysis of cold solution of ammonium sulfate in sulfuric acid results in formation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_persulfate&quot; rel=&quot;nofollow&quot;&gt;ammonium persulfate&lt;/a&gt; $\ce{(NH4)2S2O8}$, where anion has structure $\ce{ (O3SOOSO3)^{2-}}$ Similiarly, electrolysis of very cold solutions of carbonates produces various &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxocarbonate&quot; rel=&quot;nofollow&quot;&gt;percarbonates&lt;/a&gt;. A synthetic method exists, that is based on electrolysis of carboxylic acid. The anions on oxidation form $\ce{RCOO\cdot}$, that further loose $\ce{CO2}$ and resulting radicals pairs, resulting in net reaction $\ce{2RCOO- \rightarrow R2 + 2CO2 + 2 e-}$ However, in general, formation of peroxocompounds requires low temperatures, usually below zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact mechanism of the process highly depends on the nature of the electrodes, but likely includes formation and transformation of surface particles on the electrode, making reasoning difficult and electrode-specific. Since commonly anodes used have oxide coating, it can be more-or-less safely assumed, that the process includes dissociative adsorption of water with subsequent oxidation of adsorbed (hydro)oxide atoms.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-11-10T14:24:18.367" LastActivityDate="2014-11-10T14:24:18.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19313" PostTypeId="1" AcceptedAnswerId="19315" CreationDate="2014-11-09T13:41:08.240" Score="6" ViewCount="3794" Body="&lt;p&gt;Given structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSRF4.png&quot; alt=&quot;My Try&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my go at it. The longest carbon chain has 10 Carbon atoms, so the root -- &lt;code&gt;decane&lt;/code&gt;. Next: let's take the substituent in position 3, we get &lt;code&gt;3 ethyl&lt;/code&gt;. And finally, we get &lt;code&gt;5-(2, 2-Dimethylbutyl)&lt;/code&gt;. Adding them all up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;5-(2, 2-Dimethylbutyl)-3-ethyldecane&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution in my book is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KA6Za.png&quot; alt=&quot;The Solution Given In My Book&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the answer given is: &lt;code&gt;5-(2-Ethylbutyl)-3, 3-dimethyldecane&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is: Why is the main chain taken by me incorrect?&lt;/p&gt;&#xA;" OwnerUserId="6845" LastEditorUserId="275" LastEditDate="2014-12-02T20:23:13.650" LastActivityDate="2021-02-28T04:01:07.500" Title="Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19315" PostTypeId="2" ParentId="19313" CreationDate="2014-11-09T14:50:48.093" Score="11" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44 SENIORITY ORDER FOR PARENT STRUCTURES&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A parent structure is defined (P-15.1) as a parent hydride, for example, methane, a functionalized parent hydride, for example, cyclohexanol, or a functional parent compound, for example, acetic acid. For choice of the preferred IUPAC name based on a senior parent structure, see P-45.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P-44.3&lt;/strong&gt; SENIORITY OF ACYCLIC CHAINS (THE PRINCIPAL CHAIN)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In an acyclic compound consisting of individual atoms, alike or different (an acyclic chain), the chain on which the nomenclature and numbering is based is called the 'principal chain'. When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The principal chain:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) has the greater number of skeletal atoms;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P-45 SELECTION OF PREFERRED IUPAC NAMES&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P-45.0&lt;/strong&gt; INTRODUCTION&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of&#xA;  the criteria in this section will generate the preferred IUPAC name.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P-45.2.1&lt;/strong&gt; The preferred IUPAC name is based on the senior parent structure that has&#xA;  the maximum number of substituents cited as prefixes (other than 'hydro/dehydro') to&#xA;  the parent structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In accordance with P-44.3 (b), the principal chain is a decane.&#xA;It could be either “5-(2-ethylbutyl)-3,3-dimethyl&lt;strong&gt;decane&lt;/strong&gt;” or “5-(2,2-dimethylbutyl)-3-ethyl&lt;strong&gt;decane&lt;/strong&gt;”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of P-45.2.1, the preferred principal chain corresponds to “5-(2-ethylbutyl)-3,3-dimethyldecane”, which has a principal chain with three substituents (5-, 3-, and 3-), whereas the principal chain of “5-(2,2-dimethylbutyl)-3-ethyl” has only two substituents (5- and 3-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FdYLu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="402" LastEditDate="2017-04-11T04:06:02.627" LastActivityDate="2017-04-11T04:06:02.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19316" PostTypeId="1" AcceptedAnswerId="19317" CreationDate="2014-11-09T15:27:31.217" Score="-2" ViewCount="3263" Body="&lt;p&gt;Question: name the IUPAC name of the following structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yBl0P.png&quot; alt=&quot;The compound&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's what I could come up with is this: we start from the first carbon atom from the extreme left and end with the 2nd substitute while moving right (considering the first C atom at the extreme left to be in 1st position). After that, we have the first substituent in the 2nd position and we get &lt;code&gt;2 ethyl&lt;/code&gt; and then we have '3 methyl' and the root will be &lt;code&gt;pentane&lt;/code&gt;. So, we get &lt;code&gt;2-ethyl, 2-methyl pentane&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here's the answer given in my book &lt;code&gt;3, 4-Dimethylhexane&lt;/code&gt;. Any idea where I've gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="6845" LastActivityDate="2016-11-23T13:04:41.193" Title="Why is the structure of 3, 4-Dimethylhexane given as follows?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19317" PostTypeId="2" ParentId="19316" CreationDate="2014-11-09T15:32:47.390" Score="3" Body="&lt;p&gt;Yes, you started at the wrong atom. You should start on the ethyl group since it will give you a longer chain.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-09T15:32:47.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19318" PostTypeId="1" AcceptedAnswerId="19379" CreationDate="2014-11-09T16:34:37.253" Score="14" ViewCount="20140" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are acetals stable to bases and nucleophiles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could it be due to electronic effects? Lone pairs on the two oxygen atoms create an unfavorable environment for the approaching nucleophiles and bases. Or could it also be steric factors? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-15T13:24:23.110" LastActivityDate="2019-02-03T04:05:33.490" Title="Why are acetals stable to bases and nucleophiles?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;stability&gt;&lt;ethers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19319" PostTypeId="1" CreationDate="2014-11-09T16:48:03.317" Score="4" ViewCount="14320" Body="&lt;p&gt;According to the book for a binary compound, first we assign the element with greater electronegativity its oxidation number (oxygen always -2 except in peroxides). So, we have &#xA;-2 x2 =-2. So, each oxygen would have an oxidation number of -2. The book says the answer is -1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of, is since barium is an alkaline metal, it has an oxidation number of +2, so each oxygen would be -1. But, it contradicts itself because in the rules. It says for a binary compound, &quot;the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element.&quot;&lt;/p&gt;&#xA;" OwnerUserId="9695" LastEditorUserId="9961" LastEditDate="2015-10-26T03:19:32.230" LastActivityDate="2015-10-26T03:19:32.230" Title="What is the oxidation number of the oxygen in BaO2" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19320" PostTypeId="1" AcceptedAnswerId="19328" CreationDate="2014-11-09T17:31:30.687" Score="0" ViewCount="3793" Body="&lt;p&gt;Let's say the ICP-OES software gives me a concentration of 186,49 ppm (mg/L) of Mn. I diluted the acid digested sample to 14.2 mL. I weighed in 10.17 mg of the powder I analyzed. I need to calculate the actual concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite logical in my head what I need to do: multiply by diluted volume and divide by inweight. However, I don't understand my units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I multiply 186.49 mg/L with 0.0142 L, I get 22,648 mg. When I then divide this by 10.17 mg, I end up with 0.260. But this is unitless. How do I know what unit I have? I thought I converted the w/V concentration to w/w by this procedure, but the units does not follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it so, that I end up with a ratio (not sure what kind of ratio)? Then I &lt;em&gt;think&lt;/em&gt; I understand. 0.26 would then represent 26 %, meaning my sample contains 26 % Mn. But how do I know that this is in w/w, and not w/V as I started with. I just don't get the units to add up.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2014-11-17T04:10:26.397" LastActivityDate="2016-06-13T19:50:42.387" Title="ICP-OES data treatment: correct for dilution and inweight, the units" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19321" PostTypeId="2" ParentId="19318" CreationDate="2014-11-09T17:36:03.217" Score="3" Body="&lt;p&gt;In both acid-catalyzed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YYGFW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and base-catalyzed acetal formation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6TexV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the first step involves the alcohol (or alkoxide) behaving as a nucleophile and attacking the electrophilic carbonyl carbon.  At the end of the sequence, when the acetal has been formed, the nucleophilic alcohol has been transformed into an ether, but the nucleophilic center, the &quot;alcoholic&quot;  oxygen, remains intact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where would a second nucleophile attack?  The second nucleophile would be electrostatically repulsed by the still nucleophilic acetal oxygen.  It could attack an adjacent carbon in an $\ce{S_{N}2}$ process, but under basic conditions the only possible leaving group, alkoxide, is not a very favorable leaving group.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-09T17:36:03.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19322" PostTypeId="1" AcceptedAnswerId="19330" CreationDate="2014-11-09T17:36:50.490" Score="21" ViewCount="69981" Body="&lt;p&gt;In the context of automobile repair work, assume a metal surface has rust on it. Does it matter if all of the rust is removed before the metal surface is painted? If it does matter, then why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that at least some will claim you can do this, e.g. the company &lt;a href=&quot;http://www.krylon.com/how-to/faqs/product/can-i-spray-paint-over-rust.jsp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Krylon&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I've tried to figure this out myself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the rust-covered surface area will affect how well the paint will stick to the surface. Maybe this alone can be a problem in terms of durability and sealing the metal from the atmosphere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is governed by a more complex process, what process(es) should be read into further? However, I also assume that (as the aforementioned link suggests) that making the surface as smooth as possible with loose rust and debris removed will yield better results -- i.e. results that approximate that of having no rust at all there? Will it never be as good as simply having all rust removed? I know I'm asking in a way that is difficult to quantify, but &quot;how good&quot; will it get, compared to having all the rust removed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After studying the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow noreferrer&quot;&gt;rust&lt;/a&gt; I also imagine that from the dehydration equilibria we have that even a small area of rust can:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe(OH)2 &amp;lt;=&amp;gt; FeO + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even assuming the rust is properly sealed from the atmosphere it appears it can create new &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; under this protective layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we need oxygen for the key reaction to occur:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{O2 + 4e^- + 2H2O -&amp;gt; 4OH^-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this &lt;a href=&quot;http://science.howstuffworks.com/question445.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;how stuff works&lt;/a&gt; I have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How exactly is the water broken down into hydrogen and oxygen? From a chemistry textbook (Masterton &amp;amp; Hurley) I have:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, water does not spontaneously decompose to the elements by the reverse of the reaction referred to above&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2O(l) -&amp;gt; 2H2(g) + O2(g)}$$&lt;/span&gt; nonspontaneous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But electrolysis can certainly create these elements. But can it happen in this environment? Sealed under a surface of paint? What does the chemistry look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.chemicalformula.org/rust&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical formula&lt;/a&gt; I have that we can get hydroxide from:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{4e^- + 2H2O(l) + O2(g) -&amp;gt; 4OH^-(aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However still, from where does new oxygen get introduced into the system, assuming the atmosphere is actually completely isolated from this internal system taking place under the surface of the paint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it matter, assume the environment is approximately 1-atmosphere pressure and that the temperature varies between -10 to 40 degrees Celsius.&lt;/p&gt;&#xA;" OwnerUserId="6790" LastEditorUserId="402" LastEditDate="2020-01-27T22:02:06.620" LastActivityDate="2021-04-22T21:06:00.927" Title="Painting metal, does it matter if all rust is removed before?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19323" PostTypeId="1" AcceptedAnswerId="20440" CreationDate="2014-11-09T18:08:09.103" Score="6" ViewCount="31509" Body="&lt;p&gt;What is the definition of major, minor, and trace elements in a sample? I am unable to find this anywhere on the internet. Is it something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Major elements are those &gt; 0.1 %&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minor elements are those &gt; 100 ppm &amp;lt; 0.1 %&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trace elements are those &amp;lt; 100 ppm&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="73" LastEditDate="2014-11-09T21:24:43.900" LastActivityDate="2014-12-08T10:15:39.733" Title="Definition of major, minor, and trace elements in analytical chemistry" Tags="&lt;analytical-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19324" PostTypeId="2" ParentId="19322" CreationDate="2014-11-09T18:15:04.810" Score="4" Body="&lt;p&gt;Most paints will prevent oxygen and water to come into contact with the metal. Usually, rust under paint will allow the paint to break off. Some paints, like &lt;a href=&quot;http://www.rustoleum.com&quot; rel=&quot;nofollow&quot;&gt;Rust-Oleum&lt;/a&gt;, can be applied over rust, but most cannot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, with rust, you usually have an uneven surface which shows through the paint and gives an impression of a poorly done paint job.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-11-10T01:00:56.910" LastActivityDate="2014-11-10T01:00:56.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19325" PostTypeId="2" ParentId="19276" CreationDate="2014-11-09T18:24:14.200" Score="4" Body="&lt;p&gt;Although serotonin and dimethyltryptamine (DMT) have similar chemical structures (see Jori's answer), serotonin is not transported across the blood brain barrier (BBB) (&quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Serotonin&quot; rel=&quot;nofollow&quot;&gt;serotonin does not cross the blood–brain barrier&lt;/a&gt;&quot;).  On the other hand, not only does DMT cross the BBB, it is one of the few compounds that is &lt;strong&gt;actively transported&lt;/strong&gt; across the BBB.  While most compounds that cross the BBB do so by being lipid soluble (relatively small non-polar compounds) some molecules are assisted across the barrier through a variety of mechanisms.  Some of the known assistive paths include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;glucose carrier: transports glucose and several other sugars across the BBB&lt;/li&gt;&#xA;&lt;li&gt;phenylalanine carrier: transports a number of neutral amino acids across the BBB&lt;/li&gt;&#xA;&lt;li&gt;arginine carrier: transports several cationic amino acids across the BBB&lt;/li&gt;&#xA;&lt;li&gt;lactate carrier: transports several monocarboxylic acids across the BBB&lt;/li&gt;&#xA;&lt;li&gt;adenosine transporter: transports purine nucleosides as well as uridine&lt;/li&gt;&#xA;&lt;li&gt;organic cation transporter: transports a variety of cationic compounds, including monoamine neurotransmitters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are more classes and sub-classes within each.  The following two references provide nice, readable overviews&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC539320/&quot; rel=&quot;nofollow&quot;&gt;Small Molecular Drug Transfer across the Blood-Brain Barrier via Carrier-Mediated Transport Systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/jcbfm/journal/v32/n11/full/jcbfm2012126a.html&quot; rel=&quot;nofollow&quot;&gt;Drug transport across the blood–brain barrier&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't know which of the above transporters is active in the case of DMT, but I strongly suspect it is the &quot;organic cation transporter&quot;.  Here are some further references that relate specifically to DMT active transport across the BBB&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Takahashi T, Takahashi K, Ido T, Yanai K, Iwata R, Ishiwata K,&#xA;Nozoe S (1985) 11C-labeling of indolealkylamine alkaloids and&#xA;the comparative study of their tissue distributions. Int J Appl&#xA;Radiat Isot 36:965–969&lt;/li&gt;&#xA;&lt;li&gt;Yanai K, Ido T, Ishiwata K, Hatazawa J, Takahashi T, Iwata R,&#xA;Matsuzawa T (1986) In vivo kinetics and displacement study of&#xA;a carbon-11-labeled hallucinogen, N, N-[11C]dimethyltryptamine.&#xA;Eur J Nucl Med 12:141–146&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They were found in this article, &quot;&lt;a href=&quot;http://bitnest.ca/external.php?id=%257DbxUgPZC%251A%2507%2503%257D%257B%257F%2503%2507IYQ%2502TWv%257Dd%2515Um%257DA&quot; rel=&quot;nofollow&quot;&gt;Bufotenine: Toward an Understanding&#xA;of Possible Psychoactive Mechanisms&lt;/a&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  We've all read fairy tales where kissing a frog turns the frog into a prince.  When the skin of the Bufo toad (Colorado river toad, sonoran desert toad and other members) is licked, a psychedelic experience results.  It is not illegal in the US to own the toad, although the Bufotenine is a controlled substance!  Anyhow, just wondering if this, or something similar, is what the fairy tales are based on.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-09T18:55:02.083" LastActivityDate="2014-11-09T18:55:02.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19326" PostTypeId="1" CreationDate="2014-11-09T19:01:41.983" Score="3" ViewCount="373" Body="&lt;p&gt;A homework question asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The following reaction was carried out using ethanol or &#xA;  water as solvents, but gave poor yields in both cases. &#xA;  Suggest why this might be the case and how these &#xA;  problems could be overcome.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZW8ze.png&quot; alt=&quot;Deprotection hydrogenolysis&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've gathered that this is the deprotection step of a dipeptide synthesis reaction (Phe-Ser). The palladium catalyst and hydrogen gas remove the Cbz group from the n-terminal and the benzyl ester from the c-terminal, with a carbamic acid intermediate that spontaneously decarboxylates to the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand why ethanol or water would give poor yields as a solvent, since I've seen multiple references where they use exactly those as solvents for this reaction (or methanol, or THF). The only hint I could find was a suggestion that using acetic acid may be necessary when deprotecting a Cbz group from an amine, since protonating the nitrogen facilitates the reaction. A couple of references said that an aliphatic amine could poison the catalyst and require the addition of dilute acid to protonate the amine...I'm lost on why or how the poisoning occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nudge in the right direction would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="4945" LastEditDate="2019-04-25T14:20:27.560" LastActivityDate="2019-04-25T14:20:27.560" Title="Why would ethanol or water give poor yields in a peptide deprotection hydrogenolysis?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19328" PostTypeId="2" ParentId="19320" CreationDate="2014-11-09T19:15:26.290" Score="1" Body="&lt;p&gt;I think you have a typo in your calculation (also, using both dots and commas as the decimal point is rather confusing), but you're on the right track. The unfortunate thing about elemental analysis is that %, ppm, ppb, ppt, etc. get misused in a confusing way. These units are supposed to be used for only dimensionless ratios, the units you start with, ppm is actually mg/L which represents mass concentration, not mass ratio. Your calculations are correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(186.49 \frac{\mathrm{mg}}{\mathrm{L}}\right)\left(0.0142\mathrm{\ L}\right) = 2.648 \mathrm{\ mg}$$&#xA;gives the mass of the Mn in your sample, so when you divide by the sample mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2.648 \mathrm{\ mg}}{10.17\mathrm{\ mg}}=0.260$$&#xA;you get the mass fraction of Mn in your sample, and you know it's a mass fraction because you have mass units in both the numerator and the denominator, leaving you dimensionless. You can express it just like that, or you can convert it to 26.0% (and this is a dimensionless ratio, so it actually makes sense to use % here), or else express it as the mass of Mn in an arbitrary sample mass: 26.0 g/100 g.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-09T19:15:26.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19330" PostTypeId="2" ParentId="19322" CreationDate="2014-11-09T20:33:32.053" Score="10" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How exactly is the water broken down into hydrogen and oxygen?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That's extreme oversimplification. The reaction common for active metals (not including iron) is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Mg + 2 H2O -&amp;gt;  Mg(OH)2 + H2 ^ }$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Iron does not react this way. However, it does react with acids, say, with hydrochloric acids&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe + 2 HCl(aq) -&amp;gt; FeCl2(aq) + H2 ^}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reaction is quite slow even for concentrated nonoxidizing acids. However, in presence of oxygen and in diluted acids two subsequent reactions occur, the one noted above is followed by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{4FeCl2 + 10H2O + O2 = 4Fe(OH)3 v + 8HCl }$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The second reaction effectively removes iron from solution, significantly fastening the overall reaction.  However, even without acid present, even in slightly basic conditions iron slowly reacts with water in presence of free oxygen, forming rust. The process is hindered only at pH above 10 or at absence of free oxygen. Since natural water is usually slightly acidic due to presence of dissolved carbon dioxide, iron always rust on contact with water and atmosphere. This, actually, is a reason why concrete layer over steel frame is regulated: concrete always contains water, and surface layers quickly looses basicity due to reaction with atmospheric carbon dioxide, so steel near surface of concrete quickly rusts, increasing in volume and tearing the concrete from inside.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the context of automobile repair work, assume a metal surface has&#xA;rust over it. Does it matter if all of the rust is removed before the&#xA;metal surface is painted? If it does matter, then why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, it does. Rust is weak, so any paint over it will hold weak. As long as the paint film over rusty part is broken, water and atmosphere contacts the metal and start to erode it, resulting in more paint film be destroyed. So it is critical to remove all rust from the surface being painted. Abrasive blasting, if you have the equipment, is probably the best way. While some paints may be applied over rust, it should be a very thin layer, and even then I would prefer traditional paint over cleaned surface anyway.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-08-03T09:04:12.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19332" PostTypeId="1" AcceptedAnswerId="19335" CreationDate="2014-11-09T21:02:21.580" Score="2" ViewCount="109" Body="&lt;p&gt;My professor wrote an electron configuration for carbon as: 1s&lt;sup&gt;2&lt;/sup&gt; (sp&lt;sup&gt;3&lt;/sup&gt;)&lt;sup&gt;4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it was just 1s&lt;sup&gt;2&lt;/sup&gt; sp&lt;sup&gt;3&lt;/sup&gt; where did the &lt;sup&gt;4&lt;/sup&gt; come from? &lt;/p&gt;&#xA;" OwnerUserId="9708" LastEditorUserId="69830" LastEditDate="2021-05-20T07:02:00.413" LastActivityDate="2021-05-20T07:02:00.413" Title="Why is there an exponent 4 after the brackets in sp3?" Tags="&lt;orbitals&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19334" PostTypeId="1" AcceptedAnswerId="19352" CreationDate="2014-11-09T21:09:31.167" Score="2" ViewCount="2007" Body="&lt;p&gt;Why does a collision between an excited helium molecule ($\ce{He^*}$) and a ground-state helium atom create $\ce{He2^*}$?&lt;/p&gt;&#xA;" OwnerUserId="9709" LastEditorUserId="16683" LastEditDate="2016-11-29T10:38:15.713" LastActivityDate="2016-11-29T10:44:20.077" Title="How can an excited helium dimer be formed?" Tags="&lt;molecular-orbital-theory&gt;&lt;noble-gases&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19335" PostTypeId="2" ParentId="19332" CreationDate="2014-11-09T21:43:50.253" Score="7" Body="&lt;p&gt;The electron configuration of carbon is canonically written as 1s&lt;sup&gt;2&lt;/sup&gt;2s&lt;sup&gt;2&lt;/sup&gt;2p&lt;sup&gt;2&lt;/sup&gt;. What you now can do is invoke &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;noreferrer&quot;&gt;orbital hybridization&lt;/a&gt; and combine the electrons in the 2s and 2p orbitals. Then you get sp&lt;sup&gt;3&lt;/sup&gt; hybrid orbitals with a total of 4 electrons in them. The 1s electrons stay where they are. To mark the new hybrid orbitals as separate, you put brackets around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then end up with an electronic configuration of &lt;strong&gt;1s&lt;sup&gt;2&lt;/sup&gt;(2sp&lt;sup&gt;3&lt;/sup&gt;)&lt;sup&gt;4&lt;/sup&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-11-09T21:43:50.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19337" PostTypeId="1" CreationDate="2014-11-09T23:05:12.330" Score="1" ViewCount="3109" Body="&lt;p&gt;What if the concentration of an unknown is higher than the range of concentration used when you were doing an Absorption v. Concentration graph? How can you figure out the concentration of an unknown solution, basically, if it falls above the range? I have literally no idea how to go about doing it.&lt;/p&gt;&#xA;" OwnerUserId="9512" LastActivityDate="2014-11-10T19:05:47.727" Title="How to determine concentration of an unknown solution using Beer's Law?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19338" PostTypeId="1" CreationDate="2014-11-10T00:02:50.990" Score="3" ViewCount="816" Body="&lt;p&gt;Heating a protein/nucleic acid will disrupt inter-molecular and intra-molecular forces in the tertiary structure. It will also interfere with the shape of the active site if the protein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this apply with cooling also?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="69830" LastEditDate="2021-05-04T09:20:21.507" LastActivityDate="2021-05-04T09:20:21.507" Title="Denaturing of Proteins and Nucleic Acids - Effect of Temperature (Heating vs Cooling)" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;heat&gt;&lt;proteins&gt;&lt;dna-rna&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19340" PostTypeId="1" AcceptedAnswerId="19341" CreationDate="2014-11-10T01:27:24.073" Score="3" ViewCount="4807" Body="&lt;p&gt;Not all may be favorable to this project, but I will explain what I am trying to do. I work at home, and instead of walking a moderate distance to the bathroom and loosing my focus, I've been, at times, peeing in a 3 Quart Poland Springs water bottle. If you take offense at this, please do not continue reading except to be helpful in the scientific goal. I know this subject won't suit many types of people, so just ignore it if that is your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed first of all that urine is not at all as sterile as people say that it is. The rate of growth of bacteria is relatively slow, but as a precaution, I found the need to use additional measures to prevent the growth of bacteria. I settled on the following method: I have two bottles and I add to each bottle about enough salt as can be soluble in the urine and sometimes maybe a little more. The one bottle then fills up throughout the day and is emptied, washed, and refilled with salt. The salt helps to kill the bacteria which would be lingering in the empty bottle. The next day, the bottle stays empty and the other is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add that I discovered that the bacteria (without the salt) does not usually grow unless the bottle is left with urine for two days. After this, however, that same bottle (without the salt) would retain the bacteria and immediately grow, if used again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system works relatively well, so long as it is done every day. It will even withstand 2 days with only moderate growth. (If I should leave it by mistake for longer it can get ugly). Nevertheless, I am still looking to improve upon this. One reason is that, if I drink less water or relieve myself normally, the bottle does not fill in one day. I am looking for someone with knowledge of chemistry to help me find a substance that can be added to this solution which fits a number of common sense criteria. I will also add a list of the substances that I have tried or already considered.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Necessary qualities&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Safe to be handled and mistakenly touched. Implied that it will not eat through the plastic bottle. I can obtain a suitable glass bottle only with difficulty.&lt;/li&gt;&#xA;&lt;li&gt;Something that will not kill a dog or a human through trace amounts, should they drink from the (clean) toilet bowl. For example we put ammonia in toilets, flush it, no problem.&lt;/li&gt;&#xA;&lt;li&gt;Very important - something that will not react with the urine to produce gas, thereby causing the bottle to explode or expand. Something that will not (implied) produce a distinctively offensive smell. An example of this would be chlorine.&lt;/li&gt;&#xA;&lt;li&gt;(Obvious) That it will be suitable to flush down the toilet.&lt;/li&gt;&#xA;&lt;li&gt;Something that is not too expensive, physically large, or heavy in the quantity that I intend to use it. An example of this would be diluted vinegar.&lt;/li&gt;&#xA;&lt;li&gt;Something not involving much preparation, easy to obtain, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Substances considered&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vinegar: A moderate amount of vinegar did not have any effect. More than this would be difficult to obtain and store. Even at 5x the strength, I think it would still require a large amount, and this is not sold in stores that I visit.&lt;/li&gt;&#xA;&lt;li&gt;Chlorine/Bleach: I understand that this substance reacts with (things in) the urine and produces gas. It would produce a distinctive pool odor and I believe it can also cause burns.&lt;/li&gt;&#xA;&lt;li&gt;Ammonia: Same issues as chlorine - it would produce an offensive gas that burns my eyes.&lt;/li&gt;&#xA;&lt;li&gt;Acetic acid: similar to vinegar, except dangerous to handle. I don't even know that it would work.&lt;/li&gt;&#xA;&lt;li&gt;L-ascorbic acid (Vitamin C): probably dangerous to handle in a concentrated form and I don't know that it would work. Basically the same as vinegar. I know that it is used as a preservative, so it came on this list.&lt;/li&gt;&#xA;&lt;li&gt;Alcohol - might work, smell might be tolerable. I would need to know more about what concentration and amount to buy/use.&lt;/li&gt;&#xA;&lt;li&gt;Hydrochloric Acid - no&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a list of preservatives on &lt;a href=&quot;https://en.wikipedia.org/wiki/Preservative&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, and one that just now caught my eye was Sodium Propionate ($20/lb on eBay). Might not be a bad idea; maybe someone knows about it. I would much prefer something that could be bought in a store. Since this urine won't ever be consumed by humans, we should have more options than common preservatives - maybe I/we could push the PH range out to one extreme to inhibit all growth. Someone more knowledgeable could tell me how/if that could work without becoming dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I think would be something that stops the bacteria growth completely. Next best would be something that gives me an extra day or two if I need it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="9712" LastEditorUserId="9712" LastEditDate="2014-11-10T14:52:56.760" LastActivityDate="2019-01-14T02:25:01.817" Title="Storage of Urine" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;alcohols&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19341" PostTypeId="2" ParentId="19340" CreationDate="2014-11-10T02:12:11.107" Score="3" Body="&lt;p&gt;Sodium &lt;a href=&quot;http://www.ebay.com/itm/like/160721609539?lpid=82&quot; rel=&quot;nofollow&quot;&gt;Propionate&lt;/a&gt; or Sodium &lt;a href=&quot;http://www.artfire.com/ext/shop/product_view/chemistryconnection/2092766/potassium_sorbate_preservative_1_pound/supplies/handmade_supplies/bath___beauty&quot; rel=&quot;nofollow&quot;&gt;Sorbate&lt;/a&gt; are both suitable for inhibiting bacteria growth and would not pose any problem (unless consumed in large quantities). The metal ion (calcium, potassium, sodium) associated with the organic acid is not very important. I'm not sure how much you would need to use, but a teaspoon should be sufficient. If you search for &quot;food grade ...&quot;, you will be able to shop from your browser.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-10T02:12:11.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19342" PostTypeId="2" ParentId="19340" CreationDate="2014-11-10T03:03:05.417" Score="3" Body="&lt;p&gt;I use potassium metabisulfite, sodium sorbate, or sodium benzoate as preservatives in winemaking.  I have to think it would work here as well.  All three should be readily available from a homebrewing or winemaking shop, and they are fairly cheap.  They can also be purchased online fairly easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazon sells sodium benzoate for $14.00 per pound, which should be month's worth of preservative.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-11-10T03:03:05.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19344" PostTypeId="1" AcceptedAnswerId="19346" CreationDate="2014-11-10T03:24:40.347" Score="-1" ViewCount="461" Body="&lt;p&gt;Whenever a nuclear transmutation happens, do we get a new element or do we have the same element with chemical properties of a different element?&#xA;Isn't it very much against the intuition?&lt;/p&gt;&#xA;" OwnerUserId="9713" LastActivityDate="2014-11-10T05:11:17.840" Title="Do elements actually transmute in nuclear reactions?" Tags="&lt;nuclear&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19346" PostTypeId="2" ParentId="19344" CreationDate="2014-11-10T05:11:17.840" Score="4" Body="&lt;p&gt;By definition, a &lt;i&gt;&lt;a href=&quot;http://www.britannica.com/EBchecked/topic/602997/transmutation&quot; rel=&quot;nofollow&quot;&gt;nuclear transmutation&lt;/a&gt;&lt;/i&gt; involves transformation of one element into another. This means the element has undergone some sort of change in the nucleus (either by nuclear reaction or nuclear decay), hence a change in atomic weight and/or number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not this goes against &lt;i&gt;intuition&lt;/i&gt; depends upon what experiences one's intuition is based on.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-11-10T05:11:17.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19348" PostTypeId="1" CreationDate="2014-11-10T09:57:38.027" Score="2" ViewCount="1463" Body="&lt;p&gt;The textbook colour reaction of Marquis reagent (mixture cc. $\ce{H2SO4}$ + formaldehyde, optionally with some methanol to slow down polymerization) with MDMA (3,4-methylenedioxy-methamphetamine, the party drug &quot;ecstasy&quot;) is a gradually appearing dark purple/blue colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently three different samples were examined with marquis. The samples were taken in different countries, at different times. They looked a bit different too, one was light brown, the other two were clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLC results suggested they are the same compound with $R_f$ values that of MDMA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction with Mecke reagent was also MDMA-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction with Marquis started OK, with fizzing (a sign of a HCL salt) a light and then dark purple colour appeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After maybe a minute or two, the dark colour started to fade, it turned into light purple. As the colour faded even more, it turned green, and then to light yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test was repeated with two different reagents and all three samples, the results were the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found no documented nor anecdotal record of marquis turning green/yellow in the presence of MDMA after a few minutes.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this normal? Should Marquis turn green/yellow after a few minutes in the presence of MDMA?&lt;/li&gt;&#xA;&lt;li&gt;If this is really unusual, then what could possible cause fading &amp;amp; color change?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9583" LastEditorUserId="5456" LastEditDate="2014-11-10T11:25:09.220" LastActivityDate="2014-11-10T11:25:09.220" Title="Odd color reaction with Marquis and MDMA" Tags="&lt;drugs&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19349" PostTypeId="1" AcceptedAnswerId="19357" CreationDate="2014-11-10T11:00:35.463" Score="1" ViewCount="16914" Body="&lt;p&gt;first of all sorry if I have some mistakes in my &quot;chemistry words&quot;, I'm Iranian and I know the Persian terms but only some of the English ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose we have an element (x) with the atomic number of 24, so the electron configuration would be something like this : &#xA;$_{24}x : 1s^2\ \ 2s^2 \ \ 2p^6\ \ 3s^2\ \ 3p^6\ \ 3d^{5}\ \ 4s^1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is : in the element above, how many electrons are there in the subshell of l=2 (which is the subshell of d) and how many are there in the l=0 subshell (which is the subshell of s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I just took an exam and this exact question was in the exam, my answer was 5 electrons for 'd' and 7 electrons for 's'&lt;/p&gt;&#xA;" OwnerUserId="9714" LastEditorUserId="5456" LastEditDate="2014-11-10T11:24:54.070" LastActivityDate="2017-09-17T04:39:26.357" Title="how many electrons are in the second subshell of this element?" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19351" PostTypeId="2" ParentId="19340" CreationDate="2014-11-10T11:11:19.470" Score="4" Body="&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In this answer I will present to you several ways of optimizing your method, quickly summarized:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;UV irradiation&lt;/li&gt;&#xA;&lt;li&gt;Better Sanitization of the Bottles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Wait, shouldn't urine be sterile?&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I noticed first of all that urine is not at all as sterile as people say that it is.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has long been a myth that urine is sterile, but it was &lt;a href=&quot;https://www.sciencenews.org/blog/gory-details/urine-not-sterile-and-neither-rest-you&quot; rel=&quot;nofollow&quot;&gt;debunked this year&lt;/a&gt; in a study. There are in fact low amounts of bacteria present in urine. That being said, a high bacterial load in urine is an indication of an infection in the urinary tract and should not be left unchecked.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How do we know &lt;em&gt;what&lt;/em&gt; is growing in there?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I don't know how you checked what kinds of microorganisms are developing in the bottled urine, it might also be more fungal instead of bacterial. This might also affect the way to counter the problem, so without knowing exactly what is growing in there, the solutions might be worthless. For now I consider bacteria as the sole contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us proceed to the interesting part.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;UV Irradiation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's start off fancy. If you continue with your workflow of intermediate urine storage in bottles, it might be worth looking into solutions involving UV irradiation as a way to reduce the amount of microorganisms in the bottle. Something as simple as a UV diode screwed into the cap might make a huge difference. Maintenance will be low, but the higher initial cost could put you off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always worth considering long-term solutions.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Reduce Contamination of Bottle&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I don't know how you prepare the bottles prior to peeing in them, but you might well reduce the problem by simply sanitizing the bottle properly. One way would be to boil the bottle in hot water for &lt;a href=&quot;http://textbookofbacteriology.net/control.html&quot; rel=&quot;nofollow&quot;&gt;about 30 minutes&lt;/a&gt;. However, this is quite time consuming and thus defeating the purpose of the exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method would be to rinse it well with (warm) water and diluted ammonia (which you then can flush down the toilet) and then swirling some sanitizing solution (think of the commonly available hand sanitizers) around the bottle, even leaving that in there. An advantage of this method is that hand sanitizers usually contain fungicides and viricides, so you kill several birds with one stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if that already is sufficient to guarantee a two-day period where bacterial growth is inhibited enough that you will not observe it, but it might be well worth checking out, as hand sanitizers are widely available and should keep the cost of maintenance low.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="3846" LastEditDate="2014-11-11T18:37:52.927" LastActivityDate="2014-11-11T18:37:52.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19352" PostTypeId="2" ParentId="19334" CreationDate="2014-11-10T11:26:20.243" Score="4" Body="&lt;p&gt;The electronic configurations are $\mathrm{(1s)^2}$ for $\ce{He}$ and $\mathrm{(1s)^1(2s)^1}$ for $\ce{He^*}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put this information in an energy diagram and see what the molecular orbitals look like, especially how they are filled. (Further hint: Filled-unfilled orbital interactions are generally favorable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached a graphic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qiQru.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qiQru.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a bond order of $\frac{1}{2}(2-1+1) = 1$, and therefore $\ce{He2^*}$ is stable with respect to $\ce{He + He^*}$. The excited dimer is known as an &lt;em&gt;excimer&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside: The relative energy levels of the molecular orbitals are very hard to get right out of the blue, so take them with a grain of salt. As was kindly pointed out by &lt;a href=&quot;https://chemistry.stackexchange.com/users/4945/martin&quot;&gt;Martin&lt;/a&gt;, the excitation of helium will initially form a singlet excited configuration, depicted above. This should be able to relax to a triplet excited state.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2016-11-29T10:44:20.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19353" PostTypeId="1" CreationDate="2014-11-10T11:32:59.090" Score="1" ViewCount="61" Body="&lt;p&gt;I have performed an element analysis, and I I have calculated the amount of each metal in my sample, given in milligrams. Let's say I weighed in 10 mg of the powder. I want to see how much of that I have accounted for by the major components in my sample (those &gt; 1-2 %).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my metals were not free metals in my sample, but rather as various oxides, I need to convert back to the most common oxides. This is the idea:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If I have 1 mg $\ce{Mn}$ in my sample, how many milligrams would that be if I&#xA;  have $\ce{MnO2}$ instead?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should I use the number of atoms? Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 0.002648 g of $\ce{Mn}$ = 4.82×10&lt;sup&gt;-5&lt;/sup&gt; moles = 2.90×10&lt;sup&gt;19&lt;/sup&gt; atoms. I then assume I have 2.90×10&lt;sup&gt;19&lt;/sup&gt; molecules of $\ce{MnO2}$. Going backwards, this gives me 4.815×10&lt;sup&gt;-5&lt;/sup&gt; moles = 4.187 mg. Would this be the correct way of doing it?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="39539" LastEditDate="2017-02-14T21:02:26.847" LastActivityDate="2017-02-14T21:02:26.847" Title="How report metal amounts as amount of their respective metal oxides?" Tags="&lt;analytical-chemistry&gt;&lt;concentration&gt;&lt;mole&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19354" PostTypeId="1" CreationDate="2014-11-10T12:34:42.820" Score="1" ViewCount="1154" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Choose the two correct options from A-F, why isn't it commercially suitable to dissolve $\ce{SO3}$ directly in water to give concentrated sulfuric acid instead of using the contact process? The major reaction of the contact process are &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{S + O2 -&amp;gt; SO2}$$ &#xA;  $$\ce{2 SO2 + O2 &amp;lt;=&amp;gt;2 SO3}$$&#xA;  $$\ce{SO3 + H2SO4 -&amp;gt; H2S2O7}$$&#xA;  $$\ce{H2S2O7 + H2O -&amp;gt; 2H2SO4}$$&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A. To reduce the hazards of spillage&lt;/li&gt;&#xA;  &lt;li&gt;B. Because the density of the product is too high&lt;/li&gt;&#xA;  &lt;li&gt;C. To minimise transport costs of large volumes&lt;/li&gt;&#xA;  &lt;li&gt;D. Because the last step is too expensive.&lt;/li&gt;&#xA;  &lt;li&gt;E. Because the last step of the process is too exothermic.&lt;/li&gt;&#xA;  &lt;li&gt;F. An aerosol of the sulfuric acid rapidly fills the container.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is C and F, but I think it should be A and D. Is it because to make gas react with $\ce{H2O}$, too much pressure is needed and it will be expensive? Can someone explain why each choice is false and true?&lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="194" LastEditDate="2017-11-03T20:59:15.427" LastActivityDate="2017-11-03T20:59:15.427" Title="Why isn't it commercially suitable to dissolve sulfur trioxide in water to produce sulfuric acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19355" PostTypeId="1" AcceptedAnswerId="19362" CreationDate="2014-11-10T12:45:23.010" Score="0" ViewCount="2903" Body="&lt;p&gt;How do I balance this reaction using half-redox equations? &#xA;$$\ce{2SO2 + O2-&amp;gt;2SO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: $\ce{SO2 + H2O -&amp;gt; SO3 + 2e- + 2H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: $\ce{H2O + 2e- + O2 -&amp;gt; SO3 + 2H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total: $\ce{SO2 + 2H2O + O2 -&amp;gt; 2SO3 + 4H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT it's false. What did I do wrong?&lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="3846" LastEditDate="2014-11-10T12:50:39.037" LastActivityDate="2014-11-10T16:21:23.780" Title="How can I balance the redox reaction of sulfur dioxide with oxygen?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19357" PostTypeId="2" ParentId="19349" CreationDate="2014-11-10T14:40:30.327" Score="2" Body="&lt;p&gt;You're answer is correct. The s-shell has 2 electrons in the 1s subshell, 2 electrons in the 2s subshell, 2 electrons in the 3s subshell and 1 electron in the 4s subshell. The d-shell has 5 electrons in the 3d subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The element with atomic number 24 is chromium, who's electron configuration can be shown diagramatically as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mcT3w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the 3d shell is 'half-filled' before the 4s subshell is completely filled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromium is one of the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations?rq=1&quot;&gt;'exceptions to the predicted electron configurations'&lt;/a&gt; when applying Madelung's rule for &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantum numbers and electron configuration&lt;/a&gt;. The 'Aufbau' principle (represented in diagram below) can be used to determine the electron structure of relatively simple atoms, by hand, as long as one remembers the anomalies (such as chromium and copper). The 'anomaly' is typically 'explained' by stating that the a 'half-filled' or 'completely filled' d-subshell is more stable and will therefore happen before 'completely filling' the 4s subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I2X5a.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modern quantum computational chemistry, various numerical methods (such as Dirac Hyper Hartree–Fock method and density functional theory) have been devised to iteratively solve the Schrodinger equation for more accurate calculation of electronic structures of larger atoms and molecules.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-11-10T14:40:30.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19358" PostTypeId="1" AcceptedAnswerId="19376" CreationDate="2014-11-10T14:45:24.653" Score="1" ViewCount="22205" Body="&lt;p&gt;I have a question that asks if the buret tip is not filled before the experiment what would happen (effect on molarity for acid/base titration). The way I'm thinking about it, it's not like you're putting a pre measured solution into it. You fill it as high as you fill it and you can't put anything in without getting the tip filled up too, so I don't see where that will change anything. Am I thinking about that right?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="865" LastEditDate="2021-05-22T21:05:05.363" LastActivityDate="2021-05-22T21:05:05.363" Title="Filling a buret tip before an experiment" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19362" PostTypeId="2" ParentId="19355" CreationDate="2014-11-10T16:21:23.780" Score="1" Body="&lt;p&gt;WELL I think you put something erroneous in your reduction formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will never get SO3 out of just O2 and water. Instead of SO3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction formla will be 4e- + O2 + 4H+ --&gt; 2 H2O  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply oxidation formula by 2 and add it to new reduction formula and you got your answer&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2014-11-10T16:21:23.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19363" PostTypeId="2" ParentId="19354" CreationDate="2014-11-10T16:41:03.703" Score="0" Body="&lt;p&gt;A is false because even if you are doing it by contact process the hazard of spillage would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is false because the product in both processes are the same. I don't know that there is a problem with high density material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is true because sulfur is solid and SO3 is gas and gases can occupy a large volume when same no. of moles are taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D is false because if the 2nd and 3rd step have same condition requiring high pressure to force its equilibrium forward according to Le Chattelier Principle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E is false because      SO3(g) + H2O(l) ---&gt; H2SO4(aq), ......ΔH = -227.8 kJ &#xA;                        S(s)+O2(g)--&gt;SO2(g), ......ΔH = -296.8 kJ ( even more exothermic )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F is true as it is only other option left given that two answers have to be selected and also it is a fact which might be explainable, but I cannot explain with tools in my hand at present.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="402" LastEditDate="2016-12-04T21:01:46.523" LastActivityDate="2016-12-04T21:01:46.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19364" PostTypeId="2" ParentId="19211" CreationDate="2014-11-10T16:46:53.970" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two sign conventions: (1) the more common one states that a positive&#xA;  electron affinity value represents energy release when an electron is&#xA;  added to an atom; (2) the other states that a negative electron&#xA;  affinity represents a release of energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recommended values for these electron affinities, in the units&#xA;  commonly employed in introductory texts and with the sign convention&#xA;  used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba,&#xA;  respectively. The endothermic electron affinities often quoted for Be&#xA;  and Mg are also too large and should be reported simply as &quot;&gt;0&quot;. An&#xA;  argument for a return to the original sign convention for the electron&#xA;  affinity is presented in this paper.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed074p123&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ed074p123&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OP: I know that the electron will be settled in the 3d orbital &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, as explained in the above article, the electron occupies 4p.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OP: but still, why would energy be released?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments.  The reference &lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/p92-209?journalCode=cjp&quot; rel=&quot;nofollow&quot;&gt;Contributions to the electron affinity of calcium and scandium&lt;/a&gt; may be the best source for theoretical models that do and don't predict a slight energy release.  According to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0370157304000316&quot; rel=&quot;nofollow&quot;&gt;Atomic negative ions: structure, dynamics and collisions&lt;/a&gt;, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-.  Early calculations that considered only valence electrons predicted that Ca- was unstable.  &lt;a href=&quot;http://journals.aps.org/pra/abstract/10.1103/PhysRevA.62.052518&quot; rel=&quot;nofollow&quot;&gt;Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+&lt;/a&gt; presents detailed calculations of energy levels.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-11-10T19:37:00.037" LastActivityDate="2014-11-10T19:37:00.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19365" PostTypeId="1" CreationDate="2014-11-10T16:47:37.990" Score="12" ViewCount="5783" Body="&lt;p&gt;In our course on physical chemistry, which involves MOT, we have been taught that in the LCAO approach, the wave function for a molecule … say hydrogen ion ($\ce{H2+}$), can be approximated by a linear combination of two atomic orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I could gather from the very little I know about variational method is that this linear combination is going to be then optimised to give the minimum energy to find the ground state of the molecule in question. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if we are simply using the linear combination as a test/guess function for a variational method, why should we restrict ourselves to only (say $+$ and $-$) two combinations?&#xA;That is, how does the rule &quot;Number of molecular orbitals = Number of atomic orbitals constituting them&quot; arise? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why is there the restriction in form of the rule that &quot;Only atomic orbitals of close energies and proper symmetries combine to give MOs&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9349" LastEditorUserId="7475" LastEditDate="2015-12-15T22:40:02.647" LastActivityDate="2015-12-16T01:54:59.440" Title="The &quot;rules&quot; for LCAOs in Molecular Orbital Theory" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19366" PostTypeId="2" ParentId="19354" CreationDate="2014-11-10T17:43:14.253" Score="2" Body="&lt;p&gt;Part of the difficultly with this question is that it is asking about what is commercially viable as opposed to what is merely possible. It is, of course, quite possible to dissolve SO$_3$ in water, but doing so creates a runaway chemical reaction. So let's go through each of the points and explain why they don't make &lt;em&gt;commercial&lt;/em&gt; sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) The contact process produces SO$_3$, as well as several other noxious gasses anyway, so they is no reduced risk as regards spillage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) The density of the product is totally irrelevant, as you get the same product in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) Sulphur is a solid at standard temperature and pressure (STP), so one can transport a large amount using a standard container and transport vehicle. As a gas at STP, one can transport far less (many millions of moles less) SO$_3$ at STP for the same size container, so an unfeasibly large container would be needed unless a highly pressurized or very low temperature container is used. Either method is considerably more expensive, so less commercially viable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D) The last step of the contact process is to react oleum (H$_2$S$_2$O$_7$) with water. This is in fact relatively safe to do if done slowly and water is extremely cheap, so the process is not expensive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(E) The reaction of oleum and water is exothermic. However, in practice water is added relatively slowly to the oleum (both of which are controllable liquids), which controls the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(F) As said earlier, the reaction of water with SO$_3$ gas cannot be easily controlled. It runs away with itself, and the water starts to boil. This throws up lots of small droplets (an aerosol) of conc. sulfuric acid. This is highly dangerous, can increase the pressure in the reaction chamber, and also makes it difficult to cheaply extract the product from the reaction chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, on economic grounds, the answer is indeed C and F.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastActivityDate="2014-11-10T17:43:14.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19367" PostTypeId="2" ParentId="19358" CreationDate="2014-11-10T17:53:47.370" Score="0" Body="&lt;p&gt;I'm assuming that the question is talking about leaving air in the tip of the buret once you've filled it. In that case you would actually be adding less solution than you would have measured because the first half mL or so would go into filling the tip. &lt;/p&gt;&#xA;" OwnerUserId="5234" LastActivityDate="2014-11-10T17:53:47.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19368" PostTypeId="2" ParentId="19334" CreationDate="2014-11-10T17:59:37.983" Score="1" Body="&lt;p&gt;In any chemical collision process, one of the most important physical parameters to consider is the angular momentum. Throughout a simple chemical reaction, the total angular momentum of the system is preserved. In the case of a simple diatomic reaction, we need to conserve both the spin and the orbital angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the ground state of orthoheium He (1s)$^2$. This atom has two spin paired s-electrons. Therefore the total spin is 0 and the total orbital angular momentum is also 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider now the ground state of parahelium, He(1s)(2s). This atom should have (in the simple case) have two same spin electrons in each of the 1s and 2s orbitals. Therefore the total spin is 1, and the orbital angular momentum is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need to create a final molecular orbital picture with orbital angular momentum 0 (easy: all $\sigma$ orbitals) and a total spin of 1. This can only be achieved if our final molecule also has two unpaired, aligned spins. The lowest energy configuration possible is therefore (1$\sigma_g$)(1$\sigma_g$)(1$\sigma_u^*$)(2$\sigma_g$), which is indeed the excited state helium dimer He$_2^*$.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastEditorUserId="9720" LastEditDate="2014-11-21T08:32:04.520" LastActivityDate="2014-11-21T08:32:04.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19369" PostTypeId="2" ParentId="19365" CreationDate="2014-11-10T18:45:13.627" Score="12" Body="&lt;p&gt;You are right to be questioning the validity of this method, and I congratulate you for doing so. This is actually an extremely important skill, that differentiates the best students from the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variational method, like many methods from physical chemistry, is a method of approximation (a model) to what really happens. There exist a whole hierarchy of methods for computing the molecular orbitals of molecules (which are themselves models, being the stationary states of the Schrödinger equation), including at the top end post-Hartree-Fock theory and density functional theory. These methods provide quantitative information about molecular orbitals, but also require serious computer power (which is generally unavailable to undergraduate students). However, all of the basic physics can be explored and understood with simpler models that can be solved on a few sheets of paper, such as LCAO theory, hence why we teach them to undergraduates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As regards to your question, we do indeed optimise a linear combination. We start off by &lt;em&gt;assuming&lt;/em&gt; that the total molecular orbital wavefunction can be approximated using a linear set of atomic orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\Psi\rangle = c_1|\phi_1\rangle + c_2|\phi_2\rangle + \cdots + c_n|\phi_n\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then need to find the coefficients. The variational principle (also known as the Rayleigh-Ritz method) states that the coefficients that give the best approximation to the wavefunction will minimize the energy, given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{E} = \frac{\langle\Psi|\hat{H}|\Psi\rangle}{\langle\Psi|\Psi\rangle}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the second part of your question involves how to compute the two terms in this fraction. Without going into massive mathematical detail, the LCAO method can be recast into a matrix problem rather than an integral problem using the atomic orbitals as a form of basis vector. In the simplest case (Hückel theory) we assume that they are normalized, such that the denominator is always 1 or 0 (see later). The problem now is mostly how to determine the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, each element of the Hamiltonian matrix is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variational principle applied to this matrix implies that the optimized energies are the eigenvalues of the Hamiltonian matrix, and the coefficients are given by the eigenvectors. Since an $n$ x $n$ matrix has precisely $n$ eigenvalues, this implies the &quot;number of molecular orbitals&quot; rule. Notice that the reason for the +/- combinations is by our own design (LCAO). We could have picked more complex trial functions by assumption, but this would make computation far more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It most certainly is true that in order for there to be a significant interaction, two orbitals must be close in energy. The detailed reasons are complex, but essentially it comes down to the size of the $\langle\phi_i|\hat{H}|\phi_j\rangle$. Orbitals that are far apart in energy have small values of this term hence interact weakly. This doesn't necessarily mean they can't form molecular orbitals, however, but these effects are negligible and are not important in understanding the chemistry, so are neglected in simple models (although are often included in some of the most complex modern methods).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that only atomic orbitals of the same symmetry give molecular orbitals is the overlap integral $\langle\phi|\phi\rangle$. This is the total sum (integral) of the product of the AOs. This is zero for different symmetries. This must be non-zero to give an MO. Consider a $\mathrm{2p_z}$-orbital and a $\mathrm{1s}$-orbital for instance (for illustration purposes only!). The s-orbital is spherically symmetric, with all points the same sign. The $\mathrm{p_z}$-orbital has a &quot;dumbbell&quot; shape, with equal areas of different signs above and below the $xy$-plane. Therefore, the total sum of the product is zero. No overlap = no interaction.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastEditorUserId="16683" LastEditDate="2015-12-16T01:54:59.440" LastActivityDate="2015-12-16T01:54:59.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19370" PostTypeId="1" AcceptedAnswerId="19377" CreationDate="2014-11-10T18:57:08.040" Score="7" ViewCount="5014" Body="&lt;p&gt;Why is the methyl radical planar? The VSEPR theory would predict an angle between 120° and 109.5°, while it is actually 120°.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="41328" LastEditDate="2021-03-07T09:59:52.540" LastActivityDate="2021-03-07T10:02:42.397" Title="Why is the methyl radical planar?" Tags="&lt;molecular-structure&gt;&lt;radicals&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19371" PostTypeId="2" ParentId="19337" CreationDate="2014-11-10T19:05:47.727" Score="3" Body="&lt;p&gt;This can be an issue in many calibration experiments, and falls into the broader and extremely important field of &quot;design of experiments&quot; within the field of chemometrics. Clearly, your range of calibration data is not quite sufficient for your application, but you have several options available to you, each with pros and cons, some of which have been mentioned in the comments&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Go back to the calibration curve, and take some standard measurements at higher/lower concentrations than you did. Pros: this will be the most accurate method, and will help you to avoid this problem with future measurements. Cons: takes time and requires more measurements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As commented, extrapolate from your model. Pros: extremely quick. Cons: likely to be inaccurate unless you are 100% sure Beer's Law holds for your sample at all concentrations. There is also no way of estimating the errors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As commented, dilute your sample with a known quantity of solvent and remeasure. Then multiply your answer by the dilution factor. Pros: Fairly accurate, takes less time than performing more measurements for your model. Cons: Depending upon your lab skills and precision of your equipment, errors could start to propagate and become larger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="9720" LastActivityDate="2014-11-10T19:05:47.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19372" PostTypeId="2" ParentId="19319" CreationDate="2014-11-10T19:07:02.730" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;oxygen always -2 except in peroxides&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This exception is in effect here, because $\ce{BaO2}$ is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Barium_peroxide&quot; rel=&quot;nofollow&quot;&gt;peroxide&lt;/a&gt; consisting of $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Moreover, there are even more exceptions to the rule of thumb cited above, for example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Superoxide&quot; rel=&quot;nofollow&quot;&gt;superoxide&lt;/a&gt; radical anion $\ce{O2-}$ with a fractional oxidation numer of $-\frac{1}{2}$, or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dioxygenyl&quot; rel=&quot;nofollow&quot;&gt;dioxygenyl&lt;/a&gt; cation $\ce{O2+}$ with a formal oxidation number of $+\frac{1}{2}$ for each oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxide&quot; rel=&quot;nofollow&quot;&gt;peroxides&lt;/a&gt;, &quot;simple ionic compounds&quot; of oxygen are &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxide&quot; rel=&quot;nofollow&quot;&gt;oxides&lt;/a&gt;, which contain the anion $\ce{O^{2-}}$. The charge of the oxide anion is $2-$, so the oxidation number of oxygen in those compounds is -II.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2014-11-13T18:02:50.110" LastActivityDate="2014-11-13T18:02:50.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19373" PostTypeId="2" ParentId="19358" CreationDate="2014-11-10T19:14:01.413" Score="0" Body="&lt;p&gt;Yea, but in the lab we didn't measure any amount ahead of time, just filled it to the markings on the buret that we wanted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sory for the  annonymous post as an answer, away from my comp and coudn't remember my password to log in on this one.)&lt;/p&gt;&#xA;" OwnerUserId="9722" LastActivityDate="2014-11-10T19:14:01.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19374" PostTypeId="1" AcceptedAnswerId="19375" CreationDate="2014-11-10T19:17:07.447" Score="5" ViewCount="256" Body="&lt;p&gt;What am I actually changing when I click 631G(d) or 631G(d,p), etc. when I choose a &quot;basis&quot; in restricted Hartree Fock calculations? What is the difference between using an alternative basis?  Many thanks. &lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="5017" LastEditDate="2014-11-10T19:56:53.740" LastActivityDate="2014-11-10T19:56:53.740" Title="Gaussian Calculations - What is the Basis?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19375" PostTypeId="2" ParentId="19374" CreationDate="2014-11-10T19:42:04.127" Score="8" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;basis set&lt;/a&gt; is used to describe the molecular orbital wave functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest reading through a number of links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Basis_set_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/m_basis_sets.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Gaussian manual - Basis Sets&lt;/a&gt; - a nice description of the different basis sets, if a bit technical at times.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2492/naming-of-quantum-chemistry-basis-sets?rq=1&quot;&gt;Naming of quantum chemistry basis sets&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/16537/5017&quot;&gt;GAMESS Crash Course&lt;/a&gt; - applies to Gaussian or other packages&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-11-10T19:42:04.127" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19376" PostTypeId="2" ParentId="19358" CreationDate="2014-11-10T19:43:41.930" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You fill it as high as you fill it and you can't put anything in without getting the tip filled up too, &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No I don't think it will work like that. Usually you have the stock or valve closed at the bottom. Consequently, the tip will be empty until you bleed the buret. Further, you will often bleed (allow the titrant to flow) a buret while thumping the tip to remove any air bubbles that may form directly under the stop-cock when your titrant flows initially through the tip. Once you have titrant in the tip and the tip is devoid of air bubbles you will then close the valve and fill the buret to the desired level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point you can better determine just how much you have used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measure marks on the buret itself take into account what is in the tip below the stop cock. But if you do not bleed the tip initially, filling it with your titrant then your first addition will be a bit off because the tip does not fill entirely with liquid before you allow some to drip through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csudh.edu/oliver/demos/buretuse/buretuse.htm&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a good write-up on titrations with burets.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A bubble in the nozzle of a buret will produce an inaccurate volume reading if the bubble escapes during a titration. Bubbles may be large and visible as shown above left or so small as not to be seen, above center. During a titration such small bubbles begin to move in the direction of the nozzle but may remain in place even though there is a moderate flow of titrant (above right). Even when the buret valve is wide open some bubbles remain in place until you take your eyes off them. Then they sneak through the nozzle and ruin your titration. Also if you let a an air bubble stay because you think it will remain and then later it leave to the tip you will have to start over because you have no idea what&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9403" LastActivityDate="2014-11-10T19:43:41.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19377" PostTypeId="2" ParentId="19370" CreationDate="2014-11-10T19:51:09.183" Score="5" Body="&lt;p&gt;This question unfortunately illustrates the significant shortcomings of the VSEPR model. This model is based upon classical electrostatic arguments. The molecular geometry will seek to maximize the distances between the repulsive electrons in the valence (outer) shell, that are completely localized in their chemical bonds and their lone pairs/singly occupied molecular orbitals.&lt;/p&gt;&#xA;&lt;p&gt;This model works well in some cases. However, when quantum mechanics comes into play, it falls pretty flat, as it totally neglects (the really important) quantum effects. A better way of finding the molecular geometry is to find the geometry that &lt;em&gt;minimizes the total energy&lt;/em&gt; of the molecular orbitals. In this case, quantum mechanical calculations find that a set of three &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized molecular orbitals arising from a &lt;span class=&quot;math-container&quot;&gt;$D_\mathrm{3h}$&lt;/span&gt; symmetry has a lower total energy than breaking the symmetry to form the pyramidal &lt;span class=&quot;math-container&quot;&gt;$C_\mathrm{3v}$&lt;/span&gt; geometry.&lt;/p&gt;&#xA;&lt;p&gt;However, this does not mean that all &lt;span class=&quot;math-container&quot;&gt;$\ce{\overset{.}{C}X3}$&lt;/span&gt; radicals follow this pattern. Put an extremely electronegative substituent on in place of &lt;span class=&quot;math-container&quot;&gt;$\ce{H},$&lt;/span&gt; e.g. in &lt;span class=&quot;math-container&quot;&gt;$\ce{CF3},$&lt;/span&gt; and the situation changes. The lowest energy geometry is now the pyramidal one.&lt;/p&gt;&#xA;&lt;p&gt;Sadly, quantum chemistry doesn't always care to follow general rules.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastEditorUserId="41328" LastEditDate="2021-03-07T10:02:42.397" LastActivityDate="2021-03-07T10:02:42.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19378" PostTypeId="1" AcceptedAnswerId="19383" CreationDate="2014-11-10T20:13:56.870" Score="2" ViewCount="2573" Body="&lt;p&gt;Methyl chloride belongs to the $C_{3\mathrm{v}}$ point group. This is because it has a $C_3$ axis down the $\ce{Cl-C}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However what I do not understand is how it has three vertical mirror planes? It has three planes down each of the $\ce{C-H}$ bonds, but don't these bisect a bond making them dihedral planes? Although I'm wrong I can't fathom why it isn't a $C_{3\mathrm{d}}$ point group. I have used a molecular viewer to visualise it but I still get the vertical mirror planes bisecting a bond...&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="41328" LastEditDate="2017-08-06T04:23:17.433" LastActivityDate="2017-08-06T04:23:17.433" Title="Symmetry C3V For MeCl" Tags="&lt;symmetry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19379" PostTypeId="2" ParentId="19318" CreationDate="2014-11-10T20:54:39.737" Score="10" Body="&lt;p&gt;Keep in mind that in these reversible reactions of carbonyl compounds, the mechanism for the forward and reverse reactions must go through the same intermediates. The result is that if you can't make a particular functional group under particular conditions, that functional group can't be destroyed under the same conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrates (as shown in the second reaction in &lt;a href=&quot;https://chemistry.stackexchange.com/a/19321/5026&quot;&gt;ron's answer&lt;/a&gt;) and hemiacetals can be formed/decomposed under either acidic or basic conditions. In the hemiacetal formation under basic conditions, an alkoxide (methoxide below) attacks the carbonyl. Protonation gives the hemiacetal. In order to proceed to the acetal, the alkoxide would have to displace the hydroxide in an S&lt;sub&gt;N&lt;/sub&gt;2 reaction. Hydroxide is not a good enough leaving group for S&lt;sub&gt;N&lt;/sub&gt;2 reaction, so it is impossible to form the acetal under basic/nucleophilic conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YHk7S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YHk7S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hemiacetals can be decomposed to a carbonyl and alcohol under basic conditions. Deprotonation of the hydroxyl starts the mechanism, and the resulting anionic oxygen can kick out an alkoxide as a leaving group to give the carbonyl compound. Note that in this carbonyl chemistry hydroxide and alkoxides are competent leaving groups in an anionic elimination mechanism, such as the one here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u1SXD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u1SXD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an acetal, without the free hydroxyl group, alkoxide/hydroxide has nothing to deprotonate and there are no suitable leaving groups for SN2 reaction. There are no reasonable pathways for a nucleophile or base to react with the acetal.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="5026" LastEditDate="2018-03-15T14:11:56.520" LastActivityDate="2018-03-15T14:11:56.520" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19380" PostTypeId="2" ParentId="19378" CreationDate="2014-11-10T20:56:47.360" Score="0" Body="&lt;p&gt;If I remember my symmetry correctly then a dihedral plane by definition is a vertical plane that bisects a perpendicular $C_2$ axis. Since we have a perpendicular $C_2$ axis then I think that would mean you label it $D_n$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case there are no perpendicular $C_2$ axes so I think for that reason they are just vertical planes until such time that you they bisect a perpendicular $C_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what would be the case if you had $C_n$ on the principal axis and then $m$ perpendicular $C_2$ axes, where $m\neq n$, maybe someone could shed some light? My guess would be that it wouldn't be $D$ but would qualify for the $\sigma_d$ plane. &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="5017" LastEditDate="2014-11-11T03:25:43.717" LastActivityDate="2014-11-11T03:25:43.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19381" PostTypeId="2" ParentId="19249" CreationDate="2014-11-10T22:05:46.387" Score="4" Body="&lt;p&gt;The problem with carbon capture is not so much that we lack the technology to do it, it's more that we lack the technology to do it in a way that is economically feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take your example - &lt;a href=&quot;http://www.eia.gov/tools/faqs/faq.cfm?id=667&amp;amp;t=6&quot; rel=&quot;nofollow&quot;&gt;according to the US Energy Information Adminstration&lt;/a&gt;, it takes about 1.09 lbs of coal to generate 1 kWh of electricity. A typical coal-fired plant might produce an average of 500 MW throughout each day (this is a rough estimate), which works out to roughly 12,000 MWh, or 12,000,000 kWh. So that's something like 12 million pounds of coal &lt;em&gt;each day&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that coal is 100% carbon (close but not quite), then it's about 37.8 mole C per pound of coal. That's 453.6 million moles of carbon &lt;em&gt;each day&lt;/em&gt; being converted to $\ce{CO2}$, and since the combustion of carbon follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C + O2 -&amp;gt; CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that means a &lt;strong&gt;single medium sized-coal plant&lt;/strong&gt; produces roughly 454 &lt;em&gt;million&lt;/em&gt; moles  of $\ce{CO2}$ each day. By now you are probably getting a sense of the scale of this problem. Since sodium peroxide and carbon dioxide have a 1:1 ratio in the absorption equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Na2O2 + 2CO2 -&amp;gt; 2Na2CO3 + O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will need an equal number of moles of sodium peroxide. That works out to about &lt;strong&gt;78 million pounds per day.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alibaba &lt;a href=&quot;http://www.alibaba.com/showroom/sodium-peroxide.html&quot; rel=&quot;nofollow&quot;&gt;lists sodium peroxide at around \$400-500 per ton&lt;/a&gt;, so that means it would cost you (at a minimum) &lt;strong&gt;\$15.6 million dollars per day&lt;/strong&gt; to scrub the $\ce{CO2}$ emissions from one medium-sized power plant. The real costs would be much higher because of the additional equipment and operating expenses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electricity &lt;a href=&quot;http://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a&quot; rel=&quot;nofollow&quot;&gt;currently sells for about \$0.10 to \$0.20 per kWh&lt;/a&gt; in the US, and around half of that is usually for production, with the rest being delivery. That means our 500 MW coal plant is going to be bringing in something like (at the high side) \$1.2 M each day. As you can see, the costs for scrubbing with sodium peroxide far exceed the potential revenue for the plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are other options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is to reuse the sodium peroxide, although the costs of regenerating it might exceed just buying more, and you would still need to find a place to store the $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is to use a cheaper/more effective absorbent - &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_dioxide_scrubber&quot; rel=&quot;nofollow&quot;&gt;according to Wikipedia&lt;/a&gt;, amine-gas scrubbing is the most promising technology so far, although I am sure it faces the same kinds of cost scaling problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another more popular approach is &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_capture_and_storage&quot; rel=&quot;nofollow&quot;&gt;carbon capture&lt;/a&gt;, where carbon dioxide is basically just pumped into underground reservoirs in the hopes that it will stay there. While this is probably going to turn out to be the most economical approach, it has the risk of releasing large amounts of $\ce{CO2}$ accidentally, as well as having unknown effects on the local geological formations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another option is to use photosynthesis to scrub carbon dioxide. This has the advantage of being very cheap in comparison to other methods, but the downside is that the rate at which plants and algae can consume $\ce{CO2}$ is much slower than the rate at which we produce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a big problem, and all of the proposed solutions so far have been very expensive to implement - which is the main reason the fossil fuels industry opposes them.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-11-10T22:05:46.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19382" PostTypeId="2" ParentId="19258" CreationDate="2014-11-10T22:30:20.620" Score="6" Body="&lt;p&gt;Check if the green condensate is &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O3}$&lt;/span&gt;. &lt;a href=&quot;http://nitrogen.atomistry.com/nitrogen_trioxide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;It has been reported to be blue-green.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the presence of water vapor and oxygen, the nitrogen dioxide that you are trying to obtain is in equilibrium with nitric acid and nitrogen trioxide, as explained &lt;a href=&quot;http://nitrogen.atomistry.com/nitrogen_tetroxide.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As reported in the Journal of Industrial and Engineering Chemistry, volume 12, number 6, page 531-538 (1920), condensation of gaseous nitrogen dioxide in a spiral condensor at -5 °C in the presence of oxygen and water vapor resulted in liquid with distinct upper and lower layers, the upper layer being a concentrated nitric acid layer (essentially no water) and the lower layer being a diluted nitric acid layer (27 %) water.&lt;/p&gt;&#xA;&lt;p&gt;So the reason you are getting side products is likely that water vapor and oxygen are also present and causing equilibrium reactions that yield nitric acid, nitrous acid and nitrogen trioxide.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CCXIV.—The dissociation of gaseous nitrogen trioxide, Bernard Mouat Jones,  J. Chem. Soc., Trans., &lt;strong&gt;1914&lt;/strong&gt;, 105, 2310-2322, DOI: &lt;a href=&quot;https://pubs.rsc.org/en/content/articlelanding/1914/ct/ct9140502310/unauth#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1039/CT9140502310&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="17368" LastEditDate="2021-05-28T05:21:03.533" LastActivityDate="2021-05-28T05:21:03.533" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19383" PostTypeId="2" ParentId="19378" CreationDate="2014-11-10T23:45:52.190" Score="6" Body="&lt;p&gt;You've correctly identified the $C_3$ axis that contains the $\ce{C-Cl}$ bond.  Methyl chloride also has 3 planes of symmetry that &lt;strong&gt;contain&lt;/strong&gt; the $C_3$ axis and each of the $\ce{C-H}$ bonds.  Since these planes contain the $C_3$ axis they are referred to as $\sigma_{v}$ planes of symmetry.  There are no other symmetry elements in methyl chloride so it belongs to point group $C_{3v}$ (n $\sigma v$ planes plus a $C_n$ axis).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no $C_{nd}$ point group.  There is a $D_{nd}$ point group which contains a $C_n$ axis, n $\sigma_{v}$ planes (so far methyl chloride fits the description), but also n $C_2$ axis.  Methyl chloride does not contain any $C_2$ axes, so it does not belong to the $D_{nd}$ point group.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="2174" LastEditDate="2014-12-15T00:23:15.193" LastActivityDate="2014-12-15T00:23:15.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19385" PostTypeId="1" CreationDate="2014-11-11T00:41:20.943" Score="1" ViewCount="134" Body="&lt;p&gt;More or less I am trying to figure out how this reaction occurred, so I can reproduce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a plate I had an excess amount of sugar with a fair amount of grapefruit juice which was stuck to the plate. I filled the plate with water and left it for a little over 24 hours. When I finally came back to do the dishes, I poured the contents of the plate into the sink only to find that the bottom of the plate was covered in this slime very similar to clear gelatin. This slime was about as thick as honey, but was not sticky in any way, and was completely clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could have caused this?&lt;/p&gt;&#xA;" OwnerUserId="9724" LastActivityDate="2014-11-11T00:41:20.943" Title="Grapefruit and sugar gelatin like substance" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19386" PostTypeId="1" CreationDate="2014-11-11T02:37:46.093" Score="3" ViewCount="4661" Body="&lt;p&gt;The title says it all. Reasons that I can supply include: increased nuclear charge increasingly catches up in terms of influence to the increasing shielding and proof by contradiction in that if the space remained the same there would be negligible nuclear pull on the valence electrons and so this must not be the case. What is a comprehensive explanation for why this phenomenon is so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="5017" LastEditDate="2014-11-11T03:27:53.200" LastActivityDate="2014-12-11T04:10:18.097" Title="Why do principal energy levels in an atom get closer together as n increases?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19388" PostTypeId="2" ParentId="19340" CreationDate="2014-11-11T02:56:32.097" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I work at home, and instead of walking a moderate distance to the bathroom and loosing my focus, I've been, at times, peeing in a 3 Quart Poland Springs water bottle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So instead of walking into the bathroom... you just pee into the Poland Springs water bottle next to you? And then you decide to leave it next to you the next few days?!?!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All jokes aside, it depends on what you want to do with the sample. If it is strictly storage and you want to disinfect and neutralize bacteria while keeping odors at a minimum, probably the easiest and most efficient method would be either to raise or decrease the pH to sufficient levels such that bacteria growth is inhibited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Urine contains a lot of compounds that are excreted from the body. Protens, sugars, nitrites, ketones, bilirubin, urobilinogen, erythrocytes, leuckocytes, etc, etc. &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmedhealth/PMH0010405/&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, a concern that is immediately raised is disposal of such acidic or basic solution. Since you are working at home, you will generally not have designated or specific areas for such waste disposal. The easiest method is to dilute the urine solution with water sufficently such that the solution is effectively neutralized and is ~7.0 pH. You can check the pH of the solution by using Litmus paper or just standard pH paper which is relatively cheap to come by. Once you reach the desired pH limit, you can then pour the solution after &quot;walking the moderate distance to the bathroom&quot; and dispose of it in the toilet or sink, whatever floats your boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to do that and instead want to hang on to the urine/acidic/or/basic/solution, you can order a carboy. &lt;a href=&quot;https://en.wikipedia.org/wiki/Carboy&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt; Needless to say, it can be quite difficult to pour into the carboy depending on the original container you have. Then you can send it off to your local hazardous chemical disposal facility. Yay. They might wonder why your carboy's are all filled with copious amounts of urine. Be prepared to answer honestly so they can in fact, properly dispose of such material since it is biological in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Collection of Urine&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Urine is prone to collecting bacteria and other cellular contamination. Yes, the urine that comes out of your bladder is sterile. Your urethra, I don't know about you, may in fact not sterile. Common procedures are to stick a sterile catheter first up in through your urethra and clean it, although, that is in fact highly unpleasant, and is very painful. The most easiset way to collect a more or less &quot;clean&quot; sample &lt;em&gt;midstream clean catch specimen&lt;/em&gt;. First pee into one container for a few seconds. Hold and clench, and then pee into the container where you wish to collect your urine. Then hold and clench again, pee the remainder back in the first container. Ok, I'm wondering what you do about washing your hands... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Container&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your container, must, must, must, must, be sterile. Every time you collect, you must use a new container. Can I say that again? It must be sterile. Otherwise it would defeat the purpose... so yea. The next question is, how long do you wish to store your urine next to you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common containers are urine collection containers. These are generally the cups you pee in when you have a physical at your doctors. These samples are usually analyzed within the first few hours of collection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to store for a longer period of time, say 24 hrs, a urine collection container is more suitable. This is a 3L container and is amber. The amber color is necessary since analytes such as porphyrins and urobilinogen are photosensitive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preservation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common 24-hour preservatives are hydrochloric acid, boric acid, acetic acid and toluene.&lt;a href=&quot;https://www.bd.com/vacutainer/labnotes/Volume14Number2/&quot; rel=&quot;nofollow&quot;&gt;(3)&lt;/a&gt; A lot of the literature is pointing towards boric acid as the preservative. UCSF medical center has a nice protocol which you can follow here. &lt;a href=&quot;http://labmed.ucsf.edu/labmanual/mftlng-mtzn/dnld/speccol-URINE.pdf&quot; rel=&quot;nofollow&quot;&gt;(4)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boric acid is pretty cool since it's powdery and not so soluble in room temperature as other acids. Yea, get the tablet's since they are way easier to deal with than the powdered form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other methods include preservation at $4^\circ$C for first 24-hr. Then deep freeze at $-80^\circ$C for longer term storage. Those refrigerator's cost tens-of-thousands so yea... dry ice might work.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2014-11-11T02:56:32.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19390" PostTypeId="2" ParentId="19338" CreationDate="2014-11-11T03:10:02.813" Score="4" Body="&lt;p&gt;Complimentary nucleic acids are held together mainly by hydrogen bonding, so I'm not sure if they are able to be cold denatured, but cold denaturation is a well known phenomenon for proteins. &lt;a href=&quot;http://www.physics.mcgill.ca/~grant/Papers/DIAS09_cryo.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a decent review of the processes involved. In short, most proteins' tertiary structure is dominated by hydrophobic interactions and the energy barrier to unfolding a protein is generally quite small. The stabilizing effect of the hydrophobic interactions decreases with temperature and at a certain point, is insufficient to keep the protein folded. For many proteins, the cold denaturation temperature is below 0 °C, so it's difficult to study. Generally, studies on these proteins are done by adding chemical denaturants or changing the pressure of the system to shift this temperature higher, having the risk of conflating this deliberate denaturation with the temperature effect. Of course, not all proteins are stabilized this way and things like metallothioneins probably don't denature in this way as their tertiary structure is maintained by coordinating metal ions.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-11T03:10:02.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19391" PostTypeId="2" ParentId="19386" CreationDate="2014-11-11T03:50:29.057" Score="1" Body="&lt;p&gt;The energy levels for an atom are similar to a &quot;particle in a box&quot;, except that the walls have a different shape. When you analyzed the &quot;particle in a box&quot; situation, you used a rectangular box. An atom is spherical, so the walls of the atom open outwards at the same rate as the sphere increases in surface area. You should have noticed that the energy levels for a &quot;particle in a box&quot; decreased in distance at higher energy levels. The surface area of the sphere brings the energy levels even closer.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-11T03:50:29.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19393" PostTypeId="2" ParentId="8345" CreationDate="2014-11-11T04:22:21.743" Score="4" Body="&lt;p&gt;I saw this in a &quot;related questions&quot; link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that &lt;strong&gt;&lt;em&gt;yes&lt;/em&gt;&lt;/strong&gt;, you can use quantum mechanics to define atomic radii (e.g., van der Waals radii). Multiple people have done so with different schemes. I believe the most recent is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/jp8111556&quot; rel=&quot;nofollow&quot;&gt;&quot;Consistent van der Waals Radii for the Whole Main Group&quot; &lt;em&gt;J. Phys. Chem. A&lt;/em&gt; &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;113&lt;/em&gt;, 5806–5812&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar efforts have been made for covalent bonding radii using quantum mechanics.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-11T04:22:21.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19394" PostTypeId="1" AcceptedAnswerId="32681" CreationDate="2014-11-11T07:53:08.973" Score="2" ViewCount="3264" Body="&lt;p&gt;What resonance structure is the most abundant in sulfur trioxide: the one with one double and two single, the one with two double and one single or the one with all three double bonds? Is it actually correct to write an expanded octet when it is not necessary?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-06-09T23:21:22.270" Title="Sulfur trioxide structure" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19395" PostTypeId="1" CreationDate="2014-11-11T08:19:29.047" Score="0" ViewCount="248" Body="&lt;p&gt;What is the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb^2+}$&lt;/span&gt; in a lake that has a pH of &lt;span class=&quot;math-container&quot;&gt;$6.0$&lt;/span&gt; and is in equilibrium with &lt;span class=&quot;math-container&quot;&gt;$\ce{PbO2(s)}$&lt;/span&gt; and atmospheric oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{PbO2(s) + 4H+ + 2e- &amp;amp;-&amp;gt; Pb^2+ + 2H2O} &amp;amp;&#xA;\log K &amp;amp;= 49.2&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent so much time to solve this, but I could not, my final answer does not appear to make sense:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \mathrm{pH} &amp;amp;= 6 &amp;amp;&#xA;  \implies&amp;amp;&amp;amp; [\ce{H+}] &amp;amp;=10^{-6}\\&#xA;  \log K      &amp;amp;= 49.2 &amp;amp;&#xA;  \implies&amp;amp;&amp;amp; K &amp;amp;= 10^{49.2}\\&#xA;  K &amp;amp;= \frac{[\ce{Pb^2+}]}{[\ce{H+}]^4} &amp;amp;&#xA;  \implies&amp;amp;&amp;amp; 10^{49.2} &amp;amp;=\frac{[\ce{Pb^2+}]}{(10^{-6})^4} \\&#xA;  &amp;amp;&amp;amp;\implies&amp;amp;&amp;amp; [\ce{Pb^2+}] &amp;amp; =10^{25.2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my mistake?&lt;/p&gt;&#xA;" OwnerUserId="9728" LastEditorUserId="23561" LastEditDate="2018-10-22T23:47:05.597" LastActivityDate="2018-10-22T23:47:05.597" Title="How to calculate the concentration of lead(II) liberated from lead(IV) oxide under acidic conditions?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;redox&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-11-05T14:47:13.233" ContentLicense="CC BY-SA 4.0" />
  <row Id="19397" PostTypeId="1" CreationDate="2014-11-11T13:01:30.703" Score="1" ViewCount="163" Body="&lt;p&gt;In the reaction of chlorine gas with water, Hydrogen chloride and nascent oxygen is produced.The reaction of any coloured substance with this nascent oxygen makes it colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true?if yes, how?  &lt;/p&gt;&#xA;" OwnerUserId="1886" LastEditorUserId="1886" LastEditDate="2014-11-11T13:39:04.620" LastActivityDate="2014-11-11T17:23:20.677" Title="Coloured compounds on reacting with oxygen" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19398" PostTypeId="1" CreationDate="2014-11-11T16:01:49.247" Score="0" ViewCount="1443" Body="&lt;p&gt;I want to produce a solution, which contains $0.02~\mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ and $~0.2~\mathrm{mol~L}^{-1}~\mathrm{NH}_3$. The following substances are given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;copper(II) sulfate pentahydrate&lt;/li&gt;&#xA;&lt;li&gt;$100~\mathrm{mL~H}_2\mathrm{O}$&lt;/li&gt;&#xA;&lt;li&gt;diluted ammonia solution with a concentration of $2~\mathrm{mol~L}^{-1}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to determine the required volume ($V$) of $\mathrm{NH}_3$ and the mass ($m$) of $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to calculate $m$. However, to calculate $V$, I solved the following system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{0.2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V+0.1~\mathrm{L}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $V = 0.01111~\mathrm{L}$. Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $m=555~\mathrm{mg}$ correct?&lt;/p&gt;&#xA;" OwnerUserId="9733" LastEditorUserId="9733" LastEditDate="2014-11-12T17:14:09.443" LastActivityDate="2014-12-18T14:19:34.297" Title="Calculating mass and volume of two different substances required for one solution" Tags="&lt;homework&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19399" PostTypeId="2" ParentId="19397" CreationDate="2014-11-11T17:23:20.677" Score="1" Body="&lt;p&gt;The feature that makes colored compounds colored is the one that makes the compounds easily oxidized. When the compound is oxidized, something does not exist anymore. I'm not giving you a straight answer since you will experience a nice enlightenment when you figure that out. Take a look at the structure of beta-carotene, for example. Where would the oxygen go and what would the structure look like after that?&lt;/p&gt;&#xA;" OwnerUserId="4452" LastActivityDate="2014-11-11T17:23:20.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19400" PostTypeId="1" AcceptedAnswerId="19412" CreationDate="2014-11-11T17:27:14.000" Score="6" ViewCount="1849" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  The benzyloxycarbonyl group (CBZ) is widely used to protect amines, especially in aminoacid chemistry. A procedure to synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate is given in the description below. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Glycine ($0.1~\mathrm{mol}$) was dissolved in aqueous sodium hydroxide ($50~\mathrm{mL}$ of a $2~\mathrm{M}$ solution) &#xA;  and the solution was cooled in an ice bath. Benzyl chloroformate ($1.2~&#xA;\mathrm{eq}$) and aqueous &#xA;  sodium hydroxide ($25~\mathrm{mL}$ of a $4~\mathrm{M}$ solution) were added dropwise simultaneously to the &#xA;  glycine solution at $0°\mathrm{C}$ dropwise over thirty minutes. The mixture was stirred for an extra &#xA;  ten minutes, then warmed to room temperature. The solution was extracted twice with &#xA;  ether. The combined organic layers were acidified to $\mathrm{pH}=1$. The precipitate was filtered, &#xA;  washed with a little cold water and dried to give carbobenzoxyglycine ($18.5~\mathrm{g}$), m.p. $119-120°\mathrm{C}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Relative to glycine, how many equivalents of sodium hydroxide are used in this reaction? Explain why this number was used.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) Can sodium carbonate ($\ce{Na2CO3}$) be used in place &#xA;  of sodium hydroxide? Explain why it can or cannot be used.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) What inorganic by-product will be formed (write its chemical name)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(d) The procedure given includes a serious mistake, which would result in a yield of $0\%$. &#xA;  What is the mistake, and how might it be corrected?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) I mentioned $1~\mathrm{eq}$ of sodium hydroxide as sodium hydroxide reacts with glycine in a 1:1 ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) I mentioned that sodium carbonate cannot be used in place of sodium hydroxide as sodium Carbonate is not basic enough to catalyst the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) Sodium chloride &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) I mentioned that during the process when the combined organic layers were acidified to $\mathrm{pH}=1$, it could destroy the product by hydrolysing it hence giving a yield of $0\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure whether my answers are correct though, could anyone explain? Also, are there any resources on the web that contain practice questions of a similar nature to this? &lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="46306" LastEditDate="2018-03-03T13:40:13.397" LastActivityDate="2018-03-03T14:17:11.790" Title="Synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19402" PostTypeId="1" AcceptedAnswerId="19422" CreationDate="2014-11-11T18:44:45.557" Score="4" ViewCount="1996" Body="&lt;p&gt;The textbook I got from my teacher is shown to be too general. Part of the fun of learning chemistry, for me, comes from knowing the derivation of equations. But most of the derivations are missing out from the textbook, like the derivation of kinetic molecular equation and real gas equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be fine to recommend a college level textbook as long as it does not include calculus.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="5017" LastEditDate="2015-04-30T18:27:42.530" LastActivityDate="2015-04-30T18:27:42.530" Title="Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry?" Tags="&lt;reference-request&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19403" PostTypeId="1" CreationDate="2014-11-11T19:47:24.107" Score="2" ViewCount="858" Body="&lt;p&gt;I worry that the electronegative N will stabilise keto form so much that equilibrium will be so far towards keto it would be unreasonable to call the enolisiation process a factor&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yYskO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yYskO.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9739" LastEditorUserId="69830" LastEditDate="2021-05-04T09:36:42.343" LastActivityDate="2021-05-04T09:36:42.343" Title="Can ketones next to a nitrogen hetero atom tautomerise?" Tags="&lt;organic-chemistry&gt;&lt;equilibrium&gt;&lt;tautomer&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19404" PostTypeId="2" ParentId="19395" CreationDate="2014-11-11T19:47:27.637" Score="2" Body="&lt;p&gt;I think that the problem here is one of notation. I'm pretty sure that the logarithm intended to be used is the natural logarithm, &lt;span class=&quot;math-container&quot;&gt;$\log_e(K)$&lt;/span&gt; also represented as &lt;span class=&quot;math-container&quot;&gt;$\ln(K)$&lt;/span&gt;, and not the base 10 one. Otherwise your method is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use this, you get a much more reasonable answer (work it out). It also makes sense that the question was worded in this way, as the equations in which one tends to use the equilibrium constant &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; actually use its natural logarithm, for example the Gibbs's free energy and the Nernst equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this reaction is also part of the reduction process at the cathode of a lead-acid battery.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastEditorUserId="23561" LastEditDate="2018-10-22T23:39:54.147" LastActivityDate="2018-10-22T23:39:54.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19405" PostTypeId="2" ParentId="19403" CreationDate="2014-11-11T20:46:35.290" Score="2" Body="&lt;p&gt;With a nitrogen adjacent to a carbonyl moiety, the functional group is an amide, and the tautomer involving a $\ce{C=N}$ bond is called an imidic acid. The amide is usually vastly more stable and typically predominates by many orders of magnitude. This can be rationalized by several considerations. First, the two tautomers differ by the exchange of a $\ce{C=O}$ bond with a $\ce{C=N}$ bond and an $\ce{N-H}$ with an $\ce{O-H}$, but the $\ce{C=O}$ bond is particularly strong, making the amide enthalpically more stable. Second, if you consider the resonance structures (i.e., mesomers) available to the two respective tautomers, the amide has a contributing structure in which a negative formal charge is placed on the more electronegative oxygen atom, whereas the corresponding structure in the imidic acid has the charge placed on a less electronegative nitrogen atom. Granted, resonance is not a real physical process, but examination of these structures gives some idea of the distribution of electron density in the frontier orbitals, which too points to favorable increase in stability for the amide. You can apply similar logic to the case of tautomers involving a $\ce{C=C}$ bond, where the ultimate conclusion should be the same (and the difference in stability between tautomers even greater).&lt;/p&gt;&#xA;" OwnerUserId="305" LastEditorUserId="305" LastEditDate="2014-11-11T20:52:14.813" LastActivityDate="2014-11-11T20:52:14.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19406" PostTypeId="1" AcceptedAnswerId="19415" CreationDate="2014-11-11T21:05:57.693" Score="1" ViewCount="1100" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q4ZF8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminescence is defined as the amount of light emitted by a atom. But what confounds me of this graph is the fact that all the peaks are of the same height. In addition, there are no peaks in between two peaks.&#xA; &lt;strong&gt;why are the peaks of the same height?&lt;/strong&gt;&#xA;And&lt;strong&gt;why are there only six peaks, if photons can be emitted having the wave length of say, 110nm?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure those are ultraviolet rays.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-11-12T01:35:37.367" LastActivityDate="2014-11-12T03:27:32.717" Title="How to interpret a luminescence intensity vs wavelength graph?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19408" PostTypeId="2" ParentId="19400" CreationDate="2014-11-11T21:43:17.410" Score="3" Body="&lt;p&gt;Under neutral conditions glycine exists as a zwitterion as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gaf1A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form, there is no &quot;amine&quot; available to react with the carboxybenzyl (or benzyloxycarbonyl, or Cbz, or Z) protecting group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 1 equivalent of sodium hydroxide (&quot;dissolved in aqueous sodium hydroxide (50 mL of a 2 M solution)&quot;) was used to convert all of the glycine zwitterion to the glycinate anion.  In the anion, we do have a free amine group that can react with Cbz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The $\mathrm{p}K_\mathrm{a}$ of the ammonium group in the glycine zwitterion is around 9.8 (see &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/1biochem/amino2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for a nice discussion of the acid-base equilibria involved with amino acids), so, as you noted, a strong base is needed to convert the ammonium ion to the amine.  Sodium carbonate is not sufficiently basic to accomplish this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) After adding sodium hydroxide to the glycine zwitterion, we form sodium glycinate and water.  After the Cbz is added it reacts with the free amino group to produce the Cbz protected derivative of glycine plus hydrochloric acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8viS2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrochloric acid and the sodium cation react to produce sodium chloride as the inorganic by-product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) the mistake involves the &lt;strong&gt;simultaneous&lt;/strong&gt; addition of Cbz and additional sodium hydroxide (&quot;Benzyl chloroformate (1.2 equiv.) and aqueous sodium hydroxide ($\pu{25 mL}$ of a $\pu{4 M}$ solution) were added dropwise simultaneously&quot;).  For one thing, enough sodium hydroxide was added earlier to convert all of the glycine zwitterion to the glycinate anion, there is no need for additional sodium hydroxide.  Second, sodium hydroxide and Cbz will react immediately to produce the mono-sodium salt of benzylformate.  There won't be enough Cbz left to react completely with the amino acid.  Once the free amine has been generated the Cbz should be added separately in a second step to react with the free amine group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, not aware of any websites with similar practice questions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-03T14:17:11.790" LastActivityDate="2018-03-03T14:17:11.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19409" PostTypeId="2" ParentId="19403" CreationDate="2014-11-11T23:11:40.397" Score="4" Body="&lt;p&gt;If I understand your question correctly, I think you're asking if an amide can enolize like a ketone.  The answer to that question is &lt;strong&gt;yes&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nvUAl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enol content of most organic compounds is very low and special methods must be used to measure the enol content.  Perhaps the question becomes, &quot;&lt;strong&gt;is there more or less enol content&lt;/strong&gt; in an amide than in an ester, for example&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide.  But above we also drew resonance structures showing how the nitrogen can also stabilize the enol.  We can draw similar structures for both an ester and its enol as well.  Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly.  So as a first approximation I might guess that the enol content in an ester and an amide would be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nlf9Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 &lt;a href=&quot;http://etheses.dur.ac.uk/5282/1/5282_2719.PDF?UkUDh:CyT&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wiRsF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-11T23:11:40.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19410" PostTypeId="1" AcceptedAnswerId="19413" CreationDate="2014-11-11T23:13:49.060" Score="1" ViewCount="775" Body="&lt;p&gt;I am asked whether dimethyl malonate or methyl acetoacetate decarboxylate when hit with acid, water, and extreme temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is both because both are esters, and esters, upon being hit by such conditions, return to their respective carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, both carboxylic acids can undergo decarboxylation because both resulting carboxylic acids are beta-keto acids. Technically, dimethyl malonate becomes malonic acid, which can still undergo decarboxylation (but malonic acid isn't of the beta-keto class; rather the dicarboxylic acid class). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acid and heat treatment of methyl acetoacetate initially gives 3-oxobutanoic acid but this can decarboxylate under the heat as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my analysis correct? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-11-11T23:28:07.857" LastActivityDate="2014-11-11T23:59:09.200" Title="Decarboxylation under acid, water, and extreme heat" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19411" PostTypeId="1" AcceptedAnswerId="19431" CreationDate="2014-11-11T23:39:18.480" Score="1" ViewCount="324" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the complete, acid-catalyzed hydrolysis products (with application of heat) of diphenyl malonate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me, they are phenol, carbon dioxide, and acetic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aqueous acid and heat conditions indicate to me that we are returning from the carboxylic acid derivative (ester) to the carboxylic acid. The mechanism involves protonation of the carbonyl oxygen and nucleophilic attack by water. This ejects alcohol (well, in this case, phenol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat induces the malonic acid intermediate to decarboxylate and this results in the production of carbon dioxide and acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my analysis correct?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-04-14T05:24:59.090" LastActivityDate="2018-04-14T05:24:59.090" Title="Acid-catalyzed hydrolysis of diphenyl malonate" Tags="&lt;organic-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19412" PostTypeId="2" ParentId="19400" CreationDate="2014-11-11T23:46:47.113" Score="5" Body="&lt;p&gt;I agree with Ron's answer except for d. Using an extra equivalent of NaOH is the right procedure since you need to &quot;mop up&quot; the HCl that is produced when $\ce{CbzCl}$ reacts with the amino group. What is not correct is this:&#xA;When you wash the reaction mixture with ether you get rid of any impurities and your product should stay in the aqueous phase. So you have to acidify the &lt;strong&gt;aqueous&lt;/strong&gt; layer and not the ether one. pH 1 should be fine for the Cbz unless you leave it there and go away for the weekend.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-11-11T23:46:47.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19413" PostTypeId="2" ParentId="19410" CreationDate="2014-11-11T23:59:09.200" Score="2" Body="&lt;p&gt;Yes, your analysis is correct. This is a common sequence in the malonate synthesis of carboxylic acids and acetoacetate synthesis of methyl ketones. After taking advantage of the acidic, doubly alpha-proton and using the resulting enolate as a nucleophile, the ester(s) are hydrolyzed with acid. A carboxylic acid with a carbonyl in the beta-position can be thermally decarboxylated to give an enol, which typically tautomerizes to the keto form.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-11T23:59:09.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19414" PostTypeId="1" AcceptedAnswerId="19425" CreationDate="2014-11-12T01:04:58.193" Score="3" ViewCount="3196" Body="&lt;p&gt;I'm wondering if this is possible because could there not be a situation in which the reactants were pure solids/liquids and would therefore not contribute to the value of k but still contain more total moles that the product side?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="189" LastEditDate="2014-11-12T02:55:12.220" LastActivityDate="2014-11-12T18:11:10.973" Title="If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19415" PostTypeId="2" ParentId="19406" CreationDate="2014-11-12T02:57:29.723" Score="2" Body="&lt;p&gt;This is a partial spectrum of the luminescence from a hydrogen atom, the graph presents only the &lt;strong&gt;Lyman series&lt;/strong&gt; of emissions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a hydrogen atom is excited (absorbs energy), the electron can be promoted to the n=2, 3, 4, 5...infinite level.  This excited electron can eventually emit its excess energy and return to the n=1, 2, 3, 4...etc. state.  All returns to the n=1 state (ground state) are termed the Lyman series,  all returns to the n=2 state are termed the Balmer series and so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXhLe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the hydrogen atom, the wavelength of these various emmissions is given by the Rydberg formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the Lyman series this equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${1 \over \lambda} = R \left( {1} - {1 \over n^2} \right) \qquad$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Lyman_series_.28n.E2.80.B2_.3D_1.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;six lines can be seen&lt;/a&gt; in the ultraviolet.  After the 5th line (n=6 -&gt; n=1), all other transitions are so closely spaced that they appear as one line (the sixth line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|}\hline&#xA;n &amp;amp; λ (nm) \\ \hline&#xA;2 &amp;amp; 122 \\ \hline&#xA;3 &amp;amp; 103 \\ \hline&#xA;4 &amp;amp; 97.3 \\ \hline&#xA;5 &amp;amp; 95.0 \\ \hline&#xA;6 &amp;amp; 93.8 \\ \hline&#xA;\infty &amp;amp; 91.2 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the relative intensities, I'm not sure why they are all the same in your graph.  The intensities do depend upon how many electrons are excited to the various excited states.  If for some reason all excited states were created in a manner such that they were equally populated, then something like your spectrum would result.  Usually the relative intensities differ and are used much like a fingerprint.  For example, the relative intensities of the Lyman series from the hydrogen atoms in one galaxy differ from those in another galaxy due to the difference in temperature of the two galaxies.  My guess is that your data was artificially created (&lt;em&gt;e.g&lt;/em&gt;. the title in your graph says &quot;Model&quot;).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-11-12T03:27:32.717" LastActivityDate="2014-11-12T03:27:32.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19416" PostTypeId="2" ParentId="19414" CreationDate="2014-11-12T03:07:44.037" Score="3" Body="&lt;p&gt;The equilibrium constant does only contain the activities/concentrations of the species that are actually available to the reaction, e.g. in solution it would only contain species that are actually dissolved in the solvent or that are present at the interface between the solution and another phase. So, solids would not enter the equation with all there moles but only with the small percentage that is acutally available to the reaction: if some of the solid is dissolved and those dissolved species take part in the reaction then it is those (and only those) molecules that enter into the equation for the equilibrium constant; in the case of a heterogeneous reaction the solid can only enter the equation via its available surface sites. So, the bulk of a solid would not contribute to the equilibrium constant in a reaction taking part in a liquid phase that shares and interface with the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, concerning the main question: If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? This can certainly be the case if you already have some of the products present and keep their concentrations constant. Look at the example reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  \ce{A + B &amp;lt;=&amp;gt;&amp;gt; C + D}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shall have an equilibrium constant of $K_{c} = 10$. Let's assume you keep the products at a concentration of 1 mol/L each, so $[\ce{C}] = 1 \, \mathrm{mol}/\mathrm{L}$ and $[\ce{D}] = 1 \, \mathrm{mol}/\mathrm{L}$. The equation for the equilibrium constant is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  K_{c} = \frac{[\ce{C}] [\ce{D}]}{[\ce{A}] [\ce{B}]}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you've already fixed the product $[\ce{C}] [\ce{D}]$ at $1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}$ the product of your reactants must fulfill &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  [\ce{A}] [\ce{B}] = \frac{[\ce{C}] [\ce{D}]}{K_{c}} = 0.1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you use only very little $\ce{B}$, say $[\ce{B}] = 0.01 \, \mathrm{mol}/\mathrm{L}$, your concentration of $\ce{A}$ in the solution will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  [\ce{A}] = \frac{[\ce{C}] [\ce{D}]}{K_{c} [\ce{B}]} = 10 \, \mathrm{mol}/\mathrm{L}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you conduct this reaction in a vessel with a volume of 1 L you'll find a total of 2 mol of products and 10.01 mol of reactants in your reaction mixture, i.e. more reactants than products.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-11-12T18:11:10.973" LastActivityDate="2014-11-12T18:11:10.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19417" PostTypeId="1" CreationDate="2014-11-12T03:13:29.700" Score="3" ViewCount="118" Body="&lt;p&gt;I have been given a formula on the topic of alloys and temperature, however am unsure specifically what this equation calculates. The equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xC_{1} + (1 - x) C_{2} = C_{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that the C's will be given (and possibly represent Specific Heat?) - however exactly what is this equation to represent? What does solving for $x$ achieve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the letters used may vary, however our teacher has provided no more information due to sickness. I have provided all the information I know.&lt;/p&gt;&#xA;" OwnerUserId="9539" LastEditorUserId="189" LastEditDate="2014-11-12T03:58:46.977" LastActivityDate="2015-01-11T05:20:50.477" Title="What is this equation for?" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19418" PostTypeId="2" ParentId="19417" CreationDate="2014-11-12T03:29:38.453" Score="1" Body="&lt;p&gt;I fear that without additional information it will be difficult to provide you with a definitive answer. The $x$ usually represents the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mole_fraction&quot; rel=&quot;nofollow&quot;&gt;mole fraction&lt;/a&gt; and in the context of a 2-component alloy this seems plausible. The $C$s might indeed be specific heat constants, i.e. $C_{3}$ would be the specific heat of the alloy and $C_{1}$ and $C_{2}$ would be the specific heats of the components making up the alloy, whereby component 1 would have a mole fraction of $x$  and component 2 would have a mole fraction of $(1-x)$. For a mixture of ideal gases such a formula would be correct but I doubt that it will work well with alloys in the real world. But if this came up in the context of a school lesson then the approximation might be ok.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-11-12T03:29:38.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19421" PostTypeId="1" CreationDate="2014-11-12T04:58:36.867" Score="4" ViewCount="447" Body="&lt;p&gt;I'm asked to find $\Delta G$ for the isothermal compression of $\ce{N2(g)}$ (real gas) at $0\ \mathrm{^\circ C}$ if the gas expands from $P=1\space\mathrm{atm}$ to $P=1000 \space\mathrm{atm}$. The fugacity coefficients are $\phi=0.9996$ and $\phi=1.84150$ respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused about two things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My professor makes a distinction between $\phi^*$ and $\phi$, the first one being the fugacity coefficient for a pure gas and the second for a gas in a mixture. Are they the same in this case or did he make a mistake since there's not a mixture to begin with? (Not sure how Lewis—Randall rule can relate here since, again, there's no mixture to apply the relation between the two coefficients.)&lt;/li&gt;&#xA;&lt;li&gt;I already calculated the Gibbs free energy if the gas were an ideal one, I got $\Delta G=15.688\ \mathrm{kJ\over mol}$. Since $G^E=RT\ln(\gamma)$, and I have the Gibbs energy for an ideal gas, I could find the Gibbs energy for the real gas when I get a value for $G^E$; but again, in order to find $\gamma$ I need both coefficients, the one for a mixture and the one for a pure gas.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone please let me know what am I doing wrong? Or if my approach is far from the right path? Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that the data that was given corresponded to the values of fugacity coefficients for the pure gas, making it all simpler since $G^E=RT\ln(\phi)$, so I got $G^E_{N_2}=17.075765\ \mathrm{kJ\over mol}$ and then $\Delta G^R=1.387765\ \mathrm{kJ\over mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is that according to those two definitions of $G^E$, I get that $\gamma=\phi$, which I know it's not right, how's it possible?&lt;/p&gt;&#xA;" OwnerUserId="2212" LastEditorDisplayName="user7951" LastEditDate="2015-11-09T15:28:07.623" LastActivityDate="2015-11-09T15:28:07.623" Title="How to find Gibbs energy for a real gas" Tags="&lt;gas-laws&gt;&lt;free-energy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19422" PostTypeId="2" ParentId="19402" CreationDate="2014-11-12T06:48:26.643" Score="4" Body="&lt;p&gt;&quot;College Chemistry - An Introductory Textbook of General Chemistry&quot; by Linus Pauling (2nd Ed) is available as a free download (pdf &amp;amp; other formats) from &lt;a href=&quot;https://archive.org/details/CollegeChemistry&quot; rel=&quot;nofollow&quot;&gt;https://archive.org/details/CollegeChemistry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-11-12T06:48:26.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19424" PostTypeId="2" ParentId="17174" CreationDate="2014-11-12T13:54:12.683" Score="3" Body="&lt;p&gt;The rate of the third reaction affects the rate of the first reaction, i.e. they are competitive reactions and in that case the rds method can't be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of the first reaction is $r=-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t$, but the change of the concentration of the peroxydisulfate depends on the third reaction, too:&#xA;$$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2014-11-18T05:58:47.167" LastActivityDate="2014-11-18T05:58:47.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19425" PostTypeId="2" ParentId="19414" CreationDate="2014-11-12T14:41:28.230" Score="2" Body="&lt;p&gt;Yes, even if the equilibrium constant is much greater than 1, there can be almost all reactants and very little products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wood pile can just sit outside (in an oxygen containing atmosphere) all winter, but until you stick the wood in the fireplace and supply activation energy almost no reactants will be converted to product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the reactants are mixed in the same phase, the same is true.  Mix methane and air in a Bunsen burner or gas stove, and nothing noticeable happens until you provide a spark or flame.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-12T14:41:28.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19426" PostTypeId="2" ParentId="19258" CreationDate="2014-11-12T15:30:28.253" Score="4" Body="&lt;p&gt;The colorless area may be &lt;a href=&quot;https://en.wikipedia.org/wiki/Dinitrogen_tetroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt;&lt;/a&gt; which is in equilibrium with &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; is colorless and has a greater density and therefore is found at the bottom and green compound might be &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O3}$&lt;/span&gt; which is in equilibrium with &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1vfwX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vfwX.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; :&lt;br /&gt;&#xA;Boiling point of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; is 21.69 °C, so the temperature at which you perform experiment is also critical. Moreover, there is high probability that &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; might be present in liquid state too.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="17368" LastEditDate="2021-05-28T05:17:17.353" LastActivityDate="2021-05-28T05:17:17.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19427" PostTypeId="1" CreationDate="2014-11-12T15:31:25.057" Score="2" ViewCount="399" Body="&lt;p&gt;I think it would be interesting and illuminating to calculate the carbon footprint (per carat) of diamond. The answer would undoubtedly be different for artificial (industrial grade) diamonds vs. natural ones. Is anyone out there familiar enough with either natural diamond mining and processing or the high-pressure process for making artificial diamonds to venture an answer to this question? Both surely involve expenditure of a considerable amount of energy.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-11-12T15:31:25.057" Title="What is the carbon footprint of a diamond?" Tags="&lt;materials&gt;&lt;solid-state-chemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ClosedDate="2018-10-18T20:42:23.843" ContentLicense="CC BY-SA 3.0" />
  <row Id="19428" PostTypeId="2" ParentId="18923" CreationDate="2014-11-12T15:35:05.953" Score="2" Body="&lt;p&gt;With osmium tetroxide, no it does not appear that dihydroxylation and acetalization can be done simultaneously. Usually dihydroxylation is carried out with catalytic osmium tetroxide with a stoichiometric amount of N-methylmorpholine N-oxide (NMO). The latter regenerates the Os(VIII) species necessary to carry out the dihydroxylation and results in the formation of N-methylmorpholine, which is a weak base. Without an acid available for catalysis, the acetalization does not occur. The sequence can be accomplished in &quot;one-pot&quot; however. In practice, you can monitor the progress of the dihydroxylation reaction. When it is deemed complete, add catalytic acid directly to the same reaction mixture, which allows the acetal formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.tetlet.2012.02.045&quot; rel=&quot;nofollow&quot;&gt;Reference: 10.1016/j.tetlet.2012.02.045&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the overall transformation has been accomplished with mCPBA, but it's not clear if happens without any further intervention. In the referenced paper below, it's not entirely clear to me when the epoxide opens, because the procedure specifies that the epoxidation step is carried out with exclusion of water. It's likely that the epoxide is stable until addition of water, which opens the epoxide to the diol and then the acetalization can occur, catalyzed by the meta-chlorobenzoic acid byproduct from the epoxidation. In some cases, they simply isolate the epoxyketone as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/ejoc.201101087&quot; rel=&quot;nofollow&quot;&gt;Reference: 10.1002/ejoc.201101087&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-12T15:35:05.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19430" PostTypeId="2" ParentId="19055" CreationDate="2014-11-12T16:03:16.480" Score="1" Body="&lt;p&gt;We essentially deal with two different trends here and I think the answer lies within analyzing these trends. First we have that the ionization energy obviously increases with increased nuclear charge/atomic number. Secondly we have that with a higher magnetic quantum number (the s-, p-, d- and f-orbitals) the energy of the orbital increases and therefore that the ionization energy decreases. The decrease in ionization energy that the higher magnetic quantum number introduces is larger than the increase that the single extra positive charge introduces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should also take Hund's rule in account, which is favorable for a lower ionization energy of $\ce{Be}$ compared to $\ce{B}$. This factor is however relatively small compared to the above ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To verify some of these trends yourself, you can take a look at &lt;a href=&quot;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&quot; rel=&quot;nofollow&quot;&gt;this table of ionization energies&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-11-12T16:03:16.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19431" PostTypeId="2" ParentId="19411" CreationDate="2014-11-12T16:26:50.780" Score="1" Body="&lt;p&gt;If the question is &quot;What are the products of hydrolysis of diphenyl malonate?&quot;, then my answer would be 2 equivalents of phenol and one equivalent of malonic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the question is &quot;What are the products of diphenyl malonate when heated in an acid aqueous solution?&quot;, then my answer would be 2 equivalents of phenol, one equivalent of acetic acid, and one equivalent of carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see it sort of as a semantic question, since the hydrolysis of an ester gives an alcohol and a carboxylic acid, as you suggest. I don't consider decarboxylation to be a hydrolysis reaction, literally defined as cleavage of bonds by addition of water. Water isn't incorporated into the products of the decarboxylation step.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-12T16:26:50.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19432" PostTypeId="1" AcceptedAnswerId="19463" CreationDate="2014-11-12T16:58:56.640" Score="9" ViewCount="387" Body="&lt;p&gt;I'm an undergraduate Chemistry student and I'm currently writing my Final Dissertation. &lt;br/&gt; I've written a form of the Born-Oppenheimer approximation with the only initial assumption of parametric dependence on nuclear coordinates. Let me know  what do you think about it. Thanks in advance for any help you are able to provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given molecular system, within the non-relativistic approximation, the molecular Schrodinger equation can be written as $$ \hat{\mathcal{H}}_{\mathrm{tot}}\Psi\left(\mathbf{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}\Psi\left(\mathbf{R},\,\mathbf{r}\right)\quad\left(1\right) $$ where $$\hat{\mathcal{H}}_{\mathrm{tot}}=\hat{\mathcal{T}}_{\mathrm{e}}\left(\mathbf{r}\right)+\hat{\mathcal{T}}_{\mathrm{N}}\left(\mathbf{R}\right)+\hat{\mathcal{V}}_{\mathrm{ee}}\left(\mathbf{r}\right)+\hat{\mathcal{V}}_{\mathrm{NN}}\left(\mathbf{R}\right)+\hat{\mathcal{V}}_{\mathrm{eN}}\left(\mathbf{R},\mathbf{r}\right)=\hat{\mathcal{T}}_{\mathrm{N}}+\hat{\mathcal{H}}_{\mathrm{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Without loss of generality, if there exists a complete basis $ \left\{ \Psi_{k}\left(\mathbf{R},\,\mathbf{r}\right)\right\} &#xA;$ , the molecular wave function $ \Psi\left(\mathbf{R},\,\mathbf{r}\right) $ can be expanded as $$ \Psi\left(\mathbf{R},\,\mathbf{r}\right)={\displaystyle \sum_{k}}c_{k}\Psi_{k}\left(\mathbf{R},\,\mathbf{r}\right) $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The nuclei in a molecule are heavy enough that, to a good approximation, they behave as classical particles. Thus, it is completely meaningful to talk about nuclear configurations, or in other words to assume a parametric dependence from nuclear coordinates.&#xA;$$&#xA;\hat{\mathcal{H}}_{\mathrm{tot}}\Psi\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}\left(\mathrm{R}\right)\Psi\left(\mathrm{R},\,\mathbf{r}\right)&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a given complete basis set $ \left\{\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\} $, the molecular wavefunction is determined only by the coefficients $ c_{k} $. Thus, the coefficients $ c_{k} $&#xA;depends on nuclear configuration:&#xA;$$&#xA;\Psi\left(\mathrm{R},\,\mathbf{r}\right)={\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\quad\left(2\right)&#xA;$$&#xA;Now by substituting eq.(1) into eq.(2)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c}&#xA;{\displaystyle \sum_{k}}\left(\hat{\mathcal{T}}_{\mathrm{N}}+\hat{\mathcal{H}}_{\mathrm{e}}\right)c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\\&#xA;{\displaystyle \sum_{k}}\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\hat{\mathcal{H}}_{\mathrm{e}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{\mathrm{tot}}{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)&#xA;\end{array} $$ &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;First term of eq.(3)&lt;/strong&gt;&#xA;$$&#xA;\begin{array}{c}&#xA;{\displaystyle \sum_{k}}\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ \nabla_{\mathrm{N}_{a}}\cdot\nabla_{\mathrm{N}_{a}}\left[c_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right]\right\} =\\&#xA;=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ \nabla_{\mathrm{N}_{a}}\cdot\left[c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\right]\right\} =\\&#xA;=-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}^{2}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)+\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\nabla_{\mathrm{N}_{a}}^{2}c_{k}\left(\mathrm{R}\right)+2\left[\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\nabla_{\mathrm{N}_{a}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right]\right\} &#xA;\end{array} $$&#xA;Multiplying from the left by $ \Psi_{j}^{\star}\left(\mathrm{R},\,\mathbf{r}\right) $ and integrating over electron coordinates, remembering that $ \left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle =\delta_{jk} $&#xA;$$&#xA;\begin{array}{c}&#xA;-{\displaystyle \sum_{a}}\dfrac{\hbar^{2}}{2M_{a}}\nabla_{\mathrm{N}_{a}}^{2}c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle +2\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \right\} =\\&#xA;=\hat{\mathcal{T}}_{\mathrm{N}}c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle +2\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \right\} &#xA;\end{array}&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Second term of eq.(3)&lt;/strong&gt; &lt;br/&gt; Multiplying by $\Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)$ and integrating over electron coordinates&#xA;$$&#xA;{\displaystyle \sum_{k}}c_{k}\left(\mathrm{R}\right)\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle =\mathcal{E}_{k}\left(\mathrm{R}\right)c_{k}\left(\mathrm{R}\right)&#xA;$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Final form of the molecular electronic function&lt;/strong&gt; &#xA;$$&#xA;\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)c_{k}\left(\mathrm{R}\right)-{\displaystyle \sum_{a,k}}\dfrac{\hbar^{2}}{2M_{a}}\left\{ C_{2a,jk}c_{k}\left(\mathrm{R}\right)+2C_{1a,jk}\nabla_{\mathrm{N}_{a}}c_{k}\left(\mathrm{R}\right)\right\} =\mathcal{E}_{\mathrm{tot}}c_{k}\left(\mathrm{R}\right)&#xA;$$ &lt;br/&gt; where&#xA;$$&#xA;\begin{array}{c}&#xA;C_{1a,jk}=\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}^{2}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle \\&#xA;C_{2a,jk}=\left\langle \Psi_{j}\left(\mathrm{R},\,\mathbf{r}\right)\left|\nabla_{\mathrm{N}_{a}}\right|\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)\right\rangle &#xA;\end{array}&#xA;$$ &lt;br/&gt;&#xA;By neglecting the coupling terms&#xA;$$&#xA;\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)c_{k}\left(\mathrm{R}\right)=\mathcal{E}_{\mathrm{tot}}c_{k}\left(\mathrm{R}\right)&#xA;$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within this picture, the $c_{k}\left(\mathrm{R}\right)$&#xA;  behave, in a natural way, as a nuclear eigenfunction.&#xA;$$&#xA;\begin{array}{c}&#xA;\left(\hat{\mathcal{T}}_{\mathrm{N}}+\mathcal{E}_{k}\left(\mathrm{R}\right)\right)\Phi_{k}\left(\mathrm{R}\right)=\mathcal{E}_{\mathrm{tot}}\Phi_{k}\left(\mathrm{R}\right)\\&#xA;\hat{\mathcal{H}}_{\mathrm{e}}\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)=\mathcal{E}_{k}\left(\mathrm{R}\right)\Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;While computing the &lt;strong&gt;Second term of eq.(3)&lt;/strong&gt;, I assumed $ \Psi_{k}\left(\mathrm{R},\,\mathbf{r}\right)$&#xA;as an eigenfunction of the electronic Hamiltonian. How can this assumption be demonstrated?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you think that the &lt;strong&gt;Assumption 3&lt;/strong&gt; is correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="7692" LastEditDate="2014-11-13T11:33:19.103" LastActivityDate="2014-11-13T13:23:30.453" Title="Born-Oppenheimer Approximation reloaded" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19433" PostTypeId="1" CreationDate="2014-11-12T17:52:06.123" Score="6" ViewCount="9019" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Dpkq.gif&quot; alt=&quot;Iodine Pentafluoride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this lewis diagram, Iodine has 12 valence electrons? How is the possible? Only thing I can think of is that I'm mistaken in counting the covalent bonds as two valence electrons each. Do they only count as one in this scenario? &lt;/p&gt;&#xA;" OwnerUserId="8138" LastEditorUserId="15751" LastEditDate="2019-01-04T10:36:07.093" LastActivityDate="2019-01-04T10:36:07.093" Title="How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19434" PostTypeId="2" ParentId="17174" CreationDate="2014-11-12T18:02:58.453" Score="2" Body="&lt;p&gt;In this answer, I will attempt to explain using chemical principles, rather than detailed mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been said already, reactions 1 and 3 both consume $\ce{S2O8^{2-}}$. Now, for the rate determining step to dominate the decay of this species, there would necessarily be very little consumption of it in any other steps. IE, there would be little consumption of it in reaction 3, implying that there is very little supply of the $\ce{CO2^-}$ radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, unfortunately for the complexity of the kinetics, there is a rapid interconversion of $\ce{SO4^-}$ to $\ce{CO2^-}$ by reaction with formate, which in turn rapidly reacts with more $\ce{S2O8^{2-}}$ to reform the $\ce{SO4^-}$ which can react again with formate, you get the picture. So, although there will only be a very small concentration of $\ce{SO4^-}$ and $\ce{CO2^-}$ present (by virtue of the slow reaction in step 1), the sheer relative speed of formate induced interconversion makes these other steps important. Of course, the more formate we have, the more $\ce{CO2^-}$ we can liberate (to a point, hence the square root), and the faster we can remove $\ce{S2O8^{2-}}$. This is, in fact, a catalytic cycle within the greater reaction scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what is meant by the steady state approximation. As the concentrations of $\ce{SO4^-}$ and $\ce{CO2^-}$ are always very small, we can assume that their total rate of change across all reactions in the scheme is (extremely close to) zero. This allows us to write a set of simultaneous equations from which we can try to determine the (often quite complex) rate law by estimating the &quot;steady state concentrations&quot; of the low concentration species in terms of the other species of interest.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastEditorUserId="4945" LastEditDate="2014-11-18T06:03:17.070" LastActivityDate="2014-11-18T06:03:17.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19436" PostTypeId="2" ParentId="19433" CreationDate="2014-11-12T18:20:02.380" Score="1" Body="&lt;p&gt;Ah, iodine pentafluoride. This is a lovely example of a molecule that doesn't follow simple Lewis theory. You will find that this is often the case for heavier elements, where electrons from multiple subshells can contribute to the bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of iodine, the 5p and 5s subshells and the 4d subshells are actually quite close in energy to each other, and are ALL valence subshells. So, iodine can use all of these electrons in chemical bonding: 2 in the 5s, 5 in the 5p and up to 10 in the 4d, although it only needs to use 4 of its 4d electrons to do the bonding in this molecule. Hence it can have more than 8 electrons involved in its bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, the molecule has sp$^3$d$^2$ hybridized bonding, but in simpler terms, iodine (roughly!) forms one normal covalent bond to a F atom, four dative covalent bonds to 4 F atoms, leaving one lone pair remaining. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might therefore expect from valence shell electron repulsion theory (VSEPR), there are 4 equivalent F atoms lying in a plane, with one vertically above the plane with a slightly shorter bond length.&lt;/p&gt;&#xA;" OwnerUserId="9720" LastActivityDate="2014-11-12T18:20:02.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19437" PostTypeId="2" ParentId="19433" CreationDate="2014-11-12T19:37:09.583" Score="11" Body="&lt;p&gt;Explanations of bonding and hybridization in non-metals that involve d-orbitals have generally fallen out of favor for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Usually d-orbitals are considerably higher in energy than the s- and p-orbitals they would interact with leading to poor mixing, and&lt;/li&gt;&#xA;&lt;li&gt;d-orbitals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A reasonable hybridization alternative involves what is termed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypercoordinated bonding&lt;/a&gt;; where 3 center, 4 electron bonds are involved.  Such bonding involves p-orbitals with overlap geometry and molecular orbitals as pictured in the following diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7cmy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{IF5}$, the molecule has the following geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8NcZi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the hypercoordinated bonding approach, one would say that the 4 identical &quot;equatorial&quot; fluorines (they're not quite equatorial, we'll return to that later) and 2 p-orbitals on iodine would participate in this type of bonding.  Each of the two iodine p-orbitals contains two electrons from iodine and bonds to two fluorines which each contribute one electron.  We wind up with two linear hypercoordinate bonds at right angles to each other that lie in a plane.  Each of these bonds is comprised of 3 atoms ($\ce{F-I-F}$) and contains 4 electrons as described in the MO diagram above.  The only orbitals remaining on iodine are an s- and a p-orbital.  These two orbitals mix in traditional fashion (like the carbon in acetylene) to produce two $\ce{sp}$ orbitals.  these two orbitals are arranged in a linear fashion (again, like acetylene) and are perpendicular to the plane containing the two 3 atom - 4 electron hypercoordinate bonds.  One of these $\ce{sp}$ orbitals contains an iodine lone pair of electrons and the other forms a bond to fluorine using one iodine electron and one fluorine electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the asymmetry introduced by having a lone pair on one side of the molecule and a fluorine on the other side, our 4 &quot;equatorial&quot; $\ce{I-F}$ bonds distort from planarity tilting away from the electrostatically repulsive lone pair as illustrated in our figure (above) showing the geometry of $\ce{IF5}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this bonding scenario the central iodine is sharing 4 electrons in each of the two hypercoordiante bonds (total=8) and two electrons in the apical $\ce{I-F}$ bond; plus there is a lone pair on the iodine.  Therefor the iodine is neutral (7-[(8+2)/2]-2 = 0), but is it surrounded by an octet of electrons?  The answer is, yes.  Look again at the MO diagram for the hypercoordinate 3 center - 4 electron bond.  Notice that the nonbonding orbital ($\Psi_2$) contains 2 electrons, but these electrons are centered entirely on the fluorine atoms, not the central iodine.  Therefore, around iodine we have 2 (lone pair) plus 2 (apical $\ce{C-F}$ $\ce{sp}$ hybridized bond) plus 2 times the two $\Psi_1$ electrons in our hypercoordinate bonds, yielding a total of 8 electrons around the iodine.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-12T19:37:09.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19438" PostTypeId="1" CreationDate="2014-11-12T20:44:39.297" Score="3" ViewCount="2214" Body="&lt;p&gt;I want to retrieve cryotubes from the bottom of a liquid nitrogen freezer without taking all the freezer racks out and draining the liquid nitrogen. I think I can accomplish this using a tool like &lt;a href=&quot;http://www.homedepot.com/p/GrabEasy-Grabber-and-Retriever-PF0401/202505170&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; but I am not sure how well the tool will function at temperatures as low as -160 Celsius. Also, this particular tool is probably only about half as long as it needs to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming I can find a longer version of such a tool, &lt;strong&gt;how can I determine whether or not I need to be concerned about using to retrieve cryotubes from a liquid nitrogen freezer?&lt;/strong&gt; My main worry is that parts of the tool may become brittle and break or that the claw's ability grab may somehow become compromised. I am not concerned about the metal becoming excessively cold because I will be wearing gloves intended for such work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a chart or other resource that I can use to get a sense of metal's shear strength at cold temperatures?&lt;/p&gt;&#xA;" OwnerUserId="7209" LastEditorUserId="7209" LastEditDate="2014-11-12T23:10:09.313" LastActivityDate="2014-11-13T01:14:50.057" Title="How will liquid nitrogen affect a steel instrument's ability to function?" Tags="&lt;thermodynamics&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19440" PostTypeId="1" CreationDate="2014-11-12T21:38:32.207" Score="2" ViewCount="1319" Body="&lt;p&gt;Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could happen if I would strongly boil it.&#xA;Would the OH- react into the carboxyl groups to form a diol?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastActivityDate="2014-11-12T21:56:15.057" Title="Why does in the hydrolysis of acetylsalicylic acid with NaOH the solution has to be simmered gently?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19441" PostTypeId="2" ParentId="19440" CreationDate="2014-11-12T21:56:15.057" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What could happen if I would strongly boil it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mechanistically, nothing different will occur.  The same reaction that occurs at lower temperature will still occur at higher temperature.  The only difference will be the &lt;strong&gt;rate&lt;/strong&gt; of the reaction; the reaction will go faster as you increase the reaction temperature.  In a classroom setting the procedure might involve simmering rather than boiling because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The reaction is usually followed by making uv-vis absorbance measurements every 15, or so, minutes.  Depending upon the reaction temperature and the concentration of base, the reaction might proceed too fast at boiling to allow for convenient measurements (&lt;em&gt;e.g&lt;/em&gt;. the reaction might be complete in 10 minutes).&lt;/li&gt;&#xA;&lt;li&gt;Safety reasons, if something goes wrong (the flask cracks) a warm basic solution won't burn the student as bad as a boiling basic solution would.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-12T21:56:15.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19442" PostTypeId="2" ParentId="19438" CreationDate="2014-11-12T22:54:02.960" Score="2" Body="&lt;p&gt;Metals become brittle as low temperature (and that's why you cool your Parr bomb in a dry ice bath, never in liquid nitrogen when you condense gases in).  But the worst thing that can happen is that a claw might break off, which shouldn't cause any damage to equipment.  Why not try in a small dewar first.  Liquid nitrogen is cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good alternative to racks of storage boxes are storage canes, such as &lt;a href=&quot;http://www.thermoscientific.com/en/product/storage-canes.html&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-11-13T01:14:50.057" LastActivityDate="2014-11-13T01:14:50.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19447" PostTypeId="1" CreationDate="2014-11-13T02:22:07.520" Score="4" ViewCount="116" Body="&lt;p&gt;A pretty rudimentary question, but I'm reviewing the concept of energy diagrams and activation energy and I'm curious to know how chemists find the energy of reactants and of a transition state in a laboratory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kinds of instruments are used ? As well, aren't transition states only something we can imagine ? So how is it that chemistry are able to find their energy in a lab ? &lt;/p&gt;&#xA;" OwnerUserId="8260" LastActivityDate="2014-12-13T03:43:34.653" Title="How is the energy of reactants in a chemical reaction found ?" Tags="&lt;free-energy&gt;&lt;transition-state-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19449" PostTypeId="1" CreationDate="2014-11-13T02:54:17.783" Score="5" ViewCount="4031" Body="&lt;p&gt;I'm handling 0.12N of sulfuric acid for a science experiment that I am carrying out for a school project. I hear about all these terrible effects if it's ingested, inhaled, or gets in contact in skin. But how dangerous is it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes. However, I do not have a respirator and wondering if I should get one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would it be safe to handle this without a respirator?&lt;/strong&gt; I will not be spraying the sulfuric acid directly in the air. However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. &lt;strong&gt;Will it still be dangerous even spraying a low concentration in the air?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity?&lt;/strong&gt; Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa? I would really appreciate the advice that you can provide me. &lt;/p&gt;&#xA;" OwnerUserId="9697" LastActivityDate="2014-12-16T02:24:32.313" Title="What are the dangers of handling 0.12N sulfuric acid?" Tags="&lt;safety&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19450" PostTypeId="2" ParentId="19447" CreationDate="2014-11-13T03:20:43.060" Score="2" Body="&lt;p&gt;Energy itself is usually not measured, but rather &lt;em&gt;change&lt;/em&gt; in energy ($\Delta E$) is the quantity of practical importance. One may be interested in change in free energy as well, $\Delta G$, changes in enthaly $\Delta H$, and even change in entropy $\Delta S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common way to evaluate the increased energy of transition state ($\Delta E^{\ddagger}$) above the reactant is via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrhenius_equation#Arrhenius_plot&quot; rel=&quot;nofollow&quot;&gt;Arhennius Plot&lt;/a&gt;.  Here, one evaluates the rates of reaction ($k$) at a set of different temperatures ($T$ in Kelvin).  It is fairly easy to show that a plot of the $\ln(k)$ vs. $1/T$ gives a line whose slope is related to $\Delta H^{\ddagger}$, and where the intercept is related to $\Delta S^{\ddagger}$. (The ideal gas constant is in here too, but it's easy to accomodate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There is an analogous plot for energy change for a reaction, which is called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Van_.27t_Hoff_plot&quot; rel=&quot;nofollow&quot;&gt;van't Hoff plot&lt;/a&gt;, but it generally requires both reactant and product to be in equilibrium such that the concentrations of both are observable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially the enthalpy of reaction can be found by methods such as bomb calorimetry.  This measures the heat that is liberated upon reaction.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-11-13T03:26:37.257" LastActivityDate="2014-11-13T03:26:37.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19452" PostTypeId="1" CreationDate="2014-11-13T03:34:59.477" Score="2" ViewCount="661" Body="&lt;p&gt;Can the protons of two elements be swapped in a chemical reaction? &lt;/p&gt;&#xA;" OwnerUserId="9754" LastActivityDate="2014-11-13T03:48:11.190" Title="Can protons be swapped in a chemical reaction?" Tags="&lt;acid-base&gt;&lt;electrons&gt;&lt;elements&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19453" PostTypeId="2" ParentId="19452" CreationDate="2014-11-13T03:48:11.190" Score="2" Body="&lt;p&gt;Roughly speaking, no.   This would amount to nuclear fission/fusion, and is generally not considered &quot;chemistry&quot; in any colloquial sense of the term, but rather physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, chemical reactions consider nuclei to be tightly bound by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;strong force&lt;/em&gt;&lt;/a&gt;, and immutable over the course of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: there is an ill-defined border between physics and chemistry, and so one could argue that I am wrong about this.  Indeed, &lt;em&gt;alchemists&lt;/em&gt; had/have been trying to e.g. convert lead to gold for centuries, and this is related to the reaction that you propose.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2014-11-13T03:48:11.190" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19454" PostTypeId="1" AcceptedAnswerId="19468" CreationDate="2014-11-13T04:01:47.027" Score="5" ViewCount="1924" Body="&lt;p&gt;My first question is: What exactly happens on a point situated on a line? Or on a point such as E in this figure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KVymT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does point E mean that the $\delta , \gamma + \delta , \delta + \epsilon , \gamma + \epsilon $ phases all coexist here? If so, then why would it be considered a eutectoid invariant point where three phases coexist if in reality there are more than three phases coexisting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question is: How can I apply the tie line and lever rules on a point on a point situated on top of a line, if I want to find Liquid percentage Solid Percentage, and maybe concentrations of one component in the liquid phase and solid phase? Is it undefined on top of the line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7iGaO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, regarding the nature of the lever rule, is it some sort of empirical rule without any real theoretical foundation? At first glance, I can't believe it's actually accurate. Why is it true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and sorry if my questions are unclear...&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="73" LastEditDate="2014-11-13T06:07:21.977" LastActivityDate="2014-11-13T15:04:12.177" Title="Boundary lines in phase diagrams and the lever rule" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19455" PostTypeId="2" ParentId="19134" CreationDate="2014-11-13T04:39:02.673" Score="5" Body="&lt;p&gt;The EPR experiment measures the interaction of unpaired spin density ($\rho(\alpha)$) &lt;em&gt;at the nuclei&lt;/em&gt;.  This is quantified by an equation called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fermi_contact_interaction&quot; rel=&quot;noreferrer&quot;&gt;Fermi contact interaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have some prior knowledge of &quot;where to put the dot&quot; in your structure (highly localized), then generally you can assume that the EPR pattern is split most by that nucleus.  Splitting patterns are determined by standard rules in accord with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Zeeman_effect&quot; rel=&quot;noreferrer&quot;&gt;Zeeman effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus must have a non-zero nuclear spin to be observable. Certain nuclei such as hydrogen-1 and many transition metals have spin. It is very common that isotopic impurities (e.g. carbon-13) appear though the most prevalent isotope is EPR-silent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are other complications, such as nuclear quadrupole moment and broadening due to exchange of chemical sites.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional issue is that the spin density is generally spread over the molecule, and interacts with multiple nuclei.  It turns out to be easier to measure the spectrum, and then rationalize it given the particular structure and nuclear composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accurate prediction of EPR spectra is a rather difficult problem, as quantum chemical calculations generally don't emphasize basis function placement at the nucleus, as they tend to focus on the &quot;valence&quot; and not the cusp-like nature of the radial distribution.  (Here, Slater-type orbitals would be better in principle than Gaussian-type, but STOs of sufficient number and of high-enough quality of Hamiltonian are virtually impossible).  Finally, solvent and powder effects can render the gas-phase ab initio calculations virtually irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EPR simulation/interpretation of fine structure is much more difficult than NMR spectra, because in NMR spectra, one has the chemical shift to conveniently move splitting patterns out of the way.  In EPR, the splittings are often &quot;piled&quot; on top of each other, and make it necessary to use computer simulation to obtain precise fits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific example, I would identify the nuclei with spin (probably Cu, certainly phosphorus-31, and then hydrogen-1).  Then, identify where you think the unpaired density will be, and then take the nuclear spin + 1 and that will be the &quot;guesstimated&quot; isotropic spectrum.  The pattern will follow some variant Pascal's triangle if I=1/2, otherwise, you have to figure it out by writing out the number of possible configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have not even started on the characteristics of triplet spectra, as here the electrons interact.  One thing you can hope for is that all of your electrons are paired (singlet), and so it's EPR silent.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2014-11-13T05:03:59.510" LastActivityDate="2014-11-13T05:03:59.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19456" PostTypeId="1" AcceptedAnswerId="19458" CreationDate="2014-11-13T04:51:41.373" Score="0" ViewCount="146" Body="&lt;p&gt;I'm thinking in the following experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A)&lt;/strong&gt; Two (non mixable (like water and oil)) liquids are going to lie in a container, in the extremes of the container there are two electrodes, those electrodes will have a potential difference $\Phi$. Just as the figure $1.a)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for  some interesting liquids that mix, change of color when a voltage difference is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZzDLy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B)&lt;/strong&gt; The next one is really similar, now there are some solid particle present, analogously I look for a mixture that has different properties when a voltage difference is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a physics student that want to study different properties of mixtures when applying different voltage differences, and maybe study the diffusion of the solid particles when they mix, if the diffusion coefficient is grater, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with the liquids that has this property is really  appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross posted &lt;a href=&quot;https://physics.stackexchange.com/questions/146463/mixing-liquids-when-applying-a-voltage&quot;&gt;Physics stack Exchange&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9755" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-11-13T05:59:27.390" Title="Mixing liquids when applying a voltage" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19458" PostTypeId="2" ParentId="19456" CreationDate="2014-11-13T05:59:27.390" Score="2" Body="&lt;p&gt;I won't comment on B for the moment, but liquid-liquid interface electrochemistry is quite a vibrant field. There's a review of it &lt;a href=&quot;http://www.researchgate.net/publication/222670534_Electrochemistry_at_liquidliquid_interfaces_methodology_and_potential_applications/links/09e4150f467e1aab95000000ztOOmxIPWfw&amp;amp;bvm=bv.79189006,d.aWw&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, if you're interested. The simplest interaction involves ions: if one applies a potential across the interface of sufficient strength and polarity, the energy barrier of transferring a given ion from one phase to the other can be overcome. If you do this with something like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solvatochromism&quot; rel=&quot;nofollow&quot;&gt;solvatochromic&lt;/a&gt; dye, it will change colour as it moves from one phase to the other. (It would have to be something that doesn't partition into the second phase easily on its own, of course) One could also have some reagent in one phase that reacts with the ion being pushed across the interface that produces some coloured product. (iodide-starch, perhaps?) There are also voltage-sensitive dyes, though I don't have any experience with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to note is that neither water nor oils and other immiscible solvents are generally very conductive. To keep the cell resistance down to a reasonable level, it will be necessary to dissolve electrolytes in both. Things like Tetrabutylammonium hexafluorophosphate are commonly used as electrolyte in non-aqueous electrochemistry.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-13T05:59:27.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19459" PostTypeId="1" CreationDate="2014-11-13T10:03:16.940" Score="2" ViewCount="137" Body="&lt;p&gt;In an Evans aldol reaction, the enolate is typically generated by addition of a boron Lewis acid (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{(n-Bu)2BOTf}$&lt;/span&gt;) and an amine, which acts as a weak base. I'd like to know whether there's any difference between &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Triisobutylamine, &lt;span class=&quot;math-container&quot;&gt;$\ce{(i-Bu)3N}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-Diisopropylethylamine (Hünig's base), &lt;span class=&quot;math-container&quot;&gt;$\ce{(i-Pr)2NEt}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Triethylamine, &lt;span class=&quot;math-container&quot;&gt;$\ce{Et3N}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;in the selectivity of the enolate formation and subsequent aldol reaction. I'm thinking about the different &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; values (1&gt;2&gt;3) and the different steric demand (1&gt;2&gt;3). Probably the base with the highest &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; will be kinetically fastest. Also, the &lt;em&gt;cis&lt;/em&gt;-enolate should be favoured as it avoids a destabilising &lt;em&gt;syn&lt;/em&gt;-pentane interaction between the terminal methyl group and the auxiliary. However, do all of the bases lead to the same ratio of &lt;em&gt;cis&lt;/em&gt;/&lt;em&gt;trans&lt;/em&gt;-enolate formation (and hence diastereoselectivity in the subsequent aldol reaction)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9NSCi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9NSCi.png&quot; alt=&quot;Generation of enolate for Evans aldol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2018-10-14T22:47:03.077" LastActivityDate="2018-10-14T22:47:03.077" Title="Does the amine used to generate the enolate affect the selectivity of an Evans aldol reaction?" Tags="&lt;organic-chemistry&gt;&lt;stereoselectivity&gt;&lt;aldol-reaction&gt;&lt;enolate-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19463" PostTypeId="2" ParentId="19432" CreationDate="2014-11-13T13:23:30.453" Score="4" Body="&lt;p&gt;Coincidentally, I'm also currently writing my dissertation and I do also have section discussing Born-Oppenheimer approximation. :) My description is quite similar to yours, so I do not see any critical issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to your question, we do not assume that $\Psi_k(R,r)$ is an an eigenfunction of the electronic Hamiltonian, we know that for sure. If you look at equation (2) then essentially what is going on here is the following. Any well-behaved function of two independent variables $\Psi(R,r)$ can always be expanded over the complete set of functions $\{ g_{i}(R) \}$ and $\{ h_{j}(r) \}$ in the following way&#xA;$$&#xA;\Psi(R,r) = \sum\limits_{i} \sum\limits_{j} c_{ij} g_{i}(R) h_{j}(r) \, ,&#xA;$$&#xA;or by defining&#xA;$$&#xA;c_{j}(R) = \sum\limits_{i} c_{ij} g_{i}(R) \, ,&#xA;$$&#xA;as&#xA;$$&#xA;\Psi(R,r) = \sum\limits_{j} c_{j}(R) h_{j}(r) \, .&#xA;$$&#xA;Now as a complete set $\{ h_{j}(r) \}$ of functions of electronic coordinates we can choose the set of electronic wave functions $\Psi_k(R,r)$ which are by definition eigenfunctions of electronic Hamiltonian. The fact that these function carry parametric dependence on nuclear coordinates does not matter here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, this expansion is correct. For more on that story check &lt;a href=&quot;https://physics.stackexchange.com/questions/64320/expanding-two-variable-function-fx-y-over-the-complete-sets-g-ix&quot;&gt;this question at Physics.SE&lt;/a&gt; I asked about a year ago. &lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-11-13T13:23:30.453" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19465" PostTypeId="1" AcceptedAnswerId="19469" CreationDate="2014-11-13T13:45:36.337" Score="1" ViewCount="31145" Body="&lt;p&gt;I learnt at school that when sulfur dioxide gas, $\ce{SO2}$, is combined with potassium permanganate, $\ce{KMnO4}$, the purple solution will become colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens in the solution to cause this to happen?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="16683" LastEditDate="2017-11-12T16:50:23.343" LastActivityDate="2017-11-12T17:02:02.403" Title="Why does potassium permanganate turn colourless when reacted with sulfur dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19466" PostTypeId="2" ParentId="18715" CreationDate="2014-11-13T14:27:06.260" Score="3" Body="&lt;p&gt;I know this does not answer what you specifically asked for, but it may be a good way to get quick analyses of coins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to get a quick scan with an X-Ray Fluorescence gun (XRF gun), if you want to use a modern, non-destructive analytical technique. It is common for firms and certain types of stores to have one such gun, for example jeweller stores keep such guns to check the gold or silver purity, or toy stores may keep a gun to check for dangerous elements in imported toys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you could talk yourself into getting a free scan for your school project. They may have pre-installed metal calibration programs, and a scan takes 30 seconds. if your coin is plated with some metal, and the rest of the coin is some other metal-alloy, you can see this with an XRF gun if you scratch your coin on one side before analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;XRF is a nice technique, and you should be able to have a good idea of how it works (in principle) with theory from high school chemistry or physics. Modern chemical analysis is not &quot;wet chemistry&quot;, or flame tests; it is instrumental analysis (although wet techniques are still used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, see if you can find a local store or university or someone who has an XRF gun. That would give you the proportions (%) of the major components and a standard deviation of the analysis (error or uncertainty).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="5334" LastEditDate="2014-11-13T14:33:17.887" LastActivityDate="2014-11-13T14:33:17.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19467" PostTypeId="1" AcceptedAnswerId="19474" CreationDate="2014-11-13T14:53:36.020" Score="4" ViewCount="8002" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the reaction&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$ \ce{2N2O5(g) -&amp;gt; 4NO2 + O2(g)} $$&lt;/span&gt;&#xA;  the rate law is: &#xA;  &lt;span class=&quot;math-container&quot;&gt;$$ \frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k[\ce{N2O5}] $$&lt;/span&gt;&#xA;  At &lt;span class=&quot;math-container&quot;&gt;$\pu{300 K}$&lt;/span&gt;, the half-life is &lt;span class=&quot;math-container&quot;&gt;$\pu{2.50E4 s}$&lt;/span&gt; and the activation energy is &lt;span class=&quot;math-container&quot;&gt;$\pu{103.3 kJ/mol}$&lt;/span&gt;.&lt;br&gt;&#xA;  What is the rate constant at &lt;span class=&quot;math-container&quot;&gt;$\pu{350 K}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know there is something fishy about the rate law, but I can't make sense of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\frac{ln2}{k} &amp;amp;= \pu{2.50E4}\\&#xA;k &amp;amp;= \pu{2.773E-5}\\&#xA;\frac{\pu{2.773E-5}}{k_2} &amp;amp;= \frac{A\cdot\exp\left\{\frac{-103300}{8.314\times300}\right\} }{ A\cdot\exp\left\{\frac{-103300}{8.314\times350}\right\}}\\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding &lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt; from this gives me a weird value: &lt;span class=&quot;math-container&quot;&gt;$k_2 = 0.0103$&lt;/span&gt;.&#xA;The answer for this question is &lt;span class=&quot;math-container&quot;&gt;$\pu{7.47E-8 s^-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="9760" LastEditorUserId="4945" LastEditDate="2020-02-04T14:59:07.987" LastActivityDate="2020-02-04T15:06:18.283" Title="How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19468" PostTypeId="2" ParentId="19454" CreationDate="2014-11-13T15:04:12.177" Score="5" Body="&lt;p&gt;Lots of questions wrapped in one, but they do all come down to understanding what is represented on a phase diagram.  Up front, I'll suggest Porter and Easterling's classic text on 'Phase Transformation in Metals and Alloys' - it has lots of good info for lovers of thermodynamics and phase transitions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a system like Cu-Zn that you have pictured above, there are a number of different phases observed.  You could imagine that at a given temperature you could construct a free energy curve for each phase as a function of composition. At each composition, only one will have the lowest free energy, and one might think you could identify the phase diagram that way. However, a straight line across the G vs. composition plot is special - it represents where the chemical potentials of both elements are constant.  So a line that is tangent to two free energy curves represents a range of compositions where the elements in the two phases have the same chemical potential.  This then is how you construct the phase diagram - find the minimum free energy across composition, then find the common tangents (if any). An example is shown here: &lt;img src=&quot;https://i.stack.imgur.com/lUeqY.jpg&quot; alt=&quot;Phase Diagram Visualization&quot;&gt;&#xA;The top plot is the free energy curves for all the phases, the bottom plot is the phase boundaries determined by the common tangents.  The dashed vertical lines up top are where the common tangent points are, coming down to their position on the phase diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the $\delta$, $\gamma$, $\epsilon$ are specific phases.  In regions with just one of them listed, that is the only phase stable at the temperature and concentration.  Regions such as $\delta + \gamma$ are regions where both are stable. There relative amount has to follow the lever rule, which is just a linear variation across the two-phase region from, say, all $\gamma$ right on the boundary of $\gamma$ with $\gamma + \delta$, to all $\delta$ on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal lines are there mostly to emphasize when a new phase appears or disappears.&#xA;So, at point E on you diagram above, there is one specific temperature and composition at which $\delta$, $\gamma$, and $\epsilon$ are in equilibrium.  Go up in temperature, and it wants to be all $\delta$; come down in temperature and you enter the two-phase $\gamma$-$\delta$ region with the relative abundance from the lever rule. At P, the horizontal line is to point out that $\epsilon$ disappears into the $\delta$-liquid region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at the eutectoid E, there is one point where the three phases coexist (Gibbs phase rule). Moving away from that point moves into single or two-phase regions. Everywhere along the horizontal line through E is $\gamma$+$\epsilon$, except at exactly point E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the lever rule, that follow directly from coming along the common tangent - the ratios of the two phases have to vary linearly or else you are deviating from the common tangent.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-11-13T15:04:12.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19469" PostTypeId="2" ParentId="19465" CreationDate="2014-11-13T15:04:13.363" Score="4" Body="&lt;p&gt;This is a redox reaction. The permanganate ion (purple) is reduced to manganese(II) ion which is a very pale pink, or almost colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4- + 8H+ + 5e- -&amp;gt; Mn^2+ + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while sulfur dioxide is oxidized to the sulfate ion, which is colourless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO2 + 2H2O -&amp;gt; SO4^2- + 4H+ + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the solution is decolourised. The overall equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2MnO4- + 5SO2 + 2H2O -&amp;gt; 2Mn^2+ + 5SO4^2- + 4 H+}$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16683" LastEditDate="2017-11-12T17:02:02.403" LastActivityDate="2017-11-12T17:02:02.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19470" PostTypeId="2" ParentId="13885" CreationDate="2014-11-13T15:05:14.760" Score="3" Body="&lt;p&gt;First of all, after degasing by heating you have to be sure that no $\ce{CaCO_3}$ is formed, so you can minimise this problem by transfering the water in another flask.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;More accurate would be to start directly with demineralised water which you can buy cheap at filling stations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Secondly, you shouldn't wait to much while titrating, otherwise the water is taking up $\ce{CO_2}$ from the air.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You should use a bulb which doesn't contain much air (which contains $\ce{CO_2}$ and can interact as well. You can use a round flask and directly connect it with a stopper.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Third, Phenolphtalein is a good indicator because you titrate a weak acid with a strong base.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make sure, you don't use too much Phenolphtalein.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Last but not least be sure that the $\ce{CO_2}$ from flask doesn't go away in the air or into the NaoH of the burette.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;To cool will help, because $\ce{CO_2}$ is more soluble in cold water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to be a super-purist, do a potentiometric measurement and plot the point of change using a simple voltmeter.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7201" LastEditDate="2014-11-13T17:52:50.527" LastActivityDate="2014-11-13T17:52:50.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19471" PostTypeId="2" ParentId="19467" CreationDate="2014-11-13T15:12:11.927" Score="2" Body="&lt;p&gt;When using Arrhenius equation: you have to multiply the activation energy by &lt;span class=&quot;math-container&quot;&gt;$1000$&lt;/span&gt;, because it must be in &lt;span class=&quot;math-container&quot;&gt;$\pu{J}$&lt;/span&gt; and not &lt;span class=&quot;math-container&quot;&gt;$\pu{kJ}$&lt;/span&gt;.&#xA;you must also divide by &lt;span class=&quot;math-container&quot;&gt;$RT$&lt;/span&gt; and not by &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; as you did.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2020-02-04T15:06:18.283" LastActivityDate="2020-02-04T15:06:18.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19473" PostTypeId="2" ParentId="4667" CreationDate="2014-11-13T15:45:04.557" Score="1" Body="&lt;p&gt;Colored compounds of transition elements are associated with partially filled (n-1)d orbitals. Tthe transition metal ions containing unpaired d-electrons undergoes an electronic transition from one d-orbital to another. During this d-d transition process, the electrons absorb certain energy from the radiation and emit the remainder of energy as colored light. The color of ion is complementary of the color absorbed by it. hence, colored ion is formed due to d-d transition which falls in the visible region for all transition elements. &lt;/p&gt;&#xA;" OwnerUserId="9762" LastEditorUserId="402" LastEditDate="2017-04-02T20:38:05.237" LastActivityDate="2017-04-02T20:38:05.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19474" PostTypeId="2" ParentId="19467" CreationDate="2014-11-13T16:45:44.507" Score="2" Body="&lt;p&gt;I arrived at the same &lt;span class=&quot;math-container&quot;&gt;$k_{\pu{300 K}}$&lt;/span&gt; as you did. I find it a little weird that there's a 3 order of magnitude decrease in rate constant for a &lt;span class=&quot;math-container&quot;&gt;$\pu{50 K}$&lt;/span&gt; increase in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per the note below, the original equation I put had a mathematical error. The below should be correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &amp;amp;= \exp\left\{\frac{E_\mathrm{a}}{R\left(\frac{1}{T_{\pu{300 K}}} - \frac{1}{T_{\pu{350 K}}}\right)}\right\}\\&#xA;\frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &amp;amp;= \exp\left\{\frac{103000}{8.314\left(\frac{1}{300} - \frac{1}{350}\right)}\right\}=371.14&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take &lt;span class=&quot;math-container&quot;&gt;$k_{\pu{300 K}} = \pu{2.773E-5}$&lt;/span&gt; and multiply by that factor above (as you should) you get your answer, if you take &lt;span class=&quot;math-container&quot;&gt;$k_{\pu{300 K}}$&lt;/span&gt; and divide by the factor above, you get the book's answer, which is where I believe their mistake is. &lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="4945" LastEditDate="2020-02-04T15:05:30.377" LastActivityDate="2020-02-04T15:05:30.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19475" PostTypeId="1" AcceptedAnswerId="19485" CreationDate="2014-11-13T17:22:29.533" Score="12" ViewCount="1901" Body="&lt;p&gt;How many different organic structures (from the pure theoretical viewpoint) can be drawed with only 4 (exact) carbon atoms and with/without hydrogen? Polycyclic compounds and bridged compounds are also allowed. &lt;strong&gt;Please, name them all as well!&lt;/strong&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark: I have drawed the chemical graphs in a piece of paper (the skeletons) and I get 37 graphs (including linear with simple, double and triple bonds, and planar and nonplanar compounds), without geometric isomerism (cis/trans). Is that OK?&lt;/p&gt;&#xA;" OwnerUserId="3857" LastEditorUserId="19616" LastEditDate="2015-10-02T16:04:46.147" LastActivityDate="2016-05-25T12:43:27.317" Title="Hydrocarbons with only 4 carbon atoms" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19476" PostTypeId="1" AcceptedAnswerId="19518" CreationDate="2014-11-13T17:35:49.403" Score="2" ViewCount="211" Body="&lt;p&gt;I am referring to rocks in this instance. Individual crystals grow together in an igneous rock as it solidifies. I know that the $\ce{SiO4}$ tetrahedron is the basic building unit of silicate minerals, and metal cations attach to ensure electroneutrality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am wondering, however, is what force keeps the crystals together (as in bound to one another) in the solid? For comparison, sediment grains are held in place by a silica or calcite cement in fragmental rocks, but what stops individual crystals from breaking free in crystalline solids?&lt;/p&gt;&#xA;" OwnerUserId="9763" LastEditorUserId="8083" LastEditDate="2016-02-02T11:41:22.953" LastActivityDate="2016-02-02T11:41:22.953" Title="Binding forces in crystalline solids" Tags="&lt;crystal-structure&gt;&lt;rocks&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19477" PostTypeId="1" AcceptedAnswerId="19513" CreationDate="2014-11-13T18:33:01.830" Score="10" ViewCount="1815" Body="&lt;p&gt;I'm Having a real struggle trying to understand symmetry representations matrices and character tables. Is there anyone who would be kind enough to help me out here? I understand it is probably a little broad but nevertheless, I will list some areas that if anyone could shed some light on I would be very appreciative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is an irreducible representation? My book just starts going into $2\times 2$ matrices like:&#xA;\begin{equation}&#xA;\left( \begin{array}{cc}&#xA;1 &amp;amp; 0\\&#xA;0&amp;amp; 1\end{array}&#xA;\right)&#xA;\end{equation}&#xA;Saying that we can represent the effect of symmetry operations as such! When the operation actually affects the molecule by changing an atom then the diagonal of the matrix swaps round seemingly !!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Leading on from the first question, what is a reducible representation? Why the distinction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is a basis? My book introduces them via vectors down each of the cartesian axes and then a &quot;set&quot; of them together (why) to give four basis's for a $C_2v$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I can use the reduction formula to see how many irreducible representations are contained within a particular reducible representation, but what am I doing!&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="402" LastEditDate="2017-01-22T10:17:19.907" LastActivityDate="2017-01-22T10:17:19.907" Title="Symmetry representations" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19478" PostTypeId="1" CreationDate="2014-11-13T19:46:21.203" Score="1" ViewCount="73" Body="&lt;p&gt;Pyrrolidine ($2~\mathrm{mol}$) is slowly added to $164~\mathrm{ml}$ conc. $\ce{HCl}$ in $200~\mathrm{g}$ ice water to adjust the $\ce{pH}$ to $3-4$. When it is mixed, the $\ce{pH}$ is $14$, how do I lower the $\ce{pH}$ to $3-4$? I cannot add more $\ce{HCl}$ because it only calls for $164~\mathrm{ml}$.&lt;/p&gt;&#xA;" OwnerUserId="9765" LastEditorUserId="4945" LastEditDate="2014-11-13T23:18:26.617" LastActivityDate="2014-11-13T23:18:26.617" Title="How do I (further) lower the pH of a pyrrolidine solution?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19480" PostTypeId="2" ParentId="19261" CreationDate="2014-11-13T19:50:22.773" Score="2" Body="&lt;p&gt;Your proposed mechanism looks reasonable. I think that acetic acid primarily serves as an acidic solvent for the reaction. However, in the presence of concentrated sulfuric acid, $\ce{CH3CO2H}$ can be protonated by $\ce{H2SO4}$, like in the &lt;a href=&quot;http://www.chemguide.co.uk/physical/catalysis/esterify.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;first step of this esterification&lt;/a&gt;. The resulting resonance-stabilized cation is strongly acidic ($\mathrm{p}K_\mathrm{a} = -6.1$ according to &lt;a href=&quot;http://books.google.de/books?id=iVr0HWkAdS8C&amp;amp;pg=PA31&amp;amp;lpg=PA31&amp;amp;dq=protonated%20acetic%20acid%20pka&amp;amp;source=bl&amp;amp;ots=W-BnYnwvJ2&amp;amp;sig=uTs_TyUGscu3_Uub9Clb0SR0n4A&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=zwZlVNGKAue7ygO074CYBw&amp;amp;ved=0CDgQ6AEwAw#v=onepage&amp;amp;q=protonated%20acetic%20acid%20pka&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;), and should readily protonate the alcohol. This would be a possibility to include acetic acid into the mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CO2H + H2SO4 &amp;lt;=&amp;gt; CH3C(OH)2^{+} + HSO4-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PyGk6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.chemguide.co.uk/physical/catalysis/esterify.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source: ChemGuide&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the first step, the alcohol can either be protonated by $\ce{H2SO4}$ or $\ce{CH3C(OH)2^{+}}$, and in the end, the eliminated proton can either be picked up by $\ce{HSO4-}$ or $\ce{H2O}$.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-06-25T05:51:29.983" LastActivityDate="2018-06-25T05:51:29.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19481" PostTypeId="1" CreationDate="2014-11-13T21:20:14.860" Score="6" ViewCount="489" Body="&lt;p&gt;Total rewrite as was a panic-stricken 2am question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I wrote on an application that VSEPR theory is a reason of Cis-Trans isomerism (specifically in fatty acids however I doubt that would make any difference). Although in my head it seemed to work at the time, I appreciated about a week after that I was actually thinking of the wrong type of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can VSEPR and/or the electron configurations in an unsaturated fatty acid explain how cis and trans isomerism arises?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I'm aware, the Pi bond prevents (resists?) rotation around the C=C link, but I feel as if I'm missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the geometry caused by the repulsion of H's and alkyl electrons on either side of the C=C bond affect the isomers differently depending on whether or not they are Cis or Trans?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One of the below posters also highlighted a good question, does anybody have any insight on this?: &#xA;&quot;I think the main question is whether there's a preference for cis or trans isomers in a fatty acid based on steric/VSEPR considerations&quot; &#xA;(Thank you for your response)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus related question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any preference for a specific stereoisomer in the dehydrogenation of a fatty acid's alkyl chain? If so, what are they and why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9767" LastEditorUserId="7475" LastEditDate="2015-10-23T17:00:44.323" LastActivityDate="2019-05-09T20:29:44.190" Title="Explaning Cis &amp; trans isomers through VSEPR" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;stereochemistry&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19482" PostTypeId="1" AcceptedAnswerId="19533" CreationDate="2014-11-13T22:23:56.960" Score="2" ViewCount="575" Body="&lt;p&gt;To generalize the problem, I will just say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5 gram of impure ethanol (or any other type of combustible material) is placed inside a bomb calorimeter where volume is constant. If the combustion produces a certain amount of energy, how should I find the mass in gram of pure ethanol contained in that quantity of impure ethanol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is what I will do:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;find the standard molar enthalpy of ethanol. &lt;/li&gt;&#xA;&lt;li&gt;write a combustion equation. &lt;/li&gt;&#xA;&lt;li&gt;find the energy produced in the combustion of one mole of ethanol. &lt;/li&gt;&#xA;&lt;li&gt;then divide that amount of ethanol by the amount of energy stated in problem. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To find how many mole of pure ethanol, convert the mole into gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are two things that make me think that this way of solving the problem is not correct:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the quantity of impure ethanol is a mixture of pure ethanol and another unknown material. What if this unknown material produce part of the energy stated in the problem?&lt;/li&gt;&#xA;&lt;li&gt;the standard molar enthalpy is measure under standard conditions (1 atm, 25 °C). However, the condition inside the bomb calorimeter is unknown. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4805" LastEditorDisplayName="user7951" LastEditDate="2018-03-19T18:29:18.160" LastActivityDate="2018-03-19T18:29:18.160" Title="Is this a valid method to find percentage of pure ethanol in an impure sample?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19483" PostTypeId="2" ParentId="19478" CreationDate="2014-11-13T23:18:20.987" Score="2" Body="&lt;p&gt;A &lt;em&gt;pH&lt;/em&gt; of ~ 4 would correspond to a solution of pyrrolidinium chloride.  Use less pyrrolidine or more hydrochloric acid to arrive at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commercial concentrated hydrochloric acid is 37 % (w/w), this would be a concentration of ~ 12 M, and 164 ml of that is slightly less than 2 mol HCl.  It's very strange that you are seeing a basic pH.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-13T23:18:20.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19485" PostTypeId="2" ParentId="19475" CreationDate="2014-11-14T00:00:01.230" Score="21" Body="&lt;p&gt;Final update, all earlier edits incorporated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Groundrules&lt;/strong&gt;: Considering compounds with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;only carbon and hydrogen&lt;/li&gt;&#xA;&lt;li&gt;only 4 bonds to carbon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are &lt;strong&gt;37&lt;/strong&gt; isomers without considering &lt;em&gt;trans&lt;/em&gt; isomers; &lt;strong&gt;49&lt;/strong&gt; when &lt;em&gt;trans&lt;/em&gt; isomers are included.  Also, many of these compounds seem extremely unstable and therefore unlikely to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note to self&lt;/strong&gt;: check back in 20 years and see how many of the unlikely ones have been detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOSSS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Names by row:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;butane, isobutane&lt;/li&gt;&#xA;&lt;li&gt;but-1-ene, but-2-ene, 2-methylpropene&lt;/li&gt;&#xA;&lt;li&gt;buta-1,3-diene, buta-1,2-diene, buta-1,2,3-triene&lt;/li&gt;&#xA;&lt;li&gt;but-1-yne, but-2-yne, but-1-ene-3-yne, buta-1,3-diyne&lt;/li&gt;&#xA;&lt;li&gt;methylcyclopropane, 2-methylcyclopropene, 1-methylcyclopropene, methylenecyclopropane, methylenecyclopropene, methlycyclopropadiene&lt;/li&gt;&#xA;&lt;li&gt;methylcyclopropyne, methlenecyclopropyne&lt;/li&gt;&#xA;&lt;li&gt;cyclobutane, cyclobutene, cyclobuta-1,2-diene, cyclobuta-1,3-diene, cyclobutatriene, cyclobutatetraene, cyclobutyne, cyclobutenyne, cyclobutadiyne&lt;/li&gt;&#xA;&lt;li&gt;bicyclobutane, bicyclobut-1(3)-ene, bicyclobut-1(2)-ene, bicyclobuta-1,3-diene, bicyclobuta-1,2-diene&lt;/li&gt;&#xA;&lt;li&gt;tetrahedrane, tetrahedrene, tetrahedradiene&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-01T20:49:47.013" LastActivityDate="2015-04-01T20:49:47.013" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="19486" PostTypeId="2" ParentId="16792" CreationDate="2014-11-14T00:07:29.673" Score="3" Body="&lt;p&gt;This didn't get any answers specific to proteins as I was looking for - I don't think this site has a big biochem. community - though I've found what may be the answer. For any other bioscientists reading this, from what I've read it seems to not be the case, but rather a product of current experimental resolution in measuring protein folding transition states.&lt;/p&gt;&#xA;&lt;p&gt;A lecturer told me that if there was non-zero lifetime the species at the top of the free energy barrier would make it a non-two state system, and with the proliferation of potential protein conformations (into an 'ensemble' to be sampled as per Wolynes's energy landscape theory) this would increase still further, depleting concentrations on either side of the barrier, paradoxically slowing the reaction... &lt;span class=&quot;math-container&quot;&gt;$\ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(or\  words\  to\  that\  effect)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In &lt;a href=&quot;http://dx.doi.org/10.1002/pro.2539&quot; rel=&quot;nofollow noreferrer&quot;&gt;a recent issue of &lt;em&gt;Protein Science&lt;/em&gt;&lt;/a&gt;, Weikl and Paul describe &amp;quot;Decoupling and temporal ordering of conformational changes and binding events for small binding transition times&amp;quot;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In general, both conformational changes and binding/unbinding events are thermally activated processes that require the crossing of free-energy barriers. A characteristic feature of thermally activated processes is that the actual time for crossing the free-energy barrier, the &amp;quot;transition time,&amp;quot; is significantly smaller than the dwell times in the states befre and affter the barrier crossing. In single-molecule experiments that probe conformational changes of proteins, the transition times are &lt;strong&gt;typically beyond experimental resolution&lt;/strong&gt;, and the conformational changes appear as &amp;quot;sudden jumps&amp;quot; between two or more conformational states (&lt;a href=&quot;http://dx.doi.org/10.1038/nchembio.1213&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kim 2013&lt;/a&gt;; &lt;a href=&quot;http://dx.doi.org/10.1016/j.sbi.2007.12.003&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schuler 2008&lt;/a&gt;; &lt;a href=&quot;http://dx.doi.org/10.1126/science.1215768&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Chung 2012&lt;/strong&gt;&lt;/a&gt;; &lt;a href=&quot;http://dx.doi.org/10.1016/j.sbi.2012.11.007&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zoldak 2013&lt;/a&gt;; &lt;a href=&quot;http://dx.doi.org/10.1146/annurev-biophys-051013-022754&quot; rel=&quot;nofollow noreferrer&quot;&gt;Woodside 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To me this indicates the 'jump' is not actually instantaneous, but an artefact of experimental limitations. Zero lifetime is however permitted as a simplication in such a setup. The suggestion is though that it would be in reality finite, and open to determination with better methods.&lt;/p&gt;&#xA;&lt;p&gt;Chung (2012) go into such a [statistical, &lt;a href=&quot;http://www.cecam.org/workshop-4-1023.html?presentation_id=11425&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gopich-Szabo maximum likelihood&lt;/a&gt;] method in an entire paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Theory predicts that folding mechanisms are heterogeneous, so that an individual unfolded molecule can self-assemble to form its iologically active, folded structure by means of many different sequences of conformational changes. The distribution of these folding pathways can now be calculated from atomistic molecular dynamics simulations. Information on pathway distributions from experiments must come from measurements on single molecules, because only average propertiew are obtained in experiments on the large ensemble of molecules in bulk experiments. A single-molecule, equilibrium protein folding-unfolding trajectory is illustrated in Fig. 1, as monitored by FRET spectroscopy, and its relation the free-energy barrier as it crosses between the folded and unfolded states is shown.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSLdR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The most interesting part of the trajectory is contained in what appears to be an instantaneous jump between the two states, called the transition path&lt;/strong&gt;, which contains all of the information on the mechanism of folding and unfolding. The first step toward observing transition paths in protein folding, which we report here, is the determination of its average duration (transition-path time) for a fast-folding, all-β protein shown to be two-state in ensemble studies, as well as a markedly reduced upper bound compared with our previous study for α/β protein GB1 (the B1 Ig-binding domain of protein G from &lt;em&gt;Streptococcus&lt;/em&gt;. In contrast to a rate coefficient, which measures the frequency of a transition, the transition-path time is the duration of a successful barrier-crossing event.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The paper defines &amp;quot;instantaneous&amp;quot; according to the limit of a two state mathematical model, &lt;span class=&quot;math-container&quot;&gt;${\lim\limits_{\tau_s \rightarrow 0}L(\tau_s)}$&lt;/span&gt;, &amp;quot;i.e. it occurs faster than the shortest photon interval&amp;quot;, and goes on to compare the likelihood (L) plots for cases of instantaneous vs. finite transition path time, to try and assess what might be going on beneath this limit of experimental resolution.&lt;/p&gt;&#xA;&lt;p&gt;They found a significant peak in the likelihood function in the case of finite transition-path time — &amp;quot;Fig. 4B at 16 (±3) μs. (The error is the standard deviation obtained from the curvature of the peak.)&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sfkwY.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;They also found that a fast- and a slow-folding protein take almost the same time to fold when folding actually happens.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Our determination of an average transition-path time is a first step toward the goal of obtaining information on the distribution of folding pathways from measurements of interdye distance versus time trajectories during transition paths. However, the result of this first step by itself has turned out to be extremely interesting. Folding involves a complex and intricate rearrangement of a polypeptide chain to form a unique structure, yet the time for this nontrivial self-assembly process is almost the same for two proteins with different topologies and vastly different folding rates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Of biochemical interest, &amp;quot;&lt;em&gt;the diffusion coefficient of the model depends on the roughness of the underlying energy landscape and could therefore differ substantially among proteins.&lt;/em&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Googling the Gopich/Szabo maximum likelihood method I found a workshop presentation given by Chung at ETH Zurich, which explicitly states that &amp;quot;transition-paths for barrier crossings have &lt;strong&gt;never&lt;/strong&gt; been observed experimentally for any molecular system in solution. Because it is a single molecule property, even determining the average transition-path time is challenging&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;I'm still not 100% sure I've landed at the right conclusion from all this — there's no explicit statements in these papers that the species in the &amp;quot;&lt;em&gt;barrier top population&lt;/em&gt;&amp;quot; (a phrase to recognise that the transition state may actually be an ensemble) have a finite lifetime, rather that &lt;em&gt;the &amp;quot;transition path&amp;quot; traversing their formation&lt;/em&gt; has a finite lifetime.&lt;/p&gt;&#xA;&lt;p&gt;Personally, I think figures 2B and 2C are trying to infer an answer to this question: whether the TS species lifetime is finite or zero/'instantaneous', and as such, I think the fact that it's the former may have disproven the zero-lifetime TS species model (though given how recently it became possible to do so, 0-lifetime TS species may continue to be described).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: reading a 2003 paper from prominent protein scientist Martin Karplus, &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0065323303660080/pdfft?md5=195da97680662d3bd61d06050082235e&amp;amp;pid=1-s2.0-S0065323303660080-main.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catalysis and Specificity in Enzymes: A Study of Triosephosphate Isomerase and Comparison with Methyl Glyoxal Synthase&lt;/a&gt; within &lt;a href=&quot;http://www.sciencedirect.com/science/journal/00653233/66&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adv Prot Chem&lt;/a&gt;, I'm not so sure if I am right to equate non-zero transition trajectory with non-instantaneous TS species: Karplus describes the former in a consistent manner with the &amp;quot;instantaneous barrier located between the transition state and the products state and vice versa&amp;quot;. In the excerpt below he describes crossing the barrier as having a finite time scale:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5orw.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In that case, still would appreciate an answer from a protein chemist!&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-11-14T00:51:52.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19487" PostTypeId="1" CreationDate="2014-11-14T00:22:22.437" Score="0" ViewCount="692" Body="&lt;p&gt;In an aqueous solution, Nickel oxyhydroxide is oxidised with a 1:1 ratio of aqueous Sodium hydroxide. My goal is to produce Nickel oxyhydroxide ($\ce{NiOOH}$). &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong Base: $\ce{NaOH + H2O \leftrightharpoons Na+ + OH-}$ &lt;br&gt;&#xA;Add that to some stirred $\ce{Ni(OH)2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ Ni(OH)2 + OH- \leftrightharpoons NiOOH + H2O + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make $\ce{Na (s)}$?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ Na+ + e- -&amp;gt; Na (s) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, $$\ce{Na(s) + H2O \leftrightharpoons NaOH }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That just brings everything back to the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the real question is: will the initial oxidation of the $\ce{Ni(OH)2}$  happen? If not, how can I force it? I initially thought of just grounding the solution, as odd as that sounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm a 12th grade high school student. &lt;/p&gt;&#xA;" OwnerUserId="9771" LastEditorUserId="4356" LastEditDate="2014-11-14T23:44:57.443" LastActivityDate="2014-12-15T04:18:25.027" Title="What happens when oxidizing Ni(OH)2 with a strong base?" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19488" PostTypeId="1" AcceptedAnswerId="19490" CreationDate="2014-11-14T02:04:55.733" Score="12" ViewCount="41883" Body="&lt;p&gt;In my latest chem lab the objective was to create a primary standard of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; and use it to determine the concentration of sulfuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;first part&lt;/strong&gt; of the lab was determine the molarity of the &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; solution through a series of titrations.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A sample of KHP (abbreviated form of &lt;span class=&quot;math-container&quot;&gt;$\ce{KHC8H4O4}$&lt;/span&gt;) was placed into a flask with approximately &lt;span class=&quot;math-container&quot;&gt;$\pu{25 ml}$&lt;/span&gt; of water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phenolphthalein was added to the flask as the indicator. &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; was then titrated into the flask with a burette. From multiple titrations of this sort I was able to calculate the molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;.    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below I have included part of my table and calculations (Note: &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mol}$&lt;/span&gt; of KHP is equal to &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; in this experiment. If I have made any mistakes please tell me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trial 1:&lt;/strong&gt;&lt;br&gt;&#xA;Mass of KHP in flask &lt;span class=&quot;math-container&quot;&gt;$= \pu{0.5108 g}$&lt;/span&gt;&lt;br&gt;&#xA;Volume of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; added to flask &lt;span class=&quot;math-container&quot;&gt;$ = \pu{21.73 ml}$&lt;/span&gt;&lt;br&gt;&#xA;&lt;strong&gt;Calculation&lt;/strong&gt; of molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; for trial 1:&lt;br&gt;&#xA;Molar mass of KHP &lt;span class=&quot;math-container&quot;&gt;$ =  \pu{204.23 g/mol}$&lt;/span&gt;&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.5108 g}/\pu{204.23 g/mol} = \pu{0.002501 mol}$&lt;/span&gt; of KHP, which is equal to &lt;span class=&quot;math-container&quot;&gt;$\pu{0.002501 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;.&lt;br&gt;&#xA;Molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH} = \pu{0.002501 mol}/\pu{0.02173 L} = \pu{0.1151 M}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did 3 other trials like this (in total 4) and calculated the average molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; to be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1159 M}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;second half&lt;/strong&gt; of the lab is the part I had trouble with.&lt;br&gt;&#xA;We were given a sample of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; with an unknown concentration. I took &lt;span class=&quot;math-container&quot;&gt;$\pu{10 ml}$&lt;/span&gt; of this &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; and mixed it with &lt;span class=&quot;math-container&quot;&gt;$\pu{100 ml}$&lt;/span&gt; of distilled water. This new diluted solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; (I will refer to it as solution 2 now) was the solution used in the trials to determine the molarity. So &lt;span class=&quot;math-container&quot;&gt;$\pu{25 ml}$&lt;/span&gt; of solution 2 was added to a flask with a few drops of phenolphthalein. A titration using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; (the same &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; as used in the previous section) was performed.       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My task is to now figure out the concentration of the &lt;em&gt;original&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; solution. I have tried 2 different methods. The first method I attempted seems so flawed I didn't bother to put it on (it didn't even make sense to me). Each method seems incorrect and have yielded drastically different results. Below I have provided a sample of my table and one of my attempts to solve for the molarity of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net ionic equation of this procedure is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2SO4 +2NaOH &amp;lt;=&amp;gt; Na2SO4 + 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trial 1:&lt;/strong&gt;&#xA;Volume of diluted acid (solution 2) in flask: &lt;span class=&quot;math-container&quot;&gt;$\pu{25.00 ml}$&lt;/span&gt;&lt;br&gt;&#xA;Volume of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; added to flask: &lt;span class=&quot;math-container&quot;&gt;$\pu{23.81 ml}$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt 1&lt;/strong&gt; at finding molarity:&lt;br&gt;&#xA;Moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; added to flask: &lt;span class=&quot;math-container&quot;&gt;$\pu{0.02381 L} \cdot \pu{0.1159 M} = \pu{0.0027596 mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount of of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$0.0027596/2 = \pu{0.0013798 mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; (The 2 came from the net ionic equation above)&lt;br&gt;&#xA;Molarity of diluted &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; (solution 2): &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0013798 mol}/ \pu{0.025 L}= \pu{0.054172 M}$&lt;/span&gt;&lt;br&gt;&#xA;(I may be using the wrong volume, is it possible that I have to add the &lt;span class=&quot;math-container&quot;&gt;$\pu{25 ml}$&lt;/span&gt; to the &lt;span class=&quot;math-container&quot;&gt;$\pu{23.81 ml}$&lt;/span&gt; and divide by &lt;span class=&quot;math-container&quot;&gt;$\pu{0.04881 L}$&lt;/span&gt;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;C_1V_1 &amp;amp;= C_2V_2\\&#xA;C_1&amp;amp;=?\\&#xA;V_1&amp;amp;= \pu{0.01 L}\\&#xA;C_2&amp;amp;= \pu{0.054172 M}\\&#xA;V_2&amp;amp;=\pu{0.1 L}\\&#xA;\text{Therefore:}\\&#xA;C_1&amp;amp;=(\pu{0.054172 M})\cdot(\pu{0.1 L})/(\pu{0.01 L})\\&#xA;C_1&amp;amp;=\pu{0.54172 M}\\&#xA;\end{align}&lt;/span&gt;&#xA;Molarity of original/stock &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9355" LastEditorUserId="61578" LastEditDate="2019-12-08T00:20:10.063" LastActivityDate="2020-12-06T13:09:36.363" Title="How do I calculate the concentration of sulphuric acid by a titration experiment with sodium hydroxide?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;concentration&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19489" PostTypeId="1" AcceptedAnswerId="19492" CreationDate="2014-11-14T03:07:18.970" Score="10" ViewCount="11252" Body="&lt;p&gt;What is the structure of carbon fiber? Wikipedia doesn't give much insight. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The atomic structure of carbon fiber is similar to that of graphite, consisting of sheets of carbon atoms (graphene sheets) arranged in a regular hexagonal pattern, the difference being in the way these sheets interlock. Graphite is a crystalline material in which the sheets are stacked parallel to one another in regular fashion. The intermolecular forces between the sheets are relatively weak Van der Waals forces, giving graphite its soft and brittle characteristics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia description begs the question of &lt;em&gt;how&lt;/em&gt; exactly are the sheets of graphene interlocked. Are there pi-stacking interactions? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2015-04-02T18:31:06.010" LastActivityDate="2017-12-18T16:17:43.193" Title="Structure of carbon fiber" Tags="&lt;materials&gt;&lt;aromatic-compounds&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19490" PostTypeId="2" ParentId="19488" CreationDate="2014-11-14T03:31:31.687" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Molarity of diluted &lt;span class=&quot;math-container&quot;&gt;$\ce {H2SO4}$&lt;/span&gt; (solution 2): &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$&lt;/span&gt;&lt;br&gt;&#xA;  (I may be using the wrong volume, is it possible that I have to add the &lt;span class=&quot;math-container&quot;&gt;$\pu{25 ml}$&lt;/span&gt; to the &lt;span class=&quot;math-container&quot;&gt;$\pu{23.81 ml}$&lt;/span&gt; and divide by &lt;span class=&quot;math-container&quot;&gt;$\pu{0.04881 L}$&lt;/span&gt;?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, you used the correct volume since you want to know the concentration of the &lt;span class=&quot;math-container&quot;&gt;$\pu{25 mL}$&lt;/span&gt; that you added to the flask.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.54172 M}$&lt;/span&gt; - Molarity of original solution of &lt;span class=&quot;math-container&quot;&gt;$\ce {H2SO4}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="61578" LastEditDate="2019-12-07T23:49:17.250" LastActivityDate="2019-12-07T23:49:17.250" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19491" PostTypeId="1" CreationDate="2014-11-14T03:38:50.090" Score="1" ViewCount="455" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a massless piston that can move freely in a horizontal tube that is closed at both ends. To the left, there is a sample of helium gas at $300~\mathrm{K}$ and $4~\mathrm{atm}$. To the right, there is a sample of an unknown gas $\ce{X}$ at $500~\mathrm{K}$ and $2~\mathrm{atm}$. At equilibrium, the piston is located at 1/9 the length of the entire piston from the right side. The molecules of $\ce{X}$ are homonuclear. What is most likely the identity of $\ce{X}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(From Chemistry Olympiad)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused by the equilibrium part. I tried $PV = n\mathcal{R}T$ to find the ratio of moles. But without molar mass, how do you solve for it? How would you solve it?&lt;/p&gt;&#xA;" OwnerUserId="9531" LastEditorUserId="4945" LastEditDate="2014-11-14T05:51:58.323" LastActivityDate="2014-11-14T05:51:58.323" Title="How to determine an unknown gas in a tube with a massless piston at equilibrium conditions?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19492" PostTypeId="2" ParentId="19489" CreationDate="2014-11-14T03:50:41.477" Score="8" Body="&lt;p&gt;The key difference between graphite and carbon fibers is the degree (or range) of ordering.  In graphite there are large, extensive planes of carbon stacked one atop another resulting in long-range order. In the fibers, there are small ribbons of graphite-like (it is not all carbon, due to the way the fibers are prepared there is still some nitrogen present in the molecular array) material oriented along the fiber axis and there is only short range order.  As a consequence graphite with its extensive 3 dimensional planar array of ordered carbon atoms is crystalline.  Whereas the fibers, being ordered over a much smaller range,  are amorphous.  The amorphous nature of the fiber allows the smaller ribbons to bend and fold, and mechanically interlock with one another.  This mechanical interlocking is what serves to increase the fiber's strength.  See &lt;a href=&quot;http://web.mit.edu/3.082/www/team2_f01/chemistry.html&quot;&gt;here&lt;/a&gt; for a very nice and concise description of how the fibers are prepared.  The pictures in the link make clear how the short range ordering in the fibers results in a highly folded and interlocked structure.  Further heating of the fiber will convert it to graphite.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-14T03:50:41.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19493" PostTypeId="2" ParentId="18997" CreationDate="2014-11-14T04:40:24.257" Score="1" Body="&lt;p&gt;After learning more about environmental chemistry I got this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On 16 september 1987, 95 countries accepted not to use ODS(Ozone Depleting Substances)  after signing Mortal Protocol. Later other countries also accepted it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that United Nation have decided to celebrate 16 september as &quot;Ozone protection day&quot; at international level to spread awareness.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2014-11-14T04:40:24.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19494" PostTypeId="1" CreationDate="2014-11-14T06:33:09.397" Score="4" ViewCount="78119" Body="&lt;p&gt;Recently I came across the fact: &lt;em&gt;Metal oxides basic and non-metals' acidic or neutral.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for it, partly my query is solved, as I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal oxides or $\ce{O2-}$ forms $\ce{OH-}$ after hydrolysis in the following reaction so metal oxides are basic in aqeous solutions.&#xA;$$\ce{O2- + 2H2O-&amp;gt;4OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this might happen as metal form ionic compunds and non-metals form covalent and some of them may not even react tus neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got only a little $5\%$ of the exact reason, therefore I wish if someone could explain the whole process/fact. I considered Arhenius concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; My question is partly answered &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1215/why-does-polonium-form-acidic-oxides&quot;&gt;here&lt;/a&gt;, but is not satisfactory.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-10-09T22:10:00.270" Title="Why are metal oxides basic and non-metals' acidic or neutral" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19495" PostTypeId="1" AcceptedAnswerId="19496" CreationDate="2014-11-14T07:47:24.023" Score="12" ViewCount="24522" Body="&lt;p&gt;I was washing my cup with hot water (without soap) and upon it nearing my nose, there was some sort of 'smell' (I lack a better word)- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the 'smell' was different compared to when I was drinking just plain cold water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, does water have a smell? &lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorUserId="12860" LastEditDate="2015-10-30T17:16:22.337" LastActivityDate="2015-10-30T17:16:22.337" Title="Does water have a smell?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;smell&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19496" PostTypeId="2" ParentId="19495" CreationDate="2014-11-14T09:43:32.263" Score="17" Body="&lt;p&gt;Water in its pure form, i.e. $\ce{H2O}$ does not have a smell, or at least no smell that we can distinguish because the receptors in our nose (and mouth) are continuously exposed to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you smell are dissolved gases and other volatile impurities. The nature of these chemicals will vary mainly by location/source of the water, and might also be a bit influenced by your specific local plumbing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in Iceland near Myvatn there is a lot of volcanic activity and the water from the tap (in particular the warm water because it can dissolve more) smells like rotten eggs due to the sulfur. Another example is that in many countries the water will smell a bit like chlorine, because chlorine is used to keep the water safe to drink (kills germs).&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-11-14T09:43:32.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19497" PostTypeId="1" CreationDate="2014-11-14T09:59:09.013" Score="0" ViewCount="1230" Body="&lt;p&gt;I am currently taking a chemistry course about acids and bases. One of the things I am a bit confused about, is the acid and base neutralization reaction, and how to balance it. For Example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH + HCl -&amp;gt; NaCl + H2O}$$ How do you write the ionic and net ionic equations for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: My main problem here is that my teacher requires us to first write the ionization of the $\ce{HCl}$ into $\ce{H3O+}$ and $\ce{Cl-}$. This is when I get confused, because I don't know how to balance the equations with the $\ce{H3O+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I had another question in my exam, which I got wrong because of the balancing. Can you please tell me the correction of it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the neutralization reaction between nitric acid and calcium hydroxide. Write the balanced neutralization reaction and the ionic and net ionic reactions of it:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9774" LastEditorUserId="7581" LastEditDate="2015-02-17T20:44:03.770" LastActivityDate="2015-02-17T20:44:03.770" Title="Balancing Ionic and Net Ionic equations" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19500" PostTypeId="1" AcceptedAnswerId="19505" CreationDate="2014-11-14T13:10:33.300" Score="4" ViewCount="3764" Body="&lt;p&gt;At my school, my teacher tell us to extract sodium from normal salt ($\ce{NaCl}$) but without using electrolysis of molten sodium chloride. So is there a way to do that?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2015-05-03T09:32:55.483" LastActivityDate="2018-05-27T08:40:23.910" Title="Extracting sodium from salt" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19501" PostTypeId="2" ParentId="19497" CreationDate="2014-11-14T13:27:49.160" Score="1" Body="&lt;p&gt;As $\ce{NaOH}$ and  $\ce{HCl}$ are strong base and acid respectively, they are completely dissociated in water. The salt $\ce{NaCl}$ is a  strong electrolyte and completely dissociated in water. So, the ionic equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(Na^+ +OH^- )}+\ce{(H^+ +Cl^- )}\rightarrow \ce{(Na^+ +Cl^- )}+\ce{H2O}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the ions $\ce{Na^+}$ and$\ce{Cl^-}$ do not combine and remain in the solution as spectator ions. Their role is to keep the solution electrically neutral, the net ionic equation is:&#xA;$\ce{OH^- }+\ce{H^+ }\rightarrow \ce{H2O}  $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we have to notice that the ion $\ce{H^+}$ is not stable in water, it combines with $\ce{H2O}  $ to form the ion hydronium $\ce{H_3O^+}$ . We usually write the complete dissociation of $\ce{HCl }$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl} +\ce{H_2O }\rightarrow$ $\ce{Cl^-} +\ce{H_3O^+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the ionic equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(Na^+ +OH^- )}+( \ce{H_3O^+}$$ +\ce{Cl^- })\rightarrow \ce{(Na^+ +Cl^- )}+2\ce{H2O}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the net ionic equation is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{OH^- }+\ce{H_3O^+ }\rightarrow 2\ce{H2O}  $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your exam questions:&#xA;$2\ce{HNO_3} + \ce{Ca(OH)_2 }\rightarrow \ce{Ca(NO_3)_2} + 2\ce{H_2O}$&#xA; Now you can try to write the ionic and the net ionic equation to see if you understand what I have explained.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2014-11-14T13:52:09.500" LastActivityDate="2014-11-14T13:52:09.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19502" PostTypeId="1" AcceptedAnswerId="19838" CreationDate="2014-11-14T13:28:15.880" Score="4" ViewCount="8354" Body="&lt;p&gt;In my science class, I have been taught that silica (silicon dioxide/sand) has three properties, namely&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It does not react with water. &lt;/li&gt;&#xA;&lt;li&gt;It does not decompose with heat. &lt;/li&gt;&#xA;&lt;li&gt;It does not react with acid. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However I think that if it does not react with acid, will it react with bases? I am just curious because if it does react, why does my textbook not say anything about that? &lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="81" LastEditDate="2016-08-11T17:33:45.123" LastActivityDate="2018-02-13T05:27:45.793" Title="Does silicon dioxide react with alkalis?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19504" PostTypeId="2" ParentId="19494" CreationDate="2014-11-14T14:06:16.147" Score="0" Body="&lt;p&gt;Metal oxides like $\ce{CaO}$ and $\ce{Na2O}$ react with water and give hydroxides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaO + H2O &amp;lt;=&amp;gt; Ca(OH)2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2O + H2O &amp;lt;=&amp;gt; 2 NaOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While non-metal oxides like $\ce{CO2}$ and $\ce{SO2}$ react with water and give acids:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3 &amp;lt;=&amp;gt; H+ + HCO3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO2 + H2O &amp;lt;=&amp;gt; H2SO3 &amp;lt;=&amp;gt; H+ + HSO3-}$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorDisplayName="user7951" LastEditDate="2015-04-22T08:34:14.567" LastActivityDate="2015-04-22T08:34:14.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19505" PostTypeId="2" ParentId="19500" CreationDate="2014-11-14T14:09:47.090" Score="4" Body="&lt;p&gt;I don't think it is safe or realistic to do that without extensive training and equipment, but yes, historically sodium metal has been made by chemical reduction of sodium salts at high temperature and condensing sodium vapor produced.  For example, sodium carbonate was heated with charcoal.  Even though equillibrium does not favor sodium metal, the fact that the small fraction of sodium produced is removed takes advantage of Le Chatelier's principle to keep making more sodium.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-14T14:09:47.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19507" PostTypeId="1" CreationDate="2014-11-14T16:07:37.167" Score="2" ViewCount="88" Body="&lt;p&gt;Any idea where to buy chlorinated paraffin for lab application? We need very viscous analog of $\ce{CHCl_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we think about synthesizing it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a lot of places where I could buy 5 tons:) But we need only few liters for the rest of our life.&lt;/p&gt;&#xA;" OwnerUserId="4589" LastEditorUserId="16072" LastEditDate="2015-05-09T03:04:04.273" LastActivityDate="2015-05-09T03:04:04.273" Title="Where can we buy chlorinated paraffin for lab application?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-10-02T16:51:57.497" ContentLicense="CC BY-SA 3.0" />
  <row Id="19508" PostTypeId="1" AcceptedAnswerId="19511" CreationDate="2014-11-14T16:23:05.170" Score="7" ViewCount="26228" Body="&lt;p&gt;Why methyl group is 2,4-directing? I am an A level student and this isn't required at this level. I am looking for a simple explanation which a high school student can understand.&lt;/p&gt;&#xA;" OwnerUserId="6791" LastEditorUserId="5026" LastEditDate="2018-05-15T15:07:10.680" LastActivityDate="2018-05-15T15:07:10.680" Title="Why methyl group is 2,4-directing?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19509" PostTypeId="2" ParentId="19487" CreationDate="2014-11-14T16:53:14.410" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make &lt;br&gt;&#xA;  $\ce{Na (s)}$ ?: $\ce{Na+ + e- -&amp;gt; Na (s)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reaction is, you might be knowing, the reverse of one of &quot;the best&quot; known favourable reaction, as the data suggests:&#xA;$$\ce{Na-&amp;gt;Na+ +e-}\tag{$E^\circ=2.71\;V$}\\&#xA;2.71\;V\equiv-261.475\;\text{kJ/mol}\equiv1.0938\times10^{50}(\text{rate})$$ &#xA;Just imagine the reverse reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But then, &#xA;  $\ce{Na(s) + H2O -&amp;gt; NaOH }$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This then would never actually then happen. Intersetingly see &lt;a href=&quot;https://www.youtube.com/watch?v=ODf_sPexS2Q&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-15T03:35:04.280" LastActivityDate="2014-11-15T03:35:04.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19510" PostTypeId="2" ParentId="19497" CreationDate="2014-11-14T17:13:58.390" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the neutralization reaction between nitric acid and calcium hydroxide. Write the balanced neutralization reaction and the ionic and net ionic reactions of it&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For simplicity consider this:&#xA;$$\ce{HNO3}\tag{nitric acid, ionic}$$$$\ce{Ca(OH)2}\tag{calcium hydroxide, ionic}$$&#xA;So, they'll break into their constituent ions:&#xA;$$\ce{HNO3&amp;lt;=&amp;gt;H+ +NO3-}\\\ce{Ca(OH)2&amp;lt;=&amp;gt;Ca^2+ +OH-}$$&#xA;Now $\small\ce{H+}$ exists as $\small\ce{H3O+}$ and after balancing by adding water to both sides and doing $\small\ce{H2O +H+\sim H3O+}$.Now multiply the first reaction by 2 after we notice the two hydroxide ions in the decomposition of calcium hydroxide, since we want in an integral multiple, the reaction $\small\ce{H+ +OH- -&amp;gt;H2O\equiv H3O+ + OH- -&amp;gt;2H2O}$&#xA;$$\begin{array}{}\ce{&amp;amp;2HNO3 +2H2O&amp;amp;&amp;lt;=&amp;gt;&amp;amp;2NO3-&amp;amp;+&amp;amp;2H3O+}\\\ce{&amp;amp;Ca(OH)2&amp;amp;&amp;lt;=&amp;gt;&amp;amp;Ca^2+ &amp;amp;+ &amp;amp;2OH-}\\\hline\ce{&amp;amp;2HNO3 +Ca(OH)2&amp;amp;-&amp;gt;&amp;amp;\underbrace{Ca(NO3)2}_{Ca^2+ + NO3- &amp;lt;=&amp;gt; Ca(NO3)2}&amp;amp;+}&amp;amp;\underbrace{\ce{2H2O}}_{4-2=2}\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-14T17:13:58.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19511" PostTypeId="2" ParentId="19508" CreationDate="2014-11-14T17:15:03.740" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why methyl group is 2,4-directing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When toluene undergoes electrophilic aromatic substitution the products are primarily the ortho and para isomers; usually only a small amount of the meta isomer is produced.  In order to explain this observation there are two effects to consider, inductive and resonance effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductively the methyl group releases electron density into the benzene ring.  This is because the methyl group, being $\ce{sp^3}$ hybridized, is less electronegative than the $\ce{sp^2}$ hybridized aromatic carbon.  The carbon with more s-character is more electronegative, electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower its energy.  If inductive arguments were most important we would expect the ortho isomer to predominate followed by the meta and para isomers in that order.  This is not observed, the meta isomer is only a minor isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance arguments are based on the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;noreferrer&quot;&gt;hyperconjugation&lt;/a&gt; involving the $\ce{C-H}$ bonds in the methyl group.  The following figure shows some of the possible hyperconjugative resonance structures for toluene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l0eph.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional resonance structures can be drawn using the other methyl $\ce{C-H}$ bonds as well.  As the resonance structures show, electron density is increased at the ortho and para positions, not the meta.  This hyperconjugative resonance effect makes the ortho and para carbons more susceptible to electrophilic attack.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperconjugation explains how alkyl groups can be electron releasing in a resonance sense.  Hyperconjugation explains the mild activating effect of an alkyl group in electrophilic aromatic substitution and its ortho-para directing effect.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-14T17:15:03.740" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19512" PostTypeId="2" ParentId="16086" CreationDate="2014-11-14T17:28:33.593" Score="4" Body="&lt;p&gt;Ammonia is reportedly not toxic in bacteria even at molar conentrations (PMID 16604417).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, yeast, plants, and animals (presumably, all eukaryotes) are unable to tolerate high intracellular ammonia concentrations.  The best explanation for this phenomena I've seen is a hand-waving assertion that intracellular ammonia taxes overtaxes the cell's ability to maintain membrane gradients (PMID 7764110).  However, that isn't a particularly detailed or satisfying response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with your assertion that there's a paucity of literature on the subject - we probably don't know the answer yet!&lt;/p&gt;&#xA;" OwnerUserId="9780" LastActivityDate="2014-11-14T17:28:33.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19513" PostTypeId="2" ParentId="19477" CreationDate="2014-11-14T17:37:15.163" Score="14" Body="&lt;p&gt;Let's start with the basics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;&lt;em&gt;representation&lt;/em&gt;&lt;/strong&gt; is a set of matrices that fulfill the multiplication table for the point group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I might be able to use a set of 3x3 matrices, since of course I can define any of the symmetry operations over Cartesian (x,y,z) space. e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \left(&#xA;\begin{array}{ccc}&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 1 \end{array}&#xA; \right) \;&#xA;i = &#xA;\left(&#xA;\begin{array}{ccc}&#xA;-1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; -1 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; -1 \end{array}&#xA; \right) &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But usually, I can fulfill the multiplication table with much simpler (smaller) matrices.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An &lt;strong&gt;&lt;em&gt;irreducible representation&lt;/em&gt;&lt;/strong&gt; is a set of matrices that are as small as possible.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;&lt;em&gt;reducible representation&lt;/em&gt;&lt;/strong&gt; is a set of matrices that can be decomposed into a linear combination of irreducible representations. This can occur by multiple techniques, including block-diagonalization.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The distinction exists, because for chemical point groups, there are usually a finite number of irreducible representations (given in character tables), but of course there are an &lt;em&gt;infinite&lt;/em&gt; number of reducible representations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power of symmetry and group theory is that &lt;strong&gt;&lt;em&gt;any&lt;/em&gt;&lt;/strong&gt; function of a molecule, like orbitals, vibrations, etc. can be described using a representation. Thus, we usually label orbitals and vibrations by their corresponding irreducible representation.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;I'll have to hunt down my copy of Ogden, but usually people use the Cartesian axes or $p_z$, $p_x$, and $p_y$ to illustrate the different irreducible representations. That is, a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Core/Physical_and_Theoretical_Chemistry/Group_Theory/Group_Theory%3A_Theory&quot;&gt;basis&lt;/a&gt; is some function (like the Cartesian axes or an atomic orbital) used to generate a representation. We look at how the basis is transformed under the different symmetry operations and write down the matrix or characters from those transformations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I try not to use that term because students find it confusing and different guides and texts use subtly different definitions for it.)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;The reduction formula (or algorithm) is a way of decomposing the reducible representation (which is really a big matrix) into a combination of the different irreducible representations, but only needing to act on the &lt;strong&gt;&lt;em&gt;characters&lt;/em&gt;&lt;/strong&gt; of the representations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That is, I don't need to actually use the reducible matrices themselves. Using the reduction formula, I can get the linear combination of irreducible representations only using multiplication of characters of the representations. It's much easier to do the reduction formula than to figure out how to transform a large reducible matrix.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="283" LastEditDate="2017-01-22T06:07:54.757" LastActivityDate="2017-01-22T06:07:54.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19516" PostTypeId="2" ParentId="19495" CreationDate="2014-11-14T18:26:47.857" Score="0" Body="&lt;p&gt;As stated by others, you're smelling whatever substances are in solution or suspension in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason you may detect a difference between cold and hot water is that some homes have a water softener that replaces dissolved mineral salts with sodium or potassium salts. It's usually only used on the warm water line, because the main advantage is how it feels when bathing. &quot;Soft&quot; water may be less pleasant to drink than &quot;hard&quot; water. I never did like the taste of it.&lt;/p&gt;&#xA;" OwnerUserId="9784" LastActivityDate="2014-11-14T18:26:47.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19517" PostTypeId="1" AcceptedAnswerId="147821" CreationDate="2014-11-14T18:53:23.317" Score="12" ViewCount="3592" Body="&lt;p&gt;Bromine, chlorine and iodine can all be sealed in a glass container for display without the elements reacting with the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you try to seal fluorine in glass I believe it will react and fog the surface making it harder to see the gas, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way, if at all possible for long term containment of fluorine for display?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="23561" LastEditDate="2019-04-21T18:32:50.923" LastActivityDate="2021-03-20T22:44:46.947" Title="Is there a way to contain fluorine gas for long term so that it can be visually observed?" Tags="&lt;safety&gt;&lt;reactivity&gt;&lt;elements&gt;&lt;containers&gt;" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19518" PostTypeId="2" ParentId="19476" CreationDate="2014-11-14T19:01:00.347" Score="1" Body="&lt;p&gt;Converting to answer: While the energy to introduce a grain boundary into a single crystal depends on the relative orientation of the two resulting grains, there will be some atomic rearrangement (strains or reconstruction), and you will end up with a number of bonds across the grain boundary. The boundaries are usually weaker than the bulk crystal, which is why they preferentially fracture along the grain boundary. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-11-14T19:01:00.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19519" PostTypeId="2" ParentId="19495" CreationDate="2014-11-14T19:22:24.220" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;there was some sort of 'smell' (I lack a better word)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So there was a sensation in your nose that you partly (but not confidently) identified as smell. It might not have been olfactory, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert, but it seems likely to me that in addition to any impurities in the water that you really can smell, you're also detecting a sudden change in humidity (especially with the hot water) and/or temperature. Strictly speaking you can't &quot;smell&quot; water, but that's by definition of smell. You can nevertheless in the right circumstances detect water via the mucous membranes of your nose. As you inhale, the air passing over the membranes dries them (or doesn't) by a differing amount. Aside from the fact that you might falsely perceive this sensation as a smell, how dry the membranes are does affect your perception of anything you can smell in your environment. So a change in humidity can result in a change to what you smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the acoustics of the cup could be detectable (nose, ears, close together, attached to the rigid/resonating boney structure of your skull, you're putting your nose at the entrance to an enclosed space), and could be interpreted as &quot;like a smell&quot; if a smell is what you're expecting from your nose. Although I don't think that accounts for a difference between hot and cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this could well be a biology/psychology question as well as a chemistry one ;-)&lt;/p&gt;&#xA;" OwnerUserId="9786" LastEditorUserId="9786" LastEditDate="2014-11-14T19:27:25.303" LastActivityDate="2014-11-14T19:27:25.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19521" PostTypeId="1" CreationDate="2014-11-14T20:18:14.853" Score="5" ViewCount="2125" Body="&lt;p&gt;I have found that because the oxygen is double bonded to carbon, it takes about $\mathrm{1500\ ^\circ C}$ to break it down from $\ce{CO2}$ to $\ce{C}$ and $\ce{O}$. But, I am curious, would it take less heat energy to break $\ce{CO2}$ down to $\ce{CO}$ and $\ce{O}$?&lt;/p&gt;&#xA;" OwnerUserId="9788" LastEditorUserId="4356" LastEditDate="2014-11-14T22:31:27.363" LastActivityDate="2014-11-14T22:31:27.363" Title="How much heat is needed to break CO2 down to CO?" Tags="&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19522" PostTypeId="2" ParentId="19521" CreationDate="2014-11-14T21:54:02.630" Score="5" Body="&lt;p&gt;Heating CO2 at atmospheric pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1000K it is still essentially all CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 2000K: 98% CO2, 1.4% CO, 0.7% O2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 3000K: 44% CO2, 36% CO, 16% O2, 4% O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 5000K: 50% CO, 50% O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;only at even higher temperature does significant atomic C appear (9% at 6000K).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF00567372#&quot; rel=&quot;noreferrer&quot;&gt;Thermal decomposition of carbon dioxide in an argon plasma jet&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-14T21:54:02.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19523" PostTypeId="1" AcceptedAnswerId="19524" CreationDate="2014-11-14T22:28:49.973" Score="2" ViewCount="5503" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A buffer with a pH of $\mathrm{4.79}$ contains $\mathrm{ 0.39\ M}$ of sodium benzoate and $\mathrm{0.10\ M}$ of benzoic acid. What is the concentration of $\mathrm{[\ce{H} ]}$ in the solution after the addition of $\mathrm{0.058\ mol}$ of $\ce{HCl}$ to a final volume of $\mathrm{1.6\ L}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thought is that&#xA;$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]}{[acid]})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we added $\mathrm{0.058\ mol}$ of $\ce{HCl}$, the concentration of both acid and base decreased by $\frac{.058}{1.6}$, so I take that ratio and apply it.  I found $\mathrm{pK_a}$ by just applying what's given.  Not sure where I am going wrong.&lt;/p&gt;&#xA;" OwnerUserId="9791" LastEditorUserId="4356" LastEditDate="2014-11-14T22:47:21.373" LastActivityDate="2014-12-16T17:39:55.617" Title="Buffer Solution - Sodium Benzoate" Tags="&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19524" PostTypeId="2" ParentId="19523" CreationDate="2014-11-14T23:09:06.260" Score="4" Body="&lt;p&gt;In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$&#xA;From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2014-11-14T23:15:03.570" LastActivityDate="2014-11-14T23:15:03.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19525" PostTypeId="2" ParentId="19319" CreationDate="2014-11-15T00:23:10.677" Score="1" Body="&lt;p&gt;Oxidation numbers are an extremely useful form of chemical accounting, but should not be considered as having any actual reality. For instance fractional oxidation numbers are perfectly possible, but you cannot have half an electron. Anoxidation number of +7 is assigned to chlorine in perchlorates, but the idea that you could possibly strip 7 electrons off a chlorine atom is unthinkable (add up the first 7 ionisation potentials). It is the second part of the quote that you give is misleading and in many cases incorrect. The point to be born in mind first is that the oxidation number of a given element in a compound is the charge it would have &lt;em&gt;IF&lt;/em&gt; the compound was entirely ionic. The sign of the charge is based on relative electronegativity of the bonding atoms. Group I and II metals (except in very rare cases) are assigned charges of +1 and +2 respectively. Oxygen is most commonly -2, but in the case of peroxides it is assigned an oxidation number of -1 in order that the combined oxidation numbers come to zero in this case or to the total charge on the group of atoms if they happened to form an ion.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-15T00:23:10.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19527" PostTypeId="2" ParentId="19449" CreationDate="2014-11-15T02:33:59.453" Score="0" Body="&lt;p&gt;Q1) For the first question, we can calculate the molarity of the sulphuric acid.(Molarity of acid × volume of acid = molarity of the dilute acid × the volume of water)&#xA;0.12N × 0.5ml = molarity of dilute acid × 500ml&#xA;Molarity of dilute acid = 0.12N × 0.5ml / 500ml&#xA;                        = 0.00012N&#xA;Thus, you can see that the molarity of the acid is quite low, meaning the ph value is nearly at 4 or 5. I can say that is not necessary to wear the respirator but remember to prepare something base or alkaline for cases like the acid start to burn the gloves or your skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2) Yes. You must add the drop of acid into the water and not vice versa. It is because if you instantly add water to acid, the solution will form an exothermic reaction and release a lot of heat, causing the acid to boil and splatter itself everywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you add drops of acid to water. The heat produce is not enough to cause the solution to boil. Hence, it is must more safer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that it will help you on experiment.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="4934" LastEditDate="2014-12-16T01:26:12.343" LastActivityDate="2014-12-16T01:26:12.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19528" PostTypeId="1" CreationDate="2014-11-15T02:57:47.027" Score="1" ViewCount="21974" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The heat of neutralization of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl(aq)}$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{-55.9 kJ/mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; produced. If &lt;span class=&quot;math-container&quot;&gt;$\pu{50ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.6M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{25.15^\circ C}$&lt;/span&gt; is added to &lt;span class=&quot;math-container&quot;&gt;$\pu{25ml}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.79M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{26.34^\circ C}$&lt;/span&gt; in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)&lt;/p&gt;&#xA;&lt;p&gt;I need a detailed explanation to understand! I'm not sure which one is the initial temperature? &lt;span class=&quot;math-container&quot;&gt;$\pu{25.15^\circ C}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\pu{26.34^\circ C}~\ce{HCl}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9794" LastEditorUserId="69830" LastEditDate="2021-05-19T06:24:02.630" LastActivityDate="2021-05-19T06:27:13.607" Title="How to calculate the temperature from heat of neutralisation?" Tags="&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19529" PostTypeId="1" AcceptedAnswerId="19532" CreationDate="2014-11-15T03:02:29.510" Score="3" ViewCount="1347" Body="&lt;p&gt;I learnt that the chemical symbol for baking soda and baking powder is NaHCO3. However, when I do a simple experiment which is put vinegar into both substances. I can get different observations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I put vinegar into baking soda, I get a fast and fuzzy reaction. But when I put vinegar into baking powder, instead of a fast and fuzzy reaction, I get a slow and bubbly reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is what properties do each of the both substances have to make different reactions with vinegar although their chemical symbols are the same?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:17:57.973" LastActivityDate="2015-10-30T17:17:57.973" Title="Difference between baking soda and baking powder" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19530" PostTypeId="1" AcceptedAnswerId="19537" CreationDate="2014-11-15T03:06:16.240" Score="1" ViewCount="1706" Body="&lt;p&gt;Which oxygen atom is more nucleophilic in carboxylic acids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/teach/ch610bnotes/ch17.htm&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;this page&lt;/strong&gt;&lt;/a&gt;, it says the carbonyl oxygen is more nucleophilic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I also learned in class that $\mathrm{sp^3}$ atoms are more reactive than $\mathrm{sp^2}$ atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in an $\mathrm{S_n1}$ reaction where the solvent is acetic acid, which oxygen attacks?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="14" LastEditDate="2014-11-20T16:10:08.397" LastActivityDate="2014-11-20T16:10:08.397" Title="SN1 Reaction Involving Carboxylic Acids" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19531" PostTypeId="2" ParentId="19529" CreationDate="2014-11-15T03:10:02.870" Score="3" Body="&lt;p&gt;Baking soda and baking powder are not the same.&lt;br&gt;&#xA;Baking soda is (almost 100%) sodium bicarbonate ($\ce {NaHCO3}$).&lt;br&gt;&#xA;Baking powder is a mixture of sodium bicarbonate, and some acids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Baking_powder&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;: &lt;code&gt;Typical formulations (by weight) call for 30% sodium bicarbonate, 5-12% monocalcium phosphate, and 21-26% sodium aluminium sulfate.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you mix baking powder with water (in a dough), it will cause bubbles to form. Nothing will happen with baking soda.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-15T03:10:02.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19532" PostTypeId="2" ParentId="19529" CreationDate="2014-11-15T03:18:03.263" Score="4" Body="&lt;p&gt;As LDC3 says, Baking soda is simply sodium bicarbonate, $\ce{NaHCO3}$. When one adds an acid, the following reaction occurs: $\ce{HCO3- + HA -&amp;gt; H2CO3 + A-}$ and the carbonic acid releases carbon dioxide: $\ce{H2CO3 -&amp;gt; CO2 + H2O}$ (these are both equilibrium reactions, but as you can observe when you add vinegar, a lot of gas is produced and the gas is generally lost, pushing the equilibrium towards making carbon dioxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baking powder is a mixture of sodium bicarbonate and one or more weak acids. When dry, no reactions occur, but when wet, the components dissolve and can react. The observed differences when mixing vinegar with baking powder probably result from the fact that there is less sodium bicarbonate in a given volume of baking powder than baking soda, and baking powder often contains starch, which can influence the properties of the added liquid, giving the bubbles a different appearance.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-15T03:18:03.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19533" PostTypeId="2" ParentId="19482" CreationDate="2014-11-15T03:21:39.570" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What if this unknown material produce part of the energy stated in the problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be stated in the problem that the amount of heat that the unknown material produced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The standard molar enthalpy is measured under standard conditions (1 atm, 25°C). However, the condition inside the bomb calorimeter is unknown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We are given some information, the final temperature of the system. But you are right in that we do not know the pressure, and the specific heat of the remaining items is difficult to calculate.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="5026" LastEditDate="2018-03-19T07:00:33.790" LastActivityDate="2018-03-19T07:00:33.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19535" PostTypeId="1" AcceptedAnswerId="19539" CreationDate="2014-11-15T04:01:32.573" Score="5" ViewCount="3382" Body="&lt;p&gt;The element yttrium is called a rare earth element, yet &lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table_(large_version)&quot; rel=&quot;noreferrer&quot;&gt;periodic tables&lt;/a&gt; label it as a transition metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that?&lt;br&gt;&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="9795" LastActivityDate="2014-12-04T07:39:31.390" Title="Yttrium -- rare earth or transition metal?" Tags="&lt;elements&gt;&lt;rare-earth-elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19536" PostTypeId="1" CreationDate="2014-11-15T04:53:41.410" Score="3" ViewCount="1108" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A buffer with $\mathrm{pH} = 4.88$ was prepared by dissolving $\pu{0.10 mol}$ of benzoic acid ($K_\mathrm{a} = \pu{6.3E-5}$) and $\pu{0.50 mol}$ of sodium benzoate in sufficient pure water to form $\pu{1.00 L}$ solution. To a $\pu{70.0  mL}$ aliquot of this solution was added $\pu{2.00 mL}$ of $\pu{2.00 M}$ $\ce{HI}$ solution. What was the $\mathrm{pH}$ of the new $\pu{72.0 mL}$ solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Henderson–Hasselbalch equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4.88 = -\log(6.3 \cdot 10^{-5}) + \log \left( \frac{0.10-x}{x} \right),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \pu{0.0173 mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount of conjugate base in $\pu{70.0 mL}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.0173 \times 70/1000 = \pu{1.211e-3 mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount of $\ce{HI}$ added is $\pu{4e-3 mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck here having more $\ce{H+}$ than my base. &lt;/p&gt;&#xA;" OwnerUserId="9760" LastEditorUserId="5026" LastEditDate="2018-02-23T02:37:53.210" LastActivityDate="2018-02-23T02:37:53.210" Title="Finding pH of a buffer solution on addition of hydrogen iodide to it" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19537" PostTypeId="2" ParentId="19530" CreationDate="2014-11-15T06:55:02.210" Score="2" Body="&lt;p&gt;The carbonyl oxygen is more nucleophilic. You can draw a resonance structure of the carboxylic acid that places a formal negative charge on that oxygen which rationalizes its greater nucleophilicity. Also, after it adds to the carbocation, the resulting intermediate will be resonance stabilized. If the other oxygen attacked, no resonance structures would be possible.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-15T06:55:02.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19539" PostTypeId="2" ParentId="19535" CreationDate="2014-11-15T09:06:47.753" Score="7" Body="&lt;p&gt;Comparing &quot;transition elements&quot; to &quot;rare earth elements&quot; (REE) is a bit like comparing apples and oranges. It all boils down to the different ways in which the elements can be classified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a look at the periodic table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ep2xr.png&quot; alt=&quot;the rare earth elements (REE)&quot;&gt;&#xA;&lt;sup&gt;(modified from &lt;a href=&quot;http://www.texample.net/tikz/examples/periodic-table-of-chemical-elements/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rare earth elements are marked with a red outline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are all rare earth elements are technically transition elements?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, transition elements are the elements in groups 3 to 12 of the periodic table. The rare earth elements, in their widest sense, are Sc, Y and the lanthanides, La to Lu. You can see that if you group the lanthanides as one element, they fit nicely below Y in group 3 of the periodic table. The lanthanides are occasionally called the &quot;inner transition elements&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is yttrium a rare earth element?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are chemical and historical to the inclusion of yttrium in the rare earth elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, yttrium was among the first elements to be isolated from the REE-bearing minerals of the Ytterby mine in Sweden. What was though as first to be &quot;yttria&quot; was later shown to contain also the rest of the heavy REE. It just took a longer time to purify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemically, yttrium commonly occurs with the other REE in the same mineral deposits around the world. You probably know the the chemical properties of the lanthanide series vary continuously from La to Lu. Yttrium also follows that pattern, and in terms of chemical properties it sits very nicely somewhere between Dy and Ho (that's why it's associated in mineral deposits with the heavy REE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yttrium is commonly used in the same high-tech applications as the other REE: it is commonly alloyed with other REE for speciality materials. It is also commonly used in combination with the other REE in lasers and various detectors in electron microscopy and analytical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A different look&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a look at this figure, taken from here: &lt;a href=&quot;https://earthscience.stackexchange.com/a/2951/725&quot;&gt;https://earthscience.stackexchange.com/a/2951/725&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXCuD.png&quot; alt=&quot;trace elements&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that most transition elements, as they appear in nature, usually have a cationic charge between 60 and 80 pm. The REE (including Y) are somewhat larger than that, resulting in slightly different properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be interested in reading the answers to the following questions, also here on Stack Exchange:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earthscience.stackexchange.com/q/310/725&quot;&gt;What are rare earths and why do they cluster near alkaline magmatism?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/18656/8083&quot;&gt;Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-12-04T07:39:31.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19540" PostTypeId="1" AcceptedAnswerId="19608" CreationDate="2014-11-15T09:50:23.313" Score="13" ViewCount="21249" Body="&lt;p&gt;Take the example of a copper and zinc galvanic cell, connected by a salt bridge of $\ce{KNO3}$. I understand how the reactions will result in positive and negative charges, and that the ions of the salt will migrate to maintain neutrality. But why is the salt needed in the first place? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one beaker we will have an excess of zinc ions (resulting in positive charge) and the other with an excess of copper ions (resulting in negative charge). If we just have a bridge, perhaps just soaked in water, wouldn't the ions, if not purposefully then randomly, travel across the bridge and maintain neutrality? What is the purpose of the salt then, if it works without a salt?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="73" LastEditDate="2014-11-17T01:21:14.263" LastActivityDate="2015-06-11T06:22:36.620" Title="Why the salts in a salt bridge?" Tags="&lt;electrochemistry&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19541" PostTypeId="1" AcceptedAnswerId="19553" CreationDate="2014-11-15T10:02:33.470" Score="19" ViewCount="12515" Body="&lt;p&gt;Today in my thermodynamics lecture, my teacher told me that it is not possible to find the absolute value of internal energy and so we have to calculate the change in internal energy.&lt;/p&gt;&#xA;&lt;p&gt;So, my question is why isn't it possible to find the absolute value of internal energy? Is this statement true?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="95133" LastEditDate="2021-04-18T13:05:54.747" LastActivityDate="2021-04-19T02:36:22.207" Title="Reason for inability to find the absolute value of internal energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="19542" PostTypeId="2" ParentId="19540" CreationDate="2014-11-15T11:07:48.667" Score="2" Body="&lt;p&gt;This is an interesting question with two possible answers that I can see. If you are talking in terms of a galvanic cell used as a source of electrical energy, the current supplied by the cell has to be matched by the current flowing between the two half cells. As pure water is very nearly a perfect insulator due to its extremely low degree of ionisation, the internal resistance of the cell would make it impossible to supply any useful quantity of electical power. The use of an ionic solution overcomes this problem and the higher the concentration of ions the better, hence saturated solutions. However, if your question is referring to cells for measuring electrode potentials, I am not so sure of the answer. In principle pure water should do, as you suggested because its slight degree of ionisation means that it does conduct electricity very slightly and this should be all that is needed since electrod potentials should be measured when no current is flowing. I am only sorry that I did not think of this question when I had access to the apparatus needed to try it out.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-15T11:07:48.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19543" PostTypeId="1" CreationDate="2014-11-15T11:08:47.417" Score="3" ViewCount="17228" Body="&lt;p&gt;I am having real trouble understanding the first law of thermodynamics and I think the question, below, highlights most of my issues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) (a) A bomb calorimeter was calibrated by igniting a $0.825\ \mathrm{g}$ sample of benzoic acid in the presence of excess oxygen. The heat given out made the temperature of the calorimeter rise by $1.940\ \mathrm{K}$ from $298\ \mathrm{K}$. The internal energy of combustion of benzoic acid is $-3251\ \mathrm{kJ\ mol}^{-1}$. Calculate the heat capacity of the calorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) In two separate experiments in the same apparatus, $0.498\ \mathrm{g}$ of fumaric and $0.509\ \mathrm{g}$ of maleic acid (both of RMM 116) were ignited and gave temperature rises of $0.507\ \mathrm{K}$ and $0.528\ \mathrm{K}$ respectively. Use the heat capacity of the calorimeter, calculated in (a), to calculate (i) the molar internal energy of combustion, (ii) the molar enthalpy of combustion, and (iii) the molar enthalpy of formation of fumaric and maleic acids. Comment on the difference. The standard enthalpy of formation of water is $285.8\ \mathrm{kJ\ mol}^{-1}$ and of $\ce{CO2}$: $-393.5\ \mathrm{kJ\ mol}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (a) was fine - the heat capacity at a constant pressure = change in internal energy/temperature change. Then divide by number of moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (b) is where I struggle conceptually. I got the answer correct but only by following lecture notes. Sticking with just fumaric acid because there's no point talking about both. It's easy to find the molar change in internal energy using the constant volume heat capacity ($-1338\ \mathrm{kJ\ mol}^{-1}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to find the molar enthalpy change, the change in $H = U + pV$. but since $\ce{pV=nRT; H=U+nRT}$ where n depends on the change in the number of gaseous moles from products to reactants - in this case +1. So $H = U + RT$ giving $-1336\ \mathrm{kJ\ mol}^{-1}$ (less exothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly: Why are the values negative. If the system is the fumaric acid/products of reaction then surely there is an increase in internal energy because the particles will be moving faster after the bonds have broken. (Am I right in thinking that the system is the chemical)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly: Why is the enthalpy value less exothermic? (I have a very poor grasp for what enthalpy is especially in a situation like this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with answering my questions or helping me with this concept would be greatly appreciated. Also, do you know any good books/videos I could watch to help me undertand?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="35555" LastEditDate="2017-03-20T12:47:44.853" LastActivityDate="2017-03-20T12:47:44.853" Title="Difference between internal energy of combustion and enthalpy of combustion?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19546" PostTypeId="2" ParentId="19481" CreationDate="2014-11-15T12:11:01.587" Score="2" Body="&lt;p&gt;Answered my own bonus question (I believe): In case anybody was interested,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trans seems to be favoured due to the methyl group on either side of a Cis bond repelling eachother (due to to both the methyl groups wanting to retain their tetrahedral structure, hydrogens are repelling one another and hence distorting things), and hence the trans is more energetically favourable.&lt;/p&gt;&#xA;" OwnerUserId="9785" LastActivityDate="2014-11-15T12:11:01.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19547" PostTypeId="1" AcceptedAnswerId="19616" CreationDate="2014-11-15T13:39:52.567" Score="1" ViewCount="16880" Body="&lt;p&gt;I saw this experiment many times but still can't explain why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is first prepare two solutions. The first one is to dissolve potassium permanganate crystal into distilled water to form potassium permanganate (VII) while the second one is to put sodium hydroxide and glucose (sugar) into distilled water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I pour the second solution into the first one. My observation is the purple solution will slowly turn into blue, then slowly turn into green and lastly turn into yellowish orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what really happen in the solution?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="5026" LastEditDate="2018-06-28T04:44:21.880" LastActivityDate="2020-02-27T19:48:57.717" Title="Potassium permanganate (VII) + glucose + sodium hydroxide =?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19548" PostTypeId="2" ParentId="19482" CreationDate="2014-11-15T14:10:10.593" Score="0" Body="&lt;p&gt;That is impossible. If your compound is not pure and you do not know the amount of impurity and what kind of impurity, you cannot possibly calculate the amount of ethanol present. Unless we assume that the impurity is small or incombustible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't use a bomb calorimeter in this case because you want to work with enthalpy, not internal energy. An adiabatic flame calorimeter would be more suitable. If you need to use a bomb calorimeter, you can convert between $\Delta U$ and $\Delta H$ by tracking the pressure in the bomb. You can of course control the conditions inside the bomb at the start of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard ethalpy of combustion can still be used, provided that the end temperature of the products of the chemical of physical process are not much higher than the starting temperature of the reactants. A large enough reservoir of heat absorbent or higher $C_v$ (calorimeter constant) should ensure this, but not too large or too high, else you won't be able to accurately read of the temperature difference (you see what the advantage is of an accurate thermometer in this case).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-15T14:21:06.380" LastActivityDate="2014-11-15T14:21:06.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19550" PostTypeId="2" ParentId="19543" CreationDate="2014-11-15T14:41:23.567" Score="2" Body="&lt;p&gt;Combustion releases energy from the system usually as heat and/or work to the surroundings, that's why the standard energy of combustion has a negative sign. The bomb is the system in this case (alternatively you can see the chemical as the system, but you will need to include the gaseous products in your system too, changing the volume of the system, which I find a less preferable viewpoint).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value for the enthalpy will be lower since energy is trapped in the form of increased pressure introduced by the reactions products (mainly by $\ce{CO_2}$) in the bomb. This pressure can be used to do work on the surroundings (in an open system this will be usually on the atmosphere). If you determine the final pressure, you can calculate your $\Delta U$ back ($\Delta U = \Delta H - (p_{end}-p_{atm})V_{bomb}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What book do you use for the lectures?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-15T17:02:02.637" LastActivityDate="2014-11-15T17:02:02.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19551" PostTypeId="2" ParentId="19517" CreationDate="2014-11-15T15:20:26.673" Score="12" Body="&lt;p&gt;As @JonCuster mentions, some materials are pretty resistant to fluorine gas at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But out of curiosity, I checked &lt;a href=&quot;http://theodoregray.com/periodictable/Elements/009/index.s7.html#sample5&quot; rel=&quot;noreferrer&quot;&gt;Theodore Gray's website&lt;/a&gt;. He's made an effort to have a &quot;periodic table&quot; table with actual elements. He also has a &lt;a href=&quot;http://periodictable.com/theelements/&quot; rel=&quot;noreferrer&quot;&gt;beautiful book&lt;/a&gt;, called (naturally) &lt;em&gt;The Elements.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Almost anything placed in the path of a stream of fluorine gas will spontaneously burst into flame. This includes things like, oh, say glass, steel, and other things not normally thought of as flammable. It is, therefore, fairly difficult to have a sample of it in an element collection. I used to have a statement here that there was no transparent container that could hold fluorine for any length of time, but I was corrected by a man who has figured out how to do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's more &lt;a href=&quot;http://theodoregray.com/periodictable/Elements/009/index.s7.html#sample5&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; but I'll summarize. Basically, the problem with glass is the $\ce{Si-O-H}$ bonds, which will auto-catalytically react with $\ce{F2}$ to give $\ce{HF}$, and &lt;em&gt;that&lt;/em&gt; etches the glass. So with a high quality quartz tube, a lot of annealing, and a fluorocarbon grease to coat the glass and minimize reactivity, it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, I always had the idea that fluorine gas was a light blue-green. Instead, it's brownish.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-15T15:20:26.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19552" PostTypeId="1" AcceptedAnswerId="19570" CreationDate="2014-11-15T16:04:31.123" Score="3" ViewCount="389" Body="&lt;p&gt;What would be the reaction if melting iron is put in normal water? Will water be chemically changed?&lt;/p&gt;&#xA;" OwnerUserId="2504" LastEditorUserId="5026" LastEditDate="2018-04-14T05:27:00.717" LastActivityDate="2018-04-14T05:28:36.667" Title="What would be the reaction if melting iron is put in normal water?" Tags="&lt;water&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19553" PostTypeId="2" ParentId="19541" CreationDate="2014-11-15T16:17:29.730" Score="13" Body="&lt;p&gt;In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to define reference points, which we can use to compare our products with. Usually this &quot;zero&quot; point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2016-01-07T08:56:23.757" LastActivityDate="2016-01-07T08:56:23.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19554" PostTypeId="1" CreationDate="2014-11-15T16:28:30.320" Score="5" ViewCount="14196" Body="&lt;p&gt;That $\ce{C=C}$ bonds are more stable at more substituted carbon  atoms is a common argument used throughout &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot; rel=&quot;nofollow&quot;&gt;Clayden et al.&lt;/a&gt;, but is never actually explained (at least not in the first 21 chapters). They do explain however, why the HOMO is increased in energy for more substituted double bonds, through the argument of electron donation by means of hyperconjugation of neighboring groups, increasing the electrophilicity of the $\ce{C=C}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this somewhat counterintuitive. Usually when something is more reactive, it is less stable (I understand that of course by kinetic pathways this does not always hold, e.g. with $\ce{I-}$ as nucleophile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can both observations be incorporated in a consistent explanation in this case?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="186" LastEditDate="2015-10-02T16:29:07.223" LastActivityDate="2015-10-02T16:29:07.223" Title="Why are double bonds at more substituted carbon more stable?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;stability&gt;&lt;hyperconjugation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19555" PostTypeId="1" CreationDate="2014-11-15T16:33:28.720" Score="0" ViewCount="691" Body="&lt;p&gt;If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be 2.3 mol NaOH?&lt;/p&gt;&#xA;" OwnerUserId="8256" LastActivityDate="2014-11-15T16:58:09.930" Title="Limiting reactant and percent yield" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2014-11-15T23:28:57.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="19556" PostTypeId="2" ParentId="19475" CreationDate="2014-11-15T16:45:06.657" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many different organic structures (from the pure theoretical&#xA;  viewpoint) can be drawed with only 4 (exact) carbon atoms and&#xA;  with/without hydrogen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We could make strict rules like each carbon has exactly 4 bonds and get a specific answer, but this is not reality.  There can be lone pair electrons and unpaired electrons.  The octet rule is not stictly followed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C_4}$ actually has been observed and is linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;:C=C=C=C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C_4}$ has been the subject of numerous theoretical and experimetal papers because of its possible occurrence in nebulae. It has been debated whether a singlet linear, triplet linear or rhombic (kite shape) state is the lowest energy state.  &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0009261400005765&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/S0009261400005765&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither the linear nor rhombic states follow the naive rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear $\ce{C_4H}$ has been observed both in the lab and outer space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact according to the University of Kohn lists &lt;a href=&quot;http://www.astro.uni-koeln.de/cdms/molecules&quot; rel=&quot;nofollow&quot;&gt;Molecules in Space&lt;/a&gt;, linear $\ce{C_4H}$ is one of only 61 molecules and molecular ions found in extragalactic space and of only 190 found in the interstellar medium or circumstellar shells as of 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{C_4H_2}$ linear &lt;a href=&quot;http://en.wikipedia.org/wiki/Diacetylene&quot; rel=&quot;nofollow&quot;&gt;butadiyne&lt;/a&gt; in known.  Cyclobutatriene, cyclobutenyne and tetrahedrene have been ruled out theoretically as not represtenting any actual potential energy local minimum, while similar to $\ce {C_4}$, structures having lone pair or unpaired electrons and not following the octet rule (such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbene&quot; rel=&quot;nofollow&quot;&gt;carbenes&lt;/a&gt;) have been calculated to represent actual minima.  See the following references for theoretical cyclic $\ce{C_4H_2}$ structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.540020211/pdf&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/jcc.540020211/pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/jo060698k&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/pdf/10.1021/jo060698k&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/jo000941u&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/pdf/10.1021/jo000941u&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In outer space, not only has the usual HCCCCH isomer been found but also $\ce{H2CCCC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://cdsads.u-strasbg.fr/abs/1991PASJ...43..607K&quot; rel=&quot;nofollow&quot;&gt;Observations of cumulene carbenes, H2CCCC and H2CCC, in TMC-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-05-25T12:43:27.317" LastActivityDate="2016-05-25T12:43:27.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19557" PostTypeId="2" ParentId="19555" CreationDate="2014-11-15T16:57:50.337" Score="0" Body="&lt;p&gt;$$\ce{Na + H2O-&amp;gt;NaOH + \frac12H2}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced?&#xA;  Would it be 2.3 mol NaOH?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No cause the limiting reagent here is water which is in lesser amount, since both reactants have same stoichiometric coefficients, so the product will be formed in accordance with amount of water, i.e. 1.95 moles of $\ce{NaOH}$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-15T16:57:50.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19558" PostTypeId="2" ParentId="19555" CreationDate="2014-11-15T16:58:09.930" Score="1" Body="&lt;p&gt;The first step is to figure out a reaction equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na + H2O -&amp;gt; NaOH + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to balance this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Na + 2H2O -&amp;gt; 2NaOH + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every 2 moles of water used, 2 moles of sodium hydroxide are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2014-11-15T16:58:09.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19559" PostTypeId="1" AcceptedAnswerId="19566" CreationDate="2014-11-15T17:07:34.353" Score="6" ViewCount="583" Body="&lt;p&gt;The periodic table tells us that there are 6 protons in a carbon atom. Is there a way to verify this first-hand? Or are we just expected to believe it unquestioned?&lt;/p&gt;&#xA;" OwnerUserId="9802" LastActivityDate="2014-11-15T19:43:26.260" Title="How can we confirm the number of protons in an atom?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19562" PostTypeId="1" CreationDate="2014-11-15T17:37:01.497" Score="3" ViewCount="398" Body="&lt;p&gt;Typically the antisymmetric stretch in IR spectroscopy is higher than the symmetric stretch for a given functional group. For example for $\ce{NO2}$ the antisymmetric stretch falls at $\sim 1530\ \mathrm{cm^{-1}}$ compared to $1350\ \mathrm{cm^{-1}}$ for the symmetric stretch. Likewise $\ce{NH2}$ has a similar pattern ($3400\ \mathrm{cm^{-1}}$ antisymmetric and $3300\ \mathrm{cm^{-1}}$ symmetric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in some cases this pattern is inverted. e.g. $\ce{[HCO]+}$ has an symmetric stretch at $3896\ \mathrm{cm^{-1}}$ and an antisymmetric stretch at $2814\ \mathrm{cm^{-1}}$. How can this be explained?&lt;/p&gt;&#xA;" OwnerUserId="9804" LastEditorDisplayName="user7951" LastEditDate="2015-06-12T21:48:42.730" LastActivityDate="2015-06-12T21:48:42.730" Title="Assign symmetric and antisymmetric stetches in IR spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19563" PostTypeId="2" ParentId="19536" CreationDate="2014-11-15T18:27:43.093" Score="1" Body="&lt;p&gt;Amount of acid added: &#xA;$$0.002\ \mathrm L \times 0.2\ \mathrm{mol/L} = 0.004\ \mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will neutralize $0.004\ \mathrm{mol}$ of the benzoate: $$0.50\ \mathrm{mol} - 0.004\ \mathrm{mol} = 0.496\ \mathrm{mol}$$ of benzoate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will increase benzoic acid by $0.004\ \mathrm{mol}$: $$0.10\ \mathrm{mol} + 0.004\ \mathrm{mol} = 0.104\ \mathrm{mol}$$ of benzoic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] = K_\mathrm a\left(\frac{n(\ce{HA})}{n(\ce{A-})}\right)$$ (yes, it's supposed to be $[\ce{HA}]/[\ce{A-}]$, but volume cancels out leaving amount of substance so long as your numbers leave you inside the titration curve between the starting $\mathrm{pH}$ and the equivalence point $\mathrm{pH}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] = 6.3 \times 10^{-5} \left(\frac{0.104\ \mathrm{mol}}{0.496\ \mathrm{mol}}\right) = 1.32 \times 10^{-5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\log\left(1.32 \times 10^{-5}\right) = \mathrm{pH} = 4.88$$&#xA;(actually $4.8794$, so you have a small change in $\mathrm{pH}$, as you would expect with a buffer with so much of the conjugate base in solution, but not enough to make a significant difference.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not at all clear why people are married to the Henderson–Hasselbalch equation for buffer problems. The equilibrium expression covers just about any situation I can think of, there's no formula to remember if you know how to do an equilibrium expression.&lt;/p&gt;&#xA;" OwnerUserId="9806" LastEditorDisplayName="user7951" LastEditDate="2016-11-09T13:37:08.023" LastActivityDate="2016-11-09T13:37:08.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19564" PostTypeId="1" AcceptedAnswerId="19951" CreationDate="2014-11-15T18:39:27.890" Score="3" ViewCount="925" Body="&lt;p&gt;Is there a reason radioactive decay is exponential, as in after an extremely long time some of the original compound remains, or is this just an empirical model? Also, why does $\ce{^{238}U}$ have a much longer half life than $\ce{^{210}Pb}$, for example?&lt;/p&gt;&#xA;" OwnerUserId="7641" LastEditorUserId="9961" LastEditDate="2016-06-23T15:48:55.810" LastActivityDate="2016-11-26T16:16:35.453" Title="Why does radioactive decay occur?" Tags="&lt;kinetics&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19566" PostTypeId="2" ParentId="19559" CreationDate="2014-11-15T19:43:26.260" Score="4" Body="&lt;p&gt;1) While not easy, it is possible to obtain mass-spectrum of full range of ions, including fully ionized. By number of peaks with Z/E corresponding as X, X/2, X/3 ... X/n it is possible to ensure that an element has exactly n electrons and protons. It is still difficult to ensure full ionization of heavier atoms, so the method is not applicable for heavier elements. This, however, is the only direct method I can think of. Should work for carbon, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Various X-ray-derived spectral methods. While not providing direct evidence, they do provide information on electronic shell structure, that can be compared with theoretical calculations. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-15T19:43:26.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19568" PostTypeId="1" AcceptedAnswerId="21811" CreationDate="2014-11-15T21:39:36.340" Score="11" ViewCount="4969" Body="&lt;p&gt;I'm learning about electronegativity in atoms and how their difference affects the type of bond they make. I know that no difference makes a non-polar covalent bond, a difference less than 2.1 is a polar covalent bond, and a difference above 2.1 is an ionic bond. But what kind of bond forms if the difference is exactly 2.1?&lt;/p&gt;&#xA;" OwnerUserId="9808" LastEditorUserId="5017" LastEditDate="2014-12-16T20:23:28.150" LastActivityDate="2014-12-16T20:23:28.150" Title="What happens if the electronegativity difference is exactly 2.1?" Tags="&lt;bond&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19569" PostTypeId="1" CreationDate="2014-11-15T21:46:46.983" Score="7" ViewCount="1502" Body="&lt;p&gt;I have run some tests with a PEM fuel cell and the voltage we have found was $1.44\ \mathrm{V}$. Given that the standard potential for a fuel cell is $1.229\ \mathrm{V}$ what is causing the difference in our measurement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially we provide power to the fuel cell so it can generate oxygen and hydrogen into separate containers. These containers are submerged in water so there is water pressure keeping them contained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All measurements were done in standard conditions, except that the pressure of each container is slightly different (hydrogen displaced twice the amount of water since there are twice the moles of atoms present). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to calculate this using the Nernst equation since the conditions of both the oxygen and hydrogen are slightly different (each gas is of a different concentration in its respective container since there is different water pressure and water vapour pressure present).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell} = 1.229\ \mathrm{V} - \frac{RT}{2F}\cdot\ln\left(\frac{[\ce{H2O}]}{[\ce{H2}][\ce{O2}]^{0.5}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $[\ce{H2}]$ I am using the molar concentration of the gas in its containing cylinder (I believe this accounts for water vapour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also do the same thing for $[\ce{O2}]$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what to use for $[\ce{H2O}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer I am getting is $1.24\ \mathrm{V}$. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorDisplayName="user7951" LastEditDate="2015-05-13T21:05:56.857" LastActivityDate="2021-03-31T19:25:54.570" Title="Nernst equation for a fuel cell when gases are stored separately" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19570" PostTypeId="2" ParentId="19552" CreationDate="2014-11-15T22:35:30.433" Score="4" Body="&lt;p&gt;Unlike the other metals we usually see in the reactivity series, such as calcium, magnesium, zinc, etc., hot iron does not form iron hydroxide, instead it forms iron oxide. The reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3Fe + 4H2O &amp;lt;=&amp;gt; FeO.Fe2O3 + 4H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is similar to magnesium and zinc as they react only with hot water (not cold water). But the reaction of iron with hot water is less vigorous than that of magnesium and zinc.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="5026" LastEditDate="2018-04-14T05:28:36.667" LastActivityDate="2018-04-14T05:28:36.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19571" PostTypeId="2" ParentId="19554" CreationDate="2014-11-15T23:28:37.197" Score="6" Body="&lt;p&gt;Let's start by providing some useful background information that will help answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,3-Butadiene is more stable than 2 separate ethylene molecules due to the extended conjugation possible in the diene.  We can illustrate this by drawing simple resonance structures for the diene that can't be drawn for ethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ORC5L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Huckel calculations show the same thing, 1,3-butadiene is more stable than 2 individual ethylenes.  Here is the Huckel MO diagram for ethylene.  The two electrons in ethylene have a net stabilization of 2$\beta$ compared to an electron in an isolated p-orbital.  4 electrons in 2 ethylenes would have an overall stabilization energy of 4$\beta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4PnSe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Huckel diagram for 1,3-butadiene.  The 4 electrons have a net stabilization of [(2 x 0.62) + (2 x 1.62)] = 4.48$\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDmOD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the HOMO in 1,3-butadiene is higher in energy then the HOMO for ethylene, &lt;strong&gt;overall&lt;/strong&gt; 1,3-butadiene is more stable than 2 isolated ethylenes by 0.48$\beta$.  This is because the MO below the HOMO is stabilized much more (0.62 $\beta$ compared to ethylene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now let's return to your question&lt;/strong&gt;.  Hyperconjugation is a reasonable way to explain why more highly substituted double bonds are more stable. Just like in 1,3-butadiene, when we delocalize electrons over a larger number of atoms, which is what hyperconjugation does, an overall stabilization of the molecule should result.  And, as the hyperconjugated resonance structure shows, electron density will shift from the alkyl group to the double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FBxz0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the HOMO may be raised in energy, (at least some of) the other occupied MO's below the HOMO will be lowered in energy (just like in 1,3-butadiene) and &lt;strong&gt;overall&lt;/strong&gt; a net stabilization of the molecule will result.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-15T23:28:37.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19572" PostTypeId="1" AcceptedAnswerId="19595" CreationDate="2014-11-15T23:39:35.287" Score="8" ViewCount="6322" Body="&lt;p&gt;I just can't seem to get the correct answer to this symmetry question in my book. &quot;Introduction to molecular symmetry J.S.Ogden.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the reducible representations of both the $\ce{B-Cl}$ bond and the $\ce{B-B}$ bonds in $\ce{B_2Cl_4}$ in the point group of $D_\mathrm{2d}$.&lt;br&gt;&#xA;The correct answer for both is $\Gamma_{\ce{B-Cl}}=A_1+B_1+E$ and $\Gamma_{\ce{B-B}}=A_1$.&lt;br&gt;&#xA;My reducible representations are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Gamma_{\ce{B-Cl}}=4,0,0,0,2$ for the $\ce{B-Cl}$ bond in the order $E, 2S_4, C_2, 2C', 2\sigma_\mathrm{d}$. &lt;/li&gt;&#xA;&lt;li&gt;$\Gamma_{\ce{B-B}}= 1,0,1,0,1$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly either of these are not correct but honestly I can't see why! Your help is much appreciated! (I worked them out based on the number of bonds that remain unshifted during the respective symmetry operation. I have also made a little model for myself in Gaussian but it didn't really help.&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="4945" LastEditDate="2014-11-19T11:06:15.087" LastActivityDate="2015-04-24T10:13:05.283" Title="How to get the reducible representations of the B-B and the B-Cl bond of B2Cl4" Tags="&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19573" PostTypeId="2" ParentId="19568" CreationDate="2014-11-15T23:39:54.337" Score="13" Body="&lt;p&gt;Based on the the answer to this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/9222/why-electronegativity-difference-greater-than-1-7-are-ionic&quot;&gt;Why electronegativity difference greater than 1.7 are ionic?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a very nice graph of &lt;strong&gt;percent ionic character&lt;/strong&gt; as function of electronegativity difference for some common binary compounds. Here ionic character can be thought of as the degree of charge separation across the covalent bond. Compounds with 50% or more ionic character are considered &quot;ionic&quot;. This division of classes of compounds is &lt;strong&gt;arbitrary.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ikx1x.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Image source: &lt;a href=&quot;http://www.chem.ufl.edu/~itl/4411/lectures/lec_13.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Florida Chemical Bonding page&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-11-15T23:39:54.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19574" PostTypeId="1" CreationDate="2014-11-16T01:05:33.987" Score="3" ViewCount="270" Body="&lt;p&gt;It's the last one, where 1,4-dimethylcyclohexane breaks apart into... something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SXhdSl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the workup I have. I want to put out there that I'm not concerned about getting the optimal yield yet. Just making the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free radical addition of $\ce{Br}$ to the 2 position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removal of that $\ce{Br}$ and the formation of a double bond via E2 using tBuO-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking apart the structure via ozonolysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, no matter where I place the double bond, I can't get the correct configuration of double bonded Oxygens (ketones and aldehydes, right?). So, I must be doing something wrong. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="9810" LastEditorDisplayName="user10259" LastEditDate="2015-05-14T00:51:49.550" LastActivityDate="2015-05-14T00:51:49.550" Title="Designing synthetic route to dicarbonyl compound" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19576" PostTypeId="2" ParentId="19574" CreationDate="2014-11-16T01:17:30.880" Score="3" Body="&lt;p&gt;Radical bromination will occur at the 1-position (tertiary carbon). Your strategy from there is correct. Elimination using a small base such as hydroxide will give 1,4-dimethylcyclohex-1-ene. Ozonolysis with reductive workup gives the product.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-16T01:17:30.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19578" PostTypeId="2" ParentId="19574" CreationDate="2014-11-16T01:25:53.417" Score="6" Body="&lt;p&gt;Everything you said is correct except for the first step.  Free radical bromination is reasonably selective, tertiary positions are about 20 time more reactive than secondary carbons, so the &lt;strong&gt;1&lt;/strong&gt;-bromo, 1,4-dimethyl compound is formed.  Everything's a go from there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4u5K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-16T01:25:53.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19579" PostTypeId="1" CreationDate="2014-11-16T02:44:15.990" Score="5" ViewCount="1364" Body="&lt;p&gt;Two part question for you. Let's preface this with I'm in organic chemistry 1, so we're just learning the basics of these reactions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are neopentyl halides completely inert to any reaction through any pathway, $\mathrm{S_N1, S_N2, E1, E2}$? For $\mathrm{S_N1/E1}$, even after solvolysis of the halide, is there too much steric strain there for anything to attack the carbocation? Even a small base like $\ce{MeO-}$? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a methyl iodide/bromide undergo $\mathrm{S_N2}$ with any basically any nucleophile since there's practically no steric hindrance? Even a bulky compound that's not very lewis basic like tButanol? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9810" LastEditorUserId="7475" LastEditDate="2015-10-09T16:06:47.457" LastActivityDate="2015-10-09T16:06:47.457" Title="SN2: Neopentyl halides and methyl halides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19593" PostTypeId="1" AcceptedAnswerId="19740" CreationDate="2014-11-16T10:58:50.967" Score="4" ViewCount="36302" Body="&lt;p&gt;I always get confusing why the appearance of normal sugar (the white one), brown sugar and red sugar are different although they are sugar. Is it because they each have different molecular formulae or different number of bond or maybe the brown and red sugar have some transition metal inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't. So, I wish someone will help on this question. &lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:35:28.693" LastActivityDate="2018-03-28T10:55:24.040" Title="Difference between white sugar, brown sugar and red sugar" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;color&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19595" PostTypeId="2" ParentId="19572" CreationDate="2014-11-16T13:25:20.220" Score="8" Body="&lt;p&gt;There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about switching to another book as it seems to raise more questions than it answers. But I'm not working regularly with group theory, so my analysis of the problem might be faulty, but maybe someone else can support or refute my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, here is my analysis:&#xA;First, make yourself familiar with the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group. Have a look at &lt;a href=&quot;https://www.youtube.com/watch?v=f1uDLQl6re8&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; for example. Now, I looked at the effect each symmetry operation has on the bonds I'm looking at. For that I will use the following depiction of $\ce{B2Cl4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QP39W.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\ce{B-B}$ bond I used an arrow to indicate the direction of the bond for better emphasis as I will need that soon. In the following I will assume that you are familiar with the way characters are related to your reducible representation and how you get those characters from analyzing the changes of your basis vectors (in this case the bonds) under the symmetry operations of the group since you seem to have already used it to determine the reducible representations in your question.&#xA;Furthermore, I use the following notation: The reducible representation of the $\ce{B-Cl}$ bonds will be denoted as $\Gamma_{\ce{B-Cl}}$ and the character of the symmetry operation $C_{2}$ for this reducible representation will be denoted as $\Gamma_{\ce{B-Cl}}(C_{2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, let's see what the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group do to this structure, so that we can get the reducible representations:&#xA;The $E$ operation does nothing to the structure, i.e. all four $\ce{B-Cl}$ bonds and the one $\ce{B-B}$ bond stay where they are. Thus you get a character of $4$ for $\Gamma_{\ce{B-Cl}}(E)$ and a character of $1$ for $\Gamma_{\ce{B-B}}(E)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we come to the improper axis $S_{4}$. It rotates along the long the $\ce{B-B}$ bond and mirrors perpendicular to it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z92yv.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(S_{4}) = 0$. Since the mirror plane is perpedicular to the $\ce{B-B}$ bond the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(S_{4}) = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we look at the principle $C_{2}$ rotation along the $\ce{B-B}$ bond. It converts your molecule to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWqta.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}) = 0$. But as the rotation is conducted along the $\ce{B-B}$ bond axis the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the two $C_{2}^{\prime}$ axes which are perpendicular to $C_{2}$. One disects the dihedral angle between the dark red $\ce{Cl}$ and the cyan $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other disects the dihedral angle between the red $\ce{Cl}$ and the blue $\ce{Cl}$ - its effect on the structure is shown in the right picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jgcvR.png&quot; height=&quot;130&quot;&gt;&#xA;$\qquad \qquad \qquad$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9OYng.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}^{\prime}) = 0$. But now the rotation is conducted along an axis perpedicular to the $\ce{B-B}$ bond and so the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}^{\prime}) = -1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally there are the two $\sigma_{\mathrm{d}}$ planes which lie along the $C_{2}$. The one plane contains the dark red and the red $\ce{Cl}$ while disecting the angle betweeen the  cyan and the blue $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other contains the cyan and the blue $\ce{Cl}$ while disecting the angle betweeen the dark red and the red $\ce{Cl}$ - its effect on the structure is shown in the right picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ious9.png&quot; height=&quot;130&quot;&gt;&#xA;$\qquad \qquad \qquad$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mRPcx.png&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case either the red and dark red $\ce{Cl}$ atoms or the blue and cyan ones remain unchanged while the other $\ce{Cl}$ atoms change their places. So, two $\ce{B-Cl}$ bonds keep their location and direction which gives $\Gamma_{\ce{B-Cl}}(\sigma_{\mathrm{d}}) = 2$.&#xA;The $\ce{B-B}$ bond axis lies within the mirror plane and so the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(\sigma_{\mathrm{d}}) = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in conclusion you end up with the reducible representations (I kept the order of the symmetry elements you chose, i.e. $E$, $2 \, S_4$, $C_2$, $2 \, C_{2}^{'}$, $2 \, \sigma_{d}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Gamma_{\ce{B-Cl}} &amp;amp;= 4, 0, 0, 0, 2 \\&#xA;  \Gamma_{\ce{B-B}}  &amp;amp;= 1, -1, 1, -1, 1 \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and they reduce to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \Gamma_{\ce{B-Cl}} &amp;amp;= A_1 + B_2 + E \\&#xA;  \Gamma_{\ce{B-B}}  &amp;amp;= A_2 \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see both results differ from the ones your book provides.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2015-04-24T10:13:05.283" LastActivityDate="2015-04-24T10:13:05.283" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19596" PostTypeId="2" ParentId="19593" CreationDate="2014-11-16T14:18:21.047" Score="0" Body="&lt;p&gt;Molasses is collected from sugarcane when it is spun. It is dark brown, and when dried, makes brown sugar. Some varieties of sugarcane are red, producing red sugar. Further processing of the molasses will remove the color and produce white sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Processing of sugarcane:&lt;br&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sugarcane#Processing&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://scamsweet.com/how-its-made/&quot; rel=&quot;nofollow&quot;&gt;SweetScam&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-16T14:18:21.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19597" PostTypeId="2" ParentId="14860" CreationDate="2014-11-16T14:21:52.273" Score="3" Body="&lt;p&gt;You are right that $\ce{HBr}$ and $\ce{HI}$ are preferred over $\ce{HCl}$ for this reaction because $\ce{Br-}$ and $\ce{I-}$ are better nucleophiles than $\ce{Cl-}$, for the reason you have provided. Actually, the hydrolysis proceeds significantly slower with hydrochloric acid, and reasonable rates are only achieved with hydrobromic and hydroiodic acid (&lt;a href=&quot;http://www.3rd1000.com/chem301/chem301w.htm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). And while all of the acids level to the hydronium ion in water, the $\ce{H3O+}$ concentration will be higher in aqueous solutions of $\ce{HBr}$ and $\ce{HI}$, as both are available in higher concentrations than hydrochloric acid (&lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/acid-str.htm&quot; rel=&quot;nofollow&quot;&gt;57&lt;/a&gt; and &lt;a href=&quot;http://www.commonorganicchemistry.com/Common_Reagents/Hydrobromic_Acid/Hydrobromic_Acid.htm&quot; rel=&quot;nofollow&quot;&gt;48 wt.-%&lt;/a&gt; versus &lt;a href=&quot;http://www.commonorganicchemistry.com/Common_Reagents/Hydrochloric_Acid/Hydrochloric_Acid.htm&quot; rel=&quot;nofollow&quot;&gt;38 wt.-%&lt;/a&gt; for $\ce{HCl}$).&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-11-16T14:21:52.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19598" PostTypeId="1" CreationDate="2014-11-16T14:25:34.890" Score="0" ViewCount="11262" Body="&lt;p&gt;As titled, I wish to know if glycerol/glycerin is soluble in ethanol (not soluble, slightly soluble, very soluble)&lt;/p&gt;&#xA;" OwnerUserId="9817" LastActivityDate="2015-02-17T20:42:47.903" Title="Is glycerol/glycerin soluble in ethanol?" Tags="&lt;homework&gt;&lt;solubility&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19599" PostTypeId="1" AcceptedAnswerId="19792" CreationDate="2014-11-16T15:06:43.237" Score="2" ViewCount="784" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/3U2j6o2&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/3U2j6o2.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says there is no reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ROH group is resonance-stabilized with the benzene-like group. Its conjugate base is stable. There is an inductive effect of the cloud of electrons that make the oxygen more electronegative. However, resonance effects are always more important than inductive effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is there no reaction?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-11-20T23:35:46.810" Title="Reaction an ROH-resonance-stabilized Group with Hydroxide/Acid-Base Reactions" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19600" PostTypeId="1" CreationDate="2014-11-16T17:11:41.173" Score="0" ViewCount="65" Body="&lt;p&gt;I am searching for a chemical compound which meets these five conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;slightly soluble in water&lt;/li&gt;&#xA;&lt;li&gt;layered strutcture&lt;/li&gt;&#xA;&lt;li&gt;ionic conductor&lt;/li&gt;&#xA;&lt;li&gt;thermochromism&lt;/li&gt;&#xA;&lt;li&gt;lead must be a part of it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9820" LastEditorUserId="46306" LastEditDate="2018-03-04T14:58:24.723" LastActivityDate="2018-03-04T14:58:24.723" Title="A chemical compound with these conditions?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-03-04T18:44:29.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="19601" PostTypeId="2" ParentId="19579" CreationDate="2014-11-16T17:29:47.770" Score="4" Body="&lt;p&gt;Carey-Sundberg has some answers (vol. A, chapter 4).  The text is a bit advanced for you at the present stage, but you should give it a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives as reference JACS 1975, 97, 3694.  That paper has rate constants for nucleophilic substitution of compounds of type $\ce{RCH2Br}$, with $\ce{R=H, Me, Et, i-Pr, t-Bu }$ with $\ce{LiCl}$ in acetone at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rate constants in 10&lt;sup&gt;5&lt;/sup&gt; s&lt;sup&gt;-1&lt;/sup&gt;:&#xA;$\ce{R=H}$: 600, $\ce{R=Me }$: 9.9, $\ce{R=Et}$: 6.5, $\ce{R=i-Pr}$: 1.5, $\ce{R=t-Bu }$: 2.6*10&lt;sup&gt;-4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleophilic substitution does take place in neopentyl halides, but at a rate of five orders of magnitudes less than in less hindered compounds.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-16T17:29:47.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19604" PostTypeId="1" AcceptedAnswerId="20556" CreationDate="2014-11-16T17:53:13.283" Score="1" ViewCount="262" Body="&lt;p&gt;I'm very confused as to where the center of symmetry is for the PbO  (litharge form). I know that it is in the P4/nmm space group, and that it's centrosymmetric, but beyond that I'm clueless. Thank you. &lt;/p&gt;&#xA;" OwnerUserId="9709" LastActivityDate="2014-12-10T23:33:31.530" Title="Where is the center of symmetry in the PbO crystal?" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19606" PostTypeId="1" CreationDate="2014-11-16T19:33:08.727" Score="0" ViewCount="3515" Body="&lt;p&gt;I can't find a method to balance a reaction: the heating of potassium dichromate, which forms potassium dichromate (III), dichromate oxide (III) and oxygen. &#xA;I can't add water products as $\ce{H+}$, $\ce{{}^{-}OH}$ because we are not in solution. &#xA;So, how can I balance it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who oxidises in this redox? The oxygen inside the salt? The oxygen of $\ce{H2O}$ which is in the air?&lt;/p&gt;&#xA;" OwnerUserId="9824" LastEditorUserId="4945" LastEditDate="2014-11-17T02:33:52.937" LastActivityDate="2016-11-09T20:35:19.613" Title="How to find the balanced reaction equation for the decomposition of potassium dichromate in non aqueous conditions?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19608" PostTypeId="2" ParentId="19540" CreationDate="2014-11-16T20:18:28.253" Score="13" Body="&lt;p&gt;The purpose of the salt bridge is to prevent the two half-cell solutions from mixing. It is possible to make a really bad galvanic cell by putting both half-cells in a single solution, but they rapidly self discharge as the oxidizing agent ($\ce{Cu^+}$ in your example) can diffuse through the solution and react directly with the other electrode (Zn in the example), with the electrons being transferred directly instead of going through the external circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a &quot;bridge&quot; of just water soaked paper is similar to having both half-cells in the same container but poses some issues. First, the diffusion of ions through the bridge is going to be very slow, increasing the cell resistance, second, the self-discharge problem above will occur, and third, because different ions diffuse at different rates, a junction potential develops as the two containers diffuse into one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a salt bridge's job is to not only maintain electroneutrality between the two containers, but to do so without allowing the solutions to mix. An inert salt is used to prevent any reactions from occurring between the salt and the solutions, and things like $\ce{KCl}$ are commonly used for analytical cells as potassium and chloride ions have very similar diffusion coefficients, minimizing junction potential.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-16T20:18:28.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19611" PostTypeId="2" ParentId="19606" CreationDate="2014-11-16T21:23:46.057" Score="5" Body="&lt;p&gt;I think that part of your difficulty lies in incorrect naming of the products which are simply potassium chromate(VI) and chromium(III) oxide together with oxygen. The equation then becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce {4K2Cr2O7 -&amp;gt; 4K2CrO4 + 2Cr2O3 + 3O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="7475" LastEditDate="2016-11-09T20:35:19.613" LastActivityDate="2016-11-09T20:35:19.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19612" PostTypeId="2" ParentId="2851" CreationDate="2014-11-16T21:55:47.187" Score="0" Body="&lt;p&gt;I sympathise with your problem using electron half equations when one of the reactants produces two different products. For this reason and the fact that (as the previous answer points out) there are no ionic compounds invoved, I prefer to use oxidation number half equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Oxidation half equation:&lt;/strong&gt; $\ce{H2  -&amp;gt;  2H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen oxidation numer goes up 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reduction half equation&lt;/strong&gt;: $\ce{NO -&amp;gt; NH3}$&lt;br&gt;&#xA;Nitrogen's oxidation number goes down by 5 (+2 to -3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is unchanged (-2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor to balance the oxidation numbers $5~\ce{H2}$s require $2~\ce{NO}$s and the equation becomes: $$\ce{5H2 + 2NO  -&amp;gt; 2NH3 + 2H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="5026" LastEditDate="2018-02-25T03:18:28.110" LastActivityDate="2018-02-25T03:18:28.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19613" PostTypeId="1" CreationDate="2014-11-16T23:54:15.143" Score="0" ViewCount="6457" Body="&lt;p&gt;What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in an acidic solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{5NO2-   +   2MnO4-   +   6H+   -&amp;gt;   5NO3-   +  2Mn^{+2}  +  3H2O}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that right?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="4231" LastEditDate="2014-11-17T01:21:45.870" LastActivityDate="2014-12-17T03:50:36.567" Title="What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in acidic solution?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19614" PostTypeId="1" AcceptedAnswerId="19624" CreationDate="2014-11-17T00:21:47.583" Score="1" ViewCount="299" Body="&lt;p&gt;If you have an ice cube in water and the ice cube does not melt after reaching equilibrium with the water, does the water have temperature 0 degrees Celsius?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="16683" LastEditDate="2016-11-09T11:36:56.637" LastActivityDate="2016-11-09T11:36:56.637" Title="Thermodynamics of Solid-Liquid" Tags="&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19615" PostTypeId="1" CreationDate="2014-11-17T00:41:56.810" Score="6" ViewCount="3154" Body="&lt;p&gt;The Wikipedia structure is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQAiw.png&quot; width=&quot;240&quot; height=&quot;168&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the empty bond at far-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Journals on the Web&quot; has:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OSLVH.gif&quot; width=&quot;202&quot; height=&quot;96&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CH3 added at far-right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rxlist.com shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VM8f0.gif&quot; width=&quot;236&quot; height=&quot;130&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an extra H going into the page! Is one of these incorrect, or are they all consistent but w/ some implied bonds or something?&lt;/p&gt;&#xA;" OwnerUserId="9829" LastEditorUserId="7448" LastEditDate="2019-09-13T02:29:00.707" LastActivityDate="2019-09-13T02:29:00.707" Title="Why are there different chemical structures for epinephrine (adrenaline)?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19616" PostTypeId="2" ParentId="19547" CreationDate="2014-11-17T00:43:22.910" Score="4" Body="&lt;p&gt;This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastEditorUserId="84629" LastEditDate="2020-02-27T19:48:57.717" LastActivityDate="2020-02-27T19:48:57.717" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19618" PostTypeId="1" AcceptedAnswerId="20558" CreationDate="2014-11-17T00:47:00.183" Score="2" ViewCount="694" Body="&lt;p&gt;I am a student of masters study with a background only in electrical engineering, but I have studied Miller indices and crystallographic planes and directions, personally, to help myself understand the process of silicon etching which is required in my current course of study. But, I am unable to understand the following written in my text book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;When Silicon is etched anisotropically, i.e. depending on crystal direction, and a {1 0 0} surface direction is used, cavities will result which have the same {1 0 0} orientation at the bottom and are determined laterally by {1 1 1} surfaces. The reason is that {1 1 1} surfaces are etched off using a minimal etch rate and this way becomes the most resistant&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Could someone please help me understand this concept.&lt;/p&gt;&#xA;" OwnerUserId="9830" LastEditorUserId="5017" LastEditDate="2014-12-10T23:59:44.110" LastActivityDate="2014-12-11T00:18:23.660" Title="How is the crystal direction important for etching silicon surfaces?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19619" PostTypeId="2" ParentId="19615" CreationDate="2014-11-17T00:47:22.110" Score="7" Body="&lt;p&gt;They are all consistent. With line diagrams, all bonds to &quot;nothing&quot; are actually implicit bonds to carbon (this is why the benzene ring and carbon tail can be drawn without $\ce{C}$ in each of your diagrams). In addition, hydrogens are added implicitly until each carbon is saturated. So they are all the same structure.&lt;/p&gt;&#xA;" OwnerUserId="7821" LastActivityDate="2014-11-17T00:47:22.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19620" PostTypeId="2" ParentId="19615" CreationDate="2014-11-17T00:48:24.827" Score="15" Body="&lt;p&gt;There are often shortcuts taken in drawing complicated molecules, particularly if there's not any interesting chemistry going on at those functional groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{-CH3}$ is implied at the end of the &quot;stick,&quot; just like how the carbon atoms aren't explicitly labeled in the benzene ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogens are rarely explicitly drawn for the same reasons, given that you can infer the hybridization by inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Skeletal_formula&quot;&gt;Skeletal formula&lt;/a&gt; for more details.  Briefly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon atoms are usually depicted as line ends or vertices with the assumption that all carbons have a valence of 4 and carbon-hydrogen bonds, usually not shown explicitly, are assumed to complete each C valence. A skeletal formula shows the skeletal structure or skeleton of a molecule, which is composed of the skeletal atoms that make up the molecule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="73" LastActivityDate="2014-11-17T00:48:24.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19621" PostTypeId="1" CreationDate="2014-11-17T02:53:09.700" Score="6" ViewCount="733" Body="&lt;p&gt;Both carbon and silicon expose a lot of metallic properties like metallic lustre, electrical and heat conductivity, etc, so often considered metalloids. The both form alloys with other metals such as iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the usual silicon carbide is not an alloy at all. It is tranparent, electrical insulator etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder whether there is a different carbon-silicon compound that has more metallic properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon (graphite):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WHMwu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q4F2j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WK5sq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EEGmz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xlLQu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicon carbide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BMH2G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qv82B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="9831" LastEditDate="2014-11-17T20:55:05.077" LastActivityDate="2021-02-03T02:55:17.167" Title="Are there carbon-silicon alloys?" Tags="&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19622" PostTypeId="2" ParentId="19613" CreationDate="2014-11-17T02:56:39.623" Score="1" Body="&lt;p&gt;Your suggestion is certainly part of the picture, but as is usually the case with nitric acids, a number of side reactions can and often do take place depending on concentration and temperature. In acid solutions, for example, nitrite spontaneously decompose giving nitric oxide together with dinitrogen tetroxide. My feeling is that under optimum conditions your suggested reaction would predominate&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-17T02:56:39.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19623" PostTypeId="1" AcceptedAnswerId="19649" CreationDate="2014-11-17T03:05:06.480" Score="2" ViewCount="149" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TiQOq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that sodium borohydride doesn't work particularly well on resonance-stabilized carbonyl groups (i.e. esters) but what about this carbonyl in the center of the above picture? This appears to be somewhat resonance-stabilized ... why would sodium borohydride still work  here? Sodium borohydride is a nucleophilic reagent. On the other hand the second resonance contributor I've drawn is pretty crappy; yes - it's part of an aromatic system - but it's also a carbocation. And the R' group is inductively donating (although I suspect this is a fairly minor point). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it just work in the above synthesis but not with a yield comparable to something such as sodium borohydride and formaldehyde? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dNTzH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-11-17T05:54:03.170" LastActivityDate="2014-11-17T17:31:23.780" Title="Resonance and reductions with sodium borohydride" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19624" PostTypeId="2" ParentId="19614" CreationDate="2014-11-17T05:45:04.313" Score="2" Body="&lt;p&gt;Yes, when ice is present at equilibrium with liquid water (at 1 atm), the system is at 0 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you continue to remove heat from the water (such as in your freezer), the temperature of the system will remain at 0 °C until all the liquid (water) becomes solid (ice) after which, continuing to remove heat will result in a drop of temperature below 0 °C. Conversely, if heat is added to the system, the temperature of the system will remain at 0 °C until all of the solid (ice) has melted into liquid, after which, the temperature will increase above 0 °C. The amount of (latent) heat required to melt ice is called its &lt;em&gt;enthalpy of fusion&lt;/em&gt; (334 J/g). This energy is converted into intermolecular bonds (a change in internal or potential energy of the system) rather than a change in kinetic energy (that is, a temperature change).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at a much lower pressure (0.006 atm) and slightly higher temperature (0.01 °C), pure water will be in three states or &lt;em&gt;phases&lt;/em&gt; (gas, liquid and ice). This is known as the &lt;em&gt;triple point&lt;/em&gt; of water, as illustrated in the phase diagram of water, shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PgHVN.jpg&quot; alt=&quot;phase diagram of water&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorDisplayName="user7951" LastEditDate="2016-11-09T11:25:44.597" LastActivityDate="2016-11-09T11:25:44.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19625" PostTypeId="1" CreationDate="2014-11-17T00:47:14.217" Score="3" ViewCount="165" Body="&lt;p&gt;I am looking for a process in which reduces &lt;a href=&quot;http://en.wikipedia.org/wiki/Ilmenite&quot; rel=&quot;nofollow&quot;&gt;ilmenite&lt;/a&gt; to pure titanium with the highest yield possible. We are to use a process of our choice or one that we have created in conjunction with &lt;a href=&quot;http://www.factsage.com/&quot; rel=&quot;nofollow&quot;&gt;FactSage&lt;/a&gt; to calculate the yield of titanium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the materials we have available are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pure Metals&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Al, Ca, Mg, Na, Si}$ &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Pure Non-metals&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Ar, C, Cl2, H2, N2, O2, S }$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Molecules&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH4, CO, CO2, HCl, H2O, H2S, NH4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a few process that come to mind such as reduction using Magnesium, $\ce{H2SO4}$, high temperature $\ce{CO2}$, but any help on which would be the most efficient combined with the highest yield would be great. &lt;/p&gt;&#xA;" OwnerDisplayName="Eric Tenuta" LastEditorDisplayName="user7951" LastEditDate="2015-10-26T21:36:00.727" LastActivityDate="2015-10-26T21:36:48.017" Title="How to efficiently reduce of ilmenite to pure titanium?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19627" PostTypeId="1" AcceptedAnswerId="19641" CreationDate="2014-11-17T06:18:45.473" Score="1" ViewCount="68" Body="&lt;p&gt;Can we actually control catalytic hydrogenation by introducing specific molar quantities of hydrogen gas? My prof insists on no. No we cannot use a certain equivalency of hydrogen gas to stop at a desired product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet in the book I keep seeing examples of specific molar quantities of hydrogen gas being used ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4qwyj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-11-17T19:25:25.103" LastActivityDate="2014-11-17T19:25:25.103" Title="Equivalencies and catalytic hydrogenation" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19628" PostTypeId="1" AcceptedAnswerId="19644" CreationDate="2014-11-17T06:24:11.590" Score="2" ViewCount="811" Body="&lt;p&gt;Are enones resistant to hydrogenation? My solutions manual thinks so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KCj8l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't recall discussion regarding enones in my text. Nonetheless, could the resonance stabilization of the enone be preventing its hydrogenation? In that case, why is the alkene hydrogenated; both the carbonyl and the alkene are resonance-stabilized to some extent! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="186" LastEditDate="2015-10-02T16:29:44.297" LastActivityDate="2015-10-02T16:29:44.297" Title="Enones and hydrogenation" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19629" PostTypeId="1" AcceptedAnswerId="19632" CreationDate="2014-11-17T08:40:52.893" Score="1" ViewCount="4244" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The transition of an electron from a higher level to a lower level results in the emission of a photon of wavelength $350.0\ \mathrm{nm}$. If the energy of the higher level is $-3.24\times10^{-19}\ \mathrm J$, calculate the energy of the lower level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By using $E = h\nu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find $E = 5.679\times10^{-28}\ \mathrm J$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use this value to solve the qus?  &lt;/p&gt;&#xA;" OwnerUserId="9760" LastEditorDisplayName="user7951" LastEditDate="2015-12-14T14:15:05.320" LastActivityDate="2015-12-14T14:15:05.320" Title="Calculating energy level from wavelength" Tags="&lt;physical-chemistry&gt;&lt;electrons&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19631" PostTypeId="1" AcceptedAnswerId="19637" CreationDate="2014-11-17T10:14:07.750" Score="78" ViewCount="21958" Body="&lt;p&gt;Recently, I am learning the production of soluble and insoluble salts. My friend and I have done this experiment at the school lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wanted to taste them to see whether they are salty are not. The teacher luckily stopped us from doing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So without tasting them, I would really like to know whether all salts are salty.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="1677" LastEditDate="2017-11-06T23:58:19.490" LastActivityDate="2017-11-06T23:58:19.490" Title="Do all salts taste salty?" Tags="&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;&lt;safety&gt;&lt;taste&gt;" AnswerCount="6" CommentCount="7" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19632" PostTypeId="2" ParentId="19629" CreationDate="2014-11-17T13:12:10.687" Score="3" Body="&lt;p&gt;All you need are two equations and two constants (speed of light and Planck constant):&#xA;\begin{eqnarray}&#xA;\mathrm{E} &amp;amp; = &amp;amp; h\nu\\&#xA;h &amp;amp; = &amp;amp; \mathrm{6.62606957 \cdot 10^{−34}   J\cdot s}\\&#xA;\mathrm{c} &amp;amp; = &amp;amp; \lambda\nu\\&#xA;\mathrm{c} &amp;amp; = &amp;amp; 299792458\ \mathrm{m\cdot s}^{-1}\\&#xA;\end{eqnarray}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these data, the energy of a photon with $\lambda = 350\ \mathrm{nm}$ calculates to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{E} = \frac{6.62606957 \cdot 10^{-34} \cdot 2.99792458 \cdot 10^8}{350 \cdot 10^{-9}} \cdot \frac{\mathrm{J \cdot s \cdot m}}{\mathrm{s \cdot m}} = 0.05676 \cdot 10^{-17} \mathrm{J} = 5.676 \cdot 10^{-19} \mathrm{J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the energy difference to the higher level.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-11-17T18:02:20.193" LastActivityDate="2014-11-17T18:02:20.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19634" PostTypeId="2" ParentId="19631" CreationDate="2014-11-17T13:20:43.227" Score="11" Body="&lt;p&gt;The short answer is no, as suggested in the various comments. Not being a biologist I cannot give any underlying theory to explain this, but from personal and reported tasting I can give some examples: Copper sulphate, iron(II) sulphate, aluminium sulphate among others are very bitter. One you could try yourself would be sodium carbonate (washing soda) or sodium bicarbonate (baking soda). Interestingly Lead acetate as mentioned in one of the comments has a sweet taste which makes it particularly dangerous due to its toxicity. Even potassium chloride (used as a substitute for salt to reduce sodium intake) tastes quite different.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-17T13:20:43.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19635" PostTypeId="2" ParentId="19631" CreationDate="2014-11-17T13:21:43.203" Score="36" Body="&lt;p&gt;Saltiness is perceived when alkali metal enter taste buds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Taste&quot;&gt;wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Saltiness is a taste produced primarily by the presence of sodium&#xA;  ions. Other ions of the alkali metals group also taste salty, but the&#xA;  further from sodium the less salty the sensation is. The size of&#xA;  lithium and potassium ions most closely resemble those of sodium and&#xA;  thus the saltiness is most similar. In contrast rubidium and cesium&#xA;  ions are far larger so their salty taste differs accordingly.[citation&#xA;  needed] The saltiness of substances is rated relative to sodium&#xA;  chloride (NaCl), which has an index of 1. Potassium, as potassium&#xA;  chloride - KCl, is the principal ingredient in salt substitutes, and&#xA;  has a saltiness index of 0.6. Other monovalent cations, e.g. ammonium,&#xA;  NH4+, and divalent cations of the alkali earth metal group of the&#xA;  periodic table, e.g. calcium, Ca2+, ions generally elicit a bitter&#xA;  rather than a salty taste even though they, too, can pass directly&#xA;  through ion channels in the tongue, generating an action potential.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="1362" LastEditDate="2014-11-17T13:59:05.140" LastActivityDate="2014-11-17T13:59:05.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19636" PostTypeId="2" ParentId="19600" CreationDate="2014-11-17T13:58:28.437" Score="2" Body="&lt;p&gt;I will satisfy 4 of the five conditions, but you will have to research the fifth. Lead(II) iodide has a layered structure, is slightly soluble in water, is ionic (but only weakly). I hope that this helps.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-17T13:58:28.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19637" PostTypeId="2" ParentId="19631" CreationDate="2014-11-17T14:02:05.573" Score="67" Body="&lt;p&gt;No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Salts with hydrolysing cation (various alums) are acidic in addition to other notes.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-11-17T22:52:15.357" LastActivityDate="2014-11-17T22:52:15.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19638" PostTypeId="1" CreationDate="2014-11-17T14:05:32.157" Score="9" ViewCount="547" Body="&lt;p&gt;Following wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cases of systemic toxicity from accidental swallowing are rare, and attempted suicide via intravenous injection does not appear to result in systemic toxicity,[1] though it is still causes damage by physicaly blocking blood vessels both at the site of injection and the lungs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shouldn't both inhalation and IV injection of elemental mercury produce the same result, namely, mercury in the bloodstream?&lt;/p&gt;&#xA;" OwnerUserId="9839" LastActivityDate="2014-12-17T20:53:49.013" Title="Why is mercury much more toxic when inhaled than injected?" Tags="&lt;biochemistry&gt;&lt;toxicity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19640" PostTypeId="2" ParentId="19638" CreationDate="2014-11-17T14:44:31.167" Score="3" Body="&lt;p&gt;I guess the answer is that mercury atoms stick with each other very well (e.g. high surface tension in comparison to most other liquids $\gamma_{293K}^{\ce{Hg}} = 0.47 \text{ N/m}$, where  $\gamma_{293K}^{\ce{H2O}} =  0.073 \text{ N/m}$ &lt;a href=&quot;http://scipp.ucsc.edu/~haber/ph5B/bubble.pdf&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt;) and that this continues to happen in the blood stream. Such clumps of mercury are not able to pass the blood brain barrier (BBB) effectively, especially in comparison to finely divided dissolved vapor, which is transported trough the BBB by the formation of a complex with &lt;a href=&quot;https://en.wikipedia.org/wiki/Cysteine&quot; rel=&quot;nofollow&quot;&gt;cysteine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-17T14:52:51.320" LastActivityDate="2014-11-17T14:52:51.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19641" PostTypeId="2" ParentId="19627" CreationDate="2014-11-17T15:26:39.973" Score="3" Body="&lt;p&gt;Using the equipment that I have and working on the scale that I tend to work on (&amp;lt;1 mol typically), I can't deliver only one equivalent of hydrogen gas. We can measure the amount of hydrogen consumed as the reaction proceeds. The problem you have shown could be interpreted that way, i.e. 1 mol is consumed but more may be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On industrial scale it may be possible to deliver a more accurate amount of hydrogen. I'll leave it to others that might have more experience on large scale to confirm or deny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are alternative sources of hydrogen that can more easily be delivered accurately on small scale. Either ammonium formate or cyclohexadiene can be used to generate hydrogen in the presence of palladium. The former liberates ammonia and carbon dioxide as byproducts, while the latter forms benzene as byproduct.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-17T15:26:39.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19642" PostTypeId="2" ParentId="18559" CreationDate="2014-11-17T15:28:15.213" Score="2" Body="&lt;p&gt;Citric acid has three pKs: 3.1, 4.8 and 5.4.  If you add 3 equivalents of lye, you will get essentially only trisodium citrate.  If you add a lesser amount, you will have a mixture of the various citrates (at least from microscopic point of view), as well as some remaining citric acid if you add less than one equivalent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One &quot;equivalent&quot; being one mole of lye per mole of citric acid. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-17T15:28:15.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19643" PostTypeId="1" AcceptedAnswerId="19694" CreationDate="2014-11-17T15:30:14.457" Score="16" ViewCount="11671" Body="&lt;p&gt;I understand that tap water contains minute amounts of deuterium oxide (heavy water). Is there a simple way to separate, concentrate the $\ce{D2O}$?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="4231" LastEditDate="2014-11-17T15:31:42.140" LastActivityDate="2018-01-22T10:09:07.803" Title="Is there a simple way to separate deuterium oxide from tap water?" Tags="&lt;water&gt;&lt;isotope&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19644" PostTypeId="2" ParentId="19628" CreationDate="2014-11-17T15:37:55.800" Score="2" Body="&lt;p&gt;I read your question as asking why the C-C pi bond is reduced while the C-O pi bond is not. It appears to be possible to change the selectivity of hydrogenation based on the catalyst used (&lt;a href=&quot;http://www.mpi-halle.de/mpi/publi/pdf/8236_08.pdf&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;), but for simple palladium on carbon, reduction of alkenes (even conjugated to carbonyls) is much faster than an aldehyde or ketone. Coupled with complementary hydride reagents, which are selective for carbonyls, synthetic chemists can achieve the desired reduction without worrying about reaction of other reducible groups.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-17T15:37:55.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19645" PostTypeId="1" AcceptedAnswerId="19648" CreationDate="2014-11-17T15:55:06.650" Score="2" ViewCount="574" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDBVi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a much different approach from the solutions manual. My synthesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) H2 and Lindlar's catalyst to stop at the alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Oxidative cleavage to form benzoic acid. Works here because we have a benzylic hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Addition of excess allyl alcohol with catalytic concentration of acid and heat. A Fischer esterification step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also one fewer step than what my solution's manual has. The solutions manual went up to the acid chloride and down to the ester, which is a nice step (avoids the heat). Does mine also work on paper? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-05-14T16:56:37.540" Title="Creation of ester from acetylene derivative" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19646" PostTypeId="1" CreationDate="2014-11-17T16:20:27.797" Score="3" ViewCount="6853" Body="&lt;p&gt;It seems that if a carbocation intermediate is formed through solvolysis, then the product(s) are going to always be a pretty even mix of S&lt;sub&gt;N&lt;/sub&gt;1 and E1 products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any conditions that greatly favor one mechanism? I thought that a very bulky solvent - like &lt;em&gt;t&lt;/em&gt;-$\ce{BuOH}$ - would favor E1. However, the carbocation is $\ce{sp^2}$ hybridized, so there still wouldn't be a lot of steric hindrance to prevent S&lt;sub&gt;N&lt;/sub&gt;1, right?&lt;/p&gt;&#xA;" OwnerUserId="9841" LastEditorUserId="5026" LastEditDate="2018-04-14T05:31:26.383" LastActivityDate="2018-04-14T05:32:55.677" Title="What conditions would make SN1 products the major products over E1 products, and vice versa?" Tags="&lt;organic-chemistry&gt;&lt;nucleophilic-substitution&gt;&lt;elimination&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19648" PostTypeId="2" ParentId="19645" CreationDate="2014-11-17T16:54:51.157" Score="5" Body="&lt;p&gt;No, your approach doesn't give the right product. There's a carbon missing between the benzene ring and the carbonyl carbon. See the difference in the products below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly what your solutions manual suggested, but I've included a conservative approach through phenylacetaldehyde. As you suggest, it's not absolutely necessary to go through an acid chloride to make an ester from a carboxylic acid. The Fischer esterification will work. Also, for many carboxylic acid to ester conversions, you can simply react the carboxylic acid with thionyl chloride and the desired alcohol all in the reaction together, rendering it a one-step transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3gAM8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-17T16:54:51.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19649" PostTypeId="2" ParentId="19623" CreationDate="2014-11-17T17:31:23.780" Score="2" Body="&lt;p&gt;Sodium borohydride is definitely strong enough to reduce an acetophenone derivative. It is likely that the resonance stabilization does reduce the reactivity of the ketone. It's not often that a sodium borohydride reduction requires 35 hours. In the cases that I have run, usually less than one hour is sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some quick searching of whether an isolated ketone could be reduced in the presence of an acetophenone, and I did not find an adequate example of a case where only the isolated ketone was reduced. That says to me that the reactivity difference is not that great. Perhaps someone else has come across this though.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-17T17:31:23.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19650" PostTypeId="1" CreationDate="2014-11-17T17:36:04.970" Score="-2" ViewCount="1852" Body="&lt;p&gt;I came across this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quantity of electrical charge that brings about the deposition of 4.5 g $\ce{Al}$ from $\ce{Al^{+3}}$ at cathode will also produce the following volume (STP) of $\ce{H2}$ (g) from $\ce{H+}$ at cathode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A)22.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B)11.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C)5.6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D)2.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use faradays first law:&#xA;For $\ce{Al^{+3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.5-=(27*c)/ (96500*3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c=96500/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{H2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m=c/(96500*2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m=0.25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence V=2.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use 2nd law, we get mass of $\ce{H2}$ produced as 0.5&#xA;Hence, V=5.6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then what was the mistake&#xA;(Please reply if I have made calculation mistake, I generally do that)&lt;/p&gt;&#xA;" OwnerUserId="9689" LastEditorUserId="4231" LastEditDate="2014-11-18T03:41:21.417" LastActivityDate="2021-04-26T07:53:36.217" Title="faraday laws of electrolysis" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19651" PostTypeId="1" CreationDate="2014-11-17T18:04:41.920" Score="2" ViewCount="112" Body="&lt;p&gt;If holes are made in dry walls is there a substance that can be sprayed in the walls to eliminate cigarette  odor?&lt;/p&gt;&#xA;" OwnerUserId="9845" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:53:28.887" LastActivityDate="2015-10-30T16:53:28.887" Title="Can you eliminate cigarette smoke odor between walls?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19652" PostTypeId="2" ParentId="19650" CreationDate="2014-11-17T18:25:44.967" Score="1" Body="&lt;p&gt;I am not sure what your answer was (2.8 or 5.6dm3) and therfore am not sure what the mistake is. I suggest an alternative application of Faraday's law which strikes me as simpler. The law states that the same electrical charge will liberate the same mumber of moles of different atoms divided by the charge on their ions.&#xA;In this case the No. of moles of aluminium = 4.5/27 = 1/6&#xA;therefore No. of moles of H atoms = 3*1/6 (because Al has three times the charge of H+ ions)&#xA;therfore No. of moles of H2 molecules = 3/2*1/6&#xA;= 1/4. therefore volume of H2 at STP =22.4/4 = 5.6 dm3. which is one of the answers that you quoted.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-17T18:25:44.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19654" PostTypeId="2" ParentId="19651" CreationDate="2014-11-17T18:56:30.660" Score="3" Body="&lt;p&gt;Getting rid of odor is a difficult task in general, particularly since molecules stick to surfaces and often stick &lt;em&gt;in&lt;/em&gt; surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a good reason that &lt;a href=&quot;http://en.wikipedia.org/wiki/Febreze&quot; rel=&quot;nofollow&quot;&gt;&quot;Febreze&quot;&lt;/a&gt; works well. It contains a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclodextrin&quot; rel=&quot;nofollow&quot;&gt;cyclodextrin&lt;/a&gt; ring that binds many odor-causing molecules. So the molecules are still there, but bound in a form you can't smell anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I remember an anecdote that they had to develop special fragrances to work with Febreze, but I can't find that anywhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect spraying paint primer inside the walls could work. This is often used to fix smoke damage from fires.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-17T18:56:30.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19655" PostTypeId="2" ParentId="19631" CreationDate="2014-11-17T19:08:28.943" Score="5" Body="&lt;p&gt;As mentioned, while not all salts taste salty, many do, and not just salts with sodium ions. If you want to try another edible salt, try to get ahold of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_chloride&quot; rel=&quot;nofollow&quot;&gt;ammonium chloride&lt;/a&gt;, which is used (both as a powder and otherwise) in sweets in the Nordic countries, the Netherlands and Germany. Its taste can definitely be described as salty, but not the same kind of salty as sodium chloride.&lt;/p&gt;&#xA;" OwnerUserId="9847" LastEditorUserId="-1" LastEditDate="2014-11-18T14:25:05.550" LastActivityDate="2014-11-18T14:25:05.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19658" PostTypeId="2" ParentId="19528" CreationDate="2014-11-17T20:48:39.987" Score="1" Body="&lt;p&gt;First, you need to recognize what the limiting reagent is.  Because NaOH and HCl react in a 1:1 ratio, and there is less HCl (&lt;span class=&quot;math-container&quot;&gt;$\pu{25mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.79M}$&lt;/span&gt; is fewer moles than &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.6M}$&lt;/span&gt;), HCl is the limiting reagent.&lt;/p&gt;&#xA;&lt;p&gt;Second, given the amount of limiting reagent, how much heat will be released (or absorbed).  &lt;span class=&quot;math-container&quot;&gt;$\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Third, you need to approximate that the solution has the heat capacity of water, which is &lt;span class=&quot;math-container&quot;&gt;$\pu{4.18 kJ/K L}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average.  So here: &lt;span class=&quot;math-container&quot;&gt;$$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So finally calculate the amount the temperature of &lt;span class=&quot;math-container&quot;&gt;$\pu{75 ml}$&lt;/span&gt; of &amp;quot;water&amp;quot; will increase when &lt;span class=&quot;math-container&quot;&gt;$\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$&lt;/span&gt; of heat is added, and add this to &lt;span class=&quot;math-container&quot;&gt;$\pu{25.55 ^\circ C}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="69830" LastEditDate="2021-05-19T06:27:13.607" LastActivityDate="2021-05-19T06:27:13.607" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19659" PostTypeId="2" ParentId="19621" CreationDate="2014-11-17T21:32:01.843" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://www.ioffe.ru/SVA/NSM/Semicond/SiC/thermal.html&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; you can find a phase diagram for $\ce{Si/C}$ system. It does not have zones with homogeneous non-stohiometric solids. So, there is not thermodinamically stable Si/C isomorphic alloys. However, since the liquid likely to be homogeneous, it is likely for fast cooled liquid to form amorphous alloys. Indeed, google search provides plenty of links for &lt;a href=&quot;https://www.google.ru/webhp?q=amorphous%20silicon-carbon%20alloys#newwindow=1&amp;amp;safe=off&amp;amp;q=amorphous%20silicon-carbon%20alloys&quot; rel=&quot;noreferrer&quot;&gt;amorphous $\ce{Si/C}$ alloys&lt;/a&gt;. However, it is also possible for cooling liquid to form polycrystalline alloy, say, with $\ce{SiC}$ and $\ce{C}$ grains, similar to pig iron.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-17T21:32:01.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19660" PostTypeId="1" CreationDate="2014-11-17T22:03:43.480" Score="3" ViewCount="71" Body="&lt;p&gt;Phase diagrams of various compounds and systems usually include only positive pressure. But I wonder whether it is possible to somehow investigate them under negative pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there materials that change phase if mechanically stretched? For instance given that metallic form is sometimes achieved at high pressure, can a negative pressure applied to a metal produce a non-metalic form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe the effect of negative pressure may be achieved by electrically uniformly charging the material?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastActivityDate="2014-11-18T06:13:17.237" Title="Are there attempts to investigate phase states at negative pressure?" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19661" PostTypeId="2" ParentId="19646" CreationDate="2014-11-17T22:46:34.473" Score="3" Body="&lt;p&gt;In a practical way, you would have to consider each particular case, there are tons of reasons that could make a reaction to yield mostly E1 or S&lt;sub&gt;N&lt;/sub&gt;1 product (it will be always a mixture, that can vary from 50:50 to 0.01:99,9, I hope I'm making myself clear, they are always competing reactions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are several factors that contribute to make one of the two more favorable than the other, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The kind of the reactant employed: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use hydrogen halides, ($\ce{HCl, HBr, HI}$) you will mostly obtain S&lt;sub&gt;N&lt;/sub&gt;1 products.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you use acids like $\ce{H2SO4, H3PO4~or~TsOH}$ you will mostly obtain E1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is this?&lt;/strong&gt; The reagents/acids of the first case, are very good nucleophiles. The reagents lisetd on the second case, are awful nucleophiles.&#xA;A good nucleophile is key for a substitution. But for elimination you only need something that can abstract a product in the alpha position to the leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, very strong and small bases, can favour an elimination through E2 mechanism, with almost no substitution taking place. E1 mechanisms is usually observed only under mild basic conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvents, like you say, also can favour one type or another:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In some cases, solvent acts a base for E1 reactions. (i.e. $\ce{EtOH -&amp;gt; EtOH2+}$). Bulkiness of the solvent can cause both E1 and S&lt;sub&gt;N&lt;/sub&gt;1 (in case that the solvent itself is the nucleophile, which usually happens) to be slower: it has to approach the compound that undergoes the reaction anyway. Actually is easier to approach a hydrogen than a carbon, so probably will favour S&lt;sub&gt;N&lt;/sub&gt;1 over E1, BUT, do NOT take this as a determining factor at all. The alpha carbon from which you need to abstract the proton might be more or less hindered.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solvents with low nucleophility also decreases the % of substitution reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both E1 and S&lt;sub&gt;N&lt;/sub&gt;1 are favoured in protic solvents, in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, you will always have both products, but yes, there are factors that can be modified in order to maximize one of the two reactions. There are some general considerations, like the ones I posted, but in practice, each reaction is a world.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5026" LastEditDate="2018-04-14T05:32:55.677" LastActivityDate="2018-04-14T05:32:55.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19662" PostTypeId="2" ParentId="19631" CreationDate="2014-11-17T23:50:13.907" Score="16" Body="&lt;p&gt;permeakra is quite right with his counterexamples of salts that don't taste purely salty, but I'd like to expand on why. We haven't fully identified and elucidated the receptors involved in taste, but they can be broadly classed into tastes people are familiar with: salty, sweet, sour, bitter, and umami.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Salty and sour receptors are both known to be mostly dominated by ion channels, where salty receptors allow small cations to pass and sour receptors are sensitive to pH in a similar fashion. The other receptors are g-protein coupled receptors. The trick with taste (and smell, for that matter) is that there is not a single receptor type for each possible chemical one can taste, i.e. there's not a receptor entirely selective towards $\ce{Na+}$. The way it works is that there are many types of taste receptors that are somewhat selective to specific classes of molecules and a given molecule may induce a response in many different receptors. The brain takes all the data from all these receptors; combines them; integrates information from the nose, eyes, ears, tactile nerves in the mouth, etc.; and then produces the sensation of taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, though both $\ce{Na+}$ and $\ce{K+}$ will stimulate some of the same saltiness receptors, they also stimulate other receptors and in different amounts, making them taste different. In general, small cations can stimulate the ion channel receptors for saltiness, so many salts do taste salty to a certain extent, but many also stimulate other receptors. Because what we perceive as taste comes from the interaction of all these different receptors (as well as other senses), one salt may taste very salty and another quite bitter, despite both activating saltiness receptors; the responses are not independent (which is why it's possible to mask overly salty soup or whatever by adding sugar).&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-17T23:50:13.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19663" PostTypeId="1" AcceptedAnswerId="19666" CreationDate="2014-11-17T23:52:13.877" Score="2" ViewCount="5611" Body="&lt;p&gt;Soap and shampoo are both cleaning agents in our daily life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is soap in solid form and shampoo is in a liquid form, though? Is it because of the chemical components inside them or another factor? Which cleaning agent has a higher efficiency in cleaning grease from the body? &lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="73" LastEditDate="2014-11-18T01:01:08.223" LastActivityDate="2020-09-14T06:47:49.197" Title="Soap versus Shampoo" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19664" PostTypeId="2" ParentId="19302" CreationDate="2014-11-17T23:56:33.410" Score="1" Body="&lt;p&gt;Silane is the equivalent of methane for silicon, $\ce{SiH4}$. &#xA;Silicon containing organic derivatives of many kinds exist.&#xA;Generally speaking, organosilanes are those in which $\ce{Si-C}$ bonds exist, like for example, tetramethylsilane ($\ce{TMS}$, used as a reference on $\ce{{}^{1}H~NMR}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkoxysilanes are those compounds that contains an organic group bonded to a silicon but not directly, through an oxygen atom instead, like so: $\ce{R-OSi(Me)3}$, common structure on silyl enol ethers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A surface is hydrophobic if it's repulsions towards water overcome its attractions, and this is the case for many silicon compounds, but not all of them. For example, glass (silicon dioxide, $\ce{SiO2}$) is hydrophilic, due to the possibility of formation of hydrogen bonds between the glass oxygen and water molecules.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4945" LastEditDate="2014-11-18T04:45:04.640" LastActivityDate="2014-11-18T04:45:04.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19665" PostTypeId="2" ParentId="13591" CreationDate="2014-11-18T00:32:15.703" Score="5" Body="&lt;p&gt;I do not think that an entropy decrease is a requirement to form a bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about a &quot;normal&quot; case: $\ce{HCl}$ (a gas) and $\ce{NH3}$ (another gas) react together to form $\ce{NH4Cl}$, which is a solid. This is an example of bond formation where entropy clearly decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now think what would happen if two liquid or solid compounds react together to form a gaseous compound. An increase of entropy has taken place. Can't think of a concrete example, but I think this might be the case of, for example, an esterification between a liquid alcohol and a liquid carboxylic acid that form an ester where, due to the lose of $\ce{-OH}$ groups, intermolecular interactions are lower and it can result with the product molecule being more volatile than the original compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about the number of micro states in which the compounds can exist, and gas formation from solids or liquids importantly increases entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS: These are just some thoughts, really, I'm no physical-chemist.)&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5026" LastEditDate="2018-03-26T01:46:22.273" LastActivityDate="2018-03-26T01:46:22.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19666" PostTypeId="2" ParentId="19663" CreationDate="2014-11-18T02:30:20.730" Score="5" Body="&lt;p&gt;Well, whether these products are liquids or solids typically has more to do with the additional ingredients than the actual cleansing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soap, chemically speaking, is a surfactant composed of the salt of a fatty acid, typically lauric, palmitic or stearic acids. It's typically produced by adding a strong hydroxide base to liquefied plant or animal fat, which liberates the fatty acids from the triglyceride via hydrolyzation, where they then bind with the free cations from the base forming an amalgam of salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shampoo, or more specifically the detergents present in it, are similar, but produced by a different process, to create a surfactant that is less likely to bond to the mineral compounds in hard water, and so is more effective than soap when using it with tap water from most regions. Sodium lauryl sulfate and sodium laureth sulfate are the two frontrunners here; the former is very powerful, and doesn't foam as much, so it's typically used for laundry detergent. The latter is gentler on human skin (less irritating) and so is typically used for various hand and body cleansers including shampoos. The two split about 50-50 in dish soaps, with a third surfactant, lauramine oxide, also making an appearance as a heavy-hitting degreaser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these substances act in a similar way; they have a hydrophilic (water-attracting) side, and a hydrophobic (water-repelling) side. When dissolved in water (which happens slowly unless the surfactants are already dissolved), the hydrophobic side is attracted to anything that isn't water, and so it will loosely bind via various van der Waals forces to mineral compounds and organics such as those found in dirt, grease, grass stains, oil, lipstick etc. These molecules end up surrounding small particles of these substances on all sides in a bid to &quot;shelter&quot; their hydrophobic end, forming a &quot;micelle&quot;. The hydrophilic side exposed to the water, then, allows the micelle to float away in the water and resists the substance in the core of the micelle reattaching to the surface it had been on (dishes, clothes, skin, hair etc), so it can be easily rinsed away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the form in which a detergent is most useful typically determines whether its sold or packaged as a liquid or solid. Overall, its state is most directly determined by the amount of water in the product. If you've ever had a housemate that never wipes off or closes the nozzle on a bottle of dish detergent, you will have learned that these &quot;liquid&quot; soaps don't have much trouble being solid, either. The liquid products sometimes have an alcohol that helps keep the water from evaporating (and some, like cetyl alcohol, are what make certain products opaque as well as thickening them to give the appearance of a cream), but if marketing wants the product to be a solid, all the chem-engineers have to do is run the mix through a warm (160-250*F) oven to evaporate excess water, leaving behind a pliable solid that can be pressed into bars or minced into soap flakes such as for laundry powders. As far as efficacy in any given form, as I said, liquids tend to disperse in wash water more easily than solids (which is good in the case of a bar soap, not do good with a dishwasher or laundry detergent), though the powder forms are designed to dissolve well, usually through use of some sort of &quot;filler&quot; that spaces out the active ingredients and dissolves easily (and also allows for &quot;everyday&quot; units of measure, so it feels like you're adding enough soap to the water to get your clothes or dishes clean, when in reality a tablespoon of the concentrated active ingredients is more than enough as evidenced by the little gel packs).&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2014-11-18T02:54:29.387" LastActivityDate="2014-11-18T02:54:29.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19668" PostTypeId="1" CreationDate="2014-11-18T04:04:42.617" Score="5" ViewCount="850" Body="&lt;p&gt;My question has to do with some combustion related definitions. When the term &lt;em&gt;complete combustion&lt;/em&gt; is used, is it assumed that the combustion is stoichiometric (reactants combine perfectly), but dissociation does not occur, such that only carbon dioxide and water is the product? If so, does the term &lt;em&gt;stoichiometric combustion&lt;/em&gt; mean the same thing except dissociation effects do occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Minor species of combustion as a result of dissociation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9VVAG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="8117" LastEditDate="2014-11-19T15:40:53.903" LastActivityDate="2014-11-19T15:40:53.903" Title="Is There a Difference in the Definition of Complete and Stoichiometric Combustion?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19669" PostTypeId="2" ParentId="19660" CreationDate="2014-11-18T06:13:17.237" Score="3" Body="&lt;p&gt;I think there is a little misconception here.&#xA;There are no negative absolute pressures. When someone talks about negative pressures, they are referring to values under atmospherical pressure, this is,between 0 and 1 atm.&#xA;There are many ways of creating this vacuum, and of course, that matter has been studied under these conditions, for example, this kind of negative pressures are applied to get the boiling point of substances reduced, so thermolabile compounds can be vaporized (or the solvent which contains them) under its decomposition temperature.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-18T06:13:17.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19670" PostTypeId="1" AcceptedAnswerId="19804" CreationDate="2014-11-18T07:14:11.627" Score="4" ViewCount="287" Body="&lt;p&gt;The setting of this problem is on the moon of Jupiter: Io.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that chlorine exists in two isotopes: $\ce{{}^{35}_{17}Cl}~(33.3\%)$ and $\ce{{}^{37}_{17}Cl}~(66.7\%)$, what is the mass spectrum of $\ce{Cl3+}$?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I know now is that there are 3 peaks in the mass spectrum of $\ce{Cl2+}$ because there are 3 possible combinations ($\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{35}_{17}Cl}$ / $\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$ / $\ce{{}^{37}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$). Therefore, there are 4 possible combinations and 4 peaks for $\ce{Cl3+}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know how to make a mass spectrum graph with only this much information.&lt;/p&gt;&#xA;" OwnerDisplayName="user9862" LastEditorUserId="5026" LastEditorDisplayName="user9862" LastEditDate="2018-03-09T06:19:36.023" LastActivityDate="2018-03-09T06:19:36.023" Title="How to find the mass spectrum of Cl3+?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19673" PostTypeId="1" CreationDate="2014-11-18T10:58:02.323" Score="8" ViewCount="971" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3qO5F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3qO5F.png&quot; alt=&quot;Bromination of 8-chloroimidazo[1,2-a]pyrazine in acetic acid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to rationalise the regioselectivity of bromination of 8-chloroimidazo[1,2-&lt;em&gt;a&lt;/em&gt;]pyrazine, as depicted in the scheme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left ring kind of looks like a pyridazine. If it's anything like a pyridine it'll be bad at electrophilic substitution without activation, so bromination at C–5 or C–6 is out. I guess it has to be one of the carbons in the five-membered ring. But why is it C–3 and not C–2? I am trying to draw out resonance structures and reason that way.&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-09-15T13:29:27.063" LastActivityDate="2018-09-15T13:48:25.633" Title="Regioselectivity of electrophilic aromatic substitution in imidazo[1,2-a]pyrazine" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19674" PostTypeId="1" AcceptedAnswerId="19678" CreationDate="2014-11-18T12:14:48.467" Score="0" ViewCount="1448" Body="&lt;p&gt;The order of filling of molecular orbitals $\ce{O2}$ by MOT is this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OOYQO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from it, how can I deduce that there is one sigma and one pi bond in an oxygen molecule ?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastActivityDate="2014-11-18T12:31:48.867" Title="Molecular Orbital Theory and No. of bonds" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19676" PostTypeId="1" AcceptedAnswerId="19683" CreationDate="2014-11-18T12:19:48.323" Score="3" ViewCount="162" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;We have &lt;span class=&quot;math-container&quot;&gt;$\pu{10 L}$&lt;/span&gt; of solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COONa}$&lt;/span&gt; with concentration &lt;span class=&quot;math-container&quot;&gt;$C = \pu{1 M}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 10$&lt;/span&gt;. Then we add &lt;span class=&quot;math-container&quot;&gt;$\pu{90 L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;. What will be the pH of the new solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;My work:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \ce{CH3COONa &amp;amp;&amp;lt;=&amp;gt; CH3COO- + Na+}\\&#xA;  \ce{CH3COO- + H2O &amp;amp;-&amp;gt; CH3COOH + OH-}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We have &lt;span class=&quot;math-container&quot;&gt;$\pu{1 M } \ce{CH3COO-}$&lt;/span&gt; and let &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; denote the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;. If the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of the solution is &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt;, it's implied that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pOH}=4\iff-\log_{10} x=4 \iff x=\pu{10^{-4} M}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I know I have to use that the amounts of substances didn't change, so &lt;span class=&quot;math-container&quot;&gt;$$C_1V_1 = C_2V_2,$$&lt;/span&gt;&#xA;but I don't know what values I have to put in there. Is the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; I found my &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9867" LastEditorUserId="4945" LastEditDate="2020-09-02T13:59:55.773" LastActivityDate="2020-09-02T14:06:12.313" Title="Finding pH after dilution of acetic acid" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19678" PostTypeId="2" ParentId="19674" CreationDate="2014-11-18T12:31:48.867" Score="3" Body="&lt;p&gt;Overall bond order equals number of electrons that occupy bonding orbitals minus numer of electrons that occupy antibonding orbitals divided by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can apply the same concept indivudually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For s molecular orbitals, both electron pairs &quot;cancel out&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For p molecular orbitals, you have 2 electrons in sigma bonding orbitals and none in sigma antibinding orbitals, which means (2-0)/2 = 1 sigma bond.&#xA;Also, you have 4 electrons within pi bonding orbitals and only 2 electrons within pi antibonding orbitals, which means (4-2)/2 = 1 pi bond.&#xA;This yields the result that you expected.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-18T12:31:48.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19680" PostTypeId="2" ParentId="19262" CreationDate="2014-11-18T12:56:37.200" Score="3" Body="&lt;p&gt;The answer is you cannot do it almost for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain $\ce{Ca}$ metal from $\ce{Ca (+II)}$, you need to perform a reduction process. For this purpose, you would need a &lt;strong&gt;very&lt;/strong&gt; strong reducing agent, because calcium is very close to the bottom in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ellingham_diagram#Thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ellingham diagram&lt;/a&gt; (used in metallurgy for this same purpose: obtain metals from its salts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will need to perform an electrolysis of molten calcium chloride, which is not cheap, and by any reason not doable at home or at an standard lab (just a quick view of calcium salts melting point will give you a hint about why is this...)&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5026" LastEditDate="2018-04-12T04:12:39.853" LastActivityDate="2018-04-12T04:12:39.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19683" PostTypeId="2" ParentId="19676" CreationDate="2014-11-18T13:47:42.353" Score="3" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH_3COO- + H_2O&amp;lt;=&amp;gt;CH3COOH + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Approximating that all the &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is from this equilibrium, &lt;span class=&quot;math-container&quot;&gt;$[\ce{CH3COOH}] = [\ce{OH-}]$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  [\ce{OH-}]_\text{initial} &amp;amp;= [\ce{CH3COOH}]_\text{initial} \\&#xA;    &amp;amp;= \pu{10^{-4}M}\\&#xA;  \frac{[\ce{OH-}]_\text{initial}[\ce{CH3COOH}]_\text{initial}}{[\ce{CH3COO-}]_\text{initial}[\ce{H2O}]_\text{initial}} &#xA;  &amp;amp;= \frac{[\ce{OH-}]_\text{final}[\ce{CH3COOH}]_\text{final}}{[\ce{CH3COO-}]_\text{final}[\ce{H2O}]_\text{final}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then you can choose to approximate &lt;span class=&quot;math-container&quot;&gt;$[\ce{H2O}]$&lt;/span&gt; as a constant value and omit it from both sides (or not if you need great accuracy).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  [\ce{CH3COO-}]_\text{initial} &amp;amp;= \pu{1 M}\\&#xA;  [\ce{CH3COO-}]_\text{final} &amp;amp;= \pu{\frac{10}{100} M}\\&#xA;  x &amp;amp;= [\ce{OH-}]_\text{final} = [\ce{CH3COOH}]_\text{final}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solve for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2020-09-02T14:06:12.313" LastActivityDate="2020-09-02T14:06:12.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19686" PostTypeId="2" ParentId="19631" CreationDate="2014-11-18T15:11:53.873" Score="5" Body="&lt;p&gt;Ionic forms of some organic compounds that you will perfectly know, which can also be considered salts, have sweet taste, like sodium saccharin or sodium cyclamate, which are used as &quot;artificial&quot; sweeteners. The latter is considered to be about 50 times sweeter than sugar itself.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-18T15:11:53.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19687" PostTypeId="1" CreationDate="2014-11-18T15:29:47.783" Score="18" ViewCount="1869" Body="&lt;p&gt;Engines burning aluminium and liquid oxygen have been investigated for use on the moon because there is plenty of both elements in the soil - about 13% and 40% of highland soil, respectively. Propellant thus has the potential to be sourced from the moon directly, but it isn't practical unless a refining process can be found that uses the moon's environment to extract pure aluminum efficiently using local resources. A viable way to extract the oxygen is known already, using a &lt;a href=&quot;http://www.uapress.arizona.edu/onlinebks/ResourcesNearEarthSpace/resources08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyrolysis process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of taking advantage of the moon's environment to do this effectively? You have at your disposal:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A hard vacuum&lt;/li&gt;&#xA;&lt;li&gt;A month-long day that is strong sunlight for two weeks then a two week night - it is estimated that a furnace using concentrated sunlight could reach 3000 K.&lt;/li&gt;&#xA;&lt;li&gt;1/6 Earth gravity&lt;/li&gt;&#xA;&lt;li&gt;No moisture - chemicals are reduced&lt;/li&gt;&#xA;&lt;li&gt;Powdery soil that is loose to a depth of 10 or 15 cm &lt;/li&gt;&#xA;&lt;li&gt;Some soil samples from Apollo 16 were a quarter alumina by weight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A source of fuel in space is a prerequisite for a larger ongoing human presence there. A rocket launched from Earth can't manage to be more than about 5% payload. Shipping fuel from Earth to any off-world base by rocket would be prohibitively expensive (even if the rocket is reusable). This is why there has been so much interest in sources of water ice in space - it can be electrolysed into liquid hydrogen and liquid oxygen, which makes excellent propellant. But you have to go to the asteroid belt to get it (there are indications there might be significant reserves in craters at the moon's poles in permanent shade, but this is inconclusive). Once there is a way to refuel in space, all sorts of things become possible that are beyond our reach right now. An option that can compete with the price of mining asteroids has business potential. All such undertakings are going to require vast initial outlays that won't pay off for a decade or more. The process being considered here only has to meet that standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is a partner to a question at SE's Space Exploration site - &lt;a href=&quot;https://space.stackexchange.com/questions/5939/how-feasible-is-it-to-use-aluminum-and-liquid-oxygen-as-future-propellant-source&quot;&gt;How feasible is it to use aluminum and liquid oxygen as future propellant sourced from the moon?&lt;/a&gt;. I realized it was largely a chemistry question and so decided to ask about it here. There are two references in the other question: one from &lt;a href=&quot;http://www.wickmanspacecraft.com/lsp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wickman Spacecraft &amp;amp; Propellant&lt;/a&gt;, and one from the &lt;a href=&quot;http://www.asi.org/adb/06/09/03/02/095/al-o-propellants.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Artemis Data Book&lt;/a&gt;. I'm afraid neither discusses the chemistry and I haven't found anything else. The &lt;a href=&quot;http://arc.aiaa.org/action/doSearch?AllField=in%20situ%20aluminum%20propellant&quot; rel=&quot;nofollow noreferrer&quot;&gt;AIAA&lt;/a&gt; has several papers on it in their catalogue.&lt;/p&gt;&#xA;" OwnerUserId="9853" LastEditorUserId="23561" LastEditDate="2019-04-15T01:07:34.713" LastActivityDate="2019-04-15T01:07:34.713" Title="What possibilities are there for refining aluminium on the moon?" Tags="&lt;metallurgy&gt;&lt;extraction&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19688" PostTypeId="1" CreationDate="2014-11-18T15:38:23.060" Score="1" ViewCount="172" Body="&lt;p&gt;I was wondering how to write down the deterministic rate equation for a bimolecular reaction with similar particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{A -&amp;gt;[k_+] 2B}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2B -&amp;gt;[k_-] A}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the rate equations for the above reactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{\frac{\delta A}{\delta t}= - k_{+} A + k_{-} B^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathrm{\frac{\delta B}{\delta t}= 2 k_{+} A - 2 k_{-} B^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have included a factor of 2 in the second equation to conserve $\mathrm{A + \frac{B}{2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand the reasoning for this factor of 2 in the second equation.&lt;/p&gt;&#xA;" OwnerUserId="9876" LastEditorUserId="194" LastEditDate="2017-04-01T03:02:49.477" LastActivityDate="2017-04-01T03:02:49.477" Title="How do I write the deterministic rate equation for bimolecular reaction with similar particles?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19689" PostTypeId="2" ParentId="16260" CreationDate="2014-11-18T15:52:56.717" Score="7" Body="&lt;p&gt;When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  T_2=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_{p}}\right]T_1&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I will come to the derivation:&#xA;You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \delta Q = 0 \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that from the first law of thermodynamics you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d} U = \delta Q + \delta W = \delta W &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $U$ is the internal energy and $W$ is the work. &#xA;Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is $p_2$ in your equation, then this work is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \delta W = - p_{2} \mathrm{d} V \qquad \text{where:} \qquad p_{2} = \text{const.} \neq f(V)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, together with the first law of thermodynamics this yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d} U = - p_{2} \mathrm{d} V \ .  \tag{1}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  \mathrm{d} U &amp;amp;= \underbrace{\frac{\partial U}{\partial T}}_{= \, C_{V}} \mathrm{d} T + \underbrace{\frac{\partial U}{\partial V}}_{= \, \Pi \, = \, 0} \mathrm{d} V \\&#xA;  \mathrm{d} U &amp;amp;= C_{V} \mathrm{d} T \tag{2}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T$ is the temperature, $C_{V}$ is the heat capacity at constant volume, and $\Pi$ is the internal pressure, which is equal to zero for an ideal gas. &#xA;Other relations that hold for ideal gases are the ideal gas law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  p V = n R T&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the amount of the gas and $R$ is the universal gas constant, &#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  c_{p} - c_{V} = R  \tag{3}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_{p} = \frac{C_{p}}{n}$ and $c_{V} = \frac{C_{V}}{n}$ are the molar heat capacities at constant pressure and constant volume, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you just have to combine equations $(1)$ and $(2)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - p_{2} \mathrm{d} V = C_{V} \mathrm{d} T \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;integrate the result (under the assumption that $C_{V}$ is temperature independent)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - p_{2} \int^{V_2}_{V_1} \mathrm{d} V &amp;amp;= C_{V} \int^{T_2}_{T_1} \mathrm{d} T \\&#xA;  - p_{2} \left( V_2 - V_1 \right) &amp;amp;= C_{V} \left( T_2 - T_1 \right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use the ideal gas law $V = \frac{n R T}{p}$ to substitute $p$ for $V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - n R p_{2} \left( \frac{T_2}{p_2} - \frac{T_1}{p_1} \right) &amp;amp;= C_{V} \left( T_2 - T_1 \right) \\&#xA;  - n R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &amp;amp;= C_{V} \left( T_2 - T_1 \right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;introduce the molar heat capacity via $C_{V} = n c_{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  - R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &amp;amp;= c_{V} \left( T_2 - T_1 \right) \ ,&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and separate $T_1$ and $T_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &amp;amp;= T_2 \left( c_{V} + R \right) \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you can use equation $(3)$ to get the desired result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &amp;amp;= T_2 \bigl(\underbrace{c_{V} + R}_{= \, c_{p}} \bigr) \\&#xA;  T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &amp;amp;= T_2 c_{p} \\&#xA;  \Rightarrow \ T_2 &amp;amp;=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_p}\right]T_1 \ .&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="4945" LastEditDate="2014-11-20T02:28:24.113" LastActivityDate="2014-11-20T02:28:24.113" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19691" PostTypeId="2" ParentId="16093" CreationDate="2014-11-18T17:26:40.800" Score="0" Body="&lt;p&gt;I dimly remember a paper contributing to the controversy if nerve cells in the brain continue to divide in adulthood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment relied on the 14-C spike (I think it was carbon and not tritium) caused by above-ground nuclear testing.  The authors extracted DNA (which we all know is synthesized during cell division) from brains preserved before the Sixties.  It turns out that no excess 14-C was incorporated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, in its own way, radiocarbon dating can be used to gauge the age of a living being.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-18T17:26:40.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19692" PostTypeId="1" CreationDate="2014-11-18T18:14:36.973" Score="12" ViewCount="8420" Body="&lt;p&gt;The only reason I can think of is that the lone pairs in the oxygen in water are in lower energy orbital relative to the nitrogen in ammonia due to increased effective nuclear charge in oxygen. However, as far as I am aware, ammonia is SIGNIFICANTLY more basic than water. Surely the orbital energy cannot account for this.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4325" LastEditDate="2014-11-18T23:46:54.127" LastActivityDate="2018-09-19T15:13:45.903" Title="Why is ammonia so much more basic than water?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19694" PostTypeId="2" ParentId="19643" CreationDate="2014-11-18T18:56:40.427" Score="15" Body="&lt;p&gt;It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrolysis of water strongly favors &quot;H&quot; being converted to hydrogen gas rather than &quot;D&quot;, by a factor of about 8 to 1 (depending upon the electrodes).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://kb.osu.edu/dspace/handle/1811/3183&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$&lt;/a&gt; Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method.  The article says the process is suitable for &quot;those who have occasion to prepare or concentrate heavy water for research, demonstration or&#xA;advanced student laboratory use&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains.  This increased the original concentration of deuterium by a factor of 12.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level.  They used three stages.  Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7475" LastEditDate="2017-08-29T04:09:27.637" LastActivityDate="2017-08-29T04:09:27.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19696" PostTypeId="2" ParentId="19687" CreationDate="2014-11-18T19:44:11.460" Score="9" Body="&lt;p&gt;My theory was to use electromagnetic separation, aka the Calutrons used in the Manhattan Project.  Fine powder feedstock (check), into an arc discharge to vaporize and ionize (lots of good designs for high volume ion sources out there), electrostatic acceleration, then electromagnets to mass separate.  Power with a solar panel array.  No moving parts (except for the powder feed).  No vacuum systems needed.  And you separate all of the elements at once. Now, it only works during sunlight, but during the lunar night you could do maintenance and collect the separated elements.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-11-18T19:44:11.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19697" PostTypeId="1" AcceptedAnswerId="19776" CreationDate="2014-11-18T20:36:04.317" Score="1" ViewCount="105" Body="&lt;p&gt;$\mathrm{Z=33,\ Z=37,\ Z=9,\ Z=31}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we shouldn't look at the periodic table, I found their groups and periods, what should I do next?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="5026" LastEditDate="2018-03-04T02:46:27.583" LastActivityDate="2018-03-04T02:48:22.427" Title="How do I list these elements according to their radius?" Tags="&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19698" PostTypeId="1" CreationDate="2014-11-18T20:41:43.803" Score="0" ViewCount="241" Body="&lt;p&gt;$\mathrm{Z= 32}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^{10}\ 4s^2p^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me it belongs the $\mathrm{IV\ B}$ group since it has the $\mathrm{d}$ completed, but it belongs to $\mathrm{IV\ A}$. But we know that in the $\mathrm{A}$ group only the $\mathrm{s}$ and $\mathrm{p}$ elements belong. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="4325" LastEditDate="2014-11-18T23:45:06.340" LastActivityDate="2014-11-18T23:45:06.340" Title="Which group does germanium belong to?" Tags="&lt;inorganic-chemistry&gt;&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19699" PostTypeId="2" ParentId="19698" CreationDate="2014-11-18T20:56:24.557" Score="2" Body="&lt;p&gt;According to the IUPAC, the is not such thing as groups IV A and B anymore.&#xA;All the elements you are analyzing belong to the 14th group of the periodic table, each one in one period, from the 2nd to the 6th.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to your other question, you need to look for a trend on the elements of the group.&#xA;What happens as you go down the group?&#xA;I think you've already figured out that the number of electrons around the nucleus significantly increases, and this has an important effect on the radius of the atom.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-18T20:56:24.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19700" PostTypeId="1" CreationDate="2014-11-18T21:50:59.090" Score="3" ViewCount="116" Body="&lt;p&gt;Hi I am new to laser pump-probe spectroscopy and want to understand the concepts better. At the moment I am struggling to understand the following: When one takes absorption measurements with lasers in a pump-probe manner, the spectra that are collected need to be normalized. What does the Normalization mean and how does one do it? Also, why are relative changes important as opposed to absolute changes (in absorption) because I think they should give the same information?&lt;/p&gt;&#xA;" OwnerUserId="9881" LastEditorUserId="9961" LastEditDate="2015-10-23T17:24:48.447" LastActivityDate="2016-07-20T16:41:56.853" Title="Femtosecond spectroscopy" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19701" PostTypeId="2" ParentId="19692" CreationDate="2014-11-18T23:50:50.327" Score="7" Body="&lt;p&gt;When it comes to comparing ammonia and water, the acidity or basicity can be explained because of the central atoms electronegativity. When comparing atoms of the same periodic row, the more electronegative an atom is, the less it will want to donate an electron pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronegativity values from the Pauling Scale for $\ce{N}$ and $\ce{O}$ are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N}=3.04$$&#xA;$$\ce{O}=3.44$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So based off of these values we can see that nitrogen is less electronegative, it less strongly it holds onto its electrons, and thus it is a better electron donor. This allows the addition of a $\ce{H}$ to be more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic reasoning behind this is that, as a general periodic trend, as you move across the periodic table, from left to right, the nucleus is increasing, which increases it's charge, producing a stronger attraction to its valence electrons.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-11-19T00:44:05.217" LastActivityDate="2014-11-19T00:44:05.217" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19702" PostTypeId="1" AcceptedAnswerId="19704" CreationDate="2014-11-19T00:25:02.893" Score="13" ViewCount="80994" Body="&lt;p&gt;I learned these two words in school but still I don't understand what the difference is between them. From my understanding, molarity and concentration means the number of moles in a certain solution. So, what is the difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorDisplayName="user7951" LastEditDate="2015-08-11T13:17:48.110" LastActivityDate="2019-03-14T02:39:53.667" Title="What is the difference between molarity and concentration?" Tags="&lt;concentration&gt;&lt;terminology&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19703" PostTypeId="2" ParentId="19702" CreationDate="2014-11-19T00:37:56.913" Score="4" Body="&lt;p&gt;Concentration is the ratio of the amount of solute per amount of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molarity is a unit of concentration that specifically relates the number of moles of a solute per liter of solution.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-19T00:37:56.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19704" PostTypeId="2" ParentId="19702" CreationDate="2014-11-19T00:42:09.253" Score="9" Body="&lt;p&gt;Concentration is any measurement of the quantity of a solute that is present per unit of solution (in general, there are exceptions like moLality, which measures the number of moles of solute per kilogram of solvent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molarity is just one way to express concentrations, which, more specifically, is the number of moles of solute per liter of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can give values for concentrations in many different ways, like for example, grams of solute per liter of solution, percentages (grams of solute divided by the grams of solution multiplied by 100), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a wide range of ways of measuring concentrations, including molarity &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Concentration&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T00:42:09.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19705" PostTypeId="2" ParentId="19697" CreationDate="2014-11-19T01:10:50.953" Score="0" Body="&lt;p&gt;I try this question before. First you find out the electron configuration and the group and period for each substances&#xA;(I will name each substances for clearing confusion A=33, B=37, C=9, D=31)&#xA;Electron arrangement&#xA;A= 2.8.8.15 (Group 4, Period 15)&#xA;B= 2.8.8.18.1 (Group 5, Period 1)&#xA;C= 2.7 (Group 2, Period 17)&#xA;D= 2.8.8.13 (Group 4 ,Period 13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have two ways, either you can sketch the table out since the numbers are small or you can do it my way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the radius of a substance increase from right to left and from up to down. Meaning the smaller the group number and the period number the smaller the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find the substances in an ascending sequences according to their radius. First, we will look at their group number. C has the lowest group number so it has the smallest radius, then we can see that A and D have the same number of group, thus we now check on the period number. We can see that D is smaller than A. Meaning, D has a bigger radius than A . Lastly check back the group number and we have B with the highest group number so its radius is the biggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can get the sequences which is C→A→D→B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the desending sequences , do it backwards. Hope that it will help&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2014-11-20T06:37:49.693" LastActivityDate="2014-11-20T06:37:49.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19706" PostTypeId="2" ParentId="19668" CreationDate="2014-11-19T01:43:09.607" Score="1" Body="&lt;p&gt;Stoichiometric, theoretical or complete combustion, is the idealized process where a hydrocarbon is completely reduced to carbon dioxide and water, consuming oxygen (or, if other elements are present, they get all also oxidated, like sulphur compounds yielding sulphur dioxide, but as you can see, not to its maximum oxidation state, sulphur trioxide is not produced in real combustions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is called stoichiometric for obvious reasons, stoichiometric amounts of the hydrocarbon and oxygen / water carbon dioxide are consumed / generated. It is supposed to be a complete reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is also called theoretical, because it does not take place like that in real life. In practice, long hydrocarbon chains do not break down completely as efficiently as it's expected, subproducts are obtained, not only carbon dioxide is formed, but also carbon monoxide, unburned elemental carbon and like in every chemical reaction, it does not proceed to completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there is not enough oxygen (not at least an stoichiometric amount), incomplete combustion takes place, yielding almost exclusively carbon monoxide (toxic stuff).&#xA;In some cases, such as the fuels combustion processes, another process called pyrolysis, occurs before combustion. In this incomplete combustion, products of pyrolysis remain unburnt and important contaminants are produced, like the product of partial oxidation of ethanol can produce acetaldehyde, which is really toxic (Considerably small LD50) and very very flammable (flash point under 0 Celsius degrees).&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T01:43:09.607" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19707" PostTypeId="1" CreationDate="2014-11-19T01:46:06.753" Score="6" ViewCount="3577" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fuGeT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lh8bK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to reduce an aryl nitro group.&#xA;I understand that these reactions are radical reactions by the metal, but I don't understand&#xA;why alcohols are used as the solvent. Are they used as a source of protons, or just because they are good solvents?&lt;/p&gt;&#xA;" OwnerUserId="9874" LastEditorUserId="10185" LastEditDate="2015-06-20T09:54:31.813" LastActivityDate="2015-06-20T09:54:31.813" Title="Role of Alcohols(EtOH,MeOH) in nitro reduction used for Metal(Sn,Fe)" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;solvents&gt;&lt;nitro-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19708" PostTypeId="2" ParentId="19707" CreationDate="2014-11-19T02:01:59.167" Score="2" Body="&lt;p&gt;I do not believe that ethanol is mandatory. You surely need a proton source, but that's why those reactions are performed in protic acidic media, like your first example, with HCl, or what I think were the original conditions of Bechamp reduction, using acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably ethanol makes a good solvent for phenol rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;a href=&quot;http://gallica.bnf.fr/ark:/12148/bpt6k347830/f188.image&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; is the Bechamp original paper. I couldn't manage to find it in English. I'm not French speaker either, but just in case you are curious :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I'm concerned, nowadays catalytic hydrogenation is used to perform this reaction (at least at industrial scale)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the mechanism, I'm sorry but I can't give you a solid statement on this.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T02:01:59.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19709" PostTypeId="2" ParentId="19707" CreationDate="2014-11-19T02:03:59.330" Score="6" Body="&lt;p&gt;LHM...Look at this mechanism that I posted as an answer to an earlier SE Chem &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15648/nitrobenzene-reduction-with-tin-sn-catalyst-why-is-c-h-bond-cleavage-preferre&quot;&gt;question/answer&lt;/a&gt; relating to the reduction of nitrobenzene by metals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8kK8F.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most reactions that occur on metal surfaces, many of the mechanistic details are not fully understood (see @Martin's comments in the earlier post).  Nonetheless, this mechanistic scheme shows several steps where a proton ($\ce{H+}$) is required.  These proton transfers are necessary to complete the reduction process.  This is why these reactions are typically run in solvents, like ethanol, that have available protons that can be donated in order to complete the reaction...ron&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-11-19T02:03:59.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19710" PostTypeId="1" AcceptedAnswerId="19765" CreationDate="2014-11-19T04:06:28.277" Score="3" ViewCount="4027" Body="&lt;p&gt;Assume I have a salt solution with the same concentration as the dead sea and I put a iron nail into the solution (assume it sinks). Will it rust? If yes, why? If no, why not?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="5026" LastEditDate="2018-04-15T02:03:10.413" LastActivityDate="2018-04-15T02:03:39.030" Title="Will an iron nail sunk in a highly concentrated salt solution rust?" Tags="&lt;metal&gt;&lt;corrosion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19711" PostTypeId="1" CreationDate="2014-11-19T05:27:10.243" Score="6" ViewCount="3381" Body="&lt;p&gt;I am putting 100 g of boric acid, 400 g of water and trying to crystallize at low temperature and letting cool slowly, however it recrystallizes as needle like texture. I need soft large crystals like mica. Can someone help out? &lt;/p&gt;&#xA;" OwnerUserId="9744" LastEditorUserId="4325" LastEditDate="2015-07-01T11:08:22.880" LastActivityDate="2018-01-11T17:46:06.967" Title="How to recrystallize boric acid into flakes?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19712" PostTypeId="1" CreationDate="2014-11-19T00:59:12.113" Score="1" ViewCount="136" Body="&lt;p&gt;Why does adding swimming pool salt to a freshwater swimming pool cause a white sediment to build up and stick to the floor of the pool?&lt;/p&gt;&#xA;" OwnerDisplayName="malclose" LastEditorUserId="4945" LastEditDate="2014-11-19T08:12:35.440" LastActivityDate="2014-11-19T08:12:35.440" Title="What happens when adding swimming pool salt to a freshwater pool?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19713" PostTypeId="1" AcceptedAnswerId="19714" CreationDate="2014-11-19T08:20:29.180" Score="1" ViewCount="2135" Body="&lt;p&gt;Recently, I am interested with home materials and I discover that the chemical symbol for MSG or monosodium glutamate is $\ce{C5H8NO4Na}$.&#xA;My problem is shouldn't it be $\ce{NaC5H8NO4}$ as we always write the metal first before the non-metal?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="4945" LastEditDate="2014-11-19T08:32:06.693" LastActivityDate="2014-11-19T09:08:12.323" Title="What is the correct chemical formula for monosodium glutamate?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19714" PostTypeId="2" ParentId="19713" CreationDate="2014-11-19T09:08:12.323" Score="4" Body="&lt;p&gt;It should, but in organic molecules salts, often the counterion is written latter, to indicate its position on the molecule acting like so, for example, in sodium acetate, $\ce{CH3CO2Na}$, this would be something like a &quot;semicondensated&quot; formula.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T09:08:12.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19715" PostTypeId="1" CreationDate="2014-11-19T10:44:17.180" Score="7" ViewCount="19683" Body="&lt;p&gt;I know that it has something to do with HOMOs and LUMOs (I can't what the HOMO/LUMO for AlCl3 is, though). &lt;/p&gt;&#xA;" OwnerUserId="9709" LastActivityDate="2018-12-30T13:58:28.063" Title="Why does AlCl3 dimerise to Al2Cl6 at low temperatures?" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19716" PostTypeId="1" CreationDate="2014-11-19T11:42:54.607" Score="3" ViewCount="228" Body="&lt;p&gt;In the Brønsted-Lowry theory, strong acids $\ce{HA}$ dissociate using water and react into hydronium ions and the conjugate bases $\ce{A-}$. However I never understood in the strong acids case what stops the products of the reaction(ions) into reforming chemical bonds (why the reaction is one-way) and what enables them when the reagents aren't strong acids. How is this explained macrocosmically and microscopically?&lt;/p&gt;&#xA;" OwnerUserId="9867" LastEditorUserId="51553" LastEditDate="2018-04-05T11:30:50.307" LastActivityDate="2018-04-05T13:40:26.467" Title="Why do strong acids dissociate (almost) fully on a molecular level?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19718" PostTypeId="1" CreationDate="2014-11-19T12:57:55.117" Score="0" ViewCount="708" Body="&lt;p&gt;What is the mass of 150 ml of 40% sucrose, and what is the mass of 150 ml of 20% sucrose? I don't know exactly how to figure it out without using a device to tell me the mass. All I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams.&lt;/p&gt;&#xA;" OwnerUserId="9891" LastEditorUserId="5160" LastEditDate="2014-11-19T13:11:47.217" LastActivityDate="2014-11-19T13:21:33.133" Title="Mass of sucrose solutions" Tags="&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19719" PostTypeId="1" CreationDate="2014-11-19T13:01:31.917" Score="3" ViewCount="1260" Body="&lt;p&gt;I understand that hybridisation occurs due to the atom wanting to exist in the most stable state possible, and the most stable state has the lowest energy. What I don't understand, however, is say, in CH4, the bond angle is 109°18', so how does this particular value mean that the atoms and bonds in methane are farthest from each other? And if they are indeed farthest from each other, how does that mean the CH4 molecule's in the lowest energy state?&lt;/p&gt;&#xA;" OwnerUserId="9892" LastActivityDate="2014-11-19T15:20:33.183" Title="What is lowest energy state, how does an atom gain it and what does it have to do with hybridization?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;stability&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19720" PostTypeId="2" ParentId="19718" CreationDate="2014-11-19T13:11:02.010" Score="2" Body="&lt;p&gt;If 10ml of 40% sucrose is 15.7 grams, then the mass of 150 ml of 40% sucrose is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 15 \times 15.7 grams$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can not determine the mass of 150 ml of 20% sucrose from the information given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could calculate from this sucrose solution density table:&#xA;&lt;a href=&quot;http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html&quot; rel=&quot;nofollow&quot;&gt;http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the table shows the premise &quot;I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams&quot; to be incorrect.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-11-19T13:21:33.133" LastActivityDate="2014-11-19T13:21:33.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19721" PostTypeId="2" ParentId="19716" CreationDate="2014-11-19T13:38:52.010" Score="1" Body="&lt;p&gt;Deprotonation is always a reversible reaction. Hence why, even strong acids, have $\mathrm{p}K_\mathrm{a}$ values (which rely on the presence of equilibria).&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5026" LastEditDate="2018-04-05T13:40:26.467" LastActivityDate="2018-04-05T13:40:26.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19722" PostTypeId="1" CreationDate="2014-11-19T13:52:21.563" Score="10" ViewCount="1063" Body="&lt;p&gt;I am not a Chemist, but I took enough undergraduate Chemistry classes to understand the basic properties of s, p and d orbitals and how the behaviour of electrons contributes to different kinds of chemical bonds and elemental/molecular properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stopped before we got to actual quantum mechanics - my maths wouldn't have been good enough to stand it. It still isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was always fascinated as to why none of the lecturers would ever tell us about f-orbitals. When asked, they muttered dark hints that they were significantly different from lower electron shells, and that explaining why was not only a waste of time but might spoil our existing understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I left university 20 years ago, but I was suddenly blindsided by remembering this question out of nowhere. And now I want to know the answer. Google searches reveal a lot of maths that I can't follow. So can someone explain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are some of the basic properties of elements with electrons in f-orbitals, and how do they differ from lower orbital positions?&lt;/li&gt;&#xA;&lt;li&gt;I assume this is related to the fact that most (all?) elements with f-orbitals manifest as metals, so how and why? And why do some elements without f-orbitals display metallic properties?&lt;/li&gt;&#xA;&lt;li&gt;Why were my tutors so unwilling to try and explain this to non-Chemists who were supposed to stop at their appreciation of lower orbitals and basic covalent bonds?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7928" LastEditorUserId="7928" LastEditDate="2014-11-19T14:07:57.870" LastActivityDate="2014-11-19T22:00:44.790" Title="Properties of f-orbitals" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19723" PostTypeId="2" ParentId="5130" CreationDate="2014-11-19T14:03:41.160" Score="0" Body="&lt;p&gt;Both composed substances and mixtures have more than type of atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are some difference between them both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)composed substances=mix chemically, mixture=mix physically&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composed substances are usually mix together through chemical process just like burning, titration and redox reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While mixture are mix together through physical process like dissolving and even putting together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Composed substances and mixture properties&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composed substances have different properties with it original atoms.&#xA;i.e iron sulphide do not attarct to magnet although iron is magnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While mixture have the same properties with its original atoms.&#xA;i.e. a pile of iron filling and sulphur powder. The iron will still attarct to magnet and the sulphur don't&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)Bond between atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In composed substances, the atoms are bonded with either ionic bond or covelent bond. i.e the hydrogen and oxygen atoms in a water molecule have colevent bonds between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mixture, the atoms do not bond together. i.e oxygen gas, nitrogen gas, carbon dioxide gas in the air do not have bond between them as they flows freely in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are also a lot more differences between these both such as melting point, existing states of matter, conductivity of electricity and heat etc. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2014-11-19T14:03:41.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19724" PostTypeId="2" ParentId="19715" CreationDate="2014-11-19T14:58:53.820" Score="4" Body="&lt;p&gt;There are certain stable compounds where the octet rule is not obeyed, that is, their valence&#xA;shells contain more or less than 8 electrons. The compounds containing less than 8 electrons in the valence shells are called &lt;strong&gt;electron deficient compounds&lt;/strong&gt; such as $\ce{AlCl3}$, where the aluminum atom has 6 valence electrons. It has been suggested that two aluminum atoms complete their octet by accepting a pair of electrons from chlorine atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cy0x0.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that is why $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ at low temperatures.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7448" LastEditDate="2016-01-23T19:22:07.853" LastActivityDate="2016-01-23T19:22:07.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19725" PostTypeId="2" ParentId="19719" CreationDate="2014-11-19T15:20:33.183" Score="2" Body="&lt;p&gt;VSEPR stands for Valence Shell Electron Pair Repulsion.  It's a complicated acronym, but it means something that's not difficult to understand. Basically, the idea is that covalent bonds and lone pair electrons like to stay as far apart from each other as possible under all conditions (to minimize electrostatic repulsions of different pairs of electrons) .  This is because covalent bonds consist of electrons, and electrons don't like to hang around next to each other much because they have the same charge. This, of course, allows the molecule to be in its lowest energy state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This VSEPR thing explains why molecules have their shapes.  If carbon has four atoms stuck to it (as in methane), these four atoms want to get as far away from each other as they can.  This isn't because the atoms necessarily hate each other, it's because the electrons in the bonds hate each other.  That's the idea behind VSEPR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one problem with the whole VSEPR thing is that if you have four things stuck to carbon, for example, there are no orbitals that want to get 109.5 degrees apart from each other (109.5 degrees corresponds to the geometric maximum distance the atoms can get apart).  After all, s-orbitals go in a complete sphere (360 degrees) and p-orbitals are 90 degrees apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens instead of using s- or p- orbitals is that when covalent bonds are formed, the s- and p- orbitals mix to form something called hybrid orbitals.  &quot;Hybrid&quot; just means &quot;mixture of two different things&quot;, and that's exactly what a hybrid orbital is.  When three p-orbitals with 90 degree angles combine with one s-orbital with 360 degrees, they average to form four sp3 orbitals with 109.5 degree bond angles.  Depending on the numbers of s- and p-orbitals that mix, you can get a bunch of different bond angles.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-19T15:20:33.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19726" PostTypeId="2" ParentId="19688" CreationDate="2014-11-19T16:13:21.633" Score="1" Body="&lt;p&gt;There is really only one independent equation here because  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dB}{dt} = -2\frac{dA}{dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;second equation&quot; in the OP is just -2 times the first equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every molecule of A destroyed creates 2 molecules of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every 2 molecules of B destroyed creates 1 molecule of A.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-19T16:13:21.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19727" PostTypeId="2" ParentId="17008" CreationDate="2014-11-19T16:29:47.220" Score="5" Body="&lt;p&gt;After consulting with a professor, the answer seems obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two negative charges repel each other when the free cyclodextrin molecule is in solution, resulting in an &quot;open&quot; conformation that has the carboxylate groups facing &quot;out&quot;. The thioether chains interact with the rocuronium as it is drawn into the large open face of the molecule. As the positively charged nitrogen moves into place, the carboxylate groups &quot;close&quot;, and lock the drug into place. &lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2014-11-19T16:29:47.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19729" PostTypeId="1" AcceptedAnswerId="19732" CreationDate="2014-11-19T16:50:31.133" Score="3" ViewCount="152" Body="&lt;p&gt;I watched &lt;a href=&quot;http://youtu.be/6ZBwluyR2Tc&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; where a lightbulb is suspended in a solution of HF.  The result is that the HF dissolves the glass, but it seems like the part of glass on the surface reacted faster than fully submerged parts.  Here is an image of the lightbulb after the experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iidvv.png&quot; alt=&quot;After Experiment Lightbulb&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass breaks at approximately the 2:50 mark, and it looks like at that point the bottom portion was cut from a clean line.  Perhaps the larger portion dissolved afterwards or was damaged during cleanup, I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reason why the glass closer to the surface would react faster than more submerged parts?  Why would the very tip not dissolve first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that the specific line happened to be the thinnest glass.  I find it unlikely that the specific line is the thinnest part.  Wouldn't it be much more likely for there to be random holes in the submerged portion from the thinnest part of the glass?&lt;/p&gt;&#xA;" OwnerUserId="9897" LastEditorUserId="485" LastEditDate="2014-11-19T17:55:33.013" LastActivityDate="2014-11-19T18:00:44.117" Title="Why did HF disolve this lightbulb at the surface rather than more submerged points?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19731" PostTypeId="1" CreationDate="2014-11-19T17:54:42.207" Score="2" ViewCount="1338" Body="&lt;p&gt;I've asked a similar question before, but now I want to find out the real rule. Pleas tell me if my interpretation is correct: Cd, Ag , Au (increasing ray order). This happens because although we know that period and group effect the radius length, most important one is the group. So firstly we should list the elements according to their group and then list them in the same group... For example Zn, Cd (same group), then Pd, Pt, (same group), then Ph, ir (same group). From our lesson we know that from Cd to Hg the ray doesn't change but I think when we have to list them we should consider that it grows a little. Are these statements right? If not, please help me clear the concept. I'm not a native speaker of english, so I may not express correctly in some occasions&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="9878" LastEditDate="2014-11-19T18:00:59.363" LastActivityDate="2017-02-21T22:38:12.470" Title="How to list elements according to their radius? For example Ag, Cd, Au Can you check the accuracy of these sentences I've created?" Tags="&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19732" PostTypeId="2" ParentId="19729" CreationDate="2014-11-19T18:00:44.117" Score="6" Body="&lt;p&gt;Near surface of the liquid the glass is a bit hotter. All glass of the bulb gains comparable amount of heat energy from tungsten wire, but parts above the liquid are cooled only by air, so they are much hotter. Part of this heat is transferred into slightly submerged part, so it is the hottest part that is under liquid, and all reactions are quickened by increase of temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this exact experiment is a bit ambiguous, as glass here is under extreme thermal stress, and shattering near surface of the liquid may be addressed to that.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-19T18:00:44.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19734" PostTypeId="1" AcceptedAnswerId="19741" CreationDate="2014-11-19T18:33:32.993" Score="2" ViewCount="9714" Body="&lt;p&gt;I read something recently that said &quot;BN has been observed in the gas phase, is paramagnetic, and has a vibrational frequency lower than N2. These properties can be explained by the molecular orbital diagram of BN&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I drew such a diagram, I failed to see how it described any of these properties, particularly paramagnetism, because there were no lone pairs of electrons. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="9709" LastActivityDate="2014-11-20T04:55:46.687" Title="How can BN be paramagnetic?" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19735" PostTypeId="1" CreationDate="2014-11-19T18:54:31.553" Score="9" ViewCount="13524" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\delta S\ge\frac{\delta q}T$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it ($\mathrm{\frac{\delta q}{T}}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what is the significance of this inequality?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5531" LastEditDate="2014-11-20T14:01:47.293" LastActivityDate="2017-09-14T07:01:29.147" Title="What does the Clausius Inequality mean in thermodynamics? (entropy)" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19736" PostTypeId="1" CreationDate="2014-11-19T19:20:21.070" Score="0" ViewCount="689" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uuWAP.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;and I understand it belongs to the 13 th group, but my teacher wants to express it as IIIA, so how should I convert it in my mind?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastActivityDate="2014-11-19T19:50:33.573" Title="Show which group the element with Z=31 belongs" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19738" PostTypeId="2" ParentId="19731" CreationDate="2014-11-19T19:47:37.340" Score="1" Body="&lt;p&gt;This is the radius general trend on the periodic table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHJ7C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason that explains how it changes is the nuclear effective charge experimented by the electrons when atracted by the nucleos. Briefly speaking, the more electrons you add as you go down a group (in different energy levels), the lower the nuclear effective charge that the electrons will experiment, so the electron denisity will be less &quot;compresed&quot;, resulting in a larger radius.&#xA;Going from the right to the left, you add one by one, a new electron that experiments almost the same nuclear effective charge, so it will be atracted with a similar force, and also, you are adding another proton to the nucleus, so the overall atraction will be higher, the electron density will decrease and the atomic radius will decrease.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T19:47:37.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19739" PostTypeId="2" ParentId="19736" CreationDate="2014-11-19T19:50:33.573" Score="4" Body="&lt;p&gt;This is the old periodic table in which are still used these groups. You should avoid its use, but if your teacher needs you to learn them anyway, here they are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YUOrR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T19:50:33.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19740" PostTypeId="2" ParentId="19593" CreationDate="2014-11-19T19:50:38.300" Score="5" Body="&lt;p&gt;The white sugar you refer to as &quot;normal&quot; sugar is mostly pure sucrose, which forms translucent whitish crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White sugar doesn't start out white --- it's actually refined from a plant extract (primarily from sugarcane) that is a mixture of sucrose, glucose, fructose, and some minerals including calcium, iron, and magnesium. The unrefined extract is &quot;raw&quot; sugar, and these minerals give it a brown color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown sugars are partially-refined sugars which still have some of the other components besides sucrose. When it's fully refined, the appearance is white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we refine the raw sugar, the remainder is referred to as molasses -- a dense, dark brown syrupy mixture of sucrose, glucose, fructose, and some minerals which give it the signature dark brown color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red sugarcane doesn't produce red sugar as @LDC3 stated. &quot;Red sugar&quot; is actually refined sugar with a red pigment added to it... unless you're talking about what the Chinese refer to as &quot;red sugar&quot; -- which is just their name for brown sugar!&lt;/p&gt;&#xA;" OwnerUserId="60" LastEditorUserId="60" LastEditDate="2014-11-19T20:14:36.547" LastActivityDate="2014-11-19T20:14:36.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19741" PostTypeId="2" ParentId="19734" CreationDate="2014-11-19T19:57:06.040" Score="5" Body="&lt;p&gt;This is the general MO diagram you need to fill with the valence electrons of BN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VYw1a.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boron has 3 valence electrons, and nitrogen has 5 valence electrons, this makes 8 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to start filling the orbitals from those with lowest energy to those with higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 2 electrons on $\sigma_\mathrm{2s}$, two electrons on $\sigma_\mathrm{2s}^*$, two electrons on $\sigma_\mathrm{2p}$. You have now 2 electrons left, and two orbitals of the same energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you need to follow Hund's rule, which states that  if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs. So you will have to put one electron on the $\pi_x$ orbital and another one on the $\pi_y$ orbital.&#xA;So you end up with 2 umpaired electrons, and paramagnetism of the molecule is explained.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4945" LastEditDate="2014-11-20T04:55:46.687" LastActivityDate="2014-11-20T04:55:46.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19742" PostTypeId="2" ParentId="19735" CreationDate="2014-11-19T20:45:01.920" Score="8" Body="&lt;p&gt;Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $$\mathrm{\delta w \geq \delta w_{rev} \Rightarrow \delta w-\delta w_{rev}\geq 0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and according to the First Law of Thermodynamics &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\delta U = \delta q + \delta w=\delta q_{rev}+\delta w_{rev}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $$\mathrm{\delta q_{rev}-\delta q=\delta w-\delta w_{rev} \geq 0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using the definition of entropy yields &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\delta S=\cfrac{\delta q_{rev}}{T} \geq \cfrac{\delta q}{T}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $\mathrm{\delta q=0}$ (since we have an isolated system no heat can flow into it), produces &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\cfrac{\delta q_{rev}} {T} \geq 0 \Rightarrow \delta S \geq 0}$$&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-20T11:28:26.977" LastActivityDate="2014-11-20T11:28:26.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19743" PostTypeId="2" ParentId="19722" CreationDate="2014-11-19T22:00:44.790" Score="4" Body="&lt;p&gt;I think there are multiple reasons contributing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the atomic orbitals of the f-block and their properties and interactions are complicated. You have a lot of electrons, and they are far away from the nucleus, on many differently oriented orbital lobes. (well, that is the visualisation at least, in reality atomic orbitals are described by quantum mechanical wavefunctions, and those get more complicated as well for the f orbitals)&lt;br&gt;&#xA;Since you have a lot of electrons repelling each other, and on top of that there are relativistic effects coming into play, it is rather difficult to do quantum chemical calculations with these elements to determine their energy levels, etc.&#xA;A lot of &quot;rules&quot; (and approximations) valid for low atomic number elements stop working.&lt;br&gt;&#xA;Overall, heavy elements have special properties, that cannot be explained by the simple rules and approximations used for light elements. As you have said there is a lot of difficult math in the background. The rules and visualisations are just there to be an  easy to grasp abstraction of the underlying physics. The side effect of this is that you need more and more exceptions as the system gets more complicated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, they are usually rare and expensive elements.(actinides are also radioactive) They have no known biological role, and usually only have few significant uses. So usually one does not encounter them unless one is working in some niche field.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2014-11-19T22:00:44.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19744" PostTypeId="1" CreationDate="2014-11-19T23:03:26.930" Score="4" ViewCount="4350" Body="&lt;p&gt;I am working on a project in my lab, and we are analyzing capsaicin using HPLC. I am trying to find an appropriate wavelength for the HPLC using a UV-Vis spectrometer. When analyzing the pure capsaicin standard dissolved in methanol, I of course use methanol as a blank. However, I got two peaks, one around 230nm and one at 280nm. Since the system automatically subtracts the methanol, I can't understand why I am getting two peaks. &#xA;Anyway, 230 yielded the maximum absorbance, so I used that in my HPLC. I didn't get great results and decided to read some literature on the project. I saw that most other people had a UV-Vis spectrum like mine, but they decided to use 280nm. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="9902" LastEditorUserId="418" LastEditDate="2014-11-20T06:02:08.837" LastActivityDate="2014-11-20T06:02:54.037" Title="UV-Vis Spectra: Multiple peaks for one component" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19745" PostTypeId="2" ParentId="19744" CreationDate="2014-11-19T23:52:15.367" Score="2" Body="&lt;p&gt;Many organic compounds give more than one maximum peak when its UV-Vis spectra is analyzed. Each peak correspond to a electron transition from a ground state to an excited state, and more than one different transitions (with different energy, and therefore, different wavelenght) are allowed.&#xA;To determine the concentration of a compound, in principle, any fairly strong peak would be valid, but depending on the experimental conditions, sometimes one gives a stronger signal, and other times that one gives a worst signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak employed on your determinations can also be selected based on your instrument conditions, i.e., if one of the peaks is close to your measurement limits, you may prefer to use another, or if one of them has a better shape, etc.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-19T23:52:15.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19747" PostTypeId="1" AcceptedAnswerId="19798" CreationDate="2014-11-20T00:17:59.740" Score="3" ViewCount="981" Body="&lt;p&gt;I have good background knowledge on Quantum Chemistry calculations but my knowledge on Solvation Models is not that amazing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times I read a paper that uses TD-DFT, they also assign some solvent to their calculations. Some papers use Dielectric-PCM and some others use Capacitive-PCM. I know that D-PCM assumes the solvent as Dielectric media and C-PCM assumes as Capacitive Field and I have some very basic knowledge on them but what I can not understand is when we use which?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example what is the criteria for selecting DPCM or CPCM for Methanol solvent?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorDisplayName="user7951" LastEditDate="2016-12-20T19:57:19.870" LastActivityDate="2016-12-20T19:57:19.870" Title="What is the difference between Dielectric and Capacitive Polarizable Continuum Model?" Tags="&lt;quantum-chemistry&gt;&lt;solvents&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19748" PostTypeId="1" AcceptedAnswerId="19750" CreationDate="2014-11-20T00:20:38.373" Score="8" ViewCount="43822" Body="&lt;p&gt;My teacher asked me this question a few days ago. &quot;How to define hot and cold?&quot; And I said, &quot;The higher the temperature, the hotter you are. The lower the temperature the colder you are.&quot; But my teacher said, &quot;That's not 100 percent correct.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you really define hot and cold?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="189" LastEditDate="2014-11-20T02:52:24.037" LastActivityDate="2017-02-05T18:09:25.723" Title="How are hot and cold defined?" Tags="&lt;temperature&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19749" PostTypeId="1" AcceptedAnswerId="19753" CreationDate="2014-11-20T00:25:26.837" Score="3" ViewCount="152" Body="&lt;p&gt;In detail, what I really mean is which electron in which orbital is being taken when a oxidation happen? Let me give a example: chromate ion(II)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two electrons are taken but from which orbital:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/khb64.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4945" LastEditDate="2014-11-20T02:34:24.317" LastActivityDate="2014-11-20T03:17:31.600" Title="Which electrons will be taken when oxidations take place and how does the resulting orbital diagram look like?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19750" PostTypeId="2" ParentId="19748" CreationDate="2014-11-20T00:29:33.823" Score="6" Body="&lt;p&gt;The concept of temperature arises from the zeroth law of thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other. Basically, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system C, system A and system B are in thermal equilibrium with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, many hypothetical scales for temperatures are stablished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's said than one body has higher temperature than another when there is a heat exchange from the former to the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But hot and cold are just &lt;strong&gt;sensations or feelings defined by people&lt;/strong&gt;. Something is called hot when it's at higher temperature than our body (~37 °C), so a positive heat exchange will take place on contact. The opposite for cold (heat transference from our body to the cold object).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you are right, the higher the temperature of an object, the hotter it will be, but the concept hot is just to define the thermal perception.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-20T00:29:33.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19751" PostTypeId="1" AcceptedAnswerId="19779" CreationDate="2014-11-20T00:32:39.563" Score="0" ViewCount="115" Body="&lt;p&gt;When they say that we have a {1 0 0} single crystal wafer, I know that the surface normal to the wafer is oriented in the &amp;lt;1 0 0&gt; family direction, but if I draw a line on the wafer, parallel to my primary flat, does it indicate a crystallographic direction family too? Like &amp;lt;1 0 0&gt; or &amp;lt;1 0 1&gt;. I mean it's obvious that it cannot be &amp;lt;1 1 1&gt; due to the FCC lattice of Si(I am making the assumption that the thin plane of the primary flat is at exact right angles to the wafer surface, if that angle is allowed to be 54 degrees, it can also be a &amp;lt;1 1 1&gt; ), but it can surely be directed at none of the crystallographic directions at all, so the question is, if I draw a line on the wafer surface parallel to the primary flat, will it be directed towards the &amp;lt;1 0 0&gt; or the &amp;lt;1 0 1&gt; family, or does the manufacturing process just slice an arbitrary flat out of the Si ingot before slicing it into wafers?&lt;/p&gt;&#xA;" OwnerUserId="9830" LastActivityDate="2014-11-20T14:19:06.337" Title="crystalline structure of Si wafers used for MEMS" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19752" PostTypeId="2" ParentId="19748" CreationDate="2014-11-20T00:33:26.297" Score="4" Body="&lt;p&gt;Hot and cold are subjective terms that can vary from person to person. This means that what I think is warm, you might think is &lt;em&gt;cold&lt;/em&gt;, meaning there is no true value for these definitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were a reference point however, we could that boiling water is &lt;em&gt;hotter&lt;/em&gt; than you, and ice is &lt;em&gt;colder&lt;/em&gt;. But what about something more subtle like the difference between hot water for green tea (maybe 180 $^\circ$ F ) and coffee (about 198-201 $^\circ$F)? They are both hot to you, but to the coffee, green tea would be cold. Get what I'm saying?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-11-20T02:49:04.633" LastActivityDate="2014-11-20T02:49:04.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19753" PostTypeId="2" ParentId="19749" CreationDate="2014-11-20T00:36:28.823" Score="4" Body="&lt;p&gt;For transition metals (specially for those on the second and third rows), like chromium, or iron, which electrons are taken away in oxidation is not straightforwardly deduced. The ones that will be lost are those which are higher in energy and lead to a more stable situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For iron, which is $\mathrm{[Ar]3d^6 4s^2}$, the higher energy electrons are $\mathrm{4s^2}$, and those that will be removed on oxidation, leading to a $\mathrm{d^6}$ configuration for iron (II).&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4356" LastEditDate="2014-11-20T03:17:31.600" LastActivityDate="2014-11-20T03:17:31.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19756" PostTypeId="1" CreationDate="2014-11-20T04:52:12.233" Score="1" ViewCount="150" Body="&lt;p&gt;I'm wondering if the 1&lt;sup&gt;st&lt;/sup&gt; generation Grubbs catalyst can be used to do ring closing metathesis on a non-terminal alkene.&lt;/p&gt;&#xA;" OwnerUserId="9865" LastEditorUserId="7475" LastEditDate="2015-10-16T17:15:44.757" LastActivityDate="2015-10-16T17:15:44.757" Title="Can the 1st generation Grubbs Catalyst be used for ring closing metathesis on non-terminal alkenes?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19757" PostTypeId="1" CreationDate="2014-11-20T05:59:06.530" Score="1" ViewCount="118" Body="&lt;p&gt;Firstly, my question only applies to transition metal compounds with more than one hydroxide group, otherwise it would be more obvious. &lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reason I'm asking is in an effort to oxidise some nickel dihydroxide into nickel oxyhydroxide for practical use in an electrode. I've only come across somewhat laborious methods (any help in this would be appreciated) so I've considered the oxidation like in a NiMH cell (the second equation).&lt;/p&gt;&#xA;&lt;h3&gt;With nickel in a basic solution:&lt;/h3&gt;&#xA;&lt;p&gt;My question could stem from the following situation; If I stir a 2:1 ratio of NaOH:Ni(OH)2 together I would assume the following reaction would be favoured:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce {4 Ni(OH)2 + 8OH- -&amp;gt; 1O2 + 2 Ni2O3 + 8H2O}$$&lt;/span&gt;&#xA;But, if I stir a 1:1 ratio of NaOH:Ni(OH)2 and ground the solution:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce {Ni(OH)2 + OH- -&amp;gt; NiOOH + H2O + e-}$$&lt;/span&gt;&#xA;*I may or may not be misssing some electrons...I'm just starting out in electro-chemistry. &lt;br&gt;&#xA;Is this assumption true? Why or why not? Am I missing the bigger picture? Cue Gibbs? &lt;br&gt; I did try it, but with the setup I had, I couldn't check for gaseous oxygen, and unfortunately both Ni2O3 and NiOOH are black and powdery.&lt;/p&gt;&#xA;&lt;h3&gt;Similarly with iron in an acidic solution:&lt;/h3&gt;&#xA;&lt;p&gt;In insufficient acid, which one is what would happen and why?:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce {Fe(OH)3 + H+ -&amp;gt; [Fe(OH)2]+ + H2O}$$&lt;/span&gt; vs.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce {Fe(OH)3 + H+ -&amp;gt; Fe^{3+} + 3 H2O + excess \,\,\, Fe(OH)3}$$&lt;/span&gt;&#xA;&lt;br&gt;&#xA;I guess the question boils down to; &lt;b&gt;How do reduction / oxidation potentials change with multiple individual reduction / oxidations?&lt;/b&gt; &lt;br&gt;&#xA;I hope everything makes sense. If not I will gladly try to explain my question further. &lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thank you for your help!&lt;/p&gt;&#xA;" OwnerUserId="9771" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-12-21T18:44:08.027" Title="Is a metal hydroxide fully or partly oxidised in insufficient base?" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19758" PostTypeId="2" ParentId="19744" CreationDate="2014-11-20T06:02:54.037" Score="4" Body="&lt;p&gt;Different functional groups may show different absorptions and even a single functional group may exhibit two signals in your UV-VIS spectrum, e.g. corresponding to $\pi,\pi^*$ or $n,\pi^*$ transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While HPLC grade methanol is typically much cheaper than acetonitrile, it might not be the best choice here. You might want to have a look at the &lt;strong&gt;cutoff wavelengths&lt;/strong&gt; of different solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd definitely give measurements at $\lambda = $ 280 nm a try.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-20T06:02:54.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19760" PostTypeId="2" ParentId="19748" CreationDate="2014-11-20T06:35:45.363" Score="4" Body="&lt;p&gt;I finally find out the real answer. It is actually the amount of heat absorb or release which determine how hot or cold you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this answer by a simple experiment. The experiment just simply putting one in cold water and the other one in hot water. Then, put both hands in warm water. You will find the the hand with the hot water will feel cold and the other will feel hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is because &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The temperature of the cold water is lower than the warm one. So, when we put the hand in the warm from the cold water, our hand will absorb heat, causing the hand to feel hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The temperature of the hot water is higher than the warm one. So, when we put the hand in the warm from the hot water, our hand will release heat; causing  the hand to feel cold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can conclude that the more heat it absorbs, the more hotter it is. Whereas the more heat it releases, the colder it is.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2014-11-20T06:35:45.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19761" PostTypeId="1" CreationDate="2014-11-20T07:05:14.687" Score="7" ViewCount="450" Body="&lt;p&gt;I have a large loquat tree that overhangs a pond. The loquat seeds apparently contain cyanide of some sort, so I'm trying to find the effect the seeds have when they fall in the pond.&#xA;Over time I imagine the seeds would break down.&#xA;What happens to the cyanide? Does it escape into the air, or remain in the water?&lt;/p&gt;&#xA;" OwnerUserId="9908" LastEditorUserId="7475" LastEditDate="2015-10-30T17:03:09.060" LastActivityDate="2015-10-30T17:03:09.060" Title="Cyanide from seeds breaking down in a pond" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19762" PostTypeId="2" ParentId="19761" CreationDate="2014-11-20T09:10:43.753" Score="6" Body="&lt;p&gt;Loquat seeds contain &lt;em&gt;Amygdaline&lt;/em&gt;, which is also found in apricot kernels. &lt;em&gt;Amygdaline&lt;/em&gt; belongs to the group of &lt;strong&gt;cyanogenic glycosides&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these compounds, a potentially cyanide releasing aglycon, such as mandelic nitrile, is bound to a carbohydrate (= a &quot;sugar&quot;). These compounds are quite abundant in nature; other known examples are &lt;em&gt;Prunasin&lt;/em&gt; from the japanese bush cherry or &lt;em&gt;Sambunigrin&lt;/em&gt; from elderberry seeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, cyanide is released by enzymatic cleavage of the glycoside, that is when the plant material is digested by animals. Consequently, some of the plants (or at least some parts of them) are known to be toxic. In other cases, the glycosides do degrade in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hydrogen cyanide is released to the pond, I'd assume it stays there - $\ce{HCN}$ is highly soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I wouldn't be very concerned. Seeds aren't like paper bags full of toxic material. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It will likely take a while until the seeds break up to release &lt;em&gt;Amygdaline&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;To my knowledge, &lt;em&gt;Amygdaline&lt;/em&gt; is rather stable in water, hydrogen cyanide isn't released until a sugar-splitting enzyme or acid acts on the molecule.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-20T09:10:43.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19763" PostTypeId="1" CreationDate="2014-11-20T09:16:28.220" Score="3" ViewCount="1138" Body="&lt;p&gt;My answer: The inert pair effect in $\ce{Pb}$ causes it to pull back the electrons, resulting in polarisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher's answer: An ionisation state of $+4$ is too difficult to achieve, and it is not gained even when 4 chlorine atoms get reduced, as the later ionisation enthalpies of lead (or any metal) are high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is, if we don't think of the inert pair effect, we are generalising this, and moreover, high ionisation energies of lead should be explained by the inert pair effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which explanation do you think is better? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I would like to add that my teacher pointed out that the inert pair effect is an outdated concept. Is it so? If yes, why is it rejected?&lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorUserId="4945" LastEditDate="2014-11-20T12:50:35.967" LastActivityDate="2014-11-20T13:47:44.377" Title="Why is PbCl₄ covalent?" Tags="&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19764" PostTypeId="1" AcceptedAnswerId="19769" CreationDate="2014-11-20T10:40:18.710" Score="5" ViewCount="1640" Body="&lt;p&gt;A very famous experiment goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{KNO3}$&lt;/span&gt;, molten wax and sugar are mixed and later burnt; the resulting mixture produces quite a lot of smoke. &#xA;&lt;p&gt;&#xA;After hearing about the experiment, I threw in the idea, &quot;What would happen if &lt;span class=&quot;math-container&quot;&gt;$\ce{NaNO3}$&lt;/span&gt;, or other alkaline compounds were used instead?&quot;. However, the teacher wants us to be sure that the building doesn't explode. So here I am. Has this experiment been attempted before? What would happen if it was used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://styropyro.com/tutorials/smokebomb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a link with the smoke bomb experiment.&lt;/p&gt;&#xA;" OwnerUserId="9910" LastEditorUserId="15751" LastEditDate="2018-09-30T15:41:58.190" LastActivityDate="2018-09-30T15:41:58.190" Title="Can I replace potassium nitrate with sodium nitrate in a smoke bomb?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19765" PostTypeId="2" ParentId="19710" CreationDate="2014-11-20T10:46:28.413" Score="1" Body="&lt;p&gt;Rust is iron oxide, $(\ce{Fe2O3.nH2O})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It forms from iron(III)hydroxide made when iron reacts with oxygen in the presence of water. The chemical reaction can be summarized as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4Fe(s) + 6H2O +3O2 -&amp;gt;4Fe(OH)3(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, water 'naturally' contains some dissolved oxygen (8-14%), depending upon the source of the oxygen (photosynthesis from underwater plants, diffusion from the atmosphere) balanced against oxygen demands (such as fish breathing and decomposition of organic matter). The temperature of the water also affects the solubility of oxygen (gases are less soluble at higher temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ocean (which is typically 3.5% salt) typically has 20% less dissolved oxygen in it than freshwater. This is because the water molecules are attracted to the dissolved ions from the salt (&lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation&quot; rel=&quot;nofollow noreferrer&quot;&gt;solvation&lt;/a&gt;), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby driving dissolved oxygen out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the salinity of the Dead Sea is about 290 (that's 29% salt!), which is over 8 times the salinity of the ocean!  At such high levels of salinity, most of the oxygen is '&lt;a href=&quot;http://en.wikipedia.org/wiki/Salting_out&quot; rel=&quot;nofollow noreferrer&quot;&gt;salted out&lt;/a&gt;' of the water, which means that the levels of dissolved oxygen are very low, especially below the surface, so an iron nail totally immersed in water with 29% sodium chloride (salt) will not rust.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="5026" LastEditDate="2018-04-15T02:03:39.030" LastActivityDate="2018-04-15T02:03:39.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19766" PostTypeId="2" ParentId="19763" CreationDate="2014-11-20T11:10:36.383" Score="0" Body="&lt;p&gt;In $\ce{PbCl4}$ the lead is in $\ce{Pb^{IV}}$ state and $\ce{Cl}$ is having a considerable, not very large - but enough to be polarised easily, and the high charge density on lead polarizes the anion much more efficiently, so that the electron density is shifted and partially shared, in effect the compund is much covalent.You would like to see &lt;a href=&quot;http://en.wikipedia.org/wiki/Fajans%27_rules&quot; rel=&quot;nofollow&quot;&gt;Fajan's rule&lt;/a&gt;, if you didn't knew them before.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="189" LastEditDate="2014-11-20T12:38:26.967" LastActivityDate="2014-11-20T12:38:26.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19767" PostTypeId="2" ParentId="19763" CreationDate="2014-11-20T11:16:54.187" Score="3" Body="&lt;p&gt;Halogenides with metals in high oxidation state are rarely truly ionic on their own, as a strong positive charge on the cation must be stabilized. To do so, many metal-halogen contacts are required, and the metal cation is usually too small to coordinate many halogen ions. So, usually higher halogenides are either molecules with highly polar bonds or (di/oligo/poly)mers with highly polar bonds where some halogen atoms participate in two metal-halogen bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of $\ce{PbCl4}$ &lt;a href=&quot;http://doi.org/10.1021/jp102856v&quot; rel=&quot;nofollow&quot;&gt;NBO analysis&lt;/a&gt; reveals -0.4 NBO charges on chlorine atoms, so it is a highly polar bond, even though it is a molecular compound. So, I would say that due to unfavorable ionic radiuses, $\ce{Pb^{4+}}$ is incapable to form enough contacts to stabilize effectively, forming instead a tetrahedral polar-covalent molecule. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-20T11:16:54.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19768" PostTypeId="1" AcceptedAnswerId="19771" CreationDate="2014-11-20T11:21:11.973" Score="2" ViewCount="286" Body="&lt;p&gt;In fuel-rich combustion, my understanding is that enough fuel reacts and consumes all of the oxygen present. The combustion products $\ce{CO2}$ and $\ce{H2O}$ undergo dissociation until equilibrium is established and one of these is $$\ce{CO2 -&amp;gt; CO + 1/2O2}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the graph below, the equilibrium mole fraction of combustion products are plotted vs. equivalence ratio. Why is there a sudden rise in $\ce{[CO]}$ as we move into fuel rich combustion? Is this due to the drop in $\ce{[O2]}$ such that $\ce{[CO]}$ must increase to compensate for equilibrium to be established?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kfVYo.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="4945" LastEditDate="2014-11-25T06:23:54.010" LastActivityDate="2014-11-25T06:23:54.010" Title="Why is CO a major species only in fuel-rich combustion?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19769" PostTypeId="2" ParentId="19764" CreationDate="2014-11-20T11:26:02.937" Score="13" Body="&lt;p&gt;In pyrotechnics, $\ce{NaNO3}$ is normally avoided because it is very hygroscopic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With dried $\ce{NaNO3}$, the effect should be identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the usual safety rules apply!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-11-20T11:26:02.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19770" PostTypeId="2" ParentId="19763" CreationDate="2014-11-20T11:29:30.980" Score="5" Body="&lt;p&gt;The high covalent characted is just explained by the fact of $\ce{Pb^{4+}}$ being very polarizing and $\ce{Cl-}$ fairly polarizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://chemistry.stackexchange.com/q/8717/189&quot;&gt;inert pair effect&lt;/a&gt; would give an explanation about why it is difficult to reach that oxidation state, and why it has high tendency to get reduced, but there is no need to require to this concept to explain the covalent character of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your last question, I do not see why the inner pair concept should be rejected. Is just another model that can explain in a simple way properties or behaviors of certain elements. There are better models and orbital calcualtions that can also do the job? Of course, but that is no reason for avoiding the use of the model. Models that are still useful to explain stuff easily, can be employed even if more accurate ones do exist. One good example is ligand field theory, which is still useful even when molecular orbitals theory is more accurate and can still explain the same (and many more) things than the former.&#xA;Even though, a theorem or model HAS to be rejected if a new one is found that explains properties better and is incompatible with the old one.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-11-20T13:47:44.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19771" PostTypeId="2" ParentId="19768" CreationDate="2014-11-20T11:43:34.483" Score="4" Body="&lt;p&gt;I think it is quite simple: the higher the proportion of the compound that will burn in the mixture, the more oxygen will be required to a complete combustion. If you have a constant amount of oxygen (in the reactor), and more is requires while increasing the amount of fuel, the complete combustion will gradually decrease, and the ratio $CO/CO_2$ produced will increase as a result of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it can be seen on the graphic, from the point that the equivalent ratio of fuel and oxygen is larger than one, the $CO$ starts being formed in a considerable proportion, and over a ratio of 1:3, starts being the major product.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-20T11:43:34.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19773" PostTypeId="2" ParentId="19735" CreationDate="2014-11-20T12:00:01.003" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually it is not greater, &lt;em&gt;but greater than or equal to&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta S\ge\frac{\delta q}T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If heat is provide irreversibly, the term on LHS is greater otherwise in case of reversible process it is equal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also what is the significance of this inequality?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This inequality when rearranged becomes:&#xA;$$TdS-dq\ge0$$&#xA;For constant volume or pressure it becomes(recall $q_v=U,q_p=H$):&#xA;$$TdS-U\ge0\\TdS-H\ge0$$&#xA;both of which say the common equations,(for a spontaneous change):&#xA;$$dA=U-TdS\le0\\dG=H-TdS\le0$$ &lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-20T12:00:01.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19774" PostTypeId="5" CreationDate="2014-11-20T12:26:10.510" Score="0" Body="&lt;p&gt;Tautomers are constitutional isomers of (usually organic) compounds that readily interconvert into each other. The process of this interconversion is called tautomerization, which is a heterolytic molecular rearrangement. This process can quite frequently be very rapid (comp. &lt;a href=&quot;https://goldbook.iupac.org/T06253.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;goldbook&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The IUPAC goldbook describes &lt;a href=&quot;https://goldbook.iupac.org/T06252.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;tautomerism&lt;/a&gt; as an isomerism of the general form &lt;span class=&quot;math-container&quot;&gt;$$\ce{G-X-Y=Z &amp;lt;=&amp;gt; X=Y-Z-G},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\ce{X, Y, Z}$&lt;/span&gt; are typically any of the &lt;span class=&quot;math-container&quot;&gt;$\ce{C, H, O, S}$&lt;/span&gt; elements and &lt;span class=&quot;math-container&quot;&gt;$\ce{G}$&lt;/span&gt; becomes &lt;a href=&quot;https://goldbook.iupac.org/E01965.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrofuge&lt;/a&gt; or &lt;a href=&quot;https://goldbook.iupac.org/N04246.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;nucleofuge&lt;/a&gt;. The most common case is, when the electrofuge is a proton, &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, is also known as prototropy. Very common are the tautomerization of &lt;a href=&quot;https://goldbook.iupac.org/I03269.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(iso-)cyanic acids&lt;/a&gt;:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{H-C#N &amp;amp;~&amp;lt;=&amp;gt;~ {}^{\ominus}C#N^{\oplus}-H} \tag{1}\\&#xA;\ce{O=C=N-H &amp;amp;~&amp;lt;=&amp;gt;~ N#C-O-H}\tag{2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The case of &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; it is also known as dyadic tautomerism, where the proton migrates between neighbouring atoms.&lt;/p&gt;&#xA;&lt;p&gt;Other very well known cases are &lt;a href=&quot;http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism&quot; rel=&quot;nofollow noreferrer&quot;&gt;keto-enol&lt;/a&gt; and lactam-lactim &lt;a href=&quot;http://en.wikipedia.org/wiki/Tautomer&quot; rel=&quot;nofollow noreferrer&quot;&gt;tautomerism&lt;/a&gt;:&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZyjVF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZyjVF.png&quot; alt=&quot;keto-enol tautomerism&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;(graphics source: &lt;a href=&quot;https://goldbook.iupac.org/T06252.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;goldbook&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2020-12-23T21:45:30.687" LastActivityDate="2020-12-23T21:45:30.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19775" PostTypeId="4" CreationDate="2014-11-20T12:26:10.510" Score="0" Body="Tautomers are constitutional isomers of (usually organic) compounds that readily interconvert into each other. This tag is also suitable for tautomerization, the isomerization by which tautomers are interconverted. It may synonymously specified for questions about tautomerism." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-11-20T12:26:10.510" LastActivityDate="2014-11-20T12:26:10.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19776" PostTypeId="2" ParentId="19697" CreationDate="2014-11-20T12:42:45.413" Score="3" Body="&lt;p&gt;We have these elements:(just for ease, I'm naming them, you can just avoid it)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$&#xA;\begin{array}\\&#xA;Z=33&amp;amp;\text{Arsenic}&amp;amp;\ce{As}\\ &#xA;Z=37&amp;amp;\text{Rubidium}&amp;amp;\ce{Rb}\\ &#xA;Z=9&amp;amp;\text{Fluorine}&amp;amp;\ce{F}\\&#xA;Z=31&amp;amp;\text{Galium}&amp;amp;\ce{Ga}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, let's see the periodic table (the useful region, thanks to Martin) :&#xA;$$&#xA;\begin{array}{ll|cccccccccccccccccc}&#xA;&amp;amp;\text{Gr.} &amp;amp;1 &amp;amp;2 &amp;amp;13  &amp;amp;15  &amp;amp;17 \\&#xA;\text{Per.}&amp;amp;\\\hline&#xA;2&amp;amp;&amp;amp;. &amp;amp; .&amp;amp; .&amp;amp; .&amp;amp; \color{red}{\ce{\tiny^9F}}\\&#xA;4&amp;amp;&amp;amp; .&amp;amp; .&amp;amp;  \color{red}{\large\ce{^{31}Ga}}&amp;amp; \color{red}{\small\ce{^{33}As}}&amp;amp; .\\&#xA;\hline&#xA;5&amp;amp;&amp;amp; \color{red}{\huge\ce{^{37}Rb}}\\&#xA;\end{array}\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;While going down or towards left, size increases. Equivalent, is to say the size is larger towards bottom left corner&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It wouldn't be now difficult to conclude that size is as follows:&#xA;$$37&amp;gt;31&amp;gt;33&amp;gt;9$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5026" LastEditDate="2018-03-04T02:48:22.427" LastActivityDate="2018-03-04T02:48:22.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19777" PostTypeId="1" CreationDate="2014-11-20T14:08:13.823" Score="3" ViewCount="99" Body="&lt;p&gt;I am trying to make &quot;films&quot; of polyethylene glycol by melting them, pouring them into a mold and then cooling the mold down. The plan is to do this with different molecular weights. The problem is that depending on the molecular weight, the crystallinity will be different. Is there a way of ensuring that the crystallinity of the polymer is the same for all of them?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="4356" LastEditDate="2014-11-20T18:50:56.560" LastActivityDate="2014-12-20T20:20:32.500" Title="How do I achieve same crystallinity when cooling polymers?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19778" PostTypeId="1" AcceptedAnswerId="19780" CreationDate="2014-11-20T14:11:53.037" Score="5" ViewCount="12395" Body="&lt;p&gt;I just learnt about acids and bases at class.I noticed that my textbook mentions the term &quot;alkali/alkaline&quot; and sometimes it mentions &quot;base/basic&quot;. As far as I know, alkalis and bases are metal oxides hydroxides. But why does my textbook make use of &lt;em&gt;two&lt;/em&gt; words that mean the same thing instead of just &lt;em&gt;one&lt;/em&gt; of them? Do these terms (alkali and base) mean something different?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="16683" LastEditDate="2017-01-30T15:32:43.053" LastActivityDate="2019-05-24T07:22:30.997" Title="Alkalis and Bases" Tags="&lt;acid-base&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19779" PostTypeId="2" ParentId="19751" CreationDate="2014-11-20T14:19:06.337" Score="1" Body="&lt;p&gt;The flats on Si wafers are not arbitrary.  The largest, or primary flat, is perpendicular to the &amp;lt;110&gt; orientation in the plane of the wafer. If it exists, the secondary flat helps identify the wafer orientation and doping type (and can easily vary based on wafer maker and the company ordering the wafers). All modern wafers (anything 200mm or bigger, almost all 150mm) are (100) and boron doped, so the secondary flat isn't used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary flat has a specific orientation since the chips need to be aligned along the &amp;lt;110&gt; crystallographic orientations so the wafer can be diced up.  This usually uses a diamond scribe and careful breaking rather than sawing, because the saw is wider than a scribe line (so fewer chips/wafer) and dirtier .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you wafer has &amp;lt;001&gt; pointing up, and &amp;lt;110&gt; in plane towards the flat.  &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2014-11-20T14:19:06.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19780" PostTypeId="2" ParentId="19778" CreationDate="2014-11-20T14:42:28.687" Score="11" Body="&lt;p&gt;These words are in many cases interchangeably used: an &lt;em&gt;alkaline&lt;/em&gt; solution/compound is the same as a &lt;em&gt;basic&lt;/em&gt; solution/compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can specify &quot;alkali&quot; to refer to those bases that are soluble in water and that are salts of alkaline or alkaline earth metals too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkali is like a subset of bases which fulfills these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sodium hydroxide is a base, and also is an alkali metal salt, and is also soluble in water, so it is also an alkali.&lt;/li&gt;&#xA;&lt;li&gt;Ammonia is a base, soluble in water, but not a salt, so it is not an alkali.&lt;/li&gt;&#xA;&lt;li&gt;Magnesium hydroxide is a base, is a salt of an alkaline earth metal, but it is not soluble in water, so it is not an alkali.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5354" LastEditDate="2015-09-12T17:00:28.043" LastActivityDate="2015-09-12T17:00:28.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19781" PostTypeId="1" AcceptedAnswerId="19786" CreationDate="2014-11-20T15:02:34.267" Score="2" ViewCount="1784" Body="&lt;p&gt;I assume spontaneous separation to ions of salt solution (or acid) requires energy to occur, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does that energy come from? Is it from the heat of the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if someone devised a way to draw potential energy from those separated ions, could he make this process continuous (assuming solution would draw heat from its surroundings), or would it eventually stop to work?&lt;/p&gt;&#xA;" OwnerUserId="9913" LastEditorUserId="4945" LastEditDate="2014-11-25T06:25:02.167" LastActivityDate="2014-11-25T06:25:02.167" Title="Does dissociation require energy to occur?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19782" PostTypeId="2" ParentId="19781" CreationDate="2014-11-20T15:30:15.750" Score="2" Body="&lt;p&gt;When the ions of an ionic compound disolve, do not stay isolated. They form new bonds with the solvents. This makes the overall process spontaneus when a salt is soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissociation of course requires energy. But bond formation between ions and solvent molecules release energy.&#xA;This can result on a possitive (endothermic disolution, heat is absorbed in the process, like for example, $\ce{KI}$ in water. Bonds formed release less energy than bonds broken) or negative disolution enthalpy (exothermic disolution, like the one for $\ce{H_2SO_4}$ and water. Bonds formed -hydrogen bonds, stronger that ion-dipole interactions-, release more energy that is consumed by bonds broken).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that if you want to know if a process is or not spontaneous, you do not have to take into acount just enthalpy, but Gibbs free energy, which is $\Delta G= \Delta H + T\Delta S$, which also takes into account entropic factors and the influence of temperature.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="14" LastEditDate="2014-11-20T15:34:18.387" LastActivityDate="2014-11-20T15:34:18.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19783" PostTypeId="1" CreationDate="2014-11-20T16:21:43.493" Score="4" ViewCount="321" Body="&lt;p&gt;Why are all the $\ce{C-O}$ bonds in $\ce{CO_3^2-}$ equal but not those in $\ce{H2CO3}$?&lt;/p&gt;&#xA;" OwnerUserId="9915" LastEditorUserId="16683" LastEditDate="2015-10-23T16:54:29.663" LastActivityDate="2015-10-23T16:54:29.663" Title="What is 'equality' of chemical bonds?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19784" PostTypeId="1" AcceptedAnswerId="19820" CreationDate="2014-11-20T16:27:25.393" Score="2" ViewCount="1041" Body="&lt;p&gt;I've been looking for a while and I still can't get a straight answer. I just need to know the lowest CAS Registry Number on record and the corresponding chemical. &lt;/p&gt;&#xA;" OwnerUserId="9917" LastEditorUserId="5026" LastEditDate="2018-02-22T11:11:54.263" LastActivityDate="2018-02-22T11:12:57.887" Title="Lowest CAS Registry Number on record" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19785" PostTypeId="2" ParentId="19783" CreationDate="2014-11-20T16:32:38.183" Score="4" Body="&lt;p&gt;For $\ce{CO_3^2-}$ you have the following possible resonance structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufxVD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure bellow, is the one which represents better the real ionic molecule, in which all the bonds are equal. You can delocalize the negative charge through all the ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{H2CO3}$ the structure is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gN5vB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, in this case, bonds are fixed, and there is no possible resonance delocalization, so you have two single bonds (less strong) C-O and one double bond C=O (stronger). For $\ce{CO_3^2-}$ you have that all bond lenths the same showing an intermediate situation between single and double bonds.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-20T16:32:38.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19786" PostTypeId="2" ParentId="19781" CreationDate="2014-11-20T17:12:53.960" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I assume spontaneous separation to ions of salt solution (or acid)&#xA;  requires energy to occur, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it requires energy to occur.  This energy is usually referred to as the activation energy.  It is a measure of the energy barrier that must be surmounted for the process to occur.  In the case of salts ionizing in solution, the barrier is usually (but not always) quite small.  Salts that dissolve and give off heat (exothermic) will have a lower barrier than salts that require energy input (endothermic) to dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBkxn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;image source: &lt;a href=&quot;http://web.campbell.edu/faculty/nemecz/323_lect/enzyme_mech/mech_chapter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.campbell.edu/faculty/nemecz/323_lect/enzyme_mech/mech_chapter.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where does that energy come from? Is it from the heat of the solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The energy comes from the surrounding environment.  The surrounding molecules have a thermal energy that can be transferred through collisions with the reactants.  The reactants will convert to products once enough energy has been imparted to them to cross the barrier.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, if someone devised a way to draw potential energy from those&#xA;  separated ions, could he make this process continuous&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you're asking if the energy withdrawal can be made continuous.  If so, then the answer is &quot;yes&quot; - at least until all of the starting material has been consumed and converted to product.  This is done today, examples would include batteries, gasoline, power generators in general, etc.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-20T17:12:53.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19787" PostTypeId="2" ParentId="19777" CreationDate="2014-11-20T18:04:38.960" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way of ensuring that the crystallinity of the polymer is&#xA;  the same for all of them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First you need a method to assess the degree of crystallinity.  Samples of the same material that differ in degree of crystallinity should show different light transmission intensities due mainly to increased light scattering in the more crystalline samples.  If you have access to a spectrophotometer you could insert the films and measure the light transmission.  If you don't have access to a spectrophotometer, a &quot;poor-man's&quot; approach would be to use visual comparison of the transmitted light (use a black piece of paper and a pencil beam flashlight in a dark room).  It is important that the film thicknesses all be the same.  If there is variability in your film thickness, then cast thicker films so that the percentage variation will be smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cooling your sample more quickly, you will tend to preserve more of the amorphous nature found in the solution.  Conversely, cooling more slowly or annealing your sample will tend to increase the degree of crystallinity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjust your rate of cooling on each of the films until they all produce the same transmission intensity, then the degree of crystallinity should be similar in your films.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-20T18:04:38.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19789" PostTypeId="1" CreationDate="2014-11-20T20:11:26.243" Score="3" ViewCount="1343" Body="&lt;p&gt;I am studying the synthesis of a drug &lt;a href=&quot;http://en.wikipedia.org/wiki/Pregabalin&quot; rel=&quot;nofollow&quot;&gt;pregabalin&lt;/a&gt;, for which both chemical and bio-catalysts are used. Using a biocatalyst is way better than chemical route, as explained &lt;a href=&quot;https://www.dbu.de/media/130508051211a525.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. In fact, The invention of the bio-catalyst route got 2006 IchemE award in Green Chemistry and Engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in knowing is this: are there any disadvantages of using bio-catalysis in this reaction? (Some general disadvantages can be seen &lt;a href=&quot;http://www.answers.com/Q/What_are_disadvantages_of_using_a_catalyst_on_and_industrial_reaction&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) Suppose you are in a debate and you have to favor the chemical side, what would be your arguments?&lt;/p&gt;&#xA;" OwnerUserId="9921" LastEditorUserId="73" LastEditDate="2014-11-20T22:46:10.533" LastActivityDate="2014-12-18T23:13:17.920" Title="What are the specific disadvantages of using a bio-catalyst (called Lipolase) in the synthesis of the drug pregabalin?" Tags="&lt;synthesis&gt;&lt;catalysis&gt;&lt;drugs&gt;&lt;enzymes&gt;&lt;green-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19791" PostTypeId="2" ParentId="19756" CreationDate="2014-11-20T21:29:53.517" Score="3" Body="&lt;p&gt;Yes, it appears that that ring closing metathesis of non-terminal alkenes has been performed using the &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/579726?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;1st generation Grubbs catalyst&lt;/a&gt; several times:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/C3CC43250E&quot; rel=&quot;nofollow&quot;&gt;Asymmetric syntheses of enantiopure C(5)-substituted transpentacins&#xA;via diastereoselective Ireland–Claisen rearrangements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ol3029548&quot; rel=&quot;nofollow&quot;&gt;Stereoselective Synthesis of Deuterium-Labeled (2S)-Cyclohexenyl&#xA;Alanines, Biosynthetic Intermediates of Cinnabaramide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.catcom.2007.12.004&quot; rel=&quot;nofollow&quot;&gt;Ru-catalyzed metathesis of octadienylether xyloside&lt;/a&gt; &lt;a href=&quot;http://dx.doi.org/10.1021/ol0709735&quot; rel=&quot;nofollow&quot;&gt;Synthesis&#xA;of (−)-Agelastatin A by [3.3] Sigmatropic Rearrangement of Allyl&#xA;Cyanate&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1139/v06-095&quot; rel=&quot;nofollow&quot;&gt;A chiral auxiliary cleavable by ring-closing alkene&#xA;metathesis — Efficient synthesis of chiral nonracemic&#xA;cycloalkenes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.tet.2004.07.103&quot; rel=&quot;nofollow&quot;&gt;Conversion of chiral unsaturated cyanohydrins into&#xA;chiral carba- and heterocycles via ring-closing metathesis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://doi.dx.org/10.1021/jo035371n&quot; rel=&quot;nofollow&quot;&gt;Total Synthesis of Pancratistatin Relying on the [3,3]-Sigmatropic&#xA;Rearrangement&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/jo981745e&quot; rel=&quot;nofollow&quot;&gt;Stereoselective Chelate-Controlled Addition of Grignard Reagents to Unsaturated Medium-Ring Heterocycles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-11-20T21:29:53.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19792" PostTypeId="2" ParentId="19599" CreationDate="2014-11-20T21:32:33.407" Score="2" Body="&lt;p&gt;The conjugated base of the depicted alcohol is not resonance-stabilized. The negative charge cannot be delocalized into the aromatic ring because the alcohol group is not a &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenol&quot; rel=&quot;nofollow noreferrer&quot;&gt;phenolic&lt;/a&gt; OH group, whose oxygen lone pairs are conjugated to the aromatic $\pi$ electrons, like shown below. Instead, it is attached to an $sp^3$-hybridized benzylic carbon which is not part of the conjugated system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xc25W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might have mistaken the depicted compound with &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Naphthol&quot; rel=&quot;nofollow noreferrer&quot;&gt;2-naphtol&lt;/a&gt;, whose $\ce{pK_{a}}$ (9.51) is much lower than that of aliphatic alcohols and which can therefore be deprotonated by $\ce{NaOH}$. Aliphatic alcohols typically have $\ce{pK_{a}}$ values of about 16-18 (&lt;a href=&quot;http://as.vanderbilt.edu/chemistry/Rizzo/Chem220b/Chapter_17.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), so there will be no deprotonation by sodium hydroxide.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2014-11-20T21:32:33.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19795" PostTypeId="2" ParentId="19599" CreationDate="2014-11-20T23:35:46.810" Score="1" Body="&lt;p&gt;These would be the resonant structures of your deportonated compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zsJ5H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there will be no delocalization at al... If you try to move the electrons just a bit, you obtain a very unstable structure, which will not contribute to the resonance hybrid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible to deportonate it with a stronger base, like NaH, or LDA.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-20T23:35:46.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19796" PostTypeId="2" ParentId="19673" CreationDate="2014-11-21T00:45:58.633" Score="6" Body="&lt;p&gt;You are on the right track. Pyridine and similar nitrogen analogs of benzene are deactivated with respect to the electrophilic aromatic substitution reaction. You are also right that a resonance structure can help explain the regioselectivity of the reaction on the five-membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrophilic attack at C–2 (bottom line) directly gives an intermediate with non-aromatic resonance structure &lt;strong&gt;2a&lt;/strong&gt;. Moving electrons around, we are able to draw an aromatic resonance structure &lt;strong&gt;2b&lt;/strong&gt;, but this structure also contains a non-octet atom (carbocation), repulsive neighboring positively charged atoms, and charge separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y6aJi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y6aJi.png&quot; alt=&quot;Electrophilic substitution at C–2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, electophilic attack at the observed site gives an intermediate &lt;strong&gt;3&lt;/strong&gt; that maintains aromaticity in the six-membered ring and complete octets in all atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5OKhj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5OKhj.png&quot; alt=&quot;Electrophilic substitution at C–3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can predict that the intermediate arising from attack of the electrophile at C–3 would be more stable than the corresponding C–2 substitution intermediate. Since this step should be endothermic, the rate of this step will be influenced by the stability of the resulting intermediate (Hammond's postulate). From there, we can proceed through the usual addition/elimination mechanistic sequence to get to the product.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2018-09-15T13:48:25.633" LastActivityDate="2018-09-15T13:48:25.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19797" PostTypeId="1" CreationDate="2014-11-21T01:38:57.140" Score="5" ViewCount="228" Body="&lt;p&gt;I’ve stumbled across a rather peculiar recipe for a allegedly ‘health’ potion in a computer game, composed of an unspecified amount of each of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rust (iron oxide)&lt;/li&gt;&#xA;&lt;li&gt;saltpetre (potassium nitrate)&lt;/li&gt;&#xA;&lt;li&gt;oil of vitriol (sulphuric acid)&lt;/li&gt;&#xA;&lt;li&gt;aqua vitæ (ethanol)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To make things clear: I’m not suicidal, I do in no way intent to actually &lt;em&gt;mix&lt;/em&gt; that stuff (the initial tagging was maybe a bit misleading), but I’m interested in what &lt;em&gt;would&lt;/em&gt; happen if someone actually would.  &lt;em&gt;Drinking&lt;/em&gt; such a &lt;s&gt;potion&lt;/s&gt; vile concoction is of course out of the question because of the sulphuric acid alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know from highschool chemistry (and a generous helping of Wikipedia), saltpetre and sulphuric acid would be used to make explosives by way of nitrosulphuric acid.  Oversimplifying, by adding cellulose or tuluol one would end up with nitrocellulose or TNT, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I lack the terminology to go further than that.  What would be the result of adding alcohol to that, and then iron oxide, or the other way round?  Is there some known mixture that would fit, however roughly, along those lines?  From the context of where that formula came from that wouldn’t surprise me.  It may of course as well be that the formula was simply chosen because of it’s obvious unhealthyness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this clarifies the intent of my question.  The idea may have come from a pseudo-fantasy/alchemy setting, but I’m interested in the ‘real’ chemistry behind it.&lt;/p&gt;&#xA;" OwnerUserId="9928" LastEditorUserId="9928" LastEditDate="2014-11-21T20:48:29.993" LastActivityDate="2015-08-23T09:19:18.527" Title="How ‘lively’ would this ‘health’ potion be?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;safety&gt;&lt;explosives&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19798" PostTypeId="2" ParentId="19747" CreationDate="2014-11-21T02:26:49.510" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short answer: It matters most that you are using a modern solvation model than which one&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly longer answer: If you have the time, try a few solvation models, compare with experiment and see which gives you the best results. In my experience, modern models do fairly well for TDDFT calculations. Be careful about the atomic radii used and whether the computational program generates a reasonable solvation surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind that not every program supports both D-PCM and C-PCM. Many packages just implement &lt;em&gt;one&lt;/em&gt; solvation model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of excellent reviews on &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizable_continuum_model&quot; rel=&quot;nofollow&quot;&gt;polarizable continuum solvation models&lt;/a&gt;. As you're probably aware, they are a really efficient way to handle molecular solvation in calculations, since the details of the solvent are largely ignored. Instead, the solvent is treated as a dielectric environment, and you handle polarization between the molecule and the solvent self-consistently. That is, the solute molecule is treated at normal atomistic detail, creates a solvation shell (usually treated by a set of overlapping spheres) and the electrostatics of the molecule influences the solvent, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've never seen &quot;C-PCM&quot; as a &quot;capacitive&quot; PCM. Usually people refer to the &quot;C&quot; as a conductor-style model. That is, in a C-PCM model, the dielectric constant is set to $\infty$ but then you re-scale the charges on the molecule later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more than just these techniques. The main problem is that it's very hard to bench-mark solvation energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some great recent reviews:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1007/s00214-004-0582-3&quot; rel=&quot;nofollow&quot;&gt;&quot;Thirty years of continuum solvation chemistry: a review, and prospects for the near future.&quot; &lt;em&gt;Theoretical Chemistry Accounts&lt;/em&gt; (&lt;strong&gt;2004&lt;/strong&gt;) v. 112(4) pp. 184-203&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/cr9904009&quot; rel=&quot;nofollow&quot;&gt;&quot;Quantum Mechanical Continuum Solvation Models.&quot; &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;105&lt;/em&gt;(8), pp 2999–3094&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ct800029c&quot; rel=&quot;nofollow&quot;&gt;&quot;Perspective on Foundations of Solvation Modeling: The Electrostatic Contribution to the Free Energy of Solvation&quot; &lt;em&gt;J. Chem. Theory Comput.&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, 4 (6), pp 877–887&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ar800019z&quot; rel=&quot;nofollow&quot;&gt;&quot;A Universal Approach to Solvation Modeling&quot; &lt;em&gt;Acc. Chem. Res.&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, 41 (6), pp 760–768&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/wcms.56&quot; rel=&quot;nofollow&quot;&gt;&quot;The COSMO and COSMO-RS solvation models&quot; &lt;em&gt;WIREs Comput Mol Sci&lt;/em&gt;, &lt;strong&gt;(2011)&lt;/strong&gt; 1: 699–709&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/wcms.1086&quot; rel=&quot;nofollow&quot;&gt;&quot;Polarizable continuum models.&quot; &lt;em&gt;WIREs Comput Mol Sci&lt;/em&gt; (&lt;strong&gt;2012&lt;/strong&gt;) 2: 386–404.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-11-21T14:19:49.877" LastActivityDate="2014-11-21T14:19:49.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19799" PostTypeId="1" AcceptedAnswerId="19801" CreationDate="2014-11-21T03:24:58.997" Score="0" ViewCount="3343" Body="&lt;p&gt;I know sulfuric acid is a strong acid and acetic acid is a weak acid. However when I write them in their chemical symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfuric acid: $\ce{H2SO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid: $\ce{CH3COOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that acetic acid has 4 atoms of hydrogen whereas sulfuric acid has only 2 atoms of hydrogen. So, shouldn't acetic acid be stronger than sulfuric acid?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="402" LastEditDate="2017-05-24T05:00:28.087" LastActivityDate="2017-05-24T05:00:28.087" Title="Strong and weak acid" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19801" PostTypeId="2" ParentId="19799" CreationDate="2014-11-21T04:26:11.547" Score="3" Body="&lt;p&gt;Looking at the structures of those two acids might help. In both cases, the acidic hydrogens are part of hydroxyl groups. Because of polarity and the inductive effect, the O-H bond is weaker than C-H bonds which are very strong. As a result, it's nearly impossible for Carbon to lose a hydrogen without some kind of elimination reaction, whereas the O-H bond can autoprotolyze with other O-H bonds (even water). The weakness of the O-H bond is the basis for the acidity of those hydrogens. C-H bonds are essentially non-acidic, having pKa values around 45-50.&lt;/p&gt;&#xA;" OwnerUserId="4813" LastActivityDate="2014-11-21T04:26:11.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19802" PostTypeId="2" ParentId="19799" CreationDate="2014-11-21T04:45:00.680" Score="2" Body="&lt;p&gt;Sulfuric acid is a strong acid; when placed in water, it completely dissociates into $H^+$ and $HSO_4^-$. While acetic acid dissociates into $H^+$ and $CH_3COO^-$; but being a weak acid, some of the molecules remain as $CH_3COOH$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between a strong and a weak acid is not necessarily their level of acidity; it is the degree to which they dissociate. Also as the previous answerer said, due to the nature of the bond of the 3 hydrogen molecules in acetate, we wouldn't consider it an acid because they don't break into $H^+$ ions as easily.&lt;/p&gt;&#xA;" OwnerUserId="9932" LastEditorUserId="402" LastEditDate="2017-05-24T04:59:47.083" LastActivityDate="2017-05-24T04:59:47.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19804" PostTypeId="2" ParentId="19670" CreationDate="2014-11-21T06:07:57.623" Score="5" Body="&lt;p&gt;Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with &lt;a href=&quot;http://en.wikipedia.org/wiki/Pascal%27s_triangle&quot; rel=&quot;noreferrer&quot;&gt;Pascal's triangle&lt;/a&gt;, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:&#xA;\begin{array}{lrr}\hline&#xA;   &amp;amp; \text{m/z} &amp;amp;\text{abundance}/\%\\\hline&#xA;\ce{[{}^{35}_{17}Cl_3]^{+}}                &amp;amp; 105   &amp;amp;  3.7  \\&#xA;\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  &amp;amp; 107   &amp;amp; 22.2  \\&#xA;\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  &amp;amp; 109   &amp;amp; 44.4  \\&#xA;\ce{[{}^{37}_{17}Cl_3]^{+}}                &amp;amp; 111   &amp;amp; 29.7  \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following snowball diagram should further demonstrate this:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Z12Xo.png&quot; alt=&quot;snow ball diagramm&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-11-21T06:07:57.623" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19805" PostTypeId="1" AcceptedAnswerId="19807" CreationDate="2014-11-21T07:21:47.700" Score="4" ViewCount="2955" Body="&lt;p&gt;I have learned about metal and their reactivity series. However, why metals have different reactivity? What are the factors of different reactivity rate for metal? All I can think of is the number of electrons which causing it. &lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2019-01-19T05:57:43.453" Title="Reactivity series of metal" Tags="&lt;reactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19806" PostTypeId="2" ParentId="19784" CreationDate="2014-11-21T07:54:21.227" Score="2" Body="&lt;p&gt;&quot;Carotenes&quot; has a CAS Registry Number of 36-88-4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that chemical is &quot;too vague&quot;, &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sigma/n1875?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow noreferrer&quot;&gt;Naphthol AS-BI $\beta$-D-glucuronide&lt;/a&gt; has a CAS Registry Number of 37-87-6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to semi-independent source &lt;a href=&quot;http://books.google.com/books?id=ahbUc5M0I5AC&amp;amp;pg=PA6#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Chemist's Desk Reference 2nd ed. (2010) pg 6 sec 1.1.1.5&lt;/a&gt;, CAS 35-66-5 DOES seem to be the lowest, but how do we know &quot;Benzacridine, 7,​9-​dimethyl- (9CI)&quot; exist? Cl- and methyl- are occupying the same position at the aromatic ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &quot;Carotenes&quot; (CAS 36-88-4) by the name itself does not seem to exist as a single compound but a set of compounds.&lt;/p&gt;&#xA;" OwnerUserId="9933" LastEditorUserId="5026" LastEditDate="2018-02-22T11:12:57.887" LastActivityDate="2018-02-22T11:12:57.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19807" PostTypeId="2" ParentId="19805" CreationDate="2014-11-21T08:23:20.453" Score="5" Body="&lt;p&gt;I suppose that with reactivity you mean a metal's tendency to become oxidized.&#xA;There are several reasons and it depends on the metals you are analyzing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are referring to alkaline or alkaline earth metals, reactivity increases as you go down the group. They all have a large oxidation potential since it's relatively easy to remove an electron (or two) of its orbital. As you move down the group, atoms are larger and the valence electrons will be less atracted by the nucleus. Therefore, they are easier to remove and the metal will be more reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For d-block transition metals (most of them are considered &quot;transition&quot; metals) it's a whole different story. There are many factors that contribute to its reactivity and it's harder to specify general trends. Many of them (especially the ones on the second and third transition rows) can be oxidized to different oxidation states depending on the conditions. You need to look at their electron configuration and reduction potentials to start rationalizing its properties. Depending upon the electron configuration, they will be able to lose (or even gain) less or more electrons and with different ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know your background with chemistry, so I won't write a more detailed explanation but if you clarify that you have a more concrete interest in any explanation, just ask.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4325" LastEditDate="2014-11-21T18:30:40.647" LastActivityDate="2014-11-21T18:30:40.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19808" PostTypeId="1" CreationDate="2014-11-21T08:34:01.300" Score="9" ViewCount="5141" Body="&lt;p&gt;If pH is defined as the concentration of hydrogen ions in solution, then how can a ‘neutralized’ solution (defined as having an equal amount of hydrogen and hydroxide ions) have a pH other than 7?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Neutralization_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; writes: “In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in solution.” In my mind, an equivalent amount of hydrogen ions and hydroxide ions means a pH of 7 (in water), but Wikipedia carries on to write: “The pH of the neutralized solution depends on the acid strength of the reactants.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/8048/shouldnt-the-ph-at-the-equivalence-point-always-be-7&quot;&gt;This question&lt;/a&gt; doesn’t seem to answer my question.&lt;/p&gt;&#xA;" OwnerUserId="9934" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-04-24T22:57:36.017" Title="The pH of a neutralized solution" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19809" PostTypeId="1" AcceptedAnswerId="19811" CreationDate="2014-11-21T09:00:18.553" Score="1" ViewCount="874" Body="&lt;p&gt;I know that $\ce{CO_2}$ and $\ce{O_2}$ are slightly soluble in water ($\ce{H_2O}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my understanding, when carbon dioxide dissolves in water, it will react to form carbonic acid just like the equation below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O + CO2 -&amp;gt; H2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my problem now is, I can't find the equation for oxygen.&lt;br&gt;&#xA; Will they form hydroxide ions or something else?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="13339" LastEditDate="2015-05-18T08:40:36.370" LastActivityDate="2015-05-18T08:40:36.370" Title="Oxygen dissolve in water" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19810" PostTypeId="2" ParentId="19797" CreationDate="2014-11-21T09:00:40.067" Score="1" Body="&lt;p&gt;A glass of that would just kill you on the right proportions, not much of a health potion...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Honestly, I was worried about you when I first saw the title, the list and the home-experiment tag)&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-21T09:00:40.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19811" PostTypeId="2" ParentId="19809" CreationDate="2014-11-21T09:07:39.537" Score="2" Body="&lt;p&gt;Carbon dioxide is not stable in water in its original form, so it undergoes an acid-base reaction to give the carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, oxygen is stable and soluble in water in its $\ce{O_2}$ form, so it will stay like that, no reaction other than solubilization. Besides, it could react with organic matter or other compounds in water, oxidizing them, and getting reduced to $\ce{OH^-}$&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="13339" LastEditDate="2015-05-18T08:40:19.390" LastActivityDate="2015-05-18T08:40:19.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19812" PostTypeId="1" CreationDate="2014-11-21T09:17:34.963" Score="3" ViewCount="16852" Body="&lt;p&gt;I would like to know the difference between carboxylate ions carboxyl groups? An example is is $\ce{COOH}$ and $\ce{COO}$.&lt;/p&gt;&#xA;" OwnerUserId="9935" LastEditorUserId="4356" LastEditDate="2014-11-21T23:26:45.143" LastActivityDate="2019-03-29T11:22:10.413" Title="What is the difference between carboxylate ions and groups?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19814" PostTypeId="2" ParentId="19808" CreationDate="2014-11-21T10:49:16.760" Score="11" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;br&gt;&#xA;The neutral point is not the same as the equivalence point.&lt;br&gt;&#xA;A neutral aqueous solution at room temperature, $25~^\circ\mathrm{C}$ and standard pressure $1~\mathrm{atm}$, has always $\ce{pH}=7$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you look for credible sources of definitions of neutral solutions on the internet, most likely you will find something along the lines of this:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neutral solution&lt;br&gt;&#xA;  (Science) Has a $\ce{pH}$ level of $7$: a solution in which the concentration of hydrogen ions and hydroxide ions are equal (&lt;a href=&quot;http://www.biology-online.org/dictionary/Neutral_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;biology-online.org&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or even worse:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Neutral Solution Definition:&lt;br&gt;&#xA;  An aqueous solution with a $\ce{pH}$ of $7.0$ $(\ce{[H^{+}]} = 1.0 \times 10^{-7}~\mathrm{M})$. (&lt;a href=&quot;http://chemistry.about.com/od/chemistryglossary/a/neutralsoldef.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemistry.about.com&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second one especially is negligent of a lot of features.&lt;br&gt;&#xA;Unfortunately the IUPAC does not provide an actual definition of a neutral solution. There are two possible definitions, which come to the same conclusion (compare the above), given the same external conditions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The lazy definition:&lt;br&gt;&#xA;The concentrations of hydronium and hydroxide ions are identical. $$c(\ce{OH-})= c(\ce{H3O+})$$&lt;/li&gt;&#xA;&lt;li&gt;In terms of $\ce{pH}, \ce{pOH}, \mathrm{p}K_w$ one arrives at a more complete side.&lt;br&gt;&#xA;Pure water is a neutral solution. It's autoprotolysis provides &#xA;\begin{align}&#xA;\ce{2H2O &amp;amp;~&amp;lt;=&amp;gt;~ OH- + H3O+}.\tag{1}\\&#xA;\end{align}&#xA;Now we can formulate the &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilibrium constant&lt;/a&gt; and furthermore assume that autoprotolysis is small compared to the overall concentration of water.&#xA;\begin{align}&#xA;  K_c &amp;amp;= \frac{c(\ce{OH-})\cdot c(\ce{H3O+})}{c^2(\ce{H2O})}\\&#xA;  K_w &amp;amp;= c(\ce{OH-})\cdot c(\ce{H3O+})\tag{2}\\&#xA;\end{align}&#xA;Equation $(1)$ also provides the lazy definition. $$c(\ce{OH-})= c(\ce{H3O+})\tag3$$&#xA;We plug that into $(2)$ and we arrive at&#xA;$$c(\ce{H3O+})=\sqrt{K_w}\tag{4}$$&#xA;Consider the actual &lt;a href=&quot;http://goldbook.iupac.org/P04524.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition&lt;/a&gt; of &#xA;$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( m(\ce{H+}) \gamma_m(\ce{H+}) / m^\circ\right)\tag{5}$$ rewritten using concentrations&#xA;$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( c(\ce{H+}) \gamma_c(\ce{H+}) / c^\circ\right)\tag{5a}$$ &#xA;The identities used here are $a(\ce{H+})$ for the activity of a proton in aqueous solution, $\ce{H+ (aq)}$ and we will consider $\ce{H+ (aq) = H3O+}$ to be the same. Now we are going to assume, that the activity coefficient is $\gamma_c(\ce{H+})\approx 1$ for very diluted systems. Using the &lt;a href=&quot;http://goldbook.iupac.org/S05909.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard concentration&lt;/a&gt; $c^\circ$ we do not have to care about units.&#xA;$$\ce{pH} = −\lg\left( \frac{c(\ce{H+})}{c^\circ}\right)\tag{5b}$$&#xA;Now we can substitute in $(4)$ and we will have a nice definition of the $\ce{pH}$ of a neutral solution.&#xA;$$\ce{pH} = −\lg\left( \frac{\sqrt{K_w}}{c^\circ}\right)\tag{6}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The second formulation includes a very crucial point. The autoprotolysis $(1)$ is temperature, $T$, dependent, which is obvious following the definition of the &lt;a href=&quot;http://goldbook.iupac.org/S05915.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard equilibrium constant&lt;/a&gt;&#xA;$$ K^\circ = \exp\left\{\frac{-\Delta G^\circ}{\mathcal{R}T}\right\}\approx K_c.$$&#xA;Therefore also the ion product of water is temperature dependent, $K_w(T)$. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water#Dependence_on_temperature.2C_pressure_and_ionic_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; actually has a couple of values for the $\mathrm{p}K_w$. Reaction $(1)$ is endothermic. A higher temperature means providing energy, means a higher $\mathrm{p}K_w$. For example, in the human body, $37~^\circ\mathrm{C}$, this value is slightly higher than for the one we usually assume as the &quot;neutral point&quot; $\mathrm{p}K_w(25~^\circ\mathrm{C})=14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $(6)$ is more correct in the following form,&#xA;$$\ce{pH}(T) = −\lg\left( \frac{\sqrt{K_w(T)}}{c^\circ}\right)\tag{6a},$$&#xA;the neutral $\ce{pH}(T)$ therefore changes with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another note is that it is also pressure dependent, but that might take it a little too far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalence_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;equivalence point&lt;/a&gt; is usually defined for an arbitrary chemical reaction between an acid and a base, and it refers to the situation where there are stoichiometric quantities of acid and base are present.&#xA;$$\ce{AH + B &amp;lt;=&amp;gt;[\ce{H2O}] A^- + HB+}$$&#xA;This means, that at the equivalence point the following statement holds:&#xA;$$c_\text{initial}(\ce{HA}) = c_\text{final}(\ce{HB+}).$$&#xA;Therefore the $\ce{pH}$ at the equivalence point will only be governed by the reactions of $\ce{HB+}$ and might not be neutral.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The issue of solvent dependency is addressed by Fred Senese quite concise on &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/what-is-pH.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;antoine.frostburg.edu&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;pH is often used to compare solution acidities. For example, a solution of pH 1 is said to be 10 times as acidic as a solution of pH 2, because the hydrogen ion concentration at pH 1 is ten times the hydrogen ion concentration at pH 2. This is correct as long as the solutions being compared both use the same solvent. You can't use pH to compare the acidities in different solvents because the neutral pH is different for each solvent. For example, the concentration of hydrogen ions in pure ethanol is about 1.58 × 10-10 M, so ethanol is neutral at pH 9.8. A solution with a pH of 8 would be considered acidic in ethanol, but basic in water!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="41556" LastEditDate="2017-04-24T22:57:36.017" LastActivityDate="2017-04-24T22:57:36.017" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19815" PostTypeId="2" ParentId="19812" CreationDate="2014-11-21T11:38:35.003" Score="5" Body="&lt;h3&gt;Carboxyl&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The carboxyl group &lt;span class=&quot;math-container&quot;&gt;$\ce{-COOH}$&lt;/span&gt; is an organic functional group consisting of a carbon atom double bonded to an oxygen atom and single bonded to a hydroxyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hb21K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hb21K.png&quot; alt=&quot;Carboxyl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds with a carboxyl group are called carboxylic acids (e.g. acetic acid, &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2COOH}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Carboxylate&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A carboxylate &lt;span class=&quot;math-container&quot;&gt;$\ce{-COO^-}$&lt;/span&gt; is a salt or ester of a carboxylic acid. For example, acetic acid reacts with sodium bicarbonate to form sodium acetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds with this group have the carboxy- prefix or -ate suffix (e.g. sodium acetate &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COONa}$&lt;/span&gt;). When these salts are ionised (e.g. dissolved in water), the anion with the carboxylate group is called a carboxylate ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carboxylate ion is the negative ion (anion) formed as the conjugate base when a carboxylic acid loses a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{R-COOH -&amp;gt; R-COO^- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being its conjugate base, sodium acetate mixed with acetic acid makes a useful buffer solution for many biochemical reactions where a 'mildly acidic' pH of 4-6 is desired.  &lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="41328" LastEditDate="2019-03-29T11:22:10.413" LastActivityDate="2019-03-29T11:22:10.413" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19818" PostTypeId="1" CreationDate="2014-11-21T13:31:35.860" Score="11" ViewCount="4597" Body="&lt;p&gt;My question is, how do we predict the products of a cross-Cannizzaro disproportionation in presence of a concentrated base? i.e, How to judge which of the aldehydes (both don't have any alpha hydrogen) is reduced and which one is oxidized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are asked &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-O-C6H4-CHO   + HCHO -&amp;gt;[OH^- (conc)] ?}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case: if we see acc to RDS, which is hydride transfer, we must see which compound is better hydride donor. Then it will be $\ce{CH3-O-C6H4-CHO}$. This will get oxidized, and $\ce{HCHO}$ reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case: if we see which is a better nucleophilic substrate, $\ce{HCHO}$ will get oxidised, and $\ce{CH3-O-C6H4-CHO}$, reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my problem is, which of the above cases is correct? i.e, do we see which one of the substrates is better $\ce{H-}$ donor, or which one of them is a better nucleophilic substrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please advise which case is correct, and how to approach such problems.&lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorUserId="7475" LastEditDate="2016-10-19T16:29:39.817" LastActivityDate="2021-03-16T16:29:28.337" Title="How to predict the outcome of a crossed Cannizzaro reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19819" PostTypeId="2" ParentId="19502" CreationDate="2014-11-21T14:03:36.017" Score="1" Body="&lt;p&gt;Now, I realize that silicon dioxide is a very stable compound. However, there is an exception which is it will react react with a thing- Hydrofluoric acid ($\ce{HF}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is undergo like the equation below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4HF(g) + SiO2(s) -&amp;gt; SiF4(g) + 2H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only $\ce{HF}$ can react with sand ($\ce{SiO2}$), others chemical can't as sand is really that stable.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="17368" LastEditDate="2017-02-01T02:43:56.293" LastActivityDate="2017-02-01T02:43:56.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19820" PostTypeId="2" ParentId="19784" CreationDate="2014-11-21T14:26:50.660" Score="9" Body="&lt;p&gt;I work at CAS and recently answered a similar question on Twitter (@CASChemistry). In fact, there are 8 pre-50-00-0 CAS Registry Numbers in our database:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;35-66-5&lt;/li&gt;&#xA;&lt;li&gt;35-67-6&lt;/li&gt;&#xA;&lt;li&gt;36-51-1&lt;/li&gt;&#xA;&lt;li&gt;36-88-4&lt;/li&gt;&#xA;&lt;li&gt;37-71-8&lt;/li&gt;&#xA;&lt;li&gt;37-82-1&lt;/li&gt;&#xA;&lt;li&gt;37-87-6&lt;/li&gt;&#xA;&lt;li&gt;38-26-6&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the lowest CAS Registry Number on record is &lt;strong&gt;35-66-5&lt;/strong&gt; for &quot;Benzacridine, 7,​9-​dimethyl- (9CI)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9938" LastEditorUserId="5026" LastEditDate="2018-02-22T11:12:20.767" LastActivityDate="2018-02-22T11:12:20.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19821" PostTypeId="2" ParentId="19818" CreationDate="2014-11-21T15:12:12.797" Score="1" Body="&lt;p&gt;The Cannizzaro reaction, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position. The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that your fist compound is an isomer of methoxybenzaldehyde, but the other compound, &quot;hcho&quot; is... Formaldehyde?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methoxybenzaldehyde would disproportionate in their corresponding carboxylic acid and alcohol if a base were present, which is not, so it will not undergo Canizzaro disproportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if formaldehyde can undergo Canizzaro reaction to yield methanol plus formic acid, but a base would be surely needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you talk about a crossed Cannizzaro reaction, in which an aldehyde is used in combination with a more valuable chemical. In this variation, the reductant is formaldehyde, which is oxidized to sodium formate, and the other aldehyde is transformed into the correspondent alcohol. But you also need base.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-21T15:12:12.797" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19822" PostTypeId="1" CreationDate="2014-11-21T16:00:39.620" Score="-1" ViewCount="1825" Body="&lt;p&gt;This is about the ideal gas equation $$pV=nRT$$ to plot ${p}\over{T}$ against $T$, where ${nR}\over{V}$ is constant. How would this graph look like?&lt;/p&gt;&#xA;" OwnerUserId="7198" LastEditorUserId="4945" LastEditDate="2018-12-18T18:34:11.637" LastActivityDate="2018-12-18T18:38:48.560" Title="What is the temperature dependency of p/T in the ideal gas?" Tags="&lt;gas-laws&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2015-02-25T13:33:27.337" ContentLicense="CC BY-SA 4.0" />
  <row Id="19823" PostTypeId="1" CreationDate="2014-11-21T16:51:14.100" Score="1" ViewCount="85" Body="&lt;p&gt;Cation exchange capacity is often regarded as the single best summary statistic for soil fertility.  Years ago, I learned how to measure it myself, in a soil lab.  I used compulsive barium exchange, whereby I knocked off all of the cations using barium, and then I somehow (I forget) got all the barium off, before measuring how much barium is in solution.  It makes sense as a means of measuring CEC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm now in a situation where I need hundreds of soil samples analyzed, primarily for CEC.  The lab with the best price does the CEC measurements by IR spectroscopy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:  how can you measure the potential of a soil to exchange cations using IR spectroscopy?  It's based on correlations, right?  Under what conditions does this method break down?  I assume the prediction error is non-constant over the multivariate space comprising soil properties.  In what regions of this space should I expect greater bias and/or variance in the prediction, and what would the sign of such bias be?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in simpler terms, when do compulsive barium exchange and IR spectroscopy agree and when do they disagree?&lt;/p&gt;&#xA;" OwnerUserId="1570" LastActivityDate="2014-11-21T16:51:14.100" Title="Cation exchange capacity via IR spectroscopy? Is it as good as barium exchange? What is the intutition on how it works?" Tags="&lt;analytical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19825" PostTypeId="2" ParentId="19822" CreationDate="2014-11-21T17:08:44.080" Score="2" Body="&lt;p&gt;A straight line parallel to temperature axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be of the form &lt;span class=&quot;math-container&quot;&gt;$y = c$&lt;/span&gt;,  where &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is a constant, mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if you take any value of &lt;span class=&quot;math-container&quot;&gt;$\frac{p}{T}$&lt;/span&gt; it will always give same value, because &lt;span class=&quot;math-container&quot;&gt;$\frac{nR}{V}$&lt;/span&gt; is a constant.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="4945" LastEditDate="2018-12-18T18:38:48.560" LastActivityDate="2018-12-18T18:38:48.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19826" PostTypeId="2" ParentId="19822" CreationDate="2014-11-21T17:25:52.493" Score="2" Body="&lt;p&gt;Firstly, rearrange the equation to get &lt;span class=&quot;math-container&quot;&gt;$\frac{p}{T}$&lt;/span&gt; in the left hand side:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{p}{T} = \frac{nR}{V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the question says, &lt;span class=&quot;math-container&quot;&gt;$\frac{nR}{V}$&lt;/span&gt; is always a constant. So, &lt;span class=&quot;math-container&quot;&gt;$\frac{p}{T}$&lt;/span&gt; is also a constant. In other words, I double the temperature, I double the pressure as well, still giving me &lt;span class=&quot;math-container&quot;&gt;$\frac{p}{T}$&lt;/span&gt;, to make sure &lt;span class=&quot;math-container&quot;&gt;$\frac{nR}{V}$&lt;/span&gt; is always constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one were to draw graph, because is always one value, even if we change the value of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, this has to mean that a graph of &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{p}{T}\right)$&lt;/span&gt; against &lt;span class=&quot;math-container&quot;&gt;$(T)$&lt;/span&gt;, will give a straight line graph with gradient &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;, and intersecting the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis at magnitude &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{nR}{V}\right)$&lt;/span&gt;, since &lt;span class=&quot;math-container&quot;&gt;$\frac{p}{T}$&lt;/span&gt; remains constant.&lt;/p&gt;&#xA;" OwnerUserId="9926" LastEditorUserId="4945" LastEditDate="2018-12-18T18:37:39.347" LastActivityDate="2018-12-18T18:37:39.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19827" PostTypeId="2" ParentId="19805" CreationDate="2014-11-21T17:33:06.010" Score="3" Body="&lt;p&gt;There are different factors that affect the re-activity of any element. These are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The atomic radius, or the distance of the outermost shell from the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;The nuclear charge (atomic number), because the greater the number of protons, the greater the pull on the outermost shell.&lt;/li&gt;&#xA;&lt;li&gt;The electron shielding due to the repulsion of electrons from the inner shells, pushing the outermost shells away from the nucleus.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some factors can 'overpower' others depending on the position in the Periodic Table.&lt;/p&gt;&#xA;" OwnerUserId="9926" LastActivityDate="2014-11-21T17:33:06.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19828" PostTypeId="2" ParentId="19808" CreationDate="2014-11-21T17:40:04.760" Score="0" Body="&lt;p&gt;Simple contradiction is salt hydrolysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutralization Reaction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{CH3COOH +NaOH-&amp;gt;CH3COONa + H2O}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now the following reaction happen due to low acidity of the acid, thus the reversible reaction a little favored:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{CH3COO- + H2O-&amp;gt;CH3COOH + OH-}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though the reaction reaches equivalence/neutralization point, the solution will be little basic, as even though none, hydronium or hydroxyl ions, is in excess but one may become in excess with the advent of salt hydrolysis(kinda in situ).&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-21T17:40:04.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19829" PostTypeId="2" ParentId="19822" CreationDate="2014-11-21T17:43:17.233" Score="2" Body="&lt;p&gt;See the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$PV = nRT \implies \frac PT = \frac{nR}{V} = \text{const.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So whatever you do with &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P/T$&lt;/span&gt; won't change, so it would be a constant horizontal line.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2018-12-18T18:34:50.583" LastActivityDate="2018-12-18T18:34:50.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19830" PostTypeId="2" ParentId="19818" CreationDate="2014-11-21T17:51:01.090" Score="3" Body="&lt;h2&gt;Canizzaro Reaction&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A +B-&amp;gt;[\text{conc. OH}^-]A' +B'}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If A has more partial positive charge, equivalently better hydride donor or more reactive, it will reduce and B will oxidise. Exception is formaldehyde where it will always get oxidised to formate as it's hydrate is more stable in aqueous medium.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="56849" LastEditDate="2021-03-16T16:29:28.337" LastActivityDate="2021-03-16T16:29:28.337" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19832" PostTypeId="2" ParentId="19757" CreationDate="2014-11-21T18:16:48.113" Score="1" Body="&lt;p&gt;What you are worried seems to be in your comment, which I'm explaining[maybe not]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{NiOOH + H2O + e- -&amp;gt; Ni(OH)2 + OH-}	\quad E^{\circ}=+0.49\\&#xA;\ce{Ni2O3(s) + 6H+ + 2e- -&amp;gt; 2Ni^2+ + 3H2O}	\quad E^{\circ}=+1.753\\$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose if you know about electrode potential you might be able to understand now,If you wish I might explain it, if you can't on your own.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-21T18:16:48.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19834" PostTypeId="1" AcceptedAnswerId="50788" CreationDate="2014-11-21T19:09:43.010" Score="5" ViewCount="161" Body="&lt;p&gt;The ozonides of $\ce{Cs}$, $\ce{Rb}$, $\ce{K}$ are well known and relatively stable, but there is little mention of the ozonides of $\ce{Na}$, $\ce{Li}$. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="7475" LastEditDate="2016-05-07T10:50:15.423" LastActivityDate="2016-05-07T10:50:15.423" Title="Why are the ozonides of heavier elements more stable?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19835" PostTypeId="2" ParentId="19799" CreationDate="2014-11-21T20:05:42.840" Score="2" Body="&lt;p&gt;The pKa of an acid tells you how strong the acid is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pKa=-log([H^+][A^-]/[HA])=-log(Ka)$ where $A$ is the conjugate base of the acid (the species left when the proton dissociates). A low value for $pKa$ gives the stronger acid since a high value for $Ka$ indicates that the equilibrium $\ce{HA&amp;lt;=&amp;gt;H+ + A-}$ lies well to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of factors that make an acid's equilibrium position lie far to the right. By far the most important is the stability of the conjugate base ($A^-$). In your example, the reason $\ce{H2SO4}$ so much more acidic than $\ce{CH3COOH}$ because is can form more resonance structures than ethanoic acid so it distributes the negative charge better. Also the inductive effect of the alkyl group destabilizes the negative conjugate base. There are more reasons but this is the most important. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-11-21T20:05:42.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19836" PostTypeId="1" AcceptedAnswerId="19841" CreationDate="2014-11-21T20:13:12.543" Score="8" ViewCount="17631" Body="&lt;p&gt;Xenon tetroxide is tetrahedral despite having double bonds. Is it something to do with $d$ orbitals? Also things like sulfur trioxide (trigonal planar with three double bonds but the sulfur is hypervalent). Is it possible for the $\pi$ bond to be formed from overlap of $d$ orbitals from the central atom with $p$ orbitals of the oxygen?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="9961" LastEditorDisplayName="user15489" LastEditDate="2018-07-01T19:37:54.107" LastActivityDate="2018-07-01T19:37:54.107" Title="How can compounds like XeO4 form tetrahedral structures despite having double bonds?" Tags="&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19838" PostTypeId="2" ParentId="19502" CreationDate="2014-11-21T20:32:30.810" Score="6" Body="&lt;p&gt;It really does react:&#xA;\begin{align}&#xA;\ce{4HF + SiO2 &amp;amp;-&amp;gt; SiF4 + 2 H2O}\\&#xA;\ce{SiF4 + 2 HF &amp;amp;-&amp;gt; H2[SiF6]}\\&#xA;\ce{SiO2 + 2 NaOH &amp;amp;-&amp;gt; Na2SiO3 + H2O}\\&#xA;\ce{SiO2 + 2 F2 &amp;amp;-&amp;gt; SiF4 + O2}\\&#xA;\ce{SiO2 + 2 C &amp;amp;-&amp;gt; Si + 2 CO}\\&#xA;\ce{(2,1)Na2O + (1,1)SiO2 &amp;amp;-&amp;gt; Na4SiO4/Na2SiO3}\\&#xA;\ce{SiO2 + Si &amp;amp;-&amp;gt; 2 SiO}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2018-02-13T05:27:45.793" LastActivityDate="2018-02-13T05:27:45.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19839" PostTypeId="2" ParentId="19836" CreationDate="2014-11-21T20:53:32.923" Score="0" Body="&lt;p&gt;Xenon is $[\ce{Kr}]4d^{10}5s^25p^6$ which, in its excited state, becomes $[\ce{Kr}]4d^{10}5s^15p^35d^4$. Now the $s$ and $p$ mix to form the hypothetical $sp^3$ hybridised orbitals which form $\sigma$-bonds with each electron in $p$ orbital of oxygen. The rest $4$ electrons in $d$ orbital form $\pi$-bonds with each electron in $p$ orbital of oxygen. (Oxygen has two unpaired electrons in $p$ orbital.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{SO3}$, sulfur excites to $[Ne]3s^13p^33d^2$ where $sp^2$ forms $\sigma$-bonds and the remaining three orbitals, the $\pi$-bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\pi$-bonds are ignored in determining the shape of the molecule. The $\pi$-bonds shorten the bond lengths but do not affect the shape. This approach explains the $\sigma$-bonding and the shape of the molecule, but the explanation of $\pi$-bonding is unsatisfactory. This is better explained by MOT (Molecular Orbital Theory.)&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="402" LastEditorDisplayName="user15489" LastEditDate="2017-05-22T04:22:17.183" LastActivityDate="2017-05-22T04:22:17.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19840" PostTypeId="2" ParentId="19797" CreationDate="2014-11-21T20:57:58.677" Score="2" Body="&lt;p&gt;A concoction &lt;a href=&quot;http://www.apifishcare.com/product.php?id=648&quot; rel=&quot;nofollow&quot;&gt;quite like this&lt;/a&gt; (with Fe-EDTA complex instead of iron sulphate) is sold as aquarium fertilizer.  Dosed in small amounts it would at least promote plant growth...&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-21T20:57:58.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19841" PostTypeId="2" ParentId="19836" CreationDate="2014-11-21T21:15:52.423" Score="15" Body="&lt;p&gt;$\ce{XeO4}$ has a tetrahedral structure as you noted, but $d$-orbitals are not involved in explaining its geometry or hybridization. The central xenon atom in $\ce{XeO4}$ is best described as simply $\ce{sp^3}$ hybridized.  We have 8 xenon electrons so we need 4 orbitals to put them in.  The 4 equivalent orbitals are formed by mixing the $s$- and $p$-orbitals (just as we do with carbon) to form the equivalent $\ce{sp^3}$ orbitals pointing towards the vertices of a tetrahedron (which minimizes electron-electron repulsions between the 4 bonds). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule can be represented by several resonance structures and I've drawn two of them (A and B) below.  I've only drawn the lone pairs on one of the oxygen atoms for each resonance structure, but they are the same for all oxygens in each resonance structure.  So the resonance structures tell us that the xenon-oxygen bonds in $\ce{XeO4}$ are some mix of single and double bond character.  In resonance structure A, the xenon and oxygens are neutral; while in resonance structure B, the xenon has a charge of +4 and each oxygen has a charge of -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AAkKy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AAkKy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{Xe-O}$ bonds in $\ce{XeO4}$ are very weak. In resonance structure A the pi overlap is very poor due to the different atomic sizes of xenon and oxygen. In resonance structure B we just have a simple single bond connecting the oxygen and xenon and it contains 2 electrons, both from xenon and shared by oxygen, oxygen does not contribute any electrons to this bond. The single bond resonance structure B probably contributes more to the description of this molecule than resonance structure A. The bonds are so weak that $\ce{XeO4}$ is unstable and explosively decomposes around -36 C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary we can say that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the xenon in $\ce{XeO4}$ is $\ce{sp^3}$ hybridized&lt;/li&gt;&#xA;&lt;li&gt;the $\ce{Xe-O}$ bond is a partial single-double bond with the single bond probably being a more accurate description, and &lt;/li&gt;&#xA;&lt;li&gt;the $\ce{Xe-O}$ bonds are extremely weak because of the poor pi overlap in resonance contributor A; and oxygen does not contribute any electrons to the xenon-oxygen bond in resonance contributor B.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditorDisplayName="user15489" LastEditDate="2017-01-31T01:31:34.530" LastActivityDate="2017-01-31T01:31:34.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19842" PostTypeId="1" CreationDate="2014-11-21T21:29:41.753" Score="3" ViewCount="4554" Body="&lt;p&gt;I 've a soft question on the terminology of compressibility factor. It is termed compressible factor because, its value determines how any reals gas is compressible with respect to an ideal gas according as its value-less than or greater than 1. If $Z&amp;gt;1$ then $\frac{pV}{nRT}&amp;gt;1$ and so $p&amp;gt;\frac{nRT}{V}$ which means it is more compressible than ideal gas. Because for ideal gas it is $p = \frac{nRT}{V}$. Simillar for negative deviation. But the question arises why we have not taken $V$ on LHS in these equations and inequations. I mean why does scientists doesn't named this as something like &quot;volumetric factor&quot; (actually can't find any specific word, but it is not important, hope readers can understand my question.) I mean something like Volume is greater than of ideal gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another one is we always have graphs of $Z \rightarrow p$ to understand the deviation in real gases from ideal gases. (I'm posting a graph for instance.)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/w35BT.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;br&gt;&#xA;Why we don't take $T$ or $V$ on the X-axis to understand this concept.&lt;/p&gt;&#xA;" OwnerDisplayName="user9942" LastActivityDate="2014-11-23T19:51:40.093" Title="Compressible Factor" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19843" PostTypeId="1" CreationDate="2014-11-21T23:02:21.697" Score="5" ViewCount="8335" Body="&lt;p&gt;All the carbon and oxygen atoms are sp2 hybridised and can have a p orbital in the correct plane yet only conjugation of the COO- group occurs. I can't think why. I can't draw resonance structures for the negative charge being delocalised around the ring but if you look at it from the orbital perspective (sp2 hybridisation) it seems clear that this conjugation should occur. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-11-21T23:25:54.147" Title="Why can't the conjugate base of benzoic acid be stabilized by conjugation with the aromatic ring?" Tags="&lt;acid-base&gt;&lt;orbitals&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19844" PostTypeId="2" ParentId="19843" CreationDate="2014-11-21T23:25:54.147" Score="4" Body="&lt;p&gt;Here are some of the resonance structures we can draw for the carboxylate anion of benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v0Y4j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that we can't delocalize the negative charge from the oxygen into the benzene ring.  Even though there is continuous p-orbital overlap between the oxygen and benzene carbons, there is just no way to push the electrons so that a negative charge winds up in the benzene ring. However, as the resonance structures show, we can move electrons the other way and place a second negative charge on the oxygens and leave a positive charge delocalized around the ortho and para positions in the benzene ring.  This is why the carboxylate anion of benzoic acid will not react with electrophiles (electrophilic aromatic substitution) at those positions in the aromatic ring.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-21T23:25:54.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19845" PostTypeId="2" ParentId="485" CreationDate="2014-11-22T01:42:52.683" Score="3" Body="&lt;p&gt;Just browsing through this &quot;Unanswered Question&quot; now (in late 2014). I hope it's not too late?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In protein structure validation, Ramachandran outliers are a strong, but not absolute, indicator that the backbone confirmation of your protein structure deviates from all the high resolution structures known to date, and hence need your attention during the progress of refinement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the phi-psi angles (backbone conformation) all fall into the &quot;allowed/favoured&quot; regions on the Ramachandran plot, your structure model is probably, but not necessarily, esp in the case of low resolution, correct. Hence there are other weaker criteria in protein structure validation, and Cbeta deviation is just one of them. In the light of high resolution structures out there, it turns out that Cbeta also has a favoured range of dihedral angles, suggesting that any misfit between your structure and the &quot;probable&quot; side chain-backbone conformation also deserves your attention during structure refinement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MolProbity, the justification and validation procedure are described &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/12557186&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9933" LastActivityDate="2014-11-22T01:42:52.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19846" PostTypeId="1" AcceptedAnswerId="19850" CreationDate="2014-11-22T03:09:14.597" Score="0" ViewCount="2918" Body="&lt;p&gt;I have trouble to predict the effect of decreasing pressure and increasing temperature on the reaction $\ce{N2 + 2O2 &amp;lt;=&amp;gt; 2NO2}$. &lt;br /&gt;Increasing temperature caused this reaction to shift in the forward direction and decreasing pressure brought no change. &lt;br/&gt;I know increasing temperature favour endothermic reaction and decreasing temperature favour exothermic reaction. I also know on increasing pressure the system will shift to the direction where less moles are present.&lt;br /&gt; Please tell me how to use this information to predict the effect for increasing temperature and decreasing pressure.&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="10185" LastEditDate="2014-12-24T14:03:21.433" LastActivityDate="2014-12-24T14:03:21.433" Title="Trouble to predict the effect of pressure and temperature for a reaction in equilibrium" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19848" PostTypeId="1" AcceptedAnswerId="19855" CreationDate="2014-11-22T03:52:03.267" Score="0" ViewCount="171" Body="&lt;p&gt;Does the presence of UV light, or any radiation for that matter, affect the process of fermentation? I'm not sure, but I seem to recall somebody telling me before that yeast will not function properly, but I can't seem to find it anywhere. Thanks for any help. &lt;/p&gt;&#xA;" OwnerUserId="9799" LastActivityDate="2014-11-22T05:46:03.823" Title="Fermentation in the presence of UV light" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19849" PostTypeId="2" ParentId="19305" CreationDate="2014-11-22T04:26:29.803" Score="5" Body="&lt;p&gt;I would like to clarify this Q &amp;amp; A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Fe(II) and Fe(III) are found in metalloproteins that are essential to life. Fe(II) can serve as an electron donor, and Fe(III) an electron acceptor in life.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;So is iron the only element that can do that?&quot; No, for electron transfer, many biological transition metals (Co, Ni, Cu...) can also participate in redox reactions in life (but not Mg(II) in chlorophyll)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, Fe(II) in hemoglobin is not a good example for &quot;electrons donor/acceptor&quot;. Although the oxidative state of Fe(II) in heme changes slightly when binding to O2 or CO, it never oxidizes to Fe(III). (If it does, it loses its ability to bind O2). The ferrous ion in hemoglobin is important in oxygen transport.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;So is iron the only element that can do that?&quot; No, for oxygen transport, copper ion in haemocyanin can also transport oxygen in the blood of spiders, crabs and octopuses. When oxygenated, they are blue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;What is it about being able to gain and lose electrons make it essential to life?&quot; Because biological redox reactions (e.g. cellular respiration) are essential to life. Each biological metal is important in its own specific biological function. Their roles are the product of evolution. For oxygen transport, neither heme nor iron is indispensable for that function in other life forms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Iron is abundant in earth crust mainly as iron(III) oxides. However, iron is NOT abundant in bioavailability. Ferric and Ferrous hydroxides are so poorly soluble in the environment that many species have developed sophisticated iron acquisition systems such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Siderophore&quot; rel=&quot;noreferrer&quot;&gt;siderophore&lt;/a&gt; to fight for this &quot;scarce&quot; element as an essential nutrient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9933" LastActivityDate="2014-11-22T04:26:29.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19850" PostTypeId="2" ParentId="19846" CreationDate="2014-11-22T05:06:46.650" Score="0" Body="&lt;p&gt;When I posted this question I had no idea how changing each factor bring change in equilibrium. Then I bombarded my brain with everything I know about equilibrium and I found a fairly easy way to understand this.&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Change in concentration &lt;/b&gt; &lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two vessels A and B connected to each other through a thin hollow pipe through the middle of the vessel. Let vessel A be reactant and vessel B be product. The vessels are filled with water a little above the pipe. If I add water to vessel A  water flow from A to B. Thus increasing concentrated of A caused the water in vessel A to flow to B; which in other words is the forward reaction. &#xA;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Change in temperature&lt;/b&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a reaction which is in equilibrium.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Decreasing the temperature of reaction in equilibrium.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that decreasing the temperature favour exothermic reaction&#xA; and exothermic reaction are favoured by low temperature. Exothermic mean giving out heat which remove the deficiency of temperature to bring it back to normal in equilibrium.&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Increasing the the temperature of reaction in equilibrium&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know that increasing the temperature favour endothermic reaction or endothermic reaction are favoured by high temperature. Endothermic means to absorb heat energy which causes the excess heat energy ( from high temperature) to bring it back to normal in equilibrium&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Effect of pressure &lt;/b&gt;&lt;br /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing pressure play a significant role in gaseous reaction which are accompanied by change in moles. On increasing pressure of the system equilibrium will shift to the direction where number of moles decrease or where pressure decrease (because pressure is directly proportional to the number of moles). If pressure on the system is increased equilibrium shift in the direction in which decrease in number of moles take place.&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the second time I answered on stack exchange so please let me know my mistake if any. If I had known how to create the diagrams I usually find in other answers I would have added them.&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="8231" LastEditDate="2014-11-22T05:29:49.457" LastActivityDate="2014-11-22T05:29:49.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19852" PostTypeId="1" AcceptedAnswerId="19856" CreationDate="2014-11-22T05:42:23.573" Score="0" ViewCount="9103" Body="&lt;p&gt;What is the chemical composition of petrol? What are the molecular available in petrol? Some time which is marked with 95% octane. So what is the remain element occupied 5%?&lt;/p&gt;&#xA;" OwnerUserId="9946" LastEditorUserId="30836" LastEditDate="2020-03-31T15:34:46.593" LastActivityDate="2020-03-31T15:34:46.593" Title="Chemical composition of petrol" Tags="&lt;molecules&gt;&lt;petrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19853" PostTypeId="1" AcceptedAnswerId="19861" CreationDate="2014-11-22T05:42:37.783" Score="-2" ViewCount="747" Body="&lt;p&gt;I recently came across this question, I have no idea about IR:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following vibrational modes show no IR absorption bands?&lt;/p&gt;&#xA;&lt;p&gt;A) Symmetric CO2 stretch&lt;/p&gt;&#xA;&lt;p&gt;B) Antisymmetric CO2 stretch&lt;/p&gt;&#xA;&lt;p&gt;C) Symmetric S=C=O stretch&lt;/p&gt;&#xA;&lt;p&gt;D) Antisymmetric S=C=O stretch&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-19T12:02:37.377" Title="Which vibrational modes of CO2 and CSO do not lead to IR absorption bands?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-19T15:14:10.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="19854" PostTypeId="1" CreationDate="2014-11-22T05:44:58.883" Score="-2" ViewCount="585" Body="&lt;p&gt;I have no idea to which concept this belongs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The widest range over which electronic excitations in organic compounds occur, is&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) 200 nm – 780 nm&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;B) 220 nm– 500nm &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;C) 250 nm– 700 nm&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;D) 290 nm – 1000nm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="73" LastEditDate="2014-11-22T05:48:02.787" LastActivityDate="2014-11-22T08:22:06.273" Title="How can we measure electronic excitation in some &quot;nm&quot;(nanometres)?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19855" PostTypeId="2" ParentId="19848" CreationDate="2014-11-22T05:46:03.823" Score="2" Body="&lt;p&gt;Depends on the wavelength. Fermentation is basically metabolic activity by yeast and bacteria. UV-C (&amp;lt;280 nm) is pretty good at killing a lot of micro-organisms. Some species are more resistant than others, but it's safe to say that exposure to this kind of light will probably interfere with fermentation. Depending on the medium and how the light is applied (thickness of material, power, etc.), it may have more or less of an effect depending on how much light reaches the yeast or bacteria.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-22T05:46:03.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19856" PostTypeId="2" ParentId="19852" CreationDate="2014-11-22T06:06:38.040" Score="9" Body="&lt;p&gt;The composition of petrol, also called 'gasoline' or 'gas' in the USA, despite it being a liquid at room temperature, contains a complex mixture of hydrocarbon compounds selected for use in internal combustion engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;95% octane&quot; refers to the 'octane rating', 'octane number' or 'anti-knock index' of the fuel, not the amount of 'octane' in the fuel (since 'octane' is a specific organic compound in the group of alkanes).  The octane number / rating is an 'anti-knock index', based on an arbitrary scale indexed to a liquid mixture of iso-octane (2,2,4-trimethylpentane and n-heptane). Whilst different countries have different standards for anti-knock index, in the USA, the American Society for Testing and Materials (ASTM) defines two popular indices called the 'research octane number' (RON) and motor octane number (MON). Regardless if the specific stanard, the general principle is that the higher the octane number, the more the fuel can be 'compressed' before 'auto-igniting'. Auto-ignition is not desirable in petrol engines as it causes 'knocking' of the pistons which greatly shortens their life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What petrol (gasoline) actually contains is a blend of refinery products, including reformate (high in aromatic hydrocarbons, low in alkenes), olefinic hydrocarbons, alkylate (high in paraffinic hydrocarbons) and isomerate (high in branched isomers of pentane and hexane). Specific blends vary from manufacturer. Manufacturerers even offer a range of blends at the bowser (gas station) including 10% ethanol and 'premium' blends for high performance engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to '&lt;a href=&quot;http://www.utexas.edu/research/ceer/che341/Projects/octane%20number.pdf&quot; rel=&quot;noreferrer&quot;&gt;Development of a Detailed Gasoline Composition-Based Octane Model&lt;/a&gt;' for more information.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="4934" LastEditDate="2014-11-22T15:13:42.937" LastActivityDate="2014-11-22T15:13:42.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19857" PostTypeId="1" CreationDate="2014-11-22T06:20:10.420" Score="4" ViewCount="7573" Body="&lt;p&gt;Why do square planar coordination compounds of type $\ce{[Mabcd]}$ not show optical activity, although they contain 4 different ligands (i.e. chiral central metal atom)?&lt;/p&gt;&#xA;" OwnerUserId="9947" LastEditorUserId="5026" LastEditDate="2018-04-23T01:28:28.913" LastActivityDate="2018-07-30T10:53:24.960" Title="Why are square planar coordination compounds with four different ligands optically inactive?" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;coordination-compounds&gt;&lt;symmetry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19858" PostTypeId="1" AcceptedAnswerId="19871" CreationDate="2014-11-22T07:01:22.163" Score="3" ViewCount="213" Body="&lt;p&gt;I know when electrolysis on a concentration aqeuous solution which contains halide ions($Cl^-$ , $Br^-$ , $I^-$). The halide ions is discharged although $OH^-$ is more easily to discharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, electrolysis of concentration aqueous sodium chloride. The cation $Cl^-$ and $OH^-$ will attract to the anode. The $Cl^-$ ions will be discharged although according to the electrochemical series that the $OH^-$ will be more easily discharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do that happen?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2014-11-22T13:08:15.490" LastActivityDate="2014-11-22T13:41:37.243" Title="Electrolysis on aqeuous solution" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19859" PostTypeId="1" AcceptedAnswerId="19864" CreationDate="2014-11-22T07:03:34.997" Score="8" ViewCount="1052" Body="&lt;p&gt;What exactly does fluorination of some compound do in a biological sense? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the following things about what fluorine does for drugs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fluorine withdraws electron density and can make an acid stronger. &lt;/li&gt;&#xA;&lt;li&gt;Fluorine substitution can increase lipophilicity - I don't understand this. Lipids are non-polar; I can't see why adding fluorine would generally decrease the polarity of some molecule ... &lt;/li&gt;&#xA;&lt;li&gt;Carbon-fluorine bonds are exceptionally stable and resist change.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, can anyone provide some concrete examples of fluorine influencing the behavior of some drug? I could only find very general points on the internet.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-11-22T10:20:48.280" Title="Why do many pharmaceuticals contain fluorine?" Tags="&lt;biochemistry&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19860" PostTypeId="2" ParentId="19854" CreationDate="2014-11-22T08:22:06.273" Score="1" Body="&lt;p&gt;The question is asking what wavelengths of light can cause electronic excitation in organic compounds, i.e. what energy photons correspond to the energy difference between the ground and excited states. They're not the same in all compounds (which is why different compounds have different absorbance spectra), but there is a range of energies in which most of the electronic transitions in organic compounds occur.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-22T08:22:06.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19861" PostTypeId="2" ParentId="19853" CreationDate="2014-11-22T09:27:50.000" Score="5" Body="&lt;p&gt;A) Symmetric $\ce{CO2}$ stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular vibrational mode is considered IR active, if there is a change of &lt;em&gt;transition dipole moment&lt;/em&gt;. $\ce{CO2}$ is linear, and the symmetric stretching mode means both oxygens move in/outward concurrently.&#xA;You can see an illustration of the symmetric and antisymmetric stretching modes of $\ce{CO2}$ here: &lt;a href=&quot;https://www.youtube.com/watch?v=TMLnUmbLwUI&amp;amp;t=1m50s&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=TMLnUmbLwUI&amp;amp;t=1m50s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the symmetric mode, the overall center of charge stays the same, thus no change in dipole moment, thus the mode is not IR active.&lt;/li&gt;&#xA;&lt;li&gt;For the antisymmetric mode, the center of charge shifts, thus the dipole moment changes and the mode is IR active. This is also true of the $\ce{S=C=O}$ antisymmetric stretch.&lt;/li&gt;&#xA;&lt;li&gt;The $\ce{S=C=O}$ symmetric stretching mode is also IR active, since $\ce{S}$ and $\ce{O}$ are two different atoms, with different partial charges. As they vibrate, the total center of charge moves.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9949" LastEditorDisplayName="user15489" LastEditDate="2015-04-19T12:02:37.377" LastActivityDate="2015-04-19T12:02:37.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19862" PostTypeId="2" ParentId="19842" CreationDate="2014-11-22T09:43:11.773" Score="3" Body="&lt;p&gt;The compressibility factor measures the ratio of actual volume of the gas (under the given conditions of temperature and pressure) to the volume the gas must have occupied if it were an ideal gas (under the same temperature and pressure). This ratio tells us how much compressible is a gas (more compressible if the volume is lesser) by a given pressure applied on it. At this point, we tend to take the temperature as constant and assume the volume is a function of pressure alone. Then, suppose changing the pressure and observe how the volume changes for the gas. This is what the graph in your question shows. It illustrates the variation in compressibilities of different real gases with pressure. The term as well as the graph makes sense in that they show the willingness of a given real gas to get compressed upon subjecting them to various pressures.&lt;/p&gt;&#xA;" OwnerUserId="9950" LastActivityDate="2014-11-22T09:43:11.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19863" PostTypeId="2" ParentId="7683" CreationDate="2014-11-22T10:04:27.587" Score="6" Body="&lt;p&gt;I analysed the problem in different manner . Basicity implies tendency of protonation. In the first case methyl group exert steric hindrance and avoid protonation which in turn makes it less basic as compared on aniline. On the other hand in second case let us consider the acidic strength of ortho substituted and non substituted nearly equal but due to steric hindrance again tdency to loose electrons is more in unsubstituted making it less acidic as compared to substituted.Thus overall it is supporting acidic nature.&lt;/p&gt;&#xA;" OwnerUserId="9951" LastActivityDate="2014-11-22T10:04:27.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19864" PostTypeId="2" ParentId="19859" CreationDate="2014-11-22T10:20:48.280" Score="15" Body="&lt;p&gt;A definitive reference is:&#xA;&quot;The Many Roles for Fluorine in Medicinal Chemistry&quot; J. Med. Chem., &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;51&lt;/em&gt; (15), 4359–4369.&#xA;In reference to the bullet points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fluorine obviously affects $\ce{p}{K_{a}}$ and can make an acid stronger but also a base weaker which could be advantageous when optimising a drug candidate (lead) molecule.  &lt;/li&gt;&#xA;&lt;li&gt;I guess here they refer to $\ce{-CF_3}$ which can increase lipophilicity  &lt;/li&gt;&#xA;&lt;li&gt;That's one of the most useful roles of fluorine. A lot of times you may have a very potent and selective molecule, but when you test it &lt;em&gt;in vivo&lt;/em&gt; let's say in rats, it has no effect because it gets metabolised i.e. chemically modified to an inactive molecule by the rats liver (and other mechanisms). This can maybe prevented by changing to an $\ce{C-F}$ bond. For example electron rich aromatic sites are prone to metabolic oxidation. You can block this by fluorination of this site, exchanging the proton for a fluorine.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Have a look at the paper on top, you will find a new world of information.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2014-11-22T10:20:48.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19865" PostTypeId="1" AcceptedAnswerId="19866" CreationDate="2014-11-22T09:08:15.513" Score="0" ViewCount="121" Body="&lt;p&gt;Normally octane percentage is mentioned on petrol (95% octane). How is this defined? Is it percentage of chemical or some other parameters?&lt;/p&gt;&#xA;" OwnerUserId="9946" OwnerDisplayName="sugunan" LastEditorUserId="30836" LastEditDate="2019-04-14T11:43:48.657" LastActivityDate="2019-04-14T11:43:48.657" Title="How is the octane percentage defined?" Tags="&lt;fuel&gt;&lt;petrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19866" PostTypeId="2" ParentId="19865" CreationDate="2014-11-22T10:37:30.240" Score="3" Body="&lt;p&gt;See the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Octane_rating&quot; rel=&quot;nofollow&quot;&gt;octane rating&lt;/a&gt;. 95% octane is the same as an octane rating of 95, and it means the petrol has the same knocking characteristics as a mixture of 95% iso-octane and 5% heptane.&lt;/p&gt;&#xA;" OwnerUserId="1470" OwnerDisplayName="John Rennie" LastActivityDate="2014-11-22T10:37:30.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19867" PostTypeId="1" CreationDate="2014-11-22T12:26:08.413" Score="4" ViewCount="2159" Body="&lt;p&gt;My mom is using a stone like substance as a dye for hair recommended by a herbalist. That stone becomes black when exposed to air. Do you know which compound is this?&lt;/p&gt;&#xA;" OwnerUserId="9952" LastEditorUserId="7159" LastEditDate="2014-11-22T16:32:20.067" LastActivityDate="2020-01-18T01:28:27.697" Title="stone which turns to black on exposing to atmosphere?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19868" PostTypeId="1" AcceptedAnswerId="19869" CreationDate="2014-11-22T12:37:07.987" Score="7" ViewCount="19413" Body="&lt;p&gt;Is benzene carbaldehyde a correct name for Benzaldehyde? I'm (always) getting confused between common and IUPAC names. If the above name is correct then is it a common one or IUPAC? And why don't we use it more often? &lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="5026" LastEditDate="2018-02-04T03:57:48.390" LastActivityDate="2018-02-04T03:57:48.390" Title="When is the carbaldehyde suffix correct?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19869" PostTypeId="2" ParentId="19868" CreationDate="2014-11-22T13:24:32.447" Score="4" Body="&lt;p&gt;Both are IUPAC names, but used in different situations. For a 1 carbon aldehyde attached to a benzene, the one that must be used is benzaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aldehydes ($\ce{R-CHO}$) take the suffix &quot;-al&quot;. If other functional groups are present, the chain is numbered such that the aldehyde carbon is in the &quot;1&quot; position, unless functional groups of higher precedence are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a prefix form is required, &quot;oxo-&quot; is used (as for ketones), with the position number indicating the end of a chain: $\ce{CHOCH2COOH}$ is 3-oxopropanoic acid. If the carbon in the carbonyl group cannot be included in the attached chain (for instance in the case of cyclic aldehydes), the prefix &quot;formyl-&quot; or the suffix &quot;-carbaldehyde&quot; is used: $\ce{C6H11CHO}$ is cyclohexanecarbaldehyde. &lt;strong&gt;If an aldehyde is attached to a benzene and is the main functional group, the suffix becomes benzaldehyde.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this guy would be cyclohex-2-ene-1-carbaldehyde:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9mf50.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="9850" LastEditDate="2014-11-22T13:29:57.727" LastActivityDate="2014-11-22T13:29:57.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19871" PostTypeId="2" ParentId="19858" CreationDate="2014-11-22T13:41:37.243" Score="3" Body="&lt;p&gt;Let's explain this using an example. Consider a cell in which an aqueous solution of sodium chloride is electrolyzed. The $\ce{Na^+}$ ions migrate toward the negative electrode and the $\ce{Cl^-}$ ions migrate toward the positive electrode. But indeed at the anode, there are  two substances that can be oxidized at the anode: $\ce{Cl^-}$ ions and &lt;strong&gt;water molecules&lt;/strong&gt;.&#xA;$$\ce{2Cl^-   \rightarrow Cl_2 + 2e^-}$$ $$\ce{2H_2O   \rightarrow O_2 + 4H^+ + 4e^-}$$ &#xA;The standard-state potentials for these half-reactions are $E^0_{ox}= -1.36 V$ for the first couple and  $E^0_{ox}= -1.23 V$ for the second couple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, it would seem easier to oxidize water ($E^0_{ox}= -1.23 V$) than $\ce{Cl^-}$ ions ($E^0_{ox}= -1.36 V$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% $\ce{NaCl}$ by mass, which significantly decreases the potential required to oxidize the $\ce{Cl^-}$ ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as &lt;strong&gt;overvoltage&lt;/strong&gt;, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to $\ce{O_2}$ gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-22T13:41:37.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19872" PostTypeId="2" ParentId="19867" CreationDate="2014-11-22T15:32:38.493" Score="7" Body="&lt;p&gt;There is a compound called cerussite or 'natural white lead' which turns to black 'lead sulphide' when it reacts with hydrogen sulphide in the atmosphere. It was commonly used as make up by women in medieval times and is sometimes still used as a &lt;a href=&quot;http://www.naturalpigments.com/cerussite-natural-lead-white.html&quot; rel=&quot;noreferrer&quot;&gt;white pigment in 'lead paint'&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we all now know lead is toxic to many organs and tissues including the heart, bones, intestines, kidneys, and reproductive and nervous systems, as well as causing potentially permanent learning and behavior disorders, particularly in children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, it's use has diminished over time, except perhaps amongst New Age crystal 'healers' who claim it can &quot;infuse your aura with light&quot; and &quot;bring amazing change into your life&quot;, not only through a lowering of general intelligence, but, it can also help &lt;a href=&quot;http://www.rsc.org/chemistryworld/news/2010/january/14011002.asp&quot; rel=&quot;noreferrer&quot;&gt;prevent bacterial infections&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all this, you may be surprised to know that another toxic lead-based white crystalline substance called &lt;strong&gt;lead acetate&lt;/strong&gt; &lt;a href=&quot;http://www.fda.gov/Cosmetics/ProductsIngredients/Products/ucm143066.htm&quot; rel=&quot;noreferrer&quot;&gt;is approved by the FDA for use in 'progressive' hair dyes&lt;/a&gt;, where it reacts with the sulphur in hair proteins to slowly turn grey hair black, over several applications of course, so 'no-one' will notice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvRv7.jpg&quot; alt=&quot;image of lead(II) acetate crystals, courtesy of Wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;image of lead(II) acetate crystals, courtesy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead%28II%29_acetate&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: In case you were wondering why lead-based products are still used, given that there are many alternatives out there, many of which are labelled 'non-toxic', 'natural' and 'organic'...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the &lt;a href=&quot;http://www.cancer.gov/cancertopics/factsheet/Risk/hair-dyes&quot; rel=&quot;noreferrer&quot;&gt;National Cancer Institute reports&lt;/a&gt; that there are over 5000 different chemicals used in hair dyes, including toxic coal-tars, aromatic amines and phenols. Some of these are also carcinogenic (cancer-causing agents) and can be absorbed through the skin or inhaled. Whilst the industry has been working towards eliminating some of these chemicals since the late 1970's, people who regularly began using hair dyes before 1980 may be at increased risk of developing non-Hodgkin lymphoma (a form of blood cancer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://monographs.iarc.fr/ENG/Monographs/vol99/index.php&quot; rel=&quot;noreferrer&quot;&gt;International Agency for Research on Cancer (IARC)&lt;/a&gt; has said that &quot;whilst workplace exposure as a hairdresser or barber is probably carcinogenic to humans&quot;, personal hair dye use is “not classifiable as to its carcinogenicity to humans,” based on &quot;a lack of evidence from studies in people&quot;.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-11-22T15:32:38.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19874" PostTypeId="2" ParentId="19857" CreationDate="2014-11-22T16:57:29.967" Score="9" Body="&lt;p&gt;Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Chirality_%28chemistry%29#Symmetry&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, results in the same molecule[...] For tetrahedral molecules, the molecule is chiral if all four substituents are different.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So for 4-coordinated carbon atoms which are tetrahedral having four &amp;quot;ligands&amp;quot; actually establishes optical activity. But in the case of a square planar metal complex you always have a mirror plane, namely the one made up by the central atom and the ligands:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qo59Y.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;And as a result you don't get optical activity. But it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2014-11-22T16:57:29.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19875" PostTypeId="1" AcceptedAnswerId="19880" CreationDate="2014-11-22T17:33:32.800" Score="1" ViewCount="318" Body="&lt;p&gt;I know that the proton will attack the HOMO, but I'm having difficulty imagining how this will happen.&lt;/p&gt;&#xA;" OwnerUserId="9709" LastEditorUserId="23561" LastEditDate="2016-07-25T16:12:58.740" LastActivityDate="2016-07-25T16:12:58.740" Title="How is ethanenitrile protonated?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19876" PostTypeId="1" CreationDate="2014-11-22T17:44:20.667" Score="2" ViewCount="457" Body="&lt;p&gt;I am struggling with the concept of entropy but might be on the verge of understanding it. The Clausius inequality, $$\mathrm{d}S \geq \mathrm{d}q/T,$$ is true for all spontaneous changes with the equality for reversible changes. But what if a change is not spontaneous or is that impossible i.e a change cannot be forced. Any help with answering this question or to help conceptualize and understand the entropy would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-01-22T05:20:36.323" LastActivityDate="2015-01-22T05:20:36.323" Title="If dS ≥ dq/T for a spontaneous change is dS ≤ dq/T for a non-spontaneous change and is that even possible?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19877" PostTypeId="1" AcceptedAnswerId="19879" CreationDate="2014-11-22T17:54:09.863" Score="2" ViewCount="1923" Body="&lt;p&gt;Why are many exothermic reactions, like Haber process, conducted at high temperature, thus  their unfavourable equilibrium position?&lt;/p&gt;&#xA;" OwnerUserId="9733" LastActivityDate="2014-11-22T18:42:02.767" Title="Exothermic reactions at high temperature and the chemical equilibrium" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19878" PostTypeId="2" ParentId="19876" CreationDate="2014-11-22T18:23:31.103" Score="2" Body="&lt;p&gt;As you perfectly say, Clausius inequality states that $$\mathrm{d}s\geq \mathrm{d}q/T.$$ This has to be true for any process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reversible processes, what happens is that $\mathrm{d}s= \mathrm{d}q/T$, and for non-reversible processes, $\mathrm{d}s &amp;gt; \mathrm{d}q/T$. If for a given process $\mathrm{d}s&amp;lt; \mathrm{d}q/T$, it would not fulfill Clausius inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the situation where the system is isolated from its surroundings. In this case, $\mathrm{d}q_\mathrm{(sys)} = 0$ , and thus the Clausius relation implies $\mathrm{d}S\geq0$, so this tells us that the entropy of an isolated system cannot decrease in the course of a spontaneous change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the universe is itself an isolated system, this result shows that we can use entropy as the signpost of spontaneous change. Processes are only spontaneous if they cause an increase in the total entropy of the universe.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4945" LastEditDate="2015-01-22T02:19:56.023" LastActivityDate="2015-01-22T02:19:56.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19879" PostTypeId="2" ParentId="19877" CreationDate="2014-11-22T18:42:02.767" Score="3" Body="&lt;p&gt;There are two competing problems in the Haber Bosch process: the equilibrium position and the reaction rate. The reaction: $\ce{N_2 + 3H_2 -&amp;gt; 2NH3}\ (\Delta H = -92.4 \mathrm{\ kJ/mol})$ is exothermic and the equilibrium is actually favourable at room temperature, but the rate is very slow due to the large energy barrier to breaking the triple bond in $\ce{N2}$. To increase the reaction rate, a catalyst is used at high temperature. As you say, the high temperature pushes the equilibrium towards the reactants (by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;nofollow&quot;&gt;Van 't Hoff equation&lt;/a&gt;) so there is a competition between yield and rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, high pressure is also used to increase the yield (since the products side of the equation has fewer molecules), and like many industrial processes, it's run as a recirculating process. At manageable pressures and temperatures, they may only get 10 or 15% yield after one pass, but the ammonia is removed and the unreacted gases passed back into the reactor, giving quite high overall yield.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-22T18:42:02.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19880" PostTypeId="2" ParentId="19875" CreationDate="2014-11-22T19:52:10.537" Score="4" Body="&lt;p&gt;The structure of ethanenitrile is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C#N:}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualitatively, we can draw some conclusions about the molecular orbitals in the structure. If covalent bonds result in in the involved electrons being lower in energy, then which pair of electrons in the structure has the highest energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are looking at nitriles, you probably already know that both lone pairs and pi bonds are good proton acceptors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider what the difference is if the lone pair is the base or if a pi bond is the base:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lone Pair&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C#N + H+ -&amp;gt; H3C-C#N+-H}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pi Bond&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which atom will the proton end up attached to? Which of the two following carbocations is more stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C#N + H+ -&amp;gt; H3C-C+=NH}\ \  or\ \  \ce{H3C-CH=N+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... what is the relationship between the two conjugate acids drawn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C#N+-H}\ \ and\ \ \ce{H3C-C+=NH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the HOMO?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-11-22T19:52:10.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19881" PostTypeId="1" CreationDate="2014-11-22T20:09:17.003" Score="4" ViewCount="1563" Body="&lt;p&gt;Maybe this is just not included in my course, but I still wonder how to calculate the binding energy of an electron .at a certain electron orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively,The ionization energy of an element should vary depending on which orbital the valence electron is in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I only know how to calculate ionization energy using principle level.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-11-23T15:39:48.163" LastActivityDate="2018-12-21T10:01:52.217" Title="A way to calculate orbital energy of an electron" Tags="&lt;electronic-configuration&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19883" PostTypeId="1" AcceptedAnswerId="19894" CreationDate="2014-11-22T22:16:29.483" Score="7" ViewCount="609" Body="&lt;p&gt;In a paper that I'm reading, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The presence of chlorine atoms in polyvinyl chloride make it flame resistant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and I'm just wondering why the presence of chlorine atoms will make it flame resistant. To the best of my knowledge, burning something is just providing a material with enough activation energy such that it can react with oxygen to produce products such as carbon dioxide and water. If so why does the chlorine prevent this process in any way?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="5026" LastEditDate="2018-05-06T16:21:13.593" LastActivityDate="2018-05-06T19:59:25.727" Title="How does the presence of chlorine atoms in PVC make it flame resistant?" Tags="&lt;polymers&gt;&lt;combustion&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19884" PostTypeId="2" ParentId="15125" CreationDate="2014-11-22T22:52:37.900" Score="4" Body="&lt;p&gt;As for acidity of boric acid you can find on &lt;a href=&quot;https://en.wikipedia.org/wiki/Boric_acid&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Boric_acid&lt;/a&gt; that it dissociates in both ways, but is very weak. However it's acidic enough that it's only a matter of time until it would become almost completely deuterised if dissolved in D2O. It happens because in mixture there would be much more deuterium then normal hydrogen. The acid is too weak to capture or detect its anions. Maybe kinetics of deuterisation would give us some info about properties of its anions.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-03-06T08:44:22.243" LastActivityDate="2015-03-06T08:44:22.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19886" PostTypeId="1" CreationDate="2014-11-22T23:49:18.340" Score="6" ViewCount="808" Body="&lt;p&gt;Been learning about catalytic cracking of long chain hydrocarbons into shorter chain hydrocarbons, in which the catalyst zeolite is used so that the temperatures needed for the cracking are much lower (about $500\,^\circ\text{C}$ instead of $700\text{-}1000\,^\circ\text{C}$). The question is, how exactly does the zeolite help with the cracking process? Everywhere I look doesn't give a detailed answer, just that zeolite is porous, riddled with holes that can trap molecules. The only clue I have to the process I suppose is this video&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=nTse5_wXup4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=nTse5_wXup4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows an animation of some hydrocarbons whacking against a tunnel-like zeolite into smaller chain carbons. But to me the video can't be that accurate, because the whole thing seems out of proportion, the hydrocarbons are made from hydrogen and carbon which are pretty small atoms, whereas the zeolite is made from aluminium or silicon which are much larger. Even if this is the actual way it works, why not just have a plane surface to hit the hydrocarbons off to break them then? (Perhaps the rate of collision is much smaller? I'm just pulling at straws here.) Any light shed will be helpful, thanks.&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="16683" LastEditDate="2015-10-02T16:39:22.183" LastActivityDate="2018-03-04T02:57:50.437" Title="How do zeolites catalyse the cracking of hydrocarbons?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19887" PostTypeId="1" AcceptedAnswerId="19889" CreationDate="2014-11-23T00:12:42.253" Score="2" ViewCount="4316" Body="&lt;p&gt;So today we performed an experiment, where we mixed cyclohexene with Bromine water. The equation was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/NqG2Cgx.png&quot; alt=&quot;here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to make sense. The Bromine water we were using was decoloured as evidence that this equation took place, whereas when we mixed cyclohexane with Bromine water nothing happened (to show how double bonds are more 'reactive'). But what I don't understand is why the double bond was broken. I mean, wasn't it happy just being the way it was? Doesn't it still take energy to break the double bond, even if it is less than the energy required to break a single bond? It isn't like it WANTS to take the bromine? (or is it?) The only place where this energy could have come from in my mind is either the vibrational energy due to temperature, or else the kinetic energy from pouring it in? (may be taking too much of a physics approach). If so, doesn't that mean that if I got the cyclohexane, and poured it in from a great height, will it react with the Bromine water? Or is the energy difference between double bonds and single bonds so great that it can only perform under special circumstances? (i.e. under UV light)&lt;/p&gt;&#xA;" OwnerUserId="9799" LastActivityDate="2016-07-06T11:47:28.543" Title="Why do double bonds break" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19889" PostTypeId="2" ParentId="19887" CreationDate="2014-11-23T01:16:59.937" Score="6" Body="&lt;p&gt;Keep in mind, that the reason reactions occur spontaneously is because there's an overall driving force: $\Delta G = \Delta H - T\Delta S$. Since there are fewer molecules on the right hand, we can guess that $\Delta S$ isn't driving this reaction. ($G$ is Gibbs free energy, $S$ is entropy, $H$ is enthalpy, and $T$ is temperature.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then your question boils down to &quot;why is this reaction favorable for $\Delta H$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even before I look up the bond enthalpies for each of the bonds involved, I have an intuition. Yes, we're breaking the double bond in cyclohexene. We're also breaking the $\ce{Br-O}$ bond in $\ce{Br-O-H}$. So that's two bonds that are broken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right -- that costs energy, so there's clearly something &lt;em&gt;more&lt;/em&gt; energetically favorable on the other side. Well, in the product, I'm forming a $\ce{C-Br}$ and a $\ce{C-OH}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the net reaction breaks two bonds, and forms two bonds. As you say, the general intuition is that the &quot;double&quot; part of a double bond isn't quite as strong as the standard $\sigma$ single bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look up the &lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot; rel=&quot;nofollow&quot;&gt;bond enthalpies&lt;/a&gt; also through &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond-dissociation_energy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C=C}$ (vs. $\ce{C-C}$ which remains in the product): ~146 kcal/mol vs. ~84 kcal/mol = ~62 kcal/mol&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Br-O}$ = ~56 kcal/mol (I had to &lt;a href=&quot;http://www.webelements.com/bromine/bond_enthalpies.html&quot; rel=&quot;nofollow&quot;&gt;look this one up&lt;/a&gt;, so it's a bit iffy)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C-Br}$ = ~69 kcal/mol&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C-O}$ = ~85 kcal/mol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Net $\Delta H \approx$ -36 kcal/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if some of the bond enthalpies are off a bit, it's clearly energetically favorable. It has nothing to do with how the species are mixed, or really the temperature. (Well, my guess is that $\Delta S$ is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; favorable here, since two molecules form one, so high temperature might make the reaction less favorable from a $\Delta G$ standpoint.) Nor do you need UV light or anything like that -- the difference in bond strengths will push this reaction.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="7475" LastEditDate="2015-09-21T23:32:00.463" LastActivityDate="2015-09-21T23:32:00.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19890" PostTypeId="2" ParentId="19881" CreationDate="2014-11-23T01:27:16.787" Score="3" Body="&lt;p&gt;You are correct that the ionization energy of an electron depends on the orbital. &lt;strong&gt;There is no easy formula,&lt;/strong&gt; because the electrons in atoms are governed by quantum mechanics and for multi-electron atoms, the equations are un-solvable algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are some reasonable tabulations, which have been verified by calculations and by spectroscopy, e.g., &lt;a href=&quot;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&lt;/a&gt; (from a quick Google search).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ultraviolet photoelectron spectroscopy and x-ray photoelectron spectroscopy can be used to eject electrons from various materials, and you can thus identify elemental composition (and other properties) from the known orbital energy levels&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-11-23T01:27:16.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19891" PostTypeId="2" ParentId="19883" CreationDate="2014-11-23T01:44:16.780" Score="-1" Body="&lt;p&gt;My intuitive answer is that Cl has a relatively high electronegativity, so it pulls electrons towards itself, making it harder for the oxygen to react with the molecule. Reactions generally have an oxidizer which attracts electrons, and a reducer that supplies electrons. Oxidation of an organic compound usually means the carbons supply electrons to incoming oxygen atoms, making polarized and thus stable bonds. The bonds with the Chlorine atoms are themselves polarized, making the chlorine somewhat negative and the rest of the molecule slightly positive, making it less eager to give up electrons to an oxygen atom, meaning a resulting bond with oxygen would be less polarized and less stable....&lt;/p&gt;&#xA;" OwnerUserId="9962" LastActivityDate="2014-11-23T01:44:16.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19893" PostTypeId="2" ParentId="19883" CreationDate="2014-11-23T03:36:49.403" Score="0" Body="&lt;p&gt;Combustion reactions are radical reactions, and the chlorine atoms formed during decomposition of the polymer act as radical traps.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-23T03:36:49.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19894" PostTypeId="2" ParentId="19883" CreationDate="2014-11-23T06:08:13.433" Score="2" Body="&lt;p&gt;PVC is not truly flame resistant, it does decompose on relatively gentle heating (around 150 °C) However, the process produces significant volumes of hydrogen chloride, that is not flammable and isolates the plastic from the flame. This makes makes PVC &lt;strong&gt;self-extinguishing&lt;/strong&gt; or fire-retardant. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user7951" LastEditDate="2018-05-06T19:59:25.727" LastActivityDate="2018-05-06T19:59:25.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19895" PostTypeId="2" ParentId="13952" CreationDate="2014-11-23T06:32:42.537" Score="3" Body="&lt;p&gt;A company in the USA called &lt;a href=&quot;http://voltreepower.com/&quot; rel=&quot;nofollow&quot;&gt;Voltree Power&lt;/a&gt; has gone one step further to harvest electricity generated directly from live trees and other plants!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems trees are able to provide somewhere between 20 and a few hundred millivolts as well as some current, derived from a difference in pH between the trees and the soil, maintained by the trees metabolic processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science behind this technology was &lt;a href=&quot;http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0002963&quot; rel=&quot;nofollow&quot;&gt;published in 2008 by researches at MIT&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The company claims to have &lt;a href=&quot;http://www.reuters.com/article/2009/06/25/idUS230976+25-Jun-2009+BW20090625&quot; rel=&quot;nofollow&quot;&gt;demonstrated the technology&lt;/a&gt; by developing the first wireless sensor network powered exclusively by trees to detect forest fires! They are also developing the technology to monitor humidity and temperatures in the forest using electrical power harvested from the trees themselves and transmit the data using low-power wiremess mesh technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another collaborative group from University of Michigan, MIT &amp;amp; UC Berkeley &lt;a href=&quot;http://scitation.aip.org.proxy.library.adelaide.edu.au/content/aip/journal/apl/95/1/10.1063/1.3157144&quot; rel=&quot;nofollow&quot;&gt;published their research&lt;/a&gt; in 2009 into micro-power electricity scavenged from evaporative flows within the microchannels in trees.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-11-23T12:53:12.773" LastActivityDate="2014-11-23T12:53:12.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19896" PostTypeId="1" CreationDate="2014-11-23T06:57:54.053" Score="6" ViewCount="3514" Body="&lt;p&gt;Learning about biodegradable polymers and non-biodegradable polymers. What exactly sets them apart though? Here's the formula from wikipedia for the biodegradable PHB:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Poly-%28R%29-3-hydroxybutyrat.svg/213px-Poly-%28R%29-3-hydroxybutyrat.svg.png&quot; alt=&quot;here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here for polypropylene, it's non-biodegrable counterpart&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/f9/Polypropylen.svg/104px-Polypropylen.svg.png&quot; alt=&quot;here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, biodegradable just means that bacteria will degrade it into other constituent parts. Is it just that bacteria is a picky eater and won't eat certain polymers? But they seem to be made of similar elements, and the bonds between are generally similar. Any help is helpful.&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="21642" LastEditDate="2020-09-07T08:52:58.190" LastActivityDate="2020-09-07T08:52:58.190" Title="What makes something biodegradable?" Tags="&lt;biochemistry&gt;&lt;polymers&gt;&lt;enzymes&gt;&lt;green-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19897" PostTypeId="1" AcceptedAnswerId="19901" CreationDate="2014-11-23T07:46:24.970" Score="19" ViewCount="44212" Body="&lt;p&gt;To dehydrate ethanol one uses concentrated sulfuric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H5OH -&amp;gt;[\text{conc.} H2SO4] C2H4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but to go in the reverse direction, dilute sulfuric acid is used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H4 + H2O -&amp;gt;[\text{dil.} H2SO4] C2H5OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is one concentrated, and one dilute? For that matter, how does the sulfuric acid even help in the first place?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="16683" LastEditDate="2017-11-12T16:47:37.530" LastActivityDate="2017-11-12T16:48:14.117" Title="Why concentrated sulfuric acid for dehydration and dilute for hydration?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19898" PostTypeId="1" CreationDate="2014-11-23T08:06:49.893" Score="2" ViewCount="919" Body="&lt;p&gt;I found out there are two definitions of relative atomic mass $A_\mathrm r$.&lt;br&gt;&#xA;First definition is $A_\mathrm r$ is the mass of 1 atom of an element relative to 1/12 the mass of carbon-12 atom. It can be found in any chemistry school textbook.&lt;br&gt;&#xA;Second definition is what I just found and dont understand. It states $A_\mathrm r$ is the mass of 1 mole of atoms relative to 1 mole of carbon-12 atoms.&lt;br&gt;&#xA;Can you kindly explain second definition with examples?&lt;/p&gt;&#xA;" OwnerUserId="9965" LastEditorDisplayName="user7951" LastEditDate="2016-11-21T18:16:03.580" LastActivityDate="2016-11-23T00:09:39.103" Title="Definition of relative atomic mass" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19899" PostTypeId="1" AcceptedAnswerId="80359" CreationDate="2014-11-23T09:25:15.517" Score="4" ViewCount="181" Body="&lt;p&gt;I understand that entropy is defined as $\mathrm{d}S=\mathrm{d}q/T$ for a reversible change but I fail to see why $-G/T$ and $-A/T$ can be thought of as 'disguised' entropies. This is a question on my problem sheet at university and my tutor has hinted: &quot;You need to think about the total entropy change for a process&quot;. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2014-11-25T04:43:16.757" LastActivityDate="2017-07-25T03:13:30.933" Title="How can -G/T and -A/T be thought of as 'disguised' entropies?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19900" PostTypeId="1" AcceptedAnswerId="19907" CreationDate="2014-11-23T11:32:14.057" Score="0" ViewCount="5457" Body="&lt;p&gt;I've tried this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LT7LT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But its hows that the valence of Fe in normal state is 4, when in the periodic table it isn't. Why does this happen? Can you show me the answer with configuration formula in the states with energy?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="9878" LastEditDate="2014-11-23T12:02:47.130" LastActivityDate="2014-11-23T13:50:26.290" Title="Find the valences of Z=26" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19901" PostTypeId="2" ParentId="19897" CreationDate="2014-11-23T11:48:14.780" Score="13" Body="&lt;p&gt;Think of the reaction as an equilibrium equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2OH &amp;lt;=&amp;gt;[H2SO4] CH2=CH2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using Le Châtelier's Principle, we can deduce Dissenter's comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In concentrated sulfuric acid there is little to no water (note that we also start off with little to no ethylene). The equilibrium shifts to produce products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In dilute sulfuric acid there is lots of water (and little to no ethanol). The equilibrium shifts to produce reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the role of sulfuric acid, it acts as a proton source to enable the loss of leaving group in the forward direction and the electrophilic addition in the reverse direction. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2017-11-12T16:48:14.117" LastActivityDate="2017-11-12T16:48:14.117" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19902" PostTypeId="1" CreationDate="2014-11-23T12:19:16.527" Score="1" ViewCount="58" Body="&lt;p&gt;Does that happen because 3p already has a bond?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastActivityDate="2015-09-03T17:54:29.883" Title="Why V can't have the valence 7 by taking an electron from the 3p orbital and putting it into 3d?" Tags="&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19903" PostTypeId="1" AcceptedAnswerId="20053" CreationDate="2014-11-23T12:52:31.107" Score="4" ViewCount="1452" Body="&lt;p&gt;The Wikipedia article about &lt;a href=&quot;https://en.wikipedia.org/wiki/Urushibara_nickel&quot; rel=&quot;nofollow&quot;&gt;Urushibara nickel catalyst&lt;/a&gt; is rather short and vague. What is the mechanism for its preparation, and furthermore, in what sense does it differ from Raney nickel (i.e. when should you prefer the one over the other)?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2014-11-27T17:40:53.383" LastActivityDate="2015-11-29T17:17:35.897" Title="Differences between Urushibara nickel and Raney nickel" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19904" PostTypeId="1" AcceptedAnswerId="19911" CreationDate="2014-11-23T13:12:22.913" Score="11" ViewCount="29208" Body="&lt;p&gt;In the azide ion, there is a symmetrical arrangement of three nitrogen atoms. If I've been told to figure out the hybridization of each nitrogen atom, what's the best way to go about doing this? My initial guess was that the central nitrogen was sp hybridized and then the other two were sp&lt;sup&gt;2&lt;/sup&gt;, but I can't seem to justify my guess.&lt;/p&gt;&#xA;" OwnerUserId="9966" LastEditorUserId="4945" LastEditDate="2017-09-28T11:46:40.063" LastActivityDate="2017-09-28T11:46:40.063" Title="What is the hybridization of the nitrogen in the azide ion?" Tags="&lt;theoretical-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19905" PostTypeId="2" ParentId="863" CreationDate="2014-11-23T13:38:15.200" Score="1" Body="&lt;p&gt;Nanotechnology offers enormous opportunities for improving a huge range of electrochemical processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of platinum as a catalyst is been widespread throughout the electrochemical industry, helping reduce the activation energy of reactions such as in hydrogen and methanol fuel cells. However, platinum is a rare and expensive metallic element. Given that the reactions only take place on the surface of the catalyst, the use of &lt;a href=&quot;http://www.nature.com/nchem/journal/v6/n8/full/nchem.2001.html&quot; rel=&quot;nofollow&quot;&gt;nanoparticles of platinum reduces the amount of platinum needed&lt;/a&gt; to achieve the same surface area, thereby achieving the same performance at a fraction of the cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting application is in &lt;a href=&quot;http://www.formatex.info/energymaterialsbook/book/512-522.pdf&quot; rel=&quot;nofollow&quot;&gt;lowering the operating temperature in solid oxide fuel cells&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good reference on &lt;a href=&quot;http://books.google.com.au/books?id=U0G9BAAAQBAJ&quot; rel=&quot;nofollow&quot;&gt;Nanotechnology in Electrocatalysis for Energy&lt;/a&gt;, which covers background fundamentals and some applications, can be read on Google Books for free.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-11-23T15:04:47.393" LastActivityDate="2014-11-23T15:04:47.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19906" PostTypeId="2" ParentId="19898" CreationDate="2014-11-23T13:42:26.810" Score="2" Body="&lt;p&gt;These are all mathematically identical. Here’s why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{mass of one atom}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}}\\=\frac{\text{mass of one atom}\times N_\mathrm{A}}{\frac1{12}\text{ mass of one atom of }\ce{^12C}\times N_\mathrm{A}}\\=\frac{\text{mass of one mole of substance}}{\frac1{12}\text{ mass of one mole of }\ce{^12C}}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="7475" LastEditDate="2016-11-23T00:09:39.103" LastActivityDate="2016-11-23T00:09:39.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19907" PostTypeId="2" ParentId="19900" CreationDate="2014-11-23T13:50:26.290" Score="3" Body="&lt;p&gt;Hint:&#xA;$$_{26}Fe\equiv1s^22s^22p^63s^23p^64s^23d^6\equiv [Ar]\ce{\underbrace{ ^ v }_{4s^2} \underbrace{ ^ v ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }$$&#xA;Think now what will be the configuration after losing electrons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spoiler:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;! $$Fe^{2+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^ v ,\;^  ,\;^  ,\;^  ,\;^ }_{3d^6}=[Ar]\underbrace{ ^ }_{4s^2} \underbrace{ ^  ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;! $$Fe^{3+}\to[Ar]\ce{\underbrace{ - }_{4s^2} \underbrace{ ^  ,\;^  ,\;^  ,\;^  ,\;^   }_{3d^6} }&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-23T13:50:26.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19908" PostTypeId="1" AcceptedAnswerId="19909" CreationDate="2014-11-23T15:55:31.833" Score="4" ViewCount="758" Body="&lt;p&gt;I did an NMR sample to my salt p-carboxybenzyl triphenylphosphonium bromide. I got a doublet at 5.26, and according to every article it belongs to the $CH_2$ next to $P^+$, but how come I didn't get a singlet for that?&lt;/p&gt;&#xA;" OwnerUserId="9967" LastEditorUserId="4356" LastEditDate="2014-12-03T00:12:20.443" LastActivityDate="2014-12-03T00:12:20.443" Title="Why is there a doublet instead of a singlet in this Wittig reaction salt configuration NMR?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19909" PostTypeId="2" ParentId="19908" CreationDate="2014-11-23T16:12:27.490" Score="6" Body="&lt;p&gt;Phosphorous exists entirely as $\ce{^31P}$, and $\ce{^31P}$ has a magnetic moment of 1/2 (just like a proton). Therefore, a single phosphorous will split any protons coupled to it into a doublet - as you observed. If you looked at the corresponding $\ce{^31P}$ NMR spectrum you would see that the phosphorous signal would be split into a triplet by the two protons.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="470" LastEditDate="2014-11-23T20:03:47.503" LastActivityDate="2014-11-23T20:03:47.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19910" PostTypeId="2" ParentId="2766" CreationDate="2014-11-23T16:18:03.480" Score="0" Body="&lt;p&gt;I think that if you want a orbit, you should ask NASA, because the biggest ever rocket homemade reached &quot; just&quot; 56 kilometers. For NASA space starts at 100 km. A orbit asks for multiple stages with several tons of fuel, and the lowest orbit is higher than the edge of space, so its almost impossible to send a rocket into orbit.&lt;/p&gt;&#xA;" OwnerUserId="9968" LastActivityDate="2014-11-23T16:18:03.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19911" PostTypeId="2" ParentId="19904" CreationDate="2014-11-23T17:44:58.433" Score="13" Body="&lt;p&gt;Here are the resonance structures we can draw for the azide anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i2L5Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Resonance structure A has an octet on each nitrogen and just one unit of formal&#xA;charge on each nitrogen&lt;/li&gt;&#xA;&lt;li&gt;Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens.  Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the &quot;true&quot; representation of the azide anion, than resonance structure A&lt;/li&gt;&#xA;&lt;li&gt;Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less.  In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B).  The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options.  For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene.  The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in.  This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know.  Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (&lt;em&gt;e.g&lt;/em&gt;. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-19T15:54:34.263" LastActivityDate="2015-06-19T15:54:34.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19912" PostTypeId="1" AcceptedAnswerId="20099" CreationDate="2014-11-23T18:06:01.217" Score="3" ViewCount="336" Body="&lt;p&gt;Suppose methane ($CH_4$) and ethane ($C_2H_6$) in a container. The specific gravity is $r_g=0.6$. I think this means $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$. Now I am trying to understand the question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; what is the proportion of gases?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution according to my teacher:&lt;/strong&gt; $C_{CH_4}=0.1$ and $C_{C_2H_6}=0.9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eyR2h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt; $C_{CH_4}=24.2\%$ and $C_{C_2H_6}=75.8\%$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Values&lt;/strong&gt; (Curtis Phase Behaviour book p.162)&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;ol&gt;&#xA;    &lt;li&gt;&lt;p&gt;$M_{CH_4}=16.04 \frac{lbm}{lbm-mol}$&lt;/p&gt;&lt;/li&gt;&#xA;    &lt;li&gt;&lt;p&gt;$M_{C_2H_6}=30.07 \frac{lbm}{lbm-mol}$ &lt;/p&gt;&lt;/li&gt;&#xA;    &lt;/ol&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;What is $x_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{C_2C_6}}=\frac{n_{CH_4}}{n}$?&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;ol&gt;&#xA;    &lt;li&gt;&lt;p&gt;By Amagat's law and Dalton's law &lt;a href=&quot;http://www.departments.bucknell.edu/chem_eng/cheg200/CinChE_Manual/Ch06/examples/idgas.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;,&#xA;    I know $\frac{n_j}{n}=\frac{P_j}{P}=\frac{V_j}{V}$.&lt;/p&gt;&lt;/li&gt;&#xA;    &lt;li&gt;&lt;p&gt;Specific gravity $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$ where $\rho=\frac{m}{V}$ and constant $V$ so&#xA;    $0.6=\frac{m_{CH_4}}{m_{C_2H_6}}$ so $m_{CH_4}=0.6m_{C_2H_6}$.&lt;/p&gt;&lt;/li&gt;&#xA;    &lt;li&gt;&lt;p&gt;$PV=nRT\rightarrow n=\frac{PV}{RT}$ where constant $\frac{V}{RT}$ so&#xA;    $x_{CH_4}=\frac{P_{CH_4}}{P_{CH_4}+P_{C_2C_6}}=\frac{P_{CH_4}}{P}=_{(1)}\frac{n_{CH_4}}{n}=\frac{V_{CH_4}}{V}$.&lt;/p&gt;&lt;/li&gt;&#xA;    &lt;li&gt;&lt;p&gt;By $M_i=\frac{n_i}{m_i}$ so $n_i=M_i m_i$ where $n_{CH_4}=M_{CH_4}m_{CH_4}$&lt;/p&gt;&lt;/li&gt;&#xA;    &lt;/ol&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;so&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\frac{n_{CH_4}}{n}=\frac{M_{CH_4}m_{CH_4}}{M_{CH_4}m_{CH_4}+M_{C_2H_6}m_{C_2H_6}}=_{(2)}\frac{M_{CH_4}0.6m_{C_2H_6}}{M_{CH_4}0.6m_{C_2H_6}+M_{C_2H_6}m_{C_2H_6}}=\frac{0.6M_{CH_4}}{M_{CH_4}0.6+M_{C_2H_6}}=0.242455$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\rightarrow\frac{n_{C_2H_6}}{n}=1-\frac{n_{CH_4}}{n}=0.757545$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where the results are not 0.1 and 0.9.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="966" LastEditorUserId="966" LastEditDate="2014-11-23T18:28:22.567" LastActivityDate="2014-11-30T12:51:03.533" Title="Proportion of gases in Ideal Gas Mixture?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19913" PostTypeId="1" CreationDate="2014-11-23T18:59:55.913" Score="4" ViewCount="18344" Body="&lt;p&gt;Surely G would always be positive and thus would make it impossible to make diamond. Clearly this is not the case. But how? Would an increase or decrease in temperature aid it's formation? I think a decrease in temperature would favor diamond but wouldn't this contradict le chatilier's principle?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2014-11-23T19:14:29.347" LastActivityDate="2016-01-22T03:23:41.023" Title="Since the formation of diamond from Graphite is endothermic and the entropy decreases for this process, how is it possible to make diamond?" Tags="&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19915" PostTypeId="1" CreationDate="2014-11-23T19:04:13.830" Score="2" ViewCount="116" Body="&lt;p&gt;How might one determine the # of molecules in an unknown sample? If the sample is a gas,  we might use $PV=nRT$ to estimate # of molecules. If the sample is a solid or liquid and the chemical formula is known, we simply need to measure the mass. But can we determine the # of molecules without first determining the chemical formula?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastActivityDate="2014-11-23T19:57:11.170" Title="Determine # of molecules in an unknown sample" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19916" PostTypeId="1" CreationDate="2014-11-23T19:17:14.100" Score="3" ViewCount="535" Body="&lt;p&gt;How might one determine whether Ni is more dense than Cu using only the atomic masses of Ni and Cu, and the fact that Ni has a simple cubic unit cell, and Cu has a face-centered unit cell. You are not given the radii of each element. I feel like this question cannot be answered with the given information only, considering as Ni and Cu have very close densities (0.06 g/cm^3 difference).&lt;/p&gt;&#xA;" OwnerUserId="9972" LastActivityDate="2016-02-23T03:08:41.780" Title="Density of Ni vs Cu using unit cell" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19917" PostTypeId="2" ParentId="19913" CreationDate="2014-11-23T19:27:21.780" Score="4" Body="&lt;p&gt;Diamond is significantly denser than graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;atomic densities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1.14  \times 10^{23}$ cm$^{-3}$ for graphite&lt;/li&gt;&#xA;&lt;li&gt;$1.77  \times 10^{23}$ cm$^{-3}$ for diamond&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This suggests that &lt;strong&gt;higher pressure&lt;/strong&gt; would favor the formation of diamond.  Indeed, as the following phase diagram shows, diamond is the most stable allotrope of solid carbon at high pressures.  Temperature has little effect, as long as the pressure is high, diamond is favored over graphite at all temperatures up to the point of liquefication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3rgka.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-23T19:27:21.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19919" PostTypeId="2" ParentId="19913" CreationDate="2014-11-23T19:30:22.200" Score="4" Body="&lt;p&gt;If you have a look at this &lt;a href=&quot;http://dao.mit.edu/8.231/CarbonPhaseDia.htm&quot; rel=&quot;nofollow&quot;&gt;phase diagram&lt;/a&gt;, you can see the main condition at which diamond is the favourable structure: high pressure. The Gibbs free energy is dependant on pressure, in addition to temperature: $$dG = -SdT + Vdp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little math and assuming the temperature is constant, the gibbs energy moving from one pressure to another: $$G(p_2)=G(p_1)+\int_{p_1}^{p_2}Vdp$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, one method of making synthetic diamonds relies on high pressure and high temperature: carbon dissolved in a molten metal precipitates out onto small seed diamonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really necessary to calculate anything to rationalize why diamond would be the favoured allotrope at high pressure—diamond has a much higher density than graphite. At high pressure, the side of the equation with less volume will be favoured by Le Chatelier's principle.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-23T19:30:22.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19920" PostTypeId="2" ParentId="19913" CreationDate="2014-11-23T19:50:56.653" Score="1" Body="&lt;p&gt;Thermodynamics gives two possible solutions: you make $\Delta G$ positive through appropiate choice of reaction conditions (i.e. high pressure), or you couple the diamond-formation reaction with another reaction, so that the &lt;em&gt;total&lt;/em&gt; $\Delta G$ becomes positive.  There are processes for gas-phase deposition of diamond films on surfaces, and those don't depend on high pressure.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-23T19:50:56.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19921" PostTypeId="1" CreationDate="2014-11-23T19:50:57.293" Score="2" ViewCount="2242" Body="&lt;p&gt;Consider the derivation of the ideal gas law from kinetic theory presented here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_theory&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Kinetic_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some questions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This derivation assumes the container is a cube, but the law holds for containers of any shape, right?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivation assumes that the particle intercepts a give side with frequency $2L/v_x$. Why can we make this assumption? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="16683" LastEditDate="2016-09-30T09:22:06.847" LastActivityDate="2016-09-30T09:22:06.847" Title="Ideal gas law derivation from kinetic theory" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19922" PostTypeId="2" ParentId="19842" CreationDate="2014-11-23T19:51:40.093" Score="0" Body="&lt;p&gt;I look at it in a different way; not about volume but rather the relationship between molar volume and 'ideal' molar volume, $\frac{V}{n}$  and $\frac{RT}{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $PV = ZnRT$ write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{V}{n} = Z\frac{RT}{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the compressibility factor Z is called 'compressibility' factor is that it relates these two quantities like a 'spring' constant. It relates how many molecules you can actually 'squeeze' or rather  'compress' into a volume at some particular state of temperature and pressure.      For Z &gt; 1 the number of molecules, n is reduced for the same volume, temperature and pressure than when Z = 1.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2014-11-23T19:51:40.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19923" PostTypeId="2" ParentId="19915" CreationDate="2014-11-23T19:57:11.170" Score="2" Body="&lt;p&gt;There are attempts to redefine the mol through measuring spheres of monocrystalline silicon.  If you have the macroscopic dimension of your sample and the dimensions of your unit cell (from x-ray diffraction) you know how many atoms if silicon are there.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-23T19:57:11.170" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19924" PostTypeId="2" ParentId="19921" CreationDate="2014-11-23T21:34:01.267" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This derivation assumes the container is a cube, but the law holds for containers of any shape, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting law holds for any shape of container. However, the details of the derivation depend on the geometry. A cube is easy, so we use it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it holds for a mixture of any &lt;strong&gt;ideal gases&lt;/strong&gt; - the key is that only the kinetic energy of the gas molecules matters. Heavier molecules would be moving slower at the same temperature, but they would exert the same force (and therefore pressure) on the walls.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The derivation assumes that the particle intercepts a give side with frequency 2L/vx. Why can we make this assumption?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This assumption works for only one particular case. If you notice the third sentence at the beginning of the pressure derivation (emphasis is mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a gas molecule collides with the wall of the container &lt;strong&gt;perpendicular&lt;/strong&gt; to the x coordinate axis and bounces off in the opposite direction &lt;strong&gt;with the same speed&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are looking at the special case of a collision at a 90 degree angle, which bounces off the opposite wall. In that case, it will take the particle t = L/V time to reach the opposite wall, and t = L/V time to bounce back - so the total is 2L/V. This illustration may help explain it better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zzyYe.png&quot; alt=&quot;Illustration of average particle wall collision time. The first image shows a particle at one wall at the initial time, at the opposite wall at L/V, and back at the original wall at time 2L/V&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this might sound like an unreasonably strict condition, but the reason that it works is that the component velocity distribution for each particle is symmetrical (it's a normal distribution), which means that for every particle that impacts the wall at 95 degrees (for example) there is another that impacts 85 degrees. As a result, the &lt;em&gt;average&lt;/em&gt; angle of impact is 90 degrees for each wall, and therefore the average collision time is given by looking at 90 degree impacts for each component of the velocity vector.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-11-23T21:34:01.267" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19925" PostTypeId="1" AcceptedAnswerId="19954" CreationDate="2014-11-23T22:10:13.823" Score="1" ViewCount="238" Body="&lt;p&gt;For an elementary reaction like $$\ce{A + B -&amp;gt; AB},$$ $$K_{c} = \frac{\ce{[AB]}}{\ce{[A][B]}}$$ and $$\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A][B]}.$$ If I let the reaction reach equilibrium, and at this point, without altering temperature, suck away $\ce{A}$ and add $\ce{B}$ at the same rate (so $-\mathrm{d}\ce{[A]}/\mathrm{d}t = \mathrm{d}\ce{[B]}/\mathrm{d}t$) and alter both concentrations slightly, once I am finished, will the system still be in equilibrium or try to shift? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the reaction is not an elementary reaction such that $\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A]}^n\ce{[B]}^m$?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="4945" LastEditDate="2014-11-25T02:56:25.680" LastActivityDate="2014-11-25T02:56:25.680" Title="Can there be more than one combination of reactant/product concentration for a specific equilibrium constant?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19929" PostTypeId="1" AcceptedAnswerId="19989" CreationDate="2014-11-24T02:16:43.040" Score="0" ViewCount="465" Body="&lt;p&gt;I am trying to rank the concentration of $\ce{PO4^3-}$, $\ce{NO3^-}$, and $\ce{Na+}$ that remain in a reaction. The reaction occurs when 100mL of 1.0M $\ce{Na3PO4}$ solution is mixed with 100mL of 1.0M $\ce{AgNO3}$ soln. I have written the equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na3PO4(aq) + 3AgNO3(aq) -&amp;gt; 3NaNO3(aq) + Ag3PO4(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation I find that there are 0.3mol of $\ce{Na+}$ by 0.1L * 1.0M * 3mol Na, and &#xA;0.1mol of $\ce{PO4^3-}$ by: 0.1L * 1.0M * 1mol $\ce{PO4^3-}$ ,&#xA;and finally 0.3mol of $\ce{NO3-}$ by 0.1L * 1.0M * 1mol $\ce{NO3-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By these calculations, the concentration of the nitrate ion and sodium ion should by equal. However, the correct answer is that the most concentrated ion is sodium followed by nitrate followed by phosphate. Could someone please explain where I'm going wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7612" LastEditorUserId="7920" LastEditDate="2014-11-24T02:40:44.317" LastActivityDate="2014-11-25T16:21:15.560" Title="Aqueous Chemistry: Ranking Ion Concentration" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19930" PostTypeId="1" CreationDate="2014-11-24T03:19:38.440" Score="5" ViewCount="335" Body="&lt;p&gt;I was recently sorting some &lt;a href=&quot;https://en.wikipedia.org/wiki/New_Orleans_Mardi_Gras#Beads&quot; rel=&quot;nofollow&quot;&gt;Mardi Gras beads&lt;/a&gt;. These are cheap plastic beads on strings, coated in cheap metallic paint. While sorting the beads into different colors, I noticed that the purple beads seem to have suffered more fading than the other colors. This is an effect I have noticed in the past: if the beads sit in the sun for a day or two, the colors fade rapidly. I'm guessing that the cause of that is that the UV radiation from the sun is breaking the molecules used for the pigments in the paints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was that I had a pile of beads exposed to sunlight, and the purple beads in that pile had been at the top of the pile. But I noticed this happening in a few different piles that I had stored in different places and for different lengths of time. I have also noticed that the red beads and the purple/pink beads show some fading, though not as much as the purple beads. The yellow, silver, and green beads show almost no fading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current hypothesis is that the manufacturers use a pigment in the red and purple beads that is more susceptible to UV damage than the other pigments. So my question is: what is the chemical in these purple and red paints that is breaking down? (I guess there's also an implicit question here: is my thinking here chemically plausible?) &lt;/p&gt;&#xA;" OwnerUserId="323" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T15:54:28.117" LastActivityDate="2015-06-27T15:54:28.117" Title="Is purple paint more susceptible to UV damage than other colors?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19933" PostTypeId="1" AcceptedAnswerId="19947" CreationDate="2014-11-24T04:51:38.680" Score="2" ViewCount="344" Body="&lt;p&gt;I learned that when $PbO$ is hot, it is orange. When $PbO$ is cold, it is yellow. I know this's called thermochromism. But how do that happen? Is it because the ions inside doing something?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2014-11-24T15:49:46.977" Title="How thermochromism happen?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19934" PostTypeId="1" CreationDate="2014-11-24T06:54:42.883" Score="4" ViewCount="368" Body="&lt;p&gt;I want to synthesize an amide using tetrahydroisoquinoline and an unsaturated carboxylic acid which is light sensitive. I tried using DCC as a coupling agent and DCM as the solvent at room temperature, but it didn't work for me. The tetrahydroisoquinoline is in hydrochlorated form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to do acid extraction prior to synthesis, or I can use the hydrochlorated form itself? Please suggest a good method along with the conditions for the reaction to be carried out.&lt;/p&gt;&#xA;" OwnerUserId="9977" LastEditorUserId="402" LastEditDate="2019-05-18T08:45:55.270" LastActivityDate="2019-05-18T08:45:55.270" Title="Could you suggest me a good coupling method for amide synthesis between tetrahydroisoquinoline and a carboxylic acid?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19937" PostTypeId="1" AcceptedAnswerId="20554" CreationDate="2014-11-24T08:04:09.250" Score="2" ViewCount="438" Body="&lt;p&gt;How do you calculate the nuclear spin multiplicity of radicals such as $\ce{N2H+}$? How does the net charge on the molecular species effect its overall nuclear spin?&lt;/p&gt;&#xA;" OwnerUserId="8056" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:46:32.517" LastActivityDate="2015-10-23T16:46:32.517" Title="Nuclear spin multiplicity of radicals" Tags="&lt;physical-chemistry&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19938" PostTypeId="1" AcceptedAnswerId="19944" CreationDate="2014-11-24T08:41:32.607" Score="8" ViewCount="3731" Body="&lt;p&gt;So I understand than the lanthanide contraction is due to poor shielding of the 4f electrons which decreases the radius. &#xA;However, if Im not mistaken the relativistic effects lead to a contraction of the core orbitals (s,p) while the f and d orbitals expand. So what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right?&#xA;They usually say relativistic effects are one of the reasons for the lanthanide contraction&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5086" LastEditorUserId="5017" LastEditDate="2015-05-15T15:41:46.117" LastActivityDate="2017-06-15T23:29:26.870" Title="Relativistic effects and lanthanide contraction" Tags="&lt;quantum-chemistry&gt;&lt;rare-earth-elements&gt;&lt;relativistic&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19939" PostTypeId="1" CreationDate="2014-11-24T10:06:20.900" Score="6" ViewCount="7510" Body="&lt;p&gt;What is the equation and conditions for the synthesis of phenols? Would it work if we added $\ce{NaOH}$ to benzene and then refluxed?&lt;/p&gt;&#xA;" OwnerUserId="9980" LastEditorUserId="10185" LastEditDate="2016-05-31T13:11:23.313" LastActivityDate="2021-04-15T19:14:44.273" Title="How are phenols made?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;&lt;phenols&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19940" PostTypeId="2" ParentId="19939" CreationDate="2014-11-24T12:26:34.740" Score="9" Body="&lt;p&gt;That reaction will definitely not yield a phenol.&#xA;Phenols are not usually prepared directly from benzene, but from benzene derivatives, such as toluene or chlorobenzene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation of toluene, as developed by Dow Chemical:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H5CH3 + 2 O2 → C6H5OH + CO2 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrolysis of chlorobenzene, using base or steam (Raschig–Hooker process):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H5Cl + H2O → C6H5OH + HCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there are processes that allow us to obtain it directly from benzene, but these are rarely used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct oxidation of benzene with nitrous oxide, a potentially &quot;green&quot; process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H6 + N2O → C6H5OH + N2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most widely used process for industrial production (95%) involves the partial oxidation of cumene (isopropylbenzene) via the Hock rearrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C6H5CH(CH3)2 + O2 → C6H5OH + (CH3)2CO}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Source of the reactions: Wikipedia English page for phenol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="10185" LastEditDate="2015-06-17T13:53:27.780" LastActivityDate="2015-06-17T13:53:27.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19941" PostTypeId="2" ParentId="19939" CreationDate="2014-11-24T13:44:44.087" Score="1" Body="&lt;p&gt;To add to the other answer: there is an other method to synthesize phenols from benzene, namely through the benzyne intermediate. First you can react benzene with $\ce{Br2}$, in the presence of a strong Lewis acid like $\ce{AlCl3}$, to form bromobenzene. $\ce{NaNH2}$, a really strong base ($\ce{pK_{a}}=38$; compared to $\ce{NaOH}$ with $\ce{pK_{a}}=13$), can be used to remove a proton ortho to the bromo-group, where the inductive effect is largest. Then bromide is eliminated, yielding benzyne. Benzyne is incredibly unstable (very deformed &quot;$\pi$ bond&quot;) and is electrophilic enough for $\ce{OH-}$ to attack, giving the phenol after protonation by the formed $\ce{NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this only happens under the most vigorous conditions where $\ce{NaOH}$ and $\ce{NaNH2}$ are fused together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct electrophilic addition of an alcohol group to benzene is not possible, because there is no reliable source of $\ce{OH+}$.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="10185" LastEditDate="2015-06-17T13:12:46.580" LastActivityDate="2015-06-17T13:12:46.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19942" PostTypeId="1" AcceptedAnswerId="19945" CreationDate="2014-11-24T13:54:15.650" Score="5" ViewCount="77" Body="&lt;p&gt;I suspect they do, but could not find data on this. (I mean at temperatures near those used in the distillation of liquid air.)&lt;/p&gt;&#xA;" OwnerUserId="5277" LastEditorUserId="4231" LastEditDate="2015-02-28T18:15:31.060" LastActivityDate="2015-02-28T18:15:31.060" Title="Do liquid O₂ and N₂ form ideal solutions?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19943" PostTypeId="2" ParentId="19939" CreationDate="2014-11-24T14:43:03.443" Score="2" Body="&lt;p&gt;There are many ways to synthesize phenols.  The &quot;best&quot; way depends on what aromatic compound you have available as a starting material.  This &lt;a href=&quot;http://www.organic-chemistry.org/synthesis/C1O/phenols.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;web page&lt;/a&gt; lists many different methods for preparing phenols; if you find a method that looks interesting, you can click on the example and obtain more detailed information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very nice, general method for the synthesis of substituted phenols is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sandmeyer_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandmeyer reaction&lt;/a&gt;.  In this reaction you start with an aromatic amine.  The amine is diazotized to produce a diazonium salt.  When the salt is mildly heated it eliminates nitrogen producing a reactive phenyl cation.  When the diazonium salt is decomposed in the presence of water, phenols are produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fC5KG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-24T14:43:03.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19944" PostTypeId="2" ParentId="19938" CreationDate="2014-11-24T15:11:44.063" Score="9" Body="&lt;p&gt;The lanthanide contraction is caused by two effects&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals&lt;/li&gt;&#xA;&lt;li&gt;relativistic effects&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanthanide_contraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; presents a nice discussion of the lanthanide contraction.  The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what I do not understand how do they match, one of them decreases the&#xA;  radius and the other one should expand it as the f orbitals expand&#xA;  right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637&quot;&gt;this SE Chem post&lt;/a&gt; for a more detailed description).  In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius.  A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-11-24T15:11:44.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19945" PostTypeId="2" ParentId="19942" CreationDate="2014-11-24T15:23:13.253" Score="4" Body="&lt;p&gt;It's not perfectly ideal.  Try to get access to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0021961475900944&quot; rel=&quot;nofollow&quot;&gt;Excess enthalpies of the liquid mixtures nitrogen + oxygen...&lt;/a&gt; for quantitative data.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-11-24T16:19:26.157" LastActivityDate="2014-11-24T16:19:26.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19946" PostTypeId="1" AcceptedAnswerId="19992" CreationDate="2014-11-24T15:27:36.927" Score="1" ViewCount="1049" Body="&lt;p&gt;I was reading around about the different solid states of materials and I got to a point where I have reached the multi-component solids. I was not able to find a major difference between the cocrystal and the eutectic mixture from their definitions. Do you know of any fundamental difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question in another way: if you had a mixture of 2 solids that bind to each other somehow, how can you know whether this is a cocrystal or a eutectic mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks,&lt;/p&gt;&#xA;" OwnerUserId="1787" LastEditorUserId="7920" LastEditDate="2014-11-24T17:02:41.040" LastActivityDate="2014-11-25T17:15:21.973" Title="Cocrystals vs Eutectic mixtures" Tags="&lt;solid-state-chemistry&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19947" PostTypeId="2" ParentId="19933" CreationDate="2014-11-24T15:49:46.977" Score="2" Body="&lt;p&gt;In the case of lead oxide and many other inorganic thermochromic materials, the color change is a result of a phase change that slightly alters the distance separating the inorganic ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ordered materials, the energy levels present in atoms and molecules are replaced by energy bands.  Said differently, in materials with extended order there are so many energy levels clustered around one another at a certain energy point and separated from one another by such small amounts of energy, that the distinct energy levels of the molecules &quot;fuzz&quot; out into a band.  Instead of a highest occupied and lowest unoccupied molecular orbital (HOMO and LUMO), a valence band and a conduction band results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O3dqS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as the HOMO-LUMO separation will determine where in the energy spectrum an organic molecule will absorb light, the energy separation between the valence band and the conduction band (called the band gap) will determine where a material with extended order will absorb light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of lead oxide and other crystalline materials, slight changes in the crystal structure (phase change) that can be brought about by heating or cooling the material, will change the size of the band gap.  This change in the energy separation between the valence band and conduction band will result in a change in the optical (and electronic) properties of the material.  If the change in optical absorption occurs in the visible region, then the thermochromic properties of the material are apparent to the eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://amlett.com/uploads/1851516943.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to a nice paper that starts simply, but then goes into greater depth, on this subject.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-24T15:49:46.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19949" PostTypeId="2" ParentId="19896" CreationDate="2014-11-24T16:18:07.287" Score="3" Body="&lt;p&gt;The difference between biodegradable polymers and 'regular' polymers is the bonds that make up the backbone. The backbone is the part of the polymer that extents the brackets. Anything attached to that chain is a side group and is less important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two examples you show here are very dissimilar. The polypropylene is only made up of Carbon-Carbon bonds. These bonds are very stable: that holds for chemical and 'biological' stability. The PHB how ever bonds monomers with an Ester functional group (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ester&quot; rel=&quot;nofollow&quot;&gt;Ester&lt;/a&gt;). This group is far more reactive that a C-C bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactivity of the groups the polymer is made up of, is very, very important. That being said, I cannot claim that all poly-esters are biodegradable. Depending on the groups close to the esters and the enzymes (probably from bacteria) involved, the bond might be cleaved, or might be inaccessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped! &lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-11-24T16:18:07.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19950" PostTypeId="2" ParentId="19946" CreationDate="2014-11-24T17:02:07.287" Score="1" Body="&lt;p&gt;Simple: consider a system whose components are fully miscible in solution and upon solidifying separates into two phases.  At the eutectic point both solid phases melt at the same temperature.  At other compositions you'd see liquid phase and one solid phase.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-24T17:02:07.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19951" PostTypeId="2" ParentId="19564" CreationDate="2014-11-24T17:04:54.453" Score="7" Body="&lt;p&gt;The simplified version of why radiation occurs is a balance between attraction of the protons and neutrons in the nucleus and the positive charge of the protons repulsing each other. Get the balance right, and you end up with a stable isotope, get it wrong, and the isotope is unstable and will decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the exponential decay: a nucleus falling apart is stochastic process, meaning it is only dictated by chance! When you take an unstable nucleus, after it's half-life there is a 50% chance that it will have decayed. Then, you can add more nuclei the chance of half of them decaying after it's half life is far, far larger than all of them being stable. The larger amount you start of with, the better this theory describes the decay. If you then measure a vast amount of atoms (a mole for example) the curve you find will describe an exponential decay.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditDate="2016-11-26T16:16:35.453" LastActivityDate="2016-11-26T16:16:35.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19952" PostTypeId="1" AcceptedAnswerId="19955" CreationDate="2014-11-24T17:16:58.967" Score="2" ViewCount="150" Body="&lt;p&gt;Does reactivity of a compound depend on kinetic factors, thermodynamic factors, or both? i.e. how is reactivity defined? For example, out of a) 2-bromopropane and b) 4-bromobutan-2-one, which is more reactive to alcoholic KOH? Does the answer account for kinetic factors, or the fact that b) forms a conjugated ketone, and is hence more reactive?&lt;/p&gt;&#xA;" OwnerUserId="9984" LastActivityDate="2014-11-24T18:34:10.360" Title="Define &quot;reactivity&quot;" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19953" PostTypeId="2" ParentId="19930" CreationDate="2014-11-24T18:04:37.673" Score="4" Body="&lt;p&gt;I would also agree with you and the comment from ssavec. The only point that I would make is that it shouldn't be because of the colour of the pigment only the stability of the particular dye it contains (almost certainly organic). In fact the colour of the paint should make it less susceptible to fading as it is reflecting colours at the violet end of the spectrum (subject to your definition of purple) rather than absorbing them.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-24T18:04:37.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19954" PostTypeId="2" ParentId="19925" CreationDate="2014-11-24T18:28:37.493" Score="3" Body="&lt;p&gt;A single &quot;elementary reaction&quot; does not involve an equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There must be a forward reaction and a reverse reaction to have an equilibrium.  This is at least two elementary reactions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant,  [A][B] is maximal when [A] = [B].  Deviation from this maximum will shift the equilibrium away from the product and toward the reactants.  In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-11-24T19:36:42.710" LastActivityDate="2014-11-24T19:36:42.710" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19955" PostTypeId="2" ParentId="19952" CreationDate="2014-11-24T18:34:10.360" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does reactivity of a compound depend on kinetic factors, thermodynamic&#xA;  factors, or both?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reactivity of a compound depends on kinetic factors, &lt;strong&gt;but&lt;/strong&gt;... The &quot;but&quot; arises because thermodynamic factors can be a part of the kinetic factors.  Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following figure shows the conversion of a &quot;Reactant&quot; to a &quot;Product&quot;.  The product is more stable so the reaction is exothermic, heat is released.  The rate of the reaction is determined by the height of the barrier (the activation energy, $E_{a}$) that a reactant molecule must pass over in order to become the product.  The higher the barrier, the slower the reaction.  Since the barrier height is contained in the kinetic equation describing the speed of a reaction  (for example, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arrhenius equation&lt;/a&gt;)  we say that reactivity is determined by kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQYop.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now look at the reverse reaction.  Suppose you had a bottle of &quot;Product&quot; and heated it to convert it to &quot;Reactant&quot;.  This reaction is endothermic, heat must be supplied.  The free energy difference between product and reactant ($\Delta G^o)$ is a thermodynamic factor telling us whether the product or reactant is more stable.  In the reverse reaction, the barrier that we must surmount is higher than it was for the forward reaction.  The barrier for the back reaction is given as $E_{a} + \Delta G^o$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the rate or reactivity depends on the barrier height which is a kinetic phenomenon, we can see how the free energy difference ($\Delta G^o$), a thermodynamic quantity, can be viewed as a factor influencing the barrier height. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;out of a) 2-bromopropane and b) 4-bromobutan-2-one, which is more&#xA;  reactive to alcoholic KOH? Does the answer account for kinetic&#xA;  factors, or the fact that b) forms a conjugated ketone, and is hence&#xA;  more reactive?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1-bromobutane might make for a better comparison, in this case the bromine is attached to a terminal carbon as in 4-bromobutan-2-one, and both molecules contain 4 carbons - the only difference is that one is a ketone.  In any case, the answer is based on kinetic factors.  I don't know which one will react faster, but here's my thinking. They should have similar activation energies, but since the ketone reacts to produce a more stabilized conjugated $\alpha,\beta$-unsaturated ketone, the reaction of the bromo-ketone will be thermodynamically more exothermic than the bromo-alkane.  This thermodynamic difference will reduce the height of the kinetic barrier for the bromo-ketone and make this reaction faster.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-24T18:34:10.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19956" PostTypeId="2" ParentId="19934" CreationDate="2014-11-24T18:48:38.570" Score="4" Body="&lt;p&gt;A SciFinder search gave 200 or so examples of coupling tetrahydroisoquinoline with an unsaturated carboxylic acid (including benzoic acids and the like). Surprisingly dicyclohexylcarbodiimide (DCC) was not one of the reagents used, although &lt;em&gt;N&lt;/em&gt;-(3-dimethylaminopropyl)-&lt;em&gt;N&lt;/em&gt;′-ethylcarbodiimide (EDC) was used in more than 50 cases. Other common amide coupling agents including PyBOP, BOP, HBTU, and HATU were reported. Another related strategy through a mixed carbonate/anhydride by activating the acid with isobutylchloroformate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if you have to isolate the free base of your amine first, but you should certainly include a base to liberate the active nucleophile. If you haven't tried it yet, add an equivalent of triethylamine or diisopropylethylamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aapptec, a supplier of reagents and building blocks for peptide synthesis, has a nice compilation of the different amide coupling reagents that are used:&#xA;&lt;a href=&quot;http://www.aapptec.com/coupling-reagents-i-148.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aapptec coupling reagents&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="41328" LastEditDate="2019-05-17T20:07:01.297" LastActivityDate="2019-05-17T20:07:01.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19960" PostTypeId="1" AcceptedAnswerId="19961" CreationDate="2014-11-24T20:54:48.467" Score="15" ViewCount="1963" Body="&lt;p&gt;Why do the d orbitals have the following notations: $xy, yz, xz, z^2$ and $ x^2-y^2$? What do they represent in their wave-functions?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2014-11-26T09:31:29.643" LastActivityDate="2017-02-21T16:22:59.937" Title="Why do the d orbitals have these notations?" Tags="&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19961" PostTypeId="2" ParentId="19960" CreationDate="2014-11-24T22:20:02.217" Score="13" Body="&lt;p&gt;The notation is shorthand for the hydrogen atom $\ell = 2$ wavefunctions in real form, in cartesian coordinates.  In other words, the Schrodinger equation is solved in spherical coordinates, in terms of spherical harmonics, which involve complex numbers; however, linear combinations of degenerate (equal energy) solutions are also solutions.  This permits the wavefunction solutions to be expressed using real numbers only. These real solutions can be transformed to cartesian coordinates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z^2$ represents that the wavefunction is proportional to $(3z^2-r^2)/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xy$ represents that the wavefunction is proportional to $xy/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$yz$ represents that the wavefunction is proportional to $yz/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$xz$ represents that the wavefunction is proportional to $xz/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 - y^2$ represents that the wavefunction is proportional to $(x^2 - y^2)/r^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://openmopac.net/manual/real_spherical_harmonics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table of Real Spherical Harmonic Functions&lt;/a&gt; for the a list of functions in both spherical coordinates and cartesian coordinates, for s through f orbitals.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2017-02-21T16:22:59.937" LastActivityDate="2017-02-21T16:22:59.937" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19962" PostTypeId="1" AcceptedAnswerId="19980" CreationDate="2014-11-25T00:39:08.167" Score="2" ViewCount="2184" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many grams of water would you add to $\pu{1.00 kg}$ of $\pu{1.44m}$ $\ce{CH3OH (aq)}$ to reduce the molality to $\pu{1.00 m}$ $\ce{CH3OH}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried $M_1V_1=M_2V_2$. I have tried converting it to kg of $\ce{CH3OH}$ and subtracting it from the $\pu{1.00kg}$ to get kg of $\ce{H2O}$. All of which have worked to no avail.&lt;/p&gt;&#xA;" OwnerUserId="9987" LastEditorUserId="5026" LastEditDate="2018-03-03T10:43:25.267" LastActivityDate="2018-03-03T10:43:25.267" Title="Reduction of molality of methanol" Tags="&lt;concentration&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19963" PostTypeId="1" CreationDate="2014-11-25T01:51:11.130" Score="2" ViewCount="1057" Body="&lt;p&gt;I need to determine the purity of a solution and we need to discuss about the  'theoretical' refractive index of the substance that we are supposed to have.&lt;/p&gt;&#xA;" OwnerUserId="9989" LastEditorUserId="69830" LastEditDate="2021-04-28T08:36:43.430" LastActivityDate="2021-04-28T08:36:43.430" Title="Is the refractive index of a solution of four substances the median of the four substances' refractive index?" Tags="&lt;physical-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19964" PostTypeId="1" AcceptedAnswerId="19972" CreationDate="2014-11-25T02:46:17.383" Score="2" ViewCount="6374" Body="&lt;p&gt;I always mix up the definition of nuclear fission and nuclear fusion. So, what is nuclear fission and nuclear fusion, and an example for each?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="4945" LastEditDate="2014-11-25T06:11:26.370" LastActivityDate="2020-01-27T19:30:45.777" Title="What is the difference between nuclear fission and nuclear fusion?" Tags="&lt;nuclear&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19965" PostTypeId="1" AcceptedAnswerId="19969" CreationDate="2014-11-25T03:00:42.667" Score="0" ViewCount="39" Body="&lt;p&gt;I have a few more questions of this type for an interview. ( University work )&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="9990" LastActivityDate="2014-11-25T05:55:30.197" Title="Is it relevant to pursue a master degree or a doctorate when applying to a job in the petroleum field?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-11-25T11:39:40.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="19966" PostTypeId="1" CreationDate="2014-11-25T03:11:09.567" Score="0" ViewCount="2307" Body="&lt;p&gt;Why do we not need $\ce{H2SO4}$ as an acid catalyst for the di-nitration of &lt;a href=&quot;http://en.wikipedia.org/wiki/1,2-Dimethoxybenzene&quot; rel=&quot;nofollow noreferrer&quot;&gt;veratrole&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s9EDP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the 2nd nitro group go to positions 4 or 5, following the directing effects of the methoxy-group? Doesn't the first nitro group, that has been substituted onto the compound, have any effect on its position?&lt;/p&gt;&#xA;" OwnerUserId="9991" LastEditorUserId="10185" LastEditDate="2015-06-28T15:38:46.593" LastActivityDate="2015-06-28T15:38:46.593" Title="Why is nitration of veratrole specific in these positions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19967" PostTypeId="2" ParentId="19964" CreationDate="2014-11-25T04:04:17.540" Score="5" Body="&lt;p&gt;Nuclear fission involves the splitting of a heavy nucleus into two lighter fragments plus the release of energy.  As an example here is one route by which $\ce{^235U}$, a fissile material, can split apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^{235}_{92}U + ^{1}_{0}n -&amp;gt; ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + 200 MeV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the 3 neutrons generated in the fission process, they are part of the chain reaction; they allow the reaction to propagate at an even faster rate as they go on to react with 3 other $\ce{^{235}_{92}U}$ nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear fusion involves the fusing together of nuclei to form a heavier element.  Again, energy is released in the process.  Here is an example of a fusion reaction involving the combination of 2 deuterium nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^{2}_{1}H + ^{2}_{1}H -&amp;gt; ^{4}_{2}He + ^{1}_{0}n + 17$.$6 MeV}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T04:04:17.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19968" PostTypeId="2" ParentId="19963" CreationDate="2014-11-25T04:29:26.310" Score="6" Body="&lt;p&gt;There is no single equation that accurately predicts the refractive index of a mixture.  This &lt;a href=&quot;http://downloads.hindawi.com/journals/jchem/2007/485378.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt; discusses 9 approaches to estimating the refractive index of mixtures and provides the formula for each method. The simplest method is the Arago-Biot approach.  It is reasonably reliable and is described by the following formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = n_1 \phi_1 + n_2\phi_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the predicted refractive index of the mixture, $n_x$ is the refractive index of the $x^{th}$ component, and $\phi_x$ is the volume fraction of the $x^{th}$ component.  The volume fraction of the first component is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_1 = x_1 V_1 / \Sigma~ x_i V_i$$&#xA;Here $x_i$ is the mole fraction of the $i^{th}$ component and $V_i$ is the molar volume (the molecular weight divided by the density) of the $i^{th}$ component. The more ideal the liquids, the more accurate the predictions. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T04:29:26.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19969" PostTypeId="2" ParentId="19965" CreationDate="2014-11-25T05:37:52.690" Score="0" Body="&lt;p&gt;Professional careers in the petroleum industry can be very competitive, so a higher qualification such as masters degree could make you a more attractive candidate for a scientific or engineering position. Many companies nowadays specify post-graduate qualifications as necessary or desirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A PhD is generally preferred for research positions.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-11-25T05:55:30.197" LastActivityDate="2014-11-25T05:55:30.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19972" PostTypeId="2" ParentId="19964" CreationDate="2014-11-25T11:12:51.990" Score="4" Body="&lt;p&gt;Fusion is an increase in the number of protons in the nucleus by fusing two nuclei. You know fusing means joining together, right? So that can help you remember which is which. &quot;Confusion&quot; comes from the same root word, and happens when you mix things up together in your mind - the &quot;fusion&quot; in &quot;confusion&quot; is the bringing things together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fission is a decrease in the number of protons by splitting the nucleus: &quot;fission&quot; means a split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ron's given you the classic examples for fission and for fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either process can absorb energy, or release it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron is the key element here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For elements with a lower proton count (i.e. a lower atomic number) than iron, fusion releases energy, but fission absorbs energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas, for elements with a higher proton count (i.e. a higher atomic number) than iron, fusion absorbs energy, and fission releases energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(it's not quite that simple, otherwise we might find a way to fuse two lighter-than-iron nuclei to release energy and create a heavier-than-iron nucleus which we then split again back to the two original nuclei and release even more energy. That's impossible)&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2014-11-25T11:27:11.080" LastActivityDate="2014-11-25T11:27:11.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19975" PostTypeId="1" CreationDate="2014-11-25T12:59:31.033" Score="5" ViewCount="2346" Body="&lt;p&gt;Occasionally, I work with lauryl tryptose broth for growing coliform.  Being a chemical engineer, I suspected at the start that tryptose is a type of carbohydrate because of the -ose syffix.  However, a search for more info on the web did not provide even a hint of what tryptose really is. Anybody there who knows what tryptose is?  I'm particularly interested in its molecular structure.&lt;/p&gt;&#xA;" OwnerUserId="9997" LastActivityDate="2014-12-04T13:23:20.580" Title="What is tryptose" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19976" PostTypeId="2" ParentId="19966" CreationDate="2014-11-25T13:10:35.603" Score="1" Body="&lt;p&gt;When there are activating and deactivating groups on an aromatic ring, the direction of the electrophilic aromatic substitution is generally controlled by the activating groups.&#xA;Think that activating groups improve the substitution in certain positions by increasing its electron density. On the other hand, deactivating groups favor one position over another, not by activating it, but deactivating it less than the other positions. So, the directing effect of a deactivating group is masked by the directing effect of the activating ones, which is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You most likely don't need the sulphuric acid due to the presence of two activating groups. In contrast, as an example, last nitration step of trinitrotoluene is so disfavored that it has to be carried out in boiling sulphuric acid.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-11-25T13:10:35.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19977" PostTypeId="1" AcceptedAnswerId="19987" CreationDate="2014-11-25T13:30:10.797" Score="1" ViewCount="1878" Body="&lt;p&gt;How can we find the bond angle between 3 atoms in a compound?&#xA;Please specify an equation which can be used for all compounds.&lt;/p&gt;&#xA;" OwnerUserId="9775" LastEditorUserId="4231" LastEditDate="2014-11-25T19:04:09.423" LastActivityDate="2019-10-30T10:36:30.487" Title="Determine the bond angle in a compound" Tags="&lt;bond&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2014-11-25T16:16:13.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="19978" PostTypeId="1" AcceptedAnswerId="19979" CreationDate="2014-11-25T13:31:26.107" Score="7" ViewCount="1492" Body="&lt;p&gt;If I look up the IUPAC name of &lt;a href=&quot;http://en.wikipedia.org/wiki/Ascorbic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ascorbic acid&lt;/a&gt;, I find this: &lt;strong&gt;(5&lt;i&gt;R&lt;/i&gt;)-[(1&lt;i&gt;S&lt;/i&gt;)-1,2-Dihydroxyethyl]-3,4-dihydroxyfuran-2(5&lt;i&gt;H&lt;/i&gt;)-one&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am used to numbering of molecules in a way like the &lt;code&gt;3,4-dihydroxyfuran&lt;/code&gt; part in this molecule name. But now what I want to know is: what do the 5R, 1S and 5H parts signify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the 5H I came up with this: &lt;a href=&quot;http://en.wikipedia.org/wiki/Furan&quot; rel=&quot;nofollow noreferrer&quot;&gt;furan&lt;/a&gt; has two double bonds, so the H might signify that this is not the case any more, because if there is an extra H at the fifth position, this would be impossible. This can be seen in the IUPAC name of &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Furanone&quot; rel=&quot;nofollow noreferrer&quot;&gt;2-furanone&lt;/a&gt;: &lt;code&gt;5H-furan-2-one&lt;/code&gt;. I don't know though if this assumption is correct, also because in &lt;code&gt;5H-furan-2-one&lt;/code&gt; the 5H is before the furan part and in &lt;code&gt;3,4-dihydroxyfuran-2(5H)-one&lt;/code&gt; it is after (shouldn't it be &lt;code&gt;2(3,4-dihydroxy-5H)-furanone&lt;/code&gt; anyway?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact is that I can't find an explanation for the 5R, nor the 1S (as I see no sulfur). Could anyone explain me these weird molecule numberings?&lt;/p&gt;&#xA;" OwnerUserId="9999" LastEditorUserId="544" LastEditDate="2018-10-04T08:35:04.580" LastActivityDate="2018-10-04T08:35:04.580" Title="What do the numbers in the preferred IUPAC name for ascorbic acid mean?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19979" PostTypeId="2" ParentId="19978" CreationDate="2014-11-25T14:26:47.147" Score="11" Body="&lt;p&gt;The numbering of &lt;em&gt;(R)&lt;/em&gt;-5-[&lt;em&gt;(S)&lt;/em&gt;-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5&amp;#8203;&lt;em&gt;H&lt;/em&gt;)-one has four parts:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1) Numbering of the furan ring&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aA85S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2) Numbering of the ethyl side chain&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/13yAE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3) Stereodescriptors &lt;em&gt;R&lt;/em&gt; and &lt;em&gt;S&lt;/em&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1lYr7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;4) Italicized element symbols&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The italic element symbol &lt;em&gt;H&lt;/em&gt; denotes indicated or added hydrogen.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2014-11-25T14:26:47.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19980" PostTypeId="2" ParentId="19962" CreationDate="2014-11-25T14:40:26.437" Score="2" Body="&lt;p&gt;If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective of the problem is to prepare a solution of molality $1 \ce{m}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56&#xA;\ce{g}$ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solution already contains $955.95\ce{g}$ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-25T14:40:26.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19981" PostTypeId="2" ParentId="19975" CreationDate="2014-11-25T14:47:45.463" Score="1" Body="&lt;p&gt;Tryptose is not a sugar or any particular compound, so there is no corresponding molecular structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tryptose is an animal-derived mixture that includes water soluble peptides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bdbiosciences.com/documents/Tryptose.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.bdbiosciences.com/documents/Tryptose.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bd.com/ds/technicalCenter/brochures/br_1_222880_academia.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.bd.com/ds/technicalCenter/brochures/br_1_222880_academia.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-11-25T14:47:45.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19982" PostTypeId="2" ParentId="19975" CreationDate="2014-11-25T14:54:33.797" Score="4" Body="&lt;p&gt;Tryptose is not a single compound, therefore it does not have a unique molecular structure.  As you are aware, tryptose is a nutrient broth used for growing coliform. It is a &lt;strong&gt;mixture&lt;/strong&gt; of smaller proteins produced by using trypsine as the enzyme to digest  larger proteins. You might compare it to tryptone which also uses the enzyme trypsine in the digesting process, but only casein is digested.  There is also peptone which uses a  mixture of pankreatin and papain to enzymatically digest proteins.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T14:54:33.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19983" PostTypeId="1" AcceptedAnswerId="20566" CreationDate="2014-11-25T14:56:49.847" Score="3" ViewCount="64" Body="&lt;p&gt;We have been learning a bit about molecular orbitals in class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we have always considered things under a quasistatic approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are some effects that can only be explained by not using the quasistatic approximation ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5591" LastActivityDate="2014-12-11T04:47:12.097" Title="Non stationary effects of molecular orbitals" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19984" PostTypeId="1" CreationDate="2014-11-25T15:10:32.780" Score="12" ViewCount="1207" Body="&lt;p&gt;I want to clean up some old equipment that was used to visualize nucleic acids with ethidium bromide. It's a UV table, built from glass and metal (stainless steel, I assume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethidium bromide is generally considered scary in a biological lab (probably because most stuff there is pretty harmless), but there seems to be a lot of controversy about the actual danger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be an appropriate way to clean this? I'd simply wipe the whole thing with paper towels and some ethanol and dispose of them in the chemical waste. Would that be sufficient?&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="4356" LastEditDate="2015-05-15T21:24:15.550" LastActivityDate="2016-10-06T10:02:33.990" Title="What would be an appropriate way to clean up equipment contaminated with ethidium bromide?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19985" PostTypeId="1" CreationDate="2014-11-25T15:18:39.100" Score="3" ViewCount="2200" Body="&lt;p&gt;I was asked to compare the reaction between sodium hydride and water and sodium borohydride and water. I know that the former acts more violently than the latter, but apparently, I must use a curly arrow mechanism to show this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the HOMO for the latter reaction is the oxygen lone pairs, and the LUMO is the B-H bond, but I'm having difficulty moving on from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What attacks what first?&lt;/p&gt;&#xA;" OwnerUserId="9966" LastEditorUserId="402" LastEditDate="2019-05-21T00:03:12.697" LastActivityDate="2019-05-21T00:03:12.697" Title="Curly arrow mechanism for reaction between sodium borohydride and water?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="19986" PostTypeId="1" CreationDate="2014-11-25T15:44:55.423" Score="1" ViewCount="2405" Body="&lt;p&gt;Some liquid water is present inside a sealed flask at room temperature. Is water evaporating, and is the system at equilibrium? First of all, can water evaporate in a sealed flask at room temperature? If so, is the reaction happening a reversible reaction? Any help would be appreciated. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="4093" LastActivityDate="2014-11-25T15:49:36.310" Title="Will room-temperature water in a sealed flask evaporate?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19987" PostTypeId="2" ParentId="19977" CreationDate="2014-11-25T15:45:39.083" Score="7" Body="&lt;p&gt;In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{\lambda_{i}}$&lt;/span&gt; represents the hybridization index of the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-i}$&lt;/span&gt; bond (the hybridization index is the square root of the bond hybridization) and &lt;span class=&quot;math-container&quot;&gt;$\ce{\theta_{ij}}$&lt;/span&gt; represents the &lt;span class=&quot;math-container&quot;&gt;$\ce{i-C-j}$&lt;/span&gt; bond angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lRXef.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at some examples.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the case of methane, each &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; hybridized and the hybridization index of each &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\sqrt3$&lt;/span&gt;. Using Coulson's theorem we find that the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-C-H}$&lt;/span&gt; bond angle is 109.5 degrees.&lt;/li&gt;&#xA;&lt;li&gt;In the case of the methyl carbocation (&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3^{+}}$&lt;/span&gt;) the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds are &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized and the hybridization index of each &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\sqrt2$&lt;/span&gt;.  In this case, using Coulson's theorem we find that the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-C-H}$&lt;/span&gt; bond angle is 120 degrees.&lt;/li&gt;&#xA;&lt;li&gt;Finally, in the case of acetylene the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\ce{sp}$&lt;/span&gt; hybridized and the hybridization index of each &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond is &lt;span class=&quot;math-container&quot;&gt;$\sqrt1$&lt;/span&gt;.  In this case, using Coulson's theorem we find that the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C-H}$&lt;/span&gt; bond angle is 180 degrees.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Conversely, if the bond angle is known, the hybridization can be determined.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2019-10-30T10:36:30.487" LastActivityDate="2019-10-30T10:36:30.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19988" PostTypeId="2" ParentId="19986" CreationDate="2014-11-25T15:49:36.310" Score="4" Body="&lt;p&gt;Even at room temperature, some of the water molecules in the liquid will have sufficient energy to 'escape' and become vapour. Because the flask is sealed however, some of the water vapour molecules will return to the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, an equilibrium will be reached when the rate at which water molecules which escape the liquid is equal to the rate at which water molecules in the vapour enter the liquid. At this point, the pressure reached by the water vapour is known as the vapour pressure of water (at the given temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'reaction' is not completely reversible since some heat is lost to the surrounding environment through the wall of the flask.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-11-25T15:49:36.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19989" PostTypeId="2" ParentId="19929" CreationDate="2014-11-25T16:21:15.560" Score="1" Body="&lt;p&gt;Let's look at the overall reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na3PO4(aq) + 3AgNO3(aq) -&amp;gt; 3NaNO3(aq) + Ag3PO4(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net ionic equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PO4^{3-}(aq) + 3Ag^+(aq) -&amp;gt;  Ag3PO4(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial concentration of phosphate and silver ions taking into account the dilution are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:&#xA;${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:&#xA;${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$ &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-25T16:21:15.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19990" PostTypeId="2" ParentId="19962" CreationDate="2014-11-25T16:21:37.777" Score="-2" Body="&lt;p&gt;Firstly, 1.44 m of CH3OH yields a solution of 1.046 L, (where 1.44 m means (1.44*32=46.08) grams of CH3OH in 1 Kg of H2o, so solution weight will be 1.046 Kg) but we have a 1 Kg solution, so we say:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1.046 Kg solution contains 46.08 g of CH3OH&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1.000 Kg solution contains    x  g of CH3OH &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;x=CH3OH grams in your solution = 46.08×1000/1046.08 = 44.05 g  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;and water weight in your solution is 1000−44.05=955.95 g &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;now, to prepare the new solution (1 m) we say:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1 kg of water contains 32 g of CH3OH &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;y Kg of water contains 44.05 g of CH3OH &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;y= weight of water required for new solution =44.05×1000/32=1376.56g &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;then 1376.56 - 955.95 = 420.6 g  is water that should be added to the 1.44 m solution to have the 1 m solution .&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: 1 Kg almost equals 1 L (for water)&lt;/p&gt;&#xA;" OwnerUserId="9985" LastEditorUserId="9985" LastEditDate="2014-11-25T16:41:05.880" LastActivityDate="2014-11-25T16:41:05.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19991" PostTypeId="1" CreationDate="2014-11-25T17:03:12.707" Score="4" ViewCount="12929" Body="&lt;p&gt;For example $\ce{Fe}\ (Z=26)$ Short formula: $\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^6\ 4s^2}$. If we make the electronic formula we get 4 single electrons ($s=1/2$), which means that the valence in normal state is 4, but in the periodic table it is $2/3/6$. How can I find its valence?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="7475" LastEditDate="2016-12-30T23:03:57.660" LastActivityDate="2016-12-30T23:16:42.200" Title="How can I derive the preferred oxidation states of a transition metal from its electronic configuration?" Tags="&lt;transition-metals&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19992" PostTypeId="2" ParentId="19946" CreationDate="2014-11-25T17:15:21.973" Score="3" Body="&lt;p&gt;That explains a eutectic but does not answer the question, which is a good one for solid-state chemists. &#xA;A cocrystal is a crystalline, multiple component solid with a distinct stoichiometry (molar ratio) between components where a single component is neutral overall(can exist alone without a counter ion, typically non-ionic but a zwitterion counts as a single neutral component) and the components interact by non-covalent interactions that can be partially ionic and has distinctive crystalline characterization data, i.e. to be crystalline a solid the mixed solid phase needs to have X-Ray or electron diffraction data that distinguishes it from its starting components.&#xA;So, the eutectic phase of 2 solids, or the solid mixture with the lowest melting pt, can be a cocrystal, but a cocrystal is not necessarily a eutectic. A solid mixture with no regular repeating crystal lattice and no consistent molar ratio (stoichiometry) is called a solid solution. So a eutectic phase could be a cocrystal, or a salt, or a mix of those, or physical mix (solid solution).  &lt;/p&gt;&#xA;" OwnerUserId="10004" LastActivityDate="2014-11-25T17:15:21.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19993" PostTypeId="1" CreationDate="2014-11-25T17:29:14.157" Score="13" ViewCount="10546" Body="&lt;p&gt;My book mentions that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Delocalization of $\pi$ electrons makes a compound stable as the process of delocalization involves loss in internal energy&lt;/strong&gt;. This is the synopsis of resonance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how does shifting of $\pi$ electrons let the body lose internal energy? What is the cause?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While studying resonance, I observed that one structure possesses charge while in other no charge. But in the final structure, there is, according to the book, partial charge present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg.&#xA;&lt;strong&gt;In phenol, three resonating structures contain '+' charge on $\ce{O}$ . Thus, in the final structure, there is a slightly positive charge on oxygen and also, small negative charge on the main body.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="42741" LastEditDate="2017-05-19T04:54:35.077" LastActivityDate="2020-09-10T07:21:34.243" Title="Why does delocalization of π electrons bring stability?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;formal-charge&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19994" PostTypeId="2" ParentId="19991" CreationDate="2014-11-25T17:58:18.700" Score="1" Body="&lt;p&gt;Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-25T17:58:18.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19995" PostTypeId="1" AcceptedAnswerId="20001" CreationDate="2014-11-25T18:18:13.563" Score="9" ViewCount="1533" Body="&lt;p&gt;Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/Dimethyl_sulfoxide#Medicine&quot;&gt;says&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;Use of DMSO in medicine dates from around 1963, when an Oregon Health &amp; Science University Medical School team, headed by Stanley Jacob, discovered it could penetrate the skin and other membranes without damaging them and could carry other compounds into a biological system.&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this work? Is there a term for this property? Can DMSO transport molecules through other barriers? What are the limitations on which compounds DMSO can transport? Which other compounds share this property with DMSO?&lt;/p&gt;&#xA;" OwnerUserId="5021" LastActivityDate="2014-11-25T19:30:12.077" Title="How does dimethyl sulfoxide transport other molecules through skin?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19996" PostTypeId="1" CreationDate="2014-11-25T18:28:25.013" Score="2" ViewCount="72" Body="&lt;p&gt;How can you determine the possible rotational states of a molecule? &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1429218126&quot; rel=&quot;nofollow&quot;&gt;Atkins Physical Chemistry book&lt;/a&gt; says that for linear molecules there are two possible rotational states, but what is the general method for finding the amount of rotational states for other relatively more complex molecules, say for example $\ce{H2O}$ and $\ce{CH4}$?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2015-10-23T16:54:37.930" LastActivityDate="2015-10-23T16:54:37.930" Title="How to determine rotational states of a molecule?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19997" PostTypeId="2" ParentId="19993" CreationDate="2014-11-25T18:42:02.737" Score="3" Body="&lt;p&gt;In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-11-25T18:42:02.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19998" PostTypeId="1" CreationDate="2014-11-25T18:43:28.020" Score="3" ViewCount="1091" Body="&lt;p&gt;Which of the following has greater entropy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a.  $\ce{CH3COOH~ (aq)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b.  $\ce{CH3COO^{-}~ (aq) + H+~(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both in the same state, and have the same number of particles.. so how do I figure this out? My instinct says a, because the products will be more &quot;random&quot;, but I'm not completely sure.. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4231" LastEditDate="2014-11-25T18:49:35.023" LastActivityDate="2014-11-25T18:58:05.387" Title="Which has a greater entropy?" Tags="&lt;equilibrium&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19999" PostTypeId="2" ParentId="19996" CreationDate="2014-11-25T18:46:36.177" Score="3" Body="&lt;p&gt;Molecules have only 3 (or 2 if linear) rotational modes.  A molecule with n atoms has 3n-6 vibrational states, 3n-5 if it is linear.  The &quot;6&quot; or &quot;5&quot; that is subtracted represents the 3 translational modes that all molecules can move in (x, y, and z directions) plus the 3 rotational modes (rotation about the x, y and z axes).  The 3 rotational modes reduce to 2 if the molecule is linear since you can't observe rotation about its molecular axis (the axis that contains the $C_{\infty}$ symmetry element).  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T18:46:36.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20000" PostTypeId="2" ParentId="19998" CreationDate="2014-11-25T18:58:05.387" Score="1" Body="&lt;p&gt;In the gas phase I would suspect that the entropies would be equal as both systems have the same number of translational, rotational and vibrational modes (18).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you specifically ask about the liquid phase system, since you show that both &quot;a&quot; and &quot;b&quot; are solvated in aqueous media. In this case, the ionized molecule (b) has charges on both ions and this will cause increased ordering of the solvent molecules surrounding the ions, more so than neutral molecule &quot;a&quot;.  This increased ordering of the system in case &quot;b&quot; will reduce the entropy of the system.  Consequently system &quot;a&quot; will have greater entropy.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T18:58:05.387" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20001" PostTypeId="2" ParentId="19995" CreationDate="2014-11-25T19:30:12.077" Score="8" Body="&lt;p&gt;DMSO interacts with the outer membrane of skin, the stratum corneum.  The stratum corneum is a typical lipid bilayer and DMSO can solvate the exposed hydrophilic lipid head.  This results in a weakening of the lateral forces between the individual phospholipid molecules.  The weakened lateral forces result in a widening of the channel between the phospholipid molecules and the widened channel allows small molecules to pass through the lipid layer.  After the DMSO is removed the lipid bilayer restores itself back to normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCGgS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it is agreed that small molecules can pass through the DMSO-affected bilayer, there is some evidence that larger molecules such as viral agents might also pass through.  There is also some controversy over the effect of DMSO on the blood-brain barrier.  Here is a &lt;a href=&quot;http://www.dmso.org/articles/information/herschler.htm&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to a nice review of DMSO's pharmacological effects, including transport of larger molecules and transport across the blood-brain barrier.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-25T19:30:12.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20003" PostTypeId="1" AcceptedAnswerId="20075" CreationDate="2014-11-25T21:10:13.243" Score="12" ViewCount="2287" Body="&lt;p&gt;I am rather newb in GAMESS field but finally, learned to make input files for most molecular simulations and could even use TD-DFT for excited states (special thanks to Geoff Hutchison for his help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to start simulating simple chemical reactions. I looked at the GAMESS manual and think I need to make transition states and find least energy pathways but have no idea how to make input file for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a simple input file example for a simple reaction?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="402" LastEditDate="2017-04-01T00:20:43.003" LastActivityDate="2017-04-01T00:20:43.003" Title="How to simulate chemical reactions in GAMESS?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20004" PostTypeId="1" AcceptedAnswerId="20015" CreationDate="2014-11-25T21:26:07.510" Score="-1" ViewCount="62" Body="&lt;p&gt;In the reaction &#xA;$$\ce{3 C2H2(g) &amp;lt;=&amp;gt; C6H6(g)}$$&#xA;will the ratio of $\ce{[C2H2(g)]}$ to $\ce{[C6H6(g)]}$ at equilibrium be 3:1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the mole ratio related to the equilibrium (or lack thereof) of the reaction? How do I know if 3:1 is indeed the ratio?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4231" LastEditDate="2014-11-25T22:53:00.913" LastActivityDate="2014-11-26T05:15:18.820" Title="In the following reaction, will the ratio between substances be 3:1?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20005" PostTypeId="1" AcceptedAnswerId="20028" CreationDate="2014-11-25T21:31:26.583" Score="2" ViewCount="133" Body="&lt;p&gt;The following equilibrium occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NOCl(g) &amp;lt;=&amp;gt; 2 NO(g) + Cl2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gaseous mixture of $\ce{NOCl}$, $\ce{NO}$ and $\ce{Cl2}$ is put in a container. After a few minutes it is found that two moles of $\ce{NOCl}$ react for every three moles of products which react. Is the mixture at equilibrium? Why? In other words, how does the mole ratio relate to whether or not the reaction is at equilibrium? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4231" LastEditDate="2014-11-25T22:50:23.690" LastActivityDate="2014-11-26T17:07:49.813" Title="Is this reaction at equilibrium?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20006" PostTypeId="1" CreationDate="2014-11-25T23:16:48.923" Score="2" ViewCount="1829" Body="&lt;p&gt;This is for a recycling project, I am looking to recover brass pins from electronic connectors made of glass reinforced Polybutylene Terephthalate (PBT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to find a chemical which would be able to dissolve Polybutylene Terephthalate (or PBT, a common type of plastic) so that i can separate the brass pieces inside, without dissolving them too. I think that if it dissolves PBT the glass reinforcement will become dust so it won't be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far my search has been unsuccessful, either it was acids which were too strong and would also dissolved] the brass, or common solvent which had no effect on the PBT (tested Acetone). Even 30% Caustic soda doesn't attack the PBT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a method with high temperature methanol under pressure into a closed stainless steel shape, but it's way too complex for me to make. Similarly chemicals which are too cancerogenous are not worth it (at least given that i am not a specialist capable of handling the most dangerous ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has an answer or some ideas i'd love to hear it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="10008" LastActivityDate="2014-11-26T04:05:24.273" Title="Which chemicals could be used to dissolve Polybutylene Terephthalate (PBT), but not dissolve brass?" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20008" PostTypeId="1" AcceptedAnswerId="20017" CreationDate="2014-11-26T00:26:39.067" Score="0" ViewCount="5273" Body="&lt;p&gt;Latex is a natural resource and the elasticity can be increased by heating it with sulfur. My question is why they are have different elasticity than that of latex? I mean like rubber gloves have a good elasticity while rubber bands not really elastic. Is it cause by the amount of sulfur that we heat?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="16683" LastEditDate="2017-04-09T23:55:01.447" LastActivityDate="2017-04-09T23:55:01.447" Title="Elasticity of latex" Tags="&lt;polymers&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20009" PostTypeId="1" AcceptedAnswerId="20012" CreationDate="2014-11-26T01:38:29.150" Score="26" ViewCount="20253" Body="&lt;p&gt;I've asked a similar question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9803/what-happens-to-water-under-high-pressures-without-possibility-of-escape&quot;&gt;here&lt;/a&gt; but the answer given shows the behaviour of water under general conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what the behaviour of water is like as pressures increase towards infinity without being able to escape it's confinement.. i.e. a ball of water at the core of a galactic mass.. maybe this question is more for theoretical physics since we can't really measure or experiment?&lt;/p&gt;&#xA;" OwnerUserId="5117" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2016-11-12T13:48:51.727" Title="Hypothetical: What happens to water as pressure increases to infinity?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;water&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="10" ClosedDate="2015-07-01T12:17:46.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="20010" PostTypeId="2" ParentId="20005" CreationDate="2014-11-26T02:10:39.553" Score="0" Body="&lt;p&gt;When a chemical system is at equilibrium the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium expression is a ratio of the &lt;em&gt;conentrations&lt;/em&gt; of the products raised to their stoichiometric coefficients over the concentrations of the reactants raised to their stoichiometric coefficients. So, by knowing the equilibrium constant for the reaction and the concentrations of the species you can tell in what direction the reaction is going to proceed. &lt;/p&gt;&#xA;" OwnerUserId="4696" LastActivityDate="2014-11-26T02:10:39.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20011" PostTypeId="2" ParentId="20009" CreationDate="2014-11-26T02:48:25.137" Score="32" Body="&lt;p&gt;The phase diagram for water (reported in your link and reproduced below) is a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B888M.jpg&quot; alt=&quot;phase diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram shows us that at pressures around 1 terapascal (about 10 million atmospheres) ice is a solid, at least up to 400 C. It has been predicted (&lt;a href=&quot;http://militzer.berkeley.edu/papers/ice26.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference_1&lt;/a&gt;, &lt;a href=&quot;http://arxiv.org/pdf/1106.1941.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference_2&lt;/a&gt;) that at higher pressures, somewhere between 1.5 and 6 tera pascals, solid ice will undergo an insulator to metal transition and display properties typically associated with metals (band structure, electrical conduction, etc.). That's around 15-60 million atmospheres.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="9961" LastEditDate="2016-11-12T13:48:51.727" LastActivityDate="2016-11-12T13:48:51.727" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20012" PostTypeId="2" ParentId="20009" CreationDate="2014-11-26T03:47:31.517" Score="34" Body="&lt;p&gt;Ron's answer gives us a good idea of what might happen in terms of &quot;normal&quot; chemistry, but if you really mean &quot;increases to infinity&quot;, some very strange stuff happens. Suffice it to say, it doesn't stay water after a certain point. The intense temperatures created by the compression will cause the water to break apart, eventually no longer even having oxygen atoms due to nuclear reactions. Because we're talking about an externally applied pressure, the Chandrasekhar limit doesn't apply, so there is a point at which electrons and protons combine (when the &lt;a href=&quot;http://en.wikipedia.org/wiki/Degenerate_matter&quot;&gt;electron degeneracy pressure&lt;/a&gt; is overcome) and a mass of neutrons remains. Neutrons themselves also have a degeneracy pressure (though we don't have good models to predict the exact pressure that has to be overcome). From here, we don't know what happens with as much certainty, but the formation of quark matter has been predicted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, we reach a singularity. We can think of this as all the matter we had before being compressed into an infinitesimal volume with infinite density and our applied pressure ceases to mean anything. If we started with enough water, this would behave much like any other black hole, though &lt;a href=&quot;http://en.wikipedia.org/wiki/Micro_black_hole&quot;&gt;micro black holes&lt;/a&gt; are hypothesized to have some special properties.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-26T03:47:31.517" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20014" PostTypeId="2" ParentId="20006" CreationDate="2014-11-26T04:05:24.273" Score="1" Body="&lt;p&gt;PBT is a thermoplastic and its melting point is only 223 °C. Depending on its composition, brass doesn't melt until 900 °C or so. High enough heating may change the structural properties of the brass, but perhaps there is a temperature to which you can soften the plastic enough to detach the pins without damaging them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, brass is easily attacked by many chemicals and PBT is resistant to many solvents, but something like hot DMSO or N-methyl-2-pyrrolidone may swell the polymer enough to separate the parts.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-26T04:05:24.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20015" PostTypeId="2" ParentId="20004" CreationDate="2014-11-26T05:15:18.820" Score="7" Body="&lt;p&gt;No. The location of the equilibrium is dependent on many factors, most prominently temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://goldbook.iupac.org/C01023.html&quot;&gt;chemical equilibrium&lt;/a&gt; appears to have a static composition, which is of course not true. The point in question is where the rates of the reaction equal in both directions. The equilibrium can be described by the &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot;&gt;equilibrium constant&lt;/a&gt; $K$, which is defined as the product of partial pressure $p_{\ce{B}}$, concentration $c_{\ce{B}}$, [...], chemical activity $a_{\ce{B}}$, in general $x_{\ce{B}}$ of the reactants and products, with respect to their stoichiometric number $\nu_{\ce{B}}$.&#xA;$$ K_x = \prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}}$$&#xA;The standard equilibrium constant $K^\circ$ is given through the standard reaction Gibbs energy $\Delta_r G^\circ$ and the thermodynamic temperature $T$, $\mathcal{R}$ is the gas constant.&#xA;$$K^\circ = \exp\left\{\frac{-\Delta_r G^\circ}{\mathcal{R}\cdot T}\right\}$$&#xA;Since all equilibrium constants are proportional to each other it follows for a constant temperature, $T = \mathrm{const.}$, that the location is dependent on the Gibbs energy:&#xA;$$ \prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}} \propto \exp\left\{-\Delta_r G^\circ\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In your case $$\ce{3 C2H2(g) &amp;lt;=&amp;gt; C6H6(g)}$$ you would probably choose a pressure based approach $$K_p = \frac{p(\ce{C6H6})}{p^3(\ce{C2H2})} \propto \exp\left\{-\Delta_r G^\circ\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can see the temperature and pressure dependency in the &lt;a href=&quot;http://goldbook.iupac.org/G02629.html&quot;&gt;original definition&lt;/a&gt;:&#xA;\begin{align}&#xA;G(p,T) &amp;amp;= H - TS\\&#xA;H(p,T) &amp;amp;= U + pV\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will only know the product ratio by very accurate calculations or empirical measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the system in question, I believe it has been done and the results may be found in &lt;a href=&quot;https://ia601508.us.archive.org/21/items/jresv37n2p95/jresv37n2p95_A1b.pdf&quot;&gt;W. J. Taylor, &lt;em&gt;et. al.&lt;/em&gt;, J. Res. Natl. Bur. Stand., Vol. 37, No. 2, p. 95&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-11-26T05:15:18.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20017" PostTypeId="2" ParentId="20008" CreationDate="2014-11-26T05:31:35.963" Score="4" Body="&lt;p&gt;The 'elasticity' of a material is property which describes how it changes shape in response to a tensile stress. Specifically, when a material is placed under a tensile stress, it experiences some sort of elongation (strain). If removing the stress results in a return to the original size, the material is described as 'elastic' over this range of stress (note: under sufficiently high stress, all materials will eventually 'yield' to permanent deformation and ultimately 'break').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 'engineering materials terms' the ratio of tensile stress (in MPa) to strain (in mm) within its elastic region is called the 'modulus of elasticity' of the material (aka 'elastic modulus' or 'Young's modulus').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of latex rubber, it has an elastic modulus of about 0.1 GPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural rubber is derived from a milky suspension found in the milky sap of some plants such as the hevea tree. The sap which contains a polymer of isoprene called 2-methyl-1,3-butadiene, as well as a small percentage of proteins and acetone-soluble materials such as resins, fatty acids, and sterols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the isoprene consists of the long-chained stereoisomer, cis-1,4-polyisoprene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oh1Qb.gif&quot; alt=&quot;cis-1,4-polyisoprene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'cis-' indicates that substituent groups are on the same side of the C=C double bond, as shown above. The high cis- content (~99%) of natural rubber latex is what gives its elastic properties. Natural rubber in this form is a thermoplastic, that is, it is soft at higher temperatures but becomes hard and brittle when cooled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vulcanization involves heating the natural rubber with sulfur at $140-180~\mathrm{^\circ C}$ which encourages the formation of cross-links between the long-chained polymers, resulting in a 'thermosetting' elastopolymer. Other additives are also sometimes used, such as N-phenyl-P-naphthylamine antioxidants to reduce deterioration by atmospheric oxygen or ozone, or 'accelerators' such  2-mercaptohenzothiazole, used to speed up the reaction or allow it to occur at lower temperatures. Reinforcing agents such as carbon or silica can also be introduced to added the 'stiffness' and tensile strength as well as resistance to abrasion. Softeners such as fatty acids, wood resins, pine oils or coal tar byproducts are also sometimes added. Finally, colouring agents and fillers (barium sulfide, calcium carbonate) are also sometimes used for aesthetic or to reduce the cost, but sometimes at the expense of reduced strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by combining small amounts of other chemicals to latex, chemists are able to adjust the properties of the rubber in subtle ways to achieve the most desirable mechanical characteristics. &lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="402" LastEditDate="2017-04-09T18:21:27.563" LastActivityDate="2017-04-09T18:21:27.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20018" PostTypeId="2" ParentId="4045" CreationDate="2014-11-26T10:31:30.877" Score="1" Body="&lt;p&gt;Thermosets are liquids or soft solids which irreversibly 'set' or 'cure' when heated. This 'setting' produces covalently  bonded cross-linkages between adjacent molecular chains, resulting in in their mechanical rigidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermoplastics too are formed from linear polymer chains, but their molecules typically have very long chains with high molecular weight, with very few cross-linkages. The weaker van der Waals bonds which link adjacent molecules in thermoplastics are easily broken when heated, allowing the chains to 'slide' past each other, thereby giving them the property of 'softening' when heated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does a big molecule break down before the van der Waals forces&#xA;  break?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case of polytetrafluoroethylene (PTFE), we have a very long, linear polymer chain of carbons covalently bonded with fluorine. PTFE is quite unique because it doesn't have any hydrogen atoms. Instead, it has these large highly electronegative fluorine atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXO4L.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds within the PTFE molecule are very strong (C-C bonds are 607kJ/mole and C-F bonds 552kJ/mole, much stronger than the typical C-H bonds found in other plastics such as polyethylene).  In effect, the PTFE molecule is like a 'rod' of carbons bonded to a surrounding 'blanket' of fluorine atoms (arranged in a slight helical twist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the van der Waals bonds which form between adjacent the adjacent linear 'rod' molecules of PTFE are much weaker than the C-C and C-F bonds within the molecules, these van der Waals forces are able to combine in an additive manner, so the long straight &lt;a href=&quot;http://hasyweb.desy.de/science/annual_reports/2003_report/part1/contrib/46/11170.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PTFE 'rods' are able to be 'packed' together in a hexagonal arrangement&lt;/a&gt;, a bit like pencils in a box. This results in PTFE having a relatively high density and melting point of $327^o C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when PTFE melts, its extremely long molecular chain have a high probability of entanglement, resulting in a extremely high melt viscosity. Although it's melt viscosity varies with the shear stress applied and temperature, in the temperature range between $360^o C$ and $380^o C$ and shear stresses of $0.1MPa$, commercial samples of the granular polymer display viscosities of the order of $10^{10}Pa.s$. This is exceptionally high compared with most other engineering thermoplastic melts. In fact, at atmospheric pressure, it hardly seems to 'melt' in the conventional sense of 'becoming a flowing liquid', although it becomes slightly softer, it forms a near-clear gel with &lt;a href=&quot;http://www.youtube.com/watch?v=sUKUyoiv8ag&quot; rel=&quot;nofollow noreferrer&quot;&gt;very high elasticity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even below it's melting point, PTFE will start to decompose at $260^o C$, releasing toxic compounds through sublimation. When heated above $360^o C$ thermal decomposition is more rapid and the PTFE degrades noticeably. As a consequence, PTFE can not be processed by many 'conventional' thermoplastic techniques, such as injection molding. Instead, it is processed in a finely powdered or granular form, using modified powder metalurgical techniques such as cold-molding, sintering, compression and extrusion, to fabricate rods, tapes, wire, tubing, sheeting and other profiles which are typically suitable for further machining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above $480^o C$ PTFE thermally decomposes in air to release a mixture of vapours and particulates, including carbonyl fluoride, hydrogen fluoride, tetrafluoroethylene and other fluorinated hydrocarbon products, including a small amount of some highly toxic species such as perfluoroisobutylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise nature of the reactions responsible for this decomposition is still subject to research, but it appears likely that it is initiated by random-chain incisions in the PTFE chain, followed by depolymerization. &lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-11-27T09:13:10.887" LastActivityDate="2014-11-27T09:13:10.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20019" PostTypeId="1" AcceptedAnswerId="20032" CreationDate="2014-11-26T10:56:55.687" Score="3" ViewCount="3180" Body="&lt;p&gt;I'd like to ask,  how to make permanent ink, that will be invisible when used, but after like 30 minutes it will appear (preferably dark color)? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="10017" LastEditorUserId="7920" LastEditDate="2014-11-26T20:19:51.907" LastActivityDate="2015-03-16T09:37:05.667" Title="How to make permanent ink that appears after some time?" Tags="&lt;everyday-chemistry&gt;&lt;color&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20021" PostTypeId="1" AcceptedAnswerId="20027" CreationDate="2014-11-26T12:13:18.763" Score="5" ViewCount="862" Body="&lt;p&gt;How many types of ice exist and what is their nomenclature? I know that the ice on Earth is $I_h$, but why so? &lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-11-26T16:58:34.750" Title="Nomenclature for different types of ice" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20023" PostTypeId="2" ParentId="20021" CreationDate="2014-11-26T14:16:15.290" Score="2" Body="&lt;p&gt;There are different types of ice, depending on its cristalline structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess would be that Ice $I_h$ and $I_c$ are named directly after their structure, which is respectively &lt;strong&gt;h&lt;/strong&gt;exagonal and &lt;strong&gt;c&lt;/strong&gt;ubic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbering, which makes Ice I, Ice II, Ice III and so on till Ice XV, seems to be arbitrary, probably ordered in sequence as they have been discovered. This also tells you about the number of different types of ice known: 16 (or 19 if we considering &lt;em&gt;amorphous ice&lt;/em&gt; in its three forms).&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="1532" LastEditDate="2014-11-26T16:26:13.823" LastActivityDate="2014-11-26T16:26:13.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20024" PostTypeId="2" ParentId="20019" CreationDate="2014-11-26T14:23:09.603" Score="0" Body="&lt;p&gt;I do not think there is a precise answer to your question, but I would use some form of pigment/dye in a &lt;em&gt;precursor&lt;/em&gt; state which is able to react (slowly) with air, either oxygen or water vapour, to produce the coloured form of the pigment/dye. Given the type of reaction, especially if it is a form of oxidation, this could be irreversible, therefore making the ink permanent.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-26T14:23:09.603" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20027" PostTypeId="2" ParentId="20021" CreationDate="2014-11-26T16:25:44.393" Score="6" Body="&lt;p&gt;The first thing to note about ice water is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All ice is not crystalline and not all crystalline water is ice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appear to be &lt;a href=&quot;http://www.lpi.usra.edu/meetings/scssi2008/pdf/9014.pdf&quot; rel=&quot;nofollow&quot;&gt;fifteen crystalline phases of ice&lt;/a&gt;, named 'Ice 1' to 'Ice 15' using Roman numerals $I$ to $XV$. There also appears to be &lt;a href=&quot;http://homepage.uibk.ac.at/~c724117/publications/loerting11-pccp.pdf&quot; rel=&quot;nofollow&quot;&gt;three 'polyamorphic' states&lt;/a&gt;:.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$LDA$ : Low Density Amorphous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$HDA$ : High Density Amorphous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$VHDA$ : Very High Density Amorphous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All crystalline ice is tetrahedrically coordinated (where the oxygen atoms are in fixed positions relative to each other but the hydrogen atoms may or may not be disordered but obeying the 'ice rules'), and the polymorphs differ primarily in H-bond angles and next-nearest neighbour separations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice I encompasses all ice found on naturally on earth (in a stable form).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is usually split into two phases called $I_h$ and $I_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hexagonal ice (ice $I_h$) is 'ordinary ice', the form of all natural snow and ice on Earth. This is evidenced in the six-fold symmetry in ice crystals grown from water vapor (that is, snow flakes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cubic ice (ice $I_c$) is a solid form of water that can form in high clouds. Ice Ic is metastable, and irreversibly converts to ice $I_h$. Both forms contain similar layers of hexagonal rings, but differs in the way these layers are connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on their structure and properties of ice, check out '&lt;a href=&quot;http://web.archive.org/web/20091007073915/http://atom.me.gatech.edu/zhut/Courses/Courses_HarvardCollection/caiwei/phasesofice.pdf&quot; rel=&quot;nofollow&quot;&gt;the Many Phases of Ice&lt;/a&gt;'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martin Chaplin maintains a fairly thorough and well referenced online resource on water, which includes a comprehensive section on the &lt;a href=&quot;http://www1.lsbu.ac.uk/water/ice_phases.html&quot; rel=&quot;nofollow&quot;&gt;ice phases&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, Google Books has the definitive '&lt;a href=&quot;http://books.google.ca/books?id=oC941a8lXWIC&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;the Physics of Ice&lt;/a&gt;' by Petrenko &amp;amp; Whitworth available online.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-11-26T16:58:34.750" LastActivityDate="2014-11-26T16:58:34.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20028" PostTypeId="2" ParentId="20005" CreationDate="2014-11-26T17:07:49.813" Score="3" Body="&lt;p&gt;The information you provided (&quot;two moles of NOCl react for every three moles of products which react&quot;) confirms that the reaction occurs as depicted, but is insufficient to establish that the system is at equilibrium. The important missing information is not the relative proportions of these three species in the mixture, but whether those proportions are changing (system is not at equilibrium) or staying the same (system is at equilibrium). &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-11-26T17:07:49.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20030" PostTypeId="1" AcceptedAnswerId="20031" CreationDate="2014-11-26T17:21:53.070" Score="2" ViewCount="222" Body="&lt;p&gt;These days there are a number of fancy balloons available, some of them are very beautiful. As always, they loose their gas overtime and don't rise anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of those new type balloons appear to have been made of plastic and are quite easily re-usable but then you can't be going to the balloon guy every time you want a refill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any home made techniques that can let you fill those up (so they float again) and which generate a gas which is safe to handle at home without any protective gear? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: This question is not motivated by any urge to save extra money, it's just that I find so many of them lying at home useless when they can be filled up again. Looking for something that's simple to make if possible.&lt;/p&gt;&#xA;" OwnerUserId="10025" LastActivityDate="2014-11-26T17:41:33.340" Title="Any home made tricks to fill up a baloon that floats?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20031" PostTypeId="2" ParentId="20030" CreationDate="2014-11-26T17:41:33.340" Score="3" Body="&lt;p&gt;There are some options, not all easily available or safe: &lt;strong&gt;Helium&lt;/strong&gt; is the most used and it most certainly makes them float, being safe at the safe time. However, it's not an &lt;em&gt;home-made&lt;/em&gt; trick. &lt;strong&gt;&lt;strike&gt;Methane&lt;/strike&gt;&lt;/strong&gt; is more easily available (in many cities, methane is piped into homes for domestic heating and cooking purposes) but it's very unsafe for such an use, as it could explode. &lt;strong&gt;Hot air&lt;/strong&gt; is also used (think of hot-air balloons) but I wouldn't know how to make it last for a long time as the air would cool down to room temperature very quickly. Lastly, there's &lt;strong&gt;&lt;strike&gt;hydrogen&lt;/strike&gt;&lt;/strong&gt; which could also be produced with some reagents available at home, but it's as dangerous as methane. &lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-26T17:41:33.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20032" PostTypeId="2" ParentId="20019" CreationDate="2014-11-26T18:26:01.417" Score="4" Body="&lt;p&gt;Depending on what you want to do (what do you want to do?) you might be looking for &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron_gall_ink&quot; rel=&quot;nofollow&quot;&gt;gall ink&lt;/a&gt;.  Briefly, the chemistry works like this: the recipe consists of soluble polyphenols (typically &lt;a href=&quot;http://en.wikipedia.org/wiki/Gallic_acid&quot; rel=&quot;nofollow&quot;&gt;gallic acid&lt;/a&gt;) and a Fe(II) salt.  The preparation must be stored in an air-tight container and even then decomposes over time.  As soon as it is applied to paper and exposed to air, the Fe(II) is oxidized to Fe(III), which forms insoluble deeply coloured complexes with your polyphenol.  The permanent colour takes about a day to develop.  That's why often a non-permanent dye is added so people can see what they are writing.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-11-26T22:29:41.457" LastActivityDate="2014-11-26T22:29:41.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20034" PostTypeId="2" ParentId="20019" CreationDate="2014-11-26T19:53:41.313" Score="1" Body="&lt;p&gt;You are looking for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vat_dye&quot; rel=&quot;nofollow&quot;&gt;Vat dye&lt;/a&gt;.  Speed of color appearance may vary, so further research according to your requirement is needed. &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2014-11-26T19:53:41.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20036" PostTypeId="2" ParentId="20009" CreationDate="2014-11-26T20:46:51.733" Score="3" Body="&lt;p&gt;This could be equivalent to the compression generated by a black hole, hopelessly water as any other fluid is compressed to a high density solid state&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2014-11-26T20:46:51.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20038" PostTypeId="2" ParentId="11134" CreationDate="2014-11-26T22:22:42.203" Score="2" Body="&lt;p&gt;Regarding &lt;strong&gt;Q1&lt;/strong&gt;, &lt;em&gt;Henry's law&lt;/em&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At a constant temperature, the amount of a given gas that dissolves in&#xA;  a given type and volume of liquid is directly proportional to the&#xA;  partial pressure of that gas in equilibrium with that liquid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;A1&lt;/strong&gt; is answered by the definition itself: the system should be at equilibrium.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A2&lt;/strong&gt; is straightforward: if the gas reacted, you would not have a constant pressure as a certain (if not all) quantity of it would be removed from the system&lt;sup&gt;[&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Dissolving_Gases_In_Liquids%2C_Henry&amp;#39;s_Law&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;]&lt;/sup&gt;. This would also keep your system from being at equilibrium, defying the previous assumption.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A3&lt;/strong&gt; and &lt;strong&gt;A4&lt;/strong&gt; both depend on the fact that – as the concentration or the partial pressure rise – the system deviates from an ideal model where the solute is&#xA;infinite-diluted and in such system &lt;em&gt;Raoult's Law&lt;/em&gt; should be used&#xA;instead&lt;sup&gt;[&lt;a href=&quot;http://wikis.lib.ncsu.edu/index.php/CH_431/Lecture_14#Henry.27s_law&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;]&lt;/sup&gt;. So you need a low concentration and partial pressure of your gas.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To answer to &lt;strong&gt;Q2&lt;/strong&gt;, &lt;em&gt;Henry's Law&lt;/em&gt; can be applied to any solvent, whether liquid, vapour or solid&lt;sup&gt;[&lt;a href=&quot;http://www.chemengr.ucsb.edu/~ceweb/courses/che128/pdf/090733%20Avoid%20Common%20Pitfalls.pdf&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;], [&lt;a href=&quot;http://pubs.acs.org.ezproxye.bham.ac.uk/doi/abs/10.1021/ed081p1647&quot; rel=&quot;nofollow&quot;&gt;4&lt;/a&gt;]&lt;/sup&gt;. It has been used, within certain limits, to study absorption of gas in zeolites&lt;sup&gt;[&lt;a href=&quot;http://dx.doi.org/10.1016/S1381-1169(98)00021-1&quot; rel=&quot;nofollow&quot;&gt;5&lt;/a&gt;]&lt;/sup&gt; or hydrogels, but I couldn't find any precise experimental data on it. Only that in such systems it is only a limited law, adequate at low mole fractions, but useful for practical purposes when high precision is not required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Dissolving_Gases_In_Liquids%2C_Henry&amp;#39;s_Law&quot; rel=&quot;nofollow&quot;&gt;Henry's Law, UC Davis University ChemWiki&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;a href=&quot;http://wikis.lib.ncsu.edu/index.php/CH_431/Lecture_14#Henry.27s_law&quot; rel=&quot;nofollow&quot;&gt;North Carolina State University, CH 431 / Lecture 14&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;&lt;sup&gt;3&lt;/sup&gt; &lt;a href=&quot;http://www.chemengr.ucsb.edu/~ceweb/courses/che128/pdf/090733%20Avoid%20Common%20Pitfalls.pdf&quot; rel=&quot;nofollow&quot;&gt;Francis L. Smith and Allan H. Harvey (September 2007). &quot;Avoid Common Pitfalls When Using Henry's Law&quot;.&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;&lt;sup&gt;4&lt;/sup&gt; &lt;a href=&quot;http://pubs.acs.org.ezproxye.bham.ac.uk/doi/abs/10.1021/ed081p1647&quot; rel=&quot;nofollow&quot;&gt;Rosenberg, Robert M., Peticolas, Warner L., Journal of Chemical Education, Volume 81, 11, pp 1647-1652&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;&lt;sup&gt;5&lt;/sup&gt; &lt;a href=&quot;http://dx.doi.org/10.1016/S1381-1169(98)00021-1&quot; rel=&quot;nofollow&quot;&gt;Eric G. Derouane, Zeolites as solid solvents, Journal of Molecular Catalysis A: Chemical, Volume 134, 1–3, pp 29-45&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-26T22:22:42.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20039" PostTypeId="1" AcceptedAnswerId="20040" CreationDate="2014-11-27T00:13:50.387" Score="7" ViewCount="8611" Body="&lt;p&gt;A bonus question would be, if so, why? Is it because it decomposes and hence the expiry date is calculated out?&lt;/p&gt;&#xA;" OwnerUserId="7787" LastEditorUserId="5017" LastEditDate="2014-11-27T01:22:08.287" LastActivityDate="2018-07-08T16:07:08.453" Title="Do chemicals have an expiry date?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20040" PostTypeId="2" ParentId="20039" CreationDate="2014-11-27T00:38:53.527" Score="8" Body="&lt;p&gt;They do indeed. It could be for different reasons: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A chemical might undergo chemical changes due to different storage conditions: it could react with other chemicals present in the air or in the environment. This could change its nature (reaction to create other compounds), properties and/or purity.&lt;/li&gt;&#xA;&lt;li&gt;A chemical might undergo physical changes as well: the temperature and pressure vary, as well as other conditions like humidity, light, electromagnetic fields. This could change it as stated above.&lt;/li&gt;&#xA;&lt;li&gt;Certain chemicals are simply unstable and decay or degradate over time to something else, without the need of externals input / changes. This could affect the chemical as stated above.&lt;/li&gt;&#xA;&lt;li&gt;Certain chemical suffers transpiration losses.&lt;/li&gt;&#xA;&lt;li&gt;Human factors, like contaminations, spillings, loose caps, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, it might be noticed that these reasons could be divided to produce two different dates: &lt;em&gt;shelf life&lt;/em&gt; vs &lt;em&gt;expiration date&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-27T00:38:53.527" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20042" PostTypeId="1" AcceptedAnswerId="20043" CreationDate="2014-11-27T10:04:05.520" Score="2" ViewCount="889" Body="&lt;p&gt;I started to learn Chemistry with something like Hydrogen having 1 electron in the first shell, Oxygen having 2 electron in the first shell and 6 electron in the second shell... But I don't really understand how scientists know it? How do they know there are how many shells and how many electron per shell? Is that something visible? Or just an imagination theory or they using a method to measure? &lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="7448" LastEditDate="2015-04-07T14:36:33.867" LastActivityDate="2015-04-07T14:36:33.867" Title="Electron shell, how do they know it?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20043" PostTypeId="2" ParentId="20042" CreationDate="2014-11-27T10:41:46.400" Score="5" Body="&lt;p&gt;Electrons are not &lt;em&gt;actually&lt;/em&gt; separated in shells. That is a simplified model to explain the atom's structure when people first start approaching Chemistry or Physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, electrons' movements around the nucleus are governed by complex equations (some of them not even resolvable so approximations are needed) to form what is called an &lt;strong&gt;electron cloud&lt;/strong&gt;. This cloud is confined in proximity of the nucleus by the electromagnetic force. Every electron is defined by its wave equation, which tells us the probability that the electron appears to be at a particular location when its position is measured. These equations allows us to define regions in space where the electron spend the most part of its time around the nucleus, and they can be visualised as atomic orbitals, as shown below. Orbitals can have one or more ring or node structures, and they differ from each other in size, shape and orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/un0YT.jpg&quot; alt=&quot;Atomic orbitals&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each orbital (which is defined by a different equation) has different associated parameters, called quantum numbers, and a particular energy. These properties allow us to divide them into levels and sublevels, to form a structure that resembles multiple shells one inside each other. But as you can see from what I explained the structure is in fact far more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This knowledge has been gained over time both from the mathematical and physical analysis of different equations (like Schröndiger equation) and by experimental data collected by the means of different techniques. Starting from Thomson and Rutherford the model of the atomic structure has seen many changes (Bohr, Moseley, Van der Broek, de Broglie, etc.) and it has come closer to a realistic representation of the actual atom. Take a look at their experiments to understand how it has been studied. However, they often involve particle physics and are not really straightforward to understand.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="1532" LastEditDate="2014-11-27T10:55:57.520" LastActivityDate="2014-11-27T10:55:57.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20044" PostTypeId="2" ParentId="20009" CreationDate="2014-11-27T11:17:53.630" Score="4" Body="&lt;p&gt;For example here &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_VII&quot; rel=&quot;noreferrer&quot;&gt;ICE VII&lt;/a&gt; - phase of water on &lt;a href=&quot;http://en.wikipedia.org/wiki/Super-Earth&quot; rel=&quot;noreferrer&quot;&gt;Super-Earthes&lt;/a&gt; - how behaves itself water under big pressure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ice VII is a cubic crystalline form of ice. It can be formed from&#xA;  liquid water above 3 GPa by lowering its temperature to room&#xA;  temperature, or by decompressing (D2O) ice VI below 95 K.&#xA;  ©Wikipedia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10035" LastEditorUserId="10035" LastEditDate="2014-11-27T11:52:10.240" LastActivityDate="2014-11-27T11:52:10.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20045" PostTypeId="1" AcceptedAnswerId="20051" CreationDate="2014-11-27T12:27:39.823" Score="0" ViewCount="187" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have volume $V$ of gas $\ce{A}$ and an equal volume $V$ of gas $\ce{B}$ at room temperature and pressure. I mix, ignite, and find I have a gas (mixture of $\ce{A}$ and product $\ce{C}$) at room pressure, $325~ \mathrm{^\circ C}$, and total volume $3V$. What was the stoichiometry of the reaction?&lt;br&gt;&#xA;  \begin{align}&#xA; \text{a)}&amp;amp;&amp;amp;\ce{2A + B &amp;amp;-&amp;gt; C}\\&#xA; \text{b)}&amp;amp;&amp;amp;\ce{A + 2B &amp;amp;-&amp;gt; 2C}\\&#xA; \text{c)}&amp;amp;&amp;amp;\ce{2A + B &amp;amp;-&amp;gt; 2C}\\&#xA; \text{d)}&amp;amp;&amp;amp;\ce{A – 2B &amp;amp;-&amp;gt; C}\\&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is $\text{b)}$ but I can't seem to get the logic.&lt;br&gt;&#xA;$V$ vol of $\ce{A}$ plus $V$ vol of $\ce{B}$ reacts to give $2V$ vol of $\ce{C}$. How did we find stoichiometry from this? What about some unreacted $\ce{A}$ ?&lt;/p&gt;&#xA;" OwnerUserId="10036" LastEditorUserId="5026" LastEditDate="2018-02-22T11:14:05.417" LastActivityDate="2018-02-22T11:14:05.417" Title="How can I find the stoichiometry of a reaction from the volumes of the reactants and products given?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20046" PostTypeId="1" CreationDate="2014-11-27T12:41:17.870" Score="2" ViewCount="32" Body="&lt;p&gt;The anode region is on the iron where the following reaction occurs: &#xA;$$\ce{Fe(s) -&amp;gt; Fe^2+(aq) + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and at the cathode region, the following reaction occurs&#xA;$$\ce{O2(g) + 4H+(aq) + 4e- -&amp;gt; 2H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the textbook does not explicitly state what the cathode region is. Can someone provide an answer?&lt;/p&gt;&#xA;" OwnerUserId="7069" LastEditorDisplayName="user7951" LastEditDate="2015-06-16T14:57:07.357" LastActivityDate="2015-06-16T14:59:32.817" Title="Cathode region in corrosion" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20048" PostTypeId="1" CreationDate="2014-11-27T13:29:20.590" Score="2" ViewCount="1753" Body="&lt;p&gt;Is there any material that you mix/alloy with copper that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;will result in something that behaves similar to plastic (that is at room temperature it is solid, at maybe 500K or similar it can be easily formed/3D printed)&lt;/li&gt;&#xA;&lt;li&gt;When heated even more (700K?) it will make the the included copper melt, and dissolves itself, leaving only solid copper.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this even theoretically possible? The main goal is to significantly reduce the temperatures involved, 700K is a ballpark number here. Also it doesn't need to be very thick if this makes things easier.&lt;/p&gt;&#xA;" OwnerUserId="8005" LastEditorUserId="8005" LastEditDate="2014-11-27T16:59:22.210" LastActivityDate="2021-05-26T15:40:14.243" Title="Is there a material that reduces copper melting point and completely dissolves?" Tags="&lt;melting-point&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20049" PostTypeId="2" ParentId="20046" CreationDate="2014-11-27T13:35:54.997" Score="3" Body="&lt;p&gt;I would assume that the cathode region is a source of $\ce{O2(g)}$, which is probably the &lt;strong&gt;air&lt;/strong&gt; in contact with the iron. Therefore the reacting should occur at the interface iron/air. &lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="10185" LastEditDate="2015-06-16T14:59:32.817" LastActivityDate="2015-06-16T14:59:32.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20050" PostTypeId="1" CreationDate="2014-11-27T13:51:07.097" Score="2" ViewCount="543" Body="&lt;p&gt;An experiment was carried out involving $\alpha$-bromoheptaldehyde dimethyl acetal. The experimental details are given below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution of enol acetate  (15.6 g, 17.7 mL) in chloroform (20 mL) was cooled&#xA;  to 0°C. A solution of bromine (16 g, 5.1 mL) in chloroform (5 mL) was added&#xA;  slowly with stirring with the rate of addition being controlled to keep the&#xA;  reaction temperature below 10°C. After the completion of addition, anhydrous&#xA;  methanol (60 mL) was added to the mixture and it was stirred for 48 hours.&#xA;  The mixture was diluted with water (200 mL) and ether (50 mL). The organic&#xA;  layer was separated, washed with water (100 mL) and with 5% aqueous&#xA;  sodium carbonate (100 mL). The organic layer was dried using sodium sulfate and&#xA;  filtered. The solvents were removed by distillation under reduced pressure.&#xA;  The residual oil was distilled under reduced pressure in the presence of a&#xA;  small amount of sodium carbonate. The fraction boiling at about 118°C at 17&#xA;  mmHg was collected to give bromo acetal as a colourless oil (19.5 g). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) Do you think that this reaction should be done using a drying tube, under&#xA;  nitrogen or open to the air? Explain your answer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) Without doing TLC or any other form of spectroscopy or chromatography,&#xA;  how would you be able to tell if sufficient Bromine had been added?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) Will the organic layer be the upper or the lower layer? How can you&#xA;  quickly check this experimentally?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(d) Sodium carbonate is added just before the final distillation. Why?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) The procedure calls for removal of the organic solvents by “distillation&#xA;  under reduced pressure”. Which one piece of laboratory apparatus would you&#xA;  use for this?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(f) Suggest two serious and different hazards associated with this experiment. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) The organic layer will be at the bottom. Since it is know that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) To remove any excess acid that might be present in the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) Rotatory evaporator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(f) Bromine is toxic if inhaled, while methanol is a carcinogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure of my answers though. Especially part (c) as I can't decide whether the aqueous or organic layer will be at the top. Could anyone explain this.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="5026" LastEditDate="2018-03-01T06:14:50.067" LastActivityDate="2018-03-01T06:18:10.050" Title="Procedures in experimental chemistry" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20051" PostTypeId="2" ParentId="20045" CreationDate="2014-11-27T14:21:53.563" Score="0" Body="&lt;p&gt;In a problem like this I think that the best way to answer the question is to set up the equations and try all the answers. However, in this case we can infer from the text that $B$ need to be the limiting reacting, as all of it reacts but some of $A$ is left unreacted. The only answers providing this are &lt;strong&gt;answer b&lt;/strong&gt; and &lt;strong&gt;&lt;strike&gt;answer d&lt;/strike&gt;&lt;/strong&gt;. However, the latter contains a minus sign which doesn't makes sense chemically, so let's focus on the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute the moles involved in the reaction using the ideal gas equation $PV = nRT$. Let's assume the volume is $1 \mathrm{L}$, we can calculate the amount of moles for both $A$ and $B$ (as they have the same volume and are present in the same conditions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \dfrac{PV}{RT} = 40.88 \mathrm{mmol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stoichiometry is the one proposed in &lt;strong&gt;answer b&lt;/strong&gt; ($\ce{A + 2B -&amp;gt; 2C}$) &#xA;then the whole quantitity of $B$ will react with half the quantity of $A$ to give the same amount of moles of compound $C$, which will be $n = 40.88 \mathrm{mmol}$. Half of $A$ will be left unreacted, which is $n = 20.44 \mathrm{mmol}$. Using the aforementioned equation we can now compute their volumes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \dfrac{nRT}{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is $V_a = 1 \mathrm{L}$ and $V_c = 2 \mathrm{L}$, which in total is $V_{tot} = 3 \mathrm{L}$, exactly three times more than the original volume.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="1532" LastEditDate="2014-11-28T19:32:08.910" LastActivityDate="2014-11-28T19:32:08.910" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20052" PostTypeId="1" AcceptedAnswerId="20058" CreationDate="2014-11-27T14:24:25.923" Score="8" ViewCount="2214" Body="&lt;p&gt;Say, optically pure 2-bromobutane undergoes $\mathrm{S_N2}$ reaction with an alkali to form 2-butanol. The configuration about chiral carbon will now be inverted, but is it necessary that the sign of optical rotation will be opposite to that of substrate? i.e. if optical rotation of optically pure 2-bromobutane is $+x^\circ$, will optical rotation of the product be $- y^\circ$?    &lt;/p&gt;&#xA;" OwnerUserId="9984" LastEditorUserId="7475" LastEditDate="2015-10-09T15:39:11.203" LastActivityDate="2018-01-06T19:20:17.477" Title="Does inversion of configuration necessitate reversed optical rotation?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20053" PostTypeId="2" ParentId="19903" CreationDate="2014-11-27T14:51:13.610" Score="5" Body="&lt;p&gt;You can find information about the preparation of &lt;strong&gt;Urushibara nickel&lt;/strong&gt; (U-Ni) in the original paper by Urushibara&lt;sup&gt;[&lt;a href=&quot;http://dx.doi.org/10.1246%2Fbcsj.25.280&quot;&gt;1&lt;/a&gt;]&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the differences with Raney nickel (R-Ni), surface characterization by means of XPS&lt;sup&gt;[&lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1979/F1/F19797502027#!divAbstract&quot;&gt;2&lt;/a&gt;]&lt;/sup&gt; had shown that R-Ni catalysts contain metallic aluminium in the activated phase as well as oxidized aluminium, while U-Ni catalysts contain metallic zinc in the activated phase in addition to oxidized zinc. Also, U-Ni catalysts can be described as supported nickel catalysts, while R-Ni catalysts are shown to have the well known skeletal structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activity of U-Ni catalysts is very similar to R-Ni catalysts, both in nature of the reaction catalysed and in the rate of catalysis. One of the advantage of U-Ni is their simplicity of preparation, while R-Ni catalytic synthesis requires a long time. Despite this, U-Ni catalysts are also not very sensitive to impurities, so they can be prepared from commercial chemicals of ordinary grade, and they can be stored for a long time without decreasing the activity. Another advantage is the safer handling, since U-Ni catalysts are not inflammable upon contact with air, in contrast to Raney catalysts which are pyrophoric. This allows the catalyst to be recovered and regenerated. However, it should be noted that U-Ni catalyst durability is not satisfactory during the reaction, but this is also the case of R-Ni catalysts, so there is not really a big difference when it comes to practical use, maybe slightly supporting the use of U-Ni vs R-Ni as the former is less expensive and easy to produce and regenerate. The only real disadvantage of U-Ni catalysts is that for some particular reductions, they show less activity than R-Ni catalysts, favouring one reaction over another, for example U-Ni catalysts particularly favours the partial hydrogenation of acetylenic compounds to ethylenes&lt;sup&gt;[&lt;a href=&quot;https://www.erowid.org/archive/rhodium/chemistry/urushibara.html&quot;&gt;3&lt;/a&gt;]&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;a href=&quot;http://dx.doi.org/10.1246%2Fbcsj.25.280&quot;&gt;Urushibara, Yoshiyuki. &quot;A New Method of Catalytic Hydrogenation&quot;. Bulletin of the Chemical Society of Japan 25 (4): 280–280&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;sub&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1979/F1/F19797502027#!divAbstract&quot;&gt;Y. Okamoto, Y. Nitta, T. Imanaka and S. Teranishi, J. Chem. Soc., Faraday Trans. 1, 1979, 75, 2027&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;sub&gt;&lt;sup&gt;3&lt;/sup&gt; &lt;a href=&quot;https://www.erowid.org/archive/rhodium/chemistry/urushibara.html&quot;&gt;Kazuo Hata, “New Hydrogenation Catalysts: Urushibara Catalysts” (1971), pp 235-237&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-27T14:51:13.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20054" PostTypeId="1" CreationDate="2014-11-27T15:18:49.333" Score="2" ViewCount="813" Body="&lt;p&gt;I need a good book that explains the basics of chemistry in a lucid and importantly, systematic manner. Most books on this subject are not comprehensive and assume that the reader is well acquainted with certain topics. Also, most chemistry books have references to later chapters. And later chapters, references to earlier ones. This makes it very confusing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a book that assumes (effectively) no prior knowledge and builds up the basics needed to understand chemistry (structural theory, acid-base theory, electrode potentials) to a level adequate to understand undergraduate level organic and inorganic chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I need this book is because frequently, while studying inorganic and to a certain extent organic chemistry, I am unable to understand the rationale behind a certain reaction. And most textbooks do not take the effort of explaining it. So, chemistry, for me,  just reduces to memorizing a set of reactions. I think solid basics should remedy that. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2014-11-27T16:47:43.877" Title="Good book for foundational chemistry" Tags="&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-11-28T02:14:33.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="20055" PostTypeId="2" ParentId="20050" CreationDate="2014-11-27T16:03:40.467" Score="1" Body="&lt;p&gt;I believe that when you say &quot;a solution of enol acetate&quot; you are refering to a solution of the corresponding enol-acetal of hetanal? You should use full names of the compounds used, if not, it might be confusing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My opinions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.&lt;br&gt;&#xA;  It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see a reason you would need any inert conditions for this experiment. I think you will be fine if you perform it open to the air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?&lt;br&gt;&#xA;  (b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed. The end point of the reaction is reached when the calculated amount is absorbed and the bromine is no longer decolorized.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?&lt;br&gt;&#xA;  The organic layer will be at the bottom. Since it is how that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is wrong. The organic layer IS the one which is will be a mixture formed by chloroform, unreacted enol acetate and ether. The other phase will be the one composed mainly by water and methanol.&#xA;Methanol+water solution will probably be less dense than chloroform+ether+enolacetate solution, and therefore, organic layer will be the upper one. I cannot tell for sure, since you have mixtures in high proportions of compounds with quite different densities, but and easy way to tell is:&#xA;- Separate both phases using an extraction funnel. Put some tap water on the separated phase you suspect is the aqueous phase. It dissolves? That is the aqueous phase. It does not dissolve and forms a two-phased system? It was the organic phase, extract it again.&#xA;You can double check adding water to the other phase you extracted, and see that the opposite effect should be observed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(d) Sodium carbonate is added just before the final distillation. Why?&lt;br&gt;&#xA;  To remove any excess acid that might be present in the mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yeah. The acetal must be free of acid; otherwise decomposition takes place during distillation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?&lt;br&gt;&#xA;  Rotatory evaporator&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. Rotovap is your guy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(f) Suggest two serious and different hazards associated with this experiment.&lt;br&gt;&#xA;  Bromine is toxic if inhaled, Methanol is a carcinogen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bromine will definitely be your top priority concern. Even more if you have to prepare the solution from pure bromine. Also, it has a very strong unpleasant smell, which gives it the name it has (from Greek: βρῶμος, brómos, meaning &quot;strong-smelling&quot; or &quot;stench&quot;) &#xA;It is corrosive very toxic, volatile (fumes a lot, MUST be handled on a fumehood).&#xA;Chloroform is also toxic; Heptanal is volatile and flammable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methanol, even though you should not drink it wouldn't be of an important concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up, some of the reactants are of unpleasant odor or have lachrymatory effects it is best to use a good hood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also a reference: Organic Syntheses, Coll. Vol. 3, p.127 (1955); Vol. 29, p.14 (1949).&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5026" LastEditDate="2018-03-01T06:18:10.050" LastActivityDate="2018-03-01T06:18:10.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20056" PostTypeId="2" ParentId="20048" CreationDate="2014-11-27T16:15:20.033" Score="1" Body="&lt;p&gt;The usual metal 3D-printing processes usually involve metal powder and a plastic binder.  After printing, the green part is sintered.  See &lt;a href=&quot;http://www.exone.com/en/materialization/what-is-digital-part-materialization/explanation-technology&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-11-27T16:15:20.033" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20057" PostTypeId="2" ParentId="20054" CreationDate="2014-11-27T16:30:49.163" Score="2" Body="&lt;p&gt;For the concepts underlying General Chemistry find that &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/006042995X&quot; rel=&quot;nofollow&quot;&gt;Huheey's book&lt;/a&gt; gives very lucid explanations.  As far as Organic Chemistry goes, vol. A of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0387683461&quot; rel=&quot;nofollow&quot;&gt;Carey-Sundberg's opus&lt;/a&gt; might be what you are asking for.  What level of text are we looking at?&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-11-27T16:47:43.877" LastActivityDate="2014-11-27T16:47:43.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20058" PostTypeId="2" ParentId="20052" CreationDate="2014-11-27T16:45:43.703" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;if optical rotation of optically pure 2-bromobutane is some (+X)&#xA;  degrees, will optical rotation of the product be some (-Y) degrees?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, there is no correlation between the configuration at a chiral carbon and the sign of optical rotation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example (from the Aldrich catalogue), the following chiral bromo compound, &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/468347?lang=en&amp;amp;region=US&quot; rel=&quot;noreferrer&quot;&gt;(R)-(+)-2-Bromo-3-methylbutyric acid&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nNHLt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has an optical rotation of [α]22/D +21°, c = 37 (c is the sample concentration in grams/100 ml solution) in benzene.  The inverted hydroxyl analogue, &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/379093?lang=en&amp;amp;region=US&quot; rel=&quot;noreferrer&quot;&gt;(S)-(+)-2-Hydroxy-3-methylbutyric acid&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WlVfm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also rotates light in the (+) direction.  Its optical rotation is [α]20/D +19°, c = 1 in chloroform.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-27T16:45:43.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20059" PostTypeId="1" CreationDate="2014-11-27T16:56:12.367" Score="1" ViewCount="1353" Body="&lt;p&gt;i made a synthesis of this material in the lab and now i got the H-NMR specta , and by litreture i know that i got the right product.&#xA;im having trouble to match the hydrogens to the peaks,i tried to look at the molecule in it chair form, i know that the 2 hydrogens(A) attached to the cl cl group are in the highest ppm than the other 4 (B), but on the last 4 i got 2 different multiplets, which is a match to which hydrogens? i added the  spectrum&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tV2SM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ufks.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9967" LastActivityDate="2014-12-31T14:04:51.873" Title="7,7-dichlorobicyclo[4.1.0]heptane NMR solving" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20061" PostTypeId="1" AcceptedAnswerId="20064" CreationDate="2014-11-27T17:08:25.093" Score="31" ViewCount="60625" Body="&lt;p&gt;Obviously it depends on the context whether you would call a particular species a nucleophile or a base but are the two terms largely synonymous or is there a difference?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2017-03-09T13:23:26.367" Title="What's the difference between a nucleophile and a base?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20062" PostTypeId="1" AcceptedAnswerId="20063" CreationDate="2014-11-27T17:48:21.017" Score="0" ViewCount="1048" Body="&lt;p&gt;If you were given a question like below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the mass in grams of a sample of $\ce{Fe2(SO4)3}$ that&#xA;  contains $3.59 \times 10^{23}$ sulfate ions, $\ce{SO4^{2−}}$ ? The molar mass&#xA;  of $\ce{Fe2(SO4)3}$ is $399.91 \, \text{g}/\text{mol}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would you go about solving a question like that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused about how to use only the quantity of sulfate ions to find the mass in grams of the whole sample, given it's grams per mole. &lt;/p&gt;&#xA;" OwnerUserId="10042" LastEditorUserId="4990" LastEditDate="2014-11-27T22:25:31.663" LastActivityDate="2014-11-27T22:25:31.663" Title="How would you find the mass in grams, if you only know the number of particles of a portion of the formula?" Tags="&lt;molecules&gt;&lt;ions&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-06T01:06:57.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="20063" PostTypeId="2" ParentId="20062" CreationDate="2014-11-27T18:58:15.480" Score="4" Body="&lt;p&gt;The formula $\ce{Fe2(SO4)3}$ tells us that 3 molecules of $\ce{SO4^{2-}}$ exist for every $\ce{Fe2(SO4)3}$ molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\ce{1 Fe2(SO4)3 molecule}}{\ce{3 SO4^{2-} molecule}} * 3.59*10^{23} \ce{SO4^{2-} molecule}$ can be used to find the number of molecules of ferric sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then use $\frac{1 \ce{mole}}{6.022*10^{23} \ce{molecule}}$ to find the moles of ferric sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then use the given information to solve for grams of ferric sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write a comment if there's any ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2014-11-27T18:58:15.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20064" PostTypeId="2" ParentId="20061" CreationDate="2014-11-27T19:46:50.143" Score="27" Body="&lt;p&gt;The two are related, in that most nucleophiles are (Lewis) bases and vice versa. Some good nucleophiles are also strong bases, e.g. $\ce{HO-}$. However, a species can be a good nucleophile and a weak base, e.g. $\ce{I-}$; or a species can be a weak nucleophile and a strong base, e.g. $\ce{t-BuO-}$. How can we separate this behavior?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nucleophilicity is a kinetic phenomenon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophile#Nucleophilicity&quot; rel=&quot;noreferrer&quot;&gt;Nucleophilicity&lt;/a&gt; is most often defined based on the relative rate of the reactions of nucleophiles with a standard substrate in a standard solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a standard reaction might look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3I -&amp;gt;[Nu-][H2O] CH3Nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleophilicity will be related to the relative rate constant of reaction with the nucleophile (relative to the rate constant of the reaction with water $\equiv 1$).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Basicity is a thermodynamic phenomenon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basicity is based on the position of equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B + HSol &amp;lt;=&amp;gt; BH+ + Sol-}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2016-11-25T17:59:56.680" LastActivityDate="2016-11-25T17:59:56.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20065" PostTypeId="2" ParentId="20061" CreationDate="2014-11-27T19:51:45.643" Score="3" Body="&lt;p&gt;There is a difference indeed: &lt;strong&gt;basicity&lt;/strong&gt; is a particular kind of &lt;strong&gt;nucleophilicity&lt;/strong&gt;. A &lt;strong&gt;nucleophile&lt;/strong&gt; is a chemical species that donates an electron pair to an electrophile. A nucleophile can also be called a &lt;strong&gt;base&lt;/strong&gt; when this donation occurs towards a particular electrophile, which is an hydrogen ion (a proton). &lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-27T19:51:45.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20066" PostTypeId="1" CreationDate="2014-11-27T22:05:04.737" Score="2" ViewCount="285" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A cell consists of a copper electrode $\ce{Cu}$, in contact with $1.800\cdot10^{-2}~\mathrm{M}$ cupric ion and a platinum electrode, $\ce{Pt}$, in contact with $1.400\cdot10^{-1}~\mathrm{M}$ ferrous ion and $3.700\cdot10^{-2}~\mathrm{M}$ ferric ion in $1~\mathrm{M}$ $\ce{HCl}$.&lt;br&gt;&#xA;  What is the potential of cell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would have an idea of what to do with this question if it had a concentration for copper in the cupric half reaction; I would just use the nernst equation, rearranged to solve for $E$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^\circ - \frac{\mathrm{R}T}{n\mathrm{F}}\log Q$$&#xA;Then plug in the concentration values for the cupric half-reaction into $Q$ to get the $E$ value for this half reaction. I'm assuming that the half reaction taking place here is:&#xA;\begin{align}&#xA;\ce{Cu^{2+} + 2e^- &amp;amp;-&amp;gt; Cu^0} &amp;amp;E^\circ &amp;amp;= +0.337~\mathrm{V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other half reaction taking place would be:&#xA;\begin{align}&#xA;\ce{Fe^{3+} + e^{-} &amp;amp;-&amp;gt; Fe^{2+}} &amp;amp;E^\circ &amp;amp;= +0.771~\mathrm{V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I would plug the concentration values, along with their standard potentials, into the rearranged nernst equation to get the $E$ values for each half, then subtract the larger value from the smaller. I already did this, assuming that the concentrations of $\ce{Cu^0}$ and $\ce{Cu^{2+}}$ were the same, giving a logarithm of $0$ and making $E = E^\circ$. The other equation remained the same. I got an answer of $0.399~\mathrm{V}$, which is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm not sure how the concentration for $\ce{HCl}$ factors into this question, because the half reaction for $\ce{Fe}$ doesn't have any hydrogens in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me a hand with this one?&lt;/p&gt;&#xA;" OwnerUserId="9604" LastEditorUserId="5026" LastEditDate="2018-02-15T14:21:06.350" LastActivityDate="2018-02-15T14:21:06.350" Title="Half cell method of voltage calculation in an electrochemical cell" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20071" PostTypeId="2" ParentId="20066" CreationDate="2014-11-28T04:03:22.337" Score="2" Body="&lt;p&gt;Think about what state $\ce{Cu^0}$ is in. $\ce{Cu^2+}$ is dissolved in the solution, but is this true of the unionized copper? Try to write out the half-reactions with the physical states included. What activity (effective concentration) do solids have in an equilibrium expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, though there are some reactions that could happen involving the acid, and I don't know whether they actually contribute to the cell potential in this case, the way the question is written, it looks like the two half-cells you have are the significant ones.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-28T04:03:22.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20074" PostTypeId="1" CreationDate="2014-11-28T06:25:15.780" Score="1" ViewCount="369" Body="&lt;p&gt;Find $[\ce{H3O+}]$ of 0.100 M $\ce{NaHCO3}$ ($K_{\ce{H2CO3},1}=4.45×10^{-7}$, $K_{\ce{H2CO3},2}=4.7×10^{-11}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work is: $$\ce{HCO3- + H2O&amp;lt;=&amp;gt;H2CO3 + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial: $[\ce{HCO3-}]$ = 0.100 M, $[\ce{H2CO3}]$ = 0, $[\ce{OH-}]$ = 0&lt;br&gt;&#xA;Change: $[\ce{HCO3-}]$ = -X, $[\ce{H2CO3}]$ = +X, $[\ce{OH-}]$ = +X&lt;br&gt;&#xA;Equilibrium: $[\ce{HCO3-}]$ = 0.100 - X, $[\ce{H2CO3}]$ = X, $[\ce{OH-}]$= X  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{b2}=\frac{Kw}{K_{a1}} = 2.25×10^{-8}$&lt;br&gt;&#xA;$K_{b2} &amp;lt; 10^{-3}: 0.100 - x = x$&lt;br&gt;&#xA;$K_{b2} = \frac{x^2}{0.100}$&lt;br&gt;&#xA;$X = [\ce{OH-}] = 4.47×10^{-5}$ so $[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$, $[\ce{H+}] = 2.11×10^{-10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the second equation is &#xA;      $$\ce{HCO3- + H2O&amp;lt;=&amp;gt;CO2 + H3O+}$$&#xA;For this equation $K_{a2}=4.7×10^{-11}$&lt;br&gt;&#xA;So when I compared $K$ value between both of these equation, I will get $K_{b2} &amp;gt;&amp;gt; K_{a2}$ (more than $10^3$ times). So I think increase $[\ce{H3O+}]$ which occur or $[\ce{CO2}]$ can be ignored. So my answer $[\ce{H3O+}]$ is equal to $2.11×10^{-10}$ M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; the solution in my book is:&lt;br&gt;&#xA;From the formula:&lt;br&gt;&#xA;$[\ce{H3O+}]= \sqrt{\dfrac{K_1 K_2 × [\ce{NaHCO3}] + k_1 k_w}{[\ce{NaHCO3}] +K_1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{H3O+}]=4.6×10^{-9}$ (solution answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I approximately second equation false or $K$ value is false or another thing?&lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="12860" LastEditDate="2015-11-13T20:19:22.767" LastActivityDate="2015-11-13T20:19:22.767" Title="Find H3O+ of weak acid salt solution which compose of solute that can gain and give protons" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20075" PostTypeId="2" ParentId="20003" CreationDate="2014-11-28T09:03:09.343" Score="12" Body="&lt;p&gt;As in geometry optimization, you are searching for stationary point on the potential energy surface (PES). Not for local minimum, but for saddle point, therefore in GAMESS, you specify &lt;code&gt;RUNTYP=SADPOINT&lt;/code&gt;. You would also need the correct (non-guess) Hessian matrix, which you can calculate separately with &lt;code&gt;RUNTYP=HESSIAN&lt;/code&gt;. In transition state (TS), you should have zero gradients (stationary point) and one imaginary eigenvalue of Hessian (the reaction coordinate in TS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For working example, see e.g. &lt;a href=&quot;http://molecularmodelingbasics.blogspot.de/2009/08/finding-transition-state-sn2-reaction.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;SN2 reaction&quot;&gt;SN2 transition state&lt;/a&gt;. In principle you need very good guess at TS geometry, which you obtain by scanning along some chosen coordinate. The point highest in energy should be reasonably similar to the TS, and you optimize to saddle point starting from there. Already at the guess geometry you must have one imaginary Hessian eigenvalue (resembling the reaction you are looking for, not methyl rotation somewhere else).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But beware, finding the transition states is one of the most difficult topics in computational chemistry and you are not guaranteed to find one, no matter how hard you try. One of the biggest problem is solvation, as you often wish to find reaction barrier in solvent (where majority of chemistry happens), but you can calculate only in vacuum. Just think of how carefully the workbench chemist finds the correct solvent for given reaction, in computer you have only very crude approximations to it (and most often just vacuum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it is great fun and rewarding to find the transition states. For best results, use the simplest theory available, as it is better to have smooth and well behaving PES, even though it is slightly wrong, so it is no harm to start with HF or BLYP (or B3LYP). To get some ideas, start with simple reactions, i.e. where no charged species are involved, so electrocyclic reactions are probably best choice, e.g. Diels-Alder. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="194" LastEditDate="2017-03-29T04:07:31.080" LastActivityDate="2017-03-29T04:07:31.080" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20076" PostTypeId="1" AcceptedAnswerId="20081" CreationDate="2014-11-28T09:14:59.313" Score="17" ViewCount="10284" Body="&lt;p&gt;I learned the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + H+ -&amp;gt; H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I know $\ce{H+}$ and $\ce{H3O+}$ really mean the same thing. But I am confused as to when I should use $\ce{H+}$ and when I should use $\ce{H3O+}$? Is any one of them a more accurate representation?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="16683" LastEditDate="2016-05-02T22:45:12.700" LastActivityDate="2016-05-02T22:46:09.793" Title="How should the hydrated proton be represented in chemical equations?" Tags="&lt;acid-base&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20077" PostTypeId="2" ParentId="20076" CreationDate="2014-11-28T11:38:28.377" Score="9" Body="&lt;p&gt;There is no real difference between the two notations. As you know, $\ce{H+}$ cannot exist on its own in solution as it gets hydrated to $\ce{H3O+}$, but in most cases it can be used when writing equations or working with reactions. To my knowledge, $\ce{H3O+}$ is slightly preferred when dealing with acid/base reactions,  e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + HCl -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while $\ce{H+}$ is used more – for simplicity – with redox reactions, e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + 2H+ -&amp;gt; Zn^{2+} + H2}$$&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="5160" LastEditDate="2015-10-26T16:23:04.890" LastActivityDate="2015-10-26T16:23:04.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20079" PostTypeId="1" AcceptedAnswerId="20121" CreationDate="2014-11-28T14:38:31.217" Score="3" ViewCount="113" Body="&lt;p&gt;Why is the usual percent concentration of a concentrated acid's solution similar to its molar mass? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                FW (amu)     mass %&#xA;sulfuric        98.08        98%&#xA;nitric          63.00        68%&#xA;hydrochloric    36.46        36%&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there some reason for this similarity or is it coincidence?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="275" LastEditDate="2014-11-29T00:12:38.893" LastActivityDate="2014-11-29T13:28:26.147" Title="Why is the maximal percent concentration of acids similar to their molar mass" Tags="&lt;acid-base&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20081" PostTypeId="2" ParentId="20076" CreationDate="2014-11-28T16:08:06.150" Score="25" Body="&lt;p&gt;In addition to entropid's answer, let's remember why we invoke the hydronium ion $\ce{H3O+}$ in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use $\ce{H3O+}$ as a shorthand for $\ce{H+(aq)}$, which looks more like protonated water clusters of the generic formula $\ce{H+.(H2O)_{$n$}}\equiv \ce{H_{$2n+1$}O_{$n$}+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost ten years ago, a &lt;a href=&quot;http://www.sciencemag.org/content/308/5729/1765.full&quot; rel=&quot;noreferrer&quot;&gt;very interesting paper&lt;/a&gt; appeared in &lt;em&gt;Science&lt;/em&gt; that examined the structure of these clusters ($n=2,3,4,5,6,7,8$). The &lt;a href=&quot;http://www.sciencemag.org/content/308/5729/1765.full&quot; rel=&quot;noreferrer&quot;&gt;linked article&lt;/a&gt; has been made freely available by &lt;em&gt;Science&lt;/em&gt;, but you need to register for a free account at sciencemag.org.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the key points of the article was the determination of the inner structure of these clusters. For $n=1$, you have $\ce{H3O+}$, the hydronium ion, which we are reasonably familiar with. When $n=2$, you have $\ce{H5O2+}$, which has a different structure: the proton is evenly shared between two water molecules: $\ce{[H2O\bond{...}H\bond{...}OH2]+}$. Higher order clusters have these two structures, termed the &quot;Eigen&quot; and &quot;Zundel&quot; ions, respectively, at their cores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study used a combination of theory (MP2/aug-cc-pDVZ) and experiment (photodissociation vibrational spectroscopy). They found the for $n=2,6,7,8$, the clusters have a Zundel core, while the Eigen core exists for $n=3,4,5$. Whoa! And it gets more complicated as you increase the number of &lt;a href=&quot;http://www.sciencemag.org/content/304/5674/1137.full&quot; rel=&quot;noreferrer&quot;&gt;water molecules in the clusters&lt;/a&gt; (might be behind a pay wall) - though it turns out the Eigen ion is slightly more common in larger clusters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solutions, we can have an extensive dynamic network of clusters always in flux. In which case, we have a mix of solvated Eigen $\ce{H3O+}$ ions representing protons closely associated with water molecules and Zundel $\ce{H5O2+}$ ions representing protons in transit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we use $\ce{H3O+}$ to keep things simple, and also to satisfy the needs of Brønsted-Lowry acid-base theory, in which every acid-base reaction needs a &lt;strong&gt;proton-donor&lt;/strong&gt; and a &lt;strong&gt;proton-acceptor&lt;/strong&gt;. Thus, the &quot;dissociation&quot; of acids in water is not really a dissociation, but an ionization reaction - an acid-base reaction in which water is the base and hydronium is the conjugate acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA + H2O &amp;lt;=&amp;gt; A- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it is only appropriate to use $\ce{H3O+}$ in aqueous solution. In other solvents, the &quot;proton&quot; has a different structure. In general, the proton is attached to a basic site on the solvent molecule. For example, in methanol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + CH3OH &amp;lt;=&amp;gt; Cl- + CH3OH2+}$$ &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2016-05-02T22:46:09.793" LastActivityDate="2016-05-02T22:46:09.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20082" PostTypeId="1" AcceptedAnswerId="20126" CreationDate="2014-11-28T17:29:52.563" Score="0" ViewCount="169" Body="&lt;p&gt;I think everyone might be aware that +6 is considered the most stable form of sulfur. But I think why this reaction happens very slowly and at such a high temperature and even the help of a catalyst is required, but in almost all redox reactions sulfur goes to +6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(controversial where +6 is forms very forcefully):&#xA;$$\ce{SO2 +O2-&amp;gt;[\ce{Pt/V2O5}][\text{high temp.}]SO3}$$&#xA;And also does it reduces easily (controversial where +6 is reduced easily):&#xA;$$\ce{P +SO3-&amp;gt;P4O10 +SO2}$$&#xA;The reactions in my favour are(where stable +6 is formed):&#xA;$$\ce{SO2 +Cr2O7^2- -&amp;gt;SO4^2- +Cr^3+}$$&#xA;$$\ce{BaSO3- -&amp;gt;[\text{warming/$\ce{Br2-}$water}][\ce{/HNO3/H2O2/$\text{dil.}$HCl}]BaSO4 }$$&#xA;$$\ce{[AgSO3]- -&amp;gt;[\text{boil}]Ag + SO4^2- +SO2}$$&#xA;$$\ce{Ag2SO3+H2O -&amp;gt;Ag + SO4^2- +H+}$$&#xA;$$\ce{SO3^2- +MnO4- -&amp;gt;Mn^2+ +SO4^2- }$$&#xA;$$\ce{PbSO3-&amp;gt;[\text{boil}]PbSO4}$$&#xA;$$\ce{S +HNO3-&amp;gt;[\text{boil}]SO4^2-}$$&#xA;etc.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="402" LastEditDate="2016-12-04T21:00:09.690" LastActivityDate="2016-12-04T21:00:09.690" Title="Is +6 really the most stable form of sulfur?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20083" PostTypeId="1" AcceptedAnswerId="20097" CreationDate="2014-11-28T17:44:42.467" Score="0" ViewCount="206" Body="&lt;p&gt;I couldn't find which of these is correct and even someone I asked said there's no general covincement on it, but I'm unsure whether he's correct?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Hmqhs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reproduced the second structure a little wrong maybe, but arrangement and bonds are correct(I think there should be no charge on oxygen).&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4945" LastEditDate="2014-11-29T06:20:40.993" LastActivityDate="2014-11-29T06:20:40.993" Title="What is the structure of Cl₂O₆?" Tags="&lt;inorganic-chemistry&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20084" PostTypeId="1" CreationDate="2014-11-28T17:44:56.613" Score="5" ViewCount="15894" Body="&lt;p&gt;What induces  electrons to leave the atoms around the part of the zinc metal in a solution to travel up along that zinc metal and above the solution, through the wire and to the copper metal and reduce the copper ions in that solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a result of the positive metal ions in a 'sea of electrons' being attracted to the negative ions of the solution making the electrons surrounding them leave, are the electrons repelled by the electrons in the solution or do these electrons 'sense' the copper ions on the other side waiting to be reduced and flows towards them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just not sure what prompts them to leave in the first place if the positive copper ions are not directly in contact with the zinc.&lt;/p&gt;&#xA;" OwnerUserId="10057" LastActivityDate="2015-01-28T04:03:33.750" Title="Why do Electrons leave the Zinc in a Galvanic Cell" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20085" PostTypeId="1" CreationDate="2014-11-28T17:50:42.143" Score="0" ViewCount="286" Body="&lt;p&gt;We know that when we are trying to recognize iodate, we use sodium bisulphite to the solution and with the help of starch paper recognize the evolved $\ce{I2}$.But when I studied the books I found actually Iodide is formed? What's the solution to this controversy?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-29T10:24:26.660" Title="Controversy over products of iodate and bisulfite?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20086" PostTypeId="1" AcceptedAnswerId="20098" CreationDate="2014-11-28T18:04:32.667" Score="0" ViewCount="1223" Body="&lt;p&gt;Why doesn't unstable odd electron compound &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; dimerize to &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O2}$&lt;/span&gt;? Why doesn't this structure of ozonide dimerize? But why then does &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; dimerizes to &lt;span class=&quot;math-container&quot;&gt;$\ce{B2H6}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2Cl6}$&lt;/span&gt;, both with something like bridged bonds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1iwgw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="9961" LastEditDate="2019-10-24T20:27:17.833" LastActivityDate="2019-10-25T04:40:36.470" Title="Why don't unstable odd electron species dimerize so that they become stable?" Tags="&lt;inorganic-chemistry&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20088" PostTypeId="1" CreationDate="2014-11-28T18:20:14.530" Score="0" ViewCount="776" Body="&lt;p&gt;We know that $\ce{P4O10}$ is a dehydrating reagent and I might list on 100's of reaction of it's dehydrating nature, mainly orgainc elimination reaction and few inorganic oxoacid dehydrations, but it is acting like a hydrating agent, cf. these:&#xA;$$\ce{PCl5 +P4O10-&amp;gt;POCl3}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Phosphorus pentachloride also forms $\ce{POCl3}$ by reaction with water, but this reaction is less easily controlled...&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-28T19:20:34.980" Title="A dehydrating agent acting as an hydrating agent?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20089" PostTypeId="1" CreationDate="2014-11-28T18:24:17.637" Score="-9" ViewCount="4825" Body="&lt;p&gt;I have two oppsing information, one that it forms a protective layer of $\ce{PbO}$ and doesn't reacts,other that it forms lead nitrate.Can someone clarify. &lt;strong&gt;NB the acid is cc.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2016-04-28T06:19:55.800" Title="What happens in reaction of Lead and concentrated Nitric Acid?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-11-08T16:20:01.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="20090" PostTypeId="2" ParentId="20084" CreationDate="2014-11-28T18:34:53.833" Score="0" Body="&lt;p&gt;If you have studied semiconductors and the method of conductivity by them, mainly extrinsic semiconductors, you would recall that the movement of holes is relative and equivalently considered as the hole moving but actually the electrons are moving, in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the solution, copper reduces via the neighboring electrons, a very rare but favorable probability that happens similarly with zinc but with the difference that is looses some electrons in its surroundings.slowly the process starts involving electrons at far distances and suddenly when both processes approach each other, a kind of link is created which is mutually favorable and the process proceeds further more easily and now involving almost all atoms which was earlier limited to a very few atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole process is also a particular cause of the random motion of solvent and other particles like Brownian motion. Note that random always means something that involves very small factors (do you know the butterfly effect? ) so that it is almost impossible but not exactly impossible to predict the outcome, not does it means something that is luck.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-28T18:34:53.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20091" PostTypeId="1" AcceptedAnswerId="20114" CreationDate="2014-11-28T18:43:18.227" Score="1" ViewCount="2690" Body="&lt;p&gt;We know anode=oxidation=loss of electrons and cathode=reduction=gain of electrons but in the photoelectric effect the electrons are gained at the anode and lost at the cathode of the discharge tube?&#xA;&lt;strong&gt;References:&lt;/strong&gt; &lt;a href=&quot;http://goldbook.iupac.org/A00370.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anode&lt;/a&gt;,&#xA;&lt;a href=&quot;http://goldbook.iupac.org/C00905.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cathode&lt;/a&gt;, &lt;a href=&quot;http://goldbook.iupac.org/O04362.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxidation&lt;/a&gt;, &lt;a href=&quot;http://goldbook.iupac.org/R05222.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reduction&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0Y9em.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-11-29T04:04:44.787" LastActivityDate="2015-11-19T23:17:54.087" Title="Is the definition of anode and cathode failing?" Tags="&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20093" PostTypeId="2" ParentId="20088" CreationDate="2014-11-28T18:47:16.737" Score="3" Body="&lt;p&gt;Reaction with water is not always a hydration reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydration reaction is one in which all of the atoms of the water molecule get added to the reactant to form a product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MX + H2O -&amp;gt; MH2XO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H4 + H2O}\overset{\ce{H+}}{\longrightarrow}\ce{C2H6O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{P4O10 + 6H2O -&amp;gt; 4H3PO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of water with phosphorous pentachloride is more properly a substitution or partial metathesis reaction. The hydrogen atoms in water are not incorporated into the product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PCl5 + H2O -&amp;gt; POCl3 + 2HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is difficult to control because $\ce{POCl3}$ is reactive with water also, &#xA;and under the right conditions can yield phosphoric acid or even $\ce{P4O10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{POCl3 + 3H2O -&amp;gt; H3PO4 + 3HCl}$$&#xA;$$\ce{4POCl3 + 6H2O -&amp;gt; P4O10 + 12HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of $\ce{P4O10}$ and $\ce{PCl5}$ is thus also a partial metathesis. It is more obvious when you balance the equation. The oxygen and chlorine atoms are rearranging between the phosphorous atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{P4O10 + 6PCl5 -&amp;gt; 10POCl3}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4231" LastEditDate="2014-11-28T19:20:34.980" LastActivityDate="2014-11-28T19:20:34.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20094" PostTypeId="1" CreationDate="2014-11-28T18:55:20.270" Score="-6" ViewCount="1893" Body="&lt;p&gt;In the given question, the solution (according to the book) says that the electron withdrawing group will facilitate hydride transfer (not mentioning how so) and therefore, said the answer to be (D). But should not (C) be the answer, as the electron donating $\ce {-OCH3}$ group creates a &quot;pushing&quot; effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yOOTD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9984" LastEditorUserId="5026" LastEditDate="2018-02-22T13:10:26.787" LastActivityDate="2018-02-22T13:10:26.787" Title="Determine the best hydride donor" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2018-02-22T17:50:55.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="20096" PostTypeId="2" ParentId="20091" CreationDate="2014-11-28T19:40:39.030" Score="1" Body="&lt;p&gt;The definition of &lt;strong&gt;anode&lt;/strong&gt;, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Anode&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An &lt;strong&gt;anode&lt;/strong&gt; is an electrode through which positive electric charge flows &lt;em&gt;into&lt;/em&gt; a polarized electrical device. The flow of charge is an electric current.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The confusion here arises from the fact that you should not look at the chamber where the photoelectric effect occurs but at the circuit instead. This way, you will notice that the &lt;strong&gt;anode&lt;/strong&gt; is the one which is losing electrons (accumulated by the photoelectric effect onto it). These electrons flow through the circuit to the &lt;strong&gt;cathode&lt;/strong&gt;, which is gaining electrons.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="1532" LastEditDate="2014-11-28T19:46:36.477" LastActivityDate="2014-11-28T19:46:36.477" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20097" PostTypeId="2" ParentId="20083" CreationDate="2014-11-28T20:52:57.900" Score="6" Body="&lt;p&gt;Dichlorine hexoxide has different structures depending upon whether it is in the gas or solid phase. In the gas phase, the molecule adopts structure (a) (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic00283a009&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 1&lt;/a&gt;, &lt;a href=&quot;http://books.google.com/books?id=Mtth5g59dEIC&amp;amp;pg=PA463&amp;amp;lpg=PA463&amp;amp;dq=dichlorine%20hexoxide%20gas%20phase%20structure&amp;amp;source=bl&amp;amp;ots=tEVPfCPZNw&amp;amp;sig=IANaKmFXtF1w-3Z8GL-p9N4iNmU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=ar54VMyxK9CUsQS88YKIAg&amp;amp;ved=0CC4Q6AEwAg#v=onepage&amp;amp;q=dichlorine%20hexoxide%20gas%20phase%20structure&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference 2&lt;/a&gt;).  It is a relatively unstable molecule having a half-life measured in minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8M9uM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solid phase it exists as the dark-red ionic material as shown in (b) and is often referred to as chloryl perchlorate.  Both forms are very strong oxidizing agents and react explosively with organic compunds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-28T20:52:57.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20098" PostTypeId="2" ParentId="20086" CreationDate="2014-11-28T22:01:49.073" Score="8" Body="&lt;p&gt;Odd electron species do dimerize.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nitric oxide dimerizes, but only at &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitric_oxide#Reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;low temperature&lt;/a&gt; (and probably high pressure}.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nitrogen dioxide does dimerize. In fact, this is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_dioxide#Basic_thermal_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;well known property&lt;/a&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; (orange-brown) is in equilibrium with &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; (colorless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2NO2 &amp;lt;=&amp;gt; N2O4}$$&lt;/span&gt;&#xA;The position of this equilibrium is dependent on temperature and pressure (like all gas equilibria), and so it makes a nice demonstration of gas properties.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ozone does not dimerize. Dimerization would lead to &lt;span class=&quot;math-container&quot;&gt;$\ce{O6}$&lt;/span&gt;, which likely is a 6-membered ring of all oxygen atoms. While this structure violates no rules, the oxygen-oxygen single bond is not stable. This instability of the peroxide bond makes it both &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxide#Properties_and_applications_of_organic_peroxides&quot; rel=&quot;nofollow noreferrer&quot;&gt;useful&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Peroxide#Safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;dangerous&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So why does &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; dimerize readily, &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; dimerize under duress, and &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt; not dimerize at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The extra electron in the nitrogen oxides is found in an antibonding orbital. Conversion to the dimer gets that electron into a bonding orbital. There are no antibonding electrons in &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Borane dimerizes because the boron atom is electron deficient - it has an empty p orbital. This species is so electron deficient that it will form bonding interactions with nearly any electron pairs, including those already in sigma bonds. (More properly the B-H-B bonds are three-center-two-electron bonds). Aluminum compounds dimerize the same way.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So why does &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt; dimerize and &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt; does not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ozone has no electron deficient atoms. Count the electrons. All atoms have an octet, and while one has a positive formal charge, the compound as a whole is not electron deficient.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7475" LastEditDate="2019-10-25T04:40:36.470" LastActivityDate="2019-10-25T04:40:36.470" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="20099" PostTypeId="2" ParentId="19912" CreationDate="2014-11-28T22:36:49.740" Score="4" Body="&lt;p&gt;Your interpretation of specific gravity is incorrect: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Specific gravity&lt;/strong&gt; (SG) refers to the ratio between a given density and that of a reference.  Usually it is defined &lt;strong&gt;for liquids with respect to water and gasses with respect to air&lt;/strong&gt; so in this case&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;$0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$ &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;where $n_i$ are proportions by mole factor. This is the incorrect assumption &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;$0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Gasses are always a little dicier since the reference is much more important with changes in temperature and pressure affecting the resulting density.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The density form of the ideal gas laws goes like: &lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;$\rho = \text{mass / volume = molecular weight * pressure / constant / temperature}$.  &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;Therefore for constant temperature and pressure, the specific gravity&#xA;  is equal to the ratio of molecular weights.  This follows because&#xA;  $\frac{\rho_1}{\rho_2} = \frac{MW_1}{MW_2}$ at constant conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equations and Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;$M_{\text{air}} = 28.97$ (I got $28.84$ using the simple $0.21*32+0.79*28$ and this agrees with  &lt;a href=&quot;http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;)&lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;$\text{SG}=0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$ &lt;/p&gt;&#xA;    &#xA;    &lt;p&gt;$1=n_{CH_4}+n_{C_2H_6}$&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;so by the first equation and the second equation the density of the mixture is $0.6 * 28.97 = 17.38$. The third equation holds only if $0.9*16 + 0.1*30 = 17.4$ so&lt;/p&gt;&#xA;  &#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;methane is 90% and ethane is 10%.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9982" LastEditorUserId="966" LastEditDate="2014-11-30T12:51:03.533" LastActivityDate="2014-11-30T12:51:03.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20100" PostTypeId="1" CreationDate="2014-11-28T22:42:20.463" Score="22" ViewCount="5826" Body="&lt;p&gt;Why does hydroquinone possess a non-zero dipole moment? The $\ce{OH}$ groups present at para positions on the benzene ring should cancel the effect of each other... if there is a plane change then exactly when do atoms in a molecule change planes?&lt;/p&gt;&#xA;" OwnerUserId="10060" LastEditorUserId="16683" LastEditDate="2015-10-02T16:55:44.827" LastActivityDate="2021-05-26T05:49:30.903" Title="Non-zero dipole moment of hydroquinone" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20101" PostTypeId="2" ParentId="5176" CreationDate="2014-11-28T23:56:11.940" Score="2" Body="&lt;p&gt;Reactivity of these acids is very different, triflic acid shows mainly acidic properties, while chlorosulfonic is also strong oxidant and &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrophile&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrophile&lt;/a&gt;, and also &lt;a href=&quot;http://pubs.rsc.org/en/content/ebook/978-0-85404-498-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;chlorinating and dehydrating agent&lt;/a&gt;, which makes it unsuitable for using instead of triflic in your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not a good idea to even use benzene as solvent for chlorosulfonic acid, because it can get chlorosulfonated (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27437/chlorosulfonation-mechanism&quot;&gt;this question&lt;/a&gt;) and substrates in these reactions are more reactive than benzene. In condensation reaction water is created and $\ce{HSO3Cl}$ would gradually hydrolyse, even if it worked properly.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2016-06-14T20:20:55.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20102" PostTypeId="2" ParentId="20100" CreationDate="2014-11-29T00:32:43.583" Score="12" Body="&lt;p&gt;According to &lt;a href=&quot;https://doi.org/10.1021/ja01218a051&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1945,&lt;/strong&gt; &lt;em&gt;67&lt;/em&gt; (2), 322–324&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The preferred conformation for hydroquinone is a flat molecule with the hydrogens in the plane.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="95133" LastEditDate="2021-05-26T05:49:30.903" LastActivityDate="2021-05-26T05:49:30.903" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20103" PostTypeId="2" ParentId="19541" CreationDate="2014-11-29T00:35:32.960" Score="8" Body="&lt;p&gt;I also heard that and it's not exactly true, you could find total value using mentioned earlier $E=mc^2$, but it shouldn't be called absolute, as mass-energy depends on frame of reference - masses measured between &lt;a href=&quot;https://en.wikipedia.org/wiki/Inertial_frame_of_reference&quot; rel=&quot;noreferrer&quot;&gt;inertial&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-inertial_reference_frame&quot; rel=&quot;noreferrer&quot;&gt;non-inertial frames of reference&lt;/a&gt; are different. Also, as pointed in comments, you can't account for &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero-point_energy&quot; rel=&quot;noreferrer&quot;&gt;zero point energy&lt;/a&gt; of &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_state&quot; rel=&quot;noreferrer&quot;&gt;vacuum&lt;/a&gt; this way, although I wouldn't call it a problem as this energy is the same when the object of measurement is absent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still if you weighed something and multiplied this mass by $c^2$ you would get its total energy. However it would be a very rough estimate, &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence&quot; rel=&quot;noreferrer&quot;&gt;one gram is equivalent to about 90 terajoules of energy&lt;/a&gt;, which is really large amount. Even if you would measure mass with great precision, and mass would be very small, it would still be a rough estimate.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-01-07T01:20:26.007" LastActivityDate="2016-01-07T01:20:26.007" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20105" PostTypeId="1" CreationDate="2014-11-29T01:34:51.103" Score="6" ViewCount="5014" Body="&lt;p&gt;On the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyrophosphate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; about pyrophosphate, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The group (pyrophosphate) is also called diphosphate or dipolyphosphate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for example, why is ADP apparently called adenosine diphosphate but not adenosine pyrophosphate? The same for geranyl pyrophosphate (GPP) instead of geranyl diphosphate?  In which context the terms diphosphate and pyrophosphate are used?&lt;/p&gt;&#xA;" OwnerUserId="7972" LastEditorUserId="5026" LastEditDate="2018-04-07T15:59:21.517" LastActivityDate="2018-04-07T18:28:03.673" Title="When are the terms diphosphate and pyrophosphate used?" Tags="&lt;nomenclature&gt;&lt;terminology&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20106" PostTypeId="2" ParentId="20105" CreationDate="2014-11-29T01:54:08.173" Score="2" Body="&lt;p&gt;Pyrophosphate came about from heating inorganic phosphates, with the loss of water to create new compounds. Since ADP and GDP are organic compounds, diphosphate is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;a href=&quot;http://dictionary.reference.com/browse/pyro-&quot; rel=&quot;nofollow&quot;&gt;Dictionary&lt;/a&gt; definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemistry. a combining form used in the names of inorganic acids, indicating that the acid's water content is intermediate between that of the corresponding ortho- (more water) and meta- (least water) acids ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-29T01:54:08.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20107" PostTypeId="1" AcceptedAnswerId="20108" CreationDate="2014-11-29T02:25:24.607" Score="1" ViewCount="489" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kGWQe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ive been struggling to solve this dehydrohalogenataion excercise.&#xA;How does the E2 mechanism work in this case and how is it affected by the stereoselectivity and antiperiplanar position? How do I get to the transition state and how does the chair conformations for the cyclohexane ring look like?&lt;/p&gt;&#xA;" OwnerUserId="10062" LastEditorUserId="10062" LastEditDate="2014-11-30T13:16:18.477" LastActivityDate="2014-11-30T13:16:18.477" Title="How to solve 1-deuterio-3-methylcyclohexene dehydrohalogenation?" Tags="&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20108" PostTypeId="2" ParentId="20107" CreationDate="2014-11-29T03:21:37.587" Score="5" Body="&lt;p&gt;As you noted, the chlorine and the other group being eliminated must be antiperiplanar.  In the case of the chair form of cyclohexane, this means that the chlorine and other group being eliminated must be trans-diaxial.  In the figure below I've drawn the 2 possible conformations (interconverted by a chair flip) for your deuterated cyclohexane.  Only in the top conformation is the chlorine axial, and the hydrogen on the adjacent (non-methylated) carbon is also positioned trans and axial.  Only this hydrogen can be eliminated as it is the only one that is axial and antiperiplanar to the chlorine.  Therefore, the only product produced when this material is treated with base and an E2 occurs, is the pictured 1-deuterio, 3-methylcyclohexene.  The hydrogen on the methylated carbon cannot be eliminated since it is never axial when the chlorine is axial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uvSw0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-11-29T03:21:37.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20112" PostTypeId="2" ParentId="20085" CreationDate="2014-11-29T07:44:30.737" Score="3" Body="&lt;p&gt;Not much controversy here, IMO. There is more than one reaction involved, with slightly more complex kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As so often, Wikipedia already knows, so I will just quote the relevant part of the following article: &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodine_clock_reaction&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Iodine_clock_reaction&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Iodate variation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An alternative protocol uses a solution of iodate ion (for instance potassium iodate) to which an acidified solution (again with sulfuric acid) of sodium bisulfite is added.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In this protocol, iodide ion is generated by the following slow reaction between the iodate and bisulfite:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{   IO3^{−} + 3 HSO3^{−} -&amp;gt; I^{−} + 3 HSO4^{−}}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{   IO3^{−} + 5 I^{−} + 6 H+ -&amp;gt; 3 I2 + 3 H2O}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, the iodine is reduced immediately back to iodide by the bisulfite:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{   I2 + HSO3^{−} + H2O → 2 I^{−} + HSO4^{−} + 2 H+}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10050" LastEditorUserId="189" LastEditDate="2014-11-29T10:24:26.660" LastActivityDate="2014-11-29T10:24:26.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20113" PostTypeId="1" AcceptedAnswerId="20128" CreationDate="2014-11-29T08:33:00.597" Score="1" ViewCount="6083" Body="&lt;p&gt;Determine which of the following pairs of reactants will result in a spontaneous reaction at 25 celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pb2+(aq) + Cu(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ag+(aq) + Br-(aq)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li+(aq) + Cr(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe3+(aq) + Cd(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the Above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using using tables like this &lt;a href=&quot;http://ch302.cm.utexas.edu/images302/Electrochemistry_Reduction_Potentials.jpg&quot; rel=&quot;nofollow&quot;&gt;http://ch302.cm.utexas.edu/images302/Electrochemistry_Reduction_Potentials.jpg&lt;/a&gt; to use the equation Ecell = Ecath - Eanode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping anyone could answer the question, and explain why each of them could be, or could not be spontaneous. (I know that an Ecell being positive results in spontaneity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pb2+(aq) + Cu(s) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;.34-(-.13) = .47 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ag+(aq) + Br-(aq) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;.80 - (.76) = .04 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Li+(aq) + Cr(s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-.90-(-3.05) = 2.15 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe3+(aq) + Cd(s) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-.04 - (-.81) = 77 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the Above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, they're all spontaneous. I'm wondering if I grabbed the wrong values from the above table. The question asks about ONE spontaneous reaction, from my calculations, they're all spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@MichaelD.M.Dryden&lt;/p&gt;&#xA;" OwnerUserId="10067" LastEditorUserId="10067" LastEditDate="2014-11-29T09:29:57.047" LastActivityDate="2014-11-29T18:31:37.973" Title="Electrochemistry (Spontaneous Reactions)" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20114" PostTypeId="2" ParentId="20091" CreationDate="2014-11-29T08:59:12.150" Score="6" Body="&lt;p&gt;entropid's answer is correct, but I'll try to express it differently. As Bard et al. state in Electrochemical Dictionary (1), the concept of anode and cathode is related only to the direction of electron flow: The anode is the electrode at which oxidation occurs (in an electrochemical cell) and electrons flow from electrolyte to electrode and the cathode is where reduction occurs and electrons flow from electrode to electrolyte. In a galvanic cell with a zinc anode, the electrons are flowing away from the interface in the electrode. Similarly, in a phototube, electrons emitted by the cathode are collected by the anode and again, electrons are moving away from the interface in the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the electrons are actually coming from isn't important, just what direction they flow with respect to the electrode interface. Electrons moving from interface towards electrode = anode, electrons moving from electrode towards interface = cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008, pp 31.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-29T08:59:12.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20115" PostTypeId="1" CreationDate="2014-11-29T09:27:10.417" Score="3" ViewCount="739" Body="&lt;p&gt;I know of 3 different definitions for acids and bases including Arrhenius, Brønsted–Lowry and Lewis. Can anyone tell me what is the common feature among these 3, that we have to label the compounds under those definitions as acids or bases? Is there any defining physical or chemical feature that guides this? &lt;/p&gt;&#xA;" OwnerUserId="10061" LastEditorUserId="83758" LastEditDate="2020-05-26T09:13:55.810" LastActivityDate="2020-05-26T09:13:55.810" Title="A defining feature for acids / bases" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20116" PostTypeId="1" CreationDate="2014-11-29T09:44:55.473" Score="2" ViewCount="292" Body="&lt;p&gt;The work done as a ion is solvated is given by: $-z^2e^2/8\pi\epsilon r$ but since the top line can be written $Q^2$, for simplicity I will say the equation is $-Q^2/8\pi\epsilon r$. Is this derived from integrating potential with respect to charge $Q$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e    $$w=\int_Q^0 VdQ$$ where $$V=Q/4 \pi \epsilon r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this would give the correct units and equation but is this an acceptable derivation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a, sort of, side question why can't I write: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\triangle w=\int_Q^0 VdQ$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that work is an inexact differential but I fail to understand the argument regarding paths - surely if you change the charge of a particle from one value to another, you must be doing the same amount of work no matter how you do it. This argument must be wrong but I don't know why and why does this mean I cant write $\triangle w$? I know you can do this for things like entropy and enthalpy because they are state functions - I just don't understand the difference. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2014-11-29T17:06:51.997" LastActivityDate="2014-11-29T17:06:51.997" Title="How to derive the energy of solvation equation?" Tags="&lt;energy&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20117" PostTypeId="1" AcceptedAnswerId="21658" CreationDate="2014-11-29T10:54:42.630" Score="2" ViewCount="256" Body="&lt;p&gt;For wave-like (quantum coherent) energy transfer in networks (eg. propagation of excitation in photosynthetic protein complexes of algae or FMO complex in plants) 2D electronic (photon echo?) spectroscopy is used (eg. in &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1001/1001.5108.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article). I have read an article (unfortunately I can't cite it here) about how it works, but I didn't understand it. I know that the off-diagonal peaks in the spectrum mean some kind of coherence in the network but I don't know why. I also don't know what kind of coherence it means. What are meant by electronic eigenstates? What are the so called 'quantum beatings'? It would be very much appreciated if somebody would be so kind as to make these thing clear.&lt;/p&gt;&#xA;" OwnerUserId="10069" LastActivityDate="2014-12-13T04:49:34.533" Title="How is 2D spectroscopy able to show quantum coherent transport through networks?" Tags="&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20118" PostTypeId="1" CreationDate="2014-11-29T11:49:16.067" Score="5" ViewCount="680" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why the hydrogen gas should be bubbled continuously through the electrolyte in a hydrogen electrode?  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;To keep the solution saturated with hydrogen gas&lt;/li&gt;&#xA;  &lt;li&gt;To clear the platinum electrode&lt;/li&gt;&#xA;  &lt;li&gt;To gate the highest electrode potential&lt;/li&gt;&#xA;  &lt;li&gt;To minimize loss of current&lt;/li&gt;&#xA;  &lt;li&gt;To speed up the process&lt;/li&gt;&#xA;  &lt;li&gt;To minimize the cost&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my opinion:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This is false because the solution doesn't need to be saturated.&lt;/li&gt;&#xA;&lt;li&gt;Maybe true because if the platinum electrode is full with hydrogen gas which isn't bubbled continuously, there will not be enough space on platinum electrode to let another $\ce{H2}$ gas dissociate.&lt;/li&gt;&#xA;&lt;li&gt;This is false because it doesn't relate.&lt;/li&gt;&#xA;&lt;li&gt;Maybe true because if the hydrogen gas isn't bubbled continuously, $\ce{H2}$ will not dissociate continuously and will not transfer electrons. So there will be lack of current through this electric cell.&lt;/li&gt;&#xA;&lt;li&gt;Maybe true because if there's lack of current, it means that the process is too slow.&lt;/li&gt;&#xA;&lt;li&gt;False because this choice doesn't relate.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But the question asks to choose just one choice. What choice is the best? Do my reasons for each choice make sense?&lt;/p&gt;&#xA;" OwnerUserId="9686" LastEditorUserId="5026" LastEditDate="2018-02-27T13:20:52.747" LastActivityDate="2018-02-27T13:20:52.747" Title="Hydrogen electrode" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20119" PostTypeId="2" ParentId="8355" CreationDate="2014-11-29T12:06:10.477" Score="0" Body="&lt;p&gt;In dilute solutions, small particles are formed, which can suspend in the solution. As solute amount increases the particles grow bigger. At saturation, large particles are formed and precipitate from the solution.&lt;/p&gt;&#xA;" OwnerUserId="10070" LastActivityDate="2014-11-29T12:06:10.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20120" PostTypeId="2" ParentId="6035" CreationDate="2014-11-29T12:50:17.463" Score="2" Body="&lt;p&gt;white phosphorous allotrope (P_4) is a deltahedron (tetrahedron) consisting of four triangles fused by mean of each side. It doesn't bear that much angular strain, cause the measurements done on 2nd-row elements-based compounds (i.g. hydrocarbons and related) does not necessarily include 3rd-row and heavier elements. This also holds in non cyclic compounds. While the H-O-H in water is close to theoretically predictable 109,5°, the H-S-H angle in hydrogen sulfide is very close to a right angle, showing little or no hybridization to have occurred.&#xA;Moreover the rules with closed shell molecules and non-electron deficient ones, appear to be inadequate to descrive electron deficient compounds, spanning from closo-boranes (very often deltahedranes) to metal clusters&#xA;P.S. sorry for my english :)&lt;/p&gt;&#xA;" OwnerUserId="10071" LastActivityDate="2014-11-29T12:50:17.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20121" PostTypeId="2" ParentId="20079" CreationDate="2014-11-29T13:22:42.853" Score="2" Body="&lt;p&gt;Because of its solubility.&#xA;More concentrated solutions of, for example, nitric acid or hydrochloric acid, will result in fuming solutions where acid concentration evaporate out until the solution reaches an equilibrium.&#xA;Do not get confused. You can actually dissolve more than 370g of hydrochloirc acid in 630 mL of water.&#xA;In effect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility in water (Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;823 g/L (0 °C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;720 g/L (20 °C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;561 g/L (60 °C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we talk about a commercial solution, which should stay at the same concentration in time, you need to find a compromise between most quantity of acid possible in solution and less lose of acid due to vaporization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that pure hydrogen chloride is a gas at room temperature. If you continue increasing its concentration, you end up with man equilibrium shifted towards gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as you can see, if you increase the temperature, solubility decreases, so if you heat a 50% aqueous solution of hydrochloric acid, you would end up not knowing what quantity of the reactant are you actually using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitric acid is less volatile, and sulphuric acid is much less volatile, so you can have the latter almost in a pure form (98% is not an aqueous solution anymore! is just sulphuric acid with some water and other trace compounds)&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="9850" LastEditDate="2014-11-29T13:28:26.147" LastActivityDate="2014-11-29T13:28:26.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20122" PostTypeId="2" ParentId="6035" CreationDate="2014-11-29T13:43:43.703" Score="1" Body="&lt;p&gt;The best example for relatively stable 3 membered rings are the epoxides. They are very reactive, yes, but they can be isolated and stored, and then used in synthetic routes, like those sold by sigma aldrich:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=20203736&quot; rel=&quot;nofollow&quot;&gt;http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=20203736&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably wouldn't have guessed that all of those guys above did exist, would you? :-D &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also understand its stability trough the molecular orbital theory. It turns out that molecular orbitals of these 3 membered rings, like cyclopropanes or epoxides has quite similar molecular orbitals compared to those of alkenes, so their reactivity is somehow similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, square, penta-, hepta-, octa-....gonal molecules do exist!&#xA;You would be surprised about some macrocyclic compounds being stable like those over here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/patents/US6200254&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/US6200254&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="9850" LastEditDate="2014-11-29T13:49:07.220" LastActivityDate="2014-11-29T13:49:07.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20123" PostTypeId="1" AcceptedAnswerId="20132" CreationDate="2014-11-29T13:56:56.553" Score="3" ViewCount="4565" Body="&lt;p&gt;Baking soda (powder) will react when it comes into contact with a solution of citric acid crystals dissolved in a liquid medium such as water. &lt;strong&gt;But why doesn't baking soda react when mixed with a solid form of an acid (citric acid crystals)?&lt;/strong&gt; Why is water needed for a reaction to occur? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baking Soda (Solid) + Citric Acid Solution (Liquid) = Fizzy Reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baking Soda (Solid) + Citric Acid Crystals (Solid) = No Reaction (??)&lt;/p&gt;&#xA;" OwnerUserId="9802" LastActivityDate="2014-11-29T19:47:18.677" Title="Why doesn't baking soda react with the solid form of an acid?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20124" PostTypeId="1" AcceptedAnswerId="20133" CreationDate="2014-11-29T13:59:45.647" Score="0" ViewCount="15838" Body="&lt;p&gt;If I understand correctly, solubility of organic compounds depends mainly on the presence of polar groups plus putting in the factor of the number of carbon atoms in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting that in mind I do not understand why Glycine, an amino acid with two polar groups, the amine and the carboxyl groups, is considered non-polar and insoluble in water .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O5zSr.gif&quot; alt=&quot;Glycine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10061" LastEditorUserId="7475" LastEditDate="2016-06-06T22:23:33.707" LastActivityDate="2016-06-06T22:23:33.707" Title="Polarity of amino acids" Tags="&lt;solubility&gt;&lt;polarity&gt;&lt;amino-acids&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20125" PostTypeId="2" ParentId="20124" CreationDate="2014-11-29T15:05:33.237" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycine&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; and my experience, glycine is soluble in water and is a polar molecule. The reference you saw, uses polar and non-polar to mean if there is another group other than the carboxylic acid and the amine group.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solubility in water   24.99 g/100 mL (25 °C)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-11-29T15:05:33.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20126" PostTypeId="2" ParentId="20082" CreationDate="2014-11-29T15:50:26.253" Score="2" Body="&lt;p&gt;I think it's a matter of &lt;strong&gt;activation energy&lt;/strong&gt; and reaction kinetics. Although the product is much more thermodynamically stable than the reagent(s), the reaction path could feature a high activation energy which requires those harsh conditions and even a catalyst to make it happen. As shown in the graph below, despite having a favourable difference of free energy ($\Delta G &amp;lt; 0$), the activation energy $E_a$ required to go from reactants to products could be not negligible and hard to overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hm9FQm.jpg&quot; alt=&quot;Activation energy diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catalysts allows the reaction to happen more easily by lowering the free energy of the transition state (by stabilising it). The activation energy is also reduced by the high temperature, according to the Arrhenius equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = A e^{{-E_a}/{RT}}$$&lt;/p&gt;&#xA;" OwnerUserId="1532" LastActivityDate="2014-11-29T15:50:26.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20127" PostTypeId="1" CreationDate="2014-11-29T16:16:12.247" Score="22" ViewCount="4138" Body="&lt;p&gt;According to the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_equation&quot;&gt;Chemical Equations&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symbols are used to differentiate between different types of reactions. To denote the type of reaction:&lt;br&gt;&#xA;  &quot;$=$&quot; symbol is used to denote a stoichiometric relation.&lt;br&gt;&#xA;  &quot;$\rightarrow$&quot; symbol is used to denote a net forward reaction.&lt;br&gt;&#xA;  &quot;$\rightleftarrows$&quot; symbol is used to denote a reaction in both directions.&lt;br&gt;&#xA;  &quot;$\rightleftharpoons$&quot; symbol is used to denote an equilibrium.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is a &quot;reaction in both directions&quot; and an &quot;equilibrium&quot; any different? Aren't they supposed to be identical? &lt;/p&gt;&#xA;" OwnerUserId="8234" LastActivityDate="2014-12-15T05:40:56.027" Title="What is the difference between &quot;reaction in both directions&quot; and &quot;equilibrium&quot;?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20128" PostTypeId="2" ParentId="20113" CreationDate="2014-11-29T18:31:37.973" Score="1" Body="&lt;p&gt;For $\ce{Pb^2+_{(aq)} + Cu_{(s)}&amp;lt;=&amp;gt; Pb_{(s)} + Cu^2+_{(aq)}}$, your potentials are the wrong way around. The cathode is the electrode at which the reduction occurs so it's lead minus copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}&amp;lt;=&amp;gt; Ag_{(s)} + Br2_{(l)}}$, I don't know where you got .76 V for bromide, but that's not the right number. Also, consider the $K_{sp}$ of $\ce{AgBr}$ and think about whether a non-redox reaction is occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{3Li^+_{(aq)} + Cr_{(s)}&amp;lt;=&amp;gt; 3Li_{(s)} + Cr^3+_{(aq)}}$, your potentials are backwards again and the potential for $\ce{Cr}$ is wrong. Consider how easily elemental lithium is oxidized; just putting it in water will evolve hydrogen and oxidize the lithium, so anything that reduces the lithium cation would have to be a phenomenal reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}&amp;lt;=&amp;gt; 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, you have the right order, but the wrong potential for $\ce{Cd}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're having trouble keeping the order straight, it may help to think of it this way:&#xA;$$E_{\mathrm{cell}}=E_{\mathrm{reduction}} + E_{\mathrm{oxidation}}$$ where $E_{\mathrm{oxidation}}=-E_{\mathrm{reduction}}$. It's the same equation, but instead of keeping track of anode and cathode, the half-cell where reduction is happening will match what the table says and you use the potential as-is (since it's a table of &lt;em&gt;reduction&lt;/em&gt; potentials), and the half-cell where oxidation is happening will be backwards from that in the table, so you negate the potential.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-11-29T18:31:37.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20129" PostTypeId="1" CreationDate="2014-11-29T18:31:40.327" Score="-1" ViewCount="63990" Body="&lt;p&gt;I have $0.0585$ moles of $\ce{Al^{3+}}$, and need to find the number of actual ions. I thought I just divided by $6.022\times 10^{23}$ but I get $9.7\times 10^{-26}$, and the book says it should be $3.5\times 10^{22}$. &lt;/p&gt;&#xA;" OwnerUserId="10073" LastEditorUserId="275" LastEditDate="2014-11-29T19:15:03.930" LastActivityDate="2021-02-13T07:25:34.467" Title="number of ions present from moles" Tags="&lt;ions&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2021-02-19T03:05:37.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="20130" PostTypeId="2" ParentId="20127" CreationDate="2014-11-29T18:56:58.610" Score="10" Body="&lt;p&gt;As far as I know, they are used indistinguishably and interchangeably. &#xA;One or another could be chosen to emphasize whether you are talking about chemical equilibrium or just trying to show that a reaction can take place in both directions depending on the conditions (which can be understood as completely the same...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw the $⇄$ symbol used to specify some conditions for the reaction to take place in one direction and other kind of conditions for the reaction to occur in the other direction, and $⇌$ if conditions are the same for both reactions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZtMQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First case is not an equilibrium, but it's true that the reaction can go both directions using the appropriated reagents in each case. But that could not be expressed using $⇌$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reaction is an equilibrium, both reactions are happening continuously under the same conditions. &#xA;It also could be expressed using $⇄$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the only difference I can tell. Like Martin pointed out, IUPAC does not give more hints about it, so the question cannot be answered for 100% sure.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="8234" LastEditDate="2014-12-06T15:31:51.123" LastActivityDate="2014-12-06T15:31:51.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20131" PostTypeId="2" ParentId="20129" CreationDate="2014-11-29T19:31:43.073" Score="6" Body="&lt;p&gt;You should multiply by Avogadro's number $6.022\times 10^{23}$ instead of dividing by it. &#xA;To make sure you always use a conversion factor (Avogadro's number is a conversion factor between moles and numbers) correctly, you need to think about units. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 1 - What are the units of the answer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You want to find the number of ions. The units of the answer is &lt;strong&gt;number&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 2 - What are the units of the quantities you were given?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were given 0.0585 moles of $\ce{Al^{3+}}$. The units you are given is &quot;moles&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 3 - What are the units of the conversion factor?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Avogadro's number (with units) is $6.022\times 10^{23} \text{ mol}^{-1}$. The units of Avogadro's number are &quot;number per mole&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 4 - Set up the problem so that units cancel and you are left with only the units you need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multiplying by Avogadro's number&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\require{cancel} \cancel{\text{mol}}\times \dfrac{\text{number}}{\cancel{\text{mol}}}=\text{number}$$&#xA;The &quot;moles&quot; units cancel, leaving you with &quot;number&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dividing by Avogadro's number&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\text{mol}}{\frac{\text{number}}{\text{mol}}}=\text{mol}\times \dfrac{\text{mol}}{\text{number}}=\dfrac{\text{mol}^2}{\text{number}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units do not cancel appropriately, leaving us with units that do not match the desired outcome. This approach is an application of dimensional analysis commonly called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dimensional_analysis#The_factor-label_method_for_converting_units&quot; rel=&quot;noreferrer&quot;&gt;factor-label method&lt;/a&gt;. With practice and conscientious use of this method, you should not make unit conversion errors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Step 5 - Check to see if your answer makes sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$0.0585$ moles is about $\frac{1}{20}$ of a mole (or one half of one tenth). Your answer should be about $\frac{1}{20}$ of Avogadro's number.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-11-29T19:31:43.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20132" PostTypeId="2" ParentId="20123" CreationDate="2014-11-29T19:47:18.677" Score="5" Body="&lt;p&gt;How do you know a reaction is not occurring when you mix the two solids? The only cue we have that a reaction occurs in the solution state is the formation of bubbles of escaping carbon dioxide. If there is no liquid to displace, would we be able to tell if carbon dioxide was being produced? Yes, but not with the naked eye.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The real issue is that the solid state reaction is much, much slower.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To produce carbon dioxide, the bicarbonate ion needs to come in contact with an acidic proton on citric acid. This is unlikely in the solid state since there is very little kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solution, there is more kinetic energy, so the molecules are more likely to encounter each other. Additionally, the proton transfer can propagate quickly through the hydrogen bond network in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCO3- \bond{...} H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-Cit}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-11-29T19:47:18.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20133" PostTypeId="2" ParentId="20124" CreationDate="2014-11-29T20:07:10.760" Score="1" Body="&lt;p&gt;Most amino acid molecules are polar in the sense that they have polar functional groups. Even the &quot;least polar&quot; amino acids like leucine and phenylalanine are likely soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we also classify amino acids by the polarity of their side chains. When incorporated into proteins, clusters of amino acids with nonpolar side chains can create pockets of hydrophobicity in an otherwise aqueous environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glycine is the only amino acid without a side chain. As such, it is not appropriate to classify its side chain as &lt;a href=&quot;http://en.wikipedia.org/wiki/Amino_acid#mediaviewer/File:Amino_Acids.svg&quot; rel=&quot;nofollow&quot;&gt;polar or nonpolar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Polar acidic side chain:             aspartic acid, glutamic acid&#xA;Polar basic side chain:              arginine, histidine, lysine, &#xA;Other polar side chain:              asparagine, glutamine, serine, threonine&#xA;Nonpolar weakly acidic side chain:   cysteine, tyrosine&#xA;Other nonpolar side chain:           alanine, isoleucine, leucine, methionine,&#xA;                                     phenylalanine, proline, tryptophan, valine&#xA;No side chain:                       glycine&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-11-29T20:07:10.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20134" PostTypeId="2" ParentId="20115" CreationDate="2014-11-30T02:04:09.787" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr version:&lt;/strong&gt; You can only define an acid or a base in context of a reaction. Individual compounds are not acids by themselves. They are acids in reactions in which they lose protons (or gain electron pairs).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Lewis definition is the broadest acid-base definition. Each of the others are specific subsets.&lt;/p&gt;&#xA;&lt;p&gt;The Arrhenius definition is the simplest:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Acid = something that creates &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions in water.&lt;/p&gt;&#xA;&lt;p&gt;Base = something that creates &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions in water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus any compound that can ionize in or react with water, like &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{HCO2H}$&lt;/span&gt; is an acid and any compound that contains hydroxide ions or reacts with water to produce them, like &lt;span class=&quot;math-container&quot;&gt;$\ce{NaNH2}$&lt;/span&gt;, is a base.&lt;/p&gt;&#xA;&lt;p&gt;However, the Arrhenius definition breaks down in non-aqueous media.&lt;/p&gt;&#xA;&lt;p&gt;The Brønsted-Lowry definition is broader and more inclusive.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Acid = proton donor - All Arrhenius acids are Brønsted-Lowry acids&lt;/p&gt;&#xA;&lt;p&gt;Base = proton acceptor - All Arrhenius bases are Brønsted-Lowry bases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The Brønsted-Lowry definition allows us to extend our discussion of acids and bases to, say, solutions in methanol &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3OH}$&lt;/span&gt;. Note that the Brønsted-Lowry definition requires a proton transfer reaction to be occurring to properly identify acids. For example, water is amphiprotic - it can react as either an acid or a base. Many other compounds are similar. There is no absolute &amp;quot;this is an acid always&amp;quot;. Instead, you have &amp;quot;this compound is an acid in this reaction&amp;quot;. For example, water is an acid in the first reaction and a base in the second.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H- -&amp;gt; H2  + OH-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O + H2SO4 -&amp;gt; H3O+ + HSO4-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Lewis definition is broader and more inclusive still. It allows extension of acids and bases into aprotic media and situations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Acid = electron pair acceptor - all Brønsted-Lowry acids are Lewis acids&lt;/p&gt;&#xA;&lt;p&gt;Base = electron pair donor - all Brønsted-Lowry bases are Lewis bases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Again, the Lewis definition requires a reaction to occur in order to classify compounds.&lt;/p&gt;&#xA;&lt;p&gt;However, in all cases, there are compounds that behave as acids much more frequently than they act as bases, and vice versa. For example, &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; is almost always an acid, but not exclusively. The first step of the mechanism of &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitration#Aromatic_nitration&quot; rel=&quot;noreferrer&quot;&gt;aromatic nitration&lt;/a&gt; is considered to be a proton transfer step between sufluric acid and nitric acid in which &lt;strong&gt;nitric acid is the base&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HOSO3H + HONO2 &amp;lt;=&amp;gt; HOSO3- + O2NOH2+}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-11-30T02:04:09.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20137" PostTypeId="2" ParentId="20127" CreationDate="2014-11-30T03:37:22.997" Score="8" Body="&lt;p&gt;A reaction in both directions may be equilibrating, but is not necessarily equilibrating (like if the system is being excited), and it certainly isn't necessarily at equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if one puts &lt;em&gt;pure&lt;/em&gt; ammonia gas in a container, it will be far from equilibrium with respect to its decomposition products (nitrogen and hydrogen gas) but it will spontaneously proceed in the following direction:&#xA;$2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3\mathrm{H}_2$.&#xA;Some time after that, the rate of the reverse reaction becomes non-negligible;&#xA;$\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$,&#xA;so that both reactions simultaneously occurring can be summarized as:&#xA;$2\mathrm{NH}_3 \rightleftarrows \mathrm{N}_2 + 3\mathrm{H}_2$. That doesn't mean the system's at equilibrium, just that the forward and reverse rates are non-negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the equilibrium condition (so that the concentration of all species remains constant over time) applies, and the forward and reverse rates become equivalent. That's the distinction that's specified with the double harpoon arrows:&#xA;$2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the harpoon arrows are most often used to specify reaction rates, so the equilibrium condition, where the forward and reverse rates balance, is marked using them. &lt;/p&gt;&#xA;" OwnerUserId="912" LastActivityDate="2014-11-30T03:37:22.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20139" PostTypeId="1" CreationDate="2014-11-30T12:37:18.353" Score="0" ViewCount="152" Body="&lt;p&gt;I want to make composite with HDPE and wood, but when I melt it I have problem to mix them, because this polymer have high viscosity.&lt;/p&gt;&#xA;" OwnerUserId="10079" LastActivityDate="2014-12-30T19:58:39.077" Title="Can I change viscosity of an HDPE polymer?" Tags="&lt;materials&gt;&lt;polymers&gt;&lt;temperature&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20140" PostTypeId="1" AcceptedAnswerId="20141" CreationDate="2014-11-30T12:56:09.460" Score="4" ViewCount="205" Body="&lt;p&gt;I have a question with Equilibrium Constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for a reaction in equilibrium $aA +bB \rightleftharpoons cC +dD$&#xA;$$K_c= \frac{[C]^c[D]^d}{[A]^a[B]^b}$$&#xA;and $K_p$ is just using the partial pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking through my PChem1 tutorials and I saw that they wrote for &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{1/2 N2 + 3/2 H2 &amp;lt;=&amp;gt; NH_3}$$&#xA;$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that a represents activity. However what i don't get is what follows:&#xA;$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}=\frac{\left(\frac{n(NH_3)RT}{p_oV}\right)}{\left(\frac{n(N_2)RT}{p_0V}\right)^\frac12\left(\frac{n(H_2)RT}{p_oV}\right)^\frac32}=\frac{n(NH_3)}{n(N_2)^\frac12n(H_2)^\frac32}\left(\frac{p_oV}{RT}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cant' figure out how $a(NH_3)=\frac{n(NH_3)RT}{p_oV}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It obviously has something to do with the ideal gas equation $pV=nRT$ but I don't get how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;additional context:&#xA;The question asks for the equilibrium composition starting with 1 mol each of $\ce{N_2}$, $\ce{H_2}$ and $\ce{NH_3}$ in a 1L reaction vessel at 298K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s. I get that I am to use extent of reaction to solve this.&lt;/p&gt;&#xA;" OwnerUserId="9581" LastEditorUserId="194" LastEditDate="2017-04-01T01:09:14.293" LastActivityDate="2017-04-01T01:09:14.293" Title="Where does this equation for the activity in the gas phase equilibrium between N₂, H₂ and NH₃ come from?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20141" PostTypeId="2" ParentId="20140" CreationDate="2014-11-30T13:24:20.823" Score="5" Body="&lt;p&gt;The activity of a molecule $\ce{M}$ as a function of its concentration in the solution can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  a_{\ce{M}} = \frac{\gamma_{\ce{M}} c_{\ce{M}}}{c^0} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c^0$ is the standard concentration and $\gamma_{\ce{M}}$ is the (dimensionless) activity coefficient (see also: &lt;a href=&quot;https://chemistry.stackexchange.com/a/10047/189&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity#Activity_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;).&#xA;If you have a sufficiently dilute solution so that non-ideal contributions become very small, the activity coefficient can be approximated by 1, i.e. $\gamma_{\ce{M}} \approx 1$.&#xA;So, what you're left with for the activity under these conditions is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;  a_{\ce{M}} \approx \frac{c_{\ce{M}}}{c^0} \ ,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the concentration is defined by $c = \frac{n}{V}$ and with the ideal gas law $\frac{n}{V} = \frac{p}{RT}$ you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  a_{\ce{M}} &amp;amp;\approx \frac{c_{\ce{M}}}{c^0} = \frac{\frac{n_{\ce{M}}}{V}}{\frac{n^0}{V}} \\&#xA;  &amp;amp;= \frac{\frac{n_{\ce{M}}}{V}}{\frac{p^{0}}{RT}} = \frac{n_{\ce{M}} R T}{p^{0} V}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p^{0}$ is the standard pressure.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-11-30T13:24:20.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20143" PostTypeId="2" ParentId="20094" CreationDate="2014-11-30T14:35:26.833" Score="0" Body="&lt;p&gt;I would like to say that you must be knowing that the best hydride donor is almost always the entity with most partial positive charge, which is most in $\bf D$ due to strong electron withdrawing group $\bf \ce{-NO_2}$ both via inductive and resonance effects,. Remember $\bf\ce{-OCH3}$ in $\bf C$ is actually an electron donating group via resonance and electron withdrawing group via inductive effect of which former dominates largely. I think the rest two are pretty simple.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-12-12T08:21:04.547" LastActivityDate="2014-12-12T08:21:04.547" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20144" PostTypeId="2" ParentId="20118" CreationDate="2014-11-30T14:44:57.957" Score="1" Body="&lt;p&gt;Hydrogen gas must be bubbled into the solution so that they are able to oxidize by loosing electrons which they can't just in air. Platinum electrode needs not be cleaned. Potential depends upon concentration of hydrogen, by &quot;Nerst&quot; equation. If hydrogen was there or not, current can never be lost, speed of process is kinetics and in no way related to thermodynamics, Cost, rubbish.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-11-30T14:44:57.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20145" PostTypeId="2" ParentId="6876" CreationDate="2014-11-30T15:29:45.980" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are tertiary alkyl halides obtained as major than secondary or primary?&#xA;  It seems that hydrogen from tertiary carbon can be easily replaced by halogen than from secondary or primary carbon. If it is so, why is hydrogen from tertiary carbon can be easily replaced than from secondary or primary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mechanism involve:&#xA;$$\ce{X_2-&amp;gt;[\Delta][\text{or UV-light}]2X.}\\\ce{X.+RH-&amp;gt;HX +R.}\\\ce{R.+X2-&amp;gt;RX +X.}$$&#xA;where last two steps keep on repeating cyclically.&#xA;The difficult step is step is as follows which decide the rate of the mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yUmgil.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative rates of primary, secondary, tertiary alkyl halides depends on two factors: probability and Activation Energy.Probability factor for a typical case of propane is $3^\circ:2^\circ:1^\circ=0:6:2$(directly equal to number of hyrogens of that type.)&#xA;The activation energy of $\ce{R-H + X.-&amp;gt;R. +H-X}$ in KCAL/MOL are:&#xA;$$\begin{array}{c|c|c}\ce{R&amp;amp;X=Cl&amp;amp;X=Br\\CH3&amp;amp;4&amp;amp;18\\$1^\circ$&amp;amp;1&amp;amp;13\\$2^\circ$&amp;amp;0.5&amp;amp;10\\$3^\circ$&amp;amp;0.1&amp;amp;7.5}\end{array}$$&#xA;The more stable the radical, more easily it is formed.Factors that stabilize the free radical tend to stabilize the incipent free radical in the transition state:&#xA;$$\ce{-C-H+X. -&amp;gt; [C^{\delta.}\bond{...}H\bond{...}X^{\delta.}] -&amp;gt; -C.+H-X}$$&#xA;A certain factor (delocalization of the odd electron) causes the energy difference between radicals.&#xA;&lt;strong&gt;Stability of free radicals:&lt;/strong&gt;$3^\circ&amp;gt;2^\circ&amp;gt;1^\circ&amp;gt;\ce{CH3}$&#xA;&lt;strong&gt;Ease of formation of free radicals:&lt;/strong&gt;$3^\circ&amp;gt;2^\circ&amp;gt;1^\circ&amp;gt;\ce{CH3}$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is fluorine and iodide not considered in case of free radical halogenation (as you can see, the first two lines of the above passage extracted from the book, mentions only chlorine and bromine)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Iodination is reversible but it may be carried out in the presence of an oxidising agent such as $\ce{HIO3,HNO3,HgO,}$etc., which destroys the hydrogen iodide as it is formed and so drives the reaction to the right:&#xA;$$\ce{CH4 +I2&amp;lt;=&amp;gt;CH3I +HI\\5HI +HIO3-&amp;gt;3I2 +3H2O}$$&#xA;Iodines are more conviniently prepared by treating the chloro- or bromo- derivatives with $\ce{NaI}$ in methanol or acetone solution:&#xA;$$\ce{RCl+NaI-&amp;gt;[\text{acetone}]RI+NaCl}$$ This reaction is possible because $\ce{NaI}$ is soluble in methanol or acetone, whereas sodium chloride and bromide are not. This reaction is known as &lt;strong&gt;Finkelstein&lt;/strong&gt; or &lt;strong&gt;Conant-Finkelstein reaction.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct Fluorination is usually explosive; special conditions are necessary for the preparation of the fluorine derivatives of the alkanes.One such example is Swarts reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is tertiary alkyl halide not produced in the above example? Even if we think it might be produced in less amount, it should have been greater than primary alkyl or secondary alkyl halide, is n't (because of the order)? It seems the reason for tertiary alkyl halide not being in the equation is because of either, it might not have formed or it might have formed in lesser amount than primary or secondary. If it is either any case, it would violate the order mentioned in the book.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See carefully there's no tertiary alkylhalide possible, because no tertiary hydrogen is available. As radicals do not rearrange, the structure never changes!&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="24052" LastEditDate="2016-03-13T20:14:12.180" LastActivityDate="2016-03-13T20:14:12.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20146" PostTypeId="1" CreationDate="2014-11-30T15:31:36.067" Score="6" ViewCount="6277" Body="&lt;p&gt;I know that certain elements like plutonium and certain isotopes of caesium are radioactive but what gives them this ability? If it's their electron/proton/neutron arrangement, what makes that arrangement give radioactive properties and how?&lt;/p&gt;&#xA;" OwnerUserId="10081" LastEditorUserId="9961" LastEditDate="2016-06-23T15:48:01.553" LastActivityDate="2018-05-16T18:09:17.617" Title="What makes elements radioactive?" Tags="&lt;theoretical-chemistry&gt;&lt;elements&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20147" PostTypeId="1" CreationDate="2014-11-30T16:07:50.100" Score="1" ViewCount="314" Body="&lt;p&gt;I did a reaction to form $\ce{(NPCl2)3}$. The reagents I used were:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;phosphorus pentachloride $\ce{PCl5}$&lt;/li&gt;&#xA;&lt;li&gt;ammonium chloride $\ce{NH4Cl}$&lt;/li&gt;&#xA;&lt;li&gt;chlorobenzene - the solvent&lt;/li&gt;&#xA;&lt;li&gt;phosphorus oxychloride $\ce{POCl3}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The reaction is $\ce{3 PCl5 + 3 NH4Cl -&amp;gt; (NPCl2)3 + 12HCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the chlorobenzene is the solvent.&#xA;But what role does the phosphorus oxychloride $\ce{POCl3}$ have in this reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it some sort of catalyst? This reaction also needs to be performed under more acidic conditions, so it is the phosphorus oxychloride that creates the acidic conditions?&lt;/p&gt;&#xA;" OwnerUserId="10082" LastEditorUserId="5026" LastEditDate="2018-04-15T02:12:53.753" LastActivityDate="2018-04-15T02:12:53.753" Title="What is the role of POCl₃ in the reaction to form (NPCl2)3 (hexachlorophosphazene)?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20148" PostTypeId="2" ParentId="20139" CreationDate="2014-11-30T17:32:57.273" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;If you use an extruder to prepare your composite material, try to use HDPE with a melt flow index: MFI of 0.1 g/10 min.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-11-30T17:32:57.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20149" PostTypeId="1" AcceptedAnswerId="20156" CreationDate="2014-11-30T17:41:38.833" Score="9" ViewCount="13379" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The hydrogen bonds in solid HF can be best represented as:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CzAgU.png&quot; height=&quot;200&quot; alt=&quot;Various depictions of hydrogen bonding in HF&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is supposed to be &lt;strong&gt;(c)&lt;/strong&gt;, but I don't understand why this is so. Does the strength of hydrogen bonding depend on the &lt;span class=&quot;math-container&quot;&gt;$\ce{F-H\bond{...}F}$&lt;/span&gt; or the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-F\bond{...}H}$&lt;/span&gt; bond angles?&lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="16683" LastEditDate="2018-12-24T01:43:05.183" LastActivityDate="2019-01-08T16:50:53.990" Title="How can the hydrogen bonds in solid HF be best represented?" Tags="&lt;molecular-structure&gt;&lt;halides&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20152" PostTypeId="2" ParentId="20149" CreationDate="2014-11-30T18:06:25.093" Score="7" Body="&lt;p&gt;Generally hydrogen bonds &lt;span class=&quot;math-container&quot;&gt;$\ce{A-H\bond{...}B}$&lt;/span&gt; can be taken to be approximately &lt;strong&gt;linear&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the highest bonding energy for the hydrogen bond is obtained when the two electronegative atoms (fluorine here) are approximately in line with the electropositive hydrogen atom in between (&lt;span class=&quot;math-container&quot;&gt;$\ce{O\bond{...}H-O}$&lt;/span&gt; in water, &lt;span class=&quot;math-container&quot;&gt;$\ce{F\bond{...}H-F}$&lt;/span&gt; in hydrogen fluoride). Deviations from linearity will reduce the bonding energy rapidly. It follows that the &quot;best&quot; hydrogen bond is &quot;close&quot; to linear.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16683" LastEditDate="2018-12-24T01:46:17.670" LastActivityDate="2018-12-24T01:46:17.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20153" PostTypeId="2" ParentId="20147" CreationDate="2014-11-30T18:38:54.513" Score="4" Body="&lt;p&gt;If your goal was to prepare hexachlorophosphazene $\ce{(NPCl2)_3}$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wu4lU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then I'm not sure why $\ce{POCl3}$ was added.  The synthesis involving $\ce{POCl3}$ certainly works, but hexachlorophosphazene can be prepared more simply by heating phosphorous pentachloride and ammonium chloride in chlorobenzene.  Careful sublimation affords the desired product, pure and in high yield.  This route is more economical and less toxic since it avoids the use of $\ce{POCl3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{POCl3}$ is typically added to the reaction when one is interested in preparing higher polymers of $\ce{(NPCl2)_{n}}$.  In this case, use of $\ce{POCl3}$ allows you to prepare and isolate $\ce{Cl3P=N-POCl2}~(1)$ in high yield and purity (ref. 1).  The purity is important in obtaining high molecular weight polymers of $\ce{(NPCl2)_{n}}$.  Condensation of $(1)$ to produce hexachlorophosphazene typically occurs when reaction temperatures are maintained well below $\pu{200^\circ C}$, while high molecular weight polymers are produced at temperatures over $\pu{200^\circ  C}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Synthesis and Characterizations of Poly(organophosphazenes) By Roger De Jaeger, Mario Gleria. See &lt;a href=&quot;http://books.google.com/books?id=rW8cUInCRaoC&amp;amp;pg=PA29&amp;amp;lpg=PA29&amp;amp;dq=%28NPCl2%293,%20pocl3&amp;amp;source=bl&amp;amp;ots=BWbuhcxgbE&amp;amp;sig=MfbAQQMic72IsvLxowh37QsJ-G0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=iVR7VLPPE_LIsQTZj4Jg&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=(NPCl2)3%2C%20pocl3&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 26 here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-04-15T02:11:30.760" LastActivityDate="2018-04-15T02:11:30.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20154" PostTypeId="1" CreationDate="2014-11-30T18:49:38.917" Score="1" ViewCount="58" Body="&lt;p&gt;I'm very interested in making homemade droplet lens, I want to know what home element can I use to cook them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a video showing the process, with no further information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=NQLWS5j8gB4&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=NQLWS5j8gB4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a research paper about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.opticsinfobase.org/boe/fulltext.cfm?uri=boe-5-5-1626&amp;amp;id=284161&quot; rel=&quot;nofollow&quot;&gt;http://www.opticsinfobase.org/boe/fulltext.cfm?uri=boe-5-5-1626&amp;amp;id=284161&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10087" LastActivityDate="2014-11-30T18:49:38.917" Title="What home chemical can I use to make droplet lens?" Tags="&lt;everyday-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20156" PostTypeId="2" ParentId="20149" CreationDate="2014-11-30T20:03:03.283" Score="12" Body="&lt;p&gt;At &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_fluoride&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; you find this structure (drawn by &lt;a href=&quot;https://commons.wikimedia.org/wiki/User:Benjah-bmm27&quot; rel=&quot;noreferrer&quot;&gt;Benjah-bmm27&lt;/a&gt; on &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Hydrogen-fluoride-solid-2D-dimensions.png&quot; rel=&quot;noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQVLE.png&quot; height=&quot;150&quot; alt=&quot;Hydrogen bonding in HF (Wikimedia Commons)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer &lt;strong&gt;(c)&lt;/strong&gt; is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration &lt;strong&gt;(c)&lt;/strong&gt; is preferred over &lt;strong&gt;(a)&lt;/strong&gt; can basically be explained via the VSEPR theory: a fluoride ion in HF is surrounded by 3 electron pairs and 1 H–F bond: those 4 &quot;ligands&quot; should roughly be arranged tetrahedrally – approximately, not exactly, because electron pairs need more space than bonding electrons – around the F atom, and this leads to the zigzag chains from answer &lt;strong&gt;(c)&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-12-24T01:49:19.043" LastActivityDate="2018-12-24T01:49:19.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20157" PostTypeId="1" CreationDate="2014-11-30T21:49:13.713" Score="-2" ViewCount="46335" Body="&lt;p&gt;Does a radio wave or gamma radiation have a faster speed? I know that all light travels at &lt;span class=&quot;math-container&quot;&gt;$\pu{3E8 m s-1}$&lt;/span&gt;, but does that include these forms of electromagnetic radiation? Or do radio waves travel at &lt;span class=&quot;math-container&quot;&gt;$700~\mathrm{nm}$&lt;/span&gt; and gamma rays &lt;span class=&quot;math-container&quot;&gt;$400~\mathrm{nm}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10026" LastEditorUserId="4945" LastEditDate="2018-10-08T14:48:42.920" LastActivityDate="2018-10-08T14:56:27.943" Title="Is the speed of gamma rays equal to the speed of radio waves?" Tags="&lt;physical-chemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ClosedDate="2016-01-02T17:39:47.633" ContentLicense="CC BY-SA 4.0" />
  <row Id="20159" PostTypeId="1" AcceptedAnswerId="20168" CreationDate="2014-11-30T22:28:25.313" Score="3" ViewCount="309" Body="&lt;p&gt;I was reading about the integrated rate law. However I have problems to follow the solution.&#xA;I have an equilibrium reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B&amp;lt;=&amp;gt;[{k_{on}}][{k_{off}}]AB}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with forward and back reaction. I approximate the measurement to a pseudo first order reaction. The complex can be measured. &#xA;B can be calculated: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{B}]\overset{\text{def}}{=}[\ce{B_0}]-[\ce{AB}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the equilibrium is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = k_\mathrm{on} \cdot [\ce{A_0}] \cdot ([\ce{B_0}]-[\ce{AB}])-k_\mathrm{off} \cdot [\ce{AB}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be expressed also like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = k_\mathrm{on} \cdot \ce{[A_0]} \cdot \ce{[B_0]} - (k_\mathrm{on} \cdot A_0 - k_\mathrm{off})F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the differential equation I simplified to this: &#xA;\begin{align}&#xA;\frac{\mathrm{d}F(t)}{\mathrm{d}t} &amp;amp;= c_1 - c_2 \cdot F(t)\\&#xA;c_1 &amp;amp;= k_\mathrm{on} \cdot A_0 \cdot B_0\\&#xA;c_2 &amp;amp;= k_\mathrm{on} \cdot A_0 - k_\mathrm{off}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far the theory was easy and I am sure everything is correct, but now I am stuck. The solution of this differential equation would be straightforward: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(t) = \frac{c_1}{c_2}+k\cdot \mathrm{e}^{-(c_2\cdot t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution to fit and simulate the kinetic should be (in the simplified writing). This equation is published in &lt;a href=&quot;http://afm1.pharm.utah.edu/pnscourse/Anal_Biochem_1995.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://afm1.pharm.utah.edu/pnscourse/Anal_Biochem_1995.pdf&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(t) = \frac{c_1\cdot (1-\mathrm{e}^{-(c_2\cdot t)})}{c_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really would like to know, where is my mistake and how the derivations of this formula has to be. &lt;/p&gt;&#xA;" OwnerUserId="7082" LastEditorUserId="4945" LastEditDate="2017-04-15T03:39:49.617" LastActivityDate="2017-04-15T03:39:49.617" Title="Stuck with derivation of the integrated rate equation for a pseudo first order equilibrium reaction" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20162" PostTypeId="2" ParentId="20157" CreationDate="2014-12-01T00:21:33.343" Score="0" Body="&lt;p&gt;All light travels at the same speed in vacuum. What makes different kinds of light different is their wavelength. The electromagnetic spectrum gives you the range of wavelengths that the different kinds of light have. So, &lt;span class=&quot;math-container&quot;&gt;$\pu{400 - 700 nm}$&lt;/span&gt; would be the visible light range. Basically, the light is moving in a particular direction at speed &lt;span class=&quot;math-container&quot;&gt;$\pu{3x10^8 m/s}$&lt;/span&gt;, while the electric field is moving up and down and the magnetic field is moving left and right, both perpendicular to the direction the light is moving and perpendicular to each other. If the electric field went up as high as it can go, then went as low as it can go, and then came back up, the distance the light traveled during this time will be the wavelength, and the time it took will be the period (&lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;), and the amount of times it happens in one second is the frequency(&lt;span class=&quot;math-container&quot;&gt;$1/t$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;). More energy makes the electric field oscillate faster up and down, so the wavelength is shorter. The speed of propagation remains the same. &lt;/p&gt;&#xA;" OwnerUserId="5366" LastEditorUserId="4945" LastEditDate="2018-10-08T14:56:27.943" LastActivityDate="2018-10-08T14:56:27.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20163" PostTypeId="1" AcceptedAnswerId="21582" CreationDate="2014-12-01T00:23:14.767" Score="14" ViewCount="1613" Body="&lt;p&gt;Currently I am looking for the most accurate calculation method for a simple non-conjugate molecule consist of C, N, O and H. Normally I would try each method for a similar and known molecule then compare them with the experimental results but I don't have anything to compare this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I need to predict a method for ground state and excited state calculations. While I was searching I kept coming across the terms &quot;long-range&quot; and &quot;short-range&quot;. In the original publication of CAM-B3LYP for example, the authors claim that CAM-B3LYP provides additional long-range corrections, the weak part of B3LYP. What does it suppose to mean to me? I was unable to find any definitions of these terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any additional tips about choosing a method are appreciated, also.&lt;/p&gt;&#xA;" OwnerUserId="1060" LastEditorUserId="5017" LastEditDate="2014-12-11T16:22:53.007" LastActivityDate="2014-12-13T01:28:42.600" Title="What do short-range and long-range corrections mean in DFT methods?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20164" PostTypeId="2" ParentId="20157" CreationDate="2014-12-01T00:48:01.470" Score="3" Body="&lt;p&gt;In a vacuum all light travels the same speed no matter what the wavelength or frequency. Gamma radiation has a smaller wavelength and larger frequency, radio waves a smaller frequency and larger wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavelength &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; and frequency &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; are inversely proportional. As &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; gets bigger, &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; gets smaller. But if you multiply them together they always equal a constant, the speed of light &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\lambda \cdot ν = c$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] do radio waves travel at &lt;span class=&quot;math-container&quot;&gt;$\pu{700 nm}$&lt;/span&gt; [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The value &lt;span class=&quot;math-container&quot;&gt;$\pu{700 nm}$&lt;/span&gt; is the measure of wavelength; meter is a measure of distance; speed would be meters per second.&lt;/p&gt;&#xA;" OwnerUserId="10092" LastEditorUserId="4945" LastEditDate="2018-10-08T14:52:51.570" LastActivityDate="2018-10-08T14:52:51.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20165" PostTypeId="1" CreationDate="2014-12-01T00:51:42.257" Score="2" ViewCount="14280" Body="&lt;p&gt;I am trying to interpret an IR spectrum where I have $\ce{(NPCl2)3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $\ce{P-N}$, $\ce{P=N}$ and $\ce{P-Cl}$ bonds and it is cyclic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find or does anyone know what the general ranges for these stretches are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a website that told me $\ce{P=N}$ is around $1170 \, \mathrm{cm}^{-1}$.&lt;/p&gt;&#xA;" OwnerUserId="10082" LastEditorUserId="189" LastEditDate="2014-12-02T16:25:41.387" LastActivityDate="2014-12-02T19:28:34.923" Title="Where can I find the infrared band positions for phosphorus compounds?" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20166" PostTypeId="2" ParentId="10226" CreationDate="2014-12-01T01:06:20.343" Score="2" Body="&lt;p&gt;Yes, covalently bonded substances become metals in sufficiently high pressures, which vary for different compounds. Unfortunately we still don't know much about really high pressures because they are difficult to generate - you can find details about metallic &lt;a href=&quot;https://en.wikipedia.org/wiki/Metallic_hydrogen&quot; rel=&quot;nofollow&quot;&gt;hydrogen&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Solid_oxygen&quot; rel=&quot;nofollow&quot;&gt;oxygen&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-01-26T17:56:26.693" LastActivityDate="2015-01-26T17:56:26.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20168" PostTypeId="2" ParentId="20159" CreationDate="2014-12-01T02:44:56.103" Score="5" Body="&lt;p&gt;You have made no mistake. Your solution and the solution from the paper are practically identical. The solution from the paper is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  F(t)=c_1 \frac{1−e^{−c_2 t}}{c_2}=\frac{c_1}{c_2} - \frac{c_1}{c_2} e^{−c_2 t}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is exactly the form your equation has, i.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  F(t)=\frac{c_1}{c_2} + K e^{−c_2 t} \qquad \text{with} \qquad K = - \frac{c_1}{c_2} \ .&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration constant $K$ is not defined the way it is but fixed via the boundary conditions of the reaction. In this case you get $K = - \frac{c_1}{c_2}$ by requiring that at the start of the reaction, i.e. $t=0$, the reactants have not yet reacted with each other and there is no product $\ce{AB}$ present initially, i.e. $F(t\!=\!0) = 0$. This leads to the desired result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;  F(t\!=\!0) \overset{!}{=} 0 &amp;amp;=\frac{c_1}{c_2} + K e^{0} \qquad \Rightarrow \quad K=- \frac{c_1}{c_2} \ .&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-12-01T02:44:56.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20169" PostTypeId="1" AcceptedAnswerId="20172" CreationDate="2014-12-01T03:24:41.143" Score="0" ViewCount="215" Body="&lt;p&gt;I'm looking to set up a small chemistry lab in my garage for doing demos. I think it would be good for a freshman student I'm mentoring and also as a hobby. Does anyone know where I can buy apparatus (like flasks, pipettes, heaters) and chemicals?&lt;/p&gt;&#xA;" OwnerUserId="10095" LastActivityDate="2014-12-01T08:18:59.270" Title="Chemistry Experiments at Home" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2014-12-01T12:12:37.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="20170" PostTypeId="1" CreationDate="2014-12-01T04:36:07.297" Score="5" ViewCount="2225" Body="&lt;p&gt;This is a continuation of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15272/&quot;&gt;Is tropone aromatic?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dt5oc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dt5oc.png&quot; alt=&quot;Resonance forms of tropone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at the resonance form that does have the carbonyl double-bond, &lt;strong&gt;1a&lt;/strong&gt;. Can this resonance structure also be considered aromatic? Would the electrons from the C=O double bond also be delocalized through extended conjugation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, this looks like a very similar situation to styrene, when we have an alkenyl sidechain attached to a benzene ring. Apparently, for alkenyl side-chains, Huckel's number is not affected, leaving an already aromatic compound to stay aromatic even though more electrons come into the conjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of tropone, would the C=O double bond interfere with electron counting for Hückel's rule?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="16683" LastEditDate="2018-01-29T22:19:58.900" LastActivityDate="2018-01-29T22:34:57.120" Title="Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;aromaticity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20172" PostTypeId="2" ParentId="20169" CreationDate="2014-12-01T08:18:59.270" Score="1" Body="&lt;p&gt;If you live close to a research university, I would recommend you to look up if they have a surplus store. Often times when professors retire a lot of the lab supplies get sent to surplus, where they are sold very cheap. You can also build equipment out of basic stuff you can find at any thrift shop (pressure cookers, blenders, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for chemicals, it really depends on what you are interested in doing. Metals are all around you. You can take some pennies, dilute hydrogen peroxide, any carboxylic acid (vinegar, citric acid, amino acid supplements, etc.), and stick the mixture into a microwave for a few seconds. The copper will form complexes with the carboxylic acid you add. Different acids will give you different colors. Leave them standing for a bit and you will get crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get to know the plants around you, you will have access to a huge array of naturally occurring small molecules and enzymes. Some palm trees have a high concentration peroxidases, onions and garlic undergo some pretty cool chemistry which results in the formation of pigments (blue, green, pink), there's pH indicators, capsaicin, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buying pure chemicals for organic synthesis is usually tricky, and for many of them you need to be affiliated with a research institute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several YouTube channels with chemistry experiments which mentions where they obtained the chemicals. I recommend 'NurdRage'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="5366" LastActivityDate="2014-12-01T08:18:59.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20175" PostTypeId="2" ParentId="20059" CreationDate="2014-12-01T13:55:47.977" Score="1" Body="&lt;p&gt;I would start by looking at the cis rather than trans isomer as the starting point for interpreting the NMR spectrum. It is much more likely to be the product you got from this reaction. Also, be sure to identify all the chemically distinct hydrogens in your structures.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-12-01T13:55:47.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20176" PostTypeId="2" ParentId="20100" CreationDate="2014-12-01T15:26:29.407" Score="5" Body="&lt;p&gt;If you look at the two possible planar structures of hydroquinone, one will have both hydroxyl hydrogens on the same side of the molecule (with a nonzero net dipole moment since the two OH dipoles are additive as vectors), whereas the other will have them on opposite sides (and a zero dipole moment since the OH dipoles will cancel each other out).&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-12-01T15:26:29.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20177" PostTypeId="2" ParentId="20089" CreationDate="2014-12-01T15:32:34.863" Score="3" Body="&lt;p&gt;According to the CRC Handbook of Chemistry and Physics 41st edition, lead &quot;dissolves&quot; in nitric acid. As is the case with other unreactive metals like copper and silver, nitric acid acts as both an acid and an oxidizing agent, and dissolves the metal with formation of the metal nitrate and evolution of nitrogen dioxide.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-12-01T15:32:34.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20178" PostTypeId="1" AcceptedAnswerId="20183" CreationDate="2014-12-01T16:24:12.413" Score="-1" ViewCount="549" Body="&lt;p&gt;Lysine is one of the most important amino acids. IUPAC name it as &lt;strong&gt;2,6 - diaminohexanoic acid&lt;/strong&gt; . But what are its resonance forms?? And can resonance tell why it is basic??? Please help.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="865" LastEditDate="2015-10-02T16:33:21.180" LastActivityDate="2015-10-02T16:33:21.180" Title="Resonance structures of lysine and cause of its basic nature" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20179" PostTypeId="1" AcceptedAnswerId="20342" CreationDate="2014-12-01T16:25:34.847" Score="0" ViewCount="198" Body="&lt;p&gt;I've got x-ray powder diffraction (XRPD) data on bendamustine HCl from &lt;a href=&quot;http://www.google.com/patents/WO2009120386A2?cl=en&quot; rel=&quot;nofollow&quot;&gt;a patent&lt;/a&gt; and I want to know how to construct a 3D model using this data with software. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2014-12-05T04:34:38.117" LastActivityDate="2014-12-05T05:15:43.637" Title="Is it possible to construct a 3D molecular model using x-ray powder diffraction data?" Tags="&lt;crystal-structure&gt;&lt;x-ray-diffraction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20181" PostTypeId="1" AcceptedAnswerId="20188" CreationDate="2014-12-01T16:56:38.730" Score="-1" ViewCount="770" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;There can be many resonating forms for a compound. It is to be noted that they are not in any equilibrium to each other. &lt;strong&gt;Each and every structure will contribute to the ultimate hybrid, according to their stability.&lt;/strong&gt; The most important point is that &lt;strong&gt;resonating hybrid is not a mixture of the resonating structures, but rather an intermediate of them.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is what my book writes. But if it is intermediate, it must contain at least some properties of each and every structure. Now, one structure has no charge, while other structure has one positive charge and one negative charge. In the final structure, it still contains both the charges. So, how can it be an intermediate as it got no contribution from the first?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorDisplayName="user7951" LastEditDate="2020-09-26T15:26:05.183" LastActivityDate="2020-09-26T15:26:05.183" Title="Do each and every resonant structures contribute to a resonance hybrid?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20182" PostTypeId="2" ParentId="20179" CreationDate="2014-12-01T16:58:38.870" Score="3" Body="&lt;p&gt;What do you hope to achieve?  It is possible to obtain accurate interatomic distances from powder data (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/om9700122&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is an example, the structures of several alkali metal cyclopentadienides), but the method is non-routine, and the data-parameter ratio is low, which means that you cannot refine many parameters and must put much information in through appropiate restraints.  You still need accurate peak shape data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want the general shape of the molecule to illustrate Wikipedia, you are better off with &lt;a href=&quot;http://openbabel.org/wiki/Obgen&quot; rel=&quot;nofollow&quot;&gt;obgen&lt;/a&gt; from the OpenBabel project.  It usually gives good results. &lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-12-01T16:58:38.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20183" PostTypeId="2" ParentId="20178" CreationDate="2014-12-01T17:03:42.730" Score="0" Body="&lt;p&gt;The group that constitutes the side chain is nothing but a primary amine. That explains the high pKa of ~ 10.5 (compare propylamine (pka ~10.9).  No need to invoke resonance.  Are you perhaps thinking of arginine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jUkI8.png&quot; alt=&quot;Lysine&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="7920" LastEditDate="2014-12-01T17:26:04.293" LastActivityDate="2014-12-01T17:26:04.293" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20185" PostTypeId="1" AcceptedAnswerId="81565" CreationDate="2014-12-01T18:09:59.067" Score="18" ViewCount="4344" Body="&lt;p&gt;Lactones ($\mathrm pK_\mathrm a \sim 25$) tend to be more acidic than esters ($\mathrm pK_\mathrm a \sim 30$). I know that the lactone is fixed in a (&lt;em&gt;E&lt;/em&gt;)-conformation, due to the ring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rDtWj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rDtWj.png&quot; alt=&quot;Conformers of esters&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that in normal esters, the (&lt;em&gt;Z&lt;/em&gt;)-conformer is favoured based on an anomeric effect, namely the donation of the $\mathrm{sp^2}$ lone pair on the ester oxygen into the C–O σ* orbital (image taken from Clayden, Greeves &amp;amp; Warren, &lt;em&gt;Organic Chemistry&lt;/em&gt;, 2nd ed. (OUP), p 805):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SOXDJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SOXDJ.jpg&quot; alt=&quot;Stereoelectronic stabilisation of (Z)-conformer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this effect responsible for the increased acidity? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="16683" LastEditDate="2017-08-21T16:49:24.770" LastActivityDate="2017-08-21T19:36:03.580" Title="Why are lactones more acidic than esters?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;esters&gt;&lt;stereoelectronics&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20186" PostTypeId="1" AcceptedAnswerId="20187" CreationDate="2014-12-01T19:13:09.697" Score="0" ViewCount="199" Body="&lt;p&gt;I have a recipe for a rubber compound and am trying to understand some of the ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones I do not recognize are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound A&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;NSR 20 &lt;/li&gt;&#xA;&lt;li&gt;Accel CS  &lt;/li&gt;&#xA;&lt;li&gt;Accel DPG&lt;/li&gt;&#xA;&lt;li&gt;Anitiox Q&lt;/li&gt;&#xA;&lt;li&gt;Antioz PMB  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compound B&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SMR 20&lt;/li&gt;&#xA;&lt;li&gt;Accel NS&lt;/li&gt;&#xA;&lt;li&gt;Accel DPG&lt;/li&gt;&#xA;&lt;li&gt;PVI&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compound C&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Orox HR&lt;/li&gt;&#xA;&lt;li&gt;Orac CS&lt;/li&gt;&#xA;&lt;li&gt;Orac TMTD&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there a place to look these up on the web?&lt;/p&gt;&#xA;" OwnerUserId="10102" LastEditorUserId="10102" LastEditDate="2014-12-01T19:19:26.950" LastActivityDate="2014-12-02T12:30:37.640" Title="Chemicals in a Rubber Compound - NSR 20, Accel CS, Accel DPG, Anitiox Q, Antioz PMB" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20187" PostTypeId="2" ParentId="20186" CreationDate="2014-12-01T20:16:34.760" Score="0" Body="&lt;p&gt;NSR 20 and SMR 20 are natural rubber (Nigerian Standard Rubber and &lt;a href=&quot;http://www.klpk.com.my/content/display/article/site:klpk-param:standard_malaysian_rubber-link:7546&quot; rel=&quot;nofollow&quot;&gt;Standard Malaysian Rubber&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Accel&quot; and &quot;Orac&quot; mean accelerator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accel DPG is diphenyl guanidine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Antiox&quot; and &quot;orox&quot; mean antioxidant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Antioz&quot; means antiozonant &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Orac CS&quot; is likely N-cyclohexyl-2-benzothiazole sulphenamide &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Orac TMTD&quot; is likely Tetramethylthiuram disulfide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;PVI&quot; may refer to pre-vulcanization inhibitor&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-02T12:30:37.640" LastActivityDate="2014-12-02T12:30:37.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20188" PostTypeId="2" ParentId="20181" CreationDate="2014-12-01T21:18:33.437" Score="1" Body="&lt;p&gt;You can understand the relationship between the resonating forms for a compound and the ultimate hybrid by comparing it with the relationship between the different projections of a vector on the three coordinate axes and the vector itself. None of the projection can lonely represent the vector. But all of them can define the vector precisely in the space. Each projection contributes in defining the modulus of the vector according to its own value.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-01T21:18:33.437" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20189" PostTypeId="1" AcceptedAnswerId="28942" CreationDate="2014-12-01T22:00:25.660" Score="8" ViewCount="19443" Body="&lt;p&gt;Why does the methyl group at the ortho-position stabilize the carboxylate ion? I recognize that there will be a steric clash between the carboxyl group and the methyl group but I don't understand why this would help increase acidity. If there was any resonance occurring then this steric clash would DESTABILIZE the anion and make it LESS acidic given that the carboxylate group would not be in the same plane as the ring and thus delocalization, which would stabilize negative charge, would be impossible. Clearly there is no conjugation with the aromatic ring in &quot;benzoic acid based compounds&quot; from the carboxylate group. So why does having a methyl group at the ortho-position increase acidity? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2015-10-21T14:26:52.787" LastActivityDate="2018-03-09T18:13:06.617" Title="Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ClosedDate="2016-12-01T17:35:28.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="20190" PostTypeId="1" CreationDate="2014-12-01T22:06:22.600" Score="13" ViewCount="10275" Body="&lt;p&gt;When salicylic acid deprotonates (losing the proton from the carboxyl group) it forms a hydrogen bond with between the oxygen in the carboxylate anion and the hydrogen in the alcohol group. However, surely there will be a steric clash between these groups which would surely be make more significant by the long hydrogen bond? Could the hydrogen bond still form with the carboxyl group twited out of the plane? Also, why doesn't it hydrogen bond intramolecularly before deprotonation which would disfavor the loss of a proton? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="9961" LastEditDate="2015-10-02T17:01:30.483" LastActivityDate="2017-10-20T08:50:39.120" Title="Why does hydrogen bonding in salicylic acid make it more acidic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20191" PostTypeId="1" AcceptedAnswerId="20192" CreationDate="2014-12-01T22:13:12.453" Score="5" ViewCount="25232" Body="&lt;p&gt;If I were to attempt to nitrate Pyridine (essentially benzene but with one carbon replaced with Nitrogen) would I get substitution at the Ortho-, Meta- or para position? Also why is this? I can only think of an inductive effect that the Nitrogen might have but I can't think of more factors and I also can't think whether it will be directing towards o-,m- or p- nor why. Can you help?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-12-02T15:01:35.223" Title="Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20192" PostTypeId="2" ParentId="20191" CreationDate="2014-12-01T23:16:53.363" Score="9" Body="&lt;p&gt;The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3olk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons.  Below are the resonance structures we can draw for pyridine.  They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f7QyQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the resonance structures that we can draw for electrophilic attack at the ortho&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtLgu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meta&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vbDPs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and para positions,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Je6o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen.  Therefor, meta-substitution is preferred in electrophilic attack on pyridine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7YGjU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So electrophilic attack on pyridine will produce meta-substituted pyridines.  Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation.  The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred.  In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SPF1L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-02T15:01:35.223" LastActivityDate="2014-12-02T15:01:35.223" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20193" PostTypeId="1" AcceptedAnswerId="20194" CreationDate="2014-12-02T00:38:19.453" Score="0" ViewCount="250" Body="&lt;p&gt;In the following image, $\mathrm{CH}_2$ is connected to benzene rings with a $-\equiv-$ like symbol. What's the meaning of this symbol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqgwI.gif&quot; alt=&quot;Organic chemistry powered double-barrier resonant tunneling diode&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="189" LastEditDate="2014-12-02T00:47:37.253" LastActivityDate="2014-12-02T00:47:37.253" Title="What's the meaning of the $-\equiv-$-like bond symbol?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20194" PostTypeId="2" ParentId="20193" CreationDate="2014-12-02T00:44:36.263" Score="3" Body="&lt;p&gt;This is a triple bond between two $\ce{C}$ atoms, i.e. $\ce{-C#C -}$. It is drawn that way because the carbon atoms in a triple bond are $\mathrm{sp}$-hybridised and thus exhibit a linear arrangement.&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2014-12-02T00:44:36.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20195" PostTypeId="1" CreationDate="2014-12-02T01:46:17.510" Score="3" ViewCount="1094" Body="&lt;p&gt;Doing a grade 9 experiment comparing the alkalinity of antacid tablets.&#xA;Using bromothymol blue indicator.&#xA;Dissolved tablets in both tap water and distilled water and in all situations antacid came out  yellow in with bromothymol blue indicator ... so the antacid is an acid?&#xA;I was intending to have students compare brands by titrating with HCl and comparing the volume of HCl used, but have nowhere to go as the antacid is already an acid?&#xA;Can anyone explain what is wrong? &lt;/p&gt;&#xA;" OwnerUserId="10105" LastActivityDate="2015-01-01T03:06:54.120" Title="Antacid tablet is acid?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20196" PostTypeId="1" CreationDate="2014-12-02T01:58:33.520" Score="3" ViewCount="5577" Body="&lt;p&gt;(Favorable means how exothermic a reaction is, more favorable= more exothermic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some abnormalities in the trends of how favorable the electron affinity is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Al to Cl, the &#xA;Electronegativity is increasing, but the energy releasing from the electron affinity of P is smaller than that of Si. Why? Is it because of how the electrons are arranged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many determinants of how favorable electron affinity is. For example, oxygen and sulfur. Generally, we will think oxygen to have a more favorable electron affinity, due to its stronger electronegativity. Both have the same electrons config, except for the number of shells. But sulfur appears to have a more favorable electron affinity because of its larger size. The smaller size of oxygen atom produces more electron-electron repulsion when one or more electrons are added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I ask for is a list of determinants. I tried to find one but couldn't. &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="16683" LastEditDate="2017-07-04T12:29:11.527" LastActivityDate="2017-07-04T12:29:11.527" Title="What determines how favorable (exothermic) the electron affinity of a certain element is?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20197" PostTypeId="1" CreationDate="2014-12-02T01:59:40.043" Score="0" ViewCount="9991" Body="&lt;p&gt;Specific heat is the amount of heat required to change the temperature of 1 gram of a substance by 1 Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the specific heat of water in the solid phase is 2.059 J/gC and 4.184 J/gC in the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, what is the specific heat of copper in all three of its phases (liquid, solid and gas) in J/gC st standard pressure? How would one figure this out?&lt;/p&gt;&#xA;" OwnerUserId="10106" LastEditorUserId="10106" LastEditDate="2014-12-02T02:09:41.710" LastActivityDate="2014-12-02T07:53:29.583" Title="What is the specific heat of copper through its phases?" Tags="&lt;homework&gt;&lt;heat&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20198" PostTypeId="1" AcceptedAnswerId="20202" CreationDate="2014-12-02T02:14:39.703" Score="0" ViewCount="1442" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Write the combustion equation of the organic compound $\ce{C5H10O}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea at all how to write the equation. Should I add $\ce{O2}$ to make it $$\ce{C5H10O + O2 → ... ?}$$&lt;/p&gt;&#xA;" OwnerUserId="10107" LastEditorUserId="5026" LastEditDate="2018-02-22T02:15:10.247" LastActivityDate="2018-02-22T02:15:10.247" Title="Write the combustion equation of an organic compound" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20199" PostTypeId="2" ParentId="20195" CreationDate="2014-12-02T02:17:35.860" Score="1" Body="&lt;p&gt;You can do your experiment. The active ingredient in most antacids is bicarbonate, and bicarbonates are amphoteric: they can be both acids and bases. In most cases, bicarbonate is a weak base, so despite your bromothymol blue outcome, bicarbonate is a &lt;em&gt;much&lt;/em&gt; weaker acid than $\ce{HCl}$, and you should be able to do your titration. &lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-02T02:17:35.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20200" PostTypeId="1" AcceptedAnswerId="20210" CreationDate="2014-12-02T02:34:19.830" Score="1" ViewCount="477" Body="&lt;p&gt;In my book, it is written:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book cites example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carboxylate ion which contains negative monopolar charge is in fact more stable than the corresponding carboxylic acid which contains both negative and positive charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what's the reason?? What is going on?? Please help me with a math-free explanation.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="9961" LastEditDate="2015-10-16T17:24:35.300" LastActivityDate="2015-10-16T17:24:35.300" Title="Why is bipolar resonating structure less stable than monopolar structure?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20201" PostTypeId="2" ParentId="19239" CreationDate="2014-12-02T03:00:43.333" Score="0" Body="&lt;p&gt;Subtract the mass of $\ce{N2}$ from the mass of NXOY to get 14.3g. &#xA;This means there is 14.3g is from the $\ce{O2}$ so you can find the moles of $\ce{O2}$ and use the coefficients to solve for the ratio between $\ce{N2}$ and $\ce{O2}$ which will give you X and Y.&lt;/p&gt;&#xA;" OwnerUserId="9931" LastEditorUserId="4356" LastEditDate="2014-12-02T08:25:39.630" LastActivityDate="2014-12-02T08:25:39.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20202" PostTypeId="2" ParentId="20198" CreationDate="2014-12-02T03:04:30.670" Score="1" Body="&lt;p&gt;combustion is a reaction using O2 to burn the other reactant and will result in CO2 and H2O&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {\_\_\ C5H10O + \_\_\ O2 -&amp;gt; \_\_\ CO2 + \_\_\ H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then just balance&lt;br&gt;&#xA;so..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C5H10O + 7O2 -&amp;gt; 5CO2 + 5H2O}$&lt;/p&gt;&#xA;" OwnerUserId="9931" LastEditorUserId="9931" LastEditDate="2014-12-02T05:22:04.177" LastActivityDate="2014-12-02T05:22:04.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20203" PostTypeId="2" ParentId="20190" CreationDate="2014-12-02T03:06:24.803" Score="7" Body="&lt;p&gt;If the COO(-) group on salicylic acid's conjugate base didn't hydrogen bond, then wouldn't the COO(-) therefore be more inclined to pick up a proton and return the entire molecule to salicylic acid? In other words H-bonding is a stabilizing interaction. You can view H-bonding as diluting the negative charge density in the COO(-) group. This in turn makes proton abstraction less of a favorable move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure of salicyclic acid's conjugate base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LLSjx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-02T03:06:24.803" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20205" PostTypeId="2" ParentId="20157" CreationDate="2014-12-02T05:34:09.570" Score="-1" Body="&lt;p&gt;All electromagnetic waves will travel at around the same speed in a vacuum. The differences are negligible so we just refer to the speed as &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, or the speed of light which is &lt;span class=&quot;math-container&quot;&gt;$\pu{3.00E8 m/s}$&lt;/span&gt;.&lt;br&gt;&#xA;The &lt;span class=&quot;math-container&quot;&gt;$\pu{700 nm}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{400 nm}$&lt;/span&gt; you are referring to is probably referring to wavelength which is usually measured in &lt;span class=&quot;math-container&quot;&gt;$\pu{nm}$&lt;/span&gt; because it is so small. A wavelength of &lt;span class=&quot;math-container&quot;&gt;$\pu{700 nm}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{400 nm}$&lt;/span&gt; will place those waves in the visible light spectrum because the visible light runs from &lt;span class=&quot;math-container&quot;&gt;$\pu{390 nm}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{700 nm}$&lt;/span&gt;. Frequency also plays a big role and it is usually measured in &lt;span class=&quot;math-container&quot;&gt;$\pu{Hz}$&lt;/span&gt; or hertz which is just another name for &lt;span class=&quot;math-container&quot;&gt;$\pu{1/s}$&lt;/span&gt;, because it is measuring how many times the wave cycles per second. Wavelength is inversely related to frequency. Gamma rays have one of the highest frequencies and smallest wavelengths in the electromagnetic spectrum while radio waves are at the opposite end with low frequency large wavelengths.&lt;/p&gt;&#xA;" OwnerUserId="9931" LastEditorUserId="4945" LastEditDate="2018-10-08T14:55:04.060" LastActivityDate="2018-10-08T14:55:04.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20207" PostTypeId="1" AcceptedAnswerId="20212" CreationDate="2014-12-02T06:28:28.667" Score="11" ViewCount="34477" Body="&lt;p&gt;I am studying nitrogen bases and their structures and roles in nucleotide, polynucleotides and nucleic acids formation. In the structures of the nitrogen bases like Adenine, Guanine, Cytosine and Thymine, the numbering is baffling me. Why are the atoms in these molecules numbered so? Can someone please specify the rules for numbering these four molecules? I am a biologist and I'm not so good at chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wehrr.jpg&quot; alt=&quot;What is the reason behind this numbering system? Why is a particular nitrogen given number 1? Why not the other in the same ring?&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10109" LastEditorUserId="4945" LastEditDate="2015-01-21T07:18:26.640" LastActivityDate="2020-12-05T14:49:51.777" Title="What is the reason behind the numbering of nitrogen bases?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;dna-rna&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20208" PostTypeId="2" ParentId="20197" CreationDate="2014-12-02T07:12:08.640" Score="1" Body="&lt;p&gt;The NIST Webbook is generally a good place to look for thermochemistry data. You can see &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?Name=copper&amp;amp;Units=SI&amp;amp;cTG=on&amp;amp;cTC=on&amp;amp;cTP=on&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that there are data for all three phases. It looks a bit daunting, but because heat capacity is not constant with temperature, it's expressed as a Shomate equation. Basically, you find the equation that starts with $C^°_p$ and substitute the parameters from the table underneath along with the desired temperature and you can calculate the heat capacity for any temperature within the range shown at the top of the table. (note that these are molar heat capacities and you'll have to convert if you want it in terms of mass)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, you can just click &lt;code&gt;View table&lt;/code&gt; to get computed values at certain temperatures, which will save you the calculation.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-02T07:12:08.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20209" PostTypeId="2" ParentId="20197" CreationDate="2014-12-02T07:53:29.583" Score="0" Body="&lt;p&gt;The heat capacity of a material is a measure of the heat energy (in J) that the material (of mass M in g or moles) absorbs or releases for every unit rise or fall in temperature (in °C or K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HHe2e.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst the heat capacity of most materials will generally be different at different temperatures and pressures, as they undergo internal structural changes with temperature, the heat capacity is generally quoted at a given temperature and pressure. These are typically quoted as 'standard conditions' (as defined by the various standard authorities around the world). For example, the National Institute of Standards and Technology (NIST) defines 'standard conditions' as 20°C and 1atm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, copper is a solid below the temperature of 1358 K ​(1085°C) and has a specific heat capacity of 0.386 J/g.K or 24.5 J/mol.K (at 20°C and 1 atm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various empirical formulae exist which allow one to calculate the specific heat capacity of copper at other temperatures, based on measurements taken in various experiments and fitting the data to various curves such as cubic splines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd263.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nist.gov/data/PDFfiles/jpcrd263.pdf&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is a liquid between the temperature of 1358K and 2835 K ​(2562°C) with true specific heat capacity of 0.572 J/g/K or 36.33 J/mol.K at a temperature of 1400K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More accurate values of specific heat are given by suitable curve-fitting formulae, based on experimental data. (See: &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02755998&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/article/10.1007%2FBF02755998&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above it's boiling point of 2835 K ​(2562°C) copper vapourises. Copper gas has specific heat capacity of 25.14 (at 3000K) or at other temperatures, given by the Shomate Equation: $c_p(t)=A+Bt+Ct^2+Dt^3+Et^{-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t$ is the  temperature in Kelvin / 1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = -80.48635$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B= 49.35865$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=-7.578061$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=0.404960$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=133.3382$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7440508&amp;amp;Type=JANAFG&amp;amp;Table=on#JANAFG&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7440508&amp;amp;Type=JANAFG&amp;amp;Table=on#JANAFG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd263.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nist.gov/data/PDFfiles/jpcrd263.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-12-02T07:53:29.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20210" PostTypeId="2" ParentId="20200" CreationDate="2014-12-02T08:37:08.970" Score="2" Body="&lt;p&gt;In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvris.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;resonance stabilization&quot; refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram soure: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5366" LastActivityDate="2014-12-02T08:37:08.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20211" PostTypeId="1" CreationDate="2014-12-02T10:45:42.920" Score="7" ViewCount="49870" Body="&lt;p&gt;I am trying to figure out how the reactivity series comes about. My understanding is that elements with a higher electronegativity will be more reactive than elements with a lower electronegativity, and that elements with a low ionization energy will be more reactive than elements with a high ionization energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a chart of electronegativity (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativities_of_the_elements_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://mychemist.files.wordpress.com/2008/07/pauling-scale.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows electronegativity decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a chart of ionization energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.dartmouth.edu/~genchem/0405/spring/6glueck/images/IE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows ionization energy decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither electronegativity nor ionization energy can predict both reactivity trends, so what is going on?&lt;/p&gt;&#xA;" OwnerUserId="10112" LastEditorUserId="5026" LastEditDate="2018-05-16T12:25:53.373" LastActivityDate="2019-09-25T02:23:52.177" Title="How can I relate the reactivity series to electronegativity and ionization energy?" Tags="&lt;inorganic-chemistry&gt;&lt;reactivity&gt;&lt;electronegativity&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20212" PostTypeId="2" ParentId="20207" CreationDate="2014-12-02T11:57:55.277" Score="11" Body="&lt;p&gt;ACDLabs maintains a nice summary of the IUPAC nomenclature rules with examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heterocyclics (pyrimidine): &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_702.htm&quot;&gt;http://www.acdlabs.com/iupac/nomenclature/79/r79_702.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fused heterocycles (purine): &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm&quot;&gt;http://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, apply the following rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Number your rings so that the nitrogens end up with the lowest number combination. Thus pyrimidines have (1,3).&lt;/li&gt;&#xA;&lt;li&gt;Number your rings so that other functional group have lowest possible numbers. Hence pyrimidine is numbered to the exocyclic carbonyl groups are (2,4).&lt;/li&gt;&#xA;&lt;li&gt;In a fused system, numbering should prefer (in this order): ring with more nitrogens, rings with other heteroatoms, larger rings, nitrogen atom closer to ring junction.&lt;/li&gt;&#xA;&lt;li&gt;Number toward the farther ring junction from where you started. Do not number junction positions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Number 3 is fine and all, but purine seems to have a numbering scheme that does not match these rules, and &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_958.htm&quot;&gt;it is supported by IUPAC&lt;/a&gt;. Likely, this represents a historical numbering pattern that predates the IUPAC rules. Everyone was using it and so it was adopted. You will likely just need to memorize it (&lt;a href=&quot;http://en.wikipedia.org/wiki/Purine&quot;&gt;or look it up when you need to&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2014-12-02T11:57:55.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20213" PostTypeId="1" AcceptedAnswerId="24267" CreationDate="2014-12-02T13:21:09.290" Score="1" ViewCount="190" Body="&lt;p&gt;Everytime I walk by the balloon stall, I know that they use helium to make the balloons float in the air. But my problem is why when I leave the balloon aside, the balloon sinks to the bottom. Is there anything happen to the  helium inside?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:15:35.017" LastActivityDate="2015-10-30T17:15:35.017" Title="Balloon from float to sink?" Tags="&lt;everyday-chemistry&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20214" PostTypeId="1" AcceptedAnswerId="20229" CreationDate="2014-12-02T15:01:36.123" Score="10" ViewCount="11809" Body="&lt;p&gt;Everyone is familiar with the very common, hexagonal ice. Bit there are many other forms of ice - what do they look like? If you had big blocks of ice in all the different known phases next to a block of hexagonal ice, what obvious differences would you notice?&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2014-12-02T23:40:15.947" Title="What do different forms of ice look like?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20216" PostTypeId="2" ParentId="20165" CreationDate="2014-12-02T15:54:21.737" Score="2" Body="&lt;p&gt;A good resource for IR and Raman bands is &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470093072&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Infrared and Raman Characteristic Group Frequencies: Tables and Charts&quot; by George Socrates&lt;/a&gt;. There you can find a chart containing the positions and intensities of bands observed in the infrared spectra of phosphorus compounds (explanation: &quot;w&quot;, &quot;m&quot;, &quot;s&quot;  specify the observed intensity of the bands means whereby &quot;w&quot; stands for &quot;weak&quot;, &quot;m&quot; for &quot;medium&quot;, and &quot;s&quot; for &quot;strong&quot;; additionally the prefix &quot;v&quot; can be given meaning &quot;very&quot;, e.g. &quot;vs&quot; means &quot;very strong&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f5Kt9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mrgN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically you find the following numbers for the different bands you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/REZa4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7KEI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another comment on $\ce{P-Cl}$ bands:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compounds containing the $\ce{P-Cl}$ bond have a medium-to-strong absorption&#xA;  at $605 - 435$ $\mathrm{cm}^{-1}$ due to the $\ce{P-Cl}$ stretching vibration.&#xA;  The position of the band due to the $\ce{P-X}$ ($\ce{X}$=$\ce{F}$ or $\ce{Cl}$) stretching vibration is affected by the oxidation state of the phosphorus atom. 'In the presence of&#xA;  more than one halogen atom directly attached to the phosphorus atom, two&#xA;  peaks are observed due to the asymmetric and symmetric stretching vibrations&#xA;  respectively. Difluorides of the type $\ce{-P(O)F2}$ absorb at $930-895$ $\mathrm{cm}^{-1}$ and $890-870$ $\mathrm{cm}^{-1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2014-12-02T19:28:34.923" LastActivityDate="2014-12-02T19:28:34.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20217" PostTypeId="1" AcceptedAnswerId="20221" CreationDate="2014-12-02T16:35:51.600" Score="3" ViewCount="5854" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why does $\ce{O-}$ show +I effect?&#xA;On the other hand why does $\ce{OH-}$ show -I effect?&#xA;both oxygens are electron rich and are negatively charged. my teacher says that charge on the species is a headache for the species ,so according to me both the oxygens should show +I effect, that is donating electrons tendency. Plus they have lone pairs too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can we extend the similar explanation for the halides? they have 3 lone pairs so they should also not take up more electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If carbon is double bonded to oxygen then is there any inductive effect taking place over there?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Organic is confusing me. Help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="4231" LastEditDate="2015-01-02T14:48:10.590" LastActivityDate="2015-01-02T14:48:10.590" Title="Inductive effect in Organic Chemistry" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ClosedDate="2015-10-09T21:41:03.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="20218" PostTypeId="1" CreationDate="2014-12-02T17:20:18.627" Score="5" ViewCount="1442" Body="&lt;p&gt;There’s an excellent question on Stack Exchange regarding the radiolysis of water &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8929/what-is-a-product-of-water-ionization/8932#8932&quot;&gt;here&lt;/a&gt; but I have one simple outstanding question that perhaps someone might now; in radiobiology, a common DNA damaging free radial is caused by the radiolysis of water. This reaction is something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + $h\nu$ -&amp;gt; H2O+ + e- }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O+ + H2O -&amp;gt;  H3O+ + OH. }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which produces a hydroxyl radical which can prove fatal to cells. I follow the logic here, but given the massive energy of the ionizing radiation, how come a cleaving reaction like the following isn’t common? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + $h\nu$ -&amp;gt; H. + OH. }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve seen the energy quoted to break water this way at around $\pu{493.4 kJ/mol}$, which is way below the energy ($h\nu$) of an x-ray photon, even after compton scattering. Is there a reason why this reaction isn’t valid or common?&lt;/p&gt;&#xA;" OwnerUserId="1781" LastEditorUserId="5026" LastEditDate="2018-04-15T02:14:55.727" LastActivityDate="2018-04-15T02:14:55.727" Title="Reactions in water radiolysis - producing hydroxyl radicals" Tags="&lt;radicals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20220" PostTypeId="1" CreationDate="2014-12-02T17:34:52.810" Score="5" ViewCount="11249" Body="&lt;p&gt;For the 4f inner transition metals, +3 is the most common oxidation state (OS). Why is the +3 OS of cerium considered more stable than +4, at which it attains noble gas configuration?&lt;/p&gt;&#xA;" OwnerUserId="9310" LastEditorUserId="4945" LastEditDate="2014-12-04T12:36:09.433" LastActivityDate="2018-04-09T07:14:55.493" Title="Why is the +3 oxidation state of Cerium more stable than +4?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;&lt;rare-earth-elements&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20221" PostTypeId="2" ParentId="20217" CreationDate="2014-12-02T17:36:03.333" Score="2" Body="&lt;p&gt;Lone pairs can sometimes be delocalized. An acceptable rule of thumb is that if an atom possesses lone pairs and is connected to an sp2 atom, then the two atoms participate in resonance. So consider phenol. Oxygen has lone pairs. It's connected to an sp2 carbon. And so yes, the oxygen can participate in resonance with the sp2 carbon it's connected to, and also the other sp2 carbons in the aromatic ring. We can ascertain the fact that every atom is phenol is sp2 by noting that phenol is &lt;em&gt;planar&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Og6SY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a negative formal charge makes delocalization a (more) favorable move as this decreases the electron (charge) density on one atom. Spreading out electron density between or among multiple atoms generally lessens reactivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your question is why would oxygen exhibit a +I or electron-donating effect despite its high electronegativity. Well, it can participate in resonance donation as explained above; see phenol and its conjugate base, the phenoxide ion, for examples. Examining both in terms of electrophilic aromatic substitution also reveals that the oxygen atom in both is strongly electron &lt;em&gt;donating&lt;/em&gt;. True, oxygen can withdraw electron density via sigma bonds (inductive effect) but doing so is unfavorable especially in the phenoxide ion in which the oxygen &lt;em&gt;already&lt;/em&gt; bears a negative formal charge. As your teacher explained, bearing charge isn't something that is very favorable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in general, you're going to find that resonance effects outweigh inductive effects. Hence the reason why an sp2 oxygen connected to another sp2 atom generally exhibits a +I effect. Consider dimethyl malonate and methyl acetoacetate . Which is more acidic? From inductive effects alone we'd say dimethyl malonate because the oxygen is highly electronegative as opposed to an sp3 carbon and &lt;em&gt;should&lt;/em&gt; withdraw &lt;em&gt;more&lt;/em&gt; electron density via the inductive effect. From a resonance-standpoint we'd say the opposite; the oxygen in the ester donates electron density via resonance more than a methyl group can inductively donate electron density. And you'd have to realize that the resonance-standpoint is more important in evaluating the acidities of the two weak acids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common exception to resonance&gt;inductive effect in organic chemistry is an sp2 atom connected to a halogen. Yes, the halogen has lone pairs and these can be delocalized into the adjacent p-orbital. But the halogens (except possibly fluorine) are much bigger than many other atoms, and therefore the magnitude of resonance donation is limited. So the great electronegativity of the halogens trumps their magnitude resonance donation, and we can see this in the context of electrophilic aromatic substitution again; halogens &lt;em&gt;are&lt;/em&gt; ortho-para directors (delocalize the lone pairs in the Lewis structures to see why) but are &lt;em&gt;also&lt;/em&gt; mild deactivators. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-03T01:56:45.623" LastActivityDate="2014-12-03T01:56:45.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20222" PostTypeId="2" ParentId="20218" CreationDate="2014-12-02T17:59:21.273" Score="2" Body="&lt;p&gt;H (atomic hydrogen radical) is in fact formed.  It is 9% of the radial species formed by radiolysis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS-46.pdf&quot; rel=&quot;nofollow&quot;&gt;Reactivity of the Hydroxyl Radical in Aqueous Solution&lt;/a&gt; and references cited therein for further information. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-02T17:59:21.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20223" PostTypeId="1" AcceptedAnswerId="20225" CreationDate="2014-12-02T18:48:27.000" Score="4" ViewCount="332" Body="&lt;p&gt;I am trying to solve this mechanism for a disfavoured 5-endo-trig cyclisation to generate the disfavoured product (4) from starting material (1). The favoured 5-exo-trig is given by (2), which I have got a mechanism for. My efforts take me to intermediate (3) as shown below, but I'm not too sure on how to generate the desired product without reagent reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lwl1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="4945" LastEditDate="2014-12-05T04:29:36.820" LastActivityDate="2014-12-05T04:29:36.820" Title="What is the reaction mechanism of a 5-endo-trig cyclization?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20225" PostTypeId="2" ParentId="20223" CreationDate="2014-12-02T19:34:27.127" Score="5" Body="&lt;p&gt;Your mechanism to the 5-endo-trig product (4) is very close.  Notice however that when you do the initial cyclization, the intermediate formed has a &lt;strong&gt;positive charge on nitrogen&lt;/strong&gt; along with the negative charge on oxygen (as you showed).  This cyclized intermediate then just needs to move a proton from the nitrogen to oxygen to form the neutral enol which then converts to the (ester) carbonyl - which is your desired product.  No other reagents are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a3bqc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-02T19:34:27.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20226" PostTypeId="1" AcceptedAnswerId="20227" CreationDate="2014-12-02T19:36:41.333" Score="2" ViewCount="1261" Body="&lt;p&gt;I first conducted the following reactions, each in separate test tubes:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaBr}$&lt;/span&gt; + &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; + &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{KI}$&lt;/span&gt; + &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; + &lt;span class=&quot;math-container&quot;&gt;$\ce{KIO3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;...with water and dichloroethane added to each as well.&lt;/p&gt;&#xA;&lt;p&gt;(In the experiment, I put a small quantity of sodium bromide in one test tube and a similar quantity of potassium iodide in the other test tube. I then added 1 mL of water, 3 drops of 9M sulfuric acid, and 2 mL of dichloroethane to each. I then added a small quantity of sodium bromate to the first test tube and a similar quantity of potassium iodate to the other.)&lt;/p&gt;&#xA;&lt;p&gt;Then, I removed the aqueous layer from each test tube and added &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; to the dichloroethane layer in each.&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;(#1) &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; added to dichloroethane&lt;/li&gt;&#xA;&lt;li&gt;(#2) &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; added to dichloroethane&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;So far, I have the following equations written:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaBrO3 + 5NaBr + 3H2SO4 -&amp;gt; 3Na2SO4 + 3H2O + 3Br2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{KIO3 + 5KI + 3H2SO4 -&amp;gt; 3K2SO4 + 3H2O + 3I2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{3Br2 + 6OH^{-} -&amp;gt; 5Br^{-} + BrO3^{-} + 3H2O}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{3I2 + 6OH -&amp;gt; 5I^{-} + IO3^{-} + 3H2O}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Please check my work. If without error, I just need to convert these to equations to net ionic.&lt;/p&gt;&#xA;" OwnerUserId="10118" LastEditorUserId="69830" LastEditDate="2021-05-16T06:02:22.837" LastActivityDate="2021-05-16T06:02:22.837" Title="Disproportionation Reactions of the Halogens: Writing balanced net ionic equations for four reactions. See details" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20227" PostTypeId="2" ParentId="20226" CreationDate="2014-12-02T20:37:00.600" Score="2" Body="&lt;p&gt;(answer to original question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, you are not on the right track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reactions 3 and 4, a hint would be &quot;disproportionation&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-03T01:22:16.087" LastActivityDate="2014-12-03T01:22:16.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20228" PostTypeId="2" ParentId="20220" CreationDate="2014-12-02T20:43:08.573" Score="1" Body="&lt;p&gt;Are you sure $\ce {Ce}$ (III) is really considered more stable than $\ce {Ce}$ (IV)? Since the most common cerium based compound is its dioxide so $\ce {Ce}$ (IV) is actually more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, $\ce{Ce}$ has 4 electrons in its outer orbital, thus leaving only one unpaired electron on the $\ce{6s}$ orbital doesn't sound really stable to me.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="5026" LastEditDate="2018-03-04T11:26:31.523" LastActivityDate="2018-03-04T11:26:31.523" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20229" PostTypeId="2" ParentId="20214" CreationDate="2014-12-02T20:44:42.207" Score="15" Body="&lt;p&gt;As you know, ice can exist in at least 15 or so solid phases. Of these, the first phase, $Ice$ $I$ (ice-one) is the only phase found naturally on earth, mostly in the form of $Ice$ $I_h$ hexagonal crystal as seen in glaciers and snowflakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hacTc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The visible appearance of ice (eg: transparency and clarity) depend also upon the level of purity of the crystals formed. For example, air can get trapped within inclusions of the ice, making it translucent or opaque white, as shown in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4z6eT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cubic form $Ice$ $I_c$ is metastable so is rarely found, generally in high clouds in the upper atmosphere, where it is formed at ambient pressure by the condensation of water vapour at very low temperatures (-80°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most other forms of crystalline ice (designated as $Ice$ $II$ to $XV$ are metastable, being produced in the laboratory by the application of high pressures (above 200MPa) and/or low temperatures (below 72K), resulting of denser packing of the water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $Ice$ $II$ (see image below) is formed by compressing 'normal' $Ice$ $Ih$ to 200MPa at temperature of 198K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bgbay.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SEM image above is of $Ice$ $I$ decorated grain boundaries in fine-grained $Ice$ $II$. The $Ice$ $I$ stands in raised relief relative to the $Ice$ $II$ due to volumetric expansion to the lower-density phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientists believe this form of ice can be found on the interior of some large icy moons orbiting Jupiter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The macroscopic appearance of the 'exotic' forms of ice depends upon many variables such as crystal grain size. The laboratory production of these 'novel' forms of ice to ensure proper formation of fine crytallite grains (less than 10 microns) is time consuming, requiring careful control of temperature and cycling of very high pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $Ice$ $IV$ is formed by heating high-density amorphous ice at a slow (0.4K/min) rate from 145K and at a constant pressure of 0.81GPa. It forms a rhombohedral crystal (see image (a) below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eTNBH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, $Ice$ $V$ is formed from liquid water at 500MPa by lowering its temperature to 253K. It has a more complex lattice structure and forms monoclinic crystals (see image (b) above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Square snow&quot; or $Ice$ $VI$ is formed from liquid water by compressing it at 1.1GPa (10,850atm) and lowering its temperature to 270K. Its unit cell forms tetragonal crystals with all water molecules hydrogen bonded to four others, resulting in the formation of dendritic crystals (approx 45 microns across, see image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOrZ8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more images from the laboratory, see: &lt;a href=&quot;http://web.mit.edu/wbdurham/www1/short%20htmls/more%20ice%20photos.html&quot; rel=&quot;noreferrer&quot;&gt;http://web.mit.edu/wbdurham/www1/short%20htmls/more%20ice%20photos.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.lsbu.ac.uk/water/ice_phases.html#ice&quot; rel=&quot;noreferrer&quot;&gt;Water structure and science, Ice phases, known ices&lt;/a&gt;, Martin Chaplin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/wbdurham/www1/papers/Kubo-etal-2006-Science.pdf&quot; rel=&quot;noreferrer&quot;&gt;Grain size-sensitive creep in Ice II&lt;/a&gt;, Kubo, Durham, Stern &amp;amp; Kirby, Science, Vol 311 (2006).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.jstage.jst.go.jp/article/jshpreview1992/7/0/7_0_1132/_pdf&quot; rel=&quot;noreferrer&quot;&gt;Visual observations of crystal morphologies and melting of Ice IV at pressures up to 1GPa&lt;/a&gt;, Chou and Haselton, Review of High Pressure Science Technology Vol 7 (1998).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pnas.org/content/104/22/9178.full&quot; rel=&quot;noreferrer&quot;&gt;Dynamic pressure-induced dendritic and shock growth of ice IV&lt;/a&gt;, Lee, Evans &amp;amp; Yoo, Proceedings of the National Academy of Sciences, Vol 104, No. 22.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-12-02T23:40:15.947" LastActivityDate="2014-12-02T23:40:15.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20231" PostTypeId="1" AcceptedAnswerId="20235" CreationDate="2014-12-02T21:06:10.297" Score="5" ViewCount="12144" Body="&lt;p&gt;Does this route work to synthesize cyclopentane from cyclohexane?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Free-radical halogenation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Elimination with strong base to form cyclohexene.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ozonolysis to make di-aldehyde.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Oxidation to make di-carboxylic acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Esterification with ethanol, acid, and heat. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alpha-proton elimination with ethoxide ion. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Intramolecular attack of other carbonyl carbon; loss of ethoxide leaving group. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Acid, high heat, water to return to carboxylic acid. Expulsion of ethanol. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat for decarboxylication. Loss of carbon dioxide. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not illustrated: removal of carbonyl using Wolff-Kishner and Clemmensen. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think a few modifications I could make are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Formation of methyl ester using diazomethane. This would not involve heat and acid. But it would involve something that is potentially explosive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alternately, make an acid chloride and then go down to the ester if I wanted to avoid the heat involved with Fischer esterification. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HGZOx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-05-05T16:47:46.123" LastActivityDate="2020-02-11T16:14:49.253" Title="How to synthesize cyclopentane from cyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20232" PostTypeId="1" CreationDate="2014-12-02T21:15:05.460" Score="3" ViewCount="331" Body="&lt;p&gt;Q: A student standardizes sodium hydroxide by weighing out 1.207 grams of potassium hydrogen iodate, dissolving the compound in water and titrating to a suitable endpoint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the volume to reach the endpoint is 32.80 mL, what is the concentration of sodium hydroxide?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The molar mass of PHI is $389.75 \text{g/mol}$. if 1.207g is added to the titrand beaker, then there are:&#xA;$$\frac{1.207\text{g}}{389.75\text{g/mol}} = 0.0031\text{mol of KHI in the titrand beaker}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NaOH reacts with KHI in a 1:1 molar ratio, and 32.8mL of NaOH was added to reach the endpoint. This means that 0.0031mol of NaOH was added. So, the concentration of NaOH is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.0031\text{mol}}{0.0328\text{L}} = 0.0945\text{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is coming up as a wrong answer in my book. Can anyone explain where I'm going wrong?&lt;/p&gt;&#xA;" OwnerUserId="9604" LastActivityDate="2014-12-02T21:15:05.460" Title="Titration: Standardizing NaOH using PHI" Tags="&lt;titration&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20234" PostTypeId="2" ParentId="20231" CreationDate="2014-12-02T21:55:52.213" Score="2" Body="&lt;p&gt;Your route looks good until the last step. After decarboxylation, the intermediate will be cyclopentanone. So all that will be left to do is reduce the ketone to an alkane (you probably know a couple of options).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-02T21:55:52.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20235" PostTypeId="2" ParentId="20231" CreationDate="2014-12-02T22:58:33.240" Score="6" Body="&lt;p&gt;Why not just use the Aldol condensation on the dialdehyde produced from ozonolysis and save some steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zJm0V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-02T22:58:33.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20236" PostTypeId="1" AcceptedAnswerId="20240" CreationDate="2014-12-02T23:18:28.237" Score="1" ViewCount="32279" Body="&lt;p&gt;I have the question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A mixture of $\ce{KClO3}$ and $\ce{KCl}$ with a mass of 0.950g was heated to produce $\ce{O2}$. After heating, the mass of residue was 0.700g. Assuming all the $\ce{KClO3}$ decomposed to $\ce{KCl}$ and $\ce{O2}$, calculate the mass percent of $\ce{KClO3}$ in the original mixture.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took the amount of $\ce{O2}$, converted it to moles $\ce{O2}$, then divided it to moles of just O, then divived by 3, bcause for every 3 O (for the 3 in the $\ce{KClO3}$) there is a mole of $\ce{KClO3}$ so I had moles of $\ce{KClO3}$. Then I converted that to grams of $\ce{KClO3}$, but thaat ended up as more then the original mixture's mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I even thinking about this the right way?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="4356" LastEditDate="2014-12-03T02:46:44.893" LastActivityDate="2015-12-19T17:05:29.127" Title="How do I calculate the mass percent in the original mixture?" Tags="&lt;stoichiometry&gt;&lt;mole&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20237" PostTypeId="1" AcceptedAnswerId="20555" CreationDate="2014-12-02T23:42:23.147" Score="2" ViewCount="2444" Body="&lt;p&gt;Does anyone have any tips on how to approach these types of questions?&#xA;A specific example would be: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw all constitutional isomers of $\ce{C2H4O}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can get the easier ones like vinyl alcohol, but it's hard to get ethylene oxide and it takes a long time for me to get acetaldehdye. And how would I know if I got all of the possible constitutional isomers?&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="4356" LastEditDate="2014-12-03T03:04:48.080" LastActivityDate="2014-12-10T23:05:12.590" Title="How do I approach drawing all of the possible constitutional isomers?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20238" PostTypeId="1" AcceptedAnswerId="20345" CreationDate="2014-12-03T00:09:14.010" Score="0" ViewCount="109" Body="&lt;p&gt;So my question is, how can something exist a gas. Is it when the pressure exerted by the gas is greater than the atmospheric pressure, then it is a gas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also how do manometers work, how can you measure the pressure of a gas if the atmospheric pressure is greater?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://images.duckduckgo.com/iu/?u=http%3A%2F%2Fwps.prenhall.com%2Fwps%2Fmedia%2Fobjects%2F3311%2F3391331%2Fimag1002%2FAAAUAXZ0.JPG&amp;amp;f=1&quot; alt=&quot;Example Image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Taking a look at the example above, I understand that the pressure&#xA;  exerted by the gas can be greater than the atmospheric pressure,&#xA;  because only then you have a gas. But what I don't understand is why a&#xA;  substance exists as a gas when the atmospheric pressure is greater?&#xA;  Shouldn't the atmospheric pressure crush the substance to a solid or&#xA;  liquid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know this is &lt;em&gt;a lot&lt;/em&gt;, but please help.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2014-12-03T22:04:36.800" LastActivityDate="2014-12-05T12:31:21.973" Title="What conditions can cause a material to exist as a gas?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20240" PostTypeId="2" ParentId="20236" CreationDate="2014-12-03T00:28:09.927" Score="2" Body="&lt;p&gt;Looks close. Your math looks sound, but the reasoning is messy. $\ce{O2}$ is the only oxygen here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KClO3 -&amp;gt; 2KCl + 3O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might have skipped writing out that reaction — it helps &lt;strong&gt;a lot.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $0.700~\mathrm{g}$ of residue remained, that means $0.950\ \mathrm{g} - 0.700\ \mathrm{g} = 0.250\ \mathrm{g}$ of $\ce{O2}$ left the container (which means that $0.250\ \mathrm{g}$ of $\ce{O2}$ was made!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the number of moles, similar to what you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.250\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{32\ \mathrm{g}} = 0.00781\ \mathrm{mol}\ \ce{O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we have no reason to divide by 2 — $\ce{O2}$ is the only way (well, at least for this purpose) oxygen can exist in nature. Treat it as &quot;oxygen&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply a mole ratio, now. There were $\ce{2 KClO3}$ molecules used for every $\ce{3 O2}$ molecules we got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.00781\ \mathrm{mol}\ \ce{O2} \times \frac{2~\mathrm{mol}\ \ce{KClO3}}{3~\mathrm{mol}~\ce{O2}} = 0.00521\ \mathrm{mol}\ \ce{KClO3}$ which we had at the beginning. Note that we can do this because it says all of it reacted &lt;em&gt;completely.&lt;/em&gt; If only $95~\%$ of it reacted, we would multiply by $0.95$, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the grams of $\ce{KClO3}$ expended in reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.00521\ \mathrm{mol} \cdot \frac{122.55\ \mathrm{g}}{1\ \mathrm{mol}} = x~\mathrm{g}\ \ce{KClO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your percent is pretty easy from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x\ \mathrm{g}}{0.950\ \mathrm{g}} \times 100~\% = y~\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I'm sure you can figure out $x$ and $y$. (Remember to find $x$ to 3 significant digits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leave a comment if I wasn't clear on something!&lt;/p&gt;&#xA;" OwnerUserId="7641" LastEditorUserId="7475" LastEditDate="2015-12-19T17:05:29.127" LastActivityDate="2015-12-19T17:05:29.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20242" PostTypeId="1" AcceptedAnswerId="20249" CreationDate="2014-12-03T02:03:23.820" Score="9" ViewCount="10381" Body="&lt;p&gt;A lot of people rationalize the acidity of phenol by saying that resonance is responsible for much of phenol's acidity as opposed to aliphatic alcohols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7c0P5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UDBJw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this image suggests that in fact the inductive effect is responsible for a great deal of phenol's additional acidity. As we can see in the picture immediately above, it seems that the sp$^2$ carbon neighboring to the oxygen is pulling most of the weight here; that the resonance contributors with charges on carbons are minor contributors. The additional resonance contributors for phenol seem to be doing almost nothing here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, who's right? Resonance&gt;inductive effect in phenol or the other way around? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The argument I have summarized above is presented here at this website entitled &lt;strong&gt;&quot;I Judge People By Their Grammar and Knowledge of Phenol&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the chemistry in the paper was good, as The Chem Blog has noted,&#xA;  the authors’ lack of attention to detail was borderline disrespectful.&#xA;  I expected more from one of the most storied labs in synthetic organic&#xA;  chemistry. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It turns out that an inductive effect—not a resonance effect—is the&#xA;  predominant reason for the increased acidity of phenol relative to&#xA;  aliphatic alcohols. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For those interested, these data come from Evans’ Chem 206 lecture&#xA;  notes (Lecture 20, restricted access), where the point is hammered&#xA;  home in glorious detail. Professor Evans’ PowerPoint slides should be&#xA;  framed and displayed in the Smithsonian.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Addressing jerepierre's point&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\ce{Phenol + acetone~enolate \leftrightharpoons phenoxide + acetone ~enol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the pKa values we know that the above equilibrium lies mostly to the right (according to jerepierre). Given this, we can conclude that the reactants are less stable in the acid-base sense than the products; acetone enolate is less stable than phenoxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenol is also less stable than acetone enol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see how this contradict's the blog article. It does contradict ron's point about being unable to use the provided pKa values to compare the relative stabilities of the acids. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="10185" LastEditDate="2016-05-30T21:40:14.687" LastActivityDate="2016-05-30T21:40:14.687" Title="Inductive vs resonance effects and the acidity of phenol" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20247" PostTypeId="1" AcceptedAnswerId="20260" CreationDate="2014-12-03T04:45:11.630" Score="13" ViewCount="3188" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4qs3a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the phenoxide ion aromatic? I read that technically it isn't aromatic because it violates some arcane rule, but I can't find any references in the literature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see why it couldn't be aromatic, since it's &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;flat/planar&lt;/li&gt;&#xA;&lt;li&gt;has a system of conjugated p-orbitals. &lt;/li&gt;&#xA;&lt;li&gt;has 4n+2 pi electrons: 2 from the oxygen lone pair that is delocalized and n = 0. &lt;/li&gt;&#xA;&lt;li&gt;is cyclic. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-13T10:33:48.527" LastActivityDate="2018-05-05T02:30:11.013" Title="Is the phenoxide ion aromatic?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;aromaticity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20249" PostTypeId="2" ParentId="20242" CreationDate="2014-12-03T05:14:42.883" Score="4" Body="&lt;p&gt;The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A-H &amp;lt;=&amp;gt; A^- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that we are comparing the &lt;strong&gt;relative&lt;/strong&gt; stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion].  Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability.  In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in his edit Dissenter wrote, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given this, we can conclude that the reactants are less stable in the acid-base sense than the products&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissenter further posits,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's where I disagree.  We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products.  Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol?  Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$?  How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-10-09T16:30:29.300" LastActivityDate="2015-10-09T16:30:29.300" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="20250" PostTypeId="1" AcceptedAnswerId="20266" CreationDate="2014-12-03T05:28:56.963" Score="1" ViewCount="35" Body="&lt;p&gt;How significant is $\ce{NO$_x$}$ formation from $\ce{O2}$ (reactant) and $\ce{N2}$ decomposition from simply heating the two as opposed to from $\ce{CO2}$, $\ce{H2O}$, (and $\ce{N2}$) dissociation? In other words, which of the two pathways for $\ce{NO$_x$}$ formation is more significant (I imagine this will vary with equivalence ratio):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{O}$ and $\ce{N}$ radical from $\ce{N2}$ and $\ce{O2}$ dissociation&lt;/li&gt;&#xA;&lt;li&gt;$\ce{O}$ radical from $\ce{CO2}$ and $\ce{H2O}$ dissociation with $\ce{N}$ radical&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My feeling is that the second becomes more significant as equivalence ratio increases (more fuel means more $\ce{O2}$ is consumed in combustion).&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="194" LastEditDate="2017-04-01T02:59:46.603" LastActivityDate="2017-04-01T02:59:46.603" Title="How significant is air decomposition on formation of nitrogen oxides?" Tags="&lt;physical-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20251" PostTypeId="5" CreationDate="2014-12-03T05:49:21.910" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2014-12-03T05:49:21.910" LastActivityDate="2014-12-03T05:49:21.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20252" PostTypeId="4" CreationDate="2014-12-03T05:49:21.910" Score="0" Body="Electromagnetic radiation (EM radiation, EMR, or light) is a form of energy released by electromagnetic processes. In physics, all EMR is referred to as light, but colloquially light often refers exclusively to visible light, or collectively to visible, infrared, and ultraviolet light.&#xD;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-12-03T11:55:15.817" LastActivityDate="2014-12-03T11:55:15.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20253" PostTypeId="1" AcceptedAnswerId="20269" CreationDate="2014-12-03T06:36:51.343" Score="8" ViewCount="794" Body="&lt;p&gt;This question was posed by my biochemistry professor at the end of a class and then never answered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've gathered is that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It has something to do with the denaturing of a protein, since we were on the subject of denaturing that lecture.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It most likely involves gluten, because that's the main protein in wheat. And, from personal experience of making glue, gluten is very sticky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer might involve hydrophilic/hydrophobic interactions, because those interactions are disturbed during the denaturing process.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="73" LastEditDate="2014-12-03T12:09:02.583" LastActivityDate="2014-12-03T15:33:54.437" Title="Why does pasta become sticky after being cooked?" Tags="&lt;biochemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20254" PostTypeId="2" ParentId="18629" CreationDate="2014-12-03T07:04:51.980" Score="1" Body="&lt;p&gt;$$E = \frac{h\cdot c}{\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\ \mathrm J = 6.24150934 \times 10^{18}\ \mathrm{eV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you have &#xA;$$E = mc^2 = \frac{h\cdot c}{\lambda}$$&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorDisplayName="user7951" LastEditDate="2016-11-10T15:08:35.407" LastActivityDate="2016-11-10T15:08:35.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20256" PostTypeId="2" ParentId="19003" CreationDate="2014-12-03T07:46:50.847" Score="3" Body="&lt;p&gt;Please stay away from sodium hydroxide, it's way too nasty to use in non-lab setup. You can handle it safely in dry conditions, but in contact with water, it becomes sticky and causes bad burns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a substitute, I suggest $\ce{Ca(OH)_2}$, readily available as slaked lime. You can structure it into relatively high-surface sculptures and it solidifies into common plaster upon binding enough $\ce{CO_2}$. It is often used for this purpose due to its low price, low toxicity and ease of handling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation, I would just put some vertically fixed mesh soaked in very viscous suspension of calcium hydroxide. In the high humidity conditions you'd have no problems with normal drying, so once it solidifies it has done its job and you can just hit the mesh against the ground to crumble the plaster away and put on new layer. It is advisable to use gloves and goggles, but it is not as necessary as it would be in the case of sodium hydroxide (look at all the masons around).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="402" LastEditDate="2016-10-11T04:36:24.790" LastActivityDate="2016-10-11T04:36:24.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20257" PostTypeId="2" ParentId="10098" CreationDate="2014-12-03T09:25:27.563" Score="5" Body="&lt;p&gt;All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius.&#xA;\begin{align}&#xA;Q &amp;amp;= mc\Delta T\\&#xA;\end{align}&#xA;Now you can form the equations for each of the problem, while substituting $\Delta T$ with a temperature range, being $x$ the final temperature the whole system will end up on. Also note, that the iron will be cooled down, while the water will be heated. (I am using a different approach than you.&#xA;\begin{align}&#xA;Q_\mathrm{loss} &amp;amp;= m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x]\\&#xA;Q_\mathrm{gain} &amp;amp;= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transferred heat has to equal $$Q_\mathrm{gain} = Q_\mathrm{loss}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this you can solve for $x$.&#xA;\begin{align}&#xA;m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}[T(\ce{Fe})-x] &amp;amp;= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}[x-T(\ce{H2O})]\\&#xA;%m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe})-m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x &amp;amp;= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x-m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})\\&#xA;% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&amp;amp;= m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}x+m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}x\\&#xA;% m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})&amp;amp;= [m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})]\cdot{}x\\&#xA;x &amp;amp;=\frac{m(\ce{Fe})\cdot{}c(\ce{Fe})\cdot{}T(\ce{Fe}) +m(\ce{H2O})\cdot{}c(\ce{H2O})\cdot{}T(\ce{H2O})}{ m(\ce{H2O})\cdot{}c(\ce{H2O})+m(\ce{Fe})\cdot{}c(\ce{Fe})}\\&#xA;%&#xA;x &amp;amp;=\frac{30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}\cdot{}417~\mathrm{K} +40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}\cdot{}293~\mathrm{K}}{ 40~\mathrm{g}\cdot{}4.184~\mathrm{\frac{J}{gK}}+30~\mathrm{g}\cdot{}0.449~\mathrm{\frac{J}{gK}}}\\&#xA;x &amp;amp;= \frac{5616.99~\mathrm{J}+49036.48~\mathrm{J}}{167.36~\mathrm{\frac{J}{K}}+13.47~\mathrm{\frac{J}{K}}}\\&#xA;x &amp;amp;= \frac{54653.47}{180.83}~\mathrm{K} =302.24~\mathrm{K}\\&#xA;x &amp;amp;\approx 29~\mathrm{^\circ{}C}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-03T09:25:27.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20258" PostTypeId="2" ParentId="20247" CreationDate="2014-12-03T09:46:07.757" Score="12" Body="&lt;p&gt;Even when the molecule is present in a neutral form, as a phenol, the oxygen will still be close to a $\ce{sp^2}$ shape with the $\ce{p}_z$ orbital partially overlapping the $\ce{p}_z$ orbital of the carbon bound to it. The proton NMR of phenol clearly shows characteristic aromatic peaks, demonstrating that the ring is aromatic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hückel's rule states that a fully conjugated ring will be aromatic if it has $4n+2$ electrons, and antiaromatic if it has $4n$ electrons, where $n$ is an integer $\geq 0$. The rule is a generalization that comes from how the bonding and nonbonding orbitals are filled as electrons are added to an isolated conjugated ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In analyzing the resonance of the phenol, the Hückel rule applies only to the phenyl ring. It gives no understanding of how the hydroxide atom affects the system because the oxygen atom is not within the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydroxide will affect the electron distribution around the ring. This animation shows it nicely with resonance structures:&lt;br&gt;&#xA;&lt;img src=&quot;https://chemistry.boisestate.edu/richardbanks/organic/phenol.gif&quot; alt=&quot;https://chemistry.boisestate.edu/richardbanks/organic/phenol.gif&quot;&gt;&lt;a href=&quot;https://chemistry.boisestate.edu/richardbanks/organic/phenol.gif&quot;&gt;(source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonance pictures make it easy to see, but it is important to keep in mind that molecules do not actually have resonance structures. A more accurate way of seeing it would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{p}_z$of $\ce{O}$ is delocalized into $\pi^{*}$ of $\ce{C1}$, weakining $\ce{C1-C2}$ and $\ce{C1-C6}$ bonds, increasing the density on $\ce{C2}$ and $\ce{C6}$ by kicking the electrons in the bonding orbitals closer to them. The $\ce{p}_z$ electrons of $\ce{C2/C6}$ move into the $\pi^*$ of $\ce{C3/C5}$, which in turn results in an increased density on $\ce{C4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is still an MO simplification, and to get the accurate picture you would simply have to do the calculations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still possible for an aromatic ring current to flow through the ring in the presence of a magnetic field, as we can see from NMR, but perhaps it feels more like a roller coaster now.&lt;/p&gt;&#xA;" OwnerUserId="5366" LastEditorUserId="4945" LastEditDate="2014-12-03T11:02:25.633" LastActivityDate="2014-12-03T11:02:25.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20260" PostTypeId="2" ParentId="20247" CreationDate="2014-12-03T10:55:20.600" Score="14" Body="&lt;p&gt;I agree absolutely with Max, and while I was still running the calculations, he already provided the answer.&lt;br&gt;&#xA;The goldbook provides a nice definition for &lt;a href=&quot;http://goldbook.iupac.org/A00441.html&quot; rel=&quot;noreferrer&quot;&gt;aromatic&lt;/a&gt; compounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the traditional sense, 'having a chemistry typified by benzene'.&lt;/li&gt;&#xA;  &lt;li&gt;A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.&lt;/li&gt;&#xA;  &lt;li&gt;The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore not all aromatic compounds have to strictly full-fill the &lt;a href=&quot;http://goldbook.iupac.org/H02867.html&quot; rel=&quot;noreferrer&quot;&gt;Hückel (4n + 2) rule&lt;/a&gt;, but all that obey that rule are aromatic. Phenols certainly behave like benzene and are therefore aromatic, please note that aromaticity is an experimentally observed feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the aromaticity from the $\pi$ type orbitals, that are in both cases quite similar and extensively delocalised. The calculations were done at the DF-BP86/def2-SVP level of theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene exhibits a very high symmetry $D_\mathrm{6h}$, therefore the last two of the shown orbitals are degenerate.&lt;br&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/TlSRe.jpg&quot; alt=&quot;pi orbitals of benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The restriction is not present in the phenolate, as the symmetry is only $C_\mathrm{2v}$.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qrdWY.jpg&quot; alt=&quot;pi orbitals of phenolate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also due to the missing symmetry constraint the neutral phenolic radical, $\ce{H5C6O.}$, and the phenolic cation, $\ce{[H5C6O]^{\oplus}}$ would be aromatic.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-03T10:55:20.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20262" PostTypeId="1" AcceptedAnswerId="20270" CreationDate="2014-12-03T11:52:47.177" Score="4" ViewCount="646" Body="&lt;p&gt;Because aquaponics is hydroponics with fish &amp;amp; bacteria, options for adjusting pH are limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primarily people use phosphoric acid, hydrochloric acid, and nitric acid (heavily diluted).  A less popular but easy to acquire material is vinegar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using it in a very alkaline solution (100 ppm calcium carbonates, 8.4 pH) and noticed that the pH would drop by a small amount before returning to it's original state.  Anecdotally, other aquaponic hobbyists have noticed a similar effect that is not replicated by the other acids. What is happening that the pH would lower temporarily and then rise back up over several hours?&lt;/p&gt;&#xA;" OwnerUserId="10126" LastEditorUserId="275" LastEditDate="2014-12-03T11:57:56.797" LastActivityDate="2014-12-03T16:13:54.953" Title="Explanation for behavior of acetic acid in aquaponic/hydroponic systems?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20265" PostTypeId="1" AcceptedAnswerId="20267" CreationDate="2014-12-03T14:00:35.633" Score="4" ViewCount="1317" Body="&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Consider a container of volume $ 5.0$  L that is divided into two compartments of&#xA;    equal size. In the left compartment there is nitrogen at $1.0$ $atm$ and $25 °C$; in the&#xA;    right compartment there is hydrogen at the same temperature and pressure. What&#xA;    will happen when the partition is removed?       &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;p&gt;$A) $The entropy decreases, and the free energy decreases.&lt;br&gt;&#xA;  $B)$ The entropy increases, and the free energy decreases.&lt;br&gt;&#xA;  $C) $The entropy increases, and the free energy increases.&lt;br&gt;&#xA;  $D) $The entropy decreases, and the free energy increases. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Logic tells that upon removing the partition, randomness increases and hence entropy increases. I am confused about free energy. First law of thermodynamics has to be applied , I think. But I can't seem to get the right direction.&lt;/p&gt;&#xA;" OwnerUserId="10036" LastEditorUserId="23561" LastEditDate="2019-03-08T21:47:31.053" LastActivityDate="2019-03-08T21:56:53.133" Title="What will happen to the entropy and free energy of the gasses when the partition is removed?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20266" PostTypeId="2" ParentId="20250" CreationDate="2014-12-03T14:11:54.500" Score="1" Body="&lt;p&gt;In situations where NOx originiates form $\ce{N2}$, such as lightning or a high temperature furnace, the &lt;a href=&quot;http://www.arc.vt.edu/ansys_help/flu_th/flu_th_nox_thermal.html&quot; rel=&quot;nofollow&quot;&gt;Zeldovich mechanism&lt;/a&gt; dominates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 -&amp;gt; O +O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O + N2 -&amp;gt; NO +N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + N-&amp;gt; NO +O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a combusting fuel actual contains nitrogen, other mechanisms can be involved.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-03T15:15:52.993" LastActivityDate="2014-12-03T15:15:52.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20267" PostTypeId="2" ParentId="20265" CreationDate="2014-12-03T15:18:29.423" Score="4" Body="&lt;p&gt;A Spontaneous process is characterized by an increase in the total entropy (for both system and surroundings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spontaneous processes are characterized by a decrease in free energy (analogous to the decrease in gravitational potential energy occurring for a ball rolling downhill).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-03T15:18:29.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20269" PostTypeId="2" ParentId="20253" CreationDate="2014-12-03T15:33:54.437" Score="3" Body="&lt;p&gt;After cooking, pasta has a lot of free surface starch ready to bond with things. This is why we rinse well after cooking to remove extra starch and cool down the pasta to prevent it from producing more surface starch.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-03T15:33:54.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20270" PostTypeId="2" ParentId="20262" CreationDate="2014-12-03T16:13:54.953" Score="4" Body="&lt;p&gt;The source of the acid shouldn't matter much in adjusting the pH by a small amount.  What is the source of the calcium carbonate?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is dissolved calcium carbonate only, as you might find in hard water, then this is strange.  The solution is buffered by the combination of dissolved carbon dioxide and carbonate minerals, so its pH might be difficult to change, but once it has been changed it should stay where you put it, more or less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have solid calcium carbonate (cement, limestone gravel, some pottery (notably not terracotta) or marble chips, perhaps), then what is happening makes a lot more sense.  You add acetic acid, lowering the pH and consuming some of the carbonate from the solution.  There is a large reservoir of carbonate available though, in the form of the carbonate minerals, and that eventually redissolves and the solution becomes more basic with time.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-12-03T16:13:54.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20271" PostTypeId="1" AcceptedAnswerId="20272" CreationDate="2014-12-03T16:38:41.590" Score="0" ViewCount="420" Body="&lt;p&gt;In $\ce{SO5^{2-}}$, can somebody tell me which orbitals are used to form the pi bonds in this anion.&#xA;Please tell the hybridization.&#xA;And the lewis structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well I tried and the answers I got was sp3 &amp;amp; the pi bonds are formed using d orbitals.I just need to know whether I'm correct&lt;/p&gt;&#xA;" OwnerUserId="13243" LastEditorUserId="13243" LastEditDate="2014-12-04T02:18:33.337" LastActivityDate="2014-12-04T11:09:31.260" Title="What is the bonding picture in Peroxomonosulfate?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20272" PostTypeId="2" ParentId="20271" CreationDate="2014-12-03T17:00:27.177" Score="0" Body="&lt;p&gt;Here is already the Lewis structure &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/37QfQ.png&quot; alt=&quot;Here is already the Lewis structure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2014-12-03T17:00:27.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20273" PostTypeId="1" CreationDate="2014-12-03T17:08:03.183" Score="0" ViewCount="291" Body="&lt;p&gt;Its electronic configuration is $4d^15s^2$ so in normal state it has valence=1 and in energetic state, valence=2 (because our teacher said we count only the electrons in the outer shell firstly) and valence =3. First question: is it true what my teacher said? It was for Z=22 example, where she said it has the valence 2 and  4). Second question &lt;strong&gt;when we form the oxide we should use only the valence in the energetic states and not the one in the normal state?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="4945" LastEditDate="2014-12-04T09:45:57.150" LastActivityDate="2014-12-04T09:45:57.150" Title="Form the oxides of the element Z=39" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20274" PostTypeId="1" AcceptedAnswerId="20277" CreationDate="2014-12-03T17:08:45.410" Score="8" ViewCount="10493" Body="&lt;p&gt;I've read that nonpolar bonds are stronger than polar bonds. If it is true, why $\ce{C-H}$ bond entalpy ($\pu{413 kJ/mol}$) is more than $\ce{C-C}$ ($\pu{348 kJ/mol}$)? I'm very confused at the moment.&lt;/p&gt;&#xA;" OwnerUserId="4163" LastEditorUserId="5026" LastEditDate="2018-03-17T05:01:16.743" LastActivityDate="2018-03-17T05:23:29.430" Title="Which bond - C-C or C-H - is stronger?" Tags="&lt;bond&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20275" PostTypeId="1" CreationDate="2014-12-03T17:11:47.007" Score="2" ViewCount="7313" Body="&lt;p&gt;I am doing some exam preparation and I came across this Question which I am finding challenging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;List from least polar to most polar, the following molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naphthalene, Phenol, 3-aminophenol and chlorobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I know that Naphthalene is least polar as the benzene ring is very stable however, the confusion arises with chlorobenzene as the substituent Chlorine is very electronegative hence, it is strongly inductively electron-withdrawing thus is less polar than a simple benzene ring. So which is less polar, chlorobenzene or naphthalene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also which is most polar? 3-aminophenol or phenol? Phenol has a lone pair on oxygen in the oH group which can donate electrons to the ring, although with 3-aminophenol, both the NH2 and OH donate electrons, so wouldn't 3-aminophenol be more electron rich and hence more polar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give the order of the molecules from least to most polar, explaining the order itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="9464" LastActivityDate="2015-01-10T14:19:21.217" Title="List in order of increasing polarity?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20277" PostTypeId="2" ParentId="20274" CreationDate="2014-12-03T17:24:28.697" Score="2" Body="&lt;p&gt;The &quot;strength&quot; of a bond can't be evaluated just based on its enthalpy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta H = \sum \Delta H_\text{bonds broken} - \sum \Delta H_\text{bonds created} $$&#xA;the strength of your bond will depend on the reactive nearby the bond. If the molecule is likely to create bonds with higher enthalpy after the bond's breakdown, then the enthalpy variation will be negative. The process will then be spontaneous and the initial bond will breaks &quot;easily&quot;.&lt;br&gt;&#xA;There is also some steric consideration, the hydrogen atom is smaller, the bond is then more easy to be &quot;attacked&quot;.&lt;br&gt;&#xA;Finally, for a polar bond, the electronic cloud is delocalised closer to one atom rather than the other (according to their electronegativity) so you have &quot;less perturbation to add&quot; to fully remove the covalent electron from the less electronegative atom.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="9961" LastEditDate="2016-01-08T15:25:29.780" LastActivityDate="2016-01-08T15:25:29.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20278" PostTypeId="1" AcceptedAnswerId="20280" CreationDate="2014-12-03T18:29:30.830" Score="3" ViewCount="4829" Body="&lt;p&gt;What is a water blank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing some research and I am suppose to prepare a solution. I should mix 10g of soil with 95ml of water blank in order to create a soil suspension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is I am not English and I have no idea what a water blank is. I have a feeling the type of water is pretty important here though. (can't find what it is on the internet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help please. A simple definition/description will do.&lt;/p&gt;&#xA;" OwnerUserId="10131" LastActivityDate="2017-09-18T03:43:49.287" Title="What is a water blank?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20279" PostTypeId="2" ParentId="20275" CreationDate="2014-12-03T18:40:24.667" Score="0" Body="&lt;p&gt;I guess that 3- amino phenol is more polar than phenol because both OH and NH2 groups have a (+)-M-effect (donate electrons to the ring) so amino group will support electron donation to the ring.&lt;/p&gt;&#xA;" OwnerUserId="9985" LastActivityDate="2014-12-03T18:40:24.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20280" PostTypeId="2" ParentId="20278" CreationDate="2014-12-03T19:15:22.023" Score="4" Body="&lt;p&gt;Generally a blank means no sample and just solution. So it would not contain the 10g of soil in this case. I think for soil analysis it's used as a control so you can have something to compare to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some links I found by doing an internet search.&#xA;&lt;a href=&quot;http://asl.irri.org/lims/index.jsp?page=about&amp;amp;feature=soilanalysis&quot; rel=&quot;nofollow&quot;&gt;http://asl.irri.org/lims/index.jsp?page=about&amp;amp;feature=soilanalysis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.soiltest.vt.edu/PDF/lab-procedures.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.soiltest.vt.edu/PDF/lab-procedures.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastActivityDate="2014-12-03T19:15:22.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20281" PostTypeId="1" AcceptedAnswerId="20290" CreationDate="2014-12-03T20:24:30.003" Score="0" ViewCount="688" Body="&lt;p&gt;What are the &lt;em&gt;functional&lt;/em&gt; differences between alkyl halides alkyl tosylates? Both halides and tosylates are good leaving groups, yes. But is there chemistry that one will do and the other won't? Etc.? I'm not worried about their relative leaving group abilities. My prof treats the two as equivalent but I can't help but wonder if this is an oversimplification. Wikipedia doesn't point out any differences. The only difference I can think of is that an alkyl tosylate is an ester ... and this opens up the world of ester chemistry - i.e. hydrolysis with hot concentrated acid. Where as I can't see alkyl halides reacting in hot concentrated acid. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5591" LastEditDate="2015-10-09T17:29:26.110" LastActivityDate="2015-10-09T17:29:26.110" Title="Differences between alkyl halides and alkyl tosylates" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20283" PostTypeId="1" CreationDate="2014-12-03T21:01:11.960" Score="18" ViewCount="9011" Body="&lt;p&gt;What's the strongest known organic acid? Is it maybe &lt;a href=&quot;https://en.wikipedia.org/wiki/Trifluoromethanesulfonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;trifluoromethanesulfonic acid&lt;/a&gt; or tautomer of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pentacyanocyclopentadiene&quot; rel=&quot;nofollow noreferrer&quot;&gt;pentacyanocyclopentadiene&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since the question was reactivated I thought about formalising it. I can think of a few categories of compounds: neutral (actually created or only studied &lt;em&gt;in silico&lt;/em&gt;) and cationic (also two options). Second Dave's answer and ANM's one do the trick for two groups, but two other options remain (still prefer answers without boro/silico/etc. stuff).&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2018-03-30T22:15:28.020" LastActivityDate="2020-04-20T19:09:48.667" Title="What's the strongest known organic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20284" PostTypeId="1" AcceptedAnswerId="20287" CreationDate="2014-12-03T21:28:11.163" Score="2" ViewCount="3351" Body="&lt;p&gt;I came across this reading a journal article. Here I have the context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here we report for the first time the synthesis of $\ce{TiO2}$ NPs&#xA;  on a layered $\ce{MoS2}$/graphene (MG) hybrid for use in&#xA;  photocatalytic H2 production. It is shown that the activity of&#xA;  the $\ce{TiO2}$ NPs is significantly enhanced by the presence of this&#xA;  layered MG cocatalyst. In this case, ethanol was used as a&#xA;  &lt;strong&gt;sacrificial agent&lt;/strong&gt;, as it is a sustainable and renewable source and&#xA;  showed very good performance with this photocatalyst;&#xA;  however, other &lt;strong&gt;sacrificial agents&lt;/strong&gt; (e.g., glycerol) can be used&#xA;  to make this strategy feasible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Google results came up with similar instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that it would be a chemical consumed in a reaction, but one that doesn't contribute to the structure of the product. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-12-03T22:41:22.630" Title="What is a sacrificial agent?" Tags="&lt;synthesis&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20285" PostTypeId="1" AcceptedAnswerId="20297" CreationDate="2014-12-03T21:34:23.867" Score="5" ViewCount="2656" Body="&lt;p&gt;Some substances like copper sulfate for example have vivid colors. But why is water transparent? Does it not emit any visual light from the electromagnetic spectrum?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorDisplayName="user7951" LastEditDate="2015-06-16T20:28:31.347" LastActivityDate="2018-10-30T12:24:07.870" Title="Why is water transparent?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20286" PostTypeId="2" ParentId="7216" CreationDate="2014-12-03T22:41:17.583" Score="-7" Body="&lt;p&gt;Great info above, I'm glad you took time to look into H2! The initial ? here was basically the difference between Alkaline water &amp;amp; H2 Water.&#xA;Alkalinity can be easily manipulated by the addition of alkaline minerals although this does not increase the molecular Hydrogen content. Much research on Hydrogen gas as a powerful antioxidant and its endogenous antioxidant enzyme enhancing mechanism (see chart #4 in the 2nd study posted below) exists today. Hydrogen is unique in that it has the rare ability to make it past the BBB / Blood Brain Barrier making especially useful for Neuro-degenerative &amp;amp; neurodevelopmental conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium sticks can create H2 water in small amounts. Most common today is the process of ionization or Electrolysis more commonly recognized in scientific terms as &lt;strong&gt;&lt;em&gt;ERW / Electrolyzed-Reduced Water made by quality ionizers.&lt;/em&gt;&lt;/strong&gt; Some call it Kangen Water, a trade name given to the ionized water by one particular Japanese manufacturer whose machines are recognized as home use medical devices.  Here are just 2 of such studies of which the 2nd study continues on to explain the potential of molecular H2 in various prominent disease conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285010/&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285010/&lt;/a&gt; and &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660246/&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660246/&lt;/a&gt; and because of the great importance this ERW has had in my child's life as it greatly reduced her seizure activity, Autism &amp;amp; related issues which is why I study it.&lt;/p&gt;&#xA;" OwnerUserId="10137" LastEditorUserId="10137" LastEditDate="2014-12-03T22:51:29.343" LastActivityDate="2014-12-03T22:51:29.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20287" PostTypeId="2" ParentId="20284" CreationDate="2014-12-03T22:41:22.630" Score="3" Body="&lt;p&gt;&quot;Sacrificial agent&quot; usually refers to something variable, but with little value, that can be used to produce something specific or carry out a specific function. In this case, an alcohol (ethanol or glycerol) is sacrificed to act as a source of hydrogen gas, while the titanium species is a catalyst that is activated by light to allow generation of hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-03T22:41:22.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20288" PostTypeId="1" CreationDate="2014-12-03T23:53:46.653" Score="2" ViewCount="60" Body="&lt;p&gt;I saw different colors of fire burning different metals. But how does this work? The Balmer series requires electrons to drop back down to n = 2. But, not all alkali metals have empty space in their second principle energy level. Does one electron, from the second principle energy level, just jump to somewhere of energy and come back down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My description is kinda confusing, but all I'm  saying is, where does that space in the second energy level come from? &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="418" LastEditDate="2015-01-08T08:30:17.483" LastActivityDate="2015-01-08T08:30:17.483" Title="How can I interpret the colour of flames in terms of spectral series?" Tags="&lt;electronic-configuration&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20289" PostTypeId="1" AcceptedAnswerId="20292" CreationDate="2014-12-03T23:59:12.077" Score="11" ViewCount="5010" Body="&lt;p&gt;Consider two acetic acid derivatives, one with a protonated carbonyl oxygen and one with a protonated hydroxyl oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument for the one with the protonated hydroxyl oxygen being more acidic is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This structure has a lone pair on its protonated hydroxyl oxygen and the oxygen is adjacent to an $sp^2$ carbon. Therefore we can draw an extreme resonance structure in which the hydroxyl oxygen has a +2 formal charge. No such &quot;extreme&quot; structure exists for the one with the protonated carbonyl oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can definitely draw this resonance structure, but its contribution is likely very small and bordering on insignificant. I can't see how delocalization of electrons away from a highly electronegative, positively charged oxygen can help stabilize it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore: Which is more acidic? The argument that the protonated hydroxyl one should be more acidic seems to be flawed at best. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7448" LastEditDate="2015-10-02T16:56:56.087" LastActivityDate="2016-03-22T18:15:38.187" Title="More acidic - protonated carbonyl or protonated hydroxyl?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20290" PostTypeId="2" ParentId="20281" CreationDate="2014-12-04T01:02:13.410" Score="2" Body="&lt;p&gt;Both can be leaving groups. In general, TsO- are better leaving groups, thats why they are often specifically used as so. Halides can be used in many more reactions, like preparation of organometallic compunds.&#xA;TsO- more common function is to transform an alcohol group in a good leaving group. Formation of an halide, usually is not the way to go in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But their basic reactivity is very similar.&#xA;If you dive into more complex reactivity, you can complicate things to a very very large extent...&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="9850" LastEditDate="2014-12-04T01:26:26.190" LastActivityDate="2014-12-04T01:26:26.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20292" PostTypeId="2" ParentId="20289" CreationDate="2014-12-04T01:35:51.527" Score="9" Body="&lt;p&gt;You have outlined the correct reasoning but drawn the wrong conclusion. The protonated carbonyl structure (top left) is stabilized by resonance (bottom left), giving two equivalent resonance structures. Although the protonated hydroxyl (top right) has a resonance structure that can be drawn (bottom right), it should be considered a negligible contributor since there is a double positive charge on a very electronegative atom that would be sp2 hybridized (also toss in the charge separation). Since the protonated hydroxyl is not stabilized by resonance, it is more acidic, i.e., it is more willing to give away the proton (and keep the electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZ9Y7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-12-04T16:14:40.577" LastActivityDate="2014-12-04T16:14:40.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20293" PostTypeId="2" ParentId="20289" CreationDate="2014-12-04T01:47:56.603" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://cactus.dixie.edu/smblack/chem2310/summary_pages/pKa_chart.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; table, protonated ketones have a pK&lt;sub&gt;a&lt;/sub&gt; of ~ -7.3, and the pK&lt;sub&gt;a&lt;/sub&gt; of the hydronium ion is -1.76.  I think the proper argument is that the electronegativity of oxygen in acetone is greater than in water.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-12-04T01:47:56.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20295" PostTypeId="1" CreationDate="2014-12-04T02:36:54.257" Score="0" ViewCount="242" Body="&lt;p&gt;I made $10~\mathrm{M}$ of $\ce{NaOH}$ solution by diluting $40~\mathrm{g}$ of its pellets in distilled water. I'm not sure if it's really $10~\mathrm{M}$ or at least close enough, but I don't think standardize can be done in such concentration. I've been careful enough in making the solution such as measuring the mass in low humidity room and closing it with plastics/glass. &lt;/p&gt;&#xA;" OwnerUserId="10140" LastEditorUserId="4945" LastEditDate="2014-12-04T03:12:15.410" LastActivityDate="2014-12-04T08:59:53.093" Title="Do we need to standardize 10M of NaOH?" Tags="&lt;inorganic-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20296" PostTypeId="2" ParentId="20295" CreationDate="2014-12-04T03:03:53.253" Score="2" Body="&lt;p&gt;This is a highly concentrated base, which is very corrosive. I would assume, that its concentration is quite unstable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will, for example, quite rapidly react with carbon dioxide from the air:&lt;br&gt;&#xA;$$\ce{NaOH + CO2 -&amp;gt; NaHCO3}\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it will corrode glass containers, dissolving minerals like magnesium and calcium from it:&#xA;$$\ce{Mg^{2+} + 2NaOH -&amp;gt; Mg(OH)2 v + 2Na+ }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A PE bottle should be used to avoid that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you worked in a very clean and low humidity environment, the surface of your pallets will be corroded since sodium hydroxide is hygroscopic and will further react according to $(1)$, even when stored carefully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the concentration you obtain by weighing the pallets can only be quite approximate. If you need to know the exact concentration of your base, you have to calibrate it versus a &lt;a href=&quot;http://en.wikipedia.org/wiki/Primary_standard&quot; rel=&quot;nofollow&quot;&gt;primary standard&lt;/a&gt;, like dried &lt;a href=&quot;http://en.wikipedia.org/wiki/Succinic_acid&quot; rel=&quot;nofollow&quot;&gt;succinic acid&lt;/a&gt;, before and after the intended use. &#xA;In your case standardisation will indeed be not easy to perform, since your concentration is very high. Most likely there are uncertainties to the procedure that will likely leave you with similar error bars than not doing the calibration.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-04T03:03:53.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20297" PostTypeId="2" ParentId="20285" CreationDate="2014-12-04T03:11:32.417" Score="10" Body="&lt;p&gt;Water is not, in fact, colourless, but is slightly blue. As you can see below, in the visible range, there is slightly more absorption toward the long wavelengths, so more blue light makes it through than redder light. However, the absorption is very low in the visible range so the blue colour is faint and not apparent unless looking through a significant thickness of water. The reason things like $\ce{Cu^2+}$ solutions appear so vividly blue is because they also absorb red light, but very strongly, so a good deal more blue light than red passes through the solution. (see &lt;a href=&quot;http://www.inds.co.uk/howto/howto_images/Absorbance_Spectra_of_Mountain_Berry_Blast_PowerAde_and_Copper_Sulphate.png&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure exactly what you mean by transparent here, but if you mean why isn't water opaque, because water doesn't absorb much visible light (as described above), and doesn't contain any non-uniformities (like particles or the fat globules in milk) that can cause scattering, photons can travel through water in a straight line and most make it through without being absorbed, making water colourless (when looking through a small quantity) and transparent, much as air is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rBUUF.gif&quot; alt=&quot;water absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Water_absorption_coefficient_large.gif&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-12-04T03:17:24.357" LastActivityDate="2014-12-04T03:17:24.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20298" PostTypeId="2" ParentId="5544" CreationDate="2014-12-04T03:41:20.927" Score="3" Body="&lt;p&gt;I have done this demonstration in class to talk about gas laws, but not to talk about the composition of the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is happening is the candle inside the glass is obviously hot and the flame with heat up the surrounding air.  As temperature of a gas increases, the volume of the gas increases.  This is why you might see some bubbles coming out of the bottom of the glass if you don't have a good seal.  The candle will continue to burn until the oxygen is consumed from the air.  Then when the candle goes out the temperature of the air will decrease.  As the temperature decreases the gas will contract, which causes a vacuum inside of the glass (or an area of low pressure).  Water will move from high pressure to low pressure because the pressure of the air is higher on the water outside of the glass than the pressure on the water inside of the glass.&lt;/p&gt;&#xA;" OwnerUserId="10143" LastActivityDate="2014-12-04T03:41:20.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20299" PostTypeId="1" CreationDate="2014-12-04T04:31:40.040" Score="4" ViewCount="166" Body="&lt;p&gt;I bought azide-labeled primer for click chemistry, but it didn't work. So I want to prove the existence of azide group. How could I achieve this? I found a method to add &lt;code&gt;PPh3&lt;/code&gt; to Kaiser test reagents, but the concentration of the primer is too low (the protocol needs the solution be higher than 0.1M) and I cannot see the reaction.&lt;/p&gt;&#xA;" OwnerUserId="10145" LastActivityDate="2015-03-17T05:50:26.553" Title="How to detect azide group on a primer?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20300" PostTypeId="2" ParentId="20170" CreationDate="2014-12-04T04:44:52.577" Score="2" Body="&lt;p&gt;Maybe &lt;a href=&quot;https://chemistry.stackexchange.com/a/16759/189&quot;&gt;my answer&lt;/a&gt; concerning the question why aniline is considered aromatic might help a bit. There I describe that groups that are cross-conjugated with the aromatic ring are not included into the &quot;Hückel-count&quot; and give some arguments why this is so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar argument would apply here: The keto group is cross-conjugated with the ring's π-system but in a different way than the amino group is in aniline because the carbonyl carbon is part of the ring. So, the role of the keto group's π system is only to act as a bridge between the $\ce{C=C}$ π-orbitals so that the ring is &quot;closed&quot;. Thus, you wouldn't count the keto group's electrons but only the $\ce{C=C}$ π-electrons. And this would lead to the conclusion that your cyclic ketone will be aromatic if it really is flat (that will have to be checked by NMR or other methods) because there are 6 π-electrons in the ring.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="16683" LastEditDate="2018-01-29T22:19:44.433" LastActivityDate="2018-01-29T22:19:44.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20301" PostTypeId="1" CreationDate="2014-12-04T05:04:40.130" Score="6" ViewCount="11951" Body="&lt;p&gt;The dipole moment of ethyl chloride is 2.10 D (&lt;a href=&quot;https://doi.org/10.1016/0022-2860(90)80313-9&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Mol. Struct.&lt;/em&gt; &lt;strong&gt;1990,&lt;/strong&gt; &lt;em&gt;216&lt;/em&gt;, 9–26&lt;/a&gt;), and that of vinyl chloride is 1.42 D (&lt;a href=&quot;https://doi.org/10.1063/1.1700901&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;1960,&lt;/strong&gt; &lt;em&gt;32&lt;/em&gt; (1), 205–209&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case? I would have thought that hyperconjugation would lead to vinyl chloride possessing a larger dipole moment instead.&lt;/p&gt;&#xA;" OwnerUserId="10146" LastEditorUserId="16683" LastEditDate="2019-07-20T19:33:46.497" LastActivityDate="2019-07-20T19:34:24.800" Title="Why is the dipole moment of ethyl chloride larger than that of vinyl chloride?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;dipole&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20303" PostTypeId="2" ParentId="20288" CreationDate="2014-12-04T09:15:15.333" Score="4" Body="&lt;p&gt;The Balmer series relates to the electronic transitions in the hydrogen spectrum. In other elements it only requires that the energy difference between different excited states corresponds to a wavelength in the visible spectrum. These will clearly differ depending on the effective nuclear charge of the elemnt concerned. &lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-12-04T09:15:15.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20304" PostTypeId="2" ParentId="20288" CreationDate="2014-12-04T09:21:02.263" Score="1" Body="&lt;p&gt;Just to complete the previous answer, the Balmer serie isn't the only desexcitation serie producing visible light spectrum for atom nucleus.&lt;br&gt;&#xA;But if you are speaking about the Hydrogen, for instance you also have the Paschen serie for electron down to the $n = 3$ level.&lt;br&gt;&#xA;So it's the same for other elements, there is no special need for an electron to jump down on the precise level $n = 2$ to produce a photon within the visible spectra during its desexcitation. &lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2014-12-04T09:21:02.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20305" PostTypeId="2" ParentId="20273" CreationDate="2014-12-04T09:37:55.773" Score="1" Body="&lt;p&gt;I am not sure that I understand your question, but in terms of establishing the valency in this case the key point is that in d block elements the energy level of the outer s electrons are very similar to that of the d electrons in the shell below, so the valency in this case will be 3.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2014-12-04T09:37:55.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20306" PostTypeId="2" ParentId="20170" CreationDate="2014-12-04T09:41:32.573" Score="4" Body="&lt;p&gt;In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me shortly remind you of the key part of the &lt;a href=&quot;http://goldbook.iupac.org/A00441.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC definition for aromaticity&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the traditional sense, 'having a chemistry typified by benzene'.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I will therefore follow the same approach as in my answer to &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/20247/4945&quot;&gt;Is phenoxide aromatic?&lt;/a&gt;&quot; in comparing the orbital pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2aHi7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2aHi7.png&quot; alt=&quot;Pi orbitals of benzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/od3c3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/od3c3.png&quot; alt=&quot;Pi orbitals of cycloheptatrienone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kQv2T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kQv2T.png&quot; alt=&quot;LUMO of cycloheptatrienone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2018-01-29T22:18:48.943" LastActivityDate="2018-01-29T22:18:48.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20307" PostTypeId="2" ParentId="20271" CreationDate="2014-12-04T11:09:31.260" Score="3" Body="&lt;p&gt;The peroxomonosulfate is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule#Criticism&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypercoordinated molecule&lt;/a&gt;, in which the sulfur appears to exceed Lewis' rule of two, better known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;octet rule&lt;/a&gt;. In former times it was believed that d orbitals would contribute to bonding. This is &lt;em&gt;not&lt;/em&gt; the case.&lt;br&gt;&#xA;Bonding is better represented by &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;three equivalent $\ce{S-O}$ single bonds&lt;/li&gt;&#xA;&lt;li&gt;sulfur having a formal charge of +2&lt;/li&gt;&#xA;&lt;li&gt;these three oxygens having a formal charge of -1&lt;/li&gt;&#xA;&lt;li&gt;one $\ce{S-O}$ single bond that leads to the peroxyl oxygen&lt;/li&gt;&#xA;&lt;li&gt;this oxygen has a formal charge of zero&lt;/li&gt;&#xA;&lt;li&gt;one $\ce{O-O}$ single bond&lt;/li&gt;&#xA;&lt;li&gt;and the terminal oxygen having a formal charge of -1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This bonding picture is much more in accordance with quantum chemical calculations and an NBO analysis on the DF-BP86/def2-SVP level of theory. The involvement of d orbitals is minimal (&amp;lt;1%).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FvokP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy, that this molecule pretty much has the same bonding around the sulphur as in sulphate. It can therefore similarly be described. Have a look at my rant &lt;a href=&quot;https://chemistry.stackexchange.com/a/13896/4945&quot;&gt;against hypervalency&lt;/a&gt; or the &lt;a href=&quot;https://chemistry.stackexchange.com/q/13657/4945&quot;&gt;bonding picture of the $\ce{PO4^{3-}}$ anion&lt;/a&gt;, which is isoelectronic to sulphate.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-12-04T11:09:31.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20308" PostTypeId="1" CreationDate="2014-12-04T12:17:26.163" Score="7" ViewCount="144" Body="&lt;p&gt;I am looking for an easy to calculate model system in which I could see a ligand metal charge transfer. Basically I am looking for a model complex, that might have the metal/ligand in different oxidation states: $\ce{ M^{III}(Lig)}$ versus $\ce{M^{II}(Lig^{\bullet +})}$ or $\ce{M^{IV}(Lig^{\bullet -})}$. I am calling it a charge transfer since I am not sure how else to call it.&lt;br&gt;&#xA;I am also quite aware, that the description of these systems is necessarily a multireference case, so ordinary DFT methods will fail. (At least broken symmetry is necessary.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle I was thinking about an iron heme complex that might exhibit that feature.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uQ6Jg.jpg&quot; alt=&quot;iron(II)heme vs iron(III)heme.-&quot;&gt;&lt;br&gt;&#xA;However, I have the feeling, that this kind of system is already pretty demanding to calculate the electronic structure. Since I would like to use it as an exercise and for a tutorial achieving alternative occupations I would prefer a much simpler case. It should eventually treated with broken symmetry DFT and CAS/MR methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of such a system or would suggest one. It does not have to be a real one. Purely hypothetical is just as fine.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2018-11-16T13:59:38.830" LastActivityDate="2018-11-16T13:59:38.830" Title="What is a good (small) model system for ligand metal charge transfer practises?" Tags="&lt;computational-chemistry&gt;&lt;electronic-configuration&gt;&lt;multi-reference&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20309" PostTypeId="1" CreationDate="2014-12-04T12:55:34.583" Score="22" ViewCount="19010" Body="&lt;p&gt;I know there are issues of formulation and industrial processing that make it advantageous to produce many amine containing drugs as their salts rather than as freebases. And if giving the drug intravenously, the increased water solubility is clearly an advantage. But if giving orally, is there not sufficient HCl in the stomach to have this occur in vivo? Is it just that some of these drugs are too non polar to even go into solution in the stomach acid? Or do we just need to avoid neutralizing too much of the stomach acid with a drug not intended for that purpose?&lt;/p&gt;&#xA;" OwnerUserId="10148" LastEditorUserId="41890" LastEditDate="2017-11-27T18:15:37.207" LastActivityDate="2018-02-11T22:03:58.027" Title="Why formulate drugs as HCl salts when HCl is in stomach acid?" Tags="&lt;acid-base&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20310" PostTypeId="2" ParentId="19975" CreationDate="2014-12-04T13:23:20.580" Score="1" Body="&lt;p&gt;Thanks guys.  After nine months of wondering, i now realise that tryptose is not a carbohydrate but a mixture of polypeptides (or [nonfunctional] proteins?) of varying length (about 2 to 80 or so, which i estimated using the molecular weight distribution chart from the Bacto Tryptose pdf and using a rough average of 120Da for the amino acids).&lt;/p&gt;&#xA;" OwnerUserId="10149" LastActivityDate="2014-12-04T13:23:20.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20312" PostTypeId="2" ParentId="20309" CreationDate="2014-12-04T14:09:15.013" Score="13" Body="&lt;p&gt;Typically, a neutral drug species containing an amine nitrogen will have a higher melting hydrochloride salt that may be better than the &quot;free base&quot; in that it is easier to crystallize (and hence purify) for use in a drug formulation. But many types of salts are used besides the hydrochloride - it all depends on which salt gives good crystals and also works well from a physiological standpoint. For oral administration in the typical quantities used for drug doses, the counterion used probably will not make much if any difference, as it will be outnumbered by chloride and other species present in digestive juices. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-12-04T14:09:15.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20314" PostTypeId="1" AcceptedAnswerId="20316" CreationDate="2014-12-04T14:53:36.657" Score="4" ViewCount="2654" Body="&lt;p&gt;I have a bowl containing cheap wine and an immersible fish tank heater. The heater warms the wine to 30 deg C. The bowl is located in a large closed bucket, where it sits on a support. The bucket is sited out of doors in winter when the air temperature is Ca 1 deg C. Evaporation occurs and the vapour condenses on the inner surface of the bucket. Droplets of condensate run down and form a pool of liquid on the floor of the bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: what can be predicted about the composition of the condensate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #2 What effect is achieved by raising or lowering the temperature of the wine on the composition of the condensate. In plain language what temperature of the wine will give me the highest concentration of alcohol in the condensate? How can the strength of the alcohol be measured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #3 What is the effect of different air temperatures outside of the bucket (on the composition of the condensate)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #4 Why is this method not commonly used to concentrate alcohol?&lt;/p&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="10151" LastEditDate="2014-12-04T15:24:12.423" LastActivityDate="2014-12-19T20:49:54.287" Title="Can you concentrate alcohol by evaporating the aqueous solution at room temp and collecting on a cold surface?" Tags="&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20315" PostTypeId="1" AcceptedAnswerId="20321" CreationDate="2014-12-04T15:14:01.813" Score="34" ViewCount="1740" Body="&lt;p&gt;I am a journal typesetter, and recently I came to a problem that I'm not sure I know to solve. The question is how to properly typeset the &quot;NOx&quot; abbreviation for nitrogen oxides. Which is typographically correct:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathrm{NO}_x$ (because $x$ is a numeric variable); or&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{NO}_{\mathrm{x}}$ (because all names of chemical compounds are strictly upright)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2107" LastEditorDisplayName="user7951" LastEditDate="2016-12-19T17:06:06.300" LastActivityDate="2018-10-04T08:42:42.943" Title="Proper typography of &quot;NOx&quot;" Tags="&lt;molecules&gt;&lt;notation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20316" PostTypeId="2" ParentId="20314" CreationDate="2014-12-04T15:19:46.963" Score="1" Body="&lt;p&gt;Yes you can do that, this technique is called 'distillation' (it requires a minimum difference of 10 degrees celsius to seperate two liquids with different boiling points, for efficiency), I don't know what temeperature the fish tank heater runs at, but you should know that ethanol i,e. what you are trying to separate evaporates at 37 degrees celsius same as normal body temperature, and the temperature of the cooling plate should be &amp;lt; 27 degrees &lt;strong&gt;for efficiency&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit -&lt;br&gt;&#xA;yes, exactly the boiling point of ethanol is 78.3 degrees but ethanol is highly volatile and evaporates very fast at 37 degrees (without taking much water vapour with it) as far as my observation is concerned(with room temperature around, 1 atm pressure). the values I mentioned are the optimal conditions with experimental background. your goal as I can percieve is to avoid as much water as possible, if you take the condenser temperature to below 10 degrees the water vapour present in air will also start to condense at a good rate&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorDisplayName="user10153" LastEditDate="2014-12-05T13:45:29.907" LastActivityDate="2014-12-05T13:45:29.907" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="20317" PostTypeId="2" ParentId="20309" CreationDate="2014-12-04T15:22:38.470" Score="15" Body="&lt;p&gt;Another reason is that many drugs which are conjugated to hydrochloride salts is they are amines, and amines by themselves often stink. So people might be turned off from taking drugs that smell terrible (fishy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, note that even though a label might say something such as &quot;fexofenadine HCl&quot; that doesn't mean there is HCl in the drug. Fexofenadine is an amine, and amines are somewhat basic - and more than basic enough for a strong acid such as HCl to &lt;em&gt;completely&lt;/em&gt; protonate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if this is what you had in mind for the structure of fexofenadine, it isn't the most correct structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NuHkh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a much better structure. The nitrogen in the ring has been protonated and it now bears a positive formal charge. Chloride ion is the counter ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hzleW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-05T13:56:47.027" LastActivityDate="2014-12-05T13:56:47.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20318" PostTypeId="1" CreationDate="2014-12-04T15:27:24.527" Score="4" ViewCount="148" Body="&lt;p&gt;I am designing an experiment that involves immersing pieces of pipe lining in a mixture of &lt;em&gt;tert&lt;/em&gt;-butyl mercaptan (2 %) and white mineral oil (98 %) at 230 °C (the standard temperature for the test). If I was using pure &lt;em&gt;tert&lt;/em&gt;-butyl mercaptan I would be concerned with it exploding, due to its auto ignition temperature being 250 °C. However, if the white mineral oil contains only 2 % &lt;em&gt;tert&lt;/em&gt;-butyl mercaptan would the risk of autoignition be present at 230 °C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure Mineral oil has vapour pressure of less than 1mm Hg. Tert butyl mercaptan has vapour pressure of 303 mm Hg. For the tert butyl mercaptan (2%) in mineral oil (98%) mixture I calculated that the partial pressure for tert butyl mercaptan would be 6.06 mm Hg and for mineral it would be 0.098 mm Hg. Therefore, the composition of gas above the surface of the mixture would be tert butyl mercaptan 98% and mineral oil 2%. If this is the case then the auto ignition temperature of pure tert butyl mercaptan of 250 °C would apply to the resultant gas and it would not be safe operate at 230 °C. Is this right?&lt;/p&gt;&#xA;" OwnerUserId="10155" LastEditorUserId="23561" LastEditDate="2019-08-02T18:14:36.073" LastActivityDate="2019-08-02T18:14:36.073" Title="Auto ignition of tert-butyl mercaptan mixed with mineral oil" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;combustion&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20320" PostTypeId="2" ParentId="20301" CreationDate="2014-12-04T16:04:21.000" Score="0" Body="&lt;p&gt;Don't forget that dipole moment depends not only on electronegativity differences but also bond lengths. All other factors held equal, the longer the bond, the greater the dipole moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would dipole moment depend on bond length? Consider an extreme example in which, say, sodium ion were in exactly the same place as fluoride ion. Obviously the two differ tremendously in electronegativities. However, there cannot be a dipole moment; there are no &lt;em&gt;poles&lt;/em&gt; of opposite charge since the two are directly on top of each other. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-04T16:04:21.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20321" PostTypeId="2" ParentId="20315" CreationDate="2014-12-04T16:19:59.757" Score="32" Body="&lt;p&gt;I quote the &lt;a href=&quot;http://www.iupac.org/home/publications/e-resources/nomenclature-and-terminology/quantities-units-and-symbols-in-physical-chemistry-green-book.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Green Book&lt;/a&gt; by IUPAC, 2&lt;sup&gt;nd&lt;/sup&gt; printing (2008), section 1.6, enumeration item 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The overall rule is that symbols representing physical quantities or variables are italic, but symbols representing units, mathematical constants, or labels, are roman. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As such, the correct way to write it is $\ce{NO}_x$, because $x$ is a variable. Please note that it is only the sum formula for the compound and not its name. Since the element names still are in roman (upright) font, everything is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for a formula like $\ce{C_{$n$}H_{$2n+2$}}$, which is typeset exactly like that in the &lt;a href=&quot;http://goldbook.iupac.org/PDF/A00222.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gold Book&lt;/a&gt; (hat tip to &lt;a href=&quot;https://chemistry.stackexchange.com/users/4945/martin&quot;&gt;@Martin&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here at StackExchange, you can type it as &lt;code&gt;$\ce{NO_$x$}$&lt;/code&gt; ($\ce{NO_$x$}$) or &lt;code&gt;$\ce{NO_x}$&lt;/code&gt; ($\ce{NO_x}$).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-02-17T21:19:20.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20322" PostTypeId="2" ParentId="20301" CreationDate="2014-12-04T16:23:16.073" Score="7" Body="&lt;p&gt;I see two explanations for why the dipole in vinyl chloride is smaller than ethyl chloride.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider a resonance structure where the chlorine donates electrons to the alkene. This puts a formal positive charge on chlorine: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1CqKO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1CqKO.png&quot; alt=&quot;Resonance forms of vinyl chloride&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a very minor contributor given the electronegativities and the mismatch in orbital size. However, this is not possible in ethyl chloride at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The more likely explanation is that an sp&lt;sup&gt;2&lt;/sup&gt;-hybridized carbon is more electronegative than an sp&lt;sup&gt;3&lt;/sup&gt;-hybridized carbon. Since the electronegativity of chlorine will be closer to the electronegativity of the carbons in vinyl chloride than the carbons of ethyl chloride, the C–Cl bond will be less polarized in vinyl chloride, giving a smaller dipole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2019-07-20T19:34:24.800" LastActivityDate="2019-07-20T19:34:24.800" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="20323" PostTypeId="1" CreationDate="2014-12-04T17:52:06.730" Score="2" ViewCount="1129" Body="&lt;p&gt;Bismuth(III), if I am not mistaken, has only two valence electrons in the 6s orbital. However it is able to form 3 bonds with hydroxide. This goes against my intuition. How is it able to form 3 bonds with 3 hydroxide molecules with only 2 valence electrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;my best guess is that i am mistaken in thinking that the bismuth(III) is forming 3 covalent bonds when in actually it is forming ionic bonds. As a result, bismuth does not need to offer/share 3 electrons to each hydroxide, but simply take 3. Is this correct? &lt;/p&gt;&#xA;" OwnerUserId="8138" LastEditorUserId="8138" LastEditDate="2014-12-04T17:57:10.133" LastActivityDate="2014-12-04T19:28:47.380" Title="How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide?" Tags="&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20325" PostTypeId="2" ParentId="20323" CreationDate="2014-12-04T19:03:53.337" Score="4" Body="&lt;p&gt;There isn't a unique sturcture for bismuth hydroxide, but ions such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Bi6O4(OH)4]^{6+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://cdn-pubs.acs.org/doi/pdfplus/10.1021/ic000022m&quot; rel=&quot;nofollow&quot;&gt;Solvation of the Bismuth(III) Ion by Water...&lt;/a&gt; Inorg. Chem. 2000, 39, 4012-4021 as a starting point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-04T19:28:47.380" LastActivityDate="2014-12-04T19:28:47.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20326" PostTypeId="1" AcceptedAnswerId="20328" CreationDate="2014-12-04T20:06:21.950" Score="9" ViewCount="1297" Body="&lt;p&gt;The molar masses of $\ce{N2}$ is 28.0134 g/mol, for $\ce{CO}$ it's 28.0101 g/mol (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=compare%20N2%20and%20CO&quot; rel=&quot;nofollow&quot;&gt;WolframAlpha computation&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, they do have the same mass in atomic units (28). My question is, where does this small discrepancy come from? Different binding energies in the respective nuclei? I know that oxygen is a &quot;double-magic&quot; nucleus from an atomic physics POV, that is, it's binding energy per nucleon is higher than what would be expected. But since normally, you equate mass with energy, shouldn't $\ce{CO}$ be heavier then?&lt;/p&gt;&#xA;" OwnerUserId="10160" LastEditorUserId="4231" LastEditDate="2014-12-05T19:20:09.197" LastActivityDate="2014-12-05T19:20:09.197" Title="Where does the molar mass difference between N₂ and CO come from?" Tags="&lt;physical-chemistry&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20327" PostTypeId="2" ParentId="20326" CreationDate="2014-12-04T20:26:12.953" Score="5" Body="&lt;p&gt;There are three factors to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://www.chem.ualberta.ca/~massspec/atomic_mass_abund.pdf&quot;&gt;natural abundance isotope composition&lt;/a&gt; of each element.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Carbon isn't all carbon 12, there is 1.1% Carbon 13.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen isn't all oxygen 16, etc. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The neutron-proton ratio of each isotope. The mass of a neutron is greater than the mass of a proton plus electron. Isotopes having a higher neutron-proton ratio relative to carbon 12 (which is 1:1) will be relatively heavier per nucleon. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The binding energy of each isotope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-04T20:37:05.720" LastActivityDate="2014-12-04T20:37:05.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20328" PostTypeId="2" ParentId="20326" CreationDate="2014-12-04T20:53:08.313" Score="11" Body="&lt;p&gt;The differences in molecular mass stem from two sources:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy&quot;&gt;Nuclear binding energy&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition of the atomic mass unit and Avogadro's number (and thus the g/mol), is the mass of one atom of the carbon-12 isotope $\ce{^{12}_6 C}$. 1 amu = $\frac{1}{12}$ the mass of one atom of $\ce{^{12}_6 C}$ and the g/mol unit is based on the definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro_constant&quot;&gt;Avogadro's Number&lt;/a&gt;, which is the number of $\ce{^{12}_6 C}$ atoms in 12 grams of $\ce{^{12}_6 C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the differences in binding energies in the various nuclei, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Monoisotopic_mass&quot;&gt;monoisotopic mass&lt;/a&gt; of individual isotopes are not integers. For example, the mass of $\ce{^{16}_8 O}$ is &lt;a href=&quot;http://www.webelements.com/oxygen/isotopes.html&quot;&gt;15.99491463 amu&lt;/a&gt; (and not 16). The monoisotopic masses of the most common isotopes of oxygen, nitrogen, and carbon is shown below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;           mass number   % absundance    isotope mass        &#xA;Carbon     12            98.930          12 (defined)&#xA;Nitrogen   14            99.632          14.003074&#xA;Oxygen     16            99.757          15.99491463&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Isotopes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally, molecular masses (unless otherwise specified) are average molecular masses composed of average molecular masses of the elements based on their isotopes. For example, even though the most common isotope of carbon has a mass defined as 12 amu, there is another naturally occurring isotope $\ce{^{13}_6 C}$ with 1.07% natural abundance, leading to carbon having a non-integer average atomic mass of 12.01 (to 4 significant figures). There are two isotopes of nitrogen, and three isotopes of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            Average atomic mass&#xA;Carbon      12.011&#xA;Nitrogen    14.007&#xA;Oxygen      15.999&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, $\ce{CO}$ has a molecular mass of $12.011 + 15.999 = 28.010$ and $\ce{N2}$ has a molecular mass of $14.007+14.007=28.014$. Even their monoisotopic masses are different:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    average molecular mass       monoisotopic mass&#xA;dinitrogen          28.014                       28.006148&#xA;carbon monoxide     28.010                       27.99491463&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2014-12-05T02:47:44.323" LastActivityDate="2014-12-05T02:47:44.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20329" PostTypeId="2" ParentId="20315" CreationDate="2014-12-04T20:56:09.217" Score="12" Body="&lt;p&gt;The American Chemical Society recommends $\ce{NO}_x$: page&amp;nbsp;191 of Appendix&amp;nbsp;10-2, &lt;em&gt;Abbreviations, Acronyms, and Symbols&lt;/em&gt; of their style guide (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/bk-2006-STYG.ch010&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it's just in a list of abbreviations and I've not managed to find any definitive explanation of &lt;em&gt;why&lt;/em&gt; it should be typeset that way, elsewhere in the style guide. However, &lt;a href=&quot;https://chemistry.stackexchange.com/a/20321/7412&quot;&gt;tschoppi's explanation&lt;/a&gt; is what I would have said, too.&lt;/p&gt;&#xA;" OwnerUserId="7412" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-12-04T20:56:09.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20330" PostTypeId="2" ParentId="1012" CreationDate="2014-12-04T22:09:00.883" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a simple way to get the circular dichroism of a molecule from&#xA;  it's structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt;  &lt;strong&gt;Yes&lt;/strong&gt;, for many molecules if you know the molecule's structure, then you can predict the shape of the &lt;strong&gt;optical rotary dispersion&lt;/strong&gt; (ORD) or &lt;strong&gt;circular dichroism&lt;/strong&gt; (CD) curve.  Conversely, and perhaps more importantly, if you know the shape of the ORD or CD curve, then you can predict the molecule's structure, in particular, the absolute configuration at chiral centers in the molecule. There are many empirical techniques, such as &lt;a href=&quot;http://faculty.concordia.ca/muchall/chem325/2boctantrule.pdf&quot; rel=&quot;noreferrer&quot;&gt;the &quot;Octant&quot; rule&lt;/a&gt;.  These methods are based on numerous experimental observations that have been codified into sets of rules for different classes of compounds.  These empirical rules can be used to determine the absolute configuration of cyclic ketones, conjugated dienes, alpha, beta-unsaturated ketones, twisted biphenyls as well as other systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt; When polarized light is passed through a chiral sample and the light is absorbed, there are two physical phenomenon that can be measured.  One is called the &lt;strong&gt;circular birefringence&lt;/strong&gt; and it measures the difference in the refractive indices ($\ce{n_{R}, n_{L}}$) of the sample when a beam of right and left circularly polarized light is passed through the sample.  When this difference is plotted against wavelength an &lt;strong&gt;ORD&lt;/strong&gt; curve results.  The other parameter is called &lt;strong&gt;circular dichroism&lt;/strong&gt; and it measures the difference in the molar extinction coefficients ($\ce{\epsilon_{R}, \epsilon_{L}}$) when right and left circularly polarized light passes through the sample.  When this difference is plotted against wavelength a &lt;strong&gt;CD&lt;/strong&gt; curve results.  The dependence of these two phenomenon on wavelength is termed the &lt;strong&gt;Cotton effect&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following figure shows the uv-vis absorption (blue line), ORD curve (green line) and CD curve (red line) for the two enantiomers of camphor sulfonic acid.  Note how the ORD and CD curves for one enantiomer are mirror images of the same curves for the other enantiomer.  It is the carbonyl chromophore's $\ce{n-&amp;gt;\pi^{\ast}}$ absorption that is responsible for the curves seen in this image.  In ORD spectra there are typically 2 maxima, one positive and one negative.  The positive maxima is a &quot;peak&quot;, while the negative maxima is termed a &quot;trough&quot;. If, when moving from longer to shorter wavelength, a peak is encountered first, then the ORD curve is said to be positive (+).  Conversely, if a trough is encountered first, then the ORD curve is negative (-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x7AOF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.photophysics.com/tutorials/circular-dichroism-cd-spectroscopy/2-chiral-molecules&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromophores can be divided into two groups, symmetric or inherently dissymmetric.  A carbonyl, nitro and olefin are examples of groups that, by themselves, are symmetric.  They may be perturbed by asymmetry on other atoms around them, but alone, by themselves, they possess no inherent asymmetry.  On the other hand, groups such as helicenes, twisted biphenyls and unsaturated ketones possess a twist or handedness; such molecules are inherently dissymmetric.  The intensity of an ORD or CD curve from a molecule containing an inherently dissymmetric chromophore is usually much greater than the peak intensity from a molecule containing a symmetric chromophore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that information as background, we can now apply these concepts to the examination of real molecules.  One of the most studied classes of molecules where these concepts have been applied are the substituted cyclohexanones.  In the early 60's  Djerassi, Moffitt, Woodward, Moscowitz and Klyne (some of the greats!) developed an empirical set of rules to predict the absolute configuration of chiral centers in substituted cyclohexanones.  The rule became known as the &quot;Octant&quot; rule and came about because of the work they were doing with complex natural products that often contained a substituted cyclohexanone ring.  They would run a reaction on a pure enantiomer of a cyclohexanone containing natural product and wanted a method to quickly determine if a chiral carbon in the molecule retained its absolute configuration or if the configuration was inverted in the reaction.  After comparing ORD curves from countless cyclohexanone containing natural products, they were able to compile their observations into the Octant rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As its name implies, the Octant rule divides the region around a cyclohexanone carbonyl into 8 regions or octants.  As the picture below (left part of figure) illustrates, the octants are generated by bisecting the carbonyl group with 3 planes.  Two of the planes contain the carbonyl group.  One divides the carbonyl into top and bottom regions and the other divides the carbonyl into right and left regions.  The third plane actually passes through the middle of the carbonyl and is perpendicular to the 2 other planes.  This third plane divides the molecule into front and back regions.  There are a few molecules that do have substituents in the front region, but most molecules do not, so we can usually simplify our analysis by only considering the 4 rear octants.  Each octant is also assigned a plus or minus symbol telling us whether we would expect a substituent in the octant to contribute in a positive or negative manner to the ORD or CD curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfnSo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=octant%20rule&amp;amp;hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;biw=1280&amp;amp;bih=829&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=W82AVO35K671iQLF9ICQBQ&amp;amp;ved=0CAYQ_AUoAQ&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure on the right shows how a cyclohexanone would be placed in the octant diagram and how its projection appears as a rectangle when sighted down the carbonyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the case of 3-methyl cyclohexanone.  First let’s draw the possible conformers of both S- and R-3-methylcyclohexanone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/klRM5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let’s take the 2 possible conformations for R-3-methylcyclohexanone (axial and equatorial methyl group) and project them onto the octant diagram.  We see that the equatorial conformation has a positive contribution to the Cotton effect, while the axial conformer has a negative contribution.  At equilibrium, the molecule exists predominantly in the conformation with the methyl group equatorial, therefore we expect the positive Cotton effect from this conformer to outweigh the negative contribution to the Cotton effect from the axial conformer.  Therefore, we expect a positive ORD and CD curve for R-3-methylcyclohexanone.  Similar analysis leads to the expectation of a negative ORD and CD curve for the S enantiomer.  Indeed, experimentally it is found that the ORD curve for R-3-methylcyclohexanone is positive and negative for S-3-methylcyclohexanone.  The full name for the R enantiomer is R-(+)-3-methylcyclohexanone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6GoXr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=octant%20rule&amp;amp;hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;biw=1280&amp;amp;bih=829&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=W82AVO35K671iQLF9ICQBQ&amp;amp;ved=0CAYQ_AUoAQ&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail on the 3-cyclohexanone example, as well a nice, concise review of ORD &lt;a href=&quot;http://faculty.concordia.ca/muchall/chem325/2boctantrule.pdf&quot; rel=&quot;noreferrer&quot;&gt;can be found here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Books have been written on this subject so there is much that I have left out.  For those who would like to explore further here are a few phrases that, in addition to the previous link, can help you get started:  &lt;a href=&quot;https://www.google.com/search?q=ord%20curve&amp;amp;hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;biw=1280&amp;amp;bih=829&amp;amp;source=lnms&amp;amp;sa=X&amp;amp;ei=28mAVMCpA-abigLG3oGoAQ&amp;amp;ved=0CAcQ_AUoAA&amp;amp;dpr=1#hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;q=cotton%20effect&quot; rel=&quot;noreferrer&quot;&gt;Cotton effect&lt;/a&gt;, &lt;a href=&quot;https://www.google.com/search?q=ord%20curve&amp;amp;hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;biw=1280&amp;amp;bih=829&amp;amp;source=lnms&amp;amp;sa=X&amp;amp;ei=28mAVMCpA-abigLG3oGoAQ&amp;amp;ved=0CAcQ_AUoAA&amp;amp;dpr=1#hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;q=ORD%20biological%20applications&quot; rel=&quot;noreferrer&quot;&gt;ORD biological applications&lt;/a&gt;, and &lt;a href=&quot;https://www.google.com/search?q=ord%20curve&amp;amp;hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;biw=1280&amp;amp;bih=829&amp;amp;source=lnms&amp;amp;sa=X&amp;amp;ei=28mAVMCpA-abigLG3oGoAQ&amp;amp;ved=0CAcQ_AUoAA&amp;amp;dpr=1#hl=en&amp;amp;gl=us&amp;amp;authuser=0&amp;amp;q=magnetic%20circular%20dichroism&quot; rel=&quot;noreferrer&quot;&gt;magnetic circular dichroism&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-12-24T03:34:40.633" LastActivityDate="2015-12-24T03:34:40.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20331" PostTypeId="1" AcceptedAnswerId="20332" CreationDate="2014-12-04T22:24:02.890" Score="3" ViewCount="1932" Body="&lt;p&gt;I've been trying to determine how a particular waterproofing material, which claims to be &quot;water based&quot;, can still be hydrophobic. First of all, the product in question is &lt;a href=&quot;http://www.cadeco.com/productfiles/er.pdf&quot; rel=&quot;nofollow&quot;&gt;EPRO ECOLINE-R&lt;/a&gt;, which also claims to be an asphalt emulsion. My first guess is that the curing process causes a change in some way to become hydrophobic once applied. But I just can't wrap my head around how a product, which is designed to resist water absorption, can also be water based? I can find no information on the exact chemical makeup, other than &quot;petroleum hydrocarbons&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10163" LastEditorUserId="4356" LastEditDate="2014-12-04T22:43:41.920" LastActivityDate="2014-12-04T22:49:41.427" Title="How can a waterproofing material be &quot;water-based&quot;?" Tags="&lt;water&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20332" PostTypeId="2" ParentId="20331" CreationDate="2014-12-04T22:38:50.300" Score="3" Body="&lt;p&gt;One might need to have more information about the product you've posted to say what occurs in the specific example. But, the general concept could be similar to Latex (water-based) paint. Where the water proofing chemical is dispersed by an emulsion, and once the water evaporates, the proofing material is restructured by its, now unhindered, intermolecular forces. These forces might be strong enough, once the proper structure exists, to resist water coming back and trying to interrupt the interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Paint#Binder_.28or_film_former.29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; explanation for my example of latex paint.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term &quot;latex&quot; in the context of paint in the USA simply means an aqueous dispersion; latex rubber from the rubber tree is not an ingredient. These dispersions are prepared by emulsion polymerization. Such paints cure by a process called coalescence where first the water, and then the trace, or coalescing, solvent, evaporate and draw together and soften the binder particles and fuse them together into irreversibly bound networked structures, so that the paint cannot redissolve in the solvent/water that originally carried it. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-12-04T22:49:41.427" LastActivityDate="2014-12-04T22:49:41.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20333" PostTypeId="1" AcceptedAnswerId="20404" CreationDate="2014-12-05T00:55:02.787" Score="7" ViewCount="46068" Body="&lt;p&gt;Doing an exchange term so I've missed some classes that the others have attended, and unfortunately we're getting contradictionary explanations, so I thought I'd try and ask you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be extremely thankful if someone could explain the effect of conjugation of a carbonyl group on the resonance frequency in IR spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our 'assignment' is basically to figure out if it's supposed to be lower or higher than an unconjugated carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="10166" LastEditorUserId="4356" LastEditDate="2014-12-05T01:44:20.077" LastActivityDate="2020-06-23T14:07:45.880" Title="What is the effect of conjugation of a carbonyl group in IR spectroscopy?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20334" PostTypeId="1" AcceptedAnswerId="20335" CreationDate="2014-12-05T01:00:44.773" Score="2" ViewCount="215" Body="&lt;p&gt;When I went to the supermarket, I noticed that some of the product say sunscreen and some of them say sunblock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question split into two part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What are the different chemical substances use in both item? (In general)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)What is more effective to prevert sunburn? (In general) &lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2014-12-06T00:46:04.867" Title="Sunscreen or sunblock" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20335" PostTypeId="2" ParentId="20334" CreationDate="2014-12-05T01:24:26.590" Score="3" Body="&lt;p&gt;Giving advice on preventing sunburns may fall under health advice, which is not permitted on this site. For advice, you should ask a medical professional.&lt;/p&gt;&#xA;&lt;p&gt;The following information is from &lt;a href=&quot;http://www.melanomafoundation.org/prevention/facts.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Melanoma Foundation&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;What is the difference between sunscreen and sunblock?&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sunscreens can be classified into two major types: chemical and physical. &lt;strong&gt;Chemical sunscreens&lt;/strong&gt; contain special ingredients that &lt;strong&gt;act as filters and reduce&lt;/strong&gt; ultraviolet radiation penetration to the skin. These sunscreens often are colorless and maintain a thin visible film on the skin. These &lt;strong&gt;sunscreens usually contain UVB absorbing chemicals and more recently contain UVA absorbers as well&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Physical Sunscreens, most often referred to as sunblocks&lt;/strong&gt;, are products containing ingredients such a titanium dioxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{TiO2}$&lt;/span&gt;) and zinc oxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{ZnO}$&lt;/span&gt;) &lt;strong&gt;which physically block&lt;/strong&gt; ultraviolet radiation (UVR). Sunblocks provide broad protection against both UVB and UVA light. They can be cosmetically unacceptable to many people, because they are often messy, visible and do not easily wash off. However, some new zinc oxide products are available in brightly colored preparations which are popular with young people. The amount of sun protection these sunblocks provide, while potentially high, cannot be quantified in the same manner as sunscreen SPFs. Physical sunscreen is recommended for individuals who have unusual sensitivity to UVR. Most recently on the sun protection scene is sun-protective clothing designed to block UVA and UVB radiation. The effective SPF is greater that 30.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;When should you use a sunscreen?&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sunscreens should be used daily if you are going to be in the sun for more than 20 minutes.&lt;/strong&gt; Most people will receive this amount of sun exposure while performing routine activities. They can be applied under makeup. There are many cosmetic products available today that contain sunscreens for daily use because sun protection is the principal means of preventing premature aging and skin cancer. Sunscreen used on a regular basis actually allows some repair of damaged skin. Because the sun's reflective powers are great - 17 percent on sand and 80 percent on snow - don't reserve the use of these products for only sunny summer days. Even on a cloudy day 80 percent of the sun's ultraviolet rays pass through the clouds. Skiers beware, ultraviolet radiation increases 4 percent for every 1,000-foot increase in altitude.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To add to this last part, sunblock manufacturers usually include instructions with their products on the proper way to use them.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2014-12-05T01:40:30.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20336" PostTypeId="1" CreationDate="2014-12-05T02:26:10.517" Score="4" ViewCount="983" Body="&lt;p&gt;I observe some elements like $\ce{Cr}$ and $\ce{Cu}$. Some of them have strange oxidation numbers that don't really fit the purpose of making the element as stable as possible. Just Chromium, for example, it has one valence electron in 4s, the oxidation tells that it is gonna lose 2 electrons. Which obviously breaks the stability of half filled 3p orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this can be explained by the fact that $\ce{Cr}$ has weak electronegativity, then what about $\ce{Cu}$? It is among the most electronegative transition metals, but still might lose two electrons despite having fully filled 4d orbitals.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="5456" LastEditDate="2014-12-05T11:10:26.220" LastActivityDate="2014-12-05T11:10:26.220" Title="Some abnormalities in the Periodic Table" Tags="&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20337" PostTypeId="1" CreationDate="2014-12-05T03:20:29.573" Score="5" ViewCount="235" Body="&lt;p&gt;It's possible to identify a solution with several compounds* without the coupled liquid chromatography (LC) before to isolate the compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the university where I study, the LC is broken, so all we can do is inject the solution directly on the mass spectrometer. I wonder if I will be able to identify the metabolites? I don't need quantitative analysis, I just want to know what's in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) To be precise: A bacterial culture, that has gone under organic solid phase extraction, freeze dried and resuspended in methanol.&lt;/p&gt;&#xA;" OwnerUserId="10167" LastEditorUserId="4945" LastEditDate="2014-12-05T04:14:05.417" LastActivityDate="2014-12-23T18:10:20.567" Title="Is is possible to do mass spectrometry without coupled liquid chromatography?" Tags="&lt;organic-chemistry&gt;&lt;mass-spectrometry&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20338" PostTypeId="1" AcceptedAnswerId="20358" CreationDate="2014-12-05T03:35:44.193" Score="26" ViewCount="728" Body="&lt;p&gt;It's pretty hard to believe in around 4.5 billion years of this planet's existence, no snowflake pattern was repeated, and that's not including potential earth-like planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a proof to the structure or is the probability even more statistically insignificant than I thought, or what?&lt;/p&gt;&#xA;" OwnerUserId="7641" LastEditorUserId="12860" LastEditDate="2015-10-08T20:49:39.800" LastActivityDate="2015-10-08T20:49:39.800" Title="Why is it that every snowflake is unique?" Tags="&lt;everyday-chemistry&gt;&lt;crystal-structure&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20339" PostTypeId="5" CreationDate="2014-12-05T04:42:49.483" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;nofollow&quot;&gt;X-ray diffraction&lt;/a&gt; is a method to obtain the &lt;a href=&quot;/questions/tagged/crystal-structure&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'crystal-structure'&quot; rel=&quot;tag&quot;&gt;crystal-structure&lt;/a&gt; of a compound. It is generally applied to a single crystal to obtain the positions of the nuclei. It is a reliable tool to generate 3D structures of compounds, hence it is possible to measure interatomic distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the close relationship, this tag is also appropriate for x-ray &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;nofollow&quot;&gt;powder diffraction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-12-05T04:42:49.483" LastActivityDate="2014-12-05T04:42:49.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20340" PostTypeId="4" CreationDate="2014-12-05T04:42:49.483" Score="0" Body="X-ray diffraction is a method to obtain the crystal structure of a compound. This tag is also appropriate for x-ray powder diffraction." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2014-12-05T04:42:49.483" LastActivityDate="2014-12-05T04:42:49.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20342" PostTypeId="2" ParentId="20179" CreationDate="2014-12-05T05:15:43.637" Score="3" Body="&lt;p&gt;It is possible to reconstruct molecular structure from powder X-ray data. It is called the Rietveld method:&#xA;&lt;a href=&quot;http://home.wxs.nl/~rietv025/&quot; rel=&quot;nofollow&quot;&gt;http://home.wxs.nl/~rietv025/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.aps.anl.gov/Xray_Science_Division/Powder_Diffraction_Crystallography/&quot; rel=&quot;nofollow&quot;&gt;http://www.aps.anl.gov/Xray_Science_Division/Powder_Diffraction_Crystallography/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it requires high quality, high resolution data (meaning generally synchrothrone powder data), and it is an involved process which requires some experience. Most probably you don't have that quality data at hand, you don't have time to learn and figure out yourself try-and-error and most probably you don't need the result either. If you need some general insight how the molecule look like, I would join to Abel Friedman and suggest using simple calculation techniques. Molecular mechanics and other low-cost methods can give very good structure with minimal effort. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2014-12-05T05:15:43.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20344" PostTypeId="1" AcceptedAnswerId="21892" CreationDate="2014-12-05T07:31:50.787" Score="8" ViewCount="3203" Body="&lt;p&gt;I have heard a few claims about water-fueled cars. Being a student of Engineering I totally understand that water-fueled cars (using electrolysis to generate Hydrogen) are nothing more than a hoax, unless  you are giving electricity from an outer source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have heard quite knowledgeable people that water &lt;strong&gt;hybrid&lt;/strong&gt; cars are quite a reality and you can save up to 60% on your fuel costs using some kind of kits. I have difficulty agreeing with this claim as well. I think inclusion of another step (hydrolysis) in the energy cycle of car power system will only make it less efficient. Reason being you can achieve at most as much energy from making a bond (burning Hydrogen) as much you spent on breaking it (Electrolysis). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me understand if I am missing some point? Or I am correct?&lt;/p&gt;&#xA;" OwnerUserId="10169" LastEditorUserId="10169" LastEditDate="2014-12-05T07:37:44.710" LastActivityDate="2015-05-08T17:15:15.790" Title="Do water hybrid cars really exist?" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20345" PostTypeId="2" ParentId="20238" CreationDate="2014-12-05T12:31:21.973" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Generally, when a substance is at &gt; 0 K (0 Kelvin or -273°C) the atoms/molecules making up that substance possess kinetic energy, which means that each atom/molecule are [almost] constantly moving.  If that atom/molecule is in a vacuum, it would move in [almost] a straight line.  However, in the ordinary matter that we encounter everyday, atoms/molecules exist in macroscopic phases which are composed of a humongous number of particles.  Which means that each particle experiences frequent collisions with neighbouring particles, making the trajectory of each particle to appear zig-zag like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The higher the temperature in the vicinity of an atom/molecule, the higher the kinetic energy of the particle, and the faster that particle's motion becomes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, depending on the properties of the atoms/molecules in the substance, each atom/molecule experiences electrostatic forces (either attractive or repulsive) exerted by neighboring atoms/molecules.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At the interface between a liquid and a gas (e.g., water in a tumbler sitting on a table in a room at sea level), the surface of the liquid is constantly being bombarded by gaseous particles that have much higher kinetic energies than the molecules of the water in the tumbler.  This bombardment imparts enough kinetic energy to some of the water molecules at the surface.  Some of the bombarded molecules gain enough kinetic energy to enable them to leave the water surface to become gaseous water molecules.  Some of the bombarding particles (oxygen molecules, nitrogen molecules, water vapour molecules, etc.) from the air lose enough kinetic energy so that they become stuck into the liquid surface.  &lt;strong&gt;At equilibrium&lt;/strong&gt; (maximum humidity saturation of the air in the room), the number of water molecules leaving the liquid surface is equal to the number of molecules entering the liquid surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is the net result of factors 2, 3 and 4 that determines the phase of a substance.  For example, substances like water, which are composed mainly of polar molecules (molecules possessing an asymmetrical distribution of charge around it, making it behave like a bipolar/tripolar/multipolar magnet), have relatively high melting and boiling points.  Below the melting point, the molecules exist in a [3D] crystalline lattice.  Each molecule are in constant motion but are at the same time held together by electrostatic forces (hydrogen bonds).  Since in a solid, the kinetic energy or velocity of each molecule is not high/fast enough, those particles cannot overcome/escape from the electrostatic attraction exerted by their immediate neighbours.  Therefore, the molecules stay stuck in the crystal lattice, exhibiting their kinetic energy as vibrational motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the melting point is reached, the kinetic energy of the molecules starts to overcome the electrostatic attraction from neighbouring particles, allowing them to move away from their neighbours.  This is the liquid state, in which the kinetic energy of the particles are enough to prevent the particles from being stuck in one place, allowing them to move about, making the phase fluid; however, the kinetic energies are not enough to allow the particles to move farther away from each other.  In the liquid state, the distances between neighbouring particles are almost the same as that in a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a liquid to &lt;strong&gt;become a gas&lt;/strong&gt;, two factors must be overcome: (a) the intermoleculer forces in the liquid, and (b) the pressure above the liquid surface.  At the boiling point, the kinetic energies of the particles become high enough to overcome (a) and (b) to enable the particles to move farther away from each other, and we get a gaseous phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factor (a) is overcome, for example, by heating the liquid in the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can appreciate the effect of factor (b) when you go mountain-climbing (5000 feet or higher).  Take a thermometer with you to the mountain top.  You will observe that water at high altitudes boil at temperatures lower than 212°F (100°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to experience the effect of factor (b) is through the following experiment:  Get a 50-mL (1 to 2-oz) syringe, remove the needle, draw water to about 25% capacity of the syringe barrel, plug the inlet with your index finger, then with your other hand, pull the plunger until it is almost at its maximum position.  You should see the water in the syringe boil without the application of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be a simple question but a full answer can be appreciated only with some knowledge of statistical thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the pressure of a gas is the macroscopic result of a humongous number of particles impinging upon a surface (the diaphragm of a barometer, the surface of a manometer liquid, the walls of a container, etc.).&lt;/p&gt;&#xA;" OwnerUserId="10174" LastActivityDate="2014-12-05T12:31:21.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20346" PostTypeId="2" ParentId="20283" CreationDate="2014-12-05T13:20:50.730" Score="21" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8q5m5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &quot;&lt;a href=&quot;http://dx.doi.org/10.1002%2Fanie.200460005&quot; rel=&quot;noreferrer&quot;&gt;The Strongest Isolable Acid&lt;/a&gt;&quot; Angewandte Chemie International Edition 43 (40): 5352–5355, carborane acids are the strongest, with $\ce {H(CHB11Cl11)}$ being the strongest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending upon ones preferred definition of &quot;organic&quot; this might not be considered organic, but the derivative $\ce{H(CHB11Me5Br6)}$ is strong enough to fully protonate benzene.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H(CHB11Me11)}$ is stronger than $\ce{H(CHB11Me5Br6)}$ and only slightly weaker than $\ce {H(CHB11Cl11)}$, so if $\ce {H(CHB11Cl11)}$ is disregarded as no being organic, then $\ce{H(CHB11Me11)}$ is strongest.  If $\ce{H(CHB11Me11)}$ still isn't organic enough, then protonated benzene or a protonated alkane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.200701136/abstract&quot; rel=&quot;noreferrer&quot;&gt;later article&lt;/a&gt; and the article in Nicolau's comment report $\ce {H(CHB11F11)}$ and $\ce {H(CEtB11F11)}$ which may be even stronger than the above) &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-05T14:14:10.520" LastActivityDate="2014-12-05T14:14:10.520" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20347" PostTypeId="1" AcceptedAnswerId="20348" CreationDate="2014-12-05T16:08:08.180" Score="2" ViewCount="1958" Body="&lt;p&gt;Which route actually occurs? &lt;strong&gt;Is it really true that the carboxylic acid never forms under basic conditions?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not just form the -COOH and deprotonate it afterward?? Or is there a higher activation every barrier to the second route? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that carboxylic &lt;em&gt;acids&lt;/em&gt; don't stick around long in basic solutions but that doesn't necessarily mean they don't exist in basic solutions &lt;em&gt;at all&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XlFfR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-05T16:37:57.803" LastActivityDate="2014-12-12T08:11:46.107" Title="Hydrolysis under basic conditions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20348" PostTypeId="2" ParentId="20347" CreationDate="2014-12-05T16:23:15.857" Score="4" Body="&lt;p&gt;I find the top mechanism unlikely for two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the top mechanism, the proton removed in the first step should be way less acidic than a typical hydroxyl since the intermediate is already negatively charged. Going to a di-anion should be very far uphill, and therefore, slow. If that di-anion was ever formed, the next step would be exceedingly fast&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the bottom mechanism, the first step should be fast. It's intramolecular and entropically favored. The second step is instantaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't imagine any text book includes the top mechanism. Check your book for saponification of esters.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-05T16:23:15.857" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="20349" PostTypeId="1" CreationDate="2014-12-05T16:46:09.657" Score="3" ViewCount="43" Body="&lt;p&gt;I have a question about whether a particular situation qualifies as a rate limiting step. I have a catalyst that under normal conditions gets poisoned when products accumulate on the catalyst surface. I interpret that as the product desorption as being a rate limiting step, but is that true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought that that a rate limiting step means that the rate of the reaction equals the rate of that step, but in reality what I see is that the initial rate in the case where the product sticks and when you can desorb the product is the same, but just that you get an abrupt halt in reactivity once the catalyst is full of product. Conversely, by tweaking the conditions where i can cause the product to desorb it just will keep that initial rate up for 10,000 times longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that still mean that product desorption is a RLS or is there a better way to describe this poisoning event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7614" LastActivityDate="2014-12-05T16:46:09.657" Title="Does product induced poisoning qualify as a rate limiting step?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20350" PostTypeId="2" ParentId="17106" CreationDate="2014-12-05T17:06:14.147" Score="8" Body="&lt;p&gt;Graphene in its pristine form, a single sheet of pure sp2 hybridized benzenes, is (relatively speaking) one of the least chemically reactive materials known because of several reasons: one, its enormous size (we keep making larger and larger single &quot;molecules&quot;, now thanks to &lt;a href=&quot;http://ibmresearchnews.blogspot.com/2013/11/exfoliating-wafer-scale-graphene-down.html#fbid=PC2h-KrQJ8v&quot; rel=&quot;nofollow noreferrer&quot;&gt;IBM up to 4 inches&lt;/a&gt;!). Secondly, its physical hindrance: you can only do bottom/top down chemistry, granted this isn't that big of an issue but it does greatly reduce using large compounds. Last is it's stableness: this thing is crazy stable because of its gigantic p-butts and other effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long story short is that you don't have any functional groups or easy points of entry. The way to think about it is &quot;what does graphene look like and what makes those other molecules reactive?&quot;. The answer is...benzene and really any other hydrocarbon with no functional groups. The difference there is their size. But the same rules apply! You can still oxidize graphene, at the edges for example, much like you would benzene. And in the past years, we've seen even more radical methods, the most common being Hummer's Oxidizing, and some other clever ones like a &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ange.201101371/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claisen Rearrangement directly onto the surface of graphene&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: after Greg's comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't apologize! This'll get the discussion going, I was hoping to see more answers other than mine. I think you're right in terms of absolute stability but my statement is a phenomenon harder to describe than mere size. From experience, we noted that on very large graphene flakes, our functionalization of them would distort the sheets, but not break them, creating lots of hindrances and thereby decreasing its reactivity; it basically went from sheet shape to a jumbled up ball. However, on smaller (much smaller) sized flakes, the individual molecules (though hindered and flexed in a similar way) would, at the same concentrations of reactants, get functionalized more, ratio of functional groups to carbons on the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an indirect effect of the massive size of the graphene sheets. So I guess you're right in that size isn't exactly a stability factor but rather, again, it's all about hinderance. However, I wanted to point it out because the massive size of such a flat molecule is quite spectacular and does yield some interesting behavior we don't normally think of. &lt;/p&gt;&#xA;" OwnerUserId="10175" LastEditorUserId="402" LastEditDate="2017-04-01T00:54:39.570" LastActivityDate="2017-04-01T00:54:39.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20351" PostTypeId="1" CreationDate="2014-12-05T17:16:10.177" Score="5" ViewCount="57019" Body="&lt;p&gt;How to prepare $\mathrm{0.1M}$ $\ce{I_2}$ solution.&lt;br&gt;&#xA;I have found in books that it should be prepared in $\ce{KI}$ solution but none of them explains why? Please explain in detail not just the procedure.&lt;/p&gt;&#xA;" OwnerUserId="5004" LastEditorUserId="4231" LastEditDate="2014-12-05T19:13:56.813" LastActivityDate="2016-03-03T09:19:24.970" Title="Preparation of iodine solution" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20352" PostTypeId="2" ParentId="20351" CreationDate="2014-12-05T18:54:37.233" Score="13" Body="&lt;p&gt;Since $\ce{ I_2}$ is a non-polar covalent molecule, it does not ionize in water. It  cannot also be soluble in the polar water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KI}$, which is a polar, ionic compound, will ionize and dissolve in water. When KI dissolves in water, it ionizes to $\ce{K^+}$ and $\ce{I^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{I^-}$ will react with $\ce{ I_2}$ to form the complex ion $\ce{I_3^-}$. $\ce{I_3^-}$ being negatively charged will dissolve in water.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-05T18:54:37.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20353" PostTypeId="2" ParentId="20308" CreationDate="2014-12-05T19:08:16.420" Score="0" Body="&lt;p&gt;you can try $\ce{Cu^{2+}--Cys}$ vs $\ce{Cu^+--Cys}$, much smaller and it is a model of $\ce{Cu-S}$ clusters in eg $\ce{Cu}$-binding proteins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in these systems too, maybe we can share information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goedele&lt;/p&gt;&#xA;" OwnerUserId="10177" LastEditorUserId="4231" LastEditDate="2014-12-05T19:23:53.860" LastActivityDate="2014-12-05T19:23:53.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20355" PostTypeId="2" ParentId="20308" CreationDate="2014-12-05T22:30:19.433" Score="3" Body="&lt;p&gt;AFAIK, permanganate ion is colored because of intense line, corresponding to oxygene-manganese charge transfer. This is noted in wikipedia article on charge-transer complex, but, unfortunately, without quote, and I'm unable to come with one right now&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Aha. &lt;a href=&quot;http://doi.org/10.1021/jp0456990&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; there is a MO diagram on $\ce{MnO4-}$ ion and extensive discussion of (mis)understanding of its spectra, including statement I made above.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-12-05T22:39:29.287" LastActivityDate="2014-12-05T22:39:29.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20357" PostTypeId="1" AcceptedAnswerId="20360" CreationDate="2014-12-06T00:10:30.547" Score="4" ViewCount="3035" Body="&lt;p&gt;Recently, I saw a video about sodium hydride and it saw it was made up with $\ce{Na+}$ ions and $\ce{H-}$ ions. My  problem is: aren't all hydrogen ions positive? And in what conditions can $\ce{H-}$ occur?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="275" LastEditDate="2014-12-06T01:46:33.883" LastActivityDate="2015-06-27T15:22:44.817" Title="Why is the hydride ion negative?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20358" PostTypeId="2" ParentId="20338" CreationDate="2014-12-06T00:24:34.050" Score="22" Body="&lt;p&gt;&lt;strong&gt;It turns out that 2 identical snowflakes have been observed, but...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two Identical Snowflakes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although when we think &quot;snowflake&quot; we usually picture an object with 6-fold radial symmetry, snowflakes actually come in many different shapes (&lt;a href=&quot;http://www.nasa.gov/pdf/183517main_snowcrystals.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference 1&lt;/a&gt;).  Many sites report that in 1988 an NCAR researcher found 2 identical snowflakes of the hollow columnar variety (&lt;a href=&quot;http://en.wikipedia.org/wiki/Snowflake&quot; rel=&quot;noreferrer&quot;&gt;reference 2&lt;/a&gt; - see &quot;Uniqueness&quot; section, &lt;a href=&quot;http://www.proquestk12.com/curr/snow/snow395/snow395.htm#snow&quot; rel=&quot;noreferrer&quot;&gt;reference 3&lt;/a&gt;).  This pattern is much simpler than the 6-fold radial design, so the odds of finding 2 identical flakes are vastly improved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to confine the question to snowflakes of the 6-fold radial variety, then an argument can be made that 2 identical snowflakes will not be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G9KgJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXESh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the vapor pressure difference between ice and water, vapor is pushed from water towards the ice (hence, the ice grows) in an ice-water droplet falling through the atmosphere.  As it falls through different temperature zones, the rate of ice crystal growth will change leading to different dendritic patterns along the radial arms.  Further, due to perturbations caused by temperature and wind gradients across a single crystal, each radial arm will have a slightly different pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical argument against identical (6-fold radial) snowflakes is, as Jon Custer suggested in his comment, based on factorial probabilities.  The argument goes something like this.  If you have 10 different pictures to put in an album you could arrange them in 10! (10 factorial) or 3,628,800 different ways.  If you look at a snowflake under magnification, there are easily over a hundred different features that are identifiable.  In this case there are over 100! or greater than $10^{158}$ possible arrangements of these identifiable features.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Observable_universe#Matter_content_.E2.80.94_number_of_atoms&quot; rel=&quot;noreferrer&quot;&gt;Estimates&lt;/a&gt; are that there are around $10^{80}$ atoms in the universe.  Each visible feature in a snowflake contains a large number of water molecules and an even larger number of atoms.  Hence, the number of possible arrangements is far, far greater than the number of available atoms, making it extremely improbable that a duplicate of a complex snowflake would be created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're still not convinced, we can add another level of probabilistic protection.  Remember that there are isotopes of hydrogen and oxygen.  The probability of finding two 6-fold radial snowflakes with the same dendritic pattern &lt;strong&gt;and&lt;/strong&gt; the same isotopic substitution pattern is even closer to zero.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-08T02:44:56.653" LastActivityDate="2014-12-08T02:44:56.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20359" PostTypeId="2" ParentId="20334" CreationDate="2014-12-06T00:46:04.867" Score="3" Body="&lt;p&gt;One thing to add, the U.S. Food and Drug Administration (FDA) recently changed the labeling requirements for sunscreens.  As part of it the word &quot;sunblock&quot; will no longer be allowed.  You'll still find some sunscreen's using the term &quot;sunblock&quot; and this is usually because they were manufactured before the labeling changes went into effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going forward everything will be labeled &quot;sunscreen&quot; whether it is a mineral or chemical sunscreen.  So to tell in the future you'll need to look at the &quot;Active Ingredients&quot; in a sunscreen (the FDA requires sunscreens to list all their ingredients).  If you see only zinc and/or titanium there then it's a mineral sunscreen.  Any other ingredient listed under &quot;Active Ingredients&quot; is a chemical UV filter as currently 17 ingredients (2 minerals and 15 chemicals) are approved by the FDA as sunscreen filters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something you definitely want to look for on a sunscreen's label is the term &quot;Broad Spectrum&quot;.  According to the new FDA regulations only sunscreen's that protect against both UVA and UVB rays can be labeled as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested, here's a bit more about the new sunscreen labeling laws from the Skin Cancer Foundation: &quot;&lt;a href=&quot;http://www.skincancer.org/prevention/sun-protection/sunscreen/labels&quot; rel=&quot;nofollow&quot;&gt;New Sunscreen Labeling Rules Issued by FDA&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other interesting thing is that some UV filtering ingredients protect against the full UVA and UVB spectrum while others protect against only parts.  This is why some sunscreens use multiple active ingredients. If you're interested, we have bit more about that here &quot;&lt;a href=&quot;http://www.blockislandorganics.com/Blog/post/2013/11/06/UVA-UVB-and-UVC-Rays-What-They-Are-and-How-Sunscreen-Protects-You.aspx&quot; rel=&quot;nofollow&quot;&gt;UVA, UVB and UVC Rays: What They Are and How Sunscreen Protects You&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10179" LastActivityDate="2014-12-06T00:46:04.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20360" PostTypeId="2" ParentId="20357" CreationDate="2014-12-06T01:52:12.547" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And in what conditions can $\ce{H^{−}}$ occur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any phase. The hydride ion however doesn't stick around for long in water since it's &lt;em&gt;such&lt;/em&gt; a strong base. First, hydrogen isn't that electronegative to start with, so tacking on an extra electron isn't helping with its stability. Second, combination with an hydrogen proton, $\ce{H+}$, is highly entropically favorable because the product, $\ce{H2}$, is a gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $pK_a$ of $\ce{H2}$ has been estimated to be ~35 so that speaks a lot to the strength of the hydride ion as a base. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorDisplayName="user15489" LastEditDate="2015-06-27T15:22:44.817" LastActivityDate="2015-06-27T15:22:44.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20361" PostTypeId="1" CreationDate="2014-12-06T02:40:18.120" Score="2" ViewCount="1132" Body="&lt;p&gt;How would you determine the size of a polyatomic ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when determining which ion is bigger, $\ce{NH2-}$ or $\ce{I-}$, what would you use to determine the size difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\ce{I-}$ would be much smaller than $\ce{I2-}$ (since the $\ce{e-}$ are not held as tightly in the $\ce{I2-}$ causing the repulsion to increase)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what basis do we use to determine size of a nonelemental atom like the polyatomic ion or a more complex molecule for that matter?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T15:47:24.087" LastActivityDate="2015-05-02T15:47:24.087" Title="How to determine relative size of polyatomic ions?" Tags="&lt;ions&gt;&lt;atomic-radius&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20362" PostTypeId="1" AcceptedAnswerId="20369" CreationDate="2014-12-06T07:50:19.863" Score="13" ViewCount="17550" Body="&lt;p&gt;Hyperconjugation stabilizes carbocations and that makes sense because electrons are given to the empty p orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does it stabilize alkenes? Can molecular orbital theory be used to explain it?&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="402" LastEditDate="2017-05-02T23:19:39.533" LastActivityDate="2017-09-06T03:41:24.550" Title="The effect of hyperconjugation on the stability of alkenes with MO theory" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;orbitals&gt;&lt;stability&gt;&lt;hyperconjugation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20363" PostTypeId="1" CreationDate="2014-12-06T08:36:56.393" Score="2" ViewCount="1048" Body="&lt;p&gt;I want to run a reaction that needs diethyl ether as anti solvent in the last step (comes from articles).&#xA;But I can't use it due to its harmful effects on the body. &#xA;Is there any other solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reaction is about substitution of $\ce{-OH}$ of lignin with &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.80021.html&quot; rel=&quot;nofollow&quot;&gt;BiBB&lt;/a&gt; in THF and in other one in $\ce{EtOAc}$ as solvent. &lt;/p&gt;&#xA;" OwnerUserId="10181" LastEditorUserId="16683" LastEditDate="2017-04-03T15:21:44.467" LastActivityDate="2017-04-03T15:21:44.467" Title="Substitute solvent for diethyl ether?" Tags="&lt;safety&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20364" PostTypeId="1" CreationDate="2014-12-06T08:50:35.510" Score="0" ViewCount="147" Body="&lt;p&gt;Why doesn’t the energy of a body doubles when temperature in degrees Celsius is doubled?&#xA;I know that it doubles in case of kelvin scale.&lt;/p&gt;&#xA;" OwnerUserId="10182" LastEditorDisplayName="user7951" LastEditDate="2015-07-01T13:40:43.270" LastActivityDate="2015-07-01T13:40:43.270" Title="Effect on kinetic energy of a body with change in temperature" Tags="&lt;energy&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20365" PostTypeId="2" ParentId="20364" CreationDate="2014-12-06T09:28:29.197" Score="4" Body="&lt;p&gt;You might find it interesting to read up on the differences between the different temperature scales. The kelvin scale effectively measures the vibration (average kinetic energy of molecules) so it makes sense that a doubling in the kelvin means a doubling in kinetic energy. On the other hand the Celsius scale is defined by the melting and boiling points of water (quite arbitrary).&lt;/p&gt;&#xA;" OwnerUserId="7813" LastActivityDate="2014-12-06T09:28:29.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20366" PostTypeId="1" AcceptedAnswerId="20371" CreationDate="2014-12-06T11:31:26.177" Score="2" ViewCount="2657" Body="&lt;p&gt;$\ce{Cu}^{2+}$ has nine d-electrons, regardless of the ligand field strength it will have one free electron (so it is paramagnetic. Are $\ce{Cu}^{2+}$ octahedral complexes high spin or low spin? &lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7412" LastEditDate="2014-12-06T14:10:44.633" LastActivityDate="2014-12-06T14:10:44.633" Title="Cu2+ octahedral complexes" Tags="&lt;coordination-compounds&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20368" PostTypeId="1" AcceptedAnswerId="20378" CreationDate="2014-12-06T12:04:59.580" Score="2" ViewCount="18108" Body="&lt;p&gt;How is it possible to know whether a particular solvent will dissolve in a particular kind of solvent, i.e. organic, polar, protic...?&#xA;Are there any indicators to look for in molecules or elements that allow one to predict what kind of solvent should be used to dissolve it?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="189" LastEditDate="2014-12-06T12:21:22.440" LastActivityDate="2014-12-06T18:31:17.260" Title="How to predict the solubility of an organic compound in different kinds of solvents?" Tags="&lt;aqueous-solution&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ClosedDate="2014-12-08T18:33:58.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="20369" PostTypeId="2" ParentId="20362" CreationDate="2014-12-06T12:12:34.710" Score="15" Body="&lt;p&gt;Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8juF.png&quot; height=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kcdem.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="402" LastEditDate="2017-05-02T23:19:57.097" LastActivityDate="2017-05-02T23:19:57.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20370" PostTypeId="1" CreationDate="2014-12-06T12:50:58.747" Score="9" ViewCount="3075" Body="&lt;p&gt;When limestone is buried and put under intense heat and pressure, why does it turn to marble instead of thermally decomposing into calcium oxide and carbon dioxide?&lt;/p&gt;&#xA;" OwnerUserId="10184" LastEditorUserId="8083" LastEditDate="2014-12-06T19:38:36.300" LastActivityDate="2014-12-06T19:38:36.300" Title="Why doesn't limestone thermally decompose instead of turning to marble?" Tags="&lt;physical-chemistry&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20371" PostTypeId="2" ParentId="20366" CreationDate="2014-12-06T14:10:02.667" Score="4" Body="&lt;p&gt;There are not two possibilities for low spin and high spin complex for a d9 metal ion.&#xA;The only cases where you can have both is for d4, d5, d6 and d7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy. &#xA;You need extra energy for doing two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$&lt;/li&gt;&#xA;&lt;li&gt;Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the complexes where $\Delta_O$ is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite. &#xA;But this only applies for d4, d5, d7 and d8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about d1 d2 and d3:&#xA;It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each $T_2g$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now d8 d9 and d10.&#xA;There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals.&#xA;You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital?&#xA;This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin.&#xA;The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2014-12-06T14:10:02.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20372" PostTypeId="1" AcceptedAnswerId="20374" CreationDate="2014-12-06T15:33:43.817" Score="2" ViewCount="25918" Body="&lt;p&gt;I was studying the types of stereo-isomerism E-Z, Cis-Trans and then I come across &lt;strong&gt;syn-anti stereo-isomerism&lt;/strong&gt; this one is giving me a bit of fight in understanding, especially &lt;strong&gt;where lone pairs are taken into consideration&lt;/strong&gt;.How do classify the molecules with lone pairs in consideration as syn and anti Please help me out here!&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2020-12-11T02:21:16.177" Title="Stereo Isomerism - syn/anti" Tags="&lt;molecules&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20374" PostTypeId="2" ParentId="20372" CreationDate="2014-12-06T17:12:16.833" Score="5" Body="&lt;p&gt;Use of the terms &quot;syn&quot; and &quot;anti&quot; to describe the geometry about double bonds is no longer encouraged, but you'll still come across it, particularly in older articles.  Syn and anti are identical to Z(usammen) and E(ntgegen) and were often used to describe the geometry about carbon-nitrogen double bonds. In such cases, the lone pair of electrons is given the lowest priority and the sequence rule applied as usual.  For example, in the following N-methyl imines, the imine on the left has the highest priority group attached to the carbon in the double bond (ethyl) on the same side as the highest priority group attached to the nitrogen in the double bond  (the methyl has higher priority than the lone pair). Therefore the molecule is the syn isomer (but would now be called the Z isomer).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSpte.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same reasoning, the molecule on the right is the anti of E isomer.  Again, the key point is that the lone pair has a lower priority than any atom.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-06T17:12:16.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20375" PostTypeId="1" CreationDate="2014-12-06T17:33:10.230" Score="5" ViewCount="27225" Body="&lt;p&gt;Recently, while reading about isomers I learned that cis isomers are more stable than trans isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I referred many books as well as searched around the internet but I didn't find a suitable explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="7190" LastEditorUserId="402" LastEditDate="2015-11-30T03:46:56.313" LastActivityDate="2016-12-28T05:54:07.667" Title="Stability of cis vs trans isomers?" Tags="&lt;geometrical-isomerism&gt;&lt;isomers&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20376" PostTypeId="1" AcceptedAnswerId="20379" CreationDate="2014-12-06T17:44:27.927" Score="3" ViewCount="986" Body="&lt;p&gt;Is bicyclo[1.1.0]butane chiral? I think it isn't, but if one of the hydrogen atoms bound to the tertiary or secondary carbon atoms is replaced with e.g. chlorine, the formed compound will be chiral, right?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-12-06T18:30:53.927" Title="Is bicyclo[1.1.0]butane chiral?" Tags="&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20377" PostTypeId="1" AcceptedAnswerId="20384" CreationDate="2014-12-06T17:47:59.990" Score="7" ViewCount="159" Body="&lt;p&gt;It has been repeatedly emphasized that proton transfers &quot;involving&quot; three or four membered ring transition states are &lt;em&gt;not&lt;/em&gt; favorable and should not be drawn as the primary mechanism for any reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with regard to a recent discussion involving $\ce{H3CSH}$ it was noted that the rate of proton tunneling in the anionic form $\ce{H3CS-}$ was likely extremely high. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/18724/carbon-sulfur-bond-lengths-resonance-effects-or-lack-thereof&quot;&gt;Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Is this a valid rule of thumb: avoid intramolecular proton transfers unless the transition state has 5 or 6 members in its ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What's more significant with regard to proton transfer - proton tunneling or physical proton transfer? Or is the answer an &quot;it depends&quot;? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-12-06T18:56:07.237" Title="Significance of Proton Tunneling" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20378" PostTypeId="2" ParentId="20368" CreationDate="2014-12-06T18:30:16.593" Score="3" Body="&lt;p&gt;There are some general guesses one can make from looking at the structure but the &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jps.21922/abstract&quot; rel=&quot;nofollow&quot;&gt;Abraham solvation equation&lt;/a&gt; is commonly used to estimate the solubility of a compound in a given organic solvent:&#xA;$$\log P_s = c + e E + s S + a A + b B + v V$$&#xA;This equation relies on a set of descriptors to characterize the properties of solvents and solutes to give the water/solvent partition coefficient $P_s$ (the ratio of the solubilities in water and solvent, under certain assumptions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p,e,s,a,b,v$ are coefficients that describe the solvent. Abraham and coworkers have collected coefficients for 85 different solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other variables are descriptors of the solute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the excess molar refractivity, which is a measure of the polarizability of a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S$ is the solute dipolarity/polarizability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ is the overall (summation) hydrogen bond acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$ is the overall (summation) hydrogen bond basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ is the McGowan characteristic volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at the paper if you want to know exactly what these descriptors are, but their values can be obtained through various experimental and computational means, and combined with the solubility in water, used to estimate the solubility of a given compound in an organic solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, you don't have to do all this by hand as several websites, like &lt;a href=&quot;http://showme.physics.drexel.edu/onsc/models/solutesolventprediction.php&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, that offer lookup and computation where you only have to specify a compound and pick a solvent from a list.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-06T18:30:16.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20379" PostTypeId="2" ParentId="20376" CreationDate="2014-12-06T18:30:53.927" Score="2" Body="&lt;p&gt;Bicyclo[1.1.0}butane is an achiral molecule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RP0zj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any molecule containing a plane of symmetry must be achiral and bicylco[1.1.0]butane has 2 planes of symmetry.  One plane bisects the molecule and passes through the 2 tertiary carbons, the other plane bisects the molecule and contains the 2 secondary carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even replacing one of the secondary or tertiary hydrogens with a substituent does not make the molecule chiral.  Such a substitution will remove one plane of symmetry, but not the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UsW7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above bicylco[1.1.0]butane where we have substituted a chlorine for one of the tertiary hydrogens, the plane containing the 2 bridgehead carbons and the chlorine still remains.  However, if you replace one of the tertiary hydrogens and one of the secondary hydrogens, the resultant disubstituted-bicylco[1.1.0]butane will be chiral since both symmetry planes have been destroyed and no other plane or axis of symmetry exists.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-06T18:30:53.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20380" PostTypeId="2" ParentId="20368" CreationDate="2014-12-06T18:31:17.260" Score="1" Body="&lt;p&gt;In general, organic compounds tend to dissolve well in solvents which have similar properties to themselves. So polar molecules tend to dissolve well in polar solvents, apolar molecules in apolar solvents. The solubility is affected strongly by the type of intermolecular forces between the two molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example benzene and hexane are highly miscible because they are both apolar and interact via Van der Waals forces. In general molecules with large hydrocarbon chains or rings will be soluble in apolar solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big indicator to look for in molecules is the presence of highly polarised bonds, in particular N-H, F-H and O-H bonds which are able to engage in hydrogen bonding. Protic solvents (those with N-H, F-H and O-H bonds) will form H-bonds with other protic molecules leading to them tending to be very soluble. H-bonding can also occur between protic hydrogens and other highly negatively charged atoms such as the oxygen atom in a carbonyl group and so this leads to protic solvents generally being good at dissolving any significantly polar molecule. For example propanone is soluble in water due to the hydrogen bonding between the water hydrogens and the carbonyl oxygen in propanone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar aprotic solvent such as dimethylsulfoxide (DMSO) are generally miscible with polar protic and many apolar solvents and are similarly able to dissolve a wide range of molecules. After consulting several data sources it seems that DMSO and other similar polar aprotic solvents are miscible with all but the most apolar hydrocarbons and I would assume that they will also dissolve many different organic solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general when looking at molecules to decide which solvent to dissolve them in you need to evaluate what the primary intermolecular interactions will be with that molecule (In particular H-bonding or Van der Waals) and then choose a solvent which has similar properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't talked about entropy at all in this answer because I am not entirely confident that I can explain all of the relevant factors that affect it but for dissolving crystalline solids in particular that is also something that you have to take into account but I will leave that to someone more knowledgeable that me to answer.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-06T18:31:17.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20381" PostTypeId="2" ParentId="20370" CreationDate="2014-12-06T18:34:54.733" Score="9" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not hot enough, and the pressure is too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;intense heat and pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key word is pressure rather than heat. Let's look at the phase diagram of the CaO and CO&lt;sub&gt;2&lt;/sub&gt; compositional system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NKU2E.png&quot; alt=&quot;Calcium oxide and carbon dioxide phase diagram&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;(Calculated and plotted using &lt;a href=&quot;http://www.perplex.ethz.ch/&quot; rel=&quot;noreferrer&quot;&gt;Perple_X&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure limestone and marble are composed of calcium carbonate: CaCO&lt;sub&gt;3&lt;/sub&gt;. In air, the temperature required for calcium carbonate decomposition is 898 °C (considering simple equilibrium and ignoring any &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_carbonate#Calcination_equilibrium&quot; rel=&quot;noreferrer&quot;&gt;vapour pressure effects&lt;/a&gt;). As pressure increases, the temperature requirement for this equilibrium decomposition increases very rapidly. It increases so fast that in fact, at above 80 bar, the calcium carbonate will melt into a liquid (call it calcium carbonate magma or lava) instead of decomposing to lime and carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While 80 bar may seem a lot for us humans (it is higher than the &lt;a href=&quot;http://www.wolframalpha.com/input/?i=80%20bar&quot; rel=&quot;noreferrer&quot;&gt;pressure exerted by a 45 kg woman wearing stiletto heels&lt;/a&gt;), it is nothing compared to the pressure below us, deep in the Earth. Marble forms at a wide pressure range, but for this discussion let's say that it's somewhere between 2 and 10 kbar. That's 2000 to 10000 bar! Way off the scale of the phase diagram. Also, marble forms in the temperature range of 300 to 800 °C. Again, it's not even enough to melt the calcium carbonate, let alone decompose it to gas.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-06T18:34:54.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20382" PostTypeId="1" AcceptedAnswerId="20389" CreationDate="2014-12-06T18:38:38.620" Score="3" ViewCount="167" Body="&lt;p&gt;Washing up liquid is wet and it is added to water. Why does it not come in a half size package and you add the water yourself? In the same way washing powder is?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:12:04.060" LastActivityDate="2015-10-30T17:12:04.060" Title="Why isn't washing up liquid dry" Tags="&lt;everyday-chemistry&gt;&lt;surfactants&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20383" PostTypeId="2" ParentId="20375" CreationDate="2014-12-06T18:54:27.290" Score="6" Body="&lt;p&gt;As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration.&#xA;For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference between the two methyl groups either side of the double bond. However, there are exceptions to this rule which I couldn't find a quick explanation for. The rule applies to both cyclic and acyclic compounds although there are some obvious cyclic exceptions such as cyclic alkenes, where the cis isomer is more stable, especially for smaller rings, due to the large ring strain caused by the orientation of the trans isomer. &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/sterisom.htm#isom1&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; has a good illustration of this:&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-11-29T17:10:45.810" LastActivityDate="2015-11-29T17:10:45.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20384" PostTypeId="2" ParentId="20377" CreationDate="2014-12-06T18:56:07.237" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;a) Is this a valid rule of thumb: avoid intramolecular proton&#xA;  transfers unless the transition state has 5 or 6 members in its ring?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b) What's more significant with regard to proton transfer - proton&#xA;  tunneling or physical proton transfer? Or is the answer an &quot;it&#xA;  depends&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends, at temperatures around absolute zero, tunneling can become significant.  At ambient temperatures or above, physical proton transfer is much more common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the methyl mercaptan anionic example you cite, to the extent that the various carbon-sulfur double bond resonance structures contribute, I would expect a lowering of the barrier around carbon anion planarization - and consequently a lowering of the barrier to inversion at the carbon atom in the anion.  Therefore, it is not clear to me that tunneling of the sulfur bound proton would be preferred to carbon inversion in order to bring about the syn-anti interconversion, especially since any presumed experiments to generate the anion would not be performed near absolute zero.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-06T18:56:07.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20385" PostTypeId="1" AcceptedAnswerId="20499" CreationDate="2014-12-06T19:16:51.000" Score="4" ViewCount="874" Body="&lt;p&gt;The energetics of the mechanism of saponification were previously under debate, but I will admit, I misrepresented a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the revised mechanism of saponification proposed by someone other than myself. It is the one on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previously suggested mechanism is in the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook mechanism is the bottom one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it does seem that if the mechanism were concerted then proton abstraction would be much more plausible from a thermodynamic standpoint. If the second mechanism were akin to scaling Mt. Everest, then the first one looks to me to be more like climbing the rock wall at the gym. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with the same reactant - an ester - but in the second mechanism we are going through a dianion intermediate. In the first mechanism however we are simply going through a concerted transition state en route to the stable carboxylate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, again, I ask - which is (kinetically) correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BlunM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="30836" LastEditDate="2020-04-12T15:37:49.110" LastActivityDate="2020-04-12T15:37:49.110" Title="Concerted mechanism of saponification" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;saponification&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20386" PostTypeId="2" ParentId="20375" CreationDate="2014-12-06T19:38:55.603" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Contrary to the generally accepted belief, many cis isomers of olefinic compounds are more stable than their trans isomers. The stable cis form of substituted olefins such is 1–fluoro–1, 3–butadiene and l,4–difluoro–1,3–butadiene as being due to intramolecular van der Waals forces between the substituents &lt;strong&gt;[ Angew. Chem. , 75 , 793 (1963)  ]&lt;/strong&gt;. These forces are strong enough to hold the molecules in a cis–preferred configuration despite steric hindrance effects; these effects have been the basis for assuming that the trans form of such molecules is the more stable. But when the substituent atoms are so bulky that they overcome the intramolecular van der Waals forces, then the trans form becomes the more favored structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;cis cycloalkenes in general are more stable than their trans isomers. The trans double bond causes strong twisting of the ring. Because of the resulting high ring strain small trans cycloalkenes have not been observed and cis isomers show considerable ring strain. However, the latter are sufficiently stable in order to exist. To form a trans isomer the cycloalkene ring must contain at least eight carbons. The energy difference between cis- and trans-cyclooctene is approximately  $38.5\ce{kJ.mol^{-1}}$. Eventually, trans isomers become more stable than cis isomers once the ring contains more than eleven carbons. For more details, please see also: &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/cyclische_stereoisomere.vlu/Page/vsc/en/ch/2/oc/stoffklassen/systematik_struktur/cyclische_verbindungen/carbocyclisch/cycloalkene/physikalische_eigenschaften.vscml.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/cyclische_stereoisomere.vlu/Page/vsc/en/ch/2/oc/stoffklassen/systematik_struktur/cyclische_verbindungen/carbocyclisch/cycloalkene/physikalische_eigenschaften.vscml.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K7nZT.gif&quot; alt=&quot;enter image description here&quot;&gt;   &lt;img src=&quot;https://i.stack.imgur.com/ZCUGA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-03-31T06:38:17.143" LastActivityDate="2015-03-31T06:38:17.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20387" PostTypeId="1" AcceptedAnswerId="20393" CreationDate="2014-12-06T20:27:46.043" Score="2" ViewCount="5029" Body="&lt;p&gt;$$\ce{MnO4^2- (aq) -&amp;gt; MnO2(s) + MnO4^-(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I balance this in basic conditions? I know we need to separate this into two half reactions but how do we do this if there is only one reactant?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="5026" LastEditDate="2018-04-15T02:19:39.493" LastActivityDate="2018-04-15T02:20:00.500" Title="How to balance the disproportionation reaction of manganate anion in basic conditions?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20388" PostTypeId="2" ParentId="19204" CreationDate="2014-12-06T21:11:25.810" Score="2" Body="&lt;p&gt;The strengths of oxidizing and reducing agents are indicated by their &lt;strong&gt;&lt;em&gt;standard electrode potentials&lt;/em&gt;&lt;/strong&gt;. Let's consider  standard potentials of some redox couples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E^0(\ce{Li^+/Li})= -3.04V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E^0(\ce{K^+/K})= -2.92V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E^0(\ce{Cu^2+/Cu})= +0.34V$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ E^0(\ce{F_2/F^-})= +2.87V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values above are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16683" LastEditDate="2016-04-05T22:17:08.493" LastActivityDate="2016-04-05T22:17:08.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20389" PostTypeId="2" ParentId="20382" CreationDate="2014-12-06T22:09:37.393" Score="4" Body="&lt;p&gt;There's no technical reason why you couldn't make a powdered dishwashing detergent and I suspect that someone probably does sell it, just as there are liquid, tablet, and powdered automatic dishwasher detergents and laundry detergents. It does seem more convenient to use a liquid detergent for washing dishes in the sink. A liquid doesn't need time to dissolve, is easy to dispense without a scoop and if you're only washing a couple of dishes, you can simply squirt a small amount directly on a dish or use one of those brushes with a detergent reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen any dishwashing detergent that had a full ingredients list, but the one I have says &quot;contains anionic surfactants and enzymes&quot;. Anionic surfactant typically implies sodium dodecyl sulfate, which is itself a solid, usually found as a powder. The enzymes, while they could be freeze-dried, could retain more activity if stored in solution. Likewise, the fragrances often added to these detergents may be more suited to a liquid form.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-06T22:09:37.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20390" PostTypeId="1" AcceptedAnswerId="20472" CreationDate="2014-12-07T00:11:53.690" Score="1" ViewCount="871" Body="&lt;p&gt;I need a little help with one of the problems below. Any help on the problem below will be appreciated. I don't want to sound like an idiot, I will also provide my work.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solid $\ce{NaHCO3}$ decomposes to form water vapor, carbon dioxide and sodium carbonate.&#xA;  $$\ce{NaHCO3 -&amp;gt; H2O + CO2 + Na2CO3}$$&#xA;  If $2.00\ \mathrm g$ of $\ce{NaHCO3}$ decompose at $500\ \mathrm{^\circ C}$ and $700\ \mathrm{mmHg}$, how many &lt;b&gt;liters&lt;/b&gt; of $\ce{CO2}$ should be produced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\frac{750\ \mathrm{Torr}}{760\ \mathrm{Torr}}=0.99\ \mathrm{atm}}$$&#xA;$$pV=nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2.00\ \mathrm g\ \ce{NaHCO3} \cdot \frac{1\ \mathrm{mol}\ \ce{NaHCO3}}{84.02\ \mathrm g} \cdot \frac{1\ \mathrm{mol}\ \ce{CO2}}{2\ \mathrm{mol}\ \ce{NaHCO3}} \cdot \frac{22.4\ \mathrm L \ce{CO2}}{1\ \mathrm{mol}\ \ce{CO2}} = 0.287\ \mathrm L\ \ce{CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.99\ \mathrm{atm}\cdot0.287\ \mathrm L=n\cdot773\ \mathrm K\cdot0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$$&#xA;$$n = 0.004159\ \mathrm{mol}\ \ce{CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.004159\ \mathrm{mol}\ \ce{CO2} \cdot\frac{22.4\ \mathrm L}{1\ \mathrm{mol}\ \ce{CO2}} = 0.09317\ \mathrm{L}\ \ce{CO2}$$&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="865" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T21:04:36.650" LastActivityDate="2021-05-22T21:04:36.650" Title="Gas Stoichiometry &amp; Ideal Gas Law" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20391" PostTypeId="1" AcceptedAnswerId="20392" CreationDate="2014-12-07T00:12:05.893" Score="11" ViewCount="1238" Body="&lt;p&gt;If the probability is so low that a &lt;a href=&quot;https://en.wikipedia.org/wiki/Radium&quot;&gt;radium&lt;/a&gt; atom will decay into &lt;a href=&quot;https://en.wikipedia.org/wiki/Radon&quot;&gt;radon&lt;/a&gt; at any given time (the half-life is over 1600 years), then there will be a low amount of radon produced, granted it will be produced for a long time. If this is so, how can radon possibly be dangerous to us?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9961" LastEditDate="2016-06-23T15:46:06.547" LastActivityDate="2016-06-23T15:46:06.547" Title="If radium has such a long half-life, how can radon possibly be a threat to us?" Tags="&lt;elements&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20392" PostTypeId="2" ParentId="20391" CreationDate="2014-12-07T00:41:00.020" Score="16" Body="&lt;p&gt;There are two factors here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Radon is a gas and can thus enter the body very easily, through inhalation.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{^222Rn}$ and its first 4 decay products are &lt;em&gt;extremely&lt;/em&gt; radioactive, with halflives of miliseconds to a few days, releasing alpha and beta radiation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When the two are combined, we have a problem. Even if the quantity is low, inhaling radon places an extremely radioactive substance at very close proximity to the cells in one's lungs. Normally, alpha and beta radiation are relatively harmless as they cannot penetrate tissue very well, but now they have direct access. All it takes is one mutation in the wrong place to cause lung cancer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an idea of what a difference these make, the volumetric activity which is one guideline for needing to mitigate radon accumulation in a house is around 800 $\mathrm{Bq/m^3}$, which corresponds to only 141 fg of radon per cubic metre of air. This isn't like a uranium mine or something, but at this level, a certain increased risk of cancer is expected. Apparently &lt;a href=&quot;http://web.archive.org/web/20100528010149/http://deainfo.nci.nih.gov//advisory/pcp/pcp08-09rpt/PCP_Report_08-09_508.pdf&quot;&gt;some study&lt;/a&gt; found a 50% increased risk at a much lower level ~150 $\mathrm{Bq/m^3}$, too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;N.B. a becquerel (Bq) is a reciprocal second—1 Bq means one atom of the radioactive substance disintegrates every second.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-07T00:41:00.020" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20393" PostTypeId="2" ParentId="20387" CreationDate="2014-12-07T02:19:45.303" Score="1" Body="&lt;p&gt;You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^{2-}(aq) -&amp;gt; MnO2 (s) + MnO4^{-}(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -&amp;gt; MnO2 + 4OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation: $\ce{2(MnO4^{2-} -&amp;gt; MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -&amp;gt; 2MnO4^{-} + 2e^{-} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine both, and the final balanced equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3MnO4^{2-} + 2H2O -&amp;gt; MnO2 + 2MnO4^{-} + 4OH-}$&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="5026" LastEditDate="2018-04-15T02:20:00.500" LastActivityDate="2018-04-15T02:20:00.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20394" PostTypeId="1" AcceptedAnswerId="20403" CreationDate="2014-12-07T04:11:22.243" Score="4" ViewCount="1002" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwPwX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions manual said that enamine B with its roughly sp2 methyl group can force the nitrogen out of the plane and this messes around with the conjugation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the fact there is a significant degree of resonance donation by the nitrogen, which places a significant partial negative charge on carbon. In enamine A, this carbon is secondary. In enamine B, this carbon is tertiary. Extra inductive donation from the methyl group destabilizes the product. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-07T14:55:01.030" Title="Rationalizing enamine stabilities" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20395" PostTypeId="1" AcceptedAnswerId="20402" CreationDate="2014-12-07T09:32:22.323" Score="8" ViewCount="1023" Body="&lt;p&gt;When you add salt to water and stir, after a while the salt will no longer dissolve. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="10189" LastActivityDate="2014-12-07T14:23:10.383" Title="What Causes there to be a Maximum Amount of Salt which can be Dissolved in Water?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;water&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20396" PostTypeId="1" AcceptedAnswerId="20415" CreationDate="2014-12-07T09:59:32.540" Score="5" ViewCount="9387" Body="&lt;p&gt;As the title suggests, what is the difference between the conjugate and the inductive effect in organic chemistry.&lt;/p&gt;&#xA;" OwnerUserId="10189" LastEditorUserId="186" LastEditDate="2015-10-02T16:32:56.390" LastActivityDate="2016-07-24T21:23:27.470" Title="What is the Difference Between the Conjugate Effect and the Inductive Effect?" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20397" PostTypeId="1" CreationDate="2014-12-07T08:42:40.523" Score="8" ViewCount="7943" Body="&lt;p&gt;I just did a little water electrolysis experiment at home by immerging two aluminum electrodes in water with a small amount of table salt. The voltage of the battery was 6 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What surprised me is that I was able to collect a much bigger volume of gas at the electrode connected to the - than to the + (about a factor 10 by eye). I would have expected a ratio of 1/2 (2 times more hydrogen than oxygen).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why did I get less gas than expected at the electrode connected to the +?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Was the oxygen consumed by oxydising the electrode? (It appeared quite damaged at the end.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can I get a ratio of 1/2?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10193" OwnerDisplayName="Tony" LastEditorUserId="296" LastEditDate="2014-12-09T15:40:43.137" LastActivityDate="2021-01-17T13:26:15.113" Title="Ratios of produced gases in water electrolysis" Tags="&lt;electrochemistry&gt;&lt;water&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20398" PostTypeId="2" ParentId="20397" CreationDate="2014-12-07T09:55:27.677" Score="0" Body="&lt;p&gt;It looks like chlorine from table salt, rather than oxygen from water, is created at the positive electrode (I believe the reason is given in &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection&lt;/a&gt; ), and the chlorine reacts with aluminum (&lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=3673.0&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalforums.com/index.php?topic=3673.0&lt;/a&gt; ). &lt;/p&gt;&#xA;" OwnerUserId="10195" OwnerDisplayName="akhmeteli" LastActivityDate="2014-12-07T09:55:27.677" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20399" PostTypeId="1" AcceptedAnswerId="20401" CreationDate="2014-12-07T02:54:24.430" Score="12" ViewCount="196" Body="&lt;p&gt;In general, if you were to try make something hard in a liquid medium (especially water) it is quite difficult to make the material solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things such as mollusks though, have no problem generating hard shells in a wet medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly are the physics behind this formation? &lt;/p&gt;&#xA;" OwnerUserId="10192" OwnerDisplayName="gansub" LastEditorUserId="30836" LastEditDate="2019-04-14T09:21:12.190" LastActivityDate="2019-04-14T09:21:12.190" Title="In the wet medium of an ocean, how does a hard shell form?" Tags="&lt;water&gt;&lt;chemical-biology&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20401" PostTypeId="2" ParentId="20399" CreationDate="2014-12-07T11:13:00.750" Score="13" Body="&lt;p&gt;Carbon dioxide from the atmosphere (or from decaying matter in the ocean) reacts with water to form carbonic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{ CO2 + H2O -&amp;gt; H_2CO_3 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this reacts with calcium ions to form calcium carbonate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{H2CO3 + Ca^{2+} -&amp;gt; CaCO3 + 2H+ }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of calcium carbonate is about &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium_carbonate&quot; rel=&quot;noreferrer&quot;&gt;$13~\mathrm{mg \over L}$&lt;/a&gt;, so if the concentration of calcium carbonate is greater than this the excess will precipitate out as solid calcium carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shell-forming organisms actively absorb calcium from the water around them, so they are able to increase the concentration of calcium carbonate to above $13~\mathrm{mg\over L}$ and precipitate the excess to form their shells. They can get the carbon dioxide from the water around them or from their own metabolism. The actual details of shell formation is far more complex than this as it's controlled by processes with the cells of the organism rather than just being uncontrolled precipitation. I'm not sure how well the details are understood even today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, this is why shelly fauna are not keen on the acidification of the oceans that results from increased atmospheric carbon dioxide. The solubility of calcium carbonate is strongly $\mathrm{pH}$-dependent and rises sharply as the water gets more acid. Given that the shell is usually intended to stop other animals eating you, having your shell dissolve is generally not good for life expectancy.&lt;/p&gt;&#xA;" OwnerUserId="1470" OwnerDisplayName="John Rennie" LastEditorUserId="11367" LastEditDate="2016-12-12T20:29:52.983" LastActivityDate="2016-12-12T20:29:52.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20402" PostTypeId="2" ParentId="20395" CreationDate="2014-12-07T14:23:10.383" Score="6" Body="&lt;p&gt;When a salt dissolves in water, there is a competition between the electrostatic forces that keep the anion and cation together and the ion-dipole forces that solvate the ions.  One could write this equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{AB + q H2O \leftrightarrows A(H2O)_{m}+ + B(H2O)_{n}-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $q=m+n$. Since this is an equilibrium, we can write an equilibrium constant, which is oftentimes referred to as the solubility product: $$\ce{K_{sp}=[A(H2O)_{m}+] * [B(H2O)_{n}^{-}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very soluble salts will have a large, but finite, $K_{sp}$.  The magnitude of the solubility product can be rationalized by comparing the energy required to break the electrostatic bonds (the lattice enthalpy) and the energy released by forming solvated ions (solvation enthalpies).  One can use a Born-Haber cycle to predict whether the dissolution of a salt will be endothermic or exothermic, and we would need a similar approach to address entropy changes before we could get a complete, quantitative understanding of the solubility.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-12-07T14:23:10.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20403" PostTypeId="2" ParentId="20394" CreationDate="2014-12-07T14:42:33.483" Score="5" Body="&lt;p&gt;Conjugation effects are not very important in this case. &#xA;The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are exceptions, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ueSIa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4945" LastEditDate="2014-12-07T14:55:01.030" LastActivityDate="2014-12-07T14:55:01.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20404" PostTypeId="2" ParentId="20333" CreationDate="2014-12-07T15:03:41.590" Score="8" Body="&lt;p&gt;It would shift the IR peak to lower wavenumbers. You can rationalise this by understanding the bonding in the carbonyl group; in essence, by conjugating the carbonyl, the $\ce{C=O}$ bond is weakened, which shifts the IR peak down. You can think of it in terms of, if there is little or no conjugation, the electrons are firmly based in the $\ce{C=O}$ bond. As soon as we introduce some resonance structure, or conjugation, now those electrons can delocalise over 3-4 atoms or even more. This weakens the $\ce{C=O}$ bond, moving the IR peak down to lower wavenumbers. &lt;/p&gt;&#xA;" OwnerUserId="10196" LastEditorDisplayName="user15489" LastEditDate="2015-05-05T10:21:26.083" LastActivityDate="2015-05-05T10:21:26.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20405" PostTypeId="1" AcceptedAnswerId="20413" CreationDate="2014-12-07T15:22:58.013" Score="41" ViewCount="18050" Body="&lt;p&gt;Why does the electron-donating inductive effect (+I) of the isotopes of hydrogen decrease in the order $\ce{T} &amp;gt; \ce{D} &amp;gt; \ce{H}$?&#xA;(where T is Tritium and D is Deuterium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google has nothing to offer. Does it have to do anything with mass, as the order implies?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="81509" LastEditDate="2020-01-13T17:45:23.367" LastActivityDate="2020-01-13T17:45:23.367" Title="Inductive effect of hydrogen isotopes" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;isotope&gt;&lt;inductive-effect&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="20406" PostTypeId="1" CreationDate="2014-12-07T15:35:30.360" Score="1" ViewCount="290" Body="&lt;p&gt;For example if my hands were put in a HCl solution how does the protons in the solution which in high amounts will interfere/react with my hand(skin-&gt;proteins..) and cause it to burn...or how protons react with a plastic bag and cause it to melt.&lt;/p&gt;&#xA;" OwnerUserId="10197" LastEditorUserId="4325" LastEditDate="2014-12-07T16:06:00.243" LastActivityDate="2014-12-07T16:27:12.307" Title="how protons(H+) react with matter and cause the solution to be acid?" Tags="&lt;ph&gt;&lt;solutions&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20407" PostTypeId="2" ParentId="20406" CreationDate="2014-12-07T16:27:12.307" Score="2" Body="&lt;p&gt;Well HCl won't desolve all plastics. In case of your hand there are complex proteins involved containing a variety of elements e.g, carbon, nitrogen, oxygen..., when your hand comes in contact of a proton, the proton is a electrophile and seeks negative charge centers, where ever on a protein it finds the negative charge center e.g, on oxygen atom in a hydroxyl(-OH) group in some amino acid in some protein in your hand, it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water, because the ion H20^+ isn't stable and the newly formed Oxygen Hydrogen bond isn't easy to break, so the part breaks off as water. millions of such reactions or even trillions happen eventually corroding off the hand or al large part of hand exposed to protons in a mobile medium. &#xA;An acid burn isn't actually a burn but corrosion of proteins, and it doesn't either melt things away rather disintegrates them.&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2014-12-07T16:27:12.307" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20408" PostTypeId="2" ParentId="10944" CreationDate="2014-12-07T16:31:59.557" Score="8" Body="&lt;p&gt;My apologies for bumping an old question; however I'm not sure that the accepted answer provides an adequate answer to the stated question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Has anyone been taught to assign oxidation states this way? 2) If not, what do you think of this method?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part 1, I can say  personally that I use this method in teaching a 3rd year Inorganic Chemistry course.  It is covered in Rayner-Canham's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1429218142&quot;&gt;Descriptive Inorganic Chemistry&lt;/a&gt; textbook (Chapter 8 of the 5th edition).  After teaching this method for a number of years, I have anecdotal evidence that students get a better understanding of what the oxidation state is, and when it is suitable to use this type of electron bookkeeping instead of formal charge (for example).  In addition to relying on a periodic trend as opposed to a list of rules, this method helps to explain the differences in the oxidation numbers of inequivalent atoms in ions/molecules (such as sulfur in $\ce{S2O3^2-}$), which can't be done with the list-of-rules approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part 2 of the question, the biggest drawback of using the electronegativity approach to determining oxidation numbers is that it requires a knowledge of lewis dot structures.  With the current (American) General Chemistry track, students learn about oxidation numbers prior to drawing structures, and therefore must be taught using the list-of-rules approach.  I suspect a shuffling of the content to allow students to use the electronegativity approach would have some pedagogical benefits; instead of sitting at the bottom of &lt;a href=&quot;http://ww2.odu.edu/educ/roverbau/Bloom/blooms_taxonomy.htm&quot;&gt;Bloom's Taxonomy&lt;/a&gt; with remembering a list of rules, we can move up to applying the electronegativity trends to analyze the degree of electron deficiency of an atom has in an ion or molecule.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-12-07T16:31:59.557" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20412" PostTypeId="2" ParentId="20363" CreationDate="2014-12-07T16:58:15.010" Score="3" Body="&lt;p&gt;It is strange that you are concerned about diethyl ether (which has been used as an anesthetic), but you are not concerned about THF. From the &lt;a href=&quot;http://www.usp.org/sites/default/files/usp_pdf/EN/USPNF/generalChapter467Current.pdf&quot; rel=&quot;nofollow&quot;&gt;US Pharmacopoeia&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Permitted Daily Exposure&lt;br&gt;&#xA;  Tetrahydrofuran 7.2 mg/day&lt;br&gt;&#xA;  Diethyl Ether 50 mg/day&lt;br&gt;&#xA;  Ethyl Acetate 50 mg/day&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you dry the product thoroughly (above 120°C), you should not have any problems with residual solvents.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-07T16:58:15.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20413" PostTypeId="2" ParentId="20405" CreationDate="2014-12-07T17:39:14.100" Score="31" Body="&lt;p&gt;Yes, it has a lot to do with mass.  Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see &quot;Note&quot; below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a &lt;strong&gt;shorter&lt;/strong&gt; (and stronger) $\ce{C-D}$ bond length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net effect is that the shorter bond with deuterium increases the electron density at carbon, &lt;em&gt;e.g.&lt;/em&gt; deuterium is inductively more electron donating than protium towards carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Bohr Radius Detail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron.  However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary.  Given the mass difference between the electron and nucleus, this is generally a reasonable assumption.  However, in reality the nucleus does move too.  It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation for the Bohr radius becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_\text{heavy} \lt r_\text{light}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2017-12-24T15:34:25.287" LastActivityDate="2017-12-24T15:34:25.287" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20414" PostTypeId="1" CreationDate="2014-12-07T20:12:25.193" Score="10" ViewCount="6270" Body="&lt;p&gt;In class we learnt that the London forces become stronger as the relative molecular mass increases. Not just as in organic chemistry but in things like the halogens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I understand, the London forces are just temporary dipoles forming indicating that this would have nothing to do with the molecular mass but instead the number of electrons because there will be more electrons to form the dipole. This would mean that the trend would be stronger London Forces with more electrons instead of increasing molecular mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can that explain the slightly higher boiling points of heavy water (deuterium oxide ~101˚C) than normal water? &lt;/p&gt;&#xA;" OwnerUserId="10202" LastEditorUserId="5026" LastEditDate="2018-03-03T05:04:24.350" LastActivityDate="2020-09-14T11:13:01.193" Title="Why is the boiling point of heavy water more than that of normal water?" Tags="&lt;periodic-trends&gt;&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20415" PostTypeId="2" ParentId="20396" CreationDate="2014-12-07T21:29:01.337" Score="3" Body="&lt;p&gt;Conjugate effect (or delocalization) is an effect in which molecular orbitals (MOs) are conjugated to new molecular orbitals that are more delocalized and therefore generally lower in energy (the amount of MOs stays the same of course). The electrons can move freely in these new extended orbitals. As an example consider this MO diagram of butadiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/223kM.jpg&quot; alt=&quot;MO diagram of butadiene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the lowest energy MO is the MO with the least amount of nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive effect on the contrary does not increase delocalization but just pulls electrons towards the inductive (polarized) atom, this, for example, makes lone pairs less available decreasing the nucleophility of a molecule and increase its acidity. A lot of the time this inductive effect will be achieved trough a $\ce{C-X}$ (fluorine and chlorine) bond and you can observe its inductive effect simply by looking at the acidity of a series of more and less substituted molecules. Compare for example trifluoroacetic acid (pKa of 0.23) with regular acetic acid (pKa of 4.76).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not exactly sure though what effect induction has on the size of a MO (but it won't match the effects of conjugation for sure). It is clear however, that the electron distribution will shift toward the spatial direction of the inductive atom.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-07T21:29:01.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20417" PostTypeId="1" CreationDate="2014-12-07T21:36:57.913" Score="3" ViewCount="4775" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gywrS.png&quot; alt=&quot;Comparing acidic strength between 5 alcohols and carboxylic acids&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The correct answer is E&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I do not understand why?&#xA;Shouldn't (v) have the lowest &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; since it does not have a carboxyl group making it a strong acid? How do we determine the strength of acids?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="83758" LastEditDate="2020-05-26T15:40:56.083" LastActivityDate="2020-05-26T15:40:56.083" Title="How to Determine Acid Strength?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20418" PostTypeId="1" AcceptedAnswerId="82399" CreationDate="2014-12-07T22:16:12.390" Score="6" ViewCount="165" Body="&lt;p&gt;Some proteins do not unfold, or only partially unfold due to chaotropic agents such as guanidinium ions and urea. What are some reasons for this?&lt;/p&gt;&#xA;" OwnerUserId="10207" LastActivityDate="2017-09-09T18:12:32.240" Title="What would make guanidinium ions and urea more effective denaturants of proteins in some cases, but not in others?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20420" PostTypeId="1" AcceptedAnswerId="20434" CreationDate="2014-12-07T23:29:36.997" Score="5" ViewCount="1263" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G9gWw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left, there is a region for $\ce{MgO(ss)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told such regions exist because stuff are not miscible. However, does that mean the composition axis on the bottom is a lie? Namely, because it says ≠100% $\ce{MgO}$ and the region signals pure $\ce{MgO}$. &lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="4231" LastEditDate="2014-12-08T19:30:00.747" LastActivityDate="2014-12-08T19:30:00.747" Title="Phase diagrams (weird region)" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20421" PostTypeId="1" CreationDate="2014-12-08T00:36:26.910" Score="5" ViewCount="421" Body="&lt;p&gt;Synthesis of cyclopentadienyl-bis-triphenylphosphine-ruthenium chloride $\ce{RuCl(PPh3)2(\eta^5-C5H5)})$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolved $\ce{PPh3}$, $\ce{RuCl3.3H2O}$ and cyclopentadiene in ethanol and heated under reflux to prepare it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reducing agent and what is the product formed on oxidation of this reducing agent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ruthenium is being reduced but i don't know what is being oxidised. &lt;/p&gt;&#xA;" OwnerUserId="10082" LastEditorUserId="1389" LastEditDate="2015-01-07T01:46:42.647" LastActivityDate="2015-06-08T17:43:39.837" Title="Reducing agent in preparation of CpRuCl(PPh3)2" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20422" PostTypeId="2" ParentId="20421" CreationDate="2014-12-08T00:45:14.453" Score="1" Body="&lt;p&gt;You'd think it's the ethanol solvent that acts as the reducing agent here, and the byproduct is acetaldehyde.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-12-08T00:45:14.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20423" PostTypeId="1" CreationDate="2014-12-08T00:47:08.883" Score="4" ViewCount="95" Body="&lt;p&gt;If I add a small amount of chloroform to a bunch of water, it falls to the bottom as a glob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I shake the container vigorously (being carefully not to spill any of the dangerous substance!), the globs of chloroform break apart into many tiny globules. These globules immediately sink to the bottom and collect together by the action of gravity the moment I stop agitating my mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I watch this pile of globules for some time, I observe that they tend to merge after several seconds, perhaps even a minute or more for some. I suspect that if I reached in with a needle and prodded the globules, they would merge faster. Also, I suspect that if I were to tabulate how long each individual globule survives before merging, I would obtain a half-life curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What governs the merging of these globules (I'm guessing surface tension)? Why don't they instantly merge? Why do a few little globules sometimes persist for many minutes, stuck to the side of a much larger globule?&lt;/p&gt;&#xA;" OwnerUserId="5385" LastActivityDate="2014-12-08T04:00:41.680" Title="What governs the merging of globs of inmiscible liquid?" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20424" PostTypeId="2" ParentId="20417" CreationDate="2014-12-08T00:50:41.390" Score="3" Body="&lt;p&gt;First of all, let's make something crystal clear: The more acidic is a compound, they smaller will be its $pK_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate the acidity of these compounds you need to analyze the stability of the corresponding conjugate base of each one, in effect, the stability of the same compound but without a proton and a negative charge on an oxygen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you have two kinds of compounds: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two alcohols: V and II&lt;/li&gt;&#xA;&lt;li&gt;Two carboxylic acids: I, III and IV&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From those compounds, alcohols will always be less acidic than carboxylic acid. Why? Because the negative charge left by a proton leaving the molecule can be delocalizaed by resonance troughout the carbonyl, but in the case of the alcohols it will be localized on the only oxygen atom, which makes the conjugate base less stable, and therefore, the compound less acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, from the two alcohols, V will be more acidic, because the presence of the chlorine atoms substract charge from the carbon they are bond to (chlorine is more electronegative than carbon), and helps stabilizing the negative charge that is left in the molecule after the proton is substracted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regard to the carboxylic acids, you can use a similar reasoning for the carboxylic acid.&#xA;The less acidic is the one in which R is just a hydrocarbon chain, which has no effect in contrast to I, which has 1 chlorine atom that stabilizes the negative charge, and to III, which has 3 chlorine atoms that stabilize the negative charge of the conjugate base even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the order presented in option E is justified.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="4945" LastEditDate="2014-12-08T14:12:57.587" LastActivityDate="2014-12-08T14:12:57.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20425" PostTypeId="1" AcceptedAnswerId="20429" CreationDate="2014-12-08T02:35:37.137" Score="8" ViewCount="1568" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Formalin is a $37\%~\mathrm{W/W}$ solution of methanal in water. What mass of water must be added to $250~\mathrm{g}$ of methanal to make formalin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's what I've done so far. It's wrong though. According to my book answer is $430~\mathrm{g}$ $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Cw}{w} = \frac{\mathrm{m\ solute}}{\mathrm{m\ solution}} \cdot 100\%$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$37 \% = \mathrm{\frac{m\ solute}{250} \cdot 100\ \%}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\frac{37}{100} = \frac{m\ solute}{250}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{0.37 \cdot 250 = m\ solute}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{m\ solute = 92.5\ g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I went wrong? Am I completely off or at least close? &lt;/p&gt;&#xA;" OwnerUserId="10209" LastEditorUserId="13339" LastEditDate="2015-06-14T14:16:07.887" LastActivityDate="2020-01-01T15:57:56.380" Title="How to find the mass of the solute given a concentration as a mass fraction?" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20426" PostTypeId="1" AcceptedAnswerId="20433" CreationDate="2014-12-08T03:23:37.400" Score="2" ViewCount="22466" Body="&lt;p&gt;I learnt that everything is made up of atoms even paper! After that, I was thinking what are the elements that made paper? Is there any chemical symbol for paper in general?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="865" LastEditDate="2015-10-30T17:09:42.380" LastActivityDate="2017-06-17T09:54:37.450" Title="Elements of paper" Tags="&lt;everyday-chemistry&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20427" PostTypeId="2" ParentId="20423" CreationDate="2014-12-08T04:00:41.680" Score="5" Body="&lt;p&gt;Emulsion chemistry is actually rather complicated, but I'll try to give a conceptual description. Whenever you have a surface dividing two phases, there is an energy associated with creating a certain amount of surface, expressed as a surface energy density with dimensions of energy divided by area. This represents the energy cost of disrupting the intermolecular interactions of each phase at the interface, e.g. for your water/chloroform system, water can't hydrogen bond across the interface so it takes energy to break all the bonds required to make the surface. The larger the surface area, the more energy it takes. Logically, you can see that the lowest energy state is the one in which the interfacial area is minimized, which is why in the absence of external forces like gravity, liquids assume spherical shapes which give the smallest surface area for a given volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have many small droplets, there is a large surface area to volume ratio which is a high energy state, so it is energetically favourable for the droplets to form larger droplets. However, the reason it doesn't occur instantaneously is that an energy barrier can exist between the two states, for various reasons (electrostatic repulsion is a big one). In most cases, smaller droplets are lost first because the energy barrier to coalescing is surface area dependant. Larger droplets are more stable. Depending on the liquids, different separation mechanisms can occur with different kinetics. (coalescence, Ostwald ripening) I can't give a definite answer, but I'm guessing in your system you would not see an exponential decay-type half-life as that assumes a stochastic process where the probability is always the same. Since the stability depends on surface area, this isn't true in this case.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-08T04:00:41.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20428" PostTypeId="2" ParentId="20425" CreationDate="2014-12-08T05:31:23.843" Score="2" Body="&lt;p&gt;You have a couple of problems. In your equation, you've written the mass of the solution is 250 g, but there is 250 g of methanal and the total mass of the solution is 250 g + $m_{\ce{H2O}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the solution is 37% methanal not 37% water so the you've written the equation, you're solving for the wrong thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These should help you get on track:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\\&#xA;f_\mathrm{meth} &amp;amp;= \frac{m_\mathrm{meth}}{m_\mathrm{solution}}\\&#xA;m_\mathrm{solution}&amp;amp;=m_\mathrm{meth}+m_\mathrm{\ce{H2O}}\\&#xA;\%_\mathrm{meth} &amp;amp;= f_\mathrm{meth}\times100\%&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-08T05:31:23.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20429" PostTypeId="2" ParentId="20425" CreationDate="2014-12-08T05:34:55.120" Score="5" Body="&lt;p&gt;You have been given a &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_fraction_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;mass fraction&lt;/a&gt;, $w_{\ce{CH2O}}=37\%~\mathrm{w/w}$, and the mass of the solute, $m_{\ce{CH2O}}=250~\mathrm{g}$. You calculated the mass of the solution wrong.&lt;br&gt;&#xA;The mass fraction is defined as&#xA;$$w_i = \frac{m_i}{m}\tag1$$&#xA;with $m$ being the total mass and therefore&#xA;$$m = \sum_i m_i.$$&#xA;The latter is also the mass of your solution, the sum of the solute and the solvent.&#xA;$$m=m_{\ce{H2O}}+m_{\ce{CH2O}} = m_{\ce{H2O}} + 250~\mathrm{g}$$&#xA;Substitute what you know into $(1)$ and transform:&#xA;\begin{align}&#xA;w_{\ce{CH2O}} &amp;amp;= \frac{m_{\ce{CH2O}}}{m_{\ce{H2O}}+m_{\ce{CH2O}}}\\&#xA;0.37 &amp;amp;= \frac{250~\mathrm{g}}{m_{\ce{H2O}} + 250~\mathrm{g}}\\&#xA;m_{\ce{H2O}} &amp;amp;= \frac{250~\mathrm{g}}{0.37}-250~\mathrm{g} = 426~\mathrm{g}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="7920" LastEditDate="2014-12-08T06:03:29.447" LastActivityDate="2014-12-08T06:03:29.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20430" PostTypeId="2" ParentId="20414" CreationDate="2014-12-08T05:40:39.730" Score="1" Body="&lt;p&gt;I think in water and D2O the main factor is not Van Der Waals but rather hydrogen bonding. So the boiling point of D2O is higher than H2O not because of the London forces, but because of the difference in electronegativity between the hydrogen and the water, creating an electrostatic attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This begs the more interesting question: why is deuterium's electronegativity different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to clarify, electronegativity is the relative attraction an atom has for an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deuterium has a lower electronegativity than hydrogen, i.e it wants to give away its electron more. This is because the extra neutron increases the size of the nucleus and I think partially reduces the effect of the positive charge. Since deuterium has a lower electronegativity, there is a greater electronegativity difference between the deuterium and the oxygen, resulting in a stronger hydrogen bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stronger hydrogen bond = stronger intermolecular forces = greater boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think van der Waals is relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR:   In this case, hydrogen bonding is more important than London forces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above explanation is dodgy, I am still not sure to what extent it is true. There are many conflicting sources on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MUCH MORE INTUITIVE EXPLANATION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heavy water is heavier, right? So it requires more kinetic energy (i.e more heat) to escape the liquid and vapourise. Hence it has a higher boiling point&lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="402" LastEditDate="2017-04-18T19:01:26.843" LastActivityDate="2017-04-18T19:01:26.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20431" PostTypeId="1" AcceptedAnswerId="20432" CreationDate="2014-12-08T06:16:43.883" Score="2" ViewCount="4675" Body="&lt;p&gt;If $\ce{NaCl}$ crystallizes in an fcc lattice, its coordination number should be 12. Then why do we say that $\ce{NaCl}$ has a coordination number of 6?&#xA;Also, if $\ce{CsCl}$ crystallizes in a simple cubic lattice, why is it's coordination number 8 and not 6, as for other simple cubic lattices?&lt;/p&gt;&#xA;" OwnerUserId="9407" LastEditorUserId="4231" LastEditDate="2014-12-08T19:31:35.487" LastActivityDate="2016-09-03T06:40:57.290" Title="Coordination numbers of ionic and intermetallic compunds" Tags="&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20432" PostTypeId="2" ParentId="20431" CreationDate="2014-12-08T06:49:43.207" Score="4" Body="&lt;p&gt;&lt;strong&gt;NaCl&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NaCl is not monoatomic. When discussing coordination number of cations in ionic compounds, the number refers to the anions the cations is surrounded by (or bonded to, or touching).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RBm7s.png&quot; alt=&quot;NaCl structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue Na&lt;sup&gt;+&lt;/sup&gt; cations are bonded to six other Cl&lt;sup&gt;-&lt;/sup&gt; anions: top, bottom, left, right, forward, backwards (in this image). They are not bonded and do not touch the other Na&lt;sup&gt;+&lt;/sup&gt; cations in the diagonals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CsCl&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CsCl is not a &quot;simple&quot; simple cubic. Each of the different ions, Cs&lt;sup&gt;+&lt;/sup&gt; and Cl&lt;sup&gt;-&lt;/sup&gt; by themselves are simple cubic, but they are interpenetrating to form something that looks like a body centered cubic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zax8u.png&quot; alt=&quot;CsCl structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mono atomic simple cubic will have coordination number of six, but this is an ionic compound. Thus, each cation or anion that sits in the center of the cube is bonded to the eight anions or cations (respectively) that sit in the vertices of the cube.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="8083" LastEditDate="2014-12-08T20:46:32.977" LastActivityDate="2014-12-08T20:46:32.977" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20433" PostTypeId="2" ParentId="20426" CreationDate="2014-12-08T06:58:05.740" Score="11" Body="&lt;p&gt;There is no chemical symbol for paper since it is not an element but rather a mixture of several different compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that paper is made out of trees. Thus, paper is made mostly out of organic compounds: that is carbon, hydrogen and oxygen (C, H and O). Paper also contains non organic materials to improve its properties. These may be chalk (CaCO&lt;sub&gt;3&lt;/sub&gt;) and kaolin clay (Al&lt;sub&gt;2&lt;/sub&gt;Si&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;(OH)&lt;sub&gt;4&lt;/sub&gt;). Titanium oxide (TiO&lt;sub&gt;2&lt;/sub&gt;) is also commonly used in paper for bleaching it.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorDisplayName="user7951" LastEditDate="2017-06-17T09:54:37.450" LastActivityDate="2017-06-17T09:54:37.450" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20434" PostTypeId="2" ParentId="20420" CreationDate="2014-12-08T07:10:26.790" Score="9" Body="&lt;p&gt;There is nothing weird in here. What you've been told is wrong, or maybe you have misunderstood it. Regions like this exist &lt;em&gt;because&lt;/em&gt; stuff are miscible. (ss) stands for &lt;em&gt;solid solution&lt;/em&gt;, meaning that the solid crystal structure can accommodate some amount of the other component (Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram is a bit misleading, because naming something as MgO(ss) is not entirely correct. Adding the (ss) implies that it is not pure MgO. A more correct way to name the fields in the diagram would be to rename MgO to &quot;periclase&quot; and MgAl&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;4&lt;/sub&gt; to &quot;spinel&quot;, which are the mineral names for these compounds. They will not imply pure compositions of the solids, and conceptually allow the solid solutions to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some applications of the lever rule: For example, take a look at 10% Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; and 2000 degrees. You basically have periclase, which in pure form as the composition MgO. However, it has 10% of Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; in it. It is still periclase though. Let's take a more complicated example: At 40% Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; and 1800 degrees. You have two solid phases, periclase and spinel in roughly equal proportions. The periclase has 10% Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; in it, and the spinel has 65% Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; in it. Note that it's less than the ideal spinel should have!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice that Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; in the diagram does not have (ss) in it. This means that it's always going to be Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;, with no MgO in it at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curves that separate solid solution (ss) fields are appropriately called solvi (singular: solvus). They mark phase changes that are purely in the solid state, without any melting. For example, let's say you are at 70% Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; and 2000 degrees. You have only one phase: a spinel. By cooling down the system, you will cross the solvus line and periclase will be added to the system, on the expense of spinel. You will be able to microscopically see it as patches or stripes or something like that of periclase in the spinel. &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Various_eutectic_structures.png&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia has an image&lt;/a&gt; to demonstrate that (although it's built for eutectic compositions, the morphology is similar).&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="8083" LastEditDate="2014-12-08T18:45:50.900" LastActivityDate="2014-12-08T18:45:50.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20435" PostTypeId="1" CreationDate="2014-12-08T07:10:35.090" Score="1" ViewCount="541" Body="&lt;p&gt;Why does $30~\mathrm{mL}$ of $\ce{C3H7OH}$ burn in 1 second (inside a water jug) compared to a sixth of the amount: $5~\mathrm{mL}$ of $\ce{C3H7OH}$ (on a plate) burn in 180 seconds? Why does the smaller amount take longer to burn the propanol?&lt;/p&gt;&#xA;" OwnerUserId="10211" LastEditorUserId="4945" LastEditDate="2014-12-08T08:41:54.007" LastActivityDate="2014-12-08T08:53:02.760" Title="Why does a smaller amount of propanol burn slower than a much larger amount?" Tags="&lt;everyday-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20436" PostTypeId="1" CreationDate="2014-12-08T08:04:23.507" Score="1" ViewCount="92" Body="&lt;p&gt;A few years ago I saw this ultralight weight hydrophobic material and I have been trying to discover it's identity. I doubt it was aerogel since it was brought in by a high school student, but I cant say for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a transparent whitish material that was very light weight but absorbed ridiculous amounts of water, expanding as it absorbed the water. They were in small pellets, approximately 2mm or smaller randomly shaped. When water is poured on it each individual absorbed large amounts of water and expanded, and each pellet became somewhat gel like (in the sense that each individual pellet was kind of squishy). Unless obscene amounts of liquid was added all it took was a quick shake to more or less absorb every drop of water. Even when squished the material generally didn't start releasing the liquid. If you ended add enough water that the material couldn't absorb anymore, it would float on the water. If left out for a few days the water would evaporate, the material would shrink, and the dry material was somewhat squishy (less so than when wet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's off topic for me to ask something like this then how could I go about identifying this with a list of physical properties.&lt;/p&gt;&#xA;" OwnerUserId="5149" LastActivityDate="2014-12-08T08:04:23.507" Title="Identifying an aerogel like material or similar alternatives" Tags="&lt;identification&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20438" PostTypeId="2" ParentId="20435" CreationDate="2014-12-08T08:36:13.577" Score="1" Body="&lt;p&gt;For &quot;burning&quot; you need oxygen because actually when a compound is burning it's what is called an oxidation following this reaction  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H7OH + 5O2 -&amp;gt; 3CO2 + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with propanol called the combustible and oxygen is called the comburant. Both are required as well as a power source to start the combustion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a plate you have the air surrounding the compound then you have practically an endless amount of oxygen available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reaction will stop when all the Propanol will be consumed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in a jug, this time the limiting reactive is the oxygen because you have less oxygen supply. Then the combustion will stop when all the oxygen will be consumed in the jug.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up, in the jug the oxygen will be consumed faster than the propanol in the plate, that's why the combustion last longer in the plate even if you have more combustible.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="10119" LastEditDate="2014-12-08T08:53:02.760" LastActivityDate="2014-12-08T08:53:02.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20439" PostTypeId="1" AcceptedAnswerId="21687" CreationDate="2014-12-08T09:16:38.187" Score="11" ViewCount="1942" Body="&lt;p&gt;I will have some samples of azobisisobutyronitrile and I want to perform Barton-McCombie reaction. The problem is that the reagent is quite old and I need to know whether it has not decomposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any kind of test that I could perform? I was thinking if benzophenone in THF would not become blue after addition of fresh AIBN which would indicate the presence of the radicals. But I am not sure if such a test would be a good idea.&lt;/p&gt;&#xA;" OwnerUserId="7272" LastEditorUserId="9961" LastEditDate="2015-10-11T16:56:58.567" LastActivityDate="2016-06-14T19:53:19.357" Title="How to check if azobisisobutyronitrile (AIBN) is still active?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;radicals&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20440" PostTypeId="2" ParentId="19323" CreationDate="2014-12-08T10:15:39.733" Score="4" Body="&lt;p&gt;There is no definite classification of what are the major, minor and trace elements are. While the numbers you mentioned are indeed the rule of thumb, they are rather loosely applied. It depends on field (geochemistry, solid state physics, analytical chemistry). There are however several guidelines.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Major elements are the elements that define the material in question. Some examples are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gold and silver in electrum,&lt;/li&gt;&#xA;&lt;li&gt;Yttrium, aluminum and oxygen in YAG,&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen and oxygen in the ocean.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you change the major elements, you are essentially changing the material. If you replace the hydrogen with potassium then instead of water you will get potash, which is not the same material and has vastly different properties (although this example might be a bit extreme).&lt;br&gt;&#xA;Major elements are usually measured in percentage (either mass or molar), and are commonly above 1% of the chemical composition of the material.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Trace elements are elements that occur in such small concentrations that they do not change the essence of what a material is. Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Copper and platinum in electrum. It's still a gold-silver alloy, just with some other metals. The properties (such as colour) may change slightly.&lt;/li&gt;&#xA;&lt;li&gt;Neodymium, erbium, etc. in YAG. YAG is used to generate lasers, and doping it with trace amounts of other rare earth elements may change certain properties of the laser (wavelength). However, it's still a YAG.&lt;/li&gt;&#xA;&lt;li&gt;Almost every other element other than H, C, Cl and Na in ocean water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Trace elements usually have concentrations of below 0.1%. In this case it's easier to measure them in units of ppm (parts per million, this mostly refers to mass). Remember: 1% = 10000 ppm. In extreme cases ppb (parts per billion) may be the appropriate unit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Minor elements are everything in between. Technically, this means things between 1% and 0.1%. You might think of sodium and chlorine in ocean water as an example of minor elements (even though they're both slightly above 1%). To be honest, other than a mention in introductory textbooks about analytical chemistry and geochemistry, I rare see this term used in the professional literature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-08T10:15:39.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20444" PostTypeId="1" AcceptedAnswerId="20445" CreationDate="2014-12-08T12:40:10.420" Score="2" ViewCount="5413" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At room temperature, $298~\mathrm{K}$, the diffusivity of carbon in iron is &#xA;  $9.06\cdot 10^{-26}\frac{m^2}{s}$.&lt;br&gt;&#xA;  At $573~\mathrm{K}$ the diffusivity is $1.66\cdot 10^{-15}\frac{m^2}{s}$. What is the activation energy, $Q$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am wondering if my solution is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=D_0\exp\left\{\frac{-Q}{RT}\right\}$ and $D_0$ is constant (is this correct), thus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle \frac{D_{298}}{\exp\left\{-Q/(RT_{298})\right\}}=\frac{D_{573}}{\exp\left\{-Q/(RT_{573})\right\}} \implies Q\approx 122~\mathrm{\frac{kJ}{mol}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very greatful if someone could check my math here as well, if the logic is correct, since this was a test problem.&lt;/p&gt;&#xA;" OwnerUserId="8200" LastEditorUserId="5026" LastEditDate="2018-07-04T02:11:22.280" LastActivityDate="2018-07-04T02:11:22.280" Title="How to calculate the activation energy of diffusion of carbon in iron?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20445" PostTypeId="2" ParentId="20444" CreationDate="2014-12-08T13:05:47.457" Score="1" Body="&lt;p&gt;Maybe I'm wrong but my calculation has given another result:&#xA;\begin{align}&#xA;D_{298} &amp;amp;= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\&#xA;D_{573} &amp;amp;= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, going through the $\ln$ form, you get &#xA;\begin{align}&#xA;\ln\{D_{298}\} &amp;amp;= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\&#xA;\ln\{D_{573}\} &amp;amp;= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\&#xA;\ln\{D_{298}\} - \ln\{D_{573}\} &amp;amp;= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="16683" LastEditDate="2018-07-03T21:26:26.477" LastActivityDate="2018-07-03T21:26:26.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20447" PostTypeId="1" AcceptedAnswerId="20450" CreationDate="2014-12-08T14:05:33.050" Score="5" ViewCount="9561" Body="&lt;p&gt;Do you agree with the stereocenter with the $\ce{-CF3}$ and $\ce{-CN}$ substituents in the product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rLDf2.png&quot; alt=&quot;Diels Alder&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Conjugation/The_Diels-Alder_Cycloaddition&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that since $\ce{-CF3}$ is a better electron withdrawing group than $\ce{-CN}$, it would be placed opposite the diene, giving the product in which $\ce{-CF3}$ points up and $\ce{-CN}$ points down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I believe my proposal above is wrong, because probably the substituent with the highest electron-withdrawing property, the $\ce{-CF3}$ should be placed &lt;em&gt;endo&lt;/em&gt; because of secondary orbital interactions. Therefore, I expect the product above should be the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQXX4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Conjugation/The_Diels-Alder_Cycloaddition&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try to do the endo-product of the following reaction I came to the other &lt;em&gt;syn&lt;/em&gt; product with absolute stereochemistry (&lt;em&gt;S&lt;/em&gt;,&lt;em&gt;R&lt;/em&gt;) instead of the (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;S&lt;/em&gt;)-product shown here. So I'm wondering, why we get &lt;em&gt;endo&lt;/em&gt; selectivity in the first reaction and &lt;em&gt;exo&lt;/em&gt; selectivity here instead.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2016-01-04T14:20:38.917" LastActivityDate="2016-12-10T16:24:53.997" Title="Endo rule and stereoselectivity in the Diels-Alder reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20449" PostTypeId="1" AcceptedAnswerId="20492" CreationDate="2014-12-08T16:30:21.813" Score="11" ViewCount="2495" Body="&lt;p&gt;Do compounds of radioactive elements show a different behaviour? What happens when the radioactive part of the given compound decays? e.g. Radium chloride; what happens to it when radium decays?&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="9961" LastEditDate="2017-04-01T15:55:03.360" LastActivityDate="2018-06-06T07:48:19.350" Title="Does radioactivity affect chemical reactions?" Tags="&lt;reaction-mechanism&gt;&lt;radioactivity&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20450" PostTypeId="2" ParentId="20447" CreationDate="2014-12-08T17:13:53.427" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you agree with the stereocenter with the CF3/CN in the product?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, I would have expected the other isomer, the one with the $\ce{CN}$ and $\ce{NO2}$ groups pointing down in your drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rationale:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When exo and endo isomers can form in the Diels-Alder reaction, there is often a preference for formation of the endo isomer.  Although this endo preference is often referred to as the &quot;endo rule&quot;, it is not a &quot;rule&quot;, just a general, and rather weak, preference for endo product formation.  In fact, in most cases both isomers are formed, just more of the endo isomer (see figure below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation most often advanced for the endo preference is &quot;secondary orbital interactions.&quot; The following diagram illustrates the two possible transition states (TS) in the reaction of cyclopentadiene and an $\alpha,\beta$-unsaturated carbonyl compound.  One TS leads to the exo product, the other TS to endo.  Note how the endo TS has the unsaturated carbonyl bond tucked under the double bonds in the cyclopentadiene ring; the exo TS does not.  It is speculated (1) that the overlap of the carbonyl unsaturation with the diene unsaturation somehow stabilizes the endo TS making endo product preferred.  The accompanying table illustrates how weak the endo preference can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8YVMZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your top example, this endo preference for a TS with overlapping unsaturation would lead me to suspect that the product with the unsaturated nitro and cyano groups (the $\ce{CF3}$ group is not unsaturated) oriented away from the 3-carbon bridge and tucked towards the double bond would be preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, keep in mind that the endo preference applies to cases were the Diels-Alder reaction is run under kinetic control.  As the following example shows, the exo product is generally thermodynamically preferred (less steric crowding). So if the reaction is run under conditions (high temperature) where the initially formed kinetic (endo) product reverts to starting material, eventually the thermodynamically favored product (exo) will be formed eliminating the kinetic preference for the endo product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NYqa1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) see &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.23500/abstract&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for an alternate explanation for the endo preference&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-12-10T16:24:53.997" LastActivityDate="2016-12-10T16:24:53.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20451" PostTypeId="2" ParentId="20447" CreationDate="2014-12-08T17:39:21.583" Score="6" Body="&lt;p&gt;With respect to the second reaction, the endo/exo selectivity is not observable, since there is no substitution at the 1- or 4-positions on the diene. The difference between the product you predicted (S,R) and the shown (R,S) is that they are enantiomers. The product of two achiral reactants must either be achiral or a racemic mixture.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-08T17:39:21.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20454" PostTypeId="1" CreationDate="2014-12-08T19:08:56.233" Score="3" ViewCount="131" Body="&lt;p&gt;Is this best answered through quantum theory, Boltzmann statistics, or what? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a reaction energy diagram, only some of the reactants go over to the side of the products, and then get stuck there. So the progression of the reaction can be thought of in probabilities - what are the chances that something will go to products, and what are the chances the products will revert back. So you'd use the Arrhenius equation here. Could you dig deeper and find a quantum mechanical reason for this?&lt;/p&gt;&#xA;" OwnerUserId="10207" LastEditorUserId="9961" LastEditDate="2015-10-23T17:40:41.483" LastActivityDate="2015-10-23T17:40:41.483" Title="What is the mathematical foundation for the probabilistic nature of chemistry?" Tags="&lt;physical-chemistry&gt;&lt;statistical-mechanics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20455" PostTypeId="2" ParentId="7070" CreationDate="2014-12-08T19:22:16.040" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T2GQx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the fields marked by &lt;strong&gt;x&lt;/strong&gt; are solid and &lt;strong&gt;y&lt;/strong&gt; is liquid, these are the points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;a&lt;/strong&gt;: eutectic point. Cooling will turn the single liquid into two solids. The point is between the composition of the two solids involved.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;b&lt;/strong&gt;: eutectoid point. Cooling will turn the single solid into two solids.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;c&lt;/strong&gt;: peritectic point. Cooling from immediately to the left will turn the solid into a liquid and a solid. The point is outside the composition of the two solids involved.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;d&lt;/strong&gt;: congruent point: A solid will melt at this point to a liquid of the same composition.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;e&lt;/strong&gt;: incongruent melting points. A solid will melt to a liquid and a solid. Note that the left one is intimately associated with the peritectic point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are no peritectoid points in this diagram. A peritectoid point is similar to a peritectic point, but with all phases solid.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-08T19:22:16.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20456" PostTypeId="2" ParentId="10331" CreationDate="2014-12-08T19:51:18.050" Score="5" Body="&lt;p&gt;The HOH bond angle for an isolated water molecule is accurately known to be 104.5°  (104.52°±0.05° from &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/24/6/10.1063/1.1742731&quot; rel=&quot;nofollow&quot;&gt;Rotation‐Vibration Spectra of Deuterated Water Vapor J. Chem. Phys. 24, 1139&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the liquid phase, values of 105.5° (calculated) and 106° (experimental) are reported in &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/111/8/10.1063/1.479638&quot; rel=&quot;nofollow&quot;&gt;Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas phase water dimer (two otherwise isolated water molecules hydrogen bonded to each other) has also been considered and an angle of 104.7° is reported in the above reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ice Ih (common form of ice), values ranging from 106° to 108° have been reported. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempting to explain the trend, the intermolecular hydrogen bonding in the liquid and solid phases weakens the intramolecular covalent O-H bonds, and from a VSPER approach causes the electrons of the O-H bond to be more like the lone pair electrons.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-08T20:08:47.990" LastActivityDate="2014-12-08T20:08:47.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20460" PostTypeId="1" CreationDate="2014-12-08T21:11:42.773" Score="1" ViewCount="4718" Body="&lt;p&gt;Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what I'm meant to put for this.&lt;/p&gt;&#xA;" OwnerUserId="10224" LastEditorUserId="189" LastEditDate="2014-12-08T23:32:27.727" LastActivityDate="2014-12-08T23:32:27.727" Title="Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20461" PostTypeId="2" ParentId="20460" CreationDate="2014-12-08T22:10:58.277" Score="2" Body="&lt;p&gt;Carbonate contamination in oxides compounds often happens after a certain time and air exposure because of the $\ce {CO2}$ in the air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your magnesium oxide it follows the reaction  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {MgO + CO2 -&amp;gt; MgCO3}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\ce {MgCO3}$ being the magnesium carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution to prevent this kind of polution is to keep the oxide in a dessicator or to treat them in a furnace at a temperature around to $500 °C$ wich is the decomposition temperature of most of the carbonate compounds.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2014-12-08T22:10:58.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20464" PostTypeId="1" AcceptedAnswerId="20468" CreationDate="2014-12-08T23:06:13.393" Score="0" ViewCount="30" Body="&lt;p&gt;I have a quick question in Stoich and Gasses, I however could not get any far, any answer will be appreciated. Thank You.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A sample of $\ce{Pb(NO3)2}$ is heated, yielding $\ce{O2, NO2\ gas,\ and\ solid\ PbO}$. A&#xA;  $\ce{293 \ ml}$ gas sample, measured at $\ce{200C}$ and $\ce{1\ atm}$, is collected. What&#xA;  was the mass of the $\ce{Pb(NO3)2}$ sample?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I feel like in this situation I need to have only one gas so I can apply the &lt;strong&gt;Ideal Gas Law&lt;/strong&gt; to solve and find the mass of $\ce{Pb(NO3)2}$. What should I do there are two gasses and I don't know the volumes or the masses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Pb(NO3)2 -&amp;gt; O2 + 4NO2 + 4PbO}$$&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2014-12-08T23:39:13.713" LastActivityDate="2014-12-09T00:28:18.063" Title="Quick Stoich &amp; Gasses" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20465" PostTypeId="2" ParentId="12256" CreationDate="2014-12-08T23:17:24.440" Score="3" Body="&lt;p&gt;When &quot;doing&quot; science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, let's proceed with&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;A Theoretical Model&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Listed below are some standard reduction potentials I've gathered from the &lt;a href=&quot;http://hbcpnetbase.com/&quot; rel=&quot;nofollow&quot;&gt;Handbook of Chemistry and Physics&lt;/a&gt;, 95&lt;sup&gt;th&lt;/sup&gt; ed. (probably paywalled for you):&#xA;$$ \begin{align}&#xA;&amp;amp;&amp;amp;E^0_\text{red} / \mathrm{V}\\&#xA;\hline&#xA;\ce{Fe^{2+} + 2e-  &amp;amp; &amp;lt;=&amp;gt; Fe} &amp;amp; -0.447 \\&#xA;\ce{Fe^{3+} + 3e-  &amp;amp; &amp;lt;=&amp;gt; Fe} &amp;amp; -0.037 \\&#xA;\ce{Cu^{2+} + 2e-  &amp;amp; &amp;lt;=&amp;gt; Cu} &amp;amp; 0.342 \\&#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions:&#xA;$$ \begin{align}&#xA;E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &amp;amp;= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\&#xA;E^0_\text{(Fe3)} &amp;amp;= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V} &#xA;\end{align} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Confirmation With Experimental Results&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3):&#xA;$$ \varepsilon = \frac{\Delta x}{x} = 41.4\% $$&#xA;This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the experimental results in our pocket, we can now move on to the final part:&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Conclusion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, most of the calculations I did with an iPython Notebook, which you can view &lt;a href=&quot;http://nbviewer.ipython.org/gist/tschoppi/ee8cd08569e2073174be&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. I might have messed something up, you never know.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-12-08T23:17:24.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20466" PostTypeId="2" ParentId="20449" CreationDate="2014-12-08T23:26:02.593" Score="0" Body="&lt;p&gt;In addition to radioactive decay, which would, in the case of radium, give radon and an alpha particle (helium nucleus) that will strip 2 electrons from chloride ions or anything else nearby, the radioisotope will have a different mass from the stable isotope. This in turn may affect its reaction rate . The so-called isotope effect is largest in the case of hydrogen (tritium being the radioactive and three times heavier isotope, and it reacts slower in reactions involving the C-H bond.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2014-12-08T23:26:02.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20467" PostTypeId="1" CreationDate="2014-12-09T00:03:07.693" Score="2" ViewCount="279" Body="&lt;p&gt;&quot;Bonds will form only if the potential energy will be lower when the&#xA;atoms are bonded than when they are separate.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this statement true? If so, how do endothermic reactions exist? Can't you add energy to a system to make the bonds form albeit unfavorably?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="4642" LastEditDate="2014-12-09T00:18:48.130" LastActivityDate="2014-12-09T14:41:49.433" Title="Bond Potential Energy Lowering" Tags="&lt;bond&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20468" PostTypeId="2" ParentId="20464" CreationDate="2014-12-09T00:28:18.063" Score="1" Body="&lt;p&gt;The equation of the reaction provides you with the composition of the gas liberated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You assume ideal gas, which means that the gas particles are indistinguishable. This means the reaction equation could look like &#xA;$$ \ce{2Pb(NO3)2 -&amp;gt; 5X + 4PbO} $$&#xA;where 5 moles of gas are liberated for every 2 moles of lead nitrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reforming the ideal gas equation to $n=pV/(RT)$ will lead you to the numerical value. The amount of lead nitrate will then be $$n_{\ce{Pb(NO3)2}} = \frac{2n}{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that calculating the mass is a trivial task.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2014-12-09T00:28:18.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20469" PostTypeId="1" CreationDate="2014-12-09T00:45:57.753" Score="7" ViewCount="3167" Body="&lt;p&gt;I was wondering if starch can be broken up into individual glucose units. Because on a site I saw, it said that through acid hydrolysis starch can be broken up into amylose and amylopectin. However, as I researched amylopectin I found out it is branched and had a different bond than amylose so just to simplify everything:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can starch be broken up into individual glucose units and through what processes? &lt;/p&gt;&#xA;" OwnerUserId="10228" LastEditorDisplayName="user7951" LastEditDate="2016-12-30T23:33:38.967" LastActivityDate="2016-12-31T02:13:08.090" Title="Can I break starch down into glucose units?" Tags="&lt;biochemistry&gt;&lt;polymers&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20470" PostTypeId="1" AcceptedAnswerId="20473" CreationDate="2014-12-09T00:55:34.160" Score="6" ViewCount="1025" Body="&lt;p&gt;This is a question from my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine a step-wise $S_N1$ mechanism for the overall reaction:&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/HqVvN.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution shows the following carbocation rearrangement after Bromine leaves in the first step:&lt;img src=&quot;https://i.stack.imgur.com/UyQkB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the carbocation rearrange like this, going from a tertiary allylic carbocation to a secondary allylic carbocation? I thought tertiary carbocations are more stable and thus favored to be the major product? I considered Zaitsev's Rule in which the more substituted alkene is the major product, but I thought that only applies in elimination reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clarification is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="10227" LastEditorUserId="31313" LastEditDate="2017-04-01T17:24:42.837" LastActivityDate="2020-05-13T06:33:09.483" Title="Carbocation rearrangement in alkene addition reaction" Tags="&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20471" PostTypeId="2" ParentId="20467" CreationDate="2014-12-09T01:00:04.640" Score="0" Body="&lt;p&gt;They only form of the overall energy of the process is lower. You can form a less stable bond if you put in energy either through the formation of another bond which is more stable or through energy in form of light or electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: you can electrochemical my form aluminum from aluminum oxide. &lt;/p&gt;&#xA;" OwnerUserId="7614" LastActivityDate="2014-12-09T01:00:04.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20472" PostTypeId="2" ParentId="20390" CreationDate="2014-12-09T01:03:02.747" Score="1" Body="&lt;p&gt;I do not like to post an answer for my own question, however I feel the need to clarify my mistakes ton anyone else. So here is the answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Calculate the moles of $\ce{NaHCO3}$&lt;/li&gt;&#xA;&lt;li&gt;Then I need to plug in the moles to $\ce{PV = nRT}$&lt;/li&gt;&#xA;&lt;li&gt;After finding the volume of $\ce{NaHCO3}$, now I plug it in a dimensional analysis to figure out the volume of $\ce{CO2}$. That is all and the answer would be $\ce{.77\ Liters\ CO2}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2014-12-09T20:35:15.177" LastActivityDate="2014-12-09T20:35:15.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20473" PostTypeId="2" ParentId="20470" CreationDate="2014-12-09T01:23:13.150" Score="6" Body="&lt;p&gt;The mechanism shown is not a carbocation rearrangement. Those two structures are resonance structures, which means that the chloride anion can attack at either of the carbons that bear a positive charge. Because there are two different alkene products that are possible, the more substituted alkene will be favored. Strictly speaking, you're right, Zaitsev's Rule doesn't apply here, because this isn't an elimination reaction. However, the reasoning is the same: the most substituted, most stable alkene is the favored product.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-09T01:23:13.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20474" PostTypeId="1" AcceptedAnswerId="20476" CreationDate="2014-12-09T01:45:15.143" Score="6" ViewCount="731" Body="&lt;p&gt;I bake a loaf of bread containing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;500 g strong flour (1700 kcal) &lt;/li&gt;&#xA;&lt;li&gt;40 g butter (300 kcal)&lt;/li&gt;&#xA;&lt;li&gt;24 g yeast (80 kcal)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a total of 2080 kcal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many calories does my finished loaf contain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that at most, some of the energy in the yeast will be used during the proving process but I realised I know very little about the actual processes that take place during proving and baking.&lt;/p&gt;&#xA;" OwnerUserId="10229" LastEditorDisplayName="user7951" LastEditDate="2018-12-09T18:20:52.630" LastActivityDate="2018-12-09T18:20:52.630" Title="Does baked bread have the same amount of calories as its ingredients?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20476" PostTypeId="2" ParentId="20474" CreationDate="2014-12-09T02:01:20.210" Score="6" Body="&lt;p&gt;Yes, the yeast does consume sugar to produce the bubbles in the dough, but I don't think it would be more than a few grams. During the baking of the bread, some water and alcohol evaporates and proteins are denatured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denaturing proteins does not change the calories. Since the alcohol that was produced is lost, then the calories are reduced. Water doesn't have any calories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer to the question in the title: there are fewer calories in the baked bread than in the ingredients. Unfortunately, it cannot be predicted, the bread needs to be tested.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-09T02:01:20.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20478" PostTypeId="1" CreationDate="2014-12-09T03:22:12.233" Score="4" ViewCount="4949" Body="&lt;p&gt;I'm wondering if the hydrogen stolen during an $E_1$ reaction has to be antiperiplanar/anticoplanar like the hydrogen in an $E_2$ reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I'd say no, because the carbocation is flat so there's less steric hindrance than an E2 reaction, but I want to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, we could use 3-methyl-2-bromobutane. Once that carbocation forms and shifts, does it matter which hydrogen is stolen?&lt;/p&gt;&#xA;" OwnerUserId="10231" LastEditorUserId="7448" LastEditDate="2015-10-09T16:22:20.800" LastActivityDate="2015-10-09T16:22:20.800" Title="Does the hydrogen taken in an E1 reaction have to be antiperiplanar?" Tags="&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20479" PostTypeId="1" CreationDate="2014-12-09T04:42:37.767" Score="1" ViewCount="454" Body="&lt;p&gt;Refer to the mechanism posted on this site, if you will: &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-3.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-3.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the alcohol attacks the sulfur center, the pi electrons in the oxygen sulfur bond are taken by the oxygen. Then pyridine sucks up the proton. Afterwards the oxygen donates those pi electrons back and the chlorine anion leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the chlorine anion just leave? What's up with the oxygen shuffling electrons around?&lt;/p&gt;&#xA;" OwnerUserId="10231" LastEditorUserId="4945" LastEditDate="2014-12-09T10:58:23.427" LastActivityDate="2015-01-08T11:31:42.260" Title="Why is the halogenation of an alcohol a multistep process?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20480" PostTypeId="1" CreationDate="2014-12-09T05:24:32.287" Score="8" ViewCount="404" Body="&lt;p&gt;The town of Sweetwater, Missouri is known for its mineral resources, particularly rich lead sulfide (galena) deposits. I am curious to find the origin of the name &quot;Sweetwater&quot; and if it has any connection with lead compounds that at one time or another leached into the local water resources (I believe lead acetate tastes sweet). In any event, I can't seem to find any history or such a connection through internet research.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="865" LastEditDate="2015-07-30T14:19:23.547" LastActivityDate="2015-07-30T14:19:23.547" Title="Is Sweetwater, Missouri, named due to water contamination with lead compounds?" Tags="&lt;water&gt;&lt;toxicity&gt;&lt;geochemistry&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20481" PostTypeId="2" ParentId="20479" CreationDate="2014-12-09T05:34:18.820" Score="1" Body="&lt;p&gt;If the chlorine left straight away I.e after step 1 in the mechanism you linked to then you would have three charges on the structure - charges on two oxygens and a charge on sulfur. Doesn't seem too stable to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead what happens after step 1 is a proton transfer. Proton transfers are generally rapid. This proton transfer is also thermodynamically favored as it relieves the oxygen of its positive formal charge. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-09T05:34:18.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20482" PostTypeId="2" ParentId="20478" CreationDate="2014-12-09T08:32:17.587" Score="0" Body="&lt;p&gt;It actually does not matter whether or not it is antiperiplanar, because once the carbocation is formed, there is no sense of antiperiplanar. The two carbon atoms are in the same plane now. For a planar carbocation, all arrangements are equivalent(i.e. no sense of stereoisomerism). It does not matter which side the base attacks on. Also, since the bond angle in an $sp^2$ system is greater than that of an $sp^3$ system, the steric resistance to the base is less, which is the point you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, steric factors are not much significant in eliminations where protons are involved, because only a proton is required to be abstracted, and the incoming nucleophile(acting as a base ) is not required to attack or attach to a site.&lt;/p&gt;&#xA;" OwnerUserId="3863" LastEditorUserId="3863" LastEditDate="2014-12-09T13:11:43.047" LastActivityDate="2014-12-09T13:11:43.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20483" PostTypeId="2" ParentId="6314" CreationDate="2014-12-09T08:39:18.110" Score="1" Body="&lt;p&gt;Sulfurated potash (which is on that list under brown to black) does come in a gel form, it could be that.&lt;/p&gt;&#xA;" OwnerUserId="10232" LastActivityDate="2014-12-09T08:39:18.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20485" PostTypeId="2" ParentId="20479" CreationDate="2014-12-09T11:12:52.587" Score="0" Body="&lt;p&gt;As a general rule $S_{N}^{2}$ reaction do not happen at at an '$sp^{2}$' center (check here for help on hybridization: &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;). Unlike in an '$sp^{3}$' carbon, there is no likely transition state to accommodate the 'nucleophile' and 'nucleofuge'. Instead of this 'concerted' reaction, an addition-elimination takes place. Like @Dissenter already stated, the proton transfer reaction is favourable and fast, driving the reaction to completion.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-12-09T11:12:52.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20486" PostTypeId="1" AcceptedAnswerId="20548" CreationDate="2014-12-09T11:20:35.260" Score="10" ViewCount="698" Body="&lt;p&gt;For example, the cyclopentadienyl rings in ferrocene $\ce{Fe(}\eta^5 \ce{-C5H5)2}$ are often drawn with delocalized electron circles in them instead of explicit pi bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RAJ62.png&quot; alt=&quot;ferrocene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ferrocene&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article for ferrocene&lt;/a&gt; provides the following SMILES string: [cH-]1cccc1.[cH-]1cccc1.[Fe+2]. In this SMILES string, the charge on the cyclopentadienyl ligands are explicitly attached to carbon atoms in the ring.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can explicit or implicit delocalization be stored in a SMILES string? For example, would c1cccc1.c1cccc1.Fe also be an acceptable SMILES string for ferrocene that can be parsed and understood?&lt;/li&gt;&#xA;&lt;li&gt;Does the localized version of the SMILES string provide enough information for the delocalization to be reconstructed when the string is parsed back into a structure? If not, how can this be done?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="542" LastEditorUserId="275" LastEditDate="2014-12-09T15:06:14.860" LastActivityDate="2014-12-11T16:58:46.337" Title="What would be SMILES notation for a compound with delocalized bonding?" Tags="&lt;organic-chemistry&gt;&lt;notation&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20487" PostTypeId="1" AcceptedAnswerId="20488" CreationDate="2014-12-09T11:41:07.877" Score="2" ViewCount="105" Body="&lt;p&gt;PDB: &lt;a href=&quot;http://www.rcsb.org/pdb/explore/explore.do?structureId=1HIG&quot; rel=&quot;nofollow noreferrer&quot;&gt;1HIG&lt;/a&gt;, when opened using Accelrys DS Visualizer, Avogadro and Jmol produces a series of dots in space and cannot be displayed as 3D ball and stick models. Is there any way to overcome this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the output from Accelrys DS Visualizer 4.0:&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gqZ9p.jpg&quot; alt=&quot;This is the output from Accelrys DS Visualizer 4.0&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the output of Avogadro:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0d8MZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the output of Jmol:&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7vjBT.png&quot; alt=&quot;This is the output of Jmol&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2014-12-09T11:50:49.293" Title="How do I make PDB 1HIG's data usable for producing ball-and-stick models?" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20488" PostTypeId="2" ParentId="20487" CreationDate="2014-12-09T11:50:49.293" Score="4" Body="&lt;p&gt;This specific PDB file only contains the coordinates of all the $\ce{C_{\alpha}}$ atoms of the protein. It simply does not contain enough information to display an actual 3D structure of the protein. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The file does contain secondary structure assignments, so it should probably be possible to create a cartoon display of the protein, but that would require more than a simple PDB viewing program. But there is no way to create a ball-and-stick model with the data contained in that file.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2014-12-09T11:50:49.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20489" PostTypeId="2" ParentId="8509" CreationDate="2014-12-09T13:43:25.630" Score="7" Body="&lt;p&gt;O-protonated phenols direct in meta, according to an &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF03155678#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;article on that topic&lt;/a&gt;. As it occurs, phenols can be, depending on acid, protonated on oxygen, but also on carbon, and even diprotonated (but not twice on the same atom), when most powerful &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;mixtures of HF and SbF5&lt;/a&gt; are used. O-protonated forms can be brominated, hydroxylated and alkilated. C-protonated forms don't posses such reactivity and are essentially &lt;a href=&quot;https://en.wikipedia.org/wiki/Meisenheimer_complex&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meisenheimer's complexes&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2018-12-10T20:49:03.997" LastActivityDate="2018-12-10T20:49:03.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20491" PostTypeId="2" ParentId="20467" CreationDate="2014-12-09T14:41:49.433" Score="1" Body="&lt;p&gt;Two different factors play a role when considering a reaction. The energy of the bonds etc. (enthalpy) and the entropy. Together, these two components make up what is called the 'Gibbs Free Energy'. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change in Gibbs free energy ($\Delta{}G$) of a system must be $&amp;lt; 0$ for a reaction to be spontaneous. From this it is possible to see that a reaction where the entropy increases drastically, but the enthalpy increases. In this case a reaction can be endothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A case of terminology: The 'more correct' term (my teacher always said) is exergonic or endergonic. What this means is that the reaction has a $\Delta{}G &amp;lt; 0$ or $\Delta{}G &amp;gt; 0$ respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Cohorent mentioned before, it is also possible to have a system that is larger than just the chemical reaction (adding a battery for example). Meaning that the reaction is maybe endergonic, but a different energy source is used to drive the reaction. Note then, that the $\Delta{}G$ for the entire system is still below zero.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-12-09T14:41:49.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20492" PostTypeId="2" ParentId="20449" CreationDate="2014-12-09T14:57:18.020" Score="2" Body="&lt;p&gt;During the nuclear reaction, the atomic number Z (and therefore chemical identity) changes (&lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_transmutation&quot; rel=&quot;nofollow noreferrer&quot;&gt;transmutation&lt;/a&gt;).&#xA;In your example, $\ce{RaCl_2 -&amp;gt; \alpha^{2+} + RnCl_2^{2-} -&amp;gt; He + RnCl_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radon being noble gas would react further, e.g. $\ce{RnCl_2 -&amp;gt; Rn + Cl_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the decaying atom would be a transition metal with ligands, it's properties  should not change so drastically so that the ligands will stay attached.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-12-09T14:57:18.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20493" PostTypeId="1" CreationDate="2014-12-09T15:39:20.523" Score="10" ViewCount="5111" Body="&lt;p&gt;Why is benzoic acid a stronger acid than 3-aminobenzoic acid? Obviously at &lt;em&gt;meta&lt;/em&gt; position, only negative inductive effect will operate. But, how does that make it a weaker acid than benzoic acid?&lt;/p&gt;&#xA;" OwnerUserId="10234" LastEditorUserId="5026" LastEditDate="2018-04-20T16:21:19.133" LastActivityDate="2018-04-20T16:23:25.577" Title="Why is benzoic acid a stronger acid than 3-aminobenzoic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20494" PostTypeId="1" CreationDate="2014-12-09T15:48:04.530" Score="0" ViewCount="450" Body="&lt;p&gt;Question 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound A reacts according to the following hypothetical equation and has a molecular weight of 48.36 g/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 A (s) +B (aq) ––&gt; 2 C (aq) ∆H° = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sample of A, weighing 0.152 g reacts in a flask containing 250.00 g of water and the water temperature increases from 24.85 °C to a temperature of 26.26 °C. Calculate ∆H° for the reaction as written in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: -1408 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For which of the following reactions is ∆H° = ∆H°f, the heat of formation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. C (s) + 2 F2(g) ––&gt; CF4 (g) ∆H° = – 221.0 kJ &#xA;ii. H(g) + Br (g) ––&gt; HBr (g) ∆H° = –366.2 kJ &#xA;iii. 2 C(s) + H2(g) + 3 Cl2(g) ––&gt; 2 CHCl3 (g) ∆H° = –268.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: i only&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, could someone explain exactly what delta H means in terms of Thermochemistry. I would like to have a better understanding of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first question I attempted to use the q=m x s x (delta) T for the water and I attempted to use stochiometry for substance A but I always receive the wrong answer. Could someone help me on where to start with these types of problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(250) x (4.184) x (26.26-24.85)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 1474.86 kJ&lt;/p&gt;&#xA;" OwnerUserId="10235" LastActivityDate="2015-01-08T18:36:00.037" Title="Thermochemistry problems?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20495" PostTypeId="1" AcceptedAnswerId="20506" CreationDate="2014-12-09T16:12:18.980" Score="5" ViewCount="7296" Body="&lt;p&gt;Why can't an enolate anion successfully displace a halogen from a benzene? I am told that the geometry is incorrect; that such S&lt;sub&gt;N&lt;/sub&gt;2 substitutions take place on $\mathrm{sp^3}$ hybridized (i.e. tetrahedral) substrates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reasoning is circular. So what's a better reason? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of one reason but it is not satisfactory; it's still a bit circular. S&lt;sub&gt;N&lt;/sub&gt;2 = backside attack; how would one attack a planar bromobenzene? From within the ring? Seems like there would be a lot of possibly unfavorable interactions with the p-orbitals for something to get &lt;em&gt;within&lt;/em&gt; the ring. Also this doesn't explain why substitution reactions with enolate anions don't take place at other $\mathrm{sp^2}$ carbon atoms - i.e. bromoethene. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-04-03T02:10:34.267" LastActivityDate="2018-04-03T02:10:34.267" Title="Why can't an enolate anion successfully displace a halogen from a benzene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20496" PostTypeId="2" ParentId="20478" CreationDate="2014-12-09T16:21:42.423" Score="1" Body="&lt;p&gt;In an E2 reaction, there is a strong preference for the hydrogen to be coplanar with the leaving group and pointed in an anti-direction to the leaving group. In an E1 reaction, the hydrogen does not have to be anti-periplanar to the leaving group because in the E1 reaction the deprotonation event happens in a separate mechanistic step, after the leaving group has left. So there's no need to define the relative orientation of the hydrogen and leaving group in the E1 reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for the E2 reaction, remember that the anti-periplanar arrangement is preferred because it places the C-H bonding orbital anti to the C-LG antibonding orbital. In the transition state, those are the orbitals that reorganize to create the C-C pi bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the E1 reaction, there is no C-LG antibonding orbital, but there is an empty p-orbital on the carbocation. The C-H bond should be parallel to the p-orbital to create a similar (but strictly speaking, not anti-periplanar) alignment of orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pxGP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2014-12-09T16:30:36.010" LastActivityDate="2014-12-09T16:30:36.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20497" PostTypeId="2" ParentId="20495" CreationDate="2014-12-09T16:36:27.223" Score="2" Body="&lt;p&gt;When attacking a benzene ring for an SN2 reaction, I think your explanation is good enough. The $\ce{C-X}$ antibonding orbital lies in the plane and inside the ring. The nucleophile would be forced to approach through the rest of the atoms of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for other $\mathrm{sp^2}$ hybridized organohalides, there was a good explanation recently on this site: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14482/why-do-sn1-and-sn2-reaction-not-occur-at-sp2-centres&quot;&gt;Why do SN1 and SN2 reaction not occur at sp2 centres? &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-03-13T20:55:52.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20499" PostTypeId="2" ParentId="20385" CreationDate="2014-12-09T16:54:38.920" Score="2" Body="&lt;p&gt;The question borders on a semantic argument. In the third mechanism (textbook), we would assume that the deprotonation event is very fast. The first mechanism (concerted) is basically saying that the deprotonation is so fast that it cannot be distinguished from the alkoxide leaving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that acid-base reactions are fast. We also know that an anionic oxygen 'kicking out' an alkoxide is fast, at least relative to nucleophilic attack on a carbonyl (the rate determining step for any of these reactions AND the related trans-esterification is nucleophilic attack on a carbonyl to give the tetrahedral intermediate). What we're trying to answer is how fast is fast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my argument for the third mechanism. While acid-base reaction is fast, the top mechanism is essentially an E2 reaction, which does have a stereoelectronic preference for the hydrogen to be anti-periplanar to the leaving group. I have no doubt that conformation is readily accessible (the C-O bond will be rotating freely). The step is also bimolecular. In the third mechanism, an alkoxide kicking out a leaving group has no stereoelectronic preference. There are no alternate conformations to even consider. The C-O bond may be rotating, but it doesn't matter: there is always an electron pair arranged properly to kick out the alkoxide. Furthermore, the step is unimolecular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is speculation, because I have never seen an alternative to the standard textbook saponification mechanism before. To be honest I never considered the concerted mechanism, and although I think it is a reasonable mechanism, I still prefer the textbook mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-09T16:54:38.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20500" PostTypeId="2" ParentId="20494" CreationDate="2014-12-09T17:21:11.403" Score="1" Body="&lt;p&gt;H is enthalpy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition H = U + PV (where U is internal energy, P is pressure and V is volume).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 1, your work says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(250) x (4.184) x (26.26-24.85)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;= 1474.86 kJ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because you failed to show units in your work, you are off by a factor of 1000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the main reason your answer is different from the desired answer is you have calculated the amount of heat released by 0.152 g reacting, whereas the desired answer comes from calculating the heat released by 3 moles of A reacting, considering change in enthalpy of the system to be negative when heat is released to the surroundings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer ii is wrong because the monoatomic forms of hydrogen and bromine are not the standard states of the elements.  Answer iii is wrong because the equation is written for 2 moles being formed rather than 1 mole. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-09T17:59:19.287" LastActivityDate="2014-12-09T17:59:19.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20502" PostTypeId="1" AcceptedAnswerId="20534" CreationDate="2014-12-09T18:06:41.183" Score="0" ViewCount="233" Body="&lt;p&gt;I've run into trouble on this question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl-}$. Assume $\Delta_\mathrm rS$ is independent of $T$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know the reaction is&#xA;$$\ce{AgCl +e- &amp;lt;=&amp;gt; Ag(s) + Cl- (aq)}$$&#xA;$$E^\circ=0.222$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left cell is a hydrogen electrode making the equation&#xA;$$\ce{1/2H2 + AgCl(s) -&amp;gt; Ag(s) + Cl- (aq) + H+(aq)}$$&#xA;forming a nernst equation of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=E^\circ-\frac{RT}{vF}\ln\left(\frac{\alpha_{\ce{Ag(s)}}\ \alpha_{\ce{Cl-}}\ \alpha{\ce{H+}}}{\alpha_{\ce{AgCl(s)}}\ \alpha_{\ce{H_2}}^{1/2}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which reduces to&#xA;$$E=E^\circ-\frac{RT}{vF}\ln\left(\alpha_{\ce{Cl-}} \alpha_{\ce{H^{+}}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $R$, $T$, $v$ and $F$ but I don't understand where to begin to solve for $\alpha_{\ce{Cl-}} \text{and}\ \alpha_{\ce{H}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the activities? &lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorDisplayName="user7951" LastEditDate="2015-10-08T00:02:12.987" LastActivityDate="2015-10-08T00:02:12.987" Title="Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="20504" PostTypeId="1" AcceptedAnswerId="20509" CreationDate="2014-12-09T18:20:17.467" Score="1" ViewCount="438" Body="&lt;p&gt;I've been teaching myself chemistry, so any help is greatly appreciated.  I've been reading an online tutorial that claims the two orbitals that merge in Aluminum trihydride are  1 orbital of 2s and 2 orbitals of 2p.  This confuses me, because I would expect the orbitals of the electron configuration would form a hybrid with one another, that is, Al electron configuration is [Ne]3s^2 3p^1.  Therefore, why do the lower orbitals form a hybrid? Also, what is the rule in general to determine which orbitals will form hybrids?&lt;/p&gt;&#xA;" OwnerUserId="10237" LastEditorUserId="10237" LastEditDate="2014-12-09T19:20:02.047" LastActivityDate="2014-12-09T20:32:51.057" Title="Determine which orbitals will form hybrids with one another" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;hybridization&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20505" PostTypeId="2" ParentId="20449" CreationDate="2014-12-09T18:52:48.130" Score="9" Body="&lt;p&gt;I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay &lt;em&gt;can&lt;/em&gt; cause damage to crystalline solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the host crystal. The crystal can then be polished and etched to reveal these so called &quot;fission tracks&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iHkxQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;sup&gt;(source: &lt;a href=&quot;https://pangea.stanford.edu/research/groups/thermochronology/&quot; rel=&quot;noreferrer&quot;&gt;Stanford Fission Track Thermochronology Laboratory&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating of a damaged crystal will cause the damage to &quot;heal&quot; and the tracks to disappear. Thus, radioactive decay can indirectly induce chemical reactions. This routinely occurs in the deep Earth. It also has some very useful applications in geochronology, which you can read about in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fission_track_dating&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article about fission tracks&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-09T18:52:48.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20506" PostTypeId="2" ParentId="20495" CreationDate="2014-12-09T19:14:24.657" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;explain why substitution reactions with enolate anions don't take place at other sp2 carbon atoms &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As explained in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14482/why-do-sn1-and-sn2-reaction-not-occur-at-sp2-centres/14483#14483&quot;&gt;this earlier answer&lt;/a&gt;, $\ce{S_{N}2}$ reactions do occur at $\ce{sp^2}$ carbons, they're just higher energy and consequently, less common.  So the question becomes, &quot;&lt;strong&gt;why is this pathway higher in energy?&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $\ce{S_{N}2}$ reaction involves &quot;backside&quot; approach to the carbon undergoing substitution, steric considerations play a role.  However, the attacking reagent could alter its approach by 15 degrees or so lessening steric interactions (e.g. in benzene the attacking nucleophile could approach at an angle of 15 degrees above the plane of the benzene) and not increasing the activation energy too much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to explain the high energy associated with $\ce{S_{N}2}$ attack at a vinylic carbon is as follows.  Consider the ethylenic double bond as a bent bond system, that is, we have a two-membered ring (instead of a traditional sigma-pi double bond) and each carbon in the ring is $\ce{sp^{4.3}}$ hybridized.  This bent bond description is equivalent to the alternate pi bond description in all regards including electronic (btw, it makes the reason [strain] for the high heat of combustion of olefins more palpable). If possible, build a model to help visualize the situation.  As the attacking nucleophile approaches and begins to interact with one of the ring carbons, that carbon will attempt to rehybridize to an $\ce{sp^2}$ carbon (just like in the traditional case of $\ce{S_{N}2}$ attack on an $\ce{sp^3}$ carbon).  In this case we're forcing the already highly strained bond angle in our two-membered ring to try and open further - towards 120 degrees. Hence, this is a high-energy pathway that is not often observed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, this same concept can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state.  It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons in higher energy p orbitals, not offset by the energy lowering of the electrons shifting from $\ce{sp^2}$ bonds to $\ce{sp}$ bonds) thereby increasing the energy required to reach the transition state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point to consider is that since the $\ce{S_{N}2}$ reaction proceeds with inversion of configuration, such a reaction in a ring system would lead to the formation of a trans-double bond.  This would be an extremely high energy process in the case of a small ring system such as benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $\ce{sp^2}$ hybridized carbon is more electronegative than $\ce{sp^3}$ hybridized carbon due to the increased s-character.  This will increase the electron density around an $\ce{sp^2}$ carbon and, as a nucleophile approaches, the increased electron repulsion will make this a higher energy pathway that nucleophilic attack at a less electron rich $\ce{sp^3}$ carbon.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-12-13T16:55:30.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20507" PostTypeId="2" ParentId="10022" CreationDate="2014-12-09T19:19:19.693" Score="2" Body="&lt;p&gt;You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$&#xA;You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is definitely a mistake in your textbook's solutions.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-09T19:19:19.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20509" PostTypeId="2" ParentId="20504" CreationDate="2014-12-09T20:02:47.120" Score="0" Body="&lt;p&gt;Let's take your questions in reverse order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the rule in general to determine which orbitals will form&#xA;  hybrids?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;general&quot; rule is that you mix one atomic orbital for each bond you want to form.  In methane, carbon has 4 valence electrons and is forming 4 bonds so it mixes 4 atomic orbitals (1 2s and 3 2p) generating 4 $\ce{sp^3}$ hybrid orbitals, and it places 1 electron in each orbital.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of aluminum trihydride, aluminum has 3 valence electrons and will form 3 bonds, one to each of the 3 hydrogen atoms.  Therefore, aluminum must mix 3 atomic orbitals (1 3s and 2 3p) generating 3 $\ce{sp^2}$ orbitals and places 1 of its electrons in each of these orbitals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why do the lower orbitals form a hybrid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 2s orbital is relatively close in energy to a 2p orbital and the same applies for a 3s and 3p orbital.  The closer in energy 2 orbitals are, the more effective is their mixing.  Therefore, 3s and 3p orbitals can mix (form new hybrid orbitals) effectively.  The reason orbitals mix in the first place is to form new orbitals that have greater directionality and can therefore form stronger bonds.  After all, we could form 3 bonds to aluminum trihydride by just bonding to the 3s orbital and separately to 2 3p orbitals.  Two of these bonds would be oriented 90 degrees from one another and the third bond would be fluxional, moving around the surface of a (3s) sphere. But hybridizing the aluminum so that we can form 3 $\ce{sp^2}$ orbitals and use them for bonding with hydrogen produces a more stable molecule and one with the observed geometry (3 bonds in a plane with a 120 degree angle between them).  If hybridization doesn't produce stronger bonds and a molecule that is more stable, then the molecule wouldn't hybridize. $\ce{PH3}$ is an example of such a molecule, the central phosphorous atom remains unhybridized.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-09T20:32:51.057" LastActivityDate="2014-12-09T20:32:51.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20511" PostTypeId="1" AcceptedAnswerId="20513" CreationDate="2014-12-10T00:23:43.453" Score="14" ViewCount="44344" Body="&lt;p&gt;If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I'm thinking: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{(H^+)_2O^{2-}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="9795" LastEditorUserId="7412" LastEditDate="2014-12-10T14:48:37.973" LastActivityDate="2017-09-25T18:23:52.853" Title="Why isn't water an ionic compound?" Tags="&lt;bond&gt;&lt;water&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20512" PostTypeId="2" ParentId="16040" CreationDate="2014-12-10T00:37:44.700" Score="1" Body="&lt;p&gt;I found an article &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3536963/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; about PM6/7 which might be useful.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2014-12-10T00:37:44.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20513" PostTypeId="2" ParentId="20511" CreationDate="2014-12-10T00:46:02.410" Score="25" Body="&lt;p&gt;First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are &quot;in between&quot; states like &lt;em&gt;polar covalent&lt;/em&gt;, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-10T00:46:02.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20514" PostTypeId="2" ParentId="20454" CreationDate="2014-12-10T01:49:43.727" Score="2" Body="&lt;p&gt;Actually, a huge amount of the probablistic nature of chemistry is simply the Law of Large Numbers (LLN). A mole of any substance is approximately $6\times10^{23}$ molecules, which is 600 &lt;em&gt;sextillion&lt;/em&gt; molecules! The LLN kicks in at quantities &lt;strong&gt;much&lt;/strong&gt; smaller than the sextillion level. Put simply, there are &lt;em&gt;so&lt;/em&gt; many interactions occurring in any macro-level reaction that these are inevitably going to converge to the theoretical mean by the LLN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your question adds in the issue of activation energy, but I still think it fits this model, because activation energy tends to produce a sharp difference: too low, and little reaction happens; but high enough, and we mostly go to completion. Once activation energy is high enough, the LLN applies.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-10T01:49:43.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20516" PostTypeId="1" AcceptedAnswerId="20519" CreationDate="2014-12-10T02:57:36.220" Score="0" ViewCount="535" Body="&lt;p&gt;At school, I learnt about the metal and their flame colour. Here's my observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K^+}$- Lavender&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ba^2+}$- Yellow-Green&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ca^2+}$- Yellow-Red&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Li^+}$- Flame Red&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na^+}$- Bright Orange&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^2+}$- Blue-Green&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is why do the each metal has its own flame colour?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="4356" LastEditDate="2014-12-10T03:12:30.527" LastActivityDate="2014-12-10T05:57:17.237" Title="Why they have different flame colour?" Tags="&lt;ions&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20517" PostTypeId="2" ParentId="4063" CreationDate="2014-12-10T05:26:13.423" Score="2" Body="&lt;p&gt;In your example, the hydrogen (+1) bonded to the hydroxide (-1) will balance out to a net charge of 0. Because there is no overall charge, the pH will be neutral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, pure, neutral water can behave as a Lewis Base (lone pair donor) due to the lone pairs present on oxygen. For example, water behaves as a Lewis Base and Al&lt;sup&gt;3+&lt;/sup&gt; as a Lewis Acid (electron acceptor) to form a hexaaquaaluminum(III) complex. That is: Al&lt;sup&gt;3+&lt;/sup&gt; with 6 water molecules surrounding it in a cubic fashion.&lt;/p&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2014-12-10T05:26:13.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20519" PostTypeId="2" ParentId="20516" CreationDate="2014-12-10T05:57:17.237" Score="3" Body="&lt;p&gt;First, remember that electrons cannot exist between principle energy levels (distances). In a flame, electrons from lower principle levels (n=1,2,3... etc) will be knocked up to higher principle levels from the absorbed energy. Electrons do not like to stay in this high energy state and will travel back down to their ground state, and in the process release that stored energy in the form of radiation (often visible light). The frequency of the radiation (color of the light) depends upon how big the fall back to ground state was. Different metals have different attainable energy levels, and so will display different colors when burned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some elements only have a few possible emission colors, while other elements are a mixture of dozens of colors ! This principle helps us tell what the composition of stars are, as well as allows us to test for metal contents in food products or other applications. &lt;/p&gt;&#xA;" OwnerUserId="10246" LastActivityDate="2014-12-10T05:57:17.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20520" PostTypeId="2" ParentId="20454" CreationDate="2014-12-10T07:17:45.183" Score="1" Body="&lt;p&gt;If you want to address only chemical reactions, than best answer is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Statistical_mechanics&quot; rel=&quot;nofollow&quot;&gt;statistical mechanics&lt;/a&gt;, which exercises Maxwell-Boltzmann statistics, exactly as you suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any reasonable chemical experiments, you are dealing with enormously large ensembles, &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;nofollow&quot;&gt;Avogadro number&lt;/a&gt; is $6 \times 10^{23}$ and extremely long times (compared to all motions accessible to molecules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can follow quasi-equilibrium reasoning, leading to &lt;a href=&quot;https://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;nofollow&quot;&gt;transition state theory&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, this is not enough and you want to go further and include coupling between the motions, which leads you to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0521140714&quot; rel=&quot;nofollow&quot;&gt;Reaction Dynamics&lt;/a&gt; (no wiki article yet, but this is an excellent book).&#xA;This approach was needed to describe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Crossed_molecular_beam&quot; rel=&quot;nofollow&quot;&gt;crossed molecular beam experiment&lt;/a&gt; (Nobel prize 1986).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2014-12-10T07:17:45.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20523" PostTypeId="1" AcceptedAnswerId="20528" CreationDate="2014-12-10T14:21:25.870" Score="-2" ViewCount="2869" Body="&lt;p&gt;What is the proper Lewis structure for carbon suboxide, $\ce{C3O2}$?&lt;/p&gt;&#xA;" OwnerUserId="10243" LastEditorUserId="4945" LastEditDate="2018-08-16T16:34:47.290" LastActivityDate="2018-08-16T16:54:46.317" Title="What is the proper Lewis structure of carbon suboxide?" Tags="&lt;bond&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ClosedDate="2014-12-10T19:46:02.213" ContentLicense="CC BY-SA 4.0" />
  <row Id="20524" PostTypeId="2" ParentId="20523" CreationDate="2014-12-10T14:26:01.817" Score="0" Body="&lt;p&gt;Well, looks like you haven't been looking for that long time on the internet since the answer is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_suboxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article about Carbon Suboxide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G4Zwx.png&quot; alt=&quot;Lewis&amp;#39;s structure of the $\ce {C302}$ compound&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2014-12-10T14:26:01.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20528" PostTypeId="2" ParentId="20523" CreationDate="2014-12-10T15:08:50.677" Score="5" Body="&lt;p&gt;According to &quot;Electronic structure of carbon suboxide&quot;:&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed, the results of these electron population analyses&#xA;  are consistent with the following resonance forms contributing&#xA;  to the ground state of $\ce{C3O2}$:&#xA;  \begin{align}&#xA;\mathrm{A}&amp;amp;: \ce{O=C=C=C=O} \\&#xA;\mathrm{B}&amp;amp;: \ce{O=C=C^\ominus-C#O^\oplus} \\&#xA;\mathrm{C}&amp;amp;: \ce{^{\oplus}O#C-C^\ominus=C=O}\\&#xA;\mathrm{D}&amp;amp;: \ce{O=C=C^\ominus-C^\oplus=O}\\&#xA;\mathrm{E}&amp;amp;: \ce{O=C^\oplus-C^\ominus=C=O}\\&#xA;\mathrm{F}&amp;amp;: \ce{O=C=C=C^\oplus-O^\ominus}\\&#xA;\mathrm{G}&amp;amp;: \ce{^{\ominus}O-C^\oplus=C=C=O}\\&#xA;\mathrm{H}&amp;amp;: \ce{^{\oplus}O#C-C^{\ominus\ominus}-C#O^\oplus}&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1] John F. Olsen and Louis Burnelle, &lt;em&gt;J. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1969,&lt;/strong&gt; &lt;em&gt;73&lt;/em&gt; (7), 2298–2304. DOI: &lt;a href=&quot;https://doi.org/10.1021/j100727a033&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1021/j100727a033&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2018-08-16T16:54:46.317" LastActivityDate="2018-08-16T16:54:46.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20530" PostTypeId="2" ParentId="20502" CreationDate="2014-12-10T16:51:52.040" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl−}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reactions would be:&#xA;$$\ce{AgCl +e- -&amp;gt;Ag +Cl-}\tag{$E_1$ V}$$&#xA;$$\ce{2H+ +e- -&amp;gt;H2}\tag{0 V}$$&#xA;Adding:&#xA;$$\ce{AgCl +H_2-&amp;gt;Ag +Cl- +2H+}\tag{$E_1 V$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard potential would be:&#xA;$$E=E^\circ -\frac{RT}{\nu F}\log\ce{\frac{[Cl- ][H+]}{P_{H2}^2}}\quad (E^\circ =E_1,\nu=1)$$&#xA;H&#xA;Now you have $E_1$, also you must have $\ce{[Cl- ],[H+ ]}$ beforehand, there's no other way, other than considering the fact that (standard) Hydrogen electrode is maintained at $\text{1M}$ and $\text{1 atm}$, and to maintain charge $\ce{[Cl- ] =[H+ ]}$.In that case, &#xA;$$E=0.222-\frac{8.314\times 310}{96500}\ln(1)=0.222$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just for info (from Wikipedia):&#xA;$$\begin{array}{c|c}\text{Temperature}&amp;amp;\text{Potential} E^\circ\\&#xA;^\circ\text{K}	&amp;amp;\text{V versus SHE at the same temperature}\\\hline&#xA;298	&amp;amp;0.22233\\&#xA;333	&amp;amp;0.1968\\&#xA;398	&amp;amp;0.1330\\&#xA;423	&amp;amp;0.1032\\&#xA;\end{array}$$&#xA;Also when you consider $\Delta S\ne\text{ constant}$:(A.J. Bard, R. Parson, J. Jordan, &quot;Standard Potentials in Aqueous Solution&quot;, Marcel Dekker, Inc., 1985.)&#xA;$$E^\circ=0.23695 - 4.8564\times10^{−4}t - 3.4205\times10^{−6}t^2 - 5.869 \times 10^{−9}t^3$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-12-10T18:53:29.050" LastActivityDate="2014-12-10T18:53:29.050" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="20531" PostTypeId="2" ParentId="20495" CreationDate="2014-12-10T17:18:29.587" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't an enolate anion successfully displace a halogen from a benzene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you know enolate is the conjugate base of enol which is a tautomeric form of ketone. Thus being an electron rich molecule it basically acts like a nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gZStR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aryl halides are extremely less reactive towards nucleophilic&#xA;substitution reactions due to the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Resonance effect&lt;/strong&gt; : In haloarenes, the electron pairs on halogen&#xA;atom are in conjugation with π-electrons of the ring and the&#xA;following resonating structures are possible.&#xA;$\ce{C—Cl}$ bond acquires a partial double bond character due to&#xA;resonance. As a result, the bond cleavage in haloarene is difficult&#xA;than haloalkane and therefore, they are less reactive towards&#xA;nucleophilic substitution reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Difference in hybridisation of carbon atom in $\ce{C—X}$ bond&lt;/strong&gt;: In&#xA;haloalkane, the carbon atom attached to halogen is $\mathrm{sp^3}$&#xA;hybridised while in case of haloarene, the carbon atom attached&#xA;to halogen is $\mathrm{sp^2}$-hybridised.&#xA;The $\mathrm{sp^2}$ hybridised carbon with a greater s-character is more&#xA;electronegative and can hold the electron pair of $\ce{C—X}$ bond&#xA;more tightly than $\mathrm{sp^3}$-hybridised carbon in haloalkane with&#xA;less s-chararcter. Thus, $\ce{C—Cl}$ bond length in haloalkane is&#xA;177pm while in haloarene is 169 pm. Since it is difficult to&#xA;break a shorter bond than a longer bond, therefore, haloarenes&#xA;are less reactive than haloalkanes towards nucleophilic&#xA;substitution reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kMDB1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Instability of phenyl cation&lt;/strong&gt;: In case of haloarenes, the phenyl&#xA;cation formed as a result of self-ionisation will not be stabilised&#xA;by resonance and therefore, SN1 mechanism is ruled out.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Because of the possible repulsion&lt;/strong&gt;, it is less likely for the electron&#xA;rich nucleophile to approach electron rich arenes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4231" LastEditDate="2015-03-13T20:55:59.270" LastActivityDate="2015-03-13T20:55:59.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20533" PostTypeId="1" AcceptedAnswerId="20543" CreationDate="2014-12-10T17:33:25.907" Score="0" ViewCount="1722" Body="&lt;p&gt;Consider the molecule representation below. What is the systematic name of that molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WV9ME.png&quot; alt=&quot;molecule&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it possibly be &lt;em&gt;3,4-hex-2,6-anebenzene&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9677" LastEditorDisplayName="user7951" LastEditDate="2015-06-09T19:59:05.943" LastActivityDate="2015-06-09T19:59:05.943" Title="What is the systematic name of a &quot;double benzene&quot;?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20534" PostTypeId="2" ParentId="20502" CreationDate="2014-12-10T17:42:34.360" Score="1" Body="&lt;p&gt;Here's the answer from the key:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}{2}\frac{1}{2}\ce{H2 &amp;amp;-&amp;gt; H+}+e^-  \tag{0V}\\&#xA;\\ \ce{AgCl(s) +e^- &amp;amp;-&amp;gt; Ag(s) +Cl-} \tag{0.22V}\\\\&#xA;&amp;amp;E^o =0.22V\ \text{at 298 K} \\\\&#xA;\ce{2AgCl+H2 &amp;amp;-&amp;gt;Ag(s)+2HCl(aq)} \\\\&#xA;\Delta_rS &amp;amp;= \frac{vF(E^0_1-E^o_2)}{T_1-T_2} \\\\&#xA;\Delta_rS &amp;amp;=S^o_m(\ce{Ag(s)}) +S^o_m(\ce{2HCl(aq)}) - S^o_m(\ce{H2}) -S^o_m(\ce{2AgCl})\\&#xA;&amp;amp;=(2\cdot 42.55+2\cdot 56.5-130.68-2\cdot 96.2) \\ &amp;amp;= -124.98 \\ -124.98 &amp;amp;= \frac{1 \cdot 9.6485\ x \ 10^4 (0.22-E^o)}{298-310} \\ 0.22-\frac{1499.66}{9.6485x10^4} &amp;amp;=&#xA; E^o \\\\ E^o_{(310)} &amp;amp;= 0.204V \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks anyway for the attempts&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-12-16T08:34:08.943" LastActivityDate="2014-12-16T08:34:08.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20535" PostTypeId="2" ParentId="20493" CreationDate="2014-12-10T17:54:37.907" Score="6" Body="&lt;p&gt;The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.2$. The $\mathrm{p}K_\mathrm{a}$ of aniline (aniline's conjugate acid really) is $4.8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two macroscopic $\mathrm{p}K_\mathrm{a}$s of 3-aminobenzoic acid (aniline-3-carboxylic acid) are $3.1$ and $4.8$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the microscopic $\mathrm{p}K_\mathrm{a}$s of two groups in a molecule are similar, the macroscopic $\mathrm{p}K_\mathrm{a}$s are not reflective of one particular group protonating/deprotonating. The neutral form will include both a zwitterionic and an non-zwitterionic form in the case of 3-aminobenzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the groups are close in space within the molecule, there will be electrostatic interaction. If the amino group is protonated, it facilitates deprotonation of the carboxylic acid group. If the carboxylic acid group is deprotonated, it facilitates protonation of the amino group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition:&lt;/strong&gt; There is a very relavent article: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo00178a024&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbon-13 NMR determination of acid-base tautomerization equilibriums&lt;/a&gt; &lt;em&gt;J. Org. Chem.&lt;/em&gt;, &lt;strong&gt;1984&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt; &lt;em&gt;(4)&lt;/em&gt;, pp 691–696.  It reports that the neutral species is 59% zwitterion and 41% molecular (non-zwitterion).&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-04-20T16:23:25.577" LastActivityDate="2018-04-20T16:23:25.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20536" PostTypeId="2" ParentId="20533" CreationDate="2014-12-10T17:58:03.847" Score="2" Body="&lt;p&gt;As @ron already said in his comment, the systematic name of the naphtalene is Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene since you can start the carbon numerotation from 2 chemically equivalent carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find all the references &lt;a href=&quot;http://en.wikipedia.org/wiki/Naphthalene&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.chem.qmul.ac.uk/iupac/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="10119" LastEditDate="2014-12-10T18:16:56.773" LastActivityDate="2014-12-10T18:16:56.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20543" PostTypeId="2" ParentId="20533" CreationDate="2014-12-10T18:44:09.370" Score="-5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I1NqV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical Formula:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{C10H8}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Common Name:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Napthalene&lt;/strong&gt; (also accepted by IUPAC as per 1993 recommendations)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IUPAC Name: (refer numbering in figure)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene&lt;/strong&gt; or &lt;strong&gt;Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For a general guideline:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start with one of the bridgehead carbons and number it 1.&lt;/li&gt;&#xA;&lt;li&gt;Proceed round the longest chain of carbons to the second bridgehead.&lt;/li&gt;&#xA;&lt;li&gt;Number the second bridgehead carbon and continue on round the next longest chain of carbons back towards the first bridgehead carbon.&lt;/li&gt;&#xA;&lt;li&gt;Pass over the first bridgehead carbon (it already has the number 1) and along the shortest chain of carbons to the second bridgehead carbon again.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The generic name for compounds such as this is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;bicyclo[x.y.z]alkane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;where x, y, and z are the numbers of intervening carbons on the three paths between the two bridgehead carbons cited in decreasing numerical order, and alk refers to the total number of carbons in the ring systems.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2014-12-10T19:17:44.050" LastActivityDate="2014-12-10T19:17:44.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20544" PostTypeId="1" AcceptedAnswerId="21591" CreationDate="2014-12-10T18:51:08.707" Score="2" ViewCount="3821" Body="&lt;p&gt;From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling &quot;together&quot;). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-12T11:57:55.770" Title="Magnetic moments of tetrahedral Cobalt (II) and (III) complexes" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20545" PostTypeId="2" ParentId="20414" CreationDate="2014-12-10T19:26:22.387" Score="6" Body="&lt;p&gt;The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a good explanation of the physics behind these effects on &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="5026" LastEditDate="2018-03-03T05:05:31.700" LastActivityDate="2018-03-03T05:05:31.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20547" PostTypeId="1" AcceptedAnswerId="20552" CreationDate="2014-12-10T20:13:04.737" Score="9" ViewCount="149" Body="&lt;p&gt;I just read &lt;a href=&quot;http://blogs.sciencemag.org/pipeline/archives/2014/12/10/a_new_route_to_spiro_heterocycles&quot; rel=&quot;noreferrer&quot;&gt;this In the Pipeline post&lt;/a&gt; and I was slightly confused by a statement on the use of tin. Lowe reports on &lt;a href=&quot;http://dx.doi.org/10.1021/ja511232b&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;, which describes a synthetic route to spiro heterocycles using tin compounds, and caveats the finding with the comment that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No one's crazy about using tin, but the transformation is too useful to pass up.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is the use of tin undesirable in synthetic chemistry?&lt;/p&gt;&#xA;" OwnerUserId="1327" LastEditorUserId="16683" LastEditDate="2017-04-08T23:28:24.460" LastActivityDate="2019-03-26T16:14:34.623" Title="What's wrong with using tin in medicinal chemistry?" Tags="&lt;metal&gt;&lt;toxicity&gt;&lt;organometallic-compounds&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20548" PostTypeId="2" ParentId="20486" CreationDate="2014-12-10T20:17:08.540" Score="8" Body="&lt;p&gt;There are general problems representing delocalized bonding and organometallic compounds in cheminformatics (generally) and SMILES in specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote the &lt;a href=&quot;http://www.opensmiles.org/opensmiles.html&quot; rel=&quot;noreferrer&quot;&gt;Open SMILES standard&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This simple mental model [of a connection table between atoms using valence bonds] has little resemblance to the underlying quantum-mechanical reality of electrons, protons and neutrons, yet it has proved to be a remarkably useful approximation of how atoms behave in close proximity to one another. However, the valence model is an imperfect representation of molecular structure, and the SMILES language inherits these imperfections. Chemical bonds are often tautomeric, aromatic or otherwise fractional rather than neat integer multiples. Delocalized bonds, bond-centered bonds, hydrogen bonds and various other inter-atom forces that are well characterized by a quantum-mechanics description simply don’t fit into the valence model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other interesting cases are &lt;a href=&quot;http://en.wikipedia.org/wiki/Agostic_interaction&quot; rel=&quot;noreferrer&quot;&gt;agostic interactions&lt;/a&gt;, 3-center 2-electron bonds (e.g., &lt;a href=&quot;http://en.wikipedia.org/wiki/Diborane&quot; rel=&quot;noreferrer&quot;&gt;diborane&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several suggestions for advancing beyond the typical connection-table valence bond view, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ci00027a001&quot; rel=&quot;noreferrer&quot;&gt;&quot;Yet Another Representation of Molecular Structure&quot; &lt;em&gt;J. Chem. Inf. Comput. Sci.&lt;/em&gt;, &lt;strong&gt;1995&lt;/strong&gt;, 35 (5), pp 787–802&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ci9704423&quot; rel=&quot;noreferrer&quot;&gt;&quot;Overcoming the Limitations of a Connection Table Description:  A Universal Representation of Chemical Species&quot; &lt;em&gt;J. Chem. Inf. Comput. Sci.&lt;/em&gt;, &lt;strong&gt;1997&lt;/strong&gt;, 37 (4), pp 705–714&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That both papers were published 17+ years ago should be some indication of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chemists like the valence bond view of molecules, despite the limitations&lt;/li&gt;&#xA;&lt;li&gt;How complicated bonding can be in non-organic chemistry&lt;/li&gt;&#xA;&lt;li&gt;How far cheminformatics needs to go to properly handle organometallic and non-organic compounds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By the way... anyone interested in implementing such efforts should let me know.. Many people would be happy to help. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should point out that there are some practical solutions to representing such things. For example Alex Clark suggested recording zero-order bonds (and sometimes dummy atoms):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ci200488k&quot; rel=&quot;noreferrer&quot;&gt;&quot;Accurate Specification of Molecular Structures: The Case for Zero-Order Bonds and Explicit Hydrogen Counting&quot; &lt;em&gt;J. Chem. Inf. Model.&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, 51 (12), pp 3149–3157&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-12-11T16:58:46.337" LastActivityDate="2014-12-11T16:58:46.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20552" PostTypeId="2" ParentId="20547" CreationDate="2014-12-10T22:20:01.033" Score="8" Body="&lt;p&gt;Organotin compounds are rather toxic.  They are also persistent in the environment and have a long biological half-life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that trialkyltin byproducts from your reaction are difficult to separate from the product.  In the lab, this is painful and (usually) involves multiple columns, but when you want to get a drug past the FDA into the clinic you have to meet purity standards for heavy metal contamination.  To a medicinal chemist, a transformation that won't make it into pilot plant and production is useless.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastEditorUserId="23561" LastEditDate="2019-03-26T16:14:34.623" LastActivityDate="2019-03-26T16:14:34.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20553" PostTypeId="1" CreationDate="2014-12-10T22:43:01.393" Score="4" ViewCount="143" Body="&lt;p&gt;An x-ray computed tomography scanner has been used in an attempt to look at the pore space of a limestone rock sample (a carbonate, $\ce{CaCO3}$) saturated with a $\mathrm{12\ wt\%}$ sodium iodide ($\ce{NaI}$) solution as the x-ray dopant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that the sodium iodide would absorb the x-rays more so than the grain matrix of the rock and thus the fluid in the pore space of the rock can be seen, by contrast in x-ray absorption. Therefore, the primary entity that is measured through CT scanning is x-ray absorption, defined as the attenuation coefficient.  This is determined both by mass density and elemental composition of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this doped fluid wasn't able to be seen. Next tests will use a chalk (composed ~90% or greater of $\ce{CaCO3}$). Typically for tests of this nature sodium iodide is preferred because its interaction with sorts of clay is similar to that of sodium chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the highest attainable concentration of a sodium iodide solution?&lt;/li&gt;&#xA;&lt;li&gt;It is known that a solution of barium chloride has been used an x-ray dopant. Would this solution be compatible with chalk, i.e. non-reactive with chalk?  What would its highest attainable concentration be?&lt;/li&gt;&#xA;&lt;li&gt;What other possible solutions can be used as x-ray dopants?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;NIST has &lt;a href=&quot;http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;values of mass attenuation coefficients by element&lt;/a&gt; and provides a very useful &lt;a href=&quot;http://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;tool for looking up the specific attenuation coefficient for mixtures&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My quest is to understand what x-ray dopant solutions are possible and what their highest concentrations achievable are. Then to rule out ones that are reactive with the rock.&lt;/p&gt;&#xA;" OwnerUserId="6956" LastEditorUserId="23561" LastEditDate="2018-08-13T03:17:25.643" LastActivityDate="2018-08-13T03:17:25.643" Title="Best x-ray dopant?" Tags="&lt;aqueous-solution&gt;&lt;analytical-chemistry&gt;&lt;geochemistry&gt;&lt;minerals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="20554" PostTypeId="2" ParentId="19937" CreationDate="2014-12-10T22:59:10.180" Score="1" Body="&lt;p&gt;The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on a basic level, the net charge does not affect the overall nuclear spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The somewhat longer answer is that the nuclear spin states depend on the local chemical environment. This means that we can &quot;see&quot; different proton nuclear spins based on the molecular structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's are unpaired electrons, &lt;a href=&quot;http://en.wikipedia.org/wiki/Paramagnetic_NMR_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;paramagnetic NMR&lt;/a&gt; requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two main effects in paramagnetic NMR are pseudo-contact and contact shifts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-10T22:59:10.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20555" PostTypeId="2" ParentId="20237" CreationDate="2014-12-10T23:05:12.590" Score="5" Body="&lt;p&gt;I approach questions like this by considering all possibilities of arrangements of the non-hydrogen atoms. Obviously, there's C-C-O, C-O-C, and then the cyclic ethylene oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I make sure I've matched bond orders to get the hydrogen count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I get vinyl alcohol, acetaldehyde, and ethylene oxide. H2C-O-CH2 would only work for a biradical, which is probably not expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The systematic fashion then is to arrange the &quot;heavy&quot; atoms, first in all linear forms, then all cyclic forms, then all two-ring forms, etc. Of course with large formulas, the number of isomers goes up exponentially.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-10T23:05:12.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20556" PostTypeId="2" ParentId="19604" CreationDate="2014-12-10T23:33:31.530" Score="3" Body="&lt;p&gt;The best thing to do is load up the CIF at the Crystallography Open Database (COD), e.g.&#xA;&lt;a href=&quot;http://www.crystallography.net/1010977.html?cif=1010977&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crystallography.net/1010977.html?cif=1010977&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JSmol used for the visualization lets you visualize the symmetry and adjust the packing. In this case, look for the inversion center, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UoApF.png&quot; alt=&quot;inversion center&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inversion center itself is the yellow ball. But it's probably easier to see it in the inverted Pb centers I've circled in yellow. You can see that there's an exact inversion center there.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-10T23:33:31.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20557" PostTypeId="2" ParentId="19202" CreationDate="2014-12-10T23:41:14.693" Score="3" Body="&lt;p&gt;Neutral detergents are not neccesarily at pH 7, and I know of no legal restriction (at least in the EU) on labelling detergents as neutral based on pH. For example, &lt;a href=&quot;http://www.google.com/patents/EP2596090A1?cl=en&quot; rel=&quot;nofollow&quot;&gt;this patent&lt;/a&gt; claims protection for a set of neutral detergent formulae with pH of &quot;from about 5.5 to about 8.5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a commercial product is Proctor and Gamble's &quot;Tide&quot;, which has a pH of 8.1-8.5 according to &lt;a href=&quot;http://www.pg.com/productsafety/msds/professional_line/professional_line/PG_ProLine_Tide_Professional_Laundry_Detergent.pdf&quot; rel=&quot;nofollow&quot;&gt;their safety data.&lt;/a&gt; This is advertised as &quot;100%&quot; neutral ! &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-10T23:41:14.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20558" PostTypeId="2" ParentId="19618" CreationDate="2014-12-11T00:18:23.660" Score="2" Body="&lt;p&gt;Certain types of &lt;strong&gt;&lt;em&gt;wet&lt;/em&gt;&lt;/strong&gt; etching agents have differential reactivity on different crystal surfaces. Consider that the Si-Si distance and coordination environment will be quite different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start with an isotropic etch - it just eats silicon. Well, if you started with a pinprick (i.e., just a point and not a gap in the mask) you'd etch out a hemisphere. Normally, we etch through gaps in the mask, so you'll get rounded sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an &lt;a href=&quot;http://en.wikipedia.org/wiki/Etching_(microfabrication)#Anisotropic_wet_etching_.28Orientation_dependent_etching.29&quot; rel=&quot;nofollow&quot;&gt;anisotropic wet etch&lt;/a&gt; you can get flat, angular side-walls defined by crystal planes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For instance, potassium hydroxide (KOH) displays an etch rate selectivity 400 times higher in &amp;lt;100&gt; crystal directions than in &amp;lt;111&gt; directions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;OK, so let's consider that etch through a gap in a mask into a &amp;lt;100&gt; surface. As we go down 400 atomic layers of Si, we'll clearly have some etching along the side-walls (e.g., an &quot;undercut&quot;). But also consider that depending on the size of the feature for the etch, you'll go down into the &amp;lt;100&gt; but not completely uniformly. Different wet anisotropic etches will have different selectivity rates between &amp;lt;100&gt; and &amp;lt;111&gt; and thus you get different aspect ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, &lt;a href=&quot;https://www.memsnet.org/mems/processes/etch.html&quot; rel=&quot;nofollow&quot;&gt;dry etching&lt;/a&gt; (e.g., RIE, plasma) can give more vertical side-walls.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-11T00:18:23.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20559" PostTypeId="2" ParentId="19202" CreationDate="2014-12-11T00:24:33.147" Score="2" Body="&lt;p&gt;I want to add some imformation to J.LS answer. According to U.S. General Services Administration, neutral detergents are produced with strong alkalis and weak acids. They are the most common soapless detergent and are usually alkylbenzenesulfonates. So, neutral detergent is not that neutral after all.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2019-05-05T03:27:06.433" LastActivityDate="2019-05-05T03:27:06.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20560" PostTypeId="1" AcceptedAnswerId="21567" CreationDate="2014-12-11T01:37:27.980" Score="4" ViewCount="1114" Body="&lt;p&gt;I had an interesting event occur recently: I stuck a screwdriver in a potted plant, for it is the only way that I am able to unlock a certain door. It was left in there for about a month before I needed it again, and, when I took it out, it had what I thought to be rust. I am doubting my original assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for a fact that it is &lt;em&gt;not&lt;/em&gt; dirt itself, as I vigorously attempted to clean it with suds (though I'm not entirely sure whether or not that was the best approach...).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Does iron &lt;code&gt;a)&lt;/code&gt; oxidize in dirt and &lt;code&gt;b)&lt;/code&gt; if a), does it oxidize &lt;em&gt;faster&lt;/em&gt; in dirt?&lt;/p&gt;&#xA;" OwnerUserId="10264" LastActivityDate="2014-12-11T08:07:18.740" Title="Does metal oxidize (faster) in dirt?" Tags="&lt;metal&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20561" PostTypeId="1" AcceptedAnswerId="20565" CreationDate="2014-12-11T02:01:32.710" Score="2" ViewCount="4719" Body="&lt;p&gt;As most of you know, this is the diagram for nitrate ion: &lt;img src=&quot;https://i.stack.imgur.com/DaQWz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To draw this, i have to make all bonds with electrons from nitrogen, nitrogen only has five valence electron so ONE of the bonds consists of one electron from nitrogen and the other from the MYSTERIOUS electron, that comes out of no where. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book tells me that's if simply how you draw Lewis diagram for N, O, F, Cl. When the oxidation is positive, take away electron and when it is negative, add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any covalent compound, bonding atoms share their electron to complete octet.( yeh, there are exceptions) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, N and O share their electron. But when I count the number of valence electrons, a negative charge means a extra valence electron. NOW, where does this extra valence electron come from?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-12-14T17:56:48.700" LastActivityDate="2014-12-14T17:56:48.700" Title="Where does the extra electron come from in the Lewis Dot Electron structure of say for example nitrate ion?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20562" PostTypeId="1" CreationDate="2014-12-11T02:19:33.873" Score="-2" ViewCount="1694" Body="&lt;p&gt;Can anyone kindly help me to name this compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GjsHC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjsHC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the above compound &lt;strong&gt;3-Ethyl-1,1-dimethylcyclohexane&lt;/strong&gt; or &lt;strong&gt;1-Ethyl-3,3-dimethylcyclohexane&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7760" LastEditorDisplayName="user7951" LastEditDate="2018-02-14T12:00:26.717" LastActivityDate="2018-02-14T12:00:26.717" Title="Which is the correct numbering of locants for ethyldimethylcyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20563" PostTypeId="2" ParentId="20562" CreationDate="2014-12-11T02:24:55.677" Score="0" Body="&lt;p&gt;3-Ethyl-1,1-dimethylcyclohexane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the IUPAC rules, the correct name will have the lowest numbers for the substituents. Therefore, 1,1, and 3 is lower than 3,3 and 1.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2018-01-02T14:24:48.513" LastActivityDate="2018-01-02T14:24:48.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20565" PostTypeId="2" ParentId="20561" CreationDate="2014-12-11T04:03:21.867" Score="6" Body="&lt;p&gt;Unfortunately your Lewis structure for the nitrate ion, $\ce{NO3-}$, is wrong. You did not include any charges and as it is an ion there is at least one. Let's first have a look at how an ion may be formed and then hopefully you see, that there is no mystery at all about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example ordinary table salt, which has the chemical formula $\ce{NaCl}$. If you dissolve it in water the lattice will be broken and the ions will by hydrated and are now able to move more freely.&#xA;$$\ce{NaCl_{(s)} -&amp;gt; Na+{}_{(aq)} + Cl- {}_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same happens when you dissolve potassium nitrate, which releases the nitrate ion:&#xA;$$\ce{KNO3_{(s)} -&amp;gt; K+{}_{(aq)} + NO3- {}_{(aq)}}$$&#xA;Here you can see, that the nitrate ion has an extra electron which it has taken from the potassium. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_bonding&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic bonding&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Lewis structure of the nitrate ion you first have to count all the valence electrons of the involved atoms:&#xA;\begin{array}{lclr}&#xA;1 &amp;amp;\times &amp;amp;\ce{N} &amp;amp; 5\\&#xA;3 &amp;amp;\times &amp;amp;\ce{O} &amp;amp; 6\\&#xA;1 &amp;amp;\times &amp;amp;\text{charge} &amp;amp; 1\\\hline&#xA;  &amp;amp;       &amp;amp;              &amp;amp; 24\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Lewis structure you have to distribute 24 electrons in a fashion, that you do not violate the rule of two, which is better known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;octet rule&lt;/a&gt;. Therefore your structure may include a nitrogen oxygen double bond, a positive formal charge on nitrogen and a negative formal charge at each of the singly bound oxygens.&lt;br&gt;&#xA;Another representation is that all oxygens are singly bound and therefore the nitrogen has a $+2$ formal charge and each of the oxygens has a negative formal charge. The problem with this structure is that nitrogen does not have a full octet. Nevertheless, it is a strong contributor to the actual electronic structure. The bonding situation is much more complicated than can be expressed through a Lewis structure. The concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;resonance structures&lt;/a&gt; has to at least be included. Better in this case would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular orbital theory&lt;/a&gt; and or &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;valence bond theory&lt;/a&gt;. The molecule itself is a y-aromatic system, the charge is delocalised through the extensive $\pi$ system. The blue structure is an attempt of showing this feature.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Th3Jh.jpg&quot; alt=&quot;resonance structures of nitrate&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-11T04:03:21.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20566" PostTypeId="2" ParentId="19983" CreationDate="2014-12-11T04:47:12.097" Score="3" Body="&lt;p&gt;First off, let's define our terms. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Quasistatic_approximation&quot; rel=&quot;nofollow&quot;&gt;quasistatic approximation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;More generally, the quasistatic approximation can be understood through the idea that the sources in the problem change sufficiently slowly that the system can be taken to be in equilibrium at all times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the question boils down to &quot;are there ever any quantum mechanical behaviors that are non-equilibrium?&quot; Or. more precisely, are there fast changes where the system is not really in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, as noted in the comments, there's the Born-Oppenheimer approximation which, at least in my opinion, reflects a quasi-static view of electronic structure. That is, the nuclei move slowly relative to the electrons, so despite vibrations, etc. the electrons can be considered in equilibrium with nuclear positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the Born-Oppenheimer approximation fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poster suggested the question was about other non-static behavior. In that case, you fall into time-dependent quantum mechanics, e.g. time-dependent Hartree Fock (i.e., RPA) or &lt;a href=&quot;http://en.wikipedia.org/wiki/Time-dependent_density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;TD-DFT&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Linear-response TDDFT can be used if the external perturbation is small in the sense that it does not completely destroy the ground-state structure of the system. In this case one can analyze the linear response of the system. This is a great advantage as, to first order, the variation of the system will depend only on the ground-state wave-function.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are there cases where linear response breaks and the variation in the system doesn't depend only on the ground-state wave-function. Yes. Consider the effects of large perturbations, like large electric fields, &lt;a href=&quot;http://en.wikipedia.org/wiki/Nonlinear_optics&quot; rel=&quot;nofollow&quot;&gt;nonlinear optical&lt;/a&gt; effects (two photo absorption), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another fairly easy example, consider photo-ionization. That is, remove an electron from a system. In the quasi-static approximation, you'd have something like &lt;a href=&quot;http://en.wikipedia.org/wiki/Koopmans%27_theorem&quot; rel=&quot;nofollow&quot;&gt;Koopmans' theorem&lt;/a&gt; with frozen ground-state orbitals. In reality, there's orbital relaxation even though the ionization process is very fast.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-11T04:47:12.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20567" PostTypeId="2" ParentId="11048" CreationDate="2014-12-11T05:04:10.393" Score="4" Body="&lt;p&gt;It depends, because for covalent there are two types of bonds, network or molecular, or as I have also heard it be called, polar covalent and non polar covalent. But, network covalent consists of a vast network among the atoms and each one are connected, and they mostly made up of one element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a diamond for example, it is only made up of carbon, but since the atoms are connected to each other and have not bonding between molecules, like something such as salt, which is an ionic bond, it is harder to break. However, if it were a molecular covalent bond, then the whole story is different, because they tend to be very weak bonds and easily broken like sugar or otherwise known as glucose, sucrose doesn't matter, it is still a covalent molecular bond because they have molecules while as a diamond is technically one big molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, since sugar has multiple the bond between each other molecule is weaker than the bonds between the elements themselves, then it is really weak.&lt;/p&gt;&#xA;" OwnerUserId="10267" LastEditorUserId="5456" LastEditDate="2014-12-11T10:17:47.357" LastActivityDate="2014-12-11T10:17:47.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21567" PostTypeId="2" ParentId="20560" CreationDate="2014-12-11T08:07:18.740" Score="6" Body="&lt;p&gt;Iron can corrode faster in soil than in air, particularly if partly immersed in the soil. Moisture and electrolyte are retained in the soil, making it a good anode. Also, microbes in the soil can accelerate corrosion.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2014-12-11T08:07:18.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21568" PostTypeId="1" CreationDate="2014-12-11T08:43:50.553" Score="7" ViewCount="290" Body="&lt;p&gt;I'm experimenting with water electrolysis and wanted to use non-toxic materials, so I settled on pure graphite rods for both electrodes and Epsom salt (magnesium sulfate) as the electrolyte.  The system works reasonably well, but upon further reading I found that some of the $\ce {O2}$ is likely reacting with the graphite and being converted into $\ce {CO}$ and $\ce {CO2}$ at the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much $\ce {O2}$ am I probably losing to $\ce {CO}$ and $\ce {CO2}$ ?  Just a small percentage, or a majority of the potential $\ce {O2}$ production?  How can I tell?  I'm capturing the resulting mixed gas product in soap bubbles and igniting it but it seems to burn somewhat less vigorously than I was expecting, which makes me think there is rather more $\ce {H2}$ and less $\ce {O2}$ than optimal.&lt;/p&gt;&#xA;" OwnerUserId="11268" LastEditorUserId="10119" LastEditDate="2014-12-11T09:26:49.457" LastActivityDate="2014-12-11T09:26:49.457" Title="How much CO2 is created during water electrolysis with a graphite anode?" Tags="&lt;electrolysis&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21569" PostTypeId="1" CreationDate="2014-12-11T10:26:48.863" Score="2" ViewCount="172" Body="&lt;p&gt;I have searched Google for this but cannot seem to find an explanation. My question is as follows - what is the Chemistry/Biochemistry behind cis and trans fatty acids resulting in &quot;Bad&quot; LDL and &quot;Good&quot; HDL cholesterol? Is one synthesised favourably with cis/trans fatty acids as precursors? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am under the impression that LDL results in reduced cell permeability (how?), but other than that I am struggling to find information.&lt;/p&gt;&#xA;" OwnerUserId="9785" LastEditorUserId="15410" LastEditDate="2015-05-07T10:46:14.137" LastActivityDate="2015-05-07T10:46:14.137" Title="What is the Chemistry behind &quot;good&quot; and &quot;bad&quot; cholesterol &amp; their relation to Cis/Trans fatty acids?" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21570" PostTypeId="1" AcceptedAnswerId="21573" CreationDate="2014-12-11T10:30:14.413" Score="3" ViewCount="623" Body="&lt;p&gt;In chemistry lab we perform experiments to find the unknown solution. I just want to produce some fumes in lab. We are only allowed to use few common dilute solution like $\ce{NH_4Cl}$, $\ce{NH_4OH},$ $\ce{HCl}$, and few more which are generally allowed to students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to form fumes in lab? &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="418" LastEditDate="2014-12-11T12:14:38.743" LastActivityDate="2014-12-11T12:22:34.877" Title="How to produce fumes in chemistry lab?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21571" PostTypeId="1" AcceptedAnswerId="21572" CreationDate="2014-12-11T11:12:00.510" Score="-1" ViewCount="2730" Body="&lt;p&gt;Today I created $1~\mathrm{M}$ aqueous solution of $\ce{CaCl2}$ and realized that the water before the mixture had a temperature of 23.4 Celsius degrees. Then I added the $\ce{CaCl2}$ and the temperature started growing until it reached 27.5 Celsius degrees. So we have an exothermic reaction. What does that actually mean and how does it work in my example? By the book, an exothermic reaction occurs, when the energy stored in reactants is smaller than the energy stored in products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have (with my opinion):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCl2 + H2O &amp;lt;=&amp;gt; Ca^{2+} + 2Cl^{-} + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From where did that energy come from? Is it just released from the fission of $\ce{CaCl2}$ reactant or from the fusion of some molecules of $\ce{CaCl2}$ again in products? Also, is it possible to have in the products side $\ce{CaCl2}$ molecules or only ions of $\ce{Ca}$ and $\ce{Cl}$?&lt;/p&gt;&#xA;" OwnerUserId="7800" LastEditorUserId="4945" LastEditDate="2014-12-11T11:33:23.180" LastActivityDate="2014-12-11T11:33:23.180" Title="Where does the energy in the exothermic reaction of CaCl2 with water come from?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21572" PostTypeId="2" ParentId="21571" CreationDate="2014-12-11T11:16:23.283" Score="1" Body="&lt;p&gt;Actually, exothermic means that energy is released so that the initial energy is higher than the final one. That's say the contrary of your assumption.  This energy is contained in the atoms states themself and in the bondings. So when you break the bonds to form another compound, you change the energy state of your atoms, releasing energy if the final state is more stable (less energetic). Since in your reaction it's the case, the reaction happens then spontaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you should keep in mind that nothing is lost, everything is transformed, this extra energy remaining is transformed in heat. &lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2014-12-11T11:16:23.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21573" PostTypeId="2" ParentId="21570" CreationDate="2014-12-11T12:22:34.877" Score="2" Body="&lt;p&gt;Aqueous ammonia and hydrochloric acid are solutions of gases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;strong&gt;fume hood&lt;/strong&gt; place two beakers with aliquots of these two solutions (one in each beaker, &lt;strong&gt;don't mix&lt;/strong&gt;) side to side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Did I mention the fume hood?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll see the formation of solid ammonium chloride from the reaction of the two gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl(g) + NH3(g) -&amp;gt; NH4Cl(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't forget to clean up the mess!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-11T12:22:34.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21574" PostTypeId="1" CreationDate="2014-12-11T12:42:31.063" Score="2" ViewCount="3621" Body="&lt;p&gt;Can anybody share the symmetry labels for acetonitrile (&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CN}$&lt;/span&gt;), please?&#xA;I mean to assign the symmetry labels to orbitals of acetonitrile&lt;/p&gt;&#xA;" OwnerUserId="11270" LastEditorUserId="69830" LastEditDate="2021-05-07T06:14:53.543" LastActivityDate="2021-05-07T06:14:53.543" Title="Symmetry representations for acetonitrile (CH3CN)" Tags="&lt;physical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;symmetry&gt;&lt;group-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21575" PostTypeId="2" ParentId="20275" CreationDate="2014-12-11T13:21:33.393" Score="1" Body="&lt;p&gt;I would expect that the amino phenol is more polar than the phenol. This due to the fact that $\ce{NH2}$ donates electrons to the Benzene ring, making the difference in charge between the Benzene Carbon and the Oxygen stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Chlorine atom is less electro negative than oxygen (if I'm not mistaken) so that would mean: Naphatlene &amp;lt; Chlorobenzene &amp;lt; Phenol &amp;lt; Amino Phenol.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-12-11T13:21:33.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21576" PostTypeId="1" CreationDate="2014-12-11T13:57:41.343" Score="2" ViewCount="610" Body="&lt;p&gt;I did not completely understand the mechanism of Cannizzaro's reaction, especially the rate determining step i.e. the hydride transfer step, and the effect of electron withdrawing substituents on this step.&#xA;Like, I was taught that the intermediate which donates hydride, would donate hydride more readily when electron donating substituents were attached to carbon atom with the outgoing hydrogen atom, as they would increase electron density on that carbon, and therefore create a &quot;pushing&quot; effect on the outgoing hydride. But in all the questions I have attempted, the answer says that electron WITHDRAWING substituents facilitate hydride transfer.&#xA;Please tell how the electron withdrawing substituents affect the hydride transfer, and why. Also, please explain how to determine the products of a crossed Cannizzaro's reaction. Do we determine them using the concepts of the RDS,or otherwise?  &lt;/p&gt;&#xA;" OwnerUserId="9984" LastEditorUserId="9984" LastEditDate="2014-12-14T15:26:48.253" LastActivityDate="2014-12-14T15:26:48.253" Title="Please explain the mechanism of Cannizzaro's reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21577" PostTypeId="2" ParentId="7096" CreationDate="2014-12-11T14:14:12.217" Score="2" Body="&lt;p&gt;The wikipedia page on &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Phosphorescence&quot; rel=&quot;nofollow&quot;&gt;Phosphoresence&lt;/a&gt; has a red pigment. Calcium Sulfide. I can imagine that the lower energy (higher wavelength, lower energy) is more stable because the triplet is not as high in state energy as lower wavelength emitters. Because of the stability the time to emit is longer and the intensity lower.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditDate="2014-12-11T14:20:14.523" LastActivityDate="2014-12-11T14:20:14.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21578" PostTypeId="1" CreationDate="2014-12-11T14:19:14.273" Score="1" ViewCount="225" Body="&lt;p&gt;The following passage has been extracted from the book &quot;Organic Chemistry-10th edition by Solomon and Frhyle&quot;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The growth of living things from microbes to elephants rests on organic reactions, and organic reactions provide the energy that drives our muscles and &lt;em&gt;our thought processes&lt;/em&gt;&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the above statement true that, organic reactions provide energy that drives our thought processes? Is there any other process by which we are given energy to drive our thought processes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you all agree with the statement, then when a person is called &quot;dead&quot; (or after a while being called dead), I hope the reactions will cease (because he can't perform &quot;thought processes&quot;), isn't it?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these all points are agreed, is their a possibility of repairing a dead body by restarting the reactions immediately?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that, it is not as easy as I said, there is a complexity, but I believe we all can do some research &quot;together&quot; and try to do a &quot;little&quot; contribution to save ourselves.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can we repair a dead body?  &lt;/p&gt;&#xA;" OwnerUserId="2328" LastActivityDate="2015-08-02T13:59:36.357" Title="Can we repair a dead body?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-08-02T21:08:49.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="21579" PostTypeId="2" ParentId="21574" CreationDate="2014-12-11T15:28:23.193" Score="4" Body="&lt;p&gt;It depends a bit on how you handle the $\ce{CH3}$ group. Usually, you assume free rotation, so it's effectively a sphere. In that case, $\ce{CH3CN}$ is linear and asymmetric. Since the two ends are different, that places it in the $C_{\infty v}$ point group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, though, we don't assign molecular orbitals using $C_{\infty v}$ or $D_{\infty h}$ representations. It's hard to do the math when using an infinite rotation axis, so we usually use $C_{2v}$ and $D_{2h}$ instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, you'll want to use the irreducible representations from $C_{2v}$ for acetonitrile.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-12-11T17:14:02.043" LastActivityDate="2014-12-11T17:14:02.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21580" PostTypeId="1" AcceptedAnswerId="21587" CreationDate="2014-12-11T15:47:52.513" Score="1" ViewCount="145" Body="&lt;p&gt;&lt;br/&gt;&#xA;first of all I'm not not sure to be in the right stack exchange site, but I'm a material engineering student and this year I have a project: choose a material for a container filled with Sodium hypochlorite, my university has given us CES Edupack, that contains a lot of infomartion about mechanical and physical properties, but nothing about material stability.&lt;br/&gt;&#xA;Since we haven't done any lecture about materials stability/decomposition, I'm here to ask if there is someone that could enlight me about how to proceed to evaluate the stability of a material.&lt;br/&gt;The only thing that I came across while thinking about it is to avoid materials(polymers) that has got active groups with Sodium hypochlorite, is it a good approch?&#xA;&lt;br/&gt;&lt;br/&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;br/&gt;&#xA;Maybe my question is a little dispersive so I make it clear here: my main concert is more general: how do I study the stability of sodium hypochlorite inside the container? What should I check to be sure that the container doesn't &quot;melt&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11271" LastEditorUserId="11271" LastEditDate="2014-12-11T19:09:27.307" LastActivityDate="2014-12-11T19:09:27.307" Title="How to study Sodium hypochlorite stability with materials" Tags="&lt;materials&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21581" PostTypeId="1" AcceptedAnswerId="21583" CreationDate="2014-12-11T16:08:42.847" Score="5" ViewCount="9467" Body="&lt;p&gt;In hydrolysis/nucleophilic substitution of haloalkanes, the bond enthalpy indicates the rate of hydrolysis (e.g. the $\ce {C-I}$ bond is weaker than the $\ce {C-Cl}$ bond so in hydrolysis the alkane with the $\ce {C-I}$ bond will break more quickly and react with the $\ce {OH-}$ ions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in free radical substitution reactions, if you have a molecule like $\ce {CCl2F2}$ , then is the $\ce {C-F}$ bond stronger or weaker than the $\ce {C-Cl}$ bond (so which will require less energy to break)? &lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="7475" LastEditDate="2016-12-22T19:22:12.750" LastActivityDate="2018-03-31T13:11:11.217" Title="Is a carbon-fluorine bond stronger than a carbon-chlorine bond?" Tags="&lt;bond&gt;&lt;enthalpy&gt;&lt;halides&gt;&lt;covalent-compounds&gt;&lt;radicals&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21582" PostTypeId="2" ParentId="20163" CreationDate="2014-12-11T16:53:04.597" Score="14" Body="&lt;p&gt;There are many good review articles, but I really like the freely available &lt;a href=&quot;http://dft.uci.edu/doc/g1.pdf&quot; rel=&quot;noreferrer&quot;&gt;&quot;The ABC of DFT&quot; by Kieron Burke (and friends)&lt;/a&gt; and a recent &lt;a href=&quot;http://dft.uci.edu/pubs/PGB14.pdf&quot; rel=&quot;noreferrer&quot;&gt;review article&lt;/a&gt; that tries to keep to understandable language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is treating the Coulomb interactions. Conventional DFT is not &quot;asymptotically correct.&quot; That is, at long range, the 1/r behavior is not met, and dispersion interactions (e.g., van der Waals forces) are not properly handled either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in Neon atom (from ABC of DFT by Burke):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vm6G3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of course is that most DFT methods start from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Local-density_approximation&quot; rel=&quot;noreferrer&quot;&gt;local density approximation (LDA)&lt;/a&gt; which only considers the electron density at a particular (local) point. LDA indicates that the long-range behavior falls off exponentially instead of 1/r:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{xc}^{\mathrm{LDA}}(\mathbf{r}) = \frac{\delta E^{\mathrm{LDA}}}{\delta\rho(\mathbf{r})} = \epsilon_{xc}(\rho(\mathbf{r})) + \rho(\mathbf{r})\frac{\partial \epsilon_{xc}(\rho(\mathbf{r}))}{\partial\rho(\mathbf{r})}\ $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some recent efforts have used &quot;range separated&quot; functionals, e.g., &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{r} = \frac{1-g(r)}{r} + \frac{g(r)}{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first (short-range) term can be handled by the generalized gradient approximation (GGA) used by basically all modern functionals. The second term is usually handled by Hartree-Fock (i.e., exact exchange), and there's some function g(r) which smoothly scales between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think modern range-separated hybrid functionals are quite good, but choosing a functional depends a bit on the property that you want to predict. Thermodynamics? Excitation energies? Oxidation/reduction potentials from orbital eigenvalues? etc.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-11T16:53:04.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21583" PostTypeId="2" ParentId="21581" CreationDate="2014-12-11T17:00:32.950" Score="9" Body="&lt;p&gt;For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rationalized by considering the poorer orbital overlap between the $\ce {Cl}$ bonding orbital (3 s/p character, more diffuse than 2s/p) and the $\ce {C}$ $\mathrm{sp^3}$ hybrid (predominantly $\text{2p}$ character) than that between the $\ce {F}$ bonding orbital (2 s/p character) and the $\ce {C}\ \mathrm{sp^3}$ hybrid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some data on typical BDE values in $\ce {F/Cl}$ compounds can be found &lt;a href=&quot;http://lectka.chemistry.jhu.edu/assets/Bond-Strength-Files/C-X-bond-str.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="7475" LastEditDate="2016-12-22T19:23:33.797" LastActivityDate="2016-12-22T19:23:33.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21584" PostTypeId="1" AcceptedAnswerId="21710" CreationDate="2014-12-11T17:56:09.653" Score="4" ViewCount="1169" Body="&lt;p&gt;I am an A-level student going into university and I am looking for a source of organic synthesis problems and answers that is freely available on the internet. The difficulty level I am looking for is something like first year undergraduate level. Any links would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-14T20:11:10.473" Title="Organic synthesis problems" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21587" PostTypeId="2" ParentId="21580" CreationDate="2014-12-11T18:50:34.017" Score="1" Body="&lt;p&gt;Sodium Hypochlorite  is soluble in water and reacts with all organic solvents. Sodium hypochlorite is extensively used in house-holds as a bleaching agent and it is a chemical compound with formula $\ce{NaClO}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Household sodium hypochlorite bleach was introduced to Americans in 1909 and sold in steel containers, then in glass bottles. In the early 1960s, the introduction of the plastic jug brought a cheaper, lighter, and nonbreakable packaging alternative. It reduced transportation costs and protected the safety of workers involved in its shipping and handling. Additionally, the thick plastic did not permit ultraviolet light to reach the bleach, which improved its chemical stability and effectiveness. In recent years, how-ever, plastic containers have become an environmental concern because of the time it takes the material to decompose in a landfill. Many companies that depend on plastic packaging, including bleach manufacturers, have begun to reduce the amount of plastic in their packaging or to use recycled plastics. In the early 1990s, Clorox introduced post-consumer resins (PCR) in its packaging. The newer bottles are a blend of virgin high-density polyethylene (HDPE) and 25% recycled plastic, primarily from clear milk jug-type bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read more: &lt;a href=&quot;http://www.madehow.com/Volume-2/Bleach.html&quot; rel=&quot;nofollow&quot;&gt;http://www.madehow.com/Volume-2/Bleach.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-11T18:50:34.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21588" PostTypeId="2" ParentId="7629" CreationDate="2014-12-11T19:02:27.717" Score="1" Body="&lt;p&gt;In proteins, amino acid residues of histidine have pKas near 6 and tyrosine side chains near 10. The protonated form of histidine is positive, and there can be a substantial fraction protonated at physiological pH in a given protein depending upon the exact pK value.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NMR has been used to determine the pKas of particular residues in particular proteins, with interest in deviation from the usual values. In metalloenzymes, the oxidation state of a metal ion can be coupled to the protonation state a side chain.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably what the lecturer wanted you to think about, which pKas would be interesting to try to observe by NMR? not which ones change aromaticity necessarily. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-11T19:29:17.220" LastActivityDate="2014-12-11T19:29:17.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21589" PostTypeId="1" AcceptedAnswerId="22055" CreationDate="2014-12-11T19:18:52.320" Score="0" ViewCount="4538" Body="&lt;p&gt;I am unsure how to go about solving this question. I know how to get the osmotic pressure, but where do I go from there?&lt;/p&gt;&#xA;&lt;p&gt;I can't use the gas law since the protein is not a gas.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{2.38 g}$&lt;/span&gt; of a protein is dissolved in &lt;span class=&quot;math-container&quot;&gt;$\pu{100 mL}$&lt;/span&gt; of water at &lt;span class=&quot;math-container&quot;&gt;$\pu{25^\circ C}$&lt;/span&gt;. The solution rises to a height of &lt;span class=&quot;math-container&quot;&gt;$\pu{8.5 cm}$&lt;/span&gt; in an osmometer (a device for measuring osmotic pressure). What is the molar mass, in grams per mole, of the protein?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Data&lt;/strong&gt;: assuming the water ls pure, then the osmotic pressure &lt;span class=&quot;math-container&quot;&gt;$\pi = \rho gh$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the height of the water, &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density of water (&lt;span class=&quot;math-container&quot;&gt;$\pu{1 g/cm^3}$&lt;/span&gt;),and &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is the acceleration due to gravity (&lt;span class=&quot;math-container&quot;&gt;$\pu{9.81 m/s^2})$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;UPDATE: I've spoken with a masters student (my TA) about this and it turns out that the way to solve this is to use an osmotic pressure equation which is identical to the ideal gas law, but only because it is an approximation. If this question is unlocked I would be happy to share my answer.&lt;/p&gt;&#xA;" OwnerUserId="3926" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-08T09:51:19.160" Title="How to calculate the molar mass of a protein from its osmotic pressure?" Tags="&lt;homework&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21590" PostTypeId="1" AcceptedAnswerId="21597" CreationDate="2014-12-11T19:50:32.920" Score="2" ViewCount="10427" Body="&lt;p&gt;In the reaction of tin (IV) iodide in acetone and water, the solution turns a urine yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply due to the hydrolysis of tin (IV) iodide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SnI4 + 2H2O -&amp;gt; SnO2 + 4HI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why is the acetone needed though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the reaction with $\ce{KI}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SnI4 + KI -&amp;gt; ?}$$ &#xA;I understand it forms iodine, but what happens to $\ce{Sn}$ and $\ce{K}$ - do they react with the acetone for instance?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="5026" LastEditDate="2018-04-15T02:22:23.333" LastActivityDate="2018-04-15T02:22:38.843" Title="Reactions of tin (IV) iodide with potassium iodide and water" Tags="&lt;inorganic-chemistry&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21591" PostTypeId="2" ParentId="20544" CreationDate="2014-12-11T20:13:26.533" Score="3" Body="&lt;p&gt;The question isn't really clear exactly what complexes you are referring to.  However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin  while the 3.87 value is for high spin (3/2).  I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those are not the reasons, see &lt;a href=&quot;http://pubs.rsc.org/en/content/articlehtml/1968/qr/qr9682200457&quot; rel=&quot;nofollow&quot;&gt;Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments&lt;/a&gt; Quarterly Review Chemical Society volume 22, pp 457-498.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-12T11:57:55.770" LastActivityDate="2014-12-12T11:57:55.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21592" PostTypeId="1" CreationDate="2014-12-11T20:29:35.947" Score="1" ViewCount="477" Body="&lt;p&gt;On the basis of its interaction with haemoglobin, explain the toxicity of carbon monoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that ligand substitution is involved and carboxyhaemoglobin is formed meaning that the haem group in the haemoglobin molecule cannot form a ligand with the oxygen molecule, but i would like to know more in depth in terms of its chemistry, why this is the case.&lt;/p&gt;&#xA;" OwnerUserId="9464" LastActivityDate="2014-12-12T12:59:03.770" Title="The toxicity of carbon monoxide in terms of its interaction with haemoglobin" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21593" PostTypeId="1" AcceptedAnswerId="21601" CreationDate="2014-12-11T20:33:37.080" Score="7" ViewCount="11906" Body="&lt;p&gt;$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$ &#xA;The above is an exact differential. Firstly what does this mean physically? (I hear about the fact that a change in G is &lt;strong&gt;in&lt;/strong&gt;dependent of the &quot;path&quot; but I don't understand this argument because I don't know what is meant by this path. Also how would something be &lt;strong&gt;dependent&lt;/strong&gt; on the path?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be shown that $\mathrm{d}G$ can be written:&#xA;$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by comparing coefficients:&#xA;$$\left(\frac {\partial G}{\partial p}\right)_T=V$$&#xA;It is useful to integrate this to see how $G$ varies with $p$:&#xA;$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$&#xA;$V$ can be expressed in terms of $p$ using the equation $V=\frac{n\mathcal{R}T}p$ Where $n\mathcal{R}T$ are constants. The above equation becomes:&#xA;$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$&#xA;Integrating the LHS of the above equation is the crux of my question (I don't understand how that can be integrated). I am not sure whether the following step is correct mathematically, please explain why this is correct/incorrect. I cancel out the $\mathrm{d}p$'s on the LHS:&#xA;$$\int_{p_1}^{p_2}\mathrm{d}G=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$&#xA;Then the result is:&#xA;$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$&#xA;Or...&#xA;$$\Delta G=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$&#xA;This result is correct but I don't understand the left hand side. I've been told that the reason it's possible to write &quot;$G(p_2)-G(p_1)$&quot; or &quot;$\Delta G$&quot; is because $G$ is a state function/exact differential but, again, I don't understand this. If $G$ was not a state function/exact differential how would the result change?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2014-12-12T06:29:15.283" LastActivityDate="2014-12-12T06:48:02.867" Title="How to derive the pressure dependency for the Gibbs free energy?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21596" PostTypeId="2" ParentId="19687" CreationDate="2014-12-11T20:44:05.920" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://www.researchgate.net/publication/47393851_Electrorefining_of_aluminum_alloy_in_ionic_liquids_at_low_temperatures&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ionic liquid refining of aluminum&lt;/a&gt; is a promising solution which uses 3 kWh of electricity per kg instead of around 15 kWh for existing methods. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorDisplayName="user7951" LastEditDate="2019-01-19T23:18:04.440" LastActivityDate="2019-01-19T23:18:04.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21597" PostTypeId="2" ParentId="21590" CreationDate="2014-12-11T20:52:54.117" Score="5" Body="&lt;p&gt;If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SnI4 + 2H2O -&amp;gt; SnO2 + 4HI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SnO2 + 2HI -&amp;gt; I2 + SnO + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\ce{KI}$, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lugol%27s_iodine&quot; rel=&quot;nofollow noreferrer&quot;&gt;yellow-brown solution&lt;/a&gt; of $\ce{KI3}$ will be formed, which contains the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triiodide&quot; rel=&quot;nofollow noreferrer&quot;&gt;triiodide&lt;/a&gt; anion $\ce{I3-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I2 + KI -&amp;gt; KI3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (&lt;a href=&quot;http://mhchem.org/222/pdfLabs222/KineticsPartI.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)2CO + I2 -&amp;gt; CH3C(O)CH2I + HI}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5026" LastEditDate="2018-04-15T02:22:38.843" LastActivityDate="2018-04-15T02:22:38.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21598" PostTypeId="1" CreationDate="2014-12-11T20:58:03.833" Score="7" ViewCount="82" Body="&lt;p&gt;Ionic liquids have been shown &lt;a href=&quot;http://www.researchgate.net/publication/47393851_Electrorefining_of_aluminum_alloy_in_ionic_liquids_at_low_temperatures&quot;&gt;capable of reducing aluminum&lt;/a&gt; trichloride to solid aluminum at 3kw-hrs per kg of Al(s) produced instead of the current ~15kw-hr/kg. However, this would seem to neglect the fact that bauxite is aluminum trioxide not trichloride. This is no small detail and leads me to wonder if it is an insurmountable hurdle to the widescale adoption. Has this technology been implemented (or slated for implementation) on an industrial scale? If not, what holds it back? &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4945" LastEditDate="2014-12-15T08:30:00.890" LastActivityDate="2014-12-15T08:30:00.890" Title="Have Ionic Liquids been used to refine aluminum on an industrial scale?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21599" PostTypeId="2" ParentId="21584" CreationDate="2014-12-11T21:46:56.050" Score="1" Body="&lt;p&gt;Janice G. Smith's textbook &quot;Organic Chemistry&quot;, while unfortunately expensive, has a good interactive workbook online which is useful even without the companion text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find it &lt;a href=&quot;http://highered.mheducation.com/sites/0072397462/student_view0/index.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-11T21:46:56.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21601" PostTypeId="2" ParentId="21593" CreationDate="2014-12-12T01:00:32.343" Score="7" Body="&lt;p&gt;This question has a lot of parts! I hope I manage to answer most but probably edits will be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does this mean physically?&lt;/strong&gt;&lt;br&gt;&#xA;This is simply an expression that shows the contributions to the change in the Gibb's free energy, which varies with both pressure and temperature (i.e., $G$ is a function of $p$ and $T$). However, it is possible to separate the exact differential of $\mathrm{d}G$ into two terms using the mathematical expression of partial derivatives, first term is the variation of $G$ with pressure at constant temperature, and the second term is the variation of $G$ with temperature at constant pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does it mean for $\Delta G$ to be independent of the path? (same question: what does it mean for $\Delta G$ to be a state function?)&lt;/strong&gt;&lt;br&gt;&#xA;You can calculate a change in $G$, or equivalently, $\Delta G$, for any reaction. So for example for a reaction like a combustion process for glucose &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12O6~(s) + 3 O2~(g) -&amp;gt; 6 CO2~(g) + 6 H2O~(g),}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can calculate $\Delta G$ for this reaction at any particular temperature/pressure that you would like. And in fact, if you don't do the combustion directly but somehow do it via completely dissociated gaseous atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12O6~(s) + 3 O2~(g) -&amp;gt; 6 C~(g) + 18 O~(g) + 12 H~(g) -&amp;gt; 6 CO2~(g) + 6 H2O~(g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will still end up with the same $\Delta G$ for the overall reaction. Obviously if you calculate the $\Delta G$ for each individual step/arrow in the gaseous atoms case it will be very different, but they will combine to give the same overall $\Delta G$ as the direct combustion. In other words, $\Delta G$ won't be able to tell you whether you have done the combustion directly or done it laboriously via gaseous atoms, it just tells you what is the difference between the two states, first state being glucose + oxygen, second state being carbon dioxide + water. $\Delta G$ for glucose that you have eaten and combined with oxygen you breathe in and breathe out carbon dioxide + water would still the same $\Delta G$ as for combustion, even though now it is all done through cells, Kreb's cycle and a lot of biochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would something be dependent on the path?&lt;/strong&gt;&lt;br&gt;&#xA;As you can see from my example above, the amount of effort you need to make $$\ce{C6H12O6~(s) + 3 O2~(g) -&amp;gt; 6 CO2~(g) + 6 H2O~(g)}$$ happen can differ a lot, and for each of the cases, what you &quot;get back&quot; from doing that reaction is different, such as heat released when burning glucose vs keeping your cellular biochemistry going with respiring glucose. In physics/physical chemistry, this is formally &quot;work done&quot;. Work is dependent on path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little unsure whether you went from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T \tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac {\partial G}{\partial p}\right)_T=V \tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the right way, so I am going to go through it again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant $T$, $\mathrm{d}T=0$. So $(1)$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \mathrm{d}G=V\mathrm{d}p \tag3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which after rearranging is equivalent to $(2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration then should go from $(3)$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{p_1}^{p_2}\mathrm{d}G=\int_{p_1}^{p_2}V\mathrm{d}p \tag4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why what you wrote look a little weird  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is that you usually don't just cancel partial differentials like you did. Whereas $(3)$ is an exact differential, which you integrate over in $(4)$, and that is OK. Unfortunately I am not sure whether what you wrote is wrong, it is just something I would not write myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the derivation is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final question. I believe you are fine with mathematically going from $\int_{p_1}^{p_2}\mathrm{d}G$ to $G(p_2)-G(p_1)$, however, if you think about this result physically, $\Delta G$ does not depend on any other $p_3, p_4, p_5...p_n$ that the system might pass through on the way from $p_1$ to $p_2$. If $G$ is not a state function, then you would expect to have to consider at the very least all these different points of $p$ that you might have gone through, and different ways of doing the reaction passing through different $p$ would change $G$ even if the initial and final $p$ are the same. In that case, an integral like what we have written here would be a very poor way of finding $\Delta G$ because we are missing the contribution from all the intervening $p_n$, and we would have to use some different mathematical tools. In fact, I believe we would have failed to derive the master equation $(1)$ much earlier on, so the rest of the derivation would simply not follow.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastEditorUserId="4945" LastEditDate="2014-12-12T06:48:02.867" LastActivityDate="2014-12-12T06:48:02.867" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21602" PostTypeId="2" ParentId="20553" CreationDate="2014-12-12T01:07:29.097" Score="2" Body="&lt;p&gt;This is just me shooting bull because I don't know what has been done, but in the distant past, mineralogists used concentrated solutions of heavy metal salts for density determination by flotation.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Clerici_solution&quot; rel=&quot;nofollow&quot;&gt;Clerici's solution&lt;/a&gt; is a solution of equal parts thallium malonate and thallium formiate in water.  That may just be the ticket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, those with access to appropiate protective gear (i.e. fume hood and proper gloves) may consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_iodide&quot; rel=&quot;nofollow&quot;&gt;methyl iodide&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-12-12T01:07:29.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21603" PostTypeId="1" CreationDate="2014-12-12T01:12:07.873" Score="2" ViewCount="285" Body="&lt;p&gt;I have to say I have never really understood the half reaction method. What bothers me is that it seems to be too artificial. Too many little rules to remember!&#xA;Why don't we use the mass conservation law to justify the process and use sistematically a set of linear equations to do the balances? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="11278" LastActivityDate="2015-01-11T06:20:52.897" Title="Half reaction method justification" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21604" PostTypeId="1" AcceptedAnswerId="21619" CreationDate="2014-12-12T01:45:56.733" Score="2" ViewCount="1000" Body="&lt;p&gt;Why don't we usually use diatomic fluorine in electrophilic aromatic substitution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because fluorine is a potent oxidizer and will react with the carbons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might happen if we reacted benzene with diatomic fluorine in the presence of some Lewis acid such as boron trifluoride? Would we get the expected product: monosubstituted benzene? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-12T08:22:05.790" Title="Electrophilic aromatic substitution with fluorine" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21605" PostTypeId="2" ParentId="18401" CreationDate="2014-12-12T02:53:22.430" Score="4" Body="&lt;p&gt;This article &lt;a href=&quot;http://www.oocities.org/capecanaveral/campus/5361/chlorate/graphite/indgra.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; indicates that carbon/graphite electrodes are attacked both chemically (oxygen gets converted to $\ce {CO2}$, removing carbon from the electrode) and mechanically (gas bubbles form in pores in the surface and break it apart).  Pencil leads fall apart quickly because of the clay content.  A &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000RC0SZ8&quot; rel=&quot;nofollow&quot;&gt;high-quality graphite electrode&lt;/a&gt; will last far longer but will also eventually decay.  The rate of erosion goes up rapidly with increased voltage across the electrodes.  In my personal experience doing water electrolysis I got almost no mechanical erosion at 10V but obvious mechanical erosion at 24V.&lt;/p&gt;&#xA;" OwnerUserId="11268" LastActivityDate="2014-12-12T02:53:22.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21606" PostTypeId="2" ParentId="21603" CreationDate="2014-12-12T04:41:00.477" Score="2" Body="&lt;p&gt;The concept of the half reaction has nothing to do with balancing redox reactions; how one arrives at a balanced reaction is not really important. The interesting thing about redox reactions is that electrons do not necessarily have to be transferred directly between the oxidized and reduced species. In many types of electrochemical cells, the anode and cathode are isolated from one another by a salt bridge and the species in each half-cell never approach one another. In techniques such as potentiometry, often one cares only about one half of the reaction, e.g. in a combination pH electrode, the response from the half reaction that contains hydronium ions produces the analytical signal and the the other half is a reference electrode reaction whose job is to simply produce a constant potential. In things like voltammetry, one is often only concerned about the half reaction occurring at the working electrode and the half reaction that balances it at the counter electrode isn't analytically significant (and may in fact be more than one different reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two half-reactions need not necessarily occur at the same place and one half reaction can generally be substituted without affecting the other, it makes sense to think of them separately. This lets us do things like think about different lithium batteries. There are many types and they all use the $\ce{Li -&amp;gt; Li+ + e-}$ oxidation half-reaction, but it's coupled with different reduction half-reactions based on cobalt, iron, etc.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-12T04:41:00.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21609" PostTypeId="1" CreationDate="2014-12-12T05:41:46.597" Score="1" ViewCount="480" Body="&lt;p&gt;What is the mechanism of a 1,2-hydride shift? I read that there is an aromatic intermediate, in that the intermediate obeys Hückel's rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Wikipedia doesn't offer any further explanation. Can someone with more knowledge elaborate on the transition state of a 1,2-hydride shift? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-01-29T21:50:13.733" LastActivityDate="2018-01-29T21:50:13.733" Title="What is the mechanism of a 1,2-hydride shift?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;pericyclic&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21612" PostTypeId="1" CreationDate="2014-12-12T05:58:34.840" Score="1" ViewCount="302" Body="&lt;p&gt;I recently did an experiment on protein unfolding using a Perkin Elmer fluorescence spectrometer. The protein was $\alpha$-chymotrypsin. I was given &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed069p852&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; for my procedure. I am now trying to make sense of the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing temperature was used as the denaturant. Temperature and absorbance were plotted, resulting in an essentially sigmoidal curve describing unfolding. The paper gives the following equation for determining an equilibrium constant for the experiment.&#xA;$$K_u=\frac {X_N-X}{X-X_D}=\frac {[P_D]_{eq}}{[P_N]_{eq}}$$ Where $K_u$ is the equilibrium constant, $X_N$ is the absorbance for the native state, $X_D$ is the absorbance for the denatured state, and $X$ is any given absorbance. $P_D$ is the denatured protein and $P_N$ is the native protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper says that the free energy of unfolding can be described by the familiar equation: $$\Delta G = -RT\ln(K_u)$$&#xA;I'm fairly sure that I've calculated $K_u$ at each temperature recorded, but I'm not quite sure whether I use each $K_u$ as a point for another plot, or average them, or something else. If I use a Van't Hoff plot to graph each point of $\ln(K_u)$ vs. $\frac {1}{T}$, will the slope and intercept be $\frac{-\Delta H}{R}$ and $\frac{\Delta S}{R}$ respectively? I think so due to the rearrangement described &lt;a href=&quot;http://www.roanestate.edu/faculty/chemistry/chemistryslideshows/vantHoffII/vantHoffII.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, and perhaps I've answered my own question as I attempted to ask it. Nevertheless, I would appreciate having my understanding of this calculation confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;I've plotted $\ln(K_u)$ vs. $\frac {1}{T}$ and the slope of $\ln(K_u)$  vs. $\frac{1}{T}$ is not linear as had been expected. Linear regression analysis gives a trendline with an $R^2$ value of 0.87. Should I separate the data so as to graph $\ln(K_u)$ vs. $\frac {1}{T}$ for points leading up to the sudden unfolding and also for points once unfolding becomes dramatic? This would give me two values for the enthalpy of unfolding--one before it unfolds much and the other for when unfolding is happening quickly. If I do so, will the difference between those enthalpies be the overall enthalpy of unfolding? I'm not sure what the significance of those two values would be.&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4813" LastEditDate="2014-12-16T04:21:10.130" LastActivityDate="2014-12-16T04:21:10.130" Title="Finding the thermodynamics of protein unfolding from temperature and absorbance using fluorescence spectroscopy?" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21613" PostTypeId="1" AcceptedAnswerId="22003" CreationDate="2014-12-12T06:00:12.340" Score="7" ViewCount="6376" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This species is a derivative of benzene, with a single fluorine atom&#xA;  attached. Its melting point is -44 °C, which is lower than that of&#xA;  benzene, indicative of the remarkable effect of fluorination on the&#xA;  intermolecular interactions as seen throughout organofluorine&#xA;  chemistry. In contrast, the boiling points of PhF and benzene differ&#xA;  by only 4 °C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this true? Wikipedia doesn't elaborate on why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand the melting point is the point at which the solid and liquid forms are in equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling point is the point at which the liquid and gas phases are in equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that intermolecular forces play an integral role in melting and boiling point. However, why does it seem that the strength of the intermolecular forces in the solid phase is stronger than that in the liquid phase for fluorobenzene? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="10185" LastEditDate="2015-10-02T16:39:40.700" LastActivityDate="2020-07-19T19:14:31.857" Title="Melting and boiling points of benzene and fluorobenzene" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;intermolecular-forces&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21617" PostTypeId="2" ParentId="19687" CreationDate="2014-12-12T08:00:33.763" Score="5" Body="&lt;p&gt;I don't think this is going to be a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the Al and O existing in the Moon's soil are not there are free elements, but are tightly bound in minerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth, Al is extracted from bauxite - this is naturally occurring Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;. This extraction process is extremely energy intensive. In fact, they transport the bauxite mined elsewhere to Iceland because the electricity in there is so cheap, instead of doing it on the spot of the mine, or nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Al and O that occur on the Moon are mostly in the form of the mineral anorthite plagioclase: CaAl&lt;sub&gt;2&lt;/sub&gt;Si&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;8&lt;/sub&gt;. Extracting the Al from there would be even harder! You first have to extract the Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; out of the CaAl&lt;sub&gt;2&lt;/sub&gt;Si&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;8&lt;/sub&gt;, and only then you can get the Al out. This mineral is not unique to the Moon. In fact, anorthite, albite (NaAlSi&lt;sub&gt;3&lt;/sub&gt;O&lt;sub&gt;8&lt;/sub&gt;) and all minerals in between are the most common minerals in Earth's crust, and we still don't use them as a source for Al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, burning Al as a fuel would not be a good idea because the end result is a solid, not a gas.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-12T08:00:33.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21618" PostTypeId="2" ParentId="20347" CreationDate="2014-12-12T08:11:46.107" Score="1" Body="&lt;p&gt;The reaction proceeds by a mechanism named $B_{AC_2}$ (Base-Acyl Bond-2 molecularity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate is given by: (note that it is order 2 and molecularity 2) $$r=k\ce{[ester][OH- ]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\bf rds$ is the attack of hydroxide ion. Since the last step is an acid-base reaction, it is unidirectional and reverse esterification is not possible under general conditions (I think you understand what I mean by this). Thus relative rates are also given which are well explained by this mechanism (kinda proof, considering $\bf rds$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cv2FP.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/U44nr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-12T08:11:46.107" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21619" PostTypeId="2" ParentId="21604" CreationDate="2014-12-12T08:22:05.790" Score="3" Body="&lt;p&gt;Fluorine reacts so rapidly with benzene that aromatic fluorination requires special conditions&#xA;and special types of apparatus. Even then, it is difficult to limit the reaction to monofluorination&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-12T08:22:05.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21622" PostTypeId="1" CreationDate="2014-12-12T10:23:46.477" Score="1" ViewCount="814" Body="&lt;p&gt;I want to know about the dipole moment of this molecule. I think the molecule has no dipole moment, but a book says me that it has a dipole moment. &lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorDisplayName="user7951" LastEditDate="2015-05-15T08:24:22.457" LastActivityDate="2015-05-15T10:44:06.510" Title="What is the dipole moment of the most stable conformer of 1,1,2,2-tetrafluoroethane?" Tags="&lt;organic-chemistry&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21623" PostTypeId="1" CreationDate="2014-12-12T10:37:14.193" Score="5" ViewCount="2328" Body="&lt;p&gt;We know that all Lewis acids are electrophiles but I have a confusion whether the reverse is also true. Electrophiles are molecules that are in search of electrons while Lewis acids are short of electrons. So can't all electrophiles act as Lewis acids?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="7448" LastEditDate="2015-10-09T16:05:21.257" LastActivityDate="2019-02-21T12:18:29.533" Title="Why can't all electrophiles act as Lewis acids?" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21624" PostTypeId="2" ParentId="21623" CreationDate="2014-12-12T11:03:10.840" Score="0" Body="&lt;p&gt;Lewis acids are electron acceptors, the slight asymmetry in the definition arises because accepting means you have to move electrons into a Lewis acid, i.e. H+ and AlCl3. On the other hand the carbonyl carbon is electrophilic, but when it accepts two electrons, it rejects two others. Hence, it is not a Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrophilic - desires electrons&#xA;Lews Acidis   - desires and is able to take on additional electrons&lt;/p&gt;&#xA;" OwnerUserId="11280" LastActivityDate="2014-12-12T11:03:10.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21625" PostTypeId="2" ParentId="21623" CreationDate="2014-12-12T11:03:31.137" Score="3" Body="&lt;p&gt;The definition of &lt;a href=&quot;https://goldbook.iupac.org/L03508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis acids&lt;/a&gt; states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A molecular entity (and the corresponding chemical species ) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/LOqwE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOqwE.png&quot; alt=&quot;example lewis adduct&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  &lt;sub&gt;(source: &lt;a href=&quot;https://goldbook.iupac.org/SVG/L03508.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;iupac.org&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In principle, since all electrophiles can be electron acceptors, they can usually act as Lewis bases, too. However, sometimes sterically demanding moieties stabilise the electrophile, so that it is impossible to form a &lt;a href=&quot;https://goldbook.iupac.org/L03510.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis adduct&lt;/a&gt;. These compound are then one part of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Frustrated_Lewis_pair&quot; rel=&quot;nofollow noreferrer&quot;&gt;frustrated Lewis pair&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2019-02-21T12:18:29.533" LastActivityDate="2019-02-21T12:18:29.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21626" PostTypeId="1" CreationDate="2014-12-12T11:50:58.873" Score="4" ViewCount="926" Body="&lt;p&gt;I  have synthesized 1-phenylazo-2-naphthol in my laboratory tutorial and have to analyze it. I've done a thin layer chromatography (TLC) of it:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solvent: ethanol&lt;/li&gt;&#xA;&lt;li&gt;eluent: toluene&lt;/li&gt;&#xA;&lt;li&gt;stationary phase: silica gel plate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;My only problem is, I need an theoretical value for comparison but haven't found any source.&lt;/p&gt;&#xA;" OwnerUserId="9748" LastEditorUserId="4945" LastEditDate="2020-09-08T13:48:46.613" LastActivityDate="2020-09-08T13:50:28.097" Title="Are there any sources for tabulated Rf-values, especially 1-phenylazo-2-naphthol?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21627" PostTypeId="2" ParentId="21592" CreationDate="2014-12-12T12:46:09.580" Score="3" Body="&lt;p&gt;Carbon Monoxide and dioxygen both reversibly bind to the iron atom in hemoglobin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two molecules compete for the same binding site.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if both $\ce{O2}$ and $\ce{CO}$ are present, some of each will bind, but for a given amount of $\ce{O2}$, less $\ce{O2}$ will bind if $\ce{CO}$ is present. The affinity of the Fe for $\ce{CO}$ is much greater than for $\ce{O2}$, so a lower concentration of $\ce{CO}$ results in a given saturation level of hemoglobin compare to $\ce{O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same would be true for cyanide which is isoelectronic with carbon monoxide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete explanation needs to go into the cooperation between the four hemoglobin subunits.  $\ce{CO}$ bound to two of the four subunits causes $\ce{O2}$ to being too tightly bound in the other subunits.  That the oxygen can not be released from the hemoglobin, is another aspect of toxicity. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-12T12:59:03.770" LastActivityDate="2014-12-12T12:59:03.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21628" PostTypeId="1" AcceptedAnswerId="62509" CreationDate="2014-12-12T13:32:59.757" Score="4" ViewCount="1364" Body="&lt;p&gt;My teacher likes to use household materials to do experiments. Today, we did an experiment that showed the reaction of vinegar and eggshell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we were doing the analysis of the experiment, I realized that my teacher used the term acetic acid ($\ce{CH3COOH}$) instead of ethanoic acid. So, my questions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the accepted usage for the terms acetic acid and ethanoic acid?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which formula is the accepted representation of acetic acid, $\ce{CH3COOH}$ or $\ce{C2H4O2}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="7475" LastEditDate="2016-11-05T19:18:54.443" LastActivityDate="2016-11-11T15:31:12.867" Title="Acetic acid or ethanoic acid - trivial names vs systematic names and correct formula representation" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21629" PostTypeId="2" ParentId="18924" CreationDate="2014-12-12T14:16:11.857" Score="1" Body="&lt;p&gt;The issue with some of the proposed reactions of the above posts is activation energy, if its measured in Gibbs free energy, should also subject to entropy, since ΔG=ΔH-TΔS. However the wikipedia article on activation energy (as of this writing) suggests when reaction rates &lt;em&gt;decrease&lt;/em&gt; with activation energy, because the molecules were prepared to react at room temperature. The only way for the activation energy equation to work would be through would be for negative activation energy. Reactions like this are simple: i.e. two things smash together and will combine if the potential energy is larger than their momentum. But say for instance you heat this reaction up. You've accelerated these particles (that is, heated them, since heat can be considered kinetic energy) the reactants are so fast and have so much momentum they simply bounce off each other, and so the  becomes meaningless.&#xA;As an aside, you can have 'cheating' occur in normal reactions with hydrogen which exhibits quantum tunneling (because its so tiny). So you can have product form from a reaction while the reactants &lt;em&gt;are way below the activation energy&lt;/em&gt; expected. So a smartass rejoinder to your professor's comment would be &quot;sure, cliffs may not exist, but slippery slopes sure do!&quot;&lt;/p&gt;&#xA;" OwnerUserId="11281" LastActivityDate="2014-12-12T14:16:11.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21630" PostTypeId="1" AcceptedAnswerId="21641" CreationDate="2014-12-12T14:25:25.270" Score="2" ViewCount="128" Body="&lt;p&gt;In etching of mono silicon, why does the etching process around the corners differ from the etching process around the flat edges? What brings about this difference and is there some way that MEMS engineers could quantify the process around the corners?&lt;/p&gt;&#xA;" OwnerUserId="9830" LastActivityDate="2014-12-12T17:54:03.120" Title="Etching around corners" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21631" PostTypeId="1" CreationDate="2014-12-12T15:14:26.013" Score="4" ViewCount="2330" Body="&lt;p&gt;According to the definition of dative bonding electrons are shared which are given by one of the atoms, so if the electrons are shared so how come there is charge on the elements in the dative bonding?&lt;/p&gt;&#xA;" OwnerUserId="11282" LastEditorUserId="9961" LastEditDate="2015-07-27T18:45:49.817" LastActivityDate="2019-12-22T12:41:12.433" Title="Why there is charge on the elements, in the dative bonding?" Tags="&lt;bond&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21632" PostTypeId="2" ParentId="21628" CreationDate="2014-12-12T15:21:09.277" Score="6" Body="&lt;p&gt;While I agree that the systematic name ethanoic acid makes a lot more sense, the name acetic acid has been used for so long that it's the accepted IUPAC name and is used almost universally. Both refer to the same compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular formulas like $\ce{C2H4O2}$ are often seen when doing elemental analysis and the like, but give no structural information. $\ce{CH3COOH}$ makes it clear that it's an acid and shows the carbonyl group.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="-1" LastEditDate="2015-08-03T12:24:21.110" LastActivityDate="2015-08-03T12:24:21.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21633" PostTypeId="2" ParentId="21631" CreationDate="2014-12-12T15:31:14.573" Score="5" Body="&lt;p&gt;To begin with, the term &quot;dative bond&quot; is not used anymore. It has been replaced by &quot;coordinate bond&quot; : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have an example of this kind of bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l2x0q.png&quot; alt=&quot;Ammonium ion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this kind of bond occurs when an atom is giving two electrons from itself to form the bond. The typical coordinate bond is the $\ce {N}$ already bonded with $\ce {3H}$ forming a bond with a proton. Since the proton has no electron, the $\ce {N}$ atom has to fully give one of his electron from his remaining doublet to form the bond. But to respect the charge equality, you had a positive charge at the begining, no electrons were added or removed from the system, so the positive charge should remains on the final molecular edifice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, where is located this positive charge ? One could start thinking this charge is located on the $\ce {N}$ atom since it is giving an electron. But actually, the $\ce {N}$ atom has a greater electronegativity than the $\ce {H}$ atom, that's say its hability to attract electrons is greater. Therefore the positive charge remaining should be equaly distributed over all the $\ce {4H}$ since they are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, altogether, the net charge of the edifice will be $+1$.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="11619" LastEditDate="2017-11-30T17:16:26.883" LastActivityDate="2017-11-30T17:16:26.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21634" PostTypeId="2" ParentId="21631" CreationDate="2014-12-12T15:31:22.210" Score="2" Body="&lt;p&gt;It's &lt;a href=&quot;https://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;nofollow&quot;&gt;formal charge&lt;/a&gt; - it often doesn't mean there are full charges like in &lt;a href=&quot;https://en.wikipedia.org/wiki/Zwitterion&quot; rel=&quot;nofollow&quot;&gt;zwitterions&lt;/a&gt; but dative bonds are usually strongly polarised.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2014-12-12T15:31:22.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21635" PostTypeId="1" AcceptedAnswerId="21651" CreationDate="2014-12-12T15:54:32.723" Score="2" ViewCount="346" Body="&lt;p&gt;Why is it that Boron does not form Diborane with water rather than $\ce{B2O3}$?&lt;/p&gt;&#xA;" OwnerUserId="11283" LastEditorUserId="4231" LastEditDate="2014-12-12T16:31:03.437" LastActivityDate="2014-12-12T21:04:09.077" Title="Why doesn't Boron form Diborane with water?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21636" PostTypeId="1" AcceptedAnswerId="21639" CreationDate="2014-12-12T16:39:47.700" Score="0" ViewCount="48" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/j6Jl7I1&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/j6Jl7I1.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://imgur.com/HuPG4ZZ&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/HuPG4ZZ.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this reaction make sense? Mechanistically speaking, it does not for me. Is this a special case, or just a typo in the book?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-12-12T17:07:08.410" Title="Can HBr add to a non-sp2 hybridized carbon?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21638" PostTypeId="1" AcceptedAnswerId="21724" CreationDate="2014-12-12T17:02:09.447" Score="11" ViewCount="2422" Body="&lt;p&gt;Would it be possible to reduce a α,β-unsaturated nitro compound to a saturated amine using for example $\ce{NaBH4}$? I know these kind of compounds will be reduced by hydrogenation using a $\ce{Pt/C}$ catalyst, $\ce{LiAlH4}$ or freshly prepared Raney Nickel, but I'm looking for a reducing agent that is easier to handle than those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book from which I'm learning, &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot;&gt;Clayden et al.&lt;/a&gt;, does not really go into much detail on reduction of nitro compounds or unsaturated alkanes, other than that hydrogenation is a good approach to both and that the nitro group is one of the most easily reduceable functional groups. But how about a α,β unsaturated compounds then? How would I asses whether some reducing agent is able to reduce both simultaneously?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="4945" LastEditDate="2015-08-13T09:21:15.150" LastActivityDate="2015-08-13T09:21:15.150" Title="Reduction of α,β-unsaturated nitro compounds" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;nitro-compounds&gt;&lt;amines&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21639" PostTypeId="2" ParentId="21636" CreationDate="2014-12-12T17:07:08.410" Score="3" Body="&lt;p&gt;There is a carbocation rearrangement. Basically, the H can produce two secondary carbocations. One of those secondary carbocations can do a 1,2-hydride shift, which produces a tertiary carbocation. This carbocation, being tertiary, will react much faster, resulting in being in the major product. So, yes, HBr can add to a non-sp2 carbon.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-12-12T17:07:08.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21640" PostTypeId="2" ParentId="15172" CreationDate="2014-12-12T17:25:58.207" Score="4" Body="&lt;p&gt;I don't understand what is the big deal with bonds in acids or esters in higher oxidation states - I always thought of them as of single &lt;a href=&quot;https://en.wikipedia.org/wiki/Dipolar_bond&quot; rel=&quot;nofollow&quot;&gt;dipolar bonds&lt;/a&gt;. They shouldn't be written as double bonds. It's an archaism for me - now chlorine oxoanions aren't considered hypervalent. Even &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;nofollow&quot;&gt;hypervalency&lt;/a&gt; of molecules like &lt;a href=&quot;https://en.wikipedia.org/wiki/Iodine_heptafluoride&quot; rel=&quot;nofollow&quot;&gt;iodine heptafluoride&lt;/a&gt; has been criticized.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-10-20T16:28:33.997" LastActivityDate="2015-10-20T16:28:33.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21641" PostTypeId="2" ParentId="21630" CreationDate="2014-12-12T17:54:03.120" Score="3" Body="&lt;p&gt;What happens is that convex (outside) corners get severely undercut in wet etching. On a &amp;lt;100&gt; cut wafer, the edges leading up to the corner reveal &amp;lt;111&gt; planes which are resistant to etching, but right at the corner, where the two planes intersect, etching reveals a &amp;lt;110&gt; plane, which is etched much more quickly. The difference in etch rates has to do with the number of &quot;dangling&quot; bonds—faces with fewer Si-Si bonds at the surface etch faster. The reason why concave corners don't undercut, despite having the same &amp;lt;111&gt; plane intersection is that there are no dangling bonds in an inside corner so it doesn't etch any faster than the &amp;lt;111&gt; planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be noted that adding a small amount of surfactant to the etchant can dramatically decrease undercutting by adsorption to the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mnsl-journal.com/content/pdf/2213-9621-1-1.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; (fortunately open-access) article explains the phenomenon quite nicely.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-12T17:54:03.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21644" PostTypeId="1" AcceptedAnswerId="21648" CreationDate="2014-12-12T19:25:31.960" Score="0" ViewCount="3223" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two metal samples, $\mathrm{X}$ and $\mathrm{Z}$, of the same mass and initially at $\pu{25 ^\circ C}$, are heated so that each metal receives the same amount of thermal energy, which metal will have the highest final temperature?&lt;br&gt;&#xA;  Specific heat capacities: &#xA;  $c(\mathrm{X}) = \pu{0.35 J//g*K}$,&#xA;  $c(\mathrm{Z}) = \pu{0.895 J//g*K}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does the specific heat capacity matter in the final temperature or not?&lt;/p&gt;&#xA;" OwnerUserId="11286" LastEditorUserId="4945" LastEditDate="2017-11-22T06:08:10.127" LastActivityDate="2017-11-22T06:13:45.487" Title="How to determine which of two metals will have the higher temperature after applying the same amount of thermal energy?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-11-11T14:32:08.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="21648" PostTypeId="2" ParentId="21644" CreationDate="2014-12-12T20:23:42.333" Score="1" Body="&lt;p&gt;By definition, the heat capacity is the energy required for a body to be warmed by $\pu{1 K}$. For the specific heat capacity it's the same but normalized by the mass. The higher the heat capacity is, the more energy will be required to warm it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal $\mathrm{X}$ having the lowest heat capacity will have the higher temperature if it receives the same amount of energy than the metal $\mathrm{Z}$.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="4945" LastEditDate="2017-11-22T06:13:45.487" LastActivityDate="2017-11-22T06:13:45.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21651" PostTypeId="2" ParentId="21635" CreationDate="2014-12-12T21:04:09.077" Score="7" Body="&lt;p&gt;I assume the hypothetical reaction you are asking about is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3H2O + 2B -&amp;gt; B2H6 + 3/2O2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;under standard conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard enthalpy of formation of water is −285.8 kJ/mol, whereas that of diborane is 35.4 kJ/mol; this reaction would therefore be very endothermic and hence unfavourable. You're breaking up strong O-H bonds and B-B bonds to form B-H bonds (including two 3c2e bonds, which are very weak) and weak O-O bonds. In contrast the formation of boron trioxide by reaction of boron and water is &lt;strong&gt;very&lt;/strong&gt; exothermic, and has been considered in potential applications of boron as a rocket fuel. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-12T21:04:09.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21655" PostTypeId="2" ParentId="20163" CreationDate="2014-12-13T01:28:42.600" Score="6" Body="&lt;p&gt;Geoff basically answers your question. As the tip about choosing a method: if you are looking for an accuracy beyond 1 kcal/mol, then you should consider wave-function based quantum chemistry methods, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Coupled_cluster&quot;&gt;coupled-cluster&lt;/a&gt; methods. CCSD(T) method would be very appropriate for ground state calculations, while you can use EOM-CCSD methods for the excited state calculations. However, these methods have a much worse scaling compared to DFT, so depending on the size of your molecule and the computational sources you have, these methods might be out of reach. You may want to check this &lt;a href=&quot;http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.79.291&quot;&gt;review&lt;/a&gt; on coupled-cluster methods.&lt;/p&gt;&#xA;" OwnerUserId="10220" LastActivityDate="2014-12-13T01:28:42.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21657" PostTypeId="1" AcceptedAnswerId="21660" CreationDate="2014-12-13T03:20:10.283" Score="4" ViewCount="1787" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A white solid is either $\ce{Na2O}$ or $\ce{Na2O2}$. A piece of red&#xA;  litmus paper turns white when it is dipped into a freshly made aqueous&#xA;  solution of white solid. &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Identify the substance and explain with the help of balanced&#xA;  equation.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Explain what would happen to the red litmus if the white solid&#xA;  were the other compound.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the solution and theoretical knowledge, I know the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{Na2O2 + 2 H2O &amp;amp;-&amp;gt; 2NaOH + H2O2}\\&#xA;\ce{Na2O + H2O &amp;amp;-&amp;gt; 2NaOH}&#xA;\end{align}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aqueous solution of $\ce{Na2O}$ will convert red litmus into blue (basic property), so, the substance is $\bf{not}$ $\ce{Na2O}$ for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, how will it be $\ce{Na2O2}$? Why does the aqueous solution of sodium peroxide turns red litmus into white?&lt;/p&gt;&#xA;" OwnerUserId="3794" LastEditorUserId="5026" LastEditDate="2018-07-12T02:34:06.077" LastActivityDate="2019-02-01T12:22:09.690" Title="Why does the aqueous solution of sodium peroxide turns red litmus into white?" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21658" PostTypeId="2" ParentId="20117" CreationDate="2014-12-13T04:49:34.533" Score="0" Body="&lt;p&gt;I'm going to focus on one question here. You really have &lt;em&gt;several&lt;/em&gt; and it'd be better if you asked these as separate questions on the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, 2D spectroscopy involves using multiple pulses, just like multi-dimensional NMR experiments. There are a variety of techniques but in all cases, you're sending some sort of &lt;strong&gt;&lt;em&gt;pump&lt;/em&gt;&lt;/strong&gt; pulse which excites the system (e.g., electronic or vibrational levels) and then &lt;strong&gt;&lt;em&gt;probing&lt;/em&gt;&lt;/strong&gt; the system over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting data is inherently two-dimensional. You know the absorption of the system along one axis, and the emission (or second absorption) along another axis. Often, you can watch this 2D data over time, giving you even more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how does this show coherence? Let's say you have some complex system. You take a UV/Vis electronic absorption spectra and you see two peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bOkqr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now normally, you'd think of this system as having two distinct absorptions. Maybe it's a metal complex and you see a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Electronic_Spectroscopy/Selection_Rules_for_Electronic_Spectra_of_Transition_Metal_Complexes/Metal_to_Ligand_and_Ligand_to_Metal_Charge_Transfer_Bands&quot; rel=&quot;nofollow noreferrer&quot;&gt;ligand-metal or metal-ligand charge transfer band&lt;/a&gt; and a d-d transition at longer wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But perhaps instead, some of the intensity in a peak comes from energy transfer. This can particularly be the case in biology or materials systems, where there are multiple photosynthetic reaction centers with ever-so-slightly different energies. So you excite one center, but then energy dumps into another one (e.g., at lower energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using two-dimensional spectroscopy, that will show up as an off-diagonal peak. You'll excite at one wavelength, but it fluoresces at another one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZH9dO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's an example of coupling between oscillators (either vibrational or electronic). Now there are multiple ways that energy transfer could occur. Usually, this happens essentially as a diffusive random walk - all phase information is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, the wave packet transfers from one state to another, but retains its phase. That's &lt;strong&gt;&lt;em&gt;coherent&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2D photon echo spectra, you'll see oscillations in the peak heights when coherent transfer occurs, thus &quot;beats.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-13T04:49:34.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21660" PostTypeId="2" ParentId="21657" CreationDate="2014-12-13T06:34:54.230" Score="10" Body="&lt;p&gt;Your analysis seems right and you remember correctly that colour change of litmus of litmus paper will indicate the pH value of the solutions into which it is dipped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, unless the dye (litmus) is irreversible transformed to another compound which no longer shows a pH dependent colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the experiment states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A piece of red litmus paper turns white [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With other words: &lt;strong&gt;Litmus was bleached&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;strong&gt;bleaching&lt;/strong&gt; typically is the result of an &lt;strong&gt;oxidative&lt;/strong&gt; transformation of a (former) dye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the only conceivable &lt;strong&gt;oxidant&lt;/strong&gt; in solution is &lt;strong&gt;hydrogen peroxide&lt;/strong&gt;. Consequently, the initial compound in question must have been $\ce{Na2O2}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-13T06:34:54.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21661" PostTypeId="1" AcceptedAnswerId="21669" CreationDate="2014-12-13T08:57:37.983" Score="4" ViewCount="6103" Body="&lt;p&gt;Why can't esters or acids or their derivatives exhibit tautomerism? I mean if they did, then there would be a conjugation between lone pair electrons of oxygen and pi electrons adding to their stability. Then why don't they exhibit similar tautomerism?&lt;/p&gt;&#xA;" OwnerUserId="7897" LastActivityDate="2014-12-13T16:13:41.880" Title="Why do only ketones exhibit keto-enol tautomerism?" Tags="&lt;tautomer&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21662" PostTypeId="1" AcceptedAnswerId="21689" CreationDate="2014-12-13T13:30:52.093" Score="13" ViewCount="6017" Body="&lt;p&gt;Why is osmium so dense despite there being heavier elements after it in the periodic table?&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="10185" LastEditDate="2015-09-18T18:02:52.363" LastActivityDate="2017-02-15T19:46:12.823" Title="Why is osmium the densest known element?" Tags="&lt;physical-chemistry&gt;&lt;elements&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21663" PostTypeId="1" AcceptedAnswerId="21686" CreationDate="2014-12-13T14:24:15.530" Score="2" ViewCount="311" Body="&lt;p&gt;I saw &lt;a href=&quot;http://global.rakuten.com/en/store/gppro/item/1417035/?s-id=borderless_browsehist_en&quot; rel=&quot;nofollow&quot;&gt;these tiles&lt;/a&gt; and &lt;a href=&quot;http://global.rakuten.com/en/store/gppro/item/1417039/?s-id=borderless_recommend_item_en&quot; rel=&quot;nofollow&quot;&gt;these as well&lt;/a&gt; while surfing the net, they obiviously glow when their electrons are exited to a higher state and then de-exite over a long period to give off visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is which chemical might get exited by visible light and them emit back visible light for a long time.&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2014-12-13T21:16:45.393" Title="What type of compound/element might be in these glowing tiles" Tags="&lt;energy&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21664" PostTypeId="1" CreationDate="2014-12-13T14:25:02.203" Score="1" ViewCount="39" Body="&lt;p&gt;I need a good resource that explains the concept of chemical equivalence and its various application in physical chemistry. I have not found anything worthwhile from a Google search. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2014-12-13T14:25:02.203" Title="Good resource for chemical equivalence" Tags="&lt;physical-chemistry&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21665" PostTypeId="1" CreationDate="2014-12-13T14:25:18.913" Score="3" ViewCount="12514" Body="&lt;p&gt;Why does dimethyl ether have a bond angle bigger than 109.5 degrees, in contradiction with VSEPR theory?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5026" LastEditDate="2018-04-20T14:25:54.830" LastActivityDate="2019-08-22T16:29:07.087" Title="Why does dimethyl ether have a bond angle bigger than 109.5 degrees?" Tags="&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21669" PostTypeId="2" ParentId="21661" CreationDate="2014-12-13T15:57:12.543" Score="5" Body="&lt;p&gt;Esters, amides and acids do exhibit tautomerization to the enol form, but the enol content in these compounds is just much less than that found with aldehydes and ketones. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The keto-enol process is an equilibrium; to understand where an equilibrium lies, we must compare the relative stabilities of the products (enol) and reactants (carbonyl compound). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare some resonance structures for an ester and a ketone and their corresponding enols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dPkx1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ester is stabilized by resonance structures such as A, whereas in the case of the ketone, the analogous resonsnce structure A' contributes, but not nearly as much; so ester carbonyls are stabilized compared to ketone carbonyls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the enols, the ester resonance structure B is not particularly favorable due to the negative charge on carbon and positive charge on oxygen, just the reverse of what we would want based on electronegativities.  At least in the ketone case resonance structure B' has a positive charge on hydrogen, so resonance structure B' contributes more than B (this is why carbon-carbon double bond stability increases with increasing alkylation).  As a result the enol resulting from the ketone is stabilized compared to the enol resulting from the ester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just where the carbonyl-enol equilibrium lies will depend on the relative stabilities of the carbonyl compound and the enol.  As we've just discussed, the ester carbonyl is stabilized relative to the ketone carbonyl, this will push the equilibrium further to the carbonyl side for the ester.  Also, the enol from the ketone is somewhat more stabilized than the enol from the ester, this will push the equilibrium to the enol side for the ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both factors, starting material (the carbonyl compound) stability and product (enol) stability will work to increase the enol content for ketones and reduce the enol content for esters.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-13T16:13:41.880" LastActivityDate="2014-12-13T16:13:41.880" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21670" PostTypeId="1" CreationDate="2014-12-13T16:12:57.687" Score="1" ViewCount="163" Body="&lt;p&gt;In an oxygen radical (produced by the break down of O3 for instance), there are 2 unpaired electrons in the outermost subshell which make the molecule unstable. Why don't these electrons pair up which would make the atom more stable (I think)?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastActivityDate="2014-12-13T16:59:07.927" Title="Why don't electrons pair up in an oxygen radical?" Tags="&lt;electronic-configuration&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21673" PostTypeId="2" ParentId="21670" CreationDate="2014-12-13T16:59:07.927" Score="1" Body="&lt;p&gt;Pairing up of electrons in this case is unfavourable due to increased inter-electron repulsion; the average spatial separation of the electrons is higher in the triplet state where the two electrons are unpaired. Regular oxygen molecules also exist in a triplet state and have unpaired valence electrons; they are said to be biradicals.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-13T16:59:07.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21674" PostTypeId="2" ParentId="21622" CreationDate="2014-12-13T17:37:13.427" Score="3" Body="&lt;p&gt;The dipole moment is 2.45 Debye. The molecule is dipolar as you have asymmetrical electron distribution around both the carbon atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/jp001913i&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Complete Structure of Gauche 1,1,2,2-Tetrafluoroethane Determined by Microwave&#xA;Spectroscopy&lt;/em&gt;, &lt;strong&gt;&lt;em&gt;J. Phys. Chem. A&lt;/em&gt;&lt;/strong&gt;, 2000, 104, 9489-9493.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="10256" LastEditDate="2015-05-15T10:44:06.510" LastActivityDate="2015-05-15T10:44:06.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21675" PostTypeId="2" ParentId="18924" CreationDate="2014-12-13T17:49:32.473" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're asking too much. I can answer you in short:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear Decay Reactions occur when the nuclear force$^1$ between nucleons is not able to hold the nucleons together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as it can't hold large nucleons, the nucleus starts emitting particles or following special processes  to reduce mass or nucleons and reach a stable zone, the choice of process depends on where the nucleon is as per a graph $^3$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;$^1$.one of the four fundamental forces of nature. If there was no nuclear force, everything would repel apart. The reason being, if there was only electrostatic force between protons, the whole nucleus$^{1a}$  would repel away. So another &quot;nuclear force&quot;, acts into play, which is blind to the charges, and acts according to another intrinsic property$^{1b}$ called color$^{1c}$ . &lt;br&gt;So now when nucleons are at the order of 1 fermi$^{1d}$ , the force is attractive and stops acting very (very) quickly (dying).&lt;br&gt;&#xA;&lt;sup&gt;$^{1a}$ actually, only protons, to be specific&lt;br&gt;&#xA;$^{1b}$ like mass, charge, etc.&lt;br&gt;&#xA;$^{1c}$ and involves variety of particles called gluons, mesons, hyper-mesons, quarks, etc.&lt;br&gt;&#xA;$^{1d}$ named after great scientist, $1$ fm$=10^{-15}$ m&lt;br&gt;&lt;/sup&gt;&#xA;$^{2}$$\beta^-,\beta^+,\ce{_2^4He}(\alpha)$, K-capture etc.&lt;br&gt;&#xA;$^3$ which predicts that is no. of protons = no. of neutrons, then nucleus is stable ,generally, only for lighter nucleus(es) and for heavier ones it bends a little upward like a parabola ($y=x^2$)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-13T17:49:32.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21676" PostTypeId="2" ParentId="21665" CreationDate="2014-12-13T17:58:02.920" Score="5" Body="&lt;p&gt;In case of $\ce{Me2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two methyl groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other contradiction too like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of $\ce{(SiH3)2O}$, the huge bond angle increase is mainly due to back bonding and partly due to steric repulsion.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-13T17:58:02.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21677" PostTypeId="1" CreationDate="2014-12-13T17:59:32.850" Score="0" ViewCount="511" Body="&lt;p&gt;A $4~\mathrm{L}$ sample of gas is cooled from $67~\mathrm{°C}$ to a temperature at which its volume is $2.6~\mathrm{L}$. What is this new temperature? Assume no change in pressure of the gas.&lt;/p&gt;&#xA;" OwnerUserId="11296" LastEditorUserId="7448" LastEditDate="2017-03-28T17:57:17.157" LastActivityDate="2017-03-28T17:57:17.157" Title="A 4.00 L sample of gas is cooled from 67 °C to a temperature at which its volume is 2.60 L" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-12-13T21:30:21.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="21678" PostTypeId="2" ParentId="21677" CreationDate="2014-12-13T18:01:07.133" Score="2" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt; Use Ideal gas equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\large PV=nRT$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given pressure is constant, can you tell the final temperature from this relation?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-13T18:01:07.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21683" PostTypeId="2" ParentId="18924" CreationDate="2014-12-13T19:07:42.800" Score="1" Body="&lt;p&gt;The reaction between two ions to form a molecule has no activation energy as they are attracted to each other and therefore no force, and no energy is required to bring them together. e.g &#xA;$\ce{H+ + H- -&amp;gt; H2}$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-13T19:07:42.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21684" PostTypeId="1" AcceptedAnswerId="21746" CreationDate="2014-12-13T19:22:00.967" Score="13" ViewCount="15962" Body="&lt;p&gt;This also raises questions that I have about the Haber Process which produces ammonia ($\ce{NH3}$) from molecular nitrogen ($\ce{N2}$) and hydrogen ($\ce{H2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard multiple times that bond between diatomic nitrogen is one of the strongest bonds in nature due to the fact that it is a triple covalent bond that fills the valence shells of both atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that at high temperatures it is possible to break this bond, but I don't understand why the resulting Nitrogen atoms wouldn't simply return to their previous bonds as the temperature cooled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I read that lightning can result in this reaction:  $\ce{N2 + O2 -&amp;gt; 2NO}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would the atoms not return to their original bonds since they would be more stable in that manner? Is bonding indiscriminate at high energy levels? Completely random and dependent on luck?&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:03.923" LastActivityDate="2021-03-21T20:09:23.917" Title="Why does N₂ react with O₂ to Form NO at high temperatures?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;bond&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21685" PostTypeId="1" AcceptedAnswerId="22411" CreationDate="2014-12-13T21:12:20.577" Score="7" ViewCount="6067" Body="&lt;p&gt;As my understanding, there should be an oxidation-reduction reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;    \ce{2 Fe^3+ + Cu → 2Fe^2+ + Cu^2+}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I always see the process using &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; to etch copper, but I never see people using &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2(SO4)3}$&lt;/span&gt; to do this. Is it because such reaction does not occur, or is it because &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; is acidic in aqueous solution?&lt;/p&gt;&#xA;" OwnerUserId="11298" LastEditorUserId="41328" LastEditDate="2019-01-21T16:04:53.333" LastActivityDate="2019-01-21T16:06:16.263" Title="Does iron(III) sulfate react with copper?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21686" PostTypeId="2" ParentId="21663" CreationDate="2014-12-13T21:16:45.393" Score="3" Body="&lt;p&gt;Glow in the dark materials are materials that exhibit phosphorescence. The most common materials used for this effect are zinc sulfide and strontium aluminate, doped with various elements that affect the colour of luminescence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorescence is similar to fluorescence, where an electron is excited to a higher energy state, but instead of dropping back immediately by emitting a photon, an intersystem crossing event occurs where the electron's spin flips resulting in a triplet state (S=0 to S=1 in the figure). Because the electron has the same spin as the other electron it left behind in the ground state, the transition to the ground state is forbidden and thus occurs much more slowly. In practice, a given material usually exhibits both fluorescence and phosphorescence. Materials that are very phosphorescent have a high probability of intersystem crossing so that when light is shined on them, a large population builds up in triplet states that will release light slowly. What makes a given material have a high intersystem crossing probability is pretty complicated, but it tends to be higher in molecules with large atoms—even things like fused silica glasses are phosphorescent, but very weakly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r0Md8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorescence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-13T21:16:45.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21687" PostTypeId="2" ParentId="20439" CreationDate="2014-12-13T23:07:59.773" Score="8" Body="&lt;p&gt;If azobisisobutyronitrile (AIBN) underwent homolytic cleavage over time, do you expect the resulting cyanoprop-2-yl radicals would survive as free radicals for a long period of time? If not, what would be the decomposed products?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkali metals reduce benzophenone in THF to form blue diphenylketyl radical anions. I'm not sure azobisisobutyronitrile (AIBN) or cyanoprop-2-yl radical can do that.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe you can test the structural integrity of AIBN by 1H-NMR, IR or Raman spectroscopy?&#xA;The -N=N- stretching vibration of a symmetrical trans-azo compound like AIBN is &quot;IR-forbidden&quot; but absorbs in 1576cm-1 region of the Raman spectrum (1). If it decomposes, perhaps you can see the decomposed products in the mass spectrum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe you can try to activate the subject AIBN as a radical initiator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Perhaps one can observe a minute amount of nitrogen gas upon homolytic cleavage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. AIBN can be used as radical initiator in radical polymerization. Once activated as cyanoprop-2-yl radical, it readily attacks olefinic double bond of vinyl monomers. Perhaps you can test such reaction and see if some polymerization occurs using mass spectroscopy?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;R.M. Silverstein and F.X. Webster: Spectrometric Identification of Organic Compounds (6th ed.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9933" LastActivityDate="2014-12-13T23:07:59.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21688" PostTypeId="1" CreationDate="2014-12-13T23:29:11.147" Score="4" ViewCount="186" Body="&lt;p&gt;If I have $0.01~\mathrm{M}$ $\ce{HBr}$ in $0.09~\mathrm{M}$ $\ce{KBr}$, why would the presence of $\ce{Br-}$ ions from the $\ce{KBr}$ not reduce the dissociation of $\ce{HBr}$ based on Le Chatelier's principle? Or would it?&lt;/p&gt;&#xA;" OwnerUserId="2549" LastEditorUserId="16683" LastEditDate="2015-11-13T20:34:06.730" LastActivityDate="2016-05-27T07:08:31.500" Title="Common ion effect and acids" Tags="&lt;acid-base&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21689" PostTypeId="2" ParentId="21662" CreationDate="2014-12-14T00:03:37.387" Score="14" Body="&lt;p&gt;The density of an element is related to how many atoms can be placed in a given volume &lt;strong&gt;and&lt;/strong&gt; the weight of the nuclei.  Therefore, the smaller the atomic radius of an atom &lt;strong&gt;and&lt;/strong&gt; the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation.  This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The small atomic radius can be attributed to the following factors&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;f orbitals are very diffuse and therefore result in poor screening of the electrons further out.  In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Due to osmium's high atomic number, relativistic effects come into play.  These relativistic effects have been discussed previously on SE Chem for &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637&quot;&gt;gold&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13667/why-does-mercury-have-low-melting-and-boiling-points/13671#13671&quot;&gt;mercury&lt;/a&gt;.  A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts.  Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus).  Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The orbital contraction caused by these two effects leads to a small atomic radius for osmium.  Therefore, short metal-metal bonds result.  This is reflected in the &lt;strong&gt;small unit cell volume for osmium (27.96 cubic angstroms)&lt;/strong&gt;.  For comparison the unit cell volume for &lt;strong&gt;lead is 121.3 cubic angstroms!&lt;/strong&gt; As a result, more osmium atoms can pack into a given volume compared to other elements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-04-16T14:48:49.607" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21692" PostTypeId="2" ParentId="20439" CreationDate="2014-12-14T01:20:18.967" Score="2" Body="&lt;p&gt;I don't think that reaction with benzophenone is good idea - as &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzophenone&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article about benzophenon&lt;/a&gt; says it turns blue when reacted with  alkali metals - becoming ketyl anion. Those anions can trap other radicals but are too aggressive to discern whether your is AIBN good. You could maybe use &lt;a href=&quot;https://en.wikipedia.org/wiki/DPPH&quot; rel=&quot;nofollow&quot;&gt;DPPH&lt;/a&gt; - heat it in solution with AIBN and look if it reacts. Or maybe try if it can initialize another, simpler reaction like polimerisation, or maybe analyse it chromatographically - there are lots of options.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-06-14T19:53:19.357" LastActivityDate="2016-06-14T19:53:19.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21694" PostTypeId="1" CreationDate="2014-12-14T03:03:59.720" Score="0" ViewCount="316" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/IH0vUlt&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/IH0vUlt.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we are given this problem, there is a certain problem. We have a strong base, but hydrogen on the carbon of the phenyl. E2 is impossible. Also, SN2 is not possible since we have a hindered base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only product we could get is SN1. My teacher put up the answer in handwritten black in the above picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However can SN1 reactions happen when there is a strong base in the solution? Would that strong base be the nucleophile or the solvent tert-butyl alcohol be the nucleophile?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-12-14T17:32:13.607" Title="Strong Base combined with Sp2 carbons in Nucleophilic Substitution and Beta-Elimination Reactions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21696" PostTypeId="1" AcceptedAnswerId="21738" CreationDate="2014-12-14T13:08:23.267" Score="8" ViewCount="1087" Body="&lt;p&gt;Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Fe(II) is oxidised to Fe(III) ions as well, which are brown in colour, so shouldn’t the solution turn brown instead of colourless? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen these observations experimentally in the lab but still did not understand. I know I’m probably thinking about it the wrong way so any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7481" LastEditorDisplayName="user7951" LastEditDate="2015-05-13T09:25:09.663" LastActivityDate="2015-05-30T20:22:24.563" Title="Where does the Fe(III) ion go?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21697" PostTypeId="2" ParentId="21696" CreationDate="2014-12-14T13:48:41.390" Score="-1" Body="&lt;p&gt;&lt;strong&gt;This answer mentions about colours of aqueous iron solutions and how to test them (only)&lt;/strong&gt;&#xA;$$\ce{10 FeSO4 + 2 KMnO4 + 8 H2SO4 -&amp;gt; 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron(II) salts are usually very pale green in aqueous solution and iron(III) salts are yellow in aqueous solution. You can test what form of iron is present by adding thiocyanate, which gives red precipitate exclusively with iron(III): Place a drop of test solution on a spot plate place and add one drop of 1 per cent ammonium thiocyanate solution. A deep-red colouration may appear. Nitrites should be avoided in acid solution as they form nitrosyl thiocyanate $\ce{NOSCN}$, which yields a red colour, disappearing on heating similar to that with iron(III)&#xA;$$\ce{Fe^3+ + 3 SCN- -&amp;gt;Fe(SCN)3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And iron(II) can be detected most reliably with 2,2'-bipyridine reagent where deep-red complex bivalent cation $\ce{[Fe(C10H8N2)3]^2+}$ is formed in mineral acid solution. Test a drop of the faintly acidified test solution with one drop of the reagent on a spot plate; a red colouration may be obtained.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-05-30T20:22:24.563" LastActivityDate="2015-05-30T20:22:24.563" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21698" PostTypeId="1" AcceptedAnswerId="21700" CreationDate="2014-12-14T14:04:48.447" Score="1" ViewCount="684" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Compound P(&lt;span class=&quot;math-container&quot;&gt;$\ce{C7H6O}$&lt;/span&gt;) on refluxing with aqueous ethanolic &lt;span class=&quot;math-container&quot;&gt;$\ce{KCN}$&lt;/span&gt; solution for 1 hour, forms Q(&lt;span class=&quot;math-container&quot;&gt;$\ce{C14H12O2}$&lt;/span&gt;). Q on oxidation gives R(&lt;span class=&quot;math-container&quot;&gt;$\ce{C14H10O2}$&lt;/span&gt;). R on refluxing with aqueous &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; solution followed with acidification forms an &lt;span class=&quot;math-container&quot;&gt;$\alpha-$&lt;/span&gt;hydroxyacid S (&lt;span class=&quot;math-container&quot;&gt;$\ce{C14H12O3}$&lt;/span&gt;). What are the structures of the compounds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;My effort:&lt;/h3&gt;&#xA;&lt;p&gt;P has DBE (double bond equivalent) equal to 8, so 2 benzene rings (not possible with 7 carbons) or some cyclic or many double bonds might be necessary. Now some enolate reaction might be happening with &lt;span class=&quot;math-container&quot;&gt;$\ce{KCN}$&lt;/span&gt;, like  Aldol or Perkin or Knoevnagel or Stobbe or Claisen or Benzion etc. Oxidation doesn't affect the formula so it might be becoming an isomer (that too a functional one). After that a Cannizaro might be the thing.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-16T01:40:20.387" Title="Identify structure of compounds from their reactions and molecular formula" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;carbonyl-compounds&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21699" PostTypeId="1" AcceptedAnswerId="21701" CreationDate="2014-12-14T14:47:11.467" Score="2" ViewCount="2214" Body="&lt;p&gt;From the diagrams I've seen $\ce{HNO3}$ has a dative covalent bond so why doesn't it have a positive charge?&lt;/p&gt;&#xA;" OwnerUserId="8176" LastEditorUserId="4231" LastEditDate="2014-12-14T17:19:19.853" LastActivityDate="2015-05-21T13:30:39.230" Title="Why doesn't nitric acid have a positive charge?" Tags="&lt;acid-base&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21700" PostTypeId="2" ParentId="21698" CreationDate="2014-12-14T16:37:31.743" Score="5" Body="&lt;p&gt;You weren't far from resolution. The substrate is &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzaldehyde&lt;/a&gt;, first reaction &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzoin_condensation&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzoin condensation&lt;/a&gt;, so Q is &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzoin&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzoin&lt;/a&gt;. It is later oxidated to benzil - R ($\ce{C14H10O2}$) (not ($\ce{H12}$) - in oxidation formula must change). Finally benzil undergoes &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzilic_acid_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzilic acid rearrangement&lt;/a&gt;, so S is benzilic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how one could get to the solution... Compounds having 6-8 carbons and highly unsaturated are usually benzene derivatives. In this case benzaldehyde is only common thing with this formula (it was unlikely to be for example tropone). First reaction must be dimerisation, next oxidation, and hydration. &lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2017-07-31T20:27:59.637" LastActivityDate="2017-07-31T20:27:59.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21701" PostTypeId="2" ParentId="21699" CreationDate="2014-12-14T17:20:08.267" Score="5" Body="&lt;p&gt;One side of the $\ce{HNO3}$ molecule &lt;em&gt;does&lt;/em&gt; have a positive charge, but this is neutralized by the negative charge resulting from the Oxygen atom with only one bond to the nitrogen. These charges cancel leaving an overall neutral molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZyN9.png&quot; alt=&quot;nitric acid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2234" LastEditorUserId="13339" LastEditDate="2015-05-21T13:30:39.230" LastActivityDate="2015-05-21T13:30:39.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21702" PostTypeId="2" ParentId="21598" CreationDate="2014-12-14T17:26:25.947" Score="5" Body="&lt;p&gt;I think you are misunderstanding the article linked in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process described in the article, the starting material is A360 aluminum metal alloy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make very pure aluminum metal from impure aluminum metal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is an electrolysis process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Al(anode) + 7AlCl_4^{−} -&amp;gt; 4Al2Cl_7^{−} + 3e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Al2Cl_7^{−} + 3e- -&amp;gt; Al(cathode) + 7AlCl_4^{−}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existing industry techniques referred to by the article as requiring ~15kw-hr/kg are also for refining impure aluminum metal to yield high purity aluminum.  This number is not referring to purification from bauxite.  The &quot;three layer process&quot; is one of the existing refining techniques, and it does involve ionic liquids, but the electrolyte is liquid only at high temperature.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-14T17:26:25.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21703" PostTypeId="2" ParentId="21694" CreationDate="2014-12-14T17:32:13.607" Score="1" Body="&lt;p&gt;The strong base is capable of being a nucleophile in this sort of situation; deprotonating tert-butyl alcohol will increase its nucleophilicity. In an SN1 reaction the overall rate is only dependent on the loss of a leaving group, so a strong base/poor nucleophile will react via SN1 as long as there are not plausible E1 pathways (e.g. sp3 bonded hydrogen alpha to carbocation). &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-14T17:32:13.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21704" PostTypeId="2" ParentId="21688" CreationDate="2014-12-14T17:43:48.910" Score="7" Body="&lt;p&gt;The $\ce{Br-}$ would make less $\ce{HBr}$ dissociate, But $\ce{HBr}$ is such a strong acid ($\mathrm{pKa}$ is $\mathrm{-9}$) that still only a trivial amount would be molecular.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-05-26T20:07:16.493" LastActivityDate="2016-05-26T20:07:16.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21705" PostTypeId="1" CreationDate="2014-12-14T18:43:53.893" Score="2" ViewCount="1373" Body="&lt;p&gt;Let's take Iron for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hgPiH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see there are 4 desolate electrons in normal state. Can you tell me which electron moves to make the valence 2 and then 3. We have learnt to find the valence only using the configuration formula. &lt;/p&gt;&#xA;" OwnerUserId="9878" LastActivityDate="2021-04-30T12:33:30.773" Title="How to find the valence of transition elements?" Tags="&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21706" PostTypeId="1" AcceptedAnswerId="21728" CreationDate="2014-12-14T19:31:27.973" Score="-4" ViewCount="164" Body="&lt;p&gt;Thermo-gravimetrc analysis, often refered to as TGA, measures the mass of an unknown &lt;strong&gt;solid&lt;/strong&gt; substance as it heats up, of course for the elemental analysis of the very substance.  When the substance is heating up, due to special chemical or physical reactions, the mass is decreasing. e.g.: Dehydration. Anyhow, we know that in special circumstances the mass of the substance will increase. When? &lt;em&gt;That&lt;/em&gt; is the question.&#xA;( This is not quantum physics! I am talking about rest mass, if you had wondered. As a result, the increase will have nothing to do with Einstein's relativity formulas.) The diagram below is an example of some normal polymer tested in normal conditions, no relations with the exceptions!                                                                                                                                                                           &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xwo8F.png&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CBnbs.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;update: There also is a special type of gas that is blown above the surface of the substance.  The system is designed in a way that doesn't allow the gas to &quot;interfere&quot; with the measurements. And I just got lucky to find another substance experimented. I estimate it is going to be a liquid alkanol, maybe ethanol. Diagram two displays the experiment's results.&#xA;I know it seems a little bit vague, but all the info that I provided at above is all we have got to answer the question.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="23561" LastEditDate="2019-03-11T22:20:16.703" LastActivityDate="2019-03-11T22:20:16.703" Title="When will the mass of the reactant increase while being heated up?" Tags="&lt;analytical-chemistry&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21707" PostTypeId="2" ParentId="21705" CreationDate="2014-12-14T19:41:07.287" Score="1" Body="&lt;p&gt;The two 4s electrons are lost to form the $\ce{Fe^2+}$ ion and then the spin paired 3d electron is lost to form the $\ce{Fe^3+}$ ion. As a general rule the 4s electrons are lost before the 3d (at least for period 4 elements) and spin paired electrons are lost before unpaired ones.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-14T19:41:07.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21708" PostTypeId="2" ParentId="21705" CreationDate="2014-12-14T19:46:30.037" Score="2" Body="&lt;p&gt;For the 2nd and 3rd rows for 4b-7b groups the highest oxidation state equal to the group number is preferable over all else and for the 1st row for 4b-7b groups the highest oxidation state is equal to the group number.&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, for all other information you'll need case-by-case approach (i.e. plainly remember it).&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="-1" LastEditDate="2021-04-30T12:33:30.773" LastActivityDate="2021-04-30T12:33:30.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21709" PostTypeId="1" AcceptedAnswerId="23898" CreationDate="2014-12-14T19:59:19.417" Score="12" ViewCount="13830" Body="&lt;p&gt;Coordinate covalent bonds are bonds on which both electrons from one atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does this happen? Some may think it is because one of the bonding atoms have strong electronegativity. But experimental evidence suggests otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, carbon monoxide. The EN of carbon is 2.6, and oxygen 3.4. This may lead you to think that carbon will donate a pair of electron, but on Princeton.edu:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon monoxide (CO) can be viewed as containing one coordinate bond and two &quot;normal&quot; covalent bonds between the carbon atom and the oxygen atom. This highly unusual description illustrates the flexibility of this bonding description. Thus in CO, carbon is the electron acceptor and oxygen is the electron donor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, in some cases like NO2, it does follows that the less electronegative atom domates electron. But why??&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-12-14T20:43:01.030" LastActivityDate="2018-11-13T14:06:39.517" Title="Why does coordinate covalent bond form?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21710" PostTypeId="2" ParentId="21584" CreationDate="2014-12-14T20:03:00.537" Score="5" Body="&lt;p&gt;Search for the latest IChO tests and preparatory problems. You can find quite a few interesting problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some tests:&#xA;&lt;a href=&quot;http://www.iuventa.sk/files/documents/5_icho/past%20competition%20tasks/volume3-icho41-45.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.iuventa.sk/files/documents/5_icho/past%20competition%20tasks/volume3-icho41-45.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf&lt;/a&gt; (problems 18-21)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html&quot; rel=&quot;noreferrer&quot;&gt;http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html&lt;/a&gt; (problems 18-26)&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2014-12-14T20:11:10.473" LastActivityDate="2014-12-14T20:11:10.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21715" PostTypeId="2" ParentId="20439" CreationDate="2014-12-15T01:55:26.437" Score="4" Body="&lt;p&gt;You'd think that the main decomposition product is tetramethylsuccinyl dinitrile (and perhaps also some peroxidic species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;sup&gt;13&lt;/sup&gt;C NMR the quaternary carbons of both products should have distinct chemical shifts, and assuming relaxation times are similar (they should be) this should give a rough idea about the purity of the product.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2014-12-15T01:55:26.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21717" PostTypeId="2" ParentId="21574" CreationDate="2014-12-15T03:53:18.227" Score="6" Body="&lt;p&gt;I am afraid I disagree with Geoff's assignment of acetonitrile to the point group $C_{\infty v}$ and &quot;approximation&quot; with $C_{2v}$, so I will make another attempt to answer this question. The way I have structured this answer, the resources I used, and the level of detail included follow &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16305/symmetry-labels-for-orbitals/16306&quot;&gt;a similar question on this site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Molecule structure&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First you need to know the geometrical structure of your compound. This can be found easily from &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetonitrile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. The image is reproduced here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/02VL6.png&quot; alt=&quot;Structure of acetonitrile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Assigning the point group&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Then you need to find the point group of acetonitrile. This can be done by following flow charts, one of which is available &lt;a href=&quot;http://symmetry.otterbein.edu/common/images/flowchart.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. While it may be a little difficult at the beginning, with some practise you can look at a molecule and determine its point group accurately, so it is worth working through the flow chart for a range of molecules until you are confident with this. There are other arrangements of this flow chart, but all flow charts should give you the same result, so use the one provided in your course if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the flow chart I have linked to, we go through these steps (if you have a flow chart organised differently, you will encounter the same steps, but likely in a slightly different order - check this is the case!):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;molecule is not linear&lt;/li&gt;&#xA;&lt;li&gt;molecule contains only one $C_3$ axis&lt;/li&gt;&#xA;&lt;li&gt;molecule DOES contain a proper rotation axis which is $C_3$ (rotation over methyl group)&lt;/li&gt;&#xA;&lt;li&gt;the highest rotation axis is $C_3$ but there are no perpendicular $C_2$ axes&lt;/li&gt;&#xA;&lt;li&gt;the molecule does NOT contain a horizontal reflection plane ($\sigma_n$). This can sometimes be confusing -- in general, a horizontal plane is perpendicular to the highest rotation axis.&lt;/li&gt;&#xA;&lt;li&gt;the molecules DOES contain 3 vertical reflection planes ($\sigma_v$).&lt;/li&gt;&#xA;&lt;li&gt;therefore the point group is $C_{3v}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;&quot;Orbitals&quot;&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In your question, you said that you need to &quot;to assign the symmetry labels to orbitals of acetonitrile&quot;. This is a little unclear, because you can mean that you want to assign symmetry labels to atomic orbitals of acetonitrile, or molecular orbitals of acetonitrile. I am going to assume that you need the molecular orbitals here. But how do we find the molecular orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular orbitals are constructed by making combinations of atomic orbitals. Which means that, a logical next step is to consider all the valence atomic orbitals (VAOs) on all the atoms of acetonitrile.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Atomic orbitals and symmetry sets&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To help us talk about VAOs, I have put numbers on all the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5yAvj.png&quot; alt=&quot;Structure of acetonitrile with numerical labels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon, atoms 1 and 2: $2s$, $2p_x$, $2p_y$, $2p_z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen, atom 3: $2s$, $2p_x$, $2p_y$, $2p_z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen, atoms 4, 5 and 6: $1s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we try to tackle these types of problems, it is common to construct symmetry sets of atomic orbitals that would transform in the same way under the operations of the point group of the molecule. You will also notice that you cannot use any of the operations of the group to transform certain atomic orbitals on the same atom into each other (cannot transform $2s$ to $2p$, cannot transform $2p_x$ to $2p_z$, etc), so these should be considered as separate sets. The $p_x$ and $p_y$ orbitals are mixed by the rotations and reflections of the group, so we should consider them as one symmetry set.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In acetonitrile, there is only one nitrogen, so we have to consider as separate sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2s$&lt;/li&gt;&#xA;&lt;li&gt;$2p_z$&lt;/li&gt;&#xA;&lt;li&gt;$2p_x$, $2p_y$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Carbon 1 and carbon 2, though both carbons, cannot be transformed from one to another, and your chemical intuition should also tell you they are different carbons because they have different bonds attached to them, so they are also in separate sets of their own. The three hydrogen atoms, on the other hand, are related by rotation (a $C_{3v}$ rotation on the principal axis, so they can form one set of 3 hydrogen 1s orbitals.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Assigning symmetry labels to each set of valence atomic orbitals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The usual trick at this point is to simply read off the orbitals that have corresponding symmetry labels, e.g. the $p_x$ orbital would transform as $x$ on the character table. This definitely works for any atom that lies on the origin point of the point group, but it is not completely obvious where you would take the origin point for putting a (right handed) axis system on acetonitrile. The &quot;middle&quot; of this molecule would appear to be somewhere between the two carbons, and that may seem a little awkward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that in all cases, the choice of origin and axis system is actually arbitrary, however you should pick the most convenient one. Now, for acetonitrile you can pick any point on the principal rotation axis ($C_{3}$) and it would work, because that is an inherent feature of the $C_{3v}$ point group. If this is not obvious, I suggest looking at the answer &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16305/symmetry-labels-for-orbitals/16306&quot;&gt;I referenced previously&lt;/a&gt;, and also working through any other examples that you can find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily for us, $C_{3v}$ has quite a small character table (i.e. it does not have many symmetry elements), so we can work through it and check. The character table of the point group $C_{3v}$ is shown below and can also be found &lt;a href=&quot;http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=403&amp;amp;option=4&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The character tables for many common point groups can be found &lt;a href=&quot;http://symmetry.jacobs-university.de/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5fTP.png&quot; alt=&quot;Character table for $\C_{3v}$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$s$ orbitals that lie on the principal axis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now let's consider the $2s$ orbital or carbon atom 1. It should be fairly obvious that it transforms back to itself under all the operations of the point group (unchanged under identity $E$, any of the $C_3$ rotations, any of the $\sigma_v$ reflections), and so it transforms as $A_1$. It should also become very obvious that this is also true for the $2s$ orbitals of carbon atom 2 and nitrogen atom 3, even though these atoms are not symmetry-related, their atomic orbitals can have the same symmetry under the point group. This will be important when we come to construct molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$p$ orbitals that lie on the principal axis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For p orbitals, we need to impose an axis system. Conventionally, the principal axis is taken as the $z$ direction. You have to follow these conventions when doing symmetry analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the $2p_z$ orbital or carbon atom 1, which lies along the principal axis. This again transforms as $A_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $2p_x$ and $2p_y$ orbitals are a little tricky. They remain identical under the identity operator $E$, giving a character of 2. For the reflection operator $\sigma_v$, one of the three $\sigma_v$ planes would keep one identical while inverting the other one (giving zero), and the other two $\sigma_v$ planes give characters of zero, so the overall character for $3 \sigma_v$ is zero. The $C_{3v}$ is the difficult one, where you have to put vectors on the axis, rotate them, work out the geometry with trigonometry, form the $2 \times 2 $ matrix and work out the trace (i.e. the character) is -1. Or you can use the fact that the character tables show (x,y) transform as the irreducible representation $E$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the same symmetry labels apply for all the atomic orbitals for all atoms that lie on the principal axis.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Set of hydrogen $1s$ orbitals that do not lie on the principal axis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For these hydrogen $1s$ orbitals we examine explicitly how they behave under all operations of the point group.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E$: 3  (all three $1s$ remain unchanged from themselves)&lt;/li&gt;&#xA;&lt;li&gt;$C_v$: 0 (all three change and do not remain at the same position)&lt;/li&gt;&#xA;&lt;li&gt;$\sigma_v$: 1 (for any one operation, only 1 of the orbitals remain the same, the rest change)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a reducible representation. You need to reduce it to find the symmetry labels for the hydrogen $1s$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Symmetry labels for all valence atomic orbitals&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For the whole molecule of acetonitrile, the valence atomic orbitals give the irreducible representation $4 A_1 \oplus 4 E$. From the discussion above you should be able to deduce the exact symmetry label for any atomic orbital on any atom of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Molecular orbitals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You know that you need to generate as many molecular orbitals as atomic orbitals, and only the atomic orbitals of the same symmetry can mix. Molecular orbitals use lower case letters, so you expect two types of symmetry labels based on the irreducible representations of the VAOs. You should be able to say how many molecular orbitals of each type you expect, remembering that the irreducible representation $E$ is two-dimensional and indicates a degenerate pair of VAOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetry analysis cannot give you the energy of these molecular orbitals, however, so you cannot know how they are ordered relative to one another.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2014-12-15T03:53:18.227" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21723" PostTypeId="1" AcceptedAnswerId="21731" CreationDate="2014-12-15T07:24:13.293" Score="1" ViewCount="27" Body="&lt;p&gt;What is the most abundant substances which I may use like coupling agent for Wood plastic composite with HDPE polymer? I don't want maleated polyolefin because they are not so abundant in my country.&lt;/p&gt;&#xA;" OwnerUserId="11309" LastActivityDate="2014-12-15T13:08:18.310" Title="Abundant substances which I may use like coupling agents" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21724" PostTypeId="2" ParentId="21638" CreationDate="2014-12-15T09:03:46.133" Score="9" Body="&lt;p&gt;I'm not aware of a $\ce{NaBH4}$-based reduction of nitroalkenes to saturated amines, such as in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vL8fT.png&quot; alt=&quot;nitroalkene-to-sat.amine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however possible to reduce nitroalkenes to saturated hydroxylamines using $\ce{BH3*THF}$ in the presence of catalytic amounts of $\ce{NaBH4}$. (&lt;a href=&quot;http://dx.doi.org/10.1021/jo00201a030&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a chance to perform the desired transformation in one step via catalytic hydrogenation at normal pressure: &lt;strong&gt;go for it!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handling of these reactions is nothing to be afraid of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, I performed hydrogenations in a standard 2- or 3-neck flasks and used ordinary balloons (think childrens' birthday) as gas storage. &lt;strong&gt;Advice&lt;/strong&gt;: Stuff a piece of rubber tube into the inlet of the balloon and secure it with teflon tape. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either attach this to the glas olive of a valve attached to your flask &lt;/li&gt;&#xA;&lt;li&gt;or take a plastic syringe, cut off the base, stuff the the plastic cylinder into the rubber tube. Fill the balloon and attach a needle. Hydrogen may now be fed to your reaction mixture through a rubber septum cap.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still doubt that the intended reduction by $\ce{NaBH4}$ is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however another established route from $\alpha,\beta$-unsaturated nitroalkenes to amines which is worth to be mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condensation products of aromatic aldehydes and nitroalkenes are reduced by iron in hydrochloric (or acetic) acid to the corresponding arylalkanones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PhtPq.png&quot; alt=&quot;reduction of arylalkanone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These ketones can be converted to primary amines by &lt;strong&gt;reductive amination&lt;/strong&gt; in methanol (or ethanol) in the presence of (excess) ammonia and, of course, hydrogen over Raney Nickel or Urushibara Nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kcz6C.png&quot; alt=&quot;reductive amination&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the course of this reaction, the imine intermediate is &lt;em&gt;in situ&lt;/em&gt; reduced to the desired primary amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting amines are valuable precursors in the synthesis of natural products: Their conversion to imines or amides and subsequent &lt;em&gt;Pictet-Spengler&lt;/em&gt; or &lt;em&gt;Bischler-Napieralski&lt;/em&gt; cyclisations furnish tetrahydro- and 3,4-dihydroisoquinolines.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-12-17T20:53:10.703" LastActivityDate="2014-12-17T20:53:10.703" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21726" PostTypeId="2" ParentId="46" CreationDate="2014-12-15T10:32:55.510" Score="3" Body="&lt;p&gt;I would advise those who doubt in the existance of ionic solids to google ionic liquids which are widely commercialised.  They probably solidify at sub-zero temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie101653n?journalCode=iecred&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ie101653n?journalCode=iecred&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many pharmaceutical salts can be obtained in the solid amorphous state by spray-drying, freeze-drying or milling.&lt;/p&gt;&#xA;" OwnerUserId="11311" LastActivityDate="2014-12-15T10:32:55.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21727" PostTypeId="2" ParentId="21706" CreationDate="2014-12-15T12:02:54.900" Score="4" Body="&lt;p&gt;The only &quot;special &quot; circumstance I can think of is when the polymer can actually absorb or adsorb moisture or just any gas when it is heated and thus &quot;increasing&quot; the mass somehow. &lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="73" LastEditDate="2014-12-17T01:17:04.507" LastActivityDate="2014-12-17T01:17:04.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21728" PostTypeId="2" ParentId="21706" CreationDate="2014-12-15T12:15:16.633" Score="7" Body="&lt;p&gt;&lt;strong&gt;There's nothing from nowhere:&lt;/strong&gt; Heating of a polymer &lt;em&gt;in vacuum&lt;/em&gt; will never lead to an increase in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of an acidic catalyst and water (or alkanols, etc.), however, polyesters might partly undergo cleavage and/or transesterification with a net mass uptake. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-15T12:15:16.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21729" PostTypeId="2" ParentId="19449" CreationDate="2014-12-15T12:47:42.573" Score="5" Body="&lt;p&gt;0.12 Normal = 0.06 Molar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparision, concentrated sulphuric acid is 18 Molar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the 0.12 N sulphuric acid is over 99% water.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would it be safe to handle this without a respirator? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the vapor pressure of sulphuric acid is low, and it has already been diluted with considerable water, so there is no need for a respirator.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Goggles are the most important safety equipment.  An eye wash station and shower are also standard lab safety equipment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gloves are good because liquids will not pass through them.  Ordinary clothing isn't really helpful because if you did spill something on that part of your body, the substance would absorb into the cloth and you would just be delayed in washing that part of your body by having to take off the clothes.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pronged contact may cause some damage, but nothing comparable to the concentrated acid.  If you spill some on yourself, stay calm, and wash with soap and water. If you spill on your clothes, take the wet clothes off first, then wash with soap and water.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is always a good rule to keep in mind.  0.12N is dilute enough that little heat will be released from 0.5mL, but just follow the rule.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you are diluting another 1000 fold.  The solution will be around pH 4, similar to some rain water.  I would still want you to wear goggles and not spray in the direction of anyone.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-15T13:56:24.353" LastActivityDate="2014-12-15T13:56:24.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21730" PostTypeId="1" AcceptedAnswerId="21732" CreationDate="2014-12-15T13:00:22.137" Score="0" ViewCount="29" Body="&lt;p&gt;We know that higher the carbon, the more is hydrophobic the compound nature is, Still when we take a compound like sugar , a hand of sugar, it readily dissolves in water? can someone explain me reg this  &lt;/p&gt;&#xA;" OwnerUserId="11312" LastActivityDate="2014-12-15T13:10:50.343" Title="How does higher carbonyl compounds disolve in water?" Tags="&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21731" PostTypeId="2" ParentId="21723" CreationDate="2014-12-15T13:08:18.310" Score="1" Body="&lt;p&gt;If you don'y want to use maleated polyolefin (the best coupling agent for this composite), you can use oxidized polyethylene, ethylene propylene rubber, ethylene-co-glycidyl methacrylate or polyolefin elastomer.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-15T13:08:18.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21732" PostTypeId="2" ParentId="21730" CreationDate="2014-12-15T13:10:50.343" Score="1" Body="&lt;p&gt;Sugar contains a lot of hydroxyl groups. So due to hydrogen bonds, sugar can readily dissolves in water.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-15T13:10:50.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21733" PostTypeId="1" AcceptedAnswerId="21734" CreationDate="2014-12-15T13:13:46.320" Score="5" ViewCount="1927" Body="&lt;p&gt;Water dissociates into the hydroxide and hydronium ions but, why does this happen if the hydroxide ion is such a poor leaving group. Alcohols don't undergo $\ce{S_{N}1}$ reactions because the hydroxide ion doesn't dissociate so why does it happen in water?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2014-12-17T04:19:29.687" LastActivityDate="2014-12-17T04:19:29.687" Title="If the hydroxide ion is such a poor leaving group then why does water dissociate?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21734" PostTypeId="2" ParentId="21733" CreationDate="2014-12-15T13:27:26.130" Score="7" Body="&lt;p&gt;Only 1 in 10,000,000 water molecules dissociate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HI, HBr and HCl fully dissociate, and I-, Br-, and Cl- and are much better leaving groups. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-15T13:27:26.130" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21735" PostTypeId="2" ParentId="21589" CreationDate="2014-12-15T13:28:01.877" Score="-1" Body="&lt;p&gt;It seems to me that you should use &lt;a href=&quot;https://en.wikipedia.org/wiki/Osmotic_pressure#Morse_equation&quot; rel=&quot;nofollow&quot;&gt;Morse equation&lt;/a&gt; $\mathrm{\Pi=iMRT}$ - from your data you should get  $\mathrm{M}$ - molarity and then molar mass of the protein. You can skip $\mathrm{i}$ - it's not a problem if it's a dimer or oligomer.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-08-09T23:43:12.147" LastActivityDate="2016-08-09T23:43:12.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21737" PostTypeId="1" AcceptedAnswerId="21739" CreationDate="2014-12-15T14:16:53.260" Score="8" ViewCount="177" Body="&lt;p&gt;State functions such as &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; only depend on the state of the system and are not dependent on the &quot;path&quot; that took the system to that state (which would be the case for work, for example, which is not a state function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G=\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p+\left(\frac{\partial G}{\partial T}\right)_p\mathrm{d}T$$&lt;/span&gt;&#xA;Consequently, by comparing coefficients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=\left(\frac{\partial G}{\partial p}\right)_T$$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$$-S=\left(\frac{\partial G}{\partial T}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just taking the equation involving &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; now to save time and space:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\int_{p_1}^{p_2}V\mathrm{d}p=\int_{p_1}^{p_2}\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the perfect gas equation and integrating leaves the result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is independent of the path taken to get to the final state, why shouldn't I use the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=V(p_2-p_1)?$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="41328" LastEditDate="2019-12-11T07:50:44.367" LastActivityDate="2019-12-11T07:52:32.997" Title="Why is it necessary to consider infinitesimal changes in p,V,T for H,U and G given that they're state functions?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21738" PostTypeId="2" ParentId="21696" CreationDate="2014-12-15T14:46:21.950" Score="7" Body="&lt;p&gt;In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hexaaqua iron(III) has a high spin d5 electron configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the d-d transitions are spin forbidden. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.public.asu.edu/~jpbirk/qual/qualanal/iron.html&quot;&gt;Hexaaqua iron(III) is nearly colorless.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-15T16:47:08.267" LastActivityDate="2014-12-15T16:47:08.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21739" PostTypeId="2" ParentId="21737" CreationDate="2014-12-15T15:08:14.933" Score="7" Body="&lt;p&gt;It is important to make clear any requirement that a variable be constant for each equation you write.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G = V\,\mathrm{d}p,$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you would need to assume temperature is constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to go from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}G = V\,\mathrm{d}p$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G = V(p_2 - p_1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you would need to assume volume is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If temperature and volume are both constant, the pressure is constant also.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your final equation is only saying: when temperature, pressure and volume are constant, there is no change in Gibbs free energy. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2019-12-11T07:49:26.617" LastActivityDate="2019-12-11T07:49:26.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21740" PostTypeId="1" AcceptedAnswerId="21747" CreationDate="2014-12-15T15:35:18.437" Score="2" ViewCount="570" Body="&lt;p&gt;OK. I'm working in computational chemistry and the problem that bothers me is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a molecule, I know that it has a central asymmetric atom with some parity which is either &quot;Anticlockwise&quot; or &quot;AntiAntiClockwise&quot; (or a really stupid way to say clockwise). I have the molecule drawn on a 2D canvas (they are in one plane). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this example you could imagine 4 atoms. I have to specify which atoms lie in a different plane (3D perspective) and I can't seem to understand why is the parity important in that case? I'm using the SMILES &lt;code&gt;O[C@@](F)(Cl)Br&lt;/code&gt; notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZaOSs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZaOSs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused about this concept and it seems like the internet doesn't offer me a lot of information about it. So please tell me how is parity related to the plane the atom lies in? How can I map the edges and dashes to the SMILES  of this structure?&lt;/p&gt;&#xA;" OwnerUserId="11315" LastEditorUserId="41328" LastEditDate="2017-08-06T08:28:29.823" LastActivityDate="2017-08-06T08:28:29.823" Title="Confused about parity of atom and chirality" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;cheminformatics&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21742" PostTypeId="2" ParentId="21740" CreationDate="2014-12-15T16:38:54.500" Score="3" Body="&lt;p&gt;In SMILES notation, the order of the remaining three atoms corresponds to clockwise order when viewed in the direction looking from the first listed atom toward the carbon center.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have written $\ce{O[C@@](F)(Cl)Br}$.  Looking in the direction from the O atom toward the C atom, the clockwise order of the remaining atoms is F, Cl, Br.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could write the other enantiomer as $\ce{O[C@@](F)(Br)Cl}$.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5017" LastEditDate="2014-12-15T18:56:57.590" LastActivityDate="2014-12-15T18:56:57.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21743" PostTypeId="1" AcceptedAnswerId="21745" CreationDate="2014-12-15T17:31:36.843" Score="2" ViewCount="148" Body="&lt;p&gt;Solvents can be categorized as polar and protic. I understand that to be protic there had to be a significantly polar bond involving a hydrogen atom. However, how is the a hydrogen bond different in any way to another dipole-dipole interaction and therefore, what's the need for the distinction between polar and protic solvents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering the hydration enthalpy $\triangle_{hyd} G^{\theta}$ it is not sufficient to simply use the Born equation:&#xA;$$\triangle_{solv} G^{\theta}=-\frac{N_Ae^2z^2}{8 \pi \epsilon_0}(1-\frac{1}{\epsilon_r})$$&#xA;BECAUSE this only considers Coulombic interactions (i.e hydrogen bonds are not solely coulombic interactions). My textbook seems to suggest that hydrogen bonds are somehow different and more significant. My question is why? Surely there should be no distinction between different polar bonds; what makes $\alpha^{\delta-}-H^{\delta +}$ special?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="9961" LastEditDate="2015-10-02T17:01:23.730" LastActivityDate="2015-10-02T17:01:23.730" Title="Why is hydrogen bonding more significant than any other interaction between dipoles?" Tags="&lt;solvents&gt;&lt;dipole&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21744" PostTypeId="2" ParentId="21626" CreationDate="2014-12-15T17:33:12.297" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{f}$&lt;/span&gt; is sensitive to particular stationary phase, up to having different &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{f}$&lt;/span&gt; on different plates.  Traditionally, it is done comparing &lt;span class=&quot;math-container&quot;&gt;$R_\mathrm{f}$&lt;/span&gt; of freshly acquired sample and obtained previously in a different manner reference.&lt;/p&gt;&#xA;&lt;p&gt;However, I agree with Klaus, use melting point of characterisation if possible.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2020-09-08T13:50:28.097" LastActivityDate="2020-09-08T13:50:28.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21745" PostTypeId="2" ParentId="21743" CreationDate="2014-12-15T17:57:20.803" Score="3" Body="&lt;p&gt;Hydrogen bonds often have a significant covalent component. This can be seen in many aspects of the hydrogen bond: it is directional, unlike Coulombic interactions, and the bond length is generally significantly shorter than predicted by Coulombic models. There is great variation in the covalent character of hydrogen bonds depending on their environment, with e.g. bond lengths between 2.2 and 4.0 Angstroms known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other intermolecular interactions have significant covalent components; there is nothing special about hydrogen per se but it is the most important example.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="10256" LastEditDate="2014-12-27T09:37:52.277" LastActivityDate="2014-12-27T09:37:52.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21746" PostTypeId="2" ParentId="21684" CreationDate="2014-12-15T17:58:35.373" Score="12" Body="&lt;p&gt;$\Delta G = \Delta H - T \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the $\ce{N2 + O2 -&amp;gt; 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to decompose to nitrogen and oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that NO is unstable at room temperature tells us nothing about the &lt;em&gt;rate&lt;/em&gt; of the decomposition reaction.  In fact there was an &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j150561a030&quot;&gt;interesting 40 year study&lt;/a&gt; showing very little decomposition of NO sealed in glass tubes over that time period.  The authors' calculations show that without a catalyst the timescale of decompsition could be $10^{29}$ years!&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-15T17:58:35.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21747" PostTypeId="2" ParentId="21740" CreationDate="2014-12-15T18:55:44.683" Score="4" Body="&lt;p&gt;This is really a cheminformatics problem. It's a reflection of the chirality or atomic stereochemistry about the atom you mention.&lt;/p&gt;&#xA;&lt;p&gt;There are multiple ways of specifying chiral information in 2D and 3D.&lt;/p&gt;&#xA;&lt;p&gt;As mentioned by &lt;a href=&quot;https://chemistry.stackexchange.com/a/21742/5017&quot;&gt;Dave&lt;/a&gt; SMILES uses a clockwise and counter-clockwise from left to right in the string.&lt;/p&gt;&#xA;&lt;p&gt;In 2D, we usually use wedge and hash notation. The problem is that some users use wedges and hashes for graphical appearance and not for stereochemical information (e.g., pseudo-3D look). So file formats also add atom parity to indicate the &lt;em&gt;actual&lt;/em&gt; stereochemistry of atoms.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;parity&amp;quot; idea has to do with the relationship between the atoms as they are ordered in the file, vs. the stereochemistry around the central atom. As indicated in the other answer, these are enantiomers, even though they use the same &amp;quot;@@&amp;quot; SMILES stereochemical indicator:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{O[C$@@$](F)(Cl)Br}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{O[C$@@$](F)(Br)Cl}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Note that the atom ordering is different, so the &amp;quot;parity&amp;quot; will be different. For your example, &lt;span class=&quot;math-container&quot;&gt;$\ce{O[C$@@$](F)(Cl)Br}$&lt;/span&gt; I get &amp;quot;1&amp;quot; for the MDL parity.&lt;/p&gt;&#xA;&lt;p&gt;From the MDL specification (Appendix A: Stereo Notes):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Number the atoms surrounding the stereo center with 1, 2, 3, and 4 in order of increasing atom number (position in the atom block) (a hydrogen atom should be considered the highest numbered atom, in this case atom 4). View the center from a position such that the bond connecting the highest-numbered atom (4) projects behind the plane formed by atoms 1, 2, and 3.&lt;/p&gt;&#xA;&lt;p&gt;Sighting towards atom number 4 through the plane (123), you see that the three remaining atoms can be arranged in either a clockwise or counterclockwise direction in ascending numerical order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XsoKc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;For more details, you might want to look at the &lt;a href=&quot;http://openbabel.org/dev-api/group__stereo.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Open Babel documentation&lt;/a&gt;, &lt;a href=&quot;http://opensmiles.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;OpenSMILES&lt;/a&gt; or the &lt;a href=&quot;http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;MDL specification&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-03-13T20:14:20.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21749" PostTypeId="1" AcceptedAnswerId="21807" CreationDate="2014-12-15T20:12:32.150" Score="5" ViewCount="5979" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;List the following alkanes in order of increasing boiling point. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A. $\ce{CH3(CH2)4CH3}$&lt;br&gt;&#xA;  B. $\ce{(CH3)2CHCH(CH3)2}$&lt;br&gt;&#xA;  C. $\ce{CH3CH2CH(CH3)CH2CH3}$  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Answer: $C&amp;lt;B&amp;lt;A$, low to high.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is it that B has not a lower boiling point than C? Is it not more branched? I believe branching decreases surface area, leading to less intermolecular interactions, and to a higher boiling point.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorDisplayName="user10259" LastEditDate="2015-05-14T00:52:26.853" LastActivityDate="2015-05-14T00:52:26.853" Title="Does more branching lead to lower boiling point in alkanes?" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21750" PostTypeId="1" CreationDate="2014-12-15T20:15:55.390" Score="2" ViewCount="70" Body="&lt;p&gt;Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate? Alternatively, is there a hygroscopic substance that will absorb water and expand into a dense/tough non-permeable mass? This is a material to seal off a small inner diameter pipe once water intrudes into it&lt;/p&gt;&#xA;" OwnerUserId="11318" LastEditorUserId="5017" LastEditDate="2014-12-15T21:01:14.260" LastActivityDate="2015-05-19T13:13:12.893" Title="Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21751" PostTypeId="2" ParentId="20314" CreationDate="2014-12-15T20:58:24.550" Score="0" Body="&lt;p&gt;The answer to my principal question is that you can concentrate an aqueous ethanol solution by &quot;room temperature distillation&quot;. I don't have the right equipment to be able to determine what variation in output quantity and strength might be obtained by varying the temperature of the source solution and the condenser surface. But I can state that the alcohol concentration in the condensate under the conditions I used finally was ca 18% coming from the source solution which was 11%. What I like about the process is that little energy is used for heating(30 watts) and no water is wasted for cooling as in conventional distillation.&#xA;Further experiment - from a wash of sugar and beer yeast using the method above I obtained a distillate which measured 36% on my $3 Russian alcohol hygrometer. &lt;/p&gt;&#xA;" OwnerUserId="10151" LastEditorUserId="10151" LastEditDate="2014-12-19T20:49:54.287" LastActivityDate="2014-12-19T20:49:54.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21752" PostTypeId="1" CreationDate="2014-12-15T21:08:27.317" Score="1" ViewCount="92" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/nfezhDX&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/nfezhDX.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there is a mistake. If we go back to the mechanisms, E2 does not go through a carbocation. So, there should not be any impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E1 does however. So, E1 should be circled. Is there a missing point or argument in my logic.&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-12-15T21:08:27.317" Title="In which beta-elimination reaction do the structures of the carbocations formed influence the reactivity?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21753" PostTypeId="2" ParentId="21749" CreationDate="2014-12-15T22:01:43.820" Score="4" Body="&lt;p&gt;$A$ n-hexane, boiling point 69 degrees C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B$ 2,3 dimethyl butane, boiling point 58 degrees C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ 3-methyl pentane, boiling point 63 degrees C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B&amp;lt;C&amp;lt;A$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-15T22:01:43.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21754" PostTypeId="2" ParentId="16086" CreationDate="2014-12-15T22:41:18.477" Score="0" Body="&lt;p&gt;One process I can think of is that it's toxic through alkalosis and a high enough dose can overwhelm homeostasis.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2014-12-15T22:41:18.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21755" PostTypeId="2" ParentId="21638" CreationDate="2014-12-15T22:54:29.180" Score="3" Body="&lt;p&gt;There are numerous reports on transfer hydrogenation, where hydrogen is transferred from strongly reducing agent onto double bond or other groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, Raney nickel was utilized in reduction of both nitrogroups to amines and alkenes to alkanes, &lt;a href=&quot;https://www.erowid.org/archive/rhodium/chemistry/cth.nitro2amine.rani-hcooh.html&quot; rel=&quot;nofollow&quot;&gt;using formic acid for the former&lt;/a&gt;  &lt;a href=&quot;http://rua.ua.es/dspace/bitstream/10045/12847/4/proofs2.pdf&quot; rel=&quot;nofollow&quot;&gt;and isopropanol for the latter&lt;/a&gt;. I guess, that combined process is also likely to work.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="7920" LastEditDate="2014-12-19T00:35:22.740" LastActivityDate="2014-12-19T00:35:22.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21756" PostTypeId="1" AcceptedAnswerId="21780" CreationDate="2014-12-15T23:34:04.453" Score="14" ViewCount="1411" Body="&lt;p&gt;I wonder why microwave has more &quot;energy&quot; to heat things up than visible light does. Microwave has a much longer wavelength, therefore should supposedly has less energy than visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does wavelength and frequency really have to do with how fast a certain type of electromagnetic radiations give an object heat, or it is something to do with how a microwave oven is built?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4356" LastEditDate="2014-12-16T17:41:16.230" LastActivityDate="2020-10-16T08:56:28.993" Title="Why does microwave heat up things so much more quickly than visible light?" Tags="&lt;everyday-chemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21757" PostTypeId="2" ParentId="21638" CreationDate="2014-12-15T23:44:38.680" Score="3" Body="&lt;p&gt;$\ce{NaBH4}$ will only reduce the $\ce{RCH=CHNO2}$ to $\ce{RCH2CHNO2}$. An additional reducing agent, such as $\ce{Zn/HCl}$, can reduce the nitro group to the amine. A catalyst + $\ce{NaBH4}$ could do the whole thing, but you'll have to look at the literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalytic hydrogenation, using $\ce{Pd/C}$ under moderate conditions (around 5 bar $\ce{H2}$, 30-90 °C), can reduce $\ce{RCH=CHNO2}$ directly to  $\ce{RCH2CHNH2}$.&lt;/p&gt;&#xA;" OwnerUserId="11320" LastEditorUserId="418" LastEditDate="2015-05-21T12:28:21.353" LastActivityDate="2015-05-21T12:28:21.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21758" PostTypeId="2" ParentId="21684" CreationDate="2014-12-16T00:01:14.703" Score="1" Body="&lt;p&gt;The Haber process usually uses metal catalysts to help the bonds break. It runs at extremely high temperatures and pressures, enough to cause worry that it will burst steel vessels. At these conditions breaking $\ce{N2}$ and $\ce{H2}$ can and does happen. It's true that the nitrogen-nitrogen bond is broken, but once that happens ammonia forms quickly.&lt;/p&gt;&#xA;" OwnerUserId="11320" LastEditorUserId="4231" LastEditDate="2015-02-27T17:18:59.660" LastActivityDate="2015-02-27T17:18:59.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21759" PostTypeId="2" ParentId="16086" CreationDate="2014-12-16T00:07:00.243" Score="1" Body="&lt;p&gt;Ammonia will harm you like any other base. Although it's not directly caustic, it will cause inflammation and general &quot;corrosion&quot; of whatever membranes it encounters, such as those in your lungs when you get a good drag of ammonia gas. Upon contact with your body it becomes NH4OH and the hydroxide hydrolyzes amides in proteins.&lt;/p&gt;&#xA;" OwnerUserId="11320" LastActivityDate="2014-12-16T00:07:00.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21760" PostTypeId="2" ParentId="19449" CreationDate="2014-12-16T00:19:21.697" Score="1" Body="&lt;p&gt;That dilute of $\ce{H2SO4}$ won't remotely burn your skin. You can wear goggles just in case you spray it in your face for some reason.&lt;/p&gt;&#xA;" OwnerUserId="11320" LastEditorUserId="4231" LastEditDate="2014-12-16T02:24:32.313" LastActivityDate="2014-12-16T02:24:32.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21761" PostTypeId="2" ParentId="21733" CreationDate="2014-12-16T00:29:27.463" Score="1" Body="&lt;p&gt;Water dissociating into H+ and OH- isn't the same concept as free OH- leaving. A more accurate comparison would be forming NaOH by dropping Na metal in water, because in that case, OH- leaves water to become part of a new molecule as opposed to just ionizing in solution for a second before it becomes full water again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohols are very capable of SN1. Water is the leaving group as it is obviously protonated first. Benzyl alcohol to benzyl chloride, t-butanol to t-butyl bromide, etc.&lt;/p&gt;&#xA;" OwnerUserId="11320" LastActivityDate="2014-12-16T00:29:27.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21762" PostTypeId="1" AcceptedAnswerId="21769" CreationDate="2014-12-16T00:40:27.917" Score="0" ViewCount="48" Body="&lt;p&gt;The total area of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 g}$&lt;/span&gt; sample of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; powder is measured to be &lt;span class=&quot;math-container&quot;&gt;$\pu{1620 cm2}$&lt;/span&gt;. If the powder consists of tiny cubic crystals, what is the average size of these crystals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm studying for a physical chemistry final and am not sure how to do this practice question.&lt;br&gt;&#xA;I know that the density is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \frac{m}{A},\label{one}\tag1$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the surface area, so do I just find &#xA;&lt;span class=&quot;math-container&quot;&gt;$$A=\frac{m}{\delta}$$&lt;/span&gt;&#xA;to find the average &quot;size&quot;?&lt;br&gt;&#xA;Or do I use \eqref{one} and then find out what the unit cell structure of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; powder is and use that to estimate what the surface area of the crystal is? &lt;/p&gt;&#xA;" OwnerUserId="2386" LastEditorUserId="4945" LastEditDate="2018-10-26T12:24:00.060" LastActivityDate="2018-10-26T12:31:11.793" Title="How to calculate the average size of cubic crystals of KCl powder?" Tags="&lt;physical-chemistry&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-12-16T05:45:48.737" ContentLicense="CC BY-SA 4.0" />
  <row Id="21764" PostTypeId="1" AcceptedAnswerId="21786" CreationDate="2014-12-16T00:47:12.500" Score="4" ViewCount="20258" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBouc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is more stable according to its charge. The other one does not satisfy as much as the first does the formal charge rule but has a complete octet.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2014-12-16T10:59:09.833" Title="Which Lewis Structure for BeCl2 is more commonly seen?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21769" PostTypeId="2" ParentId="21762" CreationDate="2014-12-16T02:00:17.187" Score="1" Body="&lt;p&gt;For a cube, the surface area is given by the formula &lt;span class=&quot;math-container&quot;&gt;$6a^2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the length of one side. The volume is given by &lt;span class=&quot;math-container&quot;&gt;$a^3$&lt;/span&gt;. Surface area per volume is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{6a^2}{a^3}= \frac{6}{a}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the total surface area to be &lt;span class=&quot;math-container&quot;&gt;$\pu{1620 cm^2}$&lt;/span&gt;. We can calculate the total volume from the mass of the &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; and its density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  V &amp;amp;= \frac{m}{\rho}\\&#xA;  \rho(\ce{KCl}) &amp;amp;= \pu{1.98 g//cm^3}\\&#xA;  V &amp;amp;= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\&#xA;    &amp;amp;= \pu{0.3 cm^3}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An average crystal should have the same ratio as the total. So we set the two equal and solve for the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \frac{6}{a} &amp;amp;= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\&#xA;  a &amp;amp;= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\&#xA;  a &amp;amp;= \pu{9E-4 cm}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1389" LastEditorUserId="4945" LastEditDate="2018-10-26T12:31:11.793" LastActivityDate="2018-10-26T12:31:11.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21770" PostTypeId="1" CreationDate="2014-12-16T02:00:39.013" Score="1" ViewCount="96" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the activity series (table of standard reduction potentials), indicate whether or not the following reactions would occur in the direction written.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't worry about what the activity series is, I don't even know what it is so it isn't important. I'm more confused on the procedure and how to find the answer. Would you just find the oxidation states, then find the half reactions according to what the oxidation states are and find the voltage according to the reduction potentials sheet, then be able to tell whether it was spontaneous therefore whether it would run in the direction it was written?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few of the first questions, the easy ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag + KNO3 → AgNO3 + K}$$&#xA;$$\ce{Zn + AgNO3 → Ag + Zn(NO3)2}$$&#xA;$$\ce{Li + H2O → LiOH + H2}$$&#xA;$$\ce{Cu + FeSO4 → Fe + CuSO4}$$ &lt;/p&gt;&#xA;" OwnerUserId="11322" LastEditorUserId="5026" LastEditDate="2018-02-19T12:59:26.797" LastActivityDate="2018-02-19T12:59:26.797" Title="Determining the direction of chemical reactions based on the activity series" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21771" PostTypeId="2" ParentId="163" CreationDate="2014-12-16T02:02:32.890" Score="1" Body="&lt;p&gt;I say the reason why the electron affinity of fluorine is not as negative as chlorine and that of O is not as negative as S is because of the electron repulsions in the small compact atoms keep the added electrons from being tightly bound as we might expect. . .&lt;/p&gt;&#xA;" OwnerUserId="11323" LastEditorUserId="25206" LastEditDate="2016-04-29T05:19:41.713" LastActivityDate="2016-04-29T05:19:41.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21772" PostTypeId="1" AcceptedAnswerId="21775" CreationDate="2014-12-16T02:43:45.197" Score="7" ViewCount="809" Body="&lt;p&gt;I’ve attempted to solve this problem with percent composition, but it says the answer is wrong. Is it a $pV=nRT$ problem, and if so, can I use the units as they are given?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A $\pu{0.02867 g}$ sample of gas occupies $\pu{10.0 ml}$ at $\pu{290.0 K}$ and $\pu{1.10 atm}$. Upon further analysis, the compound is found to be $38.734 \%~ \ce{C}$ and $61.266\%~\ce{F}$. What is the molecular formula of the compound?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I keep on getting $\ce{CF}$ for the answer via percent composition.&lt;/p&gt;&#xA;" OwnerUserId="9822" LastEditorUserId="5026" LastEditDate="2018-02-25T08:17:12.520" LastActivityDate="2018-02-25T08:23:34.197" Title="Determine the molecular formula of the compound using percentage composition" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21773" PostTypeId="1" AcceptedAnswerId="21800" CreationDate="2014-12-16T02:59:09.000" Score="17" ViewCount="2030" Body="&lt;p&gt;Traditionally, hydrogen bonding has been defined to only include interactions between a positively polarized hydrogen and three period-two elements: nitrogen, oxygen, and fluorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why was this definition proposed? A common question is why can't chlorine, by the aforementioned definition, hydrogen bond? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have anything to do with hard and soft acid-base theory? Hydrogen is a hard acid, and chlorine is a comparatively soft base ... and the period two elements are all relatively hard? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2018-03-31T15:25:16.593" LastActivityDate="2020-04-16T15:09:25.803" Title="Is hydrogen bonding generally defined to include only three period two elements?" Tags="&lt;elements&gt;&lt;hydrogen-bond&gt;&lt;hsab&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21774" PostTypeId="1" AcceptedAnswerId="21833" CreationDate="2014-12-16T03:06:18.820" Score="12" ViewCount="7386" Body="&lt;p&gt;I guess if we look at this problem from a &quot;conjugate&quot; perspective then the conjugate of a covalent bond is two elements with electrons lying around. On the other hand, with two electrostatically attracted things, the lack of a bond leaves two separate, charged particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that still doesn't answer the question; why are electrons less stable than ions? I guess in the case of an ion though there are still electrons but also a nucleus the stabilize the electrons. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2019-06-06T15:16:21.403" Title="Why is covalent bonding stronger than electrostatic attraction?" Tags="&lt;bond&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21775" PostTypeId="2" ParentId="21772" CreationDate="2014-12-16T03:47:20.307" Score="10" Body="&lt;p&gt;You're actually on the right track. Looking at the percent composition, you've correctly identified that the ratio of $\ce{C}$ to $\ce{F}$ atoms is 1:1, however, you cannot assume that the formula is just $\ce{CF}$ (which isn't a known compound), it could be any compound with that ratio, $\ce{C2F2}$, $\ce{C3F3}$, $\ce{C4F4}$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to narrow it down is to use the gas data. You are right that you need to use the ideal gas law. If you look at the equation, you have everything except the number of moles, so you're interested in finding $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=\frac{PV}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the units, that depends on what units you have in $R$. If you have $R$ in $\mathrm{J/(mol\cdot K)}$, then $P$ needs to be in $\mathrm{Pa}$ and $V$ in $\mathrm{m^3}$, if it's in $\mathrm{L\cdot atm/(mol\cdot K)}$, then $P$ should be in $\mathrm{atm}$ and $V$ in $\mathrm{L}$. You can either convert what you have to match $R$ or convert $R$ to match your variables; it doesn't really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the number of moles, you can calculate the molar mass of the compound from the sample mass:&#xA;$$m_\mathrm M=\frac{m_\mathrm s}{n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, since we know the ratio of $\ce{C}$ to $\ce{F}$ is 1:1, the molar mass will be some multiple of the mass of $\ce{CF}$ which is $ 31.0091\ \mathrm{g/mol}$, so just divide the molar mass by that and you'll know $n$ in $\ce{C_nF_n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while not a sure way if fake compounds are possible answers, not all compounds with formula $\ce{C_nF_n}$ are known ($\ce{CF}$, for instance), so you can narrow down your choice of answer.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user7951" LastEditDate="2018-02-25T08:23:34.197" LastActivityDate="2018-02-25T08:23:34.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21776" PostTypeId="1" CreationDate="2014-12-16T04:01:20.693" Score="3" ViewCount="3330" Body="&lt;p&gt;Why is it that adding protons has a greater effect than electron-electron repulsion on periodic trends like atomic radius and ionization energy (assuming # of shells constant)? It seems that if protons and electrons have the same charge and electrons are actually closer to the electrons in question (e.g. for ionization energy), that their effect would be the same? &lt;/p&gt;&#xA;" OwnerUserId="11324" LastActivityDate="2014-12-16T05:56:18.693" Title="Periodic trends: why is effect of protons greater than electrons?" Tags="&lt;electrons&gt;&lt;periodic-table&gt;&lt;periodic-trends&gt;&lt;protons&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21777" PostTypeId="1" AcceptedAnswerId="21782" CreationDate="2014-12-16T05:36:47.410" Score="1" ViewCount="191" Body="&lt;p&gt;I am studying for my final and I am given this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution of sodium phosphate is made from 10.5 g sodium phosphate in&#xA;  150 mL of water. What is the pH of this solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given: Ka1, Ka2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution says that we can simply calculate this pH with the following eqution:&#xA;$$pH=\frac{pK_{a1}+pK_{a2}}{2}$$&#xA;However this makes absolutely no sense to me, as this is the pH at the equivalence point.  What am I missing as towhy this equation can be used?&lt;/p&gt;&#xA;" OwnerUserId="9791" LastActivityDate="2014-12-16T13:12:07.877" Title="Finding pH of tri-protic acid" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21779" PostTypeId="2" ParentId="21776" CreationDate="2014-12-16T05:56:18.693" Score="3" Body="&lt;p&gt;This is because the attraction is not between each proton and it's corresponding electron, but between each electron and the nucleus. Since the number of protons increases the charge of the nucleus, but the charge of the electron stays constant, the attraction is greater. The attraction of the electron to the nucleus depends on the charge of the nucleus, and it is independent of the number of electrons in &quot;orbit&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great analogy for this is the solar system. (Gravity works very similarly to electrostatic force.) It won't matter if suddenly there is a mother earth orbiting in the outskirts of the solar system, Earth would be unaffected by that change, and we would still orbit the sun at the sae distance. However, if the sun suddenly becomes noticeably heavier, the planets will each be pulled in with a stronger force, which reduces the orbit radius. This will happen equally to each individual planet. It doesn't matter how many planets we have, The pull of the earth is only affected by the mass of the sun and the mass of earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies to the atom if you think of the nucleus as the sun and each electron as a planet. The force attracting the electron to the nucleus is only affected by the charge of the nucleus as each electron carries the same charge, and the charge is not the collective  charge of all electrons, but the individual charge. Hence all each electron perceives is the increase in the charge of the nucleus that is increasing its attraction to the nucleus. &lt;/p&gt;&#xA;" OwnerUserId="10202" LastActivityDate="2014-12-16T05:56:18.693" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21780" PostTypeId="2" ParentId="21756" CreationDate="2014-12-16T07:45:50.317" Score="11" Body="&lt;p&gt;Microwaves don't have more energy, they just resonate at the frequency that causes molecular bonds to rotate. This specifically applies to dielectric molecules, molecules like water that have electric dipoles.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Microwave_oven#Principles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia:&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Water, fat, and other substances in the food absorb energy from the microwaves in a process called dielectric heating. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, &lt;strong&gt;and therefore rotate as they try to align themselves with the alternating electric field of the microwaves.&lt;/strong&gt; Rotating molecules hit other molecules and put them into motion, thus dispersing energy. &lt;strong&gt;This energy, when dispersed as molecular vibration in solids and liquids&lt;/strong&gt; (i.e., as both potential energy and kinetic energy of atoms), &lt;strong&gt;is heat&lt;/strong&gt;. Sometimes, microwave heating is explained as a resonance of water molecules, but this is incorrect such resonances occur only at above 1 terahertz&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For reference, energy is related to wavelength by the formula:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E=\frac{hc}{\lambda}$$&lt;/span&gt;&#xA;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\lambda=\frac{c}{f}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is wavelength&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the speed of light&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is frequency&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is Plank's constant&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is energy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;You can see by the first equation that as &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; gets smaller &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; gets larger.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="15751" LastEditDate="2020-10-15T20:34:42.033" LastActivityDate="2020-10-15T20:34:42.033" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21781" PostTypeId="1" AcceptedAnswerId="21830" CreationDate="2014-12-16T08:52:10.760" Score="2" ViewCount="9159" Body="&lt;p&gt;I'm in trouble with the oxidation of NaI by $\ce{NaClO}$, the reaction should give $\ce{I+}$ which then reacts by electrophilic substitution with an aromatic complex. As of now I've come to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ ClO- (aq) + 2I- (aq) + 2H+(aq) -&amp;gt; I2 (s) + Cl- (aq) + H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't then find a way to obtain then the electrophilic $\ce{I+}$ that should react.&lt;/p&gt;&#xA;" OwnerUserId="7050" LastEditorUserId="4945" LastEditDate="2014-12-16T11:07:52.467" LastActivityDate="2014-12-19T12:15:35.150" Title="How does the oxidation of NaI by sodium hypochlorite to give I+ work?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21782" PostTypeId="2" ParentId="21777" CreationDate="2014-12-16T08:55:22.557" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;h1&gt;amphoteric&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;from the &lt;a href=&quot;http://goldbook.iupac.org/A00306.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;http://en.wikipedia.org/wiki/Amphoterism&quot; rel=&quot;noreferrer&quot;&gt;ampholytes&lt;/a&gt; like $\ce{NaH2PO4}$ the $\ce{pH}$ is independent on the concentration (in first approximation). Consider the reactions of that are possible:&#xA;\begin{align}\ce{&#xA;H2A-    + H3+O &amp;amp;~&amp;lt;=&amp;gt; H3A     + H2O  \tag1\\&#xA;H2A-    + H2O  &amp;amp;~&amp;lt;=&amp;gt; HA^{2-} + H3+O \tag2\\&#xA;HA^{2-} + H2O  &amp;amp;~&amp;lt;=&amp;gt; A^{3-}  + H3+O \tag3\\&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the first approximation, we will assume that everything happening in equation $(3)$ is negligible, as is will only very little influence the concentration of $\ce{HA^{2-}}$. Therefore we can focus on the acidity constants $K_{a_1}$ of $(1)$ and $K_{a_2}$ of $(2)$:&#xA;\begin{align}&#xA;K_{a_1}  &amp;amp;= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})}  \tag4\\&#xA;K_{a_2}  &amp;amp;= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\&#xA;\end{align}&#xA;We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels.&#xA;\begin{align}&#xA;K_{a_1} \cdot K_{a_2} &amp;amp;= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})} \cdot&#xA;  \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\&#xA;K_{a_1} \cdot K_{a_2} &amp;amp;= c^2(\ce{H3+O})\cdot  \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\&#xA;\end{align}&#xA;Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate&#xA;$$c(\ce{HA^{2-}})=c(\ce{H3A})\tag8$$&#xA;and rewrite $(7)$ as&#xA;\begin{align}&#xA;c^2(\ce{H3+O}) &amp;amp;= K_{a_1} \cdot K_{a_2}\\&#xA;c(\ce{H3+O})   &amp;amp;= \sqrt{K_{a_1} \cdot K_{a_2}}\\&#xA;\ce{pH}        &amp;amp;= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right).&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this equation only works as a first approximation. Usually the coupled equilibria are very complex and in most instances a computer needs to be involved to calculate it accurately. More information on &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution#Polyprotic_acids&quot; rel=&quot;noreferrer&quot;&gt;polyprotic acids&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5160" LastEditDate="2014-12-16T13:12:07.877" LastActivityDate="2014-12-16T13:12:07.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21784" PostTypeId="2" ParentId="21764" CreationDate="2014-12-16T09:13:36.647" Score="2" Body="&lt;p&gt;The two structures drawn are both contributing to a total distribution of the charge. These are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;Resonance&lt;/a&gt; forms. In some cases it means the electrons are equally distributed over a molecule, due to symmetry. In the case of your molecule the contribution of the two forms will most likely not be equal, but the truth lies somewhere in between the two structures you have drawn.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-12-16T09:13:36.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21785" PostTypeId="1" CreationDate="2014-12-16T10:32:53.403" Score="2" ViewCount="589" Body="&lt;p&gt;For a research project I found out last minute that I need the coefficient of friction between nylon and Ultra-high-molecular-weight polyethylene. After hours of scouring google nothing useful turned up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I actually need the maximum and not the precise value, I came here to ask what that could possibly be and if there is any literature on that. Basically I need to find out what the maximum friction could be, caused by a Dyneema (polyethylene) rope on a plastic nylon pulley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping there is some sort of &quot;rule&quot; what the maximum coefficient of friction could be for 2 smooth plastic surfaces.&lt;/p&gt;&#xA;" OwnerUserId="7961" LastActivityDate="2016-09-05T04:00:08.293" Title="Maximum coefficient of friction between nylon and polyethylene" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21786" PostTypeId="2" ParentId="21764" CreationDate="2014-12-16T10:59:09.833" Score="6" Body="&lt;p&gt;Beryllium chloride has a significant ionic contribution to its electronic structure. A quick calculation, DF-BP86/def2-SVP, reveals that the charge of beryllium is $\mathbf{q}(\ce{Be})=0.8$ and of chlorine $\mathbf{q}(\ce{Cl})=-0.4$, based on natural bond orbital analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lewis structures best describing this are depicted below. The percentage indicates how much Lewis character this model actually describes based on the before mentioned calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RMFLK.jpg&quot; alt=&quot;structures of becl2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is just an approximation and the truth lies somewhere in between, like Eljee already mentioned. For these bonding situation, &lt;a href=&quot;http://goldbook.iupac.org/R05326.html&quot; rel=&quot;noreferrer&quot;&gt;resonance&lt;/a&gt; is unavoidable. In quantum chemistry these phenomena are usually very well described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;noreferrer&quot;&gt;valence bond theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility, which is quite complementary, is looking at the structure with &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;noreferrer&quot;&gt;molecular orbital theory&lt;/a&gt;. Since the molecule has $D_\mathrm{\infty{h}}$ symmetry, you can observe $\pi$ bonds, that are delocalised over the whole molecule. And these are only depicted in the right hand side structure above. Below are the valence orbitals of $\ce{BeCl2}$, DF-BP86/def2-SVP.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7Njt1l.jpg&quot; alt=&quot;mos of becl2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see, that one Lewis structure is not enough to actually describe the bonding in that molecule well, but as a first approximation, the left hand structure above should be sufficient.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-16T10:59:09.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21787" PostTypeId="1" CreationDate="2014-12-16T13:11:22.570" Score="0" ViewCount="95" Body="&lt;p&gt;This reaction proceeds in the galvanic cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn + Pb^{2+} -&amp;gt; Zn^{2+} + Pb}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the equation for the half-reactions of this cell and the theoretical voltage produced by this cell if concentration of $\ce{Zn^{2+}}$ is $\mathrm{0.1\ mol/L}$ and concentration of $\ce{Zn^{2+}}$ is $\mathrm{10^{-4}\ mol/L}$.&lt;/p&gt;&#xA;" OwnerUserId="11329" LastEditorUserId="418" LastEditDate="2014-12-16T13:41:10.070" LastActivityDate="2014-12-16T13:48:14.593" Title="Galvanic Cells, theoritical voltage" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2014-12-16T16:02:19.657" ContentLicense="CC BY-SA 3.0" />
  <row Id="21789" PostTypeId="2" ParentId="21787" CreationDate="2014-12-16T13:48:14.593" Score="1" Body="&lt;p&gt;$$Pb^{2+} + 2e^{-}-&amp;gt;Pb  E_{0}(Pb^{2+}) = -0.1263$$&#xA;$$Zn-&amp;gt;Zn^{2+} +2e^{-}   E_{0}(Zn^{2+}) = -0.76 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$&#xA;$$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76) $$&lt;br&gt;&#xA;$$ E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small piece of advice : Tag homework and exercises on these type of questions , read through &quot;&lt;em&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/help&quot;&gt;Help Center&lt;/a&gt;&lt;/em&gt;&quot; specifically &quot;Asking&quot; . Try to attempt the question by yourself and show your own attempt of an answer. This site isnt &quot;Solve my homework site &quot; good luck cheers&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-12-16T13:48:14.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21790" PostTypeId="1" AcceptedAnswerId="21796" CreationDate="2014-12-16T14:33:20.880" Score="0" ViewCount="352" Body="&lt;p&gt;Why is it required that the solution of copper (I) tetraiodomercurate (II), which has a thermochromic property, be filtered while it is hot?&lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="4356" LastEditDate="2014-12-16T17:43:00.057" LastActivityDate="2014-12-16T17:43:00.057" Title="Why is Cu2HgI4 filtered while it is hot?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21791" PostTypeId="2" ParentId="19523" CreationDate="2014-12-16T14:41:48.263" Score="-1" Body="&lt;p&gt;The concentration of [H] ion will be $2.45x10^{-5}$ if the pH was 4.61.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="4356" LastEditDate="2014-12-16T17:39:55.617" LastActivityDate="2014-12-16T17:39:55.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21793" PostTypeId="1" AcceptedAnswerId="21795" CreationDate="2014-12-16T14:55:32.970" Score="4" ViewCount="904" Body="&lt;p&gt;What is the systematic name of this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j9UKQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j9UKQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My try: &lt;strong&gt;3-propylpentane&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11330" LastEditorUserId="41328" LastEditDate="2017-08-05T16:28:03.620" LastActivityDate="2017-08-05T16:28:03.620" Title="IUPAC name for a branched alkane" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21794" PostTypeId="1" AcceptedAnswerId="21799" CreationDate="2014-12-16T14:58:42.390" Score="5" ViewCount="10206" Body="&lt;p&gt;How does a catalytic poison reduce a catalyst's activity, such as palladium used in the hydrogenation of alkenes or any other such catalyst?&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="5026" LastEditDate="2018-04-16T01:43:24.950" LastActivityDate="2018-04-16T01:44:17.593" Title="How does a catalytic poison reduce a catalyst's activity?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21795" PostTypeId="2" ParentId="21793" CreationDate="2014-12-16T15:00:01.710" Score="14" Body="&lt;p&gt;You need to take the longest chain first, i.e. the basis of the hydrocarbon skeleton is hexane. It has a ethyl group at position 3, so the IUPAC name will be &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.11599.html&quot;&gt;3-ethylhexane&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-16T15:00:01.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21796" PostTypeId="2" ParentId="21790" CreationDate="2014-12-16T15:01:54.077" Score="3" Body="&lt;p&gt;Copper tetraiodomercurate, $\ce{Cu2HgI4}$, is typically prepared by &lt;strong&gt;heating combined suspensions&lt;/strong&gt; of copper(I) iodide, $\ce{CuI}$ and mercury(II) iodide, $\ce{HgI2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iodides are prepared by (a) reacting copper(II) sulfate, $\ce{CuSO4*5 H2O}$ with potassium iodide, $\ce{KI}$, and (b) mercury(II) nitrate, $\ce{Hg(NO3)2}$ with with potassium iodide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to avoid contamination of the final reaction product with one (or both) of these iodides, it is essential to filter off while the reaction mixture is still hot.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-16T15:01:54.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21798" PostTypeId="2" ParentId="21793" CreationDate="2014-12-16T15:13:55.880" Score="10" Body="&lt;p&gt;First of all, make sure to find the longest chain. In this case, it is 6 starting from top right carbon to bottom right carbon. Then, number the chain. You will find that the C atom bonded to the branch has the number 3.It is not 4 because the branch should have the lowest locant and so the numbering should start from top right C atom and not from the bottom one. So, the name of the compound is 3-ethyl hex-ane.&lt;/p&gt;&#xA;" OwnerUserId="11331" LastActivityDate="2014-12-16T15:13:55.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21799" PostTypeId="2" ParentId="21794" CreationDate="2014-12-16T15:16:28.280" Score="6" Body="&lt;p&gt;Basically the catalytic poison reduces the activity by slowing or inhibiting the reaction by occupying the surface of the catalyst and therefore the surface area available for the reactants reduces. You can get a general idea from the following text.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Substances which themselves are not catalysts, but when mixed in small quantities with the catalysts increase their efficiency are called as promoters or activators.&lt;/p&gt;&#xA;&lt;p&gt;Catalyst poisoning refers to the partial or total deactivation of a catalyst caused by exposure to a range of chemical compounds. Poisoning may be desirable when it results in improved selectivity (e.g.Lindlar's catalyst) but may be undesirable when the catalyst is rendered ineffective (e.g. in catalytic converters). Poisoning refers specifically to chemical deactivation, rather than other mechanism of catalyst degradation such as thermal decomposition or physical damage.&#xA;Poisoning involves compounds which bonds chemically to the active surface sites of a catalyst. This may have two effects: the total number of catalytic sites or the fraction of the total surface area that has the capability of promoting reaction always decreases, and the average distance that a reactant molecule must diffuse through the pore structure before undergoing reaction may increase. Poisoned sites can no longer accelerate the reaction with which the catalyst was supposed to catalyze. Large scale production of substances such as ammonia in the Haber–Bosch process include steps to remove potential poisons from the product stream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For more info, read &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalyst_poisoning&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article on Catalytic poisoning&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-04-16T01:44:17.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21800" PostTypeId="2" ParentId="21773" CreationDate="2014-12-16T15:20:32.173" Score="16" Body="&lt;p&gt;I don't think there is any such traditional definition requiring &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in table 7 and the discussion thereof in &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev.pc.22.100171.002023&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Bonding&lt;/a&gt; Annual Review of Physical Chemistry Vol. 22: 347-385 the hydrogen bonding in the following species are discussed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{ClHCl^-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BrHBr^-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{IHI^-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BrHCl^-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;as well as related neutral radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonds in &lt;span class=&quot;math-container&quot;&gt;$\ce{S-H}$&lt;/span&gt; containing compounds are discussed.  These are just examples, others that do not involve &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt; are also discussed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following is copied (footnotes omitted) from &lt;a href=&quot;http://www.degruyter.com/view/j/pac.2011.83.issue-8/pac-rec-10-01-02/pac-rec-10-01-02.xml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Definition of the Hydrogen Bond (IUPAC Recommendations 2011)&lt;/a&gt; Pure and Applied Chemistry Volume 83, Issue 8 (Aug 2011)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h1&gt;2. DEFINITION&lt;/h1&gt;&#xA;  &#xA;  &lt;p&gt;The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a&#xA;  molecular fragment &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; in which &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; is more electronegative than &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;, and an atom or a group of&#xA;  atoms in the same or a different molecule, in which there is evidence of bond formation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A typical hydrogen bond may be depicted as &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y-Z}$&lt;/span&gt;, where the three dots denote the bond.  &lt;span class=&quot;math-container&quot;&gt;$\ce{X–H}$&lt;/span&gt; represents the hydrogen bond donor. The acceptor may be an atom or an anion &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt;, or a fragment or a molecule &lt;span class=&quot;math-container&quot;&gt;$\ce{Y-Z}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; is bonded to &lt;span class=&quot;math-container&quot;&gt;$\ce{Z}$&lt;/span&gt;. In some cases, &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; are the same. In more specific cases, &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; are the same and &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y-H}$&lt;/span&gt; distances are the same as well leading to symmetric hydrogen bonds. In any event, the acceptor is an electron rich region such as, but not limited to, a lone pair of &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; or π-bonded pair of &lt;span class=&quot;math-container&quot;&gt;$\ce{Y-Z}$&lt;/span&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The evidence for hydrogen bond formation may be experimental or theoretical, or ideally, a combination of both. Some criteria useful as evidence and some typical characteristics for hydrogen bonding, not necessarily exclusive, are listed below, numbered E# and C#, respectively. The greater the number of criteria satisfied, the more reliable is the characterization as a hydrogen bond.&lt;/p&gt;&#xA;  &#xA;  &lt;h2&gt;2.1 List of criteria&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;For a hydrogen bond &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y-Z}$&lt;/span&gt;:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;(E1) The forces involved in the formation of a hydrogen bond include those of an electrostatic origin, those arising from charge transfer between the donor and acceptor leading to partial covalent bond formation between &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt;, and those originating from dispersion.&lt;/li&gt;&#xA;  &lt;li&gt;(E2) The atoms &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; are covalently bonded to one another and the &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; bond is polarized, the &lt;span class=&quot;math-container&quot;&gt;$\ce{H\bond{...}Y}$&lt;/span&gt; bond strength increasing with the increase in electronegativity of &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;(E3) The &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y}$&lt;/span&gt; angle is usually linear (180º) and the closer the angle is to 180º, the stronger is the hydrogen bond and the shorter is the &lt;span class=&quot;math-container&quot;&gt;$\ce{H\bond{...}Y}$&lt;/span&gt; distance.&lt;/li&gt;&#xA;  &lt;li&gt;(E4) The length of the &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; bond usually increases on hydrogen bond formation leading to a red shift in the infrared &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; stretching frequency and an increase in the infrared absorption cross-section for the &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; stretching vibration. The greater the lengthening of the &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; bond in &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y}$&lt;/span&gt;, the stronger is the &lt;span class=&quot;math-container&quot;&gt;$\ce{H\bond{...}Y}$&lt;/span&gt; bond. Simultaneously, new vibrational modes associated with the formation of the &lt;span class=&quot;math-container&quot;&gt;$\ce{H\bond{...}Y}$&lt;/span&gt; bond are generated.&lt;/li&gt;&#xA;  &lt;li&gt;(E5) The &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y-Z}$&lt;/span&gt; hydrogen bond leads to characteristic NMR signatures that typically include pronounced proton deshielding for &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt;, through hydrogen bond spin–spin couplings between &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt;, and nuclear Overhauser enhancements.&lt;/li&gt;&#xA;  &lt;li&gt;(E6) The Gibbs energy of formation for the hydrogen bond should be greater than the thermal energy of the system for the hydrogen bond to be detected experimentally.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;h2&gt;2.2 Some characteristics of hydrogen bonds&lt;/h2&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;(C1) The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{X-H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{b}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Y-Z}$&lt;/span&gt; in a given solvent correlate strongly with the energy of the hydrogen bond formed between them.&lt;/li&gt;&#xA;  &lt;li&gt;(C2) Hydrogen bonds are involved in proton-transfer reactions (&lt;span class=&quot;math-container&quot;&gt;$\ce{X-H\bond{...}Y -&amp;gt; X\bond{...}H-Y}$&lt;/span&gt;) and may be considered the partially activated precursors to such reactions.&lt;/li&gt;&#xA;  &lt;li&gt;(C3) Networks of hydrogen bonds can show the phenomenon of co-operativity, leading to deviations from pair-wise additivity in hydrogen bond properties.&lt;/li&gt;&#xA;  &lt;li&gt;(C4) Hydrogen bonds show directional preferences and influence packing modes in crystal structures.&lt;/li&gt;&#xA;  &lt;li&gt;(C5) Estimates of charge transfer in hydrogen bonds show that the interaction energy correlates well with the extent of charge transfer between the donor and the acceptor.&lt;/li&gt;&#xA;  &lt;li&gt;(C6) Analysis of the electron density topology of hydrogen-bonded systems usually shows a bond path connecting &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; and a (3,–1) bond critical point between &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt;.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="17175" LastEditDate="2020-04-16T15:09:25.803" LastActivityDate="2020-04-16T15:09:25.803" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21801" PostTypeId="1" AcceptedAnswerId="21812" CreationDate="2014-12-16T16:59:05.573" Score="1" ViewCount="107" Body="&lt;p&gt;I have heard people go so far as to say that the &quot;driving force&quot; of a reaction is the formation of the precipitate. Why is this so? Whilst the process is almost certainly exothermic (I think - please clarify) the entropic effects are surely unfavorable.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-12-16T20:00:26.390" Title="Why is a reaction more favorable if one of the products is not soluble/sparingly soluble?" Tags="&lt;thermodynamics&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21803" PostTypeId="1" CreationDate="2014-12-16T17:03:20.313" Score="1" ViewCount="134" Body="&lt;p&gt;I cannot think of a distinction - the terms seem to be used interchangeably when talking about products of Lewis acid-base reactions. However, I am concerned that I may have overlooked a subtlety. Please can you clarify?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-12-16T17:03:20.313" Title="What distinguishes a Lewis Complex from an Adduct or are they synonymous?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21804" PostTypeId="2" ParentId="21773" CreationDate="2014-12-16T17:17:41.900" Score="3" Body="&lt;p&gt;My book* has  this  graph of boiling points of various hydrogen-element compounds. $\ce{NH3}$, $\ce{H2O}$, and $\ce{HF}$ are the only ones that violate the trend, so it would make a bit of sense to call the interaction responsible only for those three and nothing else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kXw5Q.jpg&quot; alt=&quot;boiling points&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) &lt;em&gt;Chemistry The Central Science&lt;/em&gt; by Theodore E. Brown, H. Eugene H LeMay, Bruce E. Bursten, Catherine Murphy.&lt;/p&gt;&#xA;" OwnerUserId="11334" LastEditorDisplayName="user10259" LastEditDate="2015-05-20T01:20:48.173" LastActivityDate="2015-05-20T01:20:48.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21805" PostTypeId="1" AcceptedAnswerId="21816" CreationDate="2014-12-16T17:47:53.427" Score="8" ViewCount="1208" Body="&lt;p&gt;When the $pK_a$ scale is not applicable - when there is no proton/charged species transfer - what should be used to quantify the strength of an acid?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-12-17T02:59:04.390" Title="How is it possible to quantify the strength of a Lewis acid/base?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21807" PostTypeId="2" ParentId="21749" CreationDate="2014-12-16T19:14:04.017" Score="4" Body="&lt;p&gt;B has more branching and this means that molecules of B pack less efficiently and so there are fewer Van der Waals forces between them and so B has the lowest boiling point. On the other hand A is a linear molecule and so packs efficiently, has stronger VdWs forces and so has the highest boiling point. As @DavePhD's data shows $\ce{B &amp;lt; C &amp;lt; A}$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-16T19:14:04.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21808" PostTypeId="1" CreationDate="2014-12-16T19:20:26.707" Score="0" ViewCount="197" Body="&lt;p&gt;Which is the parent chain in this structure?&lt;/p&gt;&#xA;&lt;p&gt;and since all have 3 carbons so should any of them be the parent chain, if not ..&lt;/p&gt;&#xA;&lt;p&gt;then which should be and why? is it a rule in IUPAC naming?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zoqNn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zoqNn.png&quot; alt=&quot;NEOPENTYLCHLORIDE | C5H11Cl | ChemSpider&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://www.chemspider.com/ImagesHandler.ashx?id=12416&amp;amp;w=200&amp;amp;h=200&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemspider.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="15751" LastEditDate="2021-01-01T21:21:02.670" LastActivityDate="2021-01-01T21:21:02.670" Title="Which is the parent chain in Neopentyl chloride? and why?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21809" PostTypeId="2" ParentId="21808" CreationDate="2014-12-16T19:25:12.513" Score="6" Body="&lt;p&gt;A chemist would say that all 3 methyl groups are equivalent (identical by symmetry).  Therefore they all produce the same chain, and that chain is the longest chain in the molecule.  Consequently, since the methyl groups are all the same, it doesn't matter which one you pick.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-16T19:25:12.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21810" PostTypeId="1" AcceptedAnswerId="21814" CreationDate="2014-12-16T19:42:43.453" Score="34" ViewCount="38090" Body="&lt;p&gt;I've been trying to answer my (high school) daughter's questions about the periodic table, and the reactivity series, but we keep hitting gaps in my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I showed that the noble gases have a full outer shell, which is why they don't react with anything. And then over the other side of the periodic table we have potassium and sodium, which have only one electron in their outer shell, which is what makes them so reactive, and at the top of our reactivity list. (And the bigger they get, the more reactive, which is why we were not allowed to play with caesium in class...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we looked up gold, which is at the bottom of the reactivity series, and found it also has only one electron in its outermost shell (2-8-18-32-18-1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy explanation for why gold doesn't fizz like potassium when you drop it in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question could be rephrased as &quot;What properties of each element decide their ranking in the metal reactivity series?&quot; if you prefer; that was the original question we were trying to answer.)&lt;/p&gt;&#xA;" OwnerUserId="11337" LastEditorUserId="7475" LastEditDate="2017-12-16T15:13:57.837" LastActivityDate="2018-12-17T22:35:47.957" Title="Why is gold unreactive when only one electron is in the outer shell?" Tags="&lt;metal&gt;&lt;electronic-configuration&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21811" PostTypeId="2" ParentId="19568" CreationDate="2014-12-16T19:58:23.253" Score="14" Body="&lt;p&gt;As mentioned above there is no hard line between an ionic and a covalent bond, they are just two ends of a spectrum where a bond can have a certain degree of ionic or covalent character. At the lower end of this spectrum, bonds with small electronegativity differences between the atoms have very little ionic character because the charge distribution in the bond is almost symmetrical between the two atoms. As the EN difference increases the charge distribution becomes asymmetrical and this results in an increasingly polarised bond, which therefore has an increasing degree of ionic character. Eventually at the other end of the scale you have a bond where almost all of the charge is concentrated on one atom and this is generally termed to be an ionic bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to your question; what happens if the EN difference is exacly 2.1; nothing happens, you have a relatively strongly polarised bond which under some definitions might be classed as ionic and under others as strongly polar covalent but in reality it has a large degree of both ionic and covalent character.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-16T19:58:23.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21812" PostTypeId="2" ParentId="21801" CreationDate="2014-12-16T20:00:26.390" Score="4" Body="&lt;p&gt;If the reaction product is removed from the phase (liquid solution for example) in which the equilibrium reaction is occurring, significant amount of product can be obtained even if the equilibrium constant for the forward reaction is relatively small ($K&amp;lt;1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from precipitation, product could also be removed by vaporization, with subsequent condensation (distillation) to collect the product.  Anything that reduces the product concentration in the phase where the equillibrium occurs favors the forward reaction.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-16T20:00:26.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21813" PostTypeId="2" ParentId="21810" CreationDate="2014-12-16T20:13:48.050" Score="16" Body="&lt;p&gt;Relativistic effects account for gold's lack of reactivity.  Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus.  This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals.  These relativistic electrons gain mass and as a consequence, their orbits contract.  As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms.  The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to see the formulas and math behind this (it's not all that complicated) &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637&quot;&gt;see here&lt;/a&gt;.  Also note that similar arguments explain &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13667/why-does-mercury-have-low-melting-and-boiling-points/13671#13671&quot;&gt;mercury's anomalous properties&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-12-16T20:19:04.997" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21814" PostTypeId="2" ParentId="21810" CreationDate="2014-12-16T20:19:13.383" Score="22" Body="&lt;p&gt;First off, gold &lt;em&gt;does&lt;/em&gt; react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the &lt;a href=&quot;https://chemistry.stackexchange.com/a/21813/5017&quot;&gt;other answer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nice summary of some work by Jens K. Norskov can be found here:&#xA;&lt;a href=&quot;http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490&quot; rel=&quot;noreferrer&quot;&gt;http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can, of course, react gold with &lt;strong&gt;&lt;em&gt;aggressive&lt;/em&gt;&lt;/strong&gt; reagents like &lt;a href=&quot;http://en.wikipedia.org/wiki/Aqua_regia&quot; rel=&quot;noreferrer&quot;&gt;aqua regia&lt;/a&gt;, a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If properly done, the product is $\ce{HAuCl4}$ or &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloroauric_acid&quot; rel=&quot;noreferrer&quot;&gt;chloroauric acid&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-12-17T02:44:06.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21815" PostTypeId="1" AcceptedAnswerId="23816" CreationDate="2014-12-16T20:26:47.870" Score="0" ViewCount="8294" Body="&lt;p&gt;I understand that the solvents differ in their acidity but why does this affect the basicity of the solute?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5017" LastEditDate="2014-12-17T02:44:59.103" LastActivityDate="2015-01-16T04:03:21.033" Title="How is it possible to account for the fact that NH3 is a weak base in water but a strong base in HCl?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2015-01-16T07:40:17.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="21816" PostTypeId="2" ParentId="21805" CreationDate="2014-12-16T20:35:04.927" Score="9" Body="&lt;p&gt;$$\ce{A~  + :\!\!B -&amp;gt; \ \ A\!\leftarrow:\!\!B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_f = \dfrac{\ce{[A\!\leftarrow:\!\!B]}}{\ce{[A][B]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See sections 4.1 and 4.2 in the following reference for more information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf&quot; rel=&quot;nofollow&quot;&gt;http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $K_f$ is specific to an acid-base pair.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more universal scale is formed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gutmann%E2%80%93Beckett_method&quot; rel=&quot;nofollow&quot;&gt;Gutmann–Beckett method&lt;/a&gt; by observing the phosphorous NMR spectrum of triethylphosphine oxide as it interacts with a sample lewis acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Acceptor Number (AN) scale is formed with AN=0 corresponding to triethylphosphine oxide in hexane (P chemical shift of 41), and AN = 2.21 (d-41), where d is the chemical shift of triethylphosphine oxide interacting with the sample. The greater AN is, the more acidic the compound.    &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2014-12-17T02:59:04.390" LastActivityDate="2014-12-17T02:59:04.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21817" PostTypeId="1" AcceptedAnswerId="21823" CreationDate="2014-12-16T21:19:16.147" Score="7" ViewCount="20716" Body="&lt;p&gt;It's fairly easy to generate a significant yield of hydrogen gas simply by reacting zinc metal with $\ce{HCl}$ or aluminum with $\ce{NaOH}$ solution. But is there a simple and inexpensive chemical reaction that generates a significant yield of pure nitrogen gas?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="41328" LastEditDate="2017-10-10T14:45:19.460" LastActivityDate="2020-06-19T18:29:46.277" Title="Is there a simple and inexpensive way to generate nitrogen gas?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="8" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21818" PostTypeId="2" ParentId="21817" CreationDate="2014-12-16T21:33:39.317" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen#Production&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; you can generate pure nitrogen by heating metal azides, which gives out &#xA;nitrogen and the metal. Sodium azide is &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/s2002&quot; rel=&quot;nofollow noreferrer&quot;&gt;not really expensive&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2NaN_3 -&amp;gt;2Na + 3N_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you may expect, this may not be a very safe reaction as the first reaction that came to my mind when you asked for a reaction with large yield of nitrogen gas is that of TNT, though that would not give a pure yield as you have required. Azides are also explosive. Sodium azide is &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_azide#Automobile_airbags_and_airplane_escape_chutes&quot; rel=&quot;nofollow noreferrer&quot;&gt;used in car airbag and airplane escape chutes&lt;/a&gt; where this exact reaction is used to quickly inflate these safety features. The metallic sodium formed is then further reacted to less dangerous sodium compounds.  &lt;/p&gt;&#xA;" OwnerUserId="2174" LastEditorUserId="4945" LastEditDate="2019-02-19T09:02:24.270" LastActivityDate="2019-02-19T09:02:24.270" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21819" PostTypeId="2" ParentId="21774" CreationDate="2014-12-16T22:42:37.710" Score="6" Body="&lt;p&gt;Using your &quot;conjugate&quot; way of seeing two covalently bonded atoms as electrons, let's try &lt;span class=&quot;math-container&quot;&gt;$\ce{O_2}$&lt;/span&gt; which gives you &quot;&lt;span class=&quot;math-container&quot;&gt;$\ce{2O^{\bullet}}$&lt;/span&gt;&quot;. If you think about how the electrons are arranged in terms of electron configuration, you will see that &lt;span class=&quot;math-container&quot;&gt;$\ce{O^{\bullet}}$&lt;/span&gt; has &lt;span class=&quot;math-container&quot;&gt;$\ce{[He]}2s^2 2p^4$&lt;/span&gt; which is not a particularly stable electron configuration (if you only know the &quot;secondary school electron configuration&quot; you should recognise this as 2,6). Not having a full electron shell (2,8) means this is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the same with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;, you get &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;. Both of these have full shell electron configurations (&lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ne]}$&lt;/span&gt; or 2,8; &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]}$&lt;/span&gt; or 2,8,8). So the ions are indeed more stable in this sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, relative stability of ionic and covalent bonding cannot be simply compared using your &quot;conjugate&quot; method, because you need to make sure that you are making the comparison under similar conditions. For example, you cannot really say &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; is less stable than &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; because NaCl dissociates into ions in water while &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; does not, because the surrounding water molecules have a very strong dipole which will interfere much more strongly with ionic bonding (which is electrostatic) than covalent bonding in a homonuclear diatomic (which has no reason to dissociate into neutral radicals in water). You can possibly compare hardness of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; versus diamond, i.e. carbon that is purely covalently bonded, but graphite would not make such a good comparison because between the layers of graphite there are only weak Van der Waals interactions (and it is therefore softer because you are overcoming VdW interactions not covalent bonds). There are also special cases such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_liquid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic salts with a particularly low melting point&lt;/a&gt;, and there are many examples of weak covalent bonds. You also want to pick examples with few confounding factors, such as large differences in electronegativity which can give ionic contributions to covalent bonds, or large differences in size and effective charge which can give covalent contributions to ionic bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will answer your question using two examples of giant lattices that show &lt;em&gt;near&lt;/em&gt;-ideal ionic and covalent bonding: &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt; versus diamond (with acknowledgements to &lt;a href=&quot;https://chemistry.stackexchange.com/users/5017/geoff-hutchison&quot;&gt;Geoff Hutchison&lt;/a&gt; for the choice of LiF). Lithium, fluorine and carbon are all in period 2, so avoiding further complications due to differences in size such as comparing to &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;. I am &quot;taking a shortcut&quot; here by comparing boiling points, which is a bulk property that you can use for a giant lattice, because it is difficult to find a pure ionic diatomic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt; has &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_fluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;a boiling point of 1909 K&lt;/a&gt;, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;diamond sublimes at 3900 K&lt;/a&gt;. By this measure, it would appear that covalent bonding is stronger than ionic bonding in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_radius#Determination_of_ionic_radii&quot; rel=&quot;nofollow noreferrer&quot;&gt;the distance between the lithium and fluoride ions is 202 pm&lt;/a&gt;*, while &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond_length#Bond_lengths_in_organic_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;in diamond the distance between carbon atoms is 154 pm&lt;/a&gt;. In &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt;, the two oppositely charged ions are further away than the (positively charged) nuclei of any two carbon atoms in diamond that is held together by (negatively charged) electrons between them. What I am trying to point out is that, both forms of bonding involve interactions of positive and negative charges, but in covalent bonding the charges are much closer together, so the attractive electrostatic forces are much greater. This is why in this particular comparison, diamond comes out with stronger covalent bonding than the ionic bonding found in &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I indicated in the third paragraph, there are many cases where the bonding within an ionic substance can be stronger or weaker compared to a covalent substance. This is why I have emphasised that everything that is discussed is only true for this particular comparison. It will be unwise to claim that one type of bonding is overwhelmingly stronger than another, because it always depends on what exactly is being compared. Given the near infinite combinations of atoms that can give rise to bonding that is ionic, covalent, and a mixture of the two, it would be difficult to make a full catalogue of covalent and ionic substances and compare these against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*computed by taking half the distance of the cubic lattice.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastEditorUserId="30601" LastEditDate="2018-10-02T10:16:06.147" LastActivityDate="2018-10-02T10:16:06.147" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21820" PostTypeId="1" AcceptedAnswerId="21822" CreationDate="2014-12-17T00:59:39.757" Score="13" ViewCount="28856" Body="&lt;p&gt;Noble gases have full electron shells, which virtually blocks any other element from bonding with it. However, I've heard about cases where they bond to each other  - for example, helium can apparently form a dimer $\ce{He2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="11340" LastEditorUserId="16683" LastEditDate="2017-10-06T23:17:10.630" LastActivityDate="2018-04-13T23:00:30.230" Title="How do noble gases bond with themselves?" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;noble-gases&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21821" PostTypeId="1" AcceptedAnswerId="21835" CreationDate="2014-12-17T01:11:07.943" Score="5" ViewCount="179" Body="&lt;p&gt;Under the understanding of quantum chemistry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a very miniscule chance that an electron could appear inside the nucleus of an atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What could possibly happen if this &quot;phenomenon&quot; were to occur?&lt;/strong&gt;  Would it wreak havoc upon the atom and tear it apart?  Or will it bind with the protons and create a temporary vacuum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under my current research, the electron would jump out of the nucleus as soon as possible because this phenomenon should not occur for more than a fraction of a second. Though it shouldn't leave without &lt;em&gt;some&lt;/em&gt; adverse effects.&lt;/p&gt;&#xA;" OwnerUserId="11340" LastEditorUserId="7475" LastEditDate="2016-01-31T21:38:12.843" LastActivityDate="2016-01-31T21:38:12.843" Title="What effects would an electron in the nucleus have?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21822" PostTypeId="2" ParentId="21820" CreationDate="2014-12-17T02:09:36.250" Score="24" Body="&lt;p&gt;Noble gases usually do not form strong bonds between their atoms - it takes fair amount of energy to dimerise them into &lt;a href=&quot;https://en.wikipedia.org/wiki/Excimer&quot; rel=&quot;nofollow noreferrer&quot;&gt;excimers&lt;/a&gt;, but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms. On the other hand there are many &lt;a href=&quot;https://en.wikipedia.org/wiki/Noble_gas_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;stable molecules&lt;/a&gt; created by heavier noble gases (mainly xenon) with other elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in comment, you may heard about detection of so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_molecule&quot; rel=&quot;nofollow noreferrer&quot;&gt;van der Waals molecules&lt;/a&gt; of helium, which aren't &quot;true&quot; molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2018-04-13T23:00:30.230" LastActivityDate="2018-04-13T23:00:30.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21823" PostTypeId="2" ParentId="21817" CreationDate="2014-12-17T02:15:57.173" Score="13" Body="&lt;p&gt;Per Wikipedia, we have this laboratory method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also per Wikipedia, this method generates small quantities of impurities of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt;, which can be eliminated by passing the gas through potassium dichromate &lt;span class=&quot;math-container&quot;&gt;$\ce{(K2Cr2O7)}$&lt;/span&gt; dissolved in sulfuric acid &lt;span class=&quot;math-container&quot;&gt;$\ce{(H2SO4)}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr2O7^{-2}}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr2O3}$&lt;/span&gt;). But otherwise, this looks like the best approach.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="4945" LastEditDate="2019-02-19T09:04:59.447" LastActivityDate="2019-02-19T09:04:59.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21824" PostTypeId="2" ParentId="21815" CreationDate="2014-12-17T02:56:26.820" Score="0" Body="&lt;p&gt;I'm guessing because ammonium chloride $\ce{(NH4Cl)}$ is ionic and ammonium hydroxide $\ce{(NH4OH)}$ is not. You simply get more dissolution with the former.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-17T02:56:26.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21826" PostTypeId="2" ParentId="21774" CreationDate="2014-12-17T04:12:39.700" Score="-2" Body="&lt;p&gt;Short answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote Mehrdad: You're basically comparing an intermolecular force with an intramolecular force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic bonds are two charged atoms sitting next to each other with minimal overlap of their orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covalents bonds are two atoms becoming one molecule with significant changes in orbital structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not always clear cut: there are atoms with part-ionic and part-covalent character or two atoms sitting next to each other with moderate overlap of their orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good explanation (for solids): &lt;a href=&quot;https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds-i-was-taught-the-opposite&quot;&gt;Are metallic/ionic bonds weaker than covalent bonds?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrostatic bonds by definition are incapable of any significant bond formation. Bond formation involves the formation of a lower energy &quot;bonding orbital(s)&quot; out of two (or more) higher energy ordinary orbitals on adjacent atoms within the molecule. The formation of bonding orbitals gives covalent bonding orbitals much lower energy (not smaller, see diagram) than non-bonding orbitals which exist in the unbound state. This energy difference is larger than is possible for merely electrostatic attractions. Why is this the case? Because the electrons in bonding orbital are better able to counteract the positive charges of the adjacent nuclei. This makes sense because the negative charge on one molecule in an electrostatic arrangement is only able to neutralize the positive charge from its own nuclei. Bonding orbitals that are only present in covalent bonds share those electrons between two nuclei and thereby better neutralize the charges of both nuclei, which results in a lower energy for the bonded system than is possible in an ionic system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/syPpt.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Molecular_Orbital_Theory/Bonding_and_antibonding_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2014-12-19T23:06:26.240" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21827" PostTypeId="1" AcceptedAnswerId="21832" CreationDate="2014-12-17T09:51:49.827" Score="4" ViewCount="155" Body="&lt;p&gt;I was looking for an accurate definition of halogen bonding.&#xA;I was able to find quite a few good ones, but none of them would explain if a X---H intermolecular interaction would count as a form of halogen bonding. Where X is any halogen and H is a hydrogen from another molecule. Does this count as halogen or hydrogen bonding? Or maybe as both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;" OwnerUserId="1787" LastEditorUserId="9961" LastEditDate="2015-10-02T17:06:22.727" LastActivityDate="2015-10-02T17:06:22.727" Title="Halogen bond definition" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21830" PostTypeId="2" ParentId="21781" CreationDate="2014-12-17T12:59:03.333" Score="5" Body="&lt;p&gt;I think the reaction should be done in basic solution, not acidic solution.  &quot;One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was&#xA;added dropwise over 75 min at 0-3 C&quot; J. Org. Chem. 1990,55, 5287-5291.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ ClO- (aq) + I- (aq) + H2O (l)  -&amp;gt; I+ (aq) + Cl- (aq) + 2OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq)  -&amp;gt; I+ (aq) + Cl- (aq) + H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-19T12:15:35.150" LastActivityDate="2014-12-19T12:15:35.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21831" PostTypeId="1" AcceptedAnswerId="21895" CreationDate="2014-12-17T13:18:13.393" Score="6" ViewCount="262" Body="&lt;p&gt;If something is a hard acid, it tends to form ionic bonds as it is polarizing. So why is the aluminium ion a hard acid when it forms bonds with covalent character. A similar argument follows for why carbon dioxide is considered a hard acid?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="7475" LastEditDate="2015-11-13T19:04:57.100" LastActivityDate="2015-11-13T19:04:57.100" Title="Why is aluminium(III) considered a hard acid if it forms covalently bonded compounds like aluminium oxide?" Tags="&lt;acid-base&gt;&lt;bond&gt;&lt;polarity&gt;&lt;hsab&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21832" PostTypeId="2" ParentId="21827" CreationDate="2014-12-17T13:24:27.647" Score="3" Body="&lt;p&gt;I would say no, it is only a hydrogen bond, not a halogen bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a halogen bond, the halogen atom must accept electron density from the other member of the bond.  If the other member of the bond is a hydrogen atom bonded to a more electronegative element, I don't see have the halogen atom could be an acceptor of electron density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get access see &quot;&lt;a href=&quot;http://www.sciencemag.org/content/321/5891/918.short&quot; rel=&quot;nofollow&quot;&gt;Halogen Versus Hydrogen&lt;/a&gt;&quot; Science Vol. 321 no. 5891 pp. 918-919.  Footnote 7 there explains: &quot;A halogen bonding donor is a species that contains an electrophilic halogen that can become a member of a halogen bond. In the literature on halogen bonding, the reader should be alert to how the words 'donor' and 'acceptor' are used. In a complex RX---B, RX is the halogen bond donor but the electron acceptor (Lewis acid); B is the electron donor and halogen bond acceptor (Lewis base).&quot;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-17T13:24:27.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21833" PostTypeId="2" ParentId="21774" CreationDate="2014-12-17T13:58:59.327" Score="12" Body="&lt;p&gt;&lt;strong&gt;Ionic bonds can easily be as strong as covalent ones.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have &lt;em&gt;both&lt;/em&gt; strong ionic &lt;strong&gt;&lt;em&gt;and&lt;/em&gt;&lt;/strong&gt; strong covalent character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For now, I'll stick to simple cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt;, which seems a prototypical ionic bond. The interatomic distance is &lt;span class=&quot;math-container&quot;&gt;$\pu{156 pm}$&lt;/span&gt; &lt;a href=&quot;http://cccbdb.nist.gov/exp2.asp?casno=7789244&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to NIST&lt;/a&gt;. That gives a potential energy of &lt;span class=&quot;math-container&quot;&gt;$\pu{903 kJ/mol}$&lt;/span&gt;. Now, that assumes a full electron is transferred from &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt;, which isn't the case (more like &lt;span class=&quot;math-container&quot;&gt;$0.8 - 0.9$&lt;/span&gt;). This simple calculation also ignores &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Land%C3%A9_equation#Repulsive_term&quot; rel=&quot;nofollow noreferrer&quot;&gt;Born repulsion&lt;/a&gt; which is similar to the short-range Van der Waals repulsion between atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete analysis &lt;a href=&quot;http://dx.doi.org/10.1021/ed077p1076&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;77&lt;/em&gt; (8), p 1076&lt;/a&gt; gives a much better agreement. The real bond dissociation energy is &lt;span class=&quot;math-container&quot;&gt;$\pu{577 kJ/mol}$&lt;/span&gt; (&lt;a href=&quot;https://web.archive.org/web/20140724034511/https://www.nist.gov/data/nsrds/NSRDS-NBS31.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source, PDF via the Internet Archive&lt;/a&gt;). A simple spherical ion model from the paper gives the bond enthalpy as &lt;span class=&quot;math-container&quot;&gt;$\pu{608 kJ/mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;calculations based on a simple ionic model reproduce the bond energies of &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF(g)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{NaF(g)}$&lt;/span&gt; to within &lt;span class=&quot;math-container&quot;&gt;$5\%$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Bond-dissociation_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;covalent bond dissociation energies&lt;/a&gt;, the triple bond in dinitrogen is &lt;span class=&quot;math-container&quot;&gt;$\pu{945 kJ/mol}$&lt;/span&gt;. Ok, that's perhaps stronger, but &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt; only has a &quot;single&quot; charge and clearly some ionic compounds have greater charge transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More &quot;prototypical&quot; covalent bonds include &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{410 kJ/mol}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\ce{O=O}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{497 kJ/mol}$&lt;/span&gt;), both less than &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I said earlier, it's not the case of a bond being &lt;em&gt;only&lt;/em&gt; covalent or &lt;em&gt;only&lt;/em&gt; ionic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote from R. Gillespie &lt;a href=&quot;http://dx.doi.org/10.1021/ed078p1688&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2001&lt;/strong&gt;, &lt;em&gt;78&lt;/em&gt; (12), 1688&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The large atomic charges in &lt;span class=&quot;math-container&quot;&gt;$\ce{BF3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{SiF4}$&lt;/span&gt; are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of &lt;span class=&quot;math-container&quot;&gt;$613$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{565 kJ/mol}$&lt;/span&gt;, respectively. In comparison, the &lt;span class=&quot;math-container&quot;&gt;$\ce{C–C}$&lt;/span&gt; bond in ethane has a bond dissociation enthalpy of only &lt;span class=&quot;math-container&quot;&gt;$\pu{345 kJ/mol}$&lt;/span&gt;. The great strength of the &lt;span class=&quot;math-container&quot;&gt;$\ce{BF}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{SiF}$&lt;/span&gt; bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are &lt;strong&gt;both&lt;/strong&gt; strongly covalent and strongly ionic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt; under MO theory might be something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wUS05.png&quot; alt=&quot;Hypothetical MO diagram of LiF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We readily see &lt;span class=&quot;math-container&quot;&gt;$\ce{LiF}$&lt;/span&gt; has very high ionic character. For all intents and purposes, the &quot;bonding&quot; &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; orbital is localized on the &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; atom, there's very little energy difference between the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; here.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="4945" LastEditDate="2019-06-06T13:00:13.910" LastActivityDate="2019-06-06T13:00:13.910" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="21835" PostTypeId="2" ParentId="21821" CreationDate="2014-12-17T14:24:10.237" Score="7" Body="&lt;p&gt;An electron being at/in the nucleus results in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_contact_interaction&quot;&gt;Fermi Contact Interaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This interaction can be observed experimentally through NMR, EPR, and electron capture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Paramagnetic_NMR_spectroscopy&quot;&gt;NMR of paramagnetic compounds&lt;/a&gt;, chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the &quot;contact shift&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on Fermi contact, see &quot;Origin and meaning of the Fermi contact interaction&quot; Theor Chim Acta (1988) 73:173-200.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-17T18:17:13.817" LastActivityDate="2014-12-17T18:17:13.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21836" PostTypeId="1" AcceptedAnswerId="21893" CreationDate="2014-12-17T15:25:26.237" Score="7" ViewCount="3030" Body="&lt;p&gt;Well, I was doing research about the ways we can measure electronegativity and I just realized there must be an arbitrary electronegativity set for an specific element to give us the relative amounts of the other ones. Pauling used hydrogen and suggested 2.1 for its electronegativity which later was altered to 2.20. Now, the question is, why was the electronegativity of hydrogen suggested as 2.1 and why is it 2.20 right now? What are the secrets in those numbers? &lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="4945" LastEditDate="2014-12-24T05:10:12.030" LastActivityDate="2014-12-24T05:10:12.030" Title="Why is the electronegativity of hydrogen 2.20?" Tags="&lt;hydrogen&gt;&lt;history-of-chemistry&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21837" PostTypeId="1" AcceptedAnswerId="21839" CreationDate="2014-12-17T15:45:33.610" Score="4" ViewCount="1577" Body="&lt;p&gt;Our chemistry teacher encouraged us to study the history and naming of the orbitals on the web. (actually, our textbook did, but that's irrelevant to the problem) I easily could find the reason behind their naming, and what their names stood for. However, some sources stated that f as in f orbitals stands for fine, while others claimed it to be fundamental. Which of these two is correct? And why did scientists gave f orbitals that name? (as you're aware, they did it due to the f block elements' emission spectrum; my question is, what was the specification of their emission spectrum?) I'm sorry if the question looks stupid, but I can't figure it out myself.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2014-12-17T16:09:29.473" LastActivityDate="2014-12-19T20:40:38.687" Title="About the meaning of the names of f orbitals" Tags="&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21839" PostTypeId="2" ParentId="21837" CreationDate="2014-12-17T16:22:55.687" Score="6" Body="&lt;p&gt;I would have to say &quot;f&quot; overwhelmingly stands for &quot;fundamental&quot; (not &quot;fine&quot;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An early (1915) reference is &quot;Some Recent Discoveries in Spectrum Series&quot; Astrophysical Journal, vol. 41, p.323 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Within each of these three systems there are, in general, four different &lt;em&gt;types&lt;/em&gt; of series, known as &quot;principle&quot;, &quot;sharp&quot; (second subordinate), &quot;diffuse&quot; (first subordinate), and &quot;fundamental&quot; (Bergmann).  These names are not always very appropriate; for instance the &quot;sharp&quot; series is often diffuse, the &quot;fundamental&quot; may be composed of the faintest lines in the spectrum, but this is not vital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The fundamental series was discovered by Arno Bergmann in 1907, and is sometimes called the Bergmann series. It was named &quot;fundamental&quot; by William M. Hicks, I beleive in &lt;a href=&quot;http://www.jstor.org/stable/90988&quot; rel=&quot;nofollow&quot;&gt;A Critical Study of Spectral Series. Part I: The Alkalies H and He&lt;/a&gt; (1911): &quot;F is the symbol for the new series (fundamental) as P.S.D. stand for those already known&quot; (page 57) and &quot;It was my first impression that this would be the same for all the alkalies and the .987 of the earlier discussion. It would be the basis of a fundamental series for each&#xA;element, and the letter F which has been attached to it had its origin in this idea.&quot; (page 94).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The names existed before orbital theory.  There is no need to investigate f-block elements to observe excitations to f-orbitals.  Electrons of any elment can be excited to an f-orbital. In fact, Bergmann discovered the series in the sodium spectrum.  What was orginially known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sharp_series&quot; rel=&quot;nofollow&quot;&gt;sharp series&lt;/a&gt;, were transition to s-orbitals, the primary series transitions to p-orbitals, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.che.uc.edu/jensen/W.%20B.%20Jensen/Reprints/137.%20s,%20p,%20d,%20f.pdf&quot; rel=&quot;nofollow&quot;&gt;The Origin of the s, p, d, f Orbital Labels&lt;/a&gt; for additional information. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-19T20:40:38.687" LastActivityDate="2014-12-19T20:40:38.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21840" PostTypeId="1" CreationDate="2014-12-17T17:00:26.607" Score="4" ViewCount="5804" Body="&lt;p&gt;There has been a lot of news about Graphene since its discovery in 2004. And as we are all told it is a revolutionary material which is very strong, conductive and transparent; even in some cases it has revolutionized especial laws of physics (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is it about the &lt;strong&gt;structure of Graphene&lt;/strong&gt; which makes it so &lt;strong&gt;strong&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10189" LastEditorUserId="7448" LastEditDate="2014-12-31T18:51:42.390" LastActivityDate="2018-10-06T17:53:05.317" Title="Why is Graphene So Strong?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;atoms&gt;&lt;lewis-structure&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21842" PostTypeId="1" AcceptedAnswerId="21845" CreationDate="2014-12-17T19:14:04.313" Score="3" ViewCount="143" Body="&lt;p&gt;I was unsure to the largest extent about whether I should post this question in chemistry Q &amp;amp; A or bio Q &amp;amp; A until I just read that sunscreens &quot;absorb&quot; UV rays, not allowing the most of those rays from penetrating into the skin. So when this is about the combination, it must be chemistry.&lt;br/&gt; Anyhow, is there any special name for these &quot;absorbents&quot;? And how exactly, do they &quot;absorb&quot; UVA &amp;amp; UVB? (I'm not really definite if this absorbing is really a scientific expression or not, and that's why I use quotations when using the word)&lt;br/&gt; I, myself, reckon that the sunscreen products form a special compound on the skin or they make the body produce something as a shield against UV rays, but I have got no clues.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="418" LastEditDate="2014-12-17T19:47:49.327" LastActivityDate="2014-12-18T19:36:05.017" Title="How do sunscreens protect the skin from UV rays?" Tags="&lt;organic-chemistry&gt;&lt;photochemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21844" PostTypeId="1" CreationDate="2014-12-17T19:42:11.583" Score="1" ViewCount="760" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tlIh6.png&quot; alt=&quot;Electronic configuration of silver&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For silver, why don't the electrons in the 4d orbital move to the 5p orbital? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for any elements to move their electrons between $n$d and $(n+1)$p orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the f orbital is completely filled with 14 electrons, should we take it in consideration to find the valence states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Z=58, can you show me the electronic configuration and how do we form the valence state(s)?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="2174" LastEditDate="2014-12-18T16:57:05.457" LastActivityDate="2015-01-17T16:58:28.960" Title="How to find the possible valence states of elements in the lower end of the d block and the f block?" Tags="&lt;electronic-configuration&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21845" PostTypeId="2" ParentId="21842" CreationDate="2014-12-17T19:47:24.750" Score="3" Body="&lt;p&gt;Apart from titanium dioxide, the active species in sunscreens are organic compounds from different substance classes, such as cinnamates, benzophenones, benzimidazoles, and triazines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The do absorb in the 280-315 (UVB) and/or 315-380 nm (UVA) range. This &quot;uptake&quot; of energy results in an electronic excitation, the compounds undergo a transition from the $S_0$ (ground) state to an excited $S_1$ state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The excited singlet state ($S_1)$ may &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;deactivate radiationless via &lt;em&gt;internal conversion&lt;/em&gt; (&lt;strong&gt;IC&lt;/strong&gt;), the energy of the photon is released as heat or &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;undergo &lt;em&gt;intersystem crossing&lt;/em&gt; (&lt;strong&gt;ISC&lt;/strong&gt;) to a triplet state (usually $T_1)$ with different spin multiplicity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The latter state then might again loose its energy via &lt;strong&gt;IC&lt;/strong&gt; or undergo a chemical transformation, through which the suncreen might be turned to a species that not longer is able to absorb UV radiation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-17T19:47:24.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21846" PostTypeId="2" ParentId="19638" CreationDate="2014-12-17T20:53:49.013" Score="2" Body="&lt;p&gt;The simple answer is that elemental, metallic mercury isn't very soluble in either the blood or in the contents of the digestive system. If ingested, metallic mercury will simple pass through the gut with little effect on the body and with very little being absorbed into the places where it can do harm. The same is true in the blood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury causes harm when it interacts with essential enzymes in the body and makes them malfunction. For it to do this, mercury atoms or ions have to be able to reach a significant concentration in bodily fluids. There are two ways for this to happen easily. Mercury vapour is absorbed into the bodily system (but this requires some effort such as combining bad ventilation with finely divided mercury droplets and heat or bad ventilation and a long time period as early mercury chemists discovered). Many mercuric compounds, especially organic ones, are also readily absorbed (either because they are soluble, as many salts are, or because they are volatile, as many organic mercury compounds like methyl mercury are.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the worst case of human mercury poisoning (&lt;a href=&quot;http://www.wikiwand.com/en/Minamata_disease&quot; rel=&quot;nofollow&quot;&gt;the Minamata disaster&lt;/a&gt;) resulted from the release of methyl mercury into the environment where it accumulated in fish ultimately eaten by people.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-12-17T20:53:49.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21847" PostTypeId="2" ParentId="21817" CreationDate="2014-12-17T21:01:16.807" Score="10" Body="&lt;p&gt;You could try a simple two step process. Take some air, burn some carbon-rich material in it until no further burning occurs. Then pass the remaining gas through an alkaline solution to remove the carbon dioxide. The remaining gas will be more than 95% nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you get this with no expensive chemicals. &lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-12-17T21:01:16.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21848" PostTypeId="1" AcceptedAnswerId="21917" CreationDate="2014-12-17T21:18:41.340" Score="2" ViewCount="451" Body="&lt;p&gt;The heavier the atom, the more unstable it gets, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;strong&gt;not&lt;/strong&gt; true about Uranium and we know it.&#xA;I wondered why. A brief explanation stated that since more neutrons are there in the nucleus there is more nuclear force to &lt;sup&gt;238&lt;/sup&gt;U than &lt;sup&gt;235&lt;/sup&gt;U. Seems legitimate, but then that should have been true about the other radioactive species. Again, we know that's not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, either the explanation about the Uranium exception is not totally true or the principle of the heavier-atom instability (of course, exceptions occur at elements like U) is not a principle. Or maybe there is a relation between these two (nuclear force and atomic weight) which is beyond me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hunch is that the latter is true; searched some sites which had an answer to the question, but no acceptable results gained. &lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="9961" LastEditDate="2017-06-07T15:41:20.797" LastActivityDate="2017-06-07T17:21:12.953" Title="Stability variation in heavy nuclei" Tags="&lt;stability&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21849" PostTypeId="2" ParentId="21756" CreationDate="2014-12-17T21:30:25.163" Score="8" Body="&lt;p&gt;&lt;strong&gt;The key difference with microwaves is the &lt;em&gt;amount&lt;/em&gt; of energy emitted&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Microwaves don't have more &amp;quot;energy&amp;quot; than visible light &lt;em&gt;per photon&lt;/em&gt;. But this is irrelevant in the case of a microwave oven.&lt;/p&gt;&#xA;&lt;p&gt;There are two reasons for this but the first one one (as described in the existing answer) is actually a distraction: microwaves are relatively efficient at exciting molecular rotations and vibrations (especially of ions) so dumping their energy as heat in the object they are interacting with (usually your food).&lt;/p&gt;&#xA;&lt;p&gt;But this is a distraction from a more significant point: microwave ovens push out &lt;em&gt;a lot&lt;/em&gt; of microwave energy, usually a little under a kilowatt. This is about 10 times the energy radiated by a traditional incandescent (visible) lightbulb. If you could dump that much &lt;em&gt;visible&lt;/em&gt; light into a small volume and your food absorbed it, it would cook. But you are more likely to end up with a very hot box and food that wasn't cooked efficiently. Many electric heaters (radiating red and near infra-red photons) push the same sort of power as a microwave, but they radiate it over a much larger area and don't cook you because the energy is dumped over a very much larger volume. Unless you stand too close.&lt;/p&gt;&#xA;&lt;p&gt;Electric grills are basically just very powerful incandescent bulbs tuned to emit most of their radiation in the near infra red. They emit similar power to a microwave (~kW) but don't cook food as efficiently as infra red is mostly absorbed by the &lt;em&gt;surface&lt;/em&gt; of the food they heat (microwave energy penetrates more deeply allowing food to warm from the &lt;em&gt;inside&lt;/em&gt;. Interestingly very salty foods may not cook as well in a microwave as surface effects (where the radiation couples strongly with the dielectric fields caused by salty ions) absorb more of the microwave energy near the surface.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2020-10-16T08:56:28.993" LastActivityDate="2020-10-16T08:56:28.993" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="21852" PostTypeId="2" ParentId="21848" CreationDate="2014-12-18T00:42:20.733" Score="5" Body="&lt;p&gt;The principle that &quot;heavier nuclei are more stable&quot; is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested &lt;a href=&quot;http://iopscience.iop.org/1742-6596/420/1/012001/pdf/1742-6596_420_1_012001.pdf&quot;&gt;this paper&lt;/a&gt; for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-18T00:42:20.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21853" PostTypeId="1" CreationDate="2014-12-18T01:36:12.760" Score="4" ViewCount="161" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the ratio of $\ce{MbO2/Mb}$ if the solution is saturated with air?  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Details of the question&lt;/strong&gt;:&#xA;  $$&#xA;\begin{align}&#xA;\ce{Mb + O2 (aq) &amp;amp;-&amp;gt; MbO2}\\&#xA;\Delta G' &amp;amp;= -30.3~\mathrm{kJ/mol}\\&#xA;T &amp;amp;= 298~\mathrm{K}\\&#xA;\ce{pH} &amp;amp;= 7\\&#xA;\text{Henry's Constant for }\ce{O2}  &amp;amp;= 42 \cdot 10^3~\mathrm{bar}\\&#xA;\text{Content of }\ce{O2}\text{ in dry air} &amp;amp;= 21\%\\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at Solution&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically this is a two part question, as I see it. First I need to figure out the concentration of dissolved $\ce{O2}$ using Henry's Law. Then I can simply plug this value into the $$\Delta G' = -RT\ln K$$ equation where $$K = \frac{\ce{[MbO2]}}{\ce{[O2][Mb]}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&#xA;$$&#xA;\begin{align}&#xA;P &amp;amp;= K\cdot m\\&#xA;0.21~\mathrm{bar} &amp;amp;= (42 \cdot 10^3)\cdot m\\&#xA;m &amp;amp;= 5 \cdot 10^{-6} \pu{mol/kg}\\&#xA;-30.300~\mathrm{J/mol} &amp;amp;= -8.314 \cdot 298 \cdot \ln\left(\frac{\ce{[MbO2]}}{[5\cdot10^{-6}]\ce{[Mb]}}\right)&#xA;\end{align}&#xA;$$&#xA;After dividing and raising both sides to $e^x$ (to get rid of $\ln$) I get $$\frac{\ce{MbO2}}{\ce{Mb}} = 1.02.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the answer sheet says it is $21.6$. I cannot figure out why this is so. &lt;/p&gt;&#xA;" OwnerUserId="11358" LastEditorUserId="5026" LastEditDate="2018-03-01T06:13:03.517" LastActivityDate="2018-03-01T06:13:03.517" Title="Ratio of MbO2/Mb in a satured solution of dissolved oxygen from dry air" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21854" PostTypeId="1" AcceptedAnswerId="21871" CreationDate="2014-12-18T05:17:41.287" Score="0" ViewCount="2666" Body="&lt;p&gt;I think there are $\ce{C^{4-}}$ and $\ce{C^{2-}}$ existing. How can we find the charge of ions in the elements? Should we look at the group or valence?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="4945" LastEditDate="2014-12-18T11:07:06.110" LastActivityDate="2014-12-18T16:45:34.760" Title="What kind of carbon ions exist?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21855" PostTypeId="1" AcceptedAnswerId="32672" CreationDate="2014-12-18T05:18:36.140" Score="5" ViewCount="1640" Body="&lt;p&gt;I just discovered that some solid chemicals, such as $\ce{Na2SO4}$, dissolve better in cold water than hot water from &lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0m/s17-04-effects-of-temperature-and-pre.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and would like to know if there is any particular reason as to why. Is it their structure, their charge, etc.? &lt;/p&gt;&#xA;" OwnerUserId="11359" LastEditorUserId="4945" LastEditDate="2014-12-18T06:48:47.843" LastActivityDate="2015-06-10T19:52:38.233" Title="Why do some solid chemicals dissolve better in colder water?" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21856" PostTypeId="1" CreationDate="2014-12-18T06:44:55.163" Score="13" ViewCount="8291" Body="&lt;p&gt;Benzene and nitrate ion are given in my textbook as examples for the delocalization of π-electrons. Benzene, due to symmetry of its resonating structures, is simple enough. We assume that σ-electrons are localized and π-electrons are delocalized in the ring.&lt;/p&gt;&#xA;&lt;p&gt;Each carbon atom promotes one electron from its &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbital to the empty &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2p}$&lt;/span&gt; orbital. It hybridizes two &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals with the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; to form three &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; orbitals which it then uses to form three σ-bonds, two with other carbons and one with a hydrogen.&lt;/p&gt;&#xA;&lt;p&gt;The remaining unhybridized &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital then sticks out above and below the plane of the ring. The electrons from all the six unhybridized &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals of the six carbons are then delocalized above and below the plane of the ring.&lt;/p&gt;&#xA;&lt;p&gt;However, when I try to apply a similar reasoning to the nitrate anion, problems arise. The resonance Lewis structures of nitrate ion are:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3OLJM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3OLJM.png&quot; alt=&quot;resonance Lewis structures of nitrate ion&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, assuming again that only the π-electrons are delocalized, we would expect that only two electrons are delocalized (since there is only one double bond). But my textbook claims that each atom is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;-hybridized, and as in benzene, one unhybridized &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbital per atom sticks out above and below the plane of the molecule.&lt;/p&gt;&#xA;&lt;p&gt;Assuming this to be true, I presumed that two electrons would be localized and tried to arrange the remaining 22 electrons into a Lewis structure with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized atoms. I couldn't find any viable structures with only three lone pairs around each atom.&lt;/p&gt;&#xA;&lt;p&gt;What is the exact mechanism for delocalization of electrons in nitrate? Is there a general scheme to this mechanism that applies to all such similar molecules such as &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}?$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="41328" LastEditDate="2021-05-23T12:41:41.610" LastActivityDate="2021-05-23T12:41:41.610" Title="Delocalization of π-electrons in nitrate ion" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21857" PostTypeId="2" ParentId="21840" CreationDate="2014-12-18T09:17:31.927" Score="1" Body="&lt;p&gt;Imagine carbon as some sort of wool.&#xA;Graphite would be just some wool without any pressure on it.&#xA;Diamond would be a wool ball with a structure and a shape.&#xA;Graphene would be a textile knitted with the strings of wool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wool is soft, a wool ball is more durable to pressure and a textile could carry some weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine graphene as a canvas fabric knitted with the finest strings the wool has and all strings are equally long. The pressure would be splitted equally on each of this strings and you wouldn't have a weak point. You would carry an elephant with it.&lt;/p&gt;&#xA;" OwnerUserId="11362" LastActivityDate="2014-12-18T09:17:31.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21858" PostTypeId="2" ParentId="21856" CreationDate="2014-12-18T09:49:52.760" Score="16" Body="&lt;p&gt;Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp&lt;sup&gt;2&lt;/sup&gt; hybrid orbitals, giving rise to two &amp;sigma; bonds to the neighbouring carbons and one &amp;sigma; bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised &amp;pi; system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. &lt;i&gt;D&lt;/i&gt;&lt;sub&gt;6h&lt;/sub&gt;, three of them are bonding and three of them are antibonding. In the &amp;pi; system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/30E9Z.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/30E9Z.jpg&quot; alt=&quot;pi orbitals of benzene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrate ion $\ce{NO3-}$ is what I like to call a &lt;a href=&quot;https://chemistry.stackexchange.com/q/32354/4945&quot;&gt;Y-aromat&lt;/a&gt; since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene.&lt;br&gt;&#xA;The resonance structures, however, look much more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Th3Jh.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Th3Jh.jpg&quot; alt=&quot;resonance nitrate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In quite good approximation, you can regard the nitrogen as having three sp&lt;sup&gt;2&lt;/sup&gt; hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen &amp;sigma; bonds. Each oxygen has one lone pair in the in-plane p&amp;nbsp;orbital and one lone pair in the sp&amp;nbsp;hybrid. The remaining out-of-plane p&amp;nbsp;orbitals of nitrogen and oxygen build up the delocalised &amp;pi; system. This time there are only four orbitals. Because of symmetry constraints, i.e. &lt;i&gt;D&lt;/i&gt;&lt;sub&gt;3h&lt;/sub&gt;, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vqc2x.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vqc2x.jpg&quot; alt=&quot;pi system of nitrate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the carbonate anion $\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another.&lt;br&gt;&#xA;For example, the E&quot; orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\ce{O\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ovpOp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovpOp.jpg&quot; alt=&quot;Inflated orbitals contour value 0.00001&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2017-11-01T11:12:48.177" LastActivityDate="2017-11-01T11:12:48.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21859" PostTypeId="2" ParentId="21840" CreationDate="2014-12-18T10:13:25.470" Score="4" Body="&lt;p&gt;If I understood what strong means, then graphene has high strength because of the 2-D layer structure of carbon atoms. The strength can also be attributed to its &lt;strong&gt;$sp^{2}$&lt;/strong&gt; hybridisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any crystalline material will have a lattice structure. When we mean strength of a material its usually the ability to break apart the bond between two lattice atoms. But the observed strength is usually 2 or 3 orders lower than theoretically predicted strength due to &lt;strong&gt;flaws (voids, dislocations etc)&lt;/strong&gt;. But when graphene has a near flawless 2-D structure we can reach its theoretical strength. Though its is a very expensive process to actually realize &lt;strong&gt;&quot;flawless&quot;&lt;/strong&gt;. Even with flaws its strength is really high compared to steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So its essentially the &lt;strong&gt;2-D one atom layer structure&lt;/strong&gt; that can be responsible for its high strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing is though graphene has high strength and elasticity (ability to get stretched)  it is relatively brittle (fracture toughness ~4 MPa$\sqrt{m}$).&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="9961" LastEditDate="2017-01-02T22:34:28.950" LastActivityDate="2017-01-02T22:34:28.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21860" PostTypeId="1" AcceptedAnswerId="21863" CreationDate="2014-12-18T10:16:30.190" Score="13" ViewCount="1511" Body="&lt;p&gt;All of us are aware of the importance of the catalysts in bio-chemistry. For a high school learner like me, catalysts ,and therefore, enzymes play a bridge-like role that connect high school bio to high school chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, I got baffled about the two common energy profiles drawn for &lt;strong&gt;an anonymous reaction with a presence of a catalyst&lt;/strong&gt;. The question may seem rudimentary, I agree, but the possible answers to my question were too technical for me to understand.&lt;br&gt;&#xA;One of the energy profiles had several curves in the pathway for $E$, and the other one has only &lt;em&gt;one&lt;/em&gt; curve in the pathway where the catalyst should have affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i333c.png&quot; alt=&quot;This is usually how biologists draw the energy profile &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lHWH3.png&quot; alt=&quot;This is what wiki draw&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the first energy profile is about a net change in $E_\mathrm{a}$ but the second one demonstrates the reality when an exothermic reaction occurs.&#xA;&lt;br/&gt; Now, here are the questions:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are the two energy profiles different? Please be as if you're teaching thermo-chemistry to a little kid!&lt;/li&gt;&#xA;&lt;li&gt;Why do those curves occur in the latter energy profile and is there a way to know how many of those curves will occur in the presence of a catalyst?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Image credits: chemwiki.ucdavis.edu (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Modeling_Reaction_Kinetics/Temperature_Dependence_of_Reaction_Rates/The_Arrhenius_Law/The_Arrhenius_Law%3A_Activation_Energies&quot; rel=&quot;noreferrer&quot;&gt;page here&lt;/a&gt;) and wiki (&lt;a href=&quot;https://en.wikipedia.org/wiki/Energy_profile_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;page here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7475" LastEditDate="2015-12-17T23:58:18.377" LastActivityDate="2015-12-17T23:58:18.377" Title="A misunderstanding about the energy profile of reactions with a catalyst involved" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;catalysis&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21862" PostTypeId="2" ParentId="21860" CreationDate="2014-12-18T11:08:32.500" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A reaction involving more than one elementary step has one or more intermediates being formed which, in turn, means there is more than one energy barrier to overcome. In other words, there is more than one transition state lying on the reaction pathway. As it is intuitive that pushing over an energy barrier or passing through a transition state peak would entail the highest energy, it becomes clear that it would be the slowest step in a reaction pathway. However, when more than one such barrier is to be crossed, it becomes important to recognize the highest barrier which will determine the rate of the reaction. This step of the reaction whose rate determines the overall rate of reaction is known as rate determining step or rate limiting step. The height of energy barrier is always measured relative to the energy of the reactant or starting material. Different possibilities have been shown in &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/6d/Reaction_Co-ordinate_Diagrams_for_reactions_with_0%2C_1%2C_2_intermediates.png/500px-Reaction_Co-ordinate_Diagrams_for_reactions_with_0%2C_1%2C_2_intermediates.png&quot; rel=&quot;nofollow&quot;&gt;figure 6&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the link you gave on energy profile i found this. Basically there is no &lt;strong&gt;&quot;net change&quot;&lt;/strong&gt; in $E_{a}$ . In the first graph in question the curve which is at the top is the representing how the reaction proceeds without a catalyst and the one below is of the reaction with catalyst. In the second graph there are 4 intermediate reactions/transition states before the product is formed. Hence you have 4 peaks of different activation energy as shown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A catalyst basically provides alternate routes to a reaction by forming the &lt;strong&gt;&quot;transition/metastable/intermediate state&lt;/strong&gt; at a lower energy than the original reaction. You will study later that this has to do with basic Collision theory and molecular dynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given graphs a catalyst can make the &lt;strong&gt;R-&gt;P&lt;/strong&gt; through just one route where only one transition happens (as in the graph 1) or for 4(or even more) transition states (as in the graph 2).&#xA; For one set of conditions(P,T,V etc) I am guessing the number of these peaks will be unique. It is also possible that for a certain catalyst and reactant the number of peaks / intermediate states is constant.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="7448" LastEditDate="2015-12-17T10:02:37.023" LastActivityDate="2015-12-17T10:02:37.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21863" PostTypeId="2" ParentId="21860" CreationDate="2014-12-18T11:13:26.497" Score="7" Body="&lt;p&gt;The upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\ce{S -&amp;gt; P}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The activation energy is lowered&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\#$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is the same starting marterial in the presence or absence of the catalyst, the &lt;strong&gt;energy of the transition state&lt;/strong&gt; is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly this situation is described in the second graph!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catalyst &quot;reacts&quot; with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with &lt;strong&gt;different intermediates and a transition state&lt;/strong&gt; not found in in the non-catalyzed reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-18T11:13:26.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21865" PostTypeId="1" AcceptedAnswerId="21868" CreationDate="2014-12-18T14:18:16.660" Score="5" ViewCount="1944" Body="&lt;p&gt;THF is used as a solvent for Grignard reagents, but epoxides react with Grignard reagents. There is only a small difference between epoxides and THF, what causes the difference in reactivity towards Grignard reagents?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="7475" LastEditDate="2015-10-16T17:29:30.027" LastActivityDate="2015-10-16T17:29:30.027" Title="Why do Grignard reagents react with epoxides but not THF?" Tags="&lt;reactivity&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21866" PostTypeId="2" ParentId="19398" CreationDate="2014-12-18T14:19:34.297" Score="1" Body="&lt;p&gt;We have direct formula for calculating the weight of the substance as,&#xA;$$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$&#xA;$$=\frac{0.02\times 250 \times 100}{1000}$$&#xA;$$=0.5 \text{ g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to prepare 0.02 $ \mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the volume of ammonia,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial molarity = 2 M       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial volume= ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final molarity = 0.2 M       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final volume = 100 mL&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have formula, &#xA; $$ \text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$&#xA;$$\text{Initial volume} = \frac{0.2 \times 100}{2}$$&#xA;$$= 10 mL$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-12-18T14:19:34.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21867" PostTypeId="1" AcceptedAnswerId="21875" CreationDate="2014-12-18T14:20:46.413" Score="6" ViewCount="2567" Body="&lt;p&gt;What makes $\ce{R-MgX}$ a more reactive Grignard reagent than $\ce{R-Cu}$? Has it to do with solvation, bond enthalpy, electronegativity?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2017-01-06T19:30:46.890" LastActivityDate="2017-01-06T19:30:46.890" Title="What makes R-MgX a more reactive Grignard reagent than R-Cu?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;reactivity&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21868" PostTypeId="2" ParentId="21865" CreationDate="2014-12-18T14:32:16.590" Score="7" Body="&lt;p&gt;The difference between THF and epoxides is perhapse bigger than you might think. Due to the three-membered-ring of an epoxide, a lot of energy is 'stored' in ring strain. This destabilizes the stucture and is a driving force for a Grignard reaction (or any substitution on an epoxide). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The THF is a five-membered-ring, which is far more stable. This bond is simply an ether, that happens to be in a cyclic structure. Therefore it also reacts like an ether: not easily attacked by nucleophiles.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2014-12-18T14:32:16.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21869" PostTypeId="2" ParentId="21844" CreationDate="2014-12-18T14:39:04.153" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[For silver], why don't the electrons in the 4d orbital move to the 5p orbital?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case of silver, this is likely because the energy separation between the 4d and 5s orbitals are quite large (&lt;a href=&quot;https://physics.stackexchange.com/questions/72368/why-are-most-metals-gray-silver&quot;&gt;absorbing in ultraviolet region, which is why silver appears white/silver&lt;/a&gt;), so the energy separation between 4d and 5p would be even larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, when you write down the electron configuration you are writing down the &lt;strong&gt;ground state&lt;/strong&gt;, i.e. the most stable configuration. While it is possible in principle to move electrons from 4d to 5p, it will not give you the most stable configuration as moving that electron requires a large amount of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inferring the likely valence states, you can only use orbitals with unpaired electrons, and/or move the electrons that are not separated by large energy differences between different orbitals to make hybrid orbitals with unpaired electrons. In this case, as you cannot move 4d electrons to 5p (which would make some kind of spd hybrid orbital), you can only have a valence state of 1 from the unpaired electron in 5s.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for any elements [to move their electrons between $n$d and $(n+1)$p orbitals]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but only for elements nearer the top of the periodic table (and not at the very top, as you will then not have any accessible d orbitals). Period 3 and period 4 (I think) would be most likely to do this. For example, phosphorus can have a valence state of 5, and sulfur can have a valence state of 6, both by involving their accessible 3d orbital, to make hybrid $\ce{sp^3d}$ and $\ce{sp^3 d^2}$ orbitals respectively. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the f orbital is [completely] filled with 14 electrons, should we take it in consideration to find the valence states?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, no. When the f block is completely filled, the electrons in the f orbital are generally assumed to be core-like, i.e. stable, or low in energy, therefore will have a large energy gap to the other electrons in 6s, 5d and 6p (or 7s, 6d and 7p), and do not tend to participate in bonding/valency. There are possibly one or two exceptions, but I cannot recall them right now. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When Z=58, can you show me the electronic [configuration] and how do we form the valence state(s)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Z = 58 is Cerium, which has the following electronic configuration: $\ce{[Xe] 4f^1 5d^1 6s^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On paper, it will appear that due to the diffuseness of the orbitals 4f, 5d and 6s, it is probably unusual to form covalent compounds to cerium, and you would expect a valence state of 3 (if you only use the 5d and 6s electrons), or a valence state of 4 (if you also use in addition the 4f electron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cerium#Physical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;It was found the 4f is actually near the energy of 5d and 6s, giving rise to a variable electron structure and a valence state change of 3 to 4 when cerium is cooled or compressed&lt;/a&gt;. Moreover, it is possible to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Organocerium_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;organocerium compounds with a valency of 3&lt;/a&gt;. By the way, this is not something that you can deduce simply by looking at the electronic configuration. You have to find this out empirically, using experiments!&lt;/p&gt;&#xA;" OwnerUserId="2174" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2014-12-18T14:39:04.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21870" PostTypeId="1" CreationDate="2014-12-18T15:14:10.740" Score="3" ViewCount="256" Body="&lt;p&gt;I have a program which can perform density-functional calculations for atoms, given a density functional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the simplest form of exchange potential to use is one relevant for a uniform electron gas (i.e. the original Kohn-Sham exchange, proportional to $n_e^{1/3}$). A good correlation functional is also available. However when I performed calculations for some atoms the energy of the highest occupied orbital differs greatly from the ionization potential (off by a few eVs!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That the energy of the highest occupied orbital equals the ionisation potential of the neutral atom is known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Koopmans&amp;#39;_theorem&quot; rel=&quot;nofollow&quot;&gt;Koopmans' Theorem&lt;/a&gt;. I get a feeling from the literature that DFT can be quite accurate. So what's wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly when I use the Slater exchange functional, coupled with a correction suggested by Skillman (i.e. replacing the potential by $1/r$ when the overall potential drops below that value), the results improve significantly. (See the book &quot;Atomic Structure Calculations&quot; by Herman Skillman). Well, maybe I should follow this recipe, but it seems the procedure is quite ad-hoc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there any functionals which will give reasonable values for the atomic ionisation potentials? I do not want to implement methods like the optimised effective potential method since the latter is not readily generalised to a finite-temperature scenario. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="5017" LastEditDate="2014-12-19T00:05:06.893" LastActivityDate="2014-12-19T00:14:49.143" Title="DFT Calculations, Atomic Ionization Potentials -- Which Exchange-Correlation Functional to Use, to Preserve Koopmans Theorem?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;density-functional-theory&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21871" PostTypeId="2" ParentId="21854" CreationDate="2014-12-18T15:56:33.833" Score="4" Body="&lt;p&gt;Many different carbon ions exist in the gas phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0301010497003583&quot; rel=&quot;nofollow&quot;&gt;Electronic spectra of linear carbon anions&lt;/a&gt;, spectra of $\ce{C_2^-}$, $\ce{C_3^-}$... $\ce{C_10^-}$ are discussed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dianions such as $\ce{C_7^{2-}}$ have also been observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solid phase, aluminum carbide and berylium carbide have more than 50% ionic character according to the Handbook of Refractory carbides and nitrides, and correspond to $\ce{C^{4-}}$.  Calcium carbide, which could be considered a salt of acetylene, corresponds to  $\ce{C2^{2-}}$  rather than $\ce{C^{2-}}$  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-18T16:45:34.760" LastActivityDate="2014-12-18T16:45:34.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21872" PostTypeId="1" AcceptedAnswerId="21877" CreationDate="2014-12-18T16:36:58.883" Score="1" ViewCount="155" Body="&lt;p&gt;&lt;a href=&quot;https://imgur.com/p8j47Ly&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/p8j47Ly.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question, I believe there is a mistake for iii). I think that there the tertiary carbocation would be formed (more stable), which would lead to the tetra substituted alkene (even more stable). Am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastActivityDate="2014-12-18T17:40:18.480" Title="In E1 reactions, does carbocation stability have impact on the final product?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21873" PostTypeId="1" CreationDate="2014-12-18T16:39:23.510" Score="1" ViewCount="33415" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{1.50 g}$&lt;/span&gt; sample of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; is added to &lt;span class=&quot;math-container&quot;&gt;$\pu{35.0 g}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; in a styrofoam cup and stirred until dissolved. The temperature of the solution drops from &lt;span class=&quot;math-container&quot;&gt;$24.8$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{22.4 ^\circ C}$&lt;/span&gt;. Assume that the specific heat and density of the resulting solution are equal to those of water, &lt;span class=&quot;math-container&quot;&gt;$\pu{4.18 J g-1 ^\circ C-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1.00 g mL-1}$&lt;/span&gt;, respectively, and assume that no heat is lost to the calorimeter itself, nor to the surroundings.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{KCl (s) + H2O (l) -&amp;gt; KCl (aq)} \qquad \Delta H = ?$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;a) (2 points) Is the reaction endothermic or exothermic (circle the correct answer)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Endothermic&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;b) (4 points) What is the heat of solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; expressed in kilojoules per mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q_\mathrm{rxn} = -q_\mathrm{cal}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I multiplied the sample &lt;span class=&quot;math-container&quot;&gt;$\pu{1.50 g}$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$\pu{4.18 J} \cdot (-2.4) = \pu{-15.048 J}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Divided that by &lt;span class=&quot;math-container&quot;&gt;$1000 = -0.015048$&lt;/span&gt;; thus, &lt;span class=&quot;math-container&quot;&gt;$0.015048$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, my answer seems to be wrong. I know that reaction is endothermic since the temp drops, but I am wondering which values I should be using to correctly determine the &amp;quot;Heat of Solution&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2021-03-06T17:42:00.610" Title="How to calculate the heat of dissolution from a calorimeter experiment?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21875" PostTypeId="2" ParentId="21867" CreationDate="2014-12-18T16:43:49.513" Score="11" Body="&lt;p&gt;The difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55).  This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds.  The electrons in a carbon-copper bond are better shared between the two atoms, whereas the electron distribution in a carbon-magnesium bond places more electron density on carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this difference in electron density on the carbon atoms the carbon portion of a cuprate is a softer nucleophile, while the carbon portion of a Grignard is a harder nucleophile.  These difference in nucleophilicity produce different reactivity patterns for the two reagents.  For example, the softer cuprate will undergo 1,4 addition to an $\alpha,\beta$-unsaturated carbonyl compound (a softer electrophile), while the Grignard adds directly to the carbonyl (which is a harder electrophile).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-18T16:43:49.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21876" PostTypeId="2" ParentId="21873" CreationDate="2014-12-18T16:58:57.983" Score="3" Body="&lt;p&gt;You have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the water provides most of the heat for the reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total mass of the solution is 1.50g + 35.0g = 36.5g.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be multiplying  36.5g by the temperature change and heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you need to consider how many moles 1.50g KCl is. &#xA;Divide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-18T17:12:36.613" LastActivityDate="2014-12-18T17:12:36.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21877" PostTypeId="2" ParentId="21872" CreationDate="2014-12-18T17:09:56.173" Score="2" Body="&lt;p&gt;The answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aCYay.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Thanks to Bon and Yolo123 for their comments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway.  First there is a statistical factor;  there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring.  Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-18T17:40:18.480" LastActivityDate="2014-12-18T17:40:18.480" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21881" PostTypeId="1" CreationDate="2014-12-18T17:43:48.660" Score="1" ViewCount="2523" Body="&lt;p&gt;So here is my question,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; A water sample contains 9.5% $\mathrm{MgCl_2}$ and 11.7% $\mathrm{NaCl}$ (by weight). Assuming 80% ionization of each salt. Boiling point of water will be ________. ($\mathrm{K_b}$ for water $\mathrm{=0.52}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You know, this question isn't that tough to bother anyone (definitely not!). What's all confusing here is the molality of two solutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \mathrm{T_b=K_b\cdot m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Degree of dissosiation $(\mathrm{\alpha}) = 0.8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know very well that molality of a solution having only one solute is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{m=\frac{n\ (no.\ of\ moles\ of\ solute)}{y\ (mass\ of\ solvent\ in\ kg)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the solution having more than one solute? Do they have any molality (I am sure that there will surely be, because it is there in the question). If yes then what's the method of calculating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if you are solving this, here is the answer to the question (as per my answer key),&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt;377 K&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="670" LastActivityDate="2014-12-18T18:46:39.977" Title="Molality of solution having more than one solute?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21883" PostTypeId="2" ParentId="21881" CreationDate="2014-12-18T18:09:38.537" Score="1" Body="&lt;p&gt;The boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = (0.2 + (3)(0.8))(molality of $\ce{MgCl_2}$) + (0.2 + (2)(0.8))(molality of $\ce{NaCl}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find molality of $\ce{MgCl_2}$ and molality of $\ce{NaCl}$, consider a 1kg sample of the solution.  There are 95g  $\ce{MgCl_2}$, 117g $\ce{NaCl}$ and 788g water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the number of moles $\ce{MgCl_2}$ in 95g $\ce{MgCl_2}$, and divide this number by 0.788kg.  This is the molality of $\ce{MgCl_2}$ in the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat for $\ce{NaCl}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-18T18:46:39.977" LastActivityDate="2014-12-18T18:46:39.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21884" PostTypeId="1" CreationDate="2014-12-18T18:27:58.770" Score="1" ViewCount="3066" Body="&lt;p&gt;I am trying to find the pH of the following problem (Answer key is $\mathrm{pH}=10.77$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the $\mathrm{pH}$ of a $\pu{100 mL}$ solution containing $\pu{40.0 g}$ of $\ce{Na2CO3}$ after   $\pu{400 mL}$ of $\pu{1.00 M}$ $\ce{HNO3}$ has been added. Ignore  the volume  of water    produced    by  the reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach&lt;/strong&gt;:&#xA;$$\ce{Na_2CO_3~(aq)~+~2HNO_3~(aq)\rightarrow CO_2~(g)~+~H_2O~+~2NaNO_3~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with $\pu{0.377 mol}$ $\ce{Na2CO3}$ and $\pu{0.4 mol}$ $\ce{HNO3}$.  Because of the 2:1 ratio, I use up all of the $\ce{HNO3}$, and create $\pu{0.4 mol}$ $\ce{NaNO3}$ and have $\pu{0.177 mol}$ of $\ce{Na2CO3}$ left over.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to go about setting up the ice table.  Should I use that &#xA;$$\ce{NaNO3 + H2O -&amp;gt; HNO3 + Na+ + OH-}$$&#xA;where I have the initial amount of $\ce{NaNO3}$?&lt;/p&gt;&#xA;" OwnerUserId="9791" LastEditorUserId="5026" LastEditDate="2018-03-05T14:17:48.507" LastActivityDate="2018-03-05T14:17:48.507" Title="Titration of sodium carbonate and nitric acid" Tags="&lt;physical-chemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21886" PostTypeId="2" ParentId="21884" CreationDate="2014-12-18T19:21:08.010" Score="1" Body="&lt;p&gt;To solve this problem, you need to know the pKas of carbonic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pKas are 6.35 and 10.33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add 1 equivalent of acid to $\ce{Na_2CO_{3}}$, the product will be $\ce{NaHCO_{3}}$ (sodium bicarbonate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would model the final solution as 0.354 moles of $\ce{HCO_{3}^-}$ and 0.023 moles $\ce{H_2CO_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proposed answer 10.77 is incorrect.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-18T19:21:08.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21887" PostTypeId="2" ParentId="21842" CreationDate="2014-12-18T19:36:05.017" Score="0" Body="&lt;p&gt;Just to add, Zinc (like Titanium) is another UV filter that reflects UV light rather than absorbing it.  I believe ever other UV filter, at least those approved in the U.S., absorb the UV rays.&lt;/p&gt;&#xA;" OwnerUserId="10179" LastActivityDate="2014-12-18T19:36:05.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21889" PostTypeId="1" AcceptedAnswerId="21890" CreationDate="2014-12-18T21:12:33.700" Score="3" ViewCount="8243" Body="&lt;p&gt;What is the mechanism by which benzene, cyclohexene and $\ce{AlCl3}$ form cyclohexylbenzene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that cyclohexene will form a bond with Al with its double bond. However, how do we get rid of the $\ce{AlCl3}$ now attached to the previous cyclohexene after the benzene attaches to the module?&lt;/p&gt;&#xA;" OwnerUserId="7256" LastEditorUserId="4231" LastEditDate="2014-12-18T23:20:45.657" LastActivityDate="2020-02-02T22:31:48.937" Title="Friedel-Crafts Alkylation Using Alkenes: Mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21890" PostTypeId="2" ParentId="21889" CreationDate="2014-12-18T22:02:26.057" Score="6" Body="&lt;p&gt;The Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene.  The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/65m5N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene.  A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible.  In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is &quot;invisible&quot; in this case (unless we label the starting cyclohexene).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-18T22:39:46.787" LastActivityDate="2014-12-18T22:39:46.787" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21892" PostTypeId="2" ParentId="20344" CreationDate="2014-12-18T23:22:15.543" Score="6" Body="&lt;p&gt;The idea of splitting water using electrolysis and pumping it into an engine's air intake has been around for a long time. The idea is that it's supposed to increase the efficiency of burning gasoline. However, even assuming that this is true (there are many reasons why it may not be—modern gas engines are in the range of 95–98% efficient at combustion), electrolysis is very inefficient, so I'm quite dubious of any potential gains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hunting around on the internet, it's obvious that a lot of people seem to think this works (and that people who say it doesn't are oil company shills or some such), but I wasn't able to find any compelling evidence to support it. How this kind of system interacts with a modern computer controlled engine may perhaps even make efficiency worse. There doesn't seem to be much academic research into it, though a Google Scholar search brings up a lot of kooky patents. This kind of thing is pretty difficult to test well—tests need to be performed under very well controlled conditions to get accurate results and is thus always subject to confirmation bias and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find a single report done by Environment Canada on one of these products for diesel engines and they found that adding the hydrogen generator device didn't make any statistically significant change to combustion efficiency, exhaust emission rates, or fuel consumption. They calculated that the device was about 60% efficient at making hydrogen, at best, and used about 8 mL of diesel fuel every 100 km, so it doesn't really make efficiency worse to any appreciable degree, but doesn't help either. (Source: ERMD Report # 2004-032 If you search, it's on scribd, but I couldn't find any permanent sources)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, a water-fuelled car on its own or as a hybrid with gasoline or diesel don't seem to be possible. Water is a very stable compound and it takes a great deal of energy to break it into hydrogen and oxygen gas. At most, you can get the energy you put in back, and realistically, a lot will be lost along the way. This makes a solely water fuelled car a perpetual motion machine of the first kind as it violates the law of conservation of energy—one cannot burn hydrogen and oxygen to produce energy to make the same amount of hydrogen and oxygen consumed and still have energy left over to move a vehicle. A hybrid vehicle faces a similar problem. Because the part of the system that involves water consumes energy rather than contributes it to motion, the gas or diesel part has to produce more energy to make up for the difference. Given that modern gas and diesel engines burn almost all the fuel put into them, there doesn't seem to be any mechanism by which adding a small quantity of hydrogen and oxygen could produce significantly more energy.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2014-12-26T04:33:23.843" LastActivityDate="2014-12-26T04:33:23.843" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21893" PostTypeId="2" ParentId="21836" CreationDate="2014-12-18T23:28:16.133" Score="5" Body="&lt;p&gt;Interestingly, the first paper from Pauling on electronegativity &lt;a href=&quot;http://dx.doi.org/10.1021/ja01348a011&quot;&gt;&quot;The Nature of the Chemical Bond. IV. The Energy of Single Bonds and The Relative Electronegativity Of Atoms&quot;&lt;/a&gt; published in 1932 gives a different scale with H = 0.0 and F = 2.0!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, he changed to 2.1 for H and 4.0 for F, e.g. &lt;a href=&quot;http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html&quot;&gt;http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look through his &lt;a href=&quot;http://books.google.com/books?id=L-1K9HmKmUUC&amp;amp;q=electronegativity+scale#v=snippet&amp;amp;q=electronegativity%20scale&amp;amp;f=false&quot;&gt;book&lt;/a&gt;, on p. 90 we find the following text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Following these there are given the values obtained by adding 2.05 to them...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resulting numbers are then rounded to one decimal, from H = 2.05 to H = 2.1 and F = 4.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His notes are online (e.g., &lt;a href=&quot;http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-29.html&quot;&gt;here&lt;/a&gt;) but there's no obvious explanation I can find yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;&lt;em&gt;hunch&lt;/em&gt;&lt;/strong&gt; is that he started with H = 0.0, and realized that the alkali metals would be negative numbers. The &lt;a href=&quot;http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-29.html&quot;&gt;shifted scale&lt;/a&gt; gave Li ~ 1.0 and F ~ 4.0, which seem like &quot;nice&quot; numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to hunt down the reason for the move to 2.2 as the standard for hydrogen. The first reference seems to be by Huggins (&lt;a href=&quot;http://dx.doi.org/10.1021/ja01113a001&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1953&lt;/strong&gt;, 75 (17), pp 4123–4126&lt;/a&gt;) but the paper itself doesn't really explain the reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#cite_note-Allred-5&quot;&gt;Wikipedia&lt;/a&gt; suggests a paper by Allred, but that cites the Huggins paper for the 2.2 number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably more could be found with a very careful search through the papers of Pauling and Huggins, but I've never heard any clear explanations and the documents online don't give them either.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2014-12-18T23:53:11.367" LastActivityDate="2014-12-18T23:53:11.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21894" PostTypeId="2" ParentId="21709" CreationDate="2014-12-18T23:45:02.000" Score="6" Body="&lt;p&gt;Electronegativity affects &lt;a href=&quot;https://en.wikipedia.org/wiki/Dipolar_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;coordinate covalent bonds&lt;/a&gt;; indeed atoms which accept electrons should be &quot;electron hungry&quot; enough to take them. However it stems mainly from that the acceptor doesn't have complete electron shell, thus acts as &lt;a href=&quot;https://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lewis acid&lt;/a&gt; and donor acts as base.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2017-05-12T16:26:58.093" LastActivityDate="2017-05-12T16:26:58.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21895" PostTypeId="2" ParentId="21831" CreationDate="2014-12-19T00:02:35.450" Score="6" Body="&lt;p&gt;&lt;strong&gt;Short summary - HSAB theory has it's limits of applicability, and this is a good example.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recent discussion of acid-base theory center on the frontier orbital concept, e.g. W. B. Jensen, &lt;a href=&quot;https://books.google.com/books?id=RrREAAAAIAAJ&amp;amp;q=The+Lewis+Acid-Base+Concepts&amp;amp;dq=The+Lewis+Acid-Base+Concepts&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=l2mTVO23GqHbsAT12ICgBw&amp;amp;ved=0CB8Q6AEwAA&quot;&gt;&lt;em&gt;The Lewis Acid-Base Concepts&lt;/em&gt; (1980) pp. 112-155.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this concept, an &quot;acid-base&quot; reaction creates new HOMO-LUMO orbitals of the product. So it's about energy matching between the acid (empty LUMO) and base (filled HOMO) (and also symmetry and orbital overlap). It's still very much a Lewis picture, since you have the electron pair on the base and an electron acceptor on the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we react $\ce{NH3 + H+ -&amp;gt; NH4+}$ we combine the HOMO of $\ce{NH3}$ (i.e., a lone pair) with an empty LUMO on $\ce{H+}$ to create the adduct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO, this is a nice concept, since the orbital energy match between Al and O is good, you get covalent character to the resulting product. (Now you &lt;em&gt;also&lt;/em&gt; get ionic character, but that's a different question.)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-19T00:02:35.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21896" PostTypeId="1" AcceptedAnswerId="21902" CreationDate="2014-12-19T00:03:02.660" Score="5" ViewCount="303" Body="&lt;p&gt;Deoxy- means the compound misses an oxygen comparing to the name without the prefix.&lt;br&gt;&#xA;Acetate means ethanoate in IUPAC language. Many biologists still abide old rules and call it acetate. And finally, corticosterone. I just threw away thoughts and checked for the formula  online.&lt;br&gt;&#xA;What is the law behind naming &quot;corticosterone&quot;? Is it relative to corticosteroids? Please answer in a detailed way!&lt;br&gt;&#xA;Maybe it is corti+coster+one? Since I know that there exists cortisone and cortisol.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="16683" LastEditDate="2016-12-09T12:09:59.620" LastActivityDate="2019-02-22T20:23:37.860" Title="What is the origin of the naming of deoxycorticosterone acetate?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21897" PostTypeId="2" ParentId="21870" CreationDate="2014-12-19T00:14:49.143" Score="5" Body="&lt;p&gt;I think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevLett.105.266802&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Phys. Rev. Lett.&lt;/em&gt; (2010) 105, art. 266802&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.84.075144&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Phys. Rev. B&lt;/em&gt; (2011) 84, art. 075144&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically the idea is to tune the range-separation parameter $\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find a match such that Koopmans' theorem holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is an &quot;optimally-tuned&quot; range-separated hybrid DFT functional (OT-RSH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not perfect, for example, it violates size consistency: &lt;a href=&quot;http://dx.doi.org/10.1063/1.4807325&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Chem. Phys.&lt;/em&gt; (2013) 138, 204115&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-19T00:14:49.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21898" PostTypeId="1" AcceptedAnswerId="21903" CreationDate="2014-12-19T00:55:42.347" Score="28" ViewCount="15331" Body="&lt;p&gt;This is an old question that our textbook tried to answer but worsened the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many things are soluble in water. So many, that studying solutions will always require studying aqueous ones. It is true that many non-polars like waxes are not very soluble in water, yet I have &lt;strong&gt;never&lt;/strong&gt; run into a solvent as &quot;good&quot; as water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how were we answered when we asked that &quot;why is water a good solvent&quot;? They said since water is polar so the attraction between for example $\ce{O}$ and the positive ions is so much blah blah blah!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either there are &quot;great&quot; solvents like water out there or there are other things about water that make it the &lt;em&gt;master&lt;/em&gt; solvent that are beyond me.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a solvent as &quot;versatile&quot; as water?&lt;/li&gt;&#xA;&lt;li&gt;Can this kind of solvent be non-polar?&lt;/li&gt;&#xA;&lt;li&gt;If the answer to the questions above is no, what is special about water?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5668" LastEditDate="2018-01-23T18:15:54.930" LastActivityDate="2018-01-23T18:15:54.930" Title="Why is water &quot;the universal&quot; solvent?" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;solvents&gt;&lt;polarity&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21899" PostTypeId="2" ParentId="21898" CreationDate="2014-12-19T01:48:29.087" Score="11" Body="&lt;p&gt;Since water is polar as you note, and &quot;like dissolves like&quot; remains a good principle, I think the answer is that lots and lots of substances are polar. Most (ionic) salts are, for example. Even substances that are polar covalent are at least slightly polar and so somewhat soluble (and with agitation, more soluble). So you have to be really non-polar (e.g. hydrocarbons, &quot;oils&quot;) to not be soluble in water.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-19T01:48:29.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21900" PostTypeId="2" ParentId="21896" CreationDate="2014-12-19T02:26:23.407" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the law behind naming &quot;corticosterone&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no law, since the name indicate where the compounds are produced. Corticosterone and corticosteroid are synonyms for a groups of compounds (steroids) that are &lt;a href=&quot;http://en.wikipedia.org/wiki/Corticosterone&quot; rel=&quot;nofollow&quot;&gt;produced in the cortex of the adrenal glands&lt;/a&gt;. &lt;a href=&quot;http://en.wikipedia.org/wiki/Cortisone&quot; rel=&quot;nofollow&quot;&gt;Cortisone&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cortisol&quot; rel=&quot;nofollow&quot;&gt;cortisol&lt;/a&gt; refer to 2 of the corticosteroids. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-19T02:26:23.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21901" PostTypeId="1" CreationDate="2014-12-19T03:21:57.237" Score="1" ViewCount="5189" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming an ideal gas, calculate the density in $\mathrm{g/cm^3}$ of propane, $\ce{C3H8}$ at $0.000~^\circ\mathrm{C}$ and $1.000~\mathrm{atm}$. (3 significant figures) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have figured out that I need this formula:&#xA;$$\frac{n}{V} = \frac{p}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a total of $44.09562~\mathrm{g/mol}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck at this point.   &lt;/p&gt;&#xA;" OwnerUserId="7938" LastEditorUserId="5026" LastEditDate="2018-02-28T14:22:33.537" LastActivityDate="2020-03-09T06:05:55.500" Title="How do I calculate the density of propane assuming an ideal gas?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21902" PostTypeId="2" ParentId="21896" CreationDate="2014-12-19T03:49:11.407" Score="7" Body="&lt;p&gt;Naming complex chemical compounds in a straightforward systematic fashion is usually not very easy. And often there are trivial names, that existed way before this systematic naming scheme was introduced. Especially when it comes to naturally occurring compounds like steroids. The IUPAC defines &lt;a href=&quot;https://goldbook.iupac.org/S06005.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;steroids&lt;/a&gt; as the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Naturally occurring compounds and synthetic analogues, based on the cyclopenta[a]phenanthrene carbon skeleton, partially or completely hydrogenated; there are usually methyl groups at C-10 and C-13, and often an alkyl group at C-17. By extension, one or more bond scissions, ring expansions and/or ring contractions of the skeleton may have occurred. Natural steroids are derived biogenetically from triterpenoids.&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/q1V7T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1V7T.png&quot; alt=&quot;steroid&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  &lt;sub&gt;(source: &lt;a href=&quot;https://goldbook.iupac.org/SVG/S06005.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;iupac.org&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is even a quite lengthy article about the nomenclature of steroids: [1] &lt;a href=&quot;http://dx.doi.org/10.1351/pac198961101783&quot; rel=&quot;nofollow noreferrer&quot;&gt;G. P. Moss, &lt;em&gt;Pure &amp;amp; App. Chem.&lt;/em&gt; &lt;strong&gt;1989&lt;/strong&gt;, &lt;em&gt;61&lt;/em&gt; (10), 1783-1822.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name Deoxycorticosterone acetate is itself derived from a trivial name. The underlying principle compound is &lt;a href=&quot;http://en.wikipedia.org/wiki/Corticosterone&quot; rel=&quot;nofollow noreferrer&quot;&gt;corticosterone&lt;/a&gt; and its systematic steroid name is 11β,21-dihydroxypregn-4-ene-3,20-dione, see [1] Table 2. It can be derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Cortisol&quot; rel=&quot;nofollow noreferrer&quot;&gt;cortisol&lt;/a&gt;, which has an additional hydroxy group in position 17. The systematic name therefore is (11β),17,21-trihydroxypregn-4-ene-3,20-dione.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You assigned correctly, that the prefix de(s)oxy- refers to losing an oxygen containing functional group. In this case a hydroxy group at position 11. The systematic name for 11-deoxycorticosterone would therefore be 21-hydroxypregn-4-ene-3,20-dione.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the acetate refers to an acetyl ester group at position 21 (in the sidechain). Interestingly the systematic name for cortisol acetate is 21-&lt;em&gt;O&lt;/em&gt;-acetylcortisol. Since it has already a trivial name, see [1], 3S-4.3:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the steroid acid has a trivial name that already specifies the hydroxyl group that is esterified, the prefix must be &lt;em&gt;O&lt;/em&gt;-acyl-.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Applying the same logic, the systematic name for Deoxycorticosterone acetate should be 21-&lt;em&gt;O&lt;/em&gt;-acetyldeoxycorticosterone. Which makes little to no sense. A more systematic approach is described again in  [1], 3S-4.3 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Esters of steroid alcohols&lt;/em&gt;&lt;br&gt;&#xA;  Esters of steroid alcohols are named by the appropriate steroid substituent group followed by that of the acyloxy group in its anionic form. The steroid name is formed by replacing the terminal -e of the hydrocarbon name by -yl, -diyl, etc. and inserting this before the locant with a Greek letter to designate position and configuration. When necessary locants should be used with the anionic part of the name. Examples: [...]&lt;br&gt;&#xA;  The prefix form acyloxy-, with the appropriate locants, is used if there is a preferred functional group as the suffix (e.g. an acid or lactone). Examples: [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in this case since you have a ketone, which is probably more important, I would choose the second part of this rule. Hence the systematic name becomes 21-acetoxypregn-4-ene-3,20-dione.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, naming is quite a big, often confusing subject and sometimes the origins are not clear at all. And your analysis was completely correct in every way. Trivial names are often like the vocabulary of a specific language - if you don't know it, you have to look it up.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Additional fun fact&lt;/em&gt;:&lt;br&gt;&#xA;The name acetate is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Retained_name&quot; rel=&quot;nofollow noreferrer&quot;&gt;retained name&lt;/a&gt; which can be used in a &lt;a href=&quot;https://goldbook.iupac.org/S05601.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;semisystematic&lt;/a&gt; way.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2019-02-22T20:23:37.860" LastActivityDate="2019-02-22T20:23:37.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21903" PostTypeId="2" ParentId="21898" CreationDate="2014-12-19T03:54:36.923" Score="20" Body="&lt;p&gt;To directly address where the phrase comes from:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water is called the &quot;universal solvent&quot; because it dissolves more substances than any other liquid. -&lt;a href=&quot;http://water.usgs.gov/edu/solvent.html&quot; rel=&quot;noreferrer&quot;&gt;USGS&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;What are ideal qualities of solvent?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The strength of a solvent can be attributed to the strength of its intermolecular forces like london forces, dipole-dipole forces, ion-induced dipole, and hydrogen bonding. These are forces of attraction and repulsion. Solvation occurs when a molecule is surrounded by the solvent, so when there are strong intermolecular forces, stronger solvation occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LIz4N.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The hydrogen bond&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water makes use of the hydrogen bond, a type of intermolecular force experienced when hydrogen is attracted to the electronegative atoms nitrogen, oxygen, or fluorine. Hydrogen bonding is the strongest intermolecular force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2yUjk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hydrophobia&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water is a great solvent for hydrophilic molecules but hydrophobic molecules are, by definition, not easily disturbed by water. This makes a clear exception to the &quot;Universal Solvent&quot; idea. I haven't read any literature claiming it to actually be the &lt;em&gt;best&lt;/em&gt; solvent, but it does work well in many chemical situations.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-12-19T19:43:51.957" LastActivityDate="2014-12-19T19:43:51.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21904" PostTypeId="2" ParentId="21901" CreationDate="2014-12-19T04:43:46.563" Score="4" Body="&lt;p&gt;As you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\frac{n}{V}= \frac{P}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the density, we can further modify the equation as,&#xA;\begin{align}&#xA;\frac{m}{MV} &amp;amp;= \frac{P}{RT}\\&#xA;\frac{\rho}{M}&amp;amp;= \frac{P}{RT}\\&#xA;\rho &amp;amp;= \frac{PM}{RT}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just have to replace the parameters with their respective values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of temperature should be converted to Kelvin as $(0 + 273)~\mathrm{K} = 273~\mathrm{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the value of gas constant $R$ should be chosen as  $0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}}$.&#xA;\begin{align}&#xA;\rho &amp;amp;= \frac{1~\mathrm{atm} \cdot 44.09562~\mathrm{g~mol^{-1}}}{{0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}} \cdot 273~\mathrm{K}}}\\&#xA;&amp;amp;=1.9683~\mathrm{g/L}\\&#xA;&amp;amp;=0.0019683~\mathrm{g/cm^3}\\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="5026" LastEditDate="2018-02-28T14:23:39.060" LastActivityDate="2018-02-28T14:23:39.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21905" PostTypeId="1" AcceptedAnswerId="21923" CreationDate="2014-12-19T06:06:16.360" Score="1" ViewCount="367" Body="&lt;p&gt;What would be the wave function of the lowest energy molecular orbital of a hypothetical linear H3+ molecule? &#xA;According to the LCAO method, I feel the lowest energy MO will be 1s(A) + 1s(B) + 1s(C).&#xA;Where 1s(A) is the wave function of the 1s orbital of one of the Hydrogen atoms.&#xA;This has 0 nodes and has the lowest energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't this correst?&lt;/p&gt;&#xA;" OwnerUserId="11373" LastEditorUserId="11373" LastEditDate="2014-12-19T06:26:53.813" LastActivityDate="2014-12-19T14:33:55.227" Title="What are the molecular orbitals of the hypothetical linear H₃⁺ molecule?" Tags="&lt;homework&gt;&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21906" PostTypeId="1" CreationDate="2014-12-19T07:33:01.640" Score="3" ViewCount="1974" Body="&lt;p&gt;When cooking in the kitchen you usually use some oil or butter to bake your meat and potatoes in. After some weeks without cleaning (which happens every now and then) the surroundings of the stove will get a greasy, sticky layer. Since this only happens relatively close to the stove, say a 1-2 meter radius, I'm assuming that it is caused by my cooking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is: how does the grease get so far away from the stove?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that sometimes oil will spray a bit when you add something water-containing to it, but the grease is actually also in places that are not in the 'line of fire' (on top of the ventilation hood for example). Therefore, the grease must be getting airborne somehow, but I don't understand how, because evaporation of the vegetable oil or butter seems highly unlikely to me. Does anyone know what is going on?&lt;/p&gt;&#xA;" OwnerUserId="1362" LastActivityDate="2014-12-19T07:48:53.933" Title="How does grease become airborne?" Tags="&lt;everyday-chemistry&gt;&lt;vapor-pressure&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21907" PostTypeId="2" ParentId="21906" CreationDate="2014-12-19T07:48:53.933" Score="5" Body="&lt;p&gt;Let's reword it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Traces of oil and grease are distilled off at temperatures below their boiling points. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The temperature in the frying pan is above 100 °C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The food does contain significant amounts of water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This sounds like the conditions for steam distillation, except that you're collecting the mixture of water and water-insoluble material everywhere in the kitchen and not in a flask.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-19T07:48:53.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21908" PostTypeId="2" ParentId="21898" CreationDate="2014-12-19T09:57:16.080" Score="20" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Water is incredibly easy to get and work with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;In full&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water is a good solvent for polar compounds&lt;sup&gt;[citation needed]&lt;/sup&gt;, and the reasons for this are laid out pretty well by John Snow, but that's not really what makes it the universal solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, a series of other, incidental properties makes it a popular choice:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Availability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There's a &lt;em&gt;lot&lt;/em&gt; of water&lt;sup&gt;[citation needed]&lt;/sup&gt;. It &lt;em&gt;literally&lt;/em&gt; falls from the sky. This means that for most applications, it's the absolutely cheapest solvent available.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reactivity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water is a reasonably stable molecule. Few things you can dissolve in water are expected to react significantly with it. It won't readily burn or combust, even in vapour form, and its high heat capacity and prevalence (see above) means that it can effectively douse most runaway reactions that might occur, such as fires (there are times when adding water to a fire is a bad idea, especially perhaps in a lab, but even those times the problem can be solved by adding &lt;em&gt;a lot&lt;/em&gt; of water).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Toxicity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Unlike some other common solvents, water is not toxic&lt;sup&gt;[citation needed]&lt;/sup&gt;, and is not a bio- or eco-hazard. This means that unused solvent can be disposed of without special precautions, and if the solute can be disposed of trivially, the solvent can as well.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Purity and stability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water is everywhere&lt;sup&gt;[citation needed]&lt;/sup&gt;, giving it another unique property: Unlike many other solvents, water won't absorb atmospheric water vapour, decreasing its purity as a matter of course. (It'll still exchange water with the atmospheric vapour, of course, but it doesn't significantly change its make-up.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distilled 100 % water will remain more or less pure unless something is actively added to it, which, while not unique, is useful and somewhat uncommon. Water molecules will also not spontaneously decompose under most sane conditions.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Operating temperatures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water is liquid at a wide range of temperatures, stretching from 0°C to 100°C. Under laboratory conditions of ~1 atm and 20-25°C, water is a well-behaved and practical liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, while the liquid phase is available at a wide range of temperatures, the solid and gaseous states are not out of reach when using even extremely basic equipment. Cooling water to well below freezing, say -18°C, or to boiling, are both trivial activities, using equipment that is not only available in every lab, but even most homes.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;In short...&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Water isn't always the best tool for the job, but it's almost always a good or great one, at least when working with polar compounds. A common alternative to water is ethyl alcohol, which has similar polar properties and operating temperatures, but ethanol does have weaknesses in the other categories mentioned here, i.e. its reactivity, its absorption of water vapour, and its price.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="5638" LastEditDate="2014-12-20T09:23:51.277" LastActivityDate="2014-12-20T09:23:51.277" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="21910" PostTypeId="1" CreationDate="2014-12-19T10:56:53.060" Score="2" ViewCount="543" Body="&lt;p&gt;I need to search organic semiconductors for organic photovoltaics (OPV) by their HOMO, LUMO, work function at 300K (as I understand, WF cannot be derived from HOMO and LUMO because semiconductor can be differently doped and Fermi level depends on concentration of dopants) and deposition methods (such as CVD, PVD, spin-coating, electrodeposition, etc)..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some free site where I can either download a machine-readable table of needed data or make a search based on that criteria? What is it?&lt;/p&gt;&#xA;" OwnerUserId="11374" LastEditorUserId="11374" LastEditDate="2014-12-19T14:42:37.157" LastActivityDate="2014-12-19T14:42:37.157" Title="Search materials by HOMO, LUMO and work function" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;molecular-orbital-theory&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21911" PostTypeId="1" AcceptedAnswerId="21924" CreationDate="2014-12-19T11:01:52.203" Score="5" ViewCount="3495" Body="&lt;p&gt;A useful post by @Martin indicated that probably the naming of Sweetwater town is because of the sweet tasting lead compounds in it's water.&lt;br/&gt;&#xA;Then my question arose. I know that the taste of any material is dependent on which tongue buds it provokes. Now, my question is, which chemical property of an element defines or is directly related or is causing the chemical property of taste? (I deliberately avoid asking this for all chemical species to prevent very broad answers, if you think you can give not-too-long answers, my question is about all chemical species.)&lt;br/&gt;&#xA;Note that by a responsible chemical property I mean things related specifically to an atom, like e.config.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2014-12-19T14:50:26.683" Title="What defines an element's taste?" Tags="&lt;atoms&gt;&lt;taste&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21912" PostTypeId="2" ParentId="21898" CreationDate="2014-12-19T12:36:45.200" Score="16" Body="&lt;p&gt;Just to provide an alternative answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a &lt;a href=&quot;https://www.erowid.org/archive/rhodium/pdf/solvent.miscibility.pdf&quot;&gt;solvent miscibility table&lt;/a&gt; like the one linked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least miscibile solvent? Water! Water is the worst solvent. Water is immiscible with 17 out of 30 of the other listed solvents.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 solvents in the table which are miscible with all the other solvents: ethanol, acetone, tetrahydrofuran, n- &amp;amp; iso-propanol and dioxane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is water such a bad solvent? Hydrogen bonds.  Water molecules are strongly hydrogen bonded to other water molucules.  To dissolve something else, these hydrogen bonds need to be broken.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another prespective, look across the periodic table.  What does water stably dissolve? Does water dissolve gold, silver, copper, tin, lead, aluminum, and many other metals and metaloids like antimony, silicon, and germanium, &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/10408347408542677?journalCode=batc19#preview&quot;&gt;as mercury does&lt;/a&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing universal about water, except that there is a lot of it on Earth and we need it to live. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-19T19:12:08.887" LastActivityDate="2014-12-19T19:12:08.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21913" PostTypeId="2" ParentId="21911" CreationDate="2014-12-19T12:43:20.250" Score="-1" Body="&lt;p&gt;I don't know how many variables determine the taste but we recently found out that materials can have completely different properties just by the amount. We know that with different structures of atoms materials get completely different like diamond and graphit but, we know at least since the discovery of graphene that the periodic table might be much more complicated than we had imagined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every element gets completely different properties with just the quantity of atoms. If you take for example a really small amount of an element with a spoon it glows yellow, then you take with an other spoon a bigger amount and it glow red, then you take a bigger amount it glows blue. Currently it doesn't make any sense why this happens since you don't change the atomic structure of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future the periodic table will get a 3rd dimension with every element becoming different with just the amount. This is not about the structure of the atoms or something, just the amount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the colors of matrials we have the wave lenght of the light as a reference but for the taste there are to much variables since we taste with our nose and our tongue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://periodictable.com/Posters/PlaceMatBackSide3000.JPG&quot; rel=&quot;nofollow&quot;&gt;http://periodictable.com/Posters/PlaceMatBackSide3000.JPG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://periodictable.com/Posters/LetterSizeBackSide1500.JPG&quot; rel=&quot;nofollow&quot;&gt;http://periodictable.com/Posters/LetterSizeBackSide1500.JPG&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11362" LastActivityDate="2014-12-19T12:43:20.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21914" PostTypeId="2" ParentId="21910" CreationDate="2014-12-19T13:47:25.273" Score="3" Body="&lt;p&gt;Indeed, &lt;a href=&quot;http://aspuru.chem.harvard.edu&quot; rel=&quot;nofollow&quot;&gt;Prof. Alán Aspuru-Guzik&lt;/a&gt; has placed his &lt;a href=&quot;https://cepdb.molecularspace.org&quot; rel=&quot;nofollow&quot;&gt;&quot;clean energy project&quot;&lt;/a&gt; database on the web in a searchable form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, it covers ~2.3 million compounds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Use of the data requires citation of the appropriate literature sources. The data in this database is released under the &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0/&quot; rel=&quot;nofollow&quot;&gt;Creative Commons Attribution ShareAlike&lt;/a&gt; license and therefore any academic publications associated with the CEPDB should be appropriately &lt;a href=&quot;http://www.molecularspace.org/about-us/how-to-cite&quot; rel=&quot;nofollow&quot;&gt;cited&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you point out, the database only covers HOMO and LUMO orbital eigenvalues. While this may not be the most useful for experimental applications (i.e., you really want ionization potentials and first excited state energies or electron affinities) it should be a helpful &quot;first screen&quot; to find interesting lead compounds.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-19T13:47:25.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21915" PostTypeId="1" AcceptedAnswerId="21918" CreationDate="2014-12-19T13:47:32.933" Score="33" ViewCount="13542" Body="&lt;p&gt;There are two points of view for the answer of this question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The biological view, the only one that I faced during my research, states that since it can trigger perilous conditions like metabolic acidosis, cause a neurologic sequelae, etc. methanol is toxic.&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;The chemical point of view in an answer, which is the one I seek, is the reactions that occur in the body with methanol as a reactant.&lt;br/&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why should it be a reaction? Because I assume something will be toxic for humans when it reacts and disables some of their bodies' vital compounds; e.g.: Hydrogen cyanide reacts with enzymes, renders their active sites useless, and finally causes heavy reduction in bio-chemical reactions that results in quick death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: What reactions use methanol in the body that makes it toxic for us?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-04-25T13:36:16.953" LastActivityDate="2018-03-14T02:12:13.107" Title="Why is methanol toxic?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;toxicity&gt;&lt;chemical-biology&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21916" PostTypeId="2" ParentId="21898" CreationDate="2014-12-19T14:04:57.213" Score="10" Body="&lt;p&gt;It's worth noting some of the history behind the term &quot;universal solvent,&quot; and why it is used even though water isn't truly &lt;em&gt;universal&lt;/em&gt; or even necessarily the most versatile solvent available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before chemistry existed, and for that matter before science existed as the practice we would recognize today, alchemy was an important protoscience. A number of important breakthroughs and refinements came about due to the work of alchemists, and for all that it often got buried in mysticism (and, at times, fraud), alchemy began the research into the chemical properties of matter. Chemistry developed largely in reaction to alchemy, rejecting the mystical underpinnings, and applying the scientific method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within alchemy, there was a belief in the existence of a &quot;universal solvent,&quot; which would be part of the process for producing the philosopher's stone and/or transmuting lead into gold. Numerous alchemists claimed to know how to make it, while many others considered producing it to be one of their major, overarching goals, the most difficult or secret step in obtaining whatever powers they believed they could achieve through alchemy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, no chemical with the properties described for this universal solvent has ever been found. However, because of the critical use of water as the default solvent in chemical work, it has often been referred to with the title. Other answers describe better than I can why water is a particularly useful solvent.&lt;/p&gt;&#xA;" OwnerUserId="11376" LastActivityDate="2014-12-19T14:04:57.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21917" PostTypeId="2" ParentId="21848" CreationDate="2014-12-19T14:10:52.840" Score="4" Body="&lt;p&gt;If the ratio of neutrons to protons in the nucleus is too high, the nucleus is unstable to beta decay.  A neutron emits an electon and becomes a proton.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ratio of neutrons to proton is too low, the nucleus is unstable to positron decay and/or electron capture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how high/low is too high/low, and why does this occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Semi-empirical_mass_formula&quot; rel=&quot;nofollow&quot;&gt;Semi-empirical mass formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = vA -sA^{2/3} - c\frac{Z^2}{A^{1/3}} - a\frac{N-Z}{A} - \delta (A,Z) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E$ is the binding energy of the nucleus, $N$ is number of neutrons, $Z$ is number of protons, $A$ is $N + Z$, and $v$, $c$, $s$ and $a$ are empirical constants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only the $c$ (coulomb) and $a$ (asymmetry) terms involve the relative number of protons and neutrons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The asymmetry term is zero when the number of protons and neutrons are equal, and relates to the Pauli exclusion princple (neutrons and protons being fermions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except for the coulomb term (that protons electrostatically repel each other), the ideal ratio of protons to neutrons would be 1 to 1.  Electrostatic repulsion of protons explains why somewhat more neutrons than protons is favorable.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-19T14:40:16.727" LastActivityDate="2014-12-19T14:40:16.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21918" PostTypeId="2" ParentId="21915" CreationDate="2014-12-19T14:12:57.417" Score="32" Body="&lt;p&gt;Methanol isn't particularly toxic in and of itself, although it's no walk in the park.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real culprit is one of its metabolic products, methanoic acid, also known as formic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand how formic acid, present as the formate ion, is toxic, we look to &lt;a href=&quot;http://en.wikipedia.org/wiki/Methanol&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reedit:&lt;/strong&gt; The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="5638" LastEditDate="2014-12-19T22:36:08.263" LastActivityDate="2014-12-19T22:36:08.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21919" PostTypeId="2" ParentId="21915" CreationDate="2014-12-19T14:16:45.537" Score="18" Body="&lt;p&gt;The enzyme &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcohol_dehydrogenase&quot;&gt;alcohol dehydroganase&lt;/a&gt; converts the methanol to formaldehyde in the body. Formaldehyde is then converted to formic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formaldehyde can cause blindness before being converted to formic acid, while formic acid causes acidosis as Williham Totland points out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://thetruthaboutstuff.com/pdf/(116)%20Cooper%201962%20Biochemical%20Aspects%20of%20Methano%20Poisoning.pdf&quot;&gt;Biochemical Aspects of Methanol Poisoning&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-20T12:48:12.843" LastActivityDate="2014-12-20T12:48:12.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21921" PostTypeId="1" CreationDate="2014-12-19T14:28:30.570" Score="5" ViewCount="1120" Body="&lt;p&gt;Ring strain causes the epoxides to react and THF does not react because the ring strain is not large enough. Does the four membered ring react or not? Does it favor reaction with the harder or softer nucleophile perhaps - thus only reacting with some Grignard reagents?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-11-29T15:45:01.800" Title="Grignard reagents react with epoxides but dissolve in THF, what happens in oxetane (four membered ether ring)?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21922" PostTypeId="1" CreationDate="2014-12-19T14:29:12.207" Score="4" ViewCount="155" Body="&lt;p&gt;I'm currently working towards a major in both chemistry and mathematical statistics as a part of my science degree, and I got to wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will these two go hand in hand and help each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know there's the obvious applications of statistical inference to any experimentation I do, but I'm talking about less obvious relationships. Could my knowledge of solving SDEs ever help me with my experiments? What about continuous Markov processes? Will I ever even think about a discrete random variable again if I go on to get a PhD in chemistry? Also not looking for the standard &quot;well we use statistical physics to model electrons&quot; example, either (welp, unless you can offer some nice specifics). Surely there's other ways that these skills I've found can interact with each other?&lt;/p&gt;&#xA;" OwnerUserId="11377" LastEditorUserId="865" LastEditDate="2015-11-09T13:54:02.430" LastActivityDate="2015-11-09T13:54:02.430" Title="How will my studies in mathematical statistics be applied to and assist me with problems in chemistry?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21923" PostTypeId="2" ParentId="21905" CreationDate="2014-12-19T14:33:55.227" Score="2" Body="&lt;p&gt;The lowest energy MO is : $\psi_1 = (1/2)(\phi_A+ \sqrt2\phi_B+\phi_C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then comes the MO: $\psi_2 = (1/\sqrt2)(\phi_A- \phi_C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest energy MO is  $\psi_3 = (1/2)(-\phi_A+ \sqrt2\phi_B-\phi_C)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where  $\phi$ denotes the atomic orbital $1s$ on each hydrogen atom. $A$, $B$ and  $C$ denote hydrogen atoms, where  $B$ is the central one.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-19T14:33:55.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21924" PostTypeId="2" ParentId="21911" CreationDate="2014-12-19T14:43:20.703" Score="5" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the taste buds, however, the receptors look for certain specific &lt;em&gt;characteristics&lt;/em&gt; of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you go around tasting elements, you'll find two very interesting things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you'll find that you're dying from all kinds of poisonings. At a rough count, &lt;em&gt;all&lt;/em&gt; of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the alkali metals like lithium and its friends will taste bitter, also kill you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;But wait!&quot;&lt;/strong&gt;, I hear you scream, &quot;What of, say, lithium in pharmaceuticals?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastActivityDate="2014-12-19T14:43:20.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21925" PostTypeId="1" AcceptedAnswerId="21934" CreationDate="2014-12-19T15:46:20.580" Score="2" ViewCount="3526" Body="&lt;p&gt;I was doing some tests for the multiple-choice final we've got ahead. And it was on me to count the peptide bonds in an Insulin hormone with 51 aminoacids arranged in two polypeptides with 30 and 21 aminoacids. (these are not true in reality) &lt;br/&gt;the number of bonds were 49, not 50, and that means the bond between two polypeptides doesn't count as a peptide bond. Additionally, I know that polypeptide bonds make the proteins' molecular structure, as it is now. (just look at that shape.) PEPTIDE COVALENT BONDS CAN NEVER cause that kind of 3d orientation in space. So there must be some fundamental difference between those bonds. What is it?&lt;br/&gt;&#xA; My research couldn't find any results as simple things have jammed the internet.&lt;img src=&quot;https://i.stack.imgur.com/9lGbI.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="51553" LastEditDate="2018-08-02T08:12:43.453" LastActivityDate="2018-08-02T08:12:43.453" Title="The difference between peptide bonds and the bonds between polypeptides?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21926" PostTypeId="1" AcceptedAnswerId="21928" CreationDate="2014-12-19T16:04:50.603" Score="10" ViewCount="18332" Body="&lt;p&gt;The $\mathrm{pK_{aH}}$ of $\ce I^-$ is very low which indicated that it is not favourable for it to bond with a proton. However, why would it be likely to bond with any other atom (mostly carbon when considering it's nucleophilic properties).&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user5764" LastEditDate="2016-06-11T16:13:42.377" LastActivityDate="2016-06-27T23:03:17.360" Title="Why is the iodide anion a good nucleophile but a poor base?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21927" PostTypeId="2" ParentId="21925" CreationDate="2014-12-19T16:30:40.020" Score="4" Body="&lt;p&gt;The term peptide bond refers to amide bonds. The 21 amino acids in insulin's A-chain are covalently linked by 20 amide bonds. The 30 amino acids in insulin's B-chain are linked by 29 amide bonds. The two polypeptides are covalently linked by disulfide bonds, which are not amide bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3D structure of proteins is a complex topic and a field of intense research. It's worth pointing out though that even polypeptides (with no disulfides or other covalent linkages) have a more complex 3D structure than you might expect. I suggest you look into &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_helix&quot; rel=&quot;nofollow&quot;&gt;alpha-helices&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_sheet&quot; rel=&quot;nofollow&quot;&gt;beta-sheets&lt;/a&gt; as an entry into the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Took some info from the &lt;a href=&quot;http://www.betacell.org/content/articleview/article_id/8/&quot; rel=&quot;nofollow&quot;&gt;Beta Cell Consortium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2014-12-19T16:30:40.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21928" PostTypeId="2" ParentId="21926" CreationDate="2014-12-19T16:44:48.897" Score="13" Body="&lt;p&gt;You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its proton due to the weak hydrogen-iodide bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the high nucleophilicity of iodide ion &lt;em&gt;in water&lt;/em&gt; reflects the high rate of reaction between iodide ion and some electrophilic atom (other than hydrogen). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative lack of basicity of the iodide anion in water is precisely what makes it a good nucleophile. Its lack of affinity toward protons - i.e. the protons of water - physically frees $\ce{I^-}$ to attack electrophilic atoms (not limited to carbon). There is little hydrogen bonding between iodide ion and water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the relative basicity and thus strong hydrogen-bond forming capabilities of $\ce{F^-}$ makes it a poorer nucleophile in water. Rates of reaction between fluoride anion and some electrophilic atom are slowed because the fluoride anion is trapped (strongly solvated) by water molecules. And there will also be a sizable number of fluoride anions that have completely abstracted a proton from water to make $\ce{HF}$, which isn't nucleophilic at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a polar aprotic solvent, however, where there are no positively polarized hydrogens for a nucleophile to hook onto; $\ce{F^-}$ becomes a much stronger nucleophile. Actually,  $\ce{F^-}$ becomes a better nucleophile than $\ce{I^-}$, likely due to $\ce{F^-}$'s greater charge density. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2016-06-27T23:03:17.360" LastActivityDate="2016-06-27T23:03:17.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21929" PostTypeId="1" AcceptedAnswerId="21932" CreationDate="2014-12-19T17:07:05.440" Score="2" ViewCount="3857" Body="&lt;p&gt;What about oxonium ions, such as hydronium ion? Do these hydrogen bond with water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we see hydrogen bonding as a purely electrostatic phenomenon, then why not? Ammonium ion still has nitrogen bearing a partial negative charge. Of course it can interact with positively polarized hydrogens in an electrostatic way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we also know that hydrogen bonding is partially covalent. So, it seems unlikely that the ammonium ion would be able to form any additional covalent bonds. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="9961" LastEditDate="2015-10-02T16:55:54.227" LastActivityDate="2015-10-02T16:55:54.227" Title="Does the ammonium ion hydrogen bond with water?" Tags="&lt;inorganic-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21930" PostTypeId="2" ParentId="21922" CreationDate="2014-12-19T17:33:55.230" Score="7" Body="&lt;p&gt;&lt;strong&gt;Most definitely yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took multiple statistics classes as an undergraduate and they have helped me considerably. (I can argue fairly that it helped me get tenure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few suggestions from my own work:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Statistical design of multivariate experiments&lt;/strong&gt;: Many scientists think you should test one variable at a time. No. I taught a grad student (in another group) about optimization theory and proper design and three months later, a great paper came out.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Data mining and exploration&lt;/strong&gt;: Many chemistry data sets are extremely complicated, with cross-correlated factors. Understanding how to draw empirical inferences, find the most important factors, etc. is &lt;em&gt;enormously&lt;/em&gt; helpful across a wide range of chemistry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Principal component analysis&lt;/strong&gt;: An emerging area in spectroscopy and imaging is taking complicated spectra and performing PCA to help with quantifying chemical components in mixtures and following everything from cancer growth to materials design. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Multivariate regression and ANOVA&lt;/strong&gt;: My group has created numerous quick statistical screens using multivariate surrogates for complex simulations or experiments.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cross-validated analysis&lt;/strong&gt;: Again, the understanding of training and test sets, bootstrapping, etc. is very useful.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, you also mention SDEs and Markov processes. These are crucial in chemical kinetics, including areas like reaction-diffusion systems. Monte Carlo methods are also highly used in studying dynamical processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list would be enormous. Basically, I think it's a great skill to have and you'll use it throughout your career.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2014-12-19T17:33:55.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21931" PostTypeId="1" AcceptedAnswerId="21933" CreationDate="2014-12-19T17:35:04.563" Score="1" ViewCount="10724" Body="&lt;p&gt;Clearly the amide will be unreactive given the poor quality of the leaving group and poor electrophilcity due to delocalization. However, does it react with organolithiums for example?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2018-04-19T18:54:09.890" Title="Do Grignard reagents react with amides?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21932" PostTypeId="2" ParentId="21929" CreationDate="2014-12-19T17:43:50.460" Score="4" Body="&lt;p&gt;There is a detailed consideration of this issue in  &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/002228609408381Q#&quot; rel=&quot;nofollow&quot;&gt;Hydrogen Bonding of the Ammonium Ion&lt;/a&gt; Journal of Molecular Structure 345 (1995) 77-81.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the try to be clear about what they mean by &quot;hydrogen bond&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the purposes of the discussion, a hydrogen&#xA;  bond is taken to be a specific short-range interaction&#xA;  in which the N-H bonds within the&#xA;  ammonium ion are significantly affected by the&#xA;  surrounding ions, and not simply electrostatic&#xA;  attraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In gas phase calculations, they find that one water molecule hydrogen bonds to one ammonium ion (in other words, the dimer is hydrogen bonded), based upon the lengthening of the N-H covalent bond, but as you add 2, 3, and 4 water molecules, the N-H bond length returns to the free ammonium N-H bond length and the O --- H length increases.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-19T17:43:50.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21933" PostTypeId="2" ParentId="21931" CreationDate="2014-12-19T18:11:52.190" Score="3" Body="&lt;p&gt;Amides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo01286a075&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;J. Org. Chem.&lt;/em&gt;, &lt;strong&gt;1967&lt;/strong&gt;, &lt;em&gt;32&lt;/em&gt; (11)&lt;/a&gt;). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see &lt;a href=&quot;http://www.researchgate.net/profile/Sangeeta_Jha/publication/260273495_Dehydration_of_Amides_to_Nitriles_A_Review/links/0c960530724add6933000000&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; if you're interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case of a N,O-dimethylhydroxyamide, you can &lt;a href=&quot;http://en.wikipedia.org/wiki/Weinreb_ketone_synthesis&quot; rel=&quot;nofollow&quot;&gt;make ketones using either.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-19T18:11:52.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21934" PostTypeId="2" ParentId="21925" CreationDate="2014-12-19T18:17:42.243" Score="2" Body="&lt;p&gt;Aside from covalent bonds (amide and disulfide), the sturcture of a protein is determined by hydrogen bonds, &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_bridge_(protein_and_supramolecular)&quot; rel=&quot;nofollow&quot;&gt;salt bridges&lt;/a&gt;, and less specific interactions such as hydrophobic and hydrophilic effects.  Hydrophobic portions of the protein chain tend toward the interior of the folded protein and hydrophilic regions to the exterior, in aqueous solution.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-19T18:17:42.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21936" PostTypeId="1" AcceptedAnswerId="21941" CreationDate="2014-12-19T19:06:31.257" Score="1" ViewCount="6120" Body="&lt;p&gt;Few organic molecules dissolve in water but why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2016-11-07T18:06:11.740" Title="Why is water a poor solvent for organic molecules?" Tags="&lt;organic-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21937" PostTypeId="1" AcceptedAnswerId="21939" CreationDate="2014-12-19T19:07:02.377" Score="1" ViewCount="151" Body="&lt;p&gt;I just accidentally closed a webpage (and did not find it later) which described that somehow, Aspartame poisoning victims had faults in the vital proteins of their body due to the reaction of Aspartame with some aminoacids. I doubted that this is true.&lt;br/&gt; However, considering the high solubility of Aspartame in the liquids in the body and its digestibility it is feasible for it or one of the related compounds to be causing this symptom. The info that will be shared is welcome.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2014-12-19T19:41:24.177" Title="Aspartame poisoning: Can it modify aminoacids?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;toxicity&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21938" PostTypeId="2" ParentId="21936" CreationDate="2014-12-19T19:30:30.560" Score="8" Body="&lt;p&gt;Water is not a bad solvent for all organic molecules. In fact many organic molecules &lt;em&gt;are&lt;/em&gt; soluble in water. In general the solubility of a substance in a given solvent is determined by the strength of the intermolecular forces between the solvent and solute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The predominant intermolecular force in water is hydrogen bonding and so molecules with groups which can hydrogen bond  generally dissolve well in water. These groups can have acidic hydrogens of their own such as alcohol or amine groups which can h-bond to water both through their hydrogens and through the lone pairs on their oxygen and nitrogen atoms respectively. Also other groups which have lone pairs on electronegative atoms such as carbonyl groups can h-bond with water and so will generally dissolve, although not as well as alcohols or amines. All of these groups are generally termed to be hydrophilic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also hydrophobic groups which tend not to bond well with water and so do not tend to dissolve. These are generally groups which are non-polar and interact primarily through Van der Waals forces. Therefore they do not form strong interactions with water (because water primarily interacts through h-bonding) and so do not dissolve. These groups tend to be apolar, the most common example being alkyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in answer to your question, many large organic molecules with predominant hydrophobic groups do not dissolve in water whereas smaller molecules with hydrophilic groups do tend to dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This questions has a more detailed discussion of the properties of water as a solvent: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/21898/why-is-water-the-universal-solvent&quot;&gt;Why is water &amp;quot;the universal&amp;quot; solvent?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-12-19T19:30:30.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21939" PostTypeId="2" ParentId="21937" CreationDate="2014-12-19T19:41:24.177" Score="8" Body="&lt;p&gt;Aspartame is one of the most well studied food additives there is. It seems to be controversial in the same way that evolution is—scientific research is overwhelmingly on the side of the substance being safe to consume, even at very high levels, but a Google search on it brings up things like &quot;&lt;a href=&quot;http://aspartame.mercola.com&quot;&gt;Aspartame is, by Far, the Most Dangerous Substance on the Market that is Added To Foods&lt;/a&gt;&quot; (from a site trying to sell natural health products). A lot of this comes perhaps from the naturalistic fallacy: aspartame is artificial and thus unsafe. A huge number of regulatory authorities have deemed it safe to consume: FDA, Health Canada, EFSA, FSA, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a good deal of bad biochemistry thrown about: aspartame is metabolized into the two amino acids aspartic acid and phenylalanine, as well as methanol. There have been some hypotheses that aspartic acid intake could induce some amount of toxicity, but experimental evidence is lacking. Methanol, of course, is well known to be toxic. The key is, that aspartame is very sweet and only tiny amounts are added to food. So even drinking enough diet cola to hit the FDA's acceptable daily intake of aspartame (which works out to ~7.5 L), the quantity of methanol consumed is only around 300 mg, and things like fruits are larger sources of methanol in people who drink sane amounts of diet soft drinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one condition in which aspartame is known to do harm at reasonable intake is in people who have phenylketonuria (PKU). This might be what you're thinking of by &quot;faults in the vital proteins of their body&quot;. PKU is a genetic disease caused by a defective gene for the enzyme phenylalanine hydroxylase which results in an inability to metabolize phenylalanine correctly; thus, consuming phenylalanine from any source is hazardous to sufferers of the disease.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-19T19:41:24.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21940" PostTypeId="1" AcceptedAnswerId="21945" CreationDate="2014-12-19T19:52:33.887" Score="1" ViewCount="233" Body="&lt;p&gt;Specific heat plays an important role in calculating energy and thus, heat, enthalpy and etc. &lt;br/&gt;&#xA;But as I sighted the definition for calorie, I was surprised.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A calorie is the amount of heat needed to alter the temperature of 1 gram of 14.5 °C water to 15.5 °C water in the pressure of one atmosphere.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though the unit was redefined later, the question remained still. Why were there those temperatures? It was just then that I realized specific heat capacity varies with temperature differences and it is due to vibration movements of matter and internal energy ....&lt;br/&gt;&#xA;Two questions: Can there be a unit which is constant with temperature alterations that can be used instead of S.heat capacity? If so, why isn't it replacing SHC?&lt;br/&gt;&#xA;&lt;a href=&quot;http://www.eoearth.org/view/article/150863/&quot; rel=&quot;nofollow&quot; title=&quot;Link for the quote&quot;&gt;The resource for the quote&lt;/a&gt;&lt;br/&gt; &lt;a href=&quot;http://www.quora.com/Why-does-specific-heat-increase-with-temperature&quot; rel=&quot;nofollow&quot;&gt;The resource for the reason of the increase that happens to SHC with temperature rise&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2014-12-20T03:41:38.993" Title="Can we define a new unit instead of specific heat that is constant with temperature rise?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21941" PostTypeId="2" ParentId="21936" CreationDate="2014-12-19T19:53:29.397" Score="9" Body="&lt;p&gt;Pure water can form two hydrogen bonds per molecule.  The enthalpy of these hydrogen bonds is about 23 kJ/mol of water.  You have to give water a very good reason to rip apart its hydrogen bond network, and the interactions water experiences with the organic compounds often aren't enough, even considering increased entropy, to make this favorable. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-19T19:53:29.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21942" PostTypeId="2" ParentId="21915" CreationDate="2014-12-19T21:10:58.047" Score="10" Body="&lt;p&gt;I just wanted to point out that everything is toxic in the right proportion. THe problem is with the dose. I think you could drink a bit of methanol without getting particulary harmed. Do not do it, obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LD50 is very low at 5g/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;strong&gt;Nothing to do with methanol&lt;/strong&gt;, but it is a good infographic &#xA;&lt;img src=&quot;https://i.stack.imgur.com/ru5TQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.compoundchem.com/2014/07/27/lethaldoses/&quot; rel=&quot;noreferrer&quot;&gt;http://www.compoundchem.com/2014/07/27/lethaldoses/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison with ethanol, &lt;strong&gt;I believe&lt;/strong&gt; (not sure) that the oxidation product of these two compounds, ie, acetic acid and formic acid are not equally reactive. The former (metabolite of ethanol) is less reactive than the latter (metabolite of methanol). At least from an organic point of view this makes sense. But maybe there is another more biological reason.&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="73" LastEditDate="2015-01-23T00:10:57.253" LastActivityDate="2015-01-23T00:10:57.253" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21943" PostTypeId="1" CreationDate="2014-12-19T23:02:33.100" Score="0" ViewCount="1057" Body="&lt;p&gt;Any ideas on how to get $\ce{C3H8O3}$ (Glycerol) from $\ce{CH4}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory is that by reacting $\ce{CH4}$ with $\ce{Cl2}$ (or $\ce{Br2}$), getting $\ce{CH3Cl}$, then Wurtz's method to $\ce{CH3CH3}$, adding $\ce{Cl2}$ (or $\ce{Br2}$) again, getting $\ce{CH3CH2Cl}$, then reacting with $\ce{NaOH}$ to get $\ce{CH3CH2OH}$, and finally $\ce{CH3CH2OH + CO2 + H2O}$ and getting glycerol ($\ce{C3H8O3}$), but my way seems rather off to me so I need some help. &lt;/p&gt;&#xA;" OwnerUserId="11383" LastEditorUserId="7448" LastEditDate="2017-03-27T17:41:23.370" LastActivityDate="2017-03-27T17:41:23.370" Title="Synthesis of glycerol from methane?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21944" PostTypeId="2" ParentId="21943" CreationDate="2014-12-20T00:10:35.157" Score="2" Body="&lt;p&gt;Your synthesis is unlikely to become a huge commercial success ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methylchloride is a technical product, but usually made by reaction of methanol with hydrogen chloride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3OH + HCl -&amp;gt; CH3Cl + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works nicely without side products, because&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH3Cl}$ has a low solubility in water (around $\mathrm{5\,g\cdot L^{-1}}$)&lt;/li&gt;&#xA;&lt;li&gt;under standard conditions, $\ce{CH3Cl}$ is a gas (bp around $\mathrm{-28\,°C}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The radical halogenation of methane will usually lead to various halogenation products and thus is not a good idea. The same is true for the similar reaction of ethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I fail to see how the reaction of ethanol and carbon dioxide will lead to glycerol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should further take into account that there's plenty of glycerol available as a side product in the production of &lt;em&gt;biodiesel&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a commercial and ecological point of view, methane therefore is better converted to cabon monoxide and hydrogen by steam reformation. Further hydrogen may be obtained from a subsequent shift reaction and the resulting carbon dioxide may be reacted with gycerol to yield 1,2-glycerol carbonate.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-12-20T00:16:39.030" LastActivityDate="2014-12-20T00:16:39.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21945" PostTypeId="2" ParentId="21940" CreationDate="2014-12-20T03:41:38.993" Score="5" Body="&lt;p&gt;There isn't a simple redefinition that would work in all cases, and this is probably why nobody does this.  Of course, you COULD define such a unit for a given system by defining it as a function of temperature, but it would be applicable only to the specific system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, the physics behind the temperature dependence of a material's heat capacity lies in the quantum states which become available for excitation at different energies.  Consider an insulator which stores heat primarily in the lattice vibrations.  As you increase the temperature, more vibrational modes are accessible.  This is often described using the Debye model which gives the heat capacity as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_V = k(\frac{T}{\theta})^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_V$ is the heat capacity, $k$ is a constant, $T$ is the temperature in Kelvin, and $\theta$ is a constant pertinent to a specific solid.[1]  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a metal can also store a significant amount of heat in the motions of electrons since conduction electrons are not bound to specific sites.  Heat capacity for the electron 'gas' is often described like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_V' = k' \frac{T}{\theta'}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_V' \neq C_V$, $k' \neq k$ and $\theta' \neq \theta$.[2]  Depending on $\theta$ and $\theta'$ for a given material, at low temperatures (often cryogenic) the electron contribution can be the dominant mode of storing heat in the solid.  Thus, if you chose a unit which was sensible for an insulator, it wouldn't apply to a conductor (and vice versa).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even within a specific material, the heat capacity on either side of a phase transition is often different since the phase transition has changed the quantum states available in the material.  Therefore, your new unit would become discontinuous at phase transitions and would need a different mathematical expression for each segment of the heat capacity function, not to mention a different parameterization for each material.  Hence, no real gain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 106, eq. 47b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 193, eq. 37.  Note that folks don't actually use the symbol $\theta'$ in this case, but choosing so makes the explanation above clearer.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2014-12-20T03:41:38.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21947" PostTypeId="1" CreationDate="2014-12-20T10:07:25.373" Score="2" ViewCount="297" Body="&lt;p&gt;We're currently studying &quot;the chemistry of contact lenses&quot;, which is mostly polymers and gels. I just looked at the package of two pure hydrogel lenses made out of the same monomers, but they have different water content? Why?&lt;/p&gt;&#xA;" OwnerUserId="11390" LastEditorUserId="9961" LastEditDate="2015-10-26T22:24:29.637" LastActivityDate="2016-10-16T14:35:13.933" Title="Same monomers, different water content?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;polymers&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21948" PostTypeId="1" CreationDate="2014-12-20T11:16:18.227" Score="2" ViewCount="3169" Body="&lt;p&gt;A full bottle of jewelry cleaning (links product and material safety data sheet below) broke in my car and went unnoticed for a few days.  Now the car has a sulfur-like smell whenever I go into the car.  It dissipates within a few minutes, but then returns very quickly once I leave the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary chemical components are alcohol ethoxylates, and can decompose into carbon monoxide, carbon dioxide nitrogen Oxides, sulfur oxides and other toxic gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have cleaned the area and had the carpet (rear driver) shampooed from a car detailer, but still the smell remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical compound can I use to neutralize the smell permanently (enter equilibrium)?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0044B5WN6&quot; rel=&quot;nofollow&quot;&gt;Product Link (Amazon)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://connoisseurs.com/jewelry_care/liquid_dip_cleaners_silver_bath.php&quot; rel=&quot;nofollow&quot;&gt;Product Link (Manufacturer)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://connoisseurs.com/msds/MSDS-Silver-Jewelry-Cleaner.pdf&quot; rel=&quot;nofollow&quot;&gt;Material data safety sheet&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11391" LastEditorUserId="11391" LastEditDate="2014-12-20T12:06:16.810" LastActivityDate="2018-10-28T18:24:33.580" Title="Neutralizing sulfur-like smells in car" Tags="&lt;everyday-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21949" PostTypeId="2" ParentId="21948" CreationDate="2014-12-20T11:56:06.783" Score="2" Body="&lt;p&gt;I guess that it actually was a silver cleaner, not one for gold or precious stones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and &lt;strong&gt;thiourea&lt;/strong&gt;, which probably is responsible for the smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-12-20T13:11:32.550" LastActivityDate="2014-12-20T13:11:32.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21950" PostTypeId="2" ParentId="21947" CreationDate="2014-12-20T13:42:31.047" Score="2" Body="&lt;p&gt;When a polymer is produced, the final result does &lt;strong&gt;not&lt;/strong&gt; only rely on what the monomers were. For example, there is &quot;extended&quot; polystyrene and &quot;extruded&quot; polystyrene. Or there are polystyrene sulfonate chemical polymers and etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since nowadays, we need polymers to do many things for us (sheathing, sheltering, and sometimes even heat conductivity), we usually tend to modify their structure to make them more temperature-resistant, harder, stronger and ...; and that is because researches on new polymers are not always cheap or successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's explain with an example. Imagine you have $10000$ ethylene monomers that you will need to make a polymer from. Each of the PE polymers require $100$ ethylene molecules to form. Since you need this thing to be a little bit stronger, you would pressurize it, heat it, liquify it etc. So the result is a set of well-arranged polymer molecules [You can imagine lines of pupils standing in a school's yard, waiting to go to their classes] They are resistant to minor temperature rises and can be used in places where hard polymers are needed. (actually, not yet, but you are currently imagining :))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Imagine the raw &quot;plastic&quot; that is industrially produced with $10000$ monomers and the polymers are made of 100 ethylene molecules. This time it seems as if the pupils are dispersed through the school yard, though in tact with each other. You can easily conclude that this polymer is &lt;em&gt;potentially&lt;/em&gt; weaker and is more vulnerable to temperature differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first example is called &quot;extruded polyethylene&quot; and the 2nd one is called &quot;expanded polyethylene&quot;. Notice that the monomers are the same, but the result is different in as big of a range that includes usage to specific heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the only case. You remember that I mentioned PS sulfonate. Sulfur, in here, is added to polystyrene to make chain-like &quot;links&quot; to polymers. For example, imagine the lines of pupils again. These sulfonates act as some older students that hold the pupils still and do not let them escape. These chains make polystyrene sufonate ideal for being an agent for dying clothes (since it becomes resistant to dissolution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, Hydrogel monomers are the same, but as I described above, either &lt;strong&gt;the how of their production&lt;/strong&gt; or &lt;strong&gt;additional materials that modify the behavior of the Hydrogel&lt;/strong&gt; or both are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I didn't find any appropriate applicable images without license and I hope the description is not that vague. Hope I've helped.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="34388" LastEditDate="2016-10-16T14:35:13.933" LastActivityDate="2016-10-16T14:35:13.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21951" PostTypeId="1" AcceptedAnswerId="21952" CreationDate="2014-12-20T14:40:45.787" Score="20" ViewCount="9190" Body="&lt;p&gt;I know that there is the famous process of heating material and converting them to glass; but what I don't know is, what is the chemical process of the creation of glass? Is it crystallization? &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could it result in so much visual differences?&lt;br/&gt;&#xA;What is the description of such a radical process?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="23561" LastEditDate="2019-03-14T14:01:04.263" LastActivityDate="2019-03-14T22:51:06.313" Title="Why is sand, sand; and why is glass, glass; While both mainly contain Silicon Dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;geochemistry&gt;&lt;glass&gt;&lt;rocks&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21952" PostTypeId="2" ParentId="21951" CreationDate="2014-12-20T15:04:02.637" Score="26" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it crystallization?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are correct. The main difference is that sand is crystalline and glass is not—it is amorphous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main component (&gt; 95%) of common yellow sand is quartz (the mineral whose composition is SiO&lt;sub&gt;2&lt;/sub&gt;). Note that not all sand is quartz. There are white sands containing calcite (CaCO&lt;sub&gt;3&lt;/sub&gt;) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass, the type you see in your everyday life, on the other hand, is not composed of pure SiO&lt;sub&gt;2&lt;/sub&gt;. It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO&lt;sub&gt;2&lt;/sub&gt; slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO&lt;sub&gt;2&lt;/sub&gt;, this is a network of SiO&lt;sub&gt;4&lt;/sub&gt; tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO&lt;sub&gt;2&lt;/sub&gt; crystals in various structures (CT-opal for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma (&quot;glass&quot;) deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO&lt;sub&gt;2&lt;/sub&gt; into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slow crystallization of SiO&lt;sub&gt;2&lt;/sub&gt; in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO&lt;sub&gt;2&lt;/sub&gt;) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2AxJ3.jpg&quot; alt=&quot;Quartz&quot;&gt;&lt;br&gt;&#xA;&lt;sup&gt;(By Parent Géry (Own work) [Public domain], via Wikimedia Commons)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="8083" LastEditDate="2019-03-14T22:51:06.313" LastActivityDate="2019-03-14T22:51:06.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21953" PostTypeId="2" ParentId="21951" CreationDate="2014-12-20T15:14:27.370" Score="0" Body="&lt;p&gt;Take sand, melt it, let it cool and you have (sort of) glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass is an amorphous solid. It appears to be solid, but is, in reality, a viscous liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sand is essentially the same, although broken up into very small pieces. Although sand could contain quartz (crystallized) fragments, it doesn't change what we observe about the sand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could it result in so much visual differences?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Glass is clear since the sand it is made from has chemicals added to lower the melting point and to make it transparent (by removing some impurities). Sand is composed of everything in the earth, so it is not as transparent as glass. Also the small particles deflect light so it is difficult to see a pattern through sand.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-12-20T15:35:13.627" LastActivityDate="2014-12-20T15:35:13.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21954" PostTypeId="1" AcceptedAnswerId="21960" CreationDate="2014-12-20T16:09:58.427" Score="7" ViewCount="1188" Body="&lt;p&gt;My basement is quite dry (knock on wood!).  Recently I brought into the basement for the winter a table from outside.  The table had some ice stuck to it which I couldn't remove, so I simply let it melt on the floor of the basement.  After the puddle that formed evaporated, an interesting artifact was left behind along the the perimeter of the puddle.  It's a raised white line of fuzzy material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the puddle's perimeter&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VjaiC.jpg&quot; alt=&quot;The puddle and it&amp;#39;s perimeter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a close-up of the line that's left behind with a clothespin for scale&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DPJDE.jpg&quot; alt=&quot;A close-up of the line left behind, with clothespin for scale&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this efflorescence?  I did some googling, and this is what seems most likely, but I don't know much about chemistry and thought I would ask those who would actually know.  Is it something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11393" LastEditorUserId="7448" LastEditDate="2014-12-20T20:46:41.153" LastActivityDate="2014-12-22T07:03:31.743" Title="what is the white fuzz left behind on basement floor after puddle evaporated" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21956" PostTypeId="1" CreationDate="2014-12-20T16:42:34.557" Score="2" ViewCount="223" Body="&lt;p&gt;I want to understand how are catalysts discovered for a particular chemical reactions say for example $\ce{Fe,~Mo}$ for Haber's Process reaction of $\ce{N2}$ &amp;amp; $\ce{H2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions as examples with some background (story of discovery,discoverer or method) would better serve the purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this question keep it limited to Catalysts involved in Haber's Process.&lt;/p&gt;&#xA;" OwnerUserId="11378" LastEditorUserId="670" LastEditDate="2014-12-22T18:44:01.717" LastActivityDate="2015-10-04T07:39:25.153" Title="How are catalysts discovered for a particular reaction?" Tags="&lt;reaction-mechanism&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2019-12-22T23:13:03.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="21958" PostTypeId="1" CreationDate="2014-12-20T20:40:11.313" Score="2" ViewCount="2965" Body="&lt;p&gt;In the formation of 3-bromo-1-butene and 1-bromo-1-butene, from $\ce{HBr}$ and $\ce{C_4H_6}$, in a gas phase:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HBr + C_4H_6 \to 1BrC_4H_7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HBr + C_4H_6 \to 3BrC_4H_7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the number of degrees of freedom is given by: $F=C-P+2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to rationalise this but finding it quite confusing. For example, can we state that there is only 1 phase, i.e. a gas phase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the components, is the isomer of one independent of another? I'm taking there to be 3 individual components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is then 4 degrees of freedom, which seems okay, given $p$, $t$, $X$, $V$ could vary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate your help in this, thanks. &lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="189" LastEditDate="2014-12-21T12:28:16.173" LastActivityDate="2014-12-22T06:38:21.370" Title="Phase rule and degrees of freedom" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21959" PostTypeId="1" AcceptedAnswerId="21961" CreationDate="2014-12-20T20:44:19.730" Score="12" ViewCount="472" Body="&lt;p&gt;Every high school learner, in each corner of the world, faces the lesson &lt;em&gt;History of Atom&lt;/em&gt; during his courses, just as I did. We learned about s, p, d and f orbitals, though there were no signs of orbitals in molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I wondered, are there any other orbitals, simpler or more complex, than the four mentioned? Surprisingly, I learned that there are also usages for orbitals g, h, i and even k and l.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I use the word &quot;usages&quot;. Because I believe, unless something is useful, it will never enter the domain of science. Anyway, I read in Wikipedia (though not much of it I did understand) that these orbitals are used when describing and doing the measurements of molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since no element in the periodic table has enough electrons to fill even orbital g (in its base state), in cases of molecular orbitals that have a g defined in themselves, atoms must have been excited. Excitation needs energy, doesn't it? Where does this energy come from? Isn't the formation of new bonds usually exothermic?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="9961" LastEditDate="2018-10-03T23:05:33.490" LastActivityDate="2018-10-03T23:05:33.490" Title="What is the usage of orbitals more complex than f orbitals?" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;periodic-table&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21960" PostTypeId="2" ParentId="21954" CreationDate="2014-12-20T21:29:03.450" Score="4" Body="&lt;p&gt;Three pieces of information from the comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It fizzed and disappeared in vinegar.&lt;/li&gt;&#xA;&lt;li&gt;It had a yellow-orange color in blue flame.&lt;/li&gt;&#xA;&lt;li&gt;It's salty (note that it's usually a bad idea to taste random stuff you find on your floor).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The only white crystalline material that is a plausible candidate for being there and fizzes in acetic acid (ie vinegar) is calcite - $\ce{CaCO3}$. It's also a mineral that has 'retrograde solubility' meaning it dissolves more readily in cold water. This is probably how it got to the water in the first place. This is also why it precipitates on your electric kettle. It is the same stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The salty stuff that burns yellow-orange is obviously $\ce{NaCl}$, aka table salt. It also probably dissolved with the vinegar as well. Why is the salty taste not quite like table salt? Well, first of all it is mixed with calcite. But, calcite has no taste. It's possible that there is some $\ce{KCl}$ or $\ce{MgCl2}$ in there as well. They are slightly more bitter than $\ce{NaCl}$. This is the stuff they use in low sodium salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another possibility - it could be any of the epsom salt family, $\ce{MgSO4.$n$\,H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to would this be dissolved in the first place, you can only speculate. You say that it came from a table that was outside. Did people eat there? Could people had spilled salt on it? Is the airborne dust in your area contain a lot of calcite? What about your basement? Could someone have spilled some salts on the floor, maybe even years ago?&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="4945" LastEditDate="2014-12-22T07:03:31.743" LastActivityDate="2014-12-22T07:03:31.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21961" PostTypeId="2" ParentId="21959" CreationDate="2014-12-20T21:39:26.460" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Surprisingly, I learned that there are also usages for orbitals g,h,i&#xA;  and even j.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, the letter &quot;j&quot; is not used, so it is s, p, d, f, g, h, i, k, l, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example there is the article &lt;a href=&quot;http://iopscience.iop.org/0953-4075/36/24/006&quot;&gt;Microwave spectroscopy of Al I atoms in Rydberg states: D and G terms&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-20T21:39:26.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21962" PostTypeId="1" AcceptedAnswerId="23971" CreationDate="2014-12-20T22:01:20.137" Score="8" ViewCount="305" Body="&lt;p&gt;The following has been derived from the wiki page &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Extended_periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extended periodic table&lt;/a&gt;&quot; and is one of the many examples d or f block elements being used in synthesizing new heavier elements (or at least this is the pattern I figured out by searching many other examples of the nuclear reactants in the way to synthesize heavy elements):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Attempts to synthesize still undiscovered elements:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{align}&#xA;\ce{^254_99Es + ^48_20Ca &amp;amp;-&amp;gt; ^302_119Uue^*} \\&#xA;\ce{^249_97Bk + ^50_22Ti &amp;amp;-&amp;gt; ^295_119Uue + 4 ^1_0n} \\&#xA;\ce{^{nat}_68Er + ^136_54Xe &amp;amp;-&amp;gt; ^{298,300,302,303,304,306}Ubb^* -&amp;gt; no atoms} \\&#xA;\ce{^238_92U + ^66_30Zn &amp;amp;-&amp;gt; ^304_122Ubb^* -&amp;gt; no atoms}&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is, why is it like this? How did the scientists reached an agreement on which elements to collide? Was it anything especial about those atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P. S. If you are sure that there is an exception (exceptions omit rules; at least in this case), your answer will be accepted.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="41328" LastEditDate="2017-11-15T02:00:49.583" LastActivityDate="2017-11-15T02:00:49.583" Title="What is the reason behind choosing the specific elements used for the synthesis of heavier elements?" Tags="&lt;synthesis&gt;&lt;atoms&gt;&lt;nuclear-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21963" PostTypeId="1" AcceptedAnswerId="21982" CreationDate="2014-12-20T22:11:55.103" Score="4" ViewCount="1613" Body="&lt;p&gt;A professor of mine noted that when he was in school, microscopes weren't powerful enough to resolve certain things (I forgot what it was). But current microscopes are powerful enough. &#xA;Extrapolating from this trend, he hypothesized that by the time we were his age, we would be able to see atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait. Seeing atoms? Do we expect to see discrete little dots for electrons (I don't think so)? What exactly can we expect to &quot;see,&quot; if anything at all? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2016-08-14T21:58:31.903" Title="What do atoms look like?" Tags="&lt;atoms&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21965" PostTypeId="5" CreationDate="2014-12-20T22:53:48.003" Score="0" Body="&lt;p&gt;Questions tagged are related to the chemical property of taste; Either in analyzing an unknown substance or for other uses. Not recommended for identifying an unknown chemical species due to crudeness and danger.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2014-12-21T02:18:21.310" LastActivityDate="2014-12-21T02:18:21.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21966" PostTypeId="4" CreationDate="2014-12-20T22:53:48.003" Score="0" Body="Questions related to the crude analysis of a compound by identifying its taste." OwnerUserId="7448" LastEditorUserId="402" LastEditDate="2017-05-03T00:11:49.297" LastActivityDate="2017-05-03T00:11:49.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21967" PostTypeId="2" ParentId="21926" CreationDate="2014-12-20T23:20:48.220" Score="1" Body="&lt;p&gt;The bond between I and H is weak so I- is a weak base.  I- is a strong nucleophile because it is polarizable, making it faster for its orbitals to overlap with the electrophile.  Remember that basicity is a thermodynamic concept and nucleophilicity is a kinetic concept.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that I- can be added to reactions as a catalyst because it is a good nuclephile and a good leaving group.  Leaving group capability seems to be inversely related to basicity. &lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-12-20T23:20:48.220" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21968" PostTypeId="1" CreationDate="2014-12-21T04:44:52.147" Score="1" ViewCount="1834" Body="&lt;p&gt;I want to electrolize salt ($\ce{NaCl}$) water into a solution of sodium hydroxide for soap making. The salt I have has yellow prussiate of soda (sodium ferrocyanide) added as the anticaking agent. What will happen to the ferrocyanide ($\ce{Fe(CN)_6}$) molecules during the electrolysis process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be using an old battery charger to perform the electrolysis, $20\, \mathrm{V}$, $0.9\, \mathrm{A}$ or $1.8\, \mathrm{A}$ settings.&lt;/p&gt;&#xA;" OwnerUserId="11397" LastEditorUserId="189" LastEditDate="2014-12-21T12:26:17.713" LastActivityDate="2018-02-02T16:24:31.163" Title="Electrolysis of salt water into sodium hydroxide with YSP" Tags="&lt;electrolysis&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21969" PostTypeId="1" CreationDate="2014-12-21T13:21:08.223" Score="5" ViewCount="5176" Body="&lt;p&gt;Think for example of $\ce{KH}$ or $\ce{NaH}$ in enolization reactions. This remark was made in one of my textbook, but I could not remember why this was actually the case. Why can't $\ce{NaH}$ or $\ce{KH}$, being hard nucleophiles, add to carbonyl groups?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-21T22:48:01.403" Title="Why can alkali metal hydrides usually not act as nucleophiles towards carbonyl groups?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21970" PostTypeId="1" AcceptedAnswerId="21972" CreationDate="2014-12-21T15:28:46.073" Score="1" ViewCount="15206" Body="&lt;p&gt;Anything that burns &quot;easy&quot;, has a low activation energy ($E_\mathrm a$) for the burning process. Anything that burns hotter, will have a lesser enthalpy and thus, will have a more aggressive exothermic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethane is a molecule with two carbon atoms and 6 hydrogen atoms; all bonded with covalent single bonds. Ethene has the same number of carbons, but 4 hydrogen atoms; the bond between carbon atoms, in this one, is a double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had to choose a fuel, and heat was your prior factor, which one would you have chosen? I imagined that for this, the activation energy and the enthalpy comparison could lead to an answer. I could easily have looked both $E_\mathrm a$ and enthalpies up, but I didn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="16683" LastEditDate="2017-04-01T02:40:19.987" LastActivityDate="2018-06-29T03:05:07.977" Title="Ethane and ethene: which is easier to burn? Which one burns hotter? Why?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;enthalpy&gt;&lt;hydrogen&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21971" PostTypeId="2" ParentId="21970" CreationDate="2014-12-21T15:44:03.443" Score="1" Body="&lt;p&gt;Ethane will release more heat per mole but ethene should produce a higher temperature.  The extra heat released by ethane goes into heating up more water which has a high heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for which has lower activation energy, ethyne is dangerous to store because it can explode so I am assuming ethene is faster to react because of some trend.  I also worked with someone who was researching possible fuels for a ramjet and his group chose ethene because it was the best compromise between safety and speed of combustion.  I do not know why the trend ethane&gt;ethene&gt;ethyne for reaction rates is as it is though.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-12-21T15:44:03.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21972" PostTypeId="2" ParentId="21970" CreationDate="2014-12-21T15:53:58.737" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Olefins are less stable than their corresponding alkane analogues.  For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two equivalent ways to describe the orbitals in ethene.  One is the &quot;double bond&quot; description where a pi and sigma bond exist together side by side.  We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond.  It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane.  Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring.  It is easy to see that such a ring system would contain a significant amount of strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap).  Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&#xA;Let's consider the bromination of ethene and ethane&#xA;$$\begin{align}&#xA;\ce{C2H4 + Br2 &amp;amp;-&amp;gt; BrH2C-CH2Br}\\&#xA;\ce{C2H6 + Br2 &amp;amp;-&amp;gt; 2 CH3Br}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www2.ucdsb.on.ca/tiss/stretton/database/organic_thermo.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;heats of formation&lt;/a&gt; of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is &lt;strong&gt;exothermic&lt;/strong&gt; by 90.3 kJ/mol.  On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is &lt;strong&gt;endothermic&lt;/strong&gt; by 10.3 kJ/mol.  The olefinic reaction is significantly favored for the reasons discussed above.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-06-29T03:05:07.977" LastActivityDate="2018-06-29T03:05:07.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21973" PostTypeId="1" CreationDate="2014-12-21T16:06:15.850" Score="3" ViewCount="15132" Body="&lt;p&gt;I know that there are 18 electrons available for bonding in the entire molecule, and that 6 of these are used for sigma bonds. That leaves 12 left over. Which molecular orbitals best describe the locations of these electrons?&lt;/p&gt;&#xA;" OwnerUserId="9709" LastEditorUserId="4945" LastEditDate="2014-12-22T11:38:42.607" LastActivityDate="2014-12-22T11:38:42.607" Title="How many electrons are there in the pi-system of cyanogen?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21975" PostTypeId="1" CreationDate="2014-12-21T16:52:08.887" Score="2" ViewCount="335" Body="&lt;p&gt;Note: This experiment was performed as part of a teaching lab experiment which has concluded, so I would not be able to alter any experimental conditions. I was also unable to modify any of the variables used in the GC-MS experiment itself, but I may be able to provide these upon request. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like help in the interpretation of this spectrum from experienced chromatographers, I expect more information to be required other than the spectra provided, but due to my inexperience I am uncertain what they might be. Please do point them out in the comments if necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a spectrogram plot performed in Mathematica of the relevant data for octanol quantification (octanol normally elutes at approximately 5.8-5.9 minutes, and did so in the positive control samples.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GC-MS machine was not used by any other group during the 2 days, and the only samples run on it during the 2 days were either ddH2O (with or without octanol) or sterile filtered cultures of &lt;em&gt;E. coli&lt;/em&gt; in M9 growth media+glucose (with or without octanol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All images are thumbnailed, click on the link to open a full resolution image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Representative samples taken on Wednesday, consistent high levels of tailing. All samples are in ddH2O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HYLEU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;ddH2O&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HYLEUm.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yxlFg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;25mg/L octanol&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yxlFgm.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uoAUc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;100mg/L octanol&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uoAUcm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Samples taken on Friday, varying amount of tailing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xzjXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;ddH2O&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xzjXLm.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yAPcU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blank growth media&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yAPcUm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6RVGB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;100mg/L octanol in ddH2O, not tailed&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6RVGBm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i10jz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;50mg/L octanol spiked into growth media&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i10jzm.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bmMmI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;100mg/L octanol in ddH2O, tailed&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/bmMmIm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise present (at m/z 59 and 78) seemed inconsistent with the column (100% DMPS) nor any of the alternatives present &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/9783527611300.app4/summary&quot; rel=&quot;nofollow noreferrer&quot;&gt;on a list of possible column bleed spectra&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Octanol's solubility seems to be 2300mg/L &lt;a href=&quot;http://uwrl.usu.edu/www/doucette/6730/CEE%206730%20Kow%20lec7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;as seen here&lt;/a&gt;, and I don't find it very likely that the low solubility of octanol in ddH2O resulted in the tailing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could have caused the noise at m/z 59 and 78, and why was the tailing so different despite near-identical experimental conditions?&lt;/p&gt;&#xA;" OwnerUserId="11303" LastEditorUserId="11303" LastEditDate="2014-12-21T16:57:48.703" LastActivityDate="2014-12-21T16:57:48.703" Title="GC-MS unexpected peaks during 1-octanol quantification" Tags="&lt;mass-spectrometry&gt;&lt;chromatography&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21976" PostTypeId="2" ParentId="21973" CreationDate="2014-12-21T17:14:04.053" Score="5" Body="&lt;p&gt;Here is the structure of cyanogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jxuyu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each carbon has 4 valence electrons and each nitrogen has 5, for a total of ((4x2)+(5x2)) 18 electrons as you said.  Again, as you noted, 6 electrons are involved in sigma bonds, the 2 C-N sigma bonds and the one C-C sigma bond. Each nitrogen also has a lone pair of electrons and that accounts for (2x2) 4 more electrons.  That leaves 8 electrons to fill the 4 pi bonds.  Just like acetylene with 2 pi bonds, the two carbon-nitrogen triple bonds in cyanogen also have two pi bonds, with each pi bond holding 2 electrons.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-21T17:14:04.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21977" PostTypeId="2" ParentId="21969" CreationDate="2014-12-21T17:22:25.833" Score="2" Body="&lt;p&gt;The problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Usual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-21T22:48:01.403" LastActivityDate="2014-12-21T22:48:01.403" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21978" PostTypeId="2" ParentId="21958" CreationDate="2014-12-21T18:32:55.517" Score="3" Body="&lt;p&gt;Let's do it step by step.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$C = 2$. The reason is that you have two components: $\ce{HBr}$ and $\ce{C4H6}$. The third compound, $\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\ce{BrC4H7}$ can de described as $1\times\ce{HBr} + 1\times\ce{C4H6}$, it is not an independent component.&lt;/li&gt;&#xA;&lt;li&gt;$P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\ce{HBr}$ and $\ce{C4H6}$. On the right, you one one liquid (&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.4516556.html&quot; rel=&quot;nofollow&quot;&gt;?&lt;/a&gt;) phase: $\ce{BrC4H7}$. So here are the possibilities:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.&lt;/li&gt;&#xA;&lt;li&gt;The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.&lt;/li&gt;&#xA;&lt;li&gt;The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\ce{HBr}$ and $\ce{C4H6}$ combine to form liquid $\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one. &lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="8083" LastEditDate="2014-12-22T06:38:21.370" LastActivityDate="2014-12-22T06:38:21.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21979" PostTypeId="1" AcceptedAnswerId="21991" CreationDate="2014-12-21T19:59:00.460" Score="3" ViewCount="4810" Body="&lt;p&gt;Actually, my original question was: What makes a pith of pomegranate so bitter?&lt;br/&gt;&#xA;Then I decided to be more generic. You'll hardly find someone that consumes pomegranate and doesn't nag about the bitter taste the white pith has! Additionally, when I was a kid, my parents taught me not to eat the bitter peel of an orange, which I later learned that is basic. So, since the evolution of these kind of fruit trees has unanimously been altogether, the chemical species that cause the bitterness of their fruits' piths can be of the same group.&lt;br/&gt;&#xA;I have a hunch what I seek is the &quot;family&quot; of phenols, but that is nothing more than a hunch. Answer in detail please.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="42741" LastEditDate="2017-05-13T10:55:46.377" LastActivityDate="2017-05-13T10:55:46.377" Title="What compounds exist in the bitter peel of specific fruits?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21980" PostTypeId="2" ParentId="21968" CreationDate="2014-12-21T21:47:38.560" Score="3" Body="&lt;p&gt;If you want sodium hydroxide for soap making you should just buy some: it is a cheap and common chemical. This is because there are effective industrial processes to make it in vast quantities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to make it yourself is not easy and will cost you far more effort than just buying it. The least of your problems will be what happens to contaminants like the hexacyanoferrate (which is present in such low concentrations it isn't really significant). Have a look at the way industry makes bulk hydroxide and chlorine (e.g. &lt;a href=&quot;http://www.essentialchemicalindustry.org/chemicals/chlorine.html#membrane_cell&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;); it isn't as simple as just electrolysing brine. All the processes involve sophisticated ways to separate the products produced at the different electrodes. You don't get decent quantities of either product if you don't maintain this separation (one method uses flowing mercury; another expensive ion-exchange membranes). In addition to this, uncontrolled electrolysis is dangerous as it produces toxic and explosive gases. You really don't want to try it at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the most basic components for an effective reaction will cost you far more than the couple of dollars a large bottle of pure sodium hydroxide will cost, buy the chemical and ignore the electrolysis as a route to what you want. &lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-12-21T21:47:38.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21981" PostTypeId="1" AcceptedAnswerId="21986" CreationDate="2014-12-21T23:54:48.707" Score="1" ViewCount="80" Body="&lt;p&gt;When an ionic bond is formed between Na and Cl, the lone electron in Sodium's outer shell leaves, and completes Chlorines outer shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any physical characteristics that I can use as an observer to see this movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the reaction begins (as seen in this video &lt;a href=&quot;https://www.youtube.com/watch?v=oZdQJi-UwYs&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=oZdQJi-UwYs&lt;/a&gt;) a lot of light and heat is created. Is this the electron moving from one atom to the other? If not, what is causing it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondary to this. Why does the reaction need help (by either adding water in the video above, or pre-heating the Sodium before dropping it in the Chlorine gas) before it gets started?&lt;/p&gt;&#xA;" OwnerUserId="11401" LastActivityDate="2014-12-22T02:29:57.827" Title="What physical properties are manifested when an electron leaves Na and moves to Cl to form an ionic bond?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;electrons&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21982" PostTypeId="2" ParentId="21963" CreationDate="2014-12-22T00:36:23.103" Score="9" Body="&lt;p&gt;Yes, we can definitely see atoms in modern microscopes.  I'll try not to get deep into the subject of &quot;seeing&quot;, but I will note that most people say they are seeing when they use their eyes to observe objects.  That is, they are using an optical system (eye) consisting of a lens and a detector (retina) and some image processing carried out by a computer (brain).  Thus, any instrument which uses some sort of detector and a computer to turn the signals of that detector into an image we feel comfortable looking at qualifies to me as &quot;seeing in a microscope.&quot;  If you're comfortable with that then read on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My favorite microscope which is capable of seeing atoms is the transmission electron microscope (TEM).  Using a high quality TEM, we can easily resolve objects around an angstrom in size, which is the distance between atoms.  To image a single atom, you also need to have a sample which is only a single atom thick at some point.  This is tough to accomplish but can be done.  Today, images of graphene provide arguably the best &quot;single atom&quot; TEM images.  A seminal paper is &lt;a href=&quot;http://www.nature.com/nmat/journal/v10/n3/fig_tab/nmat2964_F1.html&quot;&gt;here&lt;/a&gt; (paywall).  You can also search the internet for graphene images such as &lt;a href=&quot;http://www.er.esi.nagoya-u.ac.jp/rescwe/kaiseki/photos/GrapheneHRTEM.jpg&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the links above, the images show the shapes of atoms as little blobs, and to a large extent the blobs are more a description of the microscope's ability to focus rather than the shape of the atom itself.  However, the atomic force microscope (AFM) excels at seeing the electrons at the surface of an atom.  With AFM, researchers are even able to &lt;a href=&quot;http://www.sciencemag.org/content/342/6158/611&quot;&gt;image the chemical bonds of molecules sitting on a substrate&lt;/a&gt;.  In this case, the image shows the bonds as rods and the atom centers are not really visible.  Chemistry textbooks often show the &quot;shapes&quot; of atoms by drawing the electron orbitals, so in a sense these rods are the shapes of these atoms too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This just points up that the &quot;shape&quot; of atoms is a non-trivial subject, which is very dependent on how you look at them.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2014-12-22T00:36:23.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21983" PostTypeId="1" AcceptedAnswerId="22226" CreationDate="2014-12-22T01:21:09.940" Score="19" ViewCount="4806" Body="&lt;p&gt;I'm sorry for such a stupid question, but after googling for half an hour I know two things&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are at least a zillion pages, blog-posts and articles about how to keep your fireplace window clean or how to clean it.&lt;/li&gt;&#xA;&lt;li&gt;There is not a single page which explains what I'm looking for and which doesn't make claims without supporting them with scientific facts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Therefore, here the short form of my question: Assuming I have a fireplace with a window, which looks like this after a while:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L7d8f.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Further assume I'm able to keep all other factors constant (like type of wood, temperature of the fire etc.), is there something I can apply/do to the cleaned window to prevent the building of soot for a longer time?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The reason I'm asking is &lt;a href=&quot;https://hb2cents.wordpress.com/2013/01/27/experiment-preventing-soot-build-up-on-wood-stove-glass/&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; I stumbled over, which claims that normal corn-starch can be used to &lt;em&gt;fill the pores of the glass&lt;/em&gt;. This left me with more questions&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;does each glass (especially my one) have pores with a size that starch particles can fit in? I read somewhere that starch has a diameter between 2-170 µm depending on the type.&lt;/li&gt;&#xA;&lt;li&gt;I'm making a fire where I burn wood! Why wouldn't the starch simply burn and create itself the soot?&lt;/li&gt;&#xA;&lt;li&gt;As &lt;a href=&quot;http://en.wikipedia.org/wiki/Lotus_effect&quot; rel=&quot;noreferrer&quot;&gt;we know for a while now&lt;/a&gt; making a surface really smooth doesn't mean it keeps clean. With a size of 10-20 µm of the lotus papillae, it seems to fall in the same range as starch.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;And for the diligent, here two more sub-questions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does soot building work? I mean, why does the soot stick so hard at the window? After all it's just little particles flowing around and landing on the window. (I know it's called &lt;em&gt;burning in&lt;/em&gt;, but how does the heat work on a already burned particle?)&lt;/li&gt;&#xA;&lt;li&gt;Is there some connection to the pores (if any) in the glass in this process?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&lt;p&gt;After using my fireplace one winter, I'd like to make recommendation based on my experience so far. Important is that I can easily remove the window from my fireplace for cleaning. In case the window is fixed, cleaning might become a bit messy. Let me bring it down to 3 tips:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;My current method is to use a cleaner for ceramic glass cooktops. This works like a charm. In case someone wants to know the specific product: it's a &lt;a href=&quot;http://www.dr-beckmann.de/haushaltsreinigung/fuer-oberflaechen/glaskeramik-putzstein/&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;cleaning stone for class ceramic&lt;/em&gt;&lt;/a&gt; which is probably only available in Germany but I guess something similar can be found all around the world. This thing comes with a sponge and you just have to rub the wet sponge a few times over the cleaner to apply some. Cleaning the window with this is done in less than a minute, even if there is a lot of soot.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you don't want to invest any money, the most simple and very effective cleaning method is to use the ash itself to scrub the window. Simply take water and a newspaper, if possible make the window itself wet or better rinse it with water. Take a sheet of newspaper, make it wet and apply ash directly from the fireplace itself. It's really surprising how easily the soot can be scrubbed away with this method.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Last tip: Heat! As already said in the accepted answer, &lt;em&gt;soot is basically the deposition of incomplete combustion products from a flame&lt;/em&gt;. There is no doubt and I have tested this very often: a dirty window can partially be cleaned by make a hot fire with very dry wood and a lot of air. Therefore, try to fill your fire-room with the maximum possible amount of wood, open all air-valves and let it burn for a while.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11403" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2021-01-09T16:51:22.243" Title="Fireplace window: Can one influence how fast soot is building up?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21984" PostTypeId="1" CreationDate="2014-12-22T01:32:16.917" Score="0" ViewCount="997" Body="&lt;p&gt;I have an aromatic compound that I reduced with either $\ce{Fe + HCl}$ or $\ce{Sn + HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you choose to go about extracting the compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking to dry off all the $\ce{HCl}$, redissolve in water, and react out the iron salts to Iron Hydroxide, which can be filtered out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then dry out again, dissolve in something like acetone then filter out the sodium chloride?&lt;/p&gt;&#xA;" OwnerUserId="11404" LastEditorUserId="865" LastEditDate="2015-10-16T18:04:55.310" LastActivityDate="2015-10-16T18:04:55.310" Title="How do I purify the resulting compound after a nitro- to amine-group reduction?" Tags="&lt;organic-chemistry&gt;&lt;purification&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21985" PostTypeId="2" ParentId="21831" CreationDate="2014-12-22T02:29:20.020" Score="0" Body="&lt;p&gt;The Orbital Molecular Theory explains it very well; the symmetry and orbital overlap defined by the wave functions gives a complete approach. I add some thoughts to the question qualitatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of covalent and ionic bonds must be taken carefully when talking of acidity, a property normally refereed to aqueous solution where other interactions (hydration enthalpy) take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{Al^3+}$ ion is very polarizing, and has a high charge and low ionic radius leading to high effective nuclear charge. In solution the cation must have many shells of hydration, coordinating or polarizing water molecules and attaching $\ce{OH-}$ groups in order to compensate the positive charge, thus increasing the net amount of available $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, carbon dioxide is a gas and it can be seen as a separated phase on a soda drink (bubbles); so first it must be dissolved in water. The dissolution of a $\ce{CO2 (g)}$ molecule in water increases the acidity, $\ce{[H+]}$, as can be tasted when leaving a glass of soda open in equilibrium with air for a while. I copy here the equations found at &lt;a href=&quot;http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; which explains why carbon dioxide is acidic. If you follow the link, it has a brief explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 (g) &amp;lt;=&amp;gt; CO2 (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 (aq) + H2O (l) &amp;lt;=&amp;gt; H2CO3 (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2CO3 (aq) &amp;lt;=&amp;gt; HCO3- (aq) + H+ (aq)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCO3- (aq) &amp;lt;=&amp;gt; CO3^2- (aq) + H+ (aq)}$&lt;/p&gt;&#xA;" OwnerUserId="5683" LastEditorUserId="10185" LastEditDate="2015-09-06T17:33:37.040" LastActivityDate="2015-09-06T17:33:37.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21986" PostTypeId="2" ParentId="21981" CreationDate="2014-12-22T02:29:57.827" Score="3" Body="&lt;p&gt;The light and heat are created by the exothermic nature of the reaction.  If you look up &quot;Born Haber cycle sodium chloride&quot; you will see information on why the reaction is exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction &quot;needs help&quot; because of the fact that the sodium will likely be coated with a layer of sodium oxide or hydroxide that will not allow chlorine to collide with the sodium metal.  Even then, with pure sodium metal the reaction will be faster if heated as any reaction needs activation energy to proceed.  Adding water causes an exothermic reaction that can provide activation energy for the reaction with chlorine.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-12-22T02:29:57.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21987" PostTypeId="2" ParentId="998" CreationDate="2014-12-22T05:34:57.250" Score="-4" Body="&lt;p&gt;$\ce{Br-}$ = $\ce{[Kr]}$. $\ce{Kr}$ is a noble gas and has a full  octet, meaning that the energy is SUPER low because of how stable it is. &lt;/p&gt;&#xA;" OwnerUserId="11405" LastEditorUserId="4231" LastEditDate="2014-12-22T19:26:55.387" LastActivityDate="2014-12-22T19:26:55.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21988" PostTypeId="1" CreationDate="2014-12-22T06:18:54.353" Score="5" ViewCount="5495" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CN}$, for being a strong field, would not follow Hund's rule.&lt;/li&gt;&#xA;&lt;li&gt;the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$&lt;/li&gt;&#xA;&lt;li&gt;Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be &lt;strong&gt;trigonal bipyramidal&lt;/strong&gt; while it is &lt;em&gt;not&lt;/em&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11406" LastEditorUserId="5026" LastEditDate="2018-02-27T02:40:44.010" LastActivityDate="2018-02-27T02:40:44.010" Title="Why is tetracyanonickelate square planar?" Tags="&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21990" PostTypeId="2" ParentId="21947" CreationDate="2014-12-22T09:23:27.407" Score="3" Body="&lt;p&gt;There are a few things that influence the properties of polymers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Molecular weight of the chains, i.e. how many monomers are joined in each chain, can drastically alter the polymer's properties. For example, for polyethylene, UHMWPE has extremely long chains and produces a very strong material suitable for bullet-resistant vests, whereas LLDPE has much shorter chains that produce a weaker material that is much more ductile (think about how easy it is to stretch a plastic shopping bag or the like).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another factor that affects a polymer's properties is branching. Branching is when a polymer is not made up of nice straight chains, but chains that occasionally fork off into several different chains. The difference between LLDPE and LDPE is that LDPE has a lot more branching. This keeps the chains from packing as nicely so LDPE isn't as strong.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The other big factor (other than adding another monomer to make a copolymer) is the introduction of crosslinking. Crosslinking is where different chains are connected to one another, often by addition of a molecule that can bond to a chain at both ends, e.g. rubber vulcanization crosslinks the different chains with linear sulfur chains. Natural rubber is very different than what most people think; it's a soft sticky material that doesn't hold its shape. The polyisoprene chains it's composed of aren't crosslinked and can easily slide past one another, but once crosslinked, the chains are connected and aren't free to move as much. Depending on the degree of crosslinking, you can go from something somewhat soft, like a bouncy ball, to something very rigid, like a car tire.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hydrogels are not simple linear polymers. While they're mostly made from a single monomer, small additions of related molecules are used to control how they polymerize. N.B. These apply generally to silicones, other polymers may have other branching/termination processes. If the monomer only has two sites at which it can connect to other monomers, naturally the only structure that can be formed is a linear chain. One can control the length of the chains by adding a small amount of a terminator—a monomer that can only make one connection. If a terminator reacts with a growing chain, it stops the growth as it can't connect to anything else. The more terminator added, the shorter chains are, on average. Similarly, if you add a monomer that has more than two reactive groups, branching and crosslinking will occur. Often, a non-crosslinked polymer is supplied as a liquid, then a crosslinking agent is added to harden the polymer. &lt;a href=&quot;http://www.dowcorning.com/content/discover/discoverchem/cure-systems.aspx&quot; rel=&quot;nofollow&quot;&gt;Dow Corning&lt;/a&gt; has a great primer on the actual chemistry involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes a hydrogel special is that it's a polymer &lt;em&gt;network&lt;/em&gt; that encapsulates water, i.e. it's an extensively crosslinked polymer that is hydrophilic. If the polymer was just a series of individual chains, it would just dissolve in water since nothing is holding the different chains together. When crosslinked, water can get in, but the mesh of polymer is linked together enough to hold its shape, though it may swell to accommodate the water. Depending on the degree of crosslinking, more or less water can fit within the structure. A less linked polymer can fit more water since there are fewer crosslinks resisting, but is less rigid when hydrated for the same reason (think Jell-O). A more crosslinked polymer won't absorb as much water, but will be much more rigid (like a contact lens). Contact lenses have to balance holding their shape, oxygen permeability, surface hydrophilicity, etc. so there are many reasons why, despite being based on the same primary monomer, two different brands may have a different blend of the other stuff that lets one control the hydrogel's properties.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-22T09:23:27.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21991" PostTypeId="2" ParentId="21979" CreationDate="2014-12-22T10:09:39.307" Score="9" Body="&lt;p&gt;The reason behind the bitterness of most fruits is due to presence of Tannin in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tannin is a naturally occurring polyphenol found in plants, seeds, bark, wood, leaves and fruit skins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tannin tastes bitter,dry and astringent and you can feel it specifically on the middle of your tongue and the front part of your mouth. Besides the peels of fruit, we can feel bitterness in other food products as well like,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-Tannin Foods =&#xA; Tea Leaves&#xA;Walnuts, Almonds and Nuts with Skins&#xA;Dark Chocolate&#xA;Cinnamon, Clove and other spices&#xA;Pomegranates, Grapes &#xA;Quince&#xA;Red Beans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The  bitterness and astringency are caused by tannins binding with proteins in our saliva. The chemical constituent of tannins are: gallic acid , ellagic acid and their derivatives(present in hydrolyzable tannins) , catechin and its derivative (present in condensed tannins) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oun9V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pgXr2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your specific question about the the pomegranate fruit pith, it contains ellagitannins, pelargonidin, punicalin, punicalagin, anthocyanins, cyanidin, ellagic acid which makes it bitter.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-12-22T10:09:39.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21992" PostTypeId="2" ParentId="21973" CreationDate="2014-12-22T10:36:39.263" Score="3" Body="&lt;p&gt;To visualize ron's answer, here is the valence orbital scheme of cyanogen. Note, that this is the canonical molecular orbital picture, which does not use hybridisation. From this depiction, the stated fact from Geoff is also clearly visible. Due to delocalisation the carbon-carbon bond is significantly shorter than a normal single bond. The right hand side orbitals are taken from a DF-BP85/def2-SVP calculation. (Since I forgot to add that to the picture itself, the horizontal dotted line separates unoccupied from occupied orbitals.)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tOttM.jpg&quot; alt=&quot;mo scheme of cyanogen&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2014-12-22T10:36:39.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21993" PostTypeId="2" ParentId="21988" CreationDate="2014-12-22T14:44:01.403" Score="7" Body="&lt;p&gt;This is the square planar splitting diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9P1s0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't&lt;/p&gt;&#xA;" OwnerUserId="9850" LastEditorUserId="9850" LastEditDate="2014-12-22T14:58:33.910" LastActivityDate="2014-12-22T14:58:33.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21994" PostTypeId="2" ParentId="16614" CreationDate="2014-12-22T15:30:03.170" Score="2" Body="&lt;p&gt;Aluminium and Magnesium &#xA;react vigorously with mineral acids. &#xA;A lot of heat is evolved. &#xA;Try iron/steel wire, which is cheaper and &#xA;available in any diameter. &#xA;Georg&lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2014-12-22T15:30:03.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21996" PostTypeId="1" AcceptedAnswerId="22005" CreationDate="2014-12-22T18:26:30.840" Score="8" ViewCount="360" Body="&lt;p&gt;When a battery is performing work on a charge and moves it to the higher potential, why doesn't the charge move through the battery and return to the low potential? Why does it go through the circuit?&lt;/p&gt;&#xA;" OwnerUserId="11412" LastEditorUserId="7159" LastEditDate="2014-12-22T20:48:24.723" LastActivityDate="2015-04-19T10:07:43.013" Title="Current in Battery" Tags="&lt;electricity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21998" PostTypeId="1" CreationDate="2014-12-22T19:57:12.220" Score="1" ViewCount="53" Body="&lt;p&gt;So I have a quick question about some concept I have watched in a video, it is illustrated as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B &amp;lt;=&amp;gt; C + D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what the video is saying is that if you increase $\ce{A}$ then you get the products to &lt;b&gt;increase&lt;/b&gt; and the $\ce{C + D}$ to &lt;b&gt;increase&lt;/b&gt; as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that shouldn't the products also depend on the $\ce{B}$ and not only just the $\ce{A}$. So to increase the &lt;strong&gt;products&lt;/strong&gt; don't we need to increase both the &lt;strong&gt;reactants&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's &lt;a href=&quot;https://www.khanacademy.org/science/chemistry/reaction-rates/v/le-chatelier-s-principle&quot; rel=&quot;nofollow&quot;&gt;the video for reference&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2018-09-08T08:30:41.457" LastActivityDate="2018-09-08T08:30:41.457" Title="Reaction Rates &amp; Equilibrium" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2014-12-23T00:59:40.790" ContentLicense="CC BY-SA 3.0" />
  <row Id="21999" PostTypeId="1" AcceptedAnswerId="22008" CreationDate="2014-12-22T20:22:56.200" Score="0" ViewCount="343" Body="&lt;p&gt;Is it possible that $\ce{2 OH-}$ from water could react to form $\ce{H2O2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean most autoionization forms $\ce{H3O+}$ and $\ce{OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I think it is possible that $\ce{H-}$ and $\ce{OH+}$ forms, though the amount formed of these 2 is small. This is from water acting as an acid and H+ acting as a base.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="9850" LastEditDate="2014-12-22T20:28:13.457" LastActivityDate="2016-03-12T00:09:21.980" Title="Hydrogen peroxide from water" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;ions&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22000" PostTypeId="2" ParentId="21613" CreationDate="2014-12-22T20:34:43.550" Score="6" Body="&lt;p&gt;The answer really doesn't have much to do with fluorine.  Beacall's 1928 &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/recl.19280470107/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;The melting-points of benzene derivatives&lt;/a&gt; explains on the first page that it has long been known &quot;that the more symmetrical the constitution of a benzene derivative, the greater the resistance it offers to the passage from the solid to the liquid state&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the more modern &lt;a href=&quot;http://pubs.rsc.org/en/content/articlehtml/2004/ob/b407105k&quot; rel=&quot;nofollow noreferrer&quot;&gt;Effect of molecular symmetry on melting temperature and solubility&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As a general rule, symmetrical molecules in crystalline form have higher melting temperatures and exhibit lower solubilities compared with molecules of similar structure but with lower symmetry. Symmetry in a molecule imparts a positive amount of residual entropy in the solid phase (i.e., more possible arrangements leading to the same structure). This means that the entropy of a crystal of symmetric molecules is greater than the entropy of crystal of a similar, but non-symmetric molecule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Benzene and hexafluorbenzene have almost the same melting point, while monofluoro, monochloro, monobromo benzene all have much lower melting points.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both benzene and hexafluoro-benezne have a very stable &quot;herringbone&quot; crystal structure, where the edge of one ring approaches the center of another ring in a perpendicular manner.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AaMOr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So realizing that the cyrstal structure of benzene is unusually stablized, it is understandable that the melting point of benzene is higher than monofluorobenzene, but the boiling point is not.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-23T14:48:25.870" LastActivityDate="2014-12-23T14:48:25.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22001" PostTypeId="2" ParentId="21996" CreationDate="2014-12-22T20:40:14.393" Score="1" Body="&lt;p&gt;The answer is: it will! When no salt bridge or similar conducting element is present between the two half cells, the reaction will quickly run to equilibrium. The state in which the potential force of the difference in electron affinity between the two electrodes more or less equals the Coulomb force of the opposite charges that accumulate in them. However, if you make use of a salt bridge this charge difference accumulation can be removed, since an electrolyte, which sort of completes the circuit. Usually salt bridges are constructed in a manner that prevents intermixing of the ions in both half cells, so that there is some certainty about what redox reaction is actually going on, and how. An example of a common salt bridge used in high school chemistry experiments is a filter paper soaked in a solution of potassium chloride.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-22T20:40:14.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22002" PostTypeId="2" ParentId="21999" CreationDate="2014-12-22T21:10:57.963" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible that $\ce{2 OH-}$ from water could react to form&#xA;  $\ce{H2O2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There would be two electrons left unaccounted for.  There needs to be a balanced chemical equation, then we could consider the equillbrium constant for the reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could write $\ce{2H2O -&amp;gt; H2 + H2O2}$, but it is extremely unfavorable, like water decomposing to hydrogen and oxygen gas. You could calculate the Gibbs free energy of the reaction and an equilibrium constant, but the equilibrium constant will be extremely small. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-23T16:17:27.090" LastActivityDate="2014-12-23T16:17:27.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22003" PostTypeId="2" ParentId="21613" CreationDate="2014-12-22T22:50:04.097" Score="6" Body="&lt;p&gt;The answer has &lt;strong&gt;everything&lt;/strong&gt; to do with fluorine.  Dunitz, in his early studies on fluorine interactions found that intermolecular &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H..F}$&lt;/span&gt; interactions can be solid-state structure (&lt;em&gt;e.g.,&lt;/em&gt; crystal packing) directing.  &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.200600092/abstract;jsessionid=7445E26908FBA03C847C4AE1845D0935.f01t02?deniedAccessCustomisedMessage=&amp;amp;userIsAuthenticated=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;In the case of fluorobenzene&lt;/a&gt;, the molecule adopts a solid-state structure that sacrifices long-range order so that these short-range &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H..F}$&lt;/span&gt; interactions can be preserved (also see &lt;a href=&quot;https://books.google.com/books?id=kxGzrnJC-J0C&amp;amp;pg=PA22&amp;amp;lpg=PA22&amp;amp;dq=fluorobenzene%20crystal%20structure&amp;amp;source=bl&amp;amp;ots=or7zJG6xed&amp;amp;sig=32HBuB8zqGpHxzZm1jTilpP9qlE&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=4JaYVNSKB5X-yQTp7YKoAw&amp;amp;ved=0CB0Q6AEwADgU#v=onepage&amp;amp;q=fluorobenzene%20crystal%20structure&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorobenzene&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; where it is stated that, &amp;quot;Its [fluorobenzene] melting point is &lt;span class=&quot;math-container&quot;&gt;$\pu{-44 ^\circ C}$&lt;/span&gt;, which is lower than that of benzene, indicative of the remarkable effect of fluorination on the intermolecular interactions as seen throughout organofluorine chemistry&amp;quot;). The loss of long-range order in the solid-state, as compared to benzene, produces a lower melting point in fluorobenzene &lt;span class=&quot;math-container&quot;&gt;$(\pu{-44 ^\circ C})$&lt;/span&gt; than in benzene (&lt;span class=&quot;math-container&quot;&gt;$\pu{5.5 ^\circ C}$&lt;/span&gt;). Note too that the &amp;quot;size&amp;quot; of fluorine in organic molecules is very similar to that of hydrogen (that is why it has found such widespread use in pharmaceuticals) - it is often referred to as a hydrogen &amp;quot;mimic&amp;quot; (see &lt;a href=&quot;http://www.halocarbon.com/halocarbon_media/BenefitsToAddingFluorine-cphishowdaily.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). So steric (or symmetry) arguments cannot be used to explain the packing disruption with fluorobenzene.&lt;/p&gt;&#xA;&lt;p&gt;Once the molecule is in the liquid state the intermolecular &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H..F}$&lt;/span&gt; short-range interactions are also diminished due to increased molecular motion; as a result, fluorbenzene and benzene boil relatively close to one another (&lt;span class=&quot;math-container&quot;&gt;$\pu{84 ^\circ C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{80 ^\circ C}$&lt;/span&gt;, respectively).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-07-19T19:14:31.857" LastActivityDate="2020-07-19T19:14:31.857" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22005" PostTypeId="2" ParentId="21996" CreationDate="2014-12-23T02:19:51.990" Score="7" Body="&lt;p&gt;Batteries are constructed with an internal barrier between the two electrolytes that does &lt;strong&gt;not&lt;/strong&gt; conduct electricity (i.e., electrons can't pass), but that &lt;strong&gt;does&lt;/strong&gt; allow ions to pass through.  See the figure below (taken from &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/0/04/Alkaline-battery-english.svg/2000px-Alkaline-battery-english.svg.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Wikimedia&quot;&gt;Wikimedia&lt;/a&gt;, open source), of a simple alkaline dry cell (e.g., an AA battery), in particular the part labeled &quot;ION CONDUCTING SEPARATOR.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/10zNq.png&quot; alt=&quot;Section of dry cell&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason batteries are made this way is because in order for the anode electrolyte and the cathode electrolyte to come to thermochemical equilibrium, there &lt;strong&gt;has&lt;/strong&gt; to be a transfer of &lt;strong&gt;both&lt;/strong&gt; electrons and ions.  By putting the separator in there that only conducts ions, you block equilibration of electrons except in the case where you have it connected to an external circuit.  If you conduct an imaginary experiment where you * &lt;strong&gt;SNAP&lt;/strong&gt; * your fingers and make that ion conducting separator disappear, the electrons would indeed surge from one electrolyte to the other as you describe, as fast as mass transfer would allow.  (You would then having to imagine the battery either exploding or melting, because this would make a &lt;strong&gt;&lt;em&gt;LOT&lt;/em&gt;&lt;/strong&gt; of heat.)&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2014-12-31T12:36:17.957" LastActivityDate="2014-12-31T12:36:17.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22006" PostTypeId="2" ParentId="20337" CreationDate="2014-12-23T04:26:14.713" Score="1" Body="&lt;p&gt;It all depends on the selectivity of your SPE for the class of compounds you're looking for.  If you have some confidence that the SPE has let only a few components of the bacterial lysate pass into the eluate, then you might see few enough &lt;em&gt;m/z&lt;/em&gt; signals on the MS that qualitative analysis would be possible.  If your SPE is relatively unselective, though, you will probably have a dense array of peaks that won't tell you much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It (probably?) can't hurt to try running the sample(s), though.  If you see relatively few peaks, then you could rerun the samples, spiked with known amounts of known compounds, to see whether:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A new peak shows up (the spiked chemical is absent in the sample); &lt;/li&gt;&#xA;&lt;li&gt;An existing peak gets bigger (modest indication that the chemical is present); or,&lt;/li&gt;&#xA;&lt;li&gt;No change is observed (the spiked chemical is absent in the sample or the spike was too small)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you're looking for more than a few chemicals, this promises to be a pretty drawn-out process, with lots of samples pushed through the MS. Might still be faster than waiting for the LC, though!&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2014-12-23T04:26:14.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22007" PostTypeId="1" AcceptedAnswerId="22103" CreationDate="2014-12-23T05:52:35.603" Score="14" ViewCount="2323" Body="&lt;p&gt;It does seem that there is a correlation. Both are measures of proton affinity. Heck, isn't hydrogen bonding present in the transition states of most acid base reactions (excluding those involving proton tunneling?) Also, it seems that hydrogen bonding would be subject to the same limitations as Brønsted basicity; i.e. it seems that steric hindrance in some molecule would inhibit both proton transfer &lt;em&gt;and&lt;/em&gt; limit hydrogen bonding capability. At least it seems that way to me prima facie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also why do we depict acid base reactions as concerted? Is it because the time scale is extremely small for proton transfer reactions, so who cares if it's actually stepwise or concerted? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-03T11:19:04.800" LastActivityDate="2018-03-03T11:27:45.780" Title="Does hydrogen bonding strength correspond to Brønsted basicity in a given medium?" Tags="&lt;acid-base&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22008" PostTypeId="2" ParentId="21999" CreationDate="2014-12-23T07:21:38.720" Score="5" Body="&lt;p&gt;The only &quot;hydroxyl&quot;-type species whose recombination has been observed to yield  hydrogen peroxide are &lt;strong&gt;hydroxyl radicals&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HO^. + ^.OH -&amp;gt; H2O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\gamma$, electron beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conceivable steps yielding $\ce{HO^.}$ are either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O  -&amp;gt; H^. + ^.OH}$$&#xA;or&#xA;$$\ce{H2O  -&amp;gt; H2O^{+.} + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O^{+.} -&amp;gt; H+ + ^.OH}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="24052" LastEditDate="2016-03-12T00:09:21.980" LastActivityDate="2016-03-12T00:09:21.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22010" PostTypeId="1" AcceptedAnswerId="22011" CreationDate="2014-12-23T08:55:30.343" Score="0" ViewCount="149" Body="&lt;p&gt;Does the positivity and negativity of terminals of a battery depend on the terminal where the electrons enter the external circuit?&lt;/p&gt;&#xA;" OwnerUserId="11412" LastActivityDate="2014-12-23T11:12:26.443" Title="Terminals Of An Electrochemical Cell" Tags="&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22011" PostTypeId="2" ParentId="22010" CreationDate="2014-12-23T10:39:07.827" Score="3" Body="&lt;p&gt;The positive terminal of a battery is where 'conventional current' comes out of the battery and flows through an external circuit, returning to the battery via the negative terminal. Since electrons are (by convention) negatively charged, the negative terminal of a battery is where electrons are 'released' by the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: Of course, if it is a secondary battery which is 'recharging', electrons will 'enter' the battery at the terminal marked negative.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="708" LastEditDate="2014-12-23T11:12:26.443" LastActivityDate="2014-12-23T11:12:26.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22012" PostTypeId="1" CreationDate="2014-12-23T13:26:51.230" Score="1" ViewCount="2261" Body="&lt;p&gt;I have conducted an experiment wherein I have used different concentrations of $\ce{H2SO4}$, $\ce{HNO3}$ and $\ce{H3PO4}$ as the electrolytes. I filled the acid in glass tube and used a 9V battery, wires and two graphite electrodes to complete the circuit. I also attached an ammeter to check the current variation during a time period of 1 minute.&lt;br/&gt; I observed that the current was higher with more concentrated solutions, although my query is that when I add more of these acids, the concentration of the $\ce{OH-}$ ions reduces; it is the $\ce{OH-}$ ions and the $\ce{H+}$ ions that dissociate.&lt;br/&gt; How exactly do I explain the increase in current due to increase in concentration?&lt;/p&gt;&#xA;" OwnerUserId="11419" LastEditorUserId="4231" LastEditDate="2014-12-23T16:34:26.937" LastActivityDate="2014-12-24T14:55:24.680" Title="How to explain the increase in current with the increase in the concentration of responsible ions?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22013" PostTypeId="2" ParentId="22012" CreationDate="2014-12-23T14:53:18.843" Score="3" Body="&lt;p&gt;The electrical current is carried by &lt;strong&gt;all&lt;/strong&gt; ions in the solution, including the $\ce{SO_4^{2-}}$/ $\ce{HSO_4^-}$, $\ce{NO_3^-}$, and $\ce{PO_4^{3-}}$/ $\ce{HPO_4^{2-}}$/ $\ce{H_2PO_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$.  So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system.  For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there&lt;/li&gt;&#xA;&lt;li&gt;$\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there&lt;/li&gt;&#xA;&lt;li&gt;Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode&lt;/li&gt;&#xA;&lt;li&gt;In order for the solution to maintain &lt;a href=&quot;http://www.chem1.com/acad/webtext/elchem/ec1.html&quot; rel=&quot;nofollow&quot; title=&quot;electroneutrality&quot;&gt;electroneutrality&lt;/a&gt;, something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor.  (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$&lt;/li&gt;&#xA;&lt;li&gt;Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$&lt;/li&gt;&#xA;&lt;li&gt;Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$&lt;/li&gt;&#xA;&lt;li&gt;So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode.  These concentration differences result in a net transfer of charge between the two electrodes.&lt;/li&gt;&#xA;&lt;li&gt;All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2014-12-24T14:55:24.680" LastActivityDate="2014-12-24T14:55:24.680" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22014" PostTypeId="1" AcceptedAnswerId="22041" CreationDate="2014-12-23T17:04:29.570" Score="6" ViewCount="3949" Body="&lt;p&gt;I know that the liquid crystal displays contain some class of organic compound, which is somehow used to polarise light to make contrast variations(In calculator displays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Question is What type of Organic Compound is used in this case and How does it polarise light only when stimulated by external agents such as current in LCD's &lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2014-12-24T12:53:16.467" Title="What type of Organic Substance is used in LCD" Tags="&lt;organic-chemistry&gt;&lt;liquid-crystals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22015" PostTypeId="2" ParentId="13536" CreationDate="2014-12-23T17:12:35.017" Score="4" Body="&lt;p&gt;Looking back, I don't like my old answer so I'm adding a totally different answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A specific (theoretical) example of endothermic bond formation is described in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja9938175&quot; rel=&quot;nofollow&quot;&gt;Prediction of a Metastable Helium Compound: HHeF&lt;/a&gt; J. Am. Chem. Soc. 2000, 122, 6289-6290.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As reported in table 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy to dissociate HHeF to H + He + F is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy to dissociate HNeF to H + Ne + F is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy to dissociate HHeF to HF + He is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy to dissociate HNeF to HF + Ne is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, HHeF is stabilized by being in a potential energy well, the activation energy to dissociate being relative high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors conclude: &quot;Remarkably, HHeF is also predicted to be a metastable species, which represents the first neutral compound containing a helium chemical bond.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond formation is endothermic.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-23T17:17:53.730" LastActivityDate="2014-12-23T17:17:53.730" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22016" PostTypeId="2" ParentId="13655" CreationDate="2014-12-23T17:18:19.203" Score="2" Body="&lt;p&gt;Let me try to answer the part 1. of your question.&#xA;&lt;br/&gt;The process of compacting powdered Aluminium is also called Sintering. And as you can guess, the thin oxide layer of Alumina inhibits the binding of Aluminium grains between the powder particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while sintering, we need to either:&#xA;&lt;br/&gt;1) Break part of the Alumina film or,&#xA;&lt;br/&gt;2) Reduce part of it to Aluminium,&#xA;&lt;br/&gt; in order to facilitate formation of a metallic bond between the powder particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1), we use a nitrogen atmosphere as &lt;a href=&quot;http://www.ifam.fraunhofer.de/content/dam/ifam/de/documents/dd/Publikationen/2005/2005_schubert.pdf&quot; rel=&quot;nofollow&quot;&gt;it enhances the destruction of Alumina&lt;/a&gt;.&#xA;&lt;br/&gt;For 2), we use because magnesium as a alloying element &lt;a href=&quot;http://www.ifam.fraunhofer.de/content/dam/ifam/de/documents/dd/Publikationen/2005/2005_schubert.pdf&quot; rel=&quot;nofollow&quot;&gt;because magnesium, even in small amounts, concentrates on the surface of powder particles&#xA;and supports sintering by local reduction of Alumina&lt;/a&gt;.&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, I would say that compacted Aluminium does have Alumina inside the molded object. But metallic bonds between powders are formed after partly removing the oxide film by the above mentioned methods.&lt;/p&gt;&#xA;" OwnerUserId="3976" LastActivityDate="2014-12-23T17:18:19.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22017" PostTypeId="2" ParentId="15055" CreationDate="2014-12-23T18:10:02.293" Score="5" Body="&lt;p&gt;Resonance structures of $\ce{SO2}$ and $\ce{O3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nlOkV.jpg&quot; alt=&quot;$&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the diagram $\ce{O3}$ and $\ce{SO2}$ are isoelectric and have the same shape and this accounts for their very similar electrostatic potential maps. Both molecules are polar although $\ce{SO2}$, with a dipole moment of 1.62D, is more polar than $\ce{O3}$ (0.53D). As the diagram shows, the polarity of both molecules is due to the their trigonal planar shape with a formal positive charge on the central atom.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-23T18:10:02.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22018" PostTypeId="2" ParentId="20337" CreationDate="2014-12-23T18:10:20.567" Score="2" Body="&lt;p&gt;The practical answer really depends on two things: do you know exactly what you're looking for and can your instrument do tandem MS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know exactly which metabolites you're looking for and they're present in great enough abundance, it may be possible to resolve them from the the background, depending on the resolving power of the instrument and what else is in there (if you have an ICR instrument, it may not be a problem at all, but if it's just a single quadrupole, you may be out of luck).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know exactly which metabolites you're looking for (or know that they share some common structural feature that can be knocked off by CID or the like) &lt;em&gt;and&lt;/em&gt; your instrument can do tandem MS, you're in a much better position. Tandem MS is often used to avoid needing to do a slow LC step as it's possible to identify two compounds that aren't resolved on a precursor ion scan by looking at their fragment ions: A messy primary ion scan can be dealt with if you know that your analyte has a [M-18] transition or whatever, by CID. Also, a given class of compounds often shares certain CID reactions, which can be picked up with a neutral loss scan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have tandem, but aren't sure exactly what you want to find, it's a bit more work, but it just means you need to look at product ion scans of peaks that are ambiguous and try to work out what they are.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-23T18:10:20.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22019" PostTypeId="1" AcceptedAnswerId="22026" CreationDate="2014-12-23T18:16:50.033" Score="2" ViewCount="684" Body="&lt;p&gt;Also, when is this fact useful in practice?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="194" LastEditDate="2016-11-20T23:30:46.867" LastActivityDate="2016-11-20T23:30:46.867" Title="Why do gases only transmit sound waves whose wavelength is long relative to the mean free path?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22020" PostTypeId="1" AcceptedAnswerId="22029" CreationDate="2014-12-23T19:00:06.693" Score="1" ViewCount="1421" Body="&lt;p&gt;I learned that phenol absorbs longer wavelengths than benzene because the electron cloud is bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is about Penta-1,2,3,4-tetrien (1) and 2,2,-Dimethylpropane (2)?&#xA;(1) has the largest diameter but (2) has the smallest. Which one absorbs the longer waves?&lt;/p&gt;&#xA;" OwnerUserId="11411" LastEditorUserId="4945" LastEditDate="2014-12-24T04:13:24.340" LastActivityDate="2014-12-24T04:13:24.340" Title="Is the smallest or largest diameter important for the color of a molecule?" Tags="&lt;organic-chemistry&gt;&lt;electrons&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22021" PostTypeId="1" AcceptedAnswerId="22022" CreationDate="2014-12-23T19:03:31.700" Score="2" ViewCount="1881" Body="&lt;p&gt;Melatonin (IUPAC semi-formal name: &lt;b&gt;N-acetyl-5-methoxytryptamine&lt;/b&gt;) is a hormone produced in humans, animals and some of plants. It's mainly famous for being the drug suggested to people with sleeping problems, jet lags and even insomnia (in trivial cases, still experimental). Recent studies have shown this compound's role as an anti-oxidant and a protection for mitochondrial DNA. &lt;br/&gt;&#xA;Melatonin interacts with the immune system; and is clinically proved to have anti-inflammatory effects. It can act as a radio-protective agent...&lt;br/&gt;&#xA;Hold it there. This is chemistry, not biology. First of all, how is Melatonin radio-protective? And what is in that structure that makes it radio-protective? 2nd, what reactions occur in the body, in presence of Melatonin, that cause sleepiness (resources have claimed the chemical function to be a mystery but I doubt that's true)?&#xA;&lt;br/&gt; Most of the info above is available at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Melatonin&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia: Melatonin&lt;/a&gt;.&lt;br/&gt;&lt;img src=&quot;https://i.stack.imgur.com/C2FpF.png&quot; alt=&quot;&quot;&gt;&lt;br/&gt;image credit: &lt;a href=&quot;http://cenblog.org/the-haystack/files/2011/06/melatonin.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cenblog.org/the-haystack/files/2011/06/melatonin.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7499" LastEditDate="2014-12-23T19:41:35.533" LastActivityDate="2014-12-23T19:41:35.533" Title="How does Melatonin put you to sleep? How is this compound radio-protective?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22022" PostTypeId="2" ParentId="22021" CreationDate="2014-12-23T19:40:41.593" Score="2" Body="&lt;p&gt;Melatonin is not &quot;radio-protective&quot; in the sense that it blocks radiation or anything. As you note, it's an antioxidant. One of the ways in which ionizing radiation can interact with the body is by providing enough energy to break chemical bonds, producing highly reactive radical species that can cause biological damage by reacting with various things in the body. Melatonin can easily be oxidized by many of these reactive species, preventing them from reacting with anything else. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1600-079X.2010.00847.x/abstract&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; shows the products for the reaction with radio-generated hydroxyl radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The role melatonin plays in sleep is fairly well known (though the effects of exogenous melatonin, less so). Without going into too much detail, there are melatonin receptors in the body, found in cell membranes in the brain and a few other places (G protein-coupled receptors). When these bind melatonin on the outside of the cell, their conformation changes, further activating signalling processes on the inside of the cell. The effects of almost all signalling molecules in the body work in this way. The molecules activate some kind of receptor, which then activates a sequence of other signals, eventually resulting in biological effects.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-23T19:40:41.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22023" PostTypeId="1" AcceptedAnswerId="22030" CreationDate="2014-12-23T20:30:19.263" Score="4" ViewCount="464" Body="&lt;p&gt;In &lt;a href=&quot;http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clayden et al.&lt;/a&gt; on p. 696 the following nitroaldol reaction is listed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XI1Oq.png&quot; alt=&quot;Nitroaldol reaction&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later in the same chapter they explain the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cannizzaro_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cannizzaro reaction&lt;/a&gt;. Why is the yield of this nitroaldol reaction particularly good, while a Cannizzaro reaction is also competing for the benzaldehyde? Is the nitroaldol reaction that much faster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked in some literature and they mostly seem to use nitrogen bases (cyclohexylamine, $\ce{Et3N}$, $\ce{n-BuNH2}$, methylamine, etc.) for these kind of reactions (nitroaldol reaction with aromatic electrophile). The reported yields are also typically lower. See for example the &quot;Literature method&quot; section of &lt;a href=&quot;https://www.erowid.org/archive/rhodium/chemistry/phenyl-2-nitropropene.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the synthesis of 1-phenyl-2-nitropropene on Erowid&lt;/a&gt; (which seems to be some kind of precursor to amphetamines, so there is plentiful information about it in comparison to the 1-phenyl-2-nitroethene synthesis listed in Clayden. Can the yields be so different due to the steric hindrance of the additional methyl group?). What is the benefit of these nitrogen bases over $\ce{NaOH}$? Clayden et al. only mentions them with the comment that $\ce{NaOH}$ is not really necessary since nitrogen bases are sufficiently strong for nitroaldol reaction, because the typical pKa value of short chain nitro alkyl compounds is ~10.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="10185" LastEditDate="2015-06-17T13:10:44.197" LastActivityDate="2015-06-17T13:10:44.197" Title="Cannizzaro reaction versus Nitroaldol reaction" Tags="&lt;organic-chemistry&gt;&lt;nitro-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22024" PostTypeId="1" CreationDate="2014-12-23T21:18:21.807" Score="2" ViewCount="297" Body="&lt;p&gt;Is there a formula that shows how the density of water is affected by temperature?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, with the speed of sound it can be shown how temperature will affect the speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = 331\ \mathrm{m/s} + \left(0.6\ \mathrm{\frac{m/s}{°C}}\right)T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar formula that relates the density of a liquid to temperature? (in this case, water)&lt;/p&gt;&#xA;" OwnerUserId="11423" LastEditorDisplayName="user7951" LastEditDate="2016-10-11T13:45:42.380" LastActivityDate="2016-10-11T13:57:13.933" Title="Density of water and temperature" Tags="&lt;water&gt;&lt;temperature&gt;&lt;density&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22025" PostTypeId="1" AcceptedAnswerId="22028" CreationDate="2014-12-23T21:21:29.793" Score="4" ViewCount="3389" Body="&lt;p&gt;I would like to find out how long it would take to completely turn 50 litres of water to steam. Let's assume power of 2500 W and the water is already at boil. I am not very scientific so please forgive any omissions. Thanks&lt;/p&gt;&#xA;" OwnerUserId="11422" LastEditorUserId="-1" LastEditDate="2019-04-30T21:56:07.610" LastActivityDate="2020-10-05T01:18:49.827" Title="Boil to dry 50 liters water" Tags="&lt;water&gt;&lt;heat&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22026" PostTypeId="2" ParentId="22019" CreationDate="2014-12-23T21:31:16.210" Score="4" Body="&lt;p&gt;Since sound waves are &lt;a href=&quot;https://en.wikipedia.org/wiki/Longitudinal_wave&quot; rel=&quot;nofollow&quot;&gt;longitudinal waves&lt;/a&gt; (generally in air) there needs to be some kind of interaction between neighboring &quot;air&quot; molecules (a mixture consisting mostly of $\ce{N2}$, $\ce{O2}$, $\ce{CO2}$ and $\ce{Ar}$). You can see that this is just not possible if the mean free path of the &quot;air&quot; molecules is larger than the desired wavelength. How would the &quot;air&quot; molecules interact on such distance if there are no &quot;air&quot; molecules to interact with? This interaction (i.e. sound of a particular wavelength) will only be possible if there are sufficiently many &quot;air&quot; molecules nearby to interact with, or formulated otherwise: if the mean free path is sufficiently much smaller than the desired wavelength of the sound. Of course, it is possible to make sound waves with a larger wavelength than the length of the mean free path, just not the converse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One application I can think of lies more in Physics but whatever. In the days before Einsteins theory of special relativity was developed/accepted, there was a common consensus, which was supported by Maxwell himself, that light was travelling in a special reference frame called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aether_theories&quot; rel=&quot;nofollow&quot;&gt;The Aether&lt;/a&gt;&quot;. In this reference frame the speed of light equaled $c$. We now know of course that light can travel through vacuum and that there exists no such special reference frame as The Aether. There were various problems with this Aether theory, most famously the &lt;a href=&quot;https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment&quot; rel=&quot;nofollow&quot;&gt;Michelson-Morley&lt;/a&gt; experiment that suggested that either the Aether did not exist or was moving for a weird reason with the exact same trajectory and speed as earth in space. Another problem, related to your question, was that the speed of waves in a medium is dependent on the particle density of that medium. And since (at least some) electromagnetic waves have an incredible small wavelength (the $\lambda$ of visible light for example was known), the density of this Aether medium had to be extremely high, yet it had to also weight almost nothing and be able to penetrate trough everything!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this and other facts Einstein must have concluded that the Aether theory was in fact invalid. This idea ultimately lead to the theory of special and general relativity, both of which have many spectacular and far reaching consequences.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-23T21:31:16.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22028" PostTypeId="2" ParentId="22025" CreationDate="2014-12-23T22:30:06.447" Score="16" Body="&lt;p&gt;What actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the boiling point, then all we have to do is calculate the total amount of energy needed to vaporize all the water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enthalpy of vaporization of water is 40.65 $\mathrm{kJ/mol}$, which means 40.65 kJ of heat needs to be added to turn one mole of liquid water to steam. 50 L of water is 2775 moles:&#xA;$$\mathrm{\frac{(50\ L)\left(1000\frac{g}{L}\right)}{\left(18.02\frac{g}{mol}\right)}=2775\ mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total heat we need is:&#xA;$$\mathrm{(2775\ mol)\left(40.65\frac{kJ}{mol}\right)=112.8\ MJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 watt is one joule per second, so for a 2500 W heat source:&#xA;$$\mathrm{\frac{112.8\ MJ}{2500\ J/s}=45\ ks}$$&#xA;Or, approximately 12.5 hours. In real life, it may be significantly longer for various reasons, but this is the fastest it could go.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-23T22:30:06.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22029" PostTypeId="2" ParentId="22020" CreationDate="2014-12-23T23:33:56.603" Score="3" Body="&lt;p&gt;The size of a molecule does not correlate with the color of a molecule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Color depends upon which wavelengths of light are absorbed and which are transmitted by a molecule.  Visible light typically ranges from around 390 to 700 nm.  Light with a wavelength between approximately 635 to 700 nm appears red to the human eye.  If we have a molecule that absorbs all visible light below 635 nm, the molecule will appear red; it absorbs all visible light &lt;strong&gt;except&lt;/strong&gt; for the red wavelengths, hence only the red wavelengths will be reflected or transmitted and be available for the eye to detect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What determines which wavelengths of light are absorbed by a molecule?&lt;/strong&gt;&lt;br&gt;&#xA;This is controlled by the energy separation (energy gap) between the highest energy orbital that contains an electron (the HOMO or highest occupied molecular orbital) and the lowest energy orbital that an electron can be promoted to (the LUMO or lowest unoccupied molecular orbital). If this energy gap falls in the visible range, then the molecule will appear colored to the eye when the molecule absorbs light at the wavelength that corresponds to the energy gap (&lt;em&gt;e.g.&lt;/em&gt; when a photon is absorbed that promotes an electron from the HOMO to the LUMO) .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules like 2,2-dimethylpropane contain only sigma bonds.  The energy gap in typical sigma bonds is very large meaning that it takes a lot of energy to promote an electron from the HOMO to the LUMO.  The amount of energy required typically falls in the ultra-violet region, outside the visible range; therefore these molecules are usually colorless since they transmit all visible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules containing pi bonds have smaller HOMO-LUMO energy gaps than molecules with just sigma bonds.  If there are enough pi bonds and they are conjugated then the gap can be small enough that light in the visible range can be absorbed and make the molecule appear colored.  Here is a diagram showing how the energy gap decreases as we increase the length of our conjugated pi system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ovxjz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with 1,3,5-hexatriene the light absorption at 258 nm is still outside the visible  range, so the molecule will not be colored.  But as we continue to extend the conjugation, eventually the molecule will absorb light around 400-500 nm, the blue light range.  As the molecule absorbs blue light it will appear yellow. In addition to pi conjugation, certain substituents such as the nitro group and the carbonyl, to just name two, can also serve to extend the conjugation path in a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule you asked about, penta-1,2,3,4-tetraene, looks roughly like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/prJAh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we examine it closely, we'll notice that the orbitals really look like those in two distinct (orthogonal) 1,3-butadiene molecules.  As the chart above shows, 1,3-butadiene does not absorb in the visible range.  Nonetheless, because penta-1,2,3,4-tetraene does have an extended pi system it will have a smaller HOMO-LUMO gap than 2,2-dimethylpropane and will absorb light at longer ultra-violet wavelengths (lower energy) than 2,2-dimethylpropane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-24T01:52:36.027" LastActivityDate="2014-12-24T01:52:36.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22030" PostTypeId="2" ParentId="22023" CreationDate="2014-12-23T23:59:13.823" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the nitroaldol reaction that much faster?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My guess would be that the nitroaldol is faster than the Cannizzaro.  The Cannizzaro reaction involves&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate&lt;/li&gt;&#xA;&lt;li&gt;alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)&lt;/li&gt;&#xA;&lt;li&gt;hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The nitroaldol involves&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;generation of the nitroalkyl carbanion&lt;/li&gt;&#xA;&lt;li&gt;attack by the carbanion on the carbonyl to form a tetrahedral intermediate&lt;/li&gt;&#xA;&lt;li&gt;capture of a proton by the negatively charged oxygen to produce a neutral molecule&lt;/li&gt;&#xA;&lt;li&gt;elimination of water to produce a more stable (more conjugated) molecule than we started with (&lt;em&gt;e.g.&lt;/em&gt; energetically downhill)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the yields be so different due to the steric hindrance of the&#xA;  additional methyl group?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the benefit of these nitrogen bases over $\ce{NaOH}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-23T23:59:13.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22032" PostTypeId="1" AcceptedAnswerId="22042" CreationDate="2014-12-24T03:37:04.573" Score="1" ViewCount="8932" Body="&lt;p&gt;When an alkene bonds with an H+ ion, the electron pair from the pi bond goes towards a new dative covalent bond with the hydrogen ion, leaving, on one side of the old double bond, a carbon bonded to an extra hydrogen, and on the other side, a carbocation with a positive charge. If this gave, for example, a tertiary carbocation, I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge? I understand that it has lost electrons that were previously in the pi bond, but I don't see how it has a positive charge given that it still has more electrons than protons.&lt;/p&gt;&#xA;" OwnerUserId="11426" LastEditorUserId="11426" LastEditDate="2014-12-24T13:20:51.953" LastActivityDate="2018-03-05T12:32:23.333" Title="How does a carbocation have a positive charge?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;formal-charge&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22033" PostTypeId="1" CreationDate="2014-12-24T08:38:48.883" Score="3" ViewCount="8344" Body="&lt;p&gt;Here's a quote from Petrruci General Chemistry (pg. 160):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver chloride, $\ce{AgCl}$ is an insoluble ionic compound. When $\ce{AgCl}$ dissolved in water, it is 100% dissociated into $\ce{Ag}^+\text{ and } \ce{Cl}^-$ ions; there are no $\ce{AgCl}$ pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If $\ce{AgCl}$ is insoluble, how is it dissolved in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If  $\ce{AgCl}$ is a strong electrolyte but insoluble, does it mean a  $\ce{AgCl}$ molecule is ionized, but isn't separated - i.e. the silver and chloride remains close (they're not solvated by water molecules) but each becomes and ion?&#xA;I'm confused, can someone explain what's going on?&lt;/p&gt;&#xA;" OwnerUserId="11429" LastEditorUserId="7499" LastEditDate="2014-12-24T09:03:57.840" LastActivityDate="2021-04-22T07:32:13.307" Title="How can an insoluble compound be a strong electrolyte" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22035" PostTypeId="2" ParentId="22033" CreationDate="2014-12-24T08:58:48.827" Score="5" Body="&lt;p&gt;It sounds contradictory, but soluble and insoluble are relative terms. Silver chloride's $\mathrm{K_{sp}}$ is $1.77\times 10^{-10}$, so one can generally think of it as insoluble, but actually about a milligram will dissolve in a litre of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much like the terms strong acid and weak acid, strong electrolyte and weak electrolyte refer to the dissociation of a substance in a solvent, though they include all electrolytes, not just acids. In the case of silver chloride, though little of it dissolves, what does is present only as $\ce{Ag+}$ and $\ce{Cl-}$, not solvated  $\ce{AgCl}$. In water, all salts are strong electrolytes, but in other solvents, things can be different. For example, while perchloric acid is a strong electrolyte in water, it doesn't dissociate completely in acetic so it's a weak electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: (1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-24T08:58:48.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22036" PostTypeId="1" AcceptedAnswerId="22039" CreationDate="2014-12-24T09:46:37.313" Score="2" ViewCount="35528" Body="&lt;p&gt;I'm having a difficulty understanding the following quote from &lt;a href=&quot;http://en.wikipedia.org/wiki/Dissociation_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia - Dissociation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Acetic acid is extremely soluble in water, but most of the compound dissolves into molecules, rendering it a weak electrolyte.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To what (non-ionic) molecules can acetic acid be dissolved? If the answer is $\ce{CH3COOH}$ then in what way is it extremely soluble, if it dissolved to itself?&lt;/p&gt;&#xA;" OwnerUserId="11430" LastEditorUserId="402" LastEditDate="2017-01-12T02:44:40.283" LastActivityDate="2018-04-12T13:53:27.697" Title="How is the dissolution of acetic acid that makes its aqueous solution a poor electrolyte?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22038" PostTypeId="1" AcceptedAnswerId="22045" CreationDate="2014-12-24T12:10:17.963" Score="4" ViewCount="1967" Body="&lt;p&gt;$m_\text{e}=9.1\times10^{-31}\ \mathrm{kg}$, $T=2\times10^{6}\ \mathrm{K}$ and $p=0.030\ \mathrm{Pa}$ the solution is given as: $5.4\times10^6\ \mathrm{ms^{-1}}$ but I got $9.5\times10^6\ \mathrm{ms^{-1}}$ using the following method (which does not use the value given for pressure. I simply plugged the numbers into the equation below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_\text{rms}=\sqrt{\frac{3K_\text{B}T}{m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-04-22T19:53:28.220" LastActivityDate="2016-08-22T06:19:32.813" Title="Calculate the root-mean-squared speed of free electrons in the Sun's corona given pressure, temperature and mass of an electron" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22039" PostTypeId="2" ParentId="22036" CreationDate="2014-12-24T12:10:52.050" Score="8" Body="&lt;p&gt;There are two questions being asked here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is aqueous acetic acid a weak electrolyte?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When acetic acid is dissolved in water there is an equilibrium reaction:&#xA;$$\ce{CH3COOH + H2O &amp;lt;=&amp;gt; CH3COO- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is acetic acid highly soluble in water?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-24T12:10:52.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22040" PostTypeId="1" AcceptedAnswerId="22049" CreationDate="2014-12-24T12:40:24.680" Score="2" ViewCount="440" Body="&lt;p&gt;To derive the Maxwell-Boltzmann distribution the first step requires the knowledge that the equation will take the form $$f(v)=Ke^{-\epsilon/ kT}$$ where $\epsilon$ is the sum of the kinetic energies in the directions of the x,y and z axes. I understand that there needs to be exponential decay and why there will need to a constant. Also, I understand that the eponent needs to be dimensionless but why is KT used? This seems to pop up everywhere without much of an explaination of why (as in this case).  Can you help me understand how to get this first step? Thanks. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2016-09-30T10:02:09.953" LastActivityDate="2016-09-30T10:14:23.387" Title="How to get the first step of the derivation of the Maxwell-Boltzmann distribution" Tags="&lt;physical-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22041" PostTypeId="2" ParentId="22014" CreationDate="2014-12-24T12:53:16.467" Score="4" Body="&lt;p&gt;Liquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on &lt;a href=&quot;http://www.wikiwand.com/en/Liquid_crystal&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not &quot;turned on&quot; by the field, just altered and that is enough to make a viable display technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this &lt;a href=&quot;http://www.slideshare.net/MerckChemicals/merck-chemicals-100-years-of-liquid-crystals-at-merck-3477877&quot; rel=&quot;noreferrer&quot;&gt;slideshow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/95Hm0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2014-12-24T12:53:16.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22042" PostTypeId="2" ParentId="22032" CreationDate="2014-12-24T13:39:22.183" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-24T13:39:22.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22043" PostTypeId="1" AcceptedAnswerId="22048" CreationDate="2014-12-24T13:47:10.723" Score="11" ViewCount="56392" Body="&lt;p&gt;I want to know whether increasing the concentration of salt, specifically $\ce{NaCl}$, increases or decreases the rate of rusting. There are conflicting theories that explain opposite outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Increasing concentration of salt decreases the amount of dissolved oxygen and thus    &lt;strong&gt;decreases&lt;/strong&gt; the rate of rusting&lt;/li&gt;&#xA;&lt;li&gt;Increasing concentration increases electron mobility and thus &lt;strong&gt;increases&lt;/strong&gt; rate of rusting&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to how rusting works in full detail nor have I been able to understand why there are two contradicting theories present. It seems that people write it more out of opinion than with factual basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please clarify?&lt;/p&gt;&#xA;" OwnerUserId="11431" LastEditorUserId="23561" LastEditDate="2019-03-15T15:52:37.903" LastActivityDate="2019-03-15T15:52:37.903" Title="Does concentration of salt increase or decrease rate of rusting?" Tags="&lt;electrochemistry&gt;&lt;solutions&gt;&lt;concentration&gt;&lt;corrosion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22044" PostTypeId="1" AcceptedAnswerId="22046" CreationDate="2014-12-24T14:10:34.053" Score="1" ViewCount="3387" Body="&lt;p&gt;The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2014-12-24T16:23:05.407" Title="How do derive the equation for relative velocity of gas molecules" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22045" PostTypeId="2" ParentId="22038" CreationDate="2014-12-24T14:19:19.983" Score="6" Body="&lt;p&gt;I'll go out on a limb here and say your answer is right and theirs is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the answers differ by a factor of the square root of 3, so I'm guessing they forgot to include the factor of 3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be a small &lt;a href=&quot;http://www.marcelhaas.com/docs/maxrel.pdf&quot;&gt;correction for relativistic effects&lt;/a&gt;, but not enough to explain the differences in the two answers.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-24T14:19:19.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22046" PostTypeId="2" ParentId="22044" CreationDate="2014-12-24T14:46:55.370" Score="1" Body="&lt;p&gt;Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = m_1m_2/(m_1 + m_2)$ by definition.  If $m_1 = m_2$ then $\mu = m/2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there are two different masses, there are two different mean velocities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see where the equation with the reduced mass comes from, look at appendix 1.4 of &lt;a href=&quot;http://www.cosbkup.gatech.edu/group/chem780/CHAPT1.pdf&quot; rel=&quot;nofollow&quot;&gt;Kinetic Theory of Gases&lt;/a&gt;, along with pages 1-30 to 1-34.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2014-12-24T16:23:05.407" LastActivityDate="2014-12-24T16:23:05.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22047" PostTypeId="2" ParentId="22040" CreationDate="2014-12-24T14:55:27.837" Score="1" Body="&lt;p&gt;The Maxwell-Boltzmann distribution of speeds is derived from the Boltzmann distribution, which says that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{N_i}{N_j}=e^{-(\epsilon_i-\epsilon_j)/kT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula gives the ratio of the number of particles ($N_i$, $N_j$) in states with energy $\epsilon_i$ and $\epsilon_j$. From this it is easy to the that $f(v)=Ke^{-\epsilon/ kT}$ as you noted.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-24T14:55:27.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22048" PostTypeId="2" ParentId="22043" CreationDate="2014-12-24T15:06:05.043" Score="15" Body="&lt;p&gt;In general, salt (particularly NaCl) will &lt;strong&gt;increase&lt;/strong&gt; the rate of corrosion (rusting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, consider metallic iron $\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\ce{Fe2O3}$ in the presence of oxygen $\ce{O2}$ and water $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Fe +3O2 +6H2O-&amp;gt;4Fe(OH)3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrosion (rust) is a 'redox' reaction, which means it involves reduction of oxygen into hydroxide ions and oxidation of metallic iron to iron cations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any redox reaction to take place, electrons are transferred. In this case, from the metallic iron which loses electrons to oxygen. The presence of salt (or any electrolyte) in the water accelerates the reaction because it increases the conductivity of water, effectively increasing the concentration of ions in the water and so increasing the rate of oxidation (corrosion) of the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is complicated by the fact that salt dissolved in water actually reduced the amount of dissolved oxygen present in the water. This is because the water molecules are attracted to the dissolved ions from the salt (&lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation&quot; rel=&quot;noreferrer&quot;&gt;solvation&lt;/a&gt;), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby deriving dissolved oxygen out. So if the metal is totally submerged in water, higher salt concentrations can actually reduce the rate of corrosion, if the water is not aerated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it depends upon the amount of oxygen getting into the water and onto the metal. For example, underwater plants can release oxygen in salty water which will accelerate the rate of corrosion of the submerged metal. Of course, if the metal is not completely submerged in salt water but only partially submerged, or is affected by salt water 'spray', it will be still be exposed to oxygen and will rust faster.&lt;/p&gt;&#xA;" OwnerUserId="708" LastEditorUserId="-1" LastEditDate="2018-05-19T12:56:51.663" LastActivityDate="2018-05-19T12:56:51.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22049" PostTypeId="2" ParentId="22040" CreationDate="2014-12-24T15:10:57.213" Score="2" Body="&lt;p&gt;KT comes in from requiring agreement with the ideal gas law.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &quot;section 1.5.3 $\langle v^2\rangle$ should agree with the ideal gas law&quot; and section 1.3 &quot;The functional form of the dependence of T on $\langle\varepsilon\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)&quot; in &lt;a href=&quot;http://www.cosbkup.gatech.edu/group/chem780/CHAPT1.pdf&quot; rel=&quot;nofollow&quot;&gt;Kinetic Theory of Gasses&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="19383" LastEditDate="2016-09-30T10:14:23.387" LastActivityDate="2016-09-30T10:14:23.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22053" PostTypeId="1" CreationDate="2014-12-24T17:04:51.053" Score="1" ViewCount="79" Body="&lt;p&gt;What is the specification of Martensite (displacive) transformation in ceramics comparing to the metals? I just know about Martensite transformation in metals. For example, these transformations are thermal and diffusionless. Also, the kinetic and morphology are dominant by strain energy.&lt;/p&gt;&#xA;" OwnerUserId="11435" LastActivityDate="2015-02-27T03:25:22.210" Title="Martensitic transformation in ceramics" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22055" PostTypeId="2" ParentId="21589" CreationDate="2014-12-24T17:12:39.343" Score="2" Body="&lt;p&gt;Use osmotic pressure $Π=iMRT$&lt;br&gt;&#xA;From this get your concentration and calculate molarity of solution.&#xA;By the way if you are looking more on this topic then you must consider the chapter on solutions and their colligative properties in your chemistry textbooks&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="5026" LastEditDate="2018-03-08T09:51:19.160" LastActivityDate="2018-03-08T09:51:19.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22056" PostTypeId="1" AcceptedAnswerId="22059" CreationDate="2014-12-24T17:13:53.950" Score="-2" ViewCount="15910" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following is the strongest base?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{LiOH}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{CH3Li}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{LiNH2}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{LiF}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me the answer should be $\ce{LiF}$ as it has max number of lone pairs but the answer says its 2). I am confused.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="5026" LastEditDate="2018-02-28T13:58:18.190" LastActivityDate="2018-02-28T14:05:27.407" Title="Which of the following is the strongest base?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22058" PostTypeId="1" AcceptedAnswerId="22061" CreationDate="2014-12-24T18:07:43.397" Score="15" ViewCount="3285" Body="&lt;p&gt;When it gets to Carbon-based molecules, one very possible structure when there are more than six C atoms is the hexagon; though not mostly perfect, it emphasizes that six Carbon atoms tend to bond with each other.&lt;sup&gt;&lt;code&gt;*&lt;/code&gt;&lt;/sup&gt; Why? &lt;br/&gt;&#xA;I assume this is because of the &lt;em&gt;&quot;stability&quot;&lt;/em&gt; that is attributed to the hexagon. Explainers have discussed mainly about this issue.&lt;br/&gt;&#xA;Neither me nor my teacher were convinced by the description; (Actually, all of this is because of my teacher!) since he asked that for example, why not octagon? No response.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered that if I asked this in here what ideas I am going to face.... Sorry, but I'm accustomed to doing lots of &quot;whys&quot;.&lt;br/&gt;&#xA;&lt;sup&gt;*&lt;/sup&gt;For example; look at the pic below:&lt;br/&gt;&lt;img src=&quot;https://i.stack.imgur.com/wd492m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image above, the number of hexagonal rings is twice the number of pentagonal rings and eight times more than octagonal rings. This macro-molecule is the micro of what's happening in the nature. I don't have citations; but it takes a minute or two to see how much the number of hexagonal rings is surpassing the number of other rings. Perhaps it was so obvious, nobody mentioned it.&lt;br/&gt;&#xA;Update: According to @ron, and with a little payment of thought, I reached the fact that these rings aren't necessarily geometric; the important thing is that six C atoms, instead of five or eight or whatever else number of C atoms are the majority.&lt;br/&gt;&#xA;Image credit: &lt;a href=&quot;https://inspirehep.net/record/1281242/plots&quot; rel=&quot;noreferrer&quot;&gt;Pseudopotential model for Dirac electrons in Graphene...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditorDisplayName="user6887" LastEditDate="2015-06-14T12:50:11.973" LastActivityDate="2020-10-15T16:16:35.047" Title="Why six C atoms are usually seen in cyclic compounds?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;atoms&gt;&lt;energy&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22059" PostTypeId="2" ParentId="22056" CreationDate="2014-12-24T18:09:44.813" Score="8" Body="&lt;p&gt;In order of decreasing basicity they are ranked $\ce{LiCH3}$ &gt; $\ce{LiNH2}$ &gt; $\ce{LiOH}$ &gt; $\ce{LiF}$. We are dealing with the ions $\ce{F-}$, $\ce{OH-}$, $\ce{NH2-}$ and $\ce{CH3-}$ and so the factor which influences their basicity is the stability of the ion compared to its conjugate acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability of the ions is determined by the energy of the lone pairs. Carbon has the least effective nuclear charge ($Z_{\text{eff}}$) and so the lone pair on the $\ce{CH3-}$ ion has the highest energy, making it the least stable ion and so the strongest base. The $Z_{\text{eff}}$ increases as we move across the period until we get to fluorine which has the greatest $Z_{\text{eff}}$ and so its lone pairs have the lowest energy, making $\ce{F-}$ the least basic ion. &lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-02-28T14:04:43.133" LastActivityDate="2018-02-28T14:04:43.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22060" PostTypeId="2" ParentId="22056" CreationDate="2014-12-24T18:13:59.540" Score="7" Body="&lt;p&gt;Look at the conjugate acid of each of the anions (the base). If the conjugate acid is a strong acid (low pKa), then the anion is a weak base. If the conjugate acid is a weak acid (high pKa), then the anion is a strong base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's list the anion or base, the conjugate acid and the pKa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|} \hline&#xA;\text{Base }&amp;amp; \text{Conjugate Acid}&amp;amp; \mathrm pK_\mathrm a \\ \hline&#xA;\ce{OH-} &amp;amp; \ce{H2O} &amp;amp; 15.7\\ \hline&#xA;\ce{CH3-} &amp;amp; \ce{CH4}  &amp;amp; 48 \\ \hline&#xA;\ce{NH2-} &amp;amp; \ce{NH3} &amp;amp; 38 \\ \hline&#xA;\ce{F-} &amp;amp; \ce{HF} &amp;amp; 3 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate acid with the highest pKa is methane ($\ce{CH4}$) it is the weakest acid of the group. Therefore the $\ce{CH3-}$ in $\ce{MeLi}$ is the strongest base in the group.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-02-28T14:05:27.407" LastActivityDate="2018-02-28T14:05:27.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22061" PostTypeId="2" ParentId="22058" CreationDate="2014-12-24T18:34:56.573" Score="19" Body="&lt;p&gt;&lt;strong&gt;It is all about minimizing the energy of a molecule.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis).  In this case the hexagonal geometry is adopted because all of the carbons are &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized.  The ideal geometry (lowest energy) for an &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized carbon involves three 120 degree bond angles around the carbon atom.  This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uzlD3.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Other carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions.  For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a &amp;quot;chair conformation&amp;quot; in its ground state where the C-C-C angle is around 109.5 degrees.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/maiiu.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;why not octagon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Some other regular polygon shapes can be found in molecules.  For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon.  Both of these molecules contain only &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized carbons.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ybLD.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles.  However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively.  Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What other ring shapes are preferred, low-energy minima?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We've already discussed the &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure.  &lt;span class=&quot;math-container&quot;&gt;$\ce{sp}$&lt;/span&gt; hybridized carbon is most stable when the X-C(&lt;span class=&quot;math-container&quot;&gt;$\ce{sp}$&lt;/span&gt;)-X bond angle is 180 degrees.  Clearly small or medium size rings comprised solely of &lt;span class=&quot;math-container&quot;&gt;$\ce{sp}$&lt;/span&gt; hybridized carbon cannot exist as stable entities.  In the case of &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, &lt;em&gt;ca.&lt;/em&gt; 109.47 degrees) the lower the energy of the molecule.  We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment).  If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KtUDv.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is less strain energy in adamantane (per carbon) than in cyclohexane.  The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there.  How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar?  By continuing to add more and more cyclohexane rings we build the following ring structure...&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gAbp0.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;... Diamond, where all of the &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^3}$&lt;/span&gt; carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; response to OPs comment&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;So, the lesser energy state must be the reason that after six number&#xA;of atoms, five is the common one, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Kinda.  Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WR7gs.gif&quot; alt=&quot;enter image description here&quot; /&gt;&#xA;[&lt;a href=&quot;http://ursula.chem.yale.edu/%7Echem220/chem220js/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;  Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]&lt;/p&gt;&#xA;&lt;p&gt;5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common.  We must consider the free energy of the process, enthalpy &lt;strong&gt;and&lt;/strong&gt; entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy.  In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="55119" LastEditDate="2020-09-01T00:39:38.833" LastActivityDate="2020-09-01T00:39:38.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22062" PostTypeId="2" ParentId="22024" CreationDate="2014-12-24T21:55:52.743" Score="1" Body="&lt;p&gt;In general, we can derive the expression for the density of any substance as a function of temperature using the definition of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion&quot; rel=&quot;nofollow&quot;&gt;thermal expansion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm dV}{\mathrm dT}=\alpha V\tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By differentiating $\rho=m/V$, we can show that $\mathrm d\rho/\rho = -\mathrm dV/V$. So using this and equation $\text{(1)}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d\rho}{\mathrm dT}=-\alpha V\tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating this above expression and approximating the expression,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \rho^\circ\cdot\left(1-\alpha\Delta T\right)\tag3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho^\circ$ is the initial density at the initial temperature, $\alpha$ is the volumetric thermal expansion coefficient and $\Delta T$ is the temperature.&lt;br&gt;&#xA;For water, $\alpha$ is typically &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_expansion#Thermal_expansion_coefficients_for_various_materials&quot; rel=&quot;nofollow&quot;&gt;$207\times10^{-6}\ \mathrm{K^{-1}}$&lt;/a&gt; at $20\ \mathrm{^\circ C}$. Using this, we can observe how temperature affects the density of water.&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorDisplayName="user7951" LastEditDate="2016-10-11T13:57:13.933" LastActivityDate="2016-10-11T13:57:13.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22063" PostTypeId="2" ParentId="24" CreationDate="2014-12-25T01:46:08.607" Score="2" Body="&lt;p&gt;While it is primarily constructed to perform titration and equilibrium calculations on metal coordination complexes, once you get used to its syntax &lt;a href=&quot;http://www.hyperquad.co.uk/hyss.htm&quot; rel=&quot;nofollow&quot; title=&quot;HySS&quot;&gt;HySS&lt;/a&gt; can be effectively used to simulate acid/base equilibria and titrations.  One bonus of HySS is that it also treats solids (precipitates), which can be useful for, e.g., poorly soluble phenol derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Windows-only, unfortunately.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2014-12-25T01:46:08.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22064" PostTypeId="2" ParentId="589" CreationDate="2014-12-25T06:32:51.957" Score="2" Body="&lt;p&gt;I read it had to do with dissolved gasses, that &quot;normal tasting&quot; water has the dissolved gas but freezing water you lose the gasses.  So I tried this home experiment.  I froze water then let melt and tasted it, it tasted bad with the off flavor.  Then I took out a large freezer bag and only put a small amount of the foul tasting water.  I then shook it violently for some time to mix air back into the water.  Boom the water tasted fine again.  No double blind here but seriously is it needed?&lt;/p&gt;&#xA;" OwnerUserId="11438" LastActivityDate="2014-12-25T06:32:51.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22066" PostTypeId="2" ParentId="4411" CreationDate="2014-12-25T10:48:56.363" Score="7" Body="&lt;p&gt;&lt;strong&gt;A metalloid, possibly a metal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though nobody has so far seen or been able to test a visible piece of astatine we can extrapolate its properties based on its position in the periodic table. We know that iodine, the element above astatine, has a metallic appearance when viewed under white light, is a semiconductor (in the direction of its planes), and that there is evidence of some delocalised bonding between iodine molecules in crystalline iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given its position in the periodic table astatine could reasonably be expected to show more metallic character than iodine. Consistent with this expectation, the most common oxidation state of astatine appears to be +1 whereas that of the lighter halogens is –1; and there is some evidence to suggest that astatine forms a cation in strongly acidic solutions. In 2013, based on relativistic modelling calculations, condensed astatine was predicted to be a monatomic metal with a face-centred cubic structure. Metals with this structure are noted for their ductility whereas metalloids are regarded as being brittle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If astatine does turn out to be a metal then, like other &lt;a href=&quot;https://en.wikipedia.org/wiki/Post-transition_metal&quot;&gt;post-transition metals&lt;/a&gt;, it could be expected to show appreciable non-metallic character. Trace level experiments on the chemistry of astatine show that it appears to act like a metal in some respects and like a non-metal in others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Metalloid#Astatine&quot;&gt;metalloids&lt;/a&gt; has more information about astatine and its metallic and non-metallic properties. For now, until more evidence comes in, it would be reasonable to classify astatine as a metalloid.&lt;/p&gt;&#xA;" OwnerUserId="11440" LastActivityDate="2014-12-25T10:48:56.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22067" PostTypeId="1" CreationDate="2014-12-25T12:50:37.283" Score="0" ViewCount="101" Body="&lt;p&gt;If something contains large amounts of cyclopentane and cyclohexane, what is it called? I'd like to know what is the something as well. It's a chemistry olympiad question for this year. See &lt;a href=&quot;http://www.iuventa.sk/files/documents/2_olympiady/cho/51.%20ro%C4%8Dn%C3%ADk/%C3%BAlohy%20a%20rie%C5%A1enia/dom%C3%A1ce%20kolo/ch51dkcteul15.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, Question 3-g. Just let google translate it for you. I find some questions quite weird as well, but what can you do. Otherwise its pretty good.&lt;/p&gt;&#xA;" OwnerUserId="11441" LastEditorUserId="865" LastEditDate="2015-10-02T16:44:59.807" LastActivityDate="2015-10-02T16:44:59.807" Title="If something contains large amounts of cyclopentane and cyclohexane, what is it called?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22068" PostTypeId="1" CreationDate="2014-12-25T14:07:27.687" Score="2" ViewCount="85" Body="&lt;p&gt;I've been using aluminium foil and table salt to clean my silverware for a long time without actually concerning myself with the chemistry behind it. A couple of days ago i started digging into the chemistry and found this page on wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow&quot;&gt;Reactivity series&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and when I looked closer i saw that both silver and tin where less reactive than aluminium. Does that mean that I can safely use the same technique with my tinware as with my silverware? &lt;/p&gt;&#xA;" OwnerUserId="11420" LastEditorUserId="11420" LastEditDate="2014-12-26T21:57:33.243" LastActivityDate="2014-12-26T21:57:33.243" Title="Clean tin with aluminium" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22069" PostTypeId="1" CreationDate="2014-12-25T14:38:30.823" Score="9" ViewCount="3911" Body="&lt;p&gt;This question has a accepted answer on this site already:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2479/what-happens-during-cleaning-silverware&quot;&gt;What happens during cleaning silverware?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the reason I ask the question again is because I found different &quot;recipes&quot; for the technique and also different reaction formulas in other sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;recipe&quot; given in the linked question is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Put [the silverware] into a pot with some club soda and a piece of aluminium foil and pour over a hot water.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the reaction formulas given for &quot;blackening&quot; is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag2O(s) + 2Al(s) -&amp;gt; 3Ag(s) + Al2O3(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the reaction with the aluminium foil&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -&amp;gt; 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also found the following &quot;recipe&quot; in a &lt;a href=&quot;http://www.nyteknik.se/popular_teknik/smatt_gott/article2495688.ece&quot; rel=&quot;nofollow noreferrer&quot;&gt;Swedish newspaper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Put aluminium foil in a bowl. Put some baking powder on the foil. Put in the silverware. Add more baking powder. Pour over hot water and wait.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this recipe is more or less equivalent to the one given earlier since the baking powder essentially turns the water into carbonated water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction formula for the &quot;blackening&quot; of the silver given in the newspaper is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag + sulfurcompunds + O2 -&amp;gt; Ag2S + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the reaction with the aluminium is described as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Al(s) + 3 Ag2S(s) + 6 H2O(l) - 6 Ag(s) + Al2O3(s) + 3 H2S(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The newspaper also links to &lt;a href=&quot;http://school.chem.umu.se/Experiment/P98&quot; rel=&quot;nofollow noreferrer&quot;&gt;a small piece written at a University in Sweden with regards to this question.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where they explain the blackening of the silver in the same way as the newspaper but they give a different reaction formula for the aluminium reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Al + 3 Ag2S -&amp;gt;	6 Ag(s) + Al2S3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: &lt;strong&gt;Is any one of these reaction more &quot;true&quot; (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11420" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-05-23T13:19:11.053" Title="What is the chemistry behind cleaning silverware with aluminium foil" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;silver&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22070" PostTypeId="2" ParentId="22067" CreationDate="2014-12-25T15:23:57.730" Score="1" Body="&lt;p&gt;The most likely name would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycloalkane&quot; rel=&quot;nofollow&quot;&gt;naphthenes&lt;/a&gt;. It could also be called cycloalkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;Since the reference to the question is about the coal and oil production, the answer would be naphthenes. When the coal and oil were distilled, a layer (called naphthene) was removed which was found to contain mainly cyclopentane and cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would complete the statement as:&lt;br&gt;&#xA;If a &lt;strong&gt;fraction of distillate&lt;/strong&gt; contains large amounts of cyclopentane and cyclohexane, it is called &lt;strong&gt;naphthene&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-12-25T17:43:47.943" LastActivityDate="2014-12-25T17:43:47.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22071" PostTypeId="1" AcceptedAnswerId="22074" CreationDate="2014-12-25T16:56:50.210" Score="13" ViewCount="40048" Body="&lt;p&gt;We all know the very famous esterification reaction, where a carboxylic acid and an alcohol react to give an ester and water. But why do they actually smell? What is the reason behind its aromatic character? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially thought that it must be due to the conjugation between the two oxygens present. I do know that aromaticity is decided by the conjugated lone pairs present in the compound where the compound must follow Hückels rule. But it's only valid for a cyclic and a planar compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do esters have that fruity smell? What is the chemical explanation for a smelly non planar compound?&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="7475" LastEditDate="2016-10-29T13:39:13.990" LastActivityDate="2016-11-14T20:45:05.160" Title="Why do esters actually smell?" Tags="&lt;organic-chemistry&gt;&lt;esters&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22073" PostTypeId="1" AcceptedAnswerId="22077" CreationDate="2014-12-25T18:31:50.697" Score="1" ViewCount="1371" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5meF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer has to be both A and D. Chromic acid is a strong oxidizer and will take both aldehydes and primary alcohols to carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the computer thinks otherwise. Am I just missing something really simple here?&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-04-16T01:46:09.793" LastActivityDate="2018-04-16T01:46:09.793" Title="Chromic acid oxidation of aldehydes and primary alcohols" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;organic-oxidation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22074" PostTypeId="2" ParentId="22071" CreationDate="2014-12-25T18:40:29.317" Score="8" Body="&lt;p&gt;The term aromaticity originated with the discovery of unusually stable hydrocarbons that also happened to have strong smells. Many hydrocarbons smell, but not all are aromatic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, a compound being classified as &quot;aromatic&quot; has little to nothing to do with its smell and everything to do with its electron configuration. There are also non-aromatic compounds and anti-aromatic compounds. If we try to relate these to smell as well, what would an anti-aromatic compound smell like? Bad? But isn't that a little too subjective for science? Doesn't benzene smell &quot;bad&quot; too? Then why is it aromatic (i.e. unusually resistant to chemical change?) Why not anti-aromatic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Esters smell partly because they exhibit weak intermolecular forces. This allows ester molecules to enter the gas phase and reach your nose. Esters don't exhibit intermolecular hydrogen bonding, unlike alcohols, for example. These are no strongly positively polarized hydrogens in esters to participate in hydrogen bonding. Consider for example ethyl butyrate, which smells like pineapples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9Xvt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the molecule resembles a plain aliphatic hydrocarbon! And we know these only exhibit weak van der Waals intermolecular forces. No wonder esters smell (good &lt;em&gt;and&lt;/em&gt; bad) - they're volatile and reach our noses easily! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-26T15:46:22.680" LastActivityDate="2014-12-26T15:46:22.680" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22075" PostTypeId="1" AcceptedAnswerId="23748" CreationDate="2014-12-25T18:40:29.590" Score="4" ViewCount="1985" Body="&lt;p&gt;Black tea is an amazing product of the green tea leaves. A &lt;a href=&quot;http://www.nature.com/index.html?file=/bdj/journal/v190/n7/full/4800977a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent study&lt;/a&gt; has proven that;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The anion profile for the tea solution indicated that oxalate and citrate were the major anions detected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This resulted in the fact that a $\pu{1\!\% w\!/\!v}$ solution of normal (World Blend – Tetley USA Inc; you decide it for yourself!) tea has a pH of $\rm 4.9$. Non of these pieces of detailed info were an answer to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, definitely somehow oxalate and citrate managed to dissolve into water as we put tea leaves in it; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf) &lt;/li&gt;&#xA;&lt;li&gt;How and why do they get released from tea leaf? Since you don't usually see any &quot;big&quot; differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)&lt;/li&gt;&#xA;&lt;li&gt;What makes the color of water change? Why does the color vary with pH alteration? (There must be a reaction, what is it?)&lt;br/&gt;&#xA;Pt: These questions are additional questions to the title.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="23561" LastEditDate="2018-09-06T13:46:22.540" LastActivityDate="2018-09-06T13:46:22.540" Title="How is tea affecting water's color?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22076" PostTypeId="2" ParentId="16725" CreationDate="2014-12-25T19:41:58.987" Score="3" Body="&lt;p&gt;Enthalpy of a system is a state function, meaning, it does not depend upon the path you take to reach the end state. What that means in this case is that the change in the enthalpy of combustion will be the heat required to raise the temperature of the combustion material to the new temperature and to increase the pressure. Essentially, think of it as if the combustion is occurring at STP, and you must construct a path from the beginning state to the end state which will pass that combustion at STP. In this case, that would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Initial state of reactants -&gt; STP reactants&lt;/li&gt;&#xA;&lt;li&gt;Heat of combustion at STP&lt;/li&gt;&#xA;&lt;li&gt;STP products -&gt; final state of products&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your final state will be the same as your initial state, 650.9 K/1779 kPa. The important point is that the heat in the change of the reactants to STP will be different from the change of the products back to the initial state. This difference would be reflected if you found a heat of combustion at some condition other than STP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen is a spectator to the reaction, so you can completely ignore it for the sake of reaction. However, you do need to account for raising it to the final pressure/temperature if either is changing. I'll assume you've got that under control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the issue comes in at the fact these are somewhat extreme conditions, at least in pressure. To calculate this properly would take a great deal of time and effort. That said, there may be corners you are capable of cutting in the process. The general process you need to follow is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take octane and $\ce{O2}$ from 1779 kPa to 1 atm, 650.9 K to $T_{1,\mathrm i}$.&lt;/li&gt;&#xA;&lt;li&gt;Take octane and $\ce{O2}$ from $T_\mathrm i$ to 25 °C at constant pressure (using $C_p$).&lt;/li&gt;&#xA;&lt;li&gt;Use the heat of combustion.&lt;/li&gt;&#xA;&lt;li&gt;Take $\ce{H2O}$ and $\ce{CO2}$ from 1 atm to 1779 kPa, 25 °C to $T_{2,\mathrm i}$.&lt;/li&gt;&#xA;&lt;li&gt;Take $\ce{H2O}$ and $\ce{CO2}$ from $T_{2,\mathrm i}$ to 650.9 K.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The overall enthalpy change of this process will be the heat of combustion at your conditions. Now, you'll have to decide for yourself about steps 1 and 4, as you need to be considering intermolecular interactions, and thus need to pick an equation of state (not ideal gas). I feel like I am not necessarily knowledgeable enough in this specific field to offer suggestions in that regard. I personally would just pick something like VDW, but there might be a specific standard for combustion processes. Further, only one book that I have access to right now that has the proper temperature-dependent-equation form for $C_p$, and it doesn't have one for octane. These are the ones for the other gases involved:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2}$: $C_p = 25.46 + 1.519 10^{-2} T - 0.715 10^{-5} T^2 + 1.311 10^{-9} T^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$: $C_p = 22.243 + 5.977 10^{-2} T - 3.499 10^{-5} T^2 + 7.464 10^{-9} T^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O}$: $C_p = 32.218 + 0.192 10^{-2} T + 1.055 10^{-5} T^2 - 3.953 10^{-9} T^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Chemical and Engineering Thermodynamics (Sandler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish I could help more. Hopefully this explanation at least points you in the right direction.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastEditorDisplayName="user7652" LastEditDate="2015-12-17T20:48:32.533" LastActivityDate="2015-12-17T20:48:32.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22077" PostTypeId="2" ParentId="22073" CreationDate="2014-12-25T20:06:48.277" Score="3" Body="&lt;p&gt;There are many $\ce{CrO3}$ oxidation reagents (&lt;a href=&quot;http://en.wikipedia.org/wiki/Jones_oxidation&quot; rel=&quot;nofollow&quot;&gt;Jones&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sarett_oxidation&quot; rel=&quot;nofollow&quot;&gt;Sarett&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Collins_reagent&quot; rel=&quot;nofollow&quot;&gt;Collins&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyridinium_chlorochromate&quot; rel=&quot;nofollow&quot;&gt;PCC&lt;/a&gt;, etc.), however, most are selective and stop the oxidation of a primary alcohol at the aldehyde stage.  On the other hand, the Jones reagent is relatively unselective and will oxidize a primary alcohol to the corresponding carboxylic acid - &lt;strong&gt;if&lt;/strong&gt; enough reagent is present. It takes two-thirds of an equivalent of Jones reagent to oxidize one equivalent of the alcohol to the aldehyde and an additional two-thirds equivalent to oxidize the aldehyde to the acid.  Also the Jones oxidation is run under acidic conditions (typically dilute sulfuric acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have though that A, B (the gem-diol will equilibrate with the aldehyde under acidic conditions) and D were correct answers.  It seems like the computer answer is either in error or it is a &quot;trick&quot; question (seems unlikely for computer-based learning).  The &quot;trick&quot; would be that since an excess of reagent was not specified only 1 equivalent is assumed.  Therefore there is not enough reagent to carry out two sequential oxidations; as a consequence only the aldehyde or gem-diol could be &lt;strong&gt;completely&lt;/strong&gt; oxidized with one equivalent of reagent. But then A or B is not an acceptable answer either.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-25T20:06:48.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22078" PostTypeId="2" ParentId="22071" CreationDate="2014-12-25T20:10:28.433" Score="16" Body="&lt;p&gt;Dissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why many esters smell fruity while most thiols smell rather unpleasant is a rather interesting question. Just as most sugars taste sweet, many molecules that have structural similarities also smell similar. Humans have about &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1753-4887.2004.tb00097.x/abstract&quot;&gt;350 different olfactory receptors&lt;/a&gt;. These are G protein-coupled receptors that change conformation when they bind a compound and activate a signal pathway that is eventually sent to the brain for processing. However, since we only have 350 different receptors and there are far more than 350 odoriferous compounds, there is not a one-to-one mapping of receptors to compounds—there isn't an &quot;acetone receptor&quot;. Instead, olfactory receptors have partial and overlapping selectivity. Each receptor type responds to more than one compound, thiols with a certain general shape, for instance, and different receptor types can respond to some of the same compounds. A given compound then has a more or less unique pattern of receptor activation. All this information gets sent to the brain and is processed into the perception of smell. Compounds that have similar structural properties activate many of the same receptors, so they smell similar.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-25T20:10:28.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22079" PostTypeId="1" AcceptedAnswerId="22082" CreationDate="2014-12-25T22:10:01.200" Score="3" ViewCount="8227" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQ06Q.png&quot; alt=&quot;Chemistry Question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My attempt at the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) 2 bond pairs and 2 lone pairs on oxygen atom therefore bond angle of slightly less than 109.5 degrees, around 104 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) 120 degrees bond angle therefore &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) From what I understand, the sigma-pi separation means that you can neglect interactions between the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p_z}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p_x,p_y}$&lt;/span&gt; orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2}$&lt;/span&gt; carbon would form &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybrid orbitals.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2Gk4I.jpg&quot; alt=&quot;CH2 MO Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt; carbon would form &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybrid orbitals and would then form 3 sigma bonds with the hydrogen leaving only one orbital to interact with the oxygen. I assume that the oxygen &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; orbitals do not match the energy level of the carbon and will form non-bonding orbitals. Therefore the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; sigma bond would be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3-p}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9jTf.png&quot; alt=&quot;CH3 MO Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to use this knowledge to actually draw the MO diagram? I have tried an attempt below. I can draw MO diagrams for simple diatomic molecules. And how would the MO diagram explain the observed 120 degree bond angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ZfPr.png&quot; alt=&quot;my attempt&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iv) I haven't used the MO diagram to draw these resonance structures but I have come up with two based on the octet rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6pqL8.jpg&quot; alt=&quot;Methyl vinyl ether resonance structures&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11445" LastEditorUserId="83758" LastEditDate="2020-05-04T06:27:24.307" LastActivityDate="2020-05-04T06:27:24.307" Title="Draw a simplified MO diagram for the pi system of Methyl vinyl ether" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22080" PostTypeId="2" ParentId="22007" CreationDate="2014-12-25T22:19:06.657" Score="4" Body="&lt;p&gt;There are several methods which can be used to compare the Brønsted-Lowry (proton-transfer) basicity and the hydrogen bond basicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most common defines a parameter $ \beta_{2} =  (log K_{B}^{H} + 1.1)/4.636 $, where $logK_{B}$ is calculated via the solution of  $log K^i = L^ilogK_{B}^{H} + D^i$ ($K$ being the experimental equilibirum constant, $L$ and $D$ acid-dependent parameters) over 34 reference compounds. The numerical correction results in $ \beta_{2} = 0 $ and $ \beta_{2} = 1 $ being the lower and upper limits for hydrogen bonding strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \beta_{2} $ values are not well correlated with measures of Brønsted-Lowry basicity (e.g. proton affinity); they are well correlated when comparing very divergent systems such as water vesus phenols but not when comparing members of the same family; e.g. different halo-subsituted phenols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rationalize this by considering that hydrogen bonding strength is dependent on the electronegativity of the donor atom as well as the donor lone pair LUMO/H $\sigma*$ HOMO energy-match. Increasing the electronegativity of the donor atom at constant $pK_{a}$ results in greater hydrogen bond basicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, proton transfer is not going to occur step-wise because an unsolvated proton is very unstable; loss of a proton is always associated with formation of a charged complex in solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01045a038&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1990/p2/p29900000521#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4004274/&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2014-12-25T22:19:06.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22081" PostTypeId="1" AcceptedAnswerId="22083" CreationDate="2014-12-25T23:03:31.910" Score="5" ViewCount="5022" Body="&lt;p&gt;&lt;strong&gt;1. The problem statement, all variables and given/known data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBlVp.png&quot; alt=&quot;Question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. The attempt at a solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the MO diagram for CN- is this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LUmXo.gif&quot; alt=&quot;CN- MO diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to draw the MO diagram for cyanogen. I know that the higher the bond order, the shorter the bond length. I have tried to draw a MO diagram for cyanogen by combining two of the MO diagram for CN:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KrDyZ.png&quot; alt=&quot;NCCN MO Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if this is necessary but I have drawn the following resonance structures:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BKjkE.png&quot; alt=&quot;cyanogen resonance structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that structure (I) contributes the most as structures (II) and (III) have a positive charge on the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the C-C bond in cyanogen is shorter than the &quot;usual&quot; C-C bond lengths due to delocalisation across it but I don't know how to explain this using Molecular Orbital theory.&lt;/p&gt;&#xA;" OwnerUserId="11445" LastEditorUserId="11445" LastEditDate="2014-12-26T18:48:51.007" LastActivityDate="2014-12-26T18:48:51.007" Title="Using MO theory, give an explanation for the C-C bond length in cyanogen" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22082" PostTypeId="2" ParentId="22079" CreationDate="2014-12-25T23:43:56.213" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am unsure how to use this knowledge to actually draw the MO diagram?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, chemists used Lewis structures to explain molecular properties, but eventually this method was replaced with resonance structures (like you've drawn in part iv above).  Resonance structures focus primarily upon p orbitals in a molecule and usually do not consider the sigma system.  Molecular Orbital (MO) theory was a further improvement on resonance theory.  MO theory still focuses on p orbital connectivity and typically disregards the sigma system, but it lets you view the actual patterns that the molecular orbitals generate.  It is very similar to the orbital diagram you included above, but we leave out the sigma system and focus on the p orbitals and the electrons they contain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part ii) above you conclude that the ether oxygen is $\ce{sp2}$ hybridized.  This means that there is one p orbital on oxygen that contains a lone pair of electrons (there are also 3 $\ce{sp2}$ orbitals on the oxygen, 2 are used to form sigma bonds to carbon and the third holds the second lone pair of electrons on the oxygen).  This p orbital can interact with the 2 p orbitals involved in the carbon-carbon double bond.  These 3 p orbitals can interact to produce 3 molecular orbitals (see pictures below).  These 3 molecular orbitals are produced by combining (adding and subtracting) a p orbital and the 2 ethylenic molecular orbitals (the bonding and antibonding pi orbitals for the double bond).  All of the orbitals shown in the MO diagram below are p orbitals.  We will fill the MOs with 4 electrons, 2 from the pi double bond and 2 from the oxygen p orbital that contains one of oxygen's lone pair of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have determined that we have 3 p orbitals (one on oxygen, 2 on the olefinic carbons) that can overlap and form 3 molecular orbitals.  The system we have created with 3 p orbitals from methyl vinyl ether is similar to the all carbon allyl analogue with 3 p orbitals, one p orbital on each of the 3 carbons.  Specifically your system is analogous to the ally anion with 4 pi electrons (2 from your double bond and 2 from the  oxygen lone pair).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HAvfN.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://www.chemtube3d.com/orbitalsallylanion.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on how to construct molecular orbitals in organic molecules see &lt;a href=&quot;http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2014-12-26T16:29:15.293" LastActivityDate="2014-12-26T16:29:15.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22083" PostTypeId="2" ParentId="22081" CreationDate="2014-12-25T23:50:52.030" Score="4" Body="&lt;p&gt;See my &lt;a href=&quot;https://chemistry.stackexchange.com/questions/22079/draw-a-simplified-mo-diagram-for-the-pi-system-of-methyl-vinyl-ether/22082#22082&quot;&gt;answer&lt;/a&gt; to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, your molecule is analogous to the all-carbon system, butadiene.  In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals.  Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals.  We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons.  Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons.  For the whole molecule we have 8+4+6=18 electrons as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals.  Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GdDOO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-6-1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2NJ0v.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://classes.yale.edu/chem225b/studyaids/pericyclic/pericyclic.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top).  Don't worry about the &quot;S&quot;'s and &quot;A&quot;'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons.  This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate.  Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2014-12-26T17:04:35.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22084" PostTypeId="2" ParentId="22081" CreationDate="2014-12-26T02:58:54.140" Score="3" Body="&lt;p&gt;This feature of cyanogen was already mentioned in the comment by Geoff to the Question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/21973/4945&quot;&gt;How many electrons are there in the pi-system of cyanogen?&lt;/a&gt;&quot;. (And I was very tempted to vote to close as a duplicate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your resonance structures are correct and it is noteworthy, while I might be the predominant one, the other two are still considerably influencing the electronic structure. The problem with this is, that you cannot sufficiently express delocalisation in Lewis terms. A molecular orbitals approach is necessary to understand the bonding situations. Below you can find the complete valence MO scheme of cyanogen, which I repost updated from the linked question.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PBMd9.jpg&quot; alt=&quot;MO scheme cyanogen&quot;&gt;&lt;br&gt;&#xA;You can clearly see, that the eight $\pi$ orbitals are delocalised over the complete molecule. Half of them include the bond between the central carbon atoms, i.e. the degenerate $\pi_\mathrm{u}$ orbitals.&lt;br&gt;&#xA;For a carbon carbon single bond you would expect, a bond length of $\mathbf{d}(\ce{C-C})=150~\mathrm{pm}$ and for a double bond $\mathbf{d}(\ce{C=C})=134~\mathrm{pm}$. The calculated bond length is about $\mathbf{d}(\ce{C\bond{~-}C})=138~\mathrm{pm}$ (DF-BP86/def2-SVP) and clearly between a single and a double bond. This has of course an effect on the neighbouring carbon nitrogen bonds. You would expect the triple bond to be $\mathbf{d}(\ce{C#N})=108~\mathrm{pm}$ and the double bond $\mathbf{d}(\ce{C=N})=120~\mathrm{pm}$. The calculated bond length is $\mathbf{d}(\ce{C\bond{~=}N})=118~\mathrm{pm}$ and only little shorter than a double bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covalent radii are taken from &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.200901472/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pekka Pyykkö and Michiko Atsumi, &lt;em&gt;Chem. Eur. J.&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;15&lt;/em&gt; (46), 12770-1779.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-12-26T02:58:54.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22085" PostTypeId="2" ParentId="21962" CreationDate="2014-12-26T03:43:46.103" Score="4" Body="&lt;p&gt;The stability of heavy nuclei is a function of both atomic number and atomic mass.  Therefore, an appropriate number both of protons and of neutrons in the target nucleus must be attained, or else vanishingly short lifetimes can be expected.  (See the Wikipedia article on, e.g., the &lt;a href=&quot;https://en.wikipedia.org/wiki/Island_of_stability&quot; rel=&quot;nofollow&quot; title=&quot;island of stability&quot;&gt;island of stability&lt;/a&gt;, and links therein.)  The 'source' nuclei are chosen to provide proton and neutron counts that are anticipated to yield heavy nuclei with sufficiently long lifetime so as to be measurable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it's not as simple as just choosing the right total number of protons and neutrons in your starting nuclei, as you're not guaranteed that all of the neutrons and protons will stay put in the synthesized nucleus, nor that you won't have rapid radioactive emissions that knock you away from your desired composition.  For example, in the various &lt;a href=&quot;https://en.wikipedia.org/wiki/Dubnium&quot; rel=&quot;nofollow&quot; title=&quot;syntheses of dubnium reported at Wikipedia&quot;&gt;syntheses of dubnium reported at Wikipedia&lt;/a&gt;, various numbers of neutrons were ejected as part of the nuclear synthesis 'reaction'.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2014-12-26T03:43:46.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22086" PostTypeId="1" AcceptedAnswerId="22092" CreationDate="2014-12-26T05:12:46.007" Score="1" ViewCount="9985" Body="&lt;p&gt;In the reaction:&#xA;     $$\ce {2KClO3 -&amp;gt; 2KCl + 3O2}$$&#xA;What is the volume of oxygen released under NTP conditions when $36.75~\mathrm{g}$ of $\ce {KClO3}$ is heated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find the answer this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2~\mathrm{mol}$ $\ce{KClO3}$ form $3~\mathrm{mol}$ $\ce{O2}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$0.3~\mathrm{mol}$ $\ce{KClO3}$ form $0.45~\mathrm{mol}$ $\ce {O2}$, i.e. $36.75~\mathrm{g}~\ce{KClO3} = 0.3~\mathrm{mol}~\ce{KClO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to convert moles into litre. Can we convert mass directly to volume in $\mathrm{L}$. I read somewhere that $1~\mathrm{L} = 1~\mathrm{kg}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9428" LastEditorUserId="865" LastEditDate="2021-05-22T18:54:12.013" LastActivityDate="2021-05-22T18:54:12.013" Title="How to find the volume of oxygen release by the decomposition if potassium chlorate?" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22087" PostTypeId="2" ParentId="22086" CreationDate="2014-12-26T06:03:52.103" Score="1" Body="&lt;p&gt;1 kg for 1 liter is, in fact, liquid water at room temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to use the ideal gas law $\ce {PV=nRT}$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-26T06:03:52.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22088" PostTypeId="1" CreationDate="2014-12-26T07:33:36.037" Score="1" ViewCount="597" Body="&lt;p&gt;Why does $\ce{BeH2}$ polymerize into $\ce{(BeH2)_{n}}$ in &lt;em&gt;solid state&lt;/em&gt; whereas $\ce{AlCl3}$ dimerises into $\ce{Al2Cl6}$ in the &lt;em&gt;vapour state&lt;/em&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="11450" LastEditorUserId="9961" LastEditDate="2015-10-30T18:29:26.637" LastActivityDate="2016-12-19T15:17:39.983" Title="Dimers and polymers of electron deficient compounds" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22089" PostTypeId="1" CreationDate="2014-12-26T11:01:13.773" Score="0" ViewCount="2993" Body="&lt;p&gt;Aluminum is an ideal element for usage in airports. Unlike Iron, when oxygenating, there is a protective layer formed that does not allow the rest of the Al metal to oxygenate.&lt;br/&gt;Does the protective layer $\ce{Al2O3}$ form on the surface of aluminum alloys (such as Al-Zr) as well as when it happens to Aluminum in its pure state?&lt;/p&gt;&#xA;" OwnerUserId="11453" LastEditorUserId="7581" LastEditDate="2014-12-26T12:32:28.853" LastActivityDate="2014-12-26T12:58:28.030" Title="Does the protective layer of Aluminum oxide form on the surface of aluminium alloys (such as Al-Zr) as well?" Tags="&lt;metal&gt;&lt;surface-chemistry&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22092" PostTypeId="2" ParentId="22086" CreationDate="2014-12-26T12:46:08.430" Score="0" Body="&lt;p&gt;In fact, under NTP conditions, each mole of an ideal gas occupies the volume $22.4\ce{L }$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the volume of  $0.45\ce{mol }$ of  $\ce{O_2 }$ is  $22.4\times 0.45=10.08\ce{L }$.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-26T12:46:08.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22093" PostTypeId="2" ParentId="22089" CreationDate="2014-12-26T12:58:28.030" Score="3" Body="&lt;p&gt;Short answer: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: Any alloy containing 'passivating' elements such as aluminum, which have the tendency to form a protective oxide as you describe, has the potential to also form the same sort of passivating oxide layer as the pure material.  It's not guaranteed, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure zirconium also &lt;a href=&quot;http://www.mtialbany.com/metals/zirconium/&quot; rel=&quot;nofollow&quot; title=&quot;forms a passivating oxide&quot;&gt;forms a passivating oxide&lt;/a&gt; $(\ce{ZrO2})$$^{(1)}$, so the AlZr alloy you mention is almost certain to form a similar protective oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More broadly, a major area of research at present is in the 'alumina-forming' alloys (&quot;alumina&quot; is another word for aluminum oxide, $\ce{Al2O3}$), where aluminum is included in, e.g., steels or nickel superalloys, to try to achieve this passive aluminum oxide layer on the surface in high-temperature, corrosive environments$^{(2-5)}$.  With materials like a stainless steel-plus-aluminum, the passive oxide scale only forms if the alloy has the right percent range of aluminum -- otherwise, the aluminum oxide ends up staying spread throughout the metal and not concentrating at the surface.  If the alumina doesn't end up spread evenly over the surface, you don't get nearly as good of a protective effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;a href=&quot;http://www.mtialbany.com/metals/zirconium/&quot; rel=&quot;nofollow&quot;&gt;http://www.mtialbany.com/metals/zirconium/&lt;/a&gt;&lt;br&gt;&#xA;  (2) &lt;a href=&quot;http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf&quot; rel=&quot;nofollow&quot;&gt;http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf&lt;/a&gt;&lt;br&gt;&#xA;  (3) &lt;a href=&quot;http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf&lt;/a&gt;&lt;br&gt;&#xA;  (4) &lt;a href=&quot;http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf&lt;/a&gt;&lt;br&gt;&#xA;  (5) &lt;a href=&quot;http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed&quot; rel=&quot;nofollow&quot;&gt;http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2014-12-26T12:58:28.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22095" PostTypeId="2" ParentId="19090" CreationDate="2014-12-26T15:24:03.387" Score="7" Body="&lt;p&gt;In cyclopropyl methyl carbocation, the sigma electron cloud expands outwardly due to angle strain; therefore, the valence electron cloud of $\ce{CH2+}$ is surrounded by or enveloped by the expanded electron cloud making carbocation stable.&lt;/p&gt;&#xA;" OwnerUserId="11454" LastEditorUserId="5026" LastEditDate="2018-04-23T01:10:29.663" LastActivityDate="2018-04-23T01:10:29.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22096" PostTypeId="2" ParentId="22086" CreationDate="2014-12-26T15:40:19.287" Score="0" Body="&lt;p&gt;Avogadro's law states that &quot;&lt;em&gt;equal volumes of all gases, at the same temperature and pressure, have the same number of molecules&lt;/em&gt;&quot;, ie at constant temperature and pressure, 1 mole of any gas will occupy the same volume. At NTP, this volume is ~$24dm^3$. Therefore the volume of $0.45mol$ of $O_2$ will be $0.45 × 24 = 10.8dm^3$.  $1dm^3 = 1L$ so your answer would be $10.8L$&lt;/p&gt;&#xA;" OwnerUserId="11426" LastActivityDate="2014-12-26T15:40:19.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22097" PostTypeId="2" ParentId="18857" CreationDate="2014-12-26T15:59:28.890" Score="1" Body="&lt;p&gt;The electron is attracted to the nucleus but repelled by other electrons.  In some atoms, adding an electron is an exothermic process because the potential energy of the lone electron and the atom apart are converted to kinetic energy.  In these cases the attraction of the nucleus overcomes the repulsion of the electrons.  Not all octet forming is exothermic.  The second electron affinity of oxygen is endothermic for example.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2014-12-26T15:59:28.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22098" PostTypeId="1" CreationDate="2014-12-26T16:05:33.313" Score="2" ViewCount="101" Body="&lt;p&gt;Nitrogen is the future hope of stainless steel industry; since in the production of austenitic steel it can even completely replace Nickel (which is relatively very expensive) in some cases and for another reason: It decreases the temperature at which carbides should form.&lt;br/&gt;&#xA;As a result, there is always an optimum Carbon to Nitrogen ratio.My problems are:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How is N decreasing the temperature of the formation of carbides? Is it a catalyst or something?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Resources, however, didn't mention how the N to C ratio is measured? How? Am I missing something very easy in here?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Update: I just found this &lt;a href=&quot;http://gualstainless.com/htm/img/imatges_cms/pdf/document_tecnic_2.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt; in which almost all of the info above can be found; but it's not my main reference.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-12-08T11:10:10.137" LastActivityDate="2015-12-08T11:10:10.137" Title="How is N to C ratio measured in forming austenitic stainless steel?" Tags="&lt;inorganic-chemistry&gt;&lt;energy&gt;&lt;metal&gt;&lt;applied-chemistry&gt;&lt;alloy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22099" PostTypeId="1" AcceptedAnswerId="22101" CreationDate="2014-12-26T17:26:12.020" Score="8" ViewCount="4648" Body="&lt;p&gt;In question 16 of the A-Level Chemistry 2013 exam, there is this reaction scheme:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3CH=CH2 -&amp;gt;[?] CH3-CHBr-CH2Br}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question asks for both the name of the reagent and the product. I know that the product is 1,2-dibromopropane, but am confused about the reagent. I was told that the correct answer was &quot;bromine&quot;, whereas &quot;bromine water&quot; was a wrong answer. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="11402" LastEditorUserId="16683" LastEditDate="2018-10-20T14:34:03.403" LastActivityDate="2018-10-20T14:39:45.190" Title="What is the difference between &quot;bromine&quot; and &quot;bromine water&quot; in their reaction with an alkene?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22100" PostTypeId="1" CreationDate="2014-12-26T17:29:09.093" Score="9" ViewCount="6764" Body="&lt;p&gt;According to my course of organic chemistry, para-nitro-aminobenzene has to break his aromaticity to delocalize the electrons of the amino-group in the nitro-group.&#xA;I don't really see this, when I draw the resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i62A6.jpg&quot; alt=&quot;resonance structures&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11455" LastEditorUserId="4945" LastEditDate="2015-08-14T09:48:00.563" LastActivityDate="2020-04-03T18:56:45.703" Title="Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;nitro-compounds&gt;" AnswerCount="3" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22101" PostTypeId="2" ParentId="22099" CreationDate="2014-12-26T17:36:16.203" Score="11" Body="&lt;p&gt;From &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/alkenes/halogenation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemGuide&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Using bromine water as a test for alkenes&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you shake an alkene with bromine water (or bubble a gaseous alkene&#xA;  through bromine water), the solution becomes colourless. Alkenes&#xA;  decolourise bromine water.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The chemistry of the test&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This is complicated by the fact that the major product isn't&#xA;  1,2-dibromoethane. The water also gets involved in the reaction, and&#xA;  most of the product is 2-bromoethanol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In essence, the electrophilic pi bond in the presence of bromine &lt;em&gt;and&lt;/em&gt; water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sj4je.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sj4je.png&quot; alt=&quot;Mechanism of bromohydrin formation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p3XNN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p3XNN.png&quot; alt=&quot;Mechanism of dibromide formation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; ion to do its thing unhindered. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-10-20T14:39:45.190" LastActivityDate="2018-10-20T14:39:45.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22103" PostTypeId="2" ParentId="22007" CreationDate="2014-12-26T18:32:22.963" Score="12" Body="&lt;p&gt;@Dissenter, you really need to stop asking questions like this.  Once again you point out that the emperor has no clothes!  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the mechanism of simple proton transfer is an active area of investigation.  There is not a single yes or no answer that would adequately address your questions.  But if this truth were told in the introductory texts who would continue on when the answer to seemingly simple questions leads to such a morass?  Instead, particularly in introductory circumstances, there is a gentlemen's agreement of sorts that we will simply say that the emperor's clothes are the finest we've seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at your two questions in reverse order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why do we depict acid base reactions as concerted?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In many introductory texts, and even some that are for advanced courses, proton transfer is presented as a concerted process based largely on Occam's razor (the simplest explanation is often the correct explanation).  However, enough experiments in this area have been run that I think it would be fair to say that a stepwise mechanism is now favored, but by no means mandated for all proton transfers (we'll see why some proton transfers proceed in a concerted fashion below).  The stepwise mechanism proceeds as follows: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;step 1, formation of the encounter complex $\ce{A-H}$ and $\ce{B-}$ diffuse through solution until they are close enough to form a hydrogen bond&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A-H + B- -&amp;gt; A-H..B-}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;step 2, the $\ce{A-H}$ bond elongates while the $\ce{B-H}$ bond contracts and a transition state is reached.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A-H..B -&amp;gt; [A..H..B]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition state is not necessarily linear or symmetric.  There could also possibly be an intermediate accompanying the transition state.  This depends upon your definition of an intermediate.  How long does a species need to survive in order to be called an intermediate; is it 10 vibrations, 1 vibration or what?  Obviously it is difficult to detect such short-lived molecules, but ultra-fast spectroscopy has made great strides and we may soon know whether an intermediate exists along this multi-step reaction coordinate.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;step 3, the transition state elongates in the opposite direction &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[A..H..B] -&amp;gt; A..H-B}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;step 4, the hydrogen bond is broken and the complex diffuses apart&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A..H-B -&amp;gt; A- + B-H}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does hydrogen bonding strength correlate with Brønsted basicity in a&#xA;  given medium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we throw out cases where geometry interferes with hydrogen bonding (&lt;em&gt;e.g.&lt;/em&gt; one example would be where steric factors preclude close approach or proper alignment of the species involved), then it would not be unreasonable to expect some correlation between hydrogen bond strength and the pKa difference between the two species involved.  There is certainly experimental evidence to support this view.  However, it has been found that if we step back and examine a larger range of pKa differences, the correlation breaks down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram showing how the rate of proton transfers might change as a function of the pKa difference between the two species involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U3RIN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at a small portion of the plot, the part at low pKa, we could fairly easily convince ourselves that there is a linear correlation.  But eventually as the pKa becomes large, the rate of proton transfer becomes diffusion controlled and insensitive to any further increase in the pKa.  The same concept applies to hydrogen bonding.  There could be reactions (acid-base proton transfers for example) where there is no activated process, the reaction is diffusion controlled; as a consequence we cannot expect to accurately measure or compare hydrogen bond strengths as a function of Brønsted basicity differences in such a regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary Answers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why do we depict acid base reactions as concerted?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To keep it simple in introductory classes.  It turns out that many acid-base proton transfers are not concerted, but rather involve a multi-step process.  Some other proton transfers, for example those involved in diffusion controlled reactions, are probably concerted.  This is still an active area of research.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does hydrogen bonding strength correlate with Brønsted basicity in a&#xA;  given medium?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is &quot;yes&quot; if we 1) examine only a small range of basicities and 2) look in the range of basicities where the hydrogen bonding is described as an activated process.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-03T11:27:45.780" LastActivityDate="2018-03-03T11:27:45.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22104" PostTypeId="1" CreationDate="2014-12-26T20:45:46.953" Score="4" ViewCount="96" Body="&lt;p&gt;Would it be feasible to mix fully hydrogenated oils with unsaturated oils (eg. soybean oil) to create an edible fat with the consistency of margarine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is the motivation for interesterified oils? If not, then what is the constraint that prevents this? Do the fats not form a solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand how palm oil came to replace partially hydrogenated oils in margarines in response to trans-fat bans, and what possible alternatives might be.&lt;/p&gt;&#xA;" OwnerUserId="11458" LastEditorDisplayName="user6887" LastEditDate="2014-12-27T03:38:45.310" LastActivityDate="2017-01-24T08:44:31.097" Title="Can fully-hydrogenated oils be mixed with unsaturated oils for use in food (margarines, etc)?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;solutions&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22106" PostTypeId="2" ParentId="21996" CreationDate="2014-12-23T02:14:57.137" Score="5" Body="&lt;p&gt;The charge doesn't &quot;move back through the battery&quot; because in order to do that, it would have to pass directly through the chemical reactions that put it there in the first place. It moves through the external circuit, because that's the only path available to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take it to the next level of detail, the half-reactions that occur at the two plates inside the battery are actually occurring in both directions all of the time. When the battery is open-circuit, both reactions are occurring at the same rate. The forward reactions are slowed down by the electric field created by the separation of charge, and the reverse reactions are accelerated by it, and the battery maintains equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an external path for current is created across the battery terminals, some of the charge flows through it reducing the voltage and the E-field slightly, and this allows the forward reactions within the battery to occur at a higher rate than the reverse reactions, and this is what eventually dicharges the battery as the reactants are consumed.&lt;/p&gt;&#xA;" OwnerUserId="11461" OwnerDisplayName="Dave Tweed" LastActivityDate="2014-12-23T02:14:57.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22108" PostTypeId="1" CreationDate="2014-12-26T21:56:30.010" Score="3" ViewCount="1746" Body="&lt;p&gt;I already have &lt;a href=&quot;https://chemistry.stackexchange.com/questions/22068/clean-tin-with-aluminium&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;https://chemistry.stackexchange.com/questions/22069/what-is-the-chemistry-behind-cleaning-silverware-with-aluminium-foil&quot;&gt;questions&lt;/a&gt; touching on this subject and there is a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2479/what-happens-during-cleaning-silverware&quot;&gt;third question&lt;/a&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that all the other question talk about cleaning silver with aluminum foil and carbonated water. The recipe I use though is to soak the silverware and the aluminium foil in salt water (simply mix table salt and cold water). I got the idea many years ago watching a commercial for a product called &quot;SilverStar&quot; or something like that on TV-Shop which was essentially a 0.5 cm thick aluminium plats which you were supposed to but in a bowl with saltwater and all your jewellery and it would &lt;em&gt;magically&lt;/em&gt; become polished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanations for the &quot;recipes&quot; using carbonated water are all along the lines that the acid dissolves part of the aluminum and then the aluminum ions in solution somehow react with the silveroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then is if I only use table salt and (still) water will the chemistry be more or less the same but only much slower (relying on CO&lt;sub&gt;2&lt;/sub&gt; dissolving in the water from the air) or is there something else going on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of what was happening with the salt water was that the salt water acted as a &quot;wire&quot; and the aluminium acted as the &quot;sacrificial metal&quot; in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cathodic_protection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cathodic Protection&lt;/a&gt;. I always thought that you could essentially put the aluminium in one bowl and the silver in another and connect them with a copper wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is twofold &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the purpose of the table salt in my recipe?&lt;/li&gt;&#xA;&lt;li&gt;Is there any (chemical) difference between &quot;my&quot; recipe with table salt and the more mainstream one with different flavours of carbonated water?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11420" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2014-12-26T22:10:04.513" Title="Clean silver with aluminum foil and *" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;silver&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22110" PostTypeId="2" ParentId="22108" CreationDate="2014-12-26T22:10:04.513" Score="1" Body="&lt;p&gt;Since you want to remove the tarnished silver ($\ce {Ag2O}$), you would convert it into silver metal (or reduce it). Therefore, you would oxidize the aluminum ($\ce {Al -&amp;gt; Al^3+}$). The silver and aluminum need to be in physical contact and the sodium chloride or the carbonated water is just providing the medium to transfer the ions; the aluminum does not dissolve first.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-26T22:10:04.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22113" PostTypeId="1" AcceptedAnswerId="22361" CreationDate="2014-12-27T06:53:13.993" Score="4" ViewCount="623" Body="&lt;p&gt;So I was trying out the sodium acetate trihydrate 'hot ice' experiment, and my 300ml test worked great. So, naturally, more is better right? (cringe)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a 1000ml in a pyrex beaker cooling down, and I am belatedly wondering if this is going to generate too much heat when I set it off. I'd imagine the beaker itself is safe, I don't imagine it would do anything to the granite countertop, but it is resting on a plastic cutting board, which could conceivably melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of this can I make before I'm doing something risky?&lt;/p&gt;&#xA;" OwnerUserId="11465" LastActivityDate="2015-01-04T05:24:06.967" Title="Is 1000ml of 'hot-ice' solution dangerous?" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22114" PostTypeId="1" AcceptedAnswerId="22179" CreationDate="2014-12-27T08:24:41.303" Score="13" ViewCount="8513" Body="&lt;p&gt;How do I know whether s-p mixing would occur in some diatomic heteronuclear molecules like CN, CO, NO etc?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2014-12-29T19:24:08.487" Title="How to know whether s-p mixing will happen in heteronuclear molecules?" Tags="&lt;molecules&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22115" PostTypeId="1" AcceptedAnswerId="22117" CreationDate="2014-12-27T09:51:59.963" Score="6" ViewCount="1279" Body="&lt;p&gt;I need to etch small PCB for my project, since I do this once in a blue moon I would like to try some easy approach, preferably with household items. I'm aiming at the use of acetic acid with hydrogen peroxide to aid the reaction. But this got me wondering what are some other household items usable in the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to find potassium permanganate ($\ce{KMnO4}$) which is a strong oxidizer, but I'm not sure if this still holds for water soltuion mixed with acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My other idea for strong oxidizer was sodium hypochlorite ($\ce{NaClO}$) found in bleach, but I'm not really sure how safe this is. Even though I will etch only small PCB, 1 by 2 inches, I would rather not endanger myself or anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any of the proposed ideas make sense? &lt;strong&gt;Is $\ce{KMnO4}$ or $\ce{NaClO}$ suitable oxidizer for etching PCB using acid solution?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11466" LastEditorUserId="13339" LastEditDate="2015-06-14T14:15:59.190" LastActivityDate="2015-06-14T20:19:47.280" Title="Alternative PCB etching solutions - Is KMnO4 or NaClO suitable oxidizer?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22116" PostTypeId="1" AcceptedAnswerId="22121" CreationDate="2014-12-27T10:12:54.223" Score="4" ViewCount="2374" Body="&lt;p&gt;Consider two $\ce{H}$ atoms. Since the proton in one attracts the electron in another, they attract each other, and form a covalent bond ($\ce{H-H}$). Bond forming requires energy (436 kJ/mol for an $\ce{H-H}$ bond) – where does that energy come from?&lt;br&gt;&#xA;If I put two $\ce{H}$ atoms in a container, would they form an $\ce{H-H}$ bond by themselves, or would I need to add energy to the system in order for them to bond?&lt;/p&gt;&#xA;" OwnerUserId="11430" LastEditorDisplayName="user7951" LastEditDate="2015-06-19T17:24:28.367" LastActivityDate="2016-07-04T12:37:33.973" Title="Where does the energy of formation of covalent bonds come from?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;energy&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22117" PostTypeId="2" ParentId="22115" CreationDate="2014-12-27T10:47:32.263" Score="3" Body="&lt;p&gt;If you don't want to use the classical solid etchants, such as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iron(III)chloride ($\ce{FeCl3}$) or&lt;/li&gt;&#xA;&lt;li&gt;Ammonium persulfate ($\ce{(NH4)2S2O8}$) or&lt;/li&gt;&#xA;&lt;li&gt;Sodium persulfate ($\ce{Na2S2O8}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;there's the &lt;strong&gt;mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$)&lt;/strong&gt; in various ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Typical recipes use &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;water&lt;/li&gt;&#xA;&lt;li&gt;30% $\ce{H2O2}$&lt;/li&gt;&#xA;&lt;li&gt;37% $\ce{HCl}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;in a volume ratio of 1:1:1. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hydrochloric acid can be considered a &quot;household chemical&quot;, it was/is available at DIY stores as &lt;strong&gt;muriatic acid&lt;/strong&gt; and typically used to clean brickwork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would advise against mixing concentrated (glacial) acetic acid and hydrogen peroxide solutions of higher concentrations. You might end up forming peracetic acid, which is unhealthy and tends to decompose spontaneously (explode) at higher concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's another method worth to be mentioned, although the etchant is usually not a household item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cupric chloride $\ce{CuCl2}$ in aqueous solution can be used at room temperature to remove metallic copper from the PCB too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuCl2 + Cu -&amp;gt; CuCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking in terms of transition metal waste management, this seems to be pretty nice, since the resulting copper(I) solution can be regenerated (oxidized) to copper(II) (= the etchant) using hydrochlorid acid and hydrogen peroxide. This &lt;a href=&quot;http://www.www.journalamme.org/papers_amme03/1229.pdf&quot; rel=&quot;nofollow&quot;&gt;conference proceeding&lt;/a&gt; provides further information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the permanganate is concerned, it seems that it has been used to &lt;strong&gt;desmear&lt;/strong&gt; PCBs in industry, but it has been superseeded by plasma treatment to avoid damage to the boards.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2014-12-27T11:56:59.637" LastActivityDate="2014-12-27T11:56:59.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22118" PostTypeId="1" AcceptedAnswerId="22120" CreationDate="2014-12-27T13:17:01.163" Score="3" ViewCount="3982" Body="&lt;p&gt;Here is the reaction that I didn't find the mechanism of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The conditions are zinc dust, acetic acid (solvent) at &lt;span class=&quot;math-container&quot;&gt;$\pu{11 °C}$&lt;/span&gt;&#xA;for 1 h. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reagents are sodium acetate; 2,3,4,5-tetraacetoxy-1,6-bromohexane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The product is 3-4-diacetoxy-1,5-hexadiene.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is also some evolution of gas (I don't know what it is,&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What reaction can lead to this elimination when it seems that no proton is leaving the molecule (instead of bromide) and acetate is leaving?&lt;/p&gt;&#xA;" OwnerUserId="11469" LastEditorUserId="41328" LastEditDate="2020-05-16T13:24:39.663" LastActivityDate="2020-05-16T13:24:39.663" Title="Elimination of bromide and acetate under reducing conditions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22119" PostTypeId="2" ParentId="22115" CreationDate="2014-12-27T14:22:39.930" Score="1" Body="&lt;p&gt;I’m using $\ce{CuCl2}$ for three years now. I like it a whole lot better than $\ce{FeCl2}$ – it’s cleaner, uses domestically available chemicals, and has less of a disposal problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s related to “mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$)” because that produces the same result, ultimately, though there is more than one way to get there. I have in the past tried $\ce{HCl/H2O2}$ and ended up stripping the entire board in a few seconds: but those same chemicals can be used to make $\ce{CuCl2}$ which is far more controlled. I’m not much of a chemist, but my understanding is that $\ce{Cu + H2O2 + HCl -&amp;gt; CuCl2 + H2O}$: the initial reaction eats copper fast but the result is $\ce{CuCl2}$ and the reaction with that is slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a practical guide to making and using $\ce{CuCl2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://techref.massmind.org/techref/pcb/etch/CuCl2.htm&quot; rel=&quot;nofollow&quot;&gt;http://techref.massmind.org/techref/pcb/etch/CuCl2.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11470" LastEditorDisplayName="user7951" LastEditDate="2015-06-14T20:19:47.280" LastActivityDate="2015-06-14T20:19:47.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22120" PostTypeId="2" ParentId="22118" CreationDate="2014-12-27T15:28:48.163" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What reaction can lead to this elimination when it seems that no&#xA;  proton is leaving the molecule (instead of bromide) and acetate is&#xA;  leaving?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a reductive elimination reaction.  The classic textbook example is the reductive elimination of trans-1,2-dibromocyclohexane to cyclohexene in the presence of zinc dust and acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dD6k9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is basically a redox reaction where the zinc metal is oxidized and the dibromocyclohexane is reduced.  As with many reactions involving electron transfer from metal surfaces, the mechanism of this reaction is not fully understood.  Perhaps the zinc inserts itself into the carbon-bromine bond as a first step, but in any case, electron flow proceeds in the direction shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction has found many uses and has been widely modified to include cases such as yours where a bromine and another group are eliminated to form a double bond.  &lt;a href=&quot;http://www.scielo.cl/pdf/jcchems/v54n1/art21.pdf&quot; rel=&quot;noreferrer&quot;&gt;Here is a link to a paper&lt;/a&gt; where one of the bromines is replaced by an alkoxy group and the same zinc-mediated reductive elimination occurs.  &lt;a href=&quot;https://www.jstage.jst.go.jp/article/bcsj1926/50/1/50_1_186/_pdf&quot; rel=&quot;noreferrer&quot;&gt;Here is a link to another paper&lt;/a&gt; where zinc is used to perform a reductive elimination in a series of compounds where both bromines have been replaced, one by acetate and the other by hydroxyl.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-27T15:28:48.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22121" PostTypeId="2" ParentId="22116" CreationDate="2014-12-27T15:35:07.993" Score="7" Body="&lt;p&gt;Your view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding&quot;&gt;Fundamental forces behind covalent bonding&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H&lt;sub&gt;2&lt;/sub&gt;). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO&lt;sub&gt;2&lt;/sub&gt;. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly,  require some energy input to kick off the actual reaction process.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-12-27T15:35:07.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22122" PostTypeId="1" CreationDate="2014-12-27T16:35:17.767" Score="0" ViewCount="1695" Body="&lt;p&gt;$$\ce{N1V1 = N2V2}$$&#xA; where, N is the Normality of the Acid / Base &amp;amp;&#xA; V is the volume in litres. &#xA;We ask you to give a scientific justification or derivation to the above equation&#xA;i.e. Why must the product of Normality &amp;amp; Volume be equal for Acid &amp;amp; Base ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question I found online and would greatly appreciate an answer.&lt;/p&gt;&#xA;" OwnerUserId="11472" LastEditorUserId="7448" LastEditDate="2014-12-27T17:56:00.410" LastActivityDate="2015-01-27T07:57:50.400" Title="Why should the product of the normality of the acid or base and its volume be constant?" Tags="&lt;homework&gt;&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22123" PostTypeId="2" ParentId="22122" CreationDate="2014-12-27T18:10:20.597" Score="0" Body="&lt;p&gt;First, you need to know when you can use this equation. It is used to find the concentration of the solution when the volume of solvent is changed (when diluting the solute). Since the only thing that changed is the volume of the solvent (the amount of the solute is the same before and after), you use the equation to find the new concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. 100 mL of a 0.2 N solution of sodium chloride is diluted to 0.1 N. The new volume is:&#xA;$$\ce {N_{f} \times V_{f} = N_{i} \times V_{i}}$$&#xA;$$0.1\ce {N\times V_{f} = } 0.2\ce {N \times 100mL}$$&#xA;$$\ce {V_{f} = 200mL}$$&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2014-12-27T18:10:20.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22124" PostTypeId="1" AcceptedAnswerId="22134" CreationDate="2014-12-27T18:14:21.197" Score="5" ViewCount="348" Body="&lt;p&gt;I am working on a code to preform a HF procedure and I am a bit confused about the unitary transformation of the basis. My instructions are as follows. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;specify the basis and geometry (done)&lt;/li&gt;&#xA;&lt;li&gt;Evaluate and store overlap, Kinetic energy, nuclear attraction, and electron repulsion integrals (done)&lt;/li&gt;&#xA;&lt;li&gt;diagonalize $ \mathbf{S}$ and find $\mathbf{S}^{-1/2}$ which is the unitary transformation matrix. &lt;/li&gt;&#xA;&lt;li&gt;Use the core hamiltonian to find the initial Fock matrix:&#xA;$$ \mathbf{S^{-1/2}}^{T} \mathbf{H}~\mathbf{S^{-1/2}}\equiv\mathbf{F}_0$$&lt;/li&gt;&#xA;&lt;li&gt;Diagonalize the Fock matrix to find the mo coefficients:&#xA;$$\mathbf{F}_0 \mathbf{C}^{'}_0=\mathbf{C}^{'}_0\epsilon_0$$&lt;/li&gt;&#xA;&lt;li&gt;Back transform the coefficients.&#xA;$$\mathbf{C^{'}_0}~\mathbf{S^{-1/2}}=\mathbf{C}_0$$&lt;/li&gt;&#xA;&lt;li&gt;compute the density, use the density to evaluate new fock matrix, find new coefficients repeat until convergeance.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now step 7 kinds glazes over a lot of the work but I understand that part, so It isn't important to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused by the $\mathbf{S}^{-1/2}$ matrix.  Is this literally one over the square root of each element of $\mathbf{S}$? Or is there some decomposition I should preform that I just do not understand.  I can carry out the whole procedure but without orthonormalization my results are nonsense.&lt;/p&gt;&#xA;" OwnerUserId="11473" LastEditorUserId="3798" LastEditDate="2014-12-28T02:18:41.467" LastActivityDate="2014-12-28T02:18:41.467" Title="Unitary Transformation of the HF equations" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22125" PostTypeId="1" CreationDate="2014-12-27T18:16:08.497" Score="2" ViewCount="626" Body="&lt;p&gt;What if an explosion is not able to exit the container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a pipe bomb for example; assuming that the pipe bomb was hypothetically made out of such a strong material that the explosion is not able to rip the material apart, what would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asking this because I was told: NO MATTER WHAT the explosion has to and WILL &quot;exit&quot;, but what if it just can't?&lt;/p&gt;&#xA;" OwnerUserId="11474" LastEditorUserId="194" LastEditDate="2017-04-01T01:34:27.803" LastActivityDate="2017-04-01T01:34:27.803" Title="What happens when an explosion can't lead to volume increase of the thermodynamic system?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;explosives&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22127" PostTypeId="1" CreationDate="2014-12-27T18:43:25.840" Score="-1" ViewCount="3153" Body="&lt;p&gt;maybe because the first electron has a lower energy. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorUserId="7448" LastEditDate="2014-12-28T01:23:13.333" LastActivityDate="2014-12-28T01:23:13.333" Title="Why the second ionization energy of helium is greater than the first?" Tags="&lt;electrochemistry&gt;&lt;quantum-chemistry&gt;&lt;electronic-configuration&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2014-12-28T13:45:18.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="22128" PostTypeId="1" AcceptedAnswerId="22129" CreationDate="2014-12-27T18:45:05.720" Score="3" ViewCount="2046" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Give the name and draw the graphical formula of an alkene that is an isomer of but-1-ene and that has a different carbon skeleton.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think it's 2-methylpropene, but I'm not sure whether the correct name is 2-methylprop-1-ene or 2-methylpropene.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastEditorUserId="41328" LastEditDate="2020-09-09T18:21:34.787" LastActivityDate="2020-09-20T14:42:56.807" Title="What is the proper name for the but-1-ene's isomer methylpropene?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22129" PostTypeId="2" ParentId="22128" CreationDate="2014-12-27T18:51:15.307" Score="1" Body="&lt;p&gt;When considering whether numbering is necessary you must see whether there are alternative structures which have the same basic name but different numbering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example putting the methyl group on the 1 or the 3 position of the longest carbon chain changes its name from prop- to but-. Since the only other position that the methyl group can go is the 2 position then both numbers are unnecessary and so the shortest unambiguous name is methylpropene.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2014-12-27T19:45:23.490" LastActivityDate="2014-12-27T19:45:23.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22130" PostTypeId="2" ParentId="22127" CreationDate="2014-12-27T19:04:13.153" Score="5" Body="&lt;p&gt;The second ionization energy of any element will be greater than its first ionization energy. One way to rationalize this phenomenon is by considering that the first electron is being removed from a neutral atom and the 2nd electron is being removed from a positively charged ion that is already electron deficient.  Additionally, Slater's Rules for calculating effective nuclear charge can be used to provide a semiquantitative reasoning for this observation.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2014-12-27T19:04:13.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22131" PostTypeId="1" CreationDate="2014-12-27T19:14:45.813" Score="7" ViewCount="147" Body="&lt;p&gt;Reading that the &lt;a href=&quot;http://news.sciencemag.org/people-events/2014/12/areas-watch-2015&quot;&gt;large hadron collider will be up and running with twice as much energy&lt;/a&gt; in March 2015, I was curious whether our understanding of subatomic particles has changed our understanding of chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To attempt to narrow down the question to something with a definitive answer: has the discovery of subatomic particles changed experimental chemistry?  If so, what is an example?&lt;/p&gt;&#xA;" OwnerUserId="11476" LastEditorUserId="7448" LastEditDate="2014-12-28T20:49:22.553" LastActivityDate="2014-12-30T02:40:27.627" Title="Is it likely that increased understanding of quantum physics will change our understanding of chemistry?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22132" PostTypeId="1" AcceptedAnswerId="22144" CreationDate="2014-12-27T19:31:52.660" Score="1" ViewCount="241" Body="&lt;p&gt;Skin color is one of the things one would rather not ask anything about! Only in humans, it can vary from very dark brown to pale pink. In darker-skinned people, the color is mainly due to melanin, which is produced by melanocytes. &lt;br/&gt;&#xA;There are three types of melanin: eumelanin, pheomelanin, and neuromelanin. Eumelanin is the most common; so it must be the reason of the dark skin of the dark-skinned. And that's true, since the two major groups of eumelanin are black and brown species.&lt;br/&gt;&#xA;The first ray produced by sun that comes to mind when you think about affecting the attributes of skin is UVA, since it penetrates most into the skin (Even though its energy is lesser than UVB). So, when I attach these two together I reach the fact that if UV rays do not cause free radicals they will eventually change the skin's color. (This is what is mostly believed and its effect on natural selection is impossible to hide)&lt;br/&gt;&#xA;Now, I give up. How does UV do to the skin that makes it darker? This must be very easy, huh? What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2014-12-28T16:41:19.890" LastActivityDate="2014-12-28T16:41:19.890" Title="How does UV affect skin colors in dark-skinned people?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;color&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22133" PostTypeId="1" CreationDate="2014-12-27T22:26:37.013" Score="16" ViewCount="6018" Body="&lt;p&gt;I sprinkled salt on my turkey skin and left it to dry out over night. It was covered with aluminum foil. The next morning the aluminum foil had holes ‘burnt’ in it and the turkey skin turned silver at the spots where the salt crystals had touched the aluminum foil. Was my turkey poisoned with aluminum at this point? I did not eat it to find out.&lt;/p&gt;&#xA;" OwnerUserId="11477" LastEditorUserId="11367" LastEditorDisplayName="user6887" LastEditDate="2017-04-02T23:55:35.420" LastActivityDate="2017-05-17T07:37:56.863" Title="Why did the salt on the skin of my turkey react with the aluminum foil it was touching?" Tags="&lt;safety&gt;&lt;metal&gt;&lt;food-chemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22134" PostTypeId="2" ParentId="22124" CreationDate="2014-12-27T22:41:01.983" Score="7" Body="&lt;p&gt;You are going to want to check out Szabo and Ostlund page 143, they explain all the nitty-gritty for this matrix. The math itself is on page 22 and 23. But if you don't have a copy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathbf{S}^{-1/2}$ comes from the need to orthogonalize your basis so that you get your equation (5) instead of the $\mathbf{FC}=\epsilon\mathbf{SC}$. It's one method of several to achieve this orthogonalization, and it's called symmetric orthogonalization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you compute this function of a Hermitian matrix? Well, functions on diagonal matrices are super easy; you just apply the function to each diagonal element. For non-diagonal matrices, this is not true. More to your question, you can't just take one over the square root of each element of $\mathbf{S}$. But you can if it's diagonal. And you can make any Hermitian matrix diagonal via a unitary transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick, then, is to diagonalize $\mathbf{S}$ first, apply the inverse square root to each eigenvalue, then transform back using the eigenvectors. So you're going to want to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solve $\mathbf{SU} = \mathbf{sU}$. Where $\mathbf{s}$ are the eigenvalues of $\mathbf{S}$, and $\mathbf{U}$ are the eigenvectors.&lt;/li&gt;&#xA;&lt;li&gt;For each $s$ in $\mathbf{s}$ do $s^{-1/2}$.&lt;/li&gt;&#xA;&lt;li&gt;Transform back to original basis by $\mathbf{S}^{-1/2} = \mathbf{Us}^{-1/2}\mathbf{U}^{\dagger}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="3798" LastEditDate="2014-12-27T22:53:32.440" LastActivityDate="2014-12-27T22:53:32.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22135" PostTypeId="1" AcceptedAnswerId="27308" CreationDate="2014-12-28T00:26:37.377" Score="8" ViewCount="3016" Body="&lt;p&gt;This question is one that I'm wondering whether it belongs here or on Graphic Design SE. All I want is some software that can export a PDB structure of a protein (displayed in ribbons style) or DNA double helix, or a combination of the two, to a high-quality SVG file. I know that several pieces of software for 3D molecular modelling can create raster images of PDB files quite easily (e.g., Accelrys DS Visualizer, Jmol, PyMOL, &lt;em&gt;etc.&lt;/em&gt;). But SVG files have several advantages over PNGs, now I don't mind if the colours are all bland (i.e., each shape having one colour that is not subject to any shading or other gradients; I am saying this is OK as I realize that vector images often have problems with intricate colour gradients) just so long as the end result is a decent-looking SVG file. I have tried using Avogadro for this purpose but the resulting SVG was 264 MB in size -- a little too big for my taste. If you'd like an acceptable size range it'd be &amp;lt;5 MB. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2020-03-24T23:45:18.730" Title="How does one export a PDB structure to SVG?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22136" PostTypeId="2" ParentId="14531" CreationDate="2014-12-28T01:37:30.250" Score="3" Body="&lt;p&gt;As a specialist in poison information (SPI), I would never recommend using anything other than water to decontaminate the skin or the eyes. An acid mixed with a strong alkali can cause an exothermic reaction (release of heat) which would then cause thermal burns. Neutralizing a corrosive substance (acid or alkali) that has been ingested is CONTRA-INDICATED under all circumstances, due to the potential for exothermic reactions and the release of gas. Although I could find no specific references for skin exposures, logic would indicate the same would be true in these cases as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is always the best, most easily available and cheapest means of decontamination. AND it will do no harm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of exposure, always call your local Poison Control Centre for the most appropriate recommendations.&lt;/p&gt;&#xA;" OwnerUserId="11479" LastActivityDate="2014-12-28T01:37:30.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22137" PostTypeId="1" AcceptedAnswerId="22145" CreationDate="2014-12-28T01:43:19.737" Score="15" ViewCount="9599" Body="&lt;p&gt;Over the last year or so I've heard from friends, who subscribe to holistic medicine, talk about negative ions, saying that they are good for your mental/physical health. Being the only one among them with a chemistry background, I was suspicious when i heard this. Below are examples I've seen that use negative ions in a medicinal context:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Negative Ions Purify Air&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Virtually all particles in the air have a positive charge, while negative ions have a negative charge. In which case, negative ions and particles magnetically attract to one another. When there is a high enough concentration of negative ions in the air, they will attract to floating particles in large numbers. This causes the particle to become too heavy to remain airborne. As a result, the particle will fall out of the air, and will then be collected by normal cleaning activities, such as vacuuming or dusting. - &lt;a href=&quot;http://www.ionic-balance.com/technology/what-are-negative-ions&quot;&gt;Ion-Imbalance&lt;/a&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Negative Ions from the Beach&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may have experienced the power of negative ions when you last set foot on the beach or walked beneath a waterfall. While part of the euphoria is simply being around these wondrous settings and away from the normal pressures of home and work, the air circulating in the mountains and the beach is said to contain tens of thousands of negative ions -- Much more than the average home or office building, which contain dozens or hundreds, and many register a flat zero. -&lt;a href=&quot;http://www.webmd.com/balance/features/negative-ions-create-positive-vibes&quot;&gt;WebMD&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Apparently they Help You Breath&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;People travel from all over the world to visit salt mines like the ones found in the Himalayas, Austria, and Germany. In Germany, the hot salt springs are so revered, companies often send overworked employees there to rejuvenate.&#xA;  Why are the salt mines so beneficial?&#xA;  Quite simply... the very dry, negative ion environment of a salt mine is key.&#xA;  Well... even if you're not able to travel to Asia or Europe to take advantage of salt mines, you can still take advantage of these negative ions to help improve the air you breathe. The solution involves pure Himalayan crystal salt...- &lt;a href=&quot;http://products.mercola.com/himalayan-salt/himalayan-salt-lamps.htm&quot;&gt;Mercola&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well it's not surprising that two thirds of the above sources are trying to sell products (salt lamps, air purifiers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're dealing with ions here, I figure the chemistry world must have some sort of input. What's the chemical or biochemical support, or the evidence against, for the above claims? If there are negative ions present from waterfalls/oceans/salt rocks, what ions are we talking about?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To address @dissenter, &lt;a href=&quot;http://www.ionic-balance.com/technology/tourmaline&quot;&gt;one of these sites&lt;/a&gt; claims tourmaline is a source of negative  ions...&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2014-12-28T02:29:50.913" LastActivityDate="2015-02-25T19:08:29.540" Title="Negative ions and health: pseudo science or something more?" Tags="&lt;ions&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22139" PostTypeId="2" ParentId="18413" CreationDate="2014-12-28T02:17:18.563" Score="1" Body="&lt;p&gt;As hinted in the comments by @AngusTheMan we use the equation&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;but ultimately we are looking for the &lt;span class=&quot;math-container&quot;&gt;$v_\mathrm{rms}$&lt;/span&gt;, so we'll also use&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You've been given your &lt;span class=&quot;math-container&quot;&gt;$\mathrm{KE_{avg}},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; are constants, and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass of 1 mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ne}$&lt;/span&gt;. We can set all of these equal and the solution will come from&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="41328" LastEditDate="2020-07-05T17:04:24.063" LastActivityDate="2020-07-05T17:04:24.063" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22140" PostTypeId="2" ParentId="22122" CreationDate="2014-12-28T07:45:04.800" Score="1" Body="&lt;p&gt;$$\ce{N1V1 = N2V2}$$ The equation is used basically for dilution of solution and determining one of the parameters of acid or base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of dilution of solution, the equation holds true because the no of gm equivalents of the solute remains constant in the solution even the solution is diluted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial no of gram equivalents of solute = Final no of gram equivalents of solute ....................eq (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know, $$\text{Normality} = \frac{\text{No of gram equivalents}}{\text{Volume in Litre}}$$&#xA;$$\text{No of gram equivalents}=\text{Normality}\times\text{Volume in Litre}$$&#xA;Now lets consider,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Normality =$\ce{N1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial Volume=$\ce{V1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final Normality=$\ce{N2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final Volume = $\ce{V2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then , eq (1) becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Initial Normality}\times\text{Initial Volume}=\text{Final Normality}\times\text{Final Volume}$$&#xA;$$\ce{N1V1 = N2V2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of acid and base reaction,&#xA;No of gram equivalents of acid will always react or consume equal no of gram equivalents of base and vice versa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No of gram equivalents of acid = No of gram equivalent of base .......eq (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normality of acid = $\ce{N1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of acid =$\ce{V1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normality of base =$\ce{N2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of base =$\ce{V2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then eq (2) becomes,&#xA;$$\text{Normality of acid}\times\text{Volume of acid}=\text{Normality of base}\times\text{Volume of base}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N1V1 = N2V2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of grams of a chemical element or compound equal to the weight of its chemical equivalent. The gram equivalent of a chemical element is numerically equal to the quotient resulting from dividing the element’s atomic weight by its valence. For example, the gram equivalent of iron (atomic weight 55.847) with a valence of 2 is 27.92 g and with a valence of 3, 18.62 g. The gram equivalent of acid is that amount of the acid in grams containing one gram equivalent of hydrogen capable of being displaced by metal with the formation of salt. The gram equivalent of a base is that amount of the base in grams that is necessary for its complete reaction with one gram equivalent of acid. The gram equivalent of salt is that amount of the salt in grams containing one gram equivalent of metal. The concept of gram equivalent is widely used in volumetric analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://encyclopedia2.thefreedictionary.com/Gram+Equivalent&quot; rel=&quot;nofollow&quot;&gt;http://encyclopedia2.thefreedictionary.com/Gram+Equivalent&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2014-12-28T07:45:04.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22141" PostTypeId="2" ParentId="20066" CreationDate="2014-12-28T08:10:15.533" Score="1" Body="&lt;p&gt;Don't know why but you're just making some value errors, happens to everybody once in a while:&#xA;$$\ce{2Fe^{3+} + Cu -&amp;gt; 2Fe^{2+} +Cu^{2+}}$$&#xA;$$\mathcal{E}\approx(0.771-0.337)-\underbrace{\frac{0.059}{2}}_{\large\frac{\mathcal {RT}}{\mathcal F}\ln{10}\approx0.059,\nu=2}\cdot\log_{10}\underbrace{\overbrace{\frac{0.018\cdot0.14^2}{1\cdot0.037^2}}^{\ce{[Cu^2+][Fe^2+]^2}}}_{\ce{[Cu][Fe^3+]^2}}$$&#xA;So:&#xA;$$\mathcal{E}\approx0.451\cdot$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-28T08:10:15.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22142" PostTypeId="2" ParentId="22133" CreationDate="2014-12-28T08:23:52.763" Score="22" Body="&lt;p&gt;&lt;em&gt;If you don't mind, I'll start with a more colloquial and hopfully not to lengthy explanation&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precious metals, such gold, do not corrode under &quot;household conditions&quot;, others, such as iron or aluminium do. Rusting of iron is a typical example. Here, the corroded surface often is very porous and the corrosion carries on until the whole piece of metal is turned to rust (oxidized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of aluminium, the initial corrosive layer is very tight and serves as a protection for the rest of the aluminium foil. Such a very thin and transparent protection layer often is artificially applied. The item are sold as &lt;em&gt;Eloxal&lt;/em&gt; (= electrolytically oxidized aluminium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the table salt, which is sodium chloride ($\ce{NaCl}$). Sodium chloride is hygroscopic, it &quot;attracts&quot; the water and that's why you sprinkled it on the turkey. It is also very soluble in water. When it dissolves, sodium $\ce{Na+}$ and chloride $\ce{Cl-}$ ions are formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chloride ions attach and move into the protective oxide layer of the aluminium foil and finally induce the corrosion of the aluminium metal. Aluminium ions are released to the water (= the wet surface of the turkey).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more scientific, in-depth explanation of this corrosive processes known as &lt;strong&gt;pitting&lt;/strong&gt;, you might want to have a look at this &lt;a href=&quot;http://www.sciencemadness.org/talk/files.php?pid=237561&amp;amp;aid=17616&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; or this &lt;a href=&quot;http://jes.ecsdl.org/content/161/9/C421.full&quot; rel=&quot;noreferrer&quot;&gt;review&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Was my turkey poisoned with aluminum at this point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There seem to be some crazy myths concerning the &quot;toxicity&quot; of aluminium (ions), often spread by opponents of immunization and other quacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fact is that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a healthy human body is capable to excrete aluminium ions though the kidneys&lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://www.alzheimers.org.uk/site/scripts/services_info.php?serviceID=6&quot; rel=&quot;noreferrer&quot;&gt;British Alzheimer Society&lt;/a&gt; does not see any convincing relation between the uptake of aluminium and Alzheimer's Disease&lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;http://www.krebsinformationsdienst.de/vorbeugung/risiken/mythen.php&quot; rel=&quot;noreferrer&quot;&gt;cancer information service&lt;/a&gt; of the &lt;strong&gt;dkfz&lt;/strong&gt; (Deutsches Krebsforschungszentrum, German Cancer Research Centre) refutes the aluminium induced breast cancer myth&lt;/li&gt;&#xA;&lt;li&gt;many antacids do contain aluminium hydroxide as one ingredient. The effect of &quot;poisoning&quot; is not known.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes, aluminium ions were released to the wet surface of the turkey. &lt;/li&gt;&#xA;&lt;li&gt;There's no reason to assume that the turkey is &quot;poisoned&quot;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2016-02-27T10:55:32.083" LastActivityDate="2016-02-27T10:55:32.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22144" PostTypeId="2" ParentId="22132" CreationDate="2014-12-28T14:01:54.950" Score="3" Body="&lt;p&gt;There is a journal article almost exactly about the title question, &lt;a href=&quot;http://www.nature.com/jid/journal/v124/n6/full/5602865a.html&quot; rel=&quot;nofollow&quot;&gt;Mechanisms of Skin Tanning in Different Racial/Ethnic Groups in Response to Ultraviolet Radiation&lt;/a&gt; Journal of Investigative Dermatology (2005) 124, 1326–1332.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the body of the question assumes that eumelanin changes color.  Instead, the article finds that melanin gets closer to the surface of the skin, and the degree to which this effect occurs varies with race/ethnicity.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2014-12-28T14:01:54.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22145" PostTypeId="2" ParentId="22137" CreationDate="2014-12-28T16:28:25.827" Score="16" Body="&lt;p&gt;Charge separation at waterfalls with airborne ions, resulting in a potential between the base of the waterfall and the surrounding air, is a phenomenon examined by Philipp Lenard (Über die Elektrizität der Wasserfälle, &lt;em&gt;Ann Phys (Leipz)&lt;/em&gt;, &lt;strong&gt;1892&lt;/strong&gt;, &lt;em&gt;46&lt;/em&gt;, 584–636). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect is real and more recent field studies, such as &lt;a href=&quot;http://dx.doi.org/10.5194/acp-12-3687-2012&quot; rel=&quot;noreferrer&quot;&gt;Characterization of ions at Alpine waterfalls&lt;/a&gt;, as well as studies on the formation of charged droplets, such as &lt;a href=&quot;http://www.indiana.edu/~nano/publications/2008/Charge%20Separation%20in%20the%20Aerodynamic%20Breakup%20of%20Micrometer-Sized%20Water%20Droplets.pdf&quot; rel=&quot;noreferrer&quot;&gt;Charge Separation in the Aerodynamic Breakup of Micrometer-Sized Water Droplets&lt;/a&gt; should be taken seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the sources that you cited are &lt;strong&gt;pseudoscientific marketing bullshit&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first quote links to a web shop that offers marvels like &lt;em&gt;magnetic bands&lt;/em&gt; and &lt;em&gt;ion-balance bands&lt;/em&gt;. While the silicone band probably isn't hamful to the skin, it will not emit anything. Nada, nothing, nix!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The webshop behind the third link sells &lt;em&gt;Himalayan salt lamps&lt;/em&gt;. It's a matter of taste to have a lamp housing made from unrefined sodium chloride with traces of iron in it (hence the colour), but I wouldn't expect any health effects through the emission of whatever from these lifestyle items. &lt;em&gt;Himalayan salt&lt;/em&gt; is being sold for quite a while, even for nutrition but if at all, the stuff has seen the Himalaya mountains only from a distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no halite mines in the Himalaya mountains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the material is likely to come from the &lt;strong&gt;Salt Range&lt;/strong&gt; in the Punjab province in Pakistan, or from Poland.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, it's business as usual: Scientific facts are mashed up with esoteric mumbo jumbo to sell overpriced and useless stuff to well-meaning but naive clients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Oh what a tangled web we weave,&lt;br&gt;&#xA;When first we practise to deceive!&lt;/em&gt; &#xA;-- Sir Walter Scott&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-25T19:08:29.540" LastActivityDate="2015-02-25T19:08:29.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22146" PostTypeId="2" ParentId="22125" CreationDate="2014-12-28T18:09:21.340" Score="5" Body="&lt;p&gt;I'm not an expert in thermodynamics and gas physics so there might be some errors in this calculation but here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to several sources (wikipedia included) TNT decomposes according to this equation:&#xA;$$\ce{2C7H5N3O6 -&amp;gt; 3N2 + 5H2O + 7CO + 7C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 100kg of TNT was placed in an indestructible $1m^3$ container containing an inert gas (we will say helium) at standard temperature and pressure here is a rough estimate of what might happen if it was detonated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a Hess's Law cycle we can calculate the enthalpy of decomposition of TNT from formation enthalpy data: $$\Delta_{decomp}H = 63.2*2 - 285.8*5 - 110.5*7 = -2076.1~ kJmol^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can calculate the energy released by the decomposition: $$M_r~of~TNT = 227.13~gmol^{-1}$$ $$n = \frac{100}{0.22713} = 440.3~mol$$ $$Energy~released = 440.3 * 2076.1 *1000 = 914.1~MJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to go about specific heat capacities for gas mixtures so I won't put a figure on the temperature rise that this will cause but I suspect it will be in the thousands of degrees due to the amount of energy and so the resultant pressure rise will also be very significant, probably enough to liquefy the gases. Also as @Klaus Warzecha hinted at in his comment these kind of temperatures and pressures will probably cause some interesting chemical reactions to take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2014-12-28T18:09:21.340" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22147" PostTypeId="2" ParentId="22032" CreationDate="2014-12-28T18:17:38.353" Score="3" Body="&lt;p&gt;Well I get what your problem is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is that earlier C had 4 electrons in its valence shell. Now it is bonded to 3 other atoms giving it a total of 6 electrons in its shell. The octet rule says that it should have 8. Since the other carbon took away 2 electrons this means that our carbon should have a +2 charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well this is where the confusion lies . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is whenever we calculate formal charge we use &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Formal Charge= (No.of valence electrons in unbonded state - no of lone&#xA;  pair electrons ) - (no. of bond pair electrons/2)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case the charge comes out to be (4-0) - (6/2) =+1 We usually associate half of the bond pair electrons to each atom while calculating charge. On the other hand When counting the no of electrons in valence shell we count both the electrons in the bond pair. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck :)&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="8218" LastEditDate="2018-03-05T12:32:23.333" LastActivityDate="2018-03-05T12:32:23.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22148" PostTypeId="1" AcceptedAnswerId="22155" CreationDate="2014-12-28T18:32:11.127" Score="28" ViewCount="3224" Body="&lt;p&gt;We recently learned in school that diamonds sparkle as the are very optically dense, meaning that it takes longer for light to pass through them, thus meaning that the light totally internally reflects when you look at a diamond, whereas in glass it would pass through at that angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why have we not been able to create a solid material that is so optically dense? Is it because such materials are rare, expensive or hard to produce, or because no-one has looked into it?&lt;/p&gt;&#xA;" OwnerUserId="1896" LastEditorUserId="8083" LastEditDate="2014-12-28T20:03:13.990" LastActivityDate="2014-12-28T20:03:13.990" Title="Why are there no cheap diamond equivalents?" Tags="&lt;synthesis&gt;&lt;solid-state-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22149" PostTypeId="1" AcceptedAnswerId="22200" CreationDate="2014-12-28T18:35:43.427" Score="2" ViewCount="2327" Body="&lt;p&gt;I was doing questions on organic chemistry. I came across tonnes of questions  where I was required to compare basic and acidic strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing acidic strengths is fine &amp;mdash; I remove an $\ce{H+}$ and compare the stability of the anion because my teacher told me a strong acid has a stable anion and explained the reason to me from an equilibrium point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to comparing basic strengths what do we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I add an $\ce{H+}$ to the species and compare the resulting stability by resonance, hyperconjugation and inductive effect and then can conclude the more basic one will have a stabler cation?&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really baffled.&#xA;Please do correct me if I am mistaken in my fundamentals. &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="7448" LastEditDate="2017-03-07T11:04:01.357" LastActivityDate="2019-08-25T02:55:29.150" Title="Comparing Basic strengths of different species" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22150" PostTypeId="2" ParentId="22131" CreationDate="2014-12-28T18:36:08.980" Score="3" Body="&lt;p&gt;Well, you asked for an example, I'll give you one.&lt;br/&gt;&#xA;&quot;Quantum&quot; is undeniably the most effective keyword our science has seen, it's changed our 'understanding of chemistry and physics' drastically over the past century. &lt;br/&gt;&#xA;For example, between 1900 to 1925 (years vary from source to source, but this era exists, that's what that is important), in which Bohr's model was the main focus of study (as wikipedia puts it), is known to be the era of &lt;b&gt;&quot;old quantum theory&quot;&lt;/b&gt;.&lt;br/&gt; There were series of theories, which started by Max Plank, and partially came to results with the final modifications of Sommerfeld; and ended in Erwin Schrodinger's famous model. New efforts led to new results and therefore the &quot;old quantum theory&quot; was never consistent.&lt;br/&gt;&#xA;What was all of this about? Scientists are always eager to find out the differential behavior of matter and energy. Sub-atomic particles (in first level, n, p &amp;amp; e) were discovered in order to explain the phenomenons that Roentgen saw (radioactivity of Uranium and other heavy elements in 'pitchblende'); Faraday experimented (the $\ce{SnCl2}$ solution experiment); Greek Leucippus and Democritos tried to explain; and so on. After every discovery, since the 'rules' that were discovered usually were up to a point in explaining what's around us, we change them and make new discoveries. And that was the reason of what Einstein, Bell and people like those did. They corrected the quantum for it to be what we know today.&lt;br/&gt;&#xA;So you see, the science was always about understanding; so why wouldn't a big discovery like quantization that led to quantum mechanics be able to change our direction of thought, either we are chemists, physicists or simply &lt;em&gt;humans&lt;/em&gt;?&lt;br/&gt;&#xA;Links you might find helpful: &lt;a href=&quot;https://en.wikipedia.org/wiki/Old_quantum_theory&quot; rel=&quot;nofollow&quot;&gt;About the old quantum theory in wikipedia&lt;/a&gt;; &lt;a href=&quot;http://arxiv.org/abs/0802.1366&quot; rel=&quot;nofollow&quot;&gt;About the old quantum theory 2&lt;/a&gt;; &lt;a href=&quot;http://en.wikipedia.org/wiki/Bell%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Bell's theorem&lt;/a&gt;(which linked classical quantum to quantum mechanics) and finally &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;quantum mechanics according to wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2014-12-28T18:36:08.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22151" PostTypeId="2" ParentId="22148" CreationDate="2014-12-28T19:08:35.147" Score="17" Body="&lt;p&gt;There are plenty of synthetic materials that would fall into the &quot;cheap diamond equivalent&quot; category. If we're only talking about the optical properties, cubic zirconia ($\ce{ZrO2}$) is probably the most familiar as it is often used in jewelry. It's index of refraction isn't quite as high and it's not as hard, but it's close enough that a fair amount of it is used to make jewelry. A more recent development is moissanite (either synthetic or natural $\ce{SiC}$), which is almost as hard as diamond and actually has a higher refractive index. Other synthetic materials with even higher indices like barium titanate and gallium phosphide exist, though generally their optical properties are used for things like lasers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, we have gotten quite good at making &lt;a href=&quot;http://en.wikipedia.org/wiki/Synthetic_diamond#Gemstones&quot;&gt;synthetic diamond&lt;/a&gt; itself. They are only used for a tiny fraction of diamond jewelry, though. It wouldn't surprise me to find a concerted effort to market natural diamonds as superior, given how lucrative the gem diamond market is, which might be preventing wider use for gems. Synthetic diamonds are widespread in industrial applications, however, where smaller diamonds are used to make tools and optical properties are largely irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-28T19:08:35.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22152" PostTypeId="1" AcceptedAnswerId="24270" CreationDate="2014-12-28T19:11:29.170" Score="3" ViewCount="1056" Body="&lt;p&gt;How was the pKa value (-1.7) for the oxonium ion determined? Why does it correspond to the minimum pH value accessible in water? What reaction does it represent?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7448" LastEditDate="2014-12-28T20:02:58.357" LastActivityDate="2015-03-06T20:59:34.520" Title="Acidity of oxonium ion" Tags="&lt;acid-base&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22153" PostTypeId="1" CreationDate="2014-12-28T19:17:56.957" Score="8" ViewCount="2087" Body="&lt;p&gt;A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten degree rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be?&lt;/p&gt;&#xA;" OwnerUserId="11480" LastEditorUserId="9961" LastEditDate="2016-11-10T23:02:46.010" LastActivityDate="2016-11-10T23:02:46.010" Title="Understanding and usage of Arrhenius equation" Tags="&lt;kinetics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22154" PostTypeId="2" ParentId="22152" CreationDate="2014-12-28T19:23:39.997" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How was the pKa value (-1.7) for the oxonium ion determined?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Why does it correspond to the minimum pH value accessible in water water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that an oxonium ion is simply an ion in which an oxygen carries a positive formal charge. Many oxonium ions exist. One of which is hydronium ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydronium ion corresponds to the strongest acid able to exist in water for significant periods of time because a stronger acid will be immediately deprotonated by some base in the system. In other words, hydronium ion can exist in equilibrium with water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3O+ + H2O &amp;lt;=&amp;gt; H3O+ + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of a stronger acid with water will however favor the products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$ &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-28T19:23:39.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22155" PostTypeId="2" ParentId="22148" CreationDate="2014-12-28T20:02:25.917" Score="26" Body="&lt;p&gt;There are four properties that make a diamond look like a diamond. I will compare them with the two best diamond simulants at the moment:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;cubic zirconia $\ce{ZrO2}$ - a special (&quot;cubic&quot;) form of zirconium oxide. Don't confuse it with zircon, a different mineral with the forumula $\ce{ZrSiO4}$.&lt;/li&gt;&#xA;&lt;li&gt;moissanite $\ce{SiC}$ - the name for the naturally occurring form of silicon carbide, also used for the synthetic gemstones varieties of it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So what are the differences and why none of them is an all-round replacement for a diamond?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Refractive index&lt;/strong&gt;: determines the &lt;strong&gt;brilliance&lt;/strong&gt;. This is what you were talking about when you said &quot;optical density&quot;. This basically means the amount of light the diamond reflects to the observer. If you change the refractive index, you also have to slightly change the cut of the diamond to achieve optimal brilliance, and then your &quot;diamond&quot; will not look exactly like a diamond. The refractive index of diamond is 2.42. Moissanite is higher: 2.65 and cubic zirconia is lower: 2.17.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dispersion&lt;/strong&gt;: determines the &lt;strong&gt;fire&lt;/strong&gt;. This is what you wanted to say when you mentioned that diamonds are sparkling. Fire is the property of diamonds to break the light into different wavelengths thus creating the characteristic play of colors that you get in diamonds. Here's the interesting thing: The dispersion of diamond is 0.044 and interestingly the dispersion of both cubic zirconia and moissanite is higher: 0.060 and 0.104, respectively. So theoretically, if you consider only fire (the play of colors), then they both look better than diamonds.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Birefringence&lt;/strong&gt;: (the lack of) determines the clarity. When you look through a diamond at all angles, you can clearly see what's behind. This is also true for cubic zirconia. Both minerals are &lt;em&gt;isotropic&lt;/em&gt;. On the other hand, moissanite is &lt;em&gt;birefrigent&lt;/em&gt; - it will double the image you see through the mineral, causing it to look blurred and lose the visual clarity.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hardness&lt;/strong&gt;: while not directly related to optical properties, a mineral with higher hardness will retain it's sharp cut. If the cut is damaged, the gemstone will not look as good and lose its value. Diamond has a hardness of 10, moissanite 9.25 and cubic zirconia 8.25. Note that this scale is not linear: 10 is much harder than 9 compared to 9 and 8, for example.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you can see, none of the alternatives fully satisfies the entire set of properties that make a diamond. Moissanite may be slightly better, but it's also more expensive. And remember - one of the most appealing things about diamonds is their rarity and natural occurrence. No synthetic simulant can replace that.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-28T20:02:25.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22157" PostTypeId="2" ParentId="22125" CreationDate="2014-12-28T20:57:30.413" Score="1" Body="&lt;p&gt;I did a calculation and came up with an increase in pressure of about 280 atm. You can see my calculations &lt;a href=&quot;https://drive.google.com/file/d/0B80IPeLPxBV3STM1bFFYSWdtc2s/view?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Most gas cylinders can handle 100 to 150 atm (one is as high as 400 atm), so getting a container to hold 300 atm should not be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added:&lt;br&gt;&#xA;On this &lt;a href=&quot;http://www.alspecialtygases.com/Prd_high-pressure_steel.aspx&quot; rel=&quot;nofollow&quot;&gt;page&lt;/a&gt; is list of cylinders with their working pressure (which is at least 1/2 the maximum pressure it can handle). 2000 psig is about 130 atm and 6000 psig is 400 atm.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2014-12-28T23:32:03.163" LastActivityDate="2014-12-28T23:32:03.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22158" PostTypeId="2" ParentId="22153" CreationDate="2014-12-28T21:16:10.120" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What activation energy would lead to a doubling in the reaction rate&#xA;  for a 10 degree increase in temperature, using the Arrhenius equation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You’re right it is not generally true. Let’s set up an equation to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arrhenius equation is&#xA;$$k=A\operatorname{e}^{\frac{-E_\text{a}}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2=\frac{A\operatorname{e}^{\frac{-E_\text{a}}{R\cdot(T + 10\ \mathrm{K})}}}{A\operatorname{e}^{\frac{-E_\text{a}}{RT}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln 2 = \frac{ - E_\text{a}}{R\left( T + 10\ \text{K} \right)} - \frac{ - E_\text{a}}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let’s pick a temperature around room temperature, 300 K is a nice even number, but you can re-solve the problem with any temperature you like.&#xA;$$0.69 = \frac{ - E_\text{a}}{R \cdot 310\ \text{K} } - \frac{ - E_\text{a}}{R \cdot 300\ \text{K} }$$&#xA;Using $R=8.31\ \mathrm{J/(K\ mol)}$ will produce an answer in J/mol.  Solving we find&#xA;$$E_\text{a} = 53\,568.49\ \mathrm{J/mol}$$&#xA;or,&#xA;$$53.6\ \mathrm{kJ/mol}~~(12.8\ \mathrm{kcal/mol})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So around room temperature a reaction that has an activation energy around 54 kJ/mol (or 13 kcal/mol) would show a doubling in rate for a 10 kelvin increase in temperature.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-04-17T14:28:04.970" LastActivityDate="2015-04-17T14:28:04.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22159" PostTypeId="1" CreationDate="2014-12-29T00:13:38.453" Score="-1" ViewCount="50538" Body="&lt;p&gt;The order of precedence for functional groups I have so far is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;carboxylic acid&lt;/li&gt;&#xA;&lt;li&gt;ester&lt;/li&gt;&#xA;&lt;li&gt;amide&lt;/li&gt;&#xA;&lt;li&gt;aldehyde&lt;/li&gt;&#xA;&lt;li&gt;ketone&lt;/li&gt;&#xA;&lt;li&gt;alcohol&lt;/li&gt;&#xA;&lt;li&gt;amine&lt;/li&gt;&#xA;&lt;li&gt;multiple bonds&lt;/li&gt;&#xA;&lt;li&gt;ether&lt;/li&gt;&#xA;&lt;li&gt;organic halide &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm pretty confident that from 1 to 7, I have the order of precedence right. But from 8 onwards, I'm not too sure. Could someone help verify that the list I've got is right or help me correct it?&lt;/p&gt;&#xA;" OwnerUserId="11490" LastEditorUserId="4231" LastEditDate="2014-12-29T01:18:08.190" LastActivityDate="2020-01-11T08:14:56.780" Title="Order of nomenclature precedence among these functional groups" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22160" PostTypeId="2" ParentId="22159" CreationDate="2014-12-29T00:18:12.577" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Master Organic Chemistry&lt;/a&gt;: It looks like your table is correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMeZi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2014-12-29T00:18:12.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22161" PostTypeId="2" ParentId="20084" CreationDate="2014-12-29T01:02:11.767" Score="-1" Body="&lt;p&gt;First of all, both electrodes (the copper and the zinc, in this case) are dissolving in the electrolyte. However, normally the dissolved electrode material comes out of solution just as fast. It's like trying to dig a hole in the dirt when someone else is filling it up at the exact same speed; nothing happens. Nothing happening is also called an equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second of all, when it dissolves, it becomes an ion, that is, it either gains or loses an electron. Now, some materials want to ionize different amounts than others. This &quot;want&quot; is measured in volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account these two factors, when you put any two electrodes together in an electrolyte, you can tell from a &quot;Standard Electrode Potential Table&quot; what will happen, if anything. In the case of a lemon battery, the ionic zinc wants to lose two electrons more than the copper wants to gain them. Note that one electrode must gain the electrons and one must lose them. So we have these two electrodes giving off ions into solution and also taking them back in. However, when a copper ion by chance touches the zinc metal, it transfers two electrons to the copper making metallic copper and ionic zinc. After the copper adheres to the zinc metal, it can't dissolve anymore because if it, by chance, does dissolve, the zinc metal around it instantly changes it back to metallic copper by giving it two more electrons and dissolving another zinc atom in the copper's place. Likewise, the zinc can't adhere to the copper because it would have to steal electrons from the copper, but the copper doesn't want to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to keep in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This transfer of electrons is what generates the electricity in the first place.&lt;/li&gt;&#xA;&lt;li&gt;Not all electrode pairs react.&lt;/li&gt;&#xA;&lt;li&gt;If you want more in depth information, look up &quot;standard reduction potential table&quot; in Google, as well as &quot;how to calculate the voltage of a galvanic cell&quot;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11493" LastActivityDate="2014-12-29T01:02:11.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22162" PostTypeId="1" CreationDate="2014-12-29T02:14:36.390" Score="4" ViewCount="993" Body="&lt;p&gt;Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a $\ce{Cl}$ group? In the same vein, isn't a trialkylammonium a much weaker acid than $\ce{HCl}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-NR3+ -&amp;gt; Nuc-R + NR3}$ (Stronger leaving group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-Cl  -&amp;gt; Nuc-R + Cl-}$    (Weaker Leaving group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-NR3+ -&amp;gt; Nuc-H + NR3}$    (Weaker acid, NR3 thus weaker leaving group)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-Cl -&amp;gt; Nuc-H  + Cl-}$     (Stronger acid, Cl- thus stronger leaving group)&lt;/p&gt;&#xA;" OwnerUserId="2645" LastEditorUserId="4945" LastEditDate="2014-12-29T08:52:30.980" LastActivityDate="2014-12-29T15:34:38.250" Title="Why is a trialkylamine a better leaving group, in nucleophilic reactions, than a Cl group?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22163" PostTypeId="1" CreationDate="2014-12-29T02:47:17.930" Score="2" ViewCount="200" Body="&lt;p&gt;I bought &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007B2A3VQ&quot; rel=&quot;nofollow&quot;&gt;these two 2&quot; solid steel ball bearings&lt;/a&gt; for the purpose of rolling them in my hands. Just for something to do while sitting at my computer, without those silly bells inside. (They're still a bit noisy, but it sounds much cooler than bells.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from being moderately scary (I cracked my ceramic sink while cleaning them, when I dropped one about six inches. Some shards exploded off of the underside...don't tell my wife, luckily it's not leaking), I'm wondering if there's anything I need to worry about health-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that they might have nasty lubricants on them, and that rubbing alcohol should clean it. I soaked them in alcohol for the night, then rubbed them down with an alcohol soaked washcloth, and then cleaned them with soap and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they still smell, and it makes my hand smell as well. Not that strong, but significant and yucky, and I'm wondering if it's something to be concerned about, or if that's just the nature of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I thoroughly clean them? Is the smell something to be concerned about, or should I just wash my hands when I'm done and forget about it? Is there a safe way to coat/paint them, without causing further health issues? Anything else I should know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="11495" LastActivityDate="2014-12-29T02:58:57.423" Title="Two 2&quot; solid steel bearings for stress relief hand-rolling. Safe? What's that smell? How to thoroughly clean?" Tags="&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22165" PostTypeId="2" ParentId="22162" CreationDate="2014-12-29T03:04:55.963" Score="3" Body="&lt;p&gt;Leaving group ability refers to how well a leaving group can &lt;em&gt;stay&lt;/em&gt; off whatever it came from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following prototypical nucleophilic displacement reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-LG + Nuc- &amp;lt;=&amp;gt; R-Nuc + LG-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two issues at play here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Thermodynamics. How stable is the reactant with the leaving group, $\ce{R-LG}$, and the nucleophile, $\ce{Nuc-}$? How stable are the products, $\ce{R-Nuc}$ and $\ce{LG-}$? And when examining stability, we must look at not only the reactants and products but also their environments - i.e. the solvent(s). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Kinetics. What's the &lt;em&gt;physical&lt;/em&gt; ability of the nucleophile to actually displace the leaving group? What's the &lt;em&gt;physical&lt;/em&gt; ability of the leaving group to displace the nucleophile again? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually we're taught that leaving group ability has to do with acidity and basicity. The less basic the leaving group is, the better its ability to leave. This isn't a perfect rule of thumb, but it's acceptable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the thermodynamic part of the question. If $\ce{LG-}$ weren't thermodynamically stable, then it would be inclined to attack $\ce{R-Nuc}$ and return to the status quo. (Or just grab a proton). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how capable is $\ce{LG-}$ of carrying out such an attack? Here we come to the kinetic part of the question. The bulkier the leaving group, the more likely it is to be sterically hindered and/or strongly solvated (trapped by solvent molecules). Trialkylamine  is a very genetic term, but when I think of amines with alkyl groups I think of quaternary ammonium compounds, which have alkyl groups that can contain upwards of 18 carbons (!). Such bulk affects the ability of the leaving group to line up properly with the substrate to perform nucleophilic attack. Nucleophilic attack often has specific steric demands; $\ce{SN2}$ requires &lt;em&gt;backside&lt;/em&gt; attack; the approach of attacking nucleophiles with respect to flat carbonyl groups has an angle of 104.5 degrees, etc. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a good summary of sterics and amine nucleophilicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=tP4cqrYF7nwC&amp;amp;pg=PA231&amp;amp;dq=amine+nucleophilicity+steric&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=nXOhVLiOD8iKgwSS5YPICg&amp;amp;ved=0CEUQ6AEwBA#v=onepage&amp;amp;q=amine%20nucleophilicity%20steric&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;https://books.google.com/books?id=tP4cqrYF7nwC&amp;amp;pg=PA231&amp;amp;dq=amine+nucleophilicity+steric&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=nXOhVLiOD8iKgwSS5YPICg&amp;amp;ved=0CEUQ6AEwBA#v=onepage&amp;amp;q=amine%20nucleophilicity%20steric&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2014-12-29T15:34:38.250" LastActivityDate="2014-12-29T15:34:38.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22166" PostTypeId="2" ParentId="20084" CreationDate="2014-12-29T03:09:17.990" Score="8" Body="&lt;p&gt;First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}&amp;lt;=&amp;gt;Zn^2+_{(aq)} +2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode. This excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- &amp;lt;=&amp;gt; H_{2(g)} +2H2O}$$&#xA;The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go back to the core of your question, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I'm just not sure what prompts them to leave in the first place if the positive copper      &#xA;ions are not directly in contact with the zinc.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5477/why-is-it-important-to-use-a-salt-bridge-in-a-voltaic-cell-can-a-wire-be-used?rq=1&quot;&gt;salt bridge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed078p516&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2014-12-29T03:09:17.990" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22167" PostTypeId="1" AcceptedAnswerId="22191" CreationDate="2014-12-29T08:32:31.983" Score="11" ViewCount="6823" Body="&lt;p&gt;In my textbook, it is given that the ionic product of water $K_\mathrm{w}$ remains constant even when a non-neutral solute such as an acid is added to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{w} = \ce{[H3O+][OH-]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a strong acid is added to water, the concentration of $\ce{[H3O+]}$ increases. Yet the $K_\mathrm{w}$ must remain constant. But how? How does the concentration of $\ce{[OH-]}$ decrease?&lt;/p&gt;&#xA;" OwnerUserId="9370" LastEditorUserId="41328" LastEditDate="2017-08-06T22:33:45.620" LastActivityDate="2017-08-06T22:33:45.620" Title="Why does the ionic product of water remain constant after addition of non-neutral solute?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22168" PostTypeId="1" AcceptedAnswerId="22213" CreationDate="2014-12-29T09:39:08.143" Score="0" ViewCount="121" Body="&lt;p&gt;Calculate the equilibrium constant in &lt;span class=&quot;math-container&quot;&gt;$\ce\\M ^ 3$&lt;/span&gt; of the heterogeneous system in aqueous solution:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Knowing that the solubility of the salt is &lt;span class=&quot;math-container&quot;&gt;$15.9 \ce{mg/L}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlQ0O.jpg&quot; alt=&quot;My attemps&quot; /&gt;&#xA;They are three and gave the same result.&lt;/p&gt;&#xA;" OwnerUserId="11498" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-12-17T10:34:17.157" Title="Equilibrium Constant of a Salt knowing the solubility" Tags="&lt;homework&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22169" PostTypeId="1" CreationDate="2014-12-29T11:16:14.067" Score="9" ViewCount="9964" Body="&lt;p&gt;In elementary school, there was an experiment  to show that water has dissolved air, which involved boiling some water and observing the drops at the surface of the container,  and this proved that water has air. Now, in senior high school, I was thinking about the same,  and I think this conclusion is wrong. Aren't the drops formed by the conversion of liquid water to steam due to evaporation? Or is it really the dissolved air that is escaping? What are your thoughts? &lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorUserId="11367" LastEditDate="2014-12-29T12:36:06.417" LastActivityDate="2021-01-15T05:24:31.177" Title="Experiment to prove that water has air" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;boiling-point&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22170" PostTypeId="2" ParentId="22169" CreationDate="2014-12-29T11:52:49.960" Score="0" Body="&lt;p&gt;&lt;del&gt;Your description is somewhat vague (&quot;boiling some water and observing the drops at the surface of the container&quot;), but it does seem unlikely that such an experiment demonstrates that air dissolves into water.  If by 'drops' you mean gas bubbles, then yes: the gas bubbles in boiling water are substantially all water vapor, not air.&lt;/del&gt; You are correct: any drops of liquid forming on the inner surface of the container would be from re-condensed water vapor, itself evaporated from the boiling liquid, and would have nothing to do with air having been dissolved in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would do to attempt to demonstrate that air dissolves into water would be to rig up a 2L clear plastic (&lt;strong&gt;NOT GLASS&lt;/strong&gt;) soda bottle so that I could connect it to an air compressor. (I'd have to think a bit how to accomplish this.) Then:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fill the bottle about half full with plain water&lt;/li&gt;&#xA;&lt;li&gt;Cap the bottle &amp;amp; connect to the compressor&lt;/li&gt;&#xA;&lt;li&gt;Wearing safety glasses and being &lt;strong&gt;&lt;a href=&quot;http://www.water-rockets.com/article.pl?106,0&quot; rel=&quot;nofollow&quot; title=&quot;very careful&quot;&gt;very careful&lt;/a&gt;&lt;/strong&gt; (the bottle could burst, or turn into a rocket if the cap comes loose), pressurize the bottle to 100psi (~700 kPa), &lt;/li&gt;&#xA;&lt;li&gt;Let the bottle stand, pressurized, for a while -- thirty minutes, maybe? -- periodically swirling it gently.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;CAREFULLY&lt;/strong&gt; vent the pressure over 5-10 seconds, like when opening a bottle of soda that's been shaken.&lt;/li&gt;&#xA;&lt;li&gt;If no bubbles appear, tap the side of the bottle several times.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The amount of air that dissolves in water increases with the pressure.  The hope would be that enough additional air dissolves under 100psi air, that when you release the pressure in step 5 and the amount of air the water can hold then drops back down to normal, the excess air will form a bunch of bubbles throughout the water.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2014-12-29T15:23:25.993" LastActivityDate="2014-12-29T15:23:25.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22171" PostTypeId="1" AcceptedAnswerId="22172" CreationDate="2014-12-29T12:22:47.120" Score="25" ViewCount="22379" Body="&lt;p&gt;dU, dG, dH etc are all exact differentials and the variables themselves are known as state functions because they only depend on the state of the system. However, dq and dw for example, are inexact differentials. My questions is, what does this actually mean? I've been told that exact differentials are not dependent on the path (what does 'path' mean?) but inexact differentials are. I have been told this is related to line integrals but I am not sure how. Also is this relevant?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2020-03-24T14:13:44.363" Title="Why is dU an exact differential and dq an inexact differential?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22172" PostTypeId="2" ParentId="22171" CreationDate="2014-12-29T13:22:18.417" Score="17" Body="&lt;p&gt;As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.&#xA;And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as&#xA;$$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$ &#xA;while a change in a path function $b$ can not be evaluated in such a simple way,&#xA;$$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$&#xA;And for a state function $a$ in a thermodynamic cycle&#xA;$$\oint \mathrm{d} a = 0 \, ,$$&#xA;while for a path function $b$&#xA;$$\oint \text{đ} b \neq 0 \, .$$&#xA;The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2014-12-29T13:53:33.510" LastActivityDate="2014-12-29T13:53:33.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22173" PostTypeId="2" ParentId="22171" CreationDate="2014-12-29T13:28:34.050" Score="3" Body="&lt;p&gt;Not a complete answer, but the path is exactly what it sounds like. Say you are rolling a boulder up a hill. This increases its potential energy, which can be released by rolling the boulder down the hill. But the path you take to get to the top of the hill is irrelevant, only the height you raise the boulder matters for potential energy. So if you roll it up half way, let it fall back a quarter of the way, or any such combination of forward and back, &lt;em&gt;none&lt;/em&gt; of this matters to the total change in potential energy (assuming perfect conditions, no friction, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to Wildcat's great answer, this means for state functions the endpoints of your definite integral are all that matter: you could parameterize any path you want between the endpoints and the resulting (line) integral is the same.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-29T13:28:34.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22174" PostTypeId="2" ParentId="8598" CreationDate="2014-12-29T15:02:18.990" Score="-1" Body="&lt;p&gt;The first shell can carry up to two electrons, the second shell can carry up to eight electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third shell can carry up 18 electrons, but it is more stable by carrying only eight electrons.  There is a formula for obtaining the maximum number of electrons for each shell which is given by $2n^2~\ldots$ where n is the position of a certain shell.&lt;/p&gt;&#xA;" OwnerUserId="11500" LastEditorUserId="7448" LastEditDate="2015-11-10T10:46:09.797" LastActivityDate="2015-11-10T10:46:09.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22175" PostTypeId="2" ParentId="22169" CreationDate="2014-12-29T15:53:09.573" Score="8" Body="&lt;p&gt;There are two ways that I interpret the question you're asking, and I'll try to address each in turn.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If your teacher boiled a container of water and held a piece of glass above it to collect drops of liquid water, he or she may have been trying to show that water doesn't disappear when it boils, that instead it turns to a gas that can move through the air and can be condensed back into a liquid.  Essentially, $\ce{H2O}$ is $\ce{H2O}$, regardless of the state of matter it occupies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If instead the teacher wanted you to observe the bubbles of gas forming in a container of hot or boiling liquid, he or she might have been on the right track, but either failed to convey the correct message to you or had the wrong idea in the first place.&lt;br&gt;&#xA;&lt;br&gt;If I take water from the tap and heat it strongly, bubbles will form on the inside of the vessel I'm heating it in &lt;em&gt;prior&lt;/em&gt; to it boiling.  These are bubbles of gas that are no longer soluble because the temperature of the liquid is too high (and its pressure has decreased as compared to the tap.)  The gas &lt;em&gt;is&lt;/em&gt; dissolved air (mostly carbon dioxide, I suspect) with a large portion of water vapor.  In any case, they are bubbles of gas that were dissolved but are no longer.  You can see this happen any time you boil a pot of water on the stove.  &lt;br&gt;&lt;br&gt;&#xA;Once the pot reaches boiling, any bubbles are almost exclusively gaseous water formed via vaporization.&lt;br&gt;&lt;br&gt;&#xA;An easier, if slower, way to see this is to run a cold tap until the water is as cold as it will get, then pour a glass.  As it sits (perhaps for an hour or two), bubbles will eventually form on the walls of the glass.  These eventually will dissolve again and the gas will escape through the top of the glass, so it's not something you can set up at night and come back to the next morning.  The bubbles are air that was able to dissolve in the high pressure/low temperature water in the pipes but not in the low pressure/high temperature water in the glass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2014-12-29T15:53:09.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22176" PostTypeId="1" CreationDate="2014-12-29T16:22:10.547" Score="3" ViewCount="723" Body="&lt;p&gt;I know that the initial reactants are phosphine and menthanal in an acid ($\ce{H+ A-}$) and that the first step in the reaction is the HOMO of phosphine (the lone pairs) and the LUMO of methanal (the π* orbital). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the electrons that just moved to the oxygen atom react with $\ce{H3O+}$, so the oxygen is protonated. I'm not sure what follows, though. &lt;/p&gt;&#xA;" OwnerUserId="9709" LastEditorUserId="4231" LastEditDate="2014-12-29T17:24:18.880" LastActivityDate="2014-12-29T17:24:18.880" Title="What is the mechanism for the creation of P(CH2OH)4+ cation?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22177" PostTypeId="2" ParentId="22176" CreationDate="2014-12-29T17:23:15.323" Score="2" Body="&lt;p&gt;The acid serves as a catalyst to protonate the formaldehyde oxygen.  This makes the carbonyl carbon more electrophilic, a more reactive target for the weakly nucleophilic lone pair of electrons on phosphorous.  Here is a diagram of the basic reaction mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/35Uwx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton transfer is thought to be intramolecular according to &lt;a href=&quot;http://link.springer.com/article/10.1134/S1070363211060065#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-29T17:23:15.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22178" PostTypeId="1" AcceptedAnswerId="22180" CreationDate="2014-12-29T17:31:29.507" Score="5" ViewCount="630" Body="&lt;p&gt;An experiment from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0582462363&quot; rel=&quot;noreferrer&quot;&gt;Vogel&lt;/a&gt; that I want to conduct requires &lt;em&gt;&quot;a mixture of ice and salt of about -10 °C&quot;&lt;/em&gt;. Reading trough the text it appears that I need at least 1-2 liters of this. Obviously the salt is added to lower the freezing point of the water. I guess the interaction of water with the dissolved salt trough hydrogen bonding / lone pair donation is more favorable (less energetic) than ordinary water-water interaction.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;With &lt;em&gt;&quot;salt&quot;&lt;/em&gt;, does Vogel mean just $\ce{NaCl}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since I'm a hobby chemist I do not have the same fancy machinery as they have in the lab (those cool things that generate rough ice shards and in which everybody stuffs food against all safety regulations lol). All I can use is a small refrigerator with a tiny freezer. It's easy enough to make some ice cubes, but how do I go about making these ice shards that are used in the lab? I don't think grinding the cubes is very practical as it is a tedious job that takes a lot of time in which the shards have probably already melted (also from hitting them with a pestle continuously)...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2019-12-22T12:15:05.690" Title="The use of ice water in an experiment" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22179" PostTypeId="2" ParentId="22114" CreationDate="2014-12-29T17:39:32.327" Score="8" Body="&lt;p&gt;I don't think you can know, &lt;em&gt;a pr&lt;/em&gt;iori, whether or not s-p mixing occurs without first having some experimental data.  I'll use water, a triatomic molecule, as my case in point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H $\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\ce{sp^3}$ orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 $\ce{sp^3}$ O-H sigma bonds&lt;/li&gt;&#xA;&lt;li&gt;one lone pair in a p orbital&lt;/li&gt;&#xA;&lt;li&gt;and the second lone pair in an $\ce{sp}$ orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-29T17:39:32.327" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22180" PostTypeId="2" ParentId="22178" CreationDate="2014-12-29T17:43:46.300" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.With &quot;salt&quot;, does Vogel mean just $\ce{NaCl}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yep, it's cheap and available.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;How do I go about making these ice shards that are used in the lab?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wrap the ice cubes in a towel and hit the bundle with a hammer until you get whatever size you want.  That's how we did it in the lab.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2014-12-29T17:43:46.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22181" PostTypeId="1" AcceptedAnswerId="49880" CreationDate="2014-12-29T18:06:56.183" Score="15" ViewCount="549" Body="&lt;p&gt;What is the mechanism of the attached reaction? How could nosylate do the dehydrogenate the species?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3QFtL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is from this synthesis (46th IChO preparatory problem 26):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gS9rU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="15410" LastEditDate="2016-04-21T05:44:25.520" LastActivityDate="2016-04-21T05:44:25.520" Title="Mechanism of dehydrogenation with nosylate" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22183" PostTypeId="2" ParentId="22114" CreationDate="2014-12-29T19:16:57.507" Score="7" Body="&lt;p&gt;&lt;strong&gt;Preliminary.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hybridization is a questionable model. It was originally introduced to consider organic molecules, and there it works brilliantly. However, it is hard to use outside organic chemistry with molecules with simple bounding. For example, adequate description of cyclopropane requires fractional indices in orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, hybridization process is transformation of several orbitals using linear transform. Since molecular orbitals are described as linear combinations of atomic orbitals, there is no problem to use any imaginable set of atomic orbitals as long as their linear combinations can produce all 'original' atomic orbitals. Basically, hybridization is similar to change of basis vectors in geometry: it is OK as long as the transformation follows simple rules, and it may be built to preserve values of some scalar vector expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since from mathematical viewpoint hybridization is nothing more than a special transformation of basis set, it does no change 'actual' molecular orbitals, meaning that calculations using hybridized and unhybridized orbitals must provide same result (actually, the very transformation in question is purposefully built this way). So, your question does not have a proper answer as the question itself makes wrong assumption. Hybridization does not occur, it is a model people use, it does not correspond to any real-world phenomenon &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the proper question would be: &lt;strong&gt;when hybridization model is applicable and provides valuable insight into electronic structure of considered molecule?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are considering electronic structure of isolated 2-atom molecules it is usually better to not use hybridization. It does not provide any simple conclusions for inferring preferred geometry, but leads away from more 'accurate' MO LKAO model. Considering interaction of such molecule as a ligand with metal center, however, returns some use to hybridization for simple molecules, because it simplifies geometry consideration. For example, using electron count rules it is possible to some extent to predict geometry of $\ce{NO}$ fragment, inferring preferable hybridization state of the nitrogen atom. It is also much easier to consider interactions with central atom for $\sigma sp$ lone pair  than several dinuclear molecular orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering geometry of organic and some inorganic molecules ($\ce{BCHNOF}$) with little angular stress and no hypercoordinated atoms, hybridization works well enough as long as you do not try to use it to predict one-electron properties, like ionisation potentials. However, stressed molecules like [1,1,1]-propellane or dehydroadamantane and hypercoordinated atoms, like methanium cation, require separate, case-by-case analysis to be integrated into framework of hybridization model. For most other atoms hybridization model cannot provide valuable insights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Again, hybridization does not work for 1-electron properties.&lt;/strong&gt; For example, methane, a perfect example of ideal $sp^3$ hybridization has TWO peaks in photoionization spectrum, implying it has 2 unequal sets of valence orbitals (roughly -2.25 and -1.5 eV energy levels) despite hybridization model predicting it to have 4 perfectly equal bonds. MO LKAO model, on the other hand, predicts several bonding orbitals, &lt;a href=&quot;http://www.meta-synthesis.com/webbook/40_polyatomics/polyatomics.html&quot;&gt;one significantly lower in energy than three others&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-12-29T19:24:08.487" LastActivityDate="2014-12-29T19:24:08.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22185" PostTypeId="1" AcceptedAnswerId="22190" CreationDate="2014-12-29T23:48:43.097" Score="2" ViewCount="1752" Body="&lt;p&gt;I know that the there is no difference in density in a liquid and a gas phase of a substance after the critical point. But I do want to know if this difference in density is more continuous or if it is more like a big step when reaching the critical point? I.e if we are like 0.01 Kelvin beneath the critical point of water (with the critical pressure). Is the difference in density still big at this point or is it near zero?  &lt;/p&gt;&#xA;" OwnerUserId="11507" LastEditorUserId="11507" LastEditDate="2014-12-29T23:59:39.317" LastActivityDate="2014-12-30T07:30:51.870" Title="Does the density of a liquid and gas phase of a substance(nitrogen or water) converge when reaching critical point" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;temperature&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22186" PostTypeId="2" ParentId="22131" CreationDate="2014-12-30T02:40:27.627" Score="0" Body="&lt;p&gt;I find this a strange question, because all contemporary chemical bond theory is quantum mechanical. So quantum mechanics already influences chemistry a great deal (look &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chemistry&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)! I suppose you could ask whether deeper quantum theory gives us more knowledge, and I'd guess yes! For example, look at this Periodic Videos analysis of the chemistry of seaborgium (&lt;a href=&quot;http://www.periodicvideos.com/videos/seaborgium_compound.htm&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;). We need more experimental data to test all theories, including the quantum mechanical ones.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2014-12-30T02:40:27.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22187" PostTypeId="1" AcceptedAnswerId="22246" CreationDate="2014-12-30T04:27:21.897" Score="7" ViewCount="4722" Body="&lt;p&gt;Aren't all semiconductors' atoms bond in a covalent bond? How is a &quot;covalent semiconductor&quot; different from other semiconductors?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="23561" LastEditDate="2018-07-02T16:52:33.850" LastActivityDate="2018-10-27T18:47:22.697" Title="What is a covalent semiconductor?" Tags="&lt;covalent-compounds&gt;&lt;semiconductors&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22188" PostTypeId="1" CreationDate="2014-12-30T07:17:26.293" Score="2" ViewCount="1603" Body="&lt;p&gt;why is inert pair effect specific to certain groups? Why heavier transition elements DO NOT show inert pair effect?&lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2014-12-30T07:17:26.293" Title="Why inert pair effect is seen in P- block elements only" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22189" PostTypeId="1" CreationDate="2014-12-30T07:28:30.963" Score="12" ViewCount="70490" Body="&lt;p&gt;How can we theoretically decide the acidic or basic nature of an oxide? What are the reasons/ factors which make an oxide acidic, basic, amphoteric or netural?&lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2021-02-02T00:18:15.050" Title="How to decide acidic or basic nature of oxides" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22190" PostTypeId="2" ParentId="22185" CreationDate="2014-12-30T07:30:51.870" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the density of a liquid and gas phase of a substance converge when reaching critical point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. This is exactly what happens. Let's take a look at a qualitative $P-T$ (pressure–temperature) phase diagram of an arbitrary substance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2jgvW.png&quot; alt=&quot;supercritical 1&quot;&gt;&#xA;&lt;sup&gt;source: Cansell, F. et al. Supercritical fluid processing: a new route for materials synthesis. &lt;em&gt;J. Mater. Chem&lt;/em&gt;. 9, 67–75 (1999). &lt;a href=&quot;http://dx.doi.org/10.1039/A804964E&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1039/A804964E&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, moving directly from gas to liquid involves crossing the phase boundary. However, if you're doing it in the supercritical, the change in density is continuous. Maybe you can get a better, more quantitative view with this example of $\ce{CO2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qGFJY.jpg&quot; alt=&quot;supercritical 2&quot;&gt;&#xA;&lt;sup&gt;source: &lt;a href=&quot;http://what-when-how.com/nanoscience-and-nanotechnology/microweighing-in-supercritical-carbon-dioxide-part-1-nanotechnology/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://what-when-how.com/nanoscience-and-nanotechnology/microweighing-in-supercritical-carbon-dioxide-part-1-nanotechnology/&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is also a a $P-T$ phase diagram, with equal density $(\rho)$ lines. You can see that the lines radiating from the gas–liquid boundary lines are not equal on each side. For example, 0.1 and 0.9. However, as you get close to the critical point, $\rho$ on each side of the line converges to a single value: 0.2 and 0.7, 0.3 and 0.6 and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YouTube has some terrific examples of this happening. For example: &lt;a href=&quot;http://youtu.be/GEr3NxsPTOA&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://youtu.be/GEr3NxsPTOA&lt;/a&gt;&lt;br /&gt;&#xA;What you see is a function of density - you can see the phase boundary between liquid and gas because they have differing densities. Notice in the videos how the phase slowly and gradually disappears as the density of both phases converges, effectively turning the $\ce{CO2}$ into a supercritical fluid.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-30T07:30:51.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22191" PostTypeId="2" ParentId="22167" CreationDate="2014-12-30T08:05:40.873" Score="8" Body="&lt;p&gt;$K_w$ is the equilibrium constant of this reaction:&#xA;$$\ce{2H2 O &amp;lt;=&amp;gt; H3O+ +OH-}.$$ &#xA;So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add acid to water, the above reaction equilibrium is  displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that  ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than  $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="186" LastEditDate="2016-11-07T13:03:53.900" LastActivityDate="2016-11-07T13:03:53.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22192" PostTypeId="2" ParentId="22189" CreationDate="2014-12-30T08:42:49.173" Score="18" Body="&lt;p&gt;In general, the electropositive character of the oxide's central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide. Electropositive character increases from right to left across the periodic table and increases down the column.&#xA;The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. An amphoteric oxide is one that shows both acidic and basic properties.&#xA;This trend applies only to the oxides of the individual elements in the highest oxidation states for those elements. The pattern is less clear for other oxides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define non-metal oxide acidity in terms of the acidic solutions formed in reactions with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.&lt;br&gt;&#xA;In sum, acidic oxides are oxides of non-metals, and basic oxides are oxides of metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three non-metal oxides from the upper right portion of the periodic table, &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{NO}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O}$&lt;/span&gt;, which have such low oxidation numbers for the central atom that they give neutral aqueous solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the acidity of a cation rises rapidly with charge, d-block elements that exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties. The higher the oxidation number, the more acidic the corresponding oxide. Chromium is an example of such an element. &lt;span class=&quot;math-container&quot;&gt;$\ce{CrO}$&lt;/span&gt; is basic, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr2O3}$&lt;/span&gt; is amphoteric and  &lt;span class=&quot;math-container&quot;&gt;$\ce{CrO3}$&lt;/span&gt; is acidic.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="40029" LastEditDate="2020-02-12T15:43:29.293" LastActivityDate="2020-02-12T15:43:29.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22193" PostTypeId="1" AcceptedAnswerId="22194" CreationDate="2014-12-30T10:59:34.523" Score="9" ViewCount="12787" Body="&lt;p&gt;I know the basics of Chemistry and one thing I've always wondered if it is possible for Hydrogen to give up it's one electron? I know Hydrogen is eager to share its election through covalent bonds, but is there a scenario where Hydrogen will give up its only electron?&lt;/p&gt;&#xA;" OwnerUserId="9743" LastActivityDate="2016-08-06T12:44:05.440" Title="Is it possible for Hydrogen to lose its electron?" Tags="&lt;electrons&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22194" PostTypeId="2" ParentId="22193" CreationDate="2014-12-30T12:40:24.990" Score="11" Body="&lt;p&gt;Hydrogen can lose an electron meaning it can be in the +1 oxidation state. However, just like any other cation or anion it never occurs free in condensed matter, it always is in contact with solvent and/or anions. Moreover, because of extremely small size of proton, it is an extremely powerful Lewis acid. Consequently, in common conditions proton would react with first electron pair it comes in contact with, up to and including inert gas electron pairs and covalent bond pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, hydrogen ions are quite easy to generate in electric discharge and/or under extreme heating. In fact, producing and confinement of super-hot plasma, consisting of hydrogen ions and electrons, is an area of active research for several decades and, well, producing and confinement of relatively cold plasma is not a problem. Confinement of several billions Kelvin hot plasma, however, is still a problem.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2014-12-31T08:29:59.207" LastActivityDate="2014-12-31T08:29:59.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22195" PostTypeId="1" AcceptedAnswerId="22196" CreationDate="2014-12-30T13:12:40.890" Score="51" ViewCount="57836" Body="&lt;p&gt;Today in chemistry class we were discussing Organic Chemistry. We discussed what organic compounds basically are and then I asked the teacher whether $\ce{CO_2}$ is organic or not. She told that it is as it contains carbon and oxygen with a covalent bond. I told her it can't be as it is not found in animals (naturally). I am very confused about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some good reasons to agree with either explanation. (I have searched the internet already but found no great insights as of now).&lt;/p&gt;&#xA;" OwnerUserId="11378" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-04-06T20:43:08.890" LastActivityDate="2017-04-06T20:43:08.890" Title="Is carbon dioxide organic or inorganic?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22196" PostTypeId="2" ParentId="22195" CreationDate="2014-12-30T13:50:17.023" Score="59" Body="&lt;p&gt;It is entirely arbitrary whether you call it an organic compound or not, though most would not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distinction you make that organic compounds should be found in living things is not a useful criterion. Moreover you are wrong that carbon dioxide isn't: it is both made and used by living things. Animals make it when they metabolise sugars to release energy; plants consume it when they build more complex organic molecules through photosynthesis. In fact most organic molecules are, ultimately, derived from $\ce{CO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more importantly most molecules considered organic are neither made by nor are found in living things. Chemists make new carbon compounds all the time (tens of millions in the history of chemistry) and most have never been made by animals or plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;organic/inorganic&lt;/em&gt; terminology is mostly very simple: covalent compounds containing carbon are &lt;em&gt;organic&lt;/em&gt;. The only fuzzy area is around very simple molecules like $\ce{CO2}$ where the distinction doesn't matter much. So we would not normally think of diamond or silicon carbide as organic. But we &lt;em&gt;might&lt;/em&gt; (though many would not) call calcium carbide organic because it contains a $\ce{C2}$ unit with a carbon-carbon triple bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However since the terminology is mostly very obvious and also somewhat arbitrary, it isn't worth much argument to sort out those very simple but awkward edge cases.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T21:24:58.020" LastActivityDate="2015-06-10T21:24:58.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22197" PostTypeId="2" ParentId="22195" CreationDate="2014-12-30T13:50:59.043" Score="11" Body="&lt;p&gt;Carbon dioxide is indeed strictly speaking organic, as it is a compound containing carbon. Your claim that it doesn't naturally occur in living organisms is not correct. For instance, it is the product of breathing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, some chemists do not consider it to be organic as it has no hydrogen in it- a critical component of just about every organic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to remember that the term &quot;organic&quot; is just a vague definition given by humans to some types of chemicals. Nature does not always play by our rules, especially when you're talking about borderline cases such as carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2014-12-30T13:50:59.043" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22198" PostTypeId="1" AcceptedAnswerId="149883" CreationDate="2014-12-30T14:39:23.203" Score="6" ViewCount="6067" Body="&lt;p&gt;I'm a hobby enthusiast that loves to play with electronics so my major is not in chemistry, there for I'm hoping that this forum is the right place to find aid in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched day in and day out, and when someone refers to &quot;epoxy&quot; with &quot;extreme high temperature tolerance&quot; they're generally speaking about 200-300°C, which to me isn't that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that depending on the context that is a lot, but there has to be something that is actually &quot;extreme&quot; in the sense that it can tolerate up to 1000 °C since formula 1 cars typically hit around 1000°C at the end of a break line and typically the thickness and heat dispersion of the setup just can't take more because of the limits within FIA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what is the technical term for high temperature epoxy's and what other options is there if not? &lt;/p&gt;&#xA;" OwnerUserId="5271" LastEditorUserId="10185" LastEditDate="2015-03-23T18:09:30.487" LastActivityDate="2021-04-13T07:32:47.870" Title="Are there two compound resins that tolerate over 1000°C for glass/carbon fiber?" Tags="&lt;materials&gt;&lt;temperature&gt;" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22199" PostTypeId="2" ParentId="22198" CreationDate="2014-12-30T15:44:32.067" Score="5" Body="&lt;p&gt;Polyether ether ketone (&lt;strong&gt;PEEK&lt;/strong&gt;) should be good to go up to 250 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polybenzimidazoles (&lt;strong&gt;PBI&lt;/strong&gt;s) are stable up to temperatures of 500 °C, probably more. Protective suits for fire fighters are made from this material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to withstand higher temperatures (nose cone of a Space Shuttle), &lt;strong&gt;carbon-reinforced carbon&lt;/strong&gt; can be used. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-30T15:44:32.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22200" PostTypeId="2" ParentId="22149" CreationDate="2014-12-30T16:06:35.800" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparing acidic strengths is fine - I remove an H+ and compare the stability of the anion coz my teacher told me strong acid has a stable anion and explained me the reason from equilibrium point of view.&#xA;  When it comes to comparing basic strengths what do we do ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You follow the same general idea - the stronger the base, the weaker its conjugate acid. Consider a generic base such as $\ce{B-}$ reacting with water, the acid in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B- + H2O &amp;lt;=&amp;gt; HB + HO-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate acid of $\ce{B-}$ is $\ce{HB}$. And the stronger that $\ce{HB}$ is as an acid, the more likely it is to protonate hyroxide ion and favor the left side of the equilibrium, thereby making the base weak. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I add an H+ to the species and compare the resulting stability by resonance ,hyperconjugation and inductive effect and then can I conclude the more basic one will have a stabler cation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yep, that's it. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2014-12-30T16:06:35.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22202" PostTypeId="2" ParentId="22195" CreationDate="2014-12-30T16:12:56.617" Score="29" Body="&lt;p&gt;There is no exact definition of &quot;organic&quot; compound, although you can say that organic compounds &lt;em&gt;must&lt;/em&gt; contain carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no requirement that organic compounds can be found in living things, although the name organic comes from the fact that the first compounds in this class that were discovered did come from living things. That said,  $\ce{CO2}$ &lt;em&gt;is&lt;/em&gt; in living things - all metabolic processes either consume or produce  $\ce{CO2}$ (with the exception of some &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermophile&quot;&gt;thermophiles&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, ionic and &quot;simple&quot; molecular compounds that contain carbon are &lt;strong&gt;not&lt;/strong&gt; called &quot;organic,&quot; partly for historical reasons, but also partly because the category &quot;organic&quot; is more than just a name - it describes an entire set of bond types, reaction patterns, naming rules, and structure types. This is why there is no exact, simple definition - it is hard to draw a boundary around all of that and say &quot;these other things don't belong.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking your example, $\ce{CO2}$ is not normally called &quot;organic,&quot; even though it contains carbon, it is produced and consumed in biochemical reactions, it has similar properties to other organic molecules, and organic bonding principles can describe it. The reason for this is probably because historically,  $\ce{CO2}$ was discovered and named before organic chemistry existed as a field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people say that in order to make it easy, any carbon-containing covalent compound is &quot;organic&quot; (your teacher is probably in this group). Some people say &quot;any compound containing carbon and hydrogen&quot; should be the rule. Both of these rules have exceptions, though, so they can't be used as exact definitions.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2014-12-30T16:12:56.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22204" PostTypeId="2" ParentId="22198" CreationDate="2014-12-30T16:46:17.800" Score="7" Body="&lt;p&gt;In an oxygen-containing environment, no carbon-based polymer (which includes all epoxies) that I know of can tolerate 1000 C, at least not without significant degradation in mechanical strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two reasons for this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is that the carbon-carbon bonds will start to break down in large numbers at higher temperatures, and since the carbon-carbon to carbon-oxygen reaction is exothermic, if oxygen is around there will be no going back upon cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is that even if the polymer doesn't chemically degrade, it will undergo a physical phase transition - it will melt, or at least soften. This can be compensated for by reinforcing with fiber or particles to make a composite. However, if it softens enough, a fiber composite will &quot;delaminate&quot; - the physical bonds holding the resin onto the reinforcment will come apart, and the composite will no longer be a composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ewp.rpi.edu/hartford/~ernesto/F2012/EP/MaterialsforStudents/Patel/Beyler_Hirschler_SFPE_Handbook_3.pdf&quot; rel=&quot;nofollow&quot;&gt;Here is an interesting study&lt;/a&gt; that talks about both mechanisms of degradation, and lists several systems with their correpsonding melt and thermal decomposition temperatures. Polytetrafluoroethylene has the highest thermal decomposition temperature in this study, at 775 K (502 C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-carbon polymers might be an option, although the tradeoff is that they are weaker and composite options are a lot more limited. Silicon-based polymers generally withstand higher temperatures than carbon analogues, but I wasn't able to find any that can approach 1000 C. Fluoro-polymers are another possibility - as you can see from the reference, they tend to have much higher stability in terms of thermal decomposition. There might be a fluoro-epoxy resin out there. One thing to keep in mind, though, is that the products of fluoro-polymer decomposition are particularly toxic and reactive, and so the applications where they could safely be used at high temperatures are probably very, very limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit in response to questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One question regarding the &quot;oxygen around&quot;, would this also include if layers of epoxy were applied as a extra layer ontop of the carbon or would it need to be a complete vacuum in order to prohibit oxidization?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you had a layered carbon/polymer composite, the chemical reaction with the oxygen wouldn't start at the polymer-carbon interface, it would start at the surface exposed to oxygen. However, since the reaction is exothermic, once it starts burning, it would go quickly. The carbon would burn too at these temperatures. An inert atmosphere would prevent that reaction, but it would still thermally decompose - the difference would be that hopefully the decomposition wouldn't be as exothermic, and therefore would go more slowly. Now for epoxies, they have oxygen in the polymer structure. At high temperatures, that oxygen will react and the polymer will break apart whether there is an inert atmosphere or not.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2014-12-30T17:28:26.533" LastActivityDate="2014-12-30T17:28:26.533" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22206" PostTypeId="2" ParentId="8259" CreationDate="2014-12-30T16:56:23.293" Score="4" Body="&lt;p&gt;Chromophoric bi-product of alkali hydrolysis on cellulose is called oxycellulose. It is this which will have produced the colour - like you get on old clothes. This acid will have made the cotton stiffer/brittle, but the colour will come from the alkali.&lt;/p&gt;&#xA;" OwnerDisplayName="user11519" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T11:26:50.263" LastActivityDate="2015-06-09T11:26:50.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22207" PostTypeId="1" AcceptedAnswerId="22212" CreationDate="2014-12-30T17:06:17.833" Score="1" ViewCount="194" Body="&lt;p&gt;$\lambda=\frac{KT}{\sigma p}$ is the equation given in my textbook but everywhere else I look it is different. Is it wrong? The derivation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{v_{rel}}{z}$$ where $\lambda$ is the mean free path and $z$ is the collision frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $z=\frac{\sigma v_{rel}}{KT}$ where $\sigma$ is the area where it is possible for two particles to collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting for $z$ gives the equation $\lambda=\frac{KT}{\sigma p}$. Have I gone wrong somewhere. I often see a factor of root two in the denominator but I don't see where that might come from.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4356" LastEditDate="2014-12-30T17:13:30.843" LastActivityDate="2014-12-30T18:34:55.093" Title="Is this the correct equation for the mean free path?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22208" PostTypeId="1" AcceptedAnswerId="22216" CreationDate="2014-12-30T17:39:48.273" Score="3" ViewCount="276" Body="&lt;p&gt;When a neutralisation reaction happens, for example, &lt;span class=&quot;math-container&quot;&gt;$ \pu{100 mol l^-1}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\pu{100 mol l^-1}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NH_3}$&lt;/span&gt;, why does &lt;em&gt;all&lt;/em&gt; of the base and acid get converted to salt? Why isn't there an equilibrium established between salt and acid-base? &lt;/p&gt;&#xA;" OwnerUserId="9370" LastEditorUserId="4945" LastEditDate="2018-10-22T13:12:00.640" LastActivityDate="2018-10-22T13:14:33.857" Title="Why do acids usually completely react with bases?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22210" PostTypeId="1" CreationDate="2014-12-30T18:27:06.397" Score="2" ViewCount="88" Body="&lt;p&gt;What is the effect of rolling a graphene sheet into a carbon nanotube on the lattice constant and the geometry of the graphene?&lt;/p&gt;&#xA;" OwnerUserId="11520" LastEditorUserId="7448" LastEditDate="2015-04-02T18:28:08.183" LastActivityDate="2015-04-02T18:28:08.183" Title="Distortion of graphene structure" Tags="&lt;organic-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22211" PostTypeId="1" CreationDate="2014-12-30T18:29:12.517" Score="6" ViewCount="96" Body="&lt;p&gt;I have a question. I work at a brass metal foundry where one of our waste products is $\ce{ZnO}$, which we sell. Because we have been making new alloys which contain sulfur, our $\ce{ZnO}$ has a very small sulfur impurity level and the people that we sell our $\ce{ZnO}$ to have been complaining that our $\ce{ZnO}$ makes their product yellow. Do you think this is due to the sulfur? What are the effects of very little sulfur on tons of $\ce{ZnO}$ dust? thanks!&lt;/p&gt;&#xA;" OwnerUserId="11521" LastEditorUserId="4231" LastEditDate="2015-01-30T22:36:05.487" LastActivityDate="2015-01-30T22:56:11.910" Title="What are the physical properties affected in ZnO when there is a sulfur impurity?" Tags="&lt;temperature&gt;&lt;boiling-point&gt;&lt;color&gt;&lt;photochemistry&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22212" PostTypeId="2" ParentId="22207" CreationDate="2014-12-30T18:34:55.093" Score="1" Body="&lt;p&gt;Your derivation is correct. The $\sqrt{2}$ comes from $v_{rel}=\sqrt{2}v_{mean}$, where $v_{rel}$ is the mean relative speed (relative to the other molecules in the gas) and $v_{mean}$ is just the mean speed of the molecules in the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion: note what the subscript says about the variable!&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2014-12-30T18:34:55.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22213" PostTypeId="2" ParentId="22168" CreationDate="2014-12-30T18:41:43.003" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The molar solubility &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; equals:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;According to the equilibrium: &lt;span class=&quot;math-container&quot;&gt;$\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$&lt;/span&gt;, the  equilibrium constant is:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_{sp}=4s^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By substituting the value of &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; in the above equation, we find:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_{sp}=0.034\times10^{-9}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-12-17T10:34:17.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22216" PostTypeId="2" ParentId="22208" CreationDate="2014-12-30T19:25:50.770" Score="3" Body="&lt;p&gt;Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{HCl + H3N &amp;amp;-&amp;gt; H4N+ + Cl-}&amp;amp;&#xA; \text{Extent:}&amp;amp; \approx100\%&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The products are &lt;strong&gt;ammonium ion&lt;/strong&gt; and &lt;strong&gt;chloride ion&lt;/strong&gt; - the &lt;strong&gt;conjugate acid&lt;/strong&gt; and &lt;strong&gt;conjugate base&lt;/strong&gt;, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{H4N+ + Cl- &amp;amp;&amp;lt;&amp;lt;=&amp;gt; HCl +H3N}&amp;amp;&#xA; \text{Extent:}&amp;amp; \approx0\%&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4945" LastEditDate="2018-10-22T13:14:33.857" LastActivityDate="2018-10-22T13:14:33.857" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22217" PostTypeId="1" AcceptedAnswerId="22220" CreationDate="2014-12-30T20:37:34.503" Score="4" ViewCount="1099" Body="&lt;p&gt;In our science class, we're learning about the four fundamental macromolecules in every living organism: carbohydrates, proteins, lipids, and nucleic acids. Our class definition of a macromolecule essential is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A macromolecule is a large molecule that is essential to all life, and present in all living cells. It is a polymer- a chain of monomers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Our definition of a monomer is like so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A monomer is the simplest building block of a macromolecule with the properties of that macromolecule. They can be strung together to produce a macromolecule (usually by dehydration synthesis).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would have no problem with these definitions if not for my teacher mentioning once that some monomers can also be macromolecules by themselves. Because some monomers of certain macromolecules- such as the monosaccharide glucose vs. the disaccharide sucrose or the polysaccharide amylose - can act on their own as an essential and functional carbohydrate, they are macromolecules by themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? For example, could glucose be a macromolecule by itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11524" LastEditorUserId="7448" LastEditDate="2014-12-30T20:55:06.713" LastActivityDate="2015-01-09T05:13:31.213" Title="Can a single monomer be a macromolecule? If so, when?" Tags="&lt;biochemistry&gt;&lt;molecules&gt;&lt;polymers&gt;&lt;chemical-biology&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22218" PostTypeId="2" ParentId="22195" CreationDate="2014-12-30T23:25:49.707" Score="-7" Body="&lt;p&gt;organic usually refers to chemicals made from oil coal or gas, i.e the organic compounds are made from organic carbon sources and come from natural sources. This is why CO&lt;sub&gt;2&lt;/sub&gt; is not regarded as organic.  &lt;/p&gt;&#xA;" OwnerUserId="11525" LastEditorUserId="8083" LastEditDate="2014-12-30T23:58:44.963" LastActivityDate="2014-12-30T23:58:44.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22219" PostTypeId="1" CreationDate="2014-12-30T23:58:25.007" Score="2" ViewCount="631" Body="&lt;p&gt;Can electroplating improve the texture of a steel ball bearing so that it will have a smooth surface free of scratches even a few microns wide? I'm trying to find a way to remove scratches (about 10 microns wide) from carbon-steel electrodes and have recently started looking into electroplating. I have already ruled out electropolishing and using rotary/vibratory tumblers.&lt;/p&gt;&#xA;" OwnerUserId="11526" LastActivityDate="2018-04-12T17:38:12.853" Title="Electroplating to create a scratch-free surface?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22220" PostTypeId="2" ParentId="22217" CreationDate="2014-12-31T03:27:04.893" Score="7" Body="&lt;p&gt;&lt;strong&gt;I can't think of an example where a biological monomer would be a macromolecule.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions of &lt;a href=&quot;http://en.wikipedia.org/wiki/Macromolecule&quot; rel=&quot;nofollow&quot;&gt;macromolecule&lt;/a&gt; vary, usually by molecular weight or number of monomers (repeat units). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I'd go with ~1000 Dalton for a &lt;strong&gt;&lt;em&gt;minimum&lt;/em&gt;&lt;/strong&gt;, but the original definition of 1000 atoms is a good start too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, no biological monomer, including glucose will function the same as a macromolecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider starch - a macromolecule of sugars. It doesn't dissolve as quickly as simple sugar and has different physical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify my comments.. Macromolecules or polymers are made up of monomers the way words are made up of letters. So no, a glucose molecule isn't really the same as a macromolecule, just like &quot;R&quot; is not a word.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, macromolecules can be used to make larger assemblies like microtubules, filaments, etc., much the same way that words can form sentences and paragraphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the polymer literature there's even the concept of a &lt;strong&gt;&quot;macromonomer&quot;&lt;/strong&gt; referring to a monomer that is already large in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion though, the basic constituents (monomers) are still amino acids, nucleic acids, sugars, etc. These are not macromolecules.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-01-09T05:13:31.213" LastActivityDate="2015-01-09T05:13:31.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22221" PostTypeId="2" ParentId="22219" CreationDate="2014-12-31T04:08:44.773" Score="2" Body="&lt;p&gt;In general, this doesn't really work. To level a surface by electrodeposition, material would have to deposit in recesses in the surface faster than it does onto surface features that stick out. Normally, the opposite occurs because things that stick out are more accessible for diffusion. Electropolishing works because it's doing the opposite process in solutions viscous enough to enhance the differences in diffusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to say that it's impossible to level surfaces by electrodeposition. It's been known for a &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/bbpc.19550590731/abstract&quot; rel=&quot;nofollow&quot;&gt;long time&lt;/a&gt; that certain additives to plating baths can change the deposition selectivity to this end. However, the mechanisms behind these effects are not well known and it seems to be pretty difficult to achieve consistently. Also, you won't be able to electrodeposit carbon steel, so you would have some other metal on the surface, which may or may not adhere well. I wasn't able to find any specific method for carbon steel—electropolishing seems to be much more common.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-31T04:08:44.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22224" PostTypeId="1" AcceptedAnswerId="22227" CreationDate="2014-12-31T09:04:10.497" Score="3" ViewCount="268" Body="&lt;p&gt;I know that level A type hazmat suits can protect against minor splashes, but could somebody in one survive without harm after complete immersion for a few minutes in deadly chemicals such as sodium cyanide, high concentration (100%) hydrogen peroxide, and 80% hydrochloric/nitric acid?&lt;/p&gt;&#xA;" OwnerUserId="11530" LastEditorUserId="41328" LastEditDate="2019-06-17T05:28:33.560" LastActivityDate="2019-06-18T05:06:24.193" Title="How effective is a level A hazmat suit for immersion in deadly substances?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22226" PostTypeId="2" ParentId="21983" CreationDate="2014-12-31T09:26:44.497" Score="9" Body="&lt;p&gt;I don't have a fireplace or wood stove, so I don't have a practical technique that I know works to prevent soot buildup on glass, but I can answer some of your questions about glass and soot. To start, I should add a caveat here: while many fireplaces use tempered glass windows, wood stoves generally have glass-ceramic windows to withstand the increased heat. As far as surface properties go for this, glass-ceramic isn't dramatically different from normal glass.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does each glass (especially my one) have pores with a size that starch&#xA;  particles can fit in? I read somewhere that starch has a diameter&#xA;  between 2-170 µm depending on the type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While special techniques exist for making porous glasses, normal glass is not porous and float glass (the standard technique for fabricating flat glass panes) produces quite smooth pieces even without polishing (&lt;a href=&quot;http://www.delta-technologies.com/Products.asp?C=13&quot;&gt;hundreds of nm peak-to-peak&lt;/a&gt;), so probably whole starch particles are not getting stuck in the surface of the glass and the surface roughness probably has little impact on soot adhesion in this case.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm making a fire where I burn wood! Why wouldn't the starch simply&#xA;  burn and create itself the soot?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless the flame is right up to the glass, it probably doesn't get hot enough to really burn starch (if any is even there). That site didn't compare it directly with not applying starch so it's not clear that it makes any difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does soot building work? I mean, why does the soot stick so hard&#xA;  at the window? After all it's just little particles flowing around and&#xA;  landing on the window. (I know it's called burning in, but how does&#xA;  the heat work on a already burned particle?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Soot is basically the deposition of incomplete combustion products from a flame. A wood flame appears yellow or orange because tiny particles of mostly carbon are incandescing. There is insufficient oxygen to fully combust the carbon in the fuel to $\ce{CO2}$, so the combustion results in carbon molecules in various states of oxidation from small polymers to elemental carbon. These species are really only stable as gases within the heat of the flame itself and will thus readily deposit onto cold surfaces, like the window glass. Many modern wood stoves seem to have &quot;&lt;a href=&quot;http://www.woodheat.org/maintenance.html&quot;&gt;air-wash&lt;/a&gt;&quot; systems that flow air past the inside of the window to reduce how much soot can actually reach the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is actually deposited as soot is a &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2010/CP/b924344e#!divAbstract&quot;&gt;complex mixture&lt;/a&gt;, but what is emitted from the flame is related to how much air is available and the temperature of the flame (the two are also connected with each other). If the oxygen supply is increased, more complete combustion is favoured so less carbon species that can deposit as soot are formed. The temperature that is attained influences the composition of soot particles: higher temperatures produce more aliphatic and less aromatic compounds. The practical significance is that reducing the flame intensity by limiting oxygen can greatly increase the production of soot. (and may have some effect on how well it adheres to the glass, though I wasn't able to find anything concrete to that effect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the non-scientific sources I glanced through, it seems to be well known that excessive soot buildup can be due to wet fuel or low airflow and that burning a very hot fire can help make soot easier to remove. As for the corn starch, it may indeed be leaving some deposit that hinders soot adhesion (there are commercial cleaners that claim to leave a film behind that makes soot easier to remove), the act of applying it may be removing existing soot that's too small to see but new soot adheres readily too, or it may do nothing. Perhaps you might want to try applying corn starch to half of a freshly-cleaned window to see if there is a noticeable difference.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-31T09:26:44.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22227" PostTypeId="2" ParentId="22224" CreationDate="2014-12-31T11:07:20.847" Score="3" Body="&lt;p&gt;&lt;em&gt;Note that I'm used to standard lab safety measures including gas masks and masks mith independent air supply but I never used a Hazmat level A suit in research. You might want to take the following with a grain of salt!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;Sodium cyanide &lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;($\ce{NaCN}$) is a solid with a melting point around 560 °C. While it is a solid, jumping into a pool of it shouldn't do any harm. Level A suits are like &quot;your personal bubble&quot; with independant air support. They are however not that heat resistant: Do not jump into a pool of molten sodium cyanide!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;High concentration (100%) Hydrogen Peroxide&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3% solutions of hydrogen peroxide ($\ce{H2O2}$) in water are used for wound disinfection, solutions up to 18% are used to bleach hair. (In German, the term &lt;em&gt;wasserstoffblond&lt;/em&gt; was used for &lt;em&gt;bleach blonde&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much higher concentrations or waterfree (100%) ($\ce{H2O2}$) is another league. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide can decompose into water and &lt;strong&gt;oxygen&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O2 -&amp;gt;[\textrm{cat}] H2O + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction releases some heat, may happen spontaneously and is catalyzed by some metals. Add some burnable material that can be vapourized and you have a &lt;strong&gt;rocket propellant&lt;/strong&gt;. Think &lt;strong&gt;Messserschmitt Me 163&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hazmat suit is tight, and the material withstands &quot;bleaching&quot; by the hydrogen peroxide there's still the question whether hooks, valves, visor frames or any other metal parts, that might be part of the whole gear will catalyze the decomposition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;80% hydrochloric acid&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That does not exist in water. Water won't take up that much hydrogen chloride gas under normal conditions. Concentrated hydrochlorid acid (~ 38%) should be fine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;80% nitric acid&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's close to fuming nitric acid and again a pretty strong oxidant. I'd say that you're safe, but don't take my word for it ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2014-12-31T11:07:20.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22228" PostTypeId="1" CreationDate="2014-12-31T16:27:09.877" Score="-1" ViewCount="4948" Body="&lt;p&gt;I am searching about some basics in chemistry. I was looking for the molecular structure of all periodic table element molecules.&#xA;eg: Hydrogen molecule: as &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;; structure:&lt;img src=&quot;https://i.stack.imgur.com/lwUhA.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I was able to find out several others too. But when I reach to certain molecules like &lt;span class=&quot;math-container&quot;&gt;$\ce{He}$&lt;/span&gt; I was not able to find out the structure. &lt;br/&gt;&#xA;Does anyone know from where I can get the molecular structure of all periodic table elements. And if not, could you please explain the case of Helium and other noble gases. I am very poor in chemistry. Anyone showing me a reference on how to find the answer is welcome. Thanks in advance, and Happy New Year to all... :)&lt;/p&gt;&#xA;" OwnerUserId="11536" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-04-09T18:18:42.327" Title="Molecular structure of all periodic table element molecules exceptional cases etc" Tags="&lt;molecules&gt;&lt;elements&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2016-09-20T16:41:15.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="22229" PostTypeId="2" ParentId="22211" CreationDate="2014-12-31T17:01:05.980" Score="1" Body="&lt;p&gt;Of course, if $\ce{ZnO}$ has a very small $\ce{S}$ impurity level, this could be the reason of the yellow colour of products. &#xA;As for the effects of very little sulfur on tons of $\ce{ZnO}$ dust, it all depends on the end use of $\ce{ZnO}$:The applications of zinc oxide powder are numerous. Most applications exploit the reactivity of the oxide as a precursor to other zinc compounds. For material science applications, zinc oxide has high refractive index, high thermal conductivity, binding, antibacterial and UV-protection properties. Consequently, it is added into materials and products including plastics, ceramics, glass, cement, rubber, lubricants, paints, ointments, adhesive, sealants, concrete manufacturing, pigments, foods, batteries, ferrites, fire retardants, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, if $\ce{ZnO}$ is used in rubber industry, sulfur impurities have no harm effect.&#xA;If it is used in electrical applications (as a semiconductor), sulfur will affect the electrical conductivity. &#xA;If it is used in coating and paints applications, sulfur will affect the the color ....&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2014-12-31T17:01:05.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22230" PostTypeId="1" CreationDate="2014-12-31T17:24:01.153" Score="0" ViewCount="67" Body="&lt;p&gt;In the reaction $\ce{CH_4 + 2O_2-&amp;gt;CO_2 + 2H_2O}$ the $\ce{C}$ atom is oxidized. I can clearly see the addition of $\ce{O}$ atoms, however I don't see a change in Oxidation state. In the products, $\ce{C}$ oxidation state (O.S) was +4, and in the reactants, $\ce{C}$ O.S is +4. I believed an oxidation meant an increase in oxidation state.&#xA;What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="11430" LastActivityDate="2014-12-31T17:53:59.333" Title="Oxidation of methane" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22232" PostTypeId="2" ParentId="22230" CreationDate="2014-12-31T17:53:59.333" Score="2" Body="&lt;p&gt;Actually in $\ce{CH4}$ with slightly more electronegtaive carbon it is in $-4$ state. With $\ce{CO2}$ the oxygen is more electronegative oxygen carbon is $+4$. Therefore oxidation is also defined as addition/replacement of a more electronegative element.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Electronegativities $$\begin{array}{}\bf H&amp;amp;2.2\\\bf C&amp;amp;2.55\\\bf O&amp;amp;3.44\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2014-12-31T17:53:59.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22233" PostTypeId="1" AcceptedAnswerId="22237" CreationDate="2014-12-31T17:55:11.363" Score="2" ViewCount="3202" Body="&lt;p&gt;What type of material is used in the control rod of a nuclear reactor? (Nuclear power plant)&lt;/p&gt;&#xA;" OwnerUserId="11537" LastEditorDisplayName="user7951" LastEditDate="2019-07-03T17:06:18.633" LastActivityDate="2019-07-03T17:06:18.633" Title="Nuclear power: what material is the control rod made from?" Tags="&lt;materials&gt;&lt;nuclear&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22236" PostTypeId="1" CreationDate="2014-12-31T18:26:49.003" Score="2" ViewCount="3265" Body="&lt;p&gt;I know that electronegative metals like copper or silver can electroplate out of an aqueous solution in electrolysis. However, I've also noticed $\ce{ZnSO4_(aq)}$ can electroplate out when electrolyzed, yet Zinc is more electropositive than hydrogen. This doesn't happen with other electropositive ionic solutions like $\ce{Al(C2H3O2)3_(aq)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen, and why is it limited only to Zinc?&lt;/p&gt;&#xA;" OwnerUserId="11538" LastEditorUserId="7581" LastEditDate="2014-12-31T19:11:45.510" LastActivityDate="2014-12-31T19:42:52.903" Title="Why is an electropositive metal like Zinc capable of plating out?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22237" PostTypeId="2" ParentId="22233" CreationDate="2014-12-31T18:32:18.047" Score="1" Body="&lt;p&gt;&lt;strong&gt;Control Rods:&lt;/strong&gt; They have the ability to capture the slow neutrons and can control the chain reaction at any stage. Boron and cadmium are the best absorbers of neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start fission in &lt;span class=&quot;math-container&quot;&gt;$\ce{^{238}U}$&lt;/span&gt; we need very fast neutrons having kinetic energies more than about 1 MeV. In a nuclear reactor, the probability that a neutron will cause &lt;span class=&quot;math-container&quot;&gt;$\ce{^{235}U}$&lt;/span&gt; to fission is more for slow-moving neutrons. Thus there is less probability that the fast neutrons produced in fission will cause other fission to occur and a chain reaction will start. To slow down these neutrons (to initiate more fission events) in a nuclear reactor &lt;strong&gt;moderator&lt;/strong&gt; is used which may be ordinary water or heavy water (&lt;span class=&quot;math-container&quot;&gt;$\ce{D2O}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2019-07-03T16:53:39.443" LastActivityDate="2019-07-03T16:53:39.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22238" PostTypeId="1" AcceptedAnswerId="22243" CreationDate="2014-12-31T19:27:22.403" Score="9" ViewCount="18447" Body="&lt;p&gt;My question can also be understood as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what make something electrolyte or nonelectrolyte?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this question stems mainly from my confusion with the definitions of ionizing and dissolving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolving of a ionic substance is due to ion-dipole attraction be/wn water molecules and ionic formula unit. If water molecule has enough attraction to break down the ionic compound, the the substance will dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionizing. Uh, all I know about it is...I don't really know. I think only ionic compound will ionize because it is composed of ions. So once it dissolves it basically ionize, right?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2014-12-31T19:36:17.790" LastActivityDate="2021-02-26T05:15:05.627" Title="Why do some substances ionize instead of dissolving in water?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22239" PostTypeId="5" CreationDate="2014-12-31T19:30:58.183" Score="0" Body="&lt;p&gt;One of the main courses of chemistry research in the current decade is devoted to exploring the different allotropes of carbon. Carbon can form a variety of intramolecular bonds with itself, which allows its allotropes to have very special properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most important allotropes are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Diamond&lt;/strong&gt;&lt;br&gt;&#xA;In the bulk material, every carbon atom is covalently bonded to four other carbon atoms, giving it a tetrahedral coordination sphere. It is a very hard material and therefore has many applications in industry and also as jewelry.&lt;br&gt;&#xA;From the &lt;a href=&quot;http://goldbook.iupac.org/D01671.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An allotropic form of the element carbon with cubic structure (space group  $O_{h}^{7} − F d 3 m$) which is thermodynamically stable at pressures above  $6~\mathrm{GPa}$ at room temperature and metastable at atmospheric pressure. At low pressures, diamond converts rapidly to graphite at temperatures above $1900~\mathrm{K}$ in an inert atmosphere. The chemical bonding between the carbon atoms is covalent with $\ce{sp^3}$ hybridization.&lt;br&gt;&#xA;  &lt;em&gt;Note:&lt;/em&gt;&lt;br&gt;&#xA;  There is also a hexagonal diamond-like structure of the element carbon (lonsdaleite).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Graphite&lt;/strong&gt;&lt;br&gt;&#xA;Graphite is the most common allotrope of carbon and its thermodynamically stable form at standard temperature and pressure. It is also a good electrical conductor.&lt;br&gt;&#xA;From the &lt;a href=&quot;http://goldbook.iupac.org/G02684.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An allotropic form of the element carbon consisting of layers of hexagonally arranged carbon atoms in a planar condensed ring system graphene layers The layers are stacked parallel to each other in a three-dimensional crystalline long-range order. There are two allotropic forms with different stacking arrangements, hexagonal and rhombohedral. The chemical bonds within the layers are covalent with $\ce{sp^2}$ hybridization and with a $\ce{C–C}$ distance of  $141.7~\mathrm{pm}$. The weak bonds between the layers are metallic with a strength comparable to van der Waals bonding only.&lt;br&gt;&#xA;  &lt;em&gt;Note:&lt;/em&gt;&lt;br&gt;&#xA;  The term graphite is also used often but incorrectly to describe graphite materials, i.e. materials consisting of graphitic carbon made from carbon materials by processing to temperatures greater than $2500~\mathrm{K}$, even though no perfect graphite structure is present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Graphene layer&lt;/strong&gt;&lt;br&gt;&#xA;A special form of graphite. According to the &lt;a href=&quot;http://goldbook.iupac.org/G02683.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A single carbon layer of the graphite structure, describing its nature by analogy to a polycyclic aromatic hydrocarbon of quasi-infinite size.&lt;br&gt;&#xA;  &lt;em&gt;Note:&lt;/em&gt;&lt;br&gt;&#xA;  Previously, descriptions such as graphite layers, carbon layers or carbon sheets have been used for the term graphene. Because graphite designates that modification of the chemical element carbon, in which planar sheets of carbon atoms, each atom bound to three neighbours in a honeycomb-like structure, are stacked in a three-dimensional regular order, it is not correct to use for a single layer a term which includes the term graphite, which would imply a three-dimensional structure. The term graphene should be used only when the reactions, structural relations or other properties of individual layers are discussed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Fullerene&lt;/strong&gt;&lt;br&gt;&#xA;Also known as buckyballs. All fullerenes are molecules composed entirely of carbon and forming sphere- or cage-like structures of varying size. According to the &lt;a href=&quot;http://goldbook.iupac.org/F02547.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compounds composed solely of an even number of carbon atoms, which form a cage-like fused-ring polycyclic system with twelve five-membered rings and the rest six-membered rings. The archetypal example is [60]fullerene, where the atoms and bonds delineate a truncated icosahedron. The term has been broadened to include any closed cage structure consisting entirely of three-coordinate carbon atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;see &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropes_of_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for a more complete list of carbon allotropes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This tag should be applied to questions about the any of the allotropes of carbon and their physical or chemical properties.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="402" LastEditDate="2017-05-04T02:32:27.640" LastActivityDate="2017-05-04T02:32:27.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22240" PostTypeId="4" CreationDate="2014-12-31T19:30:58.183" Score="0" Body="This tag should be applied to questions about the various allotropes of carbon and their physical or chemical properties. Most important modifications are diamond, graphite, fullerenes, graphene and nanotubes." OwnerUserId="7448" LastEditorUserId="4945" LastEditDate="2015-01-21T09:07:24.897" LastActivityDate="2015-01-21T09:07:24.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22241" PostTypeId="2" ParentId="22236" CreationDate="2014-12-31T19:42:52.903" Score="3" Body="&lt;p&gt;It's too simple to think of this situation in terms of electronegativity and it's certainly not limited to zinc. If you look at a &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow&quot;&gt;table of standard electrode potentials&lt;/a&gt;, you can see that the reduction of many metal ions like $\ce{Zn^2+,Pb^2+,Cd^2+,}\textrm{ etc.}$ have standard potentials below that of the reduction of hydronium (0 V) and should be much less favourable than making hydrogen gas if this were the only consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in real electrochemical cells, it generally takes a certain amount of extra voltage to get a certain reaction to occur, known as the overpotential. Like all chemical reactions, redox reactions have a certain activation energy separate from the energy change between the reactants and products, e.g. elemental gold is far lower in energy than a gold solution, but a gold solution will not immediately reduce spontaneously because there is an activation barrier to the process, an extra potential must be applied to make it happen at a reasonable rate. Similarly, though the standard potential for reducing hydronium is 0 V, in practice it may not be observed until much more negative applied voltages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This overpotential for hydrogen evolution is strongly dependant on the type of electrode used. For the reaction to occur, hydronium ion must adsorb to the surface. In electrodes like platinum, this interaction is stabilized so the overpotential is relatively small and it's thus quite difficult to deposit metals like zinc onto platinum. For this reason, other electrodes must be used for working with aqueous solutions at very negative potentials. Glassy carbon, mercury, and bismuth electrodes are often used for this as hydronium is not readily adsorbed to their surfaces, giving large overpotentials for hydrogen evolution. These types of electrodes may be useful down to -1 V or so and more exotic materials like boron-doped diamond electrodes, even more negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, no electrode can prevent hydrogen evolution if the potential is negative enough. Zinc is about the most easily oxidized metal that is practical to electroplate from aqueous solution. Some metals are so easily oxidized that they are basically impossible to plate in aqueous solutions which is why one generally cannot electroplate from aqueous aluminum solutions or solutions of alkali metals (things like sodium and potassium are so easily oxidized that they react violently with water to form hydrogen gas).&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-31T19:42:52.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22242" PostTypeId="1" CreationDate="2014-12-31T20:01:25.313" Score="0" ViewCount="2254" Body="&lt;p&gt;Explain the following experiments and their results. What is their chemical equations as well?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When immersing a Magnesium (Mg) chunk inside aqueous solution of Copper ions (Cu(II)), a red-brown sediment of Copper appears. Why?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When immersing a Silver (Ag) stripe inside aqueous solution of Calcium ions (Ca(II)), no chemical reaction takes place. Why?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11522" LastEditorUserId="16683" LastEditDate="2016-10-22T15:23:03.420" LastActivityDate="2016-10-22T15:23:03.420" Title="Reaction of solid metals with a solution of metal ions" Tags="&lt;homework&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22243" PostTypeId="2" ParentId="22238" CreationDate="2014-12-31T20:13:03.093" Score="9" Body="&lt;p&gt;Dissolution is simply the mixing of two phases to produce a single homogeneous phase. Ionization is a process by which a substance with no net charge is converted into one or more ions. It includes the mundane example of dissolving an ionic compound in water that dissociates into its constituent ions as well as things like forming ions by &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_ionization&quot; rel=&quot;noreferrer&quot;&gt;knocking electrons off with an electron beam&lt;/a&gt;. They're both surprisingly complex topics, but I think the question can be answered conceptually by considering a few compounds: $\ce{NaCl}$ , acetic acid ($\ce{CH3COOH}$), and glucose: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/saYm8.png&quot; alt=&quot;glucose&quot;&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All dissolve readily in water, but only two of them are electrolytes. As you note, $\ce{NaCl}$ is composed of ions and breaks apart into $\ce{Na+}$ and $\ce{Cl-}$ in solution. An electrolyte is any compound that ionizes when dissolved in a given solution, so $\ce{NaCl}$ is an electrolyte in water. Moreover, because there's is no way for a salt to be dissolved without ionizing in water (i.e. $\ce{NaCl_{(aq)}}$ isn't possible), $\ce{NaCl}$ is known as a strong electrolyte as when it dissolves, it ionizes completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH}$ also dissolves readily in water, but isn't an ionic compound. Because it's a weak acid, it exists in this equilibrium in solution:&#xA;$$\ce{CH3COOH_{(aq)} + H2O &amp;lt;=&amp;gt; CH3COO^{-}_{(aq)} + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it can lose a proton to form ions, it does ionize in solution and is thus an electrolyte, but because it's a weak acid, the reaction doesn't go to completion and some $\ce{CH3COOH_{(aq)}}$ exists in solution making acetic acid a weak electrolyte as it doesn't completely ionize in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, glucose readily dissolves in acid, but none of its protons are readily lost so it doesn't ionize at all meaning that it isn't an electrolyte in water. Without getting too much into the thermodynamics of dissolution, the reason glucose is so soluble is because the many hydroxyl groups allow it to hydrogen bond extensively with water, where the equivalent compound without them, hexanal, is only slightly soluble due to the polar carbonyl group, and hexane is essentially insoluble.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2014-12-31T20:13:03.093" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="22244" PostTypeId="1" CreationDate="2014-12-31T21:06:52.383" Score="2" ViewCount="267" Body="&lt;p&gt;If we have the dissociation's energies of hydrogen molecule $\ce{H_{2}}$($D_{0}$) and the corresponding molecule ion $\ce{H_{2}^{+}}$ ($D_{1}$) together with the first energy of ionization of hydrogen atom ($E_{I}$), how can we estimate the first energy of ionization of hydrogen molecule?&lt;/p&gt;&#xA;" OwnerUserId="11540" LastEditorDisplayName="user15489" LastEditDate="2015-05-01T13:35:30.513" LastActivityDate="2015-05-01T13:35:30.513" Title="First ionization energy of hydrogen molecule" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;energy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22245" PostTypeId="1" AcceptedAnswerId="22249" CreationDate="2015-01-01T00:09:51.740" Score="2" ViewCount="775" Body="&lt;p&gt;I've been trying to create 3D structures of unionized tyrosine (and other amino acids, it's just tyrosine is the one I'm up to atm) using the structure of Zwitterions found within PDB files and to do this I need to know what C-OXT (which is found within the carboxylate functional groups) stands for in PDB files as this will help me to decide which oxygen in the carboxylate groups should be represented with the double bond from the associated carbon and which should have the single bond. The other oxygen in the carboxylate group is just called C-O in the PDB file.&lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2015-01-01T01:44:26.923" Title="What does C-OXT stand for in PDB files?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22246" PostTypeId="2" ParentId="22187" CreationDate="2015-01-01T00:09:56.370" Score="5" Body="&lt;p&gt;&lt;strong&gt;No, most semiconductors are ionic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I've seen the term &quot;covalent semiconductor&quot; before. Wikipedia has an article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_superconductor&quot; rel=&quot;noreferrer&quot;&gt;covalent superconductors&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main point would be that &lt;em&gt;many&lt;/em&gt; if not &lt;em&gt;most&lt;/em&gt; semiconductors are inorganic solid-state compounds. Consequently, most of these (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_semiconductor_materials#Types_of_semiconductor_materials&quot; rel=&quot;noreferrer&quot;&gt;III-V semiconductors&lt;/a&gt; are ionic solids, not really covalent ones. Consider GaAs or InP, which are typically classified as ionic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the solid state electronic structure can be considered mainly composed of cations and anions (unlike organic chemistry where the electronic structure is considered mainly of neutral atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Incidentally, defining whether some of these compounds have purely ionic or purely covalent interactions is somewhat futile, since most bonding interactions are both somewhat ionic and covalent.)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-01T00:09:56.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22247" PostTypeId="1" AcceptedAnswerId="22248" CreationDate="2015-01-01T00:17:31.750" Score="2" ViewCount="133" Body="&lt;p&gt;I'd like to be able to convert zwitterions to neutral form in Avogadro if possible. How do I do this? &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="4945" LastEditDate="2015-01-06T06:33:20.173" LastActivityDate="2015-01-06T06:33:20.173" Title="How do I convert zwitterions to neutral form in Avogadro?" Tags="&lt;ions&gt;&lt;software&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22248" PostTypeId="2" ParentId="22247" CreationDate="2015-01-01T01:37:00.380" Score="1" Body="&lt;p&gt;I don't think the neutral protomer of an amino acid can be generated in Avogadro.  What &lt;em&gt;can&lt;/em&gt; be done is adjusting the protonation state for different values of pH.  That option is in the &lt;strong&gt;Build&lt;/strong&gt; menu - &lt;strong&gt;Build&lt;/strong&gt;-&gt; &lt;strong&gt;Add Hydrogens for pH&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2015-01-01T01:37:00.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22249" PostTypeId="2" ParentId="22245" CreationDate="2015-01-01T01:44:26.923" Score="2" Body="&lt;p&gt;The naming conventions for atoms in PDB file require that at the C-terminus of a protein one of the carboxylate oxygens is called &lt;code&gt;O&lt;/code&gt; and the other is called &lt;code&gt;OXT&lt;/code&gt;.  But since it's a deprotonated carboxylate group they are both chemically equivalent, and you'd expect that bond lengths are equal.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2015-01-01T01:44:26.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22250" PostTypeId="1" CreationDate="2015-01-01T04:49:52.173" Score="3" ViewCount="1008" Body="&lt;p&gt;I accidently opened a bag of silica gel and got it on my hand what will happen to my hand because now it kinda feels weird.&lt;/p&gt;&#xA;" OwnerUserId="11544" LastEditorUserId="7448" LastEditDate="2015-01-01T08:26:32.407" LastActivityDate="2015-01-02T18:48:18.637" Title="what would happen if you got a full package of wet silica gel on your hand after it tore open?" Tags="&lt;biochemistry&gt;&lt;water&gt;&lt;safety&gt;&lt;materials&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2015-01-03T07:09:51.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="22251" PostTypeId="1" CreationDate="2015-01-01T06:48:08.613" Score="1" ViewCount="1946" Body="&lt;p&gt;Recently, I came across a question: &lt;b&gt; which of the two compounds has a greater reactivity; $\ce{NH_{3}}$ or $\ce{NF_{3}}$?&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that despite the same structure the dipole moment of $\ce{NH_{3}}$  is greater because of the electronegativity difference. So, I though that the reactivity of $\ce{NF_{3}}$ should be more as it is easy to break the bond due to the high electronegativity of fluorine but the answer turned out to be $\ce{NH_{3}}$ and I can't figure out why.&lt;/p&gt;&#xA;" OwnerUserId="7249" LastEditorUserId="865" LastEditDate="2021-05-22T20:59:44.393" LastActivityDate="2021-05-22T20:59:44.393" Title="How are reactivity and dipole moment related?" Tags="&lt;reactivity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22252" PostTypeId="1" CreationDate="2015-01-01T07:01:25.883" Score="5" ViewCount="1476" Body="&lt;p&gt;This is a concept I have never really understood. I mean to say is how can we include such a thing in a theory? How can we use them if we know that they don't actually exist? Are they some sort of calculation tool?&lt;/p&gt;&#xA;" OwnerUserId="7249" LastActivityDate="2015-01-01T13:05:46.433" Title="What is it with resonance structures?" Tags="&lt;bond&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22253" PostTypeId="2" ParentId="22251" CreationDate="2015-01-01T07:03:42.450" Score="1" Body="&lt;p&gt;Yashbhatt - interesting question, but I think that this question is too vague. Reactivity is a &lt;em&gt;very&lt;/em&gt; broad term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we talking about reactivity in the context of proton transfer? In that case NH3 will be more reactive than NF3 as a proton donor (Brønsted acid) since it actually has protons to donate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of Brønsted basicity, again, NH3 will likely be more reactive than NF3 because in NF3 the electron-withdrawing fluorines reduces affinity for positive charges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of nucleophilicity, a lot of factors come into play. In a protic solvent, it's hard to tell &lt;em&gt;a priori&lt;/em&gt; whether NH3 is more or less nucleophilic than NF3. NH3 has a greater partial negative charge on its central nitrogen atom. But this may also mean greater hydrogen bonding and therefore reduced nucleophilicity versus NF3, which doesn't exhibit hydrogen bonding (due to its lack of hydrogens). &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:55:49.637" LastActivityDate="2018-03-04T02:55:49.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22254" PostTypeId="2" ParentId="22252" CreationDate="2015-01-01T08:10:56.783" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can we use them if we know that they don't actually exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Resonance contributors don't independently exist but rather a composite of the resonance contributors exists.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are they some sort of calculation tool?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So no, resonance contributors don't exist. However, examination of resonance structures can lend insight into reactivity. Why might phenol's conjugate base be unusually stable? We can rationalize it by drawing multiple resonance contributors. These contributors are stabilizing because they show the delocalization of electron density from the oxygen to various sp2 ring carbons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, we cannot do for something such as cyclohexanol, which has no resonance contributors. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-01-01T13:05:46.433" LastActivityDate="2015-01-01T13:05:46.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22255" PostTypeId="2" ParentId="22242" CreationDate="2015-01-01T10:14:38.223" Score="0" Body="&lt;p&gt;This is based on chemical reactivity series based upon electrode potential (electrochemistry topic):&#xA;$${\bf More\;Reactive}:\ce{K\;Ca\;Mg\;Al\;Zn\;Fe\;Ni\;Sn\;Pb\;H\;Cu\;Hg\;Ag\;Au\;Pt}:{\bf Less\;Reactive}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Magnesium[Mg] displaces Copper[Cu] and all other less reactive than it, but since Calcium[Ca] is more reactive than Silve[Ag], silver can't displace Calcium:&#xA;$$\ce{Mg +Cu^2+-&amp;gt; \underbrace{Cu}_{red}+Mg^2+}\\\ce{Ca +Ag+\not\!-&amp;gt; Ag+Ca^2+}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-01T10:14:38.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22257" PostTypeId="1" AcceptedAnswerId="25149" CreationDate="2015-01-01T11:17:02.107" Score="5" ViewCount="2487" Body="&lt;p&gt;What is the reason behind it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5608" LastActivityDate="2015-03-19T05:02:27.777" Title="C-H Bond Dissociation Energy varies with substitution" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22258" PostTypeId="1" CreationDate="2015-01-01T11:20:35.760" Score="8" ViewCount="983" Body="&lt;p&gt;Iron's oxidation state in $\ce{Fe3O4}$. I really can't get the logic behind this.&lt;/p&gt;&#xA;" OwnerUserId="11549" LastEditorUserId="8083" LastEditDate="2015-01-01T15:00:42.067" LastActivityDate="2017-03-04T14:58:46.207" Title="How is it possible to obtain a value of 8/3 as an oxidation state?" Tags="&lt;oxidation-state&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22259" PostTypeId="1" CreationDate="2015-01-01T11:33:11.960" Score="17" ViewCount="22662" Body="&lt;p&gt;How does resonance lower the potential energy of the molecule? Take $\ce{O3}$ as an example.&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="1389" LastEditDate="2015-01-01T18:00:43.010" LastActivityDate="2015-01-03T16:03:20.047" Title="How does Resonance stabilize a molecule?" Tags="&lt;bond&gt;&lt;resonance&gt;&lt;stability&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22261" PostTypeId="2" ParentId="22258" CreationDate="2015-01-01T12:22:00.527" Score="8" Body="&lt;p&gt;Your assumption that all iron atoms in magnetite ($\ce{Fe3O4}$) have the same oxidation state is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a mixed $\ce{Fe(II,III)}$ oxide.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2017-03-04T14:22:05.190" LastActivityDate="2017-03-04T14:22:05.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22262" PostTypeId="2" ParentId="22258" CreationDate="2015-01-01T12:24:21.157" Score="6" Body="&lt;p&gt;The iron atoms in &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetite&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;magnetite&quot;&gt;magnetite&lt;/a&gt;, $\ce{Fe3O4}$, are not all in the same oxidation state: per Michael's comment, while it is not a &lt;a href=&quot;https://en.wikipedia.org/wiki/Spinel&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;spinel&quot;&gt;spinel&lt;/a&gt;-type material, it has a spinel structure (strictly, an inverse spinel structure), with one Fe(II) center and two Fe(III) centers.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2017-03-04T14:17:53.073" LastActivityDate="2017-03-04T14:17:53.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22263" PostTypeId="2" ParentId="22252" CreationDate="2015-01-01T12:24:51.873" Score="5" Body="&lt;p&gt;Resonance structures are a useful way of visualizing and representing the bonding in a molecule - they don't actually exist, but they are a useful tool to see what the bonding and structure of a compound is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonding in a compound can sometimes be nicely represented using &lt;strong&gt;two-center, two-electron&lt;/strong&gt; bonds - for example methane. The structure and reactivity of methane can be rationalised by representing it as a carbon with four hydrogens. Each &quot;bond&quot; can be represented by having two electrons, one from carbon, and one from hydrogen, in a &lt;strong&gt;bonding pair&lt;/strong&gt;. You might draw this as a combination between an sp3 hybrid and a 1s orbital. In reality these hybrid orbitals don't exist either - they are merely a tool to simplify the bonding to &lt;strong&gt;localised bonding&lt;/strong&gt; which is easier to understand. In reality there a multiple interactions between the carbons 2s and 2p orbitals, with the 1s orbital of the hydrogen. However, it is easier, and an accurate enough &lt;strong&gt;model&lt;/strong&gt; to model methane with four localised bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonanace is useful where this idea of &lt;strong&gt;two-center, two electron&lt;/strong&gt; bonds breaks down - for example we know from experiment that the structure of benzene is a regular hexagon, with &lt;strong&gt;identical&lt;/strong&gt; carbon-carbon bonds. However the idea of &lt;strong&gt;two-center two-electron&lt;/strong&gt; bonds breaks down here - we can't assign a structure using the ideas we know - carbon makes four bonds, the shape of benzene is a regular hexagon with identical bonds. In this case we can represent benzene as an in-between structure between two structures that &lt;strong&gt;can&lt;/strong&gt; be represented as nice &lt;strong&gt;two-center, two-electron&lt;/strong&gt; bonds. When we draw benzene as a resonance structure of two rings with alternating double bonds, what we are trying to represent is that the real structure of benzene is somewhere between these two structures - it is neither of the structure, and it doesn't &lt;strong&gt;resonate&lt;/strong&gt; - ie. flick really fast between the two structures - instead it is a mixture of the two structures. In benzene the structure we are trying to represent can be understood by considering each carbon to be sp2 hybridized - again these hybrids aren't real, but a tool to understand bonding more easily. Then we have the 6 p-orbitals in the plane of the molecule. In reality these combine two give 6 molecular orbitals. These give a final structure where the bonding is identical between all six carbon atoms, a kind of one and a half bond - but the shape and idea of this delocalised bonding is very important to the reactivity of benzene and other aromatics - which is where this idea of resonance starts to break down in some ways, but the structures can still be useful, but with a proper understanding of the bonding behind the structure.&lt;/p&gt;&#xA;" OwnerUserId="4222" LastActivityDate="2015-01-01T12:24:51.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22265" PostTypeId="2" ParentId="22258" CreationDate="2015-01-01T13:13:41.790" Score="16" Body="&lt;p&gt;The formula of $\ce{Fe3O4}$ is a shorthand for $\ce{Fe^{II}Fe^{III}2O4}$, so there are two sorts of atoms of iron here in +2 and +3 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is still possible to gain truly fractional oxidation states. The simplest example probably would be potassium superoxide, $\ce{KO2}$. Here, the anion is symmetrical $\ce{O2^{-}}$ anion, so obviously both oxygen atoms are in $\ce{-1/2}$ oxidation state. There are more complicated examples as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often fractional oxidation state comes from symmetrical structures with charge, that would have unpaired electrons otherwise. However, systems with truly fractional oxidation states usually have bonding, hard to describe in terms of common Lewis structures (and with little relevance of idea of oxidation state), and mostly are considered in semi-advanced organic chemistry and advanced inorganic chemistry.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-01-01T13:13:41.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22266" PostTypeId="2" ParentId="22242" CreationDate="2015-01-01T14:04:30.610" Score="1" Body="&lt;p&gt;You can explain these experiments with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow&quot;&gt;Standard Electrode Potential&lt;/a&gt;. With these experimental values you know which element/compound releases more energy upon receiving a comparable amount of electrons. As a rule of thumb, alkalimetals and alkaline earth metals release less energy than, for example, elements from the halogen-group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this table at hand, you can predict that $\ce{Mg}$ will be oxidized and the $\ce{Cu^2+}$ reduced, because of the potential for the two following reactions:&#xA;$$\ce{Mg^2+ + 2e^- -&amp;gt; Mg(s)} \quad E^\circ=−2.372\,\text{V}$$&#xA;$$\ce{Cu^2+ + 2e^- -&amp;gt; Cu(s)} \quad E^\circ=0.337\,\text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing copper-ions will release more energy, so the following chemical reaction will apply:&#xA;$$\ce{Mg +Cu^2+-&amp;gt; Cu + Mg^2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an analogous fashion, for the second question:&#xA;$$\ce{Ag^2+ + 2e^- -&amp;gt; Ag(s)} \quad E^\circ=+1.98\,\text{V}$$&#xA;$$\ce{Ca^2+ + 2e^- -&amp;gt; Ca(s)} \quad E^\circ=−2.868\,\text{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing $\ce{Ag^2+}$ will release more energy - since silver is already in its solid state, the following reaction will &lt;strong&gt;not&lt;/strong&gt; take place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag +Ca^2+-&amp;gt; Ag^2+ + Ca}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: To see how those potentials or potential difference relate to energy see, for example, the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Nernst_Equation&quot; rel=&quot;nofollow&quot;&gt;Nernst Equation / Gibbs free energy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="11551" LastEditDate="2015-01-01T14:11:30.757" LastActivityDate="2015-01-01T14:11:30.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22267" PostTypeId="1" AcceptedAnswerId="22278" CreationDate="2015-01-01T14:48:26.070" Score="4" ViewCount="33584" Body="&lt;p&gt;The carbon is $\mathrm{sp^2}$ hybridised and is therefore planar and should also, theoretically be $120^\circ$. However, VSEPR theory suggests that the &amp;#x3c0; bond would &quot;need more space&quot; due to greater electron repulsion. As a consequence the $\ce{H-C-H}$ bond angle would be smaller. However, since the &amp;#x3c0; bond is out of the plane of the molecule does this actually happen?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="7475" LastEditDate="2016-06-02T20:34:16.257" LastActivityDate="2016-06-02T20:34:16.257" Title="What's the H-C-H bond angle in ethene?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22269" PostTypeId="1" CreationDate="2015-01-01T15:17:15.427" Score="9" ViewCount="3848" Body="&lt;p&gt;A couple of reactions of glucose have been a source of great confusion to me. One of them is it's reaction with the Brady's reagent. It's pretty debatable whether the result is positive or negative. Could someone please shed some conclusive light onto this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, Glucose itself successfully produces an oxime with hydroxylamine. In contrast, it's pentaacetate fails to do so. Why does this occur? Why can't the pentaacetate, like glucose, &quot;break out&quot; of it's cyclic form to give a positive hydroxylamine test?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I found out about the oxime test, I had overlooked the fact that glucose pentaacetate doesn't have any hemiacetal. The Brady's test though, is still open for debate. &lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="10185" LastEditDate="2015-10-02T16:35:30.313" LastActivityDate="2016-04-21T11:18:25.557" Title="Does glucose react with Brady's reagent?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22270" PostTypeId="1" CreationDate="2015-01-01T15:20:34.633" Score="6" ViewCount="2665" Body="&lt;p&gt;How did scientists find out that  $\ce{Cl2, H2, O2}$ atoms have a two-atomic molecular structure ? &lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="22102" LastEditDate="2017-02-17T19:54:12.763" LastActivityDate="2017-02-17T19:54:12.763" Title="How was the diatomic nature of many common gaseous elements originally determined?" Tags="&lt;atoms&gt;&lt;molecules&gt;&lt;elements&gt;&lt;history-of-chemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22272" PostTypeId="2" ParentId="4151" CreationDate="2015-01-01T15:31:06.707" Score="0" Body="&lt;p&gt;From what I have read, the increasing stability of carbocations is attributed to the combined inductive and hyperconjugative effects, at least in case of alkanes. In other organic molecules, the mesomeric effect can also come into play. These effects are easy to understand, so i suggest you read up on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequently, your next question deals with why more stable carbocations have a lower activation energy in their formation. In my opinion, you're getting the cause-effect mixed up. The carbocation is stable BECAUSE the energy required to form it is lower, not the other way around. And why is the energy lower? Because of the aforementioned stabilization effects!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the point of view of a 17 year old kid who's been studying organic chemistry for about 3 years, and faced the same problems as you're facing. I won't be able to explain it as ManishEarth did though, using quantum mechanics. &lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-01-01T15:31:06.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22273" PostTypeId="1" AcceptedAnswerId="22275" CreationDate="2015-01-01T15:39:41.073" Score="1" ViewCount="1875" Body="&lt;p&gt;I apologise if this seems a very simple question, and also if I am overlooking something that is very simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part of my A Level course, we have to be able to interpret mass spectrometer readings for isotopes of elements. Further on in the course, it is extended to molecules. However, quite often it has come up where a question like this arrives:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NYeyv.png&quot; alt=&quot;Example&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that relative abundance implies that all values are relative to each other, how can all of the relative abundances sum to a number that is greater than 100?&lt;/p&gt;&#xA;" OwnerUserId="7453" LastActivityDate="2019-08-04T11:53:57.447" Title="In a mass spectrum, how can the sum of relative abundances be greater than 100?" Tags="&lt;physical-chemistry&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22274" PostTypeId="2" ParentId="22270" CreationDate="2015-01-01T15:54:21.413" Score="1" Body="&lt;p&gt;The reason to form a bond is usually for stability, and so that is why some elements are diatomic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of chlorine, it is has -effectively- 7 electrons in its outer shell. The atom will be most stable when it has 8 electrons, and so it can share an electron with another chlorine atom, which will result in them both having 8 electrons in their outer shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, hydrogen has one electron in its outer shell. At this energy level, it needs 2 electrons in its outer shell, and so can share (in what's called a covalent bond) an electron with another hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen has 6 electrons in its outer shell, and so needs two more electrons to achieve stability. To do this, it forms a double covalent bond with another oxygen molecule, and so it achieves a full outer shell and becomes stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals, such as sodium and potassium, are not actually single atoms, but atoms that are joined together in a lattice. It is made up of positive metal ions surrounded by a 'sea' of delocalised electrons. This means that they are as stable as they want to be, and so do not need to form covalent bonds with other atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit After Comment&lt;/strong&gt;&#xA;Evidence for diatomic molecules can be seen in several different ways. One example is by looking at an electron density map of, for example, chlorine:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HaV1B.jpg&quot; alt=&quot;Electron Density Map of Chlorine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that there are two atoms and these are sharing electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we can look at the mass spectrum of a chlorine molecule:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/H1kSj.gif&quot; alt=&quot;Mass Spectrum of Chlorine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the two peaks around 35 are the two isotopes of chlorine: chlorine-35 and chlorine-37. Further on, near 70, we see that there are three different peaks. These peaks are where a diatomic molecule of two chlorine isotopes has been formed. The peaks are of a 35-35 molecule, a 35-37 molecule, and a 37-37 molecule, respectively. This shows that chlorine forms a diatomic molecule with two of its isotopes. We can then extend this to saying that because it would be theoretically more stable, and that it is a gas at room temperature, it is likely that gases like chlorine are diatomic in standard conditions and so a covalent compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evidence for the metallic structure which I described is effectively its characteristics. The fact that metals can conduct electricity implies that the delocalised electrons are free to move through the lattice when a potential difference is applied. Similarly, the way metals conduct heat can also be explained using this model, as the electrons move easily and so can transmit kinetic energy rapidly through the lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high melting and boiling temperatures suggest that the forces between metal atoms must be large because it takes a lot of energy to separate them. The metallic structure suggests that the lattice of positive ions are held tightly together by this electron 'glue'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, metals are malleable and ductile, which indicates that the positive metal nuclei have room to move within the sea of electrons.&lt;/p&gt;&#xA;" OwnerUserId="7453" LastEditorUserId="7453" LastEditDate="2015-01-02T09:58:24.543" LastActivityDate="2015-01-02T09:58:24.543" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22275" PostTypeId="2" ParentId="22273" CreationDate="2015-01-01T15:56:29.910" Score="7" Body="&lt;p&gt;2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11.  However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the &lt;strong&gt;normalized&lt;/strong&gt; ratio of relative abundance of these isotopes.  It is this normalized value that adds up to 100.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-01T15:56:29.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22276" PostTypeId="1" CreationDate="2015-01-01T16:12:47.297" Score="1" ViewCount="1143" Body="&lt;p&gt;One of first redox reaction I was taught in school, is probably single replacement. It is easy to predict cause all you need is a table of metal activity series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then there are more complicated or less well-known agents. Like permanganate ion, which is used commonly in determining the purity of iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For agents like this, how do you know if it is strong enough to reduce or oxidize the other elements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Is there also a table for this?)&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-01-01T16:43:03.993" Title="How to know if something will be a good oxidizing/ reducing agent? (Besides the activity of the metal)" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22277" PostTypeId="1" AcceptedAnswerId="22313" CreationDate="2015-01-01T16:24:02.673" Score="0" ViewCount="1292" Body="&lt;p&gt;Whether be it from a textbook or teacher, precipitation reactions always occur by rearranging the elements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5fIMn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, aren't there other factors you have to bare in mind when predicting such reactions? Like electronegativity: obviously the potassium ion is much more electronegative than Silver ion, so it wouldn't be that easy for potassium to split up with iodine.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4356" LastEditDate="2015-01-02T23:23:23.157" LastActivityDate="2015-01-02T23:23:23.157" Title="Why does precipitate always form so naturally?" Tags="&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22278" PostTypeId="2" ParentId="22267" CreationDate="2015-01-01T16:26:50.277" Score="5" Body="&lt;p&gt;The H-C-H bond angle in ethene is &lt;em&gt;ca.&lt;/em&gt; 117 degrees and the H-C-C angle is &lt;em&gt;ca.&lt;/em&gt; 121.5 degrees.  There are two reasons that combine to explain this angular deformation in ethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First&lt;/strong&gt;, from these bond angles and Coulson's Theorem (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666&quot;&gt;ref_1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19977/determine-the-bond-angle-in-a-compound/19987#19987&quot;&gt;ref_2&lt;/a&gt;) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these hybridization indices (the index is the exponent &quot;n&quot; in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma  electrons will tend to flow  towards this lower energy C-C bond.  Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds.  Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second&lt;/strong&gt;, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-01-02T04:46:11.027" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22279" PostTypeId="2" ParentId="22276" CreationDate="2015-01-01T16:43:03.993" Score="1" Body="&lt;p&gt;The relative strengths of oxidising and reducing agents can be inferred from their standard electrode potentials. The strongest oxidizing agents are shown from the standard electrode table.&lt;/p&gt;&#xA;" OwnerUserId="11556" LastActivityDate="2015-01-01T16:43:03.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22280" PostTypeId="2" ParentId="22251" CreationDate="2015-01-01T16:59:47.500" Score="2" Body="&lt;p&gt;The reactivity of amines is generally based on how nucleophilic they are.  In turn, their nucleophilicity is dependent upon how much of the nitrogen lone pair of electrons really exists on the nitrogen atom (&lt;em&gt;e.g.&lt;/em&gt; the lone pair electron density on nitrogen). In ammonia ($\ce{NH3}$), the lone pair electron density on nitrogen is very high making ammonia quite nucleophilic, quite reactive.  In the case of nitrogen trifluoride ($\ce{NF3}$), the &lt;strong&gt;three&lt;/strong&gt; extremely electronegative fluorine atoms inductively remove much of the electron density from nitrogen; therefore $\ce{NF3}$ is less reactive than ammonia in a nucleophilic sense.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-01T16:59:47.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22281" PostTypeId="1" AcceptedAnswerId="22309" CreationDate="2015-01-01T17:42:40.150" Score="1" ViewCount="92" Body="&lt;p&gt;What is the benefit of finding the activation enthalpy and not the rate of reaction?&lt;/p&gt;&#xA;" OwnerUserId="11557" LastEditorUserId="194" LastEditDate="2017-04-01T02:39:18.780" LastActivityDate="2017-04-01T02:39:18.780" Title="Why is activation enthalpy found and not just rate of reaction?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22284" PostTypeId="2" ParentId="22270" CreationDate="2015-01-01T18:51:10.667" Score="-1" Body="&lt;p&gt;Hydrogen, oxygen, chlorine, etc, prefer to fill their valence electron shell by sharing electrons. Energy is released in the sharing a pair of electrons then individual atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium, potassium, etc, would rather give up an electron (to have a full valance electron shell), then to share electrons in an attempt to fill up the electron shell. In the case of bonding, a little energy is used to release an electron and a lot of energy is released when another atom pulls it into its valance shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the metal atom would use energy to take an extra electron, and the release uses energy, metal atoms will not take an extra electron.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-01T18:51:10.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22285" PostTypeId="1" CreationDate="2015-01-01T19:02:30.777" Score="13" ViewCount="2222" Body="&lt;p&gt;Is there a publicly available &lt;a href=&quot;https://en.wikipedia.org/wiki/Application_programming_interface&quot; rel=&quot;noreferrer&quot;&gt;API&lt;/a&gt; that serves physical property data for organic molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/&quot; rel=&quot;noreferrer&quot;&gt;Pubchem&lt;/a&gt; does not serve properties, &lt;a href=&quot;http://www.chemspider.com&quot; rel=&quot;noreferrer&quot;&gt;ChemSpider&lt;/a&gt; closes property access to API calls, and &lt;a href=&quot;https://toxnet.nlm.nih.gov/&quot; rel=&quot;noreferrer&quot;&gt;Toxnet&lt;/a&gt; only serves html.  Im basically looking for a free programmatic interface to the kind of info in a CRC handbook. Thanks for any suggestions.&lt;/p&gt;&#xA;" OwnerUserId="11560" LastEditorUserId="41328" LastEditDate="2017-08-06T07:50:00.353" LastActivityDate="2019-11-20T09:14:16.287" Title="Public API for physical properties" Tags="&lt;organic-chemistry&gt;&lt;reference-request&gt;&lt;cheminformatics&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="22286" PostTypeId="1" AcceptedAnswerId="22342" CreationDate="2015-01-01T22:25:12.793" Score="9" ViewCount="144" Body="&lt;p&gt;I'm thinking about a quick-and-dirty approach to molecular dynamics, mostly for teaching purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a given temperature, statistical mechanics tells us molecules have translational and rotational and vibrational energy as given by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equipartition_theorem&quot;&gt;equipartition theorem&lt;/a&gt;. At a basic level, all three motions derive from movement of the individual atoms along Cartesian axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems like a very simple quick-and-dirty approximate &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_dynamics&quot;&gt;molecular mechanics&lt;/a&gt; could be implemented by &quot;shaking&quot; the individual atoms with mass-weighted random displacements. You'd need to implement some sort of force field to make sure the added displacements would fall into the correct potential energy distribution for a given molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick searching doesn't seem to find this as an approach, likely because it won't obey a particular canonical ensemble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious... Has anyone implemented such a method for qualitative visualization? What are some of the pitfalls compared to &quot;proper&quot; molecular dynamics? (e.g., presumably ensuring the proper velocities?)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-03T16:53:12.517" Title="Quick-and-Dirty Molecular Dynamics by Mass-Weighted Atom Translations?" Tags="&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22287" PostTypeId="1" CreationDate="2015-01-01T23:26:28.483" Score="2" ViewCount="6086" Body="&lt;p&gt;What is the relationship between acid dissociation constant and equilibrium constant for an acid's reaction with water?&#xA;If we have this reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH + H2O → CH3COO- + H3O+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would calculate the equilibrium constant to 0,005241 M at 25 celcius&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HxEqB.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Acid dissociation constant for acetic acid is as known 1,75 × 10$^{−5}$. How can this be true? The equilibrium constant I have calculate is about 300 times larger than the acid dissociation constant!&lt;/p&gt;&#xA;" OwnerUserId="11561" LastEditorUserId="4231" LastEditDate="2015-01-02T14:50:19.087" LastActivityDate="2018-03-29T03:19:13.030" Title="Acid dissociation constant and equilibrium constant" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22288" PostTypeId="1" CreationDate="2015-01-02T00:07:04.107" Score="3" ViewCount="88" Body="&lt;p&gt;When conducting a Kjeldahl analysis,  ammonium sulfate is formed from nitrogen which protein consist of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How in a picture of formation of ammonium sulfate there is involved no nitrogen but only some compounds containing nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhMPA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="10185" LastEditDate="2015-04-29T18:35:59.747" LastActivityDate="2015-04-29T18:35:59.747" Title="How exactly is ammonium sulfate formed in a Kjeldahl analysis?" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22290" PostTypeId="1" AcceptedAnswerId="22324" CreationDate="2015-01-02T05:45:35.227" Score="21" ViewCount="13081" Body="&lt;p&gt;The structure of ozone is traditionally depicted using two resonance forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UeQD4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UeQD4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I am wondering is instead of using such an idea to explain the structure why can't we show the structure of ozone like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dm1Y8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dm1Y8.png&quot; alt=&quot;Triangular ozone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people have told me that such a structure is not possible but they fail to mention the reason. What I have in mind is maybe electron repulsion. But I would like to know exactly what mechanism would prevent this from happening?&lt;/p&gt;&#xA;" OwnerUserId="7249" LastEditorUserId="16683" LastEditDate="2016-11-12T19:47:54.897" LastActivityDate="2020-05-15T15:37:29.543" Title="Can ozone have a triangular structure?" Tags="&lt;bond&gt;&lt;resonance&gt;&lt;structural-formula&gt;&lt;ozone&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22291" PostTypeId="2" ParentId="14417" CreationDate="2015-01-02T05:55:54.300" Score="4" Body="&lt;p&gt;S-p mixing is the primary cause of the difference in the molecular orbitals of nitrogen and oxygen, which is influenced by the initial atomic orbital energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lighter second period elements (prior to oxygen) have a relatively small difference in energy between the 2s and 2p orbitals. This allows sufficient s-p mixing to lower the energy of the σ(2s) and σ*(2s) molecular orbitals, and is energetically offset by an increase in energy of the σ(2p) and σ*(2p) molecular orbitals. In the dinitrogen molecular orbital scheme, the dashed lines are there to represent s-p mixing influencing the energy of the four molecular orbitals involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective nuclear charge increases to the right of the period, stabilizing the 2s orbital more drastically than the 2p orbital. This can be seen qualitatively in the first figure &lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/c120/molitof.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more stabilized 2s orbital does not s-p mix as effectively, due to the greater energy difference between the 2s and 2p orbitals. As nuclear charge increases, s-p mixing becomes less significant. The change of the molecular orbital ordering between nitrogen and oxygen is the manifestation of this decreased s-p mixing. In the dioxygen molecular orbital scheme the s-p mixing effect is no longer significant enough to alter the relative orbital arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information about the details of this difference can be found in most inorganic chemistry textbooks. There are good diagrams showing the gradual change in energy differences across the second row range of homonuclear diatomic molecules. &lt;a href=&quot;http://chemistry.illinoisstate.edu/standard/che362/handouts/362corrdiagram.pdf&quot; rel=&quot;nofollow&quot;&gt;This document&lt;/a&gt; highlights the progressive lowering of molecular orbital energies due to the the atomic orbital changes, and presents a correlation diagram linking internuclear separation with molecular orbital energies as well.&lt;/p&gt;&#xA;" OwnerUserId="7653" LastActivityDate="2015-01-02T05:55:54.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22294" PostTypeId="2" ParentId="22290" CreationDate="2015-01-02T09:22:29.150" Score="1" Body="&lt;p&gt;Possible Reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Electronic Repulsion&lt;/strong&gt;: In order to form a 3-member ring your electron pairs would have to bend at 60 degrees; the energy required to push naturally repulsive electrons so close together would make the molecule very unstable. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Unstability of three membered rings (angular strain):&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-02T09:22:29.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22295" PostTypeId="1" AcceptedAnswerId="22299" CreationDate="2015-01-02T09:25:39.343" Score="3" ViewCount="700" Body="&lt;p&gt;I recently came across different names for the following compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/7/71/Cyclohexanol.png&quot; alt=&quot;what is this&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the IUPAC name for this &lt;code&gt;cyclohexol&lt;/code&gt; or &lt;code&gt;cyclohexanol&lt;/code&gt;? Is the use of &lt;code&gt;cyclohexol&lt;/code&gt; as seen in a &lt;a href=&quot;http://www.google.com/patents/CN1793099A?cl=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;google patent&lt;/a&gt; and in &lt;a href=&quot;http://www.globethesis.com/?t=1101360218458010&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Chinese thesis&lt;/a&gt; accurate?&lt;/p&gt;&#xA;" OwnerUserId="5402" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.507" LastActivityDate="2015-01-02T11:41:07.593" Title="Is cyclohexol the same as cyclohexanol?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22296" PostTypeId="2" ParentId="22288" CreationDate="2015-01-02T09:28:18.620" Score="2" Body="&lt;p&gt;Ammonia is basic so:&#xA;$$\ce{NH3 +H+&amp;lt;=&amp;gt;NH4+}$$&#xA;Sulphuric acid is acidic(since it's name is itself so acidic!):&#xA;$$\ce{H2SO4&amp;lt;=&amp;gt;H+ +HSO4- &amp;lt;=&amp;gt;2H+ +SO4^2-}$$&#xA;Now the final reaction would be like:&#xA;$$\ce{2NH3 +H2SO4&amp;lt;=&amp;gt;NH3 +NH4+ +HSO4- &amp;lt;=&amp;gt;2NH4+ +SO4^2- &amp;lt;=&amp;gt;(NH4)2SO4}$$&#xA;And remeber since proteins are polymers of peptides which itself are molecules consisiting of both a carboxylic ($\ce{-COOH/-COO-}$) and amino ($\ce{-NH3/-NH4+}$) group, therefore with strong heating ammonium sulfate is expected.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-04-29T17:58:32.067" LastActivityDate="2015-04-29T17:58:32.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22297" PostTypeId="2" ParentId="22287" CreationDate="2015-01-02T09:45:48.647" Score="3" Body="&lt;p&gt;An acid dissociation constant is a particular example of an equilibrium constant.&#xA;For the reaction:&#xA;$$\ce{CH3COOH +H2O &amp;lt;=&amp;gt;CH3COO- +H3O+}$$&#xA;The equilibrium constant would be:&#xA;$$\newcommand{\ac}[1]{\text{activity of }\ce{#1}}{\bf K_{eq}}=\frac{\ac{CH3COO-}\cdot\ac{H3O+}}{\ac{CH3COOH}\cdot\ac{H2O}}$$&#xA;And the acid dissociation constant would be:&#xA;$${\bf K_a}=\ce{\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$&#xA;Now the relation between both would be:&#xA;$$\newcommand{\acc}[1]{\gamma_{\ce{#1}}}{\bf K_{eq}}={\bf K_a}\underbrace{\frac{\acc{CH3COO-}\acc{H3O+}}{\acc{CH3COOH}\acc{H2O}}}_{\Gamma_0}\frac1{\ce{[H2O]}}$$&#xA;Now:&#xA;$${\bf K_a}={\bf K_{eq}}[\ce{H2O}]\Gamma_0^{-1}$$ &#xA;So you're missing some factors.They're not exactly the same.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-02T09:45:48.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22298" PostTypeId="1" AcceptedAnswerId="22302" CreationDate="2015-01-02T10:19:49.667" Score="1" ViewCount="677" Body="&lt;p&gt;Does $\ce{PdCl2}$ follow 18-Electron rule or octet rule?&lt;br/&gt;&#xA;How should we determine this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/24290#section=Top&quot; rel=&quot;nofollow&quot;&gt;http://pubchem.ncbi.nlm.nih.gov/compound/24290#section=Top&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compounds/3494742#section=Top&quot; rel=&quot;nofollow&quot;&gt;http://pubchem.ncbi.nlm.nih.gov/compounds/3494742#section=Top&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="7448" LastEditDate="2015-01-02T12:57:02.010" LastActivityDate="2015-01-02T13:13:02.160" Title="Do Palladium(II) Chloride and Palladium (III) Chloride follow 18-Electron rule or octet rule? How to determine this?" Tags="&lt;electrons&gt;&lt;transition-metals&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22299" PostTypeId="2" ParentId="22295" CreationDate="2015-01-02T11:41:07.593" Score="6" Body="&lt;p&gt;&lt;em&gt;Note that I can only comment on the English translations!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cyclohexanol&lt;/strong&gt; is correct, cyclohexol is just wrong. This is not the only error in the abstract of the thesis. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cyclohexone is the most [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;isn't correct either. The name is &lt;strong&gt;cyclohexanone&lt;/strong&gt;!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-02T11:41:07.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22301" PostTypeId="1" AcceptedAnswerId="22311" CreationDate="2015-01-02T12:58:56.290" Score="7" ViewCount="766" Body="&lt;p&gt;I know how to get the equation from the Clapeyron equation but I have a question regarding a the integration along a phase boundary and a small step in the derivation that I will make clear when I reach that step. Firstly, the Clapeyron equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternatively, by recognizing that $\Delta G=0$ when two phases are in equilibrium; $\Delta G =\Delta H - T\Delta S$ can be rearranged to give: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S = \frac{\Delta H}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into to the the first equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta H}{T\Delta V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a solid/gas or liquid/gas phase phase boundary, it is a reasonable to approximate that $\Delta V \approx V_{m,\text{gas}}$ as $\Delta V = V_{m,\text{gas}}-V_{m,\text{condensed}}$ since $V_{m,\text{gas}} \gg V_{m,\text{condensed}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, using the equation of state for a perfect gas and substituting for $\Delta V$ in the second form of the Clapeyron equation, the following result is obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}=\frac{\Delta H}{RT^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why is $\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The following step is the step that I don’t understand&lt;/strong&gt; In my lecture handout this step is shown as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can the left hand side be integrated with respect to $p$, yet the right hand side be integrated with respect to $T$. I cannot make sense of this. Other instances in thermodynamics when integration is used like this, both sides are integrated with respect to the same variable. For example, finding how $H$ varies with $p$. I will quickly go through what I mean without much verbal explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}p=\left(\frac{\partial H}{\partial S}\right)_p \mathrm{d}S+\left(\frac{\partial H}{\partial p}\right)_S \mathrm{d}p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, for one mole of a perfect gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\left(\frac{\partial H}{\partial p}\right)_S=\frac{RT}p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{p_1}^{p_2}\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\int_{p_1}^{p_2}\frac{RT}p\mathrm{d}p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\mathrm{d}H$ at a constant $T$ (&lt;strong&gt;I think this is right – please clarify&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{p_1}^{p_2}\mathrm{d}H=RT\int_{p_1}^{p_2}\frac{1}p\mathrm{d}p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can obviously be integrated quite easily but this only illustrates my point regarding the variable that the integration is carried out with respect to. Surely it must be the same on both the RHS and the LHS of the equation?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user5764" LastEditDate="2015-11-09T16:30:38.800" LastActivityDate="2015-11-09T16:30:38.800" Title="How to find and use the Clausius-Clapeyron equation" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22302" PostTypeId="2" ParentId="22298" CreationDate="2015-01-02T13:13:02.160" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Palladium(II)_chloride&quot; rel=&quot;nofollow&quot;&gt;Palladium (II) Chloride&lt;/a&gt; follows neither the 18 electron rule nor the octet rule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, like many Pd(II), Pt(II) and Ni(II) complexes, palldium (II) chloride follows the &lt;a href=&quot;http://www.yorku.ca/stynes/Tolman.pdf&quot; rel=&quot;nofollow&quot;&gt;16-electron rule&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pd(II) generally forms square planar complexes, and in palldium (II) chloride each Cl bridges two Pd, with 4 Cl coordinating each Pd in a roughly square planar arrangement.  8 d-electrons plus 2 electrons from each of 4 sigma bonds equals 16 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never heard of &quot;Cl3Pd&quot;, which would involve the rare oxidation state Pd(III), but the 16 and 18 electron rules should only be applied to diamagnetic complexes.  Pd(III) complexes are paramagnetic.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-02T13:13:02.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22303" PostTypeId="2" ParentId="9957" CreationDate="2015-01-02T13:34:40.877" Score="5" Body="&lt;p&gt;If you are performing these computations for non-commercial purposes, one program package you might consider is &lt;a href=&quot;https://orcaforum.cec.mpg.de/portal.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;ORCA&lt;/a&gt; (free registration required).  Binaries are available for Mac, Win and Linux.  Check out, e.g., Sections 5.9.3.5, 6.14.8.8, and 6.14.8.9 in the current manual (available at the above link) to see if the 'orca_pltvib' helper module or the 'mode trajectory run' (MTR) features will do what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'orca_pltvib' generates 'canned' OpenBabel XYZ multi-geometry animations for one or more vibrational modes of a Hessian. It may not provide sufficient configurability for your needs, though, as it only creates animations for single modes, and the displacement sizes are fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MTR feature is a lot more flexible.  It is meant to compute single-point energies at all of the points in a given mode scan (which can scan multiple modes at once) but also spits out a multi-structure XYZ file for each point, Once the TS geometry and modes are calculated, if you don't really care about the energies you can probably run the MTR at a very low level of theory, and just grab the geometries.  One possible hitch with scans over multiple modes is that it only generates the XYZ files for pairs of mode displacements, and each XYZ file only contains an animation of displacements along a single mode (see p406 in the current manual for a more complete explanation).&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="4945" LastEditDate="2016-11-30T08:03:24.723" LastActivityDate="2016-11-30T08:03:24.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22304" PostTypeId="2" ParentId="22259" CreationDate="2015-01-02T13:58:52.620" Score="12" Body="&lt;p&gt;With the concept &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;Resonance&lt;/a&gt; you can describe delocalized electrons in some molecules. To go with your example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl61F.png&quot; alt=&quot;Ozone Resonance&quot;&gt;&#xA;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the ozone molecule to be frozen in one of those two states. Positive and negative charges are localized, which is generally connected to a higher energy, i.e. a more reactive species. This would be an extreme case for ozone - in reality those electrons will be delocalized all over the molecule. So, what is the reason that molecules with localized electrons/charges have a higher total energy and are, hence, less stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you remember the &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_in_a_box&quot; rel=&quot;noreferrer&quot;&gt;Particle in a box&lt;/a&gt;. I wont go into the quantum chemistry here, but one of the take-away-messages is, that the energy for particles in a infinite square well, which is a very rough approximation of the potential in which the outer electrons reside in, does decrease with $ L^{-2} $. $L$ is the size of the box in which the particle can exist. So, the more delocalized the electrons are, the less total energy in a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, to correct you question a little bit: Resonance is not the reason why molecules are more stable, it is a concept which describes the bonding in a molecule, in which electrons are delocalized, better than a single &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_structure&quot; rel=&quot;noreferrer&quot;&gt;Lewis Structure&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to have a feedback if this answer is somewhat understandable.&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="11551" LastEditDate="2015-01-03T16:03:20.047" LastActivityDate="2015-01-03T16:03:20.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22305" PostTypeId="2" ParentId="22259" CreationDate="2015-01-02T14:12:44.990" Score="4" Body="&lt;p&gt;Let us first discuss the more fundamental phenomenon of covalent bonding. How do you think a covalent bond &quot;works&quot;? Let us say we have two hydrogen atoms. We know that the nuclei of these atoms are positively charged and the electrons negatively charged. Coulombic repulsions act between the two nuclei, and between the electrons from the two atoms. Coulombic attractions act between the electrons and nuclei within the same molecule and betweem those of the two different molecules. If the two atoms are far apart from each other, attractions between the electrons of one atom and the nucleus of the other are greater than the repulsions. If the atoms get too close to each other, repulsions are greater than the attractions. At exactly some seperation, attraction balances repulsion and the atoms settle for a seperation called the &quot;bond-length&quot;, and the atoms are said to be &quot;covalent-bonded&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that ridiculously long explanation of what a covalent bond is, let us get to the essence of it. The reason why a covalent bond &quot;works&quot; is because electrons are holding atoms 'together'. Without the electrons, the atoms would not have anything to 'bind' them. In essence, two electrons are holding two atoms in place, thereby bringing stability, and causing a lowering of potential energy of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a benzene molecule, the same thing happens. The p-orbitals of the induvidual carbon atoms in benzene overlap with each other, thereby allowing the electrons in these p-orbitals to de-localise, meaning that an electron in the p-orbital of one carbon atom no longer belongs to only one atom, but instead acts to bind all other carbon atoms in the benzene ring, just as the two electrons in the $H_2$ molecule helps bind the two atoms  of hydrogen together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JE0mi.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Image credits: chemwiki.ucdavis.edu&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since the electron belongs to the orbitals of many other atoms, and not only one or two (as in the case of the H-H covalent bond), it holds all these six carbon atoms in place, and thereby acts as a bond between the six carbon atoms, which in turn brings stability to the system and lowers its potential energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appendix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why doesn't an electron in an atom fall into the nucleus ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture we often have of electrons as small objects circling a nucleus in well defined &quot;orbits&quot; is actually quite wrong. The positions of these electrons at any given time are not well-defined, but we CAN figure out the volume of space where we are likely to find a given electron if we do an experiment to look. For example, the electron in a hydrogen atom likes to occupy a spherical volume surrounding the proton. If you think of the proton as a grain of salt, then the electron is about equally likely to be found anywhere inside a ten foot radius sphere surrounding this grain, kind of like a cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weird thing about that cloud is that its spread in space is related to the spread of possible momenta (or velocities) of the electron. So here's the key point, which we won't pretend to explain here. The more squashed in the cloud gets, the more spread out the range of momenta has to get. That's called Heisenberg's uncertainty principle. Big momenta mean big kinetic energies. So the cloud can lower its potential energy by squishing in closer to the nucleus, but when it squishes in too far its kinetic energy goes up more than its potential energy goes down. So it settles at a happy medium, and that gives the cloud and thus the atom its size. (from &lt;a href=&quot;https://van.physics.illinois.edu/qa/listing.php?id=1226&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://van.physics.illinois.edu/qa/listing.php?id=1226&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="9370" LastEditorUserId="9370" LastEditDate="2015-01-03T15:56:10.227" LastActivityDate="2015-01-03T15:56:10.227" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22306" PostTypeId="2" ParentId="22290" CreationDate="2015-01-02T14:18:58.117" Score="8" Body="&lt;p&gt;Your guess about electronic repulsion is right: the repulsion between the lone pairs of electrons on the Oxygen atoms leads to intense repulsion at such close ranges leading to instability.  In comparison to this resonance is a far far more stable configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strain at the joints in the cyclic structure does its part as well because experimentally it is found to be abnormal. The strain causes instability in bond and eventually the molecule will break into constituents. The ideal angle for stability is 109 degrees and 28 minutes (I don't know why, but experiments show this is true). Angular strain can be found in water as well: to decrease that strain water collapses to an angle of 104 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ring strain or angular strain is a type of instability that exists when bonds in a molecule form angles that are abnormal (&lt;em&gt;i.e., not generally found in nature&lt;/em&gt;).&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="5354" LastEditDate="2015-09-12T17:03:30.830" LastActivityDate="2015-09-12T17:03:30.830" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22307" PostTypeId="1" AcceptedAnswerId="22310" CreationDate="2015-01-02T14:42:51.273" Score="1" ViewCount="12578" Body="&lt;p&gt;I actually know how to calculate the oxidation number for coordination complexes but some complexes like the below are giving some reistance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Cu(NH3)_2][Fe(H2O)_3]}$ (diamminecopper(?) triaquoferrate(?)) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wrote this example now and don't know wether it exists or not I am rather concerned about how do I find their oxidation numbers&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="4231" LastEditDate="2015-01-02T14:44:29.270" LastActivityDate="2020-02-02T14:58:58.500" Title="How do I find the oxidation number of a coordination complex" Tags="&lt;coordination-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22308" PostTypeId="1" AcceptedAnswerId="22316" CreationDate="2015-01-02T14:50:22.017" Score="1" ViewCount="472" Body="&lt;p&gt;Why is it said that the presence of a carbon with 4 different groups is necessary for it to exist as an enantiomeric form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take 2-fluoropropane for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I draw the structure of it in such a way so that the 2 CH3 groups are adjacent to each other (on paper ) and take its mirror image I see that I got a mirror image which is not superimposable on the original structure. My teacher told me I am only allowed to rotate it in the plane of the paper and only by 180° without lifting it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to tell me that they are a pair of enantiomers( non superimposable mirror images).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand when I draw the structure such that the 2 CH3 groups are on opposite sides of the central carbon and then take a mirror image of it , rotate it by 180° in the plane of the paper I get the original structure.&#xA;This tells me that they are not stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also where is the mirror supposed to be placed - up down left right? Any convention (here I have placed the mirror on the right)? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="9961" LastEditDate="2015-10-09T16:48:06.383" LastActivityDate="2015-10-09T16:48:06.383" Title="Stereogenic C and enantiomers" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22309" PostTypeId="2" ParentId="22281" CreationDate="2015-01-02T14:52:46.473" Score="1" Body="&lt;p&gt;One of the Factors Influencing reaction rate is given below (&lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_rate#Factors_influencing_rate_of_reaction&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Temperature: Usually conducting a reaction at a higher temperature delivers more energy into the system and increases the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy resulting in more successful collisions (when bonds are formed between reactants). The influence of temperature is described by the Arrhenius equation. As a rule of thumb, reaction rates for many reactions double for every 10 degrees Celsius increase in temperature, though the effect of temperature may be very much larger or smaller than this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reaction rate , Activation energy are connected to temperature through Arrhenius equation. We are interested in &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy&quot; rel=&quot;nofollow&quot;&gt;Activation Energy&lt;/a&gt; as it gives an idea what temperature the reaction starts. We want to know $E_{a}$ in order to manipulate reaction rate essentially. &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalysis&quot; rel=&quot;nofollow&quot;&gt;Catalysts&lt;/a&gt; or Inhibitors work essentially on this principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking at rate wont help us in varying the rates to our interest. &lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-01-02T14:52:46.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22310" PostTypeId="2" ParentId="22307" CreationDate="2015-01-02T15:04:51.443" Score="1" Body="&lt;p&gt;Think of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy! &lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-01-02T15:04:51.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22311" PostTypeId="2" ParentId="22301" CreationDate="2015-01-02T15:35:25.360" Score="6" Body="&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Integration_by_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Integration by substitution&lt;/a&gt;, see also &lt;a href=&quot;https://math.stackexchange.com/questions/216132/integrating-each-side-of-an-equation-w-r-t-to-a-different-variable&quot;&gt;this question&lt;/a&gt; and &lt;a href=&quot;https://math.stackexchange.com/questions/182344/can-anyone-explain-the-intuitive-meaning-of-integrating-on-both-sides-of-the-eq&quot;&gt;another question&lt;/a&gt; at math.SE. This is the rule we have to apply here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{x_0}^{x_1} f(u(x))u'(x)\,\mathrm dx = \int_{u(x_0)}^{u(x_1)} f(u)\,\mathrm du $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we go from this equation:&#xA;$$\frac1p \frac{\mathrm dp}{\mathrm dT}=\frac{\mathrm d\ln p}{\mathrm dT}=\frac{\Delta H}{RT^2},$$&#xA;and integrate over temperature:&#xA;$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{T_1}^{T_2} \frac{\Delta H}{RT^2}\,\mathrm dT,$$&#xA;Now we use the first equation, with $f=\frac{1}{p}$, $p=u$, and $x=T$:&#xA;$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{p_1(T_1)}^{p_2(T_2)}\frac1p\,\mathrm dp=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus question:&lt;/strong&gt; Its just the application of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;chain rule&lt;/a&gt;, namely&#xA;$$\frac {\mathrm d}{\mathrm dx}z = \frac {\mathrm dz}{\mathrm dy} \frac {\mathrm dy}{\mathrm dx} $$&#xA;which, if applied leads to&#xA;$$\frac{\mathrm d}{\mathrm dT}\ln p = \frac{\mathrm d\ln(p)}{\mathrm dp} \frac{\mathrm dp}{\mathrm dT} = \frac{1}{p}\frac{\mathrm dp}{\mathrm dT} .$$&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:09.340" LastActivityDate="2015-11-09T16:08:15.193" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22312" PostTypeId="1" AcceptedAnswerId="23871" CreationDate="2015-01-02T15:46:17.687" Score="6" ViewCount="10433" Body="&lt;p&gt;What's the mechanism of cross-linkage for a typical household 2-party epoxy mix? I was working with some epoxy the other day and I was instructed to mix equal parts of epoxy and epoxy hardener. Stoichiometry at its finest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardener contained &quot;polymercaptans amines&quot; according to the label. Google told me that a mercaptan is a broad name for a class of thiols. Makes sense; the $\ce{-SH}$ group is likely strongly nucleophilic (to better pop open the epoxide rings). Amines contain the also nucleophilic $\ce{-NH2}$ group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what are &lt;em&gt;polymercaptan amines&lt;/em&gt;? Do these contain both thiol and amine functional groups? And again, how does cross-linkage work? I presume that this involves opening the epoxide ring. But then does the resulting oxide anion attack another epoxide ring? Wouldn't we then need a 2:1 ratio of epoxide to hardener? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-01-02T19:33:39.840" LastActivityDate="2016-11-18T06:49:23.530" Title="How do epoxys cross-link?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22313" PostTypeId="2" ParentId="22277" CreationDate="2015-01-02T17:15:13.283" Score="5" Body="&lt;p&gt;The ions combine while in solution, and since the resulting compound is not soluble, it precipitates. This happens so quickly because as the precipitate forms, it takes those individual ions out of solution, thus pulling the reaction toward the products side, according to &lt;strong&gt;Le Chatelier's principle&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-02T17:15:13.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22314" PostTypeId="1" CreationDate="2015-01-02T17:20:04.820" Score="0" ViewCount="1091" Body="&lt;p&gt;I am in real confusion over Avogadro's Law..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some of the problems I did on gas laws, there are the number of moles of the gas, the temperature. and the pressure. And it expects me to find the volume of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then in other problems there everything else is given including the volume of the container ( instead of the gas itself) and then something else is expected to be found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion starts here. If any gas occupy the volume of volumes, always, then why will volume change according the combined gas law?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one of my previous post, I learned that volumes of different gases won't be proportional (meaning all of them will occupy equal space) unless the condition of constant tempt and pressure is reached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what really happen when constant tempt and pressure are reached? Do individual gases just separate themselves from one another? Because in order for their volume ratio to be equal to anything other than 1, their volume must be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;you only have to answer the question in the title, the others questions are intended to help you understand what I mean. Sorry for the possible confusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/18957/volume-of-different-gas-in-a-container&quot;&gt;a link to my previous post&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-01-03T18:21:52.200" Title="Why two ( or more) gases have proportional volumes at constant pressure and temp.? (Avogadro's Law)" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="22316" PostTypeId="2" ParentId="22308" CreationDate="2015-01-02T17:53:45.793" Score="1" Body="&lt;p&gt;When you say: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; When I draw the structure of it in such a way so that the 2 CH3 groups are &#xA; adjacent to each other (on paper ) and take its mirror image I see that I got a &#xA; mirror image which is not superimposable on the original structure.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and then:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;On the other hand when I draw the structure such that the 2 CH3 groups are on        &#xA;opposite sides of the central carbon and then take a mirror image of it , &#xA;rotate it by 180° in the plane of the paper I get the original structure.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This tells me you are drawing structures where the bonds are at right angles to each other, like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     F                              CH3&#xA;     |                              |&#xA; CH3-C-CH3        and           CH3-C-H&#xA;     |                              |&#xA;     H                              F&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'.  In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a good place to use models.  And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider using dashes and wedges, such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ciZsl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2015-01-02T18:07:27.233" LastActivityDate="2015-01-02T18:07:27.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22317" PostTypeId="2" ParentId="22314" CreationDate="2015-01-02T18:20:38.810" Score="1" Body="&lt;p&gt;So, there are many questions, and I'm not sure if I understand you 100 %:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadros Law&lt;/a&gt;, which applies for an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas&lt;/a&gt; at constant pressure and temperature:&#xA;$$\frac{V}{n} = k \quad\Rightarrow\quad \frac{V_1}{n_1}=\frac{V_2}{n_2}$$&#xA;With $k$ as a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we have the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas law&lt;/a&gt;: &#xA;$$pV = nRT \quad\Rightarrow\quad \frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is probably what your first question is about. There are problems when you have a state $1$ and a state $2$, which differ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: You have a container where everything ($p_1,V_1,T_1,n_1$) is given. Now you double the temperature - and the task is to calculate the pressure. Since the volume and amount of substance is fixed ($V_1 = V_2$, $n_1 = n_2$), you can just rearrange everything and calculate the pressure $p_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ideal gas-model, gases do not seperate. I may be misunderstanding you, but if you put 1 mol $\ce{N2}$ and 2 mol oxygen in a container, you will have different &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;partial pressure&lt;/a&gt;s, but no demixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A relevant question was &lt;a href=&quot;https://physics.stackexchange.com/questions/96089/mixing-of-ideal-gas-thermodynamic-equilibrium&quot;&gt;posted on physics.SE&lt;/a&gt;, that may help you if you have to dive deeper into this topic.&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-01-02T18:20:38.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22318" PostTypeId="2" ParentId="22250" CreationDate="2015-01-02T18:48:18.637" Score="6" Body="&lt;p&gt;Like user137 has already written in the comment section it should not be harmful unless exposed for a long time. The main charateristic is its hygroscopy, so skin may feel dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To cite from a safety data sheet (&lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=DE&amp;amp;language=EN-generic&amp;amp;productNumber=391484&amp;amp;brand=SIAL&amp;amp;PageToGoToURL=http%3A%2F%2Fwww.sigmaaldrich.com%2Fcatalog%2Fproduct%2Fsial%2F391484%3Flang%3Dde&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 FIRST AID MEASURES&#xA;4.1 Description of first aid measures&#xA;If inhaled&#xA;    If breathed in, move person into fresh air. If not breathing, give artificial respiration.&#xA;In case of skin contact&#xA;    Wash off with soap and plenty of water.&#xA;In case of eye contact&#xA;    Flush eyes with water as a precaution.&#xA;If swallowed&#xA;    Never give anything by mouth to an unconscious person. Rinse mouth with water.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Nevertheless: Always use the safety data sheets your distributor provided and read them before dealing with the chemicals, if the safety hazards are not clear.&lt;/p&gt;&#xA;" OwnerUserId="11551" LastActivityDate="2015-01-02T18:48:18.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22319" PostTypeId="2" ParentId="7426" CreationDate="2015-01-02T19:38:48.407" Score="9" Body="&lt;p&gt;Anhydrous caffeine occurs in two different polymorphic crystal forms, with a transition between the two forms occurring at &lt;span class=&quot;math-container&quot;&gt;$\pu{136 °C}.$&lt;/span&gt; The high temperature form has a melting point of &lt;span class=&quot;math-container&quot;&gt;$\pu{236 °C}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The vapor pressure of the high temperature solid form is described by the equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\log P[\pu{Pa}] = -\frac{5477}{T[\pu{K}]} + 14.395,\tag{1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and for the liquid&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\log P[\pu{Pa}] = -\frac{3918}{T[\pu{K}]} + 11.14.\tag{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, &lt;span class=&quot;math-container&quot;&gt;$\pu{173 °C}$&lt;/span&gt; is incorrect; the boiling point is more like &lt;span class=&quot;math-container&quot;&gt;$\pu{365 °C}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Griesser, U. J.; Szelagiewicz, M.; Hofmeier, U. Ch.; Pitt, C.; Cianferani, S. Vapor Pressure and Heat of Sublimation of Crystal Polymorphs. &lt;em&gt;Journal of Thermal Analysis and Calorimetry&lt;/em&gt; &lt;strong&gt;1999&lt;/strong&gt;, 57 (1), 45–60. DOI: &lt;a href=&quot;https://doi.org/10.1023/A:1010188923713&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1023/A:1010188923713&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Bothe, H.; Cammenga, H. K. Phase Transitions and Thermodynamic Properties of Anhydrous Caffeine. &lt;em&gt;Journal of Thermal Analysis&lt;/em&gt; &lt;strong&gt;1979&lt;/strong&gt;, 16 (2), 267–275. DOI: &lt;a href=&quot;https://doi.org/10.1007/BF01910688&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1007/BF01910688&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2021-03-11T07:39:44.940" LastActivityDate="2021-03-11T07:39:44.940" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22321" PostTypeId="1" CreationDate="2015-01-02T21:32:46.563" Score="3" ViewCount="804" Body="&lt;p&gt;When I was in high school, I took two chemistry courses. The first was very elementary and didn't attempt to move beyond the picture of electrons, protons, and neutrons as little balls essentially. But in the second course I took (AP Chem), already confusion existed about waves vs particles and what exactly we mean by matter. And, in retrospect, I think jumping from thinking about localized particles  to waves is hard because it turns out neither description is complete enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently started studying quantum field theory and I feel like quantum fields are a lot more intuitive than delving into the whole wave-particle duality hubbub. The idea that particles are excitations of a field seems far more natural because one can envision waves and particles as excitations. Anthony Zee's book Quantum Field Theory in a Nutshell draws an analogy between quantum fields and mattresses. But the point is that I definitely think young people can imagine fields as easily as little balls. Yet there's a huge difference in how easily these initial intuitions allow you to navigate the physical sciences. &#xA;For instance, the idea of quantum fields would naturally lead students to wonder about how classical fields work.  Furthermore, for some particles, we can't define a wave function (eg photons). This becomes confusing because the wave function, at least to me, seem like the &quot;thing&quot; that told me how the states of a piece of elementary matter is evolving. But how can something evolve if it has no wave function? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mean we have to throw out the idea of particles as little balls. The beauty of fields is that you can keep that description as a particular type of excitation. So it would complement all the tools and tricks chemists want young students to learn (eg Lewis dot structures) but give them a much clearer picture about the nature or what a particle is and ultimately I think provide a better backdrop for eventually delving deeper into the deeper questions about how quantum mechanics works exactly. This would probably help because, when they get to college, they wouldn't have to switch from an already engrained mindset. The one they would have would be more or less correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I am ignorant and missing something, but to me, although a bit bold, it might make more sense to introduce quantum fields earlier.  &lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2015-01-03T03:09:54.910" Title="Why do middle school and high school chemistry classes teach students that electrons are little balls?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="0" CommentCount="14" ClosedDate="2015-01-03T07:01:43.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="22324" PostTypeId="2" ParentId="22290" CreationDate="2015-01-03T01:53:33.257" Score="20" Body="&lt;p&gt;Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the &lt;span class=&quot;math-container&quot;&gt;$2\pi_\mathrm u$&lt;/span&gt; orbitals (don't worry about why they are named this) are doubly occupied and the two &lt;span class=&quot;math-container&quot;&gt;$2\pi_\mathrm u$&lt;/span&gt; orbitals are both singly occupied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M4Xf5.gif&quot; alt=&quot;Walsh Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: 6a&lt;span class=&quot;math-container&quot;&gt;$_1$&lt;/span&gt; and 2b&lt;span class=&quot;math-container&quot;&gt;$_1$&lt;/span&gt; have the incorrect sign on the outer p-orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6a_1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2b_1}$&lt;/span&gt; orbitals wrong, flip the central orbital). This lowers the energy of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{1a_2}$&lt;/span&gt; orbital and the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{4b_2}$&lt;/span&gt; orbital start rising in energy, eventually rising above the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6a_1}$&lt;/span&gt; orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around &lt;span class=&quot;math-container&quot;&gt;$117^\circ$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same diagram can be used for other molecules, such as &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;. Try using it to figure out why &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is linear.&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorUserId="11568" LastEditDate="2020-05-15T15:37:29.543" LastActivityDate="2020-05-15T15:37:29.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22327" PostTypeId="1" AcceptedAnswerId="22329" CreationDate="2015-01-03T02:55:20.277" Score="7" ViewCount="5736" Body="&lt;p&gt;When suggesting nitration of an aromatic compound in the synthesis of some organic molecule, it was raised that this route should be avoided as to prevent things from going &quot;ka boom.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation was not forthcoming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why do nitro groups tend to make organic molecules explosive? Is it because the $\ce{NO2}$ group really &quot;wants&quot; to be $\ce{N2}$ since a) diatomic nitrogen has a &lt;em&gt;super&lt;/em&gt; high bond strength and b) diatomic nitrogen is a gas usually and therefore conversion to diatomic nitrogen would be entropically favorable? Is it because carbon's most oxidized state is also a gas and the $\ce{C=O}$ bond isn't a wimp either - hence the reason many organic compounds are flammable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what happens first in an explosion? Does carbon become oxidized first, which then provides the activation energy necessary to decompose the nitro group? I ask this because organic carbon compounds by themselves which are flammable - i.e. toluene - aren't considered explosive, but trinitrotoluene is an explosive. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="10185" LastEditDate="2015-06-17T12:16:07.610" LastActivityDate="2015-06-17T12:16:07.610" Title="Why do nitro groups confer explosive tendencies?" Tags="&lt;organic-chemistry&gt;&lt;explosives&gt;&lt;nitro-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22329" PostTypeId="2" ParentId="22327" CreationDate="2015-01-03T04:59:49.230" Score="7" Body="&lt;p&gt;Explosives chemistry is a rather complex topic. I've heard that &lt;a href=&quot;https://books.google.ca/books?id=ATiYCfo1VcEC&amp;amp;dq=isbn:3527316566&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=w2anVJL-M4GyogT7tYCIBg&amp;amp;ved=0CB4Q6AEwAA&quot;&gt;this book&lt;/a&gt; is a good source of information about it (I haven't read it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, your intuition about the nitro group is accurate. Formation of $\ce{N2}$ is highly energetically favourable. To get an explosive, what we need is a &lt;em&gt;rapid&lt;/em&gt; reaction that produces a lot of heat and gas to both cause the obvious effects of an explosion and to propagate the reaction to other molecules of explosive. The rapid part is what separates something like TNT from toluene. The combustion of toluene is also energetically favourable, but in TNT, to create $\ce{N2}$ is not dependant on mass transport of any other species and can thus happen very rapidly, whereas the combustion of toluene is limited by how quickly oxygen is transported to it. If one were to vaporize toluene in the correct concentration in air, this transport problem goes away and an explosion can occur. TNT also has an advantage in this regard because the nitro groups provide a source of oxygen to react with the carbon and nitrogen remaining (not enough for all of it, but it helps). Explosives are often mixed with fuels or oxidizing agents to produce a more oxygen-balanced mixture for a more efficient explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what happens first in an explosion, many different reactions can occur during an explosion, but it may be helpful to consider what it takes to actually detonate TNT and think about the timescale of the reactions. TNT is a solid at room temperature and has a flash point of 163 °C making it difficult to even ignite, and while it will burn in a fire, there is no risk of explosion. For an explosion to occur, enough gas and heat has to be produced to propagate the reaction through the bulk of the material. In practice this is done using a much more sensitive explosive (other explosives like lead azide or nitroglycerine are unstable enough to be set off by heat or pressure) to produce a small shockwave that provides the activation energy to initiate a reaction as it travels through the explosive which then sustains the shockwave through the rest of the material. In a normal explosive (not a fuel-air explosive or the like), the reactions that contribute to the bulk of the explosion are limited to what the explosive is made of because the speed of the explosion is too fast for air to play much of a role initially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of TNT, the experimentally-measured time it takes for a shockwave to pass through is 100–200 fs (&lt;em&gt;no idea how one measures that&lt;/em&gt;), so any oxidation of the carbons seem unlikely to contribute much to the the initial explosion, given the only readily available source of oxygen is from the nitro groups which must presumably decompose first. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp072121s&quot;&gt;This group&lt;/a&gt; proposed a few decomposition pathways for TNT, including homolytic cleavage of the $\ce{C-NO2}$ bond, rearrangement from $\ce{C-NO2}$ to $\ce{C-ONO}$ followed by homolytic $\ce{O-NO}$ cleavage, and $\ce{C-H}$ attack from an adjacent nitro group to the methyl ring substituent, but they found that only the first was fast enough to occur during detonation, the others possible only for lower temperature thermal decomposition. This initial decomposition step is the only thing fast enough to contribute to the shockwave that sets off the rest of the TNT, while the reactions that produce the final products occur (relatively) long after the initial blast has initiated the rest of the explosive.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-03T04:59:49.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22330" PostTypeId="1" CreationDate="2015-01-03T06:30:54.137" Score="5" ViewCount="4195" Body="&lt;p&gt;Would you consider it safe to clean and reuse a 20 litre plastic drum that contained 36% Hydrochloric Acid for drinking water? If so how would you recommend cleaning it?&lt;/p&gt;&#xA;" OwnerUserId="11571" LastEditorUserId="5456" LastEditDate="2015-01-03T07:11:44.957" LastActivityDate="2020-12-27T23:59:48.960" Title="Reuse 20 litre plastic drums for drinking water?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;cleaning&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22331" PostTypeId="2" ParentId="22330" CreationDate="2015-01-03T06:51:18.907" Score="9" Body="&lt;p&gt;Regardless of what it takes to remove any traces of hydrochloric acid from the drum, the container is probably not certified as food-grade. So in addition to any risks from the acid itself (including things like heavy metals, which are common contaminants in mineral acids), the plastic is not certified to be free of toxic substances left over from manufacturing or that the plastic will not degrade to release such substances over its lifetime.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-03T06:51:18.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22332" PostTypeId="1" AcceptedAnswerId="22334" CreationDate="2015-01-03T09:12:05.577" Score="0" ViewCount="688" Body="&lt;p&gt;I do not understand why the 3rd canonical form of propenal has the oxygen atom donating an electon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen in electronagetive so instead of taking electrons why is it giving?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this why it is most unstable but how is it even formed ??&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2015-01-03T11:07:43.510" Title="Resonance in Propenal" Tags="&lt;resonance&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22333" PostTypeId="1" CreationDate="2015-01-03T10:05:40.107" Score="1" ViewCount="1727" Body="&lt;p&gt;I have to draw the 4 four isomers for the molecule  $\ce {C2H2Cl2}$  when $\ce {Br2}$ is added to it. At last the four isomers for  $\ce {C2H2Cl2Br2}$.&#xA;I have found three of them: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8VZ06.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find the last one.&lt;/p&gt;&#xA;" OwnerUserId="11561" LastEditorUserId="865" LastEditDate="2021-05-22T21:19:37.537" LastActivityDate="2021-05-22T21:19:37.537" Title="Draw the four isomers of C2H2Cl2Br2" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22334" PostTypeId="2" ParentId="22332" CreationDate="2015-01-03T11:07:43.510" Score="3" Body="&lt;p&gt;What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of propenal the $\pi$ MOs result from the overlap of four $p$ orbitals on the four atoms in the chain. The two occupied $\pi$ MOs show how most of the electron density is between C1 and C2, and C3 and the oxygen but there is also significant electron density between C2 and C3. This reflects the different positions of the double bonds shown in the resonance contributors of propenal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nfobj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCPnT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemtube3d.com/orbitalsacrolein.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemtube3d.com/orbitalsacrolein.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-03T11:07:43.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22335" PostTypeId="2" ParentId="22333" CreationDate="2015-01-03T13:07:01.377" Score="4" Body="&lt;p&gt;You have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isomer is known as &quot;meso compound&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NUcIz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-01-03T13:24:45.777" LastActivityDate="2015-01-03T13:24:45.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22336" PostTypeId="2" ParentId="22327" CreationDate="2015-01-03T13:38:47.127" Score="3" Body="&lt;p&gt;&lt;strong&gt;What may make things go kaboom without true explosion&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, whenever you have a mixture of an oxidizer and a fuel, whatever nature, there is always risk of fast combustion.  Common gunpowder is a mixture of solids, and it can go kaboom when confined. Since nitrogroups are oxidizing, while the rest of the molecule is usually reducing, many nitrocompounds are capable of combustion without external oxidizer. The process may proceed without true explosion, for example, TNT in small portions may be safely ignited. It burns with bright yellow flame producing a lot of black thick smoke. Anyway, to produce a kaboom, the compound or mixture must have a way of decomposition producing a lot of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Self-sustaining thermal decomposition, say of hydrogen peroxide, may produce large volumes &#xA;of gases, and in case the reaction mixture is confined, the build-up of pressure may eventually lead for the reaction container to... fragment violently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Detonation wave&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A true detonation is described as a process, where instead of burning front, proceeding with subsonic speed, a supersonic detonation front is observed. The speed the front travels with may be above 10 km/s in some solids and over 3 km/s in gases. This implies that  the process does not involve diffusion of active species or thermal energy, that travel usually with subsonic speed. So decomposition in detonation wave is pressure-induced, either from adiabatic heating or from direct mechanical stress. To achieve this, the compound must be able of exothermic decomposition, producing enough energy to sustain the pressure wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nitrogroup role&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commonly utilized nitro-explosives are relatively stable and often require a significant primary explosive charge to induce detonation wave. Nitrogroups, despite having positive energy of formation, are relatively durable groups, with high barrier or bond dissociation energies, double so for nitroaromatics. This means, that in detonation wave decomposition of nitrogroups mostly consumes energy, while formation of other molecules, say, water and carbon monooxide, produces energy. So, the role of nitrogroup is mostly to provide a readily available oxygen for oxidation of the rest of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other cases&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a lot of explosives (most commonly with high amount of nitrogen), that have comparatively low decomposition energy, but are extremely sensitive. Probably, ones of the most infamous would be heavy metal azides, like lead azide. There in addition to readily available exothermic decomposition route, usually some weak bonds are present in the molecule, allowing easy thermal fragmentation, following by rearrangement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, infamous diazomethane have C-N bond energy around 172 kJ/mol, while energy of $\ce{CH2=CH2}$ bond is 611 kJ/mol,  meaning that the process produces roughly twice more energy than consumes - not a lot, but the very low initial energy investment makes diazomethane to explode from a funny look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nitrogroups mostly serve as a readily available source of oxygen to oxidize the rest of the molecule&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-01-03T13:38:47.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22337" PostTypeId="2" ParentId="22259" CreationDate="2015-01-03T13:58:35.040" Score="6" Body="&lt;p&gt;Strictly speaking resonance &lt;em&gt;doesn't&lt;/em&gt; stabilise molecules. This is because resonance is merely a descriptive way for chemists to extend simple pictures of bonding to molecules where the simplest model of bonding doesn't work. We &lt;em&gt;observe&lt;/em&gt; that some molecules are more stable than the simple picture, or a different shape than the simple picture would suggest and we &lt;em&gt;explain&lt;/em&gt; that difference by adding the idea of resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem arises because, when we draw &lt;em&gt;simple&lt;/em&gt; pictures of the bonding in molecules, we assume each bond consists of two shared electrons. So we draw benzene like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4PF6b.png&quot; alt=&quot;simple benzene bonding&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture neatly accounts for all the electrons in the C-C bonds and is often good enough for understanding and tracking the electrons and bonds in reaction mechanisms. But the picture is clearly &lt;strong&gt;wrong&lt;/strong&gt; as it doesn't match the known properties of benzene: in particular &lt;em&gt;all the bonds in benzene are the same length&lt;/em&gt;. The stability of benzene is also understated compared to the (hypothetical) molecule with isolated double bond. The real reasons for this are pretty complex and involve all sorts of quantum-mechanical molecular orbital gobbledygook. But this is way too complex to be useful when drawing simple molecular pictures or describing reactions or bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid this problem and to help retain the simple picture we add the idea of &lt;em&gt;resonance&lt;/em&gt;. This allows us to use the simple picture where a line representing a bond still means two shared electrons but we have to draw more than one structure to capture the real molecular structure. We say the multiple structures &quot;resonate&quot; but this is really just shorthand for saying that the real structure is an in-between mix of the resonance structures we draw. For benzene we draw like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tEMr7.png&quot; alt=&quot;benzene resonance structures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first structure there is a single bond between atoms 1 and 2; in the second structure there is a double bond. Real benzene is a mix of the two resonance structures where all the bonds are a sort of mix of a single and double bond. In fact benzene is sometimes drawn like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hUNUG.png&quot; alt=&quot;circle benzene structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this structure is that it is is less useful most of the time than the (incorrect) structures with single and double bonds as it makes counting electrons in bonds less clear and this is annoying when trying to follow reaction mechanisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So resonance is basically a pragmatic way to retain simplistic pictures of bonding in molecules while recognising that that simplistic pictures is often a poor representation of reality. And thereby avoiding having to do complicated quantum stuff with orbitals every time you want to understand a molecular structure or reaction. &lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-01-03T13:58:35.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22338" PostTypeId="2" ParentId="17089" CreationDate="2015-01-03T14:58:43.697" Score="0" Body="&lt;p&gt;To be very honest with you, it's almost impossible to predict accurately, the solubility of compounds. At the end of the day, it's best to memorize them. Yes, there are some cases wherein a certain logic or reasoning can be used, but mostly its better to just memorize the thing. As Abel said, it comes down to a difficult interplay of lattice enthalpy, and hydration energy.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-01-03T14:58:43.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22339" PostTypeId="2" ParentId="22286" CreationDate="2015-01-03T15:01:40.163" Score="4" Body="&lt;p&gt;I use Molecular Workbench (&lt;a href=&quot;http://mw.concord.org/modeler/&quot; rel=&quot;nofollow&quot;&gt;http://mw.concord.org/modeler/&lt;/a&gt;) when I want to employ MD for teaching.  Free of charge and easy to use.  No need to reinvent the wheel.&lt;/p&gt;&#xA;" OwnerUserId="11573" LastActivityDate="2015-01-03T15:01:40.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22340" PostTypeId="2" ParentId="22269" CreationDate="2015-01-03T16:30:27.480" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxylamine&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction of $\ce{NH2OH}$ with an aldehyde or ketone produces an oxime.&lt;br&gt;&#xA;  $$\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/2,4-Dinitrophenylhydrazine&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone).&#xA;  $$\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-01-04T15:10:20.020" LastActivityDate="2015-01-04T15:10:20.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22341" PostTypeId="2" ParentId="22330" CreationDate="2015-01-03T16:37:28.253" Score="1" Body="&lt;p&gt;I imagine the container is made of polyethylene or polypropylene (it should be marked with a numeric code). If that is the case, I have a hard time accepting that there is any downside to using it for drinking water. Chemists don't want heavy metals contaminating their hydrochloric acid, so I think you are OK on that count as well.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-03T16:37:28.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22342" PostTypeId="2" ParentId="22286" CreationDate="2015-01-03T16:53:12.517" Score="8" Body="&lt;p&gt;The random displacements in your suggested method are uncorrelated from step to step.  What you are suggesting is essentially heavily damped molecular dynamics at the Brownian limit.  The limitations are obvious, you won't sample correlated, large-scale motions efficiently.  (That might be a student project right there: turn up the Langevin parameter, see when you lose rotations of phenyl rings, and ask yourself what this means.)&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2015-01-03T16:53:12.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22343" PostTypeId="2" ParentId="6791" CreationDate="2015-01-03T17:23:46.860" Score="0" Body="&lt;p&gt;inductive effect in tertiary carbocation is much more than primary or secondary carbocation since there are 3 R groups in tertinary carbon. so (-)inductie effect by R groups reduces the (+)charge on carbocation. so it will become more stable there for it is thermodynamycally favorite. so the reactivity is heigher than primary or secondary alcohol. &lt;/p&gt;&#xA;" OwnerUserId="11574" LastActivityDate="2015-01-03T17:23:46.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22344" PostTypeId="1" CreationDate="2015-01-03T18:21:35.507" Score="5" ViewCount="1168" Body="&lt;p&gt;From my conjecture, I think it's a mix between valence electron and the principal energy state, but I'm not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, C, N, O and F can only make a maximum of four bonds, as they can only have up to sp3 hybrid orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Si, P, S and Cl, they can make a maximum of nine bonds, as they can have sp3d5 hybrid orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what I think, but I'm probably incorrect in some aspect.&lt;/p&gt;&#xA;" OwnerUserId="11490" LastEditorUserId="865" LastEditDate="2015-10-30T17:44:00.290" LastActivityDate="2015-10-30T17:44:00.290" Title="What properties of an element determines the maximum number of bonds it can make?" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22345" PostTypeId="2" ParentId="22314" CreationDate="2015-01-03T18:21:52.200" Score="1" Body="&lt;p&gt;I think some of your confusion arises from the way your questions have been posed. The laws state a relationship among pressure, volume, temperature and the number of molecules of gas (which is what moles of gas tells you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason we know the volume is because the gas is in a container. You may not be told this volume in the question, but gases expand to fill whatever container they are in (unless you are doing something on a planetary scale where gravity starts to count and you aren't going to get questions involving this in basic chemistry). So if volume is the unknown, you are being asked to calculate the volume of a container. Or the volume may be known and you need to calculate another of the 4 factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth understanding the basis of the laws. They are a high level summary of the consequences of some simple behaviour which can be understood from simple principles. One is that all molecules in an ideal gas (which means anything not close to liquefying from cold or pressure) behave like small objects that don't interact. At a constant temperature and pressure a mole of any gas will occupy the &lt;em&gt;same&lt;/em&gt; volume. This means that, if we can control the pressure and temperature, we can count the molecules (the number of moles of objects  making up the gas) by measuring the volume. Since we can also weigh the gas, we know what the objects consist of, or, at least, their mass). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the molecules making up a gas don't interact, a mixture of different molecules will simply mix and we can adjust the gas equations to pretend that the components have &lt;em&gt;partial&lt;/em&gt; pressures depending on their proportions in the mixture. So air (simplifying and rounding to make the numbers easier) is 75% nitrogen and 25% oxygen. 1 atmosphere of air will contain oxygen with a partial pressure of 0.25 atmosphere and nitrogen of .75 atm. You can apply the equations to the components using these numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the principles and you will be able to work out how to apply the laws.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-01-03T18:21:52.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22346" PostTypeId="1" CreationDate="2015-01-03T18:53:15.723" Score="2" ViewCount="15288" Body="&lt;p&gt;I want to calculate the energy change when a solute is dissolved in water. I know that I can achieve this by using the equation $q=mc \Delta T$. My question is does the mass ($m$) change with the amount of solute added to the solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I had 25g of water ($c=4.18$) as my solvent and 5g of sodium chloride as my solute. Would the energy change equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q=25 \cdot 4.18 \cdot \Delta T$$&#xA;or&#xA;$$q=(25+5) \cdot 4.18 \cdot \Delta T$$&lt;/p&gt;&#xA;" OwnerUserId="11576" LastEditorUserId="4945" LastEditDate="2015-01-05T03:38:08.463" LastActivityDate="2015-02-20T22:24:47.070" Title="Which formula is correct for calculating the heat of dissolution?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;energy&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22347" PostTypeId="1" CreationDate="2015-01-03T19:43:17.427" Score="4" ViewCount="7125" Body="&lt;p&gt;In the periodic table the trend is that as the mass of each element increase's the specific heat tends to go down. This seems to be counter-intuitive because if I am not mistaken as mass increase's the amount of heat that the substance can hold goes up. So why does the specific heat of these elements go down?&lt;/p&gt;&#xA;" OwnerUserId="8167" LastActivityDate="2020-05-02T07:28:19.957" Title="Why do heavier elements have smaller specific heat values." Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;periodic-table&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22348" PostTypeId="2" ParentId="22346" CreationDate="2015-01-03T21:04:21.597" Score="0" Body="&lt;p&gt;You have to take the mass of the mixture. So, the second expression is true.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-03T21:04:21.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22350" PostTypeId="1" AcceptedAnswerId="22370" CreationDate="2015-01-03T23:24:25.117" Score="1" ViewCount="5760" Body="&lt;p&gt;I can't seem to find a very good explanation of the difference between polypeptides and polyamides. So far, I know that polypeptides are a type of polyamides, but that is about it. My guess is that the term polyamides refers both to the polymers made from amino acids and those made from diamines and dicarboxylic acids, but polypeptides are only those made of amino acids, but this is just a guess...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance! &lt;/p&gt;&#xA;" OwnerUserId="7757" LastActivityDate="2020-04-16T08:00:52.233" Title="Difference between polypeptides and polyamides" Tags="&lt;polymers&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22351" PostTypeId="2" ParentId="22350" CreationDate="2015-01-03T23:34:27.957" Score="1" Body="&lt;p&gt;Peptides (or polypeptides) are polymers of amino acids; usually created by cell enzymes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyamides usually have a carbon chain between the amides, such as nylon 6 or nylon 66. Peptides could be classified as polyamides, but are not usually included since they have their own group.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-03T23:34:27.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22354" PostTypeId="1" CreationDate="2015-01-04T03:33:35.620" Score="3" ViewCount="607" Body="&lt;p&gt;Why do molecular orbitals in solids merge to bands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: In silicon every atom is sp&lt;sup&gt;3&lt;/sup&gt; hybridised, but when I merge two of these orbitals then it yields a bonding and an antibonding MO. When a third atom binds to one of the $\ce{Si}$ atoms in this configuration, then again we should get two MOs, separated by the same energy. This way we we don't get so called bands. How do the bands form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried wikipedia and other sites but they answer in brief. &lt;/p&gt;&#xA;" OwnerUserId="11581" LastEditorUserId="4945" LastEditDate="2015-01-05T02:30:59.347" LastActivityDate="2015-01-05T02:30:59.347" Title="Why do molecular orbitals in solids merge to bands?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22355" PostTypeId="1" CreationDate="2015-01-04T04:15:52.663" Score="0" ViewCount="744" Body="&lt;p&gt;How do we name $\ce{Cl2CHCH2COOH}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is seeing branched chain our number 1 priority so we name it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1-dichloro-3-propanoic acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is more acidic $\ce{Cl2CHCOOH}$ or $\ce{Cl2CHCH2COOH}$ ? &#xA;I feel it should be the first one as there is more −I effect but not sure.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorDisplayName="user7951" LastEditDate="2015-06-09T18:30:20.200" LastActivityDate="2015-06-09T18:30:33.080" Title="Naming rules and acidity of Cl2CHCH2COOH" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;acid-base&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22356" PostTypeId="1" AcceptedAnswerId="22359" CreationDate="2015-01-04T04:49:10.483" Score="0" ViewCount="32325" Body="&lt;p&gt;Could you explain what is unit of  molar conductivity along with derivation&lt;/p&gt;&#xA;" OwnerUserId="7039" LastActivityDate="2017-09-12T01:14:33.383" Title="What is the unit of molar conductivity?" Tags="&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="3" CommentCount="0" ClosedDate="2017-09-12T02:04:21.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="22357" PostTypeId="2" ParentId="22355" CreationDate="2015-01-04T04:54:22.990" Score="3" Body="&lt;p&gt;1- 3,3-dichloropropanoic acid. Please see &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot; rel=&quot;nofollow&quot;&gt;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The first one  is more acidic, as the inductive effect, due to chlorine atom decrease fast with the distance. (So, your answer is correct!).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-01-04T05:19:06.313" LastActivityDate="2015-01-04T05:19:06.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22358" PostTypeId="2" ParentId="22354" CreationDate="2015-01-04T04:59:26.800" Score="0" Body="&lt;p&gt;This is explained by the Pauli exclusion principle, as applied to &lt;em&gt;conductors&lt;/em&gt;: quantum mechanics forbids that fermions have exactly the same spin-state and energy level. In a single atom, that means, for example that H has one s electron in the &quot;up&quot; state and anther in thew &quot;down&quot;. In conductors (and semiconductors), electrons from one atom can exchange with those of another, so they must have (slightly) different energy levels, if all spin-states are occupied, effectively splitting the single level into a &quot;band&quot; of energies. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Pauli_exclusion_principle&lt;/a&gt; for more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, this is not true for &quot;solids&quot; in general, e.g. sulfur or window glass, but is true for metals, alloys (e.g. brass) and conductive compounds.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-04T04:59:26.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22359" PostTypeId="2" ParentId="22356" CreationDate="2015-01-04T05:01:58.947" Score="8" Body="&lt;p&gt;Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.&#xA;It's unit in S.I. is: $$\ce{S.m^2.mol^{-1}}$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-04T05:01:58.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22360" PostTypeId="2" ParentId="22347" CreationDate="2015-01-04T05:10:05.743" Score="3" Body="&lt;p&gt;The more particles in a system, the more modes of vibration (and rotation), so one gram of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; has about &lt;span class=&quot;math-container&quot;&gt;$235 \times$&lt;/span&gt; more particles than one gram of U. The heat is &quot;stored&quot; in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See &lt;a href=&quot;http://farside.ph.utexas.edu/teaching/sm1/lectures/node71.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://nptel.ac.in/courses/115106076/Module%209/Module%209.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;links&lt;/a&gt; and  for more information.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="69830" LastEditDate="2020-05-02T07:28:19.957" LastActivityDate="2020-05-02T07:28:19.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22361" PostTypeId="2" ParentId="22113" CreationDate="2015-01-04T05:24:06.967" Score="5" Body="&lt;p&gt;The &lt;em&gt;heat&lt;/em&gt; generated increases directly with the amount of sodium acetate trihydrate used, but the maximum &lt;em&gt;temperature&lt;/em&gt;, ignoring surface effects, &lt;em&gt;should not change at all&lt;/em&gt;. After all, if the heat in 300 ml is used to raise the temperature of 300 ml of substance, or if 300 kg is used to raise the temperature of 300 kg, it's being spread out proportionately. No matter how much you use, it won't exceed ~58 C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, do not try this in a container that would not allow room for expansion, e.g. a glass bottle, as in crystallizing, sodium acetate trihydrate might take up more room. See warning at &lt;a href=&quot;http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html&quot; rel=&quot;noreferrer&quot;&gt;http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html&lt;/a&gt;. You &lt;em&gt;could&lt;/em&gt; safely do this on a polyethylene cutting board, though, which melts at ~100 C&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-04T05:24:06.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22362" PostTypeId="2" ParentId="22354" CreationDate="2015-01-04T05:51:10.700" Score="4" Body="&lt;p&gt;Let's explain this with your example of silicon that every atom is sp3 hybridised. when you merge two of these orbitals, then it yields bonding and antibonding MO. But  when a third atom bounds, it yields  3 MO (not four as you mentioned) bonding, antibonding and non bonding MO. The number of MO equals the number of atomic orbitals constituting them, according to LCAO theory.&#xA;The energy of the third MO is in between the nonbonding and anti-bonding MOs. When you add a forth, a fifth and so on atoms, you will have more and more MOs very close in energy and you will form a band.&#xA;This is a qualitative description about the formation of a band.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-04T05:51:10.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22363" PostTypeId="2" ParentId="6791" CreationDate="2015-01-04T08:11:53.307" Score="-4" Body="&lt;p&gt;Primary alcohols react via $S_N2$ mechanism which involves the carbocation formation&#xA;but the secondary and tertiary react via $S_N1$ &lt;strong&gt;which is elementary&lt;/strong&gt; and no carbocation is formed.&#xA;hence it appears that both primary and tertiary alcohols should react rapidly but it is not so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the $\ce{C-O}$ bond weakens as the no of electron donating groups on the alpha-carbon increases this is due to corresponding increase in negative charge developed on $\ce{O}$ atom **&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;thats why tertiary alcohols are more reactive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&#xA;further $\ce{ZnCl2}$ is a &lt;strong&gt;&lt;em&gt;Lewis acid&lt;/em&gt;&lt;/strong&gt; which coordinates with $\ce{O}$ atom and weakens the bond thus it need to be used for primary and secondary alcohols.&lt;/p&gt;&#xA;" OwnerUserId="11581" LastEditorDisplayName="user15489" LastEditDate="2015-06-26T13:51:05.417" LastActivityDate="2015-06-26T13:51:05.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22364" PostTypeId="1" AcceptedAnswerId="22365" CreationDate="2015-01-04T08:48:44.347" Score="14" ViewCount="186647" Body="&lt;p&gt;When dealing with organic reactions, &lt;strong&gt;heating under reflux&lt;/strong&gt; is often required, such as the oxidation of Toluene using acidified $\ce{KMnO4}$ and dilute $\ce{H2SO4}$ to Benzoic acid($\ce{C6H6O2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean and how does it differ from &quot;normal&quot; heating?&lt;/p&gt;&#xA;" OwnerUserId="5402" LastEditorUserId="81" LastEditDate="2015-01-04T11:43:59.557" LastActivityDate="2020-03-13T19:50:33.343" Title="What is &quot;heating under reflux&quot;?" Tags="&lt;organic-chemistry&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22365" PostTypeId="2" ParentId="22364" CreationDate="2015-01-04T08:48:44.347" Score="21" Body="&lt;p&gt;Many organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address this, heating under reflux is used. This refers to heating a solution with an attached condenser to prevent reagents from escaping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/en/thumb/5/51/Reflux_labled.svg/668px-Reflux_labled.svg.png&quot; alt=&quot;wikied image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen above, any vapor will condense on the cool surface of the attached condenser and flow back into the flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hot water bath pictured is an optional component of heating under reflux and is usually only used for particularly sensitive reactions. Also, using it limits the reaction temperature to 100 degrees Celsius. &lt;/p&gt;&#xA;" OwnerUserId="5402" LastEditorUserId="41556" LastEditDate="2020-03-13T19:50:33.343" LastActivityDate="2020-03-13T19:50:33.343" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22366" PostTypeId="1" AcceptedAnswerId="23503" CreationDate="2015-01-04T10:40:37.810" Score="12" ViewCount="32669" Body="&lt;p&gt;In 13&lt;sup&gt;th&lt;/sup&gt; group, atomic radius increases from boron to aluminium. From aluminium to gallium, atomic radii decreases. From gallium to indium, atomic radii increases. And from indium to thallium, atomic radii decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this irregular trend given is screening effect. While explaining my teacher told that due to &lt;strong&gt;d-orbital&lt;/strong&gt; screening effect decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't get why?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5026" LastEditDate="2018-05-01T03:51:37.320" LastActivityDate="2019-10-09T10:38:02.557" Title="Why does screening effect decrease due to d-orbital?" Tags="&lt;periodic-trends&gt;&lt;boron-family&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="22367" PostTypeId="1" CreationDate="2015-01-04T10:43:43.517" Score="9" ViewCount="2079" Body="&lt;p&gt;What are the rules for choosing main branch?&lt;/p&gt;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry&quot; rel=&quot;noreferrer&quot;&gt;states&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum number of multiple bonds&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum number of single bonds.&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum length.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't understand how 3 and 4 differ. Which single bonds are we counting? Only carbon-carbon single bonds in the main chain, right?&lt;/p&gt;&#xA;&lt;p&gt;Also, I heard there is a rule that says if you've two equal length chains, then choose the more substituted one. Why is it not given here?&lt;/p&gt;&#xA;&lt;p&gt;And now now imagine you've two choices for a chain, like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gWCrj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gWCrj.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.chemdoodle.com/demos/iupac-naming&quot; rel=&quot;noreferrer&quot;&gt;The chemdoodle program&lt;/a&gt; gives the name as 2-(1-Hydroxybutyl)-3-methylhexanoic acid .&lt;/p&gt;&#xA;&lt;p&gt;I simply do not understand why the methyl substituted chain is chosen instead of hydroxy substituted one and how that fits with IUPAC rules given in Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="11587" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-08-05T13:34:30.393" Title="IUPAC naming - choosing the main branch" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22368" PostTypeId="1" CreationDate="2015-01-04T11:12:55.283" Score="6" ViewCount="12304" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{align}&#xA;\ce{NO3- + 3Fe^2+ + 4H+ &amp;amp;-&amp;gt; NO + 3Fe^3+ + 2H2O}\tag{1}\\&#xA;\ce{[Fe(H2O)6]^2+ + NO &amp;amp;-&amp;gt; \underset{\text{(brown)}}{[Fe(H2O)5(NO)]^2+} + H2O}\tag{2}&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The above are the reactions given in my book [1] for the qualitative analysis of nitrate ion (formation of the brown ring complex). I think that the reactions above are incorrect. The first one shows formation of $\ce{Fe^3+}$, which should be attacked by excess of $\ce{H2O}$ to give ferrum(III) hexaaquasulphate, not ferrum(II) hexaaquasulphate. Now that should be attacked by $\ce{NO}$ as a ligand and give $\ce{[Fe(H2O)6NO]SO4}$, where the oxidation number of $\ce{Fe}$ is being $3+$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the reactions show that the $\ce{Fe}$ already present gives $\ce{[Fe(H2O)6]^2+}$ and then $\ce{NO}$ acts as a neutral ligand to form the complex. But by the reactions shown in the book, even after formation of the brown ring complex $\ce{Fe^3+}$ is still present in the solution, which according to me shouldn't be because if $\ce{Fe^3+}$ still remains in the solution then water should also have attacked it as a ligand to give $\ce{[Fe(H2O)6NO]SO4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some clarity on this. Are the reactions given in the book correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $\ce{Fe^3+}$ will be attacked by water to give $\ce{[Fe(H_2O)_6]^3+}$ but why won't $\ce{NO}$ take electrons from the oxidation of iron ($\ce{Fe^2+ -&amp;gt; Fe^3+ + e- }$), make $\ce{NO-}$ and attack the above and substitute one water molecule and give more of the brown ring complex?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;NCERT. &lt;em&gt;Chemistry: Textbook for class XII&lt;/em&gt;; National Council of Educational Research and Training: New Delhi, &lt;strong&gt;2007&lt;/strong&gt;. ISBN 978-81-7450-648-1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="41328" LastEditDate="2017-08-06T22:26:37.000" LastActivityDate="2017-08-06T22:26:37.000" Title="Are these reaction equations for the formation of the brown ring complex correct?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;erratum&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22370" PostTypeId="2" ParentId="22350" CreationDate="2015-01-04T13:12:07.903" Score="6" Body="&lt;p&gt;Your guess is correct. Polyamides are polymers where the repeating units are linked by amide bonds. Polypeptides are a specific type of polyamide where the repeating units are amino acids.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-04T13:12:07.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22371" PostTypeId="1" AcceptedAnswerId="22372" CreationDate="2015-01-04T13:53:23.513" Score="0" ViewCount="884" Body="&lt;p&gt;I am confused regarding the fact that in a &lt;em&gt;cis&lt;/em&gt; isomer which is less stable as compared to a trans isomer why does it have a higher boiling point ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to what i feel trans is more stable so the intermolecular forces are more so melting point is also more as compared to &lt;em&gt;cis&lt;/em&gt; isomer but why not the boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even generally why is boiling point in inverse proportion with stability?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="42741" LastEditDate="2017-05-13T06:42:04.907" LastActivityDate="2017-05-13T06:42:04.907" Title="Cis-Trans Isomerism" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22372" PostTypeId="2" ParentId="22371" CreationDate="2015-01-04T14:04:11.063" Score="2" Body="&lt;p&gt;I think you are confusing some things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stability is only regarding to individual molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, boiling point depends on intermolecular bonding. You can have 2 unstable molecules strongly bonded together via hydrogen bonds for instance as you can have 2 stable compounds without hydrogen so they will not have a very high boiling point because of the absence of such hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the melting point, it's another kind of force sticking molecules or atoms together. So there is no space for comparison beteween individual stability and boiling or even melting point of a whole bunch of molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x3WeI.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;img src=&quot;https://i.stack.imgur.com/DG4mE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trans on the left and Cis on the right. &#xA;Where can you have hydrogen bonds ? You need some Oxygen or chemicaly equivalent atom to make it !&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="10119" LastEditDate="2015-01-04T14:18:31.573" LastActivityDate="2015-01-04T14:18:31.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22373" PostTypeId="1" AcceptedAnswerId="84369" CreationDate="2015-01-04T14:25:09.393" Score="8" ViewCount="221" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15511/can-a-multi-species-system-oscillate-around-equilibrium&quot;&gt;this question&lt;/a&gt;, which asks if a multi-species system could oscillate around its equilibrium, I thought about the possible concentration pathways the following more narrow specified system could take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{A -&amp;gt;[\text{$k_1$}] B} $$&#xA;$$ \ce{B -&amp;gt;[\text{$k_2$}] C} $$&#xA;$$ \ce{C -&amp;gt;[\text{$k_3$}] A} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I was interested in are the ratio of those concentrations $A:B:C$. The initial conditions are $A(0)=1$, $B(0)=0$, $C(0)=0$. (For simplicity, $[\ce{A}] \equiv A$, and analogous for $B$ and $C$)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the possible concentration pathways for different values of $k_1$, $k_2$ and $k_3$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-10-21T12:03:58.647" Title="What are the possible concentration pathways for a 3-species system?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22374" PostTypeId="2" ParentId="22373" CreationDate="2015-01-04T14:25:09.393" Score="12" Body="&lt;p&gt;Of course, we can't display every pathway there is. We have to select a wide variety of rate constants to get a &quot;good map&quot; of the parameter-space. The approach I took was this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Construct the differential equations&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm dA}{\mathrm dt} = -k_1 A + k_3 C$$&#xA;$$\frac{\mathrm dB}{\mathrm dt} = -k_2 B + k_1 A$$&#xA;$$\frac{\mathrm dC}{\mathrm dt} = -k_3 C + k_2 B$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Solve them numerically for every possible combination of 0.1, 1 and 10 for each rate constant and the above stated initial conditions&lt;/li&gt;&#xA;&lt;li&gt;This leads to 27 pathways, which are plotted in a ternary diagram&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The title of the ternary diagram is $k_1:k_2:k_3$, which is 0.1 : 0.1 : 0.1 in the first one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wz1VG.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kvFnq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kvFnq.png&quot; alt=&quot;The solution.&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;(click for larger image)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorDisplayName="user7951" LastEditDate="2016-11-15T18:43:11.137" LastActivityDate="2016-11-15T18:43:11.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22375" PostTypeId="2" ParentId="22368" CreationDate="2015-01-04T14:30:30.187" Score="2" Body="&lt;p&gt;The reactions given are correct. The nitric acid oxidizes the ferrous ions into ferric ions, and itself get's reduced to NO. But this reaction doesn't happen all at once. It's not as if all the ferrous ions are oxidized to ferric in one go. The remaining ferrous ions then form the brown ring complex with NO. Furthermore, the formation of the brown ring complex itself is a redox reaction. Iron is in the rare +1 oxidation state in the complex, whereas NO is actually NO+. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, ferric ions are present in the solution. They were produced as a result of the first reaction, but are not required in the second reaction.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-01-04T14:30:30.187" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22376" PostTypeId="1" AcceptedAnswerId="22379" CreationDate="2015-01-04T16:13:15.463" Score="3" ViewCount="26140" Body="&lt;p&gt;Specifically, I was thinking of the case where the R group is a carboxylic acid functional group or an amine group. Then there would not be a chiral carbon atom. Is there a restriction on what can be an R group that I am missing?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="865" LastEditDate="2015-10-16T18:06:09.327" LastActivityDate="2015-10-16T18:06:09.327" Title="Are all amino acids except glycine chiral compounds?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chemical-biology&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22377" PostTypeId="1" AcceptedAnswerId="22447" CreationDate="2015-01-04T16:53:12.917" Score="6" ViewCount="4728" Body="&lt;p&gt;How do you apply Wade's rules to Zintl polyhedral cluster anions ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I believe $[\ce{Bi_3}]^{3-}$ is a triangle. My approach to justify would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5 electrons per each Bi atom&lt;/li&gt;&#xA;&lt;li&gt;3 electrons for the charge &lt;/li&gt;&#xA;&lt;li&gt;18 total electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$18 = 4\times3+6 \therefore$ arachno cluster: trigonal bipyramid ($n+2$ polyhedron) with two missing vertices $\therefore$ linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was a nido cluster $(4n+4\ \ce{e-})$, then it would be a tetrahedron with one missing vertex $\therefore$ a triangle. Am I mistaken about the observed geometry or is my method incorrect?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="275" LastEditDate="2015-01-07T00:03:24.027" LastActivityDate="2015-01-08T20:57:25.297" Title="Wade's rules for Zintl ions" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22378" PostTypeId="1" AcceptedAnswerId="22381" CreationDate="2015-01-04T17:18:58.013" Score="4" ViewCount="196" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UGRnp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UGRnp.jpg&quot; alt=&quot;question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it has something to do with the fact that at high temperature it interconverts and so all the coupling constants are the same hence quintet but at lower temperatures the frequency of the rotation is less than the sample rate and so there are different coupling constants for the planar and axial fluorine atoms. We have two structures to consider; 3 planar-1 axial and 2 planar-2 axial. If my line of thought it correct the P-31 NMR spectra for each of these should superimpose to give the spectrum shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7bigK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7bigK.jpg&quot; alt=&quot;answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm lost because when I draw my diagrams they don't match the spectrum shown, I end up with a much more complicated splitting pattern. &lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="83758" LastEditDate="2020-05-04T16:29:39.143" LastActivityDate="2020-05-04T16:29:39.143" Title="NMR and pseudorotation" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22379" PostTypeId="2" ParentId="22376" CreationDate="2015-01-04T17:39:02.873" Score="4" Body="&lt;p&gt;The &lt;strong&gt;alpha&lt;/strong&gt; carbon is a chiral carbon atom, with the exception of glycine which has two indistinguishable hydrogen atoms on the alpha carbon. Therefore, all alpha amino acids but glycine can exist in either of two enantiomers, called L or D amino acids, which are mirror images of each other.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-04T17:39:02.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22381" PostTypeId="2" ParentId="22378" CreationDate="2015-01-04T18:21:39.677" Score="2" Body="&lt;p&gt;You are certainly thinking in the right direction.  Fluorine strongly prefers the axial position (for reasons described &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;here&lt;/a&gt;), so at low temperature, the isomer that you've drawn on the right (2 axial and 2 equatorial fluorines) is &lt;strong&gt;the only isomer present&lt;/strong&gt;.  As a result we have an $\ce{A2B2X}$ spectrum. The 2 equivalent equatorial fluorines (spin = 1/2) split the phosphorous signal into a triplet.  Each line of this triplet is further split into a triplet by the 2 equivalent axial fluorine's. BTW, the same reasoning that explains fluorines axial preference, also explains why generally $\ce{J(P-F)_{equatorial}}$ &gt; $\ce{J(P-F)_{axial}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At room temperature, &lt;a href=&quot;http://en.wikipedia.org/wiki/Pseudorotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;pseudorotation&lt;/a&gt; is rapid (more thermal energy is available and the barrier to pseudorotation can be surmounted) and all fluorines become equivalent on the nmr timescale. Therefore, at room temperature and above a quintet is observed in the $\ce{^{31}P}$ nmr spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-01-04T20:04:26.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22382" PostTypeId="1" AcceptedAnswerId="23873" CreationDate="2015-01-04T20:26:05.493" Score="20" ViewCount="37176" Body="&lt;p&gt;Cotton is mostly cellulose, a plant saccharide. Superglue refers to a class of cyanoacrylates. What is it about the two that causes ignition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd imagine that the cellulose in cotton is in its linear form and not in its ring form since the ring-linear form interconversion only occurs in solution. Is it that the cellulose attacks the cyanoacrylate? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The three hydroxyl groups, one primary and two secondary, in each repeating cellobiose unit of cellulose are chemically reactive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, do the hydroxyl groups attack the carbonyl group in superglue? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-09T15:48:55.683" LastActivityDate="2017-05-26T02:48:09.243" Title="Why does superglue ignite cotton?" Tags="&lt;everyday-chemistry&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="22383" PostTypeId="1" AcceptedAnswerId="22434" CreationDate="2015-01-04T20:28:21.867" Score="3" ViewCount="8295" Body="&lt;p&gt;We all have seen the meniscus of a small graduated cylinder in the lab. The reasons for the occurrence of capillary action are clear, but why does it occur more profusely in smaller-rimmed containers?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2015-07-02T08:32:53.273" Title="Why does capillary action occur better in narrow tubes than in wide tubes?" Tags="&lt;intermolecular-forces&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22384" PostTypeId="1" AcceptedAnswerId="22417" CreationDate="2015-01-04T20:36:42.753" Score="1" ViewCount="108" Body="&lt;p&gt;In many textbooks and chemistry curricula, we can see the description of large molecules (particularly in the discussion of hydrocarbons and other nonpolar compounds) as having &quot;entanglement.&quot; However, I think the term is misleading, since the molecules don't really act like tangled shoelaces or two strings pulling against each other while interlocked. Can we try to clarify what actually occurs that we describe as entanglement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things I think may happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecules are very large, and thus have many sites for IMFs, leading to high viscosity.&#xA;By way of this size, the molecules are much harder pressed for movement and so take a while to move (the distinction is that in the first example the increased size begets more IMFs whereas the second causes &quot;entanglement&quot; just by the bulky molecules' shifting past each other).&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2015-01-05T16:30:12.057" Title="Entanglement of large molecules" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22385" PostTypeId="2" ParentId="22383" CreationDate="2015-01-04T21:12:28.963" Score="1" Body="&lt;p&gt;According to the Jurin's Law &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h = \frac{2\gamma\cos(\theta)}{r\rho g} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ the height of the fluid&lt;br&gt;&#xA;$\gamma$ the surface tension&lt;br&gt;&#xA;$\theta$ the contact angle with the tube&lt;br&gt;&#xA;$r$ the tube radius&lt;br&gt;&#xA;$\rho$ the liquid density&lt;br&gt;&#xA;$g$ the gravitational constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the $h$ is inversly proportional to $r$. In other words, the larger the tube, the lower the liquid climbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this law is only valid if $r &amp;lt; \lambda_\text{c}$ with $\lambda_\text{c} = \sqrt{\frac{\gamma}{\rho g}}  $&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="10185" LastEditDate="2015-01-06T15:24:51.113" LastActivityDate="2015-01-06T15:24:51.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22389" PostTypeId="1" AcceptedAnswerId="22422" CreationDate="2015-01-04T23:52:56.297" Score="0" ViewCount="19820" Body="&lt;p&gt;Say I have a hydrate, and I am going to dissolve it in water. Does the water in the hydrate affect the final concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example: If I have &lt;span class=&quot;math-container&quot;&gt;$\pu{249.69 g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4·5H2O}$&lt;/span&gt;, and dissolve it in &lt;span class=&quot;math-container&quot;&gt;$\pu{200 mL}$&lt;/span&gt; of water, will the concentration be &lt;span class=&quot;math-container&quot;&gt;$\pu{5 M}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7789" LastEditorUserId="41328" LastEditDate="2019-06-25T09:52:57.027" LastActivityDate="2019-09-10T12:13:23.730" Title="Does the water in hydrates affect final concentration?" Tags="&lt;aqueous-solution&gt;&lt;concentration&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22392" PostTypeId="2" ParentId="22368" CreationDate="2015-01-05T02:03:33.437" Score="10" Body="&lt;p&gt;According &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ic010628q&quot; rel=&quot;noreferrer&quot;&gt;Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of&#xA;Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction&#xA;Revisited&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first equation in the book, it should be understood that the ions are aqueous.  For example H+ doesn't mean there are bare protons, it is understood to mean $\ce{H_3O+}$ and Fe2+ is $\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\ce{[Fe(H2O)6]^{3+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second equation in the book, the inner coordination sphere is explicitly shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and &quot;This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$ product&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but why won't NO take electrons from the oxidation of Iron ($Fe^{2+}&#xA; \rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and&#xA;  substitute one water molecule and give more of the brown ring complex?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe.  If the NO- then left that Fe, it would destroy one instance of the brown ring complex.  Hypothetically, the Fe2+ could reduce NO via an &lt;a href=&quot;http://en.wikipedia.org/wiki/Outer_sphere_electron_transfer&quot; rel=&quot;noreferrer&quot;&gt;outer sphere electron transfer&lt;/a&gt;, and then NO- coordinate Fe3+, but this is not what the article reports occurring. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-06T20:43:20.647" LastActivityDate="2015-01-06T20:43:20.647" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="22393" PostTypeId="2" ParentId="22354" CreationDate="2015-01-05T02:19:35.110" Score="1" Body="&lt;p&gt;I'll give a quick answer first and come back tomorrow for a more complete version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, you start with two $\ce{Si}$ atoms and then a third. As you combine these atoms into pairs and molecules, chemistry stresses that the atomic orbitals combine in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's take $\ce{Si2}$. We know that for the $3p$ orbitals, some will become more stable and some will become less stable. There are 3 for each $\ce{Si}$ atom, so a total of 6 molecular orbitals due to combinations of $3p$. (Actually, the $3s$ are also involved, and there's some amount of s-p mixing, but let's ignore that for now.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Take-home message: when we combine atomic orbitals into molecules, some orbitals become more stable and some orbitals become less stable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, now let's add a third $\ce{Si}$ atom as you said. I'm not going to derive the MO diagram right now. (I'll do that tomorrow and insert the picture.) The main point is that we'll have new combinations. If I take $\ce{H3}$ as another example, I should have 3 molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I grow into a linear chain, a 2D sheet, or a 3D solid, I don't have some small number of atoms like 2, 3, 4, 20, etc. Instead, I have 1,000 or 10,000 or many, many more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, those atomic orbitals combine just like they do when you create molecular orbitals. Some become more stable, some become less stable. Yet, rather than having 3 molecular orbitals in $\ce{H3}$ from 3 atomic orbitals, we have 1,000 or 10,000 or many, many more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are so many orbitals, the energy spacing between them is infinitesimal. These are now called &lt;strong&gt;&lt;em&gt;bands&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The type of material (and the atomic composition) will dictate the energy levels of the bands and the so-called &quot;band structure&quot; (i.e., the solid-state equivalent of molecular orbital diagrams). But simply having hundreds and thousands of atoms will give you bands.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-05T02:19:35.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22394" PostTypeId="2" ParentId="22383" CreationDate="2015-01-05T04:12:55.770" Score="2" Body="&lt;p&gt;A meniscus is formed by the opposing forces of adhesion between the fluid and the walls of the container drawing the fluid up and gravity pulling the fluid down. The adhesive forces are proportional to the diameter of the tube while the gravitational effect due to the liquid's weight is proportional to the square of the diameter. This results in the greater height of the meniscus for narrower tubes. &lt;/p&gt;&#xA;" OwnerUserId="11601" LastActivityDate="2015-01-05T04:12:55.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22395" PostTypeId="1" AcceptedAnswerId="22427" CreationDate="2015-01-05T04:34:17.190" Score="17" ViewCount="11874" Body="&lt;p&gt;&quot;Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is that the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a common misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hi6cO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation (in the main chapter, not as an answer to this question) given in Chemistry: The Central Science (13th Edition) by Brown, LeMay, and Bursten (abbreviated as CTCS) is: &quot;A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in ▲ Figure 13.18.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is, I believe, the only textbook which gives this explanation for lowering in vapour pressure on adding a non-volatile solute. The other (reputed) textbooks I've read, like Zumdahl and NCERT Grade XII Chemistry, all give the explanation that CTCS calls a &quot;common misconception&quot;. So, which explanation is correct? If possible, please give an experiment also.&lt;/p&gt;&#xA;" OwnerUserId="9407" LastEditorUserId="9407" LastEditDate="2015-01-05T19:58:28.717" LastActivityDate="2017-01-15T02:54:05.070" Title="What causes the lowering of vapour pressure in volatile/nonvolatile solvent mixtures?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22397" PostTypeId="2" ParentId="22333" CreationDate="2015-01-05T07:10:01.407" Score="1" Body="&lt;p&gt;Like Lighthart already mentioned in the comments, it is important to know which compound was used in the addition reaction. Because of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Halogen_addition_reaction#Reaction_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction mechanism&lt;/a&gt;, only certain products are formed, this is very specific. LDC3 also explained that in the comments, but let me dissect this a little further.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first structure is from geminal,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the scheme below, this is reaction $(1)$ and it forms an achiral product.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the second is from cis &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is depicted in $(2)$ and it forms the meso compound described in Yomen Atassis's answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and the third is trans (or do I have cis and trans backwards). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(No you have not.)&lt;/em&gt; This is depicted in $(3)$, but in this case a racemic mixture of the (&lt;em&gt;S,S&lt;/em&gt;)- and (&lt;em&gt;R,R&lt;/em&gt;)-product is formed, depending on which carbon atom in the bromonium ion the nucleophilic attack takes place.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The only other isomer I can think of is 1,1-dibromo-2,2-dichloroethane, but it is not made from adding $\ce{Br2}$ to dichloroethene. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For completeness, this is depicted in $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could further think of another isomer, which is like the above not formed via addition, it is depicted in $(5)$. Note that these last two reactions also yield achiral products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mSLdW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-01-05T07:10:01.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22399" PostTypeId="1" AcceptedAnswerId="22401" CreationDate="2015-01-05T07:55:15.033" Score="0" ViewCount="1537" Body="&lt;p&gt;I am studying for the International Chemistry Olympiad and some of questions involve synthesis of complicated organic compounds and i have no idea how to do them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setting of the questions is usually like this: they give you a rough idea of the end product however some of the groups are replaced with R. They replace all the other intermediates with A, B, C ... They also give the reactants and catalysts for each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time the catalysts are lewis acids. I wondering if you were able to provide mechanisms and general resulting reactions of these catalyses so i could predict what the product will be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is $\ce{NaBH4}$ a lewis acid?&lt;/p&gt;&#xA;" OwnerDisplayName="user11603" LastEditorUserId="10185" LastEditDate="2015-01-05T10:01:46.407" LastActivityDate="2015-01-05T10:01:46.407" Title="Lewis Acid Catalysts" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;synthesis&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2015-01-06T01:39:58.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="22401" PostTypeId="2" ParentId="22399" CreationDate="2015-01-05T09:11:28.427" Score="1" Body="&lt;p&gt;Firstly, $\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction. &lt;img src=&quot;https://i.stack.imgur.com/VJuAT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-01-05T09:44:34.543" LastActivityDate="2015-01-05T09:44:34.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22403" PostTypeId="1" AcceptedAnswerId="22409" CreationDate="2015-01-05T11:22:13.223" Score="4" ViewCount="2353" Body="&lt;p&gt;Every example of addition polymerisation I have seen so far has used alkenes as the monomers. I was wondering if it is possible to have some other molecule. I don't think the following molecule can exist due to stability reasons, but as an example could you not use something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H-N=N-H}\,?$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely all you need is the presence of a double bond?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="41328" LastEditDate="2019-05-17T19:50:51.777" LastActivityDate="2019-05-17T19:53:57.760" Title="Can you have addition polymerisation with monomers other than alkenes?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;polymers&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22404" PostTypeId="1" AcceptedAnswerId="22405" CreationDate="2015-01-05T11:30:03.573" Score="9" ViewCount="6880" Body="&lt;p&gt;This &lt;a href=&quot;http://www.learnsomething.tips/6-incredible-chemical-reaction-gifs-6/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; shows that a gummy bear explodes when in contact with heated potassium chlorate, &lt;span class=&quot;math-container&quot;&gt;$\ce{KClO3}$&lt;/span&gt;. But what in a gummy bear creates this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do other foods (fruit, icing sugar...) react as violently with potassium chlorate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J1Spr.gif&quot; alt=&quot;Gif of gummy bear explosion&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11609" LastEditorUserId="16683" LastEditDate="2018-10-22T23:23:47.403" LastActivityDate="2018-10-22T23:25:26.150" Title="Why do gummy bears explode when added to hot potassium chlorate?" Tags="&lt;redox&gt;&lt;combustion&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22405" PostTypeId="2" ParentId="22404" CreationDate="2015-01-05T13:02:37.603" Score="11" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_chlorate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Potassium chlorate&lt;/a&gt; is a source of oxygen. After heating, it decomposes to &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{4 KClO3 → KCl + 3 KClO4}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{KClO4 → KCl + 2O2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, &lt;a href=&quot;http://en.wikipedia.org/wiki/Combustion#Chemical_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;combustion&lt;/a&gt; occurs. For example, glucose will react in this manner:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{6O2 + C6H12O6 -&amp;gt; 6CO2 + 6H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any material present which does not burn, such as &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;, the temperature will not rise as high. For gummy bears the reaction works spectacularly because they are mainly carbohydrates (&gt;70%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An apple, for example, has only ~13% carbohydrates – unless you dry it, of course. On the other hand, &lt;a href=&quot;https://www.youtube.com/watch?v=xXts1AlpR_g#t=36&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; on YouTube is an example of how sugar itself reacts violently with potassium chlorate.&lt;/p&gt;&#xA;" OwnerUserId="11551" LastEditorUserId="16683" LastEditDate="2018-10-22T23:25:26.150" LastActivityDate="2018-10-22T23:25:26.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22406" PostTypeId="1" AcceptedAnswerId="22408" CreationDate="2015-01-05T13:03:50.327" Score="2" ViewCount="372" Body="&lt;p&gt;For example, take the 2s electron in lithium - is the reason that it's ionisation energy is lower than would be expected if &quot;shielding&quot; didn't occur at all because the 1s electrons shield the attractive coulombic force from the nucleus or because the 2s electron is repelled by the 1s electrons? Is it possibly a mixture of the two? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, shielding seems to be a model to help conceptualize what is actually happening; am I wrong about this? What is actually going on?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-03-14T13:40:16.753" Title="Is shielding of electrons really best described as shielding or is it really Coulombic repulsion between electrons?" Tags="&lt;inorganic-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22407" PostTypeId="2" ParentId="18654" CreationDate="2015-01-05T13:35:40.430" Score="2" Body="&lt;p&gt;There are some good examples in &lt;a href=&quot;http://link.springer.com/article/10.1023/A:1018817617432#&quot; rel=&quot;nofollow&quot;&gt;Multiwavelength Spectrophotometric Determination of Acid Dissociation Constants&lt;/a&gt; Pharaceutical Research Volume 16, Issue 3, pp 374-381.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzonic acid (pka = 3.98), phenol (pka 9.73), phthalic acid (pkas 2.70 and 4.86), nicotinic acid (pKas 2.1 and 4.63) and p-aminosalicylic acid (pKas 1.79 and 3.58) would be good in the ~220nm-300nm range.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-05T13:35:40.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22408" PostTypeId="2" ParentId="22406" CreationDate="2015-01-05T13:59:48.257" Score="2" Body="&lt;p&gt;The shielding model works as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AeL3f.png&quot; alt=&quot;Shielding effect&quot;&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is a mixture of attraction and pushing from protons and non-valence electrons, in accord with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law#The_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Coulomb's law&lt;/a&gt;. The prediction of the effective core charge is not an easy task - you will have to use quantum chemistry methods (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hartree-Fock&lt;/a&gt;) to predict those, like its done in &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/38/11/10.1063/1.1733573&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think its best viewed as a pragmatic model to boil down the higher, not very graphic, quantum theories, to one number, the effective nuclear charge:&#xA;$$ Z_\text{eff} = Z - S$$&lt;/p&gt;&#xA;" OwnerUserId="11551" LastActivityDate="2015-01-05T13:59:48.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22409" PostTypeId="2" ParentId="22403" CreationDate="2015-01-05T14:06:01.747" Score="2" Body="&lt;p&gt;You're right, the monomer needn't be an alkene. Non-alkenes with double bonds can also form addition polymers. In fact, the monomer needn't always have a double bond either! Surprised? Take a look at this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ytVhg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ytVhg.png&quot; alt=&quot;Caprolactam polymerization&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Image source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/7295/Nylon_6_formation.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemwiki.ucdavis.edu&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="11586" LastEditorUserId="41328" LastEditDate="2019-05-17T19:52:37.580" LastActivityDate="2019-05-17T19:52:37.580" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22410" PostTypeId="1" AcceptedAnswerId="22430" CreationDate="2015-01-05T14:22:59.353" Score="3" ViewCount="11307" Body="&lt;p&gt;I understand that $k_\text{cat}$ measures the turnover number of an enzyme. This measure is therefore a quantity of molecule conversions per unit of time. I suspect that my problem is more that of a lack of maths but why is the unit expressed in $\mathrm s^{-1}$? Why not just seconds?&lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorDisplayName="user7951" LastEditDate="2015-11-16T16:32:52.107" LastActivityDate="2015-11-16T19:55:54.990" Title="Why is the units of kcat 1/s?" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;kinetics&gt;&lt;enzymes&gt;&lt;units&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22411" PostTypeId="2" ParentId="21685" CreationDate="2015-01-05T14:25:09.573" Score="5" Body="&lt;p&gt;In &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; etching of copper, the chloride ion is extremely important.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chloride ions coordinate to &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^3+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie51394a037&quot; rel=&quot;nofollow noreferrer&quot;&gt;Copper Etching in Ferric Chloride&lt;/a&gt; Ind. Eng. Chem., 1959, 51 (3), pp 288–290 for the relative concentrations of the particular chloride complexes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, if &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe2(SO4)3}$&lt;/span&gt; were used, there would only be aqueous ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the copper surface, there is only a one-electron oxidation of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu+}$&lt;/span&gt;.  As the above article explains, without coordination of chloride, the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu+}$&lt;/span&gt; would be essentially insoluble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate limiting step of the etching reaction is diffusion of ions from the surface of the copper.  Coordination of chloride provides solubility and enables diffusion.  This is the main reason that &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; is needed.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2019-01-21T16:06:16.263" LastActivityDate="2019-01-21T16:06:16.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22412" PostTypeId="2" ParentId="22410" CreationDate="2015-01-05T14:30:33.350" Score="2" Body="&lt;p&gt;As you say this represents &quot;conversions per unit of time&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical expression for &quot;per time unit&quot; is 1/(time unit), and seconds is a particular time unit.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-05T14:30:33.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22413" PostTypeId="2" ParentId="22403" CreationDate="2015-01-05T14:42:19.687" Score="3" Body="&lt;p&gt;To my knowledge, diazene (diimine, $\ce{N2H2}$) does not polymerize but simply decomposes (to hydrazine and nitrogen). Even in the polymerization azobenzene-substituted vinylethers (consider azobenzene as diazene in with both hydrogen atoms are replaced by phenyl rings), the $\ce{N=N}$ double bond does not participate in the polymerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are right, a lot of &lt;strong&gt;copolymers&lt;/strong&gt; consists of monomers in which the polymerization does not take place at $\ce{C=C}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think in the reaction of $\ce{\alpha,\omega}$-diols&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;with phosgene ($\ce{COCl2}$) to polycarbonates&lt;/li&gt;&#xA;&lt;li&gt;with diisocyanates to polyurethanes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or the reaction of dianhydrides with diamines to polyimides.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-05T14:42:19.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22414" PostTypeId="2" ParentId="22403" CreationDate="2015-01-05T14:50:06.953" Score="4" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{N2H2}$&lt;/span&gt; is a very unstable molecule, prone to disproportionation into hydrazine and nitrogen. &lt;span class=&quot;math-container&quot;&gt;$\ce{N-N}$&lt;/span&gt; bonds except one in &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; molecule are unstable and generally unfavorable. So, no there are no polymers with nitrogen as a sole component of the main chain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several classes of compounds of 3-row elements demonstrate tendency for oligomerization, namely &lt;span class=&quot;math-container&quot;&gt;$\ce{(HP)_{n}}$&lt;/span&gt; derivatives usually exist as a ring. It is possible to obtain metastable at room temperature plastic sulfur, containing linear sulfur chains terminated by free radicals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, generally when in search of easily polymerizing molecule that produce main chain containing elements other than carbon, one usually have to look into compounds producing chains with two elements. The most common and known one is formaldehyde &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=O}$&lt;/span&gt;, producing &lt;span class=&quot;math-container&quot;&gt;$\ce{HO-(CH2-O)_{n}-H}$&lt;/span&gt; chains. A less known, but interesting example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Hexachlorophosphazene&quot; rel=&quot;nofollow noreferrer&quot;&gt;hexachlorotriphosphazene&lt;/a&gt;, the source material for production of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyphosphazene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polyphosphazenes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In  general, metastable double bonds are rare for elements other than carbon (sometimes with nitrogen or oxygen partners), so usually reactions intended to produce such bonds, say, &lt;span class=&quot;math-container&quot;&gt;$\ce{Si=Si}$&lt;/span&gt; usually end with olygomer instead, typically of ring structure. Moreover, single homoelemental bonds are usually much weaker for elements other than CHNOB, so typically when a chains containing other elements is occurred, it is usually hetero-atomic chain, produced by means other than olygomerization of double-bond containing monomer. For example, an extremely wide-used family of polymers (polysiloxanes, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicone&quot; rel=&quot;nofollow noreferrer&quot;&gt;silicones&lt;/a&gt;) is produce by hydrolisys of dialkyldicholorsilanes &lt;span class=&quot;math-container&quot;&gt;$\ce{R2SiCl2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="41328" LastEditDate="2019-05-17T19:53:57.760" LastActivityDate="2019-05-17T19:53:57.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22415" PostTypeId="2" ParentId="22410" CreationDate="2015-01-05T15:29:22.733" Score="3" Body="&lt;p&gt;You were actually describing the turnover frequency of a catalyst, and frequency has the unit hertz which is 1/s. Turnover number is a dimensionless value: number of molecules of reagent that a catalytic site can convert before becoming inactivated.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-01-05T15:29:22.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="22417" PostTypeId="2" ParentId="22384" CreationDate="2015-01-05T16:30:12.057" Score="1" Body="&lt;p&gt;Well, &lt;em&gt;very&lt;/em&gt; long hydrocarbons &lt;em&gt;can&lt;/em&gt; become twisted around each other. This is definitely known in the polymer literature. Consider some branched polyethylene with ~1,000 repeat units. It's not too hard to imagine that multiple chains can become entangled in much the same way as shoelaces or strings when you mix a lot of them in a box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are also right that with large molecules, there are more sites for intermolecular interactions. Consider that there is a &quot;combination&quot; effect with lots of sites. That is, to pull two very long linear chains apart, you need to break lots of interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5jx6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that pulling one atom on the bottom away from its neighbor on the top chain will require the movement of &lt;em&gt;all&lt;/em&gt; atoms on the bottom chain. So it's slightly harder to separate two long chains, not only because there are so many interactions, but also because of the combination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, the difficulty in separating the two chains is slightly higher than simply the sum of the interactions.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-05T16:30:12.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22418" PostTypeId="2" ParentId="22344" CreationDate="2015-01-05T16:48:52.843" Score="4" Body="&lt;p&gt;Considering the number of other atoms a central atom can be bonded to, &quot;hybridization [is] irrelevant&quot; and &quot;the coordination number ... is not restricted by the number of&#xA;hybrid orbitals that the atom can form&quot;, quoting &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00273a006?journalCode=jacsat&amp;amp;quickLinkVolume=108&amp;amp;quickLinkPage=3586&amp;amp;selectedTab=citation&amp;amp;volume=108&quot; rel=&quot;nofollow&quot;&gt;On the role of d orbitals in sulfur hexafluoride&lt;/a&gt; J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples such as $\ce{CLi6}$ and $\ce{TeF8^{2-}}$ are cited and emphasis is placed upon the ionic character of the bonding by the cited reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the main limiting factor is how many atoms can simultaneously be close to the central atom, this increasing with increasing radius of the central atom and with decreasing radius of the surrounding atoms.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-05T16:48:52.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22419" PostTypeId="1" AcceptedAnswerId="22439" CreationDate="2015-01-05T18:01:28.453" Score="4" ViewCount="793" Body="&lt;p&gt;My lecture handout says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_V = \frac{\mathrm{d}q_V}{\mathrm{d}T} = \left(\frac{\partial U}{\partial T}\right)_{\!V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $C_V = \mathrm{d}q_V/\mathrm{d}T$ and that $\mathrm{d}U = \mathrm{d}q_V$, but what's the need for the partial derivative? Why can't you just substitute and get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_V = \frac{\mathrm{d}U}{\mathrm{d}T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise for the constant pressure heat capacity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_p = \frac{\mathrm{d}q_p}{\mathrm{d}T} = \left(\frac{\partial H}{\partial T}\right)_{\!p}$$&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2016-09-23T17:42:41.600" LastActivityDate="2016-09-23T17:45:48.187" Title="Expressions for constant volume and constant pressure heat capacities" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22420" PostTypeId="2" ParentId="22389" CreationDate="2015-01-05T18:13:06.957" Score="2" Body="&lt;p&gt;Firstly, copper would have to be in the 4+ oxidation state for there to be $\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity.  If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially.  So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-05T18:13:06.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22421" PostTypeId="2" ParentId="22389" CreationDate="2015-01-05T18:39:56.903" Score="2" Body="&lt;p&gt;Just to be clear, I don't think you can actually make a 5M solution of copper sulfate, but I'll ignore solubility for this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you make a solution of a given concentration from a solid you generally dissolve the solute (your copper sulfate pentahydrate) in a relatively small amount of water and then dilute it to the final volume.  Here, you would dissolve your 249g (1 mole) of copper sulfate pentahydrate in sufficient water to dissolve it but less than 200mL total, then carefully add enough water to reach a solution volume of 200mL, yielding a 5M solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water of hydration would indeed become part of the solution.  If instead you began this process with anhydrous copper sulfate and used 1 mole of that (159.61g) then you would dissolve it in enough water to make the solution but less than 200mL total then carefully add enough to reach your final solution volume and concentration, 200mL at 5M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference here would be that you would have to add 90g of additional water to prepare a solution from the anhydrous compound as compared to the pentahydrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prepare a solution by adding a fixed mass of solute to a fixed volume of water, you are never going to make the desired concentration and volume.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2015-01-05T18:39:56.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22422" PostTypeId="2" ParentId="22389" CreationDate="2015-01-05T18:57:05.443" Score="0" Body="&lt;p&gt;Of course, the water in the hydrate affects the final concentration.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Molar mass of  &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4.5H2O}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{249.5 g/mol}.$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Amount of substance is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n = \frac{345.55}{249.5} = \pu{1.385 mol}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Concentration:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$C = \frac{1.385}{0.2} = \pu{6.925 mol/L}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But, this is only an illustrative example. &lt;strong&gt;It's not real.&lt;/strong&gt; The solubility of &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4.5H2O}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{320 g/L}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{20 °C}.$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="41328" LastEditDate="2019-06-25T09:55:15.393" LastActivityDate="2019-06-25T09:55:15.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22423" PostTypeId="2" ParentId="22389" CreationDate="2015-01-05T19:05:26.213" Score="1" Body="&lt;p&gt;I realize this may not be what you are asking, but this is worth pointing out. The most common source for error in solution concentration comes from neglecting to account for the mass of the hydrate. If you weight out an amount of a hydrate, and use the molecular mass of the anhydrous, your final concentration will be erroneous. For example, 5g of $\ce{CuSO4 \dot{} 5 H2O}$ will not yield the same as 5g $\ce{CuSO4}$ (anhydrous). When diluted to a volume of 1 liter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ce{5g\,CuSO4.5H2O}}{1} \frac{\ce{1mol\,CuSO4.5H2O}}{249.685\ce{g}}\frac{1}{\ce{1L}}=0.02\ce{M\,CuSO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\frac{\ce{5g\,CuSO4}}{1} \frac{\ce{1mol\,CuSO4}}{159.6086\ce{g}}\frac{1}{\ce{1L}}=0.03\ce{M\,CuSO4}}$$&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2015-01-05T19:05:26.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22424" PostTypeId="2" ParentId="22419" CreationDate="2015-01-05T19:42:20.260" Score="3" Body="&lt;p&gt;The difference in &lt;strong&gt;internal energy&lt;/strong&gt; ($\mathrm{d}U$) is a function of many different variables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \!$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a &lt;em&gt;total derivative&lt;/em&gt;, which is not what the equation requires.&lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="16683" LastEditDate="2016-09-23T17:45:48.187" LastActivityDate="2016-09-23T17:45:48.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22425" PostTypeId="2" ParentId="22395" CreationDate="2015-01-05T19:46:17.060" Score="0" Body="&lt;p&gt;This is similar to adsorption on a surface: van der Waals and other forces (e.g. hydrogen bond) bind the solute to the solvent, just as a gas may be bound to a solid. The lowest energy state depends on the strength of the solute-solvent bonds (adhesion) and the solute-solute bonds (cohesion) as opposed to random molecular motion causing evaporation. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Van_der_Waals_force&lt;/a&gt; and &lt;a href=&quot;http://www.chem.purdue.edu/gchelp/solutions/eboil.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.purdue.edu/gchelp/solutions/eboil.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-05T19:46:17.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22426" PostTypeId="1" AcceptedAnswerId="22428" CreationDate="2015-01-05T20:09:05.263" Score="2" ViewCount="9241" Body="&lt;p&gt;What is the difference between &quot;reaction pathway&quot; and &quot;reaction mechanism&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these expressions synonyms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some textbooks authors use both without making clear to the learner if they are synonyms or not.&lt;/p&gt;&#xA;" OwnerUserId="11612" LastActivityDate="2018-04-24T06:51:34.830" Title="What is the difference between &quot;reaction pathway&quot; and &quot;reaction mechanism&quot;?" Tags="&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22427" PostTypeId="2" ParentId="22395" CreationDate="2015-01-05T20:15:46.937" Score="9" Body="&lt;p&gt;I think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Textbooks like &lt;em&gt;Chemistry: The Molecular Nature of Matter and Change&lt;/em&gt; (Silberberg, Amateis) or &lt;em&gt;Physical Chemistry for the Biosciences&lt;/em&gt; (Chang) provides the same explanation. In addition, there is a short paper/article on this topic&lt;sup&gt;[&lt;a href=&quot;http://dx.doi.org/10.1021/ed075p787&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;]&lt;/sup&gt; where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;a href=&quot;http://dx.doi.org/10.1021/ed075p787&quot; rel=&quot;noreferrer&quot;&gt;Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787&lt;/a&gt;&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1532" LastEditorUserId="1532" LastEditDate="2015-01-05T20:32:34.147" LastActivityDate="2015-01-05T20:32:34.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22428" PostTypeId="2" ParentId="22426" CreationDate="2015-01-05T20:24:40.577" Score="6" Body="&lt;p&gt;Usually with the &lt;em&gt;reaction mechanism&lt;/em&gt; the sequence of events (electron transfers) is indicated, while the term &lt;em&gt;reaction pathway&lt;/em&gt; usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-05T20:24:40.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22429" PostTypeId="1" CreationDate="2015-01-05T23:13:37.800" Score="7" ViewCount="22695" Body="&lt;p&gt;This is a Physical Chemistry Question and comes from the Atkins Physical Chemistry book.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that &lt;span class=&quot;math-container&quot;&gt;$2.0~\mathrm{mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$2.0~\mathrm{atm}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$25~^\circ\mathrm{C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$4.0~\mathrm{mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$3.0~\mathrm{atm}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$25~^\circ\mathrm{C}$&lt;/span&gt; are mixed at constant volume. Calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta_\text{mix}G$&lt;/span&gt;. What would be the value of &lt;span class=&quot;math-container&quot;&gt;$\Delta_\text{mix}G$&lt;/span&gt; had the pressures been identical initially?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are given but I don't understand how to get to &lt;span class=&quot;math-container&quot;&gt;$-9.7~\mathrm{kJ}$&lt;/span&gt; for the first part of the question (when initial pressures are different. The gases are initially at different pressures but does that make a difference. Can I still use Raoult's Law?&lt;/p&gt;&#xA;" OwnerUserId="11615" LastEditorDisplayName="user7951" LastEditDate="2019-02-27T21:39:56.203" LastActivityDate="2019-02-27T21:39:56.203" Title="Calculating the Gibbs Free Energy of Mixing" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22430" PostTypeId="2" ParentId="22410" CreationDate="2015-01-06T00:28:08.760" Score="9" Body="&lt;p&gt;The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of &quot;widgets per hour&quot;. It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of &quot;turnovers per second&quot;. It's just that the &quot;turnover&quot; is implied and is viewed as &quot;dimensionless&quot;, so it's left off, leaving you simply with &quot;s&lt;sup&gt;-1&lt;/sup&gt;&quot; as a unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that even though the &quot;turnover&quot; is left off, it's still &lt;em&gt;very&lt;/em&gt; important to know it's there, and to know what, exactly, one &quot;turnover&quot; is. In complicated reactions there is sometimes different reasonable ways to count what &quot;one turnover&quot; is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize &lt;em&gt;this&lt;/em&gt; researcher is using the formation of one molecule of this product as &quot;one turnover&quot;, but &lt;em&gt;that&lt;/em&gt; researcher is using the disappearance of one molecule of that substrate as &quot;one turnover&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time &lt;em&gt;on average&lt;/em&gt;. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that &quot;a turnover takes 5 s&quot; gives a false impression about individual turnovers in a way that saying &quot;this enzyme works at 0.2 turnovers per second&quot; doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, in the case of multi-step reactions, saying that &quot;a turnover takes 5 s&quot; may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s.  For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)&lt;/p&gt;&#xA;" OwnerUserId="11616" LastEditorUserId="11616" LastEditDate="2015-11-16T19:55:54.990" LastActivityDate="2015-11-16T19:55:54.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22431" PostTypeId="1" AcceptedAnswerId="22437" CreationDate="2015-01-06T01:58:45.990" Score="3" ViewCount="196" Body="&lt;p&gt;So in class, I've learned about Equilibrium. So it is basically when the rate of the reverse and the forward reaction are the same. So I was like that sounds pretty easy, I can understand this. And then suddenly there was a twist (there always is in class). The teacher was like: so if you increase or add a stressor, the equilibrium will shift to stop it. How can equilibrium shift, I thought it was a concept not a &quot;thing&quot; to move up or down. Is it because it shifts to maintain the Equilibrium Constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand this. If this question has already been asked then please point me to that. I have even searched online and could not find an explanation in layman's terms. &lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="69830" LastEditDate="2021-05-20T07:10:57.590" LastActivityDate="2021-05-20T07:10:57.590" Title="Why is it possible, that an equilibrium shifts?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22432" PostTypeId="2" ParentId="22368" CreationDate="2015-01-06T07:42:13.030" Score="5" Body="&lt;p&gt;Since this reaction takes place in aqueous solution, is is implied, that the reaction equations are involving solvated ions. In general both of the given equations are correct. However, I would prefer the first equation to include the highly charged metal ions as aqua complexes: &#xA;$$\ce{NO3- + 3[Fe(H2O)_6]^{2+} + 4H+  -&amp;gt; NO + 3[Fe(H2O)_6]^{3+} + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that this is a test for nitrate, and nitrite ions will interfere accordingly.&#xA;$$\ce{NO2- + [Fe(H2O)_6]^{2+} + 2H+  -&amp;gt; NO + [Fe(H2O)_6]^{3+} + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to realise, that the Iron (II) solution is given in massive excess for this test to work. Hence the newly reduced nitrogen monoxide will react with the yet unreacted excess of $\ce{Fe^{2+}}$:&#xA;$$\ce{NO + [Fe(H2O)_6]^{2+} -&amp;gt; [Fe(NO)(H2O)_5]^{2+} + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the iron hexaaqua complexes are nearly colourless, they do not interfere with this visibility of this reaction.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-01-06T07:42:13.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22434" PostTypeId="2" ParentId="22383" CreationDate="2015-01-06T09:12:12.010" Score="4" Body="&lt;p&gt;You have to look which forces act where. You have a force that tries pull the liquid up the capillary, which stems from the gain in energy due to adhesion. At the same time, however, gravity acts against this force. In equilibrium, both forces have the same value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_\text{g} = F_{\text{surf}}$$&#xA;$$ mg=\rho Vg= \rho R^2\pi h g=2\pi R\sigma_s \cos(\varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leads to the formula presented by Babounet in another answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ h = \frac{2\sigma_s\cos(\varphi)}{R\rho g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to boil it down, if you lower the radius $R$, you lower the force due to gravity $F_g$, but the force due to surface energy $F_\text{surf}$ does not fall as fast, which leads to a higher height $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hsw4S.png&quot; alt=&quot;Sketch&quot;&gt;&#xA;(&lt;a href=&quot;http://www.engineeringarchives.com/les_fm_capillaryeffect.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;, $W=F_g$ here, naming isn't perfect, $F_{\text{surf}}$ is not drawn into) &lt;/p&gt;&#xA;" OwnerUserId="11551" LastActivityDate="2015-01-06T09:12:12.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22435" PostTypeId="1" AcceptedAnswerId="22436" CreationDate="2015-01-06T10:31:31.203" Score="3" ViewCount="4110" Body="&lt;ol&gt;&#xA;&lt;li&gt;I converted benzene into toluene by Friedel-Crafts alkylation reaction.&lt;/li&gt;&#xA;&lt;li&gt;Then by nitration I got two compounds o-nitrotoluene and p-nitrotoluene.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is it possible to get only p-nitrotoluene?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5026" LastEditDate="2018-07-10T09:41:32.617" LastActivityDate="2018-07-10T09:41:32.617" Title="How to selectively convert benzene to p-nitrotoluene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22436" PostTypeId="2" ParentId="22435" CreationDate="2015-01-06T12:05:15.987" Score="7" Body="&lt;p&gt;I'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But imagine to stuff a toluene molecule into a very narrow &quot;tube&quot;. This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite (&lt;a href=&quot;http://en.wikipedia.org/wiki/ZSM-5&quot; rel=&quot;nofollow noreferrer&quot;&gt;ZSM-5&lt;/a&gt;). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. &lt;em&gt;Applied Catalysis A: General&lt;/em&gt; &lt;strong&gt;2009,&lt;/strong&gt; &lt;em&gt;353&lt;/em&gt; (1), 1–8. &lt;a href=&quot;https://doi.org/10.1016/j.apcata.2008.10.013&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/j.apcata.2008.10.013&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-07-10T09:40:55.557" LastActivityDate="2018-07-10T09:40:55.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22437" PostTypeId="2" ParentId="22431" CreationDate="2015-01-06T14:13:52.360" Score="4" Body="&lt;p&gt;Let's do a simple generic example. We have a reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2A &amp;lt;=&amp;gt;[k_1][k_{-1}] B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally we determine that the rates of the forward and reverse reactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Rate}_1=k_1[\ce{A}]^2$$&#xA;$$\text{Rate}_{-1}=k_{-1}[\ce{B}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Rate}_1=\text{Rate}_{-1}$$&#xA;$$k_1[\ce{A}]^2=k_{-1}[\ce{B}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rate constants are &lt;strong&gt;constants&lt;/strong&gt; we can combine them into the equilibrium constant to generate the law of mass action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $k_1=2k_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\ce{A}]_{eq}$ and $[\ce{B}]_{eq}$ that defines equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{2k_{-1}}{k_{-1}}=2$$&#xA;$$K_{eq}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=2$$&#xA;$$2{[\ce{A}]_{eq}^2}=[\ce{B}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, only concentrations of $\ce{A}$ and $\ce{B}$ that satisfy $2{[\ce{A}]_{eq}^2}=[\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how this works. An equilibrium situation is $[\ce{A}]=1\text{ M}$ and $[\ce{B}]=2\text{ M}$ because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{1^2}=2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rates are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(1^2)=2k_{-1}$$&#xA;$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what if $[\ce{A}]=[\ce{B}]=2\text{ M}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{2^2}=\frac{1}{2}\ne K_{eq}$$&#xA;$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(2^2)=8k_{-1}$$&#xA;$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, &lt;strong&gt;the rates are no longer equal&lt;/strong&gt;. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\ce{A}$ is consumed (slowing down the forward reaction) and $\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can even do a bit of algebra to figure out what the concentrations of $\ce{A}$ and $\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\ce{B}$, so the change in the concentration of $\ce{A}$ is $-2x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq}=2=\frac{[\ce{B}]}{[\ce{A}]^2}=\frac{2+x}{(2-2x)^2}=\frac{2+x}{4-8x+4x^2}$$&#xA;$$2(4-8x+4x^2)=2+x$$&#xA;$$8-16x+8x^2=2+x$$&#xA;$$6-17x+8x^2=0$$&#xA;$$x=0.44695..., 1.6781...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\ce{A}$. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{A}]_{eq}=2.00-2(0.44695)=1.11$$&#xA;$$[\ce{B}]_{eq}=2.00+0.44695=2.45$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-06T14:13:52.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22438" PostTypeId="1" CreationDate="2015-01-06T14:45:22.313" Score="1" ViewCount="1333" Body="&lt;p&gt;I used  $\ce{HCl}$ and $\ce{H2O2}$ to recover gold foils by dissolving all other metals. My solution I assume contains a large amount of silver. How do I drop the silver out of the  $\ce{HCl}$  and peroxide solution.)&lt;/p&gt;&#xA;" OwnerUserId="11622" LastEditorUserId="275" LastEditDate="2015-01-06T15:33:02.307" LastActivityDate="2015-01-06T15:59:31.960" Title="Separating gold and silver from gold foil using hydrochloric acid and hydrogen peroxide" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22439" PostTypeId="2" ParentId="22419" CreationDate="2015-01-06T14:48:05.050" Score="3" Body="&lt;p&gt;To complement entropid's answer, let's look at the formulation of $U$ that is slightly more useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;U &amp;amp;= q + w \\&#xA;\mathrm{d}U &amp;amp;= \mathrm{d}q + \mathrm{d}w \\&#xA;&amp;amp;= C\,\mathrm{d}T - p\,\mathrm{d}V&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant volume, $\mathrm{d}V = 0$, so $p\,\mathrm{d}V = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\mathrm{d}U &amp;amp;= C_V\,\mathrm{d}T - 0 \\&#xA;C_V &amp;amp;= \frac{\mathrm{d}U}{\mathrm{d}T}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this notation assumes that $U$ is only a function of $T$, as in $U(T)$. It is not. In this version, $U$ is a function of $V$ and $T$: $U(V, T)$. Note that we do not need $p$ as a variable, since it is dependent only on $V$ and $T$ as well - for an ideal/perfect gas $p(V,T)=\frac{nRT}{V}$. Thus, we need partial derivative notation so that we know we are not trying to take the total derivative of $U$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2016-09-23T17:45:28.657" LastActivityDate="2016-09-23T17:45:28.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22440" PostTypeId="2" ParentId="22438" CreationDate="2015-01-06T15:23:07.967" Score="3" Body="&lt;p&gt;Hot oxidizing acids, such as nitric acid od sulfuric acid, are the more canonical methods to separate gold from less rare metals.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;My solution I assume contains a large amount of silver.&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If so, shouldn't you have noticed a precipitate of silver chloride ($\ce{AgCl}$) under these conditions?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-06T15:23:07.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22441" PostTypeId="2" ParentId="22376" CreationDate="2015-01-06T15:39:58.620" Score="6" Body="&lt;p&gt;In general, there are a variety of amino acids that are not chiral (achiral).  In physiological proteins, the amino acids are generally alpha-amino acids (one carbon between the amino and carboxylic acid groups), but there are also, &lt;a href=&quot;https://www.princeton.edu/chemistry/macmillan/group-meetings/b-aminoacids-spb.pdf&quot; rel=&quot;noreferrer&quot;&gt;beta-amino acids&lt;/a&gt;, gamma-amino acids, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just limiting ourselves to alpha-amino acids, other than glycine, the most interesting example of an achiral amino acid is &lt;a href=&quot;http://www.hmdb.ca/metabolites/hmdb01147&quot; rel=&quot;noreferrer&quot;&gt;amino malonic acid&lt;/a&gt;, which has been given its own three-letter abbreviation (Ama) along with the more-common alpha amino acids.  Ama has being found in the hydrolysates of E. Coli and human proteins.  See &lt;a href=&quot;http://www.pnas.org/content/81/3/722&quot; rel=&quot;noreferrer&quot;&gt;Aminomalonic acid: identification in Escherichia coli and atherosclerotic plaque&lt;/a&gt; PNAS vol. 81 page 722.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the amino acids of protein orgin generally have an alpha H, but if this H is substituted the compound is still an alpha amino acid.  If the two side chains on the alpha carbon are the same, the compound is achiral. For example &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Aminoisobutyric_acid&quot; rel=&quot;noreferrer&quot;&gt;alpha-methyl alanine&lt;/a&gt;.  An example that occurs naturally in cranberries is &lt;a href=&quot;http://en.wikipedia.org/wiki/1-Aminocyclopropane-1-carboxylic_acid&quot; rel=&quot;noreferrer&quot;&gt;1-aminocyclopropane 1-carboxylic acid&lt;/a&gt;. The alpha amino acids with two side chains are sometimes referred to as alpha branched amino acids, and have been &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/7277509&quot; rel=&quot;noreferrer&quot;&gt;found in meteorites&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another class of achiral alpha amino acids are those with a side chain double-bonded to the alpha carbon such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Dehydroalanine&quot; rel=&quot;noreferrer&quot;&gt;dehydroalanine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-06T17:36:34.920" LastActivityDate="2015-01-06T17:36:34.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22442" PostTypeId="2" ParentId="22438" CreationDate="2015-01-06T15:59:31.960" Score="0" Body="&lt;p&gt;I would have used aqua regia but was trying to leave the foils solid. I should have used just nitric acid.I would hate to waste the dissolved metals in the HCL. Wanted to recover the silver and copper.Am I correct to assume if I added the nitric acid the silver chloride would participate.&lt;/p&gt;&#xA;" OwnerUserId="11623" LastActivityDate="2015-01-06T15:59:31.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22443" PostTypeId="1" AcceptedAnswerId="22444" CreationDate="2015-01-06T16:02:47.280" Score="3" ViewCount="253" Body="&lt;p&gt;It makes sense to me conceptually that $dU=C_VdT$ but is that always the case because I have seen that result derived thus:&#xA;$$dU=U(V,T)=(\frac{\partial U}{\partial T})_VdT+(\frac{\partial U}{\partial V})_TdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Not the typical expression for internal energy $dU=TdS-pdV$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;for a perfect gas&lt;/strong&gt; $(\frac{\partial U}{\partial V})_T=0$ so &lt;strong&gt;for a perfect gas&lt;/strong&gt; internal energy is only a function of $T$: $U=U(T)$. Thus, it follows that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=(\frac{\partial U}{\partial T})_VdT=C_VdT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is this only true for a perfect gas? I ask this because, the knowledge of this equality is required to show that entropy is a state function and I was wondering whether the proof that uses the above argument is only valid for a perfect gas or whether it's valid more generally?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will show you the proof now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=dq-pdV=C_VdT$$&#xA;$$dq=C_VdT+pdV$$&#xA;$$\frac{dq}{T}=\frac{C_V}{T}dT+\frac{nR}VdV$$ (using pV=nRT)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you differentiate appropriately you can show it's an exact differential. However, is this only valid for a perfect gas?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-10-19T11:13:41.610" Title="Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22444" PostTypeId="2" ParentId="22443" CreationDate="2015-01-06T16:27:59.933" Score="4" Body="&lt;p&gt;The statement in the title question &quot;internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature&quot; would only apply to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfect_gas#Thermally_and_calorically_perfect_gas&quot; rel=&quot;nofollow&quot;&gt;calorically perfect gas&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you instead said &quot;an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature&quot;, that is only true for a thermally perfect gas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For real gas, the heat capacities depend upon both temperature and pressure.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-06T16:27:59.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="22446" PostTypeId="1" AcceptedAnswerId="22449" CreationDate="2015-01-06T17:54:38.337" Score="3" ViewCount="261" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Water ionization becomes much less evident if the hydrogen bonds are just a few percent stronger but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found this line in some article. You can even copy/paste this line on search bar of Google. I didn't understood the latter part of this statement which is &quot;but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker&quot;. The author is indicating to which of the following three situations?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;only $\ce{H+}$ ion concentration will increase &lt;/li&gt;&#xA;&lt;li&gt;both $\ce{H+}$ and $\ce{OH-}$ will increase &lt;/li&gt;&#xA;&lt;li&gt;$\ce{H+}$ will increase and $\ce{OH-}$ will decrease that is the equilibrium constant remains same. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="9961" LastEditDate="2015-10-02T17:12:40.743" LastActivityDate="2018-10-02T12:17:31.343" Title="If hydrogen bonding in water was weaker, what happens to H+ ion concentration?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="22447" PostTypeId="2" ParentId="22377" CreationDate="2015-01-06T18:08:34.303" Score="4" Body="&lt;p&gt;Let the number of atoms be n.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of skeletal p electrons is 2n + 2, the structure is closo (a &lt;a href=&quot;http://en.wikipedia.org/wiki/Deltahedron&quot; rel=&quot;nofollow&quot;&gt;deltahedron&lt;/a&gt; with n vertices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;http://www.chemistry.illinois.edu/research/inorganic/seminar_abstracts/2002-2003/Allen.Abstract.LitSeminar.pdf&quot; rel=&quot;nofollow&quot;&gt;Excursions Beyond the Zintl Border&lt;/a&gt; and &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/10760845&quot; rel=&quot;nofollow&quot;&gt;Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Sn_5^{2-}}$, $\ce{Pb_5^{2-}}$, $\ce{Ge_5^{2-}}$ and $\ce{Ge_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Sn_9^{4-}}$ and $\ce{Ge_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Gyroelongated_square_bipyramid&quot; rel=&quot;nofollow&quot;&gt;gyroelongated square bipyramid&lt;/a&gt; with one square pyramid apex missing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Bi_4^{2-}}$ and $\ce{Sb_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, quoting Farley and Castleman's &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00189a068&quot; rel=&quot;nofollow&quot;&gt;Observation of Gas-Phase Anionic Bismuth Zintl Ions&lt;/a&gt; J. Am. Chem. Soc. 1989, 111, 2734-2735&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since bismuth and antimony are typically trivalent, anionic clusters&#xA;  of these elements very quickly exceed the maximum 2n + 8 electrons&#xA;  accountable under Wades rules for “hypo” compounds. In particular, the&#xA;  previously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and&#xA;  $\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are&#xA;  particularly intriguing due to the inability of current theories to&#xA;  account for them and lack of other corresponding metallic compounds&#xA;  with which they may be compared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now going back to the example $\ce{Bi_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The structure is isoelectronic with species known to be bent in the ground state including $\ce{O3}$, $\ce{S3}$, $\ce{S2O}$, $\ce{SO2}$ and allyl anion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Experimental determination published in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja0038732&quot; rel=&quot;nofollow&quot;&gt;Stabilization of Ozone-like [Bi3]3-&lt;/a&gt; and in &lt;a href=&quot;http://www3.nd.edu/~sevovlab/articles/HeteroatomicDeltahedralClusters.pdf&quot; rel=&quot;nofollow&quot;&gt;a more recent article&lt;/a&gt; that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Theoretical work in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/jp980004e&quot; rel=&quot;nofollow&quot;&gt;Zintl Anions Analogous to O3&lt;/a&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Though ozone and like species are found to be bent, it is also theorized that &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_ozone&quot; rel=&quot;nofollow&quot;&gt;cyclic (equilateral triangular ring) ozone&lt;/a&gt; represents another higher local potential energy minimum, and this would be the same for $\ce{Bi_3^{3-}}$ too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Bi_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal pyramid.  A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclic_ozone&quot; rel=&quot;nofollow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-08T20:57:25.297" LastActivityDate="2015-01-08T20:57:25.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="22448" PostTypeId="2" ParentId="22377" CreationDate="2015-01-06T18:46:41.157" Score="3" Body="&lt;p&gt;The wiki article contains a note&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To predict the structure of an arachno cluster, the closo polyhedron&#xA;  with n + 2 vertices is used as the starting point, and the n+1 vertex&#xA;  nido complex is generated by following the rule above; a second vertex&#xA;  adjacent to the first is removed if the cluster is composed of mostly&#xA;  small atoms, &lt;strong&gt;a second vertex not adjacent to the first is removed if&#xA;  the cluster is composed mostly of large atoms&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{Bi}$ atom is definitely a large atom, so we should  remove out non-adjacent vertexes. The only way to choose non-adjacent vertexes in trigonal bipyramid is to take axial vertexes, gaining a triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please not, that removing two adjacent vertexes in trigonal bipyramid may also lead to triangle, if one removes one equatorial and one axial position. removing two equatorial positions will lead not to triangle or line, but to an angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for $\ce{[Bi3]^{3-}}$ the amount of electrons per atom is exactly six. This means, that &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyhedral_skeletal_electron_pair_theory#6n_rules&quot; rel=&quot;nofollow&quot;&gt;6n rules&lt;/a&gt; suite situation better. 6n rule leads to ring in the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend to consider Wade's rules as general guidelines, but not a final truth, sometimes they fail spectacularly. For example, consider $\ce{[CH]6}$ cluster. Wade's rule predicts it to be a Trigonal prism. As we know, IRL this molecule prefers to be a ring, with prismane being an esoteric and a quite rare form.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-01-06T18:51:43.200" LastActivityDate="2015-01-06T18:51:43.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22449" PostTypeId="2" ParentId="22446" CreationDate="2015-01-06T18:55:07.200" Score="4" Body="&lt;p&gt;The article by Martin Chaplin 'Water's Hydrogen Bond Strength' (arXiv:&lt;a href=&quot;https://arxiv.org/abs/0706.1355&quot; rel=&quot;nofollow noreferrer&quot;&gt;0706.1355&lt;/a&gt; &lt;strong&gt;[cond-mat.soft]&lt;/strong&gt;, no evidence of peer review) is discussing how changes to the properties of water would alter life.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hydrogen bonding in water was stronger, there would be less ionization (less &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and less &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hydrogen bonding in water was weaker, there would be more ionization (more &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and more &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2018-10-02T12:17:31.343" LastActivityDate="2018-10-02T12:17:31.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22450" PostTypeId="2" ParentId="9964" CreationDate="2015-01-06T20:48:01.950" Score="3" Body="&lt;p&gt;The spectrochemical series is an ordering of relative ligand interaction strength with metal ions. It can be used to infer relative $\sigma$-donor strength, if $\pi$-interactions are not possible for those particular ligands. Examples of this type ligand with no $\pi$-interactions include $\ce{NH3}$, $\ce{CH3-}$, and $\ce{H-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most ligands will exhibit $\pi$-interactions, operating as $\pi$-donors or $\pi$-acceptors, and these interactions &lt;strong&gt;dominate&lt;/strong&gt; the ordering of the spectrochemical series. The $\sigma$-donor interactions are obscured and heavily affected by the $\pi$-interactions. Experimentally all of these effects are important, and it is difficult to separate these interactions from one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the $\sigma$-donor interaction of a $\pi$-acceptor ligand is drastically increased from some initial level. $\pi$-backbonding reduces electron density at the metal center, leading to increased $\sigma$-donation (which increases $\pi$-backbonding, and so on). This will make a given $\pi$-acceptor ligand act as a better $\sigma$-donor than would be expected. The converse effect is observed with $\pi$-donor ligands reducing $\sigma$-donor strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complexity of these interactions tempers the question of gauging $\sigma$-donor interactions only. The combined bonding effects influence one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically separating out $\sigma$-donor interactions from all $\pi$-interactions could be possible computationally, but I am unaware of any research on this specific topic. From a very general standpoint, improved metal-ligand orbital overlap will increase $\sigma$-donor strength, which may help estimate these characteristics qualitatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most inorganic textbooks should be able to provide further detail on this topic under sections on ligand field theory.&lt;/p&gt;&#xA;" OwnerUserId="7653" LastActivityDate="2015-01-06T20:48:01.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22452" PostTypeId="1" AcceptedAnswerId="23458" CreationDate="2015-01-06T23:44:24.940" Score="9" ViewCount="1973" Body="&lt;p&gt;$$\mathrm{d}U&amp;lt;T\,\mathrm{d}S-p\,\mathrm{d}V=&#xA;\left(\frac{\partial U}{\partial S}\right)_{\!V}\,\mathrm{d}S&#xA;+\left(\frac{\partial U}{\partial V}\right)_{\!S}\,\mathrm{d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a change to occur at a constant entropy and volume, the internal energy must decrease. I get that this statement isn't useful because you can't control entropy in a lab. Furthermore, given that: &#xA;$$\mathrm{d}G&amp;lt;V\,\mathrm{d}p-S\,\mathrm{d}T=&#xA;\left(\frac{\partial G}{\partial p}\right)_{\!T}\,\mathrm{d}p &#xA;+\left(\frac{\partial G}{\partial T}\right)_{\!p}\,\mathrm{d}T$$ &#xA;this shows that a spontaneous change at a constant pressure and temperature is accompanied by a decrease in Gibbs free energy – far more useful as these variables can be controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since, for a spontaneous process, $\mathrm{d}U$ must decrease at a constant entropy and volume which is expressed by $\mathrm dU&amp;lt;T\,\mathrm dS-p\,\mathrm dV$. However, expressions for $\mathrm dS$ and $\mathrm dV$ can be expressed as a function of $p$ and $T$ making this inequality a function of pressure an temperature.&#xA;\begin{align}&#xA;\mathrm{d}U &amp;amp;&amp;lt;T\,\mathrm{d}S-p\,\mathrm{d}V\\&#xA;S&amp;amp;=S(p,T)\\&#xA;V&amp;amp;=V(p,T)\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub in the total derivatives and factorize:&#xA;$$\mathrm{d}U&amp;lt;\left[T\left(\frac{\partial S}{\partial T}\right)_{\!p} &#xA;- p\left(\frac{\partial V}{\partial T}\right)_{\!p}\right]\,\mathrm{d}T &#xA;+ \left[T\left(\frac{\partial S}{\partial p}\right)_{\!T} &#xA;- p\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\,\mathrm{d}p$$&#xA;Thus, $U=U(p,T)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that for a spontaneous change at constant pressure and temperature the internal energy must decrease however this is not the case clearly. Does making the substitutions invalidate the inequality? Where is the fault in my reasoning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's the need for Gibbs free energy. Surely a spontaneous change at constant pressure and temperature is accompanied by a decrease in internal energy? There must be a fault in my reasoning, but where and why?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2016-11-13T15:32:00.003" LastActivityDate="2016-11-13T15:32:00.003" Title="Why is Gibbs free energy more useful than internal energy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="22454" PostTypeId="1" AcceptedAnswerId="23528" CreationDate="2015-01-07T01:43:39.073" Score="15" ViewCount="4793" Body="&lt;p&gt;Potassium ($\ce{K}$) seems to be the transitional alkali metal in terms of safety. Lighter than $\ce{K}$ and we can store under oil with no worries. Heavier than $\ce{K}$ and only ampules will work. $\ce{K}$ is such that even under oil it can absorb enough oxygen to form superoxides that are contact explosives. But this is the theory: does anyone have any experience with extended proper storage of potassium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask for two reasons. First, there are lots of online videos showing $\ce{K}$ under oil that the scientist then proceeds to cut, smash, and otherwise manhandle with no fear of detonation. Second, my $\ce{K}$ arrived tarnished, which I have stored under oil, and then vacuumed sealed in a bag, which has shown zero change in visual oxidation. So while I get that leaving an open jar of $\ce{K}$ under oil is problematic, I'm less clear on well stored $\ce{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd appreciate any comments from practitioners that use $\ce{K}$ regularly. What situations are genuine dangers? Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2020-07-02T20:30:13.547" Title="Practical safety of storing potassium" Tags="&lt;safety&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23450" PostTypeId="1" AcceptedAnswerId="23451" CreationDate="2015-01-07T03:55:14.063" Score="3" ViewCount="6896" Body="&lt;p&gt;Here is something that no video in youtube about electrochemistry can explain me about galvanic cells.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have $\ce{Zn}$ metal immerse in $\ce{Zn^{2+}}$ in one side,  then $\ce{Cu}$ metal immerse in $\ce{Cu^{2+}}$ in the other side. Then, if we link the $\ce{Zn}$ metal to $\ce{Cu}$ metal with some conductor wire, electrons will flow from the $\ce{Zn}$ metal to the copper wire, but this will stop quickly without a salt bridge.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do the electrons move from $\ce{Zn}$ metal through the wire to the $\ce{Cu}$ metal in the first place? What causes that?       &lt;/p&gt;&#xA;" OwnerUserId="12625" LastEditorUserId="4945" LastEditDate="2015-01-07T06:26:13.907" LastActivityDate="2018-11-11T10:22:35.507" Title="What causes electrons to move from zinc to copper?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23451" PostTypeId="2" ParentId="23450" CreationDate="2015-01-07T04:27:07.780" Score="4" Body="&lt;p&gt;There is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermocouple&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Any junction of dissimilar metals will produce an electric potential related to temperature.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, we don't care about the temperature effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what starts the reaction, nor why the zinc dissolves easily.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-07T04:27:07.780" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23452" PostTypeId="1" CreationDate="2015-01-07T06:54:56.473" Score="4" ViewCount="85" Body="&lt;p&gt;Is there somewhere where I can find an authoritative list of chemical reactions, energy requirements, known catalysts, etc.? If not, how do you know if and how to transform substance A to substance B? What references do you use?&lt;/p&gt;&#xA;" OwnerUserId="12627" LastEditorUserId="4945" LastEditDate="2015-01-07T08:31:26.323" LastActivityDate="2015-01-07T08:31:26.323" Title="Is there an authoritative list of chemical reactions/transformations?" Tags="&lt;reference-request&gt;&lt;books&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-01-07T19:35:29.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="23453" PostTypeId="1" CreationDate="2015-01-07T08:26:00.397" Score="0" ViewCount="1472" Body="&lt;p&gt;I have three different HDPE polymers with different melting index but they don't have any elasticity. Physical form of polymers is granules. Can I change glass transition temperature (Tg) and impact resistance with additives?&lt;/p&gt;&#xA;" OwnerUserId="12629" LastEditorUserId="12629" LastEditDate="2015-01-07T09:42:04.730" LastActivityDate="2015-03-08T16:22:30.510" Title="Can I change glass transition temperature (Tg) and impact resistance with additives?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23454" PostTypeId="2" ParentId="22429" CreationDate="2015-01-07T08:32:58.903" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Let’s start with the last question: Calculate $\Delta_\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\Delta_\text{mix}G$ is:&#xA;$$\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\ \text{mol}$, $x_1=\frac{2}{6}$ and $x_2=\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\Delta_\text{mix}G= 6\ \text{mol}\times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\frac{2}{6}\ln\frac{2}{6} +\frac{4}{6}\ln\frac{4}{6})= -9462\ \text{J}$$ As $\Delta_\text{mix}G&amp;lt;0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G_\text{i}=n_1(\mu_1^0 +RT\ln p_1)+n_2(\mu_2^0 +RT\ln p_2)$$ Where $n_1=2\ \text{mol}$, $n_2=4\ \text{mol}$, $p_1=2\ \text{atm}$, $p_2=3\ \text{atm}$. $$G_\text{i}=2\ \text{mol}\times(\mu_1^0 +RT\ln2)+4\ \text{mol}\times(\mu_2^0 +RT\ln3)$$&#xA;When the partition is removed, and each gas occupies the volume $V_1+V_2$.&#xA;The partial pressure of hydrogen is $$p^\prime_1=\frac{2\ \text{mol}\times RT}{V_1+V_2}$$&#xA;The partial pressure of nitrogen is $$p^\prime_2= \frac{4\ \text{mol}\times RT}{V_1+V_2} $$  On the other hand: $$V_1=\frac{2\ \text{mol}\times RT}{2\ \text{atm}}=1\frac{\text{mol}}{\text{atm}}\times RT$$ $$V_2=\frac{4\ \text{mol}\times RT}{3\ \text{atm}}=\frac{4}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ i.e.  $$V_1+V_2=\frac{7}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ &#xA;So $$p^\prime_1=\frac{2\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{6}{7}\ \text{atm}$$ $$p^\prime_2= \frac{4\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{12}{7}\ \text{atm} $$ The final Gibbs energy is $$G_\text{f}=n_1(\mu_1^0 +RT\ln p^\prime_1)+n_2(\mu_2^0 +RT\ln p^\prime_2)$$  $$G_\text{f}=2\ \text{mol}\times(\mu_1^0 +RT\ln\frac{6}{7})+4\ \text{mol}\times(\mu_2^0 +RT\ln\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G_\text{f}$ and $G_\text{i}$ &#xA;$$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6/7}{2} +4\ \text{mol}\times RT\ln\frac{12/7}{3}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6}{14} +4\ \text{mol}\times RT\ln\frac{12}{21}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\ln\frac{3}{7} +2\times\ln\frac{4}{7})= -9744\ \text{J}\approx{-9.7\ \text{kJ}}$$ &#xA;In this case, the value of $\Delta_\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope it’s clear now.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorDisplayName="user7951" LastEditDate="2015-04-16T09:32:40.240" LastActivityDate="2015-04-16T09:32:40.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23456" PostTypeId="1" CreationDate="2015-01-07T14:00:08.020" Score="11" ViewCount="16746" Body="&lt;p&gt;Why was unified atomic mass scale introduced and why was Oxygen - 16 replaced by carbon - 12 for standardizing atomic scale?&lt;/p&gt;&#xA;" OwnerUserId="11331" LastEditorUserId="12639" LastEditDate="2015-01-07T16:56:36.170" LastActivityDate="2017-08-17T15:52:30.607" Title="Why was atomic mass scale changed from Oxygen - 16 to Carbon - 12?" Tags="&lt;atoms&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23457" PostTypeId="2" ParentId="23456" CreationDate="2015-01-07T14:43:15.193" Score="13" Body="&lt;p&gt;The mass scale has changed over time, largely due to different isotopes of the &quot;baseline.&quot; Not surprisingly, there's a good &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_mass_unit&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; article on the matter.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the 20th century, until the 1960s chemists and physicists used two different atomic-mass scales. The chemists used a &quot;atomic mass unit&quot; (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (O-16, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, $\ce{^12C}$, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short, &lt;a href=&quot;http://en.wikipedia.org/wiki/John_Dalton&quot; rel=&quot;nofollow&quot;&gt;Dalton&lt;/a&gt; suggested $\ce{^1H}$ as the basis of the mass scale, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Wilhelm_Ostwald&quot; rel=&quot;nofollow&quot;&gt;Ostwald&lt;/a&gt; pushed later for $\ce{^16O}$. &lt;strong&gt;Unfortunately, no one knew about isotopes yet.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem was that physics stuck to isotopically pure O-16 for their mass scale, and chemistry used &quot;oxygen&quot; (i.e., with some O-17 and O-18 present). There's 0.028% difference between those scales. That may not sound like much, but with molecules of ~200-500 amu, it starts to add up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using C-12, it's isotopically pure (making physics happy) very close to the previous chemistry &quot;O-16 + O-17 + O-18&quot; scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.iupac.org/publications/ci/2004/2601/1_holden.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier suggested to Mattauch that the $\ce{^12C}$ = 12 mass scale be adopted because of carbon's use as a secondary standard in mass spectrometry. Also, $\ce{^12C}$ = 12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the $\ce{^16O}$ = 16 scale (which would not acceptable to chemists). Enthusiastically, Mattauch made a worldwide effort in the late 1950s to publicize the $\ce{^12C}$ = 12 scale and obtain the physicist's approval, while Wichers obtained the chemist's approval.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-01-12T16:50:19.877" LastActivityDate="2015-01-12T16:50:19.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23458" PostTypeId="2" ParentId="22452" CreationDate="2015-01-07T14:53:20.703" Score="10" Body="&lt;p&gt;Thanks for the edit - I see now what you meant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer for why we &quot;need&quot; Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does &lt;strong&gt;not&lt;/strong&gt; necessarily decrease. Internal energy is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential&quot;&gt;thermodynamic potential&lt;/a&gt; (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for $\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason we &quot;need&quot; Gibbs energy is that it also tells us how much &quot;thermodynamically useful&quot; work we can get out of a process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical Explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two inequalities that you started with only represent spontaneous process under very specific conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}U &amp;lt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of &quot;spontaneous&quot; vs. &quot;nonspontaneous.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to arrive at this expression for $\mathrm{d}U$, we start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius_theorem&quot;&gt;Clausius inequality&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}S - \frac{\mathrm{d}q}{T} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\mathrm{d}S$) plus the change in surroundings, $-\frac{\mathrm{d}q}{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}S - \frac{\mathrm{d}U}{T} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\,\mathrm{d}S \geq \mathrm{d}U$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since $\mathrm{d}U = \mathrm{d}q$ at constant $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we then assume no entropy change for the &lt;em&gt;system&lt;/em&gt;, we can also write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}U \leq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the restriction of constant $S$ and $V$ are &lt;em&gt;built-in&lt;/em&gt; into the inequality. You can substitute in other thermodynamic variables to get $\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs energy is useful because the inequality $\mathrm{d}G &amp;lt; 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="4945" LastEditDate="2015-01-08T06:44:02.363" LastActivityDate="2015-01-08T06:44:02.363" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23459" PostTypeId="1" CreationDate="2015-01-07T15:10:20.983" Score="1" ViewCount="12161" Body="&lt;p&gt;&lt;strong&gt;Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that in simple molecular substances the atoms within a molecule are held together by strong covalent bonds, but the intermolecular forces between molecules are weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see what makes giant covalent substances have stronger intermolecular forces? What makes them stronger and have higher melting points? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I understood but I think I must have gone wrong somewhere and now I am very confused. Help would be much appreciated. Also please explain simply, chemistry does not come naturally to me so I might get even more confused.&lt;/p&gt;&#xA;" OwnerUserId="12637" LastActivityDate="2015-01-07T15:48:49.720" Title="Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?" Tags="&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23460" PostTypeId="2" ParentId="23453" CreationDate="2015-01-07T15:31:37.707" Score="1" Body="&lt;p&gt;The glass transition temperature depends on the cooling rate and molecular weight distribution and &lt;strong&gt;could be influenced by additives&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most liquid plasticizers are low molecular weight organic materials with a glass transition temperature in the range of $125-225^0\text{C}$. When this kind of additives is added to an organic polymer, the result is a weight averaging of the glass transition temperature between that of the polymer and that of the plasticiser if they are miscible. The reduction of the the glass transition temperature of a polymer by the plasticiser is usually about   $5^0\text{C}$ per weight percent of plasticiser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact modifiers ( a kind of additives) can affect the impact resistance of a polymer. Impact modifiers are key additives for&#xA;increasing flexibility and impact strength&#xA;to meet physical property requirements of&#xA;rigid polymers. All impact modifiers are elastomeric or&#xA;rubbery in nature, with a lower modulus&#xA;than the host polymer. The dispersed&#xA;rubber phase acts to absorb or dissipate the&#xA;energy of impact in order to stop craze or&#xA;crack propagation. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-07T15:31:37.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23461" PostTypeId="2" ParentId="23459" CreationDate="2015-01-07T15:46:45.403" Score="1" Body="&lt;p&gt;In simple molecules the intermolecular forces are, in order of increasing strength, dispersion forces, permanent dipole interactions and hydrogen bonding (which has significant covalent character but is generally considered to be an intermolecular force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast giant covalent repeating structures such as diamond and $\ce{SiO2}$ are not molecular in the same sense as they can theoretically be infinitely large. Therefore they do not really have intermolecular forces but they are simply held together by covalent bonds between the atoms in the structure.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-07T15:46:45.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23462" PostTypeId="2" ParentId="23459" CreationDate="2015-01-07T15:48:49.720" Score="1" Body="&lt;p&gt;Molecules are made of fixed numbers of atoms joined together by covalent bonds, and can range from the very small (even down to single atoms, as in the noble gases) to the very large (as in polymers, proteins or even DNA).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The covalent bonds holding the molecules together are very strong, but these are largely irrelevant to the physical properties of the substance. Physical properties (melting points, boiling points, solubility in water,...) are governed by the &lt;strong&gt;intermolecular forces&lt;/strong&gt; - forces attracting one molecule to its neighbors - van der Waals attractions or hydrogen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular substances tend to be gases, liquids or low melting point solids, because the intermolecular forces of attraction are comparatively weak. You don't have to break any covalent bonds in order to melt or boil a molecular substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the melting or boiling point will depend on the strength of the intermolecular forces. The presence of hydrogen bonding will lift the melting and boiling points. &lt;strong&gt;The larger the molecule the more van der Waals attractions are possible&lt;/strong&gt; - and those will also need more energy to break.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-07T15:48:49.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23463" PostTypeId="1" AcceptedAnswerId="23501" CreationDate="2015-01-07T16:01:02.323" Score="4" ViewCount="8960" Body="&lt;p&gt;Consider the following steps:&#xA;\begin{align}&#xA;\mathrm{d}U &amp;amp;= \mathrm{d}q - p\,\mathrm{d}V\\&#xA;\mathrm{d}U &amp;amp;= C_V\,\mathrm{d}T\tag1\\&#xA;\end{align}&#xA;Side question: Is equation $(1)$ only true for a perfect gas or all substances?&#xA;\begin{align}&#xA;C_V\mathrm{d}T &amp;amp;= \mathrm{d}q - p\,\mathrm{d}V\\&#xA;\mathrm{d}q &amp;amp;= C_V\,\mathrm{d}T + p\,\mathrm{d}V\\&#xA;\end{align}&#xA;Divide by T:&#xA;$$\mathrm{d}S = \frac{C_V}{T}\,\mathrm{d}T + \frac pT\,\mathrm{d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof requires a substitution of $\frac pT=\frac{nR}V$ because when it is then differentiated with respect to $T$ it equates to zero and so does $\frac{C_V}{T}$ when it is differentiated with respect to $V$ which shows it is an exact differential. However, if you do not make the substitution you do not get zero so the two differentials do not equal each other. This would suggest that entropy is not a state function. Am I going wrong somewhere? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-02-27T04:44:16.963" LastActivityDate="2015-02-27T04:46:53.453" Title="How to prove that entropy is a state function?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23464" PostTypeId="1" AcceptedAnswerId="23466" CreationDate="2015-01-07T16:20:23.023" Score="2" ViewCount="4536" Body="&lt;p&gt;I get that &lt;span class=&quot;math-container&quot;&gt;$G=H-TS$&lt;/span&gt; because then: &lt;span class=&quot;math-container&quot;&gt;$$\begin{align}\mathrm dG&amp;amp;=\mathrm dH-T\,\mathrm dS-S\mathrm dT\\&amp;amp;=T\,\mathrm dS+V\,\mathrm dp-T\,\mathrm dS-S\,\mathrm dT\end{align}$$&lt;/span&gt;&#xA;Therefore, by cancelling: &lt;span class=&quot;math-container&quot;&gt;$\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$&lt;/span&gt; which is the equations for &lt;span class=&quot;math-container&quot;&gt;$\mathrm dG$&lt;/span&gt;. However, I can’t get this result from using &lt;span class=&quot;math-container&quot;&gt;$\mathrm dG=\mathrm dH-T\,\mathrm dS$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm dH=S\,\mathrm dT+V\,\mathrm dp$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm dG=\left(S\,\mathrm dT+V\,\mathrm dp\right)-T\,\mathrm dS$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This does not give me the right equation. Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="41328" LastEditDate="2020-07-17T18:57:15.430" LastActivityDate="2020-07-17T18:57:15.430" Title="Why is dG = dH − TdS?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23465" PostTypeId="2" ParentId="8625" CreationDate="2015-01-07T16:30:38.720" Score="0" Body="&lt;p&gt;Use nitration by mixed acid and then $\ce{SOCl2}$ to replace the $\ce{OH}$ of carboxylic group. Then use $\ce{NaBH4}$ to reduce acid chloride to alcohol.  &lt;/p&gt;&#xA;" OwnerUserId="12640" LastEditorUserId="4231" LastEditDate="2015-01-07T18:21:27.333" LastActivityDate="2015-01-07T18:21:27.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23466" PostTypeId="2" ParentId="23464" CreationDate="2015-01-07T16:54:40.557" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is dG=dH-TdS?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't. That would only be true at constant temperature. In general:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2016-06-19T17:44:36.697" LastActivityDate="2016-06-19T17:44:36.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23467" PostTypeId="2" ParentId="23464" CreationDate="2015-01-07T16:54:58.717" Score="6" Body="&lt;p&gt;If you write $\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$, you drop off one term of the differentiation of $G$ which is $S\mathrm{d}T$.&#xA;By definition $G = H - TS$. So, $$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S-S\,\mathrm{d}T$$&#xA;Then, you should write $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}p$ (not as you have written $\mathrm{d}H = S\,\mathrm{d}T + V\,\mathrm{d}p$ ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $$\mathrm{d}G = T\,\mathrm{d}S + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T$$&#xA; $$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16683" LastEditDate="2016-06-19T17:47:05.927" LastActivityDate="2016-06-19T17:47:05.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23468" PostTypeId="1" AcceptedAnswerId="23471" CreationDate="2015-01-07T17:03:16.407" Score="2" ViewCount="865" Body="&lt;p&gt;how should we use di-tert-butyl dicarbonate(diBOc) as solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reference, diBOc was in the form of a liquid but when I checked the bottle it was in solid state.&lt;br/&gt;&#xA;diBOc should be stored at 8-10 ℃. So how is it possible for it to be used as liquid?&#xA;Is it OK to melt it at RT? Wouldn't melting, make diBOc have structural differences?&lt;/p&gt;&#xA;" OwnerUserId="11326" LastEditorUserId="7448" LastEditDate="2015-01-07T21:44:50.963" LastActivityDate="2015-01-07T21:44:50.963" Title="how to use di-tert-butyl dicarbonate (diBOc) as a solvent?" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23469" PostTypeId="1" CreationDate="2015-01-07T17:31:01.497" Score="3" ViewCount="146" Body="&lt;p&gt;Sorry if this is an outright stupid question. I am no chemist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nitric oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2NH3 + 2NO -&amp;gt; 2 N2 + H2 + 2H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nitrogen dioxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2NH3 + 2NO2 -&amp;gt; 2N2 + H2 + 2H2O + O2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{4NH3 + 2NO2 -&amp;gt; 3N2 + 2H2 + 4H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I'm trying to find if it is practical to take two toxins and convert them into non-volatile substances. Also, of course, harvesting this hydrogen gas would also be preferred; you could just add it back into the resulting Syngas. And I suppose ammonia doesn't have to be used. It just seemed convenient and the above reactions (producing useful Hydrogen and other non-volatiles) somehow came into my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it seems for every $\ce{O}$ (or $\ce{N}$) in the $\ce{NO_{x}}$ it can make another $\ce{O2}$ (or $\ce{N2}$, respectively). Or double the ammonia, to effectively double up on the first reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its the case that one reaction is more practical than another (as it usually is). I have noticed that conversion is simple: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2NO + O2 -&amp;gt; 2NO2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;readily happens in air and at 150C it reverses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2NO2 -&amp;gt; 2NO + O2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of this leads to gasification. Gasifiers stratify the combustion process to extract producer gas (Called Syngas. It is $\ce{H2 + CO}$, and it burns clean.). If a gasifier is built a certain way, the only pollutants could be $\ce{NO_{x}}$, and since some gasifiers can use biowaste, then ammonia is not hard to reconstitute on-site in some applications. Plus, I think it might be possible to extract ammonia directly from the pyrolysis stage when using garbage in certain gasifiers. As well, I think Ammonia is in the resulting trace gases sometimes depending on the gasifier/fuel setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some resources to be had. Gasifiers also have to typically filter out sulfur and mercury from the resulting Syngas. If those can help at all here and/or could also be included in this detoxification. As well, there are usually sources of heat (up to 1100C), tar, char (hot carbon), and hot or cold ash (burnt charcoal), $\ce{H2 + CO}$, $\ce{CO2}$, and other trace gases like $\ce{CH4}$ or whatever, depending on the gasifier and fuel. These resources could be utilized if needed, but most already have a role to play. Carbon dioxide, methane and tar are used to make more Syngas. The wasted byproducts are: sulfur, mercury, cold ash, and the &quot;whatever&quot; trace gases. I believe sulfur is in the form hydrogen sulfide. There is also probably substantial unsued hot Char (hot reactive carbon), but it burns quick, and only exists in the heart of the machine, then once all the heat is extracted it turns to ash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I wonder if I could pump the $\ce{NO_{x}}$ (or other pollutants) back into the pyrolysis stage, if they would might reform into something else. Less toxic, eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I feel like I'm barking up the wrong tree. As $\ce{NO_{x}}$ gases are closely scrutinized by municipalities, any practical reduction would probably be used already and emission of $\ce{NO_{x}}$ gases into the environment wouldn't be a problem, but that is assuming way too much for humans. So, I have to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do any chemists see a way, using these resources (or other practical external ones), to detoxify the nitrogen oxide from the resulting burnt Syngas fumes?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12636" LastEditorUserId="194" LastEditorDisplayName="user15489" LastEditDate="2017-04-06T03:17:10.657" LastActivityDate="2017-04-06T03:17:10.657" Title="Is reducing nitrogen oxides with ammonia practical in a gasifier?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;redox&gt;&lt;gas-laws&gt;&lt;theoretical-chemistry&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-04-06T16:28:26.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="23471" PostTypeId="2" ParentId="23468" CreationDate="2015-01-07T18:18:12.580" Score="4" Body="&lt;p&gt;Di-tert-butyl dicarbonate melts around room temperature (m.p.=23°C).  It does not decompose at this or even slightly higher temperatures.  For example, it is typically purified by distillation under reduced pressure at temperatures up to around 65°C.  At higher temperatures it will decompose to isobutene, t-butyl alcohol and carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oN2RE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case it is OK to melt di-tert-butyl dicarbonate and use it between room temperature and up to around 40-60°C.  Just make sure that when you open the bottle and use it that you do so in a dry environment as water will decompose di-tert-butyl dicarbonate.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-07T18:18:12.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23472" PostTypeId="1" AcceptedAnswerId="23485" CreationDate="2015-01-07T19:02:00.450" Score="1" ViewCount="1231" Body="&lt;p&gt;Upon being asked to prepare a solution of a given molarity from a solution with high concentration I was confused whether or not I should factor in the volume taken from the high concentration substance into the volume of the solvent. I think that an example could illustrate what I am asking more appropriately. Lets say I have to prepare a 5 M substance in 50 ml of water. The substance is in a container that is 10 M so I must calculate the number of moles and then the amount of solute to extract from the 10 M container.  $$M = {m\over{L}} $$ $$5 M = {m\over{.05}} $$ $$moles = .25$$ given that the solution's concentration is 10 M $$10M={.25\over{L}}$$ $$L = .025$$ Now in order to achieve the 5 M in the 50 ml of water, do I pour 25 ml of water and then the 25 ml, which then adds to 50 ml? or do I pour the 50 ml of water and then the 25ml resulting in a total volume of 75 ml?&lt;/p&gt;&#xA;" OwnerUserId="7639" LastActivityDate="2015-01-07T23:53:47.507" Title="Should the volume of a solute be factored into the volume of a solvent when preparing a substance of a certain molarity?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;stoichiometry&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23473" PostTypeId="2" ParentId="23472" CreationDate="2015-01-07T19:59:46.743" Score="0" Body="&lt;p&gt;You only add the $\ce{25 mL}$ water. After all you want to dilute to a certain molarity. You start from a 10 M solution and want to go to a 5M solution. Neglecting the volume of the dissolved compound you simply add another equivalent of solvent to dilute by a factor 2, i.e. $\ce{25 mL}$ water in this case.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="10185" LastEditDate="2015-01-07T20:54:56.443" LastActivityDate="2015-01-07T20:54:56.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23474" PostTypeId="2" ParentId="22285" CreationDate="2015-01-07T20:40:11.100" Score="5" Body="&lt;p&gt;I am not sure this is what you are looking for, but there is the Padel descriptor software that calculates a whole slew of properties for compounds you throw at it. Nothing empirical though, all calculated values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here:  &lt;a href=&quot;http://www.yapcwsoft.com/dd/padeldescriptor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.yapcwsoft.com/dd/padeldescriptor/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are willing to use a tool like KNIME, there is a Padel KNIME extension available on de Padel website. It is somewhat fiddly to get working properly in KNIME without conflicts, but once you have that going you can pump millions of molecules through this pipeline.&lt;/p&gt;&#xA;" OwnerUserId="8135" LastEditorUserId="8135" LastEditDate="2017-10-02T11:44:14.643" LastActivityDate="2017-10-02T11:44:14.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23475" PostTypeId="1" AcceptedAnswerId="23477" CreationDate="2015-01-07T20:47:00.603" Score="3" ViewCount="827" Body="&lt;p&gt;We're learning about functional groups of different organic molecules, and there are some I'm confused about: &lt;strong&gt;carboxyl&lt;/strong&gt;, &lt;strong&gt;carbonyl&lt;/strong&gt;, and &lt;strong&gt;hydroxyl&lt;/strong&gt;. I was looking online but I couldn't find a direct answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the carbonyl group is a &quot;-CO&quot;, the hydroxyl group is a &quot;-OH&quot;, and the carboxyl group is a &quot;-COOH&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We learned that carbonyl groups make molecules a ketone or aldehyde, the hydroxyl group makes molecules alcohols, and the carboxyl group makes molecules carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that that a carbonyl group is just a carbon double-bonded to an oxygen, and a carboxyl group is a carbonyl group bonded to a hydroxyl group, &lt;strong&gt;then should carboxyl molecules also have all the properties of hydroxyl and carbonyl groups (such as being an alcohol), on top of its new ones?&lt;/strong&gt; Or, does it only have the properties associated with themselves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11524" LastActivityDate="2015-01-07T21:27:12.360" Title="Do molecules with carboxyl groups have properties of carboxyl's sub-functional-groups?" Tags="&lt;biochemistry&gt;&lt;molecules&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23476" PostTypeId="2" ParentId="2766" CreationDate="2015-01-07T21:06:14.330" Score="0" Body="&lt;p&gt;Ok so first of all you cant get into orbit with anything &quot;simple&quot;... Remember that as the altitude increases, the pressure decreases, the makeup of the atmosphere changes, temperature changes and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on to rocket fuel, if you want to go high, I suggest sinking some time into a hybrid engine. A hybrid engine uses a solid fuel and a liquid or gaseous oxidizer. There are some good example videos on YouTube of engines that use a hollowed acrylic rod for fuel and use either nitrous oxide or straight O2 to produce a very effective, throttle-able engine. However if you just starting out, I suggest rocket candy, its a cheap, homemade solid fuel+oxidizer mixture that is great for experimenting with. You can usually use sucrose and either potassium nitrate or sodium nitrate, I suggest the sodium nitrate as it is slightly lighter, but then again I'm a perfectionist.&lt;/p&gt;&#xA;" OwnerUserId="12646" LastActivityDate="2015-01-07T21:06:14.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23477" PostTypeId="2" ParentId="23475" CreationDate="2015-01-07T21:21:00.490" Score="3" Body="&lt;p&gt;Carboxyl groups do share some properties with hydroxy and carbonyl groups such as the ability to engage in hydrogen bonding but in terms of chemical reactivity they are very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this lies in the structure of the carboxyl group. Instead of the carbonyl and the hydroxyl groups being completely independent of each other, there is an orbital interaction between the three $p$ orbitals on the carbon and the two oxygens which creates a $\pi$ system across the three atoms that tends to stabilise the group, particularly when it is deprotonated. (I couldn't find any decent pictures of these $\pi$ orbitals but if anyone can please edit them in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious property that carboxyl groups have which hydroxyl and carbonyl groups do not display (hydroxyls do slightly but nowehere near as much as carboxyls) is acidity. This high acidity is caused by the fact that the carboxylate ion which forms when carboxyl groups donate a proton is stabilised by the electron delocalisation $\pi$ system. This can also be expressed using resonance structures as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a1Uit.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of the acidity can be seen in the $pK_a$ values of carboxylic acids compared to alcohols.$$Ethanoic~acid~pK_a = 4.74$$$$Ethanol~pK_a = 16$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remebering that $pK_a$ has a logarithmic scale we can see that carboxylic acids are over $10^{10}$ times more acidic than their respective alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent introduction to many other properties and reactions of carboxylic acids can be found here: &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-01-07T21:27:12.360" LastActivityDate="2015-01-07T21:27:12.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23478" PostTypeId="1" CreationDate="2015-01-07T21:25:23.497" Score="-1" ViewCount="518" Body="&lt;p&gt;Water is $\ce{H2O}$ in all it's states: solid(ice), liquid, and gas(water vapor).&lt;br&gt;&#xA;In your drink, when you add artificial flavoring and coloring, does it change the chemical composition of the water? What makes the flavor and color change?&lt;br&gt;&#xA;Are you still drinking $\ce{H2O}$, with dense amounts of another liquid, or do they blend together?&lt;/p&gt;&#xA;" OwnerUserId="12647" LastEditorUserId="4231" LastEditDate="2015-01-07T21:32:08.870" LastActivityDate="2015-01-09T02:09:30.737" Title="Chemical Composition Changes in Liquids" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23481" PostTypeId="1" CreationDate="2015-01-07T22:32:11.387" Score="0" ViewCount="856" Body="&lt;p&gt;This question is on the same topic as &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13541/explanation-for-the-reactions-in-a-saltwater-battery-with-zinc-and-copper-electr&quot;&gt;this one&lt;/a&gt; but more complex.&#xA;So, let's consider the cell with $\ce{Zn}$ and $\ce{Cu}$ electrodes inside the $\ce{NaCl}$ solution. &#xA;On the $\ce{Zn}$ electrode we have &#xA;$$\ce{Zn + 2Cl^- -&amp;gt; ZnCl2  + 2e^-},$$ and on the copper one &#xA;$$\ce{H3+O + e^- -&amp;gt; 1/2 H2 + H2O}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both reactions we can find standard electrode potentials, but they are given for normal conditions, and $1~\mathrm{M}$ concentration of $\ce{Zn^{2+}}$ and $\ce{H3+O}$. To calculate the EMF for non-$1~\mathrm{M}$ concentration we use the Nernst equation. In this case we have something like&#xA;$$E=E_0 - \frac{\mathcal{R}T}{n\mathcal{F}}\ln{\frac{[\ce{Zn^{2+}}]}{[\ce{H3+O}]}}.$$&#xA;But really in the initial $\ce{NaCl}$ solution we do not have any $\ce{Zn}$ ions (so their concentration is very close to zero and unknown to us). Of course, in few microseconds some $\ce{Zn}$ from electrode will be dissolved creating some finite potential. But in the experiment when putting $\ce{Zn}$ electrode into the same solution we always observe the approximately same EMF value, $\approx 0.8~\mathrm{V}$. How to predict theoretically this value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, my question is how to calculate the EMF of such cell as precisely as possible for different experimental conditions, e.g. different temperature?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made an experiment showing that the EMF of a 'cell' with $\ce{Zn}$ and $\ce{Cu}$ electrodes into $\ce{NaCl}$ solution is decreasing when heating the $\ce{NaCl}$ solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's strange, let me explain why. The Nernst equation tells us that EMF is increasing with $T$ when $[\ce{Zn^{2+}}]$ is less than $[\ce{H3+O}]$ (because $\log$ is negative), and otherwise EMF is decreasing with $T$ ($\log$ is positive). But in my opinion the concentration of $\ce{Zn}$ ions is always less than $[\ce{H3+O}]$ because we do not have $\ce{Zn}$ in the initial solution. So then logarithm is always negative so EMF is growing with $T$.&lt;br&gt;&#xA;In this case, why do I observe the EMF decreasing for $\ce{Zn-Cu}$ pair?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last question. When we talk about $\ce{Zn-Cu}$ pair we assume that zinc dissolves; and the hydrogen gas is created on the copper electrode.&#xA;But why do I have the EMF when putting into a $\ce{NaCl}$ solution the $\ce{Al}$ ($\ce{Fe, Cr, etc}$) electrodes instead of $\ce{Cu}$, and remain instant the $\ce{Zn}$ electrode? I am not sure that hydrogen is created in such pairs: all these metals have a negative standard electrode potential. Measured EMFs of all these ($\ce{Zn-Al, Zn-Fe, Zn-Cr}$) pairs are significantly different, so the second metal is important in these cases. What reactions are going on in such pairs?&lt;/p&gt;&#xA;" OwnerUserId="11589" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-01-08T05:10:03.200" Title="How to accurately calculate the electromotive force for various conditions in a salt water battery?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23482" PostTypeId="1" AcceptedAnswerId="23484" CreationDate="2015-01-07T23:08:10.677" Score="4" ViewCount="4614" Body="&lt;p&gt;So I was reading benzotriazole's synthesis is o-phenylenediamine with acetic acid and sodium nitrite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a reversible process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not even specifically to o-phenylenediamine but any derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;phenylenedaimne's (usually para) and derivatives and generally aromatic amines form various colour dyes on old colour photographic films that I like to experiment with as a hobby.&lt;/p&gt;&#xA;" OwnerUserId="11404" LastEditorUserId="7475" LastEditDate="2015-10-09T15:49:53.590" LastActivityDate="2015-10-09T15:49:53.590" Title="Benzotriazole to o-phenylenediamine" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23484" PostTypeId="2" ParentId="23482" CreationDate="2015-01-07T23:37:12.960" Score="5" Body="&lt;p&gt;Yes, a common synthesis of benzotriazole starts with o-phenylenediamine as shown in the following figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3xqF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Benzotriazole#Synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in the reaction involves straightforward diazotization of one of the amino groups (exactly like what's done in the first part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Sandmeyer_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandmeyer reaction&lt;/a&gt;) to produce the ortho-amino diazonium ion. Instead of eliminating nitrogen as in the Sandmeyer case, the molecule loses a proton and then the diazomium ion internally captures the ortho-imino group to yield the triazole product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is not reversible, the benzotriazole is very stable and the intermediate diazonium ion is relatively high energy making the back reaction unlikely under normal conditions.  The reaction will also work with substituted o-phenylenediamines as long as the substituent doesn't interfere with the diazotization step.  So for example, alkyl- or halo-substituted o-phenylenediamines will produce the corresponding alkyl- or halo-benzotriazoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; response to comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intramolecular cyclization to produce a triazole will only occur if the amino groups in the starting diamine are ortho to one another.  If m- or p-phenylenediamines or non-ortho naphthyl diamines are used, then products resulting from intermolecular azo coupling will occur. &lt;a href=&quot;https://www.bioscience.org/2013/v18/af/4093/figures.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here&lt;/a&gt; for some examples.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-08T01:10:51.087" LastActivityDate="2015-01-08T01:10:51.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23485" PostTypeId="2" ParentId="23472" CreationDate="2015-01-07T23:53:47.507" Score="4" Body="&lt;p&gt;The only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with reference to your question, you should add enough solvent to get the proper final volume.  Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL.  The added solvent might not be exactly 25 mL to make this final volume.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-01-07T23:53:47.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23486" PostTypeId="2" ParentId="22217" CreationDate="2015-01-08T00:09:22.623" Score="1" Body="&lt;p&gt;Macromolecule can be a monomer, but it's a little tricky because macromolecules are usually polymers by themeselves. I think that only common &quot;monomeric&quot; biomolecules which create oligomers, or even polymers are proteins - for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Hemoglobin&quot; rel=&quot;nofollow&quot;&gt;hemoglobin&lt;/a&gt; is present in erythrocytes as tetramer.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-01-08T00:09:22.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23487" PostTypeId="2" ParentId="23481" CreationDate="2015-01-08T03:20:24.857" Score="3" Body="&lt;p&gt;I suspect that the reason you see a decay in the cell potential of $\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\ce{Zn}$ as $\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide (&lt;a href=&quot;http://dx.doi.org/10.1016/0010-938X(85)90070-8&quot; rel=&quot;nofollow&quot;&gt;Ichiro Suzuki, &lt;em&gt;Corrosion Science&lt;/em&gt; &lt;strong&gt;1985&lt;/strong&gt;,&lt;em&gt;25&lt;/em&gt; (11), 1029-1034&lt;/a&gt;). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.&lt;/p&gt;&#xA;" OwnerUserId="12650" LastEditorUserId="4945" LastEditDate="2015-01-08T05:10:03.200" LastActivityDate="2015-01-08T05:10:03.200" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23489" PostTypeId="1" AcceptedAnswerId="23490" CreationDate="2015-01-08T06:02:02.037" Score="3" ViewCount="978" Body="&lt;p&gt;Is a heating block necessary for heating small chemistry samples? I figure it helps a ton (not to mention it stabilizes too) in getting heat to the reaction but is there anything else it does? &lt;/p&gt;&#xA;" OwnerUserId="11585" LastEditorUserId="4945" LastEditDate="2015-01-08T06:47:42.660" LastActivityDate="2015-01-08T06:47:42.660" Title="Why use a heating block?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23490" PostTypeId="2" ParentId="23489" CreationDate="2015-01-08T06:42:21.947" Score="4" Body="&lt;p&gt;&lt;em&gt;You have already mostly answered the question yourself ;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heating block, made from a material with a high thermal conductivity, helps to keep (&lt;strong&gt;a lot of&lt;/strong&gt;) samples distributed over a larger area (several vials, Eppendorf tubes, etc.) at exactly the same temperature even if the primary heat source is punctual or just a layer of heating wires at the bottom of the instrument.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-08T06:42:21.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23493" PostTypeId="2" ParentId="1302" CreationDate="2015-01-08T12:12:28.060" Score="-3" Body="&lt;p&gt;There is a formula for this too... it is = 1×[(no. Of pi bonds b/w resonating atoms of resonating structures)/(no of sigma bonds  b/w resonating atoms of resonating structures)]....for example: in benzene, hydrogen atoms do not participate in resonance so bond order = 1+(3/6) =1.5; in carbonate ion, it is 1+ 1/3=1.33 hope this helps.... &lt;/p&gt;&#xA;" OwnerUserId="12655" LastActivityDate="2015-01-08T12:12:28.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23494" PostTypeId="1" AcceptedAnswerId="23536" CreationDate="2015-01-08T13:29:09.853" Score="3" ViewCount="4868" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NL1yU.png&quot; alt=&quot;Cyclohexane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the chair conformation of cyclohexane, what are the angles of the triangles defined by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a carbon atom, an axial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?&lt;/li&gt;&#xA;&lt;li&gt;a carbon atom, an equatorial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have unsuccessfully looked for literature values of these, and my trigonometry is not up to scratch.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="4945" LastEditDate="2015-01-09T06:35:43.277" LastActivityDate="2015-01-09T06:35:43.277" Title="What are the bond angles in cyclohexane?" Tags="&lt;molecular-structure&gt;&lt;cyclohexane&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23495" PostTypeId="1" CreationDate="2015-01-08T14:05:44.523" Score="1" ViewCount="1135" Body="&lt;p&gt;During the electrolysis of $\ce{H2SO4}$, $\ce {OH-}$ ions are more reactive than $\ce {SO4^{2-} }$ ions and so they dissociate. as $\ce {OH-}$ ions dissociate, they are the negative ions that carry current. So when the concentration of $\ce {H2SO4}$ is increased, the concentration of $\ce {OH-}$ ions actually reduces, but then the current still increases.&lt;/p&gt;&#xA;" OwnerUserId="11419" LastEditorUserId="4231" LastEditDate="2015-03-09T21:50:08.870" LastActivityDate="2015-04-08T22:02:23.710" Title="in the electrolysis of H₂SO₄ solution, when the concentration is increased why does the current increase even when OH⁻ ions reduce?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23496" PostTypeId="1" CreationDate="2015-01-08T15:09:49.657" Score="3" ViewCount="394" Body="&lt;p&gt;The original question was 'Oxygen exhibits Oxidation state -2 to +2 but other elements of grp 16 exhibit only +2,+4,+6. Explain Why'. In the solution given i can't understand meaning of this line 'It doesn't have nd-orbitals &lt;em&gt;due&lt;/em&gt; to which its valency doesn't increase'. Please explain how? &lt;br&gt;&lt;br&gt;&#xA;PS:I am a chemistry noob. So please provide as much details as possible :)&lt;/p&gt;&#xA;" OwnerUserId="12651" LastActivityDate="2017-10-14T09:34:26.893" Title="How does absence of nd orbital in Oxygen affect its valency?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23497" PostTypeId="2" ParentId="23494" CreationDate="2015-01-08T15:11:21.660" Score="2" Body="&lt;p&gt;In the chair form of cyclohexane, the carbon atoms and the bonds around them are almost perfectly tetrahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the C-C-H angles will be almost exactly 109.5 degrees. (Note that while you defined the bond midpoint, the angle will be the same regardless of whether it's the midpoint of the bond or the neighboring carbon atom itself.)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-08T15:11:21.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23498" PostTypeId="2" ParentId="23496" CreationDate="2015-01-08T15:31:08.643" Score="-1" Body="&lt;p&gt;You need to review the concepts surrounding atomic shell structure and quantum numbers, and how they relate to the organization of the periodic table, I think.  Briefly, since oxygen is in the second row of the periodic table, we know that its valence electrons occupy the second orbital shell, corresponding to orbitals with principal quantum number $n=2$.  Recall also that for a given principal quantum number $n$, only sub-shells with orbital angular quantum number $l$ up to $n-1$ are allowed.  Therefore, the valence shell of osygen only includes s ($l=0$) and p ($l=1$) orbitals.  On the other hand, the other members of group 16 will by definition have higher principal quantum numbers, which means that their valence shells include d ($l=2$) orbitals.  It is these orbitals that can be used to rationalize the existence of higher oxidation states for those elements, for example by the $sp^3d$ and $sp^3d^2$ hybridization schemes in VSEPR (valence shell electron pair repulsion) theory.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that explains why heavier elements in group 16 can have higher (i.e. more positive) oxidation numbers than +2.  Your question as stated appears to have an error however, since it says that those elements &quot;exhibit only +2,+4,+6&quot; ... that's not technically correct .. for example there are compounds where sulfur has an oxidation state of -2 (e.g. $H_2S$).  However I don't think it matters for the answer to your question here.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-08T15:31:08.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23499" PostTypeId="1" AcceptedAnswerId="23508" CreationDate="2015-01-08T16:39:14.363" Score="1" ViewCount="1027" Body="&lt;p&gt;For the reaction one mole of substance $A$ in equilibrium with one mole of substance $B$ the standard reaction enthalpy is defined:&#xA;$$\triangle_rH^{\theta}(T)=H_m^{\theta}(B;T)-H_m^{\theta}(A;T)$$&#xA;However, I have two issues with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly: the standard state of a substance (denoted by the symbol $\theta$) is the &lt;strong&gt;pure&lt;/strong&gt; material at a pressure of 1 bar. How can the reaction enthalpy be standard - the reactant have to be mixed (thus are not pure) to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly: what is $H_m^{\theta}$? Surely nothing can have a molar enthalpy because only changes in enthalpy can be measured not absolute values. Am I right in thinking that is is because the enthalpy scale has no zero value?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-10T00:22:15.473" Title="If the standard state symbol means that the substance is pure (and at 1 bar) how is it possible to have a standard REACTION enthalpy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23500" PostTypeId="1" AcceptedAnswerId="23504" CreationDate="2015-01-08T16:44:43.393" Score="3" ViewCount="3011" Body="&lt;p&gt;Why does alpha halogenation of a ketone in acidic conditions happen at the more substituted alpha carbon atom, and in basic conditions at the less substituted alpha carbon atom?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2016-07-21T03:42:45.833" Title="Regioselectivity of alpha halogenation of ketones" Tags="&lt;carbonyl-compounds&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23501" PostTypeId="2" ParentId="23463" CreationDate="2015-01-08T17:03:37.383" Score="2" Body="&lt;p&gt;I don't understand what you mean by saying &quot;if you don't make the substitution you do not get zero&quot;.  The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$.  Therefore $$\frac \partial{\partial T}\left(\frac pT\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly.  Could that be where your confusion is coming from?&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="4945" LastEditDate="2015-02-27T04:46:53.453" LastActivityDate="2015-02-27T04:46:53.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23502" PostTypeId="2" ParentId="20397" CreationDate="2015-01-08T17:33:25.073" Score="2" Body="&lt;p&gt;We know from stoichiometry that you &lt;em&gt;did&lt;/em&gt; produce twice as many moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; at the negative electrode as &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; at the positive electrode. However, we don't know a few important things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was &amp;quot;more&amp;quot;, but we won't place any bets on the factor of 10.&lt;/li&gt;&#xA;&lt;li&gt;What happened to the oxygen and hydrogen after the water split? Did it react?&lt;/li&gt;&#xA;&lt;li&gt;What else was in the water that could have reacted? We know there is &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;, but maybe there are other dissolved compounds as well?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.&lt;/p&gt;&#xA;&lt;p&gt;Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into &lt;span class=&quot;math-container&quot;&gt;$\ce{Al(OH)3}$&lt;/span&gt;, then at 1 atm and 298 K it would consume roughly 137 mL of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+ + 3OH^- -&amp;gt; Al(OH)3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$PV = nRT$&lt;/span&gt; (oxygen is close enough to ideal under normal lab conditions)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Al(OH)3}$&lt;/span&gt; is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{Al2O3}$&lt;/span&gt;) is also a potential candidate, and would consume 68 mL of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; under the same conditions.&lt;/p&gt;&#xA;&lt;p&gt;So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.&lt;/p&gt;&#xA;&lt;p&gt;How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.&lt;/p&gt;&#xA;&lt;p&gt;Question 3 is also hard to answer. We know that the &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.&lt;/p&gt;&#xA;&lt;p&gt;Summary:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Why did I get less gas than expected at the electrode connected to the +?&#xA;Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How can I get a ratio of 1/2?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="17175" LastEditDate="2021-01-17T13:26:15.113" LastActivityDate="2021-01-17T13:26:15.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23503" PostTypeId="2" ParentId="22366" CreationDate="2015-01-08T17:57:11.603" Score="13" Body="&lt;p&gt;The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals.  This is because of something called &lt;strong&gt;penetration&lt;/strong&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals.  In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the &lt;em&gt;increase&lt;/em&gt; in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.&lt;br&gt;&#xA;The trend goes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\pu{82 pm}$ ($\ce{B}$) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\pu{118 pm}$ $\ce{(Al)}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\pu{126 pm}$ $\ce{(Ga)}$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\pu{144 pm \ce{(In)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[covalent radii from &lt;a href=&quot;https://www.webelements.com&quot; rel=&quot;noreferrer&quot;&gt;www.webelements.com]&lt;/a&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This effect is generally known as the &lt;strong&gt;d-block contraction&lt;/strong&gt;.  (It can be more or less pronounced depending on how you define the atomic radii.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals&lt;/em&gt;, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm).  This effect (of the filled f-subshell) is generally known as the &lt;strong&gt;lanthanide contraction&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="51553" LastEditDate="2018-03-23T17:16:36.257" LastActivityDate="2018-03-23T17:16:36.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23504" PostTypeId="2" ParentId="23500" CreationDate="2015-01-08T18:05:55.513" Score="6" Body="&lt;p&gt;The acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively.  The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol.  &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactivity_of_Alpha_Hydrogens/Alpha_Halogenation&quot;&gt;See here&lt;/a&gt; for a good picture of this mechanism.  In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol.  Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control.  The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation.  The rate determining step is removal of an alpha-proton by the base.  The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate.  Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol.  Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-08T18:05:55.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23505" PostTypeId="1" AcceptedAnswerId="23506" CreationDate="2015-01-08T18:11:05.377" Score="7" ViewCount="2845" Body="&lt;p&gt;I'm about to teach concrete to some high school children. I'm not a chemist but as I understand it the limestone cycle is this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\ce{CaCO3 &amp;amp;-&amp;gt; CaO + CO2}\tag{R1}\\&#xA;\ce{CaO + H2O &amp;amp;-&amp;gt; Ca(OH)2}\tag{R2}\\&#xA;\ce{Ca(OH)2 + CO2 &amp;amp;-&amp;gt; CaCO3 + H2O}\tag{R3}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The water has to go somewhere or the concrete will stay wet. Air drying I could understand but how can concrete set underwater? And, assuming concrete dries from the outside in, wouldn't osmosis predict more water would be drawn &lt;em&gt;into&lt;/em&gt; the concrete?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Making kilograms of calcium carbonate will take kilograms of calcium hydroxide and kilograms of carbon dioxide. How can such a large amount of carbon dioxide reach the concrete? There isn't a lot of it in water, is there? I could imagine some reaction might produce carbon dioxide in the concrete, but then wouldn't it end up frothy/honeycombed?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10112" LastEditorUserId="41328" LastEditDate="2021-05-03T14:10:40.510" LastActivityDate="2021-05-03T14:10:40.510" Title="How can concrete set underwater?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23506" PostTypeId="2" ParentId="23505" CreationDate="2015-01-08T18:31:14.270" Score="8" Body="&lt;p&gt;The chemistry you have posted is for non-hydraulic cement, which requires air drying for precisely the reasons you enumerated.  The kind of cement that sets underwater uses silicates and various oxides (primarily calcium, aluminum and iron oxides). The basic chemical process that is occurring is hydration, whereby the mixture of oxides and silicates incorporates several water molecules directly into the crystal lattice.  The fact that the setting process both uses water, and also produces a structure that is insoluble and robust to attack by water, is what makes underwater setting of hydraulic cement possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to teaching your high school class, it's probably better to stick to the lime cycle for non-hydraulic cement as far as detailed chemical reactions are concerned. One interesting point that you can tell them about hydraulic cement is that its chemistry is still not completely understood, and even the molecular crystal structure was only very recently sorted out by &lt;a href=&quot;http://cee.mit.edu/cee-in-focus/2011/spring/cement-structure&quot; rel=&quot;noreferrer&quot;&gt;MIT scientists in 2009&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="14813" LastEditDate="2015-03-06T16:28:24.167" LastActivityDate="2015-03-06T16:28:24.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23507" PostTypeId="1" AcceptedAnswerId="23529" CreationDate="2015-01-08T19:20:47.643" Score="3" ViewCount="810" Body="&lt;p&gt;When a methyl ketone is submitted to halogenation in acidic conditions with two equivalents of a halogen, the first halogenation happens at the secondary alpha carbon atom , while the second halogenation happens at the primary alpha carbon atom. Why is it so? Why doesn't it happen on the same carbon atom (the secondary one)?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-01-09T02:52:58.137" Title="Halogenation of an alpha-halogen ketone in acidic conditions" Tags="&lt;carbonyl-compounds&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23508" PostTypeId="2" ParentId="23499" CreationDate="2015-01-08T19:22:16.530" Score="1" Body="&lt;p&gt;The standard reaction enthalpy you have describe is only for a pure substance if you are talking about a standard enthalpy of formation.  Otherwise the equation wouldn't be valid for reactions that have more than one product, right?  In the specific case of a standard enthalpy of formation, you are talking about the enthalpy change associated with the creation of one mole of a pure substance &lt;em&gt;from its component elements in their standard states.&lt;/em&gt;  This gets at your second question ... the zero on the enthalpy scale is established by convention (and this is the reason why the definition of standard enthalpy of formation works) ... the standard enthalpy of formation of an element in its standard state is DEFINED to be zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can estimate the standard reaction enthalpy for any chemical reaction by subtracting the summed standard enthalpies of formation of the reactants from the summed standard enthalpies of formation of the products, taking into account the appropriate stoichiometric coefficients from the balanced chemical equation.  That process is similar to what you have written, but not precisely the same, since you omitted the stoichiometry information (and the chemical equation, for that matter).  The standard enthalpy estimate obtained will pertain to reactions carried out at the given conditions .. standard enthalpies are defined for 1 atm pressure but variable temperature, so you may need to correct tabulated standard enthalpies of formation to match the given reaction conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last point is that enthalpy is absolutely defined as a thermodynamic quantity .. it is the internal energy plus the product of pressure and volume, $H=U + pV$.  The issue is that the internal energy of a substance is hard to define absolutely .. are you going to include the binding energies of the core electrons to their atoms?  How about the strong forces holding the nuclei together?  Thus for chemical processes it almost always makes sense to deal with just changes in enthalpy, since most of the really low level stuff I mentioned doesn't change between reactants and products.  I think that may have been part of what you were getting at with your second question.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="12658" LastEditDate="2015-01-10T00:22:15.473" LastActivityDate="2015-01-10T00:22:15.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23509" PostTypeId="2" ParentId="23495" CreationDate="2015-01-08T19:37:32.750" Score="1" Body="&lt;p&gt;Near room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid.  For pure water or pure sulfuric acid the conductivity is orders of magnitude lower.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pure water, there are only $10^{-7}$M of $\ce{H+}$ and $\ce{OH-}$ to carry current.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a relatively small amount of sulfuric acid  is added, it dissociates fully to $\ce{H+}$ and $\ce{HSO4-}$.  For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4231" LastEditDate="2015-03-09T21:50:11.760" LastActivityDate="2015-03-09T21:50:11.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23512" PostTypeId="2" ParentId="23507" CreationDate="2015-01-08T21:53:14.270" Score="0" Body="&lt;p&gt;Most probably due to steric reasons and the electron withdrawing effect that the halogen has on the substituted methylene group.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-08T21:53:14.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23513" PostTypeId="2" ParentId="22217" CreationDate="2015-01-08T22:11:21.523" Score="2" Body="&lt;p&gt;If you're a little looser with your definition of monomer, then &lt;a href=&quot;http://en.wikipedia.org/wiki/Tubulin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tubulin&lt;/a&gt; fits right in. Alpha and Beta Tubulin proteins bind to each other to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Microtubule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microtubules&lt;/a&gt; which are an important component of cell's &lt;a href=&quot;http://en.wikipedia.org/wiki/Cytoskeleton&quot; rel=&quot;nofollow noreferrer&quot;&gt;cytoskeleton&lt;/a&gt; and play roles in moving cellular components around the cell, especially during cellular division. And tubulin's not the only protein to act as a monomer, there is also &lt;a href=&quot;http://en.wikipedia.org/wiki/Actin&quot; rel=&quot;nofollow noreferrer&quot;&gt;Actin&lt;/a&gt;, which produces &lt;a href=&quot;http://en.wikipedia.org/wiki/Microfilament&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microfilaments&lt;/a&gt;. I'm sure there are others, but I don't have the time to dig them up right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ekZO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2015-01-08T22:11:21.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23514" PostTypeId="2" ParentId="15277" CreationDate="2015-01-08T22:20:05.897" Score="2" Body="&lt;p&gt;&lt;strong&gt;Assumptions&lt;/strong&gt;. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Estimating the volume of the tire and cartridge&lt;/strong&gt;. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Estimating the pressure&lt;/strong&gt;. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure&#xA;with a van der Waals calculation to see if nonideality makes any difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;P_{ideal} &amp;amp;= \frac{n R T}{V}\\&#xA;          &amp;amp;= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}}) &#xA;              (298~\mathrm{K})}{1.02~\mathrm{L}}\cdot&#xA;              \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\ &#xA;          &amp;amp;\approx 129~\mathrm{psi}\\&#xA;P_{vdw}   &amp;amp;= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\ &#xA;          &amp;amp;= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})&#xA;             (298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol})&#xA;             (0.04267~\mathrm{L~mol^{-1}})}\\&#xA;          &amp;amp;~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})}&#xA;                    {(1.02~\mathrm{L})^2}\\&#xA;          &amp;amp;= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\&#xA;          &amp;amp;\approx 124~\mathrm{psi}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-13T05:50:06.710" LastActivityDate="2015-01-13T05:50:06.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23515" PostTypeId="1" CreationDate="2015-01-08T22:51:46.113" Score="0" ViewCount="319" Body="&lt;p&gt;I wonder how photosynthesis works. I read the Wikipedia page, but I'm still confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main gap of understanding is (I assume):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a photon changes the velocity of an electron. But I do not (fully) understand how a photon creates chemical reactions. Does the atom lose its electron due to this velocity increase and thereby attracting the electrons from the other atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that all? Does this imply there is always an electron lost in photosynthesis or is that electron absorbed elsewhere (in another atom)?&lt;/p&gt;&#xA;" OwnerUserId="633" LastEditorUserId="9961" LastEditDate="2016-09-24T12:40:13.910" LastActivityDate="2016-09-24T12:40:13.910" Title="How does photosynthesis work?" Tags="&lt;biochemistry&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23516" PostTypeId="2" ParentId="9546" CreationDate="2015-01-08T23:30:08.953" Score="5" Body="&lt;p&gt;The reaction for complete combustion of acetylene is&#xA;$$\rm 2C_2H_2 + 5O_2 \rightarrow 4 CO_2 + 2H_2O$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...you'll get complete combustion when acetylene and oxygen are mixed in a 2:5 mole (or volume) ratio. If you have a higher ratio than that, combustion will be incomplete. The incomplete combustion products will appear as soot and carbon monoxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is 1/5th O$_2$, so you need 5 times as much air as you'd need pure O$_2$. The optimal acetylene/air ratio should be about 2:25. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the acetylene/air ratios for each tube. The gases are both at the same temperature and pressure, so the volume ratio is equal to the mole ratio. Any ratio that's greater than 2/25 will probably produce soot, and the higher the ratio, the more soot you'd expect. Any tube that contains an appreciable amount of oxygen mixed in with the acetylene will burn quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The full tube&lt;/strong&gt; burned slow because oxygen needed to mix with acetylene at the mouth of the tube. The little bit of acetylene that burned there produced soot, because the acetylene/air ratio was probably much higher than 2/25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The tubes that were 3/4 and 1/2 full&lt;/strong&gt; had 3:1 and 1:1 acetylene/air ratios. Both ratios are fuel-rich, so soot was produced (you'd expect more soot with the 3/4 full tube than the 1/2 full tube). Both tubes burned quickly because there was a fair amount of O$_2$ mixed with the acetylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The tube that was 1/10 full&lt;/strong&gt; had a 1:9 acetylene/air ratio. 1/9 is just a bit higher than 2/25, so this tube was fuel-rich, too, but much less so than any of the other tubes. Very little soot would be expected; maybe none, if more air was drawn into the tube as the acetylene burned.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-01-08T23:30:08.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23517" PostTypeId="1" CreationDate="2015-01-08T23:44:20.373" Score="0" ViewCount="32669" Body="&lt;p&gt;I was recently doing some organic chemistry work and came across a set of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Which of the following compounds will participate in hydrogen bonding with water? For each compound, indicate whether it will be a hydrogen bond donor, hydrogen bond acceptor, or both.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have an idea of how to approach the first part of the question. Can someone help me with the first molecule: (a) $\ce{CH3CH2OCH3}$ in order of connectivity.&lt;/p&gt;&#xA;" OwnerUserId="12664" LastEditorUserId="4325" LastEditDate="2015-04-15T20:16:21.727" LastActivityDate="2015-04-15T20:16:21.727" Title="How to tell if a molecule will participate in hydrogen bonding with water?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23518" PostTypeId="2" ParentId="23517" CreationDate="2015-01-09T00:00:34.047" Score="6" Body="&lt;p&gt;I figured out the answer. If there is a lone pair of electrons belonging to nitrogen, oxygen, or fluorine, then it is possible to bond to water. Therefore, it is can accept a hydrogen. If there is a hydrogen bonded to one of the three elements listed above, then it able to donate. &lt;/p&gt;&#xA;" OwnerUserId="12664" LastActivityDate="2015-01-09T00:00:34.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23519" PostTypeId="1" AcceptedAnswerId="23522" CreationDate="2015-01-09T00:12:12.057" Score="5" ViewCount="11317" Body="&lt;p&gt;Something I know that leads to this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water dissolved in ethanol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It might sounds weird to say water as a solute. Textbooks always says that solvent is the one present in larger quantity. This leads to a contradiction with the saying that water is an universal solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be water present in smaller quantity than ethanol in a water-ethanol solution. But which is the solute and which is the solvent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to identify the solute and the solvent in a solution besides locating the word &quot;aqueous&quot;?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="41328" LastEditDate="2017-09-18T04:55:46.180" LastActivityDate="2018-06-22T08:33:46.627" Title="Can there be more solvent in a solute?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23520" PostTypeId="1" CreationDate="2015-01-09T01:11:29.983" Score="3" ViewCount="163" Body="&lt;p&gt;I wonder if anyone knows a readily available way to obtain some chemical graph complexity index (like the J index) from the SMILES or InChl representations?I'm looking for some existing implementation, e.g. in R, or OpenBabel, prior to trying to define mine if no one else exist... Thanks!&lt;/p&gt;&#xA;" OwnerUserId="12665" LastActivityDate="2015-01-09T01:11:29.983" Title="Obtaining chemical graph complexity index from SMILES or InChl" Tags="&lt;cheminformatics&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23521" PostTypeId="2" ParentId="23519" CreationDate="2015-01-09T01:27:15.040" Score="2" Body="&lt;p&gt;I think you may be emphasizing the term &quot;universal solvent&quot; a bit too hard. Water is &quot;universal&quot; as a solvent simply because lots of molecules are at least somewhat polar and so dissolve in polar water. But otherwise I'd say your textbook is right: the solvent is generally in greater quantity than the solute by convention.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-01-09T01:32:59.567" LastActivityDate="2015-01-09T01:32:59.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23522" PostTypeId="2" ParentId="23519" CreationDate="2015-01-09T01:27:52.037" Score="10" Body="&lt;p&gt;The textbook statement you gave (&quot;solvent is always the one present in larger quantity&quot;) is simply a convention that makes it easier for us to talk to each other about mixtures of substances.  We have other words for other types of mixtures.  For example, in your ethanol-water example, you are correct in stating that it doesn't make much sense to talk about solute and solvent.  So, we call these two liquids &quot;miscible&quot;.  They mix together in all proportions.  A good demonstration that captures the essence of different types of mixtures is the phase diagram of phenol-water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8UVrd.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have a feature that is called a &quot;miscibility gap&quot;.  There is a region, defined by the rainbow shape in the diagram above in which phenol and water separate into two distinct phases.  If we had pure water and added just a little bit of phenol to it at room temperature, we would find that it would dissolve into the water up to about 7 or 8 wt%. So, phenol would be the solute and water the solvent.  Adding more than this amount pushes the mixture into the miscibility gap so that now the phenol and water would mostly separate into two separate phases.  Let's say we add enough phenol to get us to 40 wt%.  If we heat the two phase mixture up to 67 C then it will suddenly become miscible.  Now only one phase exists and we can add as much phenol as we want to it without having to worry about two phases forming again.  There is no solute or solvent here.  Just a mixture of two liquids that exist in a single phase. &lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-09T01:27:52.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23523" PostTypeId="2" ParentId="23519" CreationDate="2015-01-09T01:28:03.627" Score="2" Body="&lt;p&gt;I think you may be putting too much significance on the semantics of some mnemonics that are perhaps phrased more absolutely than they are intended to be used.  That is particularly true with the phrase that &quot;water is the universal solvent&quot;, which is not really a scientifically significant statement at all.  Certainly gasoline, or mineral oil, will not dissolve in water, even in very small quantities, so why would it be considered &quot;universal&quot; in the sense you seem to be taking it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the better rule of thumb is to use the statement that the solvent is always the component present in larger quantity .. but even that is not correct in all cases.  For example in a multiphase extraction, there can be multiple solvents ... components that are polar will dissolve in the aqueous phase, so for those components, water is the solvent .. while components that are non-polar will dissolve in the non-aqueous phase, which will be the solvent for those components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final point is that for something like a 49-51% mixture of ethanol and water, the concept of a solvent becomes largely academic ... is the ethanol dissolved in the water, or is the water dissolved in the ethanol?  The answer is: both, and neither .. they are miscible liquids, so it's a two component mixture, rather than a solution.  The physical properties of a mixture depend rather strongly on the properties of both components, while the physical properties of a solution are more likely to be dominated by the properties of the solvent.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="12658" LastEditDate="2015-01-10T17:25:42.733" LastActivityDate="2015-01-10T17:25:42.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23524" PostTypeId="2" ParentId="23515" CreationDate="2015-01-09T01:40:23.440" Score="3" Body="&lt;p&gt;A better way of thinking about it, without getting too bogged down in details, is that the plant absorbs the energy of photons and uses it to drive chemical processes - ultimately the synthesis of sugar out of carbon dioxide and water - that are energetically &quot;uphill&quot;, and would not happen spontaneously without the energy from the photons.  The energy from the photons is used by the plant to move electrons from one place to another, but no electrons are actually generated or consumed in the process ... photosynthesis is a cycle, so you always end up back where you started, with some molecules of $\ce{CO2}$ and $\ce{H2O}$ having been consumed (along with a lot of light energy), and some extra molecules of sugar and $\ce{O2}$ having been produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The places where the photonic energy is used to mobilize the electrons are protein complexes called photosystem I and photosystem II, which are concentrated in the chloroplasts of green plants.  The details of how this happen are still not completely understood, but a LOT of progress has been made fairly recently using fancy equipment like laser spectrometers that can measure physicals processes that occur on timescales shorter than a nanosecond.  This has allowed researchers to map the progress of the electron as it moves from one part of the protein complex to another, following absorption of the photonic energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help at all?&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="16683" LastEditDate="2015-10-16T20:15:20.893" LastActivityDate="2015-10-16T20:15:20.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23526" PostTypeId="2" ParentId="23478" CreationDate="2015-01-09T02:09:30.737" Score="2" Body="&lt;p&gt;Not really sure what you are after here, but maybe this will help ... artificial flavors and colors are generally present at very low concentrations, dissolved in the water.  &quot;Dissolved&quot; means that the individual additive molecules are completely surrounded by water molecules, which are interacting with it to stabilize it in its dissolved form.  Since their concentration is low, each such additive-water aggregate is pretty far away from any others, so what you have is not very different from water, in a chemical sense at least.  Note also that the water molecules around a given additive molecule are not static .. they are constantly exchanging with the &quot;bulk&quot; pure water solvent around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, an additive may not be very water-soluble ... in such a case a surfactant solubilizer molecule may be added to help the additives dissolve in the water.  The surfactant molecules have one end that likes to interact with the insoluble additives (the hydrophobic end), and one end that likes to interact with water (the hydrophilic end).  They aggregate themselves into larger spherical structures called micelles, where the hydrophobic parts of the molecules are protected on the inside (along with the additive molecules), and the hydrophilic parts are on the outside, exposed to the water.  These micellar aggregates can be large enough to scatter light, and so drinks with these additive-solubilizer combinations will often appear cloudy.  An example of this is the citrus flavoring in sodas like Mountain Dew and Squirt ... in those cases, brominated vegetable oil (BVO) is added as a surfactant to dissolve the citrus flavor molecules.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Aside: BVO is not exactly good for you, and several soft drink manufacturers have said that they will eliminate it from their formulations, which means maybe I can start drinking Fresca again)&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T02:09:30.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23528" PostTypeId="2" ParentId="22454" CreationDate="2015-01-09T02:28:58.803" Score="5" Body="&lt;p&gt;I have never worked with potassium personally, but have experience with other alkali metals.   You are spot on about potassium superoxide.  I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil.  I would therefore be extremely cautious if you intend to store it in this manner.  Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms.  Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier.  The only way I would work with this metal is in an inert glovebox, period.  That is my own preference, of course, and I always tend to err on the side of caution.        &lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-09T02:28:58.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23529" PostTypeId="2" ParentId="23507" CreationDate="2015-01-09T02:52:58.137" Score="1" Body="&lt;p&gt;The same mechanism discussed in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/23500/regioselectivity-of-alpha-halogenation-of-ketones/23504#23504&quot;&gt;my answer to your earlier question&lt;/a&gt; on acid catalyzed halogenation alpha to a carbonyl still applies here.  For reference, here is the same &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactivity_of_Alpha_Hydrogens/Alpha_Halogenation&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; supplied earlier that shows the mechanism for this acid catalyzed halogenation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that earlier answer I said that the enol that &quot;produces the more stable double bond (the one that is more highly substituted)&quot; is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product.  In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond.  Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects).  On the other hand, fluorine destabilizes a double bond.  It is so electronegative that it prefers to bond with orbitals high in p character.  When forced to bond to a double bond $\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14057/what-is-the-conformer-distribution-in-monosubstituted-fluoro-bullvalene&quot;&gt;fluorobullvalene&lt;/a&gt; example for more on why fluorine &quot;hates&quot; to be on a double bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond.  If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol.  If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These differences mean that we will have much more of the enol &lt;strong&gt;without&lt;/strong&gt; the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond.  So we have more of the unsubstituted enol in the chlorine case &lt;strong&gt;and&lt;/strong&gt; the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent.  These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to steric effects, they probably don't play a significant role.  One metric commonly used to compare &quot;steric size&quot; is the cyclohexane &lt;a href=&quot;http://en.wikipedia.org/wiki/A_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;A value&lt;/a&gt;.  The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-01-09T02:52:58.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23530" PostTypeId="1" CreationDate="2015-01-09T04:20:42.690" Score="0" ViewCount="60" Body="&lt;p&gt;I am looking for a graph that shows the basic deviation of a gas from the Charle's law. The graph should be for different values of constant pressures.This would help me evaluate how gases follow ideal behavior for low pressures and high temperatures.&lt;/p&gt;&#xA;" OwnerUserId="9713" LastActivityDate="2015-01-09T04:39:21.920" Title="Is there any graph that depicts the deviation of real gases from Charle's law?" Tags="&lt;homework&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23531" PostTypeId="2" ParentId="23530" CreationDate="2015-01-09T04:39:21.920" Score="1" Body="&lt;p&gt;Check out the van der Waals equation wikipedia page for a start ... it gives an explanation of one of the most common modifications of the ideal gas law that allows for modeling of deviations from ideality in the way you describe.  This approach is quite nice in my opinion, because it includes two coefficients (a and b) that account for specific deviations for a given gas ... a scales with the strength of intermolecular interactions, and b accounts for the actual volume of the gas molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to use that information to do what you want to do regarding Charles Law.  The graph on the page shows isotherms, but you should be able to generate the isobar curves that you are after with a little bit of work.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T04:39:21.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23532" PostTypeId="2" ParentId="22454" CreationDate="2015-01-09T05:14:44.263" Score="3" Body="&lt;p&gt;The following excerpt regarding the management of alkali metals is from: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Management of time-sensitive chemicals (I): Misconceptions leading to incidents by Jim Bailey, David Blair, Lydia Boada-Clista, Dan Marsick, David Quigley, Fred Simmons and Helena Whyte&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Slow reactions of alkali metals and their alloys with oxygen to form oxides and superoxides have been well documented*. Even when these metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react. Potassium, under these circumstances, forms yellow or orange coating that can explode or catch fire upon cutting. NaK, a eutectic mixture of potassium and sodium, can undergo a similar reaction. Lithium stored under dry nitrogen can react with the nitrogen to form a nitride. Formation of the nitride is autocatalytic and can eventually lead to autoignition of the lithium. Fires of alkali metals are extremely dangerous and difficult to extinguish.&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Management of time-sensitive chemicals (II): Their identification, chemistry and management,Jim Bailey,David Blair,Lydia Boada-Clista,Dan Marsick,David Quigley,Fred Simmons,Helena Whyte: &lt;strong&gt;Chemical Health and Safety&lt;/strong&gt;, Volume 11, Issue 6, November–December 2004, Pages 17–24&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've come to the realization that if you plan to keep chemicals around you need to know all the risks and provide effective management including a plan for periodic inspection.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-01-09T05:14:44.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23533" PostTypeId="2" ParentId="23515" CreationDate="2015-01-09T05:23:15.073" Score="1" Body="&lt;p&gt;In some steps, a red light photon has enough energy (~0.5 e.v.) to raise an electron to a higher, metastable orbit (activating it), but might not actually remove the electron from the molecule (see &lt;a href=&quot;http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120072/bio13.swf::Photosynthetic+Electron+Transport+and+ATP+Synthesis&quot; rel=&quot;nofollow&quot;&gt;http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120072/bio13.swf::Photosynthetic+Electron+Transport+and+ATP+Synthesis&lt;/a&gt; (very clear), &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html&lt;/a&gt; and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-09T05:23:15.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23534" PostTypeId="1" AcceptedAnswerId="23535" CreationDate="2015-01-09T05:25:31.853" Score="19" ViewCount="26706" Body="&lt;p&gt;I recall seeing chemistry demonstrations where a significant amount of a solute is dissolved in a solvent and the solution volume barely changes. Questions regarding this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What solvent/solute combinations best demonstrate this effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it possible to wind up with less solution, in other words Is it possible to have Volume(solution) &amp;lt; Volume(solvent) + Volume(solute) ?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="9961" LastEditDate="2018-02-19T16:22:25.737" LastActivityDate="2021-03-01T07:33:50.477" Title="How does volume contraction in solvent mixing work?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23535" PostTypeId="2" ParentId="23534" CreationDate="2015-01-09T06:05:21.723" Score="12" Body="&lt;p&gt;Yes this happens fairly regularly. It is a deviation from ideal behavior associated with the property of partial molar volume.  For an ideal solution, the total volume always equals the summed volumes of the pure components, but ideal behavior neglects several important facts about real molecules, to wit, that they may have different shapes, and may also experience stabilizing or destabilizing interactions with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly easy to see how molecular shapes and sizes could affect this property. Macroscopic objects experience similar effects.  Suppose I have a bucket &quot;full&quot; of golf balls and a bucket full of sand, if I work carefully, I can pour quite a lot of sand into the bucket of golf balls because the smaller size of the sand grains allows them to fit into the gaps between the golf balls.  Another example: Suppose you have a 10 liter bucket of golf balls and a 10 liter bucket of tiny cubes packed tightly together. If I then try to pour those two buckets into a 20 liter container simultaneously, it almost certainly will not be able to fit all of the balls and cubes, because they will not pack as efficiently in the 20 L container, costing some space.  These are extreme examples, but illustrate one way that different shapes can play a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differences in intermolecular interactions can play an even bigger role.  If you are dissolving salt in water, the individual ions making up the salt crystal can be solvated very efficiently by the polar water molecules, which experience a strong attractive interaction with the ions.  (Recall that the O and H-atoms in water have partial negative and positive charges, respectively.)  This results in the solution having a significantly smaller volume than expected based on the summed volumes of the pure salt and water used to make the solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial molar volume is not too hard to measure experimentally. For something like aqueous sodium chloride it can be determined in a fairly straightforward way from precise measurements of the densities of a series of solutions with increasing concentrations.  However it is worth noting that the volume and mass measurements used for the calculation must be very precise, yielding measured densities with 5 or more significant digits.  This requires careful use of volumetric glassware and precision balances, and the temperature of the solutions must be kept constant throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For purposes of demonstrating the qualitative effect, you could just use water and ethanol (or isopropyl alcohol), which also have a significant negative excess volume. (see the first graph on the wikipedia page for partial molar volume for details).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison you could use mixtures of benzene and toluene, which behave almost ideally with respect to volume effects.  &lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="4945" LastEditDate="2015-01-09T08:11:18.027" LastActivityDate="2015-01-09T08:11:18.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23536" PostTypeId="2" ParentId="23494" CreationDate="2015-01-09T06:33:16.797" Score="3" Body="&lt;p&gt;The crystal structure shows a strangely distorted molecule with only $C_\mathrm{i}$ symmetry. Values have quite a wide range, $\angle(\ce{CCH})\approx 100-120^\circ$. See below for all values and xyz coordinates to read in with a molecular viewer. Source:&#xA; &lt;a href=&quot;http://dx.doi.org/10.1107/S0567740873002074&quot; rel=&quot;nofollow noreferrer&quot;&gt;R. Kahn, R. Fourme, D. André and M. Renaud, &lt;em&gt;Acta Cryst.&lt;/em&gt; &lt;strong&gt;1973&lt;/strong&gt;, &lt;em&gt;B29&lt;/em&gt;, 131-138&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A DF-BP86/def2-TZVPP calculation (gas phase, 0 K) gives a highl $D_\mathrm{3d}$ symmetric structure with &#xA;\begin{array}{ll}&#xA;\angle(\ce{CCH_{ax}}) &amp;amp; =109.1^\circ,\\&#xA;\angle(\ce{CCH_{eq}}) &amp;amp; =110.3^\circ,\\&#xA;\angle(\ce{CCC}) &amp;amp;=111.5^\circ,\\&#xA;\angle(\ce{H_{eq}CH_{ax}}) &amp;amp;=106.4^\circ.\\&#xA;\end{array}&#xA; Also Bond length differ very slightly (and impossibly to determine at room temperature and pressure), with &#xA;\begin{array}{ll}&#xA;\mathbf{d}(\ce{CC}) &amp;amp;=1.537~\mathrm{\mathring{A}},\\&#xA;\mathbf{d}(\ce{CH_{ax}}) &amp;amp;=1.105~\mathrm{\mathring{A}},\\&#xA;\mathbf{d}(\ce{CH_{eq}}) &amp;amp;=1.103~\mathrm{\mathring{A}}.\\&#xA;\end{array} &#xA;The tiny differences can be explained with &lt;a href=&quot;https://chemistry.stackexchange.com/q/15671/4945&quot;&gt;Bent's rule&lt;/a&gt;.&lt;br&gt;&#xA;In principle the assignment Goeff already gave you is absolutely valid, especially if you take the dynamic nature of the molecules into account that they have at standard temperature and pressure.&lt;br&gt;&#xA;Also see xyz coordinates for visualisation below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Appendix&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Crystal structure&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/N7wMK.jpg&quot; alt=&quot;molecular structure of cyhex&quot;&gt;&#xA;\begin{array}{lr}&#xA;R(1-2)         	&amp;amp;1.528 \\         &#xA;R(1-3)         	&amp;amp;1.519 \\         &#xA;R(1-4)         	&amp;amp;0.884 \\         &#xA;R(1-5)         	&amp;amp;1.146 \\         &#xA;R(2-6)         	&amp;amp;1.055 \\       &#xA;R(2-7)         	&amp;amp;0.929 \\     &#xA;R(2-10)        	&amp;amp;1.521 \\   &#xA;R(3-8)         	&amp;amp;1.100 \\ &#xA;R(3-9)         	&amp;amp;0.944 \\&#xA;R(3-11)        	&amp;amp;1.521 \\&#xA;R(10-12)       	&amp;amp;1.519 \\&#xA;R(10-13)       	&amp;amp;1.100 \\&#xA;R(10-14)       	&amp;amp;0.944 \\&#xA;R(11-12)       	&amp;amp;1.528 \\&#xA;R(11-15)       	&amp;amp;1.055 \\&#xA;R(11-16)       	&amp;amp;0.929 \\&#xA;R(12-17)       	&amp;amp;0.884 \\&#xA;R(12-18)       	&amp;amp;1.146 \\&#xA;A(2-1-3)       	&amp;amp;110.4 \\&#xA;A(2-1-4)       	&amp;amp;106.7 \\&#xA;A(2-5)       	&amp;amp;105.8          \\&#xA;A(1-2-6)       	&amp;amp;106.6          \\&#xA;A(1-2-7)       	&amp;amp;118.2          \\&#xA;A(1-2-10)      	&amp;amp;111.3          \\&#xA;A(3-1-4)       	&amp;amp;103.0          \\&#xA;A(3-1-5)       	&amp;amp;114.0          \\&#xA;A(1-3-8)       	&amp;amp;110.6          \\&#xA;A(1-3-9)       	&amp;amp;116.9          \\&#xA;A(1-3-11)      	&amp;amp;112.4          \\&#xA;A(4-1-5)       	&amp;amp;116.7          \\&#xA;A(6-2-7)       	&amp;amp;109.5          \\&#xA;A(6-2-10)      	&amp;amp;110.6          \\&#xA;A(7-2-10)      	&amp;amp;100.5          \\&#xA;A(2-10-12)     	&amp;amp;112.4          \\&#xA;A(2-10-13)     	&amp;amp;109.1          \\&#xA;A(2-10-14)     	&amp;amp;100.4          \\&#xA;A(8-3-9)       	&amp;amp;106.9          \\&#xA;A(8-3-11)      	&amp;amp;109.1          \\&#xA;A(9-3-11)      	&amp;amp;100.4          \\&#xA;A(3-11-12)     	&amp;amp;111.3          \\&#xA;A(3-11-15)     	&amp;amp;110.6          \\&#xA;A(3-11-16)     	&amp;amp;100.5          \\&#xA;A(12-10-13)    	&amp;amp;110.6          \\&#xA;A(12-10-14)    	&amp;amp;116.9          \\&#xA;A(10-12-11)    	&amp;amp;110.4          \\&#xA;A(10-12-17)    	&amp;amp;103.0          \\&#xA;A(10-12-18)    	&amp;amp;114.0          \\&#xA;A(13-10-14)    	&amp;amp;106.9          \\&#xA;A(12-11-15)    	&amp;amp;106.6          \\&#xA;A(12-11-16)    	&amp;amp;118.2          \\&#xA;A(11-12-17)    	&amp;amp;106.7          \\&#xA;A(11-12-18)    	&amp;amp;105.8          \\&#xA;A(15-11-16)    	&amp;amp;109.5          \\&#xA;A(17-12-18)    	&amp;amp;116.7          \\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C        2.329589000      3.011988000     -0.044238000&#xA;C        3.630888000      2.632672000      0.661249000&#xA;C        1.367908000      1.836044000     -0.049671000&#xA;H        2.524936000      3.113740000     -0.900292000&#xA;H        1.933922000      3.936128000      0.506802000&#xA;H        3.388552000      2.469096000      1.674854000&#xA;H        4.348450000      3.220000000      0.605369000&#xA;H        1.030575000      1.607424000      0.972471000&#xA;H        0.592407000      1.921052000     -0.581309000&#xA;C        4.247092000      1.383956000      0.049671000&#xA;C        1.984112000      0.587328000     -0.661249000&#xA;C        3.285411000      0.208012000      0.044238000&#xA;H        4.584425000      1.612576000     -0.972471000&#xA;H        5.022593000      1.298948000      0.581309000&#xA;H        2.226448000      0.750904000     -1.674854000&#xA;H        1.266550000     -0.000000000     -0.605369000&#xA;H        3.090064000      0.106260000      0.900292000&#xA;H        3.681078000     -0.716128000     -0.506802000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Computed (DF-BP86/def2-TZVPP)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HYmyL.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C       -1.270112000     -0.733300000      0.229506000&#xA;C        0.000000000     -1.466599000     -0.229506000&#xA;C       -1.270112000      0.733300000     -0.229506000&#xA;H       -1.328386000     -0.766944000      1.332131000&#xA;H       -2.167276000     -1.251277000     -0.145050000&#xA;H        0.000000000     -1.533888000     -1.332131000&#xA;H        0.000000000     -2.502555000      0.145050000&#xA;H       -1.328386000      0.766944000     -1.332131000&#xA;H       -2.167276000      1.251277000      0.145050000&#xA;C        1.270112000     -0.733300000      0.229506000&#xA;C        0.000000000      1.466599000      0.229506000&#xA;C        1.270112000      0.733300000     -0.229506000&#xA;H        1.328386000     -0.766944000      1.332131000&#xA;H        2.167276000     -1.251277000     -0.145050000&#xA;H        0.000000000      1.533888000      1.332131000&#xA;H        0.000000000      2.502555000     -0.145050000&#xA;H        1.328386000      0.766944000     -1.332131000&#xA;H        2.167276000      1.251277000      0.145050000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-01-09T06:33:16.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23537" PostTypeId="2" ParentId="23534" CreationDate="2015-01-09T06:59:17.377" Score="9" Body="&lt;p&gt;1.Volumes do not add when you mix things together because the intermolecular forces of a mixture are different than the ones in the pure substances. The way this is handled is by using a property of mixtures called the &quot;partial molar volume.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.There are a couple of reasons why volumes of solution are not additive. One important one is that some chemical reaction occurs. The second is especially important if one component is water. Liquid water has an open structure, that is, the water molecules even in the liquid form molecule sized structures of lower density than the mean. The addition of ionic and/or polar solutes can &quot;collapse&quot; this structure resulting in an increased density. To complicate matters, the solute may also have structure that may be smaller than or greater than the average density of the solid. This will result in an expansion (or contraction) when the solute is dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no good theoretical models, of a general nature, to predict or estimate whether a solvent/solute combination will have a positive or negative &quot;excess&quot; volume of solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.When $50\mathrm{mL}$ of water and $50\mathrm{mL}$ of ethanol are combined in a $100\mathrm{mL}$ graduated cylinder, the resulting mixture has a volume of $\approx 97\mathrm{mL}$.&#xA;Intermolecular interactions between water molecules in a pure sample involve hydrogen bonding, though the details of these complex interactions are not completely understood.  Some models predict transient ice-like structures within liquid water, where the local density of the sample is temporarily diminished. Ethanol and water molecules are also attracted to each other through hydrogen bonding. The ethanol likely interrupts the transient ice-like structures present in liquid water, resulting in a slight contraction of the sample.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-09T06:59:17.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23538" PostTypeId="2" ParentId="14269" CreationDate="2015-01-09T07:11:02.500" Score="3" Body="&lt;p&gt;Passing aluminium oxide through a hot enough flame will result in the formation of sapphires, ruby and corundum (amongst others) type gem stones also known as the 'Verneuil process'.&lt;br /&gt;&#xA;A lot of man made stones are made in similar way. Here is a &lt;a href=&quot;https://www.youtube.com/watch?v=Qp0u0Vp2jQU&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; to explain this.&lt;/p&gt;&#xA;" OwnerUserId="12667" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-01-09T07:36:25.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23539" PostTypeId="1" CreationDate="2015-01-09T10:55:04.727" Score="1" ViewCount="2793" Body="&lt;p&gt;What kind of compound absorption the 365nm UV light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;compounds that has benzene ring or conjugated pi bond, make a fluorescence on 254 nm UV on TLC. &#xA;what kind of moiety(functional group) make fluorescence on 365nm UV on TLC&lt;/p&gt;&#xA;" OwnerUserId="11326" LastEditorUserId="418" LastEditDate="2015-01-09T16:34:56.960" LastActivityDate="2015-05-19T14:40:31.070" Title="What kind of compound absorption the 365nm UV light on TLC" Tags="&lt;analytical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-01-10T13:55:09.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="23540" PostTypeId="1" CreationDate="2015-01-09T12:29:11.027" Score="1" ViewCount="60" Body="&lt;p&gt;If I’ve got a reactor containing PE-Polyethylene $(\ce{{C2}$H_\text{4})n$}$ (solid) and the rest is filled with nitrogen gas. The system is &lt;strong&gt;closed&lt;/strong&gt;.  Supposing we have an initial temperature of $30~\mathrm{°C}$ and gradually increase the temperature to $180~\mathrm{°C}$. We’ll observe a change in state of the PE, passing from solid to liquid, at this point the &lt;em&gt;molecular structure is the same&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If temperature is increased beyond &lt;strong&gt;$250~\mathrm{°C}$ pyrolysis&lt;/strong&gt; will occur on PE and syngas will be produced, at this point the gas molecules will have different structures compared to $(\ce{{C2}$H_\text{4})n$}$, thus &quot;we’ll not observe&quot; a change of phase from liquid to gas maintaining  the same molecular structure. Gas molecules will nave n types of molecular structures.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible by &lt;strong&gt;varying PVT&lt;/strong&gt; to have a change of phase, from liquid to gas, of PE and &lt;strong&gt;maintain the same molecular structure&lt;/strong&gt;. Just like with water-&gt;vapor? &lt;/li&gt;&#xA;&lt;li&gt;If this is not possible, why? what are the topics to study?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2085" LastEditorUserId="2085" LastEditDate="2015-01-09T18:06:01.293" LastActivityDate="2015-01-09T18:06:01.293" Title="Can state of matter from liquid to gas maintain the same molecular structure in PE or in polymers in general, like water does?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23541" PostTypeId="1" AcceptedAnswerId="23568" CreationDate="2015-01-09T12:45:28.090" Score="1" ViewCount="339" Body="&lt;p&gt;$$G=G(P,T,n)$$&#xA;$$\mathrm dG=V\,\mathrm dP-S\,\mathrm dT+\mu\,\mathrm  dn=\left(\frac{\partial G}{\partial P}\right)_{T,n}\,\mathrm dP+\left(\frac{\partial G}{\partial T}\right)_{P,n}\,\mathrm dT+\left(\frac{\partial G}{\partial n}\right)_{T,P}\,\mathrm dn$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows open systems to be considered (where $\mathrm dn$ does not equal zero). However, can enthalpy, internal energy and Helmholtz free energy also be treated in this way to allow for open systems?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2016-11-13T14:09:27.020" LastActivityDate="2016-11-13T14:09:27.020" Title="Gibbs free energy can be expressed as a function of P,T and n but are enthalpy and internal energy also (partially) functions of n?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="23542" PostTypeId="1" CreationDate="2015-01-09T13:14:02.367" Score="2" ViewCount="215" Body="&lt;p&gt;I want to calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; for hydrogenation of benzene and producing cyclohexane using MM+ Force Field Energy. I am using Hyperchem for energy calculation and optimization. I calculated the energy for cyclohexane and then I calculated the energy for benzene. But the difference of two energies is totally false. The calculated energy is &lt;span class=&quot;math-container&quot;&gt;$8~\mathrm{kcal/mol}$&lt;/span&gt; however the real value is &lt;span class=&quot;math-container&quot;&gt;$-49~\mathrm{kcal/mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Is it a true way to calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt;? If not, how can I do it ?&lt;/p&gt;&#xA;" OwnerUserId="12669" LastEditorUserId="69830" LastEditDate="2021-05-25T06:44:34.377" LastActivityDate="2021-05-25T06:44:34.377" Title="How to calculate a reaction's DeltaH using a MM forcefield calculator software?" Tags="&lt;computational-chemistry&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23543" PostTypeId="2" ParentId="18462" CreationDate="2015-01-09T13:31:56.750" Score="4" Body="&lt;p&gt;The units of a partial derivative $\displaystyle\left(\frac{\partial Y}{\partial X}\right)_Z$ are the units of $Y$ divided by the units of $X$.  The units of $Z$ don't matter.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Isothermal compressibility is $-\displaystyle\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$, so the SI units are reciprocal pressure, $\rm Pa^{-1}$ or $\rm m^2\ N^{-1}$.&lt;/li&gt;&#xA;&lt;li&gt;Heat capacity at constant volume is $ \displaystyle\left(\frac{\partial E}{\partial T}\right)_V$, so its SI units are energy/temperature or $\rm J\ K^{-1}$. When you say &lt;em&gt;specific&lt;/em&gt; heat capacity, that means per unit mass, so the specific heat capacity at constant volume is $\rm J\ kg^{-1} K^{-1}$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-09T16:46:54.867" LastActivityDate="2015-01-09T16:46:54.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23544" PostTypeId="2" ParentId="23542" CreationDate="2015-01-09T13:35:33.763" Score="6" Body="&lt;p&gt;What you are forgetting is that the reaction will also involve hydrogen.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C6H6 + 3 H2 -&amp;gt; C6H12}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In principle, the &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; should be 0, but in practice, that may not be true with a force field. The energies from molecular mechanics are seldom accurate enough to compute reaction energies. They are parameterized on a set of molecules, but the errors are usually large, both systematic and random errors.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="69830" LastEditDate="2021-05-25T06:43:17.847" LastActivityDate="2021-05-25T06:43:17.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23545" PostTypeId="1" CreationDate="2015-01-09T14:30:31.110" Score="2" ViewCount="758" Body="&lt;p&gt;A few days ago, I learned the lengthy process of salt analysis during my practicals. The process is based on deductive reasoning but is too lengthy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am thinking is, can't it be done using the fact that all molecules have unique emission spectra? I searched for it and found that it's the same thing as the flame test which was covered in my class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a flame test is only used for metal ions. So, my question is: Is there a way in which I can determine the metal ion as well as the radical part(anion part) of the salt using spectroscopic methods? What equipment would I need for it?&lt;/p&gt;&#xA;" OwnerUserId="7249" LastEditorUserId="402" LastEditDate="2017-03-16T19:43:08.833" LastActivityDate="2017-03-16T19:43:08.833" Title="Salt analysis using spectroscopy" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23546" PostTypeId="2" ParentId="22346" CreationDate="2015-01-09T14:45:39.213" Score="1" Body="&lt;p&gt;Your energy conservation equation is $$q_{rxn} + q_{soln} = 0$$ so $$q_{rxn} = -q_{soln} = -m_{soln}c_{soln}\Delta T_{soln}$$ Note the minus sign; if $\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c_{soln} \approx c_{\rm H_2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more precision, though, you'll need to measure or look up $c_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-21T21:27:13.240" LastActivityDate="2015-01-21T21:27:13.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23547" PostTypeId="1" AcceptedAnswerId="23552" CreationDate="2015-01-09T14:48:01.983" Score="11" ViewCount="483" Body="&lt;p&gt;Would a [2]catenate whose each ring consist of a different number of methylene groups only, be optically active if one hydrogen atom of a random methylene group is replaced with e.g. chlorine? I think yes, but I need a second opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compounds attached seem to be enantiomers. I don't see how, by rotation, they could be super-imposable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Od9cM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="402" LastEditDate="2020-03-01T08:28:10.637" LastActivityDate="2020-03-01T08:28:10.637" Title="Catenane optical activity" Tags="&lt;stereochemistry&gt;&lt;supramolecular-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23548" PostTypeId="1" AcceptedAnswerId="23557" CreationDate="2015-01-09T14:53:13.277" Score="1" ViewCount="207" Body="&lt;p&gt;I want to compare two type of carbons accepting OH in adamantane which I've shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfHB4.png&quot; alt=&quot;Atom positions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the blue position is preferred. But I wanna prove it with energy optimization and MM Forcefield. I thought it is related to transition state and corresponding carbocations. But MM+ ignores all charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I test it using ForceField softwares ?&lt;/p&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="10185" LastEditDate="2015-01-09T16:04:00.913" LastActivityDate="2015-01-09T17:45:47.617" Title="Comparing adamantane carbon types for accepting OH using MM ForceField" Tags="&lt;reaction-mechanism&gt;&lt;energy&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23551" PostTypeId="1" AcceptedAnswerId="23554" CreationDate="2015-01-09T15:45:16.860" Score="1" ViewCount="1302" Body="&lt;p&gt;I am really confused between them. I know that an enantiomer is an isomer having optical activity, non super-imposability of image on original configuration, same molecular formula(Correct me if I am wrong). The thing that is causing me trouble is what diffrentiates this from diasteriomerism. It would be appreciated if you provide an example to clarify like a fischer projection&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2015-01-09T17:52:25.423" Title="Enantiomerism and Diasteriomerism" Tags="&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23552" PostTypeId="2" ParentId="23547" CreationDate="2015-01-09T16:22:23.463" Score="13" Body="&lt;p&gt;If a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral (achiral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The all hydrocarbon [2]catenane will be achiral. The molecule contains a plane of symmetry; picture a plane that contains one of the rings and bisects the other ring. In the following image, you can see that the plane of the screen contains the ring on the right and bisects the ring on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gt8Vj.gif&quot; alt=&quot;A [2]catenane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say that if we replaced the circles with cyclohexane rings, because of their chair-like shape, the plane of symmetry is lost.  That is true, but you can flatten one of the cyclohexane rings and the plane of symmetry is restored.  Remember, if any thermally accessible conformation of a molecule is achiral, then the molecule is achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes it helps to build a model or use a Newman projection to help with the stereochemical analysis.  Here is such a projection for your chloro-[2]catenane.  Note that it still possesses a plane of symmetry (a plane perpendicular to the screen).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEQZe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go further and add another chlorine atom, but now to the back ring, then the molecule becomes chiral — just like 1,3-dichloroallene.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The compounds attached seem to be enantiomers. I don't see how by rotation they could be super-imposable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the bottom figure in my second drawing.  I've taken your interlocking 5- and 6-membered catenane and rotated the 6-membered ring a bit further along (a conformational isomer; remember, as I mentioned above, if a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral [achiral]).  The plane of the screen now contains the 5-membered ring and it bisects the cyclohexane ring.  The plane of the screen also contains the methyl group.  The plane of the screen is a symmetry element of this molecule; therefore this molecule is achiral.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2020-03-01T07:57:26.747" LastActivityDate="2020-03-01T07:57:26.747" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23553" PostTypeId="2" ParentId="23539" CreationDate="2015-01-09T16:34:19.387" Score="1" Body="&lt;p&gt;&lt;em&gt;Your question is almost impossible to answer - it simply is too broad.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Excitation spectra of fluorescent organic compounds typically aren't sharp lines but consist of one (or more) bands. Even if the maximum of the excitation spectrum isn't located at $\lambda = \textrm{365 nm}$, &lt;strong&gt;monochromatic&lt;/strong&gt; irradiation at that wavelength will excite &lt;strong&gt;a lot&lt;/strong&gt; of different compounds (and even compound classes) which then deactivate by fluorescence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your light source does not emit monochromatic radiation! Hand lamps used to inspect TLCs typically have coated low-pressure mercury lamps. While the maximum of the radiation density might be centred around 360 nm, the full with at half maximum is - roughly estimated - about 30 nm or more. This means that there's even more compounds that might get excited by the emission of the hand lamp.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In summary, there's no exhaustive answer to your question, but just to name some compounds that are potential candidates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NQ99f.png&quot; alt=&quot;possible structures that match UV requirements&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diphenylacetylene (&lt;strong&gt;1&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;1,6-Diphenylhexatriene (DPH, &lt;strong&gt;2&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;4',6-diamidino-2-phenylindole (DAPI, &lt;strong&gt;3&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;6,8-difluoro-7-hydroxy-4-methylcoumarin (DIFMU, &lt;strong&gt;4&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-19T14:40:31.070" LastActivityDate="2015-05-19T14:40:31.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23554" PostTypeId="2" ParentId="23551" CreationDate="2015-01-09T16:38:56.240" Score="3" Body="&lt;p&gt;Your definition of enantiomer is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple further classification might be helpful.  Let's take 2 molecules with the same molecular formula.  These molecules are either constitutional (or structural) isomers or stereoisomers.  If they have different structures (or connectivity patterns)  and are separated by an energy barrier, then they are constitutional isomers.  If they have the same structure or connectivity pattern, then they are stereoisomers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among stereoisomers there are 3 classes that we can place a pair of molecules into.  Take any two stereoisomers and compare them, they are either&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the same (2 identical molecules)&lt;/li&gt;&#xA;&lt;li&gt;mirror images of one another (enantiomers)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If they are not the same or enantiomers, then they are&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;different, and we call them diastereomers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-09T16:38:56.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23555" PostTypeId="2" ParentId="23551" CreationDate="2015-01-09T17:27:17.090" Score="1" Body="&lt;p&gt;One additional thing to add to @ron's answer is that &lt;em&gt;diastereomers&lt;/em&gt; have different chemical properties .. they can be distinguished using conventional spectroscopic techniques, or even by measuring thermodynamic parameters (at least in principle .. sometimes the differences are very small and hard to resolve).  They can also have different chemical reactivities.  Apart from their different optical activities, &lt;em&gt;enantiomers&lt;/em&gt; have identical chemical properties ... that is to say, you have to do an experiment that exploits their chiral natures to tell them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point is that diasteromers always have at least two chiral centers ... a molecule with a single chiral center will have two enantiomers.  A molecule having two chiral centers will have two enantiomers and two disasteromers.  There is a nomenclature convention that can be used to index the various choices: each stereocenter is labeled &quot;R&quot; or &quot;S&quot; depending on whether the components are arranged clockwise (R) or counter-clockwise (S) around it, according to a conventional order of preference.  In the case of a molecule with two stereo centers, the (R,R) and (S,S) stereoisomers are mirror images, and therefore are enantiomers, and the (R,S) and (S,R) stereoisomers are not mirror images, and therefore are &lt;em&gt;usually&lt;/em&gt; diastereomers. As pointed out by @ron in the comments, if the molecule is highly symmetrical, the (S,R) and (R,S) forms may actually be indistinguishable, which would be a &quot;meso&quot; compound.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="12658" LastEditDate="2015-01-09T17:52:25.423" LastActivityDate="2015-01-09T17:52:25.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23556" PostTypeId="1" AcceptedAnswerId="23560" CreationDate="2015-01-09T17:32:20.253" Score="3" ViewCount="751" Body="&lt;p&gt;Does anyone happen to know who was the first to use Hooke's law to model vibrational absorption peak frequencies/wavenumbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Hooke's law is first outlined here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R. Hooke, Potentia restitutiva, or spring, in Lectiones cutlerianae, or a Collection of lectures: physical, mechanical, geographical, and astronomical, made before the Royal Society on several occasions ... to which are added divers miscellaneous discourses, ed. by J. Martyn (John Martyn, London, 1679), p. 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'd like to know about whose idea it was to use it for spectra.&lt;/p&gt;&#xA;" OwnerUserId="12676" LastEditorUserId="7448" LastEditDate="2015-01-09T18:14:25.237" LastActivityDate="2015-02-08T19:02:38.457" Title="Hooke's Law for absorption spectra" Tags="&lt;spectroscopy&gt;&lt;history-of-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23557" PostTypeId="2" ParentId="23548" CreationDate="2015-01-09T17:45:47.617" Score="2" Body="&lt;p&gt;One thing to try would be just to build the two isomers and compare their total energies .. that would at least show which position was preferred thermodynamically (i.e. more stable).  It sounds like you are more interested in the kinetic consideration though (i.e. which reacts faster) ... in that case, looking at the relative energies of the transition states will give you some basis for rationalizing the preference.  I am not sure how good MM is at dealing with transition states, however ... you would need to rely rather heavily on your chemical intuition to find the correct structures, and even then, the bond angles and bond lengths may be distorted to such an extent that your forcefield is not well parameterized to deal with them.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T17:45:47.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23558" PostTypeId="1" CreationDate="2015-01-09T17:49:41.330" Score="1" ViewCount="867" Body="&lt;p&gt;is there a database somewhere of materials that are luminescent  when exposed to Infra Red radiation?&lt;/p&gt;&#xA;" OwnerUserId="12677" LastEditorUserId="418" LastEditDate="2015-01-10T17:30:10.953" LastActivityDate="2020-01-08T16:05:36.113" Title="Luminescent materials under IR radiation?" Tags="&lt;spectroscopy&gt;&lt;reference-request&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23559" PostTypeId="1" AcceptedAnswerId="23563" CreationDate="2015-01-09T18:33:32.647" Score="2" ViewCount="1056" Body="&lt;p&gt;I know that when &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^{2+}}$&lt;/span&gt; ions react to &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt; atoms, it becomes &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; + &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt;. My question is, why?&lt;/p&gt;&#xA;&lt;p&gt;Copper atoms normally have 29 electrons. This means that in the K-shell there are 2 electrons, in &lt;span class=&quot;math-container&quot;&gt;$L = 8$&lt;/span&gt; electrons, in &lt;span class=&quot;math-container&quot;&gt;$M = 18$&lt;/span&gt; electrons and in &lt;span class=&quot;math-container&quot;&gt;$N = 1$&lt;/span&gt; electron. (Yes I know that there is a system with s and other letters, but I haven't been taught in that way at my school). Therefore &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^{2+}}$&lt;/span&gt; should have 27 electrons, which means that the shells are assigned like this: &lt;span class=&quot;math-container&quot;&gt;$K = 2, L = 8, M = 17$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So when the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^{2+}}$&lt;/span&gt; ions react with &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;, why does it want to become &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;? Because &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; doesn't have a complete electron shell at the end. Why doesn't it want to become &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu+}$&lt;/span&gt;, because then it would have a complete shell at the end (&lt;span class=&quot;math-container&quot;&gt;$K=2, L=8, M=18$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;I know that when you take away electrons from copper, it does it in a special way. I'm 15 years old and not english so I'm really not an expert or anything, I'm just wondering why &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^{2+}}$&lt;/span&gt; would want to become &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; when it meets iron (&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="12678" LastEditorUserId="69830" LastEditDate="2021-05-04T09:04:49.503" LastActivityDate="2021-05-04T09:04:49.503" Title="Why should the redox reaction happen between the copper ions and iron atoms?" Tags="&lt;redox&gt;&lt;ions&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23560" PostTypeId="2" ParentId="23556" CreationDate="2015-01-09T18:38:36.953" Score="3" Body="&lt;p&gt;Okay, I think I found an answer. It's a set of four papers which were published consecutively in &lt;em&gt;Physical Review&lt;/em&gt; 36 (1930). Interestingly, each was submitted on June 23rd, 1930, which makes me wonder about the back story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first comes from General Motors Corporation Research Laboratories in Detroit (&lt;a href=&quot;http://en.wikipedia.org/wiki/Charles_F._Kettering&quot; rel=&quot;nofollow&quot;&gt;Kettering&lt;/a&gt;, Shutts, and Andrews). The other three papers are out of Johns Hopkins (Donald Andrews, from Chemistry, and Yates, from Mathematics). I suppose Andrews could have graduated from Hopkins and found that GMCRL was working on the same thing; or perhaps he had a dual appointment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, there appears to be &lt;a href=&quot;http://www.google.com/patents/US1797799&quot; rel=&quot;nofollow&quot;&gt;a patent suggesting that L. W. Shutts invented shock absorbers&lt;/a&gt;, so it makes immediate sense that he would propose the spring law as an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Donald Andrews (or at least Johns Hopkins) is the obvious link between GMCRL and Robert C. Yates. Did Andrews know Yates? Was someone about to be scooped? The answer to &lt;em&gt;that&lt;/em&gt; is probably beyond the scope of this question.&lt;/p&gt;&#xA;" OwnerUserId="12676" LastActivityDate="2015-01-09T18:38:36.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23561" PostTypeId="2" ParentId="23558" CreationDate="2015-01-09T18:44:00.220" Score="1" Body="&lt;p&gt;I am not aware of any materials that have this property.  There are certainly molecules that emit fluorescence in the IR, when irradiated with shorter wavelength light, but I can't think of any that would emit in the IR after being irradiated in the IR in a manner that can be called &quot;luminescence&quot;.  Luminescence describes the specific phenomenon where a photon of a particular wavelength had excited a molecule to a higher-energy electronic state, the molecule then relaxes to a lower energy state, with emission of a second photon, which is always of longer wavelength than the original photon.  Because molecular electronic transitions are typically in the UV-visible region of the spectrum, the phenomenon of luminescence is generally restricted to those spectral regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One particular issue with IR luminescence is that it would be pretty difficult to tell apart from normal thermal IR emission ... luminescence refers specifically to light that is emitted outside of the thermal spectrum of an object.  That means that an object would need to be pretty cold initially, in order to distinguish luminescence in the IR region.  And again, since the emitted wavelength is necessarily longer than that of the absorbed wavelength, a molecule that was luminescent under IR irradiation would emit deeper in the IR than where it absorbed, making all of this harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, there are certainly molecules with electronic transitions that absorb in the IR, and if you got them cold enough and used an appropriate spectrometer, you might be able to observe IR luminescence from them.  It's something that would be restricted to pretty highly specified research labs though.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T18:44:00.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23563" PostTypeId="2" ParentId="23559" CreationDate="2015-01-09T18:56:26.143" Score="2" Body="&lt;p&gt;This is a good question, but you may have gotten off on the wrong track by thinking about the electronic shell structure (although what you have written seems correct, if a bit old-fashioned).  The relevant phenomenon here is the relative abilities of the metal ions to oxidize (or remove electrons from) each other.  For your specific example, the copper ion is a better oxidizing agent than the iron atom.  Therefore it is able to take two electrons away from the iron atom, oxidizing it and leaving the iron 2+ ion behind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &lt;em&gt;why&lt;/em&gt; this happens, at your current stage of education the best answer is probably &quot;because iron is easier to oxidize than copper&quot;.  You can look up the electrochemical activity series to find out how various metals relate to each other in terms of how easy they are to oxidize.  Copper is one of the harder ones to oxidize, but if you were to mix copper metal with silver ions, you would end up with silver metal and copper ions, because silver is even harder to oxidize than copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that help?&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T18:56:26.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23564" PostTypeId="1" CreationDate="2015-01-09T19:15:46.970" Score="4" ViewCount="368" Body="&lt;p&gt;I recently got a decent stereo microscope. Besides using the scope to teach my children, I have been thinking of picking up something new to learn myself. I have been interested about chemistry for a long time but never studied it after upper secondary school. I was thinking would there be some experiments that would allow me to learn more microscopy and chemistry at the same time? I have ~10  years of experience in machine vision, signal processing, neuroscience, programming etc., so research-grade problems are welcome too.&lt;/p&gt;&#xA;" OwnerUserId="12680" LastEditorUserId="7448" LastEditDate="2015-01-09T19:47:44.543" LastActivityDate="2015-01-12T15:53:23.257" Title="Experiments in chemistry involving use of stereo microscope?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23565" PostTypeId="1" AcceptedAnswerId="23567" CreationDate="2015-01-09T20:54:36.100" Score="2" ViewCount="440" Body="&lt;p&gt;Gibbs free energy can be defined as: $$dG=VdP-SdT+\sum_i\mu_idn_i$$ where $\mu=(\frac{\partial G}{\partial n})_{P,T}$. Last term, $\sum_i\mu_idn_i$ allows for open systems to be considered (where $dn$ does not equal zero). However, this can also be used for state functions like enthalpy too:&#xA;$$dH=TdS+VdP+\sum_i\mu_idn_i$$&#xA;But my question is why can this be used for enthalpy, considering that $\mu=(\frac{\partial G}{\partial n})_{P,T}$ (i.e it's a partial derivative of $G$ not $H$). Perhaps it's clearer for me to write the total derivative out like this: $$dH=(\frac{\partial H}{\partial S})_{P,n}dS+(\frac{\partial H}{\partial P})_{S,n}dP+(\frac{\partial G}{\partial n})_{P,T}dn$$ where I have considered a single-component open system (the component may enter or leave the system)&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-09T23:34:20.383" Title="How does the chemical potential allow for open systems to be considered?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23566" PostTypeId="1" AcceptedAnswerId="23574" CreationDate="2015-01-09T22:55:03.210" Score="5" ViewCount="877" Body="&lt;p&gt;Consider a long, vertical tube of water, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W26Qk.png&quot; alt=&quot;long tube of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P_\mathrm{atm}$ is the atmospheric pressure and $P_\mathrm{bottom}$ is the pressure at the bottom of the tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube is long enough such that $P_\mathrm{atm}$ is much less than $P_\mathrm{bottom}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that water at the surface (top of the tube) will boil at a temperature such that the water's vapor pressure equals $P_\mathrm{atm}$. Let's call this temperature $T$. For example, $T=100~^\circ\mathrm{C}$ when $P_\mathrm{atm}$ is about $1\mathrm~{atm}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: At what temperature will the water at the bottom boil? Will it be higher than $T$ or equal to $T$? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References, if available, would also be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10095" LastEditorUserId="4945" LastEditDate="2015-01-13T05:19:57.240" LastActivityDate="2015-01-13T05:19:57.240" Title="Will the water at the bottom of the tube boil at the same temperature as the water at the top of the tube?" Tags="&lt;boiling-point&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23567" PostTypeId="2" ParentId="23565" CreationDate="2015-01-09T23:34:20.383" Score="1" Body="&lt;p&gt;Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total exact differential is expressible as you had originally expected:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i&#xA;$$&#xA;  The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for any confusion my initial errors may have caused you.  Hope this clears it up.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-09T23:34:20.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23568" PostTypeId="2" ParentId="23541" CreationDate="2015-01-09T23:42:54.473" Score="2" Body="&lt;p&gt;Yes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies.  (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/23565/how-does-the-chemical-potential-allow-for-open-systems-to-be-considered/23567#23567&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any. &lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-01-09T23:42:54.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23569" PostTypeId="1" AcceptedAnswerId="23571" CreationDate="2015-01-09T23:44:09.670" Score="-9" ViewCount="508" Body="&lt;p&gt;From &quot;North Carolina Measures of Student Learning:&#xA;NC’s Common Exams Chemistry&quot; (&lt;a href=&quot;http://www.dpi.state.nc.us/docs/accountability/common-exams/released-forms/highschool/science/chemistry/common-exam.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source, Number 20&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The graph below shows a cooling curve for a sample of gas that is uniformly&#xA;  cooled from $155~^\circ\mathrm{C}$.&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/Uorew.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;  Why does the freezing point of the substance occur at $–20^\circ\mathrm{C}$?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;because the latent heat energy is absorbed by the substance as it is&#xA;  converted from a liquid to a solid&lt;/li&gt;&#xA;  &lt;li&gt;because the latent heat energy is released into the air as the substance is&#xA;  converted from a liquid to a solid&lt;/li&gt;&#xA;  &lt;li&gt;because the average kinetic energy is increasing for the substance as it is&#xA;  converted from a solid to a liquid&lt;/li&gt;&#xA;  &lt;li&gt;because the average kinetic energy is decreasing for the substance as it is&#xA;  converted from a solid to a liquid &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Its really just a simple questioning and I think the answer is 3, because the average kinetic energy is increases when converted from solid to liquid. Any reason why I'm wrong and the answer says it is 2?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2018-06-12T22:34:22.350" LastActivityDate="2018-06-12T22:34:22.350" Title="Why does the freezing point of a sample gas occur at a certain temperature?" Tags="&lt;energy&gt;&lt;heat&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-05T23:02:40.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="23570" PostTypeId="2" ParentId="23558" CreationDate="2015-01-09T23:51:49.983" Score="4" Body="&lt;p&gt;There are two types of materials I'm aware of that produce light in the way I think you want: two-photon fluorescence and photon upconversion. Two-photon fluorescence is basically the same as normal fluorescence but two photons are absorbed simultaneously, emitting a single photon when returning to the ground state. &lt;a href=&quot;http://en.wikipedia.org/wiki/Photon_upconversion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photon upconversion&lt;/a&gt; similarly involves the absorption of multiple photons, but sequentially (by various mechanisms), followed by emission of a single photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two-photon fluorescence is mostly seen used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Two-photon_excitation_microscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;two-photon excitation microscopy&lt;/a&gt; and useful dyes tend to be big conjugated organic systems. Materials used for photon upconversion are generally lanthanide-doped nanoparticles and they've found many uses in applications traditionally served by quantum dots and organic fluorophores. They're both particularly useful in medical imaging as IR can penetrate tissue deeply, but &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.201005159/full&quot; rel=&quot;nofollow noreferrer&quot;&gt;upconverting nanoparticles&lt;/a&gt; have a big advantage in that two-photon fluorescence requires very high excitation intensity (pretty much just lasers) as both photons must be absorbed simultaneously, but upconversion processes are sequential and can thus occur at lower intensity. Though, for some applications, the intensity requirements for two-photon absorption are useful as one can play tricks with focusing such that only a small region receives enough excitation light to emit, making imaging small areas possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a database of compounds, Thermo-Fisher has a &lt;a href=&quot;https://www.thermofisher.com/ca/en/home/references/molecular-probes-the-handbook/technical-notes-and-product-highlights/fluorescent-probes-for-two-photon-microscopy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalogue of two-photon dyes&lt;/a&gt;, but I don't know that there are a lot of commercial manufacturers of upconverting nanoparticles, though Sigma-Aldrich formerly carried up converting nanocrystal under the Sunstone brand.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2020-01-08T16:05:36.113" LastActivityDate="2020-01-08T16:05:36.113" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23571" PostTypeId="2" ParentId="23569" CreationDate="2015-01-09T23:54:22.313" Score="5" Body="&lt;p&gt;The first thing is that it has to be going from liquid to solid because the substance is being cooled. Like tap water put in the freezer: liquid to solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that note, the average kinetic energy goes down (directly assumed because that's what temperature is, and it's going down.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leaves us with two answer choices: latent heat absorbed or latent heat released, where latent heat is the heat of a (thermodynamic system) at a constant temperature (like a phase change!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're gaining energy, particles are going to have more velocity, like those virtual diagrams of gases with fast moving balls. Since we're going to a solid, losing energy makes sense because particles can't move as fast and are trapped in a fixed location (like an ice cube, which has a definite cuboid shape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So liquid to solid, latent heat because the temperature is constant, and losing energy to become a solid - B is the answer.&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2015-01-09T23:54:22.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23572" PostTypeId="1" AcceptedAnswerId="23573" CreationDate="2015-01-10T00:30:57.393" Score="18" ViewCount="17658" Body="&lt;p&gt;People say it dissolves, but shouldn't the acid form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{NaCl + H2O -&amp;gt;[?] HCl + ???}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="194" LastEditDate="2017-08-13T17:03:54.187" LastActivityDate="2017-08-13T17:03:54.187" Title="Why doesn't HCl form when you dissolve NaCl in water?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23573" PostTypeId="2" ParentId="23572" CreationDate="2015-01-10T00:40:56.900" Score="22" Body="&lt;p&gt;If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction that you propose - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl- + H2O -&amp;gt; HCl + HO-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is highly thermodynamically unfavorable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a &lt;strong&gt;much&lt;/strong&gt; (as in almost a trillion trillion times) stronger acid than water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;even&lt;/em&gt; if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a &lt;strong&gt;lot&lt;/strong&gt; bigger). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl- + H2O  &amp;lt;&amp;lt;=&amp;gt; HCl + HO-}$ &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-01-21T03:40:47.177" LastActivityDate="2015-01-21T03:40:47.177" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23574" PostTypeId="2" ParentId="23566" CreationDate="2015-01-10T00:46:07.463" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrothermal_vent&quot; rel=&quot;nofollow&quot;&gt;Here's a cool example&lt;/a&gt; that should help illustrate the correct answer to your question.  As for &lt;em&gt;why&lt;/em&gt; it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles.  Is it harder or easier to do that when the ambient pressure is high?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_point&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.  You will need some values for $P_0$, $T_0$ and $\Delta H_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-10T00:46:07.463" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23575" PostTypeId="2" ParentId="23564" CreationDate="2015-01-10T01:19:43.193" Score="2" Body="&lt;p&gt;For starters, it might be fun to grow different kinds of crystals in shallow dishes and watch them grow in real time under a microscope. Either a low melting solid, melted and resolidified, or growing from a saturated solution. Also, looking at how a metal surface is etched by acid using a microscope would be fascinating.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-10T01:19:43.193" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23576" PostTypeId="1" CreationDate="2015-01-10T03:22:42.143" Score="2" ViewCount="1724" Body="&lt;p&gt;I wish to know why aromatic hydrocarbons like benzene are called aromatic when they can be just called circular to avoid the confusion?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="5026" LastEditDate="2018-02-26T13:22:36.793" LastActivityDate="2018-02-26T13:23:31.847" Title="Why is benzene called &quot;aromatic&quot;? Why not &quot;circular&quot; instead?" Tags="&lt;organic-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23577" PostTypeId="2" ParentId="23576" CreationDate="2015-01-10T06:06:55.453" Score="6" Body="&lt;p&gt;Aromaticity is a electronic property of a molecule, not a description of its shape. An aromatic molecule contains a circular system of conjugated π bonds and p orbitals that contains 4n+2 electrons, where n is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene is aromatic: it has 6 electrons in a circular conjugated system (4n+2=6, n=1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclooctatetraene is not aromatic: it appears to have a circular conjugated system, but it contains 8 electrons (4n+2=8 would require n=3/2). This molecule actually adopts a non-planar geometry to avoid &lt;a href=&quot;http://en.wikipedia.org/wiki/Antiaromaticity&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;antiaromaticity&lt;/em&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexane is not aromatic, even though it is circular: it does not have a conjugated electron system.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-10T06:06:55.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23578" PostTypeId="1" AcceptedAnswerId="23580" CreationDate="2015-01-10T13:37:55.840" Score="9" ViewCount="5305" Body="&lt;p&gt;According to molecular orbital theory, two atomic orbitals form two molecular orbitals analogous to waves combining constructively or destructively but how can a wave combine destructively &lt;em&gt;and&lt;/em&gt; constructively at the same time to produce two molecular orbitals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't two atomic orbitals combine to form a molecular orbital that can be BMO &lt;em&gt;or&lt;/em&gt; ABMO?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="402" LastEditDate="2017-01-14T04:18:16.317" LastActivityDate="2017-01-14T20:17:27.830" Title="Why do two atomic orbitals form two molecular orbitals?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ClosedDate="2017-09-19T21:34:44.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="23580" PostTypeId="2" ParentId="23578" CreationDate="2015-01-10T16:49:34.533" Score="6" Body="&lt;p&gt;Well, one important thing to keep in mind is that atomic and molecular orbitals are not &quot;real&quot; things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules.  They are like a set of rules that tell us all of the ways in which it is &lt;em&gt;possible&lt;/em&gt; for the electrons with a particular set of quantum numbers to behave.  For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\mathrm p_z$ orbital (quantum numbers: $n=3$, $l=1$, $m_l=0$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when you are dealing with molecular orbital theory, you are talking about the &lt;em&gt;possible&lt;/em&gt; ways in which electrons in atomic orbitals can combine in a covalent interaction.  So it's not that the two atomic orbitals combine &quot;constructively and destructively at the same time&quot;, it's just that those are the two ways in which its &lt;em&gt;possible&lt;/em&gt; for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them.  The &quot;constructive&quot; case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the &quot;destructive&quot; case, so by the aufbau principle, electrons will occupy the lower energy &quot;bonding&quot; orbital first.  In a case like $\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\ce{He2+}$), it will occupy the higher-lying &quot;anti bonding&quot; orbital.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="7475" LastEditDate="2017-01-14T20:17:27.830" LastActivityDate="2017-01-14T20:17:27.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23581" PostTypeId="2" ParentId="23558" CreationDate="2015-01-10T17:29:39.987" Score="1" Body="&lt;p&gt;&lt;em&gt;In addition to Michael D. M. Dryden's excellent answer:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freshly prepared &lt;strong&gt;graphene oxide&lt;/strong&gt; seemingly shows two-photon absorption in the IR and deactivation by fluorescence, as described in an 2014 article in &#xA;&lt;em&gt;Nature Science Reports&lt;/em&gt;. (&lt;a href=&quot;http://dx.doi.org/10.1038/srep06090&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 2D in that article shows a linear correlation of the emission intensity at $\lambda_{\textrm{em}} = \textrm{680 nm}$ with the square of the intensity of the excitation at $\lambda_{\textrm{exc}} = \textrm{1140 nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 2C shows that the emission wavelength can be tuned by variation of the excitation wavelength!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-10T17:29:39.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23582" PostTypeId="1" AcceptedAnswerId="23585" CreationDate="2015-01-10T17:47:05.107" Score="0" ViewCount="378" Body="&lt;p&gt;So I came across this question asking whether something is physical or chemical.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Table salt is composed of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt;. Is it a &lt;strong&gt;physical&lt;/strong&gt; property or a&#xA;  &lt;strong&gt;chemical&lt;/strong&gt; property?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really don't understand why the answer says its a &lt;strong&gt;chemical&lt;/strong&gt; property. I asked the others and they did not give me an understandable answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Now as I think about it, I really don't think there is an accurate answer, even if the book said it was chemical. Bad question from the book.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="23561" LastEditDate="2018-09-29T23:02:52.030" LastActivityDate="2018-09-29T23:02:52.030" Title="Is the fact that table salt is composed of Na &amp; Cl a chemical or physical property?" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23583" PostTypeId="2" ParentId="23582" CreationDate="2015-01-10T18:10:40.447" Score="0" Body="&lt;p&gt;First, table salt is composed of $\ce {NaCl}$ wich is different from $\ce {Na}$ &amp;amp; $\ce{Cl}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, actually table salt isn't exclusively composed of $\ce {NaCl}$ depending of the soure but contains also a fair amount of $\ce {KCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, composition isn't what you can call a &quot;property&quot;. I mean, it's a definition, compound containing only $\ce {NaCl}$ or other $\ce {KCl}$ is table salt that's all. &lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-01-10T18:10:40.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23584" PostTypeId="1" AcceptedAnswerId="23586" CreationDate="2015-01-10T18:21:30.620" Score="4" ViewCount="2060" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A stability model premised on Lewis acid-base covalent bonding can be proposed to rationalize the stability shown by sulfides such as CuS and Ag2S. For instance, although the oxide anion is a stronger Bronsted base than sulfide anion, sulfide anion can be (and usually is) a stronger Lewis base than oxide anion since S has a lower EN than O. Furthermore, recall that Lewis base strength depends on the Lewis acid with which the Lewis base interacts. For the sulfide anion, since sulfide anion possess empty valence 3d orbitals, Lewis acid-base interactions are often very strong if the Lewis acid is a low oxidation state metal ion with a fairly large number of valence d electrons. for such cases, sulfide-to-cation sigma coordinate covalent bond formation couples with pi-type coordinate covalent bond formation because metal ion d-electron density &quot;drifts&quot; back to empty valence 3d orbitals on sulfide anion. This results in covalent bond interactions that have substantial, strong multiple ond character and formation of a metal-sulfide lattice which is rather macromolecular in character. In sum, such as lattice must be very stable ... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1) Is the above excerpt describing pi-backbonding? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is pi-backbonding more favorable for metals in a low rather than a high oxidation state because such metals still have substantial electron density that needs to be stabilized - and preferably stabilized by something more electronegative than a metal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Metal d-orbitals overlap with p or d-orbitals of the non-metal. How does this work? I always see diagrams such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dj038.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the overlap between the metal d-orbital and non-metal p-orbital as poor as depicted? Are the d-orbitals really at a 45 degree angle relative to the p-orbital? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) How well is the above excerpt written? I feel that two improvements could be made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) The passage seems to imply that Bronsted acidity and basicity do not depend on the corresponding base or acid. Wrong implication. HCl in HBr solvent won't be a strong acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Why electron density might just &quot;drift&quot; to the non-metal could be explained explicitly. I.e. electronegativity. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-01-10T19:55:25.817" LastActivityDate="2015-01-20T00:05:42.477" Title="Stability of Sulfides - backbonding?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23585" PostTypeId="2" ParentId="23582" CreationDate="2015-01-10T18:22:24.927" Score="6" Body="&lt;p&gt;The problem rises when we are mistaken about what a property is. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The different types of matter can be distinguished through two components: composition and properties. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Properties_of_Matter&quot; rel=&quot;nofollow&quot;&gt;UCDavis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do a search about chemical property, and you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a property or characteristic of a substance that is observed during a reaction in which the chemical composition or identity of the substance is changed. &lt;a href=&quot;http://dictionary.reference.com/browse/chemical+property&quot; rel=&quot;nofollow&quot;&gt;Dictionary.com&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A chemical property is any of a materials properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. Simply speaking, chemical properties cannot be determined just by viewing or touching the substance; the substance's internal structure must be affected for its chemical properties to be investigated. However a catalytic property would also be a chemical property. &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_property&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Any &lt;strong&gt;characteristic&lt;/strong&gt; that can be determined only by changing a substance's molecular structure.&#xA;   &lt;a href=&quot;https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/introduction-to-chemistry-1/physical-and-chemical-properties-of-matter-28/physical-and-chemical-properties-of-matter-181-1817/&quot; rel=&quot;nofollow&quot;&gt;Boundless&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's a physical property?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A physical property is any property that is measurable whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its transformations or evolutions between its momentary states. Physical properties are often referred to as observables. &lt;strong&gt;They are not modal properties.&lt;/strong&gt; - &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_property&quot; rel=&quot;nofollow&quot;&gt;Physical property in wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, a physical property is what that doesn't deal with the composition of matter; but other factors, such as it's physical state. &quot;Physical&quot; isn't at its literal meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table salt is never pure $\ce{NaCl}$; because simply there's nothing pure in the nature. It's &lt;em&gt;mainly&lt;/em&gt; composed of Sodium Chloride crystals. But if, and only if, we consider it to be only made from $\ce{NaCl}$, we are talking about its chemical composition and therefore, it must be a chemical property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that sometimes there are wrong questions for us to answer, so if composition is to be a property, it &lt;strong&gt;must&lt;/strong&gt; be a chemical one.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-01-11T09:11:34.383" LastActivityDate="2015-01-11T09:11:34.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23586" PostTypeId="2" ParentId="23584" CreationDate="2015-01-10T19:03:22.800" Score="5" Body="&lt;p&gt;1) Yes and no ... the d-d orbital overlap will almost certainly have pi-orbital character (angular node along bond axis), and it could reasonably be considered &quot;backbonding&quot; because the electron density is going in the canonically &quot;unexpected&quot; direction (i.e. away from positively charged metal center).  On the other hand, chemists (particularly inorganic chemists) usually restrict the phrase &quot;pi-backbonding&quot; to refer to back donation from the d-orbitals of a neutral or ionic metal center into anti-bonding orbitals of a ligand, typically an olefin or a carbonyl ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Yes .. that's essentially correct ... d-orbitals tend to be rather diffuse, and so can overlap with lobes of diffuse orbitals on ligands (or counter-ions, as with the metal sulfides).  In traditional metal-ligand pi-backbonding, lobes the d-orbitals overlap with lobes of an unoccupied anti-bonding orbital of the ligand.  In the example given above, they overlap with lobes of unoccupied d-orbitals on the sulfur.  For a given system, unoccupied, or virtual, orbitals tend to be much more diffuse than occupied orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Those diagrams are just intended to capture the qualitative aspects of the bonding interaction, such as the symmetries of the orbitals involved.  Remember, the orbitals are 3-dimensional shapes that describe the probability of finding an electron with a particular set of quantum numbers in a given region of space.  Depictions like the one in your post are helping to visualize the orbital by assuming a fairly tight &quot;tolerance&quot; for the probability. For example, those images might depict the surfaces that enclose 75% of the total probability density.  If you expand that to 90%, or even 99%, the lobes maintain the same basic shapes but get much larger, which might make the overlap seem more significant, but could also be confusing for other reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I think it is quite well-written, but I suppose it would depend on the broader context in which it was being presented.  I guess it is presented at an inappropriately high level for an introductory chemistry textbook, for example.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastEditorUserId="12658" LastEditDate="2015-01-10T19:11:12.110" LastActivityDate="2015-01-10T19:11:12.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23587" PostTypeId="2" ParentId="23576" CreationDate="2015-01-10T19:28:26.603" Score="2" Body="&lt;p&gt;As @Kyle said aromaticity is an electronic property, not a description of shape and so circular would not be an appropriate description, particularly seeing as these molecules are not at all circular (benzene is hexagonal, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name 'aromatic' is a historical quirk. Supposedly it was used because many of the chemically aromatic substances know at the time happened to have notable aromas although wikipedia seems to &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromaticity#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;dispute this&lt;/a&gt;. Several other reputable sites like &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/organic/aromatic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/react3.htm#rx8&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; do support this view though so I'm not sure what the right answer is (or even if anybody knows).&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-02-26T13:23:31.847" LastActivityDate="2018-02-26T13:23:31.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23590" PostTypeId="1" CreationDate="2015-01-10T21:08:07.690" Score="5" ViewCount="3851" Body="&lt;p&gt;I'd like to know why does cobalt(II),  $\ce{Co^2+}$, form $\ce{[Co(H2O)6]^2+}$ complex with water while  it forms $\ce{[Co(SCN)4]^2-}$ with the thiocyanate? Why is there $\ce{6H2O}$ but only $\ce{4 SCN-}$? And is the cobalt's  coordination number 6 in these cases?&#xA;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="12690" LastEditorUserId="4945" LastEditDate="2015-01-13T04:48:30.480" LastActivityDate="2015-01-13T04:48:30.480" Title="Why does Cobalt have different coordination numbers in its complexes with water and thiocyanate?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23591" PostTypeId="1" AcceptedAnswerId="23621" CreationDate="2015-01-10T21:09:15.280" Score="7" ViewCount="10228" Body="&lt;p&gt;On a whim, I mixed together some coconut oil and drug store (3%) hydrogen peroxide and washed my hands with it. It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did any chemical reaction occur between the coconut oil and $\ce{H2O2}$ to make a new compound? My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only had high school chemistry so I could be wrong. Just my guess. Anyone know for sure?  &lt;/p&gt;&#xA;" OwnerUserId="12691" LastEditorUserId="7448" LastEditDate="2015-01-11T14:17:42.053" LastActivityDate="2016-07-27T02:38:02.173" Title="Is there a reaction happening between Coconut Oil &amp; Hydrogen Peroxide?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23592" PostTypeId="1" CreationDate="2015-01-10T21:10:30.617" Score="1" ViewCount="4125" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Of the following atoms, which has the largest first ionization energy?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{I}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I got confused between &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt;. The answer given is &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; although in the periodic table &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; is closer to Group 18 (noble gases). Why is the answer &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; then?&lt;/p&gt;&#xA;" OwnerUserId="12679" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-21T04:47:33.327" Title="Which of the following atoms has the largest first ionization energy?" Tags="&lt;inorganic-chemistry&gt;&lt;ionization-energy&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23593" PostTypeId="1" AcceptedAnswerId="23595" CreationDate="2015-01-10T21:11:37.697" Score="-1" ViewCount="239" Body="&lt;p&gt;The definition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow noreferrer&quot;&gt;beta $(\beta^-)$ decay&lt;/a&gt; is that an electron is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^131_53I -&amp;gt; ^131_54Xe + e^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;iodine&lt;/strong&gt; forms &lt;strong&gt;Xe&lt;/strong&gt; and releases &lt;strong&gt;one electron with a -1 charge only and no weight&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that if an electron is released, then there is still a proton left so shouldn't it still be iodine? Then you should gain an extra proton. I personally feel like something is missing here, Also about Positron Emission or Positron Decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know why or what?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: The &quot;duplicate&quot; question picked by many does not fully answer the question as well as bon did here.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="194" LastEditDate="2017-04-01T03:07:15.090" LastActivityDate="2021-05-01T17:17:38.063" Title="Beta Decay - Doesn't Add Up" Tags="&lt;physical-chemistry&gt;&lt;nuclear&gt;" AnswerCount="3" CommentCount="0" ClosedDate="2015-01-11T15:39:17.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="23594" PostTypeId="2" ParentId="23593" CreationDate="2015-01-10T21:15:54.240" Score="0" Body="&lt;p&gt;Essentially, a neutron is converted to a proton, releasing an electron in the form of beta radiation. That is why the mass stays constant but the number of protons increases by 1.&lt;/p&gt;&#xA;" OwnerUserId="11597" LastEditorUserId="11597" LastEditDate="2015-01-11T00:45:48.730" LastActivityDate="2015-01-11T00:45:48.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23595" PostTypeId="2" ParentId="23593" CreationDate="2015-01-10T21:30:15.800" Score="5" Body="&lt;p&gt;This is really a physics question but I will answer it anyway.&#xA;I'm not sure how much subatomic physics you know so I will give two different versions.&#xA;Simply put, a neutron decays to form a proton and electron.&#xA;$$\ce{n -&amp;gt;  p+ + e-}$$&#xA;This explains why the proton number increases by one to form Xenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More properly a down quark inside a neutron decays to form an up quark, an electron and an electron antineutrino as a result of weak nuclear force interactions.&#xA;$$\ce{d -&amp;gt; u + e- + \overline{\nu}_{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes the neutron ($\ce{udd}$) to turn into a proton ($\ce{uud}$) leading to the increase in atomic number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of beta decay is know as $\beta^{-}$ decay. There is also $\beta^{+}$ decay which is a similar process where a proton becomes a neutron, positron and an electron neutrino:&#xA;$$\ce{^23_12Mg -&amp;gt; ^23_11Na + e+ + \nu_{e}}$$&#xA;$$\ce{u -&amp;gt; d + e+ + \nu_{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Beta_decay&lt;/a&gt; $%edit$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="24052" LastEditDate="2016-07-19T18:58:26.633" LastActivityDate="2016-07-19T18:58:26.633" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23596" PostTypeId="2" ParentId="23592" CreationDate="2015-01-10T21:30:46.277" Score="-1" Body="&lt;p&gt;Ok so I thought about the question to and I thought that Bromine should have a higher first ionization energy but then if you really think about it, the answer would be oxygen. Look at it this way. Oxygen desperately needs two electrons. And Bromine needs only one more. So if you look at it, it would be harder to lose another electron when you already lost two and it would be easier to lose one more if you are Bromine. And also if you look at a basic ionization energy trend, it goes to the top right corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you were broke, would you be willing to pay to get something even though you know that you need money? No like Oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of Bromine as an investor who needs a little bit more $ for an achievement, Bromine would be more likely to sacrifice its electron or to invest in something like an investor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really bad analogies but hopefully you get the idea.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-01-10T21:30:46.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23597" PostTypeId="2" ParentId="23593" CreationDate="2015-01-10T21:34:59.230" Score="3" Body="&lt;p&gt;When we talk about an unstable atom that will decay by emitting &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; radiation, the nucleus is the thing that's unstable, right?&lt;/p&gt;&#xA;&lt;p&gt;As &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; puts it,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An unstable atomic nucleus with an excess of neutrons may undergo β− decay, where a neutron is converted into a proton, an electron and an electron antineutrino (the antiparticle of the neutrino)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can forget the anti-matter part. Just remember than in &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; decay there is a neutron from the nucleus that is being &amp;quot;devolved&amp;quot; into eminently a proton and an electron. The electron leaves the atom with high velocity and energy, but the proton remains. So, for each atom that a &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; particle is released from, we have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^A_zM -&amp;gt; ^A_{z+1}M+}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="24052" LastEditDate="2021-05-01T17:17:38.063" LastActivityDate="2021-05-01T17:17:38.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23598" PostTypeId="2" ParentId="23572" CreationDate="2015-01-10T21:50:17.607" Score="13" Body="&lt;p&gt;It does form HCl, the only question is how much? Dissociation constant of hydrochloric acid is (&lt;a href=&quot;http://depts.washington.edu/eooptic/links/acidstrength.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{Cl-}]}{[\ce{HCl}]} = \pu{1.3e6 M} \tag{1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = \pu{1e-14 M^2} \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In neutral water, $[\ce{H+}] = \pu{1e-7 M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in $\pu{1 M}$ $\ce{NaCl}$, assuming neutral pH and substituting $[\ce{H+}] = \pu{1e-7 M}$ in equation (1), we find $\pu{1e-13 M}$ $\ce{HCl}$ is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not very much, but still billions of molecules in a 1 liter sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate rigorously, instead of assuming neutral pH solution, to the above two equations you would add two more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electrical neutrality requirement:&#xA;$$\ce{[Na+] + [H+] = [Cl-] + [OH-]}$$  &lt;/li&gt;&#xA;&lt;li&gt;Number of chlorine atoms is conserved:&#xA;$$\ce{[HCl] + [Cl-] = [Na+]}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And you would have 4 equations and 4 unknowns ($\ce{[HCl]}$, $\ce{[Cl-]}$, $\ce{[H+]}$, and $\ce{[OH-]}$), with $\ce{[Na+]}$ being known from the amount of $\ce{NaCl}$ added to the water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above is a simplification of the real situation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in &lt;em&gt;Essentials of Chemistry&lt;/em&gt; by Hessler and Smith at page 327, which even though it is over 100 years old I still think is quite accurate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in an aqueous solution of sodium chloride [are] eight distinct things: the ions hydrogen, hydroxyl, sodium, and chlorine, and the undissociated substances: water, sodium chloride, sodium hydroxide and hydrochloric acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2017-08-13T16:06:38.130" LastActivityDate="2017-08-13T16:06:38.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23599" PostTypeId="2" ParentId="23592" CreationDate="2015-01-10T21:53:43.857" Score="0" Body="&lt;p&gt;The larger that atoms get, the outer electrons experience whats called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Shielding_effect&quot; rel=&quot;nofollow&quot;&gt;shielding effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons that are closer to the nucleus have a stronger attraction than the electrons further away. Because the electrons in oxygen experience less shielding, they have a stronger affinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen's stronger electron affinity contributes to the observed &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;electronegativity&lt;/a&gt; of the atom.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the most basic level, electronegativity is determined by factors like the nuclear charge (&lt;strong&gt;the more protons an atom has, the more &quot;pull&quot; it will have&lt;/strong&gt; on negative electrons) and the number/location of other electrons present in the atomic shells (the more electrons an atom has, &lt;strong&gt;the farther from the nucleus the valence electrons will be&lt;/strong&gt;, and as a result &lt;strong&gt;the less positive charge they will experience&lt;/strong&gt;—both because of their &lt;strong&gt;increased distance from the nucleus&lt;/strong&gt;, and because the other &lt;strong&gt;electrons in the lower energy core orbitals will act to shield the valence electrons&lt;/strong&gt; from the positively charged nucleus).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2015-01-10T21:53:43.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23600" PostTypeId="2" ParentId="23592" CreationDate="2015-01-11T00:07:51.903" Score="4" Body="&lt;p&gt;I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends.  The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns.  Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$?  Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information.  It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable.  You can see the periodic trends &lt;a href=&quot;http://www.iun.edu/~cpanhd/C101webnotes/modern-atomic-theory/ionization-energy.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-11T00:07:51.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23601" PostTypeId="2" ParentId="15778" CreationDate="2015-01-11T01:37:21.457" Score="2" Body="&lt;p&gt;Well, if we take a broader look, strong neutral, and weaker negatively charged electrophiles are very common in inorganic chemistry - all Lewis acids are electrophiles, and even negatively charged complexes with number of coordination lower then optimal are still Lewis acids and therefore electrophiles.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-01-11T01:37:21.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23603" PostTypeId="2" ParentId="23545" CreationDate="2015-01-11T05:44:03.293" Score="3" Body="&lt;p&gt;Two of the methods are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-11T05:44:03.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23604" PostTypeId="2" ParentId="22382" CreationDate="2015-01-11T05:54:46.450" Score="17" Body="&lt;p&gt;Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyanoacrylate#Properties&quot;&gt;Wikipedia article for cyanoacrylates&lt;/a&gt; shows the polymerization more clearly than I can easily explain in words.  The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool provides a large number of sites for the glue to cure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I read your question I was curious; I had never heard of this before.  I searched the internet and was only able to find one or two examples of people actually trying this successfully, and both of them looked dubious.  I've spilled superglue on cotton shirts and pants any number of times and while it cures almost instantly, it has never caught my shirt on fire.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to try it myself, and I soaked a cotton ball in superglue to see what would happen.  The glue cured extremely rapidly, producing very irritating (colorless) fumes, and the mass of glue and cotton got warm, but it was nowhere near hot enough to ignite - my tap runs hotter than the glue got.  There may be a particular type of glue that cures significantly more exothermically, but I'm inclined to think that this is an internet hoax, and &lt;a href=&quot;http://www.popsci.com/diy/article/2012-09/gray-matter-playing-fire&quot;&gt;Popular Science agrees&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2015-01-11T05:54:46.450" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23605" PostTypeId="1" CreationDate="2015-01-11T09:11:51.357" Score="2" ViewCount="6779" Body="&lt;p&gt;If I leave some olive oil uncovered and fully exposed to indoors air for enough time (though away from UV exposure or any other environmental factors), what will happen to the oil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20090912072857AAkcMkw&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that some of its constituents ('lighter' molecules) will start to evaporate but that heavier ones will be left behind, which will 'not quite' evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is true (it may not be), would the non-evaporative part of the oil eventually harden and assume a cake-like consistency, almost like wax or hardened animal lard stored at refridgerator temperatures?&lt;/p&gt;&#xA;" OwnerDisplayName="user1469" LastEditorDisplayName="user1469" LastEditDate="2015-01-11T20:39:13.493" LastActivityDate="2015-01-11T22:34:46.873" Title="What will eventually happen to uncovered vegetable (seed) oil exposed to the air?" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23608" PostTypeId="1" AcceptedAnswerId="23618" CreationDate="2015-01-11T13:20:57.913" Score="4" ViewCount="238" Body="&lt;p&gt;Is there such a paint / pigment that absorbs most wavelengths of light (beyond the visible light range) that can be used for solar heat generating applications?  If not, would it be possible to make it?  Ideally, this paint would absorb light from the sun in an efficient manner and become heated quickly by solar light.  It should probably at least absorb infrared, and any other wavelengths of light that produce significant heat when absorbed.  This should be far more effective than normal black paint as black paint only &quot;for sure&quot; absorbs light in the visible spectrum, and this desired paint would absorb other types of light from the sun.&lt;/p&gt;&#xA;" OwnerUserId="6806" LastEditorUserId="6806" LastEditDate="2015-01-11T13:31:15.297" LastActivityDate="2015-01-11T16:50:00.673" Title="Is it possible to make broad spectrum light absorbing paint?" Tags="&lt;everyday-chemistry&gt;&lt;spectroscopy&gt;&lt;heat&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23609" PostTypeId="1" AcceptedAnswerId="23619" CreationDate="2015-01-11T13:24:58.830" Score="10" ViewCount="1436" Body="&lt;p&gt;Entropy is often verbally described as the order/disorder of the thermodynamic system. However, I've been told that this description is a vague &quot;hand-waving&quot; attempt at describing what entropy is. For example, a messy bedroom doesn't have greater entropy than a tidy room &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why is this the case? Also, what would better describe entropy verbally? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5456" LastEditDate="2015-01-11T15:30:12.213" LastActivityDate="2021-03-25T15:21:52.753" Title="Whats the difference between entropy and the (dis)order of a system?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23610" PostTypeId="1" AcceptedAnswerId="23615" CreationDate="2015-01-11T14:39:41.507" Score="17" ViewCount="2724" Body="&lt;p&gt;I was thinking is there really bond present at microscopic level or atoms/molecules are just nearby and are connected with force which is not visible(like gravitational force between earth and sun) and we make bonds just for understanding. &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorDisplayName="user7484" LastEditDate="2015-01-11T15:32:11.377" LastActivityDate="2015-01-11T16:38:07.003" Title="Can we really see the bonds?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="23611" PostTypeId="2" ParentId="23609" CreationDate="2015-01-11T15:13:04.053" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what would better describe entropy verbally?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can also use &quot;measurement of randomness&quot; or &quot;amount of chaos&quot; or &quot; energy dispersion&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Initially in 1862,  Rudolf Clausius asserted that  thermodynamic process always &quot;admits to being reduced to the &lt;strong&gt;alteration&lt;/strong&gt; in some way or another of the arrangement of the constituent parts of the working body&quot; and that internal work associated with these alterations is quantified energetically by a measure of &lt;strong&gt;entropy&lt;/strong&gt; change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But later after few years Ludwig Boltzmann translated word &lt;strong&gt;alteration&lt;/strong&gt; from Rudolf Clausius' assertion to &lt;strong&gt;order and disorder&lt;/strong&gt; in gas phase molecular systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you see latest books they use concept of &lt;strong&gt;energy dispersion&lt;/strong&gt; instead of order or disorder to explain entropy.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy_(order_and_disorder)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also have look at &lt;a href=&quot;https://physics.stackexchange.com/a/33385/46239&quot;&gt;physics S.E.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-01-11T15:25:33.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23612" PostTypeId="2" ParentId="23609" CreationDate="2015-01-11T15:18:10.063" Score="2" Body="&lt;p&gt;I will take a crack at this although I admit that this topic sometimes confuses me as well. Here is how I like to think about entropy. Consider a box containing equal amounts of gases A and B. So, we have a fixed volume and fixed number of molecules. Let us also isolate the box from its surroundings so that it has a fixed energy as well. We have just created a microcanonical ensemble (constant NVE). The ensemble consists of every possible configuration of molecules having the same NVE. Now, if we were able to step back and take a broad view of the ensemble, we would observe that the vast majority of boxes contain a rather bland homogeneous mixture of gases A and B. In fact, they would be indistinguishable for all practical purposes. Let us count the number of boxes in this state and call the number &lt;span class=&quot;math-container&quot;&gt;$W_1$&lt;/span&gt;. Continuing with the box counting, we find that there are only a handful of distinguishable boxes left, but they are quite interesting! One box, for instance, might have all of the A molecules crowded together in one corner and all of the B molecules in another corner! This one box can be labelled &lt;span class=&quot;math-container&quot;&gt;$W_2$&lt;/span&gt;. We start to understand that some configurations of molecules will be extremely improbable because they represent such a small fraction of the total number of possible configurations. Boltzmann quantified this relationship as &lt;span class=&quot;math-container&quot;&gt;$S=k \log W$&lt;/span&gt;. If we plug &lt;span class=&quot;math-container&quot;&gt;$W_1$&lt;/span&gt; into this equation we get a high value for the entropy, &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, because &lt;span class=&quot;math-container&quot;&gt;$W_1$&lt;/span&gt; is so large. On the other hand, if we plug in &lt;span class=&quot;math-container&quot;&gt;$W_2$&lt;/span&gt; the entropy we get is very low.&lt;/p&gt;&#xA;&lt;p&gt;So, to sum up, entropy is really the measure of how likely a given system configuration is when compared to all of the possible configurations. In this sense, I would argue that you could say a messy room has a higher entropy than a clean room because there are so many more ways a room could be considered &amp;quot;messy&amp;quot; than &amp;quot;clean&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="12650" LastEditorUserId="15751" LastEditDate="2021-03-25T15:21:52.753" LastActivityDate="2021-03-25T15:21:52.753" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23613" PostTypeId="2" ParentId="23610" CreationDate="2015-01-11T15:26:21.697" Score="-2" Body="&lt;p&gt;No I don't think that is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about the &lt;strong&gt;attractions&lt;/strong&gt; &lt;em&gt;(the wanting of a full octet)&lt;/em&gt; between the atoms, to be honest we don't even have the technology to view the actual atoms itself (if we do then we are at the initial stages). So these &quot;bonds&quot; that you are talking about are just imaginary connections between atoms so people don't get confused or overwhelmed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take an ionic bond, one of the strongest bonds out there, we know it is caused by strong electrostatic attractions between the atoms in which one completely takes an electron or gives it up. If there was like a connection between them it would look almost surreal. It would make more sense if they are beside each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the covalent bond, in a covalent bond atoms are sharing electrons, in this case the electron clouds merge, this maybe hard to grasp but that is just how this works. That is why for most covalent bonds we look at the Lewis Structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a look at an &lt;strong&gt;ionic&lt;/strong&gt; &quot;bond&quot;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dYGyJ.png&quot; alt=&quot;Typical Ionic Bond&quot;&gt;&#xA;Take a look at the &lt;strong&gt;covalent&lt;/strong&gt; &quot;bond&quot;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xSUzi.jpg&quot; alt=&quot;Typical Covalent Bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the &lt;strong&gt;double&lt;/strong&gt; and &lt;strong&gt;triple&lt;/strong&gt; &quot;bond&quot;. - Bad Image&#xA;&lt;img src=&quot;https://i.stack.imgur.com/R1yDG.jpg&quot; alt=&quot;Double Bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at the above picture you can see that a &lt;strong&gt;triple&lt;/strong&gt; &quot;bond&quot; is just two more electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in reality to make these drawings of atoms easy, we have came up with a &lt;strong&gt;bond&lt;/strong&gt;. There might be &quot;bonds&quot; but one has to note that they are &lt;strong&gt;attractions&lt;/strong&gt; instead. Here I gave you a couple examples, mostly 2D but if you want to really get the idea you should take a look at the 3D structures that include everything. But you have to also note that we don't know how far our current depictions are true, I just gave you the most plausible ones, so don't believe everything as a fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it up: You Will Not See The Bonds (Not the sticks and ball kind). Though you may see the attractions. &lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-01-11T16:38:07.003" LastActivityDate="2015-01-11T16:38:07.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23614" PostTypeId="2" ParentId="22312" CreationDate="2015-01-11T16:07:01.907" Score="4" Body="&lt;p&gt;&quot;Epoxy&quot; here refers to an epoxy resin, which is comprised of oligomers that contain at least two epoxide functional groups. Hardening agents, like polyamines and polysufides, react with the resin via nucleophilic attack on the epoxide (as you desribed). Because the hardening agent has multiple nucleophilic groups, it will react with multiple epoxy oligomers. Likewise, the epoxy oligomer reacts with multiple hardening agents. This allows a complex crosslinked network to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the MSDS for a typical epoxy hardener, &quot;polymercaptan amines&quot; is listed as a &quot;blend.&quot; Likely this is just a mix of polysulfides and polyamines that was given an ambiguous name by the manufacturer. These molecules likely contain a varying number of nucleophilic functional groups.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stoichiometry at its finest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The stoichiometry is actually not straightforward here at all. It is possible that the total number of epoxide groups in the resin equals the total number of nucleophilic groups in the hardener, but that isn't necessarily the case. The reason you get to mix the parts 1:1 is because the manufacturer provides an approprite blend of hardeners to provide the desired stoichiometry, whatever it may be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But then does the resulting oxide anion attack another epoxide ring?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reaction certainly competes with attack by the hardening agent. Whether it is competitive enough to be relevant depends on the identity and available amount of the hardening agent. It may even intentionally utilized in the crosslinking process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, it is unlikely that an alkoxide anion is involved here. Often times a Lewis acid catalyst is added to facilitate crosslinking, which would provide an alkoxide-Lewis acid adduct. Alternatively, there are plenty of hydrogen bond donors available in these mixtures, so a proton would likely be associated with the epoxide during the ring opening, providing a neutral alcohol upon completion of the reaction.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-11T16:07:01.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23615" PostTypeId="2" ParentId="23610" CreationDate="2015-01-11T16:24:38.007" Score="34" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/15xJB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All credit to Zhang et al. &quot;&lt;a href=&quot;http://www.sciencemag.org/content/342/6158/611&quot;&gt;Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy&lt;/a&gt;&quot; Science Vol. 342  no. 6158  pp. 611-614.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-11T16:24:38.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23617" PostTypeId="2" ParentId="23605" CreationDate="2015-01-11T16:39:28.743" Score="3" Body="&lt;p&gt;Some of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastEditorUserId="12684" LastEditDate="2015-01-11T22:34:46.873" LastActivityDate="2015-01-11T22:34:46.873" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23618" PostTypeId="2" ParentId="23608" CreationDate="2015-01-11T16:50:00.673" Score="4" Body="&lt;p&gt;Carbon nanotube black is highly absorbent in and outside the visible spectrum, see &lt;a href=&quot;http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes&quot; rel=&quot;nofollow&quot;&gt;http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that does not make it ideal for energy conversion, as it would also need at least the following qualities to be practical.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A good conductor of heat to the heat-exchanger surface below&lt;/li&gt;&#xA;&lt;li&gt;An enduring finish, resistant to abrasion, heat and environmental degradation&lt;/li&gt;&#xA;&lt;li&gt;Economical, both in material cost and application&lt;/li&gt;&#xA;&lt;li&gt;Safety in use and in manufacture&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-11T16:50:00.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23619" PostTypeId="2" ParentId="23609" CreationDate="2015-01-11T16:59:08.723" Score="10" Body="&lt;p&gt;Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Entropy is a measure of the number of different ways a state with a particular energy can be realized&lt;/strong&gt;. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's constant and $W$ is the number of equivalent ways to distribute energy in the system. If there are many ways to realize a state with a given energy, we say it has high entropy. Often the many ways to realize a high entropy state might be described as &quot;disorder&quot;, but the lack of order is beside the point; the state has high entropy because it can be realized in many different ways, not because it's &quot;messy&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an analogy: if energy were money, entropy would be related to the number of different ways of counting it out. For example, there, there are only two ways of counting out two dollars with American paper money (2 1-dollar bills, or 1 two-dollar bill). But there are five ways of counting out two dollars using 50-cent or 25-cent coins (4 50-cent pieces, 3 50-cent pieces and 2 quarters, and so on). You could say that the &quot;entropy&quot; of a system that dealt in coins was higher than that of a system that dealt only in paper money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look the change in entropy for a reaction $\rm A\rightarrow B$, where A molecules can take on energies that are multiples of 10 energy units, and B molecules can take on energies that are multiples of 5 units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the total energy of the reacting mixture is 20 units. If we have 3 molecules of A, there are 2 ways to distribute our 20 units among energy levels with 0, 10, and 20 units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdTLE.png&quot; alt=&quot;There are 2 ways to distribute 3 A molecules with a total energy of 20 units.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have 3 molecules of B, there are 4 ways to distribute 20 units among energy levels with 0, 5, 10, 15, and 20 units:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8i76i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy of B is higher than the entropy of A because there are more ways to distribute the same amount of energy in B than in A. Therefore, $\Delta S$ for the reaction $\rm A\rightarrow B$ will be positive.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-12T15:11:39.363" LastActivityDate="2015-01-12T15:11:39.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23620" PostTypeId="2" ParentId="23605" CreationDate="2015-01-11T17:04:11.527" Score="2" Body="&lt;p&gt;Vegetable oils do lose some lighter constituents through evaporation, particularly substances responsible for aroma, but the primary degradation pathways, making it rancid, are through oxidation, hydrolization and microorganism metabolism; see &lt;a href=&quot;http://www.ivannikolov.com/how-cooking-oil-goes-rancid/&quot; rel=&quot;nofollow&quot;&gt;http://www.ivannikolov.com/how-cooking-oil-goes-rancid/&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Rancidification&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Rancidification&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some tree-nut oil, in particular, quickly develops a noticeable off-taste. BHT, BHA and other additives (often applied to packaging rather than to the food itself) help decrease oxidative rancidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, vegetable oil used for lubrication in old machine parts certainly does thicken to the consistency of wax, though I do not know how much of that is due to evaporation vs. degradation.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-11T17:04:11.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23621" PostTypeId="2" ParentId="23591" CreationDate="2015-01-11T17:08:31.603" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Did any chemical reaction occur between the coconut oil and H2O2 to make a new compound?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most likely, &lt;strong&gt;no chemical reaction occurred&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether any reaction occurred depends on the conditions you used when mixing the oil and hydrogen peroxide. How long did they mix for? What temperature was the mixture at? Did you add anything else to the mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide is used in the production of epoxidized vegetable oils. However, the reaction is performed for prolonged times at elevated temperatures with concentrated hydrogen peroxide in the presence of a catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing that you just poured the two together at room temp, then soon after used the mixture to wash your hands. Compared to the commercial reaction, you are missing time, heat, and a catalyst, so it is very unlikely that any reaction occurred.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oils are composed of triglyceride esters of fatty acids, not fatty acids themselves. So you would not see any peroxylauric acid.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It felt less greasy because it was less greasy: you added water to it (3% hydrogen peroxide is $\ce{H2O2}$ in water). You probably made an emulsion of oil and water and used that to wash your hands (kind of like using soap).&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-11T17:08:31.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23622" PostTypeId="2" ParentId="23605" CreationDate="2015-01-11T17:34:20.673" Score="2" Body="&lt;p&gt;The key to the answer is what kind of oil do you have, there are differences. In general, oils with big amount of double bonds MAY oxidize by oxygen from air. The process is radical, UV-induced, and very slow. Some other oils, with no or too little double bonds do not oxidize. and remain mostly same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of oxidation (drying) initially produces hydroperoxides, later able to serve as a source of more free radicals, leading eventually to inter-molecular linkages. I found contradicting  notions for what these linkages consist of (it is either oxygen bridges or direct links similar to polymers), but the end process if well known and used in varnishes. However, varnishes has extra components: they are usually diluted by some solvent to reduce viscosity and always contain a metal catalyst to accelerate drying of the oil, that also is usually heated for prolonged period of time, again, to increase the speed of drying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The considerations above ignore obvious process of evaporation of light impurities and possibility of bio-degradation (I'm not aware of bio-degradation processes for pure fatty oils, but there are bacteria that live in almost undiluted toluene and there are bacteria that thrive on petrol, so I'm open to this idea).&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-01-11T17:34:20.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23623" PostTypeId="1" AcceptedAnswerId="23624" CreationDate="2015-01-11T17:34:29.823" Score="0" ViewCount="558" Body="&lt;p&gt;In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?&lt;br&gt;&#xA;Really simple, I think yes to this question but teacher says no. So what is your view into this?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="4945" LastEditDate="2015-01-13T05:14:21.397" LastActivityDate="2015-01-13T15:51:22.113" Title="In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?" Tags="&lt;homework&gt;&lt;everyday-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23624" PostTypeId="2" ParentId="23623" CreationDate="2015-01-11T17:43:23.847" Score="2" Body="&lt;p&gt;Your teacher is right. For example, at room temperature, the pressure of water vapor over a pond or lake might be around 25 torr; that's less than atmospheric pressure, but the water vapor is undeniably a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe your confusion stems from a belief that gases must be produced by boiling liquids (since a liquid must have a vapor pressure equal to the external pressure before it begins to boil). But a liquid can also become a gas through evaporation, no matter what the external pressure is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;evaporation&lt;/strong&gt;, molecules escape from the liquid's surface and become a gas.&lt;br&gt;&#xA;Molecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others. This can happen at any temperature or pressure the liquid exists at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;boiling&lt;/strong&gt;, bubbles of gas form throughout the liquid. This can only happen when the temperature is high enough to make the vapor pressure equal to the external pressure.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-13T15:51:22.113" LastActivityDate="2015-01-13T15:51:22.113" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="23625" PostTypeId="2" ParentId="23623" CreationDate="2015-01-11T18:14:13.770" Score="0" Body="&lt;p&gt;If you know Dalton's law of partial pressure that &quot;in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases&quot;&#xA;And since pressure is non-negative quantity then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(total)=P1 + P2 + ... + Pn&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO the pressure of gas in atmosphere in fact, cannot be greater than atmospheric pressures . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Dalton%27s_law_of_partial_pressures.png&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/File:Dalton%27s_law_of_partial_pressures.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also I am not saying that gas stored in container cannot have pressure greater than atmosphere . The best example is your LPG cylinder.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2015-01-11T18:14:13.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23626" PostTypeId="1" AcceptedAnswerId="23639" CreationDate="2015-01-11T19:29:46.977" Score="9" ViewCount="3765" Body="&lt;p&gt;Alanine is heavier with a ($\ce{-CH3}$) than glycine which has an ($\ce{-H}$). What makes them have different $\mathrm{p}K_\text{a}(\ce{-NH3+})$ $(\mathrm{p}K_\text{a}: \text{Ala}=9.69, \text{Gly}=9.60)$ but similar $\mathrm{p}K_\text{a}(\ce{-COOH})$ $(\mathrm{p}K_\text{a}=2.34)$ when the rest of the amino acid is similar? Why does changing a hydrogen for a methyl group affect the $\ce{-NH3+}$ but not the $\ce{-COOH}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://d2vlcm61l7u1fs.cloudfront.net/media%2F993%2F99368358-7bd8-4b0e-bdac-c52c479a1f8e%2FphpX8qXnh.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are three amino acids (glycine, leucine, aspartine) which have same $\mathrm{p}K_\text{a}(\ce{-NH3+})$ but different $\mathrm{p}K_\text{a}(\ce{-COOH})$.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="5026" LastEditDate="2018-03-06T11:59:25.713" LastActivityDate="2018-03-06T11:59:25.713" Title="Why is the pKa (COOH) of alanine and glycine similar but the pKa (NH3) different?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23627" PostTypeId="1" CreationDate="2015-01-11T19:38:13.410" Score="7" ViewCount="737" Body="&lt;p&gt;Is there anyone who knows a book where I can find a detailed explanation of cell overpotential diagrams, like the ones shown here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fkXw4.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QoWE0.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7692" LastEditorUserId="7448" LastEditDate="2015-02-27T16:39:05.320" LastActivityDate="2015-03-16T16:34:42.573" Title="Cell overpotential diagrams" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;reference-request&gt;&lt;books&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23628" PostTypeId="1" CreationDate="2015-01-11T20:02:56.420" Score="0" ViewCount="61" Body="&lt;p&gt;4 * (2 + 2) = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 * 4 = 20  (rounded to 1 significant figure)&lt;br&gt;&#xA;OR&lt;br&gt;&#xA;4 * 2 + 4 * 2 = 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.0 * (0.5 + 0.5) = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.0 * 1.0 = 2.0&lt;br&gt;&#xA;OR&lt;br&gt;&#xA;2.0 * 0.5 + 2.0 * 0.5 = 2&lt;/p&gt;&#xA;" OwnerUserId="12701" LastEditorUserId="4934" LastEditDate="2015-01-11T20:14:13.803" LastActivityDate="2015-01-11T20:14:13.803" Title="Significant Figures: To expand or not to expand?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-01-11T20:50:37.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="23629" PostTypeId="2" ParentId="23628" CreationDate="2015-01-11T20:13:20.877" Score="1" Body="&lt;p&gt;In the first example, the 4(2+2), if you were ABSOLUTELY going to adhere to significant figure rules, then 20 would be the correct answer, though I don't believe many people would fault you for saying 16. In your second case, 2.0(0.5+0.5), the appropriate answer here would indeed be 2.0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, significant figures are not really valuable when referring to purely mathematical equations like you did. We use them in order to avoid making errors when it comes to the accuracy of a result, in other words, significant figures are there so that your answer is not more specific than your measurement was. In mathematics, this is not an issue, as you are not making any measurements. So, while you COULD use significant figures in a situation like this, it really is not quite the appropriate situation in which to do so.&lt;/p&gt;&#xA;" OwnerUserId="12693" LastActivityDate="2015-01-11T20:13:20.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23631" PostTypeId="1" AcceptedAnswerId="23634" CreationDate="2015-01-11T23:21:30.610" Score="2" ViewCount="143" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Solutions are prepared by dissolving a sulphate salt in water. Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion, SO42-, in a saturated solution? What would be the concentration of sulphate ion in this case?&lt;/p&gt;&#xA;&lt;p&gt;A. &lt;span class=&quot;math-container&quot;&gt;$\ce{CaSO4}$&lt;/span&gt;    &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; = 7.1E-5&lt;/p&gt;&#xA;&lt;p&gt;B. &lt;span class=&quot;math-container&quot;&gt;$\ce{SrSO4}$&lt;/span&gt;    &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; = 3.4E-7&lt;/p&gt;&#xA;&lt;p&gt;C. &lt;span class=&quot;math-container&quot;&gt;$\ce{BaSO4}$&lt;/span&gt;    &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; = 1.1E-10&lt;/p&gt;&#xA;&lt;p&gt;D. &lt;span class=&quot;math-container&quot;&gt;$\ce{RaSO4}$&lt;/span&gt;    &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; = 4.0E-11&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My first instinct was to simply look at the &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; values.. Since &lt;span class=&quot;math-container&quot;&gt;$\ce{CaSO4}$&lt;/span&gt; has the largest &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; value, it should be the most soluble, right?&lt;/p&gt;&#xA;&lt;p&gt;But then I looked at a data page referring to the solubility of common compounds in water (Please see image. According to the table, &lt;span class=&quot;math-container&quot;&gt;$\ce{SO4^2-}$&lt;/span&gt; has low solubility with &lt;span class=&quot;math-container&quot;&gt;$Ca$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Sr$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Ba$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;My question is- does the solubility depend on the &lt;span class=&quot;math-container&quot;&gt;$K_{sp}$&lt;/span&gt; value (was my first instinct right), or is the table correct?&lt;/p&gt;&#xA;&lt;p&gt;Any ideas would  be appreciated.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IIT34.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-12T11:31:21.500" Title="Which one of the compounds listed would we use in order to get the highest concentration of sulphate ion in a saturated solution?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23632" PostTypeId="1" AcceptedAnswerId="32081" CreationDate="2015-01-11T23:31:13.947" Score="4" ViewCount="227" Body="&lt;p&gt;Internal energy, enthalpy and Gibbs free energy are all units of energy. So, for any particular process where energy is lost from the system the same amount is given off to the surroundings. Am I right in thinking that a change in internal energy/free energy/enthalpy is all a measure of how the energy of the system changes in a process and as a result should be equal to each other for the same process, albeit under different conditions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me expand slightly:&#xA;$$dU&amp;lt;TdS-pdV$$ So for a spontaneous change to occur, the internal energy must decrease provided that the entropy and the volume are kept constant. The internal energy will continue to fall until equilibrium is reached. The exact amount of energy lost by the system is gained by the surroundings. &lt;strong&gt;As a side question: how exactly is the energy lost to the surroundings?&lt;/strong&gt;&#xA;$$dH&amp;lt;TdS+VdP$$ The enthalpy of the system decreases for a spontaneous change, provided that the entropy and pressure are kept constant. Energy lost by system = energy gained by surroundings.&#xA;$$dG&amp;lt;VdP-SdT$$ The Gibbs free energy of the  system decreases for a spontaneous change, provided that the temperature and pressure are kept constant. Energy lost by system = energy gained by surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given process, surely the change in internal energy (at constant S,V) be equal to the change in enthalpy (at constant P,S) and Gibbs free energy (at costant P,V), right? The above argument seems to make sense to me but enthalpy can be defined as $H=U+PV$ which would suggest that the two values would be different wouldn't it?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-05-27T21:38:22.073" Title="Are the values for enthalpy, internal energy and Gibbs free energy the same for a particular process?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23633" PostTypeId="1" AcceptedAnswerId="23768" CreationDate="2015-01-12T00:02:46.393" Score="5" ViewCount="2403" Body="&lt;p&gt;$$\mathrm dG &amp;lt; V\,\mathrm dp - S\,\mathrm dT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a process at constant pressure and temperature ($\mathrm dp = \mathrm dT = 0$), and in the absence of any non-pV work, $\mathrm dG &amp;lt; 0$. Since energy cannot be created or destroyed (and Gibbs free energy is a measure of energy) where does the energy go when this spontaneous change occurs? Likewise for internal energy: $\mathrm dU = T\,\mathrm dS - p\,\mathrm dV$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why is this even useful? I understand that it's easy to hold temperature and pressure constant in a lab but we only seem to measure enthalpy anyway using $\mathrm dH = C_p\,\mathrm dT$. If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) rather than Gibbs free energy. I suspect I am barking up the wrong tree and I am hoping that you'll be able to put me right; the only thought that I had was that the fact that Gibbs free energy muct decrease for a spontaneous change would help to predict whether a spontaneous change would occur when done in a lab - given that $p$ and $T$ will be constant; is this right?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2018-09-06T14:24:11.400" LastActivityDate="2018-09-06T14:25:33.620" Title="For a system to undergo a spontaneous change at constant temperature and pressure the Gibbs free energy must decrease, but how?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23634" PostTypeId="2" ParentId="23631" CreationDate="2015-01-12T00:52:52.823" Score="2" Body="&lt;p&gt;The $K_{sp}$ values are the way to go .. the other information you mentioned are just some convenient rules of thumb for predicting relative solubilities of compounds when you don't know the $K_{sp}$ values.  Also, the 4th choice in your list .. radium .. probably also belongs on that list, since it's the next alkali earth element after barium, but it's included because it's a radioactive element, and people don't commonly use or encounter it.&lt;/p&gt;&#xA;" OwnerUserId="12658" LastActivityDate="2015-01-12T00:52:52.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23635" PostTypeId="1" AcceptedAnswerId="23636" CreationDate="2015-01-12T01:19:37.410" Score="0" ViewCount="430" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A metal plate has an unwanted coating of $\ce{CaCO3(s)}$. You must find a way to clean this metal plate without causing any scratches or abrasion. How might you dissolve the coating?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know I should find a substance that will dissolve the coating, and then soak the plate in it. How do I determine an appropriate substance?&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4356" LastEditDate="2015-01-12T05:11:11.057" LastActivityDate="2019-07-02T18:57:37.680" Title="How could you dissolve a coating of CaCO3(s) from a metal plate?" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23636" PostTypeId="2" ParentId="23635" CreationDate="2015-01-12T01:27:09.587" Score="2" Body="&lt;p&gt;It might take some time, but I would use vinegar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acid in vinegar is strong enough to dissolve the calcium carbonate, but it is not strong enough to dissolve metals (unless the metal reacts with water).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-01-12T01:33:01.617" LastActivityDate="2015-01-12T01:33:01.617" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23637" PostTypeId="1" CreationDate="2015-01-12T02:54:51.480" Score="3" ViewCount="4523" Body="&lt;p&gt;According to Arrhenius acid/base theory, a base is a substance which when dissolved in water furnishes $\ce{OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\ce{CuO}$ is insoluble in water and does not produce $\ce{OH-}$, so why do we consider it a base?&lt;/p&gt;&#xA;" OwnerUserId="11316" LastEditorUserId="5017" LastEditDate="2015-01-12T04:49:44.050" LastActivityDate="2018-02-14T10:59:42.413" Title="why do we consider CuO as a base?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23638" PostTypeId="2" ParentId="23637" CreationDate="2015-01-12T03:55:05.707" Score="7" Body="&lt;p&gt;There are two other widely used acid/base theories that apply here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Br%C3%B8nsted%E2%80%93Lowry_acid%E2%80%93base_theory&quot; rel=&quot;noreferrer&quot;&gt;Brønsted–Lowry theory&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lewis_acids_and_bases&quot; rel=&quot;noreferrer&quot;&gt;Lewis theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Brønsted base is defined as a proton acceptor. Technically, $\ce{CuO}$ is capable of reacting with $\ce{H+}$ to form hydroxide ions and eventually $\ce{H2O}$. Although $\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\ce{CuO}$ is present as an undissolved solid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\ce{CuO}$ has lone pairs of electrons available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CuO}$ is considered a base because it fits these two descriptions.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-12T03:55:05.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23639" PostTypeId="2" ParentId="23626" CreationDate="2015-01-12T04:41:14.580" Score="7" Body="&lt;p&gt;I found the following p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; values for glycine and alanine from the CRC Handbook of Chemistry and Physics via &lt;a href=&quot;http://www.sigmaaldrich.com/life-science/metabolomics/learning-center/amino-acid-reference-chart.html&quot; rel=&quot;nofollow&quot;&gt;Sigma-Aldrich&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glycine $\ce{COOH}$: 2.34&lt;br/&gt;&#xA;Alanine $\ce{COOH}$: 2.34&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glycine $\ce{NH3+}$: 9.60&lt;br/&gt;&#xA;Alanine $\ce{NH3+}$: 9.69&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, although the values for the carboxylic acid are identical as reported, you would see a difference if you found a value reported with enough precision. The carboxylic acid is not immune to changes in the structure of the rest of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the difference between 9.60 and 9.69 is pretty small. In most practical situations, the two amino acids would behave identically with regard to the $\ce{NH3+}$ p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the magnitude of the difference is larger for the $\ce{NH3+}$ p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; values can likely be attributed to proximity. In this case, the relevant proton is 2 bonds away from the $\alpha$ carbon, whereas the proton in the $\ce{COOH}$ group is 3 bonds away.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastEditorUserId="7448" LastEditDate="2015-01-26T19:55:12.370" LastActivityDate="2015-01-26T19:55:12.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23641" PostTypeId="2" ParentId="23623" CreationDate="2015-01-12T05:17:10.887" Score="1" Body="&lt;p&gt;No. Suppose we had a container containing nothing but water.  Suppose it is only half full, so there is empty space. &lt;strong&gt;There is no  pressure from other gases.&lt;/strong&gt; If we heat this container enough, the water will boil and eventually turn completely into a gas. &#xA;In short, clearly other things influence the state of matter in everyday life besides pressure from atmospheric gases. In this example, I illustrate how water evaporating in a vacuum is determined by temperature.  I think  this is a good example because it involves just one substance and doesn't bring into question how to think about mixed gases and partial pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an extremely useful equation that connects these relationships beautifully: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PV = nRT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P stands for pressure, V for volume, T for temperature, n for the number of molecules, and R is a constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation basically says that&#xA;&lt;em&gt;any change in pressure or volume to a closed system of gas will result in a corresponding change to the temperature of the system.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, as long as $P_1V_1\neq P_2V_2$, you can always think about a change in volume or pressure as a change in temperature for a closed system. &lt;/p&gt;&#xA;" OwnerUserId="11266" LastEditorUserId="11266" LastEditDate="2015-01-12T18:53:15.113" LastActivityDate="2015-01-12T18:53:15.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23642" PostTypeId="1" AcceptedAnswerId="23648" CreationDate="2015-01-12T05:40:30.803" Score="12" ViewCount="3076" Body="&lt;p&gt;Recently, I was taught that the coefficients of the reactants in an elementary step of a reaction indicates that reactant's order. Why is this the case? I thought that the order of a reactant was not related to its coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the following termolecular reaction mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + 2B -&amp;gt; C + 2D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rate law is $k[\ce{A}][\ce{B}]^2$. Why is this the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Also, for equilibrium reactions, do we always assume the coefficients of the reactants to provide the order for those reactants? What about for equilibrium reactions involving multiple steps?&lt;/p&gt;&#xA;" OwnerUserId="9281" LastEditorUserId="30836" LastEditDate="2017-10-18T16:39:56.260" LastActivityDate="2017-10-18T16:39:56.260" Title="In the elementary steps of a reaction, why do we assume that the coefficients of a reactant correspond to its order?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23643" PostTypeId="1" AcceptedAnswerId="23656" CreationDate="2015-01-12T06:51:19.887" Score="6" ViewCount="8066" Body="&lt;p&gt;I am doing some rough calculations to test my understanding of elementary chemistry. In what follows, I calculate the volume of one mole of liquid water at about $\pu{4 ^\circ C}$ and $\pu{1 atm}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 1: Using the molar mass and density of water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of water is $\pu{18 \times 10^-3 kg mol-1}$, and the density is $\pu{1 \times 10^3 kg m-3}$. Hence, the volume should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{(\pu{1 mol})(\pu{18 \times 10^-3 kg mol-1})}{\pu{1 \times 10^3 kg m-3}} = \pu{1.8 \times 10^-5 m3}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Method 2: Using the size of an individual water molecule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we model the mole of water as a cube, since there are $6.022 \times 10^{23}$ water molecules, each side of the cube would have $8.44 \times 10^{7}$ molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Properties_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;water data page&lt;/a&gt; on Wikipedia, I used trigonometry to calculate the H–H distance as being approximately $\pu{150 \times 10^-12 m}$. I would think the dimensions do not change too much when there are many molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I find that the volume is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = [(8.44 \times 10^7) \times (\pu{150 \times 10^-12 m})]^3 = \pu{2.05 \times 10^-6 m3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is an order of magnitude smaller than Method 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this discrepancy?&lt;/p&gt;&#xA;" OwnerUserId="12707" LastEditorUserId="16683" LastEditDate="2018-09-06T14:16:24.093" LastActivityDate="2018-09-06T14:17:21.823" Title="Calculating the volume of 1 mole of liquid water" Tags="&lt;water&gt;&lt;molecules&gt;&lt;mole&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23644" PostTypeId="1" CreationDate="2015-01-12T07:15:54.543" Score="5" ViewCount="543" Body="&lt;p&gt;Temperature before, or after reaction? Temperature change? Or what is it?&lt;/p&gt;&#xA;" OwnerUserId="12711" LastActivityDate="2015-04-12T16:32:11.613" Title="What value of temperature does T represent in Gibbs Free Energy Equation?" Tags="&lt;enthalpy&gt;&lt;temperature&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23645" PostTypeId="1" AcceptedAnswerId="23653" CreationDate="2015-01-12T08:36:01.577" Score="6" ViewCount="18501" Body="&lt;p&gt;It is commonly accepted that you can hang your wet clothes outside to dry even when the winter temperature is below freezing. It is mentioned here&#xA;&lt;a href=&quot;http://grist.org/living/ask-umbra-can-i-line-dry-my-clothes-in-the-winter-too/&quot;&gt;http://grist.org/living/ask-umbra-can-i-line-dry-my-clothes-in-the-winter-too/&lt;/a&gt;&#xA;that water in the clothes will freeze to ice, then sublimate into water vapor, making the clothes dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Why does the ice in the clothing sublimate, while the ice on the roads don't sublimate? Why is there a difference between what happens to ice on the clothing versus ice on the roads?&lt;/p&gt;&#xA;" OwnerUserId="12707" LastActivityDate="2019-01-31T13:44:48.647" Title="Drying Clothes Outside Below Freezing" Tags="&lt;water&gt;&lt;sublimation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23646" PostTypeId="1" AcceptedAnswerId="23659" CreationDate="2015-01-12T08:49:31.820" Score="2" ViewCount="82" Body="&lt;p&gt;Take internal energy for example:&#xA;$$dU\leq TdS-PdV$$&#xA;Firstly, I would like to make clear that the following question/argument relies on the following statement being correct: the left hand side of the above equation represents the change in internal energy of the system and the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two equations, above, are in fact inequalities where it only becomes an equality when the system is at equilibrium (when changes are reversible). However, when  the system is not in equilibrium the inequality implies that the work done on the system by the surroundings is greater than the internal energy gained by the system; how can this be? Where does the energy difference between the energy put in and the energy gained by the system go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2015-01-12T13:56:52.640" LastActivityDate="2015-02-11T16:39:59.027" Title="What happens to energy changes in non-reversible processes?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23647" PostTypeId="2" ParentId="23644" CreationDate="2015-01-12T09:57:07.723" Score="1" Body="&lt;p&gt;We can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation:&#xA;$$\Delta G = \Delta H - \Delta TS $$&lt;/p&gt;&#xA;" OwnerUserId="12669" LastEditorUserId="7448" LastEditDate="2015-01-12T13:10:03.263" LastActivityDate="2015-01-12T13:10:03.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23648" PostTypeId="2" ParentId="23642" CreationDate="2015-01-12T10:21:56.070" Score="9" Body="&lt;p&gt;If a reaction has just one step, it's named an &lt;em&gt;elementary reaction&lt;/em&gt;. In this reaction all of reactant directly participate on reaction and effectively collide with each other. So, the order of reaction corresponds to its coefficient, but if a reaction is not a elementary reaction, there isn't any reason for colliding all reactants with each other. For example, consider the following overall reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B + D -&amp;gt; F}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the corresponding mechanism is constituted of two elementary reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + B -&amp;gt; C}$$&#xA;$$\ce{C + D -&amp;gt; F}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown above, there is no need to collide A with D or A with C ... Then, in these type of reactions coefficients are not corresponded to order of reaction . &lt;/p&gt;&#xA;" OwnerUserId="12669" LastEditorUserId="7581" LastEditDate="2015-01-12T13:40:03.913" LastActivityDate="2015-01-12T13:40:03.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23649" PostTypeId="2" ParentId="23644" CreationDate="2015-01-12T11:04:27.890" Score="1" Body="&lt;p&gt;We've got:$$G_{(p,T)}= U+pV-TS$$&#xA;Where G is Gibbs free energy, U is the thermodynamic system's internal energy, p is pressure, V is volume, T is temperature and finally, S in entropy. In a chemical reaction, we can change the equation above to have:$$\Delta{G}_{(p,T)}= \Delta{U} +p\Delta{V}- T\Delta{S}$$&#xA;Which is then &quot;interpreted&quot; to:$$\Delta{G}_{(p,T)}= \Delta{H} - T\Delta{S}$$&#xA;This is what learners like me learn during the high school. Note that Gibbs free energy (measured in here) works well when deciding about the spontaneity of processes &lt;strong&gt;that occur in constant temperatures and pressures&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you're to use the equation, know that you oblige the poor poor system to be in a constant temperature! Think about it. If you're being asked a question from things you've studied, you &lt;em&gt;will&lt;/em&gt; definitely run into phrases like &lt;code&gt;STP&lt;/code&gt; or &lt;code&gt;RTP&lt;/code&gt; in the question content. There is, however, a &quot;formula expansion&quot; to where we have variable temperatures; but of course, that wouldn't use the equations above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's still vague for you, you can comment me.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-01-12T12:23:10.730" LastActivityDate="2015-01-12T12:23:10.730" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23650" PostTypeId="1" AcceptedAnswerId="32679" CreationDate="2015-01-12T11:16:06.017" Score="8" ViewCount="4285" Body="&lt;p&gt;Is it correct to say pyridine’s non-bonding electron pair does not participate in electron resonance while pyrrole’s non-bond electron participates in electron resonance? Would therefore pyridine’s resonance not be disturbed so act the molecule can act as a monodentate ligand?&lt;/p&gt;&#xA;" OwnerUserId="12669" LastEditorUserId="16683" LastEditDate="2017-03-22T16:39:07.893" LastActivityDate="2017-03-22T16:39:07.893" Title="Why is pyridine a monodentate ligand, but pyrrole a tridentate ligand?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23651" PostTypeId="1" CreationDate="2015-01-12T11:59:51.913" Score="0" ViewCount="6797" Body="&lt;p&gt;I am doing a AP Chemistry lab in which we determine the heat capacity of a calorimeter (a concept that seems like it should depend on a lot more than just the calorimeter.) We have to calorimeters with water of different temperatures (which is tracked the whole time) but equal mass. We then mix them. We then track temperature again. We then extrapolate to the time of mixing (with a linear regression for some reason.) We look at how much energy was lost by the hot water and gained by the cold water based on all 3 lines (each calorimeter before, and the mixed calorimeter after.) Next we take &lt;strong&gt;the difference of the energy gained and energy lost at a single instant&lt;/strong&gt; divide by the temperature change of the cold calorimeter (I have no idea why the &lt;em&gt;cold&lt;/em&gt; one) and this is the calorimeters heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble understanding this, since at a single instance, the energy gained by the hot water and lost by the cold should be equal, and any difference would be error. It also seems completely random to divide by the temperature of the cold water and not the hot. What is going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Later in the lab, extrapolating to the time of mixing is used to find the energy change due to a reaction, which makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.S One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.S.S It is the teacher's first year doing this lab I think.&lt;/p&gt;&#xA;" OwnerUserId="7219" LastEditorUserId="7219" LastEditDate="2015-01-12T12:33:17.920" LastActivityDate="2017-03-11T04:01:54.853" Title="How does this method of calculating the water equivalent of a calorimeter work?" Tags="&lt;homework&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23652" PostTypeId="1" AcceptedAnswerId="23666" CreationDate="2015-01-12T12:00:30.537" Score="5" ViewCount="2183" Body="&lt;p&gt;The Born equation gives the difference in energy required to charge a particle in a vacuum and in solution which results in the work required to transfer an ion from a vacuum into solution. It is derived thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=\int_0^Q\frac{Q}{4\pi\epsilon r}dQ-\int_0^Q\frac{Q}{4\pi\epsilon_0 r}dQ$$&#xA;Where $\epsilon=\epsilon_r\epsilon_0$ ($\epsilon_r$ is the relative permittivity of the medium/solvent - a dimensionless quantity). $w$ is work done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=\frac{1}{4\pi\epsilon r}\int_0^QQdQ-\frac{1}{4\pi\epsilon_0 r}\int_0^QQdQ$$&#xA;$$w=\frac{Q^2}{8\pi\epsilon r}-\frac{Q^2}{8\pi\epsilon_0 r}$$&#xA;$$w=\frac{z^2e^2}{8\pi\epsilon r}-\frac{z^2e^2}{8\pi\epsilon_0 r}$$&#xA;So for a mole of ions the work done is equal to:&#xA;$$\frac{N_Az^2e^2}{8\pi\epsilon r}-\frac{N_Az^2e^2}{8\pi\epsilon_0 r}=\triangle_{solv}G^{\theta}$$&#xA;My textbook seems to suggest that the last step makes it Gibbs free energy of solvation rather than just work done. I have two issues with this. Firstly, why is it Gibbs free energy and not, say, enthalpy of solvation or internal energy of solvation? Secondly, why would scaling the number of ions to one mole make it into Gibbs free energy? Isn't Gibbs free energy an extensive property with units of Joules? I know that the Gibbs free energy (and enthalpy) are often given with units of $Jmol^{-1}$ but isn't this strictly &lt;strong&gt;molar&lt;/strong&gt; gibbs free energy/chemical potential?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-12T16:16:12.760" Title="Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23653" PostTypeId="2" ParentId="23645" CreationDate="2015-01-12T12:06:07.537" Score="11" Body="&lt;p&gt;As long as there is less than 100% relative humidity, the ice will sublime to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantitative information see &quot;&lt;a href=&quot;http://snobear.colorado.edu/Markw/Research/hood99.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sublimation from a seasonal snowpack at a continental,&#xA;mid-latitude alpine site&lt;/a&gt;&quot; Hydrol. Process. 13, 1781-1797 (1999).&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2019-01-31T13:44:48.647" LastActivityDate="2019-01-31T13:44:48.647" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23654" PostTypeId="2" ParentId="23651" CreationDate="2015-01-12T12:17:42.750" Score="1" Body="&lt;p&gt;The point is that heat is not just stored in the water, but also in the calorimeter sturcture.  Heat that is unaccounted for just considering the temperature and amount of water must have been taken up by the calorimeter itself.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-12T13:20:09.990" LastActivityDate="2015-01-12T13:20:09.990" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23655" PostTypeId="2" ParentId="23644" CreationDate="2015-01-12T12:49:36.147" Score="1" Body="&lt;p&gt;The temperature must remain constant so it is the value both at the start and the end. I will try to explain this simply as, from your question, I suspect that you have just been introduced to $G$ with the equation $G=H-TS$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs free energy helps to indicate whether a process/reaction will be spontaneous at conditions where the pressure and temperature remain constant. This is useful because these are the conditions that occur in lab when we do an experiment. As I am sure you know the Gibbs free energy must decrease for a spontaneous process; in other words $\Delta G&amp;lt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the equation $G=H-TS$ does not involve $\Delta G$ is only gives a relation between Gibbs free energy and enthalpy and entropy. It is clearly useful to consider changes in $G$ so lets say $G$ changes by an infinitesimally small amount: $\mathrm{d}G$ (this basically means the same as $\Delta G$ but the change is much smaller) we can change the expression so that $\mathrm{d}G$ is written in terms of infinitesimal changes in entropy temperature and enthalpy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(G+\mathrm{d}G)=(H+\mathrm{d}H)-(T+\mathrm{d}T)(S+\mathrm{d}S)$$&#xA;Expanding the brackets in the last expression:&#xA;$$G+\mathrm{d}G=H+\mathrm{d}H-TS+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$&#xA;I will now rearrange to make the following step clear:&#xA;$$\mathrm{d}G+[G=H-TS]+\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$&#xA;Notice the $G=H-TS$ in the equation (highlighted in square brackets) – this can now cancel because it’s the same as having $G$ on each side. This leaves:&#xA;$$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$&#xA;The last term is the product of two infinitesimally small changes (very very small numbers) so is basically zero and can be ignored. This leaves:&#xA;$$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S$$&#xA;Now, remember towards the start of my answer I said that this is useful at a constant temperature and pressure. Well, if temperature is constant $\mathrm{d}T=0$ so the $S\mathrm{d}T$ term vanishes leaving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}G=\mathrm{d}H-T\mathrm{d}S$$ and for larger changes: $$\Delta G =\Delta H -T\Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is crucial that you know, and hopefully now understand after my explanation, that the above equation is &lt;strong&gt;only&lt;/strong&gt; valid at a constant pressure and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short $T$ is &lt;strong&gt;the constant&lt;/strong&gt; value for temperature.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-04-12T16:32:11.613" LastActivityDate="2015-04-12T16:32:11.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23656" PostTypeId="2" ParentId="23643" CreationDate="2015-01-12T13:17:27.227" Score="8" Body="&lt;p&gt;The molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will change your answer by a factor of 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. &lt;em&gt;J. Mol. Liq.&lt;/em&gt; &lt;strong&gt;2001,&lt;/strong&gt; &lt;em&gt;90&lt;/em&gt; (1-3), 303–312. &lt;a href=&quot;https://doi.org/10.1016/S0167-7322(01)00134-9&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/S0167-7322(01)00134-9&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2018-09-06T14:17:21.823" LastActivityDate="2018-09-06T14:17:21.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23658" PostTypeId="1" CreationDate="2015-01-12T13:40:24.590" Score="3" ViewCount="6734" Body="&lt;p&gt;I have a 6 Volt sulfuric acid battery used to power a golf cart that has all the acid drained out, and the usual solution was to add more water to the battery however I am afraid that would not work this time since there is no acid left. Could I solve it by adding more acid from other batteries? or do I need to be accurate in the amount of acid added? or could I just solve it by adding water? If you could also give me an explanation as to how these batteries work and why adding water has any benefit I would appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorUserId="7639" LastEditDate="2015-01-12T13:47:12.393" LastActivityDate="2015-01-23T17:28:24.357" Title="Reviving a 6 Volt lead-sulfuric acid battery" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23659" PostTypeId="2" ParentId="23646" CreationDate="2015-01-12T13:44:17.310" Score="1" Body="&lt;p&gt;$\Delta U = q + w$ and $dU = dq + dw$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;a href=&quot;http://en.wikipedia.org/wiki/First_law_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;first law of thermodynamics&lt;/a&gt;.  It is not an inequality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dU\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The PdV term alone represents the reversible work, the TdS term represents reversible heat&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the system must lose internal energy in a spontaneous change&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-12T15:48:59.693" LastActivityDate="2015-01-12T15:48:59.693" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="23660" PostTypeId="2" ParentId="23658" CreationDate="2015-01-12T14:27:47.430" Score="1" Body="&lt;p&gt;Obviously handling sulfuric acid is hazardous.  If all the acid leaked out, I would get a new battery, otherwise the acid will leak out again and possibly hurt someone or damage the vehicle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no acid, certainly adding water will not help.  If you do add acid, the concentration of acid needs to be correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead-acid batteries do not contain pure sulphuric acid, but acid dilute with water.  The concentration of acid can increase over time due to electrolysis of the water to hydrogen and oxygen gases.  If the concentration of acid is too high (solution density above 1.19 g/ml), adding water to dilute the acid is beneficial. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-12T14:27:47.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23661" PostTypeId="1" CreationDate="2015-01-12T14:32:32.517" Score="3" ViewCount="1185" Body="&lt;p&gt;I read somewhere that atoms cannot be created. If this is true, then how did the atoms form after the Big Bang? Also, does this mean that the number of atoms in our universe has remained the same since then? Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?&lt;/p&gt;&#xA;" OwnerUserId="12718" LastEditorUserId="4945" LastEditDate="2015-01-13T05:52:50.437" LastActivityDate="2015-01-13T05:52:50.437" Title="It is said that atoms cannot be created. If so, then how did atoms get created after the Big Bang?" Tags="&lt;atoms&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="2" ClosedDate="2015-01-12T19:43:45.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="23662" PostTypeId="1" CreationDate="2015-01-12T14:46:36.003" Score="0" ViewCount="5837" Body="&lt;p&gt;Why is the &lt;strong&gt;second&lt;/strong&gt; electron affinity of oxygen greater than that of sulfur? &#xA;I think it should be like- second electron gain enthalpy of oxygen should be greater than the sulfur but electron affinity should be less than sulfur.&lt;/p&gt;&#xA;" OwnerUserId="12719" LastEditorUserId="402" LastEditDate="2016-12-04T21:01:28.660" LastActivityDate="2016-12-04T21:01:28.660" Title="Second electron affinity of sulfur and oxygen" Tags="&lt;inorganic-chemistry&gt;&lt;atoms&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23663" PostTypeId="2" ParentId="23661" CreationDate="2015-01-12T14:54:02.880" Score="8" Body="&lt;p&gt;Atoms can be created and destroyed.  For example, in nuclear fusion two or more atoms make one atom.  In nuclear fission, one atom makes two or more atoms.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how did the atoms form after the Big Bang? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The theory of this process in known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Big Bang Nucleosynthesis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ID5B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from protons and neutrons, light elements formed.  Later, in stars, nuclei up to iron formed through nuclear fusion.  Nuclei heavier than iron formed from supernovae by the &lt;a href=&quot;http://en.wikipedia.org/wiki/R-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;r-process&lt;/a&gt; and asymptotic branch giant stars by the &lt;a href=&quot;http://en.wikipedia.org/wiki/S-process&quot; rel=&quot;nofollow noreferrer&quot;&gt;s-process&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, does this mean that the number of atoms in our universe has remained the same since then? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  The number constantly changes due to fusion, fission, supernovae, creation of neutron stars, black holes and possibily &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark_star&quot; rel=&quot;nofollow noreferrer&quot;&gt;quark stars&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently, this process does not create or destroy atoms.  However, according to the theory of &lt;a href=&quot;http://en.wikipedia.org/wiki/Accelerating_universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;accelerating expansion&lt;/a&gt;, there is a senario referred to as the &quot;&lt;a href=&quot;http://io9.com/5919193/the-big-rip-theory-says-the-universe-could-end-in-tears&quot; rel=&quot;nofollow noreferrer&quot;&gt;big rip&lt;/a&gt;&quot; in which all atoms would be destroyed.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-12T18:33:38.853" LastActivityDate="2015-01-12T18:33:38.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23664" PostTypeId="1" AcceptedAnswerId="23704" CreationDate="2015-01-12T15:13:15.653" Score="4" ViewCount="692" Body="&lt;p&gt;The physics part: The part that talks about being shiny, I agree, is mainly physics. But proceed in my question, and you'll see how it's a chemistry one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know things are shiny if they either produce light, or deflect light well. Producing is not our topic in here, but deflecting is. Metals are considered shiny objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.madsci.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mad scientist&lt;/a&gt; (the site, not our fellow friend!) has provided &lt;a href=&quot;http://www.madsci.org/posts/archives/2000-08/967126944.Ph.r.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a nice explanation&lt;/a&gt; for why metals are shiny: In short, because free electrons (electrons are partially &quot;free&quot; in metals, due to the especial way of bonding) deflect more photons than usual and do not let the inner layers of atoms in the metal absorb them.&lt;img src=&quot;https://i.stack.imgur.com/lRIKC.jpg&quot; alt=&quot;&quot;&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(These three differ in shine because of impurity; not related)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals aren't the only materials that are exceptionally &quot;shiny&quot;. We are aware of many crystals that are either well-cut or well-formed, which result in &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_internal_reflection&quot; rel=&quot;nofollow noreferrer&quot;&gt;total internal refraction&lt;/a&gt; and that makes them very shiny too. There's &lt;a href=&quot;https://physics.stackexchange.com/questions/43361/why-do-diamonds-shine&quot;&gt;a related question&lt;/a&gt; in physics.SE wherein the diamond's shiny properties are questioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organic-chemistry part: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying the word &quot;oil&quot; talks about a vast variety of chemicals; therefore, in an attempt to narrow down the question, I only ask about liquid-at-RTP triacylglycerol (the famous triglycerides which exist usually in plants or fish). The chemical structure is as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3iOfS.gif&quot; alt=&quot;&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/F9NcK.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Organic molecules like these aren't metals, are they?! So since a thorough research (maybe seemingly) didn't get me answers about why these are good light reflectors I'm asking for a way that will allow me to compare the reflectivity of these liquids, primarily from their structural formula. e.g.: Are these liquids &quot;shinier&quot; than water? Why? About water, I would say that for example, Hydrogen atoms in correspondence to the Oxygen atom, make water have hydrogen bonding, which is the main cause of surface tension. Surface tension causes the surface of water to be &quot;smoother&quot; and thus the reflectivity increases. On the other hand, it can't be only about Hydrogen bonding, can it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If the answer might be too broad, you may guide me to any resource helping me to find an answer, and that's why I add the tag: reference-request.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image sources: &lt;a href=&quot;http://www.luxrender.net/wiki/images/2/2f/Shiny_yellow_specular_series.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.luxrender.net/wiki/images/2/2f/Shiny_yellow_specular_series.jpg&lt;/a&gt; ; &lt;a href=&quot;http://library.med.utah.edu/NetBiochem/FattyAcids/7_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://library.med.utah.edu/NetBiochem/FattyAcids/7_2.html&lt;/a&gt; ; &lt;a href=&quot;http://guweb2.gonzaga.edu/faculty/cronk/biochem/images/triglyceride.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://guweb2.gonzaga.edu/faculty/cronk/biochem/images/triglyceride.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-01-13T17:18:07.187" Title="Is oil shinier than water? Why?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;molecules&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23665" PostTypeId="2" ParentId="23637" CreationDate="2015-01-12T15:39:12.613" Score="5" Body="&lt;p&gt;An additional way to consider it: $\ce{CuO}$ is defined as a base in the Lux-Flood theory because it is an oxide ion donor; it donates oxide to oxide acceptors (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lux-Flood_theory&quot; rel=&quot;nofollow&quot;&gt;Lux-Flood acids&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a Lux-Flood acid-base reaction involving $\ce{CuO}$ is the high pressure reaction with carbon dioxide to give $\ce{CuCO_3}$   (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/zaac.19744100207/abstract&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="5017" LastEditDate="2015-01-12T16:56:42.287" LastActivityDate="2015-01-12T16:56:42.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23666" PostTypeId="2" ParentId="23652" CreationDate="2015-01-12T15:48:34.553" Score="5" Body="&lt;p&gt;1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$&#xA;where the equality sign holds for reversible changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's proof:&#xA;$$\begin{align*}&#xA;H &amp;amp;= U + P V&amp;amp;\text{(definition of enthalpy)}\\&#xA;dH &amp;amp;= dq + dw + d(PV) &amp;amp;\text{(because dU = dq + dw)}\\&#xA;dG &amp;amp;= dH - T dS - S dT &amp;amp;\text{(because G = H - TS)}\\&#xA;&amp;amp;=dq + dw + d(PV) - T dS -S dT &amp;amp;\\&#xA;&amp;amp;= dq + dw + d(PV) - T dS &amp;amp; \text{(when the change is isothermal)}\\&#xA;&amp;amp;= dw_{rev} + d(PV) &amp;amp;\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\&#xA;&amp;amp;= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP &amp;amp; \text{(partitioning the work)}\\&#xA;&amp;amp;= dw_{electrical,rev} + V dP &amp;amp; (dw_{expansion,rev} = - P dV)\\&#xA;&amp;amp;= dw_{electrical,rev} &amp;amp; \text{(if pressure is constant)}\\&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-12T16:16:12.760" LastActivityDate="2015-01-12T16:16:12.760" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="23668" PostTypeId="1" AcceptedAnswerId="23679" CreationDate="2015-01-12T16:13:29.483" Score="4" ViewCount="2632" Body="&lt;p&gt;What is the $\ce{OH}$ functional group in alcohol (ethanol)? My teacher told me it isn't the same as the $\ce{OH-}$ ion or hydroxide.&lt;br/&gt;&#xA;But it does donate one electron to the respective carbon chain and it's very tempting to say it's hydroxide as it fits perfectly into the role. What exactly is it?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="9961" LastEditDate="2015-01-12T17:14:22.650" LastActivityDate="2021-01-01T00:11:20.657" Title="What is the OH in alcohol?" Tags="&lt;organic-chemistry&gt;&lt;ions&gt;&lt;alcohols&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23669" PostTypeId="2" ParentId="23668" CreationDate="2015-01-12T17:24:19.913" Score="2" Body="&lt;p&gt;Well, I don’t understand what you are trying to ask. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the $\ce{OH}$ you are bothered about can be thought exactly as that of $\ce{OH}$ of water ($\ce{H2O}$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2H5-OH}$ in the same way as $\ce{H-OH}$&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T14:25:57.890" LastActivityDate="2015-04-26T14:25:57.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23670" PostTypeId="1" AcceptedAnswerId="23678" CreationDate="2015-01-12T17:28:29.847" Score="8" ViewCount="17490" Body="&lt;p&gt;One of the uses of the Tollens' reagent is to confirm the presence of a terminal triple bond in a compound. Does acetylene give a positive Tollens' test? If not, why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorDisplayName="user7951" LastEditDate="2019-05-12T08:24:18.783" LastActivityDate="2020-02-04T14:03:42.813" Title="Does acetylene give a positive Tollens' test? If not, why is this so?" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23671" PostTypeId="2" ParentId="23590" CreationDate="2015-01-12T17:32:23.543" Score="1" Body="&lt;p&gt;I don't know whether it's correct or not but what I think is due to steric factors .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the SCN- is pretty large ligand as compare to H2O .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there might be some additional stabilisation in water as ligand due to intra atomic hydrogen bonding. &lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2015-01-12T17:32:23.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23673" PostTypeId="1" CreationDate="2015-01-12T17:44:28.823" Score="3" ViewCount="4914" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bjG0c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/c/c0/NaCl_polyhedra.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this diagram of the rock salt structure ($\ce{NaCl}$) we see that both the chloride and sodium ions have the same environment. That is to say, they each have the same number of neighbours at every distance, except that the charges are opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've always been taught to consider the chloride ions as forming a face-centred cubic (fcc) [or cubic close packed (ccp)] lattice, with sodium ions filling octahedral holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why we can't think of it the other way round, with sodium ions forming the fcc lattice and chloride ions filling octahedral holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it perhaps because sodium ions are smaller than chloride ions, or something else?&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorUserId="16683" LastEditDate="2017-08-01T03:27:48.203" LastActivityDate="2017-08-01T03:27:48.203" Title="Rock salt structure: chloride lattice or sodium lattice?" Tags="&lt;crystal-structure&gt;&lt;ionic-compounds&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23675" PostTypeId="2" ParentId="17109" CreationDate="2015-01-12T17:51:01.493" Score="1" Body="&lt;p&gt;Ordinary candle wax is readily removed by Methylated Spirits, especially if is on a candle stick. Scrape the bulk off carefully and then use a soft cloth with meths.&lt;/p&gt;&#xA;" OwnerUserId="12724" LastActivityDate="2015-01-12T17:51:01.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23677" PostTypeId="1" CreationDate="2015-01-12T17:54:53.103" Score="1" ViewCount="1290" Body="&lt;p&gt;Why is ionization potential considered a periodic property???&#xA;A periodic property is a one which appears at regular property&#xA;But we see that every element has some kind of ionization potential&lt;/p&gt;&#xA;" OwnerUserId="11316" LastEditorUserId="5633" LastEditDate="2015-01-12T20:42:57.930" LastActivityDate="2019-02-19T08:26:51.167" Title="Why is ionization potential considered a periodic property?" Tags="&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23678" PostTypeId="2" ParentId="23670" CreationDate="2015-01-12T17:59:44.163" Score="18" Body="&lt;p&gt;Tollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{RC#CH + AgNO3 + NH4OH -&amp;gt; RC#CAg v + NH4NO3 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in &lt;a href=&quot;http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a PDF from www.sakshieducation.com/&lt;/a&gt; or &lt;a href=&quot;https://web.archive.org/web/20200204140114/http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;via the Internet Archive&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HC#CH + 2AgNO3 + 2NH4OH -&amp;gt; AgC#CAg v + 2NH4NO3 + 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tollen's reagent is not very stable, therefore it needs to be freshly prepared before use.  The reagent has a number of other uses in chemistry (test for aldehydes and &amp;alpha;-hydroxy ketones), anatomical staining, and silver mirroring.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Tollens&amp;#39;_reagent&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia article&lt;/a&gt; provides a nice, concise overview.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2020-02-04T14:03:42.813" LastActivityDate="2020-02-04T14:03:42.813" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23679" PostTypeId="2" ParentId="23668" CreationDate="2015-01-12T18:02:08.580" Score="8" Body="&lt;p&gt;You could call ethyl alcohol &quot;ethyl hydroxide&quot; analogously to &quot;sodium hydroxide&quot;. But it is a poor analogy, since (unlike NaOH) ethyl alcohol is not an ionic compound, does not dissociate in water to form ethyl cation and hydroxide anion, and does not raise the pH when dissolved in water.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-12T18:02:08.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23680" PostTypeId="2" ParentId="23590" CreationDate="2015-01-12T18:05:47.390" Score="3" Body="&lt;p&gt;Firstly, in water Co(II) is not exclusively $\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. &lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/v80-223&quot; rel=&quot;nofollow&quot;&gt;Canadian Journal of Chemistry, 1980, 58(14): 1418-1426&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Greenwood's &quot;Chemistry of the elements&quot; 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-12T18:05:47.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23681" PostTypeId="1" CreationDate="2015-01-12T18:11:58.060" Score="1" ViewCount="220" Body="&lt;p&gt;I truly don't understand the method for finding the net-ionic equation for a reaction. It is not clear to me how you should decide what the products are, especially when there are multiple possibilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Excess potassium cyanide solution is added to aluminum bromide solution. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first thing that came to mind for the molecular equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3KCN\+AlBr3 \to 3KBr\+Al(CN)3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But apparently the correct molecular equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no good reason why the one I thought of is incorrect. And to get the net ionic equation, the spectator ions are $\ce{K+}$ and $\ce{Br-}$. But why should I assume the solution is aqueous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chlorine gas is bubbled through dilute sodium hydroxide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well its easy enough to get the reactants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2\+NaOH \to ?}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What first comes to mind is $\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the correct molecular equation is $\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should be the thought process for arriving at the correct answer here. I am truly lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another example&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lithium oxide reacts with sulfur dioxide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I recognize this as a synthesis reaction. However how can I predict if the product is lithium sulfite or lithium sulfate?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="5026" LastEditDate="2018-02-23T01:56:25.033" LastActivityDate="2018-02-23T01:57:47.780" Title="Is there a method for finding the net-ionic equation?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23682" PostTypeId="2" ParentId="23673" CreationDate="2015-01-12T18:33:04.373" Score="7" Body="&lt;p&gt;Unit cells are just a way of describing a given crystal. You can certainly draw $\ce{NaCl}$ the other way, with $\ce{Na+}$ in the corners and it's still correct. There are a few ways of thinking about this structure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;as two interpenetrating FCC lattices&lt;/li&gt;&#xA;&lt;li&gt;as a $\ce{Cl-}$ lattice with $\ce{Na+}$ in the octahedral holes&lt;/li&gt;&#xA;&lt;li&gt;as a $\ce{Na+}$ lattice with $\ce{Cl-}$ in the octahedral holes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;They're all equivalent. The figure above actually shows both types of octahedral holes. Some might prefer to visualize the structure as the second case since, as you say, the chloride ions are bigger, but going the other way isn't wrong, nor does it describe anything different.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-12T18:33:04.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23683" PostTypeId="2" ParentId="23681" CreationDate="2015-01-12T18:44:23.617" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But apparently the correct molecular equation is $\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's not correct, what happenend to the other 3 potassium atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could form $\ce{Al2(CN)6}$ but you can't exclude $\ce{Al(CN)3}$ without further information beyond balancing of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main steps to balancing equation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Balancing of each type of atom&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Balancing number of electrons&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you've done both of those correctly, you've done all you can with balancing alone. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Above is clearly wrong because number of oxygen atoms doesn't balance&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Above is clearly wrong because number of oxygen atoms doesn't balance&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-02-23T01:57:47.780" LastActivityDate="2018-02-23T01:57:47.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23684" PostTypeId="1" AcceptedAnswerId="23691" CreationDate="2015-01-12T19:11:41.040" Score="4" ViewCount="1572" Body="&lt;p&gt;My confusion regarding the difference between the heat given off in a reaction that takes place in an electrochemical cell compared to in a beaker/appropriate container when the reactants are simple mixed, arose from the following problem question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{HgCl2~(s) + H2~(g,1~bar) &amp;amp;-&amp;gt; 2Hg~(l) + 2H+~(aq) + 2Cl- }\\&#xA;\Delta G^\circ &amp;amp;=-51.64~\mathrm{kJmol^{-1}},\\&#xA;\Delta S^\circ &amp;amp;=-61.6~\mathrm{JK^{-1}mol^{-1}},\\&#xA;\Delta H^\circ &amp;amp;=-70~\mathrm{kJmol^{-1}}.\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (a) of the question then asks: &quot;Determine how much heat is given off at $298~\mathrm{K}$ per mole of $\ce{HgCl2}$ if the cell is operated reversibly&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct approach is to use $T\,\mathrm{d}S=\mathrm{d}q$ giving a value of: $18.36~\mathrm{kJmol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then part (b) asks: &quot;Now determine the heat given off if the reaction is brought about by simply mixing the reactants&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realised that the electrical work done in the cell would now be given off as heat so I just added the value for the Gibbs free energy to the value calculated in (a) and got: $70~\mathrm{kJmol^{-1}}$. This answer is correct but my tutor noted that this value is the value for the enthalpy change which would have been a shortcut to the answer. I didn't spot this and I don't understand why this is the case so wouldn't spot it again, should another question like this come up. &lt;strong&gt;Please can you clear up why this value is the enthalpy change.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some issues with the above working (it is all correct but I don't fully understand it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, in part (a) I had to use the Clausius inequality (except it was an equality since the process was carried out reversibly) to find $\mathrm{d}q$. However, why must I use this equation because $\Delta H=q_p$ - the process was carried out at a constant pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, if the heat given off in part (b) is $70~\mathrm{kJmol^{-1}}$ doesn't this violate the Clausius inequality? I must be wrong but it seems that this must be the case since $\mathrm{d}S \geq \frac{\mathrm{d}Q}{T}$ and using the value for the entropy change ($\Delta S^\circ=-61.6~\mathrm{JK^{-1}mol^{-1}}$) and the new calculated value for the heat given off ($70~\mathrm{kJmol^{-1}}$) then dividing this by $298~\mathrm{K}$ gives a value larger than the value for the entropy change but the Clausius inequality says that the change in entropy is &lt;strong&gt;greater than or equal to&lt;/strong&gt; $\mathrm{d}q/T$. What is going on there?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-01-16T08:58:09.763" LastActivityDate="2015-01-16T14:15:46.300" Title="Heat given off from an electrochemical cell compared to mixing reactants" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;enthalpy&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23685" PostTypeId="2" ParentId="4610" CreationDate="2015-01-12T19:42:34.957" Score="3" Body="&lt;p&gt;Batteries have water-sulfuric acid solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There freezing point of water-sulfuric acid solutions varies as a function of concentration in an unusually complex way, having plural local minima and maxima.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the solid has eight identified phases:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4 * 6 H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4 * 4 H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4* 3 H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4 * 2 H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4 * H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2S2O7}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with mixtures of the various phases reaching local minimum freezing points at &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;eutectic points&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The freezing point can be as low as -71 °C near 31% (on an &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3}$&lt;/span&gt; basis) and as high as 35 °C near 90%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See [&lt;a href=&quot;https://doi.org/10.1021/ja01160a005&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] or &lt;a href=&quot;https://www.tandfonline.com/doi/pdf/10.3402/tellusb.v45i2.15588#page=4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fig. 2&lt;/a&gt; from [&lt;a href=&quot;https://doi.org/10.1034/j.1600-0889.1993.t01-1-00006.x&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/URUE6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/URUE6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gable, C. M.; Betz, H. F.; Maron, S. H. Phase Equilibria of the System Sulfur Trioxide-Water. &lt;em&gt;Journal of the American Chemical Society&lt;/em&gt; &lt;strong&gt;1950&lt;/strong&gt;, 72 (4), 1445–1448. &lt;a href=&quot;https://doi.org/10.1021/ja01160a005&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1021/ja01160a005&lt;/a&gt;.   &lt;/li&gt;&#xA;&lt;li&gt;Ohtake, T. Freezing Points of H2SO4 Aqueous Solutions and Formation of Stratospheric Ice Clouds. &lt;em&gt;Tellus B&lt;/em&gt; &lt;strong&gt;1993&lt;/strong&gt;, 45 (2), 138–144. &lt;a href=&quot;https://doi.org/10.1034/j.1600-0889.1993.t01-1-00006.x&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1034/j.1600-0889.1993.t01-1-00006.x&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2019-01-30T20:59:19.150" LastActivityDate="2019-01-30T20:59:19.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23686" PostTypeId="1" AcceptedAnswerId="23688" CreationDate="2015-01-12T19:58:26.627" Score="2" ViewCount="640" Body="&lt;p&gt;Electrolysis is used to separate compounds like Water into Hydrogen and Oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be used to induce a synthesis reaction rather than splitting reactants apart? For example, could you use electrolysis to recombine the Hydrogen and Oxygen?&lt;/p&gt;&#xA;" OwnerUserId="12728" LastEditorUserId="7448" LastEditDate="2015-01-12T20:46:27.047" LastActivityDate="2015-01-12T20:46:27.047" Title="Electrolysis used to cause a Chemical Reaction?" Tags="&lt;synthesis&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23687" PostTypeId="2" ParentId="23677" CreationDate="2015-01-12T20:02:53.703" Score="3" Body="&lt;p&gt;The ionization energy is described as the energy required to remove an electron from an atom in the gaseous phase. Apart from some exceptions, it is possible to identify a general trend in ionization energies within the periodic table: moving from left to right in a period the ionization energy increases, whereas it decreases from top to bottom in a group.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-12T20:02:53.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23688" PostTypeId="2" ParentId="23686" CreationDate="2015-01-12T20:37:47.607" Score="3" Body="&lt;p&gt;You wouldn't need to supply electric energy to get oxygen and hydrogen to recombine into water, because the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;nofollow&quot;&gt;free energy change&lt;/a&gt; for that reaction is negative. In other words, the products have a lower potential energy than the reactants, and so the reaction will proceed &quot;downhill&quot; on its own. You would need to overcome the very small activation energy, which you could do with electricity or with a flame or spark. But once it got started, you would not need to supply additional energy, and could in fact &lt;em&gt;generate&lt;/em&gt; electricity from the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basis of how hydrogen fuel cells work - hydrogen and oxygen react in a device that is set up to extract the energy that is produced in the form of an electric current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to think about it is that if you need to supply energy to make a reaction happen, then the reverse reaction will produce the same amount of energy (if we ignore losses due to inefficiency) or less (if we don't). The reverse is true as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note - electrolysis specifically refers to splitting a compound. The &quot;-lysis&quot; suffix means &quot;break apart,&quot; and &quot;electro-&quot; prefix means &quot;with electricty.&quot; In your case, you are thinking of &quot;electrochemistry&quot; - which just means &quot;using electricity with chemistry.&quot; Electrolysis is one thing you can do with electrochemistry.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-12T20:37:47.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23689" PostTypeId="1" AcceptedAnswerId="23762" CreationDate="2015-01-12T20:54:39.843" Score="0" ViewCount="89" Body="&lt;p&gt;Take, for istance, phosphorus; it has eight different oxidation numbers, but it can only form two different anhydrides, phosphoric (when the oxidation number is 5) and phosphorous (when it is 3). How do I understand which ones I can use to get anhydrides?&#xA; Or take chlorine, it also has eight oxidation numbers but if I wanted to know the corresponding hydracid, HCl, how'd I know which number to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive me if the question is trivial and written in a poor english, but I'm having trouble with both chemistry and English language classes at present!&lt;/p&gt;&#xA;" OwnerUserId="7665" LastActivityDate="2015-01-14T12:34:36.270" Title="How do I know with which oxidation numbers I can form a certain kind of compound?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23690" PostTypeId="1" CreationDate="2015-01-12T21:04:29.183" Score="1" ViewCount="68" Body="&lt;p&gt;How does this work exactly? Does it need some sort of kick in order to come together, or does it come together when put in a pressurized chamber? &lt;/p&gt;&#xA;" OwnerUserId="12728" LastEditorUserId="7475" LastEditDate="2015-10-09T16:39:24.933" LastActivityDate="2015-10-09T16:40:18.510" Title="The Combining of Carbon Dioxide and Hydrogen" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23691" PostTypeId="2" ParentId="23684" CreationDate="2015-01-12T21:16:01.030" Score="4" Body="&lt;p&gt;I think the problem here is related to what each of these terms means, in a fundamental sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{d}S$ is the entropy of the &lt;em&gt;system&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-\mathrm{d}q}{T}$ is the entropy of the &lt;em&gt;surroundings&lt;/em&gt; - the heat change of the surroundings divided by the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{d}H$ is the enthalpy of the &lt;em&gt;system&lt;/em&gt; - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving that for dq gives you the heat absorbed by the &lt;em&gt;surroundings&lt;/em&gt;, which is equal to temperature times the entropy change of the &lt;em&gt;system&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you used:&#xA;\begin{align}&#xA;\Delta G &amp;amp;= \Delta H - T\,\mathrm{d}S\\&#xA;\Delta H &amp;amp;= \Delta G + T\,\mathrm{d}S\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Edit&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To answer your last questions, and questions from the comments:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;T\,\mathrm{d}S &amp;amp;\geq \mathrm{d}q\\&#xA;(298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &amp;amp;\geq -70~\mathrm{\frac{kJ}{mol}}\\&#xA;-18~\mathrm{kJ} &amp;amp;\geq -70~\mathrm{kJ}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is true, although it is easier to see if you divide by -1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here I assume you meant &quot;enthalpy divided by temperature.&quot; The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing &quot;regular&quot; chemical reactions at constant T &amp;amp; P.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-01-16T14:15:46.300" LastActivityDate="2015-01-16T14:15:46.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23692" PostTypeId="2" ParentId="23651" CreationDate="2015-01-12T21:23:03.607" Score="2" Body="&lt;p&gt;So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and &lt;strong&gt;heat absorbed by the calorimeter&lt;/strong&gt;. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. &lt;strong&gt;It also gives the approximate temperature change for the calorimeter&lt;/strong&gt; if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$&#xA;You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-12T21:54:00.307" LastActivityDate="2015-01-12T21:54:00.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23693" PostTypeId="1" CreationDate="2015-01-12T21:40:49.887" Score="0" ViewCount="700" Body="&lt;p&gt;So I was learning about &lt;strong&gt;pH&lt;/strong&gt; and &lt;strong&gt;pOH&lt;/strong&gt; and it made sense. I looked at &lt;strong&gt;strong, weak acids &amp;amp; bases&lt;/strong&gt; and learned that they &lt;strong&gt;dissociate completely in water&lt;/strong&gt;. So then I moved on to water and learned about the autoionization of $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + H2O -&amp;gt; H3O+ +OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then I learned about the equilibrium constant and then I figured this:&#xA;$$\ce{{[H3O+][OH-]} = 1E -14}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does this mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So after that I was doing some basic pH and pOH calculations withe &lt;strong&gt;strong&lt;/strong&gt; acids and bases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the pOH of a $\ce{1E-6M} \ce{HCL}$&#xA;So here are the steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{-log[1E-6M] = 6}$$&#xA;Ok so the solution had the &lt;strong&gt;pH of 6&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was like that's nice and I proceeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{6+pOH=14}$$&#xA;$$\ce{pOH=8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then I was like wait a minute where did the $\ce{OH-}$ come from if it dissociates like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then I thought about it and this is what I came up with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since there is some $\ce{OH-}$ then there must be some water more water that autoionized into OH and that is the only place where $\ce{OH-}$ could've come from. So that means that even though $\ce{HCl}$ completely dissociated there must be more water than $\ce{HCl}$, because referring to the original equation there is $\ce{1M}$ quantities of $\ce{H2O}$ and $\ce{HCl}$ and so in the $molarity$ of the $\ce{HCl}$ given that means there was more water and less $\ce{HCl}$ and the $\ce{OH-}$ could've only come from the autoionization of that excess water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that is what I thought could be the only possible explanation of the $\ce{OH-}$ in a &lt;strong&gt;strong&lt;/strong&gt; acid. And that could only come from $\ce{H2O}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So am I right, and If I am wrong then please tell me where the $\ce{OH-}$ came from.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the $\ce{pH}$ scale only to $14$? Because if I took a $\ce{2M}$ $\ce{HCl}$ and then I get a &lt;strong&gt;super acid&lt;/strong&gt; in which the $\ce{H+}$ ions is a lot more than the $\ce{H2O}$ available to absorb them. So why is it only to $14$? Does it have to relate to the original autoionization of $\ce{H2O}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And If I totally grasped this wrong then can you please accurately tell me what $\ce{pH}$ &amp;amp; $\ce{pOH}$ and what &lt;strong&gt;autoionization of $\ce{H2O}$&lt;/strong&gt; is for a beginner and still make sense.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-01-12T22:16:39.320" LastActivityDate="2015-03-13T23:08:50.170" Title="ph &amp; pOH - Autoionization of Water" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23694" PostTypeId="2" ParentId="1091" CreationDate="2015-01-12T21:44:54.130" Score="2" Body="&lt;p&gt;&quot;Ancestral Terms&quot; are terms such as parent and grandparent terms that couple to form the final term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you see $3d^7(^4F)4s4p(^3P°) ^6F°_{9/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this means 7 3d electons couple to form a $^4F$ grandparent term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the 4s and 4p electrons couple to form a $^3P°$ parent term and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the grandparent and parent terms couple to form the final $^6F°$ term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is one of several example given at the NIST page &lt;a href=&quot;http://physics.nist.gov/Pubs/AtSpec/node09.html&quot; rel=&quot;nofollow&quot;&gt;Notations for Different Coupling Schemes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-12T21:51:54.703" LastActivityDate="2015-01-12T21:51:54.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23695" PostTypeId="2" ParentId="23690" CreationDate="2015-01-12T22:06:08.433" Score="4" Body="&lt;p&gt;I believe you are referring to the Sabatier process. $$\ce{CO2 + 4H2 -&amp;gt; CH4 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction uses a $\ce{Ni}$ catalyst and is done at $300-400\,^{\circ}{\rm C}$ and high pressures. The reaction is exothermic, $\Delta H = −165.0~\mathrm{kJ\,mol}^{-1}$, but requires initial heating in order to start the reaction off due to the activation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sabatier_reaction&quot; rel=&quot;nofollow&quot;&gt;Read more on Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="7475" LastEditDate="2015-10-09T16:40:18.510" LastActivityDate="2015-10-09T16:40:18.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23696" PostTypeId="2" ParentId="23693" CreationDate="2015-01-12T22:13:36.803" Score="1" Body="&lt;p&gt;I think you have it all right. The OH- present must come from the autolysis of the water present, which is far in excess of the HCl. If you figure the pOH of pure water would be 7, then it makes sense to think that adding 1E-06 M (a small concentration) of HCl would lower the basicity (raise the pOH from 7 to 8). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason a scale of 0-14 is used for pH is that when you get to either extreme end of the scale (close to 1M in either acid or base), the sheer concentration of ions present, relative to water, is a game changer and the system no longer behaves in a nice linear fashion. So the scale does not work well outside the 0-14 range. Conceptually, you can extend the pH scale down below zero using concentrated mineral acids or strong acids like fluorosulfonic acid, and empirically measure their effectiveness in protonating a very weak base as a proxy for pH.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-12T22:13:36.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23698" PostTypeId="1" CreationDate="2015-01-12T23:35:35.820" Score="1" ViewCount="702" Body="&lt;p&gt;In writing net ionic equations involving strong acids, one needs to know how strong acids dissociate. For example, the dissociation of $\ce{H2SO4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}\ce{&#xA;H2SO4 &amp;amp;-&amp;gt; 2H+ + SO4^{2-}\\&#xA;H_SO4 &amp;amp;-&amp;gt; H+ + HSO4^{-}\\&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is because apparently the molecular equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 + 2KOH -&amp;gt; 2H2O + K2SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has the net-ionic form $$\ce{H+ + OH- -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful if you could describe in general how I should write the dissociation of the strong acids in the net ionic equation.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="4945" LastEditDate="2015-01-13T02:32:47.250" LastActivityDate="2015-01-13T02:32:47.250" Title="How do the strong acids dissociate?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23700" PostTypeId="1" CreationDate="2015-01-13T00:17:14.683" Score="1" ViewCount="3576" Body="&lt;p&gt;I am trying to find the net-ionic equation for the reaction of potassium thiocyanate and iron(II) acetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems reasonable to me that the molecular equation is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KSCN + Fe(CH3COO)2 -&amp;gt; 2CH3COOK + Fe(SCN)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which would give the net-ionic form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SCN- + Fe^2+ -&amp;gt; Fe(SCN)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry teacher says the correct form is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4SCN- + Fe^2+ -&amp;gt; Fe(SCN)4^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I don't know what molecular equation would give rise to this. Second, why in the world is my guess, which is simpler, wrong and this one correct?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorDisplayName="user7951" LastEditDate="2015-08-17T17:10:59.733" LastActivityDate="2015-08-17T17:10:59.733" Title="Potassium thiocyanate and iron(II) acetate" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23701" PostTypeId="2" ParentId="23700" CreationDate="2015-01-13T00:40:28.017" Score="2" Body="&lt;p&gt;As best I can tell, your answer is correct. I can't find any reference for iron(II) thiocyanate in the form your teacher says is correct, and several references that say $\ce{Fe(SCN)2}$ is the correct form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For precipitation reactions, the charge will never change. You should always assume that the charges in the product will balance. There might be some cases where the product has an unexpected formula that you can't predict based on charges alone, but I can't think of any off the top of my head, and you certainly shouldn't be expected to know them at this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the best reference I could find, it's a google books link to the &lt;a href=&quot;https://books.google.com/books?id=c1rNBQAAQBAJ&amp;amp;pg=SA4-PA69&amp;amp;lpg=SA4-PA69&amp;amp;dq=iron+ii+thiocyanate&amp;amp;source=bl&amp;amp;ots=nBHC5pzvcy&amp;amp;sig=Upbact8-zBtevG43BjYbEDuLsRo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=cAe3VIurF5CRyATf5oDADw&amp;amp;ved=0CDQQ6AEwBDge#v=onepage&amp;amp;q=iron%20ii%20thiocyanate&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;CRC handbook&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-01-15T00:21:35.167" LastActivityDate="2015-01-15T00:21:35.167" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23702" PostTypeId="1" CreationDate="2015-01-13T01:00:04.480" Score="2" ViewCount="4204" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have:&#xA;$$\ce{C_{$x$}H_{$y$}O_{$z$} + O2 -&amp;gt; CO2 + H2O -&amp;gt;[\ce{CaCl2 + NaOH}] CaCl2 + NaOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anhydrous $\ce{CaCl2}$ absorbs the $\ce{H2O}$ produced, and the $\ce{CO2}$ is absorbed from $\ce{NaOH}$? So the products are $\ce{CaCl2. 2H2O}$ and $\ce{Na2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have just figured the empirical formula out, and was wondering what the final answer is so I can compare and see where I went right/wrong. I realised that instead of $\ce{Na2CO3}$, $\ce{NaHCO3}$ would be produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Working out:&lt;/strong&gt;&#xA;Via conservation of mass, the $\ce{CO2}$ and $\ce{H2O}$ must also be $0.4953~\mathrm{g}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore: conservation of mass (mass of $\ce{CO2 + H2O}$) - increase in mass of $\ce{CaCl2 + NaOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.4953 - (0.1613 + 1.0495) = 0.7155~\mathrm{g}$ &amp;lt;--- mass of $\ce{CaCl2}$ and $\ce{NaOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of $\ce{CaCl2.2H2O} = 0.7155 + 0.1613 = 0.8768~\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of $\ce{NaHCO3} = 0.7155 + 1.0495 = 1.765~\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     C:               H:              O&#xA; 1.765x(12/84):     0.8768x(4/147.1): &#xA;&#xA; (m) 0.252142857:     0.023842284:   0.4953-(0.252142857+0.023842284)= 0.219314859&#xA;&#xA; (n) m/12       :     m/1         :    m/16 &#xA;&#xA; 0.021011905 : 0.023842284     :  0.013707179&#xA;&#xA;       (Divide by lowest value)  &#xA;&#xA;( 1.53      :     1.74        :    1 ) x4&#xA;&#xA;  6         :       7         : 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore $\ce{C6H7O4}$&lt;/p&gt;&#xA;" OwnerUserId="12733" LastEditorUserId="5026" LastEditDate="2018-04-17T10:52:33.777" LastActivityDate="2018-04-17T10:52:33.777" Title="How to determine the empirical formula for terephthalic acid from its combustion products?" Tags="&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23703" PostTypeId="2" ParentId="23698" CreationDate="2015-01-13T01:00:50.993" Score="3" Body="&lt;p&gt;This is a little tricky. I would argue that only the first proton dissociates completely and should be considered &quot;strong&quot; - using the rules for writing net ionic equations, where we only dissociate &lt;strong&gt;strong electrolytes&lt;/strong&gt;, this would meant that we would write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4 + 2KOH -&amp;gt; 2H2O + K2SO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H+ + HSO4- + 2K+ + 2OH- -&amp;gt; 2H2O + 2K+ + SO_4^{2-}}$ - here only the first proton dissociates completely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canceling out spectator ions gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H+ + HSO4- + 2OH- -&amp;gt;2H2O + SO_4^{2-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H+ + SO_4^{2-} + 2OH- -&amp;gt;2H2O + SO_4^{2-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H+ + OH- -&amp;gt;H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you dissociate protons from acids as if they were regular metal cations - just &quot;pull them off&quot; and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-13T01:00:50.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23704" PostTypeId="2" ParentId="23664" CreationDate="2015-01-13T02:42:51.493" Score="2" Body="&lt;p&gt;The Fresnel equations, &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Fresnel_equations&lt;/a&gt;, are used to calculate reflectivity of transparent substances based on the &lt;strong&gt;index of refraction&lt;/strong&gt; of the liquid vs. that of air or other medium. There are complications introduced by conductivity, and &lt;em&gt;solids&lt;/em&gt; may display anisotropic reflectivity, see &lt;a href=&quot;http://www.hindawi.com/journals/isrn/2012/732973/&quot; rel=&quot;nofollow&quot;&gt;http://www.hindawi.com/journals/isrn/2012/732973/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Dispersion staining&lt;/em&gt; is a technique for staining microscopic specimens depending on the change in the index of refraction with the color of light (dispersion), see &lt;a href=&quot;http://en.wikipedia.org/wiki/Dispersion_staining&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dispersion_staining&lt;/a&gt;. This uses the shininess of substances in a liquid, often an organic immersion oil, to help distinguish or identify them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;strong&gt;metallic reflection&lt;/strong&gt; is based on a &quot;sea&quot; of electrons bouncing back an inverted wave, and differs from refraction. Liquid metals such as mercury are shiny for that reason, but I do not know if there are sufficiently conductive organic liquids that would have metallic reflection.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-01-13T17:18:07.187" LastActivityDate="2015-01-13T17:18:07.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23705" PostTypeId="1" CreationDate="2015-01-13T03:45:18.640" Score="0" ViewCount="822" Body="&lt;p&gt;In this thought experiment, let’s consider the surrounding to be air that is composed of entirely water vapor (no other species like $\ce{O2}$ and $\ce{N2}$ are present). The surrounding is infinitely large. The system in question is a pan containing a thin layer of liquid water. The liquid water is brought to its boiling point by uniform heating. This pan is open (not covered in any way) and is exposed to the surrounding water vapor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, boiling occurs when the vapor pressure of the liquid is equal to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling&quot; rel=&quot;nofollow&quot;&gt;external pressure&lt;/a&gt;. So in this case, at boiling point, the vapor pressure of the liquid $\ce{H2O}$ is equal to the external pressure of gaseous $\ce{H2O}$. Thus, the system is in a dynamic equilibrium, and the rate of vaporization is equal to the rate of condensation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the amount of liquid water will not change over time. In particular, it will not all “boil off” as steam. Is this the right conclusion? It certainly seems a little absurd.&lt;/p&gt;&#xA;" OwnerUserId="12707" LastEditorUserId="7448" LastEditDate="2015-04-02T18:46:22.140" LastActivityDate="2015-04-02T18:46:22.140" Title="Is this the right conclusion for a thought experiment?" Tags="&lt;equilibrium&gt;&lt;water&gt;&lt;boiling-point&gt;&lt;vapor-pressure&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23706" PostTypeId="2" ParentId="23705" CreationDate="2015-01-13T04:50:01.790" Score="3" Body="&lt;p&gt;This conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.&lt;/p&gt;&#xA;" OwnerUserId="12693" LastActivityDate="2015-01-13T04:50:01.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23708" PostTypeId="1" AcceptedAnswerId="23719" CreationDate="2015-01-13T05:17:06.097" Score="3" ViewCount="1201" Body="&lt;p&gt;The following equation can be used to determine the rate:&lt;br&gt;&#xA;$Rate = \frac{-\Delta [A]}{\Delta t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation can also be used:&lt;br&gt;&#xA;$Rate = k[A]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By simple substitution, $\frac{-\Delta [A]}{\Delta t} = k[A]$ should be true. Therefore, $k$ can be represented as:&#xA;$\frac{-\Delta [A]}{[A]\Delta t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, calculating the rate constant for a given reaction is obviously not that simple. Why does this equation not work then, if the mathematics of it check out? I have tried using this equation for some problems, and as I expected, they do not yield the correct answer as opposed to using the integrated rate equation.&lt;/p&gt;&#xA;" OwnerUserId="9281" LastEditorUserId="12660" LastEditDate="2015-01-13T17:18:41.193" LastActivityDate="2015-01-13T22:35:01.313" Title="Why is it not possible to determine the rate constant this way?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23709" PostTypeId="2" ParentId="23708" CreationDate="2015-01-13T08:12:03.527" Score="0" Body="&lt;p&gt;That rate constant '$\kappa$' is for &quot;instantaneous rate of reaction&quot;(irr) which you wish to calculate from &quot;average rate of reaction&quot;(arr)&#xA;$$\frac{d[A]}{dt}=r_{irr}=\lim_{\Delta t\to0 }r_{arr}=\lim_{\Delta t\to0 }\frac{\Delta [A]}{\Delta t}$$&#xA;So if the time duration is very small you would get a better approximation.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-13T08:12:03.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23711" PostTypeId="1" CreationDate="2015-01-13T08:44:44.780" Score="4" ViewCount="408" Body="&lt;p&gt;Here is a derivation I found relating the pressure P, gravity g, height h, and density $\rho$, of a liquid in a barometer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P=\frac{mg}{A}\\&#xA;V=Ah \\&#xA;P = \frac{mgh}{V}\\&#xA;\rho=\frac{m}{V}\\&#xA;P = \rho g h&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this derivation seems misguided, since $P = \frac{F}{A}$, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid? Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.&lt;/p&gt;&#xA;" OwnerUserId="12736" LastActivityDate="2015-01-14T12:27:16.647" Title="Why does a liquids height in a Barometer depend linearly on pressure?" Tags="&lt;solid-state-chemistry&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23712" PostTypeId="2" ParentId="22270" CreationDate="2015-01-13T11:15:24.687" Score="3" Body="&lt;p&gt;Here I'll only provide the reasoning for &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; diatomic structure, I think similar reasoning was to deduce that of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;The divalency of oxygen, shown by Avogadro's hypothesis.&lt;/h2&gt;&#xA;&lt;p&gt;It is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H + O \to H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This would indicate that &lt;span class=&quot;math-container&quot;&gt;$\frac12$&lt;/span&gt; gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the&#xA;atomic weight of hydrogen was defined as &lt;span class=&quot;math-container&quot;&gt;$H = 1$&lt;/span&gt;, then the respective atomic weights of hydrogen and oxygen would be &lt;span class=&quot;math-container&quot;&gt;$H = 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$O = 16$&lt;/span&gt; (the values accepted today).&lt;/p&gt;&#xA;&lt;h2&gt;The diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.&lt;/h2&gt;&#xA;&lt;p&gt;It is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2H2 + O2 \to 2H2O}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Copied almost verbatim from &lt;a href=&quot;http://www.chem.unt.edu/%7Ejimm/CHEM%201360%20Chapters%20and%20Problems/PROOF%20THAT%20OXYGEN%20IS%20DIVALENT%20and%20that%20HYDROGEN%20AND%20OXYGEN%20ARE.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2133" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-13T11:15:24.687" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23713" PostTypeId="2" ParentId="23700" CreationDate="2015-01-13T12:13:12.340" Score="2" Body="&lt;p&gt;If you add an excess of cyanide ions to a iron(II) solution you could conceivably get precipitate of $\ce{Fe(SCN)2}$ which would then dissolve to form a complex with four thiocyanate ions. However, I can't find any references to such a complex on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations would be: $$\ce{Fe^{2+}_{(aq)} + 2SCN-_{(aq)} -&amp;gt; [Fe(SCN)2]_{(s)}}$$ $$\ce{[Fe(SCN)2]_{(s)} + 2SCN^{-}_{(aq)} -&amp;gt; [Fe(SCN)4]^{2-}_{(aq)}}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-13T12:13:12.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23714" PostTypeId="1" AcceptedAnswerId="23725" CreationDate="2015-01-13T12:42:05.790" Score="6" ViewCount="117" Body="&lt;p&gt;Looking at the images in &lt;a href=&quot;http://www.sciencemag.org/content/342/6158/611&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;, I've noticed that the molecule electron density images showed as calculated via DFT look very symmetric, while those obtained by NC-AFM appear somewhat distorted. Namely, the ring at the left (the one with $\ce{OH}$ group) looks elongated, while the one at right is, on the contrary, compressed in horizontal direction, and the $\ce{OH}$ group seems to be tilted to the $\ce N$ atom. Also, the calculated image seems much more smeared, while the measured one gives some highly localized density &quot;rods&quot; for the bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a gif-combination of these two images so that the difference was easier to see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/flKt0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the B is DFT-calculated version, while F is one measured by NC-AFM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can suppose one or several of the following might be among the reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The substrate of $\ce{Cu(111)}$ wasn't taken into account when calculating(?), while in the measurement it appears to distort the molecule&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;DFT, being an approximate method, gives too simplified results&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Smear vs thin rods difference might because in the DFT case the picture is of electron density while in AFM case the image shows AFM frequency shifts, which is not necessarily the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The probe of the microscope distorts the molecule&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What are the true reasons for the differences?&lt;/p&gt;&#xA;" OwnerUserId="2133" LastEditorUserId="2133" LastEditDate="2015-01-13T12:47:13.630" LastActivityDate="2017-07-22T19:41:55.767" Title="What's the reason for discrepancy between DFT calculated image and NC-AFM-acquired one in this article?" Tags="&lt;molecules&gt;&lt;computational-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23715" PostTypeId="2" ParentId="23714" CreationDate="2015-01-13T13:03:08.527" Score="5" Body="&lt;p&gt;1) Most definitely. Adsorbed molecules often are very distorted, up to chemical reaction with support, especially when reactive surface is used. Despite common misconception, even noble metal surfaces are quite reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) DFT, while being indeed inexact (as any calclation with limited size of basis set), catches major features of electronic density very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) AFM stands for atomic force microscopy, which is not connected to electronic density directly. Now, naphthalene molecule features large $\pi$-system, which is very polarizable. Since nonvalent interatomic forces beyond electrostatic interactions usually increases with polarizability of interacting particles, I would expect $\pi$-systems  to be highlighted on the NC-AFM images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I'm not an expert on atomic-resolved microscopy, but I never read about this possibility in the articles I worked with, though the works I usually works with used STM. and SEM mostly.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-01-13T13:03:08.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23716" PostTypeId="2" ParentId="23711" CreationDate="2015-01-13T13:41:59.883" Score="1" Body="&lt;p&gt;In a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressibility of mercury in principle does make it non-linear.  The density would be &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html&quot; rel=&quot;nofollow&quot;&gt;3.8 parts per million&lt;/a&gt; higher at the air interface than at the vacuum interface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-13T13:50:21.533" LastActivityDate="2015-01-13T13:50:21.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23717" PostTypeId="2" ParentId="23515" CreationDate="2015-01-13T13:58:50.570" Score="2" Body="&lt;p&gt;No, photons of visible light don't ionize molecules during photosynthesis - electrons are transferred via &lt;a href=&quot;https://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow&quot;&gt;redox&lt;/a&gt; reactions in &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Electron_transport_chain&quot; rel=&quot;nofollow&quot;&gt;electron transport chain&lt;/a&gt; - it seems to me that fragment of the article was maybe a little misleading.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-10-16T19:51:20.947" LastActivityDate="2015-10-16T19:51:20.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23718" PostTypeId="1" AcceptedAnswerId="23723" CreationDate="2015-01-13T14:11:51.213" Score="6" ViewCount="3868" Body="&lt;p&gt;Why is the London dispersion force inversely related to distance but not to the second power but the sixth power? Why is there not an inverse square relationship as there is for Coulombic forces? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something non-Coulombic about London dispersion forces? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-04-19T10:03:25.130" Title="London dispersion forces and distance relationship" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23719" PostTypeId="2" ParentId="23708" CreationDate="2015-01-13T14:43:11.873" Score="6" Body="&lt;p&gt;The &lt;em&gt;average&lt;/em&gt; rate $\Delta[A]/\Delta t$ over time interval $\Delta t$ is only approximately equal to the true &lt;em&gt;instantaneous&lt;/em&gt; rate $\rm d[A]/dt$ at time $t$. The difference between the two &lt;em&gt;can&lt;/em&gt; be significant if the instantaneous rate changes over the chosen time interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ig8z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger the time interval, the larger the difference between $\Delta[A]/\Delta t$ and $\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not just use very small time intervals? Well, there's a limit to how small you can make $\Delta t$ if you're computing $\Delta[A]/\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals.  &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-13T22:35:01.313" LastActivityDate="2015-01-13T22:35:01.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23720" PostTypeId="1" AcceptedAnswerId="23724" CreationDate="2015-01-13T15:29:08.430" Score="1" ViewCount="116" Body="&lt;p&gt;Since in rusting (oxidation) of Iron, transfer of 4 electrons takes place is it possible to use this reaction under catalytic conditions to create a simple electric cell, even if it just gave a millivolt of potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Fe + H2O -&amp;gt; FeOH3 + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that we have complete control over the surroundings, like volume expansion, disposing hydrogen and other such factors.&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="4945" LastEditDate="2015-01-16T09:37:43.580" LastActivityDate="2015-01-16T09:37:43.580" Title="Can we make a rusting battery" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23721" PostTypeId="1" CreationDate="2015-01-13T15:30:31.427" Score="2" ViewCount="101" Body="&lt;p&gt;The best method for calibration would be a bath filled with a chemical at its melting point with a solid/liquid composition. &#xA;The constraints being that it must be hydrophobic and relatively non-corrosive and non-toxic with a melting point between (say) 20 °C and 60 °C. Needless to say, but I’ll say it anyway, the chemical must have a very precise melting point. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:56:31.220" LastActivityDate="2015-10-23T16:56:31.220" Title="Looking for a chemical to calibrate a temperature sensor" Tags="&lt;physical-chemistry&gt;&lt;materials&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23722" PostTypeId="2" ParentId="23718" CreationDate="2015-01-13T16:45:47.973" Score="7" Body="&lt;p&gt;London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some electrostatic (Coulombic) relationships:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Two Point Charges&lt;/strong&gt;: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Point Charge and Permanent Dipole&lt;/strong&gt;: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Two Permanent Dipoles&lt;/strong&gt;: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules &lt;em&gt;rotate&lt;/em&gt; then it falls off to $1/r^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_2^* = \alpha_2 E$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \alpha_2 E^2 / 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V \propto r^{-6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \frac{C}{r^6}$&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-01-14T15:41:25.410" LastActivityDate="2015-01-14T15:41:25.410" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="23723" PostTypeId="2" ParentId="23718" CreationDate="2015-01-13T16:47:26.480" Score="7" Body="&lt;p&gt;One somewhat simplified way of looking at it is that the London dispersion forces are a dipole-dipole interactions.  The interaction between two dipoles depends upon the relative orientation of the two dipoles.  Some positions are attractive and some are replusive, but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive with potential energy having an 1/$r^6$ dependence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;http://www4.ncsu.edu/~franzen/public_html/CH431/lecture/lec_29.pdf&quot; rel=&quot;nofollow&quot;&gt;Stephan Franzen's lecture&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and especially the corporate &lt;a href=&quot;http://www.ntmdt.com/spm-basics/view/intermolecular-vdv-force&quot; rel=&quot;nofollow&quot;&gt;Intermolecular Van Der Waals page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as well as Fritz London's &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1937/TF/tf937330008b#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;The General Theory of Molecular Forces&lt;/a&gt; Trans. Faraday Soc., 1937, vol. 33, pages 8-26.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for more on the 1/$r^6$ dependence of dipoles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a proper quantum mechanical treatment shows that London dispersion interaction is not always proportional to 1/$r^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, between two hydrogen atoms in 1s states, the potential energy is proportional to 1/$r^6$ but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;between a hydrogen atom in a 1s state and a hydrogen atom in a 2p state, energy is proportional to 1/$r^3$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://depts.washington.edu/chemcrs/bulkdisk/chem551A_win06/notes_van_der_Waals_Interaction.pdf&quot; rel=&quot;nofollow&quot;&gt;Complement $C_{XI}$ of Quantum Mechanics vol. 2&lt;/a&gt; by Cohen-Tannoudji et al. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in 1948 Verwey and Overbeek demonstrated experimentally that the London dispersion interaction is even weaker than 1/$r^6$ at long distance (say hundreds of Angstroms or more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Casimir and Polder soon thereafter explained with quantum electrodynamics (QED) that the dependence should be 1/$r^7$ at relatively long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a nice historical overview and QED prespective see &lt;a href=&quot;http://george.ph.utexas.edu/~dsteck/reading/casimir/milonni94.pdf&quot; rel=&quot;nofollow&quot;&gt;Some QED vacuum effects: van der waals forces&lt;/a&gt; in The Quantum Vacuum by Peter W. Milonni.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-14T14:37:35.757" LastActivityDate="2015-01-14T14:37:35.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23724" PostTypeId="2" ParentId="23720" CreationDate="2015-01-13T16:47:39.917" Score="2" Body="&lt;p&gt;Sure! A nickel-iron battery uses the oxidation of iron at the negative plate to produce about half the voltage. Though invented by Jungner, it's often called an Edison battery. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Nickel%E2%80%93iron_battery&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are very low power semiconductor devices which could be powered by an iron battery. You could make three or four small cells from iron nails and a more electronegative element, e.g. lead, to power a digital watch; even without optimization to remove layers of hydrogen and rust, it should operate for months. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some other low power ideas, see &lt;a href=&quot;https://www.enocean.com/en/energy-harvesting/&quot; rel=&quot;nofollow&quot;&gt;https://www.enocean.com/en/energy-harvesting/&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-13T16:47:39.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23725" PostTypeId="2" ParentId="23714" CreationDate="2015-01-13T17:00:40.950" Score="4" Body="&lt;p&gt;This is not really my field, but I've certainly followed it. I'd guess some combination of all of your reasons, plus one more.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I think usually the substrate is considered, but &lt;em&gt;imperfections&lt;/em&gt; and defects in the surface may not be properly considered.&lt;/li&gt;&#xA;&lt;li&gt;Perhaps, although I think this is probably less important than other causes.&lt;/li&gt;&#xA;&lt;li&gt;Exactly. DFT-computed electron density is not exactly what the AFM measures.&lt;/li&gt;&#xA;&lt;li&gt;The probe will certainly have different interactions with the molecule, and could be subtly asymmetric.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'd guess the most likely culprit is different.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In AFM (and all scanned-probe methods) the substrate is rarely perpendicular to the probe. Some minor image correction is performed to &quot;line-fit&quot; the substrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider if you take a photo of a page on your desk. The page is probably slightly rotated in the plane of the desk, but the camera is almost never exactly parallel to the desk. This results in &lt;a href=&quot;https://en.wikipedia.org/wiki/Keystone_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;keystoning&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AFM software corrects for this effect, and users can fit a bit after data-collection, but it can cause subtle distortions.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="2133" LastEditDate="2017-07-22T19:41:55.767" LastActivityDate="2017-07-22T19:41:55.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23726" PostTypeId="2" ParentId="23721" CreationDate="2015-01-13T17:08:16.533" Score="3" Body="&lt;p&gt;In the old days, you'd look it up in the CRC organic chemical tables, but now you can go to a site such as &lt;a href=&quot;http://www.colby.edu/chemistry/cmp/cmp.html&quot; rel=&quot;nofollow&quot;&gt;http://www.colby.edu/chemistry/cmp/cmp.html&lt;/a&gt; and enter the desired m.p. and constituents. Here is the result for a search for 50C (+-5C) m.p. without O, N and S so they're likely hydrophobic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P-DICHLOROBENZENE&#xA;PENTAMETHYLBENZENE&#xA;1-METHYL-4-PHENYLBENZENE&#xA;1,2,3-TRICHLOROBENZENE&#xA;1-ETHYL-4-PHENYLBENZENE&#xA;1,2-DIPHENYLETHANE&#xA;1,1-DIPHENYL-1-PROPENE&#xA;CYCLOTETRADECANE&#xA;2-BROMONAPHTHALENE&#xA;1,2,3,4-TETRACHLOROBENZENE&#xA;1,2,3,5-TETRACHLOROBENZENE&#xA;P-CHLOROIODOBENZENE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking P-DICHLOROBENZENE as an example, you would find it melts at 53.5 C and is mildly toxic, but is used in mothballs, so is a regularly available consumer item not requiring HM permitting to buy. See &lt;a href=&quot;http://www.epa.gov/airtoxics/hlthef/dich-ben.html&quot; rel=&quot;nofollow&quot;&gt;http://www.epa.gov/airtoxics/hlthef/dich-ben.html&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/1,4-Dichlorobenzene&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/1,4-Dichlorobenzene&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-13T17:08:16.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23727" PostTypeId="2" ParentId="23721" CreationDate="2015-01-13T17:26:27.263" Score="4" Body="&lt;p&gt;The standard point recommended in that range by the CIPM's ITS-90 calibration standard is the melting point of gallium at 29.7646 °C. Gallium is not known to be particularly toxic, but it can embrittle metals so is ideally used/stored in glass or plastic containers. It's used as a standard because it is relatively easy to make very pure and is stable in storage. Also, by picking an ITS-90 standard, you can ensure your thermometer is calibrated to the same standard as many other high precision thermometers. N.B. ITS-90 specifies specifically the &lt;em&gt;melting&lt;/em&gt; point of gallium, not the freezing point, unlike some of the other calibration standards.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-13T17:26:27.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23728" PostTypeId="1" AcceptedAnswerId="23731" CreationDate="2015-01-13T17:59:48.483" Score="1" ViewCount="497" Body="&lt;p&gt;I was removing old batteries from a camera the other day. One of the batteries had leaked something very bright blue. I have seen batteries leak before and typically the colors look a bit like a corroded penny - pale green to white. This time it was a beautiful bright blue; what was it? I initially thought this could have been the metal making the connection was corroding as opposed to the color of the contents of the battery. However, I am not so sure about that.&#xA;&lt;p&gt;&#xA;&lt;strong&gt;Note:&lt;/strong&gt; I believe they were double or triple A batteries and &lt;em&gt;not&lt;/em&gt; LiION.&lt;/p&gt;&#xA;" OwnerUserId="5137" LastActivityDate="2015-01-13T18:52:54.287" Title="Unusual color from leaking battery. What happened?" Tags="&lt;electrochemistry&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23731" PostTypeId="2" ParentId="23728" CreationDate="2015-01-13T18:52:54.287" Score="2" Body="&lt;p&gt;It is, as you suspected, likely a copper corrosion product. Copper compounds are usually blue or green, and are used as pigments. Malachite, &lt;a href=&quot;http://en.wikipedia.org/wiki/Malachite&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Malachite&lt;/a&gt;, is one of the likely corrosion products of an alkaline battery and copper terminals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, you may need to rinse the battery compartment and use a file to remove the corrosion from the contacts to make connection.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-13T18:52:54.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23733" PostTypeId="1" AcceptedAnswerId="23753" CreationDate="2015-01-13T20:48:54.427" Score="6" ViewCount="4592" Body="&lt;p&gt;&lt;em&gt;Firstly I would like to say that despite the fact that you may be thinking that this is a physics question and should probably be asked on that part of the website, I came across reduced mass for the first time in Chemistry - physical and inorganic - and it is used a lot in Chemistry so I thought it suitable for this part of the site.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, I will now elaborate on my question. I hear people say that reduced mass ,$\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step [1]&lt;/strong&gt; - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$&#xA;&lt;strong&gt;Step [2]&lt;/strong&gt; - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$&#xA;&lt;strong&gt;Step [3]&lt;/strong&gt; - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$&#xA;&lt;strong&gt;Step [4]&lt;/strong&gt; - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step):&#xA;$$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-01-16T09:43:42.787" LastActivityDate="2015-01-16T09:43:42.787" Title="Why does reduced mass help when talking about two body problems?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23734" PostTypeId="1" AcceptedAnswerId="23754" CreationDate="2015-01-13T20:57:23.160" Score="1" ViewCount="6333" Body="&lt;p&gt;So I understand $\ce{CH4}$ has a low solubility, but when it does dissolve, what does it form? E.g. $\ce{HCl_{(g)} + H2O -&amp;gt; H3O+_{(aq)} + Cl^{-}_{(aq)}}$. So what does $\ce{CH4 + H2O}$ form?&lt;/p&gt;&#xA;" OwnerUserId="12743" LastEditorUserId="7448" LastEditDate="2015-12-26T21:38:36.190" LastActivityDate="2015-12-26T21:38:36.190" Title="What does CH4 form when dissolved in aqueous solution?" Tags="&lt;water&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23735" PostTypeId="1" AcceptedAnswerId="23736" CreationDate="2015-01-13T21:00:47.570" Score="3" ViewCount="774" Body="&lt;p&gt;All the definitions I'm seeing give examples of crystalline solids like diamond or boron nitride. But I'm wondering if it also includes amorphous macromolecules like vulcanized rubber. I'm thinking &quot;network&quot; probably == &quot;ordered array&quot; but not sure if that's correct.&lt;/p&gt;&#xA;" OwnerUserId="12742" LastActivityDate="2015-01-14T02:23:15.287" Title="Can a network covalent solid be amorphous?" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23736" PostTypeId="2" ParentId="23735" CreationDate="2015-01-13T21:13:54.447" Score="4" Body="&lt;p&gt;The most obvious example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_carbon&quot; rel=&quot;nofollow&quot;&gt;amorphous carbon&lt;/a&gt;. As mentioned in a comment, another good example is glass (although, whether it's truly &quot;solid&quot; past the &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass_transition&quot; rel=&quot;nofollow&quot;&gt;glass transition temperature&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, a &quot;network&quot; simply means &quot;solid with covalent bonding&quot; and of course it's easy to imagine many possible amorphous network solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vulcanized rubber might qualify too, but it would depend on whether there were many non-interlinked macromolecules. Usually macromolecules would be considered &lt;em&gt;molecular&lt;/em&gt; solids due to the high number of non-bonded packing interactions. If the amount of cross-linking is high, classifying as a network solid makes sense.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-01-14T02:23:15.287" LastActivityDate="2015-01-14T02:23:15.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23737" PostTypeId="2" ParentId="23734" CreationDate="2015-01-13T21:28:29.050" Score="0" Body="&lt;p&gt;A clathrate compound is like a cage. It relies on lattice spacing much more than molecular attraction to stick together; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Clathrate_compound&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Clathrate_compound&lt;/a&gt;. Methane-water ice clathrate is comparatively stable almost to the melting point of ice, and there are huge amounts under the oceans and in Arctic permafrost;see &lt;a href=&quot;http://en.wikipedia.org/wiki/Methane_clathrate&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Methane_clathrate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-13T21:28:29.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23738" PostTypeId="2" ParentId="23705" CreationDate="2015-01-13T22:51:54.077" Score="1" Body="&lt;p&gt;You have a closed system, consisting of a certain volume and containing a certain amount of water.  At a constant temperature it will find some equilibrium consisting of water vapor, liquid water, and ice (depending on temperature -- I'll ignore the ice for now).  As you heat the pan, you are heating the entire system and thus changing the balance of water vapor to liquid water. Where that liquid water actually resides is a problem for the water to figure out - it does not have to reside long term in the pan.  Eventually, as you keep heating, you will get to a temperature of the system where the liquid will be thermodynamically unstable, and the enclosure will contain water vapor only.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, for thermodynamic equilibrium, how fast you heat/cool does not enter into it - those are kinetic questions.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-01-13T22:51:54.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23739" PostTypeId="2" ParentId="23705" CreationDate="2015-01-13T23:42:42.340" Score="1" Body="&lt;p&gt;If the water vapor you start out with is at 1 atm then that means it has to be at 100 C. It also means that the walls of the box have to be either extremely well insulated or externally heated. Otherwise, condensation will occur on the walls and the gas will cool down. Assuming that the box and pan are somehow held at 100 C then the system will be in equilibrium and nothing will happen (i.e. no boiling). However, if the pan heats up beyond 100 C or the box walls cool off then you will see boiling. In the former case, the water vapor pressure will increase to maintain equilibrium. In the latter case, the water will boil just enough to compensate for the vapor lost by condensation.&lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-13T23:42:42.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23740" PostTypeId="1" CreationDate="2015-01-14T01:52:39.343" Score="2" ViewCount="86" Body="&lt;p&gt;Consider the diffusion of natural abundance $\ce{Li^{+}}$ and $\ce{{}^{6}Li^{+}}$ through a crystalline matrix, say a spinel such as $\ce{LiMn2O4}$.  The two lithium ions will have the same ionic radius and electric charge.  They will differ only in atomic weight.  Is there any reason to believe that the lighter isotope, $\ce{{}^{6}Li^{+}}$, will diffuse through the crystal any faster than the natural abundance isotope?  When I consider the forces acting upon the ions in the crystal, I come up with coloumbic, dispersion and hard shell repulsive forces.  I cannot see how any of those would change by simply changing the mass of the ion.  Am I missing anything?&lt;/p&gt;&#xA;" OwnerUserId="12650" LastEditorUserId="4945" LastEditDate="2015-01-14T06:29:26.987" LastActivityDate="2015-01-14T06:30:53.093" Title="Is there an isotope effect for diffusion of an ion through a crystal lattice?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23741" PostTypeId="2" ParentId="4018" CreationDate="2015-01-14T02:28:44.777" Score="0" Body="&lt;p&gt;The structure $\ce{H-H-O}$ does not technically exist in any common conditions because hydrogen does not generally form two covalent bonds at once. Such a structure would require a TON of energy to be put in because the hydrogen nucleus's lone proton would need to be able to hold electrons in the $2s$ sub level and prevent them from leaving the atom's circumference. Oxygen however readily forms two covalent bonds making $\ce{H-O-H}$ a very chemically plausible and common structure. Why can oxygen form two covalent bonds? Think of it in terms of quantum mechanics, much in the way as I explained for hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="12745" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T19:03:41.367" LastActivityDate="2015-06-09T19:03:41.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23743" PostTypeId="1" CreationDate="2015-01-14T02:37:43.307" Score="3" ViewCount="24256" Body="&lt;p&gt;Rate of disappearance is given as $-\frac{\Delta [A]}{\Delta t}$ where $\ce{A}$ is a reactant. However, using this formula, the rate of disappearance cannot be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta [A]$ will be negative, as $[A]$ will be lower at a later time, since it is being used up in the reaction. Then, $[A]_{\text{final}} - [A]_{\text{initial}}$ will be negative. Therefore, the numerator in $-\frac{\Delta [A]}{\Delta t}$ will be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta t$ will be positive because final time minus initial time will be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $-\frac{\Delta [A]}{\Delta t}$ will evaluate to $(-)\frac{(-)}{(+)} = (-) \cdot (-) =(+)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we still write the rate of disappearance as a negative number. Also, if you think about it, a negative rate of disappearance is essentially a positive rate of appearance. The reactants disappear at a positive rate, so why isn't the rate of disappearance positive?&lt;/p&gt;&#xA;" OwnerUserId="9281" LastEditorUserId="5026" LastEditDate="2018-04-17T10:55:20.587" LastActivityDate="2020-05-21T02:09:14.710" Title="Why is the rate of disappearance negative?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23744" PostTypeId="2" ParentId="23743" CreationDate="2015-01-14T03:23:20.737" Score="5" Body="&lt;p&gt;Reaction rates are generally by convention given based on the formation of the product, and thus reaction rates are positive. So, for the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A-&amp;gt;B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Rate} = \frac{\Delta[\ce{B}]}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure that you get a positive reaction rate, the rate of disappearance of reactant has a negative sign:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Rate} = -\frac{\Delta[\ce{A}]}{\Delta t}=\frac{\Delta[\ce{B}]}{\Delta t}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-04-17T10:53:34.337" LastActivityDate="2018-04-17T10:53:34.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23745" PostTypeId="1" CreationDate="2015-01-14T03:47:53.823" Score="1" ViewCount="122" Body="&lt;p&gt;standart potential of $\ce{Pb}$ is -0.13 V. how we determine an optimum potential of $\ce{Pb}$ sample? and how we choose the range potential?&lt;/p&gt;&#xA;" OwnerUserId="12747" LastEditorUserId="4231" LastEditDate="2015-01-23T23:11:10.853" LastActivityDate="2015-01-23T23:11:10.853" Title="how to determine range potential of Pb in cyclic voltammetry?" Tags="&lt;electrolysis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23746" PostTypeId="2" ParentId="23743" CreationDate="2015-01-14T04:09:28.987" Score="5" Body="&lt;p&gt;When you say &quot;rate of disappearance&quot; you're announcing that the concentration is going &lt;em&gt;down&lt;/em&gt;. If you wrote a negative number for the rate of disappearance, then, it's a double negative---you'd be saying that the concentration would be going up!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've noticed, keeping track of the signs when talking about rates of reaction is inconvenient. It would be much simpler if we defined a &lt;strong&gt;single number&lt;/strong&gt; for the rate of reaction, regardless of whether we were looking at reactants or products. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do this by&#xA;a) flipping the sign on rates for reactants, so that the rate of reaction will always be a positive number, and b) scaling all rates by their stoichiometric coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have a balanced equation for the reaction $$a \mathrm{A} +  b \mathrm{B} \rightarrow c \mathrm{C} + d \mathrm{D}$$ the rate of the reaction $r$ is defined&#xA;$$ r = -\frac{1}{a}\frac{\mathrm{d[A]}}{\mathrm{d}t} =  -\frac{1}{b}\frac{\mathrm{d[B]}}{\mathrm{d}t} = \frac{1}{c}\frac{\mathrm{d[C]}}{\mathrm{d}t} = \frac{1}{d}\frac{\mathrm{d[D]}}{\mathrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets us compute the rate of reaction from whatever concentration change is easiest to measure.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-14T04:18:06.980" LastActivityDate="2015-01-14T04:18:06.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23747" PostTypeId="2" ParentId="23733" CreationDate="2015-01-14T04:36:58.137" Score="2" Body="&lt;p&gt;You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a &lt;em&gt;single fictitious particle&lt;/em&gt;. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In step 4, you're calculating the mass $\mu$ of the fictitious particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$&#xA;This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be&#xA;$$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$&#xA;where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a &lt;em&gt;second&lt;/em&gt; fictitious particle with total mass $M=m_1+m_2$ and coordinates  $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-14T14:05:03.837" LastActivityDate="2015-01-14T14:05:03.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23748" PostTypeId="2" ParentId="22075" CreationDate="2015-01-14T05:50:51.673" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both citrate and oxalate are associated with ATP production via &lt;a href=&quot;http://en.wikipedia.org/wiki/Citric_acid_cycle&quot; rel=&quot;nofollow&quot;&gt;the citric acid cycle&lt;/a&gt;. As is obvious from the name, citrate is a critical component of the process. Oxalate is a hydrolysis product of oxaloacetic acid, which is also part of the cycle. Both of these ions are present within the plant cells.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How and why do they get released from tea leaf? Since you don't usually see any &quot;big&quot; differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your tea leaves are crushed or broken in any way, this really is just simple dissolution. If you have whole tea leaves, the oxalate and citrate are migrating into and out of the plant cells. The cell wall is permeable, so the ions (or corresponding acids) can simply flow through it. The cell membrane is semipermeable, and the diffusion of the ions across it is a biologically controlled process that I honestly don't have the background to explain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What makes the color of water change? Why does the color vary with pH alteration?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pH and color are not related properties. The pH is affected by the citric acid and oxalic acid associated with the anions previously discussed, as well as other minor components that are likely present. The color comes from some of the many other compounds that are extracted into the water, which number in the hundreds, if not thousands. Some of these compounds are inherently colored - no reaction is required to produce color in the solution.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-14T05:50:51.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23749" PostTypeId="2" ParentId="23740" CreationDate="2015-01-14T05:52:33.913" Score="1" Body="&lt;p&gt;$F = ma$, so a less massive isotope should move faster, depending on the forces moving it.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11551" LastEditDate="2015-01-14T06:30:53.093" LastActivityDate="2015-01-14T06:30:53.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23750" PostTypeId="1" AcceptedAnswerId="23796" CreationDate="2015-01-14T08:21:51.547" Score="8" ViewCount="2941" Body="&lt;p&gt;The normal technique for flame tests is to dip a clean nichrome or platinum wire into a solution of the relevant salt, and observe the resulting flame colour when inserting the wire into a non-luminous Bunsen flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the species undergoing the electron transition that emits the light of characteristic frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/9857/flame-test-is-the-metal-atom-or-the-metal-ion-responsible-for-the-flame-colour/9861#9861&quot;&gt;One explanation&lt;/a&gt; suggests vaporisation and then atomisation of the sample, with the light emission occurring from the excited atom, rather than ion. &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/group1/flametests.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another&lt;/a&gt; suggests that it’s the ions undergoing the electronic transitions. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Flame_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article is somewhat vague on the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I was doing some flame tests for a student, and in the absence of an available nichrome or platinum wire I simply took a pinch of the solid salts ($\ce{NaCl}$ and $\ce{KCl}$ respectively), and threw it into a non-luminous Bunsen flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic yellow and lilac flame colours were instantly and repeatably observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the outer regions of a Bunsen flame are at less than $1000\ \mathrm{^\circ C}$, and that the boiling points of $\ce{NaCl}$ and $\ce{KCl}$ are $1413\ \mathrm{^\circ C}$ and $1420\ \mathrm{^\circ C}$ respectively (according to Wikipedia), it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the species exhibiting the electron transitions? Is it the metal ions still within the crystal lattice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I should emphasise that the flame colour was observed in the outer regions of the flame, not the super-hot tip of the inner cone. So $1000\ \mathrm{^\circ C}$, not $1500\ \mathrm{^\circ C}$ by any stretch of the imagination. And I don’t at all buy the notion that the colours were simply the glow of hot solid particles emitting black body radiation. The colours were the characteristic sodium yellow and potassium lilac, instantly recognisable by anyone that’s familiar with simple tests like these.&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-09-01T13:46:46.300" Title="What causes the colour when conducting flame tests on solid salts?" Tags="&lt;inorganic-chemistry&gt;&lt;electrons&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23751" PostTypeId="2" ParentId="23750" CreationDate="2015-01-14T08:32:39.193" Score="4" Body="&lt;p&gt;Bunsen flame is up to 1500 C hot depending on the region.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it seems unlikely that the instantly generated characteristic colours originate from individual ions or individual atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is, however, more or less generally accepted. More precisely, in some flames the color is derived from emission of excited atoms, while in other it is molecules. alkali metal colors are believed to be derived from atomic emission, while copper bluer, strontium red and barium green are emitted by monochloride molecules (well, barium green and strontium red can be, less efficiently, produced by other molecules, but professional pyrotechnic compositions are usually manufactured with chlorine donors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism of excitation, however, is a bit less clear, but it is sometimes believed to be chemical and not thermal in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid particles in flame usually produce black body radiation, which may be, depending on temperature, red, orange, yellow and whitish (at extremely high temperatures, well above any that a flame can produce it becomes bluish)&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="9961" LastEditDate="2017-09-01T13:46:46.300" LastActivityDate="2017-09-01T13:46:46.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23752" PostTypeId="1" CreationDate="2015-01-14T09:48:16.647" Score="2" ViewCount="1260" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ClfCi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ClfCi.png&quot; alt=&quot;reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that one of the products will be 2-methylpropan-2-ol, arising via a $\mathrm{S_N1}$ reaction. But what else is formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am looking for a side product that is gaseous.&lt;/p&gt;&#xA;" OwnerUserId="12748" LastEditorUserId="16683" LastEditDate="2017-09-23T17:08:07.893" LastActivityDate="2017-09-23T17:08:07.893" Title="What is the major side product of the reaction between 2-bromo-2-methylpropane and hydroxide ion?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23753" PostTypeId="2" ParentId="23733" CreationDate="2015-01-14T11:38:36.110" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;but what sort of problems?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where two bodies move in the influence of their mutual forces only(that act in pairs and net force on system is zero, so that COM(Centre Of Mass) remains at rest), no external forces! &#xA;Actually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;When two bodies are moving in translational,rotational or vibrational motion only under their mutual interaction, then for simplified analysis, we can consider one body at rest and analyze the motion of other body wrt first body and change the inertial mass of moving body to a new value called reduced mass given as.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is such that we are treating the motional parameters of one fixed(i.e. at rest and at rest forever(until something weird happens)). Suppose you are travelling in a train, when you see an oppositely coming train, it seems faster since when you observe, in your reference frame(i.e., wrt you) the speed of the train becomes $\vec v_{other-train}-\vec v_{your-train}$ Since speed of your train has opposite sign as the speed of other train, it rather becomes: $v_1-(-v_2)=v_1+v_2$ and thus speeds add up and it looks fast. Similarly in case of moving in same direction, it appears slow as speeds subtract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This often helps because sometimes problems are solved easily since one equation is always reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vibrational Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;suppose two bodies are connected by spring and vibrating, then both are constantly moving, If they are following SHM (Simple Harmonic Motion), then to calculate time period we need to account both of them, but it could be done easily if we treat one at rest and look at the other body while sitting on the body we wish to have at rest. That turns out to be:&#xA;$$T=2\pi\sqrt{\frac{\mu}k}=2\pi\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were doing it the traditional way:&#xA;$$k_1=kl/l_1=kl(m_1+m_2)/(m_2l)\\k_1=k(m_1+m_2)/m_2\\T=2\pi\sqrt{m_1/k_1}=2\pi\sqrt{m_1m_2/(k(m_1+m_2))}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Translational example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to fin the separation velocity between two planets at a distance $r$ when their distance reduces to $r/2$ using both methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotational Motion Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to find the total kinetic energy of two particles at a distance l rotating about their COM due to their mutual gravitational attraction without any external force at angular speed $\omega$ by both methods.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-14T11:38:36.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23754" PostTypeId="2" ParentId="23734" CreationDate="2015-01-14T11:40:52.303" Score="4" Body="&lt;p&gt;Methane dissolved in water is simply methane. It does not react with the water under normal conditions.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-14T11:40:52.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23755" PostTypeId="1" CreationDate="2015-01-14T11:46:02.290" Score="3" ViewCount="82" Body="&lt;p&gt;I was teaching the lime cycle to some children yesterday. I had a limestone chip (about 10g) held at the tip of a blue Bunsen flame for several minutes. It decomposed and disintegrated but only glowed very slightly, and was mostly orange with a bit of white at the tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It certainly gave off white light, but nowhere near the quantities required to light a theatre stage, which is what this stuff is famous for, not by a few orders of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried sitting it on gauze and bringing in a second Bunsen flame from the side, but this did not improve the effect. I recall seeing quite a bright light when this was shown to me as a child.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any theories on what went wrong? Any gotchas for doing this kind of experiment?&lt;/p&gt;&#xA;" OwnerUserId="10112" LastEditorUserId="194" LastEditDate="2017-04-01T03:14:19.633" LastActivityDate="2017-04-01T03:14:19.633" Title="Why might my limelight experiment have failed and how can I improve it?" Tags="&lt;experimental-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23758" PostTypeId="2" ParentId="23752" CreationDate="2015-01-14T12:02:50.423" Score="7" Body="&lt;p&gt;On addition of hydroxide to t-butyl bromide there are two competing reactions, substitution and elimination. Depending on the conditions one of these reactions will predominate but both will always be present. If you are taking the $S_{N}1$ as the main reaction then the elimination ($E1$) side product will be 2-methylpropene, which is a gas at room temperatures and pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2G2w3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-14T12:02:50.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23760" PostTypeId="2" ParentId="23755" CreationDate="2015-01-14T12:20:11.473" Score="2" Body="&lt;p&gt;Perhaps your flame wasn't hot enough. Real limelights used oxyhydrogen torches, not wimpy natural gas flames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or perhaps your limestone chip wasn't just $\ce{CaCO3}$. If it was dolomitic limestone, there was a lot of $\ce{MgCO3}$, too, and $\ce{MgO}$ doesn't do the trick like $\ce{CaO}$ does. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-01-14T12:20:11.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23761" PostTypeId="2" ParentId="23711" CreationDate="2015-01-14T12:27:16.647" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does a liquids height in a Barometer depend linearly on pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since: $$P=\frac FA=\frac {mg}{A}=\frac {\rho Vg}A=\rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, this derivation seems misguided, since P=F/A, and P is the pressure of the gas, not of the liquid, as the derivation later relates it to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No at equilibrium the pressure of both will be same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Furthermore, it doesn't make sense to me that the height of the liquid and the pressure are necessarily linearly related. Wouldn't this depend on the compressibility of the liquid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since we are assuming uniform density of liquid the mass at any place of same volume will be same, so if we add some water mass adds, so force adds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or rather, it seems like depending upon the properties of the liquid, it may be very difficult to change the height past a certain point, for example. In this sense the added energy from increasing the pressure goes into bonds or is stored in electrical interactions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Such a situation never arrives, as far as theoretical concerns are made, since we assume water as an in compressible fluid.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-14T12:27:16.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23762" PostTypeId="2" ParentId="23689" CreationDate="2015-01-14T12:34:36.270" Score="1" Body="&lt;p&gt;You need to be familiar with some of the compounds earlier, otherwise you can't tell if it exists. Remember facts come first before explanation. It also depends usually on the atom/ion to which it is bonded. for example since Fluorine is most electronegative it can form $\ce{XeF6,XeF4}$,etc. but $\ce{XeH3}$ doesn't form also oxygen forms large oxidation states as it is electronegative too and there is also more energy released as many bonds are formed (ususlly pi-bonds.)&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-14T12:34:36.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23763" PostTypeId="1" AcceptedAnswerId="23766" CreationDate="2015-01-14T13:53:13.213" Score="4" ViewCount="209" Body="&lt;p&gt;I often hear that the production process for PVC fabrics (such as those used in Gucci and Versace handbags) is very harmful to the environment. What part of the process is harmful and can this be prevented?&lt;/p&gt;&#xA;" OwnerUserId="12750" LastEditorUserId="7448" LastEditDate="2015-01-14T16:11:04.420" LastActivityDate="2015-01-14T17:21:05.103" Title="How is the process of creating PVC fabrics harmful for the environment?" Tags="&lt;synthesis&gt;&lt;safety&gt;&lt;polymers&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23764" PostTypeId="2" ParentId="23684" CreationDate="2015-01-14T14:00:25.443" Score="0" Body="&lt;p&gt;For first part since system is operated reversibly at constant temperature:&#xA;$$\newcommand{\d}{{\rm d}}&#xA;\d S=\frac{\d q_{rev}}{T}\implies \Delta S=\frac{Q_{rev}}{T}$$&#xA;This is what you have done.&#xA;For second part, Yes Gibbs energy is the maximum (in reversible process) non-expansion/additional work (here the electrical work)&#xA;$$\Delta G^\circ=-\nu FE^\circ=w_{add,max}$$&#xA;This value would be same as $\Delta H^\circ$ because from rearrangement of definition of $$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$$$$\Delta H^\circ=\underbrace{\Delta G^\circ}_{\text{non expansion work}}+\underbrace{T\Delta S^\circ}_{\text{work extracted as heat}}$$&#xA;Also it can be viewed in another way, since $\Delta H^\circ$ is &quot;the energy supplied as heat at constant pressure&quot;.Another thing to keep in mind is $\d H=q_p$ only under constant pressure and no additional work, both to be satisfied. &#xA;And for the Classius Inequality, there's some calculation error by you:&#xA;$$\frac{Q}{T}=\frac{-70000}{298}=-234.8\le-61.6(\Delta S^\circ)$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-14T14:00:25.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23765" PostTypeId="2" ParentId="23734" CreationDate="2015-01-14T15:21:32.050" Score="1" Body="&lt;p&gt;I think you are confusing dissociation with dissolution. Bonds do not need to be broken in order to dissolve a substance. The reason so many covalent compounds do disassociate in water is that water is a highly polar solvent and therefore preferentially solvates ions (provides the driving force for the disassociation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane is pratically insoluble in water, but it will dissolve at low enough concentration to give a solution of methane in water, with only van der Waals forces between the methane and water molecules.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-01-14T15:21:32.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23766" PostTypeId="2" ParentId="23763" CreationDate="2015-01-14T17:21:05.103" Score="1" Body="&lt;p&gt;I wouldn't think that the &lt;em&gt;process&lt;/em&gt; of producing PVC fabrics is that much of a problem, for reasons:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;PVC is produced by experts that have taken all the necessary precautions for all possible reductions of safety risks it may cause in the lab or in mass scale production.&lt;/li&gt;&#xA;&lt;li&gt;The process of synthesizing PVC is not a secret. If it &lt;em&gt;was&lt;/em&gt; too harmful (rather than its profitability), then companies would have been banned from producing such a thing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Phew, I thought there was something serious about PVC fabrics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There is! The problem is not the &lt;em&gt;process&lt;/em&gt;, but &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vinyl_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;VCM&lt;/a&gt;&lt;/strong&gt; and &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phthalate&quot; rel=&quot;nofollow noreferrer&quot;&gt;phthalates&lt;/a&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Issue 1: VCM&lt;/h2&gt;&#xA;&lt;p&gt;With a chemical formula of &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H3Cl}$&lt;/span&gt;, a boiling point of &lt;span class=&quot;math-container&quot;&gt;$-13.9°C$&lt;/span&gt; and an approximate molecular weight of &lt;span class=&quot;math-container&quot;&gt;$62.5 \frac{g}{mol}$&lt;/span&gt;, VCM is the raw material for PVC. According to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1475205/?page=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, the death of four workers had been approved due to VCM in 1974. (Though wikipedia says it was three workers dead) It's reported that in 1975, an annual 5-billion-pounds rate of production of VCM was confirmed.&lt;/p&gt;&#xA;&lt;p&gt;The industry had the worst ways of producing VCM, and has suffered for decades until better ways of producing it have come around. According to &lt;a href=&quot;http://www.pvc.org/en/p/vinyl-chloride-monomer-vcm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, a traditional way of producing VCM was the reaction of acetylene with Hydrochloric acid, as following: &lt;span class=&quot;math-container&quot;&gt;$$\ce{C2H2} + \ce{HCl} ~-&amp;gt;~ \ce{C2H3Cl}$$&lt;/span&gt;&#xA;It was in the presence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury%28II%29_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mercury(II) Chloride&lt;/a&gt; as a catalyst! Nothing has been done to reverse the harmful effects done to environment because of this very reaction.&lt;/p&gt;&#xA;&lt;p&gt;In the modern process, &amp;quot;polymerization&amp;quot; is known to be the most dangerous stage and the peak of the overexposure.&lt;/p&gt;&#xA;&lt;p&gt;VCM is believed to be carcinogen. Severe symptoms have been reported due to the overexposure of VCM; such as: Hepatotoxicity (liver damage and tumor growth caused by chemicals and more precisely, angiosarcoma).&lt;/p&gt;&#xA;&lt;h2&gt;Issue 2: phthalates&lt;/h2&gt;&#xA;&lt;p&gt;VCM is the fundamental raw material for the synthesis of all kinds of PVC; rigid or soft. But &lt;a href=&quot;http://en.wikipedia.org/wiki/Phthalate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plasticizers&lt;/a&gt; are being added to rigid PVC to &amp;quot;soften it up&amp;quot;; so it would be applicable for the future use as &amp;quot;PVC fabrics&amp;quot;. Phthalates are common plasticizers in the production of soft PVCs.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They are divided into two distinct groups, with very different applications, toxicological properties, and classification, based on the number of carbon atoms in their alcohol chain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Phthalates aren't covalently bonded to the matrix of PVC. Therefore, the lighter ones can leach relatively very easy, usually due to heat or strong solvents.&lt;/p&gt;&#xA;&lt;p&gt;The most common phthalate is &lt;a href=&quot;http://en.wikipedia.org/wiki/Bis(2-ethylhexyl)_phthalate&quot; rel=&quot;nofollow noreferrer&quot;&gt;diethylhexyl phthalate&lt;/a&gt;(&lt;strong&gt;DEHP&lt;/strong&gt;). Studies show that Americans are exposed to DEHP. &lt;em&gt;DEHP is believed to cause health complications in a variety of ages&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Oh, wait: Issue 3: PVC itself! But when it degrades&lt;/h2&gt;&#xA;&lt;p&gt;PVC is known for its strong linear polymer; but when it degrades (usually due to careless use or disposal) it degrades! Degradation is excessive reduction in a plastic's average molar weight. PVC will &amp;quot;microcrack&amp;quot; when it's degraded, into macroparticles that are a big threat to the environment. These particles &amp;quot;soak up&amp;quot; &lt;a href=&quot;http://www.chem.unep.ch/pops/&quot; rel=&quot;nofollow noreferrer&quot;&gt;POPs&lt;/a&gt; and afterwards are usually ingested by living things, making it a big deal of our everyday plastic pollution exposure. (as &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyvinyl_chloride#Health_and_safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; puts it)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What should I do now?! Panic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Considering the advancement of technology and our anxiety to preserve our health we humans aren't sitting around to watch ourselves getting sick! I don't mean the hard-to-come-by sites that say overexposure to PVC isn't carcinogenic, but real efforts into solving the problem. News has been spread that &lt;a href=&quot;http://en.wikipedia.org/wiki/Registration,_Evaluation,_Authorisation_and_Restriction_of_Chemicals#REACH_Authorisation_List&quot; rel=&quot;nofollow noreferrer&quot;&gt;REACH&lt;/a&gt; attempts to authorize DEHP and two other phthalates and ban their use in specific materials. DINP is another plasticizer, approved by the EU risk assesment, that will eventually replace the former plasticizers, hopefully.&lt;/p&gt;&#xA;&lt;p&gt;Scientific researches for over ten years have proven that the amount of exposure to the chemicals mentioned is far from dangerous in current conditions and uses. So sit back relaxed; it's true that these dangers exist, but they're not serious threats to a healthy person, at least not yet.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-14T17:21:05.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23767" PostTypeId="1" AcceptedAnswerId="23795" CreationDate="2015-01-14T17:23:56.640" Score="1" ViewCount="1176" Body="&lt;p&gt;I am currently in charge of purchacing fabrics from asia. My personal policy is to try and stay as enviromentally freindly as possible. Your answer to the following question would be of great help and would prevent me from buying the wrong product in massive quantities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a TPE fabric and how does it compare to PVC fabrics?&lt;/p&gt;&#xA;" OwnerUserId="12750" LastEditorUserId="4275" LastEditDate="2015-01-14T17:38:19.550" LastActivityDate="2015-01-15T17:18:56.033" Title="TPE fabric vs PVC fabrics?" Tags="&lt;polymers&gt;&lt;environmental-chemistry&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23768" PostTypeId="2" ParentId="23633" CreationDate="2015-01-14T17:49:00.160" Score="6" Body="&lt;p&gt;If you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs free energy &lt;em&gt;is&lt;/em&gt; an energy in the sense that it is the part of the energy that is &quot;free&quot; to do useful work. It does indeed have units of energy, and $\Delta G$ can be used to compute the maximum  possible nonexpansion work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;not&lt;/em&gt; an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G_\text{system} + G_\text{surroundings}$ isn't constant in a process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an isothermal process, $\Delta G= \Delta H−T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="16683" LastEditDate="2018-09-06T14:25:33.620" LastActivityDate="2018-09-06T14:25:33.620" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23769" PostTypeId="2" ParentId="23633" CreationDate="2015-01-14T18:13:27.977" Score="4" Body="&lt;p&gt;Imagine you have a divided insulated container.  On one side is helium gas and on the other side is neon gas.  Both sides are uniformly at standard temperature and pressure.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You remove the divider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gases mix, because there are greatly more microstates we would consider &quot;mixed&quot; compare to not mixed, statisitically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy (S) increased, and in the ideal gas approximation there is no change in temperature or pressure (but there is a significant change in entropy of the system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs so-called &quot;free energy&quot; decreased solely because entropy increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No energy went anywhere.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-14T18:29:06.873" LastActivityDate="2015-01-14T18:29:06.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23771" PostTypeId="2" ParentId="4399" CreationDate="2015-01-14T19:08:06.500" Score="23" Body="&lt;p&gt;Hybridization is given by the following formula:&#xA;$$H= \frac{1}{2} (V + X - C + A)$$&#xA;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$V$ = number of valence electrons in central atom&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;$X$ = number of monovalent atoms around the central atom&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;$C$ = positive charge on cation  &lt;/li&gt;&#xA;&lt;li&gt;$A$ = negative charge on anion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$H=4 \to \ce{sp^3},\;2\to \ce{sp,\;3}\to \ce{sp^2}...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.: in $\ce{NH3}$, the hybridization of $\ce{N}$ atom is: &#xA;$$H= \frac{1}{2}(5+3-0+0)=4 \to \ce{sp^3}$$&lt;/p&gt;&#xA;" OwnerUserId="12757" LastEditorUserId="5026" LastEditDate="2018-03-20T06:59:25.293" LastActivityDate="2018-03-20T06:59:25.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23774" PostTypeId="1" CreationDate="2015-01-14T19:49:50.830" Score="1" ViewCount="90" Body="&lt;p&gt;Period 2 elements like carbon, oxygen and nitrogen cannot expand their octet because they cannot make use of d orbitals in that energy level (they don't exist). However, period 3 elements like sulfur and chlorine can because they can utilize the d orbitals with the same principal quantum number. However, the the 4s orbitals are lower in energy than the 3d orbitals (from the order of filling of electrons) so why can't they be used/ why aren't they? Moreover, why can't period 2 elements use the 3s orbital? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="7448" LastEditDate="2015-01-14T20:06:31.310" LastActivityDate="2015-01-14T20:06:31.310" Title="Why are d-orbitals required/used for hypervalent molecules (where the central atom has expanded its octet)" Tags="&lt;inorganic-chemistry&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;periodic-table&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-01-14T20:54:22.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="23775" PostTypeId="2" ParentId="23633" CreationDate="2015-01-15T00:39:19.363" Score="2" Body="&lt;p&gt;I think the other answers here are pretty good, but it seems to me that the main point of confusion is over what it means for energy to be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that total energy is conserved, but you also know that it can take different forms. Gibbs Free energy is just one form of energy. When Gibbs free energy decreases, it goes somewhere - either as heat released by the system, or into increasing the entropy of the system. Those are really the only two options, because this is how Gibbs energy is defined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why is this even useful?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your thoughts at the end are correct - it is useful because it lets us determine spontaneity, and also the amount of energy available to do &quot;useful&quot; work - useful in the sense that we can control its transformation to other forms of energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We track enthalpy even though it is the thermodynamic potential for constant P &amp;amp; S systems because at constant T &amp;amp; P, in the absence of non-PV work, it is also equal to the heat produced or absorbed by the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lab that might not seem like a big deal, but if you are designing a steam engine, air conditioning system, or a chemical reactor, being able to predict the amount of heat that is absorbed or lost during a transition is very useful!&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-15T00:39:19.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23776" PostTypeId="1" AcceptedAnswerId="23784" CreationDate="2015-01-15T05:26:06.293" Score="2" ViewCount="2446" Body="&lt;p&gt;I have one gallon of commercial, surfactant free, ammonium hydroxide solution that I would like to use to prepare a concentrated solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two options as I see it, and I am still getting into the chemistry side (amateur chemistry), but am pushing ahead and want to do this even without the understanding. I am hoping the solutions will provide adequate explanation as to what is being done, and why, which will help me learn to solve problems like this in the future. Chemistry newbie, here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility at room temperature is to be assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that I have two options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Prepare a receiving flask with cold distilled water (likely in an ice bath) and then prepare a distillation flask with the %10 ammonium hydroxide solution. Slightly heat the distillation flask to the point where the ammonia gas begins to off-gas, which is funneled into the receiving flask and dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Prepare a receiving flask and distillation flask both with an amount of the 10% ammonium hydroxide solution. As above, gently heat the distillation flask to off-gas the ammonia into the receiving flask where it should dissolve, and increase the concentration beyond 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I would like a 1 liter solution containing 28% NH3(aq), what volume of cold distilled water should be placed in the receiving flask, which I ask because I assume the solvation of NH3 into water will cause the volume to increase. If I placed 1 liter of water into the flask initially, then this would result in more than 1 liter of solution. This also depends on the amount of ammonia that needs to be dissolved to reach my desired concentration, which varies from option #1 to #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to know is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1) How to determine the volume of a solution given a specific concentration of NH3(aq). This will allow me to account for the increase in volume when adding ammonia to an existing solution, in order to reach a desired concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2) How do I know when I have reach %28 NH3(aq) concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3) A procedure leading from 10% to 28%, and any advice on more technically correct approaches to this problem. Is it a problem that should be considering molarity, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4) Any quick background info that I can look up to help me understand this problem would be welcome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line: Prepare a 1 liter solution of 28% NH3(aq) from 1 gallon of 10% NH3(aq), and please provide an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help. I am a software programmer, and have delved into he world of chemistry as a hobby, and am getting confused by terms like molarity, normality, and both of those versus concentration, etc. It's maddening to say the least.&lt;/p&gt;&#xA;" OwnerUserId="12758" LastEditorUserId="12758" LastEditDate="2015-01-15T14:56:08.393" LastActivityDate="2015-01-15T15:40:06.337" Title="Prepare 1 liter of 28% NH3(aq) solution from 1 gal of %10 NH3(aq)" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23778" PostTypeId="1" CreationDate="2015-01-15T09:07:26.230" Score="3" ViewCount="458" Body="&lt;p&gt;The www is full of crackpot claims about how various food treatments alter their nutrition content and it makes finding actual science very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically interested in the vitamin C content of orange juice.&#xA;From an answer I received in Biology SE (referencing &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1365-2621.2002.tb09543.x/abstract&quot; rel=&quot;nofollow&quot;&gt;Hodgins et al. 2002&lt;/a&gt;), pasteurisation of orange juice is unlikely to reduce the vitamin C content by more than 20%. This tallies with what I have been able to find on the reduction of vitamin C in milk by pasteurisation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is twofold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) What are the processes at work which remove some vitamin C during pasteurisation? Is it just the heat? What is the mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) If I were to &lt;em&gt;boil&lt;/em&gt; orange juice for 10 minutes or so, would that remove much more vitamin C and if so, why?&lt;/p&gt;&#xA;" OwnerUserId="12761" LastActivityDate="2015-01-15T09:07:26.230" Title="How does heat and/or boiling remove vitamin C?" Tags="&lt;heat&gt;&lt;food-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23780" PostTypeId="1" AcceptedAnswerId="23794" CreationDate="2015-01-15T09:31:49.800" Score="1" ViewCount="55" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$22~\mathrm{g}$ of dry ice is placed in an an empty $600~\mathrm{ml}$ closed vessel at $298~\mathrm{K}$. Find the final pressure inside the vessel, if all $\ce{CO2}$ gets evaporated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now by applying $pV = n\mathcal{R}T$, the value of $p$ comes out to be $20.4~\mathrm{atm}$, but in the solution shown, it adds $1~\mathrm{atm}$ i.e atmospheric pressure pressure, so the answer comes out be $21.4~\mathrm{atm}$. Now why do we need to add atm pressure, the question tells us to find the total pressure inside the vessel, isn't the atmospheric pressure acting only on the container, and not on its contents?&lt;/p&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="5633" LastEditDate="2015-01-15T17:23:22.090" LastActivityDate="2015-01-15T20:09:10.383" Title="Is atmospheric pressure acting only on the contents or also on the container?" Tags="&lt;physical-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23781" PostTypeId="1" AcceptedAnswerId="23791" CreationDate="2015-01-15T10:11:20.447" Score="2" ViewCount="1716" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$48~\mathrm{l}$ of dry $\ce{N2}$ passes through $36~\mathrm{g}$ $\ce{H2O}$ at $300~\mathrm{K}$, and this results in a water loss of $1.20~\mathrm{g}$. The vapor pressure of water is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution says, that water loss is observed because of escape of water molecules with $\ce{N2}$ gas. Then it says that water vapours occupy the volume of $\ce{N2}$, i.e  $48~\mathrm{L}$, this part I am not able to understand. Why would water occupy the same volume as that of $\ce{N2}$, it isn't mentioned that these two are in some container of volume of $48~\mathrm{L}$. What is the logic behind this? Then the solution proceeds, by applying $pV=\frac{m\mathcal{R}T}{M}$. &lt;/p&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="69830" LastEditDate="2021-05-21T17:57:40.853" LastActivityDate="2021-05-21T17:57:40.853" Title="How to calculate the vapor pressure of water?" Tags="&lt;vapor-pressure&gt;&lt;ideal-gas&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23784" PostTypeId="2" ParentId="23776" CreationDate="2015-01-15T13:24:28.040" Score="2" Body="&lt;p&gt;Disclaimer: ammonia is &lt;a href=&quot;http://avogadro.chem.iastate.edu/MSDS/NH3_gas.pdf&quot; rel=&quot;nofollow&quot;&gt;extremely hazardous&lt;/a&gt; including flammable in some vapor concentrations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I know when I have reach %28 NH4OH concentration?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can measure the specific gravity and compare to the table on page 2 of &#xA;&lt;a href=&quot;http://www.tannerind.com/PDF/green-aqua-amm.pdf&quot; rel=&quot;nofollow&quot;&gt;this document:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it is important to understand that if you heat/boil/distill the 10% solution the vapor coming off is not 100% ammonia, it is a mixture of ammonia and water, but with increased ammonia concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To predict the percent ammonia in the vapor phase, you need to look at the dew point line in a table like &lt;a href=&quot;http://www-old.me.gatech.edu/energy/laura/node4.html&quot; rel=&quot;nofollow&quot;&gt;Fig. 1.2 here&lt;/a&gt;, but that one is for 16 bars or Fig. 2 of &lt;a href=&quot;http://www.academicjournals.org/article/article1379684276_Ganesha%20and%20Srinivasb.pdf&quot; rel=&quot;nofollow&quot;&gt;Ganesh et al. Journal of Mechanical Engineering Research Vol. 3. (1), pp. 25-39&lt;/a&gt; that shows many pressures including 1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the original solution is heated, the system progresses toward the upper left in Fig. 1.2/Fig.2, the liquid phase along the bubble point line and the vapor phase along the dew point line.  The concentration of ammonia in both phases progressively decreases.  Eventually, the concentration of ammonia in the vapor phase drops below the concentration of ammonia in the original solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fig. 2 of Ganesh et al. it can be seen that at 1 bar starting from 10 weight% ammonia liquid phase, the initial vapor phase concentration would be close to 28% ammonia.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that even if you added no water to the condensate, there is no way you can get 1 liter of 28% ammonia from 1 gallon of 10% ammonia in one simple distillation step.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only through fractional distillation could you get that much 28% ammonia from that little 10% ammonia.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, practically speaking, it will be difficult to avoid lossing ammonia vapors to the environment and safety needs to be considered.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-15T15:40:06.337" LastActivityDate="2015-01-15T15:40:06.337" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23785" PostTypeId="1" AcceptedAnswerId="23790" CreationDate="2015-01-15T13:42:32.827" Score="3" ViewCount="182" Body="&lt;p&gt;In Gibbs energy equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G=\Delta H-T\,\Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant pressure, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\Delta H\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we know that $T \Delta S=Q$ (reversible)&#xA;so finally &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\,\Delta S=Q\tag 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting values of Equation $\text{(1)}$ and $\text{(2)}$ in Gibbs energy equation ,&#xA;we should always get $\Delta G=0$ for a reversible reaction. ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't it so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to search it on Google; if anyone could give me a link or otherwise explain it?&lt;/p&gt;&#xA;" OwnerUserId="12719" LastEditorDisplayName="user7951" LastEditDate="2018-09-06T17:52:02.473" LastActivityDate="2019-04-13T12:14:11.823" Title="Gibbs energy and Entropy" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23786" PostTypeId="1" AcceptedAnswerId="23789" CreationDate="2015-01-15T14:50:09.727" Score="3" ViewCount="593" Body="&lt;p&gt;PTFE (Polytetrafluoroethylene) and FEP (Fluorinated ethylene propylene) have the following chemical formulas:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4TfEf.png&quot; alt=&quot;PTFE formula, from Wikipedia&quot; /&gt;&lt;img src=&quot;https://i.stack.imgur.com/AXPuu.png&quot; alt=&quot;FEP formula, from Wikipedia&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It's been years since I have taken organic chemistry, but I do remember alkenes have double bonds and alkanes have single bonds.  Neither of these compounds have double bonds.&lt;/p&gt;&#xA;&lt;p&gt;Why do these compound names contain &amp;quot;-ene&amp;quot; in them, instead of &amp;quot;Polytetrafluoroethane&amp;quot; or something?&lt;/p&gt;&#xA;" OwnerUserId="12767" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-07-05T20:47:34.250" Title="Why are PTFE and FEP named &quot;ethylene&quot; when they don't have double bonds?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2015-01-16T02:03:05.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="23787" PostTypeId="1" AcceptedAnswerId="23824" CreationDate="2015-01-15T14:57:08.623" Score="6" ViewCount="8454" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead_dioxide&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; suggests that lead dioxide is produced by this reaction:&#xA;$$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same article also suggests that lead dioxide reacts with nitric acid in the following way:&#xA;$$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand why then, in the first reaction, $\ce{PbO2}$ doesn't further react with the residual nitric acid. Is it because the quantity of the acid is regulated? Or is there some error in the reactions?&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="16683" LastEditDate="2018-04-17T23:06:42.520" LastActivityDate="2018-04-17T23:06:42.520" Title="Reaction between lead dioxide and nitric acid" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23788" PostTypeId="2" ParentId="23786" CreationDate="2015-01-15T15:01:07.550" Score="0" Body="&lt;p&gt;The compounds are polymers. The unit structure of PTFE is tetrafluoroethene. PTFE is named in accordance with the subunits from which they are polymerized. You can apply the same logic to the second one (you apply the same logic to all polymers). There are no common names for polymers just abbreviations. IUPAC rules do have polymer rules too, so you will necessarily see them adhering to the IUPAC rules. It is, however, a matter of the unit structure in brackets, not the monomer that determines the name.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="20789" LastEditDate="2016-07-05T20:47:34.250" LastActivityDate="2016-07-05T20:47:34.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23789" PostTypeId="2" ParentId="23786" CreationDate="2015-01-15T16:04:03.677" Score="8" Body="&lt;p&gt;Polymers follow a slightly different nomenclature system than you are used to in organic chemistry. Instead of naming the polymer after the monomers &lt;em&gt;as they are&lt;/em&gt;, we name them after the monomers &lt;em&gt;as they originally were&lt;/em&gt;. In other words, a polymer made from ethylene monomers is called polyethylene. Since PTFE and FEP are made from fluorinated ethylene and propylene, we just put &quot;poly&quot; in front of the monomer names. This is despite the fact that after the reaction, the bond order is reduced by one. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-15T16:04:03.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23790" PostTypeId="2" ParentId="23785" CreationDate="2015-01-15T16:08:35.570" Score="3" Body="&lt;p&gt;Reason 1: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;while $T\,\mathrm dS=\mathrm dQ$ for a reversible process, this does not mean $T\Delta S = Q$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason 2: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;reversible reaction&quot; is unrelated to &quot;reversible process&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lame Reason 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because your &quot;always&quot; statement does not repeat the &quot;constant pressure&quot; requirement assumed for $Q=\Delta H$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorDisplayName="user7951" LastEditDate="2018-09-06T17:46:26.243" LastActivityDate="2018-09-06T17:46:26.243" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23791" PostTypeId="2" ParentId="23781" CreationDate="2015-01-15T16:19:02.697" Score="2" Body="&lt;p&gt;It doesn't matter if there is a container.  I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But adding 1.2 grams of water would increase the total volume by about 1 L.  I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-15T16:27:25.723" LastActivityDate="2015-01-15T16:27:25.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23792" PostTypeId="2" ParentId="23785" CreationDate="2015-01-15T16:26:49.963" Score="2" Body="&lt;p&gt;The problem is that you are equating &quot;reversible reaction&quot; with &quot;reversible process.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are seeing when you plug equations 2 and 1 into the Gibbs Free Energy equation is that at equilibrium, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that this works is because a reversible process is one that progresses by very small movements away from equilibrium states - when we say a process is &quot;reversible&quot; what we really mean is that we are making an approximation that allows us to ignore entropy generation. When you make this approximation and apply it to the Gibbs Free Energy equation, you are forcing it to stay at equilibrium - hence, $\Delta G = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your confusion is coming from the similar names of &quot;reversible reaction&quot; and &quot;reversible process.&quot; A &lt;a href=&quot;http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics)&quot; rel=&quot;nofollow&quot;&gt;reversible proces&lt;/a&gt;s is one in which the entropy generation term is zero, but a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reversible_reaction&quot; rel=&quot;nofollow&quot;&gt;reversible reaction&lt;/a&gt; is one in which the forward and reverse rates of reaction result in a mixture of reactants and products at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a reversible &lt;strong&gt;reaction&lt;/strong&gt; can be in an equilibrium ($\Delta G = 0$) or non-equilibrium state ($\Delta G \neq 0$), but a reversible &lt;strong&gt;process&lt;/strong&gt; is always in an equlibrium state.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-15T16:26:49.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23793" PostTypeId="2" ParentId="23781" CreationDate="2015-01-15T16:35:13.637" Score="2" Body="&lt;p&gt;This is an example of using partial pressures for a mixture of ideal gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are making a big assumption - that water is an ideal gas - but once we make that assumption, the way to proceed is to treat each gas in the mixture &lt;em&gt;as if it were by itself&lt;/em&gt; in the container. As you pointed out, it isn't very obvious from the way the question is worded that we would assume that both would occupy 48 L, but, since the volume of a gas depends either on the container volume or on temperature and pressure when there is no container, it does make sense. The idea is that whatever is constraining the $\ce{N2}$ to 48 L would also constrain the $\ce{H2O}$ to the same volume, since it is a mixture of gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the partial pressure equals the vapor pressure, the solution is given by calculating the pressure of this mass of water under these conditions. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-15T16:35:13.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23794" PostTypeId="2" ParentId="23780" CreationDate="2015-01-15T16:52:29.840" Score="1" Body="&lt;p&gt;According to the way this question is worded, adding 1 atmosphere would be incorrect. The reason is that they specify an &lt;em&gt;empty&lt;/em&gt; container - if you have a rigid container and it is truly empty (high vacuum), then the absolute pressure would be zero, and the gauge pressure would be -1 atm (see this &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure_measurement#Absolute.2C_gauge_and_differential_pressures_-_zero_reference&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; for reference to these terms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the $\ce{CO2}$ sublimates, the final absolute pressure would be given by $PV=nRT$ - there is no need to add atmospheric pressure. To find the gauge pressure, you would actually &lt;em&gt;subtract&lt;/em&gt; one atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the question wants you to assume that &quot;empty&quot; means &quot;filled with air at 1 atm,&quot; but that is a guess and I would argue that it's not safe to assume that based on the way the question is written.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-01-15T20:09:10.383" LastActivityDate="2015-01-15T20:09:10.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23795" PostTypeId="2" ParentId="23767" CreationDate="2015-01-15T17:18:56.033" Score="3" Body="&lt;p&gt;This question is impossible to answer without knowing &lt;em&gt;which&lt;/em&gt; TPE you mean - &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermoplastic_elastomer&quot; rel=&quot;nofollow&quot;&gt;TPE&lt;/a&gt; is a mixture of thermoplastic elastomers, which is a huge class of compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyvinyl_chloride&quot; rel=&quot;nofollow&quot;&gt;PVC&lt;/a&gt; is polyvinylchloride, which is a specific compound. I found an MSDS here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fishersci.com/ecomm/servlet/msdsproxy?productName=AC183320010&amp;amp;productDescription=POLY&quot; rel=&quot;nofollow&quot;&gt;http://www.fishersci.com/ecomm/servlet/msdsproxy?productName=AC183320010&amp;amp;productDescription=POLY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice is to contact the supplier for each fabric you are considering and to ask them for the MSDS. In that you will see what the acute and long-term exposure effects are for people. &quot;Environmental effects&quot; is a very broad subject, and you are not very likely to find research for any proprietary blends, particulary if they don't come from the US or EU. However, sometimes MSDS list non-proprietary compounds, and you can then check the &lt;a href=&quot;http://www.epa.gov/&quot; rel=&quot;nofollow&quot;&gt;EPA website&lt;/a&gt; for those names to see if there are any reports on broad environmental effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are also interested in the environmental impact of the entire production chain (which you should be, since end-use is usually the &lt;em&gt;least&lt;/em&gt; significant part of that), then you should also ask the suppliers about their manufacturing process, whether they have any &quot;green&quot; certifications, etc. You can look up the patents for each production process (most of the time they will be patented) to see what their raw materials are, and then check the MSDS and EPA reports for those. It is also a good idea to see if you can tour their production facilities, once you know what the potentially dangerous precursors are and how they should be handled. It is easy to temporarily hide bad working conditions, but it is harder to fake treatment and containment facilities. Of course, if there isn't good regulatory oversight, there is a very good chance that the producer will take shortcuts to reduce costs, so keep in mind that what they tell you isn't necessarily going to be the truth.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-15T17:18:56.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23796" PostTypeId="2" ParentId="23750" CreationDate="2015-01-15T17:56:10.053" Score="5" Body="&lt;p&gt;There is a very detailed discussion of what cause the emission lines in flames with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; in The Origin of the Spectra (ironically by P. Darwin Foote et al.) 1922, starting around page 165.&lt;/p&gt;&#xA;&lt;p&gt;They are saying that for &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; most of the transitions are from atomic &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt;, the &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; having vaporized, and Na+ then captured an electron, and a small fraction of the transitions are from the act itself of Na+ capturing the electron.&lt;/p&gt;&#xA;&lt;p&gt;They calculate that a Na atom emits 2000 D-line quanta per second.&lt;/p&gt;&#xA;&lt;p&gt;They also calculate the fraction of Na atoms with electrons in the &lt;span class=&quot;math-container&quot;&gt;$3p$&lt;/span&gt; state is &lt;span class=&quot;math-container&quot;&gt;$\ce{10^{-11}}$&lt;/span&gt; at 1000K, rising to &lt;span class=&quot;math-container&quot;&gt;$\ce{10^{-7}}$&lt;/span&gt; at 1500K. (Note: the book says &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$2p$&lt;/span&gt;&amp;quot;, but &lt;span class=&quot;math-container&quot;&gt;$2p$&lt;/span&gt; in Fowler notation is &lt;span class=&quot;math-container&quot;&gt;$3p$&lt;/span&gt; in modern notation)&lt;/p&gt;&#xA;&lt;p&gt;Also, they say the vapor pressure of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; in a Bunsen flame is on the order of &lt;span class=&quot;math-container&quot;&gt;$\ce{10^{-6} \ mmHg}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So the color is from vaporized &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; (vapor phase &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; being molecular), that has dissociated and formed atomic &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Also in &lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.38.699&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Absorption of Light by Flames Containing Sodium&lt;/a&gt; Phys. Rev. 38, 699 (emphasis added):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The dissociation probably occurs in two or more steps. Possibly the&#xA;molecules break up into sodium and chlorine ions which then lose their&#xA;charges becoming uncharged atoms. There may also be one or more&#xA;intermediate steps in the recombination.  &lt;strong&gt;But certainly the sodium&#xA;giving the light which we are studying consists of uncharged atoms, as&#xA;is evident from its spectrum&lt;/strong&gt;. Evidence that there are comparatively&#xA;few charged sodium atoms in the flame is given by a study of the&#xA;conductivity of flames. Wilson, (p. 85) for example, estimates that&#xA;only 1.6 percent of the sodium in a fame is ionized even when the&#xA;concentration is small.&lt;/p&gt;&#xA;&lt;p&gt;It also seems reasonable to assume that at the&#xA;edge of the flame there is more complete dissociation of the salt than&#xA;in the center, and that this is the cause of the difference between&#xA;the concentration ratios obtained with light from the center and from&#xA;the edges. A fairly definite proof that there is more dissociation at&#xA;the edges is given by the appearances of a fame into which copper&#xA;chloride is being sprayed. Such a fame gives almost no green color&#xA;except at the edges, where it is a bright green. This is the more&#xA;noticeable if there is a small amount of sodium impurity in the copper&#xA;chloride. The center of the flame is then yellow with a distinct border&#xA;of green.&lt;/p&gt;&#xA;&lt;p&gt;The greater dissociation at the surface is no doubt in some way due to the greater amount of air there, but whether it is caused by a higher temperature&#xA;or by a difference in chemical action can not at present be stated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-31T08:38:30.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23798" PostTypeId="2" ParentId="23787" CreationDate="2015-01-15T20:24:33.477" Score="2" Body="&lt;p&gt;You need to consider the reaction conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead_dioxide&quot; rel=&quot;nofollow&quot;&gt;Lead Dioxide article&lt;/a&gt; is saying that in &lt;strong&gt;dilute&lt;/strong&gt; nitric acid the first reaction occurs, while in &lt;strong&gt;warm&lt;/strong&gt; nitric acid the second reaction occurs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-15T20:24:33.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23800" PostTypeId="1" AcceptedAnswerId="23801" CreationDate="2015-01-15T20:43:42.140" Score="12" ViewCount="5276" Body="&lt;p&gt;When atmospheric $\ce{CO2}$ reacts with water to form $\ce{H2CO3}$, this is called dissolution, isn't it? What term would you use for the opposite reaction when it occurs at atmospheric pressure (e.g. due to temperature change)? For a solid it would be precipitation, but here it is for a gas. &lt;/p&gt;&#xA;" OwnerUserId="12772" LastEditorUserId="23561" LastEditDate="2018-09-06T13:38:15.350" LastActivityDate="2018-09-06T13:38:15.350" Title="Is there a name for the opposite reaction to the dissolution?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23801" PostTypeId="2" ParentId="23800" CreationDate="2015-01-15T21:05:02.943" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Effervescence&quot; rel=&quot;noreferrer&quot;&gt;Effervescence&lt;/a&gt; if there are bubbles and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Degasification&quot; rel=&quot;noreferrer&quot;&gt;Degasification&lt;/a&gt; more generally speaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the term &lt;a href=&quot;http://en.wikipedia.org/wiki/Outgassing&quot; rel=&quot;noreferrer&quot;&gt;outgassing&lt;/a&gt;, but that term is broader than just gas coming out of liquid solution.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-16T03:49:13.337" LastActivityDate="2015-01-16T03:49:13.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23803" PostTypeId="1" AcceptedAnswerId="23811" CreationDate="2015-01-15T21:13:00.257" Score="2" ViewCount="249" Body="&lt;p&gt;I was reviewing first posts here, and eating a juicy and salty (table salt) tomato at the same time. Suddenly, I realized that a few table salt crystals have been poured down on my mousepad. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got curious and I rotated the laptop $90°$. I witnessed that the salt is somehow resisting to fall! Not sticked (very firmly attached) though, and the mousepad wasn't wet and didn't contain any possible chemicals that could cause the &quot;attaching&quot;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;except that apparently the mousepad is &quot;soaked&quot; with the &lt;a href=&quot;http://health.howstuffworks.com/skin-care/information/anatomy/sebum2.htm&quot; rel=&quot;nofollow&quot;&gt;sebum&lt;/a&gt; produced by the tip of my fingers. Sebum, however, is a mixture of non-polar lipids such as fatty acids, triglycerides, sterols, sterol esters, wax esters and squalene. (See &lt;a href=&quot;http://www.nature.com/jid/journal/v53/n5/full/jid1969157a.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; also)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be another reason which lies in the electrical field that my laptop produces, and this could affect a compound like table salt that is primarily ionic. (No clue how, am I missing an easy chemistry concept?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there any possible &quot;mild&quot; (as in invisible) reaction, either chemical or physical, as a reason of the phenomenon? Or is this behavior because of the composition of table salt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I'm certain about the existence of such thing, though research showed nothing relevant in the Net.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-01-16T00:30:50.343" Title="Is table salt sticking to my laptop?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23804" PostTypeId="5" CreationDate="2015-01-15T21:14:53.173" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-15T21:14:53.173" LastActivityDate="2015-01-15T21:14:53.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23805" PostTypeId="4" CreationDate="2015-01-15T21:14:53.173" Score="0" Body="Pericyclic reactions are distinguished by their cyclic transition states and concerted mechanisms." OwnerUserId="10256" LastEditorUserId="10256" LastEditDate="2015-01-15T22:54:39.803" LastActivityDate="2015-01-15T22:54:39.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23806" PostTypeId="2" ParentId="15631" CreationDate="2015-01-15T21:43:26.477" Score="9" Body="&lt;p&gt;In your comment, you say you are interested in &quot;liquid water&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values in your self-answer are for gas phase water, and the second source is citing the first source, so it is really just one source.  The value is determined from gas phase rotation spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For liquid water, the value is not as precisely known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;105.5° (calculated) and 106° (experimental) as reported in &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/111/8/10.1063/1.479638&quot;&gt;Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-15T21:43:26.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23807" PostTypeId="1" AcceptedAnswerId="23812" CreationDate="2015-01-15T22:43:15.080" Score="1" ViewCount="6529" Body="&lt;p&gt;So I know that I need to add $4~\mathrm{mL}$ of solution to this powder, which will give me a concentration of the enzyme at $25\ \mathrm{mg/mL}$. What I do not understand is how to decide how much of each of these &quot;ingredients&quot; (below) to add. I have $\ce{TrisHCl}$ and $\ce{CaCl2}$ in granular form ($M = 157\ \mathrm{g/mol}$ and $111\ \mathrm{g/mol}$, respectively) and 100 % glycerol solution. Will I have to make a large batch of Tris and $\ce{CaCl2}$ just so I can actually weigh out the needed amount?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specifications for the working solution are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10\ \mathrm{mM}\ \ce{Tris}$&lt;br&gt;&#xA;$1\ \mathrm{mM}\ \ce{CaCl2}$&lt;br&gt;&#xA;30 % glycerol&lt;br&gt;&#xA;Final $\mathrm{pH} \approx 8$ (use $\ce{HCl}$ if necessary)&lt;br&gt;&#xA;MilliQ $\ce{H2O}$ to volume&lt;/p&gt;&#xA;" OwnerUserId="12775" LastEditorDisplayName="user7951" LastEditDate="2015-10-03T00:30:19.550" LastActivityDate="2015-10-03T00:30:19.550" Title="I have 100 mg of a proteinase K lyophilized powder and I need to make it to a working concentration of 25 mg/mL" Tags="&lt;biochemistry&gt;&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23808" PostTypeId="1" AcceptedAnswerId="23809" CreationDate="2015-01-15T22:57:56.110" Score="-1" ViewCount="30198" Body="&lt;p&gt;I understand that this is the most basic knowledge of moles, however I'm still unsure - according to easy research, $\ce{O2}$ forms a mole of Oxygen. As a mole is $6.022*10^{23}$, exactly what on the periodic table can we use to find out that it's 2 Oxygen in one mole? Wouldn't it be $6.022*10^{23}$, instead of 2?&lt;/p&gt;&#xA;" OwnerUserId="12777" LastEditorUserId="12777" LastEditDate="2015-01-15T23:04:13.597" LastActivityDate="2015-01-15T23:18:04.667" Title="Why is O2 enough to form a mole of Oxygen?" Tags="&lt;molecules&gt;&lt;atoms&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-01-21T06:51:46.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="23809" PostTypeId="2" ParentId="23808" CreationDate="2015-01-15T23:15:42.250" Score="5" Body="&lt;p&gt;$\rm O_2$ is the most common form of oxygen---the periodic table won't really tell you that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro's number tells you how many particles there are in a mole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $6.022\times 10^{23}$ O atoms in a mole of O atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are $6.022\times 10^{23}$ $\rm O_2$ molecules in a mole of $\rm O_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have 2 oxygen atoms in one $\rm O_2$ molecule, there are&#xA;$2\times 6.022\times 10^{23}$ O atoms in a mole of $\rm O_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see the difference?&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-01-15T23:15:42.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23810" PostTypeId="2" ParentId="23808" CreationDate="2015-01-15T23:18:04.667" Score="3" Body="&lt;p&gt;A 'mole' is not short for a 'molecule'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'mole' is a specific quantity (number of objects) defined by Avogadro's constant $N_A=6.022\times10^{23}mol^{-1}$. So a 'mole' of oxygen molecules has $6.022\times10^{23}$ molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The '2' in $O_2$ means there are two oxygen atoms in an oxygen molecule.&lt;/p&gt;&#xA;" OwnerUserId="708" LastActivityDate="2015-01-15T23:18:04.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23811" PostTypeId="2" ParentId="23803" CreationDate="2015-01-16T00:30:50.343" Score="0" Body="&lt;p&gt;More than likely the case of your laptop became charged and was attracting the salt. Yes salt crystals can be coerced by an electrostatic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a post on a teaching site that actually uses salt (and pepper) to demonstrate electrostatic attraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.teachervision.com/electricity/lesson-plan/5787.html&quot; rel=&quot;nofollow&quot;&gt;https://www.teachervision.com/electricity/lesson-plan/5787.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-01-16T00:30:50.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23812" PostTypeId="2" ParentId="23807" CreationDate="2015-01-16T01:08:23.467" Score="0" Body="&lt;p&gt;Simply adding 4 mL of solution to 100 mg of enzyme is not an accurate way of preparing this solution. I am going to describe what to do from an analytical chemist's point of view. If you don't need this level of precision, you could do this following the same steps without volumetric glassware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you need to do is decide what volume of your 25 mg/mL solution you need. Ideally, you will have a volumetric flask corresponding to this volume. For now, let's go with the 4 mL that you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you will need to make a stock solution of your buffer. This will include all of the components except for the enzyme:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Decide what volume of buffer to make. You will need to make at least enough for your working solution, and again you will want to use a volume for which you have a volumetric flask. Also, you want to be weighing out reasonable amounts of your reagents (because weighing out &amp;lt;10 mg of something can really suck). Let's assume you're making 1 L. You're reagents are cheap anyway and the larger scale will make everything easier.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate how much $\ce{TrisHCl}$ and $\ce{CaCl2}$ you will need to obtain the desired concentrations in 1 L of solvent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Weight out the $\ce{TrisHCl}$ and $\ce{CaCl2}$ and add it to your volumetric flask.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add 300 mL glycerol to your flask using a graduated cylinder. 30% glycerol probably refers to v/v.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fill &lt;em&gt;nearly&lt;/em&gt; to volume with water. You need everything to be in solution so you can adjust the pH, but you also need to leave some room to work with to do the actual adjustment. I'd shoot for approximately 80% full.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Adjust the pH by adding concentrated $\ce{HCl}$ dropwise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fill to volume with water and thoroughly mix the solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now that you have your buffer solution, you just need to weigh out the appropriate amount of your enzyme and dissolve it in your buffer. From the example above, you would weigh out 100 mg of the enzyme (don't just assume you have 100 mg, even if the bottle says so) and dissolve it in your buffer using a 4 mL vol. flask.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*&#xA;If you don't have a 4 mL vol. flask and you actually need all 100 mg of your enzyme in this solution, you will need to mess around with dilutions. Start by by creating a 50 mg/mL (2 mL vol. flask) or 100 mg/mL (1 mL vol. flask) solution.&lt;/p&gt;&#xA;" OwnerUserId="12684" LastEditorUserId="12684" LastEditDate="2015-01-16T04:40:18.213" LastActivityDate="2015-01-16T04:40:18.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23813" PostTypeId="1" CreationDate="2015-01-16T01:37:53.187" Score="3" ViewCount="620" Body="&lt;p&gt;I was tasked with figuring out how to make some $\ce{NH4ClO4}$(ammonium perchlorate). I found that one way of making it was to react potassium perchlorate and ammonium chloride in a double displacement reaction. according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KClO4 + NH4Cl -&amp;gt; NH4ClO4 + KCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I got to work on finding ways of making the necessary reactants without having to buy them by doing additional reactions with easier to find chemicals. After some work I came up with these two reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KCl + NH4NO2 -&amp;gt; NH4Cl + KNO2}$ (to make the ammonium cloride)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KCl + HClO4 -&amp;gt; KClO4 + HCl}$ (to make the potassium perchlorate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is (the newbie part) how should I now go about these reactions? Should I be mixing the reactants together in an aqueous solution? In water? Should I be mixing them into an ionic solution in an electrolytic cell? Or hell do I just mix them together as a granulated form and apply a little heat? How do I get these reactions to occur?&lt;/p&gt;&#xA;" OwnerUserId="12646" LastEditorUserId="9961" LastEditDate="2016-01-03T14:27:59.913" LastActivityDate="2016-01-03T14:27:59.913" Title="Obtaining ammonium perchlorate via double displacement reaction" Tags="&lt;inorganic-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23814" PostTypeId="2" ParentId="23813" CreationDate="2015-01-16T02:03:48.480" Score="8" Body="&lt;p&gt;The Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_perchlorate&quot;&gt;ammonium perchlorate&lt;/a&gt; gives a hint at how this can be done. The hint is that ammonium perchlorate is curiously less soluble than sodium perchlorate, and thus with the right concentrations, you would get ammonium perchlorate to precipitate by mixing solutions of sodium perchlorate and and, say, ammonium chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4Cl(aq) + NaClO4(aq) -&amp;gt; NH4ClO4(s) + NaCl(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the best advice that can be given on how to make ammonium perchlorate is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Don't. Especially do not use heat!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ammonium perchlorate is fairly dangerous explosive that can decompose with not much provocation into a whole heck of a lot of gas for just a little bit of solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NH4ClO4(s) -&amp;gt; N2(g) + Cl2(g) + O2(g) + 4H2O(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That this decomposition produces toxic &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine#Health_effects_of_the_free_element_and_hazards&quot;&gt;chlorine&lt;/a&gt; gas should only make you want to prepare ammonium perchlorate less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safely producing, storing, and transporting this compound is going to require a lot of specialized equipment and training (which is expensive). The danger here is not of the type that may lead to funny smells and dizziness in poorly ventilated rooms. The hazards associated with the compound involve loss of life or limb and serious structural damage if you look at it funny. &lt;strong&gt;Don't.&lt;/strong&gt; For more information into the sort of destructive power ammonium perchlorate has, consider that this is the rocket fuel that got the space shuttle into low earth orbit. This stuff has the power to &lt;a href=&quot;http://en.wikipedia.org/wiki/PEPCON_disaster&quot;&gt;level industrial complexes with the forces of an earthquake&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-16T02:03:48.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23815" PostTypeId="1" AcceptedAnswerId="23823" CreationDate="2015-01-16T03:23:58.630" Score="6" ViewCount="172" Body="&lt;p&gt;I have several ground up soil samples that I need to get fractions from. The fractions are based on the &lt;a href=&quot;http://nrel.colostate.edu/projects/century5/reference/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Century Soil Model&lt;/a&gt;. Briefly, the fractions relate to the approximate time frame it takes for an organic carbon pool to be metabolized in the soil. Fast pools ≈1 year, slow pools ≈10+ years, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at several papers to try to get an idea of the chemical methods used for digesting the soil samples to get at the more stable carbon pools. For more easily digested organic carbon pools, hot water seems to work, but for the middling and more recalcitrant pools, other reagents are used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some methods mentioned include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dissolve in $\pu{0.1 M}~\ce{NaOH} + \pu{0.1 M} ~\ce{Na4P2O7}$ &lt;/li&gt;&#xA;&lt;li&gt;dissolve in $\ce{HCl}$ &lt;/li&gt;&#xA;&lt;li&gt;reflux with $\pu{6 M} ~\ce{HCl}$ &lt;/li&gt;&#xA;&lt;li&gt;&quot;stepwise chemical digestion with cold and hot $\ce{H2SO4}$ &#xA;(95 - 98% concentration)&quot; &lt;/li&gt;&#xA;&lt;li&gt;oxidation with $\ce{KMnO4}$ ($33\text{ or }\pu{333 mM}$)  (this one I see mentioned most frequently)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not a chemist. I took the required chemistry classes for my degree and passed, but that's years ago now. When I did do chemistry labs, my results hardly ever came out as they were supposed to, so my skill in this field is minimal. Besides these methods not working, and thus wasting time and money, my real concern is how potentially dangerous these chemicals are. Would someone be so kind as to give me an indication of how risky these chemicals might be? We have a functioning fume hood, will buy PPE, and be sure there are safety procedures in place, but I'm still wondering how much I should worry if a flask breaks or something spills.  &lt;/p&gt;&#xA;" OwnerUserId="12781" LastEditorUserId="4945" LastEditDate="2017-05-24T06:40:03.560" LastActivityDate="2017-05-24T06:43:16.080" Title="How dangerous are these soil organic carbon digestion methods?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23816" PostTypeId="2" ParentId="21815" CreationDate="2015-01-16T04:03:21.033" Score="1" Body="&lt;p&gt;The difference between a weak base and a strong base is that a strong base is completely converted to ions in a given solvent, whereas a weak base is not. Although this depends on the actual basicity of the solute, it is also very dependent on the solvent involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water, $\ce{NH3}$ can react to form $\ce{NH4+}$ and $\ce{OH-}$. However, most of the ammonia remains as $\ce{NH3}$. This is due to the difference in p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;s:&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4+}$ = 9.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O}$ = 15.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the ammonium ion is more acidic than neutral water, very little $\ce{NH4+}$ will be present at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the relevant p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt;s for the $\ce{HCl}$ solution are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4+}$ = 9.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$ = -8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that $\ce{HCl}$ is much more acidic than the ammonium ion, and therefore essentially all of the ammonia will be protonated when dissolved in $\ce{HCl}$. As a result, it is considered a strong base in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow&quot;&gt;Evans' p&lt;em&gt;K&lt;/em&gt;&lt;sub&gt;a&lt;/sub&gt; tables&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12684" LastActivityDate="2015-01-16T04:03:21.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23818" PostTypeId="2" ParentId="5513" CreationDate="2015-01-16T04:39:43.943" Score="-1" Body="&lt;p&gt;This arises due to the electron making transitions between the various energy levels in an atom . The electrons fall from a higher energy state to a lower energy state or vice versa emitting or absorbing wavelengths of different magnitude which fall directly on the spectrum of hydrogen thereby releasing colours of various wavelengths and creating a continuous spectrum .&lt;/p&gt;&#xA;" OwnerUserId="12782" LastActivityDate="2015-01-16T04:39:43.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23819" PostTypeId="1" AcceptedAnswerId="23821" CreationDate="2015-01-16T05:48:30.210" Score="1" ViewCount="1952" Body="&lt;p&gt;When you collect a gas over water, you have to account for the mixed in water vapor by subtracting the room pressure from water vapor pressure. However, why do you not do something similar for volume, where you subtract the total measured volume from the volume of water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can $PV = nRT$ still hold when the pressure you are using is only for the collected gas whereas the volume is for the gas + water vapor volume?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastActivityDate="2015-01-16T08:51:22.080" Title="Collection of gas over water" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23820" PostTypeId="2" ParentId="23815" CreationDate="2015-01-16T08:19:13.113" Score="0" Body="&lt;p&gt;You must be careful when working with hot and concentrated acid solutions as they are corrosive. Otherwise, the chemicals you mentioned are common ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are not used to manipulate chemicals, I advise you to wear goggles and protective gloves while preparing solutions. Work under fume hood when refluxing $\ce{HCl}$. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-16T08:19:13.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23821" PostTypeId="2" ParentId="23819" CreationDate="2015-01-16T08:51:22.080" Score="2" Body="&lt;p&gt;You are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\ce{K}$:  it's  $13.85\,\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is  $$\frac{13.85}{760}\times100=1.83\%$$&#xA;As you can see, it's negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. To calculate the vapor pressure of water, I used the first equation in &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapour_pressure_of_water&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vapour_pressure_of_water&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-16T08:51:22.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23822" PostTypeId="1" CreationDate="2015-01-16T10:45:15.703" Score="5" ViewCount="367" Body="&lt;p&gt;I've been wondering the effect of $\ce{O2}$ on soap foam in a closed, test room. The room contains a candle and soap foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say: if the room has lower and decreasing $\ce{O2}$ in its air, does the foam is effected (i.e., the soap produces lower foams)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I have very basic knowledge in Chemistry, and, excuse me if I didn't formalize the question in a good manner.&lt;/p&gt;&#xA;" OwnerUserId="12784" LastEditorUserId="194" LastEditDate="2017-09-23T02:16:13.013" LastActivityDate="2017-09-23T02:16:13.013" Title="What is the effect of oxygen on soap foam?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23823" PostTypeId="2" ParentId="23815" CreationDate="2015-01-16T10:47:38.583" Score="5" Body="&lt;p&gt;$\pu{0.1 M}~\ce{NaOH}$ is potentially corrosive when in contact with metals or human skin (see &lt;a href=&quot;http://www.chemicals.co.uk/uploads/documents/SODIUM-HYDROXIDE-0.1M-MSDS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), so I advise precautions such as wearing protective standard gloves and keep metals away from feasible damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$ and $\ce{H2SO4}$, especially when they're hot and are in higher concentrations, are very corrosive acids that are dangerous in contact with many materials. They can also release noxious gases in the reactions, but their weakness is glass. Due to reasons irrelevant, these acids can't corrode glass. I recommend extreme precaution. You can work under a &quot;glassy&quot; protection and as Yomen mentions, under a fume hood. Gloves are very important here. &lt;a href=&quot;http://kni.caltech.edu/facilities/msds/HCl.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;'s what Caltech thinks about precautions using $\ce{HCl}$ and &lt;a href=&quot;http://www.cchem.berkeley.edu/rsgrp/SOPs2013/SulfuricAcid_Sarpong.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;'s what Berekley does about sulfuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KMNO4}$ is a very strong oxidizer. It may cause fire in contact with different materials (we had a volcano experiment in elementary school where we mixed it with glycerine and the result was very exothermic), and it has biohazards such as kidney damage. &lt;a href=&quot;http://avogadro.chem.iastate.edu/MSDS/KMnO4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; it's explained more comprehensively.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="4945" LastEditDate="2017-05-24T06:43:16.080" LastActivityDate="2017-05-24T06:43:16.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23824" PostTypeId="2" ParentId="23787" CreationDate="2015-01-16T11:04:12.723" Score="2" Body="&lt;p&gt;Exact statements are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the instability of its $\ce {Pb^4+}$ cation, lead dioxide reacts with &lt;strong&gt;warm&lt;/strong&gt; acids, converting to the more stable $\ce {Pb^2+}$ state and liberating oxygen&#xA;  $$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lead dioxide is produced commercially by several methods, which include ... or reacting $\ce {Pb3O4}$ with &lt;strong&gt;dilute&lt;/strong&gt; nitric acid:&#xA;  $$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since nitric acid in warmer in first reaction, the following reaction occurs easily, which is actually a decomposition reaction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When heated, lead(II) nitrate crystals decompose to lead(II) oxide, oxygen and nitrogen dioxide.&#xA;  $$\ce{Pb(NO3)2-&amp;gt;PbO +NO2 +O2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence the oxygen might be liberated with decomposition-cum-oxidation reaction.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-01-16T11:04:12.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23826" PostTypeId="1" AcceptedAnswerId="23832" CreationDate="2015-01-16T11:25:55.720" Score="0" ViewCount="891" Body="&lt;p&gt;Is it true that under certain conditions, Mg can reduce $\ce{SiO2}$ and the latter the former? What are those conditions?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2015-01-16T11:45:42.317" LastActivityDate="2020-01-05T05:47:47.000" Title="Is it true that under certain conditions, Mg can reduce SiO2?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23827" PostTypeId="1" CreationDate="2015-01-16T11:32:31.763" Score="-4" ViewCount="1030" Body="&lt;p&gt;Everyone knows:&#xA;$$\ce{Mg +Al2O3-&amp;gt;MgO +Al}$$&#xA;This reaction is thermodynamically feasible.&#xA;Use can we this for reduction of Alumina. And produce Aluminium. Then why don't we not use this why?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="7448" LastEditDate="2015-12-17T22:24:00.117" LastActivityDate="2015-12-17T22:24:00.117" Title="Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of auminium. Why?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23828" PostTypeId="5" CreationDate="2015-01-16T11:36:43.420" Score="0" Body="&lt;p&gt;The quantum theory of atoms in molecules (QTAIM) is a concept of &lt;a href=&quot;/questions/tagged/theoretical-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'theoretical-chemistry'&quot; rel=&quot;tag&quot;&gt;theoretical-chemistry&lt;/a&gt; and &lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'computational-chemistry'&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt;. It is based on the observable electron density and was introduced by the late Richard Bader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading and sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wikipedia: &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow&quot;&gt;Atoms in Molecules&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Principle theory: &lt;a href=&quot;https://global.oup.com/academic/product/atoms-in-molecules-9780198558651?cc=jp&amp;amp;lang=en&amp;amp;&quot; rel=&quot;nofollow&quot;&gt;Bader, Richard (1994). Atoms in Molecules: A Quantum Theory. USA: Oxford University Press. ISBN 978-0-19-855865-1.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Original review: &lt;a href=&quot;http://dx.doi.org/10.1021/cr00005a013&quot; rel=&quot;nofollow&quot;&gt;Richard F. W. Bader, &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;1991&lt;/strong&gt;, &lt;em&gt;91&lt;/em&gt; (5), 893–928.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-01-16T11:36:43.420" LastActivityDate="2015-01-16T11:36:43.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23829" PostTypeId="4" CreationDate="2015-01-16T11:36:43.420" Score="0" Body="The quantum theory of atoms in molecules (QTAIM) is a concept of theoretical and computational chemistry. It is based on the observable electron density and was introduced by the late Richard Bader." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-01-16T11:36:43.420" LastActivityDate="2015-01-16T11:36:43.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23830" PostTypeId="2" ParentId="23827" CreationDate="2015-01-16T12:43:40.807" Score="4" Body="&lt;p&gt;If you compare the &lt;a href=&quot;http://www.metalprices.com/p/MagnesiumFreeChart&quot; rel=&quot;nofollow&quot;&gt;magnesium price&lt;/a&gt; (about 2$ per pound)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to the &lt;a href=&quot;http://www.metalprices.com/p/AluminumFreeChart&quot; rel=&quot;nofollow&quot;&gt;aluminum price&lt;/a&gt; (about $0.80 per pound)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see it is unprofitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even considering the elements' relative atomic weight, you would still be losing money if you did that.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-16T12:55:55.713" LastActivityDate="2015-01-16T12:55:55.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23831" PostTypeId="1" CreationDate="2015-01-16T12:59:04.813" Score="7" ViewCount="504" Body="&lt;p&gt;So instead of regular hydrogen, it would be a deuterium (still a Hydrogen). For example, instead of $\ce{HCl}$ it would be $\ce{DCl}$ where D is a deuterium.&lt;/p&gt;&#xA;" OwnerUserId="4843" LastEditorUserId="7448" LastEditDate="2015-01-16T13:08:39.017" LastActivityDate="2015-01-16T14:51:52.013" Title="Would there be a difference if deuterium is embedded instead of protium (regular hydrogen) in acids?" Tags="&lt;inorganic-chemistry&gt;&lt;atoms&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23832" PostTypeId="2" ParentId="23826" CreationDate="2015-01-16T13:08:39.647" Score="3" Body="&lt;p&gt;Yes it is true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conditions are temperature in the 650 to 850 degrees C range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;frm=1&amp;amp;source=web&amp;amp;cd=6&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CEcQtwIwBQ&amp;amp;url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DMJ3oCT_HMoE&amp;amp;ei=7Qq5VPqNJo3ksATKxYLYCg&amp;amp;usg=AFQjCNEY1LEmWs661zxDe8Mmz9uD4EyADA&quot; rel=&quot;nofollow&quot;&gt;cool looking youtube video of the reaction&lt;/a&gt;, not that that makes it true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more serious discussion see &lt;a href=&quot;https://tspace.library.utoronto.ca/bitstream/1807/25788/3/Sadique_Sarder_E_201011_MASc_thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;Production and Purification of Silicon by Magnesiothermic Reduction of Silica Fume&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/nl801759x&quot; rel=&quot;nofollow&quot;&gt;Ordered Mesoporous Silicon through Magnesium Reduction of Polymer Templated Silica Thin Films&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-16T13:08:39.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23833" PostTypeId="2" ParentId="23831" CreationDate="2015-01-16T13:31:04.500" Score="8" Body="&lt;p&gt;Yes, there would be a difference. The acids would be weaker.  Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\rm D_2O$ are weaker than they are in ordinary water, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any reaction that involves dissociation of these acids will run more slowly, too. This is called &quot;the kinetic isotope effect&quot;. You can read more about that here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot;&gt;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to ask further questions: &lt;em&gt;why&lt;/em&gt; is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask away!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, &lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/jres/73A/jresv73An3p299_A1b.pdf&quot;&gt;here's a good reference&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R. A. Robinson, M. Paabo, R. G. Bates, &quot;Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide&quot;, &lt;em&gt;J. Res. Nat. Bureau. Standards&lt;/em&gt;, &lt;strong&gt;73A(3)&lt;/strong&gt;, 1969, pp 299-308.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-16T14:51:52.013" LastActivityDate="2015-01-16T14:51:52.013" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23835" PostTypeId="1" AcceptedAnswerId="23837" CreationDate="2015-01-16T14:41:11.657" Score="4" ViewCount="14010" Body="&lt;p&gt;How do we determine the number of alpha hydrogens in a compound? Where should we start the naming from?&#xA;For example, bicyclo[4.2.0]oct-7-ene and bicyclo[4.2.0]oct-3-ene:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Compare the stabilities of the following alkenes&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PZDxi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZDxi.png&quot; alt=&quot;Structures a and b&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In both these structures, I need to determine their stabilities based on their hyperconjugating structures. I need to determine the number of alpha hydrogens it has, to determine the number of hyperconjugating structures it makes. Where do I start?&lt;/p&gt;&#xA;&lt;p&gt;I am aware that stability is directly proportional to the number of hyperconjugating structures. More the number of alpha hydrogens, greater is the number of hyperconjugating structures, and hence greater the stability.&lt;/p&gt;&#xA;" OwnerUserId="12788" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-05-21T07:35:02.847" Title="How to determine number of alpha hydrogens in a cyclic compound?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23836" PostTypeId="2" ParentId="23822" CreationDate="2015-01-16T15:04:28.973" Score="3" Body="&lt;p&gt;Hmmm. I don't know, but I'd say that if you &lt;em&gt;significantly&lt;/em&gt; lowered the percentage of $\rm O_2$ in the air, the average molecular weight of the molecules in the bubble would drop (there'd be more low-molecular weight nitrogen and less higher-molecular weight oxygen), so the rate at which gas would diffuse out of the bubble would increase, and the foam would flatten faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try blowing a foam with nitrogen, and compare it with the results of blowing a foam with air or oxygen---if I'm right, the foam blown with nitrogen will flatten a bit faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There'll be a bit more $\rm CO_2$ in the air from the candle, and that would diffuse more slowly out of the bubble (in fact, $\rm CO_2$ is used as a &quot;blowing agent&quot; to make foams, partially for that reason). But I'd think in this case that the small amount of $\rm CO_2$ produced by the candle would dissolve in the liquid, with gas dissolution rates enhanced by the lower surface tension and higher surface area of the foam.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-16T15:32:39.730" LastActivityDate="2015-01-16T15:32:39.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23837" PostTypeId="2" ParentId="23835" CreationDate="2015-01-16T15:32:13.527" Score="7" Body="&lt;p&gt;The relative stability of your two compounds is not determined by arguments based on hyperconjugation.  Rather the relative stabilities have to do with &lt;strong&gt;ring strain&lt;/strong&gt;.  Placing a double bond in smaller and smaller rings increases the angle strain in the ring system.  A double bond, being roughly &lt;span class=&quot;math-container&quot;&gt;$\ce{sp^2}$&lt;/span&gt; hybridized, would prefer to have bond angles around 120°.  This becomes difficult in 3- and 4-membered rings where the internuclear angle is 60° and 90° respectively.  Therefore molecules containing double bonds in 3- and 4-membered rings will be more strained than molecules with the double bond in a larger ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, the given solution is not the correct answer to the problem.  It is correct that structure B has more alpha (alpha to the double bond) hydrogens (4) than structure A (only 2 alpha hydrogens).  It is also true that hyperconjugation can be used to explain why alkyl groups attached to a double bond act to stabilize a double bond (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19027/inductive-effect-and-hyperconjugation-one-elephant-different-parts/19032#19032&quot;&gt;this earlier answer&lt;/a&gt; for an explanation), and that the double bond in B is more stabilized by hyperconjugation than the double bond in A.  However, hyperconjugation has a much smaller effect on the stability of these molecules than the ring strain effect.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2019-05-21T00:04:42.880" LastActivityDate="2019-05-21T00:04:42.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23839" PostTypeId="2" ParentId="4381" CreationDate="2015-01-16T16:00:41.290" Score="2" Body="&lt;p&gt;In &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/cm0486565&quot; rel=&quot;nofollow&quot;&gt;Fabrication of Microstructures by Wet Etching of Anodic&#xA;Aluminum Oxide Substrates&lt;/a&gt; Chem. Mater. 2005, 17, 4049-4052,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5% phosphoric acid is used and is selective relative to aluminum, but you need to check how it would do with respect to the other metals you mention&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, the AAO substrate was wet-etched in a 5 wt % H3PO4 solution&#xA;  at 20 °C for 8 h&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-16T16:00:41.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23840" PostTypeId="1" AcceptedAnswerId="23844" CreationDate="2015-01-16T16:11:42.973" Score="1" ViewCount="2986" Body="&lt;p&gt;Well I was studying organic chemistry, and I came across this reaction for increasing the length of the carbon chain. Upon asking my teacher about the following reaction, she said she was unsure about it. Is this a valid reaction? And if it is, can you please provide the name of the reagent and the actual mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/odbw2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7023" LastActivityDate="2015-01-16T22:45:09.540" Title="Organic chemistry - alkanes step up reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23841" PostTypeId="1" AcceptedAnswerId="23846" CreationDate="2015-01-16T16:21:20.580" Score="2" ViewCount="965" Body="&lt;p&gt;I am told that $\ce{HCN}$ is a Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand a Lewis acid possesses (an) empty orbital that is accessible to attacking nucleophiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $\ce{HCN}$ have any empty orbital(s)? So how can it behave as a Lewis acid? I know prototypical Lewis acids are for example boron trifluoride, and these have an empty p-orbital. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="12860" LastEditDate="2015-11-13T19:09:09.980" LastActivityDate="2015-11-13T19:09:09.980" Title="Hydrogen cyanide - a Lewis acid?" Tags="&lt;acid-base&gt;&lt;molecular-orbital-theory&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23842" PostTypeId="1" CreationDate="2015-01-16T16:30:00.157" Score="0" ViewCount="1823" Body="&lt;p&gt;I read that alkanes are 'colourless and odourless'. Heptane is an alkane. Why does heptane smell like heptane? Impurities?&lt;/p&gt;&#xA;" OwnerUserId="12790" LastEditorUserId="7475" LastEditDate="2015-10-09T16:26:13.130" LastActivityDate="2015-10-09T16:26:13.130" Title="Why does heptane have an odour" Tags="&lt;organic-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23843" PostTypeId="2" ParentId="23842" CreationDate="2015-01-16T16:43:18.943" Score="2" Body="&lt;p&gt;Some higher-molecular weight alkanes definitely have an odor, but they aren't as intense as most of the compound types you'll find stinking up an organic lab. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The odor threshold for hexane is 130 parts per million (ppm), with a faint peculiar odor reported.&quot; ---&lt;a href=&quot;http://www.epa.gov/ttnatw01/hlthef/hexane.html&quot; rel=&quot;nofollow&quot;&gt;http://www.epa.gov/ttnatw01/hlthef/hexane.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;[Heptane is a] colorless liquid with a mild gasoline-like odor.&quot; ---&lt;a href=&quot;http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Alkanes have relatively high odor detection threshold (37 to 50 mg/m$^3$ for dodecane) as compared to other VOC groups such as sulphides (0.0003 to 0.16 mg/m$^3$  for&#xA;dimethylsulphide), phenolics (0.022 to 4 mg/m$^3$  for phenol), and nitrogen heterocycles (0.0004 to 0.0008 mg/m$^3$  for skatole) (O'Neill and Phillips, 1992)&quot; ---&lt;a href=&quot;http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to the following paper, there are apparently at least two olfactory receptors  in rats that respond to alkanes; one accepts alkanes with areas around 3.5 nm$^2$, and another  responds to hydrocarbons with areas of about 5 nm$^2$. The authors note that response to cycloalkanes is higher than for normal alkanes, so the shape matters too.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Concanavalin A reveals olfactory receptors which discriminate between alkane odorants on the basis of size.&quot; ---&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/2803264&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/2803264&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-16T23:38:59.763" LastActivityDate="2015-01-16T23:38:59.763" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23844" PostTypeId="2" ParentId="23840" CreationDate="2015-01-16T17:31:26.623" Score="1" Body="&lt;p&gt;I think the reagent you're talking about is ketene (R=H in the figure below).  It is a linear molecule (the central carbon is $\ce{sp}$ hybridized) and very reactive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pOGAz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When mixed with an alkane and heated it dimerizes to form diketene rather than react with the alkane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Golsv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when irradiated with light, ketene will produce the reactive carbene &quot;methlylene&quot; ($\ce{CH2:}$).  When this photolysis is performed in the presence of an alkane, the methylene will insert itself into the various $\ce{C-H}$ bonds like you've drawn.  However the reaction is not synthetically useful for a number of reasons.  For example, the $\ce{C-H}$ insertion reaction is very indiscriminate and a variety of products will be formed if different $\ce{C-H}$ bonds exist in the molecule. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-16T17:31:26.623" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23845" PostTypeId="1" CreationDate="2015-01-16T17:53:31.957" Score="3" ViewCount="197" Body="&lt;p&gt;I have a slab of aluminum and one part of the slab will be exposed to a very corrosive environment. If I put a sacrificial anode elsewhere on the slab will the section that is exposed still be protected from pit corrosion? I guess I'm unsure because the pit corrosion is a result of a localized reaction and if an anode is added will it prevent the local reaction from taking place.&lt;/p&gt;&#xA;" OwnerUserId="12791" LastActivityDate="2015-01-18T03:17:27.607" Title="Sacrificial Anodes and Pit Corrosion" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23846" PostTypeId="2" ParentId="23841" CreationDate="2015-01-16T19:17:13.657" Score="2" Body="&lt;p&gt;Low-lying, empty antibonding orbitals will suffice.  The reaction of carbon dioxide and water to form carbonic acid serves as a parallel example.  Water donates one of the lone pairs on oxygen towards bond formation, water is the Lewis base in this case.  Carbon dioxide has empty, low-lying $\ce{\pi}$* orbitals that serve to accept these electrons; carbon dioxide is a Lewis acid in this example.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-16T19:17:13.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23847" PostTypeId="1" CreationDate="2015-01-16T21:39:55.223" Score="15" ViewCount="1307" Body="&lt;p&gt;Consider astronomy. Hobbyists and amateurs are today still able to make small to moderate scientific contributions to their field. Is the same true in chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there &lt;em&gt;might&lt;/em&gt; be at least two restricting factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The equipment needed to make scientific discoveries is far too expensive (e.g. hundreds of thousands of dollars/euros/whatever) for an amateur to obtain&lt;/li&gt;&#xA;&lt;li&gt;Not having a bachelor's + master's degree (so being ready for a PhD) is too limiting (&quot;too much to study alone&quot;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If there is no &quot;correct&quot; answer for the field of chemistry in general, please post an answer per bigger sub-field (e.g. organic chemistry).&lt;/p&gt;&#xA;" OwnerUserId="12794" LastEditorUserId="12794" LastEditDate="2015-01-16T22:11:18.607" LastActivityDate="2015-01-17T15:28:43.817" Title="Is it still possible for an amateur, a hobbyist, or a science buff to make relevant discoveries in chemistry today?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="0" ClosedDate="2015-01-17T17:15:58.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="23848" PostTypeId="1" AcceptedAnswerId="23853" CreationDate="2015-01-16T22:19:24.917" Score="4" ViewCount="132" Body="&lt;p&gt;A good balance is an important piece of equipment for chemistry. For home chemistry, the price of the balance is unfortunately limiting ..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should one go about picking a balance (for home chemistry; general guidelines are welcome too)? Specifically, I'm thinking whether to buy a model with 0.01g precision (up to 200g), or a one with 0.001g precision (up to 20g) (the latter is a so called &quot;diamond balance&quot; but probably does the job). Which one would serve me better?&lt;/p&gt;&#xA;" OwnerUserId="12795" LastActivityDate="2015-01-17T02:56:09.577" Title="Choosing a balance for home chemistry" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23849" PostTypeId="2" ParentId="23840" CreationDate="2015-01-16T22:45:09.540" Score="0" Body="&lt;p&gt;I would describe the reaction as a C-H insertion reaction of the carbene generated in situ from ketene. The reaction starts with an initial release of carbon monoxide to generate the highly electrophilic 6 electrons carbene species, which is then involved in a C-H insertion process. The C-H insertion can proceed through two mechanism: via singlet state or triplet state of the carbene. The carbene is a very electrophilic sp2 hybridized species with 6 electrons: a lone pair and two pairs in bond orbitals (with H atoms). Singlet carbenes have the lone pair of electrons in a nonbonding sp2 orbital and have an empty p orbital, while triplet carbenes have two unpaired electrons, one in a sp2 and the other in a p orbital. C-H insertion mediated by singlet carbenes is believed to proceed through a concerted mechanism, with the side-on approach of the carbene which provide constructive orbitals overlapping. Triplet carbene insertions should follow a two-step radical pathway, but it is generally accepted that the reaction proceed through singlet state carbenes (supported by retention of stereochemistry observed when if the C-H bond is at a stereogenic center).&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-16T22:45:09.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23850" PostTypeId="2" ParentId="23847" CreationDate="2015-01-16T23:10:41.800" Score="3" Body="&lt;p&gt;Eric Betzig, one of this year's Nobel laureates in chemistry, did much of the work for which he won the prize (superresolved fluorescence microscopy) in his living room .&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-16T23:10:41.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23851" PostTypeId="2" ParentId="23627" CreationDate="2015-01-16T23:27:53.713" Score="1" Body="&lt;p&gt;I have always thought that corrosion textbooks give some of the most concise and enlightening overviews of electrochemistry. One of my favorites is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Corrosion and Surface Chemistry of Metals&lt;/em&gt;, by D. Landolt, EPFL Press, 2007, ISBN:  0849382335&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This book will explain to you how to interpret polarization diagrams.  The emphasis, of course, is on corrosion phenomena, but the same type of reasoning applies to any electrochemical system.  The author does a great job of breaking down all the sources of overpotential at an electrode and shows many polarization diagrams along with the discussion.  It really is a great book and I highly recommend it!&lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-16T23:27:53.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23852" PostTypeId="1" CreationDate="2015-01-17T00:13:35.737" Score="4" ViewCount="75" Body="&lt;p&gt;I was looking at why it's ok for glucose to circulate through the blood while fructose is prevented from doing so by the liver.  Fructose glycates proteins more readily than glucose.  This is the opposite of what I expected as glucose is an aldose while fructose is a ketose.  It seems reasonable to expect that the aldose would be more reactive than the ketose in many reactions.  What is the mechanism of glycation of proteins?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-16T04:37:02.147" Title="What is the mechanism by which sugars glycate proteins?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23853" PostTypeId="2" ParentId="23848" CreationDate="2015-01-17T00:14:33.000" Score="2" Body="&lt;p&gt;Home chemistry can be difficult sometimes due to the expense of the essential working materials and instruments required. As a mathematician all I need is a sharp pen (which is sometimes still quite difficult to obtain!) and blank paper. I can give you the following general (personal) advice regarding purchasing chemistry lab equipment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine your budget and the needs for the experiments you want to perform. Do you really need a scale that has to weight with an accuracy of 0.001g or is 0.01g sufficient (or even 0.1g)? Ask yourself what you REALLY need for an experiment BEFORE searching for instruments. If you do it in the converse order you often buy something way more fancier (and pricier) than what you actually needed in the first place.&lt;/li&gt;&#xA;&lt;li&gt;Look at specialized chemistry equipment supply websites to orientate on what is available and in what price range certain products are (think of scales, chemicals, glassware, etc.)&lt;/li&gt;&#xA;&lt;li&gt;A lot of stuff can be bought incredibly cheap on websites where you not expect (at least I did not) people to sell lab supplies/instruments. Think of &lt;a href=&quot;http://ebay.com&quot; rel=&quot;nofollow&quot;&gt;ebay.com&lt;/a&gt; and &lt;a href=&quot;http://amazon.com&quot; rel=&quot;nofollow&quot;&gt;amazon.com&lt;/a&gt; as examples. They sometimes list items at a fifth of the usual price that would have to be paid at a specialized store.&lt;/li&gt;&#xA;&lt;li&gt;Take a look at local non-specialized stores like the construction market (you know, the place where they sell all kinds of construction material). They sometimes hide hidden (cheap) gems! As an example I was able to obtain a vacuum source (membrane pump) for &amp;lt; 100\$, which would have cost me easily &gt; 400\$ at a chemical supply website.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;About your particular question: I really recommend a 0.01g precision 200g (or 600g) scale. It brings the benefit of having relatively accurate weighting for even small samples, while still being able to weight bulk mass. Furthermore, they are most common in the lab/home environment and hence relatively cheap in comparison with professional analytical scales. If you plan to do a lot of microscale experiments (yielding a product &amp;lt; 200 mg) the analytical scale (precision of 0.001g) may suit your needs better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish you much success, think and buy wisely!&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-17T00:14:33.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23854" PostTypeId="1" CreationDate="2015-01-17T01:48:16.543" Score="-1" ViewCount="95" Body="&lt;p&gt;I am trying to figure out the hazard quotient of copper compared to the standard. The standard for drinking water is 1300 mg/l and the first area has a value of 837 µg. How do I convert this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the formula hq (hazard quotient) = contaminant (c) divided by standard (s). &#xA;or $$HQ= \frac{C}{S}$$ I am having problems with the conversion.&lt;/p&gt;&#xA;" OwnerUserId="12798" LastEditorDisplayName="user7951" LastEditDate="2015-04-17T18:51:54.227" LastActivityDate="2015-04-17T18:53:19.277" Title="The standard for drinking water is 1300 mg/l and the first area has a value of 837 µg" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23855" PostTypeId="2" ParentId="23854" CreationDate="2015-01-17T02:00:38.243" Score="1" Body="&lt;p&gt;I don't know what this standard or the &quot;first area&quot; are, but the hazard quotient is simply the ratio of potential exposure to the maximum exposure without adverse effects. You can't take the ratio of the two values you have because 1300 mg/L is a concentration and 837 µg is an absolute quantity. You need to first decide how much water your potential exposure will involve and multiply that by 1300 mg/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, 1300 mg/L of copper shouldn't be any kind of standard level in drinking water. That's more like a standard solution for AAS or ICP, is definitely bright blue, and probably contains several percent of nitric acid.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user7951" LastEditDate="2015-04-17T18:53:19.277" LastActivityDate="2015-04-17T18:53:19.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23856" PostTypeId="2" ParentId="23852" CreationDate="2015-01-17T02:52:51.910" Score="1" Body="&lt;p&gt;One mechanism of glycation is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maillard_reaction&quot; rel=&quot;nofollow&quot;&gt;Maillard Reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism starts with nucleophilic attack of the sugar carbonyl group by a side chain amino group from the protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Milliard Reaction occurs in the human body and has been implicated in various adverse health conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cornea, lens and retina are known to be damaged by this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://linkinghub.elsevier.com/retrieve/pii/S0369811409001928?via=sd&amp;amp;cc=y&quot; rel=&quot;nofollow&quot;&gt;The Maillard reaction in the human body. The main discoveries and factors that affect glycation&lt;/a&gt; Pathologie Biologie, Volume 58, Issue 3, Pages 214-219 for a recent review.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-17T04:14:09.663" LastActivityDate="2015-01-17T04:14:09.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23857" PostTypeId="2" ParentId="23848" CreationDate="2015-01-17T02:56:09.577" Score="1" Body="&lt;p&gt;Let me first say that @Jori's answer is excellent. I would only add two things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) A range of equipment is usually desirable if it is also affordable. So measures across all 3 accuracies (0.1, 0.01, and 0.001) is what you want ideally, particularly given that precision tends to decrease with absolute mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There is a trend towards green chemistry, which emphasizes minimal reactants, products, and waste whenever possible. Thus, there will be a trend towards microchemistry where small amounts of reactants are the norm, and this will necessitate higher precision balances (i.e 0.001g precision).&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-01-17T02:56:09.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23858" PostTypeId="2" ParentId="23847" CreationDate="2015-01-17T03:12:29.707" Score="1" Body="&lt;p&gt;I'm a mathematician by training, and I think math is a decent analogy to your question. The answer is that yes, amateurs could certainly produce new insights, but this rather low probability. The reason is that humanity has been applying the scientific method to all sciences for a long time, resulting in many new results. But this also means that the &quot;low hanging fruit&quot; in all sciences have been heavily picked. It's simply the case that most of the remaining problems are hard in any scientific discipline. You may still be able to find some non-researched problem, but this is increasingly unlikely. And if it is non-researched, it is likely &lt;strong&gt;very&lt;/strong&gt; hard. So this does not rule out amateurs, but it does set a high bar.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-01-17T03:12:29.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23859" PostTypeId="2" ParentId="23847" CreationDate="2015-01-17T03:22:31.367" Score="3" Body="&lt;p&gt;Back in the early days of chemistry, it was all done by &quot;amateurs&quot;. And, as far as what costly equipment is needed--go back and look at an old pre 1900 textbook like Fownes' -- it is amazing how much chemistry they got right without NMR, IR, MS, HPLC, GC, etc.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-17T03:22:31.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23861" PostTypeId="2" ParentId="23847" CreationDate="2015-01-17T04:17:04.443" Score="4" Body="&lt;p&gt;It's conceivable that somewhere out there, someone with a completely fresh point of view could have an &lt;em&gt;annus mirabilis&lt;/em&gt; in an area like theoretical chemistry, working with just a paper and pencil (or a laptop) and home-brewed equipment and resources.  Yes, for most of us, all of the low-hanging fruit is gone---but for a real giant, the whole tree might be full of low-hanging fruit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of us aren't giants, but as Newton said, we can stand on the shoulders those who are. Doing that requires some ability to search the extant literature, even for &quot;small to moderate contributions&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In addition to the obstacles you've listed above, there is a third impediment: paywalls.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a time when you could walk into a university library and sit down with the Science Citation Index and do a good job of finding research related to yours. Nowadays, you need a faculty or student university ID to do such a search and retrieve the papers you need. Without an ID at a good university, our amateur, hobbyist, or science buff will  hit paywall after paywall. Amateurs might have to pay half a day's wages or more for the pdf of &lt;em&gt;a single paper&lt;/em&gt;! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paywalls are not only a major obstacle for amateurs---they discourage professionals in the third world, and they shut out young people who are beginning their careers outside of university.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Jack Andraka's story, here: &lt;a href=&quot;http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/&quot; rel=&quot;nofollow&quot;&gt;http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-17T15:28:43.817" LastActivityDate="2015-01-17T15:28:43.817" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23862" PostTypeId="2" ParentId="23847" CreationDate="2015-01-17T07:48:54.617" Score="1" Body="&lt;p&gt;What about the Mpemba effect? Does that count as chemistry? Maybe it's more physics or chemical engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erasto Mpemba fits the definition of a total amateur. He was in  high school back then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, 1963 was half a century ago now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least in Chemical Engineering, a fair bit of innovation happens still by amateurs. Ok, not total amateurs, but plant operators, entrepreneurs, businessman, scrap dealer, contractors, foreman,  et cetra. i.e. someone outside the conventional BS-MS-PhD academic system. &lt;/p&gt;&#xA;" OwnerUserId="12803" LastActivityDate="2015-01-17T07:48:54.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23864" PostTypeId="1" CreationDate="2015-01-17T07:54:30.443" Score="2" ViewCount="86" Body="&lt;p&gt;What is the required HLB on the Griffin Scale of 94% water and 6% mineral oil for producing a stable micro-emulsion?&lt;/p&gt;&#xA;" OwnerUserId="12802" LastActivityDate="2019-09-05T20:02:00.243" Title="Required HLB on the Griffin Scale" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-09-08T17:26:30.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="23866" PostTypeId="1" CreationDate="2015-01-17T13:06:24.757" Score="1" ViewCount="675" Body="&lt;p&gt;In Gabriel Phthalimide Synthesis, why do we use $\ce{KOH}$ in first step rather than directly adding an alkyl halide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2e37Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ainpB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12807" LastEditorUserId="7448" LastEditDate="2015-01-17T14:20:11.087" LastActivityDate="2016-05-18T02:10:13.220" Title="Why don't we use an alkyl halide directly in Gabriel Phthalimide Synthesis?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23868" PostTypeId="2" ParentId="23866" CreationDate="2015-01-17T14:28:25.933" Score="3" Body="&lt;p&gt;It is used to make the conjugate base of the imide which is far more nucleophilic than the imide itself, but not too much to attack another molecule of the alkyl halide.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-01-17T14:28:25.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23869" PostTypeId="1" CreationDate="2015-01-17T14:39:28.090" Score="1" ViewCount="523" Body="&lt;p&gt;Is there a way to measure the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorophyll#Spectrophotometry&quot; rel=&quot;nofollow&quot;&gt;chlorophyll&lt;/a&gt; content of plants (from e.g. leaves) with basic chemistry equipment (at home)?&lt;/p&gt;&#xA;" OwnerUserId="12809" LastActivityDate="2015-01-17T15:08:10.563" Title="How can the chlorophyll concentration of plants be measured?" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23870" PostTypeId="2" ParentId="23869" CreationDate="2015-01-17T15:08:10.563" Score="1" Body="&lt;p&gt;To measure the total chlorophyll from leaves, probably not, since you need access to methyl isobutyl ether or pyridine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can separate the different pigments in the leaf. From &lt;a href=&quot;http://www.thenakedscientists.com/HTML/content/kitchenscience/exp/chlorophyll-chromatography/&quot; rel=&quot;nofollow&quot;&gt;The Naked Scientist&lt;/a&gt;: Cut a rectangle 1 cm wide and long enough to stand up in a small jar from a coffee filter. Place your leaf over the top of the strip and roll the coin across the strip about 2 cm from the bottom. Now add a little nail varnish remover to the bottom of your jar, half a cm is plenty. Suspend your strip of paper so the bottom end is sticking a few mm into the acetone. you can hold it up by folding the top end of the strip over forming a hook which will rest over the lip of the jar. Wait a few minutes, with any luck you should see an interesting effect. Try the experiment with a strongly colored leaf like a purple one. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-17T15:08:10.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23871" PostTypeId="2" ParentId="22312" CreationDate="2015-01-17T16:02:15.533" Score="4" Body="&lt;p&gt;&lt;strong&gt;What's the mechanism of cross-linkage for a typical household 2-party epoxy mix?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9xEvJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the polymeric epoxy backbone, two functional groups can be distinguished that are chemically active.&#xA;The hydroxyl-groups may react with phenol-/urea-/melamine-formaldehyde resins at elevated temperatures. At room temperature reaction with for instance isocyanates is possible.&#xA;The epoxy group is very reactive, reaction enthalpy is around 96 kJ/mol typically, with many chemical species. Curing chemistries using phenol(ic) hardeners is for instance employed in powder coating applications whereas anhydride cure is often used in epoxy composite applications.&#xA;At room temperature however by far the largest groups of chemicals used are amine-functional curing agents capable of cross linking epoxy resins at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mN6hp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An epoxy group reacts with ONE amine-hydrogen, i.e. one hydrogen attached directly to a Nitrogen atom (hence also amine-hydrogen equivalent weight). Note that every time that an epoxy group is opened; one Hydroxyl group is generated (adhesion !, internal catalyst !, …). Initial chemical reaction builds linear molecular weight, the molecules get longer and longer giving (s)low viscosity increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step in the curing process is the linking of the chains with each other; cross linking. In order for this to occur it is necessary that each amine molecule will have more-than-two active amine-hydrogen’s. The inter-linking of larger molecules causes a rapid increase of the observed viscosity.&#xA;Sterical hindrance can result in non-fully chemical cure which can be remedied by means of for instance post-cure or the use of plasticisers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1sSAf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-reacted epoxy groups, either deliberately generated through off-stoichiometry or caused by non-full cure can undergo “hydroxyl-cure” catalysed by tert.-amine and heat (sun-light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mGi3z.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are polymercaptan amines? Do these contain both thiol and amine functional groups?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercaptans have been used commercially for over 30 years to cure epoxies. Despite the numerous epoxy hardeners on the market, the mercaptans provide some unique properties. The main one is their ability to cure epoxies rapidly at ambient temperature. The other key property is that the mercaptans are less sensitive in their curing speed to mass of material than other hardeners. Thus, they can be used in thin-film cures. Additional factors in favor of the polymercaptans are their low toxicity compared to other hardeners, their light color, their excellent adhesion, and the ability to accelerate other amines and polyamides. &#xA;    Polymercaptans are used when the speed of cure is important. Even the “fast”-cure amine systems are slow in comparison, especially when the mass of epoxy is small and a thin film is used. This has led to the use of the mercaptans in a wide variety of applications:&#xA;Consumer and industrial adhesives, Coatings, Sealants, Industrial patch and filler kits,Floor and flooring repair, Encapsulation, and Acceleration of other epoxy systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A number of commercial mercaptan hardeners are available. They are sold with and without amine accelerators. The polymercaptan without the accelerator (Capcure 3-800) allows the formulator to have more control over the cure speed and flexibility of the product. This material is also used to accelerate amine, amidoamine and polyamide curing of epoxies. The standard five-minute cure is obtained with the accelerated Capcure® 3830-81. The fastest polymercaptan (Capcure 40 sec HV) has a gel time of 40 seconds with a 25-gram mass. There is a more water-resistant version (Capcure WR-6) for exterior uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum up polymercaptan amines contains at least one pair of $\ce{SH}$ (thiol) groups  and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an amine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does cross-linkage work?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemistry of the mercaptan reactions is fairly straightforward. The tertiary amine catalyst forms a salt with the mercaptan to generate a mercaptide anion, which is a strong nucleophile. The mercaptide will readily open the epoxy. Reaction with another mercaptan group can regenerate the mercaptide anion, as illustrated in the Figure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEocn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low activation energy for the process allows the reaction to go at or below room temperature. Since heat is not necessary for the reaction to occur, it can cure in thin films even on metal substrates. The cure rate does slow as the mass gets smaller, and the film is thinner, but the decrease is not as dramatic as when using amine-based hardeners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wouldn't we then need a 2:1 ratio of epoxide to hardener?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of the mercaptan chemistry is the formulation flexibility. At a 1:1 weight ratio, there is an excess of epoxy compared to mercaptan groups. However, the tertiary-amine accelerator will cause the epoxy to self-cure. Thus, some variation in ratio is self-compensating: &#xA;1. More epoxy will self-cure by the tertiary-amine accelerator; &#xA;2. More mercaptan will still have epoxy groups with which to react;&#xA;3. More or less tertiary amine changes the rate of cure, but not significantly the final degree of cure.&#xA;Of course, deviating greatly from the 1:1 ratio will cause problems of incomplete cure.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-17T16:02:15.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23872" PostTypeId="2" ParentId="7079" CreationDate="2015-01-17T16:44:57.127" Score="2" Body="&lt;p&gt;If you want to determine the total content of oxalate you would need to extract it from the matrix with hot HCl 1M, for instance, since it will dissolve calcium oxalate crystals. Afterwords you can precipitate oxalate again at pH=4-5 with calcium nitrate, thus forming white crystals of calcium oxalate. You can redissolve them by adding sulfuric acid and identify oxalate by taking advantage of its reducing properties - if you add potassium permanganate it will be reduced from Mn +7 (purple colored) to Mn +2 (brown colour) and oxalate will be oxidized to carbon dioxide. But of course you start with a complex matrix, the easiest thing would be to use chromatography to separate and identify oxalate.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-17T16:44:57.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23873" PostTypeId="2" ParentId="22382" CreationDate="2015-01-17T16:52:20.240" Score="11" Body="&lt;p&gt;Cyanoacrylates include methyl 2-cyanoacrylatecommonly sold under the trade names &quot;Super Glue&quot;.In general, cyanoacrylate (consists of monomers of cyanoacrylate molecules) is an acrylic resin that rapidly polymerises in the presence of water (specifically hydroxide ions), forming long, strong chains, joining the bonded surfaces together. Because the presence of moisture causes the glue to set, exposure to normal levels of humidity in the air causes a thin skin to start to form within seconds, which very greatly slows the reaction. Because of this cyanoacrylate is applied thinly, to ensure that the reaction proceeds rapidly and a strong bond is formed within a reasonable time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eX2Zk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to sum up, in order to start the reaction off some water is normally needed.. damp things stick better/quicker than dry ones and the glue goes hard faster on a humid day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cotton wool, which is made of cellulose, a polymer of sugar molecules, there are lots and lots of hydroxy (-OH or alcohol groups), which can start the reaction in the same way as the water does, only because there are lots of them they can start many more reactions at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction gives out heat, the cotton bud therefore gets hot (and as it becomes hotter so the reaction goes faster etc), and it may get hot enough to catch fire.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-17T16:52:20.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23874" PostTypeId="1" CreationDate="2015-01-17T17:07:14.677" Score="10" ViewCount="27367" Body="&lt;p&gt;What would be the best reaction to increase the carbon chain length in an alkane? Preferably, it would be by one at a time; however, if that is not possible, just generally increasing the total length of the carbon chain (such as producing pentane from propane) would be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, is there a way that I can add a methyl group to a compound in the form $\ce{R-CH3}$?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastEditorUserId="5026" LastEditDate="2018-04-17T11:03:22.367" LastActivityDate="2018-04-17T11:04:15.407" Title="Increasing the length of a carbon chain in an alkane" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;hydrocarbons&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23875" PostTypeId="2" ParentId="5095" CreationDate="2015-01-17T17:17:00.807" Score="1" Body="&lt;p&gt;I have just had a look at the paper and actually they reported that their main product was S,R: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;...treated with excess 2-lithio-1,3-dithiane in THF at -50°C; this treatment generated the tertiary alcohols 22a and 22b in 72% combined yield with 6.2:1 diastereoselectivity as predicted by Cram's model based on participation of the piperidine nitrogen atom, giving a rigid cyclic complex with the lithium compound&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound 22a: (2S)-2-[(R)-l-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound 22b: (2S)-2-[(S)-1-(1,3-Dithian-2-yl)-l-hydroxyethyl]piperidine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not find any mistake.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-17T17:17:00.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23876" PostTypeId="2" ParentId="23874" CreationDate="2015-01-17T17:39:13.827" Score="8" Body="&lt;p&gt;One method could be metathesis of alkanes. There is a nice paper about it published by &lt;a href=&quot;https://dx.doi.org/10.1126/science.276.5309.99&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. M. Basset et al., SCIENCE 1997, 276, 99-102, &quot;Metathesis of Alkanes Catalyzed by Silica-Supported Transition Metal Hydrides&quot;&lt;/a&gt; and for sure more recent ones have been published since then. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Abstract&lt;/strong&gt;&#xA;  The silica-supported transition metal hydrides $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$ and $\ce{(≡Si-O-)xM-H (M, chromium or tungsten)}$ catalyze the metathesis reaction of linear or branched alkanes into the next higher and lower alkanes at moderate temperature ($\pu{25^\circ}$ to $\pu{200^\circ C}$). With $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$, ethane was transformed at room temperature into an equimolar mixture of propane and methane. Higher and lower homologs were obtained from propane, butane, and pentane as well as from branched alkanes such as isobutane and isopentane. The mechanism of the step leading to carbon-carbon bond cleavage and formation likely involves a four-centered transition state between a tantalum-alkyl intermediate and a carbon-carbon bond of a second molecule of alkane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="5026" LastEditDate="2018-04-17T11:04:15.407" LastActivityDate="2018-04-17T11:04:15.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23877" PostTypeId="1" AcceptedAnswerId="66643" CreationDate="2015-01-17T17:43:37.767" Score="10" ViewCount="206" Body="&lt;p&gt;Many groups of bound atoms complexively bearing a (negative) charge are called &lt;em&gt;complex anions&lt;/em&gt;. Examples are $\ce{NO3-}$, $\ce{CO3^2-}$, $\ce{BO3^2-}$, $\ce{CrO4^2-}$, $\ce{Cr2O7^2-}$, $\ce{MoO4^2-}$, $\ce{WO4^2-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{VO4^3-}$ and the several, sometimes polymeric, complex anions occurring in silicate and borate minerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologise if I am asking something trivial, but I cannot find a source, except for silicates, for which I read that the $\ce{Si-O}$ bonds are basically covalent, explaining whether the bonds within complex anions are covalent in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they always covalent? If they are not covalent in general, are they in the examples I have written?&lt;/p&gt;&#xA;" OwnerUserId="6943" LastEditorUserId="41328" LastEditDate="2017-09-22T13:55:16.240" LastActivityDate="2017-09-22T14:54:31.930" Title="Bonds within complex anions" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;coordination-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23879" PostTypeId="2" ParentId="23877" CreationDate="2015-01-17T18:26:24.597" Score="3" Body="&lt;p&gt;&quot;Electronegativity&quot; is a measure of the power of an atom in a molecule to attract shared electrons to itself. If you look up the electronegativities for the atoms on either end of a bond, there's a quick and dirty way to decide for yourself how much covalent character the bonds have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the electronegativity difference between the central atom and oxygen. The smaller the difference, the more covalent character the bond has. For example, for Si-O, Si and O have Pauling electronegativities of 1.8 and 3.5, respectively, so the electronegativity difference is 1.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some textbooks set (rather arbitrary) thresholds for electronegativity differences to help general chemistry students classify bonds as non-polar covalent, polar covalent, or ionic. Pauling's original text says that if the electronegativity difference is 1.7, the bond has about 50% ionic character, so you can say an Si-O bond is about 50% ionic, 50% covalent. If the electronegativity difference is greater than 1.7, you can call it &quot;predominantly ionic&quot;; if it's less than 1.7, you can call it &quot;predominantly covalent&quot;. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="9961" LastEditDate="2017-01-19T02:25:02.670" LastActivityDate="2017-01-19T02:25:02.670" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23880" PostTypeId="2" ParentId="23874" CreationDate="2015-01-17T18:45:19.833" Score="4" Body="&lt;p&gt;Carbenes, like $\ce{CH2}$, are known to insert into everything, including, in case of absence of better substrate, carbon-hydrogen bonds. However, the process is not clean, as carbene will insert into any suitable position it can find, with no selectivity. Carbenes are usually produced by decomposition of precursors, usually ketene ($\ce{CH2=C=O}$) or diazomethane ($\ce{CH2=N=N}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case there is some group in the compounds, it is often possible to selectively substitute hydrogen, activated by the group. For example, many electronegative groups, like carbonyl group, activate nearby hydrogenes, making them quite acidic by organic chemistry standards. Produced anion can react with equivalent of carbocation. Carbonyl group may be later eliminated by Wolff–Kishner reduction (heating with hydrazine and alkaly metal hydroxide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no general way to increase chain of alkanes cleanly and selectively. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-01-18T14:27:25.207" LastActivityDate="2015-01-18T14:27:25.207" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23882" PostTypeId="1" AcceptedAnswerId="23886" CreationDate="2015-01-17T20:36:18.443" Score="3" ViewCount="1080" Body="&lt;p&gt;If I have a solution of dissolved $\ce{NaCl}$ or sugar (or some other ionic compound), is there a chemical I could add to it that would prevent crystal growth if I boil it down? I'm looking for it to remain as a very fine powder or to keep the crystals as small as possible.&lt;/p&gt;&#xA;" OwnerUserId="12815" LastEditorUserId="4231" LastEditDate="2015-01-23T23:02:38.217" LastActivityDate="2015-01-23T23:02:38.217" Title="Prevent crystal growth by adding a chemical to a solution?" Tags="&lt;recrystallization&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23883" PostTypeId="2" ParentId="23882" CreationDate="2015-01-17T22:27:08.443" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you decrease the solvent volume, the dissolved material will start to precipitate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solubility most often increases with the temperature of the solvent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Slow decrease of the temperature or slow evaporation of the solvent at constant temperature mostly leads to slow crystallization, starting from a small(er) number of nucleation centres and typically results in the formation of less but larger, well-formed crystals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Rapid decrease of the temperature with additional scratching the surface of the crystallisation vessel (beaker, etc.) with a glas rod, stirring, or sonification will typically result in precipitation of the solute in the form of smaller particles.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-17T22:27:08.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23885" PostTypeId="2" ParentId="23874" CreationDate="2015-01-17T23:34:28.780" Score="8" Body="&lt;p&gt;Homologation methods exist, but typically not for unbranched alkanes.&#xA;A typical example is the Arndt-Eistert homologation of carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple &lt;strong&gt;unbranched&lt;/strong&gt; alkanes, stepwise chain elongation is not a technical relevant process. On the contrary, petrochemical processes reach for the opposite (formation of smaller alkanes) by thermal or catalytical cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formation of long chain alkanes, although not via subsequent step up reactions, is possible by catalytic oxidative dimerization of terminal alkynes (Glaser, Hay or Eglinton coupling) and subsequent hydrogenation or by Kolbe dimerization of carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Branched&lt;/strong&gt; alkanes, such as triptane (2,2,3-trimethylbutane), which are interesting as 'anti-knocking' fuel additives, are another league. Here, the source of the methyl group to be transfered is methanol, catalysts are either zeolites or indium(III)-iodide ($\ce{InI3}$) (&lt;a href=&quot;http://dx.doi.org/10.1021/ja803029s&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-17T23:34:28.780" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23886" PostTypeId="2" ParentId="23882" CreationDate="2015-01-18T00:21:24.720" Score="6" Body="&lt;p&gt;Spray it as a fine mist into a drying chamber - the solvent will evaporate and you are left with a powder. &#xA;No chemical involved though,  but still process chemistry. &lt;/p&gt;&#xA;" OwnerUserId="10050" LastActivityDate="2015-01-18T00:21:24.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23888" PostTypeId="2" ParentId="23845" CreationDate="2015-01-18T03:17:27.607" Score="2" Body="&lt;p&gt;Generally speaking, there is no protection unless the sacrificial anode is immersed in the electrolyte. It might be better to anodize and seal the aluminum, which should protect it against most acids. There are many aluminum anti-corrosion treatments, but in any case the area under attack would have to be covered. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Anodizing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Anodizing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-18T03:17:27.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23889" PostTypeId="1" AcceptedAnswerId="23890" CreationDate="2015-01-18T04:15:54.273" Score="3" ViewCount="398" Body="&lt;p&gt;I know only as much about chemistry as I can remember from a year in college.  So, please speak slowly and use short words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to decide on an appropriate material for a heat-transfer coil that is meant to remain submerged for extended periods in salt water, within a few feet of other metal parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a solar still that will use a cooling/condensing coil to take advantage of the heat differential between the solar-heated evaporation chamber and nearby (cooler) water temperature.  It will be mounted on a sailboat.  The coiling coil will probably be about 12-18 inches long and 2-3 inches diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impulse was to use copper, but I'm not sure how this will react in a salt-water environment over time.  I know that sea life will stay off of it, since copper is the active element in many bottom paints designed to minimize subsurface growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated.  Maybe the answer is to use a sacrificial anode on the coil?  Or, maybe some other material would be more suitable for long-term use?&lt;/p&gt;&#xA;" OwnerUserId="12818" LastActivityDate="2015-06-02T10:46:32.013" Title="Good choice for heat-conductive coil submerged in sea water" Tags="&lt;thermodynamics&gt;&lt;electrolysis&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23890" PostTypeId="2" ParentId="23889" CreationDate="2015-01-18T04:33:51.437" Score="1" Body="&lt;p&gt;My advice is to use copper and protect it with a sacrificial anode.  Copper would be the choice in a non corrosive environment and if you can protect it go for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the coils will contact parts of the boat that are important than the coils should be aluminum.  Copper would cause the other metal parts to corrode faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not use rubber or plastic, something like the pipes used in drip irrigation?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-01-18T04:33:51.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23891" PostTypeId="1" CreationDate="2015-01-18T06:28:06.417" Score="0" ViewCount="499" Body="&lt;p&gt;What are possible products from these two anion analyses?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The nitrite test, reacting diphenylamine with a nitrite:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{(C6H5)2NH + NO2- -&amp;gt; ?}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The sulphite test, reacting potassium permanganate with a sulphite:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{KMnO4 + SO3^{2-} -&amp;gt; ?}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="194" LastEditDate="2017-04-08T00:47:29.703" LastActivityDate="2017-04-08T00:47:29.703" Title="Possible products from anion analyses" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2015-01-18T15:12:42.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="23892" PostTypeId="2" ParentId="23891" CreationDate="2015-01-18T07:22:34.303" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Under acidic conditions, a blue colour is observed. This results from the formation of N,N'-diphenylbenzidine. In the course of the reaction, diphenylamine is oxidized to a quinoid radical which subsequently adds to another molecule of diphenylamine. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Under slightly acidic conditions, bleaching of the solution is observed. Sulfite is oxidized to sulfate while $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-18T07:22:34.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23893" PostTypeId="1" AcceptedAnswerId="23902" CreationDate="2015-01-18T10:09:49.267" Score="2" ViewCount="4070" Body="&lt;p&gt;Explain why three ${sp^2}$ hybrid orbitals result when one s atomic orbital is blended with two p atomic orbitals.&lt;/p&gt;&#xA;" OwnerUserId="12679" LastEditorUserId="4231" LastEditDate="2015-01-18T16:41:15.073" LastActivityDate="2015-03-20T16:44:04.433" Title="Hybridization - Mixing one atomic s orbital with 2p orbitals" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23894" PostTypeId="1" CreationDate="2015-01-18T10:37:39.580" Score="3" ViewCount="921" Body="&lt;p&gt;I often find formulas of minerals where some element symbols are grouped within square brackets and separated by a vertical bar. An example can be brochantite, whose formula I find expressed as $\ce{Cu_4[(OH)_6|SO_4]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what does such a notation means, and I have made some conjectures, although I find no reference on line or in books. I suppose that the square brackets are used to enclose the set of the anions present in the compound and the vertical bar is used to separates different anions, while the cations remain outside the square brackets. That is just an intuitive supposition. Is my supposition correct? If it is not, what does the described notation mean? &lt;/p&gt;&#xA;" OwnerUserId="6943" LastEditorUserId="4945" LastEditDate="2015-01-19T02:57:44.690" LastActivityDate="2015-01-19T17:33:05.073" Title="X[Y|Z] formulas" Tags="&lt;ionic-compounds&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="23895" PostTypeId="1" CreationDate="2015-01-18T11:48:53.253" Score="2" ViewCount="196" Body="&lt;p&gt;How can I increase the signal intensity and precision when use HPLC or LC-MS/MS ? How Also prevent Contamination and Carryover during Analysis&lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2015-01-18T11:48:53.253" Title="How can I increase the signal intensity and precision?" Tags="&lt;chromatography&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23896" PostTypeId="1" AcceptedAnswerId="23897" CreationDate="2015-01-18T12:07:00.163" Score="3" ViewCount="657" Body="&lt;p&gt;I need to know whether is it possible to calculate the atomic radius according to the number of electrons and electron configuration. Or is there any way to calculate the atomic radius using common characteristics which an atom has?&lt;/p&gt;&#xA;" OwnerUserId="11535" LastEditorUserId="5456" LastEditDate="2015-01-18T13:46:33.797" LastActivityDate="2015-01-18T13:46:33.797" Title="Is it possible to calculate atomic radius with electron configuration?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23897" PostTypeId="2" ParentId="23896" CreationDate="2015-01-18T13:19:48.163" Score="3" Body="&lt;p&gt;There's no measuring the radius of a single atom, mainly because electrons are around the nuclei: &lt;strong&gt;We can't define a radius for any single atom.&lt;/strong&gt; Take a loom at the uncertainty principle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation of position $\sigma _x$ and the standard deviation of momentum $\sigma _p$ was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928: $$\sigma _x \sigma _p \geq \frac{\hbar}{2}$$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, if you want to measure an atom's radius, you have to measure the distance between two nuclei, and then divide it by two. This results in different types of radius defined for atoms. The most common are &lt;a href=&quot;http://en.wikipedia.org/wiki/Covalent_radius&quot; rel=&quot;nofollow&quot;&gt;covalent radius&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow&quot;&gt;ionic radius&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;nofollow&quot;&gt;Van der Waals radius&lt;/a&gt;. Radii trends are what that are mostly studied in undergrad chemistry, rather than the how of measuring them with advanced techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemguide is pretty informative when it gets to teaching the trends:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/atradius.html&quot; rel=&quot;nofollow&quot;&gt;chemguide: atomic radius&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-01-18T13:19:48.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23898" PostTypeId="2" ParentId="21709" CreationDate="2015-01-18T14:25:47.417" Score="15" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;We can share the women, we can share the wine&lt;/p&gt;&#xA;&lt;p&gt;We can share what we got of yours, cause we done shared all of mine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Jack Straw&amp;quot; Grateful Dead&lt;/p&gt;&#xA;&lt;p&gt;But seriously now, the most important thing in the &lt;a href=&quot;http://www.princeton.edu/%7Eachaney/tmve/wiki100k/docs/Coordinate_covalent_bond.html&quot; rel=&quot;noreferrer&quot;&gt;Princeton.edu article&lt;/a&gt; (which explains at the bottom that its content comes from Wikipedia) is&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;&lt;strong&gt;The distinction from ordinary covalent bonding is artificial&lt;/strong&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Does &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; come from a hydride and a proton or from two hydrogen atoms?  The reason a molecule is stable is independent of a particular path by which it is made.&lt;/p&gt;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt;, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt; is isoelectronic with &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt;; the reason for bonding is the same for each.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-18T15:37:07.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23899" PostTypeId="1" AcceptedAnswerId="23904" CreationDate="2015-01-18T14:33:55.390" Score="91" ViewCount="57123" Body="&lt;p&gt;My teacher didn't answer this properly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is toothpaste solid or liquid? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can't say toothpaste is a solid because solid material have a fixed shape but toothpaste doesn't. However, you can't say it's a liquid because liquids flow easily but toothpaste needs a certain force to push it out of the tube. So is it a solid or liquid? And are there any other example just like toothpaste?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="33991" LastEditDate="2016-11-21T12:47:51.950" LastActivityDate="2020-05-31T03:24:24.227" Title="Is toothpaste solid or liquid?" Tags="&lt;everyday-chemistry&gt;&lt;phase&gt;&lt;materials&gt;" AnswerCount="7" CommentCount="8" FavoriteCount="28" ContentLicense="CC BY-SA 3.0" />
  <row Id="23900" PostTypeId="1" AcceptedAnswerId="23910" CreationDate="2015-01-18T14:34:08.130" Score="5" ViewCount="924" Body="&lt;p&gt;Upon inspection of the standard electrode potentials for the half reactions for both $\ce{Cu^2+}$/$\ce{Cu+}$ and $\ce{I2}$/$\ce{2I-}$, it becomes apparent it should be the iodide ions that act as the oxidising agent, as the standard electrode potential for the iodine half-cell is much higher. &#xA;$$\begin{alignat}{2}&#xA;\ce{Cu^2+ + e- \;&amp;amp;&amp;lt;=&amp;gt; Cu+}\quad &amp;amp;&amp;amp;E_\text{cell} = +0.159\ \mathrm{V}\\  &#xA;\ce{I2 + 2e- \;&amp;amp;&amp;lt;=&amp;gt; 2I-}\quad &amp;amp;&amp;amp;E_\text{cell} = +0.54\ \mathrm{V}\\&#xA;\end{alignat}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain this, many websites that I’ve looked at suggest it’s the weak solubility of copper(I) iodide, increasing the $\ce{Cu^2+/Cu+}$ potential to around $+0.88\ \mathrm{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such websites include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.titrations.info/iodometric-titration-copper&quot; rel=&quot;nofollow&quot;&gt;http://www.titrations.info/iodometric-titration-copper&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemistry.tutorvista.com/analytical-chemistry/iodometric-titration.html&quot; rel=&quot;nofollow&quot;&gt;http://chemistry.tutorvista.com/analytical-chemistry/iodometric-titration.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I’d like to know, is how this value of $+0.88\ \mathrm{V}$ has been determined, and a little of the theory behind as to why it has increased from $+0.159\ \mathrm{V}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user12827" LastEditorDisplayName="user7951" LastEditDate="2015-05-29T19:43:24.310" LastActivityDate="2015-06-15T05:23:38.490" Title="Iodometric titration of copper – standard electrode potentials" Tags="&lt;electrochemistry&gt;&lt;solubility&gt;&lt;titration&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23901" PostTypeId="2" ParentId="23899" CreationDate="2015-01-18T15:00:11.103" Score="13" Body="&lt;p&gt;It is &lt;strong&gt;Viscous.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viscous mean &quot;having a thick, sticky consistency between solid and liquid&quot;(dictionary meaning)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more examples like tomato ketchup, honey, wax, toothpaste, etc. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To know more check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity&quot;&gt;Wikipedia&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-01-18T16:01:45.923" LastActivityDate="2015-01-18T16:01:45.923" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23902" PostTypeId="2" ParentId="23893" CreationDate="2015-01-18T15:29:21.990" Score="3" Body="&lt;p&gt;When some number of atomic orbitals are mixed together to produce hybrid atomic  orbitals, the same number of hybrid atomic orbitals must result.  So, when N atomic orbitals are mixed together, N hybrid atomic orbitals are produced.  In the case you describe, mixing one s atomic orbital with two p atomic orbitals means that three hybrid atomic orbitals must result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three hybrid orbitals will each contain 33.333% s character and 66.666% p character because that is the ratio of the one s orbital to two p orbitals that we are mixing together.  We can describe such an orbital as $\ce{s^{33.333}p^{66.666}}$, but if we reduce that notation to the lowest common denominator we get $\ce{s^1p^2}$, or the commonly used description $\ce{sp^2}$.  This simply means that this hybrid orbital is composed of one part s character to two parts p character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To summarize if we mix 3 atomic orbitals, one s and two p orbitals, then 3 $\ce{sp^2}$ atomic hybrid orbitals will result.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-18T16:35:56.910" LastActivityDate="2015-01-18T16:35:56.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23903" PostTypeId="2" ParentId="23899" CreationDate="2015-01-18T15:39:42.057" Score="43" Body="&lt;p&gt;Here's a genchem-level answer for a genchem-level question about the classification of matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toothpaste is a &lt;em&gt;sol&lt;/em&gt;: a stable suspension of tiny solid particles in a liquid. When the toothpaste dries out you can see what the solid part alone looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixtures with more than one phase often have interesting properties and behaviors that the components alone don't have; the other answers to your question touch on some of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples of sols are paints and solid-pigment inks. Again, you can see the solid part when they dry.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-18T18:59:15.173" LastActivityDate="2015-01-18T18:59:15.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23904" PostTypeId="2" ParentId="23899" CreationDate="2015-01-18T16:02:27.767" Score="103" Body="&lt;p&gt;Toothpaste is what is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Non-Newtonian_fluid&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;non-newtonian fluid&lt;/strong&gt;&lt;/a&gt;, more specifically toothpaste is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bingham_plastic&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Bingham plastic&lt;/strong&gt;&lt;/a&gt;.  This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous &lt;em&gt;alone&lt;/em&gt; is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQL04.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty.&lt;/p&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="41328" LastEditDate="2020-05-31T03:24:24.227" LastActivityDate="2020-05-31T03:24:24.227" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23905" PostTypeId="1" CreationDate="2015-01-18T16:17:16.873" Score="3" ViewCount="126" Body="&lt;p&gt;Is there any famous experiment in chemistry where digital signal processing played an important part? I don't mean using a machine that relies on such techniques (they all do) but an experiment where signals were first recorded (e.g. temperature/pressure/voltage/etc. as a function of time) and then analyzed with some sophisticated methods (e.g. time-frequency analysis with wavelets).&lt;/p&gt;&#xA;" OwnerUserId="12828" LastEditorUserId="7448" LastEditDate="2015-01-18T16:37:32.187" LastActivityDate="2015-02-23T08:02:38.777" Title="Any popular experiment in chemistry that digital signal processing played a crucial role in?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23906" PostTypeId="2" ParentId="23894" CreationDate="2015-01-18T16:41:47.227" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://old.iupac.org/reports/provisional/abstract04/RB-prs310804/Chap2-3.04.pdf&quot; rel=&quot;nofollow&quot;&gt;The IUPAC rules for punctuation marks&lt;/a&gt; in chemical formulas don't mention [|], so that notation must be peculiar to geology. They &lt;em&gt;need&lt;/em&gt; different notation; chemists write  formulas for molecules, while geologists focus more on complicated superstructures built from smaller structures that can be arranged in different ways, even within the same superstructure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.academia.edu/7540967/Brochantite_Cu4SO4_OH_6_OD_character_polytypism_and_crystal_structures&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; describes the structure of brochantite as layers and chains of copper(II) ions octahedrally coordinated with hydroxides. The octahedra in different layers and chains  are crosslinked by sulfate ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the notation $\rm [(OH)_6|SO_4]$ is an attempt to indicate this arrangement. The brackets denote the electronegative part of the structure, with what comes before the vertical bar representing directly coordinated ions, with what comes  after representing ions that aren't directly coordinated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm speculating, though. Sorry. I would have put this into a comment, but it's too long and I wanted to give you the links to think about.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-19T17:33:05.073" LastActivityDate="2015-01-19T17:33:05.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23907" PostTypeId="1" CreationDate="2015-01-18T17:35:13.733" Score="3" ViewCount="9549" Body="&lt;p&gt;In potentiometric titration between ascorbic acid and iodine, when we plot the graph of EMF with volume of iodine consumed, we get a sudden decrease in EMF at equivalance point--why? It is also negative--why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Nernst equation at equivalence point we get&#xA;$$E_{cell}=E_{STP}~~~(voltage ~at ~STP)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is its physical meaning?&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="5577" LastEditDate="2015-05-19T21:00:08.757" LastActivityDate="2015-05-19T21:00:08.757" Title="Principle of potentiometric titration" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;titration&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23909" PostTypeId="2" ParentId="23899" CreationDate="2015-01-18T22:26:46.560" Score="12" Body="&lt;p&gt;Actually, toothpaste is both. I'm no chemist, but I am pretty sure that it can be correctly classified as a &lt;strong&gt;semisolid&lt;/strong&gt;, which means exactly what you'd think. Semisolids have properties of both solids and liquids. Slime would be another example of a semisolid.&lt;/p&gt;&#xA;" OwnerUserId="12831" LastActivityDate="2015-01-18T22:26:46.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23910" PostTypeId="2" ParentId="23900" CreationDate="2015-01-18T22:34:14.553" Score="4" Body="&lt;p&gt;Let's consider the following redox couples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^{2+} + e^- -&amp;gt;Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^{2+} + e^- + I^- -&amp;gt;CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, the two redox couples  $\ce{Cu^{2+}/CuI}$ and $\ce{Cu^{2+}/Cu^+}$  are in equilibrium and that means: $$ E_1(\ce{Cu^{2+}/Cu^+})=E_2(\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E_1(\ce{Cu^{2+}/Cu^+})=E^0_1(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}]}}$$ &#xA;$$ E_2(\ce{Cu^{2+}/CuI})=E^0_2(\ce{Cu^{2+}/CuI})+\frac{RT}{F}\ln{[\ce{I^-}][\ce{Cu^2+}]}$$&#xA;We can write then:&#xA;$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}][\ce{Cu^2+}][\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{[\ce{Cu^+}][\ce{I^-}]}}$$&#xA;$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{K_{sp}(\ce{CuI})}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^0_2(\ce{Cu^{2+}/CuI})= 0.17+\frac{8.314 \times 298}{96500}\ln{\frac{1}{10^{-12}}}=0.88 \ce{V}$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="16096" LastEditDate="2015-06-15T05:23:38.490" LastActivityDate="2015-06-15T05:23:38.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23911" PostTypeId="2" ParentId="7830" CreationDate="2015-01-18T23:41:45.567" Score="2" Body="&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\ce{Na_{x}IrO2}$&lt;/span&gt; the current answer seems to be &quot;no&quot; it has not been made but it is discussed in the literature theoretically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://arxiv.org/abs/1303.3290&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kitaev-Heisenberg models for iridates on the triangular, hyperkagome, kagome, fcc, and pyrochlore lattices&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, no triangular lattice iridate with the relevant edge sharing octahedra structure is currently available; a compound of the&#xA;  type NaxIrO2 may or may not turn out to be stabilized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in &lt;a href=&quot;http://arxiv.org/pdf/1409.7820.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chiral spin liquid and topological superconductivity in the undoped and doped&#xA;triangular Kitaev-Heisenberg model&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition, a class of ABO2 (where A and B are alkali and transition metal ions, respectively) type transition metal compounds and the possible material NaxIrO2 may be potential candidates for realizing the triangular Kitaev-Heisenberg model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(But that's your paper isn't it? I don't have any information beyond what you already know, so really this is a useless answer.)&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="17368" LastEditDate="2019-05-24T12:04:18.990" LastActivityDate="2019-05-24T12:04:18.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23912" PostTypeId="1" AcceptedAnswerId="23942" CreationDate="2015-01-18T23:57:35.217" Score="3" ViewCount="108" Body="&lt;p&gt;In relation to the vaporization of THC, when smoked, is it possible that there is any active THC residue that becomes trapped inside the evaporator after the flame goes out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't some of the vapor get stuck inside because the instrument is not a hollow tube? &lt;/p&gt;&#xA;" OwnerUserId="12833" LastEditorUserId="73" LastEditDate="2015-05-22T20:07:45.187" LastActivityDate="2015-05-22T20:07:45.187" Title="Can there be active THC be trapped in the evaporator?" Tags="&lt;combustion&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23914" PostTypeId="1" CreationDate="2015-01-19T00:51:06.243" Score="2" ViewCount="671" Body="&lt;p&gt;I am trying to work out the an unknown Ni(II) complex. I don't known the geometry and I have read that broad lines on the nmr spectrum indicate large compounds with symmetric environments. Does this mean that the complex would have tetrahedral geometry as square planner would be symmetrical and octahedral too large?&lt;/p&gt;&#xA;" OwnerUserId="12834" LastActivityDate="2015-01-19T12:57:52.317" Title="What does it mean in a Ni(II)complex if the HNMR spectrum appears unbroadened and unshifted &amp; what does this tell you about the coordination geometry?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23915" PostTypeId="2" ParentId="23914" CreationDate="2015-01-19T01:02:04.757" Score="4" Body="&lt;p&gt;If the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be low spin square planar.  Tetrahedral and octahedral d8 cannot be diamagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZqVd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could also be two nickel atoms in the complex that are antiferromagnetically&#xA;coupled for a net spin of zero.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-19T12:57:52.317" LastActivityDate="2015-01-19T12:57:52.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23919" PostTypeId="1" CreationDate="2015-01-19T04:01:31.153" Score="1" ViewCount="205" Body="&lt;p&gt;$\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{MeCN}$ and $\ce{[NEt4]2[NiBr4]}$ dissolved in $\ce{H2O}$ have different UV/visible spectra in the visible region. How would they differ, for example would one of them have more peaks? Also how would the molar extinciton coefficients differ and what do these numbers tell you? &lt;/p&gt;&#xA;" OwnerUserId="12834" LastEditorUserId="485" LastEditDate="2015-03-15T08:30:48.037" LastActivityDate="2015-03-15T08:30:48.037" Title="Why would there be differences in UV/Vis spectrum?" Tags="&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23920" PostTypeId="2" ParentId="23919" CreationDate="2015-01-19T04:52:48.687" Score="0" Body="&lt;p&gt;I haven't looked for the spectrums for the compound in the different solvents, so I can only theorize what the differences would be. Since the nickel ion has the energy levels for visible light, then the differences would be due to what surrounds the nickel ion. In water, the nickel ion would be surrounded by water, where as in acetonitrile, the molecule $\ce {[NEt4]2[NiBr]}$ would stay together.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would one of them have more peaks?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, since it is the nickel atom that influences the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also how would the molar extinction coefficients differ and what do these numbers tell you?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I cannot predict the difference in the molar extinction coefficient. The coefficient would tell you how easy the transition took place in the atom. The larger the number, the easier the transition.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-01-19T04:52:48.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23921" PostTypeId="1" AcceptedAnswerId="23922" CreationDate="2015-01-19T06:00:48.093" Score="9" ViewCount="459" Body="&lt;p&gt;I'm reorganizing my garage (I do metalworking and some machining), and now I have an empty white cabinet that I'd like to dedicate to chemical stuff. I'm not a chemist, and I don't have the typical assortment of chemicals that a chem lab would have; the stuff I'm referring to is typical stuff you'd find in a garage: adhesives, lubricants, solvents, paint, etc. (household cleaners are stored elsewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to organize things by functional groups (like those listed above). Is there a convention/standard for organizing shop chemicals?&lt;/p&gt;&#xA;" OwnerUserId="12835" LastActivityDate="2015-01-28T21:53:40.267" Title="Organizing my &quot;chemical&quot; cabinet" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23922" PostTypeId="2" ParentId="23921" CreationDate="2015-01-19T10:40:04.163" Score="11" Body="&lt;p&gt;As mentioned in the comments, safety is the main priority.  Take particular heed to the warning labels and information on the chemical label - if unsure, look up the chemical details MSDS (or SDS) = (Materials) Safety Data Sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most important thing to do (and is often overlooked) is to keep an inventory of what chemicals you have.  The safety data sheets can be obtained from the manufacturers or by looking up the chemical SDS itself, or from general websites such as &lt;a href=&quot;http://www.msds.com/&quot; rel=&quot;nofollow&quot;&gt;MSDS.com&lt;/a&gt; - an example is &lt;a href=&quot;http://www.illawarrasurf.com/msds/turps.pdf&quot; rel=&quot;nofollow&quot;&gt;this one for mineral turpentine&lt;/a&gt; - these contain considerable information about storage, disposal and first aid.  (It is always a good idea to have this readily available and updated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personal protective gear is a must - gloves, safety glasses etc - usually the label or the SDS will have this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the document &lt;a href=&quot;http://www.nerc.ac.uk/about/policy/safety/procedures/guidance_chemical_storage.pdf&quot; rel=&quot;nofollow&quot;&gt;NERC Guidance on Safe Storage of Laboratory Chemicals&lt;/a&gt; (2010), has information that is pertinent to a workshop environment.  There are 3 main principles in regard to storage of chemicals explained in the document:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Segregation&lt;/strong&gt;, from the article:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key incompatibles to segregate from each other are strong acids from&#xA;  strong bases and strong oxidisers from organic or flammable materials. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The NERC document has a list in its appendix (too long for this reply)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Separation&lt;/strong&gt;, related to the segregation,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a laboratory situation adequate separation can be achieved by means of&#xA;  storage cupboards which physically divide incompatible classes of hazardous&#xA;  chemicals. The cupboards may need specific properties or provide&#xA;  separation by means of distance. They will also need to provide secondary&#xA;  containment (eg spill trays or bunded shelves) and security (eg locks / bolted&#xA;  to wall). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ventilation&lt;/strong&gt; (from personal experience, this one is often overlooked).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ventilation is often an essential requirement for safe storage of hazardous&#xA;  chemicals. Its main function is to allow dilution and extraction of vapours or&#xA;  gases that may escape / seep out from containers during storage so they no&#xA;  longer present problems from the viewpoint of noxious smell, hazardous&#xA;  personal exposure or creation of an explosive atmosphere. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also consider an appropriate fire extinguisher, note that water is probably the worst one.  &lt;a href=&quot;http://www.wormald.com.au/fire-products/fire-extinguishers/extinguisher-selection-and-fire-classification&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is an example of a fire extinguisher type guide - this guide suggests that powder based extinguishers may be best (but, make sure you study it and seek advice if unsure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of this may seem 'overkill', but it is better to have some preparedness for if anything goes wrong, than be unpleasantly surprised by it (from personal experience), also, depending on where you are, there could be a legal requirement for a degree of proper storage, handling and protection.&lt;/p&gt;&#xA;" OwnerDisplayName="user12839" LastEditorDisplayName="user12839" LastEditDate="2015-01-28T21:53:40.267" LastActivityDate="2015-01-28T21:53:40.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23923" PostTypeId="2" ParentId="23907" CreationDate="2015-01-19T11:35:23.550" Score="1" Body="&lt;p&gt;In potentiometric analysis, there are two electrodes, an indicator electrode (platinum here) where reduction takes place and reference electrode (calomel electrode) where&#xA;oxidation takes place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here the reference is not $E_{STP}$ cell but the $E_{cathode}$ that is the calomel electrode, so there is drop to $-241mV$ after the equivalance point. Hence,&#xA; $E_{cell} = E_{cathode}$&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorDisplayName="user15489" LastEditDate="2015-05-19T20:03:43.143" LastActivityDate="2015-05-19T20:03:43.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23924" PostTypeId="1" AcceptedAnswerId="23928" CreationDate="2015-01-19T12:23:54.863" Score="5" ViewCount="1240" Body="&lt;p&gt;I did some undergraduate level Chemistry as part of my degree. It left me with the distinct impression that the polarity of water is responsible for its unusual phase properties: it's the only everyday substance that can exist in all three states under easily replicable conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My daughter is learning about states of matter at school, and I thought this might be a fun &quot;extra&quot; for her homework, so I sat down to explain it to her. At which point I realised that I couldn't explain it myself. Polarity explains why water solidifies relatively easily, and why ice floats/expands - but I couldn't see/remember why it might have anything to do with transition into a gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I taught her about phase diagrams instead, as a nice visual thing to take to class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it left me with the question: does water's polarity help explain its unusual phase properties? Or is it something else? Or are those phase properties perhaps not as unusual as I seem to remember?&lt;/p&gt;&#xA;" OwnerUserId="7928" LastActivityDate="2015-01-19T17:58:51.757" Title="Why does water change phase so easily?" Tags="&lt;water&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23925" PostTypeId="2" ParentId="23924" CreationDate="2015-01-19T14:44:00.470" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vapour_pressure_of_water&quot; rel=&quot;nofollow&quot;&gt;Vapor pressure of water&lt;/a&gt; is quite small, thanks to H-bonding in the liquid. Therefore more energy is needed to evaporate (break H-bonds with neighboring molecules) a water molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you compare water to &quot;nearest&quot; non-H-bonded analog, dimethylether, $\ce{CH_3-O-CH_3}$ you'll find, that &lt;a href=&quot;http://encyclopedia.airliquide.com/encyclopedia.asp?GasID=80&quot; rel=&quot;nofollow&quot;&gt;vapor pressure of ether&lt;/a&gt; is much higher (500 kPa) than water (2.3 kPa).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-01-19T14:44:00.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23927" PostTypeId="1" AcceptedAnswerId="23937" CreationDate="2015-01-19T15:07:38.447" Score="5" ViewCount="933" Body="&lt;p&gt;I readily admit to being a pyrophiliac and can generally be found near fire when there is fire. (Fire good, tree bad.)  One of the byproducts of fire, naturally, is charcoal.  I was looking into ways to activate charcoal instead of buying it over the counter - because how cool would it be able to just make your own - but my research so far isn't very promising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, I require intense heat and very little oxygen in order to turn the home and garden variety of charcoal into activated charcoal (as one might find in Norit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to activate your own charcoal (preferably in one's backyard), and if so, does anyone have any ideas or suggestions that I can use to turn my meagre amount of (regular) charcoal into activated charcoal?&lt;/p&gt;&#xA;" OwnerUserId="12845" LastActivityDate="2016-01-02T01:51:31.020" Title="How to activate charcoal" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23928" PostTypeId="2" ParentId="23924" CreationDate="2015-01-19T15:37:25.160" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;does water's polarity help explain its unusual phase properties?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not just its polarity, but &lt;strong&gt;its ability to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;hydrogen bonds&lt;/a&gt;&lt;/strong&gt;, together with the size and shape of its molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonds give water an unusually high boiling point for a molecule its size; many molecules of similar size that can't hydrogen-bond are gases at room temperature. They also give water a relatively high heat capacity and high enthalpies of fusion and vaporization. These latter two properties mean that ice has to absorb more heat to melt and liquid water has to absorb more heat to boil than would otherwise be necessary without hydrogen bonding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water's shape lets it form 4 tetrahedrally arranged hydrogen bonds. This lets it hydrogen-bond &lt;a href=&quot;http://www1.lsbu.ac.uk/water/hexagonal_ice.html&quot; rel=&quot;nofollow&quot;&gt;into networks of chair-shaped hexagonal rings in ice&lt;/a&gt;. The voids in the center of the rings make ice less dense than water. This is unusual; the solid form of a compound is most often denser than the liquid form. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-19T17:58:51.757" LastActivityDate="2015-01-19T17:58:51.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23930" PostTypeId="2" ParentId="23899" CreationDate="2015-01-19T16:10:17.060" Score="13" Body="&lt;p&gt;Toothpaste is a colloid. just like dust particles suspended in water form suspension, colloids are much finer particles suspended in a medium. for example jelly, it is solid particles suspended finely in a liquid....and foam; that is gas suspended in a liquid. As such is deodorant spray that is liquid suspended in air(gas)&lt;/p&gt;&#xA;" OwnerUserId="12846" LastActivityDate="2015-01-19T16:10:17.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23932" PostTypeId="1" AcceptedAnswerId="23934" CreationDate="2015-01-19T17:35:36.993" Score="9" ViewCount="60816" Body="&lt;p&gt;We were doing an experiment about hydrated crystals and more precisely how to determine &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4.nH2O}$&lt;/span&gt;. After we heated the &lt;a href=&quot;http://en.wikipedia.org/wiki/Crucible&quot; rel=&quot;noreferrer&quot;&gt;crucible&lt;/a&gt; we were to cool it down using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Desiccator&quot; rel=&quot;noreferrer&quot;&gt;Desiccator&lt;/a&gt;. Then this question came to my mind: Why &lt;strong&gt;must&lt;/strong&gt; we cool it down? By doing researches in the Net I were only able to come up with these suggestions, but none seem very applicable for me:&lt;/p&gt;&#xA;&lt;h1&gt;Using Einstein's formula:&lt;/h1&gt;&#xA;&lt;p&gt;His famous equation:&lt;span class=&quot;math-container&quot;&gt;$$E = mc^2$$&lt;/span&gt;&#xA;This means that matter is able to be converted to energy. Thus, more energy should mean more mass because &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is constant. However, I argue that since &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is a very large number, normal and observable-in-normal-labs changes in energy aren't able to modify the net change of the mass dramatically (to a degree that the balance is able to detect)&lt;/p&gt;&#xA;&lt;h1&gt;A gut feeling, no backup research:&lt;/h1&gt;&#xA;&lt;p&gt;Anyone sane, and with regards to the value the balance has in the lab, would give slightest chances that the hot container would harm the surface of the balance and therefore, damage it. But chances are that I'm wrong too.&lt;/p&gt;&#xA;&lt;p&gt;So, why must we cool the crucible in order to measure the mass of the sample in it using a balance?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-09-05T14:33:11.520" Title="Why is it obligatory to cool down the container of a sample to measure its mass in a lab?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;heat&gt;" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23933" PostTypeId="2" ParentId="23932" CreationDate="2015-01-19T17:38:17.447" Score="5" Body="&lt;p&gt;Your gut feeling is right: &lt;strong&gt;Very hot crucibles &lt;em&gt;can&lt;/em&gt; cause damage to the balance&lt;/strong&gt;. However, this isn't the main reason for why chemists cool their samples down before measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are all familiar with &lt;a href=&quot;http://en.wikipedia.org/wiki/Convection_current&quot; rel=&quot;nofollow&quot;&gt;convection currents&lt;/a&gt;: When air is heated the molecules will collide more than its previous state; resulting in an increase in either volume or pressure depending on the &quot;flexibility&quot; of the environment of effect. Thus, the density of hot air decreases, and it goes up and &quot;replaces&quot; the cold air. If the process of heating the air continues, hot air will continuously go up in what we call a &quot;convection current&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system of the crucible is thermodynamically open, so in contact with air it will lose some of its heat to its surrounding gas, which is normally atmosphere air. If we put the hot crucible for measuring its mass, convection currents will occur around the crucible, where it's in contact with the surrounding air. The motion of the hot air upwards forces your crucible upwards, but in small amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balance has to be precise enough to detect these differences, though. Most balances in the labs can measure up to 0.01 or 0.001 gram, and &lt;em&gt;will sense&lt;/em&gt; the difference. Since you usually need (or are made to write) exact measurement results up to 1 miligram, these differences are going to be irking pests in measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very useful &lt;a href=&quot;http://content.bfwpub.com/webroot_pubcontent/Content/BCS_5/Harris_Exploring%20Chemical%20Analysis%205e/Lab%20Experiments/5eExpts%20for%20Web%20Nov%202011.pdf&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt; from which the answer came from. It also has useful recommendations on practical problems in the lab.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-01-19T17:38:17.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23934" PostTypeId="2" ParentId="23932" CreationDate="2015-01-19T17:39:24.310" Score="19" Body="&lt;p&gt;A container that is not at ambient temperature will generate air currents around it.  If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient.  These air currents will cause the balance to read incorrectly.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-19T17:39:24.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23935" PostTypeId="1" AcceptedAnswerId="24410" CreationDate="2015-01-19T18:22:12.913" Score="3" ViewCount="2780" Body="&lt;p&gt;Bicyclic structures like norbornane don't undergo E2 eliminations, but why is this the case? I understand that it can't do E1 as it can't form a stable carbocation (the bridgehead carbons that is) but I don't see the issue for E2. Can you explain? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2015-10-02T16:21:44.047" LastActivityDate="2015-10-02T16:21:44.047" Title="Why can't bicyclic structures undergo E2 elimination?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23936" PostTypeId="2" ParentId="23899" CreationDate="2015-01-19T18:29:50.060" Score="21" Body="&lt;p&gt;Here's the boring answer: Toothpaste is a mixture of some solids and some liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &quot;is it solid or liquid?&quot; makes sense when you're talking about a substance or a mixture with a single &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_(matter)&quot; rel=&quot;noreferrer&quot;&gt;phase&lt;/a&gt;—that is, a substance or mixture that's pretty much completely uniform throughout space. Examples of single-phase materials include pure water (a liquid), a chunk of copper (a solid), and sugar water (which is a liquid—including the sugar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some materials consist of multiple phases, and in this case, we have to ask the question &quot;is it solid or liquid?&quot; separately for each phase. Examples of multiple-phase materials include ice water (a mixture containing both a solid (ice) and a liquid (water)), vinaigrette (a mixture of oil (a liquid) and vinegar (another liquid)), and toothpaste (a mixture of several kinds of solids and liquids).&lt;/p&gt;&#xA;" OwnerUserId="12849" LastEditorUserId="12849" LastEditDate="2019-07-23T18:36:10.920" LastActivityDate="2019-07-23T18:36:10.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23937" PostTypeId="2" ParentId="23927" CreationDate="2015-01-19T19:23:15.623" Score="2" Body="&lt;p&gt;&lt;em&gt;Note that like most of the colleagues on this site with practical lab experience, I've used activated charcoal once in a while, but I never bothered to make it myself.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed that you live on the countryside with enough space around, you can make your own charcoal in a earth-covered pile or an oven - unless local environmental regulations ban that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activation, even when starting from ordinary charcoal is different. The &lt;strong&gt;gas activation&lt;/strong&gt; requires heating (and partial burning) of the charcoal at up to 1000 °C in a stream of air, carbon dioxide or water vapour. I'd rather not do it in my backyard.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-19T19:23:15.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23938" PostTypeId="1" CreationDate="2015-01-19T19:29:14.023" Score="2" ViewCount="194" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the Rate law of the reaction below?&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{4 HBr + O2 -&amp;gt; 2 H2O + 2 Br2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How do we approach finding the rate law of a reaction? Can this be found out only through experiments?&lt;/p&gt;&#xA;" OwnerUserId="12651" LastEditorUserId="83758" LastEditDate="2020-09-15T09:47:39.123" LastActivityDate="2020-09-15T09:47:39.123" Title="How is the Rate law of a reaction measured?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;reaction-order&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23939" PostTypeId="1" CreationDate="2015-01-19T20:17:52.847" Score="4" ViewCount="3471" Body="&lt;p&gt;I am required to prepare &lt;span class=&quot;math-container&quot;&gt;$\pu{120cm3}$&lt;/span&gt; of hydrogen peroxide to each of one of my students tomorrow. Some material is given to me on the method of preparing the solutions. Within the method of preparing &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt;, it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Prepare from freshly purchased &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; volume (&lt;span class=&quot;math-container&quot;&gt;$30\%~\text{w/w}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{8.3 mol/dm3}$&lt;/span&gt;) or &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; volume (&lt;span class=&quot;math-container&quot;&gt;$6\%~\text{w/w}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pu{1.7 mol/dm3}$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt;. Dilute &lt;span class=&quot;math-container&quot;&gt;$\pu{200 cm3}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; volume solution to &lt;span class=&quot;math-container&quot;&gt;$\pu{1 dm3}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; volume solution, then dilute &lt;span class=&quot;math-container&quot;&gt;$\pu{59 cm3}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; volume solution to &lt;span class=&quot;math-container&quot;&gt;$\pu{1 dm3}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 mol/dm3}$&lt;/span&gt; solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain the quoted part?&lt;/p&gt;&#xA;" OwnerUserId="12853" LastEditorUserId="41328" LastEditDate="2019-01-19T19:09:07.433" LastActivityDate="2020-12-11T14:25:23.763" Title="How to interpret the instruction of the dilution of hydrogen peroxide?" Tags="&lt;aqueous-solution&gt;&lt;concentration&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23940" PostTypeId="1" CreationDate="2015-01-19T20:23:21.107" Score="5" ViewCount="601" Body="&lt;p&gt;I am currently trying to find an addition transition state and verify it using an IRC calculation. I have found a good candidate, but the IRC calculation gives the same minimum on each side of the transition state. In my mind, this does not seem any different from a vibrational mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the TS is submerged, and I am sure this is complicating things, but it does not make sense that an IRC calculation would give the same minimum in both directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on what this signifies or how to remedy it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="12854" LastActivityDate="2015-04-20T17:31:46.230" Title="Intrinsic Reaction Coordinate calculation-- both sides of TS lead to same minimum" Tags="&lt;computational-chemistry&gt;&lt;reaction-coordinate&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23941" PostTypeId="2" ParentId="23939" CreationDate="2015-01-19T20:32:16.907" Score="8" Body="&lt;p&gt;Take &lt;span class=&quot;math-container&quot;&gt;$\pu{200 mL}$&lt;/span&gt; of the &lt;span class=&quot;math-container&quot;&gt;$30\%$&lt;/span&gt; solution, add water to a final volume of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; liter. Then you have a &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt; solution. Next, take &lt;span class=&quot;math-container&quot;&gt;$\pu{59 mL}$&lt;/span&gt; of the &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt; solution and add water until it is &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first dilution it's not exactly &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt;, because the density of &lt;span class=&quot;math-container&quot;&gt;$30\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O2}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{1.135 g/cm3}$&lt;/span&gt;. It's more like &lt;span class=&quot;math-container&quot;&gt;$6.6\%$&lt;/span&gt;, but the instructions are referring to it as &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt;. If you want exactly &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt; you should dilute by weight instead of volume.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2019-01-19T19:10:49.977" LastActivityDate="2019-01-19T19:10:49.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23942" PostTypeId="2" ParentId="23912" CreationDate="2015-01-19T20:59:07.760" Score="6" Body="&lt;p&gt;When burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the length of the &quot;object&quot; , related to the temperature gradient between the front and the mouth piece, and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the type of the &quot;filling&quot; and its adsorptive properties&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;a part of the vaporized material might condense and/or get absorbed in a colder region of the filling. Similar effects, such as the enrichment of nicotine in the stump of cigarettes have examined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however unlikely that any of the transported and reabsorbed components can be inhaled from a cold &quot;reactor&quot;. The relevant components are no longer in the vapour phase and the vapour pressure of the tetrahydrocannabinoles will be too small.&#xA;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="10256" LastEditDate="2015-02-25T15:27:17.753" LastActivityDate="2015-02-25T15:27:17.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23943" PostTypeId="2" ParentId="5057" CreationDate="2015-01-19T21:22:08.997" Score="0" Body="&lt;p&gt;I have done some work with reactions on organic surfaces, and hybrid QM/MM calculations helped me solve my problem, particularly using the ONIOM forcefield. I understand that a classical forcefield may not be the best way to treat a metal surface, however.&lt;/p&gt;&#xA;" OwnerUserId="12854" LastActivityDate="2015-01-19T21:22:08.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23944" PostTypeId="2" ParentId="23940" CreationDate="2015-01-19T21:35:28.453" Score="5" Body="&lt;p&gt;Be sure you have really obtained TS, that means stationary point, i.e. all gradients are zero. Than, you should have one imaginary frequency corresponding to the motion along the reaction coordinate. If this is true and you still get both IRCs going to same minimum, your assumptions regarding the shape of Potential Energy Surface (PES) could be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier to find the minima first, so be sure you have well optimized reactant, product, and if by any means reasonable, also intermediates. It is possible that there exist some high energy intermediates, which complicate the whole story (if you are not aware of them).&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-01-19T21:35:28.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23945" PostTypeId="1" AcceptedAnswerId="24022" CreationDate="2015-01-19T22:23:08.717" Score="4" ViewCount="445" Body="&lt;p&gt;I have searched through printed books and the Internet, but I cannot find a definitive answer on an issue that fascinates me: what forces keep the various compounds belonging to a solid solution together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I correctly understand, a rock is a solid solution, and I know that the atoms of a single mineral species are bound by its particular covalent and ionic bonds, but what keeps different minerals together in a rock? Do they mix by sharing &quot;neighbouring&quot; cations and anions? I am not sure whether intermolecular forces such as Van der Waals forces are present, but I would definitely expect stronger forces in a solid solution or between the minerals of a rock.&lt;/p&gt;&#xA;" OwnerUserId="6943" LastEditorUserId="9961" LastEditDate="2015-12-17T17:08:43.360" LastActivityDate="2016-02-04T00:11:02.743" Title="Cohesive forces in solid solutions" Tags="&lt;bond&gt;&lt;ionic-compounds&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23946" PostTypeId="2" ParentId="23932" CreationDate="2015-01-19T22:34:49.790" Score="9" Body="&lt;p&gt;The other two answers here are fine; it's true that as the hot object you're trying to weigh warms the air around it, the air will rise, creating air currents that will give you an unsteady reading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add three things they didn't mention:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The balance pan will heat up, too, causing metal parts in the balance to expand. This will also contribute to an error in your reading, and as the metal cools, the reading will change. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another contribution to the error is that the density of a hot sample is less than the density of a cold sample, so its buoyancy is different. The object will appear to gain a tiny amount of weight as it cools! &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Don't worry about $E=mc^2$ in chemistry unless you've got some serious nuclear reactions going on. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And a fourth thing: if you put a hot object into an air-tight dessicator, the air inside will contract as the object cools---and you might have some trouble getting the lid off again! &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-20T00:46:12.883" LastActivityDate="2015-01-20T00:46:12.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23947" PostTypeId="1" CreationDate="2015-01-19T23:22:27.450" Score="1" ViewCount="195" Body="&lt;p&gt;If you place a piece of paper created by a laser printer, &lt;em&gt;e.g.&lt;/em&gt; fax paper or some receipts, on top of a piece of normal paper you have drawn on with a Bic Marking 2300 permanent marker, the image transfers to the fax/receipt paper? Do any non-toxic solvents heat up as they evaporate? If so, would that explain the reaction between the paper and ink?&lt;/p&gt;&#xA;" OwnerUserId="12858" LastEditorUserId="194" LastEditDate="2017-04-01T03:25:45.803" LastActivityDate="2017-04-01T03:25:45.803" Title="Solvent interaction with thermal paper" Tags="&lt;everyday-chemistry&gt;&lt;heat&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2017-04-01T20:44:42.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="23948" PostTypeId="1" AcceptedAnswerId="24005" CreationDate="2015-01-20T00:08:59.700" Score="4" ViewCount="363" Body="&lt;p&gt;I'm trying to determine whether or not certain compounds form solid ice at certain atmospheric pressures. These pressures vary significantly, from 0.001 atm to 800 atm. I understand that there is no equation relating the freezing points of all elements, but is there one or two that relate to carbon dioxide and water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I'd like a Python function (or library) or equation where the input is atmospheric pressure or kPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, failing that, if someone could point me in the direction of a phase diagram for both compounds where I can input specific atmospheric pressure values and read a specific temperature for freezing point, that would be helpful as well.&lt;/p&gt;&#xA;" OwnerUserId="12859" LastEditorUserId="4945" LastEditDate="2015-01-20T02:09:48.470" LastActivityDate="2015-01-21T13:31:14.440" Title="Is there a pressure dependent function for freezing points of water and carbon dioxide?" Tags="&lt;physical-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23949" PostTypeId="1" AcceptedAnswerId="24013" CreationDate="2015-01-20T02:41:45.893" Score="3" ViewCount="2973" Body="&lt;p&gt;I am trying to determine the electron pair geometry (and molecular geometry) of some atoms in ibuprofen. The first one I am trying to do is the double bonded oxygen. Ibuprofen is pictured below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2XGwi.jpg&quot; alt=&quot;Structure of Ibuprofen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I counted the number of electron domains(steric number) to be 3. I counted the double bond as one and each lone pair as one. It has two lone pairs and one double bond. So this means the electron geometry should be trigonal planar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am stuck on molecular geometry. There is no molecular geometry I have found in any chart that has two lone pairs and one bond. Am I making a mistake somewhere? &lt;/p&gt;&#xA;" OwnerUserId="7612" LastEditorUserId="83758" LastEditDate="2020-05-15T04:59:54.457" LastActivityDate="2020-05-15T04:59:54.457" Title="Electron Pair Geometry in Ibuprofen (Electron Pair Geometry in Organic Compounds)" Tags="&lt;organic-chemistry&gt;&lt;molecular-structure&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23950" PostTypeId="2" ParentId="23938" CreationDate="2015-01-20T03:48:10.110" Score="3" Body="&lt;h1&gt;Preamble&lt;/h1&gt;&#xA;&lt;p&gt;For a reaction with no intermediate steps, you could derive the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rate_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;rate law&lt;/a&gt; from the reaction equation. For a generic equation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{$a$A + $b$B -&amp;gt; C}$$&lt;/span&gt;&#xA;the rate is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r = k\cdot c^{a}(\ce{A})\cdot c^{b}(\ce{B}).$$&lt;/span&gt;&#xA;Rigorously, this is only possible for &lt;a href=&quot;http://goldbook.iupac.org/E02035.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;elementary reactions&lt;/a&gt;. A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{H2 &amp;amp;-&amp;gt;[\Delta T] 2H. }\\&#xA;\ce{Br2 &amp;amp;-&amp;gt;[h\nu] 2Br. }&#xA;\end{align}&lt;/span&gt;&#xA;An example for a bimolecular reaction would be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2NO2 -&amp;gt; NO3 + NO}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Reaction mechanism of the combustion of hydrogen bromide&lt;/h1&gt;&#xA;&lt;p&gt;This seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described &lt;a href=&quot;https://chemistry.stackexchange.com/q/14704/4945&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;A quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. &lt;em&gt;(Note: These are supposed to be gas phase reactions.)&lt;/em&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}\ce{&#xA;HBr + O2 &amp;amp;-&amp;gt; HOOBr\\&#xA;HOOBr + HBr &amp;amp;-&amp;gt; 2 HOBr\\&#xA;HOBr + HBr &amp;amp;-&amp;gt; H2O + Br\\&#xA;}\end{align}&lt;/span&gt;&#xA;The following mechanism is taken from &lt;a href=&quot;http://dx.doi.org/10.1002/kin.550150404&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lilian G. S. Shum and Sidney W. Benson &lt;em&gt;Int. J. Chem. Kin.&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;15&lt;/em&gt; (4), 341-380.&lt;/a&gt; I think this is a much more suitable and credible approach to the problem. &lt;em&gt;(Note: Most of the reactions are in equilibrium.)&lt;/em&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}\ce{&#xA;HBr + O2 &amp;amp;~&amp;lt;=&amp;gt; HO2 + Br\\&#xA;HO2 + HBr &amp;amp;~&amp;lt;=&amp;gt; H2O2 + Br\\&#xA;H2O2 + 2HBr &amp;amp;~-&amp;gt;T[wall] 2H2O + Br2\\&#xA;Br + Br + M &amp;amp;~&amp;lt;=&amp;gt; Br2 + M\\\hline&#xA;4HBr + O2 &amp;amp;~&amp;lt;=&amp;gt; 2H2O + 2Br2\\&#xA;}\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately I was unable to determine what &lt;span class=&quot;math-container&quot;&gt;$\ce{M}$&lt;/span&gt; refers to in this case. It might be any inert gas used as a solvent in the reaction or the wall.&lt;/p&gt;&#xA;&lt;h1&gt;Rate law of the combustion of hydrogen bromide&lt;/h1&gt;&#xA;&lt;p&gt;As a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realize, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.&lt;/p&gt;&#xA;&lt;p&gt;However, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm{d}[\ce{H2O}]}{\mathrm{d}t} = 2k_g[\ce{HBr}][\ce{O2}].$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore it can formally be regarded as a bimolecular reaction.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="95133" LastEditDate="2020-09-15T09:34:30.400" LastActivityDate="2020-09-15T09:34:30.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23953" PostTypeId="1" AcceptedAnswerId="23956" CreationDate="2015-01-20T05:34:18.320" Score="4" ViewCount="1430" Body="&lt;p&gt;Is it possible to make fulminates such as: magnesium fulminate, copper fulminate, etc. &#xA;by the metal + $\ce{HNO3}$ + $\ce{EtOH}$ reaction?&lt;/p&gt;&#xA;" OwnerUserId="12864" LastEditorUserId="418" LastEditorDisplayName="user12839" LastEditDate="2015-02-25T20:04:47.363" LastActivityDate="2015-02-25T20:04:47.363" Title="Is it possible to make metal fulminates?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23954" PostTypeId="1" CreationDate="2015-01-20T05:54:40.130" Score="6" ViewCount="2047" Body="&lt;p&gt;Using the example of $\ce{XeF4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/TSyjpWL.png&quot; alt=&quot;http://imgur.com/a/31Q4L&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the physical explanation enforcing the symmetry of the $\ce{1b_{1g}}$ orbital on the fluorine atoms? Why isn't the symmetry of a nonbonding orbital arbitrary?  If it's going to be nonbonding anyways, why can't we, for example, have a Fluorine p-orbital arrangement facing towards Xenon with three positive p-orbitals and one negative p-orbital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To elaborate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I imagine a free Xenon atom in space, and the approach of four individual fluorine atoms, I would expect the bond formation to be randomized with respect to the orientation of the fluorine p-orbitals, and therefore for some arrangements to not be perfectly symmetrical, such as 3 positive p-orbitals, 1 negative, facing inward.  I understand bonds can't be made without symmetry between the Xenon and Fluorine orbitals; that makes physical sense because we can argue it by looking at orbital overlap that dictates bonds can only occur with appropriate symmetry.  But in a nonbonding case, such as $\ce{1b_{1g}}$ above, I don't understand why symmetry is also required.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source for pictures: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://scilearn.sydney.edu.au/fychemistry/calculators/make_mo.shtml?type=year1&amp;amp;theMolecule=xef4&quot; rel=&quot;noreferrer&quot;&gt;https://scilearn.sydney.edu.au/fychemistry/calculators/make_mo.shtml?type=year1&amp;amp;theMolecule=xef4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chem.mun.ca/homes/cmkhome/SALCS&amp;amp;MOs.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.chem.mun.ca/homes/cmkhome/SALCS&amp;amp;MOs.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12866" LastEditorUserId="4945" LastEditDate="2015-01-20T06:28:10.947" LastActivityDate="2015-01-20T16:21:27.460" Title="Why does symmetry have to be maintained in molecular orbitals?" Tags="&lt;bond&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;symmetry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23956" PostTypeId="2" ParentId="23953" CreationDate="2015-01-20T07:51:30.480" Score="8" Body="&lt;p&gt;In &lt;em&gt;The Preparatory Manual of Explosives&lt;/em&gt;  (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0578142813&quot;&gt;paperback at amazon&lt;/a&gt;), the author, Jared Ledgard, provides a procedure for the preparation of copper fulminate that can be summarized as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To 100 mL of 70% $\ce{HNO3}$, 30 g of anhydrous $\ce{Cu(NO3)2}$ is added under stirring. If the copper nitrate doesn't dissolve properly, add water dropwise until it does. (&lt;strong&gt;NOTE&lt;/strong&gt;: &lt;em&gt;Adding water to concentrated acids usually isn't the best idea!&lt;/em&gt;) The solution is stirred for 24 hrs at room temperature and then heated to roughly 80 °C. Under stirring, 120 g of 95% ethanol are added. After addition of the ethanol, the heat source is removed and the mixture is alllowed to reach room temperature. The precipitate is rapidly filtered off, washed with several hundred mL od cold water and dried in a desiccator over magnesium sulfate. The product, copper fulminate, should be stored in a desiccator over sodium sulfate in a refrigerator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To my opinion, the procedure leaves out significant information, such as the &lt;strong&gt;exact&lt;/strong&gt; temperature of the aqueous solution while adding the ethanol. It is not clear whether and how much ethanol is supposed to evaporate. While the author suggests to carry out the reaction in a beaker, a setup with a two-neck flask, reflux condenser and a dropping funnel might be a better idea.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Handling fairly concentrated nitric acid requires safety precautions: A fume hood, eye protection, gloves and a lab coat are mandatory!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reaction product is a &lt;strong&gt;primary explosive&lt;/strong&gt;! While less shock sensitive than mercury fulminate, it still will go off on impact (or upon heating above ~200 °C).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the case of these or similar sensitive materials, there is a good rule of thumb: &lt;strong&gt;If you have to ask on a Q&amp;amp;A site - Don't touch it!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-20T07:51:30.480" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="23957" PostTypeId="2" ParentId="23954" CreationDate="2015-01-20T08:05:05.003" Score="2" Body="&lt;p&gt;The orbital and geometrical symmetry are closely related. You know that $\ce{XeF_4}$ is square planar, therefore $\ce{D_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you perform some manipulation of the molecule, e.g. rotation by $\ce{90^{\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-01-20T08:05:05.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23958" PostTypeId="5" CreationDate="2015-01-20T09:58:23.187" Score="0" Body="&lt;p&gt;&lt;strong&gt;Resonance&lt;/strong&gt; as defined by in the &lt;a href=&quot;http://goldbook.iupac.org/R05326.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC GoldBook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the context of chemistry, the term refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. Resonance among contributing structures means that the wavefunction is represented by 'mixing' the wavefunctions of the contributing structures. The concept is the basis of the quantum mechanical valence bond methods. The resulting stabilization is linked to the quantum mechanical concept of 'resonance energy'. The term resonance is also used to refer to the delocalization phenomenon itself.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mesomerism&lt;/strong&gt; as defined in the &lt;a href=&quot;http://goldbook.iupac.org/M03845.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Essentially synonymous with resonance. The term is particularly associated with the picture of π-electrons as less localized in an actual molecule than in a Lewis formula. The term is intended to imply that the correct representation of a structure is intermediate between two or more Lewis formulae.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-13T17:57:41.083" LastActivityDate="2018-03-13T17:57:41.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23959" PostTypeId="4" CreationDate="2015-01-20T09:58:23.187" Score="0" Body="Resonance refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. The tag should be applied to any question that concern resonance structures or resonance energy. This tag should not be applied to questions about aromatic compounds (use the [aromatic-compounds] tag instead)" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-13T17:58:15.960" LastActivityDate="2018-03-13T17:58:15.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23960" PostTypeId="1" AcceptedAnswerId="23967" CreationDate="2015-01-20T11:08:40.687" Score="3" ViewCount="2352" Body="&lt;p&gt;How can we determine how many delocalised electrons every atom of gold contributes to the 'sea of delocalised electrons'? More generally, how can we determine the number of electrons any metal contributes? Must we have thermodynamic data to determine this?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-01-20T17:38:05.843" Title="How many delocalised electrons in gold?" Tags="&lt;bond&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23961" PostTypeId="2" ParentId="23960" CreationDate="2015-01-20T11:59:50.043" Score="-1" Body="&lt;p&gt;Valance electron of s and p orbital take part in delocalisation over the positively charged gold atoms&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2015-01-20T11:59:50.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23962" PostTypeId="2" ParentId="23954" CreationDate="2015-01-20T12:02:42.207" Score="0" Body="&lt;p&gt;Symmetry reduces the level of energy of any thing. Most of the things of nature has inbuilt symmetry with them. It is the property of nature to have different kinds of symmetry. So that it reduces the energy of the orbital or molecule&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2015-01-20T12:02:42.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23963" PostTypeId="1" AcceptedAnswerId="23964" CreationDate="2015-01-20T13:12:30.770" Score="4" ViewCount="631" Body="&lt;p&gt;The following is extract from my book-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To illustrate this, let us consider the enol contents of acetyacetone and $\alpha$ - methylacetylacetone. Although both the enols are stabilized by H bonding the enol form of $\alpha$ - methylacetylacetone is destabilized to some extend by steric repulsion due to the presence of $\alpha$ - methyl group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What kind of H bonding (sorry I know i am being stupid, the figure in my book showed a bond connecting H and oxygen and dotted line connecting another oxygen and the previous hydrogen. How can hydrogen form two bonds?) is being talked about and what is steric repulsion? How does the steric repulsion destabilize the enol form?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="8231" LastEditDate="2015-01-20T13:26:15.690" LastActivityDate="2015-01-20T14:55:26.180" Title="Dependence of steric hinderance on enol content" Tags="&lt;organic-chemistry&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23964" PostTypeId="2" ParentId="23963" CreationDate="2015-01-20T14:55:26.180" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can hydrogen form two bonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case of an enol, hydrogen does not form two full bonds.  What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed.  It's as if the partially broken OH bond and the partially made OH bond are each &quot;half a bond&quot;, so all together the electron sharing is such that there is still just one full bond to hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aBQau.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the steric repulsion destabilize the enol form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone.  Notice the steric interactions indicated by the arrows.  The situation looks similar to that seen with &lt;em&gt;cis&lt;/em&gt;-2-butene.  The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds &quot;bump&quot; into one another causing what we often call &quot;steric destabilization.&quot;  Remember that &lt;em&gt;trans&lt;/em&gt;-2-butene is more stable than &lt;em&gt;cis&lt;/em&gt;-2-butene because the &lt;em&gt;trans&lt;/em&gt; isomer does not have the adverse steric interaction present in the &lt;em&gt;cis&lt;/em&gt; isomer.  It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in &lt;em&gt;cis&lt;/em&gt;-2-butene, it destabilizes the enol.  Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-20T14:55:26.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23965" PostTypeId="1" CreationDate="2015-01-20T15:47:50.853" Score="4" ViewCount="7086" Body="&lt;p&gt;Well consider the reaction $A + B ⇌ C$&#xA;Adding inert gas to a container where this reaction is taking place will increase the pressure of the system. In accordance with Le Chatelier's Principle, the equilibrium should shift in the direction which opposes this change in pressure, i.e. in the direction where fewer number of molecules are created. In this specific reaction, it happens to be in the forward reaction. However, I have been taught that adding an inert gas at constant volume has no effect on the equilibrium. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastActivityDate="2015-01-20T19:26:03.730" Title="Dynamic equilibrium - effect of adding inert gas" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2015-01-21T03:35:52.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="23966" PostTypeId="1" AcceptedAnswerId="23972" CreationDate="2015-01-20T15:50:33.847" Score="2" ViewCount="2537" Body="&lt;p&gt;E2 eliminations occur when the leaving group and the proton are in the anti-periplanar conformation not syn-periplanar. This is beyond a merely a preference as leads to E2 reactions being stereoselective. However, when it comes to vinyl bromides for example (involving a double bond) the reaction can occur in the Z- and E- isomers (i.e both syn-periplanar and anti-periplanar). How can this be the case? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-01-21T06:48:55.567" LastActivityDate="2015-01-21T06:48:55.567" Title="Why can syn-periplanar E2 elimination happen in molecules like vinyl bromides but seemingly not in other molecules?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23967" PostTypeId="2" ParentId="23960" CreationDate="2015-01-20T15:57:59.440" Score="3" Body="&lt;p&gt;Briefly: valence orbitals on adjacent metal atoms interfere to form &quot;bands&quot; of closely spaced molecular orbitals that spread throughout the metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have N metal atoms with occupied s orbitals, they interfere to form an &quot;s band&quot; with N molecular orbitals. The lowest energy orbital is fully bonding; the orbital on the top of the band is fully antibonding. The spread of the band will be finite, and since N is a huge number, the molecular orbital energies will be &lt;em&gt;very&lt;/em&gt; closely spaced, practically continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two valence s electrons in your metal, the s band will be full (each orbital can hold up to 2 electrons, you have 2N electrons, and N orbitals in the band). If you have only one valence s electron, the s band will be half full (the first N/2 orbitals will be filled, and the remaining orbitals will be empty). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above the s band you'll have a similar p band. In gold, the p band is unoccupied, while the s band is half full (gold has 6s$^1$). However, IF the bottom of the p band overlaps with the top of the s band, the s electrons can easily move into the bonding MO's at the bottom of the p band. They can also move into any vacant orbitals in the s band that are just above the highest occupied orbital.  &lt;em&gt;In either case, the s electron will be delocalized&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out whether the valence electrons will delocalize, then, you'll have to decide what bands are present, where the highest occupied molecular orbital is, and what the width and the spacing of the bands are (does the unoccupied band overlap with the occupied band, or is there a gap between the two?)&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-20T17:38:05.843" LastActivityDate="2015-01-20T17:38:05.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23968" PostTypeId="1" AcceptedAnswerId="23980" CreationDate="2015-01-20T16:00:02.297" Score="5" ViewCount="4396" Body="&lt;p&gt;I was trying to synthesize 1-iodo-4-nitrobenzene in the lab today – however, the appearance of my product really confused me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, I added conc. sulfuric acid to water and 4-nitroaniline. &#xA;This was followed by diazotising the amine sulfate with sodium nitrite and water. (This was done at temperatures of below $10\ \mathrm{^\circ C}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I added the diazonium salt to a solution of $\ce{NaI}$ and water. A large plume/foam formed ($\ce{N2}$). After, I performed a filtration using a Büchner funnel. That’s where things went wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used water to wet the filter paper. I added in the iodonitrobenzene. A suckback occurred within a few seconds as the product was quite viscous. After resumption, the filtration went as planned. However, the product on the filter paper was a molten brown mush. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to let it dry for a further 5 minutes but once turned off it remained a sludge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any suggestions on how I could improve on this synthesis or maybe an idea of what went wrong?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="402" LastEditDate="2017-01-14T08:49:11.267" LastActivityDate="2017-11-13T00:40:22.813" Title="Synthesis of 1-iodo-4-nitrobenzene" Tags="&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="23970" PostTypeId="2" ParentId="23954" CreationDate="2015-01-20T16:21:27.460" Score="2" Body="&lt;p&gt;It's not so much that these are arbitrary arrangements or that symmetry is &quot;conserved.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take the 4 fluorine atoms as a &lt;em&gt;set&lt;/em&gt; and consider the reproducible representation of all fluorine 2p orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the $D_{4h}$ point group of $\ce{XeF4}$, the 4 fluorine orbitals consist of an $a_{1g}$, $b_{1g}$ and $e_u$ representations. (There are other representations of the out-of-plane 2p orbitals too.) Four atomic orbitals, 2x1D 1x2D representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we try combining the representations of the fluorine orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a matter of &lt;em&gt;math&lt;/em&gt;... the reason $b_1g$ is non-bonding is because there is no Xe atomic orbital with the same symmetry. If you consider the direct product of the fluorine $b_{1g}$ with any Xe orbital, you get zero - they are orthogonal and have no overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you ask why can't you have 3:1.. well, you have 4 identical fluorines. If you had $\ce{XeF3Cl}$ or something like this, you'd then have different symmetry and different orbitals.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-20T16:21:27.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="23971" PostTypeId="2" ParentId="21962" CreationDate="2015-01-20T16:57:10.417" Score="8" Body="&lt;p&gt;There are different strategies for synthesizing superheavy elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of the main strategies are referred to as &quot;hot fusion&quot; and &quot;cold fusion&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two reactions are good examples of the &quot;hot fusion&quot; strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element.  The target element is choosen on the basis of being very heavy, availible and not rapidly decaying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{48}_{20}\ce{Ca}$ is choosen for a very special reason.  This is not ordinary calcium.  The most common isotope of calcium has only 20 neutrons, but this isotope has 28!.  It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table.  $^{50}_{22}\ce{Ti}$ is also very neutron rich.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are neutron rich nuclei chosen? There are two reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, confining many protons in a small space causes electrostatic repulsion.  As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you will notice the asterisk (*) in the 1st, 3rd and 4th equations.  This denotes the &lt;a href=&quot;http://encyclopedia2.thefreedictionary.com/Compound+Nucleus&quot; rel=&quot;noreferrer&quot;&gt;compound nucleus&lt;/a&gt; which initially forms in the fusion reaction.  It is an extremely ($&amp;lt;10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state.  It's formation is not considered creating a new element.  It is critical to the strategy of forming an element that the intermediate state loss energy through neutron &quot;evaporation&quot;.   Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown.  Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/abs/1207.5700&quot; rel=&quot;noreferrer&quot;&gt;Future of superheavy element research: Which nuclei could be synthesized within the next few years?&lt;/a&gt; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.72.733&quot; rel=&quot;noreferrer&quot;&gt;The discovery of the heaviest elements&lt;/a&gt; Rev. Mod. Phys. 72, 733&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-20T19:14:08.020" LastActivityDate="2015-01-20T19:14:08.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23972" PostTypeId="2" ParentId="23966" CreationDate="2015-01-20T17:03:22.990" Score="4" Body="&lt;p&gt;Elimination from vinyl halides occurs faster when the halide and proton have a trans relationship.  When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower.  The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the &quot;cis&quot; case.  Two possible mechanistic alternatives are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or&lt;/li&gt;&#xA;&lt;li&gt;an &lt;a href=&quot;http://chem.ucr.edu/documents/faculty_neuman/chapter%209.pdf&quot; rel=&quot;noreferrer&quot;&gt;E1CB mechanism&lt;/a&gt; where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Experiments in analogous molecules suggest that the E1CB pathway is most often followed in &quot;cis&quot; elimination from vinyl halides.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-20T17:03:22.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23974" PostTypeId="2" ParentId="23965" CreationDate="2015-01-20T17:46:38.797" Score="1" Body="&lt;p&gt;At constant temperature, only changes to the concentration (technically &lt;a href=&quot;http://en.wikipedia.org/wiki/Activity_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;activity&lt;/a&gt;) of a reactant or product, shift the equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding inert gas does not change the concentration of any reactant or product, so it does not affect equilibrium.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-20T17:46:38.797" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="23976" PostTypeId="1" CreationDate="2015-01-20T18:11:40.203" Score="2" ViewCount="12265" Body="&lt;p&gt;I have read an article online that says that ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From some preliminary internet searches, I have discovered that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Isopropyl alcohol is a slightly bigger molecule than ethanol (Methanol is $\ce{CH4O}$, Ethanol is $\ce{C2H6O}$ and Isopropyl alcohol is $\ce{C3H8O}$) and is often cheaper to make. It is not as poisonous as methanol (an ounce of methanol can kill you and a third of an ounce causes blindness). So it is useful as a solvent in the same way acetone is useful as a solvent.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One final question. It would be best to use 99% pure solvents (mostly free of water), however for an ethanol replacement, I have seen a product used for ethanol fireplaces where the MSDS mentions 95% ethanol and 5% 2-propanol. Would this be an acceptable substitute for 99% ethanol? Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="12878" LastEditorUserId="7448" LastEditDate="2015-01-20T20:27:49.317" LastActivityDate="2015-09-03T10:25:51.540" Title="What is the difference between IPA and Ethanol in terms of solubility?" Tags="&lt;organic-chemistry&gt;&lt;solutions&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23977" PostTypeId="2" ParentId="23965" CreationDate="2015-01-20T18:16:40.897" Score="0" Body="&lt;p&gt;The gas added works in both directions and the fraction of pressure is the same becaussed the total pressure increses but the inert gas do not affect the features of the others and exerts the same effect. If the number of molecules do not change, the concentration remains constant, and the principle of Le chatelier sais that in case of increase the pressure the reaction take the direction where there are less molecules because the pressure force them to agregate.&lt;/p&gt;&#xA;" OwnerUserId="12843" LastActivityDate="2015-01-20T18:16:40.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23980" PostTypeId="2" ParentId="23968" CreationDate="2015-01-20T19:32:33.537" Score="7" Body="&lt;p&gt;Here are some things to focus on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How quickly\slowly did you add the diazonium salt to the NaI solution?&lt;/li&gt;&#xA;&lt;li&gt;Did you maintain the reaction temperature below 10°C during the entire addition?&lt;/li&gt;&#xA;&lt;li&gt;Did you also keep the diazonium salt cooled while adding it?&lt;/li&gt;&#xA;&lt;li&gt;After addition was complete, did you let the reaction stir at 10°C until nitrogen evolution ceased and then let the reaction slowly warm to room temp? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://myweb.brooklyn.liu.edu/swatson/Site/Laboratory_Manuals_files/Exp18.pdf&quot;&gt;Here&lt;/a&gt; is a tested lab procedure for the preparation of 4-iodonitrobenzene.  Compare it to your notes and see where you differed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping the reaction cold during the addition is key. If the diazonium salt warms up too much, the diazonium can react with itself (&lt;a href=&quot;http://en.wikipedia.org/wiki/Azo_coupling&quot;&gt;coupling&lt;/a&gt;) to give colored by-products.  The diazonium ion can also eliminate nitrogen to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example (see page 6, the Mechanism section in this &lt;a href=&quot;http://nptel.ac.in/courses/104103022/download/module7.pdf&quot;&gt;reference&lt;/a&gt;).  Both pathways can  contaminate and reduce the yield of the desired product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-20T19:59:13.787" LastActivityDate="2015-01-20T19:59:13.787" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="23984" PostTypeId="1" CreationDate="2015-01-20T21:57:54.223" Score="2" ViewCount="144" Body="&lt;p&gt;A couple of weeks ago we did a lab where we had a bunch of standard solutions of alcohol, dichromate, and sulfuric acid. We also had one unknown concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We heated them so they would all react, and the initial orange dichromate turned slightly green (the ion $\ce{Cr^{3+}}$ formed). Then we placed each reacted soln in a calorimeter which can emit light of certain frequencies. I believe we chose 635, which should have emitted orange light through the solution. the absorbance/transmittence were calculated for each standard soln until we had a line of best fit. Then the absorbance of the unknown concentration was given, so we used to line to match it to it's concentration (of alcohol). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused as to why we had to use the color orange in our emission and why a reaction was even needed in the first place. &lt;/p&gt;&#xA;" OwnerUserId="12884" LastEditorUserId="7448" LastEditDate="2015-01-21T14:57:12.243" LastActivityDate="2015-01-21T14:57:12.243" Title="Alcohol spectrophotometry" Tags="&lt;experimental-chemistry&gt;&lt;color&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23985" PostTypeId="1" AcceptedAnswerId="23986" CreationDate="2015-01-20T21:59:06.510" Score="4" ViewCount="532" Body="&lt;p&gt;I have been asked to explain the following observation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezAHS.png&quot; alt=&quot;Comparing rate of elimination of Cl from 2 dichloro bicyclic diastereomers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a number of issues with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The fast reaction must be a syn-elimination. But how can syn E2 elimination occur? If an E2 elimination is stereospecific because it must go through an anti-periplanar transition state surely syn-peri-planar elimination cannot happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How does the slow reaction even happen? The H and the Cl are never going to be in the same plane (syn or anti).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question: does the boat conformation (which the bicyclic molecule must adopt) have equatorial and axial sites like the chair conformation does?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="83758" LastEditDate="2020-05-13T03:38:19.160" LastActivityDate="2020-05-13T03:38:19.160" Title="Why does one bicyclic diastereomer react faster than the other one?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23986" PostTypeId="2" ParentId="23985" CreationDate="2015-01-20T22:22:45.463" Score="2" Body="&lt;p&gt;Yes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KqFxE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jDMu7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-01-20T22:22:45.463" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="23987" PostTypeId="1" AcceptedAnswerId="23988" CreationDate="2015-01-20T22:45:05.413" Score="4" ViewCount="2149" Body="&lt;p&gt;What does &lt;strong&gt;&quot;higher priority&quot;&lt;/strong&gt; mean exactly in the following statement from &lt;a href=&quot;http://en.wikipedia.org/wiki/E-Z_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;this wikipedia article&lt;/a&gt; (article name: &quot;E-Z notation&quot;) on naming organic compounds in IUPAC nomenclature of organic compounds?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the two groups of &lt;strong&gt;higher priority&lt;/strong&gt; are on opposite sides of the&#xA;  double bond, the bond is assigned the configuration E (from entgegen,&#xA;  German: [ɛntˈɡeːɡən], the German word for &quot;opposite&quot;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pi2LM.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/wOnMl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11402" LastActivityDate="2015-01-20T23:10:01.140" Title="What does it mean if a functional group has &quot;higher priority&quot; in terms of IUPAC naming of organic compounds?" Tags="&lt;nomenclature&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23988" PostTypeId="2" ParentId="23987" CreationDate="2015-01-20T23:10:01.140" Score="2" Body="&lt;p&gt;Whether trying to figure out if a double bond is &quot;E&quot; or &quot;Z&quot;, or if a chiral center is &quot;R&quot; or &quot;S&quot;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cahn–Ingold–Prelog priority rules&lt;/a&gt; (see the section &quot;Assignment of priorities&quot;) are used to prioritize the groups.  In the case of a double bond there are 2 substituents (a lone pair of electrons can be a substituent) at each end of the double bond.  Priorities (1 and 2) are assigned to the substituents at each end using the Cahn–Ingold–Prelog rules provided in the link above.  The first rule used to set priorities is based on the atomic number of the atoms attached to the double bond.  If there is a tie (the 2 attached atoms at one end of the double bond have the same atomic number), then we move out to the next atoms and compare these.  The process is continued until the tie is broken (eventually the tie must be broken if the molecule being analyzed is really capable of being described as &quot;E&quot; or &quot;Z&quot;).  After the priorities have been assigned, if the two priority-1 substituents at opposite ends of the double bond are on the same side of the double bond, then the molecule is the &quot;Z&quot; (zusammen, together) isomer, otherwise it is the &quot;E&quot; (entgegen, opposite) isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZwUXt.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://driverlayer.com/img/e%20z%20nomenclature/20/image?tab=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-20T23:10:01.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23989" PostTypeId="2" ParentId="23984" CreationDate="2015-01-20T23:45:48.017" Score="1" Body="&lt;p&gt;Since green is the &lt;em&gt;absorption&lt;/em&gt; of orange light, you need to supply the orange light to measure how much is absorbed. As the alcohols are oxidized to the acids, the dichromate is reduced, so you can measure how much acid has been formed. BTW, what alcohol was used? Butyl alcohol oxidation has a distinct aroma that can be detected without spectroscopy.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-20T23:45:48.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23990" PostTypeId="1" AcceptedAnswerId="24046" CreationDate="2015-01-21T00:34:40.120" Score="5" ViewCount="1108" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a &lt;em&gt;little&lt;/em&gt; knowledge of chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was talking to a sales rep from a cleaning supplies company. They sell a scrub sponge that looks and functions just like Mr. Clean Magic Eraser. The sales rep was trying to scare me by saying that magic erasers have formaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the properties of a chemical compound are different than the properties of each component. I suspected that their product is actually the same as the magic eraser so I contacted the company directly, but got a less than satisfactory answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked if their sponge was made of formaldehyde-melamine-sodium bisulfite copolymer. They did not answer yes or no. They told me their sponge is made of melamine resin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, melamine foam (a foam-like material consisting of a formaldehyde-melamine-sodium bisulfite copolymer) is a special form of melamine resin that is used mainly as an insulating and soundproofing material and more recently as a cleaning abrasive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be more knowledgeable next time this sales rep contacts me. Hence the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer? Is there a way to make a melamine resin without using formaldehyde?&lt;/p&gt;&#xA;" OwnerUserId="12885" LastActivityDate="2015-01-22T15:58:40.450" Title="Are there different types of melamine resin?" Tags="&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23992" PostTypeId="1" CreationDate="2015-01-21T02:29:33.333" Score="3" ViewCount="1971" Body="&lt;p&gt;The US defines the &quot;proof&quot; of an alcoholic beverage to be twice the alcohol per volume percentage at 60° F (about 15.5° C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone who does not have a chemistry background, I am wondering what the 60°F requirement means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a mixture of alcohol and water that is 80 proof (40% alcohol by volume at 15.5°C), what is the alcohol by volume at room temperature 20°C?&lt;/p&gt;&#xA;" OwnerUserId="12889" LastActivityDate="2015-01-21T06:38:19.823" Title="Alcoholic proof and temperature" Tags="&lt;everyday-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23993" PostTypeId="1" CreationDate="2015-01-21T04:35:27.430" Score="-2" ViewCount="1107" Body="&lt;p&gt;I know that if an atom has 4 electrons in its outer shell it will share 4 with other atoms to have a full valence shell. But what if the atom only has 3? I understand that it will likely get rid of three instead of picking up 5 more to share, but how does it get rid of the three? &#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="12892" LastActivityDate="2015-01-21T06:21:17.130" Title="How does an atom get rid of electrons?" Tags="&lt;electrons&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-27T19:26:35.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="23994" PostTypeId="2" ParentId="23993" CreationDate="2015-01-21T06:21:17.130" Score="1" Body="&lt;p&gt;Atoms with one outer electron (e.g. sodium), called alkali metals, or two outer electrons (e.g. calcium, strontium), called alkaline earths, and even the aluminum-scandium metals are not good at sharing -- they tend to lose their outer electrons given the opportunity, e.g. when a halogen is near. These form &lt;em&gt;ionic&lt;/em&gt; compounds, such as table salt, in which case the Na+ has given up one electron, which stays close to the Cl- ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements in columns closer together don't get away with outright theft, and share their outer electrons more equitably in a covalent bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it's an oversimplification, in that a bond is not necessarily fully ionic or fully covalent and may fall somewhere in between. See &lt;a href=&quot;https://www.youtube.com/watch?v=OY2_q-1t2Q8&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=OY2_q-1t2Q8&lt;/a&gt; for a video on this. &lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-21T06:21:17.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23995" PostTypeId="2" ParentId="23992" CreationDate="2015-01-21T06:30:49.730" Score="1" Body="&lt;p&gt;Briefly, since ethyl alcohol expands more rapidly with increasing temperature than does water, then measuring the mix by volume at a warmer temperature would give you less alcohol and more water, because the alcohol would be less by weight for the same volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermal Coefficient of Expansion (approximate, measured at room temperature)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohol, ethyl (ethanol):   0.00109 / K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water:&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;0.00021 / K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these constants, you can figure out the answer to your question. But note the wording: given a &lt;strong&gt;mixture&lt;/strong&gt;. If the EtOH/H2O is &lt;strong&gt;already mixed&lt;/strong&gt;, changing the temperature does &lt;strong&gt;not&lt;/strong&gt; change the concentration or proof, and you would have to separate the ingredients to measure their respective volumes. BTW the volume of the mix is slightly less than the sum of the volumes of its ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moral: if your measuring your drinks by shot-glass rather than with a scale, you'll get a weaker drink in a warm bar ;-)&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-01-21T06:38:19.823" LastActivityDate="2015-01-21T06:38:19.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="23996" PostTypeId="2" ParentId="6037" CreationDate="2015-01-21T07:32:14.287" Score="2" Body="&lt;p&gt;The temperature is highly nonlinear with sound pressure and dependent on external pressure, tensile strength of liquid etc. Cavitation, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cavitation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Cavitation&lt;/a&gt;, can produce local shock waves and temperatures of thousands of K, causing sonoluminescence. See also &lt;a href=&quot;http://www.scs.illinois.edu/suslick/sonochembrittanica.html&quot; rel=&quot;nofollow&quot;&gt;http://www.scs.illinois.edu/suslick/sonochembrittanica.html&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of that nonlinearity and all the dependencies, I do not know if any equation for calculating the peak temperatures exists, and there is still research going on to see how much of sonoluminescence is due to black-body radiation, the recombination of radicals etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, you can view a mantis shrimp with a sonoluminescent punch at &lt;a href=&quot;https://www.youtube.com/watch?v=sNzI8K-lCqo&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=sNzI8K-lCqo&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=ti2Uoc1RXuQ&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=ti2Uoc1RXuQ&lt;/a&gt;, and another species at &lt;a href=&quot;https://www.youtube.com/watch?v=QXK2G2AzMTU&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=QXK2G2AzMTU&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-01-21T07:40:08.850" LastActivityDate="2015-01-21T07:40:08.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23998" PostTypeId="5" CreationDate="2015-01-21T08:26:20.103" Score="0" Body="&lt;p&gt;Bent bonds are also known as banana bonds. These kind of bonds are usually found in small, strained, organic molecules. The simplification of the hybridisation scheme fails for this class of compounds. Therefore a distortion in the electron density can be observed. Also see a complete description of this phenomenon by K. Wiberg in &lt;a href=&quot;http://dx.doi.org/10.1021/ar950207a&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Acc. Chem. Res.&lt;/em&gt;, &lt;strong&gt;1996&lt;/strong&gt;, &lt;em&gt;29&lt;/em&gt;, 229-234&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-01-21T08:26:20.103" LastActivityDate="2015-01-21T08:26:20.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="23999" PostTypeId="4" CreationDate="2015-01-21T08:26:20.103" Score="0" Body="Also known as banana bond. These kinds of bonds are usually found in small, strained, organic molecules. The simplification of the hybridisation scheme fails for this class of compounds." OwnerUserId="4945" LastEditorUserId="402" LastEditDate="2017-05-02T15:43:10.640" LastActivityDate="2017-05-02T15:43:10.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24000" PostTypeId="5" CreationDate="2015-01-21T09:57:11.460" Score="0" Body="&lt;p&gt;A system of written symbols used to represent qualities or quantities, elements or molecules, or formulas in general. Usually a notation follows a convention or a guideline, often published or recommended by IUPAC-IUPAP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information on specifically chemical notation can be found in the &lt;a href=&quot;http://www.iupac.org/home/publications/e-resources/nomenclature-and-terminology.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronic library of the IUPAC&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.meta.stackexchange.com/q/443/5026&quot;&gt;Which symbols are written in roman (upright) font and which are italicized?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-16T18:06:31.497" LastActivityDate="2018-03-16T18:06:31.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24001" PostTypeId="4" CreationDate="2015-01-21T09:57:11.460" Score="0" Body="A system of written symbols used to represent qualities or quantities, elements or molecules, or formulas in general. Usually a notation follows a convention or a guideline, often published by IUPAC-IUPAP. This tag should be used for questions regarding the proper use of a notation or misleading or uncommon notation in the scientific literature. Do not confuse this tag with the [nomenclature] or [terminology] tags." OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2020-03-17T21:06:24.690" LastActivityDate="2020-03-17T21:06:24.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24002" PostTypeId="1" CreationDate="2015-01-21T10:19:00.383" Score="3" ViewCount="133" Body="&lt;p&gt;I have no clue whatsoever about Chemistry on a academic point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We wish to transport some goods.&#xA;Let's say we vacuum some of those goods, by doing like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We place those goods on a box, we close the box, and then we vacuum seal the box (not the goods).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we consider the goods to be vacuum protected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean, that no hight heat will affect those goods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance, and sorry for not using specific chemical terminology.&lt;/p&gt;&#xA;" OwnerUserId="12897" LastEditorUserId="418" LastEditDate="2015-01-21T12:24:41.880" LastActivityDate="2015-11-18T18:04:00.037" Title="Vacuum good inside a box?" Tags="&lt;thermodynamics&gt;&lt;food-chemistry&gt;&lt;conductivity&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24004" PostTypeId="2" ParentId="24002" CreationDate="2015-01-21T12:04:39.507" Score="3" Body="&lt;p&gt;Let's imagine that your transport box has a decently closing lid. It you place the herbs in the box, close the lid , stuff the box into a vacuum sealer bag, &lt;strong&gt;remove the air from the bag until it is shrunk around the box&lt;/strong&gt; and finally seal it, you may consider it sealed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;According to &quot;household standards&quot;, it is air-tight to the outside. No air from the environment will enter the bag, unless it is damaged.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is however still air in the box. Whatever degradation, loss of flavour, etc. may happen to the herbs during transport still can, since there is oxygen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, if you &lt;strong&gt;vaccum-seal the box&lt;/strong&gt;, or replace the air in the box with nitrogen, you will minimize (or fully prevent) degradation of the herbs.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-21T12:04:39.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24005" PostTypeId="2" ParentId="23948" CreationDate="2015-01-21T13:14:42.887" Score="4" Body="&lt;p&gt;For ice I melting pressure as a function of temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pi = 1- 0.626000 \times 10^6 (1- \theta^{-3}) + 0.197135 \times 10^6 (1- \theta^{21.2})$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pi =$ (pressure in MPa)/0.000611657 MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta =$ (temperature in kelvins)/273.16K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source is &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd477.pdf&quot; rel=&quot;nofollow&quot;&gt;International Equation of the Pressure along the Melting and along the Sublimation Curve of Ordinary Water substance&lt;/a&gt; (no paywall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For carbon dioxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{p_m}{p_n} = 1 + a_1 (\frac{T}{T_t}) + a_2 (\frac{T}{T_t})^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ is temperature in kelvins&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_t =$ 216.592K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_m$ is the melting pressure in MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_n =$ 0.51795 MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_1 =$ 1955.5390&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_2 =$ 2055.4593&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source is &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd516.pdf&quot; rel=&quot;nofollow&quot;&gt;A New Equation of State for Carbon Dioxide...&lt;/a&gt; (no paywall)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check original sources before using equations in case I made typo(s). &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-21T13:31:14.440" LastActivityDate="2015-01-21T13:31:14.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24006" PostTypeId="1" AcceptedAnswerId="24012" CreationDate="2015-01-21T14:36:23.330" Score="2" ViewCount="110" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given three solvents X, Y, Z. Order of boiling point of pure solvent X &amp;lt; Y &amp;lt; Z and same order of molal elevation constant ($K_\mathrm{b}$). Find the order of molecular mass.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I got this question in my tuition test and the solution is unclear. This question is troubling me a lot.&lt;/p&gt;&#xA;" OwnerUserId="12905" LastEditorUserId="41328" LastEditDate="2017-09-18T05:00:56.373" LastActivityDate="2017-09-18T05:00:56.373" Title="Order of molecular mass of solvent; given boiling point" Tags="&lt;solutions&gt;&lt;solvents&gt;&lt;colligative-properties&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-01-21T20:52:12.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="24007" PostTypeId="1" AcceptedAnswerId="24011" CreationDate="2015-01-21T14:50:23.740" Score="4" ViewCount="3861" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Pencils create marks by physical abrasion, leaving behind a trail of solid core material that &lt;strong&gt;&lt;em&gt;adheres&lt;/em&gt;&lt;/strong&gt; to a sheet of paper or other surface. &lt;a href=&quot;http://en.wikipedia.org/wiki/Pencil&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Normal paper is mainly composed of cellulose fibers, which were &quot;funded&quot; by the plants. And this &quot;solid core material&quot; is supposedly composed of graphite and a type of binder clay.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0tQvQ.jpg&quot; alt=&quot;Paper&quot;&gt;&#xA;&lt;a href=&quot;http://www.chm.bris.ac.uk/emuweb/schools.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Adhering&quot; is usually caused by intermolecular forces. (e.g.: when sugar adheres to wet surfaces because of Hydrogen bonding). The questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the case of a pencil mark some kind of intermolecular force too? &lt;/li&gt;&#xA;&lt;li&gt;If so, what is it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-01-21T15:10:31.860" LastActivityDate="2018-05-15T01:11:38.587" Title="How are pencil marks &quot;adhered&quot; to the paper?" Tags="&lt;everyday-chemistry&gt;&lt;intermolecular-forces&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24008" PostTypeId="1" AcceptedAnswerId="24010" CreationDate="2015-01-21T15:25:30.480" Score="2" ViewCount="13491" Body="&lt;p&gt;The following is an extract from my book-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For an aldehyde or a ketone to exhibit keto-enol tautomerism, it is essential that it must have atleast one $\alpha$-hydrogen atom. Thus, acetophenone, butan-2-one and propionaldehyde all contain $\alpha$- hydrogen atoms and hence show keto-enol tautomerism. In contrast, benzaldehyde, benzophenone etc. do not contain $\alpha$-hydrogen atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is meant by $\alpha$-hydrogen atom? How does it affect keto-enol tautomerism?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="7448" LastEditDate="2015-01-21T16:00:13.107" LastActivityDate="2015-01-21T16:03:37.337" Title="alpha-hydrogen atom and keto-enol tautomerism" Tags="&lt;organic-chemistry&gt;&lt;hydrogen&gt;&lt;tautomer&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24009" PostTypeId="2" ParentId="24008" CreationDate="2015-01-21T16:00:05.683" Score="0" Body="&lt;p&gt;An $\alpha$ hydrogen atom is bound to a carbon next to some functional group, e.g. a ketone or an aldehyde. Without this hydrogen there is no hydrogen that can be removed by enolization. Other hydrogen atoms on the carbon chain are much less acidic due to the lack of an electron stabilizing group such as a keto-group. Benzaldehyde contain no $\alpha$ hydrogen atoms because of the $\ce{sp^2}$ hybridized carbon atom. On the other hand, acetone for example has 6 $\alpha$ hydrogen atoms.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-21T16:00:05.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24010" PostTypeId="2" ParentId="24008" CreationDate="2015-01-21T16:03:37.337" Score="3" Body="&lt;p&gt;In general, the &quot;alpha&quot; hydrogen is a hydrogen attached to a carbon that is &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_and_beta_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;alpha&quot; (adjacent to) the substituent&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6NXY0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.meritnation.com/img/shared/discuss_editlive/4255118/2013_10_30_17_50_50/zhcgfjcy4021270914596835552.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, an &quot;alpha&quot; hydrogen is a hydrogen attached to the carbon alpha to the carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8rSmx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_and_beta_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following drawing represents a typical &lt;a href=&quot;http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism&quot; rel=&quot;nofollow noreferrer&quot;&gt;keto-enol equilibrium&lt;/a&gt;.  If we removed the last alpha hydrogen and replaced it with a substituent (alkyl group, phenyl group, bromine, etc.) then we could not form a double bond with that alpha carbon, it would have 4 non-removable substituents.  If we can't form a double bond that means we can't form an enol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ozB3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-21T16:03:37.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24011" PostTypeId="2" ParentId="24007" CreationDate="2015-01-21T16:04:39.207" Score="4" Body="&lt;p&gt;The graphite itself can adhere to paper fibers with &lt;a href=&quot;http://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;London forces&lt;/a&gt;; it's not a strong adhesion because the marks can be erased easily. Some of the graphite is just mechanically stuck between the paper fibers; rubbing a mark from a soft pencil can easily blur it. Enough graphite is deposited that &lt;a href=&quot;http://pubs.acs.org/cen/whatstuff/print/7942sci4.html&quot; rel=&quot;nofollow&quot;&gt;a line drawn by a pencil can conduct electricity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clay gives the lead structural integrity; an &quot;H&quot; pencil has a harder lead, and deposits less graphite; a &quot;B&quot; pencil is softer, and an &quot;F&quot; pencil can be sharpened to a finer point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my time as an art major, I made a lot of sketches using vine charcoal on newsprint. Vine charcoal has no binding clay, and the marks were easily smeared. On paper without much &quot;bite&quot;, it was easy to lift the marks off completely with a piece of kneaded rubber, and if you wanted to keep a charcoal drawing (or a soft pencil drawing) you really had to spray it with a fixative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say that it's possible that the binding clay also plays some role in binding the graphite or charcoal to the paper. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the manufacture of graphite pencils, air-classified grades of Volclay sodium &#xA;  bentonite and micronized hectorite are regularly used to bind the graphite &#xA;  compound in pencil lead. --- &lt;a href=&quot;http://www.colloid.com/isg/Application-IndustrialAndHousehold.aspx&quot; rel=&quot;nofollow&quot;&gt;American Colloid Company, Industrial and Household Applications&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bentonite and hectorite both have ionic surfaces that probably help bind it to OH groups on the surface of cellulose fibers.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-21T19:32:58.747" LastActivityDate="2015-01-21T19:32:58.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24012" PostTypeId="2" ParentId="24006" CreationDate="2015-01-21T16:20:14.307" Score="5" Body="&lt;p&gt;I can see how this question is troubling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It don't think there is a solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_b = \frac{RT_b^2M_w}{1000\Delta H_v}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T_b$ is boiling point, $\Delta H_v$ is the molar heat of vaporization, and $M_w$ is molecular weight of the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;http://www.chem.arizona.edu/~salzmanr/480a/480ants/colprop/colprop.html&quot; rel=&quot;noreferrer&quot;&gt;Colligative Properties by W. R. Salzman&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that knowing $K_b$ and $T_b$ is insufficient to find $M_w$, knowledge of $\Delta H_v$ is also required.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, from &lt;a href=&quot;http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf&quot; rel=&quot;noreferrer&quot;&gt;Table of cryoscopic and ebullioscopic constants&lt;/a&gt; you can also see there is no solution.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-21T16:29:05.487" LastActivityDate="2015-01-21T16:29:05.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24013" PostTypeId="2" ParentId="23949" CreationDate="2015-01-21T17:46:14.030" Score="3" Body="&lt;p&gt;You are right! There is no molecular geometry in any chart that has two lone pairs and one bond, because the only possible geometry with one neighboring atom is linear.&#xA;The central atom in a chemical moiety &lt;strong&gt;must have at least two neighboring atoms&lt;/strong&gt; in order to predict the geometry around this central atom (according to VSEPR model).&#xA;Let's take the case of oxygen atom of the hydroxyl group (in Ibuprofen). The oxygen atom here, as a central atom,  has the steric number: 4 ( It has two lone pairs and two simple bonds with carbon and hydrogen atoms). So, The molecular geometry around oxygen atom is bent. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-01-21T17:46:14.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24014" PostTypeId="1" AcceptedAnswerId="24017" CreationDate="2015-01-21T18:41:47.450" Score="6" ViewCount="14654" Body="&lt;p&gt;I came across the fact that alcohols cannot turn blue litmus red. The fact confuses me as they are acidic and acids do turn blue litmus solution red. What prevents alcohols from turning blue litmus solution red?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="7448" LastEditDate="2015-01-21T19:30:05.693" LastActivityDate="2018-11-26T15:35:25.240" Title="Why can't alcohols turn blue litmus red?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;alcohols&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24015" PostTypeId="1" CreationDate="2015-01-21T18:58:51.767" Score="3" ViewCount="283" Body="&lt;p&gt;The are various types of fire extinguishers. They can be based on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;water&lt;/li&gt;&#xA;&lt;li&gt;foam&lt;/li&gt;&#xA;&lt;li&gt;dry powder&lt;/li&gt;&#xA;&lt;li&gt;carbon dioxide&lt;/li&gt;&#xA;&lt;li&gt;wet chemicals&lt;/li&gt;&#xA;&lt;li&gt;halon&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Suppose one is learning chemistry at home. What type of fire extinguisher one should ideally get? It seems the dry powder option is the cheapest ($\approx 30€$), foam mid-priced ($\approx~60€$) and the one using carbon dioxide the most expensive ($\approx~150€$). I did not find solutions based on water/wet chemicals/halon yet, though I know such exist. If there's no differences in achieved safety, I would probably pick the dry powder one.&lt;/p&gt;&#xA;" OwnerUserId="12913" LastEditorUserId="7448" LastEditDate="2015-01-21T19:08:12.477" LastActivityDate="2015-01-22T17:26:44.373" Title="What kind of fire extinguisher for chemistry at home?" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24016" PostTypeId="2" ParentId="24015" CreationDate="2015-01-21T20:20:49.343" Score="5" Body="&lt;p&gt;You need to investigate the safety and handling of every chemical you use in your experiments individually. The handling and safety precautions can be very different depending on what substances exactly you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of fire, the one thing you must know before each experiment is which methods won't work or even fail catastrophically. Trying to extinguish a sodium fire with water is a really, really bad idea, and that is something you must know in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal fires are the most common kind of fire where you have to be careful about choosing the right method. Some metal fires can't even be extinguished by carbon dioxide. There are special fire extinguishers for this (class D), other methods might not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bucket of sand is the most useful method for small fires in a lab. It works for almost everything and doesn't cause a big mess. But it is obviously useless if you cause a bigger fire and can't reliably cover it with sand anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powder extinguishers cause quite a mess, you'll spend a lot of time cleaning up the room afterwards. Obviously this is a lot better than burning down your home, but I wouldn't use them for small fires you could stop with some sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foam and powder are probably the most useful general-purpose extinguishers. Carbon dioxide is a bit trickier to use from what I heard, I probably wouldn't recommend it in your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about a fire extinguisher is a good first step. Another thing I very strongly recommend is that you read up on general safety procedures in the lab. Simple things like not storing large amounts of solvent directly in your fume hood (or workplace) might prevent a small fire from becoming a dangerously large one. Knowing that you need to inactivate any reactive reagents before disposing them might save you from throwing away some oxidizing agent and accidentally setting your waste on fire. There is a lot more to chemical safety than just choosing the right fire extinguisher.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-01-21T20:20:49.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24017" PostTypeId="2" ParentId="24014" CreationDate="2015-01-21T21:17:17.143" Score="14" Body="&lt;p&gt;Litmus is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ of about 6.5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alcohols are &lt;em&gt;extremely&lt;/em&gt; weak acids, with $\mathrm{p}K_\mathrm{a}$'s (typically) around 17. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="16683" LastEditDate="2017-01-22T16:01:39.640" LastActivityDate="2017-01-22T16:01:39.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24018" PostTypeId="1" AcceptedAnswerId="24019" CreationDate="2015-01-21T21:31:22.563" Score="16" ViewCount="1159" Body="&lt;p&gt;I experienced a situation where some electrical power supplies (rated over 1kW) were exposed to nitric acid ($\mathrm{HNO_3}$) fumes at roughly 9ppm for half a day; these power supplies intermittently failed (perhaps 2 to 3 times per hour) in the presence of the fumes.  Replacing the power supplies restored proper functionality to the equipment they were powering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, I thought that perhaps there was arcing inside the power supply; however, &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitric_acid&quot;&gt;nitric acid doesn't seem to have a flash point&lt;/a&gt;.  I'm trying to come up with a reasonable explanation for why these particular components would fail.  Maybe the fumes increased atmospheric conductivity, which shorted across components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't a compelling reason from a chemistry perspective, that's also a fair response.&lt;/p&gt;&#xA;" OwnerUserId="2152" LastEditorUserId="12660" LastEditDate="2015-01-22T05:09:11.787" LastActivityDate="2015-07-13T13:00:29.130" Title="Why would 1kW power supplies fail around Nitric acid (HNO3) gas?" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24019" PostTypeId="2" ParentId="24018" CreationDate="2015-01-21T21:38:34.383" Score="22" Body="&lt;p&gt;Nitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment.&#xA;This happens with dilute nitric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3 Cu + 8 HNO3 -&amp;gt; 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;: Here are some quantitative corrosion rates:&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1002/9783527610433.chb02210&quot;&gt;Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 2&lt;/strong&gt;: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to &lt;a href=&quot;http://www.jmcproducts.com/wp-content/uploads/2012/02/comparative_analysis.pdf&quot;&gt;this reference&lt;/a&gt; you need an airflow rate of about 53 cubic feet per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 1.5 cubic meters of air per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1.5~\mathrm{m^3}\ \text{air}}{1~\mathrm{min}} \frac{1.177~\mathrm{kg}\ \text{air}}{1~\mathrm{m^3}\ \text{air}} \frac{9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{kg}\ \text{air}} = 15.9~\frac{\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which over 12 hours is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{15.9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}} \frac{60~\mathrm{min}}{1~\mathrm{hour}} (12~\mathrm{hours}) = 11.4~\mathrm{g}\ \ce{HNO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;em&gt;all&lt;/em&gt; of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$11.4~\mathrm{g}\ \ce{HNO3}\frac{1~\mathrm{mol}\ \ce{HNO3}}{63.01~\mathrm{g}\ \ce{HNO3}}\frac{3~\mathrm{mol}\ \ce{Cu}}{8~\mathrm{mol}\ \ce{HNO3}}\frac{63.546~\mathrm{g}\ \ce{Cu}}{1~\mathrm{mol}\ \ce{Cu}} = \fbox{4 g Cu }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-22T05:11:50.907" LastActivityDate="2015-01-22T05:11:50.907" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24020" PostTypeId="2" ParentId="23976" CreationDate="2015-01-21T23:18:18.170" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why? [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They hopefully didn't write about the &quot;concentration of isopropanol&quot; but mentioned that a larger amount of the solvent has to be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not uncommon that organic compounds with polar substituents are less soluble in higher alkanols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vanillin is a typical example, the solubilities are in the range of 4.15 mol/L in methanol, 2.5 mol/L in ethanol and 1.8 mol/L in 1-propanol. (Data are taken from a freely accessible collection of experiments performed by students: &lt;a href=&quot;http://onschallenge.wikispaces.com/list+of+experiments&quot; rel=&quot;nofollow&quot;&gt;OpenNotebook Science Challenge&lt;/a&gt;). I couldn't find any data for 2-propanol. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, not really. A ketone and a secondary alkanol, although looking similar on first sight, may have behave completely different as solvents. Polystyrene (foam), as an example is pretty much insoluble in isopropanol but highly soluble in acetone!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-21T23:18:18.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24022" PostTypeId="2" ParentId="23945" CreationDate="2015-01-22T00:26:27.160" Score="2" Body="&lt;p&gt;It seems you were on right path - for example common &lt;a href=&quot;https://en.wikipedia.org/wiki/Feldspar&quot; rel=&quot;nofollow&quot;&gt;feldspars&lt;/a&gt; are &lt;a href=&quot;http://%20en.wikipedia.org/wiki/Solid_solution&quot; rel=&quot;nofollow&quot;&gt;solid solutions&lt;/a&gt; and they are mixed because they share cations, while they easily substitute one another, otherwise they exsolve - become separate phases. Also you're right that &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;van der Waals forces&lt;/a&gt; aren't important here as they are ionic compounds.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-02-04T00:11:02.743" LastActivityDate="2016-02-04T00:11:02.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24024" PostTypeId="2" ParentId="19876" CreationDate="2015-01-22T02:03:38.397" Score="0" Body="&lt;p&gt;One way of thinking about this is to consider what happens when you clean a messy room. Your room is an example of an open system (i.e. heat, work and mass can move across its boundary). If you put a lot of work into your room, in the form of good old fashioned elbow grease, you end up with a nice, clean room that has a lower entropy than what it started with. However, it is important to note that the entropy of the universe, which is an isolated system that happens to contain the open system of your room, will have increased as a result of your work. This is because you could not do the work in a reversible fashion. Your body heat, friction from moving things around, etc. all produced irreversible changes in the heat flow to your surroundings. This is where the Clausius equation holds. To describe the entropy decrease in your room you have to know the equation of state that can provide the initial and final entropies. Not really practical for an abstract concept for a room, but for a fluid such as a gas or liquid it's usually clear-cut.&lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-22T02:03:38.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24025" PostTypeId="1" CreationDate="2015-01-22T02:55:06.410" Score="5" ViewCount="902" Body="&lt;p&gt;When we add vegetables to hot water (boiling), it boils the vegetables, but when we add vegetables to hot oil (boiling) it fries the vegetables. Why? &lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="4945" LastEditDate="2017-11-29T03:45:35.267" LastActivityDate="2017-11-29T03:45:35.267" Title="Why does water boil vegetables, but oil does not?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24026" PostTypeId="2" ParentId="24025" CreationDate="2015-01-22T03:00:54.170" Score="8" Body="&lt;p&gt;The maximum temperature for water boiling is around 100 Celsius and there is a great deal of water present.  Oil can reach greater temperature and the absence of water allows other reactions, such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Maillard_reaction&quot; rel=&quot;nofollow&quot;&gt;Maillard reaction&lt;/a&gt;, to occur.  That's why you never see browning in boiled food.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="7448" LastEditDate="2015-01-22T13:12:34.133" LastActivityDate="2015-01-22T13:12:34.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24027" PostTypeId="2" ParentId="23702" CreationDate="2015-01-22T03:31:22.170" Score="1" Body="&lt;p&gt;You are on the right track, but if you google &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=terephthalic%20acid&quot; rel=&quot;nofollow&quot;&gt;terephthalic acid&lt;/a&gt;, you will know that your formula is not correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One mistake you are making is using the final mass of the $\ce{CaCl2}$ and $\ce{NaOH}$ and not the change in mass. For example, let's take the $\ce{CaCl2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we don't know what the initial mass of $\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products. &#xA;$$\ce{C}_x\ce{ H}_y\ce{ O}_z\ce{ +}\left(\frac{4x+2y-z}{4}\right)\ce{ O2 -&amp;gt; }x\ce{CO2 +}\frac{y}{2}\ce{ H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass of water produced $\ce{-&amp;gt;}$ moles of water produced $\ce{-&amp;gt;}$ moles of hydrogen in the water produced (and in the original sample!) $\ce{-&amp;gt;}$ grams of hydrogen (in sample) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass of carbon dioxide produced $\ce{-&amp;gt;}$ mass of carbon (in sample)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass of oxygen in sample (by subtraction) $\ce{-&amp;gt;}$ moles of oxygen in sample &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You are doing everything correctly after this.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-22T03:31:22.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24028" PostTypeId="2" ParentId="23702" CreationDate="2015-01-22T03:53:34.240" Score="1" Body="&lt;p&gt;I am afraid I could not follow your reasoning and unfortunately can not pinpoint you to your error therefore. The result you obtain, does not resemble the empirical formula of &lt;a href=&quot;http://en.wikipedia.org/wiki/Terephthalic_acid&quot; rel=&quot;nofollow&quot;&gt;terephthalic acid&lt;/a&gt;. As a general tip: These problems are of entirely mathematical nature, so it is important to use proper equations and keep the units at all times. I am going to tackle this problem from beginning to end.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all it is important to formulate the correct reaction equation:&#xA;$$\ce{C_{$x$}H_{$y$}O_{$z$} + $\frac{\left(2x+\frac{y}{2}-z\right)}{2}$\,O2 -&amp;gt; $x$\,CO2 + $\frac{y}{2}$\,H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information about calcium chloride and sodium hydroxide is only a way to tell you the mass of the product. It is completely irrelevant to the question itself. However, the correct equation would then be:&#xA;$$\ce{C_{$x$}H_{$y$}O_{$z$} + $\frac{\left(2x+\frac{y}{2}-z\right)}{2}$\,O2 -&amp;gt;[\ce{CaCl2, NaOH}] $x$\,NaHCO3 + $\frac{y}{2n}$\,CaCl2.$n$\,H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, we need to gather all the information we are given in the question or that we know otherwise:&#xA;\begin{align}&#xA;m(\ce{C_{$x$}H_{$y$}O_{$z$}}) &amp;amp;= 0.4953~\mathrm{g} &amp;amp;M(\ce{C_{$x$}H_{$y$}O_{$z$}}) &amp;amp;= ?\\&#xA;m(\ce{H2O}) &amp;amp;= 0.1613~\mathrm{g} &amp;amp; M(\ce{H2O}) &amp;amp;= 18~\mathrm{\frac{g}{mol}}\\&#xA;m(\ce{CO2}) &amp;amp;= 1.0495~\mathrm{g} &amp;amp; M(\ce{CO2}) &amp;amp;= 44~\mathrm{\frac{g}{mol}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can go ahead and calculate the number of moles of carbon dioxide and water produced. We therefore also know the relative number of moles of carbon and hydrogen in terephthalic acid. The basic formula that needs to be applied here is $$n=\frac{m}{M}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many moles of carbon are in carbon dioxide, what is the mathematical relationship? What is it for hydrogen and water?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{C}) : n(\ce{CO2}) = 1$&lt;br&gt;&#xA; $n(\ce{H}) : n(\ce{H2O}) = 2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How many moles of carbon dioxide and water have been produced? How does that relate to carbon and hydrogen?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{CO2})=\frac{m(\ce{CO2})}{M(\ce{CO2})}= 0.0239~\mathrm{mol} \therefore n(\ce{C})= 0.0239~\mathrm{mol}$&lt;br&gt;&#xA; $n(\ce{H2O})=\frac{m(\ce{H2O})}{M(\ce{H2O})}= 0.0090~\mathrm{mol} \therefore n(\ce{H})= 0.0180~\mathrm{mol}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Calculate the mass of hydrogen and carbon in the original sample with&#xA;$$m=n\cdot M$$ and use it to find the mass of oxygen in the sample.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $m(\ce{C}) = 0.2868~\mathrm{g}$&lt;br&gt;&#xA; $m(\ce{H}) = 0.0180~\mathrm{g}$&lt;br&gt;&#xA; $m(\ce{O}) = m(\ce{C_{$x$}H_{$y$}O_{$z$}}) - m(\ce{C}) - m(\ce{H}) = 0.1906~\mathrm{g}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How many moles of oxygen are present in the original sample?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{O}) = 0.119~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you only have to relate the number of moles for each element to integer ratios and you will find the empirical formula.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $x:y:z = n(\ce{C}) : n(\ce{H}) : n(\ce{O}) = 0.0239~\mathrm{mol} : 0.0180~\mathrm{mol} : 0.119~\mathrm{mol}$&lt;br&gt;&#xA; $x:y:z = 4:3:2$&lt;br&gt;&#xA; Therefore the empirical formula is $\ce{C4H3O2}$, which is half of the sum formula $\ce{C8H6O4}$, see wikipedia for more information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-01-22T03:53:34.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24029" PostTypeId="1" CreationDate="2015-01-22T04:36:57.073" Score="7" ViewCount="5976" Body="&lt;p&gt;I need to find the heat lost of an unknown metal dropped into a calorimeter with $70~\mathrm{g}$ $\ce{H2O}$.&lt;br&gt;&#xA;The initial temperature and final temperature of the $70~\mathrm{g}$ $\ce{H2O}$ and the calorimeter are $21~^\circ\mathrm{C}$ and $34~^\circ\mathrm{C}$. I already know that the heat gained by the water is $3807.44~\mathrm{J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal's starting temperature and mass are $100~^\circ\mathrm{C}$ and $180.45~\mathrm{g}$, but that didn't help me much as I don't know the $C_{sp}$ of the unknown metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't know is how to find the capacity of the calorimeter, any thoughts?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already tried a number of solutions that didn't work:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$C_{sp} \times \ce{H2O} = q(\mathrm{calorimeter})$  &lt;/li&gt;&#xA;&lt;li&gt;$q(\ce{H2O})/m(\ce{H2O}) \times \Delta T = q(\mathrm{calorimeter})$   &lt;/li&gt;&#xA;&lt;li&gt;$q(\ce{H2O}) = q(\mathrm{calorimeter})$  &lt;/li&gt;&#xA;&lt;li&gt;I knew the unknown metal was one of three metals (lead, aluminum, copper),     so I tried finding $q(\mathrm{metal})$ using the 3 heat specific heat capacities, but it didn't work out since the data provided were only approximations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12918" LastEditorUserId="46306" LastEditDate="2018-03-03T05:03:10.640" LastActivityDate="2018-10-31T20:23:22.213" Title="How to calculate the heat capacity of a calorimeter?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24031" PostTypeId="1" AcceptedAnswerId="24032" CreationDate="2015-01-22T03:22:26.977" Score="4" ViewCount="709" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a Dumas bulb a volatile substance is introduced. After a few minutes when the liquid has evaporated, the bulb is sealed. It is known that the initial weight of the bulb with air is $12.0468~\mathrm{g}$, $12.4528~\mathrm{g}$ with the volatile substance and $350.6264~\mathrm{g}$ with water. Calculate the molar mass of the substance if all these measurements were done at $25~^\circ\mathrm{C}$ and $1~\mathrm{atm}$ of pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is $29~\mathrm{g/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've done is the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've constructed a system of two equations to know the mass of the bulb and the volume it can contain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_b+\rho_\mathrm{air}V = 12.0468~\mathrm{g}$&lt;br&gt;&#xA;$m_b+\rho_\mathrm{water}V = 350.6264~\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions are $m_b=11.6144~\mathrm{g}$ and $V=0.339012~\mathrm{m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this I can find the molar mass of the volatile substance by knowing that its true mass was $m=0.8384~\mathrm{g}$ and the volume it occupied was $0.339~\mathrm{m^3}$ and using the ideal gas equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal gas law, can be used to say&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M=\frac{m}{pV}RT=0.060~\mathrm{g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is wrong. I've done all sorts of other things and can't get $29~\mathrm{g/mol}$. Maybe this is the craziest thing I've done until now.&lt;/p&gt;&#xA;" OwnerUserId="12920" OwnerDisplayName="ben ari" LastEditorUserId="46306" LastEditDate="2018-12-10T17:30:46.940" LastActivityDate="2018-12-10T17:30:46.940" Title="How to calculate the molecular weight for a volatile substance introduced into a Dumas bulb?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24032" PostTypeId="2" ParentId="24031" CreationDate="2015-01-22T04:28:13.080" Score="1" Body="&lt;p&gt;There are a couple of things wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the mass of the volatile gas is not $0.8384\ \mathrm g$. It's the difference between $12.4528\ \mathrm g$ and $12.0468\ \mathrm g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the volume is incorrect. $11.6144\ \mathrm g + 999.97\ \mathrm{kg/m^3} \times 0.339012\ \mathrm{m^3} \ne 350.6264\ \mathrm g$&lt;/p&gt;&#xA;" OwnerUserId="4934" OwnerDisplayName="LDC3" LastEditorDisplayName="user7951" LastEditDate="2016-12-28T00:01:27.033" LastActivityDate="2016-12-28T00:01:27.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24034" PostTypeId="1" AcceptedAnswerId="24038" CreationDate="2015-01-22T08:02:44.763" Score="6" ViewCount="885" Body="&lt;p&gt;&quot;&lt;strong&gt;Gibbs Energy&lt;/strong&gt; is the &lt;em&gt;useful work&lt;/em&gt; that can be extracted from the heat of a reaction or a thermodynamic process.&quot; I understand how it predicts the feasibility of a chemical reaction , considering the &lt;em&gt;Entropy statement of the Second Law&lt;/em&gt;. But how does it relate with the &lt;em&gt;useful work&lt;/em&gt; that can be extracted? (This definition is implicitly used in my textbook in deriving from &lt;em&gt;Nernst Equation&lt;/em&gt;, the useful work that can be extracted from an electrochemical cell and I am missing the point why this actually works.) Is there an intuitive explanation? Also if this much of energy is &quot;&lt;em&gt;free&lt;/em&gt;&quot;, what happens to the energy that is &quot;&lt;em&gt;not free&lt;/em&gt;&quot;? Why can't we extract the whole energy that is released?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also I am interested to know if there is any direct relation between &lt;strong&gt;Gibbs Function&lt;/strong&gt; and &lt;strong&gt;Kelvin-Planck Statement&lt;/strong&gt;, which apparently expresses the inherent inability to convert the whole heat to useful work, or am I mixing up two very different aspects?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12922" LastEditorUserId="62732" LastEditDate="2018-11-15T16:31:55.907" LastActivityDate="2018-11-15T16:31:55.907" Title="Gibbs Free Energy : What is it trying to say actually?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24035" PostTypeId="1" AcceptedAnswerId="24041" CreationDate="2015-01-22T10:34:06.910" Score="20" ViewCount="71817" Body="&lt;p&gt;Rotten potatoes have a distinctive unpleasant odour, and I can't find any definitive answer as to what the chemical involved is. I was thinking it might be Butyric acid, from the fermentation of the starch, but according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Butyric_acid&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Butyric acid is present in, and is the main distinctive smell of, human vomit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And those puppies don't smell like vomit. They're more like rotting flesh – the reason I'm asking is that I've spent half the day looking for a dead animal that the cat brought in or something, only to discover that the horrible smell was coming from the potato box.&lt;/p&gt;&#xA;" OwnerUserId="12923" LastEditorDisplayName="user7951" LastEditDate="2019-06-02T08:14:53.577" LastActivityDate="2020-04-28T09:20:23.123" Title="What causes the really bad smell of rotten potatoes?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24036" PostTypeId="1" AcceptedAnswerId="24039" CreationDate="2015-01-22T11:49:26.007" Score="5" ViewCount="36300" Body="&lt;p&gt;I have just minutes ago been splashed with industrial grade methanol. It seemed like a small amount (1-3mL) on my skin. I already washed the affected area thoroughly with soap and water. Online I've reviewed SDS's for methanol and they all seem to say that immediate medical attention is needed, and that the person exposed can be asymptomatic for a while. Does a small splash like that warrant medical evaluation?&lt;/p&gt;&#xA;" OwnerUserId="7122" LastActivityDate="2015-12-04T13:12:34.847" Title="Methanol skin exposure serious or not?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-01-22T18:03:06.160" ContentLicense="CC BY-SA 3.0" />
  <row Id="24037" PostTypeId="2" ParentId="24036" CreationDate="2015-01-22T12:20:46.170" Score="1" Body="&lt;p&gt;For any poisonous material we have special terms defined. In your case biochemists defined &lt;a href=&quot;http://en.wikipedia.org/wiki/Permissible_exposure_limit&quot; rel=&quot;nofollow&quot;&gt;PEL&lt;/a&gt; (permissible exposure limit). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PEL is mainly important when it comes to deciding about for example &quot;will the mineral water of that source be applicable for everyday-use of humans?&quot; Many things affect the PEL for humans; such as age (kid, adult, etc.), type of exposure (ingestion, dermal exposure, breathing etc.), the medical situation of the exposed (other ailments or symptoms that can aggravate the harmful effect of a special kind of chemical) and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your profile says, you're in your thirties, and thus are considered an adult. According to &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safe-Handling/Methanol-Health-Effects.aspx&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; an ingestion of up to 500 milligrams in an adults diet is no problem, among other trivia. And according to &lt;a href=&quot;http://www.cdc.gov/niosh/docs/81-123/pdfs/0397.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, (in which you're considered to be supposedly exposed short-termly) if you didn't have headaches there couldn't be any problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Addendum &lt;/b&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were exposed to methanol in the dermal way, not ingestion. As a result, as the second reference mentions, the liver &lt;em&gt;will&lt;/em&gt; be able to turn methanol into formaldehyde and then formate. Formate is toxic. As some of this methanol will evaporate and the rest will eventually be absorbed by the skin you might not be in serious trouble. However, a specialist's advice is the best in this case. Human biology isn't really something anyone will be able to give statements about without empirical observation.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-12-04T13:12:34.847" LastActivityDate="2015-12-04T13:12:34.847" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24038" PostTypeId="2" ParentId="24034" CreationDate="2015-01-22T12:32:19.527" Score="6" Body="&lt;p&gt;You can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\Delta G = w_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be &quot;lost&quot; as heat to basically account for the observation that is essentially stated with the Clausius inequality $\Delta S \geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\Delta S = \frac{q_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-22T12:32:19.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24039" PostTypeId="2" ParentId="24036" CreationDate="2015-01-22T12:37:41.573" Score="7" Body="&lt;p&gt;&quot;[A] factory worker who spilled a gallon of methanol down his trouser leg, was dizzy on the following day, took a short nap, and woke up total blind&quot; &lt;a href=&quot;http://thetruthaboutstuff.com/pdf/(116)%20Cooper%201962%20Biochemical%20Aspects%20of%20Methano%20Poisoning.pdf&quot;&gt;Biochemical aspects of Methanol Poisoning&lt;/a&gt;, Biochemical Pharmacology, vol. 11, page 405.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So spilling methanol on your skin is serious, but the amount in your case is relatively small.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article also explains that ingestion of as little as &quot;a teaspoon&quot; (5 mL) has caused permanent blindness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the internet is no place to get medical advice.  If you are concerned you should call your local poison control center.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-22T12:54:23.853" LastActivityDate="2015-01-22T12:54:23.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24040" PostTypeId="2" ParentId="24015" CreationDate="2015-01-22T13:11:32.013" Score="1" Body="&lt;p&gt;&lt;em&gt;Admittedly, I'm not a big fan of experimenting at home - particularly when (larger volumes of) flammable liquids are involved.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Home labs typically lack huge sinks with faucets high and huge enough to get a face under it. Neither do they have a fire shower at the exits, which is standard in German university labs. This aside and adding to the answer given by &lt;strong&gt;Mad Scientist&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foam extinguishers work well even in the rare case when a person is on fire. This does however require a second person to operate it! With respect to this and other scenarios, where pain and panic prevents you from treating a mishap reasonably (acid and alkaline spills in the face, severe cuts due to breaking glas, strong burns, etc.) the golden rule for students in university labs therefore is: &lt;strong&gt;NEVER work alone!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-22T13:11:32.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24041" PostTypeId="2" ParentId="24035" CreationDate="2015-01-22T13:46:44.897" Score="22" Body="&lt;p&gt;My first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan,  dimethyl sulfide, and dimethyl trisulfide. Check it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mYxqU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: A. Kamiya, Y. Ose, &lt;a href=&quot;https://www.jstage.jst.go.jp/article/taiki1978/18/5/18_5_453/_pdf&quot; rel=&quot;noreferrer&quot;&gt;Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste&lt;/a&gt;, Journal of Japan Society of Air Pollution, 18(5), 1983, pp  453-463.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; &lt;a href=&quot;http://en.wikipedia.org/wiki/Solanine&quot; rel=&quot;noreferrer&quot;&gt;solanine is a toxic glycoalkaloid&lt;/a&gt;, is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methanethiol&quot; rel=&quot;noreferrer&quot;&gt;Methyl mercaptan&lt;/a&gt; ($\rm CH_3SH$) has an odor described as &quot;rotting cabbage&quot; by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, &quot;flatus&quot; if we're being polite). It has an odor detection threshold as low as 1 ppb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethyl_sulfide&quot; rel=&quot;noreferrer&quot;&gt;Dimethyl sulfide&lt;/a&gt; ($\rm (CH_3)_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Dimethyl_trisulfide&quot; rel=&quot;noreferrer&quot;&gt;Dimethyl trisulfide&lt;/a&gt; ($\rm CH_3SSSCH_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gases produced by rotting potatoes are quite toxic (see &lt;a href=&quot;http://www.upi.com/Top_News/World-News/2013/08/23/Rotting-potatoes-in-basement-kill-four-members-of-Russian-family/54831377268770/&quot; rel=&quot;noreferrer&quot;&gt;Rotting potatoes in basement kill four members of Russian family&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To keep the notes below in context, in my original answer I said that a plot point in &quot;The Walking Dead&quot; was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C4PQH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-24T02:00:02.340" LastActivityDate="2015-01-24T02:00:02.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24042" PostTypeId="1" CreationDate="2015-01-22T13:55:50.323" Score="2" ViewCount="78" Body="&lt;p&gt;Could you advice an electrode for electrophoretic deposition in methanol which won't spoil methanol and also wont be spoiled by methanol. I googled a bit and found that&#xA;1 copper, platinum and gold are catalysts of methanol oxidation&#xA;2 aluminum reacts with methanol&#xA;3 silicon has poor conductivity...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can you advice?&lt;/p&gt;&#xA;" OwnerUserId="11374" LastActivityDate="2015-01-22T18:01:24.340" Title="Electrode for methanol" Tags="&lt;electrochemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24043" PostTypeId="2" ParentId="24029" CreationDate="2015-01-22T14:52:11.467" Score="2" Body="&lt;p&gt;This is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\ce{H_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-22T14:52:11.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24044" PostTypeId="1" CreationDate="2015-01-22T15:35:38.460" Score="3" ViewCount="81" Body="&lt;p&gt;I'm looking for a UV curable adhesive that can be removed without peeling or burning.  Ideal removal would be some kind of solvent over a relatively short period of time. I also need it to have a viscosity of 50,000 - 70,000 cP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now I'm using &lt;a href=&quot;http://www.dymax.com/adhesives/products/724&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; but it's not ideal.  Someone has to be there to remove it.  I'm doing this with a robot and robot's aren't very good at removing it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone out there have any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Edit&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking at UV curable investment casting wax.  I think that this could be melted off.  The only problem with what I have found is the viscosity.  They are made for 3D printing and have a relatively low viscosity.&lt;/p&gt;&#xA;" OwnerUserId="12927" LastEditorUserId="12927" LastEditDate="2015-01-27T13:35:05.810" LastActivityDate="2015-01-27T13:35:05.810" Title="UV curable adhesive" Tags="&lt;solubility&gt;&lt;polymers&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24045" PostTypeId="1" AcceptedAnswerId="24051" CreationDate="2015-01-22T15:43:50.213" Score="5" ViewCount="7368" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jOapc.png&quot; alt=&quot;2,3-dibromo-3-phenylpropanoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a picture of α,β-dibromohydrocinnamic acid (2,3-dibromo-3-phenylpropanoic acid). In the lab today I carried out an elimination reaction on this molecule using potassium carbonate giving the carbonate anion as a base. I thought about the mechanism and I initially thought that the base would attack the hydrogen on the leftmost carbon atom (out of the two with the bromine atom on it) as I thought that this would be the most acidic proton of those two carbon atoms because of resonance available with the aromatic ring. However, the proton on the carboxylate group is surely the most acidic. How does this effect things in terms of the mechanism and the product?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2016-11-14T20:01:01.600" LastActivityDate="2016-11-15T13:14:16.807" Title="Elimination in α,β-dibromohydrocinnamic acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24046" PostTypeId="2" ParentId="23990" CreationDate="2015-01-22T15:58:40.450" Score="2" Body="&lt;p&gt;Certainly the are many different types of melamine resin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to make a melamine resin without using formaldehyde?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, a different aldehyde could be used such as acetylaldehyde.  Glucose, benzaldehyde, acrolein and other alternatives to formaldehyde have been studied also (see page 33 of Polymer Synthesis by  Sandler et al.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good sources of information about formaldehyde-melamine-sodium bisulfite copolymer and possible alternatives are the patent &lt;a href=&quot;http://www.google.com/patents/US4540717&quot; rel=&quot;nofollow&quot;&gt;Resilient foam based on a melamine-formaldehyde condensate&lt;/a&gt; and related patents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the introductory portion of the patent explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is true that foams based on melamine-formaldehyde condensates have been described in various patent publications, but they have not hitherto been employed industrially for heat or sound insulation in the construction industry. They are produced by stirring air into an aqueous melamine resin solution which contains an emulsifier and a curing agent. Such as described, for example, in German Pat. No. 870,027, have the serious disadvantage that they are very hard and brittle, and easily break or crumble when handled. It is alleged in French Pat. No. 1,073,642 that melamine resin foams can be produced by heating a resin powder in a mold under reduced pressure. This process, however, does not give useful resilient foams. U.S. Pat. No. 3,093,600 describes melamine resin foams which are said to have improved resilience and resistance to crazing as a result of the incorporation of triols, e.g. trimethylolpropane. However, it has been found that the resilience, and especially the recovery after compression, of such foams is inadequate for many applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then in describing what is considered new about the invention, many different alternatives are included. Such as adding aldehydes other than formaldehyde, urea, and many other substances.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-22T15:58:40.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24047" PostTypeId="1" AcceptedAnswerId="24048" CreationDate="2015-01-22T16:12:13.687" Score="8" ViewCount="511" Body="&lt;p&gt;I was doing a reaxys substructure search for the compound shown on the left below, with the 3 amines on 1 carbon. However, the closest I found were all trinitro compounds like on the right. I would have thought the nitro compounds would be less stable and likely to explode. But given the lack of triamine compounds, they must be hard to make. All I can figure is that when all three amines are protonated you end up packing a lot of postive charge in a small place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ru2Re.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I look at that nitro structure I'm pretty sure I got the charges wrong. Just ignore that.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="4945" LastEditDate="2015-08-13T09:01:46.310" LastActivityDate="2015-08-13T09:01:46.310" Title="Why are triamino compounds so rare?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;nitro-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24048" PostTypeId="2" ParentId="24047" CreationDate="2015-01-22T16:42:43.603" Score="13" Body="&lt;p&gt;You may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below.  The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5Ale.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups are attached to the same carbon atom.  In the case of such a tris-amino compound (see figure) the imino analogue of a carbonyl is generated, but since a third amino group is present, another equilibrium between the imino compound and a nitrile also exists.  Just like in the carbonyl case, the nitrile triple bond is very strong.  So when a tris-amino compound is generated an equilibrium with a very stable nitrile will be set up and the equilibrium will (usually) lie far to the nitrile side.  Most tris-amino compounds prefer to exist as nitriles.  If the amino group is fully alkylated (all of the amino hydrogens are replaced by alkyl groups), then this equilibrium cannot occur and such alkylated tris-amino compounds should be isolable.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-24T19:04:10.257" LastActivityDate="2015-01-24T19:04:10.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24050" PostTypeId="1" CreationDate="2015-01-22T17:23:31.123" Score="0" ViewCount="92" Body="&lt;p&gt;What would be the dissociation of Hydrosulfuric acid in the presence of water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that Hydrosulfuric acid or $\ce{H2S}_{(aq)}$ would form $\ce{2 H+}$ and Sulfur ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this logic, I keep getting the wrong answer :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right answer of this question is it would not dissociate at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shall appreciate some help on explaining this. Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user12915" LastEditorUserId="7448" LastEditDate="2015-01-22T17:31:29.050" LastActivityDate="2015-01-22T20:30:51.177" Title="Dissociation of different compounds?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24051" PostTypeId="2" ParentId="24045" CreationDate="2015-01-22T17:25:51.973" Score="7" Body="&lt;p&gt;Dibromohydrocinnamic acid can eliminate in a number of different ways as shown in the following diagram. I'm not sure what conditions lead to cinnamic acid, but I know it can happen.  You have recently been asking questions about E1 and E2 eliminations, so my guess is that you are running a well-known lab experiment that generates &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-bromostyrene and allows the student to study the effect of reaction conditions on the mechanism of an elimination reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/48bME.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the experiment starts with the bromination of &lt;em&gt;trans&lt;/em&gt;-cinnamic acid.  The bromination is run under conditions that lead to trans-addition of bromine so that only the &lt;em&gt;erythro&lt;/em&gt; (&lt;a href=&quot;http://users.ipfw.edu/tahmassd/chm265/fall%202006/bromination%20procedure.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;see here for diagrams and mechanism&lt;/a&gt;) or (&lt;em&gt;S&lt;/em&gt;,&lt;em&gt;R&lt;/em&gt;) [and its mirror image (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;S&lt;/em&gt;) enantiomer] dibromocinnamic acid diastereomer is formed.  The other dibromohydrocinnamic acid diastereomer, the &lt;em&gt;threo&lt;/em&gt; isomer [(&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;R&lt;/em&gt;) and its (&lt;em&gt;S&lt;/em&gt;,&lt;em&gt;S&lt;/em&gt;) enantiomer], which can only be formed by cis-addition of bromine is not produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you build a model or draw a Newman projection it will make it easier to understand what follows.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;em&gt;erythro&lt;/em&gt; [(&lt;em&gt;S&lt;/em&gt;,&lt;em&gt;R&lt;/em&gt;) and (&lt;em&gt;R&lt;/em&gt;,&lt;em&gt;S&lt;/em&gt;)] diastereomer obtained from the above bromination, running the elimination reaction in a solvent like acetone (a solvent that won't stabilize a carbonium ion as much as protic solvents like water or methanol) you drive the mechanism towards the E2 elimination side.  With your model or drawing, put the carboxylate group (the base has removed the acidic carboxylic acid proton to produce the corresponding carboxylate anion) and the bromine on the next carbon in an anti-periplanar arrangement (you've discussed anti-periplanar in your earlier questions on the E2 mechanism) and see how in a one-step, concerted elimination of both carbon dioxide and bromide you wind up with the &lt;strong&gt;&lt;em&gt;cis&lt;/em&gt;&lt;/strong&gt;-bromostyrene isomer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ran the reaction in a protic solvent (like water or methanol) that stabilizes carbonium ions, you push the reaction towards the E1 mechanism involving loss of the bromine adjacent to the benzene ring to produce a stabilized benzylic carbonium ion.  The stable carbocation ion has an appreciable lifetime and will rotate so 1) the soon-to-be-departing carboxylate group is lined up with the carbocation p-orbital on the adjacent carbon &lt;strong&gt;and&lt;/strong&gt; 2) that the benzene ring and remaining (bulky) bromine are as far apart as possible (the lowest energy conformation).  Elimination of carbon dioxide from this conformation produces &lt;strong&gt;&lt;em&gt;trans&lt;/em&gt;&lt;/strong&gt;-bromostyrene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, the student is given several test tubes and the reaction is run under a variety of conditions where solvent and reagents are changed and then the amounts of &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-bromostyrene are measured by nmr.  From this information the student can see how solvent and reagent can shift the mechanism.  See this &lt;a href=&quot;http://www.ebah.com.br/content/ABAAAA_A0AK/journal-of-chemical-education-decarboxylative-eliminationof-2-3-dibromo-3-phenylpropanoic-acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; for a bit more detail on the actual experiment along with some good Newman projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the lab should smell nice since the &lt;em&gt;trans&lt;/em&gt;-bromostyrene produces a hyacinth-like odor.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2016-11-15T13:14:16.807" LastActivityDate="2016-11-15T13:14:16.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24053" PostTypeId="2" ParentId="24025" CreationDate="2015-01-22T17:35:01.410" Score="6" Body="&lt;p&gt;Brinnb is correct, but the other big difference between boiling and frying is that vegetables already have a lot of water in them. Boiling in water doesn't really change that because everything is already surrounded in water, but frying in oil is hot enough to boil the water inside the food, so in addition to the Maillard reaction, deep frying tends to dehydrate food. This is one of the reasons the outsides of french fries get crispy, and taken to the extreme, this effect is used to dehydrate ramen noodles in the production of instant noodles.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-22T17:35:01.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24057" PostTypeId="2" ParentId="24050" CreationDate="2015-01-22T17:53:37.883" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt;&lt;/a&gt; has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.&lt;/p&gt;&#xA;&lt;p&gt;For the dissociation part (which is usually endothermic overall) it'll need more &amp;quot;ionic character&amp;quot;. And it &lt;em&gt;does&lt;/em&gt; dissociate in water, resulting in a weak acid named  sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a &lt;strong&gt;weak acid&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Addendum:&lt;/h2&gt;&#xA;&lt;p&gt;The &amp;quot;ionic character&amp;quot; plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; is weaker than for instance, &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; and that's because &lt;strong&gt;the atomic radius changes more dramatically than electronegativity difference&lt;/strong&gt;. (See &lt;a href=&quot;http://drholly.typepad.com/ask_me_a_chemistry_questi/2005/12/why_isnt_hf_a_s.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; ) But, in the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2S}$&lt;/span&gt; molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-22T20:30:51.177" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24058" PostTypeId="2" ParentId="24042" CreationDate="2015-01-22T18:01:24.340" Score="2" Body="&lt;p&gt;Glassy carbon may be a good choice. Looking at &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/elan.1140071207/abstract&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, electrooxidation of methanol doesn't seem to be possible at reasonable potentials on glassy carbon, and it's a common working electrode material, relatively inexpensive, durable, and available in many shapes and sizes. Graphite could also work, though it may need to be replaced periodically as graphite electrodes tend to be less robust.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-22T18:01:24.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24060" PostTypeId="2" ParentId="7293" CreationDate="2015-01-22T18:28:30.080" Score="4" Body="&lt;p&gt;The key is $\Delta G = -RT \ln K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-RT \ln K = \Delta H-T\Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$-R \ln K = \frac{\Delta H}{T}-\Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an exothermic reaction, $\Delta H$ is negative.  If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7475" LastEditDate="2015-10-07T13:19:30.907" LastActivityDate="2015-10-07T13:19:30.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24061" PostTypeId="1" AcceptedAnswerId="24062" CreationDate="2015-01-22T20:04:37.027" Score="2" ViewCount="1321" Body="&lt;p&gt;Since the six strong acids dissociate completely into their ions, and the reaction goes to completion, does this mean that acids such as $\ce{HI}$ and $\ce{HCl}$ have the same acidic strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that 0.10 M $\ce{HI_{(aq)}}$ and 0.10 M $\ce{HCl_{(aq)}}$ both contain 0.10 M $\ce{H3O+_{(aq)}}$, which leads me to believe they are equally strong acids. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="865" LastEditDate="2021-05-22T21:00:07.190" LastActivityDate="2021-05-22T21:00:07.190" Title="Are all strong acids the same strength?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24062" PostTypeId="2" ParentId="24061" CreationDate="2015-01-22T20:23:29.560" Score="3" Body="&lt;p&gt;There's something called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Leveling_effect&quot; rel=&quot;nofollow&quot;&gt;&quot;solvent leveling effect&quot;&lt;/a&gt;. In short, there's a lowest pKa in a particular solvent, based on the basicity of the conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water, you're limited by $\ce{OH-}$. So given the same concentration of $\ce{HI}$ and $\ce{HCl}$ in water, they will indeed have the same pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other solvents, you may have a difference between acids, depending on the pKa.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-01-22T20:23:29.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24063" PostTypeId="2" ParentId="24061" CreationDate="2015-01-22T20:23:57.247" Score="2" Body="&lt;p&gt;No, not all strong acids are the same strength&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more than just 6 strong acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acid dissociation constant ($K_a$) quantifies the &lt;a href=&quot;http://depts.washington.edu/eooptic/links/acidstrength.html&quot; rel=&quot;nofollow&quot;&gt;strength of an acid&lt;/a&gt;.  pK is negative log base 10 of $K_a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trifluoromethanesulfonic acid pk &amp;lt;-13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perchloric acid pK = &amp;lt;-9 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HI pK = -9.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HBr pK = -9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HCl pK = -6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sulfuric acid pK = -3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chloric acid pK = -1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nitric acid pK = -1&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-22T20:31:50.670" LastActivityDate="2015-01-22T20:31:50.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24066" PostTypeId="2" ParentId="24035" CreationDate="2015-01-22T23:52:55.213" Score="5" Body="&lt;p&gt;It is a combination of potato and a fungus or bacterium that causes a potato to rot and produce the offensive odor.  A major contributor to the odor is dimethyl disulfide which has been identified as a key component of the emitted volatiles (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1745-4549.1990.tb00134.x/abstract&quot; rel=&quot;noreferrer&quot;&gt;see here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VgrOk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the above reference doesn't mention it (at least not in the abstract) dimethyl disulfide is often accompanied by hydrogen sulfide, dimethyl sulfide and dimethyl trisulfide, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/21150089&quot; rel=&quot;noreferrer&quot;&gt;the trisulfide being particularly odiferous&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only is the odor of rotting potatoes extremely unpleasant, but it has been &lt;a href=&quot;http://www.parentdish.co.uk/2013/09/04/girl-8-orphaned-after-gas-from-rotting-potatoes-kills-her-entire-family/&quot; rel=&quot;noreferrer&quot;&gt;reported to be lethal&lt;/a&gt;.  Certainly &lt;a href=&quot;http://www.smithsonianmag.com/arts-culture/horrific-tales-of-potatoes-that-caused-mass-sickness-and-even-death-3162870/?no-ist&quot; rel=&quot;noreferrer&quot;&gt;eating rotten potatoes can be lethal&lt;/a&gt; (solanine poisoning).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-23T00:05:47.733" LastActivityDate="2015-01-23T00:05:47.733" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24068" PostTypeId="2" ParentId="22367" CreationDate="2015-01-23T01:14:35.810" Score="0" Body="&lt;p&gt;For point 3 carbon-carbon single bonds count. Heteroatoms would build up a functional group. However, I could not think of a scenario where point 4 is needed. It should be a scenario where two chains with an equal amount of multiple bonds and single bonds are present but different in length.&lt;br&gt;&lt;br&gt;&#xA;The rule, which you mention, in which the number of functional groups count is point 1.&lt;br&gt;&lt;br&gt;&#xA;According to what I have learned the name &quot;2-(1-Hydroxybutyl)-3-methylhexanoic acid&quot; is wrong. There are two equal chains, one has a methyl and one a hydroxyl group. According to IUPAC nomenclature of organic chemistry the group with the higher precedence is hydroxyl. Therefore, chemdoodle gave you the wrong name.&lt;br&gt;&#xA;This is a case where this sentence applies which they put above: &quot;Please give it a try and let us know if you encounter any issues.&quot;&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2015-01-23T01:14:35.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24070" PostTypeId="2" ParentId="23905" CreationDate="2015-01-23T05:43:59.263" Score="1" Body="&lt;p&gt;Many spectroscopic experiments record analog data, which is subsequently Fourier transformed to give a frequency-domain spectrum. For example, in a nuclear magnetic resonance experiment, the signal emitted by an excited nucleus is picked up as a voltage in an RF coil for 1-2 sec. This time-domain signal is processed then converted to the frequency domain by FT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for 'popular', Richard Ernst won the 1991 Nobel in Chemistry for his contributions to this field of research. &lt;/p&gt;&#xA;" OwnerUserId="12939" LastActivityDate="2015-01-23T05:43:59.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24071" PostTypeId="1" CreationDate="2015-01-23T06:44:33.570" Score="3" ViewCount="603" Body="&lt;p&gt;Chemical formula is &lt;span class=&quot;math-container&quot;&gt;$\ce{C7H6O3}$&lt;/span&gt;. The IR spectrum has a peak at &lt;span class=&quot;math-container&quot;&gt;$\pu{3238 cm-1}$&lt;/span&gt;, which I believe could be due to an alcohol, phenol, or carboxylic acid. There's an additional peak at  &lt;span class=&quot;math-container&quot;&gt;$\pu{3015 cm-1}$&lt;/span&gt;, which I believe is due to either an alkene or aromatic group. The presence of aromatic and phenol peaks makes me think there's a phenol group in this compound.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/laU8P.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What I'm stuck on are the four protons in the 7-8 ppm range. Especially the one split into a quartet. If I'm right about the IR, I just can't think of a way to get one proton with three neighbors to split with in a phenol.&lt;/p&gt;&#xA;" OwnerUserId="12942" LastEditorUserId="69830" LastEditDate="2021-05-04T08:32:42.283" LastActivityDate="2021-05-04T08:32:42.283" Title="Could I get a hint on the structure of this molecule based on HNMR spectra?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24072" PostTypeId="2" ParentId="24071" CreationDate="2015-01-23T08:46:39.940" Score="5" Body="&lt;p&gt;That is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain distortion in strongly coupled systems, the &quot;roof effect&quot; you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.&lt;/p&gt;&#xA;" OwnerUserId="7" LastEditorUserId="7" LastEditDate="2015-01-23T12:35:14.613" LastActivityDate="2015-01-23T12:35:14.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24073" PostTypeId="2" ParentId="24071" CreationDate="2015-01-23T09:06:37.550" Score="1" Body="&lt;p&gt;I once had a similar example in which benzene was used as solvent but the spectrum was not referenced to its peak at 7.16 ppm. If the quartet is really a quartet you might think about the meaning of that singlet at 7.26 ppm.&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2015-01-23T09:06:37.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24074" PostTypeId="1" AcceptedAnswerId="24075" CreationDate="2015-01-23T10:58:18.723" Score="5" ViewCount="1069" Body="&lt;p&gt;In a book I am reading, it is said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; molecule, the shared pair of electrons spend more time nearer the chlorine atom. In the &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; molecule, the shared electrons spend more time nearer the fluorine atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does &quot;spend more time&quot; mean here? Has this been experimentally validated? If yes, how? &lt;/p&gt;&#xA;" OwnerUserId="12943" LastEditorUserId="4945" LastEditDate="2020-02-18T14:22:42.917" LastActivityDate="2020-02-19T06:03:48.397" Title="What does it mean that an electron pair &quot;spends more time&quot; nearer an atom?" Tags="&lt;electrons&gt;&lt;atoms&gt;&lt;electronegativity&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24075" PostTypeId="2" ParentId="24074" CreationDate="2015-01-23T11:12:18.840" Score="4" Body="&lt;p&gt;When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as &quot;spending more time&quot;. Electrons are less described as a dot but as a smearing (electron density).&#xA;Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say &quot;the electron spends more time at F/Cl&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: &lt;a href=&quot;http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory&lt;/a&gt;&#xA;Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2015-01-23T11:12:18.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24076" PostTypeId="2" ParentId="24071" CreationDate="2015-01-23T12:31:34.840" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the IR have any stretches near ~1700? This would be a great way to rule in (or out) a carbonyl. It looks like there is one from the above graph, and you have a broad peak ~ 11 ppm, which should stand out as a very specific functional group. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mad Scientist is correct, that is not a quartet at 7 PPM, but is a doublet and a triplet partially overlapped. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here's your hint: you have TWO oxygen-containing functional groups and an aromatic system (It looks like you already knew that from your sketches in the photo). If the aromatic system contains 4H, and is there are two functional groups, it is likely a di-substituted benzene ring. Which of the three possible types (ortho, meta, para) could give rise to two doublets and two triplets in the splitting?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S. I believe the spectrum is correctly referenced. The tiny peak near 7.26 ppm is likely chloroform from the solvent - it is much smaller than the other peaks and chloroform is the solvent of choice for most organic NMR.&lt;/p&gt;&#xA;" OwnerUserId="12939" LastActivityDate="2015-01-23T12:31:34.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24077" PostTypeId="1" CreationDate="2015-01-23T12:36:04.803" Score="2" ViewCount="252" Body="&lt;p&gt;This prevent the clothes from being ruined by &lt;strong&gt;ink&lt;/strong&gt; on the paper tag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put the garments attached with this paper through the whole process of dry-cleaning,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the questions listed below &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;paper type?&lt;/li&gt;&#xA;&lt;li&gt;ink type?&lt;/li&gt;&#xA;&lt;li&gt;Is paper need to be coated?&lt;/li&gt;&#xA;&lt;li&gt;if need &gt;&gt; what is this coating technique called?&lt;/li&gt;&#xA;&lt;li&gt;printing technique (refer to &lt;a href=&quot;http://en.wikipedia.org/wiki/Printing&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Printing&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that, the &lt;strong&gt;dry-cleaning solvent&lt;/strong&gt; is not just water.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Does that sound scientific?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the more appropriate place to ask this kind of question. If there is, just let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, &#xA;pakorn&lt;/p&gt;&#xA;" OwnerUserId="12944" LastEditorUserId="12944" LastEditDate="2015-01-23T13:13:05.973" LastActivityDate="2015-01-23T13:13:05.973" Title="How paper-tag survive in the dry-cleaning process?" Tags="&lt;materials&gt;&lt;solvents&gt;&lt;cleaning&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24078" PostTypeId="1" AcceptedAnswerId="24079" CreationDate="2015-01-23T13:49:53.787" Score="11" ViewCount="18431" Body="&lt;p&gt;By what mechanism does HF proceed in dissolving glass? Why is it the only acid that has this capability? Is it because of the small size and high electronegativity of fluorine?&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastEditorUserId="23561" LastEditDate="2018-10-05T18:47:11.330" LastActivityDate="2018-10-06T00:59:20.963" Title="How does HF dissolve glass?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24079" PostTypeId="2" ParentId="24078" CreationDate="2015-01-23T13:57:35.867" Score="7" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; reacts with glass (&lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt;).&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SiO2 + 4HF -&amp;gt; SiF4 + 2H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SiF4}$&lt;/span&gt; is not a solid that consists of vertex-connected tetrahedra like &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; but is a gas at room temperature. Technically, &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; is not a solvent since in this case it reacts with the glass vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Spierings:&lt;sup&gt;[2]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HF2-}$&lt;/span&gt; ions are adsorbed on surface silanol groups, the &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; molecules on vicinal silanol groups and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as &lt;span class=&quot;math-container&quot;&gt;$\ce{\bond{#}Si-F}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{\bond{#}Si-O-SiF3}$&lt;/span&gt;. The adsorption of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HF2-}$&lt;/span&gt; increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hollemann, A. F.; Wiberg, E. &lt;em&gt;Hollemann–Wiberg: Lehrbuch der Anorganischen&#xA;Chemie,&lt;/em&gt; 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. &lt;em&gt;J. Mater. Sci.&lt;/em&gt; &lt;strong&gt;1993,&lt;/strong&gt; &lt;em&gt;28&lt;/em&gt; (23), 6261–6273. &lt;a href=&quot;https://doi.org/10.1007/BF01352182&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1007/BF01352182&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="16683" LastEditDate="2018-10-06T00:59:20.963" LastActivityDate="2018-10-06T00:59:20.963" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24080" PostTypeId="1" AcceptedAnswerId="35553" CreationDate="2015-01-23T14:33:44.567" Score="4" ViewCount="161" Body="&lt;p&gt;When I recieve a dataset from single crystal X-ray diffraction, I can determine the quality from a graph in which F2/sig(F2) is displayed in dependence of the frame.&lt;br&gt;&lt;br&gt;&#xA;But I forgot the basics... I have found that the $R_{int}$ value is calculated as such:&#xA;$$R_{int}=\frac{\Sigma|F^2-F_{mean}^2|}{\Sigma|F^2|}$$&#xA;Reference: &lt;a href=&quot;http://www.canadiancrystallography.ca/cccw/files/CIF-file%20and%20validation.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.canadiancrystallography.ca/cccw/files/CIF-file%20and%20validation.pdf&lt;/a&gt;&lt;br&gt;&#xA;But what exactly is F2/sig(F2) now? &lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="12908" LastEditDate="2015-01-23T17:39:18.920" LastActivityDate="2015-08-27T10:27:21.607" Title="How is &quot;F2/sig(F2)&quot; the data quality in SXRD?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24081" PostTypeId="1" CreationDate="2015-01-23T14:34:30.820" Score="3" ViewCount="2113" Body="&lt;p&gt;Should the concentration of liquid water be included in the $K_\mathrm{c}$ expression for the reaction&#xA;$$\ce{2 NO2_{(g)} + 7 H2_{(g)} -&amp;gt; 2 NH3_{(g)} + 4 H2O_{(l)}}$$&#xA;if water is liquid and all the other compounds are gasses?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="15410" LastEditDate="2015-05-11T01:42:37.353" LastActivityDate="2015-05-11T03:59:09.103" Title="Should the liquid be included in Kc expression?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24082" PostTypeId="1" CreationDate="2015-01-23T14:39:02.017" Score="3" ViewCount="965" Body="&lt;p&gt;I wanted to ask clarifications about a passage in Atkin's physical chemistry book, chapter 9, in the paragraph 9.2 description of equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For studying equilibrium of a reaction, where $\nu_i$ are the stechiometric coefficients it imposes $\sum_i \nu_i \mu_i=0$ with $\mu$ the chemical coefficients. At this point it writes the chemical coefficients as a function of the activities $\mu_i=\mu_i^\circ+RT\ln(a_i)$ and identifies the term $\sum_i \nu_i \mu_i^\circ$ with the standard reaction Gibbs energy. I do not understand this identification. I thought that the standard state for the definition of activities (even dependent on wether the substance is a solute or a solvent) was not the same reference state for computing the reaction Gibbs energy (and does not depend on the role played by the substance in the reaction). I'm surely getting something wrong...  &lt;/p&gt;&#xA;" OwnerUserId="12945" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T00:06:30.737" LastActivityDate="2015-10-23T00:06:30.737" Title="reference states for activities" Tags="&lt;reaction-mechanism&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24083" PostTypeId="2" ParentId="15146" CreationDate="2015-01-23T14:58:12.277" Score="1" Body="&lt;p&gt;Compounds having potassium anions are referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkalide&quot; rel=&quot;nofollow noreferrer&quot;&gt;potassides&lt;/a&gt; and are rare, usually involving K- within a cyrpt such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbdVr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja992667v&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crystalline Salts of Na- and K- (Alkalides) that Are Stable at Room Temperature&lt;/a&gt; J. Am. Chem. Soc. vol. 121, pages 10666-10667 describes preparation of K+(aza222)K- , where &quot;aza222&quot; is the compound in the above figure, but with the 6 H's replaced with methyl groups.  The compound is formed by dissolving potassium metal in methyl amine solution of the methylated aza222.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-23T15:06:13.093" LastActivityDate="2015-01-23T15:06:13.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24084" PostTypeId="1" CreationDate="2015-01-23T15:05:11.280" Score="3" ViewCount="398" Body="&lt;p&gt;Here is the other crazy question... I know that electrons revolve around nucleus in their orbits.But my question is &quot;how the orbitals are located in the orbits? Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits?? Totally confused.... &lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="865" LastEditDate="2015-10-30T17:46:09.713" LastActivityDate="2015-10-30T17:46:09.713" Title="Totally confused about the location of &quot;s,p,d,f&quot; Orbitals inside the atom" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24085" PostTypeId="2" ParentId="24002" CreationDate="2015-01-23T15:11:50.160" Score="1" Body="&lt;p&gt;After vacuum cleaning of goods which you are mentioning.. there is still a chance of entry of gases during the packing time. If you are good enough to manage the goods vacuum protected until packing then your goods will be vacuum protected if you vacuum seal the box perfectly.. &lt;/p&gt;&#xA;" OwnerUserId="12917" LastActivityDate="2015-01-23T15:11:50.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24086" PostTypeId="2" ParentId="24084" CreationDate="2015-01-23T15:26:31.660" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that electrons revolve around nucleus in their orbits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should &quot;unknow&quot; that, because that is the failed &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow&quot;&gt;Bohr Model&lt;/a&gt; of the atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The are no fixed orbits for electrons in atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr Model has been superceded by the &lt;a href=&quot;http://www.pha.jhu.edu/~rt19/hydro/node3.html&quot; rel=&quot;nofollow&quot;&gt;Schrodinger model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atom is described in terms of probabilty density: for each point in space the probability that an electron will be at that point (when the atom is in a given state such as 1s, 2s, 2p, etc).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;my question is &quot;how the orbitals are located in the orbits?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are no orbits&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shapes you seen in drawings that are spherical, dumb bell, etc. are meant to show regions of higher electron probabilty density.  For example, someone might draw a sphere on the basis that there is a 90% probablity of the electron in that orbital being within the sphere at a given moment.  However, in reality, for example in the 1s state, the electron could be anywhere in space. Other states, such as 2s, have nodes, with the electrons possibly being anywhere except at the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Electronic_Orbitals#Radial_and_Angular_Nodes&quot; rel=&quot;nofollow&quot;&gt;nodes&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So think of the shapes as showing where the electrons in that orbital are often, but remember that they can really be anywhere excpet at the nodes.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-23T15:44:20.487" LastActivityDate="2015-01-23T15:44:20.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24087" PostTypeId="2" ParentId="24081" CreationDate="2015-01-23T15:52:06.913" Score="1" Body="&lt;p&gt;No. If the water condenses it is effectively removed from the reaction equilibrium. Just as with solids that precipitate.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-23T15:52:06.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24088" PostTypeId="1" CreationDate="2015-01-23T16:01:49.313" Score="7" ViewCount="2181" Body="&lt;p&gt;Shakashiri describes the reaction of solid Ammonium Thiocyanate and solid Barium Hydroxide Octahydrate resulting in an endothermic reaction leading to subzero temperatures. I've performed this experiment and measured temperature as low as about $-10\ ^\circ\text{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there known chemical reactions that can lead to even lower temperatures?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorDisplayName="user7951" LastEditDate="2015-05-29T14:55:38.757" LastActivityDate="2015-05-29T14:55:38.757" Title="What is an example of a very low temperature endothermic reaction?" Tags="&lt;thermodynamics&gt;&lt;reaction-mechanism&gt;&lt;enthalpy&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24089" PostTypeId="1" CreationDate="2015-01-23T16:07:39.323" Score="2" ViewCount="48" Body="&lt;p&gt;Do chemicals of an average weight, such as Polyethylene Glycols, undergo granular convection (the Brazil nut effect)?&lt;/p&gt;&#xA;" OwnerUserId="2587" LastActivityDate="2015-01-23T16:07:39.323" Title="Do chemicals undergo granular convection?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24090" PostTypeId="2" ParentId="24071" CreationDate="2015-01-23T16:29:41.643" Score="2" Body="&lt;p&gt;Looking at your picture you also seem to have annotated an IR peak at around $\mathrm{1700~cm^{-1}}$ which is indicative of the carbonyl group. Together with the IR peak at $\mathrm{3238~cm^{-1}}$ and the very broad NMR peak far downfield this supports your supposition that there is a carboxylic acid present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the molecular formula you now only have one $\ce{O}$ and one $\ce{H}$ unaccounted for so it must be a phenol group. As mentioned by Mad Scientist the pseudo-quartet on the NMR spectrum is in fact an overlapping doublet and triplet. You have two doublets and two triplets in total so you should think about which of ortho, meta and para substitution would give you this splitting pattern.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-01-23T20:20:04.113" LastActivityDate="2015-01-23T20:20:04.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24091" PostTypeId="2" ParentId="24088" CreationDate="2015-01-23T17:15:23.117" Score="5" Body="&lt;p&gt;Yes there are. In Vogels famous textbook on practical organic chemistry there is a section about the cooling effect of certain salt (mixture) solutions in water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If ice is temporarily not available, advantage may be taken of the&#xA;  cooling effect attending the solution of certain salts or salt&#xA;  mixtures in water. Thus a mixture produced by dissolving 1 part of&#xA;  NH4C1 and 1 part of NaNO3 in 1-2 parts of water causes a reduction in&#xA;  temperature from 10 to — 15 °C to — 20 °C; 3 parts of NH4C1 in 10&#xA;  parts of water from 13 to -15°C; 11 parts of Na2S2O3.5H2O in 10 parts&#xA;  of water from 11 to - 8 °C; and 3 parts of NH4NO3 in 5 parts of water&#xA;  from 13 to — 13°C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Taken from Vogels 5th edition, p. 70)&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-01-23T20:04:30.697" LastActivityDate="2015-01-23T20:04:30.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24092" PostTypeId="2" ParentId="23658" CreationDate="2015-01-23T17:28:24.357" Score="2" Body="&lt;p&gt;Depending on how long the battery has been sitting empty even adding electrolyte might not help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet is to fill the battery with distilled water and put it on a charge, charge it for a couple of days if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get yourself a hydrometer (used for checking the specific gravity of your electrolyte) and check the water's gravities. After you charge the battery for awhile check them again. If the gravities start to rise, and the liquid is reasonable clear stop charging the battery and let it sit 12-24 hours. (If however the water turns cloudy/grey the battery is scrap get a new one). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the battery has sat and cooled down, empty the water into a container, and neutralize with soda ash, or baking soda. BE CAREFUL even though what you put it was water, what will come out might/will be corrosive. Dispose of properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go to your local auto parts store and buy a box of battery electrolyte its usually sold as 1.28 spg, fill the battery with electrolyte re-charge until your gravities come up, and voila, use just re-conditioned a battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be sure to wear rubber gloves / goggles or face shield and any other appropriate safety gear. Remember that when a battery charges it releases hydrogen sulfide gas which is explosive so no open flames sparks etc.... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the battery may only last your a couple of months or another year or so, and it may not even come back at all. But I have used this procedure with great success to revive old batteries....and squeeze a bit more life out of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jeff&#xA;Industrial battery specialist for 30 years.&lt;/p&gt;&#xA;" OwnerUserId="12948" LastActivityDate="2015-01-23T17:28:24.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24094" PostTypeId="1" AcceptedAnswerId="24099" CreationDate="2015-01-23T18:16:49.137" Score="4" ViewCount="167" Body="&lt;p&gt;I just have a question on how you handle polyprotic bases.&lt;/p&gt;&#xA;&lt;p&gt;Question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Show how oxalate ion can be a polyprotic base&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nVSdN.jpg&quot; alt=&quot;Monobasic&quot; /&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2WClA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2WClA.jpg&quot; alt=&quot;Dibasic&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My question is whether we stop doing the reaction until the oxalate ion has no charge or do we keep going on as below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y53cs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y53cs.jpg&quot; alt=&quot;Tribasic&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Another question I have is ;why hypochlorite ion would be a monoprotic base, rather than a polyprotic base?&lt;/p&gt;&#xA;" OwnerUserId="12949" LastEditorUserId="106377" LastEditDate="2021-05-05T16:00:02.677" LastActivityDate="2021-05-05T16:00:02.677" Title="Polyprotic acids and bases" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24095" PostTypeId="2" ParentId="24082" CreationDate="2015-01-23T18:53:55.283" Score="2" Body="&lt;p&gt;$\mu_i=\mu_i^o+RT ln(a_i)$ is the &lt;a href=&quot;http://goldbook.iupac.org/A00115.html&quot; rel=&quot;nofollow&quot;&gt;definition of activity&lt;/a&gt; ($a_i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the &lt;a href=&quot;http://goldbook.iupac.org/S05925.html&quot; rel=&quot;nofollow&quot;&gt;standard state&lt;/a&gt; for &lt;a href=&quot;http://goldbook.iupac.org/S05908.html&quot; rel=&quot;nofollow&quot;&gt;standard chemical potential&lt;/a&gt; ($\mu_i^o$) is necessarily part of the definition of activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pure liquid or solid, the standard state is the pure substance at 1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a liquid solvent, the standard state is also the pure liquid at 1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a gas, the standard state is the fictitious state of an ideal gas at 1 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^o = \sum_i \nu_i \mu_i^o$ by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no one standard state for $\Delta G^o$, because the various substances may have different standard states.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-25T20:22:44.640" LastActivityDate="2015-01-25T20:22:44.640" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24096" PostTypeId="1" AcceptedAnswerId="24097" CreationDate="2015-01-23T19:03:37.153" Score="3" ViewCount="216" Body="&lt;p&gt;How can you elegantly describe that a polymorphic substance changes its structure?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;It polymorphs&lt;/li&gt;&#xA;  &lt;li&gt;It morphs&lt;/li&gt;&#xA;  &lt;li&gt;It undergoes polymorphic transition&lt;/li&gt;&#xA;  &lt;li&gt;It experiences a morphic change&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or what would be even better terms?&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="83758" LastEditDate="2020-04-08T17:51:13.793" LastActivityDate="2020-04-08T17:51:13.793" Title="The Verb for Polymorphism" Tags="&lt;terminology&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24097" PostTypeId="2" ParentId="24096" CreationDate="2015-01-23T19:16:19.760" Score="4" Body="&lt;p&gt;It &quot;transitions from one polymorph to another&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;transitions to another polymorph&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;transition to polymorph beta&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;transforms to another polymorph&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &quot;transforms to polymorph beta&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like &quot;transitions&quot; better than &quot;transforms&quot;, but I've seen both&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-23T19:16:19.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24099" PostTypeId="2" ParentId="24094" CreationDate="2015-01-23T19:59:39.563" Score="7" Body="&lt;p&gt;If you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Piling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton&lt;sup&gt;1&lt;/sup&gt;. You should stop with the two reactions you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;1. This kind of thing &lt;em&gt;does&lt;/em&gt; happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\rm CH_3COOH_2^+$ instead of $\rm H_3O^+$, &#xA;and $\rm CH_3COO^-$ instead of $\rm OH^-$.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-23T21:26:38.147" LastActivityDate="2015-01-23T21:26:38.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24100" PostTypeId="1" CreationDate="2015-01-23T22:25:03.630" Score="8" ViewCount="949" Body="&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dew point is the temperature at which the water vapor in a sample of air at constant barometric pressure condenses into liquid water at the same rate at which it evaporates. At temperatures below the dew point, water will leave the air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am wondering what is is about the air which determines how much $\ce{H2O}$ vapour can be stored in a set volume of air at a set temperature without the water condensing back to liquid water.&#xA;Also, why is it that at higher temperatures the air can hold a greater amount of water vapour?&lt;/p&gt;&#xA;" OwnerUserId="10189" LastEditorDisplayName="user12839" LastEditDate="2015-01-24T02:16:18.077" LastActivityDate="2015-04-23T02:22:46.697" Title="What determines humidity limit / dew point of the air? - Why can air only hold a certain amount of water?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;gas-laws&gt;&lt;temperature&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24104" PostTypeId="2" ParentId="24100" CreationDate="2015-01-24T03:20:08.980" Score="6" Body="&lt;p&gt;One way to understand this is if you think of air as bunch of atoms and molecules moving around in random fashion. The speed at which they move is measured by the temperature of air. (This is true even if we are not talking about air, but we don't need to bother about that here.) So when we say higher temperature we mean that the speed at which these particles move is higher. Furthermore the faster they move the less likely they are to stick together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now water molecules like to stick to each other and when too many of them are together they &quot;condense&quot;. But if you were to increase the temperature of the air (and hence the water molecules), they end up moving around faster and less likely to stick together. This way you can add more water molecules for the same amount of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing to think about is pressure. In some sense it describes how crowded the place is. The higher the pressure, the more crowded it is. So when you increase the pressure, you increase the chances that the water molecules will stick together and hence &quot;condense&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way to think about this is through &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Çhatelier's principle&lt;/a&gt;, which states that when a system in equilibrium (especially chemical equilibrium) is disturbed it tries to adjust in a way which minimizes the disturbance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now think of a box with some water and air. If you leave it long enough the system will reach an equilibrium, wherein no more water will evaporate or condense. But say now you increase the temperature of the air, the system will try to readjust so as to reduce the change in temperature. To do this it will cause the water to evaporate (as evaporation requires/consumes heat) and thereby reducing the change in temperature.&#xA;Similarly when you increase the pressure which is similar to decreasing the volume of the box, the air tries to account for this by condensing out some of the water (since liquid water occupies less volume than gaseous water), thereby reducing the effective decrease in volume seen by the air.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You may wonder why other particles in air do not &quot;condense.&quot; They do, but only at much lower temperatures and/or higher pressures.&lt;/p&gt;&#xA;" OwnerUserId="655" LastEditorUserId="5354" LastEditDate="2015-04-23T02:22:46.697" LastActivityDate="2015-04-23T02:22:46.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24105" PostTypeId="2" ParentId="7683" CreationDate="2015-01-24T03:28:45.683" Score="10" Body="&lt;p&gt;In  the presence of bulky group that is methyl in this case causes steric hinderence making the plane the $\ce{NH3}$ out the plane thus preventing resonance which could have  helps it in delocalising the + charge thus making the conjugate base very stable but because of hindernce the + charge gets localized and makes the conjugate base unstable and so the acid weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in case of o-toluic acid if one draws the resonating structures of the conjugate base we see that benzene ring exerts +m (mesomeric) effect, which is not so in case of o-toluidine, which indeed is inversly proportional to acidic strength and so the non planarity caused due SIR now benefits the substituent by making it free from the +M effect of the ring and so increasing its acidity. Now the substituent can freely make equivalent resonance which can will support the acidity and will not be harmed by the rings +M.&lt;/p&gt;&#xA;" OwnerUserId="12961" LastEditorUserId="7448" LastEditDate="2015-01-24T14:21:39.373" LastActivityDate="2015-01-24T14:21:39.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24106" PostTypeId="1" CreationDate="2015-01-24T03:33:01.050" Score="6" ViewCount="1164" Body="&lt;p&gt;I've deduced following postulates from studying my chem books.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Slater determinants are eigenfunctions of an unperturbed atomic Hamiltonian, which contains kinetic and central potential energy parts of each electrons only, since spin orbitals constituting the determinants are originated from one-electron Hamiltonian eigenfunctions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, atomic Hamiltonian = (kinetic part of each electrons) + (central potentials between atomic nucleus and electrons) + (interelectronic potentials)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin orbit interaction and the other effects are neglected.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Slater determinants or their linear combinations are eigenfunctions of total spin angular momentum and total orbital angular momentum operator($S^2$ and $L^2$) simultaneously.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In addition to them, I would refer to information from Quantum Chemistry, 6th Ed., written by I. N. Levine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(p. 312) Since $S^2$ and $L^2$ commute with the atomic Hamiltonian and with the exchange operator, the zeroth order functions should be eigenfunctions of $S^2$ and $L^2$.&#xA;(The zeroth order functions Levine mentioned above is indicating the single or linear combinations of Slater determinants in a same configuration)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I can accept the fact that the atomic Hamiltonian, $S^2$ and $L^2$ commute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the single or linear combinations of Slater determinants in a same configuration are the zeroth order functions (eigenfunctions of the unperturbed atomic Hamiltonian), how the fact that $S^2$ and $L^2$ commute with the atomic Hamiltonian makes the Slater determinants be the eigenfunctions of the atomic Hamiltonian??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a Slater determinant corresponding to the one of the Helium-first excited states, $|1s\alpha~2s\beta|$, is an eigenfunction of the unperturbed atomic Hamiltonian(this determinant is a zeroth order function), $S^2$ and $L^2$, but not of the atomic Hamiltonian.&lt;/p&gt;&#xA;" OwnerUserId="12962" LastEditorUserId="4945" LastEditDate="2015-03-02T02:58:51.770" LastActivityDate="2015-03-02T02:58:51.770" Title="Slater determinant as an unperturbed atomic wave function" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24107" PostTypeId="1" CreationDate="2015-01-24T04:00:32.353" Score="4" ViewCount="51207" Body="&lt;p&gt;Is a triplet state or singlet state more stable? How can an atom have a long lifetime in a triplet state even though two electrons have the same spin? &lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="5754" LastEditDate="2015-02-05T17:46:03.637" LastActivityDate="2018-08-08T15:36:43.003" Title="Is a triplet state or singlet state more stable?" Tags="&lt;photochemistry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24108" PostTypeId="1" CreationDate="2015-01-24T04:23:38.910" Score="2" ViewCount="1244" Body="&lt;p&gt;An iodine ion is the larger of the two, but why if they have the same amount of electrons? Also, $\ce{Li}$ is larger than $\ce{Li+}$, don't more electrons mean larger radius?&lt;/p&gt;&#xA;" OwnerUserId="12963" LastEditorDisplayName="user12839" LastEditDate="2015-01-24T05:27:52.187" LastActivityDate="2015-01-24T11:12:59.197" Title="Iodine ion vs caesium ion size" Tags="&lt;everyday-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24109" PostTypeId="2" ParentId="24108" CreationDate="2015-01-24T05:38:34.593" Score="2" Body="&lt;p&gt;Iodine is having 53 electrons and cesium is having 55 electrons in general. Iodine ion can be made by adding or by removing electrons from it. Here you added two electrons to iodine and made it an iodine ion in which number of electrons of iodine ion became equal to the number of electrons of cesium but the composition of nucleus(number of protons) is different in both iodine ion and cesium ion. Obviously cesium is having more number of protons than iodine, so cesium nucleus can hold it's electrons more closer in cesium than iodine nucleus in an iodine ion which results in larger radius of iodine ion than cesium.. &#xA;Same number of electrons doesn't mean that their radii are same. Same logic is applicable for lithium ion and lithium.. Lithium ion is having less electrons and more protons compared to lithium.. So radius of lithium is larger than radius of lithium ion&lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="12917" LastEditDate="2015-01-24T11:12:59.197" LastActivityDate="2015-01-24T11:12:59.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24111" PostTypeId="1" CreationDate="2015-01-24T07:19:29.177" Score="5" ViewCount="2133" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Predict the order of elution for a mixture containing caffeine and benzoic acid, using a reverse phase HPLC. Explain your answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My try:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since benzoic acid is more acidic that caffeine, which means it has a lower pH, it is hence more polar. Since for HPLC, the mobile phase is the polar phase, elution of polar compounds will be favoured hence benzoic acid will be eluted first. Is my answer and explanation correct?&lt;/p&gt;&#xA;" OwnerUserId="7422" LastEditorUserId="41686" LastEditDate="2020-06-03T11:51:43.480" LastActivityDate="2020-06-03T12:16:17.013" Title="Analytical Chemistry predicting the order of elution of compounds in a reverse phase HPLC" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24112" PostTypeId="1" AcceptedAnswerId="24126" CreationDate="2015-01-24T09:50:33.137" Score="6" ViewCount="11678" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \text{Co}[{(\text{N}{\text{H}}_{3})}_{4}{\text{Cl}}_{2}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;optically active? If so, write its stereo-isomers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that for a compound to be optically active, it should rotate the plane of Plane Polarized Light, but is there a way to do this without experiment?&lt;br /&gt;&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="4854" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-24T16:17:51.893" Title="Is tetraammine dichloro cobalt (III) an optically active compound?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;stereochemistry&gt;&lt;optical-properties&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24113" PostTypeId="1" CreationDate="2015-01-24T10:06:57.513" Score="5" ViewCount="388" Body="&lt;p&gt;Some ion-selective electrodes can be expensive (starting from $500). Is there any relatively inexpensive do-it-yourself alternatives that would still give very accurate results? How does one make an ion-selective electrode?&lt;/p&gt;&#xA;" OwnerUserId="12965" LastActivityDate="2015-02-12T08:06:11.080" Title="Ion-selective electrodes" Tags="&lt;electrochemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24114" PostTypeId="1" CreationDate="2015-01-24T11:48:25.153" Score="1" ViewCount="561" Body="&lt;p&gt;Is there a name for the point where a first order rate of reaction meets a second order one.&lt;br&gt;&#xA;&lt;img src=&quot;https://imgur.com/Nxpp9G8.png&quot; alt=&quot;reactant graph&quot;&gt;&lt;br&gt;&#xA;This point highlighted above.&lt;/p&gt;&#xA;" OwnerUserId="12967" LastEditorUserId="4945" LastEditDate="2015-03-03T02:56:47.300" LastActivityDate="2015-03-03T02:56:47.300" Title="Is there a name for the intersection of the graphs of first and second order reactions?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-22T16:26:07.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="24115" PostTypeId="1" AcceptedAnswerId="24120" CreationDate="2015-01-24T12:04:58.157" Score="5" ViewCount="1462" Body="&lt;p&gt;I need to know what a redox reaction is. How could a redox reaction possibly be identified? Also, I have a confusion whether,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H -&amp;gt; H+  +  e-}$$&#xA;$$\ce{e- + Cl -&amp;gt; Cl-}$$&#xA;$$\ce{H + Cl -&amp;gt; HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a redox reaction.&#xA;Furthermore, could anyone please tell me if ionization and reduction and gaining electron and oxidation are referring to the same phenomena?&lt;/p&gt;&#xA;" OwnerUserId="11535" LastEditorDisplayName="user12839" LastEditDate="2015-01-24T13:49:17.690" LastActivityDate="2015-01-24T14:52:44.817" Title="What is actually a redox reaction?" Tags="&lt;redox&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ClosedDate="2017-05-15T03:52:09.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="24117" PostTypeId="2" ParentId="24107" CreationDate="2015-01-24T12:52:44.580" Score="8" Body="&lt;p&gt;The key issue here that we are comparing excited states. So let's begin by defining Singlet and Triplet Excited States.&#xA;Singlet and Triplet Excited States:  A singlet or a triplet can form when one electron is excited to a higher energy level.  In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired).  In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in the illustrative Figure below. Singlet, doublet and triplet is derived using the equation for multiplicity, $2S+1$, where $S$ is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up ($s = +1/2$) or spin down ($s = -1/2$).  If we were to calculated the S for the excited singlet state, the equation would be $2[(+1/2) + (-1/2)]+1 = 2(0)+1 = 1$, therefore making the center orbital in the figure a singlet state.  If the spin multiplicity for the excited triplet state was calculated, we obtain $2[(+1/2) + (+1/2)]+1 = 2(1)+1 =3$, which gives a triplet state as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PknVI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustrative Figure: Spin in the ground and excited states The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. ***For this reason, the lifetime of the triplet state is longer than the singlet state by approximately 104 seconds fold difference.***The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-01-24T13:01:56.990" LastActivityDate="2015-01-24T13:01:56.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24120" PostTypeId="2" ParentId="24115" CreationDate="2015-01-24T13:51:41.670" Score="7" Body="&lt;p&gt;A redox reaction is any reaction in which oxidation and reduction are occurring. The word redox is a portmanteau of REDuction-OXidation. So, if we can correctly identify reduction and oxidation, we can identify redox reactants. In practice, this can be a little challenging, so I will focus first on the second part of your questions (definitions) and then go back to the reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ionization is any process that produces ions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The conversion of a hydrogen atom to a hydrogen ion by loss of an electron is an ionization. &#xA;$$\ce{H -&amp;gt; H+ + e-}$$&#xA;So to is the reaction of hydrochloric acid and water, which produces hydronium ions and chloride ions.&#xA;$$\ce{HCl + H2O -&amp;gt; Cl- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxidation is a loss of electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ionization of hydrogen atoms to hydrogen ions is an oxidation.&#xA;$$\ce{H -&amp;gt; H+ + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to is the conversion of carbon monoxide to carbon dioxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2CO + O2 -&amp;gt; 2CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this type of reaction is what gave us the word oxidation - compounds are reacting with oxygen and atoms end up with more oxygen atoms bound to them. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reduction is a gain of electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ionization of chlorine atoms to chloride ions is a reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl + e- -&amp;gt; Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So too is the conversions of acetylene to ethane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H2 + 2H2 -&amp;gt; C2H6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all oxidations and reductions are ionizations. There is a simple mnemonic in English for remembering oxidation and reduction: OIL RIG (Oxidation Is Loss Reduction Is Gain).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxidation and reduction happen together. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In principle we can write an equation that shows only oxidation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H -&amp;gt; H+ + e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but oxidation cannot happen on its own. That electron that was lost needs to be picked up by something else. This reaction that shows only oxidation is called a &quot;half-reaction&quot; and needs to be coupled with a reduction half-reaction to show a valid chemical reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H -&amp;gt; H+ + e-}$$&#xA;$$\ce{Cl + e- -&amp;gt; Cl-}$$&#xA;$$\ce{H + Cl -&amp;gt;HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other examples I gave, both processes are occurring. In the oxidation of carbon monoxide, the oxygen molecules are being reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surest way to identify oxidation and reduction (and thus redox) is to compare the oxidation states of the reactants and the products. Determining oxidation states/numbers is more complex that I want to put here, so start with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxidation_state#Determining_the_oxidation_state_or_number&quot;&gt;Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the carbon monoxide example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  reactants    products&#xA;C                 +2           +4&#xA;O (bonded to C)   -2           -2&#xA;O2                0            -2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The carbon atoms are being oxidized (oxidation number increased) and the oxygen atoms in $\ce{O2}$ are being reduced (oxidation number decreases). The oxygen atom already attached to carbon did not change its oxidation number.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-24T13:51:41.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24121" PostTypeId="2" ParentId="24115" CreationDate="2015-01-24T14:03:05.143" Score="1" Body="&lt;p&gt;A redox reaction involve transfer of electrons between species and the term redox comes indeed from the combination of terms &quot;oxidation&quot; and &quot;reduction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The species which is oxidized loses electrons whereas the species which is reduced (with higher oxidation potential) gains electrons. An oxidant (oxidating agent) is a species that can oxidize other species, whereas reducing agents can be oxidized, thus transferring electrons to other species (subsequently reduced). While being reduced, the oxidation number of a species decreases and vice versa, oxidation implies an increase in the oxidation number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of everyday life redox reaction is rusting, which consists in oxidation of iron by oxygen thus forming $\ce{Fe(III)}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Fe + 3O2 -&amp;gt; 2Fe2O3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the cellular respiration, a complex pathway of oxidative degradation of sugars producing energy (stored as chemical energy in the acid anhydride bonds of ATP). Although the process is quite complex involving several enzymes and many steps, the overall reaction is quite simple: &#xA;$$\ce{C6H12O6 + 6 O2 -&amp;gt; 6 CO2 + 6 H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorDisplayName="user12839" LastEditDate="2015-01-24T14:52:44.817" LastActivityDate="2015-01-24T14:52:44.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24122" PostTypeId="2" ParentId="24112" CreationDate="2015-01-24T14:31:12.193" Score="7" Body="&lt;p&gt;The 6 ligands are at the 6 vertices of an octahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be cis and trans isomers, but neither isomer is optically active.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the trans isomer, there will be a central axis containing Cl-Co-Cl, and the 4 N atom lie in a plane of symmetry, bisecting the Cl-Co-Co line segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cis isomer, there is a 90 degree Cl-M-Cl angle, with the line segment Cl-Cl being an edge of the octahedron.  The plane perpendicularly bisecting the Cl-Cl segment is a plane of symmetry. The plane passes through the Co and 2 of the N atoms as well.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-24T14:31:12.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24124" PostTypeId="2" ParentId="24107" CreationDate="2015-01-24T14:51:22.320" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is triplet state stable or singlet state ??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends upon the spin system (atom, molecule, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamagnetic compounds have singlet ground states, so for many compounds the ground states are singlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of triplet ground states include dioxygen.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Singlet_oxygen&quot; rel=&quot;nofollow&quot;&gt;Singlet dioxygen&lt;/a&gt; is a excited state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For dihydrogen, the singlet state is the ground state, but at room temperature, there is an equilibrium concentration of 75% triplet state and 25% ground state. Near zero kelvins, the singlet state dominates, but the energy difference between the states is so small that, in accordance with Boltzmann statistics, at room temperature the states are populated in proportion to the number of microstates, the triplet having 3 and the singlet having 1. But in this spin system, the spins are proton spins not electron spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methylene_(compound)&quot; rel=&quot;nofollow&quot;&gt;Methylene&lt;/a&gt; also has a triplet ground electronic state and singlet excited state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can an atom stay long time in a triplet state though the electrons have same spin?? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, electronic &quot;triplet state&quot; doesn't exactly mean the electrons have the same spin.  The three possibilities referred to by the word &quot;triplet&quot; are 1. both spins are up 2. both spins are down and 3.  $1/\sqrt2$(up,down + down,up)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the lifetime of any excited state is not fixed even for a particular spin system.  Mechanisms for the system to transition to the ground state need to be considered.  In the gas phase the excited state may be long lived but it could be many orders of magnitude less in a liquid solvent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to give an example, for atomic helium in the triplet 1s2s state, the lifetime to decay by radiation to the 1s1s singlet state is 7859 seconds.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-24T15:38:43.883" LastActivityDate="2015-01-24T15:38:43.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24125" PostTypeId="1" AcceptedAnswerId="24130" CreationDate="2015-01-24T14:55:19.717" Score="5" ViewCount="543" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3S5e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling with the above question. It will be the conjugate base of trichloromethane that is involved. The same is true of the carboxylate. However, I have no idea about the reaction mechanisms. Please help.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-25T20:18:10.317" Title="Making three membered ring using different reactants" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24126" PostTypeId="2" ParentId="24112" CreationDate="2015-01-24T15:01:27.380" Score="8" Body="&lt;p&gt;$\ce{Co[(NH3)_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dw0s6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.tutorvista.com/inorganic-chemistry/coordination-compound.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a molecule contains a plane of symmetry it &lt;strong&gt;cannot&lt;/strong&gt; be optically active or chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cis isomer there is a plane of symmetry containing the $\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active.  With the trans isomer there are several planes of symmetry (one contains the $\ce{Co}$ atom and the four $\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-24T16:17:51.893" LastActivityDate="2015-01-24T16:17:51.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24128" PostTypeId="2" ParentId="24125" CreationDate="2015-01-24T15:10:10.377" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It will be the conjugate base of trichloromethane that is involved. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Only as the initial intermediate! $¸\ce{CHCl3 + OH- -&amp;gt;Cl3C^- + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbanion will lose chloride and yield dichlorocarbene $\ce{Cl3C- -&amp;gt; Cl- + Cl2C:}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter then adds to the $\ce{C=C}$ double bond.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The same is true of the carboxylate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No! In this case, decarboxylation (loss of $\ce{CO2}$) is the first step. The fate of the carbanion is however as described above.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-24T15:10:10.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24129" PostTypeId="2" ParentId="24125" CreationDate="2015-01-24T15:11:45.350" Score="6" Body="&lt;p&gt;I think that both reactions involve carbene as common intermediate which then is the species involved in the subsequent cyclopropanation reaction.&#xA;These are the two equations which lead to the formation of the carbene:&#xA;$$\ce{KOH + CHCl3 -&amp;gt;  :CCl2 + KCl + H2O}$$&#xA;$$\ce{Cl3CCO2Na -&amp;gt;  :CCl2 + NaCl + CO2}$$&#xA;Subsequently, the alkene undergoes cyclopropanation by dichlorocarbene.&#xA;In the second case, heat is required in order to promote decarboxylation, which is entropically favored.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="11593" LastEditDate="2015-01-25T20:18:10.317" LastActivityDate="2015-01-25T20:18:10.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24130" PostTypeId="2" ParentId="24125" CreationDate="2015-01-24T15:16:28.977" Score="6" Body="&lt;p&gt;You are correct that the first step with both reagents involves the formation of the trichloromethyl anion ($\ce{CCl3^{-}}$).  This anion can eject $\ce{Cl^{-}}$  and irreversibly form dichlorocarbene ($\ce{:CCl2}$).  &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbenes&lt;/a&gt; are very reactive intermediates, reactive enough to add to a double bond and form a strained 3-membered ring (a cyclopropane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction you've shown is said to be stereospecific.  That is, you started with cis-2-butene and wound up with a cis-dimethyl-dichlorocyclopropane; the cis olefin geometry is preserved in your cyclopropane product.  This indicates that the reaction involved singlet dichlorocarbene.  Carbenes can exist as singlets or triplets.  As the following figure indicates, singlet carbenes add stereospecifically (your case), triplets do not.  The &quot;slow spin flip&quot; step in the triplet carbene reaction is slow enough that rotation can occur about what was the olefinic bond and this rotation gives rise to cis and trans cyclopropane isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4UOph.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-24T15:23:39.677" LastActivityDate="2015-01-24T15:23:39.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24131" PostTypeId="2" ParentId="24107" CreationDate="2015-01-24T15:45:14.960" Score="0" Body="&lt;p&gt;Stability is about the energy level which is a thermodynamic factor while lifetime is a kinetic factor. That is why they are not necessarily agree with each other.&#xA;Triplet live longer simply because it is kinetically harder for the excited electron to jump back to the original orbital and at the same time flip its spinning state. It is not really connected to the stability of it. Stability is more complicated depending many other aspects.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2015-01-24T15:45:14.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24133" PostTypeId="2" ParentId="24111" CreationDate="2015-01-24T15:58:18.503" Score="3" Body="&lt;p&gt;The logic is wrong. Lower &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; (actually you mean &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p} K_\mathrm{a}$&lt;/span&gt; here) does not mean more polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more complicated to decide the polarity then you think. For reverse phase HPLC, the more polar and more soluble compound come out first. For reverse phase HPLC, it is common to have acid (TFA for example) in the eluent. It this case, caffeine can be protonated and become much more polar because it is ionic now. Solubility is another factor to consider. So if you want to predict the order, you have to provide more information.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="83758" LastEditDate="2020-06-03T12:16:17.013" LastActivityDate="2020-06-03T12:16:17.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24134" PostTypeId="1" AcceptedAnswerId="24137" CreationDate="2015-01-24T16:14:22.803" Score="5" ViewCount="891" Body="&lt;p&gt;In elementary school, I happened to ask my teacher:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is $\ce{CO}$ they talk about these days?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was then taught that Carbon monoxide &quot;competes&quot; with the oxygen in our blood (we were  taught that oxygen gets to our cells via blood at that time) and replaces it and thus, doesn't let enough O get to cells and this will be fatal in extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't do any research about it after that because I later made the implication that $\ce{CO}$ &lt;em&gt;is&lt;/em&gt; able to replace O and react with Hb instead; but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently became doubtful about my previous lessons and decided to do a research to see what really was going on:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon monoxide mainly causes adverse effects in humans by combining with hemoglobin to form carboxyhemoglobin (HbCO) in the blood. This prevents hemoglobin from releasing oxygen in tissues, effectively reducing the oxygen-carrying capacity of the blood, leading to hypoxia. Additionally, myoglobin and mitochondrial cytochrome oxidase are thought to be adversely affected. Carboxyhemoglobin can revert to hemoglobin, but the recovery takes time because the HbCO complex is fairly stable. &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_monoxide_poisoning&quot;&gt;Wikipedia, CO poisoning&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states that the problems are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO}$ is able to combine with Hb to form HbCO.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This compound is fairly stable, so we'll have a hard time removing $\ce{CO}$ from hemoglobin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Other sources, such as &lt;a href=&quot;http://science.howstuffworks.com/question190.htm&quot;&gt;How stuff works&lt;/a&gt;, &lt;a href=&quot;http://www.answers.com/Q/Why_is_CO_toxic&quot;&gt;Answers.com&lt;/a&gt; and &lt;a href=&quot;http://www.cdc.gov/co/faqs.htm&quot;&gt;CDC.gov&lt;/a&gt; have given either only more explicit explanations of wikipedia, or the explanations weren't &quot;chemical-biology&quot; enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen a single reference in which the first assumption was declined, or agreed with. Is what I'd learnt non-scientific?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="1887" LastEditDate="2015-01-24T16:31:50.220" LastActivityDate="2015-01-26T13:12:53.473" Title="CO poisoning - What I've been taught is a hoax?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24135" PostTypeId="5" CreationDate="2015-01-24T16:38:10.440" Score="0" Body="&lt;p&gt;Relates to all chemical substance that are in the solid phase; especially non-molecular solid material. Questions including this tag can be about ways of synthesizing solid material through chemical or physical reactions, or properties concerning compounds that are in solid phase.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-01-24T22:50:47.160" LastActivityDate="2015-01-24T22:50:47.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24136" PostTypeId="4" CreationDate="2015-01-24T16:38:10.440" Score="0" Body="Study of chemical species that are in solid phase, either in cases of their properties, or about ways of synthesizing them." OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-01-24T19:38:00.790" LastActivityDate="2015-01-24T19:38:00.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24137" PostTypeId="2" ParentId="24134" CreationDate="2015-01-24T16:46:17.153" Score="12" Body="&lt;p&gt;Each hemoglobin molecule has 4 subunits (4 hemes each having one iron atom). Each time a molecule (such as oxygen or CO) binds to one subunit, it changes the binding properties of the other subunits. This is referred to as positive &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cooperative_binding&quot; rel=&quot;nofollow&quot;&gt;cooperative binding&lt;/a&gt;&quot; in the case of hemoglobin, because binding one molecule makes the other sites bind more easily.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, considering hemoglobin with no CO or oxygen bound, CO competes with oxygen for the same binding sites (ligating one of the iron atoms).  CO has a binding constant 300 times lower than oxygen, meaning a factor of 300 lower CO concentration will result in an equal percent of bound protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering CO has reached a level such that 2 out of the 4 sites can have CO bound, the remaining 2 sites would still provide enough oxygen carrying capacity for a person to live if the sites were function as normal; however, the 2 remaining sites then bind oxygen too tightly and don't release it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other iron-containing proteins such as myoglobin can also bind CO and are also inhibited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.bvsde.paho.org/bvsacd/cd51/monoxide/cap05.pdf&quot; rel=&quot;nofollow&quot;&gt;Pharmacokinetics and Mechanisms of Action of Carbon Monoxide&lt;/a&gt; for more infomation.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-26T13:12:53.473" LastActivityDate="2015-01-26T13:12:53.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24140" PostTypeId="1" AcceptedAnswerId="24142" CreationDate="2015-01-24T17:16:26.683" Score="4" ViewCount="418" Body="&lt;p&gt;Each goes through a carbocation intermediate which is enormously reactive so what would promote attack towards a proton (E1) rather than towards the carbocation for substitution. Besides an increase in temperature I can't think of much.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2015-10-02T16:50:19.403" LastActivityDate="2015-10-02T16:50:19.403" Title="What makes E1 more favourable than SN1?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24142" PostTypeId="2" ParentId="24140" CreationDate="2015-01-24T17:50:43.647" Score="3" Body="&lt;p&gt;The most important factor is the nucleophilicity versus basicity of the attacking species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the trimethylcarbenium cation as a classic example:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ahjgA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sterically hindered bases such as lithium diisopropyl amide (LDA) will favour elimination because it is physically much easier for them to remove a small proton from the outside of the cation than to reach the carbon atom at the centre surrounded by three methyl groups and form a new bond there. On the other hand, small species such as cyanide and water will favour substitution because because it is physically easier for them to reach the central carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally the strength of the base influences which reaction takes place. LDA is a strong base and the methyl hydrogens are quite acidic means that the elimination reaction is much more likely. However, weak bases such as iodide and water will favour substitution because they are less basic and so they are less likely to abstract an acidic proton from one of the methyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be noted that the nature of the solvent has a strong effect both on the stability of the cation but also on the nucleophilicity of the attacking species. This page has some discussion on the effects of solvents on nucleophilicity and the website also has good discussions of elimination and substitution mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-24T17:50:43.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24144" PostTypeId="1" CreationDate="2015-01-24T19:10:44.707" Score="5" ViewCount="1219" Body="&lt;p&gt;I was asked to write a balanced equation, as well as the net ionic equation, for the formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt; by precipitation. The equations should also include state symbols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I am not familiar with the concept of precipitation. Could anybody please show me a method for constructing these equations?&lt;/p&gt;&#xA;" OwnerUserId="12970" LastEditorUserId="16683" LastEditDate="2018-12-06T23:15:57.107" LastActivityDate="2018-12-10T13:28:17.203" Title="Constructing an equation for the precipitation of a salt" Tags="&lt;ionic-compounds&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24145" PostTypeId="1" AcceptedAnswerId="24148" CreationDate="2015-01-24T19:14:30.137" Score="5" ViewCount="1554" Body="&lt;p&gt;Typically the leaving group for E1cb is poor (like -OH or -OR) but why must this be the case? The substrate appears in the rate equation so surely a good leaving group would be beneficial?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-24T23:08:28.563" Title="Why must the leaving group in E1cb be poor?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24148" PostTypeId="2" ParentId="24145" CreationDate="2015-01-24T19:40:08.017" Score="9" Body="&lt;p&gt;There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another.  For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another.  An activation energy is associated with each of these 3 reaction pathways.  Whichever pathway has the lowest activation energy will be the major pathway followed.  By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Typically the leaving group for E1cb is poor (like -OH or -OR) but why&#xA;  must this be the case?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other.  To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group.  If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms.  If we change our leaving group to one that is an extremely &lt;strong&gt;poor leaving group&lt;/strong&gt; (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes &lt;strong&gt;less competitive&lt;/strong&gt; with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism.  Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged.  Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-24T23:08:28.563" LastActivityDate="2015-01-24T23:08:28.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24149" PostTypeId="2" ParentId="24144" CreationDate="2015-01-24T19:59:19.973" Score="5" Body="&lt;p&gt;Precipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, &lt;span class=&quot;math-container&quot;&gt;$\ce{AgBr}$&lt;/span&gt;, is insoluble in water. However, it is made up of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; ions, which we can introduce in the form of soluble salts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{LiBr(aq) + AgNO3(aq) -&amp;gt; LiNO3(aq) + AgBr(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for &quot;solid&quot;. This reflects its insolubility in water and hence its tendency to &quot;precipitate&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called &lt;strong&gt;spectator ions&lt;/strong&gt;. For example, in the reaction above, only &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Br-}$&lt;/span&gt; participated in precipitation. Thus, the net ionic equation for the reaction above is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ag+(aq) + Br-(aq) -&amp;gt; AgBr(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of deriving this net ionic equation is to realise that salts such as &lt;span class=&quot;math-container&quot;&gt;$\ce{LiBr}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; do not exist as molecules in water, but rather dissociate to give their constituent ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\ce{LiBr(s) &amp;amp;-&amp;gt;[dissolution] Li+(aq) + Br-(aq)} \\&#xA;\ce{AgNO3(s) &amp;amp;-&amp;gt;[dissolution] Ag+(aq) + NO3-(aq)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;span class=&quot;math-container&quot;&gt;$\ce{AgBr}$&lt;/span&gt; is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{\underbrace{Li+(aq) + Br-(aq)}_{from LiBr} + \underbrace{Ag+(aq) + NO3-(aq)}_{from AgNO3} -&amp;gt; \underbrace{Li+(aq) + NO3-(aq)}_{from LiNO3} + \underbrace{AgBr(s)}_{does not dissociate}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectator ions which do not take part in the reaction are &lt;span class=&quot;math-container&quot;&gt;$\ce{Li+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt;, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ag+(aq) + Br-(aq) -&amp;gt; AgBr(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take this idea and apply it back to &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt;, we first need to come up with soluble sources of the ions &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt;. A possible choice is &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2CO3}$&lt;/span&gt;, although that is hardly the only option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall equation may be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaCl2(aq) + Na2CO3(aq) -&amp;gt; 2NaCl(aq) + CaCO3(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the net ionic equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ca^2+(aq) + CO3^2-(aq) -&amp;gt; CaCO3 (s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; are the spectator ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here are some web pages you might find helpful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="16683" LastEditDate="2018-12-06T23:29:20.923" LastActivityDate="2018-12-06T23:29:20.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24151" PostTypeId="1" CreationDate="2015-01-24T20:44:10.420" Score="6" ViewCount="1439" Body="&lt;p&gt;I have an enzymatic equation in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E + S\xleftrightarrow[k_{-1} = 2.6\text{ x }10^{-2} \text{s}^{-1}]{k_1 = 3 \text{ x }10^8 \text{M}^{-1}\text{s}^{-1}} EX \xleftrightarrow[k_{-2} = 3.8\text{ x }10^{-8} \text{M}^{-1}\text{s}^{-1}]{k_2 = 1.45 \text{ x }10^3 \text{s}^{-1}} E + P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the forward reaction, I have calcluated the $K_m$ value to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${K_m}_f = \frac{k_{-1}+k_2}{k_1} = 4.83\text{ x }10^{-6}\text{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the reverse, I calculated $K_m$ to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${K_m}_r = \frac{k_{-1}+k_2}{k_{-2}} = 3.82\text{ x }10^{-6}\text{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this info, I need to somehow calculate the specific activity of the enzyme in both the forward and reverse reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of one enzyme unit is 50,000 kDa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific activity is defined as the number of 'Units' per mg of protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One enzyme unit is defined as the amount of enzyme that catalyzes the formation of 1 μmol of product per min under optimal assay conditions, which are assumed for this question (saturation of substrate in forward reaction / saturation of product in reverse reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me in the right direction? I have a feeling that this is related to the $V_{max}$ term in the Michaelis-Menten equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \frac{V_{max}\text{[S]}}{K_m + \text{[S]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that $K_m = \text{[S] at}\frac{1}{2}V_{max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $$[S]_{f} \text{ at } V_{max} = 2(4.83\text{ x }10^{-6}\text{M}) = 9.66\text{ x }10^{-6}\text{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $$[S]_{r} \text{ at } V_{max} = 2(3.82\text{ x }10^{-6}\text{M}) = 7.64\text{ x }10^{-6}\text{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I dont know how to relate it to specific activity.&lt;/p&gt;&#xA;" OwnerUserId="9604" LastEditorUserId="9604" LastEditDate="2015-01-26T18:32:58.790" LastActivityDate="2015-01-26T20:47:40.457" Title="Finding specific activity of an enzyme from Km values for a reversible enzyme reaction" Tags="&lt;biochemistry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24152" PostTypeId="1" AcceptedAnswerId="24155" CreationDate="2015-01-24T21:02:19.597" Score="8" ViewCount="6447" Body="&lt;p&gt;I recently studied an article about predicting earthquakes and how correctly realizing that an increase in $\ce{Rn}$ gas is a sign for the earthquake saved a whole city. The following is derived from &lt;a href=&quot;http://en.wikipedia.org/wiki/Earthquake_prediction#Radon_emissions&quot;&gt;the wikipedia page for earthquake prediction&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are reports of spikes in the concentrations of such gases prior to a major earthquake; this has been attributed to release due to pre-seismic stress or fracturing of the rock. One of these gases is radon, produced by radioactive decay of the trace amounts of uranium present in most rock.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Radon is useful as a potential earthquake predictor because being radioactive it is easily detected, and its short half-life (3.8 days) makes it sensitive to short-term fluctuations. A 2009 review found 125 reports of changes in radon emissions prior to 86 earthquakes since 1966.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, wikipedia then attempted to prove that most of the assretions related to radon are false statements,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but all this leaves me with one question: Why radon? (Why isn't another chemical species used for identifying the danger of feasible earthquake?) Maybe things I studied were too technical for me to understand, so please be as simple as you can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Feel free to edit tags)&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2017-12-26T17:16:31.217" Title="Radon gas in earthquake prediction - Why Rn?" Tags="&lt;physical-chemistry&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24155" PostTypeId="2" ParentId="24152" CreationDate="2015-01-24T23:40:27.897" Score="10" Body="&lt;p&gt;Uranium is found in low-levels in &lt;strong&gt;all&lt;/strong&gt; rocks and soil.  Radon is a gaseous radioactive decay product of uranium.  As the uranium undergoes radioactive decay, radioactive radon is generated and &lt;strong&gt;trapped&lt;/strong&gt; in the rocks that contain the uranium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earthquake theory involving radon suggests that prior to the actual quake, there is some subterranean movement where rocks are crushed, soil is uncompacted and the trapped radon is released producing a pre-quake spike in radon concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So radon presents the following attributes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is present in all rocks and soil&lt;/li&gt;&#xA;&lt;li&gt;if a rock is broken or if soil is disturbed, radon will be released&lt;/li&gt;&#xA;&lt;li&gt;it is gaseous and air currents / thermal gradients will carry it up to the earth's surface producing a detectable plume&lt;/li&gt;&#xA;&lt;li&gt;it is radioactive, this makes detecting small amounts or small changes in radon concentration relatively routine due to well-developed and very sensitive methods for detecting and accurately measuring radioactivity.&lt;/li&gt;&#xA;&lt;li&gt;radon has a very short radioactive half-life, a bit under 4 days.  Being a gas and having a short half-life is very useful in terms of measuring radon emissions.  If a radon emission spike occurs, the gas will dissipate quickly and after about 10 half-lives (40 days) normal background levels of radioactivity will return.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-25T00:50:06.663" LastActivityDate="2015-01-25T00:50:06.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24157" PostTypeId="1" AcceptedAnswerId="24159" CreationDate="2015-01-25T01:28:50.747" Score="13" ViewCount="15697" Body="&lt;p&gt;Is a hydrogen bond considered to be a Van der Waals force?&lt;/p&gt;&#xA;" OwnerUserId="12975" LastEditorUserId="9961" LastEditDate="2015-04-15T20:15:40.763" LastActivityDate="2020-08-11T10:54:43.060" Title="Is a hydrogen bond considered to be a van der Waals force?" Tags="&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24158" PostTypeId="1" AcceptedAnswerId="24160" CreationDate="2015-01-25T01:44:42.947" Score="8" ViewCount="6543" Body="&lt;p&gt;I am looking at the effects of increasing the surface area of which the flowing hydrogen fuel is in contact with the anode in a fuel cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode the hydrogen is catalytically split into protons and electrons. The catalyst used is most commonly platinum. So, for a fixed flow rate of hydrogen, increasing the contact surface area with the anode will increase the number of hydrogen atoms in contact with the catalyst, and the duration that they are in contact with it. Therefore the number of oxidization half cell reactions will increase. But what is the mathematical relationship that governs the number of reactions that will occur? Is it just a linear relationship, i.e doubling the surface area doubles the number of reactions, or is it more complex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is impossible to get an exact mathematical relationship, what is a good approximation on how these variables will scale with each other? Also, how do chemist usually predict the number of reactions/reaction rate, and could this be applied to my example?&lt;/p&gt;&#xA;" OwnerUserId="4422" LastEditorUserId="194" LastEditDate="2017-04-08T00:53:57.887" LastActivityDate="2017-04-08T00:53:57.887" Title="Relationship between surface area of electrode and reaction rate of hydrogen in fuel cells" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;hydrogen&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24159" PostTypeId="2" ParentId="24157" CreationDate="2015-01-25T01:55:41.873" Score="15" Body="&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/V06597.html&quot;&gt;IUPAC gold book&lt;/a&gt; a van der Waals force is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-25T01:55:41.873" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24160" PostTypeId="2" ParentId="24158" CreationDate="2015-01-25T02:37:02.173" Score="6" Body="&lt;p&gt;Any galvanic cell produces its equilibrium voltage at open-circuit.  Let's say your cell is at +3 V before you allow any current to flow through it.  As soon as you throw the switch, you allow that 3 V to push a current through the cell and some external load that you connect to it. There are several resistances within the cell that act to reduce the cell voltage from the initial 3 V.  The electrolyte resistance, concentration dependent resistances at the electrodes and electron transfer resistances all contribute to the total voltage drop in the cell, and hence, effectively limit the amount of current it can generate.  Typically, one of these resistances is much larger than any of the others depending on the exact configuration and operating conditions of the cell.  It is the electron transfer resistance that you are concerned with in your question.  The traditional way of expressing the voltage drop due to electron transfer resistance is the Tafel equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta _{a}=a+b\, log\, i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a$ and $b$ are empirical constants, $i$ is the current &lt;strong&gt;density&lt;/strong&gt; and $\eta _{a}$ is what is called the activation overpotential.  We see from this expression that the amount of voltage drop in the cell due to electron transfer increases linearly with the logarithm of the current density.  It is desirable then to minimize $\eta _{a}$ for a battery or flow cell application so that we have the maximum amount of voltage available for pushing current through the load.  A very good way to do this, as you pointed out, is to simply increase the electrode area.  By doing so, we decrease the current &lt;strong&gt;density&lt;/strong&gt; which decreases $\eta _{a}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the exact mathematical relationship you seek would be the empirical Tafel expression given above, or even better, the Butler-Volmer equation which is really just a more fundamental way of expressing the Tafel relationship.  All you should need to do is locate some references on platinum polarization in aqueous solutions.  There just happen to be tons of papers on this subject.  Here is one to get you started:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sheng W. et al, &lt;em&gt;Journal of The Electrochemical Society&lt;/em&gt;, 157 11&#xA;  B1529-B1536 &lt;strong&gt;2010&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-25T02:37:02.173" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24161" PostTypeId="1" CreationDate="2015-01-25T04:22:40.807" Score="1" ViewCount="3645" Body="&lt;p&gt;The question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-Phenyl-2-Butanol can be synthesized by three different combinations of a grignard reagent and a ketone. Show each combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only think of one. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Butanone with Br-Mg-Ph&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can't think of anything else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not too sure if this will work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;phenylbutanone withe Br-Mg-Me&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now i'm stuck. i can't think of the last one.&lt;/p&gt;&#xA;" OwnerUserId="9581" LastEditorUserId="865" LastEditDate="2021-05-22T19:01:28.213" LastActivityDate="2021-05-22T19:01:28.213" Title="Synthesis of 2-Phenyl-2-Butanol from Grignard Reagent" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24163" PostTypeId="1" CreationDate="2015-01-25T06:24:48.347" Score="-5" ViewCount="496" Body="&lt;p&gt;I want to know a way to separate aluminum hydroxide into aluminum and hydroxide ions, with a chemical reaction, or another process if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have $\ce{Al(OH)3}$ and I want to separate it into $\ce{Al}$ and $\ce{{}^{-}OH}$.&lt;/p&gt;&#xA;" OwnerUserId="12977" LastEditorUserId="4945" LastEditDate="2015-01-26T03:47:28.363" LastActivityDate="2015-01-26T03:47:28.363" Title="Is it possible to separate aluminum hydroxide?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-01-26T05:29:03.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="24164" PostTypeId="1" CreationDate="2015-01-25T07:41:58.153" Score="3" ViewCount="58" Body="&lt;p&gt;Probiotics are microorganisms that are believed to provide health benefits when consumed. Water Kefir is one such probiotic drink. This has blend of bacteria, yeast (BY), sugar water and the result of action of BY on sugar water. There is alcohol. One of my problem is to find out the alcohol content in the solution. Less expensive, simpler methods/protocols are much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12978" LastEditorUserId="418" LastEditDate="2015-01-25T09:27:34.827" LastActivityDate="2015-01-25T12:01:51.520" Title="How to find the alcohol content of a probiotic liquid?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24165" PostTypeId="1" CreationDate="2015-01-25T09:10:34.703" Score="5" ViewCount="1689" Body="&lt;p&gt;We all know that the density of the nucleus is very high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclei are made up of protons and neutrons, and while protons have the same charge, they are closely packed in a nucleus. How does the repulsion between protons not break apart the nucleus?&lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="39" LastEditDate="2015-01-25T14:12:55.677" LastActivityDate="2015-01-26T22:02:02.080" Title="How are nuclei stable?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;nuclear&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24166" PostTypeId="2" ParentId="16088" CreationDate="2015-01-25T10:12:09.917" Score="1" Body="&lt;p&gt;That's a good question. In literature there are some contrasting opinions about the antioxidant power of BHA vs BHT. TBHQ (2-tert-butyl hydroquinone) is also used as antioxidant, although maybe is less common.&#xA;There is a paper from 1997 in which the antioxidant activity of BHA, BHT and TBHQ are examined with Miller's test (M. Karamac and R. Amarowicz, Grasas y Aceites, 1997, 48 (2), 83-86), and in this test BHA and BHT gave similar results, although in the initial phase of the experiment BHA seemed to have a higher antioxidant activity than BHT.&#xA;In the paper precedent studies are also reported. Quoting from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher antioxidative activity of BHA than of BHT and lower for TBHQ was&#xA;demonstrated in thin layer model system of methyl linoleate on cellulose (J. Am. Oil Chem. Soc. 52,174-178). Berner et al. (J. Am. Oil Chem. Soc. 51, 292-296) reported antioxidative index in lard emulsion 30-fold higher for BHA than for TBHQ. Yet, in active oxygen method, antioxidant index for TBHQ was 3-fold higher than for BHA and BHT. Higher antioxidative activity of TBHQ compared with BHA and BHT in model studies with soybean oil and chicken fat in emulsion versus dry oil tests was reported by Cort et al. (J. Am. Oil Chem. Soc. 52,174-178). Relative effectiveness of antioxidants-linoleic acid monolayer on silica gel for TBHQ was almost the same as for BHT, while for BHA almost 3-fold higher (J. Food Sci. 42, 1533-1535). &lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-25T10:12:09.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24167" PostTypeId="2" ParentId="24165" CreationDate="2015-01-25T10:12:45.253" Score="0" Body="&lt;p&gt;Coulomb's electrostatic repulsion between two protons is extremely large as compared to the gravitational force of attraction between them. Therefore, as you understand correctly, if Coulomb's repulsive and gravitational attractive forces are the only forces operating inside the nucleus, it cannot be stable. The stability of nucleus has been attributed to the existence of a third type of force inside the nucleus called &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_force&quot; rel=&quot;nofollow&quot;&gt;Nuclear force&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference: Modern's abc of Physics by Satish K.Gupta, Part 2, Class 12.&lt;/p&gt;&#xA;" OwnerUserId="2328" LastEditorUserId="2328" LastEditDate="2015-01-25T10:42:57.473" LastActivityDate="2015-01-25T10:42:57.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24168" PostTypeId="2" ParentId="24164" CreationDate="2015-01-25T12:01:51.520" Score="1" Body="&lt;p&gt;Complex instrumental analytics (HPLC, photometry using a $\ce{NAD+/\,NADH}$ testing kit, etc.) is out of question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the most simple approach would be the use of a hydrometer to measure the density before and after fermentation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-25T12:01:51.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24172" PostTypeId="2" ParentId="24161" CreationDate="2015-01-25T13:25:57.330" Score="5" Body="&lt;p&gt;Hint: The carbon atom bearing the OH group in the product has three substituents--phenyl, methyl, and ethyl. Any one of these groups could be introduced as a Grignard reagent, with the appropriate ketone. You have already correctly deduced one of the three possibilities.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-25T13:25:57.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24174" PostTypeId="1" CreationDate="2015-01-25T14:20:42.597" Score="3" ViewCount="334" Body="&lt;p&gt;Which Basis Set is suitable For Mercury-complex in DFT calculations? Please provide the answers with any journal reference.&lt;/p&gt;&#xA;" OwnerUserId="12979" LastActivityDate="2015-01-25T14:20:42.597" Title="Which Basis Set is suitable For Mercury-complex in DFT calculations?" Tags="&lt;density-functional-theory&gt;&lt;basis-set&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24175" PostTypeId="2" ParentId="24165" CreationDate="2015-01-25T15:49:34.183" Score="4" Body="&lt;p&gt;Protons and neutrons in a nucleus are constantly emitting and absorbing&#xA;little particles.   When one nucleon emits a little particle called a &quot;meson&quot; that another nucleon absorbs, a strong force between the two nucleons results. This is called (strangely enough) the strong nuclear force&lt;sup&gt;1&lt;/sup&gt;, and it is strong enough to counteract the powerful electrostatic repulsion of the protons in a nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The modern explanation is richer and more complex than this. Since the 70's these &quot;mesons&quot; have been understood to be made up of smaller particles called quarks and gluons, which are also the building blocks of nucleons. In fact the strong force doesn't just bind the nucleus together; it holds the quarks in protons and neutrons together, too.  The contemporary explanation of the strong interaction is in the realm of quantum chromodynamics, and really beyond the scope of chemistry. You can read more about it &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Production and destruction of the messenger mesons violates the law of conservation&#xA;of mass and energy! However, if the messenger particle has a very short&#xA;lifetime, and so exists only for a very short time within a very small space,&#xA;the particle can exist within the limitations set by the uncertainty principle.&#xA;Particles like this are called &lt;em&gt;virtual particles&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclei are very small (on the order of femtometers in radius), so the range of&#xA;the strong nuclear force must be very small. You can make a back-of-the-envelope estimate of the range as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty principle says that you can’t &lt;em&gt;exactly&lt;/em&gt; determine the position&#xA;and momentum of a very small particle simultaneously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_x \, \sigma_p\ge \frac{\hbar}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma_x$ and $\sigma_p$ are the uncertainties in the position and momentum of a particle, and $\hbar$ is the reduced Planck constant.  Given that the mass&#xA;of the meson that mediates the strong force is $m \approx 2.4\times 10^{−28}$ kg, and the uncertainty in the velocity can’t be any larger than the speed of light $c$, you can compute $m c$ as a bound on $\sigma_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then estimate $\sigma_x$, which should be related to the maximum distance that a meson can exist from the nucleon that generated it without violating&#xA;the uncertainty principle, or the universal speed limit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_x \approx \frac{\hbar}{2 m c} = 0.7\ \text{fm}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s the right order of magnitude for the range of the strong nuclear force; it's also about the distance between nucleons in the nucleus, and about 6/10ths of the classical radius of a proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;1. There is a different nuclear force called the &lt;strong&gt;weak nuclear force&lt;/strong&gt; which plays a role in fission and radioactive decay, so calling it just &quot;nuclear force&quot; might be a little ambiguous. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-26T22:02:02.080" LastActivityDate="2015-01-26T22:02:02.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24176" PostTypeId="2" ParentId="24165" CreationDate="2015-01-25T15:56:36.667" Score="5" Body="&lt;p&gt;Since the gravitational force between two protons is negligible there must be another force holding the nucleus together. This is the strong nuclear force, which as the name suggests is extremely strong but it is also extremely short range and so it's effects are only felt on the scale of nuclei and baryons. As you can see in the graph, if two protons approach each other they repel each other due to electrostatic repulsion but once they get within about $\mathrm{3~fm}$ of each other the strong nuclear force begins to become significant and at about $\mathrm{2~fm}$ it starts to outweigh the electrostatic repulsion and bind the two protons together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also interesting to note that at even shorter ranges the strong nuclear force is repulsive and so there is an equilibrium position, indicated on the graph, where the strong nuclear attraction and the electrostatic repulsion are equal in magnitude and so the net force on the protons is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the graph shown is for two protons and so in nuclei containing more than two protons the numbers will be different but the principle is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3u1dR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-25T15:56:36.667" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24177" PostTypeId="1" AcceptedAnswerId="24226" CreationDate="2015-01-25T16:27:32.520" Score="4" ViewCount="996" Body="&lt;p&gt;From what I have read, E1 largely produces the more substituted alkene because of the fact that more substituted alkenes are more stable as a result of hyperconjugation. However, in E2 the situation is different for a number or reasons. One of these reasons is steric effects of potentially large substituents on the $\beta$- carbon which would promote formation of the less substituted alkene. This effect does not seem to be mentioned for E1 but surely it must still be valid, right? Do large subtituents or bulky bases promote the formation of less substituted alkenes in E1? If not why not? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-01-26T18:48:33.473" Title="Saytzev and Hofmann elimination in E1" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24180" PostTypeId="1" AcceptedAnswerId="24184" CreationDate="2015-01-25T18:39:07.580" Score="4" ViewCount="488" Body="&lt;p&gt;As you see in the picture, a burning match in a container causes suction and this is a traditional therapeutic method to relieve bone pain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q7YhC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason? I think it should be related to gas changes (&lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; being used for combustion and etc.), but I am not sure about it.&lt;/p&gt;&#xA;" OwnerUserId="12700" LastEditorUserId="83758" LastEditDate="2020-04-08T13:36:22.540" LastActivityDate="2020-04-08T13:36:22.540" Title="Why does a matchstick burning in a container cause suction?" Tags="&lt;everyday-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24181" PostTypeId="1" AcceptedAnswerId="24185" CreationDate="2015-01-25T18:40:01.133" Score="3" ViewCount="4809" Body="&lt;p&gt;I'm doing a research project on the fascinating chemiluminescent molecule luminol, and figured it would be a good idea to start breaking down the systematic name of the chemical and linking the terms to the structure. The systematic name of luminol is 5-amino-2,3-dihydrophthalazine-1,4-dione, and I get it all apart from the &quot;dihydro&quot; - any help? Thank you very much in advance! :)&lt;/p&gt;&#xA;" OwnerUserId="12743" LastActivityDate="2017-09-24T11:52:07.593" Title="What does the prefix &quot;dihydro&quot; in the systematic name of luminol refer to?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24182" PostTypeId="1" CreationDate="2015-01-25T18:45:32.893" Score="9" ViewCount="78461" Body="&lt;p&gt;I was boiling down water in a stainless steel pot as an experiment the other day. When the water was completely vaporised, a white substance was left behind on the bottom of the pot. I tried rubbing it off the bottom, but it would not come off. I'm unsure what this substance is. What is this substance?&lt;/p&gt;&#xA;" OwnerUserId="12981" LastEditorUserId="10185" LastEditDate="2015-01-25T19:50:20.573" LastActivityDate="2017-01-11T00:26:17.473" Title="What is the white substance left behind after boiling down water" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;aqueous-solution&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24183" PostTypeId="2" ParentId="20220" CreationDate="2015-01-25T18:57:33.410" Score="0" Body="&lt;p&gt;Well, as far my syllabus is concerned about, inspite of having F° configuration in $Ce(IV)$ it reduces itself easily to $Ce(III)$ as +3 O.S is predominantly stable in inner transitional metal series.&lt;/p&gt;&#xA;" OwnerUserId="12905" LastEditorUserId="12905" LastEditDate="2015-01-25T19:04:16.313" LastActivityDate="2015-01-25T19:04:16.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24184" PostTypeId="2" ParentId="24180" CreationDate="2015-01-25T19:29:44.823" Score="4" Body="&lt;p&gt;When the container is open to the atmosphere and has the match burning in it, the temperature of the air inside is higher than the temperature of the surrounding air.  In order for pressure to be equal in the container and outside it, the density of the air in the container is less.  When the match is removed and the container placed on the person, two things happen.  One, no air can enter or leave the container, and two, the air in the container begins to cool.  When the air in the container cools its pressure drops.  The atmosphere pushes on the air in the container indirectly through the person, creating what is called suction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to drinking through a straw.  You are not sucking the liquid through the straw as much as you are removing the opposing force that prevents the atmosphere from forcing the liquid through the straw.  The opposing force being the air pressure in the straw.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2015-01-25T19:35:25.073" LastActivityDate="2015-01-25T19:35:25.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24185" PostTypeId="2" ParentId="24181" CreationDate="2015-01-25T19:36:24.027" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dB98E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left is 5-amino-2,3-dihydrophthalazine-1,4-dione and on the right is the molecule which the name is based on, phthalazine. The 2,3-dihydro refers to the fact that at the 2 and 3 positions (the two nitrogens) hydrogens have been added (hydro is the prefix for hydrogen).&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-25T19:36:24.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24186" PostTypeId="1" AcceptedAnswerId="24205" CreationDate="2015-01-25T19:41:59.600" Score="11" ViewCount="225" Body="&lt;p&gt;Recently concerned with health effects by common chemicals existent in food I've been rather busy reading article after article; and just an interesting one came around:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Acrylamide (or acrylic amide) is a chemical compound with the chemical formula $\ce{C3H5NO}$. &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrylamide&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some evidence suggests exposure to large doses can cause damage to the male reproductive glands. Acrylamide is considered a potential occupational carcinogen by U.S. government agencies. &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrylamide#Toxicity_and_carcinogenicity&quot;&gt;Still wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A series of case-control studies have investigated the relationship between dietary intake of acrylamide and the risk of developing cancers of the oral cavity, pharynx, esophagus, larynx, large bowel, kidney, breast, and ovary. These studies generally found no excess of tumors associated with acrylamide intake. &lt;a href=&quot;http://www.cancer.gov/cancertopics/factsheet/Risk/acrylamide-in-food&quot;&gt;Cancer.gov&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, though the references are generally controversial, you can conclude that &lt;strong&gt;though nothing is proven by the studies, take care&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fda.gov/Food/FoodborneIllnessContaminants/ChemicalContaminants/ucm053569.htm&quot;&gt;FDA&lt;/a&gt;: States that acrylamide &lt;em&gt;is&lt;/em&gt; a carcinogen of animals, is a human concern, but &quot;should I worry when eating high-temperature fried or roasted food?&quot; &quot;NO.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.food.gov.uk/science/acrylamide&quot;&gt;Food standards agency&lt;/a&gt;: Lab proves acrylamide a carcinogen, but real life didn't give up many victims, though this isn't a new issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along with researches done by FAO and WHO, still the carcinogenic concerns aren't very serious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a chemist, what reasons would you have provided that acrylamide is carcinogenic or not? Should I really worry about it when I eat french fries?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-20T20:58:43.747" Title="Is Acrylamide carcinogenic? Why?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;food-chemistry&gt;&lt;toxicity&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24187" PostTypeId="1" AcceptedAnswerId="24190" CreationDate="2015-01-25T20:02:26.523" Score="3" ViewCount="1222" Body="&lt;p&gt;I am wondering if it is possible to know exact amount of acid and base that we can put in together to make a buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make a 0.05 M extraction buffer using disodium hydrogen phosphate (DHP) and sodium dihydrogen phosphate (SDP). I want to make 100 mL of buffer so I worked out that I would require 0.78 g SDP and 0.88 g DHP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that it sometimes takes a lot of time by mixing one solution into the other and keep it checking with the pH meter, there has to be some mathematical way to get the exact amounts which would save time and my chemicals.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T14:32:34.303" LastActivityDate="2017-05-30T01:57:28.267" Title="Find amount of acid and base required to make buffer" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24188" PostTypeId="1" CreationDate="2015-01-25T20:10:14.897" Score="1" ViewCount="262" Body="&lt;p&gt;as far as I've understood to measure the decomposition potential of an electrolyte using a galvanic cell we just put two electrodes, connect them to a battery, and vary the external potential till when we observe a current. The potential at which the current starts to grow linearly is the decomposition potential of the electrolyte. There is not appreciable current before the decompositon potential because the electrodes are supposed to be inert and do not provide the ionic charges needed to &quot;close the circuit&quot;. Is it right? &#xA;An other question. How the role of the electrodes is taken into consideration in the experimental setup? Is it considered an offset due to an eventual nonzero cell potential? I made this question because I'd like to be sure that I'm understanding the experimental setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tirrel&lt;/p&gt;&#xA;" OwnerUserId="12945" LastActivityDate="2015-03-06T14:45:02.793" Title="offset in measuring decomposition potentials" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24189" PostTypeId="2" ParentId="24182" CreationDate="2015-01-25T20:34:42.927" Score="9" Body="&lt;p&gt;These are minerals that are naturally present in &quot;tap&quot; water. Chemically, they are most likely a mixture of calcium carbonate and magnesium carbonate, both of which are only very lightly soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you boil away the water, these dissolved calcium/magnesium carbonates remain behind and their concentrations eventually become greater than their respective solubility limits. At this point they begin to precipitate out of the solution, forming the 'scale' you see on the pot.  Interestingly, while most salts become more soluble at higher temperatures, calcium carbonate is anomalous in that its solubility decreases as temperature increases (see the accepted answer &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4094/white-powder-observed-after-boiling-water-in-electric-kettle-for-many-weeks?rq=1&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;), which accelerates the precipitation process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to get rid of them is to add a cup of white vinegar and let it sit in the pot. You might need to agitate the scale from time to time. You can boil the vinegar in the pot if you want, but it will stink! Over time, the vinegar will dissolve the precipitate, and you will see carbon dioxide bubbles forming. You can then pour out the vinegar and flush the pot with lots of water. &lt;/p&gt;&#xA;" OwnerUserId="12939" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-03-30T16:27:28.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24190" PostTypeId="2" ParentId="24187" CreationDate="2015-01-25T21:40:37.030" Score="4" Body="&lt;p&gt;Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; provides a relationship between the ratio of the two concentrations to the pH:&#xA;$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$&#xA;$$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$&#xA;$$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$&#xA;$$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$&#xA;$$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$&#xA;$$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$&#xA;$$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-25T21:40:37.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24191" PostTypeId="2" ParentId="10147" CreationDate="2015-01-25T23:37:23.727" Score="1" Body="&lt;p&gt;The potentiometric titration of mercury with dithiooxamide has been successfully applied after preliminary separation of mercury by reduction (Comprehensive Water Analysis, Volume 2, pp.186-187, Elsevier).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selective identification of mercury in waste water can be accomplished by fluorescence enhancement chemosensors (SENSOR LETTERS 2010, 8, 1–6). In this recent paper, the ﬂuorescent emission intensity of the probe is enhanced upon binding to mercury ions. The ﬂuorescence enhancement is attributed to a 1:1 complex formation between the probe and Hg(II) ion. The chemosensor can be applied for the quantiﬁcation of mercury; it shows a high selectivity toward Hg(II) ions in comparison with common metal ions.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-25T23:37:23.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24192" PostTypeId="2" ParentId="24163" CreationDate="2015-01-26T02:16:18.270" Score="1" Body="&lt;p&gt;It isn't really possible to have isolated hydroxide (-OH), because it is charged.  Only neutral compounds can be isolated, not ions.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-26T02:16:18.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24193" PostTypeId="1" AcceptedAnswerId="24214" CreationDate="2015-01-26T05:14:07.777" Score="7" ViewCount="1342" Body="&lt;p&gt;How is it possible that some insoluble compounds in water, e.g. $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ are considered strong bases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they even strong bases? How can I tell if a compound is a strong base or not?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to explain with equations and/or formulas what happens when these compounds are dissolved into water?&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="5026" LastEditDate="2018-06-30T05:24:37.770" LastActivityDate="2018-06-30T05:24:37.770" Title="How are these compounds strong bases, when they are sparingly soluble in water?" Tags="&lt;acid-base&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24194" PostTypeId="1" AcceptedAnswerId="24222" CreationDate="2015-01-26T06:37:51.810" Score="3" ViewCount="1136" Body="&lt;p&gt;How do you compare the strength of H bonds when it's the same atom on both sides of the H? Like $\ce{N-H\bond{...}N}$ vs $\ce{F-H\bond{...}F}$? Is one still stronger than the other, or are they both the same since in both cases the pull on both sides is equal?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="9961" LastEditDate="2015-04-06T21:58:53.393" LastActivityDate="2015-04-06T21:58:53.393" Title="Hydrogen bond comparison" Tags="&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24195" PostTypeId="1" CreationDate="2015-01-26T06:46:06.830" Score="9" ViewCount="60804" Body="&lt;p&gt;I have a debate with some friends who believe that saliva is just basic, but with the enzymes which make them reactive with objects in our mouth, shouldn't it be an acid in this case? &lt;/p&gt;&#xA;" OwnerUserId="12988" LastEditorUserId="7448" LastEditDate="2015-01-26T12:52:42.047" LastActivityDate="2016-07-25T04:04:10.163" Title="Is Saliva an acid or base?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24196" PostTypeId="5" CreationDate="2015-01-26T07:00:29.633" Score="0" Body="&lt;p&gt;From the IUPAC goldbook:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/P04528.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  An entity of a material system which is uniform in chemical composition and physical state. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/P04537.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;phase transition&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  A change in the nature of a phase or in the number of phases as a result of some variation in externally imposed conditions, such as temperature, pressure, activity of a component or a magnetic, electric or stress field.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples for systems of phases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aqueous solution of sodium chloride (1 phase, liquid)&lt;/li&gt;&#xA;&lt;li&gt;A block of aluminium (1 phase, solid)&lt;/li&gt;&#xA;&lt;li&gt;Air, approximated as a mixture of four parts nitrogen and one part oxygen gas (1 phase, gaseous)&lt;/li&gt;&#xA;&lt;li&gt;Water and diethyl ether (2 phases, liquid)&lt;/li&gt;&#xA;&lt;li&gt;Dry ice in air (2 phases, solid and gaseous)&lt;/li&gt;&#xA;&lt;li&gt;Water and ice (2 phases, solid and liquid)&lt;/li&gt;&#xA;&lt;li&gt;Water, ice and steam in a closed system (idealised at its triple point, 3 phases, solid, liquid and gaseous)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Examples of phase transition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boiling water (liquid to gaseous)&lt;/li&gt;&#xA;&lt;li&gt;Melting of iron (solid to liquid)&lt;/li&gt;&#xA;&lt;li&gt;sublimation of dry ice (solid to gaseous)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This tag should be applied to question regarding chemical and physical properties of phases, systems of phases and also phase transition processes.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="39539" LastEditDate="2017-05-18T10:18:09.350" LastActivityDate="2017-05-18T10:18:09.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24197" PostTypeId="4" CreationDate="2015-01-26T07:00:29.633" Score="0" Body="An entity of a material system which is uniform in chemical composition and physical state. This tag should be applied to question regarding chemical and physical properties of phases, systems of phases and also phase transition processes." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-01-26T07:00:29.633" LastActivityDate="2015-01-26T07:00:29.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24200" PostTypeId="1" CreationDate="2015-01-26T09:17:50.737" Score="2" ViewCount="1235" Body="&lt;p&gt;For example Ethyne in combustion by limited supply of air&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$     \ce{ 2C2H2 + 3O2 -&amp;gt;[\Delta] 2CO2 + 2H2O + 2C } $$ vs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$     \ce{ 2C2H2 + 5O2 -&amp;gt; 4CO2 + 2H2O + Energy } $$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the limits of air availability to make these combustions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to do these experiments in lab?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="4834" LastEditDate="2015-01-26T15:00:15.883" LastActivityDate="2015-01-26T15:00:15.883" Title="Can we do combustion caused by limited supply of air in a lab? How?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;combustion&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24201" PostTypeId="1" AcceptedAnswerId="24202" CreationDate="2015-01-26T11:00:32.523" Score="3" ViewCount="1147" Body="&lt;p&gt;When using hydrazine as a reduction agent to reduce graphene oxide, which is the product hydrazine oxidises to?&lt;/p&gt;&#xA;" OwnerUserId="7163" LastActivityDate="2015-01-26T14:04:29.000" Title="Hydrazine as a reduction agent" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24202" PostTypeId="2" ParentId="24201" CreationDate="2015-01-26T11:31:20.533" Score="1" Body="&lt;p&gt;&lt;em&gt;I'm sorry for the wrong answer initially given; you might consider to unaccept this one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether this has ever been examined in detail. It is likely that nobody bothered to collect the (volatile) reaction products of hydrazine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a procedure provided by the group of &lt;strong&gt;Rodney Ruoff&lt;/strong&gt; (&lt;a href=&quot;http://dx.doi.org/10.1016/j.carbon.2007.02.034&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;), the reduction was performed in an open flask equipped with a reflux condenser. An aqueous suspension GO was first sonicated (to support exfoliation) until it became clear, then hydrazine hydrate was added and the mixture refluxed at 100 °C for 24 hours. During this time, the reduction product precipitated as a black solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No remarks on what actually happened to the hydrazine were given in this article.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-01-26T14:04:29.000" LastActivityDate="2015-01-26T14:04:29.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24203" PostTypeId="2" ParentId="24201" CreationDate="2015-01-26T11:38:32.230" Score="-1" Body="&lt;p&gt;The decomposition reaction of hydrazine ($\ce{N2H4}$) is as below:&#xA;$$\ce{N2H4 \rightarrow N2H2 + H2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forming $\ce{H2}$ will reduce the graphene oxide and hydrazine itself oxidises to $\ce{N2H2}$.&lt;/p&gt;&#xA;" OwnerUserId="12979" LastEditorUserId="7448" LastEditDate="2015-01-26T12:09:07.327" LastActivityDate="2015-01-26T12:09:07.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24204" PostTypeId="1" AcceptedAnswerId="24207" CreationDate="2015-01-26T12:23:33.197" Score="6" ViewCount="13223" Body="&lt;p&gt;If electrons have a negative charge and protons have a positive charge&#xA;then how come neutrons have zero charge without consisting of protons and electrons?&lt;/p&gt;&#xA;" OwnerUserId="12993" LastEditorUserId="9961" LastEditDate="2017-05-10T19:57:07.903" LastActivityDate="2017-05-10T19:57:07.903" Title="Why neutrons are neutral?" Tags="&lt;neutrons&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ClosedDate="2017-05-10T21:20:06.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="24205" PostTypeId="2" ParentId="24186" CreationDate="2015-01-26T12:38:42.440" Score="9" Body="&lt;p&gt;Acrylamide is a Michael acceptor. All Michael acceptors are potentially carcinogenic because DNA can act as a Michael donor. This Michael reaction can damage the DNA, which can ultimately develop into cancer. Of course how well a certain Michael acceptor reacts with DNA depends on its chemical structure (i.e. the carbon chain and other function groups as well as its conformation), and hence not every Michael acceptor is equally carcinogenic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you worry about eating too much French fries I would first be concerned with my caloric  and sodium intake. For that there is solid evidence that it greatly reduces human health if taken in excess.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-20T20:58:43.747" LastActivityDate="2015-02-20T20:58:43.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24206" PostTypeId="2" ParentId="24200" CreationDate="2015-01-26T12:46:21.753" Score="2" Body="&lt;p&gt;Hmm, I see a couple of problems in here:&lt;/p&gt;&#xA;&lt;h2&gt;Energy in the reaction is misunderstood:&lt;/h2&gt;&#xA;&lt;p&gt;In general, breaking bonds is endothermic, and making them is exothermic. In any combustion reaction we have bonds broken (&lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;'s for example) and bonds created (forming &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;, in this instance). The overall change in the energy defines a reaction to be endo- or exothermic; and in the case of combustion (any type) it's exothermic. If you want to write &amp;quot;heat&amp;quot; and &amp;quot;energy&amp;quot; in both sides of the reactions you face, it's going to look a little awkward. Note that in the biochemical reactions in which you see terms like &amp;quot;energy&amp;quot; and such in only the side of reactants or the products (e.g.: ATP energy releases) it's because they want to put emphasis on that particular intake or release of energy.&lt;/p&gt;&#xA;&lt;h2&gt;The equation you've provided for incomplete combustion is incomplete:&lt;/h2&gt;&#xA;&lt;p&gt;I don't see any balanced equations in there, do you? And, I can't get the reason for why you've written two &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;s.&lt;/p&gt;&#xA;&lt;p&gt;However, your second equation is about &lt;strong&gt;the complete combustion of ethyne&lt;/strong&gt;, and is balanced and correct.&lt;/p&gt;&#xA;&lt;p&gt;As for your answer, take a look at what &lt;a href=&quot;http://en.wikipedia.org/wiki/Combustion#Incomplete_combustion_of_a_hydrocarbon_in_oxygen&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The incomplete (partial) combustion of a hydrocarbon with oxygen produces a gas mixture containing mainly &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2, CO, H2O}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Depending on how much your burning sample will be out of reach of air, the mixture will vary in weight percent of the products of the reaction.&lt;/p&gt;&#xA;&lt;p&gt;In a lab, you normally will get incomplete combustion reactions, for reasons:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your sample is hardly pure hydrocarbon.&lt;/li&gt;&#xA;&lt;li&gt;The air in the lab isn't in wild currents and overall, is &amp;quot;stale&amp;quot; in simple words.&lt;/li&gt;&#xA;&lt;li&gt;The hydrocarbon doesn't get enough &amp;quot;requirements&amp;quot; for combustion and ignites incompletely.&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JBj9P.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://sciencelearn.org.nz/var/sciencelearn/storage/images/contexts/fire/sci-media/images/burning-showing-incomplete-combustion/181659-5-eng-NZ/Burning-showing-incomplete-combustion_full_size_landscape.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ouYn.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bsquaredfutures.com/pluginfile.php/273/mod_imscp/content/1/lo_4-1-scorm/lo_4-1-scorm/images/bunsen_flames.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A good sign (or alarm) of the incomplete combustion is the orange hue of fire, especially if it's accompanied by a visible &amp;quot;smoke&amp;quot;. If the experiment requires complete combustion, we'll make use of burners such as Bunsen burner.&lt;/p&gt;&#xA;&lt;p&gt;So, let's summarize: &lt;strong&gt;The incomplete combustion is the normal thing happening in the burning of ethyne. You usually do some modifications if you're to have a complete combustion reaction. When the examiner claims &amp;quot;with contact with adequate supplies of air&amp;quot;, s\he only means that a complete combustion takes place, though this is rare in real life.&lt;/strong&gt; &lt;em&gt;Plus, ethyne is not a very good hydrocarbon to start with, because as &lt;a href=&quot;http://www.cdc.gov/niosh/npg/npgd0008.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;CDC&lt;/a&gt; mentions, it has a potentially negative effect on the nervous system, causing headaches, asphyxia etc., as well as painful frostbites.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-01-26T14:24:50.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24207" PostTypeId="2" ParentId="24204" CreationDate="2015-01-26T13:18:35.060" Score="9" Body="&lt;p&gt;Protons and neutrons are both &lt;a href=&quot;http://en.wikipedia.org/wiki/Baryon&quot; rel=&quot;noreferrer&quot;&gt;baryons&lt;/a&gt;, subatomic particles composed of three quarks. &lt;a href=&quot;http://en.wikipedia.org/wiki/Quark&quot; rel=&quot;noreferrer&quot;&gt;Quarks&lt;/a&gt; are smaller fermionic particles that make up protons, neutrons, and other &lt;a href=&quot;http://en.wikipedia.org/wiki/Hadron&quot; rel=&quot;noreferrer&quot;&gt;hadrons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons are &lt;a href=&quot;http://en.wikipedia.org/wiki/Lepton&quot; rel=&quot;noreferrer&quot;&gt;leptons&lt;/a&gt;, fermions that are not composed of quarks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quarks have mass, spin, charge, and (confusingly) color charge. Color charge is a quantum property that quarks have, and all hadrons need to have a net zero color charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton is composed of three quarks: two up quarks and one down quark. Each up quark has a spin of $+\frac{1}{2}$, a mass of $2.3~\text{MeV/c}^2$, and a charge of $+\frac{2}{3}$. Each down quark has a spin of $-\frac{1}{2}$, a mass of $4.8~\text{MeV/c}^2$, and a charge of $-\frac{1}{3}$. The total charge of the proton is $2(+\frac{2}{3})+(-\frac{1}{3})=+1$. The total spin of the proton is $2(+\frac{1}{2})+(-\frac{1}{2})=+\frac{1}{2}$. The masses of the quarks do not add up because most of the proton's mass is in its confinement energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do the same analysis for a neutron, which is composed of one up quark and two down quarks. The charge of the neutron is $(+\frac{2}{3})+2(-\frac{1}{3})=0$. The total spin of a neutron is $(\frac{1}{2})+2(-\frac{1}{2})=-\frac{1}{2}$. Again the masses do not add up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons and neutrons are different because the are composed of different quarks. However... enough physics. Now for the nuclear chemistry. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interconversion of protons, neutrons, and electrons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proton is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton#Stability&quot; rel=&quot;noreferrer&quot;&gt;stable particle&lt;/a&gt; with a lifetime at least $2.1\times 10^{29}$ years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, protons can be converted into neutrons (and electron neutrinos) by electron capture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{p+ + e- -&amp;gt; n +\nu}_e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requires an investment of energy because the rest mass of the neutron is higher than the rest mass of a proton by $1.29~\text{MeV/c}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free neutrons are &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron#Beta_decay_and_the_stability_of_the_nucleus&quot; rel=&quot;noreferrer&quot;&gt;not stable&lt;/a&gt;, undergoing beta decay to a proton, electron, and electron antineutrino with a halflife of about 10 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{n -&amp;gt; p+ + e- +\bar{\nu}}_e$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, and fortunately, most neutrons bound in nuclei are stable. The protons and neutrons in a nucleus can be treated quantum mechanically analogously to electrons from several atoms bound in a molecule. There are not a collection of independent particles that do not interact. The nucleons occupy quantum mechanical energy levels. Neutrons can only decay if the proton it would become can occupy a lower energy level, which is usually not possible except for those radioisotopes which tend to undergo beta decay, thanks to the Pauli Exclusion Principle! Thus, the bound neutron is stabilized.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2015-01-26T18:35:35.823" LastActivityDate="2015-01-26T18:35:35.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24208" PostTypeId="1" AcceptedAnswerId="24415" CreationDate="2015-01-26T13:24:24.607" Score="7" ViewCount="229" Body="&lt;p&gt;When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? The sponge should still contain dish soap an hour later.&lt;/p&gt;&#xA;" OwnerUserId="12995" LastEditorUserId="7448" LastEditDate="2015-01-26T13:55:58.953" LastActivityDate="2015-01-31T15:35:54.957" Title="The sponge isn't foaming up, why?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;surfactants&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24209" PostTypeId="2" ParentId="24195" CreationDate="2015-01-26T13:47:40.823" Score="5" Body="&lt;p&gt;I can count four reasons that say saliva pH isn't the same among humans:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As people age, the composition of saliva changes in their body; this is easily explainable with the fact that the body isn't performing the actions the same way as you grow up or become old.&lt;/li&gt;&#xA;&lt;li&gt;People have different eating habits; this will put the body in impression of producing especial kinds of enzymes and reduce\increase the production of certain factors in the saliva.&lt;/li&gt;&#xA;&lt;li&gt;People aren't the same when it gets to taking care of the insides of their mouth. You can be brushing your teeth twice as much as a close friend of yours does.&lt;/li&gt;&#xA;&lt;li&gt;Human digestive system isn't always performing its duties consistently and without &quot;ups&quot; and &quot;downs&quot; during the day, which has an important effect on enzymes produced.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And a sea of other reasons....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if we are to answer this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Saliva is a watery substance located in the mouths of animals, secreted by the salivary glands. Human saliva is 99.5% water, while the other 0.5% consists of electrolytes, mucus, glycoproteins, enzymes, and antibacterial compounds such as secretory IgA and lysozyme. &lt;a href=&quot;http://en.wikipedia.org/wiki/Saliva&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saliva, wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll also say that according to &lt;a href=&quot;http://www.naturalhealthschool.com/acid-alkaline.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, a healthy range of pH of saliva varies between 6.5 and 7.5 during the day, but is optimally between 6.5 and 6.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QP5Uy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;a href=&quot;http://altered-states.net/barry/update178/DigestiveTract.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, saliva is a very weak acid, or sometimes a very weak base. Considering something like that an acid or a base is like saying that the water we drink is a base and the water of the raindrops that fall from the sky of a peaceful country is acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saliva is really important for it's being a &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution&quot; rel=&quot;nofollow noreferrer&quot;&gt;buffer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="4356" LastEditDate="2015-01-26T21:21:40.110" LastActivityDate="2015-01-26T21:21:40.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24210" PostTypeId="2" ParentId="24195" CreationDate="2015-01-26T14:04:44.537" Score="3" Body="&lt;p&gt;The pH value of resting whole saliva is slightly acidic, which varies between pH 5.75 and 7.05, and it increases with increasing flow rate up to pH 8. Besides flow rate, the pH also depends on the concentration of salivary proteins, bicarbonate and phosphate ions that have considerable buffering capacity. Bicarbonate concentration is ∼5–10 mM/L in resting conditions, and it may increase up to 40–60 mM/L with stimulation, whereas phosphate concentration is ∼4-5 mM/L in saliva rather independently from the flow rate. Besides bi-carbonate and phosphate, a significant amount of other ions are present to maintain the slightly hypotonic osmolarity of saliva. The most important ions are sodium (1–5 mM/L resting; 100mM/L stimulated), chloride (5 mM/L resting; 70 mM/L stimulated), potassium (15mM/L resting; 30–40 mM/L stimulated), and calcium (1.0 mM/L resting; 3–4 mM/L stimulated).Many other ions such as ammonium, bromide, copper,fluoride,iodide, lithium, magnesium, nitrate, perchlorate, thiocyanate, and so on can be found in the whole saliva in lower concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And appropriate ph is set up for optimum activity of enzyme especially during digestion process.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/imeXd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-01-26T14:04:44.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24211" PostTypeId="1" CreationDate="2015-01-26T14:28:10.077" Score="7" ViewCount="5415" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A 2.35 mole sample of an ideal gas, for which &lt;span class=&quot;math-container&quot;&gt;$C_{\mathrm{m},v}=3R/2$&lt;/span&gt; initially at &lt;span class=&quot;math-container&quot;&gt;$\pu{27 ^\circ C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1750 kPa}$&lt;/span&gt;, undergoes a two stage transformation. For each of the stages described in the following list, calculate the final pressure as well as &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta U$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt;. Also calculate &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta U$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; for the entire process.&lt;/p&gt;&#xA;&lt;p&gt;(a) The gas is expanded isothermally and reversibly until the volume triples.&lt;br /&gt;&#xA;(b) Beginning at the end of the first stage, the temperature is raised to &lt;span class=&quot;math-container&quot;&gt;$\pu{105 ^\circ C}$&lt;/span&gt; at constant volume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since the process is isothermal, &lt;span class=&quot;math-container&quot;&gt;$\Delta H = 0$&lt;/span&gt;. In order to calculate the final pressure. I use the formula,&#xA;&lt;span class=&quot;math-container&quot;&gt;$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$&lt;/span&gt;,&#xA;where &lt;span class=&quot;math-container&quot;&gt;$x = C_{\mathrm{m},p}/C_{\mathrm{m},v}$&lt;/span&gt;.&#xA;Hence &lt;span class=&quot;math-container&quot;&gt;$x = 5/3$&lt;/span&gt;. I then substitute this into the formula to get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_\mathrm{i}V_\mathrm{i}^{5/3} = P_\mathrm{f}V_\mathrm{f}^{5/3}.$$&lt;/span&gt;&#xA;Since the volume triples, &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{f}/V_\mathrm{i}=3$&lt;/span&gt;, hence manipulating the formula I got&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_\mathrm{i}/P_\mathrm{f}=3^{5/3}$$&lt;/span&gt;&#xA;and since &lt;span class=&quot;math-container&quot;&gt;$P_\mathrm{i}=\pu{1750 kPa}$&lt;/span&gt;, I substitute this into the above equation to get &lt;span class=&quot;math-container&quot;&gt;$P_\mathrm{f}$&lt;/span&gt;, which answers the first part of the question.&lt;/p&gt;&#xA;&lt;p&gt;However I don't seem to get the right answer, hence I could not continue doing the question. Could anyone explain?&lt;/p&gt;&#xA;&lt;p&gt;Also, am I right to say that we use&#xA;&lt;span class=&quot;math-container&quot;&gt;$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$&lt;/span&gt;&#xA;for an reversible adiabatic process only.&#xA;&lt;span class=&quot;math-container&quot;&gt;$w = -nRT\ln\left(T_\mathrm{f}/T_\mathrm{i}\right)$&lt;/span&gt; for an isothermal process and&#xA;&lt;span class=&quot;math-container&quot;&gt;$\Delta U = nC_{\mathrm{m},v}(T_\mathrm{f} - T_\mathrm{i}) = -P_{\text{external}}(V_\mathrm{f} - V_\mathrm{i})$&lt;/span&gt;&#xA;for a non-reversible adiabatic process.&lt;/p&gt;&#xA;&lt;p&gt;Im rather confused as to decide under which conditions should I use each of these three formulas.&lt;/p&gt;&#xA;" OwnerUserId="8019" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2021-05-05T03:41:43.683" Title="How to calculate the pressure of an ideal gas that undergoes isothermal expansion?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24214" PostTypeId="2" ParentId="24193" CreationDate="2015-01-26T15:12:53.737" Score="10" Body="&lt;p&gt;The typical definition of strong base that you will see at an introductory level is that there is complete dissociation to hydroxide ions, for example that NaOH completely dissociates into Na+ and OH-.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a terrible definition because reactions never go to completion.  The definition contradicts the concept of chemical equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{[Na+(aq)][OH-(aq)]}{[NaOH(aq)]} = K_B = ~4M$ (from table discussed below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even dissolved aqueous NaOH is not all dissociated.  It can exist as ion pairs or molecules in solution without being dissociated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_B$ is quantification of how strong or weak the base is.  Strong base could be defined as $K_B$ exceeding a certain value. Let us say $K_B&amp;gt;1$ and therefore $pK_B&amp;lt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the strength of the base only refers to the equilibrium between the undissoicated dissolved species and the dissociated ions.  It is independent of how much or how little solid dissolves. Some hydroxides, like NaOH and KOH have high solubilities and some have low solubilities, but $K_B$ is a different equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quantitative data on the $K_B$ values of hydroxides, see:&#xA;&lt;a href=&quot;http://pac.iupac.org/publications/pac/pdf/1969/pdf/2002x0133.pdf&quot; rel=&quot;noreferrer&quot;&gt;DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION&lt;/a&gt;&#xA;Look up the corresponding metal ion alphabetically, and a $pK_B$ value is given.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, keep in mind that for each of $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ there will be two $K_B$ values, since there are two hydroxides that can dissociate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $pK_B$ values for the &lt;em&gt;second&lt;/em&gt; hydroxide dissociating are Mg 2.6, Ca 1.4, Sr 0.8, Ba 0.6 becoming stronger bases as you go down the column in the periodic table, increasing radius stablizing the charge of the ion.  The first $pK_B$ would be even lower, but I don't see any quantitative values.  It would be so low that it would be difficult to measure.  So, yes these are strong bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as how to tell, other than experimentally, how strong a base of a hydroxide is, there is a good lecture &lt;a href=&quot;http://www.mpilkington.com/Lecture_22.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution&lt;/a&gt; by Pilkington.  As you go from left to right across the periodic table (excluding the noble gases), the hydroxides progress from being strong bases to being acids. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-27T18:06:06.790" LastActivityDate="2015-01-27T18:06:06.790" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24215" PostTypeId="2" ParentId="4040" CreationDate="2015-01-26T15:17:07.997" Score="3" Body="&lt;p&gt;Not a database, more a toolset, but there is &lt;a href=&quot;http://www.knime.org/&quot; rel=&quot;nofollow&quot;&gt;KNIME&lt;/a&gt; in wich you can make workflows that do all kinds of things.&#xA;With the right plugins you can also work with molecules in various ways, such as reading and writing, descriptors, manipulation, reactions, etc.&lt;/p&gt;&#xA;" OwnerUserId="8135" LastActivityDate="2015-01-26T15:17:07.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24216" PostTypeId="1" AcceptedAnswerId="24223" CreationDate="2015-01-26T15:26:32.537" Score="2" ViewCount="1038" Body="&lt;p&gt;I was going to ask whether there are software that could be used to predict the products of any given chemical reaction. However, I then noticed these two earlier questions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/2808/chemical-software-for-solving-reactions?rq=1&quot;&gt;Chemical software for solving reactions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/8351/software-for-predicting-chemical-reactions?rq=1&quot;&gt;Software for predicting chemical reactions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where it is said such predictions are too difficult to make. What makes the prediction difficult? &lt;/p&gt;&#xA;" OwnerUserId="12997" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-04-01T00:51:27.280" Title="Why is predicting products of chemical reactions difficult?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-01-26T18:18:58.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="24218" PostTypeId="1" AcceptedAnswerId="24221" CreationDate="2015-01-26T15:44:52.917" Score="8" ViewCount="2490" Body="&lt;p&gt;I read somewhere that the temperature of an atom is not defined. The definition of temperature is only for larger systems. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="7448" LastEditDate="2015-01-26T15:57:14.210" LastActivityDate="2015-01-26T22:16:06.400" Title="Temperature of an atom" Tags="&lt;thermodynamics&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24221" PostTypeId="2" ParentId="24218" CreationDate="2015-01-26T16:21:58.907" Score="12" Body="&lt;p&gt;Thermodynamic functions are strictly defined only for macroscopic systems (systems that have an essentially infinite number of atoms). &lt;strong&gt;You can't apply them to individual atoms because that would be confusing large-scale averages with individual microscopic values&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an analogy: the average speed of cars on a stretch of highway might be 55 mph, but it's possible that no individual car is traveling at that exact speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another analogy: the average global temperature is rising over long periods of time, but if I look at the thermometer outside my window, the local temperature is falling over a short period of time... I can't say that my local temperature drop disproves global warming, because weather is a different thing from climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you knew the speed of a single atom. Now, for a macroscopic ideal gas, the &lt;em&gt;average&lt;/em&gt; speed of molecules $\overline{v}$ is $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}}$$ It would be wrong for you to plug in the speed of your single atom and solve for its temperature, because that equation was derived by assuming that you had a macroscopic system (with a certain speed distribution over many molecules); it &lt;em&gt;only&lt;/em&gt; applies to average velocities, not individual velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This applies to other thermodynamic functions, too.&lt;/strong&gt; Let's look at a simple chemical example. For a macroscopic crystal, the Gibbs free energy might be written as $$G = Ng(P,T)$$ where $N$ is the number of atoms and $g$ is the Gibbs free energy per atom. Little $g$ is a function of pressure and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can we say that little $g$ is the Gibbs free energy of a single atom? &lt;strong&gt;No.&lt;/strong&gt; It's the average Gibbs free energy per atom for a huge number of atoms. To see the difference,  consider a small cluster of atoms from that same crystal, the previous equation will need some correction terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G = Ng(P,T) + a(P,T)N^{2/3} + b(T)\ln{N} + c(P,T)$$&#xA;where the &quot;a&quot; term is a surface free energy and the last couple of terms might come from free energy contributed by things like rotation of the cluster (Source: T. L. Hill, &lt;em&gt;Thermodynamics of Small Systems&lt;/em&gt;, Dover, 1962.) You can't say that $g(P,T)$ is the Gibbs free energy of a single atom in the cluster because of those correction terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;TL;DR: Thermodynamic functions are averages over very large numbers of particles; random fluctuations are tiny compared to the mean values. But with small systems, those fluctuations become quite important, and there are additional effects that cause macroscopic averages to poorly describe microscopic systems.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-26T18:24:40.837" LastActivityDate="2015-01-26T18:24:40.837" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24222" PostTypeId="2" ParentId="24194" CreationDate="2015-01-26T16:50:09.297" Score="4" Body="&lt;p&gt;In some cases, such as [F-H-F]- (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bifluoride&quot; rel=&quot;nofollow&quot;&gt;biflouride&lt;/a&gt;), there is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Symmetric_hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;symmetric hydrogen bond&lt;/a&gt;, with both H-F bonds equally strong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, just because the two atoms bonding to the H are the same doesn't mean it is a symmetric hydrogen bond.  For example &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_dimer&quot; rel=&quot;nofollow&quot;&gt;water dimer&lt;/a&gt; does not have a symmetrical hydrogen bond, the H is closer to one O atom than the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be cases where the hydrogen being exactly in the middle is a potential energy minimum, and other cases where the middle is not a local minimum but there are two minima offset from the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://pac.iupac.org/publications/pac/pdf/2009/pdf/8104x0571.pdf&quot; rel=&quot;nofollow&quot;&gt;Symmetry of hydrogen bonds in solution&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-26T16:50:09.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24223" PostTypeId="2" ParentId="24216" CreationDate="2015-01-26T17:16:34.150" Score="5" Body="&lt;p&gt;Because you need would need to solve the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dirac_equation&quot;&gt;Dirac equation&lt;/a&gt; (relativistic Schrodinger equation) for all the particles in the system to get a completely accurate result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even solving the time independent non-relativistic Schrodinger equation for $\ce{H2+}$ in isolation, where there is only one electron and two protons involves approximating the protons as fixed relative to each other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine trying solving a system of $10^{23}$ molecules each having multiple nuclei, and dozens or even hundreds of electrons, and including time dependence!&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-26T17:21:45.570" LastActivityDate="2015-01-26T17:21:45.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24224" PostTypeId="2" ParentId="24218" CreationDate="2015-01-26T17:47:47.483" Score="4" Body="&lt;p&gt;While generally temperature in textbooks is connected with the speed of atoms/particles, strictly speaking not the velocity, but the distribution of velocity what counts. If you microscopically observe a bunch of particles, their velocity can be whatever as your frame of observation can be any inertia system. When you want to calculate temperature, you choose  reference system which has the same velocity as the center of mass of those particles, and then you calculate the average velocities form that reference. The average will be small, if the distribution of velocities is sharp, and will be large if the distribution is broad. The calculated temperature can be low even for fast particles e.g. in a gas flow, if all the particles moving in the same directions, same velocity!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's apply this to one atom: you don't have a distribution of different velocities! Once you adjusted your reference point to the center of mass, ie the particle, it has zero velocity and meaningless to look for a distribution of different velocities. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-01-26T17:47:47.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24226" PostTypeId="2" ParentId="24177" CreationDate="2015-01-26T18:48:33.473" Score="4" Body="&lt;p&gt;In the E2 reaction, the size of the base can affect the selectivity of the deprotonation, because the steric interaction of a large base with a hindered proton will go over a higher transition state than the interaction of a large base with a less-hindered proton. This is kinetic control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any practical case where the selectivity of an E1 reaction can be controlled by the size of the base. The reason is that the deprotonation of the carbocation intermediate will have a transition state that is very close in energy to the carbocation itself. I agree with you that the size of the base should have some influence on the transition state energy, but in practice, it seems that it has less influence than the stability of the resulting alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that often E1 is run under equilibrating conditions, where even if a less-substituted (less-stable) alkene is formed, it can be protonated to reform the carbocation, and the material will funnel to the most stable product. This is thermodynamic control.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-01-26T18:48:33.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24227" PostTypeId="1" CreationDate="2015-01-26T19:28:17.947" Score="2" ViewCount="1201" Body="&lt;p&gt;Our $\ce{Cl2}$ liquifier is of the following type:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxNks.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is entirely made of mild steel (shell as well as tubes plus the connecting pipelines). At shell side there is Freon (R-12) with its refrigeration cycle (not shown here); at the tube side there is dried &amp;amp; compressed $\ce{Cl2}$ gas from compressor section (~2bar max). One must realise that drying of $\ce{Cl2}$ is done physically via passing it through packed glass wool columns &amp;amp; then washing it in $\ce{H2SO4}$ scrubbers &amp;amp; then again through glass wool columns...Compression is done via $\ce{H2SO4}$ liquid ring compressors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem we face is excess corrosion of tubes which thus lose their thickness &amp;amp; may in time compromise...For instance every 2-3 weeks we open up this liquifier &amp;amp; wash the tubes (via high pressure demi/RO water) &amp;amp; we found quiet an amount of dark-green/brown sludge indicating ineffectiveness of our drying section. Lab test are yet to be done so that I can clearly say what this sludge is made of. From the WIKI what I understand the compounds of such colour can either be oxides or chlorides of Iron because sulphates of iron are yellow in colour...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell what I can do is increase the number of drying towers or increase the number of glass wool columns, but the most active measure I am thinking of cathodic protection via impressed currents. Have any user here applied impressed currents to a S&amp;amp;T heat exchanger &amp;amp; how can it be done? I would love any literature on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note our liquifier can be electrically isolated using gaskets &amp;amp; insulated teflon bolts on any flange fitting.&lt;/p&gt;&#xA;" OwnerUserId="7583" LastEditorUserId="4231" LastEditDate="2015-02-27T17:19:33.593" LastActivityDate="2015-02-27T17:19:33.593" Title="Cathodic protection of tubes in Shell &amp; Tube Heat Exchanger" Tags="&lt;electrochemistry&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24228" PostTypeId="2" ParentId="24151" CreationDate="2015-01-26T20:30:31.853" Score="2" Body="&lt;p&gt;If the enzyme is staturated with substrate, the reaction rate will be $k_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1450 moles per second, per mole of enzyme.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;87,000 moles per minute, per mole of enzyme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;87,000,000,000 micromoles per minute, per mole of enzyme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based upon &quot;50,000 kDa&quot; (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 87/50 = 1.74 units per milligram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for when product is staturated, use $k_{-1}$ instead of $k_2$.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-26T20:47:40.457" LastActivityDate="2015-01-26T20:47:40.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24229" PostTypeId="1" CreationDate="2015-01-26T20:51:59.150" Score="13" ViewCount="4956" Body="&lt;p&gt;In the house on my own and an article popped up on my feed with the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While the Government has insisted the chemical is safe, cadmium is&#xA;  recognised as a cause of lung cancer and during the Second World War&#xA;  was considered by the Allies as a chemical weapon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proceeding paragraphs are about zinc cadmium sulphide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theguardian.com/politics/2002/apr/21/uk.medicalscience&quot;&gt;full article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a flawed argument, essentially inferring that any given molecule can be said to be toxic just because one of it's component parts is toxic but I can't think of a good example to back this up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good example of a common compound that is ingested often (daily?) but contains a particularly toxic element if that element were ingested in some way on it's own?&lt;/p&gt;&#xA;" OwnerUserId="13002" LastEditorUserId="10185" LastEditDate="2015-01-27T20:26:00.047" LastActivityDate="2020-04-24T15:55:21.360" Title="Are there any good examples of commonly ingested molecules that contain particular toxic individual elements?" Tags="&lt;molecules&gt;&lt;elements&gt;&lt;toxicity&gt;" AnswerCount="8" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24230" PostTypeId="1" CreationDate="2015-01-26T20:53:41.390" Score="6" ViewCount="1996" Body="&lt;p&gt;In one of the exams I saw it was written that dehydration of 1-methyl-cyclobutanol yields cyclobutylmethylene (double bond outside the ring) which is the anti Zaitzev product, instead of methylcyclobutene. &#xA;Why?&#xA;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="13003" LastActivityDate="2015-01-27T22:53:59.120" Title="Dehydration of methylcyclobutanol" Tags="&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24231" PostTypeId="1" AcceptedAnswerId="24233" CreationDate="2015-01-26T21:25:02.557" Score="10" ViewCount="43753" Body="&lt;p&gt;I understand the fact that ionic compounds are good conductors of electricity in molten state. But why aren't they good conductors in solid state. Cannot ions vibrate about their mean position and transfer electricity in the same way as they transfer heat? &lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2019-12-23T15:47:42.113" Title="Why are ionic compounds bad conductors of electricity in solid state?" Tags="&lt;ionic-compounds&gt;&lt;conductivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24232" PostTypeId="2" ParentId="24230" CreationDate="2015-01-26T21:32:19.403" Score="2" Body="&lt;p&gt;Assumption: The dehydration is probably occurring under acidic conditions.  The hydroxyl group is protonated and then eliminates water to generate the methylcyclobutyl carbocation pictured below, an E1 reaction mechanism is likely involved.  As you note, the carbocation can eliminate a proton from two different types of adjacent carbons; a methylene group (there are two) in the ring or the methyl group outside the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pDW1R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terms like &quot;Zaitzev&quot; and &quot;Hofmann&quot; elimination generally are used in discusing E2 eliminations. In the E1 case at hand, elimination to form the cyclobutene is statistically favored (there are 4 alpha-methylene hydrogens and only 3 methyl hydrogens), but disfavored energetically.  Cyclobutene is &lt;em&gt;ca.&lt;/em&gt; 4 kcal/m less stable than cyclobutane due to increased ring strain from placing the double bond inside a 4-membered ring.  Instead, the lower energy Hofmann product, with the double bond outside the ring, is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, a 1.4 kcal/m difference in product stabilities results in a factor of 10 shift in an equilibrium between the two products at room temperature.  In other words, if product A was 1.4 kcal/m less stable than product B, and A and B were in equilibrium, then we would expect a 10:90 A:B product ratio at equilibrium.  In E1 acid catalyzed dehydration reactions the final products are usually in equilibrium.  Since the exo-methylenecyclobutane is roughly 4 kcal/m more stable than the methylcyclobutene isomer, we might expect them to be present in something like a (4/1.6 = 2.5) 99.5 : 0.5 ratio, respectively.  An introductory textbook might well say that the cycobutene isomer is &quot;not present.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information contained in @Martin 's post strongly suggests that methylenecyclobutane is less stable than 1-methylcyclobut-1-ene.  The model compounds (cyclobutane and cyclobutene) I used in the above analysis were apparently inadequate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, this leaves unanswered the OP's question about why elimination from 1-methyl-cyclobutan-1-ol occurs towards the methyl group to produce methylenecyclobutane.  Since the OP didn't specify reaction conditions I assumed (above) that the reaction was run under E1 conditions.  From the relative stabilities Martin has provided, we now see that an E1 mechanism would be expected to produce the &quot;wrong&quot; product, 1-methylcyclobut-1-ene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of an E1 process, let's examine the possible outcomes under E2 (simultaneous elimination of the leaving group and hydrogen, no carbocation intermediate) conditions.  E2 eliminations occur with the leaving group and the hydrogen atom in an &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch05/ch5-6-1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;anti-periplanar&lt;/a&gt; configuration.  If we use Newman projections and examine the possible conformations that would favor elimination into the ring, we see that an anti-periplanar arrangement of the hydroxyl group and the adjacent ring methylene hydrogens (with the constraint of a cyclobutane ring) is not possible (figure A).  However, the Newman projection for elimination towards the methyl group to create methylenecyclobutane does have an anti-periplanar conformation accessible.  Hence, &lt;strong&gt;if the reaction proceeds through an E2 mechanism, we would only expect the Hofmann product, methylenecyclobutane to form.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the OP will add the reaction conditions shown on the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qD6Ly.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-27T22:53:59.120" LastActivityDate="2015-01-27T22:53:59.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24233" PostTypeId="2" ParentId="24231" CreationDate="2015-01-26T21:44:17.063" Score="15" Body="&lt;p&gt;Electric charge is transferred by physically moving charged particles around. In the case of an electric current moving through a wire (for example), the electrons are moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ionic compound, the ions are locked in place. They can move around a little bit, but there is not much translational motion - the ions stay in their places on the crystal lattice. In addition, the ions are &quot;happy&quot; with the number of electrons that they have. The ions formed in the first place by giving up or accepting electrons in order to minimize the overall potential energy of the system. If an anion were to transfer an electron back to a cation (for example) the energy of the system would increase, and so in general, transfer of electrons after the compound has formed is not favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solution or in a molten state, the ions themselves can move around - they become the charge carriers. In a solid, the ions can't move, and so electricity cannot be easily transferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned heat transfer - heat is the transfer of the kinetic energy of atoms and molecules. Heat can still be transferred (in some cases quite easily) in an ionic solid because, as you said, ions can vibrate about a mean position. When this happens they bump into their neighbors, which spreads the kinetic energy around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, ionic compounds don't conduct electricity very well because the charge carriers can't move through the crystal. They can conduct heat because the kinetic energy itself is the &quot;heat carrier&quot; - it can be transferred without moving ions too far from their mean positions.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-01-26T21:44:17.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24234" PostTypeId="2" ParentId="24229" CreationDate="2015-01-26T22:07:11.003" Score="33" Body="&lt;p&gt;Table salt! What would be worse than putting sodium (it can spontaneously combust if you get it wet) and chlorine (used as a war gas in WWI) all over your food?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's water which &lt;em&gt;always&lt;/em&gt; contains some &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; (an active ingredient in &lt;a href=&quot;http://en.wikipedia.org/wiki/Drano&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drano&lt;/a&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; (that stuff eats through metal, man). Although those aren't elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hiding inside every &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; molecule are two deadly oxygen atoms; once freed, they'll wreak havoc on just about anything they come into contact with.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorDisplayName="user7951" LastEditDate="2020-04-24T09:18:00.340" LastActivityDate="2020-04-24T09:18:00.340" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24235" PostTypeId="2" ParentId="24218" CreationDate="2015-01-26T22:16:06.400" Score="2" Body="&lt;p&gt;I'll just rephrase @Greg's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamics distinguishes heat and mechanical work. From a microscopic point of view, heat is the &lt;em&gt;unordered&lt;/em&gt; movement of atoms/molecules, while mechanical work is the &lt;em&gt;ordered&lt;/em&gt; movement (all the atoms of the piston move at the same velocity = same speed &amp;amp; direction; free energy). As Greg says, the actual movement is decomposed into those two components: a macroscopic translation and an unordered part where the sum of all velocities is zero. This second part constitues heat (in terms of total energy of all those microscopic movements), and temperature (distribution of speed = |velocity|).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single atom has only one velocity, so we cannot decide which part of it is ordered and which is unordered. We can thus speak of the kinetic energy and velocity of the atom, but not of distinguishable free energy and heat, nor of temperature.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2015-01-26T22:16:06.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24236" PostTypeId="1" CreationDate="2015-01-26T22:29:22.103" Score="3" ViewCount="90" Body="&lt;p&gt;Suppose I take three bases in a solution, $\ce{NaOH}$, $\ce{Na2CO3}$ and $\ce{NaHCO3}$, and I titrate this solution with $\ce{HCl}$. Since the strengths of the bases are different, does it mean that $\ce{HCl}$ would first react with the strongest base completely and then start reacting with the next strongest base? &lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="10185" LastEditDate="2015-01-27T12:37:20.250" LastActivityDate="2015-01-27T12:37:20.250" Title="Titration of bases against HCl" Tags="&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24237" PostTypeId="2" ParentId="24227" CreationDate="2015-01-26T22:55:18.810" Score="3" Body="&lt;p&gt;Cathodic protection requires an electrolyte to work. You could just pass current through the tubes but how is it going to return to the power supply? If there is no electrolyte to conduct ions it cannot provide any protection to your system. Have you considered replacing your heat exchanger with glass or glass-coated metal? If the capital costs of this are prohibitive, I would think your best bet would be to try drying the gas some more, as you mentioned. Do you happen to know the water content of the gas going into the exchanger now?&lt;/p&gt;&#xA;" OwnerUserId="12650" LastActivityDate="2015-01-26T22:55:18.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24239" PostTypeId="1" CreationDate="2015-01-27T00:00:23.407" Score="2" ViewCount="25694" Body="&lt;p&gt;Given the reaction $\ce{NaCl~(s) &amp;lt;=&amp;gt; Na+~(aq) + Cl^{-}~(aq)}$. $\ce{NaCl}$ is added to water and equilibrium is reached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the reaction shift to the left/right/not shift at all if more water is then added?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that adding water would decrease the concentrations of $\ce{Na+}$ and $\ce{Cl-}$, and therefore the reaction would shift to the right. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="13006" LastEditorUserId="4945" LastEditDate="2015-01-27T04:15:43.127" LastActivityDate="2015-01-27T11:47:39.997" Title="What are the effects of adding water to a system at equilibrium?" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24240" PostTypeId="1" CreationDate="2015-01-27T00:08:45.450" Score="1" ViewCount="27906" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You prepare $1.0~\mathrm{L}$ of a $0.25~\mathrm{M}$ acetic acid solution with a final $\ce{pH}$ of $6.0$. What are the molar concentrations of all relevant acetic acid species ($[\ce{HA}]$ and $[\ce{A-}]$) given that the $K_a$ for acetic acid is $1.74 \cdot 10^{-5}~\mathrm{M}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am getting confused with this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the $\ce{pH}$ is given, I know what the $[\ce{H^+}]$ is.&#xA;So now when I try to do the ICE table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l|ccc}&#xA;               &amp;amp; \ce{HA} &amp;amp;  \ce{H+}  &amp;amp;  \ce{A-} \\\hline&#xA;\text{Initial} &amp;amp; 0.25    &amp;amp;  0        &amp;amp;  0       \\&#xA;\text{Change}  &amp;amp;  -x     &amp;amp;  +x       &amp;amp;  +x      \\&#xA;\text{End}     &amp;amp; 0.25 -x &amp;amp;  +x       &amp;amp; +x       \\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from here I begin to assume $[\ce{H^+}] = [\ce{A^-}]$, which I am not sure about.&#xA;Then I set $x = 10^{-6.0} = 1\cdot 10^{-6}$ so I get $[\ce{A^-}] = 1\cdot 10^{-6}~\mathrm{M}$ and $[\ce{HA}] = 0.24999~\mathrm{M}$ which I think is incorrect, and to even further ensure my that it's incorrect, when I attempt to check the $K_a$ value with this, it does not match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second approach:&#xA;$[\ce{HA}] = 0.25~\mathrm{M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this I determine the $\mathrm{p}K_a$ from the $K_a$ which turns out to be $4.759$, which indicates that there should be more $[\ce{A-}]$ than $[\ce{HA}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now use the Henderson–Hasselbalch equation:&#xA;\begin{align}&#xA;    6.0   &amp;amp;= 4.759 + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\\&#xA;   17.40  &amp;amp;= \frac{[\ce{A-}]}{[\ce{HA}]}\\&#xA;   17.40  &amp;amp;= \frac{[\ce{A^-}]}{0.25~\mathrm{M}}\\&#xA;[\ce{A-}] &amp;amp;= 4.35~\mathrm{M}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel more confident about my second answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me out with this particular problem and perhaps tell me procedure I should use as well as what the correct answer should come out to be and why?&lt;/p&gt;&#xA;" OwnerUserId="13005" LastEditorUserId="865" LastEditDate="2021-05-22T19:00:37.200" LastActivityDate="2021-05-22T19:00:37.200" Title="How to calculate the concentration of all relevant species in a buffer of a given pH?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24241" PostTypeId="2" ParentId="24239" CreationDate="2015-01-27T00:26:54.970" Score="1" Body="&lt;p&gt;I think I may understand your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding water to $\ce{NaCl~(s)}$ causes it to ionize into $\ce{Na+}$ and $\ce{Cl-}$, now if you think about Le Chatelier's Principle, considering the equation: $\ce{NaCl~(s) + H2O -&amp;gt; Na+ + Cl-}$, the equilibrium will shift to the right, and the concentrations of $\ce{Na+}$ and $\ce{Cl-}$ will increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you think about it realistically, since the salt is ionized with water, the more water you add, the more $\ce{Na+}$ and $\ce{Cl-}$ you will get, now if this was considered, it is essentially increasing the concentration of the ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, usually when dealing with water, the system is considered to be in water. So you may see water on both sides of the reaction, in which case adding water would have no effect on equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="13005" LastEditorUserId="4945" LastEditDate="2015-01-27T04:18:23.400" LastActivityDate="2015-01-27T04:18:23.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24242" PostTypeId="2" ParentId="24229" CreationDate="2015-01-27T00:31:19.310" Score="8" Body="&lt;p&gt;A few things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There is a saying that &quot;the dose makes the poison.&quot; This reflects the fact that most substances (even water!) become toxic at a high enough dosage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) As to your question about individual elements, surprisingly many of them are highly problematic &lt;em&gt;in elemental form&lt;/em&gt;. But very reactive elements (Group 1, 2, 16 (some), and 17) are often not present in their elemental form in nature, and once reacted, they are quite benign (as Fred Senese noted about table salt). These are the best examples for your specific question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The other problematic elements tend to be so-called &quot;heavy metals&quot;, and they can be problematic because they can substitute for lighter metals in the same group and interfere with biological processes (generally slows them down). For example, both zinc and iron are necessary for health, but cadmium and mercury (same group and zinc) are very toxic, and the same goes for ruthenium and osmium (similar to iron). This goes to your example: these heavier metals can often be problematic regardless of the form.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-01-27T00:31:19.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24243" PostTypeId="2" ParentId="24240" CreationDate="2015-01-27T00:53:09.990" Score="3" Body="&lt;p&gt;The value given for the pH of the solution seems to be incorrect*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the reaction is $\ce{HA &amp;lt;=&amp;gt; H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your chart, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{l|ccc}&#xA;               &amp;amp; \ce{HA} &amp;amp;  \ce{H+}  &amp;amp;  \ce{A-} \\\hline&#xA;\text{Initial} &amp;amp; 0.25    &amp;amp;  0        &amp;amp;  0       \\&#xA;\text{Change}  &amp;amp;  -x     &amp;amp;  +x       &amp;amp;  +x      \\&#xA;\text{End}     &amp;amp; 0.25 -x &amp;amp;  +x       &amp;amp; +x       \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it follows that &#xA;$$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this for $x$:&#xA;\begin{align}&#xA;x^2 &amp;amp;= 1.74\cdot 10^{-5}(0.25-x)\\&#xA;x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &amp;amp;= 0\\&#xA;x &amp;amp;= 0.00207698~\mathrm{M}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] &amp;lt; 4.35~\mathrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.&lt;/p&gt;&#xA;" OwnerUserId="13006" LastEditorUserId="4945" LastEditDate="2015-01-27T03:32:23.527" LastActivityDate="2015-01-27T03:32:23.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24244" PostTypeId="1" AcceptedAnswerId="24256" CreationDate="2015-01-27T03:03:45.683" Score="4" ViewCount="604" Body="&lt;p&gt;I first want to note that I'm a layman attempting to gain an understanding of metabolism by reading resources on the internet with only a little understanding of chemistry. I've hit a snag:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation of lactate into pyruvate reduces NAD+ to NADH. Subsequent glyconeogenesis can then turn two pyruvate into a new glucose molecule at the cost of 6 ATP. However, the NADH is presumably used to generate ATP via oxidative phosphorylation producing ~2.5ATP, making an overall reaction of 2 lactate -&gt; 1 glucose cost only ~1ATP. This clashes with other things I have read - that glyconeogenesis costs 3x the energy glycolysis produces - as this would seem to discount the reduction of NAD+ to NADH in glycolysis. I feel I am clearly misunderstanding something, but I'm not sure where.&lt;/p&gt;&#xA;" OwnerUserId="13011" LastActivityDate="2015-02-21T13:30:16.933" Title="What happens to the NADH produced by lactate oxidation?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24245" PostTypeId="1" AcceptedAnswerId="24251" CreationDate="2015-01-27T03:56:29.797" Score="18" ViewCount="2688" Body="&lt;p&gt;I was never taught in school to use matrices to balance my equation in stoichiometry. But I recently started to review my chemistry and, as I have taken linear algebra, I was suddenly struck by the idea this would be a really efficient way to balance equations. It might save me from making a lot of errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are matrices a common tool for chemists when balancing equations? Does anyone have a good reference for learning about this? &lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2018-12-04T01:22:15.530" Title="Are matrices a good way to do stoichiometry?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24246" PostTypeId="1" CreationDate="2015-01-27T04:37:10.040" Score="3" ViewCount="34422" Body="&lt;p&gt;As part of a pre-lab exercise in Chemistry, we went through the necessary calculations to identify an unknown metal in a metal carbonate. The method was gas evolution by means of hydrochloric acid (see equation below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HCl + M2CO3 -&amp;gt; MCl + CO2 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unknown metal was Alkali, and we had three possibilities: Lithium, Sodium and Potassium as &lt;span class=&quot;math-container&quot;&gt;$\ce{M}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using data given to us, we first converted grams of released &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; to moles. Then here was the part I was confused on-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the molar mass of the metal carbonate, we took the mass of the sample used (the metal carbonate) and divided it by moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; released, which was just calculated. The rest here on out was a little algebra. What is the reasoning behind dividing the mass of the sample by released &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;? What makes it &quot;legal&quot;? Yes, in the balanced equation, the ratio of &lt;span class=&quot;math-container&quot;&gt;$\ce{M2CO3}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; was &lt;span class=&quot;math-container&quot;&gt;$1:1$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13012" LastEditorUserId="83758" LastEditDate="2020-05-28T14:51:45.853" LastActivityDate="2020-05-28T14:51:45.853" Title="How to find the molar mass of an unknown metal carbonate through a gas evolution reaction?" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24247" PostTypeId="2" ParentId="24236" CreationDate="2015-01-27T04:59:56.903" Score="3" Body="&lt;p&gt;This is perhaps not the best example, since hydroxide ion will react with bicarbonate ion to form carbonate ion, and also as you titrate carbonate ion note that it will form bicarbonate again. But, generally, yes, the strongest base will be consumed first as you titrate. The titration curve will tell you the amounts of the bases present and the pKas of their conjugate acids.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-01-27T04:59:56.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24248" PostTypeId="2" ParentId="24246" CreationDate="2015-01-27T06:10:10.710" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{HCl + M2CO3 -&amp;gt; MCl + CO2 + H2O}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since you already figured out, that there is a one-to-one ratio of carbonate to carbon dioxide, you can simply write&#xA;$$n(\ce{CO2}) = n(\ce{M2CO3}).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the formula for the Molar mass,&#xA;$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{M2CO3})},$$&#xA;you can simply substitute one for the other, hence&#xA;$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{CO2})}.$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-01-27T06:10:10.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24250" PostTypeId="1" AcceptedAnswerId="24252" CreationDate="2015-01-27T06:36:34.110" Score="4" ViewCount="223" Body="&lt;p&gt;I am searching for the carbon nanotubes for very low cost. I have one idea, which is to use graphite (such as that found in pencils.) Can it be converted to carbon nanotubes? If so, what would be the procedure for this?&lt;/p&gt;&#xA;" OwnerUserId="13015" LastEditorUserId="15410" LastEditDate="2016-11-03T06:40:25.570" LastActivityDate="2019-11-21T21:12:04.113" Title="Producing carbon nanotubes from graphite" Tags="&lt;nanotechnology&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24251" PostTypeId="2" ParentId="24245" CreationDate="2015-01-27T07:10:03.993" Score="22" Body="&lt;p&gt;Sure, most of the equation-balancing Javascript calculators on the web use that method. It's overkill to balance a single equation that way, but if you're automating the process or if you have a lot of equations to balance and check it can be convenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you search on &quot;balancing equations matrix&quot; you'll find many simple matrix inversion  approaches for balancing equations that fail if you don't have a single unique solution.  Here's one that's more general: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L. R. Thorne, &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1110/1110.4321.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Innovative Approach to Balancing Chemical-Reaction Equations: A&#xA;Simplified Matrix-Inversion Technique for Determining The Matrix Null&#xA;Space&lt;/a&gt;, &lt;em&gt;Chem. Educator&lt;/em&gt; 2010, &lt;strong&gt;15&lt;/strong&gt;, 304–308 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that coefficients in balanced equations are found in the nullspace of the formula composition matrix (that is, the coefficients are solutions &lt;span class=&quot;math-container&quot;&gt;$\bf x$&lt;/span&gt; of the matrix equation &lt;span class=&quot;math-container&quot;&gt;$\bf C x = 0$&lt;/span&gt;, if &lt;span class=&quot;math-container&quot;&gt;$\bf C$&lt;/span&gt; is the formula composition matrix). For example, if I want to find a balanced equation that relates &lt;span class=&quot;math-container&quot;&gt;$\rm HCl$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm K_2Cr_2O_7$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm FeCl_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm CrCl_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm H_2O$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rm FeCl_3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\rm KCl$&lt;/span&gt;, I can build a formula composition table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{matrix} &amp;amp; \ce{HCl} &amp;amp; \ce{K2Cr2O7} &amp;amp; \ce{FeCl2} &amp;amp; \ce{CrCl3} &amp;amp; \ce{H2O} &amp;amp; \ce{FeCl3} &amp;amp; \ce{KCl}\cr H &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0\cr Cl &amp;amp; 1 &amp;amp; 0 &amp;amp; 2 &amp;amp; 3 &amp;amp; 0 &amp;amp; 3 &amp;amp; 1\cr K &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1\cr Cr &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\cr O &amp;amp; 0 &amp;amp; 7 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\cr Fe &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0\end{matrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find the nullspace vectors using the multistep procedure outlined in that paper, or I can do it very simply in &lt;a href=&quot;http://maxima.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxima&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C : matrix([1,0,0,0,2,0,0],&#xA;           [1,0,2,3,0,3,1],&#xA;           [0,2,0,0,0,0,1],&#xA;           [0,2,0,1,0,0,0],&#xA;           [0,7,0,0,1,0,0],&#xA;           [0,0,1,0,0,1,0])$&#xA;nullspace(C);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{span}\left( \begin{bmatrix}14\cr 1\cr 6\cr -2\cr -7\cr -6\cr -2\end{bmatrix}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which corresponds to the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl\ =\ 14HCl + K2Cr2O7 + 6FeCl2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As chemists, we'd write that equation reversed, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{ 14HCl + K2Cr2O7 + 6FeCl2\ =\ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few more nails you can pound with your linear algebra hammer: you can determine the minimum number of chemical equations required in a reaction mechanism, or discover how many equations in a reaction mechanism are stoichiometrically independent. For example, consider the following halogen/hydrogen reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{rcl}&#xA;\rm H_2 + X_2&amp;amp;\xrightarrow{k_1}&amp;amp;\rm 2HX\\&#xA;\rm H_2 + X_2&amp;amp;\xrightarrow{k_2}&amp;amp;\rm 2X + H_2\\&#xA;\rm H_2 + X_2&amp;amp;\xrightarrow{k_3}&amp;amp;\rm 2H + X_2\\&#xA;\rm H + X_2&amp;amp;\xrightarrow{k_4}&amp;amp;\rm X + HX\\&#xA;\rm H_2 + X&amp;amp;\xrightarrow{k_5}&amp;amp;\rm H + HX\\&#xA;\rm HX + H&amp;amp;\xrightarrow{k_6}&amp;amp;\rm X + H_2\\&#xA;\rm H_2 + HX&amp;amp;\xrightarrow{k_7}&amp;amp;\rm X_2 + H_2\\&#xA;\rm 2HX &amp;amp;\xrightarrow{k_8}&amp;amp;\rm X_2 + H_2\\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make a &lt;em&gt;formula matrix&lt;/em&gt; for the five species in the reactions above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix}=\begin{pmatrix}2 &amp;amp; 0\cr 1 &amp;amp; 0\cr 1 &amp;amp; 1\cr 0 &amp;amp; 1\cr 0 &amp;amp; 2\end{pmatrix}\begin{pmatrix}H\cr X\end{pmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any set of &lt;span class=&quot;math-container&quot;&gt;$S-C$&lt;/span&gt; stoichiometrically independent equations is a &quot;complete set&quot; of reactions, where &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is the number of species and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is the number of independent equations for the conservation of elements (components). There are five species (&lt;span class=&quot;math-container&quot;&gt;$S=5$&lt;/span&gt;) and the rank of the formula matrix is 2, so the system has two components (&lt;span class=&quot;math-container&quot;&gt;$C=2$&lt;/span&gt;). There must be &lt;span class=&quot;math-container&quot;&gt;$S - C = 5 - 2 = 3$&lt;/span&gt; stoichiometrically independent reactions for the mechanism to be complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the reactions above constitute a complete set? We can write a stoichiometric matrix like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{pmatrix}-1 &amp;amp; 0 &amp;amp; 2 &amp;amp; 0 &amp;amp; -1\cr 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 2 &amp;amp; -1\cr -1 &amp;amp; 2 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\cr 0 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1\cr -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 0\cr 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 0\cr 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; -2 &amp;amp; 1\cr 1 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 1\end{pmatrix} \begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix} = \begin{pmatrix}0\cr 0\cr 0\cr 0\cr 0\cr 0 \cr 0 \cr 0\end{pmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rank of the stoichiometric matrix is 3, so the eight equations above are equivalent to three independent equations. The mechanism is complete. Whether it's &lt;em&gt;correct&lt;/em&gt; is of course a different matter. Stoichiometric constraints don't determine a mechanism; they limit the number of reactions in a particular reaction channel, but you can have more than one channel (more than one mechanism running side-by-side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this useful---why would you want to know what the number of independent stoichiometric equations (&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;) are? Well, according to Gibb's rule of stoichiometry, &lt;span class=&quot;math-container&quot;&gt;$R \le S-C$&lt;/span&gt;. If you have &lt;span class=&quot;math-container&quot;&gt;$R &amp;lt; S-C$&lt;/span&gt;, that tells you that you have a &lt;em&gt;restricted equilibrium&lt;/em&gt;: the components must be constrained by conditions other than stoichiometric equations. The restricted equilibrium composition will depend on which species are present initially in that case. There are also experimental design implications, since in a single experiment you can get only &lt;span class=&quot;math-container&quot;&gt;$S-C$&lt;/span&gt; reaction rates. You may have to figure out how to favor one channel over another in separate experiments to get all of the rates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR: If you're looking for applications of linear algebra in chemistry, chemical kinetics is one playground; there are many others. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2018-12-04T01:22:15.530" LastActivityDate="2018-12-04T01:22:15.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24252" PostTypeId="2" ParentId="24250" CreationDate="2015-01-27T09:26:16.077" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;Graphite does not contain CNTs, a simple extraction therefore is not possible.&lt;/li&gt;&#xA;&lt;li&gt;Graphite seemingly can be converted to CNTs by a &quot;simple&quot; oxidative method published by &lt;em&gt;DW Lee&lt;/em&gt; and &lt;em&gt;JW Seo&lt;/em&gt; on &lt;a href=&quot;http://arxiv.org/pdf/1007.1062v4.pdf&quot; rel=&quot;noreferrer&quot;&gt;arxiv.org&lt;/a&gt; in 2010.&lt;/li&gt;&#xA;&lt;li&gt;The authors claim to have used commercially available graphite powder (45 µm, Aldrich) of high purity (&gt;99.995%), but did not provide any information on the yields.&lt;/li&gt;&#xA;&lt;li&gt;Pencils typically contain a mixture of graphite powder and various amounts of clay as a binder.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-01-27T09:26:16.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24253" PostTypeId="2" ParentId="24229" CreationDate="2015-01-27T09:39:46.287" Score="2" Body="&lt;p&gt;As already mentioned in Fred's answer, &lt;strong&gt;Sodium (and other group 1 and 2 metals)&lt;/strong&gt; are dangerous in elemental form, but are flammable and corrosive rather than toxic. But the &lt;strong&gt;halogens&lt;/strong&gt; in elemental form are, as he says, much more toxic than the corresponding anions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another element which is very important for life in the form of phosphates but rather toxic in elemental form is &lt;strong&gt;phosphorus&lt;/strong&gt;. My high school chemistry teacher claimed that phosphorus poisoning could be diagnosed by glow in the dark vomit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carbon&lt;/strong&gt;, which is another extremely common element essential for life, will do you no good at all if inhaled in the form of coal dust.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, it's worth noting that toxic heavy metals like lead and mercury are often far worse in their compounds than as the free elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various reasons for this, one of which is bioavailability. Lead pipes were once  commonly used and only really a hazard when they corroded and thus dissolved in the water they were carrying. (BTW corrosion products of copper are also pretty bad for you.) Even if you eat lead shot, most of it will pass through your system unchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the organolead compounds (tetraethyl lead etc) that were once used in gasoline as an antiknock were extremely toxic required special handling procedures. The combustion products were much less toxic (though still a health hazard.)&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2015-01-27T09:39:46.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24254" PostTypeId="2" ParentId="24230" CreationDate="2015-01-27T11:19:43.593" Score="3" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt;&lt;br&gt;&#xA;The thermodynamic product of the dehydration reaction of 1-methylcyclobutan-1-ol is 1-methylcyclobut-1-ene, &lt;em&gt;not&lt;/em&gt; methylenecyclobutane. This is supported by [1],[3] and my own calculations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some details&lt;/strong&gt;&lt;br&gt;&#xA;I was unable to find any reference to why the dehydration of 1-methylcyclobutan-1-ol should yield methylenecyclobutane (&lt;strong&gt;3&lt;/strong&gt;) as the main product. From what I could gather up until now, the major product is 1-methylcyclobut-1-ene (&lt;strong&gt;1&lt;/strong&gt;). I also tried finding suitable mechanisms and/or protonation (hydration) energies for these two molecules according to the following scheme, to compare their relative thermodynamic stability to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k8hDg.jpg&quot; alt=&quot;protonation/hydration scheme of 1 and 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wiberg and Hao report the enthalpy difference between &lt;strong&gt;3&lt;/strong&gt; and &lt;strong&gt;1&lt;/strong&gt; to be only $\Delta H = -0.54~\mathrm{kcal/mol}$. They obtain this value from equilibrating &lt;strong&gt;1&lt;/strong&gt; and &lt;strong&gt;3&lt;/strong&gt; over sodium on alumina at various temperatures.&lt;sup&gt;[1]&lt;/sup&gt;  This would translate to a ratio of &lt;strong&gt;1&lt;/strong&gt;:&lt;strong&gt;3&lt;/strong&gt; of about 1.0 : 0.4.&lt;sup&gt;[2]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lajunen and Himottu quite similarly report, that &lt;strong&gt;1&lt;/strong&gt; is more stable than &lt;strong&gt;3&lt;/strong&gt; by about $4.3~\mathrm{kJ/mol} \approx 1~\mathrm{kcal/mol}$. They obtained this value via equilibrium constants measurements in perchloric acid.&lt;sup&gt;[3]&lt;/sup&gt;  This would in return translate to a ratio of &lt;strong&gt;1&lt;/strong&gt; : &lt;strong&gt;3&lt;/strong&gt; of about 1.0 : 0.2. It is also reported, that the transition states of protonation are quite similar, but lower for &lt;strong&gt;1&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also conducted some quantum chemical calculations, using DF-BP86/def2-SVP and the more sophisticated MP2/def2-TZVPP. These are in reasonable agreement with the previous findings. The enthalpy differences, approximated at $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$, are $1.4~\mathrm{kcal/mol}$ and $1.3~\mathrm{kcal/mol}$ respectively. This gives a &lt;strong&gt;1&lt;/strong&gt;:&lt;strong&gt;3&lt;/strong&gt; ratio of about 1.0 : 0.1. (More values, as well as the proton affinities (PA),&lt;sup&gt;[4]&lt;/sup&gt; yielding &lt;strong&gt;2&lt;/strong&gt;, are given in the table at the end.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not surprising, that these structures are quite similar in energy. They both contain a four membered ring. In &lt;strong&gt;1&lt;/strong&gt; two $\ce{sp^2}$ centres are present in the ring, forcing planarity. As a result of the shorter double bond, two angles are slightly larger than $90^\circ$, while two are forced smaller.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nJYBHm.jpg&quot; alt=&quot;bond angles of 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;3&lt;/strong&gt;, only one $\ce{sp^2}$ centre is present. However, the dihedral angle is a mere $18.4^\circ$, the out-of-plane angle is $26.8^\circ$, respectively. This is apparently not enough to significantly reduce ring strain. Because of symmetry constraints, i.e. $C_\mathrm{s}$, this arrangement forces three inner ring angles to be larger than $90^\circ$.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q6acBm.jpg&quot; alt=&quot;bond angles of 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of ring-strain, or &lt;a href=&quot;http://goldbook.iupac.org/S06037.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain&lt;/a&gt; in general, is not well defined and it cannot be measured directly. There are relative methods, to obtain this virtual quantity, sometimes involving hypothetical molecules. There are different reference systems and as far as I am aware none of them are considered the ultimative choice. In this sense, any arguments can be made. It is therefore also almost impossible to derive ring strain from just a pen-and-paper exercise, especially when the molecules are that similar. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/jo00017a022&quot; rel=&quot;nofollow noreferrer&quot;&gt;K. B. Wiberg and S. Hao, &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;1991&lt;/strong&gt;, &lt;em&gt;56&lt;/em&gt;, 5108-5110.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/14617/4945&quot;&gt;How the ratio is determined&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.3891/acta.chem.scand.41a-0252&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. Lajunen and M. Himottu, &lt;em&gt;Acta Chem. Scand.&lt;/em&gt; &lt;strong&gt;1987&lt;/strong&gt;, &lt;em&gt;41A&lt;/em&gt;, 252-258.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/10042/4945&quot;&gt;How proton affinity is defined&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Appendix&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{llrr}\hline&#xA;  &amp;amp; &amp;amp; \mathsf{\pmb1} &amp;amp; \mathsf{\pmb3} \\\hline&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \Delta E_\mathrm{el}    &amp;amp; 0.0   &amp;amp; 1.3   \\&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \Delta E_\mathrm{o}     &amp;amp; 0.0   &amp;amp; 1.6   \\&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \Delta H                &amp;amp; 0.0   &amp;amp; 1.4   \\&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \Delta G                &amp;amp; 0.0   &amp;amp; 1.6   \\&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \text{Ratio} (\Delta H) &amp;amp; 1.0   &amp;amp; 0.1   \\&#xA;\text{DF-BP86/def2-SVP} &amp;amp; \text{PA}               &amp;amp; 200.2 &amp;amp; 198.8 \\\hline&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \Delta E_\mathrm{el}    &amp;amp; 0.0   &amp;amp; 1.1   \\&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \Delta E_\mathrm{o}     &amp;amp; 0.0   &amp;amp; 1.5   \\&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \Delta H                &amp;amp; 0.0   &amp;amp; 1.3   \\&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \Delta G                &amp;amp; 0.0   &amp;amp; 1.7   \\&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \text{Ratio} (\Delta H) &amp;amp; 1.0   &amp;amp; 0.1   \\&#xA;\text{MP2/def2-TZVPP}   &amp;amp; \text{PA}               &amp;amp; 399.3 &amp;amp; 398.0 \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-01-27T11:25:12.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24255" PostTypeId="2" ParentId="24239" CreationDate="2015-01-27T11:47:39.997" Score="2" Body="&lt;p&gt;The reaction $$\ce{NaCl~(s) &amp;lt;=&amp;gt;[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chloride&quot; rel=&quot;nofollow&quot;&gt;sodium chloride&lt;/a&gt; are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}&amp;lt;c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.&lt;br&gt;&#xA;If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-01-27T11:47:39.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24256" PostTypeId="2" ParentId="24244" CreationDate="2015-01-27T11:53:52.277" Score="5" Body="&lt;p&gt;What you have read about the direct energy costs/benefits of glycolysis and gluconeogenesis  is not wrong. Glycolysis produces 2 ATP and gluconeogenesis consumes 6 ATP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net reaction for coversion of glucose $\ce{C6H12O6}$ into pyruvate $\ce{C3H3O3-}$ by &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycolysis#Overview&quot; rel=&quot;nofollow&quot;&gt;glycolysis&lt;/a&gt; is (note that I am keeping track of the charges and the magnesium counterion/cofactor on ADP/ATP):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H12O6 + 2NAD+ + 2[MgADP]- + 2HPO4^{2-} -&amp;gt;}$$&#xA;$$\ce{ 2C3H3O3- + 2NADH + 2[MgATP]^{2-} + 2H+ + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For conversion of pyruvate back to glucose by &lt;a href=&quot;http://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb1/part2/gluconeo.htm&quot; rel=&quot;nofollow&quot;&gt;gluconeogenesis&lt;/a&gt; (note that gluconeogenesis usually uses 4ATP and 2GTP - for simiplicity let's consider that a GTP has about the energy equivalence to an ATP):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C3H3O- + 2NADH + 4[MgATP]^{2-} + 2[MgGTP]^{2-} + 6 H2O -&amp;gt;}$$&#xA;$$\ce{C6H12O6 + 2NAD+ + 4[MgADP]- + 2[MgGDP]- + 6HPO4^{2-} +2H+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you compare the two processes directly, in absence of the source/fate of any of the components, you see that gluconeogenesis consumes 6 ATP/GTP and 2 NADH while glycolysis produces 2ATP and 2NADH. The NADH consumption and production is balanced out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would we compare these two directly and not worry about coupled processes? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Not all organisms undergo oxidative phosphorylation. Most do, but not all.&lt;/li&gt;&#xA;&lt;li&gt;More importantly, in the context of your question, NADH &lt;a href=&quot;http://en.wikipedia.org/wiki/Nicotinamide_adenine_dinucleotide#Functions&quot; rel=&quot;nofollow&quot;&gt;can do other things&lt;/a&gt;. Not all NADH goes to oxidative phosphorylation.&lt;/li&gt;&#xA;&lt;li&gt;Most importantly, the direct comparison is carbon balanced with the same organic species: glucose and pyruvate. If we try to include all possible variations and coupled processes we find ourselves eventually trying to do a general chemistry level stoichiometric and thermochemical analysis of an entire life-form. I'd rather not. There are lots of feed stocks into gluconeogenesis besides lactate and pyruvate. An amino acid or lipid that can be catabolized into acetyl-CoA can be fed into gluconeogenesis. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As for lactate...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;True, pyruvate can be generated from lactate $\ce{C3H5O3-}$, which also produces NADH:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H5O3- + NAD+ -&amp;gt; C3H3O3- + NADH + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with lactate, we have no net generation of NADH in gluconeogenesis. However, we are also no longer comparing the interconversion of the same carbon species. Lactate is produced from pyruvate (consuming NADH), so we are back to square one. Including lactate in the comparison instead of pyruvate leads to no net NADH production/consumption.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2015-02-21T13:30:16.933" LastActivityDate="2015-02-21T13:30:16.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24258" PostTypeId="1" CreationDate="2015-01-27T13:39:32.043" Score="1" ViewCount="824" Body="&lt;p&gt;We all know what is a wave function. There is a technique to convert non normalized wave function to normalized wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of non normalized wave function integration of square of wave function with respect to 'dt' in a length 'dx' will be any constant other than 1. In case of normalized wave equation the probability of finding will be maximum i.e. 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of converting a non-normalized wave function to a normalized wave function is known as Normalization technique. In this technique we multiply the wave function with some constant and makes it equal to 1 (condition for a normalized wave function). That means after conversion the probability of finding electron is becoming maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we multiply a wave function with any constant it's nature doesn't change. So, my question is after normalization of the wave function, the nature of the wave doesn't changes and how come the probability of finding electron becomes maximum i.e. 1???&lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="9961" LastEditDate="2015-01-27T22:43:29.483" LastActivityDate="2015-01-27T22:43:29.483" Title="Is probability of finding electron changing here?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24262" PostTypeId="2" ParentId="24258" CreationDate="2015-01-27T14:08:18.847" Score="0" Body="&lt;p&gt;Normalization means that the integral of the wave function times its complex conjugate over all space is 1.  It has nothing to do with maximizing anything.  It is only a way of saying &quot;there is one electron (or other particle)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if I say &quot;there is an 80% chance Vamsi is at school, 40% chance Vamsi is at home, 50% chance Vamsi it at work, and a 30% chance Vamsi is shopping&quot; it doesn't make any sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I say &quot;there is an 40% chance Vamsi is at school, 20% chance Vamsi is at home, 25% chance Vamsi it at work, and a 15% chance Vamsi is shopping&quot; it does make sense.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-27T14:27:23.400" LastActivityDate="2015-01-27T14:27:23.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24264" PostTypeId="2" ParentId="24258" CreationDate="2015-01-27T15:30:23.710" Score="3" Body="&lt;p&gt;I swear I read the question many times, but I still not quite sure what it is about.  So I will try to clarify some things hoping that they are relevant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First off, &lt;strong&gt;any complex multiple of a wave function&lt;/strong&gt; indeed &lt;strong&gt;describes exactly the same state&lt;/strong&gt; of a system. This directly follows from the postulate of quantum mechanics which states that the expectation value of an observable $A$ represented by a self-adjoint operator $\hat{A}$ is given as follows,&#xA;$$&#xA;\langle A \rangle&#xA;=&#xA;\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,&#xA;$$&#xA;where $\Psi$ is the wave function. It is indeed easy to show that given $\Psi$ we can multiply it by any complex number $c$ and this would not change the expectation value of any observable,&#xA;$$&#xA;\langle A \rangle&#xA;=&#xA;\frac{\langle c \Psi | \hat{A} | c \Psi \rangle}{\langle c \Psi | c \Psi \rangle} &#xA;=&#xA;\frac{c^{*} c \langle \Psi | \hat{A} | \Psi \rangle}{c^{*} c \langle \Psi | \Psi \rangle}&#xA;=&#xA;\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,&#xA;$$&#xA;since $c^{*} c$ in numerator and denominator perfectly cancels each other. Also note that the following definition of the expectation value&#xA;$$&#xA;\langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle \, ,&#xA;$$&#xA;holds true only for a normalized wave function $\Psi$ and trivially follows from the above mentioned more general definition.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Secondly, for the simplest case of one particle in one spatial dimension, if one would like to interpret the wave function $\Psi(x,t)$ as a &lt;strong&gt;probability amplitude&lt;/strong&gt; that the particle is at $x$, and consequently, it square modulus, $ \left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t)$ as the &lt;strong&gt;probability density&lt;/strong&gt; that the particle is at $x$, he/she &lt;strong&gt;have to normalize&lt;/strong&gt; it as follows,&#xA;$$&#xA;\int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1 \, .&#xA;$$&#xA;This is required because &lt;a href=&quot;http://en.wikipedia.org/wiki/Probability_axioms#Second_axiom&quot; rel=&quot;nofollow&quot;&gt;one of the axioms&lt;/a&gt; of the probability theory requires that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the probability that some elementary event in the entire sample space&#xA;  will occur is 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sample space is just a set of all possible outcomes in an experiment, which in our case is the set of events of finding a particle at any possible position, i.e anywhere between $-\infty$ and $+\infty$. Thus, in our case the probability of finding the particle anywhere between $-\infty$ and $+\infty$ have to be 1, and consequently, the wave function has to be normalized. Otherwise its interpretation as a probability amplitude would be inconsistent with the probability theory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hope it helps.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-01-27T15:30:23.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24266" PostTypeId="2" ParentId="24229" CreationDate="2015-01-27T17:33:48.270" Score="1" Body="&lt;p&gt;Soap is another good example, if you include corrosive materials. And while you're not ingesting it, you're likely absorbing some amount via your skin on a regular basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the necessary components of soapmaking is lye, or sodium hydroxide, which is so alkaline that it will eat through flesh. It also has an exothermic reaction with water, to the point that it can be explosive if you pour the water onto lye instead of the lye into the water (the former mixes it too fast, while the latter allows for a slower mix and more time for things to settle a little bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, due to the chemical reaction that lye undergoes with fatty acids, you're left with an inert and safe compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, sodium hydroxide is often used on pretzels before baking, but the (much more subtle) reaction with the dough renders it nonhazardous, as the end result is a change in the pH of the surface of the dough, which gives you that nice brown, chewy exterior. (Sodium bicarbonate or sodium carbonate can be used for this purpose, as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borax is another one, though human exposure is generally skin or indirect contact. Boric Acid, one of the chemical components of Borax, is toxic in both acute exposure (LF50 = 1-20g/kg in humans, depending on age), and chronic exposure (as little as 32mg/kg for chronic lethal dose). However, Borax, itself, is a salt of Boric Acid, and so, doesn't carry the same toxicity unless broken down with something like hydrochloric acid. In fact, Borax is a great detergent for all sorts of household cleaning purposes (laundry, dishes, all-purpose cleaning).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Baking soda (sodium bicarbonate) -- most commonly used as a leavening agent in baked goods -- is the product of a reaction of sodium chloride, ammonia, and carbon dioxide in water. Its cousin, washing soda (sodium carbonate) can be produced using sodium hydroxide. Which one you get from that initial reaction depends on how much carbon dioxide you add to the mix.&lt;/p&gt;&#xA;" OwnerUserId="13029" LastActivityDate="2015-01-27T17:33:48.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24267" PostTypeId="2" ParentId="20213" CreationDate="2015-01-27T18:41:48.687" Score="4" Body="&lt;p&gt;When the balloon is fully inflated with helium its density (mass/volume) is less than that of the air around it, allowing it to float. Over time, some of the helium will &quot;leak&quot; out of the balloon. As the helium is released the mass of the balloon decreases slightly, but the volume decreases at a much faster rate, resulting in a balloon with a density greater than that of the air around it, which is why it falls to the ground.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastActivityDate="2015-01-27T18:41:48.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24268" PostTypeId="2" ParentId="24229" CreationDate="2015-01-27T19:43:52.300" Score="3" Body="&lt;p&gt;I want to clarify that, although as the other answers demonstrate many elements are harmless in some but not all possible compounds, &lt;em&gt;cadmium is not like that&lt;/em&gt;.  Cadmium is a heavy metal; heavy metals should be assumed toxic in &lt;em&gt;all&lt;/em&gt; their forms.  This is because they easily &lt;em&gt;substitute&lt;/em&gt; for lighter elements in the same column of the periodic table (for cadmium, that would be zinc) and many of those lighter elements are critically important to normal biochemistry.  (Zinc, for instance, is used in key enzymes for aerobic respiration, protein synthesis, and DNA replication.  Substituting cadmium will make many if not all of those malfunction.)  For the same reason, heavy metals are easily absorbed by the body (the digestive system is actively trying to take up as much zinc as possible) and not easily gotten rid of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some heavy metal compounds &lt;em&gt;are&lt;/em&gt; so unreactive and/or insoluble that they're harmless, but that's the exception rather than the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the microorganisms mentioned in that article were probably orders of magnitude more dangerous than the cadmium compound.&lt;/p&gt;&#xA;" OwnerUserId="4921" LastActivityDate="2015-01-27T19:43:52.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24269" PostTypeId="2" ParentId="24230" CreationDate="2015-01-27T19:59:54.960" Score="1" Body="&lt;p&gt;Under acid conditions, cyclobutylmethylene (methylenecyclobutane) isn't even the second most abundant product experimentally.  Instead:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;45% 1-methyl-1-cyclobutene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;35% isoprene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20% methylenecyclobutane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01571a032&quot; rel=&quot;nofollow&quot;&gt;The Rate of Hydration of Methylenecyclobutane and the Effect of Structure on&#xA;Thermodynamic Properties for the Hydration of Small Ring Olefins&lt;/a&gt; J. Am. Chem. Soc., 1957, 79 (14), pp 3724–3729&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution 0.9 M in 1-methyl-1-cyclobutanol and&#xA;  0.3 M in $\ce{HClO4}$ was heated at reflux temperature&#xA;  for several hours, during which time a stream of&#xA;  nitrogen was swept through the system and the exit&#xA;  vapors were passed into a liquid nitrogen trap. The&#xA;  olefin sample collected was analyzed by infrared&#xA;  spectra in a 10-cm. gas cell at a pressure of 4.9 cm.&#xA;  The analysis indicated the decomposition product&#xA;  to be a mixture of approximately 45% 1-methyl-1-cyclobutene,&#xA;  35%, isoprene and 20% methylenecyclobutane.&#xA;  The formation of isoprene readily is&#xA;  pictured in the dehydration of 1-methyl-1-cyclobutanol&#xA;  in terms of a Whitmore [Beta]-cleavage of a C-C&#xA;  bond of the cyclobutane ring.&#xA;  An estimate of the relative thermodynamic stability&#xA;  of the endo and exo isomers may be made&#xA;  from the present results. Assuming the composition&#xA;  of the dehydration product given above to provide&#xA;  a rough measure of the relative rates of formation&#xA;  of the two olefins from aqueous carbinol solution&#xA;  at 25 [degrees C] one may calculate from this and the&#xA;  relative hydration rates, kp, that the free energy of&#xA;  formation of gaseous 1-methyl-1-cyclobutene is on&#xA;  the order of 1,5000 cal./mole more negative than&#xA;  that for methylenecyclobutane. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-27T20:26:48.327" LastActivityDate="2015-01-27T20:26:48.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24270" PostTypeId="2" ParentId="22152" CreationDate="2015-01-27T21:55:15.223" Score="5" Body="&lt;p&gt;There is a nice little article in the Journal of Chemical Eduction &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed083p1465&quot; rel=&quot;nofollow&quot;&gt;Negative pH Does Exist&lt;/a&gt; pointing out examples of negative pHs in water, for example pH = -3.6 in mine water in California.  So -1.7 is not the lowest possible pH in water. Keep in mind that pH is defined in terms of hydrogen ion activity rather than concentration.  This is emphasized in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed083p752&quot; rel=&quot;nofollow&quot;&gt;Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+]&lt;/a&gt;.  At high acid concentation, hydrogen ion &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow&quot;&gt;activity&lt;/a&gt; greatly deviates from hydrogen ion concentration.  For example, in 16 molar HCl, hydrogen ion activity is 678! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And see &lt;a href=&quot;http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1488&amp;amp;context=usgsstaffpub&quot; rel=&quot;nofollow&quot;&gt;Negative pH and Extremely Acidic Mine Waters from Iron Mountain, California&lt;/a&gt; (no paywall), which describes pH values even below -4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00790a076&quot; rel=&quot;nofollow&quot;&gt;Hydronium ion activity in strongly acidic media. Remarkable agreement between independent estimates&lt;/a&gt; Am. Chem. Soc. vol. 95, pp 3055–3057, 70% sulfuric acid is found to have a pH of -9.8 (negative 9.8)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, ignoring activity and the true definition of pH, considering that pure water has a concentration of 55M, if all the water molecules were instead hydronium ions, -log(55) = -1.7.  Then in articles like &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed063p473&quot; rel=&quot;nofollow&quot;&gt;Who Knows the Ka Values of Water and the Hydronium Ion?&lt;/a&gt; this value is attributed to the pKa of hydronium, with the supposed justification that:&#xA;$K_a = \frac{[\ce{H3O+}][\ce{H2O}]}{[\ce{H3O+}]} = [\ce{H2O}]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article &lt;a href=&quot;http://pubs.rsc.org/en/content/articlepdf/1998/AN/A705491B&quot; rel=&quot;nofollow&quot;&gt;New point of view on the meaning and on the values of&#xA;Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water&lt;/a&gt; is a serious consideration of the true meaning of the pKa of hydronium, and points out that the -1.7 value for the pKa of hydronium has no justification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, pH can be below -1.7 and the pK of hydronium isn't really -1.7.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-06T20:59:34.520" LastActivityDate="2015-03-06T20:59:34.520" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24272" PostTypeId="1" CreationDate="2015-01-28T01:25:19.883" Score="0" ViewCount="1317" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the boiling point of an aqueous solution that has a vapor pressure of &lt;span class=&quot;math-container&quot;&gt;$23.0~\text{Torr}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$25~^\circ\text{C}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(P^\circ_{\ce{H2O}}=23.78~\text{Torr};~K_\mathrm b= 0.512~^\circ\mathrm{C\ kg/mol})$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To solve this I used the equation &lt;span class=&quot;math-container&quot;&gt;$\Delta T_\mathrm b =m\times K_\mathrm b$&lt;/span&gt;. I calculated a molality of &lt;span class=&quot;math-container&quot;&gt;$1.88247\ \mathrm{mol/kg}$&lt;/span&gt; which I then used to calculate a boiling point of &lt;span class=&quot;math-container&quot;&gt;$-0.9638\ \mathrm{^\circ C}$&lt;/span&gt;. This answer is apparently wrong and I don't understand why.&lt;/p&gt;&#xA;" OwnerUserId="12959" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-11-14T23:21:04.197" Title="Calculating boiling point of a solution" Tags="&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24273" PostTypeId="2" ParentId="24272" CreationDate="2015-01-28T01:43:27.687" Score="3" Body="&lt;p&gt;If molality is 1.88, then the &lt;em&gt;change&lt;/em&gt; in boiling point is -0.96 degrees C, not the boiling point itself.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-28T01:43:27.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24274" PostTypeId="1" AcceptedAnswerId="24277" CreationDate="2015-01-28T01:57:39.810" Score="8" ViewCount="2194" Body="&lt;p&gt;In the AMC series &lt;strong&gt;Breaking Bad&lt;/strong&gt; , Walter (aka Heisenberg) smuggles in a bag of mercury fulminate under the guise of crystal methamphetamine and threatens to blow the room up if he doesn't get his money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does fulminated mercury really have the translucent white/blue crystal appearance as was shown in this scene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the compound that sensitive that throwing it against the wall would cause detonation?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="7448" LastEditDate="2015-08-30T09:35:14.587" LastActivityDate="2015-08-30T09:35:14.587" Title="Does mercury fulminate really look like crystal methamphetamine?" Tags="&lt;crystal-structure&gt;&lt;optical-properties&gt;&lt;explosives&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24275" PostTypeId="1" AcceptedAnswerId="24307" CreationDate="2015-01-28T02:35:03.700" Score="14" ViewCount="20502" Body="&lt;p&gt;There are all kinds of different dinstinctions in the internet, and I'd like to see what you guys thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I've been asked if the heat capacity of an ideal gas es independent of temperature. I've said no, even though it's practically invariant under small ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Namely the result $C_V/n = \frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What really is an ideal gas? What's its difference with a perfect gas? Is my answer to the question I was posed correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorDisplayName="user7951" LastEditDate="2016-09-23T19:15:25.993" LastActivityDate="2021-04-23T06:45:03.520" Title="What's the difference between perfect and ideal gas?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="7" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24277" PostTypeId="2" ParentId="24274" CreationDate="2015-01-28T02:42:42.503" Score="9" Body="&lt;p&gt;Mercury fulminate $\ce{[Hg(CNO)2]}$ can have a white, gray or greenish-brown color depending upon purity and dryness (it is often kept moistened to prevent accidental decomposition).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does fulminated mercury really have the translucent white/blue crystal&#xA;  appearance as was shown in this scene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've never seen a picture where it appears translucent or where it has a bluish color.  The pictures I've seen always show it as an opaque powder.  I suspect that it could be crystallized, but that's probably never done as it would further increase its sensitivity to detonation (&lt;em&gt;e.g&lt;/em&gt;. if a crystal cracked, or rubbed against another crystal, the friction, &lt;a href=&quot;http://en.wikipedia.org/wiki/Triboelectric_effect&quot;&gt;triboelectrification&lt;/a&gt;, etc. could cause the material to explode).  Here are some pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VtE2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lEyY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWzKL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the compound that sensitive that throwing it against the wall would&#xA;  cause detonation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.today.com/popculture/mythbusters-proves-breaking-bads-walt-needs-some-more-schooling-6C10904583&quot;&gt;Here is a link&lt;/a&gt; to an examination of the Breaking Bad &quot;detonation by throwing&quot; experiment (with video).  The conclusion is that it can detonate when thrown, but 1) a larger amount of material than what Walt had, and 2) a much harder throw (greater force, greater shock) would have been required to cause detonation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-28T02:42:42.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24278" PostTypeId="1" AcceptedAnswerId="24280" CreationDate="2015-01-28T02:50:04.027" Score="7" ViewCount="11395" Body="&lt;p&gt;I'm finding information in Google with the keyword (C2H5OH + K) to find the reaction between $\ce{C2H5OH}$ and $\ce{K}$ and found very little information about it. Just 2-3 related results with contradictory pieces of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.webqc.org/balance.php?reaction=K+%2B+C2H5OH+%3D+KOH+%2B+C2H5&quot; rel=&quot;nofollow noreferrer&quot;&gt;The first one gave&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{K + C2H5OH -&amp;gt; KOH + C2H5},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://in.answers.yahoo.com/question/index?qid=20140103231122AA62uyX&quot; rel=&quot;nofollow noreferrer&quot;&gt;the second one gave&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2C2H5OH + 2K -&amp;gt; 2C2H5OK + H2},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://forum.xumuk.ru/index.php?showtopic=17664&quot; rel=&quot;nofollow noreferrer&quot;&gt;the third one gave&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{C2H5OH + KOH -&amp;gt; C2H5OK + H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no trusted public sources so I'm very baffled. I think that $\ce{C2H5OH}$ will react with $\ce{K}$ to form $\ce{C2H5OK}$ because it looks like the reaction with $\ce{Na}$. But I'm not sure and just want to find a trusted source. Can anyone confirm the true result for this reaction?&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="194" LastEditDate="2017-04-08T01:08:56.503" LastActivityDate="2017-04-08T01:08:56.503" Title="Does ethanol react with potassium to form potassium ethanolate and hydrogen gas?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24279" PostTypeId="2" ParentId="24275" CreationDate="2015-01-28T02:50:15.130" Score="3" Body="&lt;p&gt;Simply put, an ideal gas follows the ideal gas law. In thermodynamics and statistical mechanics, many physical relations are derived using the ideal gas law, sometimes accounting for Van der Waals forces or other non-negligible effects, depending on the conditions. The way I understand it, a perfect gas is an ideal gas, but is never treated as having non-interacting particles. I don't think the distinction is terribly important unless you plan on doing some thesis or dissertation in the fields of physical chemistry or condensed matter physics...&lt;/p&gt;&#xA;" OwnerUserId="13008" LastActivityDate="2015-01-28T02:50:15.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24280" PostTypeId="2" ParentId="24278" CreationDate="2015-01-28T03:03:55.857" Score="10" Body="&lt;p&gt;You can view water as the simplest alcohol, and indeed, water reacts with all of the Group 1 metals to form hydrogen gas and the corresponding metal hydroxide according to the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2M + 2HOH -&amp;gt; 2MOH + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{M}$ is the metal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reaction occurs with simple alcohols to produce the corresponding metal alkoxide and hydrogen gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2M + 2ROH -&amp;gt; 2MOR + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase the hydrocarbon portion of the alcohol, or reduce accessibility to the $\ce{OH}$ portion of the alcohol (&lt;em&gt;e.g.&lt;/em&gt; make the alcohol less like water), you slow down the rate of the reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/inorganic/group1/reacth2o.html&quot;&gt;Here is a nice link&lt;/a&gt; that compares and discusses the relative reactivities of the Group 1 metals with water. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-28T03:03:55.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24282" PostTypeId="2" ParentId="22053" CreationDate="2015-01-28T03:17:36.370" Score="1" Body="&lt;p&gt;Take a look at &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0079642500000050&quot; rel=&quot;nofollow&quot;&gt;Kelly, P. M., Francis Rose, L.R., The martensitic transformation in ceramics — its role in transformation toughening, Progress in Materials Science, 2002, Vol.47(5):463–557, doi:10.1016/S0079-6425(00)00005-0&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, a transformation is martensitic if it occurs from a collective shear motion of the atoms where each atomic displacement is less than the distance between atoms.  There is no diffusion (atoms moving to other lattice sites). The definition is the same as in a metal, but in a ceramic you have multiple atom types in the crystal, that's all.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2015-01-28T03:17:36.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24283" PostTypeId="2" ParentId="19218" CreationDate="2015-01-28T03:37:00.153" Score="4" Body="&lt;p&gt;In 3D you will see spots not cones from a single crystal.  If you Google for &quot;XRD spot pattern&quot; images, you will probably find yourself enlightened about how many different shapes, and forms diffraction patterns can take.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Debye-Scherrer rings are one special case which occurs when you have a diffraction pattern from a very fine powder.  In such a powder you are summing the diffraction patterns from thousands to millions of individual crystallites.  In this case, you can imagine a large number of spot patterns rotated randomly around the beam axis so each spot projects into a cone.  Each individual crystallite, however will produce a spot pattern, and there is never anything but a spot pattern from a single crystallite.  (Sometimes spots can be smeared when there is strain in the crystal, or when the crystal is too small to produce high quality diffraction, but these are really cases of imperfect crystals.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see diffractions from rays transmitted through the crystal.  However, X-rays tend to be absorbed within a few microns to millimeters in most materials so that limits the size of crystal you can use for transmission X-ray diffraction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I noticed you said &quot;Two parallel and in-phase x-rays...&quot;  I just want to note that diffraction occurs with a single photon.  A single photon can scatter, and produce constructive/destructive interference with ITSELF.  You need to do this with many photons to see a full interference pattern, however, and this is usually the case in experimental situations.  If this is new to you, then take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;nofollow&quot;&gt;Young's double slit experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this clarifies things a bit.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditDate="2015-01-28T04:14:57.617" LastActivityDate="2015-01-28T04:14:57.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24284" PostTypeId="2" ParentId="24229" CreationDate="2015-01-28T04:25:41.120" Score="1" Body="&lt;p&gt;The obvious example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Roxarsone&quot; rel=&quot;nofollow&quot;&gt;roxarsone&lt;/a&gt;, an arsenic based growth hormone commonly found on the shelves in chicken. We are assured that it is non toxic. I guess the body handles toxic elements differently if they are bound to other elements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iatp.org/documents/arsenic-widespread-in-chicken-testing-finds&quot; rel=&quot;nofollow&quot;&gt;http://www.iatp.org/documents/arsenic-widespread-in-chicken-testing-finds&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13037" LastEditorUserId="4945" LastEditDate="2015-01-28T06:18:48.517" LastActivityDate="2015-01-28T06:18:48.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24285" PostTypeId="1" CreationDate="2015-01-28T05:27:24.497" Score="7" ViewCount="357" Body="&lt;p&gt;First of all, I have zero chemical background, but I seek for an easy explanation about Nano Zinc Oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that the name of &quot;Nano Zinc Oxide&quot; couldn't possibly exist. Instead, it should be coined as &quot;Nano Zinc Chelate&quot;, since an Oxide particle is quite big and Nano is so small, relatively. It sounds very contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is, of course, Nano Zinc Oxide, but I don't understand what he really means and  I feel enthusiastic to understand it.&lt;/p&gt;&#xA;" OwnerUserId="13038" LastEditorUserId="7448" LastEditDate="2015-01-28T20:07:24.897" LastActivityDate="2015-04-21T07:12:34.210" Title="What is Nano Zinc Oxide?" Tags="&lt;inorganic-chemistry&gt;&lt;materials&gt;&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24286" PostTypeId="2" ParentId="24285" CreationDate="2015-01-28T06:16:39.903" Score="2" Body="&lt;p&gt;Nano zinc oxide is a real material used in sunscreens and looks like &lt;a href=&quot;https://www.flickr.com/photos/fei_company/4790479142/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.  It is just little crystals of zinc oxide which are about 30 nm across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chelate refers to something which is produced by a chemical process of &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Chelation&quot; rel=&quot;nofollow&quot;&gt;chelation&lt;/a&gt;.  Chelation means that a metal atom has more than one chemical bond to surrounding organics.  (That is kind of a bad definition, but I'm trying to keep the lingo simple.)  Chelation can be &lt;a href=&quot;https://www.infona.pl/resource/bwmeta1.element.elsevier-3dbd34bb-146e-39d1-be89-b6f4cce12c61&quot; rel=&quot;nofollow&quot;&gt;used to produce zinc oxide&lt;/a&gt;, but I don't think this is the process used for manufacturing most industrial zinc oxide.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2015-01-28T06:16:39.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24287" PostTypeId="1" AcceptedAnswerId="24288" CreationDate="2015-01-22T01:18:23.840" Score="3" ViewCount="54" Body="&lt;p&gt;When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? The sponge should still contain dish soap an hour later.&lt;/p&gt;&#xA;" OwnerUserId="12995" OwnerDisplayName="blackcornail" LastActivityDate="2015-01-28T06:18:37.687" Title="Why don't sponges foam up an hour later?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24288" PostTypeId="2" ParentId="24287" CreationDate="2015-01-22T04:07:20.090" Score="6" Body="&lt;p&gt;Dish soap traditionally contained foaming agents such as sodium laureth sulfate which is chemically stable over long periods of time. But more recently environmentally conscious soap companies are turning to foaming agents that have a shorter half-life. These chemicals tend to break down sooner - perhaps just by being exposed to air. This helps reduce foaming in the settling basins of wasterwater treatment plants, or storm drains that enter the ocean and it takes less water to rinse away the soap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read the ingredients of the soap you are using and research to see which are the foaming agents, and what the stable half-life is.&lt;/p&gt;&#xA;" OwnerUserId="9411" OwnerDisplayName="docscience" LastActivityDate="2015-01-22T04:07:20.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24289" PostTypeId="1" AcceptedAnswerId="24297" CreationDate="2015-01-28T01:23:50.667" Score="2" ViewCount="63" Body="&lt;p&gt;For example, in hexagonal closed-packed (HCP) structure, the atoms may be arranged in ABAB stacks or ACAC stacks. Can these two be considered enantiomers? Do they possess different solid state properties?&lt;/p&gt;&#xA;" OwnerUserId="2593" OwnerDisplayName="syockit" LastEditorUserId="9961" LastEditDate="2015-01-28T12:46:04.503" LastActivityDate="2015-01-28T13:20:48.693" Title="Is there anything akin to molecular chirality for crystalline?" Tags="&lt;crystal-structure&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24290" PostTypeId="2" ParentId="24289" CreationDate="2015-01-28T03:31:49.090" Score="1" Body="&lt;p&gt;Typically, different ways of arranging atoms in a solid are referred to as phases. For a single composition, a material may have different phases depending on temperature or pressure. Different crystal structures are not referred to as enantiomers. For an HCP metal, there is no difference an ABAB stacking and ACAC stacking, other than the naming. If the stacking is ABCABC, then you have a face-centered cubic structure (FCC), not an HCP. The best way to alter solid-state properties such as resistivity, susceptibility, etc., is to change the temperature or introduce defects into the system.&lt;/p&gt;&#xA;" OwnerUserId="13008" OwnerDisplayName="Matt Holt" LastActivityDate="2015-01-28T03:31:49.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24292" PostTypeId="2" ParentId="13679" CreationDate="2015-01-28T09:35:52.377" Score="5" Body="&lt;p&gt;I have found an old paper that might answer your question, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Effect of Molecular Weight Distribution on Polyethylene Film Properties&quot;&#xA;P. J. Perrono, P. B. Lederman, &lt;em&gt;Polymer Egineering and Science&lt;/em&gt;, &lt;strong&gt;1972&lt;/strong&gt;, 12, 5, 340-345 (DOI: 10.1002/pen.760120505).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of polyethylene evaluated in this paper as a function of the molecular weight include film gloss, haze, tear resistance and film impact strength.&#xA;This study shows that broadening the molecular weight distribution increases haze and reduces film gloss. Further, it was shown that a linear relationship exists between film gloss and external haze. Both values are measures of surface irregularities in the film which are affected by the drawing characteristics of the polymer. A broader molecular weight distribution results in increased impact strength as measured by the Dart Drop Impact Test. This is, it is believed, a result of the increase in long chain branching of the higher molecular weight fractions of the polymer which cause a higher degree of molecular weight entanglement at the branch sites. In contrast the tear strength is reduced as the molecular weight distribution broadens because of the low molecular weight fraction in the broad spectrum material which tend to decrease resistance to tear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-28T09:35:52.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24294" PostTypeId="2" ParentId="24229" CreationDate="2015-01-28T12:22:53.633" Score="0" Body="&lt;p&gt;This may be one of the simplest molecules to answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barium sulphate is widely used, in the form of an emulsion in water, for the &quot;barium meal&quot; X-ray procedure for diagnosing problems in the digestive tract. Barium is highly toxic, as are most of its compounds, but the sulphate is insoluble in water and stomach acid, so does not pass into the bloodstream and kill the patient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barytes, the mineral, is slightly impure barium carbonate (oftem pink colour) and is found in most parts of the world where mineralisation is present, e.g. in and adjacent to veins of metals such as lead and copper ores. It is safe to handle, and obviously has not been dissolved by groundwater over millions of years, demonstrating that it is insoluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ingest almost any other barium compound, and you die. Handle with extreme care!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and as a quick edit in response to another post above, barium is indeed a heavy metal. I agree that almost all heavy metals (exceptions such as gold and platinum) are toxic, however there will probably be a few of their compounds which, like barium sulphate, can't pass into the body fluids due to lack of solubility, and as such will be non-toxic.&lt;/p&gt;&#xA;" OwnerUserId="13047" LastActivityDate="2015-01-28T12:22:53.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24295" PostTypeId="1" CreationDate="2015-01-28T12:35:01.690" Score="2" ViewCount="1345" Body="&lt;p&gt;Looking at a system in which phosphoric acid is concentrated in an evaporator which uses a constant recycle stream with a heat exchanger in this stream to heat the acid so that water evaporates out of the acid when it re-enters the main stream, How can I find the Q needed to remove a certain amount of water from the stream in order to concentrate the acid to a specific level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given the amount of water evaporated, both concentrations, and the flow rate of both the entering acid stream and the exiting acid stream. I have also been told that you cannot simply use $Q=mC_p\Delta T$, because this is inacurate for phosphoric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or sources you could link me to would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13048" LastEditorUserId="10185" LastEditDate="2015-01-28T14:38:47.833" LastActivityDate="2015-03-08T13:22:30.833" Title="Heat energy of phosphoric acid evaporation" Tags="&lt;heat&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24296" PostTypeId="2" ParentId="738" CreationDate="2015-01-28T13:13:03.497" Score="6" Body="&lt;p&gt;I totally agree with George that &lt;a href=&quot;https://chemistry.stackexchange.com/questions/738/what-is-a-neutral-atom#comment1424_738&quot;&gt;&quot;neutral atom&quot; is close to nonsense&lt;/a&gt;. Indeed, IUPAC recommends the following definition for atom:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/A00493.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;atom&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Smallest particle still characterizing a chemical element. It&#xA;  consists of a nucleus of a positive charge ($Z$ is the proton number and&#xA;  $e$ the elementary charge) carrying almost all its mass (more than&#xA;  99.9%) and $Z$ electrons determining its size.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As one can see, for an atom the proton number, also known as atomic number, which is the number of protons in the atomic nucleus, is the same as the number of electrons. Consequently, any atom is (electrically) neutral by definition, as George mentioned in his comment.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-01-28T13:13:03.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24297" PostTypeId="2" ParentId="24289" CreationDate="2015-01-28T13:20:48.693" Score="4" Body="&lt;p&gt;There are some crystal structures where the nature of the crystal structure creates chirality and therefore optical activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trivial example (important because of its history: it was involved in the discovery of optical activity and molecular chirality) is tartaric acid. But the crystal is chiral because the molecule is chiral so that might not be what yyou were looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are examples of crystals where the structure as a whole is chiral but the building blocks are not. Quartz, for example, is formed from a network of linked SiO&lt;sub&gt;4&lt;/sub&gt; tetrahedra (not molecules but a giant interlinked network where each oxygen is shared between two silicons). The overall crystal structure, though, is chiral due to small shifts in the rotation of the linked tetrahedra (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartz&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for a detailed explanation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a crystal analogy to molecular chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a few crystal structures can do this and the ones you use as examples can't.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-01-28T13:20:48.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24298" PostTypeId="2" ParentId="24295" CreationDate="2015-01-28T13:36:36.113" Score="1" Body="&lt;p&gt;You need to know the &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_of_vaporization&quot; rel=&quot;nofollow&quot;&gt;latent heat of vaporization&lt;/a&gt; of the solution, which would be the same as that of water when phosphoric acid concentration is low.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Fig. 10 of &lt;a href=&quot;http://www.potashcorp.com/media/PCS_phos_acid_manual.pdf&quot; rel=&quot;nofollow&quot;&gt;this bulletin&lt;/a&gt; for latent heat of vaporization as a function of phosphoric acid concentration, but keep in mind this is only strictly valid at the stated temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are varying phosphoric acid concentration (concentrating phosphoric acid) you would need to integrate heat of vaporization as a function of phosphoric acid concentration over the appropriate concentration range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as phosphoric acid concentration gets very high, it is no longer valid to consider the evaporating solution to be only water, it will increasingly contain phosphoric acid as well. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-28T13:36:36.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24299" PostTypeId="1" AcceptedAnswerId="33059" CreationDate="2015-01-28T14:06:53.677" Score="22" ViewCount="31865" Body="&lt;p&gt;Are there yellow metals besides gold? Are there metals with color different than yellow and grey/silver or reddish?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="9961" LastEditDate="2016-11-12T01:33:48.067" LastActivityDate="2020-10-19T02:44:48.717" Title="Are there colorful metals besides gold and copper?" Tags="&lt;metal&gt;&lt;color&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24301" PostTypeId="1" CreationDate="2015-01-28T14:38:53.793" Score="2" ViewCount="778" Body="&lt;p&gt;Well, I think $\ce{CaI2}$ is the same as $\ce{BeH2}$ and $\ce{AlCl3}$ the same as $\ce{NH3}$, but I'm not sure since $\ce{I}$ and $\ce{Cl}$ have their own pair electrons and I don't know if this affects the structure...&lt;/p&gt;&#xA;" OwnerUserId="12759" LastEditorUserId="10185" LastEditDate="2015-01-28T16:19:26.717" LastActivityDate="2015-01-28T16:35:32.140" Title="What is the geometric formula of CaI2 and AlCl3" Tags="&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24303" PostTypeId="2" ParentId="24301" CreationDate="2015-01-28T15:10:57.850" Score="1" Body="&lt;p&gt;$\ce{CaI2}$ is mostly ionic in the solid and aqueous phases.  In the gas phases it is a linear molecule.  See &lt;a href=&quot;http://pubs.rsc.org/en/content/articlepdf/2002/ce/b201975m&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ionic or covalent? Can some first hints be derived from the solid state&#xA;structures of alkaline earth metal halide adducts?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AlCl3}$ has an octaheral structure in the solid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the liquid phase:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zL2s5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trigonal planar monomers also exist at high temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Aluminium_chloride&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="10185" LastEditDate="2015-01-28T16:35:32.140" LastActivityDate="2015-01-28T16:35:32.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24304" PostTypeId="1" CreationDate="2015-01-28T15:30:00.023" Score="2" ViewCount="62" Body="&lt;p&gt;What is the best way of preparing an aqueous solution containing 5 mM &#xA;$\ce{CdSO4}$, 1 mM $\ce{SeO2}$, 0.5 M $\ce{Na2SO4}$ and adjusting pH to 3 by adding 0.5 M $\ce{H2SO4}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the conventional way of doing it (in industrial chemistry or simply just the labs)?&lt;/p&gt;&#xA;" OwnerUserId="13053" LastEditorUserId="7448" LastEditDate="2015-01-28T20:01:26.647" LastActivityDate="2015-01-28T20:01:26.647" Title="Preparing CdSe electrodeposition bath" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24305" PostTypeId="1" AcceptedAnswerId="24306" CreationDate="2015-01-28T16:59:54.013" Score="4" ViewCount="892" Body="&lt;p&gt;I did not know there are products such as a magnetic stirrer with a hot plate, enabling one to apply heat and stirring to a solution with relative ease. Did I understand correctly, that device needs a magnetic bar to be added to the solution the device to work? Does that pose any limits when it can be used (if the magnet would react with something..) How much do these devices increase productivity?&lt;/p&gt;&#xA;" OwnerUserId="13057" LastActivityDate="2015-01-28T18:10:45.393" Title="Magnetic stirrer with hot plate" Tags="&lt;thermodynamics&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24306" PostTypeId="2" ParentId="24305" CreationDate="2015-01-28T18:10:45.393" Score="2" Body="&lt;p&gt;Most of the things to be told about magnetic stirring are already in Rons comment. A rotating magnetic field is induced by a rotating permanent magnet inside the hot plate magnetic stirrer. This field drives the magnet inside your reaction flask. To ensure chemical inertness in most situations the magnet is coated with PTFE (Teflon) or protected by a thin layer of glass. Magnetic stirring can replace traditional mechanical stirring is most situations and has many benefits over it: it does not require an additional inlet to the reaction flask, which is very handy for reaction under reduced pressure and is also more easily adjustable to different scales due to the possibility to replace the magnet in the reaction flask. Two exceptions are 1) very large/very viscous reactions or reactions in which excessive amounts of solid are present 2) reactions that involve ferromagnetic compounds that interfere with the rotational movement of the magnet inside the flask (as in Urushibara nickle catalyst preparation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ia600403.us.archive.org/26/items/TextbookOfPracticalOrganicChemistry5thEd/VogelPracticalOrganicChemistry5thEditionnewfoundV.pdf&quot; rel=&quot;nofollow&quot;&gt;Vogel&lt;/a&gt; has a section on reaction heating and stirring in which this and more is discussed extensively. See pages 71-80.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-28T18:10:45.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24307" PostTypeId="2" ParentId="24275" CreationDate="2015-01-28T18:23:24.193" Score="10" Body="&lt;p&gt;An ideal gas is the same as a perfect gas. Just different naming. The usual name for such gases (for which is assumed that the particles that make up the gas have no interaction with each other) is ideal gas, perfect gas is what such a gas is named in Atkins physical chemistry book. Personally I like the perfect gas naming better as it illustrates the perfect nature of the assumptions made about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simple systems (like monoatomic gases) when we can assume a perfect/ideal gas, $C_{V,m}$ is independent of temperature. For real gases this is certainly not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $C_{V,m}=\frac{3}{2}R$ only holds for monoatomic gases.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-01-28T18:23:24.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24308" PostTypeId="2" ParentId="24299" CreationDate="2015-01-28T18:46:53.323" Score="13" Body="&lt;p&gt;There is no &lt;em&gt;element&lt;/em&gt; other than gold and copper with similar colors; see &lt;a href=&quot;http://periodictable.com/Properties/A/Color.html&quot;&gt;http://periodictable.com/Properties/A/Color.html&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many metal &lt;em&gt;alloys&lt;/em&gt; that are red and golden, such as phosphor bronze, which can have a golden hue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, some &lt;em&gt;metallic compounds&lt;/em&gt; may be shiny yellow. Freshly sliced potassium is silvery, but soon acquires an oxide/nitride coat that may look yellow (perhaps due to interference of light?). Of course, &quot;fool's gold&quot;, FeS2, was often confused with the element. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrite&quot;&gt;http://en.wikipedia.org/wiki/Pyrite&lt;/a&gt; and &lt;a href=&quot;http://www.mindat.org/arphotos/371h319-0205976001300052845.jpg&quot;&gt;Image of pyrite crystal&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-01-28T18:46:53.323" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24310" PostTypeId="5" CreationDate="2015-01-28T19:34:40.447" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-28T19:34:40.447" LastActivityDate="2015-01-28T19:34:40.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24311" PostTypeId="4" CreationDate="2015-01-28T19:34:40.447" Score="0" Body="Questions relating to radioactive substances and the chemistry of radioactivity. Also, use [rare-earth-elements] if the question is about their radioactivity specifically." OwnerUserId="10185" LastEditorUserId="23561" LastEditDate="2018-08-11T20:03:48.213" LastActivityDate="2018-08-11T20:03:48.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24312" PostTypeId="1" AcceptedAnswerId="24313" CreationDate="2015-01-28T19:34:43.920" Score="1" ViewCount="1154" Body="&lt;p&gt;The notebook is a Dell Inspiron 6400 and the coating/paint is the silver on the palmrest and back cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says 99,4% on the container of the isopropyl alcohol.&lt;/p&gt;&#xA;" OwnerUserId="13059" LastEditorUserId="10185" LastEditDate="2015-01-28T20:01:32.340" LastActivityDate="2015-01-28T20:01:32.340" Title="Will isopropyl alcohol dissolve the coating/paint of a notebook?" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;&lt;metal&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24313" PostTypeId="2" ParentId="24312" CreationDate="2015-01-28T19:44:49.873" Score="3" Body="&lt;p&gt;Unfortunately you won't really know until you try it because there doesn't appear to be any info about what they used as a coating. Having said that, isopropanol isn't a superbly strong organic solvent so while it may remove some of the coating, it's not likely to immediately eat through your laptop like how acetone might. Still, follow user137's advice and test it on a very small area before you try anything else, and give it time before you go using it everywhere else. If you don't see any damage and you decide to go through with whatever it is you want to do (I assume cleaning it), just make sure you wipe it off with a damp cloth (water) followed by a dry cloth as soon as you get the results you want to minimize any possible damage.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastActivityDate="2015-01-28T19:44:49.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24314" PostTypeId="1" CreationDate="2015-01-28T19:48:53.507" Score="14" ViewCount="10655" Body="&lt;p&gt;I am a high school student and for my chemistry lab, I did an electrolyis experiment on how the changing concentration of electrolyte (copper sulfate) will change the rate at which the copper is deposited on the cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis was that as the concentration is increased, so will the rate of electrolysis. However, my results are completely opposite. When I use 0.1 M solution, loads of copper is deposited on the cathode but as I increase the concentration, the amount of copper deposited becomes progressively less and less. With 1 M solution, there was almost no copper being deposited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes no sense to me. Could somebody explain what is going on? Am I doing something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used both copper electrodes. I did not use distilled water, just normal tap water. I used 12 V power supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried adding a little sulfuric acid (making the pH around 3) to the copper sulfate but it did not make any difference on electrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to observe the electrolysis very closely today and noticed that there was some sort of oily liquid falling off from the anode. Also, I noticed that initially some flaky brown/black substance fell off from the anode (which may or may not be copper)and soon there was a silvery layer formed on the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I checked using Hanna checker that the overall pH of the solution was around 3 but the pH near the anode read between 9 and 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that when I used a higher concentration, the copper deposition on the cathode was brown and clearly copper. But when I used lower concentration, even though the amount of deposit was a lot more, the substance was probably not pure copper as I had assumed earlier. It was black and flaky. It was certainly different to the deposit I observed when I used higher concentration of copper sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know now what is happening? Is there anything that I can do to fix the problem?&lt;/p&gt;&#xA;" OwnerUserId="2446" LastEditorUserId="402" LastEditDate="2017-01-14T08:46:43.783" LastActivityDate="2019-07-18T13:50:42.977" Title="Does increasing concentration of copper sulfate decrease or increase its conductivity?" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;electrolysis&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24315" PostTypeId="1" AcceptedAnswerId="24319" CreationDate="2015-01-28T20:53:54.417" Score="10" ViewCount="7861" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvys5.png&quot; height=&quot;80&quot; alt=&quot;Methoxycyclohexene and methoxycyclohexane&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand why it isn't just a simple addition to the double bond. I get that there's conjugation but is this what makes things different? I don't get why methanol is the leaving group and therefore I cannot predict the other product. Also, it says that methoxycyclohexane (the same molecule without the double bond) is inert under these conditions. Why does it not undergo substitution? The oxygen could surely be protonated and then methanol would be a good leaving group which would give rise to substitution (probably S&lt;sub&gt;N&lt;/sub&gt;1). Why is this not the case?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2018-05-01T22:58:34.063" LastActivityDate="2018-05-01T22:58:34.063" Title="What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;c-c-addition&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24316" PostTypeId="1" AcceptedAnswerId="24320" CreationDate="2015-01-28T21:11:28.660" Score="4" ViewCount="874" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lycopene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lycopene&lt;/a&gt;, which is a carotenoid and a phytochemical is the main reason for why tomatoes are radio-protective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t5sJz.jpg&quot; alt=&quot;&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CIhfI.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/5/5a/Lycopene_powder.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Wikimedia&quot;&gt;Source 1&lt;/a&gt;, &lt;a href=&quot;http://0.tqn.com/d/chemistry/1/S/K/1/lyco2.gif&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Chemistry.about&quot;&gt;Source 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's cut to the chase: I was wondering if this claim of &lt;a href=&quot;http://www.nutraingredients-usa.com/Research/Cooking-tomatoes-boosts-health-benefits?utm_source=copyright&amp;amp;utm_medium=OnSite&amp;amp;utm_campaign=copyright&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt; is of scientific value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Heating will result in an increase in amount of lycopene absorbed by the body.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible that the researcher made a more &quot;every-day&quot; statement for the readers to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the statement scientifically correct?&lt;/li&gt;&#xA;&lt;li&gt;If so, why and how would such a chemical be absorbed more with the heating process?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that if you want to make the argument that this is a biology question rather than a chemistry one, I would bring up these reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The temperature in which the heated lycopene and raw tomato are going to be absorbed in the body are the same. Thus, it might be possible that heating lycopene results in a chemical reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not looking for the pattern of absorption by the body. I'm looking for possible changes or specifications of its structure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="194" LastEditDate="2017-04-01T03:34:11.447" LastActivityDate="2017-04-01T03:34:11.447" Title="Why and how does lycopene absorbance increase with heating?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;heat&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24317" PostTypeId="2" ParentId="24315" CreationDate="2015-01-28T21:16:01.293" Score="4" Body="&lt;p&gt;You're close.  You are right in thinking that the first step involves protonation of the double bond.  But which end of the double bond will you protonate?  If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation.  However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it.  Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have a carbocation bearing a methoxy group.  The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal.  Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KSZy2.jpg&quot; alt=&quot;I&amp;#39;ll add a figure in a minute.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-28T21:27:26.010" LastActivityDate="2015-01-28T21:27:26.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24318" PostTypeId="1" CreationDate="2015-01-28T21:16:16.910" Score="4" ViewCount="343" Body="&lt;p&gt;I can't seem to figure out this redox reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH3COO- + CO2 -&amp;gt; CH3CH2COO-}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the charge on the carbon in the $\ce{CO2}$ supposed to match the charge on the carbon in the $\ce{CH3COO-}$? I am supposed to build a half reaction using $\ce{H2O}$ and $\ce{H+}$ as needed, but I can't seem to figure out how many electrons are actually transferred. What are the charges on the carbon?&lt;/p&gt;&#xA;" OwnerUserId="13060" LastEditorUserId="194" LastEditDate="2017-04-08T01:14:05.130" LastActivityDate="2017-04-08T01:14:05.130" Title="Building an incomplete redox half reaction/reduction" Tags="&lt;redox&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24319" PostTypeId="2" ParentId="24315" CreationDate="2015-01-28T21:21:59.857" Score="7" Body="&lt;p&gt;Vinyl ethers are &quot;masked&quot; carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPMWb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="46306" LastEditDate="2018-05-01T12:07:51.760" LastActivityDate="2018-05-01T12:07:51.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24320" PostTypeId="2" ParentId="24316" CreationDate="2015-01-28T21:51:59.817" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Heating will result in an increase in amount of lycopene absorbed by&#xA;  the body.&quot;&#xA;  Is the statement scientifically correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is a poorly phrased statement, but it is probably generally true.  It would have been better to say something like, &quot;heating tomatoes will increase the amount of all-trans-lycopene &lt;strong&gt;available&lt;/strong&gt; for ingestion.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.researchgate.net/publication/248511927_Effect_of_heating_and_exposure_to_light_on_the_stability_of_lycopene_in_tomato_pure&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; to a full paper on the subject (click the &quot;View&quot; button near the top-right of the page).  Basically two things are going on at the same time when lycopene containg food is heated.  It is the naturally occurring, &lt;strong&gt;all-trans&lt;/strong&gt; isomer of lycopene that is thought to be beneficial to the human body. Heating lycopene causes it to isomerize to a variety of cis-isomers.  Heating in the presence of oxygen can also cause oxidation.  Both of these processes (isomerization and oxidation) reduce the potency of the lycopene in the broth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in parallel with these deleterious effects, heating also makes more lycopene available in the heated broth or extract.  This is because heating disrupts cell walls and releases lycopene from the cells.  The above link shows that, at least at moderate temperatures, the latter effect predominates and so more active all-trans-lycopene is made available upon heating tomatoes, despite the loss of some lycopene to isomerization and oxidation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-28T21:51:59.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24321" PostTypeId="2" ParentId="24314" CreationDate="2015-01-28T22:13:02.977" Score="0" Body="&lt;p&gt;Kohlrausch law states that $\Lambda_m=\Lambda_m^{\circ}-\kappa \sqrt{c}$ with $\Lambda_m = \frac{k}{c}$ (in which $k$ is the electrical conductivity) for strong electrolytes and low concentrations, which is due to the strong interaction of ions. In that light your findings are not inconsistent. I am not sure though about the magnitude of $\kappa$ and whether for the conditions (copper sulfate as electrolyte, 0.1M to 1.0M concentration) you listed the law holds. At least the point I want to make is that it is not obvious or even correct to assume that conductivity depends linearly on the concentration of the electrolyte or that there is even a positive relation at high concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is definitely not intended as a definite answer (nice tautology).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-01-28T22:22:54.310" LastActivityDate="2015-01-28T22:22:54.310" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24323" PostTypeId="1" CreationDate="2015-01-29T00:18:23.000" Score="1" ViewCount="67" Body="&lt;p&gt;I know there are volumes of International Colour Index (&lt;a href=&quot;http://www.colour-index.com/&quot; rel=&quot;nofollow&quot;&gt;homepage&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Colour_Index_International&quot; rel=&quot;nofollow&quot;&gt;wiki&lt;/a&gt;, &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02692335&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;), but I can't really extract from there for example what red colorants are compatible with both silk and polyester, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.provost-inkjet.com/resources/Interactions+of+Ink+and+Media+in+Ink+Jet+Printing.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; mentions that such compatibility can be estimated using numerical methods:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm&quot; rel=&quot;nofollow&quot;&gt;Metropolis Monte Carlo&lt;/a&gt; methods have been used to determine the surface&#xA;  interaction or binding energy for dyes to cellulose.&quot;&lt;a href=&quot;http://www2.chemistry.msu.edu/courses/cem888/cukier/Allen,Tildesley,_Computer_Simulation_of_Liquids,1991.pdf&quot; rel=&quot;nofollow&quot;&gt; [ref]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a comprehensive database for this?&lt;/li&gt;&#xA;&lt;li&gt;Is there an analytical way to determine/predict/estimate such compatibility?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-01-29T00:58:43.673" LastActivityDate="2015-01-29T00:58:43.673" Title="How to choose a colorant for a given substrate?" Tags="&lt;color&gt;&lt;surface-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24324" PostTypeId="1" CreationDate="2015-01-29T00:44:18.123" Score="4" ViewCount="211" Body="&lt;p&gt;I know that dysprosium oxide is insoluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that I need to add concentrated $\ce{HCl}$ to the compound and then evaporate the $\ce{HCl}$ off. From looking at the reaction equation, I can see that I will get dysprosium chloride and water as products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I'm able to dissolve the $\ce{Dy2O3}$, how is it that I'm supposed to evaporate off the excess $\ce{HCl}$? Do I just let it sit, or do I need to gently heat the solution? How can I tell when all the excess $\ce{HCl}$ is gone and I can start adding water to obtain the concentration of dysprosium chloride that I need?&lt;/p&gt;&#xA;" OwnerUserId="13062" LastEditorUserId="4945" LastEditDate="2015-01-29T02:18:21.510" LastActivityDate="2015-04-07T16:49:58.433" Title="How do I make a dysprosium chloride solution from dysprosium oxide?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;&lt;rare-earth-elements&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24325" PostTypeId="1" AcceptedAnswerId="24414" CreationDate="2015-01-29T01:09:03.467" Score="1" ViewCount="111" Body="&lt;p&gt;Pigments are insoluble colorants that have no affinity to the substrate. So how can they function as colorants if they &quot;can't stick&quot;? Are pigments always applied together with a &quot;binder&quot;?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-01-29T01:48:13.270" LastActivityDate="2015-01-31T14:45:37.653" Title="How pigments actually function as colorants?" Tags="&lt;color&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24326" PostTypeId="1" AcceptedAnswerId="24376" CreationDate="2015-01-29T01:54:54.933" Score="3" ViewCount="2158" Body="&lt;p&gt;I want to estimate the binding energies of various molecules. What are the analytical and numerical tools I can use for this purpose? In particular I'm interested in bonding energies of dyes to substrates (&lt;a href=&quot;http://dx.doi.org/10.1016/S0143-7208(98)00052-7&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-01-30T18:44:39.160" LastActivityDate="2015-01-30T18:44:39.160" Title="What tools can be used to estimate binding energies?" Tags="&lt;computational-chemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24327" PostTypeId="1" CreationDate="2015-01-29T03:28:53.610" Score="4" ViewCount="752" Body="&lt;p&gt;So, I read in a similar chemistry thread that you can get Mn and Mw of the polymer of the interest by comparing the result of the sample obtained from GPC/SEC with the standard known molecule weight polymer like polystyrene.&#xA;I don't understand how this works. You compare what of the polymer with what of the polystyrene to determine the molecular weight?? There has to be a constant factor when determining relative values between two things, and what is it in the case of polystyrene and the polymer of the interest?&lt;br&gt;&#xA;What do you mean by polystyrene equivalent value? What is the point of determining the relative molecular weight of a polymer based on the polystyrene equivalent value?? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="13063" LastActivityDate="2015-01-29T21:32:17.340" Title="What is the relationship between the molecular weight of a polymer and that of polystyrene?" Tags="&lt;polymers&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24330" PostTypeId="1" CreationDate="2015-01-29T03:46:28.447" Score="3" ViewCount="1241" Body="&lt;p&gt;I don't mean &lt;a href=&quot;https://en.wikipedia.org/wiki/Photographic_film&quot; rel=&quot;nofollow&quot;&gt;photographic film&lt;/a&gt;/&lt;a href=&quot;https://en.wikipedia.org/wiki/Film_base&quot; rel=&quot;nofollow&quot;&gt;film base&lt;/a&gt; but rather the A4/Letter sheets of transparent film that were used in slide projectors during lectures. For example see &lt;a href=&quot;http://www.ebay.com/itm/Inkjet-Laser-Printing-A4-Transparency-Film-20-Sheets-for-Screen-Plate-Making-/181396274376&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-01-29T14:32:20.640" LastActivityDate="2015-01-29T14:32:20.640" Title="What material is an Inkjet transparent printing film (slide) made of?" Tags="&lt;materials&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24331" PostTypeId="1" AcceptedAnswerId="24333" CreationDate="2015-01-29T04:26:30.420" Score="5" ViewCount="228" Body="&lt;p&gt;For example, silk is made of &lt;a href=&quot;https://en.wikipedia.org/wiki/Fibroin&quot;&gt;fibroin&lt;/a&gt;. But what's wool made of?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-01-29T14:08:58.727" Title="What's wool made of?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24332" PostTypeId="2" ParentId="24314" CreationDate="2015-01-29T05:17:17.863" Score="5" Body="&lt;p&gt;I'm thinking maybe at the higher concentrations your tap water is making a film of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu(OH)2}$&lt;/span&gt; on the surface of your electrode, and that is what's shutting down your electroplating. Your tap water might be mildly alkaline (tap water pH's can range from about &lt;span class=&quot;math-container&quot;&gt;$6.5$&lt;/span&gt; to as high as &lt;span class=&quot;math-container&quot;&gt;$9.5$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try adding a little sulfuric acid to your solutions. If alkaline films are the problem, that will fix it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also increase the conductivity of your solutions tremendously, because hydrogen ions are extremely mobile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of copper sulfate decreases with increasing sulfuric acid concentration, so don't add too much acid.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="69977" LastEditDate="2019-07-18T05:53:23.050" LastActivityDate="2019-07-18T05:53:23.050" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24333" PostTypeId="2" ParentId="24331" CreationDate="2015-01-29T05:20:05.837" Score="4" Body="&lt;p&gt;Unlike silk or cotton, wool does not have a homogeneous structure. Its core is considered to be made of &lt;a href=&quot;https://en.wikipedia.org/wiki/Keratin&quot; rel=&quot;nofollow&quot;&gt;keratin&lt;/a&gt;, a generic type of protein found in human hair and epidermis. A detailed PDF on Merino wool can be downloaded from &lt;a href=&quot;http://www.csiro.au/~/media/CSIROau/Divisions/CSIRO%20Materials%20Science%20and%20Engineering/WoolStructure_CMSE_PDF%20Standard.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for encouraging me, through your question, to read that PDF, which explains both the chemical and physical properties in a straight-forward fashion.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="7448" LastEditDate="2015-01-29T14:08:58.727" LastActivityDate="2015-01-29T14:08:58.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24334" PostTypeId="2" ParentId="6298" CreationDate="2015-01-29T10:26:41.210" Score="1" Body="&lt;p&gt;one more category of chemical reagents Horba misses is the oxidizers . sodium hypochlorite , or ozone , can kill many colour compounds .&lt;/p&gt;&#xA;" OwnerUserId="12960" LastActivityDate="2015-01-29T10:26:41.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24335" PostTypeId="1" CreationDate="2015-01-29T13:37:45.437" Score="-2" ViewCount="84" Body="&lt;p&gt;Please tell about the hydrogen gas storage and the pressure created in tank. what is the calorific value of hydrogen gas when it is burnt?&lt;/p&gt;&#xA;" OwnerUserId="13026" LastEditorUserId="5017" LastEditDate="2015-02-06T03:05:34.857" LastActivityDate="2015-02-06T03:05:34.857" Title="Hydrogen can be stored?" Tags="&lt;safety&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-18T17:15:10.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="24336" PostTypeId="1" AcceptedAnswerId="24354" CreationDate="2015-01-29T14:49:36.290" Score="4" ViewCount="695" Body="&lt;p&gt;Are there any distinct biochemical reactions that are affected by the ingestion of distilled water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is &quot;distilled&quot; in its literal sense.&lt;/p&gt;&#xA;" OwnerUserId="13015" LastEditorUserId="194" LastEditDate="2017-04-08T01:18:00.383" LastActivityDate="2017-04-08T01:18:00.383" Title="What are the biochemical ramifications of ingesting of distilled water?" Tags="&lt;water&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24339" PostTypeId="2" ParentId="24335" CreationDate="2015-01-29T15:56:29.043" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the calorific value of hydrogen gas when it is burnt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;286 kilojoules of energy is released per mole of hydrogen burned.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please tell about the hydrogen gas storage and the pressure created in tank. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a tank at room temperature, hydrogen is a supercritical fluid, because the critical point is 33K.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a tank, a long as the tank is strong enough, the pressure could be any value up to at least 10,000 psi, but at high enough pressure it would solidify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see &lt;a href=&quot;http://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section3.pdf&quot; rel=&quot;nofollow&quot;&gt;PROPERTIES OF HYDROGEN&lt;/a&gt; (no paywall), especially page III-E for P-V-T information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-29T15:56:29.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24342" PostTypeId="1" AcceptedAnswerId="24347" CreationDate="2015-01-29T16:44:42.073" Score="15" ViewCount="14015" Body="&lt;p&gt;What is the $K_\mathrm{a}$ of $\ce{OH^-}$ and $K_\mathrm{b}$ of $\ce{H_3O^+}$? Have these constants been determined?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2015-11-06T21:04:11.720" LastActivityDate="2015-11-06T21:07:44.747" Title="What is the Ka of OH- and Kb of H3O+?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24344" PostTypeId="2" ParentId="24327" CreationDate="2015-01-29T17:05:27.313" Score="1" Body="&lt;p&gt;You have to have something to calibrate the column with. Polystyrene is typically used since it can be made cheaply (well, relatively cheaply) with a pretty small molecular weight distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there will be differences between the true molecular weight of a polymer and what the comparison to polystyrene gives you. But does that really matter? Why do you need to know the absolute molecular weight, when a relative comparison works just fine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's similar to how we determine the mass of something. At present, there is that infamous block of Pt/Ir in Paris that is the standard kilogram. All mass measurements are made through a series of comparison to that mass.&lt;/p&gt;&#xA;" OwnerUserId="13067" LastActivityDate="2015-01-29T17:05:27.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24346" PostTypeId="2" ParentId="24342" CreationDate="2015-01-29T18:48:28.887" Score="9" Body="&lt;p&gt;In water both equilibrium constants will be extremely small; the concentrations of $\ce{O^2-}$ or $\ce{H4O^2+}$ ions in water are essentially zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Cotton and Wilkinson's &lt;em&gt;Advanced Inorganic Chemistry&lt;/em&gt;, for the hydrolysis of &lt;em&gt;solid&lt;/em&gt; oxide ions you have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O^2- (s) + H2O (l) -&amp;gt; 2 OH- (aq)} \qquad \qquad K_b &amp;gt; 10^{22}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you can never have any significant concentration of oxide ions in aqueous solution. As DavePhD points out, $K_\mathrm{a}$ for the hydroxide ion is less than $K_\mathrm{w}/K_\mathrm{b} = 10^{-36}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\ce{H3O+}$ may have some capacity to act as a base in media like $\ce{HF:SbF5}$ (see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00265a031&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt; on the role of $\ce{H4O^2+}$ in isotopic exchange reactions between hydronium ions). $\ce{H4O^2+}$ can exist in sulfolane solution, too (see &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009261487804906&quot; rel=&quot;noreferrer&quot;&gt;this paper&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{H4O^2+}$ exists in water &lt;em&gt;at all&lt;/em&gt;, it's probably going to actually be two protons bridged and hydrated by a lot of waters, rather than an actual $\ce{H4O^2+}$ ion. It will have a &lt;em&gt;much&lt;/em&gt; shorter lifetime than even a hydronium ion. The molecule can exist in theory, though, and its electronic structure has been studied for at least half a century (see &lt;a href=&quot;http://actachemscand.dk/pdf/acta_vol_18_p1719-1738.pdf&quot; rel=&quot;noreferrer&quot;&gt;Rosenfeld's 1964 paper&lt;/a&gt; including SCF calculations on $\ce{H4O^2+}$). &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="16683" LastEditDate="2015-11-06T21:07:44.747" LastActivityDate="2015-11-06T21:07:44.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24347" PostTypeId="2" ParentId="24342" CreationDate="2015-01-29T18:51:46.703" Score="8" Body="&lt;p&gt;For $\ce{O^2-}$, the $\mathrm{p}K_\mathrm{b}$ is approximately $-22$, so the $\mathrm{p}K_\mathrm{a}$ of hydroxide is about $36$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sources: &lt;a href=&quot;http://www.newworldencyclopedia.org/entry/Oxide&quot; rel=&quot;nofollow&quot;&gt;http://www.newworldencyclopedia.org/entry/Oxide&lt;/a&gt; and &lt;a href=&quot;https://www.chem.tamu.edu/rgroup/hughbanks/courses/462/lecturenotes/class6-2.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.chem.tamu.edu/rgroup/hughbanks/courses/462/lecturenotes/class6-2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hydronium, the reference Fred Senese cited &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja00265a031&quot; rel=&quot;nofollow&quot;&gt;Protonated hydronium dication...&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;suggests that at acidities of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammett_acidity_function&quot; rel=&quot;nofollow&quot;&gt;Hammet acidity function&lt;/a&gt; $H_o = -25$ to $-28$, protonated hydronium can be present.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2015-11-06T21:05:03.213" LastActivityDate="2015-11-06T21:05:03.213" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24348" PostTypeId="1" CreationDate="2015-01-29T19:14:35.340" Score="2" ViewCount="5248" Body="&lt;p&gt;(All volumes were measured using a measuring cylinder which has a &lt;span class=&quot;math-container&quot;&gt;$\pm\pu{0.5 cm3}$&lt;/span&gt; uncertainty)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a &lt;span class=&quot;math-container&quot;&gt;$\pu{50 cm3}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 M}$&lt;/span&gt; solution to begin with&lt;/li&gt;&#xA;&lt;li&gt;You then make a  &lt;span class=&quot;math-container&quot;&gt;$\pu{0.4 M}$&lt;/span&gt;  solution by mixing &lt;span class=&quot;math-container&quot;&gt;$\text{40 cm}^3$&lt;/span&gt; of the above solution with &lt;span class=&quot;math-container&quot;&gt;$\pu{10 cm3}$&lt;/span&gt; water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If I'm not mistaken, the uncertainty for the volume should be &lt;span class=&quot;math-container&quot;&gt;$\pm\pu{1.0 cm3}$&lt;/span&gt; because the second solution is the result of two measurements (and also uncertainties).&lt;/p&gt;&#xA;&lt;p&gt;My question is: what is the uncertainty for the concentration?&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="69830" LastEditDate="2021-05-11T04:46:11.527" LastActivityDate="2021-05-11T04:46:11.527" Title="Concentration uncertainty due to volume measurement with cylinder" Tags="&lt;experimental-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24349" PostTypeId="1" AcceptedAnswerId="24355" CreationDate="2015-01-29T19:45:55.583" Score="4" ViewCount="1874" Body="&lt;p&gt;If someone asks me to number the most notable &quot;textbook experiments&quot; I would definitely name &quot;the comparison of the reactions of alkali metals with water&quot; among the most favorites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We just happened to do such a reaction with small pieces (approximately $\text{7.5 g}$) of Li, Na and K. As we were supposed to write down anything that we observe, I came across an interesting phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the answer is so simple, I didn't get it through many resources. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Li piece is moving rapidly (better call it spinning) on the surface of water as it is reacting with it and a subtle explosion sound is heard. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does the motion of Li on the surface of the water in the beaker take place? Could the heat released by the reaction be causing current in the water (I don't believe so, as the possibly hot water isn't involved in any convection currents)? Info you might need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our chemistry teacher cut a non-oxidized piece of well-preserved lithium. The piece of the reactant wasn't of any distinguishable geometrical shape. After repeating the experiment for several times, I sensed a headache that as I believe is because of release of the impurities of the sample.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2017-08-15T00:03:31.190" Title="Why does lithium move on the surface of water during the reaction?" Tags="&lt;experimental-chemistry&gt;&lt;metal&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24350" PostTypeId="1" CreationDate="2015-01-29T19:49:24.780" Score="2" ViewCount="4274" Body="&lt;p&gt;What's the oxidation state of nitrogen in &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt; ion. Simple calculation depicts that it's &lt;span class=&quot;math-container&quot;&gt;$+5$&lt;/span&gt;. But second period elements never show variable oxidation states. And it's structure is so confusing to me. Some people show dative/co-ordinate bonds, whereas others show a positive charge on nitrogen. Which should be considered as correct one? &lt;/p&gt;&#xA;" OwnerUserId="14751" LastEditorUserId="4945" LastEditDate="2018-09-27T14:04:06.730" LastActivityDate="2018-09-27T14:04:06.730" Title="What is the oxidation state of nitrogen in nitrate ion and what is its structure?" Tags="&lt;bond&gt;&lt;molecular-structure&gt;&lt;oxidation-state&gt;&lt;covalent-compounds&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-01-30T12:53:20.250" ContentLicense="CC BY-SA 4.0" />
  <row Id="24352" PostTypeId="2" ParentId="24350" CreationDate="2015-01-29T20:04:40.917" Score="2" Body="&lt;p&gt;You are correct that the oxidation state is +5. The second period elements, particularly carbon, nitrogen and oxygen, do show variable oxidation states regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{NO3-}$ ion cannot be depicted using a single Lewis structure. However, its resonance structures are shown below:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9Jo6y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that the actual structure of the nitrate ion is somewhere in between the three structures shown below. It is sometimes also depicted like this to show the symmetrical nature of the ion and the distribution of charges across it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/noUMt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&lt;/a&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrate#Structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Nitrate#Structure&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-01-29T20:04:40.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24353" PostTypeId="2" ParentId="24348" CreationDate="2015-01-29T21:13:48.730" Score="3" Body="&lt;p&gt;&lt;strong&gt;1. Let's assume that you're talking about random error.&lt;/strong&gt; If you specify the error in the volume with $\pm$, this is probably the case---systematic error in this case would have a single sign; for example, if the scale marks on the side of cylinder were shifted by $\rm 0.5\ cm^3$ down from what their true values should have been, you'd have a systematic error of $-0.5\ \rm cm^3$ (the negative sign tells you that your readings are $0.5\ \rm cm^3$ lower than they should have been).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can say that the error in a sum will be &lt;em&gt;less than or equal to&lt;/em&gt; the sum of errors. It's better to say that the uncertainty in a sum is the square root of the sum of squared uncertainties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(V_1+V_2) = \sqrt{\delta(V_1)^2 + \delta(V_2)^2} = \sqrt{\rm(0.5\ cm^3)^2 + (0.5\ cm^3)^2} = 0.70_7\ \rm cm^3$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a product or quotient, you can say that the relative error will be less than or equal to the sum of the relative errors in the measurements. It's better to say that the relative error in a product or quotient is the square root of the sum of squared relative errors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xA;\delta(C_{dilute}) &amp;amp;=&amp;amp; \delta\left(\frac{C_{stock}V_1}{V_1+V_2}\right)\\&#xA;\frac{\delta(C_{dilute})}{C_{dilute}} &amp;amp;=&amp;amp; \sqrt{\left(\frac{\delta(C_{stock})}{C_{stock}}\right)^2 + \left(\frac{\delta(V_{1})}{V_{1}}\right)^2 + \left(\frac{\delta(V_{1}+V_2)}{V_{1}+V_2}\right)^2&#xA;}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...can you take it from there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Let's assume you have systematic error&lt;/strong&gt;, that is, you actually meant something like &quot;The cylinder always reads $\rm 0.5\ cm^3$ too high&quot;. You will need a definite sign on a systematic error, not a $\pm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways to proceed here. Let's take the easiest route, which is to estimate the systematic error in the concentration as the difference between the concentration and its perturbed value (the value with all of the volumes having the systematic error added in). If your systematic error in volume was $\delta V = +0.5\ \rm cm^3$, you can estimate the systematic error in the concentration as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xA; \delta{C_{dilute}} &amp;amp;\approx&amp;amp; C_{dilute}^{perturbed} - C_{dilute} \\&#xA;&amp;amp;\approx&amp;amp; \frac{C_{stock}(V_1+\delta(V)}{(V_1+\delta(V))(V_2+\delta(V))} - C_{dilute} \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign on this error will be important; it tells you whether the systematic error in volume makes the concentration higher or lower than its &quot;true&quot; value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The above calculation is just an estimate of the error. You can do a better job using derivatives here. I'll show you that if you want to see it.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-30T15:35:36.617" LastActivityDate="2015-01-30T15:35:36.617" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24354" PostTypeId="2" ParentId="24336" CreationDate="2015-01-29T21:16:57.017" Score="5" Body="&lt;p&gt;As @DavePhD has noted, drinking small amounts of distilled water does not pose any significant adverse health consequences.  However, the &lt;strong&gt;World Health Organization&lt;/strong&gt; (WHO) has studied the effects of long-term drinking of distilled water and issued a &lt;a href=&quot;http://www.who.int/water_sanitation_health/dwq/nutrientschap12.pdf&quot; rel=&quot;noreferrer&quot;&gt;report&lt;/a&gt;.  They investigated the following five areas of potential adverse human health impact:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Direct effects on the intestinal mucous membrane, metabolism and mineral homeostasis or other body functions.&lt;/li&gt;&#xA;&lt;li&gt;Little or no intake of calcium and magnesium from low-mineral water.&lt;/li&gt;&#xA;&lt;li&gt;Low intake of other essential elements and microelements.&lt;/li&gt;&#xA;&lt;li&gt;Loss of calcium, magnesium and other essential elements in prepared food.&lt;/li&gt;&#xA;&lt;li&gt;Possible increased dietary intake of toxic metals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In their conclusion, they state,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Drinking water &lt;strong&gt;should contain&lt;/strong&gt; minimum levels of certain essential&#xA;  minerals (and other components such as carbonates). Unfortunately,&#xA;  over the two past decades, little research attention has been given to&#xA;  the beneficial or protective effects of drinking water substances.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Demineralised water that has not been remineralized, or low-mineral&#xA;  content water – in the light of the absence or substantial lack of&#xA;  essential minerals in it – &lt;strong&gt;is not considered ideal drinking water.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis mine)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-29T21:45:30.100" LastActivityDate="2015-01-29T21:45:30.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24355" PostTypeId="2" ParentId="24349" CreationDate="2015-01-29T21:18:12.133" Score="4" Body="&lt;p&gt;What happens when you put lithium (or sodium or potassium) in water? What is the chemical reaction that is occurring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alkali metals react with water to produce alkali hydroxides and hydrogen gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Li(s) + 2H2O(l) -&amp;gt; 2LiOH(aq) + H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two products of the reaction of water and lithium are lithium hydroxide (which dissolves) and hydrogen. Hydrogen is not very soluble in water, and it is much less dense than air. The motion of the lithium on the surface of the water is created by the escape of the hydrogen gas, which provides upward force. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-29T21:18:12.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24356" PostTypeId="1" CreationDate="2015-01-29T21:21:44.030" Score="0" ViewCount="118" Body="&lt;p&gt;As depicted in books, $\ce{NO3-}$ and $\ce{NO2-}$ are called as nitrate and nitrite respectively. Where (-ate) suffix is used for higher oxidation state and (-ite) suffix is used for lower oxidation state (just like:- sulfate and sulfite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since Nitrogen is 2nd period element, hence it can't expand it's octet (no vacant d-orbital). So I didn't get how (-ite or -ate) suffix can be used if nitrogen shows only 1 stable oxidation state i.e; (+3). Can anyone suggest?&lt;/p&gt;&#xA;" OwnerUserId="14751" LastEditorUserId="7448" LastEditDate="2015-01-29T21:44:11.100" LastActivityDate="2015-01-30T02:30:07.670" Title="Nomenclature of inorganic compounds" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24357" PostTypeId="2" ParentId="24327" CreationDate="2015-01-29T21:32:17.340" Score="1" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Size-exclusion_chromatography&quot; rel=&quot;nofollow&quot;&gt;size exclusion chromatography (SEC)&lt;/a&gt;, the molecules elute in order of decreasing molecular weight (actually solvodynamic radius - we'll get back to that). The larger molecules (or polymer chains) do not fit into the pores in the stationary phase and thus spend more time in the mobile phase and elute faster. The smaller molecules can get stuck in the pores and thus elute slower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one of the most common detectors is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_refractometer&quot; rel=&quot;nofollow&quot;&gt;differential refractometer&lt;/a&gt;, which measures the concentration of analyte in the eluent since the refractive index $\eta$ of a dilute solution depends only on the molal concentration of the solution ($m$), the refractive index of the pure solvent $\eta_0$, and a constant $k_\eta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta=k_\eta m +\eta_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relationship helps us figure out the amount of polymer that is eluting at any given time. However, the system needs to be calibrated so that we can related elution time to specific molecular weights. That's where polystyrene standards come in. Polystyrene is chosen because it can be prepared in low-polydispersity samples of known molecular weight by carefully controlling the polymerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use &quot;polystyrene-equivalent&quot; value or &quot;relative to polystyrene standards&quot; because we know the molecular weights of the polystyrene standards.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-29T21:32:17.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24358" PostTypeId="1" CreationDate="2015-01-29T20:37:26.400" Score="2" ViewCount="593" Body="&lt;p&gt;Why don't diesel and regular car engines burn all the way to $\ce{CO2}$?&lt;/p&gt;&#xA;" OwnerUserId="9473" OwnerDisplayName="Dirk Bruere" LastEditorUserId="7448" LastEditDate="2015-01-30T12:25:34.833" LastActivityDate="2015-01-30T12:25:34.833" Title="Why do internal combustion engines only burn to carbon monoxide, not dioxide?" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24360" PostTypeId="1" CreationDate="2015-01-29T23:42:13.120" Score="9" ViewCount="3402" Body="&lt;p&gt;I'm so confused... $\ce{HClO}$ is a weak acid. then why is its conjugate base, $\ce{ClO-}$, a weak base in water? shouldn't they be inversely proportional? shouldn't $\ce{ClO-}$,  be a strong base? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{HClO}$ is a weak acid, that means it does not readily give up a proton and has a strong pull on them. So when it becomes a conjugate base, $\ce{ClO-}$, shouldn't it &lt;strong&gt;readily&lt;/strong&gt; take protons and therefore be a strong base in water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALSO,&#xA;when an acid is dissolved in water, and some of that acid dissociates into $\ce{H3O+}$, is the Kw of water negligible and you only consider the Ka when finding pH? But that confuses me because I thought you use 14, the exponent of Kw when finding pH. &lt;/p&gt;&#xA;" OwnerUserId="13077" LastEditorUserId="5456" LastEditDate="2015-01-30T06:49:30.270" LastActivityDate="2019-05-09T17:17:21.980" Title="Strong and Weak Acids/Bases" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24361" PostTypeId="2" ParentId="24360" CreationDate="2015-01-30T01:44:00.223" Score="6" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm a\times K_\mathrm b = K_\mathrm w$&lt;/span&gt;.  Some teachers tell their students that the conjugate base of a weak acid is strong but it's not true.  What is true is that the stronger the acid, the weaker the conjugate base and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes to your second question.  When calculating the pH of a solution, the hydronium ion concentration is usually controlled by the strongest acid in the solution.  This is the case in the type of problems found in lower level chem classes.  In a higher level class problems with acids close enough in strength so that both acids matter may be encountered.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5026" LastEditDate="2019-05-09T17:16:10.183" LastActivityDate="2019-05-09T17:16:10.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24362" PostTypeId="2" ParentId="24356" CreationDate="2015-01-30T02:30:07.670" Score="2" Body="&lt;p&gt;Nitrogen can have a variety of &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Amines/Properties_of_Amines/Oxidation_States_of_Nitrogen&quot; rel=&quot;nofollow&quot;&gt;oxidation states&lt;/a&gt; ranging from -3 to +5. For example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nitrogen atom in ammonia $\ce{NH3}$ has an oxidation state of -3 (each hydrogen is +1).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-2  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nitrogen atoms in hydrazine $\ce{N2H4}$ has an oxidation state of -2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The oxidation state of the nitrogen atoms in $\ce{N2}$ is 0.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nitrate and nitrite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The nitrogen atom in the nitrate ion $\ce{NO3-}$ has an oxidation state of +5. Each oxygen atom has an oxidation state of -2, and the overall charge on the ion is -1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrogen atom in nitrite ion $\ce{NO2-}$ has an oxidation state of +3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen can have such a range of oxidation states because of its electron configuration, which is $\ce{[He]}2s^2 2p^3$ (5 valence electrons). Nitrogen can &quot;gain&quot; up to three electrons or &quot;lose&quot; up to five electrons.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-01-30T02:30:07.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24363" PostTypeId="1" AcceptedAnswerId="24365" CreationDate="2015-01-30T03:45:53.970" Score="3" ViewCount="1381" Body="&lt;p&gt;The protons alpha to the oxygen in methyl propionate are more shielded than their corresponding protons in ethyl propionate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOJqt.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5DY3t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xynfu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see from the NMR data that this is the case as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My TA says this is due to the inductively withdrawing effect of the extra carbon in &lt;em&gt;ethyl&lt;/em&gt; propionate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this strikes me as blatantly wrong. The carbon should be inductively donating ... I don't see how it pulls electron density away from the specified hydrogens. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-01-30T16:17:08.120" Title="proton NMR deshielding" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24364" PostTypeId="1" AcceptedAnswerId="24371" CreationDate="2015-01-30T04:02:32.257" Score="-1" ViewCount="86" Body="&lt;p&gt;I am working on a biochemistry project, and I need to know some information about reaction rates. What is the Rate of Reaction of Acetaldehyde Dehydrogenase from Saccharomyces Cerevisiae?&lt;/p&gt;&#xA;" OwnerUserId="13078" LastActivityDate="2015-01-30T12:41:16.077" Title="What is the Rate of Reaction of Acetaldehyde Dehydrogenase from Saccharomyces Cerevisiae?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24365" PostTypeId="2" ParentId="24363" CreationDate="2015-01-30T04:09:34.103" Score="3" Body="&lt;p&gt;In your example we are comparing the electron releasing/withdrawing properties of a hydrogen substituent and a methyl substituent, and the effect this has on proton ($\ce{H_{a}}$) chemical shifts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwaQR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon (2.55) is more electronegative than hydrogen (2.2).  Therefore, the methyl substituent will be more electron withdrawing compared to the hydrogen substituent.  This means that the methyl group will remove electron density from the carbon to which it is attached, compared to the hydrogen substituent.  The fact that this carbon has lower electron density in the methyl case, is further transmitted to the protons ($\ce{H_{a}}$) attached to this carbon. Consequently the $\ce{H_{a}}$ protons will also experience lower electron density in the R=methyl case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower electron density at an nmr-active nucleus results in deshielding (downfield shift). Consequently, the $\ce{H_{a}}$ protons are deshielded in the ethyl ester (R=methyl) relative to the methyl ester (R=hydrogen), in agreement with the spectral data you've provided.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-01-30T16:17:08.120" LastActivityDate="2015-01-30T16:17:08.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24367" PostTypeId="1" AcceptedAnswerId="24369" CreationDate="2015-01-30T10:21:51.883" Score="0" ViewCount="59" Body="&lt;p&gt;If the temperature on the surface of the sun and on the surface on the earth were the same, and the temperature of the universe were uniform, would the solar panels produce energy?&lt;/p&gt;&#xA;" OwnerUserId="13082" LastEditorUserId="7448" LastEditDate="2015-01-30T12:18:57.023" LastActivityDate="2015-01-30T12:18:57.023" Title="Would solar panels produce energy if uniform temperature in the universe?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-01-30T23:55:24.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="24368" PostTypeId="2" ParentId="24358" CreationDate="2015-01-30T11:14:46.357" Score="4" Body="&lt;p&gt;I think this is sort of covered by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15995/what-really-happens-atomically-in-an-explosion&quot;&gt;What really happens atomically in an explosion?&lt;/a&gt; because the burning of the fuel air mixture is basically an explosion. The point is that there isn't a nice simple answer because the reaction is fearsomely complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually most of the fuel does burn to carbon dioxide. The proportion of carbon monoxide is quite small. In a normally tuned engine there is enough oxygen to burn all the fuel to carbon dioxide but the fuel/air mixture isn't in the cylinder long enough for the reaction to go to completion. You probably know that all modern cars are fitted with &lt;a href=&quot;http://en.wikipedia.org/wiki/Catalytic_converter&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalytic convertors&lt;/a&gt;, and what they do is to complete the reaction as the exhaust gases pass through them.&lt;/p&gt;&#xA;" OwnerUserId="1470" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-01-30T11:14:46.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24369" PostTypeId="2" ParentId="24367" CreationDate="2015-01-30T12:14:04.123" Score="2" Body="&lt;p&gt;Solar panels work on the principle of photovoltaic effect. The only criteria for a voltage difference to be affected is that the light be incident on the solar panel. The energy of the incident light needs to be around that of the band-gap.&#xA;According to the question if both sun and the solar panel were of the same temperature, a solar panel on which light rays are incident would indeed generate a voltage. The band gap on the other hand depends on the temperature of the panel. Higher the temperature, smaller the band gap. Therefore, if the light incident on the panel is energetically close enough to the band gap (which depends on the temperature) the panel will generate a voltage difference.&lt;/p&gt;&#xA;" OwnerUserId="12989" LastActivityDate="2015-01-30T12:14:04.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24370" PostTypeId="1" AcceptedAnswerId="24374" CreationDate="2015-01-30T12:24:36.477" Score="1" ViewCount="4386" Body="&lt;p&gt;I have an alcoholic solution of iodine. My goal is to extract the pure iodine without using rare or expensive compounds. How could I do this?&lt;/p&gt;&#xA;" OwnerUserId="13084" LastEditorUserId="7448" LastEditDate="2015-01-30T12:28:26.383" LastActivityDate="2020-07-06T19:05:44.507" Title="How to cheaply extract iodine from an alcoholic solution?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;extraction&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24371" PostTypeId="2" ParentId="24364" CreationDate="2015-01-30T12:41:16.077" Score="2" Body="&lt;p&gt;See &lt;a href=&quot;http://www.biochemj.org/bj/173/0787/1730787.pdf&quot; rel=&quot;nofollow&quot;&gt;Kinetics and Reaction Mechanism of Potassium-Activated Aldehyde Dehydrogenase from Saccharomyces cerevisiae&lt;/a&gt; Biochem. J. (1978) 173, 787-798&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetics depend upon the particular substrates and cofactors such as the particular aldehyde, NAD vs NADP, the cation activator (K+ or something else). A thiol is also required.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-30T12:41:16.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24372" PostTypeId="1" CreationDate="2015-01-30T13:09:44.590" Score="4" ViewCount="8253" Body="&lt;p&gt;I'd like to count the total valence electron of following neutral complex in the ionic counting (=donor-pair):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIMMb.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Electron count:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align} \ce{Cl-} : \ce{6e-} \\ \ce{CH3-}: \ce{2e-} \\ \ce{PPh3} : \ce{2e-} \end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which gives &lt;span class=&quot;math-container&quot;&gt;$\ce{10e-}$&lt;/span&gt; from these three ligands.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;1. Question: How to decide the charge of NO&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;a) if I set &lt;span class=&quot;math-container&quot;&gt;$\ce{NO+}$&lt;/span&gt; the charge of Re must be +, therefore Re(I): d6&lt;/p&gt;&#xA;&lt;p&gt;b) if I set &lt;span class=&quot;math-container&quot;&gt;$\ce{NO-}$&lt;/span&gt; the charge of Re must be 2+, therefore Re(III): d4&lt;/p&gt;&#xA;&lt;p&gt;Probably only a) gives tetragonal geometry whereas b) gives square-planar geometry, correct?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2. Question: How many electrons for NO (linear/bent)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;a) I expect linear NO is in ionic counting way a &lt;span class=&quot;math-container&quot;&gt;$\ce{2e-}$&lt;/span&gt; ligand (in covalent counting (=neutral ligand method) a &lt;span class=&quot;math-container&quot;&gt;$\ce{3e-}$&lt;/span&gt; donor).&lt;/p&gt;&#xA;&lt;p&gt;b) I expect bent NO is in ionic counting way a &lt;span class=&quot;math-container&quot;&gt;$\ce{2e-}$&lt;/span&gt; ligand (in covalent counting (=neutral ligand method) a 1e donor).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;3. Question: Total valence electron number&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I remember: When we go from linear to bent the VE number decreases with -2, therefore I expect:&lt;/p&gt;&#xA;&lt;p&gt;a) linear goes along with &lt;span class=&quot;math-container&quot;&gt;$\ce{NO+}$&lt;/span&gt;, therefore: 10 + 6 + 2 = 18 VE&lt;/p&gt;&#xA;&lt;p&gt;b) bent goes along with &lt;span class=&quot;math-container&quot;&gt;$\ce{NO-}$&lt;/span&gt;, therefore: 10 + 4 + 2 = 16 VE&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Especially with &lt;strong&gt;1. Question&lt;/strong&gt;  I feel pretty uncomfortable.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="69830" LastEditDate="2021-05-08T11:21:27.180" LastActivityDate="2021-05-08T11:21:27.180" Title="NO linear vs bent - valence electron" Tags="&lt;electrons&gt;&lt;transition-metals&gt;&lt;organometallic-compounds&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24373" PostTypeId="1" AcceptedAnswerId="24382" CreationDate="2015-01-30T13:25:47.530" Score="-4" ViewCount="529" Body="&lt;p&gt;Can a transition metal ion with a charge of 2+ form without valence shell s and p electrons?&lt;/p&gt;&#xA;" OwnerUserId="11549" LastEditorUserId="11367" LastEditDate="2016-10-20T01:20:32.097" LastActivityDate="2016-10-20T01:20:32.097" Title="Is it possible to form a divalent transition metal ion that has neither s- nor p-shell valence electrons?" Tags="&lt;periodic-table&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-01-30T19:43:44.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="24374" PostTypeId="2" ParentId="24370" CreationDate="2015-01-30T13:58:16.467" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Alcoholic solution of iodine&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What firstly comes to mind is that you meant &amp;quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tincture_of_iodine&quot; rel=&quot;nofollow noreferrer&quot;&gt;iodine tincture&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;This question must have been frequently asked, since answers are available out there with different degrees of reliability. You must first follow the safety instructions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do your experiment somewhere that the &lt;strong&gt;noxious fumes&lt;/strong&gt; won't be able to do harm to anyone. It's best to work under a fume hood.&lt;/li&gt;&#xA;&lt;li&gt;You're gonna be dealing with &lt;strong&gt;strong acids&lt;/strong&gt;. They're very &lt;strong&gt;corrosive&lt;/strong&gt; when in contact with skin, metals or anything you can imagine (except glass).&lt;/li&gt;&#xA;&lt;li&gt;Iodine tends to sublime. If not contained wisely, you'll get &lt;strong&gt;irritated skin&lt;/strong&gt;, along with damages to your eyes (See about iodine hazards &lt;a href=&quot;http://www.sciencelab.com/msds.php?msdsId=9927547&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In short, what you should do is:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pour the tincture in a beaker, or any container. The important thing to note is to have &amp;quot;control&amp;quot; over what's happening in there.&lt;/li&gt;&#xA;&lt;li&gt;Add water to the solution. The volume of the water you add should preferably be equal to that of tincture.&lt;/li&gt;&#xA;&lt;li&gt;Add &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; to the solution, &lt;span class=&quot;math-container&quot;&gt;$\approx \pu{5-7 ml}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Add bleach to the solution, preferably equal or more than that amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; you added.&lt;/li&gt;&#xA;&lt;li&gt;Wait...wait...wait. For a relative amount of time, you should leave the solution for the reactions to happen.&lt;/li&gt;&#xA;&lt;li&gt;You've past the hard part! Congratulations! Now, filter the dilute solution with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Coffee_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;coffee filter&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Additional tips:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iodine crystals sublime, so you shouldn't leave them in open air for long.&lt;/li&gt;&#xA;&lt;li&gt;You should at least be wearing appropriate eye and skin protection. If you don't want them, do not do the experiment!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wikihow.com/Extract-Iodine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikihow&lt;/a&gt; has additional tips in case of any failures.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wikihow.com/Extract-Iodine&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wikihow.com/Extract-Iodine&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sci-spot.com/Chemistry/iodine.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sci-spot.com/Chemistry/iodine.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="61578" LastEditDate="2020-07-06T19:05:44.507" LastActivityDate="2020-07-06T19:05:44.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24376" PostTypeId="2" ParentId="24326" CreationDate="2015-01-30T15:36:23.347" Score="4" Body="&lt;p&gt;The steps for predicting a binding energy is relatively straightforward in computational chemistry but the process itself can become extremely difficult depending on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The types of molecules involved (size, complexity, etc.)&lt;/li&gt;&#xA;&lt;li&gt;The desired accuracy of your final answer&lt;/li&gt;&#xA;&lt;li&gt;The environment that the molecules reside in (gas phase, aqueous phase, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For very small molecules in the gas phase, the process is easy and routine assuming that electron correlation is trivial.  Aqueous phase environments, complex/large molecules and high accuracy quickly makes the problem more difficult if not impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written up &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31/do-terminal-amides-participate-in-pi-stacking/38#38&quot;&gt;a previous answer which addresses&lt;/a&gt; how to predict an &lt;strong&gt;interaction energy&lt;/strong&gt; ($E_{\mathrm{int}}$) which is &lt;em&gt;completely different&lt;/em&gt; from a &lt;strong&gt;binding energy&lt;/strong&gt; ($E_{\mathrm{bind}}$).  However, much of the concepts will remain the same.  I will try to outline here what in fact is the difference between an interaction energy and a binding energy.  Unfortunately, 'binding energy' is tossed around a lot in the literature and consequently grossly misused.  I hope to clarify the subtle differences between these two terms in my answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will focus my discussion around an elementary system of two water molecules to make this conceptually easier.  In the gas phase, an isolated water molecule will adopt a particular geometry (at 0K and in a vacuum).  We can determine what this minimum energy configuration will be at a particular level of theory by approximating the Schrodinger equation.  The geometry that gives rise to the lowest possible energy for an isolated water molecule will be the geometry that you will observe under those conditions.  Remember that the internal coordinates of water include things like the O-H bonds, and the H-O-H angle.  We don't need to guess at what these parameters will be in the end, solving the Schrodinger equation will determine that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know exactly what an isolated water molecule looks like and what the corresponding energy of that water molecule will be.  Both of these components are equally important when it comes to determining a binding energy because (as the name suggests) we are trying to determine an energy.  So the energies that we start off with are crucially important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now consider what happens when two water molecules interact (form a hydrogen bond).  What will be the binding energy of these two water molecules (i.e. what will be the strength of this bond that forms between then?).  It is common to approximate the strength of this bond by simply computing an &lt;em&gt;interaction energy&lt;/em&gt; (its computationally easier to determine $E_{\mathrm{int}}$ rather than $E_{\mathrm{bind}}$).  How would we compute an &lt;em&gt;interaction energy&lt;/em&gt;?  In this case, you wouldn't need to figure out what geometry or energy an isolated water molecule in its lowest energy configuration.  You would simply need to determine what the geometry of two interacting water molecules (i.e. dimer) would be (and corresponding energy) &lt;em&gt;as well as the&lt;/em&gt; energy of each water molecule isolated from each other &lt;strong&gt;in the geometry that it adopts in the dimer&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll explain.  We can determine the lowest energy configuration of a single water molecule in the gas phase.  This geometry is inevitably different than a water that is not isolated... a water that is interacting with something else such as another water.  If you want to get an interaction energy, you only need...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy of the lowest energy configuration of the water dimer&lt;/li&gt;&#xA;&lt;li&gt;Energy of each water (in the water dimer) isolated from each other but only at the geometry that the water takes on within the dimer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case, you are NOT allowing each isolated water molecule to 'relax' into its lowest energy configuration.  You are simply going to neglect the contributions of this energy relaxation effect to the overall bond strength.  Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\text{int}} =  E_{\text{opt-dimer}} - (E_{\text{mon1}} + E_{\text{mon2}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where mon1 is the first water molecule in your dimer and mon2 is the second water molecule in your dimer. (opt-dimer means optimized dimer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the case of a binding energy, you MUST account for the relaxation effects that occur when you break the dimer apart and allow the waters to become isolated from each other (and everything else).  To do this, you simply carry out what I've stated above... you simply optimize each water and then take its corresponding energy.  (Determining an interaction energy means you don't have to do this extra optimization(s)... you would simply do a single point energy on each monomer).  Clearly for a water dimer, you would only do one optimization as each water will become identical when isolated.  For a heterodimer, you would have to optimize each piece separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the binding energy, you would use the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\text{bind}} =  E_{\text{opt-dimer}} - (E_{\text{opt-mon1}} + E_{\text{opt-mon2}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the problem becomes more complex when taking into account basis-set superposition error (you will need to perform counterpoise corrections though there exists a heated debate in the literature about the usefulness of this type of correction).  Also, zero-point vibrational energy corrected binding energies will require that the zero-point energy of the dimer and each monomer will need to be determined (which comes from taking the 2nd derivative of the wavefunction which is computationally demanding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as tools go, you can choose from a number of available quantum software packages.  &lt;a href=&quot;http://www.gaussian.com/g_prod/g09.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaussian 09&lt;/a&gt; is a popular choice but very expensive.  &lt;a href=&quot;http://www.q-chem.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Q-Chem&lt;/a&gt; (paid) has the benefit of having the most beautiful, gorgeous, well-laid out documentation of anything I've ever seen (makes for a good supplemental textbook to boot).  &lt;a href=&quot;http://www.nwchem-sw.org/index.php/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;NWChem&lt;/a&gt; is opensource.  Google can help you find plenty of options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick note: Basis set superposition error (BSSE) should NOT be referred to as an error as 'error' suggests deviation from the RIGHT answer. BSSE has been shown to actually make  non-corrected answers even WORSE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last important note:  When determining differences of energies (such as you do to get an interaction or binding energy), ALL energies being compared must be evaluated at the SAME LEVEL OF THEORY (method/basis set, etc.).&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-01-30T16:38:33.210" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24379" PostTypeId="1" AcceptedAnswerId="24380" CreationDate="2015-01-30T15:58:43.963" Score="0" ViewCount="49" Body="&lt;p&gt;Is there any specific name of the problem of finding equilibrium concentrations with known equilibrium constants and initial concentrations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be appropriate to call it &quot;direct equilibrium problem&quot;, by analogy with the similar problem in chemical kinetics--finding concentrations as function of time for known rate constants--that is called &quot;direct kinetic problem&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11617" LastActivityDate="2015-01-30T16:39:23.667" Title="What's the name of the problem of finding equilibrium concentrations?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24380" PostTypeId="2" ParentId="24379" CreationDate="2015-01-30T16:17:03.037" Score="1" Body="&lt;p&gt;I don't think the phrase &quot;direct equilibrium problem&quot; is used in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Multiple-equilibrium problem&quot; would be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the phrase &quot;It involves solution of the multiple equilibrium problem&quot; &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/7723031&quot; rel=&quot;nofollow&quot;&gt;J. Molecular Biology vol. 247 page 774&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or &quot;chemical equilibrium problem&quot; if you don't want to imply that there is more than one equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-01-30T16:39:23.667" LastActivityDate="2015-01-30T16:39:23.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24381" PostTypeId="1" AcceptedAnswerId="24436" CreationDate="2015-01-30T16:37:39.140" Score="5" ViewCount="63" Body="&lt;p&gt;I am currently studying reactions involving the reduction of metal oxides by carbon at high temperature to produce refractory carbides (such as $\ce{ZrC}$) following equations such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{ZrO2 + 3C -&amp;gt; ZrC + 2CO (g)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across both the following terms for this type of a reaction &lt;code&gt;carbothermic reduction&lt;/code&gt; and &lt;code&gt;carboreduction&lt;/code&gt;; they appear to be used completely interchangeably. I would like to make sure (as I have not yet found a source to say otherwise) that this is the case in chemistry terminology and not that one may be applied to an equation as above and the other to a similar reaction where the product is actually the pure reduced metal, such as that for iron ore smelting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2Fe2O3 + 3C -&amp;gt; 4Fe + 3CO2}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="13090" LastEditorUserId="194" LastEditDate="2017-04-08T01:23:31.277" LastActivityDate="2017-04-08T01:23:31.277" Title="Is the term &quot;carboreduction&quot; interchangeable with &quot;carbothermic reduction&quot;?" Tags="&lt;redox&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24382" PostTypeId="2" ParentId="24373" CreationDate="2015-01-30T16:42:10.260" Score="2" Body="&lt;p&gt;A transition element in its ground state has a configuration of $ (n-1)d^{1-10} ns^{1-2} $&lt;br&gt;&#xA;For the atom to have noble gas configuration (which is what I assume you mean) all electrons of the outer orbitals have to be stripped. An oxidation state of +2 implies that only two electrons have been stripped. The lowest number of valence electrons are possessed by the third group elements as far as transition metals are concerned. Their ground state configuration is $ (n-1)d^{1} ns^{2} $ ; removing two electrons gives $ (n-1)d^{1} ns^{0} $ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the answer to your question is no, as there has to be at least one electron in the d orbital of a transition metal in +2 state.&lt;/p&gt;&#xA;" OwnerUserId="12989" LastActivityDate="2015-01-30T16:42:10.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24383" PostTypeId="1" AcceptedAnswerId="24412" CreationDate="2015-01-30T16:45:23.613" Score="16" ViewCount="28286" Body="&lt;p&gt;Tosyl chloride is used to make a hydroxyl group into a better leaving group. However, when the reaction of tosyl chloride and an alcohol occurs, a weak base such as pyridine should be used. Why?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="16683" LastEditDate="2017-02-27T19:09:52.520" LastActivityDate="2019-10-04T09:38:01.940" Title="Why is pyridine used when making tosyl esters from alcohols?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;&lt;catalysis&gt;&lt;organosulfur-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24384" PostTypeId="2" ParentId="24383" CreationDate="2015-01-30T16:55:29.117" Score="3" Body="&lt;p&gt;The byproduct of the reaction is hydrochloric acid, and with pyridine forms the water soluble pyridinium chloride, which is readily removed in work-up.  Other preparations might use triethylamine instead of pyridine.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="16683" LastEditDate="2017-02-27T20:59:41.610" LastActivityDate="2017-02-27T20:59:41.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24385" PostTypeId="1" CreationDate="2015-01-30T17:00:16.940" Score="0" ViewCount="122" Body="&lt;p&gt;I have DBCO-functionalized resin, and the DBCO group is validated with FAM-azide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I cannot use this resin to react with FAM-Azide-labled primer, which has very low concentration (20 μM, it is already very concentrate at DNA-scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that might be caused by concentration, because the reaction between DBCO-resin and FAM-azide is also not significant at such low concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already tried long reaction time and higher temperature.&#xA;How could I get this work?&lt;/p&gt;&#xA;" OwnerUserId="10145" LastActivityDate="2015-01-30T17:00:16.940" Title="How could click reaction take place at low concentration?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24386" PostTypeId="2" ParentId="17070" CreationDate="2015-01-30T17:05:08.330" Score="3" Body="&lt;p&gt;Ron's answer addresses the empircal data, but I feel compelled to comment that your assertion that pyridine has a nitrogen who's &quot;lone pair participating in resonance&quot; is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pyridine, nitrogen is sp2 hybridized and forming a pi-bond necessarily puts the lone pair perpendicular to the ring, and so it may &lt;em&gt;NOT&lt;/em&gt; donate the lone pair via resonance.  For the sake of completeness, pyrrole (like furan, but with nitrogen), will have a lone pair that is involved in resonance, and hence be an electron rich ring, similar to what was explained by Ron.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The essence of this problem is comparing the geometry of this lone-pair.  The switch between nitrogen and oxygen is a red herring, and questions like this are asked to make sure you are able to form judgments about which factors are more strongly contributing to the observed phenomena.  In this case the resonance dominates the difference in electronegativity.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-01-30T17:05:08.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24388" PostTypeId="2" ParentId="24360" CreationDate="2015-01-30T17:30:06.950" Score="4" Body="&lt;p&gt;It's true that as the strength of an acid goes up, the strength of its conjugate base goes down. For example, hydrochloric acid (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} \approx -7$&lt;/span&gt;) is really strong and chloride anion is a really weak base. On the other side of the spectrum, an alkane (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} \approx 50$&lt;/span&gt;) is a very weak acid, but its conjugate base will deprotonate nearly anything. So at the extremes it's clear: strong acid gives weak conjugate base and weak acid gives strong conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compounds with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt;'s between 0 and 14, such as hypochlorous acid with an intermediate &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt;, both the acid and the conjugate base are weak. We call them &quot;weak&quot; acids because they don't completely dissociate in water. However, their conjugate bases are also &quot;weak&quot; because they're weaker than hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes down to just how weak do you mean by &quot;weak&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="5026" LastEditDate="2019-05-09T17:17:21.980" LastActivityDate="2019-05-09T17:17:21.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24389" PostTypeId="2" ParentId="24106" CreationDate="2015-01-30T17:39:32.137" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However if the single or linear combinations of Slater determinants in&#xA;  a same configuration are the zeroth order functions (eigenfunctions of&#xA;  the unperturbed atomic Hamiltonian), how the fact that S2 and L2&#xA;  commute with the atomic Hamiltonian makes the Slater determinants be&#xA;  the eigenfunctions of the atomic Hamiltonian??&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed &lt;em&gt;not&lt;/em&gt; an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, a Slater determinant can be used as a trial wave function in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Variational_method_(quantum_mechanics)&quot; rel=&quot;noreferrer&quot;&gt;variational procedure&lt;/a&gt;. That is the whole point of the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;noreferrer&quot;&gt;Hartree-Fock method&lt;/a&gt;&lt;/em&gt;: a Slater determinant $\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\left\langle \Phi \mid H \mid \Phi \right\rangle$ using &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%E2%80%93Condon_rules&quot; rel=&quot;noreferrer&quot;&gt;Slater rules&lt;/a&gt;, and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an &lt;em&gt;approximation&lt;/em&gt; to the ground state wave function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for &lt;a href=&quot;http://en.wikipedia.org/wiki/Configuration_interaction&quot; rel=&quot;noreferrer&quot;&gt;configuration interaction (CI)&lt;/a&gt; method.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now back to all this business with angular momentum operators. A single Slater determinant is &lt;em&gt;not necessarily&lt;/em&gt; an eigenfunction of neither $\hat{L}^2$ nor $\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\hat{L}^2$ and $\hat{S}^2$ (as well as $\hat{L}_z$ and $\hat{S}_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\hat{H}, \hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to construct our trial wave function being an eigenfunction of angular momentum operators $\hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted &lt;a href=&quot;http://en.wikipedia.org/wiki/Configuration_state_function&quot; rel=&quot;noreferrer&quot;&gt;configuration state function (CSF)&lt;/a&gt; - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that this is not always possible.&#xA;For instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\hat{S}^{2}$.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-01-30T17:39:32.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24390" PostTypeId="1" CreationDate="2015-01-30T18:02:29.597" Score="2" ViewCount="2819" Body="&lt;p&gt;What happens when you dissolve white of an egg in water?&#xA;Firstly, what's white of an egg? Is it the white part? Or implies whole of an egg? I couldn't relate much of it with colloids, either.&lt;/p&gt;&#xA;" OwnerUserId="12905" LastActivityDate="2015-01-30T18:27:51.307" Title="White of an egg" Tags="&lt;solutions&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24391" PostTypeId="1" CreationDate="2015-01-30T18:08:24.967" Score="3" ViewCount="5477" Body="&lt;p&gt;I studied cyclobutadiene as an anti-aromatic compound. But I am unable to draw the resonance structures for cyclobutadiene. Can anyone help me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the driving force for the resonance to start? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that a compound is having conjugate double bond system but not satisfying resonance structures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a compound called as anti-aromatic if it is not undergoing resonance? &lt;/p&gt;&#xA;" OwnerUserId="12917" LastEditorUserId="5456" LastEditDate="2015-01-31T06:51:39.093" LastActivityDate="2016-04-05T16:46:25.953" Title="Resonance structure of cyclobutadiene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24392" PostTypeId="2" ParentId="24383" CreationDate="2015-01-30T18:20:29.173" Score="2" Body="&lt;p&gt;&lt;strong&gt;A base&lt;/strong&gt; is needed to remove $\ce{HCl}$ produced in the reaction. A &lt;strong&gt;weak&lt;/strong&gt; base is needed because strong bases are usually strong nucleophiles as well and would react with tosil chloride themselves. A &lt;strong&gt;tertiary&lt;/strong&gt; amine and not primary/secondary is used, because amines are in general stronger nucleophiles than alcohols, but tertiary amines would give quite electrophilic products themselves. &lt;strong&gt;Pyridine&lt;/strong&gt; is used, because nitrogen in it is in $\mathrm{sp^2}$ state, resulting in significantly lowered nucleophilicity. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2017-02-27T19:02:18.013" LastActivityDate="2017-02-27T19:02:18.013" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24393" PostTypeId="2" ParentId="24390" CreationDate="2015-01-30T18:27:51.307" Score="3" Body="&lt;p&gt;The white clear liquid in the egg is nothing but albumin. It helps in supplying nutrition to the yolk and gives strength to the yolk during the formation of chicken. White part doesn't imply whole of an egg.&#xA;       The composition of that white liquid is 90% water and 10% protein. &#xA;       If you dissolve it in water, then it is nothing but a colloidal solution. We can call it as emulsion as it is 'protein in water '.&#xA;    Hope helpful.. &lt;/p&gt;&#xA;" OwnerUserId="12917" LastActivityDate="2015-01-30T18:27:51.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24394" PostTypeId="2" ParentId="24391" CreationDate="2015-01-30T19:09:03.253" Score="3" Body="&lt;p&gt;Antiaromaticity is a concept describing an explicitly fictional situation, no molecule is antiaromatic, because all molecules will react to avoid the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to the implied question is that cyclobutadiene has very little resonance because the molecule is prevented from being antiaromatic (by the nature of the universe and its physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elaboration: &#xA;To avoid this situation, butadiene is very reactive; it will undergo an intermolecular Diels-Alder reaction avoid this antiaromatic state.  When reaction is prevented something much more interesting happens.  When frozen in an argon matrix, x-ray crystallography of cyclobutadiene indicates the molecule is highly rectangular, not square.  For dienyl resonance, there is some changing of bond lengths.  In butadiene, the 2,3 bond shortens.  If this were to occur in cyclobutadiene, then there would be degeneracy in the orbitals leading to a Frost diagram that would describe an antiaromatic state.  So if reaction is prevented, the molecule distorts its own geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fictional resonance structures you would draw are double bonds on the top and bottom, and double bonds on the sides.  However, generally, one might assume there is no resonance for this molecule.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2016-04-05T16:46:25.953" LastActivityDate="2016-04-05T16:46:25.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24401" PostTypeId="1" CreationDate="2015-01-31T00:06:49.973" Score="4" ViewCount="178" Body="&lt;p&gt;I know this may be a too generic question, but here it goes anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is already know that solid water (ice) organizes its molecules in a lattice (in a variety of 16 different crystals and one amorphous phase). See, for instance, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ice&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ice&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about liquid water? Is it the case that there are a (limited) number of local structures, like stable clusters or lattices configurations, combining and dissociating over relatively short periods of time, that dominate the liquid phase, loosely resembling the organization of a solid phase (the principal difference being that the hydrogen bonds would be weaker and would break more easily because of thermal agitation)? If so, what are those frequent structures? &lt;/p&gt;&#xA;" OwnerUserId="13094" LastActivityDate="2015-01-31T11:58:51.940" Title="Structures present in liquid water" Tags="&lt;water&gt;&lt;intermolecular-forces&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24402" PostTypeId="2" ParentId="2606" CreationDate="2015-01-31T03:14:01.893" Score="40" Body="&lt;p&gt;Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in &lt;em&gt;Nature Chem.&lt;/em&gt; (Ref.1).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, &amp;quot;Coulomb explosion during the early stages of the reaction of alkali metals with water,&amp;quot; &lt;em&gt;Nature Chemistry&lt;/em&gt; &lt;strong&gt;2015&lt;/strong&gt;, &lt;em&gt;7&lt;/em&gt;, 250–254 (&lt;a href=&quot;https://doi.org/10.1038/nchem.2161&quot; rel=&quot;noreferrer&quot;&gt;https://doi.org/10.1038/nchem.2161&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13096" LastEditorUserId="61578" LastEditDate="2020-07-07T16:57:57.817" LastActivityDate="2020-07-07T16:57:57.817" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24403" PostTypeId="2" ParentId="24401" CreationDate="2015-01-31T03:31:00.287" Score="-2" Body="&lt;p&gt;There is no lattice or psuedo-lattice structure associated with water while it is the liquid state. Look at it as a medley of H2O molecules combining, breaking apart, and recombining with Hydrogen bonds as the main mediator of attraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xWRW0.png&quot; alt=&quot;Water Molecules and Their Interaction Amongst Each Other&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar molecules (like the above water molecules) have a weak, partial negative charge at one region of the molecule (the O atm) and a partial positive charge elsewhere (The H atm) These bonds that form between water molecules account for some of the unique characteristics of water. &lt;/p&gt;&#xA;" OwnerUserId="6861" LastActivityDate="2015-01-31T03:31:00.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24404" PostTypeId="1" AcceptedAnswerId="24408" CreationDate="2015-01-31T07:31:24.883" Score="4" ViewCount="560" Body="&lt;p&gt;Why must the alkyl halide be sp3 hybridized with respect to the carbon connected to the halogen for the alkylation to occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, for example, can't we use acetylide with a bromine substituted for hydrogen on one side in conjunction with aluminum tribromide to alkylate benzene? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="194" LastEditDate="2017-04-08T01:25:40.100" LastActivityDate="2017-04-08T01:25:40.100" Title="Friedel-Crafts Alkylation and hybridization" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24405" PostTypeId="1" CreationDate="2015-01-31T10:40:45.230" Score="4" ViewCount="2690" Body="&lt;p&gt;Consider a situation where we have two aqueous solutions which have the same concentration (e.g. 1 M), but the solute is different. For example, one is a solution of glucose, and the other is a solution of ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that dissolving something in water can lead to changes in boiling point (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Boiling-point_elevation&quot; rel=&quot;nofollow noreferrer&quot;&gt;elevation&lt;/a&gt;). But would the boiling points of these two solutions be affected in the same manner, given that the concentration of the solute is the same? Or will the exact boiling point depend on the chemical species that is dissolved?&lt;/p&gt;&#xA;" OwnerUserId="13098" LastEditorUserId="16683" LastEditDate="2020-05-29T18:54:38.343" LastActivityDate="2020-05-29T18:55:28.137" Title="Does the boiling point of an aqueous solution depend on what the solute is?" Tags="&lt;aqueous-solution&gt;&lt;boiling-point&gt;&lt;colligative-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24406" PostTypeId="2" ParentId="24401" CreationDate="2015-01-31T11:58:51.940" Score="4" Body="&lt;p&gt;Yes, there are various water clusters leading to local order on the scale of a few molecules holding stabilized orientation with one another for measurable timescales.  &lt;a href=&quot;http://www.pnas.org/content/98/19/10533.full.pdf&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt;(1) is a pretty good introduction to the major varieties in liquid water.  &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/1521-3773(20010518)40:10%3C1808::AID-ANIE1808%3E3.0.CO;2-1/full&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt;(2) talks more about various structures available in any phase, including the ices, which gives some pretty good context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Keutsch, Frank N., Saykally, Richard J., Water clusters: Untangling the mysteries of the liquid, one molecule at a time, 2001, PNAS, 98, 19, 10533–10540, doi:10.1073/pnas.191266498&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Ludwig, Ralf, Water: From Clusters to the Bulk, 2001, Angewandte Chemie International Edition, 40, 10, 1808-1827, doi:10.1002/1521-3773(20010518)40:10&amp;lt;1808::AID-ANIE1808&gt;3.0.CO;2-1&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2015-01-31T11:58:51.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24407" PostTypeId="2" ParentId="24405" CreationDate="2015-01-31T12:59:39.000" Score="4" Body="&lt;p&gt;No, in water the two solutions will not have the same boiling point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethanol is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Volatility_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;volatile&lt;/a&gt; liquid with boiling point lower than water, and lowers the boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water forms an &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;nofollow noreferrer&quot;&gt;azeotrope&lt;/a&gt; with ethanol.  There are positive and negative azeotropes, so when two or more volatile liquids are dissolved, you cannot easily predict whether boiling point will decrease or increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can also be cases, like adding HCl to water, where as solute is added, first the boiling point increases to a maximum (108 °C around 20% HCl) above the boiling point of water, then decrease to below the boiling point of water (48 °C for 38% HCl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glucose is non-volatile and increases the boiling point of water.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling-point_elevation&quot; rel=&quot;nofollow noreferrer&quot;&gt;boiling-point elevation formula&lt;/a&gt; can be applied to non-volatile solutes in volatile solvent.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2020-05-29T18:55:28.137" LastActivityDate="2020-05-29T18:55:28.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24408" PostTypeId="2" ParentId="24404" CreationDate="2015-01-31T13:06:14.647" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the problem, sp2 and sp hybridized carbocations are less stable than sp3. Those carbocations are difficult to form and will likely undergo rearrangement faster than attack of the aromatic ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern organic synthesis has much better ways of making sp-sp2 and sp2-sp2 carbon-carbon bonds by &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Palladium-catalyzed_coupling_reactions&quot; rel=&quot;nofollow&quot;&gt;palladium catalyzed cross-coupling&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-01-31T13:06:14.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24409" PostTypeId="2" ParentId="24082" CreationDate="2015-01-31T13:52:37.567" Score="2" Body="&lt;p&gt;So is the point of contention here &quot;how does the standard state for the activities relate to the standard state for the Gibbs free energy change&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how choice of standard state for an activity relates to the standard state chosen for the chemical potential, write the equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_i = \mu_i^\circ + R T \ln a_i = \mu_i^\circ + R T \ln \frac{\gamma_i c_i}{c_i^\circ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the standard states are consistent.The standard state is being defined so that&#xA;when $\gamma_i c_i = c_i^\circ$, $\mu_i = \mu_i^\circ$. If we're using molarities, we can choose $c_i^\circ = 1\ M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose you want to define a standard state that's different from 1 in the units you're using for concentration; e. g, suppose we have a reaction that includes $\rm H^+$ and we want to define the biological standard state ($^\oplus$) as $c_{\rm H^+}^\oplus = 10^{-7}\ M = 10^{-7} c_{\rm H^+}^\circ$, while using molarities as our concentration unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute a new chemical potential:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}\mu_{\rm H^+} &amp;amp;=&amp;amp; \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\oplus}\\&#xA;&amp;amp;=&amp;amp; \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{10^{-7} c_{\rm H^+}^\circ}\\&#xA;&amp;amp;=&amp;amp; (\mu_{\rm H^+}^\circ + 7 R T\ln 10) + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\&#xA;&amp;amp;=&amp;amp; \mu_{\rm H^+}^\oplus + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\&#xA;&amp;amp;=&amp;amp; \mu_{\rm H^+}^\oplus + R T \ln a_{\rm H^+}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now at unit activity $\gamma_i c_i = c_i^\circ$ and $\mu_i = \mu_i^\oplus$. Your chemical potential (and your Gibbs free energy changes) are now defined in terms of this new standard state. &lt;em&gt;And note that the chemical potential at this new standard state ($\mu_{\rm H^+}^\oplus$) is indeed temperature dependent&lt;/em&gt;. To convert between the two standard state Gibbs free energies in this particular case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\oplus = \Delta G^\circ + 7 \nu R T \ln 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu$ is the stoichiometric coefficient of $\rm H^+$ in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;TL;DR: The standard state you choose for the concentration at unit activity becomes the standard state for the standard state Gibbs free energy change.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-02T05:46:42.547" LastActivityDate="2015-02-02T05:46:42.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24410" PostTypeId="2" ParentId="23935" CreationDate="2015-01-31T13:55:13.897" Score="1" Body="&lt;p&gt;Elimination via a E2 mechanism cannot take place at the bridgehead carbon of a bicyclic compound due to the lack of the required trans geometry between the leaving group at the bridgehead carbon and any of the six β-hydrogens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, as you have correctly observed, a E1 mechanism has to be ruled out as well due to the strain at the bridgehead carbocation. In fact bicyclic structures prevent the tertiary bridgehead carbon becoming planar, which implies that is very high in energy because the nonplanar structure forces the cation to be an empty &quot;sp3 orbital&quot; instead of an empty &quot;p orbital&quot;. This observation was made for the first time by Bredt, therefore is generally known as &quot;Bredt's rule&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-01-31T13:55:13.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24411" PostTypeId="2" ParentId="13743" CreationDate="2015-01-31T14:14:51.393" Score="5" Body="&lt;p&gt;glycerols are the triol compound used for many purposes in pure or mixed form , but glycerine is the commercial name of glycerol, which is not pure ,which contain mostly 95% of glycerol , it can't be used when pure glycerol is required .&lt;/p&gt;&#xA;" OwnerUserId="13100" LastActivityDate="2015-01-31T14:14:51.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24412" PostTypeId="2" ParentId="24383" CreationDate="2015-01-31T14:27:07.590" Score="21" Body="&lt;p&gt;The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent nucleophile. Pyridine is, in fact, more nucleophilic than the alcohol, and it attacks the acyl chloride rapidly, forming a highly electrophilic (because of the positive charge) intermediate: &lt;em&gt;N&lt;/em&gt;-tosylpyridinium chloride. It is indeed this intermediate the actual tosylating agent which reacts with the alcohol to give the ester. Because pyridine is acting as a nucleophile to speed up the reaction, yet is unchanged by the reaction, it is a &quot;nucleophilic&quot; catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5jj4c.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5jj4c.png&quot; alt=&quot;Pyridine as a nucleophilic catalyst in tosylation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="16683" LastEditDate="2017-02-27T19:01:54.060" LastActivityDate="2017-02-27T19:01:54.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24413" PostTypeId="2" ParentId="24391" CreationDate="2015-01-31T14:42:09.607" Score="1" Body="&lt;p&gt;You &#xA;mix up resonance and (anti-)aromaticity the latter being&lt;br&gt;&#xA;interpreted today by molecular orbital symmetry. &#xA;This is unhistoric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance is a concept predating Woodward-Hoffman and Fukui. &#xA;Resonance meant that delocalizing the electrons of bond(s) &#xA;lowers the energy irrespective of the kind of orbitals. &#xA;This works for conjugated double bonds, but for aromatic &#xA;compounds it is not wrong, but insufficient. &#xA;The amount of stabilisation of eg benzene is more than &#xA;conjugation can explain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For antiaromatic compounds oldfashioned resonance is simply wrong, &#xA;and drawing &quot;resonance stuctures&quot; is misleading, &#xA;because the resonance destabilisizes the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had a friendly Laplacian demon clamping the 4 carbon &#xA;atoms of cyclobutadiene on the corners of a square with a &#xA;bond length similar to benzene, we would see that &#xA;cyclobutadiene is a triplett diradical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe somebody will synthesize a molecule where a &#xA;cyclobutadiene core is clamped in in such a manner by some &#xA;planar ring structures &quot;around&quot; some day? &lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2015-01-31T14:42:09.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24414" PostTypeId="2" ParentId="24325" CreationDate="2015-01-31T14:45:37.653" Score="3" Body="&lt;p&gt;Consider &lt;a href=&quot;http://en.wikipedia.org/wiki/Indigo_dye&quot; rel=&quot;nofollow&quot;&gt;natural indigo&lt;/a&gt;, formerly used to color blue jeans. It is insoluble; reduction converted it into water-soluble &quot;indigo white&quot;. Fabric was soaked in indigo white solution; on air oxidation, it became insoluble blue indigo again, with indigo molecules now presumably mechanically entangled in the fiber and held by van der Waals forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More often &quot;binders&quot; called &lt;a href=&quot;http://goldbook.iupac.org/M04029.html&quot; rel=&quot;nofollow&quot;&gt;mordants&lt;/a&gt; are used. They form an insoluble &quot;lake&quot; when they coordinate to the pigment; the lake can bind to fabric fibers. You can read more about the chemistry of the process &lt;a href=&quot;http://web.archive.org/web/20070814015208/stainsfile.info/StainsFile/theory/mordant.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-01-31T14:45:37.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24415" PostTypeId="2" ParentId="24208" CreationDate="2015-01-31T15:35:54.957" Score="4" Body="&lt;p&gt;Tap water contains calcium and magnesium ions; more so if the water is hard. These ions can bind to soap over time and make an insoluble soap curd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the sponge clean? If not, another reason is that the surfactants in the soap that make the foam might be forming micelles around whatever grime is in your sponge, so that there aren't many left to make the foam.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-01-31T15:35:54.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24416" PostTypeId="1" AcceptedAnswerId="24417" CreationDate="2015-01-31T15:40:18.487" Score="0" ViewCount="292" Body="&lt;p&gt;I get the concept that one can measure the rate of reaction by measuring the rate at which an equilibrium state reaches another equilibrium state after the system is perturbed (usually by heating). However, I do have an issue with it. Consider the reaction $$\ce{A &amp;lt;=&amp;gt; B}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lecture handout suggests that the rate of change of $[\ce{A}]$ is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]+k_\text{reverse}[B]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, surely it's just $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]$. The equation in the lecture handout would make sense at equilibrium because that is when $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=0$ but I think it seems to insinuate that this is always valid (without explicitly saying so). Am I right or is that equation always valid? If it is: why? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2016-11-15T20:37:31.193" LastActivityDate="2016-11-15T20:37:31.193" Title="How to make sense of the 'Relaxation Method' for measuring reaction kinetics?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24417" PostTypeId="2" ParentId="24416" CreationDate="2015-01-31T16:00:40.107" Score="3" Body="&lt;p&gt;You have two simultaneous reactions, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rcl}&#xA;\ce{A} &amp;amp;\xrightarrow{k_1}&amp;amp; \ce{B}\\[6pt]&#xA;\ce{B} &amp;amp;\xrightarrow{k_{-1}}&amp;amp; \ce{A}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did it your way, this wouldn't be the case.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorDisplayName="user7951" LastEditDate="2016-11-15T20:37:24.830" LastActivityDate="2016-11-15T20:37:24.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24419" PostTypeId="1" AcceptedAnswerId="24429" CreationDate="2015-01-31T17:13:16.140" Score="0" ViewCount="26" Body="&lt;p&gt;In my experimental work, I'm following an article. It said that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 30 nm ZnO seed layer was deposited on the glass substrates by RF sputtering of a ZnO target at room temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I haven't any tools for &quot;RF sputtering&quot;. Is there any replacement for this technique?&lt;/p&gt;&#xA;" OwnerUserId="10181" LastActivityDate="2015-01-31T19:46:57.533" Title="Replacement For RF sputtering" Tags="&lt;experimental-chemistry&gt;&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24420" PostTypeId="2" ParentId="24114" CreationDate="2015-01-31T17:14:06.840" Score="2" Body="&lt;p&gt;As far as I know there isn't a specific term for that point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a one-step first order reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm A \xrightarrow{k_1} B + C, \quad \text{rate} = k_1[A]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that can run backwards as a second order reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm B + C \xrightarrow{k_{-1}} A, \quad \text{rate} = k_{-1}[B][C]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rates of those reactions are equal, you can call that point &quot;equilibrium&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't really plot the rates separately the way you have unless you have a common concentration, e. g. plot a change in concentration $x$ where $\rm [A] = [A]_0 - x$, $\rm [B] = [B]_0 + x$, and $\rm [C] = [C]_0 + x$.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-01-31T22:16:01.870" LastActivityDate="2015-01-31T22:16:01.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24421" PostTypeId="1" AcceptedAnswerId="24423" CreationDate="2015-01-31T17:31:36.420" Score="1" ViewCount="184" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The molecule below is a simplified analog of capsaicin. Explain in which of the following options, it should be most soluble:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;aqueous HCl (pH = 1)&lt;/li&gt;&#xA;  &lt;li&gt;aqueous NaOH (pH = 14)&lt;/li&gt;&#xA;  &lt;li&gt;pure water (pH = 7)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2oYOQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2oYOQ.png&quot; alt=&quot;N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide, National Center for Biotechnology Information. PubChem Compound Database; CID=&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/606880&quot; rel=&quot;nofollow noreferrer&quot;&gt;606880&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose aqueous NaOH, because then the molecule will be completely deprotonated, which will make it more soluble. Please tell me if this is correct.&lt;/p&gt;&#xA;" OwnerUserId="12664" LastEditorUserId="4945" LastEditDate="2018-08-21T14:49:49.327" LastActivityDate="2018-08-22T11:39:43.033" Title="Solubility of N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide at different pH" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24422" PostTypeId="1" AcceptedAnswerId="24424" CreationDate="2015-01-31T17:55:42.370" Score="2" ViewCount="1001" Body="&lt;p&gt;My experiment involves a potentiometric titration, and I have used $\ce{AgNO3}$ as a titrant and KCl/KI/deionised water only as an analyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below of my data, the potential difference as recorded by a multimeter drops from 0.6V to 0.15V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NkAfY.png&quot; alt=&quot;Volume of Silver Nitrate added/cm^3 along the x axis.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes the drop in potential difference when the silver nitrate and whatever analyte (all exhibited the same pattern of a sudden drop in PD) reaches its endpoint in the titration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there a higher potential difference when using KI as analyte than KCl?&lt;/p&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="7448" LastEditDate="2015-01-31T18:04:37.203" LastActivityDate="2015-02-02T21:29:40.147" Title="Why do some salt ions in solution conduct electricity better than others?" Tags="&lt;electrochemistry&gt;&lt;ions&gt;&lt;solutions&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24423" PostTypeId="2" ParentId="24421" CreationDate="2015-01-31T18:13:57.830" Score="5" Body="&lt;p&gt;I would agree that it is most soluble in NaOH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the molecule will only deprotonate the OH proton not the NH proton.  I wouldn't give your explanation much, if any, credit because you aren't saying which proton would deprotonate, and you are implying the molecule will lose all protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, at low enough pH the N will protonate, i.e. it would gain a second proton.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2018-08-21T14:51:47.187" LastActivityDate="2018-08-21T14:51:47.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24424" PostTypeId="2" ParentId="24422" CreationDate="2015-01-31T19:01:50.537" Score="4" Body="&lt;p&gt;As you add $\rm AgNO_3$ to $\rm KCl$, solid $\rm AgCl$ forms. The silver ion you add will at first be completely consumed by the chloride. Once all of the chloride is locked up in the precipitate, the silver ion concentration will rapidly increase. The Nernst equation shows how the potential is expected to change when this happens. If you're using a silver/silver chloride electrode with a reference electrode, the indicator potential is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm E = E^\circ_{\rm AgCl,Ag} - \frac{R T}{\cal F}\ln\frac{K_{sp}}{a_{\rm Ag^+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see whether you'll get a drop or a rise in $E$, ask yourself what happens when silver concentration rises. If you've got the leads hooked up to your multimeter properly (negative lead to the silver cathode, positive lead to your anode), shouldn't you be seeing an increase in potential? What exactly are your $x$ and $y$ axes on that graph? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll get a bigger change when you form $\rm AgI$, because $\rm AgCl$ is about a thousand times more soluble than $\rm AgI$. The $\rm AgI$ will lock up relatively more ions when it forms (the $K_{sp}$'s of $\rm AgCl$ and $\rm AgI$ are $1.8\times 10^{-10}$ and $1.5\times 10^{-16}$, respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 1&lt;/strong&gt;: You say you're just using a silver electrode; the indicator potential is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm E = E^\circ_{\rm Ag^+,Ag} - \frac{R T}{\cal F}\ln\frac{1}{a_{\rm Ag^+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(but if you dip a silver wire into a chloride solution you'll get a very thin layer of AgCl and it'll begin to be a silver/silver chloride electrode (Source: G. D. Christian, &lt;em&gt;Analytical Chemistry&lt;/em&gt;, 4th ed., Wiley 1986, p. 282)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also say you don't have a true reference electrode, just a graphite electrode... the fact that its potential can change as well complicates things. Your cell potential will depend on what the oxidation process is at the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addendum 2&lt;/strong&gt;: You asked for a source for the $K_{sp}$'s. CRC Handbook 58th edition gives $K_{sp}$'s at $25^\circ$C for AgCl and AgI as $1.56\times 10^{-10}$ and $1.5\times 10^{-16}$; the 84th edition gives $1.77\times 10^{-10}$ and $8.52\times 10^{-17}$; Keller's &lt;em&gt;Basic Tables in Chemistry&lt;/em&gt; gives them as $1.0\times 10^{-10}$ and $1.0\times 10^{-16}$ for &quot;temperatures near 298 K&quot;. They're very temperature dependent and they depend on conditions like ionic strength as well, so it's not surprising that the values you're finding on the web are all over the place.  I took the values above from a genchem text that happened to be lying on my desk. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-02T21:29:40.147" LastActivityDate="2015-02-02T21:29:40.147" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24426" PostTypeId="1" AcceptedAnswerId="24431" CreationDate="2015-01-31T19:20:13.443" Score="4" ViewCount="1402" Body="&lt;p&gt;In the Friedel - Crafts acetylation of bromobenzene, how is it possible to obtain a black-coloured product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the acetic anhydride was slowly added to the reactants - $\ce{AlCl3}$, DCM, bromobenzene - a small, hard, yellow lump formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be $\ce{AlCl3}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the acetic anhydride was added, the reflux was initiated. The lump appeared to disappear, leaving behind a dark red solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the reflux was turned off, the solution appeared black. I'm trying to think of potential products, such as ortho and para bromobenzene or acetic acid or $\ce{Al(OH)3}$ but why so black?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any ideas? &lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="7721" LastEditDate="2015-02-01T13:04:45.617" LastActivityDate="2015-02-01T13:04:45.617" Title="Acetylation reaction" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24427" PostTypeId="1" AcceptedAnswerId="24620" CreationDate="2015-01-31T19:27:37.087" Score="8" ViewCount="722" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Endohedral_fullerene&quot; rel=&quot;nofollow&quot;&gt;Endohedral fullerene&lt;/a&gt; is the name given to a brand new chemical species with the following attributes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A metal, usually a transition metal, is &quot;trapped&quot; inside the fullerene, like $\ce{C60}$ and $\ce{C82}$. These transition metals are often La, Sc, Ce, Y, or, Ba and Sr etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without any &quot;binding&quot; forces, the atom couldn't be possibly trapped inside the fullerene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://juwel.fz-juelich.de:8080/dspace/bitstream/2128/908/1/1981.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; it's been discussed that the bonding is the type of &quot;donor-acceptor&quot;. Later within, it was explained that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Though there isn't empirical proof, It's believed for these species to have metallic structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there another simpler name for this &quot;donor-acceptor&quot; bonding? My thoughts lead to wierd names!&lt;/li&gt;&#xA;&lt;li&gt;How &quot;close&quot; is this type of bonding to metallic bonding?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Citation: The second article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Publisher:&lt;/em&gt;&lt;/strong&gt; [AIP]; German journal of chemical physics. Volume 115, number 15, page 7215. (15 Oct 2001) (Juelich GMBH or something; Couldn't figure the name out, maybe German fellows would help.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;DOI:&lt;/em&gt;&lt;/strong&gt; &lt;a href=&quot;http://dx.doi.org/10.1063/1.1406500&quot; rel=&quot;nofollow&quot;&gt;10.1063/1.1406500&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Authors:&lt;/em&gt;&lt;/strong&gt; R. Klingeler, G. Kann, I. Wirth, S. Eisebitt, P. S. Bechthold, M. Neeb, and W. Eberhardt (Dr. Rudiger Klingeler is the main author)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Title:&lt;/em&gt;&lt;/strong&gt; $\ce{La$@$C60}$: A metallic endohedral fullerene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the other mirror links are behind paywalls. (But this article is available in &lt;a href=&quot;http://www.researchgate.net/publication/234986449_LaC60_A_metallic_endohedral_fullerene&quot; rel=&quot;nofollow&quot;&gt;Researchgate&lt;/a&gt;, in case you've got institutions to be able to create an account)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="24052" LastEditDate="2016-03-13T20:14:05.790" LastActivityDate="2016-03-13T20:14:05.790" Title="What's the type of bonding in La@C60?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;metal&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24429" PostTypeId="2" ParentId="24419" CreationDate="2015-01-31T19:46:57.533" Score="1" Body="&lt;p&gt;If you need a high quality thin film of $\ce{ZnO}$, especially on insulating substrates, you're mostly limited to sputtering techniques, PVD (including things like pulsed laser deposition), CVD (plasma-enhanced CVD, spray pyrolysis, and atomic layer deposition are common). Unfortunately, all of these require fairly sophisticated instruments. More accessible methods include sol-gel synthesis and electrodeposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0042207X04003707&quot; rel=&quot;nofollow&quot;&gt;The sol-gel synthesis&lt;/a&gt; of $\ce{ZnO}$ only requires a spin coater and an oven that can do 600 °C. It's suitable for coating large substrates, but if you really need a layer that's only 30 nm thick, it may not be the best technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ZnO}$ can be electrodeposited onto conductive substrates from aqueous solutions. This requires a potentiostat and controlling thickness and orientation can be tricky. &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2011/JM/c1jm11225b#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;One possible method&lt;/a&gt;: $$\ce{Zn^2+ + NO3- + 2e- -&amp;gt; ZnO v + NO2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should add that if you don't have access to any of these tools, there are certainly commercial suppliers of $\ce{ZnO}$ coated glass substrates.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-01-31T19:46:57.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24430" PostTypeId="1" CreationDate="2015-01-31T19:48:15.290" Score="4" ViewCount="18381" Body="&lt;p&gt;Consider this initial-rate data at a certain temperature for the reaction described by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2NOBr(g) -&amp;gt; 2NO(g) + Br2(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{cc}\hline&#xA;\ce{[NOBr]_0 (M)} &amp;amp; \mathrm{Initial \,rate \, of \,Br_2}\, (\pu{M/s}) \\ \hline&#xA;0.600 &amp;amp;1.08 \times 10^2 \\&#xA;0.750 &amp;amp; 1.69 \times 10^2\\ &#xA;0.900 &amp;amp; 2.53 \times 10^2\\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm not understanding how to find initial rates from the data given and then use that to find the rate constants. I also don't really understand how this information can give me the orders of reactions, for example when does the number double, stay the same, or quadruple, etc.&lt;/p&gt;&#xA;" OwnerUserId="13105" LastEditorUserId="95133" LastEditDate="2021-05-13T14:57:43.080" LastActivityDate="2021-05-13T14:57:43.080" Title="Determining the value and units of the rate constant" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24431" PostTypeId="2" ParentId="24426" CreationDate="2015-01-31T20:07:33.170" Score="1" Body="&lt;p&gt;I assume &quot;DCM&quot; represents dichloromethane, if so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an old paper that describes the formation of a low molecular weight polymer from the double Friedel-Crafts alkylation of benzene by dichloromethane. That's probably what's happening in your reaction as a minor pathway.  The dissolved polymer and tar is what's producing your black solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually ran that reaction once, I wanted to prepare that polymer.  It didn't work for me, but at the end, after the aluminum chloride had been removed, I did have a dark solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g65K9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-01-31T20:07:33.170" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="24432" PostTypeId="1" CreationDate="2015-01-31T20:30:16.537" Score="5" ViewCount="2352" Body="&lt;p&gt;I have prepared the 1M water solution of magnesium acetate. Then I placed $\ce{Mg}$ and $\ce{Cu}$ electrodes in it. I predict there such reactions take place: $$\ce{Mg-2e^- \rightarrow Mg^{2+}}$$ $$\ce{H_3^+O + e^- \rightarrow\frac12 H_2 + H_2O}$$&#xA;The gas bubbles emitting on both electrodes probably confirm my prediction (I did not check is it hydrogen or not).&#xA;So I expect to obtain the potential difference about 2.3 V. But really I observe something like 1.62 V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I changed the acetate solution to a water (simple water which we drink) and observed almost same value of 1.60 V. Adding there many crystals of $\ce{Mg}$ acetate (I added them till they could not dissolve more) almost did not change the voltage!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I added much acid to the solution the potential difference became larger - up to 1.9 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have two questions: 1) why the real voltage is so different from the expected value? 2) Why changing the $\ce{Mg^{2+}}$ concentration almost does not change the voltage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may say I have dirty magnesium electrode etc. But if you search e.g. experiments with potatoes (when one puts $\ce{Mg}$ and $\ce{Cu}$ strips in potato) you can observe that everywhere it is mentioned that such $\ce{Mg-Cu}$ cell gives 1.7 V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several ideas why can it happen: a)some story with overpotentials, b)oxide film on magnesium surface.&#xA;What is the real reason?&lt;/p&gt;&#xA;" OwnerUserId="11589" LastEditorUserId="7448" LastEditDate="2015-01-31T21:21:31.830" LastActivityDate="2015-02-06T04:54:33.123" Title="Why are potentials of cells with magnesium electrode always lower than expected?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24434" PostTypeId="1" AcceptedAnswerId="24467" CreationDate="2015-01-31T21:56:18.753" Score="2" ViewCount="2103" Body="&lt;p&gt;In the first propagation step of the radical chain mechanism of alkane halogenation, why does the halogen actually abstract hydrogen from the alkane? My organic chemistry textbook so far has went into details on MOs, hybridized orbitals, and general thermodynamics of reactions but has been lacking in the by and large &lt;strong&gt;why&lt;/strong&gt; of reactions. Is the $\ce{C-H}$ bond broken by sufficient collisions from halogen free atoms and the hydrogen goes to the halogen as a result of $\ce{F}$/$\ce{Cl}$/$\ce{Br}$ being more electronegative than carbon, or what should cause this to happen?&lt;/p&gt;&#xA;" OwnerUserId="13107" LastEditorUserId="7475" LastEditDate="2015-10-07T07:28:17.590" LastActivityDate="2015-10-07T07:39:31.313" Title="What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24435" PostTypeId="2" ParentId="24430" CreationDate="2015-01-31T22:07:20.383" Score="1" Body="&lt;p&gt;The ratio of 0.9 to 0.6 is 1.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of 2.43 to 1.08 is 2.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.5^2=2.25$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial rate is proportional to [NOBr]$^2$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-31T22:07:20.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24436" PostTypeId="2" ParentId="24381" CreationDate="2015-01-31T22:19:42.543" Score="1" Body="&lt;p&gt;The term &quot;carboreduction&quot; has an &lt;a href=&quot;http://goldbook.iupac.org/CT07609.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC definition:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Process in which a metal oxide is reduced in the presence of carbon or a carbon-containing compound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So technically the term does not require heating and isn't limited to elemental carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbothermic definitely implies heating.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-01-31T22:19:42.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24438" PostTypeId="1" CreationDate="2015-01-31T23:02:03.677" Score="1" ViewCount="1408" Body="&lt;p&gt;I am further interested in knowing exactly what the requirements are for a liquid to be transparent.&lt;/p&gt;&#xA;" OwnerUserId="13108" LastActivityDate="2015-01-31T23:02:03.677" Title="Is there an example of an opaque solution which can be made transparent by adding another solute, such that no chemical reaction occurs?" Tags="&lt;solutions&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24439" PostTypeId="1" CreationDate="2015-01-31T23:15:24.900" Score="8" ViewCount="838" Body="&lt;p&gt;What is the $\mathrm pK_\mathrm a$ of cacodylic acid? I cann't seem to find one answer on the internet. I have read 1.67, 6.37, and 4.8 in my $\mathrm pK_\mathrm a$ table. What is the reason for this? Also, the problem I just did had it as 0.34.&lt;/p&gt;&#xA;" OwnerUserId="12664" OwnerDisplayName="EMathS" LastEditorUserId="5026" LastEditDate="2018-02-27T02:51:52.593" LastActivityDate="2018-02-27T04:57:50.223" Title="What is the acid dissociation constant of cacodylic acid?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24441" PostTypeId="2" ParentId="24430" CreationDate="2015-02-01T00:30:52.583" Score="3" Body="&lt;p&gt;Alright, so in the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NOBr(g) -&amp;gt; 2NO(g) + Br2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rate law is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{rate} = k[\ce{NOBr}]^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $k$'s cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2.25 = 1.5^x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}\text{rate after concentration} &amp;amp;= \left(2[\ce{NOBr}]\right)^2\\&#xA;&amp;amp;= 2^2[\ce{NOBr}]^2\\&#xA;&amp;amp;= 4[\ce{NOBr}]^2\\&#xA;&amp;amp;= 4 \times \text{rate before doubling concentration}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.&lt;/p&gt;&#xA;" OwnerUserId="13111" LastEditorDisplayName="user7951" LastEditDate="2016-11-22T19:26:33.510" LastActivityDate="2016-11-22T19:26:33.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24442" PostTypeId="2" ParentId="24439" CreationDate="2015-02-01T01:07:24.923" Score="4" Body="&lt;p&gt;I found this article (TW. Shin, K. Kim, IJ. Lee, &lt;em&gt;J. Solution Chem.&lt;/em&gt; &lt;strong&gt;1997,&lt;/strong&gt; &lt;em&gt;26&lt;/em&gt;, 379. doi: &lt;a href=&quot;https://doi.org/10.1007/BF02767677&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1007/BF02767677&lt;/a&gt;) with precise analysis of the p&lt;i&gt;K&lt;/i&gt;&lt;sub&gt;a&lt;/sub&gt; of cacodylic acid - it says that in 25&amp;nbsp;&amp;deg;C it's estimated to be 6.27. Values lower than 6 are wrong, but found often. Sometimes it's said to be a diprotic (which it isn't) with first p&lt;i&gt;K&lt;/i&gt;&lt;sub&gt;a&lt;/sub&gt; 1.57 and second 6.27 - pubchem says 1.57 citing Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 76th ed. Boca Raton, FL: CRC Press Inc., 1995-1996., p. 8-45 - it might be one of the sources of problem.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="4945" LastEditDate="2018-02-27T04:57:50.223" LastActivityDate="2018-02-27T04:57:50.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24443" PostTypeId="2" ParentId="24439" CreationDate="2015-02-01T02:43:01.927" Score="4" Body="&lt;p&gt;Cacodylic acid is amphoteric, so it has a $K_\mathrm a$ and $K_\mathrm b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See table 1 of &lt;a href=&quot;http://nepis.epa.gov/Exe/ZyNET.exe/910124WK.txt?ZyActionD=ZyDocument&amp;amp;Client=EPA&amp;amp;Index=Prior%20to%201976&amp;amp;Docs=&amp;amp;Query=&amp;amp;Time=&amp;amp;EndTime=&amp;amp;SearchMethod=1&amp;amp;TocRestrict=n&amp;amp;Toc=&amp;amp;TocEntry=&amp;amp;QField=&amp;amp;QFieldYear=&amp;amp;QFieldMonth=&amp;amp;QFieldDay=&amp;amp;UseQField=&amp;amp;IntQFieldOp=0&amp;amp;ExtQFieldOp=0&amp;amp;XmlQuery=&amp;amp;File=D%3A%5CZYFILES%5CINDEX%20DATA%5C70THRU75%5CTXT%5C00000018%5C910124WK.txt&amp;amp;User=ANONYMOUS&amp;amp;Password=anonymous&amp;amp;SortMethod=h%7C-&amp;amp;MaximumDocuments=1&amp;amp;FuzzyDegree=0&amp;amp;ImageQuality=r75g8/r75g8/x150y150g16/i425&amp;amp;Display=p%7Cf&amp;amp;DefSeekPage=x&amp;amp;SearchBack=ZyActionL&amp;amp;Back=ZyActionS&amp;amp;BackDesc=Results%20page&amp;amp;MaximumPages=1&amp;amp;ZyEntry=33&quot; rel=&quot;nofollow noreferrer&quot;&gt;this EPA report on cacodylic acid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm pK_\mathrm a$  is 6.2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm pK_\mathrm b$  is 12.4, which corresponds to the $\mathrm pK_\mathrm a$  of the conjugate acid being 1.6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, according to &quot;Weed Control&quot; by the US National Research Council at page 167:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cacodylic acid is normally monobasic, but in strong sodium hydroxide it forms the disodium salt, $\ce{(CH3)2AsO3HNa2}$, of a tribasic acid. Cacodylic acid is somewhat amphoteric, forming a hydrochloride $\ce{(CH3)2AsO2H.HCl}$ by direct action with hydrogen chloride gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, the value for the $\mathrm pK_\mathrm a$  of the conjugate acid being 1.6 was confirmed by electrophorisis is &lt;a href=&quot;http://www.pnas.org/content/75/9/4262.short&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arsoniumphospholipid in algae&lt;/a&gt; Proc. Natl. Acad. Sci. USA vol. 75 pages 4262–4264.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-02-27T02:52:53.293" LastActivityDate="2018-02-27T02:52:53.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24444" PostTypeId="1" CreationDate="2015-02-01T04:29:49.513" Score="6" ViewCount="2407" Body="&lt;p&gt;Here in USA supermarkets often sell water in plastic 1 gallon jugs. These jugs seem to all be stamped with the text &quot;HDPE&quot; (meaning that the material is High-density polyethylene) and a resin number of &quot;2&quot; on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had one of these jugs, containing distilled water, on a counter in the kitchen, and another one in the refrigerator. For some reason the water on the counter had a light plastic taste, whilst the water in the fridge did not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this taste be because of plastic leaking into the water? Or what else could cause the plastic taste?&lt;/p&gt;&#xA;" OwnerUserId="13112" LastEditorUserId="7448" LastEditDate="2015-02-01T16:57:02.823" LastActivityDate="2015-04-13T07:40:09.780" Title="Can plastic leak into water from a HDPE &quot;2&quot; plastic container at room temperature?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;plastics&gt;&lt;taste&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24446" PostTypeId="1" AcceptedAnswerId="24497" CreationDate="2015-02-01T05:39:53.620" Score="2" ViewCount="101" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Hydrochloric acid, $V=75.0~\mathrm{mL}$, $c=0.250~\mathrm{M}$, is added to $225.0~\mathrm{mL}$ of $0.0550~\mathrm{M}$ $\ce{Ba(OH)2}$ solution. What is the concentration of the excess $\ce{H+}$ or $\ce{OH-}$ ions left in this solution? (&lt;a href=&quot;https://i.stack.imgur.com/MnA5T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know the balanced equation but when considering $\ce{Ba(OH)2}$ is 'marginally soluble' which&#xA;I guess could be considered insoluble, $\ce{OH-}$ wouldn't really be in the result solution. &#xA;Am I right or at where am I wrong? thanks for reading&lt;/p&gt;&#xA;" OwnerUserId="13114" LastEditorUserId="69830" LastEditDate="2021-05-05T11:12:23.407" LastActivityDate="2021-05-05T11:12:23.407" Title="Should the relatively low solubility of a base be considered in the calculation of the pH?" Tags="&lt;acid-base&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24447" PostTypeId="1" AcceptedAnswerId="24479" CreationDate="2015-02-01T06:30:58.923" Score="1" ViewCount="1839" Body="&lt;p&gt;Which base is stronger methylamine or propylamine? I know that it depends on the strength of inductive effect of alkyl group. Am I right?&#xA;But how can I determine the difference between methyl and propyl substituent?&lt;/p&gt;&#xA;" OwnerUserId="13116" LastEditorUserId="9961" LastEditDate="2015-11-13T19:49:18.623" LastActivityDate="2015-11-13T19:49:18.623" Title="Which base is stronger methylamine or propylamine?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24449" PostTypeId="2" ParentId="24444" CreationDate="2015-02-01T07:21:29.230" Score="8" Body="&lt;p&gt;HDPE &#xA;has a molecular weight more than several hundreds of thousands, &#xA;&quot;leaking&quot; is plainly impossible. &#xA;This taste is the tase of stale water, nobody knows &#xA;what that taste is actually. &#xA;Maybe its just the taste of not cold enough water,&#xA;lacking some dissolved air, or some bacteriae &#xA;develop after contact with air and light. &#xA;(google for pseudomonas) &#xA;The only substance which ever got into the jug prior &#xA;to the water is the air (or some other gas?) used to&lt;br&gt;&#xA;inflate the jug in production. &#xA;But that cause would impair the taste of the water in &#xA;the fridge as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW why do You drink such botteled water? &#xA;What about the tap water in your town/city/village? &lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2015-02-01T07:21:29.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24450" PostTypeId="2" ParentId="24434" CreationDate="2015-02-01T07:35:58.727" Score="1" Body="&lt;p&gt;First &#xA;mistake you do is speaking of &quot;the radical chain reaction&quot;.&#xA;There are many reactions with different mechanisms which &#xA;are radical chains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re the cause: All reactions take place because there is &#xA;some energy (more specific: delta G) gain. Any step &#xA;in that radical chain has to fulfil this requirement.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditDate="2015-02-01T11:07:24.310" LastActivityDate="2015-02-01T11:07:24.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24452" PostTypeId="1" AcceptedAnswerId="24459" CreationDate="2015-02-01T09:11:51.860" Score="3" ViewCount="1251" Body="&lt;p&gt;The geometry of aluminium carbide is attached. Why isn't it completely linear (in accordance to VSEPR), but bent on the Al atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LreS5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-02-01T14:38:52.970" Title="Why is aluminium carbide bent?" Tags="&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24453" PostTypeId="1" AcceptedAnswerId="24454" CreationDate="2015-02-01T09:46:43.357" Score="0" ViewCount="298" Body="&lt;p&gt;What happens when we pass sulfur vapors through cold water? What is the reaction for it? Which type is colloid is this? And why doesn't it form in hot or lukewarm water&lt;/p&gt;&#xA;" OwnerUserId="12905" LastActivityDate="2015-02-01T10:52:21.467" Title="Formation of sulfur colloids" Tags="&lt;solutions&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24454" PostTypeId="2" ParentId="24453" CreationDate="2015-02-01T10:52:21.467" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens when we pass sulfur vapors through cold water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You get more or less coarse sulfur powder-slurry. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the reaction for it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sulfur vapor ( S2 mostly) reacts to form S8 &#xA;depending on speed of cooling, some S6 might &#xA;form temporarily&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which type is colloid is this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sometimes this might result in a &quot;colloid&quot;, &#xA;but I recommend not to use this outdated word, &#xA;because so much nonsense was written in the classical &#xA;days of &quot;colloids&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And why doesn't it form in hot or lukewarm water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Who says/writes this?&lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2015-02-01T10:52:21.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24456" PostTypeId="2" ParentId="24447" CreationDate="2015-02-01T12:25:13.897" Score="2" Body="&lt;p&gt;It does not make much of a difference. Both are short chain primary alkyl amines, so chemically much alike ($\ce{pKa} \approx 10.7$ due to the $\ce{-NH2}$ group). In what context do you want to know the answer to this question? That could help me direct my answer better.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-01T12:25:13.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24458" PostTypeId="1" CreationDate="2015-02-01T14:10:01.920" Score="1" ViewCount="180" Body="&lt;p&gt;When analyzing manganese-rich samples with X-ray powder diffraction, sample fluorescence becomes a significant effect.&#xA;Is this because the incident X-rays' energy - which, if Cu K$\alpha_1$, is approximately 8.048 keV - is getting close to the K$\alpha_1$ (approximate binding energy) of the inner-most electrons of Mn (5.899 keV)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample fluorescence from Fe and Co is even stronger, which fits with the trend mentioned above. So, is this indeed a trend? If not, how do I generalize the increase in probability for sample fluorescence?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2015-02-03T12:38:05.137" LastActivityDate="2015-03-05T13:35:37.013" Title="Does the probability of fluorescence increase when energy of incident X-rays match the electron's binding energy?" Tags="&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24459" PostTypeId="2" ParentId="24452" CreationDate="2015-02-01T14:14:11.393" Score="9" Body="&lt;p&gt;That structure is incorrect. Several online chemical distributors have similar structures posted in their entries for aluminum carbide; so does &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=24854497&quot; rel=&quot;noreferrer&quot;&gt;PubChem&lt;/a&gt;! Some sources show the structure as linear; some show it as bent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That structure basically assumes that aluminum forms primarily covalent bonds with an &quot;incomplete octet&quot; in VSEPR structures, rather like boron. But aluminum atoms are a lot larger and more metallic than boron atoms, and its compounds will have comparatively more ionic character. $\rm Al_4C_3$ isn't a molecular compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cotton &amp;amp; Wilkinson's &lt;em&gt;Advanced Inorganic Chemistry&lt;/em&gt; says that aluminum carbide &quot;reacts instantly with water to produce methane, and X-ray studies have shown it to contain discrete carbon atoms (C-C = 3.16 Å); for these reasons it is sometimes considered to be a &quot;methanide&quot;, that is, a salt containing $\rm C^{4-}$, but this is probably an oversimplification.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the structure that's posted on &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_carbide&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zeX7C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that &quot;Aluminium carbide has an unusual crystal structure that consists of two types of layers. It is based on AlC$_4$ tetrahedra of two types and thus two types of carbon atoms. One is surrounded by a deformed octahedron of 6 Al atoms at a distance of 217 pm. The other is surrounded by 4 Al atoms at 190–194 pm and a fifth Al atom at 221 pm.&quot; This is attributed to Greenwood's &lt;em&gt;Chemistry of the Elements&lt;/em&gt;, but I don't see it in the edition I have.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-01T14:38:52.970" LastActivityDate="2015-02-01T14:38:52.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24460" PostTypeId="1" AcceptedAnswerId="24469" CreationDate="2015-02-01T14:20:35.570" Score="7" ViewCount="44631" Body="&lt;p&gt;Suppose I open the drinks cabinet, and pour some of the strong stuff into a saucepan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the boiling point of ethanol is lower than water, so if I boil the mixture for a while, all the alcohol will surely evaporate long before the water does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;how quickly&lt;/em&gt; does the alcohol evaporate? Is it all gone by the time the water starts to boil? Or does some remain unless you boil the mixture for a long time? (How long?) Or maybe the alcohol is gone by the time the mixture starts to get slightly warm?&lt;/p&gt;&#xA;" OwnerUserId="8072" LastEditorUserId="13101" LastEditDate="2015-02-01T15:08:22.773" LastActivityDate="2017-10-26T21:19:54.413" Title="How quickly does ethanol evaporate?" Tags="&lt;everyday-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24461" PostTypeId="1" CreationDate="2015-02-01T14:27:08.850" Score="1" ViewCount="1308" Body="&lt;p&gt;Typically, when a cathode and anode are inserted into a solution that allows current to flow, the ions are moving with a certain kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ions were to gain kinetic energy (through transfer of thermal energy for example), will more current be able to flow, and will the potential difference increase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption for this is that increased ion kinetic energy will speed up the rate of ions being attracted to each electrode, thus causing more charge to arrive in a given period of time. As &lt;span class=&quot;math-container&quot;&gt;$I = Q/t$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; = current, &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; = charge and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; = time, would this mean an increased current?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I've performed a basic experiment in which I got a Tupperware container, stuck in some multimeter probes, and tested the difference in potential (voltage) for &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; (table salt) in solution when cold and hot, then for water cold and hot. I appreciate there are many issues with reliability, but the general results were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water cold = 0.082 V&lt;br&gt;&#xA;Water hot = 0.000 V (actually went into the negative then stabilised at 0)&lt;br&gt;&#xA;NaCl cold = 0.065 V&lt;br&gt;&#xA;NaCl hot = 0.130 V  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my real experiment involved &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; being titrated into &lt;span class=&quot;math-container&quot;&gt;$\ce{KCl}$&lt;/span&gt; (mixed with deionised water and a little &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt; to acidify) – assuming the temperature remained constant, what effect could increasing the ions' velocity have on conductivity, without heating it up (increasing kinetic energy) to alter viscosity?&lt;/p&gt;&#xA;" OwnerUserId="13101" LastEditorDisplayName="user7951" LastEditDate="2019-07-14T12:59:06.847" LastActivityDate="2019-07-14T12:59:06.847" Title="Does the speed of an ion affect the current that can flow through a solution?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;ions&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24462" PostTypeId="1" AcceptedAnswerId="24466" CreationDate="2015-02-01T14:37:15.477" Score="1" ViewCount="3473" Body="&lt;p&gt;For the reaction: &#xA;$$2A\to p$$&#xA;The rate, $v$ may be given as:&#xA;$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$&#xA;Correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating this gives:&#xA;$$-\frac12\int_{[A]_0}^{[A]}\frac{d[A]}{[A]^2}=k\int_0^tdt$$&#xA;$$\frac1{2[A]}-\frac1{2[A]_0}=kt$$&#xA;Rearranging...&#xA;$$\frac1{[A]}=\frac1{[A]_0}+2kt$$&#xA;Now substituting $t=t_{1/2}$ and $[A]=\frac{[A]_0}2$...&#xA;$$\frac2{[A]_0}=\frac1{[A]_0}+2kt_{1/2}$$&#xA;Minus $1/[A]_0$ from each side...&#xA;$$\frac1{[A]_0}=2kt_{1/2}$$&#xA;This then gives:&#xA;$$t_{1/2}=\frac1{2k[A]_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have seen in textbooks that this should be written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{1/2}=\frac1{k[A]_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I gone wrong somewhere? If so where?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know which equation is correct because when figuring out the rate constant, would the gradient be equal to $\frac{1}{2k}$ or $\frac{1}{k}$?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2015-02-01T14:47:42.910" LastActivityDate="2015-02-01T15:00:20.557" Title="Second order reaction half life equation?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24463" PostTypeId="2" ParentId="4554" CreationDate="2015-02-01T14:38:08.133" Score="9" Body="&lt;p&gt;As you go down any group, there are extra electron shells (for example Lithium has 2 shells, Francium has 7) which cause electron shielding. Because the force of attraction must extend further and through these shells to the outermost shell, the force becomes less so there is less strong bonding occuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it is easier to break these less strong bonds, and because the melting point is a measure of the point at which all the bonds holding a metallic structure together are broken, it will decrease because this point is reached a lot sooner as the bonds are not as strong.&lt;/p&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-02-01T14:38:08.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24464" PostTypeId="2" ParentId="24462" CreationDate="2015-02-01T14:54:18.160" Score="0" Body="&lt;p&gt;This is purely a mathematical answer, but shouldn't the integration produce &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/[A] - 1/[A]0 = kt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means you don't have your 2kt being taken through your working, and gives you the same as the textbooks' answer(s)?&lt;/p&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2015-02-01T14:54:18.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24465" PostTypeId="1" CreationDate="2015-02-01T14:54:59.953" Score="2" ViewCount="388" Body="&lt;p&gt;Pulse-height discrimination is a common technique to filter out unwanted background signals from the analytical signals. But how exactly does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analytical signals are not continuously detected, but detected rather in pulses. This is because the detector has a &quot;dead time&quot;, and there is a pile-up of ion-pairs (or whatever the principle of detection is) in this dead-time. The intensity of the built up signal - the pulse - is then detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulse-height discrimination is just that; a discrimination of the detected pulse-heights. However, to be able to discriminate the pulse-heights, surely the system must know what a &quot;desirable&quot; pulse-height is, or, in other words, when a detected pulse deviates from the desirable? How exactly is this done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of X-ray diffraction, we know the desirable energy of the detected radiation: in theory, this energy should be equal to the K$\alpha_1$ of the X-ray tube's anode material (since diffraction is an elastic interaction, and hence no energy lose occurs). However, the intensity of the signal has nothing to do with the energy. How do we (that is, the ones developing the analytical system) know which pulse-heights are relevant and which are irrelevant to our analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope have presented my question clearly, and I appreciate all help. Finding free references online was not very easy, in my experience.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2015-02-03T12:37:23.663" LastActivityDate="2016-03-23T04:29:40.140" Title="The Workings of Pulse-Height Discrimination" Tags="&lt;physical-chemistry&gt;&lt;analytical-chemistry&gt;&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24466" PostTypeId="2" ParentId="24462" CreationDate="2015-02-01T15:00:20.557" Score="6" Body="&lt;p&gt;What you did isn't wrong, but instead of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=-\frac{d[A]}{dt}=k[A]^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is also true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twice a constant is still as constant (a different constant of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-01T15:00:20.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24467" PostTypeId="2" ParentId="24434" CreationDate="2015-02-01T15:03:22.770" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why does the halogen actually abstract hydrogen from the alkane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Look at the two tables of &lt;strong&gt;bond energies&lt;/strong&gt; on this &lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;webpage&lt;/a&gt;.  The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds.  In the top table we see that the H-Cl bond strength is 103 kcal/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$&#xA;$$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$&#xA;$$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the top row in the bottom table again.  As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As I understand, the first propagation step in the free-radical&#xA;  monochlorination of methane is slightly endothermic but is driven to&#xA;  completion by the exothermic nature of the second propagation step&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-H + Cl{.} -&amp;gt; CH3{.} + H-Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3{.} + Cl-Cl -&amp;gt; CH3-Cl + Cl{.}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These steps are consistent with my earlier argument.  You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate &quot;back of the envelope&quot; calculation and view of what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species.  It is reactive because it doesn't have an octet of electrons.  By undergoing the above-reactions it can regain its octet and become a stable entity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram).  Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products.  So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rSp8r.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-01T20:23:46.683" LastActivityDate="2015-02-01T20:23:46.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24468" PostTypeId="1" AcceptedAnswerId="24474" CreationDate="2015-02-01T15:27:20.543" Score="0" ViewCount="167" Body="&lt;p&gt;I am having trouble wit the following question on my problem sheet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The reaction $\ce{Br- + ClO^- \to BrO^- + Cl^-}$ occurs with a rate constant of $\text{0.35 dm}^3 \text{mol}^{-1}&#xA;\text{s}^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at&#xA;298 K. For the following initial conditions calculate the time taken for the Br^-&#xA;concentration to fall to half its original value:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) $[\ce{Br-}]=[\ce{ClO-}]=0.0017$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) $[\ce{Br-}]=0.0017$ and $[\ce{ClO-}]=0.084$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure I could do this question if I knew the rate equation but it isn't given. From the units of the rate constant the reaction must be second order but how am I supposed to know whether it's second order with respect to one of the reactants (and thus zeroth order w.r.t the other) or first order with respect to both? Also, what's the significance of the concentrations being equal (there's another question that stresses that the concentrations of both the reactants are equal). &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="12660" LastEditDate="2015-02-01T18:26:00.873" LastActivityDate="2015-02-01T22:50:03.210" Title="Predicting rate equations from data" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24469" PostTypeId="2" ParentId="24460" CreationDate="2015-02-01T15:38:43.130" Score="8" Body="&lt;p&gt;Certainly water and ethanol and all mixtures thereof will completely evaporate without boiling given enough time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water and ethanol form a postive &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;noreferrer&quot;&gt;azeotrope&lt;/a&gt; that is 95.6% ethanol and 4.4% water and boils at 78.2 degrees C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you boil a mixture of water-ethanol that has less than 95.6% ethanol, the concentration of water in the liquid phase will increase, but it will only asymptotically approach pure water as all the liquid boils away.  There will not be a time when all the ethanol is gone and you are left with a finite amount of pure water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you boil a mixture of water-ethanol that has more than 95.6% ethanol, the concentration of ethanol is the liquid phase will increase.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-01T15:38:43.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24470" PostTypeId="1" AcceptedAnswerId="24491" CreationDate="2015-02-01T16:48:53.810" Score="2" ViewCount="1384" Body="&lt;p&gt;Let's say $\ce{Fe}$ reacts with $\ce{Cu^{2+}}$ ions. $\ce{Fe}$ would oxidize and therefore give electrons to $\ce{Cu^{2+}}$so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe-&amp;gt; Fe^{2+}  +2e-}$$&#xA;$$\ce{Cu^{2+} +2e^- -&amp;gt;Cu}$$&#xA;The overall reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu^{2+} +Fe -&amp;gt;Cu +Fe^{2+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is an example of corrosion, right? And when this type of corrosion happens, the anode is the electrode where oxidation happens and the cathode is the electrode where reduction happens, right?&#xA;Therefore iron ($\ce{Fe}$) is the anode where oxidation happens and Copper 2+ ions ($\ce{Cu^{2+}}$) is the cathode where the reduction happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is this all correct? Have I successfully described corrosion (of this type) and given an example of it? Or is there something wrong? I'm 15 and soon I'm having a test in chemistry, and because of my age there's no point in making it &quot;super-advanced&quot;, but if there's something I missed or did wrong please point it out, I would be grateful for that!&lt;/p&gt;&#xA;" OwnerUserId="12678" LastEditorUserId="7581" LastEditDate="2015-02-01T21:51:46.680" LastActivityDate="2015-02-01T21:51:46.680" Title="The anode and cathode when corrosion happens" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24471" PostTypeId="2" ParentId="24444" CreationDate="2015-02-01T16:49:23.037" Score="3" Body="&lt;p&gt;This should've been a comment as it only backs up Georg's answer, but it just got too long for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastic doesn't &quot;leak&quot;, scientifically. However, it can go through &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymer_degradation&quot; rel=&quot;nofollow&quot;&gt;the degradation process&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't provide enough info for anyone to decide, but if you're certain that &lt;em&gt;was&lt;/em&gt; the &quot;plastic&quot; taste (I don't have much idea how you do) it's possible it may have gone through degradation, though I can tell you 95 percent of the time it isn't. And that's simply because if the plastics were too unstable to resist minor chemical or energetic &quot;attacks&quot; (read the wikipedia article for degradation to see what I mean), they wouldn't have produced it like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the urban hoax that doing stuff like pressurizing a mineral-water bottle by freezing the water inside releases dioxins that are carcinogenic and which cause the change of the taste of water, but this, is &lt;strong&gt;not&lt;/strong&gt; true. (See &lt;a href=&quot;http://cancer.about.com/od/cancercausesfaq/f/freezingwaterbo.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg's answer pretty much sums it up, but I think it's better to add: It may also be possible that you (or anyone else, or mistakenly the company itself) put the &quot;tasty&quot; water in contact with soluble mineral salts. Thus, you drink &lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_water&quot; rel=&quot;nofollow&quot;&gt;hard water&lt;/a&gt; and swear for its taste!&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-01T16:49:23.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24472" PostTypeId="1" CreationDate="2015-02-01T17:14:13.910" Score="2" ViewCount="283" Body="&lt;p&gt;Electrons that fill orbitals always do so in such a way that the resulting structure has the lowest energy state possible, though there are anomalies like chromium and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ionization no matter which element tends to violate this rule. When electrons are ionized they are from the outermost (valence) shell. But this doesn't always produce the lowest energy state possible like the case with 3d and 4s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, all I am asking is why in that specific case, is the electron taken out of the 4s orbital first instead of the 3d orbital, since the 3d orbital is a higher energy level?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="7448" LastEditDate="2015-02-01T18:38:40.790" LastActivityDate="2015-02-01T18:38:40.790" Title="Why does ionization violate the stable or lowest energy rule in atoms?" Tags="&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24473" PostTypeId="2" ParentId="24461" CreationDate="2015-02-01T17:14:52.193" Score="1" Body="&lt;p&gt;There is a &lt;em&gt;strong&lt;/em&gt; dependence on temperature in conductivity of ionic solutions; see &lt;a href=&quot;http://www.eutechinst.com/techtips/tech-tips44.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.eutechinst.com/techtips/tech-tips44.htm&lt;/a&gt;, for example. However, thermal energy, i.e. molecular motion, is random, and does not favor current flow in the forward (or backward) direction. There are a number of factors that temperature influences:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Viscosity usually changes  inversely with temperature, so ions move more freely at higher temperature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ionic solutions have other species than OH- and H+; for example OH3+ [water + hydroxyl radical] is the &lt;em&gt;hydronium ion&lt;/em&gt;; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydronium&lt;/a&gt;. The amount of each species varies with temperature, as shown in that reference. See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Self-ionization_of_water&lt;/a&gt;. Depending on mass, hydrogen bonding etc. different species have different conductivity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-02-01T21:21:28.243" LastActivityDate="2015-02-01T21:21:28.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24474" PostTypeId="2" ParentId="24468" CreationDate="2015-02-01T17:17:15.487" Score="1" Body="&lt;p&gt;You have to assume it is first order in each reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Zero-Order_Reactions&quot; rel=&quot;nofollow&quot;&gt;Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction can't really be zero-order in a reactant, otherwise it would imply the reaction continues after there is no reactant remaining.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-01T17:17:15.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24475" PostTypeId="1" CreationDate="2015-02-01T17:20:22.857" Score="3" ViewCount="25759" Body="&lt;p&gt;This question stems from one I just asked over at &lt;a href=&quot;https://cooking.stackexchange.com/questions/54259/malva-pudding-recipe-containing-too-much-bicarb-and-vinegar&quot;&gt;Seasoned Advice&lt;/a&gt;, where my recipe tasted decidedly off because the pH was wrong.&lt;/p&gt;&#xA;&lt;p&gt;The recipe called for &lt;span class=&quot;math-container&quot;&gt;$\pu{16 g}$&lt;/span&gt; bicarb of soda (sodium bicarbonate, &lt;span class=&quot;math-container&quot;&gt;$\ce{NaHCO3}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\pu{15 g}$&lt;/span&gt; of vinegar (mine was 5% acidity, and I assumed a density of &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g ml-1}$&lt;/span&gt;). I just googled the corresponding pHs of these two, and apparently vinegar is more acidic than bicarb is basic: vinegar has a pH of 2.4–3.4 depending on concentration, and bicarb is pH 9.5.&lt;/p&gt;&#xA;&lt;p&gt;How can I calculate whether I put too much vinegar or too much bicarb into my recipe? The ratio in the recipe is almost &lt;span class=&quot;math-container&quot;&gt;$1:1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What ratio in grams do I need of bicarb vs vinegar in order for the result to be neutral? I know there are other contributing factors to the pH in the recipe but I just want to focus on these two first. Assuming the vinegar is 5% which mine was.&lt;/p&gt;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&lt;p&gt;So I found the explanation for the chemical reaction between bicarb and vinegar &lt;a href=&quot;https://answers.yahoo.com/question/?qid=20110509181311AAD4F2Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Relevant bit:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\pu{1 g}\;\ce{NaHCO3}}{\pu{84.0068 g mol-1}} \times \frac{1}{1} \times \pu{60.0522 g mol-1} = \pu{0.715 g}\;\ce{CH3COOH}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What this formula tells you is that every gram of baking soda requires \pu{0.715 g} of acetic acid. But the remaining problem is that vinegar is not pure acetic acid.  In fact, most vinegars vary from 4% to 18%.&lt;/p&gt;&#xA;&lt;p&gt;Using those numbers, you would need somewhere between &lt;span class=&quot;math-container&quot;&gt;$\pu{0.715 g}/0.04 = \pu{17.9 g}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.715 g}/0.18 = \pu{3.97 g}$&lt;/span&gt; of vinegar, for each gram of baking soda.&lt;/p&gt;&#xA;&lt;p&gt;If you're lucky you can get some vinegar with the percent acetic acid listed on the label, then you can substitute that number into the equation above to determine how much vinegar you'll need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So if I plug in my 5%, I get &lt;span class=&quot;math-container&quot;&gt;$\pu{0.715 g}/0.05 = \pu{14.3 g}$&lt;/span&gt; vinegar for &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g}$&lt;/span&gt; bicarb.&lt;/p&gt;&#xA;&lt;p&gt;Yikes! This is way more than &lt;span class=&quot;math-container&quot;&gt;$1:1$&lt;/span&gt;. But I'm not sure if it answers my question. Is the result, &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COONa + CO2 + H2O}$&lt;/span&gt;, pH neutral? If yes, then my vinegar does hardly anything to cancel out the base, is that right?&lt;/p&gt;&#xA;&lt;p&gt;Edit 2:&lt;/p&gt;&#xA;&lt;p&gt;Okay so according to Wikipedia, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_acetate&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COONa}$&lt;/span&gt; is a base&lt;/a&gt;. So even though my reaction formula shows what happens when all the bicarb/vinegar is used up, the result is not neutral (how much does the water and carbon dioxide contribute to bringing the pH down?). I need more acid to make my sodium acetate neutral (or at least neutral enough to be palatable?). This means that the needed ratio is higher than 14:1!? So almost certainly way too much bicarb?&lt;/p&gt;&#xA;" OwnerDisplayName="user13121" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-09-14T14:41:38.937" Title="What ratio of bicarb to vinegar do I need in order for the result to be pH neutral?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;ph&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24477" PostTypeId="2" ParentId="24472" CreationDate="2015-02-01T18:36:49.843" Score="3" Body="&lt;p&gt;Textbooks won't bother to dig into details. Most of them are satisfied with saying that such thing exists. They're right I guess. I'll do a &lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemguide&lt;/a&gt; approach to the answer since I find it most understandable.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/otvne.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;We draw this diagram, and then assume that it works for all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.&lt;/p&gt;&#xA;&lt;p&gt;If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, you are adding extra protons to the nucleus, and extra electrons around the nucleus.&lt;/p&gt;&#xA;&lt;p&gt;The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Going through the Aufbau principle, when you pass Ca (Z = 20) you will face Sc, the first element in which a d orbital is going to be filled. But, you must make a really important change in your train of thoughts.&lt;/p&gt;&#xA;&lt;p&gt;As chemguide explained, with change in the number of protons and electrons due to the change of Z, the way their orbitals' energies are defined changes. Till Ca, the behavior insisted that 4s has a lower energy state than 3d and thus, fills sooner. But, starting from Sc, the pattern changes: &lt;strong&gt;In Scandium, 3d has a lower energy than 4s. The behavior changed&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This change in behavior results in the current confusion of yours. This much answers the problem you have, and you can visit the page at chemguide to learn more. If you want more info, I can freely add. Hope I've helped!&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-02-01T18:36:49.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24478" PostTypeId="1" CreationDate="2015-02-01T18:37:06.047" Score="0" ViewCount="250" Body="&lt;p&gt;For the syntheses of organohalides from alcohols, what would plausible reaction mechanisms be for the three following reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{R-OH + NaBr + H2SO4 -&amp;gt; R-Br + NaHSO4 + H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{R-OH + PCl5 -&amp;gt; R-Cl + POCl3 + HCl}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{R-OH -&amp;gt;[red P][Br2 or I2] R-X}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="194" LastEditDate="2017-04-08T01:35:18.583" LastActivityDate="2017-04-08T01:35:18.583" Title="Reaction mechanisms for organohalides from alcohols" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24479" PostTypeId="2" ParentId="24447" CreationDate="2015-02-01T18:46:17.913" Score="2" Body="&lt;p&gt;Different sources quote different values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia: Methylamine $\mathrm{pK_a~= 10.64}$ and Propylamine $\mathrm{pK_a~= 10.71}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pubchem: Methylamine $\mathrm{pK_a~= 10.6}$ and Propylamine $\mathrm{pK_a~= 10.71}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{pK_a}$ is also temperature dependent and this effect may be larger than the difference in the inductive effect between the two amines in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zirchrom.com/organic.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.zirchrom.com/organic.htm&lt;/a&gt;: Methylamine $\mathrm{pK_a~= 10.63~, 25^\circ C}$ and Propylamine $\mathrm{pK_a~= 10.60~, 20^\circ C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall there seems to be a consensus that propylamine is slightly more basic than methylamine at the same temperature but there is very little difference. Theoretically this can be explained by the slightly greater inductive effect of the propyl group versus the methyl group. In an experimental context I suspect that there is virtually no difference between the behaviours of the two amines.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-01T18:46:17.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24480" PostTypeId="1" CreationDate="2015-02-01T20:20:35.820" Score="0" ViewCount="809" Body="&lt;p&gt;If I am not mistaken, distilling water does not remove all possible impurities. Chemicals with boiling points less than $100\ \mathrm{^\circ C}$ will be transferred as well. Does the usual tap water contain such (I think there is at least fluoride.)? I read that activated carbon can be used to purify water as well. How is it done in practice? Suppose I first use activated carbon and then distill the water (or in the reverse order). How much impurities there would be still left?&lt;/p&gt;&#xA;" OwnerUserId="13122" LastEditorUserId="9961" LastEditDate="2016-08-03T22:15:32.730" LastActivityDate="2016-08-03T22:15:32.730" Title="How to purify water with activated carbon?" Tags="&lt;everyday-chemistry&gt;&lt;purification&gt;&lt;distillation&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24481" PostTypeId="2" ParentId="24478" CreationDate="2015-02-01T20:30:14.610" Score="1" Body="&lt;p&gt;Nucleophilic substitutions, regardless whether proceeding via $S_N1$ or $S_N2$ require a decent leaving group. $\ce{HO-}$, however, is a lousy leaving group! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To overcome this, several methods exist:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Perform the reaction under acidic conditions, where the alkanol is protonated. Your leaving group now is $\ce{H2O}$. In addition, the corresponding anion of sulfuric acid as a bad nucleophile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;React the alkanol with an acid chloride to form an ester and replace the ester group with a halide. Consider $\ce{PCl5}$ to be an acid chloride of phosphoric acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Form the acid halide &lt;em&gt;in situ&lt;/em&gt; by reaction of phosphorous and iodine.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-01T20:30:14.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24482" PostTypeId="2" ParentId="24475" CreationDate="2015-02-01T20:32:36.960" Score="4" Body="&lt;p&gt;Your calculation is correct (14.3 grams of 5% acetic acid completely reacts with 1 g sodium bicarbonate). But you're not interested really in &quot;completely reacts&quot;; you want something like pH 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've got it just about right with your mix; you're at the endpoint, and this is a point where the pH will drop very rapidly with even a small addition of vinegar. So it would only take a few additional drops to make the pH go down to 7... but that may not be what you really want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/books/NBK50958/&quot; rel=&quot;nofollow&quot;&gt;Sodium acetate has a mild salty taste by itself&lt;/a&gt;. So I'm not sure adding extra vinegar will be worthwhile, it'll make it a bit more vinegary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I'd do: Mix vinegar and baking soda in the ratio you've figured out, and then see how that tastes. Make adjustments at that point according to taste, rather than by &quot;titrating&quot; to an arbitrary pH.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-01T22:15:57.567" LastActivityDate="2015-02-01T22:15:57.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24483" PostTypeId="1" AcceptedAnswerId="24494" CreationDate="2015-02-01T20:45:42.687" Score="-2" ViewCount="1230" Body="&lt;p&gt;I already know what catalysts do (lower activation energy requirement for a reaction), but how exactly are they able to do that on a molecular level?&lt;/p&gt;&#xA;" OwnerUserId="13123" LastEditorUserId="402" LastEditDate="2017-04-01T02:04:44.617" LastActivityDate="2017-04-01T02:04:44.617" Title="On the molecular level, how do catalysts work?" Tags="&lt;reaction-mechanism&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-06T02:48:47.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="24484" PostTypeId="2" ParentId="24480" CreationDate="2015-02-01T20:56:24.370" Score="3" Body="&lt;p&gt;The main reason activated carbon is used in water treatment is that its large surface area is useful for adsorbing volatile organic contaminants and chlorine. It is not effective at removing inorganic ions (dissolved minerals, including fluoride), but these are removed quite well by distillation. I think distillation first would make more sense—distillation will remove less volatile contaminants so that less has to be adsorbed by the carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tap water has to meet certain standards (that depend on your area) for dissolved minerals and volatile organic compounds, so depending on the end use of the water, removing the small amount of remaining contaminants may not be necessary. Formerly, double-distilled water was the standard for making highly purified water, but certain applications are more or less tolerant of certain types of contaminants, and double-distillation requires a lot of energy, so it has largely been replaced for many purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consumer water filters usually contain filters to remove particulates, ion exchange resin to remove inorganic ions, and activated carbon to improve the taste of the water, and as far as purification for laboratory use, combinations of filtration, deionization, reverse osmosis, ultrafiltration, etc. are common these days. Different processes are better at removing some types of contaminants than others, so combinations are used. Distillation of tap water followed by treatment by activated carbon will produce quite pure water by consumer standards, but might not be pure enough for say semiconductor microfabrication or biomedical applications.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-01T20:56:24.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24486" PostTypeId="1" AcceptedAnswerId="24492" CreationDate="2015-02-01T21:15:05.457" Score="20" ViewCount="10201" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Beryllium has other fancy applications. It is transparent to x-rays, so it's used in the windows of x-ray tubes, which need to be strong enough to hold a perfect vacuum, yet thin enough to let the delicate x-rays out. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The quote is from the book &lt;em&gt;The Elements&lt;/em&gt; by Theodore Gray. Why is beryllium transparent to x-rays?&lt;/p&gt;&#xA;" OwnerUserId="13122" LastActivityDate="2017-08-08T08:07:54.720" Title="Why is beryllium transparent to x-rays?" Tags="&lt;physical-chemistry&gt;&lt;elements&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24488" PostTypeId="2" ParentId="15091" CreationDate="2015-02-01T21:19:51.513" Score="2" Body="&lt;p&gt;The following is the order from lowest boiling point to highest based on the types of forces these compounds have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ - dispersion forces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3Br}$ - dispersion and dipole-dipole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3OH}$ - dispersion, dipole-dipole, and hydrogen bonding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{RbF}$ - ionic&lt;/p&gt;&#xA;" OwnerUserId="13125" LastEditorDisplayName="user10259" LastEditDate="2015-05-21T23:19:31.080" LastActivityDate="2015-05-21T23:19:31.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24489" PostTypeId="1" CreationDate="2015-02-01T21:26:14.733" Score="0" ViewCount="1465" Body="&lt;p&gt;what is the difference between 'normal' food additive maltodextrin and &quot;organic maltodextrin&quot;? Is it only the primary source (i.e.: corn, wheat etc) that is organic or is it a different process to create maltodextrin? And is maltodextrin the same as food additive 1405? &lt;/p&gt;&#xA;" OwnerUserId="13126" LastActivityDate="2015-02-01T22:39:01.257" Title="food chemistry maltodextrin vs organic maltodextrin" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24491" PostTypeId="2" ParentId="24470" CreationDate="2015-02-01T21:42:44.033" Score="2" Body="&lt;p&gt;You are right! What you have written is an example of Galvanic corrosion (Galvanic corrosion occurs when two different metals have physical or electrical contact with each other and are immersed in a common electrolyte).&#xA;You have  successfully described Galvanic corrosion and given an example of it&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-01T21:42:44.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24492" PostTypeId="2" ParentId="24486" CreationDate="2015-02-01T21:44:46.923" Score="16" Body="&lt;p&gt;The amount of X-rays absorbed by an element depends on the size of its atoms (its absorption cross-section, specifically, as affected by the size of core orbitals that contain electrons that can be excited by X-ray absorption, and the number of electrons in those orbitals) and how many atoms are packed into a given volume. Big atoms that are close together can absorb more X-rays than small atoms that are far apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beryllium has very small atoms (it has a very small electronic core, with only two electrons in it), and very low density, so it will absorb relatively few X-rays compared to other materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://pd.chem.ucl.ac.uk/pdnn/diff2/abs.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, 37 % of X-rays with a wavelength of 0.56 nm will be absorbed by beryllium window that is 23.5 mm thick. If the window were made of aluminum, it would only have to be 1.4 mm thick to absorb the same amount of X-rays, and if it were made of lead it would only have to be 13 μm thick.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorDisplayName="user7951" LastEditDate="2016-11-15T19:09:59.887" LastActivityDate="2016-11-15T19:09:59.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24493" PostTypeId="2" ParentId="24489" CreationDate="2015-02-01T21:45:22.907" Score="1" Body="&lt;p&gt;&quot;Food additive 1405&quot; refers to the process for the modification of starch; in this case the process uses enzymes to break up the starch molecules into oligosaccharides between 3 and 20 saccharides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between regular maltodextrin and 'organic maltodextrin' would be that the source was grown without synthetic pesticides and synthetic fertilizers. 'Organic maltodextrin' may also need to use natural enzymes.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-02-01T22:39:01.257" LastActivityDate="2015-02-01T22:39:01.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24494" PostTypeId="2" ParentId="24483" CreationDate="2015-02-01T22:17:49.660" Score="3" Body="&lt;p&gt;In general terms, some catalysts serve as a &quot;meeting place&quot; where the reacting species can get together more easily, e.g. $\ce{H2}$ and $\ce{O2}$ adsorbed onto a $\ce{Pt}$ catalyst; see &lt;a href=&quot;https://www.chem.umn.edu/services/lecturedemo/info/Hydrogen_and_Platinum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others may actually participate in a reaction but are regenerated afterwards, e.g. according to &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1958/tf/tf9585401861&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maggs and Sutton&lt;/a&gt;, &quot;argentous hydroperoxide&quot; is produced as an intermediate in the catalytic decomposition of $\ce{H2O2}$ by $\ce{Ag}$, and regenerates the silver ions as it decomposes.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="31313" LastEditDate="2017-04-01T01:45:19.157" LastActivityDate="2017-04-01T01:45:19.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24495" PostTypeId="1" CreationDate="2015-02-01T22:24:37.633" Score="1" ViewCount="176" Body="&lt;p&gt;Can anyone explain the difference in halogen bonding to me. I understand the explanation in (c). It appears to me, that accepting that would contradict the answers to parts (a), (b), and (c).&lt;br&gt;&#xA;Is the reason for part (a) because the first molecule does not have resonance and for (b) the $\ce{F3C-I}$ molecule has a larger inductive effect? If so, that would not make sense for (c) to be correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQKMW.jpg&quot; alt=&quot;Halogen Bond Strength&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12664" LastEditorUserId="4945" LastEditDate="2015-02-02T06:19:37.367" LastActivityDate="2015-02-16T04:58:45.423" Title="How to rationalise the difference in halogen bonding?" Tags="&lt;homework&gt;&lt;bond&gt;&lt;electronegativity&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24496" PostTypeId="2" ParentId="24468" CreationDate="2015-02-01T22:39:56.293" Score="1" Body="&lt;p&gt;Being the concentrations equal makes it easier to integrate the second order law (first in each reagents). Concentrations are the same at every t and therefore you have only one variable to take into account. The resulting integrated law will be analogous to that for a second order kinetics in one reagent. &lt;/p&gt;&#xA;" OwnerUserId="13103" LastEditorUserId="13103" LastEditDate="2015-02-01T22:50:03.210" LastActivityDate="2015-02-01T22:50:03.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24497" PostTypeId="2" ParentId="24446" CreationDate="2015-02-01T22:51:48.837" Score="2" Body="&lt;p&gt;The $K_\mathrm{sp}$ of  $\ce{Ba(OH)2}$ is $0.005$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the solubility is great enough that 0.055M  $\ce{Ba(OH)2}$ can all be in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, you are diluting the $0.055~\mathrm{M}$ solution, so the concentration will be even less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, even if there was too much  $\ce{Ba(OH)2}$ to dissolve in water, the $\ce{HCl}$ would react with the $\ce{OH-}$ that is in solution, causing more  $\ce{Ba(OH)2}$ to dissolve.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2015-02-02T04:12:08.883" LastActivityDate="2015-02-02T04:12:08.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24498" PostTypeId="2" ParentId="24275" CreationDate="2015-02-01T23:05:27.053" Score="1" Body="&lt;p&gt;P.  W.  Atkins says that it is preferable to use the term &lt;strong&gt;perfect&lt;/strong&gt; gas because the particles show no interactions. On the opposite, &lt;strong&gt;ideal&lt;/strong&gt; solutions are constituted by different particles that show the same kind of interactions among them (the toluene/benzene mixture is a good approximation). &lt;/p&gt;&#xA;" OwnerUserId="13103" LastActivityDate="2015-02-01T23:05:27.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24499" PostTypeId="1" CreationDate="2015-02-01T23:23:17.727" Score="7" ViewCount="2446" Body="&lt;p&gt;In other words, a field of chemistry that is all theoretical? I have heard of theoretical chemistry, but my chemistry professor said you can't really major in theoretical chemistry. He said the closest thing is physical chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like predicting in Chemistry, but I hate labs or experiments.&lt;/p&gt;&#xA;" OwnerUserId="13128" LastActivityDate="2018-09-24T12:11:37.313" Title="Is there a field of chemistry that doesn't do experiments?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24500" PostTypeId="1" AcceptedAnswerId="24501" CreationDate="2015-01-30T15:58:23.570" Score="10" ViewCount="2180" Body="&lt;p&gt;The following links claim that inventor Arturo Solis Herrera has created an &quot;everlasting battery&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mexiconewsnetwork.com/news/bat-gen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mexiconewsnetwork.com/news/bat-gen/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.faqs.org/patents/app/20090134007#ixzz10Sh9iLQM&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.faqs.org/patents/app/20090134007#ixzz10Sh9iLQM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The article claims that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Bat gen&lt;/strong&gt; is the name of this ever-lasting battery. It consists of a biochemical process triggered by mixing water and melanin, which results in a substance capable of separating oxygen from hydrogen (the components of a water molecule), therefore liberating energy.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The process continues as the same molecule brings together both elements turning them into water again; as a result one more energy load is triggered.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;According to Solis’ research, once this process is reached, it can continue for 100 years! Moreover, there are several ways of artificially producing the melanin found in human nails, hair and retina; two of them, which he already patented, are based on vegetables and oil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although they do not claim that the battery is perpetual, they seem to imply that mixing water and melanin can give an (oscillating?) reaction that can last for 100 years, providing cheap power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have any chemical basis? What matter is consumed?&lt;br&gt;&#xA;Would that work without the user needing to add anything to the battery or is the article just describing the process in a misleading way?&lt;/p&gt;&#xA;" OwnerUserId="13134" OwnerDisplayName="mplungjan" LastEditorUserId="13134" LastEditDate="2019-03-30T07:31:59.083" LastActivityDate="2019-03-30T07:31:59.083" Title="Does the &quot;Bat Gen&quot; claim to produce more energy than it uses?" Tags="&lt;energy&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24501" PostTypeId="2" ParentId="24500" CreationDate="2015-02-01T23:58:19.950" Score="5" Body="&lt;p&gt;I'm not sufficiently familiar with the chemistry involved to state authoritatively that this guy is full of bologna, but there are some pretty fantastic claims made in the second link.  Also, if that's a US patent application, it's a disaster.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The device is able to use any light from 900-200nm, mechanical agitation, sound, or magnetic fields (and unspecified &quot;other&quot;) as sources of power.  &lt;/li&gt;&#xA;&lt;li&gt;He has also made all of his cells inside Tupperware containers sealed by silicone caulking, so the 100 year claim is a bit sketchy.&lt;/li&gt;&#xA;&lt;li&gt;The inventor's testing methodology is pretty dubious as well.  The device was placed in sunlight for 30 minutes and it experienced a change of pH from 7.3 to 7.1, therefore it works.  Also, in his own words:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is worth saying that such variation from 0.2 to 1.0 pH unit; and the reverse reaction that occurred once the containers were placed in a dark place, was anticipated by our theoric background, &lt;strong&gt;when we performed the experiments, we known already what was going to happen, as the matter of fact, we didn't do many experiments, we only made two or three times, obtaining the results we expected.&lt;/strong&gt;  {All grammar and spelling mistakes are those of the original author(s).}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That said, the idea of a cell that can catalytically break down water into hydrogen and oxygen using sunlight and then recombine them into water, extracting the energy, is not impossible by any stretch of the imagination.  Designing a cell that is economically viable compared with silicon based devices has yet to be accomplished, to the best of my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2015-02-02T00:09:21.447" LastActivityDate="2015-02-02T00:09:21.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24506" PostTypeId="1" AcceptedAnswerId="24522" CreationDate="2015-02-02T05:19:34.837" Score="4" ViewCount="600" Body="&lt;p&gt;This question has been bothering me for some time, and I can't seem to find a good answer online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have four chemical species $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and these four react in the following ways: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}\ce{&#xA;A + A &amp;amp;-&amp;gt; B\\&#xA;A + B &amp;amp;-&amp;gt; C\\&#xA;A + C &amp;amp;-&amp;gt; D\\&#xA;A + D &amp;amp;-&amp;gt; B + C\\&#xA;}\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic reaction equations for these four species should be:&#xA;\begin{align}&#xA;\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &amp;amp;= -k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &amp;amp;= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &amp;amp;= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &amp;amp;= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k_i$ is the Arrhenius coefficient for that reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my mind, the stoichiometry does not seem to add up right. It seems that when the system undergoes $\ce{A + A -&amp;gt; B}$, the concentration of $\ce{A}$ should decrease by twice as much as a reaction like $\ce{A + B -&amp;gt; C}$. Similarly, the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, stoichiometrically, I want to write: &#xA;\begin{align}&#xA;\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &amp;amp;= -2k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &amp;amp;= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &amp;amp;= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &amp;amp;= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this wrong-headed? What conceptual issue am I missing?  &lt;/p&gt;&#xA;" OwnerUserId="13132" LastEditorUserId="4945" LastEditDate="2015-02-02T06:34:25.980" LastActivityDate="2015-02-03T12:16:47.230" Title="Reaction Kinetics and Stoichiometry -- Mass conservation" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24507" PostTypeId="1" CreationDate="2015-02-02T06:03:45.580" Score="28" ViewCount="7282" Body="&lt;p&gt;This question was asked recently in an interview and I just said that &quot;Yeah, like if the reaction involves ions or paramagnetic species&quot;. But the interviewer went on and asked me to elaborate on HOW will the reaction change, and I was stumped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I am asking here is what will happen if I allow a chemical reaction to take place in a live MRI machine?&lt;/p&gt;&#xA;" OwnerUserId="13135" LastActivityDate="2019-01-18T23:19:07.327" Title="Can magnetic fields affect a chemical reaction?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24508" PostTypeId="2" ParentId="24495" CreationDate="2015-02-02T06:15:49.687" Score="5" Body="&lt;p&gt;A halogen bond can, quite analogous to a hydrogen bond, be considered a mostly electrostatic interaction with some covalent interaction. Hence the halogen bond acceptor is the electron donating molecule, i.e. a Lewis base. The halogen (bond) takes the role as the Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (b) you compare $\ce{H3C-I}$ and $\ce{F3C-I}$. Fluorine is more electronegative than hydrogen, and the statement $$\angle(\ce{H-C-H})&amp;gt;\angle(\ce{F-C-F})$$ holds. The carbon directs mor $\ce{p}$ orbital character into the carbon fluorine bonds than in the carbon hydrogen bonds. As a consequence the $\ce{s}$ orbital character in the carbon iodine bond (from the carbon) is higher in $\ce{F3C-I}$ compared to $\ce{H3C-I}$. That causes the bond to be slightly shorter and having more electron density at the carbon. This is also making the iodine a little more electron deficient and a better electron acceptor, i.e. a stronger acid (see also &lt;a href=&quot;http://goldbook.iupac.org/BT07000.html&quot;&gt;Bent's Rule&lt;/a&gt;). The electronic interaction will therefore also be slightly stronger.&lt;br&gt;&#xA;This observed effect is also often called inductive effect, which like you correctly stated is larger in $\ce{F3C-I}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (c) you are comparing the Lewis bases $\ce{NMe3}$ and $\ce{OMe2}$ acting towards $\ce{F3C-I}$ (or $\ce{HC3-I}$, the reasoning would remain the same). Oxygen is more electronegative than nitrogen, hence the lone pair orbitals of oxygen are more compact, i.e. the maximum electron density is closer to the oxygen nucleus. This causes dimethylether a weaker electron donor. Hence the the oxygen iodine interaction in $\ce{F3C-I\bond{~}OMe2}$ can be expected to be weaker than the corresponding nitrogen iodine interaction in $\ce{F3C-I\bond{~}NMe3}$. (Just like it is stated in your sheet.)&lt;br&gt;&#xA;Additionally, since methyl groups are usually slightly electron donating, and trimethylamine has one more methyl group than dimethylether, it can be expected to be an even better electron donor, stronger base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance does not contribute to these bonding situations in a significant way. (Looking at this from a valence bond perspective would certainly exceed this format and it is not necessary for a good model explanation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last a word of caution. Oxygen (in neutral molecules) can in most cases not be described as $\ce{sp^{3}}$ hybridised. Symmetry considerations and it's electronegativity are usually the reason for this. It is better described as $\ce{sp}$ hybridised in terminal position (like in ketones), or $\ce{sp^{2}}$ hybridised in bridging positions (like in ethers).&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-02-02T06:15:49.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24509" PostTypeId="1" AcceptedAnswerId="24567" CreationDate="2015-02-02T06:50:09.030" Score="7" ViewCount="3615" Body="&lt;p&gt;Why are we able to show structures of compounds with different position of electron, but with same position of them? Shouldn't the structure become unstable due to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take benzene for the example of resonance.We know that in reality benzene has no pure single or double bond. How can this be possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond length for $\ce{C-C}$ and $\ce{C=C}$ bonds in benzene is same which is $1.39~\mathring{\mathrm{A}}$. This is due to partial single and double bond between carbon atom, which is due to resonance. Why is there a resonance? Why can't benzene be happy with three alternate single and double bond between each carbon atom?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="4945" LastEditDate="2015-02-02T10:40:03.130" LastActivityDate="2015-02-03T14:39:37.987" Title="Why is there a need for resonance?" Tags="&lt;resonance&gt;&lt;lewis-structure&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24514" PostTypeId="1" AcceptedAnswerId="24590" CreationDate="2015-02-02T07:18:41.020" Score="0" ViewCount="3421" Body="&lt;p&gt;For example, in glycolysis, $\ce{NAD+} $ is reduced to $\ce{NADH}$ and then it's more energetic. Why adding electrons to an atoms gives it more energy?&lt;/p&gt;&#xA;" OwnerUserId="11430" LastActivityDate="2018-10-17T01:37:45.830" Title="Why reduced atoms are more energetic?" Tags="&lt;redox&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24515" PostTypeId="2" ParentId="24509" CreationDate="2015-02-02T07:56:05.737" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Delocalization is characteristic of the molecular orbital theory&#xA;  concerning the structure of atoms. Rather than the lone pair of&#xA;  electrons contained in specific bonds (as in the valence-bond theory),&#xA;  the MO (molecular-orbital theory) theorizes that electrons exist in&#xA;  orbitals that are spread over the entire molecule. The MO theory&#xA;  explains molecules such as ozone and benzene, which cannot be drawn&#xA;  satisfactorily with one Lewis structure, and are therefore described&#xA;  as resonance hybrids. Molecular orbitals solve this issue through the&#xA;  concept of delocalized electrons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When orbitals overlap, many of the orbitals are in hybrid states. Like&#xA;  delocalization, hybridization occurs to promote symmetry and&#xA;  stability. Delocalized electrons are often found in covalently bonded&#xA;  molecules that alternate single and multiple (usually double) bonds.&#xA;  Also, they frequently occur in aromatic systems and in mesoionic&#xA;  systems which means that they cannot exist in merely one resonance&#xA;  structure (they can and do exist in many resonance structures).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Delocalization matters for several reasons. One, spreading electron&#xA;  densities over a great area creates greater stability for the&#xA;  molecule.   Spreading electrons creates charge distribution.&#xA;  Therefore, chemical reactions expected because of a hypothesized&#xA;  configuration may not occur, a stable product is formed more readily&#xA;  in a reaction. Delocalization of electrons also paves the way for&#xA;  conductivity while in spectroscopy, wavelenght absorbtion is also&#xA;  based on resonance stability and delocalization of electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Valence_Bond_Theory/Resonance/Delocalization_of_Electrons&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13135" LastEditorUserId="13135" LastEditDate="2015-02-03T04:01:36.060" LastActivityDate="2015-02-03T04:01:36.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24517" PostTypeId="1" CreationDate="2015-02-02T09:36:53.120" Score="1" ViewCount="238" Body="&lt;p&gt;In my book the following is written:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When an electron withdrawing atom such as a halogen is attached to the end of a carbon chain, the $\sigma$-electrons of the $\ce{C-X}$ bond are attached to or displaced towards the more electronegative halogen atom, as a result $\ce{X}$ acquires a small negative charge and $\ce{C_1}$ acquires a small positive charge. The small positive charge on the $\ce{C_1}$ in turn attracts the $\sigma$-electrons of the $\ce{C_1-C_2}$ bond towards it. As a result $\ce{C_2}$ acquires a small charge, of course smaller than on $\ce{C_1}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then as $\ce{C_2}$ acquires a small a charge shouldn't the positive charge of $\ce{C_1}$ become less than earlier? Shouldn't it be true?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...This type of a displacement of $\sigma$-electrons along a saturated carbon chain whenever an electron withdrawing group is present the end of the chain is called inductive effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is there displacement of only $\sigma$-electrons not $\pi$-electrons and why does it happen along a saturated carbon chain only?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="4945" LastEditDate="2015-02-02T10:32:00.833" LastActivityDate="2015-02-02T10:32:00.833" Title="Why is the inductive effect of σ-electrons only along saturated carbon chains?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24518" PostTypeId="1" AcceptedAnswerId="24561" CreationDate="2015-02-02T10:01:38.773" Score="20" ViewCount="7610" Body="&lt;p&gt;In my 12th standard book, the formula for entropy change is given as $\Delta S = \frac{q_\text{reversible}}{T}$. What is the importance of absorbing heat reversibly and not irreversibly? What does absorbing heat reversibly mean? Does it mean that the process should be reversible, and if it does, then how would we define change in entropy of a spontaneous process i.e an irreversible process?&#xA; Would its formula also include $q_\text{reversible}$? &lt;/p&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="5026" LastEditDate="2018-03-10T15:42:06.620" LastActivityDate="2019-12-03T09:43:55.753" Title="Formula of entropy change" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24520" PostTypeId="1" AcceptedAnswerId="25167" CreationDate="2015-02-02T11:05:29.990" Score="21" ViewCount="10072" Body="&lt;p&gt;I know for a fact that &lt;em&gt;cis&lt;/em&gt; alkenes are less stable than &lt;em&gt;trans&lt;/em&gt; alkenes due to steric repulsions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what bugs me is the fact that why systems such as 2-methylprop-1-ene, where I believe steric repulsions are greater than that in &lt;em&gt;cis&lt;/em&gt;-but-2-ene, are more stable than its &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this being given in the stability order of alkenes in Solomons and Frhyle.&#xA;I looked up Morrison and Boyd and it said the same thing. Paula Bruice said otherwise stating trans to be more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct order, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Paula Bruice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lggMk.jpg&quot; alt=&quot;Paula Bruice Pg 175&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solomons and Frhyle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/61thD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/61thD.png&quot; alt=&quot;screenshot Solomons and Fryhle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For reference, &lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/entropies-organic&quot; rel=&quot;nofollow noreferrer&quot;&gt;free energies of formation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c|c} \text{Compound} &amp;amp; \Delta G^{\circ}_{f}\text{ in kJ mol}^{-1} \\\hline \text{2-methylpropene} &amp;amp; 58.1 \\ E\text{-2-butene} &amp;amp; 63.0 \\ Z\text{-2-butene} &amp;amp; 65.9  \\\hline \end{array}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="5026" LastEditDate="2018-04-21T04:17:57.393" LastActivityDate="2021-02-13T13:07:21.533" Title="Are geminal disubstituted alkenes more stable than their cis/trans isomers?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="24522" PostTypeId="2" ParentId="24506" CreationDate="2015-02-02T12:10:40.320" Score="4" Body="&lt;p&gt;Reaction rates should be defined in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Extent_of_reaction&quot; rel=&quot;nofollow&quot;&gt;Extent of reaction&lt;/a&gt; ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$:&#xA;$$\xi = \frac{[i]}{\nu_i}$$&#xA;Therefore, considering that the first reaction should be:&#xA;$$\ce{A + A -&amp;gt; B ~=~ 2A -&amp;gt; B}$$&#xA;The corresponding reaction rates are:&#xA;$$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$&#xA;In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to. &#xA;The global rate equations are:&#xA;\begin{align}&#xA;\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &amp;amp;= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &amp;amp;= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\&#xA;\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &amp;amp;= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\&#xA;\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &amp;amp;= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is conceptually wrong to assume that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reaction scheme you described is the &quot;parallel reactions&quot; model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.&lt;/p&gt;&#xA;" OwnerUserId="13103" LastEditorUserId="13103" LastEditDate="2015-02-03T12:16:47.230" LastActivityDate="2015-02-03T12:16:47.230" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24523" PostTypeId="1" CreationDate="2015-02-02T12:32:30.543" Score="2" ViewCount="1535" Body="&lt;p&gt;In my book the following was written:-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Inductive effect is a permanent effect operating in the ground state of the organic molecules and hence is responsible for higher melting point, boiling point and dipole movement of polar compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds only?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastActivityDate="2015-03-07T18:11:01.027" Title="Effect of inductive effect on boiling point,melting point and dipole movement" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;melting-point&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24524" PostTypeId="2" ParentId="24507" CreationDate="2015-02-02T13:16:18.083" Score="16" Body="&lt;p&gt;Here is a nice review paper on the subject: Rodgers, C. T. Magnetic field effects in chemical systems. &lt;em&gt;Pure Appl. Chem.&lt;/em&gt; &lt;strong&gt;2009,&lt;/strong&gt; &lt;em&gt;81&lt;/em&gt; (1), 19–43. &lt;a href=&quot;https://doi.org/10.1351/PAC-CON-08-10-18&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1351/PAC-CON-08-10-18&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemical reactions that involve radical intermediates can be influenced by magnetic&#xA;  fields, which act to alter their rate, yield, or product distribution. These effects have&#xA;  been studied extensively in liquids, solids, and constrained media such as micelles. They  may be interpreted using the radical pair mechanism (RPM). Such effects are central to the field&#xA;  of spin chemistry of which there have been several detailed and extensive reviews. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="16683" LastEditDate="2019-01-18T23:19:07.327" LastActivityDate="2019-01-18T23:19:07.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24526" PostTypeId="2" ParentId="18437" CreationDate="2015-02-02T13:51:25.687" Score="2" Body="&lt;p&gt;$K$ is just the specific value of $Q$ for which $\Delta_\mathrm r G = 0$.&#xA;In that case:&#xA;$$\begin{align}&#xA;\Delta_\mathrm r G &amp;amp;= \Delta_\mathrm r G^\circ + RT \ln Q \\&#xA;0 &amp;amp;= \Delta_\mathrm r G^\circ + RT \ln K \\&#xA;\Delta_\mathrm r G^\circ &amp;amp;= -RT \ln K &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any system not at the equilibrium it is possible that $Q &amp;lt; K$ or $Q &amp;gt; K$. What is the &quot;forward direction&quot; is just a convention, it depends on how you decide to write the equilibrium. There is nothing special in one direction or the other. The system will &quot;move&quot; toward the direction that leads to equilibrium in both cases. In the first case the reaction will consume reactants to generate products, while in the second case the opposite will happen.&#xA;You can express this idea also in terms of the ratio $\frac Q K$. If it is &gt; 1 then the reaction will consume reactants. This is also often referred to as the &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Chatelier principle&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13103" LastEditorDisplayName="user7951" LastEditDate="2015-08-22T10:08:23.470" LastActivityDate="2015-08-22T10:08:23.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24527" PostTypeId="2" ParentId="15868" CreationDate="2015-02-02T14:06:37.637" Score="1" Body="&lt;p&gt;The rate of a reaction depends on the thermodynamics of its &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory&quot; rel=&quot;nofollow&quot;&gt;transition state&lt;/a&gt;&lt;/strong&gt;.&#xA;It would be better to look at it in terms of Gibbs activation energy. Therefore there two factors affecting the rate of a reaction:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the activation enthalpy $\Delta^‡ H $, that corresponds to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow&quot;&gt;Arrhenius activation energy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;the activation entropy $\Delta^‡ S $ that takes into account the probability of a transition state.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The whole dependency of the kinetic constant on these parameters is expressed by the Eyring and Wynne-Jones equation:&#xA;$$ k = \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T$ is the temperature, $k_\text{B}$ is the Boltzman constant and $h$ is the Planck constant.&#xA;Sometimes a reflection coefficient $\kappa$ is added to the equation to take into account some other dependencies. It is near to 1 for most of the simple reactions.&#xA;$$ k = \kappa \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$&lt;/p&gt;&#xA;" OwnerUserId="13103" LastEditorDisplayName="user7951" LastEditDate="2015-05-20T15:47:49.347" LastActivityDate="2015-05-20T15:47:49.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24528" PostTypeId="1" AcceptedAnswerId="24531" CreationDate="2015-02-02T14:27:07.713" Score="33" ViewCount="158777" Body="&lt;p&gt;I'm working on a crime story about cyanide poisoning from apple seeds. I just would like to have an idea of what processes and extraction techniques might be involved in getting cyanide from the seeds. The character is supposed to have access to a high school chemistry lab; so I was thinking equipment and tools that can be found in a typical HS lab. I prefer a detailed answer, step by step, if possible. If not then just generalized terminologies would work.&lt;/p&gt;&#xA;" OwnerUserId="13141" LastEditorUserId="30836" LastEditDate="2020-07-13T23:29:45.143" LastActivityDate="2020-07-13T23:29:45.143" Title="How do I extract cyanide from apple seeds?" Tags="&lt;organic-chemistry&gt;&lt;toxicity&gt;&lt;extraction&gt;&lt;chemistry-in-fiction&gt;&lt;chemical-weapons&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="24530" PostTypeId="2" ParentId="24528" CreationDate="2015-02-02T15:28:12.580" Score="14" Body="&lt;p&gt;Honestly, you probably wouldn't be able to get this done with just a highschool chem lab, at least not to a high purity (whether or not this is needed is debatable; in fact it could be interesting in the story because it would give a clue as to how it was produced, whereas pure cyanide wouldn't really tell the detective much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, apples don't contain the harmful version of cyanide (hydrogen cyanide) in and of themselves. They do however contain amygdalin, which can be metabolized to hydrogen cyanide. A quick search ends up with &lt;a href=&quot;http://en.wikipedia.org/wiki/Amygdalin&quot;&gt;wikipedia&lt;/a&gt; giving a brief overview of how it is isolated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Amygdalin is extracted from almonds or apricot kernels by boiling in ethanol; on evaporation of the solution and the addition of diethyl ether, amygdalin is precipitated as white minute crystals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case your orgo isn't all that great (mine is a little rusty too), what you should read from the above quote is that ethanol is used to oxydize the primary alcohol group to an aldehyde or a carboxylic acid (depending on how long you let it boil for) making it polar and so non-soluble the diethyl ether (an organic solvent) which leads it to precipitating out as a white crystal. Once it precipitates out the easiest way to isolate it is just to use a paper filter, since it should be the only solid present. Suction filtration would be optimal but that's generally not available in high schools. Lastly you could use &lt;a href=&quot;http://en.wikipedia.org/wiki/Sublimation_%28phase_transition%29&quot;&gt;sublimation&lt;/a&gt; to get a really pure final product, but again this isn't normally available in high school labs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're writing a novel I recommend you take the time to read up on it and how it is metabolized within the body. The more you understand the process the more authentic it'll feel, imo. Best of luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: You may want to consider switching to apricot kernels as opposed to apple seeds. The process is effectively the same but it's more realistic, since you'd need a LOT of apple seeds to get a decent amount of cyanide. Also, like I said my orgo is a bit rusty so if someone else could look over this that would be nice. &lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="13030" LastEditDate="2015-02-02T15:39:03.350" LastActivityDate="2015-02-02T15:39:03.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24531" PostTypeId="2" ParentId="24528" CreationDate="2015-02-02T15:59:25.077" Score="45" Body="&lt;ol&gt;&#xA;&lt;li&gt;Forget about the apple seeds, they contain about 1 to 4 mg amygdalin per gramm seeds (&lt;a href=&quot;http://dx.doi.org/10.1016/j.foodchem.2014.08.083&quot;&gt;DOI&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Instead, collect &lt;strong&gt;apricot seeds&lt;/strong&gt; during the right season, the amygdalin content varies though the year and can be as high as 5% of the dry weight of the seed (&lt;a href=&quot;http://dx.doi.org/10.2306/scienceasia1513-1874.2012.38.218&quot;&gt;DOI&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;It is probably advantagenous to break the husk with a nut cracker, a plier, etc. and cut the softer inner material to smaller pieces.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Extraction of amygdalin can be performed by&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;immersing the material in methanol and subsequent ultrasonification&lt;/li&gt;&#xA;&lt;li&gt;Soxhlet extraction with methanol&lt;/li&gt;&#xA;&lt;li&gt;reflux extraction in water in the presence of citric acid&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A comparison of the extraction methods is given &lt;a href=&quot;http://dx.doi.org/10.1093/chromsci/43.7.383&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removal of the solvent in vacuum will yield a crude material with significant amounts of amygdalin. You might want to have a look at this article from the &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1272529/pdf/westjmed00222-0011.pdf&quot;&gt;Western Journal of Medicine&lt;/a&gt; on the toxicity. Here, an $\mathrm{LD_{50}}$ of 522 mg amygdalin per kg body weight was estimated for the oral application to rats. The online resource of the &lt;a href=&quot;http://chem.sis.nlm.nih.gov/chemidplus/rn/29883-15-6&quot;&gt;U.S. National Library of Medicine&lt;/a&gt; gives a value of 405 mg/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Further information on the health risk of apricot kernels are provided by of the German &lt;a href=&quot;http://www.bfr.bund.de/cm/343/verzehr_von_bitteren_aprikosenkernen_ist_gesundheitlich_bedenklich.pdf&quot;&gt;Bundesinstitut für Risikobewertung&lt;/a&gt; (Federal Institute for Risk Assessment) and the British &lt;a href=&quot;http://cot.food.gov.uk/sites/default/files/cot/cotstatementapricot200615.pdf&quot;&gt;Committee on Toxicity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note in the a German medical journal, Deutsche Ärzteblatt, (&lt;a href=&quot;http://www.aerzteblatt.de/pdf/111/50/a2240.pdf&quot;&gt;PDF&lt;/a&gt;) describes a case where boy of four years (110 cm, 18 kg body weight) was given apricot kernels during an alternative cancer treatment. Upon additional treatment with a single dose of 500 mg amygdalin, the kid showed agitation, spasms and the eyes started to roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;I'll leave it up to your fantasy as a writer on how to apply the poison, but spicing some marzipan with it might help ;) &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-02T18:27:19.453" LastActivityDate="2015-02-02T18:27:19.453" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24532" PostTypeId="1" CreationDate="2015-02-02T17:01:48.363" Score="-1" ViewCount="1021" Body="&lt;p&gt;How much hydrochloric acid (maximum) should I use to do electrolysis of water, and how much &#xA;amount of voltage should I use to get the maximum of hydrogen and oxygen?&lt;/p&gt;&#xA;" OwnerUserId="13144" LastEditorUserId="10185" LastEditDate="2015-02-02T17:52:49.113" LastActivityDate="2015-02-04T04:03:40.123" Title="Concentration of hydrochloric acid for electrolysis?" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24533" PostTypeId="1" CreationDate="2015-02-02T18:23:26.507" Score="4" ViewCount="1713" Body="&lt;p&gt;In the mechanism for the acetylation of bromobenzene, how is &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; gas produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactants are acetic anhydride, bromobenzene, &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt; and dichloromethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, 4-bromoacetophenone is made, accompanied with (as far as I know) acetic acid and the regeneration of &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then is &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remove the &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt; at the end, 3 &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; are added and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; gas is produced then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to think of ways that &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; gas is produced prior to that, however. Could a chlorine of the aluminate ion deprotonate the arenium ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could dichloromethane (solvent) donate a &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="83758" LastEditDate="2020-06-02T16:43:58.897" LastActivityDate="2020-06-02T17:07:14.110" Title="Friedel-Crafts synthesis - How is HCl produced?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;synthesis&gt;&lt;electrophilic-substitution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24534" PostTypeId="1" CreationDate="2015-02-02T18:25:52.483" Score="2" ViewCount="355" Body="&lt;p&gt;How to &lt;em&gt;partially&lt;/em&gt; hydrolyze triglycerides for &lt;strong&gt;food application&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make a mixture of mono- and diglycerides starting from triglycerides.&lt;/p&gt;&#xA;" OwnerUserId="13147" LastEditorUserId="9961" LastEditDate="2016-11-20T19:59:13.967" LastActivityDate="2016-11-20T19:59:13.967" Title="How to partially hydrolyze triglycerides?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24537" PostTypeId="2" ParentId="24533" CreationDate="2015-02-02T19:22:04.353" Score="6" Body="&lt;p&gt;You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt;, which generates &lt;em&gt;in situ&lt;/em&gt; the &quot;acylium&quot; ion (&lt;span class=&quot;math-container&quot;&gt;$\ce{H3C-C#O+}$&lt;/span&gt;) - the actual acylating agent. The second step is the electrophilic aromatic substitution, in which the electrophile is preferentially &quot;attacked&quot; by the aromatic ring at the &lt;em&gt;ortho&lt;/em&gt;- or &lt;em&gt;para&lt;/em&gt;-position (&lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-orienting effect of halides). The driving force for the reaction is the last step, which leads to restore the aromaticity and it is assisted by &lt;span class=&quot;math-container&quot;&gt;$\ce{(AcO)(Al)Cl2}$&lt;/span&gt; with liberation of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have drawn the mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNrnX.png&quot; alt=&quot;Friedel-Crafts Acylation - Mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="61578" LastEditDate="2020-06-02T17:07:14.110" LastActivityDate="2020-06-02T17:07:14.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24540" PostTypeId="2" ParentId="24532" CreationDate="2015-02-02T21:05:59.663" Score="1" Body="&lt;p&gt;You will not get oxygen if you use HCl, you will get Chlorine gas instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use sulfuric acid (like your title said before someone edited it) or NaOH as an electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.5 M would be a reasonable concentration.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-02T22:02:54.087" LastActivityDate="2015-02-02T22:02:54.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24543" PostTypeId="1" CreationDate="2015-02-03T01:02:16.897" Score="2" ViewCount="155" Body="&lt;p&gt;Agricultural grade anhydrous ammonia (liquefied gas) contains 2000 ppm $\ce{H2O}$. What would be the process to remove this water to obtain refrigeration grade (&amp;lt;100 ppm $\ce{H2O}$) anhydrous ammonia? &lt;/p&gt;&#xA;" OwnerUserId="13156" LastEditorUserId="4945" LastEditDate="2015-02-05T02:41:38.783" LastActivityDate="2015-02-05T02:41:38.783" Title="Removing trace amounts of water from agricultural grade anhydrous ammonia" Tags="&lt;everyday-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24544" PostTypeId="2" ParentId="24532" CreationDate="2015-02-03T01:05:19.753" Score="0" Body="&lt;p&gt;You need to see the concentration of acid. If it is a concentrated acid, you will get chlorine gas. But if you use diluted acid, you will get oxygen gas.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-02-03T01:05:19.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24549" PostTypeId="1" AcceptedAnswerId="24571" CreationDate="2015-02-03T05:04:57.050" Score="11" ViewCount="11787" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You don’t see carbocations on double bonds very much, and here's a good reason: compared to sp3, there is more s character in the orbitals, so the empty orbital is held more closely to the nucleus. This is destabilizing if we’re talking about positive charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.masterorganicchemistry.com/2011/03/11/3-factors-that-stabilize-carbocations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't really make sense. Positive charge arises from the lack of electrons. How does holding an empty orbital (regardless of whether that makes any physical sense) closer affect your stability if the empty orbital is ... &lt;em&gt;empty&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the exact reason sp2 (aren't most carbocations sp2 though?) and sp carbocations are generally less stable than their sp3 counterparts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also, how can we have an sp3 carbocation?&lt;/strong&gt; Huh?! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j0GJz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason I can think of is that it's energetically expensive to make an sp carbocation ... pulling electrons which are held more tightly to the nucleus away is probably harder than pulling away electrons which are not as close to the nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of any reason why the end products would be any less/more stable ... if anything an sp carbocation should be more stable than an sp2 carbocation in that in the sp carbocation the electrons are held closer to the nucleus and therefore better stabilizing the nucleus. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-06T07:06:23.730" LastActivityDate="2020-03-03T07:20:00.410" Title="Why are sp hybridized carbocations high energy?" Tags="&lt;theoretical-chemistry&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24550" PostTypeId="2" ParentId="24549" CreationDate="2015-02-03T05:40:54.097" Score="7" Body="&lt;p&gt;Let's just tackle the obvious part first: There is no sp3 carbocation. One reason for this is the same, as for why higher substituted carbocations are more stable than lower substituted. Electrons are more stable in orbitals with high s-character.&lt;br&gt;&#xA;In $\ce{\overset{\oplus}{C}(CH3)3}$ the cationic carbon is sp2 hybridised while the empty orbital is 100% p character. All s character is used in occupied orbitals.&lt;br&gt;&#xA;In $\ce{H2C=\overset{\oplus}{C}-CH3}$, one would assume from the common depiction, that the central cationic carbon is also (roughly) sp2 hybridised. My first draft of answering did also assume this. A quick DF-BP86/def2-SVP calculation revealed something different. The $\angle(\ce{CCC})$ bond angle comes out to be almost exactly $180^\circ$. The $\pi$ bond is obviously only of p character. The underlying $\sigma$ bonds can now contain the whole s character, making it sp hybridised, and as a result the p orbital of the cationic carbon atom is empty. There will be interactions with some carbon hydrogen bonds. (Hyperconjugation, see below.) This effect is most likely much less than in the prior case, causing it to be less stable. In addition to this, the molecule is much more constrained in its geometry. This linear arrangement is probably also only true for this very simple arrangement. Dependent on the moieties attached to it, a bent structure might be more stable and a different bonding angle will be observed. When that happens, there will be some s character in the empty orbital.&lt;br&gt;&#xA;In the benzyl cation, due to symmetry restrictions, $D_\mathrm{2h}$, the empty orbital must be almost exactly sp2. Hence the underlying $\sigma$ bonds will roughly take away 67% s character. The empty orbital will therefore have a very high s character, about 33%, as a result. There is almost no bond widening possible to decrease this and successively increase s character in the $\sigma$ bonds. So the s character found in empty orbitals effectively destabilised the remaining occupied orbitals. (One could also think about it as some kind of ring strain.)&lt;br&gt;&#xA;In $\ce{H3C-C#\overset{\oplus}{C}}$, symmetry is even higher, $C_\mathrm{\infty{}v}$ for the triple bond. The underlying $\sigma$ bond will have slightly more s character than in an ideal sp orbital. The empty orbital will, however, be binding a lot of s character. This is effectively destabilising the occupied orbitals again. Since the empty orbital contains the most s character in the orbital, it will also have the most destabilised $\sigma$ system of the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the more s character is in unoccupied orbitals, the less s character can be in occupied orbitals. This is destabilising these orbitals, causing the carbocation also to comparably less stable.&lt;br&gt;&#xA;The quote expresses this feature in a quite confusing way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;&lt;br&gt;&#xA;There are of course other reasons for higher substituted carbocations to be more stable than lower. One of the main effects is hyperconjugation of the $\sigma(\ce{C-H})$ bond with the empty p orbital.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-02-03T08:49:32.197" LastActivityDate="2015-02-03T08:49:32.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24552" PostTypeId="2" ParentId="24534" CreationDate="2015-02-03T07:25:47.727" Score="2" Body="&lt;p&gt;Over the last decades, partial hydrolysis of triglycerides has been examined in various contexts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this isn't my field of work and that there might hundreds of better and/or more instructive recipes, but a cursory search yielded the following freely available references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;THE DIGESTION IN VITRO OF TRIGLYCERIDES BY PANCREATIC LIPASE&lt;/em&gt;, &#xA;F. H. Mattson, and L. W. Beck,&#xA;&lt;em&gt;J. Biol. Chem.&lt;/em&gt;, &lt;strong&gt;1955&lt;/strong&gt;, &lt;em&gt;214&lt;/em&gt;, 115-125 (&lt;a href=&quot;http://www.jbc.org/content/214/1/115.full.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A stereospecific analysis of triglycerides&lt;/em&gt;, &#xA;H. Brockerhoff, Journal of Lipid Research, &lt;strong&gt;1965&lt;/strong&gt;, &lt;em&gt;6&lt;/em&gt;, 10-15&#xA;(&lt;a href=&quot;http://www.jlr.org/content/6/1/10.full.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The products from lipase-catalysed hydrolysis of bovine milkfat kill&#xA;Helicobacter pylori in vitro&lt;/em&gt;, &#xA;Cynthia Q. Sun, Charmian J. O’Connor, Alastair K. H. MacGibbon, and Anthony M. Roberton, &lt;em&gt;FEMS Immunology &amp;amp; Medical Microbiology&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt;, 235-242 &#xA;(&lt;a href=&quot;http://dx.doi.org/10.1111/j.1574-695X.2006.00185.x&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the digestion of triglycerides by &lt;strong&gt;lipases&lt;/strong&gt; in buffered solutions is a viable approach.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-03T07:25:47.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24553" PostTypeId="2" ParentId="24458" CreationDate="2015-02-03T08:04:04.523" Score="1" Body="&lt;p&gt;Short answer, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, the fluorescence signal is strongest when the exciting photon has an energy close to the excitation energy for the element absorbing the photon.  There are actually two factors coming into play:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The closer the resonance of the incident photon with the electronic orbital it is exciting, the higher the cross-section to absorb the photon.  &lt;a href=&quot;http://xdb.lbl.gov/Section1/Sec_1-5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(See here.)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The higher-Z the absorbing atom, the more likely it is that once excited, the atom will emit a fluorescence photon (as opposed to an Auger electron).  That is, it has a higher fluorescence yield.  &lt;a href=&quot;http://xdb.lbl.gov/Section1/Sec_1-3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(See here.)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the Z of the atoms are pretty close, but also in portion of the periodic table where the fluorescence yield is rapidly changing.  The chance of a fluorescence photon being emitted once excited is about 20% greater for Co than Mn (and you can look it up in the &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd136.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;literature&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-section for absorption as a function of energy for these three elements looks like this (code for the plot &lt;a href=&quot;https://gist.github.com/678df181ec1f5cd1bb46&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wsJca.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also drew a vertical line for the exciting Cu-K$\alpha$ photon at 8048 eV, and you can see that the Co curve has the highest cross-section at about 316 cm$^2$/g, and Mn has the lowest at about 270 cm$^2$/g, which means the Co cross-section is about 17% higher than the Mn.  The primary reason for the reduction in the sensitivity for the Mn edge is that the absorption cross-section drops off as $\approx\Delta E^3$.  These data were plotted using the &lt;a href=&quot;http://henke.lbl.gov/optical_constants/asf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;data files&lt;/a&gt; described by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0092640X83710132&quot; rel=&quot;nofollow noreferrer&quot;&gt;B. L. Henke, E. M. Gullikson, and J. C. Davis, Atomic Data and Nuclear Data Tables Vol. 54 No. 2 (July 1993)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both these effects are colluding to increase your fluorescence for Co vs Mn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, if you want to reduce this effect in your X-ray diffraction patterns, you can change your source target material with a heavier element such as Zr which has it's K$\alpha$ up at 15.7 keV, yet the L lines are still around 2 keV (below the transition metal K-edges) or you can use a lighter target, such as Mn.  Mn, of course, would not be able to excite its K-edge using its own K$\alpha$ fluorescence photon.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="8171" LastEditDate="2015-02-03T08:23:34.847" LastActivityDate="2015-02-03T08:23:34.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24554" PostTypeId="1" CreationDate="2015-02-03T08:21:54.783" Score="3" ViewCount="108" Body="&lt;p&gt;I'm studying Optical tweezers. I've used Polystyrene for samples. Nowadays, I've got SU-8 pillar. I wanna use that for sample, but its density is heavier than water. I can't use that. So I wanna know some liquid for replacing water.&lt;/p&gt;&#xA;" OwnerUserId="13167" LastActivityDate="2021-04-01T19:03:10.233" Title="Which solvent is good for using SU-8 particle in Optical tweezers?" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24556" PostTypeId="2" ParentId="24509" CreationDate="2015-02-03T11:12:50.003" Score="-2" Body="&lt;p&gt;One way to look at it is the &lt;strong&gt;Electron Clouds&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chemical becomes reactive when it has either excess or insufficient electron cloud. For example, in case of Ammonia $\ce {NH}_3$, Nitrogen has a lone pair of electron. Thus, it has excess of electron cloud. It can donate these electrons. Hence, Ammonia is reactive. Similarly, Aluminium is highly electropositive and has empty p-orbital. Hence, it can accept electrons and is also reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Benzene, the electrons are delocalised, that is, the electron cloud is spread over the entire ring of Benzene. Thus, there is no accumulation of charge in the ring at one place. This gives extra stability to Benzene. And every molecule wants to attain maximum stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If resonance didn't exist in Benzene, then it would react easily with many reagents. Since each of the carbons is $\ce {sp}^2$ hybridised, that is, each carbon is quite electronegative, and has the capacity to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is a better answer to your question. This is what I can think as the reason.&lt;/p&gt;&#xA;" OwnerUserId="3718" LastActivityDate="2015-02-03T11:12:50.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24559" PostTypeId="2" ParentId="24427" CreationDate="2015-02-03T12:01:59.660" Score="5" Body="&lt;p&gt;It is unlikely to find an uncharged metal atom hovering in the centre of a fullerene cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C60}$ is a good electron acceptor. This has been widely demonstrated in a lot of studies on photoinduced electron transfer in donor-acceptor dyads, where the donor was a metalloporphyrine attached to the fullerene. Here, excitation of the donor led to an intramolecular redox reaction in which the fullerene radical anion $\ce{C60^{-\cdot}}$ was formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;If you're interested in this particular topic, look for some earlier works by &lt;em&gt;Devens Gust&lt;/em&gt;, &lt;em&gt;Thomas A. Moore&lt;/em&gt; and &lt;em&gt;Ana L. Moore&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Taking that into account, full electron transfer, resulting in caged ion pair, or at least some charge transfer is conceivable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;donor-acceptor&quot; thus seems appropriate.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-03T12:16:53.287" LastActivityDate="2015-02-03T12:16:53.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24560" PostTypeId="2" ParentId="24514" CreationDate="2015-02-03T12:01:59.907" Score="0" Body="&lt;p&gt;Energy of polyelectronic atoms is hard to calculate since you have to take care of the individual energy of each electrons bu also of interaction between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify, you can assume the addition of an electron will bring the energy of the electron plus maybe some other interaction wich will rise the total energy of the atom.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-03T12:01:59.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24561" PostTypeId="2" ParentId="24518" CreationDate="2015-02-03T12:37:28.767" Score="19" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;A process is &lt;em&gt;thermodynamically&lt;/em&gt; reversible if it is essentially at equilibrium.&lt;/strong&gt; Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\text{system}} = T_{\text{surroundings}}$ if heat can be exchanged reversibly; any temperature difference will result in an &lt;em&gt;irreversible&lt;/em&gt; exchange of heat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What's special about reversible heat transfer?&lt;/strong&gt; When a tiny amount of heat $dq_{\text{rev}}$ is released by the system during a reversible process, the surroundings will absorb exactly that amount of heat, no matter what path the process takes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an &lt;em&gt;irreversible&lt;/em&gt; process, the amount of heat (and work) depends on how the system and surroundings change &lt;em&gt;during&lt;/em&gt; the process.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What's the difference between reversible and irreversible heat?&lt;/strong&gt; The heat transferred by an irreversible process $dq_{\text{irrev}}$ is less than $dq_{\text{rev}}$, where $dq_{\text{rev}}$ is the heat that would have been transferred if the process had been performed reversibly. The difference between the two is the amount of work you'd have to do on the system to reverse the effects of the irreversible process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we know that? Energy $U$ is a state function, so it'll be the same whichever way we do the process:&#xA;$$dU = dq_{\text{irrev}} + dw_{\text{irrev}} = dq_{\text{rev}} + dw_{\text{rev}}$$&#xA;so the difference between the two heats is &#xA;$$dq_{\text{rev}} - dq_{\text{irrev}} = dw_{\text{irrev}} - dw_{\text{rev}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences on either side are positive, so $$dq_{rev} &amp;gt; dq_{irrev}$$&#xA;This means that an irreversible process will always absorb less heat from the surroundings and perform less work on the surroundings than it would have if the process had been done reversibly. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Why $dS = dq_{\text{rev}}/T$?&lt;/strong&gt; With a reversible process, $$\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{system}} = -\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{surroundings}}$$&#xA;so if we define $dS \equiv dq_{\text{rev}}/T$, then we have $dS_{\text{universe}} = dS_{\text{system}} + dS_{\text{surroundings}} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what would happen in an irreversible process that took the system from state A to state B. The inequality $dq_{\text{rev}} &amp;gt; dq_{\text{irrev}}$ together with the definition of entropy in terms of reversible heats gives&#xA;$$\Delta S = S_B - S_A = \int_A^B\frac{dq_{\text{rev}}}{T} &amp;gt; \int_A^B\frac{dq_{\text{irrev}}}{T}$$ for any such process. This also implies that for any cyclic process (e. g. A to B back to A), we have $$\oint\frac{dq_{\text{irrev}}}{T} \lt 0, \ \text{and}\  \Delta S = \oint\frac{dq_{\text{rev}}}{T} = 0$$&#xA;This shows that $q_{\text{rev}}/T$ is a state function (it's zero for any cyclic process) but  $q_{\text{irrev}}/T$ is &lt;em&gt;not&lt;/em&gt; a state function. If entropy is to be a state function, then, you can't define it as simply $q/T$; you've got to use $\Delta S = q_{\text{rev}}/T$, or for the differential form $dS = dq_{\text{rev}}/T$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How could we relate the entropy change for an irreversible process to $dq_{\text{irrev}}$?&lt;/strong&gt; From the difference between $q_{rev}$ and $dq_{irrev}$ above and the definition of entropy, we have&#xA;$$dS = \frac{dq_{\text{rev}}}{T} = \frac{dq_{\text{irrev}}}{T} + \frac{dw_{\text{irrev}}-dw_{\text{rev}}}{T}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;TL;DR: There IS no TL;DR; you're asking some important questions here, and the answers need to be made carefully.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="5026" LastEditDate="2018-03-10T13:12:33.157" LastActivityDate="2018-03-10T13:12:33.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24562" PostTypeId="2" ParentId="18967" CreationDate="2015-02-03T12:38:34.163" Score="2" Body="&lt;p&gt;The rule says that for many liquids the entropy of vaporization is almost the same, around ${\rm 85\,J K^{-1}} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (partial) success of the rule is due to the fact that the entropy of a gas is so much larger than that of any liquid. Therefore, the entropy of the initial state (e.g. the liquid) usually makes no big difference.  However, if the liquid presents hydrogen bonding or any other kind of high ordered structure, its entropy will be particularly low and the entropy gain during vaporization will be larger than usual, too. &lt;/p&gt;&#xA;" OwnerUserId="13103" LastEditorUserId="7448" LastEditDate="2015-09-18T22:20:48.837" LastActivityDate="2015-09-18T22:20:48.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24563" PostTypeId="2" ParentId="24543" CreationDate="2015-02-03T13:30:40.547" Score="1" Body="&lt;p&gt;Calcium oxide can get water down to 4ppm in anhydrous ammonia according to &lt;a href=&quot;http://www.google.com/patents/US4075306&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/US4075306&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-03T13:30:40.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24567" PostTypeId="2" ParentId="24509" CreationDate="2015-02-03T14:39:37.987" Score="7" Body="&lt;p&gt;As I mentioned in my comment, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on resonance&lt;/a&gt; provides a rather nice description of the subject where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_(chemistry)#Resonance_in_quantum_mechanics&quot; rel=&quot;noreferrer&quot;&gt;Resonance in quantum mechanics section&lt;/a&gt; is of particular interest for the question asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We know that in reality benzene has no pure single or double bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now please, take a deep breathe and repeat with me: &lt;em&gt;there are no bonds&lt;/em&gt;. In reality we do have electrons and nuclei that constitutes a molecule, but a chemical bond is merely a theoretical concept introduced in the framework of a scientific theory. In other words, chemical bonds do not exist in the same sense as electrons and nuclei do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the concept of resonance was introduced by Linus Pauling in the framework of the so-called &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;noreferrer&quot;&gt;valence bond theory&lt;/a&gt;&lt;/em&gt;, according to which&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a covalent bond is formed between the two atoms by the overlap of half&#xA;  filled valence atomic orbitals of each atom containing one unpaired&#xA;  electron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, this is the way the concept of chemical bonding enters the scene in the valence bond theory: a chemical bond is used to describe overlapped atomic orbitals of two atoms in a molecule. Note that atomic orbitals also do not exist in the same sense as electrons and nuclei do, but an orbital has at least some strict mathematical and physical meaning, while a chemical bond, as introduced above, does not. Read the above quote carefully one more time: it is not even a definition! What we just say is that whenever half filled valence atomic orbitals of two atoms overlap a covalent bond is formed, but we do not say what exactly is a covalent bond. It is just something; something which is formed whenever orbitals overlap. It is a concept, purely theoretical construction. It is practically helpful in description of a reality, but no one will say that it is actually a part of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not be afraid of all this business, because it is the way science generally works. Scientific theories basically tell us how nature could be &lt;em&gt;described&lt;/em&gt;, and they do so in the language of mathematics, since it is the only language nature can unambiguously &quot;speak&quot; with us. However, it is often so tempting to force it to speak with us in a different language, in a language of our theoretical concepts (such as, for instance, the concept of a chemical bonding), no matter how fuzzy they are. This is a very dangerous road, because, after all, who are we to tell nature which language to speak? Nature will likely refuse to &quot;speak&quot; with us in that language, and that would be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is what actually happened with the valence bond theory as it was originally formulated. Indeed, all carbon-carbon bond lengths are equal in benzene, but the valence bond theory were not able to account for that simple fact. Nature refused to &quot;speak&quot; with us in the language of chemical bonding. At that point chemists had two choices: either abandon the valence bond theory or introduce another theoretical concept into it, introduce a new word in our language making it possible to communicate with nature in this language. The concept of a chemical bonding was already so tremendously useful in chemistry that the choice was actually predetermined, and in 1928 Linus Pauling introduced the concept of resonance between several valence-bond structures of a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is resonance? Is it a physical process? No way! It is just a concept, a word, used to describe&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;delocalized electrons within certain molecules or polyatomic ions&#xA;  where the bonding cannot be expressed by a single Lewis formula.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The word &quot;resonance&quot; was probably not the best choice by Pauling, since it gives an unwanted feeling that the molecule actually do oscillate back and forth between the structures, which is, as I said, not the case. But putting aside its somewhat misleading name, the concept of resonance is as good as the concept of chemical bonding and many other concepts in chemistry and other sciences.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-02-03T14:39:37.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24568" PostTypeId="1" CreationDate="2015-02-03T14:47:46.600" Score="2" ViewCount="1287" Body="&lt;p&gt;How p-phenylenediamine (PPD) is produced? PPD is used as raw material for Kevlar (Para amid Fiber).&lt;/p&gt;&#xA;" OwnerUserId="13176" LastEditorUserId="418" LastEditDate="2015-02-04T09:38:10.767" LastActivityDate="2015-02-08T08:53:08.493" Title="P-phenylenediamine Production Process" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24569" PostTypeId="1" CreationDate="2015-02-03T15:02:30.520" Score="1" ViewCount="526" Body="&lt;p&gt;Why for the Resonance effect certain group of substituents/functional group donate electron from the double bond or to a conjugated system while some withdraw electron from the double bond or to a conjugated system? What causes the groups/substituent to withdraw or donate electron causing +R and -R effect?&lt;/p&gt;&#xA;" OwnerUserId="8231" LastActivityDate="2015-02-03T18:29:56.157" Title="Why do certain groups withdraw/ donate electron to double bond or conjugated system in resonance effect?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24570" PostTypeId="2" ParentId="24568" CreationDate="2015-02-03T15:39:31.890" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;React benzene with chlorine in the presence of &lt;span class=&quot;math-container&quot;&gt;$\ce{FeCl3}$&lt;/span&gt; to yield chlorobenzene.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;According to &lt;em&gt;Ullmann's Encyclopedia of Industrial Chemistry&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Most monochlorobenzene is now produced from benzene and chlorine in continuously operated plants. Depending on the ratio of benzene to chlorine chosen, one can achieve either a low rate of benzene conversion and little dichlorobenzene&#xA;formation, or almost complete conversion of the benzene with a higher degree of dichlorobenzene formation. [...] The composition of a chlorination mixture containing the highest possible proportion of monochlorobenzene has been given as&#xA;4–5% unreacted benzene, 73% monochlorobenzene, and 22–23% dichlorobenzenes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Nitrate chlorobenzene to obtain a mixture of 2-nitrochlorobenzene and 4-nitrochlorobenzene. Separate the isomers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;In &lt;em&gt;Ullmann's Encyclopedia of Industrial Chemistry&lt;/em&gt;, we find:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nitration of chlorobenzene with mixed acid (30/56/14) typically gives an isomer&#xA;mix in 98 % yield consisting of 34–36% 2-chloronitrobenzene, 63–65 % 4-chloronitrobenzene, and only ca. 1% 3-chloronitrobenzene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Mixed acid&amp;quot; refers to a mixture of nitric acid, sulfuric acid and water.&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;React 4-nitrochlorobenzene with ammonia (Nucleophilic aromatic substitution) to obtain 4-nitroaniline.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;em&gt;Ullmann's Encyclopedia of Industrial Chemistry&lt;/em&gt;, the reaction conditions as as follows:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;4-Nitroaniline is produced in an agitated titanium reactor in 99.3 % yield by reacting 4-chloro nitrobenzene with a tenfold excess of aqueous ammonia at 175 °C and 4.2 MPa (42 bar) for 10 h.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;Ullmann's Encyclopedia of Industrial Chemistry&lt;/em&gt; cites a patent by MONSANTO (&lt;a href=&quot;https://depatisnet.dpma.de/DepatisNet/depatisnetaction=bibdat&amp;amp;docid=DE000001768518A&quot; rel=&quot;nofollow noreferrer&quot;&gt;DE1768518&lt;/a&gt;) as the relevant reference.&lt;/p&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Reduce the nitro group with iron.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-03T21:17:01.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24571" PostTypeId="2" ParentId="24549" CreationDate="2015-02-03T16:33:53.427" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are sp hybridized carbocations high energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are links to 2 earlier answers where this question has been addressed for vinylic and aryl carbocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/10509/instability-of-vinylic-and-arylic-carbocations/10516#10516&quot;&gt;link 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15409/benzene-halide-sn1-reaction/15410&quot;&gt;link 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When vinylic carbocations are generated, the empty orbital (the orbital holding no electrons) is intitially $\ce{sp^2}$ hybridized.  This carbonium ion orbital being $\ce{sp^2}$ hybridized is using some s character, the remaining orbitals around the carbonium ion carbon will therefore have less s character available for use in their bonds.  The more s character in a bond, the lower the energy of the electrons in that bond because s orbitals are lower in energy than p orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the vinylic carbocation rehybridizes such that the empty orbital becomes a p orbital, then the remaining orbitals that contain electrons will become lower in energy as they rehybridize and incorporate the s character that was initially used in the empty $\ce{sp^2}$ carbocation orbital.  Once rehybridized the carbocations overall energy has been lowered and the carbocation is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high energy of the initially generated $\ce{sp^2}$ hybridized vinylic carbocation is why it is so difficult to generate vinylic carbocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some vinylic carbocations cannot rehybridize to an $\ce{sp^2}$ carbocation carbon with a p orbital due to geometrical constraints, and they are even harder to generate.  Examples would be the phenyl carbocation and vinylic carbocations in small rings, like the cyclohexenyl carbocation (see figure below).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, how can we have an sp3 carbocation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They exist, but again, realize that since the empty orbital is not p hybridized they will be higher in energy and harder to generate then carbocations in which the empty orbital is p hybridized.  Examples of carbocations in which the empty orbital is roughly $\ce{sp^3}$ hybdridized would include bridgehead carbocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4kQrX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-02-03T22:16:49.983" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24572" PostTypeId="2" ParentId="6625" CreationDate="2015-02-03T16:53:56.750" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I right in concluding that resonance effects are stronger than&#xA;  inductive effects, if not in general, then at least for electrophilic&#xA;  substitution of substituted aryls?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No, this is not a general phenomenon.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case of electrophilic aromatic substitution of fluorobenzene.  Fluorobenzene undergoes electrophilic nitration roughly 10 times slower than benzene, yet it strongly directs ortho and para.  Nitration of fluorobenzene produces primarily ortho (13%) and para (86%) product, only a trace of meta product is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that that substitution occurs primarily at the ortho and para positions signals that the fluorine has a +R effect.  The fluorine is donating electrons through resonance into the benzene ring - but only to the ortho and para positions; resonance structures cannot be drawn that donate electron density to the meta position (this applies to both the ground state and transition state resonance structures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorine is much more electronegative than hydrogen, so inductively fluorine will remove electron density from the ring (-I effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction rate for electrophilic nitration is decelerated in fluorobenzene compared to benzene, this indicates that there is less stabilization (less electron density to stabilize the positive charge) of the transition state in the fluorobenzene reaction than in the benzene raection. If the fluorine substituent is donating electron density through resonance, but overall the electronic stabilization has decreased, then the -I effect must overwhelm the +R effect in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a continuous rebalancing of inductive and resonance effects as we move through the series benzene, toluene, dimethyl aniline, anisole and fluorobenzene.  With dimethyl aniline and anisole, the resonance effects far outweigh the inductive effects, but by the time we reach fluorine the inductive effect outweighs the resonance effect.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-03T21:24:01.463" LastActivityDate="2015-02-03T21:24:01.463" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24573" PostTypeId="2" ParentId="24569" CreationDate="2015-02-03T18:10:01.767" Score="1" Body="&lt;p&gt;Generally, it is observed that if there are lone pair(s) present on the atom of function group directly attached to to the conjugated system, then $+R$ effect is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the same atom has double bonds or dative bonds with another atom, then $-R$ effect is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; –$\ce{OH}$ shows +R due to 2 lone pairs on Oxygen.&#xA;— $\ce{NO2}$ shows -R due to one double bonds between Nitrogen and Oxygen, and one dative bond from Nitrogen to another Oxygen.&lt;/p&gt;&#xA;" OwnerUserId="12905" LastEditorUserId="7448" LastEditDate="2015-02-03T18:29:56.157" LastActivityDate="2015-02-03T18:29:56.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24574" PostTypeId="1" CreationDate="2015-02-03T18:12:50.550" Score="0" ViewCount="83" Body="&lt;p&gt;Employing the particle in 1-D box can momentum and total energy(P.E + K.E (x)) can be measured simultaneously?&lt;/p&gt;&#xA;" OwnerDisplayName="user12898" LastEditorUserId="12660" LastEditDate="2015-02-03T20:08:06.023" LastActivityDate="2015-02-03T21:47:03.170" Title="the particle in 1-D box momentum and total energy(P.E + K.E (x))" Tags="&lt;homework&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24577" PostTypeId="2" ParentId="24574" CreationDate="2015-02-03T20:07:10.043" Score="1" Body="&lt;p&gt;As this is obviously a homework problem, I'll just give you a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Observables can be measured simultaneously with arbitrarily small uncertainties if their operators commute. Compute the &lt;a href=&quot;http://vergil.chemistry.gatech.edu/notes/quantrev/node18.html&quot; rel=&quot;nofollow&quot;&gt;commutator&lt;/a&gt; of the momentum and particle-in-a-box Hamiltonian operators; if the commutator is zero, the answer to your question is &quot;yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-03T21:47:03.170" LastActivityDate="2015-02-03T21:47:03.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24578" PostTypeId="1" CreationDate="2015-02-03T20:37:41.130" Score="2" ViewCount="17681" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wAMxi.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the $\ce{HO}$ bond be written as $\ce{OH}$ and still be correct?&lt;/p&gt;&#xA;" OwnerUserId="13180" LastEditorUserId="9961" LastEditDate="2018-04-16T18:28:02.060" LastActivityDate="2018-04-16T18:28:02.060" Title="Writing atoms of functional group backwards" Tags="&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24579" PostTypeId="2" ParentId="24578" CreationDate="2015-02-03T20:41:19.183" Score="15" Body="&lt;p&gt;They mean the same thing. If the OH group on the left were written as OH, it would look like there was a carbon-hydrogen bond. By inverting the order, it makes it clear that the oxygen is bonded to the carbon and the hydrogen is bonded to the oxygen. $\ce{NH2}$ might also be written as $\ce{H2N}$ in a structure drawing if a bond was coming from the right, but the functional group is the same in both cases—the group is connected to the rest of the molecule by the nitrogen atom.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-03T20:41:19.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24580" PostTypeId="1" AcceptedAnswerId="24581" CreationDate="2015-02-03T20:42:59.290" Score="0" ViewCount="334" Body="&lt;p&gt;Are binary ionic solids like $\ce{NaCl}$ comprised of a cubic close packed structure of the anions with the cations filling interstital holes or do the cations and the anions contribute to the cubic close packed structure equally? I am really struggling with solid state chemistry - any help would be gratefully received. Also any suggestions for good learning resources on this topic would be great. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2015-02-03T22:41:30.297" LastActivityDate="2015-02-03T22:41:30.297" Title="In ionic solids such as NaCl, do the cations fill interstital holes in the cation's cubic close pack structure?" Tags="&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-04T07:01:41.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="24581" PostTypeId="2" ParentId="24580" CreationDate="2015-02-03T21:50:12.323" Score="1" Body="&lt;p&gt;The relative radii of the ions, and the stoichiometric ratio of the ions needs to be considered.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For NaCl, the radius of Na+ is 0.695 that of Cl-. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Na+ ions occupy octahedral holes in the cubic close packed lattice of Cl-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the radii are more equal (ratio greater than 0.732) then each type of ion will form a simple cubic lattice, with each ion of one lattice filing a cubic hole of the other lattice. For example, CsCl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the radii are very different (ratio less than 0.414) the smaller ions will fill tetrahedral holes in the cubic close packed lattice of the larger ions. For example ZnS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf&quot; rel=&quot;nofollow&quot;&gt;http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-03T21:50:12.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24583" PostTypeId="1" CreationDate="2015-02-04T00:23:07.407" Score="1" ViewCount="350" Body="&lt;p&gt;When boiling water at room temperature, what is the relationship energy has with the air pressure that is making the water boil cold instead of hot?&lt;/p&gt;&#xA;" OwnerUserId="13182" LastEditorUserId="7448" LastEditDate="2015-02-04T16:11:25.323" LastActivityDate="2015-02-04T16:11:25.323" Title="The relationship between energy and air pressure?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;energy&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24584" PostTypeId="1" CreationDate="2015-02-04T01:29:08.243" Score="1" ViewCount="3493" Body="&lt;p&gt;Does increasing the concentration of reactants mean the equilibrium concentration of the products will be greater (than the concentration of the reactants?)&lt;/p&gt;&#xA;" OwnerUserId="13183" LastActivityDate="2015-02-04T12:06:13.637" Title="Equilibrium Concentrations of Products/Reactants" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24585" PostTypeId="2" ParentId="24583" CreationDate="2015-02-04T02:21:05.897" Score="1" Body="&lt;p&gt;Boiling occurs when the vapor pressure of a liquid reaches ambient pressure.  In the example you asked about, water can be made to boil by removing the air in the chamber in which the vessel holding the water is kept.  When the air pressure gets low enough, there is not enough pressure to keep bubbles of water created by the water's vapor pressure from expanding.  We call this boiling.  At atmospheric pressure the water still exerts vapor pressure but the air keeps the bubbles from expanding.  The only way for water to reach the vapor phase here is for it to evaporate from the surface so we see no bubbles.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-04T02:21:05.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24586" PostTypeId="2" ParentId="24583" CreationDate="2015-02-04T02:38:42.783" Score="1" Body="&lt;p&gt;Think of it as the air pressure &lt;em&gt;holding in&lt;/em&gt; water, preventing the water in the container from reaching a lower energy state. Without pressure, the fastest-moving water molecules fly off, leaving behind slower (i.e. colder) water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As water boils, disregarding superheating, &lt;a href=&quot;http://en.wikipedia.org/wiki/Superheating&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Superheating&lt;/a&gt;, the liquid stays at the boiling point and energy from the heat source goes into breaking the bonds (e.g. hydrogen bonds) holding the liquid together. If pressure is lowered, the heat in the water can vaporize more of the liquid, removing the fastest molecules, thereby cooling the remaining liquid to the new boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing air pressure is not adding energy to the system, though work is done &lt;em&gt;against atmospheric pressure&lt;/em&gt;. If you have a piston between two cylinders, as it moves away from one side, pressure drops, but work must be done to compress the fluid on the other side.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-04T02:38:42.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24588" PostTypeId="2" ParentId="24432" CreationDate="2015-02-04T04:13:14.483" Score="2" Body="&lt;p&gt;At least one significant reason is that the voltage you &quot;expect&quot; is probably for a standard state of one mole per liter of $\ce{H+}$, i.e. at pH 0.  Neither the magnesium acetate nor the plain (tap?) water you used for the experiments come close to this pH. Also, the standard state for $\ce{H2}$ would be one bar of atmospheric pressure.  Probably this condition was not achieved either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the expected potential at any pH using the Nernst Equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half-cell reactions in your system are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Mg^0 -&amp;gt; +2e- +Mg^{+2}}$ (this has a potential of 2.372 V &lt;b&gt;not&lt;/b&gt; 2.7 V as a mistakenly wrote in a prev. version of this answer)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{2H^+ +2e- -&amp;gt; H2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The total reaction is thus $\ce{Mg^0 +2H+ -&amp;gt; Mg^{+2} + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So applying the Nernst equation to a one molar magnesium solution at pH 7 (i.e. $\ce{H+}$ concentrations of $10^{-7}$ mol/L) would give:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q$&lt;/li&gt;&#xA;&lt;li&gt;$E_\text{cell} = 2.4 \text{ V} - \frac{RT}{zF} \ln \frac{1}{(10^{-7})^2 }$&lt;/li&gt;&#xA;&lt;li&gt;$E_\text{cell} = 2.4 \text{ V} - 0.41 \text{ V} = 1.9 \text{ V}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously, this correction gives you a predicted voltage that is &lt;strike&gt;still higher than what you measure&lt;/strike&gt; closer what you measured.  &lt;strike&gt;But you reported using acid and getting a higher potential, so it's clear that the pH effect matters for your system.  But something else must be going on too.&lt;/strike&gt;  Some other possibilities for other minor effects that tweak the voltage a bit more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrogen partial pressure is unaccounted for&lt;/li&gt;&#xA;&lt;li&gt;overpotential&lt;/li&gt;&#xA;&lt;li&gt;your cell is not open but is conducting current&lt;/li&gt;&#xA;&lt;li&gt;dirty electrodes etc.&lt;/li&gt;&#xA;&lt;li&gt;other unaccounted for electrochemical reactions (copper oxidation, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-06T04:54:33.123" LastActivityDate="2015-02-06T04:54:33.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24589" PostTypeId="2" ParentId="24318" CreationDate="2015-02-04T04:25:12.613" Score="4" Body="&lt;p&gt;It helps to use &quot;fully compressed&quot; formulae so you can easily count atoms and figure out how to balance carbons, oxygens, hydrogens, and charges.  That is, instead of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH3COO- +CO2 -&amp;gt; CH3CH2COO-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;write it this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C2H3O2- +CO2 -&amp;gt; C3H5O2-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then it's easy to see that this is not a balanced equation.  Carbon atoms are the same on each side, but not oxygen or hydrogen atoms.  You need two more $\ce{O}$ atoms on the products side.  Those have to come from water since that is the only oxygen-containing species you are allowed to use.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C2H3O2- +CO2 -&amp;gt; C3H5O2- + 2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now oxygen is balanced, but hydrogen is still imbalanced.  That can be rectified by adding $\ce{H+}$ into the equation as appropriate.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C2H3O2- +CO2 +6H+ -&amp;gt; C3H5O2- + 2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now only charge is imbalanced, which can be fixed by adding electrons to complete the balanced half-reaction:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C2H3O2- +CO2 +6H+ +6e- -&amp;gt; C3H5O2- + 2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-04T04:25:12.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24590" PostTypeId="2" ParentId="24514" CreationDate="2015-02-04T05:00:36.893" Score="2" Body="&lt;p&gt;All electrochemistry is relative.  And any time you have a reduction or oxidation reaction, you are dealing with an electrochemistry problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NADH is &quot;more energetic&quot; than NAD+ because the hidden assumption is that energy levels are &lt;i&gt;relative&lt;/i&gt; to reaction with the atmosphere.  NADH can react with oxygen in the atmosphere to release energy.  But on a planet with no oxygen and lots of hydrogen in the atmosphere, NAD+ would be the &quot;more energetic&quot; molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another less exotic example is anaerobic fermentation in yeast.  NADH builds up in glycolysis, and must be converted back to NAD+ through formation of ethanol + CO2.  That process consumes energy; thus, in anaerobic environments, often times NAD+ can be thought of as a higher-energy molecule than NADH.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-04T05:00:36.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24591" PostTypeId="1" CreationDate="2015-02-04T07:14:57.127" Score="1" ViewCount="11154" Body="&lt;p&gt;I have explained how hyperconjugation leads to the directing properties of alkyl groups below.&lt;/p&gt;&#xA;&lt;p&gt;Carbon joining substituent now need only single bond. So if we remove one bond between &lt;span class=&quot;math-container&quot;&gt;$\ce{H-C}$&lt;/span&gt; along with electron from hydrogen we have a free &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ion. The bond which was removed formed a bond with carbon joining the substituent(second picture in second row). From the those two electron (on the left of the second picture in second row) we see one come from electron joining substituent(from hydrogen) and other from the carbon atom at ortho position.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How did one bond between Hydrogen and Carbon move to &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C}$&lt;/span&gt; along with one electron from hydrogen?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lge5C.png&quot; alt=&quot;Hyperconjugation in Toluene &quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we see the electron from hydrogen jumping from ortho to para position to methyl substituent. &lt;strong&gt;How does this jump actually happen?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is it that the electron( from hydrogen of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt;) on Carbon at ortho jump(&lt;strong&gt;Tell me in detail how this jump occur&lt;/strong&gt;) to meta position due to formation of double bond between the carbon at ortho and meta position. Due to the double bond, electron( from hydrogen of &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt;) on meta came to para position and so on.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How does the double bond between the carbon at ortho and meta position send electron to para position?&lt;/strong&gt;&#xA;The explanation given above is what I think. &lt;strong&gt;Is it right?&lt;/strong&gt;&#xA;In my book the following was written:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hyperconjugation involves delocalization of &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;-electron through the overlap of &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; orbital of double bond with &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; orbital of the adjacent single bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Please explain the above extract from book using the above example of hyperconjugation of toluene.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-11-19T18:54:17.567" Title="How does hyperconjugation lead to the directing properties of alkyl group?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;hyperconjugation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24592" PostTypeId="2" ParentId="24591" CreationDate="2015-02-04T08:08:35.170" Score="0" Body="&lt;p&gt;The electrons don't &quot;jump&quot; from one position to another:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;when the $\ce{H+}$ leaves the methyl group, a full p-orbital is formed on the carbon. this p orbital is on the same plane of the pi bond of the aromatic ring.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;when the p-orbital is on the same plane, the electrons can delocalize (move) along the new formed conjugated pi molecular orbital.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;so, you get 8 electrons (2 from the methyl group + 6 from the aromatic ring) on 7 carbons all on the same oriented p-orbital (or just $\pi$ bond). that means you have a total of -1 charge over this bond, so where it will be more localized? (or on which carbons there will be a negative charge?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the picture you posted are resonance structures (they describe the delocalized electrons). There are 3 resonance structures: 1. minus on the carbon of the methyl group 2. minus on orto carbon 3. minus on para carbon &#xA;The electron &quot;moves&quot; from each structure by moving the double bond.&#xA;the &quot;minus on the carbon of the methyl group&quot; will contribute to the overall structure the least, because its a primary carbon-anion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13186" LastEditorUserId="4945" LastEditDate="2015-02-04T08:33:39.970" LastActivityDate="2015-02-04T08:33:39.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24593" PostTypeId="1" AcceptedAnswerId="24594" CreationDate="2015-02-04T08:54:00.230" Score="1" ViewCount="3011" Body="&lt;p&gt;I just wanted some quick help with a lab I'm doing, based on absorption spectroscopy.  I'm given an aqueous stock solution of Yellow Dye #5 with a concentration of 54.5 micro moles. I'm asked to create a standard curve.  It gives me the diluted volume as 50 mL, and has a data table with the volume of stock solutions and absorbance values for each.  It asks me to create a standard curve (absorbance versus concentration) in excel for the Yellow Dye #5.  First off, would I just use the constant concentration of 54.5 micro moles for this standard curve graph and the absorbance values in the data table?  I'm not familiar with what a standard curve is, or if the volume of stock solution affects the concentration value in the graph. If anyone replies I'll probably comment on the right answer with a couple more, thanks for the help!  It means a lot!&lt;/p&gt;&#xA;" OwnerUserId="13187" LastEditorUserId="418" LastEditDate="2015-04-18T05:30:35.487" LastActivityDate="2015-04-18T05:30:35.487" Title="Standard curve of absorption and concentration" Tags="&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24594" PostTypeId="2" ParentId="24593" CreationDate="2015-02-04T09:27:30.563" Score="0" Body="&lt;p&gt;Supposed that &lt;em&gt;Yellow Dye #5&lt;/em&gt; refers to the FD&amp;amp;C numbering, this is a food colorant more known as &lt;strong&gt;tartrazine&lt;/strong&gt; outside the U.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard curve correlates absorbance at a particular wavelength with the concentration of the dye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed that no distortion (e.g. due to aggregation effects) is present, the relation is given by the &lt;em&gt;Lambert-Beer law&lt;/em&gt;: $$E_\lambda = \epsilon_\lambda \cdot c \cdot d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_\lambda$ is the absorbance, $\epsilon_\lambda$ is the molar absorption coefficient (specific for a particular dye at a particular wavelength $\lambda$ in a certain solvent), $c$ is the concentration of the dye and $d$ is the path length of the cuvette. &lt;em&gt;Please pay attention to the units in your calculations!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to create a standard curve, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Record the uv spectrum of the stock solution. If you have two-channel spectrometer, put a second cuvette with the neat solvent in that channel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Get a couple of volumetric flasks and dilute the stock solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Record the uv spectra for each of these solutions.&lt;/li&gt;&#xA;&lt;li&gt;Plot absorbance (probably at around 420 nm) vs concentration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-04T09:27:30.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24595" PostTypeId="2" ParentId="6944" CreationDate="2015-02-04T09:56:18.630" Score="-1" Body="&lt;p&gt;The London / van der Waals forces are weak forces but their effect depends both on the magnitude of the forces themselves and the distance over which they act. Linear molecules can align and lie parallel to each other, giving small distances of separation. This makes attraction produced by the same forces/moments greater.&lt;/p&gt;&#xA;" OwnerUserId="13188" LastEditorUserId="402" LastEditDate="2017-04-02T20:22:04.467" LastActivityDate="2017-04-02T20:22:04.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24596" PostTypeId="1" AcceptedAnswerId="24602" CreationDate="2015-02-04T10:48:40.263" Score="3" ViewCount="8352" Body="&lt;p&gt;I would like to ask which group do actinide and lanthanide belong?&#xA;I had a question in my assingment that which group has most number of elements?&#xA;if they belong to 3rd group then the 3rd group should be the answer.&#xA;Am I wrong or Right? Please Help Me.&lt;/p&gt;&#xA;" OwnerUserId="13190" LastActivityDate="2017-11-10T16:01:45.310" Title="Actinides and Lanthanides group" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24597" PostTypeId="1" AcceptedAnswerId="24610" CreationDate="2015-02-04T11:29:19.150" Score="3" ViewCount="3283" Body="&lt;p&gt;I'm a bit confused when asked to specify the chiral carbons on this sketch of Propranolol as I'm not sure how the $\ce{H3C}$ and $\ce{CH3}$ play in if they are reversed in the name like that...Does that even matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure the only chiral carbon is the one in the middle of those and the $\ce{NH}$. I've heard that you need 4 different groups/atoms surrounding the carbon for it to be chiral but how much truth is there in that, as I've found several with only 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9T9qx.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="41890" LastEditDate="2017-11-27T18:00:13.423" LastActivityDate="2017-11-27T18:00:13.423" Title="Chirality on Propranolol" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chirality&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24598" PostTypeId="2" ParentId="24597" CreationDate="2015-02-04T11:56:40.313" Score="2" Body="&lt;p&gt;$\ce{CH3}$ and $\ce{H3C}$ mean the same thing &amp;mdash; methyl &amp;mdash; But are written that way to show the C is attached directly. It is obvious, but people prefer writing it that way to show attachment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I can see only one chiral atom. These are are such that none of the groups directly attached are same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Example:&lt;/em&gt; $\ce{CH3-CH(OH)-C2H5}$  (the $\ce C_2$ is the chiral carbon).&lt;/p&gt;&#xA;" OwnerUserId="12905" LastEditorUserId="7448" LastEditDate="2016-11-03T11:50:34.713" LastActivityDate="2016-11-03T11:50:34.713" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24599" PostTypeId="2" ParentId="24584" CreationDate="2015-02-04T12:01:27.463" Score="1" Body="&lt;p&gt;Adding a reactant increases the concentration of products at equilibrium, but it doesn't necessarily mean that the product concentrations will be larger than the reactant concentrations at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a reaction $\ce{A &amp;lt;=&amp;gt; B}$. Then&#xA;$$K = \frac{[\ce{B}]}{[\ce{A}]}$$&#xA;where $[\ce{A}]$ and $[\ce{B}]$ are the concentrations at equilibrium and $K$ is a constant. The ratio $[\ce{B}]/[\ce{A}]$ is constant at equilibrium, so if you add some $\ce{A}$ to the system, some $\ce{A}$ must react and become $\ce{B}$ to maintain the ratio. The equilibrium concentration of the products will increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that increase in product concentration won't necessarily mean that the equilibrium product concentration becomes higher than that for the reactants. If $K$ is bigger than one, the equilibrium concentration of the products will be greater than that for the reactants; if it's less than one, the equilibrium concentration of the reactants is bigger.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="4945" LastEditDate="2015-02-04T12:06:13.637" LastActivityDate="2015-02-04T12:06:13.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24600" PostTypeId="1" AcceptedAnswerId="24613" CreationDate="2015-02-04T12:10:06.483" Score="2" ViewCount="317" Body="&lt;p&gt;Pauling's definition of electronegativity defines the difference of electronegativity between two elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose three elements A, B, and C. Then we can calculate $ϰ_a - ϰ_b$, $ϰ_b - ϰ_c$, and $ϰ_c - ϰ_a$ independently. Is the sum of the three values always zero? Why?&lt;/p&gt;&#xA;" OwnerUserId="13191" LastActivityDate="2015-02-04T21:48:46.547" Title="Is Pauling electronegativity well-defined?" Tags="&lt;electronegativity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24601" PostTypeId="1" CreationDate="2015-02-04T12:10:21.513" Score="3" ViewCount="832" Body="&lt;p&gt;I am learning about proton NMR and spin-spin coupling, and am confused about whether splitting occurs over an ester bond. Specifically, in the case of ethyl methanoate, HCOOCH2CH3, if I were to number the proton environments going from left to right as 1,2 and 3, would 1 and 2 cause splitting of each other?&#xA;If they would not cause splitting, then I would expect a singlet, a quartet and a triplet respectively. If they do cause splitting, then would you get a triplet, a quartet of doublets, and a triplet for 1,2 and 3 respectively? I know that with -OH and -NH the protons do not couple (although I am not sure why), so I was wondering whether the same principle may perhaps apply here...&lt;/p&gt;&#xA;" OwnerUserId="7757" LastActivityDate="2015-02-04T12:21:34.767" Title="Spin spin coupling in a proton NMR of an ester?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;spin&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24602" PostTypeId="2" ParentId="24596" CreationDate="2015-02-04T12:18:19.660" Score="7" Body="&lt;p&gt;The periodic table actually looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tJeD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...you can see that the first 14 elements in either series don't belong to any group from the main body of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could say that the lanthanides and actinides form fourteen new 2-element groups, with Lu and Lr belonging to group 3. But &lt;a href=&quot;http://www.iupac.org/publications/ci/2009/3101/1_leigh.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC does not number these groups&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGfTf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-02-04T12:18:19.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24603" PostTypeId="2" ParentId="24601" CreationDate="2015-02-04T12:21:34.767" Score="7" Body="&lt;p&gt;In the most common case you can only observe &lt;sup&gt;3&lt;/sup&gt;J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the spins are too far away from each other to cause a visible splitting, you are correct that it would result in a singlet a quartet and a triplet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of OH and NH protons, chemical exchange can play a role. This is a bit more complicated, but the simple version is that protons that can exchange quickly with the solvent are often not observable in NMR (or just as a very broad signal) and due to that also don't couple. Though in the most common cases the OH proton is usually too far away from the other protons in the molecule to show a coupling anyway.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-02-04T12:21:34.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24604" PostTypeId="1" CreationDate="2015-02-04T13:09:20.090" Score="8" ViewCount="915" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum extent of a hydride shift to form the most stable carbocation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we consider the reaction between 2-methylpropan-1-ol and $\ce{HBr}$, then of course, the carbocation formed after a hydride shift will be more stable (and lead to the major product). 3-methylbutan-1-ol will also undergo hydride shift to some extent to form the more stable carbocation, but is there a limit to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like can the number of carbons be so much that there is almost no chance to make the product with hydride shift possible? Like consider the scenario below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zbnrv.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the maximum $n$ possible here to make the product with hydride shift possible?&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="5026" LastEditDate="2018-04-12T12:03:35.573" LastActivityDate="2018-04-12T12:03:35.573" Title="What is the maximum extent of a hydride shift to form the most stable carbocation?" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24606" PostTypeId="1" AcceptedAnswerId="24614" CreationDate="2015-02-04T13:57:55.323" Score="6" ViewCount="4340" Body="&lt;p&gt;I there a &lt;em&gt;Good&lt;/em&gt; Version of a printable Periodic table Which is minimalistic and gives &lt;strong&gt;necessary&lt;/strong&gt; information only such as -&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At. Number&lt;/li&gt;&#xA;&lt;li&gt;Element Name&lt;/li&gt;&#xA;&lt;li&gt;At. weight&lt;/li&gt;&#xA;&lt;li&gt;Shell configuration &lt;/li&gt;&#xA;&lt;li&gt;color coding of various groups or Series&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="7448" LastEditDate="2015-02-04T14:45:49.057" LastActivityDate="2017-02-15T09:46:59.093" Title="Periodic table for printing" Tags="&lt;periodic-table&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24607" PostTypeId="1" AcceptedAnswerId="24609" CreationDate="2015-02-04T14:01:04.117" Score="7" ViewCount="1145" Body="&lt;p&gt;Below is a picture of the mechanism:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/F3Vfo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the mechanism but what I don't understand is why it's $\mathrm{S_N2}$. The solvent used is polar and protic, there is a tertiary carbon that would be suitable for $\mathrm{S_N1}$ as well. For what reason does this reaction occur exclusively via $\mathrm{S_N2}$?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2015-02-04T19:38:23.230" LastActivityDate="2015-02-04T19:38:23.230" Title="Why does the opening of an epoxide occur via an Sn2 like mechanism when using a methanoate ion as a nucleophile and methanol as a solvent?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24609" PostTypeId="2" ParentId="24607" CreationDate="2015-02-04T14:15:52.343" Score="4" Body="&lt;p&gt;Quotation from Clayden: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Without protonation, the epoxide oxygen is a poor leaving group, and leaves only if ‘pushed’&#xA;  by a strong nucleophile: the reaction becomes pure SN2. Steric hindrance becomes the controlling&#xA;  factor and methoxide attacks only the primary end of the epoxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-02-04T14:15:52.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24610" PostTypeId="2" ParentId="24597" CreationDate="2015-02-04T14:46:55.840" Score="5" Body="&lt;p&gt;As stated in user208322's answer, $\ce{CH3}$ and $\ce{H3C}$ are exactly the same, they're only written differently for ease of reading. As per your question about how much truth there is to having 4 groups, it's 100% truth. You &lt;em&gt;cannot&lt;/em&gt; have a chiral carbon that doesn't have 4 groups attached to it (i.e. must be sp3 hybridized). If you think you've found a chiral carbon that only has 3 groups attached to it, you're probably forgetting the unwritten hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example specifically, there is only one chiral carbon, the one attached to the secondary alcohol. Although it may look like the carbon is attached to the same group twice, you need to look further than just the first atom. Yes it is attached to two carbons, but one of those carbons is attached to a nitrogen and the other is attached to an oxygen, so the groups are different. Two groups are  only the same if you can follow it down every single atom and they match exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: In regards to your question about following the chain all the way down, realistically you only need to follow it until you see that it differs from the other chains. So for chains 1 and 3 once you get to the nitrogen and oxygen atoms you see that they're different and there's no point continuing. And the unwritten hydrogen has to do with the fact that organic chemists use shorthand in their molecular structures. Basically, carbons and hydrogens are &lt;em&gt;extremely&lt;/em&gt; common in organic molecules so they just don't write them. All the points are carbons unless otherwise labelled, and they must have 4 bonds, so where you see that the carbon is only bonded to 3 things, it is also bonded to a hydrogen as well. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Skeletal_formula&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt; for more info (specifically, &lt;a href=&quot;http://en.wikipedia.org/wiki/Skeletal_formula#Implicit_carbon_and_hydrogen_atoms&quot; rel=&quot;nofollow&quot;&gt;this section&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="13030" LastEditDate="2015-02-04T15:29:06.393" LastActivityDate="2015-02-04T15:29:06.393" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24611" PostTypeId="2" ParentId="6181" CreationDate="2015-02-04T14:51:18.500" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Osmotic pressure is measured at room temperature while other colligative properties are measured at high temperatures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;OP uses molarity while other CP uses molality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;OP is used at standard pressure i.e. 1atm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it can also measure molecular mass for a diluted solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13195" LastEditorUserId="7448" LastEditDate="2015-02-04T15:05:56.043" LastActivityDate="2015-02-04T15:05:56.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24612" PostTypeId="1" CreationDate="2015-02-04T14:52:16.103" Score="1" ViewCount="211" Body="&lt;p&gt;In the hydrolysis of propanamide ($\ce{CH3CH2CONH2}$) how do I know which molecules react with the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH3CH2CONH2 + H2O -&amp;gt; ?}&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to split the molecule. In the videos I've watched (&lt;a href=&quot;https://www.youtube.com/watch?v=MN6edmwMExw#t=612&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;) it seems as they just assume you know this.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="16683" LastEditDate="2017-04-08T01:46:46.490" LastActivityDate="2017-04-08T01:47:41.913" Title="How do I know which atoms react during hydrolysis?" Tags="&lt;organic-chemistry&gt;&lt;water&gt;&lt;carbonyl-compounds&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24613" PostTypeId="2" ParentId="24600" CreationDate="2015-02-04T14:54:44.433" Score="4" Body="&lt;p&gt;If you have a single, unique value of $\chi_i$ for each element $i$ then yes, the sum of the three values is always zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are assigning electronegativities that differ with oxidation state or chemical environment, the answer can be no (for example, if element B has a different electronegativity in an AB bond than it has in a BC bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;Pauling electronegativity&lt;/a&gt; difference is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\chi}_{a}-{\chi}_{b}=(1\ eV)^{-\frac{1}{2}}\,\sqrt{{E}_{d}\left( a,b\right) -\frac{{E}_{d}\left( a,a\right) +{E}_{d}\left( b,b\right) }{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_d(i,j)$ is the dissociation energy for a single bond between element $i$ and element $j$.&#xA;The definition assumes that you can consistently and accurately say that the square of the electronegativity difference between A and B is equal to the difference between the dissociation energy for AB and the average of the dissociation energies for AA and BB. If this isn't consistently true for each combination of three elements A, B, and C, your sum will be nonzero. Pauling fit his electronegativity values to the data to average out these inconsistencies.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-04T21:48:46.547" LastActivityDate="2015-02-04T21:48:46.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24614" PostTypeId="2" ParentId="24606" CreationDate="2015-02-04T15:00:25.777" Score="8" Body="&lt;p&gt;The &lt;a href=&quot;https://iupac.org/what-we-do/periodic-table-of-elements/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC periodic table&lt;/a&gt; (&lt;a href=&quot;http://iupac.org/cms/wp-content/uploads/2015/07/IUPAC_Periodic_Table-28Nov16.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2016-pdf&lt;/a&gt;) is good because it quantifies the accuracy of the natural abundance atomic weights, but it doesn't give electron configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.nist.gov/pml/data/periodic.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST periodic table&lt;/a&gt; has electron configuration, and probably is closest to what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dkIFU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NIST table is also good from a copyright point of view, because you can reproduce it without worrying about copyright infrigment, as &lt;a href=&quot;http://www.usa.gov/copyright.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;US government works are generally not subject to copyright protection&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2017-02-15T09:46:59.093" LastActivityDate="2017-02-15T09:46:59.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24616" PostTypeId="1" CreationDate="2015-02-04T15:51:48.360" Score="2" ViewCount="2929" Body="&lt;p&gt;We know that 1 mole is equal to $6.023\times10^{23}$ atoms. But why it is not written as $6023\times10^{20}$ ?&lt;/p&gt;&#xA;" OwnerUserId="13027" LastEditorUserId="8083" LastEditDate="2015-02-04T20:23:37.203" LastActivityDate="2015-02-05T13:55:05.100" Title="Why 6.023x10^23 is not written as 6023x10^20?" Tags="&lt;mole&gt;" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24617" PostTypeId="2" ParentId="24616" CreationDate="2015-02-04T15:54:13.457" Score="8" Body="&lt;p&gt;You could write it as either. But the former is scientific notation while the latter is &lt;em&gt;not&lt;/em&gt; considered scientific notation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In scientific notation, you move the decimal place until you have a&#xA;  number [whose absolute value is] between 1 and 10. Then you add a power of ten that tells how&#xA;  many places you moved the decimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.factmonster.com/ipka/A0876783.html&quot; rel=&quot;nofollow&quot;&gt;http://www.factmonster.com/ipka/A0876783.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-02-04T18:11:36.493" LastActivityDate="2015-02-04T18:11:36.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24618" PostTypeId="2" ParentId="24612" CreationDate="2015-02-04T16:51:21.863" Score="2" Body="&lt;p&gt;I think you mean which &lt;em&gt;atoms&lt;/em&gt; react with the water, as clearly the molecule that reacts is propanamide (it's the only non-water molecule in your question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product of the reaction is propanoic acid and ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2CONH2 + H2O -&amp;gt; CH3CH2COOH + NH3}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbonyl carbon in amides (and esters and acyl halides) has a partial positive charge because of the electron withdrawing effect of the carbonyl oxygen.  (Oxygen is very electronegative.)  This makes the carbon atom is activated toward nucleophilic attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleophilic_acyl_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia page&lt;/a&gt; has a good explanation and different mechanisms for acid vs. base conditions.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="194" LastEditDate="2017-04-08T01:47:41.913" LastActivityDate="2017-04-08T01:47:41.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24619" PostTypeId="2" ParentId="24616" CreationDate="2015-02-04T17:46:41.863" Score="12" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/24617/186&quot;&gt;Dissenter's answer&lt;/a&gt; is not quite right, since, strictly speaking, as defined there, the scientific notation will work for positive numbers only. Obviously, it can be easily fixed to work for negative as well, but the case of 0 will continue to be problematic. Just try to follow this rule to understand why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, it is the matter of definitions, but usually, &lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;scientific notation&lt;/a&gt; is defined as just a way of writing numbers in the form $a \cdot 10^b$ &lt;em&gt;without&lt;/em&gt; any restrictions on $a$ which can be &lt;em&gt;any&lt;/em&gt; real number. So, both $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ &lt;em&gt;are&lt;/em&gt; considered to be numbers written in scientific notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation#Normalized_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;normalized&lt;/em&gt; scientific notation&lt;/a&gt;, indeed,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the exponent $b$ is chosen so that the absolute value of $a$ remains at&#xA;  least one but less than ten ($1 ≤ |a| &amp;lt; 10$),&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and thus, only $6.023 \cdot 10^{23}$ is written in this notation. Normalized scientific notation  is just one special case of scientific notation. And yeah, you could not write down zero in this notation, which is kind of awkward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another special case of scientific notation is the so-called &lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_notation#Engineering_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering notation&lt;/a&gt; in which the exponent $b$ is restricted to multiples of 3. Neither $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are written in engineering notation, while $602.3 \cdot 10^{21}$ is. Although engineering notation is a special case of scientific one as defined above, is rarely called scientific notation to avoid an ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-02-04T17:46:41.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24620" PostTypeId="2" ParentId="24427" CreationDate="2015-02-04T17:51:44.567" Score="9" Body="&lt;p&gt;That the question says: &quot;Without any 'binding' forces, the atom couldn't be possibly trapped inside the fullerene&quot;, shows you are misunderstanding fullerenes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms such as Helium and $H_2$ can be confined within the interior of the fullerene.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are even reports of &lt;a href=&quot;http://www.lpi.usra.edu/meetings/impacts97/pdf/6079.pdf&quot; rel=&quot;noreferrer&quot;&gt;extraterrestrial helium trapped in fullerenes&lt;/a&gt; during astroid or comet impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helium can &lt;a href=&quot;http://www.collectionscanada.gc.ca/obj/s4/f2/dsk2/ftp03/MQ40426.pdf&quot; rel=&quot;noreferrer&quot;&gt;move around within the fullerene&lt;/a&gt;.  It is sterically confined rather than bonded.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of La, the metal ion does not remain neutral, but loses about 3 electron to the fullerene.  So, in addition to being sterically confined in the cage, there is the interaction between postively and negatively charged species.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As calculated in a later article &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.20514/abstract&quot; rel=&quot;noreferrer&quot;&gt;Computed Structure and Energetics of&#xA;La@C60&lt;/a&gt; International Journal of Quantum Chemistry, Vol 104, 272–277 (2005), the La has a charge of somewhere betwen 2+ and 3+.  In the lowest energy state, the La is off-center with respect to the fullerene, located between the center and the center of a hexagon face.  Also, the fullerene cage is distorted from icosahedral by Jahn-Teller distortion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The La ion is located 2.62 Angstroms from the nearest carbon atom (4.6 Angstroms from the furthest), when at the potential miniumun, but is expected to move within the fullerene unless the temperature is very low.  The reference also says &quot;there is obviously a relatively large negative charge on some carbon atoms of the cage&quot;, without specifying how this charge is distributed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall I would say, whereas in the case of helium and dihydrogen being confined in the fullerene is completely &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070009930.pdf&quot; rel=&quot;noreferrer&quot;&gt;because of van der Waal repulsive force of avoiding the walls&lt;/a&gt;, in the case of La, there is attractive electrostatic force (ionic interaction) between the La3+ and negatively charged carbon atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www.public.asu.edu/~baluk/reprints/399.pdf&quot; rel=&quot;noreferrer&quot;&gt;Theoretical investigation of structural and thermodynamic properties&#xA;of lanthanum carbides LaCn (n = 2–6)&lt;/a&gt; which finds strong ionic interactions between La and Cn due to charge transfer.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-05T15:57:13.997" LastActivityDate="2015-02-05T15:57:13.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24621" PostTypeId="2" ParentId="6625" CreationDate="2015-02-04T18:02:17.533" Score="6" Body="&lt;p&gt;Often, but not always, mesomeric displacement leads to a shift in prevailing over the inductive effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pi-electrons are at the peripheral orbitals. Association with the nucleus is less strong than at the sigma-electrons. The ionization potential of pi-electrons is smaller and chemical bond is more polarizable. Therefore, the dipole moment associated with the mesomeric effect can prevail over the dipole moment associated with the inductive effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are groups in which the inductive effect prevails, such as halogens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can read about hyperconjugation. Hyperconjugation can have influence to the mesomeric and inductive effects too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; Reutov  O.A., Kurts A.L., Organic Chemistry, vol. 1 of 4, ch. 2.2.2, MSU&lt;/p&gt;&#xA;" OwnerUserId="13196" LastActivityDate="2015-02-04T18:02:17.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24622" PostTypeId="1" AcceptedAnswerId="24631" CreationDate="2015-02-04T18:10:34.383" Score="4" ViewCount="2595" Body="&lt;p&gt;My professor does not believe in the hyperconjugative effect. Instead, we only consider inductive effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do both effects have the same general effect? I know for methyl groups both have the same effect; both hyperconjugation and induction can be invoked to explain the electron donating tendencies of the methyl group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any cases in which the hyperconjugative effect and the inductive effect have opposite effects? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="186" LastEditDate="2015-10-02T16:17:15.857" LastActivityDate="2015-10-02T16:17:15.857" Title="Does hyperconjugation and induction always have the same effect?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;orbitals&gt;&lt;hyperconjugation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24623" PostTypeId="1" CreationDate="2015-02-04T18:22:36.297" Score="1" ViewCount="292" Body="&lt;p&gt;Warning: This is a question from my teacher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In America hybridization isn't done, while in Europe it is done around the 10th grade. Is there any reason or advantage for doing(not doing) it as part of the school program?&lt;/p&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2015-02-04T19:57:48.133" Title="Why isn't hybridization done in America?" Tags="&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-02-04T19:50:45.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="24624" PostTypeId="2" ParentId="24623" CreationDate="2015-02-04T18:27:40.133" Score="4" Body="&lt;p&gt;In Russia hybridization of carbon learns at the beginning of the 10th grade, before the main organic chemistry. Hybridization in the atoms of metals (complexes) are studied in grade 11. These two topics are difficult for students to understand. But they can be on the exams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is easier to understand the spatial structure&lt;/li&gt;&#xA;&lt;li&gt;Explanation of coordination numbers in complexes&lt;/li&gt;&#xA;&lt;li&gt;For an explanation of the facts, when the atom forms a large number of bonds than the number of unpaired electrons in the ground state&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Difficult to understand&lt;/li&gt;&#xA;&lt;li&gt;There are many different theories&lt;/li&gt;&#xA;&lt;li&gt;Rarely used in the study follow-up program&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13196" LastEditorUserId="13196" LastEditDate="2015-02-04T19:57:48.133" LastActivityDate="2015-02-04T19:57:48.133" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24625" PostTypeId="2" ParentId="24622" CreationDate="2015-02-04T18:49:56.377" Score="0" Body="&lt;p&gt;The effect is similar to the positive mesomeric, it appears by substituting the hydrogen in the alkyl group of multiple bond. This effect is directed toward multiple bond:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CB1rj.gif&quot; alt=&quot;This effect is directed toward multiple bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperconjugation reminds positive mesomeric effect, because it gives sigma-electrons in the conjugated pi-system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G58cl.gif&quot; alt=&quot;pi-system&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive effect and hyperconjugation usually occur simultaneously. Inductive effect will undoubtedly outweigh, when the molecule is easily polarized. In addition, the spatial effect may mask the hyperconjugation effect.&#xA;In benzene homologues hyperconjugation enhances the inductive effect.&lt;/p&gt;&#xA;" OwnerUserId="13196" LastEditorUserId="13196" LastEditDate="2015-02-04T19:51:04.427" LastActivityDate="2015-02-04T19:51:04.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24626" PostTypeId="2" ParentId="24616" CreationDate="2015-02-04T19:09:04.440" Score="5" Body="&lt;p&gt;If you let the magnitude of the coefficient / significand vary outside the &lt;code&gt;[1,10)&lt;/code&gt; interval, then you have to count digits and add that to the exponent to see the magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With normalized scientific notation, you can see the magnitude at a glance.  This is WHY normalized notation is standard practice.&lt;/p&gt;&#xA;" OwnerUserId="13199" LastActivityDate="2015-02-04T19:09:04.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24627" PostTypeId="2" ParentId="24616" CreationDate="2015-02-04T19:41:24.903" Score="6" Body="&lt;p&gt;To add to all the other answers, in case you're wondering, the reason we like to use the one standard for scientific notation is because it makes comparing sizes very easy. A lot of times in making approximations you only care about the order of magnitude (this is especially true when looking at things like equilibrium constants). Having this standard essentially allows you to compare multiple values/quantities very quickly by disregarding the coefficients and just having a quick glance at their orders of magnitude. If we all used different standards, we'd have to take into account how many digits are before the decimal place.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastActivityDate="2015-02-04T19:41:24.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24628" PostTypeId="2" ParentId="24591" CreationDate="2015-02-04T19:59:48.940" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19508/why-methyl-group-is-2-4-directing&quot;&gt;Why methyl group is 2,4-directing?&lt;/a&gt; This question describes in detail why the methyl group is o,p directing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think the major point of your misunderstanding is in the interpretation of resonance structures. &lt;strong&gt;Resonance structures are not stable structures in their own right.&lt;/strong&gt;&#xA;A set of resonance structures can be used to represent bonding that cannot be represented using classic Lewis structures. The structures do not exist on their own but the true structure can be viewed as an 'average' of all the contributing resonance structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons do not jump between different atoms but are instead delocalized over all of the atoms with a slightly higher chance of being found at the ortho and para positions in the case of toluene and a slightly lower chance of being found at on the methyl carbon. This gives rise to the increased electron density of the ortho and para positions and therefore the increased negative charge and similarly the decreased electron density on the methyl carbon and the positive charge.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-02-05T12:14:11.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24630" PostTypeId="1" AcceptedAnswerId="24632" CreationDate="2015-02-04T21:05:24.903" Score="5" ViewCount="470" Body="&lt;p&gt;High school student here, I recently purchased a book titled: &quot;Solvolytic Displacement Reactions&quot; by Andrew Streitwieser published in the early 1960's. I found it at a book store and it looks like one of those incredibly valuable (in terms of knowledge) books. I tried reading the first few pages (have a basic background in chemistry) but I do not see &quot;the big picture&quot;. What are solvolytic displacement reactions, how are they used and what are they used for?&lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorUserId="194" LastEditDate="2017-09-23T02:35:20.663" LastActivityDate="2017-09-23T02:35:20.663" Title="What are solvolytic displacement reactions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24631" PostTypeId="2" ParentId="24622" CreationDate="2015-02-04T21:45:56.287" Score="5" Body="&lt;p&gt;The inductive effect occurs through overlapping sigma bonds, whereas the hyperconjugative effect occurs through overlap of a sigma bond with a p orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1be35.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many different types of hyperconjugation and hyperconjugation has been used to explain many different chemical phenomenon.  A nice overview can be found &lt;a href=&quot;https://www.princeton.edu/chemistry/macmillan/group-meetings/hyperconjugation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any cases in which the hyperconjugative effect and the&#xA;  inductive effect have opposite effects?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since inductive effects are transmitted through the sigma system, they effect should be independent of rotation around the bond(s) involved.  On the other hand, the pi type of overlap used in hyperconjugation is very dependent upon rotation about the bond(s) involved.  In the diagram above, rotation about the $\ce{C-C}$ bond by 90° shouldn't affect any inductive effect of the methyl group upon the carbocation center, but such rotation would completely destroy the overlap required for hyperconjugation to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might expect, hydrogen/deuterium isotope effects can be used to assess hyperconjugation $\ce{C-H}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: isotope effects are a subject unto themselves; a nice overview can be found &lt;a href=&quot;https://www.princeton.edu/chemistry/macmillan/group-meetings/RRK-KIE.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  See p. 12 for their application to hyperconjugation.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &quot;thought experiment&quot; where inductive effects and hyperconjugation should have different effects (I don't know if this or a similar experiment has been done).  Take the following two carbocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mJ1gt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would expect them to both have similar inductive effects upon the carbocation center.  The tris-ethyl carbocation, free to rotate about its various $\ce{C-C}$ bonds could also exist in conformations where the $\ce{C-H}$ bonds alpha to the carbocation center are correctly positioned for hyperconjugation to occur - it should show a secondary $\ce{K_{H}/K_{D} &amp;gt;1}$ since hydrogen is more electron donating than deuterium (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/20405/inductive-effect-of-isotopes-of-h/20413#20413&quot;&gt;here&lt;/a&gt; for the reasoning behind this statement).  However, the $\ce{C-H}$ bonds alpha to the carbocation center in the rigid bicycloctyl carbocation cannot adopt a conformation that would allow effective hyperconjugative overlap of these $\ce{C-H}$ bonds with the cationic center - it should show a secondary $\ce{K_{H}/K_{D} =1}$.  Different secondary isotope effects would be expected in the solvolysis of these two systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another case where inductive effects and hyperconjugation do not provide equivalent explanations relates to rotational barriers.  The origin of the rotational barrier in ethane has been a subject of investigation for a long time.  Inductive effects cannot be used to explain the barrier because, as discussed above, inductive effects are rotationally invariant.  An argument using hyperconjugation to explain the rotational barrier has been advanced and is the subject of much discussion (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;nofollow noreferrer&quot;&gt;see the &quot;Rotational barrier of ethane&quot; section&lt;/a&gt;).  Whether the hyperconjugation argument is correct remains to be proven, but it is another case where hyperconjugation can be used to explain observations that the inductive effect cannot.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-02-05T02:34:09.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24632" PostTypeId="2" ParentId="24630" CreationDate="2015-02-04T22:24:32.793" Score="5" Body="&lt;p&gt;A book could be written on this subject!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solvolysis reaction is &lt;a href=&quot;http://en.wikipedia.org/wiki/SN1_reaction&quot; rel=&quot;noreferrer&quot;&gt;an $\ce{S_{N}1}$ reaction&lt;/a&gt; where the solvent acts as the nucleophile and replaces the leaving group.  The following reaction is a solvolysis reaction run in ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jm5yv.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there are some industrial uses of solvolysis reactions and they have been used in labs and by drug companies as a step in a synthetic sequence to prepare some complex molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I suspect the biggest contribution from the study of solvolysis reactions is the impact they have had on our understanding of carbocation chemistry.  From our study of solvolysis reactions we have learned how and why some carbocations rearrange before being captured by solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KdYwp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While studying such rearrangements the solvolysis of &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Norbornyl_cation&quot; rel=&quot;noreferrer&quot;&gt;the norbornyl system was examined&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QgsxF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;norbornyl compound 1 solvolyzes several hundred times faster than compound 3 - why?&lt;/li&gt;&#xA;&lt;li&gt;if compound 1 is prepared optically active, the products are racemic - why?&lt;/li&gt;&#xA;&lt;li&gt;If compound 1 is radiolabeled, the label is scrambled in the products - why&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It took decades to answer these and other questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These investigations led to a whole new field of carbocation chemistry: the classical - non-classical carbocation debate (see the norbornyl link above for a discussion of this topic).  These investigations added a whole new section to physical organic chemistry textbooks.  What was learned from these solvolysis studies completely changed our thinking about carbocation chemistry.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-04T22:31:25.327" LastActivityDate="2015-02-04T22:31:25.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24633" PostTypeId="1" CreationDate="2015-02-05T00:14:53.617" Score="6" ViewCount="1760" Body="&lt;p&gt;I noticed that when I cook black lentils and add lemon juice to the water that's left, the water turns from black to pink!&#xA;What can be the reason for that?&lt;/p&gt;&#xA;" OwnerUserId="13202" LastEditorUserId="402" LastEditDate="2015-02-06T04:49:28.853" LastActivityDate="2015-02-06T04:49:28.853" Title="Adding lemon to lentils changes color from black to pink, why?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24634" PostTypeId="2" ParentId="24604" CreationDate="2015-02-05T00:39:59.653" Score="2" Body="&lt;p&gt;Advances in Physical Organic Chemistry, volume 19, discusses the relative transition state barriers for 1,2-, 1,3-,1-4-, and 1,5-hydride shifts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific cations considered are 2,3 dimethyl-2-butyl, 2,4 dimethyl-2-pentyl, 2,5 dimethyl-2-hexyl and 2,5 dimethyl-2-heptyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, they report that whereas the barrier for 1,4 shift in 2,5 dimethyl-2-hexyl is 12 kcal/mole, this decreases dramatically to less than 6 kcal/mole for 1,5 shift in 2,6 dimethyl-2-heptyl  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference draws the transition states as rings, with the H being transferred as a bridging member of the ring, a five membered ring for the 1,4 shift and a six member ring for the 1,5 shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference also makes a comparison to intermolecular hydride shift between isobutane and t-butyl cation, which has a barrier of 3.6 kcal/mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the reference does not discuss the possibility of longer distance than 1,5 shifts, it seems that because the ends of a long ion can meet each other, long distance (in terms of number of bonds) shifts are possible because through space the distance is short.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-05T00:39:59.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24635" PostTypeId="1" CreationDate="2015-02-05T00:40:41.650" Score="4" ViewCount="455" Body="&lt;p&gt;Raw Tahini seems to me like a non-Newtonian substance. When I mix it fast, it gets tougher. When I mix it slowly, it gets softer.&#xA;Is there any truth to that?&lt;/p&gt;&#xA;" OwnerUserId="13202" LastEditorUserId="11583" LastEditDate="2015-02-05T03:40:03.510" LastActivityDate="2015-02-05T03:40:03.510" Title="Non-Newtonian substance - Tahini" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24636" PostTypeId="2" ParentId="24606" CreationDate="2015-02-05T01:58:51.750" Score="4" Body="&lt;p&gt;The tables at &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ptable.com&lt;/a&gt; might well fill the bill. Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tnwwI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, this is the table I use as the background for my lab computer.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-02-05T01:58:51.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24637" PostTypeId="1" CreationDate="2015-02-05T02:15:58.873" Score="3" ViewCount="4325" Body="&lt;p&gt;Is there a liquid mixture that will become a solid when exposed to air/oxygen. Preferably a spongy material that is difficult to tear.&lt;/p&gt;&#xA;" OwnerUserId="13204" LastActivityDate="2019-04-17T20:55:24.213" Title="Liquid to solid when exposed to air" Tags="&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24638" PostTypeId="1" CreationDate="2015-02-05T02:23:21.027" Score="2" ViewCount="213" Body="&lt;p&gt;I was wondering how scientists are able to handle atoms. They are very small, but surely people are able to interact with them somehow? The Large Hadron Collider is one example. Also, they try and make new elements by mixing atoms together. How are they able to handle something so small?&lt;/p&gt;&#xA;" OwnerUserId="13205" LastEditorUserId="7499" LastEditDate="2015-02-05T03:59:13.360" LastActivityDate="2015-02-05T03:59:13.360" Title="How are atoms manipulated?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24639" PostTypeId="2" ParentId="24637" CreationDate="2015-02-05T02:54:45.813" Score="3" Body="&lt;p&gt;Urethane foam from isocyanates, commonly sold at hardware stores for insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click links for &lt;a href=&quot;http://www.spraypolyurethane.org/Main-Menu-Category/Do-It-Yourself/Frequently-Asked-Questions&quot; rel=&quot;nofollow noreferrer&quot;&gt;general information&lt;/a&gt; and for &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyurethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;its chemistry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="76480" LastEditDate="2019-04-17T20:55:24.213" LastActivityDate="2019-04-17T20:55:24.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24640" PostTypeId="2" ParentId="24635" CreationDate="2015-02-05T03:10:44.920" Score="1" Body="&lt;p&gt;Yes, tahini (and cornstarch-water mixtures) are dilatant, non-Newtonian fluids, with viscosity increasing with shear strain. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Dilatant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Dilatant&lt;/a&gt; for definition and &lt;a href=&quot;https://www.youtube.com/watch?v=D-wxnID2q4A&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=D-wxnID2q4A&lt;/a&gt;, &lt;a href=&quot;https://www.youtube.com/watch?v=BleCJJAKkgw&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=BleCJJAKkgw&lt;/a&gt; for demo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tahini is much better in hummus than in a swimming pool, though. ;-)&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-05T03:10:44.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24641" PostTypeId="1" CreationDate="2015-02-05T03:14:37.970" Score="12" ViewCount="17032" Body="&lt;p&gt;I realized that the $\ce{-OH}$ group when it reacts with some metal like $\ce{Na}$ or $\ce{K}$ will replace the $\ce{H}$ element to form $\ce{-OK}$ or $\ce{-ONa}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;    \ce{H2O + Na &amp;amp;-&amp;gt; NaOH + H2}\\&#xA;    \ce{CH3OH + Na &amp;amp;-&amp;gt; CH3ONa + H2}\\ &#xA;    \ce{H2O + K &amp;amp;-&amp;gt; KOH + H2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am just wondering is that always true? Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$? Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="61578" LastEditDate="2020-08-31T17:35:03.743" LastActivityDate="2020-08-31T17:35:03.743" Title="Does NaOH react with K to form NaOK or KOH?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;molecular-structure&gt;&lt;electrolysis&gt;&lt;molecules&gt;" AnswerCount="7" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24642" PostTypeId="2" ParentId="24638" CreationDate="2015-02-05T03:33:07.650" Score="5" Body="&lt;p&gt;In general, one usually doesn't work directly with single atoms, though there are &lt;a href=&quot;http://en.wikipedia.org/wiki/IBM_(atoms)&quot;&gt;ways to do that&lt;/a&gt; as well, usually involving scanning probe microscopy (basically poking at atoms with very sharp probes to either manipulate them or just image them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms become much easier to work with when they are are ionized and can be manipulated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Lorentz_force&quot;&gt;electric or magnetic fields&lt;/a&gt; (atoms have neutral charge so the Lorentz force can't be used to control them). Even then, most techniques deal with small groups of ions rather than individual ones. Ions are relatively easy to manipulate and we can easily &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion_trap&quot;&gt;trap them&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_particle_accelerator&quot;&gt;accelerate them into beams&lt;/a&gt;, and focus them. The LHC is, of course, a very impressive (and expensive) example of this. The way it works is that packets of ions are accelerated by electric and magnetic fields to form two ion beams, which are then focused into a very small area and made to intersect, allowing the ions to collide and all the cool particle physics stuff to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manipulation of ions is also used in countless other scientific applications: semiconductor manufacturing, surface analysis, medical analysis, radioisotope dating, to name a few.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-05T03:33:07.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24643" PostTypeId="1" AcceptedAnswerId="24662" CreationDate="2015-02-05T05:02:03.100" Score="6" ViewCount="162" Body="&lt;p&gt;Many times Geometry optimization gets very time consuming ( even with access to super-computers ). I have seen some pre-optimized libraries for biomolecules like proteins that you can easily download machine optimized files from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use Avogadro,  GAMESS and McMolPlt for my quantum calculations.&#xA;I am thinking such libraries should exist for some complex chemical structures ( usable on softwares mentioned above ) but couldn't find any. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I am specifically interested in &quot; Hydroxyapatite Crystals &quot;.&#xA;Does anyone know such resources ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="865" LastEditDate="2015-10-30T17:45:46.940" LastActivityDate="2015-10-30T17:45:46.940" Title="Where to find pre-optimized files for Quantum Chemistry Softwares?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24644" PostTypeId="1" CreationDate="2015-02-05T07:02:09.930" Score="9" ViewCount="14575" Body="&lt;p&gt;I'm trying to do a thermite reaction with rust and aluminum and am trying to figure out the optimal way to get the reaction started. My understanding is that I need a temperature of around &lt;span class=&quot;math-container&quot;&gt;$\pu{3200 ^\circ F}$&lt;/span&gt; to start it. I am currently aware of two methods to achieve that:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Potassium permanganate and glycerin: But apparently this requires an ambient temperature of &lt;span class=&quot;math-container&quot;&gt;$\gt \pu{70 ^\circ F}$&lt;/span&gt;, and it's winter where I live, &lt;span class=&quot;math-container&quot;&gt;$\pu{40 ^\circ F}$&lt;/span&gt; at most some days.&lt;/li&gt;&#xA;&lt;li&gt;Magnesium strip: But this requires a blowtorch (trying to avoid buying a blowtorch) and I've read is unreliable for starting this reaction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Does anyone have any alternative suggestions? Has anyone done this before? All suggestions are welcome.&lt;/p&gt;&#xA;&lt;p&gt;I am aware of the need to do it far from anything flammable/destructible, and am aware it can damage your eyes, and am aware to keep it away from water and keep my distance. Other tips are welcome.&lt;/p&gt;&#xA;" OwnerUserId="416" LastEditorUserId="61578" LastEditDate="2020-08-30T21:32:10.653" LastActivityDate="2020-10-08T13:58:58.640" Title="Best ways to start a thermite reaction?" Tags="&lt;reaction-mechanism&gt;&lt;pyrotechnics&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24645" PostTypeId="2" ParentId="24644" CreationDate="2015-02-05T07:08:19.903" Score="2" Body="&lt;p&gt;You might want to wrap the magnesium ribbon around a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sparkler&quot; rel=&quot;nofollow&quot;&gt;sparkler&lt;/a&gt;. The latter can be ignited with an ordinary lighter.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-05T07:08:19.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24646" PostTypeId="2" ParentId="24633" CreationDate="2015-02-05T07:22:05.613" Score="5" Body="&lt;p&gt;Black lentils contains a natural &lt;a href=&quot;http://en.wikipedia.org/wiki/PH_indicator&quot; rel=&quot;nofollow&quot;&gt;pH indicator&lt;/a&gt;. These dyes change their colour depending on how acidic (or alkaline) the solution is. The dyes are water-soluble, consequently you can observe the colour change in the remaining water too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural dyes, which often belong to the class of &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthocyanin&quot; rel=&quot;nofollow&quot;&gt;anthocyanines&lt;/a&gt;, are found in other food too, such as in red cabbage.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="7448" LastEditDate="2015-02-05T13:45:12.610" LastActivityDate="2015-02-05T13:45:12.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24647" PostTypeId="1" CreationDate="2015-02-05T07:54:05.147" Score="3" ViewCount="852" Body="&lt;p&gt;Why is the pure tungsten ($\ce{W}$) melting point way higher than that of its oxides $\ce{WO3}$ or $\ce{WO2}$? And why does pure $\ce{Al}$ react differently? I.e. the oxides have a higher melting point than pure $\ce{Al}$.&lt;/p&gt;&#xA;" OwnerUserId="13206" LastEditorUserId="10119" LastEditDate="2015-02-05T11:21:18.227" LastActivityDate="2015-02-05T16:15:19.883" Title="Why is the pure tungsten (W) melting point higher than that of its oxides (either WO3 or WO2)?" Tags="&lt;melting-point&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24648" PostTypeId="2" ParentId="24644" CreationDate="2015-02-05T08:18:00.373" Score="5" Body="&lt;p&gt;By blowtorch anyway. It is cheap and extremely useful for similar experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I used $\ce{Al}+\ce{S}$ mix as a starter, that was initiated by blowtorch (note, this mix produces aluminium sulphide that stinks with hydrogen sulfide.) - it is easy, cheap and reliable starter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a magnesium ribbon can be ignited by common lighter - I did it myself many times by keeping a piece of ribbon just above visible part of the flame for about 15-30 seconds.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-02-05T08:18:00.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24649" PostTypeId="1" CreationDate="2015-02-05T11:28:55.853" Score="4" ViewCount="6203" Body="&lt;p&gt;Is it possible (at home) to extract magnesium from $\ce{MgSO4·7H2O}$, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_sulfate&quot; rel=&quot;nofollow&quot;&gt;Epsom salt&lt;/a&gt; (or, alternatively, from $\ce{MgSO4}$)?&lt;/p&gt;&#xA;" OwnerUserId="13209" LastEditorUserId="30836" LastEditDate="2019-09-08T17:19:59.470" LastActivityDate="2019-10-13T10:54:45.437" Title="Is it possible to extract magnesium from epsom salt?" Tags="&lt;home-experiment&gt;&lt;extraction&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24650" PostTypeId="2" ParentId="24633" CreationDate="2015-02-05T12:44:56.490" Score="8" Body="&lt;p&gt;In the black lentils used in Indian cooking (&lt;em&gt;Vigna mungo&lt;/em&gt;), the natural water-soluble dye that causes the color change &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=2&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCMQFjAB&amp;amp;url=http%3A%2F%2Fwww.researchgate.net%2Fprofile%2FChidan_Kumar_C_S%2Fpublication%2F254356182_Extraction_and_Evaluation_of_a_New_AcidBase_Indicator_from_Black_Gram_Husk_%2528Vigna_mungo%2529%2Flinks%2F0deec53329a490f143000000&amp;amp;ei=d2HTVOifKMmuyATE5YGQBQ&amp;amp;usg=AFQjCNFLJR3tWOOfbtrWxV5f8O4gywqImQ&amp;amp;sig2=ALbKJcs9ipsb3kcdIKFPoA&amp;amp;bvm=bv.85142067,d.aWw&quot; rel=&quot;noreferrer&quot;&gt;has been identified&lt;/a&gt; as quercitin-3-O-pentoside:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/INoZE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which belongs to a class of compounds called &lt;a href=&quot;http://en.wikipedia.org/wiki/Flavonols&quot; rel=&quot;noreferrer&quot;&gt;flavonols&lt;/a&gt;. When you pop hydrogen ions on and off some of the -OH groups, you can get a pure benzenoid form (colorless), a quinonoid form (pink), and an extensive quinonoid form (yellow). The color changes&#xA;are pH dependent, and you get pale pink at pH &amp;lt; 7, yellow at pH &gt; 7, and&#xA;colorless at pH 7. When you add your lemon juice to the lentil water, you're getting the quinonoid form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other species of black lentils probably do something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH-dependent color change occurs because when you have $-\mathrm{OH}$ groups on the rings instead of $-\mathrm{O}^-$ groups, the way electrons are confined changes; that changes the spacing between the molecule's energy levels, and changes the colors of light the molecule can absorb. (See my article &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/features/water2wine.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for more about how this works). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many plants and household materials contain molecules that can change color or otherwise indicate changes in pH (See my article &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/household-indicators.shtml&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for a list of some of them). Almost any flower, fruit, or plant part that is red, blue, or purple contains a class of chemical compounds called &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthocyanin&quot; rel=&quot;noreferrer&quot;&gt;anthocyanins&lt;/a&gt; that change color with pH. The color of a flower or fruit depends on which anthocyanins are present, the pH of the pigment-bearing tissues, and the presence of other pigments, like yellow &lt;a href=&quot;http://en.wikipedia.org/wiki/Flavones&quot; rel=&quot;noreferrer&quot;&gt;flavones&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-05T13:10:31.417" LastActivityDate="2015-02-05T13:10:31.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24651" PostTypeId="2" ParentId="23940" CreationDate="2015-02-05T12:53:00.203" Score="1" Body="&lt;p&gt;When I generate an IRC from a TS - using GAMESS (US) software - I calculate the IRC for the forward direction and the IRC for the backwards direction using MacMolPlt.  Is it possible you've stitched together two forward (or two backward) direction IRCs by accident?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I once tried to calculate and IRC from a TS but both the forward and backward IRCs went in the same direction.  I don't know why ... I got around it by cribbing someone else's code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good place for code is the GAMESS (US) user group on google groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope the above helps...&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-02-05T12:53:00.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24654" PostTypeId="2" ParentId="24643" CreationDate="2015-02-05T13:46:01.337" Score="1" Body="&lt;p&gt;You could try figshare.com for comp chem files and &lt;a href=&quot;http://www.ccdc.cam.ac.uk/pages/Home.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.ccdc.cam.ac.uk/pages/Home.aspx&lt;/a&gt; for crystal structures.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-02-05T13:46:01.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24655" PostTypeId="2" ParentId="24616" CreationDate="2015-02-05T13:49:19.393" Score="6" Body="&lt;p&gt;There are a couple of advantages to using normalized scientific notation (a number less than ten and greater than or equal to one times a power of ten):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Counting significant figures is easier&lt;/strong&gt;.  All of the digits written in the number before the power of ten are significant when you're using normalized scientific notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing it other ways isn't incorrect, but it's not as convenient. For example, if you write $1.000\times 10^3$ it's absolutely clear that you've got 4 significant figures, but if you write $1000\times 10^0$ it's not so clear; you can only say &quot;I have at least one significant digit&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Comparing orders of magnitude is easier&lt;/strong&gt;. As other answers here note, you don't have to move the decimal point around to compare the powers of ten for a pair of numbers if you're using normalized scientific notation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-05T13:55:05.100" LastActivityDate="2015-02-05T13:55:05.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24656" PostTypeId="1" AcceptedAnswerId="24677" CreationDate="2015-02-05T14:21:20.947" Score="1" ViewCount="872" Body="&lt;p&gt;What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {-&amp;gt;} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help, thanks :)&lt;/p&gt;&#xA;" OwnerUserId="13213" LastEditorUserId="13213" LastEditDate="2015-02-05T18:01:53.263" LastActivityDate="2015-02-05T19:52:09.073" Title="Limiting Reagent Stoichiometry" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24657" PostTypeId="2" ParentId="24649" CreationDate="2015-02-05T14:25:51.360" Score="1" Body="&lt;p&gt;You need molten salts of some kind or a mercury cathode to get metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.scienceforums.net/topic/36363-extracting-magnesium/&quot; rel=&quot;nofollow&quot;&gt;http://www.scienceforums.net/topic/36363-extracting-magnesium/&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13027" LastActivityDate="2015-02-05T14:25:51.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24658" PostTypeId="1" AcceptedAnswerId="24668" CreationDate="2015-02-05T14:37:18.130" Score="1" ViewCount="907" Body="&lt;p&gt;&lt;strong&gt;How to remove epoxy on clothes in the perspective of chemistry?&lt;/strong&gt;&lt;br/&gt;&#xA;More precisely in my case, how to remove the &lt;strong&gt;hardener&lt;/strong&gt; of epoxy, which consists of two components: hardener and resin.&lt;br/&gt;&#xA;The clothes is made in nylon, please take the least damage to the cloth into high priorities of consideration.&#xA;Does alcohol help?&lt;br/&gt;&#xA;Thank you very much for your reply. &lt;/p&gt;&#xA;" OwnerUserId="13214" LastEditorUserId="7448" LastEditDate="2015-02-05T18:21:16.483" LastActivityDate="2015-02-05T18:21:16.483" Title="How to remove epoxy on clothes?" Tags="&lt;solubility&gt;&lt;solvents&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24659" PostTypeId="2" ParentId="24514" CreationDate="2015-02-05T14:39:23.950" Score="0" Body="&lt;p&gt;NADH is more energetic because it holds more bonds, therefore more energy is required to keep this molecule intact. So when trying to break the bond, it would require more energy to break, and therefore it is also more stable then NAD+. NAD+ is energetically unfavorable because it is a cation. Adding a hydrogen will give it favorability, but it will also hold the molecule with more energy. &lt;/p&gt;&#xA;" OwnerUserId="13007" LastActivityDate="2015-02-05T14:39:23.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24660" PostTypeId="1" AcceptedAnswerId="69888" CreationDate="2015-02-05T14:42:07.793" Score="11" ViewCount="21648" Body="&lt;p&gt;In the MO diagram of $\ce{N2^2-}$, does s-p mixing happen to a significant degree to change the ordering of orbitals - $\pi_\mathrm{2p}$ orbital below $\sigma_\mathrm{2p}$? As it is isoelectronic with oxygen, I think that the s-p mixing shouldn't change the ordering.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2017-03-06T21:11:49.193" LastActivityDate="2017-09-12T10:01:32.133" Title="MO diagram of N2(2-)" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24661" PostTypeId="2" ParentId="24523" CreationDate="2015-02-05T14:46:24.837" Score="2" Body="&lt;p&gt;Not quite sure about melting and boiling points but things like polarity ,dipole moment and inductive effect are kind of interconnected.Inductive effect is a consequence of electronegativity differences.More inductive effect leads to more dipole moment but that also depends from one case to another.For example each of the $\ce{C-Cl}$ bonds in $\ce{CCl4}$ is highly polar but net dipole moment is 0 due to the fact that Dipole moments are added through vector addition and $\ce{CCl4}$ is tetrahedral.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="10185" LastEditDate="2015-02-05T18:01:41.173" LastActivityDate="2015-02-05T18:01:41.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24662" PostTypeId="2" ParentId="24643" CreationDate="2015-02-05T14:52:10.777" Score="11" Body="&lt;p&gt;&lt;strong&gt;Short answer: we're building one for molecules&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you don't actually want molecules. You're mentioning crystals. In that case, I suggest the &lt;a href=&quot;http://www.crystallography.net&quot; rel=&quot;noreferrer&quot;&gt;Crystallography Open Database&lt;/a&gt; or if you have access, try the commercial &lt;a href=&quot;http://webcsd.ccdc.cam.ac.uk&quot; rel=&quot;noreferrer&quot;&gt;Cambridge Crystal Structure Database&lt;/a&gt;. The COD is planning a theoretical crystal database, but I don't think it's online yet. These are experimental x-ray structures, but should be good starting structures for solid-state simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For molecules, my group is building one. It's the &lt;a href=&quot;http://pqr.pitt.edu/&quot; rel=&quot;noreferrer&quot;&gt;Pitt Quantum Repository&lt;/a&gt; There are a few related efforts that I'm aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the PQR, you can either search through the website (even on mobile). The database is keyed by InChI keys, so if you have an existing structure you can generate the InChIKey and then get the structure, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pqr.pitt.edu/mol/BQJCRHHNABKAKU-KBQPJGBKSA-N&quot; rel=&quot;noreferrer&quot;&gt;D-(-)-Morphine&lt;/a&gt; -&gt; &lt;code&gt;.mol2&lt;/code&gt; file at &lt;a href=&quot;http://pqr.pitt.edu/api/mol2/BQJCRHHNABKAKU-KBQPJGBKSA-N&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;http://pqr.pitt.edu/api/mol2/BQJCRHHNABKAKU-KBQPJGBKSA-N&lt;/code&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The long answer is that some commercial software vendors consider such libraries as violations of the license. No joke, I've planned this for years to augment &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt;, but the company behind a leading program told me that they would void my license to distribute output files in this fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the rise of community and open source quantum chemistry software, this is moot. &lt;strong&gt;Use programs that let you distribute output files.&lt;/strong&gt; What remained was:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gathering the 3D coordinates (i.e., a low-energy conformer) of X thousand/million molecules&lt;/li&gt;&#xA;&lt;li&gt;Obtaining the computer time to optimize all of them&lt;/li&gt;&#xA;&lt;li&gt;Devoting a server and disk space to serving them&lt;/li&gt;&#xA;&lt;li&gt;Having a great website to help view, browse, and search them&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-07-06T18:52:50.743" LastActivityDate="2015-07-06T18:52:50.743" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24663" PostTypeId="2" ParentId="24660" CreationDate="2015-02-05T15:32:43.510" Score="5" Body="&lt;p&gt;Well, s-p mixing only occurs when the s and p atomic orbitals are close in energy (it's a necessary but not sufficient condition). Even though $\rm N_2^{2-}$ is isoelectronic with $\rm O_2$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. I think you still &lt;em&gt;would&lt;/em&gt; have s-p mixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification: I mean to cast doubt on the conjecture that there's no s-p mixing here. As you point out in the comments, electron repulsion is a consideration too. We can't definitively say which effect is more important without looking at the actual orbital energies. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-05T21:31:18.480" LastActivityDate="2015-02-05T21:31:18.480" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24664" PostTypeId="1" AcceptedAnswerId="24690" CreationDate="2015-02-05T15:48:38.247" Score="1" ViewCount="113" Body="&lt;p&gt;One of the popular do-at-home chemistry demonstrations is placing an egg in a container of vinegar for two days. After that period most if not all of the eggshell has dissolved and the outer part of the egg has congealed giving the egg a 'rubber' like appearance and feel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So chemically the vinegar (acetic acid) reacts with the egg protein (albumen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemically speaking, what is this product?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-02-06T01:33:47.560" Title="What material is synthesized in the rubber egg demo?" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24665" PostTypeId="2" ParentId="24647" CreationDate="2015-02-05T16:15:19.883" Score="2" Body="&lt;p&gt;As far as I know,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{WO2}$ and $\ce{WO3}$ have a covalent characteristic, and thus lower melting point than the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while, $\ce{Al2O3}$ have more ionic characteristic, and thus higher melting point than the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{W}$ is a d-block metal from the $\ce{3^{rd}}$ row which make it possible to have covalent materials rather than ionic materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this article might prove this point &#xA;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jp963724z?journalCode=jpcbfk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13186" LastActivityDate="2015-02-05T16:15:19.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24666" PostTypeId="2" ParentId="24649" CreationDate="2015-02-05T16:45:01.670" Score="11" Body="&lt;p&gt;Supposed you are trying to obtain &lt;strong&gt;magnesium metal&lt;/strong&gt; from Epsom salt, &lt;span class=&quot;math-container&quot;&gt;$\ce{MgSO4.7 H2O}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;h3&gt;Is is worth the effort?&lt;/h3&gt;&#xA;&lt;p&gt;Magnesium metal in the form of chips (particle size 4-30 mesh) in decent purity (99.98%) is available from suppliers such as &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/254118?lang=de&amp;amp;region=DE&quot; rel=&quot;noreferrer&quot;&gt;Sigma-Aldrich&lt;/a&gt; for about 150 Euro per kilogram.&lt;/p&gt;&#xA;&lt;h3&gt;Can I use electrolysis to generate &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; metal?&lt;/h3&gt;&#xA;&lt;p&gt;Yes, but you wouldn't want to do that at home. An older industrial process uses electrolysis of &lt;strong&gt;molten&lt;/strong&gt; magnesium chloride, &lt;span class=&quot;math-container&quot;&gt;$\ce{MgCl2}$&lt;/span&gt; (mp. is around 700 °C). While &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; metal is formed at the cathode, chlorine gas is released at the graphite anode. Did I mention that this isn't meant to be done at home?&lt;/p&gt;&#xA;&lt;h3&gt;Can I convert &lt;span class=&quot;math-container&quot;&gt;$\ce{MgSO4}$&lt;/span&gt; to something more usable to make &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; metal?&lt;/h3&gt;&#xA;&lt;p&gt;Yes, in principle. A group of Chinese researchers examined the conversion of  &lt;span class=&quot;math-container&quot;&gt;$\ce{MgSO4}$&lt;/span&gt; to  magnesium oxide, &lt;span class=&quot;math-container&quot;&gt;$\ce{MgO}$&lt;/span&gt; by heating in the presence of natural gas at temperatures around 500 °C and outlined their results in&#xA;&lt;a href=&quot;http://www.terrapub.co.jp/journals/GJ/pdf/4502/45020097.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Geochemical Journal&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt;, 97-108&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The initial pressure by natural gas at room temperature was 3 MPa, but upon heating to the final temperature, the pressure in the reactor was as high as 12-20 MPa! This can't be done at home and it shouldn't be.&lt;/p&gt;&#xA;&lt;h3&gt;What if I would have &lt;span class=&quot;math-container&quot;&gt;$\ce{MgO}$&lt;/span&gt;? What could I do with it?&lt;/h3&gt;&#xA;&lt;p&gt;Magnesium oxide, &lt;span class=&quot;math-container&quot;&gt;$\ce{MgO}$&lt;/span&gt;, is indeed the most important starting material in the production of magnesium metal in the &lt;strong&gt;Pidgeon process&lt;/strong&gt;. Here, magnesium oxide is &lt;strong&gt;heated&lt;/strong&gt; in the presence of silicon in an evacuated reactor. Silicon is oxidized to silicon dioxide, magnesium oxide is reduced to magnesium metal. The latter boils off (bp. around 1100 °C) and is condensed. This isn't kitchen chemistry either.&lt;/p&gt;&#xA;&lt;p&gt;In summary, you're better off buying some magnesium metal from a supplier of your choice!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-09-09T01:21:10.453" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24667" PostTypeId="2" ParentId="24656" CreationDate="2015-02-05T16:59:17.100" Score="0" Body="&lt;p&gt;Generally speaking, if the problem doesn't give concentration or enough information to find concentration, you can't know whether all or only a portion of the product will dissolve.  You need to know the solubility product constant (Ksp), the final volume, and the amount of each ion to calculate the amount that will dissolve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this particular question has a complicating issue because: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2BaS2 + 2H2O -&amp;gt; Ba(OH)2 + Ba(SH)2}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-05T17:16:05.937" LastActivityDate="2015-02-05T17:16:05.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24668" PostTypeId="2" ParentId="24658" CreationDate="2015-02-05T17:19:04.117" Score="0" Body="&lt;p&gt;Lestoil, PineSol or other organic-solvent cleaner should get out the unset hardener. Once set, epoxy is hard to remove; you might try methylene chloride-based paint removers [toxic!]. Some hardeners are organic peroxides, which may bleach spots on the clothing. – &lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-05T17:19:04.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24669" PostTypeId="1" AcceptedAnswerId="24670" CreationDate="2015-02-05T17:21:54.487" Score="4" ViewCount="3549" Body="&lt;p&gt;My textbook reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a gas is adsorbed, the freedom of movement of its molecules become restricted. This amounts to decrease in the entropy of the gas after adsorption, i.e. Entropy change is negative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But now, I am unable to understand why it isn't a violation of the second law of thermodynamics, which states that the entropy change of a system can never be negative. Please guide.  &lt;/p&gt;&#xA;" OwnerUserId="1886" LastEditorDisplayName="user7951" LastEditDate="2019-12-25T08:13:50.823" LastActivityDate="2019-12-25T08:13:50.823" Title="Does adsorption violate thermodynamics?" Tags="&lt;surface-chemistry&gt;&lt;entropy&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24670" PostTypeId="2" ParentId="24669" CreationDate="2015-02-05T17:48:11.383" Score="11" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;The second law of thermodynamics&lt;/a&gt; states that the entropy of the universe always increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{d}S &amp;gt; 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in entropy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta S_\mathrm{sys} &amp;lt; 0 \qquad \Delta S_\mathrm{surr} &amp;gt; 0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$|{\Delta S_\mathrm{surr}}| &amp;gt; |{\Delta S_\mathrm{sys}}|$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because adsorption is an exothermic process and so the surroundings are heated up and therefore increase in entropy. If you consider the Gibbs energy change for adsorption it will be negative because the negative &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; term is larger than the positive &lt;span class=&quot;math-container&quot;&gt;$-T\Delta S$&lt;/span&gt; term. &lt;a href=&quot;https://web.archive.org/web/20150908221122/http://www2.chem.uic.edu/rjgordon/teaching/chem549/Entropy%20of%20adsorption.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This chem.uic.edu page&lt;/a&gt; goes through the maths of this very nicely.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="41328" LastEditDate="2019-12-25T07:59:59.423" LastActivityDate="2019-12-25T07:59:59.423" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24671" PostTypeId="1" AcceptedAnswerId="24675" CreationDate="2015-02-05T18:09:23.253" Score="8" ViewCount="4174" Body="&lt;p&gt;Shouldn't reaching an octet be any atom's &quot;goal&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've recently learned about cases that are either expanding octets, or have lesser than &quot;enough&quot; electrons for an octet abiding. e.g.:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;S in sulfur hexafluoride (Expanding octet into $\text{3d}$)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bMyhf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bMyhf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;http://moziru.com/explore/Drawn%20molecule%20sf6/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;B in boron trifluoride (It's &quot;hextet&quot;, instead of octet)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VjC4S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjC4S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://opentextbc.ca/chemistry/chapter/7-3-lewis-symbols-and-structures/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, both $\ce{SnCl2}$ and $\ce{SnCl4}$ are existent. The latter is explainable with octet, but not the former. Surprisingly, $\ce{SnCl2}$ is more stable! &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How?&lt;/li&gt;&#xA;&lt;li&gt;Why is the phenomenon happening?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/users/5160/davephd&quot;&gt;Dave&lt;/a&gt; pointed out that the octet nonabidingness is happening in the vapor phase. So I redirect the question to ask about the vapor phase; since that's what I'm looking for.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="194" LastEditDate="2018-02-04T16:23:28.000" LastActivityDate="2018-02-04T16:23:28.000" Title="Why does stannous chloride occur despite the octet rule?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;covalent-compounds&gt;&lt;lewis-structure&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24672" PostTypeId="2" ParentId="24671" CreationDate="2015-02-05T18:40:24.387" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://scitation.aip.org/content/aip/proceeding/aipcp/10.1063/1.42518&quot; rel=&quot;noreferrer&quot;&gt;Theoretical and experimental studies of the electronic structure&#xA;of SnCl2 by photoionization with synchrotron radiation&lt;/a&gt; AIP Conf. Proc. 258, 60 (1992)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 99% Cl p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 99% Cl p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 7% Sn p , 92% Cl p &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 3% Sn s, 6% Sn p , 90% Cl p &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sigma bonding 14% Sn p, 83% Cl p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sigma bonding 48% Sn s, 6% Cl s 45% Cl p&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 3% Sn p, 96% Cl s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonbonding 5% Sn s, 2% Sn p, 92% Cl s &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall it seems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a total bond order of only 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is little mixing of Sn s and p orbitals expect in the antibonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;additional references:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja01046a001&quot; rel=&quot;noreferrer&quot;&gt;A Symmetry Rule for Predicting Molecular Structures&lt;/a&gt; (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/102/9/10.1063/1.468579&quot; rel=&quot;noreferrer&quot;&gt;Metal oxidation state effect in photoionization of gas‐phase metal halides&lt;/a&gt; (which further discusses the molecular orbitals of SnCl2 discussed above)&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-05T22:27:54.863" LastActivityDate="2015-02-05T22:27:54.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24673" PostTypeId="1" CreationDate="2015-02-05T19:08:23.580" Score="5" ViewCount="30158" Body="&lt;p&gt;Whenever I look up the Lewis structure for $\ce{NH2^-}$, it shows 8 electrons. Now I understand the octet rule, but I still don't see how an electron can just appear from nowhere. There are 5 valence electrons with Nitrogen, and 2 more, one from each hydrogen. I know that formal charge is calculated by subtracting the nonbonding electrons and number of bonds (or number of electrons in bonds divided by 2), but why is the formal charge of $\ce{NH2^-}$ &quot;-1&quot;, and why the extra electron out of nowhere? Because 5 - 3-2 is still just 5&lt;/p&gt;&#xA;" OwnerUserId="13218" LastEditorUserId="4945" LastEditDate="2015-07-13T03:15:39.350" LastActivityDate="2015-07-13T03:15:39.350" Title="Why is NH2 negatively charged?" Tags="&lt;organic-chemistry&gt;&lt;ions&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24674" PostTypeId="2" ParentId="24673" CreationDate="2015-02-05T19:20:45.493" Score="9" Body="&lt;p&gt;For atoms in the first 2 rows of the periodic table the octet rule arises as a consequence of filling orbitals four orbitals which each hold 2 electrons - each of which appears as a pair of dots in your Lewis structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were you to construct the Lewis structure for neutral $\ce{NH2}$ you would indeed have 5 electrons from the nitrogen, and two from hydrogen, 7 i.e. 3 pairs and 1 extra electron on its own in a orbital.  Electrons in unpaired orbitals are however very reactive, the pairing of electrons lowers the total energy of the system.  As such, it is more favourable for the $\ce{NH2}$ molecule to remove an electron from some other molecule or atom near by and form ions e.g. $\ce{Na}$ to create $\ce{Na+NH2-}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also by creating the charges, you create an electrostatic attraction between the ions which also lowers the energy and makes the ion more stable than the neutral molecule.&lt;/p&gt;&#xA;" OwnerUserId="13217" LastActivityDate="2015-02-05T19:20:45.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24675" PostTypeId="2" ParentId="24671" CreationDate="2015-02-05T19:47:12.547" Score="5" Body="&lt;p&gt;$\ce{SF6}$ does not expand it octet by way of using 3d orbitals.  It is often taught that way, particularly in introductory classes.  A better description involves treating it as a hypercoordinated molecule.  See these earlier answers for an explanation of how hypercoordination can be applied.  When finished you should be able to apply the concept to $\ce{SF6}$.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride/19437#19437&quot;&gt;reference 1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;reference 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{BF3}$ is electron deficient and reacts rapidly with molecules containing lone pairs of electrons (amines, oxygen containing compounds, etc.) to stabilize the octet configuration.  It has also been observed that the $\ce{B-F}$ bond length is somewhat shorter than expected.  This has been explained by invoking resonance structures such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Mxy4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Xfwe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To whatever degree these types of resonance structures contribute to the true description of $\ce{BF3}$ they help explain the shortened bond length and also achieve an octet around the central boron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar explanation can be applied to $\ce{SnCl2}$, drawing a resonance structure with a $\ce{Sn=Cl}$ double bond creates an octet around the central tin atom.  To whatever degree such resonance structures contribute to the real description of $\ce{SnCl2}$, they will help satisfy the octet electronic configuration around the central tin atom.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-02-05T19:47:12.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24676" PostTypeId="1" AcceptedAnswerId="24683" CreationDate="2015-02-05T19:50:21.127" Score="11" ViewCount="4946" Body="&lt;p&gt;I may be a rudimentary fact or something really &quot;not-that-hard-to-figure-out&quot;, but&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In inorganic chemistry it is common to consider a single value of the electronegativity to be valid for most &quot;normal&quot; situations. While this approach has the advantage of simplicity, it is clear that the electronegativity of an element is not an invariable atomic property and, in particular, increases with the oxidation state of the element. &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity#Variation_of_electronegativity_with_oxidation_number&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Really? That means for example $\ce{C}$ in $\ce{CO3^{2-}}$ is more electronegative than a &quot;bare&quot; C atom? Why is that so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, I'm not able to find relevant info in the Net (and the wikipedia citation that should have hopefully got me results is behind a big bad paywall), so am leaving &quot;reference-request&quot; for the case of additional study.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-02-05T20:27:33.267" LastActivityDate="2015-10-30T05:59:36.413" Title="Why is electronegativity increasing with oxidation state increase?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;&lt;reference-request&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24677" PostTypeId="2" ParentId="24656" CreationDate="2015-02-05T19:52:09.073" Score="1" Body="&lt;p&gt;So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, &lt;strong&gt;precipitates cannot be aqueous&lt;/strong&gt;, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -&gt; post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google &quot;solubility constants&quot; and &quot;common ion effect&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastActivityDate="2015-02-05T19:52:09.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24678" PostTypeId="1" CreationDate="2015-02-05T19:58:47.620" Score="-5" ViewCount="1386" Body="&lt;p&gt;Hypothetical question: &lt;br/&gt;&#xA;Knowing that orange juice has &lt;code&gt;25%&lt;/code&gt; concentration of orange. And knowing mass of juice which is &lt;code&gt;200g&lt;/code&gt;. What is mass of oranges included in juice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that orange is a molecule, am I right that mass in 25% juice would be &lt;code&gt;25% x 200g = 50gram&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking about simple formulas, without mols on basic level. &lt;br/&gt;&#xA;If I know mass of entire liquid and know that 25% of acid is a concentration, so do I have to simply computer percentage, or rather computing for acid and rest differently? &lt;/p&gt;&#xA;" OwnerUserId="13220" LastEditorUserId="5160" LastEditDate="2015-02-06T14:25:27.200" LastActivityDate="2015-02-06T14:25:27.200" Title="Mass of &quot;orange&quot; in 25% orange juice" Tags="&lt;molecules&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-02-05T22:53:11.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="24679" PostTypeId="1" AcceptedAnswerId="24685" CreationDate="2015-02-05T20:29:08.590" Score="3" ViewCount="5885" Body="&lt;p&gt;I need to write the redox reaction from tin and a sulfuric acid solution. Here is the solution given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first half-reaction: $\ce{Sn-&amp;gt;Sn^2+\+2e^-}$&lt;br&gt;&#xA;  The second half-reaction: $\ce{2H+\+2e^--&amp;gt;H2}$&lt;br&gt;&#xA;  Adding both reactions will give you: $\ce{Sn\+2H+-&amp;gt;Sn^2+\+H2}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is the sulfuric acid solution replaced with $\ce{H^+}$?&lt;/p&gt;&#xA;" OwnerUserId="13221" LastEditorUserId="5026" LastEditDate="2018-02-23T02:03:32.163" LastActivityDate="2018-02-23T02:03:32.163" Title="Redox reaction with tin and sulfuric acid solution" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24680" PostTypeId="2" ParentId="15442" CreationDate="2015-02-05T20:49:00.977" Score="-4" Body="&lt;p&gt;The sulfuric acid brakes up the bonds of the sugar and absorbs the water from the crystallization&lt;/p&gt;&#xA;" OwnerUserId="13222" LastActivityDate="2015-02-05T20:49:00.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24682" PostTypeId="2" ParentId="24678" CreationDate="2015-02-05T21:44:02.310" Score="3" Body="&lt;p&gt;If the mass of the juice is 200g, then the mass of orange in the juice would be 50 grams.&lt;/p&gt;&#xA;&lt;p&gt;You just need to rephrase your question in terms of &amp;quot;mass&amp;quot; instead of &amp;quot;molecular mass&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Knowing that orange juice has 25% concentration of orange, and knowing mass of juice is 200g, what is mass of orange included in juice?&lt;/p&gt;&#xA;&lt;p&gt;Assuming that orange is a molecule, am I right that mass of orange in the juice would be 25% x 200g = 50 gram?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then it makes sense.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-02-05T21:49:06.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24683" PostTypeId="2" ParentId="24676" CreationDate="2015-02-05T21:53:38.247" Score="10" Body="&lt;p&gt;Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation state are more electronegative - i.e. more able to attract and stabilize electrons. High oxidation state elements have essentially been &quot;robbed&quot; of their electrons and likely bear some degree of partial positive charge ... this is &quot;good&quot; for attracting and stabilizing nearby electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely consider a lone pair - which is defined to have an electronegativity of 0. Lone pairs cannot stabilize other electrons and don't attract other electrons to themselves ... the electron-electron repulsion would be too great. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-02-05T21:58:59.327" LastActivityDate="2015-02-05T21:58:59.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24685" PostTypeId="2" ParentId="24679" CreationDate="2015-02-05T22:12:44.073" Score="7" Body="&lt;p&gt;The sulfuric acid dissociates in water: $$\ce{H2SO4 -&amp;gt; H+ + HSO4-}$$$$\ce{HSO4- -&amp;gt; H+ + SO4^{2-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When writing redox equations the spectator ions are left out because they do not participate in the redox reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it is the hydrogen ions which are oxidising the tin and the sulfate and hydrogensulfate ions play no part in the reaction so they are left out. Including them in the hydrogen half equation would give (with sulfate ions): $$\ce{2H+ + SO4^{2-} +2e- -&amp;gt; H2 + SO4^{2-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the sulfate ions are unnecessary and so they are omitted from the equation.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-07T17:23:09.703" LastActivityDate="2015-05-07T17:23:09.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24686" PostTypeId="1" AcceptedAnswerId="25147" CreationDate="2015-02-05T22:39:15.880" Score="5" ViewCount="179" Body="&lt;p&gt;Is it possible to calculate bond enthalpies using the periodic table and the molecule's geometric structure? I am interested in a way of calculating this without getting the answer from measurements or pre-existing bond enthalpies.&lt;/p&gt;&#xA;" OwnerUserId="15661" LastActivityDate="2015-03-17T20:52:26.667" Title="How to calculate bond enthalpies from first principles?" Tags="&lt;bond&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24687" PostTypeId="1" AcceptedAnswerId="24693" CreationDate="2015-02-05T22:41:31.007" Score="2" ViewCount="248" Body="&lt;p&gt;Wikipedia lists dipole-dipole, hydrogen bonding and Van der Waals under &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;nofollow&quot;&gt;intermolecular interactions&lt;/a&gt;, and hydrophobic, ionic and covalent under &lt;a href=&quot;https://en.wikipedia.org/wiki/Intramolecular_force&quot; rel=&quot;nofollow&quot;&gt;intramolecular interactions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a clear distinction between each and every one of these interactions? For example: isn't hydrogen bond a private case of VdW? aren't ionic bonds and hydrophobic interactions private cases of dipole-dipole interaction?&lt;/li&gt;&#xA;&lt;li&gt;Is there one single chart that differentiates all of these types of interaction?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-02-06T02:19:03.280" Title="Is there a clear distinction between the various intra- and intermolecular interactions?" Tags="&lt;bond&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24688" PostTypeId="2" ParentId="15511" CreationDate="2015-02-05T23:02:28.693" Score="5" Body="&lt;p&gt;The comments have already mentioned one problem with the question: the trajectory of the example system includes both exergonic (OK) and endergonic (not OK for closed systems) stretches.  Think about the total Gibbs free energy of the example with time.  The regions where the trajectory is moving away from the equilibrium point are endergonic (i.e. free energy is rising) and would not be allowed unless work is being done to the system somehow.&lt;/p&gt;&#xA;&lt;p&gt;The other problem is ambiguity in what is meant by &amp;quot;around&amp;quot; equilibrium.  Passing through the equilibrium point isn't allowed -- almost by the definition of equilibrium.  Gibbs free energy would have to rise (i.e. an endergonic process would be required) after reaching equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;Orbits &amp;quot;around&amp;quot; equilibrium that always spiral closer to it would potentially be OK.  But it depends on exactly what variables are in the orbit.  Obviously, Gibbs energy must always decrease in an equilibrating system.  Concentrations don't necessarily have to.  A good graph is this one: case A is disallowed by the second law, but case B isn't.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RRMkk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRMkk.gif&quot; alt=&quot;a is disallowed, b isn't&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.pojman.com/nlcd/Fig1.5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;pojman.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;The page where I found that graph does a good job of explaining the topic in much more detail: &lt;a href=&quot;http://www.pojman.com/nlcd/intro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;see here&lt;/a&gt;.  They present chemically feasible reaction mechanisms that lead to oscillations.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-02-24T12:00:25.857" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24690" PostTypeId="2" ParentId="24664" CreationDate="2015-02-06T01:33:47.560" Score="1" Body="&lt;p&gt;The shell dissolution reaction is the calcium carbonate in the egg being dissolved by the acetic acid, resulting in release of $\ce{CO2}$ and formation of calcium acetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + 2 CH3COOH -&amp;gt; Ca(CH3COO)2 + CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rubberization of the egg white is primarily a protein denaturation reaction.  Albumin in egg whites is a globular protein, meaning each molecule is a single chain that is tightly bundled into a ball-like (&quot;globular&quot;) form.  In this form the molecules can stay nicely solvated and the egg white is liquid.  However, the low pH from the acid (and also possibly the high concentration of calcium acetate) causes each chain to unfurl and then to bind with other neighboring chains, forming a complicated cross-linked tangle of molecules.  This tangle is no longer a liquid but is a rubbery solid-like material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the products are primarily calcium acetate, $\ce{CO2}$, and denatured albumin.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-06T01:33:47.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24691" PostTypeId="1" AcceptedAnswerId="24694" CreationDate="2015-02-06T01:58:01.200" Score="58" ViewCount="10545" Body="&lt;p&gt;I've been using Ruby to write scripts for research, but I want to get into some heavier stuff that Ruby is just too slow for. I noticed there are a few things written in C and C++, but there is an oddly large proportion of software used in computational chemistry that is written in FORTRAN (in which I have zero experience.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is FORTRAN used in computational chemistry? From what I understand, FORTRAN is kind of on the ancient side (“punchcard” old.) I was a bit shocked to find fairly recently written tutorials for FORTRAN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a sort of, &quot;this is how we've always done it,&quot; thing or is there an efficiency aspect to FORTRAN I'm overlooking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I may have FORTRAN confused with later programming languages with similar names.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2019-06-12T04:25:26.713" LastActivityDate="2020-03-04T17:16:14.373" Title="Why is FORTRAN so commonly used in computational chemistry?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="6" CommentCount="11" FavoriteCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="24692" PostTypeId="1" AcceptedAnswerId="24702" CreationDate="2015-02-06T02:15:00.900" Score="2" ViewCount="923" Body="&lt;p&gt;I would like to know the reason that some materials allow light to pass (transparent), blocks it a little (translucent), or completely reflects or absorbs it (opaque.)&#xA;I'm don't have a PhD, so answer this in a way that is easy to understand.&lt;/p&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="402" LastEditDate="2015-02-06T04:10:38.423" LastActivityDate="2015-03-11T09:59:12.273" Title="Reason behind Opaque translucent transparent" Tags="&lt;optical-properties&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24693" PostTypeId="2" ParentId="24687" CreationDate="2015-02-06T02:19:03.280" Score="2" Body="&lt;p&gt;&lt;strong&gt;No, there aren't clear distinctions, particularly between intramolecular and intermolecular interactions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a caveat that we generally divide molecules along covalent bonds. So those are strictly intramolecular, in my opinion. (Network solids are a bit tricky to classify.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go through the rest one at a time.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dipole-dipole&lt;/strong&gt;: These are usually &lt;em&gt;intermolecular&lt;/em&gt;, but certainly in larger, more flexible molecules, &lt;em&gt;intramolecular&lt;/em&gt; polar-polar interactions can change the molecular shape and preferred conformation. (Think about tying together two bar magnets on one piece of string. Certainly they'll interact.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Hydrogen bonds&lt;/strong&gt;: Yes, many of these are &lt;em&gt;intermolecular&lt;/em&gt;. However, there &lt;strong&gt;definitely&lt;/strong&gt; are &lt;em&gt;intramolecular&lt;/em&gt; hydrogen bonds:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WhS7h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Van der Waals&lt;/strong&gt;: Again, these can be &lt;em&gt;intermolecular&lt;/em&gt;. In even a moderate molecule, &lt;em&gt;intramolecular&lt;/em&gt; non-bonded van der Waals interactions can change conformations and geometries. Consider axial/equitorial preferences in cyclohexanes or other types of &lt;a href=&quot;http://en.wikipedia.org/wiki/Steric_effects&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;steric clashes&quot;&lt;/a&gt; because of big bulky groups. These are the effects of van der Waals interactions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OK, so all the intermolecular interactions are also intramolecular ones too.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Hydrophobic interactions&lt;/strong&gt;: I usually think of these as &lt;em&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Hydrophobic_interactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermolecular&lt;/a&gt;&lt;/em&gt; but similar considerations make sense as intramolecular for the reasons I outline under van der Waals interactions above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Ionic&lt;/strong&gt;: I'll just think of these as different partial charges in a molecule. Clearly electrostatic interactions occur both inside a molecule (e.g., electrostatic non-bonded interactions changing conformations and geometries) and as &lt;em&gt;intermolecular&lt;/em&gt; forces, e.g., water-cation interactions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So besides covalent interactions, which we use to divide &quot;inside&quot; and &quot;outside&quot; a molecule, all of these are both intermolecular and intramolecular interactions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I've seen a great chart. Maybe someone else can comment.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-02-06T02:19:03.280" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24694" PostTypeId="2" ParentId="24691" CreationDate="2015-02-06T02:35:29.603" Score="45" Body="&lt;p&gt;I don't think that's really true anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Fortran use is historical (i.e., early codes were developed in FORTRAN because that was the best programming language for number crunching in the 70s and 80s). Heck, the name stands for &quot;formula translation.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;http://en.wikipedia.org/wiki/Fortran&quot; rel=&quot;noreferrer&quot;&gt;Fortran&lt;/a&gt; use is because of performance. The language was designed to be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;especially suited to numeric computation and scientific computing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many times, I find chemistry coders sticking to Fortran because they know it and have existing highly optimized numeric code-bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the performance side isn't necessarily true anymore when using modern, highly optimizing C and C++ compilers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I write a lot of code in C and C++ for performance and glue a lot of things with Python. I know some quantum programs are written exclusively or primarily in C++. Here are a few open source examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.psicode.org/&quot; rel=&quot;noreferrer&quot;&gt;Psi4&lt;/a&gt; - Written in C++ and Python&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/ValeevGroup/mpqc&quot; rel=&quot;noreferrer&quot;&gt;MPQC&lt;/a&gt; - Written in C++&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/evaleev/libint&quot; rel=&quot;noreferrer&quot;&gt;LibInt&lt;/a&gt; - Written in C++ for efficient quantum integrals.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tddft.org/programs/octopus/wiki/index.php/Libxc&quot; rel=&quot;noreferrer&quot;&gt;LibXC&lt;/a&gt; - Written in C with Fortran &quot;bindings&quot; for DFT exchange-correlation functionals&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is my opinion, but my recommendation for faster performance in chemistry would be Python with some C or C++ mixed in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find I'm more efficient coding in Python, partly because of the language, partly because of the many packages, partly since I don't have to compile, and that's all important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can run Python scripts and functions in parallel, on the GPU, and even compile them, e.g. with &lt;a href=&quot;http://numba.pydata.org&quot; rel=&quot;noreferrer&quot;&gt;Numba&lt;/a&gt;. As I said, if I think performance is crucial, I'll write pieces in C or usually C++ and link to Python as needed.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="402" LastEditDate="2017-04-02T19:57:32.360" LastActivityDate="2017-04-02T19:57:32.360" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="24695" PostTypeId="2" ParentId="24691" CreationDate="2015-02-06T02:38:05.453" Score="17" Body="&lt;p&gt;FORTRAN used to have an edge when it came to speed due to having much better optimizing compilers, in part due to its relative simplicity. Now that C and C++ compilers are nearly on par (and occasionally better), other factors such as programmability come into play. However, there is a lot of legacy code out there as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work in quantum chemistry, and many of the programs that I work with have a mix but are starting to incorporate more C++. The PSI4 package even mixes Python and C++ to gain the benefits of speed and usability of each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of various programs and the language(s) they use.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ORCA - C++&lt;/li&gt;&#xA;&lt;li&gt;MPQC - C++&lt;/li&gt;&#xA;&lt;li&gt;PSI4 - C++, Python&lt;/li&gt;&#xA;&lt;li&gt;PySCF - Python&lt;/li&gt;&#xA;&lt;li&gt;Q-Chem - FORTRAN, C++&lt;/li&gt;&#xA;&lt;li&gt;CFour/ACES - FORTRAN, recently some C++&lt;/li&gt;&#xA;&lt;li&gt;NWChem - FORTRAN, C&lt;/li&gt;&#xA;&lt;li&gt;GAMESS(US) and GAMESS(UK) - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;Gaussian - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;Molpro - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;Dalton - FORTRAN, some C&lt;/li&gt;&#xA;&lt;li&gt;MRCC - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;MOLCAS/OpenMOLCAS - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;DIRAC - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;ADF - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;CASINO - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;COLUMBUS - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;CP2K - FORTRAN&lt;/li&gt;&#xA;&lt;li&gt;TURBOMOL - FORTRAN&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid_state_physics_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quantum chemistry packages&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorUserId="11568" LastEditDate="2020-03-04T17:16:14.373" LastActivityDate="2020-03-04T17:16:14.373" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24696" PostTypeId="1" AcceptedAnswerId="24703" CreationDate="2015-02-06T02:57:23.507" Score="1" ViewCount="519" Body="&lt;p&gt;How do I know when I should add an electron to a single electron? Do I just always do this? I've came across this in the textbook in the form of the amide ion $\ce{NH2}$, among others. It was shown to have 8 electrons, despite nitrogen only having 5 valence electrons, and each hydrogen only having 1. The last came from some surrounding atom to complete the outer shell and thus become more stable. So, do I always do that when there is a single electron not part of a pair? &lt;/p&gt;&#xA;" OwnerUserId="13225" LastEditorUserId="7448" LastEditDate="2015-02-06T08:12:14.867" LastActivityDate="2015-02-06T11:08:28.547" Title="How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion?" Tags="&lt;organic-chemistry&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24699" PostTypeId="2" ParentId="24691" CreationDate="2015-02-06T05:30:42.843" Score="14" Body="&lt;p&gt;FORTRAN is a less expressive language than C++, and that allowed older compilers to optimize much more efficiently.  In modern compilers, there is little to no difference in performance.  There are just a few places where FORTRAN's weaker safety guarantees allow more extreme optimizations.  You likely won't notice the difference unless you spend 6-10 years developing FORTRAN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, it is easier to do development in the newer languages because they better take advantage of modern computing capabilities.  FORTRAN is very strict in its format because it had to be compiled by computers of the mid-1970's.  You tell the computer exactly what to do in its language, rather than telling it what you want to do in a more human readable language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FORTRAN is still used for two major reasons that I can uncover:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Some problems are simple enough where the expressiveness of C++ is wasted, and you can actually make a clearer program with the simpler FORTRAN.&lt;/li&gt;&#xA;&lt;li&gt;Many programs are made by the previous generation.  FORTRAN was, in the past, simply better for computational work.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13226" LastEditorUserId="402" LastEditDate="2017-04-02T19:58:39.233" LastActivityDate="2017-04-02T19:58:39.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24700" PostTypeId="1" AcceptedAnswerId="67772" CreationDate="2015-02-06T07:01:25.747" Score="9" ViewCount="2596" Body="&lt;p&gt;I was wondering what salts of gold are soluble and which are insoluble. From what I understand, silver is soluble as silver nitrate but insoluble as most other common salts like chlorides and sulphates. How does this work with gold? I was also wondering if there are any rules for solubility of salts, or whether there is some explanation for the solubility that would help in guessing the solubility of a salt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started thinking about this because of a question I came across in which there was a mixture of gold, silver and copper to which hot concentrated nitric acid was added, and an undissolved residue of given mass remained. Then $\ce{HCl}$ was added and more precipitate was formed. The question asked you to find the percentage by mass of gold, silver and copper. Judging by the general solubility trend of these metals, I'm guessing the first precipitate was gold and the second silver, but I was also wondering if there are some solubility rules...&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="15410" LastEditDate="2015-05-07T18:21:48.727" LastActivityDate="2017-02-05T06:26:13.397" Title="Solubility of gold salt solution?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;metal&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24702" PostTypeId="2" ParentId="24692" CreationDate="2015-02-06T09:26:46.310" Score="1" Body="&lt;p&gt;Opacity or transparence is a matter of light reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8zyoD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way light is reflected or absorbed depends on severals parameters, first the composition of your material. If it contains molecules or crystals absorbing the visible light, you'll see it black, or colored with the color corresponding to the wavelenght the less absorbed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also crystallographic considerations to take care of. If your material structure is well ordered, dimension of your latice will be set so to reflect the light, then you'll see it transluscent or opaque. For something without ordered structure, you'll probably see it transparent if the visible light isn't absorbed. We'll speak about &quot;amorphous&quot; material then like glass for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit : if you feel easy with those concepts, I can go further in the explainations, but since you asked for a really vulgarized answer I haven't developed more.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="10119" LastEditDate="2015-03-11T09:59:12.273" LastActivityDate="2015-03-11T09:59:12.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24703" PostTypeId="2" ParentId="24696" CreationDate="2015-02-06T09:38:31.490" Score="2" Body="&lt;p&gt;I think I know where your confusion comes from. Firstly, I would want to note an important point:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Covalent bonds&lt;/strong&gt; consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonds_vs_Ionic_Bonds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemwiki.UCDavis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is the characteristic that makes them different from ionic bonds. The surrounding atom &lt;strong&gt;will not&lt;/strong&gt; become an ion in its full sense.&lt;/p&gt;&#xA;&lt;p&gt;I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;:&lt;/h1&gt;&#xA;&lt;p&gt;We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZTwW1.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.middleschoolchemistry.com/img/content/multimedia/chapter_4/lesson_6/covalent_bond_water_1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source 1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Follow these instructions for &amp;quot;normal&amp;quot; molecules (the ones that abide octet in a perfect sense):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: &lt;span class=&quot;math-container&quot;&gt;$$6+(2*1)=8$$&lt;/span&gt;&#xA;I prefer to call this &lt;span class=&quot;math-container&quot;&gt;$\text{n}_v$&lt;/span&gt;. V represents valence shell.&lt;/li&gt;&#xA;&lt;li&gt;Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$8+2*2=12$$&lt;/span&gt;&#xA;I prefer to call this &lt;span class=&quot;math-container&quot;&gt;$\text{n}_f$&lt;/span&gt;. F represents &amp;quot;final&amp;quot;. You can name it whatever you want.&lt;/li&gt;&#xA;&lt;li&gt;If you want to know how many electrons are bonding and are shared between atoms, subtract &lt;span class=&quot;math-container&quot;&gt;$\text{n}_v$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$\text{n}_f$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$&lt;/span&gt;&#xA;For an ease in calculation, I call this &lt;span class=&quot;math-container&quot;&gt;$\text{n}_b$&lt;/span&gt;. As you might have just figured out,&lt;/li&gt;&#xA;&lt;li&gt;If you want to know how many es are nonbonding and &amp;quot;belong&amp;quot; to a single atom, subtract &lt;span class=&quot;math-container&quot;&gt;$\text{n}_b$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$\text{n}_v$&lt;/span&gt; to get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$&lt;/span&gt;&#xA;Call this whatever you want. I rather calling it &lt;span class=&quot;math-container&quot;&gt;$\text{n}_{nb}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oPSqS.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/f/fa/Water-2D-flat.png/220px-Water-2D-flat.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source 2&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;In ions, we have to take the charge of the ion into an account. &lt;strong&gt;Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from &lt;span class=&quot;math-container&quot;&gt;$\text{n}_v$&lt;/span&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;So let's get on with this thing's lewis structure:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_v = 5+(4*1) -1=8$&lt;/span&gt; (Notice (-1) that was because of a 1+ ion charge.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_f = 8 + (4*2)=16$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_b = 16-8=8$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_{nb} = 8-8=0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as &lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/dative.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;This is an awesome tutorial about the subject&quot;&gt;dative covalent bonding&lt;/a&gt;. We get this structure in accordance to the calculations and thoughts:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iacrX.gif&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text7/Tx72/NH4P.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source 4&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Finally, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2-}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;Now this case is easily solvable through the explanations for ammonium.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_v = 5 + 2 +1=8$&lt;/span&gt; The +1 is because of the charge of the ion.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_f = 8 +2*2=12$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_b = 12-8=4$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{n}_{nb} = 8-4=4$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Put some thinking, and you should get this (ignore the other ones if you like):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YucUz.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.meta-synthesis.com/webbook/45_vsepr/lewis3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source 5&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-06T11:08:28.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24704" PostTypeId="2" ParentId="24691" CreationDate="2015-02-06T09:56:57.580" Score="31" Body="&lt;p&gt;I think it does make sense to provide a somewhat alternative view and to clarify the matter.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;FORTRAN vs. Fortran&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First off, one has to distinguish the old FORTRAN from the new Fortran, where, by convention, the name of the old language is written usually in all caps. The old FORTRAN (all the way up to FORTRAN 77) is indeed still used because of tons of legacy code, but the new Fortran (starting from Fortran 90) is used mainly because it is a very elegant and simple yet powerful and efficiently implemented language &lt;em&gt;for number crunching&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;DSL vs. GPL in general&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Note specifically, that even modern Fortran, in my opinion, is &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Domain-specific_language&quot; rel=&quot;noreferrer&quot;&gt;domain-specific language&lt;/a&gt;&lt;/em&gt; (DSL) and herein lies its relative weakness comparing to some &lt;a href=&quot;http://en.wikipedia.org/wiki/General-purpose_language&quot; rel=&quot;noreferrer&quot;&gt;general-purpose languages&lt;/a&gt; (GPL) like the mentioned C++: Fortran is specialised for a particular task (number crunching) and might be not so suitable for some related tasks (say, automated analysis of the final results, their graphical representation, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General-purpose languages, such as C++, give you more flexibility (in language features, in 3rd party libraries, etc.) so that you can solve not just the primary task (number crunching) but also the related task &lt;em&gt;using the same language&lt;/em&gt;. If, however, you choose Fortran for number crunching, you often have to use &lt;em&gt;one more language&lt;/em&gt; (e.g. Python) for these related tasks. Think about it as of using two different DSLs: one for the primary task, another one for related tasks. Of course, you could also use Python together with C++, but an experienced C++ developer (which you are supposed to be if you choose to do some number crunching in it) would not necessarily take advantage of using another language instead of his C++ beast. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fortran vs. C++ specifically&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;All this is somewhat subjective, but anyway, here are my 5 cents. Overall, &lt;em&gt;Fortran is simpler&lt;/em&gt; than C++, but (and because) &lt;em&gt;C++ is more feature-rich&lt;/em&gt;. Basically, it because Fortran is DSL, while C++ is GPL. And as I said, this is subjective to some degree, and besides, the complexity is one of the most complex things in the universe, so we could start a debate about it, but hey, just one word: templates and the discussion is in principle over. Templates make C++ the beast, but everything comes for its price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I didn't say that C++ is more powerful, since, in my opinion, &quot;more feature-rich&quot; does not always mean &quot;more powerful&quot;. It depends first of all on do you actually need these additional features in the first place or not? Seriously, do you need the level of generosity (and metaprogramming) C++ templates provides for number crunching? Not necessarily. And if so, C++ would not be more powerful than Fortran, although, it will still be more feature-rich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Templates are not the only feature of C++ that Fortran does not have. Exceptions and Standard Library - are two other noticeable features that Fortran does not have. Again, it is not so likely that you will greatly benefit from these features for number crunching. But some other tasks they might be very helpful, so C++ as GPL includes them while Fortran as DSL for number crunching does not. Think about Fortran as of a &quot;suitcase language&quot; just for number crunching and about C++ as of a &quot;trunk language&quot; for everything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also one great feature of the Fortran language which C++ still could not catch up with: modules. I mean, &lt;em&gt;real modules&lt;/em&gt;, and not ancient preprocessor machinery which irritates me more than everything else.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="402" LastEditDate="2017-04-02T19:58:04.543" LastActivityDate="2017-04-02T19:58:04.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24705" PostTypeId="1" CreationDate="2015-02-06T10:10:02.660" Score="2" ViewCount="961" Body="&lt;p&gt;I have a few questions regarding the production of pure, crystalline silicon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know Silicon usually appears in nature in the form of sand or $\ce{SiO2}$ and then is split apart and brought into the form $ \ce{Si + 2CO} $ ; &#xA;I am imagining now a hot glowing molten mass. &#xA;However then I don't understand the following methods to separate the Si from the elements you don't want to have, like iron, carbon etc.&#xA;How do you remove the impurities, why do those methods work?&lt;br&gt;&#xA;For example you use a few other chemical reaction to bring Si into several other forms like $ \ce{HSiCl_3} $ and split them apart later on. &#xA;I don't understand at the moment, how this helps to separate those materials as in: There is still this big mass with lots of different chemical connection you don't want to have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also lack an actual understanding of the Czochralski process to further purify your mass of silicon. I guess want has to accept that silicon likes to join pure crystalline silicon and adapt to its structure. However I don't see, why one is not using this method beforehand for example, if one already has raw silicon you get from splitting up $ \ce{SiO_2} $ if it works so greatly. And if it not does work so greatly, why don't you have other impurities ? &lt;/p&gt;&#xA;" OwnerUserId="13232" LastEditorUserId="4945" LastEditDate="2015-02-06T11:04:29.417" LastActivityDate="2015-02-06T11:06:27.443" Title="How does the production of pure, crystalline silicon via the Czochralsky or Siemens process?" Tags="&lt;purification&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24706" PostTypeId="1" CreationDate="2015-02-06T11:00:40.043" Score="1" ViewCount="79" Body="&lt;p&gt;I have a powder which only declares to be &quot;Protein Hydrolysate&quot;&#xA;Meaning that the all the protein has been broken down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've tried to identify the source of the powder with DNA research. But no usable DNA (parts) where found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to identify the source of this powder? (ie. chicken, pork, cow, shark, dolphins or unicorns)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarification, we failed to find any DNA in the powder. so we're looking for alternative ways to analyse it.&lt;/p&gt;&#xA;" OwnerUserId="13234" LastEditorUserId="13234" LastEditDate="2015-02-06T11:17:29.907" LastActivityDate="2015-02-06T17:05:07.047" Title="Is it possible to identify the source of a protein hydrolysate" Tags="&lt;chemical-biology&gt;&lt;identification&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-02-07T00:58:36.620" ContentLicense="CC BY-SA 3.0" />
  <row Id="24707" PostTypeId="2" ParentId="24705" CreationDate="2015-02-06T11:06:27.443" Score="4" Body="&lt;h3&gt;How does trichlorosilane $\ce{HSiCl3}$ help in the purification of silicon?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Trichlorosilane is a highly volatile compound with a boiling point of 32 °C. This means that it can be destilled off the raw material and separated from impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purified trichlorosilane can either be decomposed thermally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4 HSiCl3 -&amp;gt;[\Delta] Si + 3 SiCl4 + 2 H2}$$ or reduced with hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HSiCl3 + H2-&amp;gt; Si + 3 HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does give you rather pure silicon, but it is, at best, polycrystalline.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What's the &lt;em&gt;Czochralski&lt;/em&gt; process good for?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Note that this in't just a melting/cooling cycle! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial point in the method is the use of a rotating, &lt;strong&gt;monocrystalline seeding crystal&lt;/strong&gt;, dipped into (previously purified) molten silicon. The goal is to obtain huge, monocrystalline silicon rods of defined diameter, that can be used in the production of &lt;em&gt;wafers&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-06T11:06:27.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24708" PostTypeId="1" AcceptedAnswerId="24712" CreationDate="2015-02-06T11:44:39.223" Score="11" ViewCount="1689" Body="&lt;p&gt;I wanted to ask an online group of experts/community about how I could improve in chemistry. I study in 10th grade currently and don't get very good grades. Most of the time its a B or C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to select science stream, in which physics, chemistry and maths are compulsory. I am good in physics and maths, but what should I do in chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a hard time grasping concepts like Ionisation Energy, structure of benzene etc. and these are only going to get 100 times tougher next year! &lt;/p&gt;&#xA;" OwnerUserId="13236" LastEditorUserId="7448" LastEditDate="2015-12-17T11:19:45.503" LastActivityDate="2015-12-17T11:19:45.503" Title="How do I excel in Chemistry?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="9" CommentCount="0" FavoriteCount="3" ClosedDate="2015-02-06T19:43:11.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="24709" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T11:51:03.790" Score="2" Body="&lt;p&gt;Read a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read stackexchangechemistry :P &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try this kind of website also maybe : &lt;a href=&quot;https://www.coursera.org/&quot; rel=&quot;nofollow&quot;&gt;Massive Online Open Courses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alos you can find plenty of nice second hand chemistry book on this website : &lt;a href=&quot;http://www.abebooks.com/&quot; rel=&quot;nofollow&quot;&gt;used books&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-06T11:51:03.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24710" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T12:02:03.803" Score="1" Body="&lt;p&gt;Start from bit to bit. When you don't understand what it means, stop for a while and then read the thing again&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-02-06T12:02:03.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24711" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T12:02:48.903" Score="1" Body="&lt;p&gt;I can only speak for mathematics, though I guess, it is similar: In fact I can only stress, what Babounet said and say, that you have to really work hard... really hard until your hands hurt from writing and read until you fall asleep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also work on your mindset. It is not a bad thing to not know something, however one really has to want to understand something and if necessary take every help he/she can get. Thus I think Stackexchange is a good place for you.&lt;/p&gt;&#xA;" OwnerUserId="13232" LastActivityDate="2015-02-06T12:02:48.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24712" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T12:25:34.790" Score="23" Body="&lt;p&gt;Chemistry needs a devoting reader. You should try to love chemistry as much as you want! That'll be the ignition for lots of reading and stuff. My pointers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Love chemistry.&lt;/li&gt;&#xA;&lt;li&gt;Learn in an organized matter. (For example, do not jump from studying about covalent bonds' basics to reaction mechanism.&lt;/li&gt;&#xA;&lt;li&gt;Take notes of the most important points of what you learn.&lt;/li&gt;&#xA;&lt;li&gt;Look for open-source online tutorials and not only the books.&lt;/li&gt;&#xA;&lt;li&gt;Some chemistry concepts are related to each other. Learn the first one's basics and then read the other one's. This will build up your understanding, literally. For example, first learn about electronegativity and then study the covalent bonds' basics. This is the exception for rule 2.&lt;/li&gt;&#xA;&lt;li&gt;Try to differentiate between what's taught in schools to be true and what really is. For example, look for exceptions of the periodic trends that are taught in schools.&lt;/li&gt;&#xA;&lt;li&gt;Be imaginatively practical. Don't just say enough for the lewis structure of water, but do it about $\ce{SF6}$.&lt;/li&gt;&#xA;&lt;li&gt;Always try to go relatively higher than what your textbook has written. We have a proverb in my mother language meaning &quot;If you have 100, you do have 90.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ask what misunderstandings you have in here!&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Never look to doing a research as a waste of time.&lt;/li&gt;&#xA;&lt;li&gt;When approaching to solve a chemical problem, do it step by step.&lt;/li&gt;&#xA;&lt;li&gt;When approaching to solve a problem, don't forget to write units of measurement.&lt;/li&gt;&#xA;&lt;li&gt;When you reached an answer to the problem, ask yourself: &quot;Is this possible?&quot; If you've calculated 5 grams of water to be $\frac{18}{5}$ mol, and you know that a mole of water is 18 grams, shouldn't something be wrong here?&lt;/li&gt;&#xA;&lt;li&gt;For any definitions, uses, notions, and practical experiments you encounter, ask yourself: &quot;What is this for, either in real life, experimental sciences or theoretical chemistry?&quot;&lt;/li&gt;&#xA;&lt;li&gt;Look for online exercises, free of charge, but usually important.&lt;/li&gt;&#xA;&lt;li&gt;Follow the latest news that are being discussed everywhere in the world of chemistry.&lt;/li&gt;&#xA;&lt;li&gt;For specific concepts in chemistry, e.g.: Periodic table, doing a research about the history is very helpful.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is as far as I can count. You can ask &quot;why&quot;s in comments.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-06T12:25:34.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24713" PostTypeId="1" CreationDate="2015-02-06T12:33:39.313" Score="2" ViewCount="866" Body="&lt;p&gt;I am learning about amino acids in various $\ce{pH}$ environments and how their groups are protonated or deprotonated in these environments.&#xA;I don't need to mention a specific acid for my question, I will just use the generic form of one:&#xA;$$\ce{H2N-CH(R)-COOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{H2N}$ group has a $\mathrm{p}K_a$ of $8$, I believe, and the acid has a $\mathrm{p}K_a$ of $3.1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I drop these into a solution with $\ce{pH} = 12$, the $\ce{COOH}$ will donate it's hydrogen ion, but the $\ce{H2N}$ will not. It will only add another hydrogen ion if it is basic compared to the solution to become $\ce{H3N+}$. Why is it, that it won't donate its hydrogen ion like $\ce{COOH}$, when it is acidic compared to the solution, to become $\ce{HN-}$?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="4325" LastEditDate="2015-02-13T04:51:30.677" LastActivityDate="2016-02-16T22:58:04.250" Title="How to rationalise the protonation state of amino acids" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24714" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T12:57:22.330" Score="1" Body="&lt;p&gt;I guess you haven't started with organic....after that you will &lt;strong&gt;love&lt;/strong&gt; it. Try to develop a logic behind everything you read. Think about the concepts. Make the reaction that you were cramming &quot;obvious&quot; for yourself. It might become your strong subject, perhaps more stronger than physics and maths!&lt;/p&gt;&#xA;" OwnerUserId="3771" LastActivityDate="2015-02-06T12:57:22.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24715" PostTypeId="2" ParentId="24295" CreationDate="2015-02-06T13:02:39.870" Score="0" Body="&lt;p&gt;As it turns out, I was just confused on the question being asked. You actually can use Q=mC p ΔT to answer the question. However, instead of using it for Phosphoric acid, you use it based on the water, as you know the mass of water being evaporated out, the specific heat capacity of water at this temperature, and the change in temperature of the water. This Q required by the water will be equal to that added to the phosphoric acid by the heat exchanger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I was incorrect in stating that Q=mC p ΔT is not applicable to phosphoric acid. It is applicable, however not reccomended for this problem as you do not know the mass flow rate of the phosphoric acid through the recycle stream, nor is it easy to find the heat capacity of dilute phosphoric acid at the very high temperatures at which this evaporation would take place.&lt;/p&gt;&#xA;" OwnerUserId="13048" LastActivityDate="2015-02-06T13:02:39.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24716" PostTypeId="2" ParentId="24700" CreationDate="2015-02-06T13:21:50.353" Score="2" Body="&lt;p&gt;Here is What I know:&lt;br/&gt;&#xA;Gold(III) chloride is Highly Hygroscopic [1] and is highly soluble in water,&#xA;I don't know which salts are not soluble never researched it, &#xA;maybe Gold-Mercury amalgam is not soluble in water but its not a salt.. &lt;br/&gt;&#xA;[1] = absorbing or attracting moisture from the air.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="7448" LastEditDate="2015-02-06T14:34:48.297" LastActivityDate="2015-02-06T14:34:48.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24717" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T13:40:50.257" Score="1" Body="&lt;p&gt;I am really dismayed that none of the above answers even so much as mention laboratory work. Get thee into a laboratory and get as much hands on experience with chemistry as you can. It will give you reasons and motivations to learn the theory. Also, you may find that you have a special talent--a &quot;green thumb&quot; for making things work in the lab, even if book learning is not your strong suit. And--it is fun!&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-06T13:40:50.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24718" PostTypeId="2" ParentId="24679" CreationDate="2015-02-06T13:41:13.040" Score="-1" Body="&lt;p&gt;I doubt the answer because I'm sure you'll get $\ce{Sn_{(s)} + 2H2SO4_{(aq)} → SnSO4_{(aq)} + 2H2O_{(l)} + SO2_{(g)}}$ as your Products&lt;/p&gt;&#xA;&#xA;&lt;p&gt;anyways , The Arrhenius concept of acids and bases is as follows: an acid is a substance that when dissolved in water increases the concentration of the hydrogen ion,  $$\ce{H+}$$&#xA;When Sulphuric acid is dissolved in water it gives you $$\ce{H+}$$and $$\ce{SO4^{2-}}$$ ions.&#xA;Sulphuric acid is a VERY Strong Oxidising agent, that means that it will oxidise  your Tin&#xA;therefore we get  $\text{Sn}\rightarrow \text{Sn}^{2+}+2\text{ e}^-$&#xA;in the second part I think you have reacted your 2 electrons formed in the previous reaction with Hydrogen ion, But I doubt this part&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="10185" LastEditDate="2015-02-08T18:03:53.283" LastActivityDate="2015-02-08T18:03:53.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24719" PostTypeId="2" ParentId="24641" CreationDate="2015-02-06T13:47:24.427" Score="-4" Body="&lt;p&gt;&lt;strong&gt;$\ce{NaOH}$ reacts with $\ce{K}$ to form $\ce{KOH}$&lt;/strong&gt; &#xA;there is no compound which is called as &lt;em&gt;NaOK&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?&lt;/strong&gt;&#xA;Yes of course &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaOH + Na  -&amp;gt; Na2O +  ½ H2}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2O}$ is Sodium oxide, which is like anhydrous form of $\ce{NaOH}$. When you add water to Sodium oxide it will give you back $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note - Answer edited completely&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="61578" LastEditDate="2018-04-24T15:57:47.373" LastActivityDate="2018-04-24T15:57:47.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24720" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T14:07:37.337" Score="11" Body="&lt;p&gt;Get a chemistry set and do experiments and have fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was a kid, much younger than you, I asked my mom for 75 cents to buy a chemistry set at a yard sale in our neighborhood, but she wouldn't give it to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a friend that was a couple years older bought it and we would use it at his house. Eventually I got my own chemistry set and would constantly be looking for ways to expand it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by the time I got to high school chemistry I had intuition about chemistry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once my high school teacher said, I think quoting some else, that students today ('today' being a long time ago now) know all about atomic orbitals but don't know what color permanganate is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given that you are good at math and physics, you could only be missing an intuitive aspect of chemistry that you get from practical experiments. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-06T14:24:17.500" LastActivityDate="2015-02-06T14:24:17.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24721" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T14:22:10.637" Score="1" Body="&lt;p&gt;You can watch videos related to chemistry as a starter. Later you will developed an interest in it naturally.&lt;/p&gt;&#xA;" OwnerUserId="13190" LastActivityDate="2015-02-06T14:22:10.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24722" PostTypeId="2" ParentId="24641" CreationDate="2015-02-06T14:27:17.983" Score="-1" Body="&lt;p&gt;Having both sodium hydroxide and potassium in aqueous solution would result in the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\ce{NaOH + H2O &amp;amp;-&amp;gt; Na+ + OH- + H2O}\\&#xA;\ce{K + H2O + OH- &amp;amp;-&amp;gt; K+ + 2OH- + H2}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the potassium is going to react with the water long before anything else. The only way I can think of to get your reaction to go forward would be to have the molten forms of $\ce{NaOH}$ and $\ce{K}$, which would react like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH + K -&amp;gt; KOH + Na}$$&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="5026" LastEditDate="2018-04-21T15:49:20.083" LastActivityDate="2018-04-21T15:49:20.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24723" PostTypeId="1" AcceptedAnswerId="33585" CreationDate="2015-02-06T14:27:23.420" Score="5" ViewCount="9074" Body="&lt;p&gt;Several people have said that the key to understanding chemistry is through memorizing the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to ask if there is a simple technique to learn it, or if I just have to remember every element as it is?&lt;/p&gt;&#xA;" OwnerUserId="13190" LastEditorDisplayName="user467" LastEditDate="2015-02-07T02:13:27.203" LastActivityDate="2019-07-18T18:18:43.953" Title="Memorizing the periodic table" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;&lt;elements&gt;" AnswerCount="8" CommentCount="13" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24724" PostTypeId="2" ParentId="24713" CreationDate="2015-02-06T14:42:44.150" Score="2" Body="&lt;p&gt;Let's assume you drop the amino acid in water at $\ce{pH} = 7$.&#xA;You'll have $$\ce{ H3N^+{}-CH(R)-COO^-}.$$&#xA;Then you can go to $\ce{pH}= 12$ and the amino acid loose the proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't get rid of the second proton because the $\mathrm{p}K_a$ is defined as $$\mathrm{p}k_{a}=-\lg\frac{[\ce{NH2}]\cdot[\ce{H+}]}{[\ce{NH3+}]}.$$&#xA;This means $\ce{NH3+ -&amp;gt; NH2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could define another $\ce{p}K_a$ for $\ce{NH2 -&amp;gt; NH^{-}}$.&#xA;The $\mathrm{p}K_a$ for this would be much higher than 12 - at least around 35.&#xA;You may look at this site: &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/09/29/how-to-use-a-pka-table/&quot; rel=&quot;nofollow&quot;&gt;How to use a $\mathrm{p}K_a$ table&lt;/a&gt;&#xA;You can compare the amine $\ce{NH3}$ (without the rest of the amino acid) with your $\ce{NH2}$ group of the amino acid.&lt;/p&gt;&#xA;" OwnerUserId="13238" LastEditorUserId="16683" LastEditDate="2016-02-16T22:58:04.250" LastActivityDate="2016-02-16T22:58:04.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24725" PostTypeId="2" ParentId="24723" CreationDate="2015-02-06T14:46:40.907" Score="13" Body="&lt;p&gt;There isn't really much sense in memorizing the periodic table. The elements you often use you will know them by heart after a while. And you can always use a table when you need it for the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said if you really want to do it, mnemonics are probably the best solution to memorizing the whole table. You can find some here :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mnemonic-device.com/chemistry/periodic-table/mnemonic-device-for-the-periodic-table/&quot;&gt;https://www.mnemonic-device.com/chemistry/periodic-table/mnemonic-device-for-the-periodic-table/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or here :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Periodic_table&quot;&gt;http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Periodic_table&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13179" LastActivityDate="2015-02-06T14:46:40.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24726" PostTypeId="1" AcceptedAnswerId="24750" CreationDate="2015-02-06T14:50:17.263" Score="7" ViewCount="5566" Body="&lt;p&gt;Aqua regia has the famous property of being able to dissolve noble metals, like gold and silver. I read that this has to do with the generation of nascent chlorine. How does this happen and how does it help to dissolve such metals?&lt;/p&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="16683" LastEditDate="2016-10-02T16:54:33.627" LastActivityDate="2016-10-02T21:29:18.367" Title="Aqua regia and nascent chlorine" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24727" PostTypeId="1" AcceptedAnswerId="24733" CreationDate="2015-02-06T14:57:23.487" Score="3" ViewCount="38185" Body="&lt;p&gt;I'm reading about Lysine and doing some calculations on it's charge under various $\ce{pH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the source, Lysine is usually stated as a charged amino acid with a $\ce{NH3+}$ group. However the video I'm watching is calculating the charge at 5 pH and it writes Lysine out with $\ce{NH2}$ and goes on to say that because the solution is pH 5, then $\ce{NH2 -&amp;gt; NH3+}$ and arguments that THIS is why Lysine is charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the literature state Lysine as a charged acid, assuming it is at pH 7? Or is the video I am watching wrong and it should not gain a charge as $\ce{NH3}$ will not accept a hydrogen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus: Why are only SOME groups given a Pka value? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geneinfinity.org/images/aapks.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.geneinfinity.org/images/aapks.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html&quot; rel=&quot;nofollow&quot;&gt;http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the charge for Isoleucin for example?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="10185" LastEditDate="2015-02-06T15:07:13.053" LastActivityDate="2015-02-06T16:29:41.590" Title="Why and when is lysine charged" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24728" PostTypeId="1" AcceptedAnswerId="24732" CreationDate="2015-02-06T15:05:46.553" Score="2" ViewCount="1177" Body="&lt;p&gt;Hello here is a question that is a bit confusing I hope you can help me solve it and understand it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine the hydronium and hydroxide ion concentrations in a solution&#xA;  that is $10^{-4} M \ce{Ca(OH)2} \text{ AND } 10^{-4} M \ce{HCl}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: the &quot;and&quot; is actually capitalised in my book..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I understood from this question is that they are both in one solution. Now I thought of two ways to solve it: &#xA;A) the concentrations are equal and they are both strong which means they we're neutralise each other completely and end up being a neutral solution and both $\ce{H3O+}$ and $\ce{OH-}$ will be $10^{-7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) the calcium base has 2 $\ce{OH-}$ while HCl has 1 H+ so the base concentration is double that of the acid which means that the solution is going to be basic but I don't know how can I solve this to find the actual concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which way is right and when does an acid and base really react to become neutral?&lt;/p&gt;&#xA;" OwnerUserId="9776" LastEditorUserId="19383" LastEditDate="2015-11-13T18:58:41.887" LastActivityDate="2015-11-13T18:58:41.887" Title="Ion concentration in acid and base" Tags="&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24729" PostTypeId="1" CreationDate="2015-02-06T15:20:21.503" Score="3" ViewCount="1398" Body="&lt;p&gt;Can someone tell me how to make 0.3% ammonia water from 28-30% Ammonia hydroxide?&#xA;I am trying to make a gallon or 3.78 Liters.&lt;/p&gt;&#xA;" OwnerUserId="13240" LastActivityDate="2017-12-12T02:02:31.837" Title="How to make 0.3% ammonia water from 28-30% ammonia hydroxide" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24730" PostTypeId="1" CreationDate="2015-02-06T15:28:33.180" Score="1" ViewCount="1011" Body="&lt;p&gt;Why is the boiling point of iron higher than magnesium? &lt;/p&gt;&#xA;" OwnerUserId="13243" LastEditorUserId="13243" LastEditDate="2015-02-06T15:57:57.430" LastActivityDate="2015-02-06T16:15:49.013" Title="Comparing the strength of metallic bonds" Tags="&lt;bond&gt;&lt;metal&gt;&lt;ionic-compounds&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24731" PostTypeId="1" AcceptedAnswerId="24744" CreationDate="2015-02-06T15:31:31.570" Score="7" ViewCount="15374" Body="&lt;p&gt;I know that there are organic compounds that exhibit electroluminescence, and are those types of materials difficult to synthesize? &lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorUserId="5017" LastEditDate="2015-02-06T18:36:43.907" LastActivityDate="2018-04-06T10:12:14.630" Title="What type of organic compounds are used in OLED technology?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24732" PostTypeId="2" ParentId="24728" CreationDate="2015-02-06T15:34:39.403" Score="1" Body="&lt;p&gt;Your second (B) way of thinking is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solution there are twice as many OH- ions as H+ ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the H+ from HCL will react with half the OH- from calcium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half the OH- will remain ($10^{-4} OH-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use $K_w$ to determine the final concentration of H+.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-06T15:34:39.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24733" PostTypeId="2" ParentId="24727" CreationDate="2015-02-06T15:41:30.567" Score="2" Body="&lt;p&gt;The side chain of lysine is charged at all pH lower than 10.53 (according to the table you linked), and so it would be charged at pH 7. What happens is $\ce{NH2}$ is more or less stable, but once you start to have any decent amount of hydrogen in your solution it will donate its lone pair to become $\ce{NH3+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus answer: only the amino acids with ionizable side chains are given pKa values, because, well, they're the only ones that are ionizable. In other words, they're the only groups that will accept or donate a proton at any reasonable pH (you could theoretically ionize glutamine, but the pH would be ridiculously low, and well outside of the world of biochemists who are only really concerned with pH levels inside the body). &lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="13030" LastEditDate="2015-02-06T16:03:15.910" LastActivityDate="2015-02-06T16:03:15.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24734" PostTypeId="2" ParentId="24727" CreationDate="2015-02-06T15:56:19.770" Score="3" Body="&lt;p&gt;At pH 7 lysine has a net charge of very close to +1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carboxylic acid group is fully deprotonated (-1 charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alpha amino group is about 99% protonated (+1 charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The side chain amino group is fully protonated (+1 charge).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to look at the pK of each ionizable group, consider what fraction of the group is in each ionization state (for example using the Henderson Hasselbach equation), and add the charges of each group to get the net charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For isoleucin, at pH 7, the net charge will be almost neutral, because 7 is well above the carboxylic acid pK of 2.32 and significantly below the amino pK of 9.76.  There will be a small fractional negative charge because 7 is closer to the pK of 9.76 than to the pK of 2.32.  Note in the first linked table in the question, there is a column for &quot;pI&quot;.  The pI will be the only pH where the amino acid has a net neutral charge. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-06T16:29:41.590" LastActivityDate="2015-02-06T16:29:41.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24735" PostTypeId="2" ParentId="24727" CreationDate="2015-02-06T16:02:11.067" Score="1" Body="&lt;p&gt;In aqueous solutions amino acids are usually considered as zwitterions. &#xA;Even as solid salt they are charged.&#xA;I can't really say, why usually it is drawn uncharged. It may confuse but it isn't important when you want to know which charge it really has. Just take a look at the pH and the pka values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a solution the charge depends on the pH of the solution. If there are additional groups which can protonate or deprotonate you have to take a look at their pka.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Lysin has to amino groups and one carboxylic acid group.&#xA;The pka of the both aminogroups is much higher than your pH of 5, so both are protonated. You have two positive charges.&#xA;The carboxylic acid is deprotonated at pH 5 because its pka is 2.18. So one negative charge here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means Lysin has in total one positive charges at pH 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water you have always charges because you have a acid and base in one molecule, so they can react with each other. COOH protonates the $NH_2$ and $NH_2$ deprotonates the COOH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge of Isoleucin is an easy example because you have no side chains, just COOH and $NH_2$. You can take a given pH (maybe 5) and then take a look at the pka of both functional groups as written above. Then add the charges. &#xA;Or you can take a look at the pI. This works even with side chains. If the pH = pI you have a zwitterion with on positive and one negative charge, so the whole molecule is neutral. If your pH is higher than the pI it is mor basic and something will deprotonate so you have a negative charge.&#xA;If pH &amp;lt; pI something is protonated und you have a positive charge.&#xA;pI refers to that point there half of all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BONUS:&#xA;Just take a close look at the functional groups of the side chains. Since most times pka is refered to aqueous solutions only those groups are given a pka value which can be protonated or deprotonated in water. A Methyl group for example $R-CH_3$ can't be deprotoned in water. so why should they give a pka value? It wouldn't be helpful.&lt;/p&gt;&#xA;" OwnerUserId="13244" LastEditorUserId="13244" LastEditDate="2015-02-06T16:16:39.637" LastActivityDate="2015-02-06T16:16:39.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24736" PostTypeId="2" ParentId="24730" CreationDate="2015-02-06T16:15:49.013" Score="1" Body="&lt;p&gt;There are two main reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Metallic bonding is more prevalent for transition metal elements because there are more electrons available for delocalization (from s and d orbitals), while Magnesium has only s electrons&lt;/li&gt;&#xA;&lt;li&gt;Fe has a higher atomic mass than Mg --&gt; van der Waals forces are stronger.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13086" LastActivityDate="2015-02-06T16:15:49.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24738" PostTypeId="1" AcceptedAnswerId="24776" CreationDate="2015-02-06T16:24:00.323" Score="1" ViewCount="2163" Body="&lt;p&gt;I want to know how to compare the strength or stability of ionic compounds.&#xA;Example : $\ce{Mg(HCO3)2}$ and $\ce{NaHCO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that polarity and polarizability affect the ionic and covalent nature.&lt;/p&gt;&#xA;" OwnerUserId="13243" LastEditorUserId="10185" LastEditDate="2015-02-06T16:31:27.173" LastActivityDate="2015-02-07T15:14:08.473" Title="Comparing the nature of ionic compounds" Tags="&lt;covalent-compounds&gt;&lt;ionic-compounds&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24739" PostTypeId="2" ParentId="24726" CreationDate="2015-02-06T16:35:36.997" Score="2" Body="&lt;p&gt;Chlorine gas can react with gold powder to generate $\ce{AuCl3}$. In an aqua regia solution, the driving force is the creation of the tetrachloroaurate complex, $\ce{[AuCl4]-}$.&lt;/p&gt;&#xA;" OwnerUserId="13086" LastEditorUserId="194" LastEditDate="2016-10-02T19:32:52.623" LastActivityDate="2016-10-02T19:32:52.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24740" PostTypeId="2" ParentId="24738" CreationDate="2015-02-06T16:39:43.113" Score="0" Body="&lt;p&gt;The strength of an ionic bond is determined by two factors:&#xA;The size of the ions (the smaller the ions the stronger the bond) NaF stronger than NaCl because F- is smaller than Cl-&#xA;The charge of the ions (the higher the charge the stronger the bond) Na2O stronger than NaCl, because O is a double negatively charged ion&lt;/p&gt;&#xA;" OwnerUserId="13086" LastActivityDate="2015-02-06T16:39:43.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24741" PostTypeId="2" ParentId="24706" CreationDate="2015-02-06T17:05:07.047" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;DNA is &lt;strong&gt;not&lt;/strong&gt; a protein, as @MARamezani states. See &lt;a href=&quot;http://www.answers.com/Q/Is_DNA_a_protein&quot; rel=&quot;nofollow&quot;&gt;http://www.answers.com/Q/Is_DNA_a_protein&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a prevalence (or lack) of some amino acids in certain sources. For example, maize (&quot;corn&quot; in America) is deficient in tryptophan and lysine, so if your sample is short on those two, it is &lt;em&gt;more likely&lt;/em&gt; that it is from corn. A list of some foods and amino acid content is at &lt;a href=&quot;http://www.ljhs.sandi.net/faculty/aquesnell/Biology%20Notes/Chapter%204/4.8-4.13/Essential%20Amino%20Acids.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ljhs.sandi.net/faculty/aquesnell/Biology%20Notes/Chapter%204/4.8-4.13/Essential%20Amino%20Acids.pdf&lt;/a&gt; [there appear to be some errors, though]. According to that source, yeast may be deficient in phenylalanine, so that is another indicator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You would need an analysis of the relative abundance of each amino acid, some reference showing their abundance in various sources, &lt;em&gt;and additional analysis to calculate relative abundances in a &lt;strong&gt;mixture&lt;/strong&gt;, e.g. corn supplemented with yeast hydrolyzate.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-06T17:05:07.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24742" PostTypeId="2" ParentId="20511" CreationDate="2015-02-06T18:11:53.723" Score="13" Body="&lt;p&gt;No compound is purely ionic. Water is about 33% ionic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linus Pauling discusses the ionic character of water in his famous &quot;&lt;em&gt;The Nature of the Chemical Bond&lt;/em&gt;&quot; [1] at pages 100-101.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pauling says as a first approximation water should be considered as having contributions from four resonance structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lc}&#xA;&amp;amp;37\% \qquad &amp;amp; \ce{H-O-H} \\  &#xA;&amp;amp;24\% \qquad &amp;amp; \ce{(H-O)^-~H+} \\ &#xA;&amp;amp;24\% \qquad &amp;amp; \ce{H+~(O-H)-} \\  &#xA;&amp;amp;15\% \qquad &amp;amp; \ce{H+~O^{2-}~H+}&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, he goes on to explain that this is based upon considering each $\ce{O-H}$ bond independently, and the fact that the two $\ce{O-H}$ bonds are not independent reduces the contribution of $\ce{H+~O^{2-}~H+}$ somewhat and increases the other three contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the purely convalent form $\ce{H-O-H}$ should be considered as contributing only somewhere between 37% and 44% according to Pauling.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pauling, L. The Nature of the Chemical Bond and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry, 3rd ed.; &lt;em&gt;Cornell University Press&lt;/em&gt;, &lt;strong&gt;1960&lt;/strong&gt;. ISBN 978-0-8014-0333-0.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2017-09-25T18:23:52.853" LastActivityDate="2017-09-25T18:23:52.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24743" PostTypeId="1" AcceptedAnswerId="24749" CreationDate="2015-02-06T18:18:54.003" Score="5" ViewCount="4645" Body="&lt;p&gt;Is a C-D bond better than C-H bond?  I mean if carbon has a chance to either for a bond with deuterium or hydrogen what would it choose and why?  What if tritium competes too? &lt;/p&gt;&#xA;" OwnerUserId="7897" LastEditorUserId="7448" LastEditDate="2015-02-06T18:49:33.257" LastActivityDate="2016-07-24T05:19:54.803" Title="Would C bond with H or D?" Tags="&lt;organic-chemistry&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24744" PostTypeId="2" ParentId="24731" CreationDate="2015-02-06T18:36:32.560" Score="7" Body="&lt;p&gt;The short answer is a variety of pi-conjugated materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;em&gt;huge&lt;/em&gt; variety of compounds, both small molecules, oligomers, polymers, and organometallic species have been used. Key requirements for the emissive material is a high efficiency emission process, either fluorescence or phosphorescence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the first compounds were simply conjugated aromatic molecules and dyes. Martin Pope at NYU observed luminescence from &lt;a href=&quot;http://en.wikipedia.org/wiki/Anthracene&quot; rel=&quot;noreferrer&quot;&gt;anthracene&lt;/a&gt; (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/OLED&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) in the 60s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wide variety of conjugated dyes have been tried with varying success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, there are 2-3 different types of compounds used in an OLED:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hole-transport materials&lt;/strong&gt;: Easily oxidized and conduct positive charges (holes).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Electron-transport materials&lt;/strong&gt;: Easily reduced and conduct negative charges (electrons).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Emissive materials&lt;/strong&gt;: Holes and electrons recombine and emit light.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In many cases, these are made in a sandwich device: ITO anode with hole transport layer on top, emissive layer, then electron transport layer, then cathode. (I'm leaving out various interface agents, &quot;blocking layers,&quot; adhesion layers, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're asking about the emissive component. Basically, you want materials that will efficiently emit (fluoresce or phosphoresce) at different wavelengths / colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest real breakthrough, IMHO was from &lt;a href=&quot;http://en.wikipedia.org/wiki/Tris(8-hydroxyquinolinato)aluminium&quot; rel=&quot;noreferrer&quot;&gt;$\ce{Alq3}$&lt;/a&gt; which is an aluminum organometallic compound with remarkably bright emission. But emits green, so for displays, you need other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been out of the field for a while, but here are some examples. Chemical companies like &lt;a href=&quot;http://www.lumtec.com.tw/product_list_1.php?big_id=1&amp;amp;small_id=49&quot; rel=&quot;noreferrer&quot;&gt;lumtec&lt;/a&gt; sell various components.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Polyfluorene&quot; rel=&quot;noreferrer&quot;&gt;Polyfluorenes&lt;/a&gt;: different copolymers have tailored emission colors. Usually the &quot;bridge&quot; carbon is substituted with two alkyl substituents for solubility and processability (and to avoid aggregation).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In OLEDs, polyfluorenes are desirable because they are the only family of conjugated polymers that can emit colors spanning the entire visible range with high efficiency and low operating voltage. Furthermore, polyfluorenes are relatively soluble in most solvents, making them ideal for general applications[...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/N-Vinylcarbazole&quot; rel=&quot;noreferrer&quot;&gt;Various vinyl carbazoles&lt;/a&gt;: Similar reasons to the polyfluorenes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phosphorescent metal complexes (e.g., iridium): If you work out statistics, you get a lot of electron-hole combinations that are triplets, so phosphorescent emission can be very efficient, particularly with transition metals with a high spin-orbit coupling.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think most leading emissive materials still largely fall into those categories. Small molecule purely organic targets exist, but aren't as efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger question is usually how you intend to deposit the compounds. Vapor deposition (i.e., evaporation or sputtering) can be done with almost anything that sublimates, but if you want to do drop-casting, spin-coating, or inkjet printing, you'll want soluble compounds.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-02-07T05:18:20.290" LastActivityDate="2015-02-07T05:18:20.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24745" PostTypeId="2" ParentId="24708" CreationDate="2015-02-06T18:51:13.063" Score="2" Body="&lt;p&gt;I would actually start with a general rule of thumb for any topic: &lt;strong&gt;Don't strive to learn the content, strive to want to learn the content.  The learning will often happen on its own.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google and Wikipedia can give you all the Chemistry facts you could ever need.  In fact, more than you could ever need.  If all you do is learn facts, you'll eventually be outstripped by those who played on the playground, and then asked Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The true learning comes from being able to integrate those facts with real life, and that is a skill you develop for yourself.  There's no textbook to teach it.  Once you find uses for the chemistry facts you learn, you'll suddenly find they're a WHOLE lot easier to learn.  Those who played on the playground will suddenly be awed by how easily you digest information and how quickly you turn it into application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The path is different for every person.  However, I want to give a shout out to Dave PhD's answer of &quot;Get a chemistry set and have fun with it.&quot;  If there is one &quot;best path&quot; to learn Chemistry, that's it.  Chemistry is &lt;em&gt;so&lt;/em&gt; much more interesting when you come across a problem in life such as &quot;how do I get gum off my shoe,&quot; and find that not only do you know enough about the chemical properties of the latex in the gum to understand why it is so {bleepity-bleep} sticky, but you happen to have the chemicals on hand to break it down because you were using them in a cute color-changing experiment the other night!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ideal world, one day you'd realize you need a chemical to solve some cool problem, but you don't have it.  In a flash of inspiration, you realize that you can do a chemical reaction with stuff you do have to make the chemical you don't.  All you need is some starch to complete the reaction.  Mom doesn't have starch on hand?  Saltine crackers will do nicely, I just have to wash the salt off of it so that the salt doesn't interact.  Once everything connects in your mind, you'll find memorizing ionization energies is a breeze because you know that, one day, you'll use that information!&lt;/p&gt;&#xA;" OwnerUserId="13226" LastActivityDate="2015-02-06T18:51:13.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24746" PostTypeId="1" AcceptedAnswerId="24747" CreationDate="2015-02-06T19:04:01.990" Score="4" ViewCount="164" Body="&lt;p&gt;I am asked to find the concentration of magnesium ions in ppm for a &lt;span class=&quot;math-container&quot;&gt;$0.025\ \%$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{MgCl2}$&lt;/span&gt;. I can use this formula:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{concentration in ppm} = \frac{\text{mass solute}_\mathrm{(g)}}{\text{mass solution}_\mathrm{(g)}}\times10^6$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming &lt;span class=&quot;math-container&quot;&gt;$100\ \mathrm g$&lt;/span&gt; of solution, &lt;span class=&quot;math-container&quot;&gt;$2.5 \mathrm g$&lt;/span&gt; solute/&lt;span class=&quot;math-container&quot;&gt;$\ce{MgCl2}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$97.5\ \mathrm g$&lt;/span&gt; solvent, I get &lt;span class=&quot;math-container&quot;&gt;$\pu{95.205u}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{MgCl2}$&lt;/span&gt;, and then &lt;span class=&quot;math-container&quot;&gt;$0.026259\ \mathrm{mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{MgCl2}$&lt;/span&gt;, and thus &lt;span class=&quot;math-container&quot;&gt;$0.026259\ \mathrm{mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt;. That times &lt;span class=&quot;math-container&quot;&gt;$24.305\ \mathrm{g\ mol}^{-1}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$ \ce{Mg}$&lt;/span&gt; gives me &lt;span class=&quot;math-container&quot;&gt;$0.63823\ \mathrm g$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\frac{0.63823\ \mathrm g}{100\ \mathrm g}$&lt;/span&gt; solution gives me &lt;span class=&quot;math-container&quot;&gt;$0.00638$&lt;/span&gt;, so that multiplied by &lt;span class=&quot;math-container&quot;&gt;$10^6$&lt;/span&gt; would be &lt;span class=&quot;math-container&quot;&gt;$6382.28$&lt;/span&gt;. But the only options for the questions don't go over &lt;span class=&quot;math-container&quot;&gt;$500$&lt;/span&gt;. Like &lt;span class=&quot;math-container&quot;&gt;$64\ \mathrm{ppm}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$500\ \mathrm{ppm}$&lt;/span&gt;. Nothing even close to as high as I'm getting.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="4945" LastEditDate="2020-02-08T18:51:54.940" LastActivityDate="2020-02-08T18:51:54.940" Title="How to calculate the parts-per-million concentration from a percent concentration?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24747" PostTypeId="2" ParentId="24746" CreationDate="2015-02-06T19:14:43.233" Score="7" Body="&lt;p&gt;You're using $2.5\ \mathrm g$ of solute out of $100\ \mathrm g$ of solution, which is $2.5\ \%$, but the initial question is asking about a $0.025\ \%$ solution, which is why your answer is two orders of magnitude too large. You should be using $0.025\ \mathrm g$ of solute and $99.975\ \mathrm g$ of solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general helpful hint for answering multiple choice questions, when you notice that one of the answers has the same coefficient but is off by a couple orders of magnitude (in this case the $64\ \mathrm{ppm}$ answer) go and double check all your initial concentrations and volumes, you probably forgot to account for a &quot;$\%$&quot; or a &quot;$\mathrm k$&quot; somewhere.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorDisplayName="user7951" LastEditDate="2018-08-11T09:15:04.877" LastActivityDate="2018-08-11T09:15:04.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24748" PostTypeId="2" ParentId="6625" CreationDate="2015-02-06T19:14:49.647" Score="0" Body="&lt;p&gt;(I thought the previous answers were incomplete, mainly because previous answers have talked about halo arenes, particularly about fluorobenzene)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question reminds me of a doubt I had once. Consider the two structures shown below:&lt;img src=&quot;https://i.stack.imgur.com/XXhYL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I say these two are geometrical isomers? It sounds a pretty dumb question, but here is the thing. One could argue that whenever we show that the $-COOH$ group is meta directing, we draw resonating structures that have a double bond between the carbons of the $-COOH$ and the carbon of benzene to which the group is joined to. Therefore one could say that the rotation of the carbon is restricted and hence are geometrical isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the thing is, that the rotation of the carbon in $-COOH$ is actually free(not exactly free, since it has to be in the same plane as the rest of the benzene, since it has $sp^2$ hybridization which confines it in a plane) and the $\pi e^-$ of the $-COOH$ group remains with it and &lt;strong&gt;doesn't&lt;/strong&gt; get localized in the benzene ring. But this leads to a very natural question here- then how can we explain that $-COOH$ is meta directing? By inductive effect. Or at least that is what my teacher said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-R (or -M) is shown by a group if a more electronegative element is attached by a double bond directly then it decreases electron density at ortho and para positions wrt meta and hence the group is deactivating. Considering the case of benzoic acid, electronegativity of oxygen plays a vital role , mainly because it attracts the $\pi e^-$s of the ring and decreases electron density over the ring. Although we can say that resonance does take place (which doesn't contribute much to explain why $-COOH$ is meta directing, as explained above), but the more contributing reason here is that oxygen pulls the electron density towards itself making the ring deactivated. So therefore in this case inductive effect plays a vital role here, even more important than resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could also be explained as to why $-COO^-$ is activating by the above. Since a -ve charge over a very electronegative element (oxygen) makes it very stable and kinda &quot;calms it down&quot; from attracting more electrons, therefore oxygen won't attract the electrons of benzene so much therefore relaxing the electron density of benzene a little bit and hence make it ortho and para directing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, as we can clearly see the electronegativity of oxygen here plays a vital role in making $-COOH$ meta directing and $-COO^-$ ortho and para directing. This is the best example to disprove your conclusion because here the overall effect by the oxygen is attracting electrons demonstrates a case where inductive effect is more dominant than resonance.&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="3771" LastEditDate="2015-02-11T04:49:06.460" LastActivityDate="2015-02-11T04:49:06.460" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="24749" PostTypeId="2" ParentId="24743" CreationDate="2015-02-06T19:17:21.590" Score="7" Body="&lt;p&gt;C-H would be the weakest bond, then C-D, then C-T would be the strongest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html&quot;&gt;zero point energy of the bond&lt;/a&gt; and the fact the zero point energy is proportional to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{\frac{m_C + m_H}{m_C m_H}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero point energy is the minimum energy the quantum harmonic oscillator can have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the greatest mass isotope has the smallest zero point energy, the energy of the bond is greatest.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-06T19:17:21.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24750" PostTypeId="2" ParentId="24726" CreationDate="2015-02-06T19:39:03.317" Score="7" Body="&lt;p&gt;Aqua regia is a concentrated mixture of nitric and hydrochloric acids. The two acids react to generate soluble nitrosyl chloride, water and nascent chlorine: $$\ce{HNO3 + 3HCl -&amp;gt; NOCl + 2H2O + 2Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the gold is oxidised by nitric acid to gold(III) ions: $$\ce{Au &amp;lt;=&amp;gt; Au^{3+} + 3e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is normally highly unfavourable and so very few gold(III) ions are formed if you put gold into only concentrated nitric acid. However, the nascent chlorine is highly reactive as it has an unpaired electron in a high energy orbital and it is able to react with the gold(III) ions to form the soluble tetrachloroaurate(III) complex. This removes the gold(III) ions from the equilibrium, driving the reaction to completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall the reaction is: $$\ce{Au + 3HNO3 + 4HCl -&amp;gt; [AuCl4]- + 3NO2 + H3O+ + 2H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-06T19:39:03.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24752" PostTypeId="1" CreationDate="2015-02-06T23:10:18.963" Score="1" ViewCount="1038" Body="&lt;p&gt;If water, $\ce{H2O}$ is a covalent compound formed by sharing of electrons, why is it said (in case of fuel cells) that formation of water from hydrogen and oxygen is a redox reaction (transfer of electrons, not sharing). This would make it an ionic compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights? &#xA;Note that, the concept of water being formed as a result of redox reaction is highlighted in fuel cells (which is the basic principle of fuel cells - generating electricity by exploiting this transfer of electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how fuel cells work, and it is indeed a redox reaction; but my doubt is - can water be formed by both covalent/ ionic means?  (In textbooks, or in exams when asked to identify bonding present in water); when both covalent and ionic type can be achieved in its formation? Is both &quot;covalent&quot; and &quot;ionic&quot; correct while describing the formation of $\ce{H2O}$ molecule? &lt;/p&gt;&#xA;" OwnerUserId="13249" LastEditorUserId="4945" LastEditDate="2015-03-17T09:20:26.660" LastActivityDate="2015-03-17T09:20:26.660" Title="Why is water considered ionic in fuel cells but otherwise covalent?" Tags="&lt;bond&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;water&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24753" PostTypeId="2" ParentId="24743" CreationDate="2015-02-07T00:00:38.240" Score="0" Body="&lt;p&gt;Kinetically, hydrogen will react at the fastest rate, followed by deuterium and then tritium (given a reaction such as a carbon atom or radical plus molecular hydrogen giving a C-H bond plus an •H atom). The most quickly formed product may not necessarily be the most stable. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-07T00:00:38.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24754" PostTypeId="1" AcceptedAnswerId="24780" CreationDate="2015-02-07T00:14:42.687" Score="2" ViewCount="2613" Body="&lt;p&gt;I'm dealing with a free-energy scam on another forum. It is well-known that $\ce{LiAlH4}$ reacts explosively with water. My chemistry days are long behind me, so I thought I'd come here to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1~\text{mol}$ of $\ce{LiAlH4}$ reacted with $4~\text{mol}$ of water gives how much energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not about to do this, mind you. I was just curious about the specific numbers.&lt;/p&gt;&#xA;" OwnerUserId="13250" LastEditorUserId="7448" LastEditDate="2015-02-07T16:31:07.567" LastActivityDate="2015-02-07T16:47:35.607" Title="LiAlH4 / H2O reaction energy" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24755" PostTypeId="1" AcceptedAnswerId="24758" CreationDate="2015-02-07T01:52:11.750" Score="7" ViewCount="10578" Body="&lt;p&gt;Oxygenated water has become popular for a variety of uses recently - for health purposes internally as well as enhanced wound healing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, my dad asked me recently if I knew how it is that oxygenated water does not become hydrogen peroxide. I don't know the answer to it; but it got me thinking... How does this work? Does the oxygen not bond to the water molecule? If so, how does the bottle of water not just out gas when ever it is opened until the amount of oxygen contained in the bottle is equal to that of the room air.&lt;/p&gt;&#xA;" OwnerUserId="5137" LastActivityDate="2020-01-25T19:11:37.813" Title="Why is oxygenated water not H2O2?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24756" PostTypeId="2" ParentId="24723" CreationDate="2015-02-07T02:02:11.400" Score="7" Body="&lt;p&gt;There are a couple of elements you'll want to really get to know, but other than that there's no real reason to memorize the periodic table. To me, memorizing the periodic table sounds like a quick way to hate chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend getting to know a few things about &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrogen&lt;/li&gt;&#xA;&lt;li&gt;Carbon&lt;/li&gt;&#xA;&lt;li&gt;Nitrogen&lt;/li&gt;&#xA;&lt;li&gt;Oxygen&lt;/li&gt;&#xA;&lt;li&gt;Fluorine&lt;/li&gt;&#xA;&lt;li&gt;Sulfur&lt;/li&gt;&#xA;&lt;li&gt;Chlorine&lt;/li&gt;&#xA;&lt;li&gt;Sodium&lt;/li&gt;&#xA;&lt;li&gt;Potassium&lt;/li&gt;&#xA;&lt;li&gt;Bromine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Really the best thing to learn about those is how many bonds they make and what they tend to bond with. These are the elements that I most often see on exams and in examples since they're used in a lot of textbook reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other elements, just get a feel for where they are on the periodic table so you can quickly look them up. This is especially helpful during exams. As you're taking chemistry courses, you'll notice you'll REALLY learn how to find your way around the periodic table and start memorizing the molar masses of various elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend just looking up each element (this will take a lot of time) on Wikipedia or checking out the book, &quot;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1579128955&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Elements&lt;/a&gt;.&quot; That way, you can kind of get to know elements and if you're really wanting to nerd out, you might even find that you have favorites.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2018-01-17T05:37:54.263" LastActivityDate="2018-01-17T05:37:54.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24757" PostTypeId="2" ParentId="24723" CreationDate="2015-02-07T02:10:51.423" Score="4" Body="&lt;p&gt;I think those people meant the important ones. Try to remember the first 20 elements and some useful that you often study in school like iron, silver, bromine, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since you really want to remember the whole periodic table of elements, try searching for the &quot;asapscience new periodic&quot; song on youtube. I found it useful for remembering the elements in order.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="402" LastEditDate="2015-11-02T01:11:09.963" LastActivityDate="2015-11-02T01:11:09.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24758" PostTypeId="2" ParentId="24755" CreationDate="2015-02-07T02:17:45.407" Score="11" Body="&lt;p&gt;The oxygen is dissolved in the water, just as salt can be dissolved. It does not (to any appreciable degree) combine with water molecules to form hydrogen peroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that oxygenated water is not fizzy like soda water (CO2) is the solubility of oxygen in water is about 2% that of CO2. So the amount of bubbles formed when the cap is removed is only about 1/50 as much as for soft drinks. Or champagne.&lt;/p&gt;&#xA;" OwnerUserId="13250" LastActivityDate="2015-02-07T02:17:45.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24759" PostTypeId="2" ParentId="24755" CreationDate="2015-02-07T02:18:13.843" Score="6" Body="&lt;p&gt;I think the problem is that the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2H2O2-&amp;gt; 2H2O + O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is &lt;em&gt;spontaneous&lt;/em&gt;, meaning that it proceeds (slowly) in standard conditions with no other inputs required. So far from oxygen combining with water, you have hydrogen peroxide spontaneously decomposing into water and oxygen.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-02-07T02:18:13.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24760" PostTypeId="2" ParentId="24755" CreationDate="2015-02-07T02:33:35.813" Score="3" Body="&lt;p&gt;First you need to know what is oxygenated water. It's actually form by adding additional oxygen to the water under pressure. It means it's a mixture, not a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that a mixture is two or more substances mix together physically. However, hydrogen peroxide bonds chemically which is a compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can conclude that oxygenated cannot be classified as hydrogen peroxide. &lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-02-07T02:33:35.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24762" PostTypeId="1" AcceptedAnswerId="24777" CreationDate="2015-02-07T03:22:59.243" Score="9" ViewCount="6392" Body="&lt;p&gt;As we go down the periodic table, Group 1 alkali metals hold their single outer valence electron more loosely, and so reactivity increases. Below potassium (Na, Li), we can store the metal in oil with no difficulty. Above potassium (Rb, Cs), we need to exclude all oxygen to prevent reaction in air. Potassium (K) is the unique element where we can technically store it in oil, but it is highly likely to react with any available oxygen to form explosive peroxides. I'd like an explanation of why this is; i.e. why is the reactivity favorable for the formation of peroxides. Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2016-01-28T19:17:33.400" Title="Why does potassium form peroxides but sodium does not?" Tags="&lt;periodic-table&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24763" PostTypeId="2" ParentId="16785" CreationDate="2015-02-07T03:35:17.893" Score="4" Body="&lt;p&gt;I'm no expert nor scholar, but from what I am reading in all of these explanations, and what  I notice from the illustration, it becomes obvious...at least to me...which I feel may clarify the polarity change between the Galvanic cell and electrolytic cell for this user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As established and understood, the source of electrons and transfer of ions flows from the negative pole, (Anode) and is received by the positive pole (Cathode) (intentionally using most basic terms) the anode is negative here because the the flow originates FROM the electrolyte, into the light bulb, for which, if the terminals of the bulb were labeled, they would match the electrolyte in the other cell as it is the force coming from the bulb pushing the flow to the cell's cathode, and the cell's cathode is pulling from the bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the electrolytic cell, the &quot;electrolyte&quot; is taking the role of the light bulb of the Galvanic cell, since the electrons are being SENT TO it from the power source, and is not in itself the SOURCE of flow, but is SUBJECT TO the force from the source of flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO just as the Galvanic cell's anode sends to the light bulb, and the electrolyte is labeled like the load of the galvanic cell, and transferring its incoming negative force from the current source, and this pushes through the electrolyte like the flow FROM the light bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be easier if you note that the SOURCE of power is NOT the electrolyte and technically, the black terminal of the power supply is the TRUE anode (Sending), and the red side the TRUE Cathode, (Receiving) but when identifying the reactive substance submerged/surrounded by the electrolytic substance, the anode is giving up its ions, which then add to the Cathode which is receiving them.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the tags in the electrolytic cell are not naming the &quot;source of flow&quot;, but the reaction of the substances involved, due TO the force/flow imposed on them from the power source, but is not THE source of power, and therefore should not be labeled AS one...and there are only two options for labeling them, and since it cannot be changed at the power source it can only b changed at the point of contact with the electrolyte!   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least this is what I have come to understand by reviewing the comments and illustrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sincerely hope it helps clarify the rationale for the reversal of labels for this user and any others struggling with the concept of being due to the source of current having to be labeled as - Anode and + Cathode... forcing the object the current plays upon to be the opposite despite their poles and due to direction of flow. &lt;/p&gt;&#xA;" OwnerUserId="13253" LastEditorUserId="402" LastEditDate="2017-01-22T10:06:40.903" LastActivityDate="2017-01-22T10:06:40.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24765" PostTypeId="1" AcceptedAnswerId="24766" CreationDate="2015-02-07T07:21:46.727" Score="9" ViewCount="3231" Body="&lt;p&gt;This question was given by my teacher in class:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw the structure of 2,3-diethylheptane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I drew the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GZuBR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GZuBR.png&quot; alt=&quot;2,3-diethylheptane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now if I name the compound according to the longest chain, it comes out to be 3-methyl-4-ethyloctane! How can I properly draw these structures?&lt;/p&gt;&#xA;" OwnerUserId="13236" LastEditorUserId="16683" LastEditDate="2016-10-09T22:37:49.580" LastActivityDate="2020-04-20T22:53:47.913" Title="How can I draw 2,3-diethylheptane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24766" PostTypeId="2" ParentId="24765" CreationDate="2015-02-07T07:36:46.930" Score="14" Body="&lt;p&gt;What I usually do when I start numbering is look for branching and start numbering at the end with the closest branch. Also, keep in mind that with branching you'll want to consider numbering from one of the branches (if it'll yield a long chain.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the structure you made as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I numbered from the side with the most branching. But then I numbered it in two different ways. One way was starting with the straight chain which gave me a 7 carbon chain. Then I renumbered it (the image on the right) starting from one of the branches. This gave me an 8 carbon chain. (You could also number starting from the other branch, but that gives you 7 as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose the longest chain. Since stuff that looks like a side chain can actually be part of the main chain, seriously keep this in mind during exams. It's totally okay to number the same alkane in different ways and play around with it to find which is the longest chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-ethyl-3-methyloctane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMiZc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMiZc.png&quot; alt=&quot;4-ethyl-3-methyloctane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you're having trouble drawing them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would draw just a zig-zag pattern (like I drew above) containing the number of carbons you need. You need a heptane, so start out with a 7 carbon chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will just be your bare, skeleton structure. Then number the carbons, then add the ethyls to carbons 2 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HOWEVER, the question is flawed. You actually can't have an ethyl on carbon number 2 here, because it just makes the chain longer.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2016-10-30T03:54:01.447" LastActivityDate="2016-10-30T03:54:01.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24767" PostTypeId="2" ParentId="24765" CreationDate="2015-02-07T07:40:37.157" Score="9" Body="&lt;p&gt;First, according to the name that you were provided, 2,3-diethylheptane, you did draw the correct structure. The problem actually lies with the name: it does not follow the standard &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Alkanes&quot; rel=&quot;nofollow&quot;&gt;IUPAC nomenclature&lt;/a&gt;. While it is possible to draw the structure, as you have, it's not the correct name for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any name derived from an organic alkane structure should start with the longest chain from start to end. From your structure the &quot;2-ethyl&quot; branch is actually part of the longest chain, which then includes 6 other carbons leading to the furthest end. Now you have to deal with a methyl group and an ethyl group on that octane chain: in this case, both lie on the same side, so there is no competition for which has priority in determining the numbering. So from the end that they're both closest, the methyl is on the 3-carbon; the ethyl is on the 4-carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to get correct structures from incorrect names; the system is degenerate in that sense. However there is (ideally) only 1 correct IUPAC name for any organic structure.&lt;/p&gt;&#xA;" OwnerUserId="13257" LastEditorUserId="16683" LastEditDate="2016-10-09T22:38:32.953" LastActivityDate="2016-10-09T22:38:32.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24768" PostTypeId="1" CreationDate="2015-02-07T09:56:24.667" Score="7" ViewCount="6578" Body="&lt;p&gt;Why is 4-ethyl-1-fluoro-2-nitrobenzene (structure is in the attached picture) called so ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be called as 1-ethyl-4-fluoro-3-nitrobenzene since the &lt;em&gt;precedence of ethyl is greater than halogens&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QsRCVm.png&quot; alt=&quot;4-ethyl-1-fluoro-2-nitrobenzene&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5373" LastEditorUserId="4945" LastEditDate="2015-06-01T02:02:29.147" LastActivityDate="2019-06-07T05:23:31.437" Title="Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24769" PostTypeId="1" AcceptedAnswerId="24773" CreationDate="2015-02-07T10:35:09.483" Score="6" ViewCount="8396" Body="&lt;p&gt;I have been learning chemistry for a year or two but I still can't understand this. According to maths we should place the plus or minus sign in front of a number like -2. However, In chemistry we always put the sign behind a number when writing ions like $\ce{Ba^2+}$ ,$\ce{Cu^2+}$, $\ce{O^2-}$ etc. So, I want to know why do we write it that way?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="24052" LastEditDate="2019-11-28T19:43:49.990" LastActivityDate="2020-09-24T02:56:53.653" Title="Why Ca 2+ and not Ca +2" Tags="&lt;nomenclature&gt;&lt;ions&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24770" PostTypeId="2" ParentId="24769" CreationDate="2015-02-07T11:01:28.710" Score="7" Body="&lt;p&gt;$\ce{Ba^{2+}}$ means that you have a Barium ion with two positive charges.&#xA;so the 2 counts the positive charges and the + is not the symbol to add something, the + means only a positive charge.&lt;/p&gt;&#xA;" OwnerUserId="13261" LastEditorUserId="7448" LastEditDate="2015-02-07T16:08:50.047" LastActivityDate="2015-02-07T16:08:50.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24771" PostTypeId="1" CreationDate="2015-02-07T11:47:03.257" Score="3" ViewCount="109" Body="&lt;p&gt;I'm reading a paper with a reaction scheme where an enzyme binds loosely to inhibitor reversibly and then the Enzyme-Inhibitor complex undergoes another reversible conformational change to a tightly bound complex. My question is about the notation used (which I cannot type because I can't find the appropriate symbols). The second reaction uses the regular equilibria symbol which is fine - no problem. However, the first reaction uses a modified form of the equilibria symbol which is one line rather than two with 'half arrows' at either end. What does this symbol mean?  &lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorUserId="22623" LastEditDate="2018-01-04T22:52:49.383" LastActivityDate="2018-01-06T00:27:12.963" Title="Single line equilibria notation" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24772" PostTypeId="2" ParentId="24768" CreationDate="2015-02-07T12:00:27.287" Score="8" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(g) lowest locants for the substituent cited first as a prefix in the name;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule f takes precedence over Rule g.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.3.5&lt;/strong&gt; Lowest set of locants&lt;/p&gt;&#xA;&lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with&#xA;other locant sets, each cited in order of increasing value, has the lowest term at the first&#xA;point of difference; (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the example is named as 4-ethyl-1-fluoro-2-nitrobenzene rather than 1-ethyl-4-fluoro-3-nitrobenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m0KSo.png&quot; alt=&quot;4-ethyl-1-fluoro-2-nitrobenzene&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-03-28T17:20:57.523" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24773" PostTypeId="2" ParentId="24769" CreationDate="2015-02-07T12:59:12.410" Score="13" Body="&lt;p&gt;Chemistry has an organization called the &lt;a href=&quot;https://iupac.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Union of Pure and Applied Chemistry (IUPAC)&lt;/a&gt;, which sets standards for nomenclature (among other things). Chemists &lt;em&gt;usually&lt;/em&gt; follow its recommendations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IUPAC &lt;a href=&quot;https://iupac.org/wp-content/uploads/2016/07/Red_Book_2005.pdf#page=69&quot; rel=&quot;nofollow noreferrer&quot;&gt;says&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ionic charge is indicated by means of a right upper index, as in &lt;span class=&quot;math-container&quot;&gt;$\ce{A^n+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{A^n-}$&lt;/span&gt; (&lt;em&gt;not&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A^$+n$}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{A^$-n$}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://media.iupac.org/publications/books/gbook/IUPAC-GB3-2ndPrinting-Online-22apr2011.pdf#page=63&quot; rel=&quot;nofollow noreferrer&quot;&gt;and&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The iconic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.&lt;br&gt;&#xA;  ...&lt;br&gt;&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}$&lt;/span&gt; is commonly used in chemistry and recommended by [74 = see quote above]. The forms &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^$+3$}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{S^$-2$}$&lt;/span&gt;, although widely used, are obsolete [74], as well as the old notations &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^{+++}}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{S^=}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{S^{--}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No reason is given, but you'll find a historical account of sign conventions for charge and a defense of IUPAC's choice in &lt;a href=&quot;http://www.che.uc.edu/Jensen/W.%20B.%20Jensen/Reprints/194.%20Ionic%20Charges%20.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. He says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[IUPAC's convention] avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities. Thus, in counting apples, we say two apples, three apples, etc., not apples two, apples three – that is, the number always precedes the name of the entity being counted. Likewise, when counting charges, we should say two positive charges or three negative charges, not positive charges two or negative charges three. The IUPAC ruling was intended to make the charge number symbolism consistent with this verbal convention.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally, I think the charge should be written like any other signed number. Formulas typed from the keyboard without superscripting can be ambiguous; is &lt;code&gt;X2+&lt;/code&gt; to be interpreted as &lt;span class=&quot;math-container&quot;&gt;$\rm X^{2+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\rm X_2^+$&lt;/span&gt;? (You should type &lt;code&gt;X(2+)&lt;/code&gt; or &lt;code&gt;X2(+)&lt;/code&gt;, but few people actually do that.) But it's pretty clear what &lt;code&gt;X+2&lt;/code&gt; would mean. Putting the sign first also makes it trivial to parse the charge from a formula typed this way without the need for parentheses.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="24052" LastEditDate="2019-11-28T14:29:44.733" LastActivityDate="2019-11-28T14:29:44.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24776" PostTypeId="2" ParentId="24738" CreationDate="2015-02-07T15:14:08.473" Score="4" Body="&lt;p&gt;&amp;quot;Strength&amp;quot;. Hmm, what firstly comes to mind is you want a comparison of lattice energies (or more precisely, lattice enthalpies)&lt;/p&gt;&#xA;&lt;p&gt;Lattice energy is a [crude] way of estimating how strong the electrostatic attractions are between ions in an ionic compound. Let's take the formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; into account, as I see it's the most common &amp;quot;ionic&amp;quot; compound that teachers start teaching the lesson with.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Born-Haber cycle&lt;/a&gt; this is the reaction that results in the formation of pure table salt:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Na+(g) + Cl- (g) \rightarrow NaCl(s)}$$&lt;/span&gt;&#xA;Forming the salt crystals is exothermic and this is called the lattice enthalpy of &lt;strong&gt;formation&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;Note that some textbooks mean to define lattice energy as the energy that is obtained by the thermodynamic system in the process of turning the solid compound to its gaseous ions.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To make a long story short, the more the lattice enthalpy is, the stronger &amp;quot;bonds&amp;quot; you get.&#xA;There are two &lt;strong&gt;main&lt;/strong&gt; factors that affect the lattice formation enthalpy:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Charge of the ions&lt;/li&gt;&#xA;&lt;li&gt;Radius of the ions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Charge of the ions:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BujkL.gif&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lechart3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Look at the diagram above. &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; is known to release approximately 822 kilojoules upon formation, while &lt;span class=&quot;math-container&quot;&gt;$\ce{MgO}$&lt;/span&gt; releases 3890 KJ. And what does that mean? It means that Magnesium Oxide is way more &amp;quot;stable&amp;quot; than Sodium Chloride. So, complete the rest:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The more the charge of the ions is in an ionic compound, the _____ is its lattice enthalpy and the ______ its stability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Radius of the ions:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UElso.gif&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lechart2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you see, and could've imagined if you were on the right track, bigger ions (literally) mean weaker attraction. I wanted the answer to be fairly complete, so I did an addition to the main point that answers your question. Thus, there's no need of expanding this.&lt;/p&gt;&#xA;&lt;h2&gt;In your case:&lt;/h2&gt;&#xA;&lt;p&gt;You might as well have reached a conclusion. Mg ion has a 2+ charge, while Na has only a single electron missing. The anion is pretty much the same, &lt;span class=&quot;math-container&quot;&gt;$\ce{HCO3-}$&lt;/span&gt; (hydrogen carbonate), and its priorly covalent nature shouldn't mislead you in doubting your answer. There is a case of comparison of the charge of the ions, and I would say Magnesium ion wins.&lt;/p&gt;&#xA;&lt;p&gt;In case of additional study, you can visit:&lt;br/&gt;&#xA;&lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lattice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chemguide.co.uk/physical/energetics/lattice.html&lt;/a&gt;&lt;br/&gt;&#xA;&lt;a href=&quot;http://www.knockhardy.org.uk/sci_htm_files/08le.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.knockhardy.org.uk/sci_htm_files/08le.pdf&lt;/a&gt;&lt;br/&gt;&#xA;&lt;a href=&quot;http://www.rsc.org/learn-chemistry/resources/gridlocks/downloads/answers/Born%20Haber%20NaCl%20ans.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rsc.org/learn-chemistry/resources/gridlocks/downloads/answers/Born%20Haber%20NaCl%20ans.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Addendum&lt;/strong&gt;:&lt;br/&gt;&#xA;The charge and the radius of ions are by no means the only factors that affect the lattice enthalpy. It's just that the other factors are considered &amp;quot;trivial&amp;quot; by current textbooks.&lt;br/&gt;&#xA;The lattice enthalpy and lattice energy aren't the same. They do have close meanings, though.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-07T15:14:08.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24777" PostTypeId="2" ParentId="24762" CreationDate="2015-02-07T15:31:47.540" Score="7" Body="&lt;p&gt;If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_superoxide&quot;&gt;can explode when mixed with water, acids, organics, or powdered graphite&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton &amp;amp; Wilkinson's &lt;em&gt;Advanced Inorganic Chemistry&lt;/em&gt;). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-07T15:58:40.660" LastActivityDate="2015-02-07T15:58:40.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24778" PostTypeId="2" ParentId="24729" CreationDate="2015-02-07T16:35:19.190" Score="2" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Hint&lt;/em&gt;&lt;/strong&gt; : Ammonia water is actually diluted Ammonium Hydroxide ruled by the following equilibrium :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH3 + H2O &amp;lt;=&amp;gt; NH4+ + HO-}$$&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-07T16:35:19.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24780" PostTypeId="2" ParentId="24754" CreationDate="2015-02-07T16:47:35.607" Score="3" Body="&lt;p&gt;So you have the reaction :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{LiAlH4 + 4H2O &amp;lt;=&amp;gt; LiOH + Al(OH)3 + 4H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to get the &quot;energy&quot; given by this reaction, which is nothing but the reaction enthalpy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_rH^0(T) = \sum_{i}\nu_i.\Delta H_{i,f}^0(T)$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you get this, just look for the theorical values of each $\Delta H_f^0$ you need for the 5 compounds you have (they are tabulated) and sum them up pondered with the stoechiometric coefficients.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-07T16:47:35.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24784" PostTypeId="1" AcceptedAnswerId="24785" CreationDate="2015-02-07T17:59:36.920" Score="7" ViewCount="1181" Body="&lt;p&gt;Is it possible to chemically treat grass after &quot;harvesting&quot; it to make it digestible by humans?  I know we can &quot;eat&quot; grass, but we can't break it down in our stomachs. If there is some chemical treatment to partially break grass down into something we can digest, I see it possibly being used to end world hunger!  I know we already treat bone to make it into gelatin, and I am wondering if something similar can be done for grass.&lt;/p&gt;&#xA;" OwnerUserId="6806" LastEditorUserId="6806" LastEditDate="2015-02-07T18:06:28.090" LastActivityDate="2018-06-11T02:11:01.507" Title="Is it possible to chemically treat grass to make it digestible?" Tags="&lt;biochemistry&gt;&lt;experimental-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24785" PostTypeId="2" ParentId="24784" CreationDate="2015-02-07T18:38:07.113" Score="9" Body="&lt;p&gt;The indigestible component of grasses is &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot;&gt;cellulose&lt;/a&gt;, which is polymerized glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is possible to hydrolyze the cellulose to glucose, which is of course digestible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrolysis can be catalyzed enzymatically with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulase&quot;&gt;cellulase&lt;/a&gt;, or using acid hydrolysis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of hydrolyzing cellulose is often discussed in the context of so called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulosic_ethanol&quot;&gt;cellulosic ethanol&lt;/a&gt;, as the glucose is then fermented to ethanol subsequent to hydrolysis.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-07T18:38:07.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24786" PostTypeId="1" CreationDate="2015-02-07T18:55:54.290" Score="-2" ViewCount="345" Body="&lt;p&gt;What is pKa or Hammett's function of 2,4,6-trinitrobenzenesulfonic acid? PubChem gives pKa=0.13, but doesn't specify the solvent. Even if it's real value I doubt it's a value for water. Why? Because even aliphatic sulfonic acids are strong, and with electron withdrawing groups, even stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example this &lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v78-385&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; gives -4.0 for p-nitrosulfonic acid. Even if resonance of o-nitro groups is blocked, it doesn't negate inductive effects. Reasoning with solvation effects also seems dubious - nitro groups aren't hydrofobic at all. Is there some reliable data for this, rather obscure, acid?&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2018-07-16T15:02:17.177" LastActivityDate="2018-07-16T15:02:17.177" Title="How strong is 2,4,6-trinitrobenzenesulfonic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24787" PostTypeId="1" CreationDate="2015-02-07T19:47:42.167" Score="2" ViewCount="1156" Body="&lt;p&gt;Are there any examples of alloys or ceramics that have higher density than their densest elemental constituent?  If so, do they have unusual or difficult synthesis processes?  If not, is it provable that this is &lt;em&gt;not&lt;/em&gt; possible, or demonstrably unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since homogeneous metals are essentially crystals in their solid phase it seems plausible that, in theory, there are elements that could fit within the crystal structure without altering it to increase their density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be particularly interesting for metals that are used in applications primarily for their density: e.g., lead, tungsten, and uranium.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2015-02-07T23:07:52.320" LastActivityDate="2015-02-07T23:07:52.320" Title="Can alloys or ceramics have higher density than elemental solids?" Tags="&lt;crystal-structure&gt;&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24788" PostTypeId="2" ParentId="24786" CreationDate="2015-02-07T20:25:53.847" Score="7" Body="&lt;p&gt;The reference for the &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/11045&quot; rel=&quot;nofollow noreferrer&quot;&gt;PubChem entry for picryl sulfonic acid&lt;/a&gt; is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Serjeant, E.P., Dempsey B.; Ionisation Constants of Organic Acids in Aqueous Solution. International Union of Pure and Applied Chemistry (IUPAC). IUPAC Chemical Data Series No. 23, 1979. New York, New York: Pergamon Press, Inc., p. 40&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, yes that is the value of the pKa for picryl sulfonic acid (2,4,6-trinitrobenzenesulfonic acid) in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pKa of &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/7371&quot; rel=&quot;nofollow noreferrer&quot;&gt;benzenesulfonic acid&lt;/a&gt; is 2.554, although the reference does not specify a solvent. The pKa of &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/6395&quot; rel=&quot;nofollow noreferrer&quot;&gt;methanesulfonic acid&lt;/a&gt; is -1.86, with the same reference as picrylsulfonic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between methanesulfonic acid and the arylsulfonic acids might be the ability of water to solvate the anion. The aryl groups are less polar and more hydrophobic. They are also large. This same reason (decreased solvation of the anion) is invoked as the explanation for the difference in acidity between &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14767/steric-hindrance-and-bronsted-basicity&quot;&gt;primary, secondary, and tertiary alcohols&lt;/a&gt;. The magnitude of the effect in alcohols is a few pKa units, which appears to be of a similar scale for the same effect in sulfonic acids.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-02-08T00:03:21.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24789" PostTypeId="2" ParentId="24787" CreationDate="2015-02-07T20:46:27.020" Score="3" Body="&lt;p&gt;Yes, for both alloys and ceramics it is possible have a density that is higher than their densest elemental constituent. This is because every lattice structure of the denser elemental constituent contains &lt;strong&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_Lower&amp;#39;s_Chem1/07%3A_Solids_and_Liquids/7.8%3A_Cubic_Lattices_and_Close_Packing#5_Interstitial_void_spaces&quot; rel=&quot;nofollow&quot;&gt;interstitial void spaces&lt;/a&gt;&lt;/strong&gt;, or &lt;em&gt;holes&lt;/em&gt;, typically tetrahedral holes (surrounded by 4 atoms) or octahedral holes (surrounded by 6 atoms). These holes exist despite the crystal having an optimal packing structure, thus a maximum density for the pure solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The densest elemental solids are typically made up of larger atoms toward the bottom of the periodic table, that have larger atomic radii due to electron-electron repulsion, but which have have much higher atomic mass due to their nuclei. This produces larger interstitial voids that can fit smaller atoms without significantly disturbing the lattice constants of the larger, denser component. An increase in mass without concomitant increase in volume will produce a denser composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific examples are difficult to find, as densities are not always reported. One common material which is denser than the corresponding &quot;densest elemental constituent&quot; is steel. Steel (&lt;a href=&quot;http://hypertextbook.com/facts/2004/KarenSutherland.shtml&quot; rel=&quot;nofollow&quot;&gt;maximum density&lt;/a&gt;: 8.05 g·cm⁻³) is an alloy of &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;iron&lt;/a&gt; (density 7.874 g·cm⁻³) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon&quot; rel=&quot;nofollow&quot;&gt;carbon&lt;/a&gt; (diamond density: 3.5 g·cm⁻³), although occasionally some &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium&quot; rel=&quot;nofollow&quot;&gt;chromium&lt;/a&gt; (density: 7.19 g·cm⁻³) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel&quot; rel=&quot;nofollow&quot;&gt;nickel&lt;/a&gt; (density: 7.81 g·cm⁻³) may also be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible, although less clear-cut example is the alloy &lt;strong&gt;&lt;a href=&quot;http://pdg.lbl.gov/2007/AtomicNuclearProperties/substances/279.html&quot; rel=&quot;nofollow&quot;&gt;heavymet&lt;/a&gt;&lt;/strong&gt;, which is used in ATLAS. It is composed of nickel (9.9 atom %), copper (4.0 atom %), and tungsten (86.1 atom %). Heavymet has a density of 19.3 g·cm⁻³, which despite the less dense diluents, may be slightly larger than tungsten's density which is 19.25 g·cm⁻³. &lt;/p&gt;&#xA;" OwnerUserId="13257" LastActivityDate="2015-02-07T20:46:27.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24790" PostTypeId="1" CreationDate="2015-02-07T21:44:49.077" Score="4" ViewCount="137" Body="&lt;p&gt;Recently the &lt;em&gt;sugar glycoaldehyde&lt;/em&gt; was detected in a star system 400 light years from Earth. How exactly are molecules detected in space? I am aware that spectroscopy is used to detect them, but I don't know the details. &lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="12660" LastEditDate="2015-02-08T05:59:58.350" LastActivityDate="2015-03-10T16:39:35.967" Title="How exactly are molecules detected in space?" Tags="&lt;spectroscopy&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24791" PostTypeId="1" AcceptedAnswerId="28134" CreationDate="2015-02-07T21:48:50.700" Score="9" ViewCount="91303" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What part in the strand contributes to the overall non neutral charge?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;DNA is not isolated in the body, so what keeps it stable while being charged?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is it important for DNA to be charged?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="30836" LastEditDate="2020-04-02T16:38:16.477" LastActivityDate="2020-04-02T16:38:16.477" Title="Why is DNA negatively charged and what makes it so?" Tags="&lt;biochemistry&gt;&lt;dna-rna&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24792" PostTypeId="2" ParentId="24499" CreationDate="2015-02-07T21:54:37.693" Score="2" Body="&lt;p&gt;You are correct that it is possible to study theoretical or computational chemistry, however these typically only represent a handful of courses in a normal undergraduate chemistry curriculum. Additionally, most undergraduate chemistry curriculum are designed to instruct students broadly in the core disciplines of chemistry, hence it's usually not possible to specialize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-chemisty degrees may be an alternative. However any science undergraduate program will require laboratory-based coursework. Hands-on experience teaches us how scientific concepts work in the real works, giving us an appreciation of how difficult experimental work can be. It also functions to reinforce content from books and lectures; to teach technical skills (which are usually necessary for future employment); and to develop lab safety habits. So there really is no means to escape it. Moreover, in order to take advanced chemistry electives, prior courses which include lab sessions are generally necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are other routes to get into chemistry. I've know individuals whom have majored in mathematics, engineering, physics, and biology who either did research in chemistry or who did graduate degrees in chemistry subsequent to obtaining an undergraduate degree in those fields. However in order to obtain undergraduate degrees in those fields and to be able to transition to chemistry programs, lab experience and some senior-level chemistry courses are typically required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once in graduate school, it is possible to avoid experiments altogether: theoretical and computational chemistry graduate degrees typically do not require that the candidate engage in any hands-on experiments. Many however do require that the graduate student work as a teaching assistant in undergraduate chemistry laboratories &lt;em&gt;as a condition of funding&lt;/em&gt;. In such cases, laboratory experience is either necessary or a significant advantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while it may be possible to minimize the amount of lab work that you engage in, some will always be required. Additionally I would not recommend avoiding lab work, as it greatly enriches the study of chemistry overall. &lt;/p&gt;&#xA;" OwnerUserId="13257" LastActivityDate="2015-02-07T21:54:37.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24793" PostTypeId="1" AcceptedAnswerId="24795" CreationDate="2015-02-07T23:18:02.513" Score="4" ViewCount="131" Body="&lt;p&gt;I have made a DIY furnaces out of cement. Before I do any smelting, I would just like to take any suggestions in regards to any safety measures I should take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to only melt aluminum cans and small parts. I then plan to pour them onto a flat plane to make aluminum plates. Later I plan to make casts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My furnace is very similar to the one pictured below.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/H2bMu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to connect a hair drier to the hole (with an extension metal rod) in order to increase air flow. It will operate by burning charcoal brickets.&lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorUserId="402" LastEditDate="2015-02-08T00:55:19.730" LastActivityDate="2015-10-25T21:26:12.003" Title="What safety measures should I take when operating a backyard aluminum furnace" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24795" PostTypeId="2" ParentId="24793" CreationDate="2015-02-08T00:37:36.183" Score="4" Body="&lt;p&gt;Set this up in an area free of leaves and debris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend wearing safety gear: heat resistant goggles and fire-resistance clothing and gloves that will protect you from heat, especially when pouring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get a couple practice runs in pouring before using it so you get a feel for the best angle to hold in and learn what NOT to do. Even better, you could assemble something that would do the pouring for you. (I'm thinking of &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/b/ba/US_Navy_111021-N-ZZ999-001_Shipbuilders_Jason_West,_left,_and_Kevin_Forrest_pour_molten_pewter_into_a_mold_for_production_of_the_Quicken_Loans_Carr.jpg&quot; rel=&quot;nofollow&quot;&gt;something along the lines of this&lt;/a&gt;, but it doesn't have to be as elaborate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to scrape the paint off the cans before putting them in. Otherwise, you'll  get a lot of crud from the paint in your molten aluminum.&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2015-10-25T21:26:12.003" LastActivityDate="2015-10-25T21:26:12.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24797" PostTypeId="1" AcceptedAnswerId="24799" CreationDate="2015-02-08T03:50:03.110" Score="1" ViewCount="136" Body="&lt;p&gt;I'm confused on what a triglyceride is, from what my text book it says its a type of gylercide, then from a website it said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Glycerides can be subdivided into two categories. The first group, the neutral glycerides are nonionic and nonpolar. The second group, the phosphoglycerides contain a polar region, the phosphoryl group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which makes triglyceride not a directly under a glyceride but under one of these groups. Then it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Esterification may occur at one, two or all three positions producing monoglycerides (monoacylglycerols), diglycerides (diacylglycerols), or triglycerides (triacylglycerols). The most prevalent and most important are the triglycerides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So does that mean that triglyceride can be produced by either of the two groups phosphoglycerides and neutral glycerides. And if that's correct then how are monoglycerides and diglycerides produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.latech.edu/~deddy/chem121/Lipids.htm/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12820" LastEditorUserId="865" LastEditDate="2021-05-22T19:06:07.850" LastActivityDate="2021-05-22T19:06:07.850" Title="What is a triglyceride?" Tags="&lt;nomenclature&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24798" PostTypeId="1" CreationDate="2015-02-08T04:08:50.937" Score="2" ViewCount="6042" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$\pu{−3.49 * 10^{−17} J}$&lt;/li&gt;&#xA;  &lt;li&gt;$\pu{−1.36 * 10^{−19} J}$&lt;/li&gt;&#xA;  &lt;li&gt;$\pu{+2.18 * 10^{−18} J}$&lt;/li&gt;&#xA;  &lt;li&gt;$\pu{+1.36 * 10^{−19} J}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume that the way to do this is to start from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg formula&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right),&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="12798" LastEditorUserId="194" LastEditDate="2017-09-24T16:08:48.470" LastActivityDate="2017-09-24T16:16:17.400" Title="Calculating the ionization energy for hydrogen" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24799" PostTypeId="2" ParentId="24797" CreationDate="2015-02-08T04:30:39.060" Score="1" Body="&lt;p&gt;There is some minor argument as far as nomenclature goes for these lipids. Most sources I can find, including &lt;a href=&quot;http://goldbook.iupac.org/G02647.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC&lt;/a&gt; have glycerides as only including esters of glycerol with fatty acids. Monoglycerides have a single fatty acid ester with two free OH groups, diglycerides have two fatty acid esters, and triglycerides have three. Phosphoglycerides are &lt;a href=&quot;http://goldbook.iupac.org/P04558.html&quot; rel=&quot;nofollow&quot;&gt;defined&lt;/a&gt; separately as esterified derivatives of phosphatidic acids, so you can think of them like glycerides where for one of the substituents, instead of a fatty acid, some derivative of phosphoric acid is present. However, the way the two terms are defined, they don't encompass any common molecules, i.e. there are no molecules that are both glycerides and phosphoglicerides and neither group is a subset of the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: mono-, di-, and triglycerides are subsets of glycerides while phosphoglycerides are a different group altogether, though structurally similar.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-08T04:30:39.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24801" PostTypeId="1" AcceptedAnswerId="24803" CreationDate="2015-02-08T05:42:54.050" Score="7" ViewCount="6557" Body="&lt;p&gt;I have to distinguish between copper oxide and carbon. I thought of dilute sulfuric acid because it forms copper sulfate with copper oxide. But I don't know how carbon reacts with dilute sulfuric acid. I know how carbon reacts with &lt;em&gt;concentrated&lt;/em&gt; sulfuric acid but dilute?&lt;/p&gt;&#xA;" OwnerUserId="13264" LastEditorUserId="402" LastEditDate="2017-04-01T00:34:59.103" LastActivityDate="2017-04-01T00:34:59.103" Title="Carbon and dilute sulfuric acid" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24802" PostTypeId="2" ParentId="24798" CreationDate="2015-02-08T06:10:33.057" Score="3" Body="&lt;p&gt;I assume you refer to &lt;a href=&quot;https://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bohr's atomic model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;E_{n} &amp;amp;= \frac{2\pi^{2}me^{4}Z^{2}}{n^{2}h^{2}\left(4\pi\epsilon_{0}\right)^{2}} \\&#xA;&amp;amp;= \color{\navy}{-13.6\frac{Z^2}{n^2}\pu{ eV}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Z = 1$ for hydrogen, the above equation can be further simplified. Now just plug in the required values.&lt;/p&gt;&#xA;" OwnerUserId="13216" LastEditorUserId="194" LastEditDate="2017-09-24T16:16:17.400" LastActivityDate="2017-09-24T16:16:17.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24803" PostTypeId="2" ParentId="24801" CreationDate="2015-02-08T07:03:35.237" Score="3" Body="&lt;p&gt;Sulfuric acid is  unreactive with carbon- indeed only hot concentrated sulfuric acid reacts with carbon to produce carbon dioxide. &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2015-02-08T07:03:35.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24806" PostTypeId="1" CreationDate="2015-02-08T09:42:59.947" Score="-1" ViewCount="1224" Body="&lt;p&gt;Tell me the requirements and procedure for &lt;strong&gt;producing black smoke using chemical  reactions&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13276" LastEditorUserId="7448" LastEditDate="2015-02-08T16:10:56.497" LastActivityDate="2015-02-08T16:10:56.497" Title="How to produce black smoke using chemical reaction" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-09T09:04:04.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="24807" PostTypeId="2" ParentId="24806" CreationDate="2015-02-08T10:26:04.503" Score="3" Body="&lt;p&gt;When ignited, mixtures of potassium chlorate, charcoal and anthracene (or naphthalene) produce black smoke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, note that milling and grinding dry mixtures of potassium chlorate and charcoal is a well-known method to cause an explosion and severely harm or kill yourself!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-08T10:26:04.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24808" PostTypeId="1" AcceptedAnswerId="24810" CreationDate="2015-02-08T11:24:49.780" Score="-3" ViewCount="889" Body="&lt;p&gt;I would like to know How can i create a simple (Maybe Complex) Chemical that can be used as an explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use it to blast A regular cardboard box.  &lt;/p&gt;&#xA;" OwnerUserId="13190" LastActivityDate="2015-02-08T12:35:25.463" Title="How can I Create an explosive with everyday used chemicals?" Tags="&lt;everyday-chemistry&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-02-08T16:22:09.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="24809" PostTypeId="1" AcceptedAnswerId="24821" CreationDate="2015-02-08T12:00:11.403" Score="3" ViewCount="82" Body="&lt;p&gt;I want to synthesize 3-hydroxydihydrofuran-2-one from formaldehyde but I do not know of a method. I tried a Mannich reaction and lactonization after dehydration.&lt;/p&gt;&#xA;" OwnerUserId="13277" LastEditorUserId="1887" LastEditDate="2015-02-08T18:45:27.617" LastActivityDate="2015-02-08T18:45:27.617" Title="How do I synthesize 3-hydroxydihydrofuran-2-one from formaldehyde?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24810" PostTypeId="2" ParentId="24808" CreationDate="2015-02-08T12:28:53.293" Score="3" Body="&lt;p&gt;It depends!&lt;/p&gt;&#xA;&lt;h3&gt;Can you regrow lost organs?&lt;/h3&gt;&#xA;&lt;p&gt;If you're genetically close to a newt and can easily regrow organs lost in an explosion, you should give triacetone triperoxide (TATP) a try. Cool an equimolar mixture of acetone and 50% hydrogen peroxide to 0 °C, add some drops of concentrated sulfuric acid and let it come to room temperature &lt;strong&gt;without stirring&lt;/strong&gt;. After 24h, collect the precipitate and dry it &lt;strong&gt;carefully&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;TATP is extremely shock sensitive - one careless scratching and you're done - and has a Trauzl value similar to that of TNT!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you feel by now that you better shouldn't play with it: &lt;strong&gt;You are right!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h3&gt;What about a something less suicidal?&lt;/h3&gt;&#xA;&lt;p&gt;On contact with water, calcium carbide forms calcium hydroxide and releases ethyne:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CaC2 + 2 H2O -&amp;gt; Ca(OH)2 + C2H2 ^ }$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This might be enough to burst a sealed container. Alternatively, you might consider to ignite a mixture of ethyne and air. &lt;strong&gt;In any case, take care!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-08T12:35:25.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24811" PostTypeId="1" AcceptedAnswerId="24837" CreationDate="2015-02-08T12:56:43.867" Score="2" ViewCount="66" Body="&lt;p&gt;Can clusters of an emulsion be emulsified into another emulsion? &#xA;If so is there any common product or natural substance that would have such a structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally is there a name for this? I looked around but the question gets drowned in all emulsion topics.&lt;/p&gt;&#xA;" OwnerUserId="13278" LastEditorUserId="13278" LastEditDate="2015-02-08T22:36:30.813" LastActivityDate="2015-02-08T23:30:22.087" Title="Can an emulsion be emulsified?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24812" PostTypeId="1" CreationDate="2015-02-08T13:02:43.003" Score="4" ViewCount="1055" Body="&lt;p&gt;I have been reading the books &lt;em&gt;The Elements&lt;/em&gt; by Theodore Gray and &lt;em&gt;Stuff Matters&lt;/em&gt; by Mark Miodownik, both being very popular chemistry science books. Are there some other easy-to-read, popular but scientifically accurate chemistry books available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the question should not be &quot;primarily opinion based&quot;, since &quot;popular&quot; can be measured.&lt;/p&gt;&#xA;" OwnerUserId="13279" LastActivityDate="2015-03-27T17:10:36.663" Title="What are some popular science books in chemistry?" Tags="&lt;books&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2015-05-21T23:47:05.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="24813" PostTypeId="2" ParentId="20061" CreationDate="2015-02-08T15:55:39.230" Score="0" Body="&lt;p&gt;$\ce{H2O}$ has a $\mathrm{p}K_\mathrm{a}$ of approximately 15, giving us $\ce{OH-}$ a reasonably strong base but also good nucleophile which you'll find competing between elimination and substitution reactions. Now $\ce{H2}$ has a pka of 30 which makes its conjugate base ($\ce{H-}$) an even stronger base. $\ce{CH4}$ which has a $\mathrm{p}K_\mathrm{a}$ of 50, and its conjugate base ($\ce{CH3-}$) which is an extremely strong base, in fact it's strong enough to deprotonate any acid below it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a strong base being equal to or stronger than $\ce{OH-}$. Furthermore, the higher the $\mathrm{p}K_\mathrm{a}$ of the acid the greater the strength of the conjugate base.&lt;/p&gt;&#xA;" OwnerUserId="13270" LastEditorUserId="16683" LastEditDate="2017-01-09T08:45:23.723" LastActivityDate="2017-01-09T08:45:23.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24814" PostTypeId="2" ParentId="24790" CreationDate="2015-02-08T16:02:38.467" Score="1" Body="&lt;p&gt;The research article corresponding to the observation to which the question is referring is &lt;a href=&quot;http://www.eso.org/public/archives/releases/sciencepapers/eso1234/eso1234a.pdf&quot; rel=&quot;nofollow&quot;&gt;Detection of the simplest sugar, glycolaldehyde, in a solar-type protostar with ALMA&lt;/a&gt;. Note that the molecule is &lt;a href=&quot;http://en.wikipedia.org/wiki/Glycolaldehyde&quot; rel=&quot;nofollow&quot;&gt;glycolaldehyde&lt;/a&gt; rather than glycoaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALMA is &lt;a href=&quot;http://www.almaobservatory.org/&quot; rel=&quot;nofollow&quot;&gt;the Atacama Large Millimeter/submillimeter Array&lt;/a&gt;, a radio telescope array in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atacama_Desert&quot; rel=&quot;nofollow&quot;&gt;Atacama desert&lt;/a&gt;.  The extremely low humidity of the Atacama desert minimizes water vapor interference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This observation of glycolaldehyde in space is not the first.  It was earlier observed by &lt;a href=&quot;https://public.nrao.edu/telescopes/GBT&quot; rel=&quot;nofollow&quot;&gt;Greenbank radio telescope&lt;/a&gt;, in West Virginia, USA, for example.  The Greenbank observation is reported in &lt;a href=&quot;http://astrobiology.gsfc.nasa.gov/pdf/2000%20ref/hollis_2000.pdf&quot; rel=&quot;nofollow&quot;&gt;INTERSTELLAR GLYCOLALDEHYDE: THE FIRST SUGAR&lt;/a&gt;.  Glycolaldehyde was detected through radio signals in the 71-103 GHz range that are photons emitted when the molecule transitions from an excited rotation state to a lower energy rotational state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ALMA observation, the molecule was detected based upon radio signals near 220 GHz and 690 GHz.  These correspond to photons emitted by the molecule when transitioning from an excited rotational state to a lower energy rotational state, but are a different set of transitions from those observed at Greenbank, including some for excited vibrational states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to add more information later, there are a wide variety of spectroscopic techniques: absorbance vs. emission; rotation, vibration and electronics transitions, and many different frequency ranges.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-08T16:19:53.530" LastActivityDate="2015-02-08T16:19:53.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24816" PostTypeId="1" AcceptedAnswerId="24818" CreationDate="2015-02-08T16:21:17.750" Score="12" ViewCount="17761" Body="&lt;p&gt;How pure $\ce{NaCl}$ is typical table salt? Is it pure enough for research-level experiments in chemistry? How is $\ce{NaCl}$ purified if need to be?&lt;/p&gt;&#xA;" OwnerUserId="13283" LastEditorUserId="4231" LastEditDate="2015-02-21T18:31:20.553" LastActivityDate="2015-04-14T11:31:29.410" Title="How pure NaCl is typical table salt? How is it purified?" Tags="&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24818" PostTypeId="2" ParentId="24816" CreationDate="2015-02-08T16:30:32.347" Score="13" Body="&lt;p&gt;Table salt is often iodized (iodine is an essential micronutrient, and this is one way to introduce it into the diet in the right amount). According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodised_salt&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Edible salt can be iodised by spraying it with a potassium iodate or potassium iodide solution. 60 ml of potassium iodate, costing about US $1.15 (in 2006), is required to iodise a ton of salt. Dextrose is added as a stabilizer to prevent potassium iodide from oxidizing and evaporating. Anti-caking agents such as calcium silicate are commonly added to table salt to prevent clumping.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Elsewhere, Wikipedia mentions other anti-caking agents, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_ferrocyanide&quot;&gt;sodium ferrocyanide&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_aluminosilicate&quot;&gt;sodium aluminosilicate&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_carbonate&quot;&gt;magnesium carbonate&lt;/a&gt;. And they point out that salt may also be &quot;fortified&quot; with iron or other micronutrients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The certificate of analysis for &quot;ultra pure NaCl&quot; from &lt;a href=&quot;https://www.americanbio.com/product/ab01915&quot;&gt;American Bio&lt;/a&gt; shows TRACE amounts of &lt;a href=&quot;http://www.arltma.com/Mineral_Information/Aluminum.html&quot;&gt;aluminum&lt;/a&gt;, &lt;a href=&quot;http://www.arltma.com/Mineral_Information/Arsenic.html&quot;&gt;arsenic&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bromide&quot;&gt;bromides&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Heavy_metal_%28chemistry%29&quot;&gt;heavy metals&lt;/a&gt;, &lt;a href=&quot;http://www.arltma.com/Mineral_Information/Iron.html&quot;&gt;iron&lt;/a&gt;, &lt;a href=&quot;http://www.arltma.com/Mineral_Information/Magnesium.html&quot;&gt;magnesium&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphate&quot;&gt;phosphates&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Sulfate&quot;&gt;sulfates&lt;/a&gt;. Those impurities may well be expected in table salt, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all that, you'd think that table salt is fairly impure. Let's look at the numbers to put the amounts of these impurities in perspective. The box of Morton iodized salt in my kitchen gives me the amount of sodium per serving. &lt;em&gt;If all of the sodium is in the sodium chloride&lt;/em&gt;, I estimate the purity as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.590~\mathrm{g}~\ce{Na+}}{1.5~\mathrm{g}~\text{table salt}}\frac{1~\mathrm{mol}~  \ce{Na+}}{22.989769280~\mathrm{g}~\ce{Na+}}&#xA;\frac{1~\mathrm{mol}~\ce{NaCl}}{1~\mathrm{mol}~\ce{Na+}}\frac{58.44277~\mathrm{g}~\ce{NaCl}}{1~\mathrm{mol}~\ce{NaCl}}\cdot 100\% = \fbox{99.99% NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I'm assuming the label information is actually a bit more precise than reported (&lt;em&gt;exactly&lt;/em&gt; 1.5 g of table salt contains &lt;em&gt;approximately&lt;/em&gt; 590 mg Na). This is just a BOE (Back of the Envelope) calculation; I should round my answer to three or possibly two significant figures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact there are sometimes other sodium salts present (anti-caking estimates such as Tivolex contain sodium) so this purity is a &lt;em&gt;slight&lt;/em&gt; overestimate. One salt company (&lt;a href=&quot;http://www.wasalt.com.au/Table.html&quot;&gt;WA Salt supply&lt;/a&gt;) lists &quot;typical&quot; purities for the table salt it provides at 99.72%; this salt contains uses Tivolex. &lt;a href=&quot;http://www.mortonsalt.com/faqs/food-salt-faqs&quot;&gt;Morton Salt&lt;/a&gt; claims that their iodized table salt contains 0.04% dextrose and &quot;less than 0.5% calcium silicate&quot;, so the mg Na per serving on the label is a little bit higher than it actually is; I should get something above 99.46%, considering that additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, that's pretty pure. The assay for &quot;ultra pure&quot; NaCl linked above gives 99% to 100.5% NaCl. &lt;a href=&quot;http://www.sigmaaldrich.com/programs/research-essentials-products.html?TablePage=102880799&quot;&gt;Sigma-Aldrich&lt;/a&gt; has reagent-grade NaCl (&gt;98%), ACS grade NaCl (&gt;99.5%) and TraceSelect NaCl for doing heavy metal analysis that is more than 99.999% pure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is 99% purity &quot;pure enough for high level experiments in chemistry?&quot; There is no one answer. It depends on what those experiments are. How many significant digits do you need in your results? Are the results going to be affected by the presence of tiny amounts of &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodide&quot;&gt;iodide&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Iodate&quot;&gt;iodate&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot;&gt;glucose&lt;/a&gt;, &lt;a href=&quot;http://www.arltma.com/Mineral_Information/Calcium.html&quot;&gt;calcium&lt;/a&gt;, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Silicate&quot;&gt;silicate&lt;/a&gt;, or the other impurities listed above? Maybe, maybe not. If a specific impurity interferes, find a way to lock up, remove or otherwise neutralize its effects.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-21T16:17:51.320" LastActivityDate="2015-02-21T16:17:51.320" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24819" PostTypeId="2" ParentId="24816" CreationDate="2015-02-08T16:34:14.547" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Some table salt sold for consumption contain &lt;strong&gt;additives&lt;/strong&gt; which address a variety of health concerns, especially in the developing world. The identities and amounts of additives vary widely from country to country. &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt#Fortified_table_salt&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Table salt is hardly even close to pure.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In summary, a variety of additives is used in common table salt. In addition to potassium iodide, the additives can be categorized as either iodide stabilizers or anticaking agents. Iodide stabilizers act to prevent the oxidation of iodide to iodine, while anticaking agents keep salt free-flowing. Abstract of &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed062p1016?journalCode=jceda8&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, table salt can result in an approximation in the experiment you're conducting. You can look for &lt;a href=&quot;http://en.wikipedia.org/wiki/Kosher_salt&quot; rel=&quot;nofollow&quot;&gt;kosher salt&lt;/a&gt; though.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-08T16:34:14.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24821" PostTypeId="2" ParentId="24809" CreationDate="2015-02-08T16:50:42.907" Score="2" Body="&lt;p&gt;Starting from pyruvic acid which is readily available and not very expensive: react pyruvic acid (only one enolizable side) with a Mannich base, then react the resulting amide with $\ce{MeI}$ and base to obtain the exo-methylene product. Brominate the exo-methylene. Now cyclization will prefer the five-membered ring over the strained four-membered ring. Reduce the ketone with a reletively mild reducing agent like $\ce{NaBH4}$ and then eliminate $\ce{HBr}$ in a E1cB type reaction, yielding 3-hydroxydihydrofuran-2-one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This synthesis involves a Mannich reaction as you asked, but it can be done much simpler without it I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I'm not an expert, just an enthusiastic amateur. Before trying anything in the lab  make sure your synthesis is checked by a supervisor.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-08T16:50:42.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24823" PostTypeId="1" AcceptedAnswerId="24825" CreationDate="2015-02-08T17:19:17.327" Score="3" ViewCount="26932" Body="&lt;p&gt;If you have, for example, $\pu{10 g}$ of a solute dissolved into $\pu{100 ml}$ of a solution, and the solution density is $\pu{1.5 g ml^-1}$, can you just multiply $100\ \mathrm{ml} \times 1.5\ \mathrm{\frac{g}{ml}}$, then subtract the $\pu{10 g}$ of solute from the result to get the mass of the solvent in grams? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that work or am I trying to do do this wrong?&lt;/p&gt;&#xA;" OwnerUserId="13287" LastEditorUserId="5026" LastEditDate="2018-02-27T08:17:20.643" LastActivityDate="2018-12-03T00:17:03.207" Title="Can you find the mass of solvent with mass of solute, volume of solution, and solution density?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24824" PostTypeId="1" CreationDate="2015-02-08T17:49:36.660" Score="2" ViewCount="60" Body="&lt;p&gt;I've heard Nanotubes will replace Li-ion batteries in the future and it is much more relevant. I just wanted to know could it really replace Li-ion batteries? Is synthesizing nanotubes economic?&lt;/p&gt;&#xA;" OwnerUserId="12666" LastEditorUserId="7448" LastEditDate="2015-02-08T17:52:18.720" LastActivityDate="2015-02-08T17:52:18.720" Title="Are there economical ways of synthesizing nanotubes?" Tags="&lt;nanotechnology&gt;&lt;carbon-allotropes&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24825" PostTypeId="2" ParentId="24823" CreationDate="2015-02-08T18:05:48.427" Score="2" Body="&lt;p&gt;That works!  In most chemistry (but not nuclear reactions!),  mass is always conserved, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\text{solution} = m_\text{solvent} + m_\text{solutes}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is also equal to density times volume, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\text{solution} = \rho_\text{solution} V_\text{solution} = m_\text{solvent} + m_\text{solutes}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $m_\text{solvent}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\text{solvent} = \rho_\text{solution} V_\text{solution} - m_\text{solutes}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is exactly the calculation you did.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorDisplayName="user7951" LastEditDate="2015-05-13T19:05:31.660" LastActivityDate="2015-05-13T19:05:31.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24826" PostTypeId="1" CreationDate="2015-02-08T18:06:44.277" Score="2" ViewCount="1502" Body="&lt;p&gt;After finding a reaction &quot;Transitional State&quot; by GAMESS, and finding Hessian Eigenvalues, my project instructor said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;seems your imaginary frequency (negative eigenvalue) is not that big (it is around -13.01), which means your reaction is &lt;strong&gt;barrier-less&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I am a beginner in Quantum Chemistry, I could not figure out what is the relation between a small imaginary frequency and being barrier-less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I remember from the Arrhenius equation, being barrier-less has something to do with higher temperature leading to a slower reaction, but I could not understand its relation to imaginary frequencies.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="5026" LastEditDate="2018-04-21T05:42:53.137" LastActivityDate="2018-11-10T00:45:51.463" Title="What is a barrier-less reaction in Quantum Chemistry?" Tags="&lt;reaction-mechanism&gt;&lt;quantum-chemistry&gt;&lt;energy&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24828" PostTypeId="1" AcceptedAnswerId="24892" CreationDate="2015-02-08T20:43:16.140" Score="6" ViewCount="1236" Body="&lt;p&gt;\begin{align}&#xA;\ce{Propanol + SOCl2 &amp;amp;-&amp;gt; Expected~ alkyl~ halide~ product}\\&#xA;\ce{Propanoic~ acid + SOCl2 &amp;amp;-&amp;gt; Expected ~acid ~chloride ~product}\\&#xA;\ce{Acetone ~enol + SOCl2 &amp;amp;-&amp;gt; doesn't~ work ~in~ the ~expected ~manner}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My prof says the $\ce{SOCl2}$ only works on $\ce{sp^3}$ substrates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ... both propanoic acid and acetone enol involve an $\ce{sp^2}$ carbon bound to an alcohol group ... why does $\ce{SOCl2}$ work in one case but not in the other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it instead has to do with the nucleophilicity of the electrons involved with oxygen ... the mechanism starts with electrons associated with oxygen attacking thionyl chloride. A regular alcohol has an oxygen with lone pairs that aren't fairly &quot;free,&quot; so to speak. An carboxylic acid has resonance donation enhancing the nucleophilicity of one oxygen's lone pairs. In acetone, however, there is no resonance donation ....... (inductive donation, sure).       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So does the $\ce{sp^2}$ vs. $\ce{sp^3}$ explanation hold water?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="17368" LastEditDate="2016-01-19T12:46:11.810" LastActivityDate="2016-01-19T12:46:11.810" Title="Thionyl Chloride and enols, carboxylic acids, alcohols" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24831" PostTypeId="1" AcceptedAnswerId="24841" CreationDate="2015-02-08T21:17:10.217" Score="2" ViewCount="593" Body="&lt;p&gt;The Wikipedia article titled &lt;a href=&quot;http://en.wikipedia.org/wiki/Solubility_table&quot; rel=&quot;nofollow&quot;&gt;Solubility table&lt;/a&gt; states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The table below provides information on the variation of solubility of different substances (mostly inorganic compounds) in water with temperature, at 1 atmosphere pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does pressure have an effect of on solubility?&lt;/p&gt;&#xA;" OwnerUserId="13291" LastActivityDate="2015-02-09T01:16:18.057" Title="Why does (atmospheric) pressure have an effect on solubility?" Tags="&lt;solubility&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24832" PostTypeId="2" ParentId="24831" CreationDate="2015-02-08T21:48:57.133" Score="0" Body="&lt;p&gt;Pressure has a strong effect on the solubility of gases, of course, &quot;forcing&quot; them into solution; see &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_and_Factors_Affecting_Solubility&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_and_Factors_Affecting_Solubility&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even solution of solids in liquids are affected somewhat by pressure: for example, proteins may be denatured and made less soluble by high pressure, as in ocean deeps. See &lt;a href=&quot;http://www.desy.de/news/@@news-view?id=4561&amp;amp;lang=eng&quot; rel=&quot;nofollow&quot;&gt;http://www.desy.de/news/@@news-view?id=4561&amp;amp;lang=eng&lt;/a&gt; for information on that.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-08T21:48:57.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24833" PostTypeId="1" AcceptedAnswerId="24839" CreationDate="2015-02-08T21:59:51.597" Score="2" ViewCount="10056" Body="&lt;p&gt;Suppose I pour some sodium chlorine into water. So, what happens is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{NaCl (s) -&amp;gt; Na+ (aq) + Cl- (aq)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;meaning that the ionic bond between Na and Cl breaks up (correct this far?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, does this mean that the water &lt;em&gt;actually&lt;/em&gt; contains separate charged Na and Cl particles? So... since chlorine boils at −34.04 °C according to Wikipedia, why is there then no chlorine gas evaporating? Because it is &lt;em&gt;chlorine ions&lt;/em&gt; there, not &lt;em&gt;chlorine atoms&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I feed electrons some how into the solution, will chlorine gas start forming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, could I use this so that I pour NaCl into water and get Na and Cl ions, and then (since they are separate) add something more to create Na[something] or Cl[something]? Something that would not form if I just mixed NaCl + [something else]. I hope I am not completely lost here..&lt;/p&gt;&#xA;" OwnerUserId="13291" LastEditorUserId="194" LastEditDate="2017-04-06T01:42:52.170" LastActivityDate="2017-04-06T01:42:52.170" Title="Are Na and Cl ions from NaCl in water separate charged particles?" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24834" PostTypeId="1" AcceptedAnswerId="24848" CreationDate="2015-02-08T22:00:35.257" Score="3" ViewCount="2218" Body="&lt;p&gt;I am using ED-XRF for geochemical analysis for a geology dissertation. The ED-XRF provides a ppm reading for elements and a 2sigma error, for example 472 +- 5.11ppm Sr.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example calibration curve made for Sr comprising 37 standards has a linear regression of y=1.26x + 32.1 (R^2 0.9991), which I can easily apply to correct my unknown analyses. I need to plot my corrected readings with their error bars, so how is the 2sigma error associated with this unknown calibrated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this error value also simply get calibrated with the curve equation, or just the slope value, or is this error propagation far more complex? I have searched for a solution and keep finding an equation for uncertainty propagation with calibration curves, but this only calculates the uncertainty caused by the curve, and has no relation to the uncertainty of each analysis? So is the total error the original 2sigma error provided by the machine + calibration curve uncertainty?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="13292" LastActivityDate="2015-02-09T06:59:49.020" Title="Calibration Curve Error Propagation" Tags="&lt;analytical-chemistry&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24835" PostTypeId="2" ParentId="24816" CreationDate="2015-02-08T22:03:25.867" Score="4" Body="&lt;p&gt;Salt purity is measured on &quot;dry salt&quot; - which is to say, water does not count towards the purity. In the UK, 99% pure NaCl is used for ice-treating areas that might be subject to staining by less pure gritting methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;99.9% purity is used for food:&#xA;&lt;a href=&quot;http://www.british-salt.co.uk/Products/PureDriedVacuum.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.british-salt.co.uk/Products/PureDriedVacuum.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same purity as pharmaceutical-grade salt, except that pharm. grade also requires a moisture content of 0.01% or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For export markets, British Salt can then add the required impurities for that market's requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The British Salt company provides about 2/3 of domestic cooking salt in the UK.&lt;/p&gt;&#xA;" OwnerUserId="13293" LastActivityDate="2015-02-08T22:03:25.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24836" PostTypeId="1" CreationDate="2015-02-08T23:22:08.740" Score="1" ViewCount="658" Body="&lt;p&gt;I did a reaction and I measured the volume of gas produced against time. I need to find the rate constant of the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to plot a graph of ln(vol of gas produced) against time so that the gradient will be equal to -k so i can calculate the rate constant k. I am using the equation ln[A]t = -kt + ln[A]o&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I can't plot ln0. I timed the reaction every 5 mins and gas wasn't produced until 15mins had passed. This means that the volume of gas produced for time values 0, 5 and 10 mins is 0cm^3, but ln0 doesn't exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use my data to create a graph to calculate the rate constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results were: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time 0___5___10___15___20___25___30___35___40___45___50___55___60 ____(in mins)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vol _0___0___0_____5___10___18____22___28___36___44___51____60____70___(in ml produced) &lt;/p&gt;&#xA;" OwnerUserId="12834" LastEditorUserId="12860" LastEditDate="2015-10-30T18:25:56.847" LastActivityDate="2015-10-30T18:25:56.847" Title="Rate of Reaction Graph - Gas Produced" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24837" PostTypeId="2" ParentId="24811" CreationDate="2015-02-08T23:30:22.087" Score="2" Body="&lt;p&gt;Yes, they're called multiple emulsions (and more specifically: double, triple emulsions, etc.). Usually they're made from alternating polarity phases—water inside oil inside water, for example. There are bulk ways of making them though there isn't great control over how many droplets end up in each encompassing droplet without using microfluidic techniques. &lt;a href=&quot;http://weitzlab.seas.harvard.edu&quot; rel=&quot;nofollow&quot;&gt;David Weitz's lab&lt;/a&gt; at Harvard has some interesting work in building microfluidic devices to make multiple emulsions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if there are any common natural occurrences but the pharmaceutical and cosmetics industries seem to be the the main parties interested in this stuff as they can be used for encapsulating drugs for time release and the like.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-08T23:30:22.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24838" PostTypeId="2" ParentId="24836" CreationDate="2015-02-09T00:27:09.017" Score="2" Body="&lt;p&gt;I think the first bit of gas produced probably dissolved in the water, so you didn't see it for a little while. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can skip your three initial &quot;points&quot; with zero volume, and you're not going to have $\ln(0)$, which is what's hanging you up. Note that you did this in your report, and why!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a linear fit with the remaining points. Your y-intercept will be $\ln \mathrm{[A]}_0$ and your slope will be $-k$. It's not a terribly good fit.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-09T04:43:18.363" LastActivityDate="2015-02-09T04:43:18.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24839" PostTypeId="2" ParentId="24833" CreationDate="2015-02-09T00:33:16.100" Score="1" Body="&lt;p&gt;Most of what you wrote is correct. It is difficult to feed many electrons into the salt water because the Coulomb forces are &lt;em&gt;extremely&lt;/em&gt; strong, see &lt;a href=&quot;http://physics.bu.edu/~duffy/py106/Charge.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; and &lt;a href=&quot;https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Coulombic_Forces&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, but you can &lt;strong&gt;feed electrons in if you drain them out elsewhere&lt;/strong&gt; so there is no net gain or loss of charge. that is called an &lt;em&gt;electric current&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use that to &lt;em&gt;electrolyze&lt;/em&gt; salt water, and get chlorine as you surmise; &lt;a href=&quot;http://aquarius.umaine.edu/cgi/ed_act.htm?id=47&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA has a nice lab&lt;/a&gt;. BTW, you would expect to get Cl gas at the positive electrode, and do get it, but you &lt;em&gt;do not&lt;/em&gt; get Na metal at the negative, because hydrogen bubbles out before Na. You can electrolyze molten salt, though, to get a bit of Na (or at least you see it form and then catch fire with a yellow flame).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="194" LastEditDate="2017-04-06T01:39:59.763" LastActivityDate="2017-04-06T01:39:59.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24841" PostTypeId="2" ParentId="24831" CreationDate="2015-02-09T01:16:18.057" Score="2" Body="&lt;p&gt;It comes from a combination of &lt;a href=&quot;https://en.wikipedia.org/wiki/Henry%27s_law&quot; rel=&quot;nofollow&quot;&gt;Henry's law&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Chatelier&amp;#39;s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Châtelier's principle&lt;/a&gt;. Henry's law is&#xA;$$p=k_hc$$&#xA;where $p$ is the partial pressure of the gas, $k_h$ is the Henry's law constant and $c$ is the concentration of the gas (because liquids and solids are not affected as much as gases, if at all). This means that with a greater pressure comes a greater concentration, and with a lower pressure comes a lesser concentration, and vice versa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Châtelier's principle states that if the pressure, concentration, temperature or volume of a system changes, something else will change to compensate and swing the whole thing back into equilibrium. If there is a greater pressure, there will therefore be a grater concentration. A greater concentration means there is a greater solubility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_and_Factors_Affecting_Solubility&quot; rel=&quot;nofollow&quot;&gt;UC Davis Chem Wiki&lt;/a&gt; (I didn't see that DrMoishe Pippik had linked to it until after I posted this!) is rather helpful - much more helpful than &lt;a href=&quot;https://en.wikipedia.org/wiki/Solubility#Pressure&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, in this case.&lt;/p&gt;&#xA;" OwnerUserId="13294" LastActivityDate="2015-02-09T01:16:18.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24843" PostTypeId="1" CreationDate="2015-02-09T01:48:06.373" Score="4" ViewCount="3333" Body="&lt;p&gt;Since the electron donating property of $\ce{N}$ and $\ce{C}$ is different, there is partial charge transfer between $\ce{N}$ and $\ce{C}$ atoms in aliphatic amines, namely $\ce{R-NH2}$. What is the charge transfer direction? Can the following be true? And how large is this dipole moment?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/o83NL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, how can one determine both the direction and quantity of the charge transfer dipole moment given two common organic groups? Or what is the order of electron donating property of common groups?&lt;/p&gt;&#xA;" OwnerUserId="9507" LastEditorUserId="7475" LastEditDate="2015-10-16T17:47:15.720" LastActivityDate="2015-10-16T17:47:15.720" Title="What is the dipole moment direction of an aliphatic amine, i.e., R-NH2?" Tags="&lt;dipole&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24844" PostTypeId="1" CreationDate="2015-02-09T02:09:19.440" Score="3" ViewCount="149" Body="&lt;p&gt;Is there dipole formation upon physisorption of an aliphatic amine to an aluminum substrate? What is the charge transfer direction? What if the aluminum substrate is clean, surface oxidized or covered with -OH groups?&#xA;What if aluminum is replaced by other types of metals, such as Au and Ag?&lt;/p&gt;&#xA;" OwnerUserId="9507" LastEditorUserId="865" LastEditDate="2015-10-16T17:48:49.287" LastActivityDate="2015-10-16T17:48:49.287" Title="What is the direction of charge transfer between Aluminum and aliphatic amines physisorbed on top?" Tags="&lt;electrons&gt;&lt;electronegativity&gt;&lt;surface-chemistry&gt;&lt;amines&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24845" PostTypeId="1" AcceptedAnswerId="24852" CreationDate="2015-02-09T02:37:31.213" Score="10" ViewCount="5931" Body="&lt;p&gt;I am having a difficult time understanding what makes a buffer a buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buffers in my textbook are defined as a solution of a weak acid or base and their conjugate acid/base. So if I were to just dissolve acetic acid (a weak acid) in water why would this not be defined as a buffer. Acetic acid incompletely dissociates into an acetate ion (the conjugate base) and a Hydrogen ion. Why is it that for a base to be made the acetic acid would have to be mixed with something like Sodium acetate if acetic acid incompletely dissociates into a conjugate base in the first place? Could someone please help me understand why this is the case?&lt;/p&gt;&#xA;" OwnerUserId="7612" LastEditorUserId="5026" LastEditDate="2018-03-04T16:35:27.247" LastActivityDate="2018-03-04T16:37:32.060" Title="Are all solutions of weak acid/bases buffers?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24846" PostTypeId="1" AcceptedAnswerId="24869" CreationDate="2015-02-09T04:10:24.353" Score="11" ViewCount="58116" Body="&lt;p&gt;I have trouble understanding this particular law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Greater the valency of the flocculating ion, greater will be its coagulation power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? What is the intuition behind this?&lt;/p&gt;&#xA;" OwnerUserId="13216" LastEditorUserId="27860" LastEditDate="2019-03-25T17:53:17.343" LastActivityDate="2019-03-25T17:53:17.343" Title="Hardy Schulze law" Tags="&lt;surface-chemistry&gt;&lt;nanotechnology&gt;&lt;colloids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24848" PostTypeId="2" ParentId="24834" CreationDate="2015-02-09T06:59:49.020" Score="1" Body="&lt;p&gt;I will try to offer a detailed way to propagate errors here, but before beginning it's worth asking what your audience is likely to expect &quot;error bars&quot; to mean.  In your field are error bars usually determined simply by replicate analyses of the same sample?  Or replicate samples of the same experiment or field site?  If so, it's probably best just to stick with what your audience is likely to expect.  That said...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The theoretical best way to fit your calibration curve would be with &lt;em&gt;weighted&lt;/em&gt; linear least-squares algorithms.  In that algorithm, in addition to the {x, y} data, which in your case is $\{\ce{Sr_{known}, Sr_{meas}}\}$, you also supply weights corresponding to the uncertainty in the y values.  The weight for each y data point would be $\frac{1}{\sigma^2}$, which you could calculate from the $2\sigma$ uncertainty given by your instrumentation for that data point.  In practice, if you did this, I doubt that the parameters of your calibration would change very much at all from what you did previously.  But formally the weighted variant of least-squares is required if the uncertainty in y depends on y.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Least_squares#Weighted_least_squares&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; for more.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After doing the (weighted) least squares calibration regression, you can find the uncertainty in the &lt;em&gt;estimation&lt;/em&gt; using the calibration equation.  $\sigma_{est} = \sqrt{\frac{\sum{\left(y_{pred}-y_{meas} \right)^2}}{N}}$.  In that equation $y_{pred}$ is the calibration-predicted Sr concentration and $y_{meas}$ is the measured Sr concentration, and for you $N$, the number of data points, is 37.  This uncertainty is one part of the uncertainty you'll eventually report, and reflects the uncertainty in your calibration curve.  Based on the R-squared you reported, my guess is that this uncertainty will be low.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After all that (!), you need to estimate the other major source of uncertainty, the uncertainty in the measurement of your unknowns ($\sigma_{meas}$).  There are several OK methods for doing this, the best of which would be replicate analysis of several samples.  It might be OK to use the $1\sigma$ value calculated from the $2\sigma$ uncertainty reported by your instrument too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, combine the uncertainties with the standard equation for combining multiple sources of uncertainty $\sigma_{total} = \sqrt{(\sigma_{meas})^2 + (\sigma_{est})^2}$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-09T06:59:49.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24849" PostTypeId="1" CreationDate="2015-02-09T07:15:57.470" Score="4" ViewCount="4709" Body="&lt;p&gt;While studying the Group 14 elements, I came across the anomalous behaviour of carbon. In that there was a point:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon forms only p$\pi$-p$\pi$ type bonding while other elements in the group form p$\pi$-p$\pi$, as well as d$\pi$-p$\pi$ type bonding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am searching for molecules in which  group 14 elements form d$\pi$-p$\pi$ bonds.&lt;/p&gt;&#xA;" OwnerUserId="2382" LastEditorUserId="4945" LastEditDate="2015-02-09T07:29:15.213" LastActivityDate="2015-07-28T15:15:00.260" Title="What are examples of group 14 elements forming dπ - pπ bonding?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24850" PostTypeId="1" CreationDate="2015-02-09T08:07:53.193" Score="2" ViewCount="283" Body="&lt;p&gt;In a crystal Lattice how to find the equivalent points for 3 fold and 6 fold symmetry?&lt;/p&gt;&#xA;" OwnerUserId="4034" LastActivityDate="2015-02-09T08:07:53.193" Title="Equivalent points for 3 fold and 6 fold symmetry" Tags="&lt;crystal-structure&gt;&lt;symmetry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24851" PostTypeId="2" ParentId="24816" CreationDate="2015-02-09T10:35:19.753" Score="1" Body="&lt;p&gt;See &lt;a href=&quot;http://www.spexcertiprep.com/knowledge-base/files/AppNote_GourmetSalts.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analysis of Gourmet Salts for the Presence of Heavy Metals&lt;/a&gt; which reports analysis of table salt (noniodized), 12 gourmet salts, and reagent grade NaCl using &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductively_coupled_plasma_mass_spectrometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;inductively coupled plasma mass spectrometry&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Considering metals, other than sodium what is present, by far the greatest other metal ion present in table salt was potassium at 1000 ppm (1000 micrograms per gram of sample).  Next was calcium at 218 ppm.  Considering that pure NaCl is 393000 ppm Na, the presence of K and Ca alone brings the purity of table salt down to 99.7%&lt;/p&gt;&#xA;&lt;p&gt;There is no indication that the above study tested for carbon, oxygen, silicon or hydrogen, so levels of water, sugar such as dextrose, silicates or carbonates should be considered unknown.&lt;/p&gt;&#xA;&lt;p&gt;The Pennsylvania State University agricultural extension service &lt;a href=&quot;http://extension.psu.edu/food/preservation/faq/canning-and-pickling-salt&quot; rel=&quot;nofollow noreferrer&quot;&gt;advises against using table salt for canning&lt;/a&gt; because up to 0.04% dextrose and up to 0.5% calcium silicate are added.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.codexalimentarius.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CODEX Alimentarius&lt;/a&gt; is an international set of food standards endorsed by the World Health Organization and Food and Agriculture Organization of the Unitied Nations.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.codexalimentarius.org/download/standards/3/CXS_150e.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CODEX STANDARD FOR FOOD GRADE SALT&lt;/a&gt; is linked and requires&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The content of NaCl shall not be less than 97% on a dry matter basis, exclusive of additives.&lt;/p&gt;&#xA;&lt;p&gt;The remainder comprises natural secondary products, which are present in varying amounts depending on the origin and the method of production of the salt, and which are composed mainly of calcium, potassium, magnesium and sodium sulphates, carbonates, bromides, and of calcium, potassium, magnesium chlorides as well. Natural contaminants may also be present in amounts varying with the origin and the method of production of the salt. Copper shall not exceed 2 mg/kg&lt;/p&gt;&#xA;&lt;p&gt;The name &amp;quot;salt&amp;quot; shall have in its close proximity a declaration of either &amp;quot;Food Grade&amp;quot; or &amp;quot;Cooking Salt&amp;quot; or &amp;quot;Table Salt&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So &amp;quot;Table Salt&amp;quot; as internationally recognized can have impurities of up to 3%, plus additives (such as for anti-caking or iodine supplementing), plus moisture.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-02-09T20:05:52.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24852" PostTypeId="2" ParentId="24845" CreationDate="2015-02-09T12:43:46.400" Score="8" Body="&lt;p&gt;The term &quot;buffer&quot; implies a purpose of resisting changes in pH when small amounts of acid or base are added.  The buffer will be most effective when the pH equals the pKa of the acid.  This is when there are equal amounts of acid and conjugate base. For each pH unit the system deviates from pH=pKa, the buffer is about a factor 10 less effective.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate.  However, the pH will be far from the pKa of acetic acid, and the system will not be very effective as a buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, &quot;buffer&quot; is not a very precise term, but it implies that there are significant amounts of both the acid and the conjugate base.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-03-04T16:35:50.843" LastActivityDate="2018-03-04T16:35:50.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24853" PostTypeId="2" ParentId="16387" CreationDate="2015-02-09T13:21:28.937" Score="6" Body="&lt;p&gt;Techron is a detergent based on PEA, polyeter amine. It is a complex synthetic molecule that will basically 'loosen' and 'solve' deposits in your fuel system. They then become part of the fuel-air mixture and burn in the combustion chamber with it. Be it in the carburettor, the injectors, the valves (for indirect injection vehicles) or the combustion chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Techron fuel you buy at the Chevron or Texaco stations has only a small amount of Techron added, the rest is just gas/petrol. You can also buy the Techron in seperate high concentrate bottles (typically 12oz.) These bottles contain up to 35% of PEA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the working ingrediënt PEA is just as well used in Gum-Out, Amsoil and a bunch of other fuel system cleaners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it is a difference to fuel which doesn't have these detergent but every major gas producer (Shell, BP, Chevron, Texaco, ...) will add detergents. They ARE different detergents so they do make a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact they are just auxiliary tools to keep your engine working well. It is like a band-aid. It does help tune up your engine, but it helps a lot more to do proper maintenance, change coolant and oil regularly, make sure your radiator works well so your engine operating temp isn't too high, clean your air filter and fuel filter regularly,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these fuels does have an effect eg for emissions testing, but If you ask me. I just buy the cheapest fuel and I will add a bottle of High concentrate Techron or Gumout or another cleaner before I go to an emissions test (make sure you drive about 60 miles before turning up since the deposits have to be burnt before the test). And if you really want to use it, buy a bunch of high concentrate bottles and add them to supermarket gas, that's basically the same&lt;/p&gt;&#xA;" OwnerUserId="13302" LastActivityDate="2015-02-09T13:21:28.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24854" PostTypeId="2" ParentId="24845" CreationDate="2015-02-09T13:56:50.213" Score="9" Body="&lt;p&gt;I totally agree with @DavePhD and I want to show you how the pH dependence of the degree of acid-base ionisation provides the background of the buffering mechanism. Titration curves effectively illustrate the influence of the pH on the ionisation of weak acids and bases. The Figure below shows the change in the pH of a 0.1 M acetic acid solution on its titration with a strong base. If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate. However if you add the slightest amount of a base, for example, it will result into a steep increase in the pH (look at the beginning of the titration curve) and the solution is not a buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That initial steep increase in the pH slows down quickly, and a &lt;strong&gt;pH plateau&lt;/strong&gt; appears in the range of 4.0-5.8. As in this range $[\ce{CH3COOH}]$ is comparable to $[\ce{CH3COO-}]$ (i.e. their ratio is within a tenfold range), we will have a two-component system containing both a weak acid and a weak base—i.e. a buffer. The centre of the plateau is at 50 % titration where $[\ce{CH3COOH}]=[\ce{CH3COO-}]$. Therefore, according to the equation $\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$, the $\mathrm{pH}$ will just equal the $\mathrm{p}K_\mathrm{a}$. Following the plateau (upon the exhaustion of the buffer capacity), the pH will again increase steeply, and the top of the titration step will be reached, i.e. the point of neutralization at which the degree of titration of the acetic acid will be 100 %&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qefyC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the figure above, the region marked inside the rectangle is a buffer. Because it can provide the most efficient pH stabilization against both acidic and basic shifts. In this region, the ratio of the two components (acid and conjugated base) is one or close to one. Accordingly, a buffer is effective in the $\mathrm{pH}$ range of its $\mathrm{p}K_\mathrm{a} \pm 1$ $\mathrm{pH}$ unit&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-03-04T16:37:32.060" LastActivityDate="2018-03-04T16:37:32.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24855" PostTypeId="1" CreationDate="2015-02-09T13:59:34.180" Score="2" ViewCount="54" Body="&lt;p&gt;We often know, from e.g. titration experiments, what the average surface charge density of a surface is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these charges vary over both time and space (i.e. dynamics)? I have been unable to find much (well, any really!) literature on this.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Taking the example of oxide surfaces, you have the following processes going on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S-O- + H3O+&amp;lt;=&amp;gt;[\text{K1}]S-OH +H2O}$ #Protonation/Autoionisation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S-OH + OH- &amp;lt;=&amp;gt;[\text{K2}]S-O- +H2O}$ #Deprotonation/Protonation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S-OH + H3O+ &amp;lt;=&amp;gt;[\text{K3}]S-OH2+ +H2O}$ #Protonation/Deprotonation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know (from experimental/theoetrical calculation of the Equilibrium Constants K1, K2, K3), that you have 100 negative charges per micrometer squared (i.e we are at high pH). Is it a more realistic picture on a picosecond/nanosecond timescale to consider it being something like 10,000 charges per micrometer squared which are being regularly neutralised over time or simply 100 charges per micrometer which are fixed?&lt;/p&gt;&#xA;" OwnerUserId="13304" LastActivityDate="2015-02-09T14:19:56.417" Title="Oxide surfaces develop a charge in H2O. Do we have any, even vague, handle on how long does an indivdual charge last on avg. before being neutralised?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;computational-chemistry&gt;&lt;surface-chemistry&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24856" PostTypeId="2" ParentId="24855" CreationDate="2015-02-09T14:19:56.417" Score="1" Body="&lt;p&gt;Proton transfer reactions in water such as the ones you wrote down are diffusion controlled under ambient conditions (rate constant about 1e+12 per sec.) A snapshot in time will have a certain distribution of charges on the surface, but these are continually in flux, so any one charged site will have an exceedingly short lifetime. You could reasonably represent that system with a fixed set of charges on the surface, for some types of models (e.g., states at equilibrium).&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-09T14:19:56.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24857" PostTypeId="2" ParentId="24843" CreationDate="2015-02-09T14:31:28.870" Score="1" Body="&lt;p&gt;Since the nitrogen atom is more electronegative than the carbon atom it is bonded to, at least in the case of a simple aliphatic amine, I would venture to say the direction of the dipole will be with a slight negative charge toward nitrogen. The amine nitrogen has nonbonding electrons on it, but the adjacent carbon is saturated and cannot accept the lone pair electrons on the nitrogen, and thus nitrogen in this example will not have a partial positive charge. However, if R is aromatic, the situation will be different. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-09T14:31:28.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24858" PostTypeId="2" ParentId="24844" CreationDate="2015-02-09T14:40:05.750" Score="1" Body="&lt;p&gt;Amines could bond to the aluminum surface via a Lewis acid-base reaction between the electron rich amine nitrogen and the electron poor Al atom on the surface (perhaps formed by loss of a surface hydroxyl group as hydroxide ion). The direction of charge transfer would be from the amine nitrogen to an aluminum atom. The resulting &quot;Lewis salt&quot; would have a partial positive charge on nitrogen and a partial negative charge on aluminum (like an aluminate ion or an amine-aluminum chloride adduct).&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-09T14:40:05.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24861" PostTypeId="1" CreationDate="2015-02-09T16:37:34.510" Score="3" ViewCount="1837" Body="&lt;p&gt;The given reaction is $$\ce{NH4NO2 \rightarrow N2  +  2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I did it was&#xA;$$\ce{NH4+ \rightarrow \frac{1}{2} N2 + 4H+ + 3e-}$$&#xA;and&#xA;$$\ce{NO2- +3e- +4H+ \rightarrow \frac{1}{2} N2 + 2H2O}$$&#xA;Therefore n factor is 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book I am referring to says 6. Can someone please help?&lt;/p&gt;&#xA;" OwnerUserId="13307" LastEditorUserId="51553" LastEditDate="2018-08-24T17:03:40.417" LastActivityDate="2018-08-24T17:03:40.417" Title="n-factor for ammonium nitrite?" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24862" PostTypeId="1" AcceptedAnswerId="24864" CreationDate="2015-02-09T17:09:39.373" Score="2" ViewCount="211" Body="&lt;p&gt;Consider Oxytocin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6EM5.png&quot; alt=&quot;image of oxytocin from a cutout of a local newspaper in Denmark&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to hydrolyze Oxytocin and reveal 5 amino acids which are the result of this hydrolysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must admit I have no idea where to begin. I know that generally amino acids are on the form of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH_2-CH(R)-COOH}$&lt;/span&gt;, so perhaps looking at these double bonds to oxygen may be a starting point, as I assume they will become the acid group? But I really don't know how to move forward from there!&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="83758" LastEditDate="2020-05-03T08:16:09.460" LastActivityDate="2020-05-03T08:16:09.460" Title="Hydrolysing large molecules to reveal amino acids" Tags="&lt;organic-chemistry&gt;&lt;hydrolysis&gt;&lt;proteins&gt;&lt;amino-acids&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24863" PostTypeId="2" ParentId="24861" CreationDate="2015-02-09T17:43:38.823" Score="2" Body="&lt;p&gt;&quot;n factor&quot; of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case here.You have  ammonium ion that undergoes oxidation reaction (n-factor= 3) and  nitrite ion that undergoes reduction reaction (n-factor= 3).&#xA;So, the n-factor of ammonium nitrite is 6.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-09T17:43:38.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24864" PostTypeId="2" ParentId="24862" CreationDate="2015-02-09T18:33:42.510" Score="3" Body="&lt;p&gt;You are on the right track.  $\ce{NH_2-C(H)R-COOH}$ is the formula for an &lt;em&gt;alpha&lt;/em&gt; amino acid, one where the amino group of the molecule is attached to a carbon atom one carbon atom away from the carboxylate carbon.  This is the most common type of amino acid found in biochemistry.  (A beta amino acid would be $\ce{NH_2-C(H)R^2-C(H)R^1-COOH}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When amino acids condense into a polymer or oligomer they do so by forming peptide bonds, also called amide bonds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R^1COOH + NH2R^2 -&amp;gt; R^1CONH2R^2 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to find the places in oxytocin that would be hyrolyzed into amino acids, look for individual carbon atoms that are attached to both (i) an oxygen atom via a double bond and (ii) a nitrogen atom while (iii) not being attached to any other heteroatoms.  (Item (iii) is generally necessary to exclude other functional groups such as carbamides or carbamates, but these groups are not found in oxytocin.)  For oxytocin, your impulse to look just at carbon double-bonded to oxygen (item (i) only) is OK, but be warned that this procedure is not precise enough in general, because molecules with e.g. ketone moieties will have C=O bonds but will not be hydrolyzed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red-circled amino group is a free amino group:  the carbon attached to it is not bound to a carbonyl oxygen, so it is not a place where hydrolysis will happen.  (However, look one carbon atom to the left of C atom attached to the circled N....)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@DavePhD is right; there are more than five amide bonds in oxytocin.  I think there are a total of ten.  Eight are the result of amide bond formation between alpha amino acids and two are the result of amide bond formation between an acid side chain and unsubstituted ammonia.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-09T18:33:42.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24866" PostTypeId="2" ParentId="24836" CreationDate="2015-02-09T19:49:29.203" Score="1" Body="&lt;p&gt;Fred Senese's post is an excellent starting point.  I wanted to add that transforming data in this way is conceptually very useful but distorts parameter estimates because of unequal weighting of data point uncertainty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty of your &quot;zero&quot; data points is probably not negligible.  How exactly could you read your gas measurement apparatus?  To within 0.1 mL, or 1 mL, or ...?  Say it was 0.1 mL, then your &quot;zero&quot; data point in mL is $0 \pm 0.1 \text{ mL}$.  And your 70 mL data point is $70 \pm 0.1 \text{ mL}$.  When you transform the data by taking the logarithm, however, the uncertainty in your data is unevenly affected.  The uncertainty of $\ln \left( 0 \pm 0.1 \text{ mL} \right)$ is approximately $\ln{0.1} - \ln{0}$ which as you are discovering is infinite.  The uncertainty in log-space of your 70 mL data point is $\ln{70.1} - \ln{70}$, which is 0.0014.  So even though the uncertainty on each data point was the &quot;same&quot; before logarithms, say 0.1 mL, after transformation the uncertainty is as unequal as can be (infinity vs. 0.0014)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct non-linear fitting can fit parameters to data without needing transformations.  Here is some &lt;code&gt;R&lt;/code&gt; code to do the fitting on your data.  Because of the issue related to my comment, I had to include an offset parameter.  (An exponential equation tends to a positive value as t approaches zero, not to zero...unless there's an offset.)  This method is probably overkill for your particular problem or course, but I wanted to include it in case it's useful to other chem.se readers.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#read the data from the question into R and put it in a data frame&#xA;g &amp;lt;- as.numeric(str_split('_0___0___0_____5___10___18____22___28___36___44___51____60____70___', '_{1,5}')[[1]]); g &amp;lt;- g[!is.na(g)]&#xA;t &amp;lt;- as.numeric(str_split('0___5___10___15___20___25___30___35___40___45___50___55___60 ____', '___')[[1]]); t &amp;lt;- t[!is.na(t)]&#xA;&#xA;dat &amp;lt;- data.frame(g = g, t = t)&#xA;&#xA;# scale the data to avoid numeric problems taking the exponential of large numbers &#xA;dat$g.scaled &amp;lt;- dat$g/max(dat$g)&#xA;dat$t.scaled &amp;lt;- dat$t/max(dat$t)&#xA;&#xA;#compare to fitting result by neglecting first three zero points and taking logarithm&#xA;dat$ln.g[4:13] &amp;lt;- log(dat$g)&#xA;lm(formula = ln.g ~ t, data = dat)&#xA;&#xA;# show data&#xA;dat&#xA;        g  t   g.scaled   t.scaled&#xA;1   0  0 0.00000000 0.00000000&#xA;2   0  5 0.00000000 0.08333333&#xA;3   0 10 0.00000000 0.16666667&#xA;4   5 15 0.07142857 0.25000000&#xA;5  10 20 0.14285714 0.33333333&#xA;6  18 25 0.25714286 0.41666667&#xA;7  22 30 0.31428571 0.50000000&#xA;8  28 35 0.40000000 0.58333333&#xA;9  36 40 0.51428571 0.66666667&#xA;10 44 45 0.62857143 0.75000000&#xA;11 51 50 0.72857143 0.83333333&#xA;12 60 55 0.85714286 0.91666667&#xA;13 70 60 1.00000000 1.00000000&#xA;&#xA;#do direct nonlinear fitting&#xA;fit &amp;lt;- nls(g.scaled ~ A0*exp(k*t.scaled)+offset, &#xA;           data = dat, start = list(A0 = 1, k = 1, offset = 1))&#xA;&#xA;#do logspace fitting to linear equation by transformation and neglecting first three data points&#xA;dat$ln.g[4:13] &amp;lt;- log(dat$g[4:13])&#xA;logfit &amp;lt;- lm(ln.g ~ t, data = dat)&#xA;&#xA;#compare predicted values from fit to real data&#xA;dat$pred.scaled &amp;lt;- predict(fit)&#xA;dat$pred &amp;lt;- dat$pred.scaled * max(g)&#xA;dat$logpred[4:13] &amp;lt;- exp(predict(logfit))&#xA;require(ggplot2)&#xA;ggplot(data = dat, aes(x = t)) + geom_point(aes(y = g)) + geom_line(aes(y=pred), color = 'blue') + geom_line(aes(y=logpred), color = 'red')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jvPZB.png&quot; alt=&quot;Plot produced by ggplot command&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue curve is the direct nonlinear fit and the red curve is the linear fit of log-transformed data (dropping the first three data points).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obtained rate constant is quite different in the two cases!  For the direct nonlinear fit, the rate constant is $1.4 \text{ hr}^{-1}$, and for the log-transformed data with linear fit, the rate constant is $0.053 \text{ min}^{-1}$, or $3.2 \text{ hr}^{-1}$!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-09T19:49:29.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24867" PostTypeId="1" AcceptedAnswerId="24868" CreationDate="2015-02-09T20:20:31.470" Score="13" ViewCount="30234" Body="&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;19&quot;&gt;&#xA;  &lt;li&gt;Which of the following orbitals are degenerate in the hydrogen atom with $n = 3$?&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/PbJo4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PbJo4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;  &lt;strong&gt;A&lt;/strong&gt;. II and III only;&lt;br&gt;&#xA;  &lt;strong&gt;B&lt;/strong&gt;. I and IV only;&lt;br&gt;&#xA;  &lt;strong&gt;C&lt;/strong&gt;. I, II, and IV only;&lt;br&gt;&#xA;  &lt;strong&gt;D&lt;/strong&gt;. II, III, and IV only;&lt;br&gt;&#xA;  &lt;strong&gt;E&lt;/strong&gt;. all.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer says its &lt;strong&gt;E&lt;/strong&gt;. All of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all isn't there only 1 electron in hydrogen? And how could the $\mathrm{s}$ orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?&lt;/p&gt;&#xA;" OwnerUserId="13310" LastEditorUserId="41328" LastEditDate="2017-08-04T07:55:24.140" LastActivityDate="2017-08-04T07:55:24.140" Title="Which orbitals of the hydrogen atom are degenerate for n=3?" Tags="&lt;orbitals&gt;&lt;hydrogen&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24868" PostTypeId="2" ParentId="24867" CreationDate="2015-02-09T20:32:23.910" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First of all isn't there only 1 electron in hydrogen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;degenerate&quot; means having the same energy.  &quot;Degenerate&quot; refers to a set of orbitals.  It doesn't make sense to say one orbital is degenerate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the non-relativistic Schrodinger equation, all the orbitals for a given &quot;n&quot; are degenerate.  Energy only depends upon n. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-09-01T15:02:50.687" LastActivityDate="2016-09-01T15:02:50.687" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="24869" PostTypeId="2" ParentId="24846" CreationDate="2015-02-09T20:57:57.090" Score="11" Body="&lt;p&gt;Coagulation is a process which involves coming together of colloidal particles so as to change into large sized particles which ultimately settle as a precipitate or float on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coagulation is generally brought about by the addition of electrolytes. When an electrolyte is added to a colloidal solution, the particles of the sol take up the ions which are oppositely charged and thus get neutralized. The neutral particles then start accumulating to form particles of a larger size which settle down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity of the electrolyte which is required to coagulate a definite amount of a colloidal solution depends upon the valency of the ion having a charge opposite to that of the colloidal particles. This observation of Hardy and Schulze are known as Hardy Schulze law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be defined as:&#xA;Greater is the valency of the oppositely charged ion of the electrolyte being added, the faster is the coagulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind this is the attractive electrostatic forces between ions of opposite charges. The greater the charge of the ions, the greater is  the attractive electrostatic forces between ions:&#xA;$f=-\frac{qq'}{4\pi\epsilon_0r^2}$, the faster is the coagulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for the coagulation of negatively charged arsenious sulphide sol., trivalent cations are far more effective than divalent cations which in turn are more effective than monovalent cations. Similarly for coagulation of positively charged ferric hydroxide sol, tetravalent anions are more effective than trivalent anions which are more effective than divalent anions which in turn are more effective than monovalent anions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum amount of an electrolyte that must be added to one litre of a colloidal solution so as to bring about complete coagulation or flocculation is called the coagulation or flocculation value of the electrolyte. Thus smaller is the flocculation value of an electrolyte; greater is its coagulating or precipitating power.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-09T20:57:57.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24870" PostTypeId="1" AcceptedAnswerId="24871" CreationDate="2015-02-09T21:05:52.113" Score="0" ViewCount="762" Body="&lt;p&gt;&lt;em&gt;Semi-related to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24422/why-do-some-salt-ions-in-solution-conduct-electricity-better-than-others&quot;&gt;Why do some salt ions in solution conduct electricity better than others?&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After conducting more experiments, using deionised water as an analyte only, I found the potential difference decreases as more silver nitrate is added. This goes against what I was lead to believe that it is actually disruption in forms of &quot;pure&quot; water that actually conduct electricity (i.e. pure water should not technically conduct electricity at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vPQ6t.png&quot; alt=&quot;Potential Difference Graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph shows how the potential difference drops, from my findings. The last data points plotted are at 1cm3 of silver nitrate added; the interval is 0.05cm3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why is it that I see a massive drop in potential when even the slightest (0.50cm3) amount of silver nitrate is added?&lt;/p&gt;&#xA;" OwnerUserId="13101" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-02-10T01:29:57.697" Title="Why does the potential difference of a current drop, when silver nitrate is added to deionised water?" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;solutions&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24871" PostTypeId="2" ParentId="24870" CreationDate="2015-02-10T01:29:57.697" Score="1" Body="&lt;p&gt;The potential of an electrode in the absence or paucity of redox reactions at the interface is not predictable using the Nernst equation. This situation is known as an unpoised electrode and the electrode's measured potential is related to the formation of an electrical double layer at the interface, which behaves very much like a capacitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to measure the potential of an electrode that won't easily oxidize or reduce in deionized water, due to the very low concentration of ions, establishing the double layer is slow and easily disrupted. You can think of it as having a large solution resistance in series with the double layer capacitance. When there are few ions, the resistance is very high. On that same note, how good the voltage measurement instrument is becomes important. Any current that a voltmeter draws will induce a large voltage at the electrode (as there is a large resistance to bringing ions to the interface, excess charge at the electrode is not easily balanced)—a voltmeter with a 10 nA input bias current may induce 100–200 mV of error in a very resistive solution, so it takes very good instruments to deal with these types of measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This error is proportional to the solution resistance and even adding a tiny amount of ions to deionized water will dramatically reduce the resistance. I'm guessing that repeating the experiment with a supporting electrolyte will prevent this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other trouble here is that neither of your electrodes can be assumed to maintain a fixed potential in your solution. The potential of the silver electrode should vary according to the Nernst equation once you have an appreciable quantity of silver ions in solution, but the graphite electrode is not in equilibrium with any components of your solution and its potential is free to fluctuate as the solution composition changes. Without a supporting electrolyte, adding the silver nitrate may have a large effect on the potential of the graphite electrode.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-10T01:29:57.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24872" PostTypeId="1" CreationDate="2015-02-10T01:48:29.890" Score="1" ViewCount="37" Body="&lt;p&gt;I'm new to the subject of carbon nanotubes; However, I recently studied a little about them. But I failed to understand the gas phase functionalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, this technique is used to improve the solubility of the nanotubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I failed to understand the chemistry and physics of these processes so resumidad, considering that there have been many experiments in this field, but I failed to understand the basis of this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone could explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: What causes that the nanotubes are hydrophobic?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2015-02-10T01:48:29.890" Title="Gas-Phase Functionalization of Carbon Nanotubes" Tags="&lt;carbon-allotropes&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24873" PostTypeId="1" CreationDate="2015-02-10T01:54:09.320" Score="9" ViewCount="3640" Body="&lt;p&gt;Consider tetraaminecopper (II) ion. $\ce{[Cu(NH3)4]^2+}$ or $\ce{[Cu(NH3)4(H2O)2]^2+}$.&#xA;The copper(II) ion has the electron configuration $\mathrm{[Ar]\ 3d^9}$. How do the 4 electron pairs from the ammonia ligands form the coordination bond with the copper ion, if the 3d subshell is almost entirely occupied?&lt;/p&gt;&#xA;" OwnerUserId="9839" LastEditorUserId="7475" LastEditDate="2016-11-23T01:20:00.120" LastActivityDate="2016-11-23T01:20:00.120" Title="Why do copper (II) complexes contain so many valence electrons?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;orbitals&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24875" PostTypeId="1" AcceptedAnswerId="24876" CreationDate="2015-02-10T02:59:32.363" Score="12" ViewCount="3221" Body="&lt;p&gt;The isotope of hydrogen, tritium, is indeed radioactive resulting in $\ce{^3He}$ through beta decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what of the radioactively 'stable' isotope deuterium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are all deuterium atoms destined to be deuterium for eternity or is there any process other than radioactive decay that could lead to transition such as say $\ce{^1H}$?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="7475" LastEditDate="2015-10-12T12:42:44.053" LastActivityDate="2019-05-09T13:37:48.557" Title="On the stability of deuterium" Tags="&lt;hydrogen&gt;&lt;isotope&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24876" PostTypeId="2" ParentId="24875" CreationDate="2015-02-10T03:13:04.737" Score="12" Body="&lt;p&gt;No, deuterium is completely stable. I found the answer at &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/deuteron.html&quot; rel=&quot;noreferrer&quot;&gt;Hyperphysics&lt;/a&gt;, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decay of deuterium would be &#xA;$$\ce{D -&amp;gt; P + N + e + \bar{\nu}}_e$$&#xA;where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, for some reason, deuterium &lt;em&gt;did&lt;/em&gt; decay, it could go to &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_helium#Helium-2_.28diproton.29&quot; rel=&quot;noreferrer&quot;&gt;$\ce{^2He}$&lt;/a&gt; … which would then decay back to deuterium.&lt;/p&gt;&#xA;" OwnerUserId="13294" LastEditorUserId="7475" LastEditDate="2015-10-12T12:44:27.617" LastActivityDate="2015-10-12T12:44:27.617" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="24877" PostTypeId="1" CreationDate="2015-02-10T05:10:09.593" Score="4" ViewCount="4209" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ej9E5.jpg&quot; alt=&quot;Structure of discodermolide&quot;&gt;&lt;br&gt;&#xA;It is generally said that carbon-carbon double bonds cannot behave as stereogenic centres but the above molecule contains three such centres. Can somebody please point out all the stereogenic centres (including the double bonded carbons) and explain how and why this is so?&lt;/p&gt;&#xA;" OwnerUserId="13313" LastEditorUserId="4945" LastEditDate="2015-02-10T07:31:27.367" LastActivityDate="2015-02-10T17:20:04.983" Title="How many stereogenic centres are there in discodermolide?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-02-11T01:23:00.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="24878" PostTypeId="2" ParentId="24723" CreationDate="2015-02-10T05:58:51.157" Score="8" Body="&lt;p&gt;Trying to learn chemistry by memorizing the periodic table is pretty superficial.  I would recommend learning chemistry by doing chemical reactions.  Make plastics, mix acids and bases (correctly please), ask your self why milk curdles, try to light water on fire and then ask yourself why it doesn't work even though it is made of hydrogen and oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at calcite and notice that it splits text on a page into two.  Why is ruby red?  Why is the sky blue?  Can you make blue smoke the same way?  Everything that grabs your attention is valid, and pay enough attention to safety so that you don't blow up/poison yourself before you know enough chemistry to do things safely naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curiosity is the only way to learn chemistry, so enjoy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day you will wake up and realize you know most of the periodic table.  Then you will probably memorize the rest, or the useful pieces because you can't resist it, and that is totally fine.  As long as it doesn't take you away from melting plastic with acetone (under a fume hood.)&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorUserId="19481" LastEditDate="2015-08-28T02:00:29.483" LastActivityDate="2015-08-28T02:00:29.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24880" PostTypeId="2" ParentId="24843" CreationDate="2015-02-10T06:50:37.063" Score="4" Body="&lt;p&gt;I really do not see an easy and straight forward way of determining the dipole moment's direction and quantity. It is very much dependent on how the molecule is currently shaped, i.e. rotations of the alkyl chain and also the electron cloud. Of course you could go ahead and calculate the dipole moment from tabulated values of certain bonds (&lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS-10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;for example&lt;/a&gt;), but this can be really tedious and it might now be always right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably find tabulated values for most common compounds online. Try &lt;a href=&quot;http://cccbdb.nist.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://cccbdb.nist.gov/&lt;/a&gt; or maybe just google will do. If you are lucky enough to have access, the handbook of chemistry and physics has a large number of values. There is an &lt;a href=&quot;http://www.hbcpnetbase.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;online verison&lt;/a&gt;, but you can often find it in libraries, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the meantime, I will try to illustrate something. In the series $\ce{R-NH2}$ the simplest one is $\ce{R~=~H}$, ammonia. Here the dipole moment is straight forward isn't it? I believe it is a little more complicated than that. Unfortunately you cannot just take the charges of the atoms in the molecule, since they are not point charges. Have a look at the total electron density, you will notice, that the isosurface is more extended on top of the Nitrogen, which is expectable, since it is the area where the lone pair is supposed to be. This also coincides with the highest occupied molecular orbital (HOMO). In this special case, the dipole moment is almost a direct correlation of that. (If you would add up all dipole moments of all occupied molecular orbitals, you will also obtain the total dipole moment.)&lt;br&gt;&#xA;Below you see an isosurface of the total electron density in blue on the left, the isosurface of the homo in the middle and natural charges with the resulting dipole moment on the right. The values are taken from a DF-BP86/def2-TZVPP calculation with Gaussian09.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rtmz3.jpg&quot; alt=&quot;electron density, homo, dipole and charges of ammonia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can learn from this is, that the lone pair has a huge effect on the dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go a bit further and look at $\ce{R~=~CH3}$, methyl amine. It can be expected, that the dipole moment is somewhat smaller than in ammonia, due to the methyl group. What you can also observe, that it is dependent on how the methyl group is rotated. Of course in room temperature, the group is freely rotatable and it will result in a mean (pulsating) value for the dipole moment. However, what you can observe here is, that the direction of the dipole moment is aligned with the direction of the lone pair. As you go further down the series this generally will hold, However, the magnitude will be decreasing since your &quot;quite unpolar&quot; moiety is becoming bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OdQJb.jpg&quot; alt=&quot;homos and charge/dipole moment for methyl amine in staggered and eclipsed conformation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ethyl amine, $\ce{R~=~C2H5}$, the trend holds. But to not extend this answer too much, I only pick one representative amongst all possible conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Qxdf.jpg&quot; alt=&quot;homos and charge/dipole moment for ethyl amine, all staggered&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for propyl amine, $\ce{R~=~(CH2)2CH3}$, we can observe it analogously, with decreasing magnitude of the dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gX3pg.jpg&quot; alt=&quot;homos and charge/dipole moment for propyl amine, all staggered&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the last example I have chosen heptanyl amine, $\ce{R~=~(CH2)6CH3}$, to example the trend. I removed some of the charges to not overcrowd the image. If you have a close look at the HOMO, you will notice, that it only significantly extends to the first carbon. The situation for any aliphatic amine can be expected to be very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i11tH.jpg&quot; alt=&quot;homos and charge/dipole moment for heptanyl amine, all staggered&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have no idea, why the vector for the dipole got bigger and bigger, while the magnitude was decreasing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt;&lt;br&gt;&#xA;The dipole moment in aliphatic amines is about 1 Debye and the direction is through the nitrogen directed to the lone pair.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-02-10T06:50:37.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24881" PostTypeId="1" CreationDate="2015-02-10T07:42:18.787" Score="3" ViewCount="1419" Body="&lt;p&gt;This is a very basic question about solubility, especially solution of table salt in water and solution of sugar in water. I have heard about terms like hard water, soft water, filtering of water to remove impurities etc., but I don't understand these well enough. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13317" LastEditorUserId="21964" LastEditDate="2016-07-26T20:28:58.497" LastActivityDate="2016-07-26T20:28:58.497" Title="What is hard and soft water and how does the process of dissolution work?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;solubility&gt;&lt;water-treatment&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24882" PostTypeId="2" ParentId="24881" CreationDate="2015-02-10T08:56:19.877" Score="1" Body="&lt;p&gt;Note that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chemical reaction is a process that leads to the transformation of one set of chemical substances to another.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Salt when put into water, dissolves into ions as $\ce{Na+}$ and $\ce{Cl-}$.The positive and negative part of the ions are attracted by the oppositely charged water molecules. The ions are stable and do not undergo further reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question, external impurities can be removed by filtering if they are of considerable size. Dissolved impurities like phosphates, sulphates cannot be removed by a cloth. The hard water has to be then 'treated' to make it soft again.&lt;/p&gt;&#xA;" OwnerUserId="13216" LastActivityDate="2015-02-10T08:56:19.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24883" PostTypeId="1" CreationDate="2015-02-10T09:37:58.450" Score="3" ViewCount="3575" Body="&lt;p&gt;Whenever a galvanic cell is explained in textbooks/videos, it usually starts by showing that if you put a piece of $\ce{Zn}$ in a $\ce{Cu^{2+}}$ ion solution, the $\ce{Zn}$ will become oxidized and transfer $\ce{2e-}$ to the $\ce{Cu^{2+}}$ ion, so the $\ce{Cu^{2+}}$ becomes a neutral $\ce{Cu}$ metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this, but in a galvanic cell, the $\ce{Zn}$ never actually touches the $\ce{Cu^{2+}}$ ions, the $\ce{Zn}$ and $\ce{Cu}$ are only connected with wires. I don't understand how the $\ce{Zn}$ can be oxidized and transfer ions through the current. &lt;/p&gt;&#xA;" OwnerUserId="13318" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-03-31T22:28:19.733" LastActivityDate="2017-12-19T07:19:52.097" Title="How does zinc become oxidized when it's not in direct contact with copper ions in a galvanic cell?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24884" PostTypeId="2" ParentId="24883" CreationDate="2015-02-10T10:07:07.780" Score="3" Body="&lt;p&gt;Zinc electrode is anode and copper electrode is cathode. The two electrodes are connected by a electrically conducting wire. At anode, $\ce{Zn}$ atoms are oxidised to $\ce{Zn^2+}$ which go into the solution, and the electrons are left on the metal strip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrons produced in the oxidation half-reaction at anode are transferred through the wire to $\ce{Cu}$ electrode where they are consumed. The $\ce{Cu^2+}$ ions are reduced to metallic $\ce{Cu}$ which coats on the $\ce{Cu}$ strip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, oxidation and reduction half reactions occur at separate electrodes and the current flows through the wire. A salt bridge maintains neutrality in both the compartments by a flow of ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall cell reaction can be given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn (s) + Cu^2+ -&amp;gt; Zn^2+ + Cu (s)}$$&lt;/p&gt;&#xA;" OwnerUserId="13216" LastEditorUserId="41328" LastEditDate="2017-12-19T07:19:52.097" LastActivityDate="2017-12-19T07:19:52.097" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="24885" PostTypeId="1" CreationDate="2015-02-10T11:08:41.687" Score="6" ViewCount="1864" Body="&lt;p&gt;How can one mimic a volcano eruption using household chemicals &lt;strong&gt;safely&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've tried it in a testing tube with baking soda and vinegar. This wasn't quite as impressive as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the web we've found recipes that add food coloring (red) and dish washing liquid to the mix. I guess the latter is for some bubbling effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A volcano made from paper-mâché (classic volcano cone form, perhaps a bit steeper than natural)&lt;/li&gt;&#xA;&lt;li&gt;Center hole big enough to fit a 1.5 liter water bottle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: this is supposed to be done with kids, so please don't propose any method that would generate poisonous fumes or similar.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meanwhile tried a soda ($\ce{NaCO_3}$, i.e. not baking soda $\ce{NaHCO_3}$) solution poured into a dish washing solution, then adding a citric acid solution. It was pretty good, but a bit too quick. Vinegar seems to work better than citric acid, too. Also I used warm water to dissolve the reagents.&lt;/p&gt;&#xA;" OwnerUserId="13319" LastEditorDisplayName="user15489" LastEditDate="2015-07-05T11:26:03.220" LastActivityDate="2015-07-05T11:26:03.220" Title="Mimic a volcano eruption with household chemicals?" Tags="&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24888" PostTypeId="1" CreationDate="2015-02-10T13:47:03.507" Score="1" ViewCount="386" Body="&lt;p&gt;How is the MO diagram drawn for a tetrahedral complex with heteroligands (e.g. $\ce{[NiCl2(PPh3)2]}$) different from a tetrahedral complex with the same type of ligand? Do we consider the same &quot;a1 and t2&quot; SALCs as in the case of same ligand type tetrahedral? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="7448" LastEditDate="2015-02-10T16:27:30.707" LastActivityDate="2015-02-10T16:46:06.017" Title="MO scheme for coordination complexes (tetrahedral) - Heteroligand vs. homoligand" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24891" PostTypeId="2" ParentId="24873" CreationDate="2015-02-10T14:55:12.110" Score="8" Body="&lt;p&gt;I'll address $\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\ce{[Cu(NH3)4]++}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 9 d-electrons in the complex are in non-bonding and antibonding orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See pages 6-9 of &lt;a href=&quot;http://www.chem.mun.ca/homes/cmkhome/Chemistry%203211%20Pt3.pdf&quot;&gt;Ligand Field and Molecular Orbital Theory&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t_{2g}$ set and the higher energy, antibonding $e_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t_{2g}$ orbitals and the antibonding $e_g$ orbitals, which are mainly metal in character anyway. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-10T15:07:37.640" LastActivityDate="2015-02-10T15:07:37.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24892" PostTypeId="2" ParentId="24828" CreationDate="2015-02-10T15:57:04.753" Score="2" Body="&lt;p&gt;Obviously it's not true that thionyl chloride only works with sp3 substrates. As you say, a carboxylic acid, an sp2 hybridized substrate, can be chlorinated with thionyl chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the difference is that the carbon of an enol is not readily attacked by a nucleophile, while the carbons in the other substrates can be. Each of the reactions go through the same first step(s): addition of the hydroxyl to sulfur with chloride leaving sulfur (likely through addition/elimination, not shown below). However the next steps are different depending on the functional group of the substrate. For an alcohol, chloride attacks the activated alcohol (looks very much like a sulfonate) in an SN2 reaction to give the alkyl chloride. For a carboxylic acid, chloride attacks the carbonyl carbon, and after addition/elimination (not shown in detail), the acid chloride is produced. Starting from an enol, reaction with thionyl chloride gives an enol chloro sulfate. Unlike the sp2-carboxylic acid case, the pi bond doesn't have an effective electron sink (such as oxygen), which means that the reaction would have to be a direct SN2 displacement. As I'm sure you learned, SN2 at sp2 carbons is very slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZhmdH.jpg&quot; alt=&quot;Chlorination mechanisms&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A SciFinder search shows that other reagents, notably phosphorus pentachloride can accomplish the transformation.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-02-10T15:57:04.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24893" PostTypeId="2" ParentId="24833" CreationDate="2015-02-10T16:08:45.893" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Suppose I pour some sodium chlorine into water. So, what happens is that&lt;/p&gt;&#xA;&lt;p&gt;NaCl&lt;sub&gt;(s)&lt;/sub&gt; → Na+&lt;sub&gt;(aq)&lt;/sub&gt; + Cl-&lt;sub&gt;(aq)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;meaning that the ionic bond between Na and Cl breaks up (correct this far?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That would be completely correct in an infinitely dilute solution of NaCl.&lt;/p&gt;&#xA;&lt;p&gt;However, in real solutions of NaCl there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Ion-association&quot; rel=&quot;nofollow noreferrer&quot;&gt;ion pairs&lt;/a&gt; in addition to the separately solvated ions.&lt;/p&gt;&#xA;&lt;p&gt;There is an equilibrium between the aqueous ion pairs and the aqueous seperately solvated ions:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \ce{NaCl(aq) &amp;lt;=&amp;gt; Na+(aq) + Cl-(aq)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For a detailed consideration see &lt;a href=&quot;http://www.jh-inst.cas.cz/%7Erheyrovs/jes1996.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physical Electrochemistry of Strong Electrolytes Based on Partial Dissociation and Hydration: Quantitative Interpretation of the Thermodynamic Properties of NaCl(aq) from &amp;quot;Zero to Saturation&amp;quot;&lt;/a&gt; Journal of Electrochemical Society, Vol. 143, No. 6, June 1996, pp. 1789 - 1793. (no paywall link).&lt;/p&gt;&#xA;&lt;p&gt;In summary, in a relatively concentrated solution of NaCl, there will be about 20-30% ion pairs with the remainder being seperately solvated ions.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If I feed electrons some how into the solution, will chlorine gas start forming?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You need to remove an electron from each of two Cl- to get chlorine gas, but yes you can form chlorine gas through electrolysis of NaCl solution.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Also, could I use this so that I pour NaCl into water and get Na and Cl ions, and then (since they are separate) add something more to create Na[something] or Cl[something]? Something that would not form if I just mixed NaCl + .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You could added silver nitrate solution to an NaCl solution and get AgCl precipitate.&lt;/p&gt;&#xA;&lt;p&gt;For a video of the reaction see this University of Illinois site: &lt;a href=&quot;http://www.chem.uiuc.edu/clcwebsite/AgCI.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chem.uiuc.edu/clcwebsite/AgCI.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-10T16:08:45.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24896" PostTypeId="2" ParentId="24881" CreationDate="2015-02-10T16:39:07.273" Score="3" Body="&lt;p&gt;Hard water and soft water differ only in the presence or absence of certain impurities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_water&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Hard water:&lt;/strong&gt;&lt;/a&gt; Water with impurities in the form of minerals containing calcium and magnesium added in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Water_softening&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Soft water:&lt;/strong&gt;&lt;/a&gt; Water with these minerals removed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One thing to note is that hard water is &lt;em&gt;not&lt;/em&gt; the slightest bit dangerous. Studies have been conducted by the &lt;a href=&quot;http://www.who.int/water_sanitation_health/dwq/chemicals/en/hardness.pdf&quot; rel=&quot;nofollow&quot;&gt;World Health Organization&lt;/a&gt; and the National Academy of Sciences, and no evidence was found linking hard water to and adverse health effects - or any effects at all!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By mere filtering using a cloth or a machine, how can we get rid of minute impurities, (which make water hard) which are not even visible to naked eyes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is impossible. Many devices have been proposed, but none have been shown to work. Quoting from &lt;a href=&quot;http://ianrpubs.unl.edu/live/g1274/build/g1274.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are a variety of devices that claim to manage hard water scale using primarily magnetic or electrical technology. Manufacturers generally claim the devices utilize energy to alter the behavior of compounds or elements within the water. They do not claim water chemistry is altered. In fact, the hardness of the water before and after treatment is not changed. This complicates assessment of their performance. While protocol has been established to assess the effectiveness of ion exchange water softeners (NSF/ANSI 44), at this time, no recognized agency in the U.S. has established protocol to assess the effectiveness of physical water treatment devices. Therefore, questions remain as to their effectiveness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemicals are the only way to go, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information on hard water can be found &lt;a href=&quot;http://water.usgs.gov/owq/hardness-alkalinity.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://chemistry.about.com/cs/howthingswork/a/aa082403a.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If table salt is dissolved in water, why do we call it a solution? Don't say that a chemical reaction has taken place and some new product say $\ce{NaH2OCl}$ etc. is formed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case you're interested, the reaction is&#xA;$$\ce{NaCl + H2O \to Na+ + Cl- +H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It fits the very definition of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Solution&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a homogeneous mixture composed of only one phase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl}$ is separated into ions that separate uniformly throughout the water.&lt;/p&gt;&#xA;" OwnerUserId="13294" LastEditorUserId="13294" LastEditDate="2015-02-12T00:46:05.513" LastActivityDate="2015-02-12T00:46:05.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24897" PostTypeId="1" CreationDate="2015-02-10T16:41:07.533" Score="-9" ViewCount="835" Body="&lt;p&gt;Can we know the reaction between any two elements in the periodic table? If yes then can we know the reaction between any three or more elements in the periodic table?&lt;/p&gt;&#xA;" OwnerUserId="13325" LastEditorUserId="7448" LastEditDate="2015-02-10T17:00:21.503" LastActivityDate="2015-02-10T17:01:25.883" Title="Reaction between elements in a periodic table" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-10T17:57:26.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="24898" PostTypeId="2" ParentId="24888" CreationDate="2015-02-10T16:46:06.017" Score="1" Body="&lt;p&gt;Bis(triphenylphosphine)nickel chloride can have be tetrahedral or square planar depending on conditions. In any case the SALCs for a tetrahedral complex with heteroligands will depend on the point group of the complex. For [NiCl2(PPh3)2], the point gorup is C2v and not Td like for same ligand tetrahedral. You need to check the character tables &lt;/p&gt;&#xA;" OwnerUserId="13086" LastActivityDate="2015-02-10T16:46:06.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24899" PostTypeId="2" ParentId="24897" CreationDate="2015-02-10T17:01:25.883" Score="4" Body="&lt;p&gt;We don't even know completely the reaction of one element with itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider carbon.  There are constantly new forms of carbon being discovered.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-10T17:01:25.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24900" PostTypeId="2" ParentId="24877" CreationDate="2015-02-10T17:09:21.027" Score="4" Body="&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The confusion probably arises because of incompletely standardized definitions of at least some subset of the words &quot;stereocenter&quot;, &quot;stereogenic&quot;, &quot;stereogenic center&quot;, &quot;stereogenic unit&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;IUPAC definitions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The place to turn for authoritative definitions of these terms is IUPAC's Gold Book, as noted by @jerepierre in the comments.  Here is what it has to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;&lt;a href=&quot;http://goldbook.iupac.org/D01679.html&quot; rel=&quot;nofollow&quot;&gt;stereogenic unit&lt;/a&gt; (stereogen/stereoelement)&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;A grouping within a molecular entity that may be considered a focus of stereoisomerism.&lt;/em&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;At least one of these must be present in every enantiomer (though the&#xA;  presence of stereogenic units does not conversely require the&#xA;  corresponding chemical species to be chiral). Three basic types are&#xA;  recognized for molecular entities involving atoms having not more than&#xA;  four substituents:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A grouping of atoms consisting of a central atom and distinguishable&#xA;  ligands, such that the interchange of any two of the substituents&#xA;  leads to a stereoisomer. An asymmetric atom (chirality centre) is&#xA;  the traditional example of this stereogenic unit.&lt;/li&gt;&#xA;  &lt;li&gt;A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation&#xA;  (with a change of sign of the torsion angle) about the central bond&#xA;  leads to a stereoisomer.&lt;/li&gt;&#xA;  &lt;li&gt;A grouping of atoms consisting of a double bond with substituents which give rise to cis-trans isomerism.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, the unequally substituted C=C moieties in discodermolide are &quot;stereogenic units&quot; according to IUPAC.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Stereocenter&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this term is not defined in IUPAC's Gold Book.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Stereogenic center&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this term is not defined in IUPAC's Gold Book.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Wikipedia&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Wikipedia is not as authoritative as IUPAC but it is likely that chem.se readers will encounter wikipedia entries for these terms far more often than IUPAC's in the course of regular web searching.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Stereogenic unit&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;No Wikipedia entry&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Stereogenic center&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Redirects to &quot;stereocenter&quot;, see below.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stereocenter&quot; rel=&quot;nofollow&quot;&gt;Stereocenter&lt;/a&gt;&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A stereocenter or stereogenic center is an *atom* bearing groups such that an interchanging of any two groups leads to a stereoisomer.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This definition does not apply to C=C double bonds, because only an &lt;em&gt;atom&lt;/em&gt; can be a stereocenter, not a group of atoms.  Supporting this interpretation is a complete lack of reference to cis-trans isomerism on the rest of the page.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Usage by chemists&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Discodermolide is referred to in many sources as having 13 &quot;stereocenters&quot; or &quot;stereogenic&quot; centers.  This conforms with the Wikipedia, atom-centric definition of stereocenter.  Here are some example sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Discodermolide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's discodermolide article&lt;/a&gt;: &lt;em&gt;(+)-discodermolide has&#xA;...13 stereogenic centers...&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.scs.illinois.edu/denmark/presentations/2004/gm-2004-07_06.pdf&quot; rel=&quot;nofollow&quot;&gt;A presentation from 2004&lt;/a&gt;:&#xA;&lt;em&gt;molecule possesses: 13 stereocenters&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo034901c&quot; rel=&quot;nofollow&quot;&gt;A paper from 2003&lt;/a&gt;: Paper title: &lt;em&gt;Regulated-Stereoselective Construction of Thirteen&#xA;Stereogenic Centers...&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All these uses comport with both Wikipedia definitions of &quot;stereocenter&quot;/&quot;stereogenic center&quot;.  Chemists do not seem to use the IUPAC term &quot;stereogenic unit&quot; very much in actual practice.   I could not find any relevant pages using both the term &quot;discodermolide&quot; and the phrase &quot;stereogenic unit&quot;, for example.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Bottom line&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It is just possible to pass a thread of consistency through the eye of this needle of confusing terminology.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Discodermolide has thirteen (13) stereogenic &lt;em&gt;centers&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Each of those stereogenic centers is by definition also a stereocenter, and is also a &quot;stereogenic unit&quot;.&lt;/li&gt;&#xA;&lt;li&gt;In addition to those 13 stereocenters, there are three additional &quot;stereogenic units&quot; in discodermolide corresponding to the unequally substituted C=C bonds.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-10T17:20:04.983" LastActivityDate="2015-02-10T17:20:04.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24901" PostTypeId="1" AcceptedAnswerId="24910" CreationDate="2015-02-10T17:45:59.757" Score="14" ViewCount="1065" Body="&lt;p&gt;This question is a follow-up to a recent &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24877/how-many-stereogenic-centres-are-there-in-discodermolide/24900&quot;&gt;question on discodermolide stereochemistry&lt;/a&gt; that &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24877/how-many-stereogenic-centres-are-there-in-discodermolide/24900#24900&quot;&gt;I answered here&lt;/a&gt;.  In the course of reading IUPAC's definition of &quot;stereogenic center&quot;, I wondered if axially chiral molecules such as BINOL have a &quot;stereogenic unit&quot; according to IUPAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of &lt;a href=&quot;http://goldbook.iupac.org/S05980.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC's definition&lt;/a&gt; that may (or may not) apply to BINOL is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation (with a change of sign of the torsion angle) about the central bond leads to a stereoisomer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is the structure of (&lt;em&gt;S&lt;/em&gt;)-BINOL:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPITsb.png&quot; alt=&quot;S-BINOL&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the four-atom chain consisting of the O-bond carbons plus the transannular C-C-bound carbons qualify?  I can't quite work out what is meant by &quot;with a change of sign of the torsion angle&quot; and &quot;stable conformation&quot;.  If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored).  Is that chain no longer a stereogenic unit?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="30836" LastEditDate="2020-04-08T13:37:17.337" LastActivityDate="2020-04-08T13:37:17.337" Title="Do molecules with axial chirality have &quot;stereogenic units&quot;?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24905" PostTypeId="1" AcceptedAnswerId="24923" CreationDate="2015-02-10T18:26:28.377" Score="4" ViewCount="864" Body="&lt;p&gt;There are &lt;a href=&quot;http://faculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch100_OL/aMy_FileLec/04OL_LecNotes_Ch100/05_CompoundBonding/505_IMF_Water/505_pic/descriptionIMF.gif&quot; rel=&quot;nofollow&quot;&gt;various chemical reactions&lt;/a&gt; that can occur between materials.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which chemical reactions occur with the tongue?&lt;/li&gt;&#xA;&lt;li&gt;How come all reactions are reversible (nothing sticks to my tongue)?&lt;/li&gt;&#xA;&lt;li&gt;What's special about the tongue-water interaction that makes it tasteless?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="16683" LastEditDate="2017-04-01T02:06:31.487" LastActivityDate="2017-04-01T02:06:31.487" Title="Which chemical reactions occur between my tongue and the food I eat?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24906" PostTypeId="1" CreationDate="2015-02-10T18:29:12.050" Score="4" ViewCount="869" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The metal gallium melts when held in the hand; its melting point is $\pu{29.76 ^\circ C}$. How much energy as heat is removed from the hand when $\pu{5.00 grams}$ of gallium initially at $\pu{20.0^\circ C}$ melts? The value of $\Delta H_{fusion}$ is $5.576~\mathrm{kJ~mol^{-1}}$ and the specific heat of gallium is $0.374~\mathrm{J~g^{-1}K^{-1}}$. Take the final temperature to be $\pu{29.76 ^\circ C}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I got: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = (\pu{5.00 g})(0.374~\mathrm{J~g^{-1}K^{-1}}) (\pu{9.76^\circ C})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q=\pu{18.3 J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I reason that since $\pu{18.3 J}$ of energy in the form of heat is required to melt gallium from at an initial temperature of $\pu{20.0 ^\circ C}$ to a final of $\pu{29.76 ^\circ C}$ then this is all the energy required to do this but why did they give $\Delta H_\text{fusion}$?&lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="35555" LastEditDate="2018-02-23T22:48:59.843" LastActivityDate="2018-02-23T22:48:59.843" Title="Why did they provide the value for the enthalpy change of fusion?" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24907" PostTypeId="2" ParentId="24906" CreationDate="2015-02-10T18:39:26.393" Score="7" Body="&lt;p&gt;You calculated the energy required to change solid gallium at 20&amp;deg;C to solid gallium at 29.76&amp;deg;C.  Once solid gallium reaches 29.76&amp;deg;C, additional energy is required to melt it.  The energy required is given by the $\Delta H_{\text{fusion}}$.  So to complete the question, transform the $\Delta H$ units to a grams basis instead of a mole basis, multiply it by the weight of gallium in the problem, and then add that result to the one you already calculated.  That will be the energy required to change solid gallium at 20&amp;deg;C to &lt;em&gt;liquid&lt;/em&gt; gallium at 29.76&amp;deg;C.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2018-02-23T22:46:56.137" LastActivityDate="2018-02-23T22:46:56.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24908" PostTypeId="2" ParentId="24885" CreationDate="2015-02-10T18:48:50.460" Score="6" Body="&lt;p&gt;The 'elephant's toothpaste' experiment with hydrogen peroxide decomposition works well to produce large amounts of foam and might be suitable for what you are looking for.&#xA;Low concentration hydrogen peroxide solutions are readily available and not dangerous provided you don't get it in your eyes or drink it or anything silly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill the bottle with some hydrogen peroxide solution, add a few drops of orange food colouring and some dish washing liquid and then add something to catalyse the decomposition. You could use something like potassium iodide or manganese dioxide to give you a very rapid decomposition but if you want a slightly less violent and more sustained reaction then apparently dried yeast dissolved in warm water works quite well (yeast contains good amounts of catalase, an enzyme which catalyses the decomposition of hydrogen peroxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having never done this experiment, I can't tell you exactly what quantities to use so you should probably test it before doing it as a demonstration. However, there are some descriptions of the experiment on the internet to give you a good starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencebob.com/experiments/toothpaste.php&quot;&gt;http://www.sciencebob.com/experiments/toothpaste.php&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste&quot;&gt;http://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-10T18:48:50.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24910" PostTypeId="2" ParentId="24901" CreationDate="2015-02-10T19:36:38.863" Score="8" Body="&lt;p&gt;Yes, axially chiral compounds have a stereogenic unit. Biphenyls (such as BINOL) and allenes are common types of compounds that exhibit axial chirality. As you describe, for BINOL the stereogenic unit is the set of atoms including the carbons joining the aryl groups and one carbon further from the biaryl junction. IUPAC's description is simply meant to define a torsional angle, or thought of another way, define two planes around an axis.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it does qualify, suppose temperature is increased to the point where spontaneous racemization happens (and suppose that any side reactions or decomposition can be ignored). Is that chain no longer a stereogenic unit?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a great question that forces us to think about what we mean by 'stable' and under what conditions is something stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference to 'stable conformation' means that the two conformations cannot interconvert at ambient temperature. In BINOL, there is restricted rotation around the biaryl bond, which means that the two stereoisomers cannot interconvert under ambient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think though about a much simpler compound with four carbons in a row that can be used to define a torsional angle: butane. The gauche conformation actually meets the definition of chiral: it's mirror image is not superimposable on itself. However we don't consider butane to have a stereogenic unit (or consider butane to be a chiral compound) because those two conformations can interconvert readily at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YV9vc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So those are two ends of the spectrum if we consider room temperature: BINOL has a stereogenic element but butane does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we heat up (S)-BINOL to a temperature that allows for rotation around the biaryl bond. As you say, the sample will racemize. At that temperature, the compound no longer has a stereogenic unit because rotation is no longer restricted. Allow the sample to cool, and the stereogenic unit returns, although the sample will be racemic.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-02-10T19:36:38.863" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24911" PostTypeId="1" AcceptedAnswerId="24912" CreationDate="2015-02-10T20:34:26.633" Score="3" ViewCount="609" Body="&lt;p&gt;I am seeing a significant change in specific gravity of products upon addition of $\ce{CaCl2}$ to water. The mass and volume of what goes in doesn't match my total specific gravity of what comes out. I am curious as to why that is the case?&lt;/p&gt;&#xA;" OwnerUserId="13329" LastEditorUserId="95133" LastEditDate="2021-04-12T05:43:51.417" LastActivityDate="2021-04-12T05:43:51.417" Title="Reason for inconsistency in specific gravity of calcium choride added to water" Tags="&lt;inorganic-chemistry&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24912" PostTypeId="2" ParentId="24911" CreationDate="2015-02-10T21:06:41.677" Score="3" Body="&lt;p&gt;When you say, &quot;of products&quot;, do you really mean &quot;of the solution&quot;? If so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solvent mass and the solute mass add up to the mass of the solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But solvent volume and solute volume often &lt;em&gt;don't&lt;/em&gt; add up to the volume of the solution. This is because the forces between molecules in the solution are different from the forces between molecules in the pure solute and the pure solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your solution, the calcium ions and chloride ions attract water molecules around them. The water molecules are packed more tightly than they would be if the ions weren't there. That can increase the density of the solution relative to the total volume of the unmixed solvent and solute. Another effect is that the ions in solution are packed &lt;em&gt;less&lt;/em&gt; tightly than they are in solid $\rm CaCl_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see which effect wins by dumping some calcium chloride into a volumetric flask. Fill the flask up to the mark with water, and then mix until all the calcium chloride dissolves. The solution probably doesn't quite come up to the mark anymore, because the calcium and chloride ions have drawn the water molecules in around themselves, making the solution denser than it would be otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More detail than you may want below:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the solution $V$ is given by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = n_1 \overline{V}_1 + n_1 \overline{V}_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n_1$ and $n_2$ are moles of calcium chloride and moles of water in the solution, and $\overline{V}_1$ and $\overline{V}_2$ are the &lt;a href=&quot;http://www2.stetson.edu/~wgrubbs/datadriven/fchen/bartender/partialmolarvolumechen.html&quot; rel=&quot;nofollow&quot;&gt;partial molar volumes&lt;/a&gt; of the calcium chloride and the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These partial molar volumes are actually derivatives of the volume of the solution with respect to moles of calcium chloride ($\overline{V}_1 = \partial V/\partial n_1$) and with respect to water ($\overline{V}_2 = \partial V/\partial n_2$). They are related to each other (when one rises, the other falls) and they vary with the composition and temperature of the solution in a rather complicated way. &lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-11T12:22:41.647" LastActivityDate="2015-02-11T12:22:41.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24913" PostTypeId="1" CreationDate="2015-02-10T22:12:53.753" Score="2" ViewCount="230" Body="&lt;p&gt;I was wondering what would be the best way to calculate a rough estimate for size of a nitric oxide molecule to use in comparison to a box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking volume would be the best method, but this gave me an answer in dm3/mol which I can't use for comparing against the volume of a box.&lt;/p&gt;&#xA;" OwnerUserId="13330" LastEditorUserId="4945" LastEditDate="2017-04-10T05:56:31.103" LastActivityDate="2017-06-06T14:41:30.533" Title="How to calculate a rough estimate for the volume of a nitric oxide molecule?" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24915" PostTypeId="2" ParentId="24913" CreationDate="2015-02-10T23:04:29.753" Score="1" Body="&lt;p&gt;Take the volume per mole value (b) from a &lt;a href=&quot;http://courses.chem.indiana.edu/c360/documents/vanderwaalsconstants.pdf&quot; rel=&quot;nofollow&quot;&gt;van der waals constants table&lt;/a&gt; (0.0289 L/mol) and divide by $6.022 \times 10^{23}$ molecules per mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you a volume that you can compare to the box.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-11T01:13:24.810" LastActivityDate="2015-02-11T01:13:24.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24917" PostTypeId="1" AcceptedAnswerId="24918" CreationDate="2015-02-10T23:30:28.487" Score="9" ViewCount="42007" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Vinegar generally contains 5% acetic acid. We would expect the pH of vinegar to be approximately:&lt;br&gt;&#xA;  &lt;strong&gt;a&lt;/strong&gt;. 0&lt;br&gt;&#xA;  &lt;strong&gt;b&lt;/strong&gt;. 3&lt;br&gt;&#xA;  &lt;strong&gt;c&lt;/strong&gt;. 7&lt;br&gt;&#xA;  &lt;strong&gt;d&lt;/strong&gt;. 9&lt;br&gt;&#xA;  &lt;strong&gt;e&lt;/strong&gt;. 12&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have the key for this question, so I just want to make sure that I got it right and what I got doesn't seem to be one of the answer choices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer was 1.3, so I rounded it to 1 and there wasn't an answer choice. Here is what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log(.05/1) = -1.3010$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How else would you do it because you were already given the final concentration from the original concentration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The professor said to use this formula for calculating it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{a} = \frac{(.05x)^2}{x}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and got 3, which made no sense, any way this is possible?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2017-09-09T08:42:48.177" LastActivityDate="2017-09-09T08:42:48.177" Title="What's the pH of vinegar containg 5% acetic acid?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="4" CommentCount="12" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24918" PostTypeId="2" ParentId="24917" CreationDate="2015-02-11T00:42:56.340" Score="11" Body="&lt;p&gt;You are not given a concentration, but a percentage value. Assuming its mass percent (wt%), you first have to calculate the actual concentration of the solution like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5% Acetic acid means in a $\pu{1 kg}$ solution, you can find $0.05 \times \pu{1000 g} = \pu{50 g}$ acetic acid. With a molar mass of $\pu{60.05 g/mol}$, this is similar to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{\pu{50 g}}{\pu{60.05 g/mol}} = \pu{0.83 mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration $C = n/V$, therefore is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \frac{\pu{0.83 mol}}{\pu{0.95 L}} = \pu{0.874 mol/L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid is a weak acid. This means it does not dissociate completely and $\mathrm{p}K_\mathrm{a}$ has to be taken into account to calculate the pH. The $\mathrm{p}K_\mathrm{a}$ of acetic acid is about $4.76$. The pH is then defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log C_\mathrm{a}) = 0.5 (4.76 - \log 0.874) = 2.4$$&lt;/p&gt;&#xA;" OwnerUserId="13181" LastEditorUserId="41328" LastEditDate="2017-08-07T07:17:49.450" LastActivityDate="2017-08-07T07:17:49.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24919" PostTypeId="1" AcceptedAnswerId="24932" CreationDate="2015-02-11T01:54:59.333" Score="3" ViewCount="3175" Body="&lt;p&gt;Is there a difference between the Joule coefficient and the Joule–Thomson coefficient? Also, I am having a hard time understanding what they mean.&lt;/p&gt;&#xA;" OwnerUserId="4696" LastEditorUserId="16683" LastEditDate="2020-04-20T18:03:39.133" LastActivityDate="2020-04-20T18:04:25.517" Title="What is the difference between the Joule and Joule–Thomson coefficients?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24920" PostTypeId="2" ParentId="24917" CreationDate="2015-02-11T01:55:04.120" Score="14" Body="&lt;p&gt;Consider the choices. Since it's acetic &lt;em&gt;acid,&lt;/em&gt; you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves only one choice as the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick with a question like this is to realize that it's multiple choice for a reason. It's intended to be a conceptual problem, not a calculation problem. In general, a good rule for multiple-choice questions like this is to first see if you can solve it using general concepts; only if you can't should you reach for you calculator.&lt;/p&gt;&#xA;" OwnerUserId="13334" LastActivityDate="2015-02-11T01:55:04.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24921" PostTypeId="1" CreationDate="2015-02-11T02:05:26.310" Score="1" ViewCount="361" Body="&lt;p&gt;If $4\ \mathrm{mol}$ helium gas expand against constant pressure in isothermal process, how can calculate $W$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it $W=-nRT\ln\frac{V_2}{V_1}$ as defined by isothermal or is it $W=-p\left(V_2-V_1\right)$ as define from isobaric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is enthalpy equal to $0$ in such process?&lt;/p&gt;&#xA;" OwnerUserId="13335" LastEditorDisplayName="user7951" LastEditDate="2016-11-16T21:21:08.200" LastActivityDate="2016-11-16T21:21:08.200" Title="Helium gas expand isobaric isothermal" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24922" PostTypeId="1" CreationDate="2015-02-11T03:47:19.393" Score="2" ViewCount="873" Body="&lt;p&gt;I know that equilibrium potentials are dependent on the ratio of ion concentrations inside and outside of the cell and temperature. I also know that the equilibrium potential is reached when there is no net movement of ions across the membrane, i.e. the chemical and electrical gradients are equal. Why don't the ions move until they reach the same ratio of inside and outside concentration every time? For example, why doesn't ion &quot;x&quot; always diffuse until the ratio of inside to outside concentration is 15 to 1? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="13336" LastActivityDate="2015-02-11T05:20:39.990" Title="Why don't ions have one equilibrium potential? (Nernst equation)" Tags="&lt;electrochemistry&gt;&lt;biochemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24923" PostTypeId="2" ParentId="24905" CreationDate="2015-02-11T04:12:22.230" Score="2" Body="&lt;p&gt;1) I would guess that all of the different interactions are possible except for ionic and covalent because those are much stronger than the rest and it would be difficult for the food molecules to disassociate from the taste receptors to be used again. Enzymes in saliva break down complex food molecules to smaller and simpler ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Started explaining in 1, molecules are not usually permanently bound to the taste receptor in order for the taste receptor to be used again (another molecule being able to bind to it). Some toxins form covalent bonds to receptors, such as curare. I'm not sure what kinds of receptors taste ones are but toxins target different specific receptors. Curare binds to nicotinic acetylcholine receptors, found in muscles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) We don't taste things unless a molecule binds to a taste receptor.&lt;/p&gt;&#xA;" OwnerUserId="13336" LastActivityDate="2015-02-11T04:12:22.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24924" PostTypeId="2" ParentId="24922" CreationDate="2015-02-11T05:20:39.990" Score="1" Body="&lt;p&gt;Good question.  The answer (as I understand it) is that &lt;em&gt;if&lt;/em&gt; there were only a single type of permeable ion, then there would be only one equilibrium ratio, and it would depend on the transmembrane potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in real life, there (a) is usually more than one membrane permeable ion, and more importantly (b) the transmembrane potential is not a fully independent variable but is affected by the transmembrane permeation of all those ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine sovling a single instance of the Nernst equation when you know the electrical potential $\Delta \psi$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\frac{\Delta G}{F}=\text{ion motive force}=0=-m \Delta \psi+ RT\ln{\frac{[X_{out}]}{[X_{in}]}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you know $\Delta \psi$ (and $R$ and $T$) and you are at equilibrium (so $\Delta G$ is 0), then the ion concentration ratio will always go to the same value, as you suggested in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now suppose there are many equilibrating ions, say sodium, potassium, protons, and chloride.  And suppose you don't know $\Delta \psi$.  Now, there is a Nernst equation for each species, as well as Gauss's law that $\nabla \centerdot \bf{E} = \frac{\rho}{\epsilon}$.  And probably also total material balances like $V_{out}[X_{out}] + V_{in}[X_{in}] = N_{tot}$ for each ion.   To fully solve the system to determine equilibrium, all equations have to be solved at once!  I hope you can see that (a) doing this is complicated and (b) if parameters like the volume of the cell $V_{in}$ or even more likely the total amount $N_{tot}$ of ions in the system changes for one ion, it could completely change the ratio $\frac{[X_{out}]}{[X_{in}]}$ for other ions.  That's why in general, the ratio is not the same every time.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-11T05:20:39.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24925" PostTypeId="1" CreationDate="2015-02-11T07:25:41.997" Score="3" ViewCount="1802" Body="&lt;p&gt;I am looking for ways to easily prepare artificial sea water. The chemical composition doesn't have to exactly match that of natural sea water. On-line I can find recipes to prepare sea water. But I was wondering, can I just buy natural sea salt and dissolve it in fresh water (Until I get the required salinity ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commercial shrimp farms in my region, that use natural sea water cause enormous environmental damage. That is the primary motivation behind my interest in artificial sea water. I want to explore if artificial sea water can be a eco-friendly sustainable alternative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intention is not to breed shrimp, but just to grow the shrimp that I get from hatcheries. And yes, I am planning to get the help of people with relevant expertise in this. I just want to gain some prior knowledge and experience before I go for their help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of my present concerns are similar to those of oceanarium though on a much much smaller scale :-) Like them I also want to prepare sea water, which I can control to my exact needs. Locally we get these &quot;raw sea salt&quot; used for industrial purposes, which are claimed to be very similar to natural salts found in sea. They do not contain iodine, caking agents or any other additives etc. I am guessing probably the sea salts that oceanarium imports might be same as these (probably of much better quality).&lt;/p&gt;&#xA;" OwnerUserId="13338" LastEditorUserId="73" LastEditDate="2015-02-13T00:01:33.503" LastActivityDate="2015-02-21T14:21:44.050" Title="Can artificial sea water be prepared by mixing sea salt with water?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24926" PostTypeId="2" ParentId="24723" CreationDate="2015-02-11T07:38:36.127" Score="5" Body="&lt;p&gt;Go to &lt;a href=&quot;http://www.chemie-im-fokus.de/index_fast.htm&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;&#xA;and click on &quot;Sitemap&quot; in the almost left upper corner, then select &quot;Spiele&quot; (which means &quot;Games&quot;) and then &quot;Tetris&quot;. Have fun playing Tetris with the periodic table!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to understand chemistry is not by memorising the periodic table. It is more like the language of chemists, but not its understanding. The key to understand chemistry is logic and the realisation that chemistry is not black and white but an interplay of many factors.&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="7448" LastEditDate="2016-10-07T19:31:26.677" LastActivityDate="2016-10-07T19:31:26.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24928" PostTypeId="1" AcceptedAnswerId="24934" CreationDate="2015-02-11T09:53:13.540" Score="18" ViewCount="36997" Body="&lt;p&gt;What are the oxidation states of the two inequivalent sulfur atoms of the $\ce{S2O3^{2-}}$ molecule? For all I know is that Lewis Structure does matter in most of the cases.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="5026" LastEditDate="2018-06-12T03:31:44.557" LastActivityDate="2018-06-12T04:04:12.977" Title="Oxidation state of the sulfur atoms in the thiosulfate Ion" Tags="&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24930" PostTypeId="1" AcceptedAnswerId="24931" CreationDate="2015-02-11T10:32:33.347" Score="5" ViewCount="4535" Body="&lt;p&gt;I heard people quite often speak about retention time/retention period in Chromatography and Mass Spectrometry. I see somewhere that it is the time duration between the injection time and the target compound peak. I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)What is the purpose of defining that time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)If the compound peak is absent, how one can speak about retention time (I mean if the compound completely fragments)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)Is it has importance in mass spectrometry as well?&lt;/p&gt;&#xA;" OwnerUserId="2498" LastActivityDate="2015-02-11T11:19:03.310" Title="What is 'Retention period' in chromatography and what is its importance?" Tags="&lt;mass-spectrometry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24931" PostTypeId="2" ParentId="24930" CreationDate="2015-02-11T11:19:03.310" Score="5" Body="&lt;p&gt;To answer your three questions in series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The time is a measure of how strong the interaction of a compound is with the column used. The stronger the interaction the longer the compound will stay on the column and this is therefore a way to separate two compounds. For example one molecule with a weak interaction spending a minute on the column, the other spending 5. Collecting fractions of solvent from a liquid chromatography column will yield pure compounds. Note that the time is also a measure for the amount of 'solute' flushed over the column. Many chromatograms also report retention in column volumes of solute. I will not go into detail about that here. It might be worth a separate question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If no peak is found, no retention time can be reported. In chromatography it never (actually almost always) good when a compound fragments/degrades. That means the technique is not suited for purification or analysis. In unique cases, the degradation can be used to analyse whether or not a compound will degrade when interacting with certain materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In mass spectrometry a retention time in the machine is, as far as I am aware a 'Time of Flight'. Compounds are quickly brought into a gaseous and charged state and are pulled through a magnetic field. Depending on the weight of a compound, the time it flies, is shorter or longer. That is a way of calculating the weight of a molecule. In some cases, for example small proteins, the molecules are purposefully fragmented. The fragments can then be used to reconstruct of what amino acids the protein is made of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question! If any clarification is needed, please, do ask :)&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2015-02-11T11:19:03.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24932" PostTypeId="2" ParentId="24919" CreationDate="2015-02-11T11:53:23.090" Score="7" Body="&lt;p&gt;The Joule coefficient is &lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\partial T}{\partial V}\right)_U$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, Joule attempted to measure this value by expanding gas into an evacuated, insulated container, thus ensuring &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://tccc.iesl.forth.gr/education/local/Labs-PC-II/JT.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joule–Thomson coefficient&lt;/a&gt; is &lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\partial T}{\partial P}\right)_H$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, this is realized by expanding a flow of gas in an insulted pipe from a high pressure upstream region to a lower pressure downstream region, the two regions being seperated by a porous frit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see &lt;a href=&quot;http://astrowww.phys.uvic.ca/~tatum/thermod/thermod10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;THE JOULE AND JOULE-THOMSON EXPERIMENTS&lt;/a&gt; by Dr. J. B. Tatum, University of Victoria.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2020-04-20T18:04:25.517" LastActivityDate="2020-04-20T18:04:25.517" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24933" PostTypeId="2" ParentId="24928" CreationDate="2015-02-11T12:12:13.783" Score="5" Body="&lt;p&gt;Yes, lewis structure does matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the lewis structure of thiosulfate. &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/BvOK2.png&quot; alt=&quot;&quot;&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Thiosulfate-ion-2D-dimensions.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is oxidation state measured? It's a crude way of attributing some charge to the atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;O is more electronegative than S. You have S as the central atom, and another S &quot;around&quot; the tetrahedron structure. What would that do to each S?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The central S is expanding octet, while at the same time having a positive oxidation state due to the bonds it makes. The other S is having a negative oxidation state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What oxidation state does an O atom usually have? You can apply it to this ion as well. And we have a rule that says the sum of the oxidation numbers is equal to the charge of the ion. The &quot;noncentral&quot; S will get the same oxidation number as O atoms and then you can measure the central S atom's oxidation state by calculating all of the oxidation states to be -2.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-02-14T15:46:33.247" LastActivityDate="2015-02-14T15:46:33.247" CommentCount="23" ContentLicense="CC BY-SA 3.0" />
  <row Id="24934" PostTypeId="2" ParentId="24928" CreationDate="2015-02-11T12:16:29.727" Score="19" Body="&lt;p&gt;According to &lt;a href=&quot;http://adsabs.harvard.edu/abs/1993GeCoA..57.1619V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oxidation state of sulfur in thiosulfate and implications for anaerobic energy metabolism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;according to the currently held view, the two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation. Our results based on XANES spectroscopy indicate that the two different sulfur atoms in thiosulfate have charge densities corresponding to +5 and −1 oxidation states &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and in &lt;a href=&quot;http://dx.doi.org/10.1016/S1386-1425(98)00153-X&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using X-ray absorption to probe sulfur oxidation states in complex molecules&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the oxidation states were established for the outer and inner sulfur atoms of thiosulfate as −1 and +5, respectively. The revised values give a far better understanding of the role of thiosulfate as an energy substrate for bacteria because intra-molecular electron transfer, which generates metabolic energy with the production of sulfide and sulfate, cannot be envisaged with the previous view that the oxidation states were −2 and +6 for the outer and inner sulfur atoms of thiosulfate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although both of the above articles are by the same author. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;a href=&quot;http://przyrbwn.icm.edu.pl/APP/PDF/121/a121z2p71.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Determination of Changes in Sulfur Oxidation States&#xA;in Prostate Cancer Cells&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there is a difference between formal and apparent&#xA;  oxidation state received from XANES measurements, for&#xA;  example the apparent oxidation states of sulfur in thiosulfates,&#xA;  obtained from XANES data, are -1 and +5,&#xA;  while the formal ones are -2 and +6. Such a difference&#xA;  could occur due to the calibration curve which has been done&#xA;  on the basis of chemical compounds in which there are&#xA;  ionic bonds which isn't true in the case of thiosulfates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, it must be kept in mind that according to the current IUPAC definition, the oxidation state is not an actual property of a substance but instead a charge &quot;an atom might be &lt;strong&gt;imagined&lt;/strong&gt; to have&quot;, specifically: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Oxidation State] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion; (3) hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of -1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of -1 in peroxides, e.g. H2O2); (4) the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current &lt;a href=&quot;http://goldbook.iupac.org/O04365.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC definition of oxidation state&lt;/a&gt; doesn't specify how to apportion electrons of bonds between two atoms of the same element.  This causes various oxidation states to meet the IUPAC definition depending upon how one chooses to apportion the electrons of the S-S bond between the two S atoms.  Also, experimentally, the S-S is somewhere between a single bond and a double bond based on its length, so the number of electrons to be apportioned is also in question.  One could say the oxidation states are of the S atoms are -2, +6; -1, +5 or 0, +4 without contradicting the IUPAC definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed1005213&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expanded Definition of the Oxidation State&lt;/a&gt;&quot; there is a proposed addition to the definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonds between like atoms (having the same formal charge) are cleaved homolytically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If an S=S double bond of thiosulfate is cleaved homolytically, 0 and +4 oxidation states result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an S-S single bond of thiosulfate is cleaved homolytically, -1 and +5 oxidation states result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the article concludes 0 and +4 oxidation states even with the single bond, because when there is a single S-S bond the two S-S atoms do not have the same formal charge. They give both electrons from the S-S single bond to the central S. Below is Fig. 4 from the article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NpaL0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most recent (June 2014) information (and probably most authoritative) I can find is &lt;a href=&quot;http://www.degruyter.com/view/j/pac.2014.86.issue-6/pac-2013-0505/pac-2013-0505.xml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Toward a comprehensive definition of oxidation state (IUPAC Technical Report)&lt;/a&gt; Pure and Applied Chemistry. Volume 86, Pages 1017–1081&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this article, another definition of Oxidation State is proposed requiring &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonds between two atoms of the same element are divided equally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;with no exceptions for formal charges.  This clearly rules out the possibility of -2, +6 oxidation states for the S atoms in thiosulfate, but leaves ambiguity between-1, +5&#xA; if the S-S bond is single and 0, +4 if the S=S bond is double.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, concerning thiosulfate the article states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Structural properties of sodium thiosulfate suggest that all its terminal atoms carry some of the anion&#xA;  charge. In other words, all form ionocovalent bonds to a cation. The calculated Mulliken charges (−1.4 for&#xA;  O and −1.1 for terminal S) confirm this but, being different, they suggest unequal SS and SO bond orders. The&#xA;  SS bond distance of 2.025 Å (202.5 pm) is shorter than the single bond of 2.055 Å (205.5 pm) in crystalline&#xA;  $\ce{S8}$ or 2.056 Å (205.6 pm) in the $\ce{H_2S_2}$ gas, but substantially longer than the double bond of 1.883 Å&#xA;  (188.3 pm) in $\ce{S2O}$ or 1.889 Å (188.9 pm) in $\ce{S2}$. Given this uncertain situation, the approach chosen&#xA;  here is to draw in Fig. 24 two limiting Lewis formulas. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then they proceed to draw two Lewis formulas, similar to above, one on the left with S-S single bond and -1 and +5 oxidation states for the sulfurs and one on the right with S=S double bond and 0 and +4 oxidation states for the sulfurs, and go on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The result in Fig. 24 means that the individual OS of the two sulfur atoms in thiosulfate are ambiguous,&#xA;  and no formula in-between these two limits yields reasonable values. The formula on left provides an&#xA;  unusual yet not entirely illogical OS of −1 for the terminal sulfur, reminiscent of the oxygen OS in peroxides.&#xA;  The formula on right, while having an SS bond order that is too high, suggests OS values that at times&#xA;  are used...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article also comments that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A nominal OS is one that has been nominally adjusted against the generic OS definition in order to suit a specific purpose. That purpose may be...polarizing&#xA;  a homonuclear bond in thiosulfate to OS [of S] = +6 and −2 in order to emphasize the uniqueness of the central atom and similarity of all four ligands.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="402" LastEditDate="2017-05-24T04:58:54.467" LastActivityDate="2017-05-24T04:58:54.467" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="24935" PostTypeId="1" AcceptedAnswerId="24937" CreationDate="2015-02-11T12:45:55.547" Score="3" ViewCount="7990" Body="&lt;p&gt;My book says that from drawing the Lewis structure for the molecule below, you can conclude that it has dipole moment. Can someone please explain why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ZIDZ.jpg&quot; alt=&quot;structure of PF3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the dipole moment expresses the polarity of the bond. And that the dipole moment of a molecule is the sum of the dipole moments for each bond in the molecule.&lt;/p&gt;&#xA;" OwnerUserId="2565" LastEditorUserId="4945" LastEditDate="2020-09-02T13:45:50.863" LastActivityDate="2020-09-02T16:32:33.007" Title="Why does the phosphorus trifluoride molecule have dipole moment?" Tags="&lt;molecular-structure&gt;&lt;lewis-structure&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24936" PostTypeId="1" CreationDate="2015-02-11T13:16:17.907" Score="7" ViewCount="247" Body="&lt;p&gt;I am measuring flourescence lifetimes with a picosecond pulse laser, using the LifeSpec II with an MCP detector. Instead of a single peak rapidly decaying, as would be expected, my IRF contains two peaks. I am suspecting some sort of reflection causing this. Does anyone get what I mean here? Could add a graph soon if that might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have realized this might not be an error, but I would still like to know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5JLx.png&quot; alt=&quot;The IRF&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13341" LastEditorUserId="13341" LastEditDate="2015-02-12T10:02:47.143" LastActivityDate="2015-05-31T08:26:27.460" Title="Lifetime measurement: double peak in instrument response function" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="24937" PostTypeId="2" ParentId="24935" CreationDate="2015-02-11T13:17:50.760" Score="8" Body="&lt;p&gt;Fluorine is more electronegative than phosphorus.&lt;/p&gt;&#xA;&lt;p&gt;The phosphorus is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex.&#xA;The F-P-F angles are 96 degrees.&lt;/p&gt;&#xA;&lt;p&gt;Each F-P bond contributes to the net dipole moment, as a vector from the P to the F.  The net dipole moment is the vector sum of the vectors along the three P-F bonds.&lt;/p&gt;&#xA;&lt;p&gt;If all four atoms were in a plane (a trigonal planar geometry), there would be no net dipole moment.  But because the geometry is pyramidal, the vector sum is not zero and there is a net dipole moment.&lt;/p&gt;&#xA;&lt;p&gt;In some molecules lone pairs also contribute to the net dipole moment, in this particular case the lone pair of phosphorus is in a mostly s-like orbital, so in first approximation it can be neglected.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2020-09-02T16:32:33.007" LastActivityDate="2020-09-02T16:32:33.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24938" PostTypeId="2" ParentId="24925" CreationDate="2015-02-11T13:34:12.713" Score="5" Body="&lt;p&gt;Yes, it is possible though not easy to make artificial seawater by mixing sea salt and fresh water. An interesting example of this happens at &lt;a href=&quot;http://www.oceanario.pt/&quot; rel=&quot;noreferrer&quot;&gt;the oceanarium in Lisbon, Portugal&lt;/a&gt;, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Lisbon_Oceanarium&quot; rel=&quot;noreferrer&quot;&gt;one of the largest in Europe&lt;/a&gt;. After considerable investigation, it was found that the seawater present in massive quantities just outside the building could not be economically filtered, purified and adjusted to attend the surprisingly demanding and delicate needs of the marine organisms to be displayed. The chemical composition of seawater is actually quite complex, varying significantly with the location and time of collection. As with all life, marine life can be very sensitive to small changes in concentration of compounds, not to mention possible biological incompatibilities between the micro- and macrobiota present in seawater from different regions. The solution chosen for the Lisbon Oceanarium was to import some 16.5 metric tons of sea salt monthly from the Red Sea in Israel and mix it with local fresh water in order to replace the tank water. &lt;a href=&quot;http://www.oceanario.pt/cms/29&quot; rel=&quot;noreferrer&quot;&gt;[1]&lt;/a&gt; &lt;a href=&quot;http://www.luciamalla.com/blog/2014/04/oceanario-de-lisboa-o-melhor-aquario-mundo.html&quot; rel=&quot;noreferrer&quot;&gt;[2]&lt;/a&gt; (Couldn't find a reference in English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, you really should consult people experienced in setting up seawater aquariums before trying anything yourself. Seawater aquariums have &lt;a href=&quot;http://en.wikipedia.org/wiki/Marine_aquarium&quot; rel=&quot;noreferrer&quot;&gt;significantly tougher upkeep and maintenance requirements&lt;/a&gt; compared to fresh water aquariums; you can't just mix any old cooking sea salt into water, it has to be in the right concentration, and replacement of the tank water has to be more gradual and frequent, not to mention that the salty water will corrode the actual aquarium more quickly. A good aquarium supplies store should be able to better inform you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Now I realize you may actually be interested in cultivating shrimp rather than putting them on display. That's a whole different matter, many many times more difficult. You really need to contact some experts in marine cultivation.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-02-11T13:34:12.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24939" PostTypeId="1" AcceptedAnswerId="24940" CreationDate="2015-02-11T13:54:32.273" Score="2" ViewCount="380" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;X, Y and Z are three unknown elements whose first 5 ionization energies are given below. Which of the 3 is the most electronegative?$$&#xA;\begin{array}{|c|c|c|c|c|c|}\hline&#xA;&amp;amp;\text{IE}_1&amp;amp;\text{IE}_2&amp;amp;\text{IE}_3&amp;amp;\text{IE}_4&amp;amp;\text{IE}_5\\\hline&#xA;\ce{X}&amp;amp;577&amp;amp;1815&amp;amp;2740&amp;amp;11600&amp;amp;15000\\\hline&#xA;\ce{Y}&amp;amp;800&amp;amp;2427&amp;amp;3660&amp;amp;25026&amp;amp;32827\\\hline&#xA;\ce{Z}&amp;amp;738&amp;amp;1451&amp;amp;7733&amp;amp;10543&amp;amp;13630\\\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of the IE ratios, I concluded that X and Y are in the third column and Z is in the second column. I know that EN increases from left to right for elements in the same row. Is there anything else I should know in order to solve this?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastEditorUserId="5026" LastEditDate="2018-03-02T09:11:54.923" LastActivityDate="2018-03-02T09:42:40.273" Title="Determine most electronegative element based on successive ionization energy data" Tags="&lt;periodic-table&gt;&lt;electronegativity&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24940" PostTypeId="2" ParentId="24939" CreationDate="2015-02-11T14:23:28.487" Score="1" Body="&lt;p&gt;Let's look at what amounts of energy we get from IEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{X}: 577 \rightarrow 1815 \rightarrow 2740 \rightarrow 11600 \rightarrow 15000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After taking out three electrons, we see a first jump in IEs. Thus, there are three electrons in the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Y}: 800 \rightarrow 2427 \rightarrow 3660 \rightarrow 25026 \rightarrow 32827$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, after three electrons out, we have the first jump. The result is the same as X: Three electrons in the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Z}: 738 \rightarrow 1451 \rightarrow 7733 \rightarrow 10543 \rightarrow 13630$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the first bump in IEs in between IE2 and IE3. So, the third electron comes out of a lower shell and thus, two electrons in the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more the electrons, the more shells they will fill and the farther they'll be from the nucleus (among other lesser important factors). That means X has more electrons than Y because of relatively lower IEs, and therefore Y is more electronegative as it's closer to the fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that from left to right we get more electronegativity for the atoms, but note that&lt;br&gt;&#xA;&lt;strong&gt;though with exceptions, such as Al and Ga, from top to bottom you get lesser electronegativities.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the case of Al and Ga, the one that's more electronegative will get relatively more IE to lose electrons. So that's not what you should worry about. So let's conclude:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you see two elements that are in the same group, the one with a generally higher IE is more electronegative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5026" LastEditDate="2018-03-02T09:42:40.273" LastActivityDate="2018-03-02T09:42:40.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24941" PostTypeId="1" AcceptedAnswerId="24948" CreationDate="2015-02-11T20:11:27.100" Score="7" ViewCount="40122" Body="&lt;p&gt;This is a well-known (better said: well-discussed) question in the internet. When you look for answers for popular questions, you usually see them with a variable degree of reliability and complexity. Unfortunately, for this one, I only observed very very crude and general rules of thumb. So let's get a real answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A network solid or covalent network solid is a chemical compound (or element) in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material. In a network solid there are no individual molecules, and the entire crystal may be considered a macromolecule. Formulas for network solids, like those for ionic compounds, are simple ratios of the component atoms represented by a formula unit. &lt;a href=&quot;http://en.wikipedia.org/wiki/Network_covalent_bonding&quot; rel=&quot;noreferrer&quot;&gt;Covalent network, wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diamond and SiO$_2$ are really great examples of covalent networks-lattices. So enough with stories:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you face a new chemical formula, how would you assume it's a covalent network? (In case it is) Is it somehow done by drawing the Lewis structure? Is there a rule for this? Or is it only possible to know such thing with experimental data?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fun fact: I'm preparing for an internal chemistry olympiad. The examiners usually take all the exceptions into account. So general rules such as &quot;if it contains Si, it's a covalent network&quot; aren't really reliable for me. Better phrased, I'm looking for a rule with the least exceptions.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="16683" LastEditDate="2017-12-08T23:13:06.647" LastActivityDate="2017-12-08T23:13:06.647" Title="How to know it when I see a covalent network?" Tags="&lt;molecular-structure&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24942" PostTypeId="1" CreationDate="2015-02-11T20:59:41.657" Score="2" ViewCount="493" Body="&lt;p&gt;Considering the Arrhenius theory of acids and bases: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is every existing compound either an acid or a base? &lt;/li&gt;&#xA;&lt;li&gt;Why is a compound such as $ \ce {NH_3}$ considered a base, if it doesn't even have an oxygen atom and a base is supposed to release $ \ce {OH-}$ in water?&lt;/li&gt;&#xA;&lt;li&gt;If $X$ is a base according to Arrhenius theory, will $X$ still be a base according to the other two theories (and the same if $X$ were an acid)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;According to the Brønsted-Lowry theory:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There are no absolute acid and bases, but only acid-base reaction in which one compound acts as a base and the other as an acid. So, is every reaction of two compounds an acid-base one?&lt;/li&gt;&#xA;&lt;li&gt;Except for water, are there any other substances which in some reactions act as bases while in others as acids?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7665" LastEditorUserId="9961" LastEditDate="2015-11-13T19:40:37.360" LastActivityDate="2018-03-04T02:58:49.187" Title="Clarifications about acid-base theory" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-07-30T17:52:09.990" ContentLicense="CC BY-SA 3.0" />
  <row Id="24944" PostTypeId="2" ParentId="24942" CreationDate="2015-02-11T21:48:26.337" Score="2" Body="&lt;p&gt;A couple of thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are mainly 3 different definitions of acids and bases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Arrhenius formulation: Acid = proton donor in water,  Based = hydroxide ion donor in water&lt;/li&gt;&#xA;&lt;li&gt;The Brønsted-Lowry formulation: Acid = &lt;strong&gt;&lt;em&gt;potential&lt;/em&gt;&lt;/strong&gt; proton donor, Base = &lt;strong&gt;&lt;em&gt;potential&lt;/em&gt;&lt;/strong&gt; proton acceptor&lt;/li&gt;&#xA;&lt;li&gt;The Lewis formulation: Acid = Electron pair acceptor, Base = Electron pair donor&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The Lewis definition is the most general, followed by the BL and finally by the Arrhenius definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right, for example HCl is considered a strong acid in water, but it is very weakly acidic in glacial acetic acid. If you follow the Lewis definition a large number of reactions will be considered acid base reactions. An exception are free-radical based reactions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many compounds that are amphoteric, but it really also depends on the definition you use.&lt;/p&gt;&#xA;" OwnerUserId="13086" LastEditorUserId="5026" LastEditDate="2018-03-04T02:58:49.187" LastActivityDate="2018-03-04T02:58:49.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24945" PostTypeId="1" AcceptedAnswerId="24961" CreationDate="2015-02-11T21:54:31.930" Score="0" ViewCount="210" Body="&lt;p&gt;I want to create my own glow stick that lasts a while while ligting up the surroundings. How would I go about this? I know there are youtube videos and such. But the way they do it doesn't allow for any way of &quot;recharging&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can i make a &quot;rechargable&quot; glow stick?&lt;/p&gt;&#xA;" OwnerUserId="13348" LastEditorUserId="418" LastEditDate="2015-02-12T07:33:48.990" LastActivityDate="2015-02-12T08:04:03.757" Title="How to get a &quot;rechargable&quot; chemical reaction for a glow stick" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24946" PostTypeId="1" CreationDate="2015-02-11T22:22:43.080" Score="4" ViewCount="8592" Body="&lt;p&gt;If I had hydrogen peroxide decompose with the help of manganese dioxide and measured the time taken for the reaction to complete and the volume of oxygen collected. I will get a graph that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/quoOj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what will happen to the shape of the graph if I placed the reaction vessel in ice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the following. The graph will rise but at a less rate and less steep than in the graph above. Then it will get to the point where the reaction ceases and doesn't go to completion because there are no more collisions. &lt;/p&gt;&#xA;" OwnerUserId="13349" LastEditorUserId="5026" LastEditDate="2018-02-27T02:50:25.133" LastActivityDate="2018-02-27T02:50:25.133" Title="How does an ice bath change the rate of an reaction?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24947" PostTypeId="1" CreationDate="2015-02-11T22:27:32.980" Score="2" ViewCount="1151" Body="&lt;p&gt;In the synthesis of ethyl 2-benzyl-2-methyl-3-oxobutanoate, is the nucleophilic substituion of a benzyl (chloride) group an Sn1 reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment is conducted in ethanol (protic) and is under reflux, but most textbooks have it as an Sn2?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it then be a mixture of the two?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="7721" LastEditDate="2015-02-11T23:41:42.103" LastActivityDate="2015-02-13T09:41:33.520" Title="Dialkylation reactions" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24948" PostTypeId="2" ParentId="24941" CreationDate="2015-02-11T22:27:42.617" Score="8" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvsYs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good source of information on this topic is the website &lt;a href=&quot;http://www.meta-synthesis.com/webbook/38_laing/tetrahedra.html&quot; rel=&quot;noreferrer&quot;&gt;Tetrahedra of Structure, Bonding &amp;amp; Material Type&lt;/a&gt;, from which the above image is taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extreme example of the metal corner should be cesium, the ionic corner CsF, and the molecular corner $\ce{F2}$.  At the top is diamond carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The network covalent apex is truncated on the basis that a network covalent material requires an element with the electronegativity of carbon or less (less than or equal to 2.55).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the challenge is to distinguish a boundary between network covalent and the other three types&lt;/p&gt;&#xA;&#xA;&lt;p&gt;network covalent vs. metal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;network covalent vs. ionic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;network covalent vs. molecular &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of network covalent compounds other than diamond and silicon dioxide include silicon carbide, silicon nitride, boron nitride, aluminum phosphide, gallium arsenide, aluminum oxide.  However, anytime there is an electronegativity difference there is some ionic component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin can exist in metallic and network covalent allotropes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Selenium can exist in a network covalent allotrope or as molecular $\ce{Se8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from network covalent to molecular covalent, think of how there is diamond, graphite, nanotubes, fullerenes and even small molecules like $\ce{C4}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-12T17:48:13.853" LastActivityDate="2015-02-12T17:48:13.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24950" PostTypeId="2" ParentId="24921" CreationDate="2015-02-12T00:35:58.170" Score="2" Body="&lt;p&gt;If the helium is expanding isothermally against a constant external pressure $P_{ext}$, the work is calculated as&#xA;$$w = -P_{ext}(V_2 - V_1)$$&#xA;This process is irreversible. The first expression you gave for work is (sort of) the expression for work for a &lt;em&gt;reversible&lt;/em&gt; isothermal ideal gas expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If helium is assumed to be an ideal gas&lt;/strong&gt;, then $\Delta U = 0$ because the process is isothermal. You have $q = -w$ (and not zero) for this process. $\Delta H$ will be zero too, since $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-12T00:41:02.200" LastActivityDate="2015-02-12T00:41:02.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24951" PostTypeId="1" AcceptedAnswerId="24957" CreationDate="2015-02-12T01:10:33.467" Score="9" ViewCount="2301" Body="&lt;p&gt;Why do weak bases usually contain nitrogen? I know there are two electrons on top for a bond of $\ce{H}$ but why is it mostly nitrogen? I know there are other elemental bases too but why do I keep seeing $\ce{N}$'s mostly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something special about nitrogen in this respect?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="402" LastEditDate="2017-04-01T16:42:19.917" LastActivityDate="2017-04-01T16:42:19.917" Title="Why is it that weak bases usually contain nitrogen?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24952" PostTypeId="1" CreationDate="2015-02-12T01:35:10.783" Score="1" ViewCount="1378" Body="&lt;p&gt;I'm a little confused on something:&#xA;A buffer has a certain capacity and then runs out.&lt;br&gt;&#xA;But, if the buffer is given by: $$\ce{A- + H2O &amp;lt;=&amp;gt; HA + OH-},$$ when the $\ce{A-}$ reacts with the water, the reaction shifts to the left. So how can it &quot;run out&quot;? If $Q$ just keeps becoming $0$, shouldn't the reaction keep shifting to maintain $K$? Does the right side, perhaps, approach $0$ as it keeps decreasing in concentration to maintain $K$, but it can never become $0$ molar, because that wouldn't maintain $K$?&lt;br&gt;&#xA;I guess if $\ce{H2O}$ was &quot;aqueous&quot;/had a concentration and was therefore used in the $K$ expression, this the $K$ would always be maintained.&lt;br&gt;&#xA;But how does the buffer &quot;run out&quot; of $\ce{A-}$ to react when $\ce{H2O}$ is liquid (not aqueous)? It doesn't make sense as wouldn't the reaction always adjust to make more $\ce{A-}$ to maintain $K$?&lt;/p&gt;&#xA;" OwnerUserId="13077" LastEditorUserId="4945" LastEditDate="2015-02-12T03:26:24.610" LastActivityDate="2015-02-12T03:30:56.397" Title="What is buffer capacity and how can a buffer &quot;run out&quot;?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24953" PostTypeId="1" AcceptedAnswerId="25108" CreationDate="2015-02-12T01:42:58.010" Score="3" ViewCount="1282" Body="&lt;p&gt;I was wondering, when working in a lab in an area related to organic compounds, when and why would a chemist consider the enantiomeric excess % and how would a chemist used that information?? Same Thing goes with specific rotation, why, how, and when would a chemist use this? I was just curious because I saw this in my organic chemistry book and they relate the problems to lab work and I want to get a good grasp of tis context and it made me wondered. &lt;/p&gt;&#xA;" OwnerUserId="7130" LastEditorUserId="865" LastEditDate="2015-10-16T17:41:52.267" LastActivityDate="2016-12-07T18:27:56.883" Title="Why are enantiomeric excess % and specific rotation considered?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24956" PostTypeId="2" ParentId="24952" CreationDate="2015-02-12T03:15:23.023" Score="1" Body="&lt;p&gt;If strong acid is added to a buffer system, conjugate base removes it.  Eventually, the concentration of conjugate base falls to such a low level that its effect on pH can be neglected.  At this point, the solution behaves more like a weak acid solution.  If even more acid is added, the system behaves like a strong acid system in terms of calculating pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If strong base is added to a buffer system, conjugate acid removes it.  Eventually the concentration of conjugate acid falls to such a low level that its effect on pH can be neglected.  At this point. the solution behaves more like a weak base solution.  If even more base is added, the system behaves like a strong base system in terms of calculating pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equation you wrote, $$\ce{A- + H2O &amp;lt;=&amp;gt; HA + OH-},$$ think about the effect of adding acid.  Acid reacts with the $\ce{OH-}$, producing water and removing $\ce{A-}$ from solution.  The $\ce{OH-}$ and $\ce{A-}$ cannot be removed from solution but their concentrations can be driven to negligible levels by the addition of sufficient quantities of acid.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="4945" LastEditDate="2015-02-12T03:30:56.397" LastActivityDate="2015-02-12T03:30:56.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24957" PostTypeId="2" ParentId="24951" CreationDate="2015-02-12T04:18:50.500" Score="6" Body="&lt;p&gt;The short answer is that yes, the fact that many weak bases contain nitrogen does have to do with nitrogen itself. However, it's not really accurate to say that weak bases &quot;usually&quot; contain nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm limiting my explanation to aqueous solutions, the Brønsted-Lowry definition, and main-group compounds. A base must do one of two things: either remove a proton ($\ce{H+}$) from water or dissociate to produce something else (typically $\ce{OH-}$) that can remove a proton from water. Most compounds that produce $\ce{OH-}$ directly are ionic; even if they don't dissolve much (like alkaline earth hydroxides), these ionic compounds dissociate completely and are therefore strong bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A base that removes protons from water must have a free pair of electrons for the proton to bond to. In period 2, boron tends to form compounds that tie up all three of its electrons. Carbon compounds, likewise, tend to have no free electron pairs. Oxygen and fluorine compounds (e.g., water and hydrogen fluoride) have free electron pairs, but because of their high electronegativity these elements keep their electrons to themselves and are unlikely to grab protons from water if they already have a full octet. Nitrogen is the only element in this period that balances two factors: it has a free electron pair in many of its compounds (e.g., ammonia and organic amines) and has a low enough electronegativity that it can share those electrons with a proton to form a cation (e.g., ammonium, $\text{NH}_3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covalent compounds of elements in the third period and below tend to be &lt;em&gt;very&lt;/em&gt; weak bases, if at all. I'm not sure why, but I suspect that it's because bonds involving these elements tend to form much weaker dipoles than bonds involving nitrogen. The weaker dipole would limit their ability to forcibly pull protons off of water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, there are many ionic compounds that form basic solutions due to containing the cation of a weak acid. In these cases the base is a cation, not a molecule. Certainly some of these cations contain nitrogen, but many do not. Examples without nitrogen include any soluble carbonates and bicarbonates and any salts of carboxylic acids.&lt;/p&gt;&#xA;" OwnerUserId="13334" LastEditorUserId="7475" LastEditDate="2015-09-12T18:47:09.020" LastActivityDate="2015-09-12T18:47:09.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24958" PostTypeId="1" CreationDate="2015-02-12T04:30:19.333" Score="5" ViewCount="1011" Body="&lt;p&gt;I am looking for a software (can be freeware or purchased) to look up water and steam thermodynamic properties (e.g. density, enthalpy, entropy) preferably a macro in excel. What do you use for this application?&lt;/p&gt;&#xA;" OwnerUserId="5734" LastEditorDisplayName="user7951" LastEditDate="2016-11-20T14:32:17.707" LastActivityDate="2016-11-20T17:05:14.380" Title="Thermodynamic properties software" Tags="&lt;thermodynamics&gt;&lt;software&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24961" PostTypeId="2" ParentId="24945" CreationDate="2015-02-12T08:04:03.757" Score="3" Body="&lt;p&gt;&lt;strong&gt;Chemoluminescent&lt;/strong&gt; processes based on the reaction of diaryloxalates with hydrogen peroxide in the presence of a fluorescent dye are &lt;strong&gt;not reversible&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the dye remains unchanged, both the oxalates and the hydrogen peroxide are consumed. &quot;Recharging&quot; a device (glow stick) isn't possible. When performing the reaction in a flask, only could probably prolong the light emission by subsequently adding portions of the starting materials over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photoluminescent&lt;/strong&gt; &lt;em&gt;glow-in-the-dark&lt;/em&gt; devices based on the emission of long-lived excited triplet states can indeed be &quot;recharged&quot; by holding the device in the light again. But this has nothing to do with the reactions happening in glow sticks.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-12T08:04:03.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24962" PostTypeId="2" ParentId="24113" CreationDate="2015-02-12T08:06:11.080" Score="2" Body="&lt;p&gt;Ion selective electrodes are a group of electrodes that are very different from each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common bits are, a selective membrane, a &quot;sensing&quot; electrode and a reference electrode to measure against. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of the membrane and sensing electrode depend highly on what selective electrode you are trying to build. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an old patent on a Cl selective electrode for reference:&#xA;&lt;a href=&quot;http://www.google.com/patents/US3740326&quot; rel=&quot;nofollow&quot;&gt;http://www.google.com/patents/US3740326&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding getting accurate results with a DIY: this really depends on which electrode you are trying to build. In the Cl example, I don't see any reason why a DIY would perform worse.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-02-12T08:06:11.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24963" PostTypeId="2" ParentId="23627" CreationDate="2015-02-12T08:13:24.907" Score="2" Body="&lt;p&gt;Not sure what kind of explanation you are looking for, but &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://global.oup.com/academic/product/electrode-dynamics-9780198556909?cc=us&amp;amp;lang=en&amp;amp;&quot; rel=&quot;nofollow&quot;&gt;Electrode Dynamics, A. C. Fisher, Oxford Chemistry Press, 1996, ISBN: 019855690X&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is a great concise explanation of overpotentials among other things.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="5017" LastEditDate="2015-03-16T16:34:42.573" LastActivityDate="2015-03-16T16:34:42.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24964" PostTypeId="1" CreationDate="2015-02-12T08:15:00.520" Score="2" ViewCount="2446" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we force a reversible reaction in equilibrium to become a &quot;complete&quot; one? If so, will the equilibrium constant change?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's said about reversible reactions that they never go to completion &amp;amp; I know that's because in reversible reactions, the mixture obtained in equilibrium is more stable than the extremes (i.e. either reactant or product) But if we force the reaction to proceed in a certain direction, say in the direction of products, will it ever go to 100% completion or near 100%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question is regarding the equilibrium constant. I have read about equilibrium constant that its value doesn't change if we alter the initial concentration of reactants, which I understand why, but consider this situation:&#xA; if we force equilibrium to proceed in the direction of products by altering $p,T,V$ (only and not adding or removing any substance), the equilibrium will get disturbed and obtain a new equilibrium. Wouldn't the equilibrium constant for this new equilibrium be different from the earlier one?&lt;/p&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="5026" LastEditDate="2018-02-22T03:14:25.327" LastActivityDate="2018-02-24T12:30:36.987" Title="Can we force a reversible reaction in equilibrium to achieve 100% completion?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24966" PostTypeId="1" CreationDate="2015-02-12T09:14:26.083" Score="0" ViewCount="224" Body="&lt;p&gt;Does the reversibility of a chemical reaction make it spontaneous from both sides also? (i.e.: forward &amp;amp; backward reactions) If not then is there any relationship between $\Delta G$ of forward and backward reactions. Am I right to assume that if $\Delta G$ of one of the process is positive then $\Delta G$ for the reverse process must be negative. So at equilibrium $\Delta G = 0$ ?&lt;/p&gt;&#xA;" OwnerUserId="12762" LastEditorUserId="7448" LastEditDate="2015-02-12T13:15:38.853" LastActivityDate="2015-04-19T14:07:38.700" Title="Spontaneity of reversible chemical reactions - Both sides?" Tags="&lt;homework&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="24967" PostTypeId="1" AcceptedAnswerId="24971" CreationDate="2015-02-12T09:35:59.837" Score="1" ViewCount="1493" Body="&lt;p&gt;I'm trying to understand an observed reaction from a chem practical:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a solution of ammonium sulfate, which you add to a bovine serum albumin solution. The proteins precipitate, and the solution becomes cloudy. w/v % of $\ce{(NH4)2SO4}$ at this point is 57%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add water (in a ratio of 2 parts water: 1 part serum solution)to the mix, the process reverses, and the solution becomes clear again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be explained by the ammonium sulfate solution 'stealing' water from the proteins, which causes them to precipitate, and then the process reversing once more water is added. (Right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Does the ammonium sulfate 'steal' the water because it wants to reach equilibrium, or because it's supersaturated, or what?&lt;/p&gt;&#xA;" OwnerUserId="13356" LastEditorUserId="7448" LastEditDate="2015-02-12T10:06:38.833" LastActivityDate="2015-02-12T12:13:50.847" Title="Why does Ammonium sulfate cause proteins to precipitate?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24968" PostTypeId="1" AcceptedAnswerId="24974" CreationDate="2015-02-12T09:44:42.750" Score="2" ViewCount="237" Body="&lt;p&gt;Please consider the following figure of a concentration over time (msec) diagram with variables for an enzyme, a substrate bound enzyme and a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fTcYo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that $\ce{E + S \leftrightharpoons ES \rightharpoonup E + P}$ and also recall that the velocity constant of the respective reactions are noted $k_1 , k_2, k_3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to know is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the plot provided, is the slope of $[ES]$ equal to $k_2$ and is the slope of $[E]$ equal to $k_1 + k_3$ thereby giving some insight/explanation for $k_m$ which is used to determine the &lt;em&gt;affinity&lt;/em&gt; of an enzyme?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any more details on Michaelis Menten is greatly appreciated but please note that my level of understanding is limited on the topic.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2015-02-12T16:41:09.363" Title="Rate constant and Michaelis Menten" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;&lt;enzymes&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24969" PostTypeId="1" AcceptedAnswerId="24984" CreationDate="2015-02-12T10:06:23.423" Score="5" ViewCount="1248" Body="&lt;p&gt;I ordered a DIY spectroscopy kit from public-lab, however don't have access to a halogen lamp to calibrate the wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if it would somehow be possible to calibrate the wavelength using the iPhone camera light?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastActivityDate="2015-02-12T16:54:45.407" Title="Wavelength of iPhone 4S camera light for visible spectroscopy calibration" Tags="&lt;spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24970" PostTypeId="1" CreationDate="2015-02-12T11:14:36.527" Score="1" ViewCount="195" Body="&lt;p&gt;In electrochemistry, the Nernst equation tells us the cell potential for non-standard conditions. I read that the factor $\frac{-RT}{nF}\ln{Q}$ accounts for difference in cell potential due to differing Gibbs Free Energy due to changing temperature and concentration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference in Gibbs Free Energy is only due to entropy right? Because enthalpy is a constant? So obviously, entropy significantly interferes in the process of electrolysis.&#xA;And yet, it is ignored in virtually all spheres of chemistry. I'm beginning to doubt my chemical intuition because of entropy effects, and faith in it can only be restored if entropy really was negligible. So it is negligible or not? If not, why is it always ignored?&lt;/p&gt;&#xA;" OwnerUserId="11586" LastEditorUserId="10119" LastEditDate="2015-02-12T13:36:00.057" LastActivityDate="2015-02-12T16:50:00.783" Title="Entropy - Negligible?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;energy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24971" PostTypeId="2" ParentId="24967" CreationDate="2015-02-12T12:13:50.847" Score="2" Body="&lt;p&gt;Well, I wouldn't know how to sum it up better than &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Non-Ideal_Systems/Salting_Out&quot; rel=&quot;nofollow&quot;&gt;chemwiki.ucdavis&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To achieve this conformation the molecule folds in such a way that all of the hydrophobic parts of a molecule are aggregated together and the hydrophillic groups are left to interact with the water. In the case of proteins it is the charged amino acids that allow selective salting out to occur. Charged and polar amino acids such as glutamate, lysine, and tyrosine require water molecules to surround them to remain dissolved. In an aqueous environment with a high ionic strength, the water molecules surround the charges of the ions and proteins. At a certain ionic strength, the water molecules are no longer able to support the charges of both the ions and the proteins. The result is the precipitation of the least soluble solute, such as proteins and large organic molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you ask what the aminoacid content of bovine serum albumin is, I'll cite:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Glycine,&#xA;  illanine,&#xA;  Valine,&#xA;  Leucine,&#xA;  Isoleucineh,&#xA;  Proline,&#xA;  Phenylalanine,&#xA;  Tyrosine,&#xA;  Tryptophan,&#xA;  Serine,&#xA;  Threonine,&#xA;  Half-cystinee,&#xA;  Methionine,&#xA;  Arginine,&#xA;  Histidine,&#xA;  Lysine,&#xA;  Aspartic acid and&#xA;  Glutamic acid. &lt;a href=&quot;http://www.jbc.org/content/239/3/850.full.pdf&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-12T12:13:50.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24973" PostTypeId="1" AcceptedAnswerId="24975" CreationDate="2015-02-12T12:51:42.130" Score="4" ViewCount="1844" Body="&lt;p&gt;A soft species is typically large, polarizable and has a low charge (magnitude). The silver cation is not highly charged (this supports softness) however, it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. Consequently, shouldn't it also be less polarizable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does have a tendency to form bonds that are significantly covalent in nature which backs up the assertion that $\ce{Ag^+}$ is soft; however, I don't understand why.    &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="9961" LastEditDate="2015-10-02T23:10:32.770" LastActivityDate="2019-11-19T19:24:25.623" Title="Why is Ag+ a soft acid?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;hsab&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24974" PostTypeId="2" ParentId="24968" CreationDate="2015-02-12T13:38:47.363" Score="4" Body="&lt;p&gt;Given $\ce{E + S &amp;lt;=&amp;gt;ES -&amp;gt; E + P}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}=  k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}]  $&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the slope of $[ES]$ equal to $k_2$ ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the slope of $[E]$ equal to $k_1 + k_3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-12T13:38:47.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24975" PostTypeId="2" ParentId="24973" CreationDate="2015-02-12T13:51:27.420" Score="6" Body="&lt;p&gt;According to HSAB theory, Soft Acids have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;large atomic/ionic radius; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;low or zero oxidation state bonding;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;high polarizability;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;low electronegativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ion $\ce{Ag+}$ satisfies all the aforementioned properties. &#xA;I don't agree with you when you say it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. The effective nuclear charge has nothing to do here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.webelements.com/silver/atom_sizes.html&quot; rel=&quot;nofollow&quot;&gt;http://www.webelements.com/silver/atom_sizes.html&lt;/a&gt;: The radius of this ion is $114\ \mathrm{pm}$ when it's in 4-coordinate, tetrahedral geometry and it's $116\ \mathrm{pm}$ when  it's 4-coordinate, square-planar geometry. Please just compare these radii with the radius of  $\ce{Al^3+}$ ion : $53.5\ \mathrm{pm}$ which is considered as a hard acid.&#xA;I hope it's clear now!&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T18:36:12.767" LastActivityDate="2015-04-26T18:36:12.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24976" PostTypeId="1" CreationDate="2015-02-12T14:20:33.977" Score="3" ViewCount="425" Body="&lt;p&gt;Why does the pH rapidly increase near the equivalence point during--why does the slope of the graph of a pH curve sharply increase around this area?&lt;/p&gt;&#xA;" OwnerUserId="13077" LastActivityDate="2015-02-12T18:29:22.883" Title="pH Curves for Titrations" Tags="&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24977" PostTypeId="2" ParentId="24976" CreationDate="2015-02-12T14:35:46.350" Score="6" Body="&lt;p&gt;I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the case of  the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The pH increases slowly at first because the pH scale is &#xA;logarithmic, which means that a pH of 1 will have 10 times the &#xA;hydronium ion concentration than a pH of 2. Thus, as the &#xA;hydronium ion is initally removed, it takes a lot of base to change &#xA;its concentration by a factor of 10, but as more and more &#xA;hydronium ion is removed, less base is required to change its &#xA;concentration by a factor of 10. Near the equivalence point, a &#xA;change of a factor of 10 occurs very quickly, which is why the &#xA;graph is extremely steep at this point. As the hydronium ion &#xA;concentration becomes very low, it will again take a lot of base to &#xA;increase the hydroxide ion concentration by 10 fold to change the &#xA;pH significantly. &lt;/li&gt;&#xA;&lt;li&gt;Let's find the equation of the titration curve in the the region $0&amp;lt;V_b&amp;lt;V_b(eq)$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of moles of $\ce{H3O+}$ after adding the volume   $V_b$ of the base is $C_aV_a-C_bV_b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$&#xA;So, the titration curve in the region $0&amp;lt;V_b&amp;lt;V_b(eq)$  $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$&#xA;If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-02-12T18:29:22.883" LastActivityDate="2015-02-12T18:29:22.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24978" PostTypeId="1" CreationDate="2015-02-12T14:53:40.250" Score="6" ViewCount="3276" Body="&lt;p&gt;I feel that there are very few textbooks that treat the chemical potential of mixtures in an understandable clear way, which is why I wanted to ask here about certain things. Although I do not have a single question, but 5 of them, I think that it is merely impossible to separate them thematically, which is why I hope to meet an expert here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have an ideal gas then if we know the chemical potential for a particular $(p,T)$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu(p_2,T) \approx \mu(p_1,T) + \left(\frac{\partial \mu}{\partial p} \right)_T(p_1,T) (p_2-p_1) = \mu(p_1,T) + \frac{V}{N}(p_2-p_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for changes with respect to temperature we have&#xA;$$\mu(p,T_2) \approx \mu(p,T_1) + \left(\frac{\partial \mu}{\partial T} \right)_p(p,T_1) (T_2-T_1) = \mu(p,T_1) - \frac{S}{N}(T_2-T_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we have either a gas or a liquid (Question 1: is this true, does this law hold for both fluids?) $\ce{A}$ and put another material $\ce{B}$ into it, then we have by Raoult's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{N_A}{N_A + N_B} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: Now, is the chemical potential of the total fluid given by $A$ and $B$ then the sum $\mu_{\mathrm{A+B}} = \mu_{\mathrm{A,~with~mixed~B}}(p,T) + \mu_{\mathrm{B,~with~mixed~A}}(p,T)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine that $A$ and $B$ would be gases, would this mean that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 3 : $$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_\mathrm{A}(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 4 : And what if $\ce{A}$ and $\ce{B}$ would be two two miscible liquids? could we still use the vapour pressures to calculate the chemical potential of the liquids mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{\mathrm{A,~with~mixed~B}}(p,T) = \mu_A(p,T) + RT\ln\left( \frac{p_\mathrm{A}}{p_\mathrm{A} + p_\mathrm{B}} \right) ?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that I have a liquid $\ce{A}$ and salt $\ce{B}$ in it. Also, I have a gas layer $\ce{C}$ of $\ce{A}$ above the liquid. What exactly is now in equilibrium: $\mu_{\mathrm{A+B}} = \mu_\mathrm{C}$, or $\mu_{\mathrm{A,~with~mixed~B}} = \mu_\mathrm{C}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 5: My last question would be: Imagine that you have a liquid (let's say water) and some gaseous water phase above (according to the vapour pressure). Now, you add oxygen to this. What would this mean for the chemical potentials of the liquid/gas water?&lt;/p&gt;&#xA;" OwnerUserId="13360" LastEditorUserId="16683" LastEditDate="2015-10-23T17:03:31.720" LastActivityDate="2018-12-11T12:52:56.167" Title="Chemical potential of mixtures" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24979" PostTypeId="2" ParentId="24936" CreationDate="2015-02-12T15:21:13.313" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am suspecting some sort of reflection causing this. Does anyone get what I mean here? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The time between the two peaks looks like 0.4 nanoseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time (0.4 ns) should be compared to the dimensions of the vessel (chamber, cuvette, I don't know the right word) and the speed of light in the medium being observed (water, air, diamond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example in vacuum or gas the light would travel 0.4ns x (0.3m/ns) = 0.12m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water where the index of refraction is 1.33 more like 0.09m.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-12T15:21:13.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24981" PostTypeId="2" ParentId="24969" CreationDate="2015-02-12T15:56:38.733" Score="2" Body="&lt;p&gt;I don't know about the exact calibration setup of the spectroscopy kit, which sounds interesting, but i doubt that the cell phone lamp will be useful here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both lamps seem to emit while light, but both the ways how light is generated, as well as the emission spectra of cell phone camera light and a halogen lamp are rather different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cell phone camera lights are typically blue light emitting diodes partly covered with one one more phosphors (not, it isn't phosphorous) which absorb the blue light and emit yellow (or orange) light. The emission spectrum of the devices consists of some distict bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halogen lamps are tuned heavily incandescent lamps with a tungsten filament. Here, the spectrum is more similar to a black body radiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, I suggest to get a halogen lamp. A cheap 12V lamp with a socket (often GU4) and a power supply will probably work fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Naked lamps&quot; do emit in the UV too; Lamps with a reflector to provide a collimated beam have an additional glas cover that shield the UV part. Check the manual of the spectrometer kit for some directions. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-12T15:56:38.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="24982" PostTypeId="2" ParentId="24968" CreationDate="2015-02-12T16:41:09.363" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;On the plot provided, is the slope of [ES] equal to k2 and is the&#xA;  slope of [E] equal to k1+k3 thereby giving some insight/explanation&#xA;  for km which is used to determine the affinity of an enzyme?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As DavePhD said, no to both counts.  First note that in your system, $\ce{E_0=E +ES}$, where $\ce{E_0}$ is the total amount of enzyme added, and $\ce{E_0}$ is conserved -- enzymes, as catalysts, are created or destroyed in the reaction.  This means $\frac{d\ce{E_0}}{dt}=0$, which implies that $\frac{d\ce{E}}{dt}=-\frac{d\ce{ES}}{dt}$.  So the slope of [ES] is negative the slope of [E].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, as you study more about Michaelis Menten kinetics, you will come to appreciate that this data is (mostly) for the &lt;em&gt;un&lt;/em&gt;steady-state regime, also often called the &lt;em&gt;pre&lt;/em&gt;steady-state regime.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-12T16:41:09.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24983" PostTypeId="2" ParentId="24970" CreationDate="2015-02-12T16:50:00.783" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;This difference in Gibbs Free Energy is only due to entropy right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is mostly true.  (I think it is 100% true when &lt;em&gt;concentrations&lt;/em&gt; can be used in the reaction quotient $Q$, but am not sure if it is strictly true when solutions are not ideal and &lt;em&gt;activities&lt;/em&gt; have to be used instead.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because enthalpy is a constant? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Enthalpy is a &quot;constant&quot; only because of the way electrochemical standard states are defined.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So obviously, entropy significantly&#xA;  interferes in the process of electrolysis. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have no idea what this means.  How does it &quot;interfere&quot;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And yet, it is ignored in&#xA;  virtually all spheres of chemistry. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;False.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm beginning to doubt my chemical&#xA;  intuition because of entropy effects, and faith in it can only be&#xA;  restored if entropy really was negligible. So it is negligible or not?&#xA;  If not, why is it always ignored?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It isn't negligible and is not ignored.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-12T16:50:00.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24984" PostTypeId="2" ParentId="24969" CreationDate="2015-02-12T16:54:45.407" Score="3" Body="&lt;p&gt;According to the &lt;a href=&quot;http://publiclab.org/wiki/spectral-workbench-calibration&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt;, wavelength calibration is meant to be done with a compact fluorescent lamp (for the mercury lines), but at the bottom there's a tool for calibrating using any two known wavelengths. If the LED on the phone has two distinct lines that you can calibrate against, it may work, but something like a pair of monochromatic LEDs or laser pointers (say one blue and one red) or something will probably work better. &lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-12T16:54:45.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="24985" PostTypeId="2" ParentId="24958" CreationDate="2015-02-12T16:56:44.223" Score="1" Body="&lt;p&gt;I downloaded a macro for this many years ago and have since lost it.  It might have been this one: &lt;a href=&quot;http://www.mycheme.com/steam-tables-in-excel/&quot; rel=&quot;nofollow&quot;&gt;http://www.mycheme.com/steam-tables-in-excel/&lt;/a&gt; .  That link is worth a look either way.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-12T16:56:44.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24988" PostTypeId="1" AcceptedAnswerId="24991" CreationDate="2015-02-12T18:27:11.437" Score="0" ViewCount="1602" Body="&lt;p&gt;I would like to know the density of the compressed natural gas used for natural gas vehicles in order to complete the formula in &lt;a href=&quot;https://math.stackexchange.com/q/1141752/339790&quot;&gt;this&lt;/a&gt; Mathematics Stack Exchange post.&lt;/p&gt;&#xA;" OwnerUserId="13324" LastEditorUserId="30836" LastEditDate="2019-04-13T21:00:52.390" LastActivityDate="2019-04-13T21:00:52.390" Title="What is the density of compressed natural gas (CNG) used in vehicles?" Tags="&lt;gas-laws&gt;&lt;hydrocarbons&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-04-14T00:15:51.447" ContentLicense="CC BY-SA 4.0" />
  <row Id="24989" PostTypeId="1" AcceptedAnswerId="24994" CreationDate="2015-02-12T19:03:26.830" Score="4" ViewCount="1352" Body="&lt;p&gt;What happens to the lobe of the p-orbital during sp² hybridization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H4}$&lt;/span&gt; molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that both the carbon atoms will have a total of 3 p-orbitals and they will be involved in hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us number the carbon atoms as &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; along &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis will form a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C3}$&lt;/span&gt; bond with the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; and a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C3}$&lt;/span&gt; bond with the s-orbital of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And similarly the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; along x-axis will form a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C3}$&lt;/span&gt; bond with hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both the lobes of the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; along &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis are involved in sigma bond formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; along the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axis forms a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C0}$&lt;/span&gt; bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here also both the lobes of p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; are involved in hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let say that the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}1$&lt;/span&gt; which is along the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis forms a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C3}$&lt;/span&gt; bond with the s-orbital of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And similarly the p-orbital of &lt;span class=&quot;math-container&quot;&gt;$\ce{C}2$&lt;/span&gt; along &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis forms a &lt;span class=&quot;math-container&quot;&gt;$\unicode[Times]{x3C3}$&lt;/span&gt; bond with the s-orbital of the other hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my confusion is that here two p-orbitals along the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axes have four lobes and all of them are involved in bond formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the p-orbital which is along &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis has only one of its lobes involved in bonding. What happens to the other lobe of this p-orbital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9g8z6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13366" LastEditorUserId="46306" LastEditDate="2018-12-10T17:30:16.253" LastActivityDate="2018-12-10T17:30:16.253" Title="Chemical bonding and hybridization of organic compounds" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24990" PostTypeId="1" AcceptedAnswerId="24993" CreationDate="2015-02-12T19:08:09.747" Score="1" ViewCount="113" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nanoparticle#Functionalization&quot; rel=&quot;nofollow&quot;&gt;functionalization of nanoparticles&lt;/a&gt; (with coatings such as polymer ligands; see &lt;a href=&quot;http://what-when-how.com/wp-content/uploads/2011/03/tmp64293_thumb.jpg&quot; rel=&quot;nofollow&quot;&gt;image&lt;/a&gt;) are very common (for example, DNA-functionalizied gold nanoparticles). What's the type of interaction between nanoparticles and their coatings? And why is it more favorable than forming aggregations of the the nanoparticles?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-02-12T20:09:13.187" Title="What kind of interacion is there between a nanoparticle and its functionalizing coating?" Tags="&lt;bond&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24991" PostTypeId="2" ParentId="24988" CreationDate="2015-02-12T19:09:37.807" Score="2" Body="&lt;p&gt;The density of CNG at the pressure of 200 bars (which is optimized for use in vehicles) is 435 $\frac{kg}{m^3}$. However, it's (not usually) 175 $\frac{kg}{m^3}$ in some vehicles in some countries. Note that it differs according to your home country.&lt;br/&gt; &lt;a href=&quot;http://www.chive-ltd.co.uk/chivefuels/ngvfuels.htm&quot; rel=&quot;nofollow&quot;&gt;Source 1&lt;/a&gt; and &lt;a href=&quot;http://www.np.co.tt/our-brands/cng/&quot; rel=&quot;nofollow&quot;&gt;source 2&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-02-12T19:09:37.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24992" PostTypeId="1" AcceptedAnswerId="24997" CreationDate="2015-02-12T19:14:34.460" Score="2" ViewCount="192" Body="&lt;p&gt;Monomolecularity means no molecules which experience collision with each other taking part in reaction. So how it is possible that in monomolecular reaction only one reactant experiences collision?&lt;/p&gt;&#xA;" OwnerUserId="13365" LastEditorUserId="4945" LastEditDate="2015-10-23T16:53:50.140" LastActivityDate="2015-10-23T16:53:50.140" Title="In a monomolecular reaction, how can a reactant experience collision to react?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24993" PostTypeId="2" ParentId="24990" CreationDate="2015-02-12T19:17:28.207" Score="1" Body="&lt;p&gt;In small nanoclusters/nanocrystals/quantum clusters (my goodness, they have so many names nowadays) in the size range around 13, 55 and 144 metal atoms,&lt;sup&gt;1&lt;/sup&gt; the ligands enter a real orbital interaction (&lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/dative.html&quot; rel=&quot;nofollow&quot;&gt;dative bond&lt;/a&gt;&lt;sup&gt;2&lt;/sup&gt;). They are ligated to the metal core (&lt;a href=&quot;http://dx.doi.org/10.1002/anie.201310436&quot; rel=&quot;nofollow&quot;&gt;Puls et. al., 2014&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In DNA-ligated gold clusters, the phosphate groups are able to bond to the metal core and substitute the phosphine ligands it had before (&lt;a href=&quot;http://dx.doi.org/10.1002/anie.200250235&quot; rel=&quot;nofollow&quot;&gt;Yunping et. al., 2003&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In larger nanoparticles,&lt;sup&gt;3&lt;/sup&gt; the tendency of the particles to agglomerate is much smaller. Some of them can be measured with TEM (Transmission Electron Microscopy) without full agglomeration. This is not given for the small 13 metal atom clusters. The big particles enter interactions with their surfactants for which the main contribution is given either by van der Waals forces, or by electrostatic interactions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Those are some of the most common numbers in gold cluster chemistry, like $\text{Au}_{13}(\text{PR}_3)_{12}$ for example. 55 and 144 can be found in thiolated gold clusters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; It has covalent portions as well as ionic. It is, however, better understood if you think of coordination complexes. Only in this case, there is not only a single metal atom in the centre&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; i.e. 5 nm, 12 nm. I stated this because I am not aware of any case in which a 13-metal-atom cluster was analysed by TEM (Transmission Electron Microscopy). The conditions are simply too harsh, the ligand shell is ripped off and the little clusters experience a very strong tendency to agglomerate. You need specialised TEM equipment to acquire meaningful data. Nanoparticles of a few nanometers are less prone to show such a fast agglomeration.&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="10104" LastEditDate="2015-02-12T20:09:13.187" LastActivityDate="2015-02-12T20:09:13.187" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="24994" PostTypeId="2" ParentId="24989" CreationDate="2015-02-12T19:20:29.950" Score="3" Body="&lt;p&gt;You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only &lt;em&gt;one&lt;/em&gt; lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital &lt;em&gt;will not.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemtube3d.com/orbitalsethene.htm&quot; rel=&quot;nofollow&quot;&gt;This link&lt;/a&gt; provides a nice visualisation of the different orbitals involved in ethene bonding.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="7475" LastEditDate="2015-10-02T17:14:10.830" LastActivityDate="2015-10-02T17:14:10.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24995" PostTypeId="1" CreationDate="2015-02-12T19:28:08.767" Score="0" ViewCount="157" Body="&lt;p&gt;First of all. I am really bad at chemistry and all that. So this might or more likely will sound at least primitive... But this question has been bothering me for far too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I was reading about an atom. How all different kinds of it differ by the number of protons in its nucleus. And that gave me this utopian thought: maybe in far (like, really far) future we will have some sort of super &quot;printers&quot; which could take anything as their &quot;ink&quot; and somehow turn that anything (a brick for example) into manageable atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it probably should identify each atom, by counting its protons and edit it by &quot;deleting&quot;, &quot;adding&quot; or separating needed number of the protons and neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is... Do you think this is theoretically possible?&#xA;Are there any known ways of changing the number of protons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit: as it turns out, there is a talk going on about this kind of utopia: &lt;a href=&quot;https://www.youtube.com/watch?v=RzgVWpa4fzU&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=RzgVWpa4fzU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big thanks for your answers! I wasn't expecting to receive any answers :)*&lt;/p&gt;&#xA;" OwnerUserId="13364" LastEditorUserId="9961" LastEditDate="2016-08-21T20:57:58.597" LastActivityDate="2016-08-21T20:57:58.597" Title="&quot;Editing&quot; atom nucleus" Tags="&lt;atoms&gt;&lt;nanotechnology&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24996" PostTypeId="2" ParentId="24995" CreationDate="2015-02-12T20:31:58.823" Score="2" Body="&lt;p&gt;&lt;em&gt;Theoretically&lt;/em&gt; it's possible to create atoms from raw particles and to assemble them; science fiction posits &quot;replicators&quot; in Star Trek, &lt;a href=&quot;http://en.wikipedia.org/wiki/Replicator_%28Star_Trek%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Replicator_%28Star_Trek%29&lt;/a&gt;, and elsewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; done, a few billion or trillion atoms at a time, when matter is bombarded by protons in a cyclotron or linac. Americium, &lt;sup&gt;241&lt;/sup&gt;Am, for example, was first produced that way, and you probably have some in smoke detectors in your house, though now it is extracted from the random products of nuclear fission reactions. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Americium&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Americium&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-12T20:31:58.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24997" PostTypeId="2" ParentId="24992" CreationDate="2015-02-12T20:37:27.397" Score="2" Body="&lt;p&gt;One &lt;em&gt;reactant&lt;/em&gt;, e.g. O&lt;sub&gt;3&lt;/sub&gt;, a pure chemical species, has many &lt;em&gt;molecules&lt;/em&gt; that are in continual motion and can collide with each other (as well as with the walls of a container or with a catalyst), so you'll get some O&lt;sub&gt;2&lt;/sub&gt; generated. Or the converse, if you have UV photons colliding with pure O&lt;sub&gt;2&lt;/sub&gt; gas.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-12T20:37:27.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24998" PostTypeId="2" ParentId="24995" CreationDate="2015-02-12T20:37:36.263" Score="3" Body="&lt;p&gt;In part, your vision already exists and uses &lt;strong&gt;electrons&lt;/strong&gt; as &quot;ink&quot;. In a process kown as &lt;strong&gt;electron capture&lt;/strong&gt;, a proton &quot;swallows&quot; an electron of sufficient energy and is thus converted to a neutron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^{A}_{Z}X + e- -&amp;gt; ^{A}_{Z-1}Y + \nu_{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, the &lt;strong&gt;atomic mass number A&lt;/strong&gt; remains the same, but the &lt;strong&gt;atomic number Z&lt;/strong&gt; is decreased by one: Element &lt;strong&gt;X&lt;/strong&gt; is converted to element &lt;strong&gt;Y&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical example would be the conversion of aluminium to magnesium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^{26}_{13}Al + e- -&amp;gt; ^{26}_{12}Mg + \nu_{e}}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-12T20:37:36.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="24999" PostTypeId="2" ParentId="24964" CreationDate="2015-02-12T20:41:14.600" Score="1" Body="&lt;p&gt;The two questions are sort of related, so i'll answer to the best of my ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction in dynamic equilibrium can get close to completion depending on its equilibrium constant, which is a ratio between products and reactants at a certain temperature. No, the equilibrium constant won't change if you add more reactants in an attempt to push the reaction to completion. We refer to Le Chatelier's principle which states that if one were to add another parameter, say a reactant, the reaction will act to stabilize the stress that has been placed on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this reaction $\ce{PCl3 + Cl2 &amp;lt;=&amp;gt; PCl5}$. If I were to add more $\ce{PCl3}$, the change we would witness is a shift in equilibrium, that is to say, the system will react to consume more reactants and produce more products. What we see essentially is the system reacting to stabilize the changes we made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle isn't just limited to adding reactants or products, however. We see the system tries to stabilize changes made to pressure, temperature, and volume. If we increase pressure, how will the system act to stabilize it? Pressure is a result of the gas particles hitting the walls of the container. Thus we would see the system reduced the number of gas particles in order to reach equilibrium again. Essentially, it will shift to the side with the least amount of gas particles. Volume and pressure are obviously related in this principle. An increase in volume is a decrease in pressure and a decrease in volume is an increase in pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this reaction for example $\ce{CaCO3(s)&amp;lt;=&amp;gt;CaO(s) + CO2(g)}$. If we increase the volume (decrease the pressure), we would see the system shift to the right because it has the most moles of gas and we need more gas particles to fill the volume and stabilize the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, when changing heat we view it as a product or a reactant for exothermic and endothermic reactions respectively. An exothermic reaction has heat written as a product $\ce{A &amp;lt;=&amp;gt; B} + \text{heat}$ and an endothermic reaction has heat written as reactant. Thus, when we change the heat of an endothermic or exothermic reaction, we are changing its reactants or products and we treat it in the same way we treat a change in concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that the equilibrium constant remains constant, otherwise it would be called the equilibrium variable. We do, however, see a change in equilibrium position. &lt;/p&gt;&#xA;" OwnerUserId="13368" LastEditorUserId="5026" LastEditDate="2018-02-24T12:30:36.987" LastActivityDate="2018-02-24T12:30:36.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25000" PostTypeId="1" CreationDate="2015-02-12T20:48:55.277" Score="1" ViewCount="795" Body="&lt;p&gt;Here is the calculus way: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/LNnjD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no prior knowledge of calculus. But my teacher asks me to derive the integrated rate law for any order between one to ten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be fine for me to memorize the derivations for zero to third orders. But I can't do all ten for sure.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-02-12T22:46:01.110" Title="Is it possible to derive integrated rate law algebraically?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25001" PostTypeId="1" AcceptedAnswerId="25008" CreationDate="2015-02-12T21:01:59.700" Score="5" ViewCount="906" Body="&lt;p&gt;So let's say I have a symmetric 1,2-dihalide (the halides are identical) and I wanted to make a Grignard and I only add 1 eq. of magnesium metal ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will I get 1 eq. of asymmetric Grignards (i.e. only one side of the product has the C-MgBr bond) as my product or will:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Some of the product be &quot;double&quot; Grignards - i.e. both sides of what used to be the 1,2-dihalide now have C-MgBr bonds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) I get some E2 elimination product from Grignards robbing protons from unreacted 1,2-dihalides? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) A combination of the above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Something else entirely?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's say the above reaction works the way I want it to and we get the 1 eq. of asymmetric Grignard reagent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then react it with a carbonyl containing compound such as acetone ... do we get an epoxide? Are there even any routes to an epoxide? This suggested as a possible product ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see the epoxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvCGD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-02-12T21:25:02.457" LastActivityDate="2015-02-12T22:52:14.290" Title="Grignard formation and vicinal dihalides" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25002" PostTypeId="1" CreationDate="2015-02-12T22:26:09.050" Score="7" ViewCount="988" Body="&lt;p&gt;We know, that in the exited state, electrons of Phosphorus could go to higher states and form $\ce{dsp^3}$ hybridization. Therefore it will create five equal energy states of electrons. But my question is that, why are the angles of the bonds in $\ce{PCl5}$  90 and 120 degree instead of equally distributed? How does VSEPR theory explain the difference in the angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do the  five equal bonding pairs of electrons have a trigonal&#xA; bipyramidal arrangement, with two different internal bond angles -&#xA; 120° in the plane and 90° between the plane and the axial atoms?&lt;/p&gt;&#xA;" OwnerUserId="13369" LastEditorUserId="4945" LastEditDate="2015-02-13T04:38:52.017" LastActivityDate="2018-09-24T17:40:00.847" Title="How does VSEPR theory explain the formation of different bond angles in PCl₅?" Tags="&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25003" PostTypeId="1" AcceptedAnswerId="25014" CreationDate="2015-02-12T22:38:35.220" Score="3" ViewCount="371" Body="&lt;p&gt;Recently, I learn about bonding and my teacher only focus on ionic and covelant bond, not metallic bond. So, I go on google search but I have no idea of what is it? How electrons can suddenly appear to form the bond? How can the metal highest energy level overlap if it need to obey the octet rule&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-02-12T23:51:41.310" Title="Clarification for metallic bond" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25005" PostTypeId="2" ParentId="25000" CreationDate="2015-02-12T22:46:01.110" Score="7" Body="&lt;p&gt;Calculus is pretty much just a shorthand for algebra (plus the whole limits concept), so the answer is technically yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your teacher might be intending for you to use the pattern @Klaus Warzecha mentioned:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a function $f(x) = x^n$, the primitive integral is $F(x) = \vert\ln(x)\vert$ if $n=−1$, otherwise $F(x) = \frac{1}{n+1}x^{n+1}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in that sense, you can use algebra if you let&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]^n} \, \mathrm{d}[A] = \frac{1}{n - 1} \left[ \frac{1}{[A]^{n-1}_{\tau}} - \frac{1}{[A]^{n-1}_{0}} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n \neq 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]} \, \mathrm{d}[A] = \ln \frac{[A]_\tau}{[A]_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for $n = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this only works for single-species rate equations - if you have multiple species the integration result changes. Note that this also isn't &lt;em&gt;really&lt;/em&gt; using algebra - it's memorizing and applying one of many general integration results. Which is what you do in integral calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, it doesn't seem fair to me to expect you to know how to derive integrated rate laws if integral calculus isn't a prerequisite for your course. If it is a prereq, then now you are seeing why. If it isn't, then what your teacher is asking you to do is, in my opinion, unrealistic.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-02-12T22:46:01.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25006" PostTypeId="1" CreationDate="2015-02-12T22:46:02.933" Score="24" ViewCount="6075" Body="&lt;p&gt;In many fields of science (e.g. biology, medicine, psychology, statistics, physics), machine learning and artificial intelligence techniques are becoming more and more popular to analyze data. Is it also so in chemistry?&lt;/p&gt;&#xA;" OwnerUserId="13371" LastEditorUserId="5017" LastEditDate="2019-10-07T00:03:34.237" LastActivityDate="2019-10-07T00:03:34.237" Title="Does chemistry need machine learning?" Tags="&lt;computational-chemistry&gt;&lt;machine-learning&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="5" ClosedDate="2017-12-18T19:27:49.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="25007" PostTypeId="2" ParentId="24791" CreationDate="2015-02-12T22:49:36.827" Score="3" Body="&lt;p&gt;DNA is basically a polymer of nucleotides. These are held together by covalent bonds formed between the phosphate groups, each of which forms and ester with a hydroxyl group of the pentose of the NEXT nucleotide. This uses two of the three acidic 'OH' groups of the acid, leaving the last free to ionize. This ionization leaves a negative charge on each phosphate group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These charges allow the DNA strands to band to protein molecules which stabilize the structure, giving a nucleoprotein complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This complex helps stabilize the DNA—which as important in several ways&lt;br&gt;&#xA;– The DNA molecules are VERY long and easily broken&lt;br&gt;&#xA;– particularly when being moved etc at cell division&lt;/p&gt;&#xA;" OwnerUserId="13188" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T12:08:32.217" LastActivityDate="2015-04-02T12:08:32.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25008" PostTypeId="2" ParentId="25001" CreationDate="2015-02-12T22:52:14.290" Score="2" Body="&lt;p&gt;Taken from &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm&quot; rel=&quot;nofollow&quot;&gt;the Virtual Textbook of Organic Chemistry&lt;/a&gt;, which is maintained by &lt;a href=&quot;http://www.chemistry.msu.edu/faculty-research/emeritus-faculy/william-reusch/&quot; rel=&quot;nofollow&quot;&gt;William Reusch&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If two halogen atoms are present in a given compound, reactions with&#xA;  reducing metals may take different paths depending on how close the&#xA;  carbon-halogen bonds are to each other. If they are separated by four&#xA;  or more carbons, as in the first example below, a bis-organometallic&#xA;  compound may be formed. However, if the halogens are bonded to&#xA;  adjacent (vicinal) carbons, an elimination takes place with formation&#xA;  of a double bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He does however not provide any explanation for this behavior. It could be that this result is largely empirically based, about some aspects of the Gringard reaction mechanism there is not much known about (or consensus among chemists). I looked at Clayden et al. and Vollhardt &amp;amp; Shore and neither give a satisfying answer. If I find any in an other perhaps more advanced textbook, I'll edit this answer.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-12T22:52:14.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25009" PostTypeId="1" AcceptedAnswerId="26392" CreationDate="2015-02-12T22:54:10.373" Score="6" ViewCount="3165" Body="&lt;p&gt;Why do the Hs in (&lt;em&gt;E&lt;/em&gt;)-1-bromo-2-chloroethene split into doublets instead of staying as a singlet. Wouldn't they be isolated from each other since the double bond does not rotate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l6PUB.png&quot; alt=&quot;(E)-1-bromo-2-chloroethene&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorDisplayName="user7951" LastEditDate="2016-11-17T15:10:44.893" LastActivityDate="2017-04-02T19:56:53.070" Title="Splitting across a double bond in ¹H NMR" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25010" PostTypeId="1" CreationDate="2015-02-12T23:02:06.427" Score="1" ViewCount="525" Body="&lt;p&gt;I've been planning an experiment on how to make galvanic cells and during my research I came across an article in which chloridric acid was used, along with a soluble salt and agar-agar,  in salt bridges. &#xA;Since then, I've been wondering why chloridric acid was added to agar-agar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advence.  &lt;/p&gt;&#xA;" OwnerUserId="13370" LastActivityDate="2015-02-13T11:51:21.413" Title="Why is chloridric acid used in salt bridges?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25011" PostTypeId="2" ParentId="24964" CreationDate="2015-02-12T23:03:23.117" Score="1" Body="&lt;p&gt;To make it short and simple no you cannot because that is how the reaction works, however you can change the equilibrium constant by increasing or decreasing the temperature. One thing to point out is that adding or decreasing pressure, concentration can effect the equilibrium and make it shift to a certain direction. K will change if you make a change that the reaction cannot fix like temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one thing to note is that if we do get the reaction to go one direction then there is no concept of equilibrium and reversible reactions. There won't be a K because it will always go a certain way. You can't change the K when there isn't one. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can force it.&lt;/li&gt;&#xA;&lt;li&gt;No equilibrium constant if it is a &quot;complete&quot; reaction.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is just my insight into this with what I've learned in college chemistry.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-02-12T23:03:23.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25012" PostTypeId="1" CreationDate="2015-02-12T23:23:47.293" Score="1" ViewCount="536" Body="&lt;p&gt;I just started reading about the technique of chromatography and learning how important it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm trying to focus on whether is possible to use this technique in the geological fields (fossils and dating for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just brainstorming so any thoughts will help here.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="402" LastEditDate="2016-11-16T22:01:51.717" LastActivityDate="2016-11-16T22:01:51.717" Title="Chromatography in geology" Tags="&lt;chromatography&gt;&lt;geochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25013" PostTypeId="2" ParentId="25006" CreationDate="2015-02-12T23:45:35.960" Score="14" Body="&lt;p&gt;Yes. It's not a new thing at all in chemistry; you'll find papers on chemical AI applications going back four decades or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have massive amounts of data that must be processed quickly. For example, a diode array detector in HPLC can collect large numbers of spectra per minute at hundreds of wavelengths, and we need to use that data to distinguish and possibly identify two closely eluting compounds coming off the column. Intelligent, automated data analysis and pattern recognition is a must. A lot of &quot;artificial nose&quot; sensors have similar requirements; they can use neural nets to answer questions like &quot;are these potatoes infected with dry rot?&quot; We also use expert systems in various areas of analytical chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some nice overviews by Hugh Cartwright:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0849384125&quot;&gt;Using Artificial Intelligence in Chemistry and Biology: A Practical Guide&lt;/a&gt; (CRC Press)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://global.oup.com/academic/product/applications-of-artificial-intelligence-in-chemistry-9780198557364?cc=us&amp;amp;lang=en&amp;amp;&quot;&gt;Applications of Artificial Intelligence in Chemistry&lt;/a&gt; (Oxford University Press)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/9780470189078.ch8/summary&quot;&gt;Development and Uses of Artificial Intelligence in Chemistry&lt;/a&gt; (Reviews in Computational Chemistry, Volume 25, Wiley) &lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2015-02-12T23:45:35.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25014" PostTypeId="2" ParentId="25003" CreationDate="2015-02-12T23:51:41.310" Score="5" Body="&lt;p&gt;Metallic bonds depend on the Pauli exclusion principle to cause a single energy level to be separated into a sea of &lt;em&gt;very&lt;/em&gt; closely spaced levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider metallic Ca, with its s&lt;sup&gt;2&lt;/sup&gt; filled with two electrons of opposing spin state. You cannot have another s&lt;sup&gt;2&lt;/sup&gt; electron in the same atom. Now, add a second Ca atom that can freely interchange electrons with the first (e.g. metallic conductivity). Quantum mechanically, the electrons are indistinguishable, and &lt;strong&gt;an electron of the first atom might be found on the second one&lt;/strong&gt;, so the combination acts as a &quot;super-atom&quot; with two nuclei but a single electron cloud. To allow this without violating the Pauli exclusion principle, one s&lt;sup&gt;2&lt;/sup&gt; level &quot;moves up&quot; and the other &quot;moves down&quot;, splitting the level into two bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more atoms are added, the Fermi &quot;sea&quot; becomes populated to the extent that the energy levels merge into a continuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://fog.ccsf.cc.ca.us/~wkaufmyn/ENGN45/Course%20Handouts/15_ElectricalProps/02_ConductivityConductor.html&quot; rel=&quot;noreferrer&quot;&gt;http://fog.ccsf.cc.ca.us/~wkaufmyn/ENGN45/Course%20Handouts/15_ElectricalProps/02_ConductivityConductor.html&lt;/a&gt; and &lt;a href=&quot;http://www.chemexplore.net/sodium.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemexplore.net/sodium.htm&lt;/a&gt; for further explanation.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-12T23:51:41.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25015" PostTypeId="2" ParentId="25002" CreationDate="2015-02-12T23:52:58.190" Score="11" Body="&lt;p&gt;There are only five Platonic solids: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;tetrahedron, &lt;/li&gt;&#xA;&lt;li&gt;octahedron, &lt;/li&gt;&#xA;&lt;li&gt;cube, &lt;/li&gt;&#xA;&lt;li&gt;dodecahedron, and &lt;/li&gt;&#xA;&lt;li&gt;isocahedron.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.&lt;br&gt;&#xA;Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages &lt;a href=&quot;http://www.mathpages.com/home/kmath005/kmath005.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Min-Energy Configurations of Electrons On A Sphere&lt;/em&gt;&lt;/a&gt; (&lt;a href=&quot;https://web.archive.org/web/20180411111522/http://www.mathpages.com/home/kmath005/kmath005.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stable link via The Internet Archive&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:&#xA;&lt;a href=&quot;http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 5.5 (continued 2) &lt;em&gt;MO Theory ICl4, PCl5&lt;/em&gt;. (pdf)&lt;/a&gt; (&lt;a href=&quot;https://web.archive.org/web/20170516230216/http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stable pdf link via The Internet Archive&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2018-09-24T17:40:00.847" LastActivityDate="2018-09-24T17:40:00.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25016" PostTypeId="1" CreationDate="2015-02-13T00:42:46.097" Score="1" ViewCount="87" Body="&lt;p&gt;So I've got my data for some x-ray crystals of small molecules of Ag, Cu,... I was curios to know what the best way to determine what type of lattice each of these atoms may have?&lt;/p&gt;&#xA;" OwnerUserId="13373" LastActivityDate="2015-08-27T10:18:53.710" Title="Interpreting X-ray diffraction data" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2019-07-19T17:40:41.883" ContentLicense="CC BY-SA 3.0" />
  <row Id="25017" PostTypeId="2" ParentId="25006" CreationDate="2015-02-13T01:23:50.503" Score="23" Body="&lt;p&gt;&lt;strong&gt;The short answer is yes. Machine learning, data mining, AI and other techniques are highly useful in chemistry.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completely agree with &lt;a href=&quot;https://chemistry.stackexchange.com/a/25013/5017&quot;&gt;Fred's answer&lt;/a&gt; that lots of machine learning, expert systems and statistical analysis in chemistry goes back a long time. This is particularly true in analytical chemistry - match a mass spec or NMR or IR against a library of known compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I saw your tag &quot;computational chemistry&quot; and there are some newer applications here. Basically the question is whether you can predict properties (e.g., the heat of formation, dipole moment, etc.) that would normally require quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few groups attempting this, but one of the more successful attempts so far has come from &lt;a href=&quot;http://www.chemie.unibas.ch/~anatole/&quot; rel=&quot;noreferrer&quot;&gt;Anatole von Lilienfeld&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The researchers found that with a landscape of more than 5000 molecules, the error for predicting atomisation energies of new molecules drops below 10kcal/mol, approaching the 5kcal/mol accuracy of hybrid DFT. 'Calculating a molecule's atomisation energy using hybrid DFT would take on average one hour on a single CPU,' says von Lilienfeld. 'With machine learning, it's milliseconds.'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are more conventional approaches to use machine learning, evolutionary algorithms and the like to perform parameterization of force fields, semiempirical quantum mechanics, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you have a lot of data, machine learning techniques can be effective ways to analyze the data and use it for other purposes.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-02-13T01:23:50.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25019" PostTypeId="1" AcceptedAnswerId="25021" CreationDate="2015-02-13T03:36:47.640" Score="3" ViewCount="837" Body="&lt;p&gt;In what seemed like a good idea at the time, I used a can of compressed air to clean out my Magic Mill flour grinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture of one: &lt;a href=&quot;http://ic.pics.livejournal.com/ccdesan/4606041/68177/68177_320.jpg&quot; rel=&quot;nofollow&quot;&gt;http://ic.pics.livejournal.com/ccdesan/4606041/68177/68177_320.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It did a great job of cleaning out all the extra flour stuck to the bottom of the top part. However, what I didn't realize at the time was it also left bitterant residue, ruining a batch of flour I just made!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MSDS for the compressed air (Compucessory Power Duster) just lists diflouroethane, but it says right on the can it contains a bitterant!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bitterant is likely Denatonium (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bitterant&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Bitterant&lt;/a&gt;). Is there any household chemical that will dissolve it and the other common bitterants, or better yet, break them down into non-bitter compounds?&lt;/p&gt;&#xA;" OwnerUserId="12691" LastEditorUserId="7448" LastEditDate="2015-02-13T22:11:28.603" LastActivityDate="2015-02-15T04:42:49.917" Title="Dissolving/denaturing bitterant residue" Tags="&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25021" PostTypeId="2" ParentId="25019" CreationDate="2015-02-13T04:11:11.253" Score="1" Body="&lt;p&gt;Since probably only ppm levels of denatonium benzoate are in the compressed air, there are ppm levels or lower of denatonium in your flour grinder.  At that concentration, water should dissolve denatonium benzoate.  Is that a viable way to clean your grinder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://sentinel.antalys.net/index.php3?langue=english&amp;amp;content_menu=971857696&quot; rel=&quot;nofollow&quot;&gt;this solubility table&lt;/a&gt;, the best solvent for denatonium benzoate is methanol.  However I would &lt;strong&gt;not&lt;/strong&gt; recommend using methanol on anything you are using for food.  Ethanol is only slightly less good.  Maybe you can wash your grinder in some cheap vodka a few times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://groups.google.com/forum/#!topic/atxhs-discuss/mkuAL51Vjcg&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a random thread from Google Groups where someone claims to have identified a vendor for non-bitterized air duster cans.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-13T04:11:11.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25022" PostTypeId="1" CreationDate="2015-02-13T04:30:41.593" Score="5" ViewCount="28037" Body="&lt;p&gt;Acetone and Water are miscible, and water and diethyl ether are not. So why are acetone and diethyl ether miscible if water and acetone seem to have similar polarities?&lt;/p&gt;&#xA;" OwnerUserId="13377" LastActivityDate="2015-02-13T15:42:33.337" Title="Why is acetone and diethyl ether miscible, but not water and diethyl ether?" Tags="&lt;solubility&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25023" PostTypeId="1" CreationDate="2015-02-13T05:07:52.703" Score="1" ViewCount="22" Body="&lt;p&gt;What is the experimental procedure for applying fenton's reagent to leachate? What doses of hydrogen peroxide and ferrous sulphate required? How much contact time needs to be provided?&lt;/p&gt;&#xA;" OwnerUserId="13378" LastActivityDate="2015-02-13T05:07:52.703" Title="Fenton's reagent application to leachate" Tags="&lt;oxidation-state&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25024" PostTypeId="2" ParentId="24947" CreationDate="2015-02-13T09:41:33.520" Score="1" Body="&lt;p&gt;Both mechanisms will probably be present but $S_N2$ will dominate because the enolate anion is a strong nucleophile. $E2$ will not occur because there are no $\beta$ hydrogens available on the benzyl chloride. Despite the relative stability of the benzyl cation, especially in protic solvents, the strong nature of the enolate nucleophile and the lack of an $E2$ pathway means that $S_N2$ will probably dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a comparison of alkyl halide reactivities with different nucleophiles, see &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx3.htm#hal9&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-13T09:41:33.520" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25025" PostTypeId="2" ParentId="25010" CreationDate="2015-02-13T11:51:21.413" Score="2" Body="&lt;p&gt;Any ionic solution can function as a salt bridge. Having it acid based helps with the fast conduction paths of H+ in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical question is the leak rate. Any salt bridge will leak some amount of the ions into the solution. Good ones leak less, bad ones leak more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you cannot tolerate any acidity in your solution, don't use an acid. Similarly, if you cannot tolerate any chloride in your solution, this would be a bad choice.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-02-13T11:51:21.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25026" PostTypeId="1" AcceptedAnswerId="25037" CreationDate="2015-02-13T13:42:07.990" Score="4" ViewCount="279" Body="&lt;p&gt;I'm reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0596514921&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt;, trying to learn some basics of chemistry. One of the experiments is about purifying ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have some difficulties in deciding what kind of ethanol solution I should try to purify. The drugstores appear to typically sell denatured ethanol (so it could contain e.g. methanol?) or solutions with ethanol and isopropyl alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could potable alcohol be used? I think there's glucose in those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Doing this in my country is legal. It seems that is not the case in all parts of the world (or has not been).&lt;/p&gt;&#xA;" OwnerUserId="13381" LastEditorUserId="7448" LastEditDate="2015-02-13T21:40:38.120" LastActivityDate="2015-02-13T21:40:38.120" Title="About purifying ethanol - What type of ethanol?" Tags="&lt;home-experiment&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25027" PostTypeId="1" AcceptedAnswerId="25056" CreationDate="2015-02-13T13:49:32.843" Score="1" ViewCount="228" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One gram of Hydrogen reacts with exactly (almost) 8 grams of oxygen to produce $\ce{H2O}$. Another single gram of Hydrogen reacts with 16 grams of Oxygen to produce $\ce{H2O2}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can the ratio $\frac{m_O}{m_H}$ be determined from this information? If not, what else do chemists need to know to find this ratio?&lt;/p&gt;&#xA;" OwnerUserId="13382" LastEditorUserId="31313" LastEditDate="2017-04-01T16:40:28.180" LastActivityDate="2017-04-01T16:40:28.180" Title="Atomic mass ratios - Is the problem providing enough info?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25028" PostTypeId="1" CreationDate="2015-02-13T14:15:15.787" Score="2" ViewCount="4062" Body="&lt;p&gt;I have the SI chart in front of me. It states that a centi is 10^-2 and since they are already designated cubic meters why does 1 cubic meter equal 1000000 cubic centimeters?&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:49:37.663" LastActivityDate="2015-10-30T16:49:37.663" Title="Why does 1 cubic meter = 1000000 cubic centimeters?" Tags="&lt;units&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25029" PostTypeId="1" AcceptedAnswerId="25087" CreationDate="2015-02-13T14:28:25.240" Score="12" ViewCount="16581" Body="&lt;p&gt;My chemistry textbook keeps saying that in order for a molecule to be able to absorb infrared radiation, it has to have a change in dipole moment when the bond vibrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why that is. Some clarification would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13383" LastEditorUserId="7475" LastEditDate="2015-12-29T18:00:32.793" LastActivityDate="2017-01-16T06:35:03.577" Title="Why do molecules have to have a change in dipole moment in order for them to be IR active?" Tags="&lt;spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25030" PostTypeId="2" ParentId="25028" CreationDate="2015-02-13T14:31:25.910" Score="1" Body="&lt;p&gt;$1\ \texttt{m}^3$ is equal to $(\texttt{100 cm})^{3}$ because $1\ \texttt{m}$ is equal to $100 \ \texttt{cm} = 10^2\ \texttt{cm}$.&lt;/p&gt;&#xA;" OwnerUserId="13381" LastEditorUserId="13381" LastEditDate="2015-02-13T14:36:41.310" LastActivityDate="2015-02-13T14:36:41.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25031" PostTypeId="2" ParentId="25028" CreationDate="2015-02-13T14:33:19.170" Score="0" Body="&lt;p&gt;1 cm = $10^{-2}$ meter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e 1 m = $10^2$ cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 $m^3$ = $(10^2)^3 cm^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e 1 cubic meter = $10^6$ cubic centimeter&lt;/p&gt;&#xA;" OwnerUserId="13365" LastEditorUserId="5017" LastEditDate="2015-02-13T16:31:07.363" LastActivityDate="2015-02-13T16:31:07.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25032" PostTypeId="2" ParentId="25009" CreationDate="2015-02-13T15:16:16.310" Score="3" Body="&lt;p&gt;The hydrogens are not identical, one H is connected to a carbon directly connected to a bromine and the other to a chlorine. Since chlorine is more electronegative, the H on the carbon with Cl will be more unshielded.&lt;/p&gt;&#xA;" OwnerUserId="13086" LastEditorUserId="402" LastEditDate="2017-04-02T19:56:53.070" LastActivityDate="2017-04-02T19:56:53.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25033" PostTypeId="2" ParentId="25022" CreationDate="2015-02-13T15:42:33.337" Score="4" Body="&lt;p&gt;Acetone's dipole moment is 2.91D while that of ether is only 1.15D.  Water is a very polar substance, so acetone will dissolve in it while ether will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two substances are miscible when their intermolecular forces (IMFs) are similar enough such that the forces of attraction between molecules of different substances are similar in strength to the forces of attraction between molecules of the same substance.  If we look at the miscibility of A and B as an example, if A attracts B about as strongly as A attracts A and B attracts B the substances should be miscible.  If A and A have much stronger attraction than A and B the substances are less likely to be miscible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example it's not that some molecules of ether will not dissolves in water and vice versa, it's just that if the two liquids are in contact the water will stay with water and the ether with ether for the most part.  This is because water's strongest IMF is hydrogen bonding while ether's is dispersion with some dipole-dipole forces.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-13T15:42:33.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25034" PostTypeId="1" AcceptedAnswerId="25046" CreationDate="2015-02-13T15:57:46.717" Score="7" ViewCount="6841" Body="&lt;p&gt;In our teaching lab, we were posed with the following question as an exersize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the boiling points of two compounds differ by &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt; at atmospheric pressure, what will be the effect on the relationship by decreasing the pressure to &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mm}$&lt;/span&gt; (no number, just a general process)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I did not find much information on the subject in our lab manual or book but, but based off of some reading I've done around the internet (see below), it seems that for most binary solutions, decreasing the pressure increases the relative volatility.  However, asking some of my peers, they seems to believe that while both boiling points would decrease, the difference in their new boiling points would remain about the same (i.e. &lt;span class=&quot;math-container&quot;&gt;$\pu{\approx 50 ^\circ C}$&lt;/span&gt;), To make matters slightly more confusing, using a pressure-temperature nomograph, picking arbitrary numbers, I found that the relative volatility decreased (see Figure 1).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Who is more correct?  Does decreasing the system pressure increase the relative volatility or does it stay the same?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the relative volatility increases when system pressure is decreased, why is this the case?  I would actually guess the opposite, because as you decrease the pressure towards &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; (a vacuum), it would seem that all boiling points would approach &lt;span class=&quot;math-container&quot;&gt;$\pu{0 K}$&lt;/span&gt;, thus be getting closer together.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the relative volatility increases when system pressure is decreased, does that mean it is always more efficient to perform fractional distillation under a partial vacuum?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQtt0.png&quot; alt=&quot;two nomographs&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;Figure 1&lt;/strong&gt;  Two nomographs measured via &lt;a href=&quot;http://www.sigmaaldrich.com/chemistry/solvents/learning-center/nomo-assets.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pressure-Temperature Nomograph Interactive Tool by Sigma-Aldrich&lt;/a&gt; decreasing the system pressure to &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mm}$&lt;/span&gt; for two arbitrary substances with &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt; boiling point disparity at &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;References:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Effect of temperature or pressure on relative volatility &lt;a href=&quot;https://web.archive.org/web/20160803132827/https://chemengineering.wikispaces.com/Relative+volatility&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemengineering.wikispaces.com via the Internet Archive&lt;/a&gt; (under &quot;Effect of temperature or pressure on relative volatility&quot;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_distillation#Industrial-scale_applications&quot; rel=&quot;nofollow noreferrer&quot;&gt;Industrial-scale applications of vacuum distillation at Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="4945" LastEditDate="2019-05-21T12:14:48.263" LastActivityDate="2019-05-21T12:27:55.233" Title="Why does decreasing the pressure of the system increase the relative volatility of a binary solution?" Tags="&lt;boiling-point&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25035" PostTypeId="2" ParentId="25002" CreationDate="2015-02-13T15:59:19.560" Score="7" Body="&lt;p&gt;Here's a proof that you can't place 5 ligands equidistant from each other and from the central atom in a &lt;strong&gt;nonplanar&lt;/strong&gt; molecule:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place two points anywhere you like on the surface of a sphere around the central atom.&#xA;They will define a great circle (an &quot;equator&quot;) that cuts the sphere in half.&lt;/li&gt;&#xA;&lt;li&gt;You have 3 points left. You can't place them all off the &quot;equator&quot; without putting two in one hemisphere and 1 in the other. &lt;em&gt;You cannot make all the points equidistant.&lt;/em&gt; The only way you can balance the number of points in each hemisphere is to put one in one hemisphere, one in the other, and one on the &quot;equator&quot; with the first two you placed. But the distance between the equatorial points will not be the same as all of the distances to points in the hemisphere.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why is it trigonal pyramidal, though? The best we can do is make subsets of the ligands equidistant. Place three equidistant points on the equator, and two at the &quot;poles&quot;. Imagine a linear shape (the axial positions) passing at right angles through a trigonal planar shape (the equatorial positions). You have a trigonal bipyramidal shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dsp&lt;sup&gt;3&lt;/sup&gt; hybridization is a futile attempt to explain 5 orbitals around the central atom by mixing the central atom's orbitals alone. In fact back in the early 1990's it was shown that this is a poor explanation (source: E. Magnusson, Hypercoordinate molecules of second-row elements: d functions or d orbitals? &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1990,&lt;/strong&gt; &lt;em&gt;112&lt;/em&gt; (22), 7940–7951. &lt;a href=&quot;https://doi.org/10.1021/ja00178a014&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja00178a014&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="4945" LastEditDate="2018-09-24T17:20:16.803" LastActivityDate="2018-09-24T17:20:16.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25037" PostTypeId="2" ParentId="25026" CreationDate="2015-02-13T16:22:32.317" Score="2" Body="&lt;p&gt;Forget about the denatured alcohol! Additives are typically chosen to render their removal difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed you have decent setup for distillation, analytics still might be a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will you determine whether your destillation was successful or whether it just was &quot;thermal transfer&quot; from one flask to another?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hydrometer to measure the density of the liquid could be helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use some cheap vodka, mix with water and try to distill the vodka off. Measure the density before and after distillation. Note the temperatures, read about strange things that can happen: &lt;strong&gt;azeotropes&lt;/strong&gt;. Cheers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want it more simple and obvious, mix the vodka with some food colourant. Proceed as before and watch the effect. Cheers again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that one usually would not digest anything that went through any lab equipment  for mixed used at a normal lab. But this is your home lab and you know what went through which labware. Anyway, keep your equipment clean and take care. Btw: Cheers!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-13T16:22:32.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25038" PostTypeId="1" CreationDate="2015-02-13T16:25:48.127" Score="5" ViewCount="613" Body="&lt;p&gt;What book would you recommend for learning paper and thin-layer chromatography? I don't have access to fancy equipment (just to these two basic techniques that are relatively cheap), so it would be nice if the book focused on them in addition to being easy to understand etc. It does not matter if advanced techniques are discussed in a chapter or two but if the book is solely about them, it would not offer me much.&lt;/p&gt;&#xA;" OwnerUserId="13381" LastActivityDate="2015-03-02T16:57:50.217" Title="A good book for learning paper and thin-layer chromatography" Tags="&lt;chromatography&gt;&lt;books&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25043" PostTypeId="1" CreationDate="2015-02-13T17:09:15.987" Score="4" ViewCount="167" Body="&lt;p&gt;I am a student in computer science, which is a field where sharing research data and source code is very very much recommended. I am looking for sites where data from chemistry experiments could be openly distributed. Do such exist? I would like to test some algorithms for analyzing certain kinds of experiments, without having to do them myself (which I easily can't). The motivation is that I think (.. or maybe hope) that I would have some insights to offer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not find many suitable tags, sorry.&lt;/p&gt;&#xA;" OwnerUserId="13390" LastEditorUserId="7448" LastEditDate="2015-02-13T19:46:55.170" LastActivityDate="2015-08-26T21:39:54.013" Title="Open data in chemistry?" Tags="&lt;experimental-chemistry&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25044" PostTypeId="2" ParentId="25028" CreationDate="2015-02-13T17:20:06.450" Score="4" Body="&lt;p&gt;Take a look at the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Metric_prefix&quot; rel=&quot;nofollow noreferrer&quot;&gt;SI prefixes&lt;/a&gt;, in particular, the following excerpt:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When units occur in exponentiation, for example, in square and cubic forms, &lt;strong&gt;the multiplication prefix must be considered part of the unit, and thus included in the exponentiation&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1\ km^2$&lt;/span&gt; means one square kilometre, or the area of a square of &lt;span class=&quot;math-container&quot;&gt;$1000\ m$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$1000\ m$&lt;/span&gt; and not 1000 square metres.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2\ Mm^3$&lt;/span&gt; means two cubic megametres, or the volume of two cubes of &lt;span class=&quot;math-container&quot;&gt;$1000000\ m$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$1000000\ m$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$1000000\ m$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$2\times10^{18}\ m^3$&lt;/span&gt;, and not 2000000 cubic metres (&lt;span class=&quot;math-container&quot;&gt;$2\times10^{6}\ m^3$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Thus, the relationship between the cubic metre and the cubic centimetre is as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1\ m^3 = (1\ m)^3 = (100\ cm)^3 = 100\ cm \times 100\ cm \times 100\ cm =  1000000\ cm^3 = 10^6\ cm^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1\ cm^3 = (1\ cm)^3 = (0.01\ m)^3 =  0.01\ m \times 0.01\ m \times 0.01\ m = 0.000001\ m^3 = 10^{-6}\ m^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For a real case, converting &lt;span class=&quot;math-container&quot;&gt;$243.7\ cm^3$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$m^3$&lt;/span&gt; step by step:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$243.7\ cm^3 = 243.7\times 1\ cm^3 = 243.7\times (1\ cm)^3 =243.7\times (0.01\ m)^3 = 243.7\times 0.01\ m \times 0.01\ m \times 0.01\ m = 243.7\times 0.000001\ m^3 = 243.7\times 10^{-6}\ m^3 = 2.437 \times 10^{-4}\ m^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is simply the way it was defined, and it is the standard everyone adheres to. More examples of SI prefix usage can be found on &lt;a href=&quot;http://www.bipm.org/en/publications/si-brochure/chapter3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; from one of the organizations which created the SI units, the BIPM.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-13T17:38:06.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25045" PostTypeId="1" AcceptedAnswerId="25048" CreationDate="2015-02-13T18:04:46.067" Score="2" ViewCount="4668" Body="&lt;p&gt;I had dissolved a bunch of aluminium foil in an aqeous solution of sodium hydroxide. The resulting solution consists mainly of sodium aluminate with very little excess sodium hydroxide.&lt;br&gt;&#xA;What I want to do is convert this back to sodium hydroxide. I am thinking electrolysis should do the trick. But, I am not completely sure if this is right. If it is, what kind of electrodes should I use? If not, why not? (What will be the product?)&lt;br&gt;&#xA;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="7513" LastEditorUserId="7448" LastEditDate="2015-02-13T21:38:48.973" LastActivityDate="2021-05-14T14:33:38.460" Title="How to convert sodium aluminate back to sodium hydroxide?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25046" PostTypeId="2" ParentId="25034" CreationDate="2015-02-13T18:08:38.113" Score="3" Body="&lt;p&gt;Let's assume that the vapor pressure of both components can be modeled with the Antoine equation.  This gives us a convenient way to address the question algebraically.  The vapor pressure of most chemicals is excellently modeled by the correct &lt;a href=&quot;http://en.wikipedia.org/wiki/Antoine_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Antoine equation&lt;/a&gt; for that chemical.  The equation is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\log_{10}{p} = A-\frac{B}{C+T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the the parameters &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; are compound specific parameters.  The equation can be solved for temperature explicitly yielding &lt;span class=&quot;math-container&quot;&gt;$T = \frac{B}{A-\log_{10}\, p} - C$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;, the equation for pure species &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and also a different pure species &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; that is higher boiling by &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;T_1 &amp;amp;= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\&#xA;T_1 &amp;amp;= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; is the boiling point of &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;.  To make the math easier, let's suppose the equation is in units of atmospheres, in which case &lt;span class=&quot;math-container&quot;&gt;$\log_{10}\, p_{\pu{1 atm}} = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;T_1 &amp;amp;= \frac{B_i}{A_i} - C_i\\&#xA;T_1 &amp;amp;= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what happens when &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;T_{0,i} &amp;amp;= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\&#xA;T_{0,j} &amp;amp;= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have four equations but &lt;em&gt;six&lt;/em&gt; unknown parameters, &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A_j$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B_i$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B_j$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C_i$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C_j$&lt;/span&gt;.  &lt;strong&gt;So in general, there is no generally valid constraint for the temperature difference &lt;span class=&quot;math-container&quot;&gt;$T_{0,i}-T_{0,j}$&lt;/span&gt;:  It could be higher, lower, or about the same as the difference at room temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we could make some further assumptions.  A good one here could be that both species in question have a constant heat of vaporization, which means that &lt;span class=&quot;math-container&quot;&gt;$C_i=C_j=0$&lt;/span&gt;.  &lt;em&gt;If&lt;/em&gt; that assumption is true, then the four equations become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;T_1 &amp;amp;= \frac{B_i}{A_i}\\&#xA;T_1 &amp;amp;= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\&#xA;T_{0,i} &amp;amp;= \frac{B_i}{A_i+3}&#xA; =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}}&#xA; =\frac{T_1}{1+\frac{3}{A_i}}\\&#xA;T_{0,j} &amp;amp;= \frac{B_j}{A_j+3}&#xA; =\frac{T_1+50 °C}{1+\frac{3}{A_j}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the temperature difference of interest &lt;span class=&quot;math-container&quot;&gt;$T_{0,j}-T_{0,i}$&lt;/span&gt; is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the formulas easy let me define new parameters &lt;span class=&quot;math-container&quot;&gt;$a_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a_j$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$a_x=1+\frac{3}{A_x}$&lt;/span&gt;.  Then the temperature difference is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we must introduce further assumptions about the Antoine &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; parameters.  They are positive and thus the &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; parameters must also be positive, and further positive &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; implies  &lt;span class=&quot;math-container&quot;&gt;$a&amp;gt;1$&lt;/span&gt;.  &lt;em&gt;If&lt;/em&gt; we further assume &lt;span class=&quot;math-container&quot;&gt;$a_j=a_i$&lt;/span&gt;, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know already that &lt;span class=&quot;math-container&quot;&gt;$a_j&amp;gt;1$&lt;/span&gt;, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is &lt;em&gt;less&lt;/em&gt; than the temperature difference of &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt; at \pu{1 atm}.  Now, even if &lt;span class=&quot;math-container&quot;&gt;$a_i$&lt;/span&gt; is not precisely equal to &lt;span class=&quot;math-container&quot;&gt;$a_j$&lt;/span&gt;, then the temperature difference will still be less than &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt; as long as the &lt;span class=&quot;math-container&quot;&gt;$a_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a_j$&lt;/span&gt; parameters are similar enough to keep &lt;span class=&quot;math-container&quot;&gt;$(a_i-a_j)T_1 &amp;lt;&amp;lt; a_i(\pu{50 ^\circ C})$&lt;/span&gt;.  It would take a very unusual choice of chemicals &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; for that to happen.  I don't think I could identify such a pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, to get to the result that I think your instructor was asking for, we required &lt;strong&gt;many assumptions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption).&lt;/li&gt;&#xA;&lt;li&gt;Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; parameters.&lt;/li&gt;&#xA;&lt;li&gt;The Antoine &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; parameters for both chemicals must be positive.  (A good assumption but it still must be made.)&lt;/li&gt;&#xA;&lt;li&gt;The chemicals have Antoine &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; parameters (and thus also &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; parameters as I defined them here) that are not too different from each other.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than &lt;span class=&quot;math-container&quot;&gt;$\pu{50 ^\circ C}$&lt;/span&gt;, but I'm confident that the result is not necessarily generally true.  It isn't a thermodynamic law or anything.  It definitely depends on several key assumptions about how vapor pressure &quot;usually&quot; or &quot;typically&quot; behaves for most chemicals.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="4945" LastEditDate="2019-05-21T12:22:21.407" LastActivityDate="2019-05-21T12:22:21.407" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25048" PostTypeId="2" ParentId="25045" CreationDate="2015-02-13T18:29:22.253" Score="4" Body="&lt;p&gt;I don't think you will have much luck recovering the hydroxide.  To do so you would need to reverse the reaction that happened when the aluminum dissolved, which was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Al + 2 NaOH + 2 H2O -&amp;gt; 2 NaAlO2 + 3 H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your idea is essentially to substitute reduction at an electrode for reduction with hydrogen gas.  Unfortunately electrochemical reduction of $\ce{Al(III)}$ is very difficult.  If your idea was feasible, it would revolutionize aluminum manufacture!  Instead of &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_smelting&quot; rel=&quot;nofollow&quot;&gt;doing electrolysis of molten cryolite at ~1000 degress with sacraficial carbon electrodes&lt;/a&gt;, aluminum manufacturers could use your process.  So if you could get it to work, you'd be rich...but I am not hopeful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that you don't necessarily need to do electrolysis to recover sodium hydroxide.    There is hope!  You could just boil down your solution so that it is supersatured with respect to aluminum hydroxide, and then wait for the aluminum hydroxide to precipitate, which would happen faster if you add seed crystals of aluminum hydroxide $\ce{Al(OH)3}$) to initiate precipitation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O + NaAlO2_{(aq)} -&amp;gt; Al(OH)3_{(s)} + NaOH_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step is part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bayer_process&quot; rel=&quot;nofollow&quot;&gt;Bayer process&lt;/a&gt; for commercial refining of aluminum ore.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-13T18:29:22.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25051" PostTypeId="2" ParentId="25043" CreationDate="2015-02-13T19:10:49.447" Score="5" Body="&lt;p&gt;There are a variety of &quot;open data&quot; sources in chemistry. In your case, you probably want &lt;a href=&quot;http://en.wikipedia.org/wiki/Open_notebook_science&quot; rel=&quot;noreferrer&quot;&gt;&quot;open notebook science.&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few practitioners in chemistry, although &lt;a href=&quot;http://prospect.rsc.org/blogs/cw/2014/07/03/open-notebook-science-memorial-symposium-for-jean-claude-bradley/&quot; rel=&quot;noreferrer&quot;&gt;Jean-Claude Bradley&lt;/a&gt; sadly passed this last year.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://usefulchem.wikispaces.com&quot; rel=&quot;noreferrer&quot;&gt;JC Bradley&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://oruopennotebookscience.wikispaces.com&quot; rel=&quot;noreferrer&quot;&gt;Andrew Lang&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://malaria.ourexperiment.org&quot; rel=&quot;noreferrer&quot;&gt;Open Source Malaria&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I believe there will be a collection of articles published in the open access &lt;em&gt;&lt;a href=&quot;http://www.jcheminf.com/&quot; rel=&quot;noreferrer&quot;&gt;Journal of Cheminformatics&lt;/a&gt;&lt;/em&gt; as a memorial to Jean-Claude, but I don't think these have cleared review yet.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-02-13T19:10:49.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25052" PostTypeId="1" CreationDate="2015-02-13T19:15:24.447" Score="1" ViewCount="4787" Body="&lt;p&gt;I believe that a more polar bond would be stronger, since it would be similar to an ionic bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this seems to be wrong when applied to the study of acidity of some compounds.&lt;/p&gt;&#xA;" OwnerUserId="13265" LastEditorUserId="402" LastEditDate="2015-02-14T22:47:51.260" LastActivityDate="2015-03-19T03:02:24.873" Title="Is there a relationship between bond strength and polarity?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;bond&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25053" PostTypeId="1" CreationDate="2015-02-13T19:57:26.047" Score="0" ViewCount="454" Body="&lt;p&gt;I have an aqueous solution of methylene blue. How do I determine the concentration of this solution, so I can duplicate it?  &lt;/p&gt;&#xA;" OwnerUserId="13393" LastEditorUserId="418" LastEditDate="2015-04-18T05:31:31.057" LastActivityDate="2015-04-18T05:31:31.057" Title="Methylene blue used as funguscide in cichlid reproduction - concentration measurement?" Tags="&lt;aqueous-solution&gt;&lt;spectroscopy&gt;&lt;concentration&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25055" PostTypeId="2" ParentId="25053" CreationDate="2015-02-13T22:03:50.553" Score="3" Body="&lt;p&gt;The concentration of an aqueous solution of methylene blue can be determined using a UV-VIS spectrometer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure the absorption at $\lambda = 664\, \mathrm{nm}$, use the molar extiction coefficient of $95000\, \mathrm{dm^{3}\cdot mol^{-1}\cdot cm^{-1}}$ &#xA;(&lt;a href=&quot;http://www.clays.org/journal/archive/volume%2036/36-3-214.pdf&quot; rel=&quot;nofollow&quot;&gt;Reference&lt;/a&gt;) and calculate the concentration using the &lt;strong&gt;Lambert-Beer&lt;/strong&gt; equation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-13T22:03:50.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25056" PostTypeId="2" ParentId="25027" CreationDate="2015-02-13T22:09:29.973" Score="2" Body="&lt;p&gt;Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2n\space \rm{atoms \space H} = 1 \space g \space H$$&#xA;$$1n \space \rm{atoms \space O} = 8 \space g \space O$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide the second equation by the first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you didn't know the formulae, then you could assume that the ratio was some integer &lt;em&gt;multiple&lt;/em&gt; of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, what you are describing is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Law_of_multiple_proportions&quot; rel=&quot;nofollow&quot;&gt;law of multiple proportions&lt;/a&gt;, and it is one of the key findings that led to the development of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_theory&quot; rel=&quot;nofollow&quot;&gt;atomic theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-02-13T22:09:29.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25057" PostTypeId="2" ParentId="25034" CreationDate="2015-02-13T22:18:18.497" Score="2" Body="&lt;p&gt;My first answer focused on the boiling point difference.  But that's not what's practically important when separating compounds.  The concentration ratio in the vapor phase of the two compounds to be separated is what is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define some variables.  The total overall pressure is &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and the temperature is &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;.  The partial pressures of components &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$p_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_j$&lt;/span&gt;.  The mole fractions of &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; in the liquid phase are &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_j$&lt;/span&gt; and in the vapor phase are &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y_j$&lt;/span&gt;.  Lastly, the vapor pressure of &lt;em&gt;pure&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$p_i^*(T)$&lt;/span&gt; and of &lt;em&gt;pure&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$p_j^*(T)$&lt;/span&gt;.  Those vapor pressures are functions of temperature, we could choose the Antoine equation for example.  I won't write the &lt;span class=&quot;math-container&quot;&gt;$(T)$&lt;/span&gt; but keep in mind vapor pressures of pure species are functions of temperature (only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity let's focus on ideal solutions, which mean's the simplest form of Raoult's law applies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag1 p_i&amp;amp;=p_i^*x_i\\&#xA;\tag2 p_j&amp;amp;=p_j^*x_j&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of mole fractions must be &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; in both vapor and liquid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag3 1&amp;amp;=x_i+x_j\\&#xA;\tag4 1&amp;amp;=y_i+y_j&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the definition of partial pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag5 p_i &amp;amp;= y_i P\\&#xA;\tag6 p_j &amp;amp;= y_j P&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are vapor pressure relationships, e.g. Antoine equations for each of &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\tag7 p_i^* &amp;amp;= p_i^*(T) = \text{some equation with parameters and T only}\\&#xA;\tag8 p_j^* &amp;amp;= p_j^*(T) = \text{some equation with diff. parameters and T only}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is eight independent equations but there are ten variables.  So there are two degrees of freedom.  Let's assume that we have an equimolar liquid, i.e. that &lt;span class=&quot;math-container&quot;&gt;$x_i=0.5$&lt;/span&gt;.  From equation 3 that means &lt;span class=&quot;math-container&quot;&gt;$x_j=0.5$&lt;/span&gt; too.  Now there is only one degree of freedom left.  The general question is what pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; maximizes the relative volatility &lt;span class=&quot;math-container&quot;&gt;$\frac{y_i}{y_j}$&lt;/span&gt;.  The math is tough because you have to substitute one Antoine equation into the other to eliminate temperature.  I assumed &lt;span class=&quot;math-container&quot;&gt;$C_i=C_j=0$&lt;/span&gt; again for Antoine parameters, and got that &lt;span class=&quot;math-container&quot;&gt;$p_j^*=K (p_i^*)^{\frac{B_j}{B_i}}$&lt;/span&gt;, where K is a proportionality constant that depends in a complex way on Antoine &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A_j$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B_i$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$B_j$&lt;/span&gt; parameters.  The other non-Antoine equations reduce to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;y_i P &amp;amp;= p_i^*/2\\&#xA;(1-y_i)P &amp;amp;= p_j^*/2&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subbing the Antoine result in for &lt;span class=&quot;math-container&quot;&gt;$p_j^*$&lt;/span&gt; and then subbing the first equation into the result gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2^{d-1}P^{d-1} = \frac{K y_i^d}{1-y_i} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That expression doesn't have an analytical solution for &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; but in theory you could find the solution numerically if you knew all the Antoine parameters.  I graphed the equation for a few different values of &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; and y_i was always increasing as P got lower and lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So that's a long-winded way of saying that &quot;relative volatility&quot; is different than the difference in boiling point temperatures.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="4945" LastEditDate="2019-05-21T12:27:55.233" LastActivityDate="2019-05-21T12:27:55.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25058" PostTypeId="2" ParentId="25053" CreationDate="2015-02-13T22:25:03.717" Score="2" Body="&lt;p&gt;If in addition to a UV-VIS spectrometer you also have access to a fluorimeter, I would recommend using fluorescence instead of absorbance.  It will be more sensitive and less prone to interference.  However, fluorimeters are more expensive and harder to come by than UV-VIS spectrometers which measure absorbance only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I measured methylene blue by fluorescence, I used 630 nm for excitation and 700 nm for emission.  I found fluorescence to be much better than absorbance at 670 nm for the same samples, but the absorbance measurement also worked.  The fluorescence properties of methylene blue are less well-known than the absorbance.  &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/20117811&quot; rel=&quot;nofollow&quot;&gt;Other papers&lt;/a&gt; often use different values for excitation/emission than what I used, but if you have a fluorimeter you can determine what's best for you.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-13T22:25:03.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25059" PostTypeId="2" ParentId="25043" CreationDate="2015-02-13T22:33:18.420" Score="4" Body="&lt;p&gt;The types of databases you are thinking of probably don't exist for most types of chemistry work. The reason is that &lt;em&gt;interpreting&lt;/em&gt; data typically isn't the most difficult and expensive part, &lt;em&gt;obtaining&lt;/em&gt; the data is. There are exceptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bioinformatics: &lt;a href=&quot;http://www.ebi.ac.uk/services&quot; rel=&quot;nofollow&quot;&gt;http://www.ebi.ac.uk/services&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Protein structure: &lt;a href=&quot;http://www.rcsb.org/pdb/home/home.do&quot; rel=&quot;nofollow&quot;&gt;http://www.rcsb.org/pdb/home/home.do&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Basic properties of small molecules: &lt;a href=&quot;http://cccbdb.nist.gov/&quot; rel=&quot;nofollow&quot;&gt;http://cccbdb.nist.gov/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Material properties: &lt;a href=&quot;http://www.matweb.com/search/search.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.matweb.com/search/search.aspx&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are just example sites, there are many others out there. What they all have in common is that they touch on areas where theory is not very well developed, but we have the ability to generate large amounts of data at relatively low cost. I'm sure there are other areas like this as well, but you aren't likely to find things like tables of reactants and product concentrations for reactions. You mention &quot;certain kinds of experiments&quot; - if you can be more specific, someone can probably point you in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, check out the &lt;a href=&quot;http://srdata.nist.gov/gateway/gateway?dblist=0&quot; rel=&quot;nofollow&quot;&gt;NIST website&lt;/a&gt; - it lists a lot of free online scientific databases. Not all are chemistry, but maybe you will find what you are looking for there.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-02-13T22:33:18.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25062" PostTypeId="1" AcceptedAnswerId="25086" CreationDate="2015-02-14T03:07:39.770" Score="1" ViewCount="370" Body="&lt;p&gt;I have tried for several hours to answer this question with no luck. I believe I may be missing a couple of steps or I am not utilizing the given information properly. Nonetheless, some guidance will be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of $\ce{N2}$ in the ocean at 25C is 445 $\ce{\mu}$M. The Henry's Law constant for $\ce{N2}$ is 0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part A:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the mass of $\ce{N2}$ in a liter of ocean water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.45 x $\ce 10^{-4}$ mol/kg = molality&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N2}$ in moles = 28.014 g/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Molality Equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molality = $\frac{amount of solute (mols)}{mass of solvent(kg)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want kilograms, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.45 x $\ce 10^{-4}$ $\frac{mol}{kg}$ = $\frac{28.014 g}{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where do I go from here?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part B:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the partial pressure of $\ce{N2}$ in the atmosphere&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use C = K x $P_(gas)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.45 x $\ce 10^{-4}$ = (0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$)x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = 0.73 atm&lt;/p&gt;&#xA;" OwnerUserId="7663" LastEditorUserId="865" LastEditDate="2021-05-22T21:05:41.690" LastActivityDate="2021-05-22T21:05:41.690" Title="How do I utilize Henry's Law Constant?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;gas-laws&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25064" PostTypeId="1" AcceptedAnswerId="25175" CreationDate="2015-02-14T05:28:41.340" Score="4" ViewCount="309" Body="&lt;p&gt;What properties of zinc chromate make it a good anticorrosion paint?  I know that stainless steel is protected by a layer of chromium oxide.  Is it because the zinc chromate doesn't flake off the metal it is protecting?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-16T10:26:50.890" Title="What property of zinc chromate makes it a good anticorrosion paint" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25065" PostTypeId="1" AcceptedAnswerId="25067" CreationDate="2015-02-14T05:37:47.160" Score="2" ViewCount="1915" Body="&lt;p&gt;I am looking for a solvent for polypyrrole. I read M-cresol and chloroform can be used, but I am not entirely sure. &lt;/p&gt;&#xA;" OwnerUserId="13396" LastEditorDisplayName="user7951" LastEditDate="2015-05-22T16:56:48.620" LastActivityDate="2015-05-22T16:56:48.620" Title="What Can Dissolve Polypyrrole?" Tags="&lt;polymers&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25067" PostTypeId="2" ParentId="25065" CreationDate="2015-02-14T05:50:43.807" Score="6" Body="&lt;p&gt;Polypyrrole is quite a rigid polymer (one of the reasons it's a conductive polymer) and it's generally not considered particularly soluble without being derivatized. It seems to be &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0379677999002672&quot;&gt;possible&lt;/a&gt; to dissolve it somewhat in polar organic solvents if it's been polymerized with a large anion like dodecylbenzenesulfonate, basically reducing the interactions of the chains of polymer, but it loses some of its conductivity when cast into a film. &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/app.20882/full&quot;&gt;One group&lt;/a&gt; has apparently managed to make it water soluble by polymerizing oxidatively with horseradish peroxidase in the presence of a sulfonated polystyrene counterion, which keeps the chain growing linearly along the length of the polystyrene, apparently allowing it to be soluble.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-14T05:50:43.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25071" PostTypeId="2" ParentId="16747" CreationDate="2015-02-14T08:18:36.827" Score="3" Body="&lt;p&gt;I know its an old question but just wanted to make a technical correction .&#xA;The final integral values deduced by you are quiet correct but the method of arriving at OS of W looks clumsy . &#xA;As Freddy says , the compound is neutral , hence sum of all OS must be zero .&#xA;So we get OS of W as zero not by&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;W is neutral so W=0&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but rather using the fact that the compound is neutral and the other elements (which are in the form of ligands) form neutral ligands .&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-02-14T08:18:36.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25073" PostTypeId="1" CreationDate="2015-02-14T09:28:44.037" Score="5" ViewCount="160" Body="&lt;p&gt;Using thin-layer chromatography, it is possible to see what pigments plant leaves contain. For example, I can test whether there are chlorophyll a or b, carotenoids, or anthocyanins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of technique I would need to quantify the amount of these pigments in the leaves?&lt;/p&gt;&#xA;" OwnerUserId="13398" LastEditorUserId="13398" LastEditDate="2015-02-14T15:54:30.990" LastActivityDate="2015-02-14T15:54:30.990" Title="How to quantify amount of pigments in plant leaves?" Tags="&lt;organic-chemistry&gt;&lt;mixtures&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25074" PostTypeId="2" ParentId="25073" CreationDate="2015-02-14T09:48:34.033" Score="3" Body="&lt;p&gt;Once you've found a suitable mobile phase in analytical TLC, you could upscale the separation and transfer the results to preparative TLC or column chromatography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of column chromatography, collect the eluted material in test tubes, check the content by analytical TLC and/or UV-VIS spectroscopy, unite identical fractions, evaporate the solvent and determine the weight of the respective fractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of preparative TLC, scratch the separated bands off the plate, elute with a solvent of your choice and proceed as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If necessary, repeat the chromatographic separations for the different compound classes with other solvent mixtures. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-14T09:48:34.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25075" PostTypeId="1" CreationDate="2015-02-14T09:49:29.037" Score="0" ViewCount="109" Body="&lt;p&gt;What would happen if I kept two copper plates and one zinc plate in a solution of dilute sulfuric acid, and joined two copper plates with wires?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The wires would connect to a single wire and be joined from that wire to the single zinc plate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the voltage? Will there be something wrong?&lt;/p&gt;&#xA;" OwnerUserId="13400" LastEditorUserId="402" LastEditDate="2017-05-24T04:59:20.617" LastActivityDate="2017-05-24T04:59:20.617" Title="Double anode in simple cell" Tags="&lt;electrochemistry&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25076" PostTypeId="1" CreationDate="2015-02-14T09:50:25.113" Score="14" ViewCount="17349" Body="&lt;p&gt;When you pour &lt;strong&gt;water&lt;/strong&gt; into a &lt;strong&gt;clean glass&lt;/strong&gt; in a small confined room with &lt;strong&gt;no or little fresh air&lt;/strong&gt; circulation, the only smells you perceive are those of the glass and the water (or the impurities from both of them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;strong&gt;expose&lt;/strong&gt; that very glass to a bit of &lt;strong&gt;fresh air&lt;/strong&gt;, and voilà, it starts to &lt;em&gt;smell different&lt;/em&gt;, a bit like the (wet) &lt;strong&gt;shells of mollusks&lt;/strong&gt; in the sea or from a sweetwater source (it is &lt;em&gt;not&lt;/em&gt; a &lt;em&gt;&quot;fishy&quot;&lt;/em&gt; smell in the sense that is smells like fish but rather one reminding of the sea, lakes or rivers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this smell with wet &lt;em&gt;glass, ceramics, porcelain&lt;/em&gt;, but apparently not really or &lt;strong&gt;not&lt;/strong&gt; at all with &lt;strong&gt;plastic&lt;/strong&gt; or &lt;strong&gt;metal&lt;/strong&gt; (maybe the smell of metal somehow obstructs perception, or maybe it alters or destroys the chemical compounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is this smell actually?&#xA;What chemicals are there causing this smell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it apparently need contact with fresh air (maybe oxygen radicals or something) to emerge? And it seems to be volatile so that the smell weakens with time (or maybe you just get used to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope at least some here know what I am talking about... if not, maybe try it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people are apparently more susceptible to this smell than others (a friend of mine is very susceptible to it while others have no idea what this fuss is all about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also works when you get the glass hot and fresh from a dishwasher, though it is recommended to try with a non-scented detergent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you absolutely have to try it with &lt;strong&gt;non-chlorinated&lt;/strong&gt; water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;I've come across two substances that may be related to the phenomenon, &lt;a href=&quot;https://en.wikipedia.org/wiki/Trimethylamine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trimethylamine&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dimethyl_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dimethyl sulfide&lt;/a&gt;, however, I don't know how they smell, so I can't really say if any of these are responsible for the odors, but from the Wikipedia articles, trimethylamine sounds more fishy than &quot;mollusky&quot; while dimethyl sulfide is described to be cabbage-like.&lt;/p&gt;&#xA;" OwnerUserId="13399" LastEditorUserId="13399" LastEditDate="2018-04-02T19:47:46.187" LastActivityDate="2018-04-02T19:47:46.187" Title="What is the smell of a glass of water in contact with fresh air?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;food-chemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25077" PostTypeId="1" CreationDate="2015-02-14T11:40:29.383" Score="6" ViewCount="688" Body="&lt;p&gt;The way kinetics is taught at the undergraduate level (Arrhenius and collision theory) chemical equilibrium is determined governed immensely by activation energy of the reaction. According to thermodynamics, however equilibrium is a function of free energy change. In a way, thermodynamics and kinetics seem to contradict each other. What insight am I missing. Is the energy in energy vs reaction coordinates supposed to be Gibbs free energy instead of enthalpy? Would this imply activation energy changes as a reaction proceeds? Is the way kinetics is taught wrong?&lt;/p&gt;&#xA;" OwnerUserId="11586" LastActivityDate="2015-02-15T12:22:23.937" Title="Kinetics vs Thermodynamics" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25079" PostTypeId="2" ParentId="25077" CreationDate="2015-02-14T15:19:42.360" Score="0" Body="&lt;p&gt;The energy referred to on an energy diagram is enthalpy.  Think of the reaction A = B, an exothermic reaction involving two gases.  The forward reaction has lower activation energy so it is faster.  In order for the reaction to be in equilibrium there has to be more B than A so that the forward and reverse rates can be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the reaction A = B + C, an exothermic reaction where all the species involved are gases.  The reaction will produce more products for 2 reasons.  One, the forward reaction is easier in terms of activation energy than the reverse reaction.  Two, the reverse reaction requires 2 species coming together rather than 1.  Both have to be at the correct angle etc.  That could be thought of as a way that entropy affects the equilibrium position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to deal with the entropy change in a reaction like A = B.  I suppose if B has greater entropy than A this will slow the reverse reaction due to a lower frequency factor A as it is more difficult for B to be in the correct configuration to react.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-14T15:19:42.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25080" PostTypeId="1" AcceptedAnswerId="25082" CreationDate="2015-02-14T16:41:14.897" Score="2" ViewCount="636" Body="&lt;p&gt;The reduction of an amide using LAH yields a primary amine . I looked up the mechanism &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Amides/Reactions_of_Amides/Conversion_of_Amides_into_Amines_with_LiAlH4&quot; rel=&quot;nofollow&quot;&gt;over here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism suggests that the final main product (amine) is produced before giving the products a water bath (hydrolyze) . &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;So why do we need to add water at the end of the reaction?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/organicprops/amides/other.html&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt;'s  last paragraph suggests that 4[H] are required for the reaction. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two [H] are used up to make the amine (according to the mechanism), so what happens to the other two nascent hydrogens?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="865" LastEditDate="2015-10-16T17:49:27.177" LastActivityDate="2015-10-16T17:49:27.177" Title="Reduction of amides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25081" PostTypeId="1" AcceptedAnswerId="25109" CreationDate="2015-02-14T17:22:58.377" Score="2" ViewCount="899" Body="&lt;p&gt;Maltose has a 1,4 glycosidic linkage between two molecules of α-D-Glucopyranose.&lt;br&gt;&#xA;My textbook suggests the final name of maltose as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4-O-(α-D-Glucopyranosyl)-D-glucopyranose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I wanted to know what does &quot;4-O-&quot; stands for ? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I dont think so it provides information about the atoms which are linked and moreover feel that &lt;a href=&quot;https://en.wikipedia.org/wiki/Oligosaccharide_nomenclature&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's&lt;/a&gt; nomenclature (not IUPAC) : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;α-D-Glucopyranosyl-(1→4)-β-D-glucopyranose&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;more detailized . &lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorDisplayName="user7951" LastEditDate="2015-07-17T08:55:16.557" LastActivityDate="2018-11-19T20:12:35.330" Title="Maltose Nomenclature" Tags="&lt;nomenclature&gt;&lt;chemical-biology&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25082" PostTypeId="2" ParentId="25080" CreationDate="2015-02-14T18:21:12.610" Score="4" Body="&lt;p&gt;You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the $\ce{LiOH}$ will all go to the aqueous layer, whereas $\ce{Al(OH)3}$ is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;chemguide&quot; page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into $\ce{Al(OH)3}$ but the other two &quot;$\ce{[H]}$&quot;s just go to making the carbonyl oxygen into water. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="10185" LastEditDate="2015-02-14T18:23:37.300" LastActivityDate="2015-02-14T18:23:37.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25083" PostTypeId="2" ParentId="25077" CreationDate="2015-02-14T18:41:44.923" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;...chemical equilibrium is governed immensely by activation energy of the reaction.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sentence is incorrect. The relative amounts of species in a chemical equilibrium are not at all influenced by the kinetics; if anything, kinetics are influenced by the chemical equilibrium (such as in the Marcus theory of electron transfer reactions). &lt;a href=&quot;http://en.wikipedia.org/wiki/Equilibrium_constant#Derivation_of_equilibrium_constant_expression&quot; rel=&quot;noreferrer&quot;&gt;A rigorous derivation of the expression for an equilibrium constant&lt;/a&gt; can be made with no direct reference to reaction kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical equilibria are what happens to systems after they've been left without interference for an &lt;em&gt;infinite time&lt;/em&gt;; any system must reach a steady state in finite time, otherwise it would essentially be a perpetual motion device. In other words, for a chemical equilibrium to be achieved, the system is given more than enough time for any finite activation energy barrier to be overcome.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-02-14T18:41:44.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25085" PostTypeId="2" ParentId="25075" CreationDate="2015-02-14T20:26:13.817" Score="1" Body="&lt;p&gt;Neglecting any resistance in the connecting wires, this situation is no different than having one large copper plate. If everything is connected together, you can't measure any voltage because the fermi levels of all the metals will be equal (at equilibrium). This is essentially shorting the terminals of a battery and the zinc will start to dissolve in the electrolyte and hydrogen will be produced at the copper electrodes. You can measure a potential difference between the copper plates and the zinc plate, if they're not shorted, but it won't be any different with one copper plate or two for an open circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're actually using the cell to do work, you may get more work out of the cell with more copper area, as the larger surface area will result in less solution resistance and better diffusion.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-14T20:26:13.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25086" PostTypeId="2" ParentId="25062" CreationDate="2015-02-14T22:06:47.227" Score="3" Body="&lt;p&gt;Part A: Why do you want to use the molality?! You have the concentration i.e the number of moles in one liter. So the mass of nitrogen in one liter is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of moles in one liter $\times$ molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part B: It's correct.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-14T22:06:47.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25087" PostTypeId="2" ParentId="25029" CreationDate="2015-02-14T22:14:04.993" Score="4" Body="&lt;p&gt;As a molecule vibrates, if there is a fluctuation in its dipole moment, then this induces an electric field that interacts with the electric field associated with the infra red radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule, absorption occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please see &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Theory&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; on the UC Davis ChemWiki website. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="11367" LastEditDate="2015-12-29T16:43:47.483" LastActivityDate="2015-12-29T16:43:47.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25089" PostTypeId="1" AcceptedAnswerId="25090" CreationDate="2015-02-14T22:48:43.793" Score="9" ViewCount="2709" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs7Yk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above mechanism it is clear to see that the electrons from the double bond attack one of the terminal oxygens in ozone, thus meaning it's an electrophile. How can this be? If you draw the resonance structures, the negative charge is concentrated on the terminal oxygens too. I am OK with the rest of the mechanism but I would never think to draw oxygen as an electrophile. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="9961" LastEditDate="2018-11-13T17:16:22.393" LastActivityDate="2018-11-13T17:16:22.393" Title="Is ozone an electrophile in the ozonolysis reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;pericyclic&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25090" PostTypeId="2" ParentId="25089" CreationDate="2015-02-14T23:01:54.107" Score="7" Body="&lt;p&gt;Electrons are concentrated on the two terminal oxygens in ozone but remember, ozone exhibits resonance. The electrons can be delocalized across each oxygen through p-orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sxPH7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, at first if you think about the reaction in purely Coulombic terms then it doesn't make much sense. Why would electrons go to a region that already has a relative surfeit of electrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens is that by attacking the terminal oxygen, the double bond effectively displaces electrons to the central, electron-deficient oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure someone can explain this in much greater detail with reference to molecular orbital theory but I'll leave that to someone else. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-02-14T23:01:54.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25091" PostTypeId="1" AcceptedAnswerId="25183" CreationDate="2015-02-14T23:17:55.830" Score="13" ViewCount="2764" Body="&lt;p&gt;I know a lot of chemists like to quote $\mathrm{p}K_\mathrm{a}$ values and subtract them to reinforce some notion about some acid-base reaction being highly favorable or highly unfavorable or whatnot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the $\mathrm{p}K_\mathrm{a}$ of acetic acid is about 5 and the $\mathrm{p}K_\mathrm{a}$ of LDA's conjugate acid is about 36. 36 minus 5 gives us 31 &quot;zeros&quot; of favorability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\mathrm{p}K_\mathrm{a}$ values &lt;em&gt;themselves&lt;/em&gt; don't tell you what solvent you are using … can using this method &lt;strong&gt;blindly&lt;/strong&gt; lead to problems when working across solvents? And as far as I know, a lot of $\mathrm{p}K_\mathrm{a}$ values are measured in water … but in organic chemistry we don't just use water as a solvent. So does this method even have &lt;em&gt;general&lt;/em&gt; validity? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2017-09-23T17:50:39.037" LastActivityDate="2017-09-23T17:50:39.037" Title="pKa differences and reactivity" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25092" PostTypeId="1" AcceptedAnswerId="25093" CreationDate="2015-02-15T00:06:51.760" Score="5" ViewCount="793" Body="&lt;p&gt;I am trying to buy 1g of horseradish peroxidase. On sigma-aldrich, it is sold as units/mg. I am confused as to what this unit is. It comes in 50-150, 150-250, and 250-330 u/mg. The way I currently understand it is if I buy 25 KU (kilo-units?) of 50-150 u/mg, than I should have  at minimum 167 mg of the peroxidase. Is this a correct understanding? The price of 25 KU of horseradish peroxidase is $95, so it seems ridiculous to pay that for only 1 2/3g. &lt;/p&gt;&#xA;" OwnerUserId="13396" LastEditorUserId="4958" LastEditDate="2015-02-15T15:05:32.973" LastActivityDate="2015-02-15T15:05:32.973" Title="How to buy 1g of enzyme that is sold in units/mg?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25093" PostTypeId="2" ParentId="25092" CreationDate="2015-02-15T00:39:57.793" Score="5" Body="&lt;p&gt;Sigma Aldrich defines their units as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One pyrogallol unit will form 1.0 mg purpurogallin from pyrogallol in 20 sec at pH 6.0 at 20 °C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming you're doing a biological imaging experiment, these activity values tell you much more about what you're getting than mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a casual internet search it appears that HRP is pretty expensive. I would consider whether you actually need so much of it for your research. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-02-15T00:39:57.793" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25094" PostTypeId="1" AcceptedAnswerId="25096" CreationDate="2015-02-15T00:55:49.767" Score="2" ViewCount="540" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sJfGH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the bomardment of uranium-238 this is what I see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I see a uranium-238 atom with 92 protons bombarded with a neutron. The neutron then undergoes beta decay into a proton. So we then get Neptunium as the neutron decayed into a proton and hence, atomic and mass number increase by one. An electron is also released from the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the second equation of the bombardment of Plutonium-239. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plutonium-239 with 94 protons gets bombardment with 2 neutrons. These neutrons undergo beta decay to produce protons. Thus, the atomic and mass number must increase by two. The mass number, indeed, does increase by two. The atomic number does not. An electron is also released? Not sure why aswell.&lt;/p&gt;&#xA;" OwnerUserId="13406" LastEditorUserId="7448" LastEditDate="2015-02-15T12:36:22.090" LastActivityDate="2015-02-15T12:36:22.090" Title="Why doesn't the atomic number change by two in the process of production of Americium?" Tags="&lt;quantum-chemistry&gt;&lt;nuclear&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25095" PostTypeId="1" AcceptedAnswerId="25224" CreationDate="2015-02-15T01:30:10.287" Score="1" ViewCount="675" Body="&lt;p&gt;If there was a 500 mL container filled with oxygen and hydrogen so it burns completely (2H:1O), how many moles of hydrogen gas would there be if there if pressure is 101.3 kPa and the temperature is 293 K?&lt;/p&gt;&#xA;" OwnerUserId="13407" LastEditorUserId="402" LastEditDate="2015-02-17T07:45:14.797" LastActivityDate="2016-01-05T14:45:32.700" Title="Calculating moles of Hydrogen gas in multiple gas container" Tags="&lt;hydrogen&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25096" PostTypeId="2" ParentId="25094" CreationDate="2015-02-15T01:33:10.457" Score="4" Body="&lt;p&gt;What might be confusing is that it's actually two reactions. Two neutrons are captured followed by a single beta decay:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{^239_94Pu -&amp;gt;[2(n,\gamma)]^241_94Pu-&amp;gt;[\beta^-]^241_95Am}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reaction leaves the new nucleus in an excited state, so a gamma ray is emitted. The second reaction is actually quite slow with a halflife of ~14 years. It's a beta minus decay so that's why an electron is emitted.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-02-15T01:33:10.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25097" PostTypeId="1" AcceptedAnswerId="25143" CreationDate="2015-02-15T02:02:57.753" Score="-1" ViewCount="2313" Body="&lt;p&gt;'m trying to figure out why the oxidation state of the iodate ion is +5. The iodine atom has 7 electrons in its outermost shell (comprised of s and p sub-shells). Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom. This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how is the charge on the iodate ion negative when the third oxygen requires an electron?&lt;/p&gt;&#xA;" OwnerUserId="9763" LastActivityDate="2015-02-15T19:43:05.033" Title="Iodate ion and Octet rule" Tags="&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25098" PostTypeId="1" AcceptedAnswerId="25102" CreationDate="2015-02-15T02:06:58.627" Score="7" ViewCount="1402" Body="&lt;p&gt;Why does nature &quot;prefer&quot; symmetry? Why is symmetry energy-lowering? I keep hearing statements along the lines of this tossed around by chemists. Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/135671/nature-favours-symmetry&quot;&gt;https://physics.stackexchange.com/questions/135671/nature-favours-symmetry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=WkRy8b0F2HwC&amp;amp;pg=PA153&amp;amp;dq=nature+favors+symmetry&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=V__fVPvpGsuxggSptIG4AQ&amp;amp;ved=0CDAQ6AEwAw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.com/books?id=WkRy8b0F2HwC&amp;amp;pg=PA153&amp;amp;dq=nature+favors+symmetry&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=V__fVPvpGsuxggSptIG4AQ&amp;amp;ved=0CDAQ6AEwAw&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/14651/ibrf2-van-der-waals-repulsions-and-number-of-unique-bond-angles&quot;&gt;IBrF2 - van der Waals repulsions - and number of unique bond angles&lt;/a&gt; (The more symmetric conformation of the molecule is probably preferred). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Are any of the above statements true? If so, can you provide example(s)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I think the statement is false at least to some extent. Consider cyclobutadiene ... it's not a square; it's a &lt;em&gt;rectangle&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-02-15T17:46:57.003" Title="Does symmetry lower the energy of molecules?" Tags="&lt;symmetry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25099" PostTypeId="1" CreationDate="2015-02-15T02:12:55.203" Score="0" ViewCount="2874" Body="&lt;p&gt;$$\ce{Cr2O7^2-(aq) + 14H+(aq) + 6Cl-(aq) -&amp;gt; 2Cr^3+(aq) + 3I2(aq) + 7H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two chromium atoms gain 3 electrons each. They go from $0$ to $3+$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarily, 6 iodine atoms go from $-1$ to $0$, thus gaining one electron. However, my textbook says that iodine has lost electrons. How?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It went from negative to positive, i.e. from $-1$ to $0$, just like chromium from $0$ to $+3$. &lt;/p&gt;&#xA;" OwnerUserId="13406" LastEditorUserId="41328" LastEditDate="2017-08-14T13:55:02.867" LastActivityDate="2017-08-14T13:55:02.867" Title="How has iodine gained electrons?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25100" PostTypeId="1" CreationDate="2015-02-15T02:16:05.920" Score="3" ViewCount="9692" Body="&lt;p&gt;When a sample of green potassium manganate (VI), $\ce{K2MnO4}$, is added to aqueous sodium hydroxide, a brown black solid A and a purple solution of B are obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the brown black solid A is $\ce{MnO2}$, but I can't seem to write out the entire equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! Really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13408" LastEditorUserId="10185" LastEditDate="2015-02-15T15:22:54.810" LastActivityDate="2015-02-16T09:01:54.127" Title="Green Potassium Manganate and Sodium Hydroxide Equation" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25101" PostTypeId="2" ParentId="25098" CreationDate="2015-02-15T02:45:49.500" Score="7" Body="&lt;p&gt;&lt;strong&gt;Symmetry is preferred &lt;em&gt;if&lt;/em&gt; it lowers energy, but it does not always.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My high school chemistry teacher had a good rule of thumb. A chemical system will [almost] always tend towards the lowest energy configuration. (I insert &quot;almost,&quot; because there are some cases where this isn't true - usually because of entropy.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, many things are symmetric and this usually lowers the energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You raise a good counter-example. Cyclobutadiene is not square. A square cyclobutadiene would be a di-radical, so it undergoes symmetry breaking (i.e., a Jahn-Teller distortion) and becomes a rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDlHZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see here that the rectangular distortion has lower total energy - one orbital goes down in energy and becomes doubly-occupied. The other goes up in energy and is now empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what about benzene? Why is it symmetric, vs. a double-single-double.. bond alternation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in this case, the symmetric (aromatic) version is lower in energy.&#xA;(Figure, &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticity&quot; rel=&quot;noreferrer&quot;&gt;Chemwiki&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nIV15.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-02-15T02:45:49.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25102" PostTypeId="2" ParentId="25098" CreationDate="2015-02-15T02:46:48.693" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why does nature &amp;quot;prefer&amp;quot; symmetry?&lt;/p&gt;&#xA;&lt;p&gt;Why is symmetry energy-lowering?&lt;/p&gt;&#xA;&lt;p&gt;Are any of the above statements true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Things are either &amp;quot;true&amp;quot; or &amp;quot;false&amp;quot;.  The condition is binary, there is no &amp;quot;somewhat true&amp;quot;.  In that sense, the above statements are false.  There are many, many non-symmetric things in nature.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The human heart is not symmetrically positioned within the human body.&lt;/li&gt;&#xA;&lt;li&gt;The narwhal's tusk is helically grooved and projects from the left side of the jaw.&lt;/li&gt;&#xA;&lt;li&gt;You are not symmetric with the person standing next to you.&lt;/li&gt;&#xA;&lt;li&gt;Like people, most snowflakes are different one from another (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/20338/why-is-it-that-every-snowflake-is-unique/20358#20358&quot;&gt;here&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Alkane_stereochemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;staggered&lt;/a&gt; conformation (&lt;span class=&quot;math-container&quot;&gt;$\ce{D_{3d}}$&lt;/span&gt;, lower symmetry) of ethane is lower in energy than the eclipsed conformation (&lt;span class=&quot;math-container&quot;&gt;$\ce{D_{3h}}$&lt;/span&gt;, higher symmetry).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclooctatetraene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclooctatetraene&lt;/a&gt; exists in a tub shaped conformation with 4-fold symmetry, rather than planar with 8-fold symmetry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclobutane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclobutane&lt;/a&gt; exists in a puckered &lt;span class=&quot;math-container&quot;&gt;$\ce{D_{2d}}$&lt;/span&gt; geometry rather than a planar geometry with 4-fold symmetry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclohexane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclohexane&lt;/a&gt; exists in a chair conformation with a 3-fold axis of symmetry rather than a planar configuration with 6-fold symmetry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;From these examples we see that higher symmetry is not always preferred.  In those situations where a symmetric arrangement is preferred it is likely due to&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It requires less (DNA) code to create identical arms or arrangements on a living structure&lt;/li&gt;&#xA;&lt;li&gt;Electrostatic repulsions are often minimized by creating structures with higher symmetry (e.g. methane is tetrahedral rather than planar).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In those situations where the symmetric arrangement is not preferred evolutionary advantages or steric\electrostatic repulsions, hyperconjugation (ethane conformation), etc., outweigh other factors.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-15T17:46:57.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25103" PostTypeId="2" ParentId="25098" CreationDate="2015-02-15T02:51:20.477" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions&quot;&gt;Jahn-Teller theorem&lt;/a&gt; says almost opposite:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-15T02:51:20.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25104" PostTypeId="2" ParentId="25092" CreationDate="2015-02-15T04:00:06.667" Score="4" Body="&lt;p&gt;The enzyme is sold by activity, not by weight.  The different descriptions from Sigma refer to the amount of active enzyme (in units according to their definition) in 1 mg &lt;strong&gt;of the powder or lyophilized residue they sell you&lt;/strong&gt;.  You will not be buying pure HRP.  &quot;Pure&quot; HRP is a bit of a misnomer anyway, since it comes it at least seven isoforms, differing slightly in MW and glycosylation patterns, etc.  So if the U/mg number is &lt;em&gt;higher&lt;/em&gt;, the amount of &lt;em&gt;active&lt;/em&gt; enzyme per mg of residue is &lt;em&gt;higher&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most HRPs seem to have a molecular weight of around 40,000 g/mol.  If you knew the rate constant $k_{cat}$ of HRP for your substrate in your assay conditions, you could use the rate constant and the MW to calculate a &quot;theoretical&quot; activity in terms of U/mg.  My guess would be that this theoretical activity would be higher even than 250 - 330 U/mg.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the U/mg that you buy to this &quot;theoretical&quot; activity is the weight fraction of what you buy that is active enzyme.  Probably for the cheapest preparation, more than 90% of the weight is not active enzyme.  There are probably salts, water, and inactive/misfolded HRP molecules, and other protein contaminants in there to varying degrees.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-15T04:00:06.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25105" PostTypeId="2" ParentId="25019" CreationDate="2015-02-15T04:42:49.917" Score="0" Body="&lt;p&gt;After a bit of trial and error, this formula seems to work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/2 tablespoon bleach&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7.75 oz mouthwash (enough to make a cup)&lt;/p&gt;&#xA;" OwnerUserId="12691" LastActivityDate="2015-02-15T04:42:49.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25106" PostTypeId="2" ParentId="20299" CreationDate="2015-02-15T05:01:46.617" Score="1" Body="&lt;p&gt;One possibility would be to check the IR spectrum.  If there's any azide in there the azide band should show up as a strong band at around 2150 cm&lt;sup&gt;-1&lt;/sup&gt;.  The trouble is of course that this isn't quantitative, you can only prove the absence of azide, not how much is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Staudinger_reaction&quot; rel=&quot;nofollow&quot;&gt;Staudinger reaction&lt;/a&gt; might be a viable option.  PPh&lt;sub&gt;3&lt;/sub&gt; won't go up in aqueous solution, but &lt;a href=&quot;http://en.wikipedia.org/wiki/TPPTS&quot; rel=&quot;nofollow&quot;&gt;sulfonated PPh&lt;sub&gt;3&lt;/sub&gt;&lt;/a&gt; will.  One should be able to distinguish starting material and product by &lt;sup&gt;31&lt;/sup&gt;P NMR.  With a known quantity of TPPTS you can even determine % azide in the starting material.&lt;/p&gt;&#xA;" OwnerUserId="7920" LastActivityDate="2015-02-15T05:01:46.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25108" PostTypeId="2" ParentId="24953" CreationDate="2015-02-15T06:00:03.893" Score="2" Body="&lt;p&gt;Enantiomeric excess is very important concept in organic chemistry because it clearly states the purity of one isomer over the other. For example, have a look at the structure of Thalidomide which is racemic (50:50 —R:S isomers) in nature. Between the two form, one is having very important medical property where as another form is sedative. In these cases, synthesizing enantiomerically pure (&gt;99%) form is mandatory and in January 1996, the FDA announced it would consider further incentives for developing single isomer drugs, owing to their better pharmacokinetics prosperity, safety, and tolerability.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thalidomide&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Thalidomide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the reason which makes chemists to target enantimerically pure form using latest available technologies like, Chiral GC or HPLC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific optical rotation is very old concept to judge the enatiopurity of the compounds but its not accurate because it depends on various factors like concentration of the sample, solvent which we use, temperature, etc. Even small impurity in the sample gives drastic change in the rotation value. Thats why chemists adapt latest technique like HPLC of GC, nevertheless, chemists still carry out specific optical rotation to make sure that the enantiomerically excess compound has some rotation.&lt;/p&gt;&#xA;" OwnerUserId="3852" LastActivityDate="2015-02-15T06:00:03.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25109" PostTypeId="2" ParentId="25081" CreationDate="2015-02-15T06:23:47.227" Score="1" Body="&lt;p&gt;Maltose is 4-&lt;em&gt;O&lt;/em&gt;-α-D-glucopyranosyl-D-glucose.  The &quot;O&quot; in &quot;4-O&quot; indicates the attachment of atom of the substituent.  The substituent in this case is &quot;-α-D-glucopyranosyl&quot;.  It is attached to the &lt;em&gt;oxygen&lt;/em&gt; atom that is attached to carbon 4 in the base molecule.  The base molecule in this case is &quot;-D-glucose&quot;.  &quot;4-α-D-glucopyranosyl-D-glucose&quot; could be interpreted as a different structure where the &amp;alpha;-D-glucopyranosyl substituent is bound directly to &lt;em&gt;carbon&lt;/em&gt; atom number 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;base molecule&quot; of &quot;-D-glucose&quot; still has an unsubstituted anomeric OH group.  Thus, in aqueous solution, maltose will be an equilibrium mixture of at least these three isomers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;4-&lt;em&gt;O&lt;/em&gt;-α-D-glucopyranosyl-&amp;alpha;-D-glucopyranose&lt;/li&gt;&#xA;&lt;li&gt;4-&lt;em&gt;O&lt;/em&gt;-α-D-glucopyranosyl-&amp;beta;-D-glucopyranose&lt;/li&gt;&#xA;&lt;li&gt;4-&lt;em&gt;O&lt;/em&gt;-α-D-glucopyranosyl-D-glucose (open chain on the right-most &quot;glucose&quot; moiety in the name)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, it doesn't really make sense to disambiguate &amp;alpha; vs. &amp;beta; vs. open-chain forms of the &quot;right-most&quot; glucose ring in maltose, unless you need to refer to a particular conformation of maltose that matters for a certain reaction or condition.  However, the &quot;left&quot; ring is locked (in maltose) to an &amp;alpha;-D-glucopyranosyl conformation because it does not have an anomeric OH group and rapid equilibration in water does not take place.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-15T06:23:47.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25110" PostTypeId="1" CreationDate="2015-02-15T09:20:16.920" Score="2" ViewCount="2419" Body="&lt;p&gt;Is it linear, angular, triangular or tetragonal? Well, since $\ce{O}$ has 2 electron matches of his own, I think it's tetragonal, but in &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_difluoride&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; it is angular... Why? What about $\ce{CS2}$, is it linear?&lt;/p&gt;&#xA;" OwnerUserId="13411" LastEditorDisplayName="user7576" LastEditDate="2015-02-18T19:23:45.900" LastActivityDate="2015-02-18T19:23:45.900" Title="What is the geometrical structure of OF₂?" Tags="&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25111" PostTypeId="1" AcceptedAnswerId="25122" CreationDate="2015-02-15T09:28:58.310" Score="7" ViewCount="5493" Body="&lt;p&gt;What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="5026" LastEditDate="2018-06-07T05:33:02.210" LastActivityDate="2018-06-07T05:35:40.317" Title="What are the products of the reaction between fluorine and ammonia?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;halides&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25112" PostTypeId="1" AcceptedAnswerId="25135" CreationDate="2015-02-15T09:54:55.180" Score="1" ViewCount="884" Body="&lt;p&gt;What is Sn valence in this case? Because  it depends on it if it will be linear or tetragonal.I'm talking about a molecule.&lt;/p&gt;&#xA;" OwnerUserId="13411" LastEditorUserId="13411" LastEditDate="2015-02-15T11:17:14.613" LastActivityDate="2015-02-15T17:21:41.243" Title="Build the geometric structure of SnBr2" Tags="&lt;covalent-compounds&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25113" PostTypeId="1" AcceptedAnswerId="25129" CreationDate="2015-02-15T10:13:32.223" Score="8" ViewCount="1339" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7s8AX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I would like to ask why the alkyne reacts at all with the solvated electron? Surely the alkyne group is extremely nucleophilic (more so than the alkene). This, to me, seems to suggest that an alkene would be more likely to react in this manner than the alkyne. However, this is clearly not the case; why?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-02-15T15:54:43.973" Title="If Alkynes can be reduced to Alkenes using Sodium in liquid ammonia, why can't Alkenes be reduced to Alkanes?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25114" PostTypeId="1" CreationDate="2015-02-07T23:31:20.430" Score="5" ViewCount="541" Body="&lt;p&gt;I am simply curious what the theoretical arrangement of metallic hydrogen would be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://fuelcellsworks.com/news/2012/04/03/hydrogen-that-mimics-graphene/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article which gives a picture, but I don't understand what the labels mean (e.g. is H5 an isotope, some kind of charge, or just denoting the layer?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If the image is the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What the labels mean if it is correct.&lt;/p&gt;&#xA;" OwnerUserId="14497" OwnerDisplayName="Joseph Hankel" LastEditorUserId="9961" LastEditDate="2015-02-15T20:22:20.710" LastActivityDate="2015-04-11T11:57:44.160" Title="Structure of Metallic Hydrogen" Tags="&lt;crystal-structure&gt;&lt;metal&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25115" PostTypeId="1" AcceptedAnswerId="25137" CreationDate="2015-02-15T10:39:23.703" Score="13" ViewCount="17802" Body="&lt;p&gt;I have put acetone and sodium hydroxide solution (approximately $\pu{1M}~ \ce{NaOH}$) in a beaker and didn't see any colour change for a day (actually, it was transparent, no colour).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After almost 2 days, I saw a dark brown colour (no precipitate, just diffused colour). Now (after 25 days) its colour seems black, but when I take a small amount and dilute it with water, it is brown. What chemical compound has formed that has this colour? What reaction has taken place? (I doubt it is an aldol condensation)&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorUserId="5026" LastEditDate="2018-02-22T04:32:30.257" LastActivityDate="2018-02-22T04:32:30.257" Title="Color changes of a solution of acetone and sodium hydroxide" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25116" PostTypeId="2" ParentId="25110" CreationDate="2015-02-15T11:01:26.897" Score="1" Body="&lt;p&gt;You're right, if you're talking about the arrangement which take the nonbonding electrons into consideration,&#xA;Which is not the case in VSEPR geometry, &#xA;Since we have a difference in electronegativity between F and O (F is more electronegative)  ,  so we will have &lt;strong&gt;AX2 geometry which is angular.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fzI5J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13413" LastEditorUserId="7448" LastEditDate="2015-02-15T12:38:48.213" LastActivityDate="2015-02-15T12:38:48.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25117" PostTypeId="2" ParentId="20146" CreationDate="2015-02-15T11:43:50.130" Score="8" Body="&lt;p&gt;Radioactivity is not at all dependent on the atomic or molecular structure: it a purely nuclear phenomenon. The only thing that is relevant is the structure of the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleus contains only protons and neutrons and the distinguishing feature that makes nuclei radioactive is, crudely, the ratio of the two. Some arrangements are unstable for a variety of reasons a full understanding of which requires fairly advanced quantum theory. What you need to remember is that some ratios or arrangements of protons and neutrons are inherently unstable and spontaneously decay in a variety of ways to more stable nuclei. For example, &lt;sup&gt;39&lt;/sup&gt;K (which contains 19 protons and 20 neutrons) is stable but adding a neutron makes &lt;sup&gt;40&lt;/sup&gt;K which is slightly unstable and spontaneously decays in &lt;a href=&quot;http://www.wikiwand.com/en/Potassium-40&quot;&gt;several possible ways&lt;/a&gt;. You can tell this isn't a simple phenomenon, though, as &lt;sup&gt;41&lt;/sup&gt;K, with &lt;strong&gt;two&lt;/strong&gt; extra neutrons is stable again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to note that radioactive decay is essentially uninfluenced by anything outside the nucleus. For any given nucleus there is a probability that it will decay but that event occurs randomly and we can't tell when it will happen (but the rate of decay is stable and not influenced by temperature, pressure or the chemical compounds the nuclei are present in).&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-02-15T11:43:50.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25118" PostTypeId="2" ParentId="25111" CreationDate="2015-02-15T12:06:58.693" Score="1" Body="&lt;p&gt;The reaction is this&#xA;$$\ce{5 F2 {(g)} + 2 NH3 {(g)} -&amp;gt; N2F4 {(g)} + 6 HF {(g)}}$$&#xA;And it doesn’t depend upon which reactant is taken in excess; I believe it’s because fluorine is just too reactive.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="5026" LastEditDate="2018-06-07T05:35:40.317" LastActivityDate="2018-06-07T05:35:40.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25120" PostTypeId="2" ParentId="25099" CreationDate="2015-02-15T12:21:40.183" Score="1" Body="&lt;p&gt;Two points before we begin:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;+&lt;/code&gt; charge means deficiency in electrons.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;-&lt;/code&gt; charge means excess of electrons.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now lets come to the reaction you are talking about.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two chromium atoms gain 3 electrons each. They go from 0 to 3+&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Moving from neutral to +3 charge means it just lost 3 electrons. (&lt;em&gt;Electrons have negative charge.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Similarily, 6 iodine atoms go from -1 to 0, thus gaining one electron. However my textbook says that Iodine has lost electrons. How?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your textbook is right and you're wrong here. Moving from -1 charge to 0 means that an atom just lost an electron (&lt;em&gt;in other words, it lost the excess negative charge it had and now it is neutral&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1878" LastEditorUserId="1878" LastEditDate="2015-02-15T16:07:33.777" LastActivityDate="2015-02-15T16:07:33.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25121" PostTypeId="2" ParentId="25077" CreationDate="2015-02-15T12:22:23.937" Score="3" Body="&lt;p&gt;A great deal of chemistry is determined by the interaction of kinetics and thermodynamics. The world would be a dull place if we didn't have both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While thermodynamics determines what directions a reaction can go, the kinetics often determines whether the reaction can happen. Take a simple example: diamond is not the stable from of the element carbon at room temperature; graphite is. If the world were purely determined by thermodynamics all carbon would be graphite and men would be unable to impress women with the expensive shiny jewellery they buy them. But this reaction doesn't happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it doesn't is because to convert diamond into graphite we would have to have to radically restructure the chemical bonding in diamond from a tetrahedral network breaking at least one bond for every carbon atom. Breaking all those bonds requires a huge input of energy to kick off the process (thermodynamics says you will get that energy plus a little more back if you can make this happen). You can think of diamond with a whole bunch of broken carbon-carbon bonds as an &lt;em&gt;intermediate&lt;/em&gt; structure on the path from diamond to graphite. But that intermediate structure needs a great deal of energy to create and there simply isn't enough energy at room temperature to get there. The reaction will proceed at very high temperatures (and, if the pressure is high enough, diamond will be the thermodynamically favoured form). This is how the earth (or industry) creates diamonds: at high temperatures there is enough energy to break the carbon-carbon bonds and the equilibrium can be established (and, at high pressures, will favour diamond and at lower pressures graphite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of all this is that, in chemistry, what happens involves the &lt;em&gt;interplay&lt;/em&gt; of thermodynamics and kinetics. If it takes a lot of energy to get to an intermediate, the thermodynamic products cannot be reached. So being able to achieve an equilibrium depends not just on the start and end products of a reaction but the intermediate structures or compounds that you have to pass though to interconvert them. If those structures are hard to get to because you don't have enough energy around then thermodynamics is irrelevant to what will happen. you always have to think through the &lt;em&gt;mechanism&lt;/em&gt; behind the reaction and understand the energy required to create the intermediate states along the reaction path.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-02-15T12:22:23.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25122" PostTypeId="2" ParentId="25111" CreationDate="2015-02-15T13:39:58.143" Score="13" Body="&lt;p&gt;Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - &lt;a href=&quot;https://doi.org/10.1021/ja01532a066&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1959&lt;/strong&gt;, &lt;em&gt;81&lt;/em&gt; (23), 6338-6339&lt;/a&gt; - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{3F2 + 2NH3 → N2 + 6HF}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="5026" LastEditDate="2018-06-07T05:34:41.180" LastActivityDate="2018-06-07T05:34:41.180" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25123" PostTypeId="1" AcceptedAnswerId="25127" CreationDate="2015-02-15T14:36:43.963" Score="17" ViewCount="17507" Body="&lt;p&gt;It is common that adding common salt into water will lower its melting point. But, is there any way to RAISE the melting point of water? Does it lower the boiling point of water as well? Why?&lt;/p&gt;&#xA;" OwnerUserId="13415" LastActivityDate="2018-07-23T12:12:11.240" Title="How to RAISE the melting point of water?" Tags="&lt;water&gt;&lt;melting-point&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25125" PostTypeId="2" ParentId="25099" CreationDate="2015-02-15T15:17:41.843" Score="1" Body="&lt;p&gt;The mistake here is understanding what happens to the dichromate ion. Chromium has gone from the +6 to the +3 oxidation state. Its oxidation state has decreased by three therefore it has &lt;em&gt;gained&lt;/em&gt; three electrons in the process: $$\ce{Cr2O7^{2-} + 14H+ + 6e- -&amp;gt; 2Cr^{3+} + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine has gone from the -1 to the 0 oxidation state. Its oxidation state has increased by one, &lt;em&gt;losing&lt;/em&gt; one electron in the process. $$\ce{2I- -&amp;gt; I2 + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these two half equations together gives you the overall equation which you started with. $$\ce{Cr2O7^{2-} + 14H+ + 6I- -&amp;gt; 2Cr^{3+} + 7H2O + 3I2}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-15T15:17:41.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25126" PostTypeId="1" AcceptedAnswerId="25130" CreationDate="2015-02-15T15:41:50.100" Score="9" ViewCount="44125" Body="&lt;p&gt;In many sites I see that they take $\ce{C}$ with valence 4, bur our teacher takes it with valence 2. With what valence should I take $\ce{C}$?&lt;/p&gt;&#xA;" OwnerUserId="13411" LastEditorDisplayName="user15489" LastEditDate="2015-05-23T11:03:30.467" LastActivityDate="2016-07-26T15:48:33.077" Title="What is the Lewis Structure for CO?" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25127" PostTypeId="2" ParentId="25123" CreationDate="2015-02-15T15:44:42.237" Score="15" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JKHPb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/0/08/Phase_diagram_of_water.svg/569px-Phase_diagram_of_water.svg.png&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to raise the melting point of water is to increase pressure beyond about 635 MPa.  By raising pressure you could get the melting point to be even greater than the normal boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third way would be to added enough of a high melting point solute that you pass the &lt;a href=&quot;http://en.wikipedia.org/wiki/Eutectic_system&quot; rel=&quot;noreferrer&quot;&gt;eutectic point&lt;/a&gt;. In other words, generally when you add a solute to water, the melting point decreases, but only until you add enough solute that the eutectic point is reached, beyond which adding more solute increases the melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XTShP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://www.phasediagram.dk/binary/CaCl2.PNG&quot; rel=&quot;noreferrer&quot;&gt;http://www.phasediagram.dk/binary/CaCl2.PNG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the diagram, above, A is the eutectic point.  Beyond this point, adding more CaCl2 increases freezing point.  Then beyond 38% CaCl2, melting point exceeds ) degrees C.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-16T12:37:30.623" LastActivityDate="2015-02-16T12:37:30.623" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="25128" PostTypeId="1" AcceptedAnswerId="25152" CreationDate="2015-02-15T15:51:16.923" Score="10" ViewCount="30638" Body="&lt;p&gt;In many sites, I see that they put $\ce{O}$ in the middle, but our teacher said that we should put $\ce{O}$ near the extremity. Which is the best way?&lt;/p&gt;&#xA;" OwnerUserId="13411" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2015-12-09T09:10:11.820" LastActivityDate="2015-12-30T16:25:46.847" Title="What is the Lewis structure for HClO?" Tags="&lt;molecular-structure&gt;&lt;lewis-structure&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25129" PostTypeId="2" ParentId="25113" CreationDate="2015-02-15T15:54:43.973" Score="7" Body="&lt;p&gt;As shown in your diagram, when the initial electron is transferred to the alkyne a radical anion is formed; the transferred electron pairs with an electron from the alkyne to form an anion at one end of the molecule. These two electrons are in an $\ce{sp^2}$ orbital.  If you try to repeat this process with an alkene, you wind up with the paired electrons in an $\ce{sp^3}$ orbital.  An $\ce{sp^2}$ is lower in energy than an $\ce{sp^3}$ orbital since it contains more s character and an s orbital is lower in energy than a p orbital.  The electron solvated in ammonia has enough energy to transfer into a lower energy $\ce{sp^2}$ orbital, but not enough energy to transfer into a higher energy $\ce{sp^3}$ orbital.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-15T15:54:43.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25130" PostTypeId="2" ParentId="25126" CreationDate="2015-02-15T15:59:25.847" Score="10" Body="&lt;p&gt;There is resonance between various Lewis structures as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~&amp;lt;-&amp;gt;~ :\!C=O\!:: ~&amp;lt;-&amp;gt;~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;valence&quot; is not used much nowadays, and &quot;oxidation state&quot; is used but has a different meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/O04365.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC definition of oxidation state&lt;/a&gt; requires the oxygen have a -2 oxidation state except in peroxides.  All the oxidation states must add up to 0 for a neutral molecule.  Therefore the oxidation state of carbon in CO must be +2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;http://goldbook.iupac.org/V06588.html&quot; rel=&quot;noreferrer&quot;&gt;IUPAC definition of valence&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This imposes a unique valence upon each element, as explained in &lt;a href=&quot;http://www.cs.mcgill.ca/~rwest/link-suggestion/wpcd_2008-09_augmented/wp/v/Valence_%2528chemistry%2529.htm&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;, in which case valence is always 4 for carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other definitions of &quot;valence&quot; such as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The number of hydrogen atoms that can combine with an element in a binary hydride or twice the number of oxygen atoms combining with an element in its oxide or oxides&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Chemistry of the Elements by Greenwood et al.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this definition, carbon would have a valence of 2 in CO.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2015-02-16T04:31:32.530" LastActivityDate="2015-02-16T04:31:32.530" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="25131" PostTypeId="1" CreationDate="2015-02-15T16:25:09.487" Score="6" ViewCount="4413" Body="&lt;p&gt;For example consider the bromination of phenol we get the tri-substituted product (even though bromine is deactivating so I suppose it should deactivate the ring with each addition) while sulphonation essentially gives the mono-substituted product (even at high temperatures).&lt;img src=&quot;https://i.stack.imgur.com/FlmwO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13417" LastEditorUserId="10185" LastEditDate="2016-05-31T13:22:18.560" LastActivityDate="2019-01-01T05:16:46.040" Title="How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;reactivity&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25132" PostTypeId="2" ParentId="25128" CreationDate="2015-02-15T16:25:34.857" Score="5" Body="&lt;p&gt;Putting the oxygen in the middle is definitely the correct way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qy6tG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/od1cY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-15T16:25:34.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25133" PostTypeId="2" ParentId="25128" CreationDate="2015-02-15T16:37:35.547" Score="3" Body="&lt;p&gt;The lewis structure of Hypochlorous Acid: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/yN8tE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-15T16:37:35.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25134" PostTypeId="1" CreationDate="2015-02-15T17:10:07.827" Score="0" ViewCount="10719" Body="&lt;p&gt;If I have water in a sealed container heated to say 150 degrees, how do I determine the amount of pressure being generated in the container? What about for other liquids? I have searched extensively and cannot figure this out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a formula of sorts, but for example reasons let's say a 5000 ml container with 4500 ml of water in it, with the rest of the space air, heated to 150 degrees Celsius. &lt;/p&gt;&#xA;" OwnerUserId="13419" LastEditorUserId="4945" LastEditDate="2015-10-23T17:23:45.370" LastActivityDate="2016-06-24T00:07:00.127" Title="How do I calculate the pressure of a known liquid in a sealed container heated above boiling point?" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;pressure&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25135" PostTypeId="2" ParentId="25112" CreationDate="2015-02-15T17:21:41.243" Score="1" Body="&lt;p&gt;In the gas phase (molecular), SnBr2 is bent with an angle of 95 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC definition of valence is such that there is only one valence for each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valence is independent of the compound being consider.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-15T17:21:41.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25136" PostTypeId="2" ParentId="25110" CreationDate="2015-02-15T17:22:48.373" Score="5" Body="&lt;p&gt;Oxygen difluoride ($\ce{OF2}$) is analogous to water ($\ce{OH2}$) both in terms of structure and hybridization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywht6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{OF2}$ has 4 ligands around the central oxygen, 2 $\ce{O-F}$ bonds and 2 lone pairs.  Just like in water these 4 ligands assume a roughly tetrahedral geometry around the central oxygen. Again like in water, the $\ce{F-O-F}$ angle is less than the perfect tetrahedral angle (109.5°), in this case being around 103°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon disulphide ($\ce{CS2}$) is linear.  It is analogous to the 3 carbons in &lt;a href=&quot;http://en.wikipedia.org/wiki/Allene&quot; rel=&quot;nofollow noreferrer&quot;&gt;allene&lt;/a&gt;, in both compounds the central carbon is $\ce{sp}$ hybridized making the $\ce{X=C=X}$ structure linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCYGt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-15T17:22:48.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25137" PostTypeId="2" ParentId="25115" CreationDate="2015-02-15T17:23:24.910" Score="16" Body="&lt;p&gt;Under basic condtions, acetone does condense with itself, first to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Mesityl_oxide&quot; rel=&quot;noreferrer&quot;&gt;mesityl oxide&lt;/a&gt;, which is colorless, and then to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Isophorone&quot; rel=&quot;noreferrer&quot;&gt;isophorone&lt;/a&gt;, which has a yellow color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions.  That leads to bigger and more conjugated condensation products, which leads to darker and darker colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;http://www.nt.ntnu.no/users/skoge/prost/proceedings/icheap8-pres07/icheap8webpapers/102%20Sliepcevich.pdf&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt; which looks in some detail at the base-catalyzed polymerization of acetone, albeit with a different base and at high-temperature conditions in the gas phase, but I suspect a similar pathway happens, more slowly, at ambient temperatures in the liquid.  In the liquid, even larger molecules can likely form because they will stay solvated and can keep reacting (as opposed to in the gas phase where big molecules &quot;crash&quot; out of the phase as soot/gunk/etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color is likely in part from isophorone but much moreso from large polyconjugated molecules derived from isophorone containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V99Hv.png&quot; alt=&quot;Figure 2 from Sliepcevich et al.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-15T18:21:21.853" LastActivityDate="2015-02-15T18:21:21.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25138" PostTypeId="1" AcceptedAnswerId="25139" CreationDate="2015-02-15T17:47:17.317" Score="1" ViewCount="15753" Body="&lt;p&gt;I know this question is probably really rudimentary or I may just be fundamentally misunderstanding something but I would really like to just understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$n=\pu{1 mole}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c=\pu{1 moles/L}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v=\pu{1 L}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I learned this and it makes sense. However, this way, the equation doesn't make sense. It reads: &lt;span class=&quot;math-container&quot;&gt;$$\pu{moles}=\frac{\pu{moles/L}}{\pu{L}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which simplifies into &lt;span class=&quot;math-container&quot;&gt;$$\pu{moles}=\frac{\pu{moles}}{\pu{L}^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%3D%28x%2Fl%29%2Fl&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wolframalpha.com/input/?i=x%3D%28x%2Fl%29%2Fl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is obviously wrong. What am I not getting here?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="5026" LastEditDate="2019-04-25T03:02:10.273" LastActivityDate="2019-04-25T03:03:10.280" Title="How can n = c/v?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25139" PostTypeId="2" ParentId="25138" CreationDate="2015-02-15T17:50:33.477" Score="3" Body="&lt;p&gt;It should be &lt;span class=&quot;math-container&quot;&gt;$n=cv$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, your first equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{moles}=\frac{\pu{moles/L}}{\pu{L}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes&lt;span class=&quot;math-container&quot;&gt;$$\pu{moles}=\pu{\frac{moles}{L}}\times\pu{L}=\pu{moles}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2019-04-25T03:03:10.280" LastActivityDate="2019-04-25T03:03:10.280" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25140" PostTypeId="1" CreationDate="2015-02-15T18:16:42.350" Score="3" ViewCount="174" Body="&lt;p&gt;So I've got $$\ce{CaCO_3 + HCl \rightarrow CO_2 + H_2 O + CaCl_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering how I can recognize these types of reactions? I know that when I mix an acid and a base I get salt and water but in this reaction there is also some $\ce{CO2}$ which is confusing me. When can I expect this type of salt base reaction and should I expect more like these with other by products other than $\ce{CO2}$?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="10185" LastEditDate="2015-02-15T18:27:22.263" LastActivityDate="2015-02-15T20:51:08.620" Title="Carbonate + acid reactions - How to recognize these type of reactions?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25141" PostTypeId="2" ParentId="25134" CreationDate="2015-02-15T18:34:31.350" Score="0" Body="&lt;p&gt;Since steam (water vapor) is the most common working fluid in external-combustion engines, steam tables, as mentioned by Curt F above, are widely used and will show the relationship more accurately than the ideal gas law, &quot;PV=&lt;em&gt;n&lt;/em&gt;RT&quot;,  &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Ideal_gas_law&lt;/a&gt;. See &lt;a href=&quot;http://www.wolframalpha.com/examples/SteamTables.html&quot; rel=&quot;nofollow&quot;&gt;http://www.wolframalpha.com/examples/SteamTables.html&lt;/a&gt;, &lt;a href=&quot;http://www.efunda.com/materials/water/steamtable_sat.cfm&quot; rel=&quot;nofollow&quot;&gt;http://www.efunda.com/materials/water/steamtable_sat.cfm&lt;/a&gt; or &lt;a href=&quot;http://www.tlv.com/global/TI/calculator/steam-table-pressure.html&quot; rel=&quot;nofollow&quot;&gt;http://www.tlv.com/global/TI/calculator/steam-table-pressure.html&lt;/a&gt; for an online steam table.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-15T18:34:31.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25142" PostTypeId="2" ParentId="16410" CreationDate="2015-02-15T19:08:06.397" Score="2" Body="&lt;p&gt;Before starting of with my reasoning , I must tell you that your answer ( whether guessed or logically reasoned ) looks correct according to me .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that you are familiar with the Metal Reactivity Series ( if not look &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall take use of a well presented reactivity series :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://sjesci.wikispaces.com/file/view/reactivity_series1.gif/159802325/reactivity_series1.gif&quot; alt=&quot;Picture&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to crack the answer clue by clue :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal F and G react slowly with water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence F and G are somewhere at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal H displaces metal G from its chloride salt solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly H&gt;G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal E does not react with dilute acids &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;E will come somewhere at the bottom half so F,G&gt;E (from 1) and we can further say H&gt;G&gt;E (from 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal H does NOT displace metal F from its sulphate salt solution &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This gives F&gt;H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine last two clues' result to get F&gt;H&gt;G&gt;E .&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.530" LastActivityDate="2015-02-15T19:08:06.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25143" PostTypeId="2" ParentId="25097" CreationDate="2015-02-15T19:43:05.033" Score="3" Body="&lt;p&gt;First of all I strongly advise you to go through the following link as it will make my answer more clear to you .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Iodate&quot; rel=&quot;nofollow&quot;&gt;Iodate ion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bond between the iodine atom and oxygen atom is a &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Covalent&quot; rel=&quot;nofollow&quot;&gt;covalent bond&lt;/a&gt; . The electrons from the iodine atom are not given but shared .&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two double bonded oxygens ( refer the structure in my link ) share in all 4 electrons and the 3rd oxygen shares a single electron . Iodine hence has more electrons than octet rule specified 8. This is why it violates the rule .&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And how is the charge on the iodate ion negative when the third oxygen requires an electron?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third oxygen shares a single electron and hence has a negative charge on it which makes the overall molecule an anion with charge -1.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-02-15T19:43:05.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25144" PostTypeId="1" CreationDate="2015-02-15T19:43:50.850" Score="1" ViewCount="154" Body="&lt;p&gt;I'm taking chem classes, and am interested in learning a some biochem on the side, on my own. I've bee looking for a decent book, though I'm on a bit of a tight budget so not brand shiny new ones upwards of a hundred. Hopefully. Obviously I don't want something ancient just for reduced price, so two questions. How many years back is ok to use as a cutoff for publication date, and any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2015-02-15T21:46:01.307" Title="Biochem Book age and suggestions" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;books&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ClosedDate="2015-05-25T13:22:07.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="25145" PostTypeId="2" ParentId="25144" CreationDate="2015-02-15T19:48:11.537" Score="1" Body="&lt;p&gt;I had to read biochemistry by berg. I think the 7th edition is the latest one, but you can't go wrong with the sixth either. This is a good text to get a general overview of biochemistry that you can later apply to research systems in the future.&lt;/p&gt;&#xA;" OwnerUserId="13373" LastActivityDate="2015-02-15T19:48:11.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25146" PostTypeId="1" AcceptedAnswerId="25158" CreationDate="2015-02-15T20:06:21.130" Score="6" ViewCount="1044" Body="&lt;p&gt;I've had about one liter of fluorescein solution setting on the shelf for at least several years and one day noticed that it suddenly lost its characteristic green color. It's now a light yellowish-green solution that seems to have lost the property of fluorescence altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The container is a reagent bottle with a leak-proof cap so I doubt that the solution reacted with anything in the atmosphere other than maybe 100 mL of air that was in the bottle to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottle was exposed to direct and indirect sunlight on a daily basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with the decomposition of the fluorescein, and if so, what are the expected decomposition products?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="402" LastEditDate="2017-04-01T01:12:48.687" LastActivityDate="2017-04-01T01:12:48.687" Title="Does fluorescein decompose from long term exposure to light?" Tags="&lt;reaction-mechanism&gt;&lt;photochemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25147" PostTypeId="2" ParentId="24686" CreationDate="2015-02-15T20:28:11.277" Score="3" Body="&lt;p&gt;Firstly the following basic facts can be helpful :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ionic bond is stronger than covalent bond in most cases .&lt;/li&gt;&#xA;&lt;li&gt;A metal non-metal bond is more ionic.&lt;/li&gt;&#xA;&lt;li&gt;Non-metal non-metal bond is covalent in nature.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By considering &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;nofollow&quot;&gt;Valence Bond Theory&lt;/a&gt; , we can draw the fact that overlapping of larger orbitals ( usually of large atoms ) with smaller orbitals is less efficient and will result in a weaker bond .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HF&gt;HCl&gt;HBr&gt;HI is the order of their bond strengths .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because of the fact that as we go down the group , atomic size increases and so does its p-orbital which takes part in the overlapping with small s-orbital of H , making it a weaker bond .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronic configurations of various elements ( which is derived from the atomic number provided in the periodic table ) can be used to make the Lewis structures of most molecules and then the number of covalent bonds can tell comparitive enthalpies. ( more the bonds , more is the enthalpy )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also , some Periodic tables try to be more informative and provide &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Pauling_Electronegativity&quot; rel=&quot;nofollow&quot;&gt;Pauling EN&lt;/a&gt; alongside the atomic number and mass . In that case , bond polarity can be measured by calculating the EN difference . Usually , more the polarity of the covalent bonds, more they act like an ionic bond and are more stable .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though these tips can be of help in most comparisons , relying on actual values cover anomalous behaviors and exceptions which bump along our ways .&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-02-15T20:28:11.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25148" PostTypeId="2" ParentId="25144" CreationDate="2015-02-15T20:29:38.510" Score="2" Body="&lt;p&gt;I would suggest the Lehninger Principles of Biochemistry, which is one of the best in my opinion since it covers most of the key aspects in biochemistry.&#xA;It is quite expensive to buy as new, but since it is quite common you can maybe borrow it from a library. In this way you can get an impression about the book and eventually decide if it is worth buying it or not. If you are interested in having it I would buy a secondhand one and I would say it does not matter if it is not the last version, the previous one should be good as well.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastActivityDate="2015-02-15T20:29:38.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25149" PostTypeId="2" ParentId="22257" CreationDate="2015-02-15T20:40:17.480" Score="3" Body="&lt;p&gt;Nature always try to acheive least potential energy for stability . Any organic reaction which makes the product less stable is not favourable . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the C-H bonds are cleaved homolytically in all cases , the resulting stability of the free radicals will tell us about the feasibility of the reaction . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary free radical is the most unstable and to bring about its C-H bond breaking would mean increasing its energy comparitively . Hence its bond energy is the most and so its the stronger bond of the three .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tertiary free radical is the most stable while secondary lies in between them . The same explanation can be used to draw out their comparitive strength of the C-H bonds .&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="13339" LastEditDate="2015-02-17T04:04:35.760" LastActivityDate="2015-02-17T04:04:35.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25150" PostTypeId="2" ParentId="25140" CreationDate="2015-02-15T20:43:32.370" Score="3" Body="&lt;p&gt;The reaction you showed is actually a salt &quot;exchange&quot; reaction (also called salt metathesis), in which a salt is transformed into another one during the course of the reaction. In your example when calcium carbonate is treated with hydrochloric acid, calcium chloride is formed. At the same time, carbonic acid is generated and it spontaneously decomposes into carbon dioxide and water. The equilibrium is shifted towards the products due to the release of carbon dioxide gas from the reaction mixture, thus driving the reaction to completion.&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorUserId="11593" LastEditDate="2015-02-15T20:51:08.620" LastActivityDate="2015-02-15T20:51:08.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25151" PostTypeId="2" ParentId="25131" CreationDate="2015-02-15T20:46:19.027" Score="6" Body="&lt;p&gt;It is like others have already noted that phenol is very strongly &lt;em&gt;ortho-para&lt;/em&gt; directing, while bromine is only slightly deactivating (or sometimes slightly activating). It is so activating that you do not even need a Lewis acidic catalyst or elevated temperature to let the reaction go fast and in good yield. Moreover, if you want the mono substituted product you will actually need to cool the reaction down to &amp;lt; 5 °C and use the rather dangerously flammable solvent $\ce{CS_2}$, yielding 4-bromophenol in 85% yield (source: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0582462363&quot;&gt;Vogel&lt;/a&gt;). I assume the para product is formed due to the fact that there is no significant difference in electron density between ortho and para carbon atom while ortho is slightly sterically hindered. &lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-15T22:31:52.287" LastActivityDate="2015-02-15T22:31:52.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25152" PostTypeId="2" ParentId="25128" CreationDate="2015-02-15T20:52:16.643" Score="3" Body="&lt;p&gt;The issue is that $\ce{HOCl}$ will dissociate, as usually, into $\ce{H+ + OCl-}$, which can only be possiblbe because $\ce{Cl}$ is formally $\ce{Cl+}$ and not the expected $\ce{Cl-}$. If $\ce{O}$ was to be put in an extremity, one would end up with $\ce{OHCl}$, which makes no sense because $\ce{H}$ cannot make two bonds, or $\ce{OClH}$, which makes no sense because $\ce{HCl}$ exists independently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypohalous acids, like $\ce{HOCl}$, are a point of entry into the &quot;weird&quot; chemistry of halides, which at least in Euopean countries I know is not covered until late in college.&lt;/p&gt;&#xA;" OwnerUserId="13421" LastEditorUserId="7475" LastEditDate="2015-12-09T10:46:44.487" LastActivityDate="2015-12-09T10:46:44.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25153" PostTypeId="1" CreationDate="2015-02-15T21:36:08.860" Score="2" ViewCount="285" Body="&lt;p&gt;I noticed when I poured my soy milk (which I shook as per the directions on the carton) into my coffee, a large number of very fine bubbles rose to the surface. What causes this? Is there a chemical reaction involved, or is it just a physical effect?&lt;/p&gt;&#xA;" OwnerUserId="13422" LastEditorUserId="402" LastEditDate="2017-04-01T00:56:35.460" LastActivityDate="2020-04-02T19:00:13.400" Title="Interaction between soy milk and coffee producing bubbles" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25154" PostTypeId="2" ParentId="25144" CreationDate="2015-02-15T21:46:01.307" Score="1" Body="&lt;p&gt;I know you are asking for a textbook, but if you want to do some casual learning here are a few pop-sci biochemistry-focused books that I recommend giving a try:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198607830&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Oxygen&lt;/em&gt;&lt;/a&gt; by Nick Lane.  I like this book best but any of Nick Lane's books would be a good start.  This one focuses on the profound changes oxygen has had on life on Earth, from evolution to biochemistry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0300070551&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Blondes in Venetian Paintings...&lt;/em&gt;&lt;/a&gt; by Konrad Bloch.  A series of essays on completely different topics, all recounting various aspects of the Nobel-winning career of a great biochemist.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0200000020&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;The Chemicals of Life&lt;/em&gt;&lt;/a&gt; by &lt;strong&gt;Isaac Asimov&lt;/strong&gt;(!).  Sure this one is very dated but by all accounts it's extremely well written.  (I haven't read this one yet myself but it's on my list.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-15T21:46:01.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25155" PostTypeId="1" CreationDate="2015-02-15T23:06:20.917" Score="1" ViewCount="3019" Body="&lt;p&gt;I am programming a small model to simulate diffusion of $\ce{H+}$ ions in human tissue. It's quite a simplistic model based on a naive algorithm, and I'm mostly doing it to familiarize with the concept of agent-based models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My space is an $n\times n$ grid where each position $(x;y)$ hosts $h$ $\ce{H+}$ ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $\ce{pH}$ can be calculated based on $h$ as $-\log(h)$. As I want my initial $\ce{pH}$ to be stable at average human $\ce{pH}$, I'd set $h$ across the matrix to be $10^{7.5}$. (In fact, as I don't want to simulate $10^{7.5}$ protons I decided each particle will represent a bulk of $10^5$ protons and made my equation $-\log(10^5\cdot h)$. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not understand the equation, wouldn't it at some point yield a negative $\ce{pH}$ as $-\log(x) &amp;lt; 0$ for $x &amp;gt; 1$? &lt;/p&gt;&#xA;" OwnerUserId="13423" LastEditorUserId="4945" LastEditDate="2015-02-16T09:06:13.427" LastActivityDate="2015-02-16T14:06:11.530" Title="Can a negative pH exist?" Tags="&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2015-02-16T14:16:41.083" ContentLicense="CC BY-SA 3.0" />
  <row Id="25156" PostTypeId="2" ParentId="25155" CreationDate="2015-02-15T23:27:25.503" Score="3" Body="&lt;p&gt;Yes, pH can be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The official definition of pH is: negative log of H+ activity, not negative log of H+ concentration, however.  See &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed083p752?journalCode=jceda8&quot; rel=&quot;nofollow&quot;&gt;pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+]&lt;/a&gt; J. Chem. Educ., 2006, 83 (5), p 752.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By either definition pH can be negative in concentrated solutions of strong acids.  For example 5M HCl would have a negative pH.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-16T14:06:11.530" LastActivityDate="2015-02-16T14:06:11.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25157" PostTypeId="1" AcceptedAnswerId="25193" CreationDate="2015-02-15T23:58:24.527" Score="6" ViewCount="990" Body="&lt;p&gt;When white tin ($\beta$-tin) is cooled to a temperature below $13.2\ \mathrm{^\circ C}$, it creates the allotrope of gray ($\alpha$-tin), a gray, amorphous powder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powdered form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a change in appearance of the gray powder when that happens, or is it mostly unnoticeable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the conversion to gray tin change the melting point?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="10185" LastEditDate="2015-05-18T12:29:11.847" LastActivityDate="2017-08-27T12:05:49.597" Title="Can powdered beta-tin be made from raising the temperature of alpha-tin?" Tags="&lt;physical-chemistry&gt;&lt;metallurgy&gt;&lt;melting-point&gt;&lt;allotropes&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25158" PostTypeId="2" ParentId="25146" CreationDate="2015-02-16T00:20:01.603" Score="4" Body="&lt;p&gt;Yes, Fluorescein decomposes due to exposure to light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.wrc.org.za/Knowledge%20Hub%20Documents/Water%20SA%20Journals/Manuscripts/2002/04/WaterSA_2002_04_1476.pdf&quot; rel=&quot;nofollow&quot;&gt;The Conservative Behavior of Fluorescein&lt;/a&gt; for quantitative data.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-16T00:20:01.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25160" PostTypeId="1" AcceptedAnswerId="25161" CreationDate="2015-02-16T02:42:23.240" Score="0" ViewCount="668" Body="&lt;p&gt;This is my reaction, $$\ce{Cd (g) + 1/2 Te2 (g) -&amp;gt; CdTe (s)}.$$ &#xA;Considering the solid CdTe, $$K = P_{\ce{Cd}}^\mathrm{Eq}\cdot \left(P_{\ce{Te}}^\mathrm{Eq}\right)^{\frac{1}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have elementary knowledge on chemistry. I know that the equilibrium values of pressure are dependent on temperature. My question is - does increasing the partial pressure of one of the elements only changes the equilibrium pressure? If yes, for one or both?&lt;/p&gt;&#xA;" OwnerUserId="13425" LastEditorUserId="4945" LastEditDate="2015-02-16T03:48:59.877" LastActivityDate="2015-02-16T03:48:59.877" Title="Does a change in partial pressure for a gas-solid reaction, changes the equilibrium?" Tags="&lt;equilibrium&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25161" PostTypeId="2" ParentId="25160" CreationDate="2015-02-16T03:11:27.357" Score="1" Body="&lt;p&gt;The equilibrium constant will stay the same at the same temperature.  If the pressure of one component of the system in question is changed, equilibrium will be reestablished by either an change in the relative rates of the forward and reverse reactions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the partial pressure of one reactant in this system will result in an increase in the rate of the forward reaction.  Pressure of both both reactants will be reduced until equilibrium is reestablished.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-16T03:11:27.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25162" PostTypeId="1" CreationDate="2015-02-16T03:20:29.890" Score="2" ViewCount="90" Body="&lt;p&gt;Can this compound be created?&#xA;&lt;img src=&quot;https://i.imgur.com/98fzg4E.png&quot; alt=&quot;&quot;&gt;&#xA;Where the orange is Iron and the black is Carbon. &lt;/p&gt;&#xA;" OwnerUserId="13426" LastEditorUserId="4945" LastEditDate="2015-02-16T03:42:15.777" LastActivityDate="2015-02-16T11:01:35.513" Title="Is it possible for this Iron-Carbon Compound to exist?" Tags="&lt;organometallic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25163" PostTypeId="1" CreationDate="2015-02-16T03:52:38.340" Score="0" ViewCount="32238" Body="&lt;p&gt;If they were atoms, $\ce{Mg}$ should have a smaller radii. How does this change?&lt;/p&gt;&#xA;" OwnerUserId="13411" LastEditorUserId="4231" LastEditDate="2015-02-17T02:14:30.590" LastActivityDate="2015-02-17T02:14:30.590" Title="Why does Mg2+ have the ionic radii bigger than Ca2+" Tags="&lt;ionic-compounds&gt;&lt;atomic-radius&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25164" PostTypeId="2" ParentId="25163" CreationDate="2015-02-16T04:09:30.490" Score="6" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BM70.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ca2+ is bigger than Mg2+&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-16T04:09:30.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25165" PostTypeId="1" AcceptedAnswerId="25168" CreationDate="2015-02-16T04:40:16.300" Score="15" ViewCount="1047" Body="&lt;p&gt;So the prof got to Diels-Alder without discussing it in the context of molecular orbital theory. Instead we got a resonance picture of what happens... this is what he told us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) 1,3-butadiene has both an electrophilic and a nucleophilic terminal carbon atom because we can draw resonance structures showing dipolar terminal carbon atoms ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For ethene we can similarly draw a dipolar resonance structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there we're told to simply use Coulomb's law - opposite charges attract - to complete the mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dYQD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how adequate is his electrostatic explanation? Is it adequate - probably ... if one simply wants to learn it for a test. But is it conductive to further understanding of chemistry? I feel rather short-changed here since I'm paying him the big-bucks (tuition is &lt;em&gt;insane&lt;/em&gt;) and the book goes into far greater detail ... &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ETA: I asked the guy about and grilled him on the electrostatic explanation. For example I raised &lt;strong&gt;Martin's&lt;/strong&gt; excellent point about the fact that the dieneophile cannot be that polarized - otherwise wouldn't it be a liquid instead of a gas due to the strong intermolecular attractions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He instead went on a tangent on electron-withdrawing groups and how the electrostatic explanation can explain these ... what do you think? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5yJgd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2016-01-04T14:26:42.550" LastActivityDate="2016-01-04T14:26:42.550" Title="How accurate is this polar mechanism for the Diels-Alder reaction?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25166" PostTypeId="1" CreationDate="2015-02-16T06:27:22.640" Score="0" ViewCount="366" Body="&lt;p&gt;Does a compound give a specific chromatogram peak, using which we can identify it in a mixture?&lt;/p&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2015-02-16T08:05:50.103" Title="Can we use peaks in any chromatogram to identify a compound" Tags="&lt;organic-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25167" PostTypeId="2" ParentId="24520" CreationDate="2015-02-16T06:34:47.980" Score="23" Body="&lt;p&gt;This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the answer I assume that you know about &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;noreferrer&quot;&gt;hyperconjugation&lt;/a&gt; and the various contributing structures it involves.  This gives a basic idea of hyperconjugation:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SdDgq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SdDgq.png&quot; alt=&quot;hyperconjucation&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;http://images.tutorcircle.com/cms/images/44/alkenes1.png&quot; rel=&quot;noreferrer&quot;&gt;tutorcircle.com&lt;/a&gt;&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that on hyperconjugation, the other carbon of the alkene gets a negative charge on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hyperconjugation structures of &lt;strong&gt;2-methylprop-1-ene&lt;/strong&gt; and &lt;strong&gt;&lt;i&gt;trans&lt;/i&gt;-but-2-ene&lt;/strong&gt; are drawn:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sekWJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/snecI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will notice that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ \ce{C_1} $ of the former has a negative charge on it &lt;/li&gt;&#xA;&lt;li&gt;$ \ce{C_2} $ of the latter has a negative charge on it &lt;/li&gt;&#xA;&lt;li&gt;The latter's methyl group makes the negatve charge less stable due to +I (inductive effect) &lt;/li&gt;&#xA;&lt;li&gt;The former has no such destablizing effect as the carbon containing the negative charge is just connected to hydrogens&lt;/li&gt;&#xA;&lt;li&gt;As all the hyperconjugation structure have this effect applying on them, the net effect cant be ignored for the overall stability &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence the overall stability of  &lt;strong&gt;2-methylprop-1-ene&lt;/strong&gt; &gt; &lt;strong&gt;&lt;i&gt;trans&lt;/i&gt;-but-2-ene&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="5026" LastEditDate="2018-03-02T11:35:17.950" LastActivityDate="2018-03-02T11:35:17.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25168" PostTypeId="2" ParentId="25165" CreationDate="2015-02-16T07:09:10.110" Score="14" Body="&lt;p&gt;&lt;em&gt;Disclaimer: This is not a full answer. I was just too long for a comment.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein &lt;em&gt;et. al.&lt;/em&gt; concluded that a concerted pathway is the most economical.&lt;sup&gt;[1]&lt;/sup&gt; &#xA;They also included possible diradical pathways and the similar. Of course there is no uniformly applicable mechanism, it is very dependent on the substrates. Many of the reactions can be effectively described by &lt;a href=&quot;http://en.wikipedia.org/wiki/Frontier_molecular_orbital_theory&quot;&gt;frontier molecular orbital theory&lt;/a&gt;, so it is also possible to do so for the simplest &lt;a href=&quot;http://en.wikipedia.org/wiki/Diels%E2%80%93Alder_reaction&quot;&gt;Diels-Alder reaction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be many more publications out there discussing the whole mechanism in pretty exhausting detail, but for a start I thought the cited one will do just fine. Just to name some more, see [2] through [4].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrostatic interpretation is tempting, as it is simple and it explains the simplest case very well. I believe it explains it too well for the wrong reasons:&#xA; I would not expect the polarisation, i.e. induced dipole, in &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethylene&quot;&gt;ethylene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/1,3-Butadiene&quot;&gt;butadiene&lt;/a&gt; be strong enough to make this happen at all. Both are colourless gasses at room temperature. If there were more induced dipoles, they would be liquid, which means that these molecules are basically unpolarised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a publication by Gian Franco Tantardini and Massimo Simonetta it was stated, that the contribution of the polarised Valence Bond Structure is very little, see [5]. If I read it correctly, the contribution is less than 1%. This alone should suffice to disregard a mechanism purely based on electrostatic considerations. In the same publication it is stated, that ethylene has quite a significant contribution from its polarised structure (about 25%). However, these consideration would in most cases not be able to describe streospecific reaction products, that are obtained from different substrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another aspect, one must not forget, that the ground state of 1,3-butadiene is trans and it requires just a bit of energy to rotated it to the necessary cis formation, which should be about 10 times less present. So what are the odds having a well polarised ethylene hitting a polarised cis-butadiene in the right angle at the right speed. There has to be more to it...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja9601494&quot;&gt;E. Goldstein, Brett Beno, and K. N. Houk; &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1996&lt;/strong&gt;, &lt;em&gt;118&lt;/em&gt; (25), 6036–6043.&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/0040-4020(59)80072-7&quot;&gt;R.B. Woodward, and Thomas J. Katz; &lt;em&gt;Tetrahedron&lt;/em&gt;; &lt;strong&gt;1959&lt;/strong&gt;, &lt;em&gt;5&lt;/em&gt; (1), 70-89.&lt;/a&gt; Mirror at &lt;a href=&quot;http://www.znu.ac.ir/data/members/ahmadi_ebrahim/Advanced%20Organic%20Chemistry/Advanced%20Organic%20Chemistry-Problem/%DD%D5%E1%2011/19e.pdf&quot;&gt;znu.ac.ir&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja00279a018&quot;&gt;Michael J. S. Dewar, Santiago Olivella, and James J. P. Stewart; &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1986&lt;/strong&gt;, &lt;em&gt;108&lt;/em&gt; (19),  5771–5779.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/anie.196700161&quot;&gt;J. Sauer; &lt;em&gt;Angew. Chem. Int. Ed.&lt;/em&gt;, &lt;strong&gt;1967&lt;/strong&gt;, &lt;em&gt;6&lt;/em&gt; (1), 16-33. &lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.560140811/abstract&quot;&gt;Gian Franco Tantardini and Massimo Simonetta; &lt;em&gt;Int. J. Quant. Chem.&lt;/em&gt;, &lt;strong&gt;1978&lt;/strong&gt;, &lt;em&gt;14&lt;/em&gt; (S12), 131-141.&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-02-16T23:35:36.037" LastActivityDate="2015-02-16T23:35:36.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25169" PostTypeId="2" ParentId="25163" CreationDate="2015-02-16T07:43:01.150" Score="0" Body="&lt;p&gt;Mg is smaller the Ca because it is above it in the periodic table, magnesium being a period above it shows that it has one less electron shell so it's smaller, calcium has another ring of electrons, as to say, that are around it so it's radius would be bigger.&#xA;Your question is wrong.&lt;/p&gt;&#xA;" OwnerUserId="13427" LastActivityDate="2015-02-16T07:43:01.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25170" PostTypeId="2" ParentId="25166" CreationDate="2015-02-16T08:05:50.103" Score="2" Body="&lt;p&gt;Compounds have specific retention times that are used to identify their respective peaks in chromatograms. Usually these times are dependent on many factors (temperature, column type, mobile phase...), and as such some kind of pre-calibration of the machine with respect to compounds in the mixture has to be done. The area of the peak is then indicative of the amount of substance, also subject to calibration (or comparison with reference substances with known concentrations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structure elucidation of previously unknown compounds in mixtures other methods such as GC-MS are much better suited.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-02-16T08:05:50.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25171" PostTypeId="2" ParentId="25162" CreationDate="2015-02-16T08:40:29.007" Score="1" Body="&lt;p&gt;Short answer: No&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer: Assuming you had hands tiny enough to manipulate individual atoms, you could make this molecule. The amount of time it would stay together would be as close to 0 as it gets. To start iron can't accommodate that amount of electrons, and the formal charges of this molecule are extreme to say the least. &lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-16T08:40:29.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25172" PostTypeId="2" ParentId="25155" CreationDate="2015-02-16T08:55:55.207" Score="1" Body="&lt;p&gt;Like you said the $\ce{pH}$ of blood is approximately $7.5$. So since &#xA;$$ \ce{pH} = -\log[\ce{H+}] \\&#xA;\text{or}\\ &#xA;[\ce{H+}] = 10^{-\ce{pH}}$$ &#xA;this means that the proton or hydrogen ion (same thing) has a concentration of $10^{-7.5}$ not $10^{7.5}$ (notice the negative). If you're dealing with blood you'll never have to worry about negative $\ce{pH}$ values. Like DavePhD stated, negative $\ce{pH}$ values only occur with high concentrations of strong acids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I think where you're going wrong is with $-\log(10^{7.5})$. It should be $-\log(10^{-7.5})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, you are correct. $\log(x)\geq0$ for $x \geq 1$. So $-\log(x) \leq0$ for $x\geq1$, but when do you ever have proton concentrations above 1 in human biology? Practically never.&lt;/p&gt;&#xA;" OwnerUserId="13428" LastEditorUserId="4945" LastEditDate="2015-02-16T09:16:28.567" LastActivityDate="2015-02-16T09:16:28.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25173" PostTypeId="2" ParentId="25100" CreationDate="2015-02-16T09:01:54.127" Score="3" Body="&lt;p&gt;$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -&amp;gt; MnO2 +2NaMnO4 + 6KOH }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said the brown solid A is $\ce{MnO2}$ while the purple solution is due to the permanganate ion .&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-02-16T09:01:54.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25174" PostTypeId="2" ParentId="25091" CreationDate="2015-02-16T09:40:54.037" Score="3" Body="&lt;p&gt;Doing anything blindly is, of course, counter productive. If you were to make the mistake of comparing pka values in different solvents then there would be issues ,however, if you compare pka values while maintaining a constant solvent the assumption is fine for qualitative reasoning. The only counter example I could find is a 1,8-Bis(dimethylamino)naphthalene. It's pka was significantly lower in ACN and DMSO than expected. Again, just make sure you compare things with constant solvents and you're fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the chemists you speak of are just saying &quot; for a given solvent blah dissociates more so than blah&quot; which lets them see where the reaction lies at equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-16T09:40:54.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25175" PostTypeId="2" ParentId="25064" CreationDate="2015-02-16T10:26:50.890" Score="2" Body="&lt;p&gt;Long story short, if it's scratched in the presence of water(humidity) the chromium ion inside the chromate will get reduced to Cr 3+ from Cr 6+. This 3+ ion will then form a Chromium hydroxide compound,which is insoluble in water and corrosion resistant, inside the scratch. This reduction of the chromium ion first is what inhibits the oxidation of the underlying metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the full article &lt;a href=&quot;http://www.electrochem.org/dl/interface/wtr/wtr01/IF12-01-Pages34-38.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.electrochem.org/dl/interface/wtr/wtr01/IF12-01-Pages34-38.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer: Being water resistant in the first place and being capable of producing ions that can be reduced forming another corrosion resistant compound( chromium(III)oxy-hydroxide.)&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-16T10:26:50.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25176" PostTypeId="2" ParentId="25091" CreationDate="2015-02-16T10:48:41.647" Score="3" Body="&lt;p&gt;pkA Values are measured in water, so it's only an approach to think that in other solvents they behave in a similar way. The pKa values are also depending on the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a good approach is better than nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the organic solvent is a protic solvent like alcohol (an alcohol is chemically a bit like water) this approach is quit o.k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the pKa tells something about the equilibrium and the energy that is needed to move a proton to another molecule (and so to charge a molecule). If a charged molecule is stabilized as it is in a protic environment the equilibrium allows more protons to move around. &#xA;In contrast a &quot;naked&quot; small positive charge like a proton without any stabilizing environment need so much energy that it can't exist in a normal laboratory - this is already chemistry of a plasma.&lt;/p&gt;&#xA;" OwnerUserId="13429" LastActivityDate="2015-02-16T10:48:41.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25177" PostTypeId="1" AcceptedAnswerId="25212" CreationDate="2015-02-16T10:55:01.620" Score="6" ViewCount="1784" Body="&lt;p&gt;Why specifically is bromine used in the degradation reaction (Hoffmann bromamide)? Why can't the same be performed using chlorine or iodine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what makes bromine a more convenient reagent as compared to the other halogen gases? For instance, during halogenation of alkanes, we don't use fluorine due to the explosive nature of the reaction, and we don't use iodine because of the slow rate. Is the reason somewhat analogous to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought because of kinetic factors, maybe somewhat similar to how monochlorination and monobromination are much more easier than monoiodination and monofluorination&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="5026" LastEditDate="2018-06-07T05:39:21.040" LastActivityDate="2018-06-07T05:40:17.813" Title="Why is only bromine used in the degradation reaction?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25178" PostTypeId="2" ParentId="25162" CreationDate="2015-02-16T11:01:35.513" Score="0" Body="&lt;p&gt;that molecule is not stable.&#xA;There exists &#xA;(L)xFe=C(R1)(R2) carben complexes and&#xA;(L)xRu=C=C=C=C=Ru(Lx) molecules,&#xA;so their might exists also a (L)xRu=C=Ru(L)x or (L)xFe=C=Fe(L)x molecule in theory&#xA;but at least the 3 not centered Carben-C-Atom (connected to each Iron atom) have to be stabilized. &lt;/p&gt;&#xA;" OwnerUserId="13429" LastActivityDate="2015-02-16T11:01:35.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25179" PostTypeId="1" AcceptedAnswerId="25182" CreationDate="2015-02-16T11:06:32.263" Score="7" ViewCount="770" Body="&lt;p&gt;I want to create a steam bath for a reaction the proceeds well at 105&amp;#8451;.  I cannot use pure (demineralized, anyway) water for this, since it would boil at 100&amp;#8451; at STP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such I figured that maybe I could get the required temperature by mixing water with another liquid (with which water is miscible) which has a higher boiling point (e.g. vinegar at BP around 118.5&amp;#8451; or vegetable glycerin with BP at 190&amp;#8451;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the boiling point of the mixture of liquids proceed in a linear manner between the two temperatures?  E.G. would a volume of 50% water &amp;amp; 50% VG boil at 145&amp;#8451;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is being done right on the boiling point of the combined liquids, I expect to lose some of the liquid to steam and need to add more. Will the lower temperature liquid boil out first?&lt;/p&gt;&#xA;" OwnerUserId="1071" LastEditorUserId="1071" LastEditDate="2015-02-16T11:13:38.197" LastActivityDate="2015-02-16T11:32:55.977" Title="Does the boiling boint of miscible liquids follow a linear relationship?" Tags="&lt;water&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25181" PostTypeId="2" ParentId="25155" CreationDate="2015-02-16T11:27:26.283" Score="0" Body="&lt;p&gt;$\text{pH}$ can definitely be negative.&lt;br&gt;&#xA;It is defined as $$\text{pH}=-\log[\text{H}^{+}]$$&lt;br&gt;&#xA;Note that if the $[\text{H}^{+}]\geq 1\text{M}$, the $\text{pH}$ will become negative.&lt;br&gt;&#xA;Practically, this is only possible when you have an ultrapure solution of a strong acid.&lt;br&gt;&#xA;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="7513" LastActivityDate="2015-02-16T11:27:26.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25182" PostTypeId="2" ParentId="25179" CreationDate="2015-02-16T11:32:55.977" Score="2" Body="&lt;p&gt;Except if molecular composition of your two liquids are similar, the boiling point of the mixture will not proceeds linearly but like this for instance : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIByW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the lower blue curve the boiling curve. In between the two curve you'll then have a mixture of vapor and liquid and only vapor on the upper part the binary diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should find out an already drawn binary diagram for your mixture or draw it out by yourself first.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-16T11:32:55.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="25183" PostTypeId="2" ParentId="25091" CreationDate="2015-02-16T12:51:29.980" Score="8" Body="&lt;p&gt;An important case of acidity constants changing with solvent, with consequences in organic chemistry, can be found for &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed044p304&quot; rel=&quot;nofollow&quot;&gt;compounds with hydroxyls, especially water&lt;/a&gt;. It is possible to deduce that hydroxyl compounds are more acidic in &lt;a href=&quot;http://en.wikipedia.org/wiki/Protic_solvent&quot; rel=&quot;nofollow&quot;&gt;protic solvents&lt;/a&gt; compared to aprotic solvents, which is attributed to an abnormally high stabilization of the resulting conjugate base via strong $\ce{O^{-}\ ^...\  H-O}$ hydrogen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is most clear for water. Water molecules solvated by water itself have a $\mathrm{p}K_\mathrm{a}$ of 15.7 in ambient conditions, but if the unusual stabilization from hydrogen bonding were removed, they would have a $\mathrm{p}K_\mathrm{a}$ of around 28. In other words, $\ce{H2O}$ is a weaker acid by some 10 orders of magnitude in non-protic solvents compared to water, and correspondingly, its conjugate base $\ce{OH^{-}}$ is a stronger base by the same factor. This means hydroxide ions can effect quantitative deprotonation of even rather weak acids, including the hydrocarbons cyclopentadiene, indene and fluorene, in solvents such as 1,2-dimethoxyethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, compounds containing hydroxide ions have somewhat limited use as superbases in non-protic solvents because the solubility of common hydroxide bases such as $\ce{NaOH}$ or $\ce{KOH}$ is rather small. Other hydroxides such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetramethylammonium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;quaternary ammonium hydroxides&lt;/a&gt; have appreciable solubility in non-protic solvents, but they are always found in the form of hydrates. Attempts to prepare anhydrous $\ce{R_4N^+OH^{-}}$ inevitably result in decomposition, because the crystallized water solvates the hydroxide ion and tames its reactivity; once the water is removed, the bare hydroxide ion immediately attacks the quaternary ammonium cation creating an alcohol and a tertiary amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned previously, other compounds containing hydroxyls also show changes in $\mathrm{p}K_\mathrm{a}$ comparing water and non-protic solvents, though the difference tends to be smaller. For ethanol, the difference is at most about 4 units of $\mathrm{p}K$. &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2015-03-29T19:45:42.367" LastActivityDate="2015-03-29T19:45:42.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25184" PostTypeId="1" AcceptedAnswerId="25187" CreationDate="2015-02-16T12:55:01.130" Score="21" ViewCount="10513" Body="&lt;p&gt;I noticed something while cooking and was wondering what the chemical explanation is for this. White sugar has a smell, but it's different, and much fainter than that of brown sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown sugar is unrefined, so does this have something to do with why odorous chemicals are present?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these chemicals and why are they present in one, but not the other?&lt;/p&gt;&#xA;" OwnerUserId="13432" LastEditorUserId="4325" LastEditDate="2015-07-20T20:27:20.890" LastActivityDate="2015-07-20T20:27:20.890" Title="Why does brown sugar have a strong smell but white sugar doesn't?" Tags="&lt;biochemistry&gt;&lt;food-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25185" PostTypeId="2" ParentId="579" CreationDate="2015-02-16T12:59:43.587" Score="0" Body="&lt;p&gt;Semiconductor elements have half-filled outer-most shells so if we doped group VI or II elements which have 6 and 2 electrons in the outer-most shell and they form covalent bonds with each other just like Si doped to Si and that would result in a no-free-electron insulator because there is no vacancy there .&lt;/p&gt;&#xA;" OwnerUserId="13433" LastEditorUserId="7448" LastEditDate="2015-02-16T13:09:26.747" LastActivityDate="2015-02-16T13:09:26.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25186" PostTypeId="1" AcceptedAnswerId="25205" CreationDate="2015-02-16T13:59:42.370" Score="2" ViewCount="357" Body="&lt;p&gt;I'm working on an imaging project and I have no background in chemistry or chemiluminescence for that matter. This is a Computer Science project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While most of the details for it have been worked out, I'm having difficulty figuring out the scope and possibility of an important requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a luminol image, I will develop a heat-map. For now the heat map is built based upon the intensity of the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that this chemical would burn differently at various temperature and would have a specific signature of light emitted depending on it its temperature but I'm having difficulty pinpointing whether this is true, possible and backed by research (or science).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest parallel I've found is combustion analysis. The heat release rate is calculated by measuring the amount of light radiated from a flame at various wavelengths, since certain chemicals give off radiations at specific wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm this, suggest alternatives or point me int the right direction?&lt;/p&gt;&#xA;" OwnerUserId="13434" LastEditorUserId="418" LastEditDate="2015-02-16T20:47:39.913" LastActivityDate="2015-02-16T20:56:54.557" Title="Chemiluminescence - Determining Temperature from Luminol Imaging" Tags="&lt;everyday-chemistry&gt;&lt;photochemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25187" PostTypeId="2" ParentId="25184" CreationDate="2015-02-16T14:01:13.127" Score="29" Body="&lt;p&gt;Sugar is made by repeatedly boiling and cooling cane syrup (or sugar beet syrup).  After each cooling, the solution becomes supersaturated with respect to sucrose, causing sucrose to crystallize out of the &quot;mother liquor&quot; (the industrial term for the liquid solution from which crystals form). These crystals of &quot;raw&quot; sugar are heavily processed (washed in syrup, dissolved, treated with carbonate or phosphate minerals, recrystallized, and subject to ion exchange or charcoal color removal) to result in white sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, brown sugar is usually made by mixing refined white sugar with blackstrap molasses. In former times, the brown sugar was simply unrefined or unwashed crystal fraction from the 2nd or 3rd crystallization of cane juice. But precise control over the ratio of &quot;contaminating&quot; liquid to white crystalline sugar was difficult, which is why industry moved towards reconstituting brown sugar by mixing molasses and white sugar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blackstrap molasses is simply the spent &quot;mother liquor&quot; remaining after the third cycle of boiling and cooling to produce sugar crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smell of brown sugar thus comes primarily from molasses.  The smell of molasses is very complex.  Here is the abstract of a relevant &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/00021369.1976.10862142#.VOH0yVPF_XA&quot;&gt;paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The characteristic sweet-aroma components in refinery final molasses were isolated by using a combination of sensory evaluation, column chromatography and gas chromatography, and were identified by infrared spectrometry, mass spectrometry and color reactions. Identified were ethyl n-hexanoate, ethyl phenylacetate, phenylacetic acid, n-butyl benzoate, isopropyl benzoate, l(+)-pantolactone, benzoic acid, o-toluic acid, m-toluic acid, β-phenylpropionic acid, succinic anhydride, maltol, isomaltol, and 2-methyl-5-hydroxy-6-ethyl-γ-pyrone.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The aroma of the mixture of these compounds and vanillin and vanillic acid was similar to that of refinery molasses by sensory evaluation. The results indicate that these compounds are important in producing of sweet molasses aroma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2015-02-17T04:20:38.183" LastActivityDate="2015-02-17T04:20:38.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25188" PostTypeId="1" AcceptedAnswerId="26362" CreationDate="2015-02-16T14:01:17.547" Score="11" ViewCount="1966" Body="&lt;p&gt;Secondary allylic halides can be produced in high yield with the following Mitsunobu-type reaction:&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FK6UB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield being $94~\%$.&#xA;The last step in the mechanism is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oi4K5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the $\mathrm{S_N2}$ reaction favored so much over the $\mathrm{S_N2'}$ reaction?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; Magid, R. M.; Fruchey, O. S.; Johnson, W. L.; Allen, T. G. Hexachloroacetone/triphenylphosphine: a mild reagent for the regioselective and stereospecific production of allylic chlorides from the alcohols. &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;1979,&lt;/strong&gt; &lt;em&gt;44&lt;/em&gt; (3), 359–363. &lt;a href=&quot;https://doi.org/10.1021/jo01317a011&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/jo01317a011&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5026" LastEditDate="2018-03-13T07:42:17.970" LastActivityDate="2018-03-13T07:42:17.970" Title="Why is SN2 favored greatly over SN2' in this reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25190" PostTypeId="1" AcceptedAnswerId="25191" CreationDate="2015-02-16T15:27:43.840" Score="7" ViewCount="12757" Body="&lt;p&gt;I would like to, theoretically, prepare oxalic acid from acetylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that using ozonolysis would be a good idea to do so. Upon ozonolysis of acetylene, we would get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O=CH_2-H_2C=O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On oxidizing this using hydrogen peroxide, would I obtain the desired product?&lt;/p&gt;&#xA;" OwnerUserId="7023" LastEditorUserId="7475" LastEditDate="2015-10-09T16:02:16.293" LastActivityDate="2016-11-21T13:20:58.283" Title="Preparation of oxalic acid from acetylene" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25191" PostTypeId="2" ParentId="25190" CreationDate="2015-02-16T15:47:59.810" Score="12" Body="&lt;p&gt;Ozonolysis of acetylene won't produce oxalic acid.  The triple bond holding the two acetylenic carbons together will be attacked as shown below.  The reaction is complex and depending upon reaction conditions, different products can result (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ozonolysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_1&lt;/a&gt;, &lt;a href=&quot;https://sites.smu.edu/dedman/catco/research/ozone.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_2&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8fEX1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxalic acid can be produced from acetylene by reacting acetylene with alkaline $\ce{KMnO4}$ (&lt;a href=&quot;http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_3&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-C#C-H -&amp;gt;C[KMnO4/OH^-]\ HOOC-COOH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Response to OPs comment&#xA;Here is a rough idea of what is going on mechanistically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the permanganate forms a cyclic intermediate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) that produces a diol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) since a double bond is still present the reaction can be repeated to yield 2 gem-diols that can equilibrate to an alpha-diketone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) now the standard alkaline permanganate oxidation of a carbonyl occurs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) hydroxide can add to a carbonyl and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) the manganese ion complexes with the carbonyl oxygen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) the alpha-proton donates its electrons to reform the carbonyl which is now part of a carboxylic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8) repeat steps 5-7 on the other carbonyl and you have oxalic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wiuTV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T23:39:24.333" LastActivityDate="2016-03-12T23:39:24.333" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25192" PostTypeId="1" AcceptedAnswerId="25194" CreationDate="2015-02-16T16:26:24.973" Score="8" ViewCount="26691" Body="&lt;p&gt;In terms of electronegativity, from what I understand electronegativity increases going across the period, so surely this should mean that zinc less readily loses its outer shell electrons than copper? Zinc has a greater nuclear charge but the outer shell electrons are in the same shell, so should the outer shell electrons in zinc not experience a stronger attraction to the nucleus? I have a feeling that this has something to do with which sub shells the electrons are removed from in copper and zinc...&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="5354" LastEditDate="2015-10-04T18:19:13.950" LastActivityDate="2021-03-11T20:47:02.640" Title="Why is zinc more reactive than copper?" Tags="&lt;electrons&gt;&lt;metal&gt;&lt;transition-metals&gt;&lt;electronegativity&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25193" PostTypeId="2" ParentId="25157" CreationDate="2015-02-16T16:47:50.897" Score="4" Body="&lt;p&gt;$\alpha$Sn and $\beta$Sn are the two solid allotropes of Sn.  As you not, below 13C the stable phase is $\alpha$Sn, which has a diamond cubic crystal structure (like diamond, Si, and Ge) and is a semi-metal.  Above 13C, the thermodynamically stable phase is $\beta$Sn, a body-centered tetragonal crystal. So, cycling back and forth around 13C varies the thermodynamically stable phase back and forth, with the only question being the kinetics of the transformation.  The video in the comment by @JasonPatterson shows that the transformation indeed takes place reasonably quickly (unlike diamond to graphite for carbon). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for melting, if you rapidly heated $\alpha$Sn and avoided the phase transformation, you would find that the melting temperature would be lower. Using the SGTE thermodynamic data (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)), one finds that the $\alpha$Sn -&gt; liquid phase transition would occur at about 430.7K, ~75K lower than the standard $\beta$Sn -&gt; liquid melting point at 505.06K. &lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-02-16T16:47:50.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25194" PostTypeId="2" ParentId="25192" CreationDate="2015-02-16T16:54:25.467" Score="14" Body="&lt;p&gt;You have to think about the whole process.  When a metal loses electrons to make a metal ion the following happens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The metallic bonds holding the metal atoms together are broken.&lt;/li&gt;&#xA;&lt;li&gt;The metal atom loses the electrons.&lt;/li&gt;&#xA;&lt;li&gt;The resulting metal ion is hydrated.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In your analysis you are only focusing on step 2.  The enthalpy and entropy of the entire process factor into the reduction potential.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2017-12-08T00:21:27.317" LastActivityDate="2017-12-08T00:21:27.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25195" PostTypeId="1" AcceptedAnswerId="25196" CreationDate="2015-02-16T17:02:48.580" Score="0" ViewCount="54" Body="&lt;p&gt;Here is the problem I am stuck with, in particular just the first part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/reHdS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/reHdS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated the moles $4.492 * 10^{-3} \,\mathrm{mol}$ and I am quizzed with regards to the equation that occurs. First assumption I am making is that the gas released is carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?&lt;/p&gt;&#xA;" OwnerUserId="13438" LastEditorUserId="194" LastEditDate="2017-04-09T01:54:24.900" LastActivityDate="2017-04-09T01:54:24.900" Title="Using stoichiometry to split a complex mineral" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25196" PostTypeId="2" ParentId="25195" CreationDate="2015-02-16T17:56:07.147" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{CO2}$ implies $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ is 0.8167 g.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, considering the atomic weights of C, O and H, subtract the contribution of these elements from 0.8167 g to determine the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then divide by 2 to get the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{Mg_{x}Cu_{1-x}}$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then use the atomic weights of Mg and Cu to solve for $x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="194" LastEditDate="2017-04-09T01:50:52.907" LastActivityDate="2017-04-09T01:50:52.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25197" PostTypeId="1" AcceptedAnswerId="25198" CreationDate="2015-02-16T18:05:23.010" Score="0" ViewCount="1134" Body="&lt;p&gt;I thought about this while thinking about alloys and why they are better than pure metals, even though they are technically impurities. It got me wondering, are there actually any good real-life applications where having an impure substance is completely disadvantageous?&lt;/p&gt;&#xA;" OwnerUserId="13439" LastActivityDate="2015-02-16T18:24:33.613" Title="What are disadvantages of having an impurity?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25198" PostTypeId="2" ParentId="25197" CreationDate="2015-02-16T18:24:33.613" Score="1" Body="&lt;p&gt;Impurities are the UNWANTED substances that are present in the substance.. &#xA;There are Impurities in most of stuff including a lot of chemicals and it's very hard to remove them too&#xA;However If impurity exceeds the limit its completely Dangerous..&#xA;Like for example :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surface water (river or lake)&#xA;  Surface water can be exposed to acid rain, pesticide runoff, water runoff, and industrial waste.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Groundwater (private wells or public water supplies)&#xA;  Groundwater can be contaminated by disease producing pathogens, careless disposal of household wastes, and leaching from landfills and septic tanks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which can lead to Nausea&#xA;Lung irritation,Vomiting, Dizziness, Skin rashes and even Death&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Chemistry and Chemical Industry impurities might slow down the reaction or may lead to formation of unwanted bi products , Might affect the rate of Crystallization - Most common issue I think&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of articles online out there here are some for you : &lt;a href=&quot;http://www.globalresearchonline.net/journalcontents/Volume3issue2/Article%20012.pdf&quot; rel=&quot;nofollow&quot;&gt;Impurity in medicine&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.trojanbattery.com/pdf/WP_EffectOfImpurities_0612.pdf&quot; rel=&quot;nofollow&quot;&gt;Impurity in Electrical cells &lt;/a&gt;&#xA;&lt;a href=&quot;http://www.demiwater.nl/files/AWT_IMPURITIES_IN_WATER.pdf&quot; rel=&quot;nofollow&quot;&gt;Impurity in water&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.kester.com/kester-content/uploads/2013/06/EffectsofMetallicImpurities.pdf&quot; rel=&quot;nofollow&quot;&gt;Effect on Solder machine&lt;/a&gt;&#xA;you can google and find more&lt;/p&gt;&#xA;" OwnerUserId="9963" LastActivityDate="2015-02-16T18:24:33.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25200" PostTypeId="2" ParentId="9317" CreationDate="2015-02-16T18:48:36.913" Score="10" Body="&lt;p&gt;The somewhat standard reference (~1800 citations in Google Scholar) for bulk GaAs structure is a 1982 paper by J. S. Blakemore: &lt;a href=&quot;https://doi.org/10.1063/1.331665&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Appl. Phys.&lt;/em&gt; &lt;strong&gt;53,&lt;/strong&gt; R123 (1982)&lt;/a&gt;. The experimental/numerical data in it (via &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot; rel=&quot;noreferrer&quot;&gt;X-ray diffraction&lt;/a&gt;) establishing the crystal structure of bulk GaAs goes back to the 1960s though. One of the more helpful visualizations of zincblene structure (which bulk GaAs shares) can be found on &lt;a href=&quot;https://www.youtube.com/watch?v=6lCQ6LT3Esg&quot; rel=&quot;noreferrer&quot;&gt;YouTube&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however that nanowire GaAs can have &lt;a href=&quot;http://en.wikipedia.org/wiki/Wurtzite_crystal_structure&quot; rel=&quot;noreferrer&quot;&gt;wurtzite structure&lt;/a&gt; as well; here is another &lt;a href=&quot;https://www.youtube.com/watch?v=MMJ1pCHs5b4#t=92&quot; rel=&quot;noreferrer&quot;&gt;YouTube animation&lt;/a&gt;. Nanowires weren't practical in the 1980s, so this is more recent research; see for example S. Conesa Boj's Masters thesis at the University of Barcelona, &lt;a href=&quot;http://diposit.ub.edu/dspace/handle/2445/9822&quot; rel=&quot;noreferrer&quot;&gt;&quot;HRTEM assessment of Wurtzite and Zinc-Blende phases in GaAs nanowires for optoelectronic devices&quot;&lt;/a&gt;. (HRTEM is explained well enough on &lt;a href=&quot;http://en.wikipedia.org/wiki/High-resolution_transmission_electron_microscopy&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, in case you wonder how that works). In any case, the coordination numbers are still 4 for both Ga and As in this structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases (crystal structures) the ratio of the Ga:As atoms in the unit cell is the same 1:1. I've actually checked the 11th (2012) edition of Boylestad &amp;amp; Nashelsky and they have the same old explanation. My guess is that they don't want to explain polar bonds to first-year EE undegrads, so they engage in some &quot;lies to children&quot; by sticking with the simplified covalent bond model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book by Razeghi (which gives the right ratio and explanation for bulk GaAs) does call them &quot;polar covalent bonds&quot;/&quot;mixed bonds&quot; (pp. 65-66), but from its preface this book seems aimed at senior-level undergraduates. Razeghi does mention prior to that in his book (p. 30) that GaAs has the zincblende structure, and goes into some detail on that. The flattened 2D structure on p. 66 is a bit strange in view of that prior level of detail; I'm guessing he wanted to emphasize the bond polarity (on p. 66) and he thought 2D was a simpler way to show it.&lt;/p&gt;&#xA;" OwnerUserId="13440" LastEditorUserId="16683" LastEditDate="2017-08-30T14:09:26.883" LastActivityDate="2017-08-30T14:09:26.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25201" PostTypeId="2" ParentId="25163" CreationDate="2015-02-16T19:15:02.317" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRS3U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as shown in the above pic Mg2+has more smaller size then Ca2+...&#xA;I'm sorry but your question is wrong, &#xA;Why Ca2+ has bigger Ionic Radii then Mg2+? because &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No of Shells increases down the group&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;-Property of Periodic table &lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore -&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As you move down a group in the periodic table , additional layers of electrons are being added, which naturally causes the ionic radius to increase as you move down the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends_in_Ionic_Radii&quot; rel=&quot;nofollow noreferrer&quot;&gt;you can check reference this for more information&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source : &lt;a href=&quot;http://searchpp.com/ionic-radii-periodic-table/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://searchpp.com/ionic-radii-periodic-table/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9963" LastActivityDate="2015-02-16T19:15:02.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25202" PostTypeId="1" AcceptedAnswerId="25211" CreationDate="2015-02-16T19:43:00.347" Score="3" ViewCount="1207" Body="&lt;p&gt;In Cannizaro reaction when concentration of OH- is less in the rds step hydride shift occurs, though it is very difficult for hydride ion to shift yet the reaction proceeds but what is the driving force here??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher told me that this mechanism was proposed because the hydride shift occurs but there must be some reason as to why the hydride shift?&lt;/p&gt;&#xA;" OwnerUserId="5674" LastEditorUserId="1887" LastEditDate="2015-02-16T21:45:59.947" LastActivityDate="2015-09-03T15:25:44.007" Title="Cannizaro Reaction, Hydride Shift Reason?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25203" PostTypeId="1" CreationDate="2015-02-16T19:51:15.193" Score="3" ViewCount="1747" Body="&lt;p&gt;The pKa of vinegar is 4.76. If I know the  pH of the vinegar, what would be the formula for determining the percentage acid base on the relationship between pH, pKa and concentration? Can you show the formula so that I only need to plug in the ph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar. The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated.&lt;/p&gt;&#xA;" OwnerUserId="13441" LastEditorUserId="5531" LastEditDate="2015-02-17T14:51:59.910" LastActivityDate="2015-02-17T14:55:29.420" Title="Calculating vinegar strength" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25204" PostTypeId="2" ParentId="25203" CreationDate="2015-02-16T20:20:27.260" Score="2" Body="&lt;p&gt;If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c$ is the molar concentration of the vinegar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want pH explicitly in the equation, then approximating pH = -log[H+], then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example if pH is 3, $10^{-3}$ is 0.001, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = 0.059M$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-17T00:32:59.610" LastActivityDate="2015-02-17T00:32:59.610" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="25205" PostTypeId="2" ParentId="25186" CreationDate="2015-02-16T20:54:38.967" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not! The energy released from the excited state of the 3-aminophthalate dianion formed in the oxidation of luminol (or rather its dianion) by hydrogen peroxide in alkaline is released in form of light. There's little to no heat released in this chemoluminescent reaction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-16T20:54:38.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25206" PostTypeId="2" ParentId="25186" CreationDate="2015-02-16T20:56:54.557" Score="0" Body="&lt;p&gt;There is some research out there, e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/3749789&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/3749789&lt;/a&gt;, Influence of pH and temperature on luminol-dependent chemiluminescence and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pelagiaresearchlibrary.com/der-chemica-sinica/vol4-iss3/DCS-2013-4-3-161-164.pdf&quot; rel=&quot;nofollow&quot;&gt;http://pelagiaresearchlibrary.com/der-chemica-sinica/vol4-iss3/DCS-2013-4-3-161-164.pdf&lt;/a&gt;, Effect of temperature on chemiluminescence of luminol &lt;em&gt;ethyl amine&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is also a &lt;em&gt;time&lt;/em&gt; dependency, so that increased luminous output is offset by shorter duration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's an interesting use of chemiluminescence and implies use of tomography if the sample is three-dimensional. &lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-16T20:56:54.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25207" PostTypeId="1" AcceptedAnswerId="25208" CreationDate="2015-02-16T21:13:23.803" Score="3" ViewCount="325" Body="&lt;p&gt;I am currently trying to solve a problem that seems quite simple at the beginning but the complexity reveals itself soon thereafter. The problem is the following: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You mix a known mass of water with a known mass of ethanol. Then you fill up the flask to the 50 mL mark with water, but cannot weigh the contents when you're done. What is the molar fraction of ethanol in this mixture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My attempt at solving this problem is to basically calculate the initial molar fraction, and then stepwise add small virtual amounts of water. Then the new molar fraction is calculated, and somehow the excess volume (which is negative for the water-ethanol mixture) would also be included somewhere here. This is repeated until the final volume is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other attempt would be to simply neglect the excess volume. While this might be fine at the edges of the molar fractions, it would introduce a grave error towards the middle. Or is the error still negligible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these reasonable attempts or am I completely missing something here?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-02-16T22:02:03.283" Title="Solve this mole fraction calculation with incomplete information" Tags="&lt;experimental-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25208" PostTypeId="2" ParentId="25207" CreationDate="2015-02-16T21:22:09.370" Score="3" Body="&lt;p&gt;I don't think you can solve that unless you know additional information like mass of water added or volume of water added, or the density of the final solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could approximate by dividing mass of ethanol by density of ethanol to get volume of ethanol, and saying volume of water added is 50mL-(volume ethanol), but this is only an approximation because volumes are not additive when the two solutions are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error could be estimated from this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3l6G5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Partial_molar_property&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Partial_molar_property&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the data from the graph could be used to improve the accuracy of the answer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or use this density table: &lt;a href=&quot;http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-16T22:02:03.283" LastActivityDate="2015-02-16T22:02:03.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25209" PostTypeId="1" AcceptedAnswerId="25210" CreationDate="2015-02-16T21:29:15.213" Score="0" ViewCount="356" Body="&lt;p&gt;What does&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Pb(NO3)2 + KI}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;form?&lt;/p&gt;&#xA;" OwnerUserId="13444" LastEditorUserId="13444" LastEditDate="2015-05-10T07:41:56.443" LastActivityDate="2015-05-10T07:41:56.443" Title="What does Pb(NO₃)₂ + KI form?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ClosedDate="2015-05-10T10:30:30.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="25210" PostTypeId="2" ParentId="25209" CreationDate="2015-02-16T21:38:01.707" Score="5" Body="&lt;p&gt;This is just a metathesis (double replacement) reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that nitrates are soluble, and alkali metals are soluble (at least in these general cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore balancing for now, and do that last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb(NO3)2 + KI -&amp;gt; Pb^{+2} + NO3^- + K+ + I-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb^{+2} + NO3- + K+ + I- -&amp;gt; PbI2 + KNO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the $\ce{KNO3}$ is also soluble, and would dissociate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb^{+2} + NO3- + K+ + I- -&amp;gt; PbI2 + K+ + NO3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently, we can cancel out the spectator ions $\ce{K+}$ and $\ce{NO3-}$ and are left with a net ionic equation (which is easily balanced).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb^{+2} + 2I- -&amp;gt; PbI2}$&lt;/p&gt;&#xA;" OwnerUserId="7641" LastActivityDate="2015-02-16T21:38:01.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25211" PostTypeId="2" ParentId="25202" CreationDate="2015-02-16T21:45:39.400" Score="5" Body="&lt;p&gt;Hydride shift occurs because in the intermediate dianion there is no other suitable leaving group. It is either $\ce{H-}$, $\ce{R-}$ or $\ce{O^{2-}}$. Of those $\ce{H-}$ is the best leaving group. Note though that even under these conditions $\ce{H-}$ will of course not be solvated on its own (it is way too unstable for that). It is directly transferred from the dianion to the aldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction will go &quot;to the right&quot; since the hydride shift is irreversible. There is simply no mechanism to form a dianion from the alcohol and carboxylic acid.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-09-03T15:25:44.007" LastActivityDate="2015-09-03T15:25:44.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25212" PostTypeId="2" ParentId="25177" CreationDate="2015-02-16T22:00:07.560" Score="7" Body="&lt;p&gt;It isn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under alkaline conditions, the Hofmann rearrangement of primary amides to amines can be performed with chlorine too. But why work with chlorine when bromine is much more convenient. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Choosing bromine over chlorine when possible is an obvious decision for everyone with practical lab experience! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Under standard conditions, bromine is a liquid and can be added using a dropping funnel, while chlorine is a gas, which makes adding stoichiometric amounts much more tedious.&lt;/strong&gt; &#xA;&lt;hr&gt;&#xA;Hypohalites have been used too: in the presence of alkanols, rearranged carbamates are formed. Under acidic conditions, hypervalent iodine compounds have been shown to be effective.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-06-07T05:40:17.813" LastActivityDate="2018-06-07T05:40:17.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25213" PostTypeId="1" CreationDate="2015-02-16T23:23:47.803" Score="7" ViewCount="877" Body="&lt;p&gt;I know that subcritical and supercritical $\ce{CO2}$ are excellent solvents. I've seen some examples of plastics being damaged by the $\ce{CO2}$ being impregnated into the plastic at very high pressures, but in general would liquid $\ce{CO2}$ be able to dissolve plastic?&lt;/p&gt;&#xA;" OwnerUserId="13446" LastEditorUserId="4231" LastEditDate="2015-02-17T02:03:48.947" LastActivityDate="2019-04-20T17:49:50.247" Title="Can Liquid CO₂ dissolve plastic" Tags="&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25214" PostTypeId="1" CreationDate="2015-02-17T00:46:36.187" Score="8" ViewCount="9359" Body="&lt;p&gt;I'm learning about orbital quantum numbers. While checking several elements on the periodic table I noticed that there aren't any atoms that have electrons in the g or h subshells. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="13448" LastEditorUserId="16683" LastEditDate="2018-09-20T13:53:58.113" LastActivityDate="2018-09-20T13:53:58.113" Title="Why do no known atoms have electrons in the g or h subshells?" Tags="&lt;orbitals&gt;&lt;electronic-configuration&gt;&lt;atoms&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25215" PostTypeId="1" CreationDate="2015-02-17T01:03:39.583" Score="2" ViewCount="9952" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdSIO.png&quot; alt=&quot;Vinegar concentration formula&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DavePhD was kindly helping me with this problem but the site indicated the discussion had gone on too long. &lt;a href=&quot;https://chemistry.stackexchange.com/questions/25203/calculating-vinegar-strength?noredirect=1#25204&quot;&gt;Calculating vinegar strength&lt;/a&gt; I don't have enough units to start a chat so I am starting a new thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula above is for determining the concentration or percentage of acetic acid in a vinegar sample when the ph is known. The farmer will get the ph with a ph meter, do the math and calculate the percentage acidity of her vinegar. The Ka, according to the formula DavePhD gave, is .000174 because the pKa for vinegar is 4.76. The molar mass is 60.05196 g/mol. So assuming the ph is 3, could someone write out the &quot;long math&quot; for this? I need someone to walk me through this, step by step to solve for c. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time I took algebra was about 50 years ago so, while I remember some things, it is probably better to assume I don't know much :-). Let me know if you need other numbers. Dave says the &quot;M&quot; is just a place holder&lt;/p&gt;&#xA;" OwnerUserId="13441" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-04-19T00:42:49.000" Title="Calculating the concentration of acetic acid in vinegar from pH" Tags="&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25218" PostTypeId="2" ParentId="25215" CreationDate="2015-02-17T02:04:16.030" Score="1" Body="&lt;p&gt;First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation. &lt;/p&gt;&#xA;" OwnerUserId="13428" LastEditorUserId="4356" LastEditDate="2015-02-17T21:57:01.780" LastActivityDate="2015-02-17T21:57:01.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25219" PostTypeId="2" ParentId="25214" CreationDate="2015-02-17T02:07:31.993" Score="11" Body="&lt;p&gt;&lt;strong&gt;Well, the first available sub-shell for &quot;g&quot; electrons would be 5g (i.e., 1s, 2p, 3d, 4f, so 5g). Based on current trends, we'd need row 8 of the periodic table. We just haven't found those elements yet.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at it this way.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After row 1 (1s electrons only), there are 2 rows of the s and p-block elements before &quot;d&quot; opens up with $\ce{3d^1 4s^2}$&lt;/li&gt;&#xA;&lt;li&gt;After two rows of s- p- and d-block elements, the &quot;f&quot; block opens up with cerium: $\ce{4f^1 5d^1 6s^2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So we'd need two more rows of the f-block (i.e., the lanthanides and actinides) and then complete the row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, we have found element 118, so if we can synthesize a few more, we can open up the &quot;g&quot; block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there are elements that have g electrons in excited states, so g and h-orbitals are relevant to some chemistry. But the ground-state elements haven't been discovered yet.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="16683" LastEditDate="2018-09-20T13:53:36.670" LastActivityDate="2018-09-20T13:53:36.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25220" PostTypeId="2" ParentId="25214" CreationDate="2015-02-17T02:13:41.060" Score="1" Body="&lt;p&gt;Although, through a massive amount of energy, you could excite the electrons into g and h level orbitals no elements have electrons in those orbitals at the ground state. Think about how difficult it would be to have an electron in such large and excited state. You would need more protons than any of the elements we currently know about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it takes to much energy for any known elements to hold an electron in a nonexcited state in the g or h orbital.&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-17T02:13:41.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25221" PostTypeId="1" CreationDate="2015-02-17T02:21:03.303" Score="3" ViewCount="7629" Body="&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A reaction rate doubles when the temperature increases from $\pu{25^\circ C}$ to $\pu{40^\circ C}$. Calculate $E_\mathrm a$ and the frequency factor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found the activation energy to be $\pu{35.8 kJ}$ using the two point form of the Arrhenius equation. What I'm having trouble with is finding the frequency factor. I have two unknowns, $k$ and $A$, and to me it seems like this is impossible to solve without knowing what the rate constant $k$ is. All the examples in the book solves this problem graphically, but apparently you can solve this another way according to my teacher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer given for $A$ is $1.9 \times 10^6$ but what method do you use to solve for this?&lt;/p&gt;&#xA;" OwnerUserId="13450" LastEditorUserId="5026" LastEditDate="2018-02-17T15:15:58.607" LastActivityDate="2018-02-17T15:15:58.607" Title="Finding the frequency factor when raising temperature" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25223" PostTypeId="2" ParentId="25052" CreationDate="2015-02-17T02:36:55.017" Score="1" Body="&lt;p&gt;In terms of acidity, ionic(polar) bonds dissociate more easily than covalent(polar) because the electrons are shared less equally in the ionic bond leading to easier breaking apart of the molecule. An acid for with an ionic bond will dissociate more easily leading to a stronger pH than an acid with a covalent bond. &lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-17T02:36:55.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25224" PostTypeId="2" ParentId="25095" CreationDate="2015-02-17T03:36:30.487" Score="0" Body="&lt;p&gt;Using pv = nrt$$1atm\times(0.5L) = (total moles)(.08206\frac{atm\cdot L}{mol\cdot k})(293K) \\ \text{total moles} = .02079mol $$ There are 3 total moles of gas with 2 of those being hydrogen and 1 being oxygen. This means 66% of the total moles is hydrogen and 33% is oxygen. Soooooo 66% times .02079 is the amount of hydrogen moles and 33% times .02079 is the amount of oxygen. TADA!&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-17T03:36:30.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25225" PostTypeId="2" ParentId="25076" CreationDate="2015-02-17T03:55:38.883" Score="-4" Body="&lt;p&gt;The only smell you're perceiving is evaporated water molecules. Since higher temperature increases the rate of evaporation it makes sense the smell of water vapor is more prevalent when the glass is hot. Fresh air has a much larger affect on your ability to smell than it does on the rate at which water evaporates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really it's the fresh air that let's you smell the water vapor more easily not the water actually changing composition and smelling differently.&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2015-02-17T03:55:38.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25226" PostTypeId="1" AcceptedAnswerId="25227" CreationDate="2015-02-17T03:58:02.380" Score="4" ViewCount="959" Body="&lt;p&gt;I am doing a science fair project on OLEDs. I am testing whether polypyrrole can be used as an emissive layer in a 2-layer OLED. The polypyrrole I am synthesizing is water soluble, and will be spin-coated onto ITO-coated glass substrate. I also need to spin-coat on a hole-injection layer (PEDOT:PSS). I can only seem to find it in what I believe to be water-dispersed form. This will cause obvious problems since both substances are water-soluble. I read on some informal message board that I could drop droplets of water-dispersed PEDOT:PSS into a hot organic solvent with a boiling point higher than water, boil off the water, and be left with organic-solvent dispersed PEDOT:PSS. That makes sense, I am just confused about what organic solvent to use. I believe it needs to have the properties of not dissolving the PEDOT:PSS, ITO, or the polypyrrole, and allow the PEDOT:PSS to become dispersed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is &lt;a href=&quot;http://www.heraeus-clevios.com/en/_technik/productdetail_1034017.aspx?psMarketId=1280&amp;amp;psApplicationId=1036289&quot; rel=&quot;nofollow&quot;&gt;a link&lt;/a&gt; to the PEDOT:PSS I am looking at:&lt;/p&gt;&#xA;" OwnerUserId="13396" LastEditorUserId="7448" LastEditDate="2015-02-17T06:23:33.813" LastActivityDate="2015-02-17T06:23:33.813" Title="How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent?" Tags="&lt;organic-chemistry&gt;&lt;materials&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25227" PostTypeId="2" ParentId="25226" CreationDate="2015-02-17T04:12:54.347" Score="4" Body="&lt;p&gt;&lt;strong&gt;Even if you switch to an organic solvent, if the polypyrrole is in water, it will perturb the PEDOT:PSS layer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet would be to use a poly-alkyl pyrrole or other conjugated polymer that's soluble in something that doesn't dissolve PEDOT:PSS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you switch the solvent to DMF or DMSO and it dissolves the PEDOT:PSS. Then you spin-coat that new mixture onto ITO. Once you drop the aqueous solution of polypyrrole, it will dissolve the PEDOT:PSS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you're looking at this the wrong way. Rather than trying to change the PEDOT:PSS (which will be tricky - PSS is anionic) you should consider changing your polymer. Alternatively, get rid of PEDOT:PSS and use another hole injection layer. (PEDOT:PSS is acidic and will etch the indium out of ITO.)&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-02-17T04:12:54.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25228" PostTypeId="2" ParentId="25214" CreationDate="2015-02-17T04:32:20.017" Score="9" Body="&lt;p&gt;According to Umemoto and Saito,&lt;sup&gt;[1]&lt;/sup&gt; starting with element 126, elements would possess $\mathrm{5g}$ electrons. The calculated ground-state electronic configurations for elements 126–131 are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;element 126: $\ce{[Og] 8s^2 8p^1 6f^4 5g^1}$&lt;/li&gt;&#xA;&lt;li&gt;element 127: $\ce{[Og] 8s^2 8p^2 6f^3 5g^2}$&lt;/li&gt;&#xA;&lt;li&gt;element 128: $\ce{[Og] 8s^2 8p^2 6f^3 5g^3}$&lt;/li&gt;&#xA;&lt;li&gt;element 129: $\ce{[Og] 8s^2 8p^2 6f^3 5g^4}$&lt;/li&gt;&#xA;&lt;li&gt;element 130: $\ce{[Og] 8s^2 8p^2 6f^3 5g^5}$&lt;/li&gt;&#xA;&lt;li&gt;element 131: $\ce{[Og] 8s^2 8p^2 6f^3 5g^6}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;[1] Umemoto, K.; Saito, S. Electronic Configurations of Superheavy Elements. &lt;em&gt;J. Phys. Soc. Jpn.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;65&lt;/em&gt; (10), 3175–3179. &lt;a href=&quot;https://doi.org/10.1143/JPSJ.65.3175&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1143/JPSJ.65.3175&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2018-09-20T13:50:50.593" LastActivityDate="2018-09-20T13:50:50.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25229" PostTypeId="2" ParentId="25213" CreationDate="2015-02-17T05:20:11.743" Score="0" Body="&lt;p&gt;Depending upon the plastic polymer Liquid $\ce{CO2}$ I believe It dissolves plastic. The chance of this are low. Here is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ solvent has many desirable properties; technically speaking&#xA;it is a very ‘weak’ solvent: The polarity, dielectric&#xA;constant and dipole moment are less than that of &#xA;most conventional organic solvents. Consequently,&#xA;high-molecular-weight polymers, proteins and polar molecules are only sparingly soluble or even insoluble in $\ce{CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scientist have tried to make this more efficient by adding fluorine to the solvent,  Unfortunately, fluorinated surfactants are not a good way to solve the problem, being both expensive and harmful to the environment. This research is ongoing, however.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="7448" LastEditDate="2015-02-17T06:18:30.277" LastActivityDate="2015-02-17T06:18:30.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25231" PostTypeId="2" ParentId="25221" CreationDate="2015-02-17T05:56:17.587" Score="3" Body="&lt;p&gt;This question has no answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arrhenius equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k = A e^{-\frac{E_a}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linearized form of the Arrhenius equation is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln{k} = \ln{A} - \frac{E_a}{R}T^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation linearly relates $\ln{k}$ to $T^{-1}$: the intercept is $\ln{A}$ and the slope is $-\frac{E_a}{R}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fully define a line, we need two parameters.  This can be two completely specified points that lie on the line, or any single point on the line plus a slope for the line.  For this problem that would mean either (a) two temperatures and two rates, or (b) one temperature, one rate, and one slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the information we are given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln{k} = \ln{A} - \frac{E_a}{R}T_1^{-1}$$&#xA;$$ \ln{2k} = \ln{2} + \ln{k} = \ln{A} - \frac{E_a}{R}T_2^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way we combine those two equations will only yield an equation equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln{2} = -\frac{E_a}{R}\left(T_2^{-1} - T_1^{-1} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $\ln{k}$ and $\ln{A}$ have both cancelled out.  That is because the starting two linear equations have the same coefficients for $\ln{k}$ and $\ln{A}$ in each equation.  Similarly, the two equations $2x=y$ and $2x+2=y+2$  can't be solved for $x$ and $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem as stated gives us only a slope, but not even a single point that lies on the line.  The rate could double by going from 1,000,000 $\text{s}^{-1}$ to 2,000,000 $\text{s}^{-1}$ (a very fast reaction!) or by going from 0.1 $\text{yr}^{-1}$ to 0.2 $\text{yr}^{-1}$ (pretty slow).  There is no way to find the intercept of a line when we are given only the slope.  Thus, there is no way to solve for $A$ using the information given.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-17T05:56:17.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25232" PostTypeId="2" ParentId="25157" CreationDate="2015-02-17T06:31:21.633" Score="2" Body="&lt;p&gt;Phase changes are fickle. Much like supersaturated solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a crystal to a supersaturated solution and it will precipitate rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bring a crystal of the thermodynamically favorable phase into contact with a &quot;supersaturated&quot; phase and it may just do this: &lt;a href=&quot;https://www.youtube.com/watch?v=sXB83Heh3_c&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=sXB83Heh3_c&lt;/a&gt;&#xA;I am not sure how exactly the proceedure in the video was done, so do not take it as evidence that my statement is true. &#xA;Also, Diamond is stable well below its phase transition temperature. Surely there are other compounds like that. &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2015-02-17T06:49:53.240" LastActivityDate="2015-02-17T06:49:53.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25233" PostTypeId="1" AcceptedAnswerId="25237" CreationDate="2015-02-17T07:09:25.047" Score="2" ViewCount="23000" Body="&lt;p&gt;The half equation for the reduction of the permanganate would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{5e- + 8H+ + MnO4- -&amp;gt; Mn^2+ + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with regards to the iodide, I am very confused. I have seen some sources say that $\ce{I-}$ is oxidised to $\ce{IO3-}$ in this reaction, and others saying that it is oxidised to $\ce{I2}$.&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="5026" LastEditDate="2018-06-08T02:46:19.107" LastActivityDate="2018-06-08T02:47:43.727" Title="What is the ionic equation for reaction of potassium permanganate with potassium iodide?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25234" PostTypeId="1" AcceptedAnswerId="25240" CreationDate="2015-02-17T07:29:04.557" Score="6" ViewCount="11434" Body="&lt;p&gt;I know that the zinc half equation would be&#xA;$\ce{Zn -&amp;gt; Zn^{2+} + 2e-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am confused about the half equation for iron. I would think that since iron is completely displaced, its oxidation state goes to zero, so its half equation would be $\ce{Fe^{2+} + 2e- -&amp;gt; Fe}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my teacher said that whenever you have iron, it will always change oxidation states from 2+ to 3+ or vice-versa in a redox reaction, and will not go to zero. For instance, she gave us an example in which there was a reaction between iron(iii) chloride and zinc. I thought that the zinc would completely displace the iron so that it goes from 3+ ion to 0 in the solid iron metal. However my teacher insisted that it goes to 2+, and not to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example of iron(ii) chloride, I don't even see how it is possible for it to go to 3+ because then both zinc and iron are oxidised!&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="5026" LastEditDate="2018-06-08T02:58:36.367" LastActivityDate="2018-06-08T02:58:36.367" Title="How can the iron in the reaction of iron (ii) chloride and zinc not go to zero oxidation state?" Tags="&lt;redox&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25235" PostTypeId="2" ParentId="25110" CreationDate="2015-02-17T07:51:32.023" Score="1" Body="&lt;p&gt;The angular structure is actually a distortion produced by the bond pair electrons (between oxygen and fluorine) and lone pair electrons of oxygen since every structure tries to decrease its net potential energy they assume this shape. This doesn't happen in case of Carbon-disulphide because the sulphur atoms each make a double bond with carbon leaving no lone pair or relative high electron density on carbon atom, instead the two sulphur atoms with 2 lone airs each after forming double bonds, repel each other into a configuration of linear to minimise their relative potential energies.&lt;/p&gt;&#xA;" OwnerDisplayName="user10153" LastActivityDate="2015-02-17T07:51:32.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25236" PostTypeId="1" CreationDate="2015-02-17T08:39:08.440" Score="0" ViewCount="2323" Body="&lt;p&gt;I was going through my chemistry book (RT Morrison and Boyd) and found a statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot; the collision frequency of isopropyl and n-propyl w.r.t. Chlorine atoms is the same.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it correct because I think the size will matter in this case while both have the same weight?&lt;/p&gt;&#xA;" OwnerUserId="13455" LastEditorUserId="5017" LastEditDate="2015-04-30T18:26:48.893" LastActivityDate="2015-04-30T18:26:48.893" Title="Orientation of halogenation" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="25237" PostTypeId="2" ParentId="25233" CreationDate="2015-02-17T08:50:03.380" Score="6" Body="&lt;p&gt;How stable is $\ce{IO3-}$ in the presence of $\ce{I-}$ under acidic conditions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would end up being in an equilibrium with very few $\ce{IO3-}$ anions and lots of iodine, due to the following equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 IO3- + 12 H3O+ + 10 I- &amp;lt;=&amp;gt; 6 I2 + 18 H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-06-08T02:47:43.727" LastActivityDate="2018-06-08T02:47:43.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25238" PostTypeId="1" CreationDate="2015-02-17T09:34:46.763" Score="3" ViewCount="116" Body="&lt;p&gt;To explain my question I'll use an example: We all know that sodium displaces copper in a solution of its sulfate in the reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na (s) + CuSO4 (aq) \rightarrow Na2SO4 (aq) + Cu (s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$) along side this displacement reaction. Shouldn't at least a bit of NaOH be formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate a comprehensive response that a high school sophomore (me) would understand. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="13456" LastEditorUserId="7448" LastEditDate="2015-02-17T09:47:00.867" LastActivityDate="2015-02-17T12:47:44.373" Title="Why don't water molecules react alongside aqueous reactions?" Tags="&lt;reaction-mechanism&gt;&lt;water&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25239" PostTypeId="2" ParentId="25213" CreationDate="2015-02-17T09:46:15.770" Score="5" Body="&lt;p&gt;From the abstract of &lt;a href=&quot;http://dx.doi.org/10.1021/jp9615823&quot; rel=&quot;noreferrer&quot;&gt;Solubility of Polymers and Copolymers in Supercritical &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[…] data to temperatures of 270 °C and 3000 bar […]&lt;/p&gt;&#xA;&lt;p&gt;Over the same range of conditions, &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; cannot dissolve polyethylene, poly(acrylic acid), poly(methyl methacrylate), poly(ethyl methacrylate), polystyrene, poly(vinyl fluoride), or poly(vinylidene fluoride) […]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-17T09:46:15.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25240" PostTypeId="2" ParentId="25234" CreationDate="2015-02-17T09:51:00.097" Score="8" Body="&lt;p&gt;Generally, to predict redox reactions, we consider the Standard Reduction Potentials at $25\ ^\circ\mathrm{C}$: The bigger the difference between the potentials of two redox couples, more favorable is the reaction.&#xA;$$E^0(\ce{Fe^3+ /Fe^2+})=0.77\ \mathrm{V}$$&#xA;$$E^0(\ce{Fe^3+ /Fe})=-0.04\ \mathrm{V}$$&#xA;$$E^0(\ce{Fe^2+ /Fe})=-0.44\ \mathrm{V}$$&#xA;$$E^0(\ce{Zn^2+ /Zn})=-0.76\ \mathrm{V}$$&#xA;So, when $\ce{Fe^3+}$ reacts with $\ce{Zn}$, we get $\ce{Fe^2+}$ (not $\ce{Fe}$) and $\ce{Zn^2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your reaction: when $\ce{Fe^2+}$ reacts with $\ce{Zn}$, we get  $\ce{Fe}$ and $\ce{Zn^2+}$.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorDisplayName="user7951" LastEditDate="2015-04-11T12:08:37.770" LastActivityDate="2015-04-11T12:08:37.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="25241" PostTypeId="1" AcceptedAnswerId="26248" CreationDate="2015-02-17T09:51:52.417" Score="2" ViewCount="426" Body="&lt;p&gt;I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time. I never questioned this before, but recently I have been learning about the rate equation and the order of the reaction with respect to a reactant. Seeing the graphs of how concentration changes with respect to time when you have a different order with respect to a reactant (as shown &lt;a href=&quot;https://sites.google.com/site/kwokthechemteacher/Home/kwokthechemteacher/chemical-kinetics/calculating-order-of-reactions&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that choosing a different reactant to measure the rate of change of concentration of would give you a different rate of reaction? So how is the rate of a reaction defined?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="4356" LastEditDate="2015-02-17T21:51:14.397" LastActivityDate="2015-02-17T21:53:12.170" Title="How is the rate of a reaction defined?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25242" PostTypeId="2" ParentId="14811" CreationDate="2015-02-17T11:07:22.947" Score="5" Body="&lt;p&gt;The surface of a tent consists mostly of water-resistant polymers such as latex or plastic. These polymers are disordered molecules such as spaghetti, so they have unlike crystals a high entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gough (1805) and Joule (1859) studied the thermal behavior of rubber. They made the following observations, which are known as Gough-Joule effects:&#xA;• a taut with constant weight rubber shrinks when it is heated&#xA;• rubber emits heat when it is stretched&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cause of the rubber elasticity is fundamentally different than the elasticity of crystalline materials. In contrast to a crystal, the atoms are not regularly arranged in a rubber, but they form long chains (polymers) which are interconnected and form a network. Clearly, one can think of it as a plate of cooked spaghetti. This difference in microstructure are reflected in the mechanical properties of rubber. When the rubber network is loose, very many different configurations (configurations) of the chain between two connection points are possible - more disorder (larger entropy S).&lt;/p&gt;&#xA;" OwnerUserId="13429" LastEditorUserId="7306" LastEditDate="2016-06-29T17:51:06.287" LastActivityDate="2016-06-29T17:51:06.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="25243" PostTypeId="2" ParentId="25238" CreationDate="2015-02-17T12:12:58.467" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree that sodium metal would react with the water.  Where/how did you come up with the equation implying otherwise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you'll enjoy this video: &lt;a href=&quot;https://www.youtube.com/watch?v=X5yLtqlopTM&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=X5yLtqlopTM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-17T12:47:44.373" LastActivityDate="2015-02-17T12:47:44.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25244" PostTypeId="1" CreationDate="2015-02-17T12:40:23.337" Score="4" ViewCount="430" Body="&lt;p&gt;Here is a given chemical formula for ceramic material: &lt;br&gt;&#xA;La&lt;sub&gt;0.6&lt;/sub&gt;Sr&lt;sub&gt;0.4&lt;/sub&gt;Fe&lt;sub&gt;0.8&lt;/sub&gt;Ni&lt;sub&gt;0.2&lt;/sub&gt;O&lt;sub&gt;3-δ&lt;/sub&gt;&#xA;&lt;br&gt;&#xA;There is a letter δ. What does it stand for? I guess it is something about oxygen vacancies in crystal lattice.&lt;/p&gt;&#xA;" OwnerUserId="13385" LastEditorUserId="23561" LastEditDate="2018-07-03T00:04:45.110" LastActivityDate="2018-07-03T00:04:45.110" Title="What does delta letter mean in some ceramics chemical material formula?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;materials&gt;&lt;ceramics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26242" PostTypeId="2" ParentId="25244" CreationDate="2015-02-17T13:54:40.880" Score="5" Body="&lt;p&gt;Exactly. In your example, it is related to oxygen vacancy (oxygen deficiency) in the crystal lattice.&#xA;It can affect the physical properties of the ceramic material. Take the example of this piezoelectric ceramic material:$\ce{Eu_{0.5}Ba_{0.5}TiO_{3-\delta}} $ Oxygen-vacancy induces antiferromagnetism to ferromagnetism transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, $\delta$ denotes deviation from the stoichiometry (exces or deficiency) in a chemical formula&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-17T13:54:40.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26243" PostTypeId="2" ParentId="25236" CreationDate="2015-02-17T14:16:02.073" Score="1" Body="&lt;p&gt;The statement is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;This[collision frequency] must be the same for the two reactions, since both involve collisions of the same particles: a propane molecule and a chlorine atom.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is clear that n-propyl radical and Isopropyl radical have same carbon structure. In the next step the hydrocarbon radical reacts with a chlorine ($\ce{Cl2}$) and since chlorine is symmetrical the collision frequency cannot depend upon it. It should be easy to judge from the following structures, the very minor difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ncv6s.png&quot; width=&quot;500&quot; height=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is: Since the difficult step is the abstraction of a hydrogen and the determination of rates should be done while comparing primary, secondary or tertiary radicals, which have different stability.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-02-17T14:16:02.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26244" PostTypeId="2" ParentId="25203" CreationDate="2015-02-17T14:48:59.437" Score="1" Body="&lt;p&gt;As we know:&#xA;$$\rm pH=\frac12(pKa-\log C)\quad\mid\quad C=\frac{\rm weight\;of\;acid}{\rm Mol\;wt.*Volume of sln.}\quad\mid\quad Strength=C*Mol.\;wt.$$&#xA;Calculating:&#xA;$$\rm S=10^{pKa-2pH}*M$$&#xA;So finally:&#xA;$$\huge\fbox{$\rm S=10^{4.75^{\#}-2pH}*60.05$}$$&#xA;&lt;sub&gt;#You can as well take that 4.76&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically It is correct but some deviations are bound to occur in reality. I suggest finding out that in few samples first.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2015-02-17T14:55:29.420" LastActivityDate="2015-02-17T14:55:29.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26246" PostTypeId="1" CreationDate="2015-02-17T14:53:29.347" Score="1" ViewCount="1116" Body="&lt;p&gt;I am learning about lattice enthalpies, and my textbook talks about determining the lattice enthalpy from the standard change of enthalpy of solution and the change of enthalpy of hydration of the ions. It says that the changes in enthalpy of solution and hydration can be determined experimentally but that the lattice enthalpy cannot be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am struggling to think of how one can measure the enthalpy change when one mole of gaseous ions is dissolved in water under standard conditions. I thought one would encounter the same problem as with the lattice enthalpy, and this involves one mole of an ionic lattice being formed from its gaseous ions under standard condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that the problem with this was getting the exact amounts of gaseous ions under standard conditions.&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="865" LastEditDate="2015-10-12T03:03:08.467" LastActivityDate="2016-04-09T20:26:10.660" Title="How does one measure the change in enthalpy of hydration?" Tags="&lt;aqueous-solution&gt;&lt;enthalpy&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26248" PostTypeId="2" ParentId="25241" CreationDate="2015-02-17T15:11:03.717" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It seems to me that choosing a different reactant to measure the rate of change of concentration of reactant would give you a different rate of reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose a reaction $\rm aA+bB\to cC$, then let rate law comes out to be $\rm r=k[A]^x[B]^y$ then we would write:&#xA;$$-\frac1a\frac{\rm \delta [A]}{\rm \delta t}=-\frac1b\frac{\rm \delta [B]}{\rm \delta t}=\rm k[A]^x[B]^y$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So how is rate of reaction defined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is defined in the same way you learned but the rate law is just something we &quot;humans&quot; created to relate the &lt;strong&gt;concentration&lt;/strong&gt; to &lt;strong&gt;rate of change of concentration&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="4356" LastEditDate="2015-02-17T21:53:12.170" LastActivityDate="2015-02-17T21:53:12.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26250" PostTypeId="2" ParentId="19284" CreationDate="2015-02-17T16:02:14.750" Score="5" Body="&lt;p&gt;&lt;strong&gt;Arene&lt;/strong&gt; — Compound which contains a benzene ring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aromatic Compounds&lt;/strong&gt; — Compounds having aroma (wait, the modern definition is different, I know)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every arene is an aromatic compound but every aromatic compound need not be an arene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{arene} \subset \text{aromatic compounds}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore anything which contains a benzene ring has to be aromatic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples of arenes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lu1eu.jpg&quot; alt=&quot;list of arenes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting into aromatic compounds, the &lt;em&gt;modern criteria for aromaticity&lt;/em&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The molecule is cyclic (a ring of atoms)&lt;/li&gt;&#xA;&lt;li&gt;The molecule is planar (all atoms in the molecule lie in the same plane)&lt;/li&gt;&#xA;&lt;li&gt;The molecule is fully conjugated (p orbitals at every atom in the ring)&lt;/li&gt;&#xA;&lt;li&gt;The molecule has $4n+2$ π electrons ($n=0$ or any positive integer)*&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples (non-benzenoids):&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/guiXw.png&quot; alt=&quot;non-benzenoids&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, benzenoids (compounds which contain benzene) do match with this. Therefore, benzene derivatives are aromatic compounds.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="7475" LastEditDate="2017-10-26T04:35:05.417" LastActivityDate="2017-10-26T04:35:05.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26251" PostTypeId="1" AcceptedAnswerId="26253" CreationDate="2015-02-17T16:31:56.263" Score="3" ViewCount="2905" Body="&lt;p&gt;If you have an endothermic reaction with a negative entropy change, is it still possible to induce the reaction in some way despite the fact that the Gibbs free energy change is positive or all temperatures?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="402" LastEditDate="2017-04-01T02:26:30.887" LastActivityDate="2017-04-01T02:26:30.887" Title="Can a reaction with positive Gibbs free energy change ever occur?" Tags="&lt;energy&gt;&lt;enthalpy&gt;&lt;entropy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26252" PostTypeId="2" ParentId="25241" CreationDate="2015-02-17T16:53:53.000" Score="1" Body="&lt;p&gt;Let's take a sample reaction $\ce{\it aA + bB-&amp;gt; pP + qQ}$ with stoichiometric coefficients $a$, $b$, $p$, and $q$ for reactants $A$ and $B$ and products $P$ and $Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;rate of reaction&lt;/em&gt; is the &lt;em&gt;rate of change&lt;/em&gt; of the &lt;em&gt;extent&lt;/em&gt; of reaction.  Extent of reaction is usually given the symbol $\xi$, so reaction rate is $\frac{d\xi}{dt} = \dot{\xi}=r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction rate is $\dot{\xi}$, then the rate of change of concentrations is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d[A]}{dt}=-a \dot{\xi}$$&#xA;$$\frac{d[B]}{dt}=-b \dot{\xi}$$&#xA;$$\frac{d[P]}{dt}= p \dot{\xi}$$&#xA;$$\frac{d[Q]}{dt}= q \dot{\xi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the stoichiometric coefficients link the &lt;em&gt;rate&lt;/em&gt; of formation/consumption of the products/reactants, sort of by the definition of stoiciometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when talking about the &lt;em&gt;order&lt;/em&gt; of a reaction, what we are doing in effect is giving formulas (or partial formulas) for $\dot{\xi}$.  If the reaction is &quot;elementary&quot;, then it will be $a$th order with respect to $A$ and $b$th order with respect to $B$.  This doesn't have to be true but often is.  This means that if you change the concentration of A &lt;em&gt;while holding everything else constant&lt;/em&gt;, the change in rate will be proportional to the concentration of $A$ to the $a$th power, etc.  In math form that looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\xi} \propto [A]^a$$&#xA;$$\dot{\xi} \propto [B]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\propto$ means &quot;proportional to&quot;, which is saying that the things are equal except for some one unknown constant.  Thus those two $\propto$ &quot;equations&quot; are equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\xi} = \kappa_1 [A]^a$$&#xA;$$\dot{\xi} = \kappa_2 [B]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note that $\kappa_1$ could depend on everything except $[A]$, i.e. it could be a function of $[B]$ etc.  Similarly $\kappa_2$ could depend on $[A]$ (but not on $[B]$).  The only way for both equations above to be true is if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\xi} = k [A]^a [B]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $\kappa_1 = (k [B]^b)$ and $\kappa_2 =  (k [A]^a)$.  Now that we know the reaction &lt;em&gt;order&lt;/em&gt; with respect to each reactant, and have written it down in equation form, we can relate it back to the change in concentration of each of the reactants/products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d[A]}{dt}=-a k [A]^a [B]^b$$&#xA;$$\frac{d[B]}{dt}=-b k [A]^a [B]^b$$&#xA;$$\frac{d[P]}{dt}= p k [A]^a [B]^b$$&#xA;$$\frac{d[Q]}{dt}= q k [A]^a [B]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations clearly show the relationship between the &lt;em&gt;order&lt;/em&gt; of a reaction and the rate of change of reagent/product concentrations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By looking at the four equations above, the &lt;em&gt;rate of reaction&lt;/em&gt; will only be equal to the product or reactant concentration change when the stoiciometric coefficient of that product or reactant is 1.  And that's true regardless of what the &quot;order&quot; of the reaction is.  The reaction $\ce{\it aA + bB-&amp;gt; pP + qQ}$ doesn't &lt;em&gt;have&lt;/em&gt; to be $a$th order with respect to $a$.  It could have been that $\dot{\xi} \propto [A]^{1.7}$ or even that $\dot{\xi} \propto \frac{[A]^{0.2}}{K_m+[A]}$ or whatever.  But in chemistry, it does often happen that reactions &lt;em&gt;do&lt;/em&gt; have orders that depend straightforwardly on stoichiometry, which is why those types of reactions are studied so often.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-17T16:53:53.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26253" PostTypeId="2" ParentId="26251" CreationDate="2015-02-17T16:58:44.037" Score="5" Body="&lt;p&gt;Yes, $\Delta G = -RT\ln K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a reaction is endothermic with a negative entropy change, $\Delta G$ is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That $\Delta G$ is positive only means that $K&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G$ would need to approach infinity for $K$ to approach zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, at equilibrium, there will always be some products in principle no matter how unfavorable the reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical matter, the reaction would need to be only slightly endothermic and having only a slight negative entropy change to get a significant amount of product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the small amount of product is continuously removed from the system, as for example by precipitation from a liquid phase, or gas bubbling out of a liquid phase, or by liquid-liquid extraction, one can keep the reaction going forward.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T17:26:53.017" LastActivityDate="2015-04-26T17:26:53.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26254" PostTypeId="1" AcceptedAnswerId="26263" CreationDate="2015-02-17T17:18:37.090" Score="15" ViewCount="9612" Body="&lt;p&gt;How is it possible to heat a tiny amount (30&amp;nbsp;ml)[1] of water to a high enough temperature to make a coffee, in less than 10 seconds and possibly instantly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most heaters that I know of heat water in no less than 90 seconds (induction heater) or 3 minutes which is way too much for my purposes, and a microwave oven takes me one minute and a half to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] A single coffee is between 20 - 30 ml and my coffee machine makes 30 ml for each cup.&lt;/p&gt;&#xA;" OwnerUserId="14457" LastEditorUserId="14457" LastEditDate="2015-02-18T09:21:45.887" LastActivityDate="2016-06-05T22:13:20.883" Title="How can 30 ml of water be heated in less than 10 seconds?" Tags="&lt;water&gt;&lt;experimental-chemistry&gt;&lt;heat&gt;" AnswerCount="9" CommentCount="10" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26255" PostTypeId="2" ParentId="14920" CreationDate="2015-02-17T17:22:01.203" Score="1" Body="&lt;p&gt;A reasonable trick to get the oxidation state of both the &lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt; atoms would be to write down the ionic components of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4NO3}$&lt;/span&gt;. They are-&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NO3-}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now these ions can be considered separately to get the answer which is &lt;span class=&quot;math-container&quot;&gt;$-3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$+5$&lt;/span&gt; respectively.&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="95133" LastEditDate="2020-09-12T12:07:06.010" LastActivityDate="2020-09-12T12:07:06.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26257" PostTypeId="1" AcceptedAnswerId="26260" CreationDate="2015-02-17T17:30:24.160" Score="7" ViewCount="16011" Body="&lt;p&gt;I read this statement that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Xenon is the most soluble noble gas in water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first doubt is:&lt;br&gt;&#xA;Why does the solubility of noble gases in water increase down the group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if this is the trend, why is Radon not considered most soluble?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="7475" LastEditDate="2015-11-23T13:52:11.257" LastActivityDate="2015-11-23T13:52:11.257" Title="Solubility of Noble Gases" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;periodic-trends&gt;&lt;noble-gases&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26258" PostTypeId="1" AcceptedAnswerId="27606" CreationDate="2015-02-17T17:35:55.793" Score="2" ViewCount="643" Body="&lt;p&gt;The pseudo 1st order reaction happens when one of the reactant is in large excess and its concentration doesnt change during the reaction .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other such situation when a &lt;em&gt;certain&lt;/em&gt; higher order reaction turns out to become a pseudo first order kinetic reaction (except the situation where an N$^{th}$ order reaction has N-1 species in large excess) ?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="13409" LastEditDate="2015-03-20T11:52:25.463" LastActivityDate="2015-03-21T06:31:43.650" Title="Pseudo 1st order Kinetic reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26260" PostTypeId="2" ParentId="26257" CreationDate="2015-02-17T17:53:36.110" Score="6" Body="&lt;p&gt;Wikipedia gives the following solubilities for the noble gases in water at $25^\circ \text C$ in $\mathrm{cm^3~kg^{-1}}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{He: 8.61~~~Ne: 10.5~~~	Ar: 33.6~~~	Kr: 59.4	~~~Xe: 108.1	~~~Rn: 230}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does solubility of noble gases increase down the group?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intermolecular interactions of noble gases are in the form of London dispersion forces, which are strongly influenced by the polarizability of the atom which increases as the atomic radius increases. This can be thought of as resulting from the electrons being 'less tightly bound' to the nucleus and the atomic radius increases. Atomic radius increases down the group therefore solubility also increases. For more on polarizability see &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Intermolecular_Forces/Polarizability&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is radon not the most soluble noble gas?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radon (and other heavy radioactive elements) are often ignored when talking about trends because they are uncommon. Radon is in fact the most soluble noble gas but the half lives of its isotopes are on the order of hours or days so it is not really very useful.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-17T17:53:36.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26261" PostTypeId="1" CreationDate="2015-02-17T17:53:53.240" Score="3" ViewCount="1621" Body="&lt;p&gt;In most substituents, the mesomeric effect is considered stronger and more dominating than the inductive effect.&lt;br&gt;&#xA;A notable exception are the halogens.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason behind this has been well-explained &lt;a href=&quot;https://chemistry.stackexchange.com/a/24572/13409&quot;&gt;here&lt;/a&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I further had questions about the following: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since its about the EN, would all other electronegative elements with lone pair show this anomaly?  &lt;/li&gt;&#xA;&lt;li&gt;Are there other substituents with this exceptional behavior?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-11T01:38:53.757" Title="In which all substituents does Inductive effect overpower Mesomeric effect?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;electronegativity&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26262" PostTypeId="2" ParentId="26257" CreationDate="2015-02-17T18:00:26.610" Score="3" Body="&lt;p&gt;The solubility of radon in water has long been studied, and the Weigel equation in the name of the equation for radon's solubility in water as a function of temperature, at atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Rn(aq)]/[Rn(g)] = 0.105 + 0.405 * exp(-0.0502*T) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near room temperature, the concentration of radon in the gas phase will be about 4 times the amount in the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radon is the most soluble noble gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://diginole.lib.fsu.edu/cgi/viewcontent.cgi?article=7502&amp;amp;context=etd&quot; rel=&quot;nofollow&quot;&gt;Radon Solubility In Water As A Function Of Salinity And Temperature&lt;/a&gt; by Eric B. Lieberman for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-17T18:00:26.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26263" PostTypeId="2" ParentId="26254" CreationDate="2015-02-17T18:23:10.820" Score="25" Body="&lt;p&gt;Well, let's do some math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is:&#xA;$$\begin{align}E&amp;amp;=C_Pm\Delta K \\ &#xA;&amp;amp;=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\&#xA;&amp;amp;=8.778\mathrm{\ kJ}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how much power do we need to do this in a given time? &quot;Instant&quot; doesn't really mean anything, so let's go with 10 seconds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}P&amp;amp;=\frac{E}{t}\\&#xA;&amp;amp;=\frac{8778\mathrm{\ J}}{10\mathrm{\ s}}\\&#xA;&amp;amp;=877.8 \mathrm{\ W}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an enormous amount of power, but the trick is that it all has to go into heating the water. A good microwave outputs a fair bit more power than this, but it generally doesn't all get absorbed by such a such a small amount of water in only 10 seconds. Your best bet is probably an electric heating element directly inserted into the liquid, though I don't know if you can get a ~1000 W one small enough to sit in that much water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jon Custer notes, it's not necessary to produce all the heat at once. If you heat some kind of thermal reservoir and flow the liquid past/through it, it reduces the demands on your heat source.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Also, I just tried this with a 1200 W microwave and it only took 15 seconds. How fast do you really need this coffee?&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2015-02-17T18:34:57.840" LastActivityDate="2015-02-17T18:34:57.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26264" PostTypeId="2" ParentId="26254" CreationDate="2015-02-17T18:27:05.873" Score="9" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vo5qZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Nucleate_boiling&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Nucleate_boiling&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in the above graph, there is a local heat flux maximum for heating water, when the hot surface contacting the water is 30 degrees C above the boiling point, that is 130 degrees C at atmospheric pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to minimize time to boiling, contact a 130 degree C surface made of high thermal conductivity material (such as silver, copper or aluminum) with the water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With heat flux over 100,000 Watts/m^2, you just need to make the surface area of contact large enough to acheive any desired heating time. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-17T18:27:05.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26265" PostTypeId="1" AcceptedAnswerId="26267" CreationDate="2015-02-17T18:27:57.877" Score="4" ViewCount="2111" Body="&lt;p&gt;An acid chloride being reduced through catalytic hydrogenation is an &quot;unobserved&quot; event in the context of our class, according to my professor. Nonetheless, we are &lt;strong&gt;expected&lt;/strong&gt; to &lt;em&gt;predict&lt;/em&gt; that the hydrogenation occurs &lt;em&gt;because&lt;/em&gt; acid chlorides are very reactive in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this sounds like a case of the right answer with the wrong reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, acid chlorides are reactive. We've covered their hydrolysis with water. We've covered their role in Friedel-Crafts acylation. We realize that the $\ce{C-Cl}$ bond in an acid chloride is very weak and that the carbonyl carbon is highly electrophilic. This gives rise to most of the reaction pathways with acid chlorides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what does catalytic hydrogenation have to do with the electrophilicity of the carbonyl carbon in an acid chloride? Nothing, in my mind. Hydrogen gas and platinum metal simply work here because the $\ce{C=O}$ part of the acid chloride is flat. I don't think it has anything to do with how reactive the acid chloride is. Right? Hell, even plain jane alkenes can be hydrogenated and they aren't exactly &lt;em&gt;reactive&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="7475" LastEditDate="2015-10-16T16:48:37.940" LastActivityDate="2015-10-16T16:48:37.940" Title="Acid chlorides and reduction by hydrogenation" Tags="&lt;redox&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26266" PostTypeId="1" CreationDate="2015-02-17T18:28:55.153" Score="2" ViewCount="362" Body="&lt;p&gt;I know that in physics there is thermionic emission and the photoelectric effect. These are both method of removing electrons. I think these affects are only for metals due to their de-localized electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do they say that you can remove the valence electron of a gaseous atom? Can you not do this with atoms in their solid or liquid state, and does being a metal, non metal or metalloid affect this?&lt;/p&gt;&#xA;" OwnerUserId="14459" LastActivityDate="2015-02-17T18:28:55.153" Title="Why does the definition of ionisation energy only include gaseous atoms?" Tags="&lt;electrons&gt;&lt;energy&gt;&lt;ionization-energy&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-04-06T08:09:53.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="26267" PostTypeId="2" ParentId="26265" CreationDate="2015-02-17T19:00:51.813" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Hydrogen gas and platinum metal simply work here because the C=O part&#xA;  of the acid chloride is flat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's more than just that.  Acid chlorides are significantly more reactive towards catalytic hydrogenation than other carbonyl compounds.  They are reactive enough that a poisoned catalyst must be used along with mild reaction conditions to achieve reduction to the corresponding aldehyde, and have the reaction stop there (&lt;a href=&quot;http://en.wikipedia.org/wiki/Rosenmund_reduction&quot; rel=&quot;noreferrer&quot;&gt;Rosenmund reaction&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question becomes, &quot;why are acid chlorides so reactive towards catalytic hydrogenation.&quot;  It is thought that because the acid chloride $\ce{C-Cl}$ bond is so reactive that the palladium inserts into this bond to form intermediates such the one pictured below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1ze0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other typical carbonyl compounds can't form such an intermediate; consequently acid chlorides are noticeably more reactive towards catalytic hydrogenation. Anhydrides, where the carbonyl is roughly as reactive as an acid chloride's carbonyl towards nucleophilic attack, have not been reported (to my knowledge) to produce aldehydes under Rosenmund conditions supporting this mechanism involving a palladium intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Historical Note:&lt;/strong&gt; Prior to the application of metal hydrides to the reduction of carbonyl groups, the Rosenmund reduction was a major synthetic tool for chemists.  It was one of the few methods available for the straightforward transformation of acids, esters, amides, etc, to aldehydes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-17T21:50:40.673" LastActivityDate="2015-02-17T21:50:40.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26268" PostTypeId="1" AcceptedAnswerId="27579" CreationDate="2015-02-17T19:11:58.033" Score="6" ViewCount="907" Body="&lt;p&gt;As the compound potassium chlorate exists, is there something as $\ce{KFO3}$?&#xA;All the other halogenates are mentioned somewhere on the web but what about this?&lt;br&gt;&#xA;If no, then why?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="4945" LastEditDate="2015-03-20T06:04:59.533" LastActivityDate="2016-09-05T13:50:54.117" Title="Does the compound KFO₃ exist?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26269" PostTypeId="2" ParentId="24966" CreationDate="2015-02-17T19:51:39.197" Score="1" Body="&lt;p&gt;You may look &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy#Gibbs_free_energy_of_reactions&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the formula $\Delta G=\Delta H-T\Delta S$ and also know that if for the left reaction $\Delta H_\text{r}=c$, then for the other-way reaction $\Delta H_\text{r}=-c$ (because $\Delta H_\text{r}=\sum\Delta H_\text{f} \text{(product)}-\sum\Delta H_\text{f}\text{(reactant)}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same for the entropy change $\Delta S$: If for the left reaction $\Delta S_\text{r}=c$, then for the other-way reaction $\Delta S_\text{r}=-c$ (because $\Delta S_\text{r}=\sum\Delta S_\text{f}\text{(product)}-\sum\Delta S_\text{f}\text{(reactant)}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for the reverse of a reaction, since &lt;strong&gt;both&lt;/strong&gt; $\Delta H_\text{r}$ and $\Delta S_\text{r}$ are opposite, the whole $\Delta G_\text{r}$ will be the opposite, hence at equilibrium $\Delta G=0$, so that neither way is thermodynamically favoured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In fact&lt;/strong&gt;, how much a &lt;strong&gt;reaction&lt;/strong&gt; (warning: reversible &lt;strong&gt;reactions&lt;/strong&gt; isn’t the same as reversible &lt;strong&gt;processes&lt;/strong&gt;, not to be confused) is reversible and the direction it proceeds is determined not only from thermodynamics (I would say &lt;strong&gt;much&lt;/strong&gt; more from other factors), but from Le Chatelier’s principle (temperature, pressure, concentration), catalysts and of course activation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have seen the relationship of $\Delta G_\text{r}$ of a reaction and its reverse and the fact that at equilibrium $\Delta G=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must be noted that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G_\text{r}$, $\Delta H_\text{r}$, $\Delta S_\text{r}$ refer to “reaction” (r)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G_\text{f}$, $\Delta H_\text{f}$, $\Delta S_\text{f}$ refer to “formation” (f) (of chemical compounds and elements)&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T14:07:38.700" LastActivityDate="2015-04-19T14:07:38.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26270" PostTypeId="1" CreationDate="2015-02-17T20:11:49.917" Score="13" ViewCount="4779" Body="&lt;p&gt;I am a student of biochemistry. My organic chemistry professor was giving a lecture and he talked about halohydrin formation reactions. At one point, he said that &quot;water attacks instead of bromide because water is a better nucleophile than bromide&quot;. He repeated that five times or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classmate argued that, because water is the solvent in the reaction and is in greater proportion, it will attack faster. But that still doesn't mean water is a better nucleophile, right? Bromide itself has a negative net charge, while water has no net charge, only 2 free electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, was my professor wrong or is there something I'm missing? Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="14461" LastEditorUserId="4325" LastEditDate="2015-02-17T21:47:45.347" LastActivityDate="2017-12-05T08:39:51.717" Title="Is water a better nucleophile than bromide?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;water&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26272" PostTypeId="1" AcceptedAnswerId="26278" CreationDate="2015-02-17T20:40:54.010" Score="4" ViewCount="1713" Body="&lt;p&gt;Which is the least stable trihalide of nitrogen and why?&lt;br&gt;&#xA;I have two conflicting theories, one is that fluorine-fluorine electron repulsion will make $\ce{NF3}$ least stable and another is steric reasons of iodine which give $\ce{NI3}$ as the answer.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is correct?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="4945" LastEditDate="2015-04-21T08:07:20.133" LastActivityDate="2015-04-21T08:07:20.133" Title="Which is the least stable trihalide of nitrogen?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;halides&gt;&lt;pnictogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26273" PostTypeId="2" ParentId="19598" CreationDate="2015-02-17T20:42:47.903" Score="1" Body="&lt;p&gt;As &lt;a href=&quot;https://chemistry.stackexchange.com/users/4934/ldc3&quot;&gt;LDC3&lt;/a&gt; indicated above &lt;a href=&quot;http://www.sbioinformatics.com/design_thesis/Glycerol/Glycerol_-2520Properties&amp;amp;uses.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; it states that &quot;Glycerol is &lt;strong&gt;completely soluble&lt;/strong&gt; in water and alcohol.&quot; So, this is correct, as also indicated &lt;a href=&quot;http://www.dow.com/optim/optim-advantage/physical-properties/solubility.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the reason this happens it is most likely because of its 3 hydroxyl groups ($-OH$) and the fact that it is a polar molecule.&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-02-17T20:42:47.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26274" PostTypeId="2" ParentId="26254" CreationDate="2015-02-17T20:44:17.407" Score="6" Body="&lt;p&gt;If you want the &lt;em&gt;practical&lt;/em&gt; answer, get an &lt;a href=&quot;https://en.wikipedia.org/wiki/Instant_hot_water_dispenser&quot; rel=&quot;noreferrer&quot;&gt;instant hot water dispenser&lt;/a&gt;.  Practical, HA!  Ok, let's do some science!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what &lt;em&gt;kind&lt;/em&gt; of coffee?  At 30 ml I'm guessing you're making espresso.  I'm going to go with &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jsfa.1304/abstract&quot; rel=&quot;noreferrer&quot;&gt;92C as the optimal temperature&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I'm going to take your full 10 seconds.  The rest of the process of making espresso takes far more time so whether it's 5 or 10 or 15 seconds doesn't make a real difference to how fast you get your morning hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other answers have pointed out the amount of energy needed to bring your 30 ml up from room temperature, but you can make life easier on yourself and do what a lot of &quot;instant boil&quot; systems do and preheat.  This allows you to use less energy (for the 10 seconds anyway) and bring the water up to temperature slower.  This gives you a wider tolerance between the water being too cold and it being boiled away.  If your water is already at 50 C that will only require about 500 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To deal with thermal conductivity, a wider surface area in contact with the heating element would serve to heat the water more quickly and more evenly.  If the water is too thin, it can overheat before it is removed from the element.  Too thick and the bottom may boil before the top reaches temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need a broad, flat surface that can output at least 500 W and that we can quickly add and remove small amounts of water from.  Sounds like you need &lt;a href=&quot;http://www.electricskilletreviews.org/reviews/cuisinart-csk-150-1500-watt-electric-skillet-review&quot; rel=&quot;noreferrer&quot;&gt;an electric skillet&lt;/a&gt;!  A very clean one.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoelectric_effect&quot; rel=&quot;noreferrer&quot;&gt;Or the precision laboratory version&lt;/a&gt;.  Buy one that can heat quickly and evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing when to remove the water is a problem.  Measuring the temperature of 30 ml of water on a hot electric skillet is problematic.  I would suggest trial and error with various skillet settings, times and initial water temperatures.  You can do a lot of experimental rounds in 10 seconds.  This would also allow you to include the cooling effect of whatever vessel you're transferring it into.&lt;/p&gt;&#xA;" OwnerUserId="14462" LastEditorUserId="14462" LastEditDate="2015-02-17T20:56:15.527" LastActivityDate="2015-02-17T20:56:15.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26275" PostTypeId="1" CreationDate="2015-02-17T20:45:57.910" Score="5" ViewCount="351" Body="&lt;p&gt;Why do defects in crystal lattices of various molecules and compound occur ?&lt;br&gt;&#xA;Is it due to the continuous contact or it arises due to impurities and human error during the formation?&lt;/p&gt;&#xA;" OwnerUserId="13409" LastActivityDate="2017-06-13T01:16:15.620" Title="Crystal Lattice Defects" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26276" PostTypeId="2" ParentId="16298" CreationDate="2015-02-17T20:52:40.630" Score="2" Body="&lt;p&gt;The problem does not need any difficult approach, not even relative atomic masses, if you trust the one who made this exercise. Since borane (whatever type it is)&lt;sup&gt;*&lt;/sup&gt; consists (by exercise's hypothesis) only of hydrogen and boron and since boron is &lt;span class=&quot;math-container&quot;&gt;$85.63\%$&lt;/span&gt; &lt;strong&gt;by mass&lt;/strong&gt;, hydrogen's mass percentage would be the remaining &lt;span class=&quot;math-container&quot;&gt;$100\% - 85.63\% = 14.37\%$&lt;/span&gt;. Therefore, we have &lt;span class=&quot;math-container&quot;&gt;$\pu{6.60 g}$&lt;/span&gt; of hydrogen in this sample.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;/sup&gt;The compound &lt;a href=&quot;https://en.wikipedia.org/wiki/Borane&quot; rel=&quot;nofollow noreferrer&quot;&gt;borane&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt;, is about &lt;span class=&quot;math-container&quot;&gt;$78.1\%$&lt;/span&gt; boron per mass. However, there is a class of compounds called &lt;a href=&quot;https://en.wikipedia.org/wiki/Boranes&quot; rel=&quot;nofollow noreferrer&quot;&gt;boranes&lt;/a&gt; which also only consist of boron and hydrogen. From this class &lt;a href=&quot;https://en.wikipedia.org/wiki/Pentaborane(9)&quot; rel=&quot;nofollow noreferrer&quot;&gt;pentaborane(9)&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{B5H9}$&lt;/span&gt;, with about &lt;span class=&quot;math-container&quot;&gt;$85.7\%$&lt;/span&gt;  boron by mass would fit the exercise.&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorUserId="4945" LastEditDate="2020-09-03T09:24:11.687" LastActivityDate="2020-09-03T09:24:11.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26277" PostTypeId="2" ParentId="24928" CreationDate="2015-02-17T21:12:04.097" Score="7" Body="&lt;p&gt;I'm not unsympathetic to MAR's answer. It is logical to consider sulfur's &quot;usual&quot; oxidation state to be the same as oxygen: -2. However, the fact that the two sulfurs are bonded to each other suggests that the oxidation state should be considered differently. Below is a systematic method that gives an answer (+5/-1) that matches the experimental data nicely summarized by DavePhD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we start with disulfide dianion. Overall, the two sulfurs carry two negative charges. Similar to peroxide (as described in the IUPAC Gold Book), each sulfur is considered oxidation state -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we add an oxygen to one of those sulfurs. Oxygen is (almost) always -2. To keep the overall charge -2, that means that the newly bonded sulfur is oxidation state +1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidizing that sulfur again gives sulfurs of +3 and -1. Finally, in thiosulfate the oxidations states are +5 and -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tUXZc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="46306" LastEditDate="2018-06-12T04:04:12.977" LastActivityDate="2018-06-12T04:04:12.977" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26278" PostTypeId="2" ParentId="26272" CreationDate="2015-02-17T21:13:29.257" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_triiodide&quot; rel=&quot;nofollow&quot;&gt;Nitrogen triiodide $\ce{NI3}$&lt;/a&gt; is an extremely sensitive explosive that explodes with the slightest touch when dry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_trifluoride&quot; rel=&quot;nofollow&quot;&gt;Nitrogen trifluoride $\ce{NF3}$&lt;/a&gt; is so stable that it is a greenhouse gas global warming concern, with lifetime in the atmosphere of hundreds of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NI3}$ is unstable due to steric strain as you are saying.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-17T21:13:29.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26279" PostTypeId="2" ParentId="14811" CreationDate="2015-02-17T21:39:03.953" Score="-3" Body="&lt;p&gt;The air inside the tent heats up and expands, the tent, while not completely airtight still acts like a balloon expanding with the heated air inside it and as the air cools down it shrinks, deflating the tent/balloon. &lt;/p&gt;&#xA;" OwnerUserId="14463" LastActivityDate="2015-02-17T21:39:03.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26280" PostTypeId="2" ParentId="26270" CreationDate="2015-02-17T21:46:41.557" Score="4" Body="&lt;p&gt;Under many circumstances, I would agree that bromide is a better nucleophile than water. It's negatively charged and, due to its size, bromine is more polarizable than oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in water, the bromide anion is very stable, reducing its nucleophilicity. I suspect that bromide  is still more nucleophilic than water (even considering solvent effect), but combined with the large excess of water, it is more likely that water opens the bromonium ion in the halohydrin formation.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-02-17T21:46:41.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26281" PostTypeId="2" ParentId="26270" CreationDate="2015-02-17T21:47:12.577" Score="6" Body="&lt;p&gt;That is not true. As you noted bromine has a negative charge (and also lower electronegativity than oxygen) and is therefore the better nucleophile (even under these aqueous conditions). Halohydrins are not even usual the result of bromination in water. Special conditions are required to prevent too much $\ce{Br-}$ from attacking the bromonium cation (e.g. certain pH, adding additional reagents as $\ce{AgOH}$ or using $\ce{HOBr}$, see the section &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/addene1.htm&quot; rel=&quot;noreferrer&quot;&gt;Addition Reactions Initiated by Electrophilic Halogen&lt;/a&gt; in W. Reusch's online Organic Chemistry book). You would usually expect the dihalide product since $\ce{Br-}$ is already close to the bromonium cation as it forms.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-17T21:53:22.110" LastActivityDate="2015-02-17T21:53:22.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26283" PostTypeId="1" AcceptedAnswerId="46920" CreationDate="2015-02-17T21:59:42.387" Score="2" ViewCount="226" Body="&lt;p&gt;When choosing the ideal indicator during a titration, there are two considerations to keep in mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pH = pKa of the equivalence point&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The indicator has a small range (&amp;lt;±1) over which it can change color&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which factor is more important when deciding over two indicators to see which is best?&lt;/p&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2016-02-25T12:21:36.270" Title="pKa vs range in choice of an indicator" Tags="&lt;inorganic-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26284" PostTypeId="1" AcceptedAnswerId="26329" CreationDate="2015-02-17T22:40:24.967" Score="2" ViewCount="74" Body="&lt;p&gt;Streaming potential is an electrochemical phenomenon that a that relates&#xA;electric currents with the relative movement of solid and liquid phases in contact with&#xA;each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when a liquid moves past a solid a potential is generated. The equations for the values of the potential and current generated can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Streaming_current&quot; rel=&quot;nofollow&quot;&gt;in this Wikipedia article&lt;/a&gt;. My question is, is the reverse true? If I applied an electrical potential difference (instead of applying a pressure difference), would a pressure difference be generated?&lt;/p&gt;&#xA;" OwnerUserId="4422" LastActivityDate="2015-04-20T01:49:58.833" Title="An electric potential is generated when a liquid moves with respect to a solid (streaming potential). Is the reverse true?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26285" PostTypeId="1" AcceptedAnswerId="26388" CreationDate="2015-02-17T23:19:40.717" Score="12" ViewCount="322" Body="&lt;p&gt;When running GAMESS for finding Hessian Eigenvalues and frequencies ( using Optimization with hessian calculation at the end ) , I have seen there are times that it comes out &quot;&lt;strong&gt;without any frequencies&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think even a non-optimized system should have some vibrational properties and I can not understand what such a &quot; &lt;strong&gt;no-frequency state&lt;/strong&gt; &quot; means physically.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="865" LastEditDate="2015-10-30T17:40:04.280" LastActivityDate="2019-02-26T06:03:37.160" Title="What is the meaning of &quot;no Hessian Eigenvalue &quot;?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26286" PostTypeId="1" CreationDate="2015-02-17T23:26:07.570" Score="3" ViewCount="2214" Body="&lt;p&gt;I am intending to demonstrate the formation and combustion of nitrocellulose with my high school students, however I am concerned about the potential products of the combustion equation. I have been able to satisfy the safety requirements for the creation which I believe follows this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only equation for the combustion I have found is this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C6H7N3O11 + 9/2 O2→12CO2 + 3N2 + 7H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am concerned about the possibility of $\ce{NO2}$ gas being created instead of or in addition to $\ce{N2}$ gas, my understanding of chemistry if fairly basic so I was hoping someone could confirm whether or not the production of $\ce{NO2}$ is likely.&lt;/p&gt;&#xA;" OwnerUserId="14466" LastEditorUserId="4945" LastEditDate="2015-02-19T07:15:53.527" LastActivityDate="2015-02-19T07:15:53.527" Title="Creation and Combustion of Nitrocellulose" Tags="&lt;combustion&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26287" PostTypeId="1" CreationDate="2015-02-17T23:46:10.317" Score="0" ViewCount="68" Body="&lt;p&gt;&lt;a href=&quot;http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1548283&amp;amp;tag=1&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; describes the measurement of streaming potential using a potential divider configuration. However it is unclear where the potential difference is created. From the diagram it appears that it is created across the length of the flow channel, so at one side there is a low potential and the other a high potential. However this doesn't seem to agree with the description of streaming potential. From other sources it appears that the potential difference is across the outside of the flow channel near the surfaces, to the inner middle of the flow channel (at the point furthest away from the surface). Which is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: there isn't a tag for electro-chemistry, so I have just put this under battery chemistry. However this is not necessarily for a battery.&lt;/p&gt;&#xA;" OwnerUserId="4422" LastActivityDate="2015-02-17T23:46:10.317" Title="How is streaming potential measured?" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26288" PostTypeId="1" AcceptedAnswerId="26291" CreationDate="2015-02-17T23:50:48.257" Score="0" ViewCount="75" Body="&lt;p&gt;So normality is 'equivalents' over liters solution right? I'm having a hard time understanding exactly what an equivalent is. At first from one thing I read i thought it was just the H released from acids in the reactions but then something else started talking about ti with compounds without H, and in double/single displacement reactions. From part of what I read it sounds like a single would have 1 eq and 2 for double, but what found to read was pretty brief so I'm not fully sure. Can someone please explain these a bit better?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2015-02-18T00:48:51.680" Title="Confused on normallity" Tags="&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26289" PostTypeId="2" ParentId="26254" CreationDate="2015-02-17T23:52:13.497" Score="5" Body="&lt;p&gt;&lt;strong&gt;According to Michael Dryden's answer, it takes 877.8W to heat 30ml of water by 70C (from 20C to 90C)in 10s.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My electric shower is about 10 times that power&lt;/strong&gt; (lets call it 8.778kW for convenience.) I don't like to shower at 90C. I am more than happy if my shower heats the water about half that temperature difference, say 35C (from 5C to 40C). So I would set it to deliver 300ml x 2 = 600ml every 10 seconds. That's 3.6L/minute, so my 5 minute shower uses 18L of water. Of course, &lt;strong&gt;if I did choose to raise the temperature by 70C in the shower, I would have to restrict it down to 300ml every 10 seconds, making it a continuous version of your experiment, scaled up by a factor of 10.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time I took a shower apart, the element was inside a copper vessel of about 300ml volume. So, if you restricted the flow  to 300ml in ten seconds, your residence time in the element vessel would be... ten seconds! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by Schwern, &lt;strong&gt;your issue isn't heating it, it's being able to control the temperature accurately.&lt;/strong&gt; One method that hasn't been mentioned is dunking a calibrated lump of hot metal into it (or pouring it into a hot metal container.) But so far the best method is Michael Dryden's idea of using a microwave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industry, batch processes are only used for slow processes like brewing beer. Fast processes (such as those used in oil refining: cracking, reforming, alkylation, etc.) are done continuously. For economic reasons these processes are carried out as fast as possible in the smallest possible equipment, and continuous process is therefore the only practical way of doing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you really need to heat water that rapidly, and the reactants consumed in your experiment are cheap, I suggest you set your experiment up as a continuous process&lt;/strong&gt;, with a flow of water through it. That way you will be able to control the temperature accurately with an electric heater. A shower will probably be too powerful - try scaling everything down to something like a 25W soldering iron element (make sure it's sealed and 12V DC for safety!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can supply your equipment with water from a reservoir at the top and control the flow / residence time with a valve at the outlet at the bottom. If you want to try the same conditions with double the residence time, simply reduce both the water flow and the heater output to 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may optionally replenish the reservoir with tap water to the point of (near) overflow if you need to mix things into it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I did a conversion course from chemistry to chemical engineering, several of the practicals were set up like this. It's definitely the way to go if you want to study fast processes whether they be mixing, reaction kinetics, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding Dave's points about critical heat flux, the important thing is to make sure you have sufficient surface area for the volume being heated. I don't see it as a huge issue at your scale though. For 1kW, you need 100cm2 of heating surface for a flux of 100,000W/m2. Many newer kettles provide this (with heating over the whole bottom surface for a neater design) but older models with curly heating elements are above 100,000W/m2, approaching critical heat flux. These electric kettles 'roar' as they come to the boil due to cavitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your reagents are delicate, though, you will need to space your heaters very narrowly, or even use steam injection. I remember one particularly nasty case of electric heaters converting an aqueous polymer solution into volatile aldehydes.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2015-02-17T23:52:13.497" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26291" PostTypeId="2" ParentId="26288" CreationDate="2015-02-18T00:48:51.680" Score="2" Body="&lt;p&gt;Normality = Equivalents of solute/liters of solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalents are determined by the number of electrons exchanged (or protons in water solutions, or yet other ions in different solvents). For example, if HCl is used to neutralize an alkali, then one mole of HCl will neutralize one mole of NaOH, &lt;em&gt;but&lt;/em&gt; it takes &lt;strong&gt;two&lt;/strong&gt; moles of HCl to neutralize one mole of Ca(OH)&lt;sub&gt;2&lt;/sub&gt; or &lt;strong&gt;three&lt;/strong&gt; moles of Al(OH)&lt;sub&gt;3&lt;/sub&gt;, because they have, respectively, two and three active OH&lt;sup&gt;-&lt;/sup&gt; groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some pretty clear explanations in more detail at &lt;a href=&quot;http://environmentalchemistry.com/yogi/chemistry/MolarityMolalityNormality.html&quot; rel=&quot;nofollow&quot;&gt;Molarity, Molality and Normality&lt;/a&gt; and at &lt;a href=&quot;http://chemistry.tutorvista.com/inorganic-chemistry/normality.html&quot; rel=&quot;nofollow&quot;&gt;Normality&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there's Mel Brooks &lt;em&gt;Abby Normal&lt;/em&gt;...&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-18T00:48:51.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26292" PostTypeId="2" ParentId="25089" CreationDate="2015-02-18T01:16:27.663" Score="7" Body="&lt;p&gt;As Ben Norris mentioned in his comment, ozonolysis is better described as a pericyclic reaction. Pericyclic reactions fall outside of the usual nucleophile-electrophile paradigm. This reaction is an example of a dipolar cycloaddition, because ozone is a 1,3-dipole. Chemists have observed that electron-rich alkenes react faster with ozone than electron-poor alkenes. This implies that the alkene is providing the HOMO to the reaction and ozone is providing the LUMO. In nucleophile-electrophile reactions, the nucleophile reacts through its HOMO, and the electrophile reacts through its LUMO. So if you want to make an analogy between a dipolar cycloaddition and a nucleophile-electrophile reaction, you could say that the alkene is the 'nucleophile' and ozone is the 'electrophile.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not always obvious that the 1,3-dipole is going to provide the LUMO in a dipolar cycloaddition, but we can speculate as to why ozone does. Oxygen is a very electronegative atom, so its atomic orbitals are lower energy than carbons. Since ozone is made up of three oxygens (with low energy atomic orbitals orbitals), its pi-system will be low energy with respect to an alkene's pi-system. Furthermore, the pi-system will be even lower energy due to placing a cation on an electronegative atom. These factors contribute to the LUMO being low energy relative to an alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a first approximation, if there's a positively charged oxygen in a compound, it can be assumed that compound will be electrophilic.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-02-18T01:16:27.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26294" PostTypeId="2" ParentId="26268" CreationDate="2015-02-18T01:25:45.857" Score="0" Body="&lt;p&gt;This compound can't exist. There is no such anion such as $\ce{FO3-}$, why not ? &#xA;Because fluorine is more electronegative then oxygen so that anion just can't be formed. Therefore no $\ce{FO3-}$ to combine with $\ce{K+}$ therefore no such compound is present.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="4945" LastEditDate="2015-03-20T05:13:31.130" LastActivityDate="2015-03-20T05:13:31.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26295" PostTypeId="1" AcceptedAnswerId="26298" CreationDate="2015-02-18T01:51:12.677" Score="3" ViewCount="88" Body="&lt;p&gt;I'm getting tired of writing out the full formulas for hydrates, adding $\ce{.$x$\,H2O}$ to all formulas. I'm wondering if there's standard shorthand for these.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="4945" LastEditDate="2015-02-18T05:01:52.867" LastActivityDate="2015-02-18T05:23:47.860" Title="Is there any abbreviation for hydrates?" Tags="&lt;water&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26296" PostTypeId="1" AcceptedAnswerId="26306" CreationDate="2015-02-18T04:01:24.300" Score="3" ViewCount="10518" Body="&lt;p&gt;I'm doing a high-school assignment, and I came across a question that I didn't quite understand.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Explain how the electron configurations for the following elements do not follow the diagonal rule:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Gold&lt;/li&gt;&#xA;  &lt;li&gt;Curium&lt;/li&gt;&#xA;  &lt;li&gt;Thorium&lt;/li&gt;&#xA;  &lt;li&gt;Molybdenum&lt;/li&gt;&#xA;  &lt;li&gt;Palladium&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand, all of these elements do follow the diagonal rule. How exactly do they not follow this rule?&lt;/p&gt;&#xA;" OwnerUserId="14473" LastEditorUserId="5026" LastEditDate="2018-03-08T09:53:15.713" LastActivityDate="2018-03-08T09:54:36.557" Title="Why does the electron configuration for some elements not follow the diagonal rule?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26297" PostTypeId="2" ParentId="26286" CreationDate="2015-02-18T04:59:46.247" Score="3" Body="&lt;p&gt;Flash paper is commonly used in classroom demonstrations, I think, and it is made of nitrocellulose.  A &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed072p1133&quot; rel=&quot;nofollow&quot;&gt;journal article from 1995&lt;/a&gt; describes rapid creation of flash paper through nitration with nitric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never done this procedure myself, but looking through it the biggest safety risk to my eyes is using and handling concentrated nitric and sulfuric acids.  &lt;strong&gt;$\ce{H2SO4}$ and $\ce{HNO3}$ are very dangerous especially when concentrated.&lt;/strong&gt;  Full PPE when handling these is an absolute must.  I'm glad to hear in your question that you are already thinking about safety requirements at this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important risk is the inherent risk of lighting anything on fire.  Nitrocellulose is an energetic material.  Since it is heavily nitrated, it needs less oxygen gas to completely burn than things like wood, paper, or even gasoline: a lot of oxidizer is already built into the molecule.  This risk is best managed by only ever igniting small quantities of nitrocellulose, and of course by storing any larger quantities of the nitrocellulose far away from sources of ignition.  Full PPE is a good idea too.  Perform the combustion in a well-ventilated area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm less worried about the particular risk incomplete combustion forming dangerous levels of $\ce{NO_{x}}$, especially if you limit the combustion to small sheets of nitrocellulose at a time.  High levels of $\ce{NO2}$ are unlikely to form via incomplete combustion; if large amounts did form, since $\ce{NO2}$ is a strong oxidizer, it would react with remaining nitrocellulose to less toxic, more reduced nitrogen oxides or more likely to nitrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small amounts of $\ce{NO2}$ do form from high-temperature combustion of anything in air, but if you use small amounts of flash paper, exposure of your students to $\ce{NO2}$  is likely to be background sources in the atmosphere will vastly exceed anything produced by your demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, don't take too much safety advice from the internet.  Talking to experienced chemists in person who have done this demonstration is probably the single best thing you can do to stay safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update:&lt;/em&gt;&lt;/strong&gt; I should explicitly say that although I have never &lt;em&gt;made&lt;/em&gt; flash paper, I have burned small pieces of it on numerous occasions, never with any problems or even noticeable amounts of smoke or fumes.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-18T06:00:47.140" LastActivityDate="2015-02-18T06:00:47.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26298" PostTypeId="2" ParentId="26295" CreationDate="2015-02-18T05:23:47.860" Score="3" Body="&lt;p&gt;There is, as far as I know, no shorthand for that, but you might be able to omit it completely in a first order approximation. The closest might be adding $\ce{(aq)}$ when working in aqueous environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $\ce{.$x$\,H2O}$ to the formula usually indicates, that it is a solid and that there are $x$ water molecules enclosed in the crystal structure. It depends on what you are looking at, which kind of reaction, what you might be able to omit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when you are describing a drying process, it is not possible:&#xA;$$\ce{\overset{blue}{CuSO4.5H2O} -&amp;gt;[\Delta] \overset{white}{CuSO4} + 5H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are working in aqueous solution it is usually not important, as there is much more water around it, hence&#xA;$$\ce{\overset{blue}{CuSO4.5H2O} + Na2S -&amp;gt; \overset{black}{CuS} v + 2Na+ + SO4^{2-} + 5H2O}$$&#xA;is essentially the same as&#xA;$$\ce{CuSO4~(aq)  + Na2S~(aq) -&amp;gt;CuS~(s) v + 2Na+ ~(aq) + SO4^{2-}~(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are describing a redox reaction, it might be the same, hence the following should be fine&#xA;$$\ce{CuSO4~(aq) + Zn~(s) -&amp;gt; Cu~(s) v + Zn^{2+}~(aq) + SO4^{2-}~(aq)}$$&#xA;Often you can omit the $\ce{(aq)}$ or $\ce{(s)}$ all together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So depending on what kind of chemistry you are talking about, some information can be implied and therefore be omitted. If you add an example of what kind of reactions/ substances you are working with, I can expand my answer to take that into account.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-02-18T05:23:47.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26300" PostTypeId="1" AcceptedAnswerId="27621" CreationDate="2015-02-18T06:43:37.350" Score="2" ViewCount="6578" Body="&lt;p&gt;Consider the following hydrogen fuel cell&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oT0pu.jpg&quot; alt=&quot;fuel cell&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/Enlist/Labs/FuelCellLab/FuelCell.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://butane.chem.uiuc.edu/pshapley/Enlist/Labs/FuelCellLab/FuelCell.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode, hydrogen is oxidised (losing electrons). My first question is this: why is the anode negative if it is attracting electrons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question refers to the fact that once the hydrogen atoms are iodised, the H+ ions move through the electrolyte towards the oxygen ions at the cathode, reacting to form water. &#xA;If the anode is negative, why would positive H+ ions move away from it?  &lt;/p&gt;&#xA;" OwnerUserId="7813" LastEditorUserId="7813" LastEditDate="2015-02-19T06:50:20.423" LastActivityDate="2019-09-24T15:27:13.083" Title="Hydrogen fuel cell - why do the H+ ions move through the electrolyte" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electricity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26302" PostTypeId="2" ParentId="26285" CreationDate="2015-02-18T07:59:02.093" Score="8" Body="&lt;p&gt;I suspect the problem is pretty simple: hessian is &lt;em&gt;not&lt;/em&gt; evaluated when geometry optimisation does &lt;em&gt;not converge&lt;/em&gt; to a stationary point. It is actually mentioned in GAMESS manual (emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HSSEND = a flag to control automatic hessian evaluation at the end of a &lt;em&gt;successful&lt;/em&gt; geometry search. (default=.FALSE.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And such behaviour is quite standard. The thing is that the conventional vibrational analysis (as descibed in most texts and implemented in most programs) is valid only when the first derivatives of the energy with respect to displacement of the atoms are zero. Or, in other words, the geometry which is used for vibrational analysis must result from a previously converged geometry optimisation run (at the same level of theory, of course).&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-02-18T07:59:02.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26303" PostTypeId="1" AcceptedAnswerId="26305" CreationDate="2015-02-18T09:00:11.903" Score="8" ViewCount="3973" Body="&lt;p&gt;Can I make a buffer using two different acids/bases? For example, adding 1 mole of acetic acid and 1 mole of ammonia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information I can ever find is that I can make a buffer using &lt;em&gt;only&lt;/em&gt; an acid (or base) and its &lt;em&gt;conjugate base&lt;/em&gt; (or conjugate acid). So, carbonic acid and carbonate, $\ce{H2CO3}$ and $\ce{HCO3-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I need a conjugate acid-base pair and not two random weak acid-base pairs to make a buffer?&lt;/p&gt;&#xA;" OwnerUserId="13295" LastEditorUserId="4945" LastEditDate="2015-02-20T03:43:55.177" LastActivityDate="2016-06-15T15:33:10.293" Title="Can I make a buffer from two different weak acids or bases?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26304" PostTypeId="1" AcceptedAnswerId="26319" CreationDate="2015-02-18T10:47:50.480" Score="10" ViewCount="4249" Body="&lt;p&gt;231st page of Clayden OC 2nd edition: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unlike the geometrical isomers of alkenes, however, those of an imine&#xA;  usually interconvert quite rapidly at room temperature. The&#xA;  geometrical isomers of oximes on the other hand are stable and can&#xA;  even be separated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why are oxime geometrical isomers much more stable than the geometrical isomers of immines?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-02-18T11:49:27.143" LastActivityDate="2015-02-18T17:34:18.483" Title="Why are oxime geometrical isomers stable?" Tags="&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26305" PostTypeId="2" ParentId="26303" CreationDate="2015-02-18T12:57:25.680" Score="8" Body="&lt;p&gt;Yes, having multiple weak acids and/or bases having different pKa values would give buffering capacity in more than one pH range, or over a greater pK range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a term &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Britton-Robinson_buffer&quot;&gt;universal buffer&lt;/a&gt;&quot; which refers to a buffer having multiple pKas spaced thoughout the pH ~2-12 range to give some buffering capacity over the enter pH range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01344a025&quot;&gt;A Neutral Buffered Standard for Hydrogen Ion Work and Accurate&#xA;Titrations Which Can be Prepared in One Minute&lt;/a&gt; J. Am. Chem. Soc., 1932, 54, pp 1911–1912, ammonium aceate has the property that if it is added to pure water it produces a buffered solution of pH 7.00, accurate to a hundreth of a pH unit.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-18T12:57:25.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26306" PostTypeId="2" ParentId="26296" CreationDate="2015-02-18T13:33:50.933" Score="2" Body="&lt;p&gt;The reason of these exceptions (from the diagonal rule) is that some elements are more stable with fewer electrons in some subshells and more electrons in others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronic configuration of molybdenum: $[\ce{Kr}] \ce{4d^{5} 5s^1}$, instead of  $[\ce{Kr}] \ce{5s^24d^{4}}$ according to the diagonal rule, because a half-full $\ce{4d}$ subshell and a half full $\ce{5s}$ subshell are more stable than $\ce{4d}$ filled with four electrons and a full $\ce{5s}$ subshell.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronic configuration of Gold: $[\ce{Xe}] \ce{4f^14 5d^{10} 6s^1}$, instead of  $[\ce{Xe}] \ce{4f^14 5d^{9} 6s^2}$ according to the diagonal rule, because a full $\ce{5d}$ and half full $\ce{6s}$ subshell is more stable than $\ce{5d}$ filled with 9 electrons and a full $\ce{6s}$ subshell.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronic configuration of Palladium: $[\ce{Kr}] \ce{4d^{10}}$, instead of  $[\ce{Kr}]\ce{ 5s^24d^{8}}$ according to the diagonal rule, because a full $\ce{4d}$ orbital is more stable than $\ce{4d}$ filled with eight electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronic configuration of Curium: $[\ce{Rn}] \ce{7s^2 5f^7 6d^{1}}$, instead of  $[\ce{Rn}] \ce{7s^2 5f^8} $ according to the diagonal rule, because a half full $\ce{5f}$ orbital is more stable than $\ce{5f}$ filled with eight electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-03-08T09:54:36.557" LastActivityDate="2018-03-08T09:54:36.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26307" PostTypeId="2" ParentId="26304" CreationDate="2015-02-18T13:47:30.670" Score="0" Body="&lt;p&gt;It's a matter of energetic barrier between the $E$ and $Z$ stereoismers. It's  sufficiently low for imines. So, it's relatively difficult to isolate $E$ or $Z$ stereoismer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of oximes, the energetic barrier of interconversion between the two stereoisomers is much higher and allows the isolation of the two stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please see &lt;a href=&quot;http://chemweb.bham.ac.uk/~coxlr/Teaching/1st_Year/Carbonyl_Group/Carbonyl_pdfs/lecture%207%20%28student%20HO%29.pdf&quot; rel=&quot;nofollow&quot;&gt;http://chemweb.bham.ac.uk/~coxlr/Teaching/1st_Year/Carbonyl_Group/Carbonyl_pdfs/lecture%207%20%28student%20HO%29.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-18T13:47:30.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26308" PostTypeId="1" AcceptedAnswerId="26314" CreationDate="2015-02-18T14:09:35.283" Score="0" ViewCount="176" Body="&lt;p&gt;Say I have acetohydrazide ($\ce{CH3CONHNH2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible, that when I add an acid, an intramolecular condensation reaction occurs, forming the cyclic compound 3-methyl-1⁠&lt;em&gt;H&lt;/em&gt;-diazirine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSqEbh.jpg&quot; alt=&quot;3-methyl-1H-diazirine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in general, the mechanism of such a reaction is nucleophilic substitution, right?&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorDisplayName="user7951" LastEditDate="2016-11-17T12:00:08.183" LastActivityDate="2016-11-17T12:00:08.183" Title="Intramolecular hydrazide condensation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26309" PostTypeId="1" AcceptedAnswerId="26311" CreationDate="2015-02-18T14:10:29.953" Score="2" ViewCount="6209" Body="&lt;p&gt;There are two main types of glucose, $\alpha$-D-(+)-glucose and $\beta$-D-(+)-glucose. &#xA;Can anyone please explain what is $\alpha$,D or (+)?&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TJei5m.jpg&quot; alt=&quot;Glucose&quot;&gt;&lt;br&gt;&#xA;So (maybe by permutation and combination), how many types of glucose are there?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="4945" LastEditDate="2015-02-19T07:09:10.907" LastActivityDate="2015-02-19T07:09:10.907" Title="The meaning of different parts in the naming of glucose" Tags="&lt;biochemistry&gt;&lt;nomenclature&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26310" PostTypeId="1" CreationDate="2015-02-18T14:14:06.507" Score="5" ViewCount="822" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CuSO4 + Zn -&amp;gt; ZnSO4 + Cu}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What is the cell potential of the voltaic cell with copper and zinc electrodes if the system is at 50 degrees and the solutions are both at 1 mol? How can it be equal to the standard potential from the nernst equation? &lt;span class=&quot;math-container&quot;&gt;$$E_\text{cell} = E^\circ – \frac{RT}{nF}\ln Q$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the solutions are at 1 mol concentrations each then &lt;span class=&quot;math-container&quot;&gt;$Q = 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ln Q = 0$&lt;/span&gt; resulting in &lt;span class=&quot;math-container&quot;&gt;$E_\text{cell} = E^\circ$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But the above basically shows that concentration affects cell voltage but temperature doesn't have an effect on cell potential – which isn't the case. How can I calculate how different temperatures affect the cell voltage?&lt;/p&gt;&#xA;" OwnerUserId="14481" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-06T07:18:56.273" Title="How can Nernst equation apply in this redox reaction?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26311" PostTypeId="2" ParentId="26309" CreationDate="2015-02-18T14:56:23.460" Score="3" Body="&lt;p&gt;Glucose can exist in the open form or cyclic form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/60rKo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the open form, there are only two types, D-glucose (illustrated in Fischer projection above) and L-glucose which is the mirror image of the above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, D-glucose rotates plane polarize light clockwise, hence the symbol &quot;+&quot;, and L-glucose rotates plane polarize light counterclockwise, hence the minus symbol.  For more information on the meaning of + versus - see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cyclic form, there is an additional chiral center, so there is alpha (where 3 OH groups are on the same side of the ring) vs beta (where 2 OH groups are on each side of the ring) as ilustrated in the drawing of the question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of alpha and beta have a D and an L form, so there are 4 types in the cyclic form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, there are only two types in the open form, but 4 types in the cyclic form.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-18T15:20:58.863" LastActivityDate="2015-02-18T15:20:58.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26312" PostTypeId="2" ParentId="26310" CreationDate="2015-02-18T15:12:54.993" Score="3" Body="&lt;p&gt;The flaw in your reasoning is that you are believing that $E^\circ$ is a constant, when really it is a function of temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://web.archive.org/web/20140211124924/http://www.nist.gov:80/data/PDFfiles/jpcrd355.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard Electrode Potentials and Temperature Coefficients in Water at 298.15K&lt;/a&gt; for a linear approximation of the temperature dependence of $E^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-03-06T07:18:56.273" LastActivityDate="2018-03-06T07:18:56.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26314" PostTypeId="2" ParentId="26308" CreationDate="2015-02-18T16:14:29.680" Score="1" Body="&lt;p&gt;Your reaction product is a double bond isomer of a &lt;strong&gt;diazirine&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halogene-substituted derivatives of the latter are typically prepared via oxidation of amidines with hypochlorites (&lt;em&gt;Graham&lt;/em&gt; oxidation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating your hydrazide in an aqueous solution of a mineral acid will only turn it to the corresponding carboxylic acid.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-19T09:37:22.063" LastActivityDate="2015-02-19T09:37:22.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26315" PostTypeId="1" AcceptedAnswerId="26316" CreationDate="2015-02-18T16:37:06.077" Score="3" ViewCount="27555" Body="&lt;p&gt;I'm working on a project and I require around 300-400 degree Celsius of temperature. As of now I'm taking the normal LPG gas that is available in our house for cooking purposes for my project, but I'm not sure whether it can reach 300-400 degree Celsius or not? I can definitely use multiple burners but how much heat/temperature does 1 burner generate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm wrong because heat is counted in calorific value, but then how do we know the temperature of that calorific value? Or is there some relation which says that x amount of LPG would give x degree Celsius or maybe x calorific value? But yet again, how can I know the temperature?&lt;/p&gt;&#xA;" OwnerUserId="14487" LastEditorUserId="23561" LastEditDate="2019-04-13T17:43:41.190" LastActivityDate="2020-02-27T08:18:57.750" Title="What is the temperature of heat generated from LPG gas?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;combustion&gt;&lt;hydrocarbons&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26316" PostTypeId="2" ParentId="26315" CreationDate="2015-02-18T16:51:39.020" Score="1" Body="&lt;p&gt;LPG is mostly propane and butane both of which have adiabatic flame temperature near 1970 degrees C when burned in air.  See &lt;a href=&quot;http://www.engineeringtoolbox.com/flame-temperatures-gases-d_422.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flame Temperatures of some Common Gases&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 300-400 degrees C is certainly reachable.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorDisplayName="user7951" LastEditDate="2016-11-16T23:35:22.323" LastActivityDate="2016-11-16T23:35:22.323" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26318" PostTypeId="2" ParentId="25165" CreationDate="2015-02-18T17:04:34.253" Score="5" Body="&lt;p&gt;The mechanism of the butadiene + ethylene Diels-Alder is not wrong as drawn. It shows the concerted movement of electrons from the diene to dienophile resulting in cyclohexene. This correctly describes the bond formation. It is unusual to see the parent Diels-Alder reaction explained through those minor resonance contributors. We often draw mechanisms through minor resonance contributors to explain aspects of reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are several problems with describing the Diels-Alder reaction this way. First, this tries to force the Diels-Alder reaction into the paradigm of &quot;polar&quot; reactions, i.e. nucleophile-electrophile. Although we learn some reactions of dienes and isolated alkenes as nucleophiles (electrophilic addition of H-X, X-X), there are essentially no reactions of electronically neutral dienes/alkenes as electrophiles. One of the key lessons of the Diels-Alder reaction is that it introduces a class of reactions that are &quot;non-polar&quot; (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15604/exceptions-to-motif-nucleophile-attacks-electrophile/&quot;&gt;related question&lt;/a&gt;), the pericyclic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is how to handle substitution. The Diels-Alder reaction can create four stereocenters, which suggests that upto 16 stereoisomers can result. However, that is not what is observed. There are only specific stereochemistries of products that result from specific stereochemistries of reactants: the Diels-Alder reaction is stereospecific. Without invoking molecular orbitals, explaining this is impossible. One can learn the rules for stereochemical outcomes, but the rationale relies on the interaction of MO's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addendum to the question gives some indication of what the instructor was thinking when introducing the Diels-Alder reaction. Although the Diels-Alder reaction is not a traditional nucleophile-electrophile reaction, it does follow frontier molecular orbital theory: one of the component's HOMO interacts with the other component's LUMO. Typically, the dienophile provides the LUMO, which is the traditional role of the electrophile. An alkene conjugated to an electron withdrawing group (such as a carbonyl) is more electrophilic. It's also a more reactive dienophile. So there is a parallel between electrophilicity and dienophile reactivity both explained by having a lower LUMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true with respect to the diene component. Typically, the diene provides the HOMO, which is the traditional role of the nucleophile. A diene substituted with an electron donating group (such as a methoxy group) is more reactive toward dienophiles and electrophiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the Diels-Alder reaction is non-polar, it can have polar character. It turns out that drawing resonance structures of the reactants is very useful for predicting the regiochemistry reactions with polar components. The major product will result from bond formation of the most nucleophilic end of the diene with the most electrophilic end of the dienophile (normal electron demand DA). Those resonance structures are simply substituted versions of what the instructor started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JoIzZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-02-18T17:10:21.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26319" PostTypeId="2" ParentId="26304" CreationDate="2015-02-18T17:12:14.547" Score="6" Body="&lt;p&gt;It is the same reasoning that is used to predict relative barrier heights in &lt;strong&gt;a&lt;/strong&gt;mines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example pictured below, the imine starts with the nitrogen lone pair in &quot;more or less&quot; an $\ce{sp^2}$ orbital.  In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i0wRA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;more or less&quot; is important.  If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit more p character than $\ce{sp^2}$ (say $\ce{sp^{2.2}}$ for example), then it is already closer to the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be reduced.  On the other hand, If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit less p character (more $\ce{s}$ character) than $\ce{sp^2}$ (say $\ce{sp^{1.8}}$ for example), then it is further from the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of oximes ($\ce{X=OH}$) or other molecules where the substituent on nitrogen is electronegative (electron withdrawing), the electron density in the $\ce{N-X}$ bond is &lt;strong&gt;shifted towards&lt;/strong&gt; the electronegative atom.  Consequently there is less electron density in the nitrogen portion of this bond.  As a result the nitrogen rehybridizes. Since $\ce{s}$ orbitals are lower in energy than $\ce{p}$ orbitals, orbitals with increased $\ce{s}$ character are better able to stabilize electrons. Since there is less electron density in the orbital nitrogen is using for this bond, $\ce{s}$ character is removed from the orbital.  This &quot;extra&quot; $\ce{s}$ character is used to stabilize the orbital containing the lone pair of electrons.  The lone pair orbital now has more $\ce{s}$ character and is further away from the $\ce{p}$ hybridized transition state - the barrier to reach this transition state increases.  If our substituent is an alkyl group (X=R, an imine), the carbon is less electronegative then nitrogen, there will be increased electron density in the nitrogen orbital used to form the $\ce{N-R}$ bond.  The nitrogen will hybridize such that there is more $\ce{s}$ character in this orbital so as to better stabilize these electrons.  As a consequence, there will less $\ce{s}$ character (more $\ce{p}$ character) in the lone pair orbital.  The lone pair orbital is therefore closer to the $\ce{p}$ hybridized transition state and less energy will be required to reach the transition state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this reasoning $\ce{NF3}$ has a higher inversion barrier then $\ce{NH3}$ and oximes have a higher barrier to inversion than imines.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-18T17:34:18.483" LastActivityDate="2015-02-18T17:34:18.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26320" PostTypeId="1" CreationDate="2015-02-18T17:13:50.473" Score="6" ViewCount="1045" Body="&lt;p&gt;In medicine, there is an (old?) experiment where a subject is asked to inhale radioactive Xenon gas, and radiation counters are placed at different positions near the lungs. Using this experiment, it has been demonstrated that the (alveolar?) volume of the lungs is highest in the bottom parts, and smallest in the top parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if this experiment is still used today, since inhaling radioactive gas has the obvious radiation related risks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are some magnetic particles that would be safe to inhale? I know that for example helium does not diffuse into blood, and it does not permanently bind into anything. However, I think it does not have any suitable magnetic properties. Let's assume the detector is very very sensitive to any magnetic fields (so, a super-conducting quantum interference device).&lt;/p&gt;&#xA;" OwnerUserId="14488" LastEditorUserId="16683" LastEditDate="2016-09-30T09:49:01.780" LastActivityDate="2016-09-30T09:49:01.780" Title="Are there any magnetic particles that are safe to inhale?" Tags="&lt;magnetism&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26321" PostTypeId="2" ParentId="26320" CreationDate="2015-02-18T17:22:37.630" Score="5" Body="&lt;p&gt;$\ce{O_2}$ is paramagnetic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SvkYi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;stable&lt;/strong&gt; Xenon isotope Xe-129 has been used in MRI studies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spl.harvard.edu/archive/HypX/currentresult0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperpolarized Xenon MR Imaging of the Brain&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A radically new magnetic resonance imaging (MRI) technique, using hyperpolarized 3He and 129Xe, is being developed to produce high-contrast images of important body tissues that have resisted conventional MRI techniques. Hyperpolarization of the noble gas nuclei is achieved by spin exchange with rubidium atoms that have been optically pumped into one electronic spin state with the use of circularly polarized light from a high-power diode laser system. Hyperpolarization can increase the detectability of 3He and 129Xe up to a hundred thousand times. This technology is showing dramatic results for diagnostic imaging of the lungs, brain, and other parts of the body. Most notably, we have achieved high-resolution dynamic images of the lungs and airways using 3He, and measured cerebral perfusion in the brain using 129Xe. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See video of lung being imaged: &lt;a href=&quot;http://www.spl.harvard.edu/archive/HypX/about.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.spl.harvard.edu/archive/HypX/about.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-18T18:09:27.427" LastActivityDate="2015-02-18T18:09:27.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26322" PostTypeId="2" ParentId="25165" CreationDate="2015-02-18T19:03:00.763" Score="12" Body="&lt;p&gt;I think that a major teaching opportunity was missed.  If MO theory had previously been introduced in the course, then there is absolutely no excuse, here is another topic where it was meant to be applied.  If MO theory had not been previously introduced, then it still remains a teaching failure, here's why...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was a great opportunity to introduce a new class of reactions.  So many of the reactions discussed in introductory organic chemistry are polar.  Here was a chance to point out that there are also many non-polar reactions. In order to explain these reactions mechanistically, the concept and power of aromaticity could have been reapplied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The class of concerted reactions (variuos cycloadditions, electrocyclic reactions and sigmatropic rearrangements, etc.) should have been (at least) mentioned.  In addition to presenting the Diels-Alder reaction as a member of this class of reactions, the intriguing example of the sterospecific, thermal ring opening of cis- and trans-3,4-dimethylcyclobutene could have also been presented in passing.  This example almost always elicits a &quot;why does it open like that&quot; response and whets the student's imagination and interest in further learning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VSyDn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this wasn't the time to delve into MO theory, the concept of aromatic (4n+2 electrons) and antiaromatic (4n electrons) transition states could have been introduced.  Teaching this approach doesn't take long at all and it makes for a nice tie-in back to benzene aromaticity. This approach neatly explains why the Diels-Alder reaction occurs so readily, but the dimerization of ethylene to produce cyclobutane does not.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-18T23:54:40.993" LastActivityDate="2015-02-18T23:54:40.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26323" PostTypeId="1" CreationDate="2015-02-18T20:01:12.673" Score="4" ViewCount="969" Body="&lt;p&gt;The magnetic moment of $\ce{[Fe(H2O)5NO]^2+}$ is 3.87 Bohr magneton, which implies that there are 3 unpaired electrons. Since iron is in $+1$ oxidation state (due to charge transfer with nitrosyl group), it's electronic configuration is $[\ce{Ar}]\ \mathrm{3d^7}$. Having 3 unpaired electron is only possible if the 4th and 5th electron enter the higher energy orbital $\mathrm{e_g}$ (crystal field theory). The $\mathrm{t_{2g}}$ or the lower energy orbital has 5 electrons. Why do the 4th and 5th electron enter $\mathrm{e_g}$ despite a strong field ligand (nitrosyl group) being present?&lt;/p&gt;&#xA;" OwnerUserId="14491" LastEditorUserId="7475" LastEditDate="2017-02-20T11:06:33.683" LastActivityDate="2017-02-20T11:25:39.127" Title="How can the electronic structure of pentaaquanitrosyliron be explained?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;hybridization&gt;&lt;crystal-field-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26324" PostTypeId="2" ParentId="26323" CreationDate="2015-02-18T20:11:38.927" Score="2" Body="&lt;p&gt;As explained in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ic010628q&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of&#xA;Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction&#xA;Revisited&lt;/a&gt; Inorg. Chem. 2002, 41, 4-10, the $\ce{Fe}$ is really 3+ oxidation state.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{Fe^3+}$ is spin 5/2 and antiferromagnetically couples to spin 1 $\ce{NO-}$ to yield a S=3/2 state.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="39539" LastEditDate="2017-02-14T20:53:49.030" LastActivityDate="2017-02-14T20:53:49.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26325" PostTypeId="1" AcceptedAnswerId="26377" CreationDate="2015-02-18T21:46:40.153" Score="12" ViewCount="8175" Body="&lt;p&gt;s-p mixing in $\ce{N2}$ causes the $1\sigma _g$ orbital to decrease in energy, the $2 \sigma _g$ orbital to increase in energy (to higher than the $1 \pi _u$) yet the $1 \sigma _u$ is left unchanged by the interaction. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lack of understanding may be as a result of the fact that I am unclear as to what exactly it is that is mixing. Clearly they are the s and p orbitals but is this within the atom or is the s orbital on one atom interacting with the p on the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some clarification as to why the $1\sigma _g$ orbital to decreases in energy, the $2 \sigma _g$ orbital to increases in energy would also be much appreciated.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rTIs7l.gif&quot; alt=&quot;MO diagram of dinitrogen&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2015-02-20T13:28:51.283" LastActivityDate="2015-09-23T11:27:21.357" Title="In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26326" PostTypeId="1" CreationDate="2015-02-18T22:41:53.493" Score="5" ViewCount="248" Body="&lt;p&gt;In what we usually consider combustion your fuel and oxidizer are always neutral, but is it possible to have a similar &quot;explosive&quot; combustion process where both the fuel and oxidizer have been singly ionized? If so what might some potential fuels be, and where might more information on the topic be found? A google search doesn't turn up much (other than injecting a small amount of plasma to catalyze an otherwise non-ionized reaction.)&lt;/p&gt;&#xA;" OwnerUserId="14493" LastActivityDate="2016-12-17T00:24:33.657" Title="Is it possible to combust a plasma?" Tags="&lt;combustion&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26327" PostTypeId="1" AcceptedAnswerId="26328" CreationDate="2015-02-18T23:05:13.593" Score="1" ViewCount="79" Body="&lt;p&gt;I was wondering something, if I have the same molar fraction for two different substances for a vapor on top of a solution, will this mean that the pressure of each individual substance will be the same ? Since they have the same quantity of molecules ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has a link with Raoult's law : $p_{solvent}=x_{solvent}p^*_{solvent}$, where $p_{solvent}$ is the partial pressure above the solvent, $x_{solvent}$ is the mole fraction of solvent in the mixture, and $p^*_{solvent}$ is the vapor pressure of pure solvent.&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="13184" LastEditDate="2015-02-19T00:09:14.803" LastActivityDate="2015-02-19T00:19:13.167" Title="Pressure over a solution" Tags="&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26328" PostTypeId="2" ParentId="26327" CreationDate="2015-02-19T00:11:44.970" Score="3" Body="&lt;p&gt;If the vapor (i.e. gas) phase above the solution can be treated as an ideal gas, then yes, if two species have the same partial pressure, then they have the same mole fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;em&gt;ideal&lt;/em&gt; gas mixture, $p_i = y_i P$, where $p_i$ is the partial pressure of species $i$ and $y_i$ is the mole fraction of $i$ in the vapor phase, and $P$ is the total pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal gas assumption will be very accurate in many cases commonly encountered in chemistry, but not always.  Binary mixtures of $\ce{CO2}$ and another compound at high pressure would be one example of vapor phases that are not ideal.  Concentrated vapor of aliphatic carboxylic acids is also not very ideal.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-19T00:19:13.167" LastActivityDate="2015-02-19T00:19:13.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26329" PostTypeId="2" ParentId="26284" CreationDate="2015-02-19T00:46:39.093" Score="1" Body="&lt;p&gt;Your assumption is correct. &lt;em&gt;Electrokinetic pumps&lt;/em&gt; use a potential difference across an osmotic membrane, i.e. through narrow channels, the inverse of flow causing a streaming potential difference. There are other aspects of the phenomena such as &lt;em&gt;electrophoresis&lt;/em&gt;, &lt;em&gt;capillary osmosis&lt;/em&gt; etc.; see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electrokinetic_phenomena&quot; rel=&quot;nofollow&quot;&gt;Electrokinetic Phenomena&lt;/a&gt; a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/andrew3/Public/Papers/2004/Min/2004_Sensors%20and%20Actuators%20B_On%20the%20ef%EF%AC%81ciency%20of%20electrokinetic_Min.pdf&quot; rel=&quot;nofollow&quot;&gt;On the Efficiency of Electrokinetic Pumping...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=gWH-Hn4nBxM&quot; rel=&quot;nofollow&quot;&gt;short video demo&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-19T00:46:39.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26330" PostTypeId="2" ParentId="26326" CreationDate="2015-02-19T02:01:30.060" Score="2" Body="&lt;p&gt;In a practical sense, the answer is No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that a substance reaching a plasma stage is ionized to a point where there is not opportunity to interact with another substance (in the case of combustion by definition Oxygen to produce water and CO2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only after the matter de-ionized follows the returns to a natural substance which can undergo combustion. But by the time this happens, the substance may undergo combustion but is no longer in plasma phase.&lt;/p&gt;&#xA;" OwnerUserId="5179" LastActivityDate="2015-02-19T02:01:30.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26331" PostTypeId="2" ParentId="26326" CreationDate="2015-02-19T03:02:42.513" Score="3" Body="&lt;p&gt;If you specifically mean a hot plasma, I think it might be possible but it's a tough request. To get combustion, you'll need to start off with form of bonding present between atoms in the plasma. This requires finding a molecule with atoms so strongly bound that the bond dissociation energy is higher than the molecule's ionization energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the lowest ionization energy known (3.51 eV or about 340 kJ/mol for &lt;a href=&quot;http://en.wikipedia.org/wiki/Ditungsten_tetra%28hpp%29&quot; rel=&quot;nofollow&quot;&gt;ditungsten tetra(hpp)&lt;/a&gt;) is comparable to relatively sturdy covalent single bonds, there's not much room for negotiation. Ionization energies easily become larger than even strong covalent double bond dissociation energies, so there's a large tendency for any substance to be stripped into its constituent atoms before getting a chance to ionize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is easier if you allow the usage of &quot;cold&quot; plasmas. Plasma generators are commonly used to &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_cleaning&quot; rel=&quot;nofollow&quot;&gt;scour surfaces completely clean of any organic residues&lt;/a&gt;. I don't see why you couldn't just insert oxygen and some organic gas into such a generator, ionize them both with the strong and rapidly varying electric field inside, and cause the organic gas to combust.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-02-19T03:08:31.643" LastActivityDate="2015-02-19T03:08:31.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26335" PostTypeId="2" ParentId="26300" CreationDate="2015-02-19T06:14:47.217" Score="0" Body="&lt;p&gt;Anode as such is positive and cathode is negative.Hydrogen gets oxidized at the anode by giving an electron,so the anode becomes rich in electrons. The flow of electrons is from anode to cathode. Hence anode which is rich in electrons, compared to cathode, is represented as negative. Cathode being negative attracts the hydrogen ions, which gets reduced when it receives electrons from anode via an external circuit.&lt;/p&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2015-02-19T06:14:47.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26336" PostTypeId="2" ParentId="26300" CreationDate="2015-02-19T06:39:43.323" Score="1" Body="&lt;p&gt;In a galvanic cell, the anode is positively charged.  The cathode is negatively charged.  A good mnemonic I learned a while ago is that &lt;em&gt;anions&lt;/em&gt; go to the &lt;em&gt;anode&lt;/em&gt;, and &lt;em&gt;cations&lt;/em&gt; go to the &lt;em&gt;cathode&lt;/em&gt;.  This is because opposite electric charges attract.  So negatively charged anions are attracted to the positively charged anode.  And positive cations are attracted to the negative cathode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron itself was originally discovered as the component particles of &lt;em&gt;cathode&lt;/em&gt; rays from cathode ray tubes.  That electrons come &lt;em&gt;out&lt;/em&gt; of the cathode is a hint about cathode charge: the buildup of electrons on the cathode eventually results in the emission of an electron beam.  If electrons are building up, the cathode must be negative.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-02-19T06:48:29.963" LastActivityDate="2015-02-19T06:48:29.963" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26337" PostTypeId="2" ParentId="4547" CreationDate="2015-02-19T07:02:34.760" Score="10" Body="&lt;p&gt;Valency is a concept that describes the power of an element to bind to other elements. There is an official definition in the &lt;a href=&quot;http://goldbook.iupac.org/V06588.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC goldbook&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;valence&lt;/strong&gt;&lt;br&gt;&#xA;  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in principle it is not possible to know the valency just from the electron configuration, but you have to also know the chemistry it performs. It is also noteworthy, that according to this definition all elements have only one value for it's valency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first period, according to the definition, hydrogen is monovalent, i.e. $\ce{H2}$ or $\ce{HCl}$. Helium on the other hand is technically divalent, since $\ce{HHeF}$ was theoretically predicted, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Helium#Compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;. But it is somewhat safe to assume helium is inert and therefore zerovalent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second period, where the octet rule holds, it is the number of unpaired electrons in the valence shell, with the exception of carbon. Unfortunately this is also the period which behaves most different from the rest of the periodic table. So the best way here is to just learn them as exceptions: $\ce{NaCl}$ (mono-), $\ce{BeCl2}$ (di-), $\ce{BCl3}$ (tri-), $\ce{CCl4}$ (tetra-), $\ce{NCl3}$ (tri-),&lt;sup&gt;(1)&lt;/sup&gt; $\ce{Cl2O}$ or $\ce{H2O}$ (di-), $\ce{FCl}$ or $\ce{F2}$, $\ce{Ne}$ is zerovalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below the second row it becomes a little bit more predictable. For most main group elements it is the number of electrons in the valence shell. This obviously coincides with the highest stable oxidation state.&lt;br&gt;&#xA;Some examples may illustrate that. For the first group all binary compositions of the alkaline element $\ce{A}$ of the composition $\ce{ACl}$ are known. This is also the maximum of chlorine atoms any of these elements can bear, hence they are all monovalent.&lt;br&gt;&#xA; The same applies to the earth-alkalines, $\ce{Ea}$. Here all binary chlorides of the form $\ce{EaCl2}$ are known, hence they are divalent.&lt;br&gt;&#xA;For the triels $\ce{M^{(III)}}$, all of the chlorides of the form $\ce{M^{(III)}Cl3}$ are known. However not all of them are very stable, i.e. $\ce{TlCl3}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Thallium_halides#Trihalides&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;). So they are trivalent.&lt;br&gt;&#xA;For the tetrels, $\ce{M^{(IV)}}$, the trend continues. The binary chlorides $\ce{M^{(IV)}Cl4}$ are known, even $\ce{PbCl4}$ is quite stable below $0~^\circ\mathrm{C}$ (&lt;a href=&quot;http://en.wikipedia.org/wiki/Lead_tetrachloride#Stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;).  For &lt;a href=&quot;http://en.wikipedia.org/wiki/Flerovium#Chemical&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flerovium&lt;/a&gt; only predictions have been made, but it's believed to be very unstable in its tetravalent state due to the inert pair effect.&lt;br&gt;&#xA;For the pnictogens, $\ce{M^{(V)}}$, all binary chlorides up to and including antimony are known. However, also &lt;a href=&quot;http://en.wikipedia.org/wiki/Bismuth_pentafluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{BiF5}$ is known&lt;/a&gt;, hence all elements can be considered pentavalent.&lt;br&gt;&#xA;For the chalcogens, $\ce{M^{(VI)}}$, all binary fluorides with coordination number six are known. Most prominent is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_hexafluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;sulfur hexafluoride&lt;/a&gt;, for which a chloride counterpart exists. The elements can therefore be considered hexavalent. The tetra chlorides are also stable for all elements of this group, in that case the elements can be considered tetravalent. However, more common are compounds where the chalcogens are divalent, i.e. in the respective hydrogen compounds, $\ce{H2M^{(VI)}}$.&lt;br&gt;&#xA;The halogens $\ce{Hal}$ are just amazing, and once you have looked into &lt;a href=&quot;http://en.wikipedia.org/wiki/Interhalogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;interhalogen compounds&lt;/a&gt; you will most likely remember a lot of them. For chlorine, the interhalogens $\ce{ClF_{\{1,3,5\}}}$ have been reported, making it pentavalent. For bromine, the same holds true: $\ce{BrF_{\{1,3,5\}}}$. Only $\ce{BrCl}$ has been observed. However this makes bromine also pentavalent. For iodine the colourless gas $\ce{IF7}$ has been observed, making it the only heptavalent element of the series. For &lt;a href=&quot;http://en.wikipedia.org/wiki/Astatine#Compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;astatine&lt;/a&gt; only a few compounds are known, and the current believe is, that it is monovalent.&lt;br&gt;&#xA;For the noble gases, $\ce{Ng}$, there are a few &lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_gas_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;compounds&lt;/a&gt; known, and their valency increases throughout the group. While argon can be considered at most divalent due to the discovery of the highly unstable $\ce{HArF}$, the divalent nature of Krypton is undoubtedly proven by $\ce{KrF2}$. Xenon is the most reactive and potent noble gas and it can be considered hexavalent according to the above definition, as of $\ce{XeF6}$ does exist. However, also $\ce{XeO4}$ is known, which would make it octavalent according to a different definition.&lt;sup&gt;(2)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition metals are a whole different story and it is difficult to find a generalised rule or observation for it, since the IUPAC definition is very strict on the matter. All elements have to be looked at in detail. It will certainly also exceed the purpose of this answer.&lt;br&gt;&#xA;Just to illustrate the problem have a look at &lt;a href=&quot;http://www.webelements.com/manganese/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;manganese&lt;/a&gt;. There are various compounds known, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese%28II%29_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{MnCl2}$&lt;/a&gt; (divalent), &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese(III)_fluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{MnF3}$&lt;/a&gt; (trivalent) and even &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese(IV)_fluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{MnF4}$&lt;/a&gt; (tetravalent). Binary hydrides have not yet been reported. However, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Manganese_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxides of various composition&lt;/a&gt;, ranging from divalent to heptavalent.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The molecule $\ce{NCl5}$, pentavalent, has been theoretically suggested, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_pentafluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; and references within.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A more modern interpretation of the valency concept allows elements to have more than only one value. It is therefore always depending on the compound in which it is looked at. A very brief summary can be found on &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_%28chemistry%29#Modern_definitions&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The most characteristic chemical property of an element is its valency. There are numerous measures of valency each with its own area of usefulness and applicability. Simple definitions refer to the number of hydrogen atoms that can combine with an element in a binary hydride or to twice the number of oxygen atoms combining with an element in its oxide(s).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://books.google.co.jp/books?id=vFu9BQAAQBAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;N. N. Greenwood and A Earnshaw: Chemistry of the Elements.&lt;/a&gt; Second edition. Butterworth-Heinemann, 1997. &lt;a href=&quot;http://dx.doi.org/10.1016/B978-0-7506-3365-9.50008-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 2, p. 27.&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="44197" LastEditDate="2017-04-21T20:36:47.337" LastActivityDate="2017-04-21T20:36:47.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26340" PostTypeId="2" ParentId="26254" CreationDate="2015-02-19T08:05:54.527" Score="2" Body="&lt;p&gt;As others have shown the energy requirement is quite reasonable, the problem is delivering it to the water in that timeframe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think a workable approach could be done with a collection of plates with very narrow spaces between them that can hold your 30ml.  Make the plates out of something that's not too good a conductor.  Feed an extreme current through the plates (very low voltage, though--you'll have a big step-down transformer) and it will get hot quickly.  You'll have to figure the energy left in the plates and add it to the amount of power you are dumping through this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Careful, a second run before the plates have cooled will certainly result in boiling water.&lt;/p&gt;&#xA;" OwnerUserId="7768" LastActivityDate="2015-02-19T08:05:54.527" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26341" PostTypeId="2" ParentId="26254" CreationDate="2015-02-19T08:25:57.760" Score="2" Body="&lt;p&gt;You could try Induction cooktops. 10 seconds for 30 ml should be enough at 700w or more, if you are using a flat and wide vessel (not all utensils are induction compatible).&lt;/p&gt;&#xA;" OwnerUserId="14503" LastActivityDate="2015-02-19T08:25:57.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26342" PostTypeId="1" CreationDate="2015-02-19T09:31:36.130" Score="2" ViewCount="614" Body="&lt;p&gt;Why do lithium enolates add to an electrophile in direct addition manner, while the sodium and potassium enolates tend to add in conjugated additions? The sodium and potassium enolates are well dissociated, making oxygen a hard base (which is why these enolates participate easily in O-acylations), but still, the alpha carbon atom is made softer, to react well in conjugated additions. This seems to me like a contradiction.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-02-20T14:53:07.203" LastActivityDate="2019-01-09T10:00:50.130" Title="Alkali metal enolates" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26344" PostTypeId="2" ParentId="26342" CreationDate="2015-02-19T11:57:17.083" Score="0" Body="&lt;p&gt;Because the lithium enolates are more strong than sodium or potassium enolates. The lithium enolates addition is under kinetic control, instead the other addition is under thermodinamic control.&lt;/p&gt;&#xA;" OwnerUserId="13277" LastActivityDate="2015-02-19T11:57:17.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26345" PostTypeId="1" CreationDate="2015-02-19T12:32:31.143" Score="2" ViewCount="81" Body="&lt;p&gt;I want to know if someone has experimental results and calculations of the kinetic and thermodynamic parameters of Base Catalyzed Transesterification of pig fat, if not which is the method used to calculate them?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!! :)&lt;/p&gt;&#xA;" OwnerUserId="14490" LastActivityDate="2015-03-22T01:42:31.940" Title="kinetic and thermodynamic parameters of biodiesel production from lard" Tags="&lt;thermodynamics&gt;&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26346" PostTypeId="1" AcceptedAnswerId="26347" CreationDate="2015-02-19T12:46:16.070" Score="5" ViewCount="737" Body="&lt;p&gt;How can the &lt;strong&gt;major&lt;/strong&gt; product of the addition reaction of 2-bromo-3-chloro-2-butene with hydroiodic acid be predicted, since Markovnikov's rule fails to distinguish it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C(Cl)=C(Br)-CH3~+~HI~-&amp;gt;~?}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorUserId="5026" LastEditDate="2018-04-21T05:55:47.743" LastActivityDate="2018-04-21T05:55:47.743" Title="Addition reaction of hydroiodic acid to 2-bromo-3-chloro-2-butene" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;c-c-addition&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26347" PostTypeId="2" ParentId="26346" CreationDate="2015-02-19T13:46:23.230" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;TL;DR - Sometimes you need to do the experiment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At C2 (Bromo)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -&amp;gt; H3C-CH(Cl)-C+(Br)-CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two issues to consider are resonance and induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resonance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R2C+-Br &amp;lt;-&amp;gt; R2C=Br+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At C3 (chloro)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -&amp;gt; H3C-C+(Cl)-CH(Br)-CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We once again examine resonance and induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resonance&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R2C+-Cl &amp;lt;-&amp;gt; R2C=Cl+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Induction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                  Chlorine              Bromine&#xA;Resonance         Stronger donor        Weaker donor&#xA;Induction         Stronger withdrawer   Weaker withdrawer&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now... what else might happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a &lt;a href=&quot;http://en.wikipedia.org/wiki/Halonium_ion&quot;&gt;halonium ion&lt;/a&gt;. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in effect, you can get three products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-CHCl-CIBr-CH3}$$&#xA;$$\ce{CH3-CICl-CHBr-CH3}$$&#xA;$$\ce{CH3-CClBr-CHI-CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would then want to actually do the experiment to determine which of these is really the major product. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-02-19T13:46:23.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26348" PostTypeId="2" ParentId="19079" CreationDate="2015-02-19T13:50:55.497" Score="5" Body="&lt;p&gt;&lt;strong&gt;Enthalpy $H$ is defined so that $\Delta H = q_P$, the heat at constant pressure&lt;/strong&gt;. This is convenient for constant-pressure calorimetry---it makes it possible to track energy changes without considering volume changes in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}\Delta U &amp;amp;= q + w &amp;amp; \text{(first law)}\\&#xA;\Delta U_P &amp;amp;= q_P + w_P &amp;amp; \text{(constant pressure)}\\&#xA;\Delta U_P &amp;amp;= \Delta H + w_P &amp;amp; \text{(definition of enthalpy)}\\&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...so the difference between $\Delta H$ and $\Delta U$ can be interpreted as minus the work for a &lt;em&gt;constant pressure process&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't answer your question for a general process. If the process is &lt;em&gt;not&lt;/em&gt; constant pressure, we can still apply the definition of enthalpy, but the difference isn't minus the expansion work anymore. But what is it, then? How can we use something we've defined for a constant pressure process for all processes in general? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'd like a new thermodynamic function $H$ that includes all of the information in $U$. We have $$dU = TdS - PdV$$ so the &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_potential#Natural_variables&quot; rel=&quot;noreferrer&quot;&gt;natural variables&lt;/a&gt; of $U$ are $S$ and $V$. We want this new function $H$ to have $P$ as a natural variable in place of $V$, so it will be convenient for constant-pressure calorimetry. Consider the following plot of $U$ vs. $V$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bUwdc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the tangent line at some value of $V$ is&#xA;$$ U = \left(\frac{\partial U}{\partial V}\right)_S V + intercept$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define the new function as the intercept of the tangent line shown in the graph, it will have the properties we need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H \equiv U - \left(\frac{\partial U}{\partial V}\right)_S V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the derivative $\left(\dfrac{\partial U}{\partial V}\right)_S = -P$ from the equation for $dU$ above. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H \equiv U - (-P) V = U + PV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH = dU + d(PV) = TdS - PdV + PdV + VdP = TdS + VdP$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding $PV$ to $U$, we've transformed the information in $U$ (with natural variables $S$ and $V$) to a new function $H$ (with natural variables $S$ and $P$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This trick is called a &lt;a href=&quot;http://www.iupac.org/publications/pac/2001/pdf/7308x1349.pdf&quot; rel=&quot;noreferrer&quot;&gt;Legendre transformation&lt;/a&gt;, and it can be used to understand the definitions of other thermodynamic functions like $G$ and $A$ as well.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-19T14:52:45.610" LastActivityDate="2015-02-19T14:52:45.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26349" PostTypeId="2" ParentId="26342" CreationDate="2015-02-19T14:03:29.207" Score="1" Body="&lt;p&gt;Lithium has a very high affinity for oxygen (strong bonding interaction). For the larger (i.e. ion radius) more polarized alkali metals down the periodic table, most notably sodium and potassium, the bonding is more dissociated and hence the reaction of the enolate is more reversible. Usually the conjugated addition product is the &lt;em&gt;thermodynamic&lt;/em&gt; product, while the aldol product is the &lt;em&gt;kinetic&lt;/em&gt; product. More reversibility will increase reaction time and possibly yield. Therefore sodium and potassium bases are preferred over lithium bases when the conjugated product is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above was more or less paraphrased from Clayden et Al. Organic Chemistry textbook (p. 752, chapter 29). &lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-19T14:03:29.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26350" PostTypeId="1" AcceptedAnswerId="26382" CreationDate="2015-02-19T14:17:07.993" Score="1" ViewCount="35" Body="&lt;p&gt;I want to know if it is possible to determine a chemical compound in a non-homogeneous sample. I am asking this because we are aiming to create a device that will detect histamine level in fish without homogenizing the samples. In all the journals we've read (which are a lot), fish samples are homogenized (blended + chemically-treated) before histamine is determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a computer engineering student with basic knowledge on biochemistry. I'm researching on this because this is our thesis topic (which our adviser has &lt;s&gt;forced&lt;/s&gt; recommended to us).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help/recommendation regarding this is appreciated. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="14511" LastActivityDate="2015-02-20T00:42:09.247" Title="Determination of a chemical compound in a non-homogeneous sample" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26351" PostTypeId="1" AcceptedAnswerId="26358" CreationDate="2015-02-19T14:21:42.983" Score="2" ViewCount="1205" Body="&lt;p&gt;A complex salt is a salt where the metal and ligands do not dissociate in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eg : $\space\ce{K_4[Fe(CN)_6]}$ when put into $\space\ce{H_2O}$ dissociates as : &lt;em&gt;$\space\ce{4K^+} + \ce{[Fe(CN)_6]^{4-}}$&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{[Fe(CN)_6]^{4-}}$ is a &lt;em&gt;complex ion&lt;/em&gt; which &lt;strong&gt;cannot&lt;/strong&gt; be further dissociated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;is there any way to dissociate them&lt;/strong&gt;, either by heating or by adding something? Or will they dissociate in some other solvent? Or is there no way to dissociate them?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="5026" LastEditDate="2018-04-18T06:29:04.943" LastActivityDate="2018-04-18T07:08:08.643" Title="Are there ways to fully dissociate a complex salt?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26352" PostTypeId="1" AcceptedAnswerId="26385" CreationDate="2015-02-19T14:24:40.610" Score="12" ViewCount="276" Body="&lt;p&gt;How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x4cMo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4cMo.png&quot; alt=&quot;Overall reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proposed route involves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hydrolysis of the nitrile to form a carboxylic acid, and subsequent esterification&lt;/li&gt;&#xA;&lt;li&gt;then, a Darzens reaction with benzaldehyde&lt;/li&gt;&#xA;&lt;li&gt;reduction of the ester function with DIBALH and oxidation to an aldehyde with $\ce{H2CrO4}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there a better method?&lt;/p&gt;&#xA;" OwnerUserId="13277" LastEditorUserId="16683" LastEditDate="2017-09-03T08:28:18.063" LastActivityDate="2017-09-03T08:33:56.570" Title="How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26353" PostTypeId="1" CreationDate="2015-02-19T15:11:46.677" Score="8" ViewCount="387" Body="&lt;p&gt;Why are there no chlorate esters? Or are they just undocumented or too unstable? I've seen perchlorate esters, but google &quot;chlorate esters&quot; and only perchlorate esters come up.&lt;/p&gt;&#xA;" OwnerUserId="1948" LastEditorDisplayName="user7951" LastEditDate="2015-09-25T22:29:48.333" LastActivityDate="2015-09-26T15:17:09.830" Title="No chlorate esters? Why?" Tags="&lt;organic-chemistry&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26355" PostTypeId="1" AcceptedAnswerId="26384" CreationDate="2015-02-19T15:34:36.663" Score="4" ViewCount="6739" Body="&lt;p&gt;My teacher said that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The decrease in Gibbs free energy is the maximum non-mechanical work that can be obtained when the process occurs reversibly at constant temperature and pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But for a reversible process always &lt;span class=&quot;math-container&quot;&gt;$ΔG = 0$&lt;/span&gt;, so doesn't it contradict it? I mean he's just telling that maximum useful work in any process is zero.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="41328" LastEditDate="2019-04-13T17:08:59.787" LastActivityDate="2019-04-13T17:08:59.787" Title="Gibbs Free Energy" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26356" PostTypeId="2" ParentId="26254" CreationDate="2015-02-19T15:36:41.297" Score="2" Body="&lt;h1&gt;How to get instant coffee&lt;/h1&gt;&#xA;&lt;p&gt;There are already some good answers on how you could reasonably get a cup of coffee-temperature water within 10 seconds. However, what if you wanted to get it even quicker than that?&lt;/p&gt;&#xA;&lt;h2&gt;Realistic solution: Use a boiler&lt;/h2&gt;&#xA;&lt;p&gt;If you already have hot water, you won't need to wait for 10 seconds before your cup is ready to go. This may not be the most economical solution to get your morning dose, but if you have a coffee shop and really need to churn out cups during peak times it may be interesting to look into this.&lt;/p&gt;&#xA;&lt;h2&gt;Interesting solution: Use more power&lt;/h2&gt;&#xA;&lt;p&gt;If you don't like the idea of keeping your water hot all the time, you need something to heat it quickly. Of course you could use a huge heating element, but I suppose that it may be even quicker to use something like an explosion. You just need to make sure that no deadly chemicals end up in your cup.&lt;/p&gt;&#xA;&lt;p&gt;Other answers already indicated that the main problem with using more power, is that it is hard to control the temperature. However, this should be solved with the following steps:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;(Over) Heat some water&lt;/li&gt;&#xA;&lt;li&gt;Measure the temperature (only needed if step 1 cannot produce a fixed temperature reliably)&lt;/li&gt;&#xA;&lt;li&gt;Mix the right amount of this water/steam with cold water to get a cup ful of water at the desired temperature&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Of course it will never be truely instant, but I imagine that you could get it done in a fraction of a second.&lt;/p&gt;&#xA;" OwnerUserId="14513" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-02-19T15:36:41.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26357" PostTypeId="1" AcceptedAnswerId="26378" CreationDate="2015-02-19T16:02:34.957" Score="18" ViewCount="4012" Body="&lt;p&gt;What other, rather easy to obtain, salts are more hydrophilic than $\ce{NaCl}$? Is there a hydrophilic scale for substances like the Mohs scale of mineral hardness?&lt;/p&gt;&#xA;" OwnerUserId="14515" LastEditorUserId="9961" LastEditDate="2015-10-23T17:22:04.740" LastActivityDate="2017-01-14T08:37:50.300" Title="Salts that are more hydrophilic than NaCl" Tags="&lt;physical-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26358" PostTypeId="2" ParentId="26351" CreationDate="2015-02-19T16:02:50.987" Score="4" Body="&lt;p&gt;Let's consider the formation constant of this ion complex at $25^\circ\mathrm{C}$:&#xA;$$\ce{Fe^{2+} + 6CN- &amp;lt;=&amp;gt; [Fe(CN)6]^4-}\qquad K_\mathrm{f}=10^{+24}$$&#xA;This means that, the equilibrium is highly displaced towards the formation of the  complex ion. The ion complex is very stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we want to displace this equilibrium towards the dissociation of the complex ion, we have to add an adequate quantities of  a metallic ion or a ligand that form with $\ce{CN^{-}}$ or $\ce{Fe^{2+}}$, respectively more stable complexes.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="19383" LastEditDate="2018-04-18T07:08:08.643" LastActivityDate="2018-04-18T07:08:08.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26359" PostTypeId="2" ParentId="26357" CreationDate="2015-02-19T16:26:58.597" Score="16" Body="&lt;p&gt;The scale for degree of being hydrophilic is &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot;&gt;contact angle&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contact angle ranges from 0 to 180 degrees, 0 being the most hydrophilic and 180 being the most hydrophobic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;180 degrees means that a drop of water on the surface forms a sphere, which only contacts the solid surface at one tangential point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 degrees means that the water spreads in a thin, flat layer on the solid surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NaCl and NaF have contact angles of 0 degrees according to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0021979700973508&quot;&gt;The Significance of Interfacial Water Structure in Soluble Salt Flotation Systems&lt;/a&gt; Journal of Colloid Interface Science vol. 235, page 150, so no salts are more hydrophlic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;KCl has a contact angle of 8 degrees, and KI 25 degrees.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-20T00:32:41.493" LastActivityDate="2015-02-20T00:32:41.493" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26360" PostTypeId="2" ParentId="26355" CreationDate="2015-02-19T16:58:57.783" Score="1" Body="&lt;p&gt;The term &quot;reversible process&quot; does not mean that $\Delta G = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversible process is an imaginary or idealized process which occurs in infinitesimally small steps, such that the system very close to equilibrium throughout the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the infinite number of infinitesimally small steps would add up to an finite change.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-19T16:58:57.783" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26361" PostTypeId="1" CreationDate="2015-02-19T17:07:12.727" Score="2" ViewCount="940" Body="&lt;p&gt;I'm working on a project in which I'll be dealing with pressure vessels. As of now I've drawn/conceptualized everything on theory basis. The diagrams are made on the basis of theory, but then I came through a situation where I'll require to heat water under high pressure at around 300-350 Celsius. So then I decided to use pressure vessels and to use LPG gas for heating it (LPG has flame temperature of around 1100 Celsius, but please note that I want to heat the water kept in vessel at just 300-350 Celsius)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm stuck in thinking that how do you heat pressure vessels? Are there any special port/connections provided to heat them/the content inside them? Or are any special type of heaters used to heat them? &lt;em&gt;Would be great if you can tell everything you know about heating it&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optional to read(this could be wrong): Are we supposed to directly heat them on flame? Just like cooking your food on the gas-stove running using LPG? If this is so, then won't it take too much time because the walls of vessels would be very thick?&lt;/p&gt;&#xA;" OwnerUserId="14487" LastEditorUserId="14487" LastEditDate="2015-02-19T17:23:06.533" LastActivityDate="2015-02-19T17:23:06.533" Title="How are pressure vessels heated?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;vapor-pressure&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26362" PostTypeId="2" ParentId="25188" CreationDate="2015-02-19T17:13:42.903" Score="6" Body="&lt;p&gt;While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects.  The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there.  The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2015-02-19T17:25:42.533" LastActivityDate="2015-02-19T17:25:42.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26363" PostTypeId="2" ParentId="25165" CreationDate="2015-02-19T17:19:32.997" Score="4" Body="&lt;p&gt;This is not a full answer either, but it needs to stand alone from the previous discussions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't draw resonance structures without MO theory.  This discussion is formally impossible.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-02-19T17:19:32.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26366" PostTypeId="1" CreationDate="2015-02-19T18:40:05.520" Score="4" ViewCount="4754" Body="&lt;p&gt;For example, citric acid has three acidic protons, all of which are carboxylic acids. Despite being part of the same functional groups, they all have very different pKas. Why (and how?) is this?&lt;/p&gt;&#xA;" OwnerUserId="13183" LastEditorUserId="4325" LastEditDate="2015-02-19T18:41:35.887" LastActivityDate="2015-02-19T19:05:54.283" Title="How can pKas differ across similar protons of the same atom?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26367" PostTypeId="2" ParentId="26366" CreationDate="2015-02-19T18:58:33.543" Score="2" Body="&lt;p&gt;Let's begin this discussion with the simpler oxalic acid.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     O&#xA;     #&#xA;HO-C-C-OH&#xA;   #&#xA;   O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There are two carboxylic acid functional groups, and there are two different pKa's.  This is because double deprotonation does not happen simulateneously.  One gets deprotonated first, then the other.  After the first deprotonation, the residual molecule has a negative charge.  The result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     O&#xA;     #&#xA;HO-C-C-O~&#xA;   #&#xA;   O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is much less acidic because the presences of the carboxylate functional group changes the acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's continue the discussion with the (still simpler) alpha-chlorosuccinic acid:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Cl H O&#xA;     | | #&#xA;HO-C-C-C-C-OH&#xA;   # | |&#xA;   O H H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The acid closer to the chlorine is more acidic because the chlorine is strongly electron withdrawing, and 'pulls' the electrons out of the carboxylate function, which then 'pulls' electrons away from the hydrogen, making it more acidic.  This effect decreases with distance, so the chlorine pulls on the acid function farther away less strongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more generalized way, remote groups will modify acidity because of electron withdrawing or donating effects, but the farther away from the acid function, the less effect there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For citric acid, there are actually two different flavors of acid function, so one of the kinds will be intrinsically more acidic before deprotonation because of the remove effect.  However, because of the sequential deprotonation discussed above, all three will have different pKa's&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2015-02-19T19:05:54.283" LastActivityDate="2015-02-19T19:05:54.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26368" PostTypeId="2" ParentId="26366" CreationDate="2015-02-19T18:58:43.007" Score="6" Body="&lt;p&gt;Here is a picture of citric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UAb4u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, there are 3 carboxylic acid groups in citric acid, so there are 3 acidic protons. A chemist looking at this unionized molecule would say that the carboxylic acid group drawn on the right side of the molecule is equivalent to (the same as) the carboxylic acid group drawn on the left.  Therefore, these 2 equivalent carboxylic acid groups should have the same pKa.  The carboxylic acid group drawn in the middle of the diagram is different from the other two (it is not the same as the other two) and should therefore have a different $\ce{pKa}$.  In chemistry, groups are either the same or different.  If they are the same they must be indistinguishable in all aspects, including physical properties like $\ce{pKa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, we might expect when we treat citric acid with more and more base to see only 2 different $\ce{pKa}$'s, but that is not the case.  In actuality when we treat citric acid with base one of the two different (right and left or middle) carboxylic acid groups will ionize.  Once that happens the ionized citric acid molecule contains a negative charge.  It will be harder to ionize the molecule again and remove the second proton and place a &lt;strong&gt;second&lt;/strong&gt; negative charge on the molecule (bringing like charges together is destabilizing and costs energy).  The same argument tells us that removing the third and final proton will be even more difficult because we wind up with 3 negative charges on the molecule.  As a consequence of this increasing difficulty to remove protons, we observe 3 different $\ce{pKa}$'s.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-19T18:58:43.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26370" PostTypeId="1" CreationDate="2015-02-19T19:54:56.170" Score="3" ViewCount="644" Body="&lt;p&gt;In calcination why is the ore heated in the absence of air in a reverberatory furnace,&#xA;does the presence of oxygen affect any thing?&lt;/p&gt;&#xA;" OwnerUserId="14521" LastEditorUserId="-1" LastEditDate="2015-02-20T01:33:20.663" LastActivityDate="2015-07-26T13:18:11.137" Title="Calcination in metallurgical process?" Tags="&lt;metallurgy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26371" PostTypeId="1" CreationDate="2015-02-19T20:14:24.503" Score="5" ViewCount="888" Body="&lt;p&gt;I want to do a demonstration for students where they take a capped syringe and change the volume above a solution of seltzer water.  The amount of carbon dioxide dissolved will vary, hopefully changing pH enough to cause an indicator to change color, the more dramatic the color change, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried universal, bromcresol green, and methyl red with little success.  I can add substances to the seltzer water or change the temperature of the water.  I'd like a SAFE system I can create that will change color upon moving the plunger.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2016-04-19T23:41:29.580" Title="Which pH indicator will give a dramatic color change with small changes in the amount of carbon dioxide dissolved in water?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26372" PostTypeId="2" ParentId="26370" CreationDate="2015-02-19T20:23:38.203" Score="1" Body="&lt;p&gt;Most of the ore will be oxidized if thermally treated under air. That's why it's usual to heat it under controled atmosphere or even in vacuum.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-02-19T20:23:38.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26373" PostTypeId="1" CreationDate="2015-02-19T16:31:24.250" Score="6" ViewCount="318" Body="&lt;p&gt;I have been trying to calculate chlorite mineral formulas from microprobe data. However, all of the mineral formula spreadsheets on the internet, including those from the Open University assume 28 oxygens (&lt;a href=&quot;http://www.open.ac.uk/earth-research/tindle/AGT/AGT_Home_2010/Microprobe-2.html&quot; rel=&quot;nofollow&quot;&gt;http://www.open.ac.uk/earth-research/tindle/AGT/AGT_Home_2010/Microprobe-2.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all the formula I have ever seen for chlorite only have 18 oxygens? (see DHZ or &lt;a href=&quot;http://www.mindat.org/min-1016.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mindat.org/min-1016.html&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By then calculating my formula assuming 28 oxygens, it means I have easily more cations than the mineral structure would have sites for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for any help&lt;/p&gt;&#xA;" OwnerDisplayName="Geodude" LastEditorUserId="10185" LastEditDate="2015-02-19T20:45:49.797" LastActivityDate="2015-02-19T21:38:07.323" Title="Why do people assume 28 oxygens in chlorite mineral formula?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26374" PostTypeId="2" ParentId="26371" CreationDate="2015-02-19T20:49:13.387" Score="2" Body="&lt;p&gt;Try &lt;em&gt;neutral red&lt;/em&gt;. It's transition region is from 6.8 to 8.0 &gt; red to yellow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also check this &lt;a href=&quot;http://en.wikipedia.org/wiki/PH_indicator&quot; rel=&quot;nofollow&quot;&gt;wikipage&lt;/a&gt; for other options.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-02-19T20:49:13.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26376" PostTypeId="2" ParentId="26373" CreationDate="2015-02-19T21:19:51.967" Score="3" Body="&lt;p&gt;It doesn't mean that there are 28 oxygen atoms.  It means a charge equivalent of 28 oxide anions, considering that each oxide anion has a -2 charge and each hydroxide anion has a -1 charge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the chlorite having an formula including $\ce{O20(OH)16}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example see $\ce{Mg10Al2(Si6Al2)O20(OH)16}$ in &lt;a href=&quot;http://adsabs.harvard.edu/abs/2010EGUGA..12.7294F&quot; rel=&quot;nofollow&quot;&gt;The effect of Cr on chlorite stability and the appearance of a new Al-rich silicate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In otherwords, as explained in &lt;a href=&quot;http://www.minersoc.org/pages/Archive-CM/Volume_26/26-2-149.pdf&quot; rel=&quot;nofollow&quot;&gt;OCTAHEDRAL OCCUPANCY AND THE CHEMICAL&#xA;COMPOSITION OF DIAGENETIC (LOW-TEMPERATURE) CHLORITES&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All chlorite analyses were re-calculated as chlorite structural formulae on the basis of 28 oxygen equivalents (56 negative charges) i.e. assuming an ideal anion framework of $\ce{O20(OH)16}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-19T21:38:07.323" LastActivityDate="2015-02-19T21:38:07.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26377" PostTypeId="2" ParentId="26325" CreationDate="2015-02-19T21:36:00.290" Score="6" Body="&lt;p&gt;Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ &lt;strong&gt;molecular&lt;/strong&gt; orbitals are formed from two sets of nitrogen ($\ce{N1}$) &lt;strong&gt;atomic&lt;/strong&gt; orbitals.  Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance.  With that understanding, let's step through your diagram and questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be useful to take a look at the $\ce{N2}$ MO diagram &lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/GenChem2/A5/2.html&quot;&gt;here&lt;/a&gt;.  It's just like your diagram. but they break down the steps and it might be a bit less confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals.  If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds.  In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized.  We form this bond by&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mixing the s and  a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.  &lt;/li&gt;&#xA;&lt;li&gt;Same thing happens on the other nitrogen atom&lt;/li&gt;&#xA;&lt;li&gt;As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).&lt;/li&gt;&#xA;&lt;li&gt;The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding &lt;strong&gt;molecular&lt;/strong&gt; orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).&lt;/li&gt;&#xA;&lt;li&gt;The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now for pi bonds,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.&lt;/li&gt;&#xA;&lt;li&gt;These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom.  This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With that background, let's now look at your questions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s-p mixing in N 2   causes the 1σ g   orbital to decrease in energy,&#xA;  the 2σ g   orbital to increase in energy (to higher than the 1π u  )&#xA;  yet the 1σ u   is left unchanged by the interaction. Why is this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered).  The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Clearly they are the s and p orbitals but is this within the atom or&#xA;  is the s orbital on one atom interacting with the p on the other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully I answered this above.  I tried to distinguish between atomic and molecular orbitals.  If I haven't been clear, post back.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some clarification as to why the 1σ g   orbital to decreases in&#xA;  energy, the 2σ g   orbital to increases in energy would also be much&#xA;  appreciated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully what I wrote above clarified this as well, let me know if it didn't.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-19T21:36:00.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26378" PostTypeId="2" ParentId="26357" CreationDate="2015-02-19T21:39:15.427" Score="17" Body="&lt;p&gt;As per DavePhD's answer, the contact angle of NaCl with water is zero, so we need another measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the word you are looking for is &lt;strong&gt;hygroscopic.&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Hygroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Hygroscopy&lt;/a&gt; Substances which are so hygroscopic that they dissolve in the water they absorb from the atmosphere are said to be &lt;strong&gt;deliquescent&lt;/strong&gt;, although the property of deliquesence in itself does not directly indicate the strength of a substance as a drying agent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength would best be measured as the equilibrium vapour pressure at a certain temperature, or alternatively, the temperature required to dehydrate: i.e. the temperature at which the vapour pressure of water vapour above the hydrated form of the substance becomes equal to atmospheric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheap substances which are hygroscopic find applications as &lt;strong&gt;desiccants&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is a list of some common ones&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_desiccants&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/List_of_desiccants&lt;/a&gt;. It includes the workhorse laboratory desiccants: &lt;strong&gt;silica gel&lt;/strong&gt; for use in desiccators where products are dried under vacuum; &lt;strong&gt;molecular sieves&lt;/strong&gt;, for drying solvents; &lt;strong&gt;CaCl2&lt;/strong&gt; for drying gases; and &lt;strong&gt;MgSO4&lt;/strong&gt; and &lt;strong&gt;CaSO4&lt;/strong&gt; in their anhydrous form for a quick drying of organic solutions. I would assume (though I'm not sure) that all of these would be stronger than NaCl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unfortunately the really strong ones are missing&lt;/strong&gt;, so here's a list of mine. The weaker ones above tend to act by physical absorption, or by changing from anhydrous crystals to hydrated crystals. Stronger ones tend to react chemically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Concentrated sulfuric acid&lt;/strong&gt; Strong enough to dehydrate sugar to &quot;charcoal.&quot; One of the reasons for the affinity is the energy released when &lt;code&gt;H2SO4&lt;/code&gt; proponates &lt;code&gt;H2O&lt;/code&gt;, forming &lt;code&gt;HSO4-&lt;/code&gt; + &lt;code&gt;H3O+&lt;/code&gt;. More of a curiousity than a desiccant. Not widely used as a desiccant because of its acidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Phosphorous pentoxide&lt;/strong&gt; Reacts with water to form various different polymeric phosphoric acids, ultimately ending up with 2&lt;code&gt;P2O5&lt;/code&gt; + 6&lt;code&gt;H2O&lt;/code&gt;= 4&lt;code&gt;H3PO4&lt;/code&gt; though strength is gradually reduced as water is absorbed. Commonly used in the lab when a strong desiccant is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calcium Oxide&lt;/strong&gt; The reaction &lt;code&gt;CaO&lt;/code&gt; + &lt;code&gt;H2O&lt;/code&gt; = &lt;code&gt;Ca(OH)2&lt;/code&gt; has been proposed for large scale energy storage (the storage occurs in the reverse direction), but so far has been used only for smaller applications. As the calcium species are basic, exposure to air causes absorption of CO2, which then irreversibly degrades the material to bicarbonate and carbonate (unless heated to a much higher temperature.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sodium&lt;/strong&gt; Metallic sodium, extruded into wire using a press to increase its surface area, is the best way to ensure absolutely dry solvents.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="402" LastEditDate="2017-01-14T08:37:50.300" LastActivityDate="2017-01-14T08:37:50.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26379" PostTypeId="2" ParentId="26371" CreationDate="2015-02-19T22:38:27.453" Score="2" Body="&lt;p&gt;I once worked in a QC lab where we tested the acidity of organic esters to see if they had any free acid in them. The method was based on titration with strong base in methanol solution, using phenolphtalein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this you would need an aqueous solution of sodium acetate or bicarbonate to get the phenolphthalein into its pink range, but I can tell you from experience that if you have a titration right at its endpoint, atmospheric CO2 and a bit of swirling is all it takes to turn the phenolphtalein back to colourless. Whether you can get the CO2 back out again by sucking with a syringe is another matter. Heating would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do it this way, you're going to need to change the experiment somewhat, but you should have something that will work. Here's a simple experiment (not exactly what you want but it uses the reagents I mentioned.) &lt;a href=&quot;http://www.sciencegeek.net/Chemistry/chempdfs/QualitativeAnalysis.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencegeek.net/Chemistry/chempdfs/QualitativeAnalysis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great thing about phenolpthalein is that when it's in its colourless range it is &lt;em&gt;absolutely&lt;/em&gt; colourless (due to the lack of conjugation in the molecule) so you can put lots in for dramatic effect. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Phenolphthalein&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Phenolphthalein&lt;/a&gt; Phenolphtalein is used to test the curing of concrete by CO2, and the reaction and indicator are used in various toys.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another thought: We used to add solid CO2 pellets to water with indicator in it for fun. That's certainly a way of getting a lot of CO2 into water. If I remember rightly the indicator was methyl orange. The colour changes only while the CO2 pellet is subliming. Once the pellet has completely disappeared, the colour changes back.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2015-02-19T22:38:27.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26380" PostTypeId="1" CreationDate="2015-02-19T22:42:16.463" Score="1" ViewCount="42" Body="&lt;p&gt;I asked a question previously, and the answer was quite illuminating. The answer ushered me forth in a direction that finally made sense. I ended up finding this website, chem.latech.edu/~ramu/cmen522/homework/hw06_04a.pdf, that further explained the dynamic as well. I gathered what was necessary to be done, but wasn't quite sure how to apply it. I am currently a chemist for the oil field and am trying to find an accurate representation of percent water and CaCl2*subscript*, as well as the correct specific gravity of the complete solution. Given a 10.5 Mud weight (10.5 lb/gal), I have struggled to find volumes necessary to add and partial volume actually present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further explained - I applied the formula designated $V /cm^3 = 1001.97 + 30.2207(m^.5) − 36.7889m + 37.0838(m^1.5) − 5.7447(m^2)$ for the corrected solution volume and was really close to the correct &quot;loss&quot; of volume from the packing that the answer described. The two following formulas for partial volumes$ −36.7889 + 15.1104(m^−.5) + 55.6257(m^.5) − 11.4894m $(for CaCl2*subscript*) and $18.048 + 0.2722(m^.5) − 0.3341(m^1.5) + 0.1035(m^2) $(for water), don't add up to the total solution loss. I am a bit baffled as to why this is the case. Granted my physical chemistry knowledge is a bit dated, I would assume my excel document is correct. At the 10.5 weighted mud, my molality was 3.586. This resulted in a volume of 1105 with fluid loss and a 1183 of theoretical volume. The excess volume then should be 78.27. The formula for the solute given that molality gave me a volume of 17.63 and the solution was 35.33. The volume together is 52.95. This doesn't equal the excess volume which would suggest there was an error in one of the formulas, or that I have made a significant error on my part.&lt;/p&gt;&#xA;" OwnerUserId="13329" LastEditorUserId="13329" LastEditDate="2015-02-23T17:33:51.577" LastActivityDate="2015-02-23T17:33:51.577" Title="Weighted Drilling Fluid Comparison with CaCl₂" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26382" PostTypeId="2" ParentId="26350" CreationDate="2015-02-20T00:42:09.247" Score="3" Body="&lt;p&gt;If you analyze non-homogeneous samples, the results can't be interpreted as being representative of the whole organism (or population). So, for example, if histamine is unevenly distributed within the body of a fish, your results would vary widely depending on what part of the fish you analyzed.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-20T00:42:09.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26383" PostTypeId="2" ParentId="26345" CreationDate="2015-02-20T00:52:06.740" Score="2" Body="&lt;p&gt;I don't expect a significant energy difference per ester bond between glyceryl and methyl esters of a fatty acid. The way to drive such a reaction forward is by removing the desired product as the reaction proceeds. Then more product will form as the system seeks a new equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-20T00:52:06.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26384" PostTypeId="2" ParentId="26355" CreationDate="2015-02-20T02:03:14.173" Score="4" Body="&lt;p&gt;You said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But for a reversible process delta G is always zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not true; $\Delta G$ is always zero for a reversible process &lt;em&gt;if&lt;/em&gt; it is carried out at constant temperature and pressure and the only kind of work involved is the $p-V$ work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the only kind of work involved is $p-V$, in the above conditions, it is true that the maximum work obtained is zero. In a general case, try to replace in your notes the work $w$ with $w = w_{pV} + w_{\text{no}\,pV}$. Where $w_{pV}$ is the $p-V$ work and $w_{\text{no}\,pV}$  is the work of any other types, to get the complete history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In response to your comment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Teoretically, every thermodynamic problem can be solved by using internal energy function. Many times, in practice is convenient to use others variables. It can be done by obtaining new functions that depends of the variables that we want to. We do that by apling Legendre transformations (&lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_transformation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Legendre_transformation&lt;/a&gt;) of the internal energy function. One posibility is to get the Gibbs energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy energy function naturally takes different forms depending on the system. I mean, if there are works others than $p-V$ they must to be included to account for energy changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general form of a kind of work is: $X\cdot dx$, where $X$ is a generlized force (an intensive variable) and $x$ a generalized variable of displacement (an extensive variable), so we have for the change in internal energy ($dU$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ dU = \sum _{i=1} ^k X_i\cdot dx_i $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the energy function change when we add non $p-V$ work, its Legendre transforms corresponding to $G$ also change, so in this case:&#xA;$ G \neq H - TS$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the way for calculating $\Delta G$ depends on the works involved and the specific problem.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2015-02-21T00:15:18.797" LastActivityDate="2015-02-21T00:15:18.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26385" PostTypeId="2" ParentId="26352" CreationDate="2015-02-20T02:26:41.940" Score="7" Body="&lt;p&gt;The Darzens condensation can be carried out directly from the alpha-chloronitrile. The exact reaction needed here (benzaldehyde + chloroacetonitrile) is listed 15 times in SciFinder using different base catalysts (hydroxide or an alkoxide) and conditions (sometimes with phase-transfer catalysis). The epoxide will be formed as a mixture of stereoisomers, but often chemical yields are high (reportedly upto 100%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the epoxide in place, the nitrile can be reduced with DIBAL to give an intermediate imine, which is converted to the aldehyde after hydrolysis. This is pretty much the same strategy you outlined, but with a couple fewer steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D0mhQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D0mhQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for sources of additional synthesis problems, old UC-Berkeley exams are usually very good. Check Chem 3A and 3B &lt;a href=&quot;https://tbp.berkeley.edu/courses/chem/&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this archive&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2017-09-03T08:33:56.570" LastActivityDate="2017-09-03T08:33:56.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26387" PostTypeId="1" CreationDate="2015-02-20T02:42:28.067" Score="4" ViewCount="718" Body="&lt;p&gt;I am a little bit confused between charge distribution and HOMO/LUMO site in determining chemical reactivity. In charge distribution, the sites which are deficient and abundant in electrons are localized (e.g. in carbonyl molecules, we would expect that the carbonyl carbon to be partially positive and we would predict that a nucleophile would attack into that deficient carbon) but in HOMO/LUMO orbitals the site for reactivity differs on the orientation and sign of the orbitals. It may be in HOMO this carbon has an orbital for an overlap but in LUMO a node may appear in that site. My question is where would the nucleophile react in a species, by charge distribution or HOMO/LUMO orbital? &lt;/p&gt;&#xA;" OwnerUserId="14467" LastEditorUserId="10185" LastEditDate="2015-10-02T16:11:00.787" LastActivityDate="2015-10-02T16:11:00.787" Title="HOMO/LUMO and charge distribution" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26388" PostTypeId="2" ParentId="26285" CreationDate="2015-02-20T03:12:21.617" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the meaning of “no Hessian Eigenvalue ”?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The normal modes and frequencies are retrieved from Hessian diagonalization. By diagonalizing it you get the eigen-vectors (describing normal modes) and eigen-values (related to frequencies). If it is not done, the frequencies can not be calculated (and that's why they are not printed)&#xA;If the geometry optimization fails there is no point in the calculations above.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I think even a non-optimized system should have some vibrational properties and I can not understand what such a &amp;quot; no-frequency state &amp;quot; means physically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There is not a no-frequency state here. Trully, each single point calculation does not describe exactly a quantum state of your molecule. Why? Well, it neglect the quantum particle treatment of nucleic. We calculate the quantum state of fermions on some potential focus, and get the corresponding energy for different focus positions. The &amp;quot;movement&amp;quot; of the focus is then treated classically.&lt;/p&gt;&#xA;&lt;p&gt;The standard procedure is to get the harmonic frequencies, that is, accept the Hook Law for every normal mode. To get the physical insight we can think in a diatomic molecule (that has of course only one vibrational mode).&lt;/p&gt;&#xA;&lt;p&gt;We can use Taylor series to represent the potential energy &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; at any distance &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; around the equilibrium distance &lt;span class=&quot;math-container&quot;&gt;$r_e$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U(r) = U(r_e) + U^\prime(r-r_e)+ U^{\prime\prime}(r-r_e)^2+\cdots$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The heigh over the well is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U(r) - U(r_e) = U^\prime(r-r_e)+ U^{\prime\prime}(r-r_e)^2+\cdots$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we asume harmonic behavior, we expect that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U(r) - U(r_e) = \frac{1}{2}k (r-r_e)^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At equilibrium distance: &lt;span class=&quot;math-container&quot;&gt;$U^\prime(r-r_e) = 0$&lt;/span&gt;, so it is trivial to relate &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$U^{\prime\prime}(r-r_e)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At others &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; we can't do that because we don't know &lt;span class=&quot;math-container&quot;&gt;$r-r_e$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In such case, if it has vibrational properties, we don't know much about them. BUT, it is not always true that it has vibrational properties. Imagine a potential energy curve like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hapTR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hapTR.jpg&quot; alt=&quot;Some image&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://quantavolution.net/vol_05/figs/sb_f37.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;quantavolution.net&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the molecule is between &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, it will &amp;quot;move&amp;quot; in &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; direction and no return is expected (no vibration here). It doen't happend in the normal mode of a diatomic molecule, but it is a very common situation in poliatomic species.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-02-26T06:03:37.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26389" PostTypeId="1" AcceptedAnswerId="26394" CreationDate="2015-02-20T03:57:24.193" Score="1" ViewCount="1505" Body="&lt;p&gt;For my AP Chemistry class we were told to make a molecular diagram that was &quot;better than excellent&quot; and the rest is up to us (we could choose the molecule and how we wanted to represent it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My group chose benzoylmethylecgonine, or $\ce{C17H21NO4}$, also known as cocaine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is proving to be a difficult task as I now realize that cocaine is a pretty complicated molecule and I couldn't find much on the 3D molecular structure.  If anyone could explain its chemical structure, it would be very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1884" LastEditorUserId="30836" LastEditDate="2019-04-21T00:03:11.377" LastActivityDate="2019-04-21T00:03:11.377" Title="How to represent a 3D model of cocaine?" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;structural-formula&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26390" PostTypeId="1" CreationDate="2015-02-20T04:25:13.037" Score="3" ViewCount="304" Body="&lt;p&gt;How can I determine the calcium (Ca2+ ion) content in pudding mix given the amount of milk, number of servings made with one package, the percent of recommended daily dose of calcium per serving, and the recommended daily dose of calcium?&lt;/p&gt;&#xA;" OwnerUserId="14529" LastEditorUserId="21544" LastEditDate="2015-10-04T20:12:50.320" LastActivityDate="2015-10-04T20:12:50.320" Title="How to calculate to actual calcium content in a pudding mix" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-03-11T01:56:22.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="26391" PostTypeId="1" AcceptedAnswerId="26401" CreationDate="2015-02-20T04:43:28.853" Score="2" ViewCount="1457" Body="&lt;p&gt;How can we determine the boiling point of 1 molar aqueous solution (density 1.06 g per ml) of KBr? If I use the colligative property formula than I require molality instead of molarity. Here the density given is the one of the solution not the solvent.&lt;/p&gt;&#xA;" OwnerUserId="14530" LastEditorUserId="4945" LastEditDate="2015-02-20T06:22:54.370" LastActivityDate="2015-02-20T09:36:35.790" Title="How can I calculate molality from molarity?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26392" PostTypeId="2" ParentId="25009" CreationDate="2015-02-20T05:10:30.753" Score="5" Body="&lt;p&gt;The hydrogens are not 'isolated' - they are still vicinal protons (3-bonds away) and so the J-coupling between them is strong. The strength of a splitting interaction is primarily determined by the number of bonds in between the hydrogens involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle between them also plays a role but to a lesser extent. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Karplus_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karplus equation&lt;/a&gt; describes the three-bond J-coupling between two protons as a function of their dihedral angle. &lt;/p&gt;&#xA;" OwnerUserId="12939" LastEditorUserId="402" LastEditDate="2017-04-02T19:56:45.960" LastActivityDate="2017-04-02T19:56:45.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26394" PostTypeId="2" ParentId="26389" CreationDate="2015-02-20T05:43:22.690" Score="12" Body="&lt;p&gt;While I do not think that cocaine is a particularly interesting molecule, the crystal structure has been reported in &lt;a href=&quot;http://dx.doi.org/10.1139/v83-085&quot; rel=&quot;noreferrer&quot;&gt;Ronald J. Hrynchuk, Richard J. Barton, and Beverley E. Robertson, &lt;em&gt;Can. J. Chem.&lt;/em&gt;, &lt;strong&gt;1983&lt;/strong&gt;, &lt;em&gt;61&lt;/em&gt;, 481&lt;/a&gt;. There is also an open access paper of the XII IUCr Congress available, &lt;a href=&quot;http://journals.iucr.org/a/issues/1981/a1/00/a42811/a42811.pdf&quot; rel=&quot;noreferrer&quot;&gt;R. D. Hrynchuk, R. J. Barton and B. E. Robertson, &lt;em&gt;Acta Cryst A&lt;/em&gt;, &lt;strong&gt;1981&lt;/strong&gt;, &lt;em&gt;A37&lt;/em&gt;, C72&lt;/a&gt;. The crystal structure file has been deposited to &lt;a href=&quot;https://summary.ccdc.cam.ac.uk/structure-summary?doi=10.1139%2Fv83-085&quot; rel=&quot;noreferrer&quot;&gt;ccdc.cam.ac.uk&lt;/a&gt; and can be downloaded free of charge. However, as far as I can see, it does only contain the information about the space group, etc, but no atomic co-ordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the hydrochloride structure has also been reported in &lt;a href=&quot;http://dx.doi.org/10.1107/S0365110X6300205X&quot; rel=&quot;noreferrer&quot;&gt;E. J. Gabe and W. H. Barnes, &lt;em&gt;Acta Cryst.&lt;/em&gt;, &lt;strong&gt;1963&lt;/strong&gt;, &lt;em&gt;16&lt;/em&gt;, 796-801&lt;/a&gt;. And the cif file can be also obtained at &lt;a href=&quot;https://summary.ccdc.cam.ac.uk/structure-summary?doi=10.1107%2FS0365110X6300205X&quot; rel=&quot;noreferrer&quot;&gt;ccdc.cam.ac.uk&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick calculation on this geometry on the DF-BP86/def2-SVP level of theory, the plain xyz coordinates are attached at the end of this post. You can open them with molecular viewers such as &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt; to create your own model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following, however, has been produced with the commercial &lt;a href=&quot;http://www.chemcraftprog.com/&quot; rel=&quot;noreferrer&quot;&gt;ChemCraft&lt;/a&gt; program.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/foC1H.gif&quot; alt=&quot;rotation of cocaine hydrochloride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how far you want to go with your presentation, there are some pictures on &lt;a href=&quot;http://en.wikipedia.org/wiki/Cocaine&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;. I can also recommend checking the record of it on &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.10194104.html&quot; rel=&quot;noreferrer&quot;&gt;ChemSpider&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Geometry of cocaine hydrochloride:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;symmetry c1&#xA;Cl      -2.051693000     -0.493009000      4.153090000&#xA;C       -2.451519000     -3.565091000      6.602382000&#xA;C       -6.214292000      1.164927000      8.772468000&#xA;C       -7.187531000      2.140573000      9.037661000&#xA;C       -8.160694000      1.917991000     10.029303000&#xA;C       -8.167271000      0.711328000     10.754343000&#xA;C       -7.200746000     -0.269358000     10.489848000&#xA;C       -4.535361000     -2.199809000      5.904903000&#xA;C       -4.965503000     -2.410764000      3.577976000&#xA;C       -0.472992000     -3.311748000      5.082266000&#xA;C       -3.681531000     -2.746076000      7.063993000&#xA;C       -3.208615000     -1.681350000      8.100725000&#xA;C       -1.959949000     -0.929634000      7.627840000&#xA;C       -0.886578000     -1.913758000      7.125382000&#xA;C       -0.580604000     -3.058883000      8.128726000&#xA;C       -1.633721000     -4.157723000      7.786625000&#xA;C       -5.209882000     -1.120147000      9.257751000&#xA;C       -6.215633000     -0.043947000      9.503197000&#xA;H       -2.776079000     -4.325870000      5.870251000&#xA;H       -1.141638000     -5.101415000      7.477122000&#xA;H       -5.458897000      1.318599000      7.989093000&#xA;H       -7.190344000      3.080452000      8.463274000&#xA;H       -8.921920000      2.687588000     10.235464000&#xA;H       -8.932096000      0.535630000     11.527409000&#xA;H       -7.186414000     -1.224606000     11.036739000&#xA;H       -4.915209000     -3.233215000      2.840991000&#xA;H       -4.408859000     -1.522439000      3.213266000&#xA;H       -6.015572000     -2.140911000      3.803492000&#xA;H        0.202195000     -3.979332000      5.655364000&#xA;H       -4.361351000     -3.437105000      7.610613000&#xA;H       -1.009855000     -3.900141000      4.313786000&#xA;H        0.111577000     -2.526413000      4.564425000&#xA;H       -2.986886000     -2.247300000      9.031392000&#xA;H       -2.216795000     -0.241161000      6.793930000&#xA;H       -1.551394000     -0.325210000      8.463897000&#xA;H        0.011514000     -1.355303000      6.795935000&#xA;H        0.453167000     -3.437387000      7.998115000&#xA;H       -0.655253000     -2.701279000      9.175171000&#xA;H       -2.290875000     -4.410651000      8.642544000&#xA;N       -1.471722000     -2.634200000      5.937688000&#xA;O       -4.202728000     -0.704253000      8.439562000&#xA;O       -5.256578000     -2.245544000      9.737495000&#xA;O       -5.375360000     -1.329144000      6.012186000&#xA;O       -4.299429000     -2.894414000      4.763398000&#xA;H       -1.903686000     -1.736871000      5.153339000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2015-04-15T10:36:01.540" LastActivityDate="2015-04-15T10:36:01.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26395" PostTypeId="1" CreationDate="2015-02-20T05:43:39.653" Score="4" ViewCount="857" Body="&lt;p&gt;I am stuck on this gas question and its really stressing me out. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A compressed cylinder of gas contains $2.25 \cdot 10^3~\mathrm{g}$&#xA;  of $\ce{N2}$ gas at a pressure of $4.25 \cdot 10^7~\mathrm{Pa}$ and a temperature of&#xA;  $19.4~^\circ\mathrm{C}$. What volume of gas has been released into the atmosphere&#xA;  if the final pressure in the cylinder is $1.80 \cdot 10^5~\mathrm{Pa}$?&#xA;  Assume ideal behaviour and that the gas temperature remains unchanged.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know I have to use the ideal gas law for this.&lt;br&gt;&#xA;I've seen the solution for this problem, but something still doesn't make sense to me.&#xA;In the solution they used $$\frac{nRT}{p} = \frac{n_\mathrm{f}RT}{p_\mathrm{f}},$$ to find the final amount of nitrogen gas in moles, $n_\mathrm{f}$, remaining in the gas cylinder. But why is the initial volume and final volume the same? Since gas is being released into the atmosphere, shouldn't the final volume change?&lt;/p&gt;&#xA;" OwnerUserId="14531" LastEditorDisplayName="user7951" LastEditDate="2015-02-20T21:09:33.120" LastActivityDate="2015-02-20T21:09:33.120" Title="Why is the volume in this compress gas cylinder constant throughout?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26398" PostTypeId="1" AcceptedAnswerId="26403" CreationDate="2015-02-20T07:39:15.853" Score="9" ViewCount="262" Body="&lt;p&gt;This is probably a common question because it involves TNT precursors, but if an $\ce{NO2}$ group deactivates benzene rings toward electrophilic aromatic substitution reactions, then how can toluene be tri-nitrated? I understand that the methyl group will donate electron density and activate the ring toward the first substitution, but after two nitrations I'm sure the nitro groups slow the third substitution. So what's the trick? How does toluene get three nitro groups?&lt;/p&gt;&#xA;" OwnerUserId="11585" LastEditorUserId="11585" LastEditDate="2015-11-04T08:11:25.647" LastActivityDate="2015-11-04T08:11:25.647" Title="Aromatic ring substitution: how are MNT and DNT further nitrated if nitro groups deactivate benzene rings?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26400" PostTypeId="1" AcceptedAnswerId="26583" CreationDate="2015-02-20T09:18:55.623" Score="1" ViewCount="612" Body="&lt;p&gt;I understand why solids and liquids are not included in the $K_{eq}$ expressions. However, I'm wondering what the $K$ value for a reaction involving only solid or liquid reactants and products looks like. Theoretically, it'd be 1 no matter what since $1^n = 1$. Is this right? Also, do all such reactions go to completion?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastActivityDate="2015-02-24T21:38:47.130" Title="Solid and Liquid Equilibrium" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ClosedDate="2015-02-24T22:00:59.220" ContentLicense="CC BY-SA 3.0" />
  <row Id="26401" PostTypeId="2" ParentId="26391" CreationDate="2015-02-20T09:36:35.790" Score="4" Body="&lt;p&gt;The key here is to find the mass of the solvent. If you take one mL of the solution, it will contain $10^{-3} \mathrm{mol}$ of KBr. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of $10^{-3} \mathrm{mol}$ of KBr is  $10^{-3}\times(39+79,9)=10^{-3}\times 118.9=0.1189 \mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, the mass of one mL of the solution is $1.06 \mathrm{g}$. This means that the mass of the solvent in one mL of the solution is  $1.06 -0.1189=0.9411 \mathrm{g}=0.9411\times 10^{-3}\mathrm{kg} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molality is:&#xA;$$m=\frac{10^{-3} \mathrm{mol}}{0.9411\times 10^{-3}\mathrm{kg}}=1.063\,\, \mathrm{mol/kg} $$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-20T09:36:35.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26402" PostTypeId="1" CreationDate="2015-02-20T09:49:06.070" Score="5" ViewCount="94" Body="&lt;p&gt;Let us consider the Nernst equation $$E= E^\circ -\frac{2.303RT}{nF} \log Q$$ which once at $\pu{298.15 K}$, is the equivalent  of $$E= E^\circ - \frac{0.059}{n} \log Q .$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the part of the Nernst equation $-\frac{0.059}{n} \log Q$ address Ohmic drop? In other words, is $-\frac{0.059}{n} \log Q$ related to solution (electrolyte) resistance of an electrochemical cell?&lt;/p&gt;&#xA;" OwnerUserId="13328" LastEditorUserId="194" LastEditDate="2018-02-16T14:22:53.890" LastActivityDate="2018-02-16T14:22:53.890" Title="Is the subtrahend in the Nernst equation related to ohmic drop?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26403" PostTypeId="2" ParentId="26398" CreationDate="2015-02-20T10:08:05.323" Score="9" Body="&lt;p&gt;On a industrial scale by using highly concentrated nitric acid known as fuming nitric acid (usually &gt; 90%, but in the last step from DNT to TNT may also be anhydrous) along with oleum, which is a mixture of $\ce{SO_3}$ and sulfuric acid. Preparation in the lab may be achieved by more or less ordinary nitration at elevated temperatures. In a micro scale experiment [&lt;a href=&quot;https://edivini.pbworks.com/f/nitrazionetoluene.pdf&quot;&gt;1&lt;/a&gt;] they used two different types of nitrating acid:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitrating acid I contains concentrated nitric acid 19% v/v, concentrated sulfuric acid 58% v/v and water 23% v/v.&lt;/li&gt;&#xA;&lt;li&gt;Nitrating acid II is a mixture of concentrated nitric acid 25% v/v and oleum (30% $\ce{SO_3}$) 75% v/v.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Nitrating acid I was used for $\ce{Toluene -&amp;gt; MNT}$ and $\ce{MNT -&amp;gt; DNT}$ (65 °C). Nitrating acid II was used for $\ce{DNT -&amp;gt; TNT}$ (110 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;em&gt;&lt;a href=&quot;https://edivini.pbworks.com/f/nitrazionetoluene.pdf&quot;&gt;1&lt;/a&gt; The Stepwise Nitration of Toluene: A Multistep Microscale Synthesis Based on an Industrial Process&lt;br&gt;&#xA;Richard A. Russell, Robert W. Swltzer, and Robert W. Langmwe&lt;br&gt;&#xA;University College, University of N.S.W., Northcotl Drive, Campbell ACT. 2600, Australia&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-20T20:52:22.773" LastActivityDate="2015-02-20T20:52:22.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26404" PostTypeId="2" ParentId="26395" CreationDate="2015-02-20T10:11:16.947" Score="1" Body="&lt;p&gt;It is about the volume of the cylinder, which of course does not change. You do not also consider the volume of the escaped nitrogen gas.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-20T10:11:16.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26405" PostTypeId="1" AcceptedAnswerId="26407" CreationDate="2015-02-20T10:45:33.400" Score="9" ViewCount="32398" Body="&lt;p&gt;The Nernst equation describes the relationship between cell potential and temperature. But why does temperature affect cell potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the collision model of kinetics is not relevant to electrochemical cells, plus an increase in temperature decreases cell potential.&#xA;So why does the potential change? &lt;/p&gt;&#xA;" OwnerUserId="14535" LastActivityDate="2019-05-11T12:21:13.903" Title="Why does temperature affect cell potential?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26406" PostTypeId="2" ParentId="26402" CreationDate="2015-02-20T11:04:39.643" Score="3" Body="&lt;p&gt;In short: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nernst equation describes the thermodynamic equilibrium state. By definition, there is no current flowing through the cell. Therefore, there is no ohmic drop in any term within the Nernst equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Q term relates to the activities of the reactants and products at equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-02-20T11:04:39.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26407" PostTypeId="2" ParentId="26405" CreationDate="2015-02-20T11:15:35.130" Score="5" Body="&lt;p&gt;Depending on how you write the Nernst equation the temperature might increase or decrease the potential of the cell. This depends more on the reaction quotient than on anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature comes in to the equation as a scaling factor where RT/nF has units Volt. This essentially defines how much the voltage changes per a decade change in the reaction quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature dependence here is just a matter of the scaling factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the collision model, that is more about kinetics than thermodynamics and it is not at play for the Nernst equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nernst equation describes the thermodynamics of the electrochemical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature dependence can be explained by the temperature dependence of work-functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any electrochemical reaction by definition is an electron transfer. The electron goes from the highest occupied MO of one species to the lowest unoccupied MO of the other. Both these levels are temperature dependent. &lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="13355" LastEditDate="2015-03-30T08:11:03.577" LastActivityDate="2015-03-30T08:11:03.577" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26410" PostTypeId="1" AcceptedAnswerId="26413" CreationDate="2015-02-20T15:02:13.373" Score="5" ViewCount="510" Body="&lt;p&gt;What is the mechanism for the ozonolysis of a nitronate anion? What gas is produced in this reaction? Is there any chemoselectivity in the presence of a double C=C bond in the molecule?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-02-20T17:05:18.323" Title="Ozonolysis of nitronate" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26411" PostTypeId="1" AcceptedAnswerId="26419" CreationDate="2015-02-20T15:59:23.303" Score="1" ViewCount="128" Body="&lt;p&gt;I got this redox reaction on my exam that I didn't know how to balance.&#xA;$$Sodium~iodate+sulphur~dioxide+water\ce{-&amp;gt;}sodium~sulphate+sulphuric~acid+iodine$$, thus it would be:&#xA;$$\ce{NaIO3 +SO2 +H2O-&amp;gt;Na2SO4 +H2SO4 +I2}$$&#xA;How does one go about balancing this using the oxidation number method?&#xA;I understand that S goes from +4 to +6 losing 2 electrons and I from +5 to 0 gaining 5 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we need to consider that there's 2 S atoms on the products (thus each atom loses 1 electron)? So I guess we  need to only multiply the reactants' side sulphur with 5/2, while in other cases we need to multiply sulphur both in the reactants and the products side.&lt;/p&gt;&#xA;" OwnerUserId="14537" LastEditorUserId="14537" LastEditDate="2015-02-20T17:53:18.780" LastActivityDate="2015-02-20T20:54:51.247" Title="Reaction balance" Tags="&lt;redox&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26412" PostTypeId="2" ParentId="16082" CreationDate="2015-02-20T16:03:50.470" Score="2" Body="&lt;p&gt;Try looking at Inorganic Chemistry in Aqueous Solution by Jack Barrett around page 41 on google books or elsewhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy of hydration of Mg2+ is very negative compare to Na+.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Barrett, more highly charged ions restrict the movement of water molecules of hydration more.  Also, the smaller the radius of the ion, the more restricted the movement of the hydrating waters is.  Both the number of waters whose movement is restricted, and the degree to which they are restricted, is affected by charge and radius.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete analysis would consider both the entropy and enthalpy contributions.    &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-20T16:03:50.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26413" PostTypeId="2" ParentId="26410" CreationDate="2015-02-20T17:05:18.323" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the mechanism for the ozonolysis of a nitronate anion? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably very analogous to ozonolysis of a double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rh5FF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical synthetic procedure is detailed in &lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=cv6p0648&quot; rel=&quot;noreferrer&quot;&gt;OrgSyn&lt;/a&gt;.  Note the addition of dimethylsulfide during work-up; this is common in ozonolysis reactions as it helps reduce the ozonide and also reduces any other dangerous, highly oxidized by-products.  The discussion section of the OrgSyn article says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There are various methods available for transforming a nitro group&#xA;  into a carbonyl group, including the Nef reaction (strongly acidic),²&#xA;  permanganate oxidation of nitronate anions (basic, oxidative),³&#xA;  persulfate oxidation of nitronates (basic, oxidative),⁴ treatment with&#xA;  a mixture of organic and inorganic nitrite (neutral, oxidative),⁵ and&#xA;  treatment of either a free nitro compound or a nitronate anion with&#xA;  aqueous titanium(III) (neutral, reductive).⁶ Each method is limited,&#xA;  however, by poor yield, inconvenience, or lack of generality.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With the proviso that the substrate not contain a reactive&#xA;  carbon-carbon double bond, the present ozonolysis procedure⁷ appears&#xA;  to provide a convenient and efficient method for carrying out the&#xA;  desired &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nef_reaction&quot; rel=&quot;noreferrer&quot;&gt;Nef Reaction&lt;/a&gt; is mentioned above, it is worth taking a look at the linked article since the Nef reaction is complementary to the ozonolysis reaction of nitronates.  You may also find the &quot;Reaction mechanism&quot; section interesting.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What gas is produced in this reaction? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the above mechanism, no gas would be generated, rather $\ce{NO3^{-}}$ is produced&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any chemoselectivity in the presence of a double C=C bond in the molecule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, as mentioned in the OrgSyn reference, this approach can't be used when other double bonds are present.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-20T17:05:18.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26414" PostTypeId="2" ParentId="26411" CreationDate="2015-02-20T17:30:35.980" Score="0" Body="&lt;p&gt;Let's try to solve the question starting by iodate ion and then by periodate ion (as the post changes several times while I am answering the question).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iodate ion:&#xA;$$\ce{2IO3^-  + 10e^- + 6H2O -&amp;gt;12OH^- +I2}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ SO2 +4OH^- -&amp;gt;SO4^2- +2e^- +2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we multiply the second equation by $5$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ 5SO2 +20OH^- -&amp;gt;5SO4^2- +10e^- +10H2O}$$&#xA; Now, we add the the first and the last equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2IO3^-  + 5SO2 +8OH^- -&amp;gt;5SO4^2- +4H2O +I2}$$&#xA;Now, we can rearrange this equation:&#xA;Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:&#xA;$$\ce{2IO3^-  + 5SO2 +4H2O -&amp;gt;5SO4^2- +8H+ +I2}$$&#xA;We combine $\ce{4SO4^2- +8H+ }$ to make sulfuric acid:&#xA;$$\ce{2IO3^-  + 5SO2 +4H2O -&amp;gt;SO4^2- +4H2SO4 +I2}$$&#xA;By adding $2\ce{Na+}$ to the two sides of the equation, we find:&#xA;$$\ce{2NaIO3  + 5SO2 +4H2O -&amp;gt;Na2SO4  +4H2SO4 +I2}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Periodate ion:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, if we try to answer the question with  periodate ion:&#xA;$$\ce{2IO4^-  + 14e^- + 8H2O -&amp;gt;16OH^- +I2}$$&#xA;$$\ce{ SO2 +4OH^- -&amp;gt;SO4^2- +2e^- +2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we multiply the second equation by $7$:&#xA;$$\ce{ 7SO2 +28OH^- -&amp;gt;7SO4^2- +14e^- +14H2O}$$&#xA;Now, we add the the first and the last equation:&#xA;$$\ce{2IO4^-  + 7SO2 + 12OH^- -&amp;gt; 6H2O + I2 + 7SO4^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can rearrange this equation:&#xA;Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:&#xA;$$\ce{2IO4^-  + 7SO2 + 6H2O -&amp;gt; 6H2SO4 + I2 + SO4^2-}$$&#xA;By adding $2\ce{Na+}$ to the two sides of the equation, we find:&#xA; $$\ce{2NaIO4  + 7SO2 +6H2O -&amp;gt;Na2SO4  +6H2SO4 +I2}$$&#xA;Conclusion: I think the question must be asked with periodate ion.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-02-20T20:54:51.247" LastActivityDate="2015-02-20T20:54:51.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26416" PostTypeId="1" CreationDate="2015-02-20T18:23:09.510" Score="4" ViewCount="341" Body="&lt;p&gt;Does anybody know two liquids (or a solid and a liquid) that can react to produce a gas less dense than air (or similar) at room temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is intended to make a chemical induced reaction when a lever is pulled to produce gas and decrease an object's mass.&lt;/p&gt;&#xA;" OwnerUserId="14541" LastEditorUserId="5026" LastEditDate="2018-04-18T06:35:17.033" LastActivityDate="2018-04-18T06:40:54.220" Title="Two liquids that react to produce a gas less dense than air at room temperature" Tags="&lt;gas-laws&gt;&lt;density&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26417" PostTypeId="2" ParentId="26416" CreationDate="2015-02-20T18:33:28.970" Score="8" Body="&lt;p&gt;If the other products aren't significant, I'll recommend the all famous alkali metals' reaction with water:&#xA;$$\ce{2Li(s) + 2H2O(l) -&amp;gt; 2LiOH(aq) + H2(g)}$$&#xA;Hydrogen gas is less dense than air, both at RTP and at STP, because the &lt;a href=&quot;http://eesc.columbia.edu/courses/ees/slides/climate/table_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;main components of air&lt;/a&gt; (by mole percent) are $\ce{N2}~(78\%)$, $\ce{O2}~(20.95\%)$ and $\ce{Ar}~(0.934\%)$ - all of which have a molar mass larger than hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5026" LastEditDate="2018-04-18T06:40:54.220" LastActivityDate="2018-04-18T06:40:54.220" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26418" PostTypeId="1" CreationDate="2015-02-20T18:41:33.403" Score="1" ViewCount="631" Body="&lt;p&gt;In an adiabatic irreversible process we know that q = 0. Now change in entropy is defined as q/T. So when q is zero, why is change in entropy not zero ?&lt;/p&gt;&#xA;" OwnerUserId="14542" LastActivityDate="2015-02-20T18:49:16.517" Title="Entropy change of system in adiabatic irreversible process" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-20T20:24:26.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="26419" PostTypeId="2" ParentId="26411" CreationDate="2015-02-20T18:46:37.973" Score="0" Body="&lt;p&gt;@user14537 sent a comment asking if we can solve the exercise using only oxidation number method:&#xA;$$\ce{NaIO4  + SO2 +H2O -&amp;gt;Na2SO4  +H2SO4 +I2}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The change in oxidation number between periodate ion and iodine molecule is $7\times 2=14$ (as we have two atoms in iodine molecule).&#xA;So, the first step is to multiply $\ce{NaIO4 }$ by $2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The change in oxidation number between $\ce{SO2}$ and sulfate ion is $2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As the changes must be equal, we have to multiply $\ce{SO2}$ by $7$ and $\ce{H2SO4}$ by&#xA;$6$. (As you have already a sulfate ion in   $\ce{Na2SO4}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, as you have $12$ hydrogen atoms in the right side, you add $6$ molecules of water to the left side:&#xA;$$\ce{2NaIO4  + 7SO2 +6H2O -&amp;gt;Na2SO4  +6H2SO4 +I2}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The equation is balanced.&#xA;I hope it's clear now.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-20T18:46:37.973" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26420" PostTypeId="2" ParentId="26418" CreationDate="2015-02-20T18:49:16.517" Score="0" Body="&lt;p&gt;The change in entropy is the &lt;em&gt;reversible&lt;/em&gt; heat flow divided by $T$.  So if the process is irreversible, you can't use the irreversible process's heat flow to find the entropy change.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-20T18:49:16.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26421" PostTypeId="2" ParentId="26395" CreationDate="2015-02-20T19:43:53.750" Score="0" Body="&lt;p&gt;Gases always occupy whole volume of the container, and it's not to going to change  throughout.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2015-02-20T19:43:53.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26422" PostTypeId="2" ParentId="26411" CreationDate="2015-02-20T20:12:08.040" Score="0" Body="&lt;p&gt;$\ce{2NaIO3 + 10e^- + 12H+ -&amp;gt;2Na+ +I2 + 6H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{5SO2 +10H2O -&amp;gt;5SO4^2- +10e^- + 20H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2NaIO3 + 5SO2 + 4H2O -&amp;gt;Na2SO4 + 4H2SO4 + I2}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-20T20:12:08.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26423" PostTypeId="1" AcceptedAnswerId="26446" CreationDate="2015-02-20T20:36:52.667" Score="6" ViewCount="4913" Body="&lt;p&gt;Which group gets higher priority and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-CH_2OCH_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-CH=O}$&lt;/p&gt;&#xA;" OwnerUserId="7524" LastEditorUserId="7448" LastEditDate="2015-10-02T16:02:20.970" LastActivityDate="2015-10-02T16:02:20.970" Title="Which has higher priority (Cahn-Ingold-Prelog)?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26424" PostTypeId="1" CreationDate="2015-02-20T21:01:19.670" Score="9" ViewCount="10209" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G = \Delta H_\text{system}-T\Delta S_\text{system}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{ccc}&#xA;\hline&#xA;\text{Sign of}~ΔH &amp;amp; \text{Sign of}~ΔS &amp;amp; \text{Spontaneity} \\&#xA;\hline&#xA;+ &amp;amp; + &amp;amp; \text{The reaction is spontaneous at high temperature} \\&#xA;+ &amp;amp; - &amp;amp; \text{The reaction is never spontaneous} \\&#xA;- &amp;amp; - &amp;amp; \text{The reaction is spontaneous at low temperature} \\&#xA;- &amp;amp; + &amp;amp; \text{The reaction is always spontaneous} \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When enthalpy change is negative, the reaction is exothermic, which means it releases energy into the surroundings. If the system is losing energy, shouldn’t the entropy of the system always decrease? I understand mathematically that &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}S = \mathrm{d}Q/T$&lt;/span&gt;, and if there is heat exchange, then entropy change can be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But intuitively, if energy is taken away from the system, shouldn’t the entropy of that same system decrease? For some reason, I always thought that more energy means more entropy.&lt;/p&gt;&#xA;" OwnerUserId="9625" LastEditorUserId="41328" LastEditDate="2019-03-16T17:10:44.670" LastActivityDate="2019-03-16T17:10:44.670" Title="How can enthalpy change of a system be negative while entropy change is positive?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;entropy&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26425" PostTypeId="2" ParentId="26424" CreationDate="2015-02-20T21:15:08.303" Score="11" Body="&lt;p&gt;For example, suppose you have a solid block of TNT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It explodes and releases much energy. $\Delta H$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5017" LastEditDate="2015-06-18T15:33:40.617" LastActivityDate="2015-06-18T15:33:40.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26426" PostTypeId="1" AcceptedAnswerId="26460" CreationDate="2015-02-20T21:32:51.010" Score="5" ViewCount="746" Body="&lt;p&gt;I've hit some trouble with these calculations and need help sorting out confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lab we synthesized $\ce{YBa2Cu3O7}$, per the lab manual $\ce{Y, Ba, O}$ have the usual charges of +3, +2, and -2 respectively. Spectroscopic studies show that no copper (III) centers are present in the material, but rather there are missing electrons from the copper-oxygen bonds. For the purpose of the titration though, the missing electrons are thought of as coming from the copper center and we're assuming there are copper III centers present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used idodine to tritrate, and there is the assumption that copper is doing the oxidizing. The text states, &quot;Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant equations are &#xA;$$\begin{align}&#xA;\ce{Cu^{3+} + 2I- &amp;amp;-&amp;gt; Cu+ +I2}\\&#xA;\ce{Cu^{2+} +I- &amp;amp;-&amp;gt; Cu+ + \frac{1}{2}I2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've calculated the theoretical OS to be &#xA;$$\ce{Y} (3^+ )+\ce{Ba} (2^+ * 2)+\ce{O} (2^-*7)= \frac{7}{3 \ \ce{Cu}}\ce{Cu}^{2.3+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to calulate the theoretical number of equivalents of $\ce{I-}$ oxidized by 1 g of the sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be correct for the oxidation state? If so would that mean that the total equivalents of Iodine are an addition of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ (the two equations)? Also I'm feeling brain dead, because I can't figure out the fraction of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ that would average to 2.3.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="5026" LastEditDate="2018-07-16T02:22:19.910" LastActivityDate="2018-09-14T15:39:48.333" Title="How to calculate the oxidation state and the number of equivalents oxidized with YBa₂Cu₃O₇" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26427" PostTypeId="2" ParentId="26387" CreationDate="2015-02-20T22:35:09.837" Score="1" Body="&lt;p&gt;You need &lt;em&gt;both&lt;/em&gt; an empty LUMO for the electrophile with an in phase HOMO for the nucleophile at the site of attack AND the proper charge distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LUMOs are critical; you can't shove electron density into more electron density, and LUMOs represent a lack of electron density.  However, the electron density for a nucleophile will be attracted to the partial positive charges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Organic chemistry normally requires you make judgments about the relative impacts of various phenomena, and this is another example.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-02-20T22:35:09.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26428" PostTypeId="1" AcceptedAnswerId="26431" CreationDate="2015-02-20T23:04:16.433" Score="3" ViewCount="5991" Body="&lt;p&gt;I'm taking a first year university chemistry course, and I am reading a lot of contradicting things. Some people say that ionic compounds that are soluble in water are &lt;em&gt;always&lt;/em&gt; electrically conductive, others say that it &lt;em&gt;maybe&lt;/em&gt; conductive.  Which one is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this problem to think about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that an unknown chemical compound exhibits the following properties:       &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;it is crystalline but shows no electric conductivity in the solid state &lt;/li&gt;&#xA;  &lt;li&gt;it melts at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{300^\circ C}$&lt;/span&gt; and decomposes at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{400^\circ C}$&lt;/span&gt;&lt;/li&gt;&#xA;  &lt;li&gt;the compound is soluble in water, the solution shows no electrical conductivity. &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;What kind of chemical bond would you expect in the&#xA;  given compound? Try to describe this type of bond in detail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything screams &quot;ionic&quot; (high melting point, solid at room temperature, crystalline structure, non-conducting in solid form) besides the fact that it doesn't conduct in water.&lt;/p&gt;&#xA;" OwnerUserId="14536" LastEditorUserId="83758" LastEditDate="2020-04-08T18:02:21.993" LastActivityDate="2020-04-08T18:02:21.993" Title="Can all water soluble ionic compounds conduct electricity?" Tags="&lt;solubility&gt;&lt;ionic-compounds&gt;&lt;conductivity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26429" PostTypeId="2" ParentId="26428" CreationDate="2015-02-21T00:09:34.013" Score="1" Body="&lt;p&gt;All soluble ionic compound conduct according to Debye-Hückel-Onsager theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of something molecular.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-21T00:16:22.250" LastActivityDate="2015-02-21T00:16:22.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26431" PostTypeId="2" ParentId="26428" CreationDate="2015-02-21T00:22:02.683" Score="4" Body="&lt;p&gt;DavePhD is right! Your material is an organic compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to be soluble in water, it should have some polar substituents. I'd however exclude carboxylic acids or phenolic $\ce{OH}$ (except maybe in the proximity of a carbonyl group) since these will partly dissociate and yield to minor conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the melting point wouldn't be that high, inositol would be a candidate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleobases thymine and uracil do show melting points in the 300 °C range and are both soluble in water.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-21T00:32:16.193" LastActivityDate="2015-02-21T00:32:16.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26433" PostTypeId="2" ParentId="26428" CreationDate="2015-02-21T00:55:58.703" Score="-2" Body="&lt;p&gt;I wouldn't say &quot;always&quot;. But yes, and by default, ionic compounds are usually going to conduct electricity. I would study electrolytes before trying to decide these things, if you know electrolytes, the answer would be obvious to you&lt;/p&gt;&#xA;" OwnerUserId="14545" LastActivityDate="2015-02-21T00:55:58.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26434" PostTypeId="1" CreationDate="2015-02-21T00:59:00.867" Score="0" ViewCount="509" Body="&lt;p&gt;A friend of mine bought a piece of furniture that had a super smelly wood varnish, so I read around online and someone said to light candles to burn the chemical fumes.  Does anybody know if there is a chemical reaction between the wood varnish fumes and the candles? they are unscented candles and I assume they can break down the hydrocarbons in the wood varnish fumes.  Are there any other harmful chemical in the wood varnish fumes that are not oxidized by the candles that cannot be smelled by humans?&lt;/p&gt;&#xA;" OwnerUserId="14546" LastEditorUserId="865" LastEditDate="2015-10-30T17:11:10.443" LastActivityDate="2015-10-30T17:11:10.443" Title="do candles burn wood varnish fumes?" Tags="&lt;reaction-mechanism&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26436" PostTypeId="1" CreationDate="2015-02-21T01:35:33.040" Score="-3" ViewCount="9888" Body="&lt;p&gt;Please explain why, thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="14547" LastEditorUserId="402" LastEditDate="2015-02-21T10:02:27.070" LastActivityDate="2015-02-21T10:02:27.070" Title="Which is more basic: aniline or pyridine?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-02-21T09:27:24.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="26437" PostTypeId="2" ParentId="26436" CreationDate="2015-02-21T02:02:20.537" Score="4" Body="&lt;p&gt;You can look at the pKb values.  It turns out pyridine is more basic.  This is difficult to predict from looking at the Lewis structures of the molecules.  What you should see is that the lone pair on pyridine is on an sp2 hybridized N.  This makes it less basic than an sp3 N.  The N in aniline is sp3 but the lone pair is in resonance with the pi electrons of the phenyl ring, lowering the basicity.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-02-21T02:02:20.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26440" PostTypeId="1" AcceptedAnswerId="26529" CreationDate="2015-02-21T07:55:34.277" Score="3" ViewCount="4170" Body="&lt;p&gt;Why do coupling constants change with temperature in NMR?&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="5017" LastEditDate="2015-02-23T05:37:02.500" LastActivityDate="2015-02-23T23:15:32.277" Title="Why do NMR coupling constants change with temperature?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26441" PostTypeId="2" ParentId="26423" CreationDate="2015-02-21T08:05:41.417" Score="4" Body="&lt;p&gt;$\ce{R-CH=O}$ has higher priority over &#xA;$\ce{R-CH_2OCH_3}$. Because, if $\ce{R}$ group is the same, then we have to compare between the next substituents: For $\ce{R-CH=O}$  you have one  carbon atom connected to 2 oxygen atoms (a double bond is equivalent to two simple bonds) and one hydrogen atom.&#xA;For $\ce{R-CH_2OCH_3}$, you have one  carbon atom connected to one oxygen atom and two hydrogen atoms. So, $\ce{R-CH=O}$ has higher priority.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-02-21T08:05:41.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26442" PostTypeId="1" CreationDate="2015-02-21T10:40:22.830" Score="0" ViewCount="319" Body="&lt;p&gt;I want to know if there is any substance, liquid or any other stuff to separate ferrofluid from oil after mixing them.&lt;/p&gt;&#xA;" OwnerUserId="14552" LastEditorUserId="5026" LastEditDate="2018-03-03T07:34:01.763" LastActivityDate="2020-04-21T16:35:18.713" Title="Which method can be applied to separate ferrofluid liquid from oil?" Tags="&lt;materials&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26443" PostTypeId="2" ParentId="26442" CreationDate="2015-02-21T11:20:43.360" Score="2" Body="&lt;p&gt;Typical ferrofluids are suspension of nano-sized magnetic particles in oil. If you add more of the same oil and mix thoroughly, separation is impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add a chemically different oil, you might be able to remove this compound from the mixture again by distillation in vacuum.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-21T11:20:43.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26445" PostTypeId="1" AcceptedAnswerId="26450" CreationDate="2015-02-21T12:57:46.103" Score="13" ViewCount="43096" Body="&lt;p&gt;According to molecular orbital theory s and p orbitals can mix if they are close enough in energy to each other. For period 2 diatomics, this occurs for $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$ and $\ce{N}$ - in the case of oxygen and fluorine the s and p orbitals are two far apart in energy. The effect can be seen by realizing that s-p mixing causes the $2\sigma_\mathrm{g}$ orbital to be higher in energy than the $1\pi_\mathrm{u}$ orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Does the mixing of the s and p orbitals occur between the orbitals of one atom (basically hybridization right?) or does the s orbital of one atom mix with the p orbital of the other (and vice versa)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has confused me because I realize that concept of hybridization is widely used and functions well so I initially believed that the mixing should occur in one atom - forming hybrid atomic orbitals - and  then bonding. However, after I thought about it some more I began to doubt that this is the case. Firstly, the s and p orbitals on one atom do not have the correct symmetry to mix effectively (the s orbital would simultaneously constructively interference with one lobe of the p orbital and destructively interfere with the other). Also, the molecular orbital diagram of carbon monoxide reveals that s-p mixing must be occurring since the  $3\sigma$ orbital is higher in energy than the $1\pi$ orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UfAoO.gif&quot; alt=&quot;Carbon Monoxide MO Diagram&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also seems to contradict the idea that the s and p orbitals mix on the same atom because in $\ce{O_2}$ there is no s-p mixing so why would oxygen mix its s and p orbitals when it's bonding with carbon.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2018-08-03T11:28:50.610" LastActivityDate="2018-08-03T11:28:50.610" Title="Is s-p mixing referring to hybridization or is it the mixing of one atoms s orbital with the other's p orbital?" Tags="&lt;molecular-orbital-theory&gt;&lt;theoretical-chemistry&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26446" PostTypeId="2" ParentId="26423" CreationDate="2015-02-21T14:13:32.477" Score="10" Body="&lt;p&gt;As mentioned by Yomen , the double bond counts for two atoms with a single bond between them . They are called ghost atoms . This makes aldehyde group at higher priority than the ether .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikedia has a very nice explanation on it :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a ghost atom that has the same atomic number as B but is not attached to anything except A. In turn, when B is replaced with a list of attached atoms, A itself is excluded in accordance with the general principle of not doubling back along a bond that has just been followed, but a ghost atom for A is included so that the double bond is properly represented from both ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A triple bond is handled in a similar manner . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia link : &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules&quot;&gt;Cahn–Ingold–Prelog priority rules&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="-1" LastEditDate="2015-04-22T08:10:38.857" LastActivityDate="2015-04-22T08:10:38.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26447" PostTypeId="2" ParentId="24925" CreationDate="2015-02-21T14:21:44.050" Score="-1" Body="&lt;p&gt;Yes. If you do however, that would likely unsaturate the solution. I don't know if certain animals need certain amounts of salt, but doing this without adding salt might kill them.  &lt;/p&gt;&#xA;" OwnerUserId="14545" LastActivityDate="2015-02-21T14:21:44.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26448" PostTypeId="2" ParentId="20511" CreationDate="2015-02-21T14:30:58.367" Score="-3" Body="&lt;p&gt;Simply put, H2O is not ionic because it is a nonmetal/nonmetal compound, meaning it's covalent. Water is an oddball in the chemistry world, but certain laws prohibit water being an ionic compound&lt;/p&gt;&#xA;" OwnerUserId="14545" LastEditorUserId="9961" LastEditDate="2017-09-24T22:11:51.390" LastActivityDate="2017-09-24T22:11:51.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26449" PostTypeId="2" ParentId="26445" CreationDate="2015-02-21T14:46:17.697" Score="2" Body="&lt;p&gt;s-p mixing is with respect to one atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that in the approximation that a molecular orbital is mathematically a linear combination of atomic orbitals, the best approximation involves a significant amount of the atom's s orbital and a significant amount of that atom's p orbital (plus linear combinations of one or more atomic orbitals from the other atom or atoms).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Firstly, the s and p orbitals on one atom do not have the correct symmetry to mix effectively (the s orbital would simultaneously constructively interference with one lobe of the p orbital and destructively interfere with the other). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This constructive and destructive interference is what actually occurs.  For example, in methane, with respect to the carbon, there is s-p mixing, with the constructive interference being in the direction of a hydrogen atom.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-21T14:46:17.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26450" PostTypeId="2" ParentId="26445" CreationDate="2015-02-21T15:25:13.510" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to molecular orbital theory s and p orbitals can mix if they&#xA;  are close enough in energy to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can make that statement a bit more general so that we can apply it to all orbitals, atomic and molecular, by saying,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to molecular orbital theory orbitals can mix if they are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;close enough in energy to each other, and&lt;/li&gt;&#xA;&lt;li&gt;have the same symmetry&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the mixing of the s and p orbitals occur between the orbitals of&#xA;  one atom (basically hybridization right?) or does the s orbital of one&#xA;  atom mix with the p orbital of the other (and vice versa)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good question, you are really thinking about this!  First off, yes, mixing of the s and p orbitals occurring between the orbitals of one atom is called hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I think of these problems I &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;first, hybridize the atomic orbitals (s and p) on an atom to generate hybrid atomic orbitals&lt;/li&gt;&#xA;&lt;li&gt;second, mix these hybrid atomic orbitals that exist on separate atoms to form a bond, a molecular orbital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But that's not how it really happens.&lt;/strong&gt;  It is important to realize that hybridization only occurs in response to some type of interaction between two atoms.  In other words, only when atoms begin to approach and interact with one another does hybridization start to occur.  When hybridization starts to occur, bonding between the atoms starts to occur at the same time.  To first hybridize and then bond (like I think of it in my mind) is an artificial construct that allows me to sort things out.  &lt;strong&gt;Hybridization on an atom and bonding between two atoms both occur simultaneously.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the s and p orbitals on one atom do not have the correct symmetry to&#xA;  mix effectively (the s orbital would simultaneously constructively&#xA;  interference with one lobe of the p orbital and destructively&#xA;  interfere with the other).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes they do.  As this picture shows we can add and subtract an s and a p atomic orbital to form two $\ce{sp}$ hybridized atomic orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aE7t0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bouman.chem.georgetown.edu/S02/lect12/lect12.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;picture source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in $\ce{O2}$ there is no s-p mixing so why would oxygen mix its s and&#xA;  p orbitals when it's bonding with Carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think they are leaving oxygen unhybridized.  So, for example, they are mixing (adding and subtracting) a carbon $\ce{sp}$ orbital with an oxygen s orbital to form the $\ce{1\sigma}$ and $\ce{2\sigma}$ molecular orbitals in your MO diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you know why s-p mixing raises the energy of the $\ce{2σ_{g}}$ orbital?&#xA;  (it lowers the $\ce{1σ_{g}}$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nice diagrams and a helpful explanation can be found &lt;a href=&quot;https://web.archive.org/web/20170712222509/http://classes.uleth.ca/201103/chem4000b/Transition%20Metal%20Review.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Basically, It has to do with the symmetry of the orbitals, like I mentioned above.  Look at your $\ce{1σ}$ orbital, when you add or subtract it with the $\ce{3σ}$ orbital, the symmetry is such that there is no constructive or destructive overlap. Therefore there is no interaction and there energies will not change.  On the other hand, the symmetries of the $\ce{2σ}$ and $\ce{3σ}$ orbitals are such that they can constructively and destructively overlap.  As a result they interact and their energies change, one is pushed up and one is pushed down.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="33773" LastEditDate="2018-08-03T11:15:43.677" LastActivityDate="2018-08-03T11:15:43.677" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26451" PostTypeId="1" CreationDate="2015-02-21T16:49:40.070" Score="-1" ViewCount="2907" Body="&lt;p&gt;An unknown element &lt;span class=&quot;math-container&quot;&gt;$\ce{Q}$&lt;/span&gt; has two unknown isotopes: &lt;span class=&quot;math-container&quot;&gt;$\ce{^60Q}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^63Q}$&lt;/span&gt;. If the average atomic mass is &lt;span class=&quot;math-container&quot;&gt;$\pu{61.5 u}$&lt;/span&gt;, what are the relative percentages of the isotopes?&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="4945" LastEditDate="2018-10-12T14:00:59.573" LastActivityDate="2018-10-12T14:14:28.113" Title="How to find the relative abundances of isotopes of an unknown element?" Tags="&lt;isotope&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-02-21T19:33:44.163" ContentLicense="CC BY-SA 4.0" />
  <row Id="26452" PostTypeId="1" AcceptedAnswerId="26467" CreationDate="2015-02-21T17:39:24.243" Score="1" ViewCount="377" Body="&lt;p&gt;I want to convert resultant hours of salt spray corrosion test to estimated values in years for real life environments such as marine. In other words, How could I predict the result of real life tests using salt spray data?&lt;/p&gt;&#xA;" OwnerUserId="14557" LastActivityDate="2015-02-23T18:10:16.867" Title="How to use salt spray data in real environments?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26453" PostTypeId="1" AcceptedAnswerId="26458" CreationDate="2015-02-21T17:40:41.883" Score="7" ViewCount="1141" Body="&lt;p&gt;I need to handle large quantities of $\ce{KOH}$; close to 1000 liters of 85 % conc $\ce{KOH}$ for about an hour. Any suggestions on how I go about doing it? Also the temperature during the process might touch 500 degrees Celsius. &lt;/p&gt;&#xA;" OwnerUserId="14558" LastEditorDisplayName="user7951" LastEditDate="2015-11-13T19:08:56.700" LastActivityDate="2017-12-24T20:25:40.107" Title="Storing large quantities of KOH" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;safety&gt;&lt;materials&gt;" AnswerCount="2" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26454" PostTypeId="2" ParentId="26451" CreationDate="2015-02-21T17:41:16.860" Score="0" Body="&lt;p&gt;Solve for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$60x + 63(1-x)=61.5$$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-21T17:41:16.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26455" PostTypeId="1" AcceptedAnswerId="26456" CreationDate="2015-02-21T17:41:36.300" Score="4" ViewCount="2126" Body="&lt;p&gt;Why is cyclopentadiene more reactive than 9-anthracenemethanol when acting as a diene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both molecules are locked in an s-&lt;em&gt;cis&lt;/em&gt; conformation, which is favourable. I'm thinking that it's due to the extensive conjugation in 9-anthracenemethanol?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="4231" LastEditDate="2015-02-21T18:35:37.217" LastActivityDate="2015-02-21T18:35:37.217" Title="Diels-Alder rate of reaction" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26456" PostTypeId="2" ParentId="26455" CreationDate="2015-02-21T17:55:06.117" Score="4" Body="&lt;p&gt;Yes, you are exactly right.  Anthracene is better described by this structure (where there are no fixed single and double bonds, but rather a delocalized aromatic structure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/On22t.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;than a structure like this (where the single and double bonds are fixed and distinct from one another).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h3Bj2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in anthracene, the aromatic character must be disrupted to some degree in order to undergo the Diels-Alder reaction.  In cyclopentadiene, there is no aromaticity to disrupt, the double bonds are basically fixed like in buta-1,3-diene (and, as you point out, locked in the required s-&lt;em&gt;cis&lt;/em&gt; conformation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final note, benzene won't undergo a Diels-Alder under normal conditions because its aromaticity is completely disrupted and lost during the process.  In anthracene the aromaticity remains in two of the rings, a much smaller price to pay.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-21T17:55:06.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26457" PostTypeId="1" CreationDate="2015-02-21T18:21:27.930" Score="1" ViewCount="30" Body="&lt;p&gt;For example, how does a gas stove heat a metal pot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A combustion reaction between the gas and the air takes place, and this releases energy due to new chemical bonds being formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the atoms in the metal pot absorb these photons, which then excites the atoms' electrons and creates kinetic energy within the atoms (thus, heat)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, are the atoms in the combustion reaction just moving around so fast that they're bumping into all the atoms of the metal pot, generating kinetic energy?&lt;/p&gt;&#xA;" OwnerUserId="8028" LastEditorUserId="4356" LastEditDate="2015-02-21T19:28:21.737" LastActivityDate="2015-02-21T19:28:21.737" Title="How do things heat one another?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;combustion&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-02-23T05:34:32.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="26458" PostTypeId="2" ParentId="26453" CreationDate="2015-02-21T18:21:38.490" Score="5" Body="&lt;p&gt;Various stainless steels resist boiling saturated aquous $\ce{KOH}$, such as DIN X6CrMo17 (ASI 434), DIN X5CrNi1911 (ASI 305).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dont't forget about a suitable and sufficiently large reservoir around the tank in case of spills, valve breaks, etc.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-21T18:21:38.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26460" PostTypeId="2" ParentId="26426" CreationDate="2015-02-21T18:46:29.583" Score="1" Body="&lt;p&gt;You forgot one relevant equation:&#xA;$\ce{Cu+ + I- -&amp;gt; CuI}$ which uses up some iodide as copper precipitates with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the fractions is easy. You know that they must add to 3 and that that charge balance must apply. I marked the fractions as $x$ for copper(II) and $y$ for copper (III) and set up this system of equations which is easy to solve: &#xA;$$\begin{align}&#xA;x+y&amp;amp;=3\\&#xA;2x+3y&amp;amp;=3\times\text{average oxidation state}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You multiply the fraction with the charge of the species and the average oxidation state by three because in one mole of the compound you have three moles of copper, so that the equation represents a charge balance.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="23561" LastEditDate="2018-08-10T19:55:40.213" LastActivityDate="2018-08-10T19:55:40.213" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26461" PostTypeId="2" ParentId="26451" CreationDate="2015-02-21T18:46:53.170" Score="1" Body="&lt;p&gt;You can reverse engineer the formula used to calculate the average atomic mass of all isotopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, carbon has two naturally occurring isotopes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lrr}&#xA;\text{Isotope} &amp;amp; &#xA;\text{Isotopic Mass $A$} &amp;amp; &#xA;\text{Abundance $p$} \\\hline&#xA;\ce{^{12}C}:   &amp;amp; \pu{12.000000 u} &amp;amp; 0.98892 \\&#xA;\ce{^{13}C}:   &amp;amp; \pu{13.003354 u} &amp;amp; 0.01108 &#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula to get a weighted average is the sum of the product of the abundances and the isotope mass:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A = \sum\limits_{i=1}^n p_i A_i$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For carbon this is:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0.989 \times 12.000 + 0.0111 \times 13.003 = 12.011$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, we can set the abundance of one isotope to &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, and the other to &lt;span class=&quot;math-container&quot;&gt;$1 - x$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$x = 0.989$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$1 - x = 0.0111$&lt;/span&gt;, &lt;em&gt;OR&lt;/em&gt; if &lt;span class=&quot;math-container&quot;&gt;$x = 0.0111$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$1 - x = 0.989$&lt;/span&gt;.&#xA;Therefore, we can simply set up an algebraic equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$A_1(x_1) + A_2(1 - x_1) = A$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know &lt;span class=&quot;math-container&quot;&gt;$A_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A_2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; in your example, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;60 x + 63(1 - x) &amp;amp;= 61.5\\&#xA;63 - 3 x &amp;amp;= 61.5\\&#xA;x &amp;amp;= \frac{-1.5}{-3} = 0.5&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the element with a mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{60 u}$&lt;/span&gt; has a &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; abundance, and the element with the mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{63 u}$&lt;/span&gt; has also a &lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; abundance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proof: &lt;span class=&quot;math-container&quot;&gt;$\pu{60 u}(0.5) + \ce{63 u}(0.5) = \pu{61.5 u}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="8028" LastEditorUserId="4945" LastEditDate="2018-10-12T14:14:28.113" LastActivityDate="2018-10-12T14:14:28.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26462" PostTypeId="1" AcceptedAnswerId="26464" CreationDate="2015-02-21T18:47:16.770" Score="0" ViewCount="11924" Body="&lt;p&gt;Why does the basicity of Group 15 hydrides ($\ce{NH_3,~ PH_3,~ etc}$) decreases down the group ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that's because of its Lewis Base nature due to the lone pair but wouldn't the more EN element be less reluctant to give the lone pair ?&lt;br&gt;&#xA; And nitrogen is most EN in that group .&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="23004" LastEditDate="2015-12-08T17:46:05.797" LastActivityDate="2015-12-08T17:46:05.797" Title="Basicity of Group 15 hydrides" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;periodic-trends&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ClosedDate="2015-02-21T23:37:54.433" ContentLicense="CC BY-SA 3.0" />
  <row Id="26464" PostTypeId="2" ParentId="26462" CreationDate="2015-02-21T19:14:02.160" Score="3" Body="&lt;p&gt;It can be explained by the very helpful &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent%27s_rule&quot; rel=&quot;nofollow&quot;&gt;Bent's rule&lt;/a&gt;. As nitrogen is more electronegative than phosphorus, there is a bigger s character of the sp3 hybrid orbitals that are used to form the bond to hydrogen than the ones of phosphorus. As a result, the sp3 hybrid orbital that holds the electron pair, has a bigger p character for nitrogen than for phosphorus, meaning that the pair is more distant from the nucleus and more susceptible to donation.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-02-21T19:14:02.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26465" PostTypeId="1" CreationDate="2015-02-21T19:38:07.210" Score="1" ViewCount="1506" Body="&lt;p&gt;Is it possible to create a container for helium that it can't permeate through, like a permanent balloon? Or, if not, what kind of material would have the lowest permeability for that application? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't just mean stuff that we could build today either, for instance, could it permeate through diamond? Like if we had molecular manufacturing or something, in the future, and built a box out of diamond, of the appropriate dimensions (like total mass vs volume inside) so that it would float, and constructed it so that there were no openings...well, say we build it almost all the way and then stop, and there's a hole in the top, and we drop some solid, frozen helium in, the amount we would need it to be filled with, and then built the rest of the box, sealing it in (all in a vacuum chamber of course, since that's how those hypothetical fabricators are typically depicted).  Would that be permeable? Atomically precise diamond lattice or whatever, would that be permeable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothetically speaking...&lt;/p&gt;&#xA;" OwnerUserId="14563" LastActivityDate="2015-02-21T22:03:21.873" Title="Permanent Balloon?" Tags="&lt;synthesis&gt;&lt;gas-laws&gt;&lt;materials&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26466" PostTypeId="2" ParentId="26465" CreationDate="2015-02-21T19:51:52.030" Score="5" Body="&lt;p&gt;Yes, this is feasible in practice for a fair definition of &quot;permanent&quot;. Monolayers (sheets a single atom thick) of graphene are impermeable to any gas, including helium. To be more precise, the bulk permeability rate for helium through graphene is very low, of the order of $10^{-15}\ mm\  g\ cm^{-2}\ s^{-1}\ bar^{-1}$. For comparison, it means a stack of graphene layers $100\ \mu m$ thick (about the thickness of paper or a human hair) will leak less helium than a $10\ cm$ thick slab of glass  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the original article which investigated gas retention of graphene monolayers specifically mentioned making &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/nl801457b&quot; rel=&quot;nofollow&quot;&gt;microscopic balloons, including helium balloons&lt;/a&gt;. I don't see why a macroscopic balloon wouldn't work, other than the difficulty involved in preparing a large flawless sheet of the material. There is currently a ton of research behind large scale production of graphene, so something of the sort may actually happen eventually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more interestingly, a closely related yet quite distinct material called graphene oxide has been shown to also create films which are essentially impermeable to helium, just like graphene, but &lt;a href=&quot;http://www.sciencemag.org/content/335/6067/442&quot; rel=&quot;nofollow&quot;&gt;water passes through the graphene oxide film unimpeded&lt;/a&gt;, as if there were no barrier at all!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-02-21T22:03:21.873" LastActivityDate="2015-02-21T22:03:21.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26467" PostTypeId="2" ParentId="26452" CreationDate="2015-02-21T19:56:31.273" Score="1" Body="&lt;p&gt;&lt;em&gt;Note that I'm not an expert in this field and wikipedia itself is not a scientifically relevant primary source&lt;/em&gt;. The latter might however be different for sources cited there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short, you seemingly better shouldn't.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the German wikipedia article on the &lt;a href=&quot;http://de.wikipedia.org/wiki/Salzspr%C3%BChtest&quot; rel=&quot;nofollow&quot;&gt;salt spray test&lt;/a&gt; has collected a lot of critical remarks from different sources. In summary, standardized (ASTM B117 or DIN EN ISO 9227) salt spray tests may be used in quality control, but &lt;strong&gt;cannot&lt;/strong&gt; be transfered to to real world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the wikipedia article, DIN EN ISO 9227 explicitly states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nur selten besteht ein direkter Zusammenhang zwischen der Beständigkeit gegen die Einwirkung von Salzsprühnebel und der Beständigkeit gegen Korrosion in anderen Medien. Die verschiedenen Faktoren, welche das Fortschreiten der Korrosion beeinflussen, können sich je nach den herrschenden Bedingungen sehr unterschiedlich auswirken. Dazu gehört z. B. auch die Bildung von Schutzschichten. Die Prüfergebnisse sollten deshalb nicht als direkter Hinweis auf die Korrosionsbeständigkeit der geprüften metallischen Werkstoffe in allen Umgebungsbedingungen betrachtet werden, in denen diese Werkstoffe verwendet werden können.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd roughly translate this to: &lt;em&gt;Only rarely, a direct correlation between the resistance towards saline fog and the corrosive influence of other media can be found. Depending on the conditions, different factors promoting corrosion may have a distinctively different impact. This includes, e.g., the buildup of protective layers.&#xA;Consequently, testing results should not be taken as a direct hint on the general corrosion stability of metallic materials under all planned conditions.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-21T19:56:31.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26469" PostTypeId="1" AcceptedAnswerId="26482" CreationDate="2015-02-21T20:54:50.330" Score="10" ViewCount="559" Body="&lt;p&gt;I'm new to computational chemistry, so perhaps this question has an obvious answer. I'm wondering what the intuitive reason for constructing Slater determinants that involve &lt;em&gt;excited states&lt;/em&gt; is, when one tries to describe correlations, starting from the Hartree-Fock picture. I understand that the single determinant in the Hartree-Fock model only describes exchange (antisymmetry of fermionic wave functions). &#xA;I am trying to understand the logic behind why constructing Slater determinants involving excited states can give us a picture of correlation. I guess I don't understand the relation between excited states and correlations (especially thinking about it from the point of view of DFT, which is interested only in the ground state, but correlations are still important). &lt;/p&gt;&#xA;" OwnerUserId="14566" LastEditorUserId="4945" LastEditDate="2020-06-02T09:51:21.463" LastActivityDate="2020-06-02T09:57:58.277" Title="What is the reason for including excited states in the wave function in post-Hartree Fock methods?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;ab-initio&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26471" PostTypeId="2" ParentId="26453" CreationDate="2015-02-21T22:32:49.060" Score="7" Body="&lt;p&gt;The first thing to clear up is – Why do you need to store 1000 L of KOH in liquid form? It is as others have pointed out, very concerning that someone who is planning to store this quantity needs to ask on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice is don't store 1000 L of liquid KOH, store 40 × 25 L of liquid KOH. This will reduce the risk considerably, and help ensure that a single leak isn't a catastrophic event, but is something you can deal with. This is a common philosophy, and is used, for example, in trailers of dangerous gases, which are organized into racks of 20 smaller gas bottles instead of one large tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep your 25 L containers off the floor in a bunded area for spill containment, and have appropriate means to deal with a spill. I'm not going to tell you the best way to handle a spill because I don't know. – You need to find out. Also, make sure the store is properly monitored to detect any problems.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, going back to my initial point: Let's accept that, for whatever reason, you need to store 1000 L of KOH (as you haven't shared with us what you are doing.) Solid KOH is much safer than liquid KOH, so why do you even need to mix up the solution all at once? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the causes of the Bhopal disaster (the worst industrial disaster ever, which killed many thousands of people due an uncontrolled venting event) was the unnecessary storage of a toxic and volatile compound (methyl isocyanate.) If the pesticide they were making had been synthesized via a different route, or even if the methyl isocyanate had been made only in the quantities needed, the size of the disaster could have been reduced or even eliminated (despite the extremely poor state of maintenance of the plant.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you mix up the KOH in small batches as you need it to reduce the risk&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the temperature in the process might touch 500 degrees Celsius&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By &quot;process&quot; I assume you mean mixing solid KOH with water, and that you do not actually need to use the KOH at this temperature. If you expect to reach these temperatures, you are doing it wrong. As can be seen from &lt;a href=&quot;https://koh.olinchloralkali.com/TechnicalInformation/Phase%20Diagram%200-80.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://koh.olinchloralkali.com/TechnicalInformation/Phase%20Diagram%200-80.pdf&lt;/a&gt; KOH solution in all compositions from 0 to 85% is liquid at below 150 °C. Note also that concentrations above 50% are solid at room temperature – so do you really mean 85% conc KOH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safe way to do this is to add the dangerous material to the safe material, particularly as water has such a high heat capacity and can lose heat by boiling. You are going to need either a lot of patience, or mechanical screw feeder, to perform the addition at a safe rate. You will also need decent stirring equipment. Also, I would expect most of the heat will be released in mixing the first 50% of the KOH, because after that you are effectively mixing KOH with KOH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if you are planning to do this, you should check the thermochemistry yourself and include it in your Risk Assessment, along with any relevant details from industry experience regarding specific hazards of mixing (spattering for example?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really hope you make sure what you doing before you do this. Alkalis are a lot worse on human flesh than acids, partly because they turn the oils in your skin to soap, and partly because the spills won't evaporate and therefore persist until something is done about them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a good idea to dye your KOH so you know where any spills are. I personally have spilt 50L of antifreeze solution in the lab (it was intentionally dyed beforehand in case of spills, and needless to say we used propylene glycol as it is much less toxic than the ethylene glycol used in car radiators.) Whatever you intend, something in your experiment will go wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also check the legal aspects in your area. Also, consult with your supplier! The fact that you need to ask on the internet instead of talking to them suggests you are up to something unusual. Would you care to share with us what it is?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT - Now you've clarified you're an engineer, one more very important point. Keep KOH away from aluminium. Alkali will dissolve the oxide layer that protects aluminium from corrosion, and unmask the true (very high) reactivity of the metal. A lot of heat and hydrogen gas will be generated by the reaction $\ce{3 KOH + Al -&amp;gt; K3AlO3 + 2/3 H2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing most stainless steels should be fine (but obviously check.) I have used equipment from the catering industry in some experiments, but stay away from stainless steel cookware that has the base thickened with aluminium. Even though it is external, it's not worth the risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that mixtures of acids and bases with water are often most corrosive at a medium concentration, so don't assume that just because a material is OK at 90% concentration, it's going to be OK at 50% concentration.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorDisplayName="user7951" LastEditDate="2017-12-24T20:25:40.107" LastActivityDate="2017-12-24T20:25:40.107" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="26474" PostTypeId="1" AcceptedAnswerId="26488" CreationDate="2015-02-22T00:53:33.583" Score="2" ViewCount="937" Body="&lt;p&gt;Please consider this picture of a peptide chain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JNMa1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some examining on the different amino acids and concluded that the chain is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lys-Leu-Gly-Ser-Citrullin (variant of arginine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would now like to calculate the charge of this peptide at neutral pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using this table to orient myself: &lt;a href=&quot;http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my train of thought:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we're at neutral pH the +1 and -1 of the mandatory amino and acid groups will cancel out for all amino acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lysine has a pKx of barely 11 which is some higher than 7 so it is basic in our solution. It means that it will accept a proton so Lysine has a charge of +1 in our respective solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leucine, Glycine and Serine have no charge on their x group so their charge is 0 and does not contribute anything to the total charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citrullin, I read, is uncharged at neutral pH, opposed to Arginine which is, so it does not contribute either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my total answer is +1. Is this correct? I am in no way 100% sure of my method but it would be great to get a confirmation that I am doing it correctly or to be corrected on my possibly faulty method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4984" LastActivityDate="2019-10-23T03:16:45.030" Title="Calculating charge of amino acids (am I doing it right?)" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26475" PostTypeId="2" ParentId="5323" CreationDate="2015-02-22T01:22:36.980" Score="2" Body="&lt;p&gt;Formic acid is thermodynamically unstable to decomposition to water and carbon monoxide: HCOOH ==&gt; CO + H2O. I assume this is what you are referring to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common use is acidification of silage to prevent smells. The fact that it decomposes makes it more environmentally friendly than alternatives. I was not aware that it was used in bee keeping to control varroa parasitic mites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lowering the temperature would reduce the decomposition rate from both a thermodynamic and kinetic point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MSDS I have looked at indicate that the bottle should be vented periodically to release the pressure built up from decomposition. However this one also says that formic acid should not be stored below its freezing point of 8C. &lt;a href=&quot;http://ionics.ca/wp-content/uploads/MSDS-Formic-Acid.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ionics.ca/wp-content/uploads/MSDS-Formic-Acid.pdf&lt;/a&gt; Whether this is just for convenience of use or for reasons or safety I do not know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, the build up of pressure seems far more of an issue than any loss of strength in the acid. In order to lose 50% of its strength, a 1L bottle of formic acid (1200g/46=26mol) would have to lose 13mol (about 312L) of carbon monoxide. That's a lot of venting! I would think a bottle should have a shelf life of 2 or 3 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did my university chemical engineering project on a formic acid plant. It is made as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. CO + Methanol (20 bar, 80C, potassium methoxide catalyst) == Methyl Formate &#xA;2. Methyl Formate + H2O (acidic autocatylisis)== Formic acid + Methanol&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From this you might conclude that formic acid is stable to decomposition at 20 bar, but that is not the case, because excess reactants and careful use of distillation are used to push both reactions in the desired direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was a long time ago, but I will see if I can find any relevant data in the decomposition reaction. However you would probably be best off just following the manufacturer's recommendations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and keep it in the original glass bottle. Many metals catalyse the decomposition. &lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="5541" LastEditDate="2015-02-22T01:32:53.333" LastActivityDate="2015-02-22T01:32:53.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26476" PostTypeId="1" AcceptedAnswerId="26491" CreationDate="2015-02-22T01:55:16.790" Score="3" ViewCount="10931" Body="&lt;p&gt;According to the &lt;em&gt;solutions manual&lt;/em&gt;, the oxidation of acetophenone to benzoic acid can be achieved through addition of chromic acid and application of heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions manual refers me to a section in the &lt;em&gt;book&lt;/em&gt; which explains that when a benzylic hydrogen is present, a strong oxidizing agent combined with heat can perform oxidative cleavage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is that I don't see any benzylic hydrogens on acetophenone. Neither does its enol tautomer have any benzylic hydrogens ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is the oxidation of acetphenone to benzoic acid possible given what the book and solutions manual has told me? &lt;strong&gt;Self-consistency aside is this even a possible reaction?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-02-22T19:00:14.680" LastActivityDate="2016-10-20T16:34:37.203" Title="Oxidation of acetophenone to benzoic acid" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26477" PostTypeId="2" ParentId="26476" CreationDate="2015-02-22T02:53:50.157" Score="3" Body="&lt;p&gt;I'd like to clear up one issue that has not been addressed in comments, but provides a weak answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are NO benzylic hydrogens in acetophenone.  So it is not possible based on what the textbook tells you.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-02-22T02:53:50.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26479" PostTypeId="1" CreationDate="2015-02-22T04:56:15.143" Score="3" ViewCount="1330" Body="&lt;p&gt;I heard somewhere (I think is was a show called &quot;How The Universe Works&quot;) that the big bang was the result of a hydrogen atom and a helium atom. Is this true? If so, where did all the other elements come from?&lt;/p&gt;&#xA;" OwnerUserId="14545" LastActivityDate="2015-02-22T08:02:14.013" Title="Where did all the elements come from?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-02-23T05:33:58.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="26480" PostTypeId="1" CreationDate="2015-02-22T05:03:36.703" Score="2" ViewCount="415" Body="&lt;p&gt;I am trying to calculate the expected volume of an aqueous solution of epsomite and NaCl (308 deg K but willing to go with 298 K as close enough, and 1 atmosphere pressure).  I realize the two may interact, but as a first approximation I'm looking for the partial molar volumes of both. I found NaCl's in a paper (Archer (1992) J Phys Chem Ref Data 21(4): 793-829), but I can't find the values for epsomite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a known quantity of salts and am trying to figure out how much solution I'll end up with after adding enough water to get a particular overall specific gravity. For the curious, I've got 600 pounds of epsomite, 50 pounds of NaCl, and I'm shooting for a specific gravity of 1.25.  Knowing the amount of water to add is straightforward and, unfortunately, not sufficient for this particular situation. I really need to calculate the total final volume after the salts have dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="14571" LastActivityDate="2015-03-05T12:03:09.063" Title="What is the partial molar volume of epsomite (magnesium sulfate heptahydrate)?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26482" PostTypeId="2" ParentId="26469" CreationDate="2015-02-22T06:54:54.940" Score="9" Body="&lt;p&gt;I think that here it is not accurate to say &quot;starting from the Hartree-Fock picture&quot;, instead of it, I find better to say &quot;starting from the Hartree-Fock method&quot;. When this method is carried out we get a lot of orbitals and that allow us to construct a lot of Slater determinants, even if, we can be only interested in one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a complete set of functions &lt;span class=&quot;math-container&quot;&gt;$\{f_i(x)\}$&lt;/span&gt;, any function &lt;span class=&quot;math-container&quot;&gt;$g(x)$&lt;/span&gt; can be thought as a linear combination&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(x) = \sum c_i f_i(x).$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a two variable function &lt;span class=&quot;math-container&quot;&gt;$g(x_1, x_2)$&lt;/span&gt;, fixing &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt; we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$g(x_1, x_2) = \sum c_i(x_2)f_i(x_1).$$&lt;/span&gt;&#xA;But as &lt;span class=&quot;math-container&quot;&gt;$\{f\}$&lt;/span&gt; is a complete set:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;c_i(x_2) &amp;amp;= \sum d_{ij} f_j(x_2) &amp;amp;\text{and}&amp;amp;&amp;amp;&#xA;g(x_1,x_2) &amp;amp;= \sum d_{ij}f_i(x_1)f_j(x_2).&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reasoning is trivially extended to a function of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-fermions system, the wave function depends of &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; spin-orbitals (&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; variables). To get a mathematic valid form, we only need to antisymmetrize the generalization of the expressions above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take in mind that none of these determinants are the true state (they are not eigenfunctions of the true hamiltonian). The same happens with the orbitals (that are eigenvectors of the Fock operator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the use of excited determinants arises naturally from a math view point. Just because we already have functions for the determinantal expansion, and we know that they are pretty good in some sense, because the are the best we can get at least for one determinant (the Hartree Fock solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not needed to think configuration interaction as a mixture of stationary quantum states (ground and excited states), as said above they truly aren't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Direct to your question:&lt;/em&gt; We can think the problem just as a decomposition of the wave function, somewhat like Fourier series, where the functions involved are the Slater determinants, just because it is convenient. But, this doesn't imply a physical meaning. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Response to comment:&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also this discussion is quite old, I would like to add that the term &quot;picture change&quot; is frequently used in relativistic quantum chemistry, when (effective) Hamiltonians are compared to each other. E.g., when approximating the four-component (4c) Dirac equation by an effective two-component (2c) formalism, one usually introduces further approximations. In this respect, &quot;starting from the Hartree-Fock picture&quot; is a correct terminology, as it implies certain approximations on the Hamiltonian. However, the better way is usually to end up with a certain picture, compared to starting from it... (&lt;a href=&quot;https://chemistry.stackexchange.com/users/42616/thefox&quot;&gt;TheFox&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree: By choosing an effective Hamiltonian one is working on a class of approximations that build up a picture for a set of analogous systems. When considering correlation we have in mind a &quot;picture&quot; of the system. Being that H-F is a mean field method, it belong to some picture, but I feel that when dealing with Coupled Cluster the H-F method is just a part of a bigger procedure consisting in finding some functions in order to obtain a solution for another picture. In such sense, while the H-F method itself can be understood as the method for obtaining a solution of the H-F picture, in this context is just part of method for solving a problem in another picture. I added the word &lt;em&gt;here&lt;/em&gt; for strictness.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="4945" LastEditDate="2020-06-02T09:57:58.277" LastActivityDate="2020-06-02T09:57:58.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26483" PostTypeId="1" AcceptedAnswerId="26515" CreationDate="2015-02-22T07:32:24.557" Score="5" ViewCount="12678" Body="&lt;p&gt;I was reading about concentration terms like molarity, molality, etc. and then I came across a question: which concentration is more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1M(molar) of solute x dissolved in water vs.&lt;/li&gt;&#xA;&lt;li&gt;1m(molal) of solute x dissolved in water &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When I Google it, I did not find exact answers but did find some related information.&#xA;My approach is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Convert 1M into molality and then compare if it is less or more than 1m. (?)&lt;/li&gt;&#xA;&lt;li&gt;The density of water is 1kg per L.&lt;/li&gt;&#xA;&lt;li&gt;When solute is added to it, density would definitely increase(?) &lt;/li&gt;&#xA;&lt;li&gt;And therefore 1L of that solution will weigh more than 1 kg.&lt;/li&gt;&#xA;&lt;li&gt;now if we calculate molality of 1M solution it would be less than 1m . (?) &lt;/li&gt;&#xA;&lt;li&gt;am I correct ??&lt;/li&gt;&#xA;&lt;li&gt;Density would always(?) Increase after adding solute &lt;/li&gt;&#xA;&lt;li&gt;**Hence 1m will always be more than 1M **&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14530" LastEditorUserId="13184" LastEditDate="2015-02-23T07:04:31.857" LastActivityDate="2017-02-24T14:07:46.717" Title="Whose concentration is more 1 molar or 1 molal?" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26484" PostTypeId="2" ParentId="26479" CreationDate="2015-02-22T07:59:13.283" Score="2" Body="&lt;p&gt;According to what i have heard and read, the after the BIG BANG, the first element was&#xA;$\ce{H}$ which fused to form $\ce{He}$ and after that the first compound was $\ce{H_3^+}$ and $\ce{HeH^+}$. Then these elements underwent fission and heavier elements were formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then stars were formed inside which we have super HOT cores where still heavier elements were formed. The highest being $\ce{Fe}$ . After wards after &lt;em&gt;Supernovae&lt;/em&gt; these elements got kind of thrown of and also planets were formed from Stars and hence we today have all these elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other elements like the &lt;em&gt;Actinoids&lt;/em&gt; were man made by nuclear reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a little!&lt;/p&gt;&#xA;" OwnerUserId="7434" LastActivityDate="2015-02-22T07:59:13.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26485" PostTypeId="2" ParentId="26479" CreationDate="2015-02-22T08:02:14.013" Score="9" Body="&lt;p&gt;This is a very broad question, so below I mention just the very key points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Big Bang&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;According to the prevailing cosmological model for the early development of the Universe, known as the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang&quot;&gt;Big Bang theory&lt;/a&gt;&lt;/em&gt;, the Universe begun at about 14 billion years from a singularity that can be thought of a state of an infinite density and temperature, and since its creation the Universe began to expand and cool.&#xA;Right after the Big Bang the Universe was immensely hot, the energy density was incredibly high and no particles existed, but the Universe was extremely rapidly expanding and cooling, so that soon enough energy was allowed to be converted into various elementary particles: quarks, leptons, bosons.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Baryogenesis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is generally assumed that initially there was an equal numbers of particles and corresponding antiparticles, but at some point an unknown physical process referred to as &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Baryogenesis&quot;&gt;baryogenesis&lt;/a&gt;&lt;/em&gt; lead to a very small excess of quarks and leptons over antiquarks and antileptons.&#xA;As the Universe continued to decrease in density and fall in temperature quarks and gluons combined to form baryons such as protons and neutrons, and the small excess of quarks over antiquarks led to a small excess of baryons over antibaryons, which in turn after mass annihilation resulted in the predominance of matter over antimatter in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Big Bang nucleosynthesis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After annihilation it was just the matter of time for the Universe to cool down to a point where remaining protons, neutrons and electrons could combine to form atoms.&#xA;Already a few minutes after the Big Bang neutrons combined with protons to form the lightest nuclei (H-1, H-2 or D, H-3, He-3, He-4, Li-6, Li-7, Be-7, and Be-8) in a process called the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis&quot;&gt;Big Bang nucleosynthesis&lt;/a&gt;&lt;/em&gt;, and roughly 378,000 years after the Big Bang the Universe cooled to the point when the formation of light neutral atoms (mostly H and He with trace amounts of Li) became energetically favored.&#xA;Eventually, most of the protons in the Universe were bound up in these neutral lightest atoms and matter was uniformly distributed in the Universe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stellar nucleosynthesis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;But the gravitational attraction of the nearby matter by slightly denser regions and the subsequent growth of such regions resulted in formation of gas clouds, stars, galaxies, and other large structures.&#xA;The formation of stars meant that the extreme conditions (temperatures and pressures) for the nuclear fusion were recreated and light elements (H, He and Li) from which stars were formed started to turn into heavier one by a process referred to as the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stellar_nucleosynthesis&quot;&gt;stellar nucleosynthesis&lt;/a&gt;&lt;/em&gt;. Stellar nucleosynthesis has its endpoint, namely, when nuclear fusion reaches nickel-56 which then undergoes radioactive decay into iron-56 the process stops. The reason is that iron-56 has one of the highest binding energies of all of the isotopes, and thus, no more energy can be extracted from an iron-56 nucleus, either by fusion or by fission. But it is not the end of the story.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Supernova nucleosynthesis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When the core of a star converts to iron-56, the nuclear reactions stop. And that is a big problem, since so far the star balanced the forces of gravity pulling its materials inward by nuclear reactions pushing them outward, but now all the reactions pushing outward stop. What happens then depends on how massive the star is. &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supergiant&quot;&gt;Supergiants&lt;/a&gt;&lt;/em&gt;, that have masses from 8 to 12 times the Sun, gravitationally collapse in &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Supernova&quot;&gt;supernova&lt;/a&gt;&lt;/em&gt; explosions accompanied by a large burst of neutrons. These neutrons are captured by heavy nuclei, such as iron-56, and the resulting neutron-saturated isotopes then undergo beta decays producing stable isotopes of elements heavier than iron.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-02-22T08:02:14.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26486" PostTypeId="1" CreationDate="2015-02-22T09:00:27.250" Score="1" ViewCount="14458" Body="&lt;p&gt;A salesman came to my house to demonstrate his water purifier machine which is a RO Purifier. To show the difference between his purifier machine and the machine I own, he took some water from my machine in a glass. Then he pour water from a bottle he bought in another glass claiming that this was a purified water from his machine (little suspicious). Then he put a small electrolysis machine (that one we used for experiments in school) on both glass and started it. After a minute, there was lot of uhh... particles looking like mud floating in my water, while in his glass, there were very few in dissolved state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he said that those particles were salts and bad impurities which my purifier can not filter out, and his purifier can, thus having few particles in his glass.&#xA;So my question is, can this method detect the impurities in water, or is it just some dirty sales tactic? Because as I remember, i was taught in school that electrolysis does not has to do anything with water, it just transfer cathode to anode or something like that (sorry for my bad chemistry knowledge :P)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me out!&lt;/p&gt;&#xA;" OwnerUserId="14574" LastEditorUserId="21964" LastEditDate="2016-07-25T19:24:51.373" LastActivityDate="2017-09-05T09:00:35.010" Title="Can water electrolysis detect the water impurities?" Tags="&lt;water&gt;&lt;electrolysis&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26487" PostTypeId="2" ParentId="26486" CreationDate="2015-02-22T09:55:48.400" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] can this method detect the impurities in water [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In principle, yes. The electrolysis can, as an example oxidize iron ions ($\ce{Fe^2+ -&amp;gt; Fe^3+}$) The latter forms a brownish, voluminous oxide hydroxide, $\ce{FeO(OH)·n H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WHO suggests an upper limit of 0,2 mg/l iron in drinking water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right that your drinking water comes from a well? Did you perform any analyses on the the water quality or consult local authorities concerning possible health risks and their prevention?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] or is it just some dirty sales tactic [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't comment on the dirty part, but salesmen need to sell. We don't know whether the reference sample was actually produced by reverse osmosis or just a sample of distilled water (which you would not want to drink over a longer time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if reverse osmosis can reduce the mineral content in your drinking water, you might want to check whether the additional costs for operation are really worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, try to obtain some information from local authorities.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-22T09:55:48.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26488" PostTypeId="2" ParentId="26474" CreationDate="2015-02-22T13:32:29.263" Score="1" Body="&lt;p&gt;I second Lighthart's &quot;Looks good to me&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's break out the &lt;a href=&quot;http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch equation&lt;/a&gt; to be sure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{pH}=\ce{p}K_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$&#xA;$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K_a$$&#xA;$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K_a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{pH}&amp;gt;\ce{p}K_a$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the pH is greater than the pKa, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{pH}-\ce{p}K_a&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\ce{pH}-\ce{p}K_a}&amp;gt;1$$&#xA;$$\dfrac{[\ce{A-}]}{[\ce{HA}]}&amp;gt;1$$&#xA;$$[\ce{A-}]&amp;gt;[\ce{HA}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For OH and SH functional groups in proteins, when $\ce{pH}&amp;gt;\ce{p}K_a$, then the group is deprotonated and has a negative charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}&amp;gt;\ce{p}K_a$, they are not protonated and neutral.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{pH}&amp;lt;\ce{p}K_a$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the pH is less than the pKa, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{pH}-\ce{p}K_a&amp;lt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e^{\ce{pH}-\ce{p}K_a}&amp;lt;1$$&#xA;$$\dfrac{[\ce{A-}]}{[\ce{HA}]}&amp;lt;1$$&#xA;$$[\ce{A-}]&amp;lt;[\ce{HA}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For OH and SH functional groups in proteins, when $\ce{pH}&amp;lt;\ce{p}K_a$, then the group is protonated and neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For amino groups at $\ce{pH}&amp;lt;\ce{p}K_a$, they are protonated and positively charged.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{pH}=\ce{p}K_a$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case you will find that:&#xA;$$[\ce{A-}]=[\ce{HA}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half of the groups are protonated and the other half are not.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-02-22T13:32:29.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26491" PostTypeId="2" ParentId="26476" CreationDate="2015-02-22T15:23:21.073" Score="5" Body="&lt;p&gt;From a SciFinder search, there are ~10 examples of oxidations of aryl methyl ketones resulting in a carboxylic acid using a chromium based reagent (usually dichromate but also chromium trioxide). This suggests the reaction is at least possible, although I think including it in a textbook is unwarranted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reaction slipped through my search that is informative (below). Notice that in the presence of a benzylic position that does contain a hydrogen, the methyl ketone is unchanged, while the other group is oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zB5FK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken together, it seems that chromium oxidation of an acetophenone to a carboxylic acid is slower than alkyl benzylic oxidation.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-02-22T15:23:21.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26492" PostTypeId="1" AcceptedAnswerId="26495" CreationDate="2015-02-22T15:35:04.120" Score="10" ViewCount="289" Body="&lt;p&gt;Why does $\ce{PCl_5}$ exists in the solid state as $\ce{[PCl_6]^{-}[PCl_4]^{+}}$? What is the problem with it that it doesn't exist in its original form in solid state? Like why isn't it just $\ce{PCl_5}$ in solid state as well?&lt;/p&gt;&#xA;" OwnerUserId="3771" LastEditorUserId="4231" LastEditDate="2015-02-22T17:12:49.350" LastActivityDate="2015-02-23T17:01:27.917" Title="What is the actual reason behind PCl₅ existing as an ionic solid?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26493" PostTypeId="1" CreationDate="2015-02-22T16:10:21.430" Score="2" ViewCount="767" Body="&lt;p&gt;I wish to find a pigment that can indicate when a battery contact is getting too hot, and all the thermochromic pigments from hobby stores change color below body temperature, at 31 °C is the highest I found. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kinds of thermochromic pigment can be used for electronic components, that would change color at 40–50 °C / 120–140 °F? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing would be easily obtainable and would give an rough indication of temperature transitions from 30 to 55 °C&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorDisplayName="user7951" LastEditDate="2015-02-22T16:40:49.507" LastActivityDate="2015-02-22T22:38:13.373" Title="Thermochromic pigment that changes at around 50 °C / 120 °F?" Tags="&lt;temperature&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26494" PostTypeId="2" ParentId="26493" CreationDate="2015-02-22T16:44:03.567" Score="1" Body="&lt;p&gt;The easiest way to obtain such a pigment is to mix a blue and red thermochromic pigment together, one that changes at 32 and another that changes at 50 dregrees. They aren't very easy to find, for the moment i have found the 32 degrees pigments on ebay, and a 45 degree pigment on aliexpress:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQrpC.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-02-22T22:38:13.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26495" PostTypeId="2" ParentId="26492" CreationDate="2015-02-22T17:46:32.800" Score="9" Body="&lt;p&gt;In the gas phase ions cannot be stabilized through solvation or lattice forces so ions are destabilized relative to non-ionic entities in the gas phase.  $\ce{NaCl}$ is a case in point, in the gas phase it exists as &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state/14177#14177&quot;&gt;an equilibrium of monomers and dimers&lt;/a&gt;.  Yet in the solid phase it exists as a highly ionic crystal.  This example demonstrates that going from a gas to a solid can dramatically affect the relative stabilities of ionic and covalent structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true in the case of $\ce{PCl_5}$.  In the gas phase it exists as a covalent compound with a trigonal bipyramid structure.  However, in the solid phase, it turns out that enthalpic affects favor the ionic tetrahedral/octahedral structure $\ce{[PCl4]^{+}[PCl6]^{-}}$.  Simply said, the ionic structure turns out to be thermodynamically more stable in the solid phase where positive and negative charges can be arranged to stabilize the entire lattice; whereas in the gas phase the positive and negative ions cannot be stabilized so the covalent form is favored since it is now lower in energy than the ionic form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason why the ionic form is preferred in the solid phase is that packing regular geometric shapes such as tetrahedra and octahedra together allows for a more stable arrangement since they can be packed closer together and in a more ordered arrangement than non-regular (kind of an odd shape) trigonal bipyramids can be arranged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment on ion stabilization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we generate an ion in the gas phase, it will have a certain energy associated with it. If we put that ion in a solution its energy will be much less.  We say the ion has been stabilized by placing it in solution.  Some solvents will stabilize the ion more than other solvents.  This relates to the dielectric constant of the solvent which is a measure of the solvent's polarity.  The higher the dielectric constant for a solvent, the more polar the solvent and the better its ability to stabilize and lower the energy of an ion dissolved in it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture showing a sodium ion dissolved in water.  Note how the negative ends of the dipoles in the water (the negatively polarized oxygen end of the water molecule) molecules congregate and orient themselves around the positive ion.  This orientation of the negative ends of multiple water molecules around a positive sodium ion is what stabilizes the ion in solution.  There is no such stabilization in the gas phase.  In the gas phase it is just a &quot;naked&quot; ion without any external stabilization, so in the gas phase an ion will have a higher energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwiIm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/enrgtop.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt; see the &quot;Solvation od Ions&quot; section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another picture showing showing (on the right) sodium chloride in a crystal lattice.  The sodium and chloride ions pack around each other to perform the same function as a solvent.  The opposite charges arrange themselves around each other to minimize electrostatic repulsion between like ions and maximize the stabilizing electrostatic attraction between oppositely charged ions.  Through these types of arrangements, the &quot;lattice&quot; can stabilize ions much like a solvent can stabilize ions.  Consequently, both solvents and crystal lattices stabilize ions and lower their energy compared to the ion's energy in the gas phase where no such stabilization exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RrHFt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/Enlist/Labs/LiquidSol/LiquidSol.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-02-23T17:01:27.917" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="26496" PostTypeId="1" AcceptedAnswerId="26498" CreationDate="2015-02-22T18:17:36.267" Score="2" ViewCount="1127" Body="&lt;p&gt;I always use &lt;strong&gt;GAMESS&lt;/strong&gt; and &lt;strong&gt;Avogadro&lt;/strong&gt; on my own laptop. Recently I installed them on our university supercomputer and started using them by remote logging-in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the laptop everything was super easy but unfortunately I can not find any documentation for using them from &lt;strong&gt;Terminal Command Line&lt;/strong&gt; ( I mean sending a job and receiving the result all from Terminal Command Line ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone knows a good documentation or tutorial for this subject.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastActivityDate="2015-02-22T20:04:36.937" Title="How to Run GAMESS and Avogadro on Command Line?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26497" PostTypeId="1" CreationDate="2015-02-22T19:51:01.147" Score="4" ViewCount="4213" Body="&lt;p&gt;After Osmium tetroxide adds to a double bond and is then hydrolysed the following species is formed:&lt;img src=&quot;https://i.stack.imgur.com/q5KbU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because Osmium tetroxide is toxic, NMO is used to transform the above back to Osmium tetroxide. NMO is:&lt;img src=&quot;https://i.stack.imgur.com/I1La8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism for this?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-10-19T04:52:49.507" Title="How does NMO reform Osmium tetroxide after reaction with an alkene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26498" PostTypeId="2" ParentId="26496" CreationDate="2015-02-22T20:04:36.937" Score="8" Body="&lt;p&gt;&lt;strong&gt;Contact the HPC helpdesk at your university or search for some instructions compiled by them.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot know whether your local cluster uses PBS (TORQUE/MAUI), SLURM or anything else to reserve cores, nodes, memory and cpu time to run jobs.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-22T20:04:36.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26499" PostTypeId="2" ParentId="26497" CreationDate="2015-02-22T20:21:08.290" Score="5" Body="&lt;p&gt;N-methyl-morpholine-N-oxide (NMO) is used to reoxidize and regenerate osmium tetroxide ($\ce{OsO4}$) in what is known as the &quot;Upjohn Dihydroxylation&quot; procedure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanistically, the redox occurs when a lone pair of electrons on the reduced $\ce{Os(VI)O3}$ are used to form a bond to the morpholine N-oxygen; the electrons from the former $\ce{N-O}$ bond are, in turn, transferred to the morpholine nitrogen.  The $\ce{OsO3}$ gains an oxygen and is oxidized back to osmium (VIII) tetroxide and the NMO is reduced to N-methylmorpholine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFj04.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/upjohn-dihydroxylation.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Osmium tetroxide is both toxic and expensive. Using the Upjohn procedure allows the reaction to be run with a catalytic amount of osmium teroxide and 1 equivalent of NMO.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-23T02:31:27.680" LastActivityDate="2015-02-23T02:31:27.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26500" PostTypeId="2" ParentId="24660" CreationDate="2015-02-22T20:54:50.230" Score="3" Body="&lt;p&gt;It don't think that $\ce{N2^{2-}}$ can exist as a free species.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N-}$ is unstable.  In other words, the electron affinity of nitrogen is negative.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create $\ce{N2^{2-}}$ you would need to bring together 2 negatively charged $\ce{N-}$ which are each already unstable due to their negative charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://casey.brown.edu/chemistry/research/LSWang/publications/84.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Experimental Search for the Smallest Stable Multiply Charged Anions in the Gas Phase&lt;/a&gt; &lt;em&gt;Physical Review Letters&lt;/em&gt; vol. 83, page 3402: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has also been concluded that no diatomic or triatomic dianions are stable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2017-03-07T12:13:45.697" LastActivityDate="2017-03-07T12:13:45.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26501" PostTypeId="1" CreationDate="2015-02-22T21:12:57.203" Score="14" ViewCount="8268" Body="&lt;p&gt;Electrons, as we all know, are incredibly small. Smaller things do tend to move faster, right? So exactly how fast considering how small they are? Also, does the electonegativity between two atoms alter the speed of the electron?&lt;/p&gt;&#xA;" OwnerUserId="14545" LastEditorUserId="5017" LastEditDate="2015-05-15T15:46:03.067" LastActivityDate="2017-05-15T03:58:23.793" Title="How fast do electrons move around the nucleus?" Tags="&lt;electrons&gt;&lt;electronegativity&gt;&lt;relativistic&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26503" PostTypeId="1" CreationDate="2015-02-22T21:35:07.887" Score="5" ViewCount="596" Body="&lt;p&gt;How is it possible for there to be a glacier floating on water but under clouds? Depending on the temperature, shouldn't there only be one state of water? &lt;/p&gt;&#xA;" OwnerUserId="14545" LastEditorUserId="4231" LastEditDate="2015-02-24T23:10:18.783" LastActivityDate="2015-02-25T08:29:18.187" Title="What allows all 3 states of H₂O to be present at the same time?" Tags="&lt;water&gt;&lt;gas-laws&gt;&lt;solid-state-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26504" PostTypeId="2" ParentId="26503" CreationDate="2015-02-22T22:14:45.497" Score="4" Body="&lt;p&gt;At the &lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_point&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;triple-point&lt;/strong&gt;&lt;/a&gt; of a substance, all three phases can be present. Since it takes additional energy to go from ice at 0 C to liquid water at 0 C, there can be a considerable overlap for the &lt;strong&gt;two&lt;/strong&gt; substances, e.g. 1 g ice with 99 g water or 99 g ice with 1 g water, but for thermodynamic equilibrium, the &lt;em&gt;triple point&lt;/em&gt; is very narrow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the first well-know American chemist, Josiah Willard Gibbs, pioneered in defining the relationship of temperature, pressure and state. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs%27_phase_rule&quot; rel=&quot;nofollow&quot;&gt; Gibbs' phase rule&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-22T22:14:45.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26505" PostTypeId="2" ParentId="26501" CreationDate="2015-02-22T22:25:36.300" Score="2" Body="&lt;p&gt;If they &lt;em&gt;actually&lt;/em&gt; moved in tight orbits, electrons would continuously radiate energy until they fell into the nucleus. Niels Bohr postulated that there were somehow stable orbitals and &quot;ignored&quot; the motion, the beginning of quantum theory (along with Einstein's work on the photoelectric effect). See &lt;a href=&quot;http://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow&quot;&gt;Bohr model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an electron &lt;em&gt;is&lt;/em&gt; accelerated (or decelerated), as opposed to staying in one orbital, it emits bremsstrahlung (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Bremsstrahlung&quot; rel=&quot;nofollow&quot;&gt;Bremsstrahlung&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorDisplayName="user7951" LastEditDate="2015-02-23T18:05:03.890" LastActivityDate="2015-02-23T18:05:03.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26510" PostTypeId="1" CreationDate="2015-02-23T02:36:09.580" Score="1" ViewCount="57" Body="&lt;p&gt;I would like to find a solution that can kill roof algae, fungus, mold, mildew effectively without using environmentally unfriendly products but don't know enough about chemistry to make a solution myself.&lt;/p&gt;&#xA;" OwnerUserId="14582" LastActivityDate="2015-02-23T09:27:03.353" Title="Eco Friendly Cleaning Solution" Tags="&lt;organic-chemistry&gt;&lt;cleaning&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2015-02-23T14:36:19.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="26515" PostTypeId="2" ParentId="26483" CreationDate="2015-02-23T06:39:30.210" Score="7" Body="&lt;p&gt;To convert from molal to molar, one generally needs to know the density of solution as well as the formula weight of solute.  Suppose there is a single solute &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in water, and that &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; has a formula weight of &lt;span class=&quot;math-container&quot;&gt;$f_x$&lt;/span&gt; g/mol.  Suppose also that the solution of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; has a density of of &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Let the molarity of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in the solution be &lt;span class=&quot;math-container&quot;&gt;$C_x$&lt;/span&gt; mol/L.  Then:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the mass concentration of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in g/L is &lt;span class=&quot;math-container&quot;&gt;$C_x f_x$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;the weight &lt;em&gt;fraction&lt;/em&gt; of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in grams of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; per &lt;em&gt;total&lt;/em&gt; gram of solution is thus &lt;span class=&quot;math-container&quot;&gt;$w_x = \frac{C_x f_x}{\rho}$&lt;/span&gt;.  The weight fraction &lt;span class=&quot;math-container&quot;&gt;$w_x$&lt;/span&gt; must be a number between 0 and 1.&lt;/li&gt;&#xA;&lt;li&gt;The weight &lt;em&gt;ratio&lt;/em&gt; of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in grams of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; per gram of water is &lt;span class=&quot;math-container&quot;&gt;$\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{w_x}{1-w_x}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The molality is simply the weight ratio divided by the formula weight, &lt;span class=&quot;math-container&quot;&gt;$\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}}$&lt;/span&gt;.  The factor of 1000 is because molality is moles per kg of solvent, not moles per g.&lt;/li&gt;&#xA;&lt;li&gt;That last expression can be written as  &lt;span class=&quot;math-container&quot;&gt;$\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{1000}{f_x} \frac{w_x}{1-w_x}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;This last formula shows that molality values will always be higher than molarity values.  Because &lt;span class=&quot;math-container&quot;&gt;$w_x$&lt;/span&gt; has to be between zero and 1, then the molality formula will divide by a smaller number than then molarity formula.  One molar water &amp;quot;dissolved&amp;quot; in water (say we are dissolving distilled water in tap water, or dissolving isotopically labeled water into regular water) has a molality of around 1.02.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;My approach is:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Convert 1M into molality and then compare if it is less or more than 1m. (?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Good approach.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The density of water is 1kg per L.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;True, but as noted in the comments, the density of solutions in water can be either higher or lower than the density of pure water.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When solute is added to it, density would definitely increase(?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not necessarily true.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;And therefore 1L of that solution will weigh more than 1 kg.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Not true, and this is the point where I think your logic mislead you.  To make one liter of a 1 molar solution, &lt;em&gt;less than&lt;/em&gt; one liter of pure water is usually required.  This is because part of the liter of solution is taken up by solute.  So not all of the liter needs to come from pure water.  And since water has a density of 1 kg/L, if we need &lt;em&gt;less than&lt;/em&gt; one liter of pure water, we need less than 1 kg of pure water.  This then means that our one mole of solute is being dissolved by &lt;em&gt;less&lt;/em&gt; than kg of solvent: thus &lt;em&gt;&lt;strong&gt;the molality of a one-molar solution is greater than one.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The neat thing is that this is true regardless of the density or formula weight of the species being considered.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-08-02T01:32:33.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26516" PostTypeId="2" ParentId="26510" CreationDate="2015-02-23T06:45:33.963" Score="3" Body="&lt;p&gt;I suggest to stay with commercially available, read-made solutions. These products are thoroughly tested and certified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They typically contain around 5% of various benzalkonium chlorides, such as N-Myristyl-N-benzyl-N,N-dimethylammonium chloride, as biocides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt; As far as the mildew is concerned: The orange ladybird (16 white spots) and the 22-spot ladybird apparently feed on mildew. If these two are naturally found in your region, get some. This is as eco-friendly as it can get to keep mildew in the garden close to a minumum.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-23T09:27:03.353" LastActivityDate="2015-02-23T09:27:03.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26517" PostTypeId="2" ParentId="5495" CreationDate="2015-02-23T07:07:09.157" Score="6" Body="&lt;p&gt;When heated to high temperatures, glycerol decomposes through dehydration to yield &lt;a href=&quot;http://en.wikipedia.org/wiki/Acrolein&quot;&gt;acrolein&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H3(OH)3 -&amp;gt; 2H2O + C3H4O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acrolein is a highly reactive unsaturated aldehyde.  It can polymerize all by itself into a yellowish material, and can probably react with any other impurities or materials in your sample, which is my guess for what caused the brown gunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point of uncertainty is the temperature at which acrolein can form.  Wikipedia says decomposition starts at 280 &amp;deg;C, which is much hotter than the temperature you reported.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if there is not large-scale conversion of glycerol to acrolein, the lower temperatures you used maybe might have been enough for low levels of acrolein to form, which could react and cause browning.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot depends on how pure your water and your glycerol were.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-23T07:07:09.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26518" PostTypeId="2" ParentId="26440" CreationDate="2015-02-23T07:26:16.630" Score="2" Body="&lt;p&gt;One reason that coupling constants vary with temperature is because the alignment of the coupled atoms changes with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in 1H NMR, proton-proton coupling between vicinal protons (HCCH) can be estimated by the &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/&quot; rel=&quot;nofollow&quot;&gt;Karplus equation&lt;/a&gt;, in which the coupling is mainly a function of dihedral angle between the two separate C-H bonds.  However, the degree to which the molecule can rotate freely around the C-C bond is a strong function of temperature.  Thus at high temperatures the observed coupling should be an average of all possible dihedral angles, while at lower temperatures where particular conformations are stable on NMR time scales, the Karplus equation can predict coupling directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just one reason for the T sensitivity.  I hope other chem.se users with more experience in NMR can provide better, more detailed answers.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-23T07:26:16.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26519" PostTypeId="2" ParentId="23905" CreationDate="2015-02-23T08:02:38.777" Score="1" Body="&lt;p&gt;How about the &lt;a href=&quot;http://en.wikipedia.org/wiki/Patch_clamp&quot; rel=&quot;nofollow&quot;&gt;patch-clamp technique&lt;/a&gt; in neurochemistry?  This technique allowed the study of ionic currents through single molecules of a voltage-gated ion channel.  It won the &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/medicine/laureates/1991/press.html&quot; rel=&quot;nofollow&quot;&gt;Nobel prize&lt;/a&gt; in 1991.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal processing required to simply observe the single-molecule opening and closing events is fairly minor.  However, processing single-molecule ion-current data for multiple experiments when inhibitors are titrated, or to compare ensemble-average kinetics to single- molecule kinetics, can be fairly complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a paper from 1998 on single-molecule enzymology; it's an extension of the patch-clamp experiments developed for the specific case of single-molecule ion channels: &lt;a href=&quot;http://www.sciencemag.org/content/282/5395/1877.full&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencemag.org/content/282/5395/1877.full&lt;/a&gt;  Read the paper for a better description of some of the signal processing involved.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-23T08:02:38.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26523" PostTypeId="1" AcceptedAnswerId="26524" CreationDate="2015-02-23T11:10:06.610" Score="4" ViewCount="1577" Body="&lt;p&gt;In my text book, it says that strong bases are alkalis because they fully dissociate in aqueous solution. But I'm not too sure of this. Can you not have a weak alkali? And can you not have a strong base which is not an alkali?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance :)&lt;/p&gt;&#xA;" OwnerUserId="7757" LastActivityDate="2015-02-23T12:42:04.890" Title="Are all strong bases alkalis?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26524" PostTypeId="2" ParentId="26523" CreationDate="2015-02-23T12:21:45.803" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are all strong bases alkalis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, quaternary amine hydroxides such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetramethylammonium_hydroxide&quot; rel=&quot;noreferrer&quot;&gt;tetramethylammonium hydroxide&lt;/a&gt; can also be strong bases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can you not have a weak alkali?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &quot;Acids and Bases&quot; by Kristi Lew, beryllium hydroxide is the one and only exception to the &quot;alkali bases are strong&quot; rule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add that many of the pKb values of the alkali hydroxides are very close to zero.  So you should only think of the them as strong in the sense of strong means pKb near or less than zero.  If you think of strong as full dissociated even in a concentrated solution then they (NaOH for example) would not be strong by that definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for the alkaline earth metal hydroxides (which have two hydroxides per metal ion), it is only the first pKb that is strong.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-23T12:42:04.890" LastActivityDate="2015-02-23T12:42:04.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26527" PostTypeId="1" AcceptedAnswerId="26553" CreationDate="2015-02-23T15:07:47.013" Score="4" ViewCount="352" Body="&lt;p&gt;Except pure alloys, are there any compounds with more metal elements in proportion of atoms than nonmetal elements in proportion of atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Aluminium oxide has 2 metal elements &amp;amp; 3 nonmetal elements. Do we have a compound with X metal elements &amp;amp; Y nonmetal elements, where Y is more then X.&lt;/p&gt;&#xA;" OwnerUserId="9616" LastEditorUserId="9616" LastEditDate="2015-02-24T01:26:19.437" LastActivityDate="2020-01-03T02:16:46.050" Title="Except pure alloys, are there any compounds with more metal elements proportion of atoms than nonmetal elements in proportion of atoms?" Tags="&lt;elements&gt;" AnswerCount="5" CommentCount="2" ClosedDate="2017-08-31T03:10:35.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="26529" PostTypeId="2" ParentId="26440" CreationDate="2015-02-23T16:15:13.817" Score="5" Body="&lt;p&gt;Let’s use vicinal 3-bond $\ce{H-C-C-H}$ proton coupling constants as an example.  &lt;strong&gt;Coupling is a through bond phenomenon&lt;/strong&gt; and as such the coupling constant will change depending if we have better (the coupling constant will increase) or worse (the coupling constant will decrease) overlap between the two $\ce{C-H}$ bonds involved.  Overlap is maximized when the two, vicinal $\ce{C-H}$ bonds are aligned in an eclipsed conformation ($\ce{H-C-C-H }$ dihedral angle =0°) or anti ($\ce{H-C-C-H }$ dihedral angle =180°)  conformation.  When the dihedral angle is ~90° the overlap of the two vicinal $\ce{C-H}$ bonds is minimized.  This thinking is embodied in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Karplus_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karplus equation&lt;/a&gt; which relates vicinal coupling constants to dihedral angle.  The following graph illustrates this dependence of the vicinal coupling constant on the dihedral angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QxV6C.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-05-3j.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this tells us that an experimentally observed vicinal coupling constant will be dependent upon the relative populations of the eclipsed, staggered and anti conformations. Said differently, the vicinal coupling constant will be an average of each of the individual coupling constants for each conformation, weighted by the relative population of that conformation. &lt;strong&gt;Since the relative conformer populations change with temperature, the coupling constant will also change with temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-23T23:15:32.277" LastActivityDate="2015-02-23T23:15:32.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26531" PostTypeId="2" ParentId="26527" CreationDate="2015-02-23T16:41:10.060" Score="-1" Body="&lt;p&gt;I'm not sure if I undestood you compeletly, but for example compounds called &lt;a href=&quot;https://en.wikipedia.org/wiki/Mixed_oxide&quot; rel=&quot;nofollow&quot;&gt;mixed oxides&lt;/a&gt; can have cations of many metals in their structure, like $\ce{Y3Al5O12}$ - consisting of yttrium and aluminum cations, and oxide anions.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-02-23T16:41:10.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26532" PostTypeId="1" AcceptedAnswerId="26533" CreationDate="2015-02-23T16:42:34.720" Score="8" ViewCount="9980" Body="&lt;p&gt;My teacher says $H$ is a state function but $\Delta H$ is not. But I think both of them should be state function. Is he correct?&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="4231" LastEditDate="2015-02-24T23:08:42.230" LastActivityDate="2015-02-24T23:08:42.230" Title="Is delta H a state function?" Tags="&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26533" PostTypeId="2" ParentId="26532" CreationDate="2015-02-23T16:51:03.757" Score="8" Body="&lt;p&gt;$\Delta H$ is a function of two states, the initial state and the final state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given final state, there can be infinite $\Delta H$ values depending upon what the inital state was.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given inital state, there can be infinite $\Delta H$ values depending upon what the final state is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\Delta H$ is not a state function.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4231" LastEditDate="2015-02-24T23:08:39.800" LastActivityDate="2015-02-24T23:08:39.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26534" PostTypeId="1" CreationDate="2015-02-23T16:52:26.470" Score="6" ViewCount="2300" Body="&lt;p&gt;Transition elements usually have high values, especially so for greater number of unpaired electrons. Why the exception here?&lt;/p&gt;&#xA;" OwnerUserId="14595" LastEditorUserId="14595" LastEditDate="2015-02-23T17:18:17.417" LastActivityDate="2015-02-23T21:46:18.920" Title="Why is enthalpy of atomization of Mn exceptionally low?" Tags="&lt;enthalpy&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26535" PostTypeId="2" ParentId="26501" CreationDate="2015-02-23T17:17:52.777" Score="16" Body="&lt;p&gt;The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the &lt;strong&gt;handy equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{V_{rel}=\frac{[Z]}{[137]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Z is the atomic number of the element under consideration and 137 is &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_units#Related_physical_constants&quot; rel=&quot;nofollow noreferrer&quot;&gt;the speed of light in atomic units&lt;/a&gt;, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Fine-structure_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;the fine structure constant&lt;/a&gt;.  Consequently a 1s electron in the hydrogen atom will travel at approximately 0.7% the speed of light. In silver (Z=47) the 1s electron will travel around 34% the speed of light, while the 1s electron in gold (Z=79) will travel at about 58% the speed of light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we get up to around silver, the electrons are traveling at relativistic speeds and this can dramatically impact properties of the atom. For example, the relativistic mass of an electron is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{m_{rel}=\frac{m_{e}}{\sqrt{1-(V_{rel}/c)^2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\ce{m_{e},~ V_{rel}~ and~ c}$ are the electron rest mass, the velocity of the electron, and the speed of light respectively.  The following figure provides a graphical representation of how the electron mass increases as the electron velocity increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zfGNz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following equation relates the ratio of the relativistic radius of the first Bohr orbit $\ce{R_{rel}}$ to the normal radius $\ce{R_{o}}$, to the relativistic velocity of the electron&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\frac{[R_{rel}]}{[R_{o}]}=\sqrt{1-(V_{rel}/c)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~ 18%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at these earlier Chem SE answers to see the interesting physical effects that atoms can exhibit when their electrons travel at relativistic speeds.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/9051/why-do-the-trends-in-reactivity-not-apply-for-francium/26631#26631&quot;&gt;francium&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637&quot;&gt;gold&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/13667/why-does-mercury-have-low-melting-and-boiling-points/13671#13671&quot;&gt;mercury&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2016-03-12T00:49:53.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26537" PostTypeId="2" ParentId="10478" CreationDate="2015-02-23T17:36:27.460" Score="3" Body="&lt;p&gt;The two rows of elements present at the bottom of the periodic table are called the 4f series or lanthanoids and 5f or actanoids. They are also called inner transition elements. With each series, the properties of component elements closely resembles each other so their chemistry becomes complicated than their corresponding lanthanoids. Actanoids have large number of oxidation states compared to lanthanoids. Actanoid are radioactive and are man-made in the laboratory. They are not completely studied. Hence they are placed separately in the periodic table.&lt;/p&gt;&#xA;" OwnerUserId="14599" LastEditorUserId="4325" LastEditDate="2015-02-23T17:43:44.040" LastActivityDate="2015-02-23T17:43:44.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26538" PostTypeId="1" AcceptedAnswerId="26542" CreationDate="2015-02-23T17:38:01.373" Score="4" ViewCount="801" Body="&lt;p&gt;Acetylsalicylic acid is a molecule in several popular drugs such as Magnyl and aspirin. It has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of 3.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm about to calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.010 M}$&lt;/span&gt; solution of acetylsalicylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read up on how to do this and in &lt;a href=&quot;https://www.youtube.com/watch?v=am5EWsbdvFI&quot; rel=&quot;nofollow noreferrer&quot;&gt;one video&lt;/a&gt; the author talks about &quot;equimolar amounts&quot;.&#xA;He says that now that we have equimolar amounts, we can write: &lt;span class=&quot;math-container&quot;&gt;$$\ce{\frac{[H+][A-]}{[HA]}} = \ce{\frac{[H+]^2}{[HA]}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which makes sense as there are equally many &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; ions. However when can I assume that an acid disassociated equally? Also, I thought that when it did this then it was called a strong acid? Why would I be using this formula at all for a strong acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can assume that acetylsalicylic acid disassociates into equal acid and base ions, then I can calculate my problem, but I'm skeptical as to assuming this is true.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="69830" LastEditDate="2020-05-02T07:34:03.287" LastActivityDate="2020-05-02T07:34:03.287" Title="How do I know that an acid and conjugate base are present in equimolar amounts?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;biochemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26540" PostTypeId="2" ParentId="26452" CreationDate="2015-02-23T18:10:16.867" Score="3" Body="&lt;p&gt;I have to agree with Klaus.  ASTM B117 is, at best, a QC procedure.  In my experience, most corrosion and coatings scientist who use B117 do so because it is very well established, but it does not correlate to a real-world measure of performance.  ASTM D5894 is a better cabinet test and there is also an SAE procedure developed for automobiles.&lt;/p&gt;&#xA;" OwnerUserId="14602" LastActivityDate="2015-02-23T18:10:16.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26542" PostTypeId="2" ParentId="26538" CreationDate="2015-02-23T18:44:15.820" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm really confused here!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No, you are not!&lt;/strong&gt; You only (temporarily) mixed up two things, both of with you have already understood separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA + H2O &amp;lt;=&amp;gt; H3O+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Position of the equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is described by the equilibrium constant $K$, which is the quotient of the rate constants for the forward and the back reaction: $K = \frac{k_{forward}}{k_{back}}$. For strong acids, of which you know that they completely dissociate, the equilibrium is on the product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Equimolarity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing out of nothing! If &lt;strong&gt;one molecule&lt;/strong&gt; of $\ce{HA}$ dissociates is water, there &lt;strong&gt;exactly one molecule&lt;/strong&gt; $\ce{H3O+}$ and &lt;strong&gt;exactly one molecule&lt;/strong&gt; of $\ce{A-}$ generated. Not more, not less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: How could another molecule of $\ce{A-}$ appear out of nothing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: It can't and it doesn't happen - not even in the LEGO universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1Eds.jpg&quot; alt=&quot;deprotonation in lego&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is just a proton transfer reaction from $\ce{HA}$ to $\ce{H2O}$; the overall number of atoms does not change!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If is however possible that some $\ce{A-}$ may come from a different source, just by adding a salt $\ce{MA}$ of the conjugate base (think sodium acetylsalicylate) to the solution of your acid. &lt;strong&gt;jerepierre has mentioned this in his comment!&lt;/strong&gt; But for the moment, this is another story.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-23T19:28:21.823" LastActivityDate="2015-02-23T19:28:21.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26543" PostTypeId="2" ParentId="26527" CreationDate="2015-02-23T18:54:40.173" Score="4" Body="&lt;p&gt;Sure, e.g. sodium aluminate, Na&lt;sub&gt;2&lt;/sub&gt;Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;4&lt;/sub&gt; (anhydrous formula), is a common chemical (it's the &quot;gray stuff&quot; on aluminum dishware cleaned in an automatic dishwasher with alkaline detergents).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-23T18:54:40.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26544" PostTypeId="1" AcceptedAnswerId="26545" CreationDate="2015-02-23T19:56:14.080" Score="3" ViewCount="82" Body="&lt;p&gt;I would post this in biology but as the question is biochemical in nature I decided this would be a more appropriate platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm asking is how exactly does a transmembrane protein bind to the cell membrane? I realize that the transmembrane alpha helix is about 20 amino acids long with mostly hydrophobic amino acids as the inside of the membrane is hydrophobic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't realize is how exactly this alpha helix binds to the membrane and sticks to it? I've tried searching on the subject but I can't seem to find the answer. I found out that the alpha helix satisfies the hydrogen bonds internally leaving no polar R groups for the membrane as most groups are hydrophobic, but that only explains how it &lt;strong&gt;doesn't&lt;/strong&gt; bind to the membrane.&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="9961" LastEditDate="2015-12-17T00:06:27.163" LastActivityDate="2015-12-17T00:06:27.163" Title="How do transmembrane protein alpha helixes bind to the membrane?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26545" PostTypeId="2" ParentId="26544" CreationDate="2015-02-23T20:35:37.377" Score="2" Body="&lt;p&gt;Well, it binds via &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrophobic_effect&quot; rel=&quot;nofollow&quot;&gt;hydrophobic interacions&lt;/a&gt; - non-polar R groups are attracted to hydrophobic interior of the membrane via &lt;a href=&quot;https://en.wikipedia.org/wiki/London_dispersion_force&quot; rel=&quot;nofollow&quot;&gt;dispersion forces&lt;/a&gt;, and also don't disrupt the structure of water as if they would outside of membrane. It's the same thing which holds molecules of the membrane together.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-09-06T23:32:10.540" LastActivityDate="2015-09-06T23:32:10.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26547" PostTypeId="1" CreationDate="2015-02-23T20:57:26.260" Score="5" ViewCount="2411" Body="&lt;p&gt;I'm studying chemical engineering at uni (Belgium), and this year (my 2nd) we have to develop and build a stirling engine. It has to be powered by a relatively safe exothermic reaction (not dangerous for bystanders, so no formation of toxic fumes,...) and the whole thing is electrically controlled but that's not important here,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come up with this so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sodium acetate crystallization&lt;/li&gt;&#xA;&lt;li&gt;Hydratation of anhydrous copper sulphate&lt;/li&gt;&#xA;&lt;li&gt;calcium chloride and water&lt;/li&gt;&#xA;&lt;li&gt;sodiumhypochlorite oxidizing sodiumsulfite&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An 'open flame' reaction is not allowed sadly enough..&#xA;I'm hoping to at least reach 70-80 degrees Celsius (158-176° Fahrenheit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances are quite big that there are better reactions but I really can't think of anything else.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks A LOT in advance everyone!&lt;/p&gt;&#xA;" OwnerUserId="14606" LastActivityDate="2019-12-26T19:45:42.260" Title="Safe yet strong exothermic reaction?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26548" PostTypeId="2" ParentId="26527" CreationDate="2015-02-23T21:31:20.757" Score="7" Body="&lt;p&gt;There are some rare but very interesting cases in which two metals react when mixed instead of forming an alloy, creating a non-metallic compound made only with metal atoms. Examples are caesium auride ($\ce{CsAu}$), caesium platinide ($\ce{Cs2Pt}$) and several barium platinides ($\ce{BaPt}$, $\ce{Ba2Pt}$, $\ce{Ba3Pt2}$, and possibly others), all of which best described as ionic salts rather than alloys. These are &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S129325580500230X&quot; rel=&quot;noreferrer&quot;&gt;a remarkable consequence of relativistic effects in heavy atoms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-02-23T21:31:20.757" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26549" PostTypeId="2" ParentId="26534" CreationDate="2015-02-23T21:40:19.673" Score="3" Body="&lt;p&gt;According to &quot;Atomic Structure and Periodicity&quot; by Jack Barrett, the reason is a combination of the $d^5$ electron configuration (half filled subshell) in the gas state and, relative to chromium, &quot;the extra electron in the metal occupying an anti-bonding region of a band, s or d&quot;. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-23T21:46:18.920" LastActivityDate="2015-02-23T21:46:18.920" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26550" PostTypeId="2" ParentId="26547" CreationDate="2015-02-23T22:32:28.817" Score="2" Body="&lt;p&gt;Here are some more options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pd-catalyzed oxidation of dilute hydrogen gas in air: $\ce{2H2 + O2 -&amp;gt; 2H2O}$  If you use a &lt;strong&gt;dilute&lt;/strong&gt; hydrogen gas mix (say 95% nitrogen and 5% hydrogen), no flaming combustion can occur in air.  But oxidation still happens, at least when catalyzed by palladium particles. This is how microbiological anaerobic chambers maintain a zero-oxygen environment.  You can probably buy some catalyst fairly cheaply from a microbiology anaerobic chamber company like Coy labs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phosphorus pentoxide hydration.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphorus_pentoxide#Applications&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says this reaction, $\ce{P4O10 + 6 H2O -&amp;gt; 4 H3PO4}$, has $\Delta H_{reac}=-177\text{ kJ/mol}$.  There is no flame and no fumes, but the reaction does create phosphoric acid, which is corrosive and can be dangerous.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A modification of your sulfite / bleach idea: use sodium dithionite instead of sodium sulfite.  Sodium dithionite ($\ce{Na2S2O4}$) is air-unstable but if you can work out a way to handle without air exposure it will be more exothermic than sulfite.  Instead of dithionite itself you could use the much easier to handle &lt;a href=&quot;http://en.wikipedia.org/wiki/Rongalite&quot; rel=&quot;nofollow&quot;&gt;Rongalite&lt;/a&gt; and hypochlorite.  Possibly, using calcium hypochlorite suspensions in water (mostly insoluble) will allow you to use higher concentrations of hypochlorite and Rongalite without the reaction rate and temperature skyrocketing (since the rate will be controlled by $\ce{Ca(OCl)2}$ dissolution).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-23T22:32:28.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26551" PostTypeId="2" ParentId="26532" CreationDate="2015-02-23T23:24:11.880" Score="2" Body="&lt;p&gt;I generally agree with DavePhD's answer but I also want to give the opposite view (!).  Everyone agrees $H$ is a state function, but energy (and thus enthalpy) has no built in zero-scale.  So really even in defining $H$, there is some built-in standard state that you are comparing the state of interest to.  For example, the standard state might be all electrons and all nuclei infinitely separated from each other (this standard is commonly used in quantum chemistry).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we could also specify other standard states. We could say the standard state is one where all electrons and all protons and neutrons are infinitely separated...but that's possibly only useful for nuclear chemistry.  We could also a standard state where all the atoms in a sample are separated into pure chemical elements.   So to compute &quot;the&quot; enthalpy $H$ of say $\ce{CH4}$ using this standard state, we first separate its atoms to $\ce{C}$ and $\ce{2H2}$, our standard state, and then see how much the enthalpy changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of how we definite the standard state, enthalpy is a state function.  But &quot;the&quot; enthalpy $H$ that we compute from the last standard state is the same as the enthalphy &lt;em&gt;change&lt;/em&gt; of formation, $\Delta H_{form}$!  So if those two things are equal, and one of them is a state function, then the other one must be a state function too!  So $\Delta H$ is also a state function.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-23T23:24:11.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26552" PostTypeId="1" AcceptedAnswerId="26560" CreationDate="2015-02-24T03:50:15.767" Score="3" ViewCount="120" Body="&lt;p&gt;I'm entirely chemically-naive and am confounded when trying to convert $\ce{NaCl}$ brine concentrations to moles per litre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is that I'm performing linear multiple regressions using published data form multiple studies. Because the studies report their brine concentrations using different units I am trying to convert all of the concentrations into molarity (moles per litre). To do that I'm using the conversion tool at &lt;a href=&quot;http://www.endmemo.com/chem/molarity.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;EndMemo&lt;/a&gt; - here's a screen shot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n7ACg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in more than one place in the literature people report different M values ... for example, below is an extract from one of the studies. Their M values don't match what I'm getting using the online converter (for example, 4 g/L $\ce{NaCl}$ is coming out as 0.068 M/L...hopefully I'm missing something very basic here?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bS5FB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1468-8123.2012.00369.x/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Broseta D., Tonnet N., and Shah V., &lt;em&gt;Geofluids&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;12&lt;/em&gt; (4), 280-294.&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I'm not finding consistency with the unit symbols for Molarity and litre (sometimes they are capitalised, sometimes they are not...are they referring to different things or is this another example of inconsistency)?&lt;/p&gt;&#xA;" OwnerUserId="14609" LastEditorUserId="14609" LastEditDate="2015-02-24T23:21:59.337" LastActivityDate="2015-02-24T23:31:09.240" Title="How can inconsistencies in these sodium chloride solutions be explained?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26553" PostTypeId="2" ParentId="26527" CreationDate="2015-02-24T04:17:53.097" Score="5" Body="&lt;p&gt;In addition to the answers by Nicolau and DrMoishe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most prominent examples are the binary oxides of the alkali metals &lt;span class=&quot;math-container&quot;&gt;$\ce{A2O}$&lt;/span&gt;: &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium oxide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A~=~Li}$&lt;/span&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium oxide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A~=~Na}$&lt;/span&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;potassium oxide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A~=~K}$&lt;/span&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubidium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;rubidium oxide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A~=~Rb}$&lt;/span&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Caesium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;caesium oxide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{A~=~Cs}$&lt;/span&gt;.&lt;br&gt;&#xA;Of course there are also higher homologs of these compounds like &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium sulfide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Li2S}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially for rubidium and caesium there are also a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Suboxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;suboxides&lt;/a&gt;, i.e. &lt;span class=&quot;math-container&quot;&gt;$\ce{A9O2}$&lt;/span&gt; and others. This is a whole class of compounds whith a rich variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also the nitrides of the alkali and earth alkali metals, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_nitride&quot; rel=&quot;nofollow noreferrer&quot;&gt;lithium nitride&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Li3N}$&lt;/span&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_nitride&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnesium nitride&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg3N2}$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphide&quot; rel=&quot;nofollow noreferrer&quot;&gt;phosphides&lt;/a&gt; known, that fit your description, e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{K3P}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{K4P3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{K5P4}$&lt;/span&gt;. These compound are also known from transition metals, like &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu3P}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{Ni5P2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the above, there are also some very common carbides, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_carbide&quot; rel=&quot;nofollow noreferrer&quot;&gt;aluminium carbide&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Al4C3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I am very certain, there are many more.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="17175" LastEditDate="2020-01-03T02:16:46.050" LastActivityDate="2020-01-03T02:16:46.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26554" PostTypeId="2" ParentId="26527" CreationDate="2015-02-24T04:18:40.110" Score="2" Body="&lt;p&gt;Petzite, $\ce{Ag3AuTe2}$, is an uncommon telluride containing mineral that seems to fit the bill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium oxide $\ce{Na2O}$ is well known, as is potassium oxide, rubidium oxide, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some one should write a parser and see if &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_inorganic_compounds&quot; rel=&quot;nofollow&quot;&gt;this list from Wikipedia&lt;/a&gt; has any more examples.  I didn't see any that weren't already mentioned here on a quick skim.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-24T04:18:40.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26556" PostTypeId="1" CreationDate="2015-02-24T06:32:05.720" Score="1" ViewCount="14608" Body="&lt;p&gt;A Diels-Alder reaction happens between 9-Anthracenemethanol (208.26 g/mol) and N-Methylmaleimide (111.10 g/mol). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using 0.069 g 9-Anthracenemethanol (0.033 mol) and 0.11g N-Methylemaleimide (0.0009 mol, limiting reagent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since DA reactions have 100% atom economy, the product has the molar mass of both products combined, 319.36 g/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying that by the moles of limiting reagent, we get 0.28 g product. &#xA;However, the mass of both starting materials combined (0.069g + 0.110g) together is not even 0.28g. What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="14612" LastEditorUserId="865" LastEditDate="2021-05-22T19:02:40.130" LastActivityDate="2021-05-22T19:02:40.130" Title="How to calculate the theoretical yield of a Diels-Alder reaction?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26557" PostTypeId="2" ParentId="26556" CreationDate="2015-02-24T06:39:36.627" Score="2" Body="&lt;p&gt;You are missing a  couple of zeros in the number of moles of your 9-anthracenemethanol.  I calculate 0.00033 mol of that reagent, which therefore becomes your limiting reagant, and I calculate a total yield of 0.105 g of product, with about 0.73 g of  N-Methylemaleimide left over.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-02-24T06:39:36.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26559" PostTypeId="2" ParentId="24314" CreationDate="2015-02-24T11:16:25.540" Score="3" Body="&lt;p&gt;Two things are very important in electrochemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Cleanliness: Your tap water contains all sorts of impurities that are probably electroactive. My recommendation, take some tap water, don't add any $\ce{CuSO4}$ and repeat the experiment. If you see any deposit, that is not copper. I won't even attempt to make any guesses as to what that deposit may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Potential control: This is very tricky in the type of setup you are using. In two electrode experiments like you are doing, the solution resistance, the distance between the electrodes and the current you are drawing all effect the actual potential your electrode experiences. With the constant 12V you are applying, the current that is drawn is also a function of the above parameters, so it is very hard to determine what the actual electrochemical potential will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamry has an excellent application note about 2,3 and 4 electrode experiments here: &lt;a href=&quot;http://www.gamry.com/application-notes/two-three-and-four-electrode-experiments/&quot; rel=&quot;nofollow&quot;&gt;http://www.gamry.com/application-notes/two-three-and-four-electrode-experiments/&lt;/a&gt; if you would like to read up on the details.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-02-24T11:16:25.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26560" PostTypeId="2" ParentId="26552" CreationDate="2015-02-24T12:20:56.907" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;example, 4 g/L $\ce{NaCl}$ is coming out as 0.68 M/L...hopefully I'm missing something very basic here&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you are off by a factor of 10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 g/L $\ce{NaCl}$ is 0.068M (0.068 moles per liter of solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the journal article, I would used the 4 grams per liter value (and other grams per liter values) and convert to moles per liter properly and not rely on the 0.08 (small capital &quot;M&quot; l-1) value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the value is critical to your research, email the corresponding author of the paper and and ask about it in a polite way.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4231" LastEditDate="2015-02-24T23:31:09.240" LastActivityDate="2015-02-24T23:31:09.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26561" PostTypeId="2" ParentId="13780" CreationDate="2015-02-24T12:58:23.520" Score="6" Body="&lt;p&gt;Argon is a quite unreactive element, however very few &lt;a href=&quot;http://en.wikipedia.org/wiki/Argon#Compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;compounds&lt;/a&gt; of it are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most popular, and currently the only neutral compound, is argon hydroflouride, $\ce{HArF}$, which was reported by L. Khriachtchev &lt;em&gt;et.al.&lt;/em&gt;.&lt;sup&gt;[1]&lt;/sup&gt;  In the same issue there is also a comment on the matter by my former supervisor G. Frenking.&lt;sup&gt;[2]&lt;/sup&gt;&lt;br&gt;&#xA;Another interesting read on the subject is the prediction of the metastable $\ce{HHeF}$, by M. W. Wong.&lt;sup&gt;[3]&lt;/sup&gt; It generally treats all compounds of the $\ce{HNgF}$ family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All calculations done so far indicate, that the molecule is linear and only kinetically stable. The highly accurate CCSD(T)/aug-cc-pV5Z level predicts, that the $\ce{HArF}$ molecule is $5.87~\mathrm{eV}$ higher in energy than its fragments $\ce{HF}$ and $\ce{Ar}$. The minimum barrier on this level of theory is a mere $0.35~\mathrm{eV}$ along the stretching coordinate of the hydrogen argon bond. The bending is predicted to be higher in energy. The reported bond distances for $\ce{H-Ar}$ and $\ce{Ar-F}$ are $133~\mathrm{pm}$ and $197~\mathrm{pm}$, respectively.&lt;br&gt;&#xA;An NBO analysis on the MP2/aug-cc-pVTZ level of theory shows a strong ionic character of the molecule, it is reported, that there is a strong $\ce{(HAr)+F- }$ charge transfer in the electronic structure of the molecule.&lt;br&gt;&#xA;Wong performed accurate calculations on the QCISD/6-311+G** level of theory and combined it with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quantum Theory of Atoms in Molecules&lt;/a&gt; approach. He calculated a positive value of the Laplacian of the electron density $\nabla^2\rho_\mathrm{b}$ for the bond critical point of the argon fluoride bond, which indicates a mainly electrostatic interaction. The contrary is the case for the argon hydrogen bond, where the Laplacian is found to be negative, i.e. charge concentration, and therefore indicating a dominantly covalent interaction.&lt;br&gt;&#xA;The following picture is a snapshot of this behaviour at the BP86/cc-pVTZ level of theory.&lt;sup&gt;[4]&lt;/sup&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gOI1u.jpg&quot; alt=&quot;Laplacian of HArF&quot;&gt;&lt;br&gt;&#xA;Solid lines are regions of charge depletion, $\nabla^2\rho_\mathrm{b}&amp;gt;0$, while dashed lines are regions of charge concentration $\nabla^2\rho_\mathrm{b}&amp;lt;0$. Red lines are zero flux surfaces, where $\nabla\rho_\mathrm{b}=0$. And red spheres indicate bond critical points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This molecule is quite interesting for many reasons, however, since it is only kinetically stable it has to be treated with fullest force of quantum chemical calculations. The calculations I did on a low level DFT method are not sufficient at all on its own, as has been investigated by J. Panek &lt;em&gt;et.al.&lt;/em&gt;.&lt;sup&gt;[5]]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hybridisation is in a wider sense the mixing of s and p orbitals;&lt;sup&gt;[6]&lt;/sup&gt; and upon further examination of the electron density this will be observed in this molecule. Most likely for the hydrogen argon covalent bond. As hybridisation is just a mathematical transformation of the original atomic orbitals there is no real interpretative value in this case. For example, the NBO analysis at the BP86/cc-pVTZ level of theory give only a negligible 3% s contribution of argon to the $\ce{H-Ar}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The VSEPR theory is only a observational concept for well-behaved molecules. It basically lacks a solid scientific basis and should not be used any more, but especially in cases like this, it must not be used.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Literature and Notes&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v406/n6798/full/406874a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonid Khriachtchev, Mika Pettersson, Nino Runeberg, Jan Lundell, and Markku Räsänen, &lt;em&gt;Nature&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;406&lt;/em&gt; 874-876&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v406/n6798/full/406836a0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gernot Frenking, &lt;em&gt;Nature&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;406&lt;/em&gt;, 836-837&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja9938175&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ming Wah Wong, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;122&lt;/em&gt; (26), 6289-6290&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;To avoid copyright issues I performed a quick calculation to illustrate the above raised point.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2002/cp/b109578a#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jarosław Panek, Zdzisław Latajka, and Jan Lundell, &lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;4&lt;/em&gt;, 2504-2510.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Hybridisation is a concept that dates back to the beginnings of &lt;a href=&quot;http://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Valence Bond (VB) Theory&lt;/a&gt;. It was originally implemented in the strict fashion of creating $s^{1/(n+1)}p^{n/(n+1)}$ or $sp^n$ hybrid orbitals for $n = \{1,2,3\}$. H. Bent later went on suggesting fractional hybridisation schemes, see &lt;a href=&quot;http://dx.doi.org/10.1021/cr60211a005&quot; rel=&quot;nofollow noreferrer&quot;&gt;Henry A. Bent, &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;1961&lt;/strong&gt;, &lt;em&gt;61&lt;/em&gt; (3), 275–311&lt;/a&gt;. Nowadays the term s-p-mixing in Molecular Orbital theory is almost synonymous with the VB theory term hybridisation.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="9961" LastEditDate="2016-07-20T23:15:39.857" LastActivityDate="2016-07-20T23:15:39.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26563" PostTypeId="1" AcceptedAnswerId="26569" CreationDate="2015-02-24T15:42:42.137" Score="7" ViewCount="13152" Body="&lt;p&gt;What will be the order of reactivity towards electrophilic substitution in case of the following compounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;benzene, ethyl benzene, isopropyl benzene, tert-butyl benzene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer at the end of the book says that ethyl benzene will be most reactive (the book doesn't explain the cause though it's a MCQ) but as according to inductive effect I think the tert-butyl group will have more +I effect on the ring. Doesn't it?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="15410" LastEditDate="2015-05-14T16:14:53.107" LastActivityDate="2015-05-14T16:14:53.107" Title="Order of electrophilic substitution" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26564" PostTypeId="1" CreationDate="2015-02-24T15:54:47.457" Score="2" ViewCount="2076" Body="&lt;p&gt;The local hardware stores actually sell chemicals in recycled beer bottles with plastic stoppers as lids. The bottle of hydrochloric acid I bought is clearly losing gas to diffusion through the plastic, or a seal that's not tight enough--it smells bad, and the beer bottle's metallic label is slowly corroding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I best seal this bottle? Some ideas are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to mold a stopper from silicone. Use a (synthetic) cork, whatever's used for homebrewing. Line the stopper with a few layers of thin polyethylene or polypropylene (from consumer materials, such as air-filled plastic bags for packing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or could I limit the diffusion by putting a plastic bag around it? This is tough to answer experimentally (without a lab) because resisting physical degradation isn't the same as resisting diffusion.&lt;/p&gt;&#xA;" OwnerUserId="14617" LastEditorUserId="14617" LastEditDate="2015-02-24T16:04:10.867" LastActivityDate="2015-02-27T18:24:11.720" Title="What stopper will stop HCl fumes?" Tags="&lt;physical-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26565" PostTypeId="2" ParentId="26564" CreationDate="2015-02-24T16:03:03.293" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;it smells bad, and the beer bottle's metallic label is slowly corroding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's, unfortunately, very common. At one of my former labs 'acid' cabinet  was corroded from within. For this reason it is recommended to keep such bottles under fume hood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, common stoppers are made from polyethylene or polypropylene. At one of my former labs concentrated hydrochloric acid was stored in stock plastic canister, similar to that used to store petrol (on the other hand, both nitric and sulfuric acids were stored in 20l glass bottles, even though the stoppers were made from same polymer).  I HIGHLY recommend to by (or order by mail) dedicated chemical bottles with caps, though. They are usually fairly cheap and pretty universal.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-02-24T16:03:03.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26567" PostTypeId="1" CreationDate="2015-02-24T16:43:59.963" Score="4" ViewCount="183" Body="&lt;p&gt;This question ultimately relates to brewing beer, but more specifically to water treatment before brewing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I can't get a water quality report, I suspect I have high levels of bicarbonate alkalinity in my tap water because boiling it causes a white precipitate (which I think is calcium carbonate) to form. I base this assumption on this formula, in the book &lt;a href=&quot;https://books.google.com/books?id=cr9Pv0gefCQC&amp;amp;printsec=frontcover&amp;amp;dq=brewing&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=KKbsVKr_FqzgsATMhIGQCw&amp;amp;ved=0CBQQ6AEwAA#v=onepage&amp;amp;q=brewing&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brewing&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ca(HCO3)2 -&amp;gt;[heat] CaCO3 + CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to sanitize some aluminum foil so I boiled it in fresh tap water for about 15 minutes. I noticed a few things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The precipitate formed much faster than it normally would without foil in it.&lt;/li&gt;&#xA;&lt;li&gt;The foil began to blacken in certain parts.&lt;/li&gt;&#xA;&lt;li&gt;I'm not positive, but the precipitate seems a bit finer in texture than it normally is.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I thought at first maybe the foil was just providing nucleation points for the carbonate to form around. But then I thought I remembered reading somewhere that aluminum competes with calcium in certain situations (I think it was regarding osteoporosis, maybe). The question I have, then, is whether or not it's feasible that aluminum could be affecting a different reaction than what I'm assuming normally occurs in this situation?&lt;/p&gt;&#xA;" OwnerUserId="13178" LastEditorUserId="39539" LastEditDate="2017-04-05T06:13:55.003" LastActivityDate="2017-04-05T06:13:55.003" Title="Boiling aluminum in high-alkalinity water" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26568" PostTypeId="2" ParentId="26563" CreationDate="2015-02-24T16:52:06.687" Score="3" Body="&lt;p&gt;Alkyl groups on benzene are &lt;em&gt;ortho-para&lt;/em&gt; directing. The $\sigma$-conjugation effect of the alkyl group is not dependent on the type of carbon that is attached to is (methyl, primary, secondary or tertiary). Only the carbon connected directly to the ring can interact with it. More substituted alkyl groups will however hinder reaction at the ortho-site lowering the total reaction speed (e.g. for t-butyl benzene essentially no ortho attack can occur, it is too sterically hindered). Therefore the predicted order of reactivity from most reactive to least reactive is: ethylbenzene, isopropylbenzene, tert-butyl benzene, benzene.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-24T16:52:06.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26569" PostTypeId="2" ParentId="26563" CreationDate="2015-02-24T17:24:38.593" Score="4" Body="&lt;p&gt;In your series, all of the alkyl benzenes will have roughly the same +I inductive effect.  Where they differ is with regard to the resonance effect.  Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9sQP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation.  In isopropyl benzene there is only one and there are none in *tert-*butylbenzene.  Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;benzene=1, toluene=24, &lt;em&gt;tert&lt;/em&gt;-butylbenzene=15.7 (&lt;a href=&quot;http://authors.library.caltech.edu/25034/23/BPOCchapter22.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;, page 1060)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest.  The expected order in your series would therefore be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ethylbenzene &gt; isopropylbenzene &gt; &lt;em&gt;tert&lt;/em&gt;-butylbenzene &gt; benzene&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-24T17:24:38.593" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26570" PostTypeId="1" AcceptedAnswerId="26575" CreationDate="2015-02-24T18:15:56.453" Score="24" ViewCount="25289" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which is the most stable free radical among the given species?&lt;br&gt;&#xA;  &lt;strong&gt;(1)&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{CH2=CH-CH2^.}$&lt;/span&gt; (allyl radical)&lt;br&gt;&#xA;  &lt;strong&gt;(2)&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H5-CH2^.}$&lt;/span&gt; (benzyl radical)&lt;br&gt;&#xA;  &lt;strong&gt;(3)&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{(H3C)3C^.}$&lt;/span&gt; (&lt;em&gt;t&lt;/em&gt;-butyl radical)&lt;br&gt;&#xA;  &lt;strong&gt;(4)&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5^.}$&lt;/span&gt; (ethyl radical)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book claims the answer to be the allyl radical, option 1. But how can we definitely say that the allylic free radical is more stable, compared to the benzyl radical (option 2) which is also resonance-stabilised, or the tertiary radical (option 3) which has more hyperconjugative donation from α-hydrogen atoms?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="16683" LastEditDate="2018-12-25T18:47:12.163" LastActivityDate="2020-01-02T05:51:57.363" Title="Comparing the stabilities of allylic, benzylic, and tertiary radicals" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;radicals&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26571" PostTypeId="1" AcceptedAnswerId="26572" CreationDate="2015-02-24T18:26:35.330" Score="7" ViewCount="348" Body="&lt;p&gt;So from the general reaction we can see that the carbonyl is reduced while the Grignard is oxidized to an alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for this hypothetical reaction (&lt;strong&gt;does this really occur?&lt;/strong&gt;) we are asked to provide a plausible mechanism. This is what I got, but the prof. wants something better - why transfer negative charge from oxygen to an $\text{sp}^3$ carbon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JVKet.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JVKetm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Click on the image to enlarge, if you wish.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are his hints:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Remember, prior to Grignard reaction the $\ce{Mg}$ center swaps out an ether&#xA;  solvent molecule for a reactant oxygen.  This interaction will lead to&#xA;  the reaction but in this case, due to steric interference a hydride is&#xA;  delivered instead as the Grignard is oxidized to an alkene.  In this&#xA;  way the Grignard reagent is almost reacting in the way $\ce{NaBH4}$ would in&#xA;  the presence of this ketone. Try to get all of this action working&#xA;  together in a single, concerted step.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So ... I get the first part just fine. The $\ce{Mg}$ center has an octet; it's solvated by two ether molecules in solution usually. Okay and in this case the carbonyl oxygen helps solvate the $\ce{Mg}$ center. And then hydride transfer occurs ... but how exactly do we get the hydride to leave without creating a carboanion? And is it really that important that everything be concerted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my second attempt ... if this is all concerted in real life then I must have hit the orbital alignment lottery! All the stars must have been aligned ... ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I see about my second attempt is that in the end the $\ce{Mg}$ center doesn't have a complete octet - perhaps I should have shown another ether molecule popping in to help $\ce{Mg}$ out with its octet fantasies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUARjm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2015-10-02T16:04:50.073" LastActivityDate="2020-04-12T15:52:37.223" Title="Mechanism of Grignard addition to carbonyls" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26572" PostTypeId="2" ParentId="26571" CreationDate="2015-02-24T18:56:06.727" Score="6" Body="&lt;p&gt;This reaction can be viewed as an example of a pericyclic (concerted) &lt;a href=&quot;http://en.wikipedia.org/wiki/Ene_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ene reaction&lt;/a&gt;.  &lt;strong&gt;The ene reaction is actually fairly common in organic chemistry&lt;/strong&gt;.   The following diagram details the mechanism as a concerted reaction with a cyclic 6 membered transition state involving 6 electrons (an &quot;aromatic&quot; transition state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ruO3H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A note on the Ene reaction...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ene reaction is really just a variation on the Diels-Alder theme.  The Diels-Alder reaction is a concerted [4 pi + 2 pi] reaction, whereas the Ene reaction is a [(2 pi + 2 sigma) + 2 pi] concerted reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZCToH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZCToH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-04-12T15:52:37.223" LastActivityDate="2020-04-12T15:52:37.223" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26573" PostTypeId="1" AcceptedAnswerId="26578" CreationDate="2015-02-24T19:12:44.737" Score="5" ViewCount="2157" Body="&lt;p&gt;I am getting a dilemma over this. Should the negative charge on the benzylic position participate in resonance with the benzene ring and make it more stable or is it the electronegativity of the $\ce{sp^2}$ hybridized carbon of the vinylic group that will make the carbanion more stable on it ? &lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="7475" LastEditDate="2015-10-09T16:10:35.217" LastActivityDate="2015-10-09T16:10:35.217" Title="Which is more stable — a carbanion on a vinylic position or a carbanion on a benzylic position?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26574" PostTypeId="1" CreationDate="2015-02-24T19:17:02.300" Score="2" ViewCount="267" Body="&lt;p&gt;I purchased commercial Ferrofluid, and when suspended, the water and/or alcohol turns into a brown sludge.  In 95% alcohol, all the metal particles separated from the &quot;tar&quot; Why, and how can I prevent this? &lt;/p&gt;&#xA;" OwnerUserId="14623" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:17:06.833" LastActivityDate="2015-10-30T17:17:06.833" Title="Separation of metal particles from fluid in ferrofluid" Tags="&lt;materials&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26575" PostTypeId="2" ParentId="26570" CreationDate="2015-02-24T19:36:55.237" Score="18" Body="&lt;p&gt;This is a tough question.  I think it might even be unfair to ask such a question on a test in non-advanced classes.  In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the &quot;real&quot; answer is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the &lt;em&gt;t&lt;/em&gt;-butyl or ethyl radicals which are not resonance stabilized.  You can see from the following diagram that you can draw two resonance structures for the allyl radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rt3xX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rt3xX.png&quot; alt=&quot;Resonance structures of allyl radical&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the benzyl radical you can draw even more resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9AYoh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9AYoh.png&quot; alt=&quot;Resonance structures of benzyl radicals&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the &quot;answer&quot; and reasoning provided &lt;a href=&quot;https://books.google.com/books?id=QV6cwXA9XkEC&amp;amp;pg=PA127&amp;amp;lpg=PA127&amp;amp;dq=which%20is%20more%20stable%20an%20allylic%20or%20benzylic%20radical&amp;amp;source=bl&amp;amp;ots=qgCrZyEymd&amp;amp;sig=jdZCw3GwXOgoe-b7Nwe1fXwscHU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=F8vsVMKtIompNq-igeAK&amp;amp;ved=0CEQQ6AEwBg#v=onepage&amp;amp;q=which%20is%20more%20stable%20an%20allylic%20or%20benzylic%20radical&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you do some thermochemical calculations (&lt;a href=&quot;http://www.saplinglearning.com/media/loudon/loudon5ech17sec02.pdf&quot; rel=&quot;noreferrer&quot;&gt;as done here&lt;/a&gt;), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-12-25T17:39:18.090" LastActivityDate="2018-12-25T17:39:18.090" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26576" PostTypeId="2" ParentId="26503" CreationDate="2015-02-24T19:37:49.140" Score="4" Body="&lt;p&gt;Unless you're at water's triple point (273.16 K, and a vapor pressure of 611.73 Pa), you must have fewer than three phases &lt;strong&gt;if all phases are in equilibrium&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's easy to have ice, water, and water vapor present simultaneously under other conditions &lt;strong&gt;if the phases are not all in equilibrium with each other.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Water has other triple points, but they involve multiple forms of ice.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-25T08:29:18.187" LastActivityDate="2015-02-25T08:29:18.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26577" PostTypeId="2" ParentId="26532" CreationDate="2015-02-24T19:49:54.940" Score="2" Body="&lt;p&gt;A state function is a &lt;strong&gt;property of a single state&lt;/strong&gt; of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A change in a state function is associated with a &lt;strong&gt;process that involves two states&lt;/strong&gt; of a system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be thinking that the change in enthalpy will be a state function because you can take any path you like between the initial and final states and still get the same enthalpy change. But that's beside the point; the enthalpy change isn't a property of a single state of the system, so it isn't a state function.&lt;/p&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12660" LastEditDate="2015-02-24T20:04:28.170" LastActivityDate="2015-02-24T20:04:28.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26578" PostTypeId="2" ParentId="26573" CreationDate="2015-02-24T20:15:12.487" Score="3" Body="&lt;p&gt;One of the more accurate methods to measure carbanion stabilities is negative ion photoelectron spectroscopy.  In this process, a carbanion is placed (generated) in the gas phase (in a high vacuum).  The gas-phase carbanion is then irradiated with light.  Each frequency of light provides a specific energy ($E=h\nu$).  The frequency of the light is varied (and the energy supplied changes as a consequence of the change in frequency) until enough energy is supplied to knocked out of one of the carbanion electrons (this is called photodetachment).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3^{-} + hν -&amp;gt; CH3• +~~ e^{-} +~ KE_{electron}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the photodetached electron is then experimentally measured and the electron affinity ($EA$) of the carbanion is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EA = hν ~~– ~~KE_{electron}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $EA$ is the energy required to remove one electron from the carbanion. The larger the electron affinity, the more tightly the electron was held.  Stable carbanions hold their electron more tightly than less stable carbanions. The electron affinity of the allyl anion was measured as 0.481 eV.  For the vinylic carbanion, the EA was measured as 0.667 eV.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that &lt;strong&gt;the vinyl carbanion is more stable than the allyl carbanion&lt;/strong&gt; suggests that &lt;strong&gt;hybridization effects outweigh resonance effects in terms of stabilizing these carbanions&lt;/strong&gt;.  That sounds reasonable and is what I would have guessed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ro.uow.edu.au/cgi/viewcontent.cgi?article=6579&amp;amp;context=scipapers&quot; rel=&quot;nofollow&quot;&gt;reference for EA measurements&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-02-24T20:27:08.157" LastActivityDate="2015-02-24T20:27:08.157" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26579" PostTypeId="1" CreationDate="2015-02-24T20:19:00.727" Score="3" ViewCount="1694" Body="&lt;p&gt;Using the usual notations of &lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{fus}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{vap}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{sub}$&lt;/span&gt;, as well as (&lt;span class=&quot;math-container&quot;&gt;$T_\mathrm l,p_\mathrm l$&lt;/span&gt;) and (&lt;span class=&quot;math-container&quot;&gt;$T_\mathrm s,p_\mathrm s$&lt;/span&gt;), which corresponds to the temperature and vapor pressure of liquid and solid phase, I have to derive the triple point temperature and pressure.&lt;/p&gt;&#xA;&lt;p&gt;My strategy is to use the Clausius-Clapeyron Relation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm dp}{\mathrm dT}= \frac{\Delta S}{\Delta V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;but the rest is history...&#xA;Could someone help me with&#xA;&lt;span class=&quot;math-container&quot;&gt;$\Delta H_\text{fus}$&lt;/span&gt; and (&lt;span class=&quot;math-container&quot;&gt;$T_\mathrm s,p_\mathrm s$&lt;/span&gt;)?&#xA;So that I know that I'm moving in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="14624" LastEditorUserId="83758" LastEditDate="2020-06-10T16:17:15.167" LastActivityDate="2020-06-10T16:17:15.167" Title="Deriving The Triple Point Temperature and Pressure as the following expression" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26580" PostTypeId="2" ParentId="26574" CreationDate="2015-02-24T21:01:51.333" Score="1" Body="&lt;p&gt;Assuming the ferrofluid is the one described at &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B008H40M72&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;product description&lt;/em&gt;&lt;/a&gt;, then it is &quot;an extremely fine powder, coated with a soap-like material called a surfactant, suspended in a mineral oil liquid base.&quot; From that description, it &lt;em&gt;might&lt;/em&gt; be possible to dilute or extend the ferrofluid with mineral (or even vegetable) oil, but one would not expect ferrofluid to be compatible with water or with alcohol. It might be possible to dilute it with other &lt;a href=&quot;http://en.wikipedia.org/wiki/Aliphatic_compound&quot; rel=&quot;nofollow&quot;&gt;aliphatic&lt;/a&gt; solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are directions for &lt;strong&gt;making your own water-based ferrofluid&lt;/strong&gt; at &lt;a href=&quot;http://voh.chem.ucla.edu/classes/Magnetic_fluids/pdf/instructorferro2.pdf&quot; rel=&quot;nofollow&quot;&gt;Synthesis of an Aqueous Ferrofluid&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-24T21:01:51.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26581" PostTypeId="2" ParentId="26567" CreationDate="2015-02-24T21:10:55.483" Score="3" Body="&lt;p&gt;It appears your creating &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_aluminates&quot; rel=&quot;nofollow&quot;&gt;calcium aluminate&lt;/a&gt; by reacting aluminum with alkaline water. Since that is probably &lt;em&gt;not&lt;/em&gt; wanted in the brew, you might use glass or &lt;em&gt;anodized&lt;/em&gt; aluminum, which is more resistant to chemical attack, and a water softener, if the calcium carbonate isn't wanted, either.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-24T21:10:55.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26583" PostTypeId="2" ParentId="26400" CreationDate="2015-02-24T21:38:47.130" Score="1" Body="&lt;p&gt;You could calculate an equilibrium constant at a given temperature for a reaction from the standard free energy change ($\Delta G^{\circ}$) for any reaction using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^{\circ} = -RT \ln K$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^{\circ}$ can be calculated for reactions using tabulated standard free energies of formation ($\Delta G_{f}^{\circ}$) using the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^{\circ} = \sum{\Delta G_{f}^{\circ} \mbox{(Products)}} - \sum{\Delta G_{f}^{\circ} \mbox{(Reactants)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction only contains solids and liquids, then K will still depend on the ratio between the product and reactant concentrations.  However, these concentrations will not change as a reaction moves towards equilibrium, which implies that the reaction quotient ($Q$) remains constant&lt;/p&gt;&#xA;" OwnerUserId="14596" LastActivityDate="2015-02-24T21:38:47.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26585" PostTypeId="1" CreationDate="2015-02-25T00:39:30.143" Score="0" ViewCount="448" Body="&lt;p&gt;What is the actual chemical composition of the grignard reagent and what is the method of preparation? Can it be prepared in college laboratories?&lt;/p&gt;&#xA;" OwnerUserId="14626" LastEditorUserId="275" LastEditDate="2015-02-25T03:55:33.860" LastActivityDate="2015-02-25T04:06:48.547" Title="What is the chemical composition of the grignard reagent?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26586" PostTypeId="1" AcceptedAnswerId="26599" CreationDate="2015-02-25T01:07:18.057" Score="-1" ViewCount="1573" Body="&lt;p&gt;What is the chemical composition of biodiesel and green diesel? How do they differ from each other?&lt;/p&gt;&#xA;" OwnerUserId="14626" LastEditorUserId="30836" LastEditDate="2020-04-05T21:00:27.623" LastActivityDate="2020-04-05T21:00:27.623" Title="What is the chemical composition of biofuels?" Tags="&lt;everyday-chemistry&gt;&lt;fuel&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-09T19:16:28.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="26589" PostTypeId="1" CreationDate="2015-02-25T02:56:12.407" Score="3" ViewCount="5853" Body="&lt;p&gt;How do I determine if a process is reversible without numeric information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice to know how to determine whether a process is reversible in order to solve problems LIKE this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A weather balloon is filled with Helium gas and released from the ground. It goes up 18km and achieves a diameter of 15m. Determine if the following values are greater than zero, less than zero, or equal to zero: ΔV, ΔP, ΔT, ΔU, ΔH, S_sys, S_surr, S_tot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my attempt at solving this problem. I believe at least half of my error comes from not knowing how to determine if this process, or any other process merely described, is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, ΔV increases because the balloon expands as the diameter gets to be large compared to what it presumably was at the time of release, even though the diameter at that time isn't mentioned.&#xA;ΔV &gt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔP &amp;lt; 0 because pressure decreases as altitude increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔT &amp;lt; 0 because temperature increases as altitude increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔU &amp;lt; 0 because ΔT &amp;lt; 0&#xA;ΔH &amp;lt; 0 because ΔT &amp;lt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that I have assumed that the external properties of T and P apply to the inside of the balloon, so I guess the system is reversible.&#xA;Stot = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ssys = -Ssurr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really know where to go from here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to argue that because the balloon actually does manage to overcome gravity, it's change in volume must be pretty substantial (making the w = -PdV be what causes the balloon to go up). Maybe it's more substantial than the change in temperature, giving a large positive term due to volume in the entropy state equation listed above.&#xA;Ssys &gt; 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the system is being looked at as the balloon itself rather than the air inside the balloon, then when the balloon isn't inflated, it has more freedom of movement than after it's fully inflated and therefore more entropy, meaning that the final entropy change of the system would be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am correct in guessing that this process is reversible AND in guessing that Ssys &gt; 0, then Ssurr &amp;lt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would rather not have all of my answers to physical chemistry problems be guesses. How can I solve this problem without guessing? How can I know if my guesses are correct or not? How can I better approach this problem? It's possible that the correct answer to this problem is due to something that I have completely overlooked. How do I not overlook considerations while doing these types of problems? &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2015-02-25T03:58:05.500" Title="How do I determine if a process is reversible without numeric information?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26590" PostTypeId="2" ParentId="26589" CreationDate="2015-02-25T03:44:52.633" Score="3" Body="&lt;p&gt;Any process is irreversible if it involves:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Friction&lt;/li&gt;&#xA;&lt;li&gt;Heat transfer across a finite temperature difference&lt;/li&gt;&#xA;&lt;li&gt;Expansion when there is a finite pressure difference&lt;/li&gt;&#xA;&lt;li&gt;Mixing of difference substances under any real circumstances&lt;/li&gt;&#xA;&lt;li&gt;Anything else where work is not being maximized&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So air friction against the balloon rising at a non-zero velocity makes the process irreversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat transfer from the balloon when the inside temperature is different from the outside temperature makes the process irreversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see: &lt;a href=&quot;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-25T03:55:53.580" LastActivityDate="2015-02-25T03:55:53.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26591" PostTypeId="2" ParentId="26589" CreationDate="2015-02-25T03:58:05.500" Score="0" Body="&lt;p&gt;This might help you understand why it is a reversible process&#xA;&lt;a href=&quot;http://how-it-looks.blogspot.com/2010/10/reversible-processes.html&quot; rel=&quot;nofollow&quot;&gt;http://how-it-looks.blogspot.com/2010/10/reversible-processes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and check out this website in order to get an understanding of how hot air balloon works&#xA;&lt;a href=&quot;http://socratic.org/questions/how-do-gas-laws-affect-hot-air-balloons&quot; rel=&quot;nofollow&quot;&gt;http://socratic.org/questions/how-do-gas-laws-affect-hot-air-balloons&lt;/a&gt;&#xA;Charles Law&#xA;chemed (.) chem (.) purdue (.) edu/genchem/topicreview/bp/ch4/gaslaws3.html&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking of whether or not something is reversible, ask yourself can the system goes back to the way it was before without exerting external work on the system.&lt;/p&gt;&#xA;" OwnerUserId="14624" LastActivityDate="2015-02-25T03:58:05.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26593" PostTypeId="2" ParentId="26585" CreationDate="2015-02-25T04:06:48.547" Score="5" Body="&lt;p&gt;A Grignard reagent is an organomagnesium halide of the generic form $\ce{RMgX}$, where $\ce{R}$ is a hydrocarbon group and $\ce{X}$ is a halide, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These reagents are produced by the reaction of magnesium metal with the corresponding organic halide in a non-protic (but coordinating) solvent, like diethyl ether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you would make phenylmagnesium bromide from bromobenzene and magnesium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PhBr(C6H5Br) + Mg -&amp;gt;[][\ce{Et2O}]PhMgBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are moisture and air sensitive, but not so much that you need to get crazy with inert atmosphere. These reagents are not stable outside of solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grignard reagents can be made in most laboratories set up for organic chemistry. In fact, they are &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=grignard%20reaction%20lab&quot;&gt;very common reactions to include in undergraduate organic chemistry lab courses&lt;/a&gt; because they are very well behaved.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-02-25T04:06:48.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26594" PostTypeId="1" CreationDate="2015-02-25T04:12:54.293" Score="1" ViewCount="1611" Body="&lt;p&gt;Is it possible to poison someone's food on a daily basis in order to give them cancer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing here is that i'm worried someone at my workplace might be poisoning other people's water (each employee has his own packs of water bottles at their desks, also, this person seems to have some mental issues). &#xA;It is possible to put chemicals in a bottle with no water leaking via a syringe. So my fear is that this bad person (this person IS bad., believe me) may have poisoned, in a daily basis for 3 or 4 weeks, our water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?&lt;/p&gt;&#xA;" OwnerUserId="14631" LastActivityDate="2016-03-08T14:19:59.913" Title="Can someone induce cancer in another person" Tags="&lt;water&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-08T17:04:33.157" ContentLicense="CC BY-SA 3.0" />
  <row Id="26596" PostTypeId="1" AcceptedAnswerId="26612" CreationDate="2015-02-25T05:23:12.600" Score="2" ViewCount="4897" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following is more stable ?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;$\ce{CH2=C(CH3)-CH=CH2}$&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;$\ce{CH3-CH=CH-CH3}$&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was going for answer 2 because it has 6 hyper conjugative effects. I found the answer given to be 1, which has 3 hyper conjugations but more resonance structures. Will the R-effect meet the deficit of 3 hyper conjugations and make 1 more stable than 2? &lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4945" LastEditDate="2015-02-25T07:07:30.307" LastActivityDate="2017-02-22T19:31:20.420" Title="Does resonance overrule hyper conjugation in terms of stability in alkenes?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ClosedDate="2017-02-23T03:38:06.500" ContentLicense="CC BY-SA 3.0" />
  <row Id="26597" PostTypeId="1" AcceptedAnswerId="26609" CreationDate="2015-02-25T06:09:19.117" Score="3" ViewCount="1480" Body="&lt;p&gt;Consider the hypothetical reaction described by the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3 A + 2 B -&amp;gt; C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reagent B was added in excess and the following time concentration data was obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{cr}&#xA;A\ [\mathrm{10^{-3}\ mol\cdot L^{-1}}] &amp;amp; \mathrm{time [s]} \\ \hline&#xA;5.0 &amp;amp; 10\\&#xA;2.9 &amp;amp; 50\\&#xA;1.9 &amp;amp; 80\\&#xA;1.1 &amp;amp; 120\\&#xA;0.7 &amp;amp; 150\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Derive the relevant integrated rate law and hence find the rate constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking that to solve this question you just sub the values into the integrate rate laws and see if it produces a linear relationship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Also could you please suggest a faster method of solving this type of question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. &lt;/p&gt;&#xA;" OwnerDisplayName="user14633" LastEditorUserId="418" LastEditDate="2015-02-25T06:43:23.233" LastActivityDate="2019-12-16T17:19:52.447" Title="How to Determine the Rate Law knowing time and concentrations" Tags="&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26598" PostTypeId="2" ParentId="26594" CreationDate="2015-02-25T06:17:31.213" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, in principle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information on known cancerogenic compounds can be found in &lt;strong&gt;CMR&lt;/strong&gt; (&lt;strong&gt;C&lt;/strong&gt;anceogenic, &lt;strong&gt;M&lt;/strong&gt;utagenic and toxic to &lt;strong&gt;R&lt;/strong&gt;eproduction) lists published by national organizations responsible for workplace safety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Germany, this would be the &lt;strong&gt;IFA&lt;/strong&gt; (&lt;em&gt;Institut für Arbeitsschutz der Deutschen Gesetzlichen Unfallversicherung&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://publikationen.dguv.de/dguv/pdf/10002/kmr_alte_bezeichnungen_2014.pdf&quot; rel=&quot;nofollow&quot;&gt;List with old nomenclature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://publikationen.dguv.de/dguv/pdf/10002/kmr_neue_bezeichnung_2014.pdf&quot; rel=&quot;nofollow&quot;&gt;List with new nomenclature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I have added these lists just for the sake of completeness!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you suspect any of the activities that you described, contact the local authorities.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-25T06:17:31.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26599" PostTypeId="2" ParentId="26586" CreationDate="2015-02-25T09:21:54.203" Score="2" Body="&lt;p&gt;&lt;strong&gt;Diesel fuel&lt;/strong&gt; is a fraction from the destillation of crude oil and consists of alkanes, cycloalkanes and aromatic hydrocarbons. In Europe, the properties of this mixture are defined in &lt;strong&gt;DIN EN 590&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This includes a certain density range, a minimum cetane number (CN) of 51, a maximum content of 8 mass-% for PAHs, a maximum sulfur content of 50 mg/kg. After distillation at tempertures up to 360 °C, the residue may not be larger than 5 vol-%. Note that diesel fuel may contain up to 7 vol-% biodiesel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Biodiesel&lt;/strong&gt; mostly is &lt;strong&gt;FAME&lt;/strong&gt;, &lt;em&gt;fatty acid methyl ester&lt;/em&gt;, produced by transesterification of triglycerides from biomass (vegetable oil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Green diesel&lt;/strong&gt;, also known as renewable diesel, is a fuel derived from biomass that does &lt;strong&gt;not&lt;/strong&gt; contain monoalkyl esters of fatty acids. This can be achieved by catalytic hydrodeoxygenation (hydrotreatment) of triglycerides or fatty acids and typically involves a cracking step too, leading to an alkane mixture.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-25T09:34:02.597" LastActivityDate="2015-02-25T09:34:02.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26600" PostTypeId="1" CreationDate="2015-02-25T10:02:07.543" Score="10" ViewCount="9067" Body="&lt;p&gt;According to modern atomic theory, the atom is the smallest particle which can take part in a chemical reaction. But during the formation of hydronium ion, $\ce{H+}$ ion reacts with $\ce{H2O}$ to form $\ce{H3O+}$ ion. Here the $\ce{H+}$ ion, which is smaller than hydrogen atom, takes part in a chemical reaction. So is that postulate invalid?&lt;/p&gt;&#xA;" OwnerUserId="14635" LastEditorUserId="4945" LastEditDate="2015-02-25T10:33:56.403" LastActivityDate="2015-02-25T17:15:20.997" Title="Is the atom the smallest particle, which takes part in chemical reactions?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26601" PostTypeId="1" AcceptedAnswerId="26602" CreationDate="2015-02-25T10:15:54.423" Score="2" ViewCount="247" Body="&lt;p&gt;Such as in the attached question I was taking into account the stability of the carbocations (by the number of resonating structures) formed after removal of Chlorine from and concluded to be the order I&gt;II&gt;III&gt;IV&gt;V the answer states the order to be I&gt;III&gt;IV&gt;V&gt;II&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can conclude what factors they are priority to. Help needed  &lt;img src=&quot;https://i.stack.imgur.com/dYDyD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14616" LastActivityDate="2015-02-25T16:44:53.690" Title="Which factor should I take into priority to determine reactivity of of Halides?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26602" PostTypeId="2" ParentId="26601" CreationDate="2015-02-25T11:01:10.723" Score="3" Body="&lt;p&gt;If you need to consider the stability of the carbocations obtained after chloride leaves, we can, as Klaus commented already, use Hückels rule, which more or less states that fully $\ce{sp^2}$ (i.e. usually $\ce{sp^2}$, as long as there is atleast one full p-orbital) hyberdized  compounds with $4n+2$ electrons are particularly stable or &lt;em&gt;aromatic&lt;/em&gt;. The same type of compounds that have $4n$ are particularly unstable, &lt;em&gt;anti-aromatic&lt;/em&gt;. Of course a carbocation has an empty p-orbital, so compunds I and III will be aromatic and compound II anti-aromatic. Compound I is fully benzenoid and also less strained than compound III and will hence be more stable, so I &gt; III. Then follows the allylic carbocation IV which is obviously more stable than its non-allylic counterpart V. And last but not least the anti-aromatic compound II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion I &gt; III &gt; IV &gt; V &gt; II, as the answer indicated.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-25T16:44:53.690" LastActivityDate="2015-02-25T16:44:53.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26603" PostTypeId="1" CreationDate="2015-02-25T11:11:11.613" Score="2" ViewCount="65" Body="&lt;p&gt;Platinum silicone is more dimensionaly stable, it shrinks less when it cures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, i just need silicone which has good compression set (small change by compression), and perhaps, once they are cured, platinum and tin catalyzed silicone become the same material - silicone elastomer. Perhaps that tin silicone, having shrunk, is more resistant to compression set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does tin silicone have better compression set? Are the highest quality o rings made from tin silicone?&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorUserId="14577" LastEditDate="2015-02-25T19:38:14.583" LastActivityDate="2015-03-27T20:36:59.610" Title="compression set of tin catalyzed silicone better than platinum?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26605" PostTypeId="2" ParentId="26600" CreationDate="2015-02-25T13:06:20.297" Score="13" Body="&lt;p&gt;Protons definitely participate in chemical reactions.  Free protons are generally not present in liquid water because a free proton is extremely reactive, but in the upper atmosphere or in other situations where the density of matter is low there can be free protons which participate in chemical reactions.&lt;/p&gt;&#xA;&lt;p&gt;However, I completely agree with Jori that an electron, which is much smaller than a proton, takes part in chemical reactions.  Gain or loss of one tiny electron can completely change the properties of a large metalloenzyme.&lt;/p&gt;&#xA;&lt;p&gt;How small is an electron vs. a proton? In terms of mass, the electron is 1836 times lighter, but what about in terms of radius. According to &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/1989/dehmelt-lecture.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hans Dehelt's Nobel Lecture&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Proton radius [is] &lt;span class=&quot;math-container&quot;&gt;$0.86 \times 10^{-15}$&lt;/span&gt;m...&lt;/p&gt;&#xA;&lt;p&gt;Today everybody “knows” the electron is an&#xA;indivisible atomon, a Dirac point particle with&lt;br /&gt;&#xA;radius R = 0....&#xA;But is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And goes on to explain that it is still possible the electron could have a radius on the order of &lt;span class=&quot;math-container&quot;&gt;$10^{-20}$&lt;/span&gt; m.&lt;/p&gt;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://gabrielse.physics.harvard.edu/gabrielse/overviews/ElectronSubstructure/ElectronSubstructure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Limit on Electron Substructure&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, photons, which have no mass, participate in chemical reactions.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-25T17:15:20.997" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26608" PostTypeId="2" ParentId="26579" CreationDate="2015-02-25T14:50:07.710" Score="4" Body="&lt;p&gt;On a phase diagram there are three 1-dimensional curves, the melting curve, the liquid vaporization curve, and the sublimation curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is sufficient to consider just two of the three curve, and determine the point at which they intersect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the limited information availible to you, (no information about how volume changes upon going from the solid to liquid phase), it is wise to choose the liquid vaporization curve and sublimation curve, rather than the melting curve, as $\Delta V$ can be approximated as the volume of gas, the volume of solid or liquid being approximately zero compare to the much larger volume of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Clapeyron equation in the question, you need to change variables from  $\Delta S$ which you don't know to $\Delta H$ which you do know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two phases are in equilibrium, such as along any of the three curves discussed above, $\Delta G$ = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = H-TS$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = dG = dH-TdS -SdT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that $T$ is constant during a phase change:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H = T \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dT}= \frac{ΔH}{TΔv}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dP}{dT}= \frac{ΔH_{vap}}{TΔv}$ and $\frac{dP}{dT}= \frac{ΔH_{sub}}{TΔv}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as mentioned above the approximation $\Delta V = V_{gas}$ can be made and the ideal gas law (or other equation of state) can be used to express $V$ as a function of $T$ and $P$.  Then (after intergrating) you have two equations with two unknowns, $T$ an $P$, the temperature and pressure at the triple point.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-27T16:18:44.297" LastActivityDate="2015-03-27T16:18:44.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26609" PostTypeId="2" ParentId="26597" CreationDate="2015-02-25T15:03:39.970" Score="5" Body="&lt;p&gt;You could plot the data to determine the order of the reaction.  I would use Microsoft Excel to do this.  Reproduce the data and then add columns that calculate the $\ln [A]$ and $\frac{1}{[A]}$.  Then plot $[A]$, $\ln [A]$ ,and $\frac{1}{[A]}$ vs. time (I would use separate plots).  The plot that gives the &quot;best&quot; linear fit (which you can assess using the $R^2$ value), corresponds to the appropriate integrated rate law.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If $[A]$ vs. time is linear, then the reaction is zero order.&lt;/li&gt;&#xA;&lt;li&gt;If $\ln [A]$ vs. time is linear, then the reaction is first order.&lt;/li&gt;&#xA;&lt;li&gt;If $\frac{1}{[A]}$ vs. time is linear, then the reaction is second&#xA;order.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="14596" LastActivityDate="2015-02-25T15:03:39.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26610" PostTypeId="1" AcceptedAnswerId="26625" CreationDate="2015-02-25T15:58:12.450" Score="10" ViewCount="22536" Body="&lt;p&gt;I came across a question which asks to work out the stability order for the following species-&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;b. &lt;img src=&quot;https://i.stack.imgur.com/s47MS.png&quot; height=&quot;80&quot; alt=&quot;benzene carbanion&quot;&gt;&lt;/p&gt;&#xA;&lt;p&gt;c. &lt;span class=&quot;math-container&quot;&gt;$\ce{R2C=CH-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can the lone pair on phenyl carbanion participate in resonance?  If yes, then how? Also, how is that more stable than the vinylic carbanion?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorDisplayName="user98979" LastEditDate="2020-09-19T07:31:50.263" LastActivityDate="2021-03-20T17:23:01.303" Title="Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26611" PostTypeId="1" AcceptedAnswerId="26614" CreationDate="2015-02-25T16:43:46.560" Score="12" ViewCount="2398" Body="&lt;p&gt;I know that there are 3 isotopes of Hydrogen. But according to this website it is showing 7 &lt;a href=&quot;http://periodictable.com/Properties/A/KnownIsotopes.html&quot;&gt;http://periodictable.com/Properties/A/KnownIsotopes.html&lt;/a&gt;&lt;a href=&quot;http://periodictable.com/Properties/A/KnownIsotopes.html&quot;&gt;Website&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13027" LastEditorUserId="418" LastEditDate="2015-02-25T16:57:13.270" LastActivityDate="2020-03-15T16:58:28.030" Title="Does hydrogen really have seven isotopes?" Tags="&lt;isotope&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26612" PostTypeId="2" ParentId="26596" CreationDate="2015-02-25T16:48:09.207" Score="4" Body="&lt;p&gt;We can hydrogenate unsaturated compounds and measure the heat given off, this is called the &lt;strong&gt;heat of hydrogenation&lt;/strong&gt;. As the following figure illustrates, if we hydrogenate the various &lt;strong&gt;butene&lt;/strong&gt; isomers they all wind up producing the same molecule, &lt;strong&gt;butane.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9ia3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Reactions_of_Alkenes/Catalytic_Hydrogenation_of_Alkenes&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, &lt;strong&gt;which butene is the most stable&lt;/strong&gt;. The heats of hydrogenation reported in the figure tell us that &lt;em&gt;trans&lt;/em&gt;-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in &lt;em&gt;trans&lt;/em&gt;-2-butene (6) compared to 1-butene (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only &lt;a href=&quot;https://books.google.com/books?id=tEjWioW2dH0C&amp;amp;pg=PA619&amp;amp;lpg=PA619&amp;amp;dq=heat%20of%20hydrogenation%20of%201%203%20butadiene&amp;amp;source=bl&amp;amp;ots=LvEVPe25-q&amp;amp;sig=O2fOOaRZDG5Bl70oUXixJEVbkO0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=TfTtVPz3CcSZNpjGg_AO&amp;amp;ved=0CDoQ6AEwCA#v=onepage&amp;amp;q=heat%20of%20hydrogenation%20of%201%203%20butadiene&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;57.1 kcal/mol&lt;/a&gt; of heat is given off, 3.5 kcal/mol less than we estimated. &lt;strong&gt;In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in &lt;em&gt;trans&lt;/em&gt;-2-butene (2.3 kcal/mol). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene &lt;strong&gt;plus&lt;/strong&gt; some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so &lt;strong&gt;answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-04-04T10:23:50.513" LastActivityDate="2015-04-04T10:23:50.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26613" PostTypeId="2" ParentId="26597" CreationDate="2015-02-25T16:51:01.037" Score="5" Body="&lt;p&gt;In addition to Tami M.'s answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to give &lt;a href=&quot;https://www.python.org/&quot; rel=&quot;noreferrer&quot;&gt;Python&lt;/a&gt;, the &lt;a href=&quot;http://ipython.org/notebook.html&quot; rel=&quot;noreferrer&quot;&gt;IPython Notebook&lt;/a&gt; and &lt;a href=&quot;http://matplotlib.org/&quot; rel=&quot;noreferrer&quot;&gt;matplotlib&lt;/a&gt; a try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;t = np.array([10, 50, 80, 120, 150]) &#xA;a = np.array([5.0, 2.9, 1.9, 1.1, 0.7])&#xA;&#xA;a_int = np.log(a)&#xA;&#xA;plt.figure()&#xA;plt.subplot(121)&#xA;plt.plot(t, a, 'ro-')&#xA;plt.xlabel('t [s]')&#xA;plt.ylabel('A')&#xA;&#xA;plt.subplot(122)&#xA;plt.plot(t, a_int, 'gs-')&#xA;plt.xlabel('t [s]')&#xA;plt.ylabel('ln(A)')&#xA;&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pIkpU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the plots are not fine-tuned and that I omitted a further fitting ;-)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-25T16:51:01.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26614" PostTypeId="2" ParentId="26611" CreationDate="2015-02-25T16:56:23.687" Score="28" Body="&lt;p&gt;Yes, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_hydrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;seven known isotopes of hydrogen&lt;/a&gt;, though only two (&lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^2H}$&lt;/span&gt;) are stable with respect to nuclear decay, and only three (&lt;span class=&quot;math-container&quot;&gt;$\ce{^1H}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{^2H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^3H}$&lt;/span&gt;) exist/can be made in enough quantities to be relevant outside of nuclear physics. All other hydrogen isotopes have extremely small half-lives. The next most stable hydrogen isotope after &lt;span class=&quot;math-container&quot;&gt;$\ce{^3H}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$t_{1/2}=12.3\ yr$&lt;/span&gt;) seems to be &lt;span class=&quot;math-container&quot;&gt;$\ce{^5H}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$t_{1/2} \sim 10^{-21}\ s$&lt;/span&gt;). Given that even the fastest chemical reactions happen in timespans of the order of &lt;span class=&quot;math-container&quot;&gt;$10^{-15}\ s$&lt;/span&gt;, we can safely discard any influence of hydrogen isotopes heavier than &lt;span class=&quot;math-container&quot;&gt;$\ce{^3H}$&lt;/span&gt; in chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Out of the currently 115 elements which have been confirmed to exist, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_nuclides#Summary_table_for_numbers_of_each_class_of_nuclides&quot; rel=&quot;nofollow noreferrer&quot;&gt;approximately 254 isotopes&lt;/a&gt; which are considered to be indefinitely stable, or have half-lives so enormous (over trillions of times greater than the age of the Universe, 13.8 billion years) that no value could yet be determined. Theoretically, &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotope#Radioactive.2C_primordial.2C_and_stable_isotopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;only 90 of those 254 isotopes&lt;/a&gt; are expected to be truly indefinitely stable (&lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_decay&quot; rel=&quot;nofollow noreferrer&quot;&gt;proton decay&lt;/a&gt; aside), but there are great difficulties in setting up experiments capable of measuring such rare decay events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of the previous 254, &lt;a href=&quot;http://en.wikipedia.org/wiki/Primordial_nuclide#List_of_34_radioactive_primordial_nuclides_and_measured_half-lives&quot; rel=&quot;nofollow noreferrer&quot;&gt;another 34 isotopes&lt;/a&gt; have half-lives which have actually been measured in laboratory, but are still large enough (&gt;50 million years) so that a small fraction of the isotope could conceivably have survived since the creation of the Earth 4.6 billion years ago, and may be of interest in geological timescales. There are also isotopes with smaller half-lives, but which are continuously replenished by the decay of heavier atoms or &lt;a href=&quot;https://en.wikipedia.org/wiki/Cosmogenic_nuclide&quot; rel=&quot;nofollow noreferrer&quot;&gt;by the action of cosmic rays&lt;/a&gt;, so they can still be found on Earth today. In all, about 339 different isotopes can be found naturally on our planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as many as &lt;a href=&quot;http://en.wikipedia.org/wiki/Table_of_nuclides_%28combined%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;3100-3300 different isotopes&lt;/a&gt; of the first 118 elements are claimed to have been detected in laboratories, most of them with very small half-lives (seconds or less). Nobody knows for sure how many elements and how many isotopes can exist, though the number is finite.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="17175" LastEditDate="2020-03-15T16:58:28.030" LastActivityDate="2020-03-15T16:58:28.030" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26615" PostTypeId="1" AcceptedAnswerId="26616" CreationDate="2015-02-25T17:30:24.687" Score="3" ViewCount="180" Body="&lt;p&gt;Why is $\ce{Pb^2+}$ present both in Group I and II of the basic radicals? &#xA;Is it related to the solubity product of its salts? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="4231" LastEditDate="2015-02-25T18:37:59.430" LastActivityDate="2015-02-25T18:37:59.430" Title="Salt analysis: Position of lead ion" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26616" PostTypeId="2" ParentId="26615" CreationDate="2015-02-25T18:13:46.553" Score="4" Body="&lt;p&gt;Yes, the Ksp for $\ce{PbCl2}$ is ~$10^{-5}$ while for $\ce{PbS}$ is ~$10^{-28}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the concentration of $\ce{Pb^2+}$ is too low to detect by dilute $\ce{HCl}$ precipitation, it is still possible it could be detected as the sulfide precipitate.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4231" LastEditDate="2015-02-25T18:37:50.173" LastActivityDate="2015-02-25T18:37:50.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26617" PostTypeId="1" CreationDate="2015-02-25T18:14:38.190" Score="0" ViewCount="4458" Body="&lt;p&gt;Came across the following question. The answer only says that option 3 is incorrect which is I agree to but dont you think option 4 is also incorrect. As per IUPAC rules isnt it the Halide that should get lowest possible number (in this case number 3 instead of 4) or is it both are same ? I am confused please help.&lt;img src=&quot;https://i.stack.imgur.com/PdTS7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4945" LastEditDate="2015-06-01T02:06:20.870" LastActivityDate="2015-06-01T14:35:31.010" Title="4-Bromo, 2,4-di-methylhexane and 3-Bromo, 3,5-di-methylhexane are the same thing?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26618" PostTypeId="2" ParentId="26617" CreationDate="2015-02-25T18:35:21.370" Score="0" Body="&lt;p&gt;According to the IUPAC organic chemistry nomenclature, alkyl groups take precedence over halide groups so (4) is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, C=C double bonds also take precedence over halides so (3) is incorrect and should be 2-bromoprop-1-ene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) is also incorrect as there are only five carbons in the chain not six so the correct name is 4-oxo&lt;strong&gt;pentan&lt;/strong&gt;oic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A table of functional group priorities can be found &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-02-25T18:35:21.370" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26620" PostTypeId="1" CreationDate="2015-02-25T19:14:47.973" Score="3" ViewCount="2695" Body="&lt;p&gt;The data sheet does not have any chemical shifts in this range for compounds containing only carbon and hydrogen&lt;/p&gt;&#xA;" OwnerUserId="14642" LastEditorUserId="418" LastEditDate="2015-02-25T19:24:36.810" LastActivityDate="2015-02-25T19:29:49.503" Title="Why do the two methylene protons in ethylbenzene have a chemical shift at 2-2.9?" Tags="&lt;spectroscopy&gt;&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26621" PostTypeId="2" ParentId="26620" CreationDate="2015-02-25T19:29:49.503" Score="4" Body="&lt;p&gt;Protons attached to a carbon that is connected to a benzene ring typically resonate in the range 2.2-2.9 ppm (see the Ar-C&lt;strong&gt;H&lt;/strong&gt; range in &lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/nmr.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the proton nmr spectrum for ethylbenzene.  The methylene protons absorb around 2.7 ppm, right in the upper end of the expected range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wlCgZ.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikibooks.org/wiki/Organic_Chemistry/Spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-02-25T19:29:49.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26622" PostTypeId="2" ParentId="26603" CreationDate="2015-02-25T20:33:50.433" Score="1" Body="&lt;p&gt;Tin catalyzed silicone lasts for a couple of years and platinum catalyzed lasts for 10 years, after which they lose their elasticity, becoming brittle and splitting. So while they may have equal properties for the first year, After a while the Platinum silicone will perform better. Using less catalyst for longer cure times also increases durability, and vacuum treating the silicone prior to casting and heat treatng it afterwards increases it's compression set performance.&lt;/p&gt;&#xA;" OwnerUserId="14577" LastActivityDate="2015-02-25T20:33:50.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26623" PostTypeId="1" CreationDate="2015-02-25T20:58:10.360" Score="2" ViewCount="137" Body="&lt;p&gt;I'm not totally sure which Stack Exchange community this would go in, but Chemistry seemed best.&#xA;I've been thinking for a long time now, that it would be cool to have a program where you could design your own atomic models, from the sub-atomic scale and up, then test them in a real situation. You could create anything from simple molecular structures to...anything.&#xA;I've looked around online, but there aren't any programs that you can build your own atomic structure. Anyone know if there &lt;em&gt;are&lt;/em&gt; any out there like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I've tried a few other programs, including Avogadro, &lt;a href=&quot;https://phet.colorado.edu/sims/build-a-molecule/build-a-molecule_en.jnlp&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, and another one by PHET.&lt;/p&gt;&#xA;" OwnerUserId="14643" LastEditorUserId="418" LastEditDate="2015-02-26T23:15:49.647" LastActivityDate="2015-04-19T23:59:14.520" Title="Molecular Assembly Program" Tags="&lt;atoms&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26624" PostTypeId="1" CreationDate="2015-02-25T21:44:27.193" Score="4" ViewCount="95" Body="&lt;p&gt;This question is about resveratrol in wine or grapes and not in some special delivery system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one is drinking some wine or eating grapes, does it matter if they keep it in their mouth a bit so the resveratrol within the wine or grapes get absorbed better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know now all nutrients can be gotten into onces body via buccal method so I am asking the experts what is the case with resveratrol from wine or grapes.&lt;/p&gt;&#xA;" OwnerUserId="14646" LastEditorUserId="46306" LastEditDate="2018-02-18T00:20:58.673" LastActivityDate="2018-02-18T00:20:58.673" Title="is the adsorption of resveratrol within wine or grapes linked with the time it was kept in your mouth?" Tags="&lt;biochemistry&gt;&lt;absorption&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26625" PostTypeId="2" ParentId="26610" CreationDate="2015-02-25T21:51:35.080" Score="8" Body="&lt;p&gt;The lone pair on the benzene is in the plane of the ring and hence orthogonal to the conjugated $\pi$-system so that it cannot interact with it. It is $\ce{sp^2}$ hybridized just like the vinylic carbanion, which also cannot resonate. Furthermore, both compounds lack an electron withdrawing group, so there is no inductive effect either. &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbanion&quot; rel=&quot;noreferrer&quot;&gt;Looking at the pKa values&lt;/a&gt; for both compounds, which usually gives a good indication about the stability of the anion, we see that there is indeed hardly any difference between them (pKa of benzene ~43 vs pKa of propene ~44 vs &lt;a href=&quot;http://bama.ua.edu/~kshaughn/ch531/531-handouts/pKa-values.pdf&quot; rel=&quot;noreferrer&quot;&gt;pKa of ethylene&lt;/a&gt; ~44).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question you only show us option (b) and (c). It would be nice if you could show the whole question including all possibilities (also as much as possible in text please, so the search engines can find it!)&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-02-25T23:11:45.610" LastActivityDate="2015-02-25T23:11:45.610" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="26626" PostTypeId="2" ParentId="6341" CreationDate="2015-02-25T21:59:37.050" Score="2" Body="&lt;p&gt;Here is a more complete solution&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9biG6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14648" LastActivityDate="2015-02-25T21:59:37.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26627" PostTypeId="2" ParentId="9051" CreationDate="2015-02-25T23:20:36.173" Score="1" Body="&lt;p&gt;It is predicted that francium is less reactive than caesium due to the speed at which the outer electron spins, adding strength to the bond between the atom's nucleus and the outer electron. While I am unable to fully comprehend quantum physics, a script of a 'Chemistry in its Element' podcast at the Royal Society of Chemistry, covers the subject about 2/3 of the way down:&#xA;&lt;a href=&quot;https://web.archive.org/web/20120730202925/http://www.rsc.org/chemistryworld/podcast/Interactive_Periodic_Table_Transcripts/Francium.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Francium (via the Internet Archive)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14649" LastEditorUserId="4945" LastEditDate="2020-01-28T18:13:35.023" LastActivityDate="2020-01-28T18:13:35.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26628" PostTypeId="1" AcceptedAnswerId="26630" CreationDate="2015-02-26T01:03:00.120" Score="1" ViewCount="1333" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The following reaction is not a complete reaction. Determine if the change is an oxidation, a reduction, or neither:&#xA;  $$\ce{CrO4^{-2} -&amp;gt; Cr2O7^{-2}}$$    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the answer to this problem is neither, but I don't have any idea as to how that answer is reached. &lt;strong&gt;Could anyone give me a hint?&lt;/strong&gt; I am familiar with oxidation, reduction, and oxidation numbers. &lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="4231" LastEditDate="2015-02-26T01:29:45.373" LastActivityDate="2015-03-12T01:41:11.217" Title="Determine if the following change is an oxidation, a reduction, or neither" Tags="&lt;electrochemistry&gt;&lt;oxidation-state&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26629" PostTypeId="2" ParentId="21984" CreationDate="2015-02-26T01:31:35.317" Score="1" Body="&lt;p&gt;It is very easy to purify the amino compounds from the reaction mixture of reduction of nitro- to amine- group. Firstly, we will do  workup after completion of the reaction by extraction method with organic solvents  either with DCM or with ehtylacetate, then organic layer wlii be separated, and then dried with anhydrous sodium sulphate and the solvent will be evaporated under reduced pressure by applying rotary evaporator. Then after making slurry by adding little amount of silica gel we will pack slurry in Column-chromatography silica gel (100-200 mesh) by using eluent as Ethylacetate:Hexane (20% to 60%),  the amino- group compond will be purify. This is so because nitro- group is less polar compared to anino- group. So, nitro- compound will come first from column chromatograhy and amino- compound will come letter in pure form.&lt;/p&gt;&#xA;" OwnerUserId="14622" LastActivityDate="2015-02-26T01:31:35.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26630" PostTypeId="2" ParentId="26628" CreationDate="2015-02-26T01:42:16.863" Score="3" Body="&lt;p&gt;Find the oxidation numbers of the substances in each compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should find the oxidation number of Cr in each compound to be +6, and the oxidation number of O to be -2. Since there is no change in the oxidation numbers, &lt;strong&gt;this reaction is neither an oxidation or reduction.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="4093" LastActivityDate="2015-02-26T01:42:16.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26631" PostTypeId="2" ParentId="9051" CreationDate="2015-02-26T02:08:32.343" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that reactivity increases down the group, but why does it not&#xA;  apply here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Francium has a &lt;span class=&quot;math-container&quot;&gt;$\ce{[Rn] 7s^1}$&lt;/span&gt; electron configuration. It is a relatively heavy element with atomic number &lt;span class=&quot;math-container&quot;&gt;$Z=87$&lt;/span&gt;. In such heavy elements, &lt;strong&gt;relativistic effects become significant&lt;/strong&gt; and impact the reactivity and other physical characteristics of the element. In francium, the inner &lt;span class=&quot;math-container&quot;&gt;$\ce{s}$&lt;/span&gt; electrons travel at roughly &lt;span class=&quot;math-container&quot;&gt;$\frac{[Z]}{[137]}$&lt;/span&gt; or &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26501/how-fast-do-electrons-move-around-the-nucleus/26535#26535&quot;&gt;64% the speed of light&lt;/a&gt;.  These relativistic speeds lead to a significant contraction in the radius of all s orbitals &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637&quot;&gt;as explained here&lt;/a&gt;. As the &lt;span class=&quot;math-container&quot;&gt;$\ce{7s}$&lt;/span&gt; orbital contracts, the &lt;span class=&quot;math-container&quot;&gt;$\ce{s}$&lt;/span&gt; electron&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;moves closer to the nucleus,&lt;/li&gt;&#xA;&lt;li&gt;is lowered in energy (because it is closer to the nucleus),&lt;/li&gt;&#xA;&lt;li&gt;becomes more shielded by other electrons (because it is closer to the nucleus),&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and, therefore, is less likely to be involved in chemical reactions.  Consequently, the reactivity of francium is significantly less than anticipated.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2020-01-28T18:07:02.567" LastActivityDate="2020-01-28T18:07:02.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26632" PostTypeId="1" CreationDate="2015-02-26T03:06:38.997" Score="2" ViewCount="6766" Body="&lt;p&gt;In this equilibrium equation, $$K = K_p\cdot (RT)^{-\Delta n}$$ what does $K$ represent in comparison to $K_p$? It seems to me that they both are equilibrium constants, so how are they different? &lt;/p&gt;&#xA;" OwnerUserId="12711" LastEditorUserId="194" LastEditDate="2018-02-16T14:17:58.410" LastActivityDate="2018-02-16T14:17:58.410" Title="What is the difference between K and Kp in the equilibrium equation?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26633" PostTypeId="2" ParentId="26632" CreationDate="2015-02-26T03:40:04.560" Score="2" Body="&lt;p&gt;K is for concentrations in Molarity. Kp is for what you're using pressures for reactions with all gases.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2015-02-26T03:40:04.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26634" PostTypeId="1" CreationDate="2015-02-26T03:40:44.307" Score="11" ViewCount="9134" Body="&lt;p&gt;My inorganic lab had us do an XRD measurement, but I've never been explained how to interpret the data.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the unit cell dimensions $a$, $b$, and $c$, for $\ce{YBa2Cu3O7}$ from the indexed X-ray powder pattern provided in [the textbook]. Explain why the crystals are nearly tetragonal in terms of the atomic structure of the compound. The following formula is useful:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\sin^2{\theta} = \frac{\lambda^2}{4}\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right)$$&#xA;  The three numbers, $hkl$, called Miller indices, indicate the direction of the scattering plane. The Miller indices can be read from the diffractogram provided. For simple reflections of the type $h00$, $0k0$, and $00l$, the value of $h$, $k$, or $l$ corresponds to &lt;em&gt;n&lt;/em&gt; in Bragg's law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;Bragg's law:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$$n\lambda = 2d\sin{\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The spectrum:&lt;/h1&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2CSxv.jpg&quot; alt=&quot;from Synthesis and Technique in Inorganic Chemistry&quot;&gt;&#xA;  (This is a scan from my textbook, while working on it I circled some items. Just ignore that.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h1&gt;My Attempt:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = 0.154~\mathrm{nm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scouring the internet I found the formula for tetragonal atomic structure: the axes $a = b \neq c$, and the axis angles $\alpha = \beta = \gamma = 90^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $c$:&#xA;$$\begin{align}&#xA;\sin^2{(22.9^\circ)} &amp;amp;= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2}+\frac{k^2}{b^2}+\frac{l^2}{c^2}\right) \\&#xA;&amp;amp;=\frac{0.154^2}{4} \left(\frac{0^2}{a^2}+\frac{0^2}{b^2}+\frac{3^2}{c^2}\right) \\&#xA;0.151 &amp;amp;= 0.0237\left(\frac{3^2}{c^2}\right) \\&#xA;c &amp;amp;= 1.188&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $a$ and $b$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sin^2{(22.9^\circ)} &amp;amp;= \frac{\lambda^2}{4} \left(\frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\right) \\&#xA;&amp;amp;= \frac{0.154^2}{4} \left(\frac{1^2}{a^2} + \frac{0^2}{b^2} + \frac{0^2}{c^2} \right) \\&#xA;0.151 &amp;amp;= 0.0237 \left(\frac{1^2}{a^2}\right) \\&#xA;a &amp;amp;= 0.156&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the right approach? If I have the right answer, what should the units be?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2018-03-08T02:28:29.190" LastActivityDate="2018-03-08T02:28:29.190" Title="How do you calculate a unit cell's dimensions from an XRD spectra?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26635" PostTypeId="1" AcceptedAnswerId="26661" CreationDate="2015-02-26T04:01:25.243" Score="3" ViewCount="1456" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{N2(g) + O2(g) &amp;lt;=&amp;gt; 2NO(g)}$, $K_c = 4.1 \cdot 10^{-4} (T=2000~^\circ\mathrm{C})$&lt;br&gt;&#xA;  What is $\ce{[NO]}$ when a mixture of $0.20~\mathrm{mol}$ $\ce{N2(g)}$ and $0.15~\mathrm{mol}$ $\ce{O2(g)}$ reach equilibrium in a $1.0~\mathrm{L}$ container at $T=2000~^\circ\mathrm{C}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When K is $\cdot10^{-4}$, you can assume the change is negligible, and just using the concentrations you can get from the given mole amount and the container volume, I get the right answer. When plugging back into, to see if it matches the give $K_c$, it's right:&lt;br&gt;&#xA;$$((0.20~\mathrm{M}~\ce{N2})(0.15~\mathrm{M}~\ce{O2}) \cdot 4.1\cdot10^{-4} K_c) = x^2.$$ Ans: $\ce{[NO]} = 0.0035~\mathrm{M}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read and heard you should be able to do it 'the hard' way, using the quadratic formula, like you would if $K_c$ wasn't to $10^{-4}$, and small enough. But I can't seem to get the right answer trying to do it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging everything, the equation I get to plug into the quadratic formula is&#xA;$$x^2 - 0.35x + 0.03,$$ then I get $0.2$ and $0.15$ as the answers. $0.2$ is too high, subtracting it from the $0.15~\mathrm{M}$ $\ce{O2}$ initial wouldn't be right. So using the $4.15$ as $x$, the change in $\mathrm{M}$, and subtraction that from the initial concentrations I have, then plugging that into the equation for $K_c$, even tried rearranging it in case I have got something mixed up, but I don't get anything close to the given $K_c$. &lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="69830" LastEditDate="2021-05-09T06:48:24.640" LastActivityDate="2021-05-09T06:48:24.640" Title="How to calculate the concentration of NO at equilibrium?" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26636" PostTypeId="2" ParentId="26632" CreationDate="2015-02-26T04:40:46.363" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot; rel=&quot;nofollow&quot;&gt;equilibrium constant&lt;/a&gt; is defined by the expression&#xA;$$K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.\tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x$ you can substitute a number of quantities, most commonly when working with solutions is &lt;a href=&quot;http://goldbook.iupac.org/A00295.html&quot; rel=&quot;nofollow&quot;&gt;(amount) concentration&lt;/a&gt; $c$ and when working with gasses often the partial pressure $p$ is used. In the latter case the equilibrium constants can be related via the &lt;a href=&quot;http://goldbook.iupac.org/I02935.html&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt; $$pV = n\mathcal{R}T.\tag2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into $(1)$ gives you&#xA;$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be transformed into&#xA;$$K_p=\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}} \prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{3a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the concentration is defined as&#xA;$$c_{\ce{B}}=\frac{n_{\ce{B}}}{V}\tag4$$&#xA;and from $(1)$ we can set up the concentration based equilibrium constant&#xA;$$K_c=\prod_{\ce{B}} c_{\ce{B}}^{\nu_{\ce{B}}}&#xA;     =\prod_{\ce{B}} \left(\frac{n_{\ce{B}}}{V}\right)^{\nu_{\ce{B}}}.\tag5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $5$ in $3$ we find&#xA;$$K_p=K_c\prod_{\ce{B}}\left(\mathcal{R}T\right)^{\nu_{\ce{B}}}.\tag{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the associativity of multiplication it follows therefore&#xA;$$K_p=K_c\cdot\left(\mathcal{R}T\right)^{\sum_\ce{B}\nu_{\ce{B}}}.\tag{7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often abbreviated with $K=K_c$ and $\sum_\ce{B}\nu_{\ce{B}}=\Delta n$ and therefore&#xA;$$K = K_p\cdot (\mathcal{R}T)^{-\Delta n}.\tag{7a}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Associative law of multiplications for exponents:&lt;/em&gt;&#xA;\begin{align}&#xA;  x^{n+m} &amp;amp;= x^n\cdot x^m\\&#xA;  x^{\sum d} &amp;amp;= \prod_d x^d&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-02-26T04:40:46.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26637" PostTypeId="2" ParentId="26628" CreationDate="2015-02-26T05:18:34.343" Score="5" Body="&lt;p&gt;As pointed out in the comments and the other answer, it is neither an oxidation, nor a reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: So, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: It is a &lt;strong&gt;condensation&lt;/strong&gt;: two molecules of an &lt;strong&gt;ortho acid&lt;/strong&gt; combine under loss of one molecule of water to yield a &lt;strong&gt;diacid&lt;/strong&gt; (&lt;strong&gt;pyro acid&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is frequently found in oxoacids, see&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H3PO4}$ (phosphoric acid) and $\ce{H4P2O7}$ (diphosphoric acid)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H3AsO4}$ (arsenic acid) and $\ce{H4AsO7}$ (diarsenic acid)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2SO4}$ (sulfuric acid) and $\ce{H2S2O7}$ (disulfuric acid)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-26T05:18:34.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26638" PostTypeId="1" AcceptedAnswerId="26640" CreationDate="2015-02-26T05:22:18.860" Score="3" ViewCount="559" Body="&lt;p&gt;What is the salt formation for iron(III) sulfate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia tells me that the production of iron(III) sulfate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 FeSO4 + H2SO4 + H2O2 → Fe2(SO4)3 + 2 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there was a difference between salt formation equation and production, because I thought it would've went something more like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 Fe(OH)3 + 3 H2SO4 → Fe2(SO4)3 + 6 H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="69830" LastEditDate="2021-05-09T06:58:33.867" LastActivityDate="2021-05-09T06:58:33.867" Title="How can iron(III) sulfate be formed?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26639" PostTypeId="1" CreationDate="2015-02-26T05:35:08.987" Score="2" ViewCount="449" Body="&lt;p&gt;I have trouble relating to the mole concept. For example in a question like the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many moles of $\ce{K}$ are in 1 mole of $\ce{KOH}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is it that in 1&amp;nbsp;mol of $\ce{KOH}$ there is 1&amp;nbsp;mol of $\ce{K}$ and &lt;em&gt;x&lt;/em&gt;&amp;nbsp;moles of $\ce{OH}$? How is it that 1&amp;nbsp;mol of $\ce{KOH}$ has 6.02&amp;times;10&lt;sup&gt;23&lt;/sup&gt; atoms, and then $\ce{K}$ has 6.02&amp;times;10&lt;sup&gt;23&lt;/sup&gt; atoms as well. In addition, $\ce{OH}$ will have 6.02&amp;times;10&lt;sup&gt;23&lt;/sup&gt; atoms, but if we add these two it's greater than the original number of atoms in 1&amp;nbsp;mol of $\ce{KOH}$?&lt;/p&gt;&#xA;" OwnerDisplayName="user14652" LastEditorUserId="4945" LastEditDate="2018-09-12T16:23:18.863" LastActivityDate="2018-09-12T16:23:18.863" Title="How to understand the concept of moles?" Tags="&lt;atoms&gt;&lt;molecules&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-02-27T04:00:42.030" ContentLicense="CC BY-SA 4.0" />
  <row Id="26640" PostTypeId="2" ParentId="26638" CreationDate="2015-02-26T05:37:53.223" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering if there was a difference between salt formation equation and production.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely! In this case, the latter means that you're trying to turn a side product (waste) of a technical product into another marketable product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to understand, why the oxidation of $\ce{FeSO4}$ is interesting, you have to realize that it is a side product in the synthesis of titanium dioxide ($\ce{TiO2}$). Here, ilmenite ($\ce{FeTiO3}$) is processed with concentrated sulfuric acid to yield $\ce{TiOSO4}$ as the valuable intermediate and $\ce{FeSO4}$ as &quot;waste&quot;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-26T05:37:53.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26641" PostTypeId="2" ParentId="26639" CreationDate="2015-02-26T05:41:14.020" Score="3" Body="&lt;p&gt;I had the same problem as well, until I sketched out the actual molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have one mole of $\ce{KOH}$, you have one mole of the collective &quot;molecule&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="4231" LastEditDate="2015-02-26T14:05:39.763" LastActivityDate="2015-02-26T14:05:39.763" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26643" PostTypeId="2" ParentId="26624" CreationDate="2015-02-26T06:04:19.600" Score="1" Body="&lt;p&gt;This article on the &lt;a href=&quot;http://dx.doi.org/10.1111/j.1749-6632.2010.05842.x&quot; rel=&quot;nofollow&quot;&gt;bioavailability or resveratrol&lt;/a&gt; seems to partly support your assumption: Keeping it in the mouth for a bit longer rather than swallowing directly might help to facilitate the uptake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solubility of resveratrol in water is apparently as low as 30 mg/L.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-26T06:04:19.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26644" PostTypeId="1" AcceptedAnswerId="26645" CreationDate="2015-02-26T07:18:28.353" Score="3" ViewCount="952" Body="&lt;p&gt;I'm stuck while finding what to use to make this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N3mlp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N3mlp.png&quot; alt=&quot;Retrosynthesis product&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first attempt was to use a triple bond as dienophile and a ring as my diene, as shown below -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gwwLw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwwLw.png&quot; alt=&quot;My attempt at the synthesis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure if this is right.&lt;/p&gt;&#xA;" OwnerUserId="14654" LastEditorUserId="83758" LastEditDate="2020-05-15T05:00:26.087" LastActivityDate="2020-05-15T05:00:26.087" Title="How to I identify the diene and dienophile through retrosynthesis of a given compound?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26645" PostTypeId="2" ParentId="26644" CreationDate="2015-02-26T08:04:32.920" Score="2" Body="&lt;p&gt;The $\ce{CO3}$ substituents are questionable. Let's assume that this is $\ce{CO2Me}$ instead!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is homework, I won't provide a full solution but:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Note that double bond patterns may isomerize (in the direction of conjugation).&lt;/li&gt;&#xA;&lt;li&gt;Alkynes can serve as dienophiles too!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pi2mn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-26T08:50:29.150" LastActivityDate="2015-02-26T08:50:29.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26646" PostTypeId="1" AcceptedAnswerId="26649" CreationDate="2015-02-26T08:08:52.577" Score="3" ViewCount="3893" Body="&lt;p&gt;Why does intramolecular hydrogen bonding make an organic (or possibly inorganic as well(?)) compound volatile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think is that it might be due to decrease in the solubility of the compound as intermolecular hydrogen bonding cannot be done after that. &lt;/p&gt;&#xA;" OwnerUserId="14530" LastEditorUserId="9961" LastEditDate="2015-10-02T17:01:37.677" LastActivityDate="2015-10-02T17:01:37.677" Title="Volatile nature due to hydrogen bonding" Tags="&lt;physical-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26647" PostTypeId="2" ParentId="26639" CreationDate="2015-02-26T08:16:05.240" Score="0" Body="&lt;p&gt;One mole of $\ce{KOH}$ means that there are Avogadro number of (6.023 × 10²³ molecules) $\ce{KOH}$ molecules and we need not consider the sum of number of constituent atoms.&#xA;For example,&#xA;1 mole of argon contains 6.023 × 10²³ argon atoms.&#xA;1 mole of ozone contains 6.023 × 10²³ ozone molecules (and also consists of 3 moles of oxygen atoms).&lt;/p&gt;&#xA;" OwnerUserId="14626" LastEditorDisplayName="user7951" LastEditDate="2015-09-16T15:45:16.963" LastActivityDate="2015-09-16T15:45:16.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26649" PostTypeId="2" ParentId="26646" CreationDate="2015-02-26T11:05:04.103" Score="4" Body="&lt;p&gt;&quot;Volatile&quot; usually refers to ease of evaporation, high vapor pressure, so it is a property of a pure substance while solubility is a property of the combination of two or more substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An intramolecular hydrogen bond makes a compound more volatile because the charge imbalances are offset internally instead of by forming an interaction with another molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interactions between different molecules of the substance would lower volatility, but intramolecular hydrogen bonds prevent the hydrogen bond donor and acceptor of one molecule from interacting with another molecule.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-26T11:05:04.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26650" PostTypeId="1" CreationDate="2015-02-22T01:39:06.307" Score="2" ViewCount="62" Body="&lt;p&gt;I have recorded an image of a large macromolecule and after several types of image analysis, have obtained a rotationally averaged image, that is not quite perfect, but appears to show 6-fold symmetry. With the origin on the axis of rotation, I compute a transform and unlike most of my colleagues, if I actually look at the amplitudes and phases of the transform. I have only one image of this molecule and no more specimen left. So my questions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What feature(s) I should see in the transform’s amplitudes and phases? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the amplitudes and phases do not conform exactly to my expectations, what would that mean? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the factors that degrade my image? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What should I do to make the transform conform to my expectations?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any ideas are welcome. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="15370" OwnerDisplayName="i_a_n" LastEditorUserId="9961" LastEditDate="2015-02-28T16:04:42.523" LastActivityDate="2015-02-28T16:04:42.523" Title="Amplitudes and phases of transform of an image" Tags="&lt;molecules&gt;&lt;symmetry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26651" PostTypeId="1" AcceptedAnswerId="26655" CreationDate="2015-02-26T11:52:41.003" Score="16" ViewCount="47597" Body="&lt;p&gt;How is the specific heat of water so high?&lt;br/&gt;&#xA;I was reading the hyperphysics article on it but could not understand it properly.&lt;/p&gt;&#xA;" OwnerUserId="10244" LastEditorUserId="10185" LastEditDate="2015-02-26T12:11:52.107" LastActivityDate="2016-08-20T16:46:05.170" Title="Why is the specific heat of water high?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26652" PostTypeId="1" CreationDate="2015-02-26T12:01:56.970" Score="4" ViewCount="109" Body="&lt;p&gt;So I was wondering if pre-heating coal will improve the efficiency of a steam boiler.&#xA;I did find that intake air pre-heater and water pre-heater do increase efficiency, but I could not find anything related to pre-heating coal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to take the hot exhaust, make a hot iron platform and keep the coal on top of it, it would heat it and maybe reduce the moisture content also?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemistry troll idea or actually feasible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: it is a traditional hand fired boiler&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coal type: Lignite&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Application: Industrial/Factory &lt;/p&gt;&#xA;" OwnerUserId="14655" LastEditorUserId="14655" LastEditDate="2015-02-26T13:09:48.947" LastActivityDate="2015-02-26T18:45:50.497" Title="Pre heat coal before burning" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26654" PostTypeId="2" ParentId="26652" CreationDate="2015-02-26T12:50:27.680" Score="5" Body="&lt;p&gt;&lt;em&gt;The update of your question changes everything!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Germany, around 25% of the base load is generated in lignite-fired power plants.&#xA;Lignite (brown coal) is awfully wet (50% water) and it is definitely a good idea to apply drying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the power plants, powdered lignite is used. Traditionally, the milled lignite is dried using the hot exhaust gas at temperatures up to 1000 °C, but modern plant like the one in Niederaußem operated the fluid bed drying at temperatures as low as 110 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, pre-drying of lignite is a good idea, if not essential, but it open to question how efficient the drying process of stacked lignite is.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-26T18:45:50.497" LastActivityDate="2015-02-26T18:45:50.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26655" PostTypeId="2" ParentId="26651" CreationDate="2015-02-26T13:17:30.880" Score="18" Body="&lt;p&gt;Water has hydrogen bonding in it. Hydrogen bonding is some kind of intermolecular force (&lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/hbond.html#top&quot; rel=&quot;noreferrer&quot;&gt;a tutorial&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page&lt;/a&gt;) that is usually seen in molecules that have &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{FH}$&lt;/span&gt; somewhere in their structure.&lt;/p&gt;&#xA;&lt;h2&gt;How does it happen?&lt;/h2&gt;&#xA;&lt;p&gt;Hydrogen atom is really small (atomic radius: About 37 pm) When it bonds with some very electronegative species, like O atom, (in the case of water), the electron is highly absorbed by the more electronegative. That results in a relatively high concentration of positive charge. The other O atom in water is attracted to H and this is the basis of hydrogen bonding.&lt;/p&gt;&#xA;&lt;h2&gt;So...?&lt;/h2&gt;&#xA;&lt;p&gt;Hydrogen bonding in water is far too weak to be compared with intramolecular bonds like covalent or ionic bonds, but it's strong enough that to be broken, it will require lots of energy. That's why water boils at &lt;span class=&quot;math-container&quot;&gt;$100 ° C$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Many know the reason of water's high specific heat to be hydrogen bonding. Partially, it's related to that: In giving heat to water, some heat is &amp;quot;spent&amp;quot; on loosening the hydrogen bonds rather than increasing water's kinetic energy. The argument is valid, and I haven't seen anyone bringing evidence as to its being wrong, but another reason, which was the one hyperphysics gave, relies on the motion of water.&lt;/p&gt;&#xA;&lt;p&gt;Expanding the topic isn't necessary, and it will add up to confusion. In short, the water molecule is nonlinear. That will result in three rotational &lt;a href=&quot;http://en.wikipedia.org/wiki/Degrees_of_freedom_%28physics_and_chemistry%29#Degrees_of_freedom_of_gas_molecules&quot; rel=&quot;noreferrer&quot;&gt;degrees of freedom&lt;/a&gt; that allows it to have an awful lots of rotational motions! (More types of motions are a cause of an additional bonus to the specific heat)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Liquid ammonia has a higher specific heat, just for the same reason.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Even a very easier reason to understand is the water's relatively low molar mass: &lt;span class=&quot;math-container&quot;&gt;$\frac{18 g}{mol}$&lt;/span&gt;. What should that mean? More moles of water per one kilogram.&lt;/p&gt;&#xA;&lt;p&gt;As a summary, the reasons are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The heated water will contribute much of the heat to loosening, bending or breaking the hydrogen bonds.&lt;/li&gt;&#xA;&lt;li&gt;Water had three rotational degrees of freedom. In addition to vibration, rotations happen a lot to water molecules. This will result in a higher heat capacity.&lt;/li&gt;&#xA;&lt;li&gt;Specific heat capacity is defined as the amount of heat required per unit &lt;strong&gt;mass&lt;/strong&gt; to increase the temperature by a degrees celsius. The relatively low molar mass of water allows more moles of it to be there in a mass unit (either kg or g)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;sub&gt;As a side note, the exceptional high specific heat is not the only weird property of water. See &lt;a href=&quot;http://www1.lsbu.ac.uk/water/thermodynamic_anomalies.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for example.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-02-26T14:44:11.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26656" PostTypeId="1" AcceptedAnswerId="26670" CreationDate="2015-02-26T13:51:13.710" Score="7" ViewCount="20748" Body="&lt;p&gt;To let you know where I'm coming from: I haven't studied chemistry since high school. I am now studying something else, but I have a dataset coming from your field and I would like some help in understanding just the very very basics of what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have is $\ce{NOx}$, nitric oxide and nitrogen dioxide in an engine exhaust and the equivalence ratio. It's from the dataset ethanol in the package SemiPar in R, and the original reference is Brinkman (1981). You'll find a figure at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I want:&lt;/strong&gt; I want to be able to, extremely briefly, explain what this is describing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Would you be able to provide a better explanation? Imagine you're trying to explain it to a five-year-old, that's about how much of this I know!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've read this far, please also help me with appropriate tags for the question. I'm not sure what to choose.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brinkman, N.D. (1981). Ethanol fuel – a single-cylinder engine study of efficiency and exhaust emissions. SAE transactions Vol. 90, No 810345, 1410–1424.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zhz5O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14658" LastEditorUserId="13184" LastEditDate="2015-02-26T21:06:21.377" LastActivityDate="2016-06-27T11:20:25.967" Title="Equivalence ratio explained" Tags="&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26657" PostTypeId="1" CreationDate="2015-02-26T14:15:11.763" Score="2" ViewCount="1848" Body="&lt;p&gt;Let's say that we have an element, say $\ce{Ca}$.&#xA;According to law, the last shell should contain only upto 8 e-.&#xA;Hence, electronic configuration of $\ce{Ca}$ would be 2,8,8,2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, according to the 2n2 rule, the M-shell can hold upto 18 e-.&#xA;So, doesn't this mean that the e- configuration of $\ce{Ca}$ can also be written as 2,8,9,1 or say, 2,8,7,3 or something? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it's not so... Why?&lt;/p&gt;&#xA;" OwnerUserId="12718" LastEditorUserId="4231" LastEditDate="2015-02-26T14:17:18.390" LastActivityDate="2015-02-26T17:26:41.853" Title="Electron Configuration of an element with no. of electrons &gt; 18" Tags="&lt;atoms&gt;&lt;electronic-configuration&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26658" PostTypeId="1" AcceptedAnswerId="26688" CreationDate="2015-02-26T14:28:48.993" Score="15" ViewCount="19181" Body="&lt;p&gt;Why is phenol more reactive towards electrophilic substitution than methoxybenzene? Isn't the lone pair on the methoxy group oxygen more available for participating in resonance into the ring than the lone pair on hydroxyl oxygen due to the +&lt;em&gt;I&lt;/em&gt; effect of the methyl group? Or is it because of the steric hindrance of the methoxy group that will favour lesser substitution at the &lt;em&gt;ortho-&lt;/em&gt; position of the benzene ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the hydrogen bonding of phenol have anything to do with this?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="10185" LastEditDate="2016-05-30T21:46:20.797" LastActivityDate="2019-03-24T14:56:16.177" Title="Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;reactivity&gt;&lt;phenols&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26660" PostTypeId="2" ParentId="26657" CreationDate="2015-02-26T15:39:01.890" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But, according to the 2n2 rule, the M-shell can hold upto 18 e-. So, doesn't this mean that the e- configuration of Ca can also be written as 2,8,9,1 or say, 2,8,7,3 or something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Experimentally, the ground state electron configuration is 2, 8, 8, 2, so there is no other correct way to write it.  2,8,9,1 and 2,8,7,3 would be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really the only way to know is through experiment or possibly very high level calculations that include relativity and quantum electrodynamics.  As atomic number increases, the problem becomes more and more complex.  It isn't necessarily true that one configuration accurately describes the ground state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for Lawrencium (element 103), for which no experimental determination has been made, you will see &lt;a href=&quot;http://www.nist.gov/pml/data/upload/Periodic_Table_nocrop_2014.pdf&quot; rel=&quot;nofollow&quot;&gt;some periodic tables that say it has a 7p electron&lt;/a&gt; (8 electrons in the P shell and 3 electrons in the Q shell) and &lt;a href=&quot;http://www.periodni.com/electronic_configuration_of_the_elements.html&quot; rel=&quot;nofollow&quot;&gt;some that say it has a 6d electron&lt;/a&gt; (9 electrons in the P shell and 2 electrons in the Q shell).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007EPJD...45..107F&quot; rel=&quot;nofollow&quot;&gt;The low-lying level structure of atomic lawrencium (Z=103): energies and absorption rates&lt;/a&gt; to appreciate how difficult it is to really calculate such information. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-26T17:26:41.853" LastActivityDate="2015-02-26T17:26:41.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26661" PostTypeId="2" ParentId="26635" CreationDate="2015-02-26T16:14:00.993" Score="2" Body="&lt;p&gt;$K_c = \frac{[\ce{NO}]^2}{[\ce{N2}][\ce{O2}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let x = [NO]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{x^2}{(0.20M-0.5x)(0.15M-0.5x)} = 0.00041$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 = 0.00041(0.25x^2 -0.175xM + 0.030M^2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x^2 + 0.000072xM -0.0000123M^2 = 0$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-27T03:55:21.383" LastActivityDate="2015-02-27T03:55:21.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26662" PostTypeId="1" CreationDate="2015-02-26T16:21:06.627" Score="4" ViewCount="1695" Body="&lt;p&gt;I mixed L Arginine with l Citrulline in Hibiscus and green tea water which was &lt;strong&gt;red&lt;/strong&gt; in color. Once the &lt;strong&gt;red&lt;/strong&gt; liquid hit the mix of powdered Aminoacids the solution changed &lt;strong&gt;green&lt;/strong&gt;. Can someone tell me why? Thanks so much for any info!&lt;/p&gt;&#xA;" OwnerUserId="14661" LastEditorUserId="7448" LastEditDate="2015-02-26T16:29:30.497" LastActivityDate="2020-10-28T00:42:25.010" Title="Why did the solution turn green?" Tags="&lt;experimental-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26663" PostTypeId="2" ParentId="26662" CreationDate="2015-02-26T17:07:55.293" Score="7" Body="&lt;p&gt;Both citrulline and arginine are nitrogen-rich, basic amino acids. A basic acid sounds confusing, but that's the way it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour change indicates a change in the pH of the solution (the tea mix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;pH indicator&lt;/strong&gt; that changes the colour is in the hibiscus flower and belongs to the class of the anthocyanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under acidic conditions, the dye exists as a deeply-red flavylium cation. At a slightly higher pH, the dye is reversibly converted to a colourless form. Here, the colour of the mixture is determined by the other ingredients. When more base (read: more amino acid) comes to action, further colour changes are conceivable and the double deprotonated form of some anthocyanes are known to be bluish-green.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="402" LastEditDate="2019-05-19T05:57:48.837" LastActivityDate="2019-05-19T05:57:48.837" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26664" PostTypeId="1" CreationDate="2015-02-26T17:44:09.953" Score="3" ViewCount="1328" Body="&lt;p&gt;So, here's the question which I'm having trouble solving :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively.  We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gave us:&#xA; $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$&#xA;Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gave us &#xA;$$x_{benz} =p_{benz} /P_{total} = 6.35 / 8.25= 0.770$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_{tol} =p_{tol} /P_{total} = 1.90 / 8.25 = 0.230$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, here's the answer to my question :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{tol} = x_{tol} * p^°_{tol} = 0,770 × 3,79  = 2,91 \text{ kPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{benz} = x_{benz} * p^°_{tol} = 0,230 × 12,7  = 2,92 \text{ kPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="13184" LastEditDate="2015-02-26T20:51:05.010" LastActivityDate="2015-02-26T21:02:16.410" Title="Applying Raoult's Law to a condensed gas mixture" Tags="&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26666" PostTypeId="1" AcceptedAnswerId="26667" CreationDate="2015-02-26T18:13:59.850" Score="1" ViewCount="207" Body="&lt;p&gt;&lt;em&gt;The story goes as follow:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I was in the electronic room I got bored and crushed a PNP transistor (Model &lt;a href=&quot;http://www.taydaelectronics.com/datasheets/A-115.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CDIL 2N2907&lt;/a&gt;) using a rotating pipe jaws just like in the picture bellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it took me some time to realize that I hear something like a sizzle or a tiny fire sound.&#xA;I looked at the crushed transistor and I saw a hole in the bent metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am almost clueless in Chemistry, but I guessed that there was some kind of reaction&#xA;with the substance when exposed to air (Normal air in Temperature room about 20C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at the datasheet of the exact part: &lt;a href=&quot;http://www.taydaelectronics.com/datasheets/A-115.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CDIL 2N2907&lt;/a&gt;&#xA;and I couldn't find any Chemistry info apart for the presence of silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that a N-doped semiconductor \ P-doped semiconductor are being used and in this case the semiconductor is silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Does (doped) silicon become toxic when exposed to air? Maybe there are other well known substances (insulator) being used with silicon when creating PNP that could react to air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the room had light (Fluorescent white light). Could it had any effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank in advance for any information on the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit :&lt;/strong&gt; The sound of sizziling came after I already crushed the transistor. Of course there was sound while crushing - of metal and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotrating pipe jaws:&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UypCZ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14662" LastEditorUserId="14662" LastEditDate="2015-02-26T19:53:34.393" LastActivityDate="2015-02-26T19:53:34.393" Title="Could PNP (Silicon) transistor be toxic when exposed to air\light?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26667" PostTypeId="2" ParentId="26666" CreationDate="2015-02-26T19:49:55.593" Score="3" Body="&lt;p&gt;The semiconductor itself is harmless, &lt;em&gt;but&lt;/em&gt;, since you heard some sizzling, it is &lt;em&gt;possible&lt;/em&gt; that a getter, such as an alkali metal, was in the metal-cased component. There is a patent for getters to be used with semiconductors, &lt;a href=&quot;http://patentimages.storage.googleapis.com/pdfs/US3259490.pdf&quot; rel=&quot;nofollow&quot;&gt;http://patentimages.storage.googleapis.com/pdfs/US3259490.pdf&lt;/a&gt;, though I've never seen one in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More likely, you heard the glass header breaking.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-26T19:49:55.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26669" PostTypeId="2" ParentId="26664" CreationDate="2015-02-26T21:02:16.410" Score="1" Body="&lt;p&gt;You have confused the &lt;em&gt;vapor&lt;/em&gt;-phase mole fraction with the &lt;em&gt;liquid&lt;/em&gt;-phase mole fraction.  The &lt;em&gt;liquid&lt;/em&gt;-phase is mole fraction in usually denoted as $x_i$ for compound $i$.  The &lt;em&gt;vapor&lt;/em&gt;-phase mole fraction of $i$ is $y_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raoult's law is that the partial pressure over an ideal solution is equal to the pure-component vapor pressure times the liquid mole fraction, $p_i=x_i*p^°_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separate from Raoult's law we have the ideal gas law. It says that that says that partial pressures are proportional to mole fraction in the &lt;em&gt;vapor phase&lt;/em&gt;, i.e. that $p_i=y_iP$.  So when you took the ratios of $p_{benz}/P$ and $p_{tol}/P$, you were solving for the vapor-phase mole fractions, i.e. for $y_{benz}$ and $y_{tol}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mistake comes in taking those $y_i$ values and treating them as $x_i$ values.  That's a mis-formulation of Raoult's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about Raoult's law and the ideal gas law for binary ideal mixtures in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/25034/why-does-decreasing-the-pressure-of-the-system-increase-the-relative-volatility/25057#25057&quot;&gt;my answer&lt;/a&gt; to a previous chem.se question.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-02-26T21:02:16.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26670" PostTypeId="2" ParentId="26656" CreationDate="2015-02-26T21:22:46.730" Score="6" Body="&lt;p&gt;You are correct.  There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$.  This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel.  For fuel that are pure octane, the combustion reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac{25}{2}O2 + C8H18 -&amp;gt; 8CO2 + 9H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means 12.5 moles of oxygen are required for every mole of octane.  For 12.5 moles of oxygen you need about 60 moles of &lt;em&gt;air&lt;/em&gt; because air is only 21% oxygen.  That works out to 1740 grams of air (~29 grams per mole).  One mole of octane is 114 grams.  So the theoretical, or &lt;em&gt;stoichiometric&lt;/em&gt; ratio of air to fuel is 1740/114 or about 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most motor fuels are not pure octane.  The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Air%E2%80%93fuel_ratio#Air.E2.80.93fuel_equivalence_ratio_.28.CE.BB.29&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;equivalence ratio&lt;/em&gt;&lt;/a&gt; is just the &lt;em&gt;actual&lt;/em&gt; air-to-fuel ratio divided by the theoretical.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You said the second part backwards.  If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel).  So incomplete combustion should happen more at lower equivalence ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the exhaust temperature will be maximized for equivalence ratios close to 1.  Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase.  Since temperature is important for NOx formation, I figured it was worth mentioning.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="31432" LastEditDate="2016-06-27T11:20:25.967" LastActivityDate="2016-06-27T11:20:25.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26671" PostTypeId="1" CreationDate="2015-02-26T21:24:06.090" Score="1" ViewCount="89" Body="&lt;p&gt;I just found a random example so I could highlight my inquiry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chemist wants to make .5L of a 20M solutionusing $\ce{NaOH}$. The solution will be made from solid $\ce{NaOH}$ and water. How many grams of&#xA;  $\ce{NaOH}$ are needed to make this solution?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case, I am finding the moles of $\ce{NaOH}$ that would equate to the number of moles in 0.5L of a 20M solution. But, is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L ? or am I adding x grams of $\ce{NaOH}$ to .5L of $\ce{H2O}$ and not considering the changes in volume?&lt;/p&gt;&#xA;" OwnerDisplayName="user14652" LastEditorUserId="418" LastEditDate="2015-02-26T23:33:23.237" LastActivityDate="2015-02-26T23:33:23.237" Title="Concentration Inquiry" Tags="&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26672" PostTypeId="2" ParentId="26623" CreationDate="2015-02-26T22:01:21.513" Score="2" Body="&lt;p&gt;You  might want to try BioVia's (formerly Accelrys) Discovery Studio Visualizer. This is free software and the &quot;gateway drug&quot; to their sophisticated suite of molecular modeling and chem/bio informatics software. It is a step above the various drawing programs out there and lets the user generate realistic 3-D models of an infinite variety of chemical structures. &lt;a href=&quot;http://accelrys.com/products/discovery-studio/visualization-download.php&quot; rel=&quot;nofollow&quot;&gt;http://accelrys.com/products/discovery-studio/visualization-download.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-02-26T22:01:21.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26673" PostTypeId="2" ParentId="26671" CreationDate="2015-02-26T22:05:15.337" Score="2" Body="&lt;p&gt;If the final volume of the solution is .5L then what is important here is how many moles/grams of the substance would be used to make the solution. In general when making a solution, the solution is raised to the final volume after the mixture. So whether .5L of $\ce{H2O}$ is used, or .4998L of $\ce{H2O}$ is used doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2015-02-26T22:05:15.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26674" PostTypeId="2" ParentId="26671" CreationDate="2015-02-26T22:09:55.730" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would be the &lt;strong&gt;exact&lt;/strong&gt; way!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molarity is defined as $\mathrm{\frac{moles_{solute}}{\mathbf{total}\ volume_{solution}}}$ &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4231" LastEditDate="2015-02-26T23:26:02.997" LastActivityDate="2015-02-26T23:26:02.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26675" PostTypeId="2" ParentId="26623" CreationDate="2015-02-26T22:57:26.827" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] create your own models, molecular structures, etc., then put them into a real situation and see how they act. This could maybe be used to create new medicines [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imagine a huge protein that switches a malfunction (on or off) to be like a &lt;strong&gt;lock&lt;/strong&gt;, while small organic compounds may be seen as possible &lt;strong&gt;keys&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;synthesize a huge library of potential &lt;strong&gt;keys&lt;/strong&gt; and test them in the lab or&lt;/li&gt;&#xA;&lt;li&gt;build a library of &lt;strong&gt;keys&lt;/strong&gt; on the computer and calculate their geometries, charge distribution, etc. using quantumchemical methods and then again use other programmes to examine the interaction between the &lt;strong&gt;keys&lt;/strong&gt; and the &lt;strong&gt;lock&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The latter method, figuring out how good the keys fit, is known as &lt;strong&gt;docking&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to have a look at the &lt;a href=&quot;http://www.ball-project.org/caddsuite&quot; rel=&quot;nofollow&quot;&gt;CADDSuite&lt;/a&gt;, developed by Oliver Kohlbacher at the University of Tübingen, but please note that these tools tend to be a tad above the entry level of simple moelcule editors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an online service, have a look at &lt;a href=&quot;http://www.swissdock.ch/&quot; rel=&quot;nofollow&quot;&gt;SwissDock&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5017" LastEditDate="2015-04-19T23:59:14.520" LastActivityDate="2015-04-19T23:59:14.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26676" PostTypeId="1" AcceptedAnswerId="26681" CreationDate="2015-02-27T00:16:43.313" Score="1" ViewCount="85" Body="&lt;p&gt;I am new to chemistry. I am looking for a gas that would de-oxidise aluminium. &#xA;Which gases would work for that and at what temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details: I need to sinter aluminium powder. But because the top layer of aluminium is oxidized when exposed to the atmosphere, I need to remove the oxide film before the sintering would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sergey&lt;/p&gt;&#xA;" OwnerUserId="14665" LastActivityDate="2015-02-27T03:23:19.767" Title="Aluminum deoxidation gas" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26677" PostTypeId="1" CreationDate="2015-02-27T01:23:22.707" Score="4" ViewCount="946" Body="&lt;p&gt;Say you have acid X for which $K_a(X)$ = 1E-7 and acid Y for which $K_a(Y)$ = 0.8E-7. In separate solutions, it would be easy to calculate $[H^+]$ because there would be only one dominant equilibrium. But in solution together, both contribute significantly to $[H^+]$, and as a result, both equilibria must be driven to the left by Le Chatelier's Principle. How do you calculate the pH of such a solution? Since both contribute to $[H^+]$, the resulting $[H^+]$ is not additive, but rather a bit less than that because of suppression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume X and Y are 0.1M and monoprotic.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="10185" LastEditDate="2015-02-27T15:34:58.387" LastActivityDate="2015-02-27T16:44:30.333" Title="How to calculate pH of a solution with two dominant equilibria?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26679" PostTypeId="1" AcceptedAnswerId="26682" CreationDate="2015-02-27T02:22:38.973" Score="0" ViewCount="2761" Body="&lt;p&gt;Took a pic of the paper I tried it on it on, has the exact problem description written at top, and what I tried. Figured that would be easier then typing all that. The answer is supposed to be 2.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lLZD2.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2015-02-27T18:32:37.563" Title="Finding Kp at another temp with /\Hrxn" Tags="&lt;equilibrium&gt;&lt;concentration&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-02-27T12:09:12.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="26681" PostTypeId="2" ParentId="26676" CreationDate="2015-02-27T03:23:19.767" Score="1" Body="&lt;p&gt;Since Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; is &lt;em&gt;very&lt;/em&gt; stable, powerful acids such as HF and chromic acid are used to remove the oxide surface. Since you're working with fine particles and wish to use a &lt;em&gt;gas&lt;/em&gt;, rather than &lt;em&gt;liquid&lt;/em&gt;, you'll need to consider what will become of the removed oxides. Gaseous HF would leave fluorides behind. Perhaps atomic hydrogen might work, but you'd have to remove the resultant water before the Al reoxidizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easier to prepare powdered aluminum &lt;em&gt;in an inert atmosphere&lt;/em&gt; from bulk aluminum. There are a number of sites showing how to grind aluminum using anything from a file to a coffee grinder; you'd need to adapt this to run in argon, perhaps with a glove-box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sintering might be difficult, since Al is so soft that pore spaces would close.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-02-27T03:23:19.767" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26682" PostTypeId="2" ParentId="26679" CreationDate="2015-02-27T04:36:13.413" Score="2" Body="&lt;p&gt;You used the wrong version of $R$.  Try using $R = 8.314 J mol^{-1} \ K^{-1}$.  &lt;/p&gt;&#xA;" OwnerUserId="14596" LastEditorUserId="14596" LastEditDate="2015-02-27T18:32:37.563" LastActivityDate="2015-02-27T18:32:37.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26683" PostTypeId="1" AcceptedAnswerId="26687" CreationDate="2015-02-27T06:42:17.800" Score="7" ViewCount="9705" Body="&lt;p&gt;I had to draw the molecular shape for $\ce{NHF2}$ as a Trigonal Planar like this&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LD8ML.png&quot; alt=&quot;NHF2 Structure Diagram &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I check to make sure it was right, I came across a source saying &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NHF2-difluoroammonia--tetrahedral, almost trigonal&#xA;  (&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20100304194348AA5Rjff&quot; rel=&quot;noreferrer&quot;&gt;Source 1&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After looking some more I found a source that now said it was &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trigonal pyramidal, Bond angle of 107 degrees Bond angle of F-N-F &gt; F-N-H &#xA;  (&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20120531081702AAfe0Mg&quot; rel=&quot;noreferrer&quot;&gt;Source 2&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone clarify which one it is so I can understand.&lt;/p&gt;&#xA;" OwnerUserId="12820" LastEditorUserId="4945" LastEditDate="2015-02-27T09:52:53.517" LastActivityDate="2017-03-06T00:08:56.157" Title="What is the molecular geometry of NHF₂?" Tags="&lt;lewis-structure&gt;&lt;vsepr-theory&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26684" PostTypeId="2" ParentId="26683" CreationDate="2015-02-27T07:26:02.713" Score="4" Body="&lt;p&gt;Assuming for a moment that VSEPR is a valid concept, &#xA;difluoramine ($\ce{NF2H}$) is represented by $\bf{AX_3E_1}$, and so is ammonia ($\ce{NH3}$). Try to walk your way down from there. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-27T07:26:02.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26686" PostTypeId="2" ParentId="26677" CreationDate="2015-02-27T08:22:42.443" Score="4" Body="&lt;p&gt;The first step is to write down the variables of interest, and the equations that relate them.  If there exactly as many (non-redundant) equations as variables, then the system might be solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the variables are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{[H+]}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[X^{-}]}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[Y^{-}]}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[HX]}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[HY]}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[OH^{-}]}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The equations are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$&lt;/li&gt;&#xA;&lt;li&gt;HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$&lt;/li&gt;&#xA;&lt;li&gt;HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$&lt;/li&gt;&#xA;&lt;li&gt;charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$&lt;/li&gt;&#xA;&lt;li&gt;X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$&lt;/li&gt;&#xA;&lt;li&gt;Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, &quot;all&quot; you have to do is solve those equations for the six unknown variables.  That isn't easy to do with pen and paper, but it can be done.  Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-27T08:22:42.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26687" PostTypeId="2" ParentId="26683" CreationDate="2015-02-27T09:50:18.563" Score="19" Body="&lt;p&gt;You can look up the molecule on &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.23575.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemspider&lt;/a&gt;, where you have a little applet for the 3D structure. Or you can download a coordinate file from &lt;a href=&quot;http://webbook.nist.gov/cgi/inchi/InChI%3D1S/F2HN/c1-3-2/h3H&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST&lt;/a&gt; and view it in a molecular viewer, like &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;. Or keep on reading for some deeper insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is trigonal pyramidal, like ammonia. I performed a DF-BP86/def2-SVP calculation on this to analyse the electronic structure a bit more.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uh24I.jpg&quot; alt=&quot;molecular structure of HNF2&quot;&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/H4NNf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;One reason why VSEPR does fail (a lot) is that lone pairs are more stable in orbitals with high s character. Another point that is often observed is that orbitals with higher p character are directed to more electronegative elements (Bent's Rule).&lt;sup&gt;[1,2]&lt;/sup&gt; As a result of that, the bond angle between electronegative elements gets smaller. This can be observed here, too. From the ideal sp³ arrangement we would expect an 109.5&amp;deg; angle, but here it is about 5&amp;deg; smaller. As a consequence also the hydrogen-nitrogen-fluorine angle has to decrease. In this case the bond angle becomes much smaller because of an intramolecular hydrogen bond.&lt;sup&gt;[3]&lt;/sup&gt;&lt;br&gt;&#xA;The NBO analysis consequently builds the same picture:&#xA;$$\begin{array}{llll}\hline&#xA;\text{type}&amp;amp;\text{main contribution}&amp;amp;\text{other contributions}\\\hline&#xA;\sigma\text{-Bd}~\ce{N-F}&amp;amp; 64\%~\ce{F}~(14\% s + 86\% p) &amp;amp; 36\%~\ce{N}~(15\% s + 85\% p) \\&#xA;\sigma\text{-Bd}~\ce{N-H}&amp;amp; 68\%~\ce{N}~(25\% s + 75\% p) &amp;amp; 32\%~\ce{H}~(100\% s) \\&#xA;\sigma\text{-LP}~\ce{N} &amp;amp; 100\%~\ce{N}~(52\% s + 48\% p) &amp;amp;  \\&#xA;\sigma\text{-LP}~\ce{F} &amp;amp; 100\%~\ce{F}~(75\% s + 25\% p) &amp;amp;  \\&#xA;\sigma\text{-LP}~\ce{F} &amp;amp; 100\%~\ce{F}~(10\% s + 90\% p) &amp;amp;  \\&#xA;\sigma\text{-LP}~\ce{F} &amp;amp; 97\%~\ce{F}~(100\% p) &amp;amp;  2\%~\ce{N}~(7\% s + 93\% p) &amp;amp; 1\%~\ce{F2}~(15\% s + 85\% p)\\\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planar (trigonal) structure on this level of theory is about 33 kcal/mol higher in energy. It is the transition state for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_inversion&quot; rel=&quot;nofollow noreferrer&quot;&gt;nitrogen inversion&lt;/a&gt;, which is therefore very slow.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes&lt;/em&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bent's rule on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Bent's rule on our network: &lt;a href=&quot;https://chemistry.stackexchange.com/q/15671/4945&quot;&gt;Utility of Bent&amp;#39;s Rule - What can Bent&amp;#39;s rule explain that other qualitative considerations cannot?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The bond length is much smaller than the sum of the corresponding van der Waals radii, suggesting a hydrogen bond. Compare &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100785a001&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. Bondi &lt;em&gt;J. Phys. Chem.&lt;/em&gt;, &lt;strong&gt;1964&lt;/strong&gt;, &lt;em&gt;68&lt;/em&gt; (3), pp 441–451&lt;/a&gt;.&lt;br&gt;&#xA;$r_{\mathrm{vdW}}(\ce{HF})=r_{\mathrm{vdW}}(\ce{H})+r_{\mathrm{vdW}}(\ce{F})= (106 + 140)~\mathrm{pm}= 246~\mathrm{pm}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2017-03-06T00:08:56.157" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26688" PostTypeId="2" ParentId="26658" CreationDate="2015-02-27T11:22:42.080" Score="8" Body="&lt;p&gt;First of all we must have to note that the difference in reactivity between anisole (methoxybenzene) and phenol is small. Small enough for the undergraduate texts I own to not mention anisole separately from phenol, just referring to it as having &quot;phenol-like reactivity&quot; (i.e. strong &lt;em&gt;ortho-para&lt;/em&gt; directing). We can more or less measure the difference in &lt;em&gt;ortho-para&lt;/em&gt; directing strength by looking at the proton NMR spectra of anisole and phenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QeGa.png&quot; alt=&quot;Anisole HNMR spectrum&quot;&gt;&#xA;&lt;a href=&quot;http://www.hanhonggroup.com/nmr/nmr_en/RB03030028.html&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7wVzr.png&quot; alt=&quot;Phenol HNMR spectrum&quot;&gt;&#xA;&lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_108-95-2_1HNMR.htm&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see they are almost identical to the point where you can ask yourself if there is any practical difference in reactivity at all. The A protons nicely reflect the ortho-para direction since they haven't shifted from benzene's 7.26 ppm signal and the B and C protons are all upfield as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I don't think steric hindrance is a big factor either since aromatic electrophilic substitution goes first trough a $\pi$-intermediate and a single methyl-group is not that bulky.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-02-27T11:22:42.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26689" PostTypeId="1" AcceptedAnswerId="26691" CreationDate="2015-02-27T13:31:38.823" Score="3" ViewCount="732" Body="&lt;p&gt;I have measured the pH in the water-soluble fraction of a powder. I weighed in a certain amount of powder and a certain amount of water, and extracted the water-soluble fraction. I then measured the pH in this extract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I report this? The pH depends on the amount of powder and the amount of water. Should I give the weighed-in amounts along with the pH-value? Or is there a way to &quot;normalize&quot; the pH-values to make them comparable for all? Just giving the pH values make no sense, as no one can compare anything to them.&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="7448" LastEditDate="2015-02-27T13:33:04.367" LastActivityDate="2015-02-27T20:15:28.490" Title="How to report pH values?" Tags="&lt;ph&gt;&lt;analytical-chemistry&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26690" PostTypeId="2" ParentId="26689" CreationDate="2015-02-27T13:43:10.327" Score="1" Body="&lt;p&gt;I would say &quot;pH of x% solution&quot; if you didn't separate the extract from the insoluble fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say &quot;pH of y:1 water extract&quot; you performed a separation.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7448" LastEditDate="2015-02-27T20:15:28.490" LastActivityDate="2015-02-27T20:15:28.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26691" PostTypeId="2" ParentId="26689" CreationDate="2015-02-27T13:58:08.167" Score="2" Body="&lt;p&gt;Will it become more clear for a reader if you use an an abbreviated, &quot;coded&quot; form to express what you have actually done? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why waste your time to encode it and waste the time of the reader to decode it (and maybe still leave space for ambiguity)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be fine with a report that precisely describes the experimental procedure and the observations.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-27T13:58:08.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26693" PostTypeId="1" CreationDate="2015-02-27T14:16:07.483" Score="2" ViewCount="57" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the following coupled reactions:&#xA;  $$\ce{glucose + O2 + H2O + glucose~oxydase -&amp;gt; gluconic~acid + H2O2}$$&#xA;  $$\ce{H2O2 + TMB(star) -&amp;gt; TMB}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Identify:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;A potential reagent or product that could act as a decoupling site leading to a falsely &lt;strong&gt;high&lt;/strong&gt; glucose concentration.&lt;/li&gt;&#xA;  &lt;li&gt;A potential reagent or product that could act as a decoupling site leading to a falsely &lt;strong&gt;low&lt;/strong&gt; glucose concentration.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: The second reaction is just hydrogen peroxide reacting with $\ce{3,3',5,5'-Tetramethylbenzidine}$, to form the more absorbent version (I can't remember if it's reduced or oxidized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment involves running the above reaction on a sample with an unknown amount of glucose, then running the sample through a spectrometer to determine the absorbance and thus the concentration of glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of is that if the sample warmed up too much you might lose some of the oxygen, resulting in a lower than expected glucose measurement. Though you'd have to lose enough oxygen for it to become the limiting reactant, which I don't see as being too plausible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find anything anywhere else and I'm stumped.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="5026" LastEditDate="2018-02-22T02:33:39.313" LastActivityDate="2018-02-22T02:33:39.313" Title="Identify &quot;decoupling sites&quot; in glucose oxydase reaction" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26694" PostTypeId="2" ParentId="26677" CreationDate="2015-02-27T14:53:26.390" Score="3" Body="&lt;p&gt;Curt's answer is correct, but as a practical matter considering the particular concentrations and ionizaiton constants in the question, we can reduce the problem to 3 equations and 3 unknowns as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{ax}=\frac{[X-][H+]}{0.1M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{ay}=\frac{[Y-][H+]}{0.1M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H+]=[X-] + [Y-]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H+]= \frac{0.1M~K_{ax}}{[H+]} + \frac{0.1M~K_{ay}}{[H+]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H+]^2 = 0.1M~K_{ax} + 0.1M~K_{ay}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H+]^2 = 1.8 \times 10^{-8}M^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[H+] = 1.34 \times 10^{-4}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pH = 3.9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how accurate are the approximations?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of zero, [OH-] is $7.46 \times 10^{-11}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really $[H+]=[X-] + [Y-] + 7.46 \times 10^{-11}M$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About a 1 part in 1,000,000 error in [H+] from neglecting the autoionization of water.  If the concentration of acids was very small instead of 0.1M, and pH closer to 7, this error could be much larger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error in approximating [HX] and [HY] as 0.1M is about 1 part in 1,000&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-02-27T16:44:30.333" LastActivityDate="2015-02-27T16:44:30.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26695" PostTypeId="1" AcceptedAnswerId="26700" CreationDate="2015-02-27T15:06:10.013" Score="1" ViewCount="2964" Body="&lt;p&gt;I'm having trouble finding out the salt formation equation for $\ce{(NH4)2C2O4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it was $\ce{NH3 + H2C2O4}$ but that doesn't seem to balance out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HC2O4-}$ seems to work, but are ions allowed in salt formation/neutralization formulas?&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="9961" LastEditDate="2018-11-14T18:13:32.790" LastActivityDate="2018-11-14T18:22:38.320" Title="Equation for salt formation for ammonium oxalate?" Tags="&lt;stoichiometry&gt;&lt;ionic-compounds&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26696" PostTypeId="2" ParentId="26695" CreationDate="2015-02-27T15:17:00.987" Score="2" Body="&lt;p&gt;$\Huge{\mathrm{(NH_4)_{\color{red}{2}}C_2O_4}}$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-27T15:17:00.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26698" PostTypeId="2" ParentId="26695" CreationDate="2015-02-27T16:14:39.293" Score="4" Body="&lt;p&gt;&lt;strong&gt;Where am I wrong?&lt;/strong&gt; First of all, note that the ammonia is dissolved in water. So there is an ion that's involved in the reaction, not the ammonia itself. And second, the charge of the ion involved requires a different coefficient for ammonia. Notice that the $\ce{H+}$ cations in Hydrogen oxalate are spectator ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of Ammonium Oxalate will be a precipitation reaction. (Or some may put it an acid-base one) Answer these in order to reach a final net ionic reaction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the ions that are present in the final product (ammonium oxalate)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The ions involved are ammonium ($\ce{NH4+}$) and oxalate ($\ce{C2O4^{2-}}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Notice the charges of those ions:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Ammonium has a 1+ charge (that's written correctly as +, but of course, I wanted to emphasize) and oxalate has a 2-.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The other ions are spectator ions. These two ions actually form the precipitate. But the final formula (where your misunderstanding rises from) is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $\ce{(NH4)2C2O4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can now write the net ionic equation and get it balanced.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-02-27T17:07:52.790" LastActivityDate="2015-02-27T17:07:52.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26699" PostTypeId="1" AcceptedAnswerId="26702" CreationDate="2015-02-27T16:53:22.903" Score="25" ViewCount="80477" Body="&lt;p&gt;For example for 4 given elements A,B,C,D, what's the meaning of $\text{AB} \cdot x \text{CD}$?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2020-07-29T15:58:25.377" Title="What is the meaning of the &quot;dot&quot; notation in chemical formulas?" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26700" PostTypeId="2" ParentId="26695" CreationDate="2015-02-27T16:57:52.710" Score="4" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{2 NH3 + H2C2O4}$&lt;/span&gt; - the reaction won't balance itself - you have to do it by putting stoichiometric coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/3083922#section=Top&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonium hydrogen oxalate&lt;/a&gt; - (NH4)HC2O4 also exists.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2018-11-14T18:22:38.320" LastActivityDate="2018-11-14T18:22:38.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26701" PostTypeId="2" ParentId="26695" CreationDate="2015-02-27T17:07:13.510" Score="2" Body="&lt;p&gt;I finally remembered that ammonium salts come from ammonia and does not produce water. I had trouble balancing because I thought water was part of the products. $$\ce{2NH3 + H2C2O4 -&amp;gt; (NH4)2C2O4}$$&lt;/p&gt;&#xA;" OwnerUserId="12987" LastActivityDate="2015-02-27T17:07:13.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26702" PostTypeId="2" ParentId="26699" CreationDate="2015-02-27T17:37:22.390" Score="23" Body="&lt;p&gt;By writing &lt;span class=&quot;math-container&quot;&gt;$\ce{AB.xCD}$&lt;/span&gt; chemists mean that there are CDs are found in the crystalline framework of AB. The most common example of this is water trapped inside the crystal structure of ionic compounds. (See &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_of_crystallization&quot; rel=&quot;nofollow noreferrer&quot;&gt;water of crystallization in wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;An example that's often taught is &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4.5H2O}$&lt;/span&gt;. See that 5 that's a representative of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;? It means that water can be bound only with intermolecular interactions, but it exists in the structure in a stoichiometric ratio.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9pjaL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9pjaL.gif&quot; alt=&quot;CuSO4 hydrate picture&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://202.116.45.198/wjhx/wjhx03/ziliaoku/xingzhi/1B/images/know_35.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, according to &lt;a href=&quot;http://www.che.uc.edu/Jensen/W.%20B.%20Jensen/Reprints/134.%20Dots.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In summary, when a dot is used to break a formula into subunits, it may signify ignorance of how the subunits are structurally related, as in our inorganic example; or it may correspond to actual significant structural subunits, as in our organic example; or it may represent the combining ratios of the binary starting materials required for the synthesis of the compound, as in our phase diagram example.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As a result, for example in direct addition reactions, where there is ignorance about the structural relation of the compounds, we use a dot.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="65872" LastEditDate="2020-07-29T15:58:25.377" LastActivityDate="2020-07-29T15:58:25.377" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26703" PostTypeId="2" ParentId="26699" CreationDate="2015-02-27T17:59:50.673" Score="16" Body="&lt;h2&gt;The use of dots in inorganic chemistry&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's take the example of copper sulfate penta-hydrate: $\ce{CuSO4. 5H2O}$.  The dot here is used essentially as an expression of&#xA;ignorance to indicate that, though the parts of the&#xA;molecule separated by the dot are bonded to one another in some fashion, the exact structural details of that interaction are not fully expressed in the resulting&#xA;formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Alfred Werner’s coordination&#xA;theory that indicates&#xA;that four of the five water molecules are actually&#xA;bonded directly to the $\ce{Cu}$ in the form of a complex&#xA;aquo ion  $\ce{[Cu(H2O)4]^2+}$, then the formula is better&#xA;expressed&#xA;as  $\ce{[Cu(H2O)4][SO4].H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the dot here signifies ignorance of how the&#xA;subunits of a formula are structurally related.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The use of dots in phase diagrams&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's take the example of the ternary phase diagram of $\ce{Na2O}$,$\ce{Al2O3}$,$\ce{SiO2}$. The formulas of&#xA;the various resulting complex sodium aluminosilicates&#xA;are all expressed in the form $\ce{a Na2O.b Al2O3.c SiO2}$&lt;br&gt;&#xA;in which it is implicitly understood that these initial starting components do not exist as such within the resulting compounds. They represent the combining ratios of the binary/ternary starting materials required for the synthesis of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please see &lt;a href=&quot;http://www.che.uc.edu/Jensen/W.%20B.%20Jensen/Reprints/134.%20Dots.pdf&quot;&gt;http://www.che.uc.edu/Jensen/W.%20B.%20Jensen/Reprints/134.%20Dots.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="10104" LastEditDate="2015-02-27T20:23:53.117" LastActivityDate="2015-02-27T20:23:53.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26704" PostTypeId="2" ParentId="26564" CreationDate="2015-02-27T18:24:11.720" Score="0" Body="&lt;p&gt;Okay, so no stopper will stop it? Right, then. I came up with an alternative solution that seems to work,  but time will tell: I put the stoppered bottle of acid next to an open wide-mouthed bottle of water and baking soda solution. These were both crammed together, within a couple of plastic bags. My idea is that any HCl that diffuses through the stopper will have ample chance to contact the water and be absorbed/neutralized before it makes its way to the air in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put my nose a foot from the bag and couldn't smell anything after 24 hours, so this homegrown hack seems to be working acceptably.&lt;/p&gt;&#xA;" OwnerUserId="14617" LastActivityDate="2015-02-27T18:24:11.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26706" PostTypeId="1" CreationDate="2015-02-27T19:13:56.713" Score="1" ViewCount="329" Body="&lt;p&gt;I have 2 grams of trenbolone acetate. I was told to add it to 20 ml of methanol. Then wait until it is fully dissolved. Then distilled water is to be added yeilding a precipitate that is supposed to still be trenbolone. Can anyone shed light on this? &#xA;Here is &lt;a href=&quot;http://www.basskilleronline.com/crystal-tren.shtml&quot; rel=&quot;nofollow&quot;&gt;the link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14674" LastEditorUserId="7448" LastEditDate="2015-02-27T19:25:32.763" LastActivityDate="2015-02-27T20:25:28.740" Title="Will trenbolone acetate in the presence of methanol yield trenbolone acetate if distilled water is subsequently added thereafter?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-02-27T20:31:44.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="26708" PostTypeId="2" ParentId="26706" CreationDate="2015-02-27T19:58:04.113" Score="1" Body="&lt;p&gt;The method will not convert trenbolone acetate to trenbolone, the ester is not cleaved under these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GUt5o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is however possible to separate the &lt;strong&gt;steroid&lt;/strong&gt; from a mixture with other, water-soluble compounds this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you are a veterinarian and use trenbolone or its acetate as a &lt;strong&gt;growth promoter&lt;/strong&gt; for cattle, you might want to consider the legal situation in your country. As far as Germany is concerned, the possession of as much as 2 g might lead to real trouble. Here, the non-punishable, non-significant amount is in the 150 mg range (for its use in sports).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-27T20:25:28.740" LastActivityDate="2015-02-27T20:25:28.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26710" PostTypeId="1" CreationDate="2015-02-27T20:34:00.400" Score="0" ViewCount="487" Body="&lt;p&gt;I wonder, if I wanted to know the molar mass of a some white crystalline solid that is a mix of $\ce{C6H12O6}$ and $\ce{C12H22O11}$, would I just do this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$170.1559 \; \ce{g/mol} + 343.2965 \;\ce{g/mol} = 522.4524 \; \ce{g} / 2 \ce{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus : $261,2262 \; \ce{g/mol}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be a correct operation ? Or it wouldn't make any physical sense ? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="19616" LastEditDate="2015-10-16T17:49:51.777" LastActivityDate="2015-10-16T17:49:51.777" Title="Molar mass of a mixture of two solids" Tags="&lt;organic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26711" PostTypeId="2" ParentId="26710" CreationDate="2015-02-27T21:41:49.290" Score="1" Body="&lt;p&gt;Sounds like a chemical mixture of sucrose and glucose.  In which case, you only talk about the molar mass of the two sugars separately.  The mass or number of moles of the mixture can be determined by knowing the ratio of sucrose and glucose in the mixture.  FYI... the molar mass of glucose is 180 g/mol (not 170).&lt;/p&gt;&#xA;" OwnerUserId="14480" LastActivityDate="2015-02-27T21:41:49.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26712" PostTypeId="1" AcceptedAnswerId="26751" CreationDate="2015-02-27T21:48:53.907" Score="17" ViewCount="40914" Body="&lt;p&gt;With nothing to do, I stared at the droplets that get condensed on the glass panel of my window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon examination, these droplets appear to be in some sorta pattern. There are big droplets as well as small ones that fill the space between the big ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQcsS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones in the middle, between rows of small droplets are usually big one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these intrigue me to think about what are that factors that contribute to this pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess: I think vapor gets condensed on dust particles thus forms those droplets in the picture. But what accounts for the different sizes of the droplets?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="7475" LastEditDate="2015-10-30T16:44:53.003" LastActivityDate="2018-04-25T06:46:05.110" Title="Why are water droplets shaped like that?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26713" PostTypeId="1" AcceptedAnswerId="26714" CreationDate="2015-02-27T22:10:35.443" Score="5" ViewCount="380" Body="&lt;p&gt;The first ingredient in a type of nail polish is &quot;mer&quot;. What is this? Is it some kind of chemical? How is it used? How does it work?&lt;/p&gt;&#xA;" OwnerUserId="14682" LastEditorUserId="5026" LastEditDate="2018-02-28T16:43:36.853" LastActivityDate="2018-02-28T16:43:36.853" Title="What is &quot;mer&quot;, the first ingredient in a type nail polish?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26714" PostTypeId="2" ParentId="26713" CreationDate="2015-02-27T23:15:17.427" Score="5" Body="&lt;p&gt;In the organic lab, sometimes we would run reactions that produced a tar-like substance in the reaction flask as an undesired by-product.  It was just a gooey polymer, and it happened often enough that this substance was just referred to as &amp;quot;&lt;strong&gt;mer&lt;/strong&gt;&amp;quot;, &lt;strong&gt;short for polymer&lt;/strong&gt;.  &amp;quot;Mer&amp;quot; was therefore a fairly common term used in many (most?) labs.&lt;/p&gt;&#xA;&lt;p&gt;Since &amp;quot;mer&amp;quot; is listed as the first ingredient in nail polish, that indicates that it is the major component in the formulation.  If you google &amp;quot;nail polish&amp;quot;, you'll find a &lt;a href=&quot;http://en.wikipedia.org/wiki/Nail_polish&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;. Its opening paragraph states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nail polish consists of an organic polymer with various additives&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I suspect they use the term &amp;quot;mer&amp;quot; to refer to whatever polymer is the main proprietary ingredient in their formulation.  The Wikipedia article also notes that&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nail polish consists of a film-forming polymer dissolved in a volatile organic solvent. Nitrocellulose that is dissolved in butyl acetate or ethyl acetate is common.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Once the volatile solvent evaporates, you are left with a hard polymer coating on your fingernail.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-02-23T02:42:28.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26715" PostTypeId="1" CreationDate="2015-02-28T00:32:46.513" Score="5" ViewCount="1021" Body="&lt;p&gt;Why are the HOMO and LUMOs often $\pi$ orbitals in carbon based compounds that contain multiple double bonds or conjugation such as ethene and butadiene?&lt;/p&gt;&#xA;" OwnerDisplayName="user14633" LastEditorDisplayName="user14633" LastEditDate="2015-02-28T21:57:05.293" LastActivityDate="2016-03-12T00:16:17.557" Title="HOMO and LUMOs of Carbon based Compounds" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26716" PostTypeId="1" AcceptedAnswerId="26740" CreationDate="2015-02-28T00:49:07.717" Score="1" ViewCount="110" Body="&lt;p&gt;I don't know what to call it I'll just provide a link. From what I can tell it is impossible, but why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemicalize.org/structure/#!mol=structureId%3A1609384473364&amp;amp;source=calculate&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalize.org/structure/#!mol=structureId%3A1609384473364&amp;amp;source=calculate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7786" LastActivityDate="2015-02-28T16:23:01.260" Title="Why isn't this a compound?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2016-06-14T19:25:43.917" ContentLicense="CC BY-SA 3.0" />
  <row Id="26717" PostTypeId="1" CreationDate="2015-02-28T01:12:03.697" Score="1" ViewCount="103" Body="&lt;p&gt;I am trying to convert a standard InChI string to a non-standard InChI string with fixed-hydrogen-layer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard InChI for Uracil: InChI=1S/C4H4N2O2/c7-3-1-2-5-4(8)6-3/h1-2H,(H2,5,6,7,8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Desired outcome: InChI=1/C4H4N2O2/c7-3-1-2-5-4(8)6-3/h1-2H,(H2,5,6,7,8)&lt;strong&gt;/f/h5-6H&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if this is at all theoretically possible without further information and if not, are there other ways to achieve this goal, e.g. using an existing database?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate goal is to convert the InChI string with hydrogen layer into a mol file to be able to map chromatogram peaks against it, so I installed OpenBabel which can do the trick, but it can not solve the aforementioned conversion between the InChIs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already found the &lt;a href=&quot;http://cactus.nci.nih.gov/chemical/structure&quot; rel=&quot;nofollow&quot;&gt;cactus website&lt;/a&gt; which allows to convert from InChI&lt;strong&gt;Key&lt;/strong&gt; (which I can generate from the standard InChI that I have) to the correct non-standard InChI (with hydrogen layer) via this URL:&#xA;&lt;a href=&quot;http://cactus.nci.nih.gov/chemical/structure/&quot; rel=&quot;nofollow&quot;&gt;http://cactus.nci.nih.gov/chemical/structure/&lt;/a&gt;{INCHIKEY_HERE}/inchi.&#xA;The problem here is that it returns multiple correct non-standard InChI strings (with fixed hydrogen layer). One of them is the one I am looking for but I don't see a way to figure out how to pick the correct one. As you can imagine I want to run this in an automated fashion for hundreds of conversions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your input.&lt;/p&gt;&#xA;" OwnerUserId="14684" LastEditorUserId="33746" LastEditDate="2016-08-13T13:43:43.070" LastActivityDate="2016-08-13T13:43:43.070" Title="Conversion of standard InChI to InChI with fixed-hydrogen-layer" Tags="&lt;cheminformatics&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="26718" PostTypeId="1" CreationDate="2015-02-28T01:49:30.703" Score="1" ViewCount="2504" Body="&lt;p&gt;So here's my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the osmotic pressure, noted in mm of $\ce{H2O}$, of a solution with 125 micrograms of vitamin B12 ($\ce{C63H88CoN14O14P}$) in 2.50 mL of $\ce{H2O}$ at 25 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;hint&quot; that comes with the problem says: Weak osmotic pressures are often measured and given in mm of $\ce{H2O}$. Use here the density of water (1,00 g/mL) and of $\ce{Hg}$ (13,6 g/mL). In this case, the constant $R$ has a value of $\pu{62,36 mm Hg * L / mol * K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;2.50 \,\text{microgram} = 0.000125 \,\text{grams}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;pi = cRT&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{(\pu{0.000125 g * 1 mol / 1355.3652 g})}{(\pu{2.50 mL / 1000})} * \pu{62.36 mm \ce{Hg} * L / mol * K * 298.15 K} = \pu{0.685890194 mm \ce{Hg}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I blocked here and I saw that the solution was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\pu{0.685890194 mm \ce{Hg} * 13.6 g/ml / 1.00 g/ml = 9.32 mm \ce{H2O}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here's two questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why did I have to use R = 62.36 and not the usual R = 8.31?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, what's happening in the conversion exactly? How was I supposed to figure out what they did to convert from mm $\ce{Hg}$ to mm $\ce{H2O}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="194" LastEditDate="2017-05-01T13:48:26.410" LastActivityDate="2017-05-01T13:48:26.410" Title="Converting mm Hg to mm H2O" Tags="&lt;osmosis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26721" PostTypeId="2" ParentId="19277" CreationDate="2015-02-28T04:42:33.517" Score="4" Body="&lt;p&gt;There is definitely a breakdown of mass action at very low numbers of molecules.  This can happen either because concentrations are tiny, or because the system of interest is tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, your idea to turn to probability is an excellent one, and is how these systems are modeled.  But the reason doesn't necessarily have to do with quantum mechanics.  It's just a consequence that matter is not really a &lt;em&gt;continuum&lt;/em&gt; that can have any concentration, but is in fact &lt;em&gt;granular&lt;/em&gt;.  At the scale of atoms, there can either be an atom, or not be an atom.  You can't have 0.5 of an atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right boundary at which mass action breaks down depends a bit on the particular system, in particular, if reactions are occurring that depend on H+.  In the less complicated case of no reactions depending on $\ce{H+}$, then the boundary between mass action being valid and not is actually fairly easy to calculate. If you know the average or expected value of $\ce{H+}$ ions in your water sphere (which you do, since you calculated it) the actually observed number of ions will follow a &lt;a href=&quot;http://en.wikipedia.org/wiki/Poisson_distribution&quot; rel=&quot;nofollow&quot;&gt;Poisson distrubtion&lt;/a&gt;.  If the mean number of atoms is $N$, then the standard deviation of atoms is $\sqrt{N}$, so the relative uncertainty in the mean number of atoms is proportional to $\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}$.  According to your calculation, $N$ is less than one, so the relative uncertainty is greater than 100%!  But if you enlarged the sphere so the expected number of ions was $N=10$, then the uncertainty would be about 32%, still pretty bad.  If you enlarged the sphere so $N=10000$, then the relative uncertainty is only 1%, meaning that mass action predictions for this system are pretty close to accurate!  Getting $N$ tht high requires a sphere about 10,000-times bigger in volume, which means the radius would have to be about 38 times bigger.  But keep in mind this result is only true for the water dissociation problem.  So each chemical system has its own scale where mass action breaks down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For systems where reactions are occurring that depend on $\ce{H+}$, this simple logic breaks down.  Complex probabilistic simulations need to be done.  Depending on the details of the reactions, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Master_equation&quot; rel=&quot;nofollow&quot;&gt;chemical master equation&lt;/a&gt; or the &lt;a href=&quot;http://en.wikipedia.org/wiki/Langevin_equation&quot; rel=&quot;nofollow&quot;&gt;chemical Langevin equation&lt;/a&gt; would have to be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One time I calculated the concentration of frogs in an aquarium was about 1 zeptomolar.  I don't think we could assume frogs followed the law of mass action!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-02-28T04:42:33.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26723" PostTypeId="1" CreationDate="2015-02-28T05:44:54.463" Score="2" ViewCount="105" Body="&lt;p&gt;For $\ce{CO (g) + H2 (g) &amp;lt;=&amp;gt; CH2O (g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with K&lt;sub&gt;c&lt;/sub&gt; = 0.068 @ 273K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;initial concentrations being $\ce{[CO]}$ = 1.25M, $\ce{[H2]}$ = 2.00M, $\ce{[CH2O]}$ = 1.00M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did $\frac{1.00}{((1.25)(2.00)} = 0.4$, which is more then K&lt;sub&gt;c&lt;/sub&gt;, so it's leaning to the left, will produce more reactants, and products will lower. So the I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1.00 - x}{(1.25 + x)(2.00 + x)} = 0.068$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplied the bottom and got $X^2 + 3.25x + 2.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1.00 - x}{X^2 + 3.25x + 2.5} = 0.068$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then tried multiplying the denominator by 0.068 and rearranging to solve for $x$ from there, but that's where I got stuck. $X$ is supposed to be 0.65.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="69830" LastEditDate="2021-05-08T18:35:30.410" LastActivityDate="2021-05-08T18:35:30.410" Title="Finding equilibrium concentrations, trouble simplifying equation" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26724" PostTypeId="2" ParentId="26718" CreationDate="2015-02-28T06:49:49.080" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You CAN use R=8.31 but it has the (standard) unit of (m3 Pa K−1 mol−1), simply convert your volume from liter (aka dm3) to meter cube (m3) and the pressure will be in Pascal (Pa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use R=62.36, which has a unit of (L mmHg K−1   mol−1), as long as your   volume is in liter, the pressure will be in mmHg&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But the question ask for mmH2O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure is simply force on an area (F/A), weight on an area (mg/A), density * volume on a area (rho*V /A) = density * height (rho*h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus P = density (Hg) * h (Hg) = density (water) * h (water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h (water) =  (13.6 / 1.00) * h(Hg)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9933" LastActivityDate="2015-02-28T06:49:49.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26725" PostTypeId="1" AcceptedAnswerId="26737" CreationDate="2015-02-28T08:02:58.497" Score="-1" ViewCount="1276" Body="&lt;p&gt;What will happen if someone uses tear gas, such as &lt;strong&gt;CS&lt;/strong&gt; $(\ce{C10H5ClN2})$ and laughing gas, $\ce{N2O}$, in auditorium? Means will people start laughing and crying together? Or there will be other effect? &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="418" LastEditDate="2015-02-28T08:37:26.843" LastActivityDate="2019-07-19T21:33:13.507" Title="Effect of tear gas and laughing gas together" Tags="&lt;toxicity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26726" PostTypeId="1" CreationDate="2015-02-28T11:00:44.060" Score="0" ViewCount="540" Body="&lt;p&gt;A student mixed 260ml of 1.2 M lead(II) nitrate with 300ml of 1.90M potassium iodide. What is the final concentration of $\ce{NO3^{-}}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the answer is 1.11M , but he used the moles of lead(II) nitrate to find out the moles of $\ce{NO3^{-}}$ ,, and I think he should had used the moles of potassium iodide since it is the limiting reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could anyone explain ?&lt;/p&gt;&#xA;" OwnerUserId="14688" LastEditorUserId="4231" LastEditDate="2015-02-28T14:49:59.230" LastActivityDate="2015-02-28T14:49:59.230" Title="calculate the final concentration of this equation" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-02-28T13:12:41.297" ContentLicense="CC BY-SA 3.0" />
  <row Id="26728" PostTypeId="1" AcceptedAnswerId="26731" CreationDate="2015-02-28T11:15:22.690" Score="5" ViewCount="10778" Body="&lt;p&gt;In the formula $\ce{(CH3)3-C-CH(CH3)-CH2-CH3}$, what do the parenthesized portions represent? What is the molecular structure, and how do I name the compound?&lt;/p&gt;&#xA;" OwnerUserId="14689" LastEditorDisplayName="user7951" LastEditDate="2016-11-18T15:27:03.117" LastActivityDate="2016-11-18T15:27:03.117" Title="What is the meaning of parentheses in a condensed molecular formula?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26729" PostTypeId="2" ParentId="26726" CreationDate="2015-02-28T11:42:12.487" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] he should had used the moles of potassium iodide since it is the limiting reactant.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?&lt;/li&gt;&#xA;&lt;li&gt;If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?&lt;/li&gt;&#xA;&lt;li&gt;Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update I&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the rocket science (solubilities at 20 °C):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{cr}&#xA;\mathrm{compound} &amp;amp; \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline&#xA;\ce{Pb(NO3)2} &amp;amp; 522\\ &#xA;\ce{KI} &amp;amp; 1430\\&#xA;\ce{KNO3} &amp;amp; 316\\&#xA;\ce{PbI2} &amp;amp; 0.76\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -&amp;gt; 2 K+ + 2 NO3- + PbI2 v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this in any way change the absolute amount of  $\ce{NO3-}$ &lt;strong&gt;in solution&lt;/strong&gt;?&lt;/li&gt;&#xA;&lt;li&gt;What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update II&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial $\ce{Pb(NO3)2}$ solution&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$ &lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;After mixing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but &lt;strong&gt;does not change the absolute amount&lt;/strong&gt; of  $\ce{NO3-}$ in solution. However, it &lt;strong&gt;changes the total volume&lt;/strong&gt; and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-02-28T14:23:33.493" LastActivityDate="2015-02-28T14:23:33.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26730" PostTypeId="1" CreationDate="2015-02-28T11:49:52.570" Score="2" ViewCount="41" Body="&lt;p&gt;Conceptual problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$A + 2B \rightarrow C + 2D$&lt;/span&gt; be a reaction under constant pressure (&lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;) and temperature (&lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;), that goes all the way towards the products. &lt;span class=&quot;math-container&quot;&gt;$\ce{A,B,C,D}$&lt;/span&gt; are ideal gases. Let &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; be the molar enthalpy change in the reaction.&lt;/p&gt;&#xA;&lt;p&gt;What is the change in the internal energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The total amount of moles in the reaction is constant throughout it, can we say that &lt;span class=&quot;math-container&quot;&gt;$\Delta U=\alpha nR\Delta T$&lt;/span&gt; is the term for the internal energy? (if so that will be 0).&lt;/p&gt;&#xA;&lt;p&gt;Another idea is that &lt;span class=&quot;math-container&quot;&gt;$PV=nRT \rightarrow V=\textrm{const}$&lt;/span&gt;, and then the internal energy equals the sun of chemical work and heat which we know.&lt;/p&gt;&#xA;&lt;p&gt;All help appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14690" LastEditorUserId="69830" LastEditDate="2021-05-21T07:47:38.220" LastActivityDate="2021-05-21T07:47:38.220" Title="Ideal gasses reaction under const temp and pressure" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26731" PostTypeId="2" ParentId="26728" CreationDate="2015-02-28T11:53:25.057" Score="8" Body="&lt;p&gt;The parentheses just have a grouping function: $\ce{(CH3)3-C -}$ represents a carbon atom to which 3 methyl groups are attached, $\ce{-CH(CH3)}-$ is a carbon atom in the chain to which one hydrogen atom and a methyl group are attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw the molecule, find the longest chain for the base name, determine the position of the substituents and you're done. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-28T11:53:25.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26732" PostTypeId="2" ParentId="26730" CreationDate="2015-02-28T11:53:29.097" Score="4" Body="&lt;p&gt;Got it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as $H=U+PV$ and $PV = nRT=const$ we ger $\Delta(PV)=0 =&amp;gt; \Delta U= \Delta H$&lt;/p&gt;&#xA;" OwnerUserId="14690" LastActivityDate="2015-02-28T11:53:29.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26735" PostTypeId="1" AcceptedAnswerId="26736" CreationDate="2015-02-28T12:51:11.890" Score="9" ViewCount="4505" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JAvoG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAvoG.png&quot; alt=&quot;(E)-Hexa-1,3-dien-5-yne&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both double and triple bonds are present in the hydrocarbon shown above, which has the name &lt;strong&gt;hexa-1,3-dien-5-yne&lt;/strong&gt;. Why is the numbering chosen such that the alkene has lower locants than the alkyne? Is there a rule which stipulates that the alkene is a more 'senior' functional group?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="16683" LastEditDate="2018-09-29T19:41:03.850" LastActivityDate="2021-02-13T02:30:11.497" Title="Should double or triple bonds have preference for low locants in IUPAC nomenclature?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26736" PostTypeId="2" ParentId="26735" CreationDate="2015-02-28T13:07:45.613" Score="8" Body="&lt;p&gt;The appropriate passage in the &lt;em&gt;2013 Blue Book&lt;/em&gt; is as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-31.1.1.1&lt;/strong&gt; The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the preferred IUPAC name for the above compound is &lt;strong&gt;hexa-1,3-dien-5-yne&lt;/strong&gt;, as the double bonds receive the lower locants (1,3) than in the alternative name hexa-3,5-dien-1-yne.&lt;/p&gt;&#xA;&lt;p&gt;Note that this preference for double bonds over triple bonds is only relevant when there is no difference between the locant set for the multiple bonds. In the following example, the locant set (1,4) is lower than (2,5) and hence hex-4-en-1-yne is the preferred name, even though it has a higher locant for the double bond:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DR2FY.png&quot; height=&quot;150&quot; alt=&quot;PIN of hex-4-en-1-yne&quot;&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2021-02-13T02:30:11.497" LastActivityDate="2021-02-13T02:30:11.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26737" PostTypeId="2" ParentId="26725" CreationDate="2015-02-28T13:40:55.853" Score="8" Body="&lt;p&gt;Admittedly, I had to smile when I first read your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the effect for people exposed to both agents is not comparable to hearing a joke and a sad story at the same time. There is no cuteness in that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flooding an enclosed space with $\ce{N2O}$ is not a good idea! $\ce{N2O}$ is an interesting drug, used for sedation in dentistry. Here, a &lt;strong&gt;mixture of $\ce{N2O}$&#xA;and oxygen&lt;/strong&gt; is fed to the patient through a nose mask, while the patient can freely exhale. Note that $\ce{N2O}$ apparently isn't metabolized, but just exhaled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feeding a heavier-than-air gas to a room without proper fresh air supply and circulation will lead to accumulation of this gas. Whether this could lead to (cerebral) hypoxia and permanent damages is open to calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of &lt;strong&gt;CS&lt;/strong&gt; in enclosed spaces is even worse. It is actually considered a chemical weapon, banned from war zones and only used in civil riot control. Oh, the irony! Note that it actually isn't a gas but a solid, dispersed either in solution or from granades. It is a strong irritant to all mucuous membranes and the skin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a densely packed auditorium full of people with burning sensation in their eyes and on their skin, crying, shouting, coughing, vomiting and running around in panic for an exit or relief and you know that this isn't the place to be!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-28T13:40:55.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26738" PostTypeId="1" CreationDate="2015-02-28T14:49:46.457" Score="-1" ViewCount="87" Body="&lt;p&gt;While studying equilibrium we get to know that there are $1*10^{-7} \ce{~~OH-}$ and $\ce{H+}$ ions in pure water. However, I think there are more undissociated molecules. &lt;/p&gt;&#xA;" OwnerUserId="14693" LastEditorUserId="7448" LastEditDate="2015-02-28T17:39:18.910" LastActivityDate="2015-02-28T17:39:18.910" Title="Which is more: dissociated ions in water or undissociated?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-01T02:38:07.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="26739" PostTypeId="2" ParentId="26738" CreationDate="2015-02-28T15:03:59.410" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] there are $1\cdot10^{-7}\ \ce{OH-}$ [&amp;hellip;] ions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you sure that this is an absolute number or rather a concentration in $\mathrm{\frac{mol}{L}}$? IN the latter case, what would be the concentration of undiscociated $\ce{H2O}$ molecules in 1 liter of water? &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-02-28T15:03:59.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26740" PostTypeId="2" ParentId="26716" CreationDate="2015-02-28T16:23:01.260" Score="4" Body="&lt;p&gt;This structure is impossible as either the molecule is planar and then the C-C bonds on the outer ring are extremely long (and therefore weak), or you distort from the sp2 geometry (120 degree bond angles) of the central ring carbons so that the outer ring is beneath the inner one, which would also require a very high energy state. The only stable circumscribed ring structures you'll find are macromolecular.  &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-02-28T16:23:01.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26741" PostTypeId="1" AcceptedAnswerId="26744" CreationDate="2015-02-28T16:55:28.383" Score="0" ViewCount="515" Body="&lt;p&gt;Will the Henderson–Hasselbalch equation be able to calculate the pH when you have an acidic solution and you add a basic solution? Will that be possible or can you only use it for adding an acid to an acidic buffer to calculate the pH, or only a base to a basic buffer? If it can, how is the pH calculated, if you add 0.025 mol of solid KOH to 1 L of an acidic acid buffer solution of pH 3.97.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2017-08-22T12:49:59.460" LastActivityDate="2017-08-22T12:49:59.460" Title="How to apply the Henderson–Hasselbalch equation when adding KOH to an acidic acid buffer?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26743" PostTypeId="1" AcceptedAnswerId="26746" CreationDate="2015-02-28T17:19:38.303" Score="15" ViewCount="11312" Body="&lt;p&gt;In the electrophilic aromatic substitution (EAS) of pyrrole, why is the alpha position preferred over the beta position so much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BEBus.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BEBus.png&quot; alt=&quot;Alpha and beta carbons of pyrrole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2017-10-01T14:02:53.330" LastActivityDate="2018-11-24T04:39:18.373" Title="Regioselectivity in electrophilic substitution of pyrrole" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26744" PostTypeId="2" ParentId="26741" CreationDate="2015-02-28T17:20:18.600" Score="4" Body="&lt;p&gt;If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See &lt;a href=&quot;http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7648/does-the-number-of-h-ions-in-solution-go-up-on-dilution-of-a-weak-acid&quot;&gt;Does the number of H+ ions in solution go up on dilution of a weak acid?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH = 14 - pOH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH = 14 + log(0.025) = 12.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-02-28T17:59:42.083" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="26745" PostTypeId="2" ParentId="19277" CreationDate="2015-02-28T17:33:32.890" Score="4" Body="&lt;p&gt;A detailed analysis of the significance of the mass action law at the limit of very few molecules can be found in the book &lt;a href=&quot;https://books.google.com/books?id=AuW6Mg4sr3UC&amp;amp;&amp;amp;pg=PA99#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Law of Mass Action&lt;/a&gt;. Section 3.7 ('Breakdown' of the law of mass action) on page 99 even specifically analyses the case of water dissociation in very small droplets, though the discussion and some of the results derived within are general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes statistical thermodynamics to prove, but as Curt points out well, it's a consequence of molecules/ions coming in discrete (integer) amounts. We're used to doing calculations in systems involving many billions of species, so we can get away with treating the number of any species $A$ as continuous, even though it's an approximation; we unconsciously equate &lt;em&gt;the most probable amount of&lt;/em&gt; $A$ with &lt;em&gt;the expected amount of&lt;/em&gt; $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the consequences of making the distinction clear is that &lt;em&gt;systems with zero molecules of a reagent or product can still be in equilibrium&lt;/em&gt;, even though a naive application of the mass action law would result in undefined equilibrium constants. What differentiates then an equilibrium constant of $10^{100}$ from $10^{200}$ (assuming the same amount of species are involved) is &lt;em&gt;how often&lt;/em&gt; the equilibrium can be found with zero molecules of the products; the higher the equilibrium constant, the less likely any given measurement will indicate the presence of at least one molecule of reactant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be an unintuitive concept to grasp, and since its derivation is somewhat involved, it can be hard to explain. However, there is a simpler case which shows up in physics, based on the granularity of light (photons) rather than the granularity of matter (atoms), which is useful as a stepping stone. I present it below:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Imagine a point light source in a perfect vacuum with a power of $100\ W$, emitting light at a certain wavelength, say $500\ nm$. Via the formula $E=\frac{hc}{\lambda}$, it is possible to calculate that the light source is therefore emitting about $2.52 \times 10^{20}$ photons per second. As is well known, light intensity falls off with the square of the distance $r$ from the source, since the photons are spread out over a spherical shell with area $4\pi r^2$. Thus, for example, at a distance of $r=100\ m$ from the source, the $2.52 \times 10^{20}$ photons per second are spread out over an area of approximately $31416\ m^2$, which means the photon flux is that distance is $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (100\ m)^2} = 2.00\times 10^{15}\  photons\ m^{-2} s^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we place a perfect photon detector with a $1\ m^2$ cross section facing the light source. At a distance of $100\ m$, as previously calculated it will detect $8.02\times 10^{15}$ photons arriving in one second. The distance between the light source and detector is then increased greatly to $10^7\ km$. At this new distance, we calculate a photon flux of $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (10^{10}\ m)^2} = 0.200\ photons\ m^{-2} s^{-1}$. Thus, the detector, with a $1\ m^2$ cross section at $10^7\ km$ away from the source, will count 0.2 photons per second. &lt;em&gt;There is no such thing as a fifth of a photon&lt;/em&gt;, so what does this number mean? Of course, this number is meant as a statistical time average. If the detector is left open for $10^5$ seconds, then we can expect it to detect $2\times 10^4$ photons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if we &lt;em&gt;force&lt;/em&gt; the detector to stay open for only one second at a time? Once again, there is no such thing as a fraction of a photon, so the detector must display an integer number of counts. How can the expected amount of 0.2 photons per second possibly be met in this case? Easy: &lt;em&gt;sometimes the detector will measure photons, and sometimes it will not&lt;/em&gt;. In fact, if the test is repeated many times, we will find that the detector will count exactly 0 photons with 81.9% probability, exactly 1 photon with 16.4% probability, exactly 2 photons with 1.6% probability, and so on, such that the probabilities obey &lt;a href=&quot;http://www.wolframalpha.com/input/?i=Poisson+distribution+mu+%3D+0.2&quot; rel=&quot;nofollow&quot;&gt;a Poisson distribution with an expected value of 0.2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In much the same way as photons are discrete, molecules are also discrete. Thus, when a reaction at equilibrium contains less than one molecule of a given species, what is actually meant is that there will be times when the system will be measured to have zero molecules, and other times more than zero. The smaller the fraction of expected molecules, the less probable it is to measure the system while containing more than zero molecules of the species.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-02-28T17:33:32.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26746" PostTypeId="2" ParentId="26743" CreationDate="2015-02-28T17:44:37.583" Score="14" Body="&lt;p&gt;The following figure shows the resonance structures we can draw to describe the intermediates produced by electrophilic at the 2- ($\alpha$) or 3- ($\beta$) positions in pyrrole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BrLrP.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/heterocy.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the intermediate produced by attack at the $\alpha$-position can be described by 3 resonance structures.  Whereas attack at the $\beta$-position produces an intermediate that can be described by only 2 resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the intermediate formed by $\alpha$ attack is more stable and the activation energy leading to it will be reduced.  Consequently, electrophilic attack at the $\alpha$ position in pyrrole is preferred.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-03-05T20:06:41.917" LastActivityDate="2017-03-05T20:06:41.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26747" PostTypeId="1" CreationDate="2015-02-28T17:48:00.907" Score="3" ViewCount="130" Body="&lt;p&gt;If I have $\ce{HCl}$ and I put it in water, it will form ions. But, if I take $\ce{HCl}$ and put in benzene, it won't. How do you know that? If I had other substances, how would I know if it forms ions or not?&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="7448" LastEditDate="2015-02-28T17:53:29.110" LastActivityDate="2015-02-28T18:08:25.070" Title="When does a solute form ions?" Tags="&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26748" PostTypeId="2" ParentId="26747" CreationDate="2015-02-28T18:08:25.070" Score="1" Body="&lt;p&gt;In water, HCl reacts to form $\ce{H3O+}$ and Cl-.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, water molecules surround the respective ions in a stabilizing orientation.&#xA;The partially negatively charged O atoms of water interact with the $\ce{H3O+}$.&#xA;The partially positively charged H atoms of water interact with the Cl-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of the above can occur in benzene because benzene is very reluctant to accept a proton and because benzene is non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, polar solvents have an ability to stabilize ions that nonpolar solvents do not.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-02-28T18:08:25.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26749" PostTypeId="1" AcceptedAnswerId="26750" CreationDate="2015-02-28T18:29:40.603" Score="28" ViewCount="2246" Body="&lt;p&gt;I know that this is a rather ambiguous question; but my question is, whenever we take water and freeze it in the freezer, it still tends to stay clear. Since snow is just frozen water, why is it white? Is it due to contents of the air - i.e. dust - that make snow this color? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is my freezer just weird?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-03-03T21:19:52.037" LastActivityDate="2015-03-03T21:19:52.037" Title="Why is snow white?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;water&gt;&lt;color&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="26750" PostTypeId="2" ParentId="26749" CreationDate="2015-02-28T18:49:57.747" Score="32" Body="&lt;p&gt;The difference between snow and ordinary ice cubes is mainly about the size of the particles. Snow is made from small, irregular crystals with many edges at a very small scale. Light is refracted or scattered by the edges (or the interface between air and the edges). Snow is white because the scattering effect of those edges dominates what happens to light shining on the snow. In a large block of ice like an ice cube there is very little refraction or scattering as, for any ray of light, there are only two edges (one when light enters the cube, one when it leaves). So transmission dominates (and there is little colour as ice only weakly absorbs visible light). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is common in many other compounds. Titanium dioxide is a transparent mineral but is used as the primary ingredient in many &quot;white&quot; products like paint. The secret is to use TiO&lt;sub&gt;2&lt;/sub&gt; particles that are just the right size to maximise the scattering at the particle edges thereby creating &quot;whiteness&quot; as all light is equally scattered.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-02-28T18:49:57.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26751" PostTypeId="2" ParentId="26712" CreationDate="2015-02-28T19:03:09.817" Score="11" Body="&lt;p&gt;From the &lt;a href=&quot;http://en.wikipedia.org/wiki/Surface_tension#Causes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on surface tension&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary &quot;wall tension&quot; of the surface layer according to Laplace's law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In short, the more surface tension is, the rounder shapes of water you get. And the opposite goes for gravitational potential energy: The lesser gravitational acceleration results in more spherical droplets of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symbol for surface tension is $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\gamma (\ce{H2O}) = 72.8~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)&lt;br/&gt;&#xA;   $\gamma (\text{mercury}) = 465~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the reason you hardly ever see mercury drops out of their spherical shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The spherical shape minimizes then necessary &quot;wall tension&quot; of the surface layer according to Laplace's law.&lt;sup&gt;[2]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh and I almost forgot: This &lt;a href=&quot;https://arxiv.org/pdf/1211.3854.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;great article&lt;/a&gt; - Antonin Marchand, Joost H. Weijs, Jacco H. Snoeijer, and Bruno Andreott, Why is surface tension a force parallel to the interface?. &lt;em&gt;Physique et Mecanique des Milieux Heterogenes&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt; - is very nice in case you wanted to do additional study in the case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;R Nave. Surface Tension. &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/surten.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;URL&lt;/a&gt; (accessed April 25th, 2018)&lt;/li&gt;&#xA;&lt;li&gt;R Nave. Surface Tension and Droplets. &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/surten2.html#c3&quot; rel=&quot;nofollow noreferrer&quot;&gt;URL&lt;/a&gt; (accessed April 25th, 2018)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5026" LastEditDate="2018-04-25T06:46:05.110" LastActivityDate="2018-04-25T06:46:05.110" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26752" PostTypeId="1" AcceptedAnswerId="26754" CreationDate="2015-02-28T21:28:09.667" Score="5" ViewCount="856" Body="&lt;p&gt;In the following examples; I don't understand why $\ce{Na_3PO_4}$ is a much more complicated process when calculating $\mu$ compared to $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$. Specifically, I don't understand how to determine which species to include in the $\mu$ equation. All of the examples I've come across just behave just using water and the original species at most. What is more confusing also, is why doesn't the $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ include the concentrations and charges of their conjugates as the last example with $\ce{Na_3PO_4}$ does.Thanks for any help!!! Here is the work: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.1000M$ of each reagent to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaH_2PO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] &amp;lt;&amp;lt; 0.1M$$&#xA;$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-1)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \frac{1}{2}\sum([0.1M](+1)^2 + [0.1M](-1)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = 0.1000M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaH_2PO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Ignoring} ~H^+ \text{and}~ OH^-. [C] &amp;lt;&amp;lt; 0.1M$$&#xA;$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [H_2PO_4^-](-2)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = \frac{1}{2}\sum([0.2M](+1)^2 + [0.1M](-1)^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu = 0.3000M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na_3PO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without using Quadratic (which is just to keep things simple): &#xA;$$[OH^-]=0.044M$$&#xA;$$[PO_4^{3-}]=0.056M$$&#xA;$$[HPO_4^{2-}]=0.044M$$&#xA;$$[H^+]=\frac{K_w}{0.044M}=2.27*10^{-13}M$$&#xA;$$\mu = \frac{1}{2}\sum([Na^+](+1)^2 + [PO_4^{3-}](-3)^2+[HPO_4^{2-}](-2)^2+[H^+](+1^2))$$&#xA;$$\mu = \frac{1}{2}\sum([0.3M](+1)^2 + [0.056M](-3)^2+[0.044M](-2)^2+[2.27*10^{-13}M](+1^2))$$&#xA;$$\mu=0.5120M$$&lt;/p&gt;&#xA;" OwnerUserId="14650" LastEditorUserId="5160" LastEditDate="2015-03-01T13:54:34.310" LastActivityDate="2015-03-01T13:54:34.310" Title="Why is ionic strength (mu) calculated differently for these 2 examples?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26753" PostTypeId="1" AcceptedAnswerId="26774" CreationDate="2015-02-28T22:16:42.880" Score="3" ViewCount="116" Body="&lt;p&gt;After heating an ampoule of tellurium around its melting point in a furnace for about a day, I've noticed drops of the element on the upper inside surface of the ampoule, as if it condensed there. I've also noticed similar behavior with other intermetallic compounds. Is it common for something like tellurium to condense and resolidify in drops when an ampoule of it is heated (like water as shown &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26712/why-does-water-vapor-takes-on-weird-shape-after-it-gets-condensed&quot;&gt;here&lt;/a&gt;)? &lt;/p&gt;&#xA;" OwnerUserId="9557" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-03-01T12:24:16.527" Title="condensation of tellurium on inside of ampoule?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26754" PostTypeId="2" ParentId="26752" CreationDate="2015-02-28T23:06:07.950" Score="3" Body="&lt;p&gt;The third pKa of phosphoric acid (going from disodium to trisodium phosphate) is 12.32.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trisodium phosphate is added to water it is a relatively strong base, beginning to titrate H20 to OH-, to a significant degree.  In other words, to be above the pKa of 12.32, the concentration of OH- needs to be at least 0.02M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other examples, since the other two pKas of phosphoric acid are 2.15 and 7.2, the pH in those examples will be about midway between 2.15 and 7.2 for the monosodium salt and about midway between 7.2 and 12.32 for the disodium salt, so the concentrations of H+ and OH- are insignificant compare to the other ions.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-01T13:30:47.767" LastActivityDate="2015-03-01T13:30:47.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26755" PostTypeId="2" ParentId="26715" CreationDate="2015-03-01T01:53:22.943" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why are the HOMO and LUMOs often π  orbitals in carbon based compounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The heat of hydrogenation of ethylene is -32.5 kcal/m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qt1Ae.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch02/ch2-7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, ethylene is 32.5 kcal/m less stable (higher in energy) than ethane and this amount of energy (heat) is given off when we hydrogenate ethylene and convert it to ethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that pi bonds are, generally, less stable (higher in energy) than sigma bonds as shown in the following diagram (just forget about the delta bond orbitals, they don't play into this discussion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TgV91.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chembio.uoguelph.ca/educmat/chm729/band/bond.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram is a pictorial answer to your question.  Since pi bonds are generally higher in energy than sigma bonds, they will become the HOMO in most molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since bonding and antibonding orbitals are arranged symmetrically about the zero energy point, whatever type of orbital is the HOMO, then the antibonding counterpart will be the LUMO.  If a $\ce{\pi}$ orbital is the HOMO, than a $\pi^{*}%$ (pi star) orbital will be the LUMO.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:16:17.557" LastActivityDate="2016-03-12T00:16:17.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26756" PostTypeId="1" CreationDate="2015-03-01T01:59:19.960" Score="2" ViewCount="7186" Body="&lt;p&gt;Why would you wash a solvent with $\ce{NaOH}$ solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does $\ce{NaOH}$ solution help water and solvent separate like $\ce{NaCl}$ solution does (salting out)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you wash with hot or cold $\ce{NaOH}$ solution/water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anybody know anything about washing a solvent please explain basics to me? Thank you&lt;/p&gt;&#xA;" OwnerUserId="14701" LastEditorUserId="8083" LastEditDate="2015-03-01T02:39:21.337" LastActivityDate="2015-03-01T02:55:41.140" Title="Washing a solvent?" Tags="&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26757" PostTypeId="2" ParentId="26756" CreationDate="2015-03-01T02:55:41.140" Score="2" Body="&lt;p&gt;In general, the term &lt;em&gt;washing&lt;/em&gt; means to remove something. For example, you have a water insoluble solvent containing some ionic contaminants, and you wash this with water to remove those water soluble ionic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the first comment suggests, washing with a strong base like $\ce{NaOH}$ suggests that you are trying to remove an acid. A strong base would remove any acid, strong or weak. The base would be in an aqueous layer, and so there would also be the removal of water-soluble compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key is that your solvent must be non-miscible in water! In order to wash, your solvent and water cannot mix, so that the water pulls out the water soluble compounds and purifies the non-water soluble solvent.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-03-01T02:55:41.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26758" PostTypeId="1" CreationDate="2015-03-01T04:09:32.237" Score="4" ViewCount="87" Body="&lt;p&gt;The exchange and correlation potentials refer to those defined in density functional theory. (See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Local-density_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Local-density_approximation&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the exchange potential as $V_{x}(n) = \partial E_x(n)/ \partial n$, when $E_x(n)$ is the exchange energy density of the whole system(&lt;strong&gt;not&lt;/strong&gt; per particle). Define a similar quantity $V_c(n)$ for the correlation contribution. My question is, are the two potentials (generally) of the same sign?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: It seems that the prediction of the ionization potential using the -ve energy of HOMO is not good (See the section &lt;em&gt;Exchange-correlation potential&lt;/em&gt; on &lt;a href=&quot;http://en.wikipedia.org/wiki/Local-density_approximation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Local-density_approximation&lt;/a&gt;). Using a simple slater exchange potential and a reasonable correlation potential (for a uniform electron gas), the result (?; not sure about its correctness; feel free to point it out if you think it's wrong) is about 0.7*13.6 eV, which is not good. Of course one need not use DFT for hydrogen; it however serves as a test case. Is there any reason to believe the results will improve as the atomic number increases? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="418" LastEditDate="2015-03-14T07:56:23.583" LastActivityDate="2015-03-14T07:56:23.583" Title="Signs of Exchange and Correlation Potentials" Tags="&lt;atoms&gt;&lt;density-functional-theory&gt;&lt;ionization-energy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26760" PostTypeId="2" ParentId="26749" CreationDate="2015-03-01T04:57:34.050" Score="4" Body="&lt;p&gt;As matt_black said, it's about the size.&#xA;If you take a tall, thick glass tube, it'll be transparent, right? It'll transmit the light falling on it. But make it thinner and thinner(think human hair) then it'll appear white, as it'll refract the light passing through it.&#xA;As the light from the sun(which is a combination of all colors) is white, snow appears white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people think that old people's hair are white, but they're not. They get their color(black, brown, yellow, red, whatever) from a pigment called melanin, and in white hair this pigment is absent, leaving only a transparent rod of hair. It's so small it appears white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the ice cube part, if you've observed, sometimes there's a ice cube that's transparent on the outside, but in the middle there's a white core. That happens because it has shattered in the middle, probably because of anomalous expansion of water.&lt;/p&gt;&#xA;" OwnerUserId="14703" LastActivityDate="2015-03-01T04:57:34.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26762" PostTypeId="1" CreationDate="2015-03-01T08:45:44.897" Score="1" ViewCount="88" Body="&lt;p&gt;What happens to the $\ce{C-H}$ bond strength in methane as the degree of halogenation is increased (e.g. from $\ce{CH4}$ to $\ce{CH3X}$ to $\ce{CH2X2}$...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the effect of varying the halogens from fluorine to iodine?&lt;/p&gt;&#xA;" OwnerUserId="14467" LastEditorUserId="16683" LastEditDate="2016-12-04T17:19:41.407" LastActivityDate="2016-12-04T17:23:30.773" Title="C-H bond energy in substituted methanes" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26763" PostTypeId="1" AcceptedAnswerId="26772" CreationDate="2015-03-01T08:49:38.657" Score="10" ViewCount="5290" Body="&lt;p&gt;I came across a question asking for the major product on chlorination of m-nitrophenol. In my opinion the deactivating nature of nitro group should control the regioselectivity, but I found out the answer to be 4-chloro-3-nitrophenol. But why is the activating nature of the hydroxyl group dominating over the deactivating nature of the nitro group? &lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4945" LastEditDate="2015-08-13T08:54:18.973" LastActivityDate="2015-08-13T08:54:18.973" Title="What will be the major product on chlorination of m-nitrophenol?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26764" PostTypeId="1" AcceptedAnswerId="26766" CreationDate="2015-03-01T09:45:51.893" Score="2" ViewCount="688" Body="&lt;p&gt;I was reading Linus Pauling's book How to Live Longer and Feel Better, and in various places in the book he discusses a correlation between increased sucrose intake and increased blood plasma cholesterol levels, as reported in a study by &lt;a href=&quot;https://paulingblog.wordpress.com/tag/milton-winitz/&quot; rel=&quot;nofollow&quot;&gt;Milton Winitz&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pauling writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... the metabolism of sucrose yields in it's first step equal amounts of glucose and fructose.  ... Fructose metabolism goes in part by a different route such that it produces acetate, which is a precursor of the cholesterol that we synthesize in our liver cells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering specifically what the metabolic pathway from sucrose to cholesterol is and given say 100 grams of sucrose, how much will be turned by the body into cholesterol.&lt;/p&gt;&#xA;" OwnerUserId="14705" LastEditorUserId="9961" LastEditDate="2015-03-01T13:16:04.723" LastActivityDate="2015-03-01T13:36:42.973" Title="Pathway from Sucrose to Cholesterol" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26765" PostTypeId="1" CreationDate="2015-03-01T10:17:39.137" Score="1" ViewCount="37" Body="&lt;p&gt;Information is scarce in the roof cleaning industry and the big guys are making a killing selling chemicals that are harmful to the person using it and everything around the home. I would like to find something safe that I can use to prevent microorganisms from ever popping back up on my customers roofs.&lt;/p&gt;&#xA;" OwnerUserId="14583" LastActivityDate="2015-03-01T10:17:39.137" Title="Microorganism preventative solution safe for using on a roof and will last in the sun" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;solutions&gt;&lt;surfactants&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26766" PostTypeId="2" ParentId="26764" CreationDate="2015-03-01T10:22:39.780" Score="2" Body="&lt;p&gt;As you have known that metabolism of sucrose yields glucose and fructose.So formed glucose undergoes glycolysis producing pyruvate or lactate(under anaerobic condition). In case of fructose, it is converted into fructose-1-phosphate and which on action of aldolase enzyme produces dihydroxyacetone phosphate and glyceraldehyde-3-phosphate which are the intermediate products of glycolysis and finally converted to pyruvate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6CYBe.gif&quot; alt=&quot;Glycolysis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end product of glycolysis i.e. pyruvate is converted into acetyl coa in presence of coenzyme A , which is the precursor for the synthesis of cholesterol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aYC9C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the synthesis of cholesterol , acetyl coa undergoes different enzymatic reactions finally forming cholesterol at the end.&lt;img src=&quot;https://i.stack.imgur.com/VASNt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-03-01T10:22:39.780" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26769" PostTypeId="1" AcceptedAnswerId="26771" CreationDate="2015-03-01T11:12:11.517" Score="27" ViewCount="12325" Body="&lt;p&gt;Ethyne (Acetylene) is used with $\ce{O2}$ for welding. My question is that why is ethene not used? What properties of ethyne make it suitable for welding?&lt;/p&gt;&#xA;" OwnerUserId="13236" LastEditorUserId="4231" LastEditDate="2015-03-01T13:56:02.667" LastActivityDate="2020-12-20T12:32:31.450" Title="Why is ethyne and not ethene used for welding?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26770" PostTypeId="1" AcceptedAnswerId="26880" CreationDate="2015-03-01T11:32:16.407" Score="6" ViewCount="786" Body="&lt;p&gt;What is the mechanisms of this reaction? How feasible is this reaction? Can it happen under relatively mild conditions if it is intermolecular?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GhoTc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="865" LastEditDate="2015-10-16T17:47:15.220" LastActivityDate="2015-10-16T17:47:15.220" Title="Amine addition to nitro group" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nitro-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26771" PostTypeId="2" ParentId="26769" CreationDate="2015-03-01T11:35:52.993" Score="25" Body="&lt;p&gt;An acetylene molecule is composed of two carbon atoms and two hydrogen atoms. The two carbon atoms are held together by what is known as a triple carbon bond. This bond is useful in that it stores substantial energy that can be released as heat during combustion.The triple bond which makes the oxy-acetylene flame the hottest of all gas flames is also responsible for two rather exceptional properties of acetylene gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is: If free gaseous acetylene is subjected to severe shock, or a source of ignition, some of the triple bonds may break, releasing enough energy to cause all the other molecules in the enclosed volume to decompose into carbon and hydrogen with explosive force. The force of such an explosion is not so great as that released by the explosion of most mixtures of acetylene and oxygen, or acetylene and air, but it is substantial, and can be withstood only by extra-heavy-wall steel tubing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other property of acetylene is that the flammability range of mixtures of air and acetylene is broader than that of any other fuel gas/air mixture. Acetylene/air mixtures can be ignited when they contain anywhere from 2.5 percent acetylene to 80 percent acetylene. Mixtures of methane (the principal component of natural gas) and air are flammable when they contain as little as 5 percent methane and not more than 15 percent methane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be suitable for welding operations, a fuel gas, when burned with oxygen, must have the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;High flame temperature.&lt;/li&gt;&#xA;&lt;li&gt;High rate of flame propagation.&lt;/li&gt;&#xA;&lt;li&gt;Adequate heat content.&lt;/li&gt;&#xA;&lt;li&gt;Minimum chemical reaction of the flame with base and filler metals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Among the commercially available Oxy Fuel welding fuel gases, acetylene (combination of hydrogen and carbon) most closely meets all these requirements. Other gases, fuel such as MAPP gas, propylene, propane, natural gas, and proprietary gases based on these, have sufficiently high flame temperatures but exhibit low flame propagation rates.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorDisplayName="user2175" LastEditDate="2015-03-01T18:12:31.063" LastActivityDate="2015-03-01T18:12:31.063" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="26772" PostTypeId="2" ParentId="26763" CreationDate="2015-03-01T11:50:42.200" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But why is the activating nature of hydroxyl group dominating over the deactivating nature of Nitro group?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the precise reason why &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Picric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Picric acid&lt;/a&gt; is formed. ( &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/phenol/ring.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Check here&lt;/a&gt; )&lt;br&gt;&#xA;The activating nature of the $\ce{-OH}$ group is strong enough to counter the deactivating nature of the nitro groups.&lt;br&gt;&#xA;Look at Ron's comment on the question over &lt;a href=&quot;https://chemistry.stackexchange.com/questions/25131/how-do-we-know-whether-we-will-get-a-mono-substituted-or-tri-substituted-product&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;Even methyl group sustains deactivating nature of nitro group in the formation of &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Trinitrotoluene&quot; rel=&quot;nofollow noreferrer&quot;&gt;TNT&lt;/a&gt;. So no wonder phenolic group, a better activator, dominates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phenolic group increases electron density on ortho para (making it o-p director) while nitro groups decrease e-density on the o-p positions (making it a meta director). When both these effects act simultaneously, the question is who wins and is the density on o-p carbons less than corresponding meta ones? (as that would decide where the electrophile will attack)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonace structures of the groups gives the solution.&lt;br&gt;&#xA;In nitro group:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cky3T.png&quot; alt=&quot;enter image description here&quot;&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three charged atoms bonded together in the $\ce{-NO_2}$ molecule. This is less table than the positive charge bearing oxygen atom in phenolic group's resonance structure: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0sJ97.png&quot; alt=&quot;enter image description here&quot;&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence phenolic group wins!!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to the given starting reactant:     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rqLbL.png&quot; alt=&quot;enter image description here&quot;&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we consider the dominating ortho-para effect of phenolic group over the meta-directing effect of nitro groups to get 4-chloro-3-nitrophenol:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GuStv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-18T05:41:10.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26773" PostTypeId="2" ParentId="26753" CreationDate="2015-03-01T11:54:06.390" Score="3" Body="&lt;p&gt;The melting point of tellurium is approximately 450 °C; hence, I guess, the hot part of the melt could have a temperature of roughly 500 °C.&#xA;At this temperature, the vapour pressure of tellurium is roughly 100 Pa (0.1 % of the normal atmospheric pressure). By way of comparison, the vapour pressure of water at room temperature is 2–3 kPa.&#xA;Thus, a small but noticeable amount of tellurium could evaporate during the period of one day if the tellurium vapour is continuously removed by condensation on colder parts of the ampoule.&#xA;The actual transport rate depends on the temperature gradient, the pressure of other gases in the ampoule, and the geometry of the ampoule.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-03-01T11:54:06.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26774" PostTypeId="2" ParentId="26753" CreationDate="2015-03-01T12:24:16.527" Score="5" Body="&lt;p&gt;This is a general phenomenon. All substances at any temperature above $0\ K$ will shed a few atoms from the solid or liquid, forming a gas however dilute. The tendency for this to happen varies with temperature, and it can be measured by determining the equilibrium pressure of a substance inside an evacuated chamber. The key concept behind this is &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;vapour pressure&lt;/a&gt;&lt;/em&gt;. There are several previous Chem.SE questions regarding vapour pressure, so you may find something useful, such as perhaps &lt;a href=&quot;https://chemistry.stackexchange.com/questions/11104/why-doesnt-a-table-sublimate-while-ice-does&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, the higher the boiling/sublimation point of a substance, the lower its vapour pressure at any given temperature (in fact, the definition of the boiling/sublimation point is the temperature where the vapour pressure equals the pressure exerted by the surroundings onto the solid/liquid substance, which is in many cases equal to $1\ bar$). Tellurium is a relatively low boiling-point solid, with a vapour pressure of $1\ bar$ at $988\ °C$. According to &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01121a059&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, the vapour pressure of tellurium at its melting point ($450\ °C = 723\ K$) is approximately $70\ Pa$. This means there is a fair amount of tellurium vapour present inside the ampoule at this temperature. It is entirely conceivable that some of this vapour could deposit in small droplets on the upper surface of the glass ampoule, outside the main body of the liquid. Note that theoretically the deposition of tellurium on the upper surface should also happen at room temperature, though the equilibrium vapour pressure of tellurium at $25 °C$ is only about $5 \times 10^{-4} Pa$, which would make the transfer of a macroscopically visible amount of tellurium an extremely slow process.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-01T12:24:16.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26775" PostTypeId="1" AcceptedAnswerId="28102" CreationDate="2015-03-01T12:46:42.993" Score="5" ViewCount="8097" Body="&lt;p&gt;I came across this question where I had to convert $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$.&#xA;If I need to do so can I just react the given compound with water to get ethane and then react it with $\ce{NH3}$.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer is no then why can't I?&lt;br/&gt; &#xA;Is there any set of principles in organic chemistry which govern as to which substances can be replaced with what? &lt;br/&gt;&#xA;Kind of like the reactivity series of metals?&lt;br/&gt;&#xA;For e.g.: If I react an alkane with a halogen to get a haloalkane and then react this back with hydrogen to get the alkane then I don't think so it would be correct to do so. And I don't think that organic chemistry works that way. &lt;br/&gt;&#xA;But how do I understand while solving a question whether I am going in the right direction. &#xA;Please help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: This is not a homework question its just a question I stumbled across while my preparation for a college entrance test. (edit the title maybe)&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="8218" LastEditDate="2015-03-01T14:06:36.657" LastActivityDate="2017-03-27T13:37:48.610" Title="Can I react a Grignard reagent (ethylmagnesium bromide) with water and then ammonia?" Tags="&lt;organic-chemistry&gt;" AnswerCount="4" CommentCount="1" ClosedDate="2017-04-01T20:43:49.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="26776" PostTypeId="2" ParentId="26762" CreationDate="2015-03-01T13:12:24.730" Score="2" Body="&lt;p&gt;According to David R. Lide, ed., &lt;em&gt;CRC Handbook of Chemistry and&#xA;Physics, 90th Edition (CD-ROM Version 2010),&lt;/em&gt; CRC Press/Taylor and Francis, Boca Raton, FL, the C-H bond dissociation energies for increasing chlorination of methane are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CH3}\right) = (439.3 \pm 0.4) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CH2Cl}\right) = (419.0 \pm 2.3) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CHCl2}\right) = (400.6 \pm 2.0) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of different halogens in H-CX&lt;sub&gt;3&lt;/sub&gt; is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CF3}\right) = (445.2 \pm 2.9) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CBr3}\right) = (399.2 \pm 8.4) \; \text{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D\left(\ce{H-CI3}\right) = (423 \pm 29) \; \text{kJ/mol}$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-01T13:17:42.570" LastActivityDate="2015-03-01T13:17:42.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26777" PostTypeId="2" ParentId="26764" CreationDate="2015-03-01T13:36:42.973" Score="1" Body="&lt;p&gt;Our body uses the so-called &quot;mevalonate&quot; pathway to synthesize cholesterol (and many more related terpenes) from acetate.  Sucrose is 12 carbon atoms, acetate is 2.  Even if we allow &lt;em&gt;both&lt;/em&gt; glucose and fructose to be converted to acetate, which is possible but not commonly attained because of the way metabolism is regulated) we will get only 4 acetate molecules, containing 8 carbon atoms.  The other 4 carbons are lost as $\ce{CO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cholesterol has 27 carbon atoms.  But the pathway for its synthesis goes through lanosterol, which has 30.  So each mole of sucrose (which gives rise to 16 carbon-moles of acetate) can make 8/30 = 26.7% mole of cholesterol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That calculation assumes that only &lt;em&gt;carbon&lt;/em&gt; is limiting, and that the emitted $\ce{CO2}$ cannot be recovered.  If you relax those assumptions, you can calculate the theoretical yield in more detail.  Here is &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs12257-010-0329-z&quot; rel=&quot;nofollow&quot;&gt;one paper&lt;/a&gt; that goes through very similar calculations for production of terpenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, some bacteria have a different pathway for terpene biosynthesis that has a higher theoretical yield, because it doesn't start with acetate, it starts with glycolytic precursors before pyruvate dehydrogenase.  This completely different pathway is called the DXP pathway (or the MEP pathway or the non-mevalonate) pathway.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-01T13:36:42.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26779" PostTypeId="2" ParentId="19150" CreationDate="2015-03-01T14:38:56.597" Score="3" Body="&lt;p&gt;I feel that the Original Poster actually needed examples of :&lt;br&gt;&#xA;$\Delta H &amp;lt; 0 ;  \Delta S &amp;gt; 0$&lt;br&gt;&#xA;since majority of the common reactions do not fall under this criteria . If not then Thomij's answer is perfect for the understanding of Gibbs Free Energy .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are a few examples :  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Hydrogen peroxide&lt;/strong&gt; is thermodynamically unstable and decomposes to form water and oxygen &#xA;$\ce{2 H_2O_2 → 2 H_2O + O_2}$&lt;br&gt;&#xA;$ \Delta H = -98.2~\mathrm{kJ\, mol^{-1}}$&lt;br&gt;&#xA;$ \Delta S = 70.5~\mathrm{kJ\, mol^{-1}K^{-1}}$   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Decomposition of Ammonium dichromate&lt;/strong&gt;&lt;br&gt;&#xA;$\ce{(NH_4)_2Cr_2O_7 → Cr_2O_3 + N_2 + 4 H_2O}$&lt;br&gt;&#xA;$ \Delta H = -1794.9.2~\mathrm{kJ\, mol^{-1}}$&lt;br&gt;&#xA;$ \Delta S &amp;gt; 0 $ due to formation of gases from the solid&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="5456" LastEditDate="2015-03-01T14:44:56.460" LastActivityDate="2015-03-01T14:44:56.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26780" PostTypeId="2" ParentId="4554" CreationDate="2015-03-01T14:49:50.087" Score="5" Body="&lt;p&gt;melting points of all group (i) elements is dependent on the strength of the metallic bond. In metallic bonding, the group (i) cations in the metallic lattice are attracted to the delocalised electrons. Down the group, the number of delocalised electrons and the charge on each cation remains the same at +1 but the cationic radius increases so the attraction between the cations and the electrons in the lattice get weaker down the group and so does the strength of the metallic bond&lt;/p&gt;&#xA;" OwnerUserId="14714" LastActivityDate="2015-03-01T14:49:50.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26781" PostTypeId="1" AcceptedAnswerId="27133" CreationDate="2015-03-01T17:40:39.517" Score="19" ViewCount="4238" Body="&lt;p&gt;I have conducted an experiment measuring the conductivity of both hydrochloric acid and sulfuric acid in solution respectively, with varying concentration. Wikipedia as well as a question on the site suggests a $\sqrt{c}$ relationship is viable. However, my data resembles this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6WtZ0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly not a simple $\sqrt{c}$. It reminds me of the inverse gamma distribution. I have tried looking at papers by Onsager on his conductivity theory, but they're unclear, and he doesn't propose a function. Are there any papers that look at accurately modelling conductivity as a function of concentration which exhibit this form? Obviously for low concentrations (e.g. 1-2 molar), a linear or $\sqrt{c}$ would suffice.&lt;/p&gt;&#xA;" OwnerUserId="5542" LastEditorUserId="4945" LastEditDate="2015-03-10T06:47:03.380" LastActivityDate="2019-10-08T11:37:16.647" Title="Conductivity as a function of acid concentration" Tags="&lt;reference-request&gt;&lt;conductivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26783" PostTypeId="1" CreationDate="2015-03-01T17:54:11.713" Score="1" ViewCount="34" Body="&lt;p&gt;The title can be a little unclear. But what I am basically saying is say we add a reactant to a reaction to push the reaction forward toward the product side, that reactant which was increased in amount/ concentration will surely less. Part of it used to produce more products to gain new equilibrium. &lt;strong&gt;But will it always be more than the original amount in the equilibrium before?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing much of advanced math, I just experimented with some simple fractions; thus far I haven't seen anything that violate this rule.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-03-01T17:54:11.713" Title="Does the added component (intended to disturb equilibrium) always ends up being more than the original amount, in the new equilibrium?" Tags="&lt;equilibrium&gt;" AnswerCount="0" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26784" PostTypeId="2" ParentId="26434" CreationDate="2015-03-01T18:10:16.667" Score="1" Body="&lt;p&gt;Smelly compounds can be destroyed by free radicals wich are formed by a burning candle. But you need a lot of candles for your apartment.&#xA;Candles were also used in bathrooms (after toilet visits) to destroy smelly compounds like $\ce{H2S, CH3-SH, CH3-S-CH3}$ ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better and safer way is it too use cyclodextrin. This compound absorbs molecules and destroys them as well. Cyclodextrin is the working stuff inside the air-improvement sprays.&lt;/p&gt;&#xA;" OwnerUserId="13429" LastEditorUserId="4231" LastEditDate="2015-03-01T18:12:47.553" LastActivityDate="2015-03-01T18:12:47.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26785" PostTypeId="1" AcceptedAnswerId="26799" CreationDate="2015-03-01T18:25:04.090" Score="4" ViewCount="151" Body="&lt;p&gt;I'm taking chemical thermodynamics this semester, and currently seeing things like the Maxwell's Relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor never uses explicitly the variable &quot;n&quot; and always works with molar quantities like $\bar{V}$ and stuff like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this simplifies the algebra a great deal, and can be useful if you have the temptation to get a &quot;molar quantity&quot;. However, I don't like it right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think theres nothing that can assure me that if I were to work with the variable $n$ in my derivations, I won't somehow get a quadratic dependence of my thermodynamic equation with $n$, or logarithmic, who knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems that there is always a linear dependence with $n$. This makes physical sense to me, however, how can I prove this to myself? Maybe this is more of a question of mathematics, but it has scientific implications as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastActivityDate="2015-03-03T21:44:35.107" Title="Will all thermodynamic equations have a linear dependence with moles?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26786" PostTypeId="2" ParentId="26781" CreationDate="2015-03-01T18:26:25.373" Score="3" Body="&lt;p&gt;Your carefully plotted results seem in line with this table from the &lt;a href=&quot;https://www.grc.com/dev/CES/TNS/Conductivity_v_Concentration.pdf&quot; rel=&quot;nofollow&quot;&gt;Foxboro Company&lt;/a&gt;. Though there are references that give a simple square-root relationship, they are clearly &lt;em&gt;wrong&lt;/em&gt; at high concentrations of many electrolytes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &lt;em&gt;practical&lt;/em&gt; guide to conductivity measurement at &lt;a href=&quot;http://www.tau.ac.il/~chemlaba/Files/Theoryconductivity.pdf&quot; rel=&quot;nofollow&quot;&gt;Conductivity Theory and Practice&lt;/a&gt;, which mentions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For some samples with high concentrations, the conductance = f (concentration) curve may show a maximum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Conductivity_%28electrolytic%29#Higher_concentrations&quot; rel=&quot;nofollow&quot;&gt;Wikipedia on Conductivity&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Both Kohlrausch's law and the Debye-Hückel-Onsager equation break down as the concentration of the electrolyte increases above a certain value. The reason for this is that as concentration increases the average distance between cation and anion decreases, so that there is more inter-ionic interaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, as you've done, actual conductivity curves are often determined experimentally, and formulae then are derived to fit the curves.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="4945" LastEditDate="2015-03-10T06:51:18.347" LastActivityDate="2015-03-10T06:51:18.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26787" PostTypeId="1" CreationDate="2015-03-01T19:55:52.250" Score="0" ViewCount="129" Body="&lt;p&gt;First of all, though I think that my question is related to this society, if you think reverse, tell me to delete the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if carbohydrate alcohol is inebriant; Can anyone reference me to some valid source?&lt;/p&gt;&#xA;" OwnerUserId="14719" LastEditorUserId="7448" LastEditDate="2015-03-01T20:13:15.390" LastActivityDate="2016-09-14T10:39:22.233" Title="Is carbohydrate alcohol inebriant?" Tags="&lt;organic-chemistry&gt;&lt;chemical-biology&gt;&lt;alcohols&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26788" PostTypeId="2" ParentId="6369" CreationDate="2015-03-01T20:01:43.367" Score="5" Body="&lt;p&gt;Aluminum carbonate, if existed, would decompose by the following reaction:&#xA;$$\ce{Al2(CO3)3 → Al2O3 + 3 CO2}$$&#xA;to $\ce{CO2}$ and $\ce{Al2O3}$, with a release of energy, even in the absence of water. Therefore, it is unstable, and probably could not be isolated under any conditions. &lt;/p&gt;&#xA;" OwnerUserId="4668" LastEditorUserId="7448" LastEditDate="2015-03-01T20:11:14.453" LastActivityDate="2015-03-01T20:11:14.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26789" PostTypeId="1" AcceptedAnswerId="26792" CreationDate="2015-03-01T20:41:28.733" Score="1" ViewCount="11493" Body="&lt;p&gt;Really confused about this, since there is carbon and two other types of atoms: should be polar, right? However, some people online say it is non-polar. Also, is it soluble in water?&lt;/p&gt;&#xA;" OwnerUserId="14721" LastEditorUserId="9961" LastEditDate="2015-03-01T23:04:25.520" LastActivityDate="2015-03-01T23:06:12.090" Title="Salicylic acid - polarity and solubility" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;polarity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26791" PostTypeId="2" ParentId="6369" CreationDate="2015-03-01T20:48:21.273" Score="1" Body="&lt;p&gt;Aluminium carbonate is a salt of a very weak acid and a weak base, so it tends to hydrolyze. Similar may be said for a much easier available aluminium sulfide: it reacts with water in the air, so it is a common tool for rebellious teenagers if they can get the materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, aluminium cation is extremely small. It means, that it forms very strong bonds with small anions (oxygen). Since carbonate is a rather big cation, it is not a best partner for aluminium, so it should release carbon dioxide very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a better understanding consider second group carbonates. Barium carbonate is very stable, calcium carbonate dissociates around $1000^oC$ and berrilum carbonate does not form. Aluminium is an even smaller cation, than berillium.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-03-01T20:48:21.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26792" PostTypeId="2" ParentId="26789" CreationDate="2015-03-01T22:49:21.793" Score="3" Body="&lt;p&gt;Well it has polar groups - carboxylic and phenolic but benzene ring reduces its polarity. One could say carbon atoms win 7:3 with oxygens, but it's still not bad :) &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/salicylic_acid#section=Solubility&quot; rel=&quot;nofollow&quot;&gt;PubChem&lt;/a&gt; say it's solubility is about 2g/L, but &lt;a href=&quot;http://chemister.ru/Database/properties-en.php?dbid=1&amp;amp;id=1777&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; gives more info - it dissolves much better in hot water and its solubility in methanol, which is quite polar, is good, and in benzene (non-polar) bad. Generally it seems to be somewhere in the middle between strongly polar and strongly non-polar compounds.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-03-01T23:06:12.090" LastActivityDate="2015-03-01T23:06:12.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26795" PostTypeId="1" CreationDate="2015-03-02T01:56:46.950" Score="0" ViewCount="320" Body="&lt;p&gt;I bought a box of A&amp;amp;H Washing Soda, the packaging of which implied that it was pure washing soda (sodium carbonate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On dissolving it in osmotically-purified water, however, I found that the water took on a brown tinge, which would seem to indicate a contamination or adulteration of some type. Also, the washing soda tended to form hard chips or clumps, which I have not seen in previous experiments doing the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there may be some kind of organic compound being used to adulterate the washing soda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine if there is an organic adulterant and remove it?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastActivityDate="2015-03-02T04:23:45.270" Title="Purifying washing soda" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26797" PostTypeId="1" CreationDate="2015-03-02T03:22:43.370" Score="0" ViewCount="2376" Body="&lt;p&gt;Acids: arachidic acid, palmitic acid, lauric acid, etc.&lt;br&gt;&#xA;Alcohols: butan-1-ol, pentan-1-ol, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The melting points of acids with $\ce{-COOH}$ at the end compared to melting points of alcohols, $\ce{-OH}$ at the end, is higher. I need to relate this to intermolecular forces. I was going to say that because of many more electrons in the acids, there are greater London dispersion forces and also more polarity (dipole dipole forces) because of the $\ce{COOH}$ at one end. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="14721" LastEditorUserId="4945" LastEditDate="2015-03-02T07:23:36.500" LastActivityDate="2015-03-02T12:48:10.900" Title="How to rationalise the difference in the melting points of acids and alcohols with inter molecular forces?" Tags="&lt;intermolecular-forces&gt;&lt;melting-point&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26798" PostTypeId="2" ParentId="15662" CreationDate="2015-03-02T03:32:15.427" Score="5" Body="&lt;p&gt;The strength of carboxylic acid can be discussed by positive inductive effect. Here, in ethanoic acid there is an electron releasing inductive effect from the alkyl group. The larger the positive inductive effect, the lesser will be the strength of an acid. Similarly in formic acid there is no alkyl group and hence no inductive effect takes place. So formic acid is a stronger acid than ethanoic acid.&lt;/p&gt;&#xA;" OwnerUserId="14730" LastEditorUserId="4945" LastEditDate="2015-03-02T06:58:59.313" LastActivityDate="2015-03-02T06:58:59.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26799" PostTypeId="2" ParentId="26785" CreationDate="2015-03-02T03:45:55.043" Score="2" Body="&lt;p&gt;Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is:&#xA;$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We choose $\lambda = 1/N$, so &#xA;$S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ N \,S(e, v, 1) = S(E,V,N)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(E,V,N) =$ $whatever$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work with molar properties you will find just:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDITED:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of. &lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2015-03-03T21:44:35.107" LastActivityDate="2015-03-03T21:44:35.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26800" PostTypeId="2" ParentId="26795" CreationDate="2015-03-02T04:23:45.270" Score="1" Body="&lt;p&gt;If there is an &lt;em&gt;organic&lt;/em&gt; contaminant in the $\ce{Na2CO3}$, you could try to rinse the sodium carbonate with an organic solvent in which $\ce{Na2CO3}$ is insoluble.  Based on &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/je60030a009&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, acetone and methanol would qualify.  Since those solvents, especially methanol, are toxic, I'm not recommending either one of those.  Based on the data for methanol and acetone, I'm guessing absolute ethanol would also be incapable of dissolving sodium carbonate, so it would probably the best (relatively) non-toxic solvent for extraction.  If you can't get absolute ethanol due to tax or beverage laws, you could probably used denatured ethyl alcohol instead.  It usually has small percentages of methanol, acetone, and/or isopropyl alcohol added to make it non-potable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you wash the sodium carbonate in this solvent, you should air dry it for several days, and then dry it in an explosion-proof drying oven at 100 - 200 &amp;deg;C.  Be very careful when combining flammable solvents with heated environments!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem could also be with the water you used, although you did say you used osmotically purified water.  The reason I think the water may be the culprit is that if the washing soda contained a brown contaminant, I would think it render the powder visibly brown.  If no brownness is evident until adding the soda to the water, it makes me think the high pH from the washing soda is precipitating minerals from the water, in particular iron.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-02T04:23:45.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26801" PostTypeId="2" ParentId="26775" CreationDate="2015-03-02T04:33:59.367" Score="3" Body="&lt;p&gt;The short answer is no, you can't.  Water will hydrolyze a Grignard reagent, destroying its reactivity.  The product of the hydrolysis, according to &lt;a href=&quot;http://www.chemguide.co.uk/organicprops/haloalkanes/grignard.html&quot; rel=&quot;nofollow&quot;&gt;this web site&lt;/a&gt; is an alkane and a basic magnesium bromide:&#xA;$$\ce{CH3CH2MgBr + H2O -&amp;gt; CH3CH3 + MgBrOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So before you ever got to reacting anything with ammonia, your reactants would be lost as ethane gas bubbles out of your reaction.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-02T04:33:59.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26802" PostTypeId="1" AcceptedAnswerId="26915" CreationDate="2015-03-02T04:55:17.953" Score="17" ViewCount="1327" Body="&lt;p&gt;I recently saw a protocol for extracting organic materials that are bound in the silica matrix formed by diatoms.  The extraction protocol used 10 molar ammonium fluoride ($\ce{NH4F}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I briefly considered using the protocol and then endeavoring to remove the ammonium fluoride (and water) by lyophilization or with a rotary evaporator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am concerned about the safety of doing this.  My first worry is that any glass equipment or vessels in the lyophilizer or rotary evaporator systems would be damaged by the fluoride.  My second worry is that even if the equipment used did not contain any glass, dangerous solutions of hydrogen fluoride (i.e. hydrofluoric acid) could be formed in the cold trap of the equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is, am I right that it is a bad idea to evaporate ammonium fluoride in glass equipment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second (but related) question is, even if glass is not used, what are the safety risks of evaporating ammonium fluoride under vacuum?  Imagining that a glass-free apparatus could be found, is there a way to do the evaporation safely?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2016-05-16T11:38:30.493" Title="Safety in handling of concentrated ammonium fluoride" Tags="&lt;safety&gt;&lt;materials&gt;&lt;equipment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26803" PostTypeId="2" ParentId="26785" CreationDate="2015-03-02T05:23:04.827" Score="2" Body="&lt;p&gt;One exception to the general rule that equations &lt;em&gt;will&lt;/em&gt; be linearly dependent on moles is if surface tension is to be modeled.  With surface tension, we have to specify the shape of a system as well as the extent.  The fundamental equation for a single-component system including surface tension would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d\bar{G} = \bar{V}dP - \bar{S}dT + \gamma d\bar{A} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is the surface tension and $d\bar{A}$ is a differential change in surface area (i.e. $\bar{A}$ is surface area per mole).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the shape of the system is constant (imagine say a sphere of water that grows bigger and bigger as we add more water molecules, but still stays perfectly spherical), then $A$ is proportional to $N^{2/3}$, so $\bar{A}$ is proportional to $N^{-1/3}$.  The exponent of 2/3 is not one, therefore the free energy is not linear in the number of moles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of cases are rarely studied in chemical thermodynamics at the undergraduate level, however.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-02T08:03:44.127" LastActivityDate="2015-03-02T08:03:44.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26804" PostTypeId="1" AcceptedAnswerId="26831" CreationDate="2015-03-02T05:23:56.027" Score="2" ViewCount="387" Body="&lt;p&gt;I am given a table that gives me the different total pressures at certain times. The decomposition of phosphine is the reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4 PH3(g) &amp;lt;=&amp;gt; P4(g) + 6 H2(g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time/pressure values are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{rr}&#xA;\mathrm{time\ [min]} &amp;amp; \mathrm{pressure\ [Torr]} \\&#xA;\hline&#xA;0 &amp;amp; 100 \\&#xA;40 &amp;amp; 150\\&#xA;80 &amp;amp; 167\\&#xA;100 &amp;amp; 172\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction takes place at 950K. I can then find the molarity of the total gas mixture at each time using $$\frac{n}{V}=\frac{P}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know how to get the pressure of the $\ce{P4}$ for each individual time. How would one go about doing this? &lt;/p&gt;&#xA;" OwnerUserId="7679" LastEditorUserId="10185" LastEditDate="2015-03-02T09:11:57.047" LastActivityDate="2015-03-02T16:49:17.090" Title="Finding molarities of individual reagents given the chemical equation and total concentration in a container" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26805" PostTypeId="1" CreationDate="2015-03-02T05:31:19.343" Score="3" ViewCount="991" Body="&lt;p&gt;From my understanding plotting the conc. over time for a 0-order reaction yield a negative linear slope. Plotting conc vs time for a first order reaction intuitively yields an exponential type slope. however, my teacher never covered the looks of a conc vs time graph for a second order reaction. What may this look like. In conjunction, what does it look like when we graph its rate over concentration? &lt;/p&gt;&#xA;" OwnerUserId="2197" LastEditorUserId="13184" LastEditDate="2015-03-02T05:39:02.213" LastActivityDate="2015-03-02T06:01:36.500" Title="Conc. vs Time Graph for Second Order Reaction" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26806" PostTypeId="2" ParentId="26805" CreationDate="2015-03-02T05:51:49.120" Score="8" Body="&lt;p&gt;Let's take a single-component second-order reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2A -&amp;gt; B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a 2nd-order reaction, the rate will be proportional to the concentration of $A$, squared:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dA}{dt} = -kA^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That equation can be integrated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dA}{A^2} == -k dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-(A^{-1}-A_0^{-1}) = -k(t-t_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solving for $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \frac{1}{k(t-t_0) + A_0^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the equation of a hyberbola.  It goes to zero &lt;em&gt;more&lt;/em&gt; slowly than a the expression for a first-order reaction, i.e. an exponential decay.  This makes sense because as $A$ gets more and more diluted as the reaction occurs, collisions between $A$ molecules become rarer and rarer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions&quot; rel=&quot;noreferrer&quot;&gt;this web page&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-02T05:51:49.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26807" PostTypeId="1" AcceptedAnswerId="40680" CreationDate="2015-03-02T05:58:31.497" Score="1" ViewCount="19060" Body="&lt;p&gt;Sodium acetate trihydrate dissolves in water to its constituent ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOAc.3H2O (s) -&amp;gt;[H2O] Na+ (aq) + OAc- (aq) + 3H2O (l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystallisation of sodium acetate from a supersaturated solution is well-known to be exothermic. Since dissolution is the reverse of crystallisation, should the enthalpy of dissolution be positive (i.e. an endothermic process)?&lt;/p&gt;&#xA;" OwnerUserId="14732" LastEditorUserId="16683" LastEditDate="2017-04-01T02:36:46.053" LastActivityDate="2017-04-01T02:36:46.053" Title="Is the dissolution of sodium acetate trihydate endothermic?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26810" PostTypeId="2" ParentId="26797" CreationDate="2015-03-02T06:35:36.467" Score="1" Body="&lt;p&gt;The ultimate goal of melting-point reasoning is to explain how intermolecular forces lead to better packing of molecules in the solid phase.&lt;br&gt;&#xA;Both alcohols and organic acids have hydrogen bonds, but organic acids have stronger hydrogen bonds due to the withdrawing carbonyl group, $\ce{COO-}$. This causes the acids to interact stronger with each other. More importantly, organic acids form dimers (usually) with their hydrogen bonds (acetic acid dimer, for example). For this reason, the packing of the acids in the solid phase is significantly better than that of alcohol; hence the higher melting point. Acetic acid: $16~^\circ\mathrm{C}$ vs. $-114~^\circ\mathrm{C}$ for ethanol.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastEditorUserId="4945" LastEditDate="2015-03-02T07:20:19.167" LastActivityDate="2015-03-02T07:20:19.167" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26811" PostTypeId="1" AcceptedAnswerId="26829" CreationDate="2015-03-02T07:23:00.600" Score="-5" ViewCount="1376" Body="&lt;p&gt;What would the Ksp of calcium hydroxide be at around room temp (66 Fahrenheit) and 1 atm of pressure? I know the Ksp value varies with temperature, but is there a notable difference from 20 degrees C to 25 degrees C? &lt;/p&gt;&#xA;" OwnerUserId="14733" LastActivityDate="2015-03-03T10:56:01.410" Title="Ksp of calcium hydroxide" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-04-08T02:46:10.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="26812" PostTypeId="2" ParentId="26802" CreationDate="2015-03-02T09:21:54.670" Score="8" Body="&lt;p&gt;&lt;em&gt;Note that I don't have first hand experience on this - I never used ammonium fluoride to release organic material from diatomeous earth, but I frequently used tetrabutylammonium fluoride (TBAF) in &lt;strike&gt;dichloromethane&lt;/strike&gt; &lt;strong&gt;THF&lt;/strong&gt; to cleave silyl ethers.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Cleaving silyl ethers with TBAF&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To a solution of 3.8 mmol of a cyclopentylsilylether in 50 mL THF at 0 °C under argon, 4.5 mL of a TBAF solution (1.0 M in THF) were added. The mixture was stirred for 10 min, allowed to come to RT and the stirred for another 10 min.&#xA;  Then, 100 mL diethyl ether were added.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The organic layer was extracted twice with each 20 mL water and then dried over sodium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, release of $\ce{HF}$ from $\ce{NH4F}$ (we are not talking about $\ce{NH4HF2 =\ NH4F*HF}$ here) is only to be expected under the following two conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;acidification&lt;/li&gt;&#xA;&lt;li&gt;heating $\ce{NH4F}$ up to its melting point - sublimation is accompanied by decomposition.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-02T14:47:14.917" LastActivityDate="2015-03-02T14:47:14.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26814" PostTypeId="1" CreationDate="2015-03-02T10:31:54.760" Score="0" ViewCount="118" Body="&lt;p&gt;At what percentage reduction or increase in Km and Vmax is an enzyme inhibitor regarded as effective?- is any percentage increase or reduction regarded as being a clinically useful level of inhibition, or would it have to be over a certain limit? &lt;/p&gt;&#xA;" OwnerUserId="14736" LastActivityDate="2015-03-02T14:54:28.737" Title="How to determine enzyme inhibitor efficacy?" Tags="&lt;enzymes&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26815" PostTypeId="2" ParentId="26814" CreationDate="2015-03-02T11:32:59.137" Score="1" Body="&lt;p&gt;If the inhibitor is a competitive inhibitor, to quantify inhibition the $K_m$  of the inhibitor ($K_I$) is determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_I$ is compared to the $K_m$ of the substrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the degree of inhibition is a function the inhibitor and substrate concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{max}$ is unaffected by a competitive inhibitor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of an inhibitor the kinetics are modified such that the enzyme has an effect $K_m$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_m(1 + [I]/K_I)$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-02T14:54:28.737" LastActivityDate="2015-03-02T14:54:28.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26816" PostTypeId="1" CreationDate="2015-03-02T12:18:03.947" Score="6" ViewCount="190" Body="&lt;p&gt;Will the nuclear testing done during the past decades or if a nuclear war happened in the present day, affect the accuracy of carbon dating in the future? (Hypothetically, we haven't developed a better dating method)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean you could just adjust the dating to the amount of Carbon-14 if there will be a sudden change, but will the fossils found in around and near the blast sites be dated differently from the unaffected areas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry not really good in science, I don't know if this should be in Physics STEX and I might be wrong in my understandings. Thanks in advance! &lt;/p&gt;&#xA;" OwnerUserId="13141" LastActivityDate="2015-05-04T13:23:57.200" Title="Nuclear Testing and Carbon Dating" Tags="&lt;radioactivity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26817" PostTypeId="1" CreationDate="2015-03-02T12:30:26.140" Score="1" ViewCount="134" Body="&lt;p&gt;This is again related to the crime story I'm working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how much &lt;strong&gt;biological detergent&lt;/strong&gt; (in kilograms) and &lt;strong&gt;water&lt;/strong&gt; (in liters) and how long it would take to dissolve &lt;span class=&quot;math-container&quot;&gt;$60 \,\rm{Kg}$&lt;/span&gt; of organic material at &lt;span class=&quot;math-container&quot;&gt;$50^\circ$&lt;/span&gt; Celsius with standard store-bought bio-detergent. Just an estimate of the time required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Would it dissolve hair?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do I need a special container as to not commit the same mistake Jesse had in &lt;em&gt;Breaking Bad&lt;/em&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13141" LastEditorUserId="30836" LastEditDate="2020-04-27T07:11:59.900" LastActivityDate="2020-04-27T07:11:59.900" Title="On biological detergent breaking down organic material" Tags="&lt;organic-chemistry&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26818" PostTypeId="2" ParentId="26797" CreationDate="2015-03-02T12:48:10.900" Score="0" Body="&lt;p&gt;&lt;strong&gt;Huy Nguyen&lt;/strong&gt; has already addressed the importance of dimers through hydrogen bonds in the solid state of alkanoic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to correlate the chain length of alkanoic acids of the form $\mathrm{{C_{n}H_{{2n+1}}-COOH}}$ with their melting points, a look at the data might be helpful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrr}&#xA;\mathrm{(trivial)\ name} &amp;amp; \mathrm{n} &amp;amp; \mathrm{mp}\\ &#xA;\mathrm{\tilde{} acid} &amp;amp;  &amp;amp; \mathrm{[°C]}\\&#xA;\hline&#xA;\mathrm{acetic} &amp;amp; 1 &amp;amp; 17\\&#xA;\mathrm{propionic} &amp;amp; 2 &amp;amp; -21\\&#xA;\mathrm{butyric} &amp;amp; 3 &amp;amp; -5\\&#xA;\mathrm{valeric} &amp;amp; 4 &amp;amp; -32\\&#xA;\mathrm{caproic} &amp;amp; 5 &amp;amp; -3.4\\&#xA;\mathrm{enathic} &amp;amp; 6 &amp;amp; -7.5\\&#xA;\mathrm{caprylic} &amp;amp; 7 &amp;amp; 16.7\\&#xA;\mathrm{pelargonic} &amp;amp; 8 &amp;amp; 12.5\\&#xA;\mathrm{capric} &amp;amp; 9 &amp;amp; 31.6\\&#xA;\mathrm{undecanoic} &amp;amp; 10 &amp;amp; 28.6\\&#xA;\mathrm{lauric} &amp;amp; 11 &amp;amp; 48.3\\&#xA;\mathrm{tridecanoic} &amp;amp; 12 &amp;amp; 41.5\\&#xA;\mathrm{myristic} &amp;amp; 13 &amp;amp; 54.4\\&#xA;\mathrm{pentadecanoic} &amp;amp; 14 &amp;amp; 31.6\\&#xA;\mathrm{palmitic} &amp;amp; 15 &amp;amp; 28.6\\&#xA;\mathrm{margarinic} &amp;amp; 16 &amp;amp; 48.3\\&#xA;\mathrm{stearic} &amp;amp; 17 &amp;amp; 41.5\\&#xA;\mathrm{nonadecanoic} &amp;amp; 18 &amp;amp; 54.4\\&#xA;\mathrm{arachic} &amp;amp; 19 &amp;amp; 75.5\\&#xA;\mathrm{behemic} &amp;amp; 21 &amp;amp; 80\\&#xA;\mathrm{lignoceric} &amp;amp; 23 &amp;amp; 84.2\\&#xA;\mathrm{montanic} &amp;amp; 27 &amp;amp; 90.9\\&#xA;\end{array}&#xA;(Source: German and english wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting melting points vs length of the alkyl substituent gives the following result:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NTPWB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-02T12:48:10.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26819" PostTypeId="2" ParentId="26817" CreationDate="2015-03-02T13:05:43.920" Score="3" Body="&lt;p&gt;In the household, hair and fat residues in the pipes are typically dissolved using strongly alkaine drain cleaners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the hints and directions given in the commentaries, namely &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4877/is-it-actually-possible-to-dispose-of-a-body-with-hydrofluoric-acid&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/3949/dissolving-organic-tissues&quot;&gt;that&lt;/a&gt; you might want to consider, at least in part and for the dramatic effect, a biological solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the preparation of skeletons for museum displays, adult insects and larvae of the &lt;strong&gt;bacon beetle&lt;/strong&gt; (&lt;em&gt;Dermestes lardarius&lt;/em&gt;) are frequently used. After gutting the carcass (I'll leave the gory details to the novellist), plan a couple of weeks for them.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-02T14:07:08.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26820" PostTypeId="1" CreationDate="2015-03-02T13:27:33.150" Score="7" ViewCount="1174" Body="&lt;p&gt;What can I predict about the &lt;span class=&quot;math-container&quot;&gt;$\ce{^{31}P}$&lt;/span&gt; spectra of &lt;span class=&quot;math-container&quot;&gt;$\ce{TiP2O7}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;To get the required information about the crystal structure of the compound&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here is a screenshot from the ICSD website:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5Jjh.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since there is only one oxidation state (+5) for &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt;, there will be only one peak. And from the Wyckoff symbol, 4c, the intensity of the signal is 4, i.e. 4 &lt;span class=&quot;math-container&quot;&gt;$\ce{P}$&lt;/span&gt; atoms in the unit cell.&lt;/p&gt;&#xA;&lt;p&gt;I can't go any further to complete the task and extract all the relevant information. I'm using SIMPSON software to simulate the spectra. Can somebody walk me step by step through this?&lt;/p&gt;&#xA;" OwnerUserId="4445" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2019-03-31T15:39:41.900" Title="How to predict the 31P NMR spectra?" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26821" PostTypeId="1" AcceptedAnswerId="26822" CreationDate="2015-03-02T13:40:43.903" Score="0" ViewCount="112" Body="&lt;p&gt;I was just reading about how Nitrogen and Oxygen combine to form Nitric oxide. &#xA;&lt;img src=&quot;https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcQ1yqvfJf_UlPknaS5h0frI6LP-MkMehrThZHiQRf-n1Y73v9Gi&quot; alt=&quot;Image result for n2 + o2 = 2no&quot;/&gt;&#xA;&lt;br&gt;&#xA;My question is, why would this reaction even occur? Isn't the valency of Nitrogen 3- and Oxygen 2-, and therefore this reaction seems to form a product that still hasn't achieved the octet electronic configuration.&lt;br/&gt;&#xA;Could I please have some help?&#xA;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Or even $\ce{N2O}$ or $\ce{NO2}$ for that matter?&lt;/p&gt;&#xA;" OwnerUserId="13177" LastEditorUserId="7448" LastEditDate="2015-03-03T16:31:52.630" LastActivityDate="2015-03-03T16:31:52.630" Title="Why would this reaction occur? Won't it be unfavourable?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-03-02T17:11:56.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="26822" PostTypeId="2" ParentId="26821" CreationDate="2015-03-02T13:51:28.997" Score="0" Body="&lt;p&gt;Think about it this way.  If a molecule of nitrogen and a molecule of oxygen were to collide with enough kinetic energy to break the bonds that hold the molecules together, and they collide at such an angle that allows their orbitals to overlap and form new bonds, then $\ce{NO}$ can be made.  The octet rule is a rule.  Whenever you see the word rule in chemistry you should think that it can be broken, some rules are broken often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is endothermic and is not spontaneous but by random chance, some $\ce{NO}$ can form in a mixture of nitrogen and oxygen.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="4231" LastEditDate="2015-03-02T13:54:55.640" LastActivityDate="2015-03-02T13:54:55.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26823" PostTypeId="2" ParentId="26816" CreationDate="2015-03-02T13:56:34.270" Score="1" Body="&lt;p&gt;If the amount of extra C14 was unknown this might become a problem. If the change in atmospheric C14 was recorded and the records survive into the future, it would be easy to adjust the dates. At present we have good data on C14 concentration over time for various locations around the world. I don't know the spatial resolution though.&lt;/p&gt;&#xA;" OwnerUserId="13341" LastActivityDate="2015-03-02T13:56:34.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26824" PostTypeId="2" ParentId="26821" CreationDate="2015-03-02T14:04:19.410" Score="1" Body="&lt;p&gt;You're right: $\ce{NO}$ is a free radical and the reaction between $\ce{N2}$ and $\ce{O2}$ can be achieved - &lt;strong&gt;supposed that enough energy is provided&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may happen through an &lt;strong&gt;electrical arc discharge&lt;/strong&gt; in the lab or, in the mesosphere (50-85km above earth) through &lt;strong&gt;ionizing cosmic rays&lt;/strong&gt;. Note however that nitric oxide react with oxygen to eventually furnish nitrogen dioxide $\ce{2 NO + O2 &amp;lt;=&amp;gt; 2 NO2}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5577" LastEditDate="2015-03-02T18:51:58.603" LastActivityDate="2015-03-02T18:51:58.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26825" PostTypeId="1" CreationDate="2015-03-02T14:12:52.397" Score="0" ViewCount="1768" Body="&lt;p&gt;In the reaction:&#xA;$$\ce{3 Zn + 2 H3PO4 -&amp;gt; Zn3(PO4)2 + 3H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the initial mass of $\ce{Zn}$ to be 13.08 g and the volume of phosphoric acid to be 0.5 $\mathrm{dm^3}$. How do I work out the final concentration of $\ce{Zn(PO4)2}$ and the volume of $\ce{H2}$ gas evolved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I work out the moles of Zn, are the moles of $\ce{Zn(PO4)2}$ just $\ce{Zn}$ moles/3?&lt;/p&gt;&#xA;" OwnerUserId="14737" LastEditorUserId="418" LastEditDate="2015-03-02T16:07:58.107" LastActivityDate="2015-03-02T16:07:58.107" Title="Calculating concentration and volume in acid metal reactions" Tags="&lt;acid-base&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-02T18:04:30.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="26826" PostTypeId="2" ParentId="26825" CreationDate="2015-03-02T14:32:52.957" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Your equation uses &lt;strong&gt;molar ratios&lt;/strong&gt;, rather than weights or volumes.&lt;/li&gt;&#xA;&lt;li&gt;If you know the weight of zinc used, you know the number of moles. From this, you can calculate the number of moles of zinc phosphate generated.&lt;/li&gt;&#xA;&lt;li&gt;Supposed that the phosphoric acid is sufficiently concentrated to react all zinc, the number of moles calculated above together with the volume of the phosphoric acid used allows to calculate the concentration of zinc phosphate. Note: &lt;strong&gt;Do not omit units in your calculations!&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;As far as the hydrogen is concerned: The number of moles produced can be calculated using the amount of zinc used. Additional knowledge of the molar volume of gases under standard conditions is helpful to figure out the volume.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-02T14:32:52.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26827" PostTypeId="1" CreationDate="2015-03-02T14:59:44.440" Score="4" ViewCount="604" Body="&lt;p&gt;The question asks: why is the difference in atomic radii between $\ce{K}$ and $\ce{Na}$ larger than the difference in atomic radii between $\ce{Cs}$ and $\ce{Rb}$? I do have some kind of notion that the gaps between shells decrease as you go down the periodic table, but why would this be the case?&lt;/p&gt;&#xA;" OwnerUserId="7917" LastEditorUserId="5017" LastEditDate="2015-03-05T18:30:55.257" LastActivityDate="2015-03-05T18:30:55.257" Title="Why does the &quot;gap&quot; between atomic radii get smaller down the periodic table?" Tags="&lt;periodic-table&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26829" PostTypeId="2" ParentId="26811" CreationDate="2015-03-02T15:15:22.890" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://dx.doi.org/10.1016/B978-0-08-023920-0.50010-1&quot; rel=&quot;nofollow&quot;&gt;IUPAC Solubility Data Series Volume 52, &lt;em&gt;Alkaline Earth Hydroxides in Water and Aqueous Solutions&lt;/em&gt; (1992)&lt;/a&gt;, the best fitting to experimental solubility values of calcium hydroxide in water was obtained with a three-parameter equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln\left({b/{\text{mol kg}}^{-1}}\right) = 86.1534 - 3492.14/({T/{\text{K}}}) - 13.7494\,\ln({T/{\text{K}}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $b$ is molality and $T$ is temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results have a standard uncertainty of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma \left( b \right) = 1.7 \times {10^{ - 4}}\;{\text{mol k}}{{\text{g}}^{ - 1}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;over the temperature range from 273.15 K to 623.15 K.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${T_1} = 20\;^\circ\text{C} = 293.15\;{\text{K}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the three-parameter equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xA;  \ln\left({{b_1}/{\text{mol kg}^{-1}}}\right) =  &amp;amp; \;86.1534 - 3492.14/({T_1}/{\text{K)}} - 13.7494\,\ln({T_1}/{\text{K)}} \\ &#xA;   =  &amp;amp; \;86.1534 - 3492.14/(293.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(293.15\;{\text{K}}/{\text{K)}} \\ &#xA;   =  &amp;amp; \; {-}3.8651 \\ &#xA;  {b_1}/{\text{mol kg}^{-1}} =  &amp;amp; \;0.02096 \\ &#xA;  {b_1} =  &amp;amp; \;0.02096\;{\text{mol kg}^{-1}} \\ &#xA;\end{aligned} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same way for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${T_2} = 25\;^\circ\text{C} = 298.15\;{\text{K}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the three-parameter equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{aligned}&#xA;  \ln\left({{b_2}/{\text{mol kg}^{-1}}}\right) =  &amp;amp; \;86.1534 - 3492.14/({T_2}/{\text{K)}} - 13.7494\,\ln({T_2}/{\text{K)}} \\ &#xA;   =  &amp;amp; \;86.1534 - 3492.14/(298.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(298.15\;{\text{K}}/{\text{K)}} \\ &#xA;   =  &amp;amp; \; {-}3.8978 \\ &#xA;  {b_2}/{\text{mol kg}^{-1}} =  &amp;amp; \;0.02029 \\ &#xA;  {b_2} =  &amp;amp; \;0.02029\;{\text{mol kg}^{-1}} \\ &#xA;\end{aligned} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, taking into account the above-mentioned uncertainty, the calculated solubility of calcium hydroxide in water is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${b_1} = \left( {{\text{0}}{\text{.02096}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T_1} = 20\;^\circ\text{C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${b_2} = \left( {{\text{0}}{\text{.02029}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T_2} = 25\;^\circ\text{C}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-03T10:56:01.410" LastActivityDate="2015-03-03T10:56:01.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26831" PostTypeId="2" ParentId="26804" CreationDate="2015-03-02T16:49:17.090" Score="2" Body="&lt;p&gt;The assumption you need to make is that at time zero, only $\ce{PH3}$ is present.  So relative to the moles at time zero, or $N_0$, you should find $\Delta N = N(t) - N_0$, the total &lt;em&gt;new&lt;/em&gt; moles formed by the reaction since it began.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction converts 4 moles of gas to 7.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pressure increases mean the reaction is moving to the right.  Every mole of pressure increase must be $\frac{6}{7}$ due to $\ce{H_2}$ and $\frac{1}{7}$ due to $\ce{P4}$ formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just find the total &lt;em&gt;new&lt;/em&gt; moles of gas and multiply by $\frac{1}{7}$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-02T16:49:17.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26832" PostTypeId="2" ParentId="25038" CreationDate="2015-03-02T16:57:50.217" Score="1" Body="&lt;p&gt;I think an entire book on TLC/paper chromatography would be quite overkill and your best bet is on a more general practical organic chemistry book. I can recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1118083393&quot; rel=&quot;nofollow&quot;&gt;The Organic Chem Lab Survival Manual: A Student's Guide to Techniques&lt;/a&gt; by Zubrick if you are only interested in technique or need a introductory text. Else &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0582462363&quot; rel=&quot;nofollow&quot;&gt;Vogel&lt;/a&gt; is of course &lt;em&gt;the&lt;/em&gt; book for practical organic chemistry. TLC is covered in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not so sure about paper chromatography, I don't believe it is often used in the lab (at least I have never worked with it). TLC is much more common.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-02T16:57:50.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26833" PostTypeId="1" CreationDate="2015-03-02T17:19:39.433" Score="3" ViewCount="286" Body="&lt;p&gt;What in the quantum nature of electrons gives rise to the octet rule? How is it related to the fact that an element cannot have an oxidation state higher/lower than the complementary to 8? &lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2016-12-22T07:50:54.060" Title="What's the relation between the octet rule and quantum exclusion and how is it related to oxidation state?" Tags="&lt;quantum-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26834" PostTypeId="1" AcceptedAnswerId="26836" CreationDate="2015-03-02T17:37:40.473" Score="7" ViewCount="1498" Body="&lt;p&gt;I notice that most optical calcite for sale seems to be from natural (mined) sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I know in World War 2, mining optical calcite was a strategic objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some reason why high-grade optical calcite cannot be lab grown?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastActivityDate="2015-07-17T07:38:26.933" Title="Possible to grow optical calcite?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26835" PostTypeId="1" CreationDate="2015-03-02T17:48:20.583" Score="2" ViewCount="11316" Body="&lt;p&gt;This is a &quot;homework&quot; question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm writing up a lab in which we synthesized 1,4-diphenyl-1,3-butadiene. We analyzed the products using TLC with 4:1 Hexane:Ethyl Acetate as the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rnz0x.png&quot; alt=&quot;cis,trans isomer&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SqEFJ.png&quot; alt=&quot;trans,trans isomer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the isomers has a larger Rf value than the other (0.83 and 0.95). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LLgMY.jpg&quot; alt=&quot;TLC plate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble understanding how to figure out which of the isomers is which on the TLC. I know more polar compounds are going to have a lower Rf, like cinnamaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that, although the all the double bonds are conjugated, the phenyl rings have a higher electron density than the rest of the molecule. Since the trans,trans isomer is essentially symmetrical, it has little or no molecular dipole. The cis,trans isomer has a small molecular dipole because it is asymmetric. Therefore the cis-trans isomer has more affinity for the stationary phase (giving it the Rf value of 0.83). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I on the right track here?&lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2015-03-02T18:49:41.660" Title="How can you predict relative Rf values of cis-trans isomers in thin layer chromatography?" Tags="&lt;organic-chemistry&gt;&lt;chromatography&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26836" PostTypeId="2" ParentId="26834" CreationDate="2015-03-02T18:11:31.827" Score="8" Body="&lt;p&gt;The crystal structure is trigonal, so has very low symmetry and high chance of twinning or other line defects forming during, say, Czochralski type growth methods.  My guess would be the good optical quality stuff sat and annealed in situ for a long time to clear most of the twins/dislocations out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Support for this comes from A.J. Gratz et al., Geochimica et Cosmochimica Acta 57 491-495 (1993) where they observed, with AFM, calcite growth occuring primarily on screw dislocations.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-03-02T18:11:31.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26837" PostTypeId="1" CreationDate="2015-03-02T18:39:36.080" Score="5" ViewCount="12605" Body="&lt;p&gt;The question is the same as the title states. In my textbook,&lt;sup&gt;[NCERT Chemistry I for Class 12 , pg no 217.]&lt;/sup&gt; the following is written&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^{2+}}$&lt;/span&gt; gets converted to &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^{3+}}$&lt;/span&gt; as the +3 oxidation state has half-filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; orbitals; thus it is a good reducing agent.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^{3+}}$&lt;/span&gt; gets converted to &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^{2+}}$&lt;/span&gt; as its +2 oxidation state has a half-filled d-subshell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now my questions are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why can't the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; argument be used for &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why can't the half-filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}$&lt;/span&gt;-orbital argument be used for &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^3+}$&lt;/span&gt; could be oxidized to &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^4+}$&lt;/span&gt;, which would have a half-filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; set. As oxidation and reduction both can take place in aqueous medium, why wouldn't &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}^3}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d^3}$&lt;/span&gt; configuration be more stable than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d^5}$&lt;/span&gt; configuration in aqueous medium?&lt;/p&gt;&#xA;&lt;p&gt;A similar observation is made in case of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^2+}$&lt;/span&gt;, where it observed that &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^2+}$&lt;/span&gt; is more powerful reducing agent than &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^2+}$&lt;/span&gt; (due to reasons explained above).&lt;/p&gt;&#xA;&lt;p&gt;If possible, I would like to see the crystal field splitting calculations that show this to be the case.&lt;/p&gt;&#xA;" OwnerUserId="7365" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-06-03T03:14:42.120" Title="Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;reduction-potential&gt;&lt;crystal-field-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ClosedDate="2020-06-03T14:21:45.300" ContentLicense="CC BY-SA 4.0" />
  <row Id="26839" PostTypeId="2" ParentId="26835" CreationDate="2015-03-02T18:49:41.660" Score="2" Body="&lt;p&gt;Well, apparently I was wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the trans,trans isomer is adsorbed more strongly by the stationary phase because it is entirely planar. The cis,trans-isomer cannot adopt a planar conformation because of steric hindrance between the two phenyl rings. The non-coplanar geometry lowers the efficiency of interactions with the stationary phase, making it more mobile.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the cis,trans isomer has a higher Rf, and the trans,trans isomer has a lower Rf. &lt;/p&gt;&#xA;" OwnerUserId="5434" LastActivityDate="2015-03-02T18:49:41.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26840" PostTypeId="1" CreationDate="2015-03-02T19:02:09.200" Score="3" ViewCount="113" Body="&lt;p&gt;What type of stereoisomers do these two complex compounds represent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RtKdm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-03-02T19:50:04.370" LastActivityDate="2015-03-02T19:50:04.370" Title="Bipyridyl complex isomers" Tags="&lt;coordination-compounds&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26841" PostTypeId="2" ParentId="26837" CreationDate="2015-03-02T19:07:53.930" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; argument be used for &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn}$&lt;/span&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the number of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; electrons is not changing upon going from &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^3+}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^2+}$&lt;/span&gt;.  In both &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^3+}$&lt;/span&gt; there are 3 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; electrons.  I think the point they are trying to make with &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^3+}$&lt;/span&gt; being half filled is that there are 3 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; electrons and no &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e_g}$&lt;/span&gt; electrons.  &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^3+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^2+}$&lt;/span&gt; both have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e_g}$&lt;/span&gt; electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make a similar argument for &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^3+}$&lt;/span&gt; being oxidized to &lt;span class=&quot;math-container&quot;&gt;$\ce{Mn^4+}$&lt;/span&gt;, which would have half filled &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}}$&lt;/span&gt; and empty &lt;span class=&quot;math-container&quot;&gt;$\mathrm{e_g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't the half-filled d-orbital argument be used for Cr?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because neither &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^3+}$&lt;/span&gt; nor &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^2+}$&lt;/span&gt; has a half filled d subshell.  &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^3+}$&lt;/span&gt; has 3 electrons while &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr^2+}$&lt;/span&gt; has 4 electrons.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="33586" LastEditDate="2019-08-22T07:43:03.797" LastActivityDate="2019-08-22T07:43:03.797" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26842" PostTypeId="2" ParentId="26840" CreationDate="2015-03-02T19:43:39.273" Score="2" Body="&lt;p&gt;These are examples of enantiomerism or optical isomerism. If you imagine a mirror placed between the two images you will see that they are mirror images of each other. If you try to rotate them to match up (easiest to do with a model, either physical or virtual) you will find that they are non-superimposable and this is what gives them their isomeric properties. The two isomers rotate plane polarised light in opposite directions (this is where the name, optical isomerism, comes from).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\Delta$ and $\Lambda$ labels distinguish the two isomers by their absolute stereochemical configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An decent explanation of this type of isomerism can be found &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Isomers/Stereoisomers/Optical_Isomers_in_Inorganic_Complexes&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-02T19:43:39.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26843" PostTypeId="1" CreationDate="2015-03-02T19:45:36.237" Score="2" ViewCount="77" Body="&lt;p&gt;I notice that Glan Thompson prisms seem to be the preferred optics for doing polarimetry and have been for at least 75 years. One instrument maker says that polaroid films can be used but they tend to degrade and warp over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems strange to me that no better substitute has been found for Glan Thompson prisms given their expense. I know at one point Polaroid was saying that Sodium Nitrate crystals (which can be grown) could potentially be used, but this effort seems to have gone nowhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why has no cheaper alternative been found? Aren't there growable crystals that have the birefringence necessary?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorUserId="30836" LastEditDate="2020-04-29T21:54:03.047" LastActivityDate="2020-04-29T21:54:03.047" Title="Replacement for optical calcite (Glan Thompson prism)?" Tags="&lt;crystal-structure&gt;&lt;materials&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26844" PostTypeId="1" CreationDate="2015-03-02T20:06:46.110" Score="0" ViewCount="983" Body="&lt;p&gt;In a book, I read two quite contradictory statements.:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HIO &amp;gt; HBrO &amp;gt; HClO}$&lt;/span&gt; wrt acidity while&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HF &amp;gt; HCl &amp;gt; HBr}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Aren't they contradictory? If not can anyone please explain the accurate reasoning?&lt;/p&gt;&#xA;" OwnerUserId="14742" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-12T09:36:22.297" Title="How actually do we decide acidity and basicity order?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26845" PostTypeId="2" ParentId="2727" CreationDate="2015-03-02T20:46:33.450" Score="3" Body="&lt;p&gt;According to &quot;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ar960147o&quot; rel=&quot;nofollow&quot;&gt;Small Multiply Charged Anions as Building Blocks in&#xA;Chemistry&lt;/a&gt;&quot; Acc. Chem. Res. 1996, 29, 497-502&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it is concluded that all isolated diatomic dianions or&#xA;  other multiply charged diatomic anions are not stable&#xA;  to loss of an extra electron or fragmentation in their&#xA;  ground electronic states&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So even $\ce{O2^{2-}}$, let alone $\ce{O2^{3-}}$ cannot exist in isolation.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-02T20:46:33.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26846" PostTypeId="1" CreationDate="2015-03-03T02:29:55.993" Score="7" ViewCount="11709" Body="&lt;p&gt;I learnt that on dilution of an electrolytic solution, it's conductivity decreases. However, there is an increase in the molar conductivity of the solution. The book explained the reason by comparing the definition of the two terms. However, it gave a hypothetical example which contradicts the statement it earlier stated. Maybe there is a printing error, but it has got me confused now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain why the conductivity of an electrolytic solution decreases but molar conductivity increases on dilution, maybe with the  help of a good hypothetical example?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastActivityDate="2016-11-27T16:40:55.740" Title="Why conductivity of electrolyte decreases on dilution, but molar conductivity increases?" Tags="&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26847" PostTypeId="1" AcceptedAnswerId="26849" CreationDate="2015-03-03T03:07:11.887" Score="5" ViewCount="10651" Body="&lt;p&gt;I forgot a bottle of 1L Orange Juice (100% orange juice with pulp) in the fridge for about a week, the bottle was previously open so air got inside as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a week the bottle got inflated and when I opened it I could hear the gas coming out. No odours, orange juice still the same colour (not sure if the same taste, I threw it away).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gas was produced and how did it happen? The air that got in seems to have played a part on it (these bottles stay in the supermarket closed for a long time without getting inflated).&lt;/p&gt;&#xA;" OwnerUserId="14744" LastEditorUserId="4945" LastEditDate="2015-03-03T11:40:52.427" LastActivityDate="2015-03-03T11:40:52.427" Title="Which gas makes Orange juice bottles swell?" Tags="&lt;gas-laws&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26848" PostTypeId="1" CreationDate="2015-03-03T04:02:14.953" Score="0" ViewCount="284" Body="&lt;p&gt;A question says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the empirical formula of an organic compound from the following composition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;34.62% C, 3.88% H, 61.50% O.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $\ce{C3H4O4}$.&#xA;It was found by using the mass (percentages) divided by the molar mass of each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why?&#xA;in this case it will be $\ce{CH3O3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in a similar question in the text book, they consider the moles of hydrogen and oxygen in calculations by multiplying by 2 after getting the moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anybody explain?&lt;/p&gt;&#xA;" OwnerUserId="14688" LastEditorUserId="418" LastEditDate="2015-03-03T05:20:52.150" LastActivityDate="2015-03-03T05:53:53.180" Title="Why not to consider hydrogen and oxygen moles to determine an empirical formula" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26849" PostTypeId="2" ParentId="26847" CreationDate="2015-03-03T04:07:09.687" Score="5" Body="&lt;p&gt;I can't imagine the gas was anything except $\ce{CO2}$ formed by the action of microorganisms contaminating the orange juice.  They degrade the organic components of the juice to power their metabolisms, and produce carbon dioxide as a byproduct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common contaminant of pasteurized fruit juice is a genus of bacteria called &lt;em&gt;Alicyclobacillus&lt;/em&gt;, but &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1111/j.1472-765X.2007.02229.x/abstract&quot;&gt;this paper&lt;/a&gt; says that &lt;em&gt;Alicyclobacillus&lt;/em&gt; bacteria do &lt;em&gt;not&lt;/em&gt; cause bottle swelling and &lt;em&gt;do&lt;/em&gt; cause a medicinal smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By searching Google for [&quot;bottle swelling&quot; &quot;orange juice&quot;], I found an English-language abstract of a &lt;a href=&quot;http://www.dissertationtopic.net/doc/1719787&quot;&gt;thesis&lt;/a&gt; from China which implies that the culprit could be a yeast.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-03T04:07:09.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26854" PostTypeId="2" ParentId="26848" CreationDate="2015-03-03T05:53:53.180" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The answer is $\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly! That's the way to do it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\ce{CH3O3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/82IkD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the &quot;natural appearance&quot; of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-03T05:53:53.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26859" PostTypeId="1" AcceptedAnswerId="26862" CreationDate="2015-03-03T11:29:46.700" Score="0" ViewCount="2557" Body="&lt;p&gt;When trying to work out what will be produced at each electrode during electrolysis, I'm confused as to which equations to use for hydrogen and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, an electrolytic cell containing aqueous nickel(II) bromide, with a nickel cathode and a platinum anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the cathode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ni^2+ + 2e- -&amp;gt; Ni}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that this does not occur because $\ce{Ni^2+}$ is a worse oxidising agent than $\ce{H+}$, which makes sense because $\ce{H+}$ is lower down on the left side of the electrochemical series meaning its reduction is more feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is therefore reduced at the cathode:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H+ + 2e- -&amp;gt; H2}\quad E^\circ=0.0\ \mathrm V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O + 2e- -&amp;gt; H2 + 2OH-}\quad E^\circ=-0.83\ \mathrm V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation, however, has water higher up in the electrochemical series. This would suggest nickel should actually  be reduced in preference to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know which of the two &quot;water-reduction&quot; equations to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is even less transparent for the anode, where bromide should be oxidised to bromine, but there are also two equations showing production of oxygen: one below and one above bromide in the series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{O2 + 4H+ + 4e- &amp;amp;-&amp;gt; 2H2O}\quad &amp;amp;E^\circ&amp;amp;=+1.23\ \mathrm V\\[6pt]&#xA;\ce{O2 + 2H2O + 4e- &amp;amp;-&amp;gt; 4OH-}\quad &amp;amp;E^\circ&amp;amp;=+0.40\ \mathrm V&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand which of these equations should be used and how you're supposed to know which to use.&lt;/p&gt;&#xA;" OwnerUserId="14748" LastEditorDisplayName="user7951" LastEditDate="2016-11-18T16:44:55.900" LastActivityDate="2016-11-18T16:44:55.900" Title="Electrolysis in aqueous solution – which equations to use to predict product at each electrode?" Tags="&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26862" PostTypeId="2" ParentId="26859" CreationDate="2015-03-03T14:35:45.087" Score="3" Body="&lt;p&gt;The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = E^\circ  + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H+ + 2e- &amp;lt;=&amp;gt; H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the Nernst equation reads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the potential $E$ depends on $\text{pH}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-03T14:42:23.577" LastActivityDate="2015-03-03T14:42:23.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26863" PostTypeId="1" CreationDate="2015-03-03T15:06:27.127" Score="12" ViewCount="500" Body="&lt;p&gt;I see the term &quot;superionic&quot; applied to high pressure water in articles like &lt;a href=&quot;http://www.nature.com/news/2005/050322/full/news050321-4.html&quot;&gt;Giant planets may host superionic water&lt;/a&gt;, but I don't understand what the term really means.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is &quot;superionic&quot; distinguished from simply &quot;ionic&quot;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the article refer to hydrogen &quot;atoms&quot;, which would imply that they are neutral, as opposed to protons or hydronium or some other actual ion?&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-08T13:39:20.950" Title="What is the meaning of &quot;superionic&quot;?" Tags="&lt;ionic-compounds&gt;&lt;astrochemistry&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26864" PostTypeId="1" CreationDate="2015-03-03T16:02:54.443" Score="2" ViewCount="579" Body="&lt;p&gt;Given borane's ($\ce{BH3}$) LUMO, which of these Lewis bases would have a more stabilizing interaction?  In acid bases that means the higher HOMO and lower LUMO equals a stronger dative bond I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which would have a stronger interaction (greater net stabilization) with borane, trimethylphosphine or trimethylamine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{P(CH3)3 ~vs~ N(CH3)3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus is less EN and it's electrons reside in a higher energy 3p orbital, but doesn't it compensate for this by placing the electron pair an orbital with more &quot;s&quot; character? Also carbon is more EN than phosphorous drawing some strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say trimethylphosphine has the HOMO just because of it's higher orbitals and lower overall EN making donating a pair of electrons more favorable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right?&lt;/p&gt;&#xA;" OwnerUserId="14753" LastEditorUserId="865" LastEditDate="2015-10-16T17:54:08.823" LastActivityDate="2015-10-16T17:54:08.823" Title="Which is More Stable with Boranes: Amines or Phosphines?" Tags="&lt;acid-base&gt;&lt;molecular-orbital-theory&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26865" PostTypeId="1" CreationDate="2015-03-03T16:03:48.770" Score="11" ViewCount="230" Body="&lt;p&gt;My question may be stupid, so please correct me if you find anything which is obviously erroneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following I will place a question mark (?) besides points/steps I consider doubtful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My observations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In density functional theory (DFT) (at zero temperature), the energy of the system $E$ is broken up as $E = \int_{V} d v \left[ e_{K}(n(\vec{r})) + e_{H}(n(\vec{r})) + e_{XC}(n(\vec{r})) \right]$, where $V$ is the system size, $e$ the energy density per unit volume, $n(\vec{r})$ the electronic density, subscript &quot;$K$&quot; for &quot;kinetic&quot;, &quot;$H$&quot; for &quot;Hartree&quot; and &quot;$XC$&quot; for &quot;exchange-correlation&quot;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the generalized gradient approximation (gga), $e$ also depends on $\nabla n(\vec{r})$. Thus write $e(n(\vec{r}), \nabla n(\vec{r}))$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&#xA;1. spherical symmetry (closed-shell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By spherical symmetry $e_{XC}(n(\vec{r}), \nabla n(\vec{r})) = e(n(r), \nabla n(r))$. Of course in writing this we are restricting the variation in density $\delta n$ in the &lt;strong&gt;radial&lt;/strong&gt; direction only. I'm doubtful about the validity of this step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, proceed as usual. The exchange-correlation potential is defined as the functional derivative of $e_{XC}$ w.r.t. $\delta n$. Now please point out if there are any mistakes in the following steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{eqnarray} \delta \int dv e_{XC}(n(r), n^{\prime}(r)) &amp;amp;=&amp;amp; \int dv \left[ \frac{\partial e_{XC}}{\partial n(r)} \delta n(r) + \frac{\partial e_{XC}}{\partial n^{\prime}(r)} \delta n^{\prime}(r) \right] \end{eqnarray}$$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n^{\prime}(r) = \partial_r n(r)$. The first term is done; consider the second term. $\delta n^{\prime} (r) = \partial_r \delta n(r)$. Thus we integrate the second term on the right hand side by parts, and obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 \pi \int_0^{r_0} dr \partial_r \left[ r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \delta n(r) \right] - 4 \pi \int_0^{r_0} dr \left[ \partial_r \left( r^2 \frac{e_{XC}}{n^{\prime}}\right) \delta n(r) \right]$. The first term is the boundary term; ignore it for the time being. The second term seems to give the correction to the exchange correlation potential due to the introduction of gradient-dependence. Note that the second term is equivalent to&#xA;$\int_0^{r_0} dr 4 \pi r^2 \left[ - \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right)\right]$, the square-bracketed term seems to give the exchange-correlation potential (?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $- \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right) = - \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} - \partial_r \frac{\partial e_{XC}}{\partial n^{\prime}}$. What I worry about is the first term, which seems to diverge at $r=0$ (but it shouldn't; or is it cancelled by a piece in the second term?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A expression for $e_{XC}$ may be the becke88 functional (exchange only) (PRA, 38, 3098). $$ \begin{eqnarray}e^{Becke}_{X} &amp;amp;=&amp;amp; e^{LDA}_{X}(n(r)) - \beta \sum_{\sigma} \int n_{\sigma}^{4/3} \frac{x_{\sigma}^2}{1+6 \beta x_{\sigma} \sinh^{-1}x_{\sigma}} d\vec{r} \end{eqnarray}$$, where $\sigma$ is the spin-index (For closed-shell system $n_{+} = n_{-} = n/2$ (?)), $\beta$ is a parameter (=0.0042 a.u.), and $x_{\sigma} = | \nabla n_{\sigma}|/n_{\sigma}^{4/3}$. In spherically symmetric system, can we change $| \nabla n_{\sigma}|$ to $|n^{\prime}(r)|$?. Now I am not sure about the following claims&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ is non-zero at the origin. (?) The reason is that, while it seems that $n^{\prime}(0)$ should be zero, it may not be. Consider helium. When solving the schroedinger eq. for the 1s state under arbitrary potential which tends to $2/r$ at $r=0$, the slope of the radial wave function (excluding the $r$-factor from the volume element) is non-zero (Consider for example the radial wave function of the hydrogen ground state, which behaves as $e^{-r}$). This means $n^{\prime}(0) \neq 0$. But then  $\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ will be non-zero at the origin and thus $- \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} $ will tend to infinity at $r=0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This should not be the case because the hartree (i.e. &quot;classical&quot; contribution by all electrons) contribution to the potential is small as $r \rightarrow 0$, and the exchange contribution should cancel the &quot;self-action&quot; in the hartree part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, sorry for the long post. Note that I might have done something very wrong, such that the whole argument does not stand. Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="7451" LastEditDate="2015-03-03T16:17:18.563" LastActivityDate="2015-03-03T16:17:18.563" Title="Density Functional Theory with Generalized Gradient Approxmations (GGA) - What Happens if the Electron Density Shows a &quot;Kink&quot; at the Ion Position?" Tags="&lt;atoms&gt;&lt;density-functional-theory&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26867" PostTypeId="2" ParentId="26610" CreationDate="2015-03-03T16:36:50.573" Score="16" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Can the lone pair on benzene carbanion participate in resonance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring.  Therefore it cannot participate in resonance.  Here is a drawing that illustrates this point.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQtGj.jpg&quot; alt=&quot;Structure of benzene anion&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;how is it more stable than vinylic carbanion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt;s.  A &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$43$&lt;/span&gt; means &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; ionized molecule exists in every &lt;span class=&quot;math-container&quot;&gt;$10^{43}$&lt;/span&gt; molecules.  Certainly difficult to measure experimentally and so a significant error is attached to these estimates. In fact, using the data from &lt;a href=&quot;http://evans.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Evan's &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; table&lt;/a&gt; we find a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$43$&lt;/span&gt; for the benzene &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; proton and a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$50$&lt;/span&gt; for an ethylenic proton, suggesting that the phenyl carbanion might be more stable than the vinyl carbanion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;An alternate way&lt;/strong&gt; to examine this question involves comparing the hybridizations of the phenyl and ethylenic &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds.  Since bonds with more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character will stabilize electrons better than bonds with less &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character (because the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-orbital is lower energy than a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;-orbital), whichever bond has more &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character should produce a more stable carbanion.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond in benzene is exactly &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized as required by the symmetry of the benzene molecule.  Experimentally, we know that the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-C-H}$&lt;/span&gt; angle in ethylene is &lt;span class=&quot;math-container&quot;&gt;$117^\circ$&lt;/span&gt;.  Using this fact along with Coulson's theorem (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/22267/whats-the-h-c-h-bond-angle-in-ethene/22278#22278&quot;&gt;reference1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19977/determine-the-bond-angle-in-a-compound/19987#19987&quot;&gt;reference2&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19977/determine-the-bond-angle-in-a-compound/19987#19987&quot;&gt;reference3&lt;/a&gt;) we can determine that the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bond in ethylene is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}^{2.2}$&lt;/span&gt;  hybridized.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized orbital in benzene has more s-character than the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}^{2.2}$&lt;/span&gt; hybridized orbital in ethylene.  This would lead us to suspect that &lt;strong&gt;the phenyl carbanion is more stable than the vinyl carbanion&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Of course, once we remove these protons we expect the resulting carbanions to undergo further rehybridization in order to further stabilize their lone pair (carbanion) electrons.  It seems like a reasonable first approximation that they would both rehybridize similar amounts and the &amp;quot;relaxed&amp;quot; phenyl carbanion would remain more stable than the &amp;quot;relaxed&amp;quot; vinyl carbanion.  But in any case, the transition state leading to the phenyl carbanion will be more stable than the transition state leading to the vinyl carbanion because of the hybridization \ &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt;-character reasons presented above.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; OP's comment on hybridization&lt;/p&gt;&#xA;&lt;p&gt;Molecules are not just &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp, sp^{2}}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized.  While the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds in methane are exactly &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized (as required by symmetry), the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt; bonds in &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3F}$&lt;/span&gt; are not (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622&quot;&gt;see here for a discussion&lt;/a&gt;).  Bonds can have any hybridization index (the superscript to the right of &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt;&amp;quot;) between zero and infinity.  &lt;strong&gt;Hybridization indices are not limited to integer values&lt;/strong&gt;, nature can mix &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}$&lt;/span&gt; orbitals in whatever ratio is required to create the most stable bonds and molecules.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2021-03-20T17:23:01.303" LastActivityDate="2021-03-20T17:23:01.303" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26868" PostTypeId="1" AcceptedAnswerId="26932" CreationDate="2015-03-03T16:39:37.220" Score="12" ViewCount="63546" Body="&lt;p&gt;The boiling point of &lt;strong&gt;&lt;em&gt;trans-but-2-ene&lt;/em&gt;&lt;/strong&gt; is lower than that of its &lt;strong&gt;&lt;em&gt;cis&lt;/em&gt;&lt;/strong&gt; isomer but the melting point of the former is higher than the later. Why is it not following the same order? Is there any factor of structural symmetry coming playing a role here?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4325" LastEditDate="2015-03-03T17:27:33.893" LastActivityDate="2018-02-10T07:06:47.933" Title="Boiling and melting point of trans- and cis-but-2-ene" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;stereochemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26869" PostTypeId="1" CreationDate="2015-03-03T18:05:21.557" Score="12" ViewCount="408" Body="&lt;p&gt;For hydrogen, other than in the 1s state, the electron wavefunctions have radial and/or angular nodes where the electron probability density is zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In helium or further atoms with more than one electron, considering correlation of electrons, what happens to these nodes?  Are there still radial and angular nodes where electron probability density is zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if helium is in a 2s2s state, is there anywhere that an electron cannot be?&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-04T17:15:16.617" LastActivityDate="2017-01-17T17:41:31.537" Title="Do multielectron atoms have nodes?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26870" PostTypeId="2" ParentId="26844" CreationDate="2015-03-03T18:16:01.627" Score="0" Body="&lt;p&gt;As Tejas said the sign of inequality in the second statement should be opposite. This is because in HF there is extensive hydrogen bonding which tends to reduce its acidic character. The reason being the high electro negativity of Fluorine. The extent of hydrogen bonding would decrease down the group due to decrease in electro negativity.&#xA;A similar explanation may also be extended to reason the first one. &#xA;Your book is definitely incorrect. &lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2015-03-03T18:16:01.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26871" PostTypeId="1" AcceptedAnswerId="26872" CreationDate="2015-03-03T18:29:52.290" Score="1" ViewCount="1152" Body="&lt;p&gt;I am trying to balance the following redox reaction, in basic solution. From what I have learned, to balance a redox reaction in basic solution you simply balance it as you normally would in acidic solution, then make sure the H+ ions are neutralized by adding OH- ions. In the example below, is my answer balanced in basic solution already? Or do I need to somehow cancel out the H present in H2O? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IjoOg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4093" LastActivityDate="2015-03-03T18:33:00.713" Title="Balancing REDOX reactions in a basic solution, when no H+ ions are present" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26872" PostTypeId="2" ParentId="26871" CreationDate="2015-03-03T18:33:00.713" Score="2" Body="&lt;p&gt;The bottom equation looks balanced to me. There are the same number of atoms of each element on each side of the equation and charge is conserved. The hydrogen in the $\ce{H2O}$ is balanced by the hydrogen in the $\ce{HAsO2}$ on the other side so it looks fine.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-03T18:33:00.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26873" PostTypeId="1" AcceptedAnswerId="26876" CreationDate="2015-03-03T18:40:02.977" Score="1" ViewCount="3947" Body="&lt;p&gt;I am trying to balance the REDOX equation shown below, but I got stuck after balancing the major atoms. &lt;strong&gt;Where can I go from here?&lt;/strong&gt; I know I need to balance the H and O atoms, and make sure the charges are balanced. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/uMpcy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second Attempt (Answer Only)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kumRD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this correct?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4093" LastEditorUserId="865" LastEditDate="2021-05-22T21:17:23.327" LastActivityDate="2021-05-22T21:17:23.327" Title="Balancing a REDOX reaction using the oxidation number method, in acidic solution" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26874" PostTypeId="1" AcceptedAnswerId="26877" CreationDate="2015-03-03T18:50:59.883" Score="1" ViewCount="447" Body="&lt;p&gt;In our chemistry curriculum we are doing these lab experiments where we are supposed to test salts for certain anions,cations and other stuff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that if I dissolved the salt in concentrated $\ce{H2SO4}$ and then heated it for a while over a flame, the evolution of light brown vapours would indicate the possibility of $\ce{NO3-}$. To confirm its presence, however, I would have to repeat the test but this time also add some copper bits to the solution and check for dark brown vapours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured the reason was that the copper reduced the $\ce{HNO3}$ gas to $\ce{NO2}$ which increased the amount of $\ce{NO2}$ gas produced, resulting in a darker shade of brown. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that why copper, which is a very weak reducing agent, used here? Why not a real reducing agent like zinc which will give more $\ce{NO2}$ even with lesser salt?&lt;/p&gt;&#xA;" OwnerUserId="13135" LastEditorUserId="7448" LastEditDate="2015-03-03T18:59:48.290" LastActivityDate="2015-03-03T19:43:04.980" Title="Why do we add copper when testing a salt for nitrate ion?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26875" PostTypeId="1" AcceptedAnswerId="26888" CreationDate="2015-03-03T19:22:43.737" Score="2" ViewCount="163" Body="&lt;p&gt;I have to etch a pattern on a plastic film similar to the ones used to protect phone screens. I want a very thin etch, one that is only visible on close inspection. How can I get that done?&lt;/p&gt;&#xA;" OwnerDisplayName="user14756" LastActivityDate="2015-03-04T00:14:18.857" Title="Etching a Screen Protector Film" Tags="&lt;materials&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26876" PostTypeId="2" ParentId="26873" CreationDate="2015-03-03T19:27:24.260" Score="8" Body="&lt;p&gt;Whenever you balance a redox reaction in acidic or basic solution, you have to account for interactions with the solution itself. The easiest way of doing this is by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Half-reaction#Half-reaction_balancing_method&quot;&gt;half-reaction method&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a key detail that I find a lot of students miss - you cannot balance a reaction like this without considering the solution, because &lt;strong&gt;the solution itself is participating&lt;/strong&gt;. Your clue that this is happening is the phrase &quot;in acidic (or basic) solution&quot; - any time you see that in a redox problem, you should be thinking &quot;half-reaction method.&quot; Another clue is that H and O appear on only one side of the equation. If you are familiar with the algebraic method of balancing equations, another clue is that the system of equations would be unsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, once you know that you need to use the half-reaction method, the next step is to use it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://chemistry.about.com/od/workedchemistryproblems/a/Balance-Redox-Reaction-Example-Problem.htm&quot;&gt;tutorial&lt;/a&gt; has a good description of the method, your book probably does as well, and I'll give a brief description here for good measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the major steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Split the reaction into two half-reactions. Then, for each reaction&lt;/li&gt;&#xA;&lt;li&gt;Balance everything except for H and O&lt;/li&gt;&#xA;&lt;li&gt;Balance O by adding $\ce{H2O}$ (water is available in solution, so this is legal)&lt;/li&gt;&#xA;&lt;li&gt;Balance H by adding $\ce{H^+}$ (it's an acidic solution - so this is also legal)&lt;/li&gt;&#xA;&lt;li&gt;Balance charge by adding electrons ($\ce{e^-}$)&lt;/li&gt;&#xA;&lt;li&gt;Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel - this is necessary because electrons are not created or destroyed in chemical reactions.&lt;/li&gt;&#xA;&lt;li&gt;Add the equations together, and if the solution is basic, add $\ce{OH^-}$ to both sides to remove all protons.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll use an example from my copy of Brown, LeMay, Bursten and Murphy's &quot;Chemistry: The Central Science&quot; to demonstrate (all phases are aqueous):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Balance the following reaction in acidic solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^- + C2O4^2- -&amp;gt; Mn^2+ + CO2}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Break the reaction into two half-reactions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^- -&amp;gt; Mn^2+}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2O4^2- -&amp;gt;CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not always easy to see how to do this. The quick way is to make sure that the elements that are not O and H are in their own equations. The proper way is to figure out what is being reduced and what is being oxidized, then split them so that the oxidation happens in one reaction and the reduction happens in the other. In this case, Mn is reduced and C is oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Balance everything except for H and O:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^- -&amp;gt; Mn^2+}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2O4^2- -&amp;gt; 2CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This step is usually pretty easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Balance O by adding $\ce{H2O}$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^- -&amp;gt; Mn^2+ + 4H2O }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2O4^2- -&amp;gt; 2CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with O because we are using water to balance it. Each time we add a water, we also add two H, so it doesn't make sense to start with H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Balance H by adding $\ce{H^+}$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{8H^+ + MnO4^- -&amp;gt; Mn^2+ + 4H2O }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2O4^2- -&amp;gt; 2CO2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that we can do this because it's in acidic solution. You are probably wondering what we do for basic solution. The answer is - use protons! There is a trick we can do at the end to &quot;fix&quot; it for basic solutions, and this is much easier than iterating between $\ce{OH^-}$ and $\ce{H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Balance charge by adding electrons:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{5e^- + 8H^+ + MnO4^- -&amp;gt; Mn^2+ + 4H2O }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2O4^2- -&amp;gt; 2CO2 + 2e^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this is easy to understand, I have found that in practice, a majority of mistakes happen here. Be careful when counting charges - remember to multiply by the coefficients, and remember that the charges don't have to add up to zero yet - they just have to have the same sum on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{10e^- + 16H^+ + 2MnO4^- -&amp;gt; 2Mn^2+ + 8H2O }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{5C2O4^2- -&amp;gt; 10CO2 + 10e^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part is also prone to error - be careful and remember to multiply every species by the coefficent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Add the equations together&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{16H^+ + 2MnO4^- + 5C2O4^2- -&amp;gt; 2Mn^2+ + 8H2O + 10CO2 }$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the electrons don't cancel out here, then &lt;em&gt;you made a mistake&lt;/em&gt; - go back and check each step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solution is basic, at this point you would add $\ce{OH-}$ to both sides of the equation until the 16 protons were eliminated. We can add anything we want as long as it's to both sides (since this is, after all, just algebra). Each $\ce{OH^-}$ will react with a proton to form one water molecule, and so the waters on the right would cancel out some of the waters on the left. The result would be a similar equation, but with $\ce{OH^-}$ instead of $\ce{H^+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try out your problem using these steps, and if you get stuck, update your question.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-03-03T19:33:42.727" LastActivityDate="2015-03-03T19:33:42.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26877" PostTypeId="2" ParentId="26874" CreationDate="2015-03-03T19:43:04.980" Score="3" Body="&lt;p&gt;This is a preliminary test, which is neither particularly selective nor sensitive.&#xA;With nitrate, brown vapours of $\ce{NO2}$ are released.&#xA;However, bromine or iodine could also release brownish or violet vapours.&#xA;Copper (or silver) is used to remove bromide and iodide by precipitation.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-03-03T19:43:04.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26880" PostTypeId="2" ParentId="26770" CreationDate="2015-03-03T20:46:58.797" Score="7" Body="&lt;p&gt;A reasonable mechanism for the transformation is shown below. Since this is under basic conditions, I've written a mechanism that avoids cationic intermediates. Deprotonating the hydrazine gives a very nucleophilic nitrogen that can attack the nitro group. Proton transfer gives a nitrogen anion, which can kick out hydroxide. Two additional proton transfer steps lead to the product, HOBt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dxtqp.jpg&quot; alt=&quot;Proposed mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A SciFinder search didn't reveal any examples of the intermolecular variant of this reaction, although I didn't find this exact intramolecular reaction either.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2015-03-04T00:43:49.793" LastActivityDate="2015-03-04T00:43:49.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26881" PostTypeId="1" CreationDate="2015-03-03T20:55:22.197" Score="2" ViewCount="28224" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.300N}\, \ce{NaOH}$&lt;/span&gt; would be needed to bring &lt;span class=&quot;math-container&quot;&gt;$\pu{16.0 mL}$&lt;/span&gt; of distilled water that has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$7.00$&lt;/span&gt; to a new &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$12.42$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is my attempt at a solution:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NaOH + H2O-&amp;gt; Na+ + OH- + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\mathrm{pH}(\text{initial}) &amp;amp;= 7.00 \\&#xA;\mathrm{pH}(\text{final}) &amp;amp;= 12.42 \\&#xA;\mathrm{pH} &amp;amp;= -\log{\ce{[H+]}} \\&#xA;\ce{[H+]}_{\text{intial}} &amp;amp;= 10^{-7} \\&#xA;&amp;amp;= \pu{10^{-7} ions} \\ \hline&#xA;V_1&amp;amp;= 0.016L \\&#xA;C_1 &amp;amp;= (0.300)(10^{-7})= 3\cdot10^{-8} N\\&#xA;\mathrm{pOH} &amp;amp;= 14-12.42  \\&#xA;[\ce{H+}]&amp;amp;= 10^{-1.58} \\  &#xA;&amp;amp;=\pu{ 0.02630 ions} \\&#xA;C_2 &amp;amp;= (0.300)(0.02630) \\&#xA;&amp;amp;=7.89\cdot10^{-3} N \\&#xA;C_1V_1&amp;amp;=C_2V_2 \\&#xA;(3\cdot10^{-8})(0.016)&amp;amp;= (7.89\cdot 10^{-3})(V_2) \\&#xA;V_2&amp;amp;= 6.08\cdot10^{-8} L \\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="14759" LastEditorUserId="95133" LastEditDate="2020-09-11T18:55:20.917" LastActivityDate="2020-09-11T18:56:46.463" Title="How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26882" PostTypeId="1" AcceptedAnswerId="26883" CreationDate="2015-03-03T22:13:05.270" Score="12" ViewCount="11821" Body="&lt;p&gt;How do I remove a stopper that has been stuck in the neck of a 100&amp;nbsp;mL round bottom flask since I forgot to wipe the neck of the flask. A salt was formed in the 100&amp;nbsp;mL flask and the stopper is just stuck like cement. Is there anyway to remove it without breaking the flask, breaking the stopper, or heating the flask (since I don't want to interfere with the product)?&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="4945" LastEditDate="2017-10-04T07:54:00.183" LastActivityDate="2018-11-16T01:24:34.997" Title="How to remove a stuck stopper from a flask?" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;&lt;equipment&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26883" PostTypeId="2" ParentId="26882" CreationDate="2015-03-03T22:22:25.787" Score="10" Body="&lt;p&gt;Is it a hollow glas stopper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, it helps to hold the flask at the body and hit the corner of the lab bench with the stopper. If that doesn't help, heat the neck of the flask carefully with a heat gun.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-03T22:22:25.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26884" PostTypeId="1" CreationDate="2015-03-03T22:35:20.043" Score="0" ViewCount="1734" Body="&lt;p&gt;Does an aqueous solution of sodium hypochlorite conduct current?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-03-03T23:47:35.623" Title="Is sodium hypochlorite solution an ionic solution - does it conduct current?" Tags="&lt;aqueous-solution&gt;&lt;ionic-compounds&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26885" PostTypeId="1" CreationDate="2015-03-03T22:47:53.987" Score="5" ViewCount="367" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Write an equation for the reaction you might expect to take place if trichloroethanoic acid, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl3CCO2H}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 0.65$&lt;/span&gt;, were added to dimethylpropanoic acid, &lt;span class=&quot;math-container&quot;&gt;$\ce{(CH3)3CCO2H}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 5.05$&lt;/span&gt;. Explain your answer in terms of the Brønsted–Lowry theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl3CCO2H}$&lt;/span&gt; is the stronger acid as it has a lower &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt;, however, how can dimethylpropanoic acid be a base?&lt;/p&gt;&#xA;" OwnerUserId="7884" LastEditorUserId="16683" LastEditDate="2020-02-06T00:32:56.687" LastActivityDate="2020-02-06T00:51:32.013" Title="What is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26886" PostTypeId="2" ParentId="26884" CreationDate="2015-03-03T23:47:35.623" Score="3" Body="&lt;p&gt;Yes, it is a solution of an ionic salt. a &lt;a href=&quot;http://www.forp.usp.br/restauradora/Trabalhos/prnaclo.html&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; lists electrical conductivity at various concentrations.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-03T23:47:35.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26887" PostTypeId="1" CreationDate="2015-03-03T23:48:12.650" Score="3" ViewCount="2893" Body="&lt;p&gt;I am confused as to how an order of a reactant (n or m) in the rate law can be negative. This means that increasing the concentration of a reactant would actually decrease its rate of disappearance. How does this make sense? Doesn't a higher concentration always allow for more collisions? Also, when the order is 0, this means that concentration has no effect on the rate of disappearance. How does this make sense chemically/at the atomic level? &lt;/p&gt;&#xA;" OwnerUserId="13077" LastActivityDate="2020-06-25T20:26:29.133" Title="Kinetics: Order of Reaction" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26888" PostTypeId="2" ParentId="26875" CreationDate="2015-03-04T00:14:18.857" Score="2" Body="&lt;p&gt;Assuming by thin, you mean a narrow line (not depth). It all depends on how thin you want to go. For a fine width line, a relatively cheap $\ce{CO2}$ laser cutter will get you down to 100–200 µm, but depending on the plastic and how thick it is, the heat produced may warp the plastic near the cut. I think most phone screen protectors are PET film. I've personally made ~200 µm features with one, and if the plastic is thin enough, you can keep the cut width pretty narrow. Things like pulsed UV lasers will give you much finer lines on plastics without thermal effects (as they tend to ablate rather than burn plastics), but they're much more expensive. If you don't have access to either, many companies will contract for this kind of work, even for small jobs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option for coarser cuts that don't have too complex of shapes is a vinyl cutter—basically a plotter with a cutting blade instead of a pen. These are quite cheap, but can't cut all shapes and resolution is on the order of a hundred µm or so, though they don't cause distortions along the edges in most materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're really serious about it, it's certainly possible to coat photoresist on most plastics and make a mask for reactive ion etching, which will get you down to 1 or 2 µm easily, but it sounds like overkill for you.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-03-04T00:14:18.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26889" PostTypeId="1" CreationDate="2015-03-04T00:47:52.513" Score="4" ViewCount="395" Body="&lt;p&gt;For thing like 8× dilution, that would be like 1 ml of a solution and then 7 ml of something like water to dilute right? Would 50 ml solution and 50 ml water be a 2× dilution? I’m a bit unsure about the terminology. If I an OD of 0.123 from a sample diluted 5×, would multiply 0.123 by 5 to get the real OD?&lt;/p&gt;&#xA;" OwnerUserId="14764" LastEditorDisplayName="user7951" LastEditDate="2015-05-14T18:59:52.723" LastActivityDate="2015-05-14T19:14:41.203" Title="Dilution terminology" Tags="&lt;solutions&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26890" PostTypeId="2" ParentId="26889" CreationDate="2015-03-04T01:00:27.823" Score="3" Body="&lt;p&gt;You are right.  The &amp;times; in 8&amp;times; refers to the fold-change in concentration between the concentrated source and diluted solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{DF = \frac{C_{stock}}{C_{working}}}$$ where $\mathrm{C_{stock}}$ and $\mathrm{C_{working}}$ are the concentrations of a stock solution and the desired final concentration, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of the working solution can be expanded:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{C_{working} = \frac{C_{stock} V_{stock}}{V_{working}} = \frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}}$$ where $\mathrm{V_{stock}}$ is the volume of the stock solution and $\mathrm{V_{diluent}}$ is the volume of diluent (e.g. pure water) added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging back into the DF equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{DF = \frac{C_{stock}}{\frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}} = \frac{V_{stock}+V_{diluent}}{V_{stock}}}$$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="4356" LastEditDate="2015-05-14T19:14:41.203" LastActivityDate="2015-05-14T19:14:41.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26891" PostTypeId="2" ParentId="26887" CreationDate="2015-03-04T01:00:40.917" Score="1" Body="&lt;p&gt;In principle, &lt;strong&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Zero-Order_Reactions&quot; rel=&quot;nofollow&quot;&gt;there are no zeroth-order reaction&lt;/a&gt;&lt;/strong&gt;. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of inverse first order reaction can be found &lt;a href=&quot;https://books.google.com/books?id=VaAgihZusxwC&amp;amp;pg=PA369&amp;amp;lpg=PA369&amp;amp;dq=inverse+first+order+kinetics+example&amp;amp;source=bl&amp;amp;ots=bEqroqm1FM&amp;amp;sig=v-18e3t-uIV1ITHtZsC94Gs9pIU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=E1P2VMuyKce1oQT9vYLIBA&amp;amp;ved=0CB0Q6AEwADgK#v=onepage&amp;amp;q=inverse%20first%20order%20kinetics%20example&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have a two-step reaction:&#xA;$$\ce{A + B &amp;lt;=&amp;gt; C}$$&#xA;$$\ce{C &amp;lt;=&amp;gt; D\ (slow)}$$&#xA;Increasing concentration of $\ce{C}$ would decrease the rate of reaction.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T08:45:23.357" LastActivityDate="2015-04-03T08:45:23.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26893" PostTypeId="1" CreationDate="2015-03-04T02:17:01.313" Score="3" ViewCount="1187" Body="&lt;p&gt;Here is a &lt;a href=&quot;http://youtu.be/XEbMHmDhq2I&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; that demonstrates the interesting nature of the triple point of cyclohexane, so I tried researching on the internet for a phase diagram, but only found data from a manufacture where the triple point was specified as 279.48 degK at 5.388 kPa. But in the video there doesn't seem to be a cooling plate. The flask appears to be at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the triple point specs for pure cyclohexane? If they match the data above, how were they able to achieve the phase transitions shown in the video at room temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also where can I find a phase diagram for pure cyclohexane on the internet?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-03-04T19:48:13.027" Title="What are the triple point pressure and temperature for pure cyclohexane?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26895" PostTypeId="1" CreationDate="2015-03-04T03:16:42.147" Score="3" ViewCount="429" Body="&lt;p&gt;I'm doing an assignment where we must research the separation of a mixture to obtain a useful final product. I thought about the separation of uranium from its ore after it has been mined, but I did not know if the ore counted as a mixture, or a compound.&lt;/p&gt;&#xA;" OwnerUserId="14766" LastEditorUserId="83758" LastEditDate="2020-09-16T05:02:44.730" LastActivityDate="2020-09-17T03:43:27.937" Title="Does Uranium ore count as a mixture?" Tags="&lt;analytical-chemistry&gt;&lt;metallurgy&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26897" PostTypeId="1" CreationDate="2015-03-04T04:33:29.573" Score="1" ViewCount="589" Body="&lt;p&gt;Catalyst provides alternative pathway to the reactants. Does it also mean that the potential energy surface is altered?&lt;/p&gt;&#xA;" OwnerUserId="14467" LastEditorUserId="4945" LastEditDate="2015-10-23T17:32:58.323" LastActivityDate="2015-10-23T17:32:58.323" Title="Does the catalyst alter the Potential energy Surface?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26898" PostTypeId="1" CreationDate="2015-03-04T04:40:46.250" Score="0" ViewCount="45" Body="&lt;p&gt;What is the function of silica, $\ce{SiO2}$ in the equation &#xA;$$\ce{CaO + SiO2 -&amp;gt; CaSiO3}$$ &#xA;  a) A basic oxide &#xA;  b) A acidic oxide &#xA;  c) A reducing agent &#xA;  d) An oxidising agent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain.&lt;/p&gt;&#xA;" OwnerUserId="14768" LastEditorUserId="418" LastEditDate="2015-03-04T04:57:28.467" LastActivityDate="2015-03-04T08:27:14.470" Title="Acid-base related chemistry" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-04T12:41:03.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="26899" PostTypeId="1" CreationDate="2015-03-04T05:00:41.997" Score="2" ViewCount="662" Body="&lt;p&gt;Will a real gas occupy all the space available for it? An ideal gas will expand to fill its container but will a real gas? And what are the changes in a real gas when it expands freely against a vacuum?&lt;/p&gt;&#xA;" OwnerUserId="7558" LastActivityDate="2015-04-03T11:01:17.137" Title="Real gas expansion" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26900" PostTypeId="1" CreationDate="2015-03-04T05:19:36.257" Score="-2" ViewCount="112" Body="&lt;p&gt;I have an old periodic table and I marked five metals from groups 1A and 2A with dots to indicate something about them. Only all this time later I can't figure out what exactly it was. I've been trying to figure it out for ages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements are: Li, Na, K, Ca, Ba. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doesn't anyone know any properties that just those elements, or maybe even just primarily those five, share that I could have been marking them for? &lt;/p&gt;&#xA;" OwnerUserId="14769" LastActivityDate="2015-03-04T08:59:00.133" Title="Trying to figure out what these elements have in common" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;periodic-table&gt;&lt;elements&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-04T12:44:36.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="26901" PostTypeId="1" CreationDate="2015-03-04T07:40:11.720" Score="1" ViewCount="811" Body="&lt;p&gt;Small amount of mercury and/or mercury vapors is harmful to human as it can cause damage to vital organs such as brain, heart and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are many electronic devices such as LCD monitor that used lamp that contain mercury for operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in what scenario will a LCD monitor that used lamp that contain mercury harmful?&lt;/p&gt;&#xA;" OwnerUserId="9401" LastActivityDate="2015-03-04T07:40:11.720" Title="At what scenario will a LCD monitor that used mercury inside their lamp be harmful" Tags="&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26902" PostTypeId="2" ParentId="26898" CreationDate="2015-03-04T08:27:14.470" Score="1" Body="&lt;p&gt;&lt;em&gt;Since this sounds like a &lt;a href=&quot;https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange/142#142&quot;&gt;homework question&lt;/a&gt;, I will not provide a full solution!&lt;/em&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Is &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; an oxidizing or reducing agent in the reaction?&lt;/h3&gt;&#xA;&lt;p&gt;Figure out the oxidation state for the &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Si}$&lt;/span&gt; in the starting materials and the product. As a rule of thumb, consider the oxidation state of (-2) for &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;. Unless peroxides are concerned, this works well.&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{Ca^{(+2)}O + Si^{(+4)}O2 -&amp;gt; Ca^{(+2)}Si^{(+4)}O3}$$&lt;/span&gt;&lt;br&gt;&#xA; Apparently, there is no change - this is not a redox reaction!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Is &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; a basic or an acidic oxide?&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How are basic and acidic oxides defined?&lt;/li&gt;&#xA;&lt;li&gt;Are there any rules (of thumb) for the oxides of metals and non-metals?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; Metal oxides are usually &lt;strong&gt;basic&lt;/strong&gt; oxides. These can be considered anhydrides of the corresponding metal hydroxides, &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO}$&lt;/span&gt; is a typical example: &lt;span class=&quot;math-container&quot;&gt;$\ce{CaO + H2O -&amp;gt; Ca(OH)2}$&lt;/span&gt;&#xA; What about &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-04T08:27:14.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26903" PostTypeId="2" ParentId="26897" CreationDate="2015-03-04T08:36:11.120" Score="1" Body="&lt;p&gt;As you said, catalysts do not alter the PES, but extend it to other dimensions, which usually provide new reaction pathways with lower reaction barriers. If you consider a surface catalyst and an impinged molecule. Considering the full systems PES, the pure molecule PES is a part of it (for large distances of the surface to the molecule). If we bring the molecule closer to the surface, we get new partial PES which have lower reaction barriers.&lt;/p&gt;&#xA;" OwnerUserId="14771" LastActivityDate="2015-03-04T08:36:11.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26905" PostTypeId="2" ParentId="26900" CreationDate="2015-03-04T08:59:00.133" Score="1" Body="&lt;p&gt;They are all member of the s block in the periodic table. It means they all have valence electron in the s orbital. &lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-03-04T08:59:00.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26906" PostTypeId="2" ParentId="26885" CreationDate="2015-03-04T09:44:33.147" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] I know that $\ce{Cl3CCO2H}$ is the stronger acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With other words: This is the &lt;em&gt;Br&amp;oslash;nsted&lt;/em&gt; donor. Show some confidence in your knowledge and draw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rXRKw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-04T09:44:33.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26907" PostTypeId="2" ParentId="26899" CreationDate="2015-03-04T10:11:41.293" Score="2" Body="&lt;p&gt;That's my answer to the question, which may not be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real gas molecules are bound by (attractive) Van der Waal's forces (VdW forces). When molecules move apart, part of their kinetic energy is converted to the potential energy (because Van der Waal's force is attractive), if no external source of energy (i.e. heat reservoir, external pressure sources/work done) is present. (i.e. there is no exchange of energy across the boundaries of the container).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, when a real gas expands into a vacuum, though there's no work done on the gas, the temperature of the gas should drop because part of the kinetic energy (which is proportional to the temperature) is converted to the potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the gas fill the whole container? Now first ignore surface effects (i.e. boundary walls of the container). We have to understand that although the molecules are &quot;bound&quot; by VdW forces, which tend to draw the molecules together, the latter tendency is balanced by the thermal motion of the gas molecules, which tends to pull the molecules apart. When the temperature is so low such that the thermal motion &quot;dies out&quot;, the molecules may settle into a lower energy state (e.g. lattice/solid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, real gases should fill the whole container.    &lt;/p&gt;&#xA;" OwnerUserId="7451" LastActivityDate="2015-03-04T10:11:41.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26908" PostTypeId="2" ParentId="26881" CreationDate="2015-03-04T10:12:33.440" Score="3" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NaOH + H2O-&amp;gt; Na+  +OH- + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\mathrm{pH}_1 = 7.00 &amp;amp;\implies\mathrm{pOH}_1 = 7 \\&#xA;\mathrm{pH}_2 = 12.42 &amp;amp;\implies\mathrm{pOH}_2 = 1.58&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;so &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]_1 = 10^{-7}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]_2 = 10^{-1.58}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, initially, you have &lt;span class=&quot;math-container&quot;&gt;$16^{-3}\times 10^{-7}$&lt;/span&gt; mol of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; in your water. Your &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; solution is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3 M}$&lt;/span&gt; so you'll bring &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3\times 10^{-3} mol}$&lt;/span&gt; each &lt;span class=&quot;math-container&quot;&gt;$\mathrm{mL}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Then, the resulting concentration will be &lt;span class=&quot;math-container&quot;&gt;$$C = \frac{n_\text{initial} + (0.3\times 10^{-3})\times x}{V_\text{initial} + x\times \pu{1mL}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You know &lt;span class=&quot;math-container&quot;&gt;$V_\text{initial}$&lt;/span&gt; to be &lt;span class=&quot;math-container&quot;&gt;$\pu{16 mL}$&lt;/span&gt;&lt;br /&gt;&#xA;You know &lt;span class=&quot;math-container&quot;&gt;$n_\text{initial}$&lt;/span&gt; to be &lt;span class=&quot;math-container&quot;&gt;$\pu{1.6\times 10^{-9} mol}$&lt;/span&gt;&#xA;And you know the target concentration to be &lt;span class=&quot;math-container&quot;&gt;$10^{-1.58}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I let you do the solving for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the volume of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; solution to be added ;) be careful with the units.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastEditorUserId="95133" LastEditDate="2020-09-11T18:56:46.463" LastActivityDate="2020-09-11T18:56:46.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26912" PostTypeId="2" ParentId="26895" CreationDate="2015-03-04T13:00:59.503" Score="2" Body="&lt;p&gt;The main uranium ore is pitchblende, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Uraninite&quot; rel=&quot;nofollow noreferrer&quot;&gt;uraninite&lt;/a&gt;, which is primarily composed of &lt;span class=&quot;math-container&quot;&gt;$\ce{UO2}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$\ce{UO2}$&lt;/span&gt; is ionically bonded together in a crystal structure and so it is definitely a compound and not a mixture. However, uraninite does contain small amounts of other compounds, such as &lt;span class=&quot;math-container&quot;&gt;$\ce{U3O8}$&lt;/span&gt; and trace amounts of lead oxides and so the mineral as a whole is a mixture of compounds.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="83758" LastEditDate="2020-09-16T05:02:53.980" LastActivityDate="2020-09-16T05:02:53.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26913" PostTypeId="2" ParentId="26895" CreationDate="2015-03-04T13:16:25.840" Score="3" Body="&lt;p&gt;Yes, uranium ore is a mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium compounds such as pitchblende (uraninite) are only present as a small part (usually $&amp;lt;1$%) of the ore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extensive purification is needed. Uranium ore purification would be a good choice for your project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.areva.com/EN/operations-677/uranium-processing-from-ore-to-yellow-cake.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.areva.com/EN/operations-677/uranium-processing-from-ore-to-yellow-cake.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iaea.org/sites/default/files/publications/magazines/bulletin/bull23-2/23204882428.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iaea.org/sites/default/files/publications/magazines/bulletin/bull23-2/23204882428.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tE6Le.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;image source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Development_Details/Approaches/ChemCases/Nuclear_Chemistry_and_the_Community/08._Uranium_Production&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemwiki.ucdavis.edu/Development_Details/Approaches/ChemCases/Nuclear_Chemistry_and_the_Community/08._Uranium_Production&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-04T13:39:47.320" LastActivityDate="2015-03-04T13:39:47.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26914" PostTypeId="1" CreationDate="2015-03-04T13:59:43.083" Score="3" ViewCount="1478" Body="&lt;p&gt;I am learning about the pH of buffer solutions, and am slightly confused by my text book's calculation. For the blood's buffer system of carbonic acid and hydrogencarbonate ions as the conjugate case, the equilibrium is $$\ce{H_2CO_3 &amp;lt;=&amp;gt; H^+ + HCO_3^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acid dissociation constant is $4.3 \times 10^{-7}~ \mathrm{mol/dm^{3}}$ and the $pH$ of healthy blood is 7.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $$\ce{\frac{[H_2CO_3]}{[HCO_3^{-}]}=\frac{[H^+]}{K_{a}}}=\frac{1}{10.8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this calculation, however my textbook then concludes that this shows that there are about 20 times more hydrogen carbonate ions than carbonic acid. Should it not be 10 times? Why was this value doubled?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="4231" LastEditDate="2015-03-04T14:34:18.603" LastActivityDate="2015-04-28T13:30:58.510" Title="Carbonic acid-hydrogencarbonate buffer system pH calculation?" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26915" PostTypeId="2" ParentId="26802" CreationDate="2015-03-04T15:37:08.180" Score="10" Body="&lt;p&gt;In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of etching at 25 degrees C is proportional to :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;according to &lt;a href=&quot;http://jes.ecsdl.org/content/118/11/1772.short&quot; rel=&quot;nofollow&quot;&gt;A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions&lt;/a&gt; J. Electrochem. Soc. vol. 118, pp. 1772-1775.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second equilibrium, it can be seen that if the fluoride concentation is 10M as in the question, the more active $\ce{HF2-}$ will be present at a concentration of about 100 times $[\ce{HF}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the two equilibrium equations together, and substituting $[\ce{F-}] = 10M$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{[\ce{H+}]}{[\ce{HF2-}]} = 1.4 \times 10^{-6} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore at $pH = log (1.4 \times 10^{-6}) = 5.9$, there will be about 1M $\ce{HF2-}$, diminishing by a factor of 10 for each pH unit increase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally, concentrated Ammonium Fluoride has a pH of 7.2 according to &lt;a href=&quot;http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf&lt;/a&gt; (which also says to avoid glass).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At pH 7.2, [$\ce{HF2-}$] is 0.05M.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, there will be a significant concentration of the most active etching species, $\ce{HF2-}$, present in 10M ammonium fluoride and I would be concerned about glassware.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-05-16T11:38:30.493" LastActivityDate="2016-05-16T11:38:30.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26916" PostTypeId="2" ParentId="26869" CreationDate="2015-03-04T15:56:29.933" Score="3" Body="&lt;p&gt;My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more non-trivial question is, whether the &lt;em&gt;total&lt;/em&gt; electron density can have nodes for the &lt;em&gt;ground state&lt;/em&gt;. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state &lt;strong&gt;has&lt;/strong&gt; to be occupied in the ground state.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="7451" LastEditDate="2015-03-18T14:59:31.330" LastActivityDate="2015-03-18T14:59:31.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26917" PostTypeId="1" AcceptedAnswerId="26918" CreationDate="2015-03-04T16:00:38.547" Score="4" ViewCount="21444" Body="&lt;p&gt;So, we know that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \ce{NaCl(aq) + AgNO3(aq) -&amp;gt; NaNO3(aq) + AgCl(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(In other words, the ions swap and a precipitate forms).&lt;/p&gt;&#xA;&lt;p&gt;However, both of those compounds seem pretty stable, since &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; cations both have a 1+ charge, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3}$&lt;/span&gt; both have a 1- charge. Since they are stable, I would assume no reaction would take place since both ions are happy with the ions they are currently bonded to (sodium and silver, respectively).&lt;/p&gt;&#xA;&lt;p&gt;However, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; is higher on the activity series than &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt;, and we know that these aqueous solutions actually contain dissociated ions, so I assume this has something to do with the double-replacement. But why is this? Why does this reaction take place, even though the original compounds appear stable at first glance?&lt;/p&gt;&#xA;&lt;h2&gt;Edit&lt;/h2&gt;&#xA;&lt;p&gt;I read &lt;a href=&quot;https://chemistry.stackexchange.com/questions/383/reaction-between-silver-nitrate-and-aluminium-chloride&quot;&gt;this question&lt;/a&gt;, and it says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The above compounds of sodium chloride and silver nitrate appear to be at an initial equilibrium, do they not?&lt;/p&gt;&#xA;&lt;h2&gt;Edit 2&lt;/h2&gt;&#xA;&lt;p&gt;The question isn't why silver chloride is a solid, but rather, &lt;strong&gt;why does this reaction take place if the initial two compounds are stable to begin with?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8028" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2020-03-17T09:49:30.603" Title="Why do aqueous solutions of sodium chloride and silver nitrate swap ions?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26918" PostTypeId="2" ParentId="26917" CreationDate="2015-03-04T16:14:01.827" Score="5" Body="&lt;p&gt;While solutions of silver nitrate and sodium chloride are in stable equilibrium on their own, once mixed the equilibrium is changed. When the salts are dissolved, you no longer have silver and nitrate or sodium and chloride ions associated with each other, you have a mixture of individual ions in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But silver chloride has very low solubility, about 1 mg can dissolve in 1 L of water. So when a silver ion and chloride ion meet, they quickly fall out of solution. So if we had an equilibrium between the dissolved ions and the small amount of &quot;soluble&quot; silver chloride, that equilibrium would be driven toward the silver chloride by the precipitation. When compounds escape the solution by precipitating or evaporating, they are removed from the equilibrium, so trying to replace them to restore equilibrium requires converting more free ions into silver chloride and pushes the reaction to completion.&lt;/p&gt;&#xA;" OwnerUserId="137" LastActivityDate="2015-03-04T16:14:01.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26919" PostTypeId="2" ParentId="26893" CreationDate="2015-03-04T16:32:55.843" Score="3" Body="&lt;p&gt;The NIST REFPROP &lt;em&gt;(Reference Fluid Thermodynamic and Transport Properties)&lt;/em&gt; program reports the triple-point temperature of cyclohexane as 279.47 K.&lt;br&gt;&#xA;The program refers to: S. G. Penoncello, R. T. Jacobsen, and A. R. H. Goodwin: “A Thermodynamic Property Formulation for Cyclohexane”, &lt;em&gt;International Journal of Thermophysics,&lt;/em&gt; Vol. 16., No. 2, 1995, 515–531.&lt;br&gt;&#xA;This publication mentions the triple-point temperature as &lt;em&gt;T&lt;/em&gt; = 279.47 K and the triple-point pressure as &lt;em&gt;p&lt;/em&gt; = 5.388 kPa.&lt;br&gt;&#xA;The given reference is: J. G. Aston, G. J. Szasz, and H. L. Fink, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;65&lt;/strong&gt;  1135 (1943).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following &lt;em&gt;p&lt;/em&gt;-&lt;em&gt;T&lt;/em&gt; diagram (made using REFPROP) shows the melting line (solid-liquid equilibrium, blue) and the vaporization line (liquid-gas equilibrium, red), which meet at the solid-liquid-gas triple point. Unfortunately, REFPROP does not include the sublimation line for cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P12n7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-04T19:48:13.027" LastActivityDate="2015-03-04T19:48:13.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26920" PostTypeId="1" CreationDate="2015-03-04T16:39:46.810" Score="3" ViewCount="1224" Body="&lt;p&gt;I was planning on doing some small scale hydrogenation practicals at atmospheric pressure (probably from Vogel) and ran into a couple of questions. I will be using the &quot;balloon method&quot; (&lt;a href=&quot;https://books.google.nl/books?id=lpv9D2hin6gC&amp;amp;pg=PA267&amp;amp;lpg=PA267&amp;amp;dq=Hydrogenation%20atmospheric&amp;amp;source=bl&amp;amp;ots=TxRz-QB3zd&amp;amp;sig=a7oVkg22dFcOz5ytrjrblI8uK2s&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=wy73VKpKyf5S0P-CgAg&amp;amp;ved=0CD4Q6AEwBA#v=onepage&amp;amp;q=Hydrogenation%20atmospheric&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;described in Procter's Advanced Practical Organic Chemistry book&lt;/a&gt;) as I consider it an easier technique for small scale reactions than constructing a Brown hydrogenation apparatus. Note that I'm working in an amateur environment (i.e. my home lab which does not have a fumehood). Now for the questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What rule of thumb can be used to determine the amount of catalyst (in these cases $\ce{Pd/C}$) needed per mole of hydrogen used (I suppose these two variables are depending on each other)?&lt;/li&gt;&#xA;&lt;li&gt;There are different types of $\ce{Pd/C}$ out there: 3%, 5% or 10% $\ce{Pd}$ by weight, with or without a variable amount of added $\ce{Pb}$, different solvents (mostly I see is water) and also there are different bases available, not just palladium on carbon. I understand that $\ce{Pb}$ is added to poison the catalyst such as with Lindlar catalyst, but if total reduction is required then why would this be necessary? Also how would I decide between 3%, 5% or 10% $\ce{Pd}$ by weight?&lt;/li&gt;&#xA;&lt;li&gt;What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of  pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.&lt;/li&gt;&#xA;&lt;li&gt;Why would it be required to flush the reaction mixture with an inert gas after the hydrogenation has completed? I understand that this is necessary because the dissolved hydrogen gas can ignite and possibly explode when, while filtering the catalyst, the catalyst spontaneously ignites (e.g. when the catalyst is not suspended enough and direct contact with the atmosphere is possible). However, why is an inert gas advised? What is wrong with ordinary air?&lt;/li&gt;&#xA;&lt;li&gt;What is the function of Celite when filtering the catalyst?&lt;/li&gt;&#xA;&lt;li&gt;How many times can the $\ce{Pd/C}$ catalyst be reused? How would you spot whether or not your catalyst has become too poisoned to be successfully used in hydrogenation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2018-05-18T21:49:30.153" Title="Hydrogenation at atmospheric pressure in amateur environment" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26921" PostTypeId="1" AcceptedAnswerId="26924" CreationDate="2015-03-04T17:51:53.360" Score="3" ViewCount="1060" Body="&lt;p&gt;Which is higher in energy? Cyclohexanone enol or cyclohex-2-en-1-ol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay from what I understand a higher pi HOMO is synonymous with more nucleophilic. So I would expect cyclohex-2-en-1-ol to have the higher pi HOMO because in the case of cyclohexanol enol there would be inductive withdrawal by the oxygen connected to one of the sp2 carbons involved in the pi bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would it be the other way around? Resonance donation by oxygen into the pi bond of the enol makes it more nucleophilic? How can we tell how the HOMO/LUMO energies split? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-03-04T19:47:18.090" LastActivityDate="2015-03-04T19:47:18.090" Title="Higher energy π HOMO" Tags="&lt;molecular-orbital-theory&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26922" PostTypeId="1" AcceptedAnswerId="26923" CreationDate="2015-03-04T18:10:04.600" Score="0" ViewCount="114" Body="&lt;p&gt;It makes no sense to me that these thermodynamic changes are given in kJ/mol. Reactions can have several compounds on both sides and its not clear what exactly the molar variable refers to…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been given a homework where this is the convention followed. My temptation is to just ignore the /mol as senseless, and proceed just considering the energy change. But I'm sure I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastActivityDate="2015-03-04T18:26:33.000" Title="Why are enthalpies of reaction or Gibbs FE change given in kJ/mol" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26923" PostTypeId="2" ParentId="26922" CreationDate="2015-03-04T18:26:33.000" Score="3" Body="&lt;p&gt;It makes perfect sense that the units KJ/mol are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you burn 2 cords of wood you get twice the heat as from 1 cord of wood.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthaply (and Gibbs free energy) of the reaction is proportional to the number of moles that react.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reaction is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3A + 2B =&gt; 6C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the change per mole would mean: per 3 moles of A = per 2 moles of B = per 6 moles of C.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-04T18:26:33.000" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="26924" PostTypeId="2" ParentId="26921" CreationDate="2015-03-04T18:37:37.000" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;from what I understand a higher pi HOMO is synonymous with more&#xA;  nucleophilic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that's generally true.  If you have a substituent on a double bond that increases the electron density in the double bond, then the double bond (usually one carbon more than the other) becomes more nucleophilic and the energy of the HOMO and LUMO are raised.  Conversely, electron withdrawing groups attached to a double bond withdraw electron density, making the double bond less nucleophilic and lowering the energies of the HOMO and LUMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing cyclohexanone enol (cyclohex-1-en-1-ol) to cyclohex-2-en-1-ol, you can draw a resonance structure for the former that increases electron density in the double bond (especially the $\beta$ carbon).  This electron donating resonance effect overwhelms the electron withdrawing inductive effect of the hydroxyl group.  Therefore, the double bond in the cyclohex-1-en-1-ol will be more nucleophilic and its HOMO and LUMO will be higher in energy than those for cyclohex-2-en-1-ol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-04T18:37:37.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26925" PostTypeId="1" AcceptedAnswerId="26927" CreationDate="2015-03-04T18:51:32.990" Score="1" ViewCount="5223" Body="&lt;p&gt;In spectroscopy, the isosbestic point is said to be where the molar absorptivities of two species is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A= \epsilon(X+Y) \times c(X+Y) \times l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reduces to $$A= \epsilon_t \times c \times l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the absorbances and the length are the same at this point, but the concentrations are not; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;why are the molar absorptivities defined to be the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is c taken to be the relative contribution, rather than the exact concentration?&lt;/p&gt;&#xA;" OwnerUserId="7721" LastEditorUserId="7448" LastEditDate="2015-03-04T19:47:52.373" LastActivityDate="2015-03-04T19:51:30.923" Title="Molar absorption coefficient - why are the molar absorptivities defined to be the same?" Tags="&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26926" PostTypeId="2" ParentId="26925" CreationDate="2015-03-04T19:12:34.757" Score="4" Body="&lt;p&gt;Actually that's not the correct definition of the &lt;a href=&quot;http://goldbook.iupac.org/I03310.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;IUPAC definition differs you know&quot;&gt;isosbestic point&lt;/a&gt;. From IUPAC goldbook:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Isosbestic point - Wavelength, wavenumber or frequency at which the total absorbance of a sample does not change during a chemical reaction or a physical change of the sample.&lt;/p&gt;&#xA;&lt;p&gt;A simple example occurs when one molecular entity is converted into another that has the same molar absorption coefficient at a given wavelength. As long as the sum of the concentrations of the two molecular entities in the solution is held constant there will be no change in absorbance at this wavelength as the ratio of the concentrations of the two entities is varied.&lt;/p&gt;&#xA;&lt;p&gt;Contrary to a widely accepted idea, the existence of an isosbestic point does not prove that the reaction is a quantitative conversion of one species into a unique other species or that an equilibrium exists between only two species. The observation of isosbestic points only indicates that the stoichiometry of the reaction remains unchanged during the chemical reaction or the physical change of the sample, and that no secondary reactions occur during the considered time range, For the reaction A + B → c C + d D + e E, with c, d, and e the percentages of the products C, D, and E, an isosbestic point will be observed at every wavelength where the condition ɛ A + ɛ B = c ɛ C + d ɛ D + e ɛ E, provided that the values of the percentages c, d, and e remain constant during the chemical reaction or the physical change. The use of the obsolete term isoabsorption point is not recommended.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-03-04T19:51:30.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26927" PostTypeId="2" ParentId="26925" CreationDate="2015-03-04T19:15:15.020" Score="4" Body="&lt;p&gt;An isosbestic point is found by maintaining the sum of the concentrations of the two (or more, though it's very unlikely) species the same and varying the individual contributions of each, and looking for the point at which the absorbance spectra intersect. This is easy for something like a pH indicator where a pH change just converts one species into the other—the sum of the two species is always the same. Under these conditions, a point where the absorbance doesn't change with the ratio of the two species can only exist if the molar absorptivities of the two species are the same, otherwise changing their ratio would change the absorbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the molar absorptivities are defined to be the same when you simplify your equation above because this is the only way an isosbestic point can exist. So in your simplified equation, c is the sum of the concentrations of both species as that is what determines the net absorbance, not the concentrations of each individual species.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-03-04T19:15:15.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26928" PostTypeId="2" ParentId="26920" CreationDate="2015-03-04T20:05:13.987" Score="2" Body="&lt;p&gt;I've adddressed a part of your question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/21638/reduction-of-%CE%B1-%CE%B2-unsaturated-nitro-compounds/21724#21724&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the catalyst is concerned, 3-5% Pd/C is fine. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the lab, I only used pressurized tanks. While commercial hydrogen generators are available - they are mostly used to generate hydrogen gas as a carrier gas for GC/MS - these things cost a fortune and really would be overkill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, pressurized tanks are pretty safe if you follow some rules: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Obviously, you can't attach the pressure regulator and operate the whole thing unless the steel cap over the main valve is removed. &lt;strong&gt;Don't start anything unless the tank is secured to the lab bench or the wall with a clamp, a chain or another sturdy contraption that prevents the tank from falling over!&lt;/strong&gt; You want hydrogenation, not a rocket ;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Get familiar with the pressure regulator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;When you're done, remove the regulator again and put the cap over the main valve. Keep the regulator at a dry and clean place. &lt;/li&gt;&#xA;&lt;li&gt;No grease on the threads!  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When operated with a bit of common sense, a small pressurized tank is the most convenient option.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the function of Celite when filtering the catalyst?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you filter off the catalyst, you'll typically use a suction filter with a base of sintered glass (fritted glass). Palladium on charcoal is a very fine powder that will clog the fritted glass. A pad of Celite in the suction will prevent this. Celite is cheap, suction filters are not!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-03-04T20:33:15.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26929" PostTypeId="1" CreationDate="2015-03-04T20:41:12.693" Score="0" ViewCount="1553" Body="&lt;p&gt;A 1.245 g sample of limestone ($\ce{CaCO3}$) is pulverized and then treated with 30 mL of 0.035 M $\ce{HCl}$. The excess acid requires 11.56 mL of 1.010 M $\ce{NaOH}$ to be neutralized. Calculate the %wt of limestone in the rock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first balanced equation would be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + 2HCl -&amp;gt; CaCl2 + H2O + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + NaOH -&amp;gt; H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out the moles of $\ce{CaCO3}$ if that's useful for this problem 1.245 g of $\ce{CaCO3}$ / 100.089 g (molar mass)= 0.0124 M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this but I don't know the next steps on how to solve the problem. &lt;/p&gt;&#xA;" OwnerUserId="14780" LastEditorUserId="865" LastEditDate="2015-10-02T15:45:01.867" LastActivityDate="2015-10-02T15:45:01.867" Title="Calculate the weight percent of limestone in a rock sample" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26930" PostTypeId="2" ParentId="26929" CreationDate="2015-03-04T21:01:54.677" Score="1" Body="&lt;p&gt;The question is a bit misleading ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there's apparently something else in the rock sample (that doesn't disturb the titration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you know for sure is: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the &lt;strong&gt;stoichiometry&lt;/strong&gt; of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;concentration and volume of the $\ce{HCl}$ &lt;strong&gt;used in excess&lt;/strong&gt; to neutralize all the $\ce{CaCO3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left &lt;strong&gt;after&lt;/strong&gt; reacting all $\ce{CaCO3}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HINT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you consider to solve the &lt;strong&gt;back titration&lt;/strong&gt; experiment by starting with the second neutralisation step? &lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-04T21:16:43.383" LastActivityDate="2015-03-04T21:16:43.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26931" PostTypeId="2" ParentId="6055" CreationDate="2015-03-04T22:18:29.883" Score="5" Body="&lt;p&gt;$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or  $\ce{SO4^{2-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normality of an acid is the concentration of the acid times the number of acidic protons per molecule.  Sulfuric acid has two acidic protons per molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normality $= \pu{1M} \times 2 = \pu{2N}$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-02-26T05:05:55.433" LastActivityDate="2018-02-26T05:05:55.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26932" PostTypeId="2" ParentId="26868" CreationDate="2015-03-04T22:25:24.317" Score="16" Body="&lt;p&gt;Yes . You are right that structural symmetry comes into play .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment .  The dipole moment enables electronic interactions which hold molecules together .  &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/10/25/3-trends-that-affect-boiling-points/&quot;&gt;This&lt;/a&gt; shows some general factors of boiling point . Also the below link to the google book covers the structural aspect of a molecule towards b.p on pg47.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For melting points , let me quote Joseph Hornback from Chapter 2.6 &lt;a href=&quot;http://www.worldcat.org/title/organic-chemistry/oclc/424076558&quot;&gt;Organic Chemistry&lt;/a&gt; pg46 (&lt;a href=&quot;https://books.google.com/books?isbn=0534389511&quot;&gt;google books&lt;/a&gt;) :   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The shape of a compound is very important and has a dramatic effect on its melting point . More symmetrical molecules pack into the crystal lattice better , allowing closer approach and larger attractive forces , resulting in higher melting points .    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Properties_of_Alkenes/Physical_Properties_of_Alkenes&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trans 2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point .   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading I suggest you visit 'Molecular Shape' over &lt;a href=&quot;http://www.brynmawr.edu/chemistry/Chem/mnerzsto/Labs/Forces_and_Factors.pdf&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="13339" LastEditDate="2015-03-06T16:51:45.513" LastActivityDate="2015-03-06T16:51:45.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26933" PostTypeId="1" AcceptedAnswerId="26938" CreationDate="2015-03-04T23:41:13.193" Score="3" ViewCount="244" Body="&lt;p&gt;I have to examine the exact material of a specific fabric (non-woven), of which I know is either viscose or cotton. I know that viscose and cotton are structurally similar, but there should be a way to tell them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the result of the microscopic examination (x 100):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/284jm.jpg&quot; alt=&quot;Viscose or cotton&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor tells me this is viscose, but I thought of cotton because of the long fibers. But I also know that cotton has a 'twisted' look under the microscope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor doesn't give any more clues, which leaves me with a bit of a problem. How can I tell the two apart? Does the picture show enough evidence for viscose (because of the linear structure)? I would like to double check this, just to be sure.&lt;/p&gt;&#xA;" OwnerUserId="3935" LastEditorUserId="10185" LastEditDate="2015-10-09T16:33:07.390" LastActivityDate="2016-05-24T09:17:42.707" Title="Is this cotton or viscose?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26935" PostTypeId="1" CreationDate="2015-03-05T00:05:14.610" Score="4" ViewCount="2869" Body="&lt;p&gt;USA pennies were nearly pure copper pre-1982.  Post 1982 they are copper clad zinc.  A common experiment in US high schools is to notch pennies with a file and add them to an HCl solution.  The old pennies remain untouched and the zinc of the new pennies dissolves, leaving the copper cladding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the copper cladding also dissolves and sometimes a considerable amount of the old penny dissolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the copper dissolve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I propose that the copper in the penny is reacting with atmospheric oxygen, forming copper oxide that dissolves in the acidic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proposal is that the formation of a complex ion between copper and chloride ion drives the equilibrium, allowing the copper to react in HCl.  This is analogous to how aqua regia dissolves gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another proposal is that the copper oxide layer on the penny reacts with copper metal in a comproportionation to form copper (I) ion that then reacts with oxygen to make copper (II) and dissolves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is right?  Why does the copper dissolve?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-03-05T05:50:48.473" Title="Why do the pennies in the pre/post 1982 penny experiment sometimes dissolve completely?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26936" PostTypeId="2" ParentId="26929" CreationDate="2015-03-05T00:07:28.557" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Calculate the amount of excess acid:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{x mol HCl = 11.56mL NaOH \times \frac{1.010 mol NaOH}{1000mL}\times \frac{1 mol HCl}{1mol NaOH} = 0.01167 mol HCl }$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Determine the amount of acid originally added:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{x mol HCl = 30 mL HCl \times \frac{0.035 mol NaOH}{1000mL}= 0.0011 mol HCl }$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Realize more acid was neutralized than was originally added.  There must be an error in the question some where.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-05T00:07:28.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26937" PostTypeId="1" CreationDate="2015-03-05T01:30:24.207" Score="3" ViewCount="1369" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;An aqueous effluent contains &lt;span class=&quot;math-container&quot;&gt;$12\ \mathrm{mol\ l^{-1}}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$10\ \mathrm{mol\ l^{-1}}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt; as solution of nitrates. Current practice is use &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; to selectively precipitate the metals.&lt;/p&gt;&#xA;&lt;p&gt;(a) Is it feasible to get 0 &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd}$&lt;/span&gt; impurities in &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; precipitate by using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; (back up your answer with adequate calculations)&lt;/p&gt;&#xA;&lt;p&gt;(b) What trade-off with respect to yield and purity do you have to accept if you use &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;(c) If consider switching the precipitating agent during the process which sequence should be used&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Attempt at answer:&lt;/p&gt;&#xA;&lt;h3&gt;Part (a)&lt;/h3&gt;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{MgF2}) = 3.7\times10^{-8}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{CdF2}) = 6.44\times10^{-3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{MgF2}) = [\ce{Mg^2+}][\ce{F-}]^2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{CdF2}) = [\ce{Cd^2+}][\ce{F-}]^2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;As the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{MgF2}$&lt;/span&gt; is smaller it is less soluble and will precipitate first. You need to add &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; to the point where &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt; will just not start precipitating. The limiting concentration is when the ion product = &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Limiting concentration for &lt;span class=&quot;math-container&quot;&gt;$\ce{MgF2}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{Mg^2+}]} = \sqrt\frac{3.7\times10^{-8}}{10} = 6.08\times10^{-5}\ \mathrm{mol\ l^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Limiting concentration for &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{CdF2})}{[\ce{Cd^2+}]} = \sqrt\frac{6.44\times10^{-3}}{12} = 0.023\ \mathrm{mol\ l^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The limiting concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt; is lower so this will precipitate first. By adding &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; up to the limiting concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd^2+}$&lt;/span&gt; no &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd}$&lt;/span&gt; will precipitate so &lt;span class=&quot;math-container&quot;&gt;$\ce{MgF2}$&lt;/span&gt; precipitate will be pure.&lt;/p&gt;&#xA;&lt;h3&gt;Part (b)&lt;/h3&gt;&#xA;&lt;p&gt;I assume you need to find the yield and purity for both methods (using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;).&#xA;To find the purity of &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt; need to know how much &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt; is left in solution at the point when &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt; starts precipitating.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$[\ce{Mg^2+}]_\text{sol} = \frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{F-}]^2} = \frac{3.7\times10^{-8}}{0.023^2} = 6.99\times10^{-8}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Purity of &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\frac{[\ce{Cd}]}{[\ce{Cd}]+[\ce{Mg2+}]_\text{sol}} \times 100\ \% = 99.99\ \%$&lt;/span&gt;&lt;br /&gt;&#xA;I assume the yield of &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt; would be &lt;span class=&quot;math-container&quot;&gt;$100$&lt;/span&gt; as you can keep adding &lt;span class=&quot;math-container&quot;&gt;$\ce{F-}$&lt;/span&gt; until it’s all precipitated out.&lt;/p&gt;&#xA;&lt;p&gt;I’m not sure if to calculate the yield of magnesium I need to do the original concentration − impurities in &lt;span class=&quot;math-container&quot;&gt;$\ce{CdF2}$&lt;/span&gt;/original concentration. If I did this then:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\frac{10-6.99\times10^{-8}}{10} \times 100\ \% = 99.99\ \%$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To compare with using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; as the precipitating agent:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{Mg(OH)2}) = 1.8\times10^{-11}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{Cd(OH)2}) = 2.5\times10^{-14}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{Mg(OH)2}) = [\ce{Mg^2+}][\ce{OH-}]^2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}(\ce{Cd(OH)2}) = [\ce{Cd^2+}][\ce{OH-}]^2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;As the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd(OH)2}$&lt;/span&gt; is smaller it is less soluble and will precipitate first.&lt;/p&gt;&#xA;&lt;p&gt;Limiting concentration for &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg(OH)2}$&lt;/span&gt;:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{Mg(OH)2})}{[\ce{Mg^2+}]} = \sqrt\frac{1.8\times10^{-11}}{10} = 1.43\times10^{-6}\ \mathrm{mol\ l^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To find the purity of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg(OH)2}$&lt;/span&gt; need to know how much &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd^2+}$&lt;/span&gt; is left in solution at the point when &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg(OH)2}$&lt;/span&gt; starts precipitating.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$[\ce{Cd^2+}]_\text{sol} = \sqrt\frac{K_\mathrm{sp}(\ce{Cd(OH)2})}{[\ce{F-}]^2} = \frac{2.5\times10^{-14}}{(1.43\times10^{-6})^2} = 0.0138\ \mathrm{mol\ l^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Purity of &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\frac{10}{10+0.0138}\times100\ \% = 99.86\ \%$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Yield of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cd}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$\frac{12-0.0138}{12}\times100\ \% = 99.86\ \%$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For part B the way the question is phrased I assume the purity or yield using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt; shouldn’t be as great as using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt; but I don’t know if I have done my calculations right as from my results it would seem they’re both accurate methods.&lt;/p&gt;&#xA;" OwnerUserId="14783" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-04-01T00:50:54.340" Title="Purity of precipitate in selective precipitation" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;solubility&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26938" PostTypeId="2" ParentId="26933" CreationDate="2015-03-05T01:42:11.203" Score="4" Body="&lt;p&gt;Yes, the picture should be sufficient to decide between viscose and cotton; see photo of cotton at &lt;a href=&quot;http://whrhs-forensic-chem-spring-2010.wikispaces.com/Fiber+Analysis?responseToken=906bf6a76a5e268a00e54c8bce6927ba&quot; rel=&quot;nofollow&quot;&gt;http://whrhs-forensic-chem-spring-2010.wikispaces.com/Fiber+Analysis?responseToken=906bf6a76a5e268a00e54c8bce6927ba&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that cotton was the &lt;em&gt;cell wall&lt;/em&gt; of a plant cell, so it retains some of the shape of the cell. (Silk, on the other hand, is excreted from spinnerets, as are man-made fibers, so it looks more like viscose).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-05T01:42:11.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26939" PostTypeId="2" ParentId="5194" CreationDate="2015-03-05T03:50:00.430" Score="-1" Body="&lt;p&gt;The outermost shell of an atom can accomodate a max of 8 electrons. It was observed that the atoms of elements,having a completely filled outermost shell show little chemical activity. In other words their combining capacity is 0. Of these inert elements, the helium atom has 2 electrons in its outermost shell and all other elements have atoms with 8 electrons in the outermost shell. Their combining capacity of the atoms of other elements,that is their tendency to react &amp;amp; form molecules with same or different elements,was thus explained as an attempt to attain a fully-filled outermost shell. &lt;/p&gt;&#xA;" OwnerUserId="14787" LastActivityDate="2015-03-05T03:50:00.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26940" PostTypeId="1" AcceptedAnswerId="26941" CreationDate="2015-03-05T05:14:08.337" Score="-1" ViewCount="118" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Suppose we had an aqueous solution of density 0.984 g cm$^{-3}$, containing 100.0 g ethanol per liter. How would we compute the mole fraction of ethanol in this solution?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My calculations so far propose that we examine a fixed volume, 1 L of the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CH$_3$CH$_2$OH: m = 100.0g | M = 46.068g mol$^{-1}$ | n = $\frac{m}{M}$ = 2.170 ... mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solution (incl. H$_2$O): m$_{tot}$ = 984 g | M$_{tot}$ = M$_{solute}$ + M$_{solvent}$ = 46.068 + 18.016 = 64.084 g mol$^{-1}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\therefore$ n$_{tot}$ = 15.35 ... mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $x$ = $\frac{n}{n_{tot}}$ = 0.1413 ... $\approx$ 0.141 (3 sig. fig.)&lt;/p&gt;&#xA;" OwnerUserId="7161" LastActivityDate="2015-03-05T05:37:30.613" Title="Calculation of Solute Mole Fraction" Tags="&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26941" PostTypeId="2" ParentId="26940" CreationDate="2015-03-05T05:37:30.613" Score="2" Body="&lt;p&gt;This is not correct because you assuming that the molecular weight rations are one to one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coorect formulation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(100 g Methano/46.068)/(100/46.068 + (984-100)/18.016)&lt;/p&gt;&#xA;" OwnerUserId="5179" LastActivityDate="2015-03-05T05:37:30.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26943" PostTypeId="2" ParentId="26935" CreationDate="2015-03-05T05:50:48.473" Score="-2" Body="&lt;p&gt;Your conclusion about formation of the Copper Oxide is correct, but not about the reaction with atmospheric oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of the Oxygen is from the hydrochloric acid solution. The solution does not mean is pure HCl , it means is a dissolve in water.  There is a redox reaction with water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave to you to find the redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="5179" LastActivityDate="2015-03-05T05:50:48.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26944" PostTypeId="1" CreationDate="2015-03-05T07:45:29.180" Score="1" ViewCount="105" Body="&lt;p&gt;When does 'H' Shows +I/-I effect and why does T,D show better +I effect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my book Hydrogen is given in both +I &amp;amp; -I effect showing group ,i was curious to know under what conditions does it shows such opposite/dual effects? Also are there any other commonly encountered groups which show both +I/-I contition to condition?&lt;/p&gt;&#xA;" OwnerUserId="14742" LastActivityDate="2015-03-05T07:45:29.180" Title="When does 'H' Shows +I/-I effect and why does T,D show better +I effect?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ClosedDate="2015-03-05T11:31:26.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="26946" PostTypeId="1" CreationDate="2015-03-05T09:27:50.217" Score="0" ViewCount="229" Body="&lt;p&gt;Suppose I have gas bottle and gas cooker. Suppose I have propane in my bottle and I suspect it is diluted with unknown gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I measure with simple instruments, how many propane I have (percentage)?&lt;/p&gt;&#xA;" OwnerUserId="14790" LastEditorUserId="10185" LastEditDate="2015-03-23T18:03:47.470" LastActivityDate="2015-03-23T18:03:47.470" Title="How to determine methane percentage at home?" Tags="&lt;home-experiment&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26947" PostTypeId="2" ParentId="26480" CreationDate="2015-03-05T11:53:18.527" Score="1" Body="&lt;p&gt;As you know, partial molar volume of given substance is the change in solution volume after one mole of this substance is added. When you add $\ce{MgSO4.7H2O}$ to water it dissolves to magnesium cation and sulfate anion but the hydration water simply mixes with the rest of water. So, I suppose that partial molar volume of the magnesium sulfate heptahydrate is the sum of partial molar volume of magnesium sulfate and seven times molar volume of water. If $\widehat{V}_i$ denotes partial molar volume of substance $i$, and $V_i$ denotes molar volume, then: $$\widehat{V}_{\ce{MgSO4.7H2O}}=\widehat{V}_{\ce{MgSO4}}+7V_{\ce{H2O}}.$$&#xA;Nevertheless, partial molar volume is function of composition and commonly quoted values are for very low concentration (adding substance to large quantity of water). If you want to obtain solution with higher concentration, finding its exact volume requires knowledge about $\widehat{V}_i$ as a function of composition and such function has to be integrated.&lt;/p&gt;&#xA;" OwnerUserId="14772" LastEditorUserId="4945" LastEditDate="2015-03-05T12:03:09.063" LastActivityDate="2015-03-05T12:03:09.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26948" PostTypeId="1" CreationDate="2015-03-05T13:05:38.063" Score="2" ViewCount="412" Body="&lt;p&gt;I have recently bought a carbon monoxide spot detector which is not working the way I intended for it. A carbon monoxide spot detector is an analog carbon-monoxide detecting patch that changes color when it comes in contact with carbon monoxide. The problem is that it changes back to it's original color whenever the carbon monoxide leaves the room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd wish for it to stay the same color it changed to after it came into contact with carbon monoxide. What should the reaction look like to fix this problem? I have copy/pasted a scholarly paper from Google to better explain what's happening inside of the patch:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Principle of the $\ce{CO}$ – Detector:&lt;/strong&gt; The palladium (II) chloride inside is reduced into metallic palladium when exposed to $\ce{CO}$. While the $\ce{CO}$ – level in air is rising, higher amount of palladium is formed. A color change of the solution from pale yellowish orange to orange or grey is followed.&#xA;  $$\ce{CO + PdCl2.2H2O → CO2 + Pd + 2HCl + H2}$$&#xA;  Afterwards, when exposing to $\ce{CO}$ – free air, copper (II) chloride oxidizes palladium back into palladium (II) chloride. In addition, copper (II) chloride is reduced to copper (I) chloride. &#xA;  $$\ce{Pd + 2CuCl2.2H2O → PdCl2.2H2O + Cu2Cl2}$$ &#xA;  Then, the regeneration of copper (II) chloride occurs. &#xA;  $$\ce{Cu2Cl2 + 2HCl + H2O + 1/2 O2 → 2CuCl2.2H2O}$$&#xA;  Thus, the solution resumes to its initial state. (the oxygen comes from the surrounding air)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14792" LastEditorUserId="5026" LastEditDate="2018-02-23T13:58:17.277" LastActivityDate="2018-02-23T13:58:17.277" Title="Carbon monoxide detector reaction" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26949" PostTypeId="1" AcceptedAnswerId="26950" CreationDate="2015-03-05T14:08:41.390" Score="12" ViewCount="48916" Body="&lt;p&gt;I came across a question asking for the major product on chlorination of 2-methylbutane. The answer in the back of the book says it will be 2-chloro-3-methylbutane. However, I thought the major product would be 2-chloro-2-methylbutane. The later is formed through a tertiary free radical with eight hyper-conjugative structures, whereas the former forms through a secondary free radical with only four hyper-conjugative structures. How is this possible?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4325" LastEditDate="2015-03-06T00:49:12.150" LastActivityDate="2021-01-28T10:32:26.960" Title="What is the major product on chlorination of 2-methylbutane?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26950" PostTypeId="2" ParentId="26949" CreationDate="2015-03-05T15:16:08.903" Score="15" Body="&lt;p&gt;Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to R. Brückner &lt;em&gt;Advanced Organic Chemistry,&lt;/em&gt; the chlorination of isopentane has four monochlorination products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22 % 2-chloro-2-methylbutane&lt;br&gt;&#xA;33 % 2-chloro-3-methylbutane&lt;br&gt;&#xA;30 % 1-chloro-2-methylbutane&lt;br&gt;&#xA;15 % 1-chloro-3-methylbutane  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical factor is responsible for this.&lt;br&gt;&#xA;If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 hydrogen atom (8 %) → 2-chloro-2-methylbutane&lt;br&gt;&#xA;2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane&lt;br&gt;&#xA;6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane&lt;br&gt;&#xA;3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In terms per hydrogen atom,&lt;/strong&gt; the experimental yield is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;22 % / 1 = &lt;strong&gt;22 %&lt;/strong&gt; → 2-chloro-2-methylbutane&lt;br&gt;&#xA;33 % / 2 = &lt;strong&gt;16.5 %&lt;/strong&gt; → 2-chloro-3-methylbutane&lt;br&gt;&#xA;30 % / 6 = &lt;strong&gt;5 %&lt;/strong&gt; → 1-chloro-2-methylbutane&lt;br&gt;&#xA;15 % / 3 = &lt;strong&gt;5 %&lt;/strong&gt; → 1-chloro-3-methylbutane  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-03-05T15:16:08.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26951" PostTypeId="1" CreationDate="2015-03-05T16:06:27.873" Score="4" ViewCount="8194" Body="&lt;p&gt;I bought a new refrigerator that makes ice cubes. The instructions said to discard the first 3 full bins of ice in order to ensure the new water line is flushed clean. I dumped the full bins in my stainless steel sink each time. After the ice melted each time a chalky residue was left in the sink. I emptied a couple of extra full bins and still was left with the chalky residue. I cleaned the sink after each dumping to be certain there wasn't any left over residue from the previous dump. What is the chalky substance? Is it something to be concerned about or not? &lt;/p&gt;&#xA;" OwnerUserId="14795" LastEditorUserId="73" LastEditDate="2015-03-06T06:46:48.847" LastActivityDate="2015-12-11T14:36:29.050" Title="Chalky residue from melted ice cubes" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26952" PostTypeId="2" ParentId="26946" CreationDate="2015-03-05T16:37:02.883" Score="1" Body="&lt;p&gt;A couple of ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you think the diluent gas is non-flammable, you should be able to use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adiabatic_flame_temperature&quot; rel=&quot;nofollow&quot;&gt;adiabatic flame temperature&lt;/a&gt; of the gas when you burn it in air to estimate the composition.  You will need to carefully adjust your stove to give the optimal fuel/air blend (that maximizes the flame temperature).  Then you will need to accurately measure the flame temperature.  If you assume a heat capacity for the unknown diluent, you can estimate its concentration in the gas by comparing the measured temperature to the theoretical adiabatic flame temperature for pure propane burning in air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a vacuum pump to empty a container of known volume, and then allow the cooking gas into the container at ~1 atm pressure.  Measure the density of the gas and compare to the density of pure propane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you think the diluent gas does not contain carbon: hook up flow meters to the gas bottle (and an air or oxygen stream) to form a torch flame with known gas flow rate.  Collect all of the the exhaust produced by the flame, pass it through a cold trap to freeze out water vapor, and then pass the cold flame exhaust over a bed of anhydrous sodium hydroxide of known initial mass.  It will absorb the carbon dioxide from the exhaust.  Measure the mass increase of the sodium hydroxide.  By comparing the moles of gas that flow into the flame to the moles of CO2 absorbed (as calculated from the mass increase) you can estimate gas purity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find a spectrometer and record the IR or Raman spectrum of the gas!  This will probably be more accurate than any of the other methods.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-05T16:37:02.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26954" PostTypeId="1" AcceptedAnswerId="26957" CreationDate="2015-03-05T17:08:51.123" Score="2" ViewCount="4296" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nr1qi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this up until the last two steps. How does $\ln(K_\mathrm{eq})$ become $2{,}303\log(K_\mathrm{eq})$ and how does $-RT\cdot2{,}303$ become $-1{,}42$?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorDisplayName="user7951" LastEditDate="2017-12-26T10:13:52.660" LastActivityDate="2017-12-26T10:13:52.660" Title="Correlation between the equilibrium constant and delta G" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26955" PostTypeId="1" CreationDate="2015-03-05T17:11:45.207" Score="5" ViewCount="1070" Body="&lt;p&gt;I am confused with the last step of clozapine synthesis:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ppDiE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the amine group attack the amide group to form the amidine in such acidic environment? Is the intermolecular reaction between an amine and amide group feasible under not too extreme conditions?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7475" LastEditDate="2015-10-16T17:36:44.637" LastActivityDate="2015-10-16T17:36:44.637" Title="Attack of amine on amide" Tags="&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26956" PostTypeId="2" ParentId="26954" CreationDate="2015-03-05T17:42:27.197" Score="4" Body="&lt;p&gt;The first step is simple:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ - RT\ln {K_{{\text{eq}}}} \approx  - RT \times 2.303 \times \log {K_{{\text{eq}}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is terrible because $\text{kcal}$ is used and the units are missing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = 310{\text{ K}}$ (my guess)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;  \Delta G =  &amp;amp; \; {-} RT \times 2.303 \times \log {K_{{\text{eq}}}} \\ &#xA;   =  &amp;amp; \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K_{{\text{eq}}}} \\ &#xA;   =  &amp;amp; \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K_{{\text{eq}}}} \\ &#xA;\end{aligned} $$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-03-05T17:48:05.707" LastActivityDate="2015-03-05T17:48:05.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26957" PostTypeId="2" ParentId="26954" CreationDate="2015-03-05T17:43:22.390" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does ln(Keq) become 2,303log(Keq) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because $e^{2.303} = 10$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x) = \log (10^{\ln(x)})$ therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x)=\log((e^{2.303})^{\ln(x)})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x)=\log (x^{2.303})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln(x)=2.303\log(x)$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how does -RT*2,303 become -1,42?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would only be true at a particular temperature in a particular set of units&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-05T18:02:04.313" LastActivityDate="2015-03-05T18:02:04.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26958" PostTypeId="1" AcceptedAnswerId="26959" CreationDate="2015-03-05T17:49:52.140" Score="1" ViewCount="1619" Body="&lt;p&gt;The common route for aspirin synthesis (described &lt;a href=&quot;http://en.wikipedia.org/wiki/Aspirin#Synthesis&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) employs acetic anhydride and salicylic acid, in acidic conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step of such a reaction mechanism is the protonation of the carbonyl oxygen of acetic anhydride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why, instead, protonation of carbonyl oxygen in salycilic acid should not occur ? Of course, such a protonation is not desired, since it won't lead to aspirin: however, can it be neglected and why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any hints.&lt;/p&gt;&#xA;" OwnerUserId="3871" LastActivityDate="2015-03-05T18:13:22.223" Title="Why, during aspirin synthesis in acidic conditions, anydride carbonyl oxygen is protonated rather than salicylic acid carbonyl oxygen ?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26959" PostTypeId="2" ParentId="26958" CreationDate="2015-03-05T18:13:22.223" Score="3" Body="&lt;p&gt;Mechanisms are explanative, and not predictive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step of the reaction mechanism is protonation of the anhydride because that is the first step of the reaction.  That just is.  If the anhydride is not protonated, step 2 simply won't happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean that the salicyclic acid is not protonated.  It just means that the protonation of salicyclic acid is not necessary to explain how acetylsalicyclate forms.  Proton exchanges are at equilibrium.  It is likely that the salycilic acid is protonated during this process.  However, that protonated salicyclic acid may then protonate acetic anhydride and proceed as indicated in your link.  For the purposes of describing the mechanism, it is not necessary to distinguish where the proton that protonates the acetic anhydride comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your question really is self-answered, where you say, roughly, 'Such a protonation does not lead to product directly'.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-05T18:13:22.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26960" PostTypeId="2" ParentId="26955" CreationDate="2015-03-05T18:19:20.680" Score="3" Body="&lt;p&gt;There reaction environment would have to be acidic to protonate the amide, the amide is not natively electrophilic enough to be attacked by the amine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible point of concern is 'won't the nuclophilic amine be protonated?'  And the answer is absolutely.  But the protonation is a reversible reaction, and at equilbrium with a catalytic amount of acid (almost always TSA is used in catalytic quantities), the proton will be exchanged with various, some of which will lead to the condensation resulting in product.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-05T18:19:20.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26961" PostTypeId="2" ParentId="26951" CreationDate="2015-03-05T18:19:26.943" Score="4" Body="&lt;p&gt;It is possible that the white material is calcium carbonate or magnesium carbonate from your water source precipitating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if vinegar dissolves the white residue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the white residue dissolves in vinegar, it is probable calcium carbonate and/or magnesium carbonate.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-05T18:19:26.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26962" PostTypeId="2" ParentId="26827" CreationDate="2015-03-05T18:19:52.253" Score="3" Body="&lt;p&gt;The first thing to remember is that the &quot;shell&quot; model is an over-simplification of quantum chemistry. A better picture is that the electrons represent &quot;clouds&quot; (i.e., electron density).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question basically boils down to &quot;why don't 5th and 6th row atoms get much larger?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics can give a relatively easy formula for the &quot;average&quot; radius of a given hydrogen orbital&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle r \rangle = a_0 n^2 \left( 1 + \frac{1}{2} \left[ 1- \frac{l(l+1)}{n^2} \right] \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $a_0$ is the Bohr radius, the expected radius of the hydrogen atom, and $n$ is the principal quantum number and $l$ is the angular quantum number (e.g., l=0 for &quot;s&quot; orbitals and n=5 and 6 for Rb and Cs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty obvious that as n gets large for s orbitals (l=0), that formula approximates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle r \rangle \approx \frac{3 a_0}{2} n^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The left side becomes smaller and smaller, so the coefficient gets closer and closer to 3/2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a &lt;em&gt;hydrogen&lt;/em&gt; atom (one electron), the 5s and 6s orbitals in fact get very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is about &lt;em&gt;atoms&lt;/em&gt;. So we have to consider many electrons. Basically, this is another representation of the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19938/relativistic-effects-and-lanthanide-contraction&quot;&gt;lanthanide contraction&lt;/a&gt; caused by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26501/how-fast-do-electrons-move-around-the-nucleus/26535#26535&quot;&gt;relativistic effects&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, in that last link, there's a formula for the atomic radius &lt;em&gt;after&lt;/em&gt; relativistic corrections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7NRul.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atomic numbers 11 and 19, the relativistic correction is 0.99 (i.e., no real change to the radius). For atomic numbers 37 and 55, it's 0.963 and 0.916, respectively. So ~9% of the &quot;ideal&quot; radius is lost with $\ce{Cs}$.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-05T18:19:52.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26963" PostTypeId="1" AcceptedAnswerId="26971" CreationDate="2015-03-05T19:32:14.483" Score="6" ViewCount="1169" Body="&lt;p&gt;I've heard that most of the weight you &quot;lose&quot; is actually carbon dioxide released when you exhale. In percent its much more than what is released in pure heat energy and feces. Does anyone have statistics on what's the split between where the energy leaves your body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many percentage of food weight is actually released through carbon dioxide when its burned in your body? For example when you eat a hamburger and fries, is all the carbon released later when you exhale? How much is heat energy? Provided it does not go into fat reserves. &lt;/p&gt;&#xA;" OwnerUserId="14797" LastEditorUserId="7448" LastEditDate="2015-03-05T19:35:32.457" LastActivityDate="2019-12-18T13:17:25.777" Title="Is most of the weight you lose actually released as carbon dioxide when you exhale?" Tags="&lt;biochemistry&gt;&lt;energy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="26967" PostTypeId="2" ParentId="26864" CreationDate="2015-03-05T23:00:45.930" Score="3" Body="&lt;p&gt;OK, first off, let's comment on the stability question. When making a Lewis adduct (and rationalizing with the frontier orbital approach), you want the resulting complex to be stable. That requires two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The HOMO of the Lewis Base ($\ce{NH3}$ or $\ce{PH3}$ here) is low (i.e., hard to remove an electron).&lt;/li&gt;&#xA;&lt;li&gt;The coupling between the acid and base orbitals is &lt;strong&gt;large&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kU79n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want those electrons to be stabilized. Either (or both) approaches can work. So to correct your statement, you don't want a &quot;high HOMO&quot; or &quot;low LUMO.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catch, of course, is that a large splitting between the Lewis acid LUMO and Lewis base HOMO can lead to low coupling (poor orbital overlap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I went and ran some quick B3LYP/6-31G(d) calculations on both $\ce{N(CH3)3}$ and $\ce{P(CH3)3}$ because I couldn't find any quick comparisons in the textbooks I use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amine has a computed orbital energy of -5.64 eV, and the phosphine has a computed orbital energy of -5.98 eV. (Standard disclaimers about orbital energies being non-physical apply.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would suggest that the phosphine would form a stronger bond with a borane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's supporting evidence here: &lt;a href=&quot;http://dx.doi.org/10.1016/j.theochem.2003.06.001&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Ab initio molecular orbital study of the substituent effect on ammonia and phosphine–borane complexes&quot; &lt;em&gt;Journal of Molecular Structure: THEOCHEM&lt;/em&gt;, (2004) 709(1–3), pp 103-107.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The complexation energies of $\ce{H3BXH_{$3−n$}F_{$n$}}~(\ce{X~$=$~N, P}; n=0–3)$ and the proton affinities of $\ce{XH_{$3–n$}F_{$n$}}$ compounds have been investigated at the G2(MP2) level of theory. The G2(MP2) results show that the phosphine complexes are more stable than the corresponding ammonia ones. ... The NBO partitioning scheme shows that the stability of the phosphine complexes was related to the hyperconjugation effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, the phosphine complexes seem more stable because there is some back-donation of electrons from the borane into the phosphine. It's not just a simple frontier orbital picture - more orbital interactions come into play.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="4945" LastEditDate="2015-03-06T03:02:37.157" LastActivityDate="2015-03-06T03:02:37.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26968" PostTypeId="1" CreationDate="2015-03-05T23:43:32.277" Score="2" ViewCount="5079" Body="&lt;p&gt;I am looking to install a water filter system and I wanted to test the water for free and total chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand from the city water report that it uses chloramine and NO chlorine (&lt;a href=&quot;http://www.sandiego.gov/water/pdf/wq13.pdf&quot; rel=&quot;nofollow&quot;&gt;test report&lt;/a&gt;)and that chloramine content averages around 4 ppm in the water supply. I purchased one of these to test it out. &lt;a href=&quot;http://www.bulkreefsupply.com/insta-test-free-total-chlorine-strips-lamotte.html&quot; rel=&quot;nofollow&quot;&gt;http://www.bulkreefsupply.com/insta-test-free-total-chlorine-strips-lamotte.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above product can measure free and total chlorine ( 0, 0.5, 1, 3, 5, 10 ) PPMs. I was surprised to find out after using these test strips that city supply has 0 PPM for both free and total chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions&#xA;1. Any idea how I can test if these test strips are indeed working?&#xA;2. Anything wrong with the way I have tested?&lt;/p&gt;&#xA;" OwnerUserId="14802" LastEditorUserId="21964" LastEditDate="2016-08-18T16:39:17.570" LastActivityDate="2017-10-04T20:44:06.210" Title="Test for chlorine and chloramine in city water supply" Tags="&lt;water&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26969" PostTypeId="1" CreationDate="2015-03-06T01:52:54.263" Score="0" ViewCount="55" Body="&lt;p&gt;I'd like to know why my $\ce{TiO2}$ sol gel did not get coated nicely on my 200 gsm fiberglass? Here my solution, 5 g of tween 80, 40.15 ml isopropanol, 3.4 ml titanium isopropoxide, 0.62 ml sulphuric acid (0.02 mol).  Beside that, my $\ce{TiO2}$ sol gel can't calcinated at 500°C.  It will turn into powders and get burned into a black colour.  What should I do to overcome this problem?&lt;/p&gt;&#xA;" OwnerUserId="14804" LastEditorUserId="4325" LastEditDate="2015-03-06T02:10:14.167" LastActivityDate="2015-03-06T11:50:46.987" Title="Sulfur Doped Titanium Dioxide" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26970" PostTypeId="2" ParentId="26963" CreationDate="2015-03-06T02:28:28.670" Score="5" Body="&lt;p&gt;Yes, nearly all of the weight you &quot;lose&quot; when your body mass goes down is lost as $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infintessimally small -- essentially zero -- weight is lost from &quot;heat&quot;.  We have Einstein's theory that $E=c^2m$, but if you plug in the numbers, the mass change from burning off a pound of fat is &lt;strong&gt;&lt;em&gt;less than a microgram&lt;/em&gt;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your body could also lose weight as water, but in the long term, if you kept doing this, you'd dry up and die from dehydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to lose the weight is to &quot;burn off&quot; fat by converting it to $\ce{CO2}$ and exhaling the $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great article which goes through the math -- and which also shows that many dieticians and doctors do not understand this point! -- is this one: &lt;a href=&quot;http://www.bmj.com/content/349/bmj.g7257&quot; rel=&quot;nofollow&quot;&gt;http://www.bmj.com/content/349/bmj.g7257&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-12T15:45:32.100" LastActivityDate="2015-03-12T15:45:32.100" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="26971" PostTypeId="2" ParentId="26963" CreationDate="2015-03-06T02:57:20.810" Score="9" Body="&lt;p&gt;Let's start with:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much is heat energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;2000 kilocalories per day, considering $E=mc^2$, is &#xA;$9.3 \times 10^{-8}$ grams&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is all the carbon released later when you exhale? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  Brainstorming, the others ways would include: feces, urine, perspiration, flatulence, shedding skin, hair and nails, mucous from nose and mouth, for women: menstruation, vaginal discharge, lactation, and giving birth, and for men ejaculation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://ceadserv1.nku.edu/longa//haiti/kids/feces_value.html&quot; rel=&quot;noreferrer&quot;&gt;The Value of Human Feces&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per day the amount of carbon excreted in feces is 21 grams and in urine is 7 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2205022/&quot; rel=&quot;noreferrer&quot;&gt;The Composition of Human Perspiration&lt;/a&gt; Proc R Soc Med. 1934 May; 27(7): 839–848, a typical amount of perspiration a day is 700mL, and per 100mL the main carbon containing compounds are 20 mg urea, 20 mg glucose and 5mg amino acids.  So clearly less than 1 gram of carbon loss per day from perspiration. If we were considering loss of salts instead of just carbon, that would be about 3 grams per day from perspiration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For menstruation, blood loss is about 40mL per cycle, and &lt;a href=&quot;http://www.waterinfo.org/resources/water-facts&quot; rel=&quot;noreferrer&quot;&gt;blood is 92% water&lt;/a&gt;, so maybe 1 gram of carbon on a heavy day of a period.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skin loss per day is 1.5 grams according to &lt;a href=&quot;http://www.centreofthecell.org/centre/?page_id=118&quot; rel=&quot;noreferrer&quot;&gt;Cells for life, cells for an instant&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A highly controlled study having a person in a chamber for 24 hours and analyzing the air found 653g of CO2 or 178g carbon &quot;&lt;a href=&quot;http://books.google.com/books?id=hrm3AAAAIAAJ&amp;amp;pg=PA177&amp;amp;dq=weight%20of%20co2%20exhaled&amp;amp;source=gbs_toc_r&amp;amp;cad=3#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;A Respiration Calorimeter with Appliances for the Direct Determination of Oxygen&lt;/a&gt;&quot; Carnigie Institution of Washington Publication 42.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a person that is not lactating or giving birth, I would estimate about 30 grams per day of carbon loss is other than from exhaling carbon dioxide.  Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 86% of carbon loss is through exhaling carbon dioxide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this will certainly vary depending upon the amount of indigestible carbon such as fiber, in ones diet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if mass loss through nitrogen and phosphorous (but not hydrogen and oxygen) were also considered the loss through breathing would come down to about 80%.  Inclusion of salts would bring this percentage down further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of water from feces, urine and respiration is about 1.8 kg per day and the total mass emission of the body is 2.6kg, according to the Carnigie source above.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-06T16:53:59.270" LastActivityDate="2015-03-06T16:53:59.270" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26972" PostTypeId="2" ParentId="26844" CreationDate="2015-03-06T03:25:41.480" Score="2" Body="&lt;p&gt;I'm not sure about the first order, but I'll try to explain the reason for the second one. The acidic strength increases as $$\ce{HF&amp;lt;HCl&amp;lt;HBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is the bond strength between the hydrogen and the halogen atom. Since the bond between $\ce F$ and $\ce H$ is very strong, the molecule does not easily lose $\ce H^+$. On the other hand, as the size of the halogen atom increases, the overlapping of the orbitals between halogen and hydrogen becomes less efficient. Thus, $\ce {Cl}$ and $\ce {Br}$ can more easily lose $\ce {H^+}$ than $\ce {F}$. Hence, the acidic strength increases as size of halogen atom increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second reason is the stability of formation of conjugate base. When an acid loses $\ce {H^+}$ ion, it forms what is called as conjugate base. Similarly, when a base loses $\ce {OH^-}$ ion, it forms a conjugate acid. In case of $\ce {HF}$, we have $$\ce{HF &amp;lt;=&amp;gt; H^+ +F- }$$&#xA;Here, $\ce {F^-}$ is the conjugate base. Now, due to small size of $\ce {F}$, it is difficult for $\ce {F-}$ to hold negative charge. There are strong repulsions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for $\ce {HBr}$, we have $$\ce{HBr &amp;lt;=&amp;gt; H^+ +Br^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {Br}$ has a greater atomic radius. It is also less electronegative than $\ce {F}$. Hence, the repulsions are less in $\ce {Br-}$. Thus, $\ce {Br^-}$ is quite stable. The more stable is the conjugate base, the more readily the acid loses $\ce {H^+}$ ion, and so, more is its acidic strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/09/22/five-key-factors-that-influence-acidity/&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; link. Maybe you'll find it useful.&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="4945" LastEditDate="2015-03-12T09:36:22.297" LastActivityDate="2015-03-12T09:36:22.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26973" PostTypeId="1" AcceptedAnswerId="26976" CreationDate="2015-03-06T03:51:48.067" Score="3" ViewCount="150" Body="&lt;p&gt;An aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$0.00$&lt;/span&gt; if &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}] = 1.00 \; \mathrm{M}$&lt;/span&gt;. I tried to see if an aqueous solution of acetic acid could have a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$0.00$&lt;/span&gt;, given that its &lt;span class=&quot;math-container&quot;&gt;$K_a$&lt;/span&gt; value is &lt;span class=&quot;math-container&quot;&gt;$1.8 \times 10^{-5}$&lt;/span&gt;. I tried finding the concentration &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt; of such a solution using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{C2H3O2^-}]}{[\ce{HC2H3O2}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then using molar mass and density conversions to determine whether such an aqueous solution exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense to say that such a solution does not exist if the amount of substance of acetic acid in 1 liter of solution actually take up more than 1 liter of volume?&lt;/p&gt;&#xA;" OwnerUserId="12745" LastEditorUserId="4945" LastEditDate="2019-11-04T10:53:07.333" LastActivityDate="2019-11-04T10:53:07.333" Title="Impossible pH For Aqueous Weak Acid Solution" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26974" PostTypeId="2" ParentId="26914" CreationDate="2015-03-06T03:56:38.183" Score="0" Body="&lt;p&gt;I would assume that it is just a typo and that it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are still not sure, try this: do the same problem again, but substitute a different weak acid (and be sure to use the new $K_a$ value!). Obviously the numbers will be different, but make sure that the procedure still makes sense and that you do not second-guess yourself here.&lt;/p&gt;&#xA;" OwnerUserId="12745" LastActivityDate="2015-03-06T03:56:38.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26975" PostTypeId="2" ParentId="4105" CreationDate="2015-03-06T04:16:08.353" Score="0" Body="&lt;p&gt;you can use also H++ server just go to this link;  &lt;a href=&quot;http://biophysics.cs.vt.edu/&quot; rel=&quot;nofollow&quot;&gt;http://biophysics.cs.vt.edu/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14805" LastActivityDate="2015-03-06T04:16:08.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26976" PostTypeId="2" ParentId="26973" CreationDate="2015-03-06T05:55:56.250" Score="6" Body="&lt;p&gt;Your reasoning is fine. But a major caveat (or additional factor) is that the pH scale itself no longer applies to concentrated solutions (say, more than 1 molar in acid) where the molarity of water falls significantly below 55M (pure water). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can extend the pH scale down below 0 by empirically measuring how effective an acid solution is at protonating a weak base, but the scale deviates from the straight line predicted by the concentration of acid present.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-06T05:55:56.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26978" PostTypeId="2" ParentId="26844" CreationDate="2015-03-06T06:19:22.017" Score="2" Body="&lt;p&gt;$$\ce{HOCl&amp;gt;HOBr&amp;gt;HOI}$$&#xA;$\mathrm{p}K_a(\ce{HOCl})=7.5$,  $\mathrm{p}K_a(\ce{HOBr})=8.6$,  $\mathrm{p}K_a(\ce{HOI})=10.6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{H-O}$ bond  in those oxo-acides ionizes more readily when the oxygen atom is bonded to a more electronegative atom. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The conjugate base $\ce{OCl-}$ is more stable than $\ce{OBr-}$. Because $\ce{Cl}$ being more electronegative, is better able to accommodate the negative charge.&lt;/li&gt;&#xA;&lt;li&gt;The conjugate base $\ce{OBr-}$ is more stable than $\ce{OI-}$. Because $\ce{Br}$ being more electronegative, is better able to accommodate the negative charge.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HI&amp;gt;HBr&amp;gt;HCl\gg HF}$$&#xA;$\mathrm{p}K_a(\ce{HF})=3.1$, $\mathrm{p}K_a(\ce{HCl})=-6.0$, $\mathrm{p}K_a(\ce{HBr})=-9.0$, $\mathrm{p}K_a(\ce{HI})=-9.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ are all strong acids, whereas $\ce{HF}$ is a weak acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As halogen size increases (as we move down the periodic table), $\ce{H-X}$  bond strength decreases and acid strength increases (a weaker bond produces a stronger acid, and vice versa).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2015-03-12T09:33:22.747" LastActivityDate="2015-03-12T09:33:22.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26979" PostTypeId="1" AcceptedAnswerId="26993" CreationDate="2015-03-06T07:44:56.897" Score="9" ViewCount="6002" Body="&lt;p&gt;In Organic Chemistry (Wade) there is a question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following compounds would give a positive Tollens' test? (Remember that the Tollens' test involves mild basic aqueous conditions.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers manual says this compound - $\ce{CH3-(CH2)3-CH(OCH3)2}$ - gives a negative result as it is an &quot;acetal stable in base: no reaction&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I would have thought this is positive. Earlier in the chapter it's taught that since the Tollens reagent removes any aldehyde present, even if at equilibrium the amount of aldehyde is very small, the Tollens reagent will nevertheless predominate due to Le Chatelier's principle (the small amount of aldehyde will be removed slowly but surely, until eventually all of the solution is converted to aldehyde and then to the tollens product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, whilst I respect that the acetal product may be by far the major product, surely a very, very small percentage of the acetal is nevertheless converted to the aldehyde, via protonation of a methoxy group, which leaves, and then attack by $\ce{H2O}$, followed by deprotonation of $\ce{H2O -&amp;gt; OH}$, and then protonation of the other methoxy group, which leaves, leaving a carbocation, which the $\ce{OH}$ attacks forming a double bond and a positive charge on the $\ce{OH}$, followed by deprotonation on the $\ce{OH}$  giving an aldehyde?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be unlikely, but according to Le Chatelier's principle, it just needs to occur at some small quantity in order to be continuously removed by the tollens reagent, and thus ends up predominating and removing all of the acetal product given enough time? Why is this not positive?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="5026" LastEditDate="2018-02-24T13:22:40.657" LastActivityDate="2018-02-24T13:22:40.657" Title="Does an acetal give a positive Tollens test?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="26980" PostTypeId="1" CreationDate="2015-03-06T10:56:07.137" Score="3" ViewCount="74" Body="&lt;p&gt;I'm performing the reaction between acetone and 2-nitrobenzaldehyde catalysed by (L)-Proline, as answered to one of my questions a while ago: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16421/simple-experiments-involving-enantiomer-synthesis&quot;&gt;Simple experiments involving enantiomer synthesis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, while the literature specified using 1 mmol of (L)-Proline in 20ml of solution, due to the cost of the materials and the budget available, I've only been able to use a fraction of that, around 0.01mmol. I've scaled down the entire experiment to 5 ml to try and compensate for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is, I've left this solution reacting for a long period of time (12+ hours) and I'm not seeing any optical activity at all. With the reduced concentrations, I'd have expected to see a lot less activity than the original literature specifies, but to see at least some indication that my reaction is working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the concentrations of (L)-Proline simply too small for any significant reaction to take place? Or can there be some other reason why my reaction isn't working? Note: I'm a chemistry 'newbie', so don't assume anything!&lt;/p&gt;&#xA;" OwnerUserId="7892" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-05-30T23:55:23.553" Title="Enantioselective organic reaction not working" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="26981" PostTypeId="2" ParentId="26969" CreationDate="2015-03-06T11:21:26.340" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Regarding why your ceramic material did not get coated nicely on the fiberglass, I think there's a problem of adherence between the fibers and the the ceramic. You need to prepare the surface of the fibers by etching for example before coating. They are often shielded.&lt;/li&gt;&#xA;&lt;li&gt;Regarding why your  sol-gel can't be calcinated at $500°\mathrm{C}$. You need to figure out the proper temperature of calcination by monitoring the phase appearance of the ceramic materials by XRD diffractometry. You start with the XRD diffractogram of your as-prepared sample(normally, it should be amorphous). Then, you treat your sample at different  temperatures within different periods of times (you have two parameters to find: Temperature and Time). Unless you are working on a published article that has addressed all these issues. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-03-06T11:50:46.987" LastActivityDate="2015-03-06T11:50:46.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26991" PostTypeId="2" ParentId="24188" CreationDate="2015-03-06T14:45:02.793" Score="1" Body="&lt;p&gt;Ok I will answer your  last question first. If you have two identical electrodes(i.e. same electrode material), the cell potential will be zero and there won't be any issues there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to use inert electrodes so that you are not measuring the dissolution of the electrode itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one part I am not sure you understand properly is the current growing. You want to vary the potential until the current raises exponentially when continuing to sweep. This will be a very sharp rise that will be hard to miss. &lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-03-06T14:45:02.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26992" PostTypeId="1" CreationDate="2015-03-06T15:22:51.417" Score="1" ViewCount="75" Body="&lt;p&gt;If I have three thermochemical reactions in a series(in order to find the enthalpy change) how can I cancel an element having not included in the reaction that's being asked about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and only mentioned in one of the reactions ( either products or reactants)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g: $$\ce{NO + O3 \rightarrow NO2 + O2}$$&#xA;     $$\ce{O3 \rightarrow 3/2 O2}$$&#xA;     $$\ce{O2 \rightarrow 2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The reaction that is being asked&lt;/strong&gt;: $$\ce{2NO2 \rightarrow 2NO + O2}$$&lt;/p&gt;&#xA;" OwnerUserId="14688" LastEditorUserId="12700" LastEditDate="2015-03-12T15:45:53.667" LastActivityDate="2015-03-12T15:45:53.667" Title="How to cancel once-mentioned element in thermochemical reactions" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26993" PostTypeId="2" ParentId="26979" CreationDate="2015-03-06T15:24:55.547" Score="9" Body="&lt;p&gt;The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids&lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt; [Evans pKa table]&lt;/a&gt;. In basic solution, water (pKa 16) is the strongest acid available to protonate the acetal. The protonated acetal will have a pKa close to protonated methanol (pKa -2). So the equilibrium for that protonation will lie to the water side by approximately $\ce{10^18}$. As Jori is alluding to in the comment, the rate moving forward is going to be exceedingly slow because the concentration of the protonated intermediate will be exceedingly low. This is where the rule of thumb that acetals are stable to base comes from. See also: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19318/why-are-acetals-stable-to-bases-and-nucleophiles/&quot;&gt;Why are acetals stable to bases and nucleophiles?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Tollen's test was performed under acidic conditions, then you would be correct that the acetal would test positive for an aldehyde. A hemi-acetal would also test positive.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2017-04-11T07:33:47.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26995" PostTypeId="1" AcceptedAnswerId="27000" CreationDate="2015-03-06T16:04:12.810" Score="3" ViewCount="692" Body="&lt;p&gt;&lt;em&gt;(This is an interdisciplinary question, between molecular chemistry, materials science &amp;amp; &lt;a href=&quot;https://diy.stackexchange.com/questions/15375/what-type-of-concrete-sealer-to-use-to-prevent-frost-heaves-in-my-driveway&quot;&gt;construction&lt;/a&gt;. I think Chemistry SE is the right fit, but please move it if needed.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silane/siloxane solutions are applied to porous building materials (e.g. brick or concrete) to repel water. Unlike a solid coating, these sealants are &quot;&lt;a href=&quot;https://www.google.com/search?q=silane+OR+siloxane+breathable&quot; rel=&quot;nofollow noreferrer&quot;&gt;breathable&lt;/a&gt;&quot;, allowing water vapor to exit instead of being trapped, which would cause slow structural damage. How does that work, molecularly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume these compounds affect surface tension in some way, reducing the ability of liquid water to permeate, without completely sealing the pores, but I don't understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optional: also, how does this sealant affects the permeability of other liquids (polar vs non?) and other gases?&lt;/p&gt;&#xA;" OwnerUserId="14813" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:17.580" LastActivityDate="2017-08-23T11:42:42.660" Title="How does &quot;breathable&quot; siloxane water sealant work?" Tags="&lt;water&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26996" PostTypeId="1" AcceptedAnswerId="27001" CreationDate="2015-03-06T16:09:51.090" Score="4" ViewCount="1882" Body="&lt;p&gt;&lt;em&gt;This question is purely out of curiosity and to help me remember the &quot;why&quot; of a mechanism to help me remember the &quot;what/how&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In organic chemistry we recently studied the  S&lt;sub&gt;n&lt;/sub&gt;2 Substitution of Primary Alcohol with PBr&lt;sub&gt;3&lt;/sub&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w4Qx0.png&quot; alt=&quot;Originally from Wikipedia PBr3_alcohol_rxn.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why is it favorable for PBr&lt;sub&gt;3&lt;/sub&gt; to act as a Lewis acid (i.e. accept electrons from the oxygen to form a bond) when it already has a full octet and no formal charge?&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="13384" LastEditDate="2015-03-09T14:25:13.723" LastActivityDate="2015-05-19T04:04:54.680" Title="Why does phosphorus tribromide act as a Lewis acid (electron acceptor)?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26997" PostTypeId="2" ParentId="26992" CreationDate="2015-03-06T16:31:42.973" Score="2" Body="&lt;p&gt;We can make use of this formula: $$\Delta H = \Sigma ~\Delta H_f~(\text{products}) - \Sigma ~\Delta H_f~(\text{reactants}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H_f$ for $\ce{O3}$. If you're interested, &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29&quot; rel=&quot;nofollow&quot;&gt;wikipedia has documented&lt;/a&gt; some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H_f = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that the $\Delta H_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H_f \ce{(NO2)} + 0) - (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from the equation above we can conclude that $$x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now write the master formula for the last reaction: $$(2 * \Delta H (\ce{NO}) + 0) - 2 * \Delta H_f (\ce{NO2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did we conclude from the last reaction? $x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H_f$ for $\ce{NO}$ &amp;amp; $\ce{NO2}$?&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-03-06T16:31:42.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="26998" PostTypeId="2" ParentId="26992" CreationDate="2015-03-06T16:41:54.660" Score="2" Body="&lt;p&gt;If you're just interested in the reaction enthalpy of the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NO2 &amp;lt;=&amp;gt; 2NO + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and one set of equations for the calculation of $\Delta H_f(\mathrm{products})$ and/or $\Delta H_f(\mathrm{reactants})$ doesn't work out, just use another one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It does not matter! The Hess Law is invariant of the path.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-06T16:41:54.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="26999" PostTypeId="1" CreationDate="2015-03-06T16:53:06.120" Score="0" ViewCount="764" Body="&lt;p&gt;I've been struggling with the various methods of purification of organic compounds. If a question comes up in my examination, asking me to explain steam distillation in my own words, I'd like to know if this applies and if this is accurate.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;i) Steam distillation is utilized to separate volatile constituents which are immiscible with water but steam volatile.&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;ii) Steam from a steam generator is passed through a heated flask which contains the liquid to be distilled.&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;iii) The volatile component separates and thusly, one obtains a mixture of steam and the volatile organic compound.&lt;/h2&gt;&#xA;&#xA;&lt;h2&gt;iv) This is then condensed and separated by means of a separatory funnel.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In steam distillation, the liquid boils when the sum of vapour pressured due to organic liquid and that due to water become equal to the atmospheric pressure. &#xA;i.e Patm=Pw+Pc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $Pc&amp;lt;P$ ; liquid vaporises at temperature lower than its boiling point. That is, if water insoluble substance and water are the components, the mixture boils at a temperature close to, but below 373K. &#xA;Example: Aniline-Water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please be highly specific. There should be no error, however minute.&#xA;Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="7448" LastEditDate="2015-03-06T17:08:37.960" LastActivityDate="2015-03-06T20:47:00.117" Title="Is the following paragraph about steam distillation accurate?" Tags="&lt;organic-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27000" PostTypeId="2" ParentId="26995" CreationDate="2015-03-06T18:11:26.853" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.dowcorning.com/content/discover/discoverchem/si-repellents.aspx&quot; rel=&quot;nofollow&quot;&gt;How silicone water repellents work&lt;/a&gt;, &#xA;&quot;Because silicones have a very open molecular structure, water vapor trapped in the substrate can easily slip through. Water molecules, however, are too large to enter.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, siloxanes' Si-O bonds are not &lt;em&gt;intrinsically&lt;/em&gt; hydrophobic, but the long molecular chains with outward-facing hydrophobic methyl groups, and thermal molecular motion, prevent liquid water from seeping through, which presents less of a barrier to water vapor and other low-molecular-weight gases.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-06T18:11:26.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27001" PostTypeId="2" ParentId="26996" CreationDate="2015-03-06T18:30:03.863" Score="3" Body="&lt;p&gt;The &lt;em&gt;why&lt;/em&gt; of a mechanism is always the best approach.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons for this attack are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{P}$ can expand its octet due to presence of vacant $d$-orbitals.  &lt;/li&gt;&#xA;&lt;li&gt;The $\ce{P-Br}$ bond is polar due to the electronegativity difference between the atoms and hence P develops partial positive charge favouring a nucleophilic attack on it.    &lt;/li&gt;&#xA;&lt;li&gt;Looking at the molecule as a whole, $\ce{Br}$ is a good leaving group so $\ce{P}$ doesn't mind a nucleophilic attack on it by the lone pair of oxygen.  &lt;/li&gt;&#xA;&lt;li&gt;&quot;Nature loves stability&quot; and  $\ce{P-O}$ bond is stronger than $\ce{P-Br}$ so that might be the driving force for this reaction mechanism.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="15410" LastEditDate="2015-05-19T04:04:54.680" LastActivityDate="2015-05-19T04:04:54.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27002" PostTypeId="1" CreationDate="2015-03-06T18:45:12.780" Score="1" ViewCount="619" Body="&lt;p&gt;I'm completely stuck on this one - I was doing an experiment in a lab and now I cannot find the molarity of an ion in a solution. I've looked everywhere on the web but after 2 hours of searching I couldn't find anything. Maybe I am looking in the wrong direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mixed three solutions: 1mL of $\ce{0.1M ~CuSO4}$,  $\ce{0.28 M ~~NH3}$ and  water.&#xA;This solution turned into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cu(H2O)4^{2+} + 4NH3 = Cu(NH3)4^{2+} + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turned the solution from light blue to dark blue-ish purple. The final mL of the mixed solutions was 25 mL.&#xA;Now I am supposed to calculate the molarity of the $\ce{Cu(NH3)4^{2+}}$ ion. I am pretty sure that I only have to calculate the molarity of $\ce{Cu^{2+}}$ but I am not entirely sure. If you guys could give me some help on this I would be very grateful.  &lt;/p&gt;&#xA;" OwnerUserId="14815" LastEditorUserId="14815" LastEditDate="2015-03-07T20:52:47.477" LastActivityDate="2015-03-07T20:52:47.477" Title="Calculating the Molarity of an Ion?" Tags="&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;&lt;ions&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27003" PostTypeId="2" ParentId="4105" CreationDate="2015-03-06T19:06:35.810" Score="3" Body="&lt;p&gt;This is a challenging problem and there are &lt;em&gt;many&lt;/em&gt; programs that do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One review was published a few years ago: &lt;a href=&quot;http://dx.doi.org/10.1021/ci900289x&quot; rel=&quot;nofollow&quot;&gt;&quot;Comparison of Nine Programs Predicting pKa Values of Pharmaceutical Substances&quot;, &lt;em&gt;J. Chem. Inf. Model.&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt; (12), pp 2801–2812&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Knowledge of the possible ionization states of a pharmaceutical substance, embodied in the pKa values [...] is vital [...] We compare nine commercially available or free programs for predicting ionization constants. [...] We compared their performances by applying them to 197 pharmaceutical substances with 261 carefully determined and highly reliable experimental pKa values from a literature source. The programs ADME Boxes 4.9, ACD/pKa DB 12.0, and SPARC 4.2 ranked as the top three with mean absolute deviations of 0.389, 0.478, and 0.651 and $r^2$ values of 0.944, 0.908, and 0.894, respectively. ACD/pKa DB 12.0 predicted all sites, whereas ADME Boxes 4.9 and SPARC 4.2 failed to predict 5 and 18 sites, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, the quantum program did poorly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's even a full book on the subject: &lt;a href=&quot;http://www.crcpress.com/product/isbn/9781466508781&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Computational Approaches for the Prediction of pKa Values&lt;/em&gt;&lt;/a&gt; by George Shields and Paul Seybold. It covers both quantum and QSAR approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the best seem to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.acdlabs.com/products/percepta/predictors/pka/&quot; rel=&quot;nofollow&quot;&gt;ACD/pKa&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.simulations-plus.com/Products.aspx?pID=13&amp;amp;lID=41&quot; rel=&quot;nofollow&quot;&gt;ADMET Predictor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://propka.org&quot; rel=&quot;nofollow&quot;&gt;Propka&lt;/a&gt; Also available as open source code: &lt;a href=&quot;https://github.com/jensengroup/propka-3.1&quot; rel=&quot;nofollow&quot;&gt;https://github.com/jensengroup/propka-3.1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemaxon.com/products/calculator-plugins/property-predictors/&quot; rel=&quot;nofollow&quot;&gt;Marvin / ChemAxon&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But there are &lt;em&gt;dozens&lt;/em&gt; out there. I'd check the review articles and try a few.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-06T19:06:35.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27005" PostTypeId="2" ParentId="26999" CreationDate="2015-03-06T20:47:00.117" Score="1" Body="&lt;p&gt;You have it right on all counts. But, to clarify...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) Steam distillation is used to separate steam-volatile organic components of a mixture from non-volatile ones (not from each other, however). It is most useful for higher boiling, relatively non-polar organic compounds that are immiscible with water.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-06T20:47:00.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27006" PostTypeId="1" CreationDate="2015-03-06T20:58:53.947" Score="0" ViewCount="1933" Body="&lt;p&gt;Can we make Gold in a Lab? Is it possible to make changes in metals that will create Gold? and How?&#xA;I have tried to search something related to this but didn't get any good results.&#xA;Thanks in Advance&lt;/p&gt;&#xA;" OwnerUserId="14817" LastEditorUserId="7448" LastEditDate="2015-03-06T21:02:41.360" LastActivityDate="2015-03-06T21:08:09.670" Title="Can we make Gold in a Lab?" Tags="&lt;atoms&gt;&lt;metal&gt;&lt;nuclear&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-03-06T22:55:13.477" ContentLicense="CC BY-SA 3.0" />
  <row Id="27007" PostTypeId="2" ParentId="27006" CreationDate="2015-03-06T21:08:09.670" Score="2" Body="&lt;p&gt;Yes, gold has been made from mercury. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &quot;&lt;a href=&quot;http://journals.aps.org/pr/abstract/10.1103/PhysRev.60.473&quot; rel=&quot;nofollow&quot;&gt;Transmutation of Mercury by Fast Neutrons&lt;/a&gt;&quot; Phys. Rev. 60, 473&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-06T21:08:09.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27008" PostTypeId="1" AcceptedAnswerId="27013" CreationDate="2015-03-06T22:24:49.333" Score="9" ViewCount="4321" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;Failure to locate stationary point, SCF has not coverged&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the most annoying message after Optimization run ( even with a good primary start geometry like UFF we ay see that) .&#xA;The only solution I know is to use the output log file for another run but it is very time consuming.&#xA;Is there any way to force GAMESS to automatically do that ?( I mean makes another automatic run on current output results ).&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="7448" LastEditDate="2015-03-07T12:51:47.363" LastActivityDate="2015-11-10T11:24:53.187" Title="How can you manage SCF convergence problems?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27011" PostTypeId="1" CreationDate="2015-03-07T02:02:56.453" Score="0" ViewCount="1438" Body="&lt;p&gt;When developing the chromatography paper, what chemical will you use to spray on the chromatography paper?&lt;/p&gt;&#xA;" OwnerUserId="14819" LastEditorUserId="402" LastEditDate="2015-11-24T01:10:42.547" LastActivityDate="2015-11-24T01:10:42.547" Title="What is sprayed on chromatography paper when developing?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-24T16:50:04.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="27012" PostTypeId="1" CreationDate="2015-03-07T02:15:48.030" Score="2" ViewCount="45" Body="&lt;ol&gt;&#xA;&lt;li&gt;I'm seeking clarification as I believe this to be the answer to why boron compounds are electron deficient.&#xA;While  bonding, the $2s^2$ electrons of Boron are excited to the $2p_y$ orbital, they engage in bonding. However, even after bonding, there exists an empty $2p_z$ which leaves an incomplete shell, thus these compounds require two more electrons and so, they are electron deficient.&#xA;(As in the case of $\ce{BCl3}$)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CO2}$ cannot put out fire caused by the reaction of saline hydrides and water as it will be converted into formate ions during the process. (although, I'd appreciate some help on chemical equations if possible)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moreover, if this is accurate, does the same logic apply to $\ce{NH3}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grade 11. &lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="1878" LastEditDate="2015-03-07T16:10:08.130" LastActivityDate="2015-03-07T16:32:27.690" Title="Am I right in giving the following reason for some questions relating to the p-block?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27013" PostTypeId="2" ParentId="27008" CreationDate="2015-03-07T02:56:25.290" Score="15" Body="&lt;p&gt;There can be &lt;em&gt;many&lt;/em&gt; reasons for the SCF to not converge.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The initial guess wasn't great, and the convergence is slow. Increasing the number of SCF iterations can help here.&lt;/li&gt;&#xA;&lt;li&gt;The SCF oscillates. This can happen if there are two orbitals close in energy. Using damping, mixing (i.e., the next iteration is 50% the old and 50% the updated iteration), Fermi broadening, etc. can help here.&lt;/li&gt;&#xA;&lt;li&gt;The SCF strategy (e.g., DIIS) is causing problems. Usually DIIS can greatly accelerate convergence, but in some hard-to-converge systems, you need a lot of iterations with another strategy (e.g., Newton-Raphson)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Without more information from your particular file, it's impossible to know how to help. Many times, you may simply want to increase the maximum number of SCF iterations and it'll work fine.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-07T02:56:25.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27015" PostTypeId="1" AcceptedAnswerId="27040" CreationDate="2015-03-07T04:54:48.313" Score="4" ViewCount="1218" Body="&lt;p&gt;I've been having trouble with reducing COD (Chemical Oxygen Demand) below approx. 500 ppm by using Fentons Reagent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 100 gm of the sample (initial COD 3000 ppm) I use 5 gm of 30% $\ce{H2O2}$ &amp;amp; 0.1 gm of $\ce{FeSO4}$. pH is adjusted to between 3 to 5. Temp. is approx. Room Temp. &amp;amp; I've tried reaction times up to 6 hrs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these sound reasonable choices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice? &lt;/p&gt;&#xA;" OwnerUserId="12803" LastEditorUserId="4231" LastEditDate="2015-03-07T14:49:29.387" LastActivityDate="2017-01-31T20:06:46.373" Title="Fentons Reagent tips: H₂O₂ + FeSO₄" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27016" PostTypeId="1" AcceptedAnswerId="27017" CreationDate="2015-03-07T06:31:37.147" Score="3" ViewCount="1136" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDTAA.png&quot; alt=&quot;Question regarding predicting the color of pH indicator in different solutions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have is above. Could someone explain why the answer is (A)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the sodium hydroxide turn the indicator blue because it is a strong base and hence it will react completely to form water,&#xA;shifting equilibrium really far to the right?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the HCl turn the indicator red for a similar reason?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;I am unsure of the sodium acetate effect&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14824" LastEditorUserId="83758" LastEditDate="2020-04-26T06:22:16.467" LastActivityDate="2020-04-26T06:22:16.467" Title="How to predict the color of a pH indicator using Le Chatelier's principle" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27017" PostTypeId="2" ParentId="27016" CreationDate="2015-03-07T06:52:18.650" Score="2" Body="&lt;p&gt;&lt;strong&gt;You're absolutely on the right track!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The keys to the question are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;to  order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a &lt;strong&gt;part&lt;/strong&gt; of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. &lt;strong&gt;Some amount&lt;/strong&gt; of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is all about the position of the equilibria.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-07T07:05:26.667" LastActivityDate="2015-03-07T07:05:26.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27018" PostTypeId="1" CreationDate="2015-03-07T07:16:28.403" Score="4" ViewCount="423" Body="&lt;p&gt;Often we're concerned with changes in thermodynamic properties during reactions, like $\Delta H_p = Q_{reaction}$ (change in enthalpy for an isobaric process). What seems odd to me is that in the derivation of this relationship,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_p = \Delta U + P\Delta V=Q-W+P\Delta V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we assume the work is given by $W=P\Delta V$, which requires that the process is reversible in the thermodynamic sense. Then the change in enthalpy simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_p=Q - P\Delta V + P\Delta V=Q$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposedly, measuring the heat transfer during a reaction at constant pressure will give the enthalpy. Even though enthalpy is a state function in general, our case requires that the process is reversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent does the restriction of thermodynamic reversibility apply to the reaction? Do we have to assume the reaction happens slowly and can be undone, leaving only an infinitesimal change in the universe? Or does it not matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if any of my points are wrong. Classical thermodynamics usually doesn't mention &lt;em&gt;kinds&lt;/em&gt; of matter, so I'm a bit confused when chemistry comes into it.&lt;/p&gt;&#xA;" OwnerUserId="14825" LastActivityDate="2015-05-31T19:06:13.823" Title="Relationship between thermodynamic reversibility and reactions" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27020" PostTypeId="1" AcceptedAnswerId="27023" CreationDate="2015-03-07T10:07:17.920" Score="5" ViewCount="33381" Body="&lt;p&gt;Is there any particular reason that peaks in proton NMR can appear with a larger integral ratio than expected, that is to say one specific proton absorption stretches over a wide ppm range? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a very broad singlet that is integrating to nearly 5x expected. The rest of the spectrum is fine in terms of integrating and the position of the peaks. This particular proton is a singlet from 9.95 to 10.25 ppm and is the alpha proton of a carbene hydrogen chloride salt. Either that or it is impure, but so far my spectrum is very clean whenever I run it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks! :)  &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2016-09-15T14:13:32.967" Title="1H NMR Broad peaks" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27021" PostTypeId="1" AcceptedAnswerId="27059" CreationDate="2015-03-07T10:29:19.377" Score="1" ViewCount="79" Body="&lt;p&gt;At $t = 0$ we have 10 moles of $\ce{NO2}$ and 10 moles of $\ce{N2O4}$ satisfying this reaction in a 0.75 liter tank (the barrier is closed at this time):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2O4 &amp;lt;=&amp;gt; 2NO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is in thermal contact with the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At time t=t0 the barrier is released and the gases are free to expend to the other tank ($V_2=1$ liter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hjbLu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is what will be the concentration of $\ce{NO2}$ when they reach equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I tried is calculating the chemical equilibrium before based on the concentration quotient $\text{K}_c$ (13.33) and then to use it as reference after the system reaches equilibrium. I got a weird result (negative number) Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastEditorUserId="865" LastEditDate="2021-05-22T18:58:51.773" LastActivityDate="2021-05-22T18:58:51.773" Title="Concentration of expanding gases" Tags="&lt;equilibrium&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27023" PostTypeId="2" ParentId="27020" CreationDate="2015-03-07T13:17:00.487" Score="4" Body="&lt;p&gt;One reason for broadening is chemical exchange.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the proton you are referring to is in exchange with another population of protons, for example from a solvent, the integrated area could represent the total population that is exchanging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordination of a paramagnetic metal ion is another source of broadening.  Integrated area should be unaffected if there is sufficient time between each scan.  However, since the paramagnetic metal ion can cause faster relaxation of nearby protons, it is possible that the integration is distorted by protons near the paramagnetic ion fully (or closer to fully) relaxing between scans, while the other protons do not.  This effect is often used intentionally to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0022236483901099&quot; rel=&quot;nofollow&quot;&gt;enhance detection of protons near a paramagnetic metal center&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-09-15T14:13:32.967" LastActivityDate="2016-09-15T14:13:32.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27024" PostTypeId="2" ParentId="27020" CreationDate="2015-03-07T14:43:39.710" Score="5" Body="&lt;p&gt;The most common process that can cause both peak broadening &lt;strong&gt;and&lt;/strong&gt; an increase in the value of a peak integral is &lt;a href=&quot;http://www.bioc.aecom.yu.edu/labs/girvlab/nmr/course/COURSE_2012/Lecture_ExchangeDynamics_2012.pdf&quot; rel=&quot;noreferrer&quot;&gt;exchange broadening&lt;/a&gt; between two different protons.  The following diagram represents a typical case where the protons in water are exchanging with the hydroxyl proton in an alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kZN1r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rate of exchange is fast on the nmr timescale, then a single, sharp peak of increased amplitude and a weighted average of the individual chemical shifts is observed.  If the rate of exchange is slow compared to the nmr timescale, then two distinct peaks, both with accurate integrals, are seen.  In the case where the rate of exchange is comparable to the timescale of the nmr experiment, then a broadened signal with integral intensity equal to the two combined signals and located at a weighted average of the two separate signals is observed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way you can test to see if a small amount of water in your sample is the culprit is as follows.  Add a small amount of $\ce{H2O}$ to your sample and see if the peak for your sample shifts to a new location and the integral value further increases.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-07T18:06:16.620" LastActivityDate="2015-03-07T18:06:16.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27025" PostTypeId="2" ParentId="7541" CreationDate="2015-03-07T15:09:48.143" Score="0" Body="&lt;p&gt;&quot;When you touch water or a water wetted material part of the water molecules are transfer to your hands (an hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be cause however in some case of parasthesias even if there are no liquid particles. In fact there are no &quot;wet receptor&quot; over your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in you skin.&quot;  Let's see, the reason that water molecules are transferred to your hands is because your hands are made of a 'water-loving' material (i.e., 'hydrophilic').  Sounds like we haven't progressed beyond Moliere (opium induces sleep because of its 'dormative' property), or Nietzsche (by reason of a 'faculty').  Or, perhaps, even Empedocles ('love' and 'strife' rule the universe).  Actually, 'wetness' doesn't seem to be a property at all, but rather the name of a &lt;em&gt;relation&lt;/em&gt; that describes a regularity without providing any explanation at all.  Why is water wet?  We have &lt;em&gt;no&lt;/em&gt; idea!&lt;/p&gt;&#xA;" OwnerUserId="14826" LastActivityDate="2015-03-07T15:09:48.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27027" PostTypeId="1" AcceptedAnswerId="27037" CreationDate="2015-03-07T16:37:47.423" Score="3" ViewCount="7341" Body="&lt;p&gt;What substance that is liquid at STP has the lowest freezing/melting point? I &lt;em&gt;think&lt;/em&gt; it's either 2-methylpentane or 3-methylpentane, but I get different numbers from different sources.&lt;/p&gt;&#xA;" OwnerUserId="14827" LastActivityDate="2015-03-08T03:19:49.927" Title="Lowest freezing point room temperature liquid?" Tags="&lt;melting-point&gt;" AnswerCount="1" CommentCount="16" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27028" PostTypeId="1" CreationDate="2015-03-07T16:54:22.437" Score="1" ViewCount="64" Body="&lt;p&gt;I read in a book that zeolites are formed by Aluminium replacing a silicon atom in Silicon dioxide. It says there is a negative charge formation. How does this happen? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14637" LastActivityDate="2015-03-08T04:56:59.850" Title="Zeolite formation" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27029" PostTypeId="2" ParentId="27028" CreationDate="2015-03-07T18:50:08.850" Score="1" Body="&lt;p&gt;There's no magic in this - just the beauty of science.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Imagine a lattice of $\ce{Si^{(+4)}O4}$ tetrahedrons ($\ce{Si}$ in the centres, $\ce{O}$ on the corners) connected through the &lt;strike&gt;edges&lt;/strike&gt; corners (= the $\ce{O}$ atoms)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replace some of the $\ce{Si^{(+4)}}$ with $\ce{Al^{(+3)}}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a result, the whole lattice becomes a huge polyanion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charges are balanced via cations, such as $\ce{H+}$, $\ce{Na+}$, etc. embedded in the pores of the lattice. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-08T04:56:59.850" LastActivityDate="2015-03-08T04:56:59.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27030" PostTypeId="1" CreationDate="2015-03-07T21:06:43.427" Score="2" ViewCount="434" Body="&lt;p&gt;I recently was told that the reason you can mix colors (blue and yellow make green because wavelengths 605nm + 450nm = green) was based on wavelength (which seems very obvious when you think about it but please humor me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of cool but I don't get it - is this based on the wavelength range that's transmitted?  If  605nm + 450nm = green then how would you find that exact wavelength of green? Would it be the average of the two, or is there some constant that I am missing?&lt;/p&gt;&#xA;" OwnerUserId="14815" LastActivityDate="2015-04-06T22:29:08.403" Title="Mixing Different Solutions with Different Wavelengths to Get Different Colors?" Tags="&lt;solutions&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27031" PostTypeId="1" CreationDate="2015-03-07T21:13:29.347" Score="1" ViewCount="1321" Body="&lt;p&gt;I am trying to isolate pure carbon in its elemental form. Are there any decomposition reactions or other easy to carry out reactions to yield solid carbon? and in the case that it is isolated, does it start reacting with other elements in the air?&lt;/p&gt;&#xA;" OwnerUserId="7639" LastActivityDate="2019-01-17T15:00:57.590" Title="What easy to conduct (at home) reactions could yield solid carbon?" Tags="&lt;home-experiment&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27032" PostTypeId="2" ParentId="27031" CreationDate="2015-03-07T21:48:34.923" Score="6" Body="&lt;p&gt;A classic reaction to yield carbon is the dehydration of sucrose (table sugar) with concentrated sulfuric acid: Put some sugar in a beaker, carefully add sulfuric acid and watch a voluminous &quot;snake&quot; of porous carbon (blow up due to the steam released during the reaction) from the beaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find lots of videos on this on youtube or other platforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is performed with concentrated sulfuric acid! If you decide that you must do this at home, protect yourself (googles, gloves, lab coat) and keep the sulfuric acid locked away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a more simple way to yield carbon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hold a spoon (with a long handle) into the flame of a candle. The black stuff that deposits on the spoon is carbon. Play with the distance for optimal results.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the reactivity of carbon is concerned: There's no danger in that! Millions of people stockpile carbon in the form of coal at home in their cellars or sheds for heating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And actually, this points to the easiest and most safe option: Don't perform any reaction at home and grind some charcoal or anthracite coal instead.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-07T22:32:10.853" LastActivityDate="2015-03-07T22:32:10.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27033" PostTypeId="2" ParentId="27030" CreationDate="2015-03-07T22:14:48.790" Score="1" Body="&lt;p&gt;You have to distinguish two sorts of colour mixing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Subtractive mixing&lt;/strong&gt; is observed when you mix two coloured solutions ot put two colored films on top of each other and shine white light through it. Here, $\mathrm{\color{yellow}{yellow}}$ and $\mathrm{\color{blue}{blue}}$ yield $\mathrm{\color{green}{green}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Additive mixing&lt;/strong&gt; is observed when you overlay two coloured light sources. In the RGB model, $\mathrm{\color{red}{red}}$ and $\mathrm{\color{green}{green}}$ yield $\mathrm{\color{yellow}{yellow}}$, the combination of $\mathrm{\color{red}{red}}$, $\mathrm{\color{green}{green}}$ and $\mathrm{\color{blue}{blue}}$ yields white.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as the wavelengths are concerned: Each &quot;colour&quot; is represented by a wavelength range, rather than an exact value. The fact that there's more than one red or green is intuitively understood - particularly by women ;) &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-07T22:14:48.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27034" PostTypeId="1" CreationDate="2015-03-07T23:31:40.380" Score="4" ViewCount="97" Body="&lt;p&gt;I have acquired the crushed magnet out of a motor I was given. If I'm correct that the magnet is $\ce{NdFeB}$ magnet, how could I separate out all the metals into a pure or oxide form? And for hard drive magnets, is there any way to remove the $\ce{Ni}$ coating without having to get nitric acid?  I don't have any proper chemistry equipment at all.&lt;/p&gt;&#xA;" OwnerUserId="14830" LastEditorUserId="402" LastEditDate="2017-04-01T01:15:23.107" LastActivityDate="2017-04-01T01:15:23.107" Title="Separating metals out of a magnet" Tags="&lt;acid-base&gt;&lt;metal&gt;&lt;metallurgy&gt;&lt;rare-earth-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27035" PostTypeId="1" CreationDate="2015-03-08T02:02:45.973" Score="0" ViewCount="275" Body="&lt;p&gt;This is the question extracted from my country high school level' exam. &#xA;So it asks to discuss the feasibility of the electrochemical cell to carry out electrolysis of an aqueous solution of sodium chloride at standard condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Al(s)/Al(aq)//Cu(aq)/Cu&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how I should start to answer this question?&lt;/p&gt;&#xA;" OwnerUserId="14831" LastActivityDate="2015-03-08T02:02:45.973" Title="The feasibility of the electrochemical cell." Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27036" PostTypeId="2" ParentId="27031" CreationDate="2015-03-08T02:40:41.057" Score="2" Body="&lt;p&gt;Why don't you just buy activated carbon? Generally it is rather pure and contains no carcinogen or poisonous substances, so it is very safe to handle, unlike most pyrolysis products.  &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-03-08T02:40:41.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27037" PostTypeId="2" ParentId="27027" CreationDate="2015-03-08T03:19:49.927" Score="4" Body="&lt;p&gt;After a bit of random searching through small molecules, I found &lt;a href=&quot;http://figshare.com/articles/Jean_Claude_Bradley_Open_Melting_Point_Datset/1031637&quot; rel=&quot;nofollow&quot;&gt;this very large dataset&lt;/a&gt; with 28,645 experimentally measured melting points. Sorting them by increasing melting point and checking the first few entries individually, it seems that &lt;em&gt;N,N&lt;/em&gt;-diethylmethylamine holds the record, with a melting point of -196°C (!) and a &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/d98203&quot; rel=&quot;nofollow&quot;&gt;boiling point of 63-65°C&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also mentioned another possibility for very low melting liquid substances at room temperature, namely ionic liquids. Even though the species present are much larger and have ionic interactions, presumably the ions pack so extremely poorly in the solid that freezing is very unfavourable.  Aldrich already offers multigram quantities of &lt;a href=&quot;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Aldrich/Brochure/al_chemfile_v5_n6.pdf&quot; rel=&quot;nofollow&quot;&gt;ionic liquids melting at -88°C&lt;/a&gt;, and ongoing research offers even lower melting alternatives, such as the ones in &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2013/cp/c2cp43066e#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; which purportedly do not crystallize until -150°C. This limit may well be improved upon in the next few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If mixtures are allowed as well as pure substances, then a combination of small organic molecules or ionic liquids will very likely reach even lower temperatures than the examples above without freezing.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-03-08T03:19:49.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27039" PostTypeId="1" AcceptedAnswerId="27042" CreationDate="2015-03-08T04:44:39.197" Score="1" ViewCount="674" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5jy9W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble with this question since I'm getting (A), when the answer says its (B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My logic was that there is a methyl group on the first carbon, and a hydroxide functional group on the second carbon, hence the name 1-methyl-2-pentanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain why I am wrong?&lt;/p&gt;&#xA;" OwnerUserId="14824" LastEditorDisplayName="user7951" LastEditDate="2018-12-27T18:03:06.717" LastActivityDate="2018-12-27T18:03:06.717" Title="Nomenclature of methyl groups in an alkanol" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27040" PostTypeId="2" ParentId="27015" CreationDate="2015-03-08T04:47:40.373" Score="2" Body="&lt;p&gt;I'm not an expert in Fenton chemistry but &lt;a href=&quot;http://www.h2o2.com/industrial/fentons-reagent.aspx?pid=143&amp;amp;name=General-Chemistry-of-Fenton-s-Reagent&quot; rel=&quot;nofollow&quot;&gt;it appears that difficulty in getting the Fenton reagent to reduce COD to below 500 ppm is not unusual.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does your particular wastewater have  chelants such as phosphates, EDTA, formaldehyde, and/or citric/oxalic acids?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Perhaps at long reaction times all the iron you add is being precipitated as iron (hydr)oxides.  Could you lower the pH further?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reaction rates with Fenton’s Reagent are generally limited by the rate&#xA;  of $\ce{.OH}$ generation (i.e., concentration of iron catalyst) and less so&#xA;  by the specific wastewater being treated. Typical $\ce{Fe:H2O2}$ ratios are&#xA;  1:5-10 wt/wt, though iron levels &amp;lt; 25-50 mg/L can require excessive&#xA;  reaction times (10-24 hours). This is particularly true where the&#xA;  oxidation products (organic acids) sequester the iron and remove it&#xA;  from the catalytic cycle. Fenton’s Reagent is most effective as a&#xA;  pretreatment tool, where CODs are &gt;500 mg/L. This is due to the loss&#xA;  in selectivity as pollutant levels decrease...In addition to free radical scavengers,&#xA;  the process is inhibited by (iron) chelants such as phosphates, EDTA,&#xA;  formaldehyde, and citric/oxalic acids. Because of the sensitivity of&#xA;  Fenton’s Reagent to different wastewaters, it is recommended that the&#xA;  reaction always be characterized through laboratory treatability tests&#xA;  before proceeding to plant scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My tip would be to increase the concentration of both $\ce{H2O2}$ and $\ce{FeSO4}$, and, if possible, to make sure the pH is 3 (or maybe even lower).  If that doesn't work, do you think iron chelators could be present in your particular wastewater sample(s)?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-08T04:47:40.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27041" PostTypeId="2" ParentId="27039" CreationDate="2015-03-08T04:51:21.790" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could someone please explain why I am wrong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the longest chain! This determines the base name: &lt;strong&gt;hexane&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Find the position of the $\ce{OH}$ substituent. Lower number wins.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-08T04:51:21.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27042" PostTypeId="2" ParentId="27039" CreationDate="2015-03-08T04:52:53.890" Score="4" Body="&lt;p&gt;Don't let the graphical layout of the molecule deceive you!  There are no branches in the carbon chain.  (That is, no carbon atom is bound to more than two carbon atoms.)  The molecule ethane is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; called &quot;methylmethane&quot;, so calling your molecule a methylpentane derivative would similarly be an error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no branches (and no rings and no double bonds etc.), and with six carbon atoms, we must be dealing with a hexane derivative.  Once you know that, you just need to apply IUPAC rules that minimize the numbering of atoms to which substituents are bound.  So answer B makes more sense than C, for that reason.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-08T04:52:53.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27044" PostTypeId="1" AcceptedAnswerId="27054" CreationDate="2015-03-08T06:04:49.377" Score="13" ViewCount="1661" Body="&lt;p&gt;Several companies are offering a service to turn ashes of human remains into diamonds.&#xA;Does anyone know any details about the chemical aspects of the process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea, but suspect that the ashes are composed mostly of phosphates and oxides. Where does the carbon hide? Any carbonate would have decomposed at the creation temperature. Or does the cremation chamber has such a high $\ce{CO2}$ partial pressure that $\ce{CO2}$ from burners is captured by the $\ce{CaO}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the carbon then extracted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, diamond is normally made out of graphite, which agrees with a statement on the &lt;a href=&quot;http://lifegemaustralia.com.au/the_process.html&quot;&gt;LifeGem website&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume they can squeeze out a little $\ce{CO2}$ by heating the ashes in an oxygen stream. Then they concentrate the $\ce{CO2}$ on a molecular sieve, use the Sabatier reaction to create $\ce{CH4}$ which they decompose on a hot metal wire. Then they dissolve that in acid and get a little bit of powder rich in $sp^2$ bonds and similar to graphite. Could one make diamonds out of that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the page mentioned above states that they first extract the carbon and then turn it into graphite. How do you turn a few mg of fine carbon powder into graphite? &lt;/p&gt;&#xA;" OwnerUserId="14832" LastEditorUserId="7448" LastEditDate="2015-03-08T13:43:30.293" LastActivityDate="2016-04-14T18:52:52.183" Title="Ashes to diamonds --- How?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27045" PostTypeId="1" CreationDate="2015-03-08T08:36:38.483" Score="1" ViewCount="1051" Body="&lt;p&gt;Can anyone tell the product of the reaction between $\ce{AlCl3}$ (aluminium chloride) and $\ce{N(CH3)3}$ (trimethylamine)?&lt;/p&gt;&#xA;" OwnerUserId="14835" LastEditorUserId="5026" LastEditDate="2018-02-23T01:58:56.100" LastActivityDate="2018-02-23T02:00:39.877" Title="Product of reaction between aluminium chloride and trimethylamine" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2018-02-24T17:12:28.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="27046" PostTypeId="1" CreationDate="2015-03-08T09:30:09.527" Score="0" ViewCount="4777" Body="&lt;p&gt;A student analysed a hydrocarbon X and found it to be containing 43g carbon and 7.2g hydrogen.The relative molecular mass of X was found to be 42.What is the molecular formula of X?&lt;/p&gt;&#xA;" OwnerUserId="14836" LastActivityDate="2016-04-21T20:13:18.030" Title="the molecular formula of an unknown hydrocarbon" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-08T19:38:49.407" ContentLicense="CC BY-SA 3.0" />
  <row Id="27047" PostTypeId="1" CreationDate="2015-03-08T09:39:48.697" Score="3" ViewCount="89" Body="&lt;p&gt;If quenching is seen as a result of running electrochemical experiments on a solution, what kind of processes may be responsible? I'm thinking about UV/Vis spectroscopy, if it makes any difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redox couple would be iodide ions, $\ce{I-/I^3-}$. I understand iodide ions to be quenching agents, so my suspicion would be that their varying concentrations would cause a variation in quenching over the course of the experiment.&lt;/p&gt;&#xA;" OwnerUserId="13341" LastEditorDisplayName="user15489" LastEditDate="2015-05-31T08:15:59.250" LastActivityDate="2015-05-31T08:15:59.250" Title="Which causes of quenching increase specifically in electrochemical experiments?" Tags="&lt;electrochemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27048" PostTypeId="2" ParentId="26863" CreationDate="2015-03-08T10:40:44.680" Score="3" Body="&lt;p&gt;Superionic water is water under extreme heat and temperature with the properties of both solids and liquids. &lt;br/&gt;&#xA;It is where the water only exists as a mixture of $\ce{H+}$ and $\ce{OH-}$ ions. At high pressures, the oxygen crystallises, leaving the hydrogen ions free - thus making superionic water highly conductive.&lt;br/&gt;&#xA;It's crystal structure is a face-centred cubic lattice, and it is thought that Uranus and Neptune hold a layer of superionic water.&lt;/p&gt;&#xA;" OwnerUserId="14837" LastEditorUserId="7448" LastEditDate="2015-03-08T13:39:20.950" LastActivityDate="2015-03-08T13:39:20.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27049" PostTypeId="2" ParentId="27046" CreationDate="2015-03-08T10:40:50.403" Score="7" Body="&lt;p&gt;$43\ \mathrm g$ of carbon for $7.2\ \mathrm g$ hydrogen.&lt;br&gt;&#xA;Molar mass of carbon is $12\ \mathrm{g/mol}$ and that of hydrogen is $1\ \mathrm{g/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there are $\frac{43\ \mathrm g}{12\ \mathrm{g/mol}} = 3.58\ \mathrm{mol} \approx 3.6\ \mathrm{mol}$ of carbon for every $\frac{7.2\ \mathrm g}{1\ \mathrm{g/mol}}=7.2\ \mathrm {mol}$ of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the empirical formula of the compound is $\ce{CH2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative molecular mass of $\ce{CH2}$ is $12+2\times1 = 14$. Given relative molecular mass is $42$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle\frac{42}{14}=3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the formula of the compound is $\ce{C3H6}$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorDisplayName="user7951" LastEditDate="2016-04-21T20:13:18.030" LastActivityDate="2016-04-21T20:13:18.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27050" PostTypeId="2" ParentId="27011" CreationDate="2015-03-08T10:56:54.637" Score="2" Body="&lt;p&gt;It depends on what you will use the chromatography paper for. Different testable substances will require different mediums. However, for a plain-simple chromatography without staining/substance identification, aluminium oxide (or silica) coated on a plate is used, however paper can also be used.&lt;/p&gt;&#xA;" OwnerUserId="14837" LastActivityDate="2015-03-08T10:56:54.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27052" PostTypeId="2" ParentId="16563" CreationDate="2015-03-08T11:27:07.067" Score="-3" Body="&lt;p&gt;When sugar dissolves into tea or coffee, the liquid transforms the sugar into a liquid so it can fit in with the liquid and slide in with the molecules. If you try to &lt;em&gt;evaporate&lt;/em&gt; the water for long enough, you will turn the sugar back into a solid.&lt;/p&gt;&#xA;" OwnerUserId="14839" LastActivityDate="2015-03-08T11:27:07.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27053" PostTypeId="1" CreationDate="2015-03-08T14:30:57.617" Score="4" ViewCount="36247" Body="&lt;p&gt;I’m really wondering how can decreasing the temperature refer to an endothermic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I faced this when I was calculating the heat of a reaction &lt;em&gt;Q&lt;/em&gt; in a constant-pressure calorimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I believe is that when something becomes cooler, then it has released some amount of heat.&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorDisplayName="user7951" LastEditDate="2015-05-10T12:33:31.693" LastActivityDate="2020-04-29T05:29:04.050" Title="How can decreasing in temperature indicate an endothermic reaction?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27054" PostTypeId="2" ParentId="27044" CreationDate="2015-03-08T14:54:00.843" Score="6" Body="&lt;p&gt;According to US published patent application &lt;a href=&quot;https://www.google.com/patents/US20030017932&quot; rel=&quot;nofollow&quot;&gt;20030017932 &quot;Cremeting; filtering carbon particles; sublimation; polishing&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The preferred process for collection begins with the oven operator positioning the body in the oven so that the head and chest area are not positioned directly underneath the main burner. This can be accomplished by positioning the body to the left or right side of the main burner, or positioning the body so that the legs and feet are underneath the main burner rather than the head and torso. Positioning the body in this manner assures that carbon will remain in the body's head area. The carbon can then be gathered by hand, or by using a metal shovel or scoop, or the like.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, in this technique (or proposed technique) the idea is to not let the normal cremation process occur, which would change much of the carbon to CO2.  It seems more like pyrolysis to charcoal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They go on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another alternative embodiment for carbon collection involves collecting carbon from pulverized cremated remains. These remains consist mostly of ash, but depending on how the cremation was performed, there may be traces of carbon particles mixed in with the ashes. The remains can be placed in a vacuum induction furnace. The furnace is heated to 2000 degrees centigrade in a vacuum ranging from 30 toir to 500 torr. Chlorine gas is injected into the furnace, and reacts with the impurities to form chlorides. The impurities leave the carbon in the form of chloride gases, and are filtered as they exit the furnace.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For more possible methods read this and other patent applications such as &lt;a href=&quot;https://www.google.com/patents/US20040031434&quot; rel=&quot;nofollow&quot;&gt;US20040031434&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In another alternative embodiment of the invention, the remains can be cremated conventionally, mixed with additional carbon from another source, and purified as described above. It is contemplated that, using this technique, a gem containing at least some of the original carbon from the cremated remains can be prepared, even if the amount of carbon present in the remains alone is insufficient to make a gemstone of desired size or type.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no way to say what actually occurs at a particular company, unless someone has first hand knowledge, which I don't.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-09T08:35:04.727" LastActivityDate="2015-03-09T08:35:04.727" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27055" PostTypeId="2" ParentId="27045" CreationDate="2015-03-08T15:00:06.120" Score="5" Body="&lt;p&gt;Here are a few tips to get you started:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lewis acids are electron pair acceptors; Lewis bases are electron pair donors. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This looks like a Lewis acid-base reaction. You mention that this &lt;em&gt;is&lt;/em&gt; an acid-base reaction but there are no easily accessible protons in either reactant. Aluminum trichloride has &lt;strong&gt;no&lt;/strong&gt; protons; trimethylamine has does have protons but these are $\ce{sp^3~C-H}$ type protons - which are extremely weak. Also, there is no path to stability; deprotonation of one of the methyl groups leaves a &lt;em&gt;localized&lt;/em&gt; negative charge on a carbon. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are other heuristics one can use. For example, note that aluminum trichloride has an electron-deficient aluminum atom (the Al doesn't have an octet). Acceptance of an electron pair &lt;em&gt;will&lt;/em&gt; provide it with an octet. This heuristic works fairly well; consider boron trichloride for example. Boron is similarly electron-deficient. And boron trichloride is also a Lewis acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the amine has a lone pair. Keeping in mind that Lewis acid-base reactions are about the movement of electrons (rather than protons) then we might reasonably suspect this to be a Lewis acid-base reaction. Lone pairs generally indicate Lewis basicity, as these are generally available for donation to a Lewis acid. Delocalization of lone pairs can weaken the lone pair's Lewis basicity. Thus, the presence of a lone pair only serves as an indicator of Lewis basicity rather than the &lt;em&gt;strength&lt;/em&gt; of Lewis basicity. And of course you can also donate bonding electrons rather than lone pairs (halogenation of ethene would be an example; the $\ce{C=C}$ double bond is broken and the electrons previously involved in that double bond are put into the antibonding orbital of some diatomic halogen)&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-02-23T02:00:39.877" LastActivityDate="2018-02-23T02:00:39.877" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="27056" PostTypeId="2" ParentId="24324" CreationDate="2015-03-08T16:25:25.417" Score="2" Body="&lt;p&gt;DyCl3 is often prepared by the &quot;ammonium chloride route,&quot; starting from either Dy2O3 or hydrated chloride or oxychloride.[2][3] or DyCl3·6H2O.[4] These methods produce (NH4)2[DyCl5]:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    10 NH4Cl + Dy2O3 → 2 (NH4)2[DyCl5] + 6 NH3 + 3 H2O&#xA;&#xA;    DyCl3·6H2O + 2 NH4Cl → (NH4)2[DyCl5] + 6 H2O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The pentachloride decomposes thermally according to the following equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    (NH4)2[DyCl5] → 2 NH4Cl + DyCl3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The thermolysis reaction proceeds via the intermediacy of (NH4)[Dy2Cl7].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treating Dy2O3 with aqueous HCl produces hydrated chloride (DyCl3·6H2O). This salt cannot be rendered anhydrous by heating. Instead one obtains an oxychloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: wikipedia&lt;/p&gt;&#xA;" OwnerUserId="13429" LastActivityDate="2015-03-08T16:25:25.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27058" PostTypeId="2" ParentId="10390" CreationDate="2015-03-08T16:39:33.717" Score="0" Body="&lt;p&gt;I had a similar, but slightly different &lt;a href=&quot;https://physics.stackexchange.com/questions/168993/rutherfords-gold-foil-experiment-can-alphas-be-deflected-by-electrons-vs-nucl&quot;&gt;question over on the physics exchange&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant to this question: Grabbing electrons would also seem to change the trajectory of the He+ (as would bumping into one of the 79 high-speed electrons per gold atom).  While my question was more about the impact this has on the scatter (as this seems important to the measurement of the size of the nucleus of the atoms), seems like deflections should give clues to pick-up of electrons as well.  &lt;/p&gt;&#xA;" OwnerUserId="14828" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-03-10T14:05:01.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27059" PostTypeId="2" ParentId="27021" CreationDate="2015-03-08T17:16:36.257" Score="2" Body="&lt;p&gt;This is a pretty straight forward equilibrium problem in disguise. If we assume the initial state represents equilibrium, we can determine the equilibrium constant by plugging in the concentrations to the law of mass action. We can get the concentrations because we have the number of moles and the volume of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2O4 &amp;lt;=&amp;gt; 2NO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c=Q_{eq}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, now we let the gas expand. Being in thermal contact with the surroundings is important, since it means the temperature will not change (which means the equilibrium constants does not change). Now we will find that we are not at equilibrium any more. We will probably find that $Q_c &amp;lt; K_c$, so equilibrium needs to shift toward $\ce{NO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can set up the ICE (initial, change, equilibrium) table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;          [N2O4]           [NO2]&#xA;I         5.71 M           5.71 M&#xA;C         -x               +2x&#xA;E         5.71 -x          5.71 + 2x&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Substitute into the law of mass action at equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_c = \dfrac{(5.71+2x)^2}{5.71-x}=13.33$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange and solve for $X$. Being a quadratic equation, you should get two roots (two solutions for $x$), and as is always the case for an equilibrium problem, one of them will be nonsense. For example, you could get one value of $x$ to be larger than the initial concentrations!&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-03-08T17:16:36.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27060" PostTypeId="1" CreationDate="2015-03-08T17:44:14.297" Score="3" ViewCount="403" Body="&lt;p&gt;I am very much confused: in a simple voltaic cell (&lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; is positively charged and is the anode, where in a Galvanic cell (&lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{ZnSO4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4}$&lt;/span&gt;),&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; is positively charged but is the cathode. How can I define it, where both are electric or voltaic cell?&lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorUserId="4945" LastEditDate="2020-09-09T13:46:10.187" LastActivityDate="2020-09-09T13:53:57.543" Title="How to correctly define anode and cathode?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27061" PostTypeId="2" ParentId="27060" CreationDate="2015-03-08T18:02:07.030" Score="3" Body="&lt;p&gt;&lt;strong&gt;Cathode&lt;/strong&gt; is the electrode where the &lt;strong&gt;Reduction&lt;/strong&gt; reaction takes place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anode&lt;/strong&gt; is the electrode where the &lt;strong&gt;Oxidation&lt;/strong&gt; reaction takes place&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-03-08T18:02:07.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27063" PostTypeId="1" CreationDate="2015-03-08T18:27:20.100" Score="1" ViewCount="272" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e8vyt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it is a Diels Alder reaction, but I don't recognize the start materials!&#xA;First I make the tertiary alcohol from carboxylic acid with organo-lithium reagent; after I try to make a Diels Alder reaction, but I don't find the right conjugated diene and dienophile.&lt;/p&gt;&#xA;" OwnerUserId="13277" LastEditorUserId="865" LastEditDate="2021-05-22T20:52:35.470" LastActivityDate="2021-05-22T20:52:35.470" Title="Synthesis of a bicyclo[2.2.2]octane from benzoic acid" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27064" PostTypeId="2" ParentId="27053" CreationDate="2015-03-08T19:26:44.843" Score="2" Body="&lt;p&gt;By definition, an &lt;em&gt;exothermic&lt;/em&gt; reaction gives off heat, and &lt;em&gt;endothermic&lt;/em&gt; reaction (e.g. dissolution of ammonium nitrate in water, as in &quot;cool-packs&quot;) &lt;em&gt;absorbs&lt;/em&gt; heat from the environment in order to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumption is based on the idea that heat &lt;em&gt;leaves&lt;/em&gt; the reactants in order to cool them, which is not the case in an &lt;em&gt;endothermic&lt;/em&gt; reaction, but &lt;em&gt;is&lt;/em&gt; the case with a heat pump. These are different mechanisms; in the endothermic reaction, heat energy is used to force the reaction to proceed, e.g. a change of state from solid to liquid.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-08T19:26:44.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27067" PostTypeId="1" AcceptedAnswerId="27068" CreationDate="2015-03-08T21:17:19.650" Score="1" ViewCount="2035" Body="&lt;p&gt;Since these two chemicals have different freezing temps (pyrrolidone is $-11~\mathrm{^\circ F}$ [$-24~\mathrm{^\circ C}$] and $\unicode[Times]{x3b3}$-butyrolactone freezes (also called melting) at $-46~\mathrm{^\circ F}$ [$-43~\mathrm{^\circ C}$]), could a person wanting to separate these just immerse them in liquid nitrogen (or something else?) to get the desired chemical to freeze, then filter off the unfrozen butyrolactone? Other routes seem quite difficult for a non chemist. Any objections or other routes? Anything you can add? Even a step by process will be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14842" LastEditorUserId="7475" LastEditDate="2015-10-16T17:18:53.883" LastActivityDate="2019-02-02T00:58:28.750" Title="how to separate gamma-butyrolactone from N-methyl-2-pyrrolidone" Tags="&lt;organic-chemistry&gt;&lt;melting-point&gt;&lt;purification&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27068" PostTypeId="2" ParentId="27067" CreationDate="2015-03-08T21:42:04.047" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Cud [sic] a person wanting to separate these just emerse [sic] them into liquid nitro [sic] or any [sic] to get the desired chemical to freeze then filter off the unfrozen butyrolacton [sic]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With liquid nitrogen this would not be possible because liquid nitrogen condenses/boils at -195 &amp;deg;C.  It would freeze both the pyrrolidone and the lactone.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-09T19:57:29.997" LastActivityDate="2015-03-09T19:57:29.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27069" PostTypeId="1" AcceptedAnswerId="27075" CreationDate="2015-03-08T22:09:39.533" Score="2" ViewCount="359" Body="&lt;p&gt;I'm learning about quantum numbers right now, and I can't figure out how to calculate the number of electrons in each shell without memorizing it. Is there a formula to find the possible numbers of electrons in the s, p, d, f, and other (theoretical) subshells? If it helps, this problem is put under the &quot;Quantum Numbers, Exclusion Principle&quot; topic.&lt;/p&gt;&#xA;" OwnerUserId="5457" LastActivityDate="2015-03-09T03:54:14.320" Title="Show that there can be 18 electrons in a &quot;g&quot; subshell" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27070" PostTypeId="2" ParentId="27044" CreationDate="2015-03-08T22:15:21.227" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Let's assume they can squeeze out a little CO2 by heating the ashes in an oxygen stream. Then they concentrate the CO2 on a molecular sieve, use the Sabatier reaction to create CH4 which they decompose on a hot metal wire. Then they dissolve that in acid and get a little bit of powder rich in sp2 bonds and similar to graphite. Could one make diamonds out of that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You ask good questions about how $\ce{CO2}$ is obtained.  But if there is a way to get $\ce{CO2}$, I doubt methanation is done en route to graphite.  At elevated temperatures, direct reduction of $\ce{CO2}$ to graphite can be accomplished by many metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it used in the ashes-to-diamonds industry, but in sample preparation for accelerator mass spectrometry, $\ce{CO2}$ is converted to graphite routinely.  &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0020708X85902510&quot; rel=&quot;nofollow&quot;&gt;Here is one paper from the 1980s&lt;/a&gt;, which uses finely divided magnesium at 550 &amp;deg;C to reduce carbon dioxide to elemental carbon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On the other hand, the page mentioned above states that they first extract the carbon and then turn it into graphite. How do you turn a few mg of fine carbon powder into graphite?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://e-reports-ext.llnl.gov/pdf/336483.pdf&quot; rel=&quot;nofollow&quot;&gt;A much later report from Lawrence Livermore National laboratory&lt;/a&gt; uses zinc for the same purpose, and shows that their protocol is usable on only a few mg of carbon.  (Note since they are interested in analytical applications, their protocol is extremely rigorous/tedious, but I would guess that for diamond manufacturing applications this level of rigor would not be required.)  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-08T22:15:21.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27071" PostTypeId="2" ParentId="27053" CreationDate="2015-03-08T22:23:36.877" Score="4" Body="&lt;p&gt;In case of endothermic reactions, products possess more energy than substrates. The energy difference (more precisely - enthalpy difference) was taken from surroundings (e.g. solvent), so the surroundings energy is lower. Thus, its temperature, which is a measure of average kinetic energy of molecules, is lower.&lt;/p&gt;&#xA;" OwnerUserId="14772" LastActivityDate="2015-03-08T22:23:36.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27072" PostTypeId="2" ParentId="27053" CreationDate="2015-03-08T22:55:52.767" Score="5" Body="&lt;p&gt;Let’s say that you have a beaker of water. You measure the water’s temperature using a thermometer, and you determine that it is 25 °C. Now you add two reactants to the beaker, and mix them together well until they are completely reacted. If the reaction between these two reactants is endothermic, then the chemicals will absorb heat from the water. If you were to now remeasure the water’s temperature, you might find that it was something like 23 °C. This is because the heat that was absorbed by the chemicals was lost by the water.&lt;/p&gt;&#xA;" OwnerUserId="13111" LastEditorDisplayName="user7951" LastEditDate="2015-05-10T12:30:42.180" LastActivityDate="2015-05-10T12:30:42.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27073" PostTypeId="1" CreationDate="2015-03-09T00:26:53.043" Score="4" ViewCount="2473" Body="&lt;p&gt;As the title says, I would like to understand chemically why this occurs. My theory behind it is that in the former case both are completely dissociated, and so at the equivalence point (and approaching it), everything is controlled increasingly by the autoionization of water, whereas in weak acids there is also a much more sizable contribution of $H^+$ by the acid itself, so the change is not as drastic. If this is correct, I'd like further explanation as to why it is, since I don't quite feel comfortable enough with it intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a duplicate. I am not asking why the strong acid curve is steep, but rather why the strong acid curve is steep &lt;strong&gt;while the weak acid one is not&lt;/strong&gt;.!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dlt.ncssm.edu/tiger/diagrams/acid-base/StrongAcid-WeakAcid.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weak acid and strong acid titration curves with addition of a strong base&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="-1" LastEditDate="2017-04-15T15:30:12.213" LastActivityDate="2017-04-15T15:30:12.213" Title="Why is the vertical part of a titration curve longer for a strong acid by a strong base than for a weak?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27074" PostTypeId="1" CreationDate="2015-03-09T03:52:44.487" Score="0" ViewCount="520" Body="&lt;p&gt;I initially guessed it was due to the formation of $\ce{H3BO3}$ but am now doubtful. Does it have anything to do with $\ce{HCl}$ formation? If the question referred to a boron halide, I could answer with that. Either the question is wrong, or my guess is. What's accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate any help.&#xA;Grade 11 Chem.&lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="14775" LastEditDate="2015-03-09T04:10:12.067" LastActivityDate="2018-10-09T13:37:11.003" Title="Why does boron fume in moist air? Is the following accurate?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27075" PostTypeId="2" ParentId="27069" CreationDate="2015-03-09T03:54:14.320" Score="4" Body="&lt;p&gt;Simple mnemonic: each orbital, s, p, d and f, has double the consecutive odd numbers of electrons, i.e. 1, 3, 5 &amp;amp; 7 --&gt; 2, 6, 10 &amp;amp; 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, from spectroscopy, s, p, d &amp;amp; f refer to splitting of lines, the &lt;em&gt;sharp&lt;/em&gt;, &lt;em&gt;principal&lt;/em&gt;, &lt;em&gt;diffuse&lt;/em&gt; and &lt;em&gt;fine&lt;/em&gt;; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow&quot;&gt;Atomic orbital&lt;/a&gt;. As Tom Lehrer stated, &quot;These are the only ones of which the news has come to Hahvahd, and there may be many others, but they haven't been discovahd.&quot; Although element 121, unbiunium, begins the theoretical g-block...&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-09T03:54:14.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27076" PostTypeId="2" ParentId="27063" CreationDate="2015-03-09T06:17:46.010" Score="1" Body="&lt;p&gt;Supposed that the synthesis does involve a Diels-Alder reaction, it is nowhere written that it is the &lt;strong&gt;last&lt;/strong&gt; step in the sequence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take another look at your product and imagine a $\ce{C=C}$ double bond somewhere. Does it now look more like a Diels-Alder product? The $\ce{C=C}$ can be turned to $\ce{C-C}$ using... (You figure that out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work your path back from here and try to identify diene and dienophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Don't look until you've figured out the diene ;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Convert acid to ester, 2 x $\ce{MeMgBr}$, Diels-Alder reaction with vinylmethyl ether, hydrogenate double bond. You didn't cheat, did you? ;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-09T07:41:14.717" LastActivityDate="2015-03-09T07:41:14.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27077" PostTypeId="1" CreationDate="2015-03-09T06:47:12.263" Score="0" ViewCount="21585" Body="&lt;p&gt;For the expansion of a gas, isothermally and reversibly, what is the sign taken?&#xA;Do I use this equation&#xA;$-nRT\ln  (Vf/Vi)$&#xA;or this one&#xA;$nRT\ln  (Vf/Vi)$ ?&#xA;Moreover, when I'm asked about the 'work done', what exactly does the work? The gas itself? The surroundings? I'm confused about this because my text says one thing and the guidebook, another.&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14775" LastActivityDate="2018-03-19T13:22:31.270" Title="What are the signs for work done by (on?) a gas during isothermal reversible compression?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27078" PostTypeId="1" CreationDate="2015-03-09T06:52:12.597" Score="7" ViewCount="878" Body="&lt;p&gt;Normally, we think of combustion as occurring only in the gas phase. I think I understand the main reason for this - in the gas phase it's very hard for a bimolecular reaction to produce a single product, which makes it more likely for radical chain reactions to propagate rather than being terminated - but I'm curious about whether it's possible for combustion to occur in the liquid state at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: suppose I mixed liquid $\ce{CH4}$ with liquid $\ce{O2}$ under high pressure and reasonably low temperature. Would they react immediately? If not, could I 'ignite' the mixture in the same way as I could in the gas phase? (Note that the combustion of $\ce{CH4}$ into $\ce{H2O}$ and $\ce{CO2}$ doesn't change the mole number, so the volume change should be relatively small.) Would the reactions that occur be similar to the reactions that occur during combustion of $\ce{CH4}$ in the gas phase, or would they be completely different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this reaction wouldn't resemble combustion, is there some other example of a self-propagating oxidation reaction in liquid or aqueous phase with similar properties to combustion?&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2015-03-09T18:42:26.423" Title="Combustion in liquid state" Tags="&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27079" PostTypeId="2" ParentId="27067" CreationDate="2015-03-09T07:29:43.263" Score="2" Body="&lt;p&gt;As Curt F. has already pointed out, liquid nitrogen is an overkill and won't help here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to try one of the following cooling mixtures - values are taken from (&lt;a href=&quot;http://www.chemieunterricht.de/dc2/tip/08_98.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.chemieunterricht.de/dc2/tip/08_98.htm&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;81 g $\ce{CaCl2*6H2O}$ + 100 g ice : -21,5 °C (this might not be enough)&lt;/li&gt;&#xA;&lt;li&gt;66 g $\ce{NaBr}$ + 100 g ice : -28 °C&lt;/li&gt;&#xA;&lt;li&gt;105 g ethanol + 100 g ice : -30 °C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to enrich NMP in ther solid fraction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-09T07:29:43.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27080" PostTypeId="2" ParentId="27060" CreationDate="2015-03-09T08:06:52.657" Score="2" Body="&lt;p&gt;Voltaic cell is a synonym for a Galvanic cell. In the mentioned system (&lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; can be charged both positively and negatively. It depends how you connect it to the power source. In a Galvanic/Voltaic cell (as Babounet mentioned):&lt;/p&gt;&#xA;&lt;p&gt;The &lt;em&gt;cathode&lt;/em&gt; is a negatively charged electrode where &lt;em&gt;reduction&lt;/em&gt; takes&#xA;place, and the &lt;em&gt;anode&lt;/em&gt; is a positively charged electrode where &lt;em&gt;oxidation&lt;/em&gt; takes place.&lt;/p&gt;&#xA;&lt;p&gt;Let's analyse the second cell. Without any additional information, I assume that no external power source is connected. This system is known as Daniell cell:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qTfZg.png&quot; alt=&quot;Wikipedia Daniell cell&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{Zn | Zn^2+ &amp;amp;|| Cu^2+ | Cu}\\&#xA;\ce{Zn - 2e- &amp;amp;-&amp;gt; Zn^2+} &amp;amp;\therefore\text{oxidation} &amp;amp;= \text{anode}\\&#xA;\ce{Cu^2+ + 2e- &amp;amp;-&amp;gt; Cu} &amp;amp;\therefore\text{reduction} &amp;amp;= \text{cathode}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The situation is nearly the same for the first cell, but this situation is used and known as Sacrificial Anode. &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; acts in the system as an anode and it is dissolved.&lt;/p&gt;&#xA;&lt;p&gt;You can have two electrodes that are both positively charged such as &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;. Standard electrode potential of both metals are positive: &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+|Ag (s)} = \pu{+0.80 V}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+ | Cu (s)} = \pu{+0.34 V}$&lt;/span&gt;, but if you connect them in a galvanic cell, one will act as cathode (&lt;span class=&quot;math-container&quot;&gt;$\ce{Au}$&lt;/span&gt;) and the other as anode (&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;As far as I can tell (based on provided informations), the copper electrode will be in both cases positively charged and therefore be a cathode. There must be a mistake in the description or in the statement.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4945" LastEditDate="2020-09-09T13:53:57.543" LastActivityDate="2020-09-09T13:53:57.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27081" PostTypeId="1" CreationDate="2015-03-09T08:30:00.833" Score="3" ViewCount="931" Body="&lt;p&gt;I am dealing with two conformations of &lt;em&gt;cis&lt;/em&gt;-1,4-dimethylcyclohexane. One conformation is di-axial, and the other one is di-equatorial. Why does di-axial have higher strain energy. I would make sense if the molecule was 1,3-dimethylcyclohexane because then it would the steric hinderance that would cause higher energy. Is it similar with 1,4-.. conformation? If its not steric repulsion, what factors cause di-axial to have higher strain energy?&lt;/p&gt;&#xA;" OwnerUserId="14844" LastEditorUserId="14844" LastEditDate="2015-03-09T08:40:50.200" LastActivityDate="2015-03-10T08:38:49.627" Title="Strain Energy in di-axial vs di-equatorial" Tags="&lt;organic-chemistry&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27082" PostTypeId="1" CreationDate="2015-03-09T09:03:56.797" Score="3" ViewCount="10279" Body="&lt;p&gt;I assumed this:&#xA;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.12491.html?rid=222625d4-811c-4b07-b263-8930fafa831a&quot; rel=&quot;nofollow&quot;&gt;http://www.chemspider.com/Chemical-Structure.12491.html?rid=222625d4-811c-4b07-b263-8930fafa831a&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be 1-phenylbut-4-ene&#xA;Where is the priority here? No source I know of tells me of where the priority of the phenyl  group lies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is wrong. The book states it is 4-PhenylBut-1ene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the priority of the phenyl group? &lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="10185" LastEditDate="2015-03-09T17:23:32.897" LastActivityDate="2015-03-09T17:23:32.897" Title="What is the priority of a phenyl group in a compound such as this?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27084" PostTypeId="2" ParentId="27082" CreationDate="2015-03-09T10:54:01.920" Score="4" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdJVJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For preferred IUPAC names, the ring is preferred to the chain.&lt;br&gt;&#xA;Therefore, the preferred IUPAC name for the given structure is &lt;strong&gt;but-3-en-1-ylbenzene.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the context may favour the chain, so that, for example, an unsaturated acyclic structure may be recognized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one double bond is present, an unbranched alkene is named by changing the ending to ‘ene’. The double bond is assigned the lower locant possible, which is placed immediately in front of the ending ‘ene’. When several structural features appear, low locants are assigned to ‘ene’ endings before detachable alphabetized prefixes.&lt;br&gt;&#xA;Therefore, a correct name for the given structure is &lt;strong&gt;4-phenylbut-1-ene.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-03-09T10:54:01.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27085" PostTypeId="2" ParentId="27081" CreationDate="2015-03-09T11:03:10.297" Score="4" Body="&lt;p&gt;&lt;strong&gt;You can't have both in a chair conformation!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the methyl groups in 1,4-dimethylcyclohexane adopt a bis-axial orientation, they are &lt;em&gt;trans&lt;/em&gt; and they stay &lt;em&gt;trans&lt;/em&gt; after ring flip:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5hosP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your 1,4-dimethylcyclohexane is &lt;em&gt;cis&lt;/em&gt;-substituted the orientation of the methyl groups is axial and equatorial, respectively:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/levxN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2015-03-10T08:38:49.627" LastActivityDate="2015-03-10T08:38:49.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27086" PostTypeId="1" CreationDate="2015-03-09T11:29:52.030" Score="25" ViewCount="25480" Body="&lt;p&gt;I was wondering whether the following order (&lt;a href=&quot;http://www.kshitij-iitjee.com/Study/Chemistry/Part2/Chapter5/50.jpg&quot; rel=&quot;noreferrer&quot;&gt;from here&lt;/a&gt;) is correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZNpky.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZNpky.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In the &lt;strong&gt;first inequality&lt;/strong&gt;, why did 9 hyperconjugations dominated resonance effect of benzene ring?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the &lt;strong&gt;last inequality&lt;/strong&gt; doesn't the 2 hyperconjugations of vinylic carbocation overpower the methyl carbocation?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In middle there is:&lt;br&gt;&#xA;&lt;strong&gt;allyl = secondary&lt;/strong&gt;     &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why is there an equality here when resonance dominates inductive effect?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="4945" LastEditDate="2017-08-01T11:23:39.950" LastActivityDate="2020-09-22T13:40:18.230" Title="Carbocation stability order" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27087" PostTypeId="1" CreationDate="2015-03-09T11:35:06.400" Score="1" ViewCount="138" Body="&lt;p&gt;I want to know about a transparent material,In which penetration of light decreases as intensity of light increases.&lt;/p&gt;&#xA;" OwnerUserId="14847" LastActivityDate="2015-03-09T21:40:11.447" Title="dependence of Intensity on penetration of light" Tags="&lt;optical-properties&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27089" PostTypeId="1" CreationDate="2015-03-09T12:46:05.223" Score="2" ViewCount="108" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AQdoX.gif&quot; alt=&quot;molecule&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this compound's IUPAC name?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://m.wolframalpha.com/input/?i=2-phenyloctane&amp;amp;x=0&amp;amp;y=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;2-phenyloctane&lt;/a&gt;.&lt;br&gt;&#xA;2) &lt;a href=&quot;http://www.chemicalize.org/structure/#!mol=2-phenyloctane&quot; rel=&quot;nofollow noreferrer&quot;&gt;octan-2-ylbenzene&lt;/a&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is using phenyl correct or is taking benzene as the base correct?  &lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="14846" LastEditDate="2015-03-09T13:12:10.637" LastActivityDate="2015-03-09T13:12:10.637" Title="IUPAC Nomenclature of Benzene+Aliphatic chain" Tags="&lt;nomenclature&gt;" AnswerCount="0" CommentCount="11" ClosedDate="2015-03-09T13:38:45.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="27090" PostTypeId="2" ParentId="6102" CreationDate="2015-03-09T13:18:32.450" Score="1" Body="&lt;p&gt;That recipe is taken from book dedicated to chemical restoration techniques: &lt;a href=&quot;http://www.allbeton.ru/wiki/%D0%A5%D0%B8%D0%BC%D0%B8%D1%8F+%D0%B2+%D1%80%D0%B5%D1%81%D1%82%D0%B0%D0%B2%D1%80%D0%B0%D1%86%D0%B8%D0%B8+%28%D0%BD%D0%B8%D0%BA%D0%B8%D1%82%D0%B8%D0%BD%29/&quot; rel=&quot;nofollow&quot;&gt;ISBN 5-7245-0274-7&lt;/a&gt;, 1990.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) AgNO3, 5g + water, 60ml. Divide to 20 and 40 ml solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 40ml of Solution(1) + NH4OH, slowly, active mixing. Dark precipitation occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Solution(2) + 20ml of Solution(1), add 5ml per 2 sec. Then add NH4OH. You obtain cloudy solution (otherwise, add more AgNO3 solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Solution (3) diluted up to 500ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can reduce Ag to obtain metalic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Refined sugar, 75g + water, 500-600 ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Solution(1) + H2SO4, 10ml, 10%. Boil for 5-6 h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Solution(2) becomes dark-yellow. Cool it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Dilute soluion(3) up to 1L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Before using dilute 10 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you mix 2 solutions Ag is reduced as precipitate: Ag(NH3)2OH -&gt; Ag + NH3 + OH-&lt;/p&gt;&#xA;" OwnerUserId="14852" LastEditorUserId="14852" LastEditDate="2015-03-09T13:30:01.283" LastActivityDate="2015-03-09T13:30:01.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27091" PostTypeId="1" AcceptedAnswerId="27109" CreationDate="2015-03-09T13:40:43.363" Score="3" ViewCount="89" Body="&lt;p&gt;A question in &lt;em&gt;Thermodynamics of Natural Systems&lt;/em&gt; by G. M. Anderson is to find the &quot;standard heat of reaction&quot; of this reaction:&#xA;$$\ce{C8H16N2O3(aq) + H2O(l) -&amp;gt; C6H13NO2(aq) + C2H5NO2(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand it, it should be a rather simple problem of finding $\Delta_\mathrm r H$.&#xA;So it should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_\mathrm r H=\Delta_\mathrm fH (\ce{C6}) + \Delta_\mathrm fH (\ce{C2}) - \Delta_\mathrm fH (\ce{C8}) - \Delta_\mathrm fH (\ce{H2O})$ (note that I shortened the organic formulae)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the data from the same textbook, it's:&#xA;$(-632077~\pu{J/mol}-513988~\pu{J/mol})-(-847929~\pu{J/mol}-285830~\pu{J/mol})=-12306~\pu{J/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer given is $−89475~\pu{J/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something? Do I need to add some terms of pressure or volume expansion? Is this not the enthalpy that I'm looking for?&lt;/p&gt;&#xA;" OwnerUserId="14853" LastEditorUserId="16683" LastEditDate="2018-02-23T16:53:18.427" LastActivityDate="2018-02-23T16:53:18.427" Title="Trying to figure out heat of reaction" Tags="&lt;physical-chemistry&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27092" PostTypeId="2" ParentId="27078" CreationDate="2015-03-09T13:49:50.483" Score="4" Body="&lt;p&gt;Your question is addressed by &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/je60002a006&quot; rel=&quot;nofollow&quot;&gt;Explosive Systems Containing Liquid Oxygen: Liquid Oxygen-Liquid Methane Mixtures&lt;/a&gt;. J. Chem. Eng. Data, 1959, vol. 4, pp. 127–131.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If liquid methane and liquid oxgygen at 90K are mixed, they are miscible in all proportions.  Mixtures over a wide range of proportions detonated with use of a blasting cap. 100 grams of mixture punched a 10cm hole in a 3/16 inch steel plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparision, the TNT control only dented the metal plate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further tests on the stoichimetric liquid methane/oxgygen mixture found that it was detonated by a .22 caliber copper bullet at 40 foot range, flame and spark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So nothing happens if the liquids are simply mixed, but a powerful explosion occurs if a stimulus is applied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid oxygen and liquid methane have also been used in rocket engines.  See &lt;a href=&quot;https://archive.org/details/nasa_techdoc_20000004538&quot; rel=&quot;nofollow&quot;&gt;Liquid Methane/Liquid Oxygen Injectors for Potential Future Mars Ascent Engines&lt;/a&gt; for example.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-09T18:42:26.423" LastActivityDate="2015-03-09T18:42:26.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27093" PostTypeId="2" ParentId="27077" CreationDate="2015-03-09T14:57:28.113" Score="1" Body="&lt;p&gt;The easiest way to remember the sign conventions is to think of the gas particles as &quot;wanting&quot; to get away from each other. To reduce the volume of the gas, you have to do work &lt;em&gt;on the gas&lt;/em&gt; - the surroundings are doing the work. In chemistry we always take the point of view of the system. If we do work &lt;em&gt;on&lt;/em&gt; the system, we add energy to it, and therefore, the sign of work is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the opposite is true as well. When the volume of the system increases, the sign of the work term must be negative - the system is losing energy and doing work on the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your equations, all of this holds for the first one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-nRT\ln  (Vf/Vi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because a &lt;em&gt;decrease&lt;/em&gt; in volume will give you a fraction in the $\rm{ln}$ term, which leads to a negative sign. The leading negative sign reverses that. The result is that any compression will have a positive work term, and any expansion will have a negative work term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how the correct sign is a result of the derivation in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Isothermal_process#Calculation_of_work&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2018-03-19T13:22:31.270" LastActivityDate="2018-03-19T13:22:31.270" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27095" PostTypeId="1" CreationDate="2015-03-09T15:26:05.853" Score="35" ViewCount="20583" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;While naming any compound, the numbering should be done such that position of substituted groups gives smallest sum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm struggling with the IUPAC naming for cyclic compounds. How exactly does this rule work?&lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="16683" LastEditDate="2016-09-25T17:14:56.890" LastActivityDate="2020-10-04T14:28:18.950" Title="IUPAC nomenclature: &quot;Smallest sum of locants&quot;?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="27096" PostTypeId="2" ParentId="27095" CreationDate="2015-03-09T15:38:25.477" Score="7" Body="&lt;p&gt;The way you have stated the rule can be misleading or even incorrect at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what you are talking about is the lowest locant rule or the first-point difference rule. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The principal functional group is given the lowest possible locant.&lt;/li&gt;&#xA;&lt;li&gt;The substituents are then assigned locants based on this numbering scheme in such a way as to give the lowest locant number at the first time there is a difference.&lt;/li&gt;&#xA;&lt;li&gt;In the event that there is no first point of difference in the locants, substituents are numbered based on alphabetization giving the first substituent the lowest possible locant number.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/orgnom/main/dontsum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;You might find this link relevant.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13216" LastEditorUserId="402" LastEditDate="2017-04-11T04:04:29.327" LastActivityDate="2017-04-11T04:04:29.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27097" PostTypeId="2" ParentId="27095" CreationDate="2015-03-09T15:48:12.177" Score="3" Body="&lt;p&gt;There are lots of sites,blogs and other online sources where you can learn about the topic in depth. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkanes/Properties_of_Alkanes/Cycloalkanes/Nomenclature_of_Cycloalkanes&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkanes/Properties_of_Alkanes/Cycloalkanes/Nomenclature_of_Cycloalkanes&lt;/a&gt;. and here is the site where you can learn by watching online &lt;a href=&quot;http://leah4sci.com/naming-organic-compounds-iupac-nomenclature/&quot; rel=&quot;nofollow&quot;&gt;http://leah4sci.com/naming-organic-compounds-iupac-nomenclature/&lt;/a&gt;   . You can get many videos in YouTube as well. So go for these sources.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4656" LastEditDate="2016-01-10T07:40:02.920" LastActivityDate="2016-01-10T07:40:02.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27098" PostTypeId="2" ParentId="27078" CreationDate="2015-03-09T18:27:24.330" Score="2" Body="&lt;p&gt;Perhaps your question is rather what would it take to have a fluidic version of a controlled &quot;burn&quot; rather than an explosion. Interesting! The necessary conditions would be introduction of one liquid (&quot;fuel&quot;) into another (&quot;oxidizer&quot;) via a nozzle, with a reaction occurring at the interface at a rate sustained by continuous introduction of more of the &quot;fuel&quot; component. Methane and oxygen react too violently - maybe you could go to fluoromethane or difluoromethane and tone it down. Or go to hydrogen peroxide as the oxidizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of work has been done on developing extremely compact, lightweight energy sources that can power an engine in extreme environments. For example, liquid lithium and sulfur have been used to power a reciprocating engine.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-09T18:27:24.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27099" PostTypeId="1" CreationDate="2015-03-09T18:52:58.800" Score="10" ViewCount="7440" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MoEBt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MoEBt.png&quot; alt=&quot;1,2,4-trimethyl-1H-imidazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that 1,2,4-trimethyl-1⁠&lt;em&gt;H&lt;/em&gt;-imidazole is aromatic. As it has four obvious pi electrons (in the double bonds), I assumed that it was anti-aromatic as opposed to aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been operating under the assumption that a nitrogen does not contribute its pi electrons to the aromaticity count for Hückel's rule unless it is attached to a hydrogen. Does the nitrogen with a methyl group here donate its pi electrons to the count?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="16683" LastEditDate="2018-01-08T15:53:13.070" LastActivityDate="2018-01-08T16:07:46.763" Title="When do lone pairs on nitrogen contribute to aromaticity?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;aromaticity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27100" PostTypeId="2" ParentId="27099" CreationDate="2015-03-09T19:17:58.980" Score="11" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BZgTO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BZgTO.png&quot; alt=&quot;1,2,4-trimethyl-1H-imidazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both nitrogens in your drawing are roughly $\mathrm{sp^2}$ hybridized, therefore they each have a p-orbital and 3 $\mathrm{sp^2}$ hybdridized orbitals. All of the p-orbitals, including those on the 3 carbon atoms in the double bonds, are perpendicular to the plane of the screen (&lt;em&gt;e.g.&lt;/em&gt; they project above and below the plane of the screen). So there are 5 p-orbitals and each contains one electron &lt;strong&gt;except&lt;/strong&gt; for the one p-orbital located on your &quot;bottom&quot; nitrogen (in green). This p-orbital contains the nitrogen lone pair, so there are two electrons in this p-orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count the p-electrons we have $1+1+1+1+2 = 6$ electrons in $5$ p-orbitals that form a continuous loop. Therefore, the molecule is aromatic ($4n+2, n=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a closer look at the &quot;top&quot; nitrogen in your drawing (in purple). Where is its lone pair? It resides in an $\mathrm{sp^2}$ orbital on this nitrogen. The $\mathrm{sp^2}$ orbital lies in the plane of the screen. This orbital is orthogonal (perpendicular) to the 5 p-orbitals; therefore these 2 lone pair electrons are separated from the electrons that make the molecule aromatic. They are not involved in the molecule's aromaticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how to approach this situation in the future:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Identify the nitrogen(s)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine if the lone pair is in a p-orbital or an $\ce{sp^2}$ orbital.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the nitrogen is in a double bond, then the lone pair is in one $\ce{sp^2}$ orbital, 1 electron is in the p-orbital and 1 electron is in each of the 2 remaining $\ce{sp^2}$ orbitals.&lt;/li&gt;&#xA;&lt;li&gt;If the nitrogen is not in a double bond, then the lone pair is in a p-orbital and 1 electron is in each of the 3 $\ce{sp^2}$ orbitals.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Count the electrons in the cyclic p-orbitals to determine aromaticity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-01-08T16:07:46.763" LastActivityDate="2018-01-08T16:07:46.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27101" PostTypeId="2" ParentId="27099" CreationDate="2015-03-09T19:18:30.063" Score="4" Body="&lt;p&gt;The nitrogen with a methyl group does indeed contribute to aromaticity in this molecule. The criteria for aromaticity under Hückel's rule are that $4n+2$ electrons reside in coplanar p-orbitals in a cyclic system. This allows the p-orbitals to overlap and form a set of π-orbitals, which are delocalised over the whole ring system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example all the atoms in the ring are sp&lt;sup&gt;2&lt;/sup&gt; hybridised. The three carbons in the ring form three σ bonds with their sp&lt;sup&gt;2&lt;/sup&gt; orbitals and then have one half-filled p-orbital perpendicular to the plane of the ring which participates in π-bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrogen with a double bond forms two σ bonds with two of its sp&lt;sup&gt;2&lt;/sup&gt; orbitals and uses the third to accommodate its lone pair. The lone pair is therefore in the plane of the ring, facing outwards, and is thus unable to participate in π-bonding. It is left with a half-filled p-orbital in the plane which contributes another electron to the π-system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrogen with the methyl group attached forms three σ bonds with its sp&lt;sup&gt;2&lt;/sup&gt; orbitals, and therefore has a full p-orbital remaining perpendicular to the plane of the ring, which can overlap with the other p-orbitals to form a π-system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In total there are six π-electrons residing in five coplanar p-orbitals. So, Hückel's rule is fulfilled and aromaticity is present.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="16683" LastEditDate="2018-01-08T15:51:08.723" LastActivityDate="2018-01-08T15:51:08.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27102" PostTypeId="2" ParentId="27099" CreationDate="2015-03-09T19:24:55.430" Score="0" Body="&lt;p&gt;Here's a crude, qualitative method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that on top, delocalizing the electrons from that particular nitrogen results in a resonance contributor that has an $\ce{sp}$ hybridized nitrogen. $\ce{sp}$ hybridization implies a linear bond angle (180 degrees). Do you think we can get anywhere close to a linear bond angle in a 5-membered ring of all things? Remember, in resonance contributors the nuclei don't move. It's just the electrons that are &quot;moving&quot; (delocalized). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bottom I try to delocalize the electrons from the other nitrogen. This is fine, as the resulting resonance contributor now has an $\ce{sp^2}$ nitrogen. Note that this does &lt;em&gt;not&lt;/em&gt; mean that the nitrogen in the ring is exactly $\ce{sp^2}$. It's likely somewhere in between $\ce{sp^3}$ and $\ce{sp^2}$. A more or less $\ce{sp^2}$ atom in a 5-membered ring is more plausible than a $\ce{sp}$ atom. So we can conclude that it's this nitrogen's lone pairs that are delocalized throughout the ring. Incidentally, this makes for 6 pi electrons. Additionally, the other Huckel rules are fulfilled (even the planar requirement; both nitrogens have $\ce{sp^2}$ character so they're more or less planar). So we can conclude this molecule is aromatic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DfIV4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DfIV4.png&quot; alt=&quot;Resonance structures of imidazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="16683" LastEditDate="2018-01-08T15:44:16.230" LastActivityDate="2018-01-08T15:44:16.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27103" PostTypeId="1" CreationDate="2015-03-09T19:31:13.800" Score="3" ViewCount="524" Body="&lt;p&gt;I was asked to solve this problem, and the answer key shows this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/GWTFuyn.png&quot; alt=&quot;molecule&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what's going on here, but it looks like some sort of a rearrangement and reduction. Is it an internal reaction between the ether and the alkene?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="10185" LastEditDate="2015-03-10T18:27:15.923" LastActivityDate="2015-03-10T18:27:15.923" Title="Identifying rearrangement reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27104" PostTypeId="2" ParentId="27103" CreationDate="2015-03-09T19:48:46.473" Score="3" Body="&lt;p&gt;This is an example of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Claisen_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claisen rearrangement&lt;/a&gt;.  It is basically the oxa-analogue of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cope_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cope rearrangement&lt;/a&gt;.  The Claisen rearrangement can occur with aromatic or non-aromatic allyl vinyl ethers.  The reaction mechanism involves a concerted, intramolecular, [3,3]-sigmatropic rearrangement.  In the aromatic case, the rearrangement is followed by a proton shift (keto-enol tautomerization) to yield the final product with the aromaticity restored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a diagram of the aromatic Claisen rearrangement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mqpco.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a diagram for the non-aromatic counterpart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6BuEg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-09T20:05:00.300" LastActivityDate="2015-03-09T20:05:00.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27105" PostTypeId="1" CreationDate="2015-03-09T20:39:20.383" Score="0" ViewCount="923" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $\pu{2.4 mol}$ of gas fill a volume of $\pu{175 mL}$, what volume, in mL, will $5.4$ additional moles of gas fill?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was a quiz question which I got wrong the first attempt, and I think I got it right this time however I would like some confirmation that it is/is not. The unnamed gas is assumed to be ideal.&#xA;$$\frac{\pu{2.4 mol}}{\pu{175 mL}} = \frac{\pu{7.8 mol}}{X}, \quad \text{or} \quad X = \frac{\pu{7.8 mol} \times \pu{175 mL}}{\pu{2.4 mol}} = \pu{568.75 mL}$$&#xA;Is this accurate?&lt;/p&gt;&#xA;" OwnerUserId="14859" LastEditorUserId="41328" LastEditDate="2017-08-05T18:03:21.850" LastActivityDate="2017-08-05T18:03:21.850" Title="Applying Avagadro's law: the relationship between moles and volume of a gas" Tags="&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27106" PostTypeId="1" CreationDate="2015-03-09T21:12:14.037" Score="0" ViewCount="3340" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Polyvinyl acetate, PVA, is a useful adhesive for gluing together articles made from wood, paper or cardboard. The monomer of PVA is ethenyl ethanoate.&lt;br&gt;&#xA;   PVA is formed from B by the process of addition polymerisation.&lt;br&gt;&#xA;  Draw a section of the PVA molecule containing at least 2 monomer molecules and identify clearly the repeating unit.&lt;br&gt;&#xA;  (&lt;a href=&quot;http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20%289701%29/9701_s11_qp_42.pdf&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;: University of Cambridge international Examinations, General Certificate of Education Advanced Level, Chemistry, Paper 4 Structured Questions, May/June 2011, Q4)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the $\ce{C=C}$ double bond break and each of the carbon atoms involved in the double bond are attached to $\ce{CH3OCO}$. so the repeating unit is $\ce{CH3OCO}$.&lt;/p&gt;&#xA;" OwnerUserId="14862" LastEditorUserId="14862" LastEditDate="2015-03-10T06:40:13.917" LastActivityDate="2015-04-09T16:10:17.117" Title="How to identify the repeating unit in a polymer?" Tags="&lt;homework&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27107" PostTypeId="2" ParentId="27087" CreationDate="2015-03-09T21:40:11.447" Score="2" Body="&lt;p&gt;In general, when light penetrates a material (water for example) its intensity decreases exponentially with increasing depth. This process is known as &lt;strong&gt;attenuation or absorption&lt;/strong&gt;. The severity of attenuation differs with the wavelength of electromagnetic radiation. In the region of visible light, the red part of the spectrum attenuates more rapidly than the shorter-wavelength blue part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what you are asking about is a non-linear optical property known under the name: &lt;code&gt;Saturable absorption&lt;/code&gt; It is a property of materials where &lt;code&gt;the absorption of light decreases with increasing light intensity&lt;/code&gt;. At sufficiently high incident light intensity, atoms in the ground state of a saturable absorber material become excited into an upper energy state at such a rate that there is insufficient time for them to decay back to the ground state before the ground state becomes depleted, and the absorption subsequently saturates. The material becomes transparent at that wavelength. Saturable absorbers are useful in laser cavities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another kind  of non-linear materials is known under the name &quot;optical power limiting materials&quot;. In these materials &lt;code&gt;the absorption of light increases with intensity&lt;/code&gt; such that beyond a certain input intensity the output intensity approaches a constant value. Such a material can be used to limit the amount of optical power entering a system. This can be used to protect expensive or sensitive equipment such as sensors, can be used in protective goggles, or can be used to control noise in laser beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please see: &lt;a href=&quot;http://en.wikipedia.org/wiki/Saturable_absorption&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Saturable_absorption&lt;/a&gt;&#xA;and &lt;a href=&quot;http://en.wikipedia.org/wiki/Two-photon_absorption&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Two-photon_absorption&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-03-09T21:40:11.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27108" PostTypeId="2" ParentId="27105" CreationDate="2015-03-09T21:46:56.067" Score="0" Body="&lt;p&gt;I think that it is right, because I resolve it at the same method&lt;/p&gt;&#xA;" OwnerUserId="13277" LastActivityDate="2015-03-09T21:46:56.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27109" PostTypeId="2" ParentId="27091" CreationDate="2015-03-09T21:53:10.683" Score="3" Body="&lt;p&gt;There is a &quot;readme&quot; note from the author that says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The answers are not guaranteed, in the sense that not only&#xA;  might I make a mistake here and there, but I might have used different sources of&#xA;  data, or the data in a database may have changed. I have not been able to check all&#xA;  the answers, or to make them all consistent with the data in Appendix B, although I&#xA;  believe that most are.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I think that this problem is an example of one of those situations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book itself works out the problem on page 267 and gets the same answer as you: −12306 J/mol. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-09T21:53:10.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27110" PostTypeId="1" CreationDate="2015-03-09T22:30:40.493" Score="0" ViewCount="134" Body="&lt;p&gt;Can anyone tell me how I can convert a rate constant derived from first order kinetics (units are $\mathrm{h^{-1}}$) into an actual rate (turnovers/h)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario is that I know my rate constant for a molecule degrading in solution and it is $0.012\ \mathrm{h^{-1}}$. I know that I have $10^{14}$ catalyst molecules in the solution. How can I translate that into a number of molecules converted  per catalyst per hour?&lt;/p&gt;&#xA;" OwnerUserId="7614" LastEditorDisplayName="user7951" LastEditDate="2015-05-12T16:11:42.257" LastActivityDate="2015-05-12T16:11:42.257" Title="Convert kinetic rate constant into a rate?" Tags="&lt;kinetics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27112" PostTypeId="1" CreationDate="2015-03-09T23:08:20.920" Score="4" ViewCount="5958" Body="&lt;p&gt;I do not understand how you can work out which element will react more readily with another element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have lithium (Li) and sodium (Na) and I am trying to work out which element will be more reactive with oxygen. How do I manage to work that out?&lt;/p&gt;&#xA;" OwnerUserId="14865" LastEditorUserId="5026" LastEditDate="2018-07-20T14:58:08.800" LastActivityDate="2018-07-20T14:59:44.833" Title="Which of sodium or lithium will be more reactive with oxygen?" Tags="&lt;inorganic-chemistry&gt;&lt;alkali-metals&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27114" PostTypeId="2" ParentId="24499" CreationDate="2015-03-09T23:30:28.620" Score="2" Body="&lt;p&gt;It is possible to not perform experiments at all and only concentrate on theoretical chemistry (at least after your undergraduate studies). What your professor said might be related to your university and his group of friends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;University:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three names of professors in theoretical chemsitry I know from our university are Hättig, Marx and Kutzelnigg: &lt;a href=&quot;http://www.theochem.ruhr-uni-bochum.de/home.en.html&quot; rel=&quot;nofollow&quot;&gt;http://www.theochem.ruhr-uni-bochum.de/home.en.html&lt;/a&gt; . The scientists working for them did not enter a laboratory after the 6th semester (final semester of bachelor studies) as far as I know. Their major is theoretical chemistry. This is what they can have highlighted on their master certificate.&#xA;(You might want to look at this website and check out the member lists. Then you have an idea how many people are working in this field.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is work for theoretical chemistry at BASF and Bayer. We heard presentations from several people of that field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lab work is not required. There are people who specialise in lab work and master it to a fascinating level. The people in theoretical chemistry receive the results from those in the laboratories and work their magic of interpretation, explanation and prediction. The lab skills of a theoretician are not needed if a professional is working on the experiments. The computer skills of an experimentalist are not needed if a professional is working on the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[To my understanding, theoretical chemistry does not make experiments obsolete but experiments cannot give a full explanation about the multitude of properties and comprehension of its scientific context. Furthermore, theoretical chemistry gives valuable guidance. Nowadays, theoretical and experimental chemistry are roughly on eye level (even though each thinks of himself as more important).]&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2015-03-09T23:30:28.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27117" PostTypeId="2" ParentId="16792" CreationDate="2015-03-10T02:12:10.320" Score="6" Body="&lt;p&gt;I think that most treatments of TS miss the point. I'll try to organize my answer stressing some points. At first, proteins have nothing special in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the definition of the transition state.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1º - What kind of state are we talking about? A quantum mechanical state? A classical one? As we need to fall in Born-Oppenheimer approximation, we are dealing (according to Copenhagen interpretation) with a kind of mix of them. We treat particles (atoms) following classical mechanics in some PES given for 'quantum' mechanics.&#xA;It is important because in classical mechanics we define a state with the positions and momentum of particles (as will do in Transition State Theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2º - In this context, we have the same situation for the definition of an equilibrium state and for the definition of a TS, just because we need to define it in terms of positions and momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3º - For a given time $t$, is almost sure that there is not even a single molecule in the exact position corresponding to the minimum of the potential energy surface, and the same is true for the TS. This is simply because the atomic positions $\mathbf{x}=\{x_1,x_2,...,x_n\}$ and $t$ correspond to a continuous variables (that have infinity possible values), that is, these two state are two points of a infinite probable points.&#xA;I mean, it is like to say, for example, what is the probability of picking  the number (say...) 0.45 when (truly) randomly someone choose a number between 0 and 1 with an uniform distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4º - So to speak about equilibrium states or TS, we need to refer them at least as &quot;a&quot; state that is truly many different states, all those with $\mathbf{x}$ values in the range $\{x_1 \pm \delta x_1,x_2\pm \delta x_2,...,x_n\pm \delta x_n\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the usefulness of the definition.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some sense it is all about probability. It essence lies in statistical physics. If we want to know how fast a reaction is, and we assume that it is carried out thought the TS, we need to know how probable (relative to something) is the occurrence of a TS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the things go like that: we define a TS as those states in some volume in phase space. Recall that knowing energy distributions and degeneracies you can get the relative amount of states. So we use the equilibrium states (defined for volumes in phase space) because we can know much of them,like the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next steps are irrelevant here, but we should recognize the &lt;strong&gt;&lt;em&gt;necessity&lt;/em&gt;&lt;/strong&gt; of defining these states by volumes and not just points in the phase space. By doing that (and recalling point 3º) we are establishing that it has no sense (physical or practical) to define TS giving them zero life time quality. (notice that momentum are finite)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the equilibrium constant&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, there is nothing wrong with call it an well defined equilibrium constant. We are always said that it validity requires equilibrium (like everything equilibrium thermodynamics), but it has few sense if we recognize that we can derive it from statistical physics. An equilibrium state just needs to satisfy a constant density in phase, so if it is recovered fast enough we are safe working with thermodynamic equations. States that doesn't change in the time .. .. is just a way of be (almost) completely safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bibliography&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not recommend the original bibliography (Eyring papers). I think a better idea would be for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variational Transition State Theory&#xA;Annual Review of Physical Chemistry&#xA;Vol. 35: 159-189 (Volume publication date October 1984)&#xA;DOI: 10.1146/annurev.pc.35.100184.001111&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other good source for some points can be:&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471400661&quot;&gt;http://www.amazon.com/Chemical-Dynamics-Papers-Advances-Physics/dp/0471400661&lt;/a&gt;&#xA;but only if you really have time to read a lot of articles...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The principles of statistical mechanics&quot; by Tolman is (imho) a book that can make you feel it intuitively (although it is rigorous enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for a &quot;short&quot; description (based in statistical physics) of the TST, I would recommend &quot;Statistical mechanics, the theory of the properties of matter in equilibrium&quot;, by Fowler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final comment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not completely sure if I address your dude. Honestly it seems to me that this kinds of confusions can be avoided thinking in terms of statistical physics, from where the answers to questions like the life time of a TS arises naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, let me know if I didnt answer your question or if something needs more details. Few years ago this theory filled me of doubts and I know that it is not easy find answers about it.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2015-03-10T02:30:39.420" LastActivityDate="2015-03-10T02:30:39.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27120" PostTypeId="1" CreationDate="2015-03-10T04:56:04.407" Score="1" ViewCount="55" Body="&lt;p&gt;I'm confused on this problem for estimating protein concentration. Upon first glance, it looks like a simple Beer's law problem. but then there's the issue that the protein is not pure. How do I account for that?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;E. coli Maltose binding protein, like alkaline Phosphatase is a periplasmic protein. Audrey, has a solution containing solublized E. coli periplasmic proteins and maltose binding protein and she just read the absorbance of the solution at $280~\mathrm{nm}$ which was $0.84$, path length $1~\mathrm{cm}$.&lt;br&gt;&#xA;  Help Audrey estimate the protein concentration of her solution. Also note Audrey knows the extinction coefficient for maltose binding protein ($\mathrm{MW} \approx43,000$) is $32,290~\mathrm{M^{-1}cm^{-1}}$.&lt;br&gt;&#xA;  (Hint: not pure protein solution.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As mentioned above I think it would just be Beer's law&#xA;\begin{align}&#xA;A &amp;amp;= \epsilon cl\\&#xA;A &amp;amp;= 0.84\\&#xA;l &amp;amp;= 1~\mathrm{cm}\\&#xA;\epsilon &amp;amp;= 32,290~\mathrm{M^{-1}cm^{-1}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $c$ gives $0.26~\mathrm{mM}$. My guess would be to say that it is less than that. How do I approach this?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4945" LastEditDate="2015-03-10T07:02:35.340" LastActivityDate="2015-03-10T07:02:35.340" Title="Estimating the protein concentration of an unpure protein solution" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27121" PostTypeId="5" CreationDate="2015-03-10T04:58:36.757" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-03-10T04:58:36.757" LastActivityDate="2015-03-10T04:58:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27122" PostTypeId="4" CreationDate="2015-03-10T04:58:36.757" Score="0" Body="The Beer–Lambert law, also known as Beer's law, the Lambert–Beer law, or the Beer–Lambert–Bouguer law relates the attenuation of light to the properties of the material through which the light is traveling. The law is commonly applied to chemical analysis measurements and used in understanding attenuation in physical optics, for photons, neutrons or rarefied gases. In mathematical physics, this law arises as a solution of the BGK equation." OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-03-10T13:12:44.710" LastActivityDate="2015-03-10T13:12:44.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27124" PostTypeId="1" AcceptedAnswerId="27144" CreationDate="2015-03-10T06:39:33.423" Score="1" ViewCount="125" Body="&lt;p&gt;Let's say you are doing a standard titration problem, for example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the pH after adding $10~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{NaOH}$ in the&#xA;  titration of $25~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{HF}$ with $0.3~\mathrm{M}$ $\ce{NaOH}$.&#xA;  The $K_a$ of $\ce{HF}$ value is $6.6 \cdot 10^{-4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most sources start by considering the reaction between $\ce{HF}$ and $\ce{OH^-}$, finding the pre-equilibrium concentrations of $\ce{HF}$ and $\ce{F^-}$ and then using Henderson-Hasselbalch or the equilibrium expression directly to find $\ce{[H^+]}$ (as seen &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantitative_Analysis/Titration/Titration_Of_A_Weak_Acid_With_A_Strong_Base&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, don't you have to consider the dissociation of $\ce{HF}$ first, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HF &amp;lt;=&amp;gt; H^+ + F^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then consider the reaction between those produced $\ce{H^+}$ and the added $\ce{OH^-}$? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?&lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorUserId="16683" LastEditDate="2016-12-04T17:32:42.167" LastActivityDate="2016-12-04T17:38:32.083" Title="Including H⁺ from the dissociation of the acid in a titration pH calculation?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27125" PostTypeId="1" CreationDate="2015-03-10T06:44:50.727" Score="1" ViewCount="21" Body="&lt;p&gt;I think it may not be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given all the pair correlation functions, it seems possible to obtain the internal energy (assuming different species interact through pair-potentials (i.e. potentials that do not involve more than two bodies)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the free energy, the &quot;naive&quot; way is to obtain the partition function and take logarithm. Yet it seems not easy (except possibly through cluster expansions? which is not exact.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="7451" LastEditDate="2015-03-10T06:52:21.013" LastActivityDate="2015-03-10T06:52:21.013" Title="Given the Pair Correlation Functions for a Multi-component Fluid, is There a Simple way to Obtain the Helmholtz Free Energy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27126" PostTypeId="1" AcceptedAnswerId="27154" CreationDate="2015-03-10T06:51:47.063" Score="1" ViewCount="102" Body="&lt;p&gt;Following &lt;a href=&quot;https://byo.com/stories/issue/item/3113-etch-your-kettle-projects&quot; rel=&quot;nofollow&quot;&gt;this plan&lt;/a&gt;, I am wondering what the byproducts of this method are? At the site with the q-tip, the solution turned a bright yellow, with the q-tip turning a dark yellow, almost red. Holding the q-tip in place for about 10 seconds results in a small amount of smoke, and bubbles around the contact point. What are these colors, smoke, and bubbles? I don't know if the kettle is steel or aluminum, it's an old Budweiser keg, if that helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case the link above dies, the method is etching metal using a vinegar and salt solution, combined with a 9 volt battery. Connecting the negative lead to a piece of metal, and using the positive lead with the solution results in a type of reverse electroplating, thus removing metal ions and etching the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure on the tags for this question, so feel free to adjust them, and the description of the etching method. &lt;/p&gt;&#xA;" OwnerUserId="14873" LastActivityDate="2015-03-10T18:13:19.893" Title="Reverse electro-plating byproducts" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27127" PostTypeId="2" ParentId="24499" CreationDate="2015-03-10T08:15:22.797" Score="1" Body="&lt;p&gt;Many researchers do no traditional lab experiments. Computational chemistry is important and a huge subject. The mathematical theory of chemistry is the Schrödinger equation, known since over 70 years, but unfortunately too difficult to solve for most molecules with more than a few atoms. Thus, an ever-expanding set of approximative methods have been developed and keep developing. One cool example would be Anatole von Lilienfeld who uses neural networks to predict molecule properties by pattern recognition, extrapolating from known molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that lab work can be avoided for a chemistry student. I suppose an alternate route would be to become a physicist then switch subject. I did that, although I actually do lab work and enjoy it now.&lt;/p&gt;&#xA;" OwnerUserId="13341" LastActivityDate="2015-03-10T08:15:22.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27128" PostTypeId="2" ParentId="27105" CreationDate="2015-03-10T10:18:28.667" Score="3" Body="&lt;p&gt;You are correct in your approach and answer. Assuming the gas to be ideal, one can apply Avogadro's Law which states that for an ideal gas $V \propto n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies $V_1/n_1 = V_2/n_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the given values one does obtain the volume as $568.75\ \mathrm{mL}$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorDisplayName="user7951" LastEditDate="2015-09-24T19:47:59.333" LastActivityDate="2015-09-24T19:47:59.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27129" PostTypeId="1" CreationDate="2015-03-10T11:44:37.763" Score="2" ViewCount="2181" Body="&lt;p&gt;I think that this is decided by the number of carbon atoms. The part which has greater the number of carbon atoms, is considered the parent chain. But this fact is not leading to the correct result always. Are there some rules that govern this fact besides the number of carbon atoms. &lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-04-20T15:59:17.677" Title="How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27130" PostTypeId="1" CreationDate="2015-03-10T13:14:31.943" Score="11" ViewCount="32590" Body="&lt;p&gt;I'm curious how to test purity of isopropyl alcohol? I simply got very curious today. I bought rubbing alcohol and it doesn't say what's the purity anywhere. It could be 70%, but it could also be 90%, or 99% or even 99.9%. I'm a curious man so I started googling but couldn't find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were a chemist, how would I test the purity?&lt;/p&gt;&#xA;" OwnerUserId="14878" LastEditorUserId="12860" LastEditDate="2015-10-20T05:06:23.840" LastActivityDate="2020-08-02T02:49:55.730" Title="Testing isopropyl alcohol purity" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;alcohols&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27131" PostTypeId="1" CreationDate="2015-03-10T13:17:34.873" Score="6" ViewCount="218" Body="&lt;p&gt;Lets say you have two times the same molecule in PDB format.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;PDB1:&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HETATM 1916  C1  EQU A1128      22.568 -34.213 -37.632  1.00 19.82           C&#xA;HETATM 1917  O1  EQU A1128      21.807 -35.267 -37.474  1.00 17.07           O&#xA;HETATM 1918  C2  EQU A1128      22.218 -33.078 -36.885  1.00 19.81           C&#xA;HETATM 1919  C3  EQU A1128      23.024 -31.912 -37.017  1.00 21.42           C&#xA;HETATM 1920  C4  EQU A1128      24.206 -31.909 -37.929  1.00 24.79           C&#xA;HETATM 1921  C5  EQU A1128      24.504 -33.085 -38.661  1.00 20.61           C&#xA;HETATM 1922  C6  EQU A1128      23.703 -34.217 -38.516  1.00 17.92           C&#xA;HETATM 1923  C10 EQU A1128      22.658 -30.748 -36.242  1.00 22.26           C&#xA;HETATM 1924  C11 EQU A1128      23.461 -29.573 -36.367  1.00 21.01           C&#xA;HETATM 1925  C12 EQU A1128      24.609 -29.558 -37.249  1.00 24.75           C&#xA;HETATM 1926  C13 EQU A1128      24.992 -30.715 -38.034  1.00 24.09           C&#xA;HETATM 1927  C16 EQU A1128      25.402 -28.301 -37.351  1.00 23.63           C&#xA;HETATM 1928  C17 EQU A1128      26.110 -28.153 -38.728  1.00 25.12           C&#xA;HETATM 1929  C18 EQU A1128      26.990 -29.384 -39.036  1.00 23.20           C&#xA;HETATM 1930  C19 EQU A1128      26.187 -30.715 -38.965  1.00 23.69           C&#xA;HETATM 1931  C24 EQU A1128      24.645 -26.944 -37.282  1.00 25.63           C&#xA;HETATM 1932  C25 EQU A1128      25.847 -25.958 -37.473  1.00 24.96           C&#xA;HETATM 1933  C26 EQU A1128      26.770 -26.799 -38.426  1.00 26.13           C&#xA;HETATM 1934  O26 EQU A1128      27.843 -26.441 -38.854  1.00 29.46           O&#xA;HETATM 1935  C27 EQU A1128      25.111 -27.931 -39.875  1.00 24.05           C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;PDB2:&lt;/h3&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HETATM    1  C10 LIG     1      22.735 -30.818 -36.280  1.00  0.00           C&#xA;HETATM    2  C5  LIG     1      24.497 -33.076 -38.614  1.00  0.00           C&#xA;HETATM    3  C11 LIG     1      23.469 -29.649 -36.412  1.00  0.00           C&#xA;HETATM    4  C6  LIG     1      23.704 -34.219 -38.496  1.00  0.00           C&#xA;HETATM    5  C2  LIG     1      22.306 -33.076 -36.899  1.00  0.00           C&#xA;HETATM    6  C3  LIG     1      23.085 -31.931 -37.022  1.00  0.00           C&#xA;HETATM    7  C4  LIG     1      24.186 -31.919 -37.892  1.00  0.00           C&#xA;HETATM    8  C12 LIG     1      24.550 -29.604 -37.290  1.00  0.00           C&#xA;HETATM    9  C13 LIG     1      24.951 -30.740 -38.009  1.00  0.00           C&#xA;HETATM   10  C1  LIG     1      22.602 -34.190 -37.652  1.00  0.00           C&#xA;HETATM   11  C26 LIG     1      26.758 -26.817 -38.473  1.00  0.00           C&#xA;HETATM   12  C19 LIG     1      26.180 -30.692 -38.934  1.00  0.00           C&#xA;HETATM   13  C25 LIG     1      25.881 -26.016 -37.526  1.00  0.00           C&#xA;HETATM   14  C18 LIG     1      27.001 -29.371 -38.996  1.00  0.00           C&#xA;HETATM   15  C24 LIG     1      24.709 -26.963 -37.185  1.00  0.00           C&#xA;HETATM   16  C16 LIG     1      25.367 -28.337 -37.396  1.00  0.00           C&#xA;HETATM   17  C17 LIG     1      26.086 -28.150 -38.747  1.00  0.00           C&#xA;HETATM   18  C27 LIG     1      25.093 -27.987 -39.938  1.00  0.00           C&#xA;HETATM   19  O26 LIG     1      27.821 -26.440 -38.923  1.00  0.00           O&#xA;HETATM   20  O1  LIG     1      21.738 -35.208 -37.538  1.00  0.00           O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In the third column you can see the atom names (&lt;code&gt;C1&lt;/code&gt;, &lt;code&gt;O20&lt;/code&gt;, etc.). They are different for both PDB files even though it is the same molecule. I would like them to have the same names for the same atoms. So instead of &lt;code&gt;C13&lt;/code&gt; and &lt;code&gt;C9&lt;/code&gt; I would like to have them both named &lt;code&gt;C13&lt;/code&gt; or &lt;code&gt;C9&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the convention to name these atoms? What method should I use to convert them to a common naming?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. I left out most of the PDB to save space&lt;/p&gt;&#xA;" OwnerUserId="14877" LastEditorUserId="5017" LastEditDate="2015-03-10T15:51:03.550" LastActivityDate="2019-02-06T20:51:21.103" Title="Consistent atom names PDB" Tags="&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27132" PostTypeId="1" AcceptedAnswerId="27136" CreationDate="2015-03-10T13:46:08.837" Score="4" ViewCount="7940" Body="&lt;p&gt;Why are thioesters relatively reactive with regard to nucleophilic attack? Prof says to wait until pchem 3 when we learn about orbital symmetry. He also said that sulfur’s d-orbitals (?!) don't have the correct symmetry to participate in resonance with the carbonyl carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait. I thought that sulfur didn’t utilize its d-orbitals. Was it previously taught that sulfur’s lone pairs were in d-orbitals? Why would sulfur even need d-orbitals when in a thioester? Can't we just explain this based on the poor overlap between carbon's 2p and sulfur's hypothetical 3p orbitals? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-03-10T21:37:46.740" LastActivityDate="2019-10-09T01:00:28.783" Title="Reactivity of thioesters with respect to nucleophilic attack" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="27133" PostTypeId="2" ParentId="26781" CreationDate="2015-03-10T13:47:34.283" Score="10" Body="&lt;p&gt;At low concentration, conductivity is proportional to concentration (a linear relationship).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each ion will have its own unique mobility, as discovered by Kohlrausch. &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; has the highest mobility. As you can see in your graph the acids have higher conductivities than the salts. &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is also highly mobile. As concentration increases, the linear relationship breaks down for two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, in infinitely dilute solution, for the strong electrolytes in the graph, there is complete dissociation into separately solvated ions. However, as concentration increases, a portion of the electrolyte exists as ion pairs.  See for example &lt;a href=&quot;http://jes.ecsdl.org/content/144/7/2380.abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equations for Densities and Dissociation Constant of NaCl(aq) at 25°C from “Zero to Saturation” Based on Partial Dissociation&lt;/a&gt; &lt;em&gt;J. Electrochem. Soc.&lt;/em&gt; &lt;strong&gt;1997&lt;/strong&gt; vol. 144, pp. 2380-2384. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the mobility of the ions that are solvated is decreased by the fact that they are no longer moving through water, but are instead moving past other ions as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have a maximum in the curves of the question, and to account for the above factors, it is necessary to subtract a term from the linear term.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a function of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Conductivity} = Ac - Bf(c),$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; are constants, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is concentration, and &lt;span class=&quot;math-container&quot;&gt;$f(c)$&lt;/span&gt; is some function of concentration. Historically, the function &lt;span class=&quot;math-container&quot;&gt;$Ac - Bc\sqrt{c}$&lt;/span&gt; was the first and most simple to use.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start by trying to fit your data to that function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More advanced treatments replace &lt;span class=&quot;math-container&quot;&gt;$\sqrt{c}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\sqrt{I}$&lt;/span&gt;, where I is ionic strength. Further advances involve higher order &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; in addition to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{I}$&lt;/span&gt;, such as &lt;span class=&quot;math-container&quot;&gt;$I \ln (I)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$I^{3/2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$I \ln (I)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; terms arise upon considering the ions as charged spheres of finite diameter, rather than a simple point charges, as explained in &lt;a href=&quot;http://www.nist.gov/data/nsrds/NSRDS-NBS331.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrolytic Conductance and Conductances of the Halogen Acids in Water&lt;/a&gt; and references cited therein. This is the reference to look at if you want the best function to fit the HCl curve in your graph as it includes numerical coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information see &lt;a href=&quot;https://books.google.com/books?id=T0XPAAAAMAAJ&amp;amp;printsec=frontcover&amp;amp;dq=inauthor:%22Olin%20Freeman%20Tower%22&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=W_P-VLqfO4arNpLKg9gB&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Conductivity of Liquids&lt;/a&gt; by Olin Freeman Tower, which although not the most recent work, has the advantage of being understandable, or &lt;a href=&quot;http://www.aslo.org/lomethods/free/2008/0489.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculating the conductivity of natural waters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a very advanced consideration, see &quot;&lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF01074870#&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrical conductance of electrolyte mixtures of any type&lt;/a&gt;&quot; &lt;em&gt;Journal of Solution Chemistry&lt;/em&gt;, &lt;strong&gt;1978&lt;/strong&gt;, vol. 7, pp. 533-548.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2019-10-08T11:37:16.647" LastActivityDate="2019-10-08T11:37:16.647" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27134" PostTypeId="2" ParentId="27129" CreationDate="2015-03-10T13:52:12.633" Score="1" Body="&lt;p&gt;If you have a simple chain with carbon atoms and an aromatic ring, the priority has the ring, not matter how the chain is long. This rule is NOT valid when you have a functional group: if it is in the linear chain, it has the priority, and the numeration starts from the carbon more near at the functional group, for example&#xA;$\ce{CH3-CH2-CO-CH2-CH2-CH3}$. The numeration starts from left, so you write hexan-3-one.&lt;/p&gt;&#xA;" OwnerUserId="13277" LastEditorDisplayName="user7951" LastEditDate="2015-03-21T15:54:58.640" LastActivityDate="2015-03-21T15:54:58.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27135" PostTypeId="2" ParentId="2446" CreationDate="2015-03-10T14:05:45.913" Score="1" Body="&lt;p&gt;In enzyme kinetics, there is a similar situation, where the symbol $E_t$ is used to indicate the &quot;total enzyme concentration&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there can be free enzyme with concentration [E], substrate bound enzyme with concentration [ES], and inhibitor bound enzyme [EI], but for the total concentration of all forms of the enzyme the symbol $E_t$ with no brackets is often used.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example page 354 of &lt;a href=&quot;https://books.google.com/books?id=V5M1bMjOTQoC&amp;amp;pg=PA353&amp;amp;dq=Total%20enzyme%20concentration%20Et&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=XPf-VLblHoi4ggT_2IPgBQ&amp;amp;ved=0CC4Q6AEwAQ#v=onepage&amp;amp;q=Total%20enzyme%20concentration%20Et&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Essentials of Chemical Reaction Engineering&lt;/a&gt; by H. Scott Fogler where he has a note about using that symbol.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-10T14:05:45.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27136" PostTypeId="2" ParentId="27132" CreationDate="2015-03-10T14:19:01.977" Score="3" Body="&lt;p&gt;According to &lt;a href=&quot;http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/L26-index.cfm?L26resource=thioesters&quot; rel=&quot;nofollow&quot;&gt;these lecture notes from Gonzaga University&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question arises over why thioesters occupy a prominent place in metabolism while oxygen esters play a relatively minor role. One answer is thermodynamic. The sulfur in the thioester linkage is less able to participate in electron delocalization through the acyl group (the usual explanation given for this is poor overlap between the 2p orbital of carbon and the 3p orbital of sulfur), and this makes the thioester bond less stable than the ester bond. Hence, the thioester bond has a more negative standard free energy of hydrolysis (−7.5 kcal/mole vs. about −5 kcal/mole for most oxygen esters). In many cases, it appears that thioesters are more reactive than oxygen esters, undergoing more facile nucleophilic displacement reactions at the acyl group. The reactivity of a thioester at the alpha carbon compares favorably with that of ketones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because resonance structure 2c [see web page - CF] makes a more significant contribution for a thioester than an analogous structure for an ester, the acyl carbon is more positive, hence more susceptible to nucleophilic attack. An attacking nucleophile would be more readily acylated by a thioester than it would be by an ester. Furthermore, the C-S bond is weaker than the C-O bond, and the thiolate (or thiol, if protonated) is a better leaving group than alkoxide (or alcohol). These factors all make a thioester in general a better acylating agent than an ester.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the answer doesn't seem to involve d orbitals explicitly, but instead the difference between 2p and 3p orbitals (as well as other factors).&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-10T14:19:01.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27137" PostTypeId="2" ParentId="27130" CreationDate="2015-03-10T14:27:08.967" Score="15" Body="&lt;p&gt;If the isopropanol mixture was bought commercially, the chances that it contains significant amounts of methanol or ethanol are fairly small, I think.  (I've never seen commercial isopropanol in the US that contains &amp;gt; 3% of these impurities, at least, but the situation may vary in other countries or markets.)&lt;/p&gt;&#xA;&lt;p&gt;If you can assume that the only other main component is water, the easiest way to estimate purity is simply by measuring the density.  100% pure isopropanol has a density at 20 °C of 0.786 g per mL.  Pure water has a density of 1.00 g/mL.  At intermediate concentrations of isopropanol, the &lt;a href=&quot;https://www.aat-corp.com/ipa-water-density-percentage-chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;density is in between&lt;/a&gt; those values.  If you can accurately measure both (a) the volume and (b) the weight of a portion of isopropanol, you should be able to measure the density.  I know my kitchen scale and kitchen measuring cups would be precise enough for this.&lt;/p&gt;&#xA;&lt;p&gt;If you &lt;strong&gt;can't&lt;/strong&gt; assume that water is the only other impurity, other methods such as NMR, chromatography, or spectroscopy would probably be required.  I think NMR would probably be the most informative (but perhaps the least accessible to the home chemist.)&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2020-06-26T10:52:29.337" LastActivityDate="2020-06-26T10:52:29.337" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27138" PostTypeId="2" ParentId="26781" CreationDate="2015-03-10T14:41:09.407" Score="4" Body="&lt;p&gt;To expound on DavePhD's nice answer, &lt;em&gt;molar&lt;/em&gt; conductivity follows &lt;a href=&quot;http://en.wikipedia.org/wiki/Molar_conductivity&quot; rel=&quot;nofollow&quot;&gt;Kolrausch's law&lt;/a&gt;, $\Lambda_m=\Lambda_m^{\circ} - \kappa \sqrt{c}$ &lt;strong&gt;at low concentrations&lt;/strong&gt;, which means that extensive conductivity follows $\Lambda_m c = c\Lambda_m^{\circ} - \kappa c \sqrt{c}$ in this domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've discovered this relationship breaks down at high concentration.  In a loose analogy to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_state#Virial_equation_of_state&quot; rel=&quot;nofollow&quot;&gt;virial equation of state for gases&lt;/a&gt;, I'd propose fitting your data to a modification of Kolrausch's law, specifically, to an expansion in $\sqrt{c}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Lambda_m=\Lambda_m^{\circ} - \kappa_1 \sqrt{c} - \kappa_2 (\sqrt{c})^2 - \kappa_3 (\sqrt{c})^3 + ...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not enough of Debye-Huckel theorist to know if this expansion has the same theoretical justification/derivation in terms of intermolecular forces that the virial equation for gases does.  But, (a) it definitely introduces extra degrees of freedom that you will need to fit your more complex data, and (b) if your data did follow Kolrausch's law (and the data for e.g. sodium chloride doesn't look too far off, at least by eye), your fit would tell you so, meaning you wouldn't have to convert or translate your fit parameters to a new system of units to compare with tabulated values for Kolrausch's law -- your fit would tell you the Kolrausch parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;UPDATE:&lt;/em&gt;&lt;/strong&gt; I dug around in the literature a bit.  Wikipedia pointed me towards &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja01494a003&quot; rel=&quot;nofollow&quot;&gt;this 1960s JACS paper&lt;/a&gt;.  There they give the Fuoss-Onsager equation, which can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Lambda_m=\Lambda_m^{\circ} - \beta_1 \sqrt{c} - \beta_2 c - \beta_3 c \ln{c}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\beta_i$ parameters are expanded in the paper to show dependence on some underlying solvent and solute properties, but that isn't important for fitting compound-specific curves to your conductivity data.  You can find papers on the Fuoss-Onsager equation and cite them if you like.  And this equation still has the advantage that is reduces to Kolrausch's law for $\beta_2 = \beta_3 = 0$.  So if that seems better to you than the expansion in $\sqrt{c}$, try that one.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-11T18:35:53.493" LastActivityDate="2015-03-11T18:35:53.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27139" PostTypeId="1" CreationDate="2015-03-10T14:50:39.153" Score="3" ViewCount="21773" Body="&lt;p&gt;Molarity (M) is defined as moles of solute per liter of solution while Normality (N) is defined as moles of reacting units per liter of solution.  Most texts use sulfuric acid as an example to demonstrate the difference: a 1 M solution of sulfuric acid is 2 N because each mole of sulfuric acid gives two moles of $\ce{H+}$ (the &quot;reacting unit&quot;). Why then is concentrated sulfuric acid labeled as 18 N when it should be 18 M and 36 N (according to my text's example above)?&lt;/p&gt;&#xA;" OwnerUserId="14879" LastEditorUserId="4325" LastEditDate="2015-03-10T15:16:25.400" LastActivityDate="2015-03-10T17:42:22.793" Title="Molarity vs Normality in labeling sulfuric acid" Tags="&lt;analytical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27140" PostTypeId="2" ParentId="27139" CreationDate="2015-03-10T15:06:50.003" Score="3" Body="&lt;p&gt;Who labels concentrated sulfuric acid as 18N?  36N sounds right to me and Sigma-Aldrich:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Concentration: The commercial reagent contains 93-98% H2SO4, the balance being water. Its effective concentration is 18 M (36N).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/258105pis.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/258105pis.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IUPAC discourages use of normality because it is ambiguous.  The number of equivalents can be ambiguous depending upon the purpose (equivalents of protons, equivalents of electrons, etc.).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following IUPAC document: &lt;a href=&quot;http://old.iupac.org/publications/analytical_compendium/Cha06sec3.pdf&quot; rel=&quot;nofollow&quot;&gt;http://old.iupac.org/publications/analytical_compendium/Cha06sec3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-10T17:42:22.793" LastActivityDate="2015-03-10T17:42:22.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27141" PostTypeId="2" ParentId="27131" CreationDate="2015-03-10T15:57:57.050" Score="4" Body="&lt;p&gt;You'd need to have some sort of ligand dictionary. For many PDB ligands, this exists, and you can look up the atom naming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the &lt;a href=&quot;http://ligand-expo.rcsb.org/&quot; rel=&quot;nofollow&quot;&gt;PDB ligand expo&lt;/a&gt; allow you to search for ligands by 3-letter code (e.g., &lt;a href=&quot;http://ligand-expo.rcsb.org/pyapps/ldHandler.py?formid=cc-index-search&amp;amp;target=HEM&amp;amp;operation=ccid&quot; rel=&quot;nofollow&quot;&gt;HEM&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.ebi.ac.uk/pdbe-srv/pdbechem/atom/list/HEM&quot; rel=&quot;nofollow&quot;&gt;PDBeChem&lt;/a&gt; also lets you find atom names based on 3-letter codes (the link is HEM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as far as I know, there isn't (yet) any sort of pattern matching that would allow you to go through PDB files and automatically perceive your ligands, sort the atoms, and name them systematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you find one, please let me know - it would be a great addition to programs like &lt;a href=&quot;http://openbabel.org/&quot; rel=&quot;nofollow&quot;&gt;Open Babel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-10T15:57:57.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27142" PostTypeId="2" ParentId="27106" CreationDate="2015-03-10T16:05:56.660" Score="3" Body="&lt;p&gt;It's not that the &quot;double bond breaks,&quot; it's that the vinyl C=C bond reacts during the radical addition polymerization to create &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyvinyl_acetate&quot; rel=&quot;nofollow&quot;&gt;polyvinyl acetate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd disagree that the repeat unit is $\ce{CH3OCO}$ since that ignores the backbone entirely. Instead, if you look at the Wikipedia article (or any other picture of polyvinyl acetate), I'd call the repeat unit something like $\ce{-(CH2CH(OC=OCH3)-)}$ or something like that.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-10T16:05:56.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27143" PostTypeId="1" CreationDate="2015-03-10T16:14:38.403" Score="6" ViewCount="962" Body="&lt;p&gt;Why does magnesium have the lowest melting point of all earth alkali metals?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-03-10T16:14:38.403" Title="Why does magnesium have the lowest melting point of all earth alkalis?" Tags="&lt;melting-point&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ClosedDate="2020-07-27T04:59:24.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="27144" PostTypeId="2" ParentId="27124" CreationDate="2015-03-10T16:18:04.803" Score="3" Body="&lt;p&gt;There are numerous approximations made when pH calculations like these are made.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a problem like this the approximations that would usually be made are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;pH = -log[H+] (as opposed to pH being -log (H+ activity) which is the true definition).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NaOH is an infinitely strong base, with NaOH completely dissociated. (really, [Na+][OH-]/[NaOH] is about 4) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The dissciation of water  can be ignored (really (Kw =[H+][OH-]= 10^-14).  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There exists a constant Ka for HF that is [H+][F-]/[HF], (whereas a true equilibrium constant would be a function of activities, rather than concentrations, of these species, and of water.  The Ka expressed as concentrations is dependent upon ionic strength for example in accordance with the Davies equation approximation).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Particular to this example, it is being assumed that the only fluorine species are HF and F- (really there is also HF2- and another equilbrium constant for it).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you can take away one or more of these approximating assumptions to try to more accurately calculate pH, but it makes the problem very difficult.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, don't you have to consider the dissociation of HF first, and then consider the reaction between those produced  H+  and the added OH-? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end of the titration (or at any other point during the titration along as the solution is homogeneously mixed) the system will be in equilibrium.  The correct approach is to consider all the species that are present and all the equilbrium constants amoung the species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the species would be H2O, H+, OH-, HF, F-, Na+, and NaOH (and if I want to be picky HF2-).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilbrium constants are Kw, Ka of HF, Kb(NaOH), (and Ka of HF2-).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need the same number of independent equations as unknowns to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[F-] + [HF] = (0.1/0.35)0.3M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Na+] + [NaOH] = (0.1/0.35)0.3M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[H+] + [Na+] = [F-] + [OH-] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[H20] = 55M (or, if that is too much of assumption [H2O] + [OH-] + [NaOH] = constant by conservation of oxygen atoms)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus the 3 equilibrium constant equations gives you 7 equations and 7 unknowns.   That would be a comprehensive consideration of the problem dropping assumptions number 2 and 3.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2016-12-04T17:29:05.720" LastActivityDate="2016-12-04T17:29:05.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27145" PostTypeId="1" CreationDate="2015-03-10T16:18:14.177" Score="2" ViewCount="73" Body="&lt;p&gt;Is there a material which can be applied to a surface that will block liquids but is dissolved by blood?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to use the liquid to close a circuit, but only if blood is present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is something I can coat on the circuit to act as a barrier preventing other liquids from reaching the circuit, but that would react to the chemical components of blood by dissolving, allowing the blood to reach and close the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One application might be in a bandage, to detect if a wound has reopened. The number of liquids in that context is pretty limited. A limited number of false positive edge cases is also acceptable&lt;/p&gt;&#xA;" OwnerUserId="14881" LastEditorUserId="14881" LastEditDate="2015-03-11T13:32:01.283" LastActivityDate="2015-03-11T13:32:01.283" Title="Is there something which will block liquids but is dissolved by blood?" Tags="&lt;water&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27146" PostTypeId="1" CreationDate="2015-03-10T16:23:59.823" Score="2" ViewCount="173" Body="&lt;p&gt;Suppose I had an acid HA, and it reaches equilibrium in water. Then I remove the hydronium ions. I don't know how I would do this, maybe adding some hydroxide ions. My question is can I do this and will I be able to eventually create a 50/50 mixture of acid and conjugate base? Is this a very good buffer?&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="4325" LastEditDate="2015-03-10T17:44:23.857" LastActivityDate="2015-04-09T18:10:30.000" Title="What happens if I have an acid/base equillibrium and remove hydronium ions until there is 50/50 of acid and conjugate base?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27148" PostTypeId="1" AcceptedAnswerId="27149" CreationDate="2015-03-10T16:35:32.013" Score="4" ViewCount="552" Body="&lt;p&gt;Why is it difficult to detect &lt;a href=&quot;http://www.hmdb.ca/spectra/nmr_one_d/1070&quot; rel=&quot;nofollow&quot;&gt;dopamine&lt;/a&gt; using $\ce{^1H}$ Magnetic Resonance Spectroscopy (MRS) in the brain (&lt;em&gt;in vivo&lt;/em&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="14882" LastEditorUserId="4231" LastEditDate="2015-03-10T17:14:57.667" LastActivityDate="2015-03-10T17:34:05.193" Title="¹H-MRS Detection of Dopamine" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27149" PostTypeId="2" ParentId="27148" CreationDate="2015-03-10T17:19:23.070" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The concentration of dopamine is low&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The concentration of other protons is very high&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The magnetic field is not homogeneous.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The brain cannot be rapidly rotated the way a laboratory solution state NMR sample is to average magnetic field inhomogenity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compare to laboratory solution state NMR, the viscosity in the brain would be high causing further line broadening.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you look at the spectra in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja305051u&quot;&gt;1H NMR Probe for in Situ Monitoring of Dopamine Metabolism and Its Application to Inhibitor Screening&lt;/a&gt; J. Am. Chem. Soc., 2012, 134 (30), pp 12398–12401, you'll see that even in tissue samples, there are too many other stronger signals for conventional 1H NMR to be successful, so C-13 labeling is used in combination with 1H NMR.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-10T17:34:05.193" LastActivityDate="2015-03-10T17:34:05.193" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27150" PostTypeId="1" AcceptedAnswerId="27177" CreationDate="2015-03-10T17:55:14.803" Score="1" ViewCount="33255" Body="&lt;p&gt;Given a 19kg Liquid Petroleum Gas (LPG) cylinder, with this specification:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LPG inside the cylinder weighs &lt;code&gt;19kg&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The internal volume of the cylinder is &lt;code&gt;44.5 litres&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The LPG is a mixture of &lt;code&gt;propane gas (60%)&lt;/code&gt; and &lt;code&gt;butane (40%)&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate the &lt;strong&gt;volume&lt;/strong&gt; of LPG contained in the cylinder. This is the volume the pressurised gas will uncompress to, if it escapes the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the Ideal Gas Law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Pressure x Volume = Number of Moles x Gas constant x Temperature&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I proceed with this problem?&lt;/p&gt;&#xA;" OwnerUserId="14884" LastEditorUserId="9961" LastEditDate="2017-06-20T20:44:44.803" LastActivityDate="2018-11-25T07:33:50.363" Title="How much gas is inside a pressurised gas cylinder?" Tags="&lt;gas-laws&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27151" PostTypeId="2" ParentId="27146" CreationDate="2015-03-10T18:00:22.747" Score="1" Body="&lt;p&gt;You can't just remove hydronium ions because the solution needs to remain electrically neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't just add hydroxide ions, but you could add sodium hydroxide until the HA/A- ratio was 50/50.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this a very good buffer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is the optimum ratio of HA and A- for buffering.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-10T18:00:22.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27152" PostTypeId="1" AcceptedAnswerId="27158" CreationDate="2015-03-10T18:05:48.713" Score="12" ViewCount="738" Body="&lt;p&gt;Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with a 0.5mW laser pointer, changes color?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be mainly for the visual effects (bartender show) so it should not require aiming the laser at the drink for more than a few seconds. The reason for choosing a 0.5mW laser is to minimize the need for eye protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not matter what color laser is used, but would this be possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, what kind of chemical(s) would be needed in the drink?&lt;/p&gt;&#xA;" OwnerUserId="13112" LastEditorUserId="4231" LastEditDate="2015-03-10T21:24:59.063" LastActivityDate="2015-03-10T21:24:59.063" Title="Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with laser, changes color?" Tags="&lt;color&gt;&lt;photochemistry&gt;&lt;alcohols&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27153" PostTypeId="1" AcceptedAnswerId="27157" CreationDate="2015-03-10T18:10:28.777" Score="8" ViewCount="16958" Body="&lt;p&gt;My textbook says 2-bromobutane is chiral, but it look like it will form superimposition mirror image&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="30836" LastEditDate="2020-04-05T20:45:23.103" LastActivityDate="2020-04-05T20:45:23.103" Title="Is 2-bromobutane chiral?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27154" PostTypeId="2" ParentId="27126" CreationDate="2015-03-10T18:13:19.893" Score="4" Body="&lt;p&gt;The yellow colour in solution is likely from iron ions from the metal being oxidized and dissolved: &#xA;$$\ce{Fe_{(s)} + 2e- -&amp;gt; Fe^{2+}_{(aq)}}$$ (Probably $\ce{Fe^3+}$, too)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas bubbles/smoke is the product of the other half of the reaction, the reduction of something in your solution. Because you have chloride in solution, chlorine gas is a possibility:&#xA;$$\ce{2Cl^{-}_{(aq)} -&amp;gt; Cl_{2(g)} + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen may also be produced depending on what you're using as an electrode on the q-tip.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-03-10T18:13:19.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27155" PostTypeId="2" ParentId="27153" CreationDate="2015-03-10T18:19:38.303" Score="3" Body="&lt;p&gt;Yes, it is chiral because there are 4 different groups on the second carbon, H, Br, methyl and ethyl.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you are looking at a planar depiction.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-10T18:19:38.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27156" PostTypeId="2" ParentId="27150" CreationDate="2015-03-10T18:25:05.550" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;LPG inside the cylinder weighs &lt;span class=&quot;math-container&quot;&gt;$\mathrm{19\,kg}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The LPG is a mixture of propane gas (60%) and butane (40%).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With other words, you know:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the total mass of the gas mixture&lt;/li&gt;&#xA;&lt;li&gt;the mass fractions of propane and butane&lt;/li&gt;&#xA;&lt;li&gt;the molar masses of both gases&lt;/li&gt;&#xA;&lt;li&gt;the molar volume of a gas under standard conditions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-10T18:25:05.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27157" PostTypeId="2" ParentId="27153" CreationDate="2015-03-10T18:29:42.400" Score="15" Body="&lt;p&gt;For a molecule to be chiral it must have &lt;strong&gt;non-superimposable mirror images&lt;/strong&gt;.  Here is a drawing of the two mirror images for 2-bromobutane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hggjD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chiral carbon atom is denoted by an asterisk.  In the case of 2-bromobutane there are 4 different substituents attached to it. The molecule is chiral, you can't pick up one of the mirror images and superimpose it on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note to think about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4 different substituents may make an atom chiral, but the molecule as a whole may still be achiral&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example meso compounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ulYYl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Meso_Compounds&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;some molecules without 4 different substituents attached to an atom may still be chiral .&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, compounds with &lt;a href=&quot;http://en.wikipedia.org/wiki/Axial_chirality&quot; rel=&quot;noreferrer&quot;&gt;axial chirality&lt;/a&gt; such as chiral allenes (1,3-dimethylallene), chiral biphenyls, chiral helicenes, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FfYqa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.med-chemistry.com/2014/09/chiral-compounds-without-asymmetric.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HZdtj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemhelper.com/unusualchirality.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0IVud.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thelab.photophysics.com/circular-dichroism/publications-alert-latest-circular-dichroism-research-2/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-10T22:17:07.037" LastActivityDate="2015-03-10T22:17:07.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27158" PostTypeId="2" ParentId="27152" CreationDate="2015-03-10T19:43:34.507" Score="8" Body="&lt;p&gt;I see a couple of problems here:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Beam diameter vs volume of the beverage&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source definitely is the better option.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Is is possible in general?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Yes, a number of compound classes undergo transformations upon iradiation and change their colour (spiropyranes, fulgides, etc.). Often, the reverse reaction can be induce thermally.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;So what?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Food dyes are strictly regulated, you can't add a photochromic dye to a drink just because the colour change looks nice in the lab. If the dye isn't approved as a food dye, there's no legal way! To my knowledge, food dyes typically do not show photochromism but are photostable. Consumers usually don't like that their food changes the colour in the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the idea sounds funny but I rather doubt that it can be realized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only option to add a dramatic effect would be the use of fluorescent and food-compatible ingedients. Think in quinine (tonic water).&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-10T19:43:34.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27159" PostTypeId="1" AcceptedAnswerId="27160" CreationDate="2015-03-10T20:41:54.787" Score="3" ViewCount="4570" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;During the formation of $\ce{Al2O3}$ from 5.4 grams of $\ce{Al}$ and enough oxygen, the temperature of 2 kg of water climbs by 20 degrees. What’s the enthalpy of formation of $\ce{Al2O3}$ (per mole) ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I tried:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Al + 3 O2 -&amp;gt; 2 Al2O3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.4 grams of $\ce{Al}$ are 0.2 moles.&#xA;If 2 kg of water climbed by 20 degrees then 40 000 calories were invested during the reaction.&#xA;To get one mole of $\ce{Al2O3}$ we need 2 moles of $\ce{Al}$ , therefore the enthalpy of formation (per mole) of $\ce{Al2O3}$ is 400 000 calories. However, the correct answer is −400 000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain the negative sign?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastEditorUserId="5026" LastEditDate="2018-02-27T02:25:52.453" LastActivityDate="2018-02-27T02:25:52.453" Title="How to determine the sign of the enthalpy of formation?" Tags="&lt;physical-chemistry&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27160" PostTypeId="2" ParentId="27159" CreationDate="2015-03-10T20:54:12.283" Score="7" Body="&lt;p&gt;$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A thermodynamic system is the content of a macroscopic volume in space, along with its walls and surroundings; it undergoes thermodynamic processes according to the principles of thermodynamics. &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_system&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it &lt;strong&gt;loses&lt;/strong&gt; heat. Thus, $\Delta H_f$ is negative.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-03-10T21:02:53.077" LastActivityDate="2015-03-10T21:02:53.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27161" PostTypeId="2" ParentId="27152" CreationDate="2015-03-10T21:12:51.833" Score="6" Body="&lt;p&gt;Klaus has provided an excellent answer.  Basically your desired approach is possible, but fraught with difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps there is another way to attack the problem.  Still, I suspect considerable testing will be required.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons mentioned by Klaus, it is necessary to use a food dye that is already approved.  Many food dyes are photobleachable.  For example, azo dyes exist as &lt;em&gt;cis-trans&lt;/em&gt; isomer pairs.  Typically, the &lt;em&gt;trans&lt;/em&gt; isomer is more stable and provides the dye color.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cyqSZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often (not always, this is why you'll need to test), when the &lt;em&gt;trans&lt;/em&gt;-dye is irradiated with laser light it will convert to the colorless (or differently colored) &lt;em&gt;cis&lt;/em&gt; isomer. This is termed &quot;photobleaching&quot; - light has &quot;bleached&quot; the color out of the dye. Generally the &lt;em&gt;cis&lt;/em&gt; isomer remains stable as long as you keep it cold. The same concept can be applied to non-azo dyes that can exist as isomeric pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vXE1C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hindawi.com/journals/ijp/2009/434897/&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy a number of food dyes that provide the color you desire and can exist as isomers (&lt;em&gt;cis-trans&lt;/em&gt; or otherwise).  Dissolve each one separately (until you find one that works) in the minimum amount of one of the ingredients in your &quot;Blue Light Special&quot;.  Irradiate the (presumably colored) solution with a general lab laser (maybe you have a chemist friend who can do this for you) -  a laser pointer probably won't have enough power.  If the experiment is successful, the color should disappear from the solution.  Keep the solution cold in your bar fridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When someone orders the &quot;Blue Light Special&quot; mix all of the other ingredients in a glass at room temperature.  Add some of the cold, colorless (photobleached) dye solution to the room temperature mix.  As the cold dye warms up it should convert back to its colored form and the drink should &lt;strong&gt;become colored right before the customer's eyes.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-10T21:21:11.400" LastActivityDate="2015-03-10T21:21:11.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27162" PostTypeId="1" AcceptedAnswerId="27163" CreationDate="2015-03-10T21:14:44.203" Score="2" ViewCount="1657" Body="&lt;p&gt;Suppose there's a spaceship that has a hole in it. Gas is expanding through the hole from the spaceship to outer space (vacuum). According to pressure volume work ($P\Delta V$) the work done by the gas is infinity. According to my homework assignment, it's $0$. Can someone please explain why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_%28thermodynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$P$ denotes the pressure inside the system and outside the system, against which the system expands; the two pressures are practically equal for&#xA;  a reversible process;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can $P$ denote two different quantities? Can't this rule be applied for situations where the pressures inside and outside the system are different?&lt;/p&gt;&#xA;" OwnerUserId="2565" LastEditorUserId="5026" LastEditDate="2018-02-17T16:07:07.080" LastActivityDate="2018-02-17T16:07:07.080" Title="Why is the work done by a gas during free expansion zero and not infinity?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27163" PostTypeId="2" ParentId="27162" CreationDate="2015-03-10T21:24:56.330" Score="2" Body="&lt;p&gt;The external pressure in space is 0 (well, depends where and not quite, but for the discussion let's say that it is). So $P \cdot \Delta V$ is also 0. Gas expanding from a hole in a spaceship is not an irreversible process. Such processes are very slow and gradual, such that the external pressure has time to equilibrate with the internal pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary - you have the most irreversible process in here. Gas leaking from a hole in a spaceship into a medium (space) that will never equilibrate with the internal pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it from an intuitive point of view. Work done by an expanding system is there because it moves other things around. In space there is nothing to move, so no work done.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2015-03-10T21:24:56.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27165" PostTypeId="1" CreationDate="2015-03-10T22:21:01.697" Score="5" ViewCount="986" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWV4z.png&quot; alt=&quot;Question in photo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get everything up until the very end. We need to identify the bond enthalpies first:&#xA;\begin{align}&#xA;\ce{C=C} &amp;amp;= 612\\&#xA;\ce{O=O} &amp;amp;= 248\\&#xA;\text{SUM of these}&amp;amp;= 860\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{C-C} &amp;amp;= 348\\&#xA;2\cdot(\ce{C-O}) &amp;amp;= 720\\&#xA;\text{SUM of these} &amp;amp;= 1068\\&#xA;\end{align}&#xA;We know that enthalpy change is = products - reactants, which is $1068 - 860 = 208$, but the problem comes here: I know a minus sign is supposed to go there, but&#xA;how do we know that? Exothermic reaction? Is it because of the oxygen, suggesting it's a combustion reaction or not?&lt;/p&gt;&#xA;" OwnerUserId="14888" LastEditorUserId="4945" LastEditDate="2018-02-27T04:46:29.107" LastActivityDate="2018-02-27T04:52:20.407" Title="How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27166" PostTypeId="1" CreationDate="2015-03-11T00:22:36.360" Score="0" ViewCount="1233" Body="&lt;p&gt;I placed a container with 1000 grams (1 kg) of water at room temperature (20 degrees Celsius) inside a vacuum at 0.03 atm pressure. The water boiled for a few seconds then froze solid.&#xA;I need to calculate how much water evaporated before freezing occurred. Anyone know a formula I could use?&lt;/p&gt;&#xA;" OwnerUserId="14890" LastEditorDisplayName="user7951" LastEditDate="2019-06-09T17:18:21.720" LastActivityDate="2019-06-09T17:18:21.720" Title="How much water evaporates in a vacuum before freezing?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27167" PostTypeId="1" CreationDate="2015-03-11T00:33:51.887" Score="6" ViewCount="2274" Body="&lt;p&gt;I'm interested in the mushroom cloud produced by a nuclear explosion: What is it made of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hWn54.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the uranium fission reaction is the following (from &lt;a href=&quot;https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/nuclear-chemistry-19/nuclear-fission-137/the-atomic-bomb-544-3501/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this web page&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{{}^{235}U + {}^1_0n -&amp;gt; {}^{95}Sr + {}^{139}Xe +2~{}^1_0n + 180~MeV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that these elements are not stable and decay, so what is left over?&lt;/p&gt;&#xA;" OwnerUserId="9755" LastEditorUserId="5354" LastEditDate="2015-04-23T02:23:23.497" LastActivityDate="2018-05-09T16:16:30.237" Title="What chemical elements does a nuclear explosion release into its mushroom cloud?" Tags="&lt;nuclear&gt;&lt;explosives&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27168" PostTypeId="2" ParentId="27150" CreationDate="2015-03-11T01:21:46.027" Score="2" Body="&lt;p&gt;The cylinder contains $\pu{19kg}$ of LPG, $\pu{60\%}$ of which is propane (presumably by mass) so we have $\pu{11.4kg}$ propane and the rest is butane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of propane $\ce{C3H8}$ is $\pu{3*12+8*1 = 44 g/mol}$, so we can work out how many moles of propane there are. Then do the same for butane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now we use the ideal gas law:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;You are not interested in the pressure inside the cylinder.&lt;/strong&gt; You are interested in what the volume is when the LPG uncompresses. &lt;strong&gt;Therefore the pressure you need to use is atmospheric pressure.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply the formula for the ideal gas law $PV=nRT$ you will find that $1$ mole of gas occupies about $\pu{24 L}$ at $\pu{25^oC}$ ($\pu{298K}$) and atmospheric pressure. That is a good shortcut to remember for cases like this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that the ideal gas law only works when the conditions are in the range where the gas behaves like an ideal gas. You know the volume of the cylinder, but you cannot use the ideal gas law to find the pressure inside the cylinder, because the gas behaves non-ideally at high pressure (It liquifies!)&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="39539" LastEditDate="2017-05-23T21:44:09.353" LastActivityDate="2017-05-23T21:44:09.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27169" PostTypeId="1" AcceptedAnswerId="27186" CreationDate="2015-03-11T03:21:57.423" Score="4" ViewCount="930" Body="&lt;p&gt;I know GAMESS is not best suited for crystal simulation but I am not ready to start learning another package as VASP. As I heard GAMESS is able to do some calculations in a &lt;strong&gt;&quot;terminated&quot;&lt;/strong&gt; crystal (I mean using an ion in the periphery instead of repeating it multiple times).  Mostly I need to calculate HOMO-LUMO gap / hessian matrix and transitional state of a surface reaction .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked in the web but couldn't find a good worked example of using GAMESS for crystals.  Is there anywhere I can find such an example ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="5017" LastEditDate="2015-03-11T13:32:18.863" LastActivityDate="2015-03-11T13:41:50.613" Title="How to run crystal simulation on GAMESS?" Tags="&lt;computational-chemistry&gt;&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27171" PostTypeId="1" CreationDate="2015-03-11T04:42:57.483" Score="1" ViewCount="8082" Body="&lt;p&gt;I've often herd that aryl halides are very less reactive towards nucleophilic substitution reactions. One of the reasons for this I found was that there is resonance effect in the compound, but how is  that responsible for it to less reactive towards S.N. reactions and in addition to resonance effect are there other factors also responsible ?&lt;/p&gt;&#xA;" OwnerUserId="5423" LastActivityDate="2015-03-11T13:56:21.823" Title="Why aryl halides are very less reactive towards nucleophilic substitution reactions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-11T16:45:34.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="27173" PostTypeId="1" CreationDate="2015-03-11T06:18:22.620" Score="0" ViewCount="1903" Body="&lt;p&gt;Can anyone explain the resonating structure of halo-benzene? How does the transfer of an electron from the halogen atom to the α-carbon and from there to ß-carbon and so on take place?&lt;/p&gt;&#xA;" OwnerUserId="5423" LastEditorUserId="4945" LastEditDate="2015-03-11T08:17:59.160" LastActivityDate="2015-03-11T15:26:35.650" Title="What is the reasoning behind the resonating structure of halo-benzene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-03-16T12:42:40.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="27174" PostTypeId="2" ParentId="27159" CreationDate="2015-03-11T07:34:17.187" Score="0" Body="&lt;p&gt;I guess you are confused by sign convention of physics and chemistry. In chemistry, aluminium oxide is the system, and the reaction releases energy, and that energy is obviously coming out of the system. Since it is a loss of energy for the system it is referred as negative sign. But in physics the sign convention is just the opposite. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="4945" LastEditDate="2015-03-11T08:35:29.613" LastActivityDate="2015-03-11T08:35:29.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27175" PostTypeId="1" CreationDate="2015-03-11T07:40:23.617" Score="6" ViewCount="363" Body="&lt;p&gt;Gold has a special property called aurophilicity. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/006042995X&quot;&gt;Inorganic chemistry by Huheey&lt;/a&gt; has some explanation, but that was not enough to clear the conception. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="4945" LastEditDate="2015-03-11T08:26:57.653" LastActivityDate="2015-03-11T08:26:57.653" Title="What is the most accepted explanation of aurophilicity?" Tags="&lt;inorganic-chemistry&gt;&lt;elements&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27176" PostTypeId="2" ParentId="4033" CreationDate="2015-03-11T07:56:56.437" Score="2" Body="&lt;p&gt;I would expect that &lt;a href=&quot;https://en.wikipedia.org/wiki/Gilding&quot; rel=&quot;nofollow&quot;&gt;gilding&lt;/a&gt; - classic manual gold plating using &lt;a href=&quot;https://en.wikipedia.org/wiki/Gold_leaf&quot; rel=&quot;nofollow&quot;&gt;gold leaf&lt;/a&gt; - works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(But as the frequency, amplitude and radius of flexing is not known, it's hard to tell)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be starting with the polished polymer surface.&lt;br&gt;&#xA;That is coated with a special glue,&lt;br&gt;&#xA;the gold leaf is just layed down on it,&lt;br&gt;&#xA;and the gold surface is polished/burnished to create the mirror surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficult part may be finding the right glue substance, which depends on environment conditions. But a standard product should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - why do I propose this ancient method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see the advantage in using an actual (microscopic) compact layer of sheet metal.&lt;br&gt;&#xA;Gold leaf is just a thin gold foil.&lt;br&gt;&#xA;But there is a special property in gold: it's very ductile. It's so ductile, that gold leaf can actually be produced roughly by taking a piece of gold and hammering it flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I would think that the gold layer will not have much problems with the flexing polymer under it.&lt;br&gt;&#xA;Another important property of gold here is the innertness of the metal, because we have the mirror layer directly exposed to air - not behind glass like in &quot;consumer mirrors&quot;. With silver, that would be a problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold makes a very good mirror for many wavelengths - you did not say which you are interested in. But if the yellow color tint of reflected visible light is a problem, one could try the same with other metals. No other metal has the level of ductility of gold, so they are possibly less suitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes me think that creating a sheet of metal first, and putting it on the surface in a second step is that any kind of metal plating where the metal is deposited on the surface will follow any roughness of the surface, which will cause irregular thickness and mechanical strength. On the other hand, the layer following the contours has a larger area, when looking closely, which shoud help with mechanical stretching.&#xA;The typical thikness of deposited metas ist thinner than a foil. It can be made as thick for sure, but this could make coating expensive in terms of processing time.&lt;/p&gt;&#xA;" OwnerUserId="5078" LastEditorUserId="5078" LastEditDate="2015-03-13T12:42:49.160" LastActivityDate="2015-03-13T12:42:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27177" PostTypeId="2" ParentId="27150" CreationDate="2015-03-11T09:14:21.877" Score="1" Body="&lt;p&gt;Ideal Gas Law only works when a gas behaves like an ideal gas. It cannot be used in this case for finding pressure inside the cylinder as LPG behaves non-ideally at high pressure (it liquefies). Thanks to @steveverrill for pointing this out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we will use Ideal Gas Law to find out what the volume of LPG will be at normal atmospheric pressure (1 atm) and a temperature of 20 degrees Celsius, when it is still a gas:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A 19kg cylinder contains 19kg of LPG.&#xA;Mass of propane                    = 60% of 19kg = 11.4kg = 11400g&#xA;Mass of butane                     = 40% of 19kg = 7.6kg  = 7600g&#xA;Molar mass of propane              = 44.1g/mol&#xA;Number of moles of propane         = 11400g / 44.1g/mol = 258.5mol&#xA;Molar mass of butane               = 58.12g/mol&#xA;Number of moles of butane          = 7600g / 58.12g/mol = 130.8mol&#xA;Total number of moles in 19kg LPG  = 258.5 + 130.8 = 389.3mol&#xA;&#xA;Pressure x Volume = Number of moles x Gas constant x Temperature&#xA;1 atm x V = 389.3mol x 0.00008205736 m3 atm K-1 mol-1 x 293.15 K&#xA;V = 9.365 m3  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, there is 9.365 cubic metres of LPG at normal atmospheric pressure and temperature of 20 degrees Celsius. &lt;/p&gt;&#xA;" OwnerUserId="14884" LastActivityDate="2015-03-11T09:14:21.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27179" PostTypeId="1" CreationDate="2015-03-11T09:58:45.057" Score="1" ViewCount="167" Body="&lt;p&gt;When dealing with different problems in thermodynamics&#xA;I noticed that some of the problem consider the number of moles in calculating the heat absorbed or released, whereas some don't and they deal with the heat,capacity,difference in temperature WITHOUT considering the number of moles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example1:       1 gram of a sample burns in .....&#xA;they divide the grams over the molar mass to get the moles, then multiply by the enthalpy and solve for the needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example2:       1 gram of a sample burns in ....&#xA;they get one one of (q,C,delta H) directly without considering the number of moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what's the difference?&lt;/p&gt;&#xA;" OwnerUserId="14840" LastActivityDate="2015-03-11T09:58:45.057" Title="when do we consider the moles in bomb calorimeter?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27180" PostTypeId="1" AcceptedAnswerId="27185" CreationDate="2015-03-11T10:09:34.863" Score="-1" ViewCount="3506" Body="&lt;p&gt;Why is Hydrogen so reactive? What makes it combustible?&lt;/p&gt;&#xA;" OwnerUserId="14893" LastEditorUserId="14893" LastEditDate="2015-03-12T03:12:43.880" LastActivityDate="2015-03-12T03:12:43.880" Title="Why is hydrogen gas so highly reactive?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-11T13:42:09.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="27183" PostTypeId="1" CreationDate="2015-03-11T11:02:17.960" Score="1" ViewCount="22" Body="&lt;p&gt;I noticed some problems consider the samples in a calorimeter as a solution thereby calculating the mass of the &lt;strong&gt;combining samples&lt;/strong&gt;, whereas in others I found that they separate the samples into different calculations by using the &lt;strong&gt;distinct mass&lt;/strong&gt; and specific heat for each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could you explain why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks|!&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="7448" LastEditDate="2015-03-11T14:02:09.967" LastActivityDate="2015-03-11T14:02:09.967" Title="when do we consider the samples in a calorimeter as a solution in calculation?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27184" PostTypeId="2" ParentId="27166" CreationDate="2015-03-11T13:33:25.817" Score="6" Body="&lt;p&gt;You will need to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the heat capacity of water $C_p$&lt;/li&gt;&#xA;&lt;li&gt;the latent heat of fusion of ice $L_f$&lt;/li&gt;&#xA;&lt;li&gt;the latent heat of vaporization of water $L_v$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;First determine how much heat must be lost to change the temperature from 20 to 0 using the heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next determine how much heat must be lost to change from 0 degree water to 0 degree ice using the latent heat of fusion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, add the two heats from above together to get a total heat loss, and use the heat of vaporization of water to calculate the amount of water evaporated that corresponds to this total heat loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The above involves an approximation that the amount of water evaporated is relatively small.  In reality, once an infinitesimal portion of water evaporates, the amount remaining that needs to cool will be infinitesimally less than one kg.  If you are expected to know and use intergral calculus for your class, you may need to take into consideration how the mass is varying thoughout the process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate more accurately, break the problem is into two parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, a mass m1 evaporates, to cool water from 20 degres to 0 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $m$ be the varible liquild mass during the process and temperature be $t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mC_pdt=L_vdm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_p\int_{20}^{0}dt = L_v \int_{1}^{1-m1}(dm/m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining mass is 1kg-m1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a mass m2 evaporates, to freeze the remaining water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_v m2 = L_f (1kg - m1 - m2)$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-12T15:21:10.900" LastActivityDate="2015-03-12T15:21:10.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27185" PostTypeId="2" ParentId="27180" CreationDate="2015-03-11T13:41:25.327" Score="2" Body="&lt;p&gt;Hydrogen is not particularly reactive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, just mixing hydrogen and oxygen gas will not cause a reaction at room temperature, but many metal elements oxidize at least on the surface in air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most reactive elements in my opinion are fluorine in the non-metals, and the alkali metals like sodium and potassium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is combustible because oxygen wants electrons and takes them from hydrogen to form water.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-11T13:41:25.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27186" PostTypeId="2" ParentId="27169" CreationDate="2015-03-11T13:41:50.613" Score="3" Body="&lt;p&gt;As you said, there are many packages capable of handling calculations of crystals, slabs, etc. with periodic boundary conditions. VASP is only one of dozens. (That's probably a separate question in its own right.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAMESS does not include treatment of periodic boundary conditions, so you would need to run a calculation on a &quot;cluster.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention wanting to treat a surface reaction, you'd use a builder (e.g., &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;) to create a large cluster (e.g., Build-&gt;Supercell). So rather than one primitive unit cell, you create maybe a 2x2x2 or 3x3x3 slab. If needed, you'll terminate dangling bonds (e.g., adding hydrogens to cap oxygens in $\ce{TiO2}$ for example.. notice the terminal &quot;uncapped&quot; atoms on the back and right of the unit cell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DivJZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, after running calculations in GAMESS, you'd need to check for convergence. So you'd want to run calculations with increasing system size and see if the properties converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this isn't as strange as it sounds. In some cases ($\ce{MgO}$ and $\ce{TiO2}$ are good examples) the properties require more advanced calculation methods like multi-configuration QM, and this may be easier to do in a quantum chemical package with a cluster, rather than a solid-state crystal code.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-11T13:41:50.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27187" PostTypeId="2" ParentId="27171" CreationDate="2015-03-11T13:56:21.823" Score="2" Body="&lt;p&gt;No, that is a very incorrect explanation for why (unsubstituted) aryl halides do not participate in nucleophilic substitution reaction. Conjugating effect between the halide substituent and the benzene ring is not responsible for this. The problem is that there is simply no available mechanism to perform the substitution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{S_{N}2}$ is impossible since  inversion would give an incredibly weird strained ring (even more weird than a bridgehead carbon that is already forbidden by Bredt's rule).&lt;/li&gt;&#xA;&lt;li&gt;$\ce{S_{N}1}$ would leave a very unstable empty $\ce{sp^2}$ orbital. This can happen, but only for the very best leaving groups, like $\ce{N2}$.&lt;/li&gt;&#xA;&lt;li&gt;There is no electron stabilizing group that is conjugated with the halide group. This renders an addition-elimination type mechanism impossible.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-11T13:56:21.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27188" PostTypeId="1" AcceptedAnswerId="103627" CreationDate="2015-03-11T14:33:53.160" Score="5" ViewCount="584" Body="&lt;p&gt;IUPAC has a system for naming organic compounds, ionic compounds, and inorganic compounds. I have seen something vaguely suggesting there is one for intermetallic compounds, such as: &lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe-Co}$$&lt;/span&gt; What is the system, if it even exists? If it doesn't exist what is the generally accepted way of naming intermetallic compounds?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="402" LastEditDate="2019-05-17T06:05:18.580" LastActivityDate="2019-05-17T06:05:18.580" Title="What is the IUPAC nomenclature for Intermetallic Compounds?" Tags="&lt;nomenclature&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27190" PostTypeId="2" ParentId="27173" CreationDate="2015-03-11T15:26:35.650" Score="3" Body="&lt;p&gt;I think the misunderstanding here (it's a common one) is over the interpretation of resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals. So electrons do not actually move from the halogen to the $\alpha$ and $\beta$ carbons but what the resonance structures represent is that, only considering mesomeric effects, there is a greater electron density on the ortho and para carbons and than on the meta carbons, the carbon attatched to the halogen and the halogen itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7PxyG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not the full picture because halogens; especially fluorine and chlorine, have a very strong inductive effect and so there is actually considerably more electron density on the halogen than on any other atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrostatic potential map for chlorobenzene:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s6NHQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-11T15:26:35.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27191" PostTypeId="1" AcceptedAnswerId="27197" CreationDate="2015-03-11T15:32:08.783" Score="0" ViewCount="1523" Body="&lt;p&gt;Specifically $\ce{KF}$ into the $\ce{K+}$ and $\ce{F-}$ ions. I can find tables on the DE into atoms, but where would I go to find the DE between ions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also given that the bond length is 2.17e-10m. Can I use that at all?&lt;/p&gt;&#xA;" OwnerUserId="14899" LastEditorUserId="10185" LastEditDate="2015-03-11T18:26:42.253" LastActivityDate="2015-03-11T19:34:09.407" Title="How do you figure out the dissociation energy for ions?" Tags="&lt;homework&gt;&lt;bond&gt;&lt;ions&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27192" PostTypeId="2" ParentId="7233" CreationDate="2015-03-11T15:57:29.233" Score="0" Body="&lt;p&gt;Because the oxygens are shared with more than one silicon,so in large crystal structure the ratio of silicon to oxygens will be 1:2 not 1:4&lt;/p&gt;&#xA;" OwnerUserId="14900" LastActivityDate="2015-03-11T15:57:29.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27194" PostTypeId="2" ParentId="27165" CreationDate="2015-03-11T16:17:28.047" Score="1" Body="&lt;p&gt;When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants.  In other words &quot;reactants - products&quot;.  This is because all bond enthalpies are positive, since energy must be put into the bond to break it.  Bond formation releases energy; the minus sign compensates for this.  In short, bonds of reactant molecule must break, and new are formed to form product molecules.  So, your answer should be negative.  &lt;/p&gt;&#xA;" OwnerUserId="14596" LastActivityDate="2015-03-11T16:17:28.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27196" PostTypeId="1" CreationDate="2015-03-11T17:03:10.707" Score="3" ViewCount="597" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/icTlb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critical points are defined (by wikipedia) as the &quot;end-point of the phase equilibrium curve&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say this diagram therefore has three critical points. But this is unnatural. I doubt its true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other curves are different from the ones with critical points because &quot;they can run to infinity&quot;. But I think thats ridiculous because its physically impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I'm interpreting PDs correctly but I think that what defines that you are on a phase in these diagrams is that if you travel along a horizontal line you meet with a phase equilibrium curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't then you were either on a single phase the whole time or on a sort of &quot;critical phase&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't like it though. For example if you travel along a horizontal line that wont intersect the L-V curve, how would you know when the LV critical phase is present? Regions are not clearly defined here, except for monoclinic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats wrong with my ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many critical points does this diagram really have and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="7448" LastEditDate="2015-03-11T17:05:00.337" LastActivityDate="2015-03-11T18:23:25.013" Title="How many critical points? Why?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27197" PostTypeId="2" ParentId="27191" CreationDate="2015-03-11T18:08:22.893" Score="0" Body="&lt;p&gt;If you are trying to find the heterolytic dissociation energy for a gaseous $\ce{KF}$ molecule then as a rough estimate you can model the ions as point charges and calculate the energy required to separate them from their equilibrium separation (the bond length) to infinity, using Coulomb's Law: $$E = -\int_{2.17\times10^{-10}}^\infty\frac{Qq}{4\pi\epsilon_0r^2}dr$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that modelling the ions as point charges is a major simplification and so this is probably not a very accurate estimate. Unfortunately I can't find any data for the actual energy so I can't say how accurate this is.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-03-11T19:34:09.407" LastActivityDate="2015-03-11T19:34:09.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27198" PostTypeId="2" ParentId="27196" CreationDate="2015-03-11T18:23:25.013" Score="6" Body="&lt;p&gt;This phase diagram has one critical point - it is where the line separating the liquid and gas comes to an end - it is now one fluid. So, you can go from liquid to gas (or vice versa) by swinging T,P around the critical point without undergoing a first order phase transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points with arrows pointing to them are triple points, which are also special points, but different from the critical point. Movement away from a triple point means that one (or more) of the phases have to undergo a first order phase transition to the stable phase(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically, these are very different things.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-03-11T18:23:25.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27200" PostTypeId="1" CreationDate="2015-03-11T19:24:34.333" Score="9" ViewCount="7131" Body="&lt;p&gt;I'm just a simple man curious about things. I've no big understanding of chemistry. I had this question today in mind. What chemicals could I use to achieve the cleanest metallic surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I could first start by using isopropyl alcohol to remove all dirt and oils. But there are other liquids such as acetone and acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were a chemist, how would you clean a metallic surface?&lt;/p&gt;&#xA;" OwnerUserId="14878" LastActivityDate="2016-03-07T13:48:02.487" Title="What are the best chemical liquids to clean metallic surfaces?" Tags="&lt;metal&gt;&lt;cleaning&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27201" PostTypeId="1" AcceptedAnswerId="27225" CreationDate="2015-03-11T19:27:20.123" Score="5" ViewCount="1412" Body="&lt;p&gt;The following reaction occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jdimq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{HOO^{-}}$ ion reacts as a nucleophile at carbon number three and consequently the reaction is conjugate addition. However, Why would a peroxide not react through direct addition. I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur (Baeyer-Villiger reaction: &lt;a href=&quot;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&lt;/a&gt;). Why must it react at the alkene?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2015-03-11T19:30:29.700" LastActivityDate="2015-03-13T23:50:46.297" Title="Why does basic peroxide react via conjugate addition rather than through direct addition?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27203" PostTypeId="1" CreationDate="2015-03-11T19:44:10.097" Score="5" ViewCount="17687" Body="&lt;p&gt;I did an experiment irradiating benzophenone in UV light and worked out the rate constant for the reaction. Then I repeated the experiment using various concentrations of a quencher and worked out their respective rate constants. Now I need to calculate the rate constant for quenching $k_\mathrm{q}$ using a Stern-Volmer plot. How would I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation I have used before for the Stern-Volmer relationship is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{I_\mathrm{f}} = \frac{1}{I_\mathrm{abs}}\left(1 + \frac{k_\mathrm{q}[\ce{Q}]}{k_\mathrm{lum}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the &lt;em&gt;x&lt;/em&gt;-axis is $[\ce{Q}]$, the concentration of the quencher, and the &lt;em&gt;y&lt;/em&gt;-axis is $1/I_\mathrm{f}$, and to work out $k_\mathrm{q}$ you calculate the gradient/&lt;em&gt;y&lt;/em&gt;-intercept = $k_\mathrm{q}/k_\mathrm{lum}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I would plot a graph of concentration of quencher $[\ce{Q}]$ on the $x$-axis against 1/rate constant on the y axis. But how would I work out $k_\mathrm{q}$?&lt;/p&gt;&#xA;" OwnerUserId="14905" LastEditorUserId="16683" LastEditDate="2016-12-14T09:38:40.343" LastActivityDate="2016-12-14T09:38:40.343" Title="Stern-Volmer plot and calculation of rate constant for quenching" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27204" PostTypeId="2" ParentId="27201" CreationDate="2015-03-11T19:55:47.270" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why would a peroxide not react through direct addition […]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Nobody says that it doesn't, but apparently that isn't a productive route: there are no products found that would result from such an attack. Consequently, this reaction, which cannot be ruled out, probably is reversible.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I know that if the double bond wasn't there it would attack the carbonyl directly and then a rearrangement would occur  […]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;True indeed, but the double bond &lt;strong&gt;is&lt;/strong&gt; there.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JlDEk.png&quot; alt=&quot;epoxidation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As a result, attack on the &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;-carbon of the enone, formation of a hydroperoxy enolate and subsequent formation of the epoxide are seemingly favoured.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-11T19:55:47.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27207" PostTypeId="2" ParentId="27201" CreationDate="2015-03-11T21:07:08.740" Score="1" Body="&lt;p&gt;First off, it may be useful to go back and take a look at this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/21867/what-makes-r-mgx-a-more-reactive-grignard-reagent-than-r-cux/21875#21875&quot;&gt;earlier question&lt;/a&gt; of yours and the answer to it. The concept of hard and soft electrophiles and nucleophiles was discussed.  Basically, it was said that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hard nucleophiles are generally small, highly charged atoms. Because&#xA;  of this they tend to be more reactive. They prefer to react with&#xA;  similar (hard) electrophiles. Their reactions are usually kinetically&#xA;  controlled with early transition states and are governed by&#xA;  electrostatic interactions. Soft nucleophiles are generally larger&#xA;  systems with a more diffuse charge. Because of this they tend to be&#xA;  less reactive (more discriminating). They prefer to react with similar&#xA;  (soft) electrophiles&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;take away&quot; is that a hard electrophile prefers to react with a hard nucleophile while a soft electrophile prefers to react with a soft nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the system at hand, the nucleophile is the hydroperoxide anion ($\ce{HOO^{-}}$).  You can draw two resonance structures to describe the anion, delocalizing the negative charge over both oxygens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H-O-O^{-} &amp;lt;-&amp;gt; H^{+}~ [O-O]^{-2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, this anion is a softer nucleophile than $\ce{HO^{-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbonyl group is a hard electrophile, while an $\ce{\alpha,\beta}$ unsaturated carbonyl, where the electron density is distributed over 4 atoms is a softer electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soft reacts with soft, so would you expect the soft hydroperoxide anion to preferrentially react with a hard (carbonyl; 1,2 addition) or soft ($\ce{\alpha,\beta}$ unsaturated carbonyl; 1,4 addition) electrophile?&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-12T14:59:03.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27208" PostTypeId="2" ParentId="27203" CreationDate="2015-03-11T21:11:46.460" Score="0" Body="&lt;p&gt;If you plot such a graph you should obtain a straight line with some slope $a$ and intercept $b$: $y=ax+b$. You can see that also Stern Volmer relation is a linear function (of concentration). Thus, $k_q$ can be calculated by equating value of slope you obtained ($a$) to $k_q/klum$. Such procedure is simple and important to understand as can be used to determine many quantities which occur in linear equations.&lt;/p&gt;&#xA;" OwnerUserId="14772" LastActivityDate="2015-03-11T21:11:46.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27210" PostTypeId="2" ParentId="27200" CreationDate="2015-03-12T00:27:06.813" Score="5" Body="&lt;p&gt;Assuming clean means anti-pathogenic, I would use Hydrogen Peroxide &quot;oxyclean&quot; since it can &quot;clean&quot; metals by removing pathogens. $\ce{H2O2}$ is marketed in various strengths. 1% dilution can kill up to 99.99% bacteria in 30 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any butyl based cleaning product is a very good degreaser but it may harm rubber products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.mbzponton.org/valueadded/other/clean.htm&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.finishing.com/133/33.shtml&quot;&gt;here&lt;/a&gt; for more.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-03-12T00:27:06.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27211" PostTypeId="2" ParentId="27167" CreationDate="2015-03-12T00:33:06.360" Score="3" Body="&lt;p&gt;According to &lt;em&gt;The Preparatory Manual of Explosives&lt;/em&gt;, the mushroom cloud is produced by the partial vacuum left after initial expansive shock wave. That implies that it is primarily the condensation of water that makes it visible.  These clouds form in high explosive as well as in nuclear detonations near the ground. See &lt;a href=&quot;https://books.google.com/books?id=J55D3HcgPuoC&amp;amp;pg=PA98#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;https://books.google.com/books?id=J55D3HcgPuoC&amp;amp;pg=PA98#v=onepage&amp;amp;q&amp;amp;f=false&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-12T00:33:06.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27212" PostTypeId="1" CreationDate="2015-03-12T00:46:05.103" Score="1" ViewCount="896" Body="&lt;p&gt;The solubility product of $\ce{AgCl}$ in water is given as $1.8\cdot10^{-10}$&#xA;Determine the solubility of $\ce{AgCl}$ in pure water and in a $0.25~\mathrm{M}$ $\ce{MgSO4}$ solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solve to the $K_\mathrm{{sp}}$ and got $1.34\cdot10^{-10}$&#xA;for Part B I was told to use the activity and am not sure if I did it right,&#xA;my new $K_\mathrm{{sp}}$ turn out to be $1.123\cdot10^{11}$&#xA;could someone help me with part B?&lt;/p&gt;&#xA;" OwnerUserId="14624" LastEditorUserId="865" LastEditDate="2021-05-22T21:24:57.297" LastActivityDate="2021-05-22T21:24:57.297" Title="How to calculate the solubility of AgCl in a MgSO₄ solution?" Tags="&lt;physical-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27214" PostTypeId="2" ParentId="26628" CreationDate="2015-03-12T01:41:11.217" Score="0" Body="&lt;p&gt;Two ways you can go about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, recall that the oxidation number of oxygen is always -2 beyond hydrogen peroxide and the elemental forms of oxygen (monoatomic, diatomic, and ozone). Then you can show that the oxidation number for chromium is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be clever (though not as efficient or just want another way to check yourself), you can attempt to balance the half-reaction like a half-reaction occurring in acidic solution. Why does that attempt show that no oxidation/reduction exists?&lt;/p&gt;&#xA;" OwnerUserId="12745" LastActivityDate="2015-03-12T01:41:11.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27215" PostTypeId="1" CreationDate="2015-03-12T01:44:17.010" Score="3" ViewCount="3560" Body="&lt;p&gt;Why does $\ce{Fe(CO)4}$ adopt a tetrahedral, as opposed to square planar, geometry? Here's my analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors for square planar:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;coordination: 4&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{Fe(0)}$: d8&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO}$: strong field ligand (gives large d-splitting)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Factors for tetrahedral:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;oxidation state: 0 (gives weak d-splitting)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO}$: relatively bulky (?) ligand&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general guidelines for determining whether or not a row 1 transition metal complex is square planar or tetrahedral (given it  has 4-coordination and d8 electrons), are the strength and bulkiness of the ligand. Generally, strong field ligands like $\ce{CO}$ and $\ce{CN^{-}}$ gives square planar complexes. In fact, $\ce{[Ni(CN)4]^{2-}}$ is square planar. Thus, the only way I could explain why $\ce{Fe(CO)4}$ would be tetrahedral would be because of its low oxidation state. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="14910" LastEditorUserId="4945" LastEditDate="2015-03-23T09:28:19.917" LastActivityDate="2015-03-23T09:28:19.917" Title="Why does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;molecular-structure&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27216" PostTypeId="1" CreationDate="2015-03-12T02:04:41.113" Score="0" ViewCount="2104" Body="&lt;p&gt;Bear with me, I'll get to the chemistry soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a float tank that I want to attach a skimmer to.  Unfortunately, the enclosure is weakened by cutting a hole out of the ABS plastic.  The weight of the salt water (1.25SG) to somehow causes a bend in the plastic just enough that getting a good seal isn't possible without additional support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The float tank consists primary of (pardon the units) 800lbs of $\ce{MgSO4}$ Heptahydrate raised to about 96 degrees Fahrenheit.  The pH is roughly 7.2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern with supporting the ABS plastic is that the Epsom salt water may corrode the metal, if it somehow isn't protected by a liner.  Would I absolutely need to have stainless steel, or could I get by with galvanized (which is much cheaper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it matters, I estimate the volume of water directly pressing on the skimmer opening to be about 9+ gallons (again at 1.25sg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for answering non-chemist,&#xA;Ben&lt;/p&gt;&#xA;" OwnerUserId="5597" LastEditorUserId="4097" LastEditDate="2015-03-12T16:54:49.850" LastActivityDate="2019-08-31T21:29:15.490" Title="Could I settile for galvanized steel over stainless for salt corrosion resistance?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27217" PostTypeId="1" AcceptedAnswerId="27218" CreationDate="2015-03-12T02:06:19.997" Score="5" ViewCount="272" Body="&lt;p&gt;I'd like some assistance in understanding the IUPAC name for tetrahedrane. I understand the basics of polycyclic ring naming however this example has me confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/etHuZ.png&quot; alt=&quot;Tetrahedrane&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8193" LastEditorUserId="4945" LastEditDate="2015-06-01T01:11:25.953" LastActivityDate="2015-06-01T01:11:25.953" Title="How to rationalise the IUPAC name for tetrahedrane?" Tags="&lt;nomenclature&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27218" PostTypeId="2" ParentId="27217" CreationDate="2015-03-12T02:22:53.687" Score="8" Body="&lt;p&gt;Let's start with bicyclobutane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/24yix.png&quot; alt=&quot;bicyclobutane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper name for this compound is bicyclo[1.1.0]butane. The compound is numbered starting at one bridgehead and working around. The numbers in brackets represent the number of atoms between the bridgeheads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, tetrahedrane, which has an extra bridge between carbon atoms 2 and 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mw01m.png&quot; alt=&quot;tetrahedrane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, we list all of the bridges in the brackets, but we have an extra bridge, and it is not between the main bridgeheads (1 and 3). We list this extra bridgehead last and then use superscript locants to indicate which ring atoms this bridge connects. This extra bridge is between carbon atoms 2 and 4, thus $0^{2,4}$: tricyclo[1.1.0.0$^{2,4}$]butane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more examples, including applying the rules to crazier ring systems, here is the &lt;a href=&quot;http://pac.iupac.org/publications/pac/pdf/1999/pdf/7103x0513.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC publication&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2015-03-12T10:33:21.750" LastActivityDate="2015-03-12T10:33:21.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27219" PostTypeId="1" CreationDate="2015-03-12T03:31:58.147" Score="5" ViewCount="92" Body="&lt;p&gt;I'm poaching some pears.  The recipe calls for bringing the liquid to a boil and adding the pears, then simmering until done, about 25 minutes.  It then calls for removing the pears from the liquid and letting both come to room temperature, then putting the liquid over the pears again and marinating overnight.  Separating the pears and the liquid will let the pears cool more quickly and presumably prevent overcooking.  I have heard a rule of thumb that every 10C in temperature doubles the reaction rate.  As the pears simmered for 25 minutes, it seems that the cooldown (even if they stayed in the liquid) will not add much to the reactions of cooking. Is there any sense to extracting the pears from the liquid?&lt;/p&gt;&#xA;" OwnerUserId="11344" LastEditorUserId="11344" LastEditDate="2015-03-12T04:28:28.700" LastActivityDate="2015-03-12T06:43:29.067" Title="Effect of temperature on cooking" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27220" PostTypeId="2" ParentId="27219" CreationDate="2015-03-12T06:43:29.067" Score="3" Body="&lt;p&gt;(&lt;em&gt;Note that I'm a rather bad cook, except in the lab!&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any sense to extracting the pears from the liquid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes! &lt;strong&gt;Yomen Atassi&lt;/strong&gt; has already explained that you actually have to consider two (distinct) processes in the recipe:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cooking the pears for a short period of will yield the right consistency. Additionally, some transformations happening during the cooking will possible break up some cells, convert some ingredients to even more tasty ones, and maybe even facilitate the later take up of other of other compounds from the liquid. If you cook for too long, the pears will be sludge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Marinating means that the compounds dissolved in the liquid will permeate the pear and add to the taste. Imagine that the pear acts like a spongue and sucks all the good stuff in from the marinade. This needs time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With a two-step procedure given in the recipe, you should get the best of both worlds and I hope that the dessert was really tasty.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-12T06:43:29.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27221" PostTypeId="1" AcceptedAnswerId="27240" CreationDate="2015-03-12T07:34:43.143" Score="1" ViewCount="5589" Body="&lt;p&gt;Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 + 2NaOH -&amp;gt; Na2O + H2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as opposed to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2 + 2NaOH -&amp;gt; Na2CO3 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in the context of fractional distillation of liquified air, before which carbon dioxide must be removed using sodium hydroxide. &lt;/p&gt;&#xA;" OwnerUserId="14911" LastEditorUserId="9961" LastEditDate="2015-10-25T22:00:34.467" LastActivityDate="2017-07-06T09:50:16.977" Title="Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27222" PostTypeId="2" ParentId="27215" CreationDate="2015-03-12T08:13:03.400" Score="6" Body="&lt;p&gt;The hypothetical $\ce{Fe(CO)4}$ complex is &lt;a href=&quot;http://en.wikipedia.org/wiki/Isoelectronicity&quot; rel=&quot;noreferrer&quot;&gt;isoelectronic&lt;/a&gt; to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.&lt;sup&gt;[1,2]&lt;/sup&gt; Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a good ligand causing a strong field and hence a strong splitting. In a tetrahedral complex there would be triply degenerated HOMO, which would only be occupied by four electrons. This arrangement cannot be more stable than a planar complex. However, the iron tetracarbonyl complex as a singular entity has not been observed (to my current knowledge). The observed structures for the compound are various dimers and trimers.&lt;sup&gt;[3]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a neutral iron carbonyl complex, but it binds five instead of only four ligands. Therefore $\ce{Fe(CO)5}$ is a trigonal bipyramidal 18 electron complex.&lt;sup&gt;[4]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observed $\ce{[Fe(CO)4]^{2-}}$ complex&lt;sup&gt;[5]&lt;/sup&gt; is isoelectronic to $\ce{Ni(CO)4}$ which is a tetrahedral 18 electron complex.&lt;sup&gt;[1,4]&lt;/sup&gt; Without going into more detail, the iron in this complex is of course $\ce{Fe^{-II}}$ and therefore $\mathrm{d^{10}}$. A square planar arrangement would no longer be more stable and a complex with a higher symmetry and more degeneracy is preferable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;References&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF00533488&quot; rel=&quot;noreferrer&quot;&gt;J. Demuynck, and A. Veillard, &lt;em&gt;Theor. chim. acta&lt;/em&gt;, &#xA;&lt;strong&gt;1973&lt;/strong&gt;, &lt;em&gt;28&lt;/em&gt; (3), 241-265.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.19995981.html&quot; rel=&quot;noreferrer&quot;&gt;ChemSpider.com&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01148a060&quot; rel=&quot;noreferrer&quot;&gt;Raymond K. Sheline, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1951&lt;/strong&gt;, &lt;em&gt;73&lt;/em&gt; (4), 1615–1618&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/om00028a082&quot; rel=&quot;noreferrer&quot;&gt;Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, &lt;em&gt;Organometallics&lt;/em&gt;, &lt;strong&gt;1993&lt;/strong&gt;, &lt;em&gt;12&lt;/em&gt; (4), 1481–1483.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00425a009&quot; rel=&quot;noreferrer&quot;&gt;Henry B. Chin, and Robert Bau, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1976&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt; (9), 2434–2439.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;&lt;br&gt;&#xA;Carbonyl is nowhere close to a bulky ligand. Apart from halides and hydrides it might as well be the smallest possible ligand.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-03-12T08:13:03.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27223" PostTypeId="2" ParentId="27203" CreationDate="2015-03-12T11:09:07.053" Score="5" Body="&lt;p&gt;Benzophenone is a notorious case where prompt fluorescence, phosphorescence and delayed fluorescence can mix. Have a look at &lt;em&gt;Modern Molecular Photochemistry&lt;/em&gt; by N.J. Turro for some general information and &lt;a href=&quot;http://dx.doi.org/10.1016/0009-2614(72)87176-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Prompt and delayed fluorescence from benzophenone&lt;/em&gt;&lt;/a&gt; by &#xA;R.E. Brown and L.A. Singer for the original reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\ce{BP &amp;amp;-&amp;gt;[h\nu]\ ^{1}BP^{\ast}}\\ &#xA;\ce{^{1}BP^{\ast} &amp;amp;-&amp;gt; BP + h\nu'\quad &amp;amp;(prompt\ fluorescence)}\\&#xA;\ce{^{1}BP^{\ast} &amp;amp;-&amp;gt;[ISC]\ ^{3}BP^{\ast} &amp;amp;(intersystem\ crossing)}\\&#xA;\ce{^{3}BP^{\ast} &amp;amp;-&amp;gt; BP + h\nu''\quad &amp;amp;(phosphorscence)}\\&#xA;\ce{^{3}BP^{\ast} &amp;amp;-&amp;gt;\ ^{1}BP^{\ast} &amp;amp;(repopulation\ of\ S_1)}\\&#xA;\ce{^{1}BP^{\ast} &amp;amp;-&amp;gt; BP + h\nu'\quad &amp;amp;(delayed\ fluorescence)}\\&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore all this (for a moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorescence quenching, i.e the observed decrease of fluorescence intensity in the presence of a quencher, may occcur through two different pathways:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Static fluorescence quenching&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;strong&gt;formation of a ground state complex between the fluorophore $\ce{F}$ a quencher $\ce{Q}$&lt;/strong&gt; decreases the concentration of $\ce{F}$, hence the concentration of the excited state $\ce{^1F^{\ast}}$ and the fluorescence intensity. &lt;strong&gt;This has no impact on the fluorescence lifetimes!&lt;/strong&gt; These are the same in the absence ($\tau_0$) and presence ($\tau$) of a quencher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\frac{\tau_0}{\tau} = 1&#xA;\] &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Dynamic fluorescence quenching&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;strong&gt;quenching occurs from the excited state $\ce{^1F^{\ast}}$&lt;/strong&gt;. With other words, the presence of a quencher introduces an additional, non-radiative pathway for the deactivation of the excited singlet state. Consequently, &lt;strong&gt;the fluorescence lifetime $\tau$ is shortened!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try to derive the &lt;em&gt;Stern-Volmer&lt;/em&gt; equation and see if that leads to some insights: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;You probably know all of this already, but I'm just in the mood...&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an experiment on dynamic fluorescence quenching, the following processes and their rate constants have to be considered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\ce{F + h\nu &amp;amp;-&amp;gt;[k_{abs}]\ ^1F^{\ast}\quad &amp;amp;\mathrm{(excitation)} } \\&#xA;\ce{^1F^{\ast} &amp;amp;-&amp;gt;[k_{ic}]\ F + heat\quad &amp;amp;(internal\ conversion)}\\&#xA;\ce{^1F^{\ast} &amp;amp;-&amp;gt;[k_{f}]\ F + h\nu'\quad &amp;amp;(fluorescence)}\\&#xA;\ce{^1F^{\ast} + Q &amp;amp;-&amp;gt;[k_{q}]\ F + Q'\quad &amp;amp;(quenching)}\\&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluorescence quantum yield $\Phi$ is given as the quotient of emitted ($I_{em}$) and absorbed ($I_{abs}$) photons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\Phi = \frac{I_{em}}{I_{abs}}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of emitted photons depends on the concentration of the fluorophore in its excited singlet state $\ce{^1F^{\ast}}$ and the rate constant $k_{f}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;I_{em} = k_{f}\cdot[^1F^{\ast}]&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of absorbed photons depends on the concentration of the fluorophore $\ce{F}$ and the rate constant $k_{abs}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;I_{abs} = k_{abs}\cdot[F]&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of absorption processes is difficult to measure, they're bloody fast. Fortunately, we don't have to! &lt;strong&gt;Under steady-state conditions, population of the excited singlet state and its deactivation balances out.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, we can express $I_{abs}$ as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;I_{abs} = k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for the fluorescence quantum yield thus is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\Phi &amp;amp;= \frac{k_{f}\cdot[^1F^{\ast}]}{k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]}\\&#xA;&amp;amp;= \color{\red}{\frac{k_{f}}{k_{f} + k_{ic} +k_{q}\cdot[Q]}}&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is pretty convenient since we don't have to think about the pesky $[^1F^{\ast}]$ anymore! &lt;strong&gt;YAY!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do perform a fluorescence measurement in the absence of the quencher,  $[\ce{Q}] = 0$, $\Phi_0$, the fluorescence quantum yield under this condition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\Phi_0= \frac{k_{f}}{k_{f} + k_{ic}}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of a quencher $\ce{Q}$ on the fluorescence is expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\color{red}{\frac{\Phi_0}{\Phi}} &amp;amp;= \frac{k_{f}}{k_{f} + k_{ic}} \cdot \frac{k_{f} + k_{ic} +k_{q}\cdot[Q]}{k_{f}}\\&#xA;&amp;amp;= \color{\red}{1+ \frac{k_{q}\cdot[Q]}{k_{f} + k_{ic}}}&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too bad that we don't know all these rate constants and using just a steady-state spectrometer, we won't be able to measure them! And for a lab chemist, there's too much stuff in the equation anyway! &lt;strong&gt;Bummer!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can argue about what influences the &lt;strong&gt;fluorescence lifetimes&lt;/strong&gt;, which, btw, can be measured!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lifetimes are pretty much the inverse of the (sum of the) rate constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the quencher, the fluorescence lifetime $\tau_0$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\tau_0 = \frac{1}{k_{f} + k_{ic}}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of a quencher, the fluorescence lifetime $\tau$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\tau = \frac{1}{k_{q}\cdot[Q] + k_{f} + k_{ic}}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we plug that in again and rearrange a bit, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\frac{\Phi_0}{\Phi} &amp;amp;= 1 + \tau_0\cdot k_q\cdot[\ce{Q}]\\&#xA;&amp;amp;= 1+ {K_{SV}}\cdot[\ce{Q}]&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We name $K_{SV}$ the Stern-Volmer constant!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Summary&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Plotting $\frac{\Phi_0}{\Phi}$ or the quotient of the areas of the fluorescence spectra vs the concentration of the quencher should give a straight line, that cuts the $y$ axis at 1.&lt;/li&gt;&#xA;&lt;li&gt;The slope of the curve is the Stern-Volmer constant $K_{SV}$.&lt;/li&gt;&#xA;&lt;li&gt;The real rate constant $k_q$ for the reaction of the excited state $\ce{^1F^{\ast}}$ with the quencher can only be obtained from $K_{SV}$ if the fluorescence lifetime $\tau_0$ has been obtained from a &lt;strong&gt;time-resolved fluorescence measurement&lt;/strong&gt;! &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2016-12-14T08:27:13.940" LastActivityDate="2016-12-14T08:27:13.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27225" PostTypeId="2" ParentId="27201" CreationDate="2015-03-12T12:03:56.530" Score="2" Body="&lt;p&gt;The reason that is given by Clayden et al. (chapter 23) is that because the peroxide has two adjecant lone pair on oxygen it exhibits an &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;alpha effect&lt;/a&gt;, which is more or less a delocalization interaction between the lone pair orbitals. For reasons not entirely known (the basis of the alpha effect is an open problem in chemistry) this results in an increase in energy of the electrons in the resulting delocalized HOMO, making it more nucleophilic. You can imagine that this delocalized HOMO is also softer than the single localized oxygen lone pair and hence it tends to react at the soft double bond instead of the hard carbonyl site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excerpt from Clayden et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zZpX2.png&quot; alt=&quot;Excerpt from Clayden et al.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-03-13T23:50:46.297" LastActivityDate="2015-03-13T23:50:46.297" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27227" PostTypeId="2" ParentId="27216" CreationDate="2015-03-12T12:53:07.147" Score="1" Body="&lt;p&gt;I'm a chemist, but this seems more like a question for an engineer. &#xA;Unfortunately I don't have some references to back up my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use galvanized steel you have a protective layer on the outside of the steel. As soon as you handle it for attaching, welding... you destroy/scratch/bruise this top layer and the underlying steel gets in contact with the salt solution which is normally aggressive towards metal. This mostly is right for e.g. chloride salts in a mix with oxygen and I'm not quite sure how corrosive MgSO4 is.&#xA;Here is a paper about corrosive currents of steel in Saltwater/cement solutions, but that`s not my speciality: &lt;a href=&quot;http://www.jmst.org/fileup/PDF/pass543.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.jmst.org/fileup/PDF/pass543.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stainless steel is more robust, because it's an alloy of steel and chromium. Therefor it's not just covered with a protective layer but saver as a whole. An I guess it's therefor recommended.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As a side note, some years ago I had a field trip to a big chemistry company with lots and lots of pipelines. We had some time there with the technical stuff in charge of teh pipelines. They explained, that the most annoying pipelines were the ones with salt water, since it was more aggressive towards the pipes than acids, mineral oils, gases etc.)&lt;/p&gt;&#xA;" OwnerUserId="14917" LastActivityDate="2015-03-12T12:53:07.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27229" PostTypeId="1" CreationDate="2015-03-12T13:20:40.557" Score="1" ViewCount="72" Body="&lt;p&gt;I have prepared hydrogel nanoparticles in aqueous solution. How to calculate concentration of hydrogel nanoparticles ?&lt;/p&gt;&#xA;" OwnerUserId="14919" LastActivityDate="2015-03-12T13:20:40.557" Title="How to calculate concentration of hydrogel nanoparticles" Tags="&lt;concentration&gt;&lt;nanotechnology&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2019-05-15T03:53:29.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="27230" PostTypeId="1" CreationDate="2015-03-12T13:30:44.027" Score="4" ViewCount="3632" Body="&lt;p&gt;&lt;a href=&quot;http://www.smbc-comics.com/index.php?id=3668&quot; rel=&quot;nofollow noreferrer&quot;&gt;I was reading todays SMBC&lt;/a&gt; which asked what happens if you make a gas of magnetic particles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pLD3C.jpg&quot; alt=&quot;Comic strip&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does happen if you make a gas of magnetic particles?&lt;/p&gt;&#xA;" OwnerUserId="14920" LastActivityDate="2015-03-12T14:10:27.887" Title="What happens if you make a gas of magnetic particles" Tags="&lt;gas-laws&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="11" ClosedDate="2015-03-12T19:22:36.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="27231" PostTypeId="1" AcceptedAnswerId="27237" CreationDate="2015-03-12T13:47:34.963" Score="0" ViewCount="1780" Body="&lt;p&gt;Loosely speaking, is the reduction potential equivalent to the ease of discharge in the context of electrolysis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. is it easier to discharge an ion with a higher reduction potential as opposed to one with a lower one?&lt;/p&gt;&#xA;" OwnerUserId="14911" LastActivityDate="2015-03-12T15:20:16.757" Title="Is reduction potential equivalent to the ease of discharge?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27233" PostTypeId="1" CreationDate="2015-03-12T14:10:13.533" Score="1" ViewCount="113" Body="&lt;p&gt;I have 20mL gas samples and I need to determine the [CO2] in ppm. I have a Li-Cor 6200 with a closed loop, so essentially I will inject a volume of gas into the loop and the change in the output is how to determine the concentration of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calibrate I'm using a standard of gas that is 5020ppm CO2, and injecting volumes of 50, 75, &amp;amp; 100 uL then recording the change in [CO2] in the closed loop.&#xA;After calibration I will inject 50uL of sample and record the change of CO2 in the closed loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I set up the conversion? It is spring break and I am having trouble finding help around campus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the average change in CO2 output from 3 different volumes of standard injected, the output changes 0.114 per uL of standard gas injected (5020ppm CO2). So, (5020/.114)=44035.0877. Then, I just multiply the change in output per uL of sample injected by 44035.0877 to get the CO2 ppm of the sample.&#xA;My first sample I inject 50uL and the output changes by 20.2, so I divide to get the change per uL; (20.2/50=0.404). Then multiply, (0.404 * 44035.0877 = 17790.18 ppm CO2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it really this simple, or is my thinking to simple?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. This is my first post, I hope I have followed the rules.&lt;/p&gt;&#xA;" OwnerUserId="14923" LastActivityDate="2015-03-12T14:10:13.533" Title="Determining [CO2] of a sample in ppm" Tags="&lt;gas-laws&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27234" PostTypeId="2" ParentId="27230" CreationDate="2015-03-12T14:10:27.887" Score="-1" Body="&lt;p&gt;If you heat up the magnet above Curie temperature (or more drastically just evaporates it), the magnetic spins will be randomly aligned and the magnetic field  is not produced anymore (at higher temperature). &lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-03-12T14:10:27.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27235" PostTypeId="1" AcceptedAnswerId="64196" CreationDate="2015-03-12T14:47:51.340" Score="5" ViewCount="20040" Body="&lt;p&gt;Mixtures of ethanol and acetone have vapor pressures that are &lt;em&gt;higher&lt;/em&gt; than predicted by Raoult's law, while mixtures of acetone and water have &lt;em&gt;lower&lt;/em&gt; vapor pressures than predicted by Raoult's law.  Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, why is there a &quot;positive deviation&quot; from Raoult's law for acetone/ethanol solutions but a &quot;negative deviation&quot; for acetone/water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't an acetone-ethanol hydrogen bond stronger than an acetone-acetone hydrogen bond? If the reason for the positive deviation of the ethanol/acetone solution is that the H bonds there are weaker than ethanol-ethanol H bonds, then there wouldn't that imply that acetone/water solutions would also have a positive deviation, as water-water H bonds must be much stronger than water-acetone, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there some other explanation?&lt;/p&gt;&#xA;" OwnerUserId="14573" LastEditorUserId="13184" LastEditDate="2016-12-08T18:05:10.377" LastActivityDate="2016-12-09T17:19:37.500" Title="the chemical mechanism for deviations from Raoult's law" Tags="&lt;solutions&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27236" PostTypeId="1" AcceptedAnswerId="27241" CreationDate="2015-03-12T14:49:15.867" Score="4" ViewCount="4428" Body="&lt;p&gt;What gas is released during dissolving solid $\ce{NaOH}$ in a high purity water? Water more than 10 M$\Omega$/cm and standard quality $\ce{NaOH}$.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="10185" LastEditDate="2015-03-18T17:50:46.740" LastActivityDate="2015-03-18T17:50:46.740" Title="What gas is released during dissolving NaOH" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;identification&gt;" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27237" PostTypeId="2" ParentId="27231" CreationDate="2015-03-12T15:20:16.757" Score="3" Body="&lt;p&gt;Easy is a big word. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower potential means it requires less energy, but that doesn't say anything about how quick the reaction will proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A low potential reaction may still have slow kinetics and therefore not be &quot;easy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are comparing ions as to which will produce more product as a result of an electrochemical reaction, you need to find something with low potential and fast kinetics.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-03-12T15:20:16.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27238" PostTypeId="2" ParentId="27236" CreationDate="2015-03-12T15:58:32.130" Score="4" Body="&lt;p&gt;That is simply air included in the evaporation/drying process. &lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditDate="2015-03-12T17:01:31.397" LastActivityDate="2015-03-12T17:01:31.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27240" PostTypeId="2" ParentId="27221" CreationDate="2015-03-12T16:21:08.740" Score="3" Body="&lt;p&gt;Sodium oxide is not a base in either the Bronsted-Lowry or Arrhenius senses.  It is technically a &lt;a href=&quot;http://en.wikipedia.org/wiki/Base_anhydride&quot; rel=&quot;nofollow&quot;&gt;base anhydride&lt;/a&gt;, meaning that it can be &lt;em&gt;hydrated&lt;/em&gt; to yield a base:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$$\ce{Na2O + H2O -&amp;gt; 2NaOH}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Carbon dioxide can similarly be viewed as an acid anhydride, meaning it can be hydrated to form an acid.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;$$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The difference is that the hydration of sodium oxide is essentially irreversible.  One program I used estimated $K_{eq}=10^{39}$ for sodium oxide hydration.  In contrast the hydration of carbon dioxide is reversible, with $K_{eq}\approx1.5\times10^{-3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction that you are wondering about, $\ce{CO2 +2NaOH⟶Na2O +H2CO3}$, is essentially the dehydration of sodium hydroxide by $\ce{CO2}$ to form sodium oxide and carbonic acid, that is, it is reaction 2 minus reaction 1.  But the equilibrium constants I just mentioned show that sodium oxide has an &quot;affinty&quot; for water that is about 42 orders of magnitude greater than carbon dioxide's affinity for water.  So that reaction is very unfavorable.  The reverse reaction would be extremely favored: Sodium oxide would be a powerful reagent for $\ce{CO2}$ removal prior to air liquefaction, at least as powerful as sodium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are curious, here is the R code I used to estimate the equilibrium constant of sodium oxide hydration, using the package &lt;code&gt;CHNOSZ&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; subcrt(c(&quot;Na2O&quot;,&quot;H2O&quot;,&quot;NaOH&quot;),c(-1,-1,2),T=25)&#xA;info.character: found H2O(liq), also available in gas&#xA;subcrt: 3 species at 298.15 K and 1 bar (wet) &#xA;$reaction&#xA;     coeff         name formula state ispecies&#xA;2031    -1 sodium oxide    Na2O    cr     2031&#xA;1       -1        water     H2O   liq        1&#xA;1157     2         NaOH    NaOH    aq     1157&#xA;&#xA;$out&#xA;      logK         G         H         S         V       Cp&#xA;1 39.01742 -53229.29 -57143.24 -13.24728 -36.04979 -40.8745&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="10185" LastEditDate="2015-03-12T19:20:57.553" LastActivityDate="2015-03-12T19:20:57.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27241" PostTypeId="2" ParentId="27236" CreationDate="2015-03-12T16:48:03.453" Score="5" Body="&lt;p&gt;I don't think any gas is released.  You are probably experiencing dust or mist that contains NaOH itself, or tiny amounts of gas phase NaOH.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is related discussion at a soapmaking website &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tikvah.com/medical/lye.html&quot;&gt;http://www.tikvah.com/medical/lye.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &quot;&lt;a href=&quot;http://europepmc.org/backend/ptpmcrender.fcgi?accid=PMC1285477&amp;amp;blobtype=pdf&quot;&gt;Pneumothorax following inhalation of caustic soda fumes&lt;/a&gt;&quot; Archives of Emergency Medicine, 1988, 5, 45-47&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-12T17:13:02.757" LastActivityDate="2015-03-12T17:13:02.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27242" PostTypeId="2" ParentId="27165" CreationDate="2015-03-12T16:56:39.407" Score="1" Body="&lt;p&gt;There is one mistake in your calculation: oxirane is not just two standard C-O bonds plus a standard C-C bond. Due to ring strain there is some extra energy in the three-member ring. &lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="5026" LastEditDate="2018-02-27T04:52:20.407" LastActivityDate="2018-02-27T04:52:20.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27243" PostTypeId="2" ParentId="7588" CreationDate="2015-03-12T17:07:34.440" Score="3" Body="&lt;p&gt;A good analogy to your proposed reaction is the formation of carbon dioxide above a solution of carbonate salts.  Like sulfur trioxide, carbon dioxide is an acid anhydride.  The acid for sulfur trioxide is sulfuric acid, $\ce{H2SO4}$.  The acid for carbon dioxide is carbonic acid, $\ce{H2CO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dry carbonate salts like sodium carbonate can be made to release $\ce{CO2}$ by heating to high temperatures.  For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_carbonate#Calcination_equilibrium&quot; rel=&quot;nofollow&quot;&gt;calcium carbonate decomposes to calcium oxide&lt;/a&gt; and $\ce{CO2}$ at around 800 &amp;deg;C to 900 &amp;deg;C.  Since such high temperatures are required, this technique does not work with aqueous solutions at ambient temperatures.   (The water would boil off way before gas is released.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, sulfur trioxide evaporation from sulfate salts would be even more disfavored at room temperature than carbon dioxide evaporation from carbonate salt solutions, because of sulfur trioxide's affinity for water is much higher than $\ce{CO2}$'s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the reaction mechanism you have drawn is not a dehydration, but a reduction of the sulfate.  That would not form sulfur trioxide, but a reduced sulfur compound, such as sulfite.  Sulfur &lt;em&gt;di&lt;/em&gt;oxide gas evaporation from above sulfite salts is probably more favored than sulfur trioxide evaporation, but that isn't saying much. Either one is still highly disfavored under neutral or alkaline pH.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-12T17:07:34.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27244" PostTypeId="1" AcceptedAnswerId="27247" CreationDate="2015-03-12T17:41:56.860" Score="4" ViewCount="110" Body="&lt;p&gt;I am attempting to create an experiment to show the effects of metastability to a public audience and get people excited about science. More specifically, I am wondering if there is any sort of liquid that can mimic the effects that water had in the book &lt;em&gt;Cat's Cradle&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book, the idea was that a shard of a more stable polymorphic crystal of ice with a high melting point could &quot;infect&quot; water. This crystal would quickly spread, solidifying oceans and destroying most life on earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic question I'm asking is as follows. Is there any public display I could make to show a liquid quickly solidifying from a seed crystal? It is preferable (but not necessary) for the materials to be reasonably inexpensive and require no special equipment, so as to allow for the experiment to be done on a grander scale. Additionally, the experiment should have a low failure rate (unlike using supercooled water, which is difficult to control in large amounts).&lt;/p&gt;&#xA;" OwnerUserId="5162" LastActivityDate="2015-03-12T18:16:51.437" Title="Is there an experiment that visibly shows the effects of metastability?" Tags="&lt;experimental-chemistry&gt;&lt;crystal-structure&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27246" PostTypeId="1" AcceptedAnswerId="27252" CreationDate="2015-03-12T17:58:20.683" Score="12" ViewCount="2413" Body="&lt;p&gt;I travel a bit so am flexible where I live, and have taken a recent interest in chemistry to try to gain an understanding of physical reality. I find learning from textbooks to be good, but want to have the comfort of applying a lot of this knowledge in my own lab. To see and feel how it works first hand. From my understanding, the west has a huge paranoia with clandestine chemistry, which is of no interest to me. However, it seems a vast quantity of common reagents and catalysts in organic chemistry books are banned in western countries due to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would someone know how it's possible to practice amateur chemistry from home, without fear of accidentally breaking the law by using some reagent I'm not supposed to? And if not, what would be the best way to gain hands-on experience like this, where I can have complete freedom to experiment with what I want to do? I find at university I'm limited to lab sessions, and if one day I gain employment with a lab I'd no doubt not be free to conduct my own experiments. I wish to emphasize, I'm financially independent from my chemical studies, so purchasing the equipment isn't the issue. It's the legality I'm concerned about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an option to do this, to pursue amateur knowledge in chemistry, or is it illegal? Is there a country in which it isn't illegal?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="4945" LastEditDate="2015-03-13T04:45:24.570" LastActivityDate="2015-11-11T16:10:41.590" Title="In which country is it allowed to practice non-clandestine amateur chemistry?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;teaching-lab&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ClosedDate="2015-11-11T16:27:42.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="27247" PostTypeId="2" ParentId="27244" CreationDate="2015-03-12T18:16:51.437" Score="7" Body="&lt;p&gt;The standard demo for this is crystallization of sodium acetate trihydrate from its melt.  The trihydrate melts at 58 &amp;deg;C.  The molten liquid is easily supercooled, after which point dumping in a few crystals of unmelted stuff results in a rapid solidification of the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several videos on YouTube demonstrate the reaction; &lt;a href=&quot;https://www.youtube.com/watch?v=7HDZI2rwyHg&quot;&gt;Nurdrage's&lt;/a&gt; is my personal favorite.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-12T18:16:51.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27248" PostTypeId="1" CreationDate="2015-03-12T18:18:39.023" Score="9" ViewCount="1019" Body="&lt;p&gt;Discussing about the pros and cons of glassware from different manufacturers for chemistry experiments easily becomes a heated debate. Well, that's how it seems if you read the chemistry related blogs, discussions, questions, etc. over the internet. How much does it really matter what kind or type of glass one is using? How big the differences are across different manufacturers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice if the answers also elaborated on the differences between laboratory and home chemistry.&lt;/p&gt;&#xA;" OwnerUserId="14924" LastEditorUserId="46306" LastEditDate="2018-02-20T10:58:43.717" LastActivityDate="2018-02-20T10:58:43.717" Title="How much does the quality of glassware matter, really?" Tags="&lt;equipment&gt;&lt;teaching-lab&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27249" PostTypeId="2" ParentId="27236" CreationDate="2015-03-12T18:28:30.040" Score="2" Body="&lt;p&gt;I think it is due to the heat of solvation of NaOH, causing the solution to evaporate. If you dissolve NaOH with a large amount of solution, the &quot;gas&quot; wouldn't be released.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastActivityDate="2015-03-12T18:28:30.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27250" PostTypeId="2" ParentId="27246" CreationDate="2015-03-12T19:43:59.677" Score="5" Body="&lt;p&gt;In the USA there is a list of 41 special federally regulated chemicals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there are two lists, &lt;a href=&quot;http://www.deadiversion.usdoj.gov/21cfr/cfr/1310/1310_02.htm&quot; rel=&quot;noreferrer&quot;&gt;list 1 having 30 chemicals&lt;/a&gt; and &lt;a href=&quot;http://www.deadiversion.usdoj.gov/21cfr/cfr/1310/1310_02.htm&quot; rel=&quot;noreferrer&quot;&gt;list 2 having 11 chemicals&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these are common lab chemicals including:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetone&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethyl ether &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium permanganate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toluene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrochloric acid &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfuric acid &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iodine &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are planning to use these, I would recommend studying &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.deadiversion.usdoj.gov/21cfr/cfr/index.html&quot; rel=&quot;noreferrer&quot;&gt;Title 21 parts 1300-1321&lt;/a&gt; to make sure you are in compliance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the &lt;a href=&quot;http://www.ecfr.gov/cgi-bin/retrieveECFR?gp=&amp;amp;SID=b2605939192d71b9ceb3d5cf5b7fd37d&amp;amp;n=sp40.28.355.d&amp;amp;r=SUBPART&amp;amp;ty=HTML&quot; rel=&quot;noreferrer&quot;&gt;Extremely Hazardous Chemical list&lt;/a&gt; that has seperate reporting requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;http://www.bnl.gov/esh/env/compliance/docs/SaraTitleList.pdf&quot; rel=&quot;noreferrer&quot;&gt;EPA List of Lists&lt;/a&gt; that tries to catalog the laws and regulations that apply to various chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a relavent article &lt;a href=&quot;https://pubs.acs.org/cen/science/86/8645sci1.html&quot; rel=&quot;noreferrer&quot;&gt;Underground Science Chemistry hobbyists face a labyrinth of local and state regulations&lt;/a&gt; in Chemical and Engineering News Vol. 86, pages 38-40.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-12T20:28:27.017" LastActivityDate="2015-03-12T20:28:27.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27251" PostTypeId="1" AcceptedAnswerId="59973" CreationDate="2015-03-12T19:52:21.297" Score="8" ViewCount="11296" Body="&lt;p&gt;What is the mechanism of the fluorescein synthesis performed by von Baeyer?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sIIYm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="1887" LastEditDate="2015-03-12T22:19:11.560" LastActivityDate="2021-01-04T20:24:43.453" Title="Fluorescein synthesis" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27252" PostTypeId="2" ParentId="27246" CreationDate="2015-03-12T20:14:34.980" Score="10" Body="&lt;p&gt;Your question is basically too broad to be answered in full detail. I suggest that if you want a more detailed answer about a certain part, you should open a question for that matter separately. That being said, I'll try to summarize my own experience as a hobby chemist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, are you actually studying chemistry? IMHO, if this is within your possibilities, it is the best way to gain experience and also do your your own research. You will have supervision and the best equipment available. You will get the space to do research in your own areas mostly during your graduate studies, but here (The Netherlands) you will also get some space as an undergraduate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following points (in no particular order):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You say that finance is not a problem, but do not underestimate the price of all the chemicals and equipment you will need to do any bit of serious chemistry. Scales, vacuum pumps, glassware, solvents, a dedicated fridge, a specially constructed cabinet to store flammables, a fume hood with explosion proof suction, etc (and I'm not even mentioning NMR, mass spectrometers, IR, all of which are available to you as student). To run any decent lab from your backyard be prepared to invest several thousands of dollars (or euros). This can be done of course a lot cheaper, but then you will be limited in your possibilities or you will have to exchange safety for money.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Almost every chemical you will ever need when doing a normal synthesis (i.e. no explosive making or pharmaceuticals) is legally available in most countries (I heard however that some states in the US are quite notorious w.r.t. hobby chemistry, like Texas). You are even able to obtain chemicals that are clearly dangerous and should not be sold to amateurs, like osmium tetroxide, sodium cyanide (not everywhere), gaseous hydrogen chloride in large 110L pressurized tanks. There are a few exceptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Common precursors and preprecursors to illicit drugs in large quantities. Most notably phenylacetone and derivatives, piperonal, piperidine, methylamine, benzaldehyde, nitroethane, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chemicals that can be used to manufacture explosives in large quantities. Examples include: nitromethane and ammonium nitrate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chemicals that are severely dangerous to the environment or can be easily abused to poison people, although oddly enough some of the most hazardous are freely available. Examples are: elemental mercury, a lot of (organo) arsenic and other heavy metal compounds, a lot of radioactive compounds, some cyanides (may be available though from mining suppliers), etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note:&lt;/em&gt; you are able to obtain most of the above mentioned chemicals (also in larger quantities) if you start a business and have it legitimately registered to the Chamber of Commerce. This business does not actually have to do anything, it can function as a facade. It still won't work for pharmaceutical precursors or very large quantities of chemicals however, but for a hobby chemist that is mostly no problem.&#xA;&lt;em&gt;Note 2:&lt;/em&gt; At least here in The Netherlands some pharmaceutical precursors and preprecursors can be legally owned and synthesized, they are just not available commercially because of the high change for abuse. These chemicals however (e.g. phenylacetone) can be used in many other legal synthesis and hence it might be useful to synthesize them yourself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Be prepared to spend a lot of time on this hobby, both studying theory and doing practical work. Independently doing (serious) chemistry is not easy and you will have to rethink and adjust your setup a lot of times. Also the possibility of failure or accidents is higher without supervision. It is &lt;strong&gt;not&lt;/strong&gt; entirely without risk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-12T20:14:34.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27253" PostTypeId="1" AcceptedAnswerId="27271" CreationDate="2015-03-12T20:41:53.797" Score="6" ViewCount="1947" Body="&lt;p&gt;My professor wanted me to set up a lab to synthesize Methyl Vinyl Ketone (AKA Butenone) from some MEK we have lying around (2-Butanone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know there are multiple ways to do this, but the tricky part is finding out a safe way to go about this as MVK has a very low flashpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfC3k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured we could do an &lt;strong&gt;acid catalyzed&lt;/strong&gt; alpha halogenation with &lt;em&gt;Bromine&lt;/em&gt; and some &lt;em&gt;Acetic Acid&lt;/em&gt;, and then follow up with an Elimation using a bulky base such as &lt;em&gt;tert-butoxide&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is I have no idea how long, how much, or how cold/warm these materials need to be for the reaction to occur.  Ideally the entire reaction would take place in an ice water bath due to the danger, but at the same time I don't want to take five hours synthesizing one compound.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know what you think, and feel free to recommend a different (but safe) route of synthesis, as our lab has plenty of chemicals to work with.  By the way, we only need to create about &lt;strong&gt;10-20ml&lt;/strong&gt; of the stuff, but more is better.&lt;/p&gt;&#xA;" OwnerUserId="14927" LastActivityDate="2015-03-13T06:49:00.410" Title="Possible Synthesis of MVK from 2-Butanone?" Tags="&lt;organic-chemistry&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27254" PostTypeId="2" ParentId="27212" CreationDate="2015-03-12T21:49:32.773" Score="1" Body="&lt;p&gt;In pure water, as you calculated, for the reaction&lt;br&gt;&#xA;$\ce{AgCl -&amp;gt; Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$&lt;br&gt;&#xA; the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:&lt;br&gt;&#xA;$\ce{Ag2SO4 -&amp;gt; 2Ag+ + SO4^2-}$ with &lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm&quot; rel=&quot;nofollow&quot;&gt;$K_\mathrm{sp} = 1.4\cdot10^{-5}$&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't. &#xA;$(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} &amp;lt; (K_\mathrm{sp})_{\ce{Ag2SO4}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorDisplayName="user7951" LastEditDate="2015-04-12T08:54:20.193" LastActivityDate="2015-04-12T08:54:20.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27255" PostTypeId="1" CreationDate="2015-03-12T22:20:32.130" Score="2" ViewCount="1154" Body="&lt;p&gt;I work in a (physics) lab in which an aluminum oxide coating is applied to the inner surface of a plasma confinement vessel (made of Cu) to provide an electrically insulating layer that can withstand high temperature environments. The product we use comes in a medium that allows it to be brushed or sprayed onto the metal surface (something like: &lt;a href=&quot;http://www.zypcoatings.com/aluminum-oxide-coatings/&quot; rel=&quot;nofollow&quot;&gt;http://www.zypcoatings.com/aluminum-oxide-coatings/&lt;/a&gt;  -- see the data sheets). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if a similar product might exist for black aluminum oxide. I've seen that black aluminum oxide powder exists, but cannot find information on a coating product that is black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to understand what the difference is between the two: Does black aluminum oxide have the same physical properties as the white variant? Is it reasonable to expect that a similar suspension or solution of black aluminum oxide in water or a different solvent could be made? What makes one black and the other white? Any information on their differences would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14928" LastActivityDate="2015-04-21T02:49:58.077" Title="In what ways (other than color) does black Aluminum Oxide differ from white?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27256" PostTypeId="1" AcceptedAnswerId="27262" CreationDate="2015-03-12T22:33:36.023" Score="13" ViewCount="718" Body="&lt;p&gt;I just learned how Slater's rules work on Wikipedia. These rules really are very simple. But the presentation of the rules seemed not very efficient. I would think there would be someway to set up and equation of some sort such that all you have to do is plug in the relevant quantum numbers and the element number and use some sort of formula to just compute Slater's rules more easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such an equation / formula? If so, I'd like to see it. It would be more convenient than memorizing the rules in verbal form. &lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2015-03-13T03:02:40.650" Title="Is there a general / simple formula for Slater's rules?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27258" PostTypeId="1" CreationDate="2015-03-13T00:22:42.550" Score="6" ViewCount="891" Body="&lt;p&gt;I've been scouring the web for a website/pdf that lists/tabulates reactions and their corresponding rate laws. Something like:&#xA;\begin{align}&#xA;\ce{A + B &amp;amp;-&amp;gt; C} &amp;amp; \text{rate}&amp;amp;=1.234[\ce{A}]^2[\ce{B}] &amp;amp;&amp;amp;\text{at } 25~^\circ\mathrm{C}\\&#xA;\ce{D + E &amp;amp;-&amp;gt; F} &amp;amp; \text{rate}&amp;amp;=5.678[\ce{D}][\ce{E}]^2 &amp;amp;&amp;amp;\text{at } 25~^\circ\mathrm{C}\\  &#xA;\ce{G + H &amp;amp;-&amp;gt; I} &amp;amp; \text{rate}&amp;amp;=9.001[\ce{G}][\ce{H}] &amp;amp;&amp;amp;\text{at } 25~^\circ\mathrm{C}\\&#xA;&amp;amp;\cdots&amp;amp;&amp;amp;\cdots&amp;amp;&amp;amp;\cdots\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a list of known rate laws for reactions?&lt;/p&gt;&#xA;" OwnerUserId="14930" LastEditorUserId="5026" LastEditDate="2018-06-09T01:37:57.363" LastActivityDate="2018-06-09T01:40:05.720" Title="Is there a list of known rate laws for reactions?" Tags="&lt;kinetics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27259" PostTypeId="1" AcceptedAnswerId="27266" CreationDate="2015-03-13T01:00:07.333" Score="3" ViewCount="86" Body="&lt;p&gt;The book &quot;Lehrbuch der anorganischen Chemie&quot; by Arnold F. Hollemann on page 47[link here][1] states that dropping 0.8g Alcohol (M = 46g/mol) in a sea or ocean would result in every litre seawater containing 8 molecules alcohol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This completely baffles me. How exactly does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;https://books.google.de/bookshttps://books.google.de/books?id=eGkvSDAqY9gC&amp;amp;pg=PA47&amp;amp;lpg=PA47&amp;amp;ots=MFR61amS0x&amp;amp;focus=viewport&amp;amp;dq=ein&quot; rel=&quot;nofollow&quot;&gt;https://books.google.de/bookshttps://books.google.de/books?id=eGkvSDAqY9gC&amp;amp;pg=PA47&amp;amp;lpg=PA47&amp;amp;ots=MFR61amS0x&amp;amp;focus=viewport&amp;amp;dq=ein&lt;/a&gt; mol alkohol in den weltmeeren&amp;amp;hl=de&amp;amp;output=html_text&lt;/p&gt;&#xA;" OwnerUserId="4434" LastActivityDate="2015-03-13T05:04:49.023" Title="How exactly do alcohol molecules dissolve in seawater?" Tags="&lt;molecules&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27262" PostTypeId="2" ParentId="27256" CreationDate="2015-03-13T03:02:40.650" Score="12" Body="&lt;p&gt;Sadly, no. It wouldn't be too hard to program a &quot;get Slater shielding&quot; script. But it wouldn't be terribly useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let's back up briefly. &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater&amp;#39;s_rules&quot; rel=&quot;noreferrer&quot;&gt;Slater's rules&lt;/a&gt; allow you to estimate the effective nuclear charge $Z_{eff}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital &quot;shell&quot; (e.g., to compare the effective nuclear charge and shielding for say $3d$ vs. $4s$ in transition metals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slater wanted to use these for quantum calculations, such that the radial wavefunction would be something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_{ns}(r) = r^{n-1}e^{-\frac{(Z-\sigma)r}{n}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rules he devised were fairly simple and produced fairly accurate predictions of things like the electron configurations of transition metals and ionization potentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, others performed better optimizations of $\sigma$ and $Z_{eff}$ using variational Hartree-Fock methods. For example, Clementi and Raimondi published &lt;a href=&quot;http://dx.doi.org/10.1063/1.1733573&quot; rel=&quot;noreferrer&quot;&gt;&quot;Atomic Screening Constants from SCF Functions.&quot; &lt;em&gt;J Chem Phys&lt;/em&gt; (&lt;strong&gt;1963&lt;/strong&gt;) &lt;em&gt;38&lt;/em&gt;, 2686–2689.&lt;/a&gt; (That article covers elements up to Kr.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMEbS.png&quot; alt=&quot;Slater shielding vs. nuclear charge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty obvious that no simple formula will cover these curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days, basis set optimization is done by computer, and I doubt many people even glance at the exponents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my mind, Slater's rules serve two main purposes at this point:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They're relatively simple to teach and learn and allow quick qualitative predictions, particularly for electron configurations in the transition metals.&lt;/li&gt;&#xA;&lt;li&gt;They emphasize that exponents for multi-electron atoms must be empirically derived.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I still teach them, particularly, for the first reason. To a lesser degree, they also reinforce the idea of effective nuclear charge and provide some semi-quantitive basis for that concept.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-13T03:02:40.650" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27263" PostTypeId="1" AcceptedAnswerId="27268" CreationDate="2015-03-13T04:42:34.870" Score="0" ViewCount="3033" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Arrange the following according to increasing &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_a$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cyclohexanol&lt;/li&gt;&#xA;&lt;li&gt;Phenol&lt;/li&gt;&#xA;&lt;li&gt;Cyclohexane carboxylic acid&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Please tell me the logic along with the solution.&#xA;I am confused between phenol and cyclohexanol because i guess phenol is resonance stabilized-the phenoxide ion&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="-1" LastEditorDisplayName="user14857" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-13T05:47:38.650" Title="What is the reasoning behind the increasing pKa for organic compounds?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-03-13T13:52:24.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="27266" PostTypeId="2" ParentId="27259" CreationDate="2015-03-13T05:04:49.023" Score="3" Body="&lt;p&gt;Here is my calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ocean has a volume of 1.3 billion cubic kilometers or 1.3e+18 cubic meters or 1.3e+21 liters. 0.8/46=0.0173 moles of ethyl alcohol x 6.02e+23 molecules per mole = 0.1047e+23 molecules, which if dissolved evenly in 1.3e+21 liters would give 10.47/1.3 or approx. 8 molecules of ethanol per liter.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-13T05:04:49.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27267" PostTypeId="1" CreationDate="2015-03-13T05:08:32.117" Score="5" ViewCount="200" Body="&lt;p&gt;Why do the covalent radii of transition series decrease at the start rapidly, then become almost constant and at the end of series begin to increase from left to right in periodic table?&lt;/p&gt;&#xA;" OwnerUserId="14932" LastEditorUserId="4945" LastEditDate="2015-03-13T06:52:13.333" LastActivityDate="2015-03-13T06:52:13.333" Title="How to rationalise the trend in the covalent radii of the transition metals?" Tags="&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;periodic-trends&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27268" PostTypeId="2" ParentId="27263" CreationDate="2015-03-13T05:47:38.650" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MMfZI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexanol is an alcohol. An alcohol contains an $\ce{OH}$ group connected to a tetrahedral carbon. &#xA;Phenol is slightly different; it contains an $\ce{OH}$ group connected to a trigonal planar carbon that is part of an aromatic ring.&#xA;cyclohexanecarboxylic acid contains a very different functional group, carboxylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexanol has a $\mathrm{p}K_a$ of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a $\mathrm{p}K_a$ of about 9 and is considered mildly acidic. cyclohexanecarboxylic acid has a $\mathrm{p}K_a$ of about 4.9; its acidity is similar to hydrofluoric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare stability of the conjugate bases of cyclohexanol, phenol and cyclohexanecarboxylic acid:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.&lt;/li&gt;&#xA;&lt;li&gt;In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.&lt;/li&gt;&#xA;&lt;li&gt;In cyclohexanecarboxylic  acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ibQpV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWEn5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please visit: &lt;a href=&quot;http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-03-13T05:47:38.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27269" PostTypeId="1" CreationDate="2015-03-13T05:51:25.017" Score="1" ViewCount="477" Body="&lt;p&gt;These two are simple ionic compounds , not complexes [if I do not consider water of hydration] then how can I rationalise the difference in colour?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-03-13T05:51:25.017" Title="why colour of CuSO4 and CuCl3 is different?" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2015-03-13T06:51:18.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="27270" PostTypeId="1" CreationDate="2015-03-13T06:22:15.103" Score="1" ViewCount="1359" Body="&lt;p&gt;I have some gas cylinders in a gas cabinet. Every time I change the cylinder, I use N2(100%v/v, 60L/min) to purge the residue gas in the pipe to a local scrubber. I want to know how many gases I should treat per purge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;follow is my customer give me the condition：&#xA;gas name                 flow rate         pressure(MPa)&#xA;F2(5%F2,&#xA;Blance Kr/Ne 95%)      max 4.4L/min         14.8&#xA;BCl3                    same↑                0.0627&lt;br&gt;&#xA;WF6                     same↑                0.04 &#xA;ClF3                    same↑                0.0588&#xA;N2                       60                   ─&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is someone give me the solution：&#xA;100% gas volume (L/purge)= 0.07(L)*[pressure (pa)/98066.5(pa)]*1(cylinder/purge)*concentration%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come?&#xA;In my thought, it use the formula P1V1=P2V2, where V2 is what we want to calculate. P1 is the gas pressure in the cylinder, P2= the pressure in the pipe.&#xA;But 0.7(L)? Why? And why P2 is not 1 atm = 101325 (pa) but use 98066.5(pa)=1kgf/cm^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the pressure(MPa) value let me confused. Generally, how much pressure is the gas cylinder for each kind of gas above? The value is closed to the pressure of a new cylinder or to a used-up cylinder?&lt;/p&gt;&#xA;" OwnerUserId="14937" LastActivityDate="2015-03-13T06:22:15.103" Title="how to calculate the gas volume per purge when I change the cylinder?" Tags="&lt;safety&gt;&lt;gas-laws&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27271" PostTypeId="2" ParentId="27253" CreationDate="2015-03-13T06:49:00.410" Score="3" Body="&lt;p&gt;As pointed out in the comments, there is no way to make it safe, as methyl vinyl ketone itself is a very nasty compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also not an expert on organic synthesis, but I found some procedures, that might help you in adopting a synthetic route. The one you suggested might work, but be aware, that you are dealing with a methyl ketone, which can undergo the &lt;a href=&quot;http://en.wikipedia.org/wiki/Haloform_reaction&quot; rel=&quot;nofollow&quot;&gt;haloform reaction&lt;/a&gt; quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in [1] they describe the bromination with bromine and absolute methanol at the methyl group for 3-methyl-2-butanone at a nice moderate temperature range 0-10&amp;nbsp;&amp;deg;C. The yield is 95%. But this is not what you want. Adding an acid as catalyst might work in your favour though, as after protonation the partial positive charge is better stabilised at the higher substituted carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In [2] they use &lt;em&gt;N&lt;/em&gt;-bromosuccinimide and aqueous hydrogen bromide to introduce the halogen in &amp;alpha; position. They convert hexanoic acid with thionylchlorid to the acid chloride and subsequently to 2-bromohaxanoylchloride. The reaction takes place in the also quite nasty carbon tetrachloride at about 85&amp;nbsp;&amp;deg;C. This might also work for your system, it's a bit warmer than you expected though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In [3] they use sulfuryl chloride in carbon tetrachloride to convert pantane-3-one to 2-chloropentane-3-one at about 45&amp;nbsp;&amp;deg;C with about 80% yield. How selective this will work with an asymmetric ketone is not clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have obtained the 3-halogeno-butan-2-one, things will not get easier. I found a patent [4], where they convert 3-chlorobutan-2-one via pyrolysis with sodium acetate at 500-550&amp;nbsp;&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find a reference for an elimination, but that should also work as long as you use a bulky base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia [5] has a different route for the synthesis, which might be a bit more economical and easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, good luck!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=CV6P0193&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Org. Synth.&lt;/em&gt; &lt;strong&gt;1976&lt;/strong&gt;, &lt;em&gt;55&lt;/em&gt;, 24.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=cv6p0190&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Org. Synth.&lt;/em&gt; &lt;strong&gt;1976&lt;/strong&gt;, &lt;em&gt;55&lt;/em&gt;, 27.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=V79P0251&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Org. Synth.&lt;/em&gt; &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;79&lt;/em&gt;, 251.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.google.com/patents/US2256149&quot; rel=&quot;nofollow&quot;&gt;J. R. Long: &lt;em&gt;Preparation of methyl vinyl ketone&lt;/em&gt;. US 2256149 A, 16. 09. 1941.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_vinyl_ketone#Production&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Methyl_vinyl_ketone#Production&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-03-13T06:49:00.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27272" PostTypeId="2" ParentId="27258" CreationDate="2015-03-13T07:21:15.553" Score="1" Body="&lt;p&gt;Integrated Rate Laws in Chemical Kinetics are very well known (&lt;a href=&quot;http://www.csus.edu/indiv/m/mackj/chem142/kinetics.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.csus.edu/indiv/m/mackj/chem142/kinetics.pdf&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good article with the list of the constants: &lt;a href=&quot;https://web.archive.org/web/20160804015722/http://www.nist.gov/data/nsrds/NSRDS-NBS67.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table of Recommended Rate Constants for Chemical Reactions Occurring in Combustion&lt;/a&gt; by Francis Westley&lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="5026" LastEditDate="2018-06-09T01:40:05.720" LastActivityDate="2018-06-09T01:40:05.720" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27273" PostTypeId="2" ParentId="27236" CreationDate="2015-03-13T07:26:47.640" Score="-2" Body="&lt;p&gt;Ammonia is possible in the small amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4+ (aq) + H2O (l) &amp;lt;=&amp;gt; NH3 (aq) + H3O+ (aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="4945" LastEditDate="2015-03-13T07:58:15.510" LastActivityDate="2015-03-13T07:58:15.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27274" PostTypeId="2" ParentId="27236" CreationDate="2015-03-13T07:55:06.147" Score="1" Body="&lt;p&gt;Thanks for the comments. Based on them, here what I observed: The thank is filled with 100 l of water then $\ce{NaOH}$ is added. If it is added slowly, no bubbles or gases were observed. We need to shorten the preparation time so we stirred it intensively. During stirring, air bubbles are introduced to the solution. After their release, a alkaline mist/water dispersion is released to air - that is what I was observed. Thanks again for your thoughts. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-03-13T07:55:06.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27276" PostTypeId="1" CreationDate="2015-03-13T10:55:52.780" Score="0" ViewCount="93" Body="&lt;p&gt;how can we know when to indicate &quot;+ or -&quot; sign in uncertainty calculations because I noticed some problems has been solved by this sign and some has not.&lt;/p&gt;&#xA;" OwnerUserId="14840" LastActivityDate="2015-03-13T12:19:17.290" Title="when do we indicate the &quot;+ or -&quot; sign in uncertainty" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-03-14T06:50:53.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="27277" PostTypeId="2" ParentId="27276" CreationDate="2015-03-13T12:06:05.937" Score="6" Body="&lt;p&gt;You either use both &quot;+&quot; and &quot;-&quot;, $\pm$, or neither in which case it is implied that you mean $\pm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually uncertainty is expressed in a way that is symmetric about the mean, such as $123\pm4$, meaning to a certain degree of confidence the true value is between 119 and 127. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you could write $123^{+5}_{-3}$ to indicate the best estimate of the true value is 123 and to a certain degree of confidence the true value is between 120 and 128.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you were highly certain that an object is a cube, and measure its edge as 100 $\pm 20$, then it might be better to express the volume of the cube as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.0^{+0.7}_{-0.5} \times 10^6$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-13T12:19:17.290" LastActivityDate="2015-03-13T12:19:17.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27278" PostTypeId="2" ParentId="9847" CreationDate="2015-03-13T13:15:49.280" Score="19" Body="&lt;p&gt;The controversy surrounding the $\mathrm{p}K_\mathrm{a}$ of hydronium mostly arises from the definition of $K_\mathrm{a}$ or lack thereof.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no IUPAC definition of $\mathrm{p}K_\mathrm{a}$ or $K_\mathrm{a}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest IUPAC defined term is the &lt;a href=&quot;http://goldbook.iupac.org/S05915.html&quot; rel=&quot;noreferrer&quot;&gt;standard equilibrium constant&lt;/a&gt;, which can be denoted as $K^\circ$ or just $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physical chemistry texts such as Levine and respected works such Bates's &lt;em&gt;&quot;Determination of pH — Theory and Practice&quot;&lt;/em&gt; define $K^\circ_\mathrm{a}$ of an acid in water as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a(\ce{A-})a(\ce{H3O+})}{a(\ce{HA})a(\ce{H2O})} \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a$ is &lt;a href=&quot;http://goldbook.iupac.org/A00019.html&quot; rel=&quot;noreferrer&quot;&gt;activity&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting that the acid is $\ce{H3O+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a(\ce{H2O})a(\ce{H3O+})}{a(\ce{H3O+})a(\ce{H2O})} = 1 \tag{2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and of course $-\log(1) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number $-1.74$ that some quote for the hydronium $\mathrm{p}K_\mathrm{a}$ comes from:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;omitting the activity of water from the denominator of the $K^\circ_\mathrm{a}$ definition (equation $(1)$); and&lt;/li&gt;&#xA;&lt;li&gt;taking the concentration of water (about $55.5~\mathrm{M}$) as the $K_\mathrm{a}$ of $\ce{H3O+}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With that, one obtains the $\mathrm{p}K_\mathrm{a}$ as $-\log (55.5) = -1.74$.  For example, Levine even has the $-1.74$ value in a figure comparing the $\mathrm{p}K_\mathrm{a}$'s of various acids, but has a footnote explaining that the value for $\ce{H3O+}$ is based upon the alternative $\mathrm{p}K_\mathrm{a}$ definition.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, revisiting the analysis that $K^\circ_\mathrm{a} =1$ is probably the most authoritative paper on this issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlepdf/1998/AN/A705491B&quot; rel=&quot;noreferrer&quot;&gt;New point of view on the meaning and on the values of Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water&lt;/a&gt;, which insists that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + H3O+ &amp;lt;=&amp;gt; H3O+ + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;does not correspond to an actual chemical process&quot; and therefore &quot;it is not legitimate&quot; to extend the concept of $K^\circ_\mathrm{a}$ to $\ce{H3O+}$ in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article goes on to say that only by studying the $K_\mathrm{a}$ of $\ce{H3O+}$ in another solvent such as ethanol can $\ce{H3O+}$ be compared to other acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in ethanol is $0.3$ and $\mathrm{p}K_\mathrm{a}$ values are $1.0 \pm 0.3$ units lower in water than in ethanol, so the article suggests a $\mathrm{p}K_\mathrm{a}$ of $-0.7$ for $\ce{H3O+}$ in water, for the purpose of comparison to other acids.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2016-09-29T06:05:24.700" LastActivityDate="2016-09-29T06:05:24.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27279" PostTypeId="1" AcceptedAnswerId="27283" CreationDate="2015-03-13T13:43:19.630" Score="4" ViewCount="1611" Body="&lt;p&gt;I'm designing an experiment and it involves testing various concentrations of solutions, and I will be testing 5 in total. I proposed to make up a stock solution, and distribute it evenly in 5 different beakers and add different amounts of water to each solution but my teacher doesn't like this method and says I should simply dilute down but I don't know what he means. So if I had a 1.00 mol/dm$^{-3}$ solution and wanted concentrations of 0.800, 0.600, 0.400 and 0.200  mol/dm$^{-3}$, how do I dilute it down? To be honest, the concentrations don't really matter, it's just the method of making them - would it be easier to dilute them by half each time? Each solution is going to be used up so I can't simply reuse them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I will need 200ml of each solution in the experiment design, and the precision of the concentrations is not very important. How can I dilute one stock solution to 4 other solutions?&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="418" LastEditDate="2015-03-14T08:13:59.763" LastActivityDate="2015-03-14T21:37:08.147" Title="Diluting down solutions method" Tags="&lt;aqueous-solution&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27281" PostTypeId="1" CreationDate="2015-03-13T13:54:04.513" Score="0" ViewCount="628" Body="&lt;p&gt;How to make &quot;Aqua Regia&quot; in the lab conditions? Which lab equipment must be used?&lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="418" LastEditDate="2015-03-13T14:12:10.110" LastActivityDate="2015-03-13T14:19:18.813" Title="How it is possible to make Aqua Regia in the lab?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27282" PostTypeId="2" ParentId="27281" CreationDate="2015-03-13T14:11:14.883" Score="5" Body="&lt;p&gt;Typically, 1 part of concentrated nitric acid is added to 3 parts of concentrated hydrochloric acid. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Personal protection (goggles, gloves, lab coat) is vital. &lt;/li&gt;&#xA;&lt;li&gt;Prepare and use it under a well-vented fumehood only - hands inside the fumehood, head outside!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comment of Crystal Lettuce is very important and points to a crucial aspect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aqua regia isn't a mixture to keep bottled on the shelf: it has to be prepared and used immediately!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-13T14:19:18.813" LastActivityDate="2015-03-13T14:19:18.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27283" PostTypeId="2" ParentId="27279" CreationDate="2015-03-13T14:18:05.070" Score="3" Body="&lt;p&gt;It depends on what degree of accuracy you need and what order of magnitude the final concentrations will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need high accuracy and concentrations are in the range you are describing, I would put a class A volumetric flask on a balance, add the appropriate weight of solute, add water to not quite the full volume, dissolve the solute, and add water to the volumetric line, seperately for each solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need concentrations much lower than what you are saying, when weighing becomes your largest source of error, then it is more accurate to make a concentrated solution, and the dilute the concentrated solution.  For example, you would use a class A TD (to deliver) volumetric pipette to measure the concentrated solution into a class A volumetric flask.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-13T14:18:05.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27285" PostTypeId="1" CreationDate="2015-03-13T14:21:52.193" Score="2" ViewCount="65" Body="&lt;p&gt;My neighbour warned me that saltpetre (potassium (sodium) nitrate) is danger in the certain conditions. Is it possible that it can explode in my garage?&lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="418" LastEditDate="2015-03-13T15:39:19.000" LastActivityDate="2015-03-13T15:39:19.000" Title="Saltpetre in the Garage" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27287" PostTypeId="1" AcceptedAnswerId="27383" CreationDate="2015-03-13T14:45:27.193" Score="5" ViewCount="5780" Body="&lt;p&gt;Are 2s or 2p orbitals more stabilized when going from left to right in the period? There are good arguments for both possible answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) 2p orbitals are more stabilized because they penetrate less in the space of the 1s orbital, so the increase of the effective nuclear charge affects them more. When plotting the 1st ionization energy with the increase of the atomic number, it can be seen that the ionization energy increases with the increase of the atomic number, but the slope is bigger for O-F-Ne (3.97eV) than for B-C-N (3.12eV). 3s and 3p orbitals aren't stabilized so much (because they both penetrate into the space of the inner orbitals), therefore it is much easier to get a high oxidation state of sulfur than of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) 2s orbitals are more stabilized because the electron spends more time closer to the nucleus, hence the orbital is more affected by the increase of the effective nuclear charge. Proof: the MO diagrams for B, C and N differ from the ones for O, F, and Ne, because s-p mixing in B, C and N happens as the 2s and 2p orbitals are similar in energy. The energy difference increases towards the right (because 2s orbital is more stabilized), so 2s and 2p can't mix efficiently in O,F and Ne. If the 2p orbital was more stabilized, then the difference between 2s and 2p would be smaller through the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really need some clarification here. &lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-03-17T17:22:05.363" Title="Effect of effective nuclear charge increase on the stabilization of 2s and 2p orbitals" Tags="&lt;orbitals&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27288" PostTypeId="1" AcceptedAnswerId="27290" CreationDate="2015-03-13T15:08:38.667" Score="5" ViewCount="177" Body="&lt;p&gt;We've learned in class that epoxide ring openings can be catalyized with acids, and I seem to recall either the professor or a classmate mentioning that they also react with bases, but some quick internet searching didn't turn up anything useful.  Do epoxide rings react with bases, and if so, what is the mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Dis)claimer: this is just for my own curiosity/learning, and not being asked as a homework/test question (yet).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="13384" LastEditorUserId="10185" LastEditDate="2015-03-13T19:45:49.580" LastActivityDate="2015-03-13T19:45:49.580" Title="Do epoxide rings react with bases?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27289" PostTypeId="2" ParentId="27285" CreationDate="2015-03-13T15:09:22.647" Score="3" Body="&lt;p&gt;Potassium and sodium nirate can be used to make black powder (gun powder), but they are not explosives by themselves. It is safe to store bags of salpetre in the garage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's salpetre forming on the brickworks, I'd rather be concerned to keep the bags closed and away from that wall: sodium nitrate is rather hygroscopic and salpetre crystals on and between the bricks indicate that the wall is wet.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-13T15:09:22.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27290" PostTypeId="2" ParentId="27288" CreationDate="2015-03-13T15:25:33.787" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do epoxied rings react with bases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they can be opened, e.g. with alkoxides.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the mechanism?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/idTak.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, this is a $S_N2$ reaction, with the typical approach of the nucleophile from the back side. The difference to a non-epoxide case is that removal of the leaving group (bond breaking, ring opening) is facilitated by the ring strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the general rules for the reactivity of alkyl-substituted centres in $S_N2$ nevertheless apply: the higher the substitution, the lower the rate. This is attributed to the &lt;em&gt;steric hindrance&lt;/em&gt; of the respective centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the regioselectivity is different under acidic conditions!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-13T17:58:04.357" LastActivityDate="2015-03-13T17:58:04.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27291" PostTypeId="2" ParentId="4033" CreationDate="2015-03-13T15:42:39.433" Score="6" Body="&lt;p&gt;There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common, commercially-available materials include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dupont.com/products-and-services/electronic-electrical-materials/flexible-rigidflex-circuit-materials/brands/pyralux-flexible-circuit/sub-brands/pyralux-copper-clad-laminates.html&quot; rel=&quot;noreferrer&quot;&gt;Pyralux&lt;/a&gt; - copper-coated polyimide from DuPont.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemistry.about.com/od/polymers/f/What-Is-Mylar.htm&quot; rel=&quot;noreferrer&quot;&gt;Mylar&lt;/a&gt; - DuPont's name for &lt;a href=&quot;http://en.wikipedia.org/wiki/BoPET&quot; rel=&quot;noreferrer&quot;&gt;biaxially-oriented polyethylene terephthalate (BoPET)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/polymers/f/What-Is-Mylar.htm&quot; rel=&quot;noreferrer&quot;&gt;Chemistry.About.com&lt;/a&gt; does mention vapor deposition of metals into Mylar, though.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-13T15:42:39.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27292" PostTypeId="1" CreationDate="2015-03-13T16:07:33.167" Score="9" ViewCount="6040" Body="&lt;p&gt;Could anybody explain the following statement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The boundary surface is a common way to represent atomic orbitals, incorporating the volume in which there is about a 90 percent probability of finding the electron at any given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what does the author means starting from &quot;incorporating the volume..&quot;&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="5017" LastEditDate="2015-03-14T18:06:42.383" LastActivityDate="2015-03-14T18:06:42.383" Title="What is an orbital boundary surface?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27293" PostTypeId="2" ParentId="27292" CreationDate="2015-03-13T16:17:35.507" Score="8" Body="&lt;p&gt;It means that the boundary surface is a closed surface or set of closed surfaces, such that there is a 90% chance that an electron is interior of the closed surface and a 10% chance that the electron is exterior of the closed surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way of specifying how an orbital is represented is &quot;equal probability boundary surface&quot; which further specifies that each point on the surface must have any equal electron probablity density, and the surface must be a closed surface or set of closed surfaces with a 90% (or other percent) chance of containing the electron at a given time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the &quot;equal probability&quot; statement there are infinite different boundary surfaces for each orbital (infinite different ways to enclose 90% of the probability density).  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-13T16:29:15.677" LastActivityDate="2015-03-13T16:29:15.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27294" PostTypeId="1" AcceptedAnswerId="28597" CreationDate="2015-03-13T17:32:41.653" Score="4" ViewCount="559" Body="&lt;p&gt;I am discussing only two types of vaporization in my question that are, evaporation and boiling. We know that evaporation is the phenomena which describes vaporization of liquid under a certain temperature. And boiling point is the temperature at which vapors of liquid exceeds their pressure by $760\ \mathrm{Torr}$ and starts escaping. The liquid doesn’t absorb more heat and utilize all of the heat to change its state and is said to boil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more details about these two kinds but there is no need to discuss them right now so we should go back to the question that asks that why vapors exerts pressure rather than diffusing in atmosphere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tk8LD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14946" LastEditorDisplayName="user7951" LastEditDate="2015-06-27T14:00:01.340" LastActivityDate="2019-03-24T00:18:17.613" Title="Why water vapors exerts pressure rather than diffusing in air before reaching the boiling point of water?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27295" PostTypeId="1" AcceptedAnswerId="27296" CreationDate="2015-03-13T17:34:08.857" Score="9" ViewCount="12511" Body="&lt;p&gt;I've been trying to find out as much as I can about peracetic acid, especially regarding its use as a sanitizer. In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Peracetic_acid#Production&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia entry&lt;/a&gt; it notes that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Peracetic acid is always sold in solution as a mixture with acetic acid and hydrogen peroxide to maintain its stability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is definitely born out by commercial examples: Loeffler Lerasept PAA (a sanitizer I've used before) lists 31% hydrogen peroxide, 17% peracetic acid, 16% acetic acid and 1% phosphonic acid among its ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, my understanding was that mixing acetic acid and hydrogen peroxide &lt;em&gt;makes&lt;/em&gt; peracetic acid. From the same Wikipedia article: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It forms upon treatment of acetic acid with hydrogen peroxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So my question is: what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid? Does the phosphonic acid inhibit this reaction? Is there something else not listed that does this? Or is it just some bit of chemistry I'm not aware of?&lt;/p&gt;&#xA;" OwnerUserId="13178" LastEditorUserId="5026" LastEditDate="2018-06-09T01:45:50.600" LastActivityDate="2020-03-26T19:08:31.547" Title="Formation of peracetic acid from acetic acid and hydrogen peroxide and its stability in their presence" Tags="&lt;organic-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27296" PostTypeId="2" ParentId="27295" CreationDate="2015-03-13T18:42:41.823" Score="6" Body="&lt;p&gt;The formation of peracetic acid from acetic acid and hydrogen peroxide is an equilibrium reaction and so in order for the peracetic acid to remain at a constant concentration as desired, acetic acid and hydrogen peroxide must also be present so that the reaction can be at equilibrium. $$\ce{CH3COOH + H2O2 &amp;lt;=&amp;gt; CH3COOOH + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what exactly keeps the excess hydrogen peroxide from reacting with the excess acetic acid to simply form more peracetic acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nothing. The hydrogen peroxide reacts with the acetic acid to form peracetic acid and water but the peracetic acid and water also react to reform hydrogen peroxide and acetic acid. At equilibrium the rates of these reactions are equal and so the concentrations of the species do not change. At room temperature the equilibrium constant is somewhere around 2.5 (see references) which indicates that significant amounts of reactants and products are present, supporting the reasoning above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper &lt;a href=&quot;http://www.researchgate.net/profile/Xuebing_Zhao/publication/51995019_Preparation_of_peracetic_acid_from_acetic_acid_and_hydrogen_peroxide_Experimentation_and_modeling/links/09e41506011ee8d08a000000.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Chin J Proc Eng&lt;/em&gt; &lt;strong&gt;2008&lt;/strong&gt; February, &lt;em&gt;8&lt;/em&gt; (1), 35–41&lt;/a&gt; has some data on the values of the equilibrium constant and the variation in concentration of peracetic acid formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another value for the equilibrium constant on page vii of &lt;a href=&quot;http://nrl.northumbria.ac.uk/2284/1/unis.melod_phd.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unis, Melod (2010) Peroxide reactions of environmental relevance in aqueous&#xA;solution. Doctoral thesis, Northumbria University&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-06-09T01:51:04.653" LastActivityDate="2018-06-09T01:51:04.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27297" PostTypeId="2" ParentId="27295" CreationDate="2015-03-13T18:43:55.290" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But, my understanding was that mixing acetic acid and hydrogen peroxide makes peracetic acid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly! But within the mixture, peracetic acid reacts back to hydrogen peroxide and acetic acid. With other words, the components are in an &lt;strong&gt;equilibrium&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\ce{CH3COOH + H2O2 &amp;lt;=&gt; CH3CO3H + H2O}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For disinfection purposes, these mixtures are sold as made. This is safe (and cheap) for both the manufacturer and the consumer. The fairly stable and safe to use, while concentrated solutions of peracetic acid tend to decompose - violently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphonic acid with a (first) pKa around two is the strongest acid within the mix. One might speculate that it is added to adjust the pH, keep more acetic acid protonated and therefore shift the reaction a bit more towards the peracetic acid.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-13T18:43:55.290" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27298" PostTypeId="2" ParentId="27294" CreationDate="2015-03-13T19:04:00.200" Score="2" Body="&lt;p&gt;Pictures are incomplete: they don't show molecules rebounding from others and pressing against all surfaces, both the container and the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal gas laws can be derived from thermodynamics; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_theory&quot; rel=&quot;nofollow&quot;&gt;Kinetic theory&lt;/a&gt;. There's a nice animated GIF there that could help you to understand this.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-13T19:04:00.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27299" PostTypeId="1" CreationDate="2015-03-13T19:16:24.973" Score="3" ViewCount="253" Body="&lt;p&gt;How can I store distilled water without introducing impurities?  I'm going to be doing so electrical studies of pure water and ice and I need a way of keeping the water pure during the experiments.  The water needs to remain relatively pure (I'm not yet entirely sure how relative yet), over at least the course of $1$ day.  What is a good container that won't be leeching any salts or other chemicals into the water?&lt;/p&gt;&#xA;" OwnerUserId="14948" LastEditorUserId="402" LastEditDate="2016-11-20T19:22:17.050" LastActivityDate="2016-11-20T19:22:17.050" Title="How to store distilled water without introducing impurities?" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27300" PostTypeId="1" AcceptedAnswerId="27500" CreationDate="2015-03-14T00:52:31.577" Score="14" ViewCount="1090" Body="&lt;p&gt;Is it possible to make a drug that liquefies heart plaque to treat heart disease without damaging other parts of the body?  If so, would the liquefied plaque be eliminated as regular fluid is through the kidneys?  I believe that if this is possible, it would definitely be worthwhile to make as it could save many lives!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard there are drugs that can dissolve a blood clot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also heard of a technique being developed in Israel that dissolves heart plaque with an ultrasonic probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like there are some efforts already in this area, but it appears they halt rather than dissolve the plaque:&#xA;&lt;a href=&quot;http://heartdisease.about.com/cs/coronarydisease/a/plaquebuster.htm&quot;&gt;http://heartdisease.about.com/cs/coronarydisease/a/plaquebuster.htm&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6806" LastEditorUserId="6806" LastEditDate="2015-03-18T01:04:47.303" LastActivityDate="2015-03-19T00:23:49.263" Title="Is it possible to make a drug that liquefies heart plaque to treat heart disease?" Tags="&lt;biochemistry&gt;&lt;solubility&gt;&lt;chemical-biology&gt;&lt;medicinal-chemistry&gt;&lt;drugs&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27301" PostTypeId="1" CreationDate="2015-03-14T01:00:18.163" Score="0" ViewCount="2078" Body="&lt;p&gt;I asked my chemistry teacher this today and didn't get an answer. It would seem to me that the equation for determining the morality of the unknown in a titration, MaVa = MbVb, would only work if the acid and the base were both strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always interpreted the equation as a way to say (moles H+)=(moles OH-), but if the acid was weak, the moles of H+ would not simply equal MaVa, because the acid wouldn't fully dissociate, and the same for a base and OH-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet the formula still works for both strong and weak acids and bases. Why?&lt;/p&gt;&#xA;" OwnerUserId="14952" LastActivityDate="2015-03-14T08:20:36.083" Title="Why is Ma*Va=Mb*Vb still valid for weak acids/bases?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-14T12:38:34.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="27302" PostTypeId="1" AcceptedAnswerId="27418" CreationDate="2015-03-14T03:05:10.863" Score="36" ViewCount="4258" Body="&lt;p&gt;I have a very very general question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In DFT functional selection , mostly people speak about the most recent ones. For example my professor always asks: &quot; which DFT Functional did you select ? &quot;  and if I say B3LYP, he says : &quot; No ! that's too old ! &quot; but if I answer: M06 , he says : &quot; hmm ... sounds promising, that's a modern functional &quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is too naive to select functionals just based on their chronologic sequence. I want to ask if there is any good and reliable criteria for functional selection. For example a criteria that says for an alkene with some specific characteristics, go for M06-L and for an alkane with other characteristics, use B97xxx family and so on..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a criteria ?   I hope this topic can become a good guideline for future reference. !&lt;/p&gt;&#xA;" OwnerUserId="7188" LastActivityDate="2020-12-22T02:58:41.870" Title="DFT Functional Selection Criteria" Tags="&lt;computational-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="27305" PostTypeId="1" CreationDate="2015-03-14T07:05:51.863" Score="0" ViewCount="99" Body="&lt;p&gt;what does it mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the portion starting from &quot; or readily give rise to...&quot;&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="7581" LastEditDate="2015-03-14T07:48:42.377" LastActivityDate="2015-03-14T07:56:17.953" Title="What does this property of transition metals mean?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-14T19:02:35.617" ContentLicense="CC BY-SA 3.0" />
  <row Id="27306" PostTypeId="2" ParentId="27305" CreationDate="2015-03-14T07:56:17.953" Score="1" Body="&lt;p&gt;Take the electronic configuration of copper (atomic number $29$) $\ce{[Ar] 3d^10 4s^1}$. According to the first portion of the definition, copper doesn't have incompletely filled subshells. But, copper readily gives rise to cation $\ce{Cu^{2+}}$ that has the electronic configuration $\ce{[Ar] 3d^9 4s^0}$ with incompletely filled d subshells. So, copper is considered as a transition metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can guess why zinc (atomic number $30$) is not considered as a transition metal.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-03-14T07:56:17.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27307" PostTypeId="2" ParentId="27301" CreationDate="2015-03-14T08:20:36.083" Score="0" Body="&lt;p&gt;We normally titrate a weak acid by a strong base, or  weak base by a strong acid, to ensure that the reaction of titration is total (quantitative). We don't titrate weak acid by a weak base (and vice-versa).&#xA;I will calculate the equilibrium constant of the titration of  weak acid $\ce{HA}$ by a strong base $\ce{OH-}$:&#xA; $$\ce{HA + OH- &amp;lt;=&amp;gt;A- + H2O}$$ The equilibrium constant is $$K=10^{(\mathrm{p}K_w -\mathrm{p}K_a)}$$ For common weak acids (acetic acid $\mathrm{p}K_a=4.7$ ): $$K&amp;gt;&amp;gt; 10^3$$&#xA;So, the equation of titration (neutralization) must be written:$$\ce{HA + OH- -&amp;gt;A- + H2O}$$&#xA;At the equivalence point: $$n(\ce{HA})=n(\ce{OH-})$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-03-14T08:20:36.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27308" PostTypeId="2" ParentId="22135" CreationDate="2015-03-14T11:25:07.017" Score="7" Body="&lt;p&gt;If you haven't found answer till now, you can use VESTA (&lt;strong&gt;V&lt;/strong&gt;isualization for &lt;strong&gt;E&lt;/strong&gt;lectronic and &lt;strong&gt;ST&lt;/strong&gt;ructural &lt;strong&gt;A&lt;/strong&gt;nalysis) package. It is a freeware and can be downloaded from &#xA;&lt;a href=&quot;http://jp-minerals.org/vesta/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jp-minerals.org/vesta/en/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is available for both Windows and Linux. It can be used to import any many structural files (pdb included) and get an output in any supported format(which includes SVG). From their website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;VESTA is a 3D visualization program for structural models, volumetric data such as electron/nuclear densities, and crystal morphologies.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if you do not strictly require SVG format them my personal favorite is &quot;QuteMol&quot;. It gives way too many options to adjust visual representation&#xA;It is an open source (citeware, i.e. you have to cite their paper for any academic publishing) molecular rendering software available from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://qutemol.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://qutemol.sourceforge.net/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can generate PNG, JPEG anf GIF(animated) images of maximum dimensions up to 8192x8192 pixels so you can use them for high resolution printing/posters etc. too. One small flaw is that it is rather ill maintained and considerable amount of bugs are there.For e.g. using default options is fine but in case you want to use customized colour scheme then their example ART(atomic rendering table, just a table of atomic symbols, their radii, and colour in RGB codes) file gives following example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;X    0       0.853  204    0    0&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;H    1    0.930  199  199  199&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;He    2  1.085  107  115  140&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Li    3  3.100    0  128  128&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However in reality you will require 2 colums of radii for programme to recognize it (core and vander waal radii). Similarly atomic symbols with single alphabet are not rendered properly!!! But all these problems have work around and results are worth it (specially for large protein molecules).&#xA;sample this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YirOv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YirOv.png&quot; alt=&quot;nano motor&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://qutemol.sourceforge.net/sidetoside/fineMotion.pdb.ao.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;sourceforge.net&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="14960" LastEditorUserId="15751" LastEditDate="2019-02-27T09:55:43.410" LastActivityDate="2019-02-27T09:55:43.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27309" PostTypeId="1" AcceptedAnswerId="27311" CreationDate="2015-03-14T11:48:43.617" Score="1" ViewCount="15992" Body="&lt;p&gt;I do not understand why and how an atom with $\ce{sp^2}$ hybridization has only one p orbital?? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in pyridine why is the lone pair of nitrogen not counted as a pi electron.  &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="4231" LastEditDate="2015-03-14T14:51:55.020" LastActivityDate="2015-03-14T14:51:55.020" Title="sp² hybridized orbital" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27310" PostTypeId="1" CreationDate="2015-03-14T13:11:33.370" Score="1" ViewCount="25719" Body="&lt;p&gt;What is the amount of energy (J or Wh) needed to boil one litre of tap-water in a regular kettle starting at room temperature (20°C)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation itself assuming ideal and standard prerequisites would be interesting as well as factoring in the non-purity of tap-water and a typical kettle performance factor.&lt;/p&gt;&#xA;" OwnerUserId="2451" LastEditorUserId="2451" LastEditDate="2015-03-14T18:18:40.177" LastActivityDate="2017-03-17T20:10:42.790" Title="Amount of Energy Needed for Boiling a Litre of Water" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-14T19:08:38.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="27311" PostTypeId="2" ParentId="27309" CreationDate="2015-03-14T13:14:49.320" Score="6" Body="&lt;p&gt;The picture below from the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_07%3A_Organic_compounds_as_acids_and_bases/Section_7.4%3A_More_on_resonance_effects_on_acidity_and_basicity#7.4B:_The_basicity_of_nitrogen-containing_groups:_aniline.2C_imines.2C_pyridine.2C_and_pyrrole&quot; rel=&quot;noreferrer&quot;&gt;UC Davis ChemWiki&lt;/a&gt; summarizes why that lone pair is not available. It also explains how pyridine can function as a base.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dxpOH.png&quot; alt=&quot;pyridine orbitals&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The lone pair in the &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; orbital is perpendicular to the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbitals and &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is &lt;strong&gt;the one in the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=N}$&lt;/span&gt; bond&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;noreferrer&quot;&gt;Conservation of Angular Momentum&lt;/a&gt;, which does apply in quantum systems.&lt;/p&gt;&#xA;&lt;p&gt;Nitrogen, in its valence shell, has four orbitals: &lt;span class=&quot;math-container&quot;&gt;$2s,2p_x,2p_y,2p_z$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; hybridization is achieved by mixing two &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbitals and one &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbital that was not mixed is still there (and is the only one still there).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We cannot have two &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbitals at an &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; hybridized atom because then we would have 5 orbitals (or we would not have &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now, let's create a general rule for aromatic heterocycles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Look at pyrrole (also from ChemWiki);&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o1GRx.png&quot; alt=&quot;pyrrole&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the structure of pyrrole, the &lt;span class=&quot;math-container&quot;&gt;$\ce{N-H}$&lt;/span&gt; bond uses the &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; orbital and the lone pair is in the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; bond uses the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbital and the lone pair is in the &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a &lt;span class=&quot;math-container&quot;&gt;$\ce{N-H}$&lt;/span&gt; bond (or some other bond to the nitrogen). This &lt;span class=&quot;math-container&quot;&gt;$\ce{N-H}$&lt;/span&gt; bond uses the &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; orbital and the lone pair uses the &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; orbital and counts toward aromaticity.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-14T13:14:49.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27312" PostTypeId="2" ParentId="27310" CreationDate="2015-03-14T13:25:49.623" Score="6" Body="&lt;p&gt;It is simple. Density of water is $1\;\frac{\text{g}}{\text{ml}}$ as we have 1 kg water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now according to the laws of calorimetry the amount of heat required&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$c$ = specific heat capacity of water, which is 1 cal/(g °C)&lt;/li&gt;&#xA;&lt;li&gt;$m$ = mass of water; i.e. 1 kg&lt;/li&gt;&#xA;&lt;li&gt;$T$ = the room temperature in °C &lt;/li&gt;&#xA;&lt;li&gt;$\Delta {h_{\text{vap}}}$ = specific enthalpy of vaporization of water; &lt;a href=&quot;https://www.thermexcel.com/english/tables/vap_eau.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;540 cal/g&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you will get the amount of energy needed (in cal) to boil a litre of water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get answer in Joule multiply the answer in cal by 4.184 J/cal. And be careful with units!&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="37235" LastEditDate="2017-03-17T20:10:42.790" LastActivityDate="2017-03-17T20:10:42.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27313" PostTypeId="2" ParentId="22406" CreationDate="2015-03-14T13:40:16.753" Score="0" Body="&lt;p&gt;Its actually Coulombic repulsion between electrons and attraction between electron and protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be said shelding if you are considering the last electron. But if u consider any electron in the lower shells the electron in the upper shells will push the electron towards the nucleus and it won't be called shilding.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2015-03-14T13:40:16.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27314" PostTypeId="1" AcceptedAnswerId="27315" CreationDate="2015-03-14T13:43:53.710" Score="6" ViewCount="2639" Body="&lt;p&gt;I fail to understand why the hybridization of oxygen in furan is $\ce{sp^2}$. I think that since there are 2 bond pairs and two lone pairs so it should be $\ce{sp^3}$ hybridized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it to do with the aromaticity of furan?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="16683" LastEditDate="2017-02-27T21:19:41.360" LastActivityDate="2017-02-27T21:20:57.570" Title="What is the hybridization of oxygen in furan?" Tags="&lt;aromatic-compounds&gt;&lt;orbitals&gt;&lt;hybridization&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27315" PostTypeId="2" ParentId="27314" CreationDate="2015-03-14T13:53:06.037" Score="5" Body="&lt;p&gt;Furan is aromatic because one of the lone pairs of electrons on the oxygen atom is delocalized into the ring, creating a $4n+2$ Hückel aromatic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steric number is sometimes incorrectly described as the sum of the number of sigma bonds and the number of lone pairs. However, it is more accurately given by the number of sigma bonds plus the number of &lt;strong&gt;localised&lt;/strong&gt; lone pairs.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="16683" LastEditDate="2017-02-27T21:20:57.570" LastActivityDate="2017-02-27T21:20:57.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27316" PostTypeId="1" AcceptedAnswerId="27329" CreationDate="2015-03-14T15:50:49.013" Score="4" ViewCount="232" Body="&lt;p&gt;I've been in a few classes now that teach about energy, and I feel they're always so bad. There is always such a rush to start using equations to get problems solved, and I often feel I don't have an intuitive grasp on energy at all. I guess this question is also about pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does pressure decrease with time? If pressure is caused by the force of gas molecules hitting the inside surface of a container, don't they transfer energy and eventually slow down enough to totally reduce pressure?&lt;/li&gt;&#xA;&lt;li&gt;Along the same lines, energy can be released or absorbed by a reaction as work or heat. If the only change that happens is through work (in something that is totally insulated - where gas is formed and pressure is created), after the pressure is created, doesn't it require continuous energy to maintain pressure against a constant force of 1 atm (or whatever it is)? Heat dissipates to all things, and that makes sense as everything gets colder in time. The same does not seem true for pressure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noobsauce&lt;/p&gt;&#xA;" OwnerUserId="14965" LastEditorUserId="7448" LastEditDate="2015-03-14T16:18:19.467" LastActivityDate="2016-03-12T00:49:58.880" Title="Thermochemistry: forms of energy &amp; pressure" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27317" PostTypeId="1" CreationDate="2015-03-14T15:52:09.407" Score="4" ViewCount="999" Body="&lt;p&gt;the question is,&lt;br&gt;&#xA;The two square planar isomers of $\ce{Pt[BrCl(PR_{3})2]}$ have a different $\ce{^{31}P}$-nmr spectra. One shows single group of lines, the other shows two distinct resonances, each one being similar to the single resonance region of the first one. Which isomer is cis and which is trans?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14966" LastEditorDisplayName="user7951" LastEditDate="2016-11-20T15:30:20.213" LastActivityDate="2016-11-20T15:30:20.213" Title="Pt[BrCl(PR₃)₂] cis, trans isomer nmr - Which is cis and which is trans?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27318" PostTypeId="2" ParentId="27317" CreationDate="2015-03-14T16:03:19.833" Score="2" Body="&lt;p&gt;Chemical shifts and multiplicity are always controlled by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Who sees what?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two centres have the same environment/neighbours or are related by a symmetry operation, they will have identical shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the phosphor atoms in the &lt;em&gt;trans&lt;/em&gt; isomer both &quot;see&quot; the same and appear as a single set of lines.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-14T16:03:19.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27319" PostTypeId="2" ParentId="27317" CreationDate="2015-03-14T16:07:43.320" Score="8" Body="&lt;p&gt;Look at the two structures drawn below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4MTDf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;trans&lt;/em&gt;-isomer each phosphorous atom is in the exact same chemical environment.  Each one has a chlorine on one side, a bromine on the other side and a phosphorous diagonally across.  Therefore only one signal is observed for the phosphorous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cis-isomer, each phosphorous is in a different chemical environment.  One phosphorous has a chlorine diagonally across from it, the other has a bromine diagonally across from it.  In this case we observe two phosphorous signals, one for each of the two different phosphorous environments.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-14T16:07:43.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27320" PostTypeId="1" AcceptedAnswerId="27322" CreationDate="2015-03-14T16:24:11.120" Score="8" ViewCount="533" Body="&lt;p&gt;Using organic chemistry I have failed to find a method that doesn't involve halogens (gasseous especially).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please ignore the enzyme mono-oxygenase, which adds $\ce{O2}$ directly to alkanes to form alcohols. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[edit] Regarding selectivity the reaction should preferably stop at the alcohol stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most promising method I have found &lt;a href=&quot;http://rutchem.rutgers.edu/~agoldman/ACS885_Chap08_CH_Functionalization_Borane_Reagents_Hartwig.pdf&quot; rel=&quot;nofollow&quot;&gt;involves Boranes&lt;/a&gt;, but only works for terminal alkanes. The termial organoborane can then &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroboration%E2%80%93oxidation_reaction&quot; rel=&quot;nofollow&quot;&gt;be converted to an alcohol&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found it using a google search inspired by the answers below: &quot;photochemistry of alkanes.&quot; &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4945" LastEditDate="2015-03-18T03:09:22.873" LastActivityDate="2015-03-18T11:36:14.117" Title="Non-halogen intermediate between alkane and alcohol under basic conditions?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27321" PostTypeId="2" ParentId="27316" CreationDate="2015-03-14T16:33:45.203" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.Does pressure decrease with time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ideal gas law tells us that&#xA;$$PV=nRT$$&#xA;or &#xA;$$\mathrm{P=\frac{[nRT]}{[V]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature, number of moles or volume does not change, then the pressure will not change.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If pressure is caused by the force of gas molecules hitting the inside&#xA;  surface of a container, don't they transfer energy and eventually slow&#xA;  down enough to totally reduce pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that's correct.  &lt;strong&gt;If energy is transferred out of the system&lt;/strong&gt;, then the temperature will fall and the pressure will decrease.  But if everything is already at the same temperature, there will be no energy flow in or out and the pressure will remain constant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;after the pressure is created, doesn't it require continuous energy to&#xA;  maintain pressure against a constant force&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, there is no requirement that energy must be supplied to maintain the pressure.  Again, if the gas, the vessel and the surroundings are all at the same temperature the pressure will not change.  However, if the entire system is not at thermal equilibrium, then heat will flow so as to put things at equilibrium and the temperature will change.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="24052" LastEditDate="2016-03-12T00:49:58.880" LastActivityDate="2016-03-12T00:49:58.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27322" PostTypeId="2" ParentId="27320" CreationDate="2015-03-14T17:08:36.840" Score="6" Body="&lt;p&gt;I think that you're asking if there is a way to convert an alkane to an alcohol without using halogens in the process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkanes are relatively unreactive and so high energy species (like free radicals) are most often used to carry out reactions on alkanes. The  reaction of alkanes with halogens that you've mentioned is an example of this free radical process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other high energy molecules that can be used to convert alkanes to alcohols include ozone and (photo-) chemically generated singlet oxygen.  Just like with the halogens, these reactive molecules show little selectivity, so if primary, secondary and tertiary hydrogens co-exist in your alkane, then a mixture of primary, secondary and tertiary alcohols result.  However in molecules such as cyclohexane, where there is only one type of hydrogen, somewhat useful results can be obtained.  I say &quot;somewhat useful&quot; because ozone and singlet oxygen can oxidize the resultant alcohol further to a carbonyl.  Here is a &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ba-1968-0077.ch060&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt; to an abstract on the use of ozone to functionalize alkanes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reagents such as potassium permanganate and chromic acid also oxidize alkanes by a free radical mechanism to produce alcohols.  Again the issues of selectivity and further oxidation remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a nice review entitled &quot;&lt;a href=&quot;http://www.scripps.edu/baran/images/grpmtgpdf/Voica_Mar_09.pdf&quot; rel=&quot;nofollow&quot;&gt;Alkane Hydroxylation&lt;/a&gt;.&quot; It covers a wide variety of reagents.  Most of the molecules depicted contain functional groups which can help improve selectivity, but there are a number of hydrocarbon examples as well.  Perhaps you'll find an example that is similar to your molecule of interest.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-14T18:36:05.273" LastActivityDate="2015-03-14T18:36:05.273" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27323" PostTypeId="2" ParentId="2547" CreationDate="2015-03-14T17:15:20.630" Score="3" Body="&lt;p&gt;I would like to add: the key for a substance to be coloured, is the excitation of electrons, because while it goes back from the excited state to the ground one, it releases photons, which we can see as any of the colours. (Niels Bohr).&lt;br/&gt;&#xA;This is also the reason why glass is colourless. The valence electrons of silicates in glass can't 'afford' the energy of light, so they don't get excited, while adding any transition metal in glass, they will get excited and therefore the glass will be coloured. The relation between absorbance of a molecule and the wavelength of light gives us the spectrum and it is pretty much important to know in order to understand the colours of any given substance.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="7448" LastEditDate="2015-03-14T17:21:38.863" LastActivityDate="2015-03-14T17:21:38.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27324" PostTypeId="2" ParentId="27248" CreationDate="2015-03-14T17:21:58.367" Score="-1" Body="&lt;p&gt;The simple glass is not recomended to use in laboratory IF you are using substances thay may emit Ultra Violet light, because the glass absorbs it and we know that that kind of light has a really high energy. Therefore quartz ( silicon dioxide ) is more suitable. And of course, radioactive elements will need special conditions.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="160" LastEditDate="2015-03-14T21:40:28.607" LastActivityDate="2015-03-14T21:40:28.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27325" PostTypeId="2" ParentId="27302" CreationDate="2015-03-14T17:44:35.647" Score="13" Body="&lt;p&gt;There are many criteria that you can follow to choose a functional, in no special order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Availability: Do you have access to a software package in which the functional is implemented? If yes, are analytic derivatives also implemented? Do you need extra dispersion corrections? Are they available?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Computational cost: Functionals in different steps of &quot;Jacob's ladder&quot; (pure, hybrids, double-hybrid) scale in different way with system size (N^3,N^4,N^5).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Accuracy of the functional for the property that you want to calculate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course it is not a good idea to select a functional based in a chronological sequence. It is true that newer functionals tend to be more accurate that old ones. BUT, it is not always the case, there are very new functionals that have high theoretical interest, but lacks of outstanding performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;¿What about B3LYP? It is true that the 'magical functional' is not the best performer today, but it is widely accepted, no one will complain about its usage. For most applications, it is very good for geometry optimization in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not a simple recipe, the best you can do is just compare the results of many functionals in many test sets. Look at the test set designed to test the properties that you need to predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, till today, for almost any application you will be ok if just read the results of this paper:&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1039/C0CP02984J&quot; rel=&quot;noreferrer&quot;&gt;Lars Goerigk and Stefan Grimme, &lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;13&lt;/em&gt;, 6670-6688.&lt;/a&gt; (DOI: 10.1039/C0CP02984J)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are too few functionals out there in which you can be interested&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="4945" LastEditDate="2015-03-16T06:22:25.440" LastActivityDate="2015-03-16T06:22:25.440" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="27326" PostTypeId="2" ParentId="27320" CreationDate="2015-03-14T17:49:10.617" Score="4" Body="&lt;p&gt;ron has already pointed out that the functionalization of alkanes often involves radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical process is the &lt;em&gt;Fenton reaction&lt;/em&gt;, typically using $\ce{FeSO4}$ and $\ce{H2O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the reactive species is the hydroxyl radical, $\color{\red}{\ce{HO\cdot}}$, which is generated via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{Fe^{2+} + H2O2 -&gt; Fe^{3+} + OH- + \color{\red}{HO\cdot}}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a variant, the &lt;em&gt;photo-Fenton reaction&lt;/em&gt;, the process is facilitated by UV/VIS irradiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irradiation in the visible range enables photoreduction of $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$, furnishing more hydroxyl radicals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{Fe^{3+} + H2O -&gt;[h\nu] Fe^{2+} + \color{\red}{HO\cdot} + H+}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon irradiation at wavelength below 400 nm, direct excitation (and homolytical cleavage) of hydrogen peroxide provides an additional source for the reactive species: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{H2O2 -&gt;[h\nu] 2\color{\red}{HO\cdot}}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, due to the lack of selectivity, this process is rarely used in synthesis and rather applied for the opposite: degradation of organic matter in wastewater treatment.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-14T18:28:12.483" LastActivityDate="2015-03-14T18:28:12.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27328" PostTypeId="1" CreationDate="2015-03-14T18:16:39.513" Score="1" ViewCount="294" Body="&lt;p&gt;The binding of oxygen by haemoglobin giving oxy-haemoglobin is partially regulated by the concentration of $\ce{H3O+}$ and dissolved in $\ce{CO2}$ in blood&#xA;$$\ce{HbO2 + H3O+ +CO2&amp;lt;=&amp;gt;H+.Hb.CO2 + O2 + H2O}$$ &#xA;If there is production of lactic acid and $\ce{CO2}$ in a muscle during excretion then is more oxygen released or is more $\ce{HbO2}$ formed?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="275" LastEditDate="2015-03-14T18:48:04.783" LastActivityDate="2015-03-14T18:48:04.783" Title="Hemoglobin equilibrium - effect of increased carbon dioxide and lactic acid" Tags="&lt;equilibrium&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27329" PostTypeId="2" ParentId="27316" CreationDate="2015-03-14T19:18:03.153" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does pressure decrease with time? If pressure is caused by the force of gas molecules hitting the inside surface of a container, don't they transfer energy and eventually slow down enough to totally reduce pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. As you know, an aerosol can remains pressurised indefinitely until it is used. The molecules do not slow down if no energy transfer occurs, because there are no smaller bodies to transfer their energy to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the start of a game of snooker/pool/billiards (whatever you play in your part of the world) the balls are arranged in a triangle at the bottom of the table. You place the white ball at the top of the table and hit it into the other balls. When the collision occurs, the white ball shares its energy (normally unevenly) with the other balls. What we see is an &lt;em&gt;increase in disorder.&lt;/em&gt; Even if we neglect friction, once this has occured it is very unlikely (impossible even) that all the other balls would stop and the white ball would have all the energy again. The way the energy of the white ball is spread between the other balls is a macroscopic display of how kinetic energy is dissipated by friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will learn later on that while energy must be conserved, disorder in a &lt;em&gt;closed&lt;/em&gt; system can only stay the same or increase; it can never decrease. Disorder is quantified by &lt;em&gt;entropy&lt;/em&gt; but the quantification is beyond the scope of your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balls on the table will soon stop moving, due to friction. What happen is that the kinetic energy is shared around in a more disordered way amongst the molecules of the balls and table. The energy is still there (the balls and table are very very slightly warmer) but it is now distributed in random movements of the molecules so that the balls are no longer moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once it gets to this level, the molecules just keep vibrating (or in the case of a gas, colliding.) There is no friction in these interactions, because there is nothing smaller than a molecule for energy to be dissipated into. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(well actually there is one thing: the molecules could shake themselves free of the solid they are in and become a liquid, shake themselves free of the liquid they are in and become a gas, or even shake themselves apart and become smaller molecules or atoms. But breaking these bonds or interactions can only occur with additional heat absorption, or by cooling of the material.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;after the pressure is created, doesn't it require continuous energy to maintain pressure against a constant force of 1 atm (or whatever it is)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;work&quot; (which is a transfer of mechanical energy) only occurs when something is moved against a force. For example, when you put your finger over the end of a bicycle pump and squeeze the plunger, your squeezing arm does work. If you release the plunger, it moves outward again, releasing the stored energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you find it difficult to maintain the force on the plunger indefinitely does not mean that any work is being done. Your body is not designed to maintain a force with perfect efficiency; energy is expended within your body to maintain that muscle contracted. But if you had a latch on the bicycle pump, you could close that and maintain the pressure indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when you lie down on an airbed, you increase the pressure inside it. You have no expenditure of energy while you are lying on the bed, yet you maintain the pressure. Similarly the aerosol can maintains its internal pressure with no expenditure of energy. &lt;/p&gt;&#xA;" OwnerUserId="5541" LastActivityDate="2015-03-14T19:18:03.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27330" PostTypeId="1" AcceptedAnswerId="27340" CreationDate="2015-03-14T19:53:09.130" Score="1" ViewCount="69" Body="&lt;p&gt;In one problem it's asking me to find the solubility of $\ce{Ca(OH)2}$ in water. In that problem it assumes the starting concentration $\ce{OH}$ to be zero. But for other problems like $\ce{BaSO4}$ in $0.1~M~~ \ce{Na2SO4}$, we account for the $\ce{SO4}$ in the $\ce{NaSO4}$. So in the first problem, why don't we worry about the $\ce{OH}$ from the self ionization of water? Is it just that extremely low of an amount? &lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="4231" LastEditDate="2015-03-14T20:14:41.600" LastActivityDate="2015-03-15T00:46:28.690" Title="Solubility determination of Ca(OH)₂ in water" Tags="&lt;solubility&gt;&lt;solutions&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27331" PostTypeId="1" AcceptedAnswerId="27334" CreationDate="2015-03-14T19:57:12.253" Score="4" ViewCount="2610" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RS6S2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphs such as the above keep coming up when talking about harmonic oscillators in a quantum mechanical sense. However, I simply cannot make sense of them. What does each line represent why are they waves and what is the parabola? Also, I see sometimes graphs such as:&lt;img src=&quot;https://i.stack.imgur.com/T8C9S.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These seem to follow on from the initial graph. It's something to do with the recovery of classical behavior but I don't understand the first graph so stand no chance of getting the second ones at high n - perhaps you could explain the relation between the first and second graph please? Thanks&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-03-15T00:50:06.590" Title="Please explain the following graph for a quantum mechanical harmonic oscillator" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27332" PostTypeId="2" ParentId="16640" CreationDate="2015-03-14T20:19:25.540" Score="12" Body="&lt;p&gt;One could melt them floating on a pool of high-boiling point denser metal, or in space where they can readily be contained. Or one could create a thick actively cooled shell and melt them inside it, melting part of the shell as well. Finally, it's probably not very practical, but one could use an air jet to keep then suspended away from other matter and then melt them with lasers or superheated air.&lt;/p&gt;&#xA;" OwnerUserId="14971" LastActivityDate="2015-03-14T20:19:25.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27333" PostTypeId="1" AcceptedAnswerId="27342" CreationDate="2015-03-14T20:59:54.023" Score="0" ViewCount="314" Body="&lt;p&gt;As is well known, water can be electrolyzed into oxygen and hydrogen gasses. However, all redox reactions can be run in reverse. If I were to electrolyze water in 2 closed containers (in order to trap the gasses) and i were to run the cell in reverse using a battery, would the hydrogen and oxygen gasses turn back into ions and, therefore, water? If so how would this work on a physical level? And what path do the ions take to get to each other?&lt;/p&gt;&#xA;" OwnerUserId="1948" LastActivityDate="2015-04-14T01:58:46.123" Title="Ionize Hydrogen and Oxygen Gas in Electrolytic Cell?" Tags="&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27334" PostTypeId="2" ParentId="27331" CreationDate="2015-03-14T21:02:14.627" Score="5" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by considering the case of 2 hydrogen atoms separated by a large distance.  If they happen to be on a trajectory that brings them closer together they can interact with one another.  If they get close enough together they can form a bond and create a hydrogen molecule.  This process is pictured in the following graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Srgg1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemicool.com/definition/morse_potential.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;potential well&quot; is often referred to as a &quot;Morse potential&quot;.  Out on the right edge of the curve (say 5 angstroms), we have our 2 separated hydrogen atoms.  As the distance between them lessens, stabilization occurs, they can fall into the well and form a hydrogen molecule.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red lines in the Morse potential represent different vibrational states of the hydrogen molecule.  Initially, the hydrogen atom is formed in one of the higher vibrational states.  It can either absorb more energy through collisions and go back to 2 dissociated atoms or it can give away some excess energy through collisions and fall to a lower vibrational level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical view of the hydrogen atom is two weights connected by a spring.  The more energy in the system, the wider the oscillation of the spring.  With lower energy the oscillations are smaller.  The energy of the spring system (y-axis) as a function of separation (x-axis) is given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Hooke&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt; as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E(x)={1\over2}kx^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really an equation that describes a parabola, so the lower part of a Morse potential can be accurately described by a parabola.  &lt;strong&gt;Your top drawing represents the lower part of the Morse potential diagram&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we have used hydrogen as our model in the above explanation, the same reasoning applies to any bond formation (and subsequently bond stretch) process.  For example, we could describe a $\ce{C-H}$ bond stretch in a completely analogous manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What does each line represent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Different vibrational levels, as explained above&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why are they waves&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Each energy state (vibrational level) is described by its own wavefunction ($\Psi_n$).  The square of the wavefunction (${\Psi_{n}^2}$) describes the probability of finding the electron at a certain point on the x-coordinate for that specific vibrational energy level. From the solution to the wavefunction we find that as we go to higher and higher energy levels, the number of nodes (a point where the wavefunction = 0) in our wave continually increases.  The square of the wavefunction produces positive values (the &quot;upside-down&quot; parabolas). We always have the same number of nodes in the wavefunction and the probability distribution.  For example, in the lowest vibrational energy level (n=0) there is no node in either the wavefunction or the probability distribution.  In the n=1 vibrational level there is one node in the wavefunction and probability distribution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the parabola?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Explained in the background section&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-15T00:50:06.590" LastActivityDate="2015-03-15T00:50:06.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27336" PostTypeId="2" ParentId="27279" CreationDate="2015-03-14T21:37:08.147" Score="2" Body="&lt;p&gt;@DavePhD already gave a good description of how to prepare the solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are possible good reasons why the teacher doesn't like your original proposed method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;distribute evenly&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;does not work in practice (unless you mean: prepare a bit more, and then pipette 5 equal amounts).  But if you pipette, there is no reason not to pipette the amounts of stock solution that directly lead to the neeeded volume for the dilutions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;beaker&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is no volumetric glassware, and in addition not the preferred glassware for mixing (though mixing with magnet stirrer is OK). Volumetric flasks are much better: you can close them and then thorough mixing is possible. If you don't have volumentric flasks, Erlenmeyers are better than beakers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would it be easier to dilute them by half each time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would lead to an exponential series of dilutions, like 1 mol/l, 0.1 mol/l, 0.01 mol/l etc. instead of the equispaced concentration series you describe. Also, serial dilution means that concentration errors build up: each dilution step adds error.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt; Ultimately I will need 200ml of each solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With volumetric flasks, you can directly prepare 250 ml of each solution. Your proposed method would lead to varying volumes. It is unlikely that the proper volumes are available as volumetric flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the needed volumnes for the volumetric flask + pipetting stock solution or the graduated cylinder version is easy. A plausibility check of this can probably be done on the fly in the lab without calculator.&lt;br&gt;&#xA;Calculating the needed additional volume for your method is somewhat more elaborate. This is of course a bad reason if it  is only about laziness. But it is a good reason if it is about doing plausibility checks while doing lab work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last possible reason: working with volumetric flasks thorough mixing is easier and the volume is anyways more precise than with a graduated cylinder. (Far more precise if you are still learning the practical skills of lab work: with the graduated cylinder you neet to get both volumes right at the first attempt. &#xA;With a pipette getting the volume right is much easier. So only the flask volume needs to be right at the first attempt - and the flask has a neck that helps with precise volumina.)&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2015-03-14T21:37:08.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27337" PostTypeId="2" ParentId="23919" CreationDate="2015-03-14T21:49:11.210" Score="1" Body="&lt;p&gt;$\ce{H2O}$ does not absorb the radiation almost at all, so it will not interfere in the absorbance of the analyte we are analysing, thats why we use water almost always as a solvent in instrumental methods of analysis.Therefore $\ce{MeCN}$ wil absorb some of the radiation so then of course the absorbance of the analyte will also change ( decrease ) and the spectrum will be different ( the peaks of the analyte will be lower ). &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="4231" LastEditDate="2015-03-14T21:51:17.450" LastActivityDate="2015-03-14T21:51:17.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27338" PostTypeId="2" ParentId="27248" CreationDate="2015-03-14T22:14:00.767" Score="4" Body="&lt;p&gt;In addition to differences in the absorption spectrum, different types of glass are more or less able to deal with temperature or mechanical shocks. But that is less about manufacturer: you'll find different types of glassware from the same manufacturer for different purposes (e.g. Fiolax and Duran are both by Schott). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @LDC already commented, some glassware is thin walled and more mechanically fragile. OTOH, thick walled glass is more likely to break with a temperature shock. So e.g. for test tubes you have to decide whether you want to have them &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;suitable for local heating =&gt; thin wall borosilicate glass&lt;/li&gt;&#xA;&lt;li&gt;suitable for centrifuge =&gt; thick wall borosilicate glass&lt;/li&gt;&#xA;&lt;li&gt;cheap =&gt; soda glass&lt;/li&gt;&#xA;&lt;li&gt;UV transmittant =&gt; quartz &lt;/li&gt;&#xA;&lt;li&gt;suitable for huge temp shocks =&gt; quartz&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If this question is about whether your borosilicate glass ware is called Duran or  Pyrex (or Jenaer or Ilmabor or Rasotherm) I don't see that much of a difference. &lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2015-03-14T22:14:00.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27339" PostTypeId="1" CreationDate="2015-03-14T23:00:37.103" Score="7" ViewCount="9000" Body="&lt;p&gt;If I determine the order of the reactants in a reaction without using a catalyst, and then use a catalyst, will the order of the reactants then be different?&lt;/p&gt;&#xA;" OwnerUserId="14974" LastEditorUserId="30836" LastEditDate="2017-10-18T18:17:27.047" LastActivityDate="2017-10-18T18:17:27.047" Title="How does a catalyst affect the rate equation?" Tags="&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27340" PostTypeId="2" ParentId="27330" CreationDate="2015-03-15T00:14:22.407" Score="2" Body="&lt;p&gt;The self ionization of $\ce{OH}$ is $10^{-7}$. If the concentration of $\ce{Ca(OH)2}$ is greater by a few magnitudes, then the self ionization should not matter much. In fact, $\ce{Ca(OH)2}$ is considered soluble, so if not in miniscule amounts, the self ionization should not matter much&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="4231" LastEditDate="2015-03-15T00:46:28.690" LastActivityDate="2015-03-15T00:46:28.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27341" PostTypeId="1" AcceptedAnswerId="29417" CreationDate="2015-03-15T00:45:47.723" Score="2" ViewCount="27772" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the process, $\ce{CH3OH(l) -&amp;gt; CH3OH(g)}$&#xA;  $\Delta G^\circ = 4.30\ \mathrm{kJ/mol}$ at 25 °C. What is the vapor pressure&#xA;  of $\ce{CH3OH(l)}$ at 25 °C in mmHg?&lt;br&gt;&#xA;  (A) 0.176 mmHg&lt;br&gt;&#xA;  (B) 14.0 mmHg&lt;br&gt;&#xA;  (C) 134 mmHg&lt;br&gt;&#xA;  (D) 759 mmHg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is from the 2012 National Chemistry Olympiad (US).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt at solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^\circ = -RT\ln K$&lt;br&gt;&#xA;$4.3\times10^3\ \mathrm{J/mol}=-8.314\ \mathrm{J/(mol\cdot K)}\cdot 298\ \mathrm{K} \cdot\ln K$&lt;br&gt;&#xA;thus&lt;br&gt;&#xA;$0.1763\ \text{(unitless)} = K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I’m supposed to know that $K$ is in units of atm and convert to mmHg accordingly. How would it know that this unitless quantity is atm and not pascal or mmHg?&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T19:58:30.453" LastActivityDate="2016-11-26T18:48:37.953" Title="Calculating Vapor Pressure from Delta G/ Kp" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27342" PostTypeId="2" ParentId="27333" CreationDate="2015-03-15T01:02:29.027" Score="1" Body="&lt;p&gt;Since the monatomic hydrogen and oxygen have combined, respectively, in H2 and O2 gases, those molecular bonds must be broken in order to get them to recombine as water and produce electricity. That is precisely what a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fuel_cell&quot; rel=&quot;nofollow&quot;&gt;fuel cell&lt;/a&gt; does. You can lease a hydrogen fuel cell powered Hyundai Tucson, if you're close to H2 filling stations, and Iceland is looking into setting up a hydrothermal-powered H2 fuel station system.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-15T01:02:29.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27346" PostTypeId="1" CreationDate="2015-03-15T01:50:23.570" Score="24" ViewCount="16234" Body="&lt;p&gt;Other isomers include 1-butene, cis 2-butene, and trans 2-butene. Why would 2-methylpropene be less in energy if there is more steric hindrance?&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="10185" LastEditDate="2015-03-30T17:14:43.250" LastActivityDate="2017-11-30T01:45:54.570" Title="Why is 2-methylpropene less in energy than its alkene counterparts?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="27347" PostTypeId="2" ParentId="27346" CreationDate="2015-03-15T02:43:59.747" Score="27" Body="&lt;p&gt;There are two things we need to understand before we can answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) More highly substituted double bonds are generally more stable than less substituted double bonds.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$  hybridized carbon in the double bond. Electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower the energy of the electrons in that orbital.  As the electrons flow from the $\ce{sp^3}$ carbon to the $\ce{sp^2}$ carbon the energy of the electrons (and therefore the energy of the molecule) is lowered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;shorthand&quot; way to illustrate this electron flow is by drawing hyperconjugated resonance structures like the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c09kw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently 2-butene (where we can draw 6 hyperconjugated resonance structures) is more stable than 1-butene (where we can only draw 2 hyperconjugated resonance structures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Carbanion stability decreases in the order primary &gt; secondary &gt; tertiary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we've just shown above, alkyl groups are electron releasing.  If we have a negative charge on a carbon (a carbanion), the last thing we want is to have more alkyl groups attached to that carbon releasing more electron density onto an already negative carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now we can return to your question&lt;/strong&gt; and try to understand why 2-methylpropene is more stable than &lt;em&gt;trans&lt;/em&gt;-2-butene.   Carbon-hydrogen hyperconjugation is at the center of the answer.  Let's compare some of the hyperconjugated resonance structures that can be drawn for these two molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hzIS2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that &lt;strong&gt;a)&lt;/strong&gt; more highly substituted double bonds are more stable &lt;strong&gt;and b)&lt;/strong&gt; that primary carbanions are more stable than secondary carbanions, we can compare hyperconjugated resonance structures A and B. Resonance structure A contains a disubstituted double (more stable) and a primary carbanion (more stable), while resonance structure B contains a monosubstituted double bond (less stable) and a secondary carbanion (less stable). Therefore, resonance structure A should be more stable than resonance structure B.  Since resonance structure A is more stable than resonance structure B, &lt;strong&gt;2-methylpropene will be more stable than trans-2-butene.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we note that &lt;em&gt;cis&lt;/em&gt;-2-butene is less stable that &lt;em&gt;trans&lt;/em&gt;-2-butene for steric reasons.  This allows us to place these $\ce{C_4}$ isomers in the following order of stability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2-methylpropene is more stable than &lt;em&gt;trans&lt;/em&gt;-2-butene which is more stable than &lt;em&gt;cis&lt;/em&gt;-2-butene, which is more stable than 1-butene.&lt;/strong&gt; If we examine the heats of combustion for these isomers, we see that our order of stability is confirmed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1cp2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://quimicaorganica.webcindario.com/Eliminacion1.pdf&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-29T23:30:50.353" LastActivityDate="2015-03-29T23:30:50.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27348" PostTypeId="1" AcceptedAnswerId="28836" CreationDate="2015-03-15T04:56:52.400" Score="4" ViewCount="998" Body="&lt;p&gt;Assume in a software like Avogadro, I put molecules A and B in a specific distance and I know they will involve in a charge transfer reaction ( as Marcus Theory outer sphere transfer ). Then I run a calculation package as Gaussian or GAMESS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to see if any charge has moved from A to B.  One approach is to measure their charges and see if they are partially charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I don't know how to find that molecular partial charge in a selected part of the system in such softwares (like I select a part of the system on the screen and see its charge )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach I found is in Avogadro: View&gt;properties &gt; atom properties and find the charge of each single atom and then add the selected atoms together , but it is very very time consuming in larger systems. I am pretty sure there must be a more effective method.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="5017" LastEditDate="2015-03-16T01:21:40.507" LastActivityDate="2015-04-16T19:24:53.827" Title="How to find molecular charge of fragments using quantum calculations?" Tags="&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27353" PostTypeId="2" ParentId="5393" CreationDate="2015-03-15T08:39:23.587" Score="0" Body="&lt;p&gt;Let it be organic or inorganic compound &#xA;More the no. of bonds between 2 atoms more the energy required to break it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because intermolecular force of attraction increases as the no. of bonds increase. So to break the bonds one needs more energy or so called more bond enthalpy.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="4945" LastEditDate="2015-03-17T09:13:52.650" LastActivityDate="2015-03-17T09:13:52.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27355" PostTypeId="2" ParentId="24752" CreationDate="2015-03-15T09:08:13.933" Score="2" Body="&lt;p&gt;Water is a covalent compound. But as we know no compound is 100% ionic or 100% covalent. Since the electronegative difference (END) between hydrogen and oxygen is very high it is a strongly polarised covalent compound therefore the electron cloud is shifted towards the oxygen atom and therefore water has a very high % ionic character.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="4945" LastEditDate="2015-03-17T09:16:54.050" LastActivityDate="2015-03-17T09:16:54.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27356" PostTypeId="1" AcceptedAnswerId="27367" CreationDate="2015-03-15T10:01:47.217" Score="3" ViewCount="4291" Body="&lt;p&gt;We were taught that nitric acid can oxidize $\ce{Cl-}$ to $\ce{Cl2}$, but sulfuric acid cannot. Is this due to its smaller size, or structure, or what?&lt;/p&gt;&#xA;" OwnerUserId="14982" LastEditorUserId="5026" LastEditDate="2018-03-12T06:16:03.370" LastActivityDate="2018-03-12T06:16:03.370" Title="Why is nitric acid a stronger oxidising agent than sulfuric acid?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27357" PostTypeId="1" CreationDate="2015-03-15T10:07:21.333" Score="8" ViewCount="15963" Body="&lt;p&gt;$\ce{HNO3}$ always reduces its anion to give nitrogen containing compounds ( $\ce{NO2}$, $\ce{NO}$, and even $\ce{N2O}$) when reacting with metals, but with $\ce{Mg}$ and $\ce{Mn}$ it releases $\ce{H2}$. Is there a specific reason for this variation?&lt;/p&gt;&#xA;" OwnerUserId="14982" LastEditorUserId="9961" LastEditDate="2018-03-18T22:28:11.347" LastActivityDate="2018-03-18T22:28:11.347" Title="Does nitric acid release hydrogen only when reacting with Mg and Mn?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27358" PostTypeId="2" ParentId="27357" CreationDate="2015-03-15T10:23:46.690" Score="-3" Body="&lt;p&gt;Because reaction of Metal+Acid= Salt + hydrogen&#xA;$$\ce{Mg +2HNO3 -&amp;gt; Mg(NO3)2 +H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg}$ has a charge of +2 while nitric acid ($\ce{HNO3}$) has a charge of -1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals always react with acids to form salt and hydrogen gas&lt;/p&gt;&#xA;" OwnerUserId="14980" LastEditorDisplayName="user15489" LastEditDate="2015-04-12T10:31:38.623" LastActivityDate="2015-04-12T10:31:38.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27360" PostTypeId="1" AcceptedAnswerId="27370" CreationDate="2015-03-15T11:25:53.837" Score="5" ViewCount="16749" Body="&lt;p&gt;Do compounds like tropolone, furan, and thiophene (which are aromatic) show resonance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jspHP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jspHP.png&quot; alt=&quot;Compounds mentioned in question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes then what are the resonance structures that can be drawn? Also, do all aromatic compounds show resonance?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="16683" LastEditDate="2017-03-05T19:41:15.040" LastActivityDate="2017-03-05T19:41:15.040" Title="Resonance structures of some aromatic compounds" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27361" PostTypeId="1" AcceptedAnswerId="27819" CreationDate="2015-03-15T11:38:34.690" Score="7" ViewCount="10492" Body="&lt;p&gt;Why is &lt;em&gt;p&lt;/em&gt;-methoxyphenol more acidic than &lt;em&gt;p&lt;/em&gt;-methylphenol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, inductive effect really doesn’t matter at para positions and &lt;em&gt;p&lt;/em&gt;-methoxyphenol should be less acidic due to the high mesomeric effect of the $\ce{-OCH3}$ group in comparison to the weak hyperconjugative effect of the $\ce{-CH3}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/4-Methoxyphenol#section=Ionicity&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;p&lt;/em&gt;-Methoxyphenol&lt;/a&gt; – $\mathrm{p}K_\mathrm{a} = 10.1$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/4-Methylphenol#section=Ionicity&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;p&lt;/em&gt;-Methylphenol&lt;/a&gt; – $\mathrm{p}K_\mathrm{a} = 10.3$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorDisplayName="user7951" LastEditDate="2015-03-28T18:40:43.797" LastActivityDate="2018-08-06T17:43:04.597" Title="Why is p-methoxyphenol more acidic than p-methylphenol?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27362" PostTypeId="2" ParentId="16327" CreationDate="2015-03-15T11:43:33.163" Score="2" Body="&lt;p&gt;Well whatever metal you add will interfere in the absorbance of Ca, because the metal will absorb some of the light too, that's why that of the Ca decreases and thats why the analyte is almost every time mixed only with water which does not absorb the light.( and ofcourse you can add a reagent to form the complex which helps inorganic compounds to absorb the light )&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-03-15T15:28:21.450" LastActivityDate="2015-03-15T15:28:21.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27363" PostTypeId="2" ParentId="6735" CreationDate="2015-03-15T11:58:51.337" Score="2" Body="&lt;p&gt;I had this problem too, and what helped me was remembering some of the preparation charachteristics for each group in periodic table, because the preparation of elements in a group is pretty much similiar.&#xA;For example in 5th group, most elements are prepared from their most important alotropic modifications, more exact from their reduction with carbon. Example quartz $\ce{SiO2 + C ~ = ~Si + CO2}$&#xA;So does the $\ce{As}$ oxide, the $\ce{Bi}$ and the $\ce{Sb}$ oxide. Also $\ce{Pb}$ (4th) is prepared by its most important modification which is galenit $\ce{PbS}$.Concentrate in those similiarities in each group and you will have no trouble remembering them. If you don't find those similiarities i can help you.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="4231" LastEditDate="2015-03-15T13:21:55.587" LastActivityDate="2015-03-15T13:21:55.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27364" PostTypeId="1" CreationDate="2015-03-15T12:24:25.420" Score="5" ViewCount="904" Body="&lt;p&gt;I understand that true (non-relative) masses of atoms are calculated with a mass spectrometer. using this relationship: Centripetal Force = Force due to magnetic field(B) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I was wondering how masses of atoms were calculated before the mass spectrometers were invented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Please don't get this mixed up with relative atomic mass. That's not what I'm talking about.)&lt;/p&gt;&#xA;" OwnerUserId="14984" LastEditorDisplayName="user10259" LastEditDate="2015-05-15T00:34:47.713" LastActivityDate="2015-05-15T00:34:47.713" Title="How were the absolute masses of atoms calculated before the mass spectrometers?" Tags="&lt;atoms&gt;&lt;mass-spectrometry&gt;&lt;history-of-chemistry&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27365" PostTypeId="2" ParentId="24278" CreationDate="2015-03-15T13:04:16.477" Score="1" Body="&lt;p&gt;When reacted with alkaline metals, alcohols act somewhat like acids, oxidizing the metal and releasing hydrogen gas while the rest of the molecule becomes an anion. In this case, the anion consists of a negatively charged alkoxy group, called an alkoxide or alcoholate ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of the reaction is indeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2C2H5OH + 2K -&amp;gt; 2C2H5OK + H2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the product is named potassium ethoxide or potassium ethanolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, the reaction $\ce{C2H5OH + KOH -&amp;gt; C2H5OK + H2O}$ does happen as well as the reverse reaction: potassium ethoxide hydrolyses into ethanol and potassium hydroxide. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{C2H5OH + KOH &amp;lt;-&amp;gt; C2H5OK + H2O}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They exist in an ionic equilibrium with hydroxide being the favored ion, because water has a higher $K_a$ and thus a higher tendency than ethanol to remain in ionic form.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt; Wurtz reaction&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;About the other reaction $\ce{C2H5OH + K -&amp;gt; C2H5^{.} + KOH}$ which is provided in the question, I'm not sure if this reaction actually occurs. However, if the hydroxyl group is replaced by a halogen atom, that is in the case of a haloalkane/alkyl halide instead of an alcohol, this reaction actually happens, and is known as the Wurtz reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CH3CH2Cl + K -&amp;gt; CH3CH2^{.} + KCl}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the ethyl radicals produced would then dimerize to form butane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2CH3CH2^{.} -&amp;gt; CH3CH2CH2CH3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the overall reaction would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{2CH3CH2Cl + 2K -&amp;gt; C4H10 + 2KCl}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction can be used to synthesize structurally symmetrical alkanes, linear or branched(e.g. using 2-chloropropane).&lt;/p&gt;&#xA;" OwnerUserId="14983" LastEditorUserId="194" LastEditDate="2017-04-08T01:02:36.013" LastActivityDate="2017-04-08T01:02:36.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27367" PostTypeId="2" ParentId="27356" CreationDate="2015-03-15T13:53:38.680" Score="3" Body="&lt;p&gt;The nitrate ion is a strong oxidant whereas the sulfate ion is a milder oxidant that only shows oxidizing properties in high concentrations. The oxidizing atom in question is the oxygen atom in the acid, not the hydrogen ion. This is because hydrogen ion is not very electrophilic, which is why non-oxidizing acids cannot dissolve copper: even copper has a higher affinity to electrons than hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nitrate ion, the molecular structure is inherently not very stable, as the nitrogen atom has to bear a positive charge to maintain the molecular structure. Besides, nitrogen is in itself a very electronegative element, so that makes the nitrate ion even more unstable, hence its oxygen atom readily reacts with other atoms or molecules that oxygen can oxidize, including chloride ions: &lt;br/&gt;&#xA;$\ce{Cl- + O -&amp;gt; OCl-}$ &lt;br/&gt;&#xA;$\ce{H+ + OCl- &amp;lt;=&amp;gt; HOCl}$ &lt;br/&gt;&#xA;$\ce{H+ + Cl- &amp;lt;=&amp;gt; HCl}$ &lt;br/&gt;&#xA;$\ce{HCl + HOCl &amp;lt;=&amp;gt; H2O + Cl2 }$&lt;/p&gt;&#xA;" OwnerUserId="14983" LastEditorUserId="402" LastEditDate="2017-04-01T00:44:03.800" LastActivityDate="2017-04-01T00:44:03.800" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27368" PostTypeId="1" AcceptedAnswerId="27379" CreationDate="2015-03-15T14:00:39.383" Score="7" ViewCount="5317" Body="&lt;p&gt;When chloral undergoes reaction with a concentrated alkali, will it undergo Cannizarro reaction to form the corresponding alcohol and acidic salt, or should it form chloroform instead by departure of trichloro carbanion instead of the hydride ion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the trichlorocarbanion is stabilized by the $-I$ effect of chlorine, so why not form that instead?&lt;/p&gt;&#xA;" OwnerUserId="7005" LastEditorUserId="7475" LastEditDate="2015-10-16T16:57:32.690" LastActivityDate="2017-07-31T15:35:40.383" Title="Cannizarro reaction of chloral" Tags="&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27370" PostTypeId="2" ParentId="27360" CreationDate="2015-03-15T14:17:19.620" Score="9" Body="&lt;p&gt;Resonance structures can be drawn for &lt;strong&gt;all&lt;/strong&gt; aromatic compounds.  Tropolone, furan and thiophene are considered to be aromatic compounds because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can draw resonance structures&lt;/li&gt;&#xA;&lt;li&gt;and they are cyclic compounds with 4n+2 (n=1) pi electrons&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here are the resonance structures we can draw for &lt;strong&gt;tropolone&lt;/strong&gt;, the positive charge can be delocalized to every carbon in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G62l6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like in benzene, where instead of drawing 2 resonance structures (top of following figure) we usually just draw the one resonance structure shown at the bottom of the figure;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zx9oT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;instead of drawing all 8 resonance structures for tropolone, we usually represent them by just drawing the one resonance structure shown at the bottom of the tropolone figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here are the resonance structures for &lt;strong&gt;furan&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9udD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for &lt;strong&gt;thiophene&lt;/strong&gt;, they are the same as furan, just replace the oxygen with a sulfur atom in the resonance structures.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-15T17:14:29.380" LastActivityDate="2015-03-15T17:14:29.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27371" PostTypeId="1" CreationDate="2015-03-15T14:21:14.600" Score="7" ViewCount="2937" Body="&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/b/b1/Nu3_carbon_dioxide.png&quot; alt=&quot;Spectrum of CO2 at 2349 cm^-1&quot;&gt;&#xA;Spectrum of $\ce{CO2}$ at $2349\space \mathrm{cm^{-1}}$   region&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some other molecules (for example $\ce{CO}$) the peak separation corresponds to $2B$ (where $B$ is the rotational constant wavenumber)&#xA;However for $\ce{CO2}$ when I calculated the peak separation and compared it to literature value, it seems approximately to equal to $4B$? Any idea why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the explanation for $\ce{CO2}$ spectrum being asymmetric? and why does the $P$ branch have larger peak separations than the $R$ branch?&lt;/p&gt;&#xA;" OwnerUserId="14988" LastEditorUserId="41556" LastEditDate="2017-11-20T16:53:05.560" LastActivityDate="2017-11-20T19:46:01.417" Title="What does the peak separation of carbon dioxide IR rotational-vibrational spectrum corresponds to and why?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27372" PostTypeId="2" ParentId="24752" CreationDate="2015-03-15T14:21:35.380" Score="0" Body="&lt;p&gt;Water is certainly a covalent compound. Being produced from electron transfer doesn't mean that it is ionic. In a fuel cell, hydrogen is first ionized: $\ce{H2 &amp;lt;=&amp;gt; 2 H+ + 2 e- }$ &lt;br/&gt;&#xA;Next, the electron moves to the other electrode, reducing oxygen while doing electric work: $\ce{O2 + 4e- -&amp;gt; 2O^2^-}$ &lt;br/&gt;&#xA;this step is the step that generates electricity, because the oxygen has a higher electrode potential than hydrogen.&lt;br/&gt;&#xA;Finally, hydrogen ions and oxide ions unite to form water: $\ce{2H+ + O^2^- -&amp;gt; H2O}$&lt;br/&gt;&#xA;in this step, it is essentially similar to the reaction of an ionic oxide with an acid. Although the electron is initially transferred to the oxygen atom, as the ions of the opposite charge merge, the hydrogen ions gain their electrons back and share them with oxygen atoms, thus forming covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that water can &lt;a href=&quot;https://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow&quot;&gt;self-ionize&lt;/a&gt; and become ionic, although these ions have only a low concentration of $10^{-7} M$ in pure water under room conditions. The presence of solutes and temperature of water affects the amount of water ionized at any given moment.&lt;/p&gt;&#xA;" OwnerUserId="14983" LastActivityDate="2015-03-15T14:21:35.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27373" PostTypeId="2" ParentId="6735" CreationDate="2015-03-15T14:42:47.587" Score="6" Body="&lt;p&gt;Let me tell you how I have managed to, with about 85-90% accuracy, been able to predict a large number of inorganic reactions. There are some prerequisites to do this. They are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. The Periodic Table:&lt;/strong&gt; You need to learn the stable oxidation states and configurations of ions. For instance, pretty much all sulphides, upon oxidation, become sulphates, because +6 is one of the most stable states of sulphur. Or, Ferrous and Ferric ions are pretty much equally stable, so much so that I like to call them toggler ions, as the switch between each other with ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Electrochemical Series:&lt;/strong&gt; This is extremely important. It was simply amazing how much i could put into context after learning about this series. I could now understand why Iron metal produces hydrogen gas upon reacting with an acid. It will help you in countless redox reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Solubility of compounds (in water)&lt;/strong&gt; This will help you to predict precipitation reactions. In lower classes, we are taught these as lousy double displacement reactions, but they are actually pretty inaccurate. A ppt reaction will occur if the initial reactants are soluble, and at least one of the products is insoluble. Le-Chatelier's principle, if you will. Also, there is usually no way to predict solubility, as it is a complex interplay of hydration and lattice enthalpies, so you HAVE to memorize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a lot of reactions have no apparent logic, so there is no way to predict them. You will have to memorize these.&#xA;For instance, in the Rasching's process to prepare hydrazine, Ammonia reacts with the hypochlorite ion to produce chloramine. This cannot be predicted with the above steps. It has to be memorized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can say is, inorganic chemistry needs 55% memorization, and 45% common sense, which includes some intrinsic feeling as to what should be the outcome of a reaction, which will develop after a lot of practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-03-15T14:42:47.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27375" PostTypeId="1" AcceptedAnswerId="27378" CreationDate="2015-03-15T15:05:04.467" Score="3" ViewCount="499" Body="&lt;p&gt;All Google searches have simply returned more info on catalase. I'm looking for a catalyst that isn't found inside living organisms which can break down $\ce{H2O2}$. Looking for any resources to look at, or just examples of other catalysts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side question: Can catalase be used for anything other than the decomposition of  $\ce{H2O2}$?&lt;/p&gt;&#xA;" OwnerUserId="14990" LastEditorUserId="5026" LastEditDate="2018-06-09T01:58:46.613" LastActivityDate="2018-06-09T01:58:46.613" Title="Reaction of hydrogen peroxide and catalase" Tags="&lt;organic-chemistry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27376" PostTypeId="2" ParentId="21820" CreationDate="2015-03-15T15:40:15.210" Score="8" Body="&lt;p&gt;The reason why noble gases form molecules is because of London forces, which are very weak. The electrons in noble gas atoms are on the move all the time, so there will be a moment when they will spend more time in an area, which generates a temporary dipole and allows for interaction with another atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An addendum regarding noble gas &lt;em&gt;compounds&lt;/em&gt; in general: the first noble gas compound synthesised was $\ce{XePtF6}$. When $\ce{[O2+][PtF6-]}$ was discovered by the oxidation of dioxygen with platinum hexafluoride, it was hypothesised that $\ce{XePtF6}$ could be made via the oxidation of xenon with platinum hexafluoride, because of the similar ionisation energies of dioxygen and xenon.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="16683" LastEditDate="2017-10-06T23:15:12.233" LastActivityDate="2017-10-06T23:15:12.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27378" PostTypeId="2" ParentId="27375" CreationDate="2015-03-15T16:09:47.613" Score="2" Body="&lt;p&gt;Manganese (IV) oxide, iron (III) chloride, and lead (IV) oxide are common examples for a catalyst that isn't found inside living organisms and which can break down $\ce{H2O2}$. Potassium iodide can also be used as a catalyst, along with other transition metal compounds, of metals such as silver or platinum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This didn't take much effort to find so I'm sure you can find more examples easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for catalase, it seems to just decompose $\ce{H2O2}$, but this is useful in many different areas. For example, decomposing $\ce{H2O2}$ in contact lens cleaning solutions, preventing foods from being oxidised in wrappers (in conjunction with glucose oxidase, another enzyme), and removing $\ce{H2O2}$ from milk before cheese is made. The decomposition reaction is also used as a test for catalase in bacteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources&lt;/strong&gt;: General use - &lt;a href=&quot;https://web.archive.org/web/20151008025225/http://www.ncbe.reading.ac.uk:80/ncbe/materials/enzymes/catazyme.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Center of Biotechnology Education&lt;/a&gt; - and more uses (including the cheese example) - &lt;a href=&quot;http://www.worthington-biochem.com/CTL/default.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Worthington Enzyme Manual&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14863" LastEditorUserId="5026" LastEditDate="2018-06-09T01:58:22.593" LastActivityDate="2018-06-09T01:58:22.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27379" PostTypeId="2" ParentId="27368" CreationDate="2015-03-15T16:52:11.610" Score="8" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cannizzaro_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cannizarro reaction&lt;/a&gt; an aldehyde with &lt;strong&gt;no alpha-hydrogens&lt;/strong&gt; is treated with $\ce{OH^-}$.  Through a series of equilibria the aldehyde goes on to disproportionate into the corresponding carboxylate and alcohol as shown in the following mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LAdBr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the only requirement for the Cannizzaro reaction to occur is that the aldehyde not have any alpha-hydrogens.  Formaldehyde and benzaldehyde are two very common aldehydes that undergo the reaction when treated with $\ce{OH^-}$.  Chloral ($\ce{CCl3CHO}$), having no alpha-hydrogens also undergoes the Cannizzaro reaction (&lt;a href=&quot;https://books.google.com/books?id=YyoUNYgBYJoC&amp;amp;pg=PA239&amp;amp;lpg=PA239&amp;amp;dq=chloral%20cannizzaro%20reaction&amp;amp;source=bl&amp;amp;ots=OUBbAA7vAr&amp;amp;sig=MXtYb4RM_6Hz4d2x2iJD_oFlKBQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=uJUFVe_cGMjsgwSExYDgDQ&amp;amp;ved=0CCgQ6AEwAzgK#v=onepage&amp;amp;q=chloral%20cannizzaro%20reaction&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key part of the Cannizzaro mechanism is shown in the second line of the above mechanism.  Here a hydrogen is transferred in a &lt;strong&gt;concerted&lt;/strong&gt; step to the second molecule of aldehyde.  Free $\ce{H^-}$ is not generated in the process, rather it is a smooth, concerted transfer of a hydrogen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: This has been proven experimentally, when the reaction is run in $\ce{D2O}$, no deuterium winds up attached to the alcohol carbon.  This shows that free $\ce{H^-}$ was never involved, but rather a concerted transfer of the aldeyhdic hydrogen takes place].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Mithoron for noting that, with chloral, the haloform reaction can apparently compete with the Cannizarro reaction.  Presumably, the pathway followed depends upon the reaction conditions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2017-07-31T15:35:40.383" LastActivityDate="2017-07-31T15:35:40.383" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27380" PostTypeId="1" AcceptedAnswerId="27381" CreationDate="2015-03-15T17:23:15.773" Score="7" ViewCount="1648" Body="&lt;p&gt;E2 reactions (&lt;a href=&quot;https://en.wikipedia.org/wiki/Dehydrohalogenation&quot; rel=&quot;nofollow&quot;&gt;dehydrohalogenation&lt;/a&gt;) only occur when the hydrogen and leaving group are in the same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of a reason that they would need to be in the same plane, so what is the reason for this strange phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="470" LastEditDate="2015-03-15T23:58:25.900" LastActivityDate="2015-03-15T23:58:25.900" Title="Why do E2 reactions (dehydrohalogenation) only occur when the hydrogen and leaving group are in the same plane?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27381" PostTypeId="2" ParentId="27380" CreationDate="2015-03-15T17:38:55.953" Score="9" Body="&lt;p&gt;When the $\ce{C-X}$ and $\ce{C-H}$ $\ce{sp^3}$bonds break, the orbitals left behind on the two carbons will rehybidize into p orbitals and form the pi bond in the forming double bond.  For these two p orbitals to form a pi bond they must be in the same plane (or very close).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is illustrated in the following figure.  This is called the anti-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~180° with respect to each other) and is the preferred geometry for E2 elimination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcqmX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mhhe.com/physsci/chemistry/carey/student/olc/graphics/carey04oc/ref/ch05eliminationreactions.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elimination can also occur from the syn-periplanar conformation (the soon-to-be p orbitals are in the same plane and arranged ~0° with respect to each other) as shown below, but this conformation is slightly less favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3dNND.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why in molecules where the $\ce{C-X}$ and $\ce{C-H}$ bonds are locked in a conformation (perhaps by an inflexible ring system) where they are oriented 90° apart, E2 elimination becomes very difficult, if not impossible.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-15T17:50:16.000" LastActivityDate="2015-03-15T17:50:16.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27382" PostTypeId="2" ParentId="27339" CreationDate="2015-03-15T17:38:57.927" Score="8" Body="&lt;p&gt;Yes. It is possible for a catalyst to change the order of a reaction since it shows different path to the reaction i.e. reaction occurs through different mechanism. In the new path, rate determining step (slow step) may involve different number of reactant molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in case of reaction catalyzed on metal surfaces, most of the times, the rate does not depend on the concentration of the reactants at higher concentration since now it is surface area that governs the rate.&lt;/p&gt;&#xA;" OwnerUserId="7266" LastActivityDate="2015-03-15T17:38:57.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27383" PostTypeId="2" ParentId="27287" CreationDate="2015-03-15T18:04:41.297" Score="4" Body="&lt;p&gt;2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything you wrote above is correct &lt;strong&gt;excep&lt;/strong&gt;t for,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2p orbitals are more stabilized because they penetrate less in the&#xA;  space of the 1s orbital&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is correct that they penetrate less, but they are &lt;strong&gt;not&lt;/strong&gt; stabilized more.  Because they penetrate less, they are stabilized less by the effective nuclear charge.  &lt;strong&gt;Electron penetration correlates with electron stabilization&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's first comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;explain then why 2nd period non-metals can't easily form higher&#xA;  oxidation states while 3p elements can?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state.  But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel.  3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons.  &lt;strong&gt;3p electrons are therefore held more loosely than 2p electrons&lt;/strong&gt;.  This leads to lower electronegativities, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/23BVg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.tutorcircle.com/inorganic-chemistry/electronegativity-trends.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lower ionization potentials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xofNY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Ionization_Energy/Ionization_Energies&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and higher oxidation states being more common for third period non-metals compared to second period non-metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's second comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3p electrons penetrate well into the space near the nucleus (&lt;a href=&quot;https://www.rose-hulman.edu/~tilstra/Engineering%20Chemistry%20I/RadialDisributionFunctions.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;)&#xA;  Therefore it&#xA;  should be easier for 2p elements to form higher oxidation states than&#xA;  3p elements, but it's not, how?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are &quot;on average&quot; further from the nucleus than 2p electrons - which is the key point.  Here is a 3-dimensional representation of the same thing.  Look how much further away the 3p electrons are from the nucleus than the 2p electrons.  Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states.  This is consistent with the electronegativity and ionization potential arguments presented in my first edit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8hb5H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's third comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the effective nuclear charge has bigger increase for 2p than 2s&#xA;  (except N-O) according to &lt;a href=&quot;http://img.docstoccdn.com/thumb/orig/118492050.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this chart&lt;/a&gt;. How so?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key here is that the s orbital is spatially symmetric, it has a spherical pattern.  An electron in an s orbital can screen outer electrons from the nucleus &lt;strong&gt;equally&lt;/strong&gt; in &lt;strong&gt;all&lt;/strong&gt; directions.  On the other hand, a p orbital is directional, it is not spherically symmetric.  An electron in the 2px orbital will &lt;strong&gt;not&lt;/strong&gt; screen outer electrons from the nucleus &lt;strong&gt;equally&lt;/strong&gt; in &lt;strong&gt;all&lt;/strong&gt; directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641).  However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape.  So when we add the second 2p electron, it &quot;feels&quot; the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You noticed the &quot;anomaly&quot; for the 2p electrons as we go from nitrogen to oxygen.  This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern.  The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -&gt; O 2p effective nuclear charge increase=0.619).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-17T17:22:05.363" LastActivityDate="2015-03-17T17:22:05.363" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="27384" PostTypeId="1" CreationDate="2015-03-15T18:07:29.553" Score="9" ViewCount="5496" Body="&lt;p&gt;On regular paper, ink is absorbed, while on photopaper ink stays at the surface and dries for a longer time. What forces act in these two different cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(see also related questions about &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24007/how-are-pencil-marks-adhered-to-the-paper&quot;&gt;pencil marks&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24323/how-to-choose-a-colorant-for-a-given-substrate&quot;&gt;colorants&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-03-16T17:21:59.363" Title="What forces hold ink on paper?" Tags="&lt;everyday-chemistry&gt;&lt;intermolecular-forces&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27385" PostTypeId="1" CreationDate="2015-03-15T19:00:02.820" Score="14" ViewCount="4524" Body="&lt;p&gt;According to Wikipedia what gives epoxy its holding power is apparently &quot;ionic&quot; interactions between the epoxy and the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I thought that epoxy was overall neutral when cured. How can it make ionic bonds with the surface it is bonded to? Or are there unprotonated alkoxides left in the epoxy which then make the bonds with the surface? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="4356" LastEditDate="2015-05-15T19:42:05.813" LastActivityDate="2015-05-15T19:46:11.513" Title="How does epoxy stick to stuff?" Tags="&lt;materials&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27386" PostTypeId="2" ParentId="27364" CreationDate="2015-03-15T20:41:37.630" Score="6" Body="&lt;p&gt;Absolute mass can be determined electrochemically, once the &lt;em&gt;coulomb&lt;/em&gt; is defined in terms of numbers of electrons, as through the Millikan oil-drop experiment, to be  $6.24×10^{18}$ electrons per coulomb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.if.ufrj.br/~moriconi/termo_fisest/avogadro.htm&quot; rel=&quot;noreferrer&quot;&gt;Determination of Avogadro's Number&lt;/a&gt; and &lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/jres/79A/jresv79An6p747_A1b.pdf&quot; rel=&quot;noreferrer&quot;&gt;Absolute  Determination  of  the  Electrochemical  Equivalent and  the  Atomic  Weight  of Zinc&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that caused confusion in the determination of atomic mass is natural isotopic separation, e.g. sulfur derived from condensation from vulcanism vs. from mineral compounds, so that there were slightly different experimental results depending on source.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorDisplayName="user467" LastEditDate="2015-03-15T21:06:00.167" LastActivityDate="2015-03-15T21:06:00.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27387" PostTypeId="1" AcceptedAnswerId="27493" CreationDate="2015-03-15T21:08:42.233" Score="2" ViewCount="39918" Body="&lt;p&gt;I have a buffer containing 0.2 M of the acid $\ce{HA}$, and 0.15 M of its conjugate base $\ce{A-}$, with a pH of 3.35. I need to find the pH after 0.0015 mol of $\ce{NaOH}$ is added to 0.5 L of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started by converting everything to moles, writing out an equation for $\ce{HA}$ and $\ce{OH-}$ reacting to make $\ce{H2O}$ and $\ce{A-}$, getting 0.0985 mol of $\ce{HA}$ left after all the $\ce{OH-}$ was used up, and 0.015 mol of $\ce{A-}$ produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At that point I got a bit stuck. I tried doing things a bunch of different ways after that, such as trying to calculate $\mathrm{p}K\mathrm{_a}$ by using the Henderson-Hasselbalch equation, but I'm not getting anywhere near the answer. I calculated a $\mathrm{p}K\mathrm{_a}$ of 3.47. I added the amount of $\ce{A-}$ made when $\ce{HA}$ reacts with $\ce{OH-}$ to the original moles of $\ce{A-}$ in the buffer (I calculated 0.18 M $\ce{A-}$), and tried putting that, the leftover moles of $\ce{HA}$, and my $\mathrm{p}K\mathrm{_a}$ value into the equation, but that wasn't right either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="865" LastEditDate="2015-08-05T15:01:47.700" LastActivityDate="2017-09-06T20:37:45.090" Title="Finding new pH after NaOH added to buffer solution" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27388" PostTypeId="1" CreationDate="2015-03-15T21:21:13.860" Score="0" ViewCount="4312" Body="&lt;p&gt;So far, maybe it's just that my course isn't advanced enough, but I have never seen anything called 'neutralization constant'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any such reaction involving an acid and a base will be complete no matter the strength of either reactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buffer works the same way. If the reaction is reversible then the buffet is not effective.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-03-15T23:07:08.433" Title="Can neutralization be reversed?" Tags="&lt;titration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27389" PostTypeId="2" ParentId="27388" CreationDate="2015-03-15T21:26:04.393" Score="1" Body="&lt;p&gt;The reverse reaction of neutralization is called hydrolysis ( X salt + H2O = acid + base ) &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-03-15T21:26:04.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27390" PostTypeId="2" ParentId="27388" CreationDate="2015-03-15T23:07:08.433" Score="0" Body="&lt;p&gt;In general under normal conditions of temperature and pressure the reverse reaction of the neutralization is not possible because of their (acids &amp;amp; bases)  tendency (release of H+ ion &amp;amp; OH- ion) to form salt water due to which there is no any constant so called &quot;neutralization constant&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as for the consideration, the reverse of the neutralization reaction is called hydrolysis. In a hydrolysis reaction a salt reacts with water to yield the acid or base:&#xA;BA + H2O → HA + BOH&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-03-15T23:07:08.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27391" PostTypeId="1" AcceptedAnswerId="27395" CreationDate="2015-03-15T23:28:38.260" Score="2" ViewCount="52" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the topic of identifying an organic compounds using peaks generated in &lt;em&gt;Mass Spectrometry&lt;/em&gt;, a rule of thumb expressed in the UK educational textbook, CGP, is that that &quot;the M peak is the one with the second highest mass/charge ratio&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the significance of this M peak anyway?&lt;/p&gt;&#xA;" OwnerUserId="11402" LastActivityDate="2015-03-16T00:57:44.890" Title="What is the significance of an organic compound's M peak?" Tags="&lt;terminology&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27392" PostTypeId="1" AcceptedAnswerId="27394" CreationDate="2015-03-15T23:43:54.887" Score="3" ViewCount="3082" Body="&lt;p&gt;Question 7 of Edexcel's June 2014 Unit 2 exam paper reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a mass spectrum of butane, C4H10, where would a peak be seen for&#xA;  the molecular ion if it had a charge of 2+?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why?&lt;/p&gt;&#xA;" OwnerUserId="11402" LastActivityDate="2015-03-16T00:16:52.477" Title="Why would the 2+ molecular ion of butane have a mass spectrum peak at 29?" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27393" PostTypeId="1" AcceptedAnswerId="27400" CreationDate="2015-03-16T00:04:43.400" Score="5" ViewCount="1970" Body="&lt;p&gt;&lt;strong&gt;Background research&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Magnesium hydroxide exists naturally as brucite (&lt;a href=&quot;http://en.wikipedia.org/wiki/Magnesium_hydroxide#Preparation&quot; rel=&quot;nofollow&quot;&gt;source: wiki&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Passing hot steam over magnesium produces magnesium oxide (source: &lt;a href=&quot;http://freeexampapers.com/?dir=A-Level/Chemistry/Edexcel/2014-Jun&quot; rel=&quot;nofollow&quot;&gt;Edexcel's June 2014 Unit 2 exam (CH02_01): question 18&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't magnesium hydroxide form when treated with hot steam, instead of magnesium oxide?&lt;/p&gt;&#xA;" OwnerUserId="11402" LastEditorUserId="9961" LastEditDate="2015-10-31T00:00:09.843" LastActivityDate="2018-04-20T01:12:18.137" Title="Why does hot steam produce MgO instead of Mg(OH)2?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27394" PostTypeId="2" ParentId="27392" CreationDate="2015-03-16T00:16:52.477" Score="6" Body="&lt;p&gt;The mass spectrometer observes the mass to charge ratio (&lt;em&gt;m&lt;/em&gt;/&lt;em&gt;z&lt;/em&gt;). Most of the time, small organic molecules like butane will be singly charged by the ionizer, which means that the mass can be read off directly since &lt;em&gt;z&lt;/em&gt; =1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If butane was somehow ionized to the +2 state, then the mass spec would observe 58/2=29, which is the mass of butane (58 g/mol) divided by the charge (+2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes particularly important for larger compounds with multiple easily ionizable groups, such as peptides. A series of peaks corresponding to different charge states can be used to identify the compound of interest.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-03-16T00:16:52.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27395" PostTypeId="2" ParentId="27391" CreationDate="2015-03-16T00:20:17.970" Score="3" Body="&lt;p&gt;&quot;M&quot; stands for &lt;strong&gt;molecular&lt;/strong&gt; ion.  It is often represented as $\ce{M, M^{+} or M^{+~.}}$. In the mass spec experiment, an electron is knocked out of some of the sample molecules.  The peak at M/q (q=1) represents the molecular weight of the entire molecule, less an electron, before it fragments to smaller ions.  From precise measurement of the molecular ion (&lt;em&gt;e.g&lt;/em&gt;. 82.34567890) you can determine the exact molecular formula of your sample compound.  This can help greatly with sample identification.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-16T00:57:44.890" LastActivityDate="2015-03-16T00:57:44.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27396" PostTypeId="2" ParentId="27348" CreationDate="2015-03-16T01:27:10.510" Score="4" Body="&lt;p&gt;Your question boils down to &quot;where are the electrons&quot; in a quantum chemical system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, is if there's any level of interaction between the two fragments, &lt;strong&gt;you don't know&lt;/strong&gt;. The charge will be delocalized across the entire A+B complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put it another way.. Let's say you have a bunch of atoms and you run a calculation and the overall charge on the molecule is +1. Where's the extra positive charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, you might make some guesses, but it's probably delocalized across multiple atoms. The same thing happens with intermolecular interactions, but to a lesser degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two approaches:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Constrained DFT: This is a method where you &lt;em&gt;force&lt;/em&gt; the charge on a particular fragment to be the way you want. You can then compare this semi-localized charge with the distributed (unconstrained) calculation and compare.&lt;/li&gt;&#xA;&lt;li&gt;Fragment MO: You can diagonalize the wavefunction onto multiple fragments that you define.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't use GAMESS much, but a quick search implies that &lt;a href=&quot;http://myweb.liu.edu/~nmatsuna/gamess/refs/howto.fmo.html&quot; rel=&quot;nofollow&quot;&gt;FMO&lt;/a&gt; is supported in GAMESS.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-16T01:27:10.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27398" PostTypeId="1" AcceptedAnswerId="27399" CreationDate="2015-03-16T03:25:26.430" Score="6" ViewCount="278" Body="&lt;p&gt;I bought a thermometer from Cole-Parmer having kerosene as the fluid, and the fluid is broken up. The fluid in the bottom part goes up to about 10 °C even though it is 20 °C in the room, then there is a gap followed by fluid between 23 °C and 25 °C, then there is no fluid going all the way to the top where there is another small stretch of fluid. There is a short black rubber sleeve on the top of thermometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean the thermometer is defective or am I using it incorrectly, not “calibrating” it or something?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorDisplayName="user7951" LastEditDate="2015-06-15T17:28:04.193" LastActivityDate="2015-06-15T17:28:04.193" Title="Thermometer fluid is separated, defective?" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27399" PostTypeId="2" ParentId="27398" CreationDate="2015-03-16T03:41:24.003" Score="6" Body="&lt;p&gt;No, it is not defective, just slightly damaged, possibly from shipping. You should call customer service at Cole-Palmer and see if they will replace (or repair) the thermometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, there are 2 ways to fix it. I haven't tried either one, so I don't know how well they work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a bulb at the top and the thermometer is low range (under 150°C), then you would heat the thermometer in an upright position until all the bubbles are pushed up into the bulb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other method is to cool the thermometer down until all the liquid is in the bottom bulb. The second method may not work if the liquid freezes before all the liquid has moved into the bulb.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-03-16T03:41:24.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27400" PostTypeId="2" ParentId="27393" CreationDate="2015-03-16T05:58:08.327" Score="9" Body="&lt;p&gt;There are two steps in the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg + H2O -&amp;gt; MgO + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MgO + H2O &amp;lt;-&amp;gt; Mg(OH)2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium point in the second step is dependent on the temperature. See file &lt;a href=&quot;http://www.nsm.buffalo.edu/courses/gly206/StableAssemblages.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Equilibrium Assemblages&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2018-04-20T01:12:18.137" LastActivityDate="2018-04-20T01:12:18.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27402" PostTypeId="2" ParentId="27384" CreationDate="2015-03-16T07:18:26.223" Score="6" Body="&lt;p&gt;There's actually two questions being asked here: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is holding ink on paper?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do some inks take longer to dry?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ink sticks to paper because it has a high affinity for the medium it's being applied to. To group things into simple categories we could say, there are hydrophobic inks (non-polar inks that don't like water) and hydrophilic inks (polar inks that do like water). As with the inks, surfaces can also fall into this category. When ink is thrown onto a surface that it doesn't like, it aggregates, clumping together to avoid all interaction with the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple illustration showing why this is unfavorable can be seen with water interacting with lipids. In the picture below, water is forming ordered cages around the hydrophobic part. These cages cause a decrease in entropy, and by the 2nd law of thermodynamics, this is unfavorable. You can think of the ink as the water, and the hydrocarbon chain of the lipid as the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jWN4r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When inks do favor their substrate (surface) they will attach via h-bonding, london dispersion forces, and/or other inter-molecular forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second question, the answer relies more on what the ink is made out of. According to the wiki on inks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ink&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ink&lt;/a&gt; can be a complex medium, composed of solvents, pigments, dyes, resins, lubricants, solubilizers, surfactants, particulate matter, fluoresces, and other materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally, the ink, which has become mobile in its solvent, will remain mobile until the solvent evaporates. So you'll find that the dry time is dependent upon the rate of evaporation. That's why the dry times can be longer for some oil-based pens. &lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-03-16T17:21:59.363" LastActivityDate="2015-03-16T17:21:59.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27403" PostTypeId="1" CreationDate="2015-03-16T07:18:31.153" Score="3" ViewCount="311" Body="&lt;p&gt;The question is as in the title. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for a one-component hard-sphere gas, there is an analytic solution in the $\vec{k}$-space (PRA, 5, 939). I know that to obtain the correct solution in $r$-space I just need to do something like a Fast Fourier Transform. The problem is that in doing so I can only obtain values for the radial function on a grid, and problem arises as to the magnitude of errors on the grid values, the errors in interpolating between the grid values, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to apologize for the fact that I rarely do fast fourier transforms, and even less estimate their errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard way (e.g fitting formula) to do this? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorUserId="7451" LastEditDate="2015-03-16T07:24:47.810" LastActivityDate="2015-03-16T07:24:47.810" Title="Radial Distribution Function for a Hard Sphere Gas in the Percus-Yevick Approximation - Is There a Quick Way to Evaluate it Numerically?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27405" PostTypeId="1" AcceptedAnswerId="27410" CreationDate="2015-03-16T07:59:48.010" Score="2" ViewCount="9314" Body="&lt;p&gt;Given the following data, how would you work out the average bond enthalpy for $\ce{C-F}$ bond. I've tried setting up the chemical equations and applying Hess's Law, but that's not getting me anywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H_\mathrm f^\circ(\ce{CF4(g)})=-680~\mathrm{kJ~mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bond enthalpy, $\ce{F2(g)}=+158~\mathrm{kJ~mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C(s) -&amp;gt; C(g)}\quad \Delta H=+715~\mathrm{kJ~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: These are the equations I used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{C(s) + 2F2(g) &amp;amp;-&amp;gt; CF4(g)}\\[6pt]&#xA;\ce{F2(g) &amp;amp;-&amp;gt; 2F-(g)}\\[6pt]&#xA;\ce{C(s) &amp;amp;-&amp;gt; C(g)}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="14994" LastEditorDisplayName="user7951" LastEditDate="2016-11-20T18:52:43.857" LastActivityDate="2017-01-19T11:54:03.320" Title="Determining the Average Bond Enthalpy for the C-F bond" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27407" PostTypeId="1" AcceptedAnswerId="27408" CreationDate="2015-03-16T08:33:45.783" Score="24" ViewCount="18429" Body="&lt;p&gt;Normal metals like sodium or Calcium have a positive charge as $\ce{Na}^+$ or $\ce{Ca}^{2+}$.&#xA;Transition metals have a loot of variable oxidation states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yesterday I read about Iridium in Wikipedia and it stated that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a &lt;strong&gt;+9&lt;/strong&gt; oxidation state has been found in $\ce{IrO_4^+}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is astounding for me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i was thinking can metals have a net negative charged ion also?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Google :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H2XlW.jpg&quot; alt=&quot;Metals USUALLY have positive charges&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is written that metals &lt;em&gt;USUALLY&lt;/em&gt; have positive charges which means there may be conditions or compounds when they have negative charges too. Is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also can Noble gasses form ions too?&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="7383" LastEditDate="2015-03-16T14:28:08.390" LastActivityDate="2020-07-30T00:40:21.023" Title="Can metals have a net negative charge" Tags="&lt;ions&gt;&lt;metal&gt;&lt;transition-metals&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27408" PostTypeId="2" ParentId="27407" CreationDate="2015-03-16T09:11:05.167" Score="25" Body="&lt;p&gt;These species usually do not exist in nature, but they can be synthesized.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Silver has been reduced in liquid ammonia to give $\ce{Ag-}$.&lt;/li&gt;&#xA;&lt;li&gt;A lot of anionic metal carbonyl complexes $\ce{M(CO)_{n}^{m-}}$ have been synthesized: &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;-1&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[V(CO)6]-}$, $\ce{[Nb(CO)6]-}$, $\ce{[Ta(CO)6]-}$, $\ce{[Mn(CO)5]-}$, $\ce{[Ir(CO)4]-}$, $\ce{[Co(CO)4]-}$, $\ce{[Rh(CO)4]-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;-2&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[Ti(CO)6]^{2-}}$, $\ce{[Zr(CO)6]^{2-}}$, $\ce{[Hf(CO)6]^{2-}}$, $\ce{[Fe(CO)4]^{2-}}$, $\ce{[Ru(CO)4]^{2-}}$, $\ce{[Os(CO)4]^{2-}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;-3&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[V(CO)5]^{3-}}$, $\ce{[Nb(CO)5]^{3-}}$, $\ce{[Ta(CO)5]^{3-}}$, $\ce{[Mn(CO)4]^{3-}}$, $\ce{[Re(CO)4]^{3-}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;-4&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[Cr(CO)4]^{4-}}$, $\ce{[Mo(CO)4]^{4-}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There's a review by &lt;em&gt;John E. Ellis&lt;/em&gt;: &lt;a href=&quot;http://dx.doi.org/10.1016/S0065-3055(08)60508-0&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Highly Reduced Metal Carbonyl Anions: Synthesis, Characteri's zation, and Chemical Properties&lt;/em&gt;&lt;/a&gt; on these compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Christian Aichinger's comment concerning $\pi$-backbonding in metal carbonyls is correct, it is important to realize that $\ce{CO}$ is a neutral ligand! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With other words: The &lt;em&gt;Mond&lt;/em&gt; process: $\ce{Ni(s) + 4CO -&amp;gt; Ni(CO)4}$, using nickel powder and carbon monoxide, is not a redox reaction and the oxidation state of the metal still is $0$ in the metal carbonyl. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="9961" LastEditDate="2015-03-19T01:53:11.070" LastActivityDate="2015-03-19T01:53:11.070" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27410" PostTypeId="2" ParentId="27405" CreationDate="2015-03-16T11:37:26.137" Score="3" Body="&lt;p&gt;Your approach to use Hess's Law is reasonable!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the enthalpy for $\ce{CF4}$ - a molecule with &lt;strong&gt;four&lt;/strong&gt; $\ce{C-F}$ bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average $\ce{C-F}$ bond enthalpy is smaller:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx  427\ \mathrm{kJ\cdot mol^{-1}}\]&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-16T11:37:26.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27411" PostTypeId="1" CreationDate="2015-03-16T11:58:55.033" Score="4" ViewCount="4167" Body="&lt;p&gt;Elements of group 6A, compared to 5A, require less ionization energy due to the paired electrons of 6A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a question arises: why does group 7A, compared to group 6A, require more ionization energy in spite of the paired electrons in group 7a too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the general trend that the ionization energy increases as we go from left to right in periods, why didn't we apply the same procedure as the first comparison?&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="10185" LastEditDate="2015-07-08T19:48:06.700" LastActivityDate="2015-07-08T19:48:06.700" Title="How adding one more electron increases the ionization energy?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27412" PostTypeId="2" ParentId="27411" CreationDate="2015-03-16T13:02:06.930" Score="5" Body="&lt;p&gt;&lt;strong&gt;It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, as you go across a row, the &lt;a href=&quot;https://chemistry.stackexchange.com/q/22406/5017&quot;&gt;shielding effect&lt;/a&gt; is fairly small &lt;a href=&quot;https://chemistry.stackexchange.com/q/4090/5017&quot;&gt;compared to the effect of increasing nuclear charge&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-03-16T13:02:06.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27413" PostTypeId="1" CreationDate="2015-03-16T13:03:10.397" Score="5" ViewCount="131" Body="&lt;p&gt;I'm trying to find a substitute of Glucono delta-lactone (GDL). This food additive is usually used to produce silk tofu. Because of its slow hydrolysis the pH decreases very smoothly, which results in a tofu(coagulated soy milk protein) very soft. At the same time it is neutral in flavour, which is a great feature for a food additive. Now, my problem is that I'm trying to find an alternative because GDL is not allowed in organic food (in EU). Does anyone know any possible alternative compatible with the allowed additives (Annex III in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2008:250:0001:0084:EN:PDF&quot; rel=&quot;nofollow&quot;&gt;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2008:250:0001:0084:EN:PDF&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, how can I know the reaction rates of these acidifiers? For example, does Lactic acid hydrolyses slowly? &#xA;Note: When I refer to fast or slow, my time reference is a minute (fast) or hours (slow).&lt;/p&gt;&#xA;" OwnerUserId="15000" LastEditorUserId="275" LastEditDate="2015-03-16T15:45:16.923" LastActivityDate="2015-03-16T15:45:16.923" Title="Slow acidification process" Tags="&lt;ph&gt;&lt;food-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27414" PostTypeId="2" ParentId="27411" CreationDate="2015-03-16T13:05:15.893" Score="-1" Body="&lt;p&gt;Elements of group 6A (also called group 16) require less ionization energy than group 5A (group 15) elements because group 5A elements have a half filled p-orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group 7A (group 17) &lt;strong&gt;DOES NOT&lt;/strong&gt; have paired electrons. Ionization energy of group 7A elements is more than that of group 6A elements due to higher effective nuclear charge on group 7A elements.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-03-16T13:05:15.893" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27415" PostTypeId="2" ParentId="27302" CreationDate="2015-03-16T13:42:14.270" Score="6" Body="&lt;p&gt;&lt;strong&gt;Short answer:&lt;/strong&gt; No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer:&lt;/strong&gt; There are too many different ways to categorize the myriad conceivable chemical species, and too many properties of potential interest, for such a 'lookup database' of functionals to be feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerous theoretical chemists are actively engaged in benchmarking of functionals for their performance in predicting particular properties of particular sets of related compounds; searching the literature for benchmark studies pertinent to the system/property you wish to examine is likely the best course of action.  &lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-03-16T13:42:14.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27416" PostTypeId="2" ParentId="27302" CreationDate="2015-03-16T13:55:15.027" Score="11" Body="&lt;p&gt;In addition to the good answers that have already been given, it is important to note that for DFT to be properly implemented, &lt;strong&gt;you must calibrate the functional(s)&lt;/strong&gt; to whatever it is you are examining.  While this is idealistic, it is far from common practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calibrating is very tedious and time intensive than just running with a single functional.  For instance, when examining systems with large dispersion components to say, an intermolecular interaction, a slew of  13 DFT functionals (and corresponding variants) were tested to find qualitative agreement against a variety of computed properties (nature of the stationary point, binding energy, geometry, etc.) and compared against the MP2 wavefunction method (see: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp502588h&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jp502588h&lt;/a&gt;).  Some functionals were clearly underperforming while quite a few functionals were in very good (qualitative) agreement amongst each other as well as MP2.  This is one approach you can use to validate your choice of functional.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to be careful about promoting one functional over another based on this type of analysis alone if experimental results are not available (or some previous calibration study) as was the case in this paper.  Experimental values were available for formaldehyde dimer but not for any of the sulfur containing systems.  The computational methods were validated by producing numbers that agree with experiment and extended to the sulfur containing systems.  But this does not mean that you will get the &lt;em&gt;same&lt;/em&gt; performance on these novel systems.  In this case, more rigorous examinations are required and the authors of this paper state that future work include implementing the CCSD(T) level of theory to verify not only the MP2 results but also the functionals used.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2015-03-16T13:55:15.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27417" PostTypeId="1" AcceptedAnswerId="27419" CreationDate="2015-03-16T14:07:39.263" Score="4" ViewCount="859" Body="&lt;p&gt;I am trying to determine the concentration of products from the kinetic reactions of wood pyrolysis. The kinetic scheme separates the wood into cellulose, hemicellulose, and lignin. For example, the kinetic reactions for just the cellulose are shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B8Zgz.png&quot; alt=&quot;kinetic reactions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $G1$ and $G2$ terms represent a group of chemical species, $LVG$ represents levoglucosan, and the $K$ terms represent the kinetic rate constants for each reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using Python (see code below) to plot the concentration profiles and mass balance of the system. As shown in the plot below, the total mass remains constant (as it should). The plot represents each component as a percent of the original wood concentration which is given as $kg/m^3$. My next step is to determine the concentration ($kg/m^3$) of each chemical species. From the plot, I have the amount of Group 1 ($G1$) produced but I would like to determine the concentrations for $H_2O$ and $Char$. For example, $5\, H_2O + 6\, Char = G1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the concentration (as $kg/m^3$) of each individual species in Groups $G1$ and $G2$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFiqp.png&quot; alt=&quot;cellulose reactions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python function for the cellulose reactions (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the activation energy units are in $\frac{kcal}{kmol}$ but the concentration of wood provided to the system is a mass basis represented as the density of the wood $\frac{kg}{m^3}$. However, when using the appropriate units for $R$ the rate constant $K$ is in units of $\frac{1}{s}$ where s is seconds. The amount of cellulose provided to the reactions is assumed to be 50% of the original wood as $0.5 * 700 \frac{kg}{m^3}$. The concentrations returned from the cellulose function &lt;code&gt;cell&lt;/code&gt; are in terms of $kg/m^3$. The plot (see above) displays the percent of the concentration relative to the original concentration of wood. For example, cellulose in the plot is calculated as $\frac{\rho_{cell}}{\rho_{wood}}*100$ where $\rho$ is the concentration in terms of $kg/m^3$.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre class=&quot;lang-python prettyprint-override&quot;&gt;&lt;code&gt;def cell(T, pw, dt, p):&#xA;    &quot;&quot;&quot;&#xA;    T = temperature, K&#xA;    pw = vector of initial wood concentration, kg/m^3&#xA;    dt = time step, s&#xA;    p = total number of time steps&#xA;    &quot;&quot;&quot;&#xA;&#xA;    # array to store species concentrations as a density, kg/m^3&#xA;    # row = chemical species&#xA;    # column = concentration at time step&#xA;    spec = np.zeros([16, p])&#xA;&#xA;    # initial cellulose concentration in wood, kg/m^3&#xA;    spec[0] = pw*0.5&#xA;&#xA;    R = 1.987   # universal gas constant, kcal/kmol*K&#xA;&#xA;    # A = pre-factor (1/s) and E = activation energy (kcal/kmol)&#xA;    A1 = 4e13;  E1 = 45000  # CELL -&amp;gt; CELLA&#xA;    A2 = 0.5e9; E2 = 29000  # CELLA -&amp;gt; products&#xA;    A3 = 1.8;   E3 = 10000  # CELLA -&amp;gt; LVG&#xA;    A4 = 4e7;   E4 = 31000  # CELL -&amp;gt; 5*H2O + 6*Char&#xA;&#xA;    # reaction rate constant for each reaction, 1/s&#xA;    K1 = A1 * np.exp(-E1 / (R * T))         # CELL -&amp;gt; CELLA&#xA;    K2 = A2 * np.exp(-E2 / (R * T))         # CELLA -&amp;gt; G2&#xA;    K3 = A3 * T * np.exp(-E3 / (R * T))     # CELLA -&amp;gt; LVG&#xA;    K4 = A4 * np.exp(-E4 / (R * T))         # CELL -&amp;gt; G1&#xA;&#xA;    # concentrations for each chemical species, kg/m^3&#xA;    for i in range(1, p):&#xA;        r1 = K1 * spec[0, i-1]&#xA;        r2 = K2 * spec[1, i-1]&#xA;        r3 = K3 * spec[1, i-1]&#xA;        r4 = K4 * spec[0, i-1]&#xA;        spec[0, i] = spec[0, i-1] - (r1+r4)*dt          # CELL&#xA;        spec[1, i] = spec[1, i-1] + r1*dt - (r2+r3)*dt  # CELLA&#xA;        spec[2, i] = spec[2, i-1] + r4*dt               # G1&#xA;        spec[3, i] = spec[3, i-1] + r2*dt               # G2&#xA;        spec[4, i] = spec[4, i-1] + r3*dt               # LVG&#xA;&#xA;    # return species array concentrations, kg/m^3&#xA;    return spec&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Python script to plot the reactions from the above cellulose function:&lt;/p&gt;&#xA;&#xA;&lt;pre class=&quot;lang-python prettyprint-override&quot;&gt;&lt;code&gt;# Modules&#xA;#------------------------------------------------------------------------------&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as py&#xA;import funcRanzi as kn&#xA;&#xA;py.close('all')&#xA;&#xA;# Parameters from Papadikis 2010a&#xA;#------------------------------------------------------------------------------&#xA;&#xA;rhow = 700  # density of wood, kg/m^3&#xA;Tinf = 773  # ambient temp, K&#xA;&#xA;# Initial Calculations&#xA;#------------------------------------------------------------------------------&#xA;&#xA;dt = 0.01   # time step, delta t&#xA;tmax = 4   # max time, s&#xA;t = np.linspace(0, tmax, num=tmax/dt)   # time vector&#xA;p = len(t)  # total number of time steps&#xA;&#xA;# Calculate Concentrations of Chemical Species&#xA;#------------------------------------------------------------------------------&#xA;&#xA;# vectors for wood, gas, tar, char concentrations as a density, kg/m^3&#xA;pw = np.zeros(len(t))   # wood &#xA;pg = np.zeros(len(t))   # gas&#xA;pt = np.zeros(len(t))   # tar&#xA;pc = np.zeros(len(t))   # char&#xA;&#xA;pw[:] = rhow    # initial wood concentration as density&#xA;&#xA;# array of chemical species concentrations as a density, kg/m^3&#xA;spec = kn.cell(Tinf, pw, dt, p)&#xA;&#xA;# concentration as percent relative to original wood, %&#xA;cell = spec[0]/rhow*100     # CELL&#xA;cella = spec[1]/rhow*100    # CELLA&#xA;g1 = spec[2]/rhow*100       # G1&#xA;g2 = spec[3]/rhow*100       # G2&#xA;lvg = spec[4]/rhow*100      # LVG&#xA;&#xA;total = cell + cella + g1 + g2 + lvg&#xA;&#xA;# Plot Results&#xA;#------------------------------------------------------------------------------&#xA;&#xA;py.rcParams['xtick.major.pad'] = 8&#xA;py.rcParams['ytick.major.pad'] = 8&#xA;py.rcParams['lines.linewidth'] = 2&#xA;py.rcParams['axes.grid'] = True&#xA;&#xA;py.figure(1)&#xA;py.plot(t, cell, label='cell')&#xA;py.plot(t, cella, label='cella')&#xA;py.plot(t, g1, label='g1')&#xA;py.plot(t, g2, label='g2')&#xA;py.plot(t, lvg, label='lvg')&#xA;py.plot(t, total, label='total')&#xA;py.title('Cellulose Reactions at T = {} K'.format(Tinf))&#xA;py.xlabel('Time (s)')&#xA;py.ylabel('Conversion (% dry basis)')&#xA;py.legend(loc='best', numpoints=1)&#xA;py.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="16683" LastEditDate="2017-08-07T15:58:11.197" LastActivityDate="2017-08-07T15:58:11.197" Title="Calculate species concentration from first-order kinetic reactions" Tags="&lt;kinetics&gt;&lt;concentration&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27418" PostTypeId="2" ParentId="27302" CreationDate="2015-03-16T14:17:23.540" Score="39" Body="&lt;h1&gt;What's up with all that magic? (A chapter formerly known as Introduction)&lt;/h1&gt;&#xA;&lt;p&gt;The hunt for for the holy grail of density functional theory (DFT) has come a long way.&lt;sup&gt;[1]&lt;/sup&gt; Becke states in the introduction of the cited paper:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the &lt;em&gt;total electron density&lt;/em&gt; alone, is so compelling it can drive one mad.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I really like this description, it points out why we use and need DFT, and as it also points out the flaws, that every computational chemist has to deal with: How can something with such a simple approach be correct?&lt;br /&gt;&#xA;Something that is often forgotten about DFT is, that &lt;em&gt;in principle&lt;/em&gt; it is correct. It's the implementations and approximations, that make it incorrect, but usable. Becke states this in the following quote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let us introduce the acronym DFA at this point for “&lt;em&gt;density-functional approximation&lt;/em&gt;.” If you attend DFT meetings, you will know that Mel Levy often needs to remind us that DFT is &lt;em&gt;exact&lt;/em&gt;. The failures we report at meetings and in papers are not failures of DFT, but failures of DFAs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I sometimes here that the abbreviation DFT is often used in the wrong context, since we are not talking about the theory itself any more, but about the implementations and approximations of it. One suggestion I heard was, that it should rather be used as &lt;em&gt;density functional technique&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;With that in mind I would like to state, that I absolutely agree with &lt;a href=&quot;https://chemistry.stackexchange.com/a/27325/4945&quot;&gt;the previous answer&lt;/a&gt; by user1420303 and it's subsequent comment by Geoff Hutchison. Since you asked for a somewhat more practical approach, I like to offer the advice I usually give to students new in the field.&lt;/p&gt;&#xA;&lt;h1&gt;Old is bad, isn't it?&lt;/h1&gt;&#xA;&lt;p&gt;Some of the functionals are now around for about thirty years. That does not make them bad, maybe even the opposite. It shows, that they are still applicable today, giving reasonable results. One of my personal favourites is the conjunction of Becke 1988 and Perdew 1986, often abbreviated as BP86.&lt;sup&gt;[2]&lt;/sup&gt; It's a pure functional which is available most modern quantum chemical packages.&lt;sup&gt;[3]&lt;/sup&gt; It performs usually well enough for geometries and reasonable well for energies for simple systems, i.e. small organic molecules and reactions.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;em&gt;magical&lt;/em&gt; functional B3LYP was one of the first hybrid functionals, and it was introduced by Gaussian's very own developers.&lt;sup&gt;[4]&lt;/sup&gt; A lot of people were surprised how well it worked and it quickly became one of the most popular functionals of all time. It combines Becke's three parameter functional B3&lt;sup&gt;[5]&lt;/sup&gt; with Lee, Yang and Parr's correlation functional.&lt;sup&gt;[6]&lt;/sup&gt; But why are we surprised it works? The answer is quite simple, it was not fitted to anything. Frish &lt;em&gt;et. al.&lt;/em&gt; just reworked the B3PW91 functional to use LYP instead of PW91. As a result, it heavily suffers or benefits from error compensation. Some even go as far as to say: “It is right for the wrong reasons.”&lt;sup&gt;[7-9]&lt;/sup&gt; Is it a bad choice? No. It might not be the best choice, but as long as you know what you are doing and you know it is not failing for your system, it is a reasonable choice.&lt;/p&gt;&#xA;&lt;h1&gt;One functional is enough, is it?&lt;/h1&gt;&#xA;&lt;p&gt;Now that we established, that old functionals are not out of fashion, we should establish something very, very important: &lt;em&gt;One is never enough.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are a few things, where it is appropriate to do most of the work with one functional, but in these cases the observations have to be validated with other methods. Often it is best to work your way up Jacob's ladder.&lt;sup&gt;[10]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;h1&gt;How do I start?&lt;/h1&gt;&#xA;&lt;p&gt;It really depends on your system and what you are looking for. You are trying to elucidate a reaction mechanism? Start with something very simple, to gain structures, many structures. Reaction mechanisms are often about the quantity of the different conformers and later about suitable initial structures for transition states. As this can get complex very fast, it's best to keep it simple. Semi-empirical methods and force fields can often shorten a long voyage. Then use something more robust for a first approach to energy barriers. I rely on BP86 for most of the heavy computing. As a modern alternative, another pure density functional, M06-L is quite a good choice, too.&lt;sup&gt;[11]&lt;/sup&gt; Some of the popular quantum chemistry suites let you use density fitting procedures, which allow you to get even more out of the computer. Just to name a few, without any particular order: &lt;a href=&quot;http://www.gaussian.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaussian&lt;/a&gt;, &lt;a href=&quot;http://www.molpro.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MolPro&lt;/a&gt;, &lt;a href=&quot;http://www.turbomole.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turbomole&lt;/a&gt;.&lt;br /&gt;&#xA;After you have developed a decent understanding of the various structures you obtained, you would probably want to take it up a notch. Now it really depends on what equipment you have at hand. How much can you afford? Ideally, more is better. At least you should check your results with a pure, a hybrid, and a meta-hybrid functional. But even that can sometimes be a stretch.&lt;sup&gt;[12]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you are doing bonding analysis, elucidation of the electronic structure, conformation analysis, or you want to know more about the spectrum, you should try to use at least five different functionals, which you later also check versus &lt;em&gt;ab initio&lt;/em&gt; approaches. Most of the times you do not have the hassle to deal with hundreds of structures, so you should focus of getting the most accurate result. As a starting point I would still use a pure functional, the worst thing that could happen is probably, that is reduces the times of subsequent optimisations. Work your way up Jacob's ladder, do what you can, take it to the max.&lt;sup&gt;[13]&lt;/sup&gt;&lt;br /&gt;&#xA;But of course, keep in mind, that some functionals were designed for a specific purpose. You can see that in the Minnesota family of functionals. The basic one is M06-L, as previously stated a pure functional, with the sole purpose of giving fast results. M06 is probably the most robust functional in this family. It was designed for a wide range of applications and is best chosen when dealing with transition metals. M06-2X is designed for main group chemistry. It comes with somewhat built in non-covalent interactions and other features. This functional (like most other though) will fail horribly, if you have multi-reference character in your system. The M06-HF functional incorporates 100% Hartree-Fock exchange and was designed to accurately calculate time dependent DFT properties and spectra. It should be a good choice for charge transfer systems. See the original publication for a more detailed description.&lt;sup&gt;[14]&lt;/sup&gt;&lt;br /&gt;&#xA;Then we have another popular functional: PBE.&lt;sup&gt;[15a]&lt;/sup&gt; In this initial publication an exchange as well as a correlation functional was proposed, both pure density functionals, often used in conjunction.&lt;sup&gt;[15b]&lt;/sup&gt; I don't know much about it's usefulness, since I prefer another quite robust variation of it: PBE0, which is a hybrid functional.&lt;sup&gt;[15c,d]&lt;/sup&gt; Because of its adiabatic connection formula, it is described by the authors as a non-empirical hybrid functional.&lt;sup&gt;[15d]&lt;/sup&gt;&lt;br /&gt;&#xA;Over the years there have been various developments, some of the are called improvement, but it often boils down to personal taste and applicability. For example, Handy and Cohen reintroduced the concept of left-right correlation into their OPTX functional and subsequently used it in combination with LYP, P86 and P91. Aparently, they work well and are now often used also as a reference for other density functionals. They went on and developed a functional analogous to B3LYP but outperforming it.&lt;sup&gt;[16]&lt;/sup&gt;&lt;br /&gt;&#xA;But these were obviously not the only attempts. Xu and Goddard III extended the B3LYP scheme to include long range effects. They claim a good description of dipole moments, polarizabilities and accurate excitation energies.&lt;sup&gt;[17]&lt;/sup&gt;&lt;br /&gt;&#xA;And with the last part in mind, it is also necessary to address long range corrections. Sometimes a system cannot be described accurately without them, sometimes they make the description worse. To name only one, CAM-B3LYP, which uses the coulomb attenuating method.&lt;sup&gt;[18]&lt;/sup&gt; And there are a couple of more, and a couple of more to come, head on over to a similar question: &lt;a href=&quot;https://chemistry.stackexchange.com/q/20163/4945&quot;&gt;What do short-range and long-range corrections mean in DFT methods?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see, there is no universal choice, it depends on your budget and on the properties you are interested in. There are a couple of theoretical/ computational chemists on this platform. I like BP86 as a quick shot and answer questions relating to MO theory with it, shameless self-promotion: &lt;a href=&quot;https://chemistry.stackexchange.com/q/10908/4945&quot;&gt;Rationalizing the Planarity of Formamide&lt;/a&gt; or &lt;a href=&quot;https://chemistry.stackexchange.com/q/14618/4945&quot;&gt;Rationalising the order of reactivity of carbonyl compounds towards nucleophiles&lt;/a&gt;. And sometimes we have overachievers like LordStryker, that use a whole bunch of methods to make a point: &lt;a href=&quot;https://chemistry.stackexchange.com/q/16342/4945&quot;&gt;Dipole moment of cis-2-butene&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h1&gt;So I picked a functional, what else?&lt;/h1&gt;&#xA;&lt;p&gt;You still have to pick a basis set. And even here you have to pick one that fits what you need. Since this answer is already way longer than I intended in the first place (Procrastination, yay!), I will keep it short(er).&lt;/p&gt;&#xA;&lt;p&gt;There are a couple of universally applicable basis sets. The most famous is probably 6-31G*. This is a nice ancient basis set that is often used for its elegance and simplicity. Explaining how it was built is easier, than for other basis sets. I personally prefer the Ahlrichs basis set def2-SVP, as it comes with a pre-defined auxiliary basis set suitable for density fitting (even in Gaussian).&lt;sup&gt;[19]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;Worth mentioning is the Dunning basis set family cc-pVDZ, cc-pVTZ, ... . They were specifically designed to be used in correlated molecular calculations. They have been reworked and improved after its initial publication, to fit them to modern computational standards.&lt;sup&gt;[20]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;The range of suitable basis sets is large, most of them are available through the &lt;a href=&quot;https://www.basissetexchange.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;basis set exchange portal&lt;/a&gt; for a variety of QC programs.&lt;/p&gt;&#xA;&lt;p&gt;Sometimes an effective core potential can be used to reduce computational cost and is worth considering.&lt;/p&gt;&#xA;&lt;h1&gt;*Sigh* What else?&lt;/h1&gt;&#xA;&lt;p&gt;When you are done with that, consider dispersion corrections. The easiest way is to pick a functional that has already implemented this, but this is quite dependent on the program of you choice (although the main ones should have this by now, it's not something brand new). However, the standalone DFT-D3 program by Stefan Grimme's group can be obtained from his website.&lt;sup&gt;[21]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Still reading? Read more! (A chapter formerly known as Notes and References)&lt;/h1&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.4869598&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axel D. Becke, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;140&lt;/em&gt;, 18A301.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;(a) &lt;a href=&quot;https://journals.aps.org/pra/abstract/10.1103/PhysRevA.38.3098&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. D. Becke, &lt;em&gt;Phys. Rev. A&lt;/em&gt;, &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;38&lt;/em&gt;, 3098-3100.&lt;/a&gt; (b) &lt;a href=&quot;http://journals.aps.org/prb/abstract/10.1103/PhysRevB.33.8822&quot; rel=&quot;nofollow noreferrer&quot;&gt;John P. Perdew, &lt;em&gt;Phys. Rev. B&lt;/em&gt;, &lt;strong&gt;1986&lt;/strong&gt;, &lt;em&gt;33&lt;/em&gt;, 8822-8824.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Unfortunately this functional is not always implemented in the same way, although the differences are pretty small. It basically boils down as to which VWN variation is used in the local spin density approximation term. Also see &lt;a href=&quot;http://dx.doi.org/10.1139/p80-159&quot; rel=&quot;nofollow noreferrer&quot;&gt;S. H. Vosko, L. Wilk, and M. Nusair, &lt;em&gt;Can. J. Phys.&lt;/em&gt;, &lt;strong&gt;1980&lt;/strong&gt;, &lt;em&gt;58&lt;/em&gt; (8), 1200-1211.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100096a001&quot; rel=&quot;nofollow noreferrer&quot;&gt;P. J. Stephens, F. J. Devlin, C. F. Chabalowski, and M. J. Frisch, &lt;em&gt;J. Phys. Chem.&lt;/em&gt;, &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;98&lt;/em&gt; (45), 11623–11627.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/98/7/10.1063/1.464913&quot; rel=&quot;nofollow noreferrer&quot;&gt;Axel D. Becke, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1993&lt;/strong&gt;, &lt;em&gt;93&lt;/em&gt;, 5648.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://journals.aps.org/prb/abstract/10.1103/PhysRevB.37.785&quot; rel=&quot;nofollow noreferrer&quot;&gt;C. Lee, W. Yang, and R. G. Parr, &lt;em&gt;Phys. Rev. B&lt;/em&gt;, &lt;strong&gt;1988&lt;/strong&gt;, &lt;em&gt;37&lt;/em&gt;, 785–789&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Unfortunately the B3LYP functional suffers from the same problems that are mentioned in [3].&lt;/li&gt;&#xA;&lt;li&gt;The failures of B3LYP are known and often well documented. Here are a few recent papers, but there are many, many more. (a) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo302156p&quot; rel=&quot;nofollow noreferrer&quot;&gt;Holger Kruse, Lars Goerigk, and Stefan Grimme, &lt;em&gt;J. Org. Chem.&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;77&lt;/em&gt; (23), 10824–10834.&lt;/a&gt; (b) &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/127/2/10.1063/1.2747249&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joachim Paier, Martijn Marsman and Georg Kresse, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;127&lt;/em&gt;, 024103.&lt;/a&gt; (c) &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2010/CC/c000677g&quot; rel=&quot;nofollow noreferrer&quot;&gt;Igor Ying Zhang, Jianming Wu and Xin Xu, &lt;em&gt;Chem. Commun.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;46&lt;/em&gt;, 3057-3070.&lt;/a&gt; (&lt;a href=&quot;https://www.researchgate.net/publication/43051014_Extending_the_reliability_and_applicability_of_B3LYP&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf via researchgate.net&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Just my two cents, that I am hiding in the footnotes: “&lt;em&gt;Pretty please&lt;/em&gt; do not make this your first choice.”&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/proceeding/aipcp/10.1063/1.1390175&quot; rel=&quot;nofollow noreferrer&quot;&gt;John P. Perdew and Karla Schmidt, &lt;em&gt;AIP Conf. Proc.&lt;/em&gt;, &lt;strong&gt;2001&lt;/strong&gt;, &lt;em&gt;577&lt;/em&gt;, 1.&lt;/a&gt; (&lt;a href=&quot;http://molphys.org/Curso_colombia/gamess/DFT/Jacob_latter_1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf via molphys.org&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/125/19/10.1063/1.2370993&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yan Zhao and Donald G. Truhlar, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;125&lt;/em&gt;, 194101.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Note, that not always full recomputations of geometries are necessary for all different functionals you apply. Often single point energies can tell you quite much how good your original model performs. Keep the computations to what you can afford.&lt;/li&gt;&#xA;&lt;li&gt;Don't use an overkill of methods, if you already have five functionals agreeing with each other and possibly with an MP2 calculation, you are pretty much done. What can the use of another five functionals tell you more?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs00214-007-0310-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Y. Zhao, N.E. Schultz, and D.G. Truhlar, &lt;em&gt;Theor. Chem. Account&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;120&lt;/em&gt;, 215–241.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;(a) &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.77.3865&quot; rel=&quot;nofollow noreferrer&quot;&gt;John P. Perdew, Kieron Burke, and Matthias Ernzerhof, &lt;em&gt;Phys. Rev. Lett.&lt;/em&gt;, &lt;strong&gt;1996&lt;/strong&gt;, &lt;em&gt;77&lt;/em&gt;, 3865.&lt;/a&gt; (b) The exchange functional was revised in &lt;a href=&quot;http://dx.doi.org/10.1063/1.476928&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matthias Ernzerhof and John P. Perdew, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1998&lt;/strong&gt;, 109, 3313.&lt;/a&gt; (c) &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/110/13/10.1063/1.478522&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carlo Adamo and Vincenzo Barone, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;110&lt;/em&gt;, 6158.&lt;/a&gt; (d) &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0009261496013735&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kieron Burke, Matthias Ernzerhof, and John P. Perdew, &lt;em&gt;Chem. Phys. Lett.&lt;/em&gt;, &lt;strong&gt;1997&lt;/strong&gt;, &lt;em&gt;265&lt;/em&gt;, 115-120.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;(a) &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/00268970010018431#.VQbGLY7LfYg&quot; rel=&quot;nofollow noreferrer&quot;&gt;N. C. Handy and A. J. Cohen, &lt;em&gt;Mol. Phys.&lt;/em&gt;, &lt;strong&gt;2001&lt;/strong&gt;, &lt;em&gt;99&lt;/em&gt;, 403-12.&lt;/a&gt; (b) &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/00268970010023435#.VQbHw47LfYg&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. J. Cohen and N. C. Handy, &lt;em&gt;Mol. Phys.&lt;/em&gt;, &lt;strong&gt;2001&lt;/strong&gt;, &lt;em&gt;99&lt;/em&gt; 607-15.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pnas.org/content/101/9/2673&quot; rel=&quot;nofollow noreferrer&quot;&gt;X. Xu and W. A. Goddard III, &lt;em&gt;Proc. Natl. Acad. Sci. USA&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;101&lt;/em&gt;, 2673-77.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0009261404008620&quot; rel=&quot;nofollow noreferrer&quot;&gt;T. Yanai, D. P. Tew, and N. C. Handy, &lt;em&gt;Chem. Phys. Lett.&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;393&lt;/em&gt;, 51-57.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;(a) &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2005/CP/b508541a&quot; rel=&quot;nofollow noreferrer&quot;&gt;Florian Weigend and Reinhart Ahlrichs, &lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;7&lt;/em&gt;, 3297-3305.&lt;/a&gt; (b) &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2006/CP/b515623h&quot; rel=&quot;nofollow noreferrer&quot;&gt;Florian Weigend, &lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 1057-1065.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The point where the use of more basis functions does not effect the calculation. For the correlation consistent basis sets, see a comment by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0009261496009177&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ernest R. Davidson, &lt;em&gt;Chem. Phys. Rev.&lt;/em&gt;, &lt;strong&gt;1996&lt;/strong&gt;, &lt;em&gt;260&lt;/em&gt;, 514-518&lt;/a&gt; and references therein. Also see &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/90/2/10.1063/1.456153&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thom H. Dunning Jr, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;1989&lt;/strong&gt;, &lt;em&gt;90&lt;/em&gt;, 1007&lt;/a&gt; as the original source.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.thch.uni-bonn.de/tc/index.php?section=downloads&amp;amp;subsection=DFT-D3&amp;amp;lang=english&quot; rel=&quot;nofollow noreferrer&quot;&gt;DFT-D3 Website&lt;/a&gt;; &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/132/15/10.1063/1.3382344&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stefan Grimme, Jens Antony, Stephan Ehrlich, and Helge Krieg,&#xA;&lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;132&lt;/em&gt;, 154104.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Have fun and good luck!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="41556" LastEditDate="2020-12-22T02:58:41.870" LastActivityDate="2020-12-22T02:58:41.870" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27419" PostTypeId="2" ParentId="27417" CreationDate="2015-03-16T14:18:26.563" Score="3" Body="&lt;p&gt;According to the diagram you wrote, &lt;code&gt;g1&lt;/code&gt; and &lt;code&gt;g2&lt;/code&gt; are just stoichiometric groupings of certain components.  Because there is no modeled back-reaction from those components to the rest of the system, you can find the component amounts by simple stoichiometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asssuming the &quot;stoichiometric&quot; coefficients in the groupings are meant as &lt;em&gt;mass&lt;/em&gt; coefficients, then the calculations would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;py.plot(t, 6*g1/(6+5)+0.61g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='char')&#xA;py.plot(t, 5*g1/(6+5)+0.83g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='water')&#xA;py.plot(t, 0.88g2/(sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='HAA')&#xA;py.plot(t, 0.2g2/sum([0.8, 0.2, 0.1, 0.25, 0.3, 0.21,...etc...]), label='glyox')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-16T16:11:02.477" LastActivityDate="2015-03-16T16:11:02.477" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="27420" PostTypeId="1" CreationDate="2015-03-16T14:40:42.573" Score="7" ViewCount="10509" Body="&lt;p&gt;Many first electron affinities are positive, indicating a favourable process, but the corresponding second electron affinities are negative. For example, the first and second electron affinities of oxygen are $+141$ and $-780~\mathrm{kJ~mol^{-1}}$ respectively. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="14840" LastEditorUserId="16683" LastEditDate="2016-05-07T17:43:35.997" LastActivityDate="2021-01-20T07:41:32.217" Title="Why does the second electron affinity has an opposite sign of the first one?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27421" PostTypeId="1" CreationDate="2015-03-16T15:33:59.043" Score="7" ViewCount="6354" Body="&lt;p&gt;Where can you find the enthalpy and Gibbs free energy of formation and entropy of compounds either as a data file or web API? Technically, I only need two since I can figure out the other one from two of them.&lt;/p&gt;&#xA;" OwnerUserId="7219" LastEditorUserId="41328" LastEditDate="2021-04-12T10:32:42.980" LastActivityDate="2021-04-12T10:32:42.980" Title="Where to find data for Gibbs energy, enthalpy, and entropy?" Tags="&lt;physical-chemistry&gt;&lt;enthalpy&gt;&lt;entropy&gt;&lt;free-energy&gt;&lt;databases&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27422" PostTypeId="2" ParentId="27420" CreationDate="2015-03-16T15:53:17.097" Score="7" Body="&lt;p&gt;Many non-metals like those in Group 16 and 17 have first electron affinities that are negative.  In other words, the following reaction is &lt;strong&gt;exothermic&lt;/strong&gt; and energy is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{X + e^{−} -&amp;gt; X^{−} + ~energy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By gaining an electron the group 17 atoms attain a complete octet of electrons and become stabilized.  By gaining an electron the Group 16 atoms are just one electron away from having a filled shell.  Further in both the Group 16 and 17 cases, the protons in the nucleus are not fully screened by the outer electrons, so adding one more electron to form an anion is electrostatically favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we try to add a second electron the reaction becomes &lt;strong&gt;endothermic&lt;/strong&gt;; adding a second electron is unfavorable and energy must be added (endothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{X^{-} + e^{−} + ~energy~ -&amp;gt; X^{2−}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to add a second electron primarily because of the increased electron-electron repulsion that occurs with the second electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since adding the first electron is exothermic, while adding the second electron is endothermic, the sign reverses when we measure the energy added or released by these two processes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-16T15:53:17.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27423" PostTypeId="2" ParentId="27421" CreationDate="2015-03-16T16:12:19.960" Score="9" Body="&lt;p&gt;NIST webbook does have a lot of data, though they are not in any kind of an API form as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webbook.nist.gov/chemistry/&quot;&gt;http://webbook.nist.gov/chemistry/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-03-16T16:12:19.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27424" PostTypeId="2" ParentId="27421" CreationDate="2015-03-16T17:21:21.870" Score="8" Body="&lt;p&gt;NIST is the best place to turn for lots of data.  However, more easily parsed, smaller datasets are available in a couple of other locations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;http://www.chnosz.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;CHNOSz&lt;/code&gt; package&lt;/a&gt; in &lt;code&gt;R&lt;/code&gt; has thermochemical data for a variety of species, mostly inorganic.  Their database is referenced back to the chemical literature.  See &lt;a href=&quot;https://chemistry.stackexchange.com/a/27240/13184&quot;&gt;an answer I gave to an old question&lt;/a&gt; for an example of how to use it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Robert Alberty's &lt;a href=&quot;http://library.wolfram.com/infocenter/MathSource/797/BasicBiochemData2.m?file_id=4428&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathematica package&lt;/a&gt; for biochemical thermodynamics has a lot of built in thermochemical data for common biochemical compounds -- relative to a biochemical standard state of constant pH.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Water-specific databases.  In case you're only interested in water, ice, and steam, there are &lt;a href=&quot;https://chemistry.stackexchange.com/a/24985/13184&quot;&gt;Excel plugins&lt;/a&gt; that let you get thermochemical data by calling functions inside Excel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There appears to be a &lt;a href=&quot;https://pythonhosted.org/thermopy/burcat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Python library called thermopy&lt;/a&gt; which contains a &lt;code&gt;burcat()&lt;/code&gt; module in which some thermochemical property data lives.  I haven't tried it out though.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The database in that python module seems to be &lt;a href=&quot;http://burcat.technion.ac.il/dir/READ.ME.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For biochemical reaction systems, there is also &lt;a href=&quot;http://equilibrator.weizmann.ac.il/&quot; rel=&quot;nofollow noreferrer&quot;&gt;eQuilibrator&lt;/a&gt; which has a web interface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-03-16T17:21:21.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27425" PostTypeId="1" AcceptedAnswerId="27430" CreationDate="2015-03-16T17:43:41.537" Score="3" ViewCount="3221" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the ratio of copper(II) ions to EDTA in a &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cu(EDTA)]^{2-}}$&lt;/span&gt; complex? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that EDTA forms up to 6 coordinate bonds with the central atom, but is that what they are asking for with &quot;ratio&quot;?&#xA;According to &lt;a href=&quot;http://www.chemthes.com/entity_datapage.php?id=3974&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;, it seems that copper actually bonds with all 6 free electron pairs.&lt;/p&gt;&#xA;" OwnerUserId="15002" LastEditorUserId="83758" LastEditDate="2020-04-02T17:50:48.333" LastActivityDate="2020-04-02T17:50:48.333" Title="Ratio of copper(II) ions to EDTA?" Tags="&lt;coordination-compounds&gt;&lt;chemical-formula&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="27426" PostTypeId="1" CreationDate="2015-03-16T17:44:27.170" Score="4" ViewCount="3814" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;From the reaction of &lt;span class=&quot;math-container&quot;&gt;$\ce{NiBr2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph2EtP}$&lt;/span&gt;, it is possible to isolate green and red crystals of &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ni(Ph2EtP)2Br2]}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;A) coordination number and geometries of the complexes?&lt;br /&gt;&#xA;B) possible isomeric structures for the green and red crystals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I know that d orbitals effect the color but I don't know how to apply it to isomers.&lt;/p&gt;&#xA;" OwnerUserId="14967" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-07-15T02:54:13.457" Title="Green and red crystals of [Ni(Ph2EtP)2Br2]" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27427" PostTypeId="1" AcceptedAnswerId="27429" CreationDate="2015-03-16T18:03:09.113" Score="2" ViewCount="614" Body="&lt;p&gt;In my next lab, I'll be doing a complexometric titration of a copper solution with EDTA and a murexide indicator. The lab guide advises to keep the pH level nearly neutral by adding a diluted $\ce{NH3}$ solution. Why is that necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that during the formation of the copper complex $\ce{[Cu(EDTA)]^{2-}}$, the solution becomes more acidic. Do we need to add the basic solution in order to make sure the murexide does not change color due to the decreasing pH level? Or is it simply so that we can revert the color change (of the formation of the copper complex) and continue the titration until the equivalence point is reached (and all copper ions are in complex form)?&lt;/p&gt;&#xA;" OwnerUserId="15002" LastEditorUserId="7448" LastEditDate="2015-03-16T18:20:39.560" LastActivityDate="2015-03-16T22:15:03.543" Title="Why should a complexometric titration be kept neutral?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27428" PostTypeId="2" ParentId="4526" CreationDate="2015-03-16T18:35:00.287" Score="3" Body="&lt;p&gt;I don't think you can use the approach you are outlining without an equation of state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $P$ and $V$ are &lt;em&gt;both&lt;/em&gt; constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$.  Actually, that assumption is probably not horrible for most liquids at &quot;normal&quot; pressures: I think it amounts to assuming the liquid is incompressible.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-16T18:35:00.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27429" PostTypeId="2" ParentId="27427" CreationDate="2015-03-16T19:10:20.123" Score="0" Body="&lt;p&gt;Your last guess is exact. This is a classical method, the equivalent point is seen by our eyes, so in our interest is for the complex to be formed so we can recognize the colour, and we should add $\ce{NH3}$ so the colour won't change immediately after the equivalent point.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="7448" LastEditDate="2015-03-16T22:15:03.543" LastActivityDate="2015-03-16T22:15:03.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27430" PostTypeId="2" ParentId="27425" CreationDate="2015-03-16T19:13:24.937" Score="2" Body="&lt;p&gt;The ratio of Cu to EDTA is the number of copper ions per EDTA ion. In this case there is one EDTA ion for every copper ion so the ratio is 1:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper bonds with six electron pairs on EDTA to form the complex shown below where the central metal species is copper(II):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tx7HQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-16T19:13:24.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27431" PostTypeId="2" ParentId="27426" CreationDate="2015-03-16T19:36:35.693" Score="6" Body="&lt;p&gt;The coordination number of $\ce{[Ni(Ph2EtP)2Br2]}$ is four;&lt;br&gt;&#xA;thus, the complex may be tetrahedral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7JOGx.png&quot; alt=&quot;tetrahedral&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or square planar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mkc9g.png&quot; alt=&quot;square planar&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(figures taken from &lt;a href=&quot;https://chemistry.stackexchange.com/a/4651/7951&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains $\ce{Ni^2+}$, which has eight d electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When filled with eight electrons, the tetrahedral structure is paramagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively small (4.4 Dq).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding square-planar structure is diamagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively large (10 Dq).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the square planar complex absorbs light of higher energy (lower wavelength) than the tetrahedral complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green crystals absorb red light, whereas the red crystals absorb green light. Thus, the red crystals absorb light of higher energy than the green crystals. Therefore, the red crystals correspond to the square-planar complex and the green crystals correspond to the tetrahedral complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, since the complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains two different ligands, there are &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; isomers of the square-planar structure.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-03-16T21:05:15.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27432" PostTypeId="2" ParentId="27426" CreationDate="2015-03-16T19:46:26.013" Score="4" Body="&lt;p&gt;It's a well known experiment (&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1970,&lt;/strong&gt; &lt;em&gt;92,&lt;/em&gt; 2691). Of course, it's possible to isolate green and red crystals $\ce{[Ni(Ph2EtP)2Br2]}$. The coordination number for both complexes is four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry of the complex with  red crystals is square planar. It's diamagnetic. Hybridization of $\ce{Ni^2+}$ is $\ce{dsp^2}$ The geometry of the complex with green crystals is tetrahedral. It's paramagnetic.&#xA;Hybridization of $\ce{Ni^2+}$ is $\ce{sp^3}$&#xA;Electron configuration of $\ce{Ni^2+}$: $\ce{[Ar] 4s^0 3d^8}$.&#xA;We can explain the formation of the red crystals as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/beuh2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we can explain the formation of the green crystals as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lHnQq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for possible isomers, there are two isomers (cis and trans) for the square planar complex since it has two kinds of ligands.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-07-15T02:54:13.457" LastActivityDate="2018-07-15T02:54:13.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27436" PostTypeId="1" AcceptedAnswerId="27438" CreationDate="2015-03-16T21:33:54.163" Score="11" ViewCount="2414" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2nx1N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling to find a mechanism for the above reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have suggested that the proton adds to one of the reacting aromatic species leaving the resonance stabilized secondary cation. This then adds to the ortho position on the second reactant (since alkyl groups are activating). I then get a positively charged intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simpler reaction I would simply remove the proton but I wonder whether the reaction from this step to the product is concerted? I couldn't see an obvious step so I deprotonated and again could find no obvious way to continue apart from a rather elaborate proton transfer from a carbon that does not possess the most acidic proton.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4325" LastEditDate="2015-03-16T22:50:18.927" LastActivityDate="2018-01-11T02:33:21.793" Title="Mechanism of styrene dimerization" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27437" PostTypeId="1" AcceptedAnswerId="27441" CreationDate="2015-03-16T21:47:51.367" Score="19" ViewCount="15606" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6iMNA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above reaction looks very similar to a conventional sulfonation reaction but if the reaction pathway is analogous to sulfonation I do not see how the $\ce{SO_2Cl^+}$ would form. Surely $\ce{Cl^-}$ would leave rather than $\ce{H_2O}$ in the formation of the electrophile? Please help me find a mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would like to approach questions, such as this one, like I would a puzzle rather than calling upon my memory. Is there a series of steps that I should follow? (i.e: 1. Identify electro/nucleophile, 2. Identify acidic protons ...).&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="5026" LastEditDate="2018-04-27T15:15:30.227" LastActivityDate="2018-04-27T17:52:23.407" Title="What is the mechanism of chlorosulfonation of benzene?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27438" PostTypeId="2" ParentId="27436" CreationDate="2015-03-16T22:04:58.257" Score="7" Body="&lt;p&gt;You're right that the first step involves protonation of the terminal carbon in the  styrene double to yield a resonance stabilized benzyl carbocation.  This carbocation then adds to another styrene molecule in the exact same fashion, reacting at the terminus of the styrene double bond to again yield a resonance stabilized benzyl carbocation.  This latter carbocation then alkylates itself (like a Friedel-Crafts alkylation).  Loss of a proton from this final carbocation yields the indicated product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SM0N6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-16T22:19:41.110" LastActivityDate="2015-03-16T22:19:41.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27439" PostTypeId="2" ParentId="27437" CreationDate="2015-03-16T22:29:25.993" Score="7" Body="&lt;p&gt;The $\ce{OH}$-group is protonated by another molecule of chlorosulphuric acid, creating a cation. This cation can be attacked by benzene under the right conditions. $\ce{H2O}$ is not a better leaving group than $\ce{Cl-}$ (look at the charge, they mostly tell a lot about the ability of the leaving group, though not always). A famous example of this reaction is of course the famous old synthesis of saccharin (Clayden et al. p. 644 &amp;amp; p. 645) giving the widely used $\ce{TsCl}$ as byproduct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XLmeT.png&quot; alt=&quot;Synthesis of saccharin - picture 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/StUsu.png&quot; alt=&quot;Synthesis of saccharin - picture 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-16T22:29:25.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27440" PostTypeId="1" CreationDate="2015-03-16T22:59:02.817" Score="0" ViewCount="6231" Body="&lt;p&gt;I've been given a problem asking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Estimate the standard enthalpy of formation of hydrazine, $\ce{N2H4(g)}$, from the following data.&#xA;  \begin{align}&#xA;\Delta{}H_\mathrm{B}(\ce{H2}) &amp;amp;= 436~\mathrm{kJ/mol}\\&#xA;\Delta{}H_\mathrm{B}(\ce{N2}) &amp;amp;= 944~\mathrm{kJ/mol}\\&#xA;\Delta{}H_\mathrm{B}(\ce{N–N}) &amp;amp;= 163~\mathrm{kJ/mol}\\&#xA;\Delta{}H_\mathrm{B}(\ce{N–H}) &amp;amp;= 388~\mathrm{kJ/mol}\\&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While I know that $\Delta{}H_\mathrm{B}(\ce{N–N})$ refers to the energy released from breaking that bond, I'm not sure what $\Delta{}H_\mathrm{B}(\ce{H2})$ or $\Delta{}H_\mathrm{B}(\ce{N2})$ refers to. Could someone please help me understand?&lt;/p&gt;&#xA;" OwnerUserId="15013" LastEditorDisplayName="user7951" LastEditDate="2015-05-13T21:17:41.930" LastActivityDate="2021-02-27T11:31:40.257" Title="How to calculate the enthalpy of formation of hydrazine from tabulated values?" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27441" PostTypeId="2" ParentId="27437" CreationDate="2015-03-16T22:59:58.810" Score="12" Body="&lt;p&gt;Rob, IMO this is a difficult problem, so don't feel bad about not &quot;seeing&quot; the solution right off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorosulfonic acid (or chlorosulfuric acid) is both difficult and hazardous to work with so it's reactions aren't discussed all that frequently.  It can also give rise to different products at different temperatures.  At higher temperatures it generates $\ce{SO3}$ which will behave as an electrophile and react with aromatic compounds to produce sulfonation products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQ3IM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.4college.co.uk/a/Cd/react.php&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At lower temperatures (&lt;strong&gt;your case&lt;/strong&gt;), chlorosulfonic acid reacts by way of the following equilibrium (who knew this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3 ClSO2(OH) &amp;lt;=&amp;gt; SO2Cl^{+} + 2 SO3Cl^{-} + H3O^{+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The equilibrium generates $\ce{SO2Cl^{+}}$ which is the electrophile of interest in your reaction&lt;/strong&gt;.  $\ce{SO2Cl^{+}}$ will react with an aromatic nucleus to produce the corresponding sulfonyl chloride by way of a typical electrophilic aromatic substitution reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to read more about the complexities of reactions involving chlorosulfonic acid see &lt;a href=&quot;https://books.google.com/books?id=roI_CwMEcHMC&amp;amp;pg=PA35&amp;amp;lpg=PA35&amp;amp;dq=chlorosulfonic%20acid,%20chlorosulfonation%20by&amp;amp;source=bl&amp;amp;ots=FeI7puOh4X&amp;amp;sig=KmCQSqv-chgvNQyb6xAyEaWwgrY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=BFsHVcanL8ergwSpooJQ&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=chlorosulfonic%20acid%2C%20chlorosulfonation%20by&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; starting around page 11.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficult part here was identifying the source of the electrophile. Looking at the final product you get an idea of what the electrophile is, but how is it produced?  Sometimes you have to do a little googling to sort things out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, IMO this was a tough one.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-17T00:17:57.813" LastActivityDate="2015-03-17T00:17:57.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27443" PostTypeId="1" AcceptedAnswerId="27445" CreationDate="2015-03-17T00:03:37.323" Score="1" ViewCount="54" Body="&lt;p&gt;I have to use oxidization number approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Co^2+ (aq) + H2SO3(aq) -&amp;gt; Co^3+(aq) + S2O3^2- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my approach so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The change in Co is 1 electron, the change is S is 2 electrons, so CO must be multiplied by two&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Co^2 + H2SO3(aq) -&amp;gt; 2 Co^3+ + S2O3^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I balance the O's, then balance the H's and I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 Co^2+ + 2H2SO3(aq) + 2H^+ -&amp;gt; 2 Co^3+ + S2O3^2- + 3H2O(l)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did it the half reaction way as well, I got the same answer but the coefficients on the Co's were 4. I suspect my problem is my assumptions about the oxidization number change at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="15015" LastEditorUserId="13184" LastEditDate="2015-03-17T00:52:40.680" LastActivityDate="2015-03-17T18:35:47.667" Title="How do I go about balancing these equations?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27445" PostTypeId="2" ParentId="27443" CreationDate="2015-03-17T07:51:47.737" Score="1" Body="&lt;p&gt;I agree with you, the right equilibration of the redox reaction is the following:&lt;br&gt;&#xA;$$\ce{4 Co^2+(aq) + 2H2SO3(aq) + 2H^+ (aq)-&amp;gt; 4 Co^3+(aq) + S2O3^2- (aq) + 3H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is  in your assumptions about the oxidization number change at the beginning, and more precisely about sulfur.&#xA;The oxidation number of  sulfur in $\ce{H2SO3}$ is four. But in thiosulfate ion, you have two inequivalent sulfur atoms: &#xA;Please see:&#xA; &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24928/oxidation-state-of-the-sulfur-atoms-in-the-thiosulfate-ion-contradiction&quot;&gt;Oxidation states of the sulfur atoms in the thiosulfate ion&lt;/a&gt;&#xA;The oxidation number is $\ce{+IV}$ for the central one, while it's zero for the terminal one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum, the variation in the oxidation number for sulfure is four, not two.&#xA;(As if you have two sulfur atoms with oxidation number of four in $\ce{2H2SO3}$. One of them changes its oxidation number to zero. while the other one stays intact).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-03-17T07:57:11.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27446" PostTypeId="1" AcceptedAnswerId="27452" CreationDate="2015-03-17T08:18:11.850" Score="14" ViewCount="4698" Body="&lt;p&gt;The mass of carbon-12 is &lt;span class=&quot;math-container&quot;&gt;$\pu{12 u}$&lt;/span&gt; by definition. However, one carbon-12 atom comprises 6 neutrons (each weighing &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0087 u}$&lt;/span&gt;), 6 protons (each weighing &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0072 u}$&lt;/span&gt;), and 6 electrons (each weighing &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0005 u}$&lt;/span&gt;), which all add up to &lt;span class=&quot;math-container&quot;&gt;$\pu{12.0894 u}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the &lt;span class=&quot;math-container&quot;&gt;$0.7\%$&lt;/span&gt; difference in mass come from?&lt;/p&gt;&#xA;" OwnerUserId="15017" LastEditorUserId="23561" LastEditDate="2018-12-15T17:32:52.550" LastActivityDate="2018-12-15T17:32:52.550" Title="Why are the masses of atoms less than the sum of their subatomic particles?" Tags="&lt;atoms&gt;&lt;mole&gt;&lt;nuclear-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27447" PostTypeId="2" ParentId="27440" CreationDate="2015-03-17T08:43:50.693" Score="1" Body="&lt;p&gt;In order to &quot;make&quot; hydrazine, $\ce{H2N-NH2}$ from hydrogen and nitrogen atoms, you have to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;break $\ce{N2}$ once, giving 2 nitrogen atoms&lt;/li&gt;&#xA;&lt;li&gt;break $\ce{H2}$ &lt;strong&gt;twice&lt;/strong&gt;, giving 4 hydrogen atoms&lt;/li&gt;&#xA;&lt;li&gt;form one $\ce{N-N}$ bond&lt;/li&gt;&#xA;&lt;li&gt;form &lt;strong&gt;four&lt;/strong&gt; $\ce{N-H}$ bonds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You have all the enthalpies that you need. Just pay attention to the coefficients and the signs.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-17T08:43:50.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27450" PostTypeId="1" AcceptedAnswerId="27469" CreationDate="2015-03-17T10:59:50.493" Score="1" ViewCount="137" Body="&lt;p&gt;I read in a biology course that you can determine the concentration of oxygen in water by adding some $\ce{NaOH}$ and then a small amount of $\ce{MgCl2}$. They say that a dark coloured precipitate is formed.&#xA;I can't see a way of writing this reaction. Can anyone help me?&lt;/p&gt;&#xA;" OwnerUserId="9297" LastEditorUserId="10185" LastEditDate="2015-03-17T11:43:14.370" LastActivityDate="2015-03-17T15:33:22.940" Title="Finding out the concentration of oxygen in water" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27452" PostTypeId="2" ParentId="27446" CreationDate="2015-03-17T12:07:53.390" Score="18" Body="&lt;p&gt;This is due to the mass-energy equivalence and a phenomenon called &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy&quot; rel=&quot;noreferrer&quot;&gt;binding energy&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.&lt;/p&gt;&#xA;&lt;p&gt;The binding energy of carbon-12 is quoted on Wikipedia as &lt;span class=&quot;math-container&quot;&gt;$\pu{92.162 MeV}$&lt;/span&gt;. Therefore we can estimate the mass defect of a carbon-12 atom, &lt;span class=&quot;math-container&quot;&gt;$\Delta m$&lt;/span&gt;, using &lt;span class=&quot;math-container&quot;&gt;$ E = (\Delta m)c^2$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The difference in the mass of carbon-12 to the mass of its constituent particles is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.08940 u}$&lt;/span&gt;, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-09-02T15:35:57.460" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27453" PostTypeId="1" AcceptedAnswerId="27568" CreationDate="2015-03-17T12:11:05.907" Score="10" ViewCount="5448" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MtQDC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find a mechanism for the above reaction but I cannot make much progress. I do have something in mind but it doesn't seem entirely plausible: First the tert-butyl group leaves (this seems wrong but tert-butyl cation is stable and thus a reasonable leaving group maybe? Also the negatively charged ring that it would leave behind would be stabilised by the EWG Bromines). The cation is then attacked by Cl and the ring is attacked by a proton (one product formed). Then a Friedel-Craft alkylation to make the other product (hence the excess of benzene to promote only mono-substitution). I doubt that is right, purely because of the first step. Can you help? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4325" LastEditDate="2015-03-17T15:53:54.673" LastActivityDate="2015-03-19T19:36:58.450" Title="Reverse Friedel-Crafts mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27454" PostTypeId="1" CreationDate="2015-03-17T12:35:49.093" Score="0" ViewCount="53" Body="&lt;p&gt;TL;DR;&lt;br&gt;&#xA;I need a compound that is harmeless to the human body, odorless, tasteless, causes some kind of visible reaction (urine coloration perhaps) AND can be bought without medical recomendation.&lt;br&gt;&#xA;LONG VERSION;&lt;br&gt;&#xA;Food stealing in my company fridge has become a serious problem. It's not only a recurring issue but a targeted one... our thief has been targeting &quot;high&quot; value, particularly coconut water. HR seems to ignore our complains... so I decided to take matters in my own hands... since I have access to timesheets, I thought that if I can scare out thief to leave work or skip it, I can narrow-it-down/point-him-out because we have about 30 people in our office...&lt;/p&gt;&#xA;" OwnerUserId="9331" LastActivityDate="2015-03-17T14:38:31.173" Title="Looking for a compound" Tags="&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-17T19:58:57.183" ContentLicense="CC BY-SA 3.0" />
  <row Id="27455" PostTypeId="1" AcceptedAnswerId="27456" CreationDate="2015-03-17T12:37:58.197" Score="2" ViewCount="767" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.rexresearch.com/hhusb/hh6thc.htm#HH62&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cannabinoid Chemistry - 6.2 Extraction&lt;/a&gt; &amp;amp; pondered over the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cannabis must be dried be(fore) it is extracted, because it is not possible to remove more than 50% of the cannabinoids from fresh material. THC-Acid is difficult to extract. If you plant(sic) to convert the THCA to THC, the &lt;em&gt;plant material should be thoroughly decarboxylated by heating it &lt;strong&gt;under nitrogen&lt;/strong&gt; at 105&amp;#8451; for 1 hour before performing a solvent extraction.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is decarboxylation done 'under nitrogen'?  My first guess would be that at 105&amp;#8451; the plant begins to oxidize (but that is just my naive guess).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also sourced this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6d83.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Effect of heating time and temperature on the THC content of an &lt;em&gt;n&lt;/em&gt;-hexane&#xA;marihuana extract &lt;br&gt;after heating on the glass surface in an open reactor.  Journal of Chromatography 520 (1990).&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which suggests that 122&amp;#8451; for 27-30 minutes will be even more effective for decarboxylation, but if the sample begins to oxidize at 105&amp;#8451; it seems even more important to get the right mixture of gases around it at 122℃.&lt;/p&gt;&#xA;" OwnerUserId="1071" LastEditorUserId="1071" LastEditDate="2015-03-18T12:47:34.123" LastActivityDate="2015-03-18T12:47:34.123" Title="Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27456" PostTypeId="2" ParentId="27455" CreationDate="2015-03-17T13:49:57.650" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] it seems even more important to get the right mixture of gases around it at 122℃&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's only important to keep the oxygen away while you are heating the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the structures: the thermal decarboxylation of the THCA isn't different from the decarboxylation of salicylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with the carboxyl group or without, (upon heating,) phenols are easily oxidized by air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the sample could be heated under vacuum too. In this case, you might want to use a cooling trap to collect volatile material. But usually, one does that anyway to protect the vacuum pump.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="1071" LastEditDate="2015-03-18T03:06:14.520" LastActivityDate="2015-03-18T03:06:14.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27457" PostTypeId="1" AcceptedAnswerId="27459" CreationDate="2015-03-17T13:50:37.273" Score="10" ViewCount="50033" Body="&lt;p&gt;Identifying reducing / non-reducing sugar been confusing me for a while now , I know that reducing sugar contain aldehyde or ketone group . It's easy to identify them in monosaccharides but this becomes confusing in case of disaccharides! &#xA;&lt;img src=&quot;https://i.stack.imgur.com/FoVM1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fail to see carbonyl group here, How can I identify reducing / non-reducing sugar by looking at structure in picture? can anyone help me out?&lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorDisplayName="user7951" LastEditDate="2016-12-22T18:17:13.933" LastActivityDate="2017-11-18T17:04:05.573" Title="How do you identify reducing / non-reducing sugar by looking at structure?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;carbonyl-compounds&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27459" PostTypeId="2" ParentId="27457" CreationDate="2015-03-17T14:07:08.410" Score="10" Body="&lt;p&gt;Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hemiacetal&quot; rel=&quot;noreferrer&quot;&gt;hemiacetal&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFpF5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you realize that a hemiacetal can equilibrate with a carbonyl (&lt;em&gt;e.g&lt;/em&gt;. it is a carbonyl in disguise), identification of reducing sugars becomes easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example from &lt;a href=&quot;http://en.wikipedia.org/wiki/Reducing_sugar&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, it is Maltose, the same as your third compound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yk9w0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the aldehyde in the open isomer drawn on the right; see how it can close to form a hemiacetal on the left.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-17T15:08:37.873" LastActivityDate="2015-03-17T15:08:37.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27460" PostTypeId="1" CreationDate="2015-03-17T14:09:01.407" Score="4" ViewCount="355" Body="&lt;p&gt;Disclaimer: Potentially more of a physics question, but this area is muddy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find 720 nm LEDs but they're proving to be very uncommon. In fact, there seems to be a gap between 660 nm up to ~730 nm. I've only been able to find one supplier of 720 nm LEDs (data sheet &lt;a href=&quot;http://www.roithner-laser.com/datasheets/led_highsingle/apg2c1-720.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). These LEDs also tend to be very low powered compared to their counterparts in the nearby range. For example, a typical 690 nm LED outputs 690 mW, whereas the 720 nm LED only ouputs 170 mW, and that's the strongest one I could find (which is no longer carried).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling says this has to do with the material being used in the LED (in this case GaAlAs). Perhaps it's very difficult to dope the material in such a way that it emits 720 nm light? I've already noticed that these LEDs are prone to burning out. Is the material just not very good for this purpose? Hopefully someone with more insight into this topic could shed some &lt;em&gt;light&lt;/em&gt; (heh) on the situation.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="10185" LastEditDate="2015-04-10T17:53:28.257" LastActivityDate="2015-04-10T17:53:28.257" Title="Why are 720 nm LEDs so uncommon?" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27461" PostTypeId="1" CreationDate="2015-03-17T14:27:13.003" Score="3" ViewCount="640" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GvVIs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture shows the first step of Oxymercuration. The electrons attack the Mercury atom but the OAc group leaves. Why is this? Hg is a transition metal with a number of available oxidation states, why cant it just accept the electrons? For similar reactions (like glycol cleavage using periodate or lead tetra acetate) how is it possible to tell whether a group will leave (retaining oxidation number) or the oxidation state changes?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-03-17T14:27:13.003" Title="Why does the OAc group leave in the first step of Oxymercuration?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27462" PostTypeId="1" AcceptedAnswerId="27463" CreationDate="2015-03-17T14:28:14.897" Score="7" ViewCount="13390" Body="&lt;p&gt;We write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{Alcohol -&gt;[\ce{KMnO4}] Acid}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alcohol forms a carboxylic acid, i.e. gains an $\ce{O}$ from $\ce{KMnO4}$. This means $\ce{KMnO4}$ is a reactant! Then why is it written above the arrow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else do we write above the arrow?&lt;/p&gt;&#xA;" OwnerUserId="13236" LastEditorUserId="7475" LastEditDate="2015-10-09T15:59:31.987" LastActivityDate="2015-10-09T15:59:31.987" Title="Why do we sometimes write reactants above the reaction arrow?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27463" PostTypeId="2" ParentId="27462" CreationDate="2015-03-17T14:36:16.077" Score="6" Body="&lt;p&gt;Things written on arrows are chemicals that are part of the reaction in some way, but not one of the reactants or products. In this case, the $\ce{KMnO4}$ is your oxidizing agent, which can be swapped out for any other strong oxidizer, and is not an inherent part of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things you could see on the arrow would be $\ce{H2O}$ or Diethyl Ether, both of which are solvents. Other things include $h\nu$ and $\Delta T$, both of which are conditions under which the reaction takes place. Lastly the most common thing you'll see are catalysts such as $\ce{Pt}$. As you  can see, all of these things are necessary to the reaction, but aren't really what we care about in the reaction/can be substituted for other, similar chemicals.&lt;/p&gt;&#xA;" OwnerUserId="13030" LastEditorUserId="4945" LastEditDate="2015-03-18T06:56:47.343" LastActivityDate="2015-03-18T06:56:47.343" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27464" PostTypeId="2" ParentId="27454" CreationDate="2015-03-17T14:38:31.173" Score="3" Body="&lt;p&gt;Urine typically is yellow, but in combination with blue dyes, a green colour can be obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noteworthy compounds are methylene blue, thymol, Cimetidine (antacid), or Mitoxanthrone (used in cancer therapy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've mentioned these just for the sake of completeness - &lt;strong&gt;I think it is irresponsible and stupid to spike food with anything that doesn't belong there.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hiding a webcam in a shelf and running zoneminder probably isn't legal either and will get you in trouble with the privacy officer, but at least that doesn't mess around with the health of anybody else.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-17T14:38:31.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27465" PostTypeId="1" AcceptedAnswerId="27468" CreationDate="2015-03-17T14:48:03.573" Score="8" ViewCount="1719" Body="&lt;p&gt;If you put a paper inside an oven vacuum packed and go beyond $250~\mathrm{^\circ C}$, will it burn? There's no oxidizer. If not, what will happen?&lt;/p&gt;&#xA;" OwnerUserId="15023" LastEditorUserId="39539" LastEditDate="2020-04-05T21:02:02.167" LastActivityDate="2020-04-06T00:18:11.520" Title="Can paper burn without oxygen or air?" Tags="&lt;combustion&gt;&lt;temperature&gt;&lt;carbohydrates&gt;&lt;pyrolysis&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27466" PostTypeId="1" AcceptedAnswerId="27467" CreationDate="2015-03-17T15:07:04.567" Score="2" ViewCount="504" Body="&lt;p&gt;I was solving some problems on chemical kinetics, then I was just struck at some ques of sequential reaction kinetics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given reactions $$\ce{O_3 + Cl \rightarrow O_2 + ClO ~~~ k_1=5.2 \times 10^9~Lmol^{-1}s^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and  $$\ce{ClO + O \rightarrow O_2 + Cl ~~~ ~~~  k_2=2.6 \times 10^{10} ~Lmol^{-1}s^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which value is closest to rate const of overall net reaction? $$\ce{O_3 + O \rightarrow 2O_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the answer given was $=5.2 \times 10^9 ~\mathrm{Lmol^{-1}s^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I am confused, how did they find the rate const of the net reaction?&lt;/p&gt;&#xA;" OwnerUserId="13138" LastEditorUserId="4231" LastEditDate="2015-03-17T15:16:32.297" LastActivityDate="2015-03-17T15:16:32.297" Title="Sequential Kinetics" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27467" PostTypeId="2" ParentId="27466" CreationDate="2015-03-17T15:14:20.807" Score="3" Body="&lt;p&gt;The rate constant of the overall reaction is determined by the rate constant of the RDS, Rate Determining Step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RDS is the one with the lowest rate constant. Hence in this case the overall rate of the reaction is $5.2 \times 10^9 Lmol^{-1}s^{-1}$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-03-17T15:14:20.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27468" PostTypeId="2" ParentId="27465" CreationDate="2015-03-17T15:32:11.800" Score="17" Body="&lt;p&gt;No, the paper will not burn without oxygen being present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper is made primarily of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cellulose&quot; rel=&quot;noreferrer&quot;&gt;cellulose&lt;/a&gt; which is a polymer of glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4U14W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you heat paper in a vacuum the cellulose simply decomposes to $\ce{H2O}$, $\ce{CO2}$, $\ce{CO}$ and carbon.  As the paper decomposes it will &quot;char&quot; or turn brown to black as the cellulose polymer degrades.  Here is a &lt;a href=&quot;http://jes.ecsdl.org/content/83/1/161.abstract&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt; to an abstract describing the process in more detail.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-17T21:44:36.247" LastActivityDate="2015-03-17T21:44:36.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27469" PostTypeId="2" ParentId="27450" CreationDate="2015-03-17T15:33:22.940" Score="3" Body="&lt;p&gt;This is the old Winkler titration, published as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Die Bestimmung des im Wasser gelösten Sauerstoffes&lt;/em&gt; by &lt;em&gt;Ludwig Wilhelm Winkler&lt;/em&gt; in &lt;em&gt;Ber. Dtsch. Chem. Ges.&lt;/em&gt;, &lt;strong&gt;1888&lt;/strong&gt;, &lt;em&gt;21&lt;/em&gt;, 2843-2854 (&lt;a href=&quot;http://gallica.bnf.fr/ark:/12148/bpt6k90715s/f46.image&quot; rel=&quot;nofollow&quot;&gt;SRC&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does it work?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add $\ce{MnSO4}$ to your water sample, then $\ce{NaOH}$ and $\ce{KI}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Under alkaline conditions, oxygen dissolved in your water sample oxidizes $\ce{Mn(II)}$ to $\ce{Mn(IV)}$.&#xA;\[\ce{2Mn^{2+} + O2 + 4 OH- -&gt; MnO(OH)2 v}\]&#xA;This is the brownish precipitate mentioned in the question.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Now, the sample is acidified (Winkler used hydrochloric acid) and the solution becomes yellowish due to the iodine formed. Under these conditions, $\ce{Mn(IV)}$ oxidizes iodide:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;\[\ce{Mn(IV) + 2 I- -&gt; Mn^{2+} + I2}\]&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-17T15:33:22.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27470" PostTypeId="1" AcceptedAnswerId="27482" CreationDate="2015-03-17T16:23:49.213" Score="3" ViewCount="106" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Clark_electrode&quot; rel=&quot;nofollow&quot;&gt;Clark Type Oxygen Electrode&lt;/a&gt; converts water, oxygen, and electrons into hydroxide via a hydrogen peroxide intermediate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{O2 + 4e^{−} + 2 H_2O -&gt; 4 OH^{−}}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the sum of:&#xA;\[&#xA;\begin{align*}&#xA;\ce{2H2O +O2 + 2e- &amp;amp;-&amp;gt; H2O2 + OH- \\&#xA;H2O2 + 2e- &amp;amp;-&amp;gt; 2 OH-}&#xA;\end{align*}&#xA;\]&#xA;Can this be reversed?&lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="4945" LastEditDate="2015-03-18T03:10:57.647" LastActivityDate="2015-03-18T03:37:44.007" Title="Formation of O₂, Water, and Electricity from Hydroxide?" Tags="&lt;electrochemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27471" PostTypeId="2" ParentId="27443" CreationDate="2015-03-17T18:35:47.667" Score="0" Body="&lt;p&gt;The problem with your answer is that the charges do not balance. There is a net charge of +6 on the left hand side and +4 on the right hand side. This can be solved by doubling both of the Co coefficients giving a net charge of +10 on each side.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-17T18:35:47.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27472" PostTypeId="1" CreationDate="2015-03-17T19:31:16.803" Score="5" ViewCount="141" Body="&lt;p&gt;I'm considering creating a program that automatically completes an ICE chart/table when the minimum given information is provided. The reason I'm posting my question here is because I need to know the possible cases of minimum information required before I can even start the program.&lt;/p&gt;&#xA;&lt;p&gt;I know you can complete the table if you're given these following sets of information:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;K and initial concentrations&lt;/li&gt;&#xA;&lt;li&gt;K and final concentrations&lt;/li&gt;&#xA;&lt;li&gt;K and initial concentrations of reactants and final concentrations of products&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I also believe there are a ton more, but I'm not sure how to complete the list.&lt;/p&gt;&#xA;" OwnerUserId="15025" LastEditorUserId="83758" LastEditDate="2020-09-30T06:27:52.423" LastActivityDate="2020-09-30T06:27:59.760" Title="What is the minimum number of given sets of information required to complete an ICE chart for an equilibrium reaction?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27474" PostTypeId="2" ParentId="7142" CreationDate="2015-03-17T21:23:14.327" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://journals.aps.org/pra/abstract/10.1103/PhysRevA.36.467&quot;&gt;Analytical Dirac-Hartree-Fock-Slater screening function for atoms (Z=1–92)&lt;/a&gt; Phys. Rev. A 36, 467 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;seems to address the deficiencies mentioned in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effective nuclear charge is expressed as a function of radius for elements through uranium and relativistic calculations are used.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1574181811000474&quot;&gt;A new set of relativistic screening constants for the screened hydrogenic model&lt;/a&gt; High Energy Density Physics vol. 7 page 169-179.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-17T21:23:14.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27475" PostTypeId="2" ParentId="27472" CreationDate="2015-03-17T21:32:55.607" Score="1" Body="&lt;p&gt;Alright I think you have the basics down, first of all I have a couple pointers for you since I have done many (&lt;em&gt;many&lt;/em&gt;) RICE charts. What I would do if I were you? Since I have some experience in programming I would just use molarities of the initial and final.&lt;/p&gt;&#xA;&lt;p&gt;You must make the user input the data, say if the user was given moles then make him/her convert it to molarity.&lt;/p&gt;&#xA;&lt;p&gt;I think you have it pretty much down, you need either the initial concentrations and K or find the K through using the initial concentrations of the products and reactants or find the initial using the product and K.&lt;/p&gt;&#xA;&lt;p&gt;I mean then again I would work through making this first part down because those are the essentials (atleast what I've learned) and then work on making updates to that program.&lt;/p&gt;&#xA;&lt;p&gt;Since I do have programming experience I would just use if statements to do the entire thing. &lt;code&gt;If this is available then solve for this.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Don't worry about the entire thing yet. Otherwise I would have to say you have pretty much everything down.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: This is not about Ka, Ksp, Kb, just for Keq. Because then you would have to work with the pH and pOH more equations.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="83758" LastEditDate="2020-09-30T06:27:59.760" LastActivityDate="2020-09-30T06:27:59.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27476" PostTypeId="1" CreationDate="2015-03-17T21:43:23.330" Score="1" ViewCount="143" Body="&lt;p&gt;This is a rather physics question but I wanted to get an insight from a chemistry perspective. I heard from some that you can beat the speed or go faster than it but is that really true. Physicists as we all know look at the theoretical possibilities but what is your insight from a chemistry-ish view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bbc.com/future/story/20121003-can-we-travel-faster-than-light&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=EPsG8td7C5k&amp;amp;t=34&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we can then how much more faster is it? I heard one view into this that goes along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you distorted space and moved it then you would be light-years away from where you started. This seemed very &lt;em&gt;dumb&lt;/em&gt; at first but as I thought about it is this anyway possible.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-03-18T14:27:45.113" Title="Can we really beat the speed of light?" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="0" ClosedDate="2015-03-18T12:16:08.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="27477" PostTypeId="1" CreationDate="2015-03-17T22:04:53.573" Score="1" ViewCount="1102" Body="&lt;p&gt;In &lt;a href=&quot;http://www.youtube.com/watch?v=xFNeN9rONKY&amp;amp;t=3m50s&quot; rel=&quot;nofollow&quot;&gt;3:50 of this video&lt;/a&gt; two FTIR spectras are shown. He explains that the % transmittance (double bond in his example) depends on the dipole moment, $\mu = \delta \cdot d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the absorbance center frequency depends on $\delta, d$, but why the &lt;strong&gt;percentage&lt;/strong&gt; of transmittance depends on $\delta, d$? A private case of this would be a symmetric bond ($\delta=0$): why the transmittance is 100%?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-03-17T22:10:39.473" LastActivityDate="2015-03-18T00:18:51.033" Title="Why the magnitude of dipole moment influences on FTIR % transmittance?" Tags="&lt;spectroscopy&gt;&lt;dipole&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27478" PostTypeId="2" ParentId="27476" CreationDate="2015-03-17T22:10:26.997" Score="2" Body="&lt;p&gt;From classical special relativity it can be shown that anything (particle, object, electromagnetic wave, Z-wave, a dishwasher, anything) that travels faster than the speed of light (in vacuum) would violate the causality principle. In relation to chemistry, this means more qualitatively that the second law of thermodynamics would be violated. Hence I would argue that from a chemistry viewpoint traveling faster than the speed of light is impossible.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-03-18T10:42:34.240" LastActivityDate="2015-03-18T10:42:34.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27479" PostTypeId="2" ParentId="27477" CreationDate="2015-03-17T23:30:42.943" Score="1" Body="&lt;p&gt;Light has both oscillating electric and magnetic fields.  In the common UV-Visible-IR spectroscopy experiments it is the interaction of the molecule with the  electric field that is being observed. A molecule is very small compared to the size of a beam of light, so it is fair to view the molecule as being immersed in or surrounded by the light.  In other words, it is fair to view the molecule as being immersed in a uniform electric field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of the interaction ($I_s$) of the light's electric field ($E$) with the molecule is dependent upon both the strength of the light's electric field and the vector description of the molecule's electric field.  The molecule's electric field is described by the spatial arrangement of the molecule's electrons.  The spatial arrangement of the molecule's electrons is described by the dipole moment ($\mu$), a vector sum.  Hence,&#xA;$$I_s = E \cdot \mu $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absorption intensity of a spectral transition (the area under the absorption curve - usually it correlates with the peak height) is given by the same equation &lt;strong&gt;except&lt;/strong&gt; that rather than use the molecular dipole moment $\mu$, one uses what is called the &lt;strong&gt;transition dipole moment&lt;/strong&gt;.  Rather than consider the ground state distribution of electrons around the molecule, one must use the rearranged distribution of electrons around the molecule that occurs when the molecule absorbs light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the transition dipole moment is always similar to the ground state dipole moment, then what the video says is correct.  This may be true for larger molecules when we are looking at an absorption that excites a $\ce{C-H}$ or $\ce{C-C}$ bond, or something similar.  In such cases, the small perturbation in electron density around the specific bond probably doesn't radically change the overall electron distribution around the entire molecule.  In smaller molecules like $\ce{CO}$ it seems possible to me that light absorption could make a significant change to the overall electron distribution and that consequently the ground state dipole moment might differ substantially from the transition dipole moment.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-18T00:18:51.033" LastActivityDate="2015-03-18T00:18:51.033" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27480" PostTypeId="2" ParentId="27476" CreationDate="2015-03-18T00:57:15.553" Score="3" Body="&lt;p&gt;There are a number of things that exceed the speed of light:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Particles in a medium such as water can enter faster than light [in that medium] but are slowed and release energy as &lt;a href=&quot;http://en.wikipedia.org/wiki/Cherenkov_radiation&quot; rel=&quot;nofollow&quot;&gt;Cherenkov radiation&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_velocity&quot; rel=&quot;nofollow&quot;&gt;phase velocity&lt;/a&gt; of a wave can exceed that of light [in that medium].&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An arbitrary point, such as the point on a pair of scissors where blades,&#xA;meet, can move faster than light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most puzzling to me: &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_entanglement&quot; rel=&quot;nofollow&quot;&gt;entangled particles&lt;/a&gt;. Einstein called this &quot;spooky action at a distance&quot;. Although each &lt;em&gt;particle&lt;/em&gt; moves at C or less, information &lt;em&gt;might be&lt;/em&gt; shared between them in an indefinitely short time; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Superluminal_communication&quot; rel=&quot;nofollow&quot;&gt;superluminal communication&lt;/a&gt;. Entanglement &lt;em&gt;does&lt;/em&gt; apply to chemistry, as do other odd quantum effects; e.g. the detection of single photons with silver halide photography as they form an interference pattern in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;nofollow&quot;&gt;double-slit experiment&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-03-18T14:27:45.113" LastActivityDate="2015-03-18T14:27:45.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27482" PostTypeId="2" ParentId="27470" CreationDate="2015-03-18T03:37:44.007" Score="2" Body="&lt;p&gt;Yes, the overall reaction can be reversed.  The reverse reaction is called the &lt;a href=&quot;http://link.springer.com/referenceworkentry/10.1007%2F978-1-4419-6996-5_407&quot; rel=&quot;nofollow&quot;&gt;oxygen evolution reaction&lt;/a&gt; or simply &lt;a href=&quot;http://en.wikipedia.org/wiki/Heterogeneous_water_oxidation&quot; rel=&quot;nofollow&quot;&gt;water oxidation&lt;/a&gt;.  Depending on the pH, you could write it as either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O -&amp;gt; O2 + 4 H+ + 4e-}$$ or&#xA;$$\ce{4 OH- -&amp;gt; O2 + 2 H2O + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse reaction happens when water is electrolyzed to make hydrogen at one electrode and oxygen at the other.  You may have seen that demo at one time or another -- &lt;a href=&quot;https://www.youtube.com/watch?v=HQ9Fhd7P_HA&quot; rel=&quot;nofollow&quot;&gt;here's a Youtube version&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is a thermodynamically powerful oxidant, which is another way of saying that it is easily reduced.  This means that the water oxidation reaction requires energy input, about 1.23 V at the chemical reference state of pH 0.  But it usually takes even more energy to effectively reverse the reaction -- notice they used a 9 V battery in the Youtube demo, so about 8 of those volts are wasted in the demo.  Developing new electrode materials that can carry out the water oxidation reaction with less energy loss is a major area of fundamental research in chemistry today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also said the Clark electrode works with $\ce{H2O2}$ as an intermediate.  Depending on the details of the particular materials used, in the water oxidation reaction hydrogen peroxide may or may not be an intermediate.  Working out the fundamental reaction steps that happen at electrode surfaces can be very difficult.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-18T03:37:44.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27483" PostTypeId="1" CreationDate="2015-03-18T03:39:18.587" Score="1" ViewCount="158" Body="&lt;p&gt;I would like to spin coat Graphene oxide on the glass slides and intend to have a thickness of 100nm/0.1micrometre. May I know what is the amount of Graphene oxide to be poured onto the glass slides, spin time, spin speed and time for it to develop to obtain the optimized and desired thickness of 100nm?&lt;/p&gt;&#xA;" OwnerUserId="15027" LastEditorUserId="7448" LastEditDate="2015-04-02T18:28:40.700" LastActivityDate="2015-04-02T18:28:40.700" Title="Spin coat of Graphene oxide on glass slides" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27485" PostTypeId="1" CreationDate="2015-03-18T07:52:45.850" Score="3" ViewCount="681" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is the acidity order true?&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COCH3} &amp;gt; \ce{CH3CO2CH3} &amp;gt; \ce{CH3CON(CH3)2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I deduced that in methyl acetate and &lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-dimethylacetamide there is a −&lt;em&gt;I&lt;/em&gt; effect of the methoxy group, &lt;span class=&quot;math-container&quot;&gt;$\ce{N(CH3)2}$&lt;/span&gt;  and a +&lt;em&gt;R&lt;/em&gt; effect, too so the attached keto-carbon develops a positive charge and the conjugate base becomes unstable due to the +&lt;em&gt;R&lt;/em&gt; effect of methoxy and &lt;span class=&quot;math-container&quot;&gt;$\ce{N(CH3)2}$&lt;/span&gt; groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my logic right?&lt;/p&gt;&#xA;" OwnerDisplayName="user15029" LastEditorUserId="41328" LastEditDate="2020-05-27T11:44:39.297" LastActivityDate="2020-05-27T11:44:39.297" Title="Acidity order of acetone, methyl acetate and N,N-dimethylacetamide" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27486" PostTypeId="1" AcceptedAnswerId="27489" CreationDate="2015-03-18T08:57:23.373" Score="5" ViewCount="319" Body="&lt;p&gt;Fructose is described to have a different metabolic pathway (a more fat-inducing one) than glucose (see: &lt;a href=&quot;http://healthyeating.sfgate.com/difference-between-sucrose-glucose-fructose-8704.html&quot; rel=&quot;nofollow&quot;&gt;http://healthyeating.sfgate.com/difference-between-sucrose-glucose-fructose-8704.html&lt;/a&gt;) as it can only be metabolized by the liver and does not impact insulin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand, is that in glycolysis, a paramount step of metabolizing glucose, it is very quickly converted to fructose-6-phosphate. Why does the body want to convert glucose into fructose?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorDisplayName="user7951" LastEditDate="2016-11-27T20:47:22.520" LastActivityDate="2016-11-27T20:47:22.520" Title="How is it that fructose has a different metabolic pathway than glucose but yet glucose is converted to fructose?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27487" PostTypeId="2" ParentId="27300" CreationDate="2015-03-18T09:26:10.530" Score="1" Body="&lt;p&gt;This drug would have to attack the cholesterol accretion or help the body do it by raising the HDL levels in relation to the LDL levels. &#xA;Statins are commonly used for this.&lt;/p&gt;&#xA;" OwnerUserId="8135" LastActivityDate="2015-03-18T09:26:10.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27488" PostTypeId="1" AcceptedAnswerId="27531" CreationDate="2015-03-18T09:40:18.927" Score="4" ViewCount="1853" Body="&lt;p&gt;Sigma bonds form due to the axial overlap of two orbitals whereas pi bonds form due to their lateral overlapping. Since the former is more effective, we can say that sigma is stronger than pi. We mean that the orbitals overlap to a further extent in sigma bonds and not as much in pi bonds(...or at least that is what I infer). To me this sounds like a circular proof/explanation, so I thought up another way to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sigma bond, the electron cloud of one atom is attracted to the nucleus of the other atom's nucleus.This happens because of the electrostatic forces of attraction between the positively charged protons in the nucleus of one atom and the negatively charged electron cloud of the other atom. The attractive force causes the electron clouds to the atoms to overlap, despite the fact that the electron clouds should repel each other.  However, the electrons start to repel each other after a certain extent of overlapping has occurred. (Because the repulsive forces between the two nuclei and the electron clouds negate the attractive forces)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a pi bond, there is no positively charged nucleus that would attract the electrons, so there must be another factor favoring the bond formation.&#xA;Since I haven't studied the quantum mechanics of bonding yet, my best guess is that the pi bond (or lateral overlapping) occurs because of the pairing of electrons.&#xA;So this means that sigma bonds are stronger because of electrostatic forces, which consequently causes more effective overlapping than pi bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct to some (if any) extent?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastActivityDate="2021-01-01T15:47:40.230" Title="Is this a possible explanation as to why sigma bonds are stronger than pi bonds?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27489" PostTypeId="2" ParentId="27486" CreationDate="2015-03-18T09:49:36.830" Score="7" Body="&lt;p&gt;The metabolic pathway you are talking about is how fructose is converted into energy and how its concentration in the blood is regulated. It is indeed true that blood fructose level does not affect blood insulin level, and this is why it is more &quot;fat-inducing&quot;, as it cannot be effectively regulated like glucose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also true that in glycolysis, glucose is converted to fructose-6-phosphate. However, this should not be confused with fructose itself. The fructose phosphate that fructolysis produced is fructose-1-phosphate, not fructose-6-phosphate.&lt;/p&gt;&#xA;" OwnerUserId="14983" LastActivityDate="2015-03-18T09:49:36.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27490" PostTypeId="2" ParentId="25134" CreationDate="2015-03-18T10:34:27.397" Score="1" Body="&lt;p&gt;&lt;em&gt;Note: for convenience, &quot;gas&quot; refers to any gas at a temperature beyond its boiling point; &quot;vapor&quot; refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of &quot;gasses&quot; can directly be calculated using the ideal gas law. Of course the partial pressure of &quot;gasses&quot; is 0Pa if there is no gas but only vapor in your container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its &lt;strong&gt;equilibrium vapor pressure&lt;/strong&gt;, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this: &lt;br/&gt;&#xA;$\log P = A- {B \over C+T}$ &lt;br/&gt;&#xA;where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture: &lt;br/&gt;&#xA;$p_i = p_i^\star x_i$ &lt;br/&gt;&#xA;where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.&lt;br/&gt;&#xA;At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:&lt;br/&gt;&#xA;$p_{total} = \sum p_i^\star x_i$&lt;/p&gt;&#xA;" OwnerUserId="14983" LastActivityDate="2015-03-18T10:34:27.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27491" PostTypeId="2" ParentId="27483" CreationDate="2015-03-18T11:05:08.513" Score="4" Body="&lt;p&gt;The procedure will vary with the size and shape of the glass, the concentration and viscosity of your graphene oxide dispersion, the solvent you're using, the wettability of your glass slide, and so on. You'll need to figure out the parameters yourself by varying the spinning conditions, measuring the resulting film thicknesses via profilometry/AFM/cross-sectional SEM, and then constructing spin curves (&lt;a href=&quot;http://nanolithography.gatech.edu/zep520acurve.html&quot; rel=&quot;nofollow&quot;&gt;examples&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-03-18T11:05:08.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27493" PostTypeId="2" ParentId="27387" CreationDate="2015-03-18T11:55:06.343" Score="2" Body="&lt;p&gt;Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2016-08-20T16:41:27.140" LastActivityDate="2016-08-20T16:41:27.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27497" PostTypeId="2" ParentId="27300" CreationDate="2015-03-18T12:50:17.717" Score="3" Body="&lt;p&gt;A good starting point to read about drugs that can potentially remove plaque is &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3459726/&quot; rel=&quot;nofollow&quot;&gt;The prevention and regression of atherosclerotic plaques: emerging treatments&lt;/a&gt; Vasc Health Risk Manag. 2012; 8: 549–561. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there is is evidence that drugs can remove plaque, but the word &quot;liquify&quot; is probably not the best description.  Also, the removed plaque would not be especially excreted by the kidneys.  The main carbon compound in urine is urea, which has 2 nitrogen atoms per carbon atom, and the plaque doesn't have that much nitrogen.  If the plaque is excreted, it would be after metabolism and the carbon in the plaque would mainly be excreted as CO2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to the cited article, rather than liquify, the term &quot;regress&quot; is used:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is now known that plaques are able to regress. Plaque reversal occurs by removal of lipids and necrotic material, endothelial repair, or halt of vascular smooth muscle cell proliferation. Several mechanisms explain this reversal, such as high-density lipoprotein cholesterol (HDL-C) action, destruction of foam cells and macrophages in lymph nodes and restoration of endothelium by neighboring cells or circulating progenitors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reference is made to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3209544/&quot; rel=&quot;nofollow&quot;&gt;An integrated approach for the mechanisms responsible for atherosclerotic plaque regression&lt;/a&gt;. Exp Clin Cardiol. 2011;16(3):77–86 for further information.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the article goes through several studies on drugs that may remove plaque.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drug treatments discussed fall into the three broad catagories of &lt;a href=&quot;http://en.wikipedia.org/wiki/Statin&quot; rel=&quot;nofollow&quot;&gt;statins&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Phosphodiesterase_inhibitors&quot; rel=&quot;nofollow&quot;&gt;phosphodiesterase inhibitors&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Thiazolidinediones&quot; rel=&quot;nofollow&quot;&gt;thiazolidinediones&lt;/a&gt; and specfically are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rosuvastatin&quot; rel=&quot;nofollow&quot;&gt;rosuvastatin&lt;/a&gt; alone and in combination with &lt;a href=&quot;http://en.wikipedia.org/wiki/Ramipril&quot; rel=&quot;nofollow&quot;&gt;ramipril&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atorvastatin&quot; rel=&quot;nofollow&quot;&gt;atorvastatin&lt;/a&gt; alone or in combination with &lt;a href=&quot;http://en.wikipedia.org/wiki/Ezetimibe&quot; rel=&quot;nofollow&quot;&gt;ezetimibe&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pitavastatin&quot; rel=&quot;nofollow&quot;&gt;pitavastatin&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cilostazol&quot; rel=&quot;nofollow&quot;&gt;cilostazol&lt;/a&gt; alone or in combination with &lt;a href=&quot;http://en.wikipedia.org/wiki/Probucol&quot; rel=&quot;nofollow&quot;&gt;probucol&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pioglitazone&quot; rel=&quot;nofollow&quot;&gt;pioglitazone&lt;/a&gt; alone or in combination with a statin&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-18T13:09:22.090" LastActivityDate="2015-03-18T13:09:22.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27498" PostTypeId="1" CreationDate="2015-03-18T13:13:35.620" Score="2" ViewCount="668" Body="&lt;p&gt;I'm preparing a chemistry show where I end the show with yellow powder bang.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powder is a mixture of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K2CO3 + KNO3 + S} $$&#xA;My question is, what's the reaction between these chemicals?&lt;/p&gt;&#xA;" OwnerUserId="15035" LastEditorUserId="13409" LastEditDate="2015-03-23T08:38:59.023" LastActivityDate="2015-04-22T09:27:20.153" Title="What is the chemistry behind the reaction of KNO₃, K₂CO₃ and Sulfur?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27499" PostTypeId="1" CreationDate="2015-03-18T13:36:56.480" Score="2" ViewCount="641" Body="&lt;p&gt;I saw a Youtube video which showed that cotton (cellulose) dissolves in deep blue solution of tetramine copper made by adding 20% ammonia in copper carbonate salt. I tried to replicate the process using $\ce{CuSO4}$ and copper chloride. None worked. Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="7448" LastEditDate="2015-03-19T23:34:05.333" LastActivityDate="2015-03-19T23:34:05.333" Title="Does tetramine copper dissolve cellulose?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27500" PostTypeId="2" ParentId="27300" CreationDate="2015-03-18T14:09:10.437" Score="9" Body="&lt;p&gt;Plaques are continually being removed from arteries by natural mechanisms within the body.  Statin use, when combined with aggressive dietary changes, can slow down the rate of plaque deposition to the point that the rate of plaque removal is actually higher than the rate of plaque deposition, and, as a consequence, plaque thinning can be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing plaques faster (plaque liquefication) has some potential, life threatening drawbacks.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We've all seen how ice cubes, when frozen together into a clump and then placed in a water bath, dissolve.  Some ice cubes break off from the main structure and float away to dissolve separately.  The same thing can happen with a large plaque deposit.  If a piece of the plaque breaks off during the liquefication process and continues to travel through the blood stream it can lead to downstream blockages, particularly in the brain (stroke).  Rapid dissolution of plaques would increase the risk for such adverse events.&lt;/li&gt;&#xA;&lt;li&gt;While plaques form more or less everywhere as relatively uniform deposits, the real concern is areas where plaques have grown to large size and begin to block arteries.  It has been found that these larger deposits are initiated by events related to damage of the arterial wall at that specific location. The damage leads to infection (that's why &lt;a href=&quot;http://www.webmd.com/heart-disease/guide/heart-disease-c-reactive-protein-crp-testing&quot;&gt;highly sensitive C-reactive protein&lt;/a&gt; [hs-crp] measurements are routinely made in conjunction with cholesterol measurements).  The body mobilizes the immune system to fight the infection; microphages and white blood cells are directed to these infection sites.  Once this happens deposition of other materials (lipids, calcium, &lt;em&gt;etc&lt;/em&gt;.) start at the infection site and abnormal plaque growth results.  Because of the 1) initial damage, 2) the infection and 3) infection fighting, the arterial wall becomes abnormally thinned at the site.  To some degree, the plaque actually serves to stabilize the thin wall preventing wall rupture and leaks.  Rapid removal of the plaque would increase the chances for the wall to rupture,  producing adverse events and increasing mortality.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So while it is likely that drugs could be designed to remove the plaque extremely fast (&lt;em&gt;e.g&lt;/em&gt;. liquefication), the risks might far outweigh the benefits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;would the liquefied plaque be eliminated as regular fluid is through&#xA;  the kidneys?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A major component of plaque is lipoprotein.  Lipoprotein is transported by blood as insoluble particles of varying density and size.  Lipoprotein itself is an assembled complex of proteins and fats.  The fats are various esters of glycerol including esterified cholesterol.  The lipoprotein transports cholesterol and fats, &lt;em&gt;via&lt;/em&gt; the bloodstream to cells where these compounds are needed for biosynthesis. After the lipoprotein has released the transported molecules it returns to the liver where it is &quot;reloaded&quot; and sent out again. Lipoproteins are not eliminated (unless damaged) from the body, but are rather recycled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a large amount of lipoprotein were suddenly released into the body, I'm sure some would be eliminated through the kidneys.  Certainly the lipoprotein receptors would sense this event and stop synthesis of new lipoprotein until systemic balance was restored. it seems likely that adverse events like kidney damage, gallstones, bile duct clogging, &lt;em&gt;etc&lt;/em&gt;. would occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since statins were first released onto the market, drug companies have been considering where to go next.  I have to believe that plaque liquefication was among the options considered.  While the reduction of plaque build up (statins) have their share of problems, I would guess that these problems are minor compared to the possible problems that can result from liquefication (discussed above).  Indeed, with the recent announcements (&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/24422577&quot;&gt;see here for example&lt;/a&gt;) of monoclonal antibodies achieving plaque reductions superior to statins, it seems that drug companies are continuing to focus on avoiding the problem in the first place, rather than fixing things (liquefication) down the road. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-19T00:23:49.263" LastActivityDate="2015-03-19T00:23:49.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27501" PostTypeId="1" AcceptedAnswerId="27511" CreationDate="2015-03-18T14:55:24.040" Score="2" ViewCount="322" Body="&lt;p&gt;If a mixture of Mn$^{2+}$, Zn$^{2+}$, Co$^{2+}$, Ni$^{2+}$ and Al$^{3+}$ salts is dissolved in water, a white dim solution is obtained. I guess that the aluminium and zinc ions hydrolysed and the respective hydroxides are formed. Am I right, or maybe one of them won't hydrolyze so much that the solution becomes dim or maybe some other cation could have hydrolysed, too?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-03-18T17:47:46.067" LastActivityDate="2015-03-18T18:23:08.377" Title="3rd analytical group of cations" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;analytical-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27502" PostTypeId="2" ParentId="27485" CreationDate="2015-03-18T14:55:37.007" Score="4" Body="&lt;p&gt;The increased acidity of the protons on the alpha-carbon in these compounds, compared to the acidity of the protons in methane, is due to stabilization of the resultant negative charge through &lt;strong&gt;resonance&lt;/strong&gt; with the adjacent carbonyl.  In the top line of the following figure we show a ketone where the proton has been removed.  We can draw 2 resonance structures to describe this carbanion.  In structure II the negative charge has been delocalized onto the electronegative oxygen atom - a very favorable and stabilizing situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aoXGH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To whatever degree we remove the carbonyl's ability to resonance stabilize that adjacent charge we will decrease the acidity of the alpha-protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of both the ester (III) and amide (VI) we can draw a third resonance structure (V and VIII respectively).  To whatever extent these resonance structures contribute to the overall description of the ester and the amide, the carbonyl will be less available to stabilize the negative charge through structures IV and VII.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that resonance structures V and VIII are significant contributors, with the amide structure VIII having a positive charge on nitrogen being more important than resonance structure V which places the positive charge on the more electronegative oxygen (less favorable).  In fact, resonance structure VIII is such a significant contributor that restricted rotation occurs about the $\ce{C-N}$ bond and we can typically observe two different R group signals in the proton-nmr spectrum of amides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as stated earlier, as the contribution of the third resonance structure increases, the contribution of the second structure (the structure that stabilizes our carbanion) will decrease making our compound less acidic.  The ketone doesn't have a third structure so it will be the most acidic structure. Structure VIII is more important than structure V, so the ester will be more acidic than the amide.  Therefore, the overall order of decreasing acidity is&#xA;$$\ce{CH3COCH3 &amp;gt; CH3CO2CH3 &amp;gt; CH3CON(CH3)2}$$&#xA;just as you noted.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-18T14:55:37.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27503" PostTypeId="1" AcceptedAnswerId="27505" CreationDate="2015-03-18T15:06:14.830" Score="3" ViewCount="634" Body="&lt;p&gt;Can somebody explain as concrete as possible, how the change of &lt;strong&gt;temperature&lt;/strong&gt; and the change of &lt;strong&gt;pH&lt;/strong&gt;  leads to deviations of this law?&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorDisplayName="user7951" LastEditDate="2015-04-18T15:06:18.347" LastActivityDate="2015-04-18T15:06:18.347" Title="Lambert Beer's law deviations" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27505" PostTypeId="2" ParentId="27503" CreationDate="2015-03-18T15:54:55.957" Score="4" Body="&lt;p&gt;A compound &lt;strong&gt;S&lt;/strong&gt; follows the Lambert-Beer law if the absorbance $E_\lambda$ &lt;strong&gt;at a particular wavelength $\lambda$&lt;/strong&gt; is proportional to the concentration $c$ of &lt;strong&gt;S&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[E_\lambda = \left(\frac{I_0}{I} \right) = \epsilon\cdot c \cdot d \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;deviation&quot; is observed when the &lt;strong&gt;real&lt;/strong&gt; concentration $c$ is different from the expected/assumed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may happen if&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;S&lt;/strong&gt; is an acid or a base: &#xA;&lt;ul&gt;&#xA;&lt;li&gt;variation of the pH leads to (de)protonation and forms another species with a completely different absorption spectrum, or at least a different $\epsilon$ at the wavelength under observation&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;S&lt;/strong&gt; undergoes a reaction to &lt;strong&gt;S'&lt;/strong&gt; (with different optical properties) at elevated temperatures (or change of pH)&lt;/li&gt;&#xA;&lt;li&gt;Decrease of temperature or change of pH leads to the formation of a turbid solution from which light is scattered into the detector &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-18T15:54:55.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27506" PostTypeId="2" ParentId="27499" CreationDate="2015-03-18T16:15:09.360" Score="1" Body="&lt;p&gt;You might want to check the pH - you would have a less basic solution using copper sulfate or chloride than you would get from copper carbonate, and that could be less effective in dissolving the cotton. Also, your particular sample of cotton might behave differently from others, such as that used in the video.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-18T16:15:09.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27508" PostTypeId="1" CreationDate="2015-03-18T17:10:06.637" Score="0" ViewCount="36" Body="&lt;p&gt;I've been taught that if you have two aqueous reactants that react together and produce heat it is exothermic. Shouldn't that be endothermic? Did my professor misspeak?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's gaining heat, that means it's pulling heat from the outside. The opposite of what happens when water freezes and releases heat (which is actually exothermic).&lt;/p&gt;&#xA;" OwnerUserId="14965" LastActivityDate="2015-03-18T17:10:06.637" Title="Why is it that reactions that cause an increase of temperature are exothermic?" Tags="&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-03-18T19:12:27.023" ContentLicense="CC BY-SA 3.0" />
  <row Id="27509" PostTypeId="2" ParentId="27460" CreationDate="2015-03-18T17:27:04.823" Score="3" Body="&lt;p&gt;I suspect the answer is that AlGaAs is an alloy with a bandgap between that of AlAs - 2.16 (which is indirect) and GaAs - 1.42 eV (which is direct). If there is enough aluminum, the resulting semiconductor is an indirect bandgap semiconductor, which means that to emit a photon, it must be coupled with a lattice phonon, greatly reducing the probability of photon emission, which would make a AlGaAs LED of short enough wavelength rather inefficient at emitting light, instead producing a lot of heat rather than light.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-03-18T17:27:04.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27510" PostTypeId="1" AcceptedAnswerId="27512" CreationDate="2015-03-18T17:57:16.880" Score="2" ViewCount="403" Body="&lt;p&gt;It can form a square planar and a tetrahedral complex but within square planar or within tetrahedral can there be isomers possible?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-03-18T22:51:16.513" Title="what are possible isomers of (meso-stilbenediamine)(isobutylenediamine) palladium (II)?" Tags="&lt;coordination-compounds&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27511" PostTypeId="2" ParentId="27501" CreationDate="2015-03-18T18:16:24.820" Score="3" Body="&lt;p&gt;Aluminium and zinc will produce a white precipitate due to formation of hydroxide. Both these hydroxides dissolve in excess acid or alkali due to their amphoteric character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manganese, nickel, cobalt hydroxides are probably soluble as they are precipitated in the 4th group as sulphides.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="13357" LastEditDate="2015-03-18T18:23:08.377" LastActivityDate="2015-03-18T18:23:08.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27512" PostTypeId="2" ParentId="27510" CreationDate="2015-03-18T22:23:18.897" Score="4" Body="&lt;p&gt;The use of the term &quot;meso&quot; in the compound name indicates that both phenyl groups are on the same side of the molecule, &quot;&lt;em&gt;cis&lt;/em&gt;&quot; to one another, not &quot;&lt;em&gt;trans&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the tetrahedral and square planar representations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsUpZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I used $\ce{Pt}$ instead of $\ce{Pd}$ as the central metal, but it makes no difference in terms of isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tetrahedral molecule has a plane of symmetry containing the &quot;top&quot; 5-membered ring, bisecting the 2 methyl groups on that ring and then bisecting the &quot;bottom&quot; 5-membered ring with the 2 &quot;meso&quot; phenyl groups. Since the tetrahedral molecule contains a plane of symmetry it is achiral and has no isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the square planar form, there is no symmetry element.  Therefore the square planar molecule is chiral and exists as an enantiomeric pair of isomers.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-18T22:51:16.513" LastActivityDate="2015-03-18T22:51:16.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27514" PostTypeId="2" ParentId="27499" CreationDate="2015-03-19T00:05:11.260" Score="3" Body="&lt;p&gt;As @iad22agp states, it is not sufficiently alkaline. Schweizer's reagent is tetraamminediaquacopper dihydroxide, $\ce{[Cu(NH3)4(H2O)2](OH)2}$. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Schweizer%27s_reagent&quot; rel=&quot;nofollow&quot;&gt;Schweizer's reagent&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;em&gt;not&lt;/em&gt; Doktor Schweizer, I presume.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-19T00:05:11.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27515" PostTypeId="2" ParentId="5568" CreationDate="2015-03-19T01:02:15.130" Score="1" Body="&lt;p&gt;Both exergonic and exothermic reactions release energy, however, the energies released have different meanings as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Exothermic reaction&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy released is just called energy &lt;/li&gt;&#xA;&lt;li&gt;Energy of reactants is greater than that of products&lt;/li&gt;&#xA;&lt;li&gt;Energy of the reaction system decreases relative to that of the    surounding, i.e. the surrounding becomes hotter.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exergonic reaction&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy released, has a special name called Gibbs energy or Gibbs free   energy&lt;/li&gt;&#xA;&lt;li&gt;Energy reactants is greater than that of the products&lt;/li&gt;&#xA;&lt;li&gt;It has nothing to do with how hot or cold reactants become. Has a more chemical meaning - it relates to the spontaneity of the reaction; thus it always means that a reaction is feasible, i.e. reaction will always happen.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary, whereas, an exergonic reaction means that a reaction is spontaneous, an exothermic reaction has nothing to do with spontaneity, but that an energy is released to the surrounding.&lt;/p&gt;&#xA;" OwnerUserId="15040" LastEditorUserId="4945" LastEditDate="2018-10-22T08:31:47.793" LastActivityDate="2018-10-22T08:31:47.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27517" PostTypeId="1" CreationDate="2015-03-19T02:21:12.120" Score="3" ViewCount="161" Body="&lt;p&gt;Alright, so Tin ions in most of it's salts have the propensity to form Tin(II,IV) Hydroxides and then precipitating out of aqueous solutions. What can I throw in the solutions to prevent this? Using Tin Fluoroborate doesn't precipitate Hydroxides, but I don't have access Fluoroborates. I really don't want very acidic solutions, either, for safety purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any complexes out there that does the job?&lt;/p&gt;&#xA;" OwnerUserId="15043" LastEditorUserId="4945" LastEditDate="2015-03-19T06:35:43.193" LastActivityDate="2015-03-19T06:35:43.193" Title="How to prevent Tin ions from precipitating as Tin(II,IV) Hydroxides" Tags="&lt;acid-base&gt;&lt;solutions&gt;&lt;precipitation&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27522" PostTypeId="1" CreationDate="2015-03-19T04:21:38.980" Score="3" ViewCount="338" Body="&lt;p&gt;I'm looking to create a light up dancing pole. This requires a &lt;strong&gt;translucent&lt;/strong&gt; material that is strong enough to support the weight and force of a human being, even when it's hollowed out to so LEDs can go inside.  The material must also be reasonably affordable, so something like diamond is out of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've discovered a couple materials during my research (please recommend any I'm leaving out):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;acrylic (I know there are acrylic poles out there, but I'm not sure&#xA;if they're able to support a human when hollowed out) &lt;/li&gt;&#xA;&lt;li&gt;polycarbonate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To take the example of polycarbonate, how much weight can it support as a function of its thickness/diameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Also, if the editors feel like this is an inappropriate question I'd just like to point out that the weight that different materials can support is a very appropriate question here.  My intent is not to scandalize, only to build a dance pole and learn some Chemistry in the process.  Thank you]&lt;/p&gt;&#xA;" OwnerUserId="15047" LastEditorUserId="15047" LastEditDate="2015-03-19T04:37:48.770" LastActivityDate="2015-03-19T04:37:48.770" Title="Chemistry of Stripper Poles: How much weight can different materials support?" Tags="&lt;materials&gt;&lt;plastics&gt;" AnswerCount="0" CommentCount="6" ClosedDate="2015-03-19T15:06:35.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="27524" PostTypeId="1" AcceptedAnswerId="27539" CreationDate="2015-03-19T05:22:07.127" Score="1" ViewCount="877" Body="&lt;p&gt;Recently I read about Hyperconjugation. But i have a doubt. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/fiaF7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the need of hyperconjugation in this structure when there is charge separation resulting in lesser stable structure.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="16683" LastEditDate="2017-03-09T22:15:18.920" LastActivityDate="2017-03-09T22:15:18.920" Title="Need of hyperconjugation?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;hyperconjugation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27526" PostTypeId="1" CreationDate="2015-03-19T05:53:04.450" Score="-1" ViewCount="357" Body="&lt;p&gt;why chelates which form 3 or 4 member rings or 6 or 7 member rings are as stable as 5 member ring chelates? &#xA;for example if we measure stability of chelates with oxygen donors, say oxalate, malonate, succinate etc oxalate chelated with metal shoes more stability. why malonate , succinate as chelating ligands are less stable?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-04-18T07:40:27.637" Title="Why member ring is favoured in chelates?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27527" PostTypeId="2" ParentId="27526" CreationDate="2015-03-19T06:54:08.407" Score="1" Body="&lt;p&gt;In its &lt;strong&gt;general&lt;/strong&gt; form, your statement on the stability of these complexes is wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Be(II)}$-malonato complexes are more stable than the corresponding oxalato complexes, cf. &lt;a href=&quot;http://dx.doi.org/10.1016/S0020-1693(97)05520-5&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Preparative, potentiometric and NMR studies of the interaction of beryllium(II) with oxalate and malonate&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-19T06:54:08.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27528" PostTypeId="1" AcceptedAnswerId="27530" CreationDate="2015-03-19T06:58:24.820" Score="1" ViewCount="3562" Body="&lt;p&gt;If a reaction is endothermic, then the reaction must take energy from the environment to occur (as far as I know), and if this is the case, then to me the reaction seems energetically unfavorable, as there is only so much energy you can get from the environment. Does this means all naturally occurring reactions, (that don't just happen once in a blue moon but at a steady rate) are exothermic? This seems pretty silly, so if not, why not?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="10185" LastEditDate="2015-03-19T18:03:40.630" LastActivityDate="2017-05-10T19:15:37.523" Title="Are all naturally occuring reactions exothermic?" Tags="&lt;energy&gt;&lt;enthalpy&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27529" PostTypeId="1" CreationDate="2015-03-19T07:02:26.693" Score="2" ViewCount="84" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose I am given following reaction in a vessel:&#xA;  $$\ce{NH4HS(s) &amp;lt;=&amp;gt; NH3(g) + H2S(g)}$$&#xA;  $P_\ce{NH3} = P_\ce{H2S} = \pu{1 atm}$ at equilibrium; if volume is doubled at this instant, then what would be $P_\ce{NH3}$ at new equilibrium? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that as soon as volume is doubled, pressure would drop to half at this instant, and $K_\mathrm{c}$ will have the same value at new equilibrium, but how to actually use this to calculate the partial pressure of ammonia at new equilibrium? &lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorUserId="41328" LastEditDate="2017-09-27T12:56:03.353" LastActivityDate="2017-09-27T12:56:03.353" Title="Partial pressure and chemical equilibirum" Tags="&lt;inorganic-chemistry&gt;&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27530" PostTypeId="2" ParentId="27528" CreationDate="2015-03-19T07:09:20.830" Score="5" Body="&lt;p&gt;It's not necessary that all naturally occurring reactions are exothermic. The definition of Gibbs free energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = \Delta H - T\Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spontaneous reaction needs to have $\Delta G&amp;lt;0$, and therefore for spontaneous endothermic reactions to occur we only need that $\Delta H &amp;lt; T\Delta S$.&lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorUserId="16683" LastEditDate="2017-05-10T19:14:30.333" LastActivityDate="2017-05-10T19:14:30.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27531" PostTypeId="2" ParentId="27488" CreationDate="2015-03-19T07:30:07.690" Score="7" Body="&lt;p&gt;This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &amp;sigma; orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a &amp;pi; orbital the situation is different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we can see that for &amp;pi; orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In very short terms:&lt;/em&gt; In &amp;sigma; orbitals the maximum electron density is on the bonding axis, while in &amp;pi; (and &amp;delta;) orbitals it is off the bonding axis.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-03-19T07:30:07.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27532" PostTypeId="1" AcceptedAnswerId="31782" CreationDate="2015-03-19T09:22:10.377" Score="35" ViewCount="3361" Body="&lt;p&gt;I realise, that this question is a stretch, but I was wondering, how would a bonding orbital be called if it was formed from two $f_{x(x^2−3y^2)}$ or $f_{y(3x^2−y^2)}$ orbitals. Have there been any suggestions on this, was it anywhere proposed or discussed? I am not arguing about the existence of such a thing, but as a thought experiment, how would it be called? Phi, $\varphi$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also throwing in a reference request here, since I am interested in any literature related to the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A $\sigma$ orbital is an orbital, that has no nodal plane between the bonding partners. It is $C_\infty$ symmetric with respect to the bonding axis (in first approximation - of course $d_{xy}$ orbitals can align in a $\sigma$ fashion, too, but for the sake of the argument, let's not consider this.) The point group of this orbital can further be assumed as $D_\mathrm{\infty h}$. It shares therefore most features of an $s$ orbital, hence $s$igma, $\sigma$.&lt;br&gt;&#xA;The most prominent example for this is probably the dihydrogen, $\ce{H2}$, molecule.&lt;/li&gt;&#xA;&lt;li&gt;A $\pi$ orbital has one nodal plane and the bonding axis of the involved atoms is part of this plane. It is therefore asymmetric, $C_1$, with respect to this plane. However, it is $C_\mathrm{s}$ symmetric with respect to the plane that is perpendicular, also sharing the bonding axis. The orbital has further the point group $C_\mathrm{2v}$. It has therefore most features of a $p$ orbital, hence $p$i, $\pi$.&lt;br&gt;&#xA;The most prominent examples for $\pi$ bonds are probably ethylene, $\ce{C2H4}$, and acetylene, $\ce{C2H2}$.&lt;/li&gt;&#xA;&lt;li&gt;A $\delta$ orbital has two nodal planes between the bonding partners. These are perpendicular to each other. The bond orbital is asymmetric with respect to both of them. As a result, it is $C_\mathrm{s}$ symmetric with respect to the bisecting mirror planes. The point group of this orbital is $D_\mathrm{2h}$. Therefore it has shares features with a $d$ orbital, hence $d$elta, $\delta$.&lt;br&gt;&#xA;One example is the $\ce{[Re2Cl8]^{2-}}$ ion, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Quadruple_bond&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's quadruple bond&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I imagine two $f$ orbitals aligning with the bonding axis being simultaneously a $C_3$ symmetry axis, giving it an overall $D_\mathrm{3h}$ point group. Since there is no greek letter starting with an $f$, I think the logical thing would be to choose $\phi$ as it sounds a like. (I cannot be the only one thinking that.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The following graphic should somewhat demonstrate what I mean:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/El7QH.jpg&quot; alt=&quot;sigma, pi, delta, and phi (?) linear combinations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit, that I have not done much research about this. It was more like an impulsive question for me to ask.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2021-01-06T18:05:59.507" Title="What would follow in the series sigma, pi and delta bonds?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;orbitals&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27533" PostTypeId="2" ParentId="27529" CreationDate="2015-03-19T09:24:21.353" Score="1" Body="&lt;p&gt;No, at new equilibrium $K_c$ would actually is the same value. This constant stands for reaction between these two compounds no matter the pressure, it is only affected by temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As ammonium hydrosulfide is a solid, the reaction is only relevant at the surface of the crystals, and the &quot;concentration&quot; of ammonium hydrosulfide at the relevant areas will remain constant. Therefore same partial pressure of both ammonia and hydrogen sulfide is needed to maintain the equilibrium. Hence, at equilibrium, the partial pressure of both ammonia and hydrogen sulfide should be the same as the initial partial pressure.&lt;/p&gt;&#xA;" OwnerUserId="14983" LastActivityDate="2015-03-19T09:24:21.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27534" PostTypeId="2" ParentId="27498" CreationDate="2015-03-19T10:18:05.557" Score="2" Body="&lt;p&gt;Here's just a guess, but I do have some confidence that I'm right:&lt;br/&gt;&#xA;$\ce{KNO3 + S -&amp;gt; K2SO4 + K2SO3 + K2S + K2O + SO2 + N2}$&lt;br/&gt;&lt;br/&gt;&#xA;$\ce{K2CO3 -&amp;gt;[\Delta] K2O + CO2}$&lt;br/&gt;&#xA;The first, unbalanced equation is the reaction between potassium nitrate and sulfur, producing quite some gas as product, and a variety of potassium compounds left behind, with the composition depending on your formulation of the powder. Of course there might be some sulfur trioxide produced, but the major oxide of sulfur should be the dioxide. In the second equation is the thermal decomposition of potassium carbonate releasing carbon dioxide. This is the part that I'm not very sure of, but it may be to release more gas to give your reaction more puff. However, IIRC the decomposition requires a very high temperature. In contrast, bicarbonate has a much lower decomposition temperature and also release more gas(carbon dioxide and water), but the products would likely be absorbed by potassium oxide to form hydroxide and carbonate, so  bicarbonate might not help much.&lt;/p&gt;&#xA;" OwnerUserId="14983" LastActivityDate="2015-03-19T10:18:05.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27535" PostTypeId="1" AcceptedAnswerId="27571" CreationDate="2015-03-19T11:27:21.523" Score="3" ViewCount="90752" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the ionic character of a bond, $\ce{A-B}$, in terms of the electronegativities of $\ce{A}$ and $\ce{B}$ ($\chi_\ce{A}$ and $\chi_\ce{B}$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have been taught that the percentage ionic character is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\text{observed value of ionic character}}{\text{calculated value of character}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I can't understand how electronegativity is used here. I couldn't find anything on the internet either.&lt;/p&gt;&#xA;" OwnerUserId="13418" LastEditorUserId="5026" LastEditDate="2018-03-23T11:38:05.840" LastActivityDate="2020-09-22T13:26:46.417" Title="Percentage ionic character when electronegativity is given" Tags="&lt;covalent-compounds&gt;&lt;electronegativity&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27536" PostTypeId="1" AcceptedAnswerId="27537" CreationDate="2015-03-19T11:49:48.267" Score="3" ViewCount="560" Body="&lt;p&gt;The title's pretty much it. At school, we're taught that ionic substances are composed of an infinitely-repeating lattice, with atoms at fixed angles from other atoms. How is this possible when I can get an individual grain of salt? Clearly if there's an individual grain, it can't be infinitely repeating.&lt;/p&gt;&#xA;" OwnerUserId="15033" LastEditorUserId="10185" LastEditDate="2015-09-26T21:13:46.580" LastActivityDate="2015-09-26T21:13:46.580" Title="How can ionic substances have an infinitely-repeating lattice structure?" Tags="&lt;crystal-structure&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27537" PostTypeId="2" ParentId="27536" CreationDate="2015-03-19T12:35:33.427" Score="9" Body="&lt;p&gt;&lt;strong&gt;To be technically accurate, it's &quot;practically&quot; infinite.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not &lt;em&gt;infinite&lt;/em&gt; since as you mention a grain of salt (or even a large crystal) has some finite size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to remember the actual atomic dimensions $10^{-10} \:\mathrm{m}$. So while it's not truly infinite, there are an incredibly large number of atoms in a crystal.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-19T12:35:33.427" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27538" PostTypeId="1" CreationDate="2015-03-19T13:44:23.867" Score="3" ViewCount="3878" Body="&lt;p&gt;I am teaching my students how to write the electronic configuration of the elements of the periodic table.&lt;/p&gt;&#xA;&lt;p&gt;For example, following the rules on how to fill the orbitals, the electronic configuration of iron is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe}:~\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(4s)^2(3d)^6}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus, its abbreviated form is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe}:~[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, I has not been easy to find the explanation on why in any periodic table it is written as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;instead of&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Same case for many other elements. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="15054" LastEditorUserId="41328" LastEditDate="2020-12-07T06:28:02.890" LastActivityDate="2020-12-07T06:39:02.337" Title="Abbreviations in electronic configurations" Tags="&lt;electronic-configuration&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27539" PostTypeId="2" ParentId="27524" CreationDate="2015-03-19T14:13:19.540" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the need of hyper conjugation in this structure when there is&#xA;  charge separation resulting in lesser stable structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think we'd all agree that the charge separated structure you've drawn is a resonance structure.  Charge has been separated so it may well be higher in energy than the first (neutral) resonance structure.  Nonetheless, to whatever extent the charge separated resonance structure contributes to the description of propene - even if it is just a small amount - it will stabilize propene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are 3 hydrogens on the methyl group you can draw three identical resonance structures, this will increase the importance of the structure.  Note too the charge separated resonance structures involve an allyl carbanion (resonance stabilized) and a proton; two very stable structures - this will further increase the weight, or importance, of the hyperconjugated structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These hyperconjugated resonance structures are really showing us that an $\ce{sp^3}$ carbon can donate electron density to an $\ce{sp^2}$ hybridized carbon.  Electron movement in this direction lowers the energy of the electron because it moves the electron from an orbital that is 25% s-character to an orbital that is 33% s-character.  The more s-character in an orbital, the lower the energy of the orbital (an s orbital is lower in energy than a p orbital).  As we stabilize the electron, we stabilize the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This hyperconjugation argument comes in handy.  It can be used to explain&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;why double bonds with more alkyl groups are more stable than double bonds with fewer alkyl groups (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/27346/why-is-2-methylpropene-less-in-energy-than-its-alkene-counterparts/27347#27347&quot;&gt;ref1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19554/why-are-double-bonds-at-more-substituted-carbon-more-stable/19571#19571&quot;&gt;ref2&lt;/a&gt;,)&lt;/li&gt;&#xA;&lt;li&gt;why carbocation stability follows the order: tertiary &gt; secondary &gt; primary (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19027/inductive-effect-and-hyperconjugation-one-elephant-different-parts/19032#19032&quot;&gt;ref1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8412/organic-chemistry-hyper-conjugation/8420#8420&quot;&gt;ref2&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;why alkyl groups are ortho-para directing in electrophilic aromatic substitution (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19508/why-methyl-group-is-2-4-directing/19511#19511&quot;&gt;ref&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-03-20T01:38:13.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27540" PostTypeId="2" ParentId="6460" CreationDate="2015-03-19T14:25:34.373" Score="16" Body="&lt;p&gt;Aromatic cations like the tropylium cation (&lt;span class=&quot;math-container&quot;&gt;$\ce{C7H7+}$&lt;/span&gt;)&lt;sup&gt;[1]&lt;/sup&gt; or the cyclopropenyl cation (&lt;span class=&quot;math-container&quot;&gt;$\ce{C3H3+}$&lt;/span&gt;)&lt;sup&gt;[2]&lt;/sup&gt; can coordinate to metals, creating &lt;a href=&quot;http://en.wikipedia.org/wiki/Sandwich_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;sandwich&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Half_sandwich_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;half-sandwich compounds&lt;/a&gt;. The positive charge is shared by all carbon atoms of the rings.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Anderson, J. E.; Maher, E. T.; Kool, L. B. Electrochemical and spectroelectrochemical properties of the titanium sandwich complexes (Cp)Ti(L) and (Cp*)Ti(L), where L is either cyclooctatetraene (C&lt;sub&gt;8&lt;/sub&gt;H&lt;sub&gt;8&lt;/sub&gt;) or the tropylium ion (C&lt;sub&gt;7&lt;/sub&gt;H&lt;sub&gt;7&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt;). &lt;em&gt;Organometallics&lt;/em&gt; &lt;strong&gt;1991,&lt;/strong&gt; &lt;em&gt;10&lt;/em&gt; (5), 1248–1254. &lt;a href=&quot;https://doi.org/10.1021/om00051a009&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/om00051a009&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lichtenberger, D. L.; Hoppe, M. L.; Subramanian, L.; Kober, E. M.; Hughes, R. P.; Hubbard, J. L.; Tucker, D. S. Electron distribution and bonding in η&lt;sup&gt;3&lt;/sup&gt;-cyclopropenyl-metal complexes. &lt;em&gt;Organometallics&lt;/em&gt; &lt;strong&gt;1993,&lt;/strong&gt; &lt;em&gt;12&lt;/em&gt; (6), 2025–2031. &lt;a href=&quot;https://doi.org/10.1021/om00030a011&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/om00030a011&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="16683" LastEditDate="2018-12-26T23:12:16.857" LastActivityDate="2018-12-26T23:12:16.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27541" PostTypeId="1" AcceptedAnswerId="27545" CreationDate="2015-03-19T14:34:11.687" Score="4" ViewCount="1554" Body="&lt;p&gt;I'm working on a science fiction short story and was wondering if there were any liquids that would be safe to bathe in while traveling on a space craft with artificial (spin-based) gravity.  In such a context hydrogen and oxygen are valuable resources not to be wasted.  While bath water could be recycled, it is not available for other uses while someone is sitting in it, so I was wondering if there are other liquids at Standard Temperature Pressure that would be safe to bathe in (potentially ingesting small amounts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of using carbon bucky-ball fullerenes in a fluid but the viscosity looks to be much too high (essentially tar-like).  I know there are waterless chemicals that can be used for cleaning (e.g. DryBath), the purpose of this is more to have a hot-tub to relax in than to disinfect and clean.  If the addition of some hydrogen and oxygen (either as H2O or in other compounds) can improve the viscosity or is necessary to form the liquid, that is acceptable as long as less than 10% of the liquid is comprised of hydrogen and oxygen.  The purpose is to save water and oxygen for air while still being able to soak in a hot-tub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately most oils contain a substantial amount of hydrogen and some oxygen, so they don't qualify unless the amount of carbon offers substantial savings in volume.  I'd prefer something without any hydrogen or oxygen if such a chemical exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first post here and I'm unfamiliar with the tags, so feel free to edit or add any you feel are appropriate.&lt;/p&gt;&#xA;" OwnerUserId="15055" LastActivityDate="2015-03-19T18:44:20.927" Title="Are there any STP liquids not containing large amounts of hydrogen and oxygen that are safe for human contact?" Tags="&lt;applied-chemistry&gt;&lt;viscosity&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27543" PostTypeId="2" ParentId="6460" CreationDate="2015-03-19T14:36:37.597" Score="13" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitroso&quot; rel=&quot;nofollow noreferrer&quot;&gt;nitroso&lt;/a&gt; ligand (&lt;span class=&quot;math-container&quot;&gt;$\ce{NO^+}$&lt;/span&gt;) is also one of the most commonly occurring positively charged ligands. &lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorUserId="16683" LastEditDate="2018-12-26T23:15:18.910" LastActivityDate="2018-12-26T23:15:18.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27544" PostTypeId="2" ParentId="27538" CreationDate="2015-03-19T14:43:08.687" Score="1" Body="&lt;p&gt;In some textbooks that I have read, I came across electron configurations that listed the 3d electrons before the 4s electrons.  I believe it is to keep the electrons with the same principal quantum number &lt;span class=&quot;math-container&quot;&gt;$(n)$&lt;/span&gt; together.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$&lt;/span&gt; in full would be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(3d)^6(4s)^2}.$&lt;/span&gt;  When I teach general chemistry, I opt to list the sub-shells in order of increasing energy instead, since it correlates with the structure of the periodic table.&lt;/p&gt;&#xA;" OwnerUserId="14596" LastEditorUserId="41328" LastEditDate="2020-12-07T06:39:02.337" LastActivityDate="2020-12-07T06:39:02.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27545" PostTypeId="2" ParentId="27541" CreationDate="2015-03-19T14:50:11.723" Score="7" Body="&lt;p&gt;Perfluorocarbons would be a possibility, $\ce{(-CF2-)_{n}}$, they contain only carbon and fluorine and are relatively low cost and widely available in large quantities.  They are inert (&lt;em&gt;e.g&lt;/em&gt;. non-reactive) and depending upon chain length and structure, you could pick one that can be warmed up nicely to a bath temperature.  They are also used industrially in degreasing operations, so they might have some cleaning ability as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some perflurocarbons are actually being investigated as &lt;a href=&quot;http://en.wikipedia.org/wiki/Blood_substitute&quot; rel=&quot;nofollow noreferrer&quot;&gt;human blood substitutes&lt;/a&gt; - they can transport and release oxygen to tissues.  Perfluorodecalin is a prime example. As shown below, it is basically a naphthalene ring in which all of the hydrogens and double bonds have been replaced with fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uZjvR.png&quot; alt=&quot;PFDecalin image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ingesting small amounts would not be a problem. In fact, these same materials have been used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_breathing&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid breathing&lt;/a&gt; experiments.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of other scientific aspects of these perfluorocarbons that could play into science fiction.  For example, the small size of the molecules (compared to say, haemoglobin) allows it to travel to smaller places that haemoglobin cannot access due to its larger size. Perhaps the perfluorcarbon could also carry small drugs to relax, rejuvenate, repair body parts, or enhance sensory experiences during the bath. Since these perfluorocarbons act as oxygen transport agents they could also be worked into the story as a bath medium that also scavenges &quot;loose&quot; oxygen.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-19T16:12:58.050" LastActivityDate="2015-03-19T16:12:58.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27546" PostTypeId="1" CreationDate="2015-03-19T14:58:52.587" Score="7" ViewCount="174" Body="&lt;p&gt;By this, I mean if I buy compressed dry air from somebody like British Oxygen how much variation in composition might I expect between batches? For example, is Argon evenly distributed globally, by height, by geography etc? Is $\ce{CO2}$ removed? Are contaminants filtered out eg $\ce{NOx}$?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorUserId="4231" LastEditDate="2015-03-19T17:43:40.127" LastActivityDate="2015-03-19T17:43:40.127" Title="How chemically homogeneous is air?" Tags="&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27547" PostTypeId="1" CreationDate="2015-03-19T15:11:52.493" Score="2" ViewCount="84" Body="&lt;p&gt;Is there any paper making alternative? What is there in trees that makes it the only thing from which paper can be made? Do we even need paper anymore with all electronic devices and stylus', etc&lt;/p&gt;&#xA;" OwnerUserId="15057" LastActivityDate="2015-03-19T17:49:54.037" Title="Paper making alternative" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27548" PostTypeId="1" CreationDate="2015-03-19T15:23:59.387" Score="1" ViewCount="1113" Body="&lt;p&gt;I understand why it decreases for most of the gases but i cannot understand why it increases after reaching a certain minimum.&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="10185" LastEditDate="2015-03-19T18:01:23.280" LastActivityDate="2015-03-19T18:01:23.280" Title="Why does the graph of compressibility factor(z) vs pressure increase after reaching the minimum?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27549" PostTypeId="2" ParentId="168" CreationDate="2015-03-19T15:33:07.903" Score="11" Body="&lt;p&gt;Epoxides are stable because, first and foremost, they are ethers. Ethers are an exceptionally unreactive functional group. Why? Because alkoxides are poor leaving groups that do not participate in many common nucleophile-electrophile reactions (e.g., SN2). Of course, there are ways to force ethers to react, such as through an SN1 pathway utilizing strong acid catalysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the ethers, epoxides are (among) the most reactive due to the ring strain, which enables chemists to perform reactions with epoxides that do not work with other ethers. For example, the reaction of an epoxide with a Grignard reagent is one of the first carbon-carbon bond forming reactions introduced in introductory organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it's for this reason, their early importance in organic chemistry education, that epoxides have a reputation as exceptionally reactive. However, most epoxides are very readily prepared, isolated through aqueous workup, and purified on silica gel with no deleterious effects. In fact, epoxides can be survive being subjected to strong nucleophiles if a more reactive group is present in the molecule. For example, an ester can be reduced with DIBAL in the presence of an epoxide, and recently I was able to carry out a synthesis that used butyllithium in the presence of an unhindered epoxide with no trouble.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-03-19T15:33:07.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27550" PostTypeId="1" CreationDate="2015-03-19T15:39:24.123" Score="1" ViewCount="805" Body="&lt;p&gt;Pressure is the magnitude (absolute value) of the force over the area, so how can that result in a negative value for pressure as would happen in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure#Negative_pressures&quot; rel=&quot;nofollow&quot;&gt;certain circumstances&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="15059" LastEditorDisplayName="user7951" LastEditDate="2016-11-21T14:11:56.710" LastActivityDate="2016-11-21T18:51:59.283" Title="How can you achieve negative pressure?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;pressure&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27551" PostTypeId="1" CreationDate="2015-03-19T15:47:54.490" Score="5" ViewCount="123" Body="&lt;p&gt;Let's say I have a sample of 2 (assumingly different) diatomics A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through spectroscopy I found the data below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For molecule A:&#xA;$$\begin{align}&#xA;B&amp;amp;\approx2.17690\ \mathrm{cm^{−1}}\\&#xA;D&amp;amp;\approx4.79000\times10^{-5}\ \mathrm{cm^{−1}}\\&#xA;I&amp;amp;=1.28590\times10^{-46}\ \mathrm{kg\ m^2}\\&#xA;\tilde ν_e&amp;amp;=928.15\ \mathrm{cm^{−1}}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For molecule B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;B&amp;amp;\approx1.51695\ \mathrm{cm^{−1}}\\&#xA;D&amp;amp;\approx7.1049\times10^{-6}\ \mathrm{cm^{−1}}\\&#xA;I&amp;amp;=1.84533\times10^{-46}\ \mathrm{kg\ m^2}\\&#xA;\tilde ν_e&amp;amp;=1401.86\ \mathrm{cm^{−1}}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through Mass Spec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$. How do I deduce which is which and further, how do I determine the bond length. [FYI, I'm thinking the former mass is the molecule NO and the latter NaF just by guessing]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that the heavier molecule will have the larger inertia value. And from the inertia equation $I=\mu R^2$ I'll derive the bond length $R$. but how do I find the $\mu$ from the actually mass of the molecule from the mass spec data? &lt;/p&gt;&#xA;" OwnerUserId="9581" LastEditorDisplayName="user7951" LastEditDate="2016-10-13T06:34:26.220" LastActivityDate="2019-06-21T21:17:43.603" Title="Finding identities of diatomics from Spectroscopy and Mass Spec Data" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27553" PostTypeId="2" ParentId="27538" CreationDate="2015-03-19T16:05:50.227" Score="3" Body="&lt;p&gt;I would recommend following the &lt;a href=&quot;http://physics.nist.gov/PhysRefData/DFTdata/configuration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;convention used by NIST&lt;/a&gt; which is to list the subshells in order of principle quantum number &lt;span class=&quot;math-container&quot;&gt;$(n),$&lt;/span&gt; and in order of angular momentum quantum number &lt;span class=&quot;math-container&quot;&gt;$(\ell)$&lt;/span&gt; for a given &lt;span class=&quot;math-container&quot;&gt;$n.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://pdg.lbl.gov/2014/reviews/rpp2014-rev-elements-electronic-struct.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ELECTRONIC STRUCTURE OF THE ELEMENTS (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The question refers to “rules on how to fill the orbitals”, but I don't think there are any rules, just memorization aids.&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;http://adsabs.harvard.edu/abs/2007EPJD...45..107F&quot; rel=&quot;nofollow noreferrer&quot;&gt;The low-lying level structure of atomic lawrencium &lt;span class=&quot;math-container&quot;&gt;$(Z = 103):$&lt;/span&gt; energies and absorption rates&lt;/a&gt; for an sample of how difficult it is to calculate the ground state configuration of an atom.  The only rule is quantum electrodynamics.&lt;/p&gt;&#xA;&lt;p&gt;Also, the configurations are really just designations or leading configurations, with the actually ground state being a mix of configuration.  See for example the above lawrencium reference or &lt;a href=&quot;http://iopscience.iop.org/1402-4896/76/5/024/&quot; rel=&quot;nofollow noreferrer&quot;&gt;A critical compilation of energy levels and spectral lines of neutral boron&lt;/a&gt;, where it explained “Configuration and&#xA;term labels have little physical meaning for highly mixed levels”. Boron ground state is &lt;span class=&quot;math-container&quot;&gt;$95\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(1s)^2(2s)^2(2p)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$4\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(1s)^2(2p)^3}.$&lt;/span&gt;  Lawrencium is calculated to be &lt;span class=&quot;math-container&quot;&gt;$86\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(7s)^2(7p)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$6\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(6d)(7s)(7p)}.$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="41328" LastEditDate="2020-12-07T06:37:16.503" LastActivityDate="2020-12-07T06:37:16.503" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27555" PostTypeId="2" ParentId="27548" CreationDate="2015-03-19T16:12:25.350" Score="2" Body="&lt;p&gt;After a certain amount of pressure, the molecules start repelling each other and this leads to a rise in compressibility factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high pressures, the van der Wall constant $b$ which represents the finite volume occupied by the gas molecules plays a greater role than the van der Wall constant $a$ which represents the attractive forces between the gas molecules.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-03-19T16:12:25.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27556" PostTypeId="2" ParentId="27550" CreationDate="2015-03-19T16:21:46.707" Score="7" Body="&lt;p&gt;The term “&lt;strong&gt;negative pressure&lt;/strong&gt;” is often used in engineering to refer to a situation in which an enclosed volume has lower pressure than its surroundings. If the a region is surrounded by more pressurized area around it would cause substances to flow inwards and thus the term &lt;strong&gt;negative pressure&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorDisplayName="user7951" LastEditDate="2016-11-21T14:11:46.117" LastActivityDate="2016-11-21T14:11:46.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27557" PostTypeId="2" ParentId="27541" CreationDate="2015-03-19T16:27:01.450" Score="1" Body="&lt;p&gt;I think we should propose something that uses some water. We cannot forget that our body is compatible with water as a solvent. The chemistry in our body is based on the chemistry of aqueous solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On the other hand, we have to take into account the  large quantities of solvents, other than water to be transported to the spacecraft for astronauts baths. The water used for my proposed solution is a secondary product of the fuel cells  actually used to generate electricity in the spacecraft. The fuel for the fuel cells are oxygen and hydrogen. And as @Ben Norris added hydrogen is the most common element in the universe and oxygen is the third most common element in the universe. So, with   the right technology, we should have them on the space craft.&lt;/strong&gt;&lt;br&gt;&#xA;I would like to suggest to use &quot;Bentonite based superabsorbent polymer composite&quot;. It's a kind of hydrogels that absorbs water and loaded up to 50% wt/wt with a clay, bentonite (aluminium phyllosilicate). This composite is taken to space as powder. When astronauts  want to take a bath, they can add some (water+fragrance) to the powder that swells and forms a gel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of bentonite as stabilizing and rheological agents is due to the colloidal structure in water. Each smectite particle is composed of thousands of submicroscopic platelets stacked in sandwich fashion with a layer of water between each. A single platelet is one nanometer thick and up to several hundred nanometers across. The faces of these platelets carry a negative charge, while edges have a slightly positive charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bentonite is largely used in cosmetics, it is bio-compatible and has excellent rheological properties to get perfect emulsions and optimize flow properties. On the other hand, the composite I'm suggesting has been formulated and tailored to absorb moderate quantities of water  and fragrance that has a soothing properties to relax astronauts.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-03-19T18:44:20.927" LastActivityDate="2015-03-19T18:44:20.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27558" PostTypeId="2" ParentId="27546" CreationDate="2015-03-19T16:50:26.900" Score="8" Body="&lt;p&gt;Oxygen/nitrogen ratios &lt;a href=&quot;http://bluemoon.ucsd.edu/publications/ralph/3_Seasonal.pdf&quot;&gt;vary seasonally&lt;/a&gt; and geographically even at sea level on the order of 200 ppm.  Oxygen concentration has also been slowly decreasing in correlation with carbon dioxide increase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using the &lt;a href=&quot;http://ccmc.gsfc.nasa.gov/modelweb/models/msis_vitmo.php&quot;&gt;MSIS-E-90 Atmosphere Model&lt;/a&gt; which allow location, date, time and height input and provides composition output.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-19T16:50:26.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27559" PostTypeId="1" AcceptedAnswerId="27560" CreationDate="2015-03-19T16:59:38.887" Score="3" ViewCount="207" Body="&lt;p&gt;Recently there has been announced a new 3d printing technology which uses UV light, from lasers or a projector, to harden slices of epoxy resin so that the printed object rises out of a pool of epoxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some information about an oxygen boundary and UV light, can someone explain the chemistry of the process?&lt;br&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=UpH1zhUQY0c&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=UpH1zhUQY0c&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14577" LastActivityDate="2015-08-06T07:56:22.987" Title="How does 3dCarbon CLIP printer work?" Tags="&lt;organic-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27560" PostTypeId="2" ParentId="27559" CreationDate="2015-03-19T17:26:01.860" Score="2" Body="&lt;p&gt;Actually, the technology isn't that new. It seems however, that the setups are becoming more affordable in the hobby market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technology is know as &lt;em&gt;stereolithography&lt;/em&gt; and is a form of light-induced polymerization (&lt;strong&gt;UV curing&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liquid monomers used have a substituent that can be activated using UV light. Out of practical considerations, UV is more convenient: polymerization of the resin will not occur under natural light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the procedure is comparable to UV curing in dental repair. There, a lamp that irradiates the whole area is used: you want a fast hardening all over the tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3D printing, this is obviously different! The light source is a laser with a narrow beam: you only want to irradiate a very small volume element at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want a rapid polymerization either: only the volume element irradiated should react!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laser beam is focussed at the boundary of the resin and different volume elements are addressed by moving the focus in the $xy$ plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than moving the (heavy) laser itself, the beam is &quot;moved&quot; using a mirror optic. This allows for a very fast adressing of reaction points within the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;UPDATE&lt;/strong&gt;&#xA;In the example linked, in the question and a comment, however, the setup is different from laser-based STL printers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irradiation from the bottom of of the resin tank through an oxygen-permeable membrane indicates indeed that the UV curing proceeds in the presence of oxygen. Anyway, if the light source would be just a lamp, one would nevertheless achieve homogenous curing of the whole layer. &#xA;But the manufacturer doesn't use the term lamp or light source. Instead, they write about a &lt;strong&gt;projector&lt;/strong&gt; and that's probably what they do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;They are not illuminating the whole layer, but are projecting masks for every slice of the object to be printed.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-20T06:26:24.473" LastActivityDate="2015-03-20T06:26:24.473" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27561" PostTypeId="1" CreationDate="2015-03-19T17:43:10.390" Score="3" ViewCount="57" Body="&lt;p&gt;What does this sentence mean - I assumed that burning vaporised something? If something is vaporised, isn't that turning it into a gas? How can you burn a gas? This is a sentence in my AS Chemistry book and I am obviously missing something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Alkanes must be vaporised before they will burn&quot;&lt;/p&gt;&#xA;" OwnerUserId="12743" LastEditorUserId="10185" LastEditDate="2015-03-19T17:59:53.250" LastActivityDate="2015-03-19T17:59:53.250" Title="&quot;Alkanes must be vaporised before they will burn&quot; - what does this mean?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27562" PostTypeId="2" ParentId="27547" CreationDate="2015-03-19T17:49:54.037" Score="1" Body="&lt;p&gt;Paper mostly consists of cellulose, form which wood is a good source.&#xA;Another interesting source is the hemp plant and it seems that hemp was indeed the major source in paper manufacturing up to the 19th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, recycling of paper probably is the most reasonable way to reduce the wood use. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do we even need paper anymore with all electronic devices and stylus'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, for different reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Think in how long old books and handwritings have lasted. Compare that with the durability of magnetic (tapes, hard disks) or optical recordings (CD, DVD, Bluray, etc.) &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Supposed that the ability to read doesn't get lost and a document is written in a well-defined alphabet, you do not need any additional technology to access the information. Who can read data from a 9 track tape nowadays (the technology is only fifty years old) or an &quot;old&quot; MFM hard drive? 8&quot; or 5.25&quot; floppy disks, anyone? ZipDrives, supposed they survived the &quot;click-od-death&quot;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most people do not use electronic signatures for documents. Papers are signed, sealed, etc. Consequently, written contracts, bills, etc. are considered more forge-proof than electronic documents. Storing them is therefore mandatory in business.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-19T17:49:54.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27563" PostTypeId="2" ParentId="27561" CreationDate="2015-03-19T17:58:30.737" Score="4" Body="&lt;p&gt;Burning an alkane means reacting it with oxygen (in air). The larger the contact surface between the alkane phase and the air, the better the reaction. That's why cars have carburators: there's no larger &quot;contact surface&quot; than between the molecules in a gas mixture.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-19T17:58:30.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27564" PostTypeId="1" CreationDate="2015-03-19T18:46:59.420" Score="2" ViewCount="2255" Body="&lt;p&gt;Why is the basicity of &lt;em&gt;para&lt;/em&gt;- &gt; &lt;em&gt;meta&lt;/em&gt;- &gt; aniline &gt; &lt;em&gt;ortho&lt;/em&gt;-&#xA;for phenylenediamine?&lt;br&gt;&#xA;($\mathrm pK_\mathrm a$: &lt;em&gt;ortho&lt;/em&gt;: 9.53, &lt;em&gt;meta&lt;/em&gt;: 9.12, &lt;em&gt;para&lt;/em&gt;: 7.96) I tried deducing it using resonance and inductive effects but got stuck.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorDisplayName="user7951" LastEditDate="2016-11-20T20:55:08.290" LastActivityDate="2016-11-20T20:55:08.290" Title="How to rationalise the basicity order of phenylenediamine?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27565" PostTypeId="1" CreationDate="2015-03-19T19:06:35.300" Score="2" ViewCount="2133" Body="&lt;p&gt;Why is $\ce{H-OH}$ more acidic than $\ce{CH3CH2-OH}$? A common answer to this question is because of the $+I$-effect of $\ce{CH3}$ the negative charge at $\ce{O}$ increases making it less acidic. But if we compare electronegativity difference, the difference between $\ce{O}$ and $\ce{H}$ is more than that between $\ce{O}$ and $\ce{C}$ so there should be even more negative charge at $\ce{O}$ in case of $\ce{H2O}$. In this way $\ce{H2O}$ should be less acidic than $\ce{CH3CH2-OH}$. This is also the reason that $+M$ (electron donating ability) of $\ce{-OH}$ (which is having more negative charge at $\ce{O}$) is greater than the $+M$ of $\ce{-OR}$. &lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="7475" LastEditDate="2015-10-16T16:43:47.930" LastActivityDate="2015-10-16T16:43:47.930" Title="Why is water more acidic than alkanols?" Tags="&lt;acid-base&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27567" PostTypeId="1" CreationDate="2015-03-19T19:35:24.023" Score="0" ViewCount="1505" Body="&lt;p&gt;I know Sorbitol, $\ce{C_6H_14O_6}$ $$\ce {C(OH)H_2 (C(OH)H)_4C(OH)H_2}$$is soluble in water however I did not have much luck finding out whether or not it is soluble in ethanol. My guess is that it isn't however I am not so sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is Sorbitol soluble in ethanol?&lt;/p&gt;&#xA;" OwnerUserId="2558" LastActivityDate="2017-03-01T04:38:42.453" Title="Does sorbitol dissolve in ethanol?" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27568" PostTypeId="2" ParentId="27453" CreationDate="2015-03-19T19:36:58.450" Score="7" Body="&lt;p&gt;This reaction proceeds through a reverse Friedel-Crafts alkylation followed by a forward Friedel-Crafts alkylation. The last step of Friedel-Crafts alkylation is deprotonation, which means that the first step of the reverse process will be protonation. Next, the carbon-carbon bond breaks, with 1,2,4,5,-tetrabromobenzene acting as a leaving group. The resulting t-butyl cation can undergo the typical electrophilic aromatic substitution steps, attack by an aromatic ring (benzene here) and deprotonation to regain aromaticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might wonder why aluminum chloride is present in the reaction mixture, since we can write a reasonable mechanism that excludes it. Likely it is present in case chloride (from HCl) traps it to form t-butyl chloride. Aluminum chloride will catalyze the reformation of t-butyl cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g3qVZ.jpg&quot; alt=&quot;Reverse Friedel-Crafts / forward Friedel-Crafts&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-03-19T19:36:58.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27569" PostTypeId="2" ParentId="27499" CreationDate="2015-03-19T20:28:07.767" Score="0" Body="&lt;p&gt;There should be no anions exept OH- present. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;made by adding 20% ammonia in copper carbonate salt.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is wrong, using CuSO4 is even wronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to precipitate Cu(OH)2 and wash the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;precipitate thoroughly, a very tedious operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a IG Farbenindustrie Patent You &lt;/p&gt;&#xA;&#xA;&lt;p&gt;throw copper pieces in  &gt; 25 % Ammonia  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(cut from electrical wire) and percolate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with oxygen or air. When using air, you have to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maintain temperature under 8 °C. This needs days &lt;/p&gt;&#xA;&#xA;&lt;p&gt;until &quot;ready&quot; but has the advantage that the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;reagent is pure without cleaning operations. &lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="105" LastEditDate="2015-03-19T21:07:34.227" LastActivityDate="2015-03-19T21:07:34.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27571" PostTypeId="2" ParentId="27535" CreationDate="2015-03-19T21:05:57.413" Score="10" Body="&lt;p&gt;Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'd like to correct the definition of percent ionic character in your question using dipole moment  $\mu$ (not &lt;em&gt;Observed value of ionic character&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated}  }}$ $\times 100 \%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details please see &lt;a href=&quot;http://wwwchem.uwimona.edu.jm/courses/CHEM1902/IC10K_MG_Fajans.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5026" LastEditDate="2018-03-23T11:39:40.867" LastActivityDate="2018-03-23T11:39:40.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27572" PostTypeId="2" ParentId="27565" CreationDate="2015-03-19T21:32:37.710" Score="5" Body="&lt;p&gt;\begin{array} {|l|r|r|}&#xA;\hline&#xA;  &amp;amp;\mathrm{pK_a~in~water} &amp;amp;\mathrm{pK_a~in~DMSO}\\&#xA;\hline&#xA;\ce{H2O} &amp;amp;15.74 &amp;amp;31.4\\&#xA;\hline&#xA; \ce{CH3CH2OH} &amp;amp;16 &amp;amp;29.8 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf&quot; rel=&quot;nofollow&quot;&gt;http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/pkatable/index.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.wisc.edu/areas/reich/pkatable/index.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="7448" LastEditDate="2015-10-09T15:52:11.183" LastActivityDate="2015-10-09T15:52:11.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27574" PostTypeId="2" ParentId="27528" CreationDate="2015-03-19T22:44:37.557" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this means all naturally occurring reactions, (that don't just&#xA;  happen once in a blue moon but at a steady rate) are exothermic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, not all naturally occurring reactions are exothermic ($\Delta H &amp;gt; 0$) or exergonic ($\Delta G &amp;gt; 0$).  For example, the reaction fundamental to our existence, photosynthesis, where carbon dioxide is converted to oxygen and sugar is both &lt;a href=&quot;http://en.wikipedia.org/wiki/Photosynthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;endothermic&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Endergonic_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;endergonic&lt;/a&gt; (see the figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{n CO2 + 2n H2O + photons → (CH2O)_n + n O2 + n H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy required to surmount the barrier is provided by photons (radiation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerous other biochemical reactions are also endothermic and endergonic. &lt;a href=&quot;http://en.wikipedia.org/wiki/Endergonic_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; mentions protein synthesis and the $\ce{Na+/K+}$ pump which drives nerve conduction and muscle contraction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-05-10T19:15:37.523" LastActivityDate="2017-05-10T19:15:37.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27575" PostTypeId="1" AcceptedAnswerId="27577" CreationDate="2015-03-19T23:51:24.233" Score="6" ViewCount="2341" Body="&lt;p&gt;Is there any difference between the phrase &quot;continuously overlapping p-orbitals&quot; and &quot;conjugated pi bonds&quot; when referring to aromatic compounds? I've heard both used and I wasn't sure if they mean the same thing.&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="7475" LastEditDate="2015-10-16T16:39:52.937" LastActivityDate="2015-10-16T16:39:52.937" Title="Difference between continuously overlapping p orbitals and conjugated pi bonds" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27577" PostTypeId="2" ParentId="27575" CreationDate="2015-03-20T01:26:00.090" Score="10" Body="&lt;p&gt;Interesting question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have &lt;strong&gt;continuously&lt;/strong&gt; overlapping p-orbitals, the p-orbitals must be arranged in a cyclic loop.  The p-orbitals in benzene are continuously overlapping.  The p-orbitals in benzene are also conjugated pi bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vv56x.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ch.ic.ac.uk/rzepa/mim/environmental/html/benzene.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In buta-1,3-diene the p-orbitals form conjugated pi bonds, however, they are not continuously overlapping - there are two &quot;end&quot; carbons that are not connected with each other, there is no continuous loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jjE8s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.science.oregonstate.edu/~gablek/CH335/Chapter14/butadiene_MOs.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's take it a step further:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuously overlapping p-orbitals are not always aromatic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: Cyclobuta-1,3-diene (the square planar structure, $\ce{D_{4h}}$) has 4 continuously overlapping p-orbitals but it is not aromatic, it is antiaromatic.  For a continuously overlapping system of p-orbitals to be aromatic, it must contain 4n+2 pi electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z1YKP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://science.uvu.edu/ochem/index.php/alphabetical/a-b/aromatic/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: Have you heard of the &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_strip&quot; rel=&quot;noreferrer&quot;&gt;Möbius strip&lt;/a&gt;? The mathematical (topological) concept has application in chemistry.  Imagine a large cyclic loop of p-orbitals where each p-orbital is twisted a few degrees from the preceeding p-orbital.  If the loop is large enough, then by the time we come back to the first p-orbital we have twisted the last p-orbital by 180°; there is a phase inversion (node) between the first and last p-orbital.  In a &lt;a href=&quot;http://en.wikipedia.org/wiki/M%C3%B6bius_aromaticity&quot; rel=&quot;noreferrer&quot;&gt;Möbius system&lt;/a&gt; the p-orbitals are continuous, but not continuously overlapping!  Möbius systems are aromatic if they contain 4n pi electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYN3J.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/mobius.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous pi systems with an even number (0, 2, 4...) of nodes (cyclobutadiene, benzene, etc) are called Hückel systems in order to differentiate then from Möbius systems where there are an odd number of nodes (1, 3, 5...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lest you think the Möbius concept is purely theoretical take a look at these two molecules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KudNt.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3ccGF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/mobius.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't this what makes chemistry fascinating!&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-20T01:26:00.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27578" PostTypeId="2" ParentId="13563" CreationDate="2015-03-20T05:43:34.613" Score="1" Body="&lt;p&gt;The &lt;strong&gt;&lt;em&gt;liquid-vapor coexistence&lt;/em&gt;&lt;/strong&gt; line of the phase diagram of water has a positive slope $\mathrm{d}P/\mathrm{d}T$. Humidity in air is a partial pressure of water vapor. If partial pressure of water is increased starting from $&amp;lt;100\%$ humidity, it will reach the vapor-liquid transition and any additional water vapor will simply condense into droplets ($=100\%$ humidity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first-order phase transition of condensation will force any water molecules added to a saturated system to condense into liquid, and so there is a high limit to the humidity at a given temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase in temperature corresponds to an increase in the saturation vapor pressure, thus at warmer temperatures there will be a higher partial pressure of water vapor in the air at a given $\%$ humidity.&lt;/p&gt;&#xA;" OwnerUserId="15065" LastActivityDate="2015-03-20T05:43:34.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27579" PostTypeId="2" ParentId="26268" CreationDate="2015-03-20T06:02:31.460" Score="12" Body="&lt;p&gt;It has been already mentioned in the comments and the other answer, that fluorine is more electronegative than oxygen. Therefore it is highly unlikely that this compound exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate the problem at hand, take a look at similar, known compound with a fluorine oxygen bond: &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypofluorous_acid&quot; rel=&quot;noreferrer&quot;&gt;hypofluorous acid&lt;/a&gt;, $\ce{HFO}$, which is actually better described as hydroxylfluoride, $\ce{HOF}$. According to Holleman and Wiberg's&lt;sup&gt;[1,2]&lt;/sup&gt; textbook the synthesis of this compound is already difficult. Low pressure fluorine gas is passed over ice at about $-40~^\circ\mathrm{C}$. Obtained is a mixture of $\ce{HF}$, $\ce{O2}$, $\ce{H2O2}$, $\ce{H2O}$ and also $\ce{HOF}$. Fractional freezing until about $-183~^\circ\mathrm{C}$ leads to a white substance, that turns pale yellow at about $-117~^\circ\mathrm{C}$, and boils at about $10-20~^\circ\mathrm{C}$. It decomposes via a radical pathway at about $25~^\circ\mathrm{C}$ and $100~\mathrm{mbar}$ with a half life of about 30 minutes to oxygen and hydrogen fluoride: $$\ce{2HOF -&amp;gt; 2HF + O2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic structure of the molecule is most likely $$\ce{HO^{\delta+}\bond{-}F^{\delta-}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Salts, $\ce{M^{+}OF^{-}}$, that are derived from this compound are not known. Covalent compounds, however, are known, i.e. $\ce{O3ClOF}$ or $\ce{O2NOF}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to imagine, that a compound with an even higher oxygen load cannot be stable. Therefore we can conclude that a halogenate like $\ce{HFO3}$, let alone its salt $\ce{M^{+}[FO3]^{-}}$, cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes and References&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Unfortunately I only have access to the German version of this textbook. I tried to excerpt and translate the important parts to the best of my knowledge.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.degruyter.com/view/product/19732&quot; rel=&quot;noreferrer&quot;&gt;Holleman, A., Wiberg, N., Wiberg, E., &lt;em&gt;et al.&lt;/em&gt; (2008). &lt;em&gt;Lehrbuch der Anorganischen Chemie&lt;/em&gt;. Berlin, Boston: De Gruyter.&lt;/a&gt; Kapitel XII 4.2. &lt;em&gt;Sauerstoffsäure des Fluors&lt;/em&gt;. S. 465.  (in german)&lt;/li&gt;&#xA;&lt;li&gt;Translation of the 101st edition available as &lt;a href=&quot;http://books.google.co.jp/books?id=Mtth5g59dEIC&quot; rel=&quot;noreferrer&quot;&gt;Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2016-09-05T13:50:54.117" LastActivityDate="2016-09-05T13:50:54.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27580" PostTypeId="2" ParentId="27564" CreationDate="2015-03-20T06:09:28.477" Score="3" Body="&lt;p&gt;As &lt;strong&gt;Mithoron&lt;/strong&gt; pointed out in a comment, we had a similar question convering the cases of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases/7685#7685&quot;&gt;2-methylaniline (&lt;em&gt;o&lt;/em&gt;-toluidine) and 2-methylbenzoic acid (&lt;em&gt;o&lt;/em&gt;-toluic acid)&lt;/a&gt; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of 1,2-diaminobenzene (&lt;em&gt;o&lt;/em&gt;-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, &lt;em&gt;o&lt;/em&gt;-phenylenediamine is significantly less basic than the &lt;em&gt;m&lt;/em&gt;- and &lt;em&gt;p&lt;/em&gt;-isomer, but a tad more basic than aniline.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-03-20T06:09:28.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27584" PostTypeId="1" AcceptedAnswerId="27611" CreationDate="2015-03-20T10:33:29.250" Score="5" ViewCount="1434" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IRfyS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IRfyS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this compound there are three cyclic and three acyclic carbons. Even the substituent comes on the 2nd position in both cases. So, should the main chain be the &lt;em&gt;cyclic&lt;/em&gt; one, since there is a rule that cyclic has a priority over acyclic?&lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="41328" LastEditDate="2017-08-06T21:07:54.873" LastActivityDate="2017-08-06T21:07:54.873" Title="IUPAC nomenclature of alicyclic compound with an equal number of cyclic and acyclic carbons" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27585" PostTypeId="1" AcceptedAnswerId="27587" CreationDate="2015-03-20T10:35:25.183" Score="2" ViewCount="3396" Body="&lt;p&gt;$$\ce{C(s) + 2 Cl2 (g)  -&amp;gt; CCl4 (g)}$$&#xA;\begin{align}  &#xA;\Delta H^\circ_\mathrm{f} (\ce{C(g)}) &amp;amp;= 121~\mathrm{kJ/mol}\\&#xA;\Delta H^\circ_\mathrm{f} (\ce{Cl(g)}) &amp;amp;= 716~\mathrm{kJ/mol}\\&#xA;\Delta H^\circ_\mathrm{f} (\ce{CCl4(g)}) &amp;amp;= -106.48~\mathrm{kJ/mol}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I already answered this, but I was a bit confused about the result, cause I don't know if there is a bond that is negative, cause am familiar with the energy given to &quot;dissociation&quot; of a bond or the &quot;formation&quot; of it , but here it's given the energy to the formation of chlorine and carbon, I don't know if that's the same as the formation of the bond or not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &amp;amp;= \frac{\Delta H^\circ_\mathrm{f}(\ce{C(g)}) + 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) - \Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}}{4} \\&#xA;\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &amp;amp;= \frac{121 + 4 \cdot 716 + 106.48}{4} \\&#xA;\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &amp;amp;= 772.87~\mathrm{kJ/mol}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Should I take the $\triangle H^\circ_\mathrm{f}(\ce{Cl(g)})$ same as $\triangle H^\circ_f(\ce{Cl-Cl})$? In that case I'm going to multiply it by $2$, not $4$. &lt;/li&gt;&#xA;&lt;li&gt;Should the  $\Delta H^\circ_\mathrm{f}(\ce{C-Cl})$ be prefixed by a minus $-$ or a plus $+$, cause I don't know if it's an energy that is needed to formulate $\ce{CCl4}$? I'm not saying that the enthalpy of the bond should be negative, but the direction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15070" LastEditorUserId="69830" LastEditDate="2021-05-13T13:54:39.223" LastActivityDate="2021-05-13T13:54:39.223" Title="How to calculate the bond energy of C-Cl?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27586" PostTypeId="1" AcceptedAnswerId="27594" CreationDate="2015-03-20T12:47:42.553" Score="8" ViewCount="1471" Body="&lt;p&gt;The group 15, 16, and 17 are called the pnictogens, chalcogens, and halogens respectively. Is there a name for the groups 13 and 14 as well?&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="5026" LastEditDate="2018-04-08T15:52:13.890" LastActivityDate="2018-04-08T15:55:15.993" Title="Is there a common name to refer to the groups 13 and 14?" Tags="&lt;periodic-table&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27587" PostTypeId="2" ParentId="27585" CreationDate="2015-03-20T13:38:04.123" Score="1" Body="&lt;p&gt;Using your Given data &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)}) +\Delta{}H^\circ_\mathrm{f} (\ce{C(g)}) -4\times\Delta H_{C-Cl}=\Delta{}H^\circ_\mathrm{f} (\ce{CCl4 (g)})$$&lt;/p&gt;&#xA;" OwnerUserId="13457" LastActivityDate="2015-03-20T13:38:04.123" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27588" PostTypeId="1" CreationDate="2015-03-20T14:06:34.110" Score="3" ViewCount="30208" Body="&lt;p&gt;What are the steps to follow in order to determine which molecule has a higher bond order by drawing the Lewis structure? For example, with $\ce{CO}$ and $\ce{CO3^{2-}}$? &lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="402" LastEditDate="2017-05-15T21:49:31.607" LastActivityDate="2017-05-16T01:53:58.257" Title="How to determine the bond order using the Lewis structure?" Tags="&lt;bond&gt;&lt;resonance&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27589" PostTypeId="1" CreationDate="2015-03-20T14:24:53.030" Score="2" ViewCount="176" Body="&lt;p&gt;We need a pure copper phthalocyanine for our experiments. Of course, we can buy it from Sigma Aldrich, but this means we'll have to wait a while. In local area we can buy only dyes which are about 98% purity, which is not enough. Is there any fast and easy method to purify it in lab?&lt;/p&gt;&#xA;" OwnerUserId="11374" LastActivityDate="2015-03-20T14:24:53.030" Title="Make pure phthalocyanines from phthalocyanine dyes" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;&lt;purification&gt;&lt;organometallic-compounds&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27591" PostTypeId="2" ParentId="27588" CreationDate="2015-03-20T14:51:43.203" Score="3" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9hoHo.png&quot; alt=&quot;Lewis Structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly draw lewis structure, then count the total number of bonds which is equal to $4$ here. Finally, count the number of bond groups between individual atoms, which is $3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond order will be equal to $\frac{4}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorUserId="402" LastEditDate="2017-05-15T21:50:34.767" LastActivityDate="2017-05-15T21:50:34.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27592" PostTypeId="2" ParentId="2495" CreationDate="2015-03-20T15:15:45.110" Score="-1" Body="&lt;p&gt;Rain water contains very low amounts of salts and other nitrates but it takes in any gas present in air. That is why acid rain occurs when the air is polluted. The air contains oxygen so people recommend the rain to be consumable as it has lots of oxygen contents in them. But distilled water does not contain &lt;em&gt;any&lt;/em&gt; minerals in them, so it is not recommended for consumption because of lack of minerals. &lt;/p&gt;&#xA;" OwnerUserId="15073" LastEditorUserId="4945" LastEditDate="2015-03-23T06:03:40.133" LastActivityDate="2015-03-23T06:03:40.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27594" PostTypeId="2" ParentId="27586" CreationDate="2015-03-20T15:37:43.887" Score="6" Body="&lt;p&gt;It seems that the carbon group is known as either the &lt;em&gt;tetrels&lt;/em&gt; or the &lt;em&gt;crystallogens&lt;/em&gt;; while the boron group is known as &lt;em&gt;triels&lt;/em&gt; or &lt;em&gt;icosagens&lt;/em&gt;. These names are very rarely used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source&lt;/strong&gt;: Wikipedia articles on &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbon family&lt;/a&gt; (also, &lt;a href=&quot;https://en.wiktionary.org/wiki/crystallogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;crystallogen&lt;/a&gt;) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Boron_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boron family&lt;/a&gt;. Although other (seemingly) respectable sites use them.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="5026" LastEditDate="2018-04-08T15:55:15.993" LastActivityDate="2018-04-08T15:55:15.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27595" PostTypeId="1" CreationDate="2015-03-20T16:05:29.700" Score="3" ViewCount="62" Body="&lt;p&gt;I am planning to use polydimethylsiloxane (PDMS) to create phantom for mimicking tissue for imaging. For this, I plan on using &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/761036?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;SYLGARD® 184&lt;/a&gt;, which is a viscous liquid. When adding other components to it, the air bubbles will be introduced. I have learned that one way (available to me) is using the negative pressure/ vacuum pump to remove bubbles / de-gas the solution. Does any one knows what negative pressure is sufficient to remove the air from the gel-like solution?&lt;/p&gt;&#xA;" OwnerUserId="15075" LastEditorUserId="4231" LastEditDate="2015-03-20T16:10:34.183" LastActivityDate="2015-03-20T16:10:34.183" Title="Silicon degassing with negative pressure" Tags="&lt;home-experiment&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27598" PostTypeId="2" ParentId="27564" CreationDate="2015-03-20T21:05:08.427" Score="3" Body="&lt;p&gt;Klaus's answer is nice, but is omits one important detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen bonding would necessarily require that the lone pair of electrons go 'out of conjugation' with the ring system, and hence there can be no donation of electron density via pi-effects as in the para-phenylenediamine case.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-20T21:05:08.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27599" PostTypeId="1" AcceptedAnswerId="27608" CreationDate="2015-03-20T21:25:38.343" Score="12" ViewCount="1415" Body="&lt;p&gt;I have looked at GAMESS and Gaussian manuals but can not find the maximum number of &lt;strong&gt;efficient&lt;/strong&gt; nodes. As I tried GAMESS on our supercomputer, I can add as many CPUs as I want but it really does not Improve the calculation time and efficacy . On the other hand just adding maximum number of CPUs is just waste of CPU-Hour time and resources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any criteria for selection of best number of CPUs for optimum resource usage ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastActivityDate="2015-03-26T16:24:41.260" Title="GAMESS and Gaussian : what is the maximum efficient number of supported nodes?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27601" PostTypeId="1" CreationDate="2015-03-21T02:19:23.003" Score="0" ViewCount="7088" Body="&lt;p&gt;I assumed it would increase, but I think I am wrong. I know that when the forces are strong, the vapor pressure would decrease. Does this same principle apply to the IM forces increasing?&lt;/p&gt;&#xA;" OwnerUserId="15080" LastActivityDate="2015-03-21T04:10:50.240" Title="When the intermolecular forces in a liquid increase, does the equilibrium vapor pressure of the liquid at a given temperature increase or decrease?" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27602" PostTypeId="1" CreationDate="2015-03-21T02:19:57.937" Score="1" ViewCount="1023" Body="&lt;p&gt;I've been asked to draw a relative energy sketch of the reactants and the two products. The reaction being examined was adding copper acetate monohydrate to glycine in two different fashions to create the two isomers cis-bis(glycinato)copper(II) monohydrate and trans-bis(glycinato)copper(II) monohydrate. I found that the cis-bis(glycinato)copper(II) monohydrate was  thermodynamically stable and trans-bis(glycinato)copper(II) monohydrate was kinetically favored since when adding heat to trans-bis(glycinato)copper(II) monohydrate it changed to the same color as cis-bis(glycinato)copper(II) monohydrate therefore cis-bis(glycinato)copper(II) monohydrate has a lower energy relative to trans-bis(glycinato)copper(II) monohydrate. The part where I'm stuck is where the reactant energy is relative to these two. I know it must be lower than the intermediate trans-bis(glycinato)copper(II) monohydrate, but where the energy level of the reactants compared to the thermodynamically stable cis-bis(glycinato)copper(II) monohydrate has me stumped, whether it is higher or lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="15081" LastActivityDate="2015-03-24T06:27:23.367" Title="Relative energies between reactants and products" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;energy&gt;&lt;free-energy&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27603" PostTypeId="2" ParentId="27601" CreationDate="2015-03-21T04:03:51.283" Score="3" Body="&lt;p&gt;Yeah your hunch is right. Higher the intermolecular forces between the liquid particles, harder it is for it to escape into the vapor phase, ie., you need more energy to convert it from liquid to the vapor &#xA;phase, in other words, higher its boiling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is harder for a liquid to escape into vapor, because it is held back into the liquid by the neighboring particles attraction, you have less vapor and hence low vapor pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in nutshell, if the intermolecular forces are increased, the equilibrium vapor pressure of the liquid will decrease.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4656" LastEditDate="2015-03-21T04:10:50.240" LastActivityDate="2015-03-21T04:10:50.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27604" PostTypeId="1" CreationDate="2015-03-21T04:31:17.310" Score="-1" ViewCount="678" Body="&lt;p&gt;The equation we need it to change to &#xA;\[\ce{FeO + CO -&gt;Fe + CO2}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three equations and delta heat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[&#xA;\begin{align*}&#xA;\ce{3Fe2O3 + CO &amp;amp;-&amp;gt; 2Fe3O4 + CO2} &amp;amp;\Delta H=-47.0\ \mathrm{kJ}\\&#xA;\ce{Fe2O3 + 3CO &amp;amp;-&amp;gt; 2Fe + 3CO2} &amp;amp;\Delta H=-25.0\ \mathrm{kJ}\\&#xA;\ce{Fe3O4 + CO &amp;amp;-&amp;gt; 3FeO + CO2} &amp;amp;\Delta H= 19.0\ \mathrm{kJ}&#xA;\end{align*}&#xA;\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the enthalpy change and how to combine the 3 equations to make the above equation? (Please show work)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&#xA;     Tiffany&lt;/p&gt;&#xA;" OwnerUserId="15082" LastEditorUserId="418" LastEditDate="2015-03-21T06:14:39.727" LastActivityDate="2015-03-21T07:14:25.653" Title="What is the enthalpy change for the following equation?" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-03-21T14:59:54.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="27605" PostTypeId="1" AcceptedAnswerId="27610" CreationDate="2015-03-21T04:55:48.437" Score="0" ViewCount="318" Body="&lt;p&gt;What form is the ionic equalibrium equation? I have problem asking for it, but I'be done so many different equalibrium equations that I'm starting to get mixed up on which one is called what... Which one is it?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastActivityDate="2015-03-21T09:28:38.347" Title="Ionic equilibrium equation" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-21T13:55:04.790" ContentLicense="CC BY-SA 3.0" />
  <row Id="27606" PostTypeId="2" ParentId="26258" CreationDate="2015-03-21T06:31:43.650" Score="3" Body="&lt;h1&gt;Example from biochemistry&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The Michaelis-Menten equation for enzyme kinetics is an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v=\frac{k_{cat}E_0S}{K_S+S}$ where $S$ is substrate concentration and $E$ in enzyme concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low $S$ ($S\ll K_S$), the reaction is 2nd order with $v=\frac{k_{cat}}{K_S}E_0S$, i.e. first order in both enzyme and in substrate.  However at high $S$ ($S\gg K_S$), the reaction is first order only in enzyme with $v=k_{cat}E_0$, and is zero-order in substrate.  The substrate concentration doesn't have to be in molar excess to other reactants.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example from process control&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In chemical engineering, nonlinear systems such as higher-order reactions occuring in a continuous process at steady-state are often modeled using linear (which you can think of as first-order reactions) systems as an approximation for developing process control techniques.  A &lt;a href=&quot;http://www.mathworks.com/help/slcontrol/ug/linearizing-nonlinear-models.html&quot; rel=&quot;nofollow&quot;&gt;MATLAB documentation page&lt;/a&gt; has some good information.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Another example, somewhat contrived&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As a last example, consider an unstable molecule $A$.  It can decompose to $B$ by two entirely separate elementary mechanisms.  The first is $\ce{A + A -&amp;gt; B + B}$.  The second is $\ce{A -&amp;gt; B}$.  At &lt;em&gt;high&lt;/em&gt; concentrations of $A$, the second-order kinetics will dominate and the decomposition of $A$ will be second order.  But &lt;em&gt;lower&lt;/em&gt; the concentration enough, and the reaction will become first order in $A$ because the rate the bimolecular reaction will be dwarfed by the unimolecular route at low $A$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-21T06:31:43.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27607" PostTypeId="2" ParentId="27604" CreationDate="2015-03-21T07:14:25.653" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;[…] how to combine the 3 equations […]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;&lt;strong&gt;Equations&lt;/strong&gt; is the keyword!&lt;/h3&gt;&#xA;&lt;p&gt;This is just an example on how to solve a system of linear equations, disguised as a chemistry problem.&lt;/p&gt;&#xA;&lt;p&gt;If the calculation with &lt;span class=&quot;math-container&quot;&gt;$\ce{FeO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;, etc. feels strange, just translate them to &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$x_4$&lt;/span&gt;, etc.&lt;/p&gt;&#xA;&lt;p&gt;The question for the enthalpy of the first reaction therefore is equivalent to finding the solution for&lt;/p&gt;&#xA;&lt;p&gt;\[y = x_3 + x_4 - x_1 - x_2 \]&lt;/p&gt;&#xA;&lt;h3&gt;How to work in the other equations?&lt;/h3&gt;&#xA;&lt;p&gt;\[\ce{3Fe2O3 + CO -&amp;gt; 2Fe3O4 + CO2} \quad\quad\Delta H=-47.0\ \mathrm{kJ}\]&lt;/p&gt;&#xA;&lt;p&gt;You have the reactants, the products and the enthalpy: translate that to a mathematical equation following the naming scheme introduced above:&#xA;\[2x_6 + x4 - x_2 - 3x_5 = -47.0\]&#xA;Do the same for the other two equations, rearrange and try to solve the equation system.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-21T07:14:25.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27608" PostTypeId="2" ParentId="27599" CreationDate="2015-03-21T07:40:34.003" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any criteria for selection of best number of CPUs for optimum&#xA;  resource usage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The criteria is basically the same as for choosing the level of theory: &lt;em&gt;benchmarking&lt;/em&gt;. Parallel scaling depends on many factors: the level of theory, the size of the molecule, the machine architecture, etc., so there is no golden rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though from my experience I can say that Gaussian 09 single point energy, geometry optimisation and frequencies calculations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;scale well for HF &amp;amp; DFT up to 16 physical cores;&lt;/li&gt;&#xA;&lt;li&gt;scale well for MP2 up to 8 physical cores;&lt;/li&gt;&#xA;&lt;li&gt;almost do not scale for CI &amp;amp; CC.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also note that Gaussian 09 is about 2 times faster for almost any kind of calculations than the previous version (Gaussian 03) and doesn't suffer from some awkwardly strange memory effects (see e.g. &lt;a href=&quot;http://biowulf.nih.gov/apps/gaussian/#memory&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-03-21T07:40:34.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27609" PostTypeId="1" AcceptedAnswerId="27613" CreationDate="2015-03-21T08:00:25.987" Score="3" ViewCount="1704" Body="&lt;p&gt;Why does the color of iodine differ when it forms a charge transfer complex with benzene (pi donor) - pink and ethanol (electron pair donor) - brown? Does the HOMO-LUMO energy differ, if so how?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="418" LastEditDate="2015-03-21T09:57:51.187" LastActivityDate="2016-07-10T20:42:39.383" Title="Iodine color change in different complexes" Tags="&lt;coordination-compounds&gt;&lt;color&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27610" PostTypeId="2" ParentId="27605" CreationDate="2015-03-21T09:28:38.347" Score="1" Body="&lt;p&gt;An ionic equilibrium equation is simply one that involves ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : $\ce {HA &amp;lt;=&amp;gt; H^+ + A^-}$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-03-21T09:28:38.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27611" PostTypeId="2" ParentId="27584" CreationDate="2015-03-21T09:55:14.300" Score="10" Body="&lt;p&gt;According to &lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;current IUPAC recommendations (2013)&lt;/a&gt;, for the preferred IUPAC name, a ring has seniority over a chain. Therefore, the given structure is a propylcyclopropane. (However, the IUPAC recommendations allow that the context may favour the chain.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two prefixes of cyclopropane (chloro- and 2-bromopropyl-) are arranged alphabetically. Thus, the name is &lt;strong&gt;1-(2-bromopropyl)-2-chlorocyclopropane.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1tW4F.png&quot; width=&quot;244&quot; alt=&quot;1-(2-bromopropyl)-2-chlorocyclopropane&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-06-14T14:51:32.850" LastActivityDate="2016-06-14T14:51:32.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27612" PostTypeId="1" CreationDate="2015-03-21T10:17:52.413" Score="1" ViewCount="73" Body="&lt;p&gt;I have to convert  a whole lot of nutrient data : $\ce{NH_4}$, $\ce{PO_4}$ and $\ce{NO_x}$ data that is in µg/L into mM (millimolars). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me an example for 26.62 $\ce{NO_x}$ µg/L? or 7.73 $\ce{PO_4}$ µg/L? or 14.39 $\ce{NH_4}$ µg/L?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a large data set in Excel to convert and send to someone and want to get it right. Any help would be really appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15083" LastEditorUserId="7448" LastEditDate="2015-03-21T10:50:14.077" LastActivityDate="2015-03-21T13:48:41.687" Title="Converting 26.62 NOx µg/L into mM (millimolars)?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27613" PostTypeId="2" ParentId="27609" CreationDate="2015-03-21T10:58:47.937" Score="6" Body="&lt;p&gt;The formation of charge transfer complexes between iodine and aromatic hydrocarbons according to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{Ar + I2 &amp;lt;=&gt; Ar\cdot I2}\] is a long known phenomenon and has been described by &lt;em&gt;Benesi&lt;/em&gt; and &lt;em&gt;Hildebrand&lt;/em&gt; in &lt;a href=&quot;http://dx.doi.org/10.1021/ja01176a030&quot;&gt;A Spectrophotometric Investigation of the Interaction of Iodine with Aromatic Hydrocarbons&lt;/a&gt;. The following data for the absorption maxima of iodine in the VIS and UV range (wavelength in nm) in different solvents are taken from that article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrrr}&#xA;\mathbf{solvent} &amp;amp; \lambda_{max}(\textrm{VIS}) &amp;amp; \lambda_{max}(\textrm{UV})\\&#xA;\hline&#xA;\textrm{benzene} &amp;amp; 500 &amp;amp; 297\\&#xA;\textrm{toluene} &amp;amp; 497 &amp;amp; 306\\&#xA;\textrm{p-xylene} &amp;amp; 495 &amp;amp; 315\\&#xA;\textrm{mesitylene} &amp;amp; 490 &amp;amp; 333\\&#xA;\hline&#xA;\textrm{trifluormethylbenzene} &amp;amp; 512\\&#xA;\textrm{n-heptane} &amp;amp; 520 &amp;amp; \\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the effect is stronger in the UV than in the VIS range and the shift correlates with the (combined) $+I$ effects of the methyl groups, while a $\ce{CF3}$ substituent disfavours then formation of a charge transfer complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethanol, on the other hand, is a protic and polar solvent, nice to stabilize charged species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might argue that under these conditions, other species than in the case of aromatic solvents exist - the solvation of triiodide $\ce{I3-}$&#xA; is certainly possible here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the coordination of iodine by ethanol - we cannot exclude interaction with several molecules of ethanol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{n\ EtOH +I2 &amp;lt;=&gt; I2\cdot(EtOH)_{n}}\] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is conceivable that disproportionation of iodine and solvation of the resulting charged species might play a role here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ I2\cdot(EtOH)_{n} + I2 &amp;lt;=&amp;gt;[\ce{EtOH}] I+\cdot(EtOH)_{n} + I3-} $$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-21T10:58:47.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27614" PostTypeId="1" AcceptedAnswerId="27620" CreationDate="2015-03-21T11:43:09.357" Score="3" ViewCount="1890" Body="&lt;p&gt;According to my textbook, all the protons in $\ce{^{12}C_6{}^{1}H_6}$ are magnetically equivalent, but those in $\ce{^{13}C_6{}^{1}H_6}$ are not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule. I do not understand this with respect to benzene as I expect Hc to couple differently to Ha than Hb to Ha and thus for them to be magnetically inequivalent irrespective of the isotope. &lt;img src=&quot;https://i.stack.imgur.com/yQ3Uw.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="4231" LastEditDate="2015-03-21T14:33:49.220" LastActivityDate="2015-03-21T16:38:52.960" Title="NMR: Magnetic and chemical equivalence in benzene" Tags="&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27617" PostTypeId="2" ParentId="27612" CreationDate="2015-03-21T13:42:10.510" Score="1" Body="&lt;p&gt;I would make it easy for yourself and convert your µg/L to g/L. You can then divide it by the formula weight to obtain the molarity (in M). After this it is just a simple conversion to mM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose in Excel you can create a new column and use a formula something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=((CELL/10&lt;sup&gt;6&lt;/sup&gt;)/Formula weight)*1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your example of phosphate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=((0.00000773)/94.97)*1000&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=0.00008139 mM // 8.139x10&lt;sup&gt;-5&lt;/sup&gt; mM&lt;/p&gt;&#xA;" OwnerDisplayName="user15085" LastEditorDisplayName="user15085" LastEditDate="2015-03-21T13:48:41.687" LastActivityDate="2015-03-21T13:48:41.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27620" PostTypeId="2" ParentId="27614" CreationDate="2015-03-21T14:32:50.987" Score="7" Body="&lt;p&gt;All of the protons in $\ce{^{12}C_6{}^{1}H_6}$ benzene are equivalent by symmetry, therefore they are &lt;strong&gt;chemical shift equivalent&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wrote, &quot;I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other &lt;strong&gt;nuclei&lt;/strong&gt; in the molecule.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the correct statement would be, for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other &lt;strong&gt;non-chemical shift equivalent nuclei&lt;/strong&gt; in the molecule.  If there are no other non-chemical shift equivalent nuclei in the molecule, then the chemical shift equivalent nuclei are also magnetically equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{^{12}C_6{}^{1}H_6}$ benzene, all of the protons are equivalent by symmetry, therefore they are chemically equivalent and since there are no &quot;different&quot; nuclei (non-chemical shift equivalent nuclei) for them to couple with, they are also magnetically equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{^{13}C_6{}^{1}H_6}$ benzene all of the protons are equivalent by symmetry so they are all chemical shift equivalent.  Since all of the carbons are $\ce{^{13}C}$ (spin=1/2), there are now &quot;different&quot; (the $\ce{^{13}C}$ nuclei are non-chemical shift equivalent) nuclei for the chemical shift equivalent protons to couple with; therefore the protons will be magnetically non-equivalent.  The spectrum is now an AA'A'A&quot;A&quot;A&quot;'XX'X'X&quot;X&quot;X&quot;' (A=proton, X=carbon-13) spectrum.  Since the protons are not magnetically equivalent (at least some of) the couplings will be observable.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-21T16:38:52.960" LastActivityDate="2015-03-21T16:38:52.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27621" PostTypeId="2" ParentId="26300" CreationDate="2015-03-21T14:54:08.083" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At the anode, hydrogen is oxidised (losing electrons). My first question is this: why is the anode negative if it is attracting electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not attracting electrons, it is attracting hydrogen atoms. The electrode binds the hydrogen and releases the electrons. The electrons are then able to flow out through the wire.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My second question refers to the fact that once the hydrogen atoms are iodised, the H+ ions move through the electrolyte towards the oxygen ions at the cathode, reacting to form water. If the anode is negative, why would positive H+ ions move away from it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once the electrons have returned to the cathode, the cathode is more negative than the anode. The positive ions move to the cathode and the negative ions move to the anode.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-03-21T14:54:08.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27622" PostTypeId="1" CreationDate="2015-03-21T14:55:35.533" Score="3" ViewCount="7372" Body="&lt;p&gt;From what I understand, adding $\ce{NaOH}$ to a solution containing transition metal ions gives transition metal hydroxide precipitates. Why is $\ce{NaOH}$ soluble but transition metal hydroxides insoluble? I have a feeling it has something to do with ligands and complexes formed, which I am familiar with.&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T14:54:19.990" LastActivityDate="2016-01-20T13:07:41.333" Title="Why are transition metal hydroxides insoluble?" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;ionic-compounds&gt;&lt;transition-metals&gt;&lt;precipitation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27623" PostTypeId="1" CreationDate="2015-03-21T15:02:23.233" Score="2" ViewCount="2108" Body="&lt;p&gt;I am very much confused: in the cell (Zn, Cu in $\ce {H2SO4}$), Cu is positively charged and is anode, where in this cell (Zn in $\ce {ZnSO4}$ and Cu in $\ce {CuSO4}$), Cu is positively charged but is Cathode. So how can I define it, where both are electric or voltaic cell. Please help me. Please see these picture for better understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CopiE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/221Tt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJ5C5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1rwq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;reference:A Text Book Of Electrical Technology,Volume 1,B.L.Theraja &lt;/p&gt;&#xA;" OwnerUserId="15089" LastEditorUserId="4934" LastEditDate="2015-03-21T16:01:06.510" LastActivityDate="2015-03-21T17:14:38.147" Title="Anode and Cathode Understanding" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27624" PostTypeId="1" CreationDate="2015-03-21T15:03:33.457" Score="1" ViewCount="2416" Body="&lt;p&gt;For instance, copper can form $\ce{[Cu(H2O)6]^2+}$ so it accepts 6 electron pairs from $\ce{H2O}$ molecules to form dative covalent bonds. However it has one electron in the 4s subshell and 10 in the 3d subshell. To which shell does it accept these electrons? Is it to the 4d subshell? And how are these accepted electrons arranged?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorDisplayName="user7951" LastEditDate="2015-04-12T08:33:19.797" LastActivityDate="2015-04-12T08:33:19.797" Title="How can transition metals form so many bonds with ligands?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-04-12T16:56:42.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="27625" PostTypeId="2" ParentId="27624" CreationDate="2015-03-21T15:11:25.030" Score="1" Body="&lt;p&gt;Copper is  in +2 oxidation state so it has configuration $\ce{[Ar] 3d^9}$. Since 4s, 4p and 4d orbitals are vacant, Cu will show $\ce{sp^3d^2}$ hybridisation and will accept these elctron pairs in its 6 $\ce{sp^3d^2}$ hybridised orbitals. Read more about VBT here &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/valence.php&quot; rel=&quot;nofollow&quot;&gt;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/valence.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2015-03-21T15:11:25.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27627" PostTypeId="1" CreationDate="2015-03-21T15:29:57.330" Score="12" ViewCount="38549" Body="&lt;p&gt;Larger value of b signifies that the size of the molecules is larger and hence the forces acting should become larger due to increased surface area.This should ensure easier liquefication. But my book says that smaller value of b corresponds to easier liquefication. Where am going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says &quot;Chlorine is more easily liquefied than Ethane because 'a' for  chlorine is greater than 'a' for ethane but 'b' for fluorine is less than 'b' for ethane. Van der Waals constant 'a' is due to force of attraction and 'b' due to finite size of molecules. Thus, greater the value of 'a' and smaller the value of b, larger the liquefaction&quot;&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="16683" LastEditDate="2016-09-30T09:34:03.910" LastActivityDate="2018-09-13T05:43:58.627" Title="Why does small value of van der Waals gas constant &quot;b&quot; ensure easier liquefication?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;equation-of-state&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27628" PostTypeId="1" AcceptedAnswerId="27643" CreationDate="2015-03-21T15:43:13.613" Score="8" ViewCount="575" Body="&lt;p&gt;I have long known that mixing bleach and peroxide produces oxygen gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOCl + H2O2 -&amp;gt; O2 + H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I only recently learned that the specific form of oxygen produced by the reaction is &lt;a href=&quot;http://en.wikipedia.org/wiki/Singlet_oxygen&quot; rel=&quot;noreferrer&quot;&gt;singlet oxygen&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOCl + H2O2 -&amp;gt; O2(a^1\Delta_{g}) + H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple chemical demonstration that can show the difference between singlet oxygen and triplet (ground state) oxygen?  Will singlet $\ce{O2}$ relight a splint &quot;faster&quot;, or decolorize a dye faster when bubbled through solution, or otherwise be distinguishable from triplet oxygen in a home lab?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="418" LastEditDate="2015-03-21T21:06:01.090" LastActivityDate="2018-07-26T17:53:07.620" Title="A unique demo for singlet oxygen?" Tags="&lt;home-experiment&gt;&lt;electronic-configuration&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27629" PostTypeId="2" ParentId="27627" CreationDate="2015-03-21T15:43:38.853" Score="4" Body="&lt;p&gt;The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written in &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1429218126&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atkins Physical Chemistry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tMXXQ.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-03-22T11:42:32.700" LastActivityDate="2015-03-22T11:42:32.700" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27630" PostTypeId="2" ParentId="27623" CreationDate="2015-03-21T15:51:30.760" Score="1" Body="&lt;p&gt;Here Cu plate is cathode and Zn plate is anode, it's written wrong in it. In the first case (Zn in  $\ce{ZnSO4}$ and Cu in $\ce{CuSO4}$) the electrodes are connected through an external circuit (You see that bulb glowing?). But what this article is trying to say is that in the cell (Zn, Cu in $\ce{H2SO4}$), if you leave the solution as it is without connecting externally, Zn having higher tendency to lose electrons, will lose electrons and these will appear on its plate surface making it negatively charged, the $\ce{Zn^2+}$ thus formed will combine with $\ce{ SO4^2-}$ and form $\ce{ZnSO4}$. The remaining two $\ce{H^+}$ ions will more towards copper plate where they will combine with the electrons on its plate and make it positive charged, releasing $\ce{H2}$. Now if a wire is connected across this setup then the electrons on Zn surface will leave it and go towards positive charged Cu plate.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2015-03-21T17:14:38.147" LastActivityDate="2015-03-21T17:14:38.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27631" PostTypeId="1" CreationDate="2015-03-21T16:05:17.743" Score="10" ViewCount="12738" Body="&lt;p&gt;Why is it that some liquid mixtures (that exhibit positive/negative deviation from Raoult's law) form azeotropic mixtures at certain compositions? What is the physical reason behind this; are there any extra forces holding the two components together? If so, please elaborate on this. &lt;/p&gt;&#xA;" OwnerUserId="15090" LastEditorUserId="41328" LastEditDate="2019-02-19T07:49:42.897" LastActivityDate="2021-04-27T20:27:29.623" Title="Reason for the formation of azeotropes" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;reference-request&gt;&lt;boiling-point&gt;&lt;vapor-pressure&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27633" PostTypeId="2" ParentId="27628" CreationDate="2015-03-21T16:43:37.707" Score="6" Body="&lt;p&gt;There is a great amount of relevant information in &lt;a href=&quot;https://doi.org/10.1021/ja01006a023&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemistry of singlet oxygen. IV. Oxygenations with hypochlorite-hydrogen peroxide&lt;/a&gt; &lt;em&gt;J. Am. Chem. Soc.,&lt;/em&gt; &lt;strong&gt;1968,&lt;/strong&gt; &lt;em&gt;90&lt;/em&gt; (4), 975–981.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lifetime of the singlet oxygen in solution will be on the order of a mircosecond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions of the formed singlet oxygen with tetramethylethylene, tetraphenylcyclopentadienone and several other compounds are described.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For tetraphenylcyclopentadienone, it says &quot;The reaction was followed by loss of the intense color of tetraphenylcyclopentadienone.&quot;, which sounds good for a demonstration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effects of temperature, concentration and solvent are discussed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solvent selection is apparently very important.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5026" LastEditDate="2018-07-26T17:53:07.620" LastActivityDate="2018-07-26T17:53:07.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27634" PostTypeId="2" ParentId="27627" CreationDate="2015-03-21T16:55:50.337" Score="4" Body="&lt;p&gt;The virial equation of state reads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = 1 + \left(b - \frac{a}{RT}\right)\frac{1}{V_\mathrm{m}} + \left(\frac{b}{V_\mathrm{m}}\right)^2 \cdots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going by this equation, it seems like increasing the value of $b$ will increase $Z$, which means repulsive forces will increase, making the gas liquefaction difficult. Also for $a = 0$ and $b = 0$, the gas becomes an ideal gas whose liquefaction is not possible, so I think what you are saying is incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question you are talking about, that why chlorine is more easily liquefied than ethane? It is because $a$ of $\ce{Cl2}$ is greater than $a$ of $\ce{C2H6}$, which means $\ce{Cl2}$ is easier to liquify. $\ce{C2H6}$, being greater in size than $\ce{Cl2}$, has a higher value of $b$. It's wrong to say &quot;the smaller the value of $b$, the larger the liquefaction&quot;. &lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="16683" LastEditDate="2015-11-02T16:40:00.787" LastActivityDate="2015-11-02T16:40:00.787" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27635" PostTypeId="1" AcceptedAnswerId="27638" CreationDate="2015-03-21T17:27:18.440" Score="13" ViewCount="7717" Body="&lt;p&gt;I've read in a book that the main factor for determining S&lt;sub&gt;N&lt;/sub&gt;2 reaction rate is steric hindrance. The lesser it is, the faster the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{KI}$ in acetone undergoes S&lt;sub&gt;N&lt;/sub&gt;2 reaction with each of $\ce{P,Q,R}$ and $\ce{S}$. The rates of reaction vary as:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkYaX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand how $\ce{S}$ is the fastest. In my opinion it should be the slowest as it is very sterically hindered. What is going on here?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; Joint Entrance Exam (JEE) 2013 India&lt;/p&gt;&#xA;" OwnerUserId="11587" LastEditorUserId="5026" LastEditDate="2018-04-04T01:28:51.403" LastActivityDate="2018-04-04T14:56:36.500" Title="Comparing SN2 reaction rates" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27636" PostTypeId="1" CreationDate="2015-03-21T17:27:49.110" Score="3" ViewCount="4597" Body="&lt;p&gt;I realise that there is a similar question here &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7304/difference-between-sodium-ion-and-a-transition-metal-ion-dissolving-in-water&quot;&gt;Difference between sodium ion and a transition metal ion dissolving in water?&lt;/a&gt; and it seems to answer my question, however I was reading about how aluminion ions form complexes in water, and aluminium is not a transition metal. So how can aluminium form complexes like transition metal ions, but sodium does not?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2016-08-28T16:44:25.553" Title="Does sodium form complexes like transition metal ions?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27637" PostTypeId="2" ParentId="27635" CreationDate="2015-03-21T18:06:53.167" Score="6" Body="&lt;p&gt;The reaction rate of S&lt;sub&gt;N&lt;/sub&gt;2 depends primarily on the leaving group involved. Since in your question, $\ce{-Cl}$ is common in all 4 options we move on to the next important criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial positive charge on carbon (or the electrophilicity of the carbon) is considered more important than the steric effect that you mentioned. &lt;a href=&quot;http://en.m.wikipedia.org/wiki/SN2_reaction#Factors_affecting_the_rate_of_the_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Look at Wikipedia's first line&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleophile attacks the more electrophilic carbon. This can be easily compared by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inductive effect&lt;/a&gt; of the substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us the correct answer as option B.&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="5026" LastEditDate="2018-04-04T14:56:36.500" LastActivityDate="2018-04-04T14:56:36.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27638" PostTypeId="2" ParentId="27635" CreationDate="2015-03-21T18:11:20.717" Score="11" Body="&lt;p&gt;There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction.  Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors.  Considering only steric factors for a moment we would order the compounds as &lt;strong&gt;P&gt;S~R&gt;Q&lt;/strong&gt;.  But now let's consider electronic factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the allyl system in compound &lt;strong&gt;R&lt;/strong&gt;, the transition state looks something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gyNRh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map:_Robert&amp;#39;s_%22Basic_Principles_of_Organic_Chemistry%22/14:_Organohalogen_%26_Organometallic_Compounds/14.03A:_2-Propenyl_(Aliylic)_Halides&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The p orbitals in the adjacent double bond overlap with the p orbital at the $\mathrm{S_{N}2}$ reaction center (remember that $\mathrm{S_{N}2}$ transition states are roughly $\ce{sp^2}$ hybridized) and stabilizes the transition state through resonance delocalization.  Note that the transition state is electron rich, there is a full unit of negative charge in the transition state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a carbonyl is placed next to the $\mathrm{S_{N}2}$ reaction center, we replace the $\ce{C=C}$ double bond in the above drawing with a $\ce{C=O}$ double bond.  Again there is overlap and resonance stabilization of the transition state.  But in the carbonyl case the stabilization is even greater than with the $\ce{C=C}$ double bond because the carbonyl carbon is positively polarized and inductively stabilizes the electron rich (remember that negative charge) transition state even more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z7Vc1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_11%3A_Nucleophilic_carbonyl_addition_reactions/Section_11.1%3A_Nucleophilic_additions_to_aldehydes_and_ketones%3A_the_general_picture&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we would expect the allyl case (&lt;strong&gt;R&lt;/strong&gt;) to be faster than the ethyl case, in your series of compounds apparently it is not faster than the methyl case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: often the allyl system will react a bit faster than the methyl analogue, but the two rates are usually close)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &lt;strong&gt;additional inductive stabilization&lt;/strong&gt; from the carbonyl allows it to react faster than the methyl compound so our final reactivity order becomes &lt;strong&gt;S&gt;P&gt;R&gt;Q&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-21T19:13:07.720" LastActivityDate="2015-03-21T19:13:07.720" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="27640" PostTypeId="2" ParentId="27636" CreationDate="2015-03-21T20:02:59.270" Score="12" Body="&lt;p&gt;Although less common than transition metal complexes, sodium does form complexes with some ligands, particularly oxygen based ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aqua complexes are formed in aqueous solution, the most common being $\ce{[Na(H2O)6]+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium forms many complexes with crown ethers, cryptands and other related ligands. For example, 15-crown-5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GrC4j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-03-21T20:02:59.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27641" PostTypeId="2" ParentId="27635" CreationDate="2015-03-21T20:05:31.683" Score="2" Body="&lt;p&gt;If you'll look at the transition state in SN2 mechanism, you'll find that there's a partial negative charge on carbon.&#xA;In option S, The leaving group is attached to a 1 degree carbon and there will be a -M by the carbonyl group making the negative charge most stable.&#xA;In P, we have a 1 degree carbon with minimum hindrance.&#xA;In R, we have a 1 degree carbon and on which the negative charge can be stabilized by resonance.&#xA;In Q, we have a 2 degree carbon means steric hindrance, further the +I effect of CH3 will make the negative charge on carbon even worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So correct answer is B.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2015-03-21T20:05:31.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27642" PostTypeId="2" ParentId="27086" CreationDate="2015-03-21T20:41:39.537" Score="2" Body="&lt;p&gt;About the first two, it's an exception to the rule that resonance dominates hyperconjugation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the last inequality, the vinyl cation is sp hybridized which means it is highly electronegative, and a positive charge on it will be very unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the middle inequality, the benzyl carbocation (with five resonance structures) is more stable than allyl carbocation (two resonance structures). You may verify this &lt;a href=&quot;https://books.google.co.in/books?id=QV6cwXA9XkEC&amp;amp;pg=PA120&amp;amp;lpg=PA120&amp;amp;dq=allyl%20cation%20and%20benzyl%20cation&amp;amp;source=bl&amp;amp;ots=qgCt_EJAmc&amp;amp;sig=SZGKF7jaIhzhxTOxs7u5JbuhY3I&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=5tQNVeXkEZeIuATt24CwCA&amp;amp;ved=0CEUQ6AEwCA#v=onepage&amp;amp;q=allyl%20cation%20and%20benzyl%20cation&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2015-03-21T20:41:39.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27643" PostTypeId="2" ParentId="27628" CreationDate="2015-03-21T20:54:15.673" Score="11" Body="&lt;p&gt;&lt;em&gt;Admittedly, I'm having a professional photochemical bias when it comes to the generation of singlet oxygen!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rather safe and simple approach - even in home experiments - is the use of air, sunlight and a sensitizer (rose bengal, methylene blue, or even chlorophyll from natural resources, such as &lt;em&gt;Spirulina&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the sensitizer is irradiated and excited to the first singlet state, undegoes intersystem crossing (ISC) and transfers the energy to ground state triplet oxygen, upon which singlet oxygen is formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{Sens &amp;amp;-&amp;gt;[h\nu]\ ^1Sens^{\ast}}\\&#xA;\ce{^1Sens^{\ast} &amp;amp;-&amp;gt;[ISC]\ ^3Sens^{\ast}}\\&#xA;\ce{^3Sens^{\ast} +\ ^3O2&amp;amp;-&amp;gt;\ Sens +\ ^1O2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prominent reaction of $\ce{^1O2}$ is the concerted, diastereoselective addition to 1,3-dienes in Diels-Alder fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LSsuH.png&quot; alt=&quot;ascaridole synthesis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A classical example for this reaction is the synthesis of the natural peroxide &lt;em&gt;ascaridole&lt;/em&gt; (&lt;strong&gt;2&lt;/strong&gt;) from $\alpha$-terpinene (&lt;strong&gt;1&lt;/strong&gt;). &lt;em&gt;G. O. Schenck&lt;/em&gt;, the former director of the Max Planck institute for Radiation Research, performed the reaction in batteries of flasks in his garden and published the results in 1944 (&lt;a href=&quot;https://doi.org/10.1007/BF01467891&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Naturwissenschaften&lt;/em&gt; &lt;strong&gt;1944,&lt;/strong&gt; &lt;em&gt;32&lt;/em&gt; (14-26), 157–157&lt;/a&gt;). Ascaridole is a natural anthelmintic (= against parasitic worms), hence the name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe not really suitable for home experiments due to the chromatographic workup envolved but worth to be mentioned is the &lt;em&gt;Schenck ene reaction&lt;/em&gt; of allylic alcohols:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uvrsc.png&quot; alt=&quot;schenck-ene-reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction products can be converted to &lt;em&gt;1,2,4-trioxanes&lt;/em&gt; by peroxyacetalization, as shown by &lt;em&gt;Griesbeck&lt;/em&gt; and coworkers (&lt;a href=&quot;https://doi.org/10.3762/bjoc.6.61&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Beilstein J. Org. Chem.&lt;/em&gt; &lt;strong&gt;2010,&lt;/strong&gt; &lt;em&gt;6&lt;/em&gt; (61)&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5UYEI.png&quot; alt=&quot;1,2,4-trioxanes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the &lt;em&gt;1,2,4-trioxane&lt;/em&gt; moiety is the key element in the natural &lt;strong&gt;anti-malaria&lt;/strong&gt; drug *artemisinin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MX9Ow.png&quot; alt=&quot;artemisinin&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-07-26T17:52:38.797" LastActivityDate="2018-07-26T17:52:38.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27645" PostTypeId="1" AcceptedAnswerId="27659" CreationDate="2015-03-22T00:09:18.710" Score="2" ViewCount="4748" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Passivation_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;Surface passivation&lt;/a&gt; is done to deal with &quot;dangling bonds&quot; (for example, $\mathrm{SiO_2}$ layer on top of a $\mathrm{Si}$ layer). But isn't the new surface will have dangling bonds on its own at the new formed surface?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-03-22T20:00:30.880" Title="How surface passivation works?" Tags="&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27646" PostTypeId="1" CreationDate="2015-03-22T00:35:48.037" Score="5" ViewCount="6514" Body="&lt;p&gt;I got this problem in school as a tool to get ready for International Chemistry Olympiad, but I have some problems with this challenge.&#xA;So, the problem is to calculate as precise as possible the Gibbs free energy for $\ce{CaCO3}$ decomposition at 1200 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\mathrm{d}G=\mathrm{d}H-T\cdot\mathrm{d}S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\mathrm{d}S$ and $\mathrm{d}H$ are temperature dependant, so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}H=\mathrm{d}H_0+\int_{T_0}^TC_p\mathrm{d}T=\mathrm{d}H_0+\mathrm{d}C_p\cdot\left(T-T_0\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}S=\mathrm{d}S_0+\ln\left(T/T_0\right)\cdot\mathrm{d}C_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But molar heat capacity at constant pressure is also temperature dependant, and the equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_p=a+b\cdot T+c/T^2$, where a, b, c coefficients for each compound at specific phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is with both solids, because this equation can not be used for them, because in 1200 K they are not gases and Einstein’s Quantum Mechanics and Debye models do not work, because Einstein’s approximates are maxed at $C_p\approx3R$, which is about 24.9, but $C_p$ at 298 K, 1 atm for both solids in tables are &gt; 24.9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the problem? I would appreciate any possible lead to mistake/ correct answer.&lt;/p&gt;&#xA;" OwnerUserId="15093" LastEditorDisplayName="user7951" LastEditDate="2015-04-18T14:43:43.700" LastActivityDate="2015-10-17T02:33:44.347" Title="Heat Capacity temperature dependence and Gibbs energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;enthalpy&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27647" PostTypeId="1" CreationDate="2015-03-22T00:40:02.570" Score="0" ViewCount="1386" Body="&lt;p&gt;I did the iodine clock reaction using sodium metabisulfite and potassium iodate. As I understand it, increasing concentration is supposed to speed up your reaction rate. I did 10 trials. The first 5 I varied the concentration of metabisulfite and kept the iodate constant. Trials 6-10 I varied the potassium iodate and kept the meta constant. I don't understand why as I increased concentration, the reaction rate became slower? For example in trial 6 I had 2 ml of sodium metabisulfite and 4 ml of potassium iodate, that only took 7 seconds. In trial 10 I had 2 ml of meta and 10 ml of potassium iodate and that took 17 seconds? Does anyone know what I did wrong? Increasing concentrations should have increased collisions and made the rate faster, but it did the opposite! Please help me figure out why things went wrong!&lt;/p&gt;&#xA;" OwnerUserId="15094" LastEditorUserId="4231" LastEditDate="2015-03-22T17:31:26.403" LastActivityDate="2015-03-22T17:31:28.927" Title="Experimental difficulty with the iodine clock reaction" Tags="&lt;experimental-chemistry&gt;&lt;kinetics&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27648" PostTypeId="2" ParentId="19735" CreationDate="2015-03-22T01:28:57.197" Score="2" Body="&lt;p&gt;The reason that there is so much confusion over this is that the usual form of the Clausius inequality appearing in most references is not completely and properly specified.  In an irreversible process, the temperature within the system is not going to be spatially uniform, so there is considerable ambiguity as to what value of the temperature T to use in calculating the integral of dq/T.  The correct value to use is the temperature &lt;em&gt;at the interface between the system and surroundings&lt;/em&gt; (at which the heat transfer is occurring), $T_I$.  So the correct form of the Clausius inequality should more properly read:&#xA;$$\Delta S \ge \int {\frac{dq}{T_I}}$$&#xA;So, for example, if the surroundings consists of a constant temperature bath, the T in the Clausius inequality is the temperature of the bath (i.e., $T_I=T_{bath}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the path between the initial and final equilibrium states of the system is reversible, then the system temperature is uniform throughout and equal to the interface temperature, such that $T_I=T$, and the equality in the Clausius relationship applies.  The integral then gives the change in entropy.  For any irreversible path between the same two thermodynamic equilibrium states, the value of the integral is less than the integral for a reversible path, $\Delta S$.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="15095" LastEditDate="2015-12-02T15:35:25.460" LastActivityDate="2015-12-02T15:35:25.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27649" PostTypeId="1" CreationDate="2015-03-22T05:00:46.653" Score="7" ViewCount="23401" Body="&lt;p&gt;The change in enthalpy of carbon in diamond form to graphite is negative. This suggests that graphite should be more stable than diamond. Is it true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought was that graphite has van der Waals forces while diamond has covalent forces so diamond should be more stable. But the melting point of graphite is more than diamond which implies that it is thermodynamically more stable. I am thoroughly confused. So which is more stable?&lt;/p&gt;&#xA;" OwnerUserId="15096" LastEditorUserId="23561" LastEditDate="2019-08-23T16:16:55.860" LastActivityDate="2019-08-23T16:16:55.860" Title="Is graphite more stable than diamond?" Tags="&lt;thermodynamics&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27650" PostTypeId="1" AcceptedAnswerId="27657" CreationDate="2015-03-22T05:00:55.697" Score="4" ViewCount="360" Body="&lt;p&gt;From &lt;a href=&quot;https://chemistry.stackexchange.com/q/27455/1071&quot;&gt;Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done &amp;#39;under nitrogen&amp;#39;?&lt;/a&gt; I confirmed that nitrogen is used to exclude oxygen - in order to prevent oxidation of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can $\ce{CO_2}$ be substituted for nitrogen, or will the presence of carbon dioxide slow down the decarboxylation reaction (which is &lt;em&gt;causing carbon dioxide to be released&lt;/em&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="1071" LastEditorUserId="5026" LastEditDate="2018-04-18T02:13:24.043" LastActivityDate="2018-04-18T02:13:59.033" Title="Can decarboxylation be done under carbon dioxide?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27652" PostTypeId="1" CreationDate="2015-03-22T05:58:04.397" Score="6" ViewCount="3260" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wXYhc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At room temperature, alpha and alpha+$\ce{Fe3C}$ are present (depending on carbon content), which is ferrite and pearlite. In the lab we polished and etched samples of carbon steel to observe the microstructure, and saw both ferrite grains AND pearlite grains. Not unlike the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KLmf9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from my understanding of the phase diagram, only one phase should be present. For any steel with less than 0.02% carbon we should see only ferrite, and for steel with &gt;0.02% carbon we should see only pearlite. &lt;strong&gt;&lt;em&gt;Though this was not the case, and rather there was a mix of pearlite and ferrite,&lt;/em&gt;&lt;/strong&gt; with the higher carbon content steel having a higher proportion of pearlite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be much appreciated if someone could explain to me what is happening here. Links to further reading would be helpful but not required.&lt;/p&gt;&#xA;" OwnerUserId="5237" LastEditorUserId="11310" LastEditDate="2015-05-02T15:34:56.073" LastActivityDate="2020-11-03T04:08:59.470" Title="Having trouble understanding the iron-carbon phase diagram" Tags="&lt;crystal-structure&gt;&lt;phase&gt;&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27653" PostTypeId="2" ParentId="1246" CreationDate="2015-03-22T07:14:31.257" Score="-1" Body="&lt;p&gt;phosphorus has 5 valence electrons.each chlorine has 7 valence electrons.phosphorus give 1 electron to each chlorine and hence all the 6 attain 8 electrons in the outer shell thus forming an ionic bond. &lt;/p&gt;&#xA;" OwnerUserId="15098" LastActivityDate="2015-03-22T07:14:31.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27655" PostTypeId="1" AcceptedAnswerId="27693" CreationDate="2015-03-22T07:38:38.450" Score="3" ViewCount="634" Body="&lt;h3&gt;Question&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a cubic lattice of &lt;span class=&quot;math-container&quot;&gt;$\ce{XYZ},$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; atoms are present at all corners except one corner which is occupied by &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atoms. &lt;span class=&quot;math-container&quot;&gt;$\ce{Z}$&lt;/span&gt; atoms are present at face centres. What is the formula of compound?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Answer&lt;/h3&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{X2YZ24}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;My solution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Let the number of atoms in a unit cell be &lt;span class=&quot;math-container&quot;&gt;$N.$&lt;/span&gt; Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;N(\ce{X}) &amp;amp;= 7 × \frac{1}{8} = \frac{7}{8}\\&#xA;N(\ce{Y}) &amp;amp;= 1 × \frac{1}{8} = \frac{1}{8}\\&#xA;N(\ce{Z}) &amp;amp;= 6 × \frac{1}{2} = 3&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the ratio would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N(\ce{X}):N(\ce{Y}):N(\ce{Z}) = \frac{7}{8}:\frac{1}{8}:3,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same thing as &lt;span class=&quot;math-container&quot;&gt;$7:1:24.$&lt;/span&gt; So, the formula is &lt;span class=&quot;math-container&quot;&gt;$\ce{X7YZ24}.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="6845" LastEditorUserId="83758" LastEditDate="2020-03-27T07:49:58.767" LastActivityDate="2020-03-27T07:49:58.767" Title="Chemical formula of a compound crystallized in a cubic lattice" Tags="&lt;solid-state-chemistry&gt;&lt;crystallography&gt;&lt;chemical-formula&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27657" PostTypeId="2" ParentId="27650" CreationDate="2015-03-22T10:35:50.503" Score="2" Body="&lt;p&gt;Because nitrogen is the standard inert gas in lab practice. If you are &quot;rich&quot; You can use argon, but carbon dioxide is seldom used, because it won't work on alkaline solutions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;or will the presence of carbon dioxide slow down the reaction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, what you presumably mean is influencing the equilibrium unfavourably, don't you? You should learn to discern kinetics and equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: not really. Whether you have a carbon dioxide atmosphere or not will shift the decomposition temperature not more than some degrees.&lt;/p&gt;&#xA;" OwnerUserId="105" LastEditorUserId="5026" LastEditDate="2018-04-18T02:13:59.033" LastActivityDate="2018-04-18T02:13:59.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27658" PostTypeId="2" ParentId="27649" CreationDate="2015-03-22T11:02:56.537" Score="11" Body="&lt;p&gt;At normal room temperature and pressure graphite is (slightly) more stable than diamond. But the melting point is not a good indicator of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Melting points are determined by the bonding structure of the solid and any potential liquid (indeed there may not be a liquid phase at some pressures). They don't necessarily tell you whether a different bonding structure is more thermodynamically stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermodynamic stability of different carbon structures depends on subtle tradeoffs among several factors in the bonding of the various solids (allotropes) than can be formed with different bonding arrangements. There is no easy way to tell which is the most thermodynamically stable other than by measurement or some very high-powered calculations about the quantum mechanics of the bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The different allotropes have radically different structures (tetrahedral in diamond, a ball of hexagons and pentagons in buckminsterfullerene and flat plates of hexagons in graphite). Changing one into another requires a lot of bond breaking which isn't going to happen much at room temperature. Interconversion is possible at high pressures and temperatures when carbon is dissolved in liquid rock or metal and the carbon allotrope that forms is more likely to be whatever is stable at that whatever the pressure is (diamond is more stable at higher pressure which is why the earth or industry needs very high pressures to create diamond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for most practical purposes, it doesn't matter than graphite is the more stable allotrope: your diamonds are not going to spontaneously turn to ash during your, or the universe's, lifetime.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2015-06-15T18:50:45.497" LastActivityDate="2015-06-15T18:50:45.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27659" PostTypeId="2" ParentId="27645" CreationDate="2015-03-22T11:19:21.697" Score="4" Body="&lt;p&gt;Surface passivation creates different chemistry on the surface than the chemistry of the bulk solid. Imagine bulk Si with a clean surface (presumably in a vacuum and freshly made). It can be thought of as having &quot;dangling bonds&quot;, though this is not necessarily the best way to think about it. Think more generally that the structure on the surface is different to the bulk structure. There may be different elements present and the bonds may be &quot;satisfied&quot;. Or they may be strained compared to the bulk material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expose the Si to, for example, O&lt;sub&gt;2&lt;/sub&gt; and the surface layer (your dangling bonds, if you like) will react to give some SiO&lt;sub&gt;2&lt;/sub&gt; just on the top layer. This new layer may be much less reactive than bulk silicon in normal air, hence can be said to passivate the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chemical and physical properties of the new layer are right, the layer will protect the bulk from further reaction: this is the definition of passivation. A similar situation occurs naturally for aluminium, for example, where the Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; layer that forms is both strong and relatively impermeable to air preventing further reaction (aluminium with the layer removed will react so rapidly with oxygen it gets warm and disintegrates). Iron, on the other hand, forms rust which doesn't protect the surface as it is flaky and weaker than the bulk metal. To passivate iron you need to add other elements which encourage the formation of stronger passivating layers (this is how many steels are made not to rust).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemical manipulation of the surface can create specific types of passivation that serve a particular purpose rather than relying on the natural reactions of air and the surface. This is common in the semiconductor industry and can also be used to generate the coloured layers on aluminium and titanium products where the goal is aesthetic as much as protective.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="10104" LastEditDate="2015-03-22T20:00:30.880" LastActivityDate="2015-03-22T20:00:30.880" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27661" PostTypeId="1" AcceptedAnswerId="27668" CreationDate="2015-03-22T12:31:58.347" Score="3" ViewCount="166" Body="&lt;p&gt;In a hydrogen atom only 1s orbital is filled. Does this mean that the energies of all other orbitals are equal? If so,why?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="10185" LastEditDate="2015-03-22T21:23:32.363" LastActivityDate="2015-03-22T21:23:32.363" Title="Are the energies of orbitals that are not filled in an atom, equal?" Tags="&lt;atoms&gt;&lt;orbitals&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27662" PostTypeId="1" AcceptedAnswerId="29651" CreationDate="2015-03-22T12:47:24.763" Score="6" ViewCount="44594" Body="&lt;p&gt;The electronic configuration of Fe is $\ce{[Ar] 3d^6 4s^2}$.&#xA;So after removing two electrons the configuration becomes:&#xA;$\ce{[Ar] 3d^6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why can't the electrons rearrange themselves to give a more stable&#xA;$\ce{[Ar] 3d^5 4s^1}$ configuration?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="25052" LastEditDate="2018-04-28T16:18:47.337" LastActivityDate="2018-06-02T23:42:30.043" Title="What is the electronic configuration of Fe(II) ion?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27663" PostTypeId="1" CreationDate="2015-03-22T13:01:49.120" Score="4" ViewCount="27643" Body="&lt;p&gt;I was reading about electrochemical cells and I really don't understand why a salt bridge is needed. Can't you just combine the solutions? I've found a lot of webpages explaining electrochemical cells but strangely enough none of them seem to find this important enough to explain. All thoughts appreciated :)&lt;/p&gt;&#xA;" OwnerUserId="15101" LastEditorUserId="10185" LastEditDate="2015-03-22T21:23:04.417" LastActivityDate="2015-03-23T11:10:46.817" Title="Why use a salt bridge at all in an electrochemical cell?" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27665" PostTypeId="1" CreationDate="2015-03-22T13:46:13.277" Score="0" ViewCount="122" Body="&lt;p&gt;My book says: frequency is considered as the only REAL characteristics of electromagnetic radiation, because it is unchangeble if the radiation comes up with some other environment, while the wavelength and the wavelength number do change. How can wavelength change and the frequency not ? The formula tells us very clear that while frequency increases, wavelength decreases and vice versa ?! &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-03-22T15:25:17.447" Title="Frequency, the only real characteristics of electromagnetic radiation ??" Tags="&lt;physical-chemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27666" PostTypeId="2" ParentId="27647" CreationDate="2015-03-22T13:54:09.203" Score="2" Body="&lt;p&gt;The theory is correct. Increasing concentrations increase the number of effective collisions per unit time and the rate increases. So accordingly even in the experiment it should happen like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in visualizing your situation, it must have been an experimental error and  not the theory. (Say there must have been less amount of $\ce{H2O2}$ in the mixture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this reaction what happens first is the conversion of $\ce{I-}$ to $\ce{I2}$ (have to add $\ce{H2O2}$). This $\ce{I2}$ will then react with metabisulphite or (even sodium thiosulphate) to produce $\ce{I-}$. Since you have added a constant value of meta in trials 6-10 there is a definite amount of $\ce{I2}$ removed in each trial by meta then the remaining $\ce{I2}$ will react with starch to give the color change.&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="4231" LastEditDate="2015-03-22T17:31:28.927" LastActivityDate="2015-03-22T17:31:28.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27668" PostTypeId="2" ParentId="27661" CreationDate="2015-03-22T14:44:07.350" Score="4" Body="&lt;p&gt;Sure, if they are unoccupied they contribute nothing to the total energy of the system. Remember that orbitals are mathematical constructs and only have meaning with regard to their energy when being physically occupied by an electron that adheres this mathematical description. The energy required to &lt;strong&gt;fill&lt;/strong&gt; such an orbital however differs (vastly) from orbital to orbital and nucleus to nucleus (see for example &lt;a href=&quot;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-22T14:44:07.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27669" PostTypeId="2" ParentId="27665" CreationDate="2015-03-22T14:48:46.463" Score="2" Body="&lt;p&gt;Recall that frequency $f$, wavelength $\lambda$ and the speed $v$ of a wave are all related as follows&#xA;$$&#xA;    v = f \lambda \, .&#xA;$$&#xA;So presumably, your book is talking about a situation when an electromagnetic wave crosses boundaries between different media. In such cases frequency indeed remains constant, while the speed of the wave, and consequently, its wavelength change.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-03-22T14:48:46.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27670" PostTypeId="2" ParentId="27665" CreationDate="2015-03-22T15:25:17.447" Score="0" Body="&lt;p&gt;Frequency of an electromagnetic wave is the characteristic of solely the source of the wave. A charge's frequency of oscillation would determine the frequency of the wave it produces. Refraction doesn't cause change in frequency, only wavelength, which results in a change in velocity as well.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastActivityDate="2015-03-22T15:25:17.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27671" PostTypeId="2" ParentId="27663" CreationDate="2015-03-22T15:49:53.890" Score="2" Body="&lt;p&gt;The use of a salt bridge is to maintain the electrical neutrality throughout the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A salt bridge will allow the movement of ions.&#xA;Electrons flow from anode to cathode.Due to this a negative charge will be developed in the cathode and a positive charge in the anode.This charge developed will oppose the electron flow (current from anode to cathode) and at one level the electron flow will be ceased.So the cell's action will be stopped.&#xA;So in order to prevent this, salt bridges are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar question in this site.Please refer that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/5477/why-is-it-important-to-use-a-salt-bridge-in-a-voltaic-cell-can-a-wire-be-used&quot;&gt;Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-03-22T16:08:20.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27672" PostTypeId="1" AcceptedAnswerId="27680" CreationDate="2015-03-22T16:15:49.770" Score="3" ViewCount="3427" Body="&lt;p&gt;Given the simple first order reaction $\ce{A -&amp;gt; P}$ derive the integrated rate law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k[\ce{A}]$$&#xA;Collect terms:&#xA;$$\frac{\mathrm d[\ce{A}]}{[\ce{A}]} = -k\,\mathrm dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the bit I need help with, the integration: &#xA;Apparently the integrated form of $\frac{\mathrm d[\ce{A}]}{[\ce{A}]}= \ln[\ce{A}]$ but I'm struggling to see exactly how. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step (usually omitted)I think should be to split the fraction before integration&#xA;$$\frac{\mathrm d[\ce{A}]}{1}\frac{1}{[\ce{A}]}=\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}=-kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}}=\int{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=\int{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to evaluate this I use the general integration law for a constant $\int{[\ce{A}]\,\mathrm dt} = [\ce{A}]t + C$ and reciprocal  $\int{\frac{1}{[\ce{A}]}=\ln[\ce{A}]+C}$. Therefore $$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=[\ce{A}]t + C_1+ \ln[\ce{A}]+C_2=\int{\frac{\mathrm d[\ce{A}]}{[\ce{A}]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is wrong because the textbooks say something different but I don't know how to correct it. Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorUserId="5026" LastEditDate="2018-04-18T02:15:12.183" LastActivityDate="2018-04-18T02:15:57.337" Title="Integration of the first order rate equation" Tags="&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27673" PostTypeId="1" CreationDate="2015-03-22T16:41:00.883" Score="-1" ViewCount="701" Body="&lt;p&gt;Do atoms combine with the nearest atoms possible because of lower delta E (less energy needed to form a stable nucleus)?&lt;/p&gt;&#xA;" OwnerUserId="15105" LastActivityDate="2015-03-22T22:54:19.063" Title="How do atoms combine?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-03-23T03:54:16.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="27674" PostTypeId="2" ParentId="27672" CreationDate="2015-03-22T17:00:38.980" Score="4" Body="&lt;p&gt;You omitted the limits for the integration and therefore miss the practical (and most important) aspect of the whole exercise! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember why all this integration is done: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is about measuring  $[\ce{A}_t]$ over time, starting with $[\ce{A_0}]$ at $t = 0$, in order to confirm a hypothesis on the rate law.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work the limits for the integration in, you'll reach a useful functional relation between concentration of the starting material and time - useful for a chemist and a biologist ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-22T17:00:38.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27675" PostTypeId="2" ParentId="27673" CreationDate="2015-03-22T17:22:58.953" Score="1" Body="&lt;p&gt;Let's say we have an atomic soup that contains atoms (what follows also applies to molecules) $\ce{A, B}$ and $\ce{C}$.  From these reactants it would be possible to form products $\ce{A-B, A-C}$ and $\ce{B-C}$.  Just which of these molecules, if any, will form depends on several factors including the $\Delta E$ that you mention.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We usually talk about a reaction coordinate, like the one shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6qvrm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sparknotes.com/chemistry/kinetics/mechanisms/section1.rhtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have our reactants on one side and products on the other, and a barrier connecting the two.  The height of the barrier is referred to as the activation energy ($E_{a}$).  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arrhenius equation&lt;/a&gt; is often used to describe the rate (k) for this process, where A is the pre-exponential factor (more on this in a minute), R is a constant and T is the temperature (°K)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = Ae^{-E_a/(RT)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back to the question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because atoms collide doesn't necessarily mean they will react to produce a product.  Sometimes they must collide from a certain direction.  Think of molecules with complex shapes, for them to react we need the other reactant to collide with a certain part of the larger molecule, maybe too from a certain angle or direction.  These geometrical factors are contained in the pre-exponential factor.  Also the collision must have enough energy to get the reactants over the energy barrier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just because molecules are close doesn't guarantee that they will react.  More important is how often molecules collide (concentration), the geometry of the collision (pre-exponential factor) and the energy (E) they carry.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-22T17:34:18.100" LastActivityDate="2015-03-22T17:34:18.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27677" PostTypeId="1" CreationDate="2015-03-22T17:54:01.493" Score="0" ViewCount="1846" Body="&lt;p&gt;Under constant pressure (and only PV work), change in enthalpy can be shown to equal change in heat. If enthalpy is not meaningful outside of constant pressure, then why not use heat instead? So I think it must be meaningful outside of constant pressure scenarios, but I couldn't figure it out myself. Can someone help?&lt;/p&gt;&#xA;" OwnerUserId="15104" LastEditorUserId="10185" LastEditDate="2015-03-22T21:22:32.107" LastActivityDate="2015-04-21T22:11:04.980" Title="What is the significance of enthalpy definition not under constant pressure?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27678" PostTypeId="1" CreationDate="2015-03-22T17:57:37.003" Score="1" ViewCount="671" Body="&lt;p&gt;I bought a phosphate buffer from Sigma/Fluka (73212-1L) containing potassium dihydrogen phosphate ($\ce{KH_2PO_4}$) and disodium hydrogen phosphate ($\ce{Na_2HPO_4}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is at pH 8.0 but they do not provide the phosphate concentration. Can I determine it from pH via titration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table shows the concentrations for a 0.2M solution:&#xA;&lt;a href=&quot;http://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#phosphate&quot; rel=&quot;nofollow&quot;&gt;http://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#phosphate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="8083" LastEditDate="2015-04-20T12:53:09.027" LastActivityDate="2015-04-20T12:53:09.027" Title="Determining concentration of phosphate buffer from pH?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27679" PostTypeId="2" ParentId="27672" CreationDate="2015-03-22T18:17:59.513" Score="7" Body="&lt;p&gt;You can't write $\int \frac{\mathrm{d}[A]}{[A]} $ as $\int \mathrm{d}[A] \int\frac{1}{[A]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac{1}{[A]}$ doesn't make any sense. Integration is essentially summation of infinite infinitely small terms and so each term must be infinitely small. $\frac{1}{[A]}$ isn't infinitely small. $\frac{\mathrm{d}[A]}{[A]}$ is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\int \frac{1}{[A]} \neq \ln([A]) + C$ instead&#xA;$\int \frac{\mathrm{d}[a]}{[A]} = \ln([A]) + C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as &lt;a href=&quot;https://chemistry.stackexchange.com/a/27674/13357&quot;&gt;Klaus Warzecha&lt;/a&gt; has stated, you forgot to put the limits.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="5026" LastEditDate="2018-04-18T02:15:57.337" LastActivityDate="2018-04-18T02:15:57.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27680" PostTypeId="2" ParentId="27672" CreationDate="2015-03-22T18:40:25.470" Score="10" Body="&lt;p&gt;Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="16683" LastEditDate="2016-11-21T19:12:54.797" LastActivityDate="2016-11-21T19:12:54.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27681" PostTypeId="1" CreationDate="2015-03-22T19:13:37.573" Score="6" ViewCount="150" Body="&lt;p&gt;I'd like someone to explain, scientifically, what makes some baby clothes' stains so hard to remove, and what the right way to treat the stains is, chemistry-wise. A lot of internet sites have solutions (pardon the pun) that are 'my experience' and anecdotal, which is better than nothing, but not always transferable (and sometimes may damage clothes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yeah... how do you get out the hard to remove orange stains (e.g. from iron, or carrot/sweet potato soup)? How about other things? Why? I'd love someone to make this the definitive guide to removing stains from baby clothes, if you can :D. (Feel free to expand beyond my stated question above so there's just a one-stop reference question here.)&lt;/p&gt;&#xA;" OwnerUserId="15109" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T18:36:13.330" LastActivityDate="2016-02-05T18:52:43.750" Title="Definitive Guide To Removing Stains From Baby Clothes" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27682" PostTypeId="2" ParentId="27677" CreationDate="2015-03-22T19:26:55.497" Score="2" Body="&lt;p&gt;There are many processes that take place at constant pressure, such as the change of phase from a liquid to a vapor for a pure substance at constant temperature.  Also, the heats of chemical reactions and heats of formation of compounds are typically cataloged at constant temperature and pressure.  So the enthalpy comes into play in both these instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first law of thermodynamics for a closed system is expressed conveniently in terms of the change in internal energy of the system.  However, for an open system (in which mass is flowing in and out) operating at steady state, the first law is more conveniently expressed in terms of the enthalpy, not the internal energy.  And, of course, all industrial applications of thermodynamics involving continuous processes are operating at steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chet&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-03-22T19:26:55.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27683" PostTypeId="1" CreationDate="2015-03-22T20:29:17.290" Score="5" ViewCount="1033" Body="&lt;p&gt;I was wondering why temperature regulation is an important factor for producing explosives. I know many are exothermic reactions but what would happen if you did not use an ice bath and allow the temperature to go too high (Considering the explosive substance has not already been formed in the reaction)? Would it explode? &lt;br/&gt;For example when mixing acetone with sulfuric acid and hydrogen peroxide to make TATP.&lt;/p&gt;&#xA;" OwnerUserId="15110" LastEditorUserId="46306" LastEditDate="2018-03-04T07:51:58.663" LastActivityDate="2018-03-04T07:51:58.663" Title="Why is temperature regulation an important factor for producing explosives?" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;explosives&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27684" PostTypeId="2" ParentId="27646" CreationDate="2015-03-22T20:32:10.003" Score="3" Body="&lt;p&gt;At constant pressure, $dG=-SdT$.  Also, the definition of G is $G=H-TS$, so $S=\frac{H-G}{T}$.  Substituting,&#xA;$$dG=\frac{G-H}{T}dT$$&#xA;This equation can be manipulated into the form:&#xA;$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$&#xA;So,&#xA;$$\frac{d(\Delta G/T)}{dT}=-\frac{\Delta H}{T^2}$$&#xA;I leave the rest up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A math error was corrected on 10/16/15&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="15095" LastEditDate="2015-10-17T02:33:44.347" LastActivityDate="2015-10-17T02:33:44.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27685" PostTypeId="1" CreationDate="2015-03-22T20:34:48.007" Score="2" ViewCount="249" Body="&lt;p&gt;I want to make an algorithm that will give me the boiling temperature of a substance as a function of pressure, the critical temperature of a substance, and the critical pressure of a substance. I believe this can be done using a combination of the Peng-Robinson Equation of state and the Clayperon or Clausius-Clayperon equation of state. The issue is that I believe this requires additional input criteria, specifically the latent heat but this function is supposed to be more general. Is there any way to work this method without the latent heat, or to calculate the latent heat from the other known values?&lt;/p&gt;&#xA;" OwnerUserId="15111" LastActivityDate="2015-03-23T13:57:14.737" Title="How can I calculate the boiling temperature of a pure substance as a function of pressure?" Tags="&lt;physical-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27687" PostTypeId="2" ParentId="27683" CreationDate="2015-03-22T21:20:39.783" Score="4" Body="&lt;p&gt;Temperature regulation is vital in producing explosives as the products are, well, &lt;em&gt;explosive&lt;/em&gt;. Worse, many of the products are temperature sensitive explosives and the reactions themselves are exothermic and can generate a lot of heat if uncontrolled. If the heat gets high, the reaction rate will increase and the product will not only be produced faster but will detonate. In the early days of explosives many experimenters paid for this knowledge with their life or their limbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result most modern experiments are done on a very small scale in very carefully controlled environments.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-03-22T21:20:39.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27688" PostTypeId="1" CreationDate="2015-03-22T21:23:55.573" Score="2" ViewCount="61" Body="&lt;p&gt;Two fluids with same reduced volume and temperature have the same reduced pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two fluids with same reduced pressure and temperature have the same compression factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What justifies these two statements and why exactly are both saying the same thing?&lt;/p&gt;&#xA;" OwnerUserId="8091" LastActivityDate="2015-03-22T21:23:55.573" Title="What is the justification behind the principle of corresponding states?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27691" PostTypeId="2" ParentId="27683" CreationDate="2015-03-22T22:56:00.713" Score="6" Body="&lt;p&gt;My advice on TATP is simple: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you're an inexperienced amateur researcher, do the synthesis on paper - and only there!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the synthesis on TATP can be performed at temperatures above 0 °C. When &lt;em&gt;Richard Wolffenstein&lt;/em&gt; accidently found in 1895 during his works on coniine that TATP had been formed in a mixture of 10% hydrogen peroxide and acetone, he did not use any acidic catalyst and had kept the mixture at room temperature for weeks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ueber die Einwirkung von Wasserstoffsuperoxyd auf Aceton und Mesityloxyd, &lt;em&gt;Ber. dtsch. chem. Ges.&lt;/em&gt;, &lt;strong&gt;1895&lt;/strong&gt;, &lt;em&gt;28&lt;/em&gt;, 2265-2269 (&lt;a href=&quot;http://dx.doi.org/10.1002/cber.189502802208&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When he had collected the precipitate and recrystallized it from diethyl ether, it exploded, which Wolffenstein attributed to to the formation of peroxides in ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the first reports on the synthesis of TATP under acidic conditions was given by &lt;em&gt;Milas&lt;/em&gt; and &lt;em&gt;Golubovic&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Studies in Organic Peroxides. XXVI. Organic Peroxides Derived from Acetone and Hydrogen Peroxide, &#xA;  &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1959&lt;/strong&gt;, &lt;em&gt;81&lt;/em&gt;, 6461-6462 (&lt;a href=&quot;http://dx.doi.org/10.1021/ja01533a033&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Btw, the experimental part of this article starts with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;These peroxides are dangerously explosive and should be handled with extreme care.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under controlled lab conditions, the decomposition of TATP in solution seemingly has been examined:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;J.C. Oxley, J.L. Smith and H. Chen, &#xA;  Decomposition of a Multi-Peroxidic Compound: Triacetone Triperoxide (TATP), &lt;em&gt;Propellants, Explosives, Pyrotechnics&lt;/em&gt;, &lt;strong&gt;2002&lt;/strong&gt;, &lt;em&gt;27&lt;/em&gt;, 209-216&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and the activation energy for this first-order reaction was calculated to be around $\mathrm{150\,kJ\cdot mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes TATP different from most of the other known explosives is &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the molecule (obviously) does not contain nitro groups. Typical detectors for explosives thus fail here. Together with the simplicity of the synthesis, this is a real problem in law enforcement!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the explosion does &lt;strong&gt;not&lt;/strong&gt; generate a lot of heat, but is an &lt;strong&gt;entropic explosion&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The latter was througly examined by &lt;em&gt;Ehud Keinan&lt;/em&gt; and coworkers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Decomposition of Triacetone Triperoxide Is an Entropic Explosion, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;127&lt;/em&gt;, 1146-1159. (&lt;a href=&quot;http://dx.doi.org/10.1021/ja0464903&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't want to go though all their calculations of a possible decomposition mechanism, the following hopefully is impressive enough:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The approximate pressure increase, assuming an&#xA;  infinitely rapid conversion that occurs in the volume of 1 mol&#xA;  of solid &lt;strong&gt;1&lt;/strong&gt; (=TATP), is of the order of 100 bar. The associated energy&#xA;  content is approximately 2500 kcal/mol. This is the origin of&#xA;  the explosive power of solid &lt;strong&gt;1&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm aware that I probably haven't fully anwered your question, but as a take-home message I can only suggest again to not attempt any experiment with this peroxide!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-22T22:56:00.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27692" PostTypeId="2" ParentId="27112" CreationDate="2015-03-23T00:55:50.317" Score="4" Body="&lt;p&gt;Reactivity of Group 1 and 2 elements increases as you go down the periodic table.&#xA;So sodium is more reactive than lithium. Sodium will react with oxygen forming $\ce{Na2O}$ (sodium oxide). Lithium forms lithium oxide $\ce{(Li2O)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, there are general trends (reactivity with acids, air, etc.) down a group or along a period.&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="5026" LastEditDate="2018-07-20T14:58:55.663" LastActivityDate="2018-07-20T14:58:55.663" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27693" PostTypeId="2" ParentId="27655" CreationDate="2015-03-23T01:25:51.023" Score="5" Body="&lt;p&gt;I would like to say your answer is right, but the question is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that a lattice is &quot;infinitely&quot; repeated units in 3 dimensional space so you should be able to expand the unit cell in x, y and z axes, indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine your cubic lattice with an &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atom in just 1 (out of 8) corners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding the unit cell in the x-axis will necessarily duplicate the &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atom in the x-axis because all expanded unit cells must be identical in composition. Similarly, expanding the unit cell in the y-axis will duplicate the &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atom in the y-axis, and expanding the unit cell in the z-axis will duplicate the &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atom in the z-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end you will find the &lt;b&gt;having one corner as &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atom will necessarily have ALL corners as &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; atoms&lt;/b&gt;. In other words, &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; IS &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt;, and the formula is necessarily &lt;span class=&quot;math-container&quot;&gt;$\ce{XZ_3}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{YZ_3}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don'k think the designer of the question saw his fatal contradiction because this very question is seen in the (mock) Joint Entrance Examination (JEE) in India.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/upSdD.png&quot; alt=&quot;cubic exam question&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$\ce{X_2YZ_24}$&lt;/span&gt; is not in the options even in the wrong question.&lt;/p&gt;&#xA;" OwnerUserId="9933" LastEditorUserId="83758" LastEditDate="2020-03-24T16:03:35.463" LastActivityDate="2020-03-24T16:03:35.463" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27694" PostTypeId="1" AcceptedAnswerId="27737" CreationDate="2015-03-23T02:32:24.027" Score="2" ViewCount="3607" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How many moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4Cl}$&lt;/span&gt; must be added to &lt;span class=&quot;math-container&quot;&gt;$\pu{2.0 L}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.10 M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; to form a buffer with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}=9$&lt;/span&gt;? Assume the addition does not change the volume of the solution significantly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The solution given is as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The equilibrium between &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; is given by-&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH3 + H2O &amp;lt;=&amp;gt; NH4+ + OH-}$$&lt;/span&gt;&#xA;We know the &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{b}$&lt;/span&gt; value and &lt;span class=&quot;math-container&quot;&gt;$\ce{[OH-]}$&lt;/span&gt; from the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[NH3]}$&lt;/span&gt; is given, thus we can solve for &lt;span class=&quot;math-container&quot;&gt;$\ce{[NH4+]}$&lt;/span&gt;. The solution given says that the final answer is given by this concentration multiplied by the total volume &lt;span class=&quot;math-container&quot;&gt;$\pu{2 L}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't follow the logic in the solution. The &lt;span class=&quot;math-container&quot;&gt;$\ce{[NH4+]}$&lt;/span&gt; we found was the &lt;em&gt;total&lt;/em&gt; concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; in the solution after adding the salt. But before adding &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4Cl}$&lt;/span&gt;, there was already some &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; in the solution (we don't know how much).&lt;/p&gt;&#xA;&lt;p&gt;Thus to find the number of moles &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4Cl}$&lt;/span&gt; we need to add, shouldn't we subtract the number of moles &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; already in the solution from the answer we got in the textbook solution?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="83758" LastEditDate="2020-09-18T16:39:54.740" LastActivityDate="2020-09-18T18:17:58.747" Title="How many moles NH4Cl must be added to NH3 to create buffer with pH=9?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;concentration&gt;&lt;buffer&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27695" PostTypeId="2" ParentId="20211" CreationDate="2015-03-23T03:04:47.717" Score="1" Body="&lt;p&gt;With less electronegativity, there is also less ionization energy. This means for metals, it is easier to lose the electrons. When it is easier to lose the electrons, the metal becomes more reactive because it uses less energy to fulfill its goal; to gain a full outer shell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nonmetals, the main goal is to gain electrons and not lose them. Thus, an increase in electronegativity and ionization energy means an increase in reactivity because the nonmetals want to hold on to their electrons and gain more vigorously to achieve the full outer shell. &lt;/p&gt;&#xA;" OwnerUserId="15117" LastActivityDate="2015-03-23T03:04:47.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27697" PostTypeId="1" CreationDate="2015-03-23T03:50:33.440" Score="4" ViewCount="1307" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QfX2L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QfX2L.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kl1RD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kl1RD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have no idea how to determine the substituents on the aromatic ring from this NMR spectrum. There is (or might be) a triplet on my NMR around &lt;span class=&quot;math-container&quot;&gt;$\pu{7.59 - 7.65 ppm}$&lt;/span&gt; which has 4 hydrogens in the same environment. If so, there must be two &lt;em&gt;ortho&lt;/em&gt; substituents on two aromatic rings but I think the 4 Hs won’t be in the same environment.&lt;/p&gt;&#xA;&lt;p&gt;I assume the substance to be the one I added here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3jWk3.jpg&quot; alt=&quot;suggested structure&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="15119" LastEditorUserId="69830" LastEditDate="2021-05-05T06:17:56.137" LastActivityDate="2021-05-05T06:17:56.137" Title="How can I deduce the structure of my compound from this NMR spectra?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27698" PostTypeId="1" AcceptedAnswerId="31922" CreationDate="2015-03-23T03:54:07.653" Score="10" ViewCount="157" Body="&lt;p&gt;In &lt;a href=&quot;https://www.facebook.com/video.php?v=977834295591048&quot; rel=&quot;nofollow noreferrer&quot;&gt;this rather shaky Facebook video&lt;/a&gt; (sorry I can't provide a better link), a science teacher takes a burning beaker of something and spills it out onto the floor.  It then proceeds to shoot fire out in all directions, and we can see embers burning themselves out, leaving black ash of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wb9rr.png&quot; alt=&quot;teacher with fire&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vWyRf.png&quot; alt=&quot;explosion!&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen teachers do similar stunts with alcohol fires, but those burn blue.  This one is much more explosive and stays yellow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there any way to tell what the teacher is using here?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5478" LastEditorUserId="402" LastEditDate="2017-04-01T01:15:13.430" LastActivityDate="2017-04-01T01:15:13.430" Title="What is involved in this firey demonstration?" Tags="&lt;identification&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27700" PostTypeId="5" CreationDate="2015-03-23T07:26:20.403" Score="0" Body="&lt;p&gt;When applied to a chemical species the term expresses a thermodynamic property. In general this is a relative concept stating that one state has a lower energy than another.&lt;br&gt;&#xA;The &lt;a href=&quot;http://goldbook.iupac.org/S05900.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC goldbook&lt;/a&gt; defines the term stable:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species $\ce{A}$ is more stable than its isomer $\ce{B}$ if  $\Delta_\mathrm{r}G^\circ &amp;gt; 0$ for the (real or hypothetical) reaction&#xA;  $$\ce{A -&amp;gt; B},$$&#xA;  under standard conditions. If for the two reactions:&#xA;  \begin{align}&#xA;\ce{P &amp;amp;-&amp;gt; X + Y}&amp;amp;&amp;amp; (\Delta_\mathrm{r} G_1^\circ)\\&#xA;\ce{Q &amp;amp;-&amp;gt; X + Z}&amp;amp;&amp;amp; (\Delta_\mathrm{r} G_2^\circ)\\&#xA;\Delta_\mathrm{r} G_1^\circ &amp;amp;&amp;gt; \Delta_\mathrm{r} G_2^\circ,\\&#xA;\end{align} $\ce{P}$ is more stable relative to the product $\ce{Y}$ than is $\ce{Q}$ relative to $\ce{Z}$. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As stability is a relative concept, this tag may also be specified for questions seeking information about the instability of certain chemical species. The &lt;a href=&quot;http://goldbook.iupac.org/ET07027.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC goldbook&lt;/a&gt; defines unstable as the opposite of stable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The opposite of stable, i.e. the chemical species concerned has a higher molar Gibbs energy than some assumed standard. The term should not be used in place of reactive or transient, although more reactive or transient species are frequently also more unstable. (Very unstable chemical species tend to undergo exothermic unimolecular decompositions. Variations in the structure of the related chemical species of this kind generally affect the energy of the transition states for these decompositions less than they affect the stability of the decomposing chemical species. Low stability may therefore parallel a relatively high rate of unimolecular decomposition.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tag should be applied to questions seeking answers with respect to the stability (or instability) of a certain chemical species, molecular entity and/or electronic structure. It must not be applied to questions about &lt;a href=&quot;/questions/tagged/reactivity&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'reactivity'&quot; rel=&quot;tag&quot;&gt;reactivity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &quot;inert&quot; involves both concepts, stability and reactivity (&lt;a href=&quot;http://goldbook.iupac.org/I03026.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC goldbook&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stable and unreactive under specified conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T07:26:20.403" LastActivityDate="2015-03-23T07:26:20.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27701" PostTypeId="4" CreationDate="2015-03-23T07:26:20.403" Score="0" Body="Applied to a chemical species, the term expresses a thermodynamic property in reference to a standard, stating that one state is lower in energy than another. The tag should be applied to questions seeking answers with respect to the stability (or instability) of a certain chemical species, molecular entity and/or electronic structure. It must not be applied to questions about the reactivity of particular chemical species." OwnerUserId="4945" LastEditorUserId="5354" LastEditDate="2015-12-07T03:12:49.170" LastActivityDate="2015-12-07T03:12:49.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27703" PostTypeId="1" AcceptedAnswerId="27715" CreationDate="2015-03-23T08:46:23.080" Score="10" ViewCount="13184" Body="&lt;p&gt;I learnt that Gibbs free energy is the maximum amount of non-expansion work.&#xA;But the phrase non expansion work confuses me. Work is defined as pressure times change in volume. If there is no expansion, how can work be done?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="10185" LastEditDate="2015-03-23T18:08:55.840" LastActivityDate="2016-12-28T08:21:53.527" Title="What is non-expansion work?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27704" PostTypeId="5" CreationDate="2015-03-23T09:05:39.730" Score="0" Body="&lt;p&gt;Applied to a chemical species, the term expresses a kinetic property in reference to another species. The IUPAC goldbook defines the term reactive:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As applied to a chemical species, the term expresses a kinetic property. A species is said to be more reactive or to have a higher reactivity in some given context than some other (reference) species if it has a larger rate constant for a specified elementary reaction. The term has meaning only by reference to some explicitly stated or implicitly assumed set of conditions. It is not to be used for reactions or reaction patterns of compounds in general. The term is also more loosely used as a phenomenological description not restricted to elementary reactions. When applied in this sense the property under consideration may reflect not only rate, but also equilibrium, constants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Closely related (and therefore the tag also applies to) is the concept of being unreactive, as defined by the IUPAC goldbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Failing to react with a specified chemical species under specified conditions. The term should not be used in place of stable, since a relatively more stable species may nevertheless be more reactive than some reference species towards a given reaction partner.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tag should be applied to questions seeking answers with respect to the reactivity (or unreactivity) of a certain chemical compound, species, molecular entity and/or functional groups. It must not be applied to questions about the &lt;a href=&quot;/questions/tagged/stability&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stability&amp;#39;&quot; rel=&quot;tag&quot;&gt;stability&lt;/a&gt; of certain chemical species.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="9961" LastEditDate="2017-06-07T22:18:30.097" LastActivityDate="2017-06-07T22:18:30.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27705" PostTypeId="4" CreationDate="2015-03-23T09:05:39.730" Score="0" Body="Applied to a chemical species, the term expresses a kinetic property in reference to another species. The tag should be applied to questions seeking answers with respect to the reactivity (or unreactivity) of a certain chemical compound, species, molecular entity and/or functional groups. It must not be applied to questions about the stability of certain chemical species." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T09:05:39.730" LastActivityDate="2015-03-23T09:05:39.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27706" PostTypeId="5" CreationDate="2015-03-23T09:32:52.440" Score="0" Body="&lt;p&gt;As defined in the first entry of the definition in the &lt;a href=&quot;http://goldbook.iupac.org/C00844.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Compounds containing the carbonyl group, C=O. The term is commonly used in the restricted sense of &lt;a href=&quot;http://goldbook.iupac.org/A00208.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;aldehydes&lt;/a&gt; and &lt;a href=&quot;http://goldbook.iupac.org/K03386.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ketones&lt;/a&gt;, although it actually includes &lt;a href=&quot;http://goldbook.iupac.org/C00852.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;carboxylic acids&lt;/a&gt; and derivatives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T04:11:55.770" LastActivityDate="2018-03-19T04:11:55.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27707" PostTypeId="4" CreationDate="2015-03-23T09:32:52.440" Score="0" Body="For questions about compounds containing the carbonyl group, C=O, the carboxylic acid group (-COOH), or its derivatives (anhydrides/esters/amides/acid halides). For questions about complexes involving carbon monoxide as a ligand use [carbonyl-complexes] instead." OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T04:11:41.237" LastActivityDate="2018-03-19T04:11:41.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27708" PostTypeId="5" CreationDate="2015-03-23T09:36:46.313" Score="0" Body="&lt;p&gt;Complexes or (coordination) compounds involving carbon monoxide as a ligand. The second part of the IUPAC goldbook definition of carbonyl compounds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metal carbonyls, in which carbon monoxide is a formal ligand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For questions about compounds containing the carbonyl group, C=O, use &lt;a href=&quot;/questions/tagged/carbonyl-compound&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged 'carbonyl-compound'&quot; rel=&quot;tag&quot;&gt;carbonyl-compound&lt;/a&gt; instead.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T09:36:46.313" LastActivityDate="2015-03-23T09:36:46.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27709" PostTypeId="4" CreationDate="2015-03-23T09:36:46.313" Score="0" Body="Complexes or (coordination) compounds involving carbon monoxide as a ligand. For compounds of the carbonyl group use carbonyl-compounds instead." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T09:36:46.313" LastActivityDate="2015-03-23T09:36:46.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27710" PostTypeId="2" ParentId="16640" CreationDate="2015-03-23T09:44:23.417" Score="22" Body="&lt;p&gt;We use an levitating furnace to heat samples of refractory ceramics up to approx $3000~^\circ\mathrm{C}$. Its for research purposes, so the samples are small (2 mm) beads. These are balanced on a jet of argon and heated with $\ce{CO2}$ lasers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a paper which talks about the technique:&lt;br&gt;&#xA;&lt;a href=&quot;http://dx.doi.org/10.1063/1.4832115&quot;&gt;D. Langstaff, M. Gunn, G. N. Greaves, A. Marsing, and F. Kargl, &lt;em&gt;Rev. Sci. Instrum.&lt;/em&gt;; &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;84&lt;/em&gt;, 124901.&lt;/a&gt; (&lt;a href=&quot;http://cadair.aber.ac.uk/dspace/bitstream/handle/2160/13128/ojpstmpstampPdf_d05T1119_RSINAK_84_12_124901_1.pdf?sequence=1&quot;&gt;Mirror&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15123" LastEditorUserId="4945" LastEditDate="2015-03-23T09:51:15.510" LastActivityDate="2015-03-23T09:51:15.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27711" PostTypeId="5" CreationDate="2015-03-23T09:45:22.390" Score="0" Body="&lt;p&gt;From the IUPAC goldbook, &lt;strong&gt;Gibbs energy (function), $G$&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Enthalpy minus the product of thermodynamic temperature and entropy.  It was formerly called free energy or free enthalpy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$G = H - TS$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T09:45:22.390" LastActivityDate="2015-03-23T09:45:22.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27712" PostTypeId="4" CreationDate="2015-03-23T09:45:22.390" Score="0" Body="Also known as Gibbs energy, it is the enthalpy minus the product of thermodynamic temperature and entropy." OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-23T09:45:22.390" LastActivityDate="2015-03-23T09:45:22.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27713" PostTypeId="1" AcceptedAnswerId="27719" CreationDate="2015-03-23T10:23:59.813" Score="2" ViewCount="299" Body="&lt;p&gt;The Gibbs energy increases from the reactants to the transition state, but according to the Second Law of Thermodynamics, entropy of the universe can only increase. Why is the transition state possible then?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-03-23T18:07:37.217" LastActivityDate="2015-03-23T18:07:37.217" Title="If Gibbs energy of activation is positive, how is the change to the transition state possible?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;&lt;transition-state-theory&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27714" PostTypeId="1" CreationDate="2015-03-23T10:39:01.397" Score="3" ViewCount="902" Body="&lt;p&gt;In &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Conversion_of_carboxylic_acids_to_alcohols_using_LiAlH4&quot; rel=&quot;nofollow&quot;&gt;reduction of carboxylic acid&lt;/a&gt; with $\ce{LiAlH_4}$ the first step is deprotonation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the mechanism of reduction of amide with $\ce{LiAlH_4}$ (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Amides/Reactions_of_Amides/Conversion_of_Amides_into_Amines_with_LiAlH4&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), the first step is addition of hydride on carbonyl group. Why doesn't the hydride ion abstract a proton from $\ce{-NH2}$? Since acid base reactions are faster than nucleophilic addition.&lt;/p&gt;&#xA;" OwnerUserId="3928" LastEditorUserId="3928" LastEditDate="2015-03-23T11:10:20.257" LastActivityDate="2020-05-01T19:13:18.137" Title="Reduction of amide with LiAlH4" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27715" PostTypeId="2" ParentId="27703" CreationDate="2015-03-23T10:42:22.643" Score="14" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Work_(thermodynamics)#Pressure-volume_work&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pressure-Volume work&lt;/a&gt; is a type of mechanical work .&lt;br&gt;&#xA;There are also &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_(thermodynamics)#Non-mechanical_forms_of_work&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-mechanical&lt;/a&gt; forms of work where in pressure and volume terms are not involved .&lt;br&gt;&#xA;The most famous example is &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrical_work&quot; rel=&quot;nofollow noreferrer&quot;&gt;electrical work&lt;/a&gt; . The work done to move electrical charges ( like motion of electrons in a conductor) comes under this category .&lt;/p&gt;&#xA;" OwnerUserId="13409" LastEditorUserId="16683" LastEditDate="2016-12-28T08:21:53.527" LastActivityDate="2016-12-28T08:21:53.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27716" PostTypeId="1" AcceptedAnswerId="27717" CreationDate="2015-03-23T10:59:58.253" Score="9" ViewCount="11665" Body="&lt;p&gt;I've seen a &quot;2.303&quot; term multiplied to $R$ in Van't Hoff equation several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Error_propagation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Error_propagation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\Delta H^\circ=2.303R\frac{\log K_1-\log K_2}{\frac1{T_2}-\frac1{T_1}}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is this number and where does it come from?&lt;/p&gt;&#xA;" OwnerUserId="14853" LastEditorDisplayName="user7951" LastEditDate="2016-09-06T10:26:58.013" LastActivityDate="2016-09-06T10:39:21.427" Title="What's the source of &quot;2.303&quot; in Van't Hoff equation?" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27717" PostTypeId="2" ParentId="27716" CreationDate="2015-03-23T11:07:14.070" Score="12" Body="&lt;p&gt;It is the multiplier used when converting the equation to use a base 10 log instead of a natural log.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes out to be the natural log of 10.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="13355" LastEditDate="2016-09-06T09:48:16.357" LastActivityDate="2016-09-06T09:48:16.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27718" PostTypeId="2" ParentId="27663" CreationDate="2015-03-23T11:10:46.817" Score="1" Body="&lt;p&gt;There are a lot of cases where you can work without a salt bridge. If the two reactions(reduction and oxidation) don't interfere with each other, you can just let the solution be in a single pot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only there to separate the two solutions and still maintain ionic contact.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-03-23T11:10:46.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27719" PostTypeId="2" ParentId="27713" CreationDate="2015-03-23T11:23:31.787" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...&lt;em&gt;according to the Second Law of Thermodynamics, entropy of the universe can only increase.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, there was a &lt;a href=&quot;https://biology.stackexchange.com/questions/28022/relationship-between-evolution-and-the-increase-of-entropy-of-earth/28026#28026&quot;&gt;nice similar question&lt;/a&gt; in bio.SE about this. &lt;strong&gt;A system is &lt;em&gt;not&lt;/em&gt; universe.&lt;/strong&gt; The &lt;em&gt;overall&lt;/em&gt; entropy of the universe increases, but that isn't true for only a particular thermodynamic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, no one said &lt;a href=&quot;http://www.masterorganicchemistry.com/2010/11/03/whats-a-transition-state/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the transition state&lt;/a&gt; is stable!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The transition state of a chemical reaction is a particular configuration along the reaction coordinate. It is defined as the state corresponding to the highest potential energy along this reaction coordinate. &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(High potential energy = lesser stability)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Laws_of_Thermodynamics/Second_Law_of_Thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;This chemwiki.ucdavis page&lt;/a&gt; is also good to help reshape your understanding of second law of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:14.547" LastActivityDate="2015-03-23T11:51:11.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27720" PostTypeId="2" ParentId="27714" CreationDate="2015-03-23T12:27:28.140" Score="2" Body="&lt;p&gt;As compared to the carboxylic acid, deprotonation of the amide is negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't recall any data for THF or 1,4-dioxane as the solvent, but &lt;em&gt;F.G. Bordwell&lt;/em&gt; has determined a lot of $pK_a$ values in DMSO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In DMSO, the $pK_a$ of acetic acid is 12.6 [&lt;a href=&quot;http://dx.doi.org/10.1021/jo00410a001&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;], while the $pK_a$ of acetamide is 25.5 [&lt;a href=&quot;http://dx.doi.org/10.1021/jo00876a042&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;].&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-23T12:27:28.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27721" PostTypeId="1" CreationDate="2015-03-23T12:27:37.430" Score="12" ViewCount="3079" Body="&lt;p&gt;This weekend I accidentally left a glass of water on my kitchen table for a few days while I was out of town. Unsurprisingly, bubbles formed on the walls.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand from the last time I attended a chemistry class quite a while ago, this happens due to water self-ionizing (at about a ratio of $10^{-14}$ if I recall correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + H2O \rightleftharpoons H3O+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for some reason, the bubbles only formed up to a certain depth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/wf9iXUc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5Ae0.png&quot; alt=&quot;Sample image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this? Does self-ionization happen uniformly in the glass? Do bubbles start forming from the top down and would they have reached the bottom if left untouched for longer?&lt;/p&gt;&#xA;" OwnerUserId="5275" LastEditorUserId="7448" LastEditDate="2015-03-23T20:26:17.857" LastActivityDate="2015-03-26T07:15:18.743" Title="Where do the bubbles in a glass of water come from?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27722" PostTypeId="2" ParentId="27721" CreationDate="2015-03-23T12:50:23.277" Score="14" Body="&lt;p&gt;Gases are less soluble in warm water than cold water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bubbles are likely from dissolved gas coming out of solution as the water warmed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bubbles are unrelated to self-ionization of water.  The self-ionization does vary somewhat with pressure, so at an extreme depth in the ocean you might need to consider what Kw is, but not in a glass of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of gases increases with pressure, so on this basis it is more likely that you would have bubble appear near the top.  However, the pressure is only 1-2% greater at the bottom of the glass compare to the top.  There may be fewer nucleation sites (such as tiny scratches) near the bottom of that particular glass.  Also, if the bubbles move up along the wall sometimes, it would be logical that there are less at the bottom.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-23T16:52:25.360" LastActivityDate="2015-03-23T16:52:25.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27724" PostTypeId="2" ParentId="27685" CreationDate="2015-03-23T13:57:14.737" Score="1" Body="&lt;p&gt;If you know the latent heat of vaporization at one temperature, you should be able to determine the latent heat of vaporization at any other temperature from Hess' law and knowledge of the heat capacities of the liquid and vapor as a function of temperature.  You may also have to include a pressure contribution as the critical point is approached.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-03-23T13:57:14.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27728" PostTypeId="1" CreationDate="2015-03-23T16:27:16.190" Score="1" ViewCount="827" Body="&lt;p&gt;This seems counter intuitive to me since ionic radius increases with increasing coordination number. Surely the solid doesn't expand under high pressure? Can you explain?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-03-24T16:38:06.163" Title="Why do ionic solids change to a structure with a higher coordination number at high pressures?" Tags="&lt;inorganic-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27729" PostTypeId="1" CreationDate="2015-03-23T17:23:24.277" Score="0" ViewCount="80" Body="&lt;p&gt;I understand how the protons and electrons move around through various carriers, shuttles and proteins, but there's this talk of electrons being in &quot;lower in energy&quot; at the end. What does that mean? How is an electron lower in energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also why do they need an electron acceptor? What would happen if there was no $\ce{O2}$ to accept the electrons? Wouldn't they just bind to something else?&lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="15410" LastEditDate="2016-05-28T05:39:59.703" LastActivityDate="2016-05-28T05:39:59.703" Title="How exactly is energy created by oxidative phosphorylation" Tags="&lt;biochemistry&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27731" PostTypeId="1" CreationDate="2015-03-23T17:54:36.980" Score="2" ViewCount="5477" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;What order of reactivity do you predict will be observed when each alkyl halide is mixed with sodium iodide in acetone?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;1-Chlorobutane&lt;/li&gt;&#xA;  &lt;li&gt;1-Bromobutane&lt;/li&gt;&#xA;  &lt;li&gt;2-Chloro-2-methylpropane&lt;/li&gt;&#xA;  &lt;li&gt;Bromobenzene&lt;/li&gt;&#xA;  &lt;li&gt;2-Chlorobutane&lt;/li&gt;&#xA;  &lt;li&gt;1-Chloro-2-butene&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that $\ce{I-}$ is the best leaving group, followed by $\ce{Br-}$ and $\ce{Cl-}$.&lt;br&gt;&#xA;For the $\ce{NaI}$ reaction, tertiary halides should react fastest and primary halides should react slowest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;&lt;p&gt;What order of reactivity do you predict will b observed when each alkyl halide is mixed with silver nitrate in ethanol?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;1-Chlorobutane&lt;/li&gt;&#xA;  &lt;li&gt;1-Bromobutane&lt;/li&gt;&#xA;  &lt;li&gt;2-Chloro-2-methylpropane&lt;/li&gt;&#xA;  &lt;li&gt;Bromobenzene&lt;/li&gt;&#xA;  &lt;li&gt;2-Chlorobutane&lt;/li&gt;&#xA;  &lt;li&gt;1-Chloro-2-butene&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the $\ce{AgNO3}$ reaction, primary halides should react fastest and tertiary halides should react slowest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which takes preference, leaving group or rank? Also, how do I rank 1-chlorobutane vs. 1-chloro-2-butene? They are both primary and the halide is $\ce{Cl-}$ for both.&lt;/p&gt;&#xA;" OwnerUserId="15132" LastEditorUserId="69830" LastEditDate="2021-05-18T06:38:32.960" LastActivityDate="2021-05-18T06:38:32.960" Title="Predicting order of nucleophilic substitution reactivity" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27732" PostTypeId="1" CreationDate="2015-02-23T08:37:39.663" Score="2" ViewCount="966" Body="&lt;p&gt;I read in Russian popular articles that it is allowed to preserve herring/fish with &lt;a href=&quot;https://en.wikipedia.org/wiki/Hexamethylenetetramine#Food_additive&quot; rel=&quot;nofollow&quot;&gt;Hexamethylenetetramine&lt;/a&gt; in Europe. Since Soviet times, I like to buy salty herring and &lt;a href=&quot;https://en.wikipedia.org/wiki/Marination&quot; rel=&quot;nofollow&quot;&gt;marinade&lt;/a&gt; it with Vinegar. The food-safety article says that &lt;strong&gt;&lt;em&gt;Hexamethylenetetramine + Vinegar = Formaldehyde&lt;/em&gt;&lt;/strong&gt;, which is a serious poison.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I easily test presence of Hexamethylenetetramine or Formaldehyde in food?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The salty herring that I buy is sold loosely, without packaging, so that I cannot simply read the ingredients.&lt;/p&gt;&#xA;" OwnerUserId="15152" OwnerDisplayName="Recognize Evil as Waste" LastEditorUserId="10185" LastEditDate="2015-03-23T18:22:17.383" LastActivityDate="2015-03-23T18:22:17.383" Title="Testing for the presence of Hexamethylenetetramine / Formaldehyde" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;analytical-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-11T17:03:28.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="27733" PostTypeId="2" ParentId="27732" CreationDate="2015-02-23T12:30:32.727" Score="2" Body="&lt;p&gt;I don't believe the premise that E239 (the E number designated to hexamethylene tetramine) is used to preserve fish within the EU is correct - in January, 2014, the European Food Safety Authority published an article in their journal about this preservative where it states quite clearly that the only permitted use for this additive is in Provolone cheese. There are concerns that over-consumers of this particular cheese &lt;em&gt;may&lt;/em&gt; accrete more levels of E239 within the body than are healthy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E239 does appear to be used in the UK for preserving fish, or at least, it is listed as one of its intended uses. Whether its actually used or not I haven't been able to determine, but given we're in the EU, technically, it shouldn't be. If I source any other research regarding this, I'll add it later, but currently, it seems it depends where you're buying your fish from in the first place, that is, the country of origin.&lt;/p&gt;&#xA;" OwnerDisplayName="Bamboo" LastActivityDate="2015-02-23T12:35:48.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27735" PostTypeId="1" CreationDate="2015-03-23T19:44:44.697" Score="4" ViewCount="1166" Body="&lt;p&gt;Compounds of silver form particularly strong bonds which is accounted for by the significant covalent character of the bonding. Furthermore, is the tendency to form covalent bonds linked to the fact that $\ce{Ag+}$ is a soft acid?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="7448" LastEditDate="2015-10-06T20:59:41.763" LastActivityDate="2015-10-06T20:59:41.763" Title="Why does Silver form bonds with covalent character?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;silver&gt;&lt;hsab&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27737" PostTypeId="2" ParentId="27694" CreationDate="2015-03-23T21:08:15.407" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; is such a weak base (&lt;span class=&quot;math-container&quot;&gt;$K_{\mathrm{b}}=1.8\times 10^{-5}$&lt;/span&gt;) that the initial concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; can be considered negligible. You can consider the initial concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt;, but the answer you will get is the same to a considerable number of significant figures.&lt;/p&gt;&#xA;&lt;p&gt;By considering the equilibrium of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; dissociating to form &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; it is easy to find the initial concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_{\mathrm{b}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;\begin{align}[\ce{NH4+}] &amp;amp;= [\ce{OH-}] \tag{1}\\&#xA;K_{\mathrm{b}}\cdot [\ce{NH3}] &amp;amp;= [\ce{NH4+}]^{2} \tag{2}&#xA;\end{align}&lt;/span&gt;&#xA;Plugging in values gives:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{NH4+}] = 1.34\times 10^{-3} (3\text{s.f.})$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which equates to &lt;span class=&quot;math-container&quot;&gt;$\pu{2.68\times 10^{-3} mol}$&lt;/span&gt; in the &lt;span class=&quot;math-container&quot;&gt;$\pu{2 L}$&lt;/span&gt; solution.&lt;/p&gt;&#xA;&lt;p&gt;If I'm right this should be orders of magnitude smaller than the textbook answer, and therefore negligible.&lt;/p&gt;&#xA;" OwnerUserId="14863" LastEditorUserId="95133" LastEditDate="2020-09-18T18:17:58.747" LastActivityDate="2020-09-18T18:17:58.747" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27738" PostTypeId="1" CreationDate="2015-03-23T22:44:08.227" Score="3" ViewCount="885" Body="&lt;p&gt;Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively is it possible to electrically purify iron using an anode and cathode?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastActivityDate="2016-03-04T09:21:13.943" Title="Obtain precipitated iron?" Tags="&lt;purification&gt;&lt;precipitation&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27739" PostTypeId="1" CreationDate="2015-03-23T23:02:42.627" Score="2" ViewCount="168" Body="&lt;p&gt;In organic chemistry I was told that the magnesium ion which is part of Grignards is coordinated to several ether molecules. The overall structure around the magnesium is &lt;em&gt;tetrahedral&lt;/em&gt;, which implies that &lt;em&gt;two&lt;/em&gt; ether molecules have coordinated to the magnesium center. The professor said something about fulfilling magnesium ion's octet with this configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In inorganic chemistry and biochemistry however I learned about hexacoordinated magnesium - i.e. $\ce{Mg(H2O)6}$. Magnesium ion tends to form octahedral complexes; this implies a coordination number of 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Does magnesium ion adopt both these configurations? Can it be found as part of tetrahedral and octahedral complexes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Which is preferred? I suspect this answer depends on a variety of factors. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-03-27T12:05:35.683" Title="Does the magnesium 2+ ion prefer an octet or more?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27740" PostTypeId="1" CreationDate="2015-03-23T23:25:19.920" Score="0" ViewCount="4158" Body="&lt;p&gt;What are the possible resonance structures for $\ce{ClO_2^-}$? Assigning one double bond to the structure makes for formal charges of $\ce{O(-1)-Cl(0)=O(0) &amp;lt;-&amp;gt; O(-1)=Cl(0)-O(-1)}$. It appears that the same set of formal charges can be achieved with two double bonds (expanding the octet on $\ce{Cl}$ to 12 electron capacity, using the d orbital). That is, $\ce{O(0)=Cl(-1)=O(0)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this feasible, and if not, why not?&lt;/p&gt;&#xA;" OwnerUserId="7161" LastEditorUserId="24052" LastEditDate="2016-03-11T23:05:05.943" LastActivityDate="2016-03-11T23:05:05.943" Title="What are the resonance structures for chlorite anion?" Tags="&lt;inorganic-chemistry&gt;&lt;resonance&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27741" PostTypeId="1" AcceptedAnswerId="27744" CreationDate="2015-03-23T23:42:35.210" Score="3" ViewCount="1785" Body="&lt;p&gt;After running a TD-DFT , we can calculate the HOMO-LUMO difference and then calculate the the excitation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is just simply pick the energies calculated in the TD-DFT and use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me the latter makes more sense as that's the reason we do TD-DFT , otherwise we could access the HOMO-LUMO difference by any other methods but I have seen papers using the first concept. I also have noticed that the energies in second concept are closer to the experimental values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when I run TD-DFT, I can see Oscillator Strength. Should I pick the energy that has the largest Oscillator Strength for calculation of the maximum wavelength ? &#xA;Here is an example. Should I pick number 8 for maximum wavelength calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ExHU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="186" LastEditDate="2015-05-12T09:14:03.893" LastActivityDate="2016-02-25T10:45:48.587" Title="Which is most reliable excitation energy: LUMO-HOMO energy or TD-DFT excitation energies?" Tags="&lt;computational-chemistry&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;&lt;td-dft&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27742" PostTypeId="1" AcceptedAnswerId="33147" CreationDate="2015-03-24T00:33:18.050" Score="6" ViewCount="836" Body="&lt;p&gt;I think the following problem in my textbook is malformed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solid &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; is placed in &lt;span class=&quot;math-container&quot;&gt;$1\,\mathrm{L}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$0.55\,\mathrm{M}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;. Find the mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; which dissolves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know the method to solve the problem. I think the problem is malformed: When we get an answer, say &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; grams &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; dissolves, but if we placed less than &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; grams solid &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt; in the solution initially, then it would be impossible for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; grams to dissolve.&lt;br&gt;&#xA;So we obviously have to make certain assumptions, but I am looking at the solution in my textbook and I don't see in which part of the solution is the assumption used (implicitly). And by the way, how do we solve the problem if we have place less than &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; grams solid &lt;span class=&quot;math-container&quot;&gt;$\ce{AgCl}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="41328" LastEditDate="2019-05-20T14:28:16.060" LastActivityDate="2019-05-20T14:28:16.060" Title="How to find the mass of dissolved AgCl when placed in a solution of NaCl?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;precipitation&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27743" PostTypeId="2" ParentId="27738" CreationDate="2015-03-24T01:11:49.337" Score="4" Body="&lt;p&gt;Sure... here's a &lt;a href=&quot;http://phys.org/news/2013-05-electrolysis-method-green-iron.html&quot; rel=&quot;nofollow&quot;&gt;an article on it&lt;/a&gt; and a &lt;a href=&quot;https://www.google.com/patents/US4134800&quot; rel=&quot;nofollow&quot;&gt;patent&lt;/a&gt; for the process. In brief,&#xA;&quot;[Donald Sadoway, the John F. Elliott Professor of Materials Chemistry at MIT] found that a process called molten oxide electrolysis could... [produce] steel as a byproduct... Sadoway's method used an iridium anode... But after more research... the MIT team identified an inexpensive metal alloy that can replace the iridium anode in molten oxide electrolysis.&quot;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-03-25T23:34:11.483" LastActivityDate="2015-03-25T23:34:11.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27744" PostTypeId="2" ParentId="27741" CreationDate="2015-03-24T01:17:58.897" Score="5" Body="&lt;p&gt;&lt;strong&gt;When you can, use TDDFT, in part because of the oscillator strengths&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to point out to my group that optical excitations consist of two things consistently (e.g., in a UV/Vis spectra). There are the excitation energies (the x-axis) and the intensity or extinction coefficient (the y-axis). To some degree, the peak widths give a bit more information (e.g., into vibrational structure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example you give above is a great example. Most of the excitations are &lt;a href=&quot;http://en.wikipedia.org/wiki/Selection_rule#Electronic_spectra&quot; rel=&quot;noreferrer&quot;&gt;forbidden&lt;/a&gt; and have no oscillator strength. I'd pick the same excitation, although I'm curious what the molecule is - usually there are &lt;em&gt;much&lt;/em&gt; higher oscillator strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right that there are papers and groups that use the HOMO and LUMO eigenvalues. While some DFT methods (e.g. optimally tuned range-separated hybrids) do provide HOMO-LUMO energies that correspond nicely with experiments, there's nothing physical about Kohn-Sham orbital eigenvalues. Moreover, in general, unfilled orbitals are not handled well by DFT methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a while since I looked for reviews, but general TDDFT excitation energies are ~0.2-0.3 eV from experimental absorption energies after correcting for solvent effects, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my group, we also do what I call a &quot;consensus&quot; model. That is, when you run a TDDFT calculation, you get the HOMO and LUMO energies anyway. Since there are slightly different random and systematic errors from each approach, you can calibrate &lt;em&gt;both&lt;/em&gt; against experiment and use a multivariate linear regression. The result decreases the average error significantly.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-03-24T01:17:58.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27746" PostTypeId="1" AcceptedAnswerId="27815" CreationDate="2015-03-24T01:59:04.483" Score="2" ViewCount="850" Body="&lt;p&gt;I've noticed that chalcogens never form binary acids, and instead end up like water, with very little acidity. Why is this? Or am I wrong and there's a counterexample, if so please give it.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="7786" LastEditDate="2015-03-25T01:40:57.453" LastActivityDate="2015-03-27T15:01:13.310" Title="Chalcogens' hydrides as acids?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27747" PostTypeId="1" AcceptedAnswerId="27758" CreationDate="2015-03-24T02:14:38.070" Score="3" ViewCount="985" Body="&lt;p&gt;A shortage of a given chemical (sodium thiopendal) has prompted Utah to allow &lt;a href=&quot;http://www.cnn.com/2015/03/23/us/utah-death-penalty-firing-squad/&quot; rel=&quot;nofollow&quot;&gt;firing squads for executions&lt;/a&gt;. This shortage is due to mostly European chemical factories &lt;a href=&quot;http://www.theatlantic.com/international/archive/2014/02/can-europe-end-the-death-penalty-in-america/283790/&quot; rel=&quot;nofollow&quot;&gt;refusing to export the drug to the US&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am against the death penalty, but can't help wondering why USA, arguably the most powerful nation in the world, can not manufacture this chemical themselves. Is it difficult to produce ?&lt;/p&gt;&#xA;" OwnerUserId="15144" LastEditorUserId="41890" LastEditDate="2017-11-27T18:02:22.213" LastActivityDate="2017-11-27T18:02:22.213" Title="Why does the US not manufacture sodium thiopental themselves?" Tags="&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27748" PostTypeId="1" AcceptedAnswerId="27752" CreationDate="2015-03-24T02:24:39.480" Score="7" ViewCount="79553" Body="&lt;p&gt;Here is a problem a made up: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have some aqueous $\ce{HCl}$. What is the maximum possible molarity of the $\ce{HCl}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do I solve this? I don't have any good ideas.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="7475" LastEditDate="2016-11-03T18:29:55.010" LastActivityDate="2019-09-14T00:40:05.327" Title="What is the maximum molarity of aqueous HCl?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27749" PostTypeId="2" ParentId="27721" CreationDate="2015-03-24T03:21:34.410" Score="4" Body="&lt;p&gt;I've always assumed that the bubbles formed because they're coming out of solution after the water pressure drops. Inside your plumbing, the pressure is high and the water can hold a lot of gas, and then after you pour it into the glass, the pressure is lower and you have a supersaturated solution, and the gas gradually comes out of solution at the nucleation sites on the glass walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggested experiment: take such a glass of water after it's been standing a few days, and pour the water into a new glass. See if new bubbles form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to try: seal the water off from the atmosphere, while leaving it at approximately atmospheric pressure. Maybe you could invert a glass full of water in a tub, taking care not to trap a bubble? I'm going to do this right now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverted glass gathers about as many bubbles as the upright glass, so the gas seems to be coming from &quot;inside&quot; the water, and not diffusing there. I didn't notice any bias toward the top of the glass at my first check after a few hours, but now, after two days, the inverted glass has some more, and larger, bubbles near the top, where they get trapped. I'm speculating that that's just gravity doing it's thing: occasionally a bubble will detach from the walls and simply float up, although I don't have the patience to see if they ever reattach to the walls rather than simply always continuing to the top.&lt;/p&gt;&#xA;" OwnerUserId="389" LastEditorUserId="389" LastEditDate="2015-03-25T17:25:24.647" LastActivityDate="2015-03-25T17:25:24.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27751" PostTypeId="2" ParentId="27729" CreationDate="2015-03-24T03:47:15.433" Score="2" Body="&lt;p&gt;&lt;strong&gt;How exactly is energy created by oxidative phosphorylation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is &quot;created&quot; by the syntesis of ATP. The ATP molecule is considered an energy carrier, because it is used in other reactions as a source of energy. The released energy comes from the loss of one or two phosphate gropus to form ADP or AMP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that energy is not created, but simply transferred. The oxidative phosphorylation is a mechanism to recreate the ATP using the energy from other sources (O&lt;sub&gt;2&lt;/sub&gt;, nutrients, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also why do they need an electron acceptor? What would happen if there was no O2 to accept the electrons?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I can give you a precise answer, but I'll try. In chemistry reactions, the system is always working towards equilibria. In general, if you take away one reagent of a reaction, there won't be a reaction at all. If there's no oxygen available, ATP won't be formed, and as a consequence the body resorts to other sources of energy, or cells will die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wouldn't they just bind to something else?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think so. The electrons aren't free particles moving around the cell, the transfer only happens if all the reagents for the reaction are complete. Enzymes are, generally speaking, very specific, and what could happen is a &quot;saturation&quot; of NADH's which wouldn't be converted back to NAD and hence the cycle would stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But there's this talk of electrons being in &quot;lower in energy&quot; at the end. What does that mean? How is an electron lower in energy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember hearing or reading that. Can you post some context or source?&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-03-24T03:47:15.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27752" PostTypeId="2" ParentId="27748" CreationDate="2015-03-24T04:01:16.640" Score="14" Body="&lt;p&gt;The HCl molecule is, at standard conditions, a gas. The reagent used in laboratories is HCl dissolved in water, which is why you'll find in the label that it is around 37% HCl in weight. The other 63% is water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering 37% as the maximum solubility of HCl, you can calculate the molarity using the solution density (1.2 g/mL) and HCl's molar mass (36.46 g/mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1000 mL of solution, you will have 1200g of weight, of which 37% is HCl: 444g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 444g of HCl you have 12.18 moles (444/36.46), which means the concentration of P.A. HCl is around 12.18 moles per liter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that because P.A. HCl is a solution saturated with a gas, vapors are likely to be released and escape the solution (which is why P.A. HCl shouldn't be handled outside fume hoods), which makes this concentration an approximation, and it should never be used to prepare a solution which you need to know the exact acid concentration.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-03-24T04:08:41.810" LastActivityDate="2015-03-24T04:08:41.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27753" PostTypeId="1" CreationDate="2015-03-24T04:43:12.963" Score="4" ViewCount="364" Body="&lt;p&gt;$$\ce{CaCO3 + 2H2O + Pb^{2+} -&amp;gt; Pb(CO3) + Ca(OH)2 + H2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this equation balance? &#xA;On the right you have lead with 2+ charge, whereas on the right its completely neutral. So my question is do you need to balance the charges in order to make the equation correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that equation is wrong, what is the right equation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to show that hydrogen gas is formed (or some other gas but I don't know what else it can be), that the solution will become more basic (proven with $\ce{Ca(OH)2}$), and that the lead has to replace the calcium.&lt;/p&gt;&#xA;" OwnerUserId="15146" LastEditorUserId="5026" LastEditDate="2018-04-22T10:14:57.100" LastActivityDate="2018-04-22T12:02:51.493" Title="Does this equation - involving calcium carbonate and a lead (II) cation - balance?" Tags="&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27754" PostTypeId="1" CreationDate="2015-03-24T05:42:55.870" Score="2" ViewCount="17594" Body="&lt;p&gt;I answered wrong on my quiz the following question on finding an equilibrium constant:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For the following reaction, at &lt;span class=&quot;math-container&quot;&gt;$1~\mathrm{bar}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$300~\mathrm{K}$&lt;/span&gt;, partial pressures of &lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O4}$&lt;/span&gt; are both &lt;span class=&quot;math-container&quot;&gt;$0.5~\mathrm{bar}$&lt;/span&gt;. If pressure is increased to &lt;span class=&quot;math-container&quot;&gt;$2~\mathrm{bar}$&lt;/span&gt; (this is the final pressure), and temperature is increased to &lt;span class=&quot;math-container&quot;&gt;$400~\mathrm{K}$&lt;/span&gt;, calculate the final composition of the system. &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt; for this reaction is &lt;span class=&quot;math-container&quot;&gt;$57.2~\mathrm{kJ/mol}$&lt;/span&gt;; assume it is independent of temperature.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{N2O4 (g) &amp;lt;=&amp;gt; 2NO2 (g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&lt;p&gt;I first found &lt;span class=&quot;math-container&quot;&gt;$K_\textrm{p}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_p (300~\mathrm{K}) = \frac{P^2(\ce{NO2})}{P(\ce{N2O4})} = (0.5^2)/0.5 = 0.5$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then I used&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln\left(\frac{K_p(400~\mathrm{K})}{K_p(300~\mathrm{K})}\right) = - \frac{\Delta H}{\mathcal{R}} \left( \frac{1}{400~\mathrm{K}} - \frac{1}{300~\mathrm{K}}\right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I solved for &lt;span class=&quot;math-container&quot;&gt;$K_\textrm{p}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{400 K} = 92695.44$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is where everything went wrong. I can not figure out how to get from &lt;span class=&quot;math-container&quot;&gt;$K_p$&lt;/span&gt; to partial pressures. All I got to is the following line:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_\text{total} = 2P(\ce{NO2}) + P(\ce{N2O4})$&lt;/span&gt;, since &lt;span class=&quot;math-container&quot;&gt;$n(\ce{NO2})$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$n(\ce{N2O4})$&lt;/span&gt; ratio is &lt;span class=&quot;math-container&quot;&gt;$2:1$&lt;/span&gt;. I assumed they exert pressure proportionally.&lt;/p&gt;&#xA;&lt;p&gt;Can some one guide me in the right direction?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Multiple choice answers:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P(\ce{N2O4(g)}) = 0.0252~\mathrm{bar}$&lt;/span&gt;; &lt;span class=&quot;math-container&quot;&gt;$P(\ce{NO2(g)}) = 1.9748~\mathrm{bar}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P(\ce{N2O4(g)}) = 0.039~\mathrm{bar}$&lt;/span&gt;; &lt;span class=&quot;math-container&quot;&gt;$P(\ce{NO2(g)}) = 1.961~\mathrm{bar}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P(\ce{N2O4(g)}) = 1.45~\mathrm{bar}$&lt;/span&gt;; &lt;span class=&quot;math-container&quot;&gt;$P(\ce{NO2(g)}) = 0.098~\mathrm{bar}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P(\ce{N2O4(g)}) = 1~\mathrm{bar}$&lt;/span&gt;; &lt;span class=&quot;math-container&quot;&gt;$P(\ce{NO2(g)}) = 1~\mathrm{bar}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15147" LastEditorUserId="69830" LastEditDate="2021-05-07T11:14:51.870" LastActivityDate="2021-05-07T11:14:51.870" Title="How to find partial pressures from a given equilibrium constant Kp?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27756" PostTypeId="2" ParentId="27753" CreationDate="2015-03-24T06:04:10.873" Score="5" Body="&lt;p&gt;Unfortunately your equation does not add up, let me briefly explain why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_carbonate&quot; rel=&quot;nofollow&quot;&gt;Calcium carbonate&lt;/a&gt; is a not very soluble salt. There is ionic binding between the calcium cation and the molecular carbonate anion. If you dissolve it in water you, the following equilibrium will establish: $$\ce{CaCO3~(s) &amp;lt;=&amp;gt;[\ce{H2O}] Ca^{2+}~(aq) + CO3^{2-}~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lead_carbonate&quot; rel=&quot;nofollow&quot;&gt;Lead carbonate&lt;/a&gt; is also a salt, it is even less soluble. Only a little bit will dissolve according to the following equilibrium:$$\ce{PbCO3~(s) &amp;lt;=&amp;gt;[\ce{H2O}] Pb^{2+}~(aq) + CO3^{2-}~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;Calcium hydroxide&lt;/a&gt; is also a salt. It is a little bit more soluble than calcium carbonate. The following equilibrium is established when put in water:$$\ce{Ca(OH)2~(s) &amp;lt;=&amp;gt;[\ce{H2O}] Ca^{2+}~(aq) + 2{}^{-}OH~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You write the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{CaCO3 + 2H2O + Pb^{2+} -&amp;gt; Pb(CO3) + Ca(OH)2 + H2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in principle you have $\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$ and $\ce{CO3^{2-}}$ on both sides of the equation. What is left is not a balanced equation, since there are two electrons missing:&#xA;$$\ce{2H2O ~\nrightarrow 2{}^{-}OH + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a source for the electrons, one way of balancing the equation is&#xA;$$\ce{CaCO3 + 2H2O + Pb -&amp;gt; Pb(CO3) + Ca(OH)2 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the insolubilities of all the species involved, it might be hard to see any reaction at all. And I have not checked if the redox reaction is favourable, with potentials.  &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-03-24T08:37:02.123" LastActivityDate="2015-03-24T08:37:02.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27757" PostTypeId="2" ParentId="27602" CreationDate="2015-03-24T06:27:23.367" Score="1" Body="&lt;p&gt;That's interesting; I just did this experiment (I'm actually writing my lab report up now) and for me the trans-bis(glycinato)copper(II) monohydrate was the thermodynamically favoured product and the cis-bis(glycinato)copper(II) monohydrate the kinetically favoured. As far as the reactant energy levels are concerned, I attached a photo of a rough graph my lab instructor gave us that illustrates the energy levels for the two pathways to the final thermodynamically favoured product. I hope that helps!! &#xA;&lt;img src=&quot;https://i.stack.imgur.com/21QAN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15149" LastActivityDate="2015-03-24T06:27:23.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27758" PostTypeId="2" ParentId="27747" CreationDate="2015-03-24T06:48:39.883" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I [&amp;hellip;] can't help wondering why USA [&amp;hellip;] can not manufacture this chemical themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;They can and they did!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the nineteen thirties, &lt;em&gt;Ernest H. Volwiler&lt;/em&gt; and &lt;em&gt;Donalee L. Tabern&lt;/em&gt; filed a couple of patents on sedative thiobarbituric acid derivatives for the Illinois-based &lt;em&gt;Abbott Laboratories&lt;/em&gt;. Among these was &lt;a href=&quot;http://patft.uspto.gov/netacgi/nph-Parser?Sect2=PTO1&amp;amp;Sect2=HITOFF&amp;amp;p=1&amp;amp;u=/netahtml/PTO/search-bool.html&amp;amp;r=1&amp;amp;f=G&amp;amp;l=50&amp;amp;d=PALL&amp;amp;RefSrch=yes&amp;amp;Query=PN/2153731&quot; rel=&quot;nofollow&quot;&gt;US 2,153,731&lt;/a&gt; for sodium thiopental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Abbott Laboratories&lt;/em&gt; produced it for a long time and when the company was restructured, the still Illinois-based split-off &lt;em&gt;Hospira&lt;/em&gt; continued the production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is not a &quot;bad compound&quot; as such and actually, it is still listed as an alternative to propofol on the &lt;a href=&quot;http://www.who.int/medicines/publications/essentialmedicines/18th_EML.pdf&quot; rel=&quot;nofollow&quot;&gt;WHO Model List of Essential Medicines&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a 2010 article in the German &lt;a href=&quot;http://www.aerztezeitung.de/panorama/article/621766/thiopental-mangel-verzoegert-hinrichtungen-usa.html&quot; rel=&quot;nofollow&quot;&gt;ÄrzteZeitung (online, 29.09.2010)&lt;/a&gt;, the FDA was aware of the thiopental shortage and a speaker of &lt;em&gt;Hospira&lt;/em&gt; told the NYT that his company is rather unlucky about the fact that this anaestetic found it's way into death row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Around this time, the production of thiopental at the facilities in Illinois was seemingly stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 2011 article in the &lt;a href=&quot;http://www.aerztezeitung.de/news/article/637976/todesstrafe-usa-gehen-giftspritzen.html&quot; rel=&quot;nofollow&quot;&gt;ÄrzteZeitung (online, 24.01.2011)&lt;/a&gt; it is stated that &lt;em&gt;Hospira&lt;/em&gt; had initially announced to continue the production at a facility in North Carolina and in Liscate, Italy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems that the production in North Carolina never started. Moreover, the Italian authorities demanded a declaration that the thiopental produced in Liscate &lt;strong&gt;would not be used for lethal injections&lt;/strong&gt;. Obviously, &lt;em&gt;Hospira&lt;/em&gt; could not comply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I think that &lt;em&gt;Hospira&lt;/em&gt; wasn't overly unlucky about the legal situation in Europe. They are a health care company and I suppose that they prefer to make profit by improving the medical situation of their customers, rather than by killing them!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-24T07:12:59.973" LastActivityDate="2015-03-24T07:12:59.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27761" PostTypeId="1" AcceptedAnswerId="27762" CreationDate="2015-03-24T09:40:05.520" Score="3" ViewCount="719" Body="&lt;p&gt;Are there any fuel-air bombs / thermobaric weapons that utilise ammonium nitrate as a fuel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that BLU-82 is often mistakenly referred to as fuel-air bomb, so I became curious as to whether or not there were any &lt;em&gt;actual&lt;/em&gt; fuel-air bombs that used ammonium nitrate. &lt;/p&gt;&#xA;" OwnerUserId="15033" LastActivityDate="2015-03-24T10:49:17.593" Title="Ammonium nitrate in fuel-air bombs" Tags="&lt;explosives&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27762" PostTypeId="2" ParentId="27761" CreationDate="2015-03-24T10:26:35.110" Score="5" Body="&lt;p&gt;Explosives that use ammonium nitrate and a fuel are mostly referred to as &lt;a href=&quot;http://en.wikipedia.org/wiki/ANFO&quot;&gt;ANFO&lt;/a&gt; explosives, you might find more detailled information by using that term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most explosives used in military (especially bombs) are based on TNT, mostly in the form of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tritonal&quot;&gt;Tritonal&lt;/a&gt;, a mixture of TNT and aluminium powder. Generally, only outdated bombs use ANFO explosives (due to their low price and ease-of-use). (See &lt;a href=&quot;http://en.wikipedia.org/wiki/BLU-82&quot;&gt;BLU-82&lt;/a&gt;, which does indeed use ammonium nitrate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANFO explosives are rather used in an industrial context, especially &lt;a href=&quot;http://en.wikipedia.org/wiki/ANFO#Industrial_use&quot;&gt;mining&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the common mixture of ammonium nitrate with nitromethane &lt;a href=&quot;http://en.wikipedia.org/wiki/ANFO#ANNM&quot;&gt;(ANNM)&lt;/a&gt;, no oxygen from the air is needed to sustain the explosion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3NH4NO3 + 2CH3NO2 -&amp;gt; 4N2 + 2CO2 + 9H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ANFO explosives carry their oxygen with them by definition (the ammonium nitrate molecule contains 3 oxygen atoms), they do not need oxygen from the surroundings and are therefore not thermobaric or fuel.air weapons.&lt;/p&gt;&#xA;" OwnerUserId="15002" LastActivityDate="2015-03-24T10:26:35.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27763" PostTypeId="2" ParentId="27761" CreationDate="2015-03-24T10:49:17.593" Score="1" Body="&lt;p&gt;I'm not aware of any FAE devices using ammonium nitrate as a fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxirane is rather common, the thermobaric hellfire (AGM-114N) uses aluminium dust and the original WWII designs of Mario Zippermayr's anti-aircraft device used coal dust. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-24T10:49:17.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27764" PostTypeId="1" AcceptedAnswerId="27766" CreationDate="2015-03-24T11:08:39.027" Score="1" ViewCount="5328" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Write equations for the reactions of each of: (a) nitric acid (b) sulfuric acid *(c) hydroiodic acid, with each of: (i) sodium hydroxide (ii) zinc oxide (iii) ammonia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the reaction of nitric acid with sodium hydroxide and zinc oxide is simple because I know that acid + base = salt + water. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\ce{HNO3(aq) + NaOH (aq) &amp;amp;-&amp;gt; NaNO3 (aq) + H2O (l)}\\&#xA;\ce{2HNO3(aq) + ZnO (s) &amp;amp;-&amp;gt; Zn(NO3)2 (aq) + H2O (l)}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes sense. In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with ammonia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HNO3(aq) + NH3 -&amp;gt; NH4NO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How am I supposed to know this equation? It's not metal + acid = salt + water. Is there word equations for &lt;em&gt;everything&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="15151" LastEditorUserId="5026" LastEditDate="2018-04-26T02:40:01.313" LastActivityDate="2018-04-26T02:40:26.730" Title="How do I write out the chemical reaction of ammonia with strong acid?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27765" PostTypeId="2" ParentId="27764" CreationDate="2015-03-24T11:22:19.347" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How am I supposed to know this equation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to &lt;em&gt;memorise&lt;/em&gt; it. In aqueous solution, $\ce{NH3}$ deprotonates a small fraction of the water to give ammonium and hydroxide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;    \ce{NH3 + H2O &amp;lt;=&amp;gt; NH4+ + OH−} \, ,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and while it is usually impossible to isolate $\ce{NH4OH}$ itself, the above mentioned ions are there in the solution. And thus, an aqueous solution of $\ce{NH3}$ is a weak base and reacts with acids in the similar way as other weak bases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I access all the word equations in chemistry so I am able to write every single equation given to me like these.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, learning general principles plus memorising all the special cases &amp;amp; exceptions is the only way.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="24052" LastEditDate="2016-03-11T23:51:49.567" LastActivityDate="2016-03-11T23:51:49.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27766" PostTypeId="2" ParentId="27764" CreationDate="2015-03-24T11:55:48.687" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't agree on the -1 for hydrogen, let's look again at that example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HNO3(aq) + ZnO(s) -&amp;gt; Zn(NO3)2(aq) + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Metal oxides&lt;/strong&gt;, such as zinc oxide, &lt;strong&gt;are basic anhydrides&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{ZnO}$ was made by heating $\ce{Zn(OH)2}$ until it loses water:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn(OH)2 -&amp;gt;[\Delta] ZnO(s) + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;$\ce{ZnO}$ is a salt of $\ce{Zn^2+}$ and $\ce{O^2-}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reaction between nitric acid and zinc oxide and just a neutralization reaction between two protons and $\ce{O^2-}$ (which, as such, isn't present in solution).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the reaction with ammonia: For this you may want to remember that there are different acid-base theories: The one that helps here is the &lt;em&gt;Bronsted-Lowry&lt;/em&gt; theory:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Acids are proton donors&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\ce{HNO3}$ definitely is that.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bases are proton acceptors&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can something else other than the $\ce{OH-}$ that you knew and the $\ce{O^2-}$ from $\ce{ZnO}$ be a proton acceptor = form a bond to a (proton)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that possible if this acceptor doesn't even have a negative charge? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Yes, otherwise I wouldn't buld up the tension like in a suspense story :D&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the proton can't provide the electrons for the bond, the acceptor has to deliver them! And that can happen in the form of an electron pair, a double filled, non-bonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore a reaction, such as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H+ + NH3 &amp;lt;=&amp;gt; NH4+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is perfectly valid. The rest of the example is just arranging the counter ions.  &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="5026" LastEditDate="2018-04-26T02:40:26.730" LastActivityDate="2018-04-26T02:40:26.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27767" PostTypeId="1" AcceptedAnswerId="27768" CreationDate="2015-03-24T12:37:51.030" Score="8" ViewCount="1956" Body="&lt;p&gt;I am writing my bachelor thesis about alternative materials for feminine hygiene products and for testing I need to know what the viscosity is of menstruation blood. Can I use water as replacement for menstruation blood because this kind of blood exists out of a big part of water? And what is the composition of menstruation blood in percentages?&lt;/p&gt;&#xA;" OwnerUserId="2657" LastActivityDate="2015-04-17T08:40:58.050" Title="What is the viscosity of menstruation blood?" Tags="&lt;viscosity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27768" PostTypeId="2" ParentId="27767" CreationDate="2015-03-24T12:51:01.677" Score="12" Body="&lt;p&gt;It is difficult to give a typical value as the composition and hence the physical properties of menstrual blood varies over the course of a woman's period. &lt;a href=&quot;https://books.google.co.uk/books?id=beenEjKmvPwC&amp;amp;pg=PA159#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;This reference&lt;/a&gt; has a good discussion which you may find useful; you can find on p. 159:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is fair to say that a large proportion of menses samples are considerably more viscous than blood or water - about four times as viscous as venous blood and 35 times as viscous as water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-03-24T12:51:01.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27770" PostTypeId="1" AcceptedAnswerId="27894" CreationDate="2015-03-24T13:34:39.677" Score="7" ViewCount="370" Body="&lt;p&gt;I notice that Gd has a second ionization potential which is significantly higher than would be expected from the general trend in the lanthanides (see &lt;a href=&quot;http://www.uv.es/~borrasj/ingenieria_web/temas/tema_1/lecturas_comp/p938.pdf&quot;&gt;this paper p. 945 for a graph&lt;/a&gt;). What is the reason for this effect ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} =&amp;gt; [Xe] 4f^{7} 5d^{1}$. I understand that the 5d is occupied to prevent inter-electron repulsion between antiparallel f electrons, and this can be used to rationalize why the correspnding Tb transition requires less energy, but I'm not sure why the potential should be significantly higer than that for Eu. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-03-27T01:26:11.093" Title="Second ionization potential of Gadolinium" Tags="&lt;electronic-configuration&gt;&lt;periodic-trends&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27773" PostTypeId="1" CreationDate="2015-03-24T14:16:47.557" Score="2" ViewCount="1259" Body="&lt;p&gt;I know how to derive the expression for the equilibrium constant starting from $\mathrm dG=0$,&#xA;$$\ln(K)=-\frac{\Delta G ^\circ}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we are not at equilibrium, $\mathrm dG$ is not necessarily zero. In this case:&#xA;$$\frac{\partial G}{\partial \epsilon} = \sum \mu_i \nu_i = \Delta G ^\circ + RT \ln \left( \prod a_i ^{\nu_i} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\nu_i$ are the stoichiometric coefficients, $\mu_i$ are the chemical potentials, $\Delta G^\circ$ is the Gibbs free energy of reaction at standard state and $a_i$ are the activities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence,&#xA;$$\frac{\partial G}{\partial \epsilon} = RT \ln \left(\frac{Q}{K} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q=\prod a_i ^{\nu_i}$ is the reaction quotient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, if we want to kown if a reaction is spontaneous, do we look at the ratio $Q/K$? If this is correct, if $Q &amp;gt; K$ then $\frac{\partial G}{\partial \epsilon} &amp;gt; 0$ and the reaction would be non-spontaneous, whereas if $Q &amp;lt; K$ then $\frac{\partial G}{\partial \epsilon} &amp;lt; 0$ and the reaction is spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, if we start with reactants only, $Q=0$ (since the activites of the products are zero because there are none) and the reaction would be spontaneous, and we would always get at least a little bit of products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, various pages in internet say that if $\Delta G ^\circ &amp;lt; 0$ then the reaction is spontaneous, and it is not if $\Delta G ^\circ&amp;gt;0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you know, then, if you start with only reactants, if a reaction is going to happen?&lt;/p&gt;&#xA;" OwnerUserId="4713" LastEditorDisplayName="user7951" LastEditDate="2016-06-18T09:28:36.473" LastActivityDate="2016-06-18T09:28:36.473" Title="Spontaneous Reaction and Gibbs Free Energy" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27774" PostTypeId="2" ParentId="27728" CreationDate="2015-03-24T14:26:43.683" Score="5" Body="&lt;p&gt;Lets put some numbers on the phase transition of NaCl from the normal NaCl structure (based on fcc, 6-fold coordinated) at atmospheric pressure, to the CsCl (based on bcc, 8-fold coordinated) structure at high pressure. &#xA;From Compounds and Alloys Under High Pressure: A Handbook, by E. Yu Tonkov (google found it) one gets that the transition occurs at 27-29GPa, with a volume change of 5%.&lt;br&gt;&#xA;The density of NaCl, taken from Wikipedia, is 2.165 gm/cc, or about 27cm$^3$/mol, or 27E-6 m$^3$/mol. &#xA;Multiply on through and you get, very roughly, the PV energy gained by converting from the less dense NaCl structure to the denser CsCl structure is 40kJ/mol - this is not an inconsequential energy gain for the Gibbs free energy.  &lt;strong&gt;Thus, the PV energy gained by a denser solid with the higher coordination plays off (and pays off) against the ionic unhappiness.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the literature there are also indications that the CsCl, and/or higher pressure phases, may be metallic in nature, so the assumption of purely ionic bonding may be in jeopardy.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2015-03-24T16:38:06.163" LastActivityDate="2015-03-24T16:38:06.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27775" PostTypeId="2" ParentId="27773" CreationDate="2015-03-24T15:25:31.967" Score="1" Body="&lt;p&gt;Your analysis is correct.  It is silly to say that a reaction is or is not spontaneous based soley on $\Delta G^{\circ}$.  If you have only reactants present and no products present, then the reaction will always be spontaneous, irrespective of $\Delta G^{\circ}$.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="13184" LastEditDate="2015-03-25T02:09:25.750" LastActivityDate="2015-03-25T02:09:25.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27777" PostTypeId="2" ParentId="27731" CreationDate="2015-03-24T16:10:21.937" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Acetone a is aprotic polar solvent which is often used as solvent for $\ce{S_{N}2}$ reactions (notably in conjunction with $\ce{NaI}$ as in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Finkelstein_reaction&quot; rel=&quot;nofollow&quot;&gt;Finkelstein reaction&lt;/a&gt;). For $\ce{S_{N}2}$ reactions steric hindrance and leaving group capability is of great importance, so primary halides will be usually easier replaced by iodine then tertiary.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{AgNO3}$ will probably react as a base forming nitric acid. Even though nitric acid is a rather strong acid this is more likely than $\ce{NO_3-}$ acting as nucleophile. Besides that, $\ce{Ag+}$ is known to have a unique selectivity for halogens and will attack them readily, forming $\ce{AgX}$ and a carbocation, hence promoting $\ce{E 1}$/$\ce{S_{N}1}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-24T16:10:21.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27778" PostTypeId="1" AcceptedAnswerId="27820" CreationDate="2015-03-24T16:32:49.773" Score="2" ViewCount="1615" Body="&lt;p&gt;The Frost diagram for sulfur shows the relative stability (in terms of cell potentials) of the possible aqueous oxidation states. What is the reason for the instability of higher oxidation states in the acidic regime and their stability in the basic regime ? Something I have always noticed but never been able to apply chemical intuition to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e5f4u.png&quot; alt=&quot;Frost diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-03-25T15:27:58.943" Title="Frost diagram and acidic/basic conditions." Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27780" PostTypeId="1" AcceptedAnswerId="27785" CreationDate="2015-03-24T18:13:45.690" Score="2" ViewCount="5661" Body="&lt;p&gt;The ultimate analysis (by weight percent) for beech wood is:&#xA;$$\ce{C}=49.05\,\%,\; \ce{H}=5.83\,\%,\; \ce{O}=45\,\%,\; \ce{N}=0.12\,\%$$&#xA;Using the ultimate analysis values, the molar mass of each element, and assuming 100 grams of wood (therefore C = 49.05 g, H = 5.83 g, etc.), a representative formula for the beech wood is:&#xA;$$&#xA;\ce{C} = 49.05/12.01 = 4.0841\; \ce{mol} \\&#xA;\ce{H} = 5.83/1.01 = 5.77\; \ce{mol} \\&#xA;\ce{O} = 45/16 = 2.81\; \ce{mol} \\&#xA;\ce{N} = 0.12/14.01 = 0.0086\; \ce{mol} \\&#xA;\ce{C}_{\,4.0841\,}\ce{H}_{\,5.77\,}\ce{O}_{\,2.81\,}\ce{N}_{\,0.0086\,}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the molecular weight of the beech wood based on the representative formula?&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="2151" LastEditDate="2015-03-26T19:22:08.483" LastActivityDate="2015-03-26T19:22:08.483" Title="Determine molecular weight of biomass based on ultimate analysis" Tags="&lt;concentration&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-03-27T01:19:12.923" ContentLicense="CC BY-SA 3.0" />
  <row Id="27781" PostTypeId="1" CreationDate="2015-03-24T18:17:09.457" Score="2" ViewCount="116" Body="&lt;p&gt;$$\ce{NaCl + H2SO4 -&amp;gt; NaHSO4 + HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above reaction happens, but why doesn't the following happen? Or does it but only when one of the reactants is in excess?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NaCl + H2SO4 -&amp;gt; Na2SO4 + 2HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reaction seems plausible to me. How can I predict whether a similar type of reaction will be like the first equation or the second equation?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2016-11-22T15:41:19.033" LastActivityDate="2016-11-22T15:41:19.033" Title="Why is only one proton transferred in the reaction between sulfuric acid and sodium chloride" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27782" PostTypeId="1" AcceptedAnswerId="27787" CreationDate="2015-03-24T18:33:15.580" Score="2" ViewCount="1874" Body="&lt;p&gt;After creating a solution of copper(II)-sulfate-pentahydrate, water and concentrated ammonia, the newly formed complex (tetraamminecopper(II)-sulfate-monohydrate) is completely dissolved in water. By adding ethanol, one is able to decrease the solubility of that complex in water and have crystals percipitates out of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the complex is not soluble in ethanol, I am having trouble understanding why adding ethanol changes anything about the solubility of the complex. Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?&lt;/p&gt;&#xA;" OwnerUserId="15002" LastActivityDate="2015-03-24T20:34:49.850" Title="Changing solubility of metal complex by adding ethanol" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27783" PostTypeId="1" CreationDate="2015-03-24T18:42:21.067" Score="2" ViewCount="630" Body="&lt;p&gt;I am finding different equations for the synthesis. My lecture handout suggests a reaction with Sodium tetrafluroborate and Sulfuric acid. This is fine but I have to remember this for an exam and that reaction looks extremely hard to remember/balance in an exam. A second way seems to be reaction of boron oxide with Calcium fluoride and Sulfuric acid. However, I cannot find consistent equations for this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia simply suggests that Boron oxide reacts with $\ce{HF}$ which is made in situ by the reaction between sulfuric acid and calcium fluoride:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B2O3 + 6HF -&amp;gt; 2BF3 + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to remember but I am reluctant to accept this because it excludes two of the reactants (sulfuric acid and calcium fluoride). Would it be acceptable to merge the formation of $\ce{HF}$ into the above equation - thus including all reactants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B2O3 + 3CaF2 + 6H2SO4 -&amp;gt; 2BF3 + 3H2O + 3Ca(HSO4)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a little tricky to balance but not too hard to remember. Is there anything wrong with the above equation? I haven't seen it anywhere but only variations on it.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2016-05-13T12:22:02.347" LastActivityDate="2016-05-13T12:22:02.347" Title="How to synthesize Boron tri-fluoride using Boron Oxide" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27784" PostTypeId="1" CreationDate="2015-03-24T19:34:06.087" Score="0" ViewCount="100" Body="&lt;p&gt;Why does $\ce{H2Se}$ have a bigger $K_\text{a2}$ than both $\ce{H2S}$ and $\ce{H2Te}$?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorDisplayName="user7951" LastEditDate="2015-03-24T20:16:56.980" LastActivityDate="2015-03-24T20:26:14.157" Title="Why does H₂Se have a bigger Ka2 than H₂Te" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27785" PostTypeId="2" ParentId="27780" CreationDate="2015-03-24T20:16:55.290" Score="2" Body="&lt;p&gt;As a general rule, it is not a good idea to treat complex materials like wood as something you can calculate a mole of. You can of course analyze the material and get elemental rates, but that does not mean they are bonded in such way that this rate forms a minimum fraction which represents the whole thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you calculate the number of moles of each atom in 100g, what you have is an estimation of the expected quantities in that sample. Undoubtedly, if you took different samples of the wood you would find different values, specially if sampling from different parts of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an exaggerated example, think about the elemental constitution of the human body, which is mainly 65% Oxygen, 18% Carbon, 10% Hydrogen and 3% Nitrogen in mass (&lt;a href=&quot;http://en.wikipedia.org/wiki/Composition_of_the_human_body&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). With that in mind you can calculate a mole ratio of these elements and create a &quot;representative formula&quot;. Obviously, though, there's no such thing as the &lt;em&gt;human molecule&lt;/em&gt;, and you can't say that someone's body is X or Y moles of &lt;em&gt;human&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, like I originally answered, if you define what the representative formula is, you can calculate it's molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce {M}=4.0841*12.0107 + 5.77*1.00794+2.81*15.9994+0.0086*14.0067 = 99.947 \ \ce{g/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing ${\ce M}$, the number of moles is ${\ce 700,000/99.947 \approx 7004}$, and the concentration of biomass is $7004\ \ce{mol/m^3}$&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-03-24T21:48:06.900" LastActivityDate="2015-03-24T21:48:06.900" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="27786" PostTypeId="2" ParentId="27784" CreationDate="2015-03-24T20:26:14.157" Score="4" Body="&lt;p&gt;The following data are from the &lt;em&gt;Rubber Book&lt;/em&gt; (CRC Handbook of Chemistry and Physics), 89th edition, 2009, p 8-40. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lll}&#xA;\mathrm{acid} &amp;amp; \mathrm{pK_{a1}} &amp;amp; \mathrm{pK_{a2}} \\&#xA;\hline&#xA;\ce{H2S} &amp;amp; 7.05 &amp;amp; 19\\ &#xA;\ce{H2Se} &amp;amp; 3.89&amp;amp; 11\\&#xA;\ce{H2Te} &amp;amp; 2.6 &amp;amp; 11\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\mathrm{pK_{a1}}$ of $\ce{H2Te}$ was determined at 18 °C, all other data at 25 °C&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-24T20:26:14.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27787" PostTypeId="2" ParentId="27782" CreationDate="2015-03-24T20:34:49.850" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would adding some organic phase like petrol ether or a non-soluble solid such as ceramic pellets do the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. But adding other organic solvents which are miscible with water would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind the &quot;trick&quot; is to reduce the polarity of the &lt;strong&gt;aqueous&lt;/strong&gt; phase. With other words, you're making water more ethanol-like.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-24T20:34:49.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27788" PostTypeId="2" ParentId="27781" CreationDate="2015-03-24T20:40:59.377" Score="3" Body="&lt;p&gt;Sodium bisulphate (pKa ~ 2 - &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfuric_acid&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Sulfuric_acid&lt;/a&gt;) is a much weaker acid that hydrochloric acid (pKa = ~ -7), which means that chloride is not a strong enough base to deprotonate sodium bisulphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However all acid pbased reactions are at equilibrium.  Roughly one part in 10^9 of the species involve sulphate anion, which for all practical purposes is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table of pKa charts: &lt;a href=&quot;http://www.lauher.com/che142/acidbase.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.lauher.com/che142/acidbase.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2015-03-24T21:52:08.457" LastActivityDate="2015-03-24T21:52:08.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27789" PostTypeId="1" CreationDate="2015-03-24T20:52:34.563" Score="11" ViewCount="14327" Body="&lt;p&gt;Is it possible to chemically convert methane into propane? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what is involved and is it economical?&lt;/p&gt;&#xA;" OwnerUserId="15157" LastEditorUserId="30836" LastEditDate="2019-04-13T14:58:27.387" LastActivityDate="2019-04-13T14:58:27.387" Title="Conversion of methane to propane" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27790" PostTypeId="2" ParentId="27789" CreationDate="2015-03-24T21:02:41.140" Score="15" Body="&lt;p&gt;Yes, it can be done by mimicking the conditions in the earth's mantle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was shown by &lt;em&gt;Alexander Goncharov&lt;/em&gt; and coworkers &lt;a href=&quot;http://dx.doi.org/10.1038/ngeo591&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt; by placing methane in a diamond anvil cell, applying a pressure of more than 2 GPa and heating everything to a temperature around 1000-1500 K with a laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it economical? Hardly, but very interesting anyway!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Kolesnikov, A.; Kutcherov, V. G.; Goncharov, A. F. Nature Geosci 2009, 2 (8), 566–570. DOI &lt;a href=&quot;http://dx.doi.org/10.1038/ngeo591&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1038/ngeo591&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="41328" LastEditDate="2017-08-08T22:27:51.980" LastActivityDate="2017-08-08T22:27:51.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27791" PostTypeId="2" ParentId="27780" CreationDate="2015-03-24T21:51:43.857" Score="2" Body="&lt;p&gt;Using your mass percentages and the molar masses of the elements in your sample, you have correctly found out how many mol of each element are in 100g of your sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you need to keep in mind that wood is more than one simple molecule but instead rather a collection of quite a number of molecules. The masses of the elements that show up in all of those molecules add up to the percentages you found by this analysis. You see now that trying to formulate this distribution as a molecular formula has its limitations. Depending on how you select your sample, you will once include maybe $\ce{1 mol}$ of molecule A and $\ce{18 mol}$ of molecule B, whereas if you take twice the amount, you will count a total of 38 molecules as one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Molx stated in his answer, if you always select the same mass for comparision, this might be a reasonably suitable unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To rephrase this: if you only had one molecule or a specific ratio of known molecules in your sample, it would be clear how the different elements distribute themselves to the different molecules. Since you do not know what molecules beech wood is made of, this molecular formula is not very useful for such a case.&lt;/p&gt;&#xA;" OwnerUserId="15002" LastActivityDate="2015-03-24T21:51:43.857" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27792" PostTypeId="2" ParentId="19226" CreationDate="2015-03-24T22:04:12.140" Score="1" Body="&lt;p&gt;This turns out to be very difficult to do for two reasons. One problem is that the voltage of a silver cell is different from that of Ni-Cd cell, so it would require a specialized, multi-cell configuration to emulate the voltage characteristics of the Ni-Cd rechargeables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is that silver cells generally have a sophisticated frame inside of them that is produced by an intricate high-temperature welding process. To duplicate this process and produce a suitable frame would require a significant amount of experimentation and work.&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorUserId="41328" LastEditDate="2019-11-03T14:10:28.397" LastActivityDate="2019-11-03T14:10:28.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27793" PostTypeId="1" CreationDate="2015-03-24T22:06:35.080" Score="1" ViewCount="425" Body="&lt;p&gt;In titrating 25 mL of a 0.2 M weak acid, HA, with 0.5 M of a strong base, say NaOH, the equivalence point is reached after 10 mL of base is added. I can calculate the pH by first setting up an ICE table to calculate the moles of A- remaining at the equivalence point, and then setting up an additional ICE table for the hydrolysis of A- to calculate the moles of OH- at the equivalence point. From that, I can determine the pOH and thus the pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: what would be the effect of adding more weak acid? Would I need to set up an ICE table with non-zero initial values for HA, OH-, and A-? If that's the case, then how do I determine the pH?&lt;/p&gt;&#xA;" OwnerUserId="15156" LastActivityDate="2015-03-24T22:06:35.080" Title="How can I determine the pH of a solution when I add more analyte after the equivalence point?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27794" PostTypeId="1" AcceptedAnswerId="27798" CreationDate="2015-03-24T23:01:26.617" Score="2" ViewCount="394" Body="&lt;p&gt;Are the thermal decomposition products of polyurea polymers known, especially of the urea group, in the absence of water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the formula&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;R-NH-C=O&#xA;     |&#xA;    HN-R&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;there appears to be no obvious candidate for a decomposition product - not enough oxygen for CO&lt;sub&gt;2&lt;/sub&gt;, not enough hydrogen for Ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas or, even better, references?&lt;/p&gt;&#xA;" OwnerUserId="15159" LastEditorUserId="8083" LastEditDate="2015-03-25T00:24:39.400" LastActivityDate="2015-03-25T02:06:22.850" Title="Thermal decomposition products of polyurea?" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27795" PostTypeId="1" CreationDate="2015-03-25T01:08:22.083" Score="9" ViewCount="7343" Body="&lt;p&gt;For the (E)-2-pentene, there was no C=C reading, about 1600. For the (Z)-2-pentene, there is a C=C. Why is there a reading on the Z but not on the E?&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="4945" LastEditDate="2015-03-31T08:47:00.413" LastActivityDate="2015-03-31T08:47:00.413" Title="In infrared spectroscopy, why is there no C=C reading for E alkenes?" Tags="&lt;spectroscopy&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27796" PostTypeId="2" ParentId="27789" CreationDate="2015-03-25T01:52:19.537" Score="10" Body="&lt;p&gt;Generally, this reaction is run in reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen reformers are used to generate hydrogen by the following reaction scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + H2O -&amp;gt; CO + 3H2}~~~\mathrm{(Endothermic, 1000C)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is frequently followed by the shift reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO + H2O -&amp;gt; CO2 + H2}~~~\mathrm{(Exothermic, 300C)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen reformers that run on heavier hydrocarbons than methane are frequently fitted with a pre-reformer in which the reaction starts at a lower temperature. In the pre-reformer the following reaction occurs, amongst others, including some that generate hydrogen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H8 + 2H2 -&amp;gt; 3CH4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this reaction is useful is debatable, but it does indeed occur because it is mildly exothermic and has no change in the number of moles of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running the reaction in the way that you want would require &lt;strong&gt;the removal of some of the hydrogen, to push the equilibrium in favour of propane.&lt;/strong&gt; This might be useful in the case of stranded gas: i.e natural gas which is economic to extract (or which is extracted as a byproduct of oil extraction) but is not economic to ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequently stranded gas is converted to methanol for shipping. Synthesis gas (a mixture of $\ce{H2}$ and $\ce{CO}$ is an intermediate step.) Methanol &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_to_liquids&quot; rel=&quot;noreferrer&quot;&gt;can be converted&lt;/a&gt; to heavier hydrocarbons such as gasoline if desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as mentioned in the same reference we can also perform a more direct conversion of natural gas to heavier hydrocarbons via the Fischer-Tropsch process (using the synthesis gas step but not the methanol step). &lt;a href=&quot;http://en.wikipedia.org/wiki/Fischer%E2%80%93Tropsch_process&quot; rel=&quot;noreferrer&quot;&gt;Main article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial oxidation of the natural gas eliminates the excess hydrogen and ensures the correct balance of of $\ce{H2}$ and $\ce{CO}$ in the synthesis gas to ensure heavier hydrocarbons are formed. We have two reactions that can vary the ratio of $\ce{H2/CO}$ in the feed to the Fischer-Tropsch process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2 + O2 -&amp;gt;  H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consumption of $\ce{H2}$ in preference to consumption of $\ce{CO}$, decreases $\ce{H2/CO}$ ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO + H2O -&amp;gt; CO2 + H2}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shift reaction, increases $\ce{H2/CO}$ ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This composition control and several other parameters enable the desired mixture of hydrocarbons to be obtained from the Fischer-Tropsch process, though a single compound such as propane is rarely if ever obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, $\ce{3CH4 -&amp;gt; C3H8 + 2H2}$ is not possible under normal conditions, but $\ce{3CH4 + O2 -&amp;gt; 2H2O + \mathrm{a~mixture~of~hydrocarbons~of~average~formula~C3H8}}$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding air to the mixture for oxidation is problematic because it adds nitrogen which must be purged away later, but I am guessing it is much cheaper than running a pressure swing absorption plant to remove excess hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pearl_GTL&quot; rel=&quot;noreferrer&quot;&gt;Here's&lt;/a&gt; an example of a gas to liquid hydrocarbon plant. I'm guessing it doesn't make propane, because heavier hydrocarbons are more valuable and easier to ship, but the principle remains the same.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="10185" LastEditDate="2015-08-18T18:27:43.990" LastActivityDate="2015-08-18T18:27:43.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27797" PostTypeId="2" ParentId="27795" CreationDate="2015-03-25T02:03:48.220" Score="12" Body="&lt;p&gt;The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment.  If we examine the $\ce{C=C}$ stretch in &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-2-butene we find that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0PNJJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://socratic.org/questions/how-can-i-draw-cis-and-trans-isomers&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;cis&lt;/em&gt;-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore &lt;strong&gt;we would expect&lt;/strong&gt; to see an ir peak for the $\ce{C=C}$ stretch in this molecule&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;trans&lt;/em&gt;-2-butene does not have a dipole moment  (0 D) and upon stretching the double bond the dipole moment will remain zero, there is no change; therefore we &lt;strong&gt;would not expect&lt;/strong&gt; to see an ir peak for the $\ce{C=C}$ stretch in this molecule&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A general rule based on these dipole moment arguments is that, for symmetrically di-substituted olefins $\ce{C=C}$ stretches in &lt;em&gt;trans&lt;/em&gt; olefins are non-existent or very weak, the absorption is much stronger in the &lt;em&gt;cis&lt;/em&gt; isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can extend this reasoning to molecules with dipole moments, if the dipole moment is very small.  &lt;em&gt;Trans&lt;/em&gt;-2-pentene does have a dipole moment but it is very small and it is less than the dipole moment in the &lt;em&gt;cis&lt;/em&gt;-isomer &lt;strong&gt;and&lt;/strong&gt; the dipole moment in the &lt;em&gt;trans&lt;/em&gt;-isomer (being smaller to begin with)  will change less upon stretching the double bond than it will for the &lt;em&gt;cis&lt;/em&gt; isomer.  Therefore we would expect a very weak absorption in &lt;em&gt;trans&lt;/em&gt;-2-pentene, while the &lt;em&gt;cis&lt;/em&gt; isomer should show a stronger absorption.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-25T16:35:46.290" LastActivityDate="2015-03-25T16:35:46.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27798" PostTypeId="2" ParentId="27794" CreationDate="2015-03-25T02:06:22.850" Score="2" Body="&lt;p&gt;Here's &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/pol.1982.170200703/abstract&quot; rel=&quot;nofollow&quot;&gt;one abstract&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The primary fragmentation processes in the thermal decomposition of polymers were studied in detail on a series of structurally related polyureas by direct pyrolysis with a mass spectrometer. Our results indicate that polyureas I–III undergo a quantitative depolycondensation process analogous to that observed for N-monosubstituted polyurethanes. The thermal decomposition of polyureas IV–VI proceeds by intramolecular hydrogen transfer processes that occur at higher temperatures with respect to depolycondensation. Polycarboxypiperazine VI is decomposed by a single-stage decomposition mechanism that leads to fragments with amino end groups and carbon oxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I took a look at the article.  Depolycondensation is accurate: most of the polyureas examined decomposed to the &quot;parent&quot; isocyanates and amines.  The temperature range examined was 150&amp;deg;C to 350&amp;deg;C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in what happens at higher temperatures, I would guess a decomposition to mixtures of CO, ammonia, water, and possibly methane and/or hydrogen.  But I didn't find any references that would support me after a quick look&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-25T02:06:22.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27799" PostTypeId="1" AcceptedAnswerId="27822" CreationDate="2015-03-25T02:24:06.620" Score="7" ViewCount="5063" Body="&lt;p&gt;While comparing the relative stabilities of acetic acid and formic acid why don't we consider the 3 possible hyper conjugative structures which would stabilize the positive charge on carbon thus making the conjugate of acetic acid more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor told me hyper conjugation does not operate here. But what is the reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="13409" LastEditDate="2015-03-25T15:29:02.463" LastActivityDate="2018-08-28T03:34:13.587" Title="Hyper conjugation in acetic acid?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27800" PostTypeId="1" CreationDate="2015-03-25T02:30:43.617" Score="2" ViewCount="2446" Body="&lt;p&gt;At the beginning of a Weak Acid/Strong Base titration curve there is a sharp decrease in $[\ce{H+}]$ and a sharp increase in pH before the ratio between the weak acid and it's conjugate base becomes low enough that we have a buffer solution.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EK0wX.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/F5wmq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening at that point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is that, given the balanced equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{&#xA;CH3COOH~(aq) + H2O(~l) &amp;lt;=&amp;gt; CH3COO- ~(aq) + H3O+~(aq)&#xA;}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of $\ce{OH-}$ would shift equilibrium to the left by Le Chatelier's principle making the newly generated $\ce{CH3COO-}$ react with $\ce{H3O+}$ to make more $\ce{CH3COOH}$, thereby increasing pH. After that, the $[\ce{H3O+}]$ is so low that the addition of the anion is not enough to keep shifting the equilibrium position to the left, so the remaining weak acid resists the changes in pH with its dissociation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make any sense?&lt;/p&gt;&#xA;" OwnerUserId="15058" LastEditorUserId="4945" LastEditDate="2015-03-25T04:17:59.303" LastActivityDate="2019-04-08T12:21:31.770" Title="What's happening at the beginning of a weak acid titration?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27802" PostTypeId="2" ParentId="27800" CreationDate="2015-03-25T04:42:40.097" Score="2" Body="&lt;p&gt;Unfortunately this is not correct. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{CH3COOH~(aq) + H2O(~l) &amp;lt;=&amp;gt; CH3COO- ~(aq) + H3O+~(aq)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have a pure acetic acid solution the equation you state governs the pH of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.&lt;br&gt;&#xA;Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated).&#xA;$$\ce{H3CCOOH + {}^{-}OH &amp;lt;=&amp;gt; H3CCOO- + H2O}$$&#xA;According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play.&#xA;$$\ce{H3CCOO- + H2O &amp;lt;=&amp;gt; H3CCOOH + {}^{-}OH}$$&#xA;This equilibrium is essentially responsible for raising the pH.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-03-25T04:42:40.097" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27803" PostTypeId="1" AcceptedAnswerId="27808" CreationDate="2015-03-25T05:28:50.517" Score="6" ViewCount="1231" Body="&lt;p&gt;I'm reading about water models and their dispersion coefficients, and going back to S. C. Wang's work according to the citation of an equation (Google books &lt;a href=&quot;https://books.google.co.uk/books?id=Apyi_FXKnSkC&amp;amp;lpg=PA236&amp;amp;ots=3R_BmUrJsx&amp;amp;dq=s%20c%20wang%20u(r)%20c6r%206&amp;amp;pg=PA236#v=onepage&amp;amp;q=s%20c%20wang%20u(r)%20c6r%206&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUpXW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see one citation's to a biographical work, and the other to a German work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowhere in the chapter I'm reading (chapter 5, ‘Resolution’, in Rowlinson’s 2002 &lt;em&gt;Cohesion: A Scientific History of Intermolecular Forces&lt;/em&gt;) does it say what $C_6$ stands for, and it doesn't seem to be a series of $C_n$, so I'm a little confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's the equation, apparently a Casimir-Polder integral&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\mathrm{DS} = ‒\sum_{n=6,8,10,...} \left(\frac{C_n}{r_n}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in something else I'm reading, but it's biochemistry and so they're skipping over actually defining these things (so I can see DFT would be density functional theory but can't actually figure out what either $E_\mathrm{DS}$ is, nor the meaning of a $C_6$ coefficient. Looking for $E_\mathrm{DS}$ is just bringing up a bunch of articles' editors in search results (&lt;em&gt;Eds.&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone enlighten me as to the latter - the meaning of a $C_6$ coefficient with respect to [London] dispersion forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My own answer to this after more reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The oscillating electrons in a molecule generate not only instantaneous dipoles but also quadrupoles and higher multipoles. It is to be expected, therefore, that the London dispersion energy is only the first term in a series expansion for the attractive energy&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;that starts in $$u(r) = −C_6r^{−6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term comes originally from the wave equation (as Rowlinson's text does eventually describe), the number six arising from a Cartesian coordinate system for each of the two interaction partners&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W34Sy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2119" LastEditorUserId="2119" LastEditDate="2015-03-25T11:47:44.470" LastActivityDate="2015-03-25T11:47:44.470" Title="Why are C6 dispersion forces named C6?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27804" PostTypeId="1" AcceptedAnswerId="27805" CreationDate="2015-03-25T06:37:39.623" Score="3" ViewCount="1590" Body="&lt;p&gt;For an experiment, I'm adding several substances (as powders) to hydrogen chloride, and measuring the temperature changes to calculate enthalpy. What is a container I could perform it in (other than a styrofoam cup) that would minimise the heat lost to the surroundings? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up using a vacuum flask of some sort, but would that be appropriate for this sort of reaction? The temperature is being measured via a temperature probe.&lt;/p&gt;&#xA;" OwnerUserId="14994" LastEditorUserId="418" LastEditDate="2015-03-25T06:59:06.777" LastActivityDate="2015-03-25T09:59:56.883" Title="How to lose as little heat as possible to the surroundings from an endothermic reaction?" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27805" PostTypeId="2" ParentId="27804" CreationDate="2015-03-25T06:57:01.697" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;What is the scale?&lt;/li&gt;&#xA;&lt;li&gt;What is the budget?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shop.faust.ch/QSRCH.HTM?UID=550055125b5d000000000000&amp;amp;KEYVAL=9.142+106&quot; rel=&quot;nofollow&quot;&gt;Reactors are available&lt;/a&gt; from 250 mL up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be cheaper to place a flask in a cut-off bottom of a suitable PE bottle and insulate it with PU foam from a DIY market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small Dewar cylinder might be an option too. &#xA;The &lt;a href=&quot;https://de.vwr.com/store/catalog/product.jsp?catalog_number=478-4102&quot; rel=&quot;nofollow&quot;&gt;model 00C&lt;/a&gt; has a volume of 100 mL and lids are available too. Usually, these Dewars are used to store liquid nitrogen and cooling mixtures (2-propanol + dry ice, etc.) so I don't know how resistant the lid is towards $\ce{HCl}$ over time. In addition, this setup isn't as tight as a flask with hollow stoppers and/or a silicone seal if that's an issue. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-25T09:59:56.883" LastActivityDate="2015-03-25T09:59:56.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27806" PostTypeId="1" CreationDate="2015-03-25T07:15:39.930" Score="10" ViewCount="15406" Body="&lt;p&gt;We will be buying methanol by a about 200 liters at a time, I was wondering what I should consider when storing it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I know that it should be &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in the shade&lt;/li&gt;&#xA;&lt;li&gt;earthed with a copper stake&lt;/li&gt;&#xA;&lt;li&gt;in a secondary plastic container&lt;/li&gt;&#xA;&lt;li&gt;wash empty containers and drain them&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I am wondering what else should be considered? How far should it be away from a house? &lt;/p&gt;&#xA;" OwnerUserId="15164" LastActivityDate="2015-05-30T13:27:20.520" Title="How to store methanol" Tags="&lt;safety&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27807" PostTypeId="1" AcceptedAnswerId="27823" CreationDate="2015-03-25T10:48:37.960" Score="1" ViewCount="292" Body="&lt;p&gt;In the uncertainty principle , whats the significance of the inequality sign ?&lt;br&gt;&#xA;Why is it not equal to over there ?&lt;br&gt;&#xA;Why do we say greater than equal to instead of just equal to ?   &lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="10185" LastEditDate="2015-03-25T13:04:21.080" LastActivityDate="2015-03-27T18:23:31.607" Title="Heisenberg's Uncertainty Principle's inequality" Tags="&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2015-03-25T22:44:49.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="27808" PostTypeId="2" ParentId="27803" CreationDate="2015-03-25T11:35:02.847" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/56/6/10.1063/1.1677610&quot;&gt;Calculation of Coefficients in the Power Series Expansion of the Long‐Range Dispersion Force between Atoms&lt;/a&gt; J. Chem. Phys. 56, 2801 (1972)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Highly accurate calculations&#xA;  of $C_6$, the coefficient of the $R^{-6}$ term, resulting from&#xA;  an induced dipole-induced dipole interaction, have&#xA;  been made for many atomic systems, while values of&#xA;  $C_8$ (dipole-quadrupole interaction) and $C_{10}$ (dipoleoctopole&#xA;  plus quadrupole-quadrupole interactions) are...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, in London dispersion interaction is usually represented as induced dipole - induced dipole interaction, the energy of which is proportional to $R^{-6}$, but there are other components of the interaction.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-25T11:35:02.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27809" PostTypeId="1" AcceptedAnswerId="27817" CreationDate="2015-03-25T12:01:06.817" Score="5" ViewCount="45" Body="&lt;p&gt;There's a &lt;a href=&quot;http://www.irishtimes.com/news/crime-and-law/courts/criminal-court/graham-dwyer-trial-jury-resumes-deliberating-verdict-1.2152615&quot; rel=&quot;nofollow&quot;&gt;murder trial ongoing in Ireland at the moment&lt;/a&gt; and one of the pieces of evidence is a spade found at the scene along with the skeletal remains. It was contended by the prosecution that this spade was belonging to the defendant but while they were similar&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;forensic evidence found the paint splatters on the spade were not the&#xA;  same as on the [defendant's] garden fence&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is: can something left out in the open be changed either chemically or structurally by the environment such that it appears to be similar but not the same? If so why would this not have been called out as a caveat by the expert forensics witness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specifics of the chemical tests weren't gone into in the media (&lt;a href=&quot;http://www.rte.ie/news/2015/0226/683012-graham-dwyer-trial/&quot; rel=&quot;nofollow&quot;&gt;sample article here&lt;/a&gt;) so I don't know how accurate it was.&lt;/p&gt;&#xA;" OwnerUserId="15169" LastEditorUserId="10185" LastEditDate="2015-03-25T13:16:17.113" LastActivityDate="2015-03-25T14:45:27.397" Title="Can weathering change the composition of something so that it appears different?" Tags="&lt;everyday-chemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27811" PostTypeId="2" ParentId="27807" CreationDate="2015-03-25T12:25:37.873" Score="4" Body="&lt;p&gt;As andynitrox said in the comments :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The HUP just gives a lower bound in the uncertainty of Δx and Δpx, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The inequality exists to show that, the uncertainty is atleast this much and in real-life cases, its much larger than this . (&lt;a href=&quot;https://physics.stackexchange.com/a/170686&quot;&gt;source&lt;/a&gt;)(provided by Mithoron)&lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-03-27T18:23:31.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27814" PostTypeId="1" CreationDate="2015-03-25T13:17:37.787" Score="1" ViewCount="2956" Body="&lt;p&gt;Can someone tell me, which of the following terms should be used or avoided? (Google shows the presence of all the candidate terms.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;synthesis route&lt;/li&gt;&#xA;&lt;li&gt;synthetic route&lt;/li&gt;&#xA;&lt;li&gt;route of synthesis&lt;/li&gt;&#xA;&lt;li&gt;synthesis pathway&lt;/li&gt;&#xA;&lt;li&gt;synthetic pathway&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Imagine for example a sentence such as &quot;A new ... was developed.&quot;&lt;/p&gt;&#xA;" OwnerUserId="12908" LastEditorUserId="7448" LastEditDate="2015-03-25T13:37:49.460" LastActivityDate="2015-04-26T12:20:54.500" Title="Chemical writing: the term &quot;synthesis route&quot;" Tags="&lt;terminology&gt;" AnswerCount="0" CommentCount="9" ClosedDate="2015-04-26T18:47:07.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="27815" PostTypeId="2" ParentId="27746" CreationDate="2015-03-25T13:18:23.067" Score="6" Body="&lt;p&gt;To start off, all chalcogens are known to form binary hydrides of the form $\ce{H2X}$ which can then behave as binary acids, however weak or strong, by the equation $\ce{H2X &amp;lt;=&amp;gt; HX^- + H+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to compare the strength of the acids. The most straightforward way of doing this is comparing gas phase acidities, which removes the complication of solvation effects. For this, we look at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_affinity&quot; rel=&quot;nofollow noreferrer&quot;&gt;proton affinities&lt;/a&gt; of the conjugate bases $\ce{HX^-}$. The definition of a substance's proton affinity is the $\Delta G$ when it is protonated in the gas phase (always an exergonic reaction for neutral species and anions), so gas phase acidity is simply the inverse reaction, with the opposite sign for the $\Delta G$. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Proton_affinity_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0168117689830769&quot; rel=&quot;nofollow noreferrer&quot;&gt;two&lt;/a&gt; sources, the gas phase acidities of the hydrogen chalcogenides are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O_{(g)} &amp;lt;=&amp;gt; HO^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta  G = +1635\  kJ\  mol^{-1},\ K^{25ºC} = 2.5\times 10^{-287}$$&#xA;$$\ce{H2S_{(g)} &amp;lt;=&amp;gt; HS^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1477\  kJ\  mol^{-1},\ K^{25ºC} = 1.2\times 10^{-259} $$&#xA;$$\ce{H2Se_{(g)} &amp;lt;=&amp;gt; HSe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1417\  kJ\  mol^{-1},\ K^{25ºC} = 4.1\times 10^{-249} $$&#xA;$$\ce{H2Te_{(g)} &amp;lt;=&amp;gt; HTe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G \simeq +1386\  kJ\  mol^{-1},\  K^{25ºC} \simeq 1.1\times 10^{-243} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, when going down the group, deprotonation becomes less endergonic, so the binary hydrides definitely become stronger acids, even if against what one might expect from electronegativity considerations. Calculating the equilibrium constants allows a more direct comparison. Don't worry that the constants are all extremely small, what matters to us is the ratio of the equilibrium constants. Thus, &lt;strong&gt;in the gas phase&lt;/strong&gt;, $\ce{H2S}$ is a whopping 28 orders of magnitude more acidic than $\ce{H2O}$, while $\ce{H2Se}$ and $\ce{H2Te}$ are 38 and 44 orders of magnitude more acidic than $\ce{H2O}$, respectively. It is certainly noteworthy that $\ce{H2O}$ is a much, much weaker acid than its heavier homologues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the data presented, we ask, &lt;em&gt;why does acidity increase down the group&lt;/em&gt;? There are two main factors presented in different sources. One of them is that the chalcogen-hydrogen bond becomes weaker for heavier chalcogens, and this is ascribed to the poor overlap between the tiny $\ce{H}\ 1s$ orbital and the increasingly larger chalcogen orbitals which weakens the covalent bond and presumably makes ionization easier. This can be seen from &lt;a href=&quot;https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the dissociation energies of the $\ce{X-H}$ bonds&lt;/a&gt; ($\ce{O-H} = 428\  kJ\  mol^{-1}$ | $\ce{S-H} = 344 \  kJ\  mol^{-1}$ | $\ce{Se-H} = 305\  kJ\  mol^{-1}$ | $\ce{Te-H} = 268\  kJ\  mol^{-1}$). The second factor is the size of the chalcogen atoms. The heavier atoms are larger, meaning they spread the same charge in the conjugate base $\ce{HX^-}$ over a larger volume, which makes reprotonation less favourable. This latter explanation seems to be the most common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to now we've been talking about acidity in the gas phase, but now let's add the solvent, specifically water. The presence of a solvent stabilizes ions immensely relative to gaseous ions (especially protons), and water is a particularly good solvent for ions, so all the dissociation equilibria will be shifted &lt;strong&gt;massively&lt;/strong&gt; to the right compared to the gas phase. Let us compare the acid dissociation constants in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O_{(l)} &amp;lt;=&amp;gt; HO^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ k_w^{25ºC} = 1.0\times 10^{-14}$$&#xA;$$\ce{H2S_{(aq)} &amp;lt;=&amp;gt; HS^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.0\times 10^{-7} $$&#xA;$$\ce{H2Se_{(aq)} &amp;lt;=&amp;gt; HSe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.3\times 10^{-4} $$&#xA;$$\ce{H2Te_{(aq)} &amp;lt;=&amp;gt; HTe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \   K_a^{25ºC} = 2.3\times 10^{-3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;All the binary chalcogen hydrides still increase in acidity after solvation when going down the group. This is not necessarily always the case. A classic example of relative acidities changing on solvation is the dimethylammonium ($\ce{Me_2NH2^+}$) and trimethylammonium ($\ce{Me_3NH^+}$) cations; in the gas phase the former is more acidic, but in aqueous solution the latter is more acidic as the $\ce{Me_3NH^+}$ ion is not as well solvated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As I mention in the comments, aqueous hydrogen telluride ($pK_a=2.6$) is about as acidic as aqueous phosphoric acid ($pK_{a_1}=2.15$). Compare also with a common weak acid such as acetic acid ($pK_a=4.76$), a borderline strong acid such as nitric acid ($pK_a \simeq -1$), and a recognized strong acid such as hydrochloric acid ($pK_a \simeq -6$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The $K_a$ of water is a little tricky, since it is both the reactant and the solvent. Its value &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9847/what-is-the-pka-of-the-oxonium-hydronium-h%E2%82%83o%E2%81%BA-ion&quot;&gt;depends on the definition used&lt;/a&gt;. I quoted the $k_w$, but many people use a $pK_a$ of 15.74 which implies $K_a = 1.8\times 10^{-16}$. The difference is not too important here. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{H2O}$ is a much stronger acid than expected from gas phase data, relative to its heavier homologues. This is ascribed to the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/25091/pka-differences-and-reactivity/25183#25183&quot;&gt;exceptionally strong aqueous solvation possible only for $\ce{OH^{-}}$&lt;/a&gt; among the listed conjugate bases, stemming from strong hydrogen bonding interactions, which helps weigh the equilibrium to the right side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-03-27T15:01:13.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27816" PostTypeId="1" AcceptedAnswerId="27818" CreationDate="2015-03-25T13:20:28.140" Score="5" ViewCount="22861" Body="&lt;p&gt;2-Chloro-3,3-dimethylbutane or 3-Chloro-2,2-dimethylbutane?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to these sources, both are correct:    &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wtt-pro.nist.gov/wtt-pro/index.html?cmp=2-chloro-3.3-dimethylbutane&quot; rel=&quot;noreferrer&quot;&gt;NIST/TRC Web Thermo Tables (WTT)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.455173.html&quot; rel=&quot;noreferrer&quot;&gt;ChemSpider&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AHhLt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AHhLt.png&quot; alt=&quot;nomenclature discrepancy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="402" LastEditDate="2017-04-11T04:10:06.673" LastActivityDate="2017-04-11T04:10:06.673" Title="What is has a higher priority in naming organic compounds, alkyl groups or halides?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27817" PostTypeId="2" ParentId="27809" CreationDate="2015-03-25T14:45:27.397" Score="2" Body="&lt;p&gt;Yes, weather can and probably will change the composition of materials and make them different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, forensic analysis usually takes that into account. There are some properties that might change because of weathering (you shouldn't compare water content, for example), but others won't. According to the sample article, the difference was found in the chemical composition of the paints.  Since details were not disclosed, one can only guess what these differences are, but some possibilites are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Presence or absence of metals in the paints&lt;/li&gt;&#xA;&lt;li&gt;Substances that wouldn't appear or disappear because of wheatering&lt;/li&gt;&#xA;&lt;li&gt;Different isotopic ratios of elements&lt;/li&gt;&#xA;&lt;li&gt;Traces of the solvent indicating they were not the same&lt;/li&gt;&#xA;&lt;li&gt;Probably others&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming the analysis were performed correctly, it is usually not hard to say two samples are different. On the other side, when the results are similar, you can hardly say they are equal, but only that they match and that no differences were found.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-03-25T14:45:27.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27818" PostTypeId="2" ParentId="27816" CreationDate="2015-03-25T15:12:27.007" Score="8" Body="&lt;p&gt;The current IUPAC recommendations (2013) read as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Numbering&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acylic compounds, low locants are assigned to them in the following decreasing order of seniority.&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(g) lowest locants for the substituent cited first as a prefix in the name;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule f takes precedence over Rule g.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the example is named as &lt;strong&gt;3-chloro-2,2-dimethylbutane&lt;/strong&gt; rather than 2-chloro-3,3-dimethylbutane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5Kpa.png&quot; alt=&quot;3-chloro-2,2-dimethylbutane&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-03-25T15:17:40.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27819" PostTypeId="2" ParentId="27361" CreationDate="2015-03-25T15:17:30.577" Score="11" Body="&lt;p&gt;First off, it is a very small difference in $\mathrm{p}K_\mathrm{a}$ that you are asking about.  In fact, I would say that the two compounds have similar acidity.  It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question we are comparing the effect of two different substituents on an equilibrium.  To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion).  If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27986/stabilisation-of-phenoxide-ion&quot;&gt;your other question&lt;/a&gt; we see that the &lt;em&gt;p&lt;/em&gt;-methoxy group destabilizes the phenoxide anion to a greater extent than the &lt;em&gt;p&lt;/em&gt;-methyl group.  Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the &lt;em&gt;p&lt;/em&gt;-methoxy substituent pushes the equilibrium more strongly to the phenol side than the &lt;em&gt;p&lt;/em&gt;-methyl group will.  This would tend to make &lt;em&gt;p&lt;/em&gt;-methoxyphenol less acidic than &lt;em&gt;p&lt;/em&gt;-methylphenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the phenol side  of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect.  The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring.  Certainly such an effect would be greater with methoxy than methyl.  If true, then we could argue that the &lt;em&gt;p&lt;/em&gt;-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making &lt;em&gt;p&lt;/em&gt;-methoxyphenol more acidic than &lt;em&gt;p&lt;/em&gt;-methylphenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect &lt;em&gt;p&lt;/em&gt;-methoxyphenol and &lt;em&gt;p&lt;/em&gt;-methylphenol to have comparable acidities &lt;strong&gt;and&lt;/strong&gt; if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_\mathrm{a}$'s similar to phenol ($\mathrm{p}K_\mathrm{a}=10$), which they do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approach used here is sound.  I think the explanation of the substituent effects on the phenoxide are also sound.  The explanation of the substituent effects on the phenol side is (admittedly) weak.  Nonetheless, this explanation does correctly explain why &lt;em&gt;p&lt;/em&gt;-methoxyphenol and &lt;em&gt;p&lt;/em&gt;-methylphenol have comparable acidities &lt;strong&gt;and&lt;/strong&gt; why these acidities are close to the acidity of phenol.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-22T04:18:58.660" LastActivityDate="2018-03-22T04:18:58.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27820" PostTypeId="2" ParentId="27778" CreationDate="2015-03-25T15:27:58.943" Score="1" Body="&lt;p&gt;Write out the half-reactions and you will see that oxidation of S almost always creates lots of protons.  So you can think of Le Chatelier's principle: if there are already lots of protons around, it will be harder to create more of them.  Here's the example for complete oxidation of sulfide to sulfate at pH ~7:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HS- + 4 H2O -&amp;gt; 8e- + 9H+ + SO4^{-2} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, under alkaline conditions, the equilibrium favors the products even more, because of the &quot;removal&quot; of the protons by the alkaline conditions.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-25T15:27:58.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27821" PostTypeId="2" ParentId="27748" CreationDate="2015-03-25T15:34:13.300" Score="3" Body="&lt;p&gt;According to &quot;&lt;a href=&quot;http://monographs.iarc.fr/ENG/Monographs/vol54/mono54-8.pdf&quot; rel=&quot;nofollow&quot;&gt;Hydrochloric acid&lt;/a&gt;&quot; IARC Monographs volume 54:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubilty of HCl in water is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;82.3 grams HCl per 100 grams water at 0 degree C &lt;/p&gt;&#xA;&#xA;&lt;p&gt;67.3 grams HCl per 100 grams of water at 30 degrees C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the density of 39.1% HCl aqueous solution is 1.20. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example at 0 degrees C, 1 liter of solution is 1200 grams and contains (82.3/182.3)(1200 grams) = 542 grams which corresponds to 14.9 moles HCl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 30 degrees C, this works out to 13.2 moles HCl.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the maximum concentration is about 15M at 0 degrees C decreasing to 13M at 30 degrees C.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For data outside this temperature range and not limited to atmospheric pressure see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01938a001&quot; rel=&quot;nofollow&quot;&gt;A STUDY OF THE SYSTEM HYDROGEN CHILORIDE AND WATER&lt;/a&gt; J. Am. Chem. Soc., 1909, 31 (8), pp 851–866&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-25T16:13:28.710" LastActivityDate="2015-03-25T16:13:28.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27822" PostTypeId="2" ParentId="27799" CreationDate="2015-03-25T15:50:10.887" Score="7" Body="&lt;p&gt;&lt;strong&gt;Hyperconjugation does play a role&lt;/strong&gt; in your example.  Acetic acid ($\mathrm{p}K_\mathrm{a}=4.75$) is a weaker acid than formic acid ($\mathrm{p}K_\mathrm{a}=3.75$).  To understand this we must consider the relative stabilities of &lt;strong&gt;both&lt;/strong&gt; the ionized and unionized acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tl1Pl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In acetic acid there is a partial positive charge on the carbonyl carbon.  Both the inductive effect and resonance effect (&lt;em&gt;e.g&lt;/em&gt;. &lt;strong&gt;hyperconjugated resonance structures&lt;/strong&gt; like B) of the methyl group act to stabilize this charge, thereby stabilizing the unionized form of acetic acid.  In formic acid, without a methyl group, similar stabilization of its unionized form is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the ionized acids.  The negative charge in the anion is delocalized as shown in resonance structure C. This negative charge on the two oxygens makes them &lt;strong&gt;much less electron withdrawing&lt;/strong&gt;.  Consequently, there is much less positive charge on the &quot;carbonyl&quot; carbon in the ionized form.  In turn, there is less need for stabilization of positive charge by the methyl group.  The methyl group does not stabilize the ionized form as much as it stabilizes the unionized form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the methyl group stabilizes the unionized form of acetic acid more than a hydrogen &lt;strong&gt;and&lt;/strong&gt; since the methyl doesn't stabilize the ionized form much more than a hydrogen, there is a higher energy barrier for acetic acid to ionize compared to formic acid.  This means that acetic acid will be a weaker acid than formic acid.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-07T14:37:38.283" LastActivityDate="2018-03-07T14:37:38.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27823" PostTypeId="2" ParentId="27807" CreationDate="2015-03-25T15:58:13.520" Score="0" Body="&lt;p&gt;In addition to my comment further above I might add that the Heisenberg Uncertainty Principle has nothing to do with an error in a measurement. The HUP does not state that we cannot measure something more accurately than this lower bound, but instead that it is inherently not determined to a better degree.&lt;/p&gt;&#xA;&lt;p&gt;To illustrate, the two properties of a one-dimensional wave packet , frequency and location, are similarily interrelated.&lt;/p&gt;&#xA;&lt;p&gt;Imagine a sharp wave packet, a peak, something like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CMT3S.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You'll have no trouble locating 'the wave packet', simply because it is fairly sharp.&#xA;Conversely, for a monochromatic (and therefore technically infinitely long wave 'packet'), it is very easy to figure out what frequency or wavelength it has:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnEB3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, trying to find out &lt;em&gt;where&lt;/em&gt; this wave is, cannot really be answered. Keeping in mind that a wave packet is essentially made up of different monochromatic waves with different frequencies and amplitudes, you'll see that the sharper a packet is, the more frequencies it contains, so it will become nearly impossible to say what frequency it has.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uL85p.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can give a lower bound for the accuracy of both measurements: the certainty in the measurement of the frequency and of the location of such a wave packet. We cannot obviously measure both at the same time with arbitrary accuracy. Note that we may very well see exactly what such a wave packet looks like, but we can still not pin-point one single frequency &lt;em&gt;and&lt;/em&gt; one single location that desribes such a packet. That is basically the essence of the Heisenberg Uncertainty Principle.&lt;/p&gt;&#xA;&lt;p&gt;Using the wave number &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and as location &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, we can describe this relation in the case of a wave packet as &lt;span class=&quot;math-container&quot;&gt;$$\Delta k \cdot \Delta x \geq 2\pi$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The case of the wave-particle duality is analog to this purely mathematical example (the factor &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt; comes from the coefficient in the Fourier transform that is used to switch between the &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;-space and the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-space).&lt;/p&gt;&#xA;&lt;p&gt;Images: &lt;a href=&quot;http://www.newtonphysics.on.ca/heisenberg/image615.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.newtonphysics.on.ca/heisenberg/image615.gif&lt;/a&gt;, &lt;a href=&quot;https://skullsinthestars.files.wordpress.com/2007/12/sinesnapshot.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://skullsinthestars.files.wordpress.com/2007/12/sinesnapshot.gif&lt;/a&gt; and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/wpac5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/wpac5.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15002" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-26T06:51:46.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27824" PostTypeId="1" CreationDate="2015-03-25T16:03:31.700" Score="3" ViewCount="4071" Body="&lt;p&gt;We observe that nitrate ion has a trigonal planar geometry according to its lewis structure and the VSEPR theory. Its resonance hybrid consists of three contributing structure -three probable double bond organisations. According to my knowledge, we would expect that pi bond to be distributed equally over the molecule and thus, the dipole moments would cancel each other out. My question is, why such an ion would dissolve in water even though it is  non-polar?&lt;/p&gt;&#xA;" OwnerUserId="11549" LastEditorUserId="7448" LastEditDate="2015-03-25T18:11:55.420" LastActivityDate="2015-03-25T18:11:55.420" Title="Polarity of the nitrate ion" Tags="&lt;solubility&gt;&lt;resonance&gt;&lt;polarity&gt;&lt;vsepr-theory&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="27825" PostTypeId="1" AcceptedAnswerId="27828" CreationDate="2015-03-25T16:09:46.957" Score="2" ViewCount="815" Body="&lt;p&gt;During hybridization of d-orbitals why the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are used up before the $d_{xy}$, $d_{yz}$ and $d_{zx}$ ?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="14616" LastEditDate="2015-03-25T17:13:08.403" LastActivityDate="2015-03-25T17:46:36.033" Title="Order of priority in hybridization of d-orbitals" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27826" PostTypeId="1" CreationDate="2015-03-25T16:12:47.290" Score="16" ViewCount="2417" Body="&lt;p&gt;I know that soft bases cleave diborane symmetrically but hard ones don't. However, what makes ammonia harder than trimethylamine? And why does hardness/softness affect this? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4325" LastEditDate="2015-11-20T21:54:55.000" LastActivityDate="2018-08-31T16:13:25.777" Title="Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?" Tags="&lt;inorganic-chemistry&gt;&lt;hsab&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27827" PostTypeId="1" AcceptedAnswerId="27831" CreationDate="2015-03-25T17:45:05.730" Score="11" ViewCount="2324" Body="&lt;p&gt;Valence electrons are associated with &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital&quot;&gt;molecular orbitals&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_hybridisation&quot;&gt;hybridizations&lt;/a&gt;. Do core electrons have molecular/hybridized orbitals, or the original &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;atomic orbitals&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-03-27T18:18:48.307" Title="Do core electrons have molecular orbitals?" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27828" PostTypeId="2" ParentId="27825" CreationDate="2015-03-25T17:46:36.033" Score="2" Body="&lt;p&gt;The rotational frame of reference which defined which axis is the z-axis is changed such that $d_{z^2}$ and $d_{x^2-y^2}$ are hybridized first is a convention / practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory (meaning mathematically), you should be able to get equivalent geometry if you used the other d orbitals for hybridzation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the donut thing is weird enough we try to fold it in so we can deal with the more normal cloverleafs as leftovers.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-25T17:46:36.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27829" PostTypeId="2" ParentId="7985" CreationDate="2015-03-25T17:49:49.930" Score="-2" Body="&lt;p&gt;Partial pressure is the pressure exerted by the individual gas in the mixture of different gases at the same temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vapour pressure is the pressure exerted by the condensed(cooling) gas by lowering the temperature.&lt;/p&gt;&#xA;" OwnerUserId="15175" LastEditorUserId="13184" LastEditDate="2015-03-25T18:19:14.730" LastActivityDate="2015-03-25T18:19:14.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27830" PostTypeId="2" ParentId="27824" CreationDate="2015-03-25T17:51:04.807" Score="2" Body="&lt;p&gt;Don't overthink it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygens have a partial negative charge in nitrate from resonance.  The protons in water have a partial positive charge from dipoles.  Positives attract negatives.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule polarity becomes an issue when discussing neutral molecules interacting with neutral molecules.  Molecular polarity is one thing, and bond polarity is another.  Transformations / arrangements / alignments of molecules such that the dipoles (the things that make a molecule or bond 'polar') align to maximize interaction are certainly important for dissolution, but when a species is a bona fide free ion, water will tend to dissolve it.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1285" LastEditDate="2015-03-25T17:58:01.223" LastActivityDate="2015-03-25T17:58:01.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27831" PostTypeId="2" ParentId="27827" CreationDate="2015-03-25T17:57:33.497" Score="5" Body="&lt;p&gt;tl;dr: If you are talking about &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow&quot;&gt;VSEPR&lt;/a&gt; for an undergraduate level class, core orbitals do not mix.  If you are talking about &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;nofollow&quot;&gt;Hartree-Fock theory&lt;/a&gt; for a graduate level thesis, core orbitals most certainly do mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to this question depends greatly on the level of theory you are looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-level Quantum Mechanics calculations will typically involve mixing of core orbitals as well as valence orbitals, and give the most realistic picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, by making the simplifying assumption that core orbitals do not hybridize (a situation which is almost assuredly untrue), one might make many useful predictions about geometry and reactivity. For example, the entire discipline of Organic Chemistry is based (in part) upon the central assumption that the carbon 1s orbital does not participate meaningfully in hybridization, and yet the discipline still provides many useful tools for discussing structure and reactivity.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="10104" LastEditDate="2015-03-25T20:09:39.673" LastActivityDate="2015-03-25T20:09:39.673" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="27832" PostTypeId="2" ParentId="27827" CreationDate="2015-03-25T17:58:43.313" Score="6" Body="&lt;p&gt;When valence atomic orbitals start to interact with orbitals on another atom or molecule (&lt;em&gt;e.g.&lt;/em&gt; bond formation), the valence electrons respond by forming molecular orbitals with the electrons on the other atom or molecule.  The core electrons also respond to this bonding interaction.  Their energy levels and spatial distributions are &lt;strong&gt;perturbed&lt;/strong&gt;.  The more deeply buried the core electrons, the smaller the perturbation - but the core electrons will also be affected by the bonding interaction.  If one were able to write out the exact wave function to describe the new molecule that was formed, the wave function would include the core electrons.  Their importance would be much less than the valence electrons, but they would be involved in describing the total bonding situation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-25T17:58:43.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27833" PostTypeId="1" AcceptedAnswerId="27834" CreationDate="2015-03-25T18:03:23.097" Score="5" ViewCount="4762" Body="&lt;p&gt;P forms $\ce{P4}$ due to its inability to form double/triple bonds. So it makes a sort of tetrahedral structure.&#xA;S however makes $\ce{S8}$ cycles w/ $sp^2$ hybridization. Why go all that way to connect the chain? Can't $\ce{S3}$ do the trick? It will still have $sp^2$. What makes $\ce{S8}$ more stable?&lt;/p&gt;&#xA;" OwnerUserId="14982" LastEditorUserId="402" LastEditDate="2017-01-14T20:33:53.383" LastActivityDate="2017-01-14T20:34:24.333" Title="Why does sulfur form S8 instead of S3?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27834" PostTypeId="2" ParentId="27833" CreationDate="2015-03-25T18:06:59.770" Score="3" Body="&lt;p&gt;The ring you propose would have severe issues with ring strain, and would quickly 'pop-open' into a $\ce{S3}$ chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other chains have similar benefits and penalties.  The larger the ring, the greater the entropic cost for formation in a growing chain model.  this is traded off with ring strain: too tight causes one kind of strian and too loose causes another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S8}$ is the 'sweet spot' for handling these competing factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as phosphorous goes, my understanding is that $\ce{P4}$ is the most reactive form of molecular phosphorous, and other allotropes are thermodynamically preferred.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="5017" LastEditDate="2015-03-26T03:44:46.947" LastActivityDate="2015-03-26T03:44:46.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27835" PostTypeId="1" AcceptedAnswerId="28148" CreationDate="2015-03-25T18:14:51.837" Score="7" ViewCount="8777" Body="&lt;p&gt;Can a hydride shift be followed by a methyl shift for stabilizing a carbocation?&#xA;In case both are possible, which will occur first?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="5026" LastEditDate="2018-03-13T06:22:37.327" LastActivityDate="2018-03-13T14:02:40.680" Title="Can hydride shift and methyl shift happen one after another in a carbocation?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27837" PostTypeId="1" AcceptedAnswerId="27855" CreationDate="2015-03-25T19:21:43.157" Score="8" ViewCount="984" Body="&lt;p&gt;The Bessemer process for making steel involves blowing air or oxygen thru hot iron. Impurities and carbon oxidize away. Why does the iron itself not oxidize?&lt;/p&gt;&#xA;" OwnerUserId="15176" LastEditorUserId="25316" LastEditDate="2016-07-07T07:47:34.623" LastActivityDate="2016-07-07T07:47:34.623" Title="Why doesn't the Bessemer process oxidize the iron?" Tags="&lt;metallurgy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27838" PostTypeId="1" CreationDate="2015-03-25T19:26:13.810" Score="2" ViewCount="29" Body="&lt;p&gt;I am reading the book &lt;em&gt;Soft Matter Physics - Synthetic and Biological Self-Assembling Materials&lt;/em&gt; from Hamley and particularly page 201. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to think how this packing occurs in case of molecules or &lt;em&gt;molecular patches&lt;/em&gt;, particularly in case of nitrates and nitrites because they have highest priority set by WHO in their drinking water guideline. &#xA;The curvature can be defined as the amount by which a geometric object deviates from being flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume that these molecules clamp together such that there is no other interactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can you approximate the mean curvature of nitrates?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="292" LastActivityDate="2015-03-25T19:26:13.810" Title="What is the mean curvature of nitrates and nitrites?" Tags="&lt;molecules&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27839" PostTypeId="1" CreationDate="2015-03-25T20:20:04.207" Score="0" ViewCount="1038" Body="&lt;p&gt;In the case of both Silicon and Oxygen, the double bond is less stable than two single bonds. Consequently, surely both should polymerize, why don't they? What accounts for the difference between Carbon and Silicon in this context?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2015-03-25T22:47:16.743" LastActivityDate="2015-03-26T21:37:23.330" Title="Why does Si2H4 readily polymerize at RTP but ethene does not (requires catalyst)?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27840" PostTypeId="1" AcceptedAnswerId="27847" CreationDate="2015-03-25T20:22:20.050" Score="4" ViewCount="505" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7NFaB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the bridged ion intermediate aromatic or not?&#xA;What is the way to quickly identify aromaticity?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="10185" LastEditDate="2015-10-09T19:16:10.170" LastActivityDate="2018-07-29T08:34:49.753" Title="Aromaticity-Bromonium Ion" Tags="&lt;ions&gt;&lt;aromatic-compounds&gt;&lt;transition-state-theory&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27841" PostTypeId="1" AcceptedAnswerId="27848" CreationDate="2015-03-25T20:25:45.497" Score="6" ViewCount="836" Body="&lt;p&gt;The energy band of valence electrons gets smaller and smaller when the lattice (bulk) atoms are more and more separated from each other, until discrete energy levels are formed, due to the isolation of the atoms from each other (see image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens with energy levels of core electrons in the bulk, as the lattice constant is changed? Do they form a continuous energy band in the first place? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Hhq1.gif&quot; alt=&quot;Diamond band structure vs. lattice constant&quot;&gt;&#xA;(&lt;a href=&quot;http://ecee.colorado.edu/~bart/ecen3320/newbook/chapter2/ch2_3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If core electrons form a continuous energy band, how come K,L,M,... shells are differentiable in xray spectroscopy? In other words, if all electrons form one continuous band, how can xray have element specificity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NdAvM.png&quot; alt=&quot;Electron shells&quot;&gt;&#xA;(&lt;a href=&quot;https://principlequantumnumber.wordpress.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-03-27T19:09:22.873" LastActivityDate="2015-03-27T19:31:06.907" Title="Do core electrons in bulk material have discrete energy levels or continuous energy band?" Tags="&lt;quantum-chemistry&gt;&lt;energy&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27842" PostTypeId="1" AcceptedAnswerId="111165" CreationDate="2015-03-25T20:32:33.680" Score="10" ViewCount="1503" Body="&lt;p&gt;What substance has the lowest &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{sp}$&lt;/span&gt; and what is its value? The lowest I could find is &lt;span class=&quot;math-container&quot;&gt;$2.6\cdot 10^{-124}$&lt;/span&gt; for cobalt(III) sulfide &lt;span class=&quot;math-container&quot;&gt;$\ce{Co2S3}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2020-07-25T08:26:08.487" LastActivityDate="2020-07-25T08:30:01.430" Title="What substance has the lowest solubility product?" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27844" PostTypeId="1" CreationDate="2015-03-25T21:12:52.410" Score="2" ViewCount="64" Body="&lt;p&gt;Suppose I eat a whole $450~ \rm{g}$ chocolate Easter rabbit. The enthalpy of combustion of chocolate is $−22.4~\rm{ kJ/g}$. One kilogram of body fat contains $39000 ~\rm{kJ}$ of energy. Walking for an hour uses up $690 ~\rm{kJ}$ of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how many hours do I have to walk to use up the energy consumed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have calculated the energy found in $450 ~\rm{g}$ of the Easter bunny by multiplying $450~ \rm{g}$ by $-22.4 ~\mathrm{kJ/g}$. This gave me $-10080~ \mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m thinking my next step would be dividing $-10080 ~\mathrm{kJ}$ by $690 ~\mathrm{kJ}$ to find the answer. But I feel this is incorrect because I don’t understand the relevance of $1~\mathrm{kg}$ of body fat containing $39000 ~ \mathrm{kJ}$.&lt;/p&gt;&#xA;" OwnerUserId="15086" LastEditorUserId="4325" LastEditDate="2015-03-27T21:28:28.107" LastActivityDate="2015-04-27T00:42:51.230" Title="Exercising to use up energy consumed" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27845" PostTypeId="1" CreationDate="2015-03-25T23:18:09.907" Score="2" ViewCount="1501" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why is &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph-CH2-COOH}$&lt;/span&gt; more acidic than &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt; although the equilibrium of ionization lies mainly backward for both of them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&lt;p&gt;In the ionized form of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph-CH2-COOH}$&lt;/span&gt; we have a &lt;span class=&quot;math-container&quot;&gt;$\ce{Ph-CH2}$&lt;/span&gt;-group donating its electron to carbonyl carbon while in case of acetic acid we have a methyl group donating its electron to the carbonyl carbon.&lt;/p&gt;&#xA;&lt;p&gt;But which one of them is stabilized more?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="69830" LastEditDate="2021-05-25T06:15:00.600" LastActivityDate="2021-05-25T06:15:00.600" Title="How to arrange carboxylic acids (phenylacetic acid and acetic acid) by increasing acidity?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27847" PostTypeId="2" ParentId="27840" CreationDate="2015-03-25T23:49:03.497" Score="3" Body="&lt;p&gt;A chemist would say that the reaction proceeds through an &quot;&lt;strong&gt;aromatic transition state&lt;/strong&gt;&quot;. The bridged, chloronium ion transition state drawn in your question involves 2 electrons.  No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in the $\ce{C-Cl}$ bond and no electrons in the carbocation p-orbital).  If a transition state contains 4n+2 electrons, then it is said to be aromatic. This case has n=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example. The top line in the figure below shows a suprafacial 1,5 hydrogen shift.  6 electrons are involved here, so the transition state is also referred to as aromatic (4n+2, n=1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jcO9o.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/special0.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt; (look far down on the page)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like the bridged halonium ion, this reaction is also very common because it too has an energetically favored (aromatic) transition state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second line in the figure depicts a suprafacial 1,3 hydrogen shift.  4 electrons are involved here, so the transition state does not fit the 4n+2 rule.  The transition state is said to be anti-aromatic.  Because it is disfavored energetically (anti-aromatic), the suprafacial 1,3 hydrogen shift does not occur (at least not via a low-energy concerted pathway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra Credit:&lt;/strong&gt; I don't want to jump to far ahead, but let me briefly say, that what I've written above applies to Hückel transition states, opposite rules apply to Möbius transition states.  See this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27575/difference-between-continuously-overlapping-p-orbitals-and-conjugated-pi-bonds/27577#27577&quot;&gt;earlier answer&lt;/a&gt;, and references therein, for an explanation of the Hückel and Möbius concepts. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-03-25T23:49:03.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27848" PostTypeId="2" ParentId="27841" CreationDate="2015-03-26T00:00:44.337" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens with energy levels of core electrons in the bulk, as the&#xA;  lattice constant is changed? Do they form a continuous energy band in&#xA;  the first place? Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using an organic example, as you move from ethylene to butadiene to hexatriene, you have 1, 2 and 3 bonding and antibonding pi molecular orbitals respectively. If you continue this chain extension and wind up with a very large polyene, the energy difference between all of the bonding (or antibonding) orbitals becomes smaller and smaller.  In addition to this compression between energy levels, each molecule is in a slightly different environment.  Some are near the surface, some deep in the bulk, some have a structural deviation or packing non-uniformity close by.  This serves to broaden the energy width of each orbital.  Between the energy broadening of an orbital and the compression of the energy separation between orbitals, one can imagine a limiting case where all of the bonding orbitals are so close to one another that one continuous band, the valence band, exists. The same would be true of the antibonding orbitals and they would merge into a continuous conduction band. The only remaining separation is the separation between the valence band and the conduction band.  This separation is referred to as the band gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, at least at normal pressures, this limiting case of a single band comprised of all bonding orbitals seems unlikely. Instead, some orbitals where the energy separation between them is large likely form separate bands.  In this case, the HOMO would still broaden significantly (as would the other molecular orbitals) and transform into the valence band, remaining distinct from the other bands at lower energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, it is difficult to probe the bulk and to understand &lt;em&gt;exactly&lt;/em&gt; just what the situation is there. X-ray spectroscopy probes to a depth of only a micron or so, so it is really examining the surface states.  Certainly at least some of the molecules at the surface retain their individual orbital identity and are not part of the band structure.  Therefore it is not surprising that XPS (since it probes only the surface) can detect individual orbitals and serve to identify materials.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-27T19:31:06.907" LastActivityDate="2015-03-27T19:31:06.907" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27849" PostTypeId="2" ParentId="27721" CreationDate="2015-03-26T01:15:36.570" Score="1" Body="&lt;p&gt;You are mixing things up. The self ionization of water is not related with bubbles at all. The ionization is an equilibrium process which goes on forever under constant conditions meaning thereby the concentration of ions does not increase under constant conditions. Also ions are produced in aqueous phase not gaseous phase&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O~(l) -&amp;gt; H3O^{+}~(aq) + OH-~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drinking water is not pure, it is a solution of many minerals and gases. Changing temperature and pressure of the surroundings can affect the solubility of the gases. Gases are less soluble at high temperature. So as water absorbs heat gases come out in form of bubbles.&#xA;About the particular depth, solubility varies inversely with pressure. As pressure increases with depth so bubbles appear only upto a particular depth.&lt;/p&gt;&#xA;" OwnerUserId="15174" LastEditorUserId="4945" LastEditDate="2015-03-26T07:15:18.743" LastActivityDate="2015-03-26T07:15:18.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27853" PostTypeId="2" ParentId="2498" CreationDate="2015-03-26T02:17:05.817" Score="2" Body="&lt;p&gt;I actually make metalpoint pencils and use pure lead for the writing material. It's obviously a bit toxic (because lead), though I'm experimenting with making custom lead alloys (possibly with a percentage of bismuth, as I have some laying around) to reduce the overall lead content. In terms of utility, however, pure lead leaves a reasonably dark and easy-to-make mark.&lt;/p&gt;&#xA;" OwnerUserId="15184" LastActivityDate="2015-03-26T02:17:05.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27854" PostTypeId="2" ParentId="27839" CreationDate="2015-03-26T02:23:28.567" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does $\ce{Si2H4}$ readily polymerize at RTP but ethene does not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{Si2H4}$, the silicon analogue of ethylene, is referred to as &lt;strong&gt;disilene&lt;/strong&gt;, a member of the silene family.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VuOvw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The silicon atoms in disilene are $\ce{sp^2}$ hybridized and the orbitals  overlap to form both a sigma and pi bond, analogous to the hybridization and bonding in ethylene. However, the $\ce{Si=Si}$ bond length is significantly over &lt;a href=&quot;http://en.wikipedia.org/wiki/Silenes&quot; rel=&quot;nofollow noreferrer&quot;&gt;2 angstroms&lt;/a&gt;, whereas the double bond length in an ethylene is much shorter at 1.34 angstroms. This much larger double bond length in silenes results in extremely poor p-orbital overlap which yields an extremely weak silicon-silicon pi bond.  While it takes energies on the order of 65-70 Kcal/mol to cause rotational &lt;em&gt;cis-trans&lt;/em&gt; isomerization about a carbon-carbon double bond, it only takes 10-20 Kcal/mol to effect similar isomerization about the silene double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the pi bond in a silene is so weak it doesn't require much energy to disrupt it making the molecule &lt;strong&gt;very&lt;/strong&gt; reactive, &lt;strong&gt;much&lt;/strong&gt; more so than an olefin.  The molecule can polymerize or react with whatever else may be around such as oxygen, water, &lt;em&gt;etc&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing the 4 hydrogen atoms in disilene with bulky groups stabilizes the molecule and allows the molecule to be isolated and characterized.  The bulky groups make it more difficult for another molecule to get close enough to react with the disilene pi bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HcOOi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, even these sterically protected molecules react with oxygen and water at temperatures not far above room temperature.  Here is a &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3409869/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to a full paper&lt;/a&gt; that discusses the preparation and reactivity of these sterically protected silenes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-26T21:37:23.330" LastActivityDate="2015-03-26T21:37:23.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27855" PostTypeId="2" ParentId="27837" CreationDate="2015-03-26T03:23:05.730" Score="7" Body="&lt;p&gt;I thought you asked a great question, one that I have wondered about myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer, as best I could determine, is that the iron does oxidize.  A good description of the Basic Oxygen Steelmaking process, today's descendent of the Bessamer process, &lt;a href=&quot;https://www.steel.org/Making%20Steel/How%20Its%20Made/Processes/Processes%20Info/The%20Basic%20Oxygen%20Steelmaking%20Process.aspx&quot; rel=&quot;noreferrer&quot;&gt;says so&lt;/a&gt; anyway.  The BOS process uses pure oxygen instead of the original Bessamer processes's air.  It reports that some iron &lt;em&gt;does&lt;/em&gt; oxidize.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation of carbon to form carbon monoxide is more thermodynamically favorable than the oxidation of iron, so it occurs preferentially.  The page I found (once again the &lt;a href=&quot;https://www.steel.org/Making%20Steel/How%20Its%20Made/Processes/Processes%20Info/The%20Basic%20Oxygen%20Steelmaking%20Process.aspx&quot; rel=&quot;noreferrer&quot;&gt;link is here&lt;/a&gt;) mentions using on the order of 1800 scf of oxygen per ton of metal.  A ton of steel is about 18 kmol of iron.  1800 scf of oxygen is about 2 kmol of oxygen.  So there is not enough oxygen to oxidize even close to all the iron in the average BOS run.  That fact, combined with the fact that carbon, silicon, and other impurities in the pig iron oxidize preferentially, mean that not much iron is oxidized during BOS or Bessamer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &quot;not much&quot; does not mean zero.  My guess is that iron oxides are more soluble in the slag than they are in molten iron and so are also removed with the slag after BOS.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-03-26T03:23:05.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27856" PostTypeId="2" ParentId="27738" CreationDate="2015-03-26T03:29:24.910" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very pure iron can be obtained by pyrolysis of iron pentacarbonyl.  Iron metal obtained in this manner is in fact called &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonyl_iron&quot;&gt;carbonyl iron&lt;/a&gt; and is commercially available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(CO)5 -&amp;gt; Fe(s) + 5 CO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing this yourself is very dangerous, however, because iron pentacarbonyl is volatile and toxic.  The decomposition reaction to produce iron requires high temperatures, and the byproduct is carbon monoxide gas, which is also notoriously toxic.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alternatively is it possible to electrically purify iron using an anode and cathode?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also possible.  I don't know about other salts of iron, but to form iron from iron &lt;em&gt;oxides&lt;/em&gt;, very high temperatures are required.  The other answer mentioned Dr. Sadoway's work.  Let me &lt;a href=&quot;http://www.nature.com/nature/journal/v497/n7449/abs/nature12134.html&quot;&gt;highlight&lt;/a&gt; a recent article of his unveiling new anode materials for electrolytic decomposition of iron oxide.  At temperatures as &quot;low&quot; as 1538&amp;deg;C, molten iron oxide can by electrolyzed to pure oxygen gas and to molten iron metal using electrochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is not an appropriate solvent for electrochemical formation of iron metal.  At potentials low enough to favor $\ce{Fe^0}$ formation from $\ce{Fe^{+2}}$, water is unstable and would be reduced to hydrogen gas.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-26T03:37:04.373" LastActivityDate="2015-03-26T03:37:04.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27861" PostTypeId="1" AcceptedAnswerId="27864" CreationDate="2015-03-26T06:27:25.620" Score="0" ViewCount="236" Body="&lt;p&gt;The problem gives that the solubility of Silver dichromate is $8.3\times10^{-3} g/100mL$. I need to find $K_{sp}$ which is supposed to be $2.8\times10^{-11}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{sp}$ is $\ce{[Ag+]^2[Cr2O7^{2-}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I changed the $8.3\times 10^{-3}$ from g/mL to mol/L and got $1.9\times 10^{-8}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver dichromate breaks into 2 Ag and one dichromate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that would be $(2 \times 1.9\times10^{-8})^2 \times ( 1.9\times10^{-8})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what I did but it's not giving me the right answer. Where am I going wrong here?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="402" LastEditDate="2016-11-12T01:31:10.483" LastActivityDate="2016-11-12T01:31:10.483" Title="How do I solve for Ksp" Tags="&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27862" PostTypeId="1" CreationDate="2015-03-26T06:27:51.660" Score="-2" ViewCount="3555" Body="&lt;p&gt;The sulphate ion has a higher number of resonance structures than the carbonate ion. Hence, it should be more stable than the carbonate ion.&#xA;But my book says that the carbonate ion is stronger than the sulphate ion. Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="7475" LastEditDate="2016-05-02T14:49:24.047" LastActivityDate="2016-06-03T18:04:24.870" Title="What is more stable: a carbonate ion or a sulphate ion?" Tags="&lt;ions&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2016-06-04T04:06:52.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="27863" PostTypeId="1" CreationDate="2015-03-26T06:36:57.867" Score="2" ViewCount="1064" Body="&lt;p&gt;A strong electrolyte is an electrolyte that completely, or almost completely, ionizes or dissociates in a solution. &#xA;My book has a question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a saturated solution of sparingly soluble strong electrolyte $\ce{AgIO3}$ the equilibrium with sets in is $\ce{AgIO3→Ag+ + IO3-}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the above statement wrong or is there anything as sparingly soluble strong electrolyte?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="4325" LastEditDate="2015-03-27T23:31:40.440" LastActivityDate="2015-03-27T23:31:40.440" Title="Can a strong electrolyte be sparingly soluble?" Tags="&lt;solubility&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27864" PostTypeId="2" ParentId="27861" CreationDate="2015-03-26T06:53:06.300" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The problem gives that the solubility of Silver dichromate is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{8.3\cdot10^{-3} g/100\,mL}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;[…]&lt;/p&gt;&#xA;&lt;p&gt;I changed the &lt;span class=&quot;math-container&quot;&gt;$8.3\cdot10^{-3}$&lt;/span&gt; from g/mL to mol/L and got &lt;span class=&quot;math-container&quot;&gt;$1.9\cdot10^{-8}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;[…] Were am I going wrong here?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Was the solubility given in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{g/100\,mL}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{8.3\cdot10^{-3} g/100\,mL = 8.3\cdot10^{-2}\,g/L = \frac{8.3}{431.72}\,mol/L = 1.92\cdot10^{-4}\,mol/L}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-26T06:58:23.937" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27865" PostTypeId="1" CreationDate="2015-03-26T07:52:53.007" Score="0" ViewCount="612" Body="&lt;p&gt;Why is chloroethene said to be a conjugated system?Is it because of p orbitals present in chlorine or something?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="10185" LastEditDate="2015-03-26T11:39:31.227" LastActivityDate="2015-03-27T11:39:16.533" Title="Why is chloroethene said to be a conjugated system?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27866" PostTypeId="2" ParentId="27865" CreationDate="2015-03-26T08:00:47.723" Score="0" Body="&lt;p&gt;Chloroethene is said to be a conjugated system because the lone pair on the chlorine atom in in conjugation with the double bond between the 2 carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.expertsmind.com/CMSImages/2061_properties%20of%20vinyl%20chloride1.png&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; image for the resonating structures.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-03-26T08:00:47.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27868" PostTypeId="2" ParentId="27845" CreationDate="2015-03-26T08:13:45.897" Score="6" Body="&lt;p&gt;The $\mathrm pK_\mathrm a$ difference between phenylacetic acid and acetic acid is around 0.5 and the trends become more obvious when other arylalkanoic acids are included in the comparison. Numbers in red are the $\mathrm pK_\mathrm a$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zk5eJ.png&quot; alt=&quot;arylalkanoic acids&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Apparently, phenylacetic acid is a stronger acid than acetic acid. Adding a further phenyl subsituent even increases the acid strength.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The effect levels off with the distance of the phenyl substituent and the carboxylate. Note that 4-phenylbutyric acid is just as strong as acetic acid. It's as if there isn't any phenyl group at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would you think that there is a $-I$ effect of the phenyl substituent on the $\ce{-COOH}$ group?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2016-12-23T20:05:37.710" LastActivityDate="2016-12-23T20:05:37.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27870" PostTypeId="1" CreationDate="2015-03-26T08:45:05.857" Score="1" ViewCount="1225" Body="&lt;p&gt;For an experiment, I'm adding several substances (as powders) to hydrochloric acid, and measuring the temperature changes to calculate enthalpy. However when adding the powders, the mixture requires some stirring to get all of the substance reacting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I am measuring temperature via a temperature probe hooked up to a data logger (to obtain a temp. against time graph), I need a way to briefly stir the mixture in a controlled way that won't affect the change in temperature (reactions are both endo and exothermic). I have tried stirring with the probe itself, but this is not ideal as the probe is pretty sensitive and friction/bumping with the side of the vessel may affect the recording. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially looked at a magnetic stirring to obtain uniform and controlled stirring, but then I realized the magnet would absorb heat and potentially increase the temperature of the mixture through kinetic energy. So I'm after a stirring method that will not affect temperature, so that the temperature change I record is solely due to the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions are welcome, because I can't think of any tool or method that would suit. &lt;/p&gt;&#xA;" OwnerUserId="14994" LastEditorUserId="39539" LastEditDate="2017-03-27T20:59:06.370" LastActivityDate="2017-03-27T20:59:06.370" Title="Stirring the reaction between hydrochloric acid and a powdered substance" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27874" PostTypeId="1" AcceptedAnswerId="27890" CreationDate="2015-03-26T09:44:02.520" Score="4" ViewCount="65" Body="&lt;p&gt;I am sitting over a TG diagram of the decomposition steps of hexaurea-chromic-chloride-trihydrate ($\ce{[Cr(CO(NH2)2)6]Cl3 \cdot 3 H2O}$) and cannot really make sense of the products it might decompose into. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kb2G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that water leaves first (-3.5%), and the next step might be (with a wide error margin) fumes of $\ce{CO2}$, $\ce{H2O}$ and $\ce{2 NH3}$. What is left is then $\ce{[Cr(CO(NH2)2)5]Cl3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not found any resources on what might happen further - the four urea ligands could decompose into $$\ce{4 CO} + \ce{5 NH3} + \ce{3/2 N2} + \ce{1/2 H}$$ and I am left with $\ce{CrCl3}$ and urea, but those steps show up last (17.9% + 17.3% + 5.4% = 40.6%), leaving the 15.5% in the middle unexplained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other decomposition could take place after water and two urea ligands are gone? Is there a way to figure that out other than by guessing or experiment?&lt;/p&gt;&#xA;" OwnerUserId="15002" LastEditorUserId="194" LastEditDate="2017-08-13T13:36:55.293" LastActivityDate="2017-08-13T13:36:55.293" Title="Decomposition products of hexaurea-chromic-chloride-trihydrate" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27875" PostTypeId="1" CreationDate="2015-03-26T10:05:34.710" Score="0" ViewCount="606" Body="&lt;p&gt;The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?&lt;/p&gt;&#xA;" OwnerUserId="15189" LastActivityDate="2015-03-26T11:49:22.363" Title="The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-26T13:21:56.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="27877" PostTypeId="2" ParentId="27833" CreationDate="2015-03-26T10:39:55.263" Score="5" Body="&lt;p&gt;Trisulfur &lt;strong&gt;does&lt;/strong&gt; exist, but is not circular, it is ozone-like and unstable at common temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that while angle strain does exist in ring form, it is not that harsh. Unlike carbon, having natural bond angle ~109 degree, for bivalent sulfur it is closer to 90 degree (sulfur exhibits little to no hybridization).  However, the S-S bond strength is only 40% of C-C bond and in trisulfur ring  all S-S bonds would have dihedral angle equal to zero, which is unfavorable because of nonbonding pairs interaction. This issue is a lot more severe than in tetraphosphorus, and leads to ring opening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information on sulfur allotropes see &lt;a href=&quot;http://en.wikipedia.org/wiki/Allotropes_of_sulfur&quot;&gt;the Wikipedia article&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="4945" LastEditDate="2015-03-26T11:04:16.933" LastActivityDate="2015-03-26T11:04:16.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27879" PostTypeId="1" AcceptedAnswerId="27925" CreationDate="2015-03-26T11:21:18.293" Score="4" ViewCount="8395" Body="&lt;p&gt;I get really confused by this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order of melting points of group 2 chlorides is: $$\ce{BaCl2 &amp;gt; SrCl2 &amp;gt; CaCl2 &amp;gt; MgCl2 &amp;gt; BeCl2}$$ This is completely understandable by Fajan's Rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the order for group 1 halides is: $$\ce{NaCl &amp;gt; KCl &amp;gt; RbCl &amp;gt; CsCl &amp;gt; FrCl &amp;gt; LiCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I don't get! Considering Fajan's Rules, $\ce{NaCl}$ is the most covalent, least cationic size, so shouldn&quot;t it have the lowest melting point?&lt;/p&gt;&#xA;" OwnerUserId="15192" LastEditorUserId="4945" LastEditDate="2015-03-26T11:28:03.500" LastActivityDate="2019-11-11T10:38:21.767" Title="How do we predict the order of melting points of group 1 and group 2 halides?" Tags="&lt;inorganic-chemistry&gt;&lt;melting-point&gt;&lt;halides&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27880" PostTypeId="1" AcceptedAnswerId="27904" CreationDate="2015-03-26T11:22:01.043" Score="15" ViewCount="25233" Body="&lt;p&gt;What is the difference between them?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think radial nodes and spherical nodes are the same, and angular and planar nodes are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blamp.sites.truman.edu/files/2012/03/Atomic-Orbitals-PowerPoint-Presentation.pdf&quot; rel=&quot;noreferrer&quot;&gt;Reference&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, how many spherical nodes are there in a $2p$ orbital?&lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="10185" LastEditDate="2015-06-20T15:27:24.237" LastActivityDate="2017-01-08T08:03:43.583" Title="Difference between radial, planar, angular and spherical nodes" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27882" PostTypeId="2" ParentId="27875" CreationDate="2015-03-26T11:49:22.363" Score="2" Body="&lt;p&gt;What you meant is known as erosion. Erosion is the chemical loss of enamel due to acid.When acid continuously attacks teeth, they cannot repair themselves and will gradually begin to turn fuzzy and dissolve. Most soft drinks contain acids such as citric, phosphoric and malic or tartaric acids. A study was done where extracted human teeth were placed in softdrinks. Within two days, the teeth became very soft and the enamel surface lost much of its calcium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brush your teeth at least twice a day with a fluoride-containing toothpaste.A mouth rinse containing fluoride can help prevent tooth decay, according to the American Dental Association. For children, sealants can be used.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-03-26T11:49:22.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27883" PostTypeId="1" CreationDate="2015-03-26T12:05:32.783" Score="2" ViewCount="3766" Body="&lt;p&gt;The basic strength is determined by the ability of an ion or molecule to accept a proton. How do I know whether RSH is more stable than ROH?&#xA;(R is an alkyl group)&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="9961" LastEditDate="2021-03-21T17:06:43.957" LastActivityDate="2021-03-21T17:06:43.957" Title="Is RS- more basic than RO-?" Tags="&lt;acid-base&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27884" PostTypeId="1" CreationDate="2015-03-26T12:25:33.123" Score="0" ViewCount="671" Body="&lt;p&gt;Experiment A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3cm metal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.5M acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 degrees temperature change&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiment B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6cm metal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1M acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much would the temperature change be, and how much faster or slower would the reaction take place in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the answer would be either &quot;the same&quot; or &quot;4x as much&quot;. I don't exactly understand the reaction yet; would the doubling of each material make the reaction 4x stronger (2x2), or would they cancel each other out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it might've been something else I haven't thought of, too.&lt;/p&gt;&#xA;" OwnerUserId="15194" LastEditorUserId="15194" LastEditDate="2015-03-26T13:55:20.863" LastActivityDate="2015-03-27T07:16:30.890" Title="What happens in an acid and metal reaction where the acid is twice as concentrated and the metal twice as much in quantity?" Tags="&lt;acid-base&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27885" PostTypeId="1" AcceptedAnswerId="27887" CreationDate="2015-03-26T12:35:21.407" Score="2" ViewCount="1309" Body="&lt;p&gt;Is the order of defined by IUPAC or we follow some special conventions to know the order of priority?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-03-26T13:00:06.207" Title="What decides the order of priority of functional groups in organic compounds?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27886" PostTypeId="1" AcceptedAnswerId="27888" CreationDate="2015-03-26T12:43:12.077" Score="5" ViewCount="246" Body="&lt;p&gt;This was in J. Huheey's book and no explanation was given&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-03-26T13:57:02.617" Title="Why macromolecule ligands such as Crown ethers form more stable complexes than open chain ligands?" Tags="&lt;computational-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27887" PostTypeId="2" ParentId="27885" CreationDate="2015-03-26T13:00:06.207" Score="3" Body="&lt;p&gt;It’s an arbitrary agreement by IUPAC, although note&#xA;that there is some correlation between the&#xA;oxidation state of the carbon and the priority&#xA;(more oxidized groups tend to be higher&#xA;priority). Generally, the functional group which shows its characteristics more dominantly even when present with other group gets a higher priority.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-03-26T13:00:06.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27888" PostTypeId="2" ParentId="27886" CreationDate="2015-03-26T13:57:02.617" Score="6" Body="&lt;p&gt;The reason for crown ethers being more stable than their equally long open chain derivatives is the difference in Gibbs Free energy. (If you are not used to these terms, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; will help I think, though it is quite, long and thorough.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point to take from the use of the Gibbs free energy in this case is this: it is made up of two parts, enthalpy ($\Delta{}H$) and entropy ($\Delta{}S$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta{}G = \Delta{}H - T \cdot{} \Delta{}S&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only the binding affinity of the (crown) ether chain is considered, only the enthalpy is relevant. The strength of the individual ether moieties is as strong in a crown as it is in a chain. The difference then, lies with the entropy. To have a crown ether form a structure around a ligand 'costs some entropy'. The same entropy difference should be considered for the chain ether, with &lt;strong&gt;added to that&lt;/strong&gt; the cost of pulling the ends of the chain together! That difference in entropy results in the difference in $\Delta{}G$ and therefore yields a more stable complex. (The ends of the crown ether are also pulled together, this cost of entropy however, has been paid in an earlier reaction when making the crown ether, and therefore has no influence here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this also from the relation between the Gibbs free energy and the equilibrium constant of the complex (In the imaginary reaction $\ce{A + B -&amp;gt; AB})$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta{}G = - RT \cdot{} Ln (K)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K = \frac{[\text{AB}]}{[\text{A}] \cdot{} [\text{B}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A higher $\Delta{}G$ leads to a higher equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helped, and I'd be happy to clarify anything further.&lt;/p&gt;&#xA;" OwnerUserId="2359" LastActivityDate="2015-03-26T13:57:02.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27889" PostTypeId="2" ParentId="27833" CreationDate="2015-03-26T14:24:35.310" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q5q2z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Trisulfur&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trisulfur&lt;/a&gt; or thiozone is significantly present in the gas phase of sulfur at high temperature and has been studied by rotational spectroscopy as reported in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja049645f&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Rotational Spectrum and Geometrical Structure of Thiozone&lt;/a&gt; J. Am. Chem. Soc., 2004, 126 (13), pp 4096–4097&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of $\ce{S3}$ is very close to isoelectronic $\ce{O3}$ (116.8 degrees), $\ce{SO2}$ (119.5 degrees), and $\ce{S2O}$ (118 degrees).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above article states that the central sulfur is $sp^2$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the ground state is bent, the cyclic form is only slightly higher in energy, about 5-10 kcal/mol higher.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is evidence that the cyclic form can exist as a ligand of a transition metal complex, as reported in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja044809d&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transition Metal Complexes of Cyclic and Open Ozone and Thiozone&lt;/a&gt; J. Am. Chem. Soc., 2005, 127 (4), pp 1278–1285.  According to this reference, early calculation incorrectly showed the cyclic form to be more stable for both ozone and thiozone &quot;due to the fact that electron correlation plays an important role in these molecules, especially in the open form&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="402" LastEditDate="2017-01-14T20:34:24.333" LastActivityDate="2017-01-14T20:34:24.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27890" PostTypeId="2" ParentId="27874" CreationDate="2015-03-26T14:33:50.440" Score="3" Body="&lt;p&gt;Using the &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs10973-013-3260-7&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; Klaus Warzecha linked to and some trial-and-error, a possible decomposition could look like this and is in accordance with the TG diagram in the question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Decomposition of $\ce{3 H2O}$ and $\ce{2 CO(NH2)2}$:&#xA;$\ce{[Cr(CO(NH2)2)6]Cl3 $\cdot$ 3 H2O -&amp;gt; [Cr(CO(NH2)2)4]Cl3 + 3 H2O \uparrow + 2 CO(NH2)2 \uparrow}$&lt;/li&gt;&#xA;&lt;li&gt;Decomposition of $\ce{2 CO(NH2)2}$ and $\ce{2 HCl}$:&#xA;$\ce{[Cr(CO(NH2)2)4]Cl3 -&amp;gt; CrH6C2N4ClO2 + 2 CO(NH2)2 \uparrow + 2 HCl \uparrow}$&lt;/li&gt;&#xA;&lt;li&gt;Decomposition of $\ce{2 CO2}$, $\ce{N2}$, $\ce{3 H2}$ and $\ce{1/2 Cl2}$ and oxidation of $\ce{Cr}$ to form $\ce{CrN2}$:&#xA;$\ce{CrH6C2N4ClO2 -&amp;gt; CrN2 + 2 CO2 \uparrow + N2 \uparrow + 3 H2 \uparrow + 1/2 Cl2 \uparrow}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given the temperature range and pressure (probably standard pressure), the formation of $\ce{CrO2}$ is unlikely.&lt;/p&gt;&#xA;" OwnerUserId="15002" LastActivityDate="2015-03-26T14:33:50.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27891" PostTypeId="2" ParentId="27884" CreationDate="2015-03-26T14:58:33.267" Score="1" Body="&lt;p&gt;Without providing more detailed specifications of the metal such as: state of the metal - powder or solid, shape of the solid form etc., the rate of the reaction will be the same, but the second reaction will be a bit faster because of the bigger temperature change (as it affects reaction rate coefficient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that you meant 3 and 6 $cm^3$ and not $cm$.&#xA;The same metal reacts in both reactions. So we can assume that the density of both metals is also the same ($\rho_1=\rho_2=\rho$). We can further assume that metals were dissolved completely and that the temperature change is actually temperature increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho=\frac{m_1}{V_1}$, $\rho=\frac{m_2}{V_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m_1}{V_1}=\frac{m_2}{V_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m_1}{3}=\frac{m_2}{6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{m_1}{1}=\frac{m_2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger piece of metal weights 2 times more than the smaller metal. We can convert weights to moles (molar mass of both metals is the same):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=\frac{m}{M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n_1}{1}=\frac{n_2}{2}$&#xA;so we can say $2*n_1=n_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 1st reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_1$ reacts with 0,5M of acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 2nd reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2*$n_1$ react with 1M solution and it is the same as $n_1$ reacts with 0,5M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the solution is the same and the specific heat capacity will be nearly the same. The reaction enthalpy corresponds to the amount of reacted material. Because two times more material reacts in the second reaction,  the temperature change will be nearly two times bigger (nearly - because of the change in specific heat capacity).&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-03-27T07:16:30.890" LastActivityDate="2015-03-27T07:16:30.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27892" PostTypeId="1" CreationDate="2015-03-26T15:06:52.490" Score="1" ViewCount="125" Body="&lt;p&gt;We have been asked to conduct an analytic project for chemistry. It is important that the project conducted hold social implications and the conclusion achieved help inform people better. Several ideas have come to my mind for such projects, such as determining amount of phosphoric acid in cold drinks, acetic acid in fruits such as guava, amount of insecticides/pesticides in vegetables/fruits in the market. However, I am still yet to find an idea which is striking. I would like to do something unique. Any suggestions? &lt;/p&gt;&#xA;" OwnerUserId="7023" LastEditorUserId="418" LastEditDate="2015-03-26T15:49:19.693" LastActivityDate="2015-03-26T16:04:26.160" Title="Analysis project to conduct" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-03-27T01:18:36.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="27893" PostTypeId="2" ParentId="27892" CreationDate="2015-03-26T15:25:14.543" Score="0" Body="&lt;p&gt;As you might have guessed: It depends!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the educational level of the study (High School, College, etc.)?&lt;/li&gt;&#xA;&lt;li&gt;What's the time frame - how many man-hours?&lt;/li&gt;&#xA;&lt;li&gt;What's the equipment of the lab?&lt;/li&gt;&#xA;&lt;li&gt;If equipment or material has to be bought: What is the budget?&lt;/li&gt;&#xA;&lt;li&gt;Is it a plain academic study or is it intended to go public with the results in order to change/improve consumer awareness?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&quot;Something unique&quot; might mean that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there are no reference data&lt;/li&gt;&#xA;&lt;li&gt;it is too &quot;academic&quot; to have social implications (or at least that it will be difficult to explain to the public that it actually has)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What about something as &quot;boring&quot; as sugar content in local soft drink brands as compared to similar products on the international market? I'm sure you'll find lots of references, such as &lt;a href=&quot;http://psasir.upm.edu.my/2370/1/Determination_of_Sugars_in_Soft_Drinks_by.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go for something that is relevant for the average Joe Bloggs. Go for something that can be bought thoughout the while time of the study, just in can you have to repeat measurements or need to cover seasonal variations. If it is a fresh, seasonal product, you might have to plan ahead and stockpile and deep-freeze material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, it is up to you to decide - planning and outlining the study is part of the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking of relevance for the average consumer: Everybody needs excrete - but you do not want to examine that. Fortunately, everybody needs to eat too and seemingly, a lot of people tend to eat too much of the wrong stuff. The recent report of the &lt;a href=&quot;http://www.health.gov/dietaryguidelines/2015-scientific-report/02-executive-summary.asp&quot; rel=&quot;nofollow&quot;&gt;US Dietary Guidelines Advisory Committee&lt;/a&gt; might give some directions what to look for.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-03-26T16:04:26.160" LastActivityDate="2015-03-26T16:04:26.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27894" PostTypeId="2" ParentId="27770" CreationDate="2015-03-26T15:44:16.660" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all the other elements Pr through Yb, there is no 5d electron, and instead an additional 4f electron. In all cases, it is the 6s electron that is being lost in the second ionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more the nuclear charge is screened, the easier it is for the 6s electron to be lost.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5d electron is less effective than a 4f electron at screening the 6s electron from the nuclear charge, because 4f electron density is concentrated much closer to the nucleus (see Fig. 2 of below reference).  Therefore it takes more energy to loss the 6s electron for gadolinium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on nuclear charge screening see &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/47/4/10.1063/1.1712084&quot; rel=&quot;nofollow&quot;&gt;Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons&lt;/a&gt; J. Chem. Phys. 47, 1300 (1967).&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2015-03-27T01:26:11.093" LastActivityDate="2015-03-27T01:26:11.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27895" PostTypeId="1" AcceptedAnswerId="27898" CreationDate="2015-03-26T15:51:48.937" Score="6" ViewCount="643" Body="&lt;p&gt;Woody biomass is composed of cellulose, hemicellulose, and lignin components. The following diagram provides representative formulas for each component:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sl16W.png&quot; alt=&quot;biomass components&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biomass can be represented by it's density, ultimate analysis, and composition. For example, beech wood can be represented by its density, $\rho = 700~\mathrm{kg/m^3}$, its ultimate analysis by weight percent such as $\ce{C}=49.05\%$, $\ce{H}=5.83\%$, $\ce{O}=45\%$, $\ce{N}=0.12\%$, and its composition by weight percent such as $\text{cellulose}=48\%$, $\text{hemicellulose}=28\%$, and $\text{lignin}=24\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following diagram represents a kinetic scheme for cellulose, &lt;strong&gt;CELL&lt;/strong&gt;, which can be represented by the formula $\ce{C6H10O5}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jDr6r.png&quot; alt=&quot;cellulose kinetic scheme&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction rates for each pathway are first-order Arrhenius equations where $A$ is the pre-factor in $\mathrm{s^{-1}}$, $E$ is the activation energy in $\mathrm{kcal/kmol}$, $T$ is the temperature, and $\mathcal{R}$ is the universal gas constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K = A\cdot \exp{\left(\frac{-E}{\mathcal{R}T}\right)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To apply the cellulose kinetic reaction to beech wood, I used the density of the wood, $700~\mathrm{kg/m^3}$ to calculate the initial concentration of cellulose as 48% of the wood density, $336~\mathrm{kg/m^3}$.  Using the initial mass concentration of cellulose, the reaction rates at a certain temperature can be applied to that concentration. This approach will give the mass concentration for each group (G1, G4, LVG). Notice that each group represents various chemical species on a mole basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the individual chemical species are represented on a mole basis, it seems like there are two approaches to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass Basis Approach&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;start with an initial mass concentration of the biomass&lt;/li&gt;&#xA;&lt;li&gt;divide that mass conc. into cellulose, hemicellulose, and lignin components based on the weight % composition&lt;/li&gt;&#xA;&lt;li&gt;apply the reaction rates on a mass basis for each component to determine the mass concentration of each group&lt;/li&gt;&#xA;&lt;li&gt;finally use the mass concentration of the group to calculate the individual chemical species (as mass or moles?) in that group&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mole Basis Approach&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;represent the biomass as moles of C, H, O, N using the ultimate analysis&lt;/li&gt;&#xA;&lt;li&gt;divide the biomass into cellulose, hemicellulose, and lignin components&lt;/li&gt;&#xA;&lt;li&gt;apply the reaction rates on a mole basis to determine the chemical species&lt;/li&gt;&#xA;&lt;li&gt;atomic balances for C, H, O, N should be checked at each step&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How do I attempt to solve this problem on a mole basis?&#xA;Or is it best to treat it initially on a mass basis then convert to moles to solve for the individual chemical species?&lt;/p&gt;&#xA;" OwnerUserId="2151" LastEditorUserId="194" LastEditDate="2017-04-06T02:28:03.743" LastActivityDate="2017-08-29T12:00:10.443" Title="Representing woody biomass concentration for kinetic reactions" Tags="&lt;kinetics&gt;&lt;concentration&gt;&lt;elemental-analysis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27897" PostTypeId="2" ParentId="27862" CreationDate="2015-03-26T16:21:00.287" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dEFPh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQrwJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lone pairs on oxygen atom repel each other .The lone pair-lone pair repulsion are minimized to a greater extent in carbonate ion than in sulphate ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence carbonate ion is more stable than sulphate ion.&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="15161" LastEditDate="2016-05-02T14:01:21.990" LastActivityDate="2016-05-02T14:01:21.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27898" PostTypeId="2" ParentId="27895" CreationDate="2015-03-26T16:49:27.493" Score="4" Body="&lt;p&gt;First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$&lt;/li&gt;&#xA;&lt;li&gt;hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$&lt;/li&gt;&#xA;&lt;li&gt;lignin: $\ce{C31H34O11}$ with $M_l = 582~\mathrm{g/mol}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using a density of $700~\mathrm{g/dm^3}$ you will find the relative masses of the three compounds to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$0.48 \cdot 700~\mathrm{g/dm^3} = 336~\mathrm{g/dm^3}$ (cellulose)&lt;/li&gt;&#xA;&lt;li&gt;$0.28 \cdot 700~\mathrm{g/dm^3} = 196~\mathrm{g/dm^3}$ (hemicellulose)&lt;/li&gt;&#xA;&lt;li&gt;$0.24 \cdot 700~\mathrm{g/dm^3} = 168~\mathrm{g/dm^3}$ (lignin)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With $n = m/M$ we now find the number of moles of each compound in the sample ($1~\mathrm{dm^3}$):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$2.074~\mathrm{mol/dm^3}$ cellulose&lt;/li&gt;&#xA;&lt;li&gt;$1.307~\mathrm{mol/dm^3}$ hemicellulose&lt;/li&gt;&#xA;&lt;li&gt;$0.289~\mathrm{mol/dm^3}$ lignin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that this is a very rough estimate. You'll probably get a better result by taking the average of all species of hemicellulose, etc. in your sample and cross-checking it to make sure it fits the ratios of the elements (weight percent).&lt;/p&gt;&#xA;" OwnerUserId="15002" LastEditorUserId="4945" LastEditDate="2015-03-27T02:19:42.127" LastActivityDate="2015-03-27T02:19:42.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27899" PostTypeId="1" AcceptedAnswerId="28097" CreationDate="2015-03-26T17:15:03.463" Score="10" ViewCount="5429" Body="&lt;p&gt;When something goes wrong with a lithium polymer battery (as are commonly used to power electric models and robots), and it &quot;catches fire&quot;, what is actually happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What gas or gasses are being released?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/gisdMQbtJqk&quot;&gt;Here's a video of such an event&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15195" LastEditorUserId="9961" LastEditDate="2015-04-01T00:38:00.550" LastActivityDate="2015-04-01T02:26:30.317" Title="What reaction is happening when a lithium polymer battery &quot;is on fire&quot;" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27900" PostTypeId="1" AcceptedAnswerId="29564" CreationDate="2015-03-26T17:17:21.133" Score="8" ViewCount="3995" Body="&lt;p&gt;I know that it corrodes copper, but does it do the same to all brass or bronze compositions?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastActivityDate="2020-05-06T08:51:28.720" Title="Does Gallium (liquid or solid) corrode all forms of brass?" Tags="&lt;inorganic-chemistry&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27901" PostTypeId="1" AcceptedAnswerId="27902" CreationDate="2015-03-26T17:28:08.630" Score="5" ViewCount="1706" Body="&lt;p&gt;As we all know that there are 5 tastes : sweet, sour, bitter, umami, salty. &#xA;So what chemicals are there that help form these tastes, in other words how do our tastebuds experience these tastes?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also can there be more tastes than these basic five. I am not saying that there should be a blend of these to make a taste but rather are there any other undiscovered tastes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe we haven't tasted them yet?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-03-26T18:27:47.857" Title="What chemicals help define &quot;taste&quot;?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27902" PostTypeId="2" ParentId="27901" CreationDate="2015-03-26T18:27:47.857" Score="10" Body="&lt;p&gt;It's not exactly true that there are 5 &quot;tastes&quot;. There are 5 generally accepted classes of gustatory transducers, each comprising many different types. The perception of taste is a product of not only the output of these transducers, but also information from the olfactory system, visual cues, textures, sound, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gustatory system itself is based (like the olfactory system) on overlapping sensors with partial selectivity. So a given compound, sucrose for instance, may stimulate several sweetness sensors (and probably even some from other classes) to varying degrees, and the particular combination of the responses is what gives sucrose its unique taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The types of molecules that stimulate the different sensors is not known clearly, but we can make some broad generalizations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sour&lt;/strong&gt; - These sensors are thought to be primarily ion channels sensitive to pH—acidic solutions produce a sour taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Salty&lt;/strong&gt; - Also thought to be mostly ion channels sensitive to small metal cations. Different ions have somewhat difference response. ($\ce{Na+}$ doesn't have the same response as $\ce{K+}$) There is some debate about the actual mechanism though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest are g-protein coupled receptors. They change shape when they bind a molecule which causes a chemical signalling cascade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sweet&lt;/strong&gt; - Not known precisely, but many compounds containing hydroxyl or carbonyl groups trigger these receptors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Umami&lt;/strong&gt; - Mostly triggered by amino acids, particularly glutamate. Inosinate and guanylate are known to enhance the effect of glutamate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bitter&lt;/strong&gt; - This one is rather strange. A very wide variety of compounds stimulate these receptors, even with very different structural features. This taste is thought to have evolved to protect us against toxic compounds as many poisonous alkaloids and such taste bitter. &lt;a href=&quot;http://en.wikipedia.org/wiki/Denatonium&quot;&gt;Denatonium&lt;/a&gt; is an example of a bitter compound (one of the most bitter known).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for whether any other classes of transducers exist, it's entirely possible. Some researchers have identified candidates for receptors for things like fats. Also, there some receptors like TRPV1 (which responds to capsaicin) are known that respond to compounds found in food, but aren't usually considered taste receptors.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-03-26T18:27:47.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27903" PostTypeId="1" CreationDate="2015-03-26T18:49:37.443" Score="4" ViewCount="288" Body="&lt;p&gt;In my previous occupation the usage of bleach--$NaClO$--I found was quite staggering, and based off of the health issues that I and other coworkers were having, I wanted to see how close we were cutting it to being over exposure limits. Taking the definition, $$ 1 \, \, ppm = \frac{1 \,\,mg \,\,solute}{1 \,\,L \,\,solvent} = \frac{1 \,\,g \,\,solute}{1 \,\,m^3 \,\,solvent} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a $ 946 \,\,mL $ bottle of solution labeled to be $.65\%$ Sodium Hypochlorite. This leads me to calculate the presence of $6.149 \,\,mL$ of the entire mixture being $NaClO$. Subtracting that from the total volume to get the volume of the solvent, I obtain $.940 \,\,L \,\,solvent$. Using the fact that the density of $NaClO$ is $ \rho =1.11 \frac{g}{mL}$, I obtained $6.83\,\,g$ or $6830\,\,mg\,\,solute$. Plugging these values in for the concentration in $ppm$ I obtain $\frac{6830\,\,mg}{.940\,\,L}= 7265.957\,\,ppm$ ($\frac{g}{m^{3}}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The American Industrial Hygiene Association only recommends $2\,\,ppm$ for short term exposure, which leads me to believe that my calculations might be in error in some way. Is this calculation correct or do I have some fundamental misunderstanding about the concept of molarity? I just don't feel like something that's $3633$ times more concentrated than the AIHA recommends wouldn't even make it off the drawing board, let alone into workplaces.&lt;/p&gt;&#xA;" OwnerUserId="15197" LastEditorUserId="10185" LastEditDate="2015-03-27T19:43:05.370" LastActivityDate="2015-03-27T20:02:54.560" Title="NaClO Concentration" Tags="&lt;safety&gt;&lt;concentration&gt;&lt;toxicity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27904" PostTypeId="2" ParentId="27880" CreationDate="2015-03-26T19:21:00.637" Score="12" Body="&lt;p&gt;The wave function $\Psi(r,\theta,\phi)$ of a one-electron (hydrogen-like) atom is seperable as the product of a radial function $R(r)$ and an angular function $Y(\theta,\phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Psi(r,\theta,\phi) =  R(r)Y(\theta,\phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $R(r_1) = 0$, there exists a radial node.  The radial node is a sphere with radius $r_1$. Therefore the terms &quot;radial node&quot; and &quot;spherical node&quot; are the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y(\theta,\phi)$ is further seperable as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Y(\theta,\phi) = P(\theta)F(\phi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either $P(\theta)$ or $F(\phi)$ is zero for a given respective angle value, there is an angular node.  However, a angular node is not necessarily a planar node.  An angular node could be a planar node or a conical node.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(\phi)$ being zero corresponds to a planar node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P(\theta)$ being zero corresponds to either a conical node or a planar node (some think of the planar case as a specical case of conical, with apex angle being 180 degrees)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, there will be $n-1$ nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l$ nodes will be angular&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n-l-1$ nodes will be radial (spherical) &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-26T19:21:00.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27905" PostTypeId="1" AcceptedAnswerId="27908" CreationDate="2015-03-26T19:21:47.403" Score="2" ViewCount="1461" Body="&lt;p&gt;How concentrated can an acid be without it being supersaturated? Is there a certain limit to how concentrated an acid or base can be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I mean concentration, I mean molarity; so how concentrated can an acid or base be? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; Can be a &lt;em&gt;strong&lt;/em&gt; or &lt;em&gt;weak&lt;/em&gt; acid/base. Because we are just looking for the amount of &lt;strong&gt;solvent&lt;/strong&gt; needed to dissolve to start the dissociation process.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; No autoprotolysis please. Acids that &lt;strong&gt;need&lt;/strong&gt; a solvent to dissociate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-03-28T22:42:32.840" LastActivityDate="2015-03-28T22:42:32.840" Title="How concentrated can an acid be?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;water&gt;&lt;solubility&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27906" PostTypeId="2" ParentId="27905" CreationDate="2015-03-26T19:35:42.080" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am talking about dissolving in water since an acid needs some sort of solvent to dissociate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it doesn't. You neglect autoprotolysis where the water-free acid itself is the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{2H2SO4 &amp;amp;&amp;lt;=&amp;gt; H3SO4+ + HSO4-}\\&#xA;\ce{3HF &amp;amp;&amp;lt;=&amp;gt; H2F+ + HF2-}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-26T19:35:42.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27907" PostTypeId="2" ParentId="27903" CreationDate="2015-03-26T19:51:20.210" Score="5" Body="&lt;p&gt;The recommended &lt;em&gt;Workplace Environmental Exposure Level&lt;/em&gt; (WEEL) refers to the concentration of airborne substances in the workplace air. This value is not applicable to the concentration in cleaning agents or other products.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-03-26T19:51:20.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27908" PostTypeId="2" ParentId="27905" CreationDate="2015-03-27T00:32:25.387" Score="8" Body="&lt;p&gt;Are you suggesting that if an acid molecule is surrounded by a sufficiently small number of water molecules, then it might display lower than 100% ionization even if it is a &quot;strong acid&quot; ($pK_a&amp;lt;-1.76$)? That is true; some acids we commonly consider to be strong, such as nitric, hydrochloric or sulphuric acid, will not completely ionize in sufficiently concentrated aqueous solutions, even if there is more than one molecule of water per molecule of acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there an acid capable of protonating water quantitatively even when there is only one molecule of water per molecule of acid? Yes, in fact there are several. Interesting proof of this is that sufficiently strong acids will form &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydronium#Solid_hydronium_salts&quot;&gt;solid crystalline salts containing the hydronium cation&lt;/a&gt; when mixed with water in a 1:1 molar ratio. The most striking examples are probably the hydronium carborane salts, such as $\ce{(H3O)^+(HCB11Cl11)^{-}}$. I recommend taking a look at &lt;a href=&quot;http://www.reedgrouplab.ucr.edu/publications/Chem%20Comm%202005%201669.pdf&quot;&gt;this fantastic article&lt;/a&gt; from the Reed group for details on the wonderful properties of carborane superacids, with a section describing its several salts when reacted with water in different proportions and environments.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-03-27T00:32:25.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27909" PostTypeId="1" CreationDate="2015-03-27T01:42:18.040" Score="2" ViewCount="95" Body="&lt;p&gt;Sulfurous acid, $\ce{H2SO3}$, is a diprotic acid.&#xA;\begin{align}&#xA;K_{a1} &amp;amp;= 1.6 \cdot 10^{-2}\\&#xA;K_{a2} &amp;amp;= 6.4 \cdot 10^{-8}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to find the concentration of $\ce{SO3^2-}$ from this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think you would need to know the concentration of at least one thing. It is like trying to solve for 3 unknowns with 2 equations!&lt;/p&gt;&#xA;" OwnerUserId="15199" LastEditorUserId="4945" LastEditDate="2015-03-27T02:27:17.770" LastActivityDate="2015-03-31T06:16:59.933" Title="Can you find [SO₃²⁻] with just the two acid dissociation constants Kₐ₁ and Kₐ₂?" Tags="&lt;acid-base&gt;&lt;concentration&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27910" PostTypeId="2" ParentId="27909" CreationDate="2015-03-27T02:08:29.797" Score="3" Body="&lt;p&gt;No, it is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, as total concentration of acid approaches zero, concentration of sulfate would also approach zero.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-27T02:08:29.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27912" PostTypeId="1" CreationDate="2015-03-27T02:35:03.373" Score="1" ViewCount="94" Body="&lt;p&gt;I am doing a research and experiment on using Graphene oxide for deicing. My purpose is to coat graphene oxide on the surface of the ship in the north and south pole where large amount of large size ice cubes and icebergs are present. So when voltage is applied to it, it can generate a large amount of heat to melt the ice but the amount of heat must not be significant enough to vaporize the water to steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, the problem is:&lt;/strong&gt; As you know graphene oxide has a very high resistance. So there must be some process or material that can enhance the conductivity of the graphene oxide. What is it? In other words, is there a process that can enhance the conductivity of graphene?&lt;/p&gt;&#xA;" OwnerUserId="15027" LastEditorUserId="4356" LastEditDate="2015-04-03T03:46:36.240" LastActivityDate="2015-04-03T04:34:39.500" Title="What is the process or material that can enhance the conductivity of the graphene oxide?" Tags="&lt;organic-chemistry&gt;&lt;materials&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27913" PostTypeId="1" AcceptedAnswerId="27957" CreationDate="2015-03-27T02:40:08.070" Score="3" ViewCount="1594" Body="&lt;p&gt;If you have a weak base XOH titrated with a strong monotonic acid, where do you start with that? Would you first have the base react with water and break up and determine how much of it does? Or just it breaking up alone, not reacting with water? Reacting just with the acid right away? I'm not sure where to start for this problem type...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the pH at the equivalence point. I have the concentrations of the acid and base, which are both equal, no volume given. &lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="7689" LastEditDate="2015-03-27T18:14:48.467" LastActivityDate="2015-03-27T20:21:56.520" Title="Weak base titrated with a strong acid" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="27914" PostTypeId="2" ParentId="27883" CreationDate="2015-03-27T05:16:43.613" Score="5" Body="&lt;p&gt;&lt;strong&gt;RobChem&lt;/strong&gt; has already pointed out in his comment that your assumption is not quite correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at $\mathrm{p}K_\mathrm{a}$ values for $\ce{ROH}$ and $\ce{RSH}$ in water from the &lt;em&gt;CRC Handbook of Chemistry and Physics&lt;/em&gt; and/or other online sources, such as &lt;a href=&quot;http://www.diederich.chem.ethz.ch/teach/pKa_Table.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf&quot; rel=&quot;nofollow&quot;&gt;that&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lrr}&#xA;\mathbf{R} &amp;amp; {\ce{\mathbf{OH}}} &amp;amp;\ce{\mathbf{SH}}\\&#xA;\hline&#xA;\ce{H} &amp;amp; 15.7 &amp;amp;7.0\\&#xA;\ce{Et} &amp;amp; 15.9 &amp;amp; 10.6\\&#xA;\ce{(H3C)3C} &amp;amp; 18.0 &amp;amp; 11.7\\&#xA;\ce{C6H5} &amp;amp; 9.9 &amp;amp; 6.6\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all the cases above, the $\mathrm{p}K_\mathrm{a}$ value for $\ce{RSH}$ is smaller than that of $\ce{ROH}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-03-28T18:43:06.393" LastActivityDate="2015-03-28T18:43:06.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27915" PostTypeId="1" AcceptedAnswerId="27918" CreationDate="2015-03-27T05:51:45.090" Score="0" ViewCount="939" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xz80T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will the products on addition reaction of 1,4-benzoquinone with HBr ?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorDisplayName="user7951" LastEditDate="2015-03-27T09:33:00.890" LastActivityDate="2015-03-30T05:05:16.407" Title="Hydro-Bromination Product 1,4-Benzoquinone" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-03-27T11:01:45.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="27918" PostTypeId="2" ParentId="27915" CreationDate="2015-03-27T07:37:40.923" Score="0" Body="&lt;p&gt;This is the mechanistic approach :&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YdMYI.jpg&quot; alt=&quot;the full step by step mechanism:&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15204" LastEditorUserId="15204" LastEditDate="2015-03-30T05:05:16.407" LastActivityDate="2015-03-30T05:05:16.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27920" PostTypeId="1" AcceptedAnswerId="27935" CreationDate="2015-03-27T10:51:44.127" Score="20" ViewCount="6652" Body="&lt;p&gt;&lt;a href=&quot;http://www.gizmag.com/scientists-create-forbidden-compounds-table-salt/30520/&quot; rel=&quot;noreferrer&quot;&gt;There&lt;/a&gt;. And &lt;a href=&quot;http://www.sci-news.com/othersciences/chemistry/science-sodium-chlorides-foundation-chemistry-01633.html&quot; rel=&quot;noreferrer&quot;&gt;there&lt;/a&gt;. Almost a year ago, a group of scientists claimed to have reached compounds of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; with weird stochiometries (&lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl3, Na3Cl, NaCl7, Na3Cl2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2Cl}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;h2&gt;What one of the articles says:&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These compounds are thermodynamically stable and, once made, remain indefinitely; nothing will make them fall apart.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl7, NaCl3, Na3Cl2}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2Cl}$&lt;/span&gt; are metals (that explains the apparent violation of electroneutrality since charge balance rules are inapplicable to metals), and only one semiconducting phase of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl3}$&lt;/span&gt; is stable in the pressure range between 250 and 480 thousand atmospheres.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At first, I thought this whole thing was bogus. But then, I faced &lt;a href=&quot;http://www.chemistryviews.org/details/ezine/6731321/Against_Chemical_Intuition_Na3Cl_and_NaCl3_Do_Exist.html&quot; rel=&quot;noreferrer&quot;&gt;another, apparently convincing, article&lt;/a&gt; and even &lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/12/new-sodium-chlorides-assault-chemical-rules&quot; rel=&quot;noreferrer&quot;&gt;one from RSC&lt;/a&gt;. Then the questions popped up:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why does this happen? (Why is such a compound more thermodynamically stable at high pressure?)&lt;/li&gt;&#xA;&lt;li&gt;How can the bonding in these be explained?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IUWOSm.jpg&quot; alt=&quot;&quot; /&gt;&#xA;(&lt;a href=&quot;http://www.chemistryviews.org/SpringboardWebApp/userfiles/chem/image/2014/2014_09/hull.jpg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-10-22T13:03:27.330" Title="Why is NaCl3 possible?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27924" PostTypeId="2" ParentId="27909" CreationDate="2015-03-27T11:08:19.600" Score="1" Body="&lt;p&gt;You can use approximations like $K_\mathrm{a1}$ is much greater than $K_\mathrm{a2}$ and that first dissociation goes to almost completion. Accurate values cannot be calculated mathematically though.&lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="4945" LastEditDate="2015-03-31T06:16:59.933" LastActivityDate="2015-03-31T06:16:59.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27925" PostTypeId="2" ParentId="27879" CreationDate="2015-03-27T11:22:17.383" Score="-1" Body="&lt;p&gt;Melting point depends on Lattice energy. NaCl has similar sized ions which make the lattice stronger. LiCl comes last since it has a lot of covalent character.&lt;/p&gt;&#xA;" OwnerUserId="15208" LastActivityDate="2015-03-27T11:22:17.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27926" PostTypeId="2" ParentId="27863" CreationDate="2015-03-27T11:26:08.837" Score="2" Body="&lt;p&gt;It means that by just adding it to water wont make it dissociate into ions(sparingly soluble) but if current is passed through its solution, it will easily dissociate and give ions. These cases look rare.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Curt.F's reasoning looks better :  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the low concentration regime where the salt is &quot;soluble&quot;, it completely dissociates, but that the solubility limit is just low.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="15208" LastEditDate="2015-03-27T17:31:49.580" LastActivityDate="2015-03-27T17:31:49.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27927" PostTypeId="2" ParentId="27844" CreationDate="2015-03-27T11:34:27.220" Score="2" Body="&lt;p&gt;As binary geek said, that information is not required in the calculations. Its probably given to make people realize that by gulping chocolate and walking for just 5-10 mins wont make you reduce your weight. Standard Satire has been employed!&lt;/p&gt;&#xA;" OwnerUserId="15208" LastActivityDate="2015-03-27T11:34:27.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27928" PostTypeId="2" ParentId="27865" CreationDate="2015-03-27T11:39:16.533" Score="3" Body="&lt;p&gt;Conjugation can be between pi-bond and another pi-bond, lone pair of electrons and a pi-bond(as in your case) and pi-bond and a vacant orbital(p or d).&lt;/p&gt;&#xA;" OwnerUserId="15208" LastActivityDate="2015-03-27T11:39:16.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27930" PostTypeId="2" ParentId="27835" CreationDate="2015-03-27T11:48:17.453" Score="11" Body="&lt;p&gt;Multiple rearrangements can happen if the resulting species is stabilized by it. In general hydride shift is more favourable over methyl shift because of its smaller size.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example proposed by James was not known to me but its truly an amazing example.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;James:&lt;br&gt;&#xA;  It’s possible for multiple hydride/alkyl shifts to occur. One amazing example is in the biosynthesis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Lanosterol&quot; rel=&quot;noreferrer&quot;&gt;lanosterol&lt;/a&gt;.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.&#xA;  For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.masterorganicchemistry.com/2012/08/15/rearrangement-reactions-1-hydride-shifts/&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Lanosterol#Biosynthesis&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vt8FB.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="15208" LastEditDate="2015-03-28T04:50:59.670" LastActivityDate="2015-03-28T04:50:59.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27932" PostTypeId="2" ParentId="27740" CreationDate="2015-03-27T12:03:17.903" Score="-3" Body="&lt;p&gt;That would be like calling it Chlorine dioxide   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NHd8t.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Chlorine_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.m.wikipedia.org/wiki/Chlorine_dioxide&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in chlorite ion, the ones you specified will only appear as stated here &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chlorite (compound) is a compound that contains this group, with chlorine in oxidation state +3.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Chlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.m.wikipedia.org/wiki/Chlorite&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="15208" LastEditDate="2015-03-27T17:27:36.053" LastActivityDate="2015-03-27T17:27:36.053" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27935" PostTypeId="2" ParentId="27920" CreationDate="2015-03-27T12:37:04.047" Score="16" Body="&lt;p&gt;Yes NaCl3 is possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the more common iodine analog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KI + I2 -&amp;gt; KI3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a comprehensive analysis see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1211/1211.3644.pdf&quot;&gt;Unexpected stable stoichiometries of sodium chlorides&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 25-48 GPa, NaCl3 is stable in the Pnma structure, which has 4 formula units in the&#xA;  unit cell. Unlike all the other new phases predicted here (which are metallic), this phase is&#xA;  a semiconductor. Its structure (Fig. 2(B)) contains almost linear asymmetric Cl3 groups.&#xA;  Bader analysis (19) shows that the middle atom in the Cl3-group is nearly neutral, with&#xA;  most negative charge on the side atoms (Table 1) and the total charge of this anion group&#xA;  being ~-0.8. This result is reminiscent of well-known triiodides and azides with I3&#xA;  - and&#xA;  N3&#xA;  - ions, and of recent calculations on hypothetical H3&#xA;  - ions (20), which were found to&#xA;  have charge configuration [H-0.81H+0.72H-0.81]-0.9. Pnma-NaCl3 still can be viewed as an&#xA;  octet compound, with covalent bonding within [Cl3]--groups and ionic bonding between&#xA;  Na+ and [Cl3]-. At 48 GPa, with the formation of a new phase of NaCl3, the octet rule&#xA;  breaks down.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This research group compressed chlorine gas between NaCl plates, while heating with lasers and observed formation of $\ce{NaCl3}$. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-27T13:34:56.857" LastActivityDate="2015-03-27T13:34:56.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27936" PostTypeId="1" AcceptedAnswerId="27946" CreationDate="2015-03-27T12:38:16.910" Score="2" ViewCount="86" Body="&lt;p&gt;Why is in the reaction of hydrogen peroxide and sodium hypochlorite, singlet oxygen obtained? Why isn't triplet oxygen produced instead, as it is thermodynamically more stable?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-03-27T19:43:34.870" LastActivityDate="2015-03-27T19:43:34.870" Title="Singlet oxygen production" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-09-30T11:53:23.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="27937" PostTypeId="1" AcceptedAnswerId="27939" CreationDate="2015-03-27T12:44:48.863" Score="2" ViewCount="120" Body="&lt;p&gt;Can I use my extech pH60 meter to measure soil pH if so how?&#xA;I included a link to the meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0012W4HW2&quot; rel=&quot;nofollow&quot;&gt;Extech pH 60&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="554" LastActivityDate="2015-03-31T23:20:03.433" Title="Can I use my extech pH 60 meter to measure soil pH if so how?" Tags="&lt;physical-chemistry&gt;&lt;biochemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27938" PostTypeId="1" AcceptedAnswerId="27949" CreationDate="2015-03-27T12:58:01.570" Score="0" ViewCount="353" Body="&lt;p&gt;Can you explain to me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are Cis-Trans, E-Z and Syn-Anti Isomerization? &lt;/li&gt;&#xA;&lt;li&gt;What is the difference between them? &lt;/li&gt;&#xA;&lt;li&gt;How many functional groups is cis-trans isomerism applicable to?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I  know mostly about cis trans and E-Z but have no idea about syn- anti isomers. Correct me if needed: I think cis and trans isomers respectively occur when identical atoms or groups are on the same or opposite side of the double bond respectively. E and Z isomers are for trans and cis in complex cases where more functional groups are involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I couldn't follow the Wikipedia Link. &lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="52" LastEditDate="2015-03-28T01:15:50.770" LastActivityDate="2015-03-28T01:15:50.770" Title="Geometrical Isomerization" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27939" PostTypeId="2" ParentId="27937" CreationDate="2015-03-27T13:51:54.703" Score="5" Body="&lt;p&gt;Take a soil sample and mix it with distilled water. The most common ratio used for soil-water pH is 1:1 soil:water.&#xA;Stir the soil and solution and wait 15 minutes to 1 hour to set the slurry.&#xA;Measure the pH (check the temperature first).&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-03-27T13:51:54.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27941" PostTypeId="1" AcceptedAnswerId="27943" CreationDate="2015-03-27T14:05:50.703" Score="1" ViewCount="1474" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The volume of certain saturated solution is greater than the sum of the volumes of the water and salt from which it is made. On increasing pressure solubility of this salt&lt;/p&gt;&#xA;&lt;p&gt;(a) Increases&lt;/p&gt;&#xA;&lt;p&gt;(b) Decreases&lt;/p&gt;&#xA;&lt;p&gt;(c) Remains unaffected&lt;/p&gt;&#xA;&lt;p&gt;(d) Can't be predicted&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The solution says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Density of product &amp;lt; Density of reactant [mass of reactant = mass of product]&#xA;So on increasing pressure equilibrium shift toward higher density solid or liquid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of &lt;strong&gt;low&lt;/strong&gt; density when pressure is increased?&lt;/p&gt;&#xA;&lt;p&gt;Is my interpretation of the Principle wrong or is the solution of the answer wrong?&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="865" LastEditDate="2021-05-22T20:57:33.417" LastActivityDate="2021-05-22T20:57:33.417" Title="Effect of pressure on solubility of salts" Tags="&lt;equilibrium&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27942" PostTypeId="2" ParentId="27941" CreationDate="2015-03-27T14:37:38.640" Score="1" Body="&lt;p&gt;It shifts towards high density and not low.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told: &#xA;  graphite &amp;lt;--&gt; diamond &#xA;  We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as: &#xA;  graphite &amp;lt;--&gt; diamond + volume &#xA;  Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source:&#xA;(&lt;a href=&quot;http://www.mcat-prep.com/forum/gs-1-physical-sciences-f8/topic2040.html&quot; rel=&quot;nofollow&quot;&gt;http://www.mcat-prep.com/forum/gs-1-physical-sciences-f8/topic2040.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="15208" LastEditDate="2015-03-27T17:21:42.457" LastActivityDate="2015-03-27T17:21:42.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27943" PostTypeId="2" ParentId="27941" CreationDate="2015-03-27T14:48:58.783" Score="2" Body="&lt;p&gt;The answer is &quot;(b) Decreases.&quot;  The reason is that the saturated solution occupies more volume than the starting reagents.  At higher pressures, creating this extra volume requires more $P dV$ work be done by the dissolving reagents.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to Le Chatelier's Principle, shouldn't equilibrium be shifted towards the side of low density when pressure is increased?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find it easier to think in terms of volumes, not densities.  High pressure should favor low volumes.  Imagine a balloon filled with saturated solution.  If we squeeze the balloon by raising the pressure on the outside of it, the balloon will &quot;want to&quot; shrink.  In the example given, the balloon can shrink only by saturated solution converting back into solid salt and liquid water, so that is what would be thermodynamically favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;: Based on discussion in the comments, I wanted to add a point about compressibility and thermodynamic stability.  The degree to which a system opposes a change in pressure is called the compressibility and is related to dV/dP for a given system, which you can find from the equation of state. In general, systems in thermodynamic equilibrium cannot have a positive dV/dP value. That is, the volume must shrink as pressure rises for mechanically stable thermodynamic systems. &lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-27T16:45:09.230" LastActivityDate="2015-03-27T16:45:09.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27944" PostTypeId="2" ParentId="24812" CreationDate="2015-03-27T14:55:32.430" Score="2" Body="&lt;p&gt;1) Curiosity: How Science Became Interested in Everything&lt;br&gt;&#xA;by Philip Ball&lt;br&gt;&#xA; (&lt;a href=&quot;https://www.goodreads.com/book/show/13650233-curiosity&quot; rel=&quot;nofollow&quot;&gt;https://www.goodreads.com/book/show/13650233-curiosity&lt;/a&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Molecules: The Elements and the Architecture of Everything&lt;br&gt;&#xA;by Theodore Gray, Nick Mann&lt;br&gt;&#xA;(&lt;a href=&quot;https://www.goodreads.com/book/show/20729580-molecules&quot; rel=&quot;nofollow&quot;&gt;https://www.goodreads.com/book/show/20729580-molecules&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="15208" LastEditDate="2015-03-27T17:10:36.663" LastActivityDate="2015-03-27T17:10:36.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27946" PostTypeId="2" ParentId="27936" CreationDate="2015-03-27T15:07:08.370" Score="5" Body="&lt;p&gt;You are starting with spin-paired starting materials (an even number of electrons, with half in the up state and half in the down state), and they react (exothermically) to create an oxygen molecule. Initially, spin is conserved so that the oxygen molecule also has to be spin paired (singlet), despite the fact that the triplet state of oxygen is thermodynamically more stable. It will then react with a substrate or eventually decay over minutes to hours (most likely thermally) to the triplet state.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-03-27T15:07:08.370" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="27948" PostTypeId="1" CreationDate="2015-03-27T16:20:51.303" Score="4" ViewCount="10081" Body="&lt;p&gt;Are methyl n-propyl ether and methyl iso-propyl ether metamers?&lt;br&gt;&#xA;It says so &lt;a href=&quot;http://chem-guide.blogspot.in/2010/04/introduction-nomenclature.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the difference must be in number of carbon atoms or the distribution?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition(&lt;a href=&quot;http://dictionary.reference.com/browse/metamerism&quot; rel=&quot;nofollow&quot;&gt;http://dictionary.reference.com/browse/metamerism&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="402" LastEditDate="2016-12-17T03:02:05.853" LastActivityDate="2017-04-16T07:40:39.523" Title="Are methyl n-propyl ether and methyl iso-propyl ether metamers?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27949" PostTypeId="2" ParentId="27938" CreationDate="2015-03-27T16:25:20.423" Score="0" Body="&lt;p&gt;&lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; are older notations that express roughly same concept of &lt;em&gt;E&lt;/em&gt; and &lt;em&gt;Z&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explicitly, E and Z have absolute meaning and are assigned based on precedent, where the Z is assigned to the case where the highest priority substituents are on the same side of the alkene/double bond and &lt;em&gt;E&lt;/em&gt; being opposite.  These notations explicitly refer to a pair wise relationship.  It is not normally said '... the group &lt;em&gt;E&lt;/em&gt; to the ...'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; refer to substituents of context, with &lt;em&gt;cis&lt;/em&gt; being substituents on the same side of the alkene/double bond, and &lt;em&gt;trans&lt;/em&gt; being opposite.  Frequently these refer to the most important nomenclaturally, but not always.  Additionally &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; are used in discussions such as '...the group &lt;em&gt;trans&lt;/em&gt; to the ...' describing a relation of one object relative to the other.  &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; are no longer considered IUPAC-correct nomenclature, but still have some usage for descriptive discussion.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-03-27T16:25:20.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27950" PostTypeId="2" ParentId="27948" CreationDate="2015-03-27T16:30:27.747" Score="6" Body="&lt;p&gt;The term metamer is equivalent to isomer.  Isomers differ in connectivites of atoms between molecules with the identical number and type of atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methyl &lt;em&gt;n&lt;/em&gt;-propoyl Ether and methyl &lt;em&gt;iso&lt;/em&gt;-propyl ether have the same numbers of carbon, oxygen and hydrogen atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I have never heard the term metamer before, I expect it is archaic.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="13184" LastEditDate="2015-03-27T16:56:11.450" LastActivityDate="2015-03-27T16:56:11.450" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27951" PostTypeId="2" ParentId="22100" CreationDate="2015-03-27T17:00:27.490" Score="4" Body="&lt;p&gt;All the resonating structures follow Huckel's Rule and so they are not losing their aromaticity.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planar Cyclic Ring   &lt;/li&gt;&#xA;&lt;li&gt;4n+2 pi-electrons (the exocyclic pi electrons are not considered)&lt;a href=&quot;https://chemistry.stackexchange.com/questions/22100/what-is-the-resonance-structure-of-nitro-aminobenzene-with-a-broken-aromaticity#comment41055_27951&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;https://chemistry.stackexchange.com/questions/22100/what-is-the-resonance-structure-of-nitro-aminobenzene-with-a-broken-aromaticity#comment41280_27951&quot;&gt;2&lt;/a&gt;.                      &lt;/li&gt;&#xA;&lt;li&gt;All atoms of the ring in conjugation  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A classic example of an aromatic cabanion is &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Cyclopentadiene#Deprotonation&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclopentadiene&lt;/a&gt; ion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives $sp^2$ and not $sp^3$. So its planar.  &lt;/p&gt;&#xA;" OwnerUserId="15208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-04-10T15:21:07.490" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="27952" PostTypeId="2" ParentId="27941" CreationDate="2015-03-27T17:12:06.417" Score="0" Body="&lt;p&gt;There was another answer here which got removed because it just contained a link. The &lt;a href=&quot;http://www.mcat-prep.com/forum/gs-1-physical-sciences-f8/topic2040.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; was also useful to me in addition to &lt;a href=&quot;https://chemistry.stackexchange.com/a/27943/13357&quot;&gt;Curt's&lt;/a&gt; answer. I'll quote the relevant portion here.&lt;/p&gt;&#xA;&lt;p&gt;The topic in discussion was over the conversion of graphite to diamond but the concept applied was the same as in this question.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;You are right to suggest that when applying Le Chatelier's Principle, it's often useful to write and equation. We are told:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{graphite &amp;lt;=&amp;gt; diamond}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We are also given that diamond has a greater density meaning that it occupies less volume than the same mass of graphite, so the balanced equation could be written as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{graphite &amp;lt;=&amp;gt; diamond + volume}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now think of it this way: if you were to add volume, you would shift to the left according to Le Chatelier's Principle. Therefore, an increase in pressure (= decrease in volume) must shift to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-27T17:12:06.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27953" PostTypeId="1" CreationDate="2015-03-27T18:01:30.457" Score="3" ViewCount="69" Body="&lt;p&gt;I’ve done absolutely no chemistry since high school.  Now, 16 years later, I’ve been asked to look at some chemical analyses and I’m having trouble!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: I’ve got a series of spectroscopy results giving the concentrations (in mg/l) of chlorine and $\ce{SO4}$ in a sample.  I want to find the molar and mass ratios of $\ce{S}/\ce{Cl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I’ve done: I convert the mg to g, then (mass) concentration into a molar concentration by dividing by the molar mass (35.5 g/mol for $\ce{Cl}$ and 96 g/mol for $\ce{SO4}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: am I right in thinking that the molar concentration of $\ce{S}$ will be the same as the molar concentration of $\ce{SO4}$?  My reasoning is that 1 mol $\ce{SO4}$ contains 1 mol $\ce{S}$ and 2 mol $\ce{O2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, to get the mass concentration of $\ce{S}$ alone, can I just multiply my concentration of $\ce{SO4}$ by the molar mass ratio, $\frac{M(\ce{S})}{M(\ce{SO4})}$?&lt;/p&gt;&#xA;" OwnerUserId="15213" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T11:48:55.613" LastActivityDate="2015-04-02T14:07:56.207" Title="Amount of substance of a molecule in a solute the same as amount of substance of constituent elements?" Tags="&lt;molecules&gt;&lt;spectroscopy&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27954" PostTypeId="1" CreationDate="2015-03-27T18:09:28.910" Score="4" ViewCount="5406" Body="&lt;p&gt;I know  that specific metals like $\ce{Pb}, \ce{Sn}$ and $\ce{Al}$ become passive in conc. nitric acid. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why does this happen only with conc. nitric acid, and not other conc. acids?  &lt;/li&gt;&#xA;&lt;li&gt;Why is the passivity specific to these metals? Do they have something in common or are there individual reasons? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14982" LastEditorUserId="24052" LastEditDate="2016-03-11T22:51:21.220" LastActivityDate="2020-05-23T15:23:31.927" Title="What causes passivity of specific metals in conc. nitric acid?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27955" PostTypeId="2" ParentId="27827" CreationDate="2015-03-27T18:13:44.190" Score="3" Body="&lt;p&gt;When the electronic structure of a molecule is represented by molecular orbitals, all electrons occupy molecular orbitals.&lt;/p&gt;&#xA;&lt;p&gt;Molecular orbitals can be approximated as linear combinations of atomic orbitals.  A complete basis set for the molecular orbitals is an infinite set of atomic orbitals.&lt;/p&gt;&#xA;&lt;p&gt;A finite set of atomic orbitals will be a less accurate approximation of the molecular orbitals.&lt;/p&gt;&#xA;&lt;p&gt;So if you represent a low-lying molecular orbital as only a linear combination of core atomic orbitals, for example the lowest energy sigma MO of NN as only linear combination of the two 1s-1s orbitals, there will be some error associated with this approximation.  And if you do not include the 1s orbitals in the representation of the the higher molecular orbitals there will be some error associated with this approximation.  However, in the cases that I know of the error is small.&lt;/p&gt;&#xA;&lt;p&gt;For example, in the following early molecular orbital SCF LCAO papers, the 1s orbital are linearly combined (mixed, hybridized) with the other atomic orbitals:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/27/2/10.1063/1.1743730&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCF LCAO MO Study of Li2&lt;/a&gt; J. Chem. Phys. 27, 369 (1957)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/23/3/10.1063/1.1742031&quot; rel=&quot;nofollow noreferrer&quot;&gt;An SCF LCAO MO Study of N2&lt;/a&gt; J. Chem. Phys. 23, 569 (1955)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/23/10/10.1063/1.1740588&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electronic Population Analysis on LCAO–MO Molecular Wave Functions. I&lt;/a&gt; J. Chem. Phys. 23, 1833 (1955)&lt;/p&gt;&#xA;&lt;p&gt;As the nitrogen paper summarizes &amp;quot;As was expected, the inner-outer-shell mixing&#xA;is extremely small&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-03-27T18:18:48.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27956" PostTypeId="2" ParentId="27903" CreationDate="2015-03-27T20:02:54.560" Score="0" Body="&lt;p&gt;0.65% NaClO is lower than normal household bleach, which about 5%.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ucfoodsafety.ucdavis.edu/files/26437.pdf&quot; rel=&quot;nofollow&quot;&gt;Guidelines for the Use of Chlorine Bleach as&#xA;a Sanitizer in Food Processing Operations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe helpful to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, in the USA, it is permissible to wash fruits and vegetables in upto 2000 ppm NaClO. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-03-27T20:02:54.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27957" PostTypeId="2" ParentId="27913" CreationDate="2015-03-27T20:21:56.520" Score="1" Body="&lt;p&gt;Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$  (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived &lt;a href=&quot;http://www.chemteam.info/AcidBase/KaKb=Kw.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the equilibrium     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the acid dissociation constant has the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps, and please correct me if I've made any errors.&lt;/p&gt;&#xA;" OwnerUserId="14863" LastActivityDate="2015-03-27T20:21:56.520" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27958" PostTypeId="2" ParentId="27909" CreationDate="2015-03-27T20:46:07.040" Score="1" Body="&lt;p&gt;Here is something I found, on weak polyprotic acids, that I think answers the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9I2Cu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then [$\ce{SO3^{2−}}$] $= 6.4 \cdot 10^{-8}$?&lt;/p&gt;&#xA;" OwnerUserId="15199" LastEditorUserId="4945" LastEditDate="2015-03-31T05:37:42.747" LastActivityDate="2015-03-31T05:37:42.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27959" PostTypeId="1" AcceptedAnswerId="27962" CreationDate="2015-03-27T21:59:50.713" Score="10" ViewCount="17472" Body="&lt;p&gt;What happens when alkynes react with sodium in ammonia (&lt;span class=&quot;math-container&quot;&gt;$\ce{Na/NH3}$&lt;/span&gt;), or when they react with sodium amide (&lt;span class=&quot;math-container&quot;&gt;$\ce{NaNH2}$&lt;/span&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two sets of conditions different?&lt;/p&gt;&#xA;" OwnerUserId="7266" LastEditorUserId="16683" LastEditDate="2018-12-13T01:03:13.040" LastActivityDate="2021-04-17T17:42:03.510" Title="Reactions of alkynes with sodium/ammonia vs sodium amide" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;hydrocarbons&gt;&lt;organic-reduction&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27960" PostTypeId="2" ParentId="22100" CreationDate="2015-03-27T22:11:01.187" Score="10" Body="&lt;p&gt;Your teacher is correct, some of the resonance structures in your figure are not aromatic.  Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic.  The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a &quot;quinone&quot; rather than an aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;o&lt;/em&gt;- and &lt;em&gt;p&lt;/em&gt;-benzoquinone are examples of quinones; they are &lt;strong&gt;not&lt;/strong&gt; aromatic compounds (&lt;a href=&quot;http://en.wikipedia.org/wiki/Quinone&quot; rel=&quot;noreferrer&quot;&gt;ref&lt;/a&gt;, note the sentence &quot;Quinones are conjugated but not aromatic&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ux1sR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, their carbon analogues, &lt;em&gt;o&lt;/em&gt;- and &lt;em&gt;p&lt;/em&gt;-&lt;a href=&quot;http://en.wikipedia.org/wiki/Xylylene&quot; rel=&quot;noreferrer&quot;&gt;xylylene&lt;/a&gt; are also non-aromatic quinoid-like compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a compound to be aromatic it must be&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cyclic&lt;/li&gt;&#xA;&lt;li&gt;contain 4n+2 pi electrons&lt;/li&gt;&#xA;&lt;li&gt;have a continuous loop of p-orbitals&lt;/li&gt;&#xA;&lt;li&gt;and you must be able to draw double bonds - &lt;strong&gt;internal to the ring&lt;/strong&gt; - from each ring carbon atom.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclophane&quot; rel=&quot;noreferrer&quot;&gt;[6]-paracyclophane&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5KBE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-28T16:52:52.087" LastActivityDate="2015-03-28T16:52:52.087" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="27962" PostTypeId="2" ParentId="27959" CreationDate="2015-03-27T22:32:20.917" Score="11" Body="&lt;p&gt;Sodium in ammonia (&lt;span class=&quot;math-container&quot;&gt;$\ce{Na/NH3}$&lt;/span&gt;) and sodium amide (&lt;span class=&quot;math-container&quot;&gt;$\ce{NaNH2}$&lt;/span&gt;) are very different sets of conditions, despite superficially looking similar!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sodium metal itself, &lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt;, is a one-electron reducing agent – it is oxidised to &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{Na/NH_3}$&lt;/span&gt; is a way of &lt;strong&gt;reducing&lt;/strong&gt; some organic compounds (including alkynes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal alkynes &lt;span class=&quot;math-container&quot;&gt;$\ce{R-C#C-R}$&lt;/span&gt; can be reduced to &lt;em&gt;trans&lt;/em&gt;-alkenes using &lt;span class=&quot;math-container&quot;&gt;$\ce{Na/NH3}$&lt;/span&gt;. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate &lt;strong&gt;4&lt;/strong&gt; can interconvert between a &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; geometry, and the &lt;em&gt;trans&lt;/em&gt; geometry is preferred as this minimises steric repulsion between &lt;span class=&quot;math-container&quot;&gt;$\ce{R^1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{R^2}$&lt;/span&gt;. When this anion is protonated, the &lt;em&gt;trans&lt;/em&gt;-alkene is formed. This is a nice synthetic procedure for the preparation of &lt;em&gt;trans&lt;/em&gt;-alkenes, complementary to Lindlar hydrogenation which produces &lt;em&gt;cis&lt;/em&gt;-alkenes from alkynes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3MWvk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MWvk.png&quot; alt=&quot;Mechanism of Na/NH3 reduction of alkynes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in sodium amide &lt;span class=&quot;math-container&quot;&gt;$\ce{NaNH2}$&lt;/span&gt;, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion &lt;span class=&quot;math-container&quot;&gt;$\ce{NH2-}$&lt;/span&gt; is a very strong base (the &lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a$&lt;/span&gt; of ammonia, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;, is 38). So &lt;span class=&quot;math-container&quot;&gt;$\mathbf{NaNH_2}$&lt;/span&gt; is a way of &lt;strong&gt;deprotonating&lt;/strong&gt; some organic compounds (including terminal alkynes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton attached to the terminal carbon in a terminal alkyne is &lt;a href=&quot;https://chemistry.stackexchange.com/q/32341/16683&quot;&gt;acidic enough&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\mathrm pK_\mathrm a \sim 25$&lt;/span&gt;) to react with sodium amide, forming the corresponding carbanion &lt;strong&gt;7&lt;/strong&gt;. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HqReP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HqReP.png&quot; alt=&quot;Deprotonation of terminal alkyne with NaNH2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-12-13T01:35:39.910" LastActivityDate="2018-12-13T01:35:39.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27964" PostTypeId="1" CreationDate="2015-03-27T23:26:06.023" Score="2" ViewCount="41" Body="&lt;p&gt;On G2:M1:V1 window, result summary is grayed out. Does anybody know how to enable so that I can see my calculation summary?&lt;/p&gt;&#xA;" OwnerUserId="15217" LastEditorUserId="7448" LastEditDate="2015-10-09T16:00:52.927" LastActivityDate="2015-10-09T16:00:52.927" Title="how do you enable result summary in gaussview" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ClosedDate="2015-10-09T16:48:31.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="27969" PostTypeId="1" CreationDate="2015-03-28T02:21:04.260" Score="3" ViewCount="60" Body="&lt;p&gt;To my understanding, dehydration is defined as a subset of a condensation reaction. A condensation defined as being two molecules joining together (in the case of loss of water, this is referred to as a dehydration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought however, if a molecule of water leaves without a condensation necessarily occurring, that this is also a dehydration. In this case, how could the dehydration be a subset of condensation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with any such molecule off the top of my head, but some easily protonated alcohol that is resonance stabilized when water leaves? Would this leaving group of water not be a dehydration, without a condensation, since no addition would occur thereafter?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="7448" LastEditDate="2015-06-22T15:22:55.457" LastActivityDate="2015-06-22T15:22:55.457" Title="Is it correct to say that dehydration isn't a subset of condensation?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27970" PostTypeId="1" AcceptedAnswerId="27974" CreationDate="2015-03-28T04:51:44.080" Score="2" ViewCount="38241" Body="&lt;p&gt;Suppose I am given some reaction in which &#xA;$$\ce{C6H12O6 -&amp;gt; CO2}$$&#xA;and I want to calculate n-factor for this reaction to ultimately calculate equivalent weight of carbon for this reaction. Since $$E=\frac{M}{n}$$&#xA;I know that n-factor is no of electron gained/lost by one atom of a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Initial Oxidation state of carbon is $$6x + 12-12=0$$ $$x=0$$ meaning &#xA;and final oxidation state of carbon is $$-4$$ Hence, n-factor of this reaction should be $4$ but it's given to be $28$. How is this possible? Can someone explain how to correctly calculate n-factor with few more examples?&lt;/p&gt;&#xA;" OwnerUserId="13457" LastEditorUserId="41328" LastEditDate="2019-09-21T19:41:32.250" LastActivityDate="2019-09-21T19:41:32.250" Title="How to calculate valence factor (n-factor) for a element?" Tags="&lt;stoichiometry&gt;&lt;oxidation-state&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="27972" PostTypeId="2" ParentId="16135" CreationDate="2015-03-28T05:36:22.320" Score="-3" Body="&lt;p&gt;We can calculate the number of structural isomers by double bond equivalent i.e by using the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1+ \sum_i \frac{n_i (v_i - 2)}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ repesents the number of atoms $i$ of a given valency, and $v_i$ represents these atoms’ valency. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C4H6}: 1+ \sum_i \frac{n_i (v_i - 2)}{2}  = 1+ \frac{4(4-2)}{2}+\frac{6(1-2)}{2}  \\&#xA;= \frac{8-6}{2} +1 = 2 $$&#xA;Thus the compound may contain two double bonds or a triple bond and there are two possible isomers.&lt;/p&gt;&#xA;" OwnerUserId="15223" LastEditorUserId="-1" LastEditDate="2016-09-19T12:32:45.933" LastActivityDate="2016-09-19T12:32:45.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27974" PostTypeId="2" ParentId="27970" CreationDate="2015-03-28T06:40:42.187" Score="12" Body="&lt;p&gt;Hint :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;n-factor of a molecule/compound is defined as the change in oxidation state &lt;strong&gt;per molecule&lt;/strong&gt;.&lt;br&gt;&#xA;You have correctly calculated the change of &lt;strong&gt;one&lt;/strong&gt; carbon atom as 4.&lt;br&gt;&#xA;But how many carbon atoms are there in the glucose molecule?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;average&lt;/strong&gt; oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. (&lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/GenChem2/B7/1.html&quot;&gt;Reference&lt;/a&gt;)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n-factor of a reaction is not defined but it is defined for a single species participating in that reaction.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;n-factor of glucose is not equal to the n-factor of the product formed in that reaction, that is, $\ce{CO_2}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="13339" LastEditDate="2015-05-18T05:29:26.717" LastActivityDate="2015-05-18T05:29:26.717" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27975" PostTypeId="1" CreationDate="2015-03-28T07:07:04.973" Score="6" ViewCount="558" Body="&lt;p&gt;I want to optimize geometry and to analyse frequencies for transition metals ($\ce{Cu}$, $\ce{Mn}$, etc.) compounds using GAMESS. I use Avogadro for preparing input files and I have a problem: one of my calculations terminates abnormally with the following message in the output file.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NO FORCE FIELD, SCF DOES NOT CONVERGE AT VIB0 POINT&lt;br&gt;&#xA;  EXECUTION OF GAMESS TERMINATED -ABNORMALLY- AT Thu Mar 26&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the actual problem and how can I overcome it?&lt;/p&gt;&#xA;" OwnerUserId="15224" LastEditorUserId="186" LastEditDate="2015-03-28T13:19:21.343" LastActivityDate="2015-03-28T13:19:21.343" Title="Abnormal termination of GAMESS in frequencies calculation" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2016-09-06T21:14:12.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="27976" PostTypeId="2" ParentId="27969" CreationDate="2015-03-28T07:27:29.033" Score="2" Body="&lt;p&gt;Alkanols are indeed a good example. Think in the E2 and E1 reactions of primary and tertiary alkanols  in the presence of sulfuric acid at elevated temperatures, yielding alkenes. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-28T07:27:29.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27980" PostTypeId="2" ParentId="16135" CreationDate="2015-03-28T12:26:17.307" Score="25" Body="&lt;p&gt;&lt;strong&gt;It isn't easy but it is an interesting research topic&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. &lt;a href=&quot;http://www.wikiwand.com/en/List_of_straight-chain_alkanes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been an interesting research topic in computational chemistry and mathematics. &lt;a href=&quot;http://prod.sandia.gov/techlib/access-control.cgi/2004/040960.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This old paper (pdf),&lt;/a&gt; for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is &lt;a href=&quot;http://oeis.org/A000602&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up &lt;em&gt;space&lt;/em&gt; in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1gnQY.png&quot; alt=&quot;the first &amp;quot;impossible&amp;quot; hydrocarbons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available &lt;a href=&quot;http://www-jmg.ch.cam.ac.uk/data/isomercount/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; if your Java settings allow it. The results are discussed in a paper available in the &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ci0497657&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of Chemical Information and Modelling&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="402" LastEditDate="2019-05-17T06:40:20.430" LastActivityDate="2019-05-17T06:40:20.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27981" PostTypeId="2" ParentId="5394" CreationDate="2015-03-28T12:30:46.050" Score="-3" Body="&lt;p&gt;In the case of ionic compounds, there is a complete transfer of electrons from one atom of the element to another atom of the another element. So, there exists a strong electrostatic force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of covalent compounds, electrons are not completely dragged, the atoms of different elements are attracted by a weaker electrostatic force. Thus, ionic compounds have atoms bound by stronger electrostatic force than in case of covalent compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in general it is easier to break covalent bonds than ionic bonds. Thus, the melting and boiling points of ionic compounds are greater than covalent compounds.&lt;/p&gt;&#xA;" OwnerUserId="15226" LastEditorUserId="16683" LastEditDate="2017-06-13T11:07:17.290" LastActivityDate="2017-06-13T11:07:17.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27982" PostTypeId="2" ParentId="4" CreationDate="2015-03-28T12:59:30.710" Score="7" Body="&lt;p&gt;It might be helpful if we defined the changes in free energy in these equations in a more precise manner.  Let's just consider an ideal gas reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G^\circ$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;State 1&lt;/strong&gt;:  Pure reactants (in separate containers) in stochiometric proportions at 1 atm pressure and temperature T &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;State 2&lt;/strong&gt;:  Pure products (in separate containers) in corresponding stochiometric proportions at 1 atm pressure and temperature T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly measure $\Delta G^\circ$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G^\circ$ for that path.  This path might involve the use of constant temperature reservoirs and semipermeable membranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;State 1&lt;/strong&gt;:  Pure reactants (in separate containers) in stochiometric proportions at specified pressures and temperature T &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;State 2&lt;/strong&gt;:  Pure products (in separate containers) in corresponding stochiometric proportions at specified pressures and temperature T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specified pressures just happen to correspond to the partial pressures of the gases in a reaction mixture at equilibrium, then $\Delta G = 0$.  For small excursions of the partial pressures from the equilibrium values,  $\Delta G$ will increase as the squares of the partial pressure increments.  That is what we mean when we say that it is at a minimum at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly measure $\Delta G$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G$ for that path.  This path might involve the use of constant temperature reservoirs, small weights to be added or removed from a piston, and semipermeable membranes.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="16683" LastEditDate="2017-02-09T18:54:52.543" LastActivityDate="2017-02-09T18:54:52.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27983" PostTypeId="1" CreationDate="2015-03-28T13:13:21.767" Score="2" ViewCount="63" Body="&lt;p&gt;I can't find any publication to deal with aggregation of colloidal metalic nanoparticles. What type of interaction(s) is/are there between aqeous metalic nanoparticles which give(s) rise to aggregation? &lt;/p&gt;&#xA;" OwnerUserId="4445" LastActivityDate="2015-03-30T04:35:24.213" Title="Interaction between metalic nanoparticles and aggregation" Tags="&lt;physical-chemistry&gt;&lt;synthesis&gt;&lt;metal&gt;&lt;materials&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27984" PostTypeId="1" CreationDate="2015-03-28T14:22:52.653" Score="1" ViewCount="84" Body="&lt;p&gt;Most resources on pH and the electrochemistry behind it that I've found are IMHO too superficial. For example, they explain that an electrode with less than 80% slope should be discarded, but not why the decrease in slope happens, or why it's a problem if you can just calibrate it. Or they explain the potential generated at the Ag/AgCl eletrode and the glass membrane potential, but not the interaction between the two - does the H+ combine with the e- from the electrode? And what happens if the current is reversed? After reading for a few days I still feel like I don't understand any of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've read a couple PDFs on nico2000.net (the 30 page Introduction to pH measurements), the 160 page book pH Measurements by C. Westcott (1978), a few manufacturer's guides to pH measurement theory and practice (by Radiohm, Emerson, Mettler) and the relevant chapters in the Analytical Chemistry books that my library had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="15101" LastActivityDate="2015-03-30T05:48:09.080" Title="What's a good resource/book on pH measurements?" Tags="&lt;electrochemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27986" PostTypeId="1" AcceptedAnswerId="27988" CreationDate="2015-03-28T14:57:14.377" Score="6" ViewCount="2513" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which one of the following substituents at para-position is &lt;strong&gt;least effective&lt;/strong&gt; in stabilizing the phenoxide ion &lt;span class=&quot;math-container&quot;&gt;$\ce{PhO-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(a)&lt;span class=&quot;math-container&quot;&gt;$\ce{-CH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(b)&lt;span class=&quot;math-container&quot;&gt;$\ce{-OCH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(c)&lt;span class=&quot;math-container&quot;&gt;$\ce{-COCH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(d)&lt;span class=&quot;math-container&quot;&gt;$\ce{-CH2OH}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to me it should be &lt;span class=&quot;math-container&quot;&gt;$\ce{-OCH3}$&lt;/span&gt; because of its mesomeric effect which will counter the mesomeric effect of &lt;span class=&quot;math-container&quot;&gt;$\ce{O-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{-CH3}$&lt;/span&gt; has a positive inductive effect but, the mesomeric effect of &lt;span class=&quot;math-container&quot;&gt;$\ce{-OCH3}$&lt;/span&gt; should be stronger and destabilize the phenoxide ion to a greater extent&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-28T16:03:52.107" Title="Stabilisation of phenoxide ion" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27987" PostTypeId="2" ParentId="5394" CreationDate="2015-03-28T15:50:52.840" Score="12" Body="&lt;p&gt;The answer relates to the strength of the interactions between the component units that make up a crystal or a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why anything is a solid at a given temperature is, crudely, that the interactions between the units that make up the solid (atoms, ions or molecules) are stronger than the amount of thermal energy available at that temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond, for example, is a solid because each unit carbon is bonded to its neighbours by a strong carbon-carbon bond that takes a great deal of energy to break. Non-polar molecular compounds, like candle wax (in reality a mixture but imagine a pure long chain hydrocarbon to keep it simple) are held together by the intermolecular forces between the molecules (often called dispersion or Van der Waals forces). These are much weaker than chemical bonds and depend on the surface area of the molecules. So small non-polar or not-very polar molecules tend to have low melting points. Hexane is a volatile liquid. The shape and symmetry of the molecule matters a bit to these interactions so toluene is a liquid down to -95°C but benzene, which is very similar to toluene but far more symmetric and slightly flatter only melts at about 6°C as the molecules pack together better encouraging stronger interactions. Polar molecules have stronger interactions because the molecular dipoles interact and this produces stronger forces than the non-polar dispersion forces. So phenol (very similar in size to toluene but far more polar) melts at about 41°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple ionic solid are made from charged ions. Table salt crystals consist of Na&lt;sup&gt;+&lt;/sup&gt; and Cl&lt;sup&gt;-&lt;/sup&gt; ions. The key interaction between these is electrostatic and this is a strong force that takes a lot of energy to break. The melting point is about 800°C. Permanent electrostatic interactions are strong especially if the components can get close (remember, it's an inverse square law). This explains why not every ionic compound is a solid. There are ionic compounds that are liquid at room temperature. These tend to consist of molecular ions that are large and sometime irregular so forcing structures where the ionic force has to act over much larger distances and, therefore, be much weaker. An example is [BMIM]PF&lt;sub&gt;6&lt;/sub&gt; which has a melting point well below room temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CPrYX.png&quot; alt=&quot;BMIMPF6&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These ionic liquids are an interesting area of research in chemistry as they sometimes enable much less wasteful chemical reactions in industry (see &lt;a href=&quot;http://www.wikiwand.com/en/Ionic_liquid&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-03-28T15:50:52.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27988" PostTypeId="2" ParentId="27986" CreationDate="2015-03-28T16:03:52.107" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which one of the following substituents at para-position is &lt;strong&gt;least&#xA;  effective&lt;/strong&gt; in stabilizing the phenoxide ion $\ce{PhO^{−}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the &lt;em&gt;p&lt;/em&gt;-methoxy group is the least stabilizing of the choices. I'm not quite sure what you mean by&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;its mesomeric effect which will counter the mesomeric effect of $\ce{O^{−}}$ &#xA;  .&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so let me add a few words of explanation for the choice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below, I've drawn a few of the resonance structures possible for &lt;em&gt;p&lt;/em&gt;-methoxyphenoxide (A-C).  The methoxy group donates electrons through resonance and resonance structure B illustrates this effect.  However, note that in resonance structure B we have placed two negative charges close to each other - this is very destabilizing.  Therefore, as the methoxy group donates electrons through resonance it destabilizes the system. In resonance structure C, the electron withdrawing inductive effect of the methoxy group is stabilizing.  However the resonance effect outweighs the inductive effect and overall the &lt;em&gt;p&lt;/em&gt;-methoxy group is strongly destabilizing in the case of phenoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structure D illustrates the &lt;em&gt;p&lt;/em&gt;-methyl case.  Methyl is very mildly electron releasing both through resonance and inductively, so it also destabilizes the phenoxide anion, but since its effect is weak, it is not as destabilizing as the &lt;em&gt;p&lt;/em&gt;-methoxy substituent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, notice in the &lt;em&gt;p&lt;/em&gt;-acetyl case we can draw resonance structure E which actually stabilizes the phenoxide anion &lt;em&gt;via&lt;/em&gt; resonance.  The &lt;em&gt;p&lt;/em&gt;-acetyl group is also inductively stabilizing in the phenoxide example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZYb76.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-03-28T16:03:52.107" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="27990" PostTypeId="1" AcceptedAnswerId="27991" CreationDate="2015-03-28T17:49:04.370" Score="11" ViewCount="2646" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Three-dimensional_examples&quot; rel=&quot;noreferrer&quot;&gt;According to Wikipedia&lt;/a&gt;, there's an infinite set of possible wavefunctions (orbitals) for the hydrogen atom:&#xA;$$\psi_{n\ell m}(r,\theta,\phi) = \sqrt {{\left (  \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^{\ell} L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^{m}(\theta, \phi )$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would an unperturbed electron ever go outside the lowest orbital?&lt;/li&gt;&#xA;&lt;li&gt;At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ru0Ob.jpg&quot; alt=&quot;Probability densities for the first few hydrogen atom orbitals&quot;&gt;&#xA;&quot;Probability densities for the first few hydrogen atom orbitals&quot; (&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Hydrogen_Density_Plots.png&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="1109" LastEditDate="2015-03-28T22:02:59.793" LastActivityDate="2017-10-23T19:49:45.853" Title="Why are so many wave functions associated with hydrogen?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27991" PostTypeId="2" ParentId="27990" CreationDate="2015-03-28T18:17:49.413" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.Would an unperturbed electron ever go outside the lowest orbital?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, there is no such thing as &quot;outside the lowest orbital&quot; because the lowest orbital has probability density at all points in space (no nodes).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes, at a given temperature electrons will be distributed amoung the energy levels in accordance with a Boltzmann distribution.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are infinite solutions none of which exceed the ionization energy (the ionization energy is only approached as n approaches infinity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a series like 0, 3/4, 8/9, 15/16, etc. where there are infinite members without reaching 1 (1 is approached in the limit of infinity).  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-03-28T18:23:17.657" LastActivityDate="2015-03-28T18:23:17.657" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="27994" PostTypeId="1" CreationDate="2015-03-28T20:34:42.503" Score="3" ViewCount="1319" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Which is the better solvent for molar mass determinations by freezing point depression, benzene or cyclohexane?&lt;/li&gt;&#xA;  &lt;li&gt;Which is the better method for molar mass determination, freezing point depression or boiling point elevation?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the correct approach for solving these?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In my opinion, resonance structures might be the key concept, but I don't know how it would help or hurt.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think boiling has a decomposing effect that we must avoid, meaning freezing the compound(s) is preferable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But, both of these statements are inadequate.&lt;/p&gt;&#xA;" OwnerUserId="11549" LastEditorUserId="194" LastEditDate="2018-02-05T03:44:50.890" LastActivityDate="2018-02-05T03:44:50.890" Title="Molecular weight determination from freezing point depression" Tags="&lt;solutions&gt;&lt;solvents&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="27995" PostTypeId="1" AcceptedAnswerId="29352" CreationDate="2015-03-28T21:01:29.543" Score="8" ViewCount="8640" Body="&lt;p&gt;What is the mechanism of the reduction of organic acids with borane? Borane reductions don't work well with electron deficient carbonyl groups. Why is that so, and how isn't the carboxyl group an electron deficient group? &lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-04-27T13:09:11.037" Title="Reduction of acids with borane" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27996" PostTypeId="2" ParentId="27994" CreationDate="2015-03-28T21:43:28.163" Score="2" Body="&lt;pre&gt;&lt;code&gt;Molar freezing point depression  &#xA;Benzene      5.12 °C/m  &#xA;Cyclohexane  20.0 °C/m  &#xA;&#xA;Molar boiling point elevation  &#xA;Benzene      2.53 °C/m  &#xA;Cyclohexane  2.79 °C/m  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would base my argument on the constants, instead of waving hands.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-03-28T21:43:28.163" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="27997" PostTypeId="1" CreationDate="2015-03-28T23:17:55.310" Score="4" ViewCount="1422" Body="&lt;p&gt;I got &lt;strong&gt;some confusion&lt;/strong&gt; going on with this explanation of Raoult's  Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what my book say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the ideal solution has a higher entropy than the pure solvent, the energy of a vapor increases as its molecules expand into a larger volume, that is, if they are at a lower pressure. Thus, solutes in a solution reduce the vapor pressure of the solvent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, I should mention both solutions have the same $\Delta S$ based on the same $\frac{\Delta H}{T}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is I am confused about the last part of this explanation about how molecules expand. First off, this is all in a closed container, otherwise there's no vapor pressure. If so, &lt;strong&gt;the volume should be the same&lt;/strong&gt; there is no larger or smaller volume. &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="15133" LastEditDate="2015-03-29T05:06:06.040" LastActivityDate="2016-07-06T15:34:21.023" Title="Raoult's Law explained with thermodynamics/ free energy" Tags="&lt;gas-laws&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27998" PostTypeId="1" AcceptedAnswerId="28058" CreationDate="2015-03-29T00:00:36.310" Score="7" ViewCount="2206" Body="&lt;p&gt;Often called &quot;hypervalent&quot;, chemicals like phosphorous pentachloride and sulfur hexafluoride are possible due to the fact that their central atoms form covalent bonds with more than four other atoms, giving rise to uncommon arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{PCl5}$, the phosphorous atom forms 5 bonds, giving rise to trigonal bipyramidal shape according to the VSEPR model. With $\ce{SF6}$, the Sulfur atom forms 6 bonds, giving rise to octahedral shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, which central atom can achieve the highest number of bonds through the use of an expanded octet, and what is its shape? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest I can think of is iodine heptafluoride, whose shape is pentagonal bipyramidal due to its fluorine atom of 7 bonds.&lt;/p&gt;&#xA;" OwnerUserId="15236" LastEditorUserId="15410" LastEditDate="2016-04-29T04:59:14.767" LastActivityDate="2016-04-29T04:59:14.767" Title="What is the highest possible expanded octet?" Tags="&lt;molecular-structure&gt;&lt;valence-bond-theory&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="27999" PostTypeId="1" AcceptedAnswerId="28000" CreationDate="2015-03-29T00:19:14.997" Score="2" ViewCount="68" Body="&lt;p&gt;According to the VSEPR model, when a molecule in formed in the shape of &lt;span class=&quot;math-container&quot;&gt;$\ce{AB_{x}}$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;like &lt;span class=&quot;math-container&quot;&gt;$\ce{IF7}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{SCl6}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt;,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;its structure is found by reducing the repulsions between peripheral atoms. This gives rise to many shapes, most of which have constant bond angles. Examples of those with constant bond angles include&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CH4}$&lt;/span&gt; with 109.5° bond angles&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; with 180° bond angles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My question is then, why do some have non-constant bond angles?&lt;/p&gt;&#xA;&lt;p&gt;The following do not&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{IF7}$&lt;/span&gt; with 72° and 90° bond angles&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{PCl5}$&lt;/span&gt; with 120° and 90° bond angles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15236" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-03-29T17:12:26.600" Title="Why do many hypervalant chemicals have non-constant bond angles?" Tags="&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28000" PostTypeId="2" ParentId="27999" CreationDate="2015-03-29T00:43:04.857" Score="3" Body="&lt;p&gt;The $\ce{IF7}$ and $\ce{PCl5}$ molecules have 7 and 5 ligands, and pentagonal and trigonal bipyramidal geometries, respectively. These geometries happen &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;simply because&#xA;  there is no geometrical arrangement which can result in five equally&#xA;  sized bond angles in three dimensions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Trigonal_bipyramidal_molecular_geometry&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no arrangement in which all the angles would be the same, except in a plane, where the repulsion would obviously be a lot higher than in the pyramids.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="5017" LastEditDate="2015-03-29T14:13:18.253" LastActivityDate="2015-03-29T14:13:18.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28001" PostTypeId="1" AcceptedAnswerId="28005" CreationDate="2015-03-29T00:59:48.717" Score="8" ViewCount="5746" Body="&lt;p&gt;All of the exams that I've sat so far, and the homework questions I've had, always ask for the reaction between a substance and a &lt;em&gt;dilute acid&lt;/em&gt;. Is there a difference in the reaction between a dilute acid and a concentrated acid when each is applied to a substance? If so, what is it, and how dilute is dilute?&lt;/p&gt;&#xA;" OwnerUserId="15033" LastEditorUserId="5026" LastEditDate="2018-02-18T06:10:15.100" LastActivityDate="2018-02-18T06:10:15.100" Title="Is there a difference in the reactions of a dilute acid and a concentrated acid?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28002" PostTypeId="1" CreationDate="2015-03-29T02:38:08.927" Score="0" ViewCount="514" Body="&lt;p&gt;I have a problem solivng this chemical problem. Any help would be highly appreciative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium Thiosulfate can be made by the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na_2CO_3 + 2Na_2S + 4SO_2 -&amp;gt;  3Na_2S_2O_3 + CO_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How many grams of $\ce{Na_2S}$ are required to react with 25.0 g of $\ce{Na_2CO_3}$ if $\ce{SO_2}$ is present in excess?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25.0 \text{ g } \ce{Na_2CO_3}\frac{1 \text{ mol } \ce{Na_2CO_3}}{105.99}\frac{2}{1}\frac{78.05}{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer from the book states: 25.9g$\ce{Na_2S}$&lt;/p&gt;&#xA;" OwnerUserId="15237" LastEditorUserId="10185" LastEditDate="2015-03-29T17:08:41.453" LastActivityDate="2015-03-29T17:08:41.453" Title="Sodium thiosulfate formation" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28003" PostTypeId="2" ParentId="28002" CreationDate="2015-03-29T03:03:14.937" Score="2" Body="&lt;p&gt;You are completely correct in your dimensional analysis and stoichiometry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;36.8g is the correct answer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Molar mass $\ce{Na2CO3 = 106g/mol }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${25g \rightarrow 25/106 = 0.236}\text{ mol } \ce{Na2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will react with ${0.236*2 = 0.472} \text{ mol } \ce{Na2S}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar mass $\ce{Na2S} = 78.0 \text{ g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.472 \text{ mol } = 0.472*78.0 = 36.8 \text{ g } \ce{Na2S}$ will be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I will now prove your book wrong&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25.9g $\ce{Na2S}$ = 0.332mol $\ce{Na2S}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.332mol $\ce{Na2S}$ reacts with 0.166mol $\ce{Na2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.166mol $\ce{Na2CO3}$ = 17.5g of $\ce{Na2CO3}$, which is different from our starting quantity of 25.0g.&lt;/p&gt;&#xA;" OwnerUserId="15236" LastEditorUserId="15236" LastEditDate="2015-03-29T03:35:13.893" LastActivityDate="2015-03-29T03:35:13.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28004" PostTypeId="1" CreationDate="2015-03-29T04:19:39.823" Score="2" ViewCount="616" Body="&lt;p&gt;I want to know why concentration of solids and liquids are taken unity in expression of equilibrium constant( heterogenous equilibrium). It is usually mentioned that the density of solids and liquids( ideally) does not change and hence concentration does not change and so rate does not depend on solids and liquids. I accept that it is true. But will not it change the actual value of $ K_c$? We use $K_c$ everywhere like calculating extent of reaction, in thermodynamics- calculation of standard gibbs free energy and so on... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the assumption of unit concentration in expression  is justified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to give you an example:&#xA; Suppose in expression of $K_c$ the concentration of water(liquid) is involved  then the concentration wil be $(1000 g/L)/(18g/mol)$ ( density/ gram molar mass) Now if we take the concentration of water(liquid) unity then definitely value of $K_c$  will change and this change will be reflected. for example: when we want to predict extent of reaction $OR$ put value of $K_c$ in different equations relating it like $G^○=-RTlnK_c$&lt;/p&gt;&#xA;" OwnerUserId="14914" LastEditorUserId="14914" LastEditDate="2015-03-30T04:18:50.797" LastActivityDate="2015-03-30T04:18:50.797" Title="The concentration is unity! Why?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ClosedDate="2015-03-29T13:24:33.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="28005" PostTypeId="2" ParentId="28001" CreationDate="2015-03-29T04:26:07.643" Score="5" Body="&lt;p&gt;While it would be easier with examples, I leave here some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be because the idea of concentration is an approximation of the concept of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;nofollow&quot;&gt;thermodynamic activity&lt;/a&gt;, which is frequently referred to as &lt;strong&gt;effective concentration&lt;/strong&gt;. Depending on the question being asked, it might be important that the solute behaves as if it was in an infinite concentration, where there's no interaction between solute molecules/ions and the concentration equals the activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mention of dilution might also be important if the reaction involves water as a reagent. Even though water usually does not receive much attention when discussing reactions, since it's the solvent with &quot;constant&quot; concentration, it is important to note that several reactions wouldn't happen in its absence. That would be the case if the acid is available in a purified form, such as a gas (like HCl) or as solid (such as benzoic acid). It's also worth mentioning that the concept of acidity is inherent to solutions, so the two above aren't acids unless diluted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above also highlights something that is noteworthy. While &lt;em&gt;diluted&lt;/em&gt; is the opposite of concentrated, it may also be said as the opposite of &quot;not-in-solution&quot;, meaning not mixtured with the solvent. This could also be a reason to mention dilute acids, to make it clear they are in solution instead of any other form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the example, I do believe it's because the reactions happen in the presence of water, but honestly, I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible reason to mention dilute acids would be in case it's necessary to consider the concentration of a medium-strength acid. In dilute solutions, those can be considered completely dissociated, which makes calculations a lot easier, if that's the case.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-03-29T17:08:48.840" LastActivityDate="2015-03-29T17:08:48.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28006" PostTypeId="1" AcceptedAnswerId="31521" CreationDate="2015-03-29T06:30:44.380" Score="11" ViewCount="6176" Body="&lt;p&gt;The standard example is $\ce{CrCl3.6H2O}$, which can have three distinctly different&#xA;crystalline compounds now known as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Cr(H2O)6]Cl3}$ (violet),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[CrCl(H2O)5]Cl .H2O}$&#xA;(blue-green), and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[CrCl2(H2O)4]Cl.2H2O}$ (dark green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this difference of ligands in octahedral system affect the colour of compound? I think it may be due to the change of splitting energy but if that so then how does it actually work?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="13357" LastEditDate="2015-03-29T11:55:33.660" LastActivityDate="2016-03-20T12:14:00.913" Title="How hydrate isomerism results difference in colours?" Tags="&lt;coordination-compounds&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28007" PostTypeId="1" CreationDate="2015-03-29T07:24:56.117" Score="5" ViewCount="1976" Body="&lt;p&gt;I searched a lot to find the products of the haloform reaction of acetylacetone (pentane-2,4-dione),  but I didn't find any results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the products of the haloform reaction of acetylacetone? &lt;/p&gt;&#xA;" OwnerUserId="15239" LastEditorUserId="5026" LastEditDate="2018-02-24T13:23:48.757" LastActivityDate="2018-02-24T13:37:48.413" Title="What are the products of the haloform reaction of acetylacetone?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28009" PostTypeId="2" ParentId="27095" CreationDate="2015-03-29T09:35:20.463" Score="43" Body="&lt;p&gt;The guideline that you have mentioned &lt;em&gt;(‘While naming any compound, the numbering should be done such that position of substituted groups gives the smallest sum.’)&lt;/em&gt; does not exist in the IUPAC recommendations.&lt;/p&gt;&#xA;&lt;p&gt;The corresponding section in the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; actually reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.3.5&lt;/strong&gt; Lowest set of locants&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.&lt;br /&gt;&#xA;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the &lt;em&gt;smallest sum of locants&lt;/em&gt; can lead to wrong results.&lt;/p&gt;&#xA;&lt;p&gt;For example, for 2,7,8-trimethyldecane, the &lt;em&gt;smallest sum of locants&lt;/em&gt; would yield the wrong name ‘3,4,9-trimethyldecane’, since &lt;span class=&quot;math-container&quot;&gt;$2+7+8=17 &amp;gt; 3+4+9=16$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BtoLn.png&quot; width=&quot;190&quot; alt=&quot;2,7,8-trimethyldecane&quot;&gt;&#xA;&lt;p&gt;Likewise, for 2,2,6,6,7-pentamethyloctane, the &lt;em&gt;smallest sum of locants&lt;/em&gt; would yield the wrong name ‘2,3,3,7,7-pentamethyloctane’, since &lt;span class=&quot;math-container&quot;&gt;$2 + 2 + 6 + 6 + 7 = 23 &amp;gt; 2 + 3 + 3 + 7 + 7 = 22$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Lt0yy.png&quot; width=&quot;170&quot; alt=&quot;2,2,6,6,7-pentamethyloctane&quot;&gt;&#xA;&lt;p&gt;For 1,6,7-trimethylnaphthalene, the &lt;em&gt;smallest sum of locants&lt;/em&gt; would yield the wrong name ‘2,3,5-trimethylnaphthalene’, since &lt;span class=&quot;math-container&quot;&gt;$1 + 6 + 7 = 14 &amp;gt; 2 + 3 + 5 = 10$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lpJGf.png&quot; width=&quot;158&quot; alt=&quot;1,6,7-trimethylnaphthalen&quot;&gt;&#xA;&lt;p&gt;For 1,6-dimethylcyclohex-1-ene, the &lt;em&gt;smallest sum of locants&lt;/em&gt; would yield the wrong name ‘2,3-dimethylcyclohex-1-ene’, since &lt;span class=&quot;math-container&quot;&gt;$1 + 6 = 7 &amp;gt; 2 + 3 = 5$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oozeW.png&quot; width=&quot;163&quot; alt=&quot;1,6-dimethylcyclohex-1-ene&quot;&gt;&#xA;&lt;p&gt;Anyway, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2020-10-04T14:28:18.950" LastActivityDate="2020-10-04T14:28:18.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28010" PostTypeId="2" ParentId="16940" CreationDate="2015-03-29T10:27:41.057" Score="0" Body="&lt;p&gt;I think these could be the reason-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not possible because $\ce{NH_3^+}$(no vacant orbital) doesn't have any space at all to get involved with the lone pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and the second part &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abel already answered that at one time only one $\ce{-NH_2}$ takes part when we determine basicity and the second $\ce{-NH_2}$ plays no role.&lt;/p&gt;&#xA;" OwnerUserId="15244" LastEditorUserId="4945" LastEditDate="2015-04-01T09:21:28.280" LastActivityDate="2015-04-01T09:21:28.280" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28013" PostTypeId="1" AcceptedAnswerId="28018" CreationDate="2015-03-29T16:01:05.213" Score="0" ViewCount="2719" Body="&lt;p&gt;I know that sulfuric acid is considered a strong acid, and the dissociation of its first &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ion is complete and of its second &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ion is partial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid is weak and all three of its &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions dissociate partially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just wanted to clarify: are dibasic and tribasic acids considered strong if their first dissociation is complete (and subsequent ones will naturally be partial)?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="41328" LastEditDate="2019-10-07T09:18:18.597" LastActivityDate="2019-10-07T09:19:55.643" Title="Confused about strong/weak dibasic and tribasic acids" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;protons&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28014" PostTypeId="1" AcceptedAnswerId="28019" CreationDate="2015-03-29T16:31:21.117" Score="11" ViewCount="849" Body="&lt;p&gt;This is rather a bunch of questions that I decided to post on Chemistry StackExchange since I thought the chemists would have the most knowledge about the chemical processes of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So recently I have came across stem cells and a lot of revolutionary medical products that can rebuild tissue and even organs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I thought that can we use these products to transplant our organs and in a sense, rebuild our own body. I have even came across making a heart and lungs as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so is there any thing that stands between having a longer life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What else do we need to have in order to prolong and in a sense immortalize ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it would be impossible to do this for everyone since our planet cannot sustain more life, but theoretically is it possible to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are you views into this?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="10185" LastEditDate="2015-03-29T17:03:21.273" LastActivityDate="2015-03-30T04:44:14.257" Title="Can we prolong life?" Tags="&lt;biochemistry&gt;&lt;medicinal-chemistry&gt;&lt;drugs&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ClosedDate="2015-03-30T06:55:21.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="28015" PostTypeId="1" AcceptedAnswerId="28025" CreationDate="2015-03-29T16:38:14.300" Score="4" ViewCount="1761" Body="&lt;p&gt;I'm going to use aluminium in my example, but this is applicable to &lt;em&gt;all&lt;/em&gt; metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium has delocalised electrons. It has aluminium cation -electron attraction. It's easier to visualize the atomic structure of aluminium in solid form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it's in the gas phase, will delocalised electrons return to the atom itself, or is there going to be an electron and cation mixture (electrons floating around)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can aluminium (or any other metal) conduct electricity as a vapour, and what would its conductance be compared to its solid form?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="87609" LastEditDate="2020-01-11T15:43:58.643" LastActivityDate="2020-01-11T15:43:58.643" Title="Structure of a metal in gas phase" Tags="&lt;metal&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28016" PostTypeId="1" AcceptedAnswerId="28023" CreationDate="2015-03-29T16:45:01.643" Score="4" ViewCount="127" Body="&lt;p&gt;Can we picture metallic bonding as an equilibrium between electrons and cations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{Al^3+ + 3e- &amp;lt;=&amp;gt; Al}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13208" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T20:44:51.793" LastActivityDate="2015-04-24T20:44:51.793" Title="Can we picture metallic bonding as an equilibrium between electrons and cations?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28017" PostTypeId="2" ParentId="28013" CreationDate="2015-03-29T16:51:44.033" Score="1" Body="&lt;p&gt;If you think about each dissociate state separately, than it's easy to answer your question. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; is a very strong acid &lt;span class=&quot;math-container&quot;&gt;$(\mathrm{p}K_\mathrm{a}\approx -3.0),$&lt;/span&gt; but &lt;span class=&quot;math-container&quot;&gt;$\ce{HSO4-}$&lt;/span&gt; is not so strong &lt;span class=&quot;math-container&quot;&gt;$(\mathrm{p}K_\mathrm{a}\approx 2.0).$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid starts with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}\approx 2.14,$&lt;/span&gt; which as as you said is not so strong, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HPO4^2-}$&lt;/span&gt; have &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}\approx 7.2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}\approx 12.3,$&lt;/span&gt; so definetly weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, we say sulfuric acid is strong because we're talking about &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4},$&lt;/span&gt; and what comes after should be mentioned separately. &lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="41328" LastEditDate="2019-10-07T09:19:55.643" LastActivityDate="2019-10-07T09:19:55.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28018" PostTypeId="2" ParentId="28013" CreationDate="2015-03-29T16:53:40.237" Score="1" Body="&lt;p&gt;I mean whenever you have a polyprotic acid/base and they say that if it is strong or weak you can only assume the first dissociation is strong or weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say like you have $\ce{H2SO4}$ and we all know that it is a strong acid. This only means that the first dissociation is strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 -&amp;gt; HSO4- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case above only the first dissociation is strong. And then you cannot and do not know if $\ce{HSO4}$ is strong as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So overall in any case we cannot assume and the data must be provided.&lt;/strong&gt;&#xA;Good Question.  &lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-03-29T16:53:40.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28019" PostTypeId="2" ParentId="28014" CreationDate="2015-03-29T17:07:36.707" Score="9" Body="&lt;p&gt;&lt;strong&gt;Ageing is a disease.  Just as treatments have been found for other diseases, ageing can be slowed or reversed as well.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not that long ago, scientists noticed that at the end of each DNA strand there were thousands of nucleotides.  Nucleotides are used for genetic coding.  That is to say, certain sequences of nucleotides instruct the body to synthesize  certain proteins when the genetic code in the DNA is being read.  DNA is really like a massive software program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, when scientists noticed all of this code at the end of a DNA strand, at first they couldn't figure out what is was for, so they thought it must be gibberish.  Later they noticed that everytime a cell divided (&lt;em&gt;e.g&lt;/em&gt;. everytime the DNA was read) a piece of this gibberish was lopped off.  Eventually it became understood that this &quot;gibberish&quot; was the basis of aging.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each lopped off piece was termed a telomere.  Each telomere contains the same nucleotide sequence (code).  There are about 15,000 telomere nucleotides at the end of the DNA strands in an embryo.  This has been shortened to about 10,000 when you are born and when it reaches a length around 5,000, the body dies.  It turns out that the telomeres play a role in stabilizing DNA strands during the copying process, when their length gets down to around 5,000 nucleotides the telomere length becomes to short to provide adequate stabilization and errors (genetic mutations) occur in the copying process and/or the cells become senescent (become inactive, go to sleep).  Errors or sleep, either way, the body stops functioning efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This telomere shortening occurs during cell division in all human cells &lt;strong&gt;except&lt;/strong&gt; for sperm and egg cells.  Sperm and egg cells are &lt;strong&gt;immortal&lt;/strong&gt;.  There is an enzyme called telomerase, it is expressed only in the sperm and egg cells.  After these cells divide a telomere is lopped off, but telomerase re-adds a telomere to the DNA in sperm and egg cells.  Hence, their telomere length never shortens - they live forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All cells contain the code to make telomerase, but that code  (or signaling) is turned off in all cells except for sperm and egg cells.  A specific protein combines with a certain portion of DNA and that is all that is required to prevent telomerase expression in a cell.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If scientists could find a way to prevent that (repressor) protein from complexing with DNA, then telomerase expression could be turned on and that cell would then be immortal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can imagine, this is a big area of research.  Here is a &lt;a href=&quot;http://www.sierrasci.com/&quot; rel=&quot;nofollow&quot;&gt;link to a company&lt;/a&gt; that has been active in this area for a long time, and here is a link to a &lt;a href=&quot;http://www.sierrasci.com/?p=telomere_basics&quot; rel=&quot;nofollow&quot;&gt;basic tutorial&lt;/a&gt; on the subject of telomeres.  Poke around the web site, the videos and other tutorials are extremely interesting.  Even aspects such as the social consequences of curing aging are discussed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enough progress has been made that you can buy products in the over-the-counter market that reduce telomere shortening, but they're very expensive at the present time.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-03-29T21:49:55.577" LastActivityDate="2015-03-29T21:49:55.577" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28021" PostTypeId="1" CreationDate="2015-03-29T19:51:09.913" Score="6" ViewCount="186" Body="&lt;p&gt;We have UV SHİMADZU 2550 and its diffuse reflectance attachment. I want to measure band gap of single crystal samples. What should I do? Which formulas should I use to calculate band gap? I don't have much sample...&lt;/p&gt;&#xA;" OwnerUserId="15249" LastActivityDate="2016-02-13T14:28:56.350" Title="How to measure band gap of powder sample with using diffuse reflectance method" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28022" PostTypeId="1" AcceptedAnswerId="42574" CreationDate="2015-03-29T21:11:17.677" Score="24" ViewCount="1946" Body="&lt;p&gt;According to Wikipedia, the conjugate acid of LDA has a pKa of about 35 - which is in line with figures I've seen elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then can such a weak acid's pKa be determined? If the determination is made in water, how is that even possible? Water would be a stronger acid by far and couldn't it easily mask the acidity of LDA's conjugate acid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is LDA's pKa value even determined in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, how are extremely strong acids' pKa values determined? Wouldn't the leveling effect come into play as one introduces, say, tosic acid into water? How then can we determine the strength of an acid stronger than hydronium ion in water? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="402" LastEditDate="2017-05-08T00:58:20.363" LastActivityDate="2017-05-08T00:58:20.363" Title="How is the pKa of extremely weak acids determined?" Tags="&lt;organic-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28023" PostTypeId="2" ParentId="28016" CreationDate="2015-03-29T21:29:12.700" Score="3" Body="&lt;p&gt;In metals, electrons are non-localized, forming a &quot;sea&quot; of electrons, rather than having them localized, as in the $\ce{Na+Cl-}$ lattice of crystalline salt. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallic_bonding&quot; rel=&quot;nofollow&quot;&gt;Metallic bonding&lt;/a&gt; for a more complete description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, of course, a matter of degree, as covalent, ionic and metallic bonding can &quot;blend&quot; from one to the other. A bond can be considered &lt;em&gt;partially&lt;/em&gt; ionic and covalent, for example; see these &lt;a href=&quot;http://www.longview.k12.wa.us/ralong/tietjen/Chemistry/Metallic.JAT.pdf&quot; rel=&quot;nofollow&quot;&gt;helpful graphics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-03-29T21:36:01.520" LastActivityDate="2015-03-29T21:36:01.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28025" PostTypeId="2" ParentId="28015" CreationDate="2015-03-29T21:52:56.767" Score="7" Body="&lt;p&gt;Metal vapors &lt;em&gt;do&lt;/em&gt; conduct electricity, as in the mercury vapor lamp (also used in fluorescent lamps), but, since the atoms are isolated or in small (molecular) clusters, &lt;strong&gt;conduction is ionic, not metallic.&lt;/strong&gt; Since electrons are far less massive (and therefore more mobile) than the heavy ions, they are responsible for the majority of the conducted current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrons are forcibly ripped away from the mercury atom (well, raised to higher levels) -- the first ionization potential is 10.39V, and its second is 18.65V, so in a mercury vapor lamp or &lt;a href=&quot;http://www.r-type.org/addtext/add092.htm&quot; rel=&quot;noreferrer&quot;&gt;mercury vapor rectifier&lt;/a&gt;, the potential drop of at least 20 volts implies more than one of the outer electrons are lost from many of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that ionized gases are &lt;em&gt;much less conductive than solid or liquid metals&lt;/em&gt;. Consider a 70 watt sodium vapor lamp, with an arc length of perhaps 10 cm and a &lt;em&gt;potential drop of ~18 volts.&lt;/em&gt; The finest commercially available wire, 40 gauge, would have a &lt;em&gt;potential drop of just ~1.3 volts&lt;/em&gt; for metallic copper wire, or &lt;em&gt;~4 volts for sodium&lt;/em&gt;, which has 1/3 the conductivity of metallic copper (if they made fine sodium wire), at the same length and current (~3 A). &#xA;See the useful &lt;a href=&quot;http://www.webexhibits.org/causesofcolor/4AC.html&quot; rel=&quot;noreferrer&quot;&gt;chart&lt;/a&gt; explaining sodium-vapor ionization .&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-03-29T22:28:57.807" LastActivityDate="2015-03-29T22:28:57.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28026" PostTypeId="1" AcceptedAnswerId="28070" CreationDate="2015-03-29T22:03:34.370" Score="7" ViewCount="514" Body="&lt;p&gt;This is what I read on &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Laws_of_Thermodynamics/Second_Law_of_Thermodynamics#Major_players_in_developing_the_Second_Law&quot; rel=&quot;nofollow&quot;&gt;Chemwiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rudolf Clausius was a German physicist, and he developed the Clausius statement, which says &quot;Heat generally cannot flow spontaneously from a material at a lower temperature to a material at a higher temperature.&quot; However, there is one exception to this statement and that is sometimes a hot material will take heat from the cold material, making the hot material warmer and the cold material colder.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemwiki does not provide a reference for the exception about heat transfer from cold material to hot material. How is this possible? What is an example of this happening?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="1887" LastEditDate="2015-03-30T23:09:17.563" LastActivityDate="2016-12-17T12:43:41.167" Title="When and why do hot material take heat from cold material?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28027" PostTypeId="1" CreationDate="2015-03-29T22:20:59.160" Score="0" ViewCount="1389" Body="&lt;p&gt;So my textbook gives me this random equation:&#xA; $\Delta U = q + w$. Then consider two conditions, under constant pressure and under constant volume. Because $w = -P\Delta V$, constant volume will make $\Delta U = q$. And constant pressure equation is basically the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is isn't heat under neither condition considered? Because it is of significance?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4231" LastEditDate="2015-03-29T23:32:55.497" LastActivityDate="2015-03-30T02:34:17.713" Title="Why heat of reaction under neither constant pressure or constant volume is considered (in my textbook, or probably somewhere else)?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28028" PostTypeId="2" ParentId="28001" CreationDate="2015-03-29T23:01:35.347" Score="1" Body="&lt;p&gt;It depends to some extent on the acid. The most important example would be sulphuric acid because in concentrated form ~98% there is not enough water present to accept the protons available so it becomes a weak acid if concentrated enough. This is not the case for conc. nitric acid (~65%) or conc. hydrochloric acid (~36%). In addition conc. H2SO4 is a powerful dehydrating agent (eg. with sucrose). It is also a strong oxidising agent eg. gives SO2 rather than hydrogen with metals. It can also be simply that the reaction could be inconveniently vigorous with too high a concentration. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="9792" LastActivityDate="2015-03-29T23:01:35.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28029" PostTypeId="1" CreationDate="2015-03-30T00:18:15.670" Score="6" ViewCount="247" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yPbtm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yPbtm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask about the rate of hydrolysis for 4-hydroxyphenyl acetate (A), &lt;em&gt;N&lt;/em&gt;-(4-hydroxyphenyl)acetamide (B), 4-hydroxyphenyl chloroacetate (C), and methyl 4-hydroxybenzoate (D). I know that hydrolysis depends on the positive charge of the carbonyl carbon and also the substituent around it (&lt;em&gt;e.g.&lt;/em&gt; electron withdrawing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, from the picture above I can see that compound C is the most reactive since it has a phenol and a chlorine as its substituents. Compound B is the least reactive due to nitrogen being the weakest electronegatively as compared to oxygen (I think?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused between compounds A and D because they are isomers of each other. The oxygen and phenol group are on the same substituent in compound A and are in two different substituent in compound D. Does this affect anything? or can I say that the strength of the electron withdrawing groups are additive, meaning compound A and D are similar in terms of reactivity? &lt;/p&gt;&#xA;" OwnerUserId="15253" LastEditorUserId="16683" LastEditDate="2016-11-30T14:06:55.477" LastActivityDate="2016-11-30T14:06:55.477" Title="Electrophilicities of carboxylic acid derivatives" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;reactivity&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28030" PostTypeId="2" ParentId="28014" CreationDate="2015-03-30T02:28:51.767" Score="3" Body="&lt;p&gt;There is no &lt;em&gt;chemical&lt;/em&gt; reason why life couldn't be prolonged at least &lt;em&gt;somewhat&lt;/em&gt;.  There are a huge variety of undesirable biochemical reactions (oxidation of proteins for instance) that cells must deal with as part of being alive, and given that &lt;em&gt;lineages&lt;/em&gt; are immortal (as the process of growing up and creating germ cells and then fusing them into a new being somehow resets the aging process) we can see that chemical immortality is entirely possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, that's what DNA is: the chemically immortal molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are all sorts of practical reasons why it may be difficult.  There is extensive research in a variety of organisms implicating all sorts of fundamental biochemistry in lifespan: insulin signaling, the EGF pathway, TOR, the unfolded protein response, and so on.  Ultimately it is not the chemistry which is the barrier: life already knows how to select and grow its way out of any undesirable side-reactions.  It is the biology: can you actually regulate yourself to do the biochemistry you need to do to live longer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to this chemistry question is: chemically speaking, it's a biology question.&lt;/p&gt;&#xA;" OwnerUserId="15254" LastActivityDate="2015-03-30T02:28:51.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28031" PostTypeId="2" ParentId="28027" CreationDate="2015-03-30T02:34:17.713" Score="2" Body="&lt;p&gt;The constant pressure and constant volume scenarios are presented when you first study thermodynamics because they present an easy way to simplify the calculation of energy changes ($\Delta U,\ \Delta H,$ etc.) based on the change of one easy to measure phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Internal Energy $U$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the change in internal energy must equal the sum of the work done $w$ and the heat transferred $q$. This is not random, it is a consequence of the Law of Conservation of Energy. If the internal energy of a system has changed then energy must be exchanged with the surroundings. There are only a few ways this can happen (heat, work, electromagnetic radiation, ...), and heat and work are the most important for &lt;em&gt;most&lt;/em&gt; thermodynamic scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\Delta U = q + w$. We need to be able to relate $q$ and $w$ to things we can measure. We calculate $q$ based on the temperature change of the system $\Delta T$ and the heat capacity of the system $c$. Thus, $q=c\Delta T$. Work is defined (in one way) as the change in volume of the system times the pressure. The negative sign is present to make sure that an expanding system is doing work on the surroundings (and thus losing energy): $w=-P\Delta V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have $\Delta U = c\Delta T -P\Delta V$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Constant Volume&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If volume is constant $\Delta V = 0$, which means that $w=0$, regardless of pressure. Thus, $\Delta U = q = c\Delta T$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Constant Pressure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This one is trickier, but we can do it &lt;strong&gt;for a gaseous system&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do not need to measure volume, since $PV=nRT$, but it is easier to measure volume than $n$, which may also be changing. If $n$ is constant, then, the equation for change in internal energy becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta U = c\Delta T - nR\Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $n$ is not constant, then we need to measure that change in volume and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the process is in another state, like in solution, $\Delta V$ might be very challenging to measure. Let's hope we can measure $\Delta n$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Enthalpy $H$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because of the challenge of measuring volume changes in solution, we invented a new thermodynamic potential $H = U + PV$. Total differentiation and whatnot gives us: &#xA;$$\Delta H = \Delta U +P\Delta V + V\Delta P = c\Delta T - p\Delta V + P\Delta V + V\Delta P$$&#xA;$$\Delta H = c\Delta T + V\Delta P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, at constant pressure, $\Delta H = q = c\Delta T$, which is easy to measure. Enthalpy is similarly challenging at constant volume as internal energy is at constant pressure. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Constant pressure and volume. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under these conditions, $\Delta U = \Delta H = q = c\Delta T$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressure and volume are not constant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Okay, so if $w=-P\Delta V$, how do we calculate work? If pressure is changing, which value of $P$ do we use? It turns out we use all of them and we need some calculus to do it. This case is often left out because many introductory textbooks consider the likelihood that many students beginning their study of chemistry may not have taken calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go. We need to replace the total change in $V$ with the infinitessimal change in $V$ at each moment $\Delta V \rightarrow  dV$. $\Delta V$ is the sum of all of the infitinessimal $dV$ values. Now, we can multiply each $dV$ with the pressure at that moment and add them up to get work (which involves integration):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w = \sum_{j=1}^n P_j dV_j=\int_{V_i}^{V_f}PdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we cannot leave well enough alone, because $P$ is changing, so we need to substitute for $P=\frac{nRT}{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w=\int_{V_i}^{V_f} \dfrac{nRT}{V}dV=nRT\dfrac{\ln V_f}{\ln V_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the above maths only work if temperature is constant. What if it's not?&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-03-30T02:34:17.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28033" PostTypeId="2" ParentId="27983" CreationDate="2015-03-30T04:35:24.213" Score="1" Body="&lt;p&gt;Metallic nanoparticles, considered as colloids, are attracted to each other by van der Waals forces. To prevent their aggregation, we normally use dispersant using electrostatic repulsion or steric hindrance to stabilize the colloids.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-03-30T04:35:24.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28034" PostTypeId="2" ParentId="28014" CreationDate="2015-03-30T04:44:14.257" Score="2" Body="&lt;p&gt;To provide a more abstract answer not particularly founded in science, have you considered that we have a finite lifetime in order to encourage us to reproduce and therefore create/promote genetic variability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus providing a far greater chance for our DNA to live on forever (albeit in a different form). This seems to already offer us immortality in some way. Unless of course global disaster strikes causing total extinction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems this is the only way to achieve true immortality as something as simple as a disease mutating could wipe us all out, variation is what allows DNA to continue to live.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately what would living forever allow us to achieve, there would be no goal in life and instead the potential for our DNA's death would increase as other reproductive organisms evolve and adapt.&lt;/p&gt;&#xA;" OwnerUserId="15258" LastActivityDate="2015-03-30T04:44:14.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28035" PostTypeId="2" ParentId="27984" CreationDate="2015-03-30T05:28:50.343" Score="1" Body="&lt;p&gt;Looks like you understand pH measurement, but you are trying to go deeper into the details of potentiometry and open circuit potential measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There I would recommend basic electrochemistry texts. Here is one simple book that is cheap, easy to read and very informational:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://global.oup.com/academic/product/electrode-potentials-9780198556848?lang=en&amp;amp;cc=us&quot; rel=&quot;nofollow&quot;&gt;Electrode Potentials; Richard G. Compton and Giles H. W. Sanders&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="13355" LastEditDate="2015-03-30T05:48:09.080" LastActivityDate="2015-03-30T05:48:09.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28036" PostTypeId="1" CreationDate="2015-03-30T07:40:58.813" Score="1" ViewCount="585" Body="&lt;p&gt;According to the Nernst equation, the electrode potential $E$ is&#xA;$$E = E^\circ - \frac{RT}{nF}\ln\frac{[\ce{M}]}{[\ce{M^{$n$+}}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is, why? Assuming a simple galvanic cell, we know that in general if the temperature is increased, there should be more collisions of the solution with the electrode, resulting in a higher accumulation of the metal ions on the electrode. It should also make it easier for the metal atoms in the electrode to leave the electrode and go into the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if temperature reaches zero, the reaction should stop altogether as there would be no more collisions and no more potential difference should be developed on the electrodes. Yet according to the Nernst equation if temperature reaches zero, maximum standard electrode potential would be reached instead. Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="15260" LastEditorDisplayName="user7951" LastEditDate="2015-05-26T19:29:04.100" LastActivityDate="2015-05-26T19:29:04.100" Title="Why does an increase in temperature decrease the electrode potential?" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-05-27T03:42:46.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="28037" PostTypeId="1" CreationDate="2015-03-30T07:46:54.307" Score="0" ViewCount="736" Body="&lt;p&gt;I am trying to find the K&lt;sub&gt;sp&lt;/sub&gt; of silver thiocyanate:&#xA;AgSCN&lt;sub&gt;(s)&lt;/sub&gt; --&gt; Ag&lt;sup&gt;+&lt;/sup&gt; &lt;sub&gt;(aq)&lt;/sub&gt;SCN&lt;sup&gt;-&lt;/sup&gt;&lt;sub&gt;(aq)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I need to first prove that the &lt;strong&gt;lower&lt;/strong&gt; [Ag+] = 1.27&lt;sup&gt;-12&lt;/sup&gt; M, but don't know how to find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I am given:&#xA;Volume of 10&lt;sup&gt;-4&lt;/sup&gt; M Ag+ solution used: 10.00ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of 1.0 M KSCN added: 50.00ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measured cell potential when an Ag(s) electrode was placed into the above mixture, and then joined to the Ag+(0.1M)/Ag(s) half cell: 0.645V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results:&#xA;Calculated lower [Ag+] from cell potential: 1.27&lt;sup&gt;-12&lt;/sup&gt; M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am suppose to use this equation for the concentration cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eVZRd.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where n is the number of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really my question is how I would go about finding the higher concentration of Ag&lt;sup&gt;+&lt;/sup&gt; and then use that to determine that &lt;strong&gt;lower&lt;/strong&gt; [Ag+] = 1.27&lt;sup&gt;-12&lt;/sup&gt;M&lt;/p&gt;&#xA;" OwnerUserId="15259" LastActivityDate="2015-03-30T07:46:54.307" Title="Concentration cell lower half cell concentration from measured cell potential" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;solubility&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28039" PostTypeId="1" CreationDate="2015-03-30T08:23:40.863" Score="2" ViewCount="37" Body="&lt;p&gt;What mass of fragrance are airborne in scented environments for example in a perfume deparment? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum g/m3 of fragrance that a human can detect, and how much is present when it evaporates from an oven, in a cookie shop, from a cut lawn, near a perfumed lady and average levels from a day of walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What range of fragrance concentrations do humans typically have to cope with.&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorUserId="14577" LastEditDate="2015-04-02T08:12:08.027" LastActivityDate="2015-04-02T08:12:08.027" Title="What mass of fragrances does a cubic meter of air contain on average?" Tags="&lt;aromatic-compounds&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-04-03T05:02:10.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="28040" PostTypeId="1" AcceptedAnswerId="28049" CreationDate="2015-03-30T08:49:35.700" Score="3" ViewCount="59" Body="&lt;p&gt;Are there examples of UV-Vis emission from charge-separated states? The cases I have encountered have only absorption. Is the recombination process of a nonradiative kind by necessity?&lt;/p&gt;&#xA;" OwnerUserId="13341" LastActivityDate="2015-03-30T13:22:13.470" Title="UV-Vis emission from charge-separated states?" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28041" PostTypeId="1" AcceptedAnswerId="28128" CreationDate="2015-03-30T09:28:44.177" Score="6" ViewCount="4042" Body="&lt;p&gt;This would mean the formation of a &amp;#x3c3; and one or two &amp;#x3c0; bonds, all coordinate. I haven't heard of this happening, like a fluorine anion or water with two lone pairs on oxygen will donate only one lone pair, but I wanted to know why it's not possible.&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="7475" LastEditDate="2016-02-05T01:05:06.430" LastActivityDate="2016-02-05T01:05:06.430" Title="Can any atom donate more than 1 lone pair as a coordinate bond to the same metal?" Tags="&lt;bond&gt;&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28042" PostTypeId="1" CreationDate="2015-03-30T09:37:54.293" Score="2" ViewCount="261" Body="&lt;p&gt;I recently noticed that whenever I washed the egg yolk stuck on some utensils the water becomes milky white.&lt;br&gt;&#xA;It has to be the egg yolk that turns white and give off the white color to water right?  &lt;/p&gt;&#xA;" OwnerUserId="5674" LastEditorUserId="7448" LastEditDate="2015-03-30T10:26:21.940" LastActivityDate="2015-03-30T10:26:21.940" Title="Why does egg yolk turn white when washed with water?" Tags="&lt;biochemistry&gt;&lt;water&gt;&lt;color&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28044" PostTypeId="1" AcceptedAnswerId="28050" CreationDate="2015-03-30T11:43:25.843" Score="3" ViewCount="16132" Body="&lt;p&gt;Why is fat nonpolar while water is polar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water there is the $\ce{H-O}$ bond and the $\ce{O}$ is more negative, we have&#xA;$\text{dipole = charge}* \text{distance}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fat molecule is like&#xA;$\ce{H-O-C-C-C-C-C-C...}$&#xA;but the carbon atoms don't matter, the polarity is still ${q*d}$, as in the water molecule, so they should have the same amount of polarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do people say fat is a nonpolar molecule?&lt;/p&gt;&#xA;" OwnerUserId="15262" LastEditorUserId="15133" LastEditDate="2015-03-30T18:16:32.333" LastActivityDate="2015-03-30T20:38:29.243" Title="Why is fat nonpolar?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28046" PostTypeId="1" AcceptedAnswerId="28071" CreationDate="2015-03-30T12:03:56.200" Score="8" ViewCount="5181" Body="&lt;p&gt;I posed &lt;a href=&quot;https://cooking.stackexchange.com/questions/56153/lime-substitute-for-nixtamalizing-corn&quot;&gt;this question&lt;/a&gt; at the Seasoned Advice SE and thought I might get some more information here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nixtamalization&quot; rel=&quot;nofollow noreferrer&quot;&gt;Traditionally prepared corn&lt;/a&gt; is cooked in a calcium hydroxide solution to make it more digestible, but when my search for $\ce{Ca(OH)2}$ hit a dead end, I began thinking: could I make a substitute with stuff I already have (sodium hydroxide, which I use for pretzel/soap making; and calcium chloride, for brewing)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought process was that, upon dissolving, sodium hydroxide plus calcium chloride would provide the same ions (though not necessarily in the same concentration) as calcium hydroxide and sodium chloride. Since I would be adding salt to this recipe at some point anyway, I didn't see any harm to this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this feasible, or am I off-base in my chemical understanding about this? More importantly, is there some chemical consideration I've failed to take into account?&lt;/p&gt;&#xA;" OwnerUserId="13178" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:30.103" LastActivityDate="2017-10-21T01:26:17.987" Title="'Emulating' calcium hydroxide" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28049" PostTypeId="2" ParentId="28040" CreationDate="2015-03-30T13:22:13.470" Score="3" Body="&lt;p&gt;There are at least cases in which charge recombination results in an excited species that deactivates via fluorescence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the ionization of a hydrocarbon ($\ce{RH}$) to yield a radical cation and a solvated electron.&#xA;\[\ce{RH -&gt;[\mathrm{e-beam}] RH^{+\cdot} + e-}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trap the electron by reducing an aromatic hydrocarbon ($\ce{ArH}$) to the coresponding radical anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \ce{ArH +e- -&gt; ArH^{-\cdot}}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge recombination between the radical cation and the radical anion can yield $\ce{ArH}$ in an excited state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{RH^{+\cdot} + ArH^{-\cdot} -&gt; RH + ArH^{\ast}} \]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the days, &lt;em&gt;Alexander D. Trifunac&lt;/em&gt; did a lot of work on that at ANL.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-03-30T13:22:13.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28050" PostTypeId="2" ParentId="28044" CreationDate="2015-03-30T14:01:49.787" Score="4" Body="&lt;p&gt;The polarity of a single bond might indeed be the same for some different molecules. When speaking of polarity of molecules however, the whole molecule should be taken into account, since the macro properties are dictated by &lt;strong&gt;all&lt;/strong&gt; the parts of a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the fatty acids: The 'head' of such a molecule is indeed polar, and in specific cases this means that the interactions and reactivity will be like polar molecules. This effect however is only contained to the head of the molecule. Fatty acids as a whole are still considered non-polar because this type of molecule will not be dissolved in water. Note that the main reason for this is not due to the interactions of water with the fatty acid. The fatty acid is just as 'happy' interacting with water as it is with other fatty acid chains. Water, on the other end, does not 'want' to interact with fatty acids, since it will have to rearrange around the fatty acid molecule to ensure the same number of hydrogen bonds are made. This is an entropic cost, that makes the total Gibbs Free Energy change unfavorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that water molecules do not rearrange the hydrogen bonds with other water molecules and the fatty acids interact with each other. Due to this effect, water not being willing to rearrange hydrogen bonds, fatty acids are considered non-polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this has helped, and if I used to much terminology or jargon, please let me know!&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="2359" LastEditDate="2015-03-30T18:42:11.860" LastActivityDate="2015-03-30T18:42:11.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28051" PostTypeId="1" AcceptedAnswerId="28358" CreationDate="2015-03-30T14:58:37.590" Score="5" ViewCount="390" Body="&lt;p&gt;A common science demonstration is to &lt;a href=&quot;https://www.youtube.com/watch?v=1PJTq2xQiQ0&quot;&gt;float a &quot;boat&quot; on sulfur hexafluoride&lt;/a&gt;.&#xA;How long would the sulfur hexafluoride stay in the container (in the experiment linked above) if the container were to be left uncovered and undisturbed at room temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm wondering what would happen if you scaled the experiment up to the size of a lake, ignoring the time it would take to fill said lake and the impossibility of obtaining that much sulfur hexafluoride.  Let's also assume that there is no wind to blow it away. Could you have a large permanent pool of sulfur hexafluoride or would it eventually dissipate and mix with the atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this change if a different dense (non-reactive) gas were used, like xenon?&lt;/p&gt;&#xA;" OwnerUserId="15266" LastActivityDate="2015-04-07T22:57:24.547" Title="How quickly does sulfur hexafluoride mix with the atmosphere?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28052" PostTypeId="2" ParentId="6369" CreationDate="2015-03-30T15:03:37.507" Score="1" Body="&lt;p&gt;aluminium form smaller cation,since carbonate ion is larger in size thus greater polarization lead to weak covalent bond hence at room temperature aluminium carbonate tend to break down.&lt;/p&gt;&#xA;" OwnerUserId="15267" LastActivityDate="2015-03-30T15:03:37.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28054" PostTypeId="1" AcceptedAnswerId="28088" CreationDate="2015-03-30T16:45:36.110" Score="5" ViewCount="40843" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the pH at the equivalence point for the titration of $\pu{0.130 M}$ methylamine ($\ce{CH3NH2}$) with $\pu{0.130 M}$ $\ce{HCl}$. The $K_\mathrm{b}$ of methylamine is ${5.0 \cdot 10^{–4}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I started with the equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + CH3NH2 &amp;lt;=&amp;gt; CH3NH3+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then I knew that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[base]}}{\ce{[acid]}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I put ${\log \left(\frac{0.130}{0.130}\right) = \log 1 = 0}$ and then added that to the $\mathrm{p}K_\mathrm{a}$, which I got from the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}} \quad \rightarrow &#xA; \quad \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after I plugged those in to get a $\mathrm{pH}$, it turned out to be wrong and then comments said that when titrated a weak base with a strong acid, the volume is doubled at equivalence point and the concentrations are halved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? I now know that my original equation was wrong and it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3NH3+ &amp;lt;=&amp;gt; H+ + CH3NH2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from there I should make an ICE table with the concentration of $\pu{0.0650 M}$. &lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="41328" LastEditDate="2017-08-07T02:17:58.307" LastActivityDate="2017-08-07T02:17:58.307" Title="Calculating pH for titration of weak base with strong acid" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28057" PostTypeId="1" AcceptedAnswerId="28064" CreationDate="2015-03-30T17:40:03.103" Score="2" ViewCount="1351" Body="&lt;p&gt;Why do most all double replacement reactions result in a precipitate and an aqueous molecule? Is it because the individual solute molecules always form stronger bonds such as how in single replacement the molecule with more energy bonds with the molecule of the opposite charge?&lt;/p&gt;&#xA;" OwnerUserId="15271" LastEditorUserId="10185" LastEditDate="2015-03-30T18:37:23.480" LastActivityDate="2018-03-28T16:10:05.210" Title="Double Replacement Reactions form Precipitates" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28058" PostTypeId="2" ParentId="27998" CreationDate="2015-03-30T17:43:18.703" Score="10" Body="&lt;p&gt;I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention &lt;em&gt;hypervalency&lt;/em&gt;, what you seem interested in is &lt;em&gt;hypercoordination&lt;/em&gt;, or even more generally, just compounds with high &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_number&quot;&gt;coordination numbers&lt;/a&gt; (hypercoordination is used specifically when the number of ligands in a compound is larger than &quot;normal&quot;). Hypercoordination and high coordination numbers are entirely independent of hypervalency or VSEPR theory altogether. Regardless of the precise electronic structure in a compound, coordination numbers can often be determined far more directly, especially when the compounds can create crystal structures for &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_crystallography&quot;&gt;x-ray crystallography&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutron_diffraction&quot;&gt;neutron diffraction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually there is little focus given to compounds with more than six ligands, as the vast majority of compounds will have atoms surrounded by six or less ligands. However, &lt;a href=&quot;http://en.wikipedia.org/wiki/Coordination_geometry#Table_of_coordination_geometries&quot;&gt;there are some representatives for higher coordination numbers&lt;/a&gt;. Some books will mention iodine heptafluoride, $\ce{IF7}$, as a seven-coordinate compound, with its pentagonal bipyramidal structure. It is still possible to add a fluoride and obtain an example of coordination number 8 in the octafluoroiodate(VII) anion, $\ce{IF8^{-}}$, which has an interesting &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_antiprism&quot;&gt;square antiprismatic geometry&lt;/a&gt; (take a cube and twist a face by 45°). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For coordination number 9, representatives can be found in the transition metal hydrides, such as the nonahydridorhenate anion, $\ce{ReH_9^{2-}}$, and the lowest energy configuration in this case is a curious &lt;a href=&quot;http://en.wikipedia.org/wiki/Tricapped_trigonal_prismatic_molecular_geometry&quot;&gt;tricapped triagonal prism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to go higher than coordination number 9 in coordination compounds? While there are examples, at the present it seems that &lt;a href=&quot;http://en.wikibooks.org/wiki/Inorganic_Chemistry/Chemical_Bonding/VSEPR_theory#Examples&quot;&gt;none of them&lt;/a&gt; contain solely &lt;a href=&quot;http://en.wikipedia.org/wiki/Denticity&quot;&gt;monodentate ligands&lt;/a&gt;, that is, individual ligands which only bind to the centre once. This is because either there would have to be a lot of crowding over the central atom, creating repulsions between the ligands, or because to allow enough space the ligands would have to stay relatively far from the central atom, making their bonds weak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if a single ligand is allowed to bond to the centre through more than atom simultaneously, then the coordination number can keep increasing without requiring the presence of too many ligands. Actinides have a very rich coordination chemistry and are capable of generating several impressive compounds, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranocene&quot;&gt;uranocene&lt;/a&gt; with coordination number 16, but it seems the current record is held by actinide elements surrounded by four cyclopentadienyl rings, each with five carbon atoms, reaching an amazing coordination number of 20 in &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/9/95/Tetrakis%28cyclopentadienyl%29thorium%28IV%29-3D-balls.png&quot;&gt;tetrakis(cyclopentadienyl)thorium(IV) ($\ce{Th}\mathrm{(\eta ^5-}\ce{C5H5}\mathrm{)_4}$&lt;/a&gt; or its uranium analogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To finalize, it's interesting to note that though there are no examples yet of compounds with coordination number 10 or above containing only monodentate ligands, &lt;a href=&quot;http://www.mathpages.com/home/kmath005/kmath005.htm&quot;&gt;their expected geometries can be calculated&lt;/a&gt; even for much higher coordination numbers (under certain assumptions). For very high coordination numbers, calculations and physical reality will likely diverge, but perhaps a few more coordination numbers with the expected geometry will be unlocked by the study of ultraheavy element chemistry. Of course, the calculations linked here don't help much to study the geometry of coordination compounds with polydentate ligands, as they rely significantly on the geometry of the ligand itself&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.200604148/abstract&quot;&gt;This very relevant article&lt;/a&gt; raises some interesting points. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Endohedral_fullerene&quot;&gt;endohedral fullerene compounds&lt;/a&gt; could be thought of as a central atom surrounded by a single ligand in the shape of a cage, so one could possibly make a case for structures with coordination numbers of 60, 70, 80 or even more. The article also calculates the possible existence of a compound containing 15 monodentate ligands, the cation $\ce{PbHe_{15}^{2+}}$, though it would be very weakly bound (as you might expect from a helium compound) and possibly restricted to the gas phase.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-03-30T18:28:56.053" LastActivityDate="2015-03-30T18:28:56.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28064" PostTypeId="2" ParentId="28057" CreationDate="2015-03-30T19:00:08.203" Score="2" Body="&lt;p&gt;Precipitation of silver chloride is a common example of a double replacement reaction. e.g. $$\ce{AgNO3~_{(aq)} + NaCl_{(aq)} -&amp;gt; AgCl_{(s)} + NaNO3~_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially all four ions are present in solution. When the solutions are mixed the silver and chloride ions bond together and precipitate, leaving the sodium and nitrate ions remaining in solution along with a small number of silver and chloride ions because the precipitation is an equilibrium so a small number of dissolved ions will still be present (thanks to @Martin for pointing this out). The reaction happens because the precipitation of silver chloride removes these ions from the solution equilibrium, driving the reaction to completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver chloride is insoluble because the $\ce{Ag-Cl}$ bond has a high degree of covalent character due to the high polarizing ability of the silver cation and the high polarizability of the chloride anion and the similar size of both ions which allows a shorter bond length and so a stronger bond requiring more energy to break apart in order to dissolve the solid. See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/410/why-is-agcl-less-soluble-than-agno%E2%82%83&quot;&gt;this&lt;/a&gt; question for more detail on this point.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-03-31T17:25:18.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28065" PostTypeId="1" AcceptedAnswerId="28066" CreationDate="2015-03-30T19:20:49.267" Score="8" ViewCount="338" Body="&lt;p&gt;Why does a dianion (such as malonate) bind cations more strongly than its equivalent anion (acetate)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations? Does the second $\ce{COO-}$ group on malonate distribute charge in a better way that can allow for stronger binding?&lt;/p&gt;&#xA;" OwnerUserId="9604" LastEditorUserId="19616" LastEditorDisplayName="user15489" LastEditDate="2015-10-23T16:45:59.570" LastActivityDate="2015-10-23T16:45:59.570" Title="Why do dianions (such as malonate) bind cations more strongly than anions?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;bond&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28066" PostTypeId="2" ParentId="28065" CreationDate="2015-03-30T19:43:10.227" Score="10" Body="&lt;p&gt;This is known as the chelate effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason why you observe this is that cations in solution have an ordered solvent shell around them, especially in polar solvents where there will be defined solvent geometry around the shell; often octahedral for metal ions in water. The formation of a complex with a bidentate ligand such as malonate is more favourable than that with a monodentate ligand such as acetate because the same number of solvent ligands are lost with half the number of ligands complexed. This results in a greater increase in &lt;strong&gt;entropy&lt;/strong&gt; - you can see this in terms of the degrees of freedom gained by the water molecules more than compensating for those lost by the new ligands.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is likely also a factor, although in most cases less important. In the case of the monodentate ligands there is a small energetic cost associated with bringing the anionic oxygen centres close to complex the metal centres. In the case of the bidentate ligand, some of this cost is &quot;already paid&quot; in the enthalpy of formation of the ligand - it is preorganized. This contribution becomes dominant for macrocyclic ligands. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T21:16:10.160" LastActivityDate="2015-05-09T21:16:10.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28067" PostTypeId="2" ParentId="28057" CreationDate="2015-03-30T19:49:32.970" Score="-3" Body="&lt;p&gt;Precipitation helps drive the reaction in one direction by removing some of the reactants from solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example of double displacement is $\ce{FeS + H2SO4 -&amp;gt; FeHSO4 + H2S}$ where the reaction is driven in the forward direction by the loss of gaseous $\ce{H2S}$.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="10185" LastEditDate="2015-03-30T21:09:32.050" LastActivityDate="2015-03-30T21:09:32.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28068" PostTypeId="2" ParentId="28044" CreationDate="2015-03-30T20:38:29.243" Score="4" Body="&lt;p&gt;While there is nothing wrong with @eljee's answer there is a more obvious reason why fats are not as polar as you assume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fats are &lt;strong&gt;not&lt;/strong&gt; long chain alcohols: they are the triglyceride esters of long chain carboxylic acids. So, rather than being terminated by a relatively polar carboxylic acid or alcohol group, they are terminated by a relatively non polar glycerol ester group. Only when saponified (which breaks the free acids from the ester) do they become the more schizophrenic long chain acids which act as surfactants because one end of the molecule is polar enough to like water while the other is fatty enough to prefer oily, non-polar environments. Hence soap!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.wikiwand.com/en/Fat&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; page summarises this well.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-03-30T20:38:29.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28069" PostTypeId="1" CreationDate="2015-03-30T21:26:38.123" Score="3" ViewCount="6211" Body="&lt;p&gt;A transition element is one which forms one or more stable ions which have partially filled d-orbitals. Manganese (Mn) could have an oxidation state of 7+, which means that it will lose all of the $\ce{3d^5}$ and $\ce{4s^2}$ electrons. So how could it make a 7+ ion in that case (which means that it will lose all of the $\ce{4s^2}$ and $\ce{3d^5}$ electrons) and still be defined as a transition element?&lt;/p&gt;&#xA;" OwnerUserId="14862" LastEditorUserId="41328" LastEditDate="2017-10-06T13:59:38.353" LastActivityDate="2017-10-06T13:59:38.353" Title="Manganese(VII) oxidation state" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28070" PostTypeId="2" ParentId="28026" CreationDate="2015-03-30T23:09:06.337" Score="6" Body="&lt;p&gt;The statement is wrong and/or poorly formulated. Heat cannot spontaneously flow from cold areas to hot areas, there are no exceptions known. It contradicts Clausius' statement of the second law and hence also the more general observation/statement that the entropy of an isolated system is increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can drain heat from cold regions and transport it to hotter regions, but this should always go paired with some other (larger) increase in entropy somewhere else, e.g. by doing work like in a Linde refrigeration. An other example is that some salts have a very high entropy of solvation and can cool water to ~ -15 °C in a lab at room temperature. These salts can be/are used in practical chemistry for cooling a reaction mixture.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-03-30T23:09:06.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28071" PostTypeId="2" ParentId="28046" CreationDate="2015-03-31T00:02:00.777" Score="6" Body="&lt;p&gt;Your idea seems mostly sound, but one thing you should consider is the solubility of calcium hydroxide (a.k.a. slaked lime).  It is fairly low.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That low solubility provides you with an opportunity.  If you have calcium chloride and sodium hydroxide, you can mix saturated solutions of both and a white precipitate should form.  The precipitate is calcium hydroxide.  You can then remove the precipitated calcium hydroxide, wash it with ice cold water, and then you'll have $\ce{Ca(OH)2}$!  A &lt;a href=&quot;http://www.youtube.com/watch?v=MdIHl2JnD-I&quot; rel=&quot;noreferrer&quot;&gt;YouTube demo&lt;/a&gt; of this idea provides some more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCl2(aq) +  2NaOH(aq) -&amp;gt; 2NaCl(aq) + Ca(OH)2(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this method is preferred because the low solubility of calcium hydroxide compared to sodium hydroxide naturally limits its alkalinity and corrosiveness.  It would be too easy to use undesirable (and possibly even dangerous) levels of NaOH if you made a mistake in how much you used for nixtamalification.  Additionally, simply mixing NaOH and calcium chloride and using the resulting milky liquid as a substitute for calcium hydroxide will have far too much salt (NaCl) present to be a good substitute.  Or conversely, if you lower the amount of NaOH and calcium chloride to where the saltiness is tolerable, you will have too little alkalinity.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-03-31T00:12:06.940" LastActivityDate="2015-03-31T00:12:06.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28072" PostTypeId="1" CreationDate="2015-03-31T00:30:32.300" Score="3" ViewCount="3489" Body="&lt;p&gt;A friend of mine sprayed &quot;Black Flag Home Defense&quot; insecticide under her&#xA;computer. It died immediately. I'm trying to figure out why, and how&#xA;fixable it might be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an MSDS, but it only says what the active ingredient is,&#xA;deltamethrin, 0.02%. Apparently everything else is proprietary and/or&#xA;non-hazardous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deltamethrin is not water-soluble, so I'm thinking that this probably&#xA;doesn't contain any water; more like nonpolar hydrocarbon sorts of&#xA;compounds. I'm thinking there's some particular nonpolar solvent that&#xA;makes up the bulk of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Information or informed speculation as to the likely solvent&#xA;involved.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For those with knowledge of electronics, information or informed&#xA;speculation about the effect of such solvents on PC boards and&#xA;electronic components.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Answers to 1 alone are also useful, as I can then ask on&#xA;electronics.se what those compounds would do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I don't have access to the computer myself. I passed on some of&#xA;the suggestions to my friend, and she will bring it in to a shop and see&#xA;how repairable the damage seems to be. Thanks for all your thoughts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE 2: I spoke with the tech who looked at it. Apparently the&#xA;computer was very dirty inside; years of dust, cigarette smoke residue,&#xA;and insects. She's not sure exactly what happened, but there seems to be&#xA;a short somewhere on the motherboard, and that may have fried the power&#xA;supply as well. Given the dirt, it was probably not in good shape even&#xA;before the incident; the heat sinks were thoroughly clogged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall conclusion: keep your computer clean! Keeping unknown chemicals&#xA;away from it is a good idea too.&lt;/p&gt;&#xA;" OwnerUserId="15279" LastEditorUserId="7448" LastEditDate="2015-10-09T16:02:36.903" LastActivityDate="2015-10-09T16:02:36.903" Title="Computer killed by insecticide; what compounds are likely?" Tags="&lt;solutions&gt;&lt;identification&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28073" PostTypeId="1" AcceptedAnswerId="28076" CreationDate="2015-03-31T00:53:35.140" Score="2" ViewCount="591" Body="&lt;p&gt;Consider a mixture made up of alcohol and water at room temperature. If I freeze this mixture, why does it change its temperature during the phase change? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the difference between the case when you have a pure substance (only water or only alcohol) and that one. How could it be different (I mean, in molecular thinking...)?&lt;/p&gt;&#xA;" OwnerUserId="15280" LastEditorUserId="4934" LastEditDate="2015-03-31T02:25:50.430" LastActivityDate="2015-03-31T03:25:48.103" Title="Why does the mixture's temperature change while phase transition?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28076" PostTypeId="2" ParentId="28073" CreationDate="2015-03-31T03:25:48.103" Score="2" Body="&lt;p&gt;The concentration of water decreases as the mixture cools, so one is left with fairly pure ice, freezing at ~273 K, and pure(r) ethanol, freezing ~159 K (well, the eutectic mixture, about 93% ethanol). Since the ratio of the components are changing as it freezes, there is no distinct freezing point (unless you &lt;em&gt;start&lt;/em&gt; with the eutectic mixture). See the &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Phase_diagram_ethanol_water_s_l_en.svg&quot; rel=&quot;nofollow&quot;&gt;phase diagram for ethanol-water mixture&lt;/a&gt;. Instead, the slush mix has a range of &lt;em&gt;solidus&lt;/em&gt; and &lt;em&gt;liquidus&lt;/em&gt; temperature. BTW, this is a method to make &lt;em&gt;apple jack&lt;/em&gt; or brandy by freezing out water from cider or wine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is noticed with solder, where it crystallizes gradually and dully, except for the sudden freezing of a eutectic mix (~63% Sn, ~37% Pb), making a shiny joint.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-03-31T03:25:48.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28078" PostTypeId="2" ParentId="28072" CreationDate="2015-03-31T03:57:06.417" Score="3" Body="&lt;h2&gt;Ingredients&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So we could probably guess what's in the insecticide based on what's used in some of their other products.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;http://whatsinproducts.com//brands/show_msds/1/10766&quot; rel=&quot;nofollow&quot;&gt;Black Flag Flying Insect Killer&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Tetramethrin - Sounds like the &quot;active ingredient&quot; here. By &quot;active ingredient&quot;, I mean poison.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-phenothrin - Another poison. This product sounds like it kills more than one insect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Petroleum Distilates - Likely a solvent. Unspecified&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquefied Petroleum Gas - Propane or butane. An aerosol propellant. Was your product from a bucket or can? If a can, then it probably has this stuff.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So what happened?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My guess is that one of the products created a &lt;a href=&quot;https://en.wikipedia.org/wiki/Short_circuit&quot; rel=&quot;nofollow&quot;&gt;short circuit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an excessive electric current limited only by the Thévenin equivalent resistance of the rest of the network and potentially causes &lt;strong&gt;circuit damage, overheating, fire or explosion&lt;/strong&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically, when something like a computer experiences a short circuit, it will stop functioning, &quot;die&quot;, once the irreversible damage is done. As specified in the above text, circuit damage is likely done, and the circuits will need to be rewired. If there was no damage, then removing any remaining substances through a solvent, followed by thoroughly drying the computer, would remove the short circuit and allow it to function normally. It's possible that if the short circuit was created by a solvent, that the solvent has already been removed by evaporation, due to common organic solvents' melting points.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-03-31T04:17:14.117" LastActivityDate="2015-03-31T04:17:14.117" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28079" PostTypeId="2" ParentId="28072" CreationDate="2015-03-31T07:18:37.230" Score="0" Body="&lt;p&gt;From the answer of John Snow I saw that it contained &lt;code&gt;Petroleum Distilates&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my years as IT tech it was absolutely forbidden to use any Petrolium/oil based product for cleaning out the computer parts, because they cause corrosion/react with the compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acidity of crude oil based products is just too high. It's also a damn good rust remover(try it, it makes your bike all sparkly new again). Even when refined it's still highly reactive(cleaning petrol, lamp oil, etc...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So spraying it in your computer where there are heaps of soft, easely corrodable metals is generally a bad idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.nl/books?id=ZDPOBQAAQBAJ&amp;amp;pg=PA254&amp;amp;lpg=PA254&amp;amp;dq=how%20does%20crude%20oil%20react%20to%20metals&amp;amp;source=bl&amp;amp;ots=4z5q6PRZTW&amp;amp;sig=LEU07viZC5QfRLe_Zb0gL42Mvzc&amp;amp;hl=nl&amp;amp;sa=X&amp;amp;ei=pkkaVceRFMrgaKqigMAM&amp;amp;ved=0CEUQ6AEwBA#v=onepage&amp;amp;q=how%20does%20crude%20oil%20react%20to%20metals&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;For your reading enjoyment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to fix it? Depends, if pieces have melted I have bad hopes for you being to fix it. You might try dousing it in IsoPropyl Alcohol to flush away any solvents, but i'm afraid the damage has been done and the chips and resistors are dead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to replace resistors if they are damaged. You can recognise those by them bulging out instead of having a flat top.&#xA;Damaged chips are recognisable by being burnt. Might need a magnifying glass to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But usually in these cases the damage is nigh invisible.&lt;/p&gt;&#xA;" OwnerUserId="15286" LastEditorUserId="4356" LastEditDate="2015-03-31T08:05:38.807" LastActivityDate="2015-03-31T08:05:38.807" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28080" PostTypeId="1" AcceptedAnswerId="28131" CreationDate="2015-03-31T07:34:11.640" Score="2" ViewCount="151" Body="&lt;p&gt;I was reading the &lt;a href=&quot;http://en.wikipedia.org/wiki/Indium#History&quot; rel=&quot;nofollow&quot;&gt;History&lt;/a&gt; section about Indium on Wikipedia when I came across the following sentence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1924, indium was found to have a valued ability to stabilize non-ferrous metals, which was the first significant use for the element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if the ability is to improve crystallinity, structural stability of alloys, resistance to corrosion, or something else.&lt;/p&gt;&#xA;" OwnerUserId="12904" LastEditorUserId="418" LastEditDate="2015-03-31T07:49:42.893" LastActivityDate="2015-04-01T22:37:32.350" Title="What exactly is Indium's ability to stabilize non-ferrous metals?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-table&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28081" PostTypeId="2" ParentId="27740" CreationDate="2015-03-31T08:04:46.177" Score="2" Body="&lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt;&lt;br&gt;&#xA;The here used methods are also only approximations and I do not claim that the provided numbers are accurate. However, they are used here to make the point, that bonding situations involving resonance are not easily explained.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Oxygen is slightly more electronegative than chlorine, therefore the bonds will always be slightly polarised towards the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following resonance scheme is based on an NBO analysis on the DF-BP86/def2-SVP level of theory. The percentage value gives an estimate how close the given conformation is to an idealised Lewis structure. From this you could also derive how much such a structure would contribute to a resonance situation (in terms of Valence Bond theory).&lt;br&gt;&#xA;What we do see here is, that the electronic structure is better described with more formal charges. The contribution of a &amp;pi; bond is best described by an ionic donor-acceptor structure, rather than breaking the octet rule at chlorine.&lt;br&gt;&#xA;Noteworthy is also that the doubly ionic donor-acceptor configuration is as equally good a lewis structure than any that breaks the octet rule.&lt;br&gt;&#xA;The worst representation is the one with a four electron three centre bond (which is also not part of the Lewis bonding picture anyway).&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yheC2.jpg&quot; alt=&quot;resonance structures of chlorite anion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/zs41nJdCdCs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Videos&lt;/a&gt; like the one mentioned in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; are easy to follow, but often they are too simplified to be correct.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-03-31T08:04:46.177" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28082" PostTypeId="5" CreationDate="2015-03-31T10:23:05.933" Score="0" Body="&lt;p&gt;This tag should be used when asking about resources to learn about a particular thing. Resources include books, videos, online textual content, journals, etc.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="13357" LastEditDate="2015-03-31T15:00:46.907" LastActivityDate="2015-03-31T15:00:46.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28083" PostTypeId="4" CreationDate="2015-03-31T10:23:05.933" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-03-31T10:23:05.933" LastActivityDate="2015-03-31T10:23:05.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28088" PostTypeId="2" ParentId="28054" CreationDate="2015-03-31T14:09:54.630" Score="3" Body="&lt;p&gt;First of all, when you titrate  a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -&amp;gt; CH3NH3+ }$$The reaction of titration, as you can see is total and quantitative. The constant of the aforementioned equilibrium is:  $$ K=\frac{K_\mathrm b}{K_\mathrm w}= 5\times 10^{10} \gg 10^3$$&#xA;According to the stoichiometry of the neutralization equation, at the equivalence point: $n_{\mathrm{acid}}=n_{\mathrm{base}}$ $$ C_{\mathrm{acid}}.V_{\mathrm{acid}}=C_{\mathrm{base}}.V_{\mathrm{base}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the initial concentration of the base equals the initial concentration of the acid, the volume of the acid at the equivalence point equals the volume of the base:  $$ V_{\mathrm{acid}}^{equiv.}=V_{\mathrm{base}}$$&#xA;You can see that the total volume is doubled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's calculate the $\mathrm{pH}$ at the equivalence point: We have a weak acid at the initial concentration $$C_{\ce{CH3NH3+}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{V_{\mathrm{base}}+V_{\mathrm{acid}}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{2V_{\mathrm{base}}}=\frac{C_{\mathrm{base}}}{2}= \frac{0.13}{2} = \pu{0.065 M}$$&#xA;It's in equilibrium with its conjugated base:&#xA;$$\ce{CH3NH3+&amp;lt;=&amp;gt;H+(aq) + CH3NH2}$$ The acid is partially dissociated:$$ [\ce{CH3NH3+}]= 0.065-x$$ $$ [\ce{CH3NH2}]=x$$ $$ [\ce{H+(aq)}]=x$$&#xA;Let's write the constant of this equilibrium:&#xA;$$K_\mathrm a= 0.2\times 10^{-10}= \frac{x^2}{0.065-x}$$&#xA;We solve this equation of second order to find $x$, the concentration of ion hydronium.&#xA;We find: $\mathrm{pH}= 5.94$ &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2017-04-08T12:29:46.143" LastActivityDate="2017-04-08T12:29:46.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28090" PostTypeId="1" AcceptedAnswerId="28176" CreationDate="2015-03-31T14:31:26.140" Score="6" ViewCount="17466" Body="&lt;p&gt;For carbonates and bicarbonates, I know that stability increases down the group, and for chlorides and fluorides, stability decreases down the group. Why does this happen?  Can someone explain this in detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am talking about S block alkali metals)&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4356" LastEditDate="2015-04-03T00:01:35.940" LastActivityDate="2015-04-03T00:33:59.187" Title="What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides?" Tags="&lt;stability&gt;&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28092" PostTypeId="1" CreationDate="2015-03-31T15:31:55.633" Score="8" ViewCount="102" Body="&lt;p&gt;I have a year-long research project next year in my senior year of high school, in which we have to make some product or prototype to solve some issue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest was in the Liquid Fluoride Thorium Reactor but considering that I do not have the access I need, I opted to invest in the heat it produces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One use as proposed by Kirk Sorensen proposed using the heat from the reactor to fuel the sulfur-iodine cycle to split water into $\ce{H2}$ and $\ce{O2}$.  Then use those combined with carbon collected from the waste of coal and fossil fuel plants and nitrogen from the atmosphere to make methanol dimethyl-ether and ammonia.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher and I have ruled out the first two, but decided that we might be able to make ammonia.  My question is whether it would be feasible and practical to use the sulfur-iodine cycle in combination with the Haber process to split water and make ammonia at a much lower temperature and pressure to ensure safety.&lt;/p&gt;&#xA;" OwnerUserId="5471" LastEditorDisplayName="user15489" LastEditDate="2015-04-30T11:06:32.150" LastActivityDate="2017-05-13T19:27:27.290" Title="Will the sulfur-iodine cycle and Haber Method work at lower pressures and temperatures?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28094" PostTypeId="1" CreationDate="2015-03-31T20:53:43.690" Score="3" ViewCount="4068" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The boiling point of an aqueous solution containing sucrose, &lt;span class=&quot;math-container&quot;&gt;$\ce{C12H22O11}$&lt;/span&gt;, is &lt;span class=&quot;math-container&quot;&gt;$101.45~^\circ\mathrm{C}$&lt;/span&gt;. Calculate the osmotic pressure of this solution at &lt;span class=&quot;math-container&quot;&gt;$35~^\circ\mathrm{C}$&lt;/span&gt;, at which the solution density is &lt;span class=&quot;math-container&quot;&gt;$\pu{1.036 g/mL}$&lt;/span&gt;. The boiling point elevation constant is &lt;span class=&quot;math-container&quot;&gt;$k_b = 0.512~^\circ\pu{C\cdot m^{-1}}$&lt;/span&gt;.&#xA;  The expected answer is&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\Pi = 37.6~\mathrm{atm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried it this way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I used &lt;span class=&quot;math-container&quot;&gt;$\Delta T_b = k_b \cdot m$&lt;/span&gt; and substituted &lt;span class=&quot;math-container&quot;&gt;$T_b = 101.45$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k_b = 0.512$&lt;/span&gt;&#xA;and got &lt;span class=&quot;math-container&quot;&gt;$m = 198.144~\mathrm{mol/kg}$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Density equal to &lt;span class=&quot;math-container&quot;&gt;$\pu{1.036 g/ml}$&lt;/span&gt; equals &lt;span class=&quot;math-container&quot;&gt;$\pu{1.036 kg/L}$&lt;/span&gt;, which means &lt;span class=&quot;math-container&quot;&gt;$\pu{1.036 kg}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1 kg} = \pu{0.965 L}$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M = \pu{198.144 mol}/\pu{0.965 L} = \pu{205.33 mol/L}$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The osmotic pressure equation &lt;span class=&quot;math-container&quot;&gt;$\Pi = MRT = 205.33 \cdot 0.08206 \cdot (35+273)&#xA;  = \pu{640 atm}$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="15293" LastEditorUserId="41556" LastEditDate="2020-05-27T19:33:29.790" LastActivityDate="2020-05-28T13:27:15.817" Title="How to calculate the osmotic pressure of a sucrose solution?" Tags="&lt;solutions&gt;&lt;mole&gt;&lt;osmosis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28095" PostTypeId="2" ParentId="28094" CreationDate="2015-03-31T21:19:48.620" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This should be the first clue that something went wrong.  198 mol of sucrose is about 67 &lt;strong&gt;&lt;em&gt;kilograms&lt;/em&gt;&lt;/strong&gt; of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose.  That can't be right.  (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you forgot to use the &lt;em&gt;change&lt;/em&gt; in boiling point temperature in your $\Delta T_b = k_b m$ equation.  The &lt;em&gt;change&lt;/em&gt; is only 1.45 &amp;deg;C, not 101.45 &amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if you plug in that value and redo your calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true, but keep in mind the definition of &lt;em&gt;molality&lt;/em&gt;: it is moles of solute per kg of &lt;em&gt;solvent&lt;/em&gt;.  However the density is obviously referring to the kg of &lt;em&gt;total solution&lt;/em&gt;.  So, it is inappropriate to find the molarity by simply dividing the molality by the density.  As I described &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26483/whose-concentration-is-more-1-molar-or-1-molal&quot;&gt;in a past answer on chem.se&lt;/a&gt;, if the &lt;em&gt;molarity&lt;/em&gt; is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density.  So you will have to invert this relationship to find the molarity from the molality.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-04-01T03:09:54.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28096" PostTypeId="2" ParentId="27937" CreationDate="2015-03-31T23:20:03.433" Score="0" Body="&lt;p&gt;Here's a great video showing how to do this&#xA;In the video below it says to do 1:2 (1 part soil 2 parts distilled water)  &lt;a href=&quot;https://youtu.be/int0FevLw8I&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/int0FevLw8I&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="554" LastActivityDate="2015-03-31T23:20:03.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28097" PostTypeId="2" ParentId="27899" CreationDate="2015-04-01T02:26:30.317" Score="5" Body="&lt;p&gt;I'll take a stab at this one, but I'm not a battery expert.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What is a lithium &quot;polymer&quot; battery?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_polymer_battery&quot;&gt;Wikipedia&lt;/a&gt; says that &quot;lithium polymer battery&quot; is now commonly used to refer to nearly any lithium ion battery that is encased in a flexible polymer shell instead of a hard metal casing.  There is a chance for confusion here because it apparently used to mean a battery with a polymeric electrolyte instead of a purely small-molecule liquid-phase electrolyte.  I'm assuming you mean the former &quot;in a polymer shell&quot; definition for the purposes of my answer.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What type of anode, cathode, and electrolyte materials?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There are many types of lithium ion batteries.  Electrolyte composition, cathode materials, and anode materials can all vary.  Perhaps the most common is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium_cobalt_oxide&quot;&gt;lithium cobalt oxide&lt;/a&gt; cathode with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite_intercalation_compound&quot;&gt;intercalated graphite&lt;/a&gt; anode.  Organic solvents must be used as electrolyte because water is not stable to the high voltages that lithium ion cells provide.  Solvents stable over a wider potential range must be used instead to dissolve salts that can carry electric current.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What happens when it's crushed:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;So for &lt;em&gt;lithium cobalt oxide batteries with liquid electrolyte&lt;/em&gt;, I was able to find &lt;a href=&quot;http://cdn.intechopen.com/pdfs-wm/10407.pdf&quot;&gt;a good source&lt;/a&gt; describing the problem of &quot;thermal runaway&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The temperature of the cell rises due to the&#xA;  chemical reactions between the organic solvent and electrode materials leading to cell failure/explosion. The chemical reaction is prompted because of the coexistence of 1)&#xA;  combustible organic solvent with inorganic salt electrolyte 2) lithiated graphite anode and 3)&#xA;  partially delithiated LixCoO2 cathode in the charged state. The graphite anode intercalate&#xA;  fairly well during the initial cycles. As the cycle proceeds, the available intercalating sites in&#xA;  the graphite slowly decreases and the lithium deposition over the surface of the anode&#xA;  increases during charging process. Such a situation prompts non-even lithium deposition or&#xA;  even dendrite growth over the graphite surface. When the battery is in the charged state the&#xA;  cathode remains in a delithiated state. Flow of current over and above the&#xA;  tolerable/standard charged state causes instability to the cathode which starts to release&#xA;  oxygen into the electrolyte. Thus released oxygen reacts exothermally with the lithium&#xA;  plated over the graphite anode and increases the temperature of the cell making the cathode&#xA;  to release oxygen further. The cell which was in the dormant state in the absence of oxygen&#xA;  and heat now becomes an explosive device in the event the cell is met with any abuse. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That describes what happens when the battery is overcharged and leaks oxygen into the electrolyte.  I suspect that when the batteries are crushed, even if they aren't overcharged, that at least two things can happen leading to similar thermal runaway: First, the lithiated graphite is air- and water- reactive, heating the now-crushed battery and eventually lighting the electrolyte solvent on fire.  Second, the cell may be shorted by the mechanical disruption, so that even without air or water contact, the voltage in the battery can discharge very quickly.  This would create large heating from ohmic resistance, also eventually igniting the elctrolyte.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-01T02:26:30.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28101" PostTypeId="2" ParentId="26775" CreationDate="2015-04-01T05:44:04.750" Score="1" Body="&lt;p&gt;Alkanes react only with very powerful oxidazing agents, like chlorine or oxygen, because they are very unreactive (due to their weak polarity).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yes, there are such principles in organic chemistry and they are based most of the times on the reactivity of the substante. (acidity, polarity, aromaticity).&lt;/p&gt;&#xA;" OwnerUserId="9297" LastActivityDate="2015-04-01T05:44:04.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28102" PostTypeId="2" ParentId="26775" CreationDate="2015-04-01T05:46:45.483" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I came across this question where I had to convert&#xA;  $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$. If I need to do so can I&#xA;  just react the given compound with water [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Always be cautious about unwanted, side reactions that may occur. Proton exchanges are one huge class of possible side reactions. Protons transfers occur readily in many circumstances because the protons is small and can be shuffled around without much to get in its way, physically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the case of your hypothetical reaction between a Grignard reagent and water - realize that Grignards are strongly basic and therefore will deprotonate water. Unfortunately protonated Grignards are no good in terms of reactivity. They just fall back apart into hydrocarbons and a basic halide salt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Further reading regarding Grignards&lt;/em&gt; - It is tempting to use Grignards to create carbon-carbon bonds, but anything that is acidic as water or more acidic will be deprotonated by the Grignard before any carbon-carbon bonds are created. When preparing Grignards, glassware must be perfectly dry. Not just wiped dry with a paper towel but heated dry. Grignards also live in ether solvents since ethers are only very weakly acidic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grignards may be used to attack aldehydes and ketones in creating carbon-carbon bonds, but realize that if the aldehyde or ketone contains a functional group that is about as acidic as water or more so - then the first equivalent of Grignard you add isn't going to the aldehyde or ketone carbon - it's going to perform a proton-transfer. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any set of principles in organic chemistry which govern as to&#xA;  which substances can be replaced with what?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, it is good to keep in mind the motif &lt;em&gt;nucleophile attacks electrophile&lt;/em&gt; in introductory organic chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15604/exceptions-to-motif-nucleophile-attacks-electrophile&quot;&gt;Exceptions to motif: nucleophile attacks electrophile&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other heuristics you can use in predicting what reactions might commence ... &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bond energies. This is mainly useful for studying radical reactions - to which the motif nucleophile attacks electrophile doesn't apply as well. Are we going from weaker to stronger bonds? That's a favorable move, enthalpically. &lt;/li&gt;&#xA;&lt;li&gt;Polarities. This falls under nucleophile attacks electrophile. &lt;/li&gt;&#xA;&lt;li&gt;Kinetics. It's good to learn the difference between thermodynamics and kinetics early in organic chemistry. Sometimes chemical entities may be thermodynamically unstable but kinetically stable. Maybe a carbon is particularly electrophilic, but it's too well &quot;hidden&quot; by neighboring groups (steric hindrance). So will a nucleophile attack this carbon? Probably not to a great extent. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-04-01T20:56:46.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28108" PostTypeId="2" ParentId="28041" CreationDate="2015-04-01T06:41:38.793" Score="7" Body="&lt;p&gt;There is a class of compounds, called carbodiphosphoranes, which formally has a divalent carbon&amp;nbsp;(0) central atom. Employed as a ligand it acts as a sigma and pi donor. This capability is also one of the major features to distinguish these compounds and derivatives from carbenes, which are sigma donors and pi acceptors. The following is reproduced from &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.201200741/abstract&quot; rel=&quot;noreferrer&quot;&gt;Thi Ai Nhung Nguyen and Gernot Frenking, &lt;em&gt;Chem. Eur. J.&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;18&lt;/em&gt;, 12733-1748&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x5LyZ.jpg&quot; alt=&quot;carboduphosphoranes and N-heterocyclic carbenes (and their wolfram carbonyl complexes)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly carbodicarbenes (substitute the phosphine with a carbene), also known as bent allenes, are also double donors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This post shall only serve as an example, I am quite certain, there are more compound with this kind of bonding motive out there. Nicolau mentioned $\ce{OsO3N^{-}}$ and other &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_nitrido_complex&quot; rel=&quot;noreferrer&quot;&gt;metal nitrido complexes&lt;/a&gt;, which he will hopefully extend in an answer.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-04-01T06:41:38.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28111" PostTypeId="2" ParentId="28090" CreationDate="2015-04-01T11:54:51.920" Score="1" Body="&lt;p&gt;By Fajan's Rule you should be getting the answer and then more electropositive metal will have more ionic character and then that will increase stability.&lt;/p&gt;&#xA;" OwnerUserId="15305" LastActivityDate="2015-04-01T11:54:51.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28112" PostTypeId="1" CreationDate="2015-04-01T12:18:03.300" Score="2" ViewCount="602" Body="&lt;p&gt;A $52\ \mathrm{g}$ piece of iron at $250\ \mathrm{^\circ C}$ is added to a mixture of $100\ \mathrm{g}$ of ice and $100\ \mathrm{g}$ of liquid water at thermal equilibrium $(0\ \mathrm{^\circ C})$. Calculate the mass of ice that melts (not all of the ice will melt, so you know the final temp of everything mixed together).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the temperature of the mixture should stay same, if is at thermal equilibrium. Because then it less willing to accept change in temperature. But I am not sure how to use heat of fusion and heat capacity of liquid water to solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ lost from iron = $Q$ gained by water + ice mixture  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(0.93\ \mathrm{mol}\ \ce{Fe})(25.1\ \mathrm{J\ mol^{-1}\ K^{-1}})(T_\text{f}-250\ \mathrm{^\circ C}) = -(200\ \mathrm{g}\ \text{mixture})(\text{what is }C\text{?})(T_\text{f}-0\ \mathrm{^\circ C})$&lt;/p&gt;&#xA;" OwnerUserId="15306" LastEditorUserId="39539" LastEditDate="2017-05-04T01:54:14.967" LastActivityDate="2017-05-04T01:54:14.967" Title="heat capacity and thermal equilibrium" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28114" PostTypeId="1" AcceptedAnswerId="28741" CreationDate="2015-04-01T13:22:11.677" Score="11" ViewCount="650" Body="&lt;p&gt;The Royal Institution seems to claim that they still possess some of Faraday's original gold colloids: &lt;a href=&quot;http://www.rigb.org/our-history/iconic-objects/iconic-objects-list/faraday-gold-colloids&quot;&gt;&quot;These liquids are some of the first examples of metallic gold colloids, made by Michael Faraday over 150 years ago.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a publication in Gold Bulletin (&lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF03215598#page-1&quot;&gt;D. Thompson, Gold Bull. 2007, 40(4), 267-269.&lt;/a&gt;), the following is stated: &quot;Faraday's original gold colloids were very stable and at least one example survived until the second World War, when it was accidentally destroyed by enemy action [...].&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do Faraday's gold colloidal solutions still exist?&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2015-04-15T05:00:37.190" Title="Do Faraday's gold colloidal solutions still exist?" Tags="&lt;history-of-chemistry&gt;&lt;colloids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28117" PostTypeId="2" ParentId="28072" CreationDate="2015-04-01T14:49:36.753" Score="1" Body="&lt;p&gt;The only reason for a short could be that by spraying a lot &#xA;of that can some moisture from air condensed due to the &#xA;refrigerating action of the spray. &#xA;But: Does the computer start again after some time has passed now? &#xA;Usually all supply voltages in Computers are short-Proof &#xA;and shut down on a short. &#xA;After dryinfg the computer should start again.&lt;/p&gt;&#xA;" OwnerUserId="105" LastActivityDate="2015-04-01T14:49:36.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28121" PostTypeId="1" CreationDate="2015-04-01T17:04:21.547" Score="1" ViewCount="80" Body="&lt;p&gt;This is a question in a problem sheet I have been set. Is it do do with the following equation: $\Lambda=\Lambda_0-a\sqrt c$? Surely charge density is proportional to concentration so therefore molar conductivity would decrease linearly with the square root of concentration (and $c$ $\alpha$ charge density?). Also, what is the physical origin of this equation; my lecture notes simply say that the ions interact with each other, is it as simple as that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part to this question asks &quot;Does the actual conductivity decrease?&quot;. Presumably by &quot;actual conductivity&quot;, he means $\kappa$ where $\kappa=\Lambda c$. I have no idea about this part so any help would be much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-01T17:04:21.547" Title="Why does the molar conductivity decrease with increasing charge density?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28122" PostTypeId="1" CreationDate="2015-04-01T18:01:55.743" Score="5" ViewCount="1200" Body="&lt;p&gt;I have calcium chloride in the form of small (1-2 mm) round white pellets, originally bought at a home-brew shop. The packaging gives no indication what hydrate it is. Since each water molecule adds ~16% of the weight of anhydrous CaCl&lt;sub&gt;2&lt;/sub&gt; this introduces a big error when using it to treat water to a specified amount (in ppm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally I figured I could throw it in the toaster-oven for a while at 350°F and weigh it every 15 minutes, to figure out how much water was driven off. Results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 min.  -  1.15 g.&#xA;15 min. -  0.92 g.&#xA;30 min. -  0.90 g.&#xA;45 min. -  0.90 g.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I originally assumed this would be anhydrous, but after &lt;a href=&quot;http://en.wikipedia.org/wiki/Calcium_chloride&quot; rel=&quot;noreferrer&quot;&gt;reading&lt;/a&gt; that calcium chloride dihydrate decomposes at 347°F, and calcium chloride monohydrate decomposes at 500°F, I'm not so sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find any other information on figuring out the hydrate online  or indicating what state calcium chloride is typically in (at least that I could understand). Am I correct in thinking I'm left with the monohydrate? Or is it actually anhydrous?&lt;/p&gt;&#xA;" OwnerUserId="13178" LastActivityDate="2017-02-18T09:16:54.473" Title="Determining which hydrate of calcium chloride I have?" Tags="&lt;water&gt;&lt;home-experiment&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28123" PostTypeId="1" AcceptedAnswerId="28130" CreationDate="2015-04-01T18:52:27.517" Score="7" ViewCount="2114" Body="&lt;p&gt;What is palindromic DNA and why isn't every complementary strand palindromic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know AGCT is palindromic, but what is an example of a strand that isn't?&lt;/p&gt;&#xA;" OwnerUserId="15312" LastEditorUserId="418" LastEditDate="2015-04-01T19:45:47.773" LastActivityDate="2015-04-01T22:34:27.127" Title="What is palindromic DNA?" Tags="&lt;dna-rna&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28124" PostTypeId="1" AcceptedAnswerId="28129" CreationDate="2015-04-01T19:02:12.777" Score="3" ViewCount="20468" Body="&lt;p&gt;Which elements of Group 15 of the periodic table have allotropes?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://periodictable.com/Properties/A/AllotropeNames.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;P, As and Sb have well-known allotropes&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P: White, Red and Black;&lt;br&gt;&#xA;As: Yellow, Black and Gray (Grey);&lt;br&gt;&#xA;Sb: White, Yellow and Black.          &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What about Nitrogen and Bismuth?&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The same &lt;a href=&quot;http://periodictable.com/Properties/A/AllotropeNames.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt; suggests that Nitrogen has an allotrope named &lt;a href=&quot;http://periodictable.com/Elements/007/data.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dinitrogen&lt;/a&gt; while Bismuth has none.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_15%3A_The_Nitrogen_Family#Nitrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemwiki&lt;/a&gt; says:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{N2}$ does not have any allotropes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and doesn't comment on Bismuth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Royal Society of Chemistry says yes for &lt;a href=&quot;http://www.rsc.org/periodic-table/element/7/nitrogen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nitrogen&lt;/a&gt; and no for &lt;a href=&quot;http://www.rsc.org/periodic-table/element/83/bismuth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bismuth&lt;/a&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/9780470145210.ch10/summary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inorganic Reactions and Methods: The Formation of Bonds to N, P, As, Sb, Bi (Part 1), Volume 7&lt;/a&gt; has an article which suggests that Bismuth allotrope exists.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://sg.answers.yahoo.com/question/index?qid=20070401182351AA9cWIv&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; not so reliable source says no for Bismuth.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia doesn't mention anything for Bismuth but for &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Nitrogen#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nitrogen&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At atmospheric pressure, molecular nitrogen condenses (liquefies) at $\pu{77 K}$ ($\pu{−195.79^\circ C}$) and freezes at $\pu{63 K}$ ($\pu{−210.01^\circ C}$) into the beta hexagonal close-packed crystal allotropic form.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unstable allotropes of nitrogen consisting of more than two nitrogen atoms have been produced in the laboratory, like $\ce{N3}$ and $\ce{N4}$. Other (as yet unsynthesized) allotropes include hexazine and octaazacubane.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to me, nitrogen has allotropes and bismuth doesn't, but the above sources contradict. Any clarifications and reasons for them showing or not showing allotropy would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorUserId="41328" LastEditDate="2017-08-06T10:23:00.517" LastActivityDate="2017-08-13T02:30:47.133" Title="Allotropy in Nitrogen and Bismuth of Group 15" Tags="&lt;inorganic-chemistry&gt;&lt;allotropes&gt;&lt;pnictogen&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28126" PostTypeId="2" ParentId="28123" CreationDate="2015-04-01T19:43:50.323" Score="3" Body="&lt;p&gt;It has nothing to do with the complementarity (of the other strand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try this to figure out whether some sequences are palindromic:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python3&#xA;def is_palindromic(seq):&#xA;    translation_table = str.maketrans('ACGT', 'TGCA')&#xA;    translation = seq.translate(translation_table)&#xA;    #print(seq, translation[::-1])&#xA;    return seq == translation[::-1]&#xA;&#xA;is_palindromic('GATTACA')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-01T19:43:50.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28128" PostTypeId="2" ParentId="28041" CreationDate="2015-04-01T20:42:58.897" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Water has 2 lone pairs on O, so can it donate both (one as a pi bond) just like those compounds? If not, why?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I posted:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;It is possible for pi-donating ligands, like $\ce{NH2^{-}}$ (amide anion), but not really possible for water (neutral, oxygen more electronegative).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pi-donation from water would lead to oxygen having +2 formal charge, thus becoming very acidic and losing proton - turning into hydroxide ligand, but I think it's the other way around; it first loses proton and then hydroxide pi-donates (becomes doubly bonded). Therefore &lt;a href=&quot;https://en.wikipedia.org/wiki/Metal%E2%80%93ligand_multiple_bond#Pi-donor_ligands&quot; rel=&quot;nofollow&quot;&gt;pi-donor ligands&lt;/a&gt;, are often anions. Also unsaturated ligands have this property.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-02-04T17:08:59.847" LastActivityDate="2016-02-04T17:08:59.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28129" PostTypeId="2" ParentId="28124" CreationDate="2015-04-01T21:12:12.970" Score="7" Body="&lt;p&gt;Bismuth certainly has other known crystal structures at elevated pressures and temperatures, at least 4 others in addition to the rhombohedral structure stable at room temperature and pressure.  One place to start would be &lt;a href=&quot;https://str.llnl.gov/str/JulAug06/Lorenzana.html&quot;&gt;an article from NIF&lt;/a&gt; on shock physics of bismuth. For a scholarly article, you might start with 'Phase Diagrams of Arsenic, Antimony, and Bismuth at Pressures up to 70 kbars', W. Klement et al., Phys Rev 131(2) 632-637 (1963).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for nitrogen, it too would appear to exhibit a variety of solid diatomic phases under pressure.  A starting point to explore might be 'High-pressure amorphous nitrogen', E. Gergoryanz et al., Phys Rev. B64 052103 (2001), which mentions a variety of phases in addition to the titular amorphous phase they explore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both Physical Review articles should be available to the general public...&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-04-01T21:12:12.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28130" PostTypeId="2" ParentId="28123" CreationDate="2015-04-01T22:34:27.127" Score="7" Body="&lt;p&gt;A palindromic stretch of DNA is a strand whose reverse complement is itself.  So &lt;code&gt;5'-AAAT-3'&lt;/code&gt; is not palindromic.  It's reverse complement is &lt;code&gt;5'-ATTT-3'&lt;/code&gt;.  Those two pieces of DNA are not identical.  However, &lt;code&gt;5'-GGATCC-3'&lt;/code&gt; is palindromic, because the reverse complement is identical.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-01T22:34:27.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28131" PostTypeId="2" ParentId="28080" CreationDate="2015-04-01T22:37:32.350" Score="0" Body="&lt;p&gt;All I did was click on the reference from the History wikipedia link above for the sentence you were talking about. The sentence comes from a paragraph of a manuscript which is found here: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed011p270&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ed011p270&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the bolded part, that was what is meant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;&lt;/strong&gt; Its[Indium's] identification as well as its name came from the beautiful indigo-blue line which it so prominently displays in its spectrum. Its commercial development came as the result of the knowledge that it should be a valuable stabilizer of non-ferrous metals &lt;strong&gt;&lt;em&gt;much as chromium is a stabilizer of the ferrous metals.&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Above is quote from: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;French, Sidney J. (1934). &quot;A story of indium&quot;. Journal of Chemical Education 11 (5): 270. Bibcode:1934JChEd..11..270F. doi:10.1021/ed011p270.&lt;/p&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2015-04-01T22:37:32.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28132" PostTypeId="1" CreationDate="2015-04-01T23:27:20.907" Score="6" ViewCount="2802" Body="&lt;p&gt;How do you use charge density ($Z/r$) as a measurement of polarising power and polarisability, and hence hard/soft acid character?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've been looking at different polarising power and polarisability and how that affects a substance in whether it is a hard or soft acid/base, but I don't understand how charge density comes into play, and how we can use it to understand the characteristics.&lt;/p&gt;&#xA;" OwnerUserId="15317" LastEditorUserId="194" LastEditDate="2018-01-23T16:32:03.743" LastActivityDate="2018-01-23T16:36:43.233" Title="How is charge density a useful measure of polarising power and polarisability?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;hsab&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28134" PostTypeId="2" ParentId="24791" CreationDate="2015-04-02T01:03:42.010" Score="21" Body="&lt;p&gt;As cibrail said, DNA is a polymer of nucleotides. They join themselves through phosphodiester bonds (a specific kind of covalent bond) that can grow to as much as millions of nucleotides.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What part in the strand contributes to the overall non neutral charge?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reason why DNA is negatively charged is the phosphate group that makes up every nucleotide (pentose + nitrogenous base + phosphate). When forming part of the phosphodiester bond, they retain 1 of 2 negative charges (the other being lost to form the other ester bond to a new pentose, that's why the bond is called &quot;phospho-di-ester&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJqx9.jpg&quot; alt=&quot;Phosphodiester bond&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lehninger 5th edition. Part of figure 25-5&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;DNA is not isolated in the body, so what keeps it stable while being charged?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The main molecules in charge of that are proteins called &lt;strong&gt;histones&lt;/strong&gt;. They have alkali features and positive charge, and so, they neutralize these negative charges. On the other hand, that's not the main point of histones. Histones play a really important key role in the regulation and compaction of DNA within the nucleus of the cell, and are important targets of diverse molecules that want to alter DNA structure and function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xnUGe.jpg&quot; alt=&quot;Histones&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lehninger 5th edition.Figure 24-26&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why is it important for DNA to be charged?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It has great importance because cell machinery is adapted to that property. If DNA wasn't negatively charged, the structural fundamentals of interactions between DNA and transcription factors would be completely different. &#xA;Thanks to the fact that it is negatively charged, it becomes spontaneously pretty straight when denaturalised, because negative charges repel themselves. It forms very wide curves.&#xA;And of course, histones wouldn't be able to join DNA and exert their key regulating functions.&lt;/p&gt;&#xA;" OwnerUserId="15318" LastActivityDate="2015-04-02T01:03:42.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28137" PostTypeId="1" AcceptedAnswerId="28170" CreationDate="2015-04-02T03:06:46.460" Score="4" ViewCount="2110" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Cellulose&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, a single cellulose molecule is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a linear chain of several hundred to many thousands of &amp;beta;(1&amp;rightarrow;4) linked&#xA;  D-glucose units.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which means a cellulose molecule is made of ~1&amp;mu;m of glucose units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471809101&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trotman, 1984&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each cotton fibre is a unicellular hair [...] Each fibre, however long&#xA;  or short it may be, consists of one single complete vegetable cell&#xA;  [...] Average fibre length 13-51 mm.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(6th ed., pages 27-33)&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/NLKcF.png&quot; alt=&quot;Cotton fibre morphology&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(source: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471809101&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trotman, page 31&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if cellulose molecules are only ~1&amp;mu;m in length, what forces keep the cellulose molecules together inside the S1,S2,S3 structures throughout the fiber? How ~1&amp;mu;m units end up giving cotton its mechanical properties?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-04-02T21:23:01.257" LastActivityDate="2015-04-02T22:33:26.887" Title="What forces keep a cotton fiber together?" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28138" PostTypeId="1" AcceptedAnswerId="28144" CreationDate="2015-04-02T03:13:35.700" Score="1" ViewCount="544" Body="&lt;p&gt;I'm exploring the &lt;a href=&quot;http://en.wikipedia.org/wiki/Platonic_hydrocarbon&quot; rel=&quot;nofollow&quot;&gt;platonic hydrocarbons&lt;/a&gt; and one of the methods I'd like to use is to compare their enthalpies of formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that carbon's standard form is a covalent lattice and I can't think of a way to efficiently obtain its electronic and thermal enthalpies without resorting to optimising a giant &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycyclic_aromatic_hydrocarbon&quot; rel=&quot;nofollow&quot;&gt;PAH&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW I'm using GaussView/Gaussian.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="186" LastEditDate="2015-04-02T09:27:22.763" LastActivityDate="2015-04-04T12:36:08.463" Title="Is there a way to efficiently estimate the enthalpy of formation for an organic molecule in computational chemistry?" Tags="&lt;thermodynamics&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28140" PostTypeId="1" AcceptedAnswerId="28164" CreationDate="2015-04-02T05:26:20.553" Score="1" ViewCount="2073" Body="&lt;p&gt;What happens to a complexometric titration of calcium carbonate (analyte) with $\ce{EDTA}$ (titrant) when no magnesium(II) chloride crystals are added to the buffer ($\ce{NH3/NH4^+ + EDTA}$) solution? In a water analysis (where water sample is the analyte and $\ce{EDTA}$ is the titrant)?&lt;/p&gt;&#xA;" OwnerUserId="7921" LastEditorUserId="4231" LastEditDate="2015-04-02T19:48:53.773" LastActivityDate="2015-04-02T19:48:53.773" Title="What happens if don't use MgCl₂ in buffer solution in Ca²⁺/Mg²⁺ titration?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28142" PostTypeId="1" AcceptedAnswerId="28152" CreationDate="2015-04-02T07:40:42.487" Score="2" ViewCount="320" Body="&lt;p&gt;So, after dissolving a good portion of my brain by watching a ton of sci-fi, I started thinking about all of the strange new materials that we will supposedly have in the future. For example, the ever-present &quot;di-lithium&quot; in Star Trek. But, as I started thinking about it, I thought it was a little ridiculous (I know science-FICTION) that all these exotic elements could exist. No matter where a species comes from in the galaxy or universe, they would be working with the exact same Table of Elements as us. Furthermore, so long as they are similar to human-life, i.e. they breathe oxygen, need water, etc., and therefore live on similar planets, wouldn't the compounds be the same as well? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we are 250 million years into the future and have discovered superluminal space-flight at 100,000,000c. Say we cruise around to inhabitable planets throughout the galaxy and universe. Upon landing on these new worlds, the elements and compounds we find would be the same that occur on Earth, right? Would we still find diamonds, rubies, uranium, iron, etc.? Or is it possible that there are combinations of the elements yet unknown? Is it for certain that we have catalogued every element and isotope that occurs naturally? I understand that new compounds could be formed, but as far as the base elements and the naturally occurring minerals, have we found them all?&lt;/p&gt;&#xA;" OwnerUserId="15216" LastEditorUserId="16683" LastEditDate="2017-02-08T18:54:12.053" LastActivityDate="2017-02-08T18:54:12.053" Title="Could we possibly discover new extraterrestrial elements and minerals?" Tags="&lt;chemistry-in-fiction&gt;&lt;astrochemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28143" PostTypeId="1" CreationDate="2015-04-02T08:16:17.130" Score="6" ViewCount="8070" Body="&lt;p&gt;I’m heating up several metal blocks as part of an experiment to determine their specific heat capacities. In a nutshell, I’m heating the blocks of known mass to 100 °C, then adding them to an insulated vessel containing water of known temperature and measuring the increase. Pretty basic stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m trying to work out how I can measure the temperature of the blocks, so that I know they’re at 100 °C. Leaving them in the boiling water for a long period of time is not ideal – I want to be sure that they are at the desired temperature that I can record.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this effectively? I thought about using an infrared thermometer, but wouldn’t that just measure the temperature on the surface of the block?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought about something to do with measuring the resistance of the metals, but is this actually accurate and easy to do? Any advice or suggestions are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14994" LastEditorDisplayName="user7951" LastEditDate="2015-04-02T09:04:05.317" LastActivityDate="2016-05-03T15:08:33.763" Title="How can I measure the internal temperature of a solid metal?" Tags="&lt;experimental-chemistry&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28144" PostTypeId="2" ParentId="28138" CreationDate="2015-04-02T09:20:37.273" Score="4" Body="&lt;p&gt;I would say &lt;em&gt;yes&lt;/em&gt;, you could estimate standard enthalpies of formation for molecules more or less accurately by application of quantum and statistical mechanics, although there is a &lt;em&gt;'but'&lt;/em&gt;: the usual procedure works for &lt;em&gt;gas-phase&lt;/em&gt; heats of formation only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus, if the standard state of the material of interest corresponds to the &lt;em&gt;condensed phase&lt;/em&gt;, you have to resort to some procedure which allows you to calculate a condensed phase heat of formation given the corresponding gas-phase one. Such procedures are based on a trivial application of the infamous &lt;a href=&quot;http://en.wikipedia.org/wiki/Hess&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hess's law&lt;/a&gt;, which for the case of solids can be written as follows&#xA;$$&#xA;   \Delta H_{\mathrm{solid}} = \Delta H_{\mathrm{gas}} - \Delta H_{\mathrm{subl}} \, ,&#xA;$$&#xA;where $\Delta H_{\mathrm{subl}}$, the heat of sublimation, is estimated usually from experiementally available data on simillar compounds by a correlation scheme. However, as with any empirically fitted scheme the reliability and the accuracy of any such approach are doubtful.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively, in many cases, it is enough to compare the gas-phase heat of formations, even if the standard state of material is solid or liquid. To calculate the gas-phase heat of formation for a substance in Gaussian all you need is to run frequencies calculations on the previously optimised geometry of a single molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like to give the gas-phase heats of formations a try, I suggest running G4 calculations first. In the route section just write&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# G4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and that would do it. Alternatively, in GaussView's Gaussian Calculation Setup window first choose &quot;Energy&quot; at the &quot;Job Type&quot; tab and then at the &quot;Method&quot; tab choose &quot;Compound...&quot; in the second combobox and &quot;G4&quot; in the third one as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMogv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-04-04T12:36:08.463" LastActivityDate="2015-04-04T12:36:08.463" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28146" PostTypeId="2" ParentId="27953" CreationDate="2015-04-02T10:09:56.087" Score="1" Body="&lt;p&gt;You are right thinking that the molar concentration of S will be the same as the molar concentration of SO4, unless there is no other molecule whith S in the sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is 1 mole of molecules of sulphate (SO4), there will be exactly 1 mole of atoms of sulfur (S), and 4 moles of atoms of oxygen (O).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, though &quot;theoretically&quot; it is correct to say there will be 2 moles of molecules of O2, it is really not correct, because there is absolutely no molecule of O2 in your sample. The amount of oxygen atoms is the same in both cases, but how they are organized is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get the mass concentration of S, say in g/l, you would just multiply the concentration of SO4 in g/l by the mass ratium of sulfur within sulphate, (sulfur mass over sulphate mass, 32/96=0.33333). Since for every sulphate gram there is &lt;strong&gt;1/3 sulfur mass&lt;/strong&gt; and &lt;strong&gt;2/3 oxygen mass&lt;/strong&gt;, 1/3 of your sulphate mass is due to sulfur.&#xA;Another strategy would be to get the concentration of sulphate, which is exactly the same as that of sulfur (1 mol of atoms of S for every mol of molecules of SO4), and multiply that by the molar mas of the sulfur atom (32).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, the &lt;strong&gt;molar ratio&lt;/strong&gt; between S and SO4 is 1, and the mass ratio is 1/3 grams of sulfur for every gram of sulphate&lt;/p&gt;&#xA;" OwnerUserId="15318" LastEditorUserId="15318" LastEditDate="2015-04-02T14:07:56.207" LastActivityDate="2015-04-02T14:07:56.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28147" PostTypeId="2" ParentId="28132" CreationDate="2015-04-02T12:41:46.123" Score="1" Body="&lt;p&gt;What does higher charge density of a cation means? It means that the positive charge on a cation is concentrated in a small volume. Small, concentrated charge can cause greater distortion in the anion and therefore they have greater polarising power, or greater power to cause attract the negative electron cloud of the anion and hence greater distortion. The greater the positive charge on a cation, the larger the charge density will be (it is directly proportional to $Z$). Greater positive charge will cause greater distortion in the electron cloud of anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do small and concentrated charge cause greater distortion? What will affect you more? A force applied on a very very small area of your body, or the same amount of force applied on whole of your body? Of course, the former one will affect us more, i.e., a small and concentrated force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the anion? When the electron cloud of the anion is small, it is closer to the nucleus and experiences a greater attraction from the nucleus, therefore it is not easy to distort it. But, if the electron cloud is large, the inward force experienced by it from the nucleus is less, hence it can be polarised much easily. The inference is that the greater the radius, the larger the polarisability of the anion will be. What does greater radius mean? The greater the radius, the smaller the charge density will be (it's inversely proportional to $r$, or proportional to $1/r$).&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="194" LastEditDate="2018-01-23T16:36:43.233" LastActivityDate="2018-01-23T16:36:43.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28148" PostTypeId="2" ParentId="27835" CreationDate="2015-04-02T13:13:54.627" Score="6" Body="&lt;p&gt;Yes it is possible. But it must fulfil some basic condition for that shift.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The empty p-orbital of the carbocation must be in plane with the $\ce{C-H}$ or the $\ce{C-C}$ bond. When the alignment of a carbon–hydrogen bond with a vacant p orbital takes place allowing for hyperconjugation, a &quot;pseudo-double-bond&quot; develops. As illustrated in Figure 5.13, this can be envisioned as a double bond with a closely associated hydrogen ion.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If, as shown in Figure 5.13, hyperconjugation results in the formation of species possessing both double-bond character and associated hydrogen ions, equilibrium-controlled migration of the associated hydrogen ion can be expected. This transformation, shown in Scheme 5.8, is known as a 1,2-hydride shift and results in the migration of a proton&#xA;  from carbon 1 to carbon 2.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;While the example illustrated in Scheme 5.8 shows equilibrium between two chemically identical carbocations, there are factors influencing the direction of these transformations when applied to more complex systems. If we consider Scheme 5.9, we notice that the positive charge migrates exclusively to the tertiary center, reflecting the increased stability of tertiary carbocations over primary carbocations. In general, where 1,2-hydride shifts are possible, rearrangement of less stable carbocations to more stable carbocations is expected.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is important to remember that hydride shifts occur much more readily than the corresponding alkyl shifts. In fact,as a general rule, alkyl shifts will not occur unless a hydride shift cannot take place. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Alkyl migrations occur when the resulting carbocation is more stable than the starting carbocation. But for hydride shift its not such a hard and fast rule.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dOO8D.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bipgj.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7GgNu.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &quot;Arrow Pushing in Organic Chemistry, An Easy Approach to Understanding Reaction Mechanisms&quot; by Daniel E. Levy; Copyright 2008 by John Wiley &amp;amp; Sons, Inc.; page 110 in chapter 6&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="5026" LastEditDate="2018-03-13T14:02:40.680" LastActivityDate="2018-03-13T14:02:40.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28149" PostTypeId="2" ParentId="28143" CreationDate="2015-04-02T13:19:51.470" Score="2" Body="&lt;p&gt;You can use ultrasound thermometry. It uses change in speed of sound under heat. AFAIK, there are industrial-grade solutions on the market.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-04-02T13:19:51.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28150" PostTypeId="1" AcceptedAnswerId="28160" CreationDate="2015-04-02T13:40:59.463" Score="0" ViewCount="2610" Body="&lt;p&gt;Hydrogen and hydroxide both exceed sodium and chlorine in terms of reduction and oxidation potential respectively.&lt;/p&gt;&#xA;&lt;p&gt;While electrolyzing a concentrated solution of aqueous NaCl, it is known that chlorine is discharged at the anode (contaminated with traces of oxygen). However, at the cathode, hydrogen is charged in place of sodium. Why is this so? I have read many explanations, but they never provide the reasoning.&lt;/p&gt;&#xA;&lt;p&gt;I suspect these particular properties of factoring in preferential discharge:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Atomic mass (and consequently weight).&lt;/li&gt;&#xA;&lt;li&gt;Magnitude of oxidation &amp;amp; reduction potentials.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14911" LastEditorUserId="14911" LastEditDate="2021-03-05T09:51:26.543" LastActivityDate="2021-03-05T09:51:26.543" Title="While electrolyzing concentrated aqueous sodium chloride, why is it that chlorine is discharged but not sodium?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28151" PostTypeId="1" CreationDate="2015-04-02T13:44:48.820" Score="1" ViewCount="242" Body="&lt;p&gt;Which orbitals are filled? Is it diamagnetic? According to CFT, $\ce{CO}$ is a strong field ligand, so will it pair the electrons and be diamagnetic?&lt;/p&gt;&#xA;" OwnerUserId="15325" LastEditorUserId="4231" LastEditDate="2015-04-02T15:39:13.263" LastActivityDate="2015-04-02T15:39:13.263" Title="Orbitals involved in Ni(CO)₄" Tags="&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;magnetism&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28152" PostTypeId="2" ParentId="28142" CreationDate="2015-04-02T13:45:53.580" Score="9" Body="&lt;p&gt;One thing which can be said with &lt;strong&gt;absolute certainty&lt;/strong&gt; is that no species  in the Universe (or even all their combined efforts) will &lt;em&gt;even come close&lt;/em&gt; to exhausting the daunting enormity of chemical space. When you drag combinatorics into a problem, you can easily stumble on massive numbers, and &lt;a href=&quot;http://www.rsc.org/chemistryworld/2014/10/pipeline-chemical-space-big&quot; rel=&quot;nofollow noreferrer&quot;&gt;this is the case here&lt;/a&gt;; it has been roughly calculated that there are on the order of $10^{60}$ &quot;chemically reasonable&quot; distinct molecules below a molar mass of $\mathrm{500\ g\ mol^{-1}}$! This means that &lt;strong&gt;almost all chemical substances which can exist, never will exist&lt;/strong&gt;, be it from natural or artificial production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how far chemistry develops, we will always be interested in investigating the chemical composition of new samples, and will always find many new compounds doing so. Most of the time we'll find the same of what we do on Earth; there's no escaping the fact that $\ce{N_2}$ and $\ce{Al2O3}$ are very stable substances, for example, and can exist in a wide range of conditions. But there will always be something new for us waiting out there. Organic substances are tremendously varied thanks to the concatenability of carbon atoms. One might think inorganic substances are far less variable, but just the class of silicate minerals is enormous, and many polymorphs of the same substances can exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, at least one area of chemistry seems far closer to being exhausted; the elements themselves. There is no possibility of us having &quot;missed&quot; any element among the first 112, and we're still filling in the gaps at higher atomic numbers, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/8073/the-last-elements-atomic-number/8077#8077&quot;&gt;but each heavier element discovered involves ever more complicated procedures and lower yields&lt;/a&gt;. Much of the experimentally determined properties of elements above $\mathrm{Z=104}$ come from the analysis of a tiny amount of atoms, as little as tens of them. Each new element is significantly more difficult to synthesize, and is less stable, which is hampering progress. We don't yet know very well how to get past $\mathrm{Z=122}$ or so even in principle, and are unsure how accessible the ultraheavy elements will prove to be. There is still great academic interest in several aspects of the elements and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26611/does-hydrogen-really-have-seven-isotopes/26614#26614&quot;&gt;their isotopes&lt;/a&gt;, but current research deals with the production of very shortlived species, which have limited applicability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all, we will always find something &lt;em&gt;new&lt;/em&gt; to explore, but whether we will always find something &lt;em&gt;strange&lt;/em&gt; depends on what you mean by the word. There are still many, many substances to find which will display impressive biochemical properties, for example, and many lethal or crippling diseases today might still become curable in the future. But there is zero chance of finding a substance which makes magic real, which unlocks faster-than-light travel, which has &lt;a href=&quot;http://en.wikipedia.org/wiki/Thiotimoline&quot; rel=&quot;nofollow noreferrer&quot;&gt;endochronic properties&lt;/a&gt; or which can be used to make overunity engines and perpetual motion devices. Ultimately, Chemistry is a form of applied quantum electrodynamics, a subsection of Physics, and quantum electrodynamics is a very developed theory &lt;a href=&quot;http://en.wikipedia.org/wiki/Precision_tests_of_QED&quot; rel=&quot;nofollow noreferrer&quot;&gt;with incredible precision&lt;/a&gt;. Thus, there is little room for surprises, and they likely will not come from a chemical system unless it is &lt;em&gt;very&lt;/em&gt; specially tailored, rather than randomly found in nature.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-04-02T13:45:53.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28153" PostTypeId="1" AcceptedAnswerId="66540" CreationDate="2015-04-02T13:57:12.667" Score="10" ViewCount="1864" Body="&lt;p&gt;I have a container with 3% hydrogen and the rest is air. How fast will the hydrogen get oxidized by air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much will the reaction speed up with increased temperature? &lt;/p&gt;&#xA;" OwnerUserId="14665" LastEditorUserId="402" LastEditDate="2017-04-01T00:22:04.483" LastActivityDate="2017-04-01T00:22:04.483" Title="Does hydrogen react with air at room temperature and pressure?" Tags="&lt;redox&gt;&lt;kinetics&gt;&lt;hydrogen&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28154" PostTypeId="1" CreationDate="2015-04-02T14:09:13.973" Score="4" ViewCount="2129" Body="&lt;p&gt;Just wondering if you fitted a cold water tank overflow pipe into a car park at the height of a car exhaust , if the carbon monoxide fumes went back up the oveflow pipe into the tank would this contaminate the water?&lt;/p&gt;&#xA;" OwnerUserId="15326" LastEditorUserId="7454" LastEditDate="2015-04-02T19:49:19.527" LastActivityDate="2015-04-02T19:49:19.527" Title="Can carbon monoxide contaminate water?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28155" PostTypeId="2" ParentId="28154" CreationDate="2015-04-02T14:41:12.027" Score="5" Body="&lt;p&gt;Contamination would occur if a sufficient quantity of carbon monoxide wouldn't just bubble through but &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;would react with water to form an unhealthy product or &lt;/li&gt;&#xA;&lt;li&gt;is dissolved in water&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both isn't really true:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{CO}$ does not react under these conditions.&lt;/li&gt;&#xA;&lt;li&gt;The solubility of $\ce{CO}$ in water at 20 °C is as low as $\mathrm{30\,mg\cdot L^{-1}}$ (&lt;a href=&quot;http://gestis-en.itrust.de/nxt/gateway.dll?f=templates$fn=default.htm$vid=gestiseng:sdbeng&quot;&gt;GESTATIS database&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-02T14:41:12.027" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28156" PostTypeId="1" CreationDate="2015-04-02T14:42:53.480" Score="1" ViewCount="40" Body="&lt;p&gt;Why is zinc oxide yellow when hot and white when cold?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it yellow because of the flame used to make it hot?&lt;/p&gt;&#xA;" OwnerUserId="15325" LastActivityDate="2015-04-02T14:42:53.480" Title="Reason behind ZnO color" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2015-04-02T15:52:29.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="28158" PostTypeId="1" AcceptedAnswerId="28162" CreationDate="2015-04-02T15:27:25.277" Score="0" ViewCount="15458" Body="&lt;p&gt;What are the products of partial hydrolysis of $\ce{XeF_6}$?&lt;br&gt;&#xA;Is $\ce{XeOF_4}$ formed or $\ce{XeO_2F_2}$ is formed?&lt;/p&gt;&#xA;" OwnerUserId="15325" LastEditorUserId="15325" LastEditDate="2015-05-02T10:28:15.050" LastActivityDate="2017-11-09T05:03:57.027" Title="Partial hydrolysis of XeF6" Tags="&lt;inorganic-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28159" PostTypeId="1" CreationDate="2015-04-02T17:00:08.777" Score="5" ViewCount="390" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Which species can't appear in the rate expression for this reaction scheme?&lt;br&gt;&#xA;  Here is the reaction scheme in steps: &#xA;  \begin{align}&#xA;\ce{A + B~(slow) &amp;amp;\rightarrow C} \\&#xA;\ce{C~(fast) &amp;amp;\rightarrow D + B}\\&#xA;\ce{D + E~(fast) &amp;amp;\rightarrow F}\\&#xA;\ce{F~(fast) &amp;amp;\rightarrow G}\\&#xA;\end{align}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers are apparently F and G, but I don't see why D and E couldn't also be valid (considering the first step is the slowest). A and B will inevitably be in the rate equation since they are reactants in the rate determining step. C produces B (alongside D) and so will also be in the rate equation as a result. However, why can't we rule out D and E of being in the rate equation, considering they don't seem to have any effect on the concentrations of A and  B. &lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="46306" LastEditDate="2018-03-07T05:05:58.477" LastActivityDate="2018-03-07T05:05:58.477" Title="Which species can't appear in the rate expression for this reaction scheme?" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28160" PostTypeId="2" ParentId="28150" CreationDate="2015-04-02T17:01:04.637" Score="7" Body="&lt;p&gt;Let's take a look at a table of &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;noreferrer&quot;&gt;standard electrode potentials&lt;/a&gt; and find the relevant half-reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xA;&amp;amp;\ce{Na+ + e- &amp;lt;=&amp;gt; Na} &amp;amp; &amp;amp;E°=-2.71\ \mathrm{V}\\&#xA;&amp;amp;\ce{2H+ + 2e- &amp;lt;=&amp;gt; H2} &amp;amp; &amp;amp;E°=0\ \mathrm{V}&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidations (reduction reactions reversed):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*}&#xA;&amp;amp;\ce{2H2O &amp;lt;=&amp;gt; O2 + 4H+ + 4e-} &amp;amp; &amp;amp;E°=-1.229\ \mathrm{V}\\&#xA;&amp;amp;\ce{2Cl- &amp;lt;=&amp;gt; Cl2 + 2e-} &amp;amp; &amp;amp;E°=-1.36\ \mathrm{V}&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you know the concentrations of the species involved, you can use the Nernst equation to figure out the actual potentials, but we can reason around it without knowing the exact numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big thing to notice is how close the potentials of each of the reductions and oxidations are to one another—the reduction reactions are separated by almost 3 V, but the oxidation reactions are very similar. This means that for reduction, it is far less energetically costly to produce hydrogen gas than to reduce sodium, and in fact, if any sodium were produced, it would instantly react with the water to produce hydrogen gas, so if there is any water around, that half-reaction will be favoured. In the real world, the exact potential hydrogen is produced depends a great deal on the electrode used (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Overpotential&quot; rel=&quot;noreferrer&quot;&gt;overpotential&lt;/a&gt;), but it's more or less impossible to electrodeposit metals with much lower reduction potentials than zinc in aqueous solutions. To actually produce metallic sodium, electrolysis of molten NaCl without any water present is the most common method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the oxidations, the story's quite different. It doesn't take much more energy to oxidize chloride than it does water (and if the chloride concentration is high, the oxidation potential will go up). This means that (subject to the effects of the electrodes on overpotential), some amount of both half-reactions will occur and because chlorine is a gas and therefore leaves the solution and the oxidations potentials are so close (leaving little energy to overcome kinetic barriers), the chlorine produced will not significantly oxidize the water.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-04-02T17:01:04.637" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28161" PostTypeId="1" CreationDate="2015-04-02T17:11:06.120" Score="1" ViewCount="60" Body="&lt;p&gt;I need to know, in terms of electronic charge - the electrophiles and nucleophiles involved - how the electronegativity of the potassium and hydrogen atoms alongside the hydroxide and H&lt;sup&gt;+&lt;/sup&gt; ions result in this reaction taking place. You could also think of this as asking why this reaction is exothermic from an organic chemistry perspective - why are the products more stable in terms of their electronic arrangement. It might be helpful to have the mechanism explained perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2K + 2H2O -&amp;gt; 2KOH + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, I have learnt before that potassium is higher up the reactivity series than hydrogen, but what does this mean on a deeper level - what factors mean that potassium is more reactive than hydrogen in terms of electron arrangement.&lt;/p&gt;&#xA;" OwnerUserId="9286" LastEditorUserId="13184" LastEditDate="2015-04-02T17:47:31.050" LastActivityDate="2015-04-02T17:47:31.050" Title="Why does this reaction take place in terms of electronic structure?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;electronic-configuration&gt;&lt;electronegativity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28162" PostTypeId="2" ParentId="28158" CreationDate="2015-04-02T17:44:02.087" Score="8" Body="&lt;p&gt;Both can be formed. It depends on the amount of water added: &#xA;\begin{align}&#xA;&amp;amp;&amp;amp;\ce{XeF6 + H2O  &amp;amp;-&amp;gt;  XeOF4 + 2HF}\\&#xA;&amp;amp;\text{or}&amp;amp;&#xA;\ce{XeF6 + 2H2O &amp;amp;-&amp;gt; XeO2F2 + 4HF}\\&#xA;&amp;amp;\text{or even}&amp;amp;&#xA;\ce{XeF6 + 3H2O &amp;amp;-&amp;gt; XeO3 + 6HF}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="4945" LastEditDate="2017-11-09T05:03:57.027" LastActivityDate="2017-11-09T05:03:57.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28163" PostTypeId="1" CreationDate="2015-04-02T17:46:25.550" Score="1" ViewCount="1535" Body="&lt;p&gt;What is the difference between diatomic molecules and elements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, If oxygen is element of a single atom, then what makes it a diatomic element? But, then I read that a single oxygen atom can bind with another element to form a compound, but it's a diatomic element with two atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please can someone enlighten some clarity on this subject. Also, it would help to see some examples to see the differences and context of how these diatomic molecules and elements interact.&lt;/p&gt;&#xA;" OwnerUserId="15330" LastEditorUserId="7448" LastEditDate="2015-04-02T17:50:39.353" LastActivityDate="2015-04-02T17:50:39.353" Title="Diatomic Molecules/Elements" Tags="&lt;atoms&gt;&lt;molecules&gt;&lt;elements&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28164" PostTypeId="2" ParentId="28140" CreationDate="2015-04-02T18:09:01.360" Score="1" Body="&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt; The Magnesium is added because the color change is easier to spot with it, even though calcium also changes the color of the solution in presence of TBT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long explanation and experiment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though your question might looks silly at first, I know it makes sense because I work with Soil Analysis, and the titrimetric determination of $\ce{Ca^2+/Mg^2+}$ is performed using the same pH 10 $\ce{NH3/NH4^+}$ buffer, with the addition of a small amount $\ce{MgSO4}$ and $\ce{EDTA}$ to the buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, and according to a &lt;a href=&quot;http://www.titrations.info/EDTA-titration-water-hardness&quot; rel=&quot;nofollow noreferrer&quot;&gt;fast online search&lt;/a&gt;, the addition of magnesium is to ensure the visibility of the end point when using Eriochrome Black T (EBT) as indicator. Supposedly, the color change wouldn't be easily spotted without the magnesium addition. This is how it's supposed to happen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering there will be no change of color in EBT only due to Calcium, a small amount of magnesium is necessary to interact with the indicator and dye the solution. The equivalent number of moles of EDTA is added to the buffer, otherwise the results would be overestimated. Upon completion of the titration, the remaming $\ce{Mg^2+}$ complexed with TBT would be chelated by EDTA and the endpoint detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is important in basically two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no magnesium in your sample.&lt;/li&gt;&#xA;&lt;li&gt;The buffer used is not pH 10, but pH 12, exactly to remove magnesium and detect only calcium.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In both cases, you need some (known) amount of magnesium to see the color change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this is that Calcium is supposed to color TBT. It's even the example given on &lt;a href=&quot;http://en.wikipedia.org/wiki/Eriochrome_Black_T&quot; rel=&quot;nofollow noreferrer&quot;&gt;TBT's Wikipedia page&lt;/a&gt;, and there's also a &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ac60099a026&quot; rel=&quot;nofollow noreferrer&quot;&gt;full article&lt;/a&gt; about the interactions of magnesium and TBT. So why would you add magnesium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your queston - which also became my own - I did a &lt;strong&gt;qualitative&lt;/strong&gt; experiment, because I'm in the lab right now. And the result was interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added small amounts of Calcium Chloride and Magnesium Sulphate to two different beakers containing purified water, and then added a few drops of TBT. I also added TBT to the purified water as a blank. This is what I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9VXGt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the $\ce{Ca^2+}$ solution did change the color of the indicator, but not nearly as much as $\ce{Mg^2+}$ did. If you think about real-sample analysis, where usually the water (or soil extract) contains other coloured materials, making sure the endpoint is easily visible is very important. As a colorblind, I already find this change in color a bit hard, and for $\ce{Ca^2+}$ alone I suppose it would be even harder without the magnesium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally: The Magnesium is added because the color change is easier to spot, even though calcium changes the color of the solution in presence of TBT.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-04-02T18:09:01.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28165" PostTypeId="1" CreationDate="2015-04-02T18:44:21.550" Score="8" ViewCount="1014" Body="&lt;p&gt;How would you write the $K_\mathrm p$ expression for the reaction: $$\ce{4 Ag(s) + O2(g) \rightarrow 2 Ag2O(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know normally it would be products over reactants, but there is a solid on the product side and they aren't included, so I don't know how you would write it?&lt;/p&gt;&#xA;" OwnerUserId="15331" LastEditorUserId="5026" LastEditDate="2018-02-24T11:24:54.833" LastActivityDate="2018-02-24T11:24:54.833" Title="Writing pressure equilibrium constant expression for a reaction with a single solid product" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28166" PostTypeId="1" CreationDate="2015-04-02T18:47:40.290" Score="1" ViewCount="9600" Body="&lt;p&gt;I am tying to determine if an acid is a liquid or aqueous solution? Please can some one enlighten me?&lt;/p&gt;&#xA;" OwnerUserId="15330" LastEditorUserId="-1" LastEditDate="2018-06-20T22:53:32.380" LastActivityDate="2018-06-20T22:53:32.380" Title="Are acids and bases always aqueous?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2015-04-03T13:10:21.277" ContentLicense="CC BY-SA 4.0" />
  <row Id="28167" PostTypeId="1" AcceptedAnswerId="28180" CreationDate="2015-04-02T20:44:39.807" Score="3" ViewCount="178" Body="&lt;p&gt;&lt;strong&gt;The setting:&lt;/strong&gt; I want to simulate protein docking and let &lt;em&gt;some&lt;/em&gt; dihedral angles vary, but in order to keep it low-dimensional I have to select those which are most likely to change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I've thought of so far:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The easiest solution would be some tool/database that simply returns a list of dihedral angles ordered by their variance (something like &lt;a href=&quot;http://www.ifm.liu.se/compchem/msi/doc/life/cerius46/qsar/theory_descriptors.html#515786&quot; rel=&quot;nofollow&quot;&gt;http://www.ifm.liu.se/compchem/msi/doc/life/cerius46/qsar/theory_descriptors.html#515786&lt;/a&gt;, but that only returns the number of &quot;meaningful&quot; rotamers)&lt;/li&gt;&#xA;&lt;li&gt;Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles.&lt;/li&gt;&#xA;&lt;li&gt;Take Ramachandran plots and try to combine all those conditional variances (if φ is …, then ψ has some amount of freedom) into a more &quot;holistic&quot; statement.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problems I encountered:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I couldn't find such a list.&lt;/li&gt;&#xA;&lt;li&gt;Searching &lt;a href=&quot;http://rcsb.org&quot; rel=&quot;nofollow&quot;&gt;rcsb.org&lt;/a&gt;, I only found about ~100 conformations (for albumin), but if I'm not mistaken the format isn't the exactly the same, so it'd be kinda hard to compare. Besides, ~100 samples won't yield good results for &gt;100 dimensions (=angles).&lt;/li&gt;&#xA;&lt;li&gt;The only approach for which data seems to be available, but it just looks like a real pain.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15084" LastActivityDate="2015-04-03T01:19:09.377" Title="How to get a protein's dihedral angles ordered by variance?" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28168" PostTypeId="2" ParentId="28166" CreationDate="2015-04-02T22:12:11.873" Score="2" Body="&lt;p&gt;There are varying definitions of&quot;acid&quot; (or &quot;base&quot;) depending on solvent system; some do not involve water at all. For example, there are &lt;a href=&quot;http://www.personal.psu.edu/the1/acidbase.htm&quot; rel=&quot;nofollow&quot;&gt;Brønsted–Lowry acids (proton donors) and Lewis acids (electron acceptors)&lt;/a&gt;, though the &lt;a href=&quot;https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Arrhenius_definition&quot; rel=&quot;nofollow&quot;&gt;Arrhenius definition&lt;/a&gt; &lt;em&gt;is&lt;/em&gt; based on &lt;em&gt;aqueous&lt;/em&gt; solutions dissociating into $\ce{H+}$ (acid) or $\ce{OH-}$ (base) ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, many acids are &lt;em&gt;not&lt;/em&gt; liquids, such as solid &lt;a href=&quot;https://en.wikipedia.org/wiki/Citric_acid&quot; rel=&quot;nofollow&quot;&gt;citric acid&lt;/a&gt;, which may be anhydrous or the monohdrate.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-02T22:12:11.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28170" PostTypeId="2" ParentId="28137" CreationDate="2015-04-02T22:33:26.887" Score="3" Body="&lt;p&gt;A glucose molecule is roughly 1 nanometer long (&lt;a href=&quot;http://mathbench.umd.edu/modules/measurement_sizes/page06.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;).  There are roughly 9,000-15,000 glucose molecules in a cellulose fiber (&lt;a href=&quot;http://www.cottoninc.com/product/NonWovens/Nonwoven-Technical-Guide/Cotton-Morphology-And-Chemistry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;, see Table 5).  If we use 10,000 as the number of glucose units, then the length of a cellulose fiber will be &#xA;$$10^{-9}~ \mathrm{m/molecule} \cdot 10^4 ~\mathrm{molecules/fiber}=10^{-5}~\mathrm{m/fiber} $$&#xA;or ~10 microns/fiber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical strength of the fiber comes from two sources.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Structural:&lt;/strong&gt; As your diagram illustrates there are many walls in the cellulose fiber.  All of these walls reinforce one another and build up the mechanical stability of the fiber. Each wall is composed of highly-organized cellulose chains.  &quot;Highly-organized&quot; is another way of saying &quot;crystalline&quot;.  Cotton has an average crystallinity around 75%.  This allows the cellulose fibers to pack closely together further increasing mechanical stability.  The cellulose fibers in the different walls are orientated in different directions, again serving to increase the fiber strength.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Chemical:&lt;/strong&gt; Here is a drawing of 2 glucose molecules from a cellulose chain.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xA0Ah.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &lt;strong&gt;each&lt;/strong&gt; glucose molecule has 3 equatorial hydroxyl groups oriented out and away from the polymer chain.  Because of the high crystallinity mentioned above, the cellulose chains are positioned very close to one another.  This means that &lt;strong&gt;thousands&lt;/strong&gt; of hydrogen bonds can form between hydroxyl groups on different chains.  As a rule of thumb, a hydrogen bond might have a strength around 5 kcal/m; if we have thousands of these bonds holding the cellulose chains together, then we have the equivalent of hundreds of normal chemical bonds holding the chains together!  These hydrogen bonds account for much of the stability of the cotton fiber.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-02T22:33:26.887" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28172" PostTypeId="1" CreationDate="2015-04-02T23:07:22.097" Score="0" ViewCount="344" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o7N1N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L7GBt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is happening in Step 2 and would you please give the reaction mechanism for it as well?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastActivityDate="2015-04-02T23:22:47.740" Title="Is Bromide anion acting as a base in reaction with ethylamine and methylbromide?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28173" PostTypeId="2" ParentId="28172" CreationDate="2015-04-02T23:22:47.740" Score="1" Body="&lt;p&gt;The bromide anion is acting as a counter anion to balance out the ammonium charge in step 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given sufficient heat, step 2 can occur, with bromide acting as a weak base.  This is likely to produce gaseous produces which would drive the equilibrium as the gasses escape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanistically, this is possible, but synthetically, it is more likely another base would be added in a typical synthesis.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-04-02T23:22:47.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28174" PostTypeId="1" AcceptedAnswerId="28191" CreationDate="2015-04-02T23:48:54.840" Score="2" ViewCount="100" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r0e4i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tsd4e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these reaction mechanisms correct for Ammonia and Bromide anions acting as a base respectively?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="7475" LastEditDate="2015-10-16T17:30:34.623" LastActivityDate="2015-10-16T17:30:34.623" Title="Are these reaction mechanisms correct?" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28176" PostTypeId="2" ParentId="28090" CreationDate="2015-04-02T23:55:28.960" Score="11" Body="&lt;h2&gt;Carbonates&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The quote from your text:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbonates of alkaline earth&#xA;  metals are insoluble in water and can be&#xA;  precipitated by addition of a sodium or&#xA;  ammonium carbonate solution to a solution&#xA;  of a soluble salt of these metals. The solubility&#xA;  of carbonates in water decreases as the atomic&#xA;  number of the metal ion increases. All the&#xA;  carbonates decompose on heating to give&#xA;  carbon dioxide and the oxide. Beryllium&#xA;  carbonate is unstable and can be kept only in&#xA;  the atmosphere of CO2. &lt;strong&gt;The thermal stability&#xA;  increases with increasing cationic size&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the stability that you are referring to is &lt;em&gt;thermal stability.&lt;/em&gt; This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What's happening to cause thermal instability?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, lets look at the carbonate ion here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p2zYe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below the illustration shows where the negative charge is likely to be concentrated (colored in red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiLb0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ipQaa.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this polarization, the carbon dioxide will become more stable and energetically favorable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How does going down a group play into this?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower &lt;strong&gt;charge density&lt;/strong&gt;. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Chloride and fluoride stability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQ359.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Information and illustrations on carbonate ions were sourced from &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/group2/thermstab.html&quot; rel=&quot;noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-04-03T00:33:59.187" LastActivityDate="2015-04-03T00:33:59.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28177" PostTypeId="2" ParentId="28165" CreationDate="2015-04-03T00:21:11.020" Score="4" Body="&lt;p&gt;The expression $\mathrm{K_p}$ is an expression of the equilibrium of partial pressures. So, it just includes the the partial pressure of the oxygen because that's the only gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{K_P} = \frac{1}{\ce{P_{O2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information about the equilibrium constant $\mathrm{K_p}$ can be found &lt;a href=&quot;http://www.chemguide.co.uk/physical/equilibria/kp.html&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-04-03T06:56:12.320" LastActivityDate="2015-04-03T06:56:12.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28180" PostTypeId="2" ParentId="28167" CreationDate="2015-04-03T01:19:09.377" Score="2" Body="&lt;p&gt;If I understand you, I can't imagine how to solve the your problem with this approach. Also I don't get the physical meaning of the proposed solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have to deal with a molecule with more than 100 rotatable bonds, you are just in problems.&#xA;You could find the rotatable bonds with some tools (see for example OpenBabel). But the problem will just start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, you can not just try one by one the $N$ angles ($\{\theta _i \}$) (by changing the angle value $n$ times and trying each of them), because for exploring all the conformational space you will just have $n^N$ points. And you just can't know a priori which are the most important angles (in case (I think not) that effectively exist the most important angles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that not every point ($\theta _1, \theta _2, \cdots,\theta _N)$ is acceptable, just because parts of the molecule can superpose with other parts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that this is not possible. The values of the rotatable angles will be just chosen by you.&#xA;Even more, suppose that you have two rotatable bonds very close each other, in a linear part of the molecule. If not a large superposition ocurrs, you will have no diference in the efect of them. But the effect of them will be very correlated. So, you can get useful info. with PCA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A possible approach&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I know very few of biology, I would make a research about if there are prefered configuration, that are much more common than others (because of biological facts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are forced to work with the entire configurational space, the computational approach that comes to my mind is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use very fast method for energy computations that allow you set the connectivity of the atoms. This will prevent changes of the covalent bonds due to superpositions.&lt;/li&gt;&#xA;&lt;li&gt;Use the Monte Carlo method, this should explore the configurational space giving you a reasonable idea of this space.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find extrange doing docking with a large protein and ¿without a ligand? If there is a ligand for binding to the protein that we are talking about, I won't take too much care of the protein. The scoring functions or energies are few accurate, I find hard to believe that just one found configuration would represent accurately the problem (considering the large degeneracies involved)&lt;/p&gt;&#xA;" OwnerUserId="14527" LastActivityDate="2015-04-03T01:19:09.377" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28181" PostTypeId="1" CreationDate="2015-04-03T02:28:56.963" Score="5" ViewCount="5664" Body="&lt;p&gt;I understand that if you rotate a molecule and it has the same configuration as its enantiomer, then it is indeed the same compound as the enantiomer (meso), and hence is achiral (despite having chirality centers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If compound is rotated and the resulting Fisher projection has exactly the same configuration after its rotation, what does that mean in terms of chirality? (I'm not talking about enantiomers here,) It will have a different enantiomer (and hence not be meso), and will be polarizable, but the compound itself has exactly the same perspective if you rotate it. Does this hold any signifance chemically or stereochemically, or is it meaningless?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example D-idaric acid (2&lt;i&gt;S&lt;/i&gt;,3&lt;i&gt;R&lt;/i&gt;,4&lt;i&gt;R&lt;/i&gt;,5&lt;i&gt;S&lt;/i&gt;)-2,3,4,5-tetrahydroxyhexanedioic acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below example of Idaric acid, the enantiomer appears to be a different compound. When the compound is rotated 180 degrees, it looks exactly the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iLt7T.png&quot; alt=&quot;Idaric acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below example of galactaric acid, when the compound is rotated 180 degrees, it is a different image. This is despite it being meso.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkVVq.png&quot; alt=&quot;Galactaric acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, being meso has a lot of chemical significance (no polarization of light). But does a compound possessing the property of being rotated 180 degrees, and having the same image (in Fischer form) present any chemical significance compared to a compound that doesn't do this?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="4945" LastEditDate="2017-12-11T05:08:22.477" LastActivityDate="2017-12-11T05:08:22.477" Title="Are molecules chiral if a rotation leads back to the same compound?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;symmetry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28182" PostTypeId="2" ParentId="28181" CreationDate="2015-04-03T03:43:25.613" Score="4" Body="&lt;p&gt;&lt;strong&gt;You can have &quot;symmetric&quot; chiral molecules like you describe.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a subtle point and your particular choice isn't chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In typical organic chemistry, instructors mention about mirror images, but leave some of the subtle details to later courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of chiral molecules and some of them look quite symmetric. Consider a pinwheel or propeller. You can rotate along the axis of the propeller and it will be symmetric, but there's a tilt/twist to the blades:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iaiHW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these two images are mirror, but there's a chiral axis and the direction of rotation is reversed. They are enantiomers. This case has three-way symmetry, but it's possible to have a &quot;two leaf&quot; pinwheel that's still chiral. Proper rotations don't make mirror images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case, D-idaric acid, is C2 symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Hhfu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-04-05T00:11:03.820" LastActivityDate="2015-04-05T00:11:03.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28183" PostTypeId="2" ParentId="27912" CreationDate="2015-04-03T04:10:07.267" Score="2" Body="&lt;p&gt;Addressing the problem of high resistance, we can identify that the resistivity derives from the oxidation of graphene, changing $sp^2$ carbons to $sp^3$ in the lattice, which increases the band gap, and creates an insulator. There are a number of answers to this question that could be correct, including simply mixing another, more conductive material in the synthesis, but since the question seems to imply that the material has already been synthesized and simply needs to be reduced, removing the oxygen, creating reduced graphene oxide (rGO). One simple approach is through thermal annealing, where GO is reduced by a heat treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0008622311008967&quot; rel=&quot;nofollow&quot;&gt;the reduction of graphene oxide here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-04-03T04:34:39.500" LastActivityDate="2015-04-03T04:34:39.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28184" PostTypeId="1" CreationDate="2015-04-03T04:21:27.867" Score="4" ViewCount="27338" Body="&lt;p&gt;I was wondering, how do I determine what metal (element) has the highest density by using the periodic table? Is it possible? &lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="5456" LastEditDate="2015-05-04T04:05:55.910" LastActivityDate="2020-08-21T16:11:41.130" Title="How to find density of element metals" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28186" PostTypeId="1" CreationDate="2015-04-03T05:19:21.957" Score="2" ViewCount="2897" Body="&lt;p&gt;The question is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For which exothermic reaction is $\Delta\,E$ more negative than&#xA;  $\Delta\,H$?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(A) $\ce{Br2(l) &amp;lt;=&amp;gt; Br2(g)}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(B) $\ce{2C(s) + O2(g) r 2CO(g)}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(C) $\ce{H2(g) + F2(g) -&amp;gt; 2HF(g)}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(D) $\ce{2SO2(g) + O2(g) r 2SO2(g)}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;r is supposed to be reaction? Either a one headed or equilibrium arrows. The &quot;correct&quot; answer is supposed to be D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using the formula $H = U + PV$ to $\Delta H = \Delta U + \Delta PV$, I found that $\Delta U$ would be less if $\Delta V$ was positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, more gaseous products should be produced. As a result, I chose B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, A is not exothermic, so would not be considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone confirm my answer or offer an explanation for my answer being wrong?&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="5026" LastEditDate="2018-02-24T11:24:07.480" LastActivityDate="2018-02-24T11:24:07.480" Title="Relating change in internal energy and to that in enthalpy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28187" PostTypeId="2" ParentId="28186" CreationDate="2015-04-03T06:43:00.747" Score="2" Body="&lt;p&gt;You are right. B is the correct choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H = \Delta E + \Delta PV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Delta E $ is more negative than $\Delta H$, $$\Delta H &amp;gt; \Delta E$$.&#xA;Hence $$\Delta PV&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta E \space $has to be negative. So option A is ruled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the other choices, B is the only one where number of gaseous moles are increasing from left to right, leading to a positive value of $\Delta V$ and hence $\Delta PV$.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorDisplayName="user7951" LastEditDate="2015-04-03T10:38:45.203" LastActivityDate="2015-04-03T10:38:45.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28190" PostTypeId="1" CreationDate="2015-04-03T10:13:49.540" Score="3" ViewCount="4548" Body="&lt;p&gt;how phosphorus forms 2 types of bonds with oxygen i.e. 5 in $\ce{P4O10}$ and 3 in $\ce{P4O6}$ when only three are needed to complete its octet?&lt;/p&gt;&#xA;" OwnerUserId="15343" LastEditorUserId="418" LastEditDate="2015-04-03T10:20:58.690" LastActivityDate="2015-05-30T17:27:36.880" Title="why does phosphorus makes 5 bonds with oxygen in P4O10 whereas 3 bonds in P4O6?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28191" PostTypeId="2" ParentId="28174" CreationDate="2015-04-03T10:26:08.433" Score="3" Body="&lt;p&gt;Except for the missing equilibrium arrow (these reaction are acid-base equilibria), they are fine. Or do you have any objections :) (what makes you doubt) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take @jerepierre's comment in account. This would probably be considered wrong if at an exam.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-04-03T20:45:03.863" LastActivityDate="2015-04-03T20:45:03.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28192" PostTypeId="2" ParentId="28181" CreationDate="2015-04-03T13:03:56.473" Score="6" Body="&lt;p&gt;The property that you are describing for D-idaric acid is called C2 symmetry. This means that when you rotate a molecule (or any object) by 180 degrees, it appears identically to the original orientation. In D-idaric acid, the C2 symmetry axis is halfway along the bond between C3 and C4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a C2 symmetry axis doesn't tell us if the compound is chiral or not. Achiral compounds have an (internal) plane of symmetry or a point of symmetry. D-Idaric acid does not meet either of these requirements and is a chiral molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galactaric acid does not have a C2 symmetry axis but does have a plane of symmetry. Galactaric acid is achiral. Since galactaric acid has stereocenters but is not chiral, we describe it as meso.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2015-04-03T14:33:39.143" LastActivityDate="2015-04-03T14:33:39.143" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28193" PostTypeId="1" AcceptedAnswerId="28196" CreationDate="2015-04-03T13:14:21.440" Score="4" ViewCount="3193" Body="&lt;p&gt;Are there any simple methods to completely remove nitrate ions from few ml of impure water sample? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method I am curious about is the following process:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add one or more reagents to water in test tube, which would crystallize nitrate ions into some solids&lt;/li&gt;&#xA;&lt;li&gt;Then filter out the solids from water such that the water is now nitrate free.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is the above process possible? If so, what are some reagents that can be used?&lt;/p&gt;&#xA;" OwnerUserId="15344" LastEditorUserId="194" LastEditDate="2017-04-06T01:31:22.110" LastActivityDate="2018-07-12T18:33:26.097" Title="How to remove nitrate from water sample?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28194" PostTypeId="1" CreationDate="2015-04-03T13:21:46.417" Score="1" ViewCount="488" Body="&lt;p&gt;I am struggling to understand ternary phase diagrams for reactor engineering. The course slides offer little to no help so I'm hoping to find an answer to this question&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the slope of the tie-lines affect on which side of the&#xA;  equilibrium line the extract/raffinate is? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand is that if the lines slope down, the extract is on the right and the raffinate is on the left. If they slope up, the sides are flipped. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidenote: does anyone know a good resource where ternary diagrams for chemical/reactor engineering are well explained?&lt;/p&gt;&#xA;" OwnerUserId="7961" LastActivityDate="2015-04-03T13:21:46.417" Title="How do tie-line slopes affect raffinate/extract envelope in ternary phase diagrams?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28195" PostTypeId="1" AcceptedAnswerId="28197" CreationDate="2015-04-03T14:01:18.333" Score="1" ViewCount="2276" Body="&lt;p&gt;I am an electronics guy with not so much knowledge of chemistry.&#xA;I would like to prepare a pH 7 buffer to test pH meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All web references I have found so far describing how to make a phosphate buffer use&#xA;disodium phosphate (Na&lt;sub&gt;2&lt;/sub&gt;HPO&lt;sub&gt;4&lt;/sub&gt;) and Monosodium phosphate (NaH&lt;sub&gt;2&lt;/sub&gt;PO&lt;sub&gt;4&lt;/sub&gt;) as ingredients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a good supply of 85% phosphoric acid and solid sodium hydroxide. Can I make a pH 7 buffer from these ingredients instead or is there a catch? &lt;/p&gt;&#xA;" OwnerUserId="15346" LastActivityDate="2015-04-03T14:41:16.833" Title="Can I make a pH 7 buffer from phosphoric acid and sodium hydroxide?" Tags="&lt;ph&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28196" PostTypeId="2" ParentId="28193" CreationDate="2015-04-03T14:37:16.927" Score="7" Body="&lt;h1&gt;Precipitation won't work&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Nitrate removal is a very tough problem in wastewater treatment.  One of the reasons is that &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot;&gt;nitrate salts of nearly every common cation are highly water soluble&lt;/a&gt;.  Unfortunately, I can't think of a way to precipitate nitrate from your water sample.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How commercial nitrate removal processes work&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Nitrate is removed from wastewater through a microbiological process called &lt;a href=&quot;http://en.wikipedia.org/wiki/Denitrification&quot;&gt;denitrification&lt;/a&gt;.  In this process, organic carbon substrates are used as food by special anaerobic bacteria.  In the absence of oxygen, the bacteria use nitrate as an electron acceptor so they can oxidize their food and obtain energy via respiration.  As an example organic food molecule, consider hexose sugar.  Then the reaction could be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{5 C6H12O6 + 24NO3- -&amp;gt; 30CO2 + 24 OH- + 18H2O + 12N2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be difficult to purify a few mL of water in this way unless you know exactly the &lt;a href=&quot;http://en.wikipedia.org/wiki/Biochemical_oxygen_demand&quot;&gt;BOD&lt;/a&gt; of the water and are an experienced microbiologist with access to cultures of the appropriate bacteria.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Another not very good option&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Another possibility would be electrochemical removal of the nitrate.  That is a topic which has been researched but which is not practices commercially to any significant extent.  Here is &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0013468605005566&quot;&gt;one paper&lt;/a&gt; where you can read about the technique.  I don't think it would be very easy for a home chemist either.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The tried and true&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;That leaves more tried and true methods.  You've simply got to distill your water!  Boil it and collect and condense the steam and you'll have &lt;a href=&quot;http://en.wikipedia.org/wiki/Distilled_water&quot;&gt;substantially purified water&lt;/a&gt; to behold and enjoy!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-03T14:37:16.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28197" PostTypeId="2" ParentId="28195" CreationDate="2015-04-03T14:41:16.833" Score="2" Body="&lt;p&gt;You should be able to do this with little trouble.  The only caveats:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Make a stock solution of phosphoric acid and also of NaOH.  When you add them to each other to make your buffer, heating could be a problem, much moreso than if you were using the mono- and di-basic sodium phosphate salts.  So you may want to cool on ice when mixing these compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can calculate approximately how much NaOH and phosphoric acid to mix in order to get a pH 7 result, but it's best to use a pH meter to verify that you've reached pH 7. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-03T14:41:16.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28199" PostTypeId="1" CreationDate="2015-04-03T15:17:29.360" Score="1" ViewCount="369" Body="&lt;p&gt;I understand $d_{xy}$ and $d_{yz}$ and those orientations but ones like $d_{z^2}$ are very strange to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain that to me?&lt;/p&gt;&#xA;" OwnerUserId="15348" LastEditorUserId="7448" LastEditDate="2015-04-03T16:18:05.550" LastActivityDate="2015-04-03T16:18:05.550" Title="Trouble understanding 5 orientations of d orbitals" Tags="&lt;orbitals&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28202" PostTypeId="1" CreationDate="2015-04-03T17:57:42.767" Score="5" ViewCount="7334" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B8OwF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B8OwF.png&quot; alt=&quot;Hept-1-en-6-yne&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preferred IUPAC name for the compound above is &lt;strong&gt;hept-1-en-6-yne&lt;/strong&gt;. I understand that the numbers are chosen in order to give the alkene the lower locant, as explained in &lt;a href=&quot;https://chemistry.stackexchange.com/q/26735/16683&quot;&gt;Should double or triple bonds have preference for low locants in IUPAC nomenclature?&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm wondering why the name has the form -en-yne instead of -yn-ene. In other words, why is it not &quot;hept-6-yn-1-ene&quot; instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if this is an IUPAC specification, is there any rationale for why en-yne should be preferred to yn-ene?&lt;/p&gt;&#xA;" OwnerUserId="15352" LastEditorUserId="16683" LastEditDate="2018-09-29T19:53:06.117" LastActivityDate="2019-11-05T05:56:11.673" Title="Why are enynes named as -en-yne and not -yn-ene?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28203" PostTypeId="1" AcceptedAnswerId="28207" CreationDate="2015-04-03T18:01:32.797" Score="2" ViewCount="124" Body="&lt;p&gt;Azide is a fairly common ion, but why aren’t there similar versions such as $\ce{P3-}$ or $\ce{B3+}$? Do they exist, is there something wrong with these ions as to make them more likely to just break apart rather than for ionic compounds.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T08:11:14.507" LastActivityDate="2015-05-03T08:11:14.507" Title="Analogies of azide?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28204" PostTypeId="1" AcceptedAnswerId="28209" CreationDate="2015-04-03T18:13:12.333" Score="4" ViewCount="450" Body="&lt;p&gt;Well my book says that on heating- amorphous solids crystalline at some temperature. As far as I know , amorphous solids have a very disordered structure compared to crystalline solids and on further heating the level of disorder should increase. So why do they become more ordered or crystalline at high temperatures or am I missing something fundamental in here? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2015-04-03T19:27:48.057" Title="Amorphous to crystalline: Opposite trend?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28205" PostTypeId="2" ParentId="28202" CreationDate="2015-04-03T18:23:13.410" Score="6" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-31.1.1.1&lt;/strong&gt; The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.&lt;br&gt;&#xA;  (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since, in names, the ending ‘ene’ always precedes ‘yne’ (with elision of the final letter ‘e’ in ‘ene’), the given compound is a ‘hept-&lt;em&gt;x&lt;/em&gt;-en-&lt;em&gt;y&lt;/em&gt;-yne’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the lowest sets of locants in ‘hept-1-en-6-yne’ and ‘hept-6-en-1-yne’ are equivalent (1, 6), preference for low locants is given to the double bond. Therefore, locant 1 is assigned to the double bond and locant 6 is assigned to the triple bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the correct name is &lt;strong&gt;hept-1-en-6-yne.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nhFkv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhFkv.png&quot; alt=&quot;hept-1-en-6-yne&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-08-24T17:18:35.047" LastActivityDate="2015-08-24T17:18:35.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28206" PostTypeId="1" CreationDate="2015-04-03T18:48:15.147" Score="9" ViewCount="581" Body="&lt;p&gt;For metal complexes with $A_{2}$ or $E$ ground state terms there is angular momentum contribution to the magnetic moment which is generally positive for more than half-filled subshells and negative otherwise, and related to the spin-only magnetic moment by &#xA;$$m_\mathrm{eff} = m_\mathrm{SO}\left(1-\frac{\alpha\lambda}{\Delta}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta$ is the ligand field splitting parameter, $\alpha$ is the orbital angular momentum degeneracy of the state and $\lambda$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.&lt;/li&gt;&#xA;&lt;li&gt;Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;: a simple way to do this is via Hund's third rule. If the d-orbitals are half or more full, the highest J state will be lowest in energy, and then the deviation for the T state will be positive ! Exactly why is probably rather subtle. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="10256" LastEditDate="2015-05-19T19:33:35.330" LastActivityDate="2015-05-19T19:33:35.330" Title="Predicting orbital angular momentum effects on magnetic moments" Tags="&lt;transition-metals&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28207" PostTypeId="2" ParentId="28203" CreationDate="2015-04-03T18:59:39.087" Score="6" Body="&lt;p&gt;Compare the covalent radii of $\ce{N}$ ($75\,\mathrm{pm}$) and $\ce{P}$ ($106\,\mathrm{pm}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus is a bit on the chubby side. Larger atom-to-atom distance means less overlap of p-orbitals in multiple bonds: $\ce{P=P}$ and $\ce{P#P}$ bonds are much weaker than similar bonds between nitrogen atoms.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-03T18:59:39.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28208" PostTypeId="2" ParentId="28203" CreationDate="2015-04-03T19:08:49.897" Score="1" Body="&lt;p&gt;Nitrogen is special in that its pi bonds are stronger than its sigma bonds. This explains why organic compounds containing nitrogen connected with only single bonds or non-resonant double bonds (especially -N=N-) tend to be unstable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost every other element has stronger sigma than pi bonds so they prefer single bonds. Azide analogues will most likely dissociate and/or polymerise.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2015-04-03T19:08:49.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28209" PostTypeId="2" ParentId="28204" CreationDate="2015-04-03T19:27:48.057" Score="4" Body="&lt;p&gt;What they say in your book is right. Take the case of polyethylene terephthalate (PET) which is a kind of polyester used in water bottles for example. It's an amorphous polymer (remember the bottles are transparent). When it is heated, the strained chains have the possibility to move and reorder themselves and the polymer becomes partially crystalline. This fact can readily be proved experimentally by heating a piece of PET (over its glass transition temperature, around $74 ^0\mathrm{C}$), then quenching the sample. It becomes white and not transparent which means it's partially crystalline.&#xA;We can monitor this transformation using a specific instrument called differential scanning calorimeter, where we can notice the formation of an exothermic peak corresponding to crystallization.  &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-03T19:27:48.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28210" PostTypeId="1" CreationDate="2015-04-03T23:22:59.920" Score="1" ViewCount="459" Body="&lt;p&gt;Why is the keto form of acetyl chloride more stable even though the enol form is stabilized by the hydrogen bonding?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorDisplayName="user7951" LastEditDate="2015-04-04T22:23:53.367" LastActivityDate="2015-04-05T12:53:41.743" Title="Relative stability of keto- enol form of acetyl chloride" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28211" PostTypeId="2" ParentId="7940" CreationDate="2015-04-04T02:04:02.463" Score="0" Body="&lt;p&gt;Solubility is maximum amount of solute that will dissolve in a given solvent at a given temperature to form saturated solution. Dissociation is a more kinetic process which depend on lattice and hydration energy.&lt;/p&gt;&#xA;" OwnerUserId="15353" LastEditorUserId="4945" LastEditDate="2015-04-04T05:10:55.663" LastActivityDate="2015-04-04T05:10:55.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28212" PostTypeId="1" AcceptedAnswerId="28213" CreationDate="2015-04-04T02:34:28.530" Score="2" ViewCount="619" Body="&lt;p&gt;At equilibrium, the rates of the forward and backward reaction become equal. In a reaction &#xA;A(s) giving B(g) + C(g)&#xA;Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected. By the time the rate of the backward reaction becomes equal to that of the forward reaction, more of A must have converted into B as compared to how much it did when there was less surface area. Since the activity of a pure solid is unity, it doesn't appear in the expression for K. So K must have increased. The equilibrium constant on the other hand depends on the ∆G standard for a reaction. The standard ∆G is unchanged, so K must be unchanged. How does one resolve these two?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-04-04T02:50:35.823" Title="Equilibrium Constant and Surface Area" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28213" PostTypeId="2" ParentId="28212" CreationDate="2015-04-04T02:50:35.823" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a reaction A(s) giving B(g) + C(g) Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is an assumption that you're making and it is thermodynamically and geometrically questionable.  If increasing the surface area of $\ce{A}$ increases the rate at which it sublimes/gasifies to form $\ce{B + C}$, it means that the reactive molecules of $\ce{A}$ are only those at the surface, in which case it is misleading to denote the reactant as $\ce{A(s)}$.  It is really $\ce{A(surface)}$ which is reacting.  Similarly, if the rate of the reverse reaction is not really dependent on the surface area and surface geometry of $\ce{A(s)}$, then it implies that very quickly molecules of $\ce{B}$ and $\ce{C}$ would condense to form a nanocrystallite of $\ce{A}$ that is one molecule big.  These nanocrystallites of $\ce{A}$ would have, collectively, nearly infinite surface area.  So the backward reaction &lt;em&gt;creates&lt;/em&gt; surface area under the assumptions you have made.  You have to account for that created area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid-phase reactions are complicated by phenomena such as crystal &lt;a href=&quot;https://en.wikipedia.org/wiki/Nucleation&quot; rel=&quot;nofollow&quot;&gt;nucleation&lt;/a&gt; and other mechanisms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_growth&quot; rel=&quot;nofollow&quot;&gt;crystal growth&lt;/a&gt;.  Treating solids as uniformly reactive materials with unit activity is inappropriate for understanding such phenomena.  Another way to say it is that the assumption that solids have unit activity and don't appear in expressions for $K$ or $\Delta G$ is an assumption, one that is inapplicable when dealing with reactions whose rates depend on surface area.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-04T02:50:35.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28214" PostTypeId="1" AcceptedAnswerId="28217" CreationDate="2015-04-04T04:22:07.473" Score="1" ViewCount="4230" Body="&lt;p&gt;My teacher told me that order of lewis basic strength is inversely proportional to electro negativity. Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion. Why this is so? Is it related to localised and delocalize lone pair?&lt;/p&gt;&#xA;" OwnerUserId="15354" LastEditorUserId="10185" LastEditDate="2015-11-13T19:32:40.990" LastActivityDate="2020-01-19T10:17:24.563" Title="Order of lewis basic strength" Tags="&lt;acid-base&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28215" PostTypeId="1" CreationDate="2015-04-04T04:42:00.990" Score="1" ViewCount="9578" Body="&lt;p&gt;According to the solutions there exist 7 different structural isomers 'containing 3 degree amines' for the molecular formula  $\ce{C_6H_{15}N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried and made a few, here are those &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KDYsw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying since an hour and all I made were the repetitions of these 5 only. Could anyone suggest me the remaining 2? &lt;/p&gt;&#xA;" OwnerUserId="11606" LastEditorUserId="865" LastEditDate="2015-10-16T18:05:21.437" LastActivityDate="2015-10-16T18:05:21.437" Title="What are all the structural isomers of a tertiary amine with the sum formula C6H15N?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28216" PostTypeId="2" ParentId="28215" CreationDate="2015-04-04T07:35:47.680" Score="2" Body="&lt;p&gt;&lt;em&gt;Please note that this is just a visualization of the comment given by &lt;strong&gt;LDC3&lt;/strong&gt;!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the two missing isomers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ZTMO.png&quot; alt=&quot;C6H15N&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-04T07:35:47.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28217" PostTypeId="2" ParentId="28214" CreationDate="2015-04-04T08:05:27.797" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My teacher told me that order of lewis basic strength is inversely proportional to electro negativity. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I agree with your teacher's guideline. Now, note that this cannot be your sole guideline. Lewis basicity (and acidicty) depend on a number of factors including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Strength of the corresponding Lewis acid or Lewis base and the strength of the resulting bonds in the Lewis adduct. &lt;/li&gt;&#xA;&lt;li&gt;Solvent effects &lt;/li&gt;&#xA;&lt;li&gt;Geometry of the interacting species &lt;/li&gt;&#xA;&lt;li&gt;Electronegativity &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you are confusing Bronsted basicity with Lewis basicity. It is true that &lt;strong&gt;in water&lt;/strong&gt; (a polar, protic solvent) chloride ion is a stronger Bronsted base than iodide ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of Lewis basicity, however, things become a lot more complicated. Iodide ion has the more readily available lone pair electrons for donation since iodide ion is less electronegative than chloride ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the strength of bonds to iodine are generally not as strong as bonds with chloride ion. This is a result of poor orbital overlap between iodide ion - which is a big atom - much bigger than many Lewis acids. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I asked my teacher today the same question and he told that since both iodide and chloride ion have delocalize lone pair both iodide and chloride ion have delocalize lone pair in their own vacant orbital therefore the electron donating tendency decreases. here we can't compare using electro negativity. And comparing them since iodide ion has more big size than chloride ion therefore the delocalization tendency is better in iodine therefore chloride ion is better lewis base. Any comments?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I mostly agree; both have electrons in their orbitals but I'm not a fan of the term &quot;delocalization&quot; - that implies that the electrons are spread across multiple nuclei - but in chloride ion and iodide ion there's only one nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there are multiple factors to consider - not just electronegativity. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-04-04T17:05:31.807" LastActivityDate="2015-04-04T17:05:31.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28218" PostTypeId="1" CreationDate="2015-04-04T09:16:25.927" Score="2" ViewCount="191" Body="&lt;p&gt;From what I understand is ingesting high amounts of sugar alcohols can lead to various effects such as abdominal pain and diarrhea, my question is, is there a certain amount of sugar alcohol if ingested cause drunkenness (even if it causes a person to be in a critical state or death) ? &lt;/p&gt;&#xA;" OwnerUserId="15355" LastActivityDate="2015-04-04T15:17:09.357" Title="Do sugar alcohols have the potential to get a person drunk?" Tags="&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28219" PostTypeId="2" ParentId="28114" CreationDate="2015-04-04T10:04:54.753" Score="10" Body="&lt;p&gt;&lt;em&gt;Please note that I haven't seen the sample myself!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Royal Institution claims that a bottle with the collidal gold solution is on display at &lt;a href=&quot;http://www.rigb.org/our-history/michael-faraday/magnetic-laboratory&quot;&gt;Michael Faraday's Magnetic Laboratory&lt;/a&gt; (see bottle at the bottom of the image) in the museum of the Royal Institution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be nitpicking and English isn't my native language, but &lt;em&gt;David Thompson&lt;/em&gt; does not explicitly state in his article in &lt;em&gt;Gold Bulletin&lt;/em&gt; that &lt;strong&gt;all samples&lt;/strong&gt; are lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;em&gt;Ryan D. Tweney&lt;/em&gt; in his article &lt;em&gt;Discovering Discovery: How Faraday Found the First Metallic Colloid&lt;/em&gt;, published in &lt;em&gt;Perspectives on Science&lt;/em&gt;, &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;14&lt;/em&gt;, 97-121 (&lt;a href=&quot;http://dx.doi.org/10.1162/posc.2006.14.1.97&quot;&gt;DOI&lt;/a&gt;) not only describes the finding of 700 surviving microscopic slides in Faraday's reconstructed magnetic with gold film samples and numbered in Faraday's hand and that the numbers correspond to the numbering in Faraday's 1856 Diary (lab journal) but continues:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It had long been known that some of Faraday's gold colloids survive from the 1856 research: Four bottles of these are on display at the Royal Institution in London [&amp;hellip;]. These few colloids are all that survive of many dozens referred to in the Diary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-07T06:38:57.233" LastActivityDate="2015-04-07T06:38:57.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28220" PostTypeId="1" CreationDate="2015-04-04T10:34:31.613" Score="1" ViewCount="478" Body="&lt;p&gt;Given a table with equivalent conductivity of certain electrolytes, how can I determine the conductivity of separate ions? I know that you can find the difference between for example $\ce{K+}$ and $\ce{Na+}$ if you know the limiting conductivity of $\ce{KOH}$ and $\ce{NaOH}$, but how do you calculate the actual value for those two ions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;I've found a different source for the table, my apologies. Let me rephrase my question. Many texts on electrolytic conductivity refer to a table containing the molar conductivity of individual ions, (&lt;a href=&quot;http://www.fondriest.com/environmental-measurements/wp-content/uploads/2014/05/DOW.pdf&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;). What I'm wondering is, how are these values obtained? Can they be calculated with data from (&lt;a href=&quot;http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/equivalent%20conductivity%20of%20electrolytes.pdf&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="15356" LastEditorUserId="13339" LastEditDate="2015-05-18T09:00:08.750" LastActivityDate="2015-05-18T09:00:08.750" Title="How to determine limiting conductivity of specific ions?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28221" PostTypeId="1" AcceptedAnswerId="28224" CreationDate="2015-04-04T11:22:53.070" Score="16" ViewCount="78845" Body="&lt;p&gt;Can someone please explain what exactly the difference between an intermediate and a transition state is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that they are formed as part of the process in converting the reactants to products in a chemical reaction. Other than that, all I know is that usually a transition state exists very briefly (if at all) whereas an intermediate can be isolated. But why? &lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="21296" LastEditDate="2017-12-09T11:53:33.503" LastActivityDate="2017-12-13T22:11:02.273" Title="Difference between intermediates and transition states" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28222" PostTypeId="1" AcceptedAnswerId="28228" CreationDate="2015-04-04T11:38:49.170" Score="8" ViewCount="22597" Body="&lt;p&gt;As explained by my instructor, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Allylic Halides form highly stable Transition States. They are stable because they are:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$sp^2$ hybridized&lt;/li&gt;&#xA;  &lt;li&gt;possess a conjugated system of pi bond and empty p orbital&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if we consider the $S_N1$ mechanism, a carbocation intermediate is stabilized by resonance. Why is the formation of the transition state more favorable than the formation of the resonance stabilized intermediate?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="4325" LastEditDate="2015-04-04T18:18:26.357" LastActivityDate="2015-04-04T18:18:26.357" Title="Why do Allylic Halides prefer SN2 reaction over SN1?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28223" PostTypeId="2" ParentId="26617" CreationDate="2015-04-04T11:44:58.033" Score="4" Body="&lt;p&gt;Both names, ‘4-bromo-2,4-dimethylhexane’ and ‘3-bromo-3,5-dimethylhexane’ are unambiguous and describe the same compound. However, only ‘4-bromo-2,4-dimethylhexane’ is in accordance with IUPAC recommendations.&lt;/p&gt;&#xA;&lt;p&gt;First of all, the &lt;em&gt;seniority order of functional classes&lt;/em&gt;, which is used to choose the functional class (here: …hexane), does not apply to &lt;em&gt;numbering of locants within the same compound class.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Concerning numbering of locants within the same compound class, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.3.5&lt;/strong&gt; Lowest set of locants&lt;/p&gt;&#xA;&lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with&#xA;other locant sets, each cited in order of increasing value, has the lowest term at the first&#xA;point of difference; (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the example is correctly named as &lt;strong&gt;4-bromo-2,4-dimethylhexane&lt;/strong&gt; rather than 3-bromo-3,5-dimethylhexane since the locant set ‘2,4,4’ is lower than ‘3,3,5’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/He9CG.png&quot; alt=&quot;4-bromo-2,4-dimethylhexane&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-04T12:14:43.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28224" PostTypeId="2" ParentId="28221" CreationDate="2015-04-04T12:53:58.840" Score="23" Body="&lt;p&gt;An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle. Intermediates may be unstable molecules (in which case they are called reactive intermediates) or highly stable molecules.&#xA;The difference between them can be better described through the energy profile diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition states are local energy maximums and have partial bonds. This might be one of the reasons why they cant be isolated as intermediates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abo58.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-04-04T12:53:58.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28227" PostTypeId="2" ParentId="28218" CreationDate="2015-04-04T15:17:09.357" Score="1" Body="&lt;p&gt;Sugar alcohols are only partially absorbed. It is very unlikely anyone will become intoxicated with sugar alcohols. Also, their effects are less likely to cause intoxications.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-04T15:17:09.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28228" PostTypeId="2" ParentId="28222" CreationDate="2015-04-04T15:27:06.410" Score="10" Body="&lt;p&gt;&lt;strong&gt;You and your instructor are both correct, it is your premise that's wrong.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your instructor's explanation about the allyl double bond stabilizing the $\mathrm{S_{N}2}$ transition state is correct. The following figure shows the stabilizing interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1TEdh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map:_Robert&amp;#39;s_%22Basic_Principles_of_Organic_Chemistry%22/14:_Organohalogen_%26_Organometallic_Compounds/14.03A:_2-Propenyl_(Aliylic)_Halides&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we remove the attacking nucleophile and the leaving group and place a positive charge in the system, then we have the $\mathrm{S_{N}1}$ allyl intermediate.  As you point out, this intermediate is resonance stabilized, and therefore, the transition state leading to it is also stabilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your premise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do Allylic Halides prefer SN2 reaction over Sn1?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not true.  There are many cases where allylic halides react preferentially by an $\mathrm{S_{N}1}$ process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;strong&gt;both&lt;/strong&gt; the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways.  We can shift from one mechanism to the other by changing reaction conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, allyl chloride reacts by an $\mathrm{S_{N}1}$ mechanism to produce allyl alcohol when we place it in a 50:50 mix of $\ce{H2O:EtOH}$ at 45°C.  If instead we run the reaction in the presence of ethoxide ion in ethanol at 45°C, the corresponding ethyl ether is formed by an $\mathrm{S_{N}2}$ process.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-04T15:27:06.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28230" PostTypeId="2" ParentId="220" CreationDate="2015-04-04T17:07:14.550" Score="0" Body="&lt;p&gt;Volatilty is the tendency of a substance to vapourize.&#xA;It is directly related to vapoure pressure of a substance(pressure exerted by the vapourse in equilibrium with the liquid at a given temperature).&#xA;more voletile substance have low BP because of weake intermolecular forces.&lt;/p&gt;&#xA;" OwnerUserId="15361" LastActivityDate="2015-04-04T17:07:14.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28231" PostTypeId="1" AcceptedAnswerId="29237" CreationDate="2015-04-04T17:15:01.177" Score="6" ViewCount="1968" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the rapid interconversion, tautomers are generally&#xA;  considered to be the same chemical compound. (&lt;a href=&quot;https://en.wikipedia.org/wiki/Tautomer&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But looking at the examples provided on that same Wikipedia article, it is obvious that different chemical reactions would take place in each case (for example enol vs. keto).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aTj9U.gif&quot; alt=&quot;Tautomer examples&quot;&gt;&#xA;(&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tautomers.gif&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-04-23T21:20:41.427" Title="Why tautomers are considered to be the same chemical compound?" Tags="&lt;isomers&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28232" PostTypeId="1" CreationDate="2015-04-04T17:22:49.323" Score="1" ViewCount="482" Body="&lt;p&gt;Looking at the examples in &lt;a href=&quot;https://en.wikipedia.org/wiki/Tautomer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tautomer on Wikipedia&lt;/a&gt;, it seems that the hydrogen needs to travel a significant distance (~20&amp;Aring;) for the interconversion to take place. What lets it happen so fast? Why interconversion is preferred over maintaining its position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aTj9U.gif&quot; alt=&quot;Tautomer examples&quot;&gt;&#xA;(&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tautomers.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2019-01-31T20:27:54.843" Title="Why interconversion in tautomers is so rapid?" Tags="&lt;tautomer&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28234" PostTypeId="1" CreationDate="2015-04-04T20:44:07.727" Score="1" ViewCount="3624" Body="&lt;p&gt;I have read that the actual definition of a strong acid is one which will fully dissociate in water, so please do not turn the problem on its head when answering. It is also my understanding that &lt;em&gt;every&lt;/em&gt; acid reaction is in fact reversible (just as with any other reaction) and yet the thing that is different about strong acid forward reactions is that they are shifted so far forward (the equilibrium constant is &lt;strong&gt;very&lt;/strong&gt; high) to the extent that hardly any of the reactants remain after reaction. The acid reactions follow the laws of chemical reaction kinetics and the reason for this is that the Gibbs free energy decrease of the acid reaction is fairly high and so according to a Maxwell-Boltzmann Curve of energy distributions, only a very small fraction of the molecules will have enough energy to actually revert back to their undissociated form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My real question is what is it about the electronic (or otherwise) structure of these compounds, for example $\ce{HCl}$, as opposed to weak acids, e.g. $\ce{CH3COOH}$, that makes them more easily dissociated in water. It might also be helpful to learn the same about strong and weak bases. I think it might have something to do with the electronegativity of the ions involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can help answer the above, please could you also explain the weak/strong conjugate rule (that if a weak/strong acid/base reacts it will produce a strong/weak conjugate base/acid respectively).&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2018-02-22T16:58:50.443" Title="What is the reason for strong acids/bases dissociating in water?" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;equilibrium&gt;&lt;ions&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28235" PostTypeId="2" ParentId="28234" CreationDate="2015-04-04T22:01:29.293" Score="3" Body="&lt;p&gt;The easy answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes down to the strength of the bonds. Or that covalent bond strength is directly related to the bond dissociation energy and thus dissociation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A little Physical Chemistry:&lt;/em&gt;&#xA;(The following is how Pauling explained this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physical chemistry, they have something called bond dissociation ($D_o$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Atkins (9th Ed 10.4): &quot;&lt;em&gt;bond dissociation energy, $D_o$, the energy required to separate the atoms to infinity or by the well depth $D_e$&lt;/em&gt;.&quot; - What is meant here by &lt;em&gt;well&lt;/em&gt; is a &lt;em&gt;potential well&lt;/em&gt; – for now just think of that as the 2D box talked about in physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets shift gears. You may be familiar with bond enthalpy (sometimes just called bond energies). A common formula used in general chemistry is $\Delta H_\mathrm{rxn} = n(\text{bonds broken}) - n(\text{bonds formed})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that $\text{bonds broken}$ or $\text{bonds formed}$ is actually referring to the bond energy or bond enthalpy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It actually works out that there is a direct relationship to bond dissociation energy and bond enthalpy; which makes sense. It is elegantly related by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H^o_\mathrm{bond} = D_o + \frac32 RT $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I guess the natural question to ask now is why do some bonds have a greater enthalpy of formation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was not well understood for a long time. In reality there is a lot of factors that can come into play. The way that Pauling initially came up with dissociation energy, $D_o$, in &lt;em&gt;The Nature of the Chemical Bond&lt;/em&gt; followed this protocol:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Spectroscopy, (I did this with FTIR for example, but there are several choices) to get a molecular potential energy curve. &lt;/li&gt;&#xA;&lt;li&gt;Extract the well depth, $D_e$&lt;/li&gt;&#xA;&lt;li&gt;Find the dissociation Energy using $D_o = D_e - \frac12 h\nu$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In modern thermo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the relationship with enthalpy, we can measure enthalpy with calorimetry and back calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the best description uses statistical thermodynamics … Which I will not get into. &lt;/p&gt;&#xA;" OwnerUserId="7957" LastEditorUserId="7475" LastEditDate="2016-10-24T19:36:08.717" LastActivityDate="2016-10-24T19:36:08.717" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28236" PostTypeId="1" CreationDate="2015-04-04T23:34:01.873" Score="3" ViewCount="6901" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Q: A buffer with a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of 4.44 contains &lt;span class=&quot;math-container&quot;&gt;$\pu{0.21 M}$&lt;/span&gt; of sodium benzoate and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.12 M}$&lt;/span&gt; of benzoic acid. What is the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{[H+]}$&lt;/span&gt; in the solution after the addition of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.050 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; to a final volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.3 L}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A: So these are the steps that I have taken for this:&lt;/p&gt;&#xA;&lt;p&gt;The reaction is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H5COOH +H2O -&amp;gt; C6H5COO- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$K_ \mathrm{a}$&lt;/span&gt; of benzoic acid is &lt;span class=&quot;math-container&quot;&gt;$6.3 × 10^{-5}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But, &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[C6H5COOH]}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We don't know the concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{[H+]}$&lt;/span&gt; so we solve for that. We then assume at equilibrium &lt;span class=&quot;math-container&quot;&gt;$\ce{[C6H5COO-]}$&lt;/span&gt; is equal to it.&lt;/p&gt;&#xA;&lt;p&gt;Then: &lt;span class=&quot;math-container&quot;&gt;$6.3×10^{-5}=\ce{\frac{[C6H5COO- ][H+]}{[C6H5COOH]}}$&lt;/span&gt; and&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{[H+]}=0.00272$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But this is wrong. I think I'm doing something wrong in my problem, but really can't figure out what.&lt;/p&gt;&#xA;" OwnerUserId="15366" LastEditorUserId="83758" LastEditDate="2020-06-10T18:27:00.347" LastActivityDate="2020-06-11T08:23:16.043" Title="Finding is the concentration of H⁺ in a benzoate buffer solution after adding HCl" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;concentration&gt;&lt;buffer&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28237" PostTypeId="1" CreationDate="2015-04-05T05:15:36.307" Score="1" ViewCount="127" Body="&lt;p&gt;There are van-der-Waals forces within non-polar molecules. I read it that is mainly due to the instantaneous movements of the electron cloud's distribution. When looking at symmetrical molecules like $\ce{CH4}$, the concept isn't hard to understand. But I have a question regarding the bioelectricity and action potential of human cell membrane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. Outside the cell membrane the ion concentration is high and inside it's low, thus it creates an instantaneous dipole effect between the cell wall. Can this be due to van-der-Waals forces?&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="4945" LastEditDate="2015-04-07T10:20:29.413" LastActivityDate="2015-04-07T10:20:29.413" Title="Is the momentary dipole in between a cell membrane due to the van-der-Waals forces?" Tags="&lt;intermolecular-forces&gt;&lt;dipole&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28239" PostTypeId="2" ParentId="28236" CreationDate="2015-04-05T08:48:10.753" Score="3" Body="&lt;p&gt;Initially, it's true that you have the following equilibrium between sodium benzoate and benzoic acid:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H5COOH + H2O &amp;lt;=&amp;gt; C6H5COO- + H3O+_\mathrm{(aq)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;When you add  &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H5COO- + H3O+_\mathrm{(aq)}  &amp;lt;=&amp;gt;  C6H5COOH + H2O }$$&lt;/span&gt;&#xA;Now, the equilibrium constant of this reaction is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K=\frac{1}{K_ a}=0.16 \times 10^5&amp;gt;10^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reaction is quantitative (total). So, we can rewrite it:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{C6H5COO- + H3O+_\mathrm{(aq)} -&amp;gt;  C6H5COOH + H2O }$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, we have to construct an ICE table:&#xA;Assuming that we didn't dilute the solution when we added &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; (this point is not clear in your question): &lt;span class=&quot;math-container&quot;&gt;$$\ce{[C6H5COOH]_0} = \pu{0.12 M}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[C6H5COO^-]_0} =  \pu{0.21 M}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]_0} = (\frac{0.050}{1.3} + 10^{-4.4}) \mathrm{M}\approx 0.0385 \mathrm{M}$$&lt;/span&gt;&#xA;As you can clearly seen &lt;span class=&quot;math-container&quot;&gt;$\ce{[H^+]_0}$&lt;/span&gt; is the lowest, and so ion hydronium is the limiting reactant.&#xA;At the end of the reaction, we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[C6H5COO^-]_\mathrm{(eq)}} = (0.21-0.0385) = \pu{0.1715 M}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[C6H5COOH]_\mathrm{(eq)}} = (0.12+0.0385) = \pu{0.1585 M}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]_\mathrm{(eq)}} = \epsilon$$&lt;/span&gt;&#xA;Substituting all these numbers in the expression of &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K = \frac{ 0.1585}{0.1715\epsilon}=0.16 \times 10^5$$&lt;/span&gt;&#xA;By solving the above equation, we find:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H^+]} = \epsilon = \pu{5.82 \times 10^{-5} M}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH} = 4.23$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="61578" LastEditDate="2020-06-11T08:23:16.043" LastActivityDate="2020-06-11T08:23:16.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28243" PostTypeId="2" ParentId="28210" CreationDate="2015-04-05T12:53:41.743" Score="1" Body="&lt;p&gt;The stability of keto-enol compounds depends on the solution they are present in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keto compounds are stabilized due to inter-molecular hydrogen bonding when present in a polar solvent (for example water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enol compounds are stable when present in non-polar solvents, in which intra-molecular hydrogen bonding stabilizes them.&lt;/p&gt;&#xA;" OwnerUserId="15031" LastActivityDate="2015-04-05T12:53:41.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28244" PostTypeId="1" AcceptedAnswerId="28269" CreationDate="2015-04-05T13:24:26.910" Score="2" ViewCount="741" Body="&lt;p&gt;My textbook says that degree of dissociation= [HA]dissociated/[HA]initial * 100%.&#xA;However, it also says that citric acid has a greater degree of dissociation than acetic acid because citric is triprotic whereas acetic is monoprotic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reasoning doesn't fit with the top definition because, more protons donated doesn't necessarily mean more acid molecules dissociated (triprotic acid vs monoprotic acid for example; for every acid molecule dissociated, triprotic acid donates more protons). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, how does one define degree of dissociation for polyprotic acids?&lt;/p&gt;&#xA;" OwnerUserId="6988" LastEditorUserId="6988" LastEditDate="2015-04-06T01:45:02.957" LastActivityDate="2021-05-28T08:17:35.727" Title="What is degree of dissociation for polyprotic acids?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28245" PostTypeId="1" CreationDate="2015-04-05T13:25:12.253" Score="1" ViewCount="94" Body="&lt;p&gt;I know that a magnetic dipole moment is given by $$\mu=\frac {-e}{2m}I$$ and that the z component of angular momentum is $$m_j\hbar.$$ However, I have also seen that angular momentum $I$ is given by $$I=\hbar\sqrt{l(l+1)}.$$ Is this also the component in the z direction or is it different? Are both expressions for angular momentum right? Which one do I use to answer the question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question was asked in a problem sheet that I have been asked to do over the vacation. See below for context of question).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/thlJC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="865" LastEditDate="2021-05-22T21:16:19.953" LastActivityDate="2021-05-22T21:16:19.953" Title="Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28246" PostTypeId="2" ParentId="16921" CreationDate="2015-04-05T14:27:02.320" Score="1" Body="&lt;p&gt;A rather quick googling shows up that there is some Longuet-Higgins theorem, which states almost what you are looking for. For instance, the abstract of &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/344/1637/147&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is proved that if the wave function of a given electronic state&#xA;  changes sign when transported adiabatically round a loop in nuclear&#xA;  configuration space, then the state must become degenerate with&#xA;  another one at some point within the loop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the paper itself it is further stated that this in turn implies that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[the corresponding] potential energy surface intersects one of another electronic state&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and that the theorem &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;makes it possible to diagnose the presence of an intersection between potential energy surfaces from the behaviour of the electronic wave function as it is transported adiabatically round a closed loop remote from the intersection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which answers both your questions, if I understand them correctly.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-04-05T14:27:02.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28247" PostTypeId="1" AcceptedAnswerId="28271" CreationDate="2015-04-05T15:03:25.300" Score="1" ViewCount="14804" Body="&lt;p&gt;What are the configurations of 1-bromo-3-methylcyclopent-1-ene and 1-nitro-2-(propan-2-yl)benzene? Are they &lt;em&gt;E&lt;/em&gt; or &lt;em&gt;Z&lt;/em&gt;? Are they &lt;em&gt;cis&lt;/em&gt; or &lt;em&gt;trans&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7qk9l.jpg&quot; alt=&quot;1-bromo-3-methylcyclopent-1-ene&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sWqyA.jpg&quot; alt=&quot;1-nitro-2-(propan-2-yl)benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the rules for these types of isomerism?&lt;/p&gt;&#xA;" OwnerUserId="13335" LastEditorDisplayName="user7951" LastEditDate="2016-11-23T00:40:42.720" LastActivityDate="2016-11-23T00:40:42.720" Title="What is the E/Z configuration of cyclic compounds?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28249" PostTypeId="2" ParentId="28247" CreationDate="2015-04-05T15:51:03.070" Score="4" Body="&lt;h3&gt;What is the meaning of &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt;?&lt;/h3&gt;&#xA;&lt;p&gt;&lt;em&gt;cis&lt;/em&gt; means &lt;em&gt;on the same side&lt;/em&gt; (or face), &lt;em&gt;trans&lt;/em&gt; means &lt;em&gt;on the opposite side&lt;/em&gt; (or face).&lt;/p&gt;&#xA;&lt;h3&gt;Can I use &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; for cylic compounds?&lt;/h3&gt;&#xA;&lt;p&gt;Yes, you can describe the relative orientation of substituents on a ring using &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt;, such as in the case of 1,3-dibromcyclohexane.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HUZZA.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Can I use &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; for alkenes?&lt;/h3&gt;&#xA;&lt;p&gt;Yes, in priciple. But you should rather use &lt;em&gt;&lt;strong&gt;E&lt;/strong&gt;&lt;/em&gt; and &lt;em&gt;&lt;strong&gt;Z&lt;/strong&gt;&lt;/em&gt;. Note that these descriptors are taken from German.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i61Fn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;E&lt;/em&gt; means &lt;strong&gt;entgegen&lt;/strong&gt; and refers to &lt;strong&gt;opposite&lt;/strong&gt; directions (or sites), while &lt;em&gt;Z&lt;/em&gt; means &lt;strong&gt;zusammen&lt;/strong&gt;, which is &lt;strong&gt;together&lt;/strong&gt;, that is &lt;strong&gt;the same&lt;/strong&gt; direction (or site).&lt;/p&gt;&#xA;&lt;h3&gt;What about the compounds in my question?&lt;/h3&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;E&lt;/strong&gt;&lt;/em&gt;/&lt;em&gt;&lt;strong&gt;Z&lt;/strong&gt;&lt;/em&gt; or &lt;em&gt;cis&lt;/em&gt;/&lt;em&gt;trans&lt;/em&gt; are not applicable here.&lt;/p&gt;&#xA;&lt;p&gt;In the first case, the &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; atom is in the ring plane, the methyl group can be above or below it. You could describe the absolute stereochemistry at the centre bearing the &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3}$&lt;/span&gt; substituent using &lt;em&gt;&lt;strong&gt;R&lt;/strong&gt;&lt;/em&gt; or &lt;em&gt;&lt;strong&gt;S&lt;/strong&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In the second case, there's no alkene at all but a disubstituted benzene. The relative orientation of the substituents on these systems can be described by assigning the proper numbers or by using the &lt;em&gt;ortho&lt;/em&gt;- &lt;em&gt;meta&lt;/em&gt;- or &lt;em&gt;para&lt;/em&gt;-prefix. For this compounds, 2-isopropylnitrobenzene or &lt;em&gt;o&lt;/em&gt;-isopropylnitrobenzene would be ok.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-05T20:02:33.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28250" PostTypeId="2" ParentId="13478" CreationDate="2015-04-05T16:21:57.107" Score="2" Body="&lt;p&gt;Simple commercially available tests contain &lt;strong&gt;formaldehyde dehydrogenase.&lt;/strong&gt; This enzyme selectively catalyzes the chemical reaction of formaldehyde:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH2O + NAD+ + H2O -&amp;gt; HCOO- + NADH + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enzymatic reaction is typically indicated by reduction of a tetrazolium salt to a coloured formazan dye.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-05T16:21:57.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28251" PostTypeId="1" CreationDate="2015-04-05T16:53:12.243" Score="4" ViewCount="12240" Body="&lt;p&gt;Polymorphism is defined as &lt;em&gt;the ability of a solid material to exist in more than one form or crystal structure&lt;/em&gt;, whereas allotropy is defined as &lt;em&gt;the property of some chemical elements to exist in two or more different forms, in the same physical state, known as allotropes of these elements&lt;/em&gt;. (From wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism... but this is always true? That is, both terms, polymorphism and allotropy refer always to crystal structures? Polymorphism definition mentions &quot;crystal structures&quot; but &quot;crystal&quot; does not appears in the definition of allotropy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&#xA;Thank you for your answer Then, in some cases allotropy is more general than polymorphism. For instance, dioxygen and ozone are allotropes but not polymorphisms. But... my question arises from my TRUE/FALSE set of questions which I have to answer, and one of them is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a non-crystal material the concept of allotropy or polymorphism does not exist. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would you answer to that? My conclusion is that the question is not clear... For polymorphism is true whereas for allotropy it is not.&lt;/p&gt;&#xA;" OwnerUserId="15371" LastEditorUserId="10185" LastEditDate="2015-05-18T12:28:55.280" LastActivityDate="2015-05-18T12:28:55.280" Title="Allotropy and polymorphism" Tags="&lt;crystal-structure&gt;&lt;allotropes&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28252" PostTypeId="2" ParentId="28251" CreationDate="2015-04-05T17:18:29.963" Score="2" Body="&lt;p&gt;In my own work, I've never used the term polymorph for anything but molecular crystals in a highly but differently ordered solid state.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, &lt;strong&gt;allotropy&lt;/strong&gt; is the &lt;em&gt;polymorphism&lt;/em&gt; of &lt;strong&gt;elements in the same state&lt;/strong&gt;. This does not mean that each solid allotrope has to be highly ordered: amorphous forms of carbon, phosphorus or selenium are considered as allotropes too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of allotropy isn't confined to the solid state: ozone is an allotrope of dioxygen.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-05T17:18:29.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28254" PostTypeId="1" AcceptedAnswerId="31875" CreationDate="2015-04-05T18:08:44.807" Score="6" ViewCount="103" Body="&lt;p&gt;How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couldn't there have been an entirely different negatively charged particle? What justified that assumption?&lt;/p&gt;&#xA;" OwnerUserId="15373" LastEditorUserId="7448" LastEditDate="2015-04-05T18:17:01.080" LastActivityDate="2015-05-23T09:13:37.387" Title="How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28255" PostTypeId="2" ParentId="28245" CreationDate="2015-04-05T19:01:09.147" Score="1" Body="&lt;p&gt;The first equation you have is for spin. Let's look at it in a operator form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ μ_B =-g \frac {eℏ}{2m}  \frac { \boldsymbol {S}}{ℏ} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and &lt;strong&gt;S&lt;/strong&gt; is the spin operator.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For orbital magnetic moment (your question) you use: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s: l=0 &#xA;p: l=1&#xA;d: l=2&#xA;f: l=3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So plugging in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$&#xA;$$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Units are $ \boldsymbol {\mu_B}$)&lt;/p&gt;&#xA;" OwnerUserId="7957" LastActivityDate="2015-04-05T19:01:09.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28256" PostTypeId="1" CreationDate="2015-04-05T20:29:17.587" Score="2" ViewCount="293" Body="&lt;p&gt;Is it possible to create a galvanic cell without a half cell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book I was reading, Illustrated Guide to Home Chemistry Experiments, the following lab (Laboratory 16.4) was suggested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;add 2 M HCl and add copper metal to one end of the beaker, with magnesium at the other end. Then, add patch cables to connect the metal plate to an LED to have physical evidence of success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't understand how useful work could be generated without half cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, we add HCl to react with the magnesium and copper, to produce ions (&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt;). Then, they can spontaneously produce work due to their eV differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am missing is how the work can be useful, as opposed to just generating heat in the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would believe that, rather than traveling though the patch cables, electrons can travel directly through the solution and produce heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading and I really appreciate any help!&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="24052" LastEditDate="2019-06-19T07:14:21.420" LastActivityDate="2019-06-20T03:14:33.863" Title="Galvanic Electrochemical Cell Without half cells" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28257" PostTypeId="1" AcceptedAnswerId="28259" CreationDate="2015-04-05T20:29:28.080" Score="1" ViewCount="969" Body="&lt;p&gt;Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately? &#xA;Normally by using $SV=constant$ I get pH will change by &lt;strong&gt;1 unit&lt;/strong&gt; (without taking common ion effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will common ion effect affect the answer? Help.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="15133" LastEditDate="2015-04-06T05:04:22.673" LastActivityDate="2015-04-12T18:54:21.213" Title="Effect of dilution on pH for HCl" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28259" PostTypeId="2" ParentId="28257" CreationDate="2015-04-05T21:10:14.250" Score="-2" Body="&lt;p&gt;Yes, you will definitely need to account for common ion effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should get a result around 6.79=pH.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="7448" LastEditDate="2015-04-12T18:54:21.213" LastActivityDate="2015-04-12T18:54:21.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28261" PostTypeId="1" CreationDate="2015-04-05T21:23:24.517" Score="4" ViewCount="6487" Body="&lt;p&gt;Why is pyridine so much more acidic than other similar imines (5.5 versus ~9, see picture)? Wouldn't delocalization of the positive charge into the aromatic system stabilize it and hence decrease acidity? Also the lone pair of pyridine is not delocalized into the perpendicular $\pi$-system and hence not any less available than the one of the projected imine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNt0x.png&quot; alt=&quot;Organic Chemistry by Clayden et al. p. 1149.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Organic Chemistry by Clayden et al. p. 1149. )&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="16683" LastEditDate="2017-02-27T15:47:00.613" LastActivityDate="2017-02-27T16:11:38.760" Title="pKaH of pyridine versus other imines" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28263" PostTypeId="1" AcceptedAnswerId="28265" CreationDate="2015-04-05T21:24:55.603" Score="7" ViewCount="2816" Body="&lt;p&gt;I recently stumbled across the substance &quot;p,p′-terphenyldicarboxylic acid&quot; in a scientific journal when recently reading about metal–organic frameworks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain the prefix p,p'- to me?&lt;/p&gt;&#xA;" OwnerUserId="15375" LastEditorDisplayName="user7951" LastEditDate="2015-12-29T17:09:35.757" LastActivityDate="2015-12-29T17:09:35.757" Title="Meaning of prefix p,p′-?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28264" PostTypeId="2" ParentId="28261" CreationDate="2015-04-05T21:30:25.183" Score="-1" Body="&lt;p&gt;One interpretation of acidity is the ability to stabilize a negative charge (since the electron is generally kept by the acid, and the proton harbors the positive charge, if we go by bronsted lowry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the negative charge is delocalized and thus stabilized. Its impact is not felt as greatly on one atom, and therefore it is more stable, not seeking a proton as much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the negative charge is better for nitrogen, an electronegative atom which can better stabilize it (but this doesn't explain the difference. I think the main difference is resonance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This website may help:&lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/heterocyclen/arom_gessaet_heterocyclen.vlu/Page/vsc/en/ch/12/oc/heterocyclen/sechsringaromat/sechsring_aromat.vscml.html&quot; rel=&quot;nofollow&quot;&gt;http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/heterocyclen/arom_gessaet_heterocyclen.vlu/Page/vsc/en/ch/12/oc/heterocyclen/sechsringaromat/sechsring_aromat.vscml.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="14975" LastEditDate="2015-04-05T21:44:41.263" LastActivityDate="2015-04-05T21:44:41.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28265" PostTypeId="2" ParentId="28263" CreationDate="2015-04-05T21:37:55.283" Score="7" Body="&lt;p&gt;I am guessing that this is the compound you are interested in:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/61CmI.png&quot; alt=&quot;p-Terphenyl-4,4′′-dicarboxylic acid&quot;&gt;&lt;br&gt;&#xA;Sigma labels this compound as p-Terphenyl-4,4′′-dicarboxylic acid (or 1,4-Bis(4-carboxyphenyl)benzene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the p,p' indicates that the 2 carboxylic acids are in the para positions on 2 of the phenyl rings.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-05T21:37:55.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28266" PostTypeId="2" ParentId="28263" CreationDate="2015-04-05T21:43:57.800" Score="6" Body="&lt;p&gt;As &lt;strong&gt;LDC3&lt;/strong&gt; has pointed out, the two &lt;em&gt;p&lt;/em&gt; stand for &lt;em&gt;para&lt;/em&gt;. But taking into account that there are three different terphenyls, &lt;em&gt;p,p'&lt;/em&gt;-terphenyldicarboxylic acid is ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w2aF6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An unambiguous name would denote both the terphenyl core, as well as the positions of the $\ce{COOH}$ substituents on the outer rings, such as in &lt;strong&gt;&lt;em&gt;p&lt;/em&gt;-terphenyl-4,4'-dicarboxylic acid&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-05T21:43:57.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28268" PostTypeId="2" ParentId="28261" CreationDate="2015-04-05T22:33:21.767" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is pyridine so much more acidic than other similar imines?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When we discuss the acidity of a compound, we are really discussing where the equilibrium between the acid and its conjugate base lies.  Therefore, when we compare acidities we are really comparing equilibria, the relative stabilities of the reactant (acid) and product (conjugate base).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following diagram illustrates the equilibrium between the iminium ion and the corresponding imine, it's conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0Zqi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't draw any good resonance structures for the imine, but we can draw 2 resonance structures to describe the iminium ion, A and B in the figure. In the pyridinium ion - pyridine case we can draw 2 resonance structures for pyridine (just like in benzene), but we can't draw any resonance structures for the pyridinium ion that delocalize the positive charge without destroying the aromaticity of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the iminium ion - imine case&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can stabilize the iminium ion through resonance - this stabilizes the iminium ion and shifts the equilibrium to the left making the iminium ion less acidic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whereas in the pyridinium ion - pyridine case&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can't draw good resonance structures that stabilize the positive charge in the pyridinium ion - this makes pyridinium ion more acidic.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both factors operate in the same direction and tend to make the pyridinium ion ($\mathrm{p}K_\mathrm{a}=5.5$) more acidic than the iminium ion ($\mathrm{p}K_\mathrm{a}$ $\sim$ 9).&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2017-02-27T16:11:38.760" LastActivityDate="2017-02-27T16:11:38.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28269" PostTypeId="2" ParentId="28244" CreationDate="2015-04-05T22:38:39.883" Score="4" Body="&lt;p&gt;I think this resembles the question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28013/confused-about-strong-weak-dibasic-and-tribasic-acids/28018#28018&quot;&gt;Confused about strong/weak dibasic and tribasic acids?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Saying that the degree of dissociation &lt;span class=&quot;math-container&quot;&gt;$(\alpha)$&lt;/span&gt; of citric acid is greater because its triprotic is not correct. The degree of dissociation should be calculated for each dissociation separately. The degree of dissociation of citric acid's first dissociation &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H8O7 &amp;lt;=&amp;gt; C6H7O7- + H+}$&lt;/span&gt; is greater than that of acetic acid, because its &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; is smaller (3.13 vs 4.76)&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;, meaning it's a stronger acid.&lt;/p&gt;&#xA;&lt;p&gt;The second value of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; for citric acid, however, will be very to close to acetic acid's, because the second dissociation has approximately the same &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; (4.76)&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;. The third &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; will be smaller, since the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of the third dissociation is greater (6.40)&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This can be seen in the common polyprotic acids dissociation graphs, like this one (&lt;a href=&quot;http://ion.chem.usu.edu/%7Esbialkow/Classes/3600/Overheads/H3A/H3A.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nNxrl.gif&quot; alt=&quot;polyprotic acids dissociation graphs&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is represented on the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-axis in a 0-1 scale. For each of the four states of the molecule, there is a different &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, and different values of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; happen simultaneously.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; values taken from Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="61578" LastEditDate="2021-05-28T08:17:35.727" LastActivityDate="2021-05-28T08:17:35.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28270" PostTypeId="1" CreationDate="2015-04-05T22:50:04.183" Score="0" ViewCount="5903" Body="&lt;p&gt;How do quantum numbers change for ions?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;A sample question:&lt;/em&gt; Give the set of four quantum numbers that could represent the electron lost to form the Rb &lt;strong&gt;ION&lt;/strong&gt; from the Rb atom. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer given is $n=5; l=0; m_l=0; m_s= +\frac12$. I chose the answer: $n=4; l=1; m_l=1; m_s= -\frac12$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an explanation. I am confused because I thought the question asked me to describe $\ce{Rb+}$ and the answer given is the same that I would choose for $\ce{Rb}$ at ground state. If it is asking me to describe the &lt;strong&gt;electron&lt;/strong&gt;, how is that technically possible if I don't know where the electron went? Sounds silly and I know I'm over thinking it. Some insight would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15377" LastEditorUserId="7475" LastEditDate="2016-12-08T21:11:30.307" LastActivityDate="2016-12-08T21:11:30.307" Title="How can I determine the quantum number of a electron lost during ion formation?" Tags="&lt;quantum-chemistry&gt;&lt;ions&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28271" PostTypeId="2" ParentId="28247" CreationDate="2015-04-06T01:26:53.127" Score="2" Body="&lt;p&gt;It's not necessary to include the &lt;em&gt;E/Z&lt;/em&gt; descriptors for these compounds, because the alternative geometries would be far too strained to be stable molecules. However, there is no reason that we can't apply the rules to these compounds and assign the &lt;em&gt;E/Z&lt;/em&gt; descriptor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case, compare at the substituents on each side of the pi-bond. For the top compound, one side of the alkene has Br compared to a carbon group, and Br is higher priority. On the other side, hydrogen is lower priority than a carbon group. The higher priority groups are trans, which makes the geometrical stereochemistry &lt;em&gt;E&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the bottom compound, we must compare two carbon groups, and the alkene containing one is higher priority. On the other side, nitro is higher priority than the carbon group. Again, the higher priority groups are trans, which means that the stereochemistry is &lt;em&gt;E&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9BGMo.jpg&quot; alt=&quot;compounds of interest&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it is unnecessary to define the &lt;em&gt;E/Z&lt;/em&gt; stereochemistry until the ring size is 8.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-04-06T01:26:53.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28272" PostTypeId="1" CreationDate="2015-04-06T03:00:15.787" Score="1" ViewCount="205" Body="&lt;p&gt;I'm studying for a chem exam and there are often questions of this form: What kind of inter-molecular interactions are important in (insert compound here). Then the possible answers are any combination of dipole-dipole, hydrogen bonding, and dispersion force.&#xA;My question is: When ever a molecule has hydrogen bonding does that mean there is also dipole-dipole? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you tell if a molecule has a dipole, for instance CH3COOH,  has one but how do you know?&lt;/p&gt;&#xA;" OwnerUserId="15379" LastActivityDate="2015-04-06T03:00:15.787" Title="How do you know what inter-molecular interactions are present?" Tags="&lt;intermolecular-forces&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28273" PostTypeId="1" AcceptedAnswerId="28280" CreationDate="2015-04-06T04:53:13.670" Score="1" ViewCount="4727" Body="&lt;p&gt;Which one of hydrogen and helium behaves more like ideal gas &lt;strong&gt;in room temperature&lt;/strong&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought hydrogen gas would show more ideal behavior in room temperature because it has lighter mass (less no of electron) although it's diatomic, so there would be less attraction.&#xA;However, I saw one question on the book and they said that the helium gas is more likely to show the ideal gas behavior in room temperature, not the hydrogen gas..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can someone please give me an idea how my thinking is wrong?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for advance&lt;/p&gt;&#xA;" OwnerUserId="15380" LastActivityDate="2015-04-06T08:57:41.163" Title="Hydrogen and Helium ideal gas behavior in room temperature" Tags="&lt;inorganic-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28275" PostTypeId="1" AcceptedAnswerId="28276" CreationDate="2015-04-06T05:45:26.587" Score="4" ViewCount="7490" Body="&lt;p&gt;I'm specifically talking about the decomposition of hydrogen peroxide by bovine catalase. The number of catalase molecule obviously doesn't change, so there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, and doesn't change the pH or temperature, then shouldn't the rate of reaction remain the same regardless of concentration, since every environmental conditions &amp;amp; the number of catalase molecule in the solution before dilution is the same after it is diluted?&lt;/p&gt;&#xA;" OwnerUserId="11490" LastActivityDate="2015-04-06T14:58:51.070" Title="When an enzyme is diluted with water, what is slowing down the rate of reaction?" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;catalysis&gt;&lt;enzymes&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28276" PostTypeId="2" ParentId="28275" CreationDate="2015-04-06T06:02:16.153" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: Why is dating between a fixed number of guys and girls more difficult on a large square packed with nuns than in a small club (without the nuns)? ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;: It's more difficult to bump into each other since the girls a more far away and there are legions of nuns in between!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right in that the absolute amount of catalase molecules does not change upon dilution, but the &lt;strong&gt;concentration&lt;/strong&gt; does!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions are about collisions of the reactants and in the case of dilutions, there are more collisions with &lt;em&gt;innocent&lt;/em&gt; bystanders than with an other reactant.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-06T08:14:56.307" LastActivityDate="2015-04-06T08:14:56.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28277" PostTypeId="1" AcceptedAnswerId="28278" CreationDate="2015-04-06T08:00:30.757" Score="0" ViewCount="150" Body="&lt;p&gt;A task requires us to experiment water quality.&#xA;How would you reduce turbidity of a clay water sample with an effective and cost-effective method? &lt;/p&gt;&#xA;" OwnerUserId="15384" LastEditorUserId="418" LastEditDate="2016-11-08T02:49:13.797" LastActivityDate="2016-11-08T02:49:13.797" Title="How to reduce the turbidity of a water sample?" Tags="&lt;experimental-chemistry&gt;&lt;filtering&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2015-04-07T07:11:21.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="28278" PostTypeId="2" ParentId="28277" CreationDate="2015-04-06T08:30:50.000" Score="1" Body="&lt;p&gt;Apart from the &lt;a href=&quot;http://dx.doi.org/10.5402/2011/632189&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; linked in a comment above, possible options are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;wait and let settle, then decant&lt;/li&gt;&#xA;&lt;li&gt;use an established flocculant, such as potassium alum $\ce{KAl(SO4)2*12H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-06T08:30:50.000" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28279" PostTypeId="1" CreationDate="2015-04-06T08:54:42.577" Score="0" ViewCount="160" Body="&lt;p&gt;I want to take a training on computational works in chemistry. It should be of a short duration. Where in Europe I can find it?&lt;/p&gt;&#xA;" OwnerUserId="15386" LastActivityDate="2015-04-06T09:21:50.850" Title="short/extensive training on computational chemistry" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-06T13:33:31.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="28280" PostTypeId="2" ParentId="28273" CreationDate="2015-04-06T08:57:41.163" Score="1" Body="&lt;p&gt;For inert gas molecules, intermolecular interactions would be by Van der Waals forces, which are based on the total volume and polarizability of the gas molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diatomic hydrogen is larger and more polarizable of the two.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-06T08:57:41.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28281" PostTypeId="1" CreationDate="2015-04-06T09:10:33.567" Score="2" ViewCount="88" Body="&lt;p&gt;For example, would a box of baking soda in a fridge lose weight as it absorbs odor? How long would it take to drop 100 g?&lt;/p&gt;&#xA;" OwnerUserId="15387" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:08:41.507" LastActivityDate="2015-10-30T17:08:41.507" Title="Is there a visible change to baking soda when it reacts with odour?" Tags="&lt;everyday-chemistry&gt;&lt;absorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28282" PostTypeId="2" ParentId="28256" CreationDate="2015-04-06T09:12:18.367" Score="0" Body="&lt;p&gt;Electrons cannot move through the solution. Only ions can. Therefore the circuit will still complete properly and work will be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar idea to the lemon battery. Two dissimilar metals will have different potentials and things will work out.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-06T09:12:18.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28283" PostTypeId="2" ParentId="28279" CreationDate="2015-04-06T09:21:50.850" Score="1" Body="&lt;p&gt;Did you check the offerings of the national chemical societies and their special interest groups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Germany, that would be the &lt;a href=&quot;https://www.gdch.de/netzwerk-strukturen/fachstrukturen/chemie-information-computer-cic.html&quot; rel=&quot;nofollow&quot;&gt;CIC - Chemie Information Computer&lt;/a&gt; group of the &lt;a href=&quot;https://www.gdch.de&quot; rel=&quot;nofollow&quot;&gt;GDCh - Gesellschaft Deutscher Chemiker&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cosmologic provides &lt;a href=&quot;http://www.cosmologic.de/support-download/workshops.html&quot; rel=&quot;nofollow&quot;&gt;TURBOMOLE workshops&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2014, the Max Planck Institute for Chemical Energy Conversion organized a &lt;a href=&quot;http://www.cec.mpg.de/downloads/summer-school-2014.html&quot; rel=&quot;nofollow&quot;&gt;Summer School&lt;/a&gt; during which a course on QM/MM with Frank Neese's ORCA was given. I don't know if they plan something similar this year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.nsccs.ac.uk/events.php&quot; rel=&quot;nofollow&quot;&gt;NSCCS&lt;/a&gt; has some upcoming events.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-06T09:21:50.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28284" PostTypeId="1" CreationDate="2015-04-06T10:15:55.627" Score="6" ViewCount="13994" Body="&lt;p&gt;I need help understanding the following statement in my book (regarding trigonal bipyramidal geometry):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Placing a lone pair in an axial position would put it at 90 degree to three other electron domains. Placing it in an equatorial put is at 90 degree to only two other domains, thus minimizing the number of strong repulsive interactions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does it mean by &quot;90 degree&quot; to two/three &quot;domains&quot;?&lt;/p&gt;&#xA;" OwnerUserId="15381" LastEditorUserId="5026" LastEditDate="2018-03-01T06:01:14.723" LastActivityDate="2018-03-01T06:01:14.723" Title="How can placing a lone pair in axial position differs from that in equatorial in number of electron domains?" Tags="&lt;bond&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28285" PostTypeId="2" ParentId="28277" CreationDate="2015-04-06T10:41:50.440" Score="1" Body="&lt;p&gt;Expensive method is to use heat to evaporate the water, unless you have cheap heat (e.g. sunlight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheap way is to push or suck it through a filter. You can buy bottles with filters in places that sell camping supplies.&lt;/p&gt;&#xA;" OwnerUserId="15388" LastActivityDate="2015-04-06T10:41:50.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28286" PostTypeId="1" CreationDate="2015-04-06T11:34:00.703" Score="4" ViewCount="1420" Body="&lt;p&gt;I have the &lt;sup&gt;1&lt;/sup&gt;H, &lt;sup&gt;13&lt;/sup&gt;C, and DEPT spectra of an unknown compound, listed in text form below, but I have also attached original images of the spectra. I know that it is an amino acid that has a cyclic structure and a relative molecular mass of 145. I'm supposed to deduce the structure, but am having some trouble with it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;H NMR (in D&lt;sub&gt;2&lt;/sub&gt;O): 4.2 ppm (1H, s), 3.85 ppm (1H, dd), 3.2 ppm (2H, m), 2.1 ppm (1H, d), 1.8 ppm (3H, m) (&lt;a href=&quot;https://i.stack.imgur.com/y2KaR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image of spectrum&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;13&lt;/sup&gt;C NMR: 174.27 ppm (quaternary), 61.92 ppm (CH&lt;sub&gt;2&lt;/sub&gt;), 54.09 ppm (CH&lt;sub&gt;2&lt;/sub&gt;), 38.69 ppm (CH/CH&lt;sub&gt;3&lt;/sub&gt;), 32.87 ppm (CH/CH&lt;sub&gt;3&lt;/sub&gt;), 28.04 ppm (CH/CH&lt;sub&gt;3&lt;/sub&gt;) (&lt;a href=&quot;https://i.stack.imgur.com/fmmaF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image of 1D &lt;sup&gt;13&lt;/sup&gt;C spectrum&lt;/a&gt;; &lt;a href=&quot;https://i.stack.imgur.com/SQ9Tx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image of DEPT spectrum&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the spectral data, I can tell that the compound doesn't have an aromatic ring, but beyond that I don't know how to proceed. Unfortunately, although it would be really helpful, I'm not able to determine the coupling constants in the &lt;sup&gt;1&lt;/sup&gt;H spectrum.&lt;/p&gt;&#xA;" OwnerUserId="15389" LastEditorUserId="16683" LastEditDate="2018-09-07T14:00:51.697" LastActivityDate="2018-09-07T14:08:41.973" Title="Identifying the structure of an amino acid from NMR spectra" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;&lt;amino-acids&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28287" PostTypeId="1" CreationDate="2015-04-06T13:13:48.990" Score="5" ViewCount="1539" Body="&lt;p&gt;I don't understand how to interpret graphs like the following:&lt;img src=&quot;https://i.stack.imgur.com/fZGnb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget the difference between the Harmonic and Morse (I think I get that), It's more what the horizontal lines mean and what the boundary line means. I know that vibrational energy is quantized and I can rationalize the shape of the Lennard-Jones potential but I don't understand how they're combined in this diagram. Please could you walk me through what's going on?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-07T01:35:43.190" Title="Lennard-Jones potential and vibrational energy level diagram explanation" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28288" PostTypeId="2" ParentId="28284" CreationDate="2015-04-06T13:55:17.200" Score="9" Body="&lt;p&gt;Here's a drawing of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9tlx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example with the axial lone pair of electrons, when they say, &quot;90 degree to three other electron domains&quot;, they just mean that the axial lone pair is situated 90° relative to 3 other substituent bonds - the 3 equatorial substituent bonds in this case.  When the lone pair is located in an equatorial position, then it has a 90° interaction with only 2 other bonds - the 2 axial substituent bonds in this case; the lone pair is 120° away from the 2 other equatorial substituents.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-06T13:55:17.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28289" PostTypeId="2" ParentId="28275" CreationDate="2015-04-06T14:58:51.070" Score="3" Body="&lt;p&gt;If you diluted the catalase, &lt;em&gt;while keeping the peroxide concentration fixed&lt;/em&gt; (by adding both water and peroxide to the solution), then you're right, the total reaction rate (measured in reactions catalyzed per second, within the whole solution) would not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by adding just water to the solution, you're diluting both the enzyme &lt;em&gt;and&lt;/em&gt; the peroxide.  The lower the peroxide concentration is, the longer it takes for each catalase enzyme unit to bump into another peroxide molecule after reacting with the previous one.&lt;/p&gt;&#xA;" OwnerUserId="283" LastActivityDate="2015-04-06T14:58:51.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28290" PostTypeId="2" ParentId="28281" CreationDate="2015-04-06T15:19:58.760" Score="2" Body="&lt;p&gt;Let us assume that baking soda works by not simply absorbing, but reacting with acidic/odiferous substances via an acid-base reaction. So, for example if you had rancid butter in the fridge and it gave off the nasty-smelling butyric acid, that acid would react with the baking soda to give sodium butyrate plus carbonic acid (which breaks down to carbon dioxide and water). Let us assume that the carbon dioxide and water would evaporate. That would be a net gain of weight since the butyric acid molecule has a bigger mass that carbon dioxide and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most odiferous substances will likely have a higher molecular weight than carbon dioxide plus water, so I believe over time the baking soda will increase in weight. This will also be true for those cases where there is no reaction - but simply absorption.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-06T15:19:58.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28291" PostTypeId="1" CreationDate="2015-04-06T15:37:07.267" Score="6" ViewCount="83" Body="&lt;p&gt;If one reads reviews on ceramic glazed cookware (esp. frying pans) one pattern becomes apparent: the pan is awesome at first, nothing sticks to it, even with little to no oil. Then, after a few months the non-stick properties more or less plummet and the pan may become worse than stainless steel pans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why that is, I've read multiple claims. Some say (or assume) that the non-stick layer erodes/dissolves/gets damaged. But to my humble knowledge ceramic is harder than steel, so that seems somewhat simplistic of an explanation. Some sources say that the coating reacts with oil when overheated (which will happen on high settings of the stove) and that residue is what dimishes the non-stickiness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That sounded more plausible to me and I wondered if that proccess can be reversed/undone in some way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first part of my question: how does a ceramic pan deteriorate? If it's not plain &quot;erosion&quot;, what can be done to restore its properties? I'm afraid to scrub if harder for the fear of scrathing it. &lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="7475" LastEditDate="2015-10-30T16:50:19.663" LastActivityDate="2015-10-30T16:50:19.663" Title="What makes ceramic coated pans fail? Is it reversible?" Tags="&lt;everyday-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-10-30T21:25:26.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="28292" PostTypeId="1" CreationDate="2015-04-06T15:49:03.493" Score="2" ViewCount="493" Body="&lt;p&gt;For hydrogen fuel cell to produce electricity, hydrogen  will need to be supplied to the anode. To split oxygen and hydrogen from water and obtain hydrogen, we need to use electrolysis. To use electrolysis, we need to use a battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why do we bother splitting up hydrogen and oxygen and then feed it into the fuel cell instead of just using battery to generate electricity? Or is it not practical to obtain hydrogen and oxygen from electrolysis and use it to generate electricity from fuel cell? Thank you&lt;/p&gt;&#xA;" OwnerUserId="15395" LastEditorUserId="4356" LastEditDate="2015-04-06T19:37:09.857" LastActivityDate="2015-04-06T19:37:09.857" Title="Hydrogen Fuel Cell and Battery" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;electrolysis&gt;&lt;hydrogen&gt;&lt;fuel&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28293" PostTypeId="2" ParentId="28292" CreationDate="2015-04-06T17:19:06.720" Score="0" Body="&lt;p&gt;It is more likely they use the mains to electrolyze the water, but hydrogen is also a byproduct in the steam reforming of crude oil, which produced &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_production&quot; rel=&quot;nofollow&quot;&gt;11 million tonnes&lt;/a&gt; of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is having to put a large number of heavy batteries into the car, or a smaller, lighter (but still reinforced) tank of hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-06T17:19:06.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28294" PostTypeId="1" AcceptedAnswerId="28295" CreationDate="2015-04-06T17:29:27.990" Score="4" ViewCount="259" Body="&lt;p&gt;Alkylation of acetone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/quiP0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is done with a strong base (LDA in this case). Is this because the protons aren't particularly acidic? Secondly it should be done at low temperatures; why is this so? -78 degrees Celsius is typical. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-06T17:59:02.837" Title="Why should alkylation of enols/carbonyl compounds be done at low temperature with a strong base?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28295" PostTypeId="2" ParentId="28294" CreationDate="2015-04-06T17:59:02.837" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why should alkylation of enols/carbonyl compounds be done at low&#xA;  temperature with a strong base?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you start with an unsymmetrical ketone, then there are two different sides of the molecule you could remove a proton from. Let's use methyl ethyl ketone as an example.  Removing a proton from the methyl group leads to what is termed the &lt;strong&gt;kinetically&lt;/strong&gt; favored anion (&lt;em&gt;e.g.&lt;/em&gt; this hydrogen is the most acidic and the corresponding enolate anion is the fastest formed anion).  Removing a proton from the methylene group on the ethyl side leads to the &lt;strong&gt;thermodynamically&lt;/strong&gt; favored anion (&lt;em&gt;e.g&lt;/em&gt; the enolate anion with the most highly substituted [most stable] double bond).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkylation of carbonyls/enols is not always done using strong base and low temperature.  Strong base and low temperature generally leads to the kinetically favored enolate anion, while weaker bases and higher temperatures generally lead to the thermodynamically favored enolate anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In your case&lt;/strong&gt;, acetone is symmetric so it doesn't really matter which side you alkylate.  &lt;strong&gt;But&lt;/strong&gt;, once you form your desired product, methyl ethyl ketone, it can react further to give polyalkylated products.  These secondary products are more stable and thermodynamically preferred.  To reduce the likelihood of thermodynamic control leading to these undesired products, the reaction is run under kinetic control - strong base and low temperature - to maximize formation of the monoalkylated product.  Here is an &lt;a href=&quot;https://chemistry.stackexchange.com/questions/17103/why-is-structure-b-disfavored/17112#17112&quot;&gt;earlier answer&lt;/a&gt; that further highlights kinetic &lt;em&gt;vs.&lt;/em&gt; thermodynamic control in alkylation reactions.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-04-06T17:59:02.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28296" PostTypeId="1" AcceptedAnswerId="28297" CreationDate="2015-04-06T18:20:17.633" Score="6" ViewCount="843" Body="&lt;p&gt;What is the product of the permanganate oxidation of 1-methylindane (1-methyl-2,3-dihydro-1⁠&lt;em&gt;H&lt;/em&gt;-indene)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5LMR.png&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two choices of benzylic protons (one proton on the carbon with the methyl group at the top of the cyclopentane, and two protons on the carbon in the cyclopentane at the bottom of the molecule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure that the ketone is added to the carbon that has two benzylic protons, and that the other carbon with a benzylic proton is untouched. Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorDisplayName="user7951" LastEditDate="2016-11-23T15:10:53.440" LastActivityDate="2016-11-23T15:10:53.440" Title="Benzylic Oxidation of Cyclic Molecules Using KMnO₄" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28297" PostTypeId="2" ParentId="28296" CreationDate="2015-04-06T18:41:59.880" Score="8" Body="&lt;p&gt;For the homogenous permanganate oxidation of aromatic side chains, as long as there is at least one benzylic hydrogen, then the side chain will be oxidized to the corresponding benzoic acid.  Here are some examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EBrgq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.masterorganicchemistry.com/reaction-guide/oxidation-of-aromatic-alkanes-with-kmno4-to-give-carboxylic-acids/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the third example, the permanganate oxidation of tetralin to phthalic acid.  It is very analogous to your molecule.  &lt;strong&gt;Your molecule would also be oxidized to phthalic acid.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-06T18:41:59.880" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28298" PostTypeId="2" ParentId="28292" CreationDate="2015-04-06T19:10:15.823" Score="1" Body="&lt;p&gt;If you have a fuel cell, only hydrogen needs to be carried with the car; the air provides all the oxygen you need. Hydrogen is an extremely light fuel, but there are some issues with current means to contain it (e.g. heavy high pressure container, or massive hydride tanks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://cleantechnica.com/2013/09/18/iceland-bullish-hydrogen-video/&quot; rel=&quot;nofollow&quot;&gt;Iceland is Bullish on Hydrogen&lt;/a&gt;, they've started to use their plentiful geothermal and hydroelectric power to electrolyze water, using the hydrogen to power cars and the oxygen byproduct for other uses.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-06T19:10:15.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28299" PostTypeId="1" CreationDate="2015-04-06T20:54:32.667" Score="8" ViewCount="263" Body="&lt;p&gt;Ins this video &lt;a href=&quot;https://youtu.be/txk-VO1hzBY?t=2391&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/txk-VO1hzBY?t=2391&lt;/a&gt; the speaker claims that $$\ce{H2 + O2 -&amp;gt; H2O}$$ involves 23 reactions and 11 species. How do you list all of that ?&lt;/p&gt;&#xA;" OwnerUserId="15398" LastEditorUserId="12860" LastEditDate="2015-10-30T19:03:32.357" LastActivityDate="2017-03-05T23:01:18.607" Title="How do you list all possible reactions and intermediates of a reaction?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28301" PostTypeId="1" CreationDate="2015-04-06T22:37:10.490" Score="1" ViewCount="119" Body="&lt;p&gt;Can someone please help me find the equation that defines the relation between &lt;em&gt;molar conductivity&lt;/em&gt; and &lt;em&gt;temperature&lt;/em&gt;. I have found some information, but it was not really that much complete.&lt;/p&gt;&#xA;" OwnerUserId="15401" LastEditorUserId="4945" LastEditDate="2015-04-13T12:53:57.103" LastActivityDate="2015-04-13T12:53:57.103" Title="What is the equation that demonstrates the relationship between molar conductivity and temperature?" Tags="&lt;concentration&gt;&lt;temperature&gt;&lt;conductivity&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28302" PostTypeId="1" CreationDate="2015-04-06T22:44:13.947" Score="4" ViewCount="1254" Body="&lt;p&gt;Well I have to do research about the HPLC. It says, that because of the pressure the stationary phase can be made out of smaller particles and the pillars can be shorter. Also the seperation of the contents is supoosed to be superior, compared to other kinds of liquid chromatographies. Can somebody explain that to me?&lt;/p&gt;&#xA;" OwnerUserId="15402" LastActivityDate="2015-04-07T20:22:40.413" Title="What does the pressure in the HPLC(High performance liquid chromatography) do?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28303" PostTypeId="2" ParentId="1020" CreationDate="2015-04-07T01:01:37.043" Score="4" Body="&lt;p&gt;I assume you are aware of the common clock-reactions, such as the iodine clock, etc. By varying concentration of reagents and pH of the solution, you could run several repetitions to generate a trend-line, and from that extrapolate the estimated pH and reagent concentration to provide you with the time delay you are looking for. The iodine clock is a two-step reaction as outlined below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first, slow reaction, iodine is produced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 + 2I- + 2H+ -&amp;gt; I2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2S2O3^2- + I2 -&amp;gt; S4O6^2- + 2I-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By decreasing pH in reaction 1 and increasing the thiosulfate level in reaction 2, you should be able to lengthen the ‘clock time’. Let me know how things turn out!&lt;/p&gt;&#xA;" OwnerUserId="15404" LastEditorDisplayName="user7951" LastEditDate="2015-04-12T12:11:09.500" LastActivityDate="2015-04-12T12:11:09.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28304" PostTypeId="1" AcceptedAnswerId="28313" CreationDate="2015-04-07T01:24:04.430" Score="3" ViewCount="2152" Body="&lt;p&gt;This is a practical question about the use of &lt;em&gt;rotary evaporators&lt;/em&gt; (&quot;rotovap&quot;). So I'm asking for answers from people who actually use the things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotovap typically uses a cold condenser to condense the solvent into a collection flask. But the distillation is typically done under weak vacuum as well. So the solvent sitting in the collection flask seem susceptible to re-vaporizing and getting sucked into the vacuum pump. Now I have a cold trap in front of my pump that should be cold enough (-50C) to trap most solvents under weak vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is that if the vacuum is weak enough to vaporize the solvent under the hot water bath, but not under the cooler ambient air temperature, then the loss to the pump should be minimal. Indeed, this is what the instructions for the rotovap suggest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question: In practice, what are the risks to losing solvent into the pump?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="418" LastEditDate="2016-10-31T08:02:26.133" LastActivityDate="2016-10-31T08:02:26.133" Title="How to prevent losing solvent into the vacuum pump in rotovap solvent collection?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28305" PostTypeId="1" CreationDate="2015-04-07T01:26:10.227" Score="0" ViewCount="560" Body="&lt;p&gt;I'm just starting OChem and I see that for a SN1 reaction, the rate of the reaction changes with the number of alkyl groups attached to the carbocation that is formed. The textbook (Klein) explains that this is due to the electron-donating nature of alkyl groups which helps to stabilize the carbocation. I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Klein, David. Organic Chemistry as a Second Language, First Semester.&lt;/p&gt;&#xA;" OwnerUserId="15405" LastActivityDate="2015-04-07T01:45:49.570" Title="What happens to a molecule when it is unstable?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28306" PostTypeId="2" ParentId="28287" CreationDate="2015-04-07T01:35:43.190" Score="2" Body="&lt;p&gt;This diagram is conceptually horrible, because it is a mix of quantum and pseudo-classical physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget for a while the vibrations. If you calculate single point energies (within Born-Oppenheimer approximation) for different internuclear values, you'll get something like blue curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the $r_e$ distance between nucleus (again, within Born-Oppenheimer approximation) you can calculate the &quot;vibrational (normally harmonic) frequency&quot; in a pseudo-classic fashion. For this frequency, you can assign energies (of a quantum harmonic oscillator having this frequency). This energy values correspond to the difference between energy values for the horizontal blue lines, and the well (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The widths of the horizontal lines have not physical meaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness, the $D_e$ would be the energy of equilibrium state if no vibrations are considered. $D_0$ when the fundamental vibrational state energy is add. The former energy is not measurable because this is not a real energy for the system.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastActivityDate="2015-04-07T01:35:43.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28307" PostTypeId="2" ParentId="28305" CreationDate="2015-04-07T01:45:49.570" Score="1" Body="&lt;p&gt;Think of a &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_06%3A_Introduction_to_organic_reactivity_and_catalysis/Section_6.2%3A_Energy_diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction-energy diagram&lt;/a&gt;.  Specifically, think of a &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/5333/image041.png?revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;two-step reaction energy diagram&lt;/a&gt; linked like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oNJpo.png&quot; alt=&quot;Two-step reaction-energy diagram&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbocation is the intermediate $\mathbf{I}$.  The relative energy for alkyl-substituted carbocations will be lower.  This also practically means that $\mathrm{TS1}$ will have a lower relative energy as well.  The fact that $\mathrm{TS1}$ and $\mathbf{I}$ are of &lt;em&gt;lower&lt;/em&gt; energy for a substituted carbocation (vs. a primary or methyl carbocations) means that it will be formed more often under the &lt;a href=&quot;http://en.wikipedia.org/wiki/Transition_state_theory#Derivation_of_the_Eyring_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;quasi-equilibrium assumption&lt;/a&gt; of transition state theory.  Basically, if the barrier is lower, then then the intermediate can form more often.  And thus the reaction can happen more often.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you are correct as far as your thought goes.  If you could somehow instantaneously create a beaker full of primary carbocations, they might well react more quickly than other carbocations.  (And that's not only via the SN1 reaction we are considering but also via other pathways such as elimination as well.)  But instantaneously creating a beaker full of primary carbocations is &lt;em&gt;way more difficult&lt;/em&gt; than creating a corresponding beaker full of secondary or tertiary carbocations.  So (referring to the diagram again) yes, &lt;strong&gt;iff&lt;/strong&gt; you have lots of $\mathbf{I}$, for $\mathbf{I}=$ primary carbocations, reactions would probably be quicker than the case where $\mathbf{I}=$ higher carbocations.  But in reality in the primary carbocation case essentially no $\mathbf{I}$ is ever generated.  So the overall reaction rate is slower.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-07T01:45:49.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28311" PostTypeId="1" CreationDate="2015-04-07T04:04:06.787" Score="-2" ViewCount="5228" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If the absorbance of one of the buffer solutions (pH = 4.623) was 0.319 and the absorbance of the basic solution was 0.625, what is the $\mathrm{p}K_\mathrm{a}$ of the indicator? Include activity coefficients in the calculations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is 4.927, but I don’t know how to approach the problem to get the right answer. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="15409" LastEditorDisplayName="user7951" LastEditDate="2015-04-08T15:16:45.970" LastActivityDate="2015-04-08T15:16:45.970" Title="How to find pKa given absorbances and pH" Tags="&lt;acid-base&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2015-04-07T10:35:58.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="28313" PostTypeId="2" ParentId="28304" CreationDate="2015-04-07T06:00:06.410" Score="2" Body="&lt;p&gt;Honestly, I never bothered to actually measure the volume of the condensed solvent in order to compare it with the original volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept described in the question, that is using a cooling trap with dry ice + acetone, is good practise and typically good enough to retain the evaporated solvent and protect the membrane pump. In rare cases, I have used a second, flat Dewar filled with dry ice + acetone on a &lt;a href=&quot;http://de.wikipedia.org/wiki/Laborboy&quot; rel=&quot;nofollow&quot;&gt;&quot;Laborboy&quot;&lt;/a&gt; to cool the condensate-collecting flask.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-07T06:00:06.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28314" PostTypeId="1" CreationDate="2015-04-07T06:29:16.200" Score="5" ViewCount="1763" Body="&lt;p&gt;I have not been able to get my head around it yet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance in the reaction between ammonia and borontrifloride why is it that a positive and a negative charge results? Is it that ammonia donates is lone pair completely? If not so, if sharing of lone pair donated by ammonia occurs why then should charges result?&lt;/p&gt;&#xA;" OwnerUserId="15411" LastEditorUserId="7475" LastEditDate="2015-06-09T22:50:22.007" LastActivityDate="2015-06-09T22:50:22.007" Title="Why do charges result on atoms when coordinate bonds are formed?" Tags="&lt;bond&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28316" PostTypeId="1" AcceptedAnswerId="28326" CreationDate="2015-04-07T06:50:45.540" Score="2" ViewCount="1009" Body="&lt;p&gt;Basic and acidic materials exist in nature, so it makes sense that our forefathers could know what happens when they react. I would like to know when and how did they learn about this. In particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. What is the earliest time in which people could witness acid-base reaction that causes fire or explosion (as in &lt;a href=&quot;https://youtu.be/X_vMr-3P0KQ&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;)?  What are the earliest written accounts of such reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. What are the earliest written accounts of neutralization reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. When were acid-base reactions intentionally used, e.g. for neutralizations or to create explosions?&lt;/p&gt;&#xA;" OwnerUserId="15413" LastEditorUserId="15413" LastEditDate="2015-04-07T07:30:15.467" LastActivityDate="2016-03-11T23:25:45.190" Title="History of acid-base reactions" Tags="&lt;acid-base&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28317" PostTypeId="2" ParentId="28311" CreationDate="2015-04-07T09:11:43.573" Score="1" Body="&lt;p&gt;The following only gives some directions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar extinction coefficient (molar absorptivity) is not necessary to correlate the absorbance and the concentration of the observed species, supposed that the same species is measured at one particular wavelength at different pH values!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the &lt;strong&gt;Lambert-Beer Law&lt;/strong&gt;, we can derive that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\frac{E_1}{E_2} = \frac{\epsilon\cdot c_1\cdot d}{\epsilon\cdot c_2\cdot d} = \frac{c_1}{c_2}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The ratio of absorbances is linear proportional to the ratio of concentrations.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For a pH of 4.623, it seems fair to assume that an acetate puffer is used here. Taking this into account, you could even estimate the ratio of $\ce{HOAc}$ and $\ce{NaOAc}$ used.&lt;/li&gt;&#xA;&lt;li&gt;The species monitored by UV is either the indicator $\ce{HInd}$ or its corresponding base $\ce{Ind-}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the absorbance is higher in basic solution, we might conclude that $\ce{Ind-}$ is observed.&#xA;\[\ce{HInd + B &amp;lt;=&gt; Ind- +BH+}\]&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{HInd}$ is typically a rather weak acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Is it safe to assume that the basic solution only contains $\ce{Ind-}$, while in buffered solution $\ce{HInd}$ is present as well?&lt;/li&gt;&#xA;&lt;li&gt;Is it furthermore conceivable that $\ce{HInd}$ and $\ce{Ind-}$ constitute a buffer system for which the pH can be calculated using the &lt;strong&gt;Hasselbalch-Henderson&lt;/strong&gt; equation?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-07T09:11:43.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28318" PostTypeId="1" AcceptedAnswerId="29253" CreationDate="2015-04-07T09:30:52.177" Score="4" ViewCount="982" Body="&lt;p&gt;While calculating the adiabatic flame temperature for the following combustion reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\Phi\, C2H6 + $3.5$\,(O2 + $3.76$\,N2) -&amp;gt; $a$\,CO2 + $b$\,CO + $d$\,H2O + $e$\,H2 + $f$\,O2 + $g$\,N2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Phi$ is the equivalence ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the product composition ratios $a$, $b$, $d$, $e$, $f$ I need to make mass balance on $\ce{C}$, $\ce{O}$ and $\ce{H}$ and then use two $K_{p}$ (Chemical equilibrium constant) relations. Then solve this system of equation simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the elementary reactions of dissociation I should use to apply $K_{p}$ equations, As I could have the following scenarios happening (regarding dissociation):&#xA;\begin{align}&#xA;\ce{CO + H2O &amp;amp;&amp;lt;=&amp;gt; CO2 + H2}\\&#xA;\ce{CO2      &amp;amp;&amp;lt;=&amp;gt; CO + $0.5$\,O2}\\&#xA;\ce{H2O      &amp;amp;&amp;lt;=&amp;gt; H2 + $0.5$\,O2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need two equations for $K_{p}$ which two should I use and why?&lt;/p&gt;&#xA;" OwnerUserId="15414" LastEditorUserId="4945" LastEditDate="2015-04-07T10:29:45.453" LastActivityDate="2016-08-17T18:28:12.760" Title="Chemical equilibrium for simultaneous dissociation reactions" Tags="&lt;equilibrium&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28319" PostTypeId="1" CreationDate="2015-04-07T10:06:21.870" Score="3" ViewCount="1170" Body="&lt;p&gt;I understand the acidity of a substance increases with electronegativity and size (valence shell space) of an atom (periodic trend) that contains a hydrogen atom (Brønsted-Lowry). However when looking at following pairs, I am not able to determine which one is a stronger acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3O+ ,\ H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, I would want to reason, the relationship between the charge and acidity/ basicity.&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="5026" LastEditDate="2018-03-03T11:22:07.587" LastActivityDate="2018-03-03T11:22:07.587" Title="What's the relationship between charge and acidity/basicity?" Tags="&lt;acid-base&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-06-10T14:22:11.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="28321" PostTypeId="5" CreationDate="2015-04-07T10:21:45.187" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T10:21:45.187" LastActivityDate="2015-04-07T10:21:45.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28322" PostTypeId="4" CreationDate="2015-04-07T10:21:45.187" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T10:21:45.187" LastActivityDate="2015-04-07T10:21:45.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28323" PostTypeId="5" CreationDate="2015-04-07T10:22:59.120" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T10:22:59.120" LastActivityDate="2015-04-07T10:22:59.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28324" PostTypeId="4" CreationDate="2015-04-07T10:22:59.120" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T10:22:59.120" LastActivityDate="2015-04-07T10:22:59.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28325" PostTypeId="1" AcceptedAnswerId="28327" CreationDate="2015-04-07T10:33:28.167" Score="8" ViewCount="7631" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following product is formed in the reaction $\ce{CH3MgBr}$ in DryIce/acid?&lt;br&gt;&#xA;  &lt;strong&gt;A&lt;/strong&gt;) $\ce{CH3COOH}$&lt;br&gt;&#xA;  &lt;strong&gt;B&lt;/strong&gt;) $\ce{CH4}$&lt;br&gt;&#xA;  &lt;strong&gt;C&lt;/strong&gt;) $\ce{CH3OH}$&lt;br&gt;&#xA;  &lt;strong&gt;D&lt;/strong&gt;) $\ce{CH3CH2CHO}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My answer is &lt;strong&gt;A&lt;/strong&gt;, since dry ice is $\ce{CO2}$ and it has one carbon atom and reactant has one carbon, so in all the product should have two carbon atoms, which is option &lt;strong&gt;A&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="15416" LastEditorUserId="41328" LastEditDate="2017-08-08T22:22:59.980" LastActivityDate="2017-12-28T14:07:27.370" Title="Organic reaction of dry ice and Grignard reagent" Tags="&lt;organic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28326" PostTypeId="2" ParentId="28316" CreationDate="2015-04-07T10:34:22.087" Score="5" Body="&lt;p&gt;There's a bit of a misconception on what actually happens in the video linked in your question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions between alkali metals, namely lithium ($\ce{Li}$) and sodium  ($\ce{Na}$) and water, diluted acid and concentrated sulfuric acid are not neutralisations, but redox reactions, that is transfer of an electron from one reaction partner to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{2H2O &amp;amp;&amp;lt;=&amp;gt; 2H+ + 2OH-\\&#xA;2Na &amp;amp;-&amp;gt; 2Na+ + 2e-\\&#xA;2H+ + 2e- &amp;amp;-&amp;gt; H2 ^}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take this three reactions as a (mathematical) equation system, with which you're probably more familiar with, and sum it up, the total reaction is&#xA;\[\ce{2Na + 2H2O -&gt; 2NaOH + 2 H2 ^} \]&#xA;To put this into words:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The metal is dissolved and forms metal hydroxide. &lt;/li&gt;&#xA;&lt;li&gt;The resilting solution is alkaline (basic), hence the name alkali metals.&lt;/li&gt;&#xA;&lt;li&gt;The reaction releases a lot of heat! &lt;/li&gt;&#xA;&lt;li&gt;Hydrogen gas is formed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the reaction releases a lot of heat, the hydrogen gas is burnt with the oxygen from the air, yielding water, This is the explosion that you noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken these vigorous reactions into account, it is not astonishing that in nature, sodium and and lithium do not exist as &lt;strong&gt;metals&lt;/strong&gt;, but in the form of various salts. It wasn't until 1807 that sodium metal was actually made in the laboratory, followed by lithium in 1818.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning back to your question on neutralisations, let's define what it actually means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;I apologize if I'm getting to colloquial here!&lt;/em&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term &lt;em&gt;neutralisation&lt;/em&gt; is typically used to describe that two effects don't sum up to result in something stronger but rather incapacitate each other. In the case of acid-base reactions, this translates to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{effective\ acid + effective\ base -&gt; something\ ineffective + water}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water, a typical example would be the reaction of of a strong acid (such as hyrochloric acid) with &lt;strong&gt;equimolar&lt;/strong&gt; amounts of strong base (such as sodium hydroxide):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{HCl + NaOH -&gt; NaCl + H2O}\] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the right amount of a strong and corrosive acid to a solution of also corrosive and harmful caustic soda a harmless solution of sodium chloride (table salt) in water.&#xA;The only side effect of the reaction is that the solution can get pretty hot, but an explosion is not to be expected here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, usable explosions do not result from neutralisation reactions but from fast combustion reactions in with large volumes of gases are formed, possibly (but not necessarily) accompanied by the release of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account that strong acids had to be made by men, I'd be surprised to find that these materials were wasted in simple neutralisation reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will however find historical accounts on the use of the acids or the bases. Nitric acid was used to separate silver from gold (&lt;em&gt;Scheidewasser&lt;/em&gt;). Calcium oxide (burnt lime), the precursor of calcium hydroxide e.g. is used to make cement. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="24052" LastEditDate="2016-03-11T23:25:45.190" LastActivityDate="2016-03-11T23:25:45.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28327" PostTypeId="2" ParentId="28325" CreationDate="2015-04-07T10:47:32.493" Score="12" Body="&lt;p&gt;You are correct. Grignard reagents $\ce{RMgX}$ (where $\ce{R}$ stands for some hydrocarbon group and $\ce{X}$ is a halogen, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch19/ch19-2-1.html&quot;&gt;react with carbon dioxide to produce carboxylic acids&lt;/a&gt; after acidic workup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{RMgX -&amp;gt;[1)\ \ce{ CO2}][2)\ \ce{ H3O+}] RCO2H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{RMgX + CO2 -&amp;gt; RCO2- MgX+}$$&#xA;$$\ce{RCO2- + H3O+ -&amp;gt; RCO2H + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general your reasoning when approaching this problem is a good one. If you are reacting two 1-carbon reagents, then you probably have a 2-carbon product.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-04-07T10:47:32.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28328" PostTypeId="1" AcceptedAnswerId="31480" CreationDate="2015-04-07T11:04:53.927" Score="6" ViewCount="8926" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Benzamide can be converted to benzylamine using     &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A) $\ce{Br2,\ KOH}$&lt;br&gt;&#xA;   B) $\ce{PCl5}$&lt;br&gt;&#xA;   C) $\ce{LiAlH4}$&lt;br&gt;&#xA;   D) $\ce{NaBH4}$     &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer:&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing agent is used so C and D options both must be correct but since its single choice answer, I do not know which is more appropriate. Any help?&lt;br&gt;&#xA;Also, what does option A and B reagents do to amides?&lt;/p&gt;&#xA;" OwnerUserId="15416" LastEditorUserId="16683" LastEditDate="2018-03-02T12:06:02.503" LastActivityDate="2018-04-09T02:38:20.313" Title="How to convert benzamide to benzylamine?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="28329" PostTypeId="1" CreationDate="2015-04-07T12:20:05.880" Score="0" ViewCount="6477" Body="&lt;p&gt;why some times 2nd shell onward it can hold 8 electrons and other times it is like they can hold K2 L8 M18 N32? when do i have to use first method(K2 L8 M8 N8) and when to use second method (k2 L8 M18)? I'm really confused.&lt;/p&gt;&#xA;" OwnerUserId="15418" LastActivityDate="2015-04-08T19:21:26.440" Title="How many electrons does each shell hold?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ClosedDate="2017-06-29T16:11:57.243" ContentLicense="CC BY-SA 3.0" />
  <row Id="28330" PostTypeId="1" CreationDate="2015-04-07T12:28:43.403" Score="2" ViewCount="58" Body="&lt;p&gt;Differential Scanning Colorimetry (DSC) is a technique used to identify Polymorphism by revealing two melting points of a pure crystalline structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DSC shows a peak associated with the formation of the crystals. Hot Stage Microscopy (HSM) shows that only one of the polymorphs forms at the crystallisation temperature, the second polymorph forms between the crystallisation temperature and the melting temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the formation of the second polymorph not accompanied with a change in energy? If not, then why would the two structures have a different melting point seeing as no energy has entered or exited the system?&lt;/p&gt;&#xA;" OwnerUserId="15417" LastEditorUserId="194" LastEditDate="2017-04-06T02:47:36.807" LastActivityDate="2017-04-06T02:47:36.807" Title="Is Polymorphism accompanied with an energy change?" Tags="&lt;energy&gt;&lt;crystal-structure&gt;&lt;differential-scanning-calorimetry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28331" PostTypeId="2" ParentId="28329" CreationDate="2015-04-07T13:40:21.420" Score="1" Body="&lt;p&gt;This is a great question that allows us to learn from the periodic table. The previous answer asked if you have learned about orbitals, and if you haven't this is probably why you are having difficulty.&#xA;If you look at the periodic table, you will see that the elements are arranged in two ways, vertically (group) and horizontally (period), there is good reason for this. Electrons can only occupy space based on mathematical probability; each of these probability functions is what we refer to as orbitals. Each orbital can only hold a maximum number of electrons ( the equation for total electrons is 4l+2 electrons where l is a quantum number starting at l=0 for s, l=1 for p etc.) in the valence shell (for 1, it can hold 2 Starting on the left side of the periodic table with Hydrogen we know that it is in group 1, which means that it has 1 electron in its valence orbital. Element number 2 (He) is a bit of a special case because although it only has 2 electrons in its valence orbital, it is full because the orbital number is 1. What I mean by this is that if you look at period 2 elements, starting with Li, you see that it falls under Hydrogen, and has 1 valence electron. This is because the 1s orbital is full with 2 electrons, and the additional electron goes into the next lowest energy orbital which is called the 2s orbital. Be has 4 electrons, 2 of them in the valence orbital, and 2 of them in the 1s orbital. When we move to B(oron) the 1s and 2s orbitals are full, and we start to fill the 2p orbital. The p orbital can hold a total of 6 (4l+2 where l = 1) electrons and so the boron atom has the following configuration (1s)2(2s)2(2p)1. What you see is that the highest energy orbital (in this case 2 (s + p) is the valence orbital and so B(oron) has 3 valence electrons, but also has 2 electrons in the much lower energy 1s orbital. Think of these orbitals like layers of an onion, where the further from the center, the greater the energy of the orbital. There are rules to how the orbitals are filled and the rules are dictated by quantum mechanics but generally speaking the orbitals are filled as follows 1s 2s 2p 3s 3p 4s 3d 4p 4d using what is know as the Madelung rule. s orbitals can hold 2 electrons, p orbitals can hold 6 electrons, d orbitals can hold 10 electrons. As you work your way from low atomic number on the period table to high you will see why it is organized the way it is. The break between H and He is because of electrons, when you move to period 2 you start filling valence shell 2, when you move to period 3 you start filling valence shell 3 and so on.  I hope this helps, and let me know if you need clarification on anything!&lt;/p&gt;&#xA;" OwnerUserId="15404" LastActivityDate="2015-04-07T13:40:21.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28332" PostTypeId="2" ParentId="20042" CreationDate="2015-04-07T13:49:47.477" Score="0" Body="&lt;p&gt;One way to think of this is to think of the total number of bonds that atoms can make. You can then use experimentation to look at the structure of the compounds, and that will tell you a lot about how the electrons are configured around the nucleus. For instance, Carbon compounds tend to take on tetrahedral shape if the bonds are saturated (single covalent), this means that carbon bonds most favourably by forming 4 bonds. If the moieties that are bound are all the same such as hydrogen, the angle of the bonds should be the same, and using x-ray crystallography for instance shows us that this is the case. We understand chemical bonding quite well, and we can deduce that each hydrogen, donates its single valence electron to a single valence electron from the carbon to form a covalent bond, the total number of valence electrons for carbon must therefore be 4. This is because the 2p shell is higher energy than the 2s shell and only the 2p electrons are free to bond. (This isn't really the case, the bonds are hybrid sp bonds but I won't get in to that right now)&lt;/p&gt;&#xA;" OwnerUserId="15404" LastActivityDate="2015-04-07T13:49:47.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28333" PostTypeId="1" AcceptedAnswerId="28382" CreationDate="2015-04-07T14:50:20.117" Score="2" ViewCount="3765" Body="&lt;p&gt;What is the electrical conductivity of pool water? I tried to find some information, unfortunately without success. I need to know it to make an underwater electronic project.&lt;/p&gt;&#xA;" OwnerUserId="15420" LastEditorUserId="4945" LastEditDate="2020-04-30T11:38:11.940" LastActivityDate="2020-04-30T11:38:11.940" Title="What is the electrical conductivity of pool water?" Tags="&lt;electrochemistry&gt;&lt;water&gt;&lt;conductivity&gt;&lt;electricity&gt;&lt;electrical-properties&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28334" PostTypeId="1" CreationDate="2015-04-07T15:59:24.850" Score="1" ViewCount="2093" Body="&lt;p&gt;In $\ce{[MnCl_6]}^{2-}$, $\ce{Mn}$ has +4 oxidation state which gives it a $\ce{3d^3}$  $\ce{4s^0}$ configuration. But how is it having 4 unpaired electrons?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4231" LastEditDate="2015-04-07T17:06:57.997" LastActivityDate="2015-04-07T17:06:57.997" Title="Why [MnCl6]^2- has got 4 unpaired electrons instead of 3?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28336" PostTypeId="2" ParentId="28206" CreationDate="2015-04-07T16:16:59.510" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to ADVANCES IN INORGANIC CHEMISTRY AND RADIOCHEMISTRY, Volume 12, page 193 &quot;The tris(bipyridal) cobalt(I) cation has a magnetic momment of 3.2-3.4 B.M. which is independent of temperature from 70K to 340K&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are two unpaired electrons, the spin only value is $\sqrt{2(2+2)} =2.83$.  So your experimental data are inconsistent with the literature, as 3.2-3.4 is greater than the spin only value, 2.83.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the deviation from the spin only value is not always positive for complexes with T ground states. The main characteristic of complexes with T ground states is that the magnetic moment is temperature dependent, because the spin-orbital coupling splits the T state into states that differ in energy by an amount close to kT.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example $\ce{Cs2VCl6}$ has a magnetic moment of 1.4 at 80K and 1.8 at 300K, with the spin only formula predicting 1.73.  So deviation can be negative, and isn't necessarily positive or negative even for the same compound, depending upon temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See section 4.3 of Dr. Mark D. Spicer's &lt;a href=&quot;http://personal.strath.ac.uk/m.d.spicer/docs/magnetism%20notes%20complete.doc&quot; rel=&quot;nofollow&quot;&gt;magnetism lecture&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-04-07T16:16:59.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28337" PostTypeId="1" CreationDate="2015-04-07T16:21:58.683" Score="-3" ViewCount="268" Body="&lt;p&gt;2.50 mL of 0.0150 M A, 3.50 mL of 0.0250 M B, and 4.00 mL of distilled water were added to a 50-mL beaker. At equilibrium the concentration of C is 2.05*10^-3M at 25degrees celsius. Calculate Kc for the following reaction.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{2A + B &amp;lt;=&gt; 3C}\]&lt;/p&gt;&#xA;" OwnerUserId="15422" LastEditorUserId="418" LastEditDate="2015-04-07T17:46:53.343" LastActivityDate="2015-04-07T17:49:17.580" Title="How to calculate an equilibrium constant from concentrations and a reaction equation" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-04-07T18:25:36.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="28338" PostTypeId="1" CreationDate="2015-04-07T16:30:10.777" Score="1" ViewCount="434" Body="&lt;p&gt;I am reading a book that basically states what is the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something I observed when I was doing ICE charts: &lt;strong&gt;when acid and base neutralize, we always assume complete neutralization.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, when we put acid in water, which can also be considered a Brønsted-Lowry base, we called it ionization of acid, and thereof acid ionization constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leveling effect happens when multiple strong acids dissolve completely and we don't know which is stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question: &lt;strong&gt;does complete neutralization always take place with a base and an acid, why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="12860" LastEditDate="2015-11-13T19:23:55.637" LastActivityDate="2015-11-13T19:23:55.637" Title="Why leveling effect is reduced when strong acid is dissolved in a weaker acid than water?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28339" PostTypeId="2" ParentId="28337" CreationDate="2015-04-07T17:42:03.370" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Start with the equation for the equilibrium constant&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For a reaction $\ce{a\cdot A + b\cdot B &amp;lt;=&amp;gt; c\cdot C}$, the equilibrium constant $K_c$ is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[K_c = \mathrm{\frac{[C]^c}{[A]^a\cdot[B]^b}}\]&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Get familiar with the concept of &lt;strong&gt;ICE&lt;/strong&gt; charts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine the concentrations of &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; for the moment when you have poured together the solutions of &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; and diluted the mixture with water. (Don't forget about the water!) &lt;/li&gt;&#xA;&lt;li&gt;Imagine that you've worked incredibly fast and no reaction has taken place! &lt;em&gt;This is the &lt;strong&gt;I&lt;/strong&gt;nitial situation&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Now have a look at the reaction equation to figure out what happens when &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; react to yield &lt;strong&gt;C&lt;/strong&gt;. Denote the change of the concentration of &lt;strong&gt;C&lt;/strong&gt; as $x$ (or $3x$) if you think that is easier to handle. How does the formation of $x$ moles of &lt;strong&gt;C&lt;/strong&gt; change the concentration of &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt;? &lt;em&gt;This the &lt;strong&gt;C&lt;/strong&gt;hange part&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Can you use the information from the &lt;strong&gt;I&lt;/strong&gt;nitial and the &lt;strong&gt;C&lt;/strong&gt;hange part to express the concentration of the reactants in the &lt;strong&gt;E&lt;/strong&gt;quilibrium, knowing that $[\ce{C}]$ is $2.05\cdot 10^{-3} \mathrm{mol\cdot L^{-1}}$?&#xA;&lt;em&gt;(Yes, that's the &lt;strong&gt;E&lt;/strong&gt; part of the &lt;strong&gt;ICE&lt;/strong&gt;)&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Feed these concentrations, actually &quot;some expressions with $x$&quot; in the equation for $K_c$ and solve.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Success ;) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-07T17:49:17.580" LastActivityDate="2015-04-07T17:49:17.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28340" PostTypeId="1" AcceptedAnswerId="28350" CreationDate="2015-04-07T18:00:22.877" Score="10" ViewCount="684" Body="&lt;p&gt;I am reading &lt;em&gt;The Selfish Gene&lt;/em&gt; by Richard Dawkins.&lt;/p&gt;&#xA;&lt;p&gt;I am on chapter two.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;He speaks of the observation of the formation of amino-acids when you simulate environmental conditions of primordial earth.&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;UV light + water + carbon dioxide + methane + ammonia + a couple of weeks time = amino-acids&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;I understand that amino acids are an organic compound that serve a lot of important functions in our bodies.&lt;/p&gt;&#xA;&lt;p&gt;Presumably, they are important to all life, because from the context I can derive, they seem to be somewhat of a precursor to life itself.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Dawkins then says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Processes analogous to these must have given rise to the 'primeval&#xA;soup'  which biologists and chemists believe constituted the seas some&#xA;three to  four thousand million years ago.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;He goes on and eventually states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;At some point a particularly remarkable molecule was formed by&#xA;accident. We will call it the Replicator.   ....&lt;/p&gt;&#xA;&lt;p&gt;Actually a molecule that makes copies of itself is not as difficult to&#xA;imagine as it seems at first, and it only had to arise once. Think of&#xA;the  replicator as a mould or template. Imagine it as a large molecule&#xA;consisting of a complex chain of various sorts of building block&#xA;molecules.  The small building blocks were abundantly available in the&#xA;soup  surrounding the replicator. Now suppose that each building block&#xA;has an  affinity for its own kind. Then whenever a building block from&#xA;out in the  soup lands up next to a part of the replicator for which&#xA;it has an affinity,  it will tend to stick there. The building blocks&#xA;that attach themselves in  this way will automatically be arranged in&#xA;a sequence that mimics that of the replicator itself. It is easy then to think of them joining up to&#xA;form a  stable chain just as in the formation of the original&#xA;replicator. This  process could continue as a progressive stacking up,&#xA;layer upon layer.  This is how crystals are formed. On the other hand,&#xA;the two chains might  split apart, in which case we have two&#xA;replicators, each of which can go  on to make further copies.&lt;/p&gt;&#xA;&lt;p&gt;A more complex possibility is that each building block has affinity&#xA;not for  its own kind, but reciprocally for one particular other kind.&lt;/p&gt;&#xA;&lt;p&gt;Then the replicator would act as a template not for an identical copy,&#xA;but  for a kind of 'negative', which would in its turn re-make an&#xA;exact copy of  the original positive. For our purposes it does not&#xA;matter whether the  original replication process was positive-negative&#xA;or positive-positive,  though it is worth remarking that the modem&#xA;equivalents of the first  replicator, the DNA molecules, use positive-&#xA;negative replication. What  does matter is that suddenly a new kind of&#xA;'stability' came into the world.  Previously it is probable that no&#xA;particular kind of complex molecule was  very abundant in the soup,&#xA;because each was dependent on building  blocks happening to fall by&#xA;luck into a particular stable configuration. As  soon as the&#xA;replicator was born it must have spread its copies rapidly  throughout&#xA;the seas, until the smaller building block molecules became  a scarce&#xA;resource, and other larger molecules were formed more and  more&#xA;rarely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;As someone with a limited understanding of Biology and Chemistry, I have a couple of questions.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We have observed that the chemical conditions outlined at the top of this question seem to yield more complex organic compounds with time. Do we know (or have theories) as to &lt;em&gt;why&lt;/em&gt; this is, or is it merely something we have observed and re-created?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The theory about replicators suggests that they are like a chain of building-block molecules each with an affinity for either its own kind, or some other kind. (positive-positive vs positive-negative). -- Which ever was the case, you could end up with a big chain of molecules that had a fractal nature, because these building blocks were freely available in the soup surrounding the replicator. -- If the chain split apart, you would get two identical copies and each part could go off and replicate further.&lt;/p&gt;&#xA;&lt;p&gt;Now, I suppose this question mirrors the first, but why is it that it is so easy to imagine the affinity relationship? What makes molecules want to stick together with other, specific molecules?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further more, what makes chains of molecules split apart (replicate)?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Can somebody elaborate further on these points, and on what Dawkins is saying here in general, so as to pander to the cravings of someone who is not particularly versed in the parlance?&lt;/p&gt;&#xA;" OwnerUserId="15424" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-03-11T21:41:02.033" Title="Explain the formation of replicators to a layman" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28342" PostTypeId="1" AcceptedAnswerId="28344" CreationDate="2015-04-07T18:43:00.597" Score="5" ViewCount="2157" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8iUfh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8iUfh.png&quot; alt=&quot;Structures in question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Huckel's rule states that an aromatic species must have $4n+2$ π-electrons. Is the last molecule also aromatic?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="16683" LastEditDate="2018-01-09T16:10:28.867" LastActivityDate="2018-01-09T16:11:30.317" Title="Are the following compounds aromatic?" Tags="&lt;organic-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28343" PostTypeId="2" ParentId="28340" CreationDate="2015-04-07T18:47:18.983" Score="6" Body="&lt;p&gt;The situation leading to the formation of amino acids was simulated by &lt;em&gt;Stanley Miller&lt;/em&gt; and &lt;em&gt;Harold Urey&lt;/em&gt; in their famous experiment and published as &lt;em&gt;A Production of Amino Acids Under Possible Primitive Earth Conditions&lt;/em&gt; in &lt;em&gt;Science&lt;/em&gt;, &lt;strong&gt;1953&lt;/strong&gt;, &lt;em&gt;117&lt;/em&gt;, 528-529 (&lt;a href=&quot;http://dx.doi.org/10.1126/science.117.3046.528&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that this formation of amino acids might have happened not just in the sea, but also in shallow pits at the shore that once in a while fell dry and were heated by sunlight. Under these conditions, particularly in the presence of insoluble salts serving as catalysts, amino acids may have undergone condensation reactions, i.e., splitting off water and forming peptide bonds. The resulting molecules are oligopeptides, small brothers of proteines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that these pools were filled with water again and that the process repeats. Between oligopeptides of fitting amino acid sequences, hydrogen bonds can form, resulting in $\beta$-sheets or similar aggregates. Upon repetition of the condensation step, free amino acids bound to one chain by hydrogen bonds might preferably undergo condensation in the other. This may be the molecular background of the replicator concept described by &lt;em&gt;Richard Dawkins&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-07T18:47:18.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28344" PostTypeId="2" ParentId="28342" CreationDate="2015-04-07T18:52:30.653" Score="10" Body="&lt;p&gt;Molecule &lt;strong&gt;1&lt;/strong&gt;, the cyclopentadienyl cation, is not aromatic. It only has 4 π electrons and does not fit the $4n+2$ rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule &lt;strong&gt;2&lt;/strong&gt;, the cycloheptatrienyl cation (or tropylium cation), is aromatic; it has 6 π electrons and fits the $4n+2$ rule with $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecule &lt;strong&gt;3&lt;/strong&gt;, azulene, is generally considered to be aromatic. It fits the $4n+2$ rule with $n=2$, and is often referred to as a &quot;nonbenzenoid&quot; aromatic compound. It can also be viewed as a fusion of two other aromatic units, the tropylium cation and the cyclopentadienyl anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4eCxW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-01-09T16:11:30.317" LastActivityDate="2018-01-09T16:11:30.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28346" PostTypeId="1" AcceptedAnswerId="28372" CreationDate="2015-04-07T20:16:40.290" Score="26" ViewCount="3598" Body="&lt;p&gt;Today, I'm looking for how to find the 2nd perturbation to the base in Rayleigh Schrödinger Perturbation Theory (RSPT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SETUP&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the 2nd order perturbation in Dirac's notation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\hat{H}^0 |n^2\rangle + \hat{V}  |n^1\rangle  = E_n^0  |n^2\rangle + E_n^1 |n^1\rangle  +  E_n^2  |n^0\rangle &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply to the left with $\langle k^0 |$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\langle k^0| \hat{H}^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = \langle k^0|  E_n^0  |n^2\rangle + \langle k^0|  E_n^1 |n^1\rangle  + \langle k^0|  E_n^2  |n^0\rangle &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hermicity of the Hamilton operator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle  + E_n^2 \langle k^0  |n^0\rangle &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cancel the third term on the right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gathering same inner products:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle = ( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle  &#xA;\end{equation}&#xA;\begin{equation}&#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle  = \langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle&#xA;\end{equation}&#xA;Replacing $| N^1 \rangle$ and $E_N^1$ with the first order perturbation to the base and to the energy:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;\begin{equation}&#xA;|n^1\rangle =\sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}&#xA;\end{equation}&#xA;    \begin{equation}&#xA;E_N^{1} = \langle n^0 | \hat{V} | n^0 \rangle&#xA;\end{equation}&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \langle k^0| \hat{V}  \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - E_n^1 \langle k^0 \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n}  \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I'm not sure if the bracket algebra was done properly not even forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then ... are another some algebra steps that I cannot figure out the properly path to find the respective coefficient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle k^0 | n^2 \rangle$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to replace it in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;|n^2\rangle =\sum_{k \neq n} |k^0 \rangle \frac{\langle k^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_k^0 ???}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, Would someone help me out to show me how to get the properly 2nd perturbation to the wave function?&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="4945" LastEditDate="2015-04-09T09:06:07.093" LastActivityDate="2016-05-04T04:44:33.733" Title="How to find the second order perturbation to wave function?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28347" PostTypeId="2" ParentId="28302" CreationDate="2015-04-07T20:22:40.413" Score="3" Body="&lt;p&gt;The pressure in HPLC systems is not an intended feature, but a consequence of how the system works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several adjustable parameters/properties of a HPLC/UPLC system that will influence on the internal pressure inside the chromatographic column. Some of those are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Solvent flow&lt;/strong&gt;:  A greater flow of solvents mean a faster chromatophy. On the other hand, faster runs also mean lower efficiency. An increase in flow leads to an increase in pressure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Column length&lt;/strong&gt;: A bigger column will improve the separation efficiency, but also increase the duration of the run. Bigger columns leads to a greater pressure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Particle diameter&lt;/strong&gt;: The lower the particle, the greater the separation efficiency, but this also leads to a greater pressure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Particle pore size&lt;/strong&gt;: This has the same effect as above, but is not the same thing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Optmizing this parameters and/or improving the technology related to them is what generally drives the research in the HPLC/UPLC development field, besides of course find the appropriate type of column, solvent gradient, etc. You can't simply make a system where the pressure is as high as possible, or some connectiong will eventually burst. On the other hand, it's not interesting to have a chromatographic run that lasts several hours, or doesn't really separate the molecules, specially when dealing with hundreds or thousands of different analytes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is much more to the above in several books and thousands of papers, including very detailed descriptions, experiments and calculations that show a much more precise relation between those and other parameters.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-04-07T20:22:40.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28348" PostTypeId="1" CreationDate="2015-04-07T20:46:51.830" Score="0" ViewCount="6299" Body="&lt;p&gt;For my chemistry lab we had to heat sodium chloride and there was a colour change and a sizzling sound. Is it a decomposition or synthesis reaction, and what is the balanced chemical equation for it?&lt;/p&gt;&#xA;" OwnerUserId="15430" LastActivityDate="2015-04-07T22:50:21.090" Title="Whats the Balanced chemical equation for heating sodium chloride and what type of a reaction is it?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-08T09:45:21.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="28350" PostTypeId="2" ParentId="28340" CreationDate="2015-04-07T21:38:24.030" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;We have observed that the chemical conditions outlined at the top of this question seem to yield more complex organic compounds with time.&#xA;  Do we know (or have theories) as to why this is, or is it merely&#xA;  something we have observed and re-created.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A recurring theme in chemistry,and all of your questions, is &quot;&lt;strong&gt;lower energy usually equates with more stable structures&lt;/strong&gt;.&quot;  Oversimplifying just a bit (see note on time to reach equilibrium below), we can say that all chemical transformations are also equilibria.  That means that our starting materials are converted to products and our products are converted back to starting materials - reactions are continually happening in both directions.  Once at equilibrium the rate of the forward reaction matches that of the back reaction and the concentrations of starting materials and products will no longer change. At equilibrium the equilibrium constant ($\ce{K_{eq}}$) describing the relative concentrations of our starting materials and products will have the following form&#xA;$$\mathrm{K_{eq}={\frac{(concentration ~products)}{(concentration~starting~ materials)}}}$$&#xA;and the equilibrium constant is related to the free energy difference between our starting materials and products by&#xA;$$\mathrm{\Delta G = -RT\ln K_{eq}}$$&#xA;These equations tell us that the more stable a structure, the more it will predominate at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take carbon, oxygen, hydrogen and nitrogen an equilibrium will be set up between these elements and the simple amino acid glycine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGg2L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;according to the following equation&#xA;$$\ce{4C + 5H2 + N2 + 2O2 -&amp;gt; 2C2H5NO2~(glycine) + heat}$$&#xA;Over 500kJ of heat is produced per mole of glycine formed.  This means that glycine is over 500kJ/mol more stable than the elements used to create it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom line:&lt;/strong&gt; If a complex molecule is more stable than a simpler molecule, then over time the higher energy &quot;simple&quot; molecule will transform into the more stable (lower energy) &quot;complex&quot; molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a word about time.  It may take a long or a short time for an equilibrium to be attained on its own.  Suffice it to say that with catalysts and\or energy, equilibria can be attained quickly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;...why is it that it is so easy to imagine the affinity relationship? What makes molecules want to stick together with other, specific molecules?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again the answer has to do with lowering the energy of the system and making things more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some molecules have what is called an &quot;active site&quot;.  It may be an odd-shaped kink or pocket in the molecule.  Let's think of this odd-shaped pocket as a lock, only molecules shaped like the key will fit.  Further, there are often other functional groups (perhaps hydroxyl groups) around the active site that will bind (hydrogen bonds perhaps) with the key and stabilize it once it is inserted into the lock. Stabilization means lowering the energy of the system.  Only the key fits into the lock and once inserted it is stabilized - energy is lowered.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Further more, what makes chains of molecules split apart (replicate)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By now you've probably guessed that the splitting apart step occurs because it lowers the energy of the system.  Perhaps once the molecule reaches a certain size it folds differently - a new lower energy shape is now possible.  A  protein that was bound to the molecule and stabilizing it can no longer bind to it due to this new folding pattern.  The protein is displaced, it is replaced by a different protein that stabilizes the molecule's new folded structure but predisposes it to splitting apart to a smaller size.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2016-03-11T21:41:02.033" LastActivityDate="2016-03-11T21:41:02.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28351" PostTypeId="2" ParentId="28348" CreationDate="2015-04-07T21:39:08.640" Score="3" Body="&lt;p&gt;My guess would be that this was rather the dissociation of &lt;strong&gt;Sodium Carbonate&lt;/strong&gt; or &lt;strong&gt;Sodium Hydrogen Carbonate&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for the dissociation of &lt;strong&gt;Sodium Carbonate&lt;/strong&gt; and &lt;strong&gt;Sodium Hydrogen Carbonate&lt;/strong&gt;, respectively, are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{Na2CO3 -&amp;gt; Na2O + CO2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{and}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{2 NaHCO3 -&amp;gt; Na2O + H2O + 2 CO2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15431" LastEditorUserId="15431" LastEditDate="2015-04-07T22:50:21.090" LastActivityDate="2015-04-07T22:50:21.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28352" PostTypeId="1" CreationDate="2015-04-07T21:46:44.127" Score="1" ViewCount="1330" Body="&lt;p&gt;I am doing a University project on Magnetically Assisted Combustion and am curious to know more about how magnetic fields may or may not affect gases such as butane and propane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I am right in saying that due to the gas molecules being covalently bonded, there are no free electrons to be influenced in anyway by an induced magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone briefly mentioned something about that depending on whether they are saturated or unsaturated but the conversation didn't go any further as he then promptly left.. so I am still unsure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could shed any light on this or has any thoughts at all I'd be very grateful for your input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of some experiments I have done was attach a neodynium magnetic arrangement around the fuel line of a gas burner and measuring heat transfer to water.  So far the results have been fairly inconclusive as I am using a camping stove which has problems sustaining a constant gas flow as well as a number of inconsistonsies with heat loss and human error etc .. but that's another issue. &lt;/p&gt;&#xA;" OwnerUserId="15433" LastActivityDate="2015-11-29T10:51:26.533" Title="Can covalently bonded hydrocarbons be influenced by a magnetic field?" Tags="&lt;gas-laws&gt;&lt;combustion&gt;&lt;covalent-compounds&gt;&lt;magnetism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28354" PostTypeId="5" CreationDate="2015-04-07T22:02:32.663" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T22:02:32.663" LastActivityDate="2015-04-07T22:02:32.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28355" PostTypeId="4" CreationDate="2015-04-07T22:02:32.663" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-07T22:02:32.663" LastActivityDate="2015-04-07T22:02:32.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28356" PostTypeId="2" ParentId="28352" CreationDate="2015-04-07T22:22:01.967" Score="2" Body="&lt;p&gt;Just because there are no unpaired electrons doesn't mean magnetic fields can't influence something. Unpaired electrons tend to lead to &lt;em&gt;strong&lt;/em&gt; interactions with magnetic fields (the dioxygen in air has two and is paramagnetic, an effect best seen in the liquid phase where a magnet will attract the cold liquid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But compounds with paired electrons also interact with magnetic fields, though the effects are usually weak. But the repulsive effect is sometimes strong enough to levitate a diamagnetic substance (see this &lt;a href=&quot;https://www.youtube.com/watch?v=IFv4VOrWecI&quot; rel=&quot;nofollow&quot;&gt;youtube video&lt;/a&gt; showing how strong magnets can levitate plates of graphite which is strongly paramagnetic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unlikely that such effects would be easy to see in gases. But you are talking about combustion and the most likely effect will not be on the neutral gas but on the flame where the gas is &lt;strong&gt;not&lt;/strong&gt; neutral but a plasma of charged particles. Charged particles or molecules will have a &lt;em&gt;much&lt;/em&gt; stronger interaction with a magnetic field though it might still require sensitive apparatus to detect how the field makes a difference.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-04-07T22:22:01.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28357" PostTypeId="1" CreationDate="2015-04-07T22:38:43.617" Score="1" ViewCount="7385" Body="&lt;p&gt;We heated rock salt for a chemistry lab and it changed from white crystally rock to brown rock and ash-like materiak on the bottom of the test tube. Also there was a sizzling\popping noise. Why did it change colour and why was there a noise? What caused this? &lt;/p&gt;&#xA;" OwnerUserId="15430" LastActivityDate="2017-10-27T14:19:34.250" Title="Why does rock salt change colour when it's heated" Tags="&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28358" PostTypeId="2" ParentId="28051" CreationDate="2015-04-07T22:39:02.547" Score="3" Body="&lt;p&gt;Up front: this depends on what you mean by &quot;stay in&quot;. It would diffuse out of the container and, given the surroundings are closed, obtain a small equilibrium concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not that I could provide an exact answer to your question (it is rather a physics question), but &lt;a href=&quot;http://dx.doi.org/10.1109/TEI.1982.298567&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; (paywalled) deals with the diffusion of the $\ce{N2-SF6}$ pair. The binary diffusion coefficient is reported to be 0.0945 cm&lt;sup&gt;2&lt;/sup&gt;/s. In their experiments they used a container of d = 0.5 m, l = 1 m, introduced a small amount (2-5%) of $\ce{SF6}$ at the bottom of the vessel and measured the time it took to obtain the equilibrium concentration of the mixture. For this setup, this was ~15 h, with a reported time constant of 4-5 hours (I believe this is the time it takes to reach a value of [local $\ce{SF6}$ concentration]/[equilibrium concentration] of 0.5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They used a lengthy formula derived from a one-dimensional model for the comparison of calculated and measured concentrations and in my opinion the measured data agreed quite well with the calculated ones.&lt;/p&gt;&#xA;" OwnerUserId="15431" LastEditorUserId="15431" LastEditDate="2015-04-07T22:57:24.547" LastActivityDate="2015-04-07T22:57:24.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28359" PostTypeId="1" CreationDate="2015-04-07T22:49:35.030" Score="7" ViewCount="328" Body="&lt;p&gt;I can think of any other way to influence the value of $\Delta G$ than (1) change the temperature (2) providing an external source to overcome the negative cell potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between $\Delta G$ and $E_{\mathrm{cell}}$ is &lt;strong&gt;$\Delta G =-nFE_{\mathrm{cell}}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between $\Delta H$, $\Delta S$ and $\Delta G$ is &lt;strong&gt;$\Delta G=\Delta H - T\Delta S$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, temperature won't do anything to change the spontaneity. But if we change $\Delta G$ through $\Delta E_{\mathrm{cell}}$ by providing external voltage in excess of the negative voltage, and though $\Delta G$ we are also able to influence enthalpy and entropy(???).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the part I am not sure about. Besides the title question, a sub-question is: &lt;strong&gt;how in this way can we change the enthalpy and entropy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="15410" LastEditDate="2015-05-07T03:07:51.580" LastActivityDate="2015-05-07T15:10:56.990" Title="Ways to change the spontaneity of a reaction" Tags="&lt;thermodynamics&gt;&lt;reduction-potential&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28360" PostTypeId="1" CreationDate="2015-04-07T22:54:14.083" Score="3" ViewCount="244" Body="&lt;p&gt;I've been dealing with assigning stereochemistry and I seem to be doing okay, but I've come across a molecule that I've found a little tricky to deal with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/olhcZsH.png&quot; alt=&quot;The molecule in question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've identified three stereocentres, highlighted below. My first question would be to confirm that this is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/JvTA7XH.png&quot; alt=&quot;stereocentres highlighted&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No. 1 is standard fare and I got that it was an R enantiomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for 2 &amp;amp; 3, does the fact that both carbons are bonded to the oxygen have an effect on the rotation of the molecule? As in, when you go through the process of rotating the lowest priority group to the back and then tracing from first priority to third, does this affect the process? It may seem a silly question, but I don't have ball &amp;amp; stick models to easily visualise it so I'm struggling a bit. I treated them as normal and got that they were both R enantiomers, but I'm wondering if I've missed a trick.&lt;/p&gt;&#xA;" OwnerUserId="15434" LastActivityDate="2015-04-07T22:54:14.083" Title="Assigning R/S stereochemistry of a cyclic molecule" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28361" PostTypeId="2" ParentId="28359" CreationDate="2015-04-08T01:14:18.837" Score="2" Body="&lt;p&gt;The Nernst Equation relates the reduction potential of a cell to the chemical reaction that is occurring in the cell, and is written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{\mathrm{cell}}=E^\mathrm{o}-\frac{RT}{nF}\ln(Q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{\mathrm{cell}}$, and thereby $\Delta G$, can be affected by changes in the concentration of the ions in either half cell.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-05-07T01:42:37.547" LastActivityDate="2015-05-07T01:42:37.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28362" PostTypeId="1" AcceptedAnswerId="28371" CreationDate="2015-04-08T02:07:45.207" Score="5" ViewCount="22328" Body="&lt;p&gt;I'm trying to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/gMkTlnI.png&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, how does one know which nitrogen is more basic? I'm assuming that for the first molecule, that it's the nitrogen without the proton, since it seems like the other nitrogen has already been protonated (and already has acted as a base.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct? How does one solve the other molecules?&lt;/p&gt;&#xA;" OwnerUserId="2613" LastActivityDate="2015-04-08T06:44:26.303" Title="How does one determine the relative basicity of a nitrogen in a molecule?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28363" PostTypeId="1" AcceptedAnswerId="28368" CreationDate="2015-04-08T03:57:16.407" Score="1" ViewCount="2798" Body="&lt;p&gt;I was asked to &quot; Use VSEPR theory to predict the most stable structure for $\ce{SF3Cl}$ and justify your answer. &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to interpret the structure as seesaw and with a lone pair on the equatorial position. However, I had chosen that the most stable structure was that with Chlorine on the equatorial position because I thought that the Chlorine-Sulfur bond length would be greater than that of Chlorine-Fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the answer key stated the Fluorine was more electronegative, and therefore would be more stable on the axial (with chlorine as equatorial).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the answer key was accurate and if so, how would I compare the effects of electronegativity and bond length? I had considered electronegativity but I had thought it would simply make the bond even shorter, increasing repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="7448" LastEditDate="2015-04-08T14:01:43.950" LastActivityDate="2015-04-08T14:01:43.950" Title="SF3Cl Vsepr Most stable structure" Tags="&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28364" PostTypeId="2" ParentId="28362" CreationDate="2015-04-08T04:25:06.360" Score="0" Body="&lt;p&gt;For amine basicity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basicity decreases with increasing s character&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basicity decreases with increasing stability, due to resonance/ delocalization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/amine1.htm&quot; rel=&quot;nofollow&quot;&gt;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/amine1.htm&lt;/a&gt; (taken from this webpage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first molecule for example, the double bonded nitrogen is less basic due to increasing s character (closer to nucleus, less likely to bind to H).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think most of your examples can be reasonably thought out with these two key points.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-08T04:25:06.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28365" PostTypeId="2" ParentId="28357" CreationDate="2015-04-08T04:31:31.517" Score="0" Body="&lt;p&gt;I'm not too sure what you are asking. If you are refering to rock salt as in sodium chloride (NaCl), I too am at a loss at how you would have created such an effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume that you have heated some random other compound since NaCl does not decompose/ react to heat so readily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the &quot;rock salt&quot; reacted to the heat and perhaps decomposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, CaCO3 (calcium carbonate) decomposes to CaO and Oxygen, perhaps forming the sizzling noise.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-08T04:31:31.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28367" PostTypeId="2" ParentId="28357" CreationDate="2015-04-08T05:10:31.450" Score="7" Body="&lt;p&gt;&quot;Rock salt&quot; contains &lt;a href=&quot;http://www.salt-partners.com/pdf/MasuzawaT.5thSaltSymp.pdf&quot; rel=&quot;noreferrer&quot;&gt;many impurities&lt;/a&gt;, in addition to its main ingredient, sodium chloride. Some trace chemicals in rock salt, such as lithium chloride or sodium carbonate, hold onto water, which might boil out, sizzling and popping, and if your specimen got hot enough, carbonates could decompose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you had used &quot;table salt&quot;, rather than rock salt, chemicals are intentionally added, such as sodium ferrocyanide (to prevent caking), which has a yellowish color and potassium iodide (as a food supplement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had you heated &lt;em&gt;pure&lt;/em&gt; sodium chloride, there should not have been a color change.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-08T05:10:31.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28368" PostTypeId="2" ParentId="28363" CreationDate="2015-04-08T06:07:46.197" Score="0" Body="&lt;p&gt;The bond lengths have little to do with the stability of the atom positioning around the central $\ce{S}$ atom. Rather, the relative sizes of the electron clouds are what should be considered when positioning these atoms. An atom of $\ce{Cl}$ has a 8 more electrons than does an atom of $\ce{F}$, so naturally its electron cloud will be larger and therefore more repulsive. The $\ce{Cl}$ should therefore be positioned in the equatorial position for the same reason you predicted that the lone pair would be present in the equatorial position--because these positions offer the least repulsion between the electron clouds of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the axial position, the 3 equatorial positions, present at 90 degree angles to the axial position, repel the electron cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an equatorial position, however, only the 2 axial positions are present at 90 degree angles to repel the electron cloud. The other 2 equatorial positions are positioned away from the equatorial bond in question, and repel its electron cloud only slightly. For this reason unbonded electron pairs and atoms with large electron clouds occupy these spaces first-- they offer the least repulsion and the most stable positioning.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-08T06:07:46.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28369" PostTypeId="2" ParentId="9537" CreationDate="2015-04-08T06:35:41.717" Score="0" Body="&lt;p&gt;Short answer is, heat is a byproduct of the chemical reaction. It is released by the reaction. It is such that we now use combustion for (most of the time) the sole purpose of getting the heat out of it. We don't want the CO2, but it comes with it (if you burn hydrocarbons).&lt;/p&gt;&#xA;" OwnerUserId="12852" LastActivityDate="2015-04-08T06:35:41.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28370" PostTypeId="1" CreationDate="2015-04-08T06:41:08.120" Score="1" ViewCount="542" Body="&lt;p&gt;So if we have an alpha halogen ketone, why is an Sn1 type mechanism so unfavorable for that species?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand. It seems like the C+ would be resonance stabilized by the carbonyl's double bond. &lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorUserId="9961" LastEditDate="2015-10-16T17:53:43.880" LastActivityDate="2015-10-16T17:53:43.880" Title="Why are alpha-carbonyl relatively less stable?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28371" PostTypeId="2" ParentId="28362" CreationDate="2015-04-08T06:44:26.303" Score="4" Body="&lt;p&gt;Use a combination of &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the &lt;em&gt;exclusion principle&lt;/em&gt;: sort out the unlikely centres with low electron density and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;paper work to figure out those nitrogen atoms for which the additional positive charge can be distributed over a larger part of the molecule without moving the proton.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SP16u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of caffeine (&lt;strong&gt;1&lt;/strong&gt;) the amide and the imide are rather unlikely centres for protonation! You have probably already ruled out the amide nitrogen in the benzamide case (your second molecule) in favour of the aniline site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an initial number of four possible centres in &lt;strong&gt;1&lt;/strong&gt;, only two remain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's see whether the positive charge resulting from protonation can be stabilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F3LV2.png&quot; alt=&quot;protonation of tertiary amine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks fair, but creates a quarternary ammonium ion for which &lt;strong&gt;no obvious charge distribution&lt;/strong&gt; seems to be available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the last possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PTLRu.png&quot; alt=&quot;protonation with charge delocalisation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you agree that this looks much more promising?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-08T06:44:26.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28372" PostTypeId="2" ParentId="28346" CreationDate="2015-04-08T07:05:29.700" Score="31" Body="&lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt; This post is some kind of a legacy post. Find the notation used in the question in the other answer. I added this proof as I was not entirely certain I understood the notation correctly. As it turned out, I did not. As a result I posted the complete derivation of RSPT up to second order, trying to guide anyone through it using a different notation.$\require{cancel}$&lt;br&gt;&#xA;&lt;em&gt;Obviously, I will come back to the initial problem only at the end, if you are impatient, just skip ahead.&lt;/em&gt;&lt;br&gt;&#xA;Please also note, that I removed the parts that became redundant when I posted the second answer. If you are interested in legacy versions, just look them up in the history.&lt;br&gt;&#xA;Also, this page is heavily loaded with $\mathcal{M}^{ath}\mathrm{J_{a}X}$, &lt;strong&gt;give it time to load completely&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In Rayleigh Schrödinger perturbation theory, we set up the Hamiltonian as $$\mathbf{H}=\mathbf{H_0}+\lambda\mathbf{H'}.\tag1$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perturbed Schrödinger equation is therefore $$\mathbf{H}\Psi=W\Psi_\mathrm{ST}.\tag2$$&#xA;You can develop this into Taylor series:&#xA;\begin{align}&#xA;W                &amp;amp;= &#xA;     \lambda^0W_0    + \lambda^1W_1    + \lambda^2W_2    + \cdots &#xA;     &amp;amp;&amp;amp;= \sum_{i=0}\lambda^i W_i\tag3\\&#xA;\Psi_\mathrm{ST} &amp;amp;= &#xA;     \lambda^0\Psi_0 + \lambda^1\Psi_1 + \lambda^2\Psi_2 + \cdots &#xA;     &amp;amp;&amp;amp;= \sum_{i=0}\lambda^i \Psi_i\tag4\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your unperturbed reference system is a complete solution denoted as &#xA;$$\mathbf{H}_0\Phi_i = E_0\Phi_i,\tag5$$&#xA;where $i\in\mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You chose your system to be intermediate normalised.&#xA;$$\langle\Psi_\mathrm{ST}|\Phi_0\rangle = 1\tag6$$&#xA;Since all your solutions of the unperturbed system are orthogonal,&#xA;$\langle\Phi_i|\Phi_j\rangle=\delta_{ij},$&#xA;the same holds for your perturbed system, hence&#xA;$$\langle\Psi_{i\neq0}|\Phi_0\rangle = 0.\tag7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can combine $(1)$ through $(4)$ and gather them according to their order.&#xA;\begin{align}&#xA;\lambda^0&amp;amp;:&amp;amp;&#xA;  \mathbf{H}_0\Psi_0 &#xA;    &amp;amp;= W_0\Psi_0\tag{8a}\\&#xA;\lambda^1&amp;amp;:&amp;amp; &#xA;  \mathbf{H}_0\Psi_1 + \mathbf{H'}\Psi_0 &#xA;    &amp;amp;= W_0\Psi_1 + W_1\Psi_0\tag{8b}\\&#xA;\lambda^2&amp;amp;:&amp;amp; &#xA;  \mathbf{H}_0\Psi_2 + \mathbf{H'}\Psi_1 &#xA;    &amp;amp;= W_0\Psi_2 + W_1\Psi_1 + W_2\Psi_0\tag{8c}\\&#xA;\lambda^n&amp;amp;:&amp;amp; &#xA;  \mathbf{H}_0\Psi_n + \mathbf{H'}\Psi_{n-1} &#xA;    &amp;amp;= \sum_{i=0}^n W_i\Psi_{n-i}\tag{8d}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do have a complete set of solutions, hence we can express perturbations as linear combinations of these.&#xA;\begin{aligned}&#xA;\Psi_0 &amp;amp;= \Phi_0 &amp;amp; &#xA;\Psi_1 &amp;amp;= \sum_i c_i \Phi_i &amp;amp; &#xA;\Psi_2 &amp;amp;=\sum_i d_i \Phi_i &amp;amp; &#xA;\dots&#xA;\end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we project equations $8$ onto $\Psi_0$. For $\mathrm{8a}$ this is trivial and it leads back to the expectation value of the energy for the unperturbed system. (In the last transformation we use the linear combinations.)&#xA;\begin{align}&#xA;\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle &#xA;    &amp;amp;= \langle\Phi_0|W_0|\Psi_0\rangle\\&#xA;\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle &#xA;    &amp;amp;= W_0\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9a}\\&#xA;W_0 &amp;amp;= \langle\Phi_0|\mathbf{H}|\Phi_0\rangle = E_0\tag{9a'}&#xA;\end{align}&#xA;The next step is similar, and everything else will follow from there analogously. For $\mathrm{8b}$ we obtain:&#xA;\begin{align}&#xA;\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle + &#xA;\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle &#xA;    &amp;amp;=   \langle\Phi_0|W_0|\Psi_1\rangle &#xA;       + \langle\Phi_0|W_1|\Psi_0\rangle\\&#xA;\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle + &#xA;\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle &#xA;    &amp;amp;= W_0\cancelto{c_0}{\langle\Phi_0|\Psi_1\rangle} +&#xA;       W_1\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\\&#xA;\cancelto{c_o\cdot E_0}{\sum_i c_i \langle\Phi_0|\mathbf{H}_0|\Phi_i\rangle} &#xA;       + \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle &#xA;    &amp;amp;= W_0\cdot c_0 + W_1\tag{9b}\\&#xA;W_1 &amp;amp;= \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle\tag{9a'}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly we can obtain the coefficients by projecting on some other function than $\Phi_0$. I will cut this a little bit shorter.&#xA;\begin{align}&#xA;\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle &#xA;     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle &#xA;    &amp;amp;= \langle\Phi_j|W_0|\Psi_1\rangle + &#xA;       \langle\Phi_j|W_1|\Psi_0\rangle\\&#xA;\cancelto{c_j\cdot E_j}{\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle} &#xA;     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle &#xA;    &amp;amp;= \cancelto{c_j\cdot E_0}{\langle\Phi_j|W_0|\Psi_1\rangle} &#xA;     + \cancelto{0}{\langle\Phi_j|W_1|\Psi_0\rangle}\tag{10b}\\&#xA;c_j &amp;amp;= \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{E_0-E_j}\tag{10b'}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is impossible to solve for $c_0$, but we still have the intermediate normalisation.&#xA;\begin{align}&#xA;0 &amp;amp;= \langle\Phi_0|\Psi_1\rangle&#xA;   = \sum_i c_i \langle\Phi_0|\Phi_i\rangle&#xA;   = c_0\cancelto{1}{\langle\Phi_0|\Phi_i\rangle} + &#xA;     \sum_{i\neq0} c_i \cancelto{0}{\langle\Phi_0|\Phi_i\rangle} = c_0\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly this applies to all zeroth coefficients, $c_0 = d_0 = \dots = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's get back to business and continue with $\mathrm{8c}$. &#xA;\begin{align}&#xA;\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle &#xA;  + \langle\Phi_0|\mathbf{H'}|\Psi_1\rangle &#xA;  &amp;amp;= \langle\Phi_0|W_0|\Psi_2\rangle &#xA;   + \langle\Phi_0|W_1|\Psi_1\rangle &#xA;   + \langle\Phi_0|W_2|\Psi_0\rangle\\&#xA;\scriptsize\cancelto{d_0\cdot E_0}{\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle} &#xA;   + \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle &#xA;  &amp;amp;= \scriptsize W_0\cancelto{d_0=0}{\langle\Phi_0|\Psi_2\rangle} &#xA;   + W_1\cancelto{c_0=0}{\langle\Phi_0|\Psi_1\rangle} &#xA;   + W_2\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9c}\\&#xA;W_2 &amp;amp;= \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle\tag{9c'}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do now insert $\mathrm{10b'}$ and get the second order correction:&#xA;\begin{align}&#xA;W_2 &amp;amp;= \sum_{i\neq0}\frac{&#xA;             \langle\Phi_0|\mathbf{H'}|\Phi_i\rangle&#xA;             \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle&#xA;            }{E_0-E_i}\tag{9c''}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the expression for the second order energy, let's do the messy stuff for the second order contribution to the wave function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Skip here for the short version, which is still very long, but that's how we roll for the second order correction to the wave function&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start by multiplying $\mathrm{8c}$ with $\Phi_j$ from the left and integrating (projection). I will include a few more steps for this problem, hopefully that will help transferring it to your notation.&#xA;\begin{align}&#xA;\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +&#xA;\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &amp;amp;=&#xA;\langle\Phi_j|W_0|\Psi_2\rangle +&#xA;\langle\Phi_j|W_1|\Psi_1\rangle +&#xA;\langle\Phi_j|W_2|\Psi_0\rangle\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will go through this term by term this time, as it is quite a long equation.&#xA;Let's start with the most simplest term, the third one on the right hand side. As you noted, it cancels.&#xA;\begin{align}&#xA;\langle\Phi_j|W_2|\Psi_0\rangle &#xA;   &amp;amp;= W_2 \cancelto{0}{\langle\Phi_j|\Psi_0\rangle }\\&#xA;   &amp;amp;= 0&#xA;\end{align} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's continue on the right hand side:&#xA;\begin{align}&#xA;\langle\Phi_j|W_1|\Psi_1\rangle &#xA;   &amp;amp;= W_1 \langle\Phi_j|\Psi_1\rangle\\&#xA;   &amp;amp;= W_1 \sum_i c_i \langle\Phi_j|\Phi_i\rangle\\&#xA;   &amp;amp;= W_1 \left(c_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} +  &#xA;         \sum_{i\neq j} c_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\&#xA;   &amp;amp;= c_j W_1&#xA;\end{align} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the same for the last contribution:&#xA;\begin{align}&#xA;\langle\Phi_j|W_0|\Psi_2\rangle &#xA;   &amp;amp;= W_0 \langle\Phi_j|\Psi_2\rangle\\&#xA;   &amp;amp;= W_0 \sum_i d_i \langle\Phi_j|\Phi_i\rangle\\&#xA;   &amp;amp;= W_0 \left(d_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} &#xA;        + \sum_{i\neq j} d_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\&#xA;   &amp;amp;= d_j E_0 &amp;amp;\mathrm{9a':}&amp;amp;(W_0 = E_0)&#xA;\end{align} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go to the left side:&#xA;\begin{align}&#xA;\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle &#xA;   &amp;amp;= \sum_i d_i \langle\Phi_j|\mathbf{H}_0|\Phi_i\rangle \\&#xA;   &amp;amp;= \sum_i d_i E_i \langle\Phi_j|\Phi_i\rangle\\&#xA;   &amp;amp;= d_j E_j \cancelto{1}{\langle\Phi_j|\Phi_j\rangle}&#xA;      + \sum_{i\neq j} d_i E_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\\&#xA;   &amp;amp;= d_j E_j&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last one can barely be simplified:&#xA;\begin{align}&#xA;\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle&#xA;  &amp;amp;= \sum_i c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\&#xA;  &amp;amp;= \cancelto{0}{c_0 \langle\Phi_0|\mathbf{H'}|\Phi_0\rangle}&#xA;   + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\ &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Avengers reassemble, rearrange, and insert the expression for $W_1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +&#xA;\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &amp;amp;=&#xA;\langle\Phi_j|W_0|\Psi_2\rangle +&#xA;\langle\Phi_j|W_1|\Psi_1\rangle +&#xA;\langle\Phi_j|W_2|\Psi_0\rangle\\&#xA;d_j E_j + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle&#xA; &amp;amp;= c_j W_1 + d_j E_0\\&#xA;d_j (E_j - E_0)&#xA; &amp;amp;= c_j \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle&#xA;   - \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\&#xA;d_j &amp;amp;=&#xA;   c_j \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)} &#xA;   - \sum_{i\neq 0} c_i \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now introduce the first order base, $c_i$ and $c_j$ from $\mathrm{10b'}$, there will be some minus shuffling as well.&#xA;\begin{align}&#xA;d_j &#xA;  &amp;amp;=&#xA;   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} &#xA;   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)} &#xA;   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}&#xA;     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}\\&#xA;  &amp;amp;=&#xA;   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} \cdot&#xA;   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(-1)(E_0 - E_j)} &#xA;   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}&#xA;     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(-1)(E_0 - E_j)}\\&#xA; d_j &amp;amp;=&#xA;   \sum_{i\neq 0} &#xA;   \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle%&#xA;         \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle%&#xA;        }{(E_0 - E_j)(E_0 - E_i)} &#xA;   -\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle%&#xA;          \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)^2} \tag{10c'}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I am officially exhausted right now. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And lastly, a small remark on your notation (I keep this paragraph for nostalgic reasons). It took me quite some time understanding it, and as we saw (in some earlier version of this answer), I was mistaken.&lt;br&gt;&#xA;While I do not mind the hat notation for operators, indicating order on top of the quantities is problematic in my opinion. To me this place is reserved for powers. I suggest you change it to one of the following: $E_{n,0},\ {}^{(0)}\!E_n,\ E^{(0)}_n$. When you have a look at the other answer, you will notice, I have decided to stick to the last option. But as I do not know which book your are using it is also not easy to recommend. I believe there are as many different notations as there are books out there. I personally preferred the book &lt;a href=&quot;http://as.wiley.com/WileyCDA/WileyTitle/productCd-0470011874.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;Introduction to Computational Chemistry, 2nd Edition&quot;by Frank Jensen&lt;/a&gt;.&lt;br&gt;&#xA;As we settled in the comments, I appreciate that you stick to Dirac's original notation. And also, notation is like Marvel and DC, you pick one and stick with it, you can still appreciate the other, but just not as much. Or a football (Americans pronounce this as &quot;soccer&quot;) club, or ...&lt;br&gt;&#xA;As long as you understand it and can explain it to someone else, you are fine.&lt;br&gt;&#xA;And now we all deserve some chocolate and some coffee and cake. Thank you for reading this post until the end.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-12-03T10:51:25.547" LastActivityDate="2015-12-03T10:51:25.547" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28379" PostTypeId="2" ParentId="28370" CreationDate="2015-04-08T09:49:33.950" Score="3" Body="&lt;p&gt;The carbon atom of the carbonyl group has a partial positive charge due to electron withdrawal by the electronegative oxygen atom. An $S_{N}1$ mechanism would result in an intermediate with a positive charge next to a partial positive charge, which would be high in energy.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-08T09:49:33.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28380" PostTypeId="1" AcceptedAnswerId="28383" CreationDate="2015-04-08T09:50:13.267" Score="14" ViewCount="31576" Body="&lt;p&gt;According to present definition torr and mm of Hg differ though slightly.&#xA;I am still confused over it. Can someone write an explicit answer differentiating between them and the reason for differentiation as earlier they were considered same. The explanation is probably related to the changes is pressure at different heights so I want the exact definitions of these units that was used earlier and what modifications have been made?&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T16:57:37.633" LastActivityDate="2017-10-05T10:48:39.100" Title="What is the difference between the units 'torr' and 'mm of Hg'?" Tags="&lt;pressure&gt;&lt;units&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28381" PostTypeId="1" AcceptedAnswerId="28529" CreationDate="2015-04-08T10:03:00.283" Score="5" ViewCount="230" Body="&lt;p&gt;My son and daughter got given gifts where you place a plastic egg in water, and the (rubber?) toy inside expands (over three days) and breaks the shell of the egg so it 'hatches'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H6CJp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7aUB3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also observed that three weeks later the toys shrank again, but grew once placed in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the chemical reaction fascinating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What is the reaction in a child's 'add water and watch the egg hatch' toy?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2016-05-27T22:18:09.753" Title="What is the reaction in a child's 'add water and watch the egg hatch' toy?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28382" PostTypeId="2" ParentId="28333" CreationDate="2015-04-08T10:58:51.427" Score="4" Body="&lt;p&gt;The pH should be around 8.4 and the conductivity in the range of 2.5 - 3 mS/cm. The lower value is appropriate for freshly prepared water. The conductivity is used for the determination of TDS (total dissolved particles).&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-04-08T11:06:13.090" LastActivityDate="2015-04-08T11:06:13.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28383" PostTypeId="2" ParentId="28380" CreationDate="2015-04-08T11:44:35.307" Score="18" Body="&lt;p&gt;The two units torr, and mm of Hg were the same until they were redefined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torr was named after the Italian Evangelista Torricelli. 1 atmosphere is $101325\ \mathrm{Pa}$. The torr is defined as $1/760$ of an atmosphere. This is equal to $133.322\overline{368421052631578947}~\mathrm{Pa}$, which periodically infinitely repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mm of Hg was defined as the pressure generated by a column of mercury one millimeter high. Since the pressure of mercury depends on temperature and gravity, it is now redefined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{0.001\ mL \cdot 13595.1~mg \cdot 9.80665\ m/s^2 = 133.322387415 ~Pa}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has caused a difference between the units of $0.000015~\%$, so unless you require extremely high precision, you can assume that $1\ \mathrm{torr} = 1\ \mathrm{mmHg}$&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="7475" LastEditDate="2017-01-25T19:55:20.667" LastActivityDate="2017-01-25T19:55:20.667" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28384" PostTypeId="1" CreationDate="2015-04-08T11:58:32.027" Score="4" ViewCount="433" Body="&lt;p&gt;I recently heard that during some period of time Purina pet food was being produced from crude oil via a set of complex chemical transformations. Which, I think, is mostly bluff and marketing. So is it possible, and have somebody had any real positive practical experience of getting amino acids from hydrocarbons?&lt;/p&gt;&#xA;" OwnerUserId="15446" LastEditorUserId="7475" LastEditDate="2015-10-16T16:36:56.990" LastActivityDate="2015-10-16T16:36:56.990" Title="Is it possible to synthesize amino acids from hydrocarbons?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28386" PostTypeId="1" AcceptedAnswerId="28388" CreationDate="2015-04-08T12:25:27.630" Score="7" ViewCount="246" Body="&lt;p&gt;The product of an aldol reaction is a $\beta$-hydroxy ketone. This can go on to eliminate to form an alkene thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/alKG4.png&quot; alt=&quot;Aldol reaction showing possible elimination product&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the species on the far left of the above diagram (the $\beta$-hydroxy ketone product of the first aldol reaction) can react through an aldol reaction again given that it has two more available protons at the $\alpha$ position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible to tell which reaction pathway will occur? Is this the issue that is referred to when the Aldol reaction is referred to as &quot;hard to control&quot;? &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="21296" LastEditDate="2017-06-06T22:11:07.100" LastActivityDate="2017-06-06T22:11:07.100" Title="What determines whether an aldol reaction will occur again or eliminate?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aldol-reaction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28387" PostTypeId="1" AcceptedAnswerId="28496" CreationDate="2015-04-08T12:37:51.430" Score="13" ViewCount="4825" Body="&lt;p&gt;As the title says, why do some element symbols contain a single letter while others contain two letters?&lt;/p&gt;&#xA;" OwnerUserId="15327" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-07-04T15:22:48.127" LastActivityDate="2020-10-27T15:14:32.963" Title="Why do some element symbols contain a single letter while others contain two letters?" Tags="&lt;nomenclature&gt;&lt;elements&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28388" PostTypeId="2" ParentId="28386" CreationDate="2015-04-08T12:57:50.283" Score="8" Body="&lt;p&gt;Partially yes. The big problem with aldol reactions in general is that there are a lot of species that can react together to form stable compounds resulting in low yield. There are numerous ways to counter this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Specific enolate equivalants, like lithium enolates, silyl enol ethers, aza enolates, enamines, etc. These are among the most important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lack of proton in $\alpha$-position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Strong thermodynamic favoring of certain products in combination with reversibility of the reaction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for your example the $\beta$-hydroxy ketone is more sterically hindered than acetone and is therefore not very likely to be the major contributor (although polymerization can still happen of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you cannot predict whether or not elimination of $\ce{H2O}$ occurs after the aldol reaction, but the likeliness can be increased by forcing certain reaction conditions.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-08T12:57:50.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28389" PostTypeId="2" ParentId="28384" CreationDate="2015-04-08T13:44:11.850" Score="5" Body="&lt;p&gt;Out of memory, I don't recall a method for the &lt;strong&gt;direct&lt;/strong&gt; synthesis of $\alpha$-amino acids from plain hydrocarbons (including alkenes and alkynes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is different for $\beta$-amino acids. The $\ce{Pd(II)}$-catalyzed reaction of alkenes with amines in the presence of carbon monoxide, known as &lt;strong&gt;aminocarbonylation&lt;/strong&gt; can furnish these amino acids or their derivates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] from crude oil via a set of complex chemical transformations&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;however, is pretty vague and may involve anything, including the formation of aldehydes and their subsequent use in a &lt;strong&gt;Strecker&lt;/strong&gt; reaction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-08T13:44:11.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28390" PostTypeId="1" AcceptedAnswerId="28394" CreationDate="2015-04-08T13:45:24.443" Score="3" ViewCount="325" Body="&lt;p&gt;I want to know how a radiofrequency pulse of known flip angle $\beta$ changes the population difference of a set of isolated spin 1/2 nuclei.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the equilibrium magnetization is related to the population difference by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$M_0= \frac{1}{2}\gamma\hbar\Delta{n_{eq}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to know how to calculate the x, y, z components of the magnetization after the pulse, and then calculate the population difference from these values. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="10256" LastEditDate="2015-04-08T14:11:42.673" LastActivityDate="2015-04-08T14:34:41.760" Title="NMR population difference from a pulse of known flip angle" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28391" PostTypeId="2" ParentId="28384" CreationDate="2015-04-08T13:45:40.817" Score="5" Body="&lt;p&gt;Well (almost) everything is possible. A better question might be whether or not it is profitable, but remember that A LOT of organic compounds used in synthesis originate from the petrochemical industry (these include other than simple alkanes as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible industrial pathway from an alkane to an amino acid could be: 1) chlorination of the alkane using a radical chain reaction 2) reaction of the chlorinated functionalized compound with ammonia to yield the alkylamine 3) chlorinate again and react with $\ce{NaCN}$ to obtain the nitrile amine 4) hydrolyze in aqueous acid which yields the desired amino acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the above procedure is highly simplified as it does neither consider stereo chemistry nor the question of how to chlorinate at the right carbon (regioselectivity), but you do get the idea that even completely unfunctionalized alkanes can be turned into complex molecules with enough manipulations. Notice that for each functional group (at least for the first) we introduce, we must do some radical chain reaction to functionalize the otherwise unreactive alkane. Chlorination of alkanes is done widely on industrial scale as it is one of the few ways to functionalize an alkane.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-08T13:45:40.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28392" PostTypeId="1" AcceptedAnswerId="28396" CreationDate="2015-04-08T14:23:24.543" Score="1" ViewCount="1463" Body="&lt;p&gt;My problem sheet asks me: &quot;Draw a mechanism for the formation of $\ce{EtCH2COOH}$ from the reaction of $\ce{MeCOCH(Et)COOEt}$ with $\ce{NaOH}$ in $\ce{EtOH}$&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has got me stumped. The mechanisms that I have attempted use the hydroxide ion as the base/nucleophile, is this right? My initial thoughts were to make the enolate by deprotonating between the caronyl groups then doing and aldol into the ketone part of another molecule of ethylacetoacetate but this just gives me a complicated molecule that I can't see reacting further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simply cannot find a way to make the product using only the reactants in the question.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-04-13T19:29:07.957" LastActivityDate="2015-04-13T19:29:07.957" Title="Mechanism for the formation of butanoic acid from ethylacetoacetate" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28393" PostTypeId="2" ParentId="17018" CreationDate="2015-04-08T14:24:02.807" Score="1" Body="&lt;p&gt;Try this equation I found in a company manual...not sure about the source but I tested it against the monographs you can find at say &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanex-TISH-Guide.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanex-TISH-Guide.aspx&lt;/a&gt;, p7&#xA;and found it to be quite good... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Percent MeOH by wt. in MeOH-Water Solution&#xA;= ((32.04 x (F8 x (0.0004745 x ((G8-32) x 0.555)+0.9894)))/(14.03 + F8 x (0.0147 x ((G8-32) x 0.555) - 9.53))) x 100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where F8 = Sp. Gr. of Soln. and G8 = Temp. of Solution in deg F&lt;/p&gt;&#xA;" OwnerUserId="15448" LastActivityDate="2015-04-08T14:24:02.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28394" PostTypeId="2" ParentId="28390" CreationDate="2015-04-08T14:34:41.760" Score="2" Body="&lt;p&gt;immediately after the pulse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_z = M_0cos(\beta )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_x = M_0sin(\beta )$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_y = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for example after a 90 degree pulse, all the magnetization has been rotated into the x-y plane, and after a 180 pulse, the magnetization is in the -z direction.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-04-08T14:34:41.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28395" PostTypeId="1" CreationDate="2015-04-08T14:52:56.683" Score="4" ViewCount="1180" Body="&lt;p&gt;First question regarding corrosion of iron: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is anodic area always located at nicks and scratches, where energy state is high? (does&quot;high [potential] energy&quot; mean unstable structure?) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqNUH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second questions(refer to the image):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why and how does $\ce{Fe^{2+}}$ move to the top of the iron piling(part near the surface)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on what I learn about conductor, there cannot charges piled up at a particular region of conductor, as opposed to insulator? Also the image is sorta misleading, the ions move through the iron piling or the electrolytes in sea water?&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4356" LastEditDate="2015-04-09T20:32:35.017" LastActivityDate="2015-04-12T09:27:10.107" Title="Corrosion of iron" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28396" PostTypeId="2" ParentId="28392" CreationDate="2015-04-08T15:01:19.950" Score="2" Body="&lt;p&gt;I thought about adding solid $\ce{NaOH}$ to ethanol and therefore used $\ce{EtO-}$ ;) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iQjwg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-08T15:01:19.950" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28397" PostTypeId="1" CreationDate="2015-04-08T15:19:09.273" Score="3" ViewCount="177" Body="&lt;p&gt;According to my organic chemistry textbook, Cis-Decalin has the two rings nearly perpendicular to one another, with the middle hydrogens cis to one another. Trans-Decalin has the ring nearly flat (planar), with the middle hydrogens trans to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, typing Cis-Decalin and Trans-Decalin into ChemBioDraw (where they appear correctly in 2D form) and copying them into ChemBio3D shows the exact opposite conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a bug with ChemBio3D, or a mistake in the textbook?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="5017" LastEditDate="2015-04-08T16:22:15.530" LastActivityDate="2015-04-08T18:36:20.150" Title="Cis/Trans-Decalin in ChemBio3D" Tags="&lt;organic-chemistry&gt;&lt;software&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28398" PostTypeId="1" AcceptedAnswerId="28400" CreationDate="2015-04-08T15:44:11.203" Score="2" ViewCount="961" Body="&lt;p&gt;I have a question about fats. This may sound cliché! Because we are in the era of &quot;trans-fat&quot; free, &quot;unsaturated&quot; food. Not to mention it's the period &lt;em&gt;where people get sicker and die faster (oh may be, the statistics weren't obvious in the past...)&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fats are important part of our diet and major source of energy (&lt;em&gt;the mighty omega-3 fatty acids&lt;/em&gt;). But, The trans-fat, it's a linear molecule, it tends to stick and stacks up, builds up bad cholesterol, so that's not good for our cardiovascular system...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean, the opposite configuration, cis-fat is healthy for our body?&#xA;If cis is what's naturally found, by default it's unsaturated fat isn't it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's that time of the year again, for a community project we (&lt;em&gt;the people who are new to biology, organic, biochemistry education&lt;/em&gt;) need to explain &lt;strong&gt;that&lt;/strong&gt; to kids of age 7 and 10. They are going to go for shopping with us and we have to pop our eye balls to make sure they don't buy the wrong stuff. Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips to doughnuts...list goes on. Saying something isn't bad not going to convince them an &lt;em&gt;atom&lt;/em&gt;! So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing and prevent any &quot;kids-melo-drama&quot;.&#xA;I am struggling with &lt;a href=&quot;https://chemistry.stackexchange.com/a/10852/7450&quot;&gt;this explaination&lt;/a&gt; and I doubt it's going to be any easier for the kids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acaloriecounter.com/breakfast-cereal.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;The breakfast cereals reference.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/blHfd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-04-10T10:50:32.823" Title="When disqualifying trans fat, are we qualifying cis-fat as healthy?" Tags="&lt;food-chemistry&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28399" PostTypeId="1" AcceptedAnswerId="28432" CreationDate="2015-04-08T16:28:06.007" Score="25" ViewCount="16431" Body="&lt;p&gt;Standard Gibbs free energy of formation of liquid water at &lt;span class=&quot;math-container&quot;&gt;$\pu{298 K}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{−237.17 kJ mol-1}$&lt;/span&gt; and that of water vapour is &lt;span class=&quot;math-container&quot;&gt;$\pu{−228.57 kJ mol-1}$&lt;/span&gt; therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O (l) -&amp;gt; H2O (g)}\qquad\Delta G =\pu{8.43{kJ mol-1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\Delta G&amp;gt;0$&lt;/span&gt;, it should not be a spontaneous process but from common observation, water &lt;em&gt;does&lt;/em&gt; turn into vapour from liquid over time without any apparent interference. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="23561" LastEditDate="2018-09-30T01:02:55.827" LastActivityDate="2018-09-30T01:02:55.827" Title="Why does water evaporate spontaneously at room temperature despite ΔG &gt; 0?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28400" PostTypeId="2" ParentId="28398" CreationDate="2015-04-08T16:34:54.420" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If cis is what's naturally found, by default it's unsaturated fat isn't it? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both cis and trans are unsaturated.  &quot;Unsaturated&quot; means having at least one C-C bond that is not a single bond (but instead a double or triple bond).  The terms &lt;a href=&quot;http://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism&quot; rel=&quot;nofollow noreferrer&quot;&gt;cis and trans&lt;/a&gt; refer to the geometry of the molecule about the double bond.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trans fat also occurs naturally in milk, for example fats of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vaccenic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;vaccenic acid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since a fatty acid can have more than one double bond, it can be both cis and trans, for example fats of &lt;a href=&quot;http://en.wikipedia.org/wiki/Rumenic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rumenic acid&lt;/a&gt; found in milk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUhas.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When disqualifying trans fat, are we qualifying cis-fat as healthy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cis-fat can also have negative health consequences.  For example, according to the review article &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0261561404000263&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dietary fat, insulin sensitivity and the metabolic syndrome&lt;/a&gt; &lt;em&gt;Clinical Nutrition&lt;/em&gt; vol. 23 pages 447-456.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;monounsaturated and ω-6 polyunsaturated fatty acids have less negative effect on insulin sensitivity than saturated fat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, even consumption of too much unsaturated, naturally occuring, cis fat can have negative health consequences, such as insulin resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but is this really true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kmqYU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've seen, cereals have little or no fat of any kind, let alone trans-fat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, maple syrup has no fat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4UNx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even potato chips don't necessarily contain any trans fat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pA0W8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The amounts of partially hydrogenated trans fat in foods has been greatly reduced in the last 10 years.  Milk and meats naturally contain trans fat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't want you telling my kids they can't drink milk or eat a hamburger.  Sure it's bad to eat too many hamburgers.  The focus should be a balanced diet.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-04-09T14:50:43.920" LastActivityDate="2015-04-09T14:50:43.920" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28401" PostTypeId="1" CreationDate="2015-04-08T17:00:49.223" Score="3" ViewCount="979" Body="&lt;p&gt;What is the decarboxylation product of the following species:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5pUYt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may seem straightforward but the product enolate involves a double bond with one end at the bridgehead carbon which is impossible/very unstable. What is the correct product and how is the product reached mechanistically. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-08T20:43:21.010" Title="Decarboxylation product of bicyclic compound" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28402" PostTypeId="1" AcceptedAnswerId="34986" CreationDate="2015-04-08T17:28:25.467" Score="7" ViewCount="3622" Body="&lt;p&gt;Having never prepared a buffer solution before, I am hoping for a little help in understanding the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.whitelabs.com/files/Alkaline_Methylene_Violet_Solution.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This method&lt;/a&gt; I plan to use calls for a $\pu{0.1M}$ glycine buffer at pH $10.6$. Some internet searching yielded &lt;a href=&quot;http://microscopy.berkeley.edu/Resources/instruction/buffers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this recipe&lt;/a&gt; (very bottom of the page) for making a glycine-$\ce{NaOH}$ buffer at $10.6$ pH, which I assume is what I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of it &lt;em&gt;seems&lt;/em&gt; to make sense to me. $\pu{0.2M}$ stock solutions of both glycine and $\ce{NaOH}$ would be prepared as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.2M NaOH solution = 0.2 * 39.9971 = ~8g NaOH/l&#xA;0.2M glycine solution = 0.2 * 75.07 = ~15g glycine/l&#xA;(Molar mass: NaOH = 39.9971; glycine = 75.07 g/mol)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And, to yield a buffer at $10.6$ pH, mix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;25ml glycine stock + 22.75ml NaOH stock + deionized water to 100ml&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where I get lost is figuring out the molar concentration of the resulting solution, since there are two different solutes. Would it be just the total number of moles present of both, something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(25ml + 22.75ml) * 0.2M / 100ml = 0.0955M ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems pretty close to the $\pu{0.1M}$ concentration called for, but I can't seem to find any confirmation of this.&lt;/p&gt;&#xA;" OwnerUserId="13178" LastEditorUserId="23561" LastEditDate="2018-09-08T05:45:27.563" LastActivityDate="2018-09-08T05:45:27.563" Title="Alkaline glycine buffer molarity" Tags="&lt;ph&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28404" PostTypeId="2" ParentId="28399" CreationDate="2015-04-08T17:35:39.067" Score="9" Body="&lt;p&gt;$\Delta G = -RT \ln K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means there for any given process an equilibrium constant can be deduced.  Given the products are 'gaseous' they will escape the system, and further drive the equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = 0.05$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied pragmatically, we define a system as 'near the puddle'.  When the water evaporates, it can either condense again, or move 'far from the puddle' and hence leaves the system.  Over time the 'moving away' portion of this process leads to what we observe humanisticly as the puddle disappearing.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorDisplayName="user7951" LastEditDate="2015-04-11T15:23:37.527" LastActivityDate="2015-04-11T15:23:37.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28405" PostTypeId="2" ParentId="28401" CreationDate="2015-04-08T17:49:19.853" Score="4" Body="&lt;p&gt;${\beta}$-keto acids typically decarboxylate very smoothly at temperatures between 100-200°C. The mechanism involves a cyclic, 6-membered transition state that first produces an enol intermediate (see figure). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vBfoz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the case of your bicyclo[2.2.1] ${\beta}$-keto acid, a bridgehead double bond would result.  As you correctly point out, such a structure would be very high energy.  Therefore &lt;strong&gt;this compound does not decarboxylate&lt;/strong&gt; at normal temperatures (&lt;a href=&quot;https://books.google.com/books?id=Zr7algbSigoC&amp;amp;pg=PA217&amp;amp;lpg=PA217&amp;amp;dq=decarboxylation%20of%20bridgehead%20keto%20acid&amp;amp;source=bl&amp;amp;ots=iUxZSPi48U&amp;amp;sig=tox5gzY6UMx3DIC_UpGStW8t3BU&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=3ZElVdv7CYrysAXAi4G4AQ&amp;amp;sqi=2&amp;amp;ved=0CCAQ6AEwAQ#v=onepage&amp;amp;q=decarboxylation%20of%20bridgehead%20keto%20acid&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;, see p. 218, scheme 6.53).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure if you heated it hot enough something would happen.  It may decarboxylate &lt;em&gt;via&lt;/em&gt; a free radical mechanism. The resultant bridgehead free radical could fragment or dimerize.  Alternately, and my preferred guess, would have the starting compound, in equilibrium with its non-bridgehead enol tautomer, undergo a reverse Diels-Alder reaction as pictured above.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-08T20:43:21.010" LastActivityDate="2015-04-08T20:43:21.010" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28407" PostTypeId="1" AcceptedAnswerId="28410" CreationDate="2015-04-08T18:22:16.697" Score="2" ViewCount="936" Body="&lt;p&gt;How do I write a figure legend for a thin layer chromatography plate drawing?&lt;/p&gt;&#xA;" OwnerUserId="15454" LastActivityDate="2015-04-08T20:22:46.170" Title="Figure Headings in Laboratory Reports" Tags="&lt;chromatography&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28409" PostTypeId="2" ParentId="28407" CreationDate="2015-04-08T18:38:29.770" Score="0" Body="&lt;p&gt;At the bottom of each spot mark your legends by alpha/numerals. Outside put the meaning of your legends. Mention solvent system used for elution.&lt;/p&gt;&#xA;" OwnerUserId="15455" LastActivityDate="2015-04-08T18:38:29.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28410" PostTypeId="2" ParentId="28407" CreationDate="2015-04-08T18:38:49.760" Score="2" Body="&lt;p&gt;Provide all the data that are necessary to reproduce the experiment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;TLC material (silica gel or aluminium oxide)&lt;/li&gt;&#xA;&lt;li&gt;composition of the mobile phase&lt;/li&gt;&#xA;&lt;li&gt;special tricks, such as running the TLC in methanol first for 5 mm to squeeze the initial spot to a very narrow band&lt;/li&gt;&#xA;&lt;li&gt;detection of the separated bands/spots:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{R_f}$ values&lt;/li&gt;&#xA;&lt;li&gt;do the spots have a colour&lt;/li&gt;&#xA;&lt;li&gt;do the show fluorescence&lt;/li&gt;&#xA;&lt;li&gt;which reagents did you use for staining&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you report on a series of measurements, it's usually good practise describe the conditions that didn't change (type of TLC plate, staining solutions, etc.) in the general section of the experimental part of the report/article/thesis.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-08T20:22:46.170" LastActivityDate="2015-04-08T20:22:46.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28411" PostTypeId="1" AcceptedAnswerId="28417" CreationDate="2015-04-08T19:17:54.473" Score="2" ViewCount="800" Body="&lt;p&gt;A &quot;reaction map&quot; in my lecture handout suggests that the product of this reaction is: &lt;img src=&quot;https://i.stack.imgur.com/TU34N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically overall it just merges two ethylacetoacetate molecules together at the central carbon. What is the mechanism for this? My thoughts are to swap the acidic proton for an Iodine atom by making the enol (in acidic conditions to avoid iodoform reaction) and react with $I_2$. Then, the enol of another molecule of EAA substitutes the Iodine to make the product. However, I don't know whether this is $S_N1$ or $S_N2$. Typically alkylation is $S_N2$ (I don't know why, please explain if you can) but in this case its a bulky secondary substituent and the solvent is protic (acidic conditions). Surely it's $S_N1$, right?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-08T20:00:58.830" Title="Reaction of ethylacetoacetate with Iodine" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28412" PostTypeId="1" AcceptedAnswerId="28774" CreationDate="2015-04-08T19:19:40.577" Score="13" ViewCount="1974" Body="&lt;p&gt;I've been searching the web for this phenomenon and surprisingly I found only one post about it. And &lt;a href=&quot;http://www.madsci.org/posts/archives/2001-06/991666551.Ch.r.html&quot;&gt;the reply&lt;/a&gt; to this &lt;a href=&quot;http://www.madsci.org/posts/archives/2001-06/991666551.Ch.q.html&quot;&gt;question about canned air&lt;/a&gt; begins with words:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't know about the shaking part (it is hard for me to see how that would make a difference)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This beginning has strongly shaken my faith, so for sure I picked up a can with deodorant and shaken it. And it &lt;em&gt;does become cooler&lt;/em&gt;. Significantly! After standing still, the can doesn't seem to become warmer, so this reaction doesn't revert itself and maybe is irreversible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I imagine it, when being shaken the average pressure in the can as a whole &lt;em&gt;can't possibly change&lt;/em&gt;. Also it's interesting how you add energy by shaking only to have more energy drained by cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So please, can you explain this strange behavior? Could you also cover the energy flow? When does the energy return back out of the can?&lt;/p&gt;&#xA;" OwnerUserId="1677" LastActivityDate="2017-07-15T03:33:37.107" Title="What causes spray cans to get cooler when shaken?" Tags="&lt;thermodynamics&gt;&lt;molecules&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28414" PostTypeId="2" ParentId="28399" CreationDate="2015-04-08T19:49:12.013" Score="12" Body="&lt;p&gt;There are two considerations about why water evaporates but only one of them has to do with the Gibbs free energy of vaporisation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the first answer correctly states, there is an equilibrium (heavily biased in favour of the liquid not the vapour) at room temperature. &lt;strong&gt;In equilibrium&lt;/strong&gt; (which is what the $\Delta G$ measures) there is enough thermal energy to put a small amount of water into the vapour phase but most water will remain liquid. This is the correct analysis in a closed vessel. The water doesn't continue to evaporate once the equilibrium is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, mostly, you don't observe water in &lt;em&gt;closed vessels&lt;/em&gt;. This explains why the water continues to evaporate. The point is that, when the vapour can be removed from the vessel containing the liquid, the normal equilibrium is not established and the water can &lt;em&gt;slowly&lt;/em&gt; evaporate (at least as long as the surrounding environment has a lower humidity than the equilibrium vapour pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In open systems the simple thermodynamic numbers don't correctly describe the long term behaviour because equilibrium isn't what happens.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-04-08T19:49:12.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28415" PostTypeId="1" CreationDate="2015-04-08T19:50:30.993" Score="1" ViewCount="28" Body="&lt;p&gt;I need to compare the hydrophilicity of those materials to compare contact angles of water with those materials' surfaces for surface tension measurement. Materials are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PTFE (Teflon), PVC, PS, PVDF, Si-Wafer-Si-SiO$_x$, Parafilm and PMMA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ordering the polarity can also work. Where can I found such a comparison? Can someone help me with the ordering. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="15457" LastEditorUserId="7448" LastEditDate="2015-04-08T20:05:02.357" LastActivityDate="2015-04-08T20:05:02.357" Title="ordering the hydrophilicity of some polymers" Tags="&lt;polymers&gt;&lt;polarity&gt;&lt;surface-chemistry&gt;&lt;surfactants&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28416" PostTypeId="1" AcceptedAnswerId="28448" CreationDate="2015-04-08T19:54:46.343" Score="2" ViewCount="354" Body="&lt;p&gt;I'm trying to figure out which form of chromium oxide forms naturally in our atmosphere, i.e., if you take a piece of elemental $\ce{Cr}$ from vacuum to a normal, room temperature environment with air, does it form $\ce{CrO}$ on the surface? Or $\ce{Cr2O3}$? Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/d-Block_Elements/Group_6%3A_Transition_Metals/Chemistry_of_Chromium#Chromium_Oxides_and_Hydroxides&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chromium (III) oxide, $\ce{Cr2O3}$ is the main oxide of chromium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems like it might be that one, but &quot;main&quot; is pretty vague. Does anyone have a more definitive answer, with a source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you!&lt;/p&gt;&#xA;" OwnerUserId="3764" LastEditorUserId="17368" LastEditDate="2017-03-25T07:42:41.400" LastActivityDate="2017-03-25T07:48:27.407" Title="Which form of chromium oxide is formed naturally in the atmosphere?" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28417" PostTypeId="2" ParentId="28411" CreationDate="2015-04-08T20:00:58.830" Score="2" Body="&lt;p&gt;Are you certain that the coupling itself isn't a radical reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond energy of an ordinary $\ce{C-I}$ bond is in the range of $210\,\mathrm{kJ\cdot\,mol^{-1}}$, which isn't a hell of a lot. Moreover, the resulting radical of an 1,3-dicarbonyl compound is nicely stabilized. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-08T20:00:58.830" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28418" PostTypeId="1" CreationDate="2015-04-08T20:11:40.427" Score="3" ViewCount="76" Body="&lt;p&gt;Consider the Robin boundary condition for the diffusion/heat equation $u_t=a(t)u_{xx}+f(x,t)$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-k(t)u_x(0,t)=h(t)u(0,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_x(0,t)+\frac{h(t)}{k(t)}u(0,t)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k(t)$ thermal conductivity and $h(t)$ heat tranfer coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Question: Is it possible that the ratio $h(t)/k(t)$ to be constant? Could anyone please help me? I have really no idea.&lt;/p&gt;&#xA;" OwnerUserId="15458" LastEditorDisplayName="user7951" LastEditDate="2016-11-23T15:48:08.663" LastActivityDate="2016-11-23T15:48:08.663" Title="Ratio between heat tranfer coefficient and thermal conductivity" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28420" PostTypeId="1" AcceptedAnswerId="28422" CreationDate="2015-04-08T20:33:33.297" Score="-3" ViewCount="87" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=sGDKxlW1ZoQ&quot; rel=&quot;nofollow&quot;&gt;Ionic Liquids separation of oil and sand.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="114" LastActivityDate="2015-04-08T20:56:53.440" Title="What types of chemicals/ionic liquid could perform this type of separation?" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-04-12T21:28:21.410" ContentLicense="CC BY-SA 3.0" />
  <row Id="28422" PostTypeId="2" ParentId="28420" CreationDate="2015-04-08T20:56:53.440" Score="2" Body="&lt;p&gt;In the video from PennState, Aron Lupinsky tells that he is working in the group of Prof. Painter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these information, a &lt;strong&gt;cursory search&lt;/strong&gt; furnishes the article&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Recovery of Bitumen from Low-Grade Oil Sands Using Ionic Liquids&lt;/em&gt; by &lt;em&gt;Phillip Williams&lt;/em&gt;, &lt;em&gt;Aron Lupinsky&lt;/em&gt; and &lt;em&gt;Paul Painter&lt;/em&gt;, published in &#xA;&lt;em&gt;Energy Fuels&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;24&lt;/em&gt;, 2172–2173&#xA;(&lt;a href=&quot;http://dx.doi.org/10.1021/ef901384s&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first page of the article, it is mentioned that a mixture of toluene and 1-butyl-2,3-dimethylimidazolinium tetrafluoroborate, $\ce{[bmmim][BF4]}$, was used for the extraction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-08T20:56:53.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28424" PostTypeId="1" AcceptedAnswerId="28425" CreationDate="2015-04-08T21:49:10.947" Score="2" ViewCount="3840" Body="&lt;p&gt;What is the final product of this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/uH9l1gS.png&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/LFeO6Lr.png&quot; alt=&quot;my solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this look correct? I'm concerned because I'm used to seeing the nitrogen removed from the product (in the cases of primary amines).&lt;/p&gt;&#xA;" OwnerUserId="2613" LastEditorUserId="95133" LastEditDate="2021-05-16T14:14:46.137" LastActivityDate="2021-05-16T14:14:46.137" Title="Predicting product of a Hofmann elimination with a cyclic amine" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28425" PostTypeId="2" ParentId="28424" CreationDate="2015-04-08T22:12:45.980" Score="4" Body="&lt;p&gt;With excess of methylamine, I wouldn't exclude additional methylation, yielding a trimethylanilinium iodide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/biQLC.png&quot; alt=&quot;alkylation of dimethylaniline&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the reaction should stop there since a further hydrogen abstraction from a $\beta$-position isn't possible.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-08T22:27:20.313" LastActivityDate="2015-04-08T22:27:20.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28426" PostTypeId="1" AcceptedAnswerId="28465" CreationDate="2015-04-09T00:45:57.587" Score="13" ViewCount="29437" Body="&lt;p&gt;I have seen videos and discussions about alcohol cracking acrylic, and waterblock manufacturers reminding users to not use alcohol in cooling systems with plexiglass acrylic, but why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question proposed as duplicate only asked for the effect, but here I already know the effect. I am asking for the reason for the effect, which is not answered in the answer to the 'duplicate' question.&lt;/p&gt;&#xA;" OwnerUserId="15462" LastEditorUserId="31313" LastEditDate="2017-04-01T01:07:38.683" LastActivityDate="2019-12-20T00:40:55.547" Title="Why does alcohol crack acrylic (plexiglass)?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;alcohols&gt;&lt;plastics&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28427" PostTypeId="1" CreationDate="2015-04-09T01:23:08.440" Score="2" ViewCount="469" Body="&lt;p&gt;Is there any difference between having the addition of 1 mole of hydrogen to 2-pentyne with a platinum catalyst compared to the addition of 2 moles of hydrogen(H2). I was under the impression that both would result in an alkane yet I have 2 problems of that nature. &#xA;would one only result in a alkene, though I thought that only happens with the poisoned catalyst.&lt;/p&gt;&#xA;" OwnerUserId="9649" LastActivityDate="2015-04-09T09:49:24.607" Title="In regards to the addition of hydrogen to alkynes, does the amount of moles of Hydrogen matter?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28428" PostTypeId="1" AcceptedAnswerId="28438" CreationDate="2015-04-09T01:35:25.803" Score="2" ViewCount="240" Body="&lt;p&gt;For the hydrolysis of $\ce{\mathit{trans\hyphen}[Co(en)2Cl2]Cl}$ (and &lt;em&gt;cis&lt;/em&gt;) we measured reaction rates via spectrophotometry. Plots of change in absorbance were used to determine the rate law and the rate for the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lab report asks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suggest another method, other than spectrophotometry, for determining the reaction rate measured in this experiment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Knowing little about measuring rates, I said one could measure the enthalpy and entropy of the reaction at a given temperature, and use those to calculate the rate constant. This is based off of the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\frac{k}{T}=-\frac{\Delta H_\text{act}}{RT}+\ln\frac{k_\mathrm{B}}{h}+\frac{\Delta S_\text{act}}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but, my professor said thermodynamics couldn’t be used to characterize kinetics. Also, reflecting on my answer, I didn’t really specify a technique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What’s another technique that could be applied here?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorDisplayName="user7951" LastEditDate="2015-04-09T14:40:23.167" LastActivityDate="2015-04-09T14:40:23.167" Title="Suggest a method, other than spectrophotometry, for determining the reaction rate" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28429" PostTypeId="1" AcceptedAnswerId="28455" CreationDate="2015-04-09T01:51:12.820" Score="3" ViewCount="881" Body="&lt;p&gt;For the hydrolysis of &lt;span class=&quot;math-container&quot;&gt;$trans-\ce{[Co(en)2Cl2]Cl}$&lt;/span&gt; (and &lt;em&gt;cis&lt;/em&gt;) we measured reaction rates via spectrophotometry. Plots of change in absorbance &lt;span class=&quot;math-container&quot;&gt;$(\ln(A-A_\infty))$&lt;/span&gt; were used to determine the rate law and the rate for the reaction. Using my data, I created the first-order plot seen below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YXREy.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The question is for the report was:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Justify assuming first-order kinetics&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My explanation was that first-order reactions can be expressed as &lt;span class=&quot;math-container&quot;&gt;$$[\ce{A}]=[\ce{A}]_0\operatorname{e}^{-kT}$$&lt;/span&gt;&#xA;and because the concentration decreases exponentially over time, we can assume its first order. Other rate laws I looked up didn’t show an exponential function, so this seemed like sound reasoning, but I was told I was wrong.&lt;/p&gt;&#xA;&lt;p&gt;What would be the justification for thinking this is first-order?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="95133" LastEditorDisplayName="user7951" LastEditDate="2021-04-12T14:17:08.143" LastActivityDate="2021-04-12T14:17:08.143" Title="Justify assuming first-order kinetics for a complex" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28430" PostTypeId="1" AcceptedAnswerId="28433" CreationDate="2015-04-09T02:17:43.990" Score="10" ViewCount="5636" Body="&lt;p&gt;When my family buys powdered cloves in bulk, we put them in plastic until we get home - which takes at very most an hour. As soon as we get home, we transfer them into glass because we have had issues with them eating through the plastic cups until they are soft and have holes in them - this takes &lt;em&gt;maybe&lt;/em&gt; a week. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know exactly what type of plastic the cups are made out of; if this helps, they have the recyclable number 6 on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, what is it in the cloves that is causing them to eat through the plastic at such a high rate? Are they really acidic? Or is it perhaps due to their oils? &lt;/p&gt;&#xA;" OwnerUserId="5137" LastEditorUserId="402" LastEditDate="2016-12-29T00:43:47.533" LastActivityDate="2019-12-02T00:55:14.007" Title="How do cloves break down plastic?" Tags="&lt;acid-base&gt;&lt;food-chemistry&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28432" PostTypeId="2" ParentId="28399" CreationDate="2015-04-09T03:48:20.343" Score="30" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, &lt;span class=&quot;math-container&quot;&gt;$$\ce{H2O(l)-&amp;gt;H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\Delta G&amp;gt;0$&lt;/span&gt;, it should not be a spontaneous process but from common observation, water &lt;em&gt;does&lt;/em&gt; turn into vapour from liquid over time without any apparent interference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Your math is correct but you left out a very important symbol from your equations.  There is a big difference between &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta G^\circ$&lt;/span&gt;.  Only &lt;span class=&quot;math-container&quot;&gt;$\Delta G^\circ$&lt;/span&gt; means the Gibbs energy change under &lt;em&gt;standard&lt;/em&gt; conditions, and as you noted in the question, the free energy values you quoted are the &lt;em&gt;standard&lt;/em&gt; gibbs free energy of water and water vapor.&lt;/p&gt;&#xA;&lt;p&gt;Whether or not something is spontaneous under standard conditions is determined by &lt;span class=&quot;math-container&quot;&gt;$\Delta G^\circ$&lt;/span&gt;.  Whether something is spontaneous under other conditions is determined by &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt;.  To find &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; for real conditions, we need to know how they differ from standard conditions.&lt;/p&gt;&#xA;&lt;p&gt;Usually &amp;quot;standard&amp;quot; conditions for gases correspond to one bar of partial pressure for that gas.  But the partial pressure of water in our atmosphere is usually much lower than this.  Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;noreferrer&quot;&gt;given by&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$G = G^\circ + RT \ln{\frac{p}{p^\circ}}$&lt;/span&gt;.  If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so &lt;span class=&quot;math-container&quot;&gt;$\ln{\frac{p}{p^\circ}}$&lt;/span&gt; would be negative infinity.  That would translate to an infinitely &lt;em&gt;negative&lt;/em&gt; -- i.e. highly spontaneous -- &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; for the water evaporation reaction.&lt;/p&gt;&#xA;&lt;p&gt;Small but not-quite zero dryness in the atmosphere would still lead to the &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; of water vapor that is more negative than liquid water.  So water evaporation is still spontaneous.&lt;/p&gt;&#xA;&lt;p&gt;Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which &lt;span class=&quot;math-container&quot;&gt;$\Delta G = 0$&lt;/span&gt; for water evaporation.  And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-09T03:48:20.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28433" PostTypeId="2" ParentId="28430" CreationDate="2015-04-09T03:59:17.990" Score="16" Body="&lt;p&gt;Most of the odor of cloves comes from volatile molecules, &lt;a href=&quot;http://en.wikipedia.org/wiki/Oil_of_clove&quot;&gt;especially eugenol&lt;/a&gt;, found in the cloves.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Eugenol&quot;&gt;Eugenol&lt;/a&gt; is a water-immiscible, oily liquid at ambient temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastic with &lt;a href=&quot;http://en.wikipedia.org/wiki/Resin_identification_code&quot;&gt;resin identifiction code #6&lt;/a&gt; is polystyrene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://sevierlab.vet.cornell.edu/resources/Chemical-Resistance-Chart-Detail.pdf&quot;&gt;Polystyrene is not compatible&lt;/a&gt; with most organic solvents, because it tends to be softened, swelled, or even completely dissolved by them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I suspect that the problem is that oil in the cloves is attacking the plastic that you use.  You could avoid the problem (mostly) by using a plastic with resin code #2 or #5 (i.e. high-density polyethylene or polypropylene) instead of #6.  But, you are probably right that glass would be better than any plastic for storing your cloves.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T03:59:17.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28435" PostTypeId="2" ParentId="28416" CreationDate="2015-04-09T05:36:59.487" Score="2" Body="&lt;p&gt;Drawing a parallel to the formation of rust (oxides of iron), in which the following &lt;a href=&quot;http://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow&quot;&gt;reactions&lt;/a&gt;:&#xA;$$\ce{Fe&amp;lt;=&amp;gt;Fe^2+ +2e-}$$&#xA;$$\ce{4Fe^2+ +O2&amp;lt;=&amp;gt;4Fe^3+ +2O^2-}$$&#xA;form both of the oxidation states of iron, reacting to eventually form both $\ce{Fe2O3}$ and $\ce{FeO}$, the two most common forms of rust. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron_oxide&quot; rel=&quot;nofollow&quot;&gt;more common&lt;/a&gt; of the two forms is $\ce{Fe2O3}$, which is likely because $\ce{Fe^3+}$ has an electron configuration of $[\ce{Ar}] 3d^5$, which is more stable than the electron configuration of $\ce{Fe^2+}$, which is $[\ce{Ar}] 3d^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely then that both $\ce{CrO}$ and $\ce{Cr2O3}$ form when $\ce{Cr}$ is exposed to the atmosphere, and also that $\ce{Cr^3+}$ has a more stable electron configuration than $\ce{Cr^2+}$, though this is not intuitive since neither half an empty or half-filled 3d orbital.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-09T05:36:59.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28437" PostTypeId="2" ParentId="28429" CreationDate="2015-04-09T06:37:18.853" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the justification for thinking this is first-order?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/KISS_principle&quot; rel=&quot;noreferrer&quot;&gt;KISS principle&lt;/a&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you start to model complex kinetics, begin with a simple model. If it fits, consider your job done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Good enough is good enough!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Imagine that you perform &lt;em&gt;therapeutic drug monitoring&lt;/em&gt;. It's not your job to elucidate the exact metabolism of a drug here, you just want to be sure that the patients receive the right dose.)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-09T06:37:18.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28438" PostTypeId="2" ParentId="28428" CreationDate="2015-04-09T07:23:32.897" Score="3" Body="&lt;p&gt;Spectroscopy is often the best method, since it allows for (almost) continuous monitoring of the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not wanted, ask yourself: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is released in the hydrolysis of the complex?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Chloride would be a good guess.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can I take samples and perform a wet chemical analysis to determine the concentration of new species?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Looking at the graph in your other related question, the answer is yes. The hydrolysis is slow enough.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which methods can I possibly use to determine the concentration $\ce{Cl-}$ in solution?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An &lt;strong&gt;argentometric&lt;/strong&gt; titration, based on the low solubility of $\ce{AgCl}$ is a good idea. What about the &lt;em&gt;Mohr&lt;/em&gt; titration, using $\ce{K2CrO4}$ as an indicator?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-09T07:23:32.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28439" PostTypeId="1" CreationDate="2015-04-09T08:50:16.773" Score="1" ViewCount="471" Body="&lt;p&gt;I want to decrease the boiling point of a compound. So I need to create an azeotrope mixture from two or three compounds. Where can I find out how make it? &lt;/p&gt;&#xA;" OwnerUserId="15468" LastEditorUserId="7448" LastEditDate="2015-04-09T09:11:15.470" LastActivityDate="2015-04-09T16:47:09.213" Title="How do I create azeotrope mixture?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ClosedDate="2018-08-12T06:24:00.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="28440" PostTypeId="1" CreationDate="2015-04-09T09:09:44.100" Score="2" ViewCount="270" Body="&lt;p&gt;Are there any simple colorimetric tests for measuring the dissolved oxygen in salt water, which do not involve titration like in Winkler method ?  i.e. Are there any chemical reagents, which when added to a salt water sample, would cause a change in the colour of the solution , giving us the measure of DO ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards.&lt;/p&gt;&#xA;" OwnerUserId="15344" LastActivityDate="2015-04-09T16:53:26.430" Title="Colorimetric test to measure dissolved oxygen in salt water" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28441" PostTypeId="2" ParentId="28439" CreationDate="2015-04-09T09:17:48.207" Score="2" Body="&lt;p&gt;Boiling point can be decrased by lowering pressure over liquide. If you insist on the azeotrop preparation, find azeotrop diagram such as: &lt;img src=&quot;https://i.stack.imgur.com/IzLT4.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or data table such as: &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope_%28data%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt; and determine the composition that must be achieved to obtain the azeotropic mixture. Then mix it accordingly.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-04-09T09:17:48.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28442" PostTypeId="2" ParentId="28427" CreationDate="2015-04-09T09:49:24.607" Score="1" Body="&lt;p&gt;Your question is not entirely clear, but I think that you ask whether or not an excess of hydrogen gas would interfere with the reduction of an alkyne to an alkane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case. A lot of the time hydrogen is just drawn into the reaction flask directly from a large reservoir (20L $\ce{H2}$ gas cylinder under pressure or a simple balloon). When the pressure in the reservoir does not change anymore (in the case of the balloon when it does not shrink anymore) it means that the reaction is finished, i.e. no more hydrogen can be added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general over-reduction cannot be prevented by limiting the amount of hydrogen present. It will just result in an awful mixture of products (alkanes/alkenes/alkynes). A poisoned catalyst is required to prevent this (e.g. Lindlar's catalyst).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-09T09:49:24.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28443" PostTypeId="1" CreationDate="2015-04-09T10:26:04.167" Score="5" ViewCount="322" Body="&lt;p&gt;I watched a ted talk (&lt;a href=&quot;https://www.youtube.com/watch?v=yzOcvINn8Iw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Luca Turin: The science of scent&lt;/a&gt;) and was wondering if anyone had any examples of how to calculate vibrational spectra of molecules (used as fragrances)?  Would love to figure out how he computed these charts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IH6kx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gwzyi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="554" LastEditorUserId="4945" LastEditDate="2015-04-10T10:38:20.200" LastActivityDate="2015-05-03T13:32:16.267" Title="How to calculate vibrational spectra of organic molecules?" Tags="&lt;computational-chemistry&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28444" PostTypeId="1" AcceptedAnswerId="28464" CreationDate="2015-04-09T10:27:27.637" Score="3" ViewCount="168" Body="&lt;p&gt;I'm looking at some data with substrate concentration on the $x$-axis and $k_\text{dep}$ on the $y$ axis. What does $k_\text{dep}$ mean? I understand that it must be a rate of some kind but what is the significance of the 'dep' part? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I'm looking at can be found in Fig. 9 in &lt;a href=&quot;https://doi.org/10.1124/mol.114.096784&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; (&lt;a href=&quot;http://molpharm.aspetjournals.org/content/87/3/430.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jSA6Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jSA6Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Essentially this is a dose response curve for a substrate and enzyme. I suspect $k_\text{dep}$ is simply the rate but why $k_\text{dep}$ and not just $k$?&lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorUserId="30836" LastEditDate="2017-10-21T12:43:02.897" LastActivityDate="2017-10-21T12:43:02.897" Title="What does &quot;dep&quot; mean?" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28446" PostTypeId="1" CreationDate="2015-04-09T12:25:29.650" Score="3" ViewCount="4907" Body="&lt;p&gt;I know that methane to methanol is a common practice, but is it possible to convert methanol back into methane, and would it be a relatively easy process to achieve?&lt;/p&gt;&#xA;" OwnerUserId="15469" LastEditorUserId="5026" LastEditDate="2018-06-28T12:45:43.163" LastActivityDate="2018-06-28T12:45:43.163" Title="Converting methanol into methane" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28447" PostTypeId="2" ParentId="28446" CreationDate="2015-04-09T12:46:34.163" Score="4" Body="&lt;p&gt;Some methanogenic miroorganisms isolated from human feces are known to digest $\ce{H2}$ and $\ce{CH3OH}$ to produce $\ce{CH4}$.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-09T12:46:34.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28448" PostTypeId="2" ParentId="28416" CreationDate="2015-04-09T13:19:56.413" Score="5" Body="&lt;p&gt;An investigation of the phase diagram reveals that the only stable chrome oxide near room temperature is $\ce{Cr2O3}$.  $\ce{Cr3O4}$ appears, but only near 2000K.  There is no stable chrome analogue to $\ce{FeO}$ on the phase diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the one stable oxide that will form under normal atmospheric conditions is $\ce{Cr2O3}$. However, note that it is easy to find mixed metal-oxygen compounds with high mutual solubility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in the thermodynamics of the $\ce{Cr-O}$ system, you could start with &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0364591695000249&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. Kowalski and P.J. Spencer, Calphad &lt;em&gt;19(3) 229-243&lt;/em&gt; (&lt;strong&gt;1995&lt;/strong&gt;)&lt;/a&gt;, which covers the $\ce{Cr-O, Fe-O}$ and $\ce{Ni-O}$ systems&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="17368" LastEditDate="2017-03-25T07:48:27.407" LastActivityDate="2017-03-25T07:48:27.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28449" PostTypeId="1" CreationDate="2015-04-09T13:38:29.163" Score="2" ViewCount="2442" Body="&lt;p&gt;When water boils into water vapor, its temperature (average kinetic energy) does not change because the inputed energy is used to break its bonds. But how does that work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in a liquid, aren’t intermolecular bonds broken and reformed constantly anyway? Why does a phase change require a time interval for breaking bonds?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastActivityDate="2016-09-17T12:46:58.620" Title="Bonding and Phase Changes?" Tags="&lt;phase&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28450" PostTypeId="1" AcceptedAnswerId="28660" CreationDate="2015-04-09T13:57:33.470" Score="3" ViewCount="2478" Body="&lt;p&gt;Very short question but maybe not a short answer !   As the probability of presence of electrons in a specific spatial position is defined as square of the wave function, I am wondering if there is a technique for molecular volume calculation based on the wavefunction.&#xA;Is there such a technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:  I could run the volume keyword in Gaussian.  This is the output for methane , but I can not find the volume in it !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/si2CE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="7188" LastEditDate="2015-04-12T06:04:38.113" LastActivityDate="2015-04-16T19:26:05.580" Title="How to calculate molecular volume from the wavefunction?" Tags="&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28451" PostTypeId="2" ParentId="28429" CreationDate="2015-04-09T14:11:17.453" Score="4" Body="&lt;p&gt;This might be an oversimplification, but my reasoning for assuming that it's first order would be because the plot of the log of the change in absorbance versus time gave a straight line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log of the concentration vs time will normally indicate a first order equation (given a straight line) but using the Beer-Lambert law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = ecl (where A is absorbance, e is the extinction coefficient, c is the concentration and l is the path length)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and assuming that the path length (l) is 1, you can see that the absorbance is proportional to the concentration. So plotting the graph of absorbance is analogous to plotting the graph of concentration, in that you should get a similar straight line and the result for the reaction constant k.&lt;/p&gt;&#xA;" OwnerUserId="15435" LastActivityDate="2015-04-09T14:11:17.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28452" PostTypeId="2" ParentId="28449" CreationDate="2015-04-09T14:12:30.820" Score="4" Body="&lt;p&gt;Temperature involves kinetic energy; however, bonds and intermolecular forces (IMF's) involve potential energy.  The potential energy of the system increases when water vaporizes, but kinetic energy does not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a liquid, for each IMF broken, another is made, resulting in a net change of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A phase change requires a time interval because it takes time to add energy to the system.      &lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="21964" LastEditDate="2016-09-17T01:42:48.133" LastActivityDate="2016-09-17T01:42:48.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28453" PostTypeId="2" ParentId="28446" CreationDate="2015-04-09T14:41:28.753" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methanation&quot; rel=&quot;noreferrer&quot;&gt;Methanation&lt;/a&gt; is a process whereby synthesis gas is converted to methane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO + 3H2 -&amp;gt; CH4 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methanol, like nearly all organic molecules, can be &lt;a href=&quot;http://www.netl.doe.gov/research/coal/energy-systems/gasification/gasifipedia/gasification-chemistry&quot; rel=&quot;noreferrer&quot;&gt;gasified&lt;/a&gt; to form synthesis gas at elevated temperatures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3OH -&amp;gt; CO + 2 H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: the gasification reaction is actually a complex balance of many types of reactions.  Often times mixtures of $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, and $\ce{CO2}$ are obtained.  The reaction above is the ideal case where complete conversion is achieved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a process for converting methanol to methane could proceed via gasification followed by methanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since methanol is made commercially from methane by essentially the reverse process, this route is unlikely to be economically viable!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T14:41:28.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28454" PostTypeId="1" AcceptedAnswerId="28459" CreationDate="2015-04-09T14:51:03.043" Score="3" ViewCount="1547" Body="&lt;p&gt;What are the main fragmentation pathways and products of a phenyl radical cation in mass spectrometry ? I notice sometimes that there's a m/z = 51 peak after the m/z = 77 for phenyl radical cation; I assume due to loss of acetylene.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-09T18:21:14.313" Title="Fragmentation of phenyl radical cations in mass spectrometry" Tags="&lt;mass-spectrometry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28455" PostTypeId="2" ParentId="28429" CreationDate="2015-04-09T14:52:19.407" Score="5" Body="&lt;p&gt;As you have explained, the reaction is a hydrolysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Co(en)2Cl2]+ + H2O -&amp;gt; [Co(en)2(H2O)Cl]^2+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you may expect a second-order rate equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_1 \cdot \left[\ce{[Co(en)2Cl2]+}\right] \cdot \left[\ce{H2O}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you are using a dilute aqueous solution. Compared to the concentration of the complex, the concentration of water is very large. Thus, the concentration of water remains approximately constant and it can be included in the rate constant, obtaining a &lt;strong&gt;pseudo-first-order&lt;/strong&gt; rate equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = -\frac{\mathrm{d}c}{\mathrm{d}t} = k_2 \cdot \left[\ce{[Co(en)2Cl2]+}\right]$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-04-09T19:37:06.247" LastActivityDate="2015-04-09T19:37:06.247" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28456" PostTypeId="2" ParentId="28443" CreationDate="2015-04-09T14:56:18.523" Score="4" Body="&lt;p&gt;The usual approach is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the equilibrium geometry (within the Born-Oppenheimer app.).&lt;/li&gt;&#xA;&lt;li&gt;Expand the energy in Taylor series till second order term. The first order will be 0 because of the minimum energy condition.&lt;/li&gt;&#xA;&lt;li&gt;Include this expression in the Hamiltonian.&lt;/li&gt;&#xA;&lt;li&gt;Generate a coordinate transformation to get normal coordinates.&lt;/li&gt;&#xA;&lt;li&gt;Relate second order derivative of the potential energy with the force constant of Hook law (Using second order term we can identify it with a harmonic oscillator). Then get the harmonic frequencies.&lt;/li&gt;&#xA;&lt;li&gt;The software employed should give also the intensity for each frequency.&lt;/li&gt;&#xA;&lt;li&gt;Plot intensity vs. frequency/energy/wave length. You'll get only &quot;points&quot;. To transform them to a continuous function you can use a Lorentzian like function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Details can be found in most computational quantum chemistry books.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2015-04-10T10:10:23.960" LastActivityDate="2015-04-10T10:10:23.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28457" PostTypeId="1" CreationDate="2015-04-09T14:57:21.820" Score="3" ViewCount="112" Body="&lt;p&gt;How can we detect polyethylene from a distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method should be mobile as I am thinking of employing it a mobile robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion or external link would be welcomed.&lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="13184" LastEditDate="2015-04-09T16:34:45.730" LastActivityDate="2015-04-09T16:34:45.730" Title="Detect Polyethylene" Tags="&lt;analytical-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28458" PostTypeId="2" ParentId="28454" CreationDate="2015-04-09T15:12:03.777" Score="2" Body="&lt;p&gt;What about $\ce{phenyl^{*+} -&amp;gt; Dewar\ benzene^{*+} -&amp;gt; cyclobutadiene^{+*} + C2H2}$? &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-09T15:41:52.350" LastActivityDate="2015-04-09T15:41:52.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28459" PostTypeId="2" ParentId="28454" CreationDate="2015-04-09T15:52:03.537" Score="2" Body="&lt;p&gt;Without more experimental details, it isn't 100% clear that the small peak is exclusively derived from the phenyl radical cation.  It also isn't clear that observed the &quot;phenyl&quot; radical cation is actually cyclic!  It could be another isomer of $\ce{C6H5^{+.}}$.  With benzilic acid as the ion source, it is possible that the small peak contains an oxygen atom, although searching for possible formulas with appropriate &quot;exact&quot; masses makes me think this isn't true (see below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://metlin.scripps.edu/metabo_info.php?molid=1297&quot; rel=&quot;nofollow&quot;&gt;fragmentation spectrum of benzoic acid&lt;/a&gt; shows similar peaks.  The &quot;exact&quot; mass of the observed ion 51.0237 Daltons, &lt;a href=&quot;http://www.chemcalc.org/mf_finder/mfFinder_em_new&quot; rel=&quot;nofollow&quot;&gt;within a milliDalton&lt;/a&gt; of the mass of $\ce{C4H3+}$.   The negative mode fragmentation spectrum of benzoic acid shows a corresponding ion at 49.0084, corresponding to $\ce{C4H1-}$ within a mDa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the existence of something like $\ce{C4H1-}$ in negative mode supports Klaus's answer of cyclic structure because it's hard to see how else such an &quot;unsaturated&quot; cluster of carbon atoms can form.  Perhaps a &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedrane&quot; rel=&quot;nofollow&quot;&gt;tetrahedrane&lt;/a&gt;-type structure is even possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to get more information on this ion's structure would be to fragment a variety of partially deuterated benzene rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;: I found &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/jp990624v&quot; rel=&quot;nofollow&quot;&gt;a paper&lt;/a&gt; talking about fragmentation of fluorobenzene, and they argue the &quot;phenyl&quot; cation fragment derived from that molecule may not be cyclic after all: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In combining these approaches, several mechanistic variants can&#xA;  be excluded, while a new, hitherto not considered pathway&#xA;  involving the formation of acyclic $\ce{C6H5+}$ is proposed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know if the same thing happens during the fragmentation of benzilic acid, but if so, then I wonder if the observed peaks could be protonated &lt;strong&gt;butadiyne&lt;/strong&gt; (positive mode) or deprotonated butadiyne (negative mode).&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-04-09T18:21:14.313" LastActivityDate="2015-04-09T18:21:14.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28460" PostTypeId="2" ParentId="28457" CreationDate="2015-04-09T16:27:25.230" Score="3" Body="&lt;p&gt;There is a growing literature on using &lt;a href=&quot;http://en.wikipedia.org/wiki/Near-infrared_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;near-infrared&lt;/a&gt; (NIR) &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperspectral_imaging&quot; rel=&quot;nofollow&quot;&gt;hyperspectral imaging&lt;/a&gt; (HSI) to detect a variety of materials, including plastics.  &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0921344912000080&quot; rel=&quot;nofollow&quot;&gt;Here's one example&lt;/a&gt; that focuses on resolving polyethylene from the very similar polypropylene using NIR HSI.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T16:27:25.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28461" PostTypeId="2" ParentId="28439" CreationDate="2015-04-09T16:47:09.213" Score="0" Body="&lt;p&gt;If you cannot find a phase diagram like the one above for methyl jasmonate, you can use the principle that a mixture of two dissimilar substances is likely to have a lower boiling point. You might try water plus methyl jasmonate, for example - it should boil slightly below 100C.  Purifying it this way is called steam distillation. Also, another way to lower the boiling point is to lower the ambient pressure - i.e. do a vacuum distillation. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-09T16:47:09.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28463" PostTypeId="2" ParentId="28440" CreationDate="2015-04-09T16:53:26.430" Score="1" Body="&lt;p&gt;Unfortunatly I don't think there is an easy colorimetric technique available.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are reagents that are strongly colored and oxygen-sensitive.  One exampleis reduced &lt;a href=&quot;http://en.wikipedia.org/wiki/Viologen&quot; rel=&quot;nofollow&quot;&gt;methylviologen&lt;/a&gt; radical.  That species can be prepared by mixing methylviologen with sodium dithionite in aqueous solution.  The one-electron reduced species is intensely blue.  It is rapidly oxidized by oxygen, leading to a loss in color.  In theory, you could use the degree of this color loss to assay DO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem in real life with applying this reagent to DO determination is that oxygen in the air will also react, very quickly.  So you would have to put your seawater sample under an inert atmosphere (and also prepare the reduced MV radical solution under an inert atmosphere).  That is very inconvenient, and could create problems where your seawater sample outgases oxygen into the anaerobic environment before or during measurement.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T16:53:26.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28464" PostTypeId="2" ParentId="28444" CreationDate="2015-04-09T18:01:48.307" Score="6" Body="&lt;p&gt;The &quot;dep&quot; probably stands for &quot;depletion&quot;: read the paper you linked to for more context. Often times, subscripts on rate constants are used to refer to which reaction in a system of multiple reactions. You'll see $k_1$ and $k_2$ for rate constants of &quot;reaction 1&quot; and &quot;reaction 2&quot;, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect in the paper, they were measuring the depletion of a substrate, and not what the substrate was converted to.  So they wouldn't know if they were measuring the rate of a single reaction $\ce{A-&amp;gt;B}$, or the &lt;em&gt;combined rates&lt;/em&gt; of multiple reactions such as $\ce{A-&amp;gt;B}$ and $\ce{A-&amp;gt;C}$.  Thus (probably) the desire to indicate $k_{depletion}$ instead of just $k$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T18:01:48.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28465" PostTypeId="2" ParentId="28426" CreationDate="2015-04-09T18:52:21.887" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.eplastics.com/Plastic/Plastics_Library/Chemical-Resistance-of-Plexiglass-Acrylic&quot; rel=&quot;noreferrer&quot;&gt;chemical resistance of plexiglass acrylic&lt;/a&gt; alcohol is not as aggressive as you mention with plexiglass acrylic.&#xA;But, we have to signal that acrylics is a big family of polymers (thermoplastics or thermosets) derived from acrylic acid, methacrylic acid or other related compounds. Plexiglass is only a member of this family.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way, in response to your question why  alcohol might cause severe crazing  and cracking?&#xA;It's simply because alcohol, as a good solvent, has the ability to attack the surface of acrylics and dissolve the polymer chains.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-09T18:52:21.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28467" PostTypeId="1" CreationDate="2015-04-09T19:26:08.943" Score="6" ViewCount="1211" Body="&lt;p&gt;Can someone explain (and preferably draw) the intermediates of this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://scontent-ord.xx.fbcdn.net/hphotos-xat1/v/t1.0-9/11138154_10153755063967785_1986665322492311719_n.jpg?oh=e829103ee0a9d033f09490900711d24e&amp;amp;oe=55B22FE2&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that it is a Claisen Reaction, and likely a condensation due to the heat, but I am not sure why there isn't an ester in the final product.&lt;/p&gt;&#xA;" OwnerUserId="2613" LastActivityDate="2015-04-10T02:59:28.367" Title="Claisen Condensation to Ketone" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28468" PostTypeId="1" AcceptedAnswerId="28470" CreationDate="2015-04-09T19:43:33.437" Score="2" ViewCount="2647" Body="&lt;p&gt;From what I understand, it is the functional group attached to a structure that determines the odor it will give off, for the most part. Which explains why two similar molecules, except for their functional groups, would smell different. So, would that make it so that two different molecules with the same functional groups would smell the same? &lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="4356" LastEditDate="2015-04-09T20:16:09.517" LastActivityDate="2015-04-09T20:43:21.137" Title="Why might two different chemical structures smell the same?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28470" PostTypeId="2" ParentId="28468" CreationDate="2015-04-09T20:43:21.137" Score="1" Body="&lt;p&gt;The molecular determinants of odor perception is a disputed area in fragrance biochemistry.  There are two basic theories: the &quot;shape&quot; hypothesis and the &quot;vibrations&quot; hypothesis.  Luca Turin is the foremost exponent of the more controversial and less widespread &quot;vibrations&quot; hypothesis, but he does a good job of explaining both theories in &lt;a href=&quot;https://www.ted.com/talks/luca_turin_on_the_science_of_scent?language=en&quot; rel=&quot;nofollow&quot;&gt;his TED talk&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the vibrational theory, since different functional groups have different vibrational modes, they should smell quite different.  An example from Luca's talk is a complex alcohol vs. the corresponding thiol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the shape theory, molecules which have similar but not identical functional groups should smell similarly, unless the protein receptors in our nose that sense odors have very exquisite shape sensitivity, enough to perceive the very slight difference in the shape of thiol vs. an alcohol in a large otherwise identical molecule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why might two different chemical structures smell the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to one theory, they would smell the same if they had similar vibrational (i.e. infrared) spectra.  In 2011, &lt;a href=&quot;http://www.pnas.org/content/108/9/3797.abstract&quot; rel=&quot;nofollow&quot;&gt;PNAS published&lt;/a&gt; is a very intriguing result that showed this was true for fruit flies in the case of some odorants.  They changed the vibrations by isotopic substitutions, a method that should leave essentially unchanged molecule's shape and its chemical reactivity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, would that make it so that two different molecules with the same functional groups would smell the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In my mind this is true if you believe the vibrational theory as well.  But it is harder to understand why it is true if you believe the shape theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the real answer may be a combination of both theories.  I'm a tepid supporter of the vibration theory myself, but I would be very interested in a chem.se answer from another expert who is less impressed with Dr. Turin's theories than me.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-09T20:43:21.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28473" PostTypeId="1" AcceptedAnswerId="28476" CreationDate="2015-04-09T21:15:28.180" Score="2" ViewCount="124" Body="&lt;p&gt;Do we know what is the wavelength of the color of &lt;a href=&quot;http://en.wikipedia.org/wiki/Melanin#Pheomelanin&quot; rel=&quot;nofollow noreferrer&quot;&gt;pheomelanin&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="15477" LastEditorUserId="16683" LastEditDate="2017-07-13T14:55:20.273" LastActivityDate="2017-07-13T14:56:33.743" Title="Color of pheomelanin?" Tags="&lt;physical-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28475" PostTypeId="1" CreationDate="2015-04-09T21:51:12.330" Score="2" ViewCount="54" Body="&lt;p&gt;What determines the uniformity? Just a fun fact I am wondering about, because I saw cyanide solutions used in electroplating.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="9961" LastEditDate="2015-04-10T11:36:41.797" LastActivityDate="2015-04-10T13:03:09.010" Title="Why is the deposition of silver more uniform from [Ag(Cl)₂]⁻ (aq) than from Ag⁺?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28476" PostTypeId="2" ParentId="28473" CreationDate="2015-04-09T21:51:39.477" Score="4" Body="&lt;p&gt;Pheomelanin has a very broad absorption spectrum in the visible, with monotonically increasing absorption as the wavelength decreases until a maximum is reached at 300 nm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that this would correspond to a dark red/brown colour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://doi.org/10.1039/C1CP20131J&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt; &lt;strong&gt;2011,&lt;/strong&gt; &lt;em&gt;13&lt;/em&gt; (20), 9119–9127&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="16683" LastEditDate="2017-07-13T14:56:33.743" LastActivityDate="2017-07-13T14:56:33.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28477" PostTypeId="2" ParentId="28418" CreationDate="2015-04-09T21:54:45.810" Score="1" Body="&lt;p&gt;The heat transfer coefficient and the thermal conductivity can often be treated as constants in practical heat transfer problems.  However, the heat transfer coefficient can change if the exterior convective flow conditions change.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-04-09T21:54:45.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28478" PostTypeId="2" ParentId="28467" CreationDate="2015-04-09T22:09:25.633" Score="6" Body="&lt;p&gt;Step 1 is a Claisen ester condensation with loss of ethoxide from the ester acting as the carbonyl part. The product is the $\beta$-ketoester (&lt;strong&gt;2&lt;/strong&gt;).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9dlNR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon acidification and heating, the ester is cleaved, the resulting $\beta$-ketoacid decarboxylates to yield the symmetrical ketone.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-09T22:09:25.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28479" PostTypeId="1" CreationDate="2015-04-09T22:26:53.730" Score="-1" ViewCount="71" Body="&lt;p&gt;At first this question might get the following response, &quot;What are you talking about?&quot;. But really all I am kind of asking is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What is the probability of us humans being disorderly?&lt;/strong&gt; Or in other&#xA;  words, I am kind of talking about the &lt;em&gt;Second Law of&#xA;  Thermodynamics&lt;/em&gt;. But just to be clear, in how many ways can we humans&#xA;  be &quot;&lt;strong&gt;ordered&lt;/strong&gt;&quot; or &quot;&lt;strong&gt;arranged&lt;/strong&gt;&quot; to still be &lt;em&gt;human&lt;/em&gt; and &lt;em&gt;living&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;em&gt;different&lt;/em&gt; question than what you usually see on Chemistry SE, but I just want to know &lt;em&gt;what is the probability of being human&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorDisplayName="user467" LastEditDate="2015-04-09T23:49:37.653" LastActivityDate="2015-04-09T23:49:37.653" Title="What is the probability of us being &quot;orderly&quot;?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28480" PostTypeId="2" ParentId="28479" CreationDate="2015-04-09T23:44:51.523" Score="3" Body="&lt;p&gt;I'm not certain that this question is appropriate for this site. However, there is an awfully large range of configurations that are human. You can have physical defects, genetic anomalies, mutations, and you are still human. That's not even mentioning all the more typical human variations like height. But strictly mathematically, there are probably many more non-viable configurations of your cells than viable, so you might say that the probability is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT! Your invocation of the Second Law seems wrong to me. This only applies in a &lt;em&gt;closed system&lt;/em&gt;, and that does not apply to the creation of a new human. &lt;em&gt;Lots&lt;/em&gt; of energy is expended in that creation, which is not at all in defiance of the Second Law because we are not in a closed system (e.g. we ingest food and convert to energy).&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-04-09T23:44:51.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28481" PostTypeId="2" ParentId="28429" CreationDate="2015-04-10T00:37:34.383" Score="3" Body="&lt;p&gt;You were told you are wrong, but &lt;strong&gt;you are right&lt;/strong&gt;. This is clearly a first order reaction, just because of the reasoning you exposed. But I will try to present you a simple mind exercise to realize why in an heuristic way (your argument is the more accurate one in my opinion, but many chemist need an inaccurate representation to feel that the concept is understood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bet that it is just a misunderstand of reaction order concept. In the case of elemental reactions in gas phase, it will be (just for statistical reasons) the sum of reactive molecules involved. BUT, here we are talking about a reaction in aq. phase, where one of the reactives is not another solute. So, thinking this hydrolysis just as molecule $A$ plus &lt;em&gt;only one&lt;/em&gt; water molecule is &lt;em&gt;not&lt;/em&gt; accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is making a field around each molecule of your reactive, so represent (in your head) it like your molecule surrounded a electric field is more accurate that only thinking in two molecules. Thinking so, the first order for this reaction arises naturally.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastEditorUserId="14527" LastEditDate="2015-04-10T10:14:36.137" LastActivityDate="2015-04-10T10:14:36.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28482" PostTypeId="2" ParentId="28467" CreationDate="2015-04-10T02:59:28.367" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone explain (and preferably draw) the intermediates of this reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ethoxide - a base comparable in strength to hydroxide ion - abstracts a proton from the alpha position of the ester. This forms an &lt;em&gt;enolate&lt;/em&gt; ion. Note that while protons alpha to a carbonyl center are mildly acidic - they are still not very acidic. Definitely not acidic enough for ethoxide to completely deprotonate. So we have &lt;em&gt;both&lt;/em&gt; enolate anion and untouched, still-protonated ester. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enolate anions are nucleophilic at both their oxygens and carbons. However, we mostly consider the reactions of enolates at their negatively charged carbons rather than their oxygens as we are interested in forming C-C bonds (as opposed to O-C bonds). Also, forming O-C bonds isn't particularly favorable versus C-C bonds - you'll have to study a bit of hard-soft acid-base theory to understand that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we have enolate anions in the system and untouched esters, which have electrophilic carbonyl carbons, we can make carbon-carbon bonds. The enolate anions can attack the carbonyl carbons, forming a tetrahedral intermediate. This intermediate collapses and ejects ethoxide ion. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-04-10T02:59:28.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28483" PostTypeId="1" CreationDate="2015-04-10T03:25:34.163" Score="7" ViewCount="1796" Body="&lt;p&gt;Using the equation $$\ce{CaSO4 (s) &amp;lt;=&amp;gt; Ca^2+ (aq) + SO4^2- (aq)},$$ and $\Delta H/\Delta S$ values, calculate $\Delta G$ at $50~^\circ\mathrm{C}$ when the solution is saturated with $\ce{Ca^2+}$ and $\ce{SO4^2-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;&#xA;$$\Delta G = \Delta G^\circ + RT \ln Q$$&#xA;I calculated $\Delta G^\circ$ at $50\ \mathrm{^\circ C}$, But I am confused as to how we know what $Q$ is if there is no reactant then does $Q$ not become $\ce{[Ca^2+][SO4^2- ]/0}$, but this is impossible.&lt;/p&gt;&#xA;" OwnerUserId="15480" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T13:37:52.343" LastActivityDate="2015-05-15T00:34:23.630" Title="Gibbs Free Energy of Solutes and Solutions" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;solutions&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28485" PostTypeId="1" AcceptedAnswerId="28491" CreationDate="2015-04-10T05:18:13.610" Score="10" ViewCount="2594" Body="&lt;p&gt;I want to make a plasma speaker. I'm worried about the amount of ozone being produced by the ionization of the air and it possibly not being a safe amount. Even if it doesn't produce a dangerous amount, I don't like the smell of ozone. I'd like this all to be continuous. I am very unfamiliar with the chemistry here but I have read that UV light, around 250nm, can cause $\ce{O_3 \to O_2 + O}$ and also $\ce{O_2 \to 2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also without any help it seems that $\ce{2O \to O_2}$ and also $\ce{O + O_2 \to O_3}$. So it seems that if I shined UV light of the right frequency on the spark I'd make a lot of $\ce{O}$ and perhaps that would cause more $\ce{O_2}$ than $\ce{O_3}$ but I'm not really sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also thinking that it might oxidize some sort of metal filter really well and I could just keep replacing those but I wasn't sure. I could also heat the filter slightly to speed up the reaction if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the UV light idea doesn't work out very well I still want a way to do this. Is the amount of ozone produced by a continuous electrical current though air harmful? Is there a way of getting rid of the ozone to make some odorless non-harmful chemical?&lt;/p&gt;&#xA;" OwnerUserId="15483" LastEditorUserId="418" LastEditDate="2015-04-10T07:36:04.677" LastActivityDate="2015-11-01T00:46:44.847" Title="How much ozone is produced by ionization of air and how turn ozone into oxygen" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;photochemistry&gt;&lt;toxicity&gt;&lt;ozone&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28489" PostTypeId="1" CreationDate="2015-04-10T07:00:11.753" Score="7" ViewCount="647" Body="&lt;p&gt;So my textbook tells me that using a malonic ester to make a quaternary alpha carbon is impossible (I get that this might be hyperbole). However, it doesn't explain why and I don't understand why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought was that after the first addition the tertiary compound would be too bulky to act as a nucleophile. However in the next section, an ethyl acetate is treated with LDA and acts as a nucleophlile 3 times to create a quaternary alpha carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also considered that you'd be making a tertiary carbon anion to make a second addition to a malonic ester's alpha carbon. However you also make a tertiary carbon anion using LDA on ethyl acetate and then alkyl halides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've added a screenshot of my textbook. Skip to the part in parenthesis if you want to read it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rdksC.png&quot; alt=&quot;Screenshot of the paragraph. Skip the part in parenthesis.&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorUserId="10185" LastEditDate="2015-09-18T18:14:52.483" LastActivityDate="2015-09-18T18:14:52.483" Title="Why can you not make a compound with a quaternary alpha carbon using malonic ester?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;esters&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28490" PostTypeId="2" ParentId="28485" CreationDate="2015-04-10T07:31:07.487" Score="7" Body="&lt;p&gt;I would say it would be much easier to try to keep the area oxygen free. Atomic oxygen (i.e. $\ce{O}$) is a highly reactive radical. This would react either with another $\ce{O}$, or an $\ce{O_2}$. It is hard to avoid this unless you can get rid of the $\ce{O_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what kind of a geometry you are thinking of, but purging the volume with $\ce{N_2}$ gas might be an easy way to achieve this.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="7475" LastEditDate="2015-11-01T00:46:44.847" LastActivityDate="2015-11-01T00:46:44.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28491" PostTypeId="2" ParentId="28485" CreationDate="2015-04-10T07:33:34.843" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Even if the UV light idea doesn't work out very well I still want a way to do this&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With some photochemical background, I suggest to forget about running such a UV lamp&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Remember that your plasma will be formed &lt;strong&gt;in air&lt;/strong&gt;. For every molecule of ozone formed, there are much more harm- and odourless $\ce{O2}$ molecules around. If you cleave $\ce{O3}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;\[\ce{O3 + O2-&gt;[\mathrm{h}\nu][\lambda\ \lt\ \mathrm{300\ nm}] [O2 + O* + O2] -&gt; O2 + O3}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the resulting oxygen atom will be captured by the next $\ce{O2}$ to form ozone again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall process will not reduce the number of ozone molecules, unless $\ce{O*}$ reacts with $\ce{N2}$, which might lead to some other unpleasant species ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these processes are &lt;strong&gt;fortunately&lt;/strong&gt; effective in the ozone to protect us from short wave UV radiation, which brings us to the next aspect of the setup.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;In order to photolyze ozone, you'll neeed a light source that emits UV at $\lambda\lt300\,\mathrm{nm}$. It's not that these light sources would not be available, but this UV range is &lt;strong&gt;not&lt;/strong&gt; that of a tanning booth lamp! Remember that the ozone layer protects our DNA from running havoc! The least you can expect from short-wave irradtion is the sun burn of your life! Consequently, you will have to shield the whole setup properly!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In summary, I absolutely like the idea of building a plasma speaker yourself and there even seem to be some DIY kits around, but I suggest to forget about the UV part, which can be much more dangerous than the ozone formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Burak Ulgut&lt;/strong&gt; brought up a very interesting point. Back in the days, plasma speakers using other gases (or gas mixtures) were tried, and even went commerical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.plasmatweeter.de/plasmatronic.htm&quot;&gt;site in German&lt;/a&gt; describes a &lt;strong&gt;helium&lt;/strong&gt; driven plasma speaker system :)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-10T07:49:28.817" LastActivityDate="2015-04-10T07:49:28.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28492" PostTypeId="2" ParentId="28489" CreationDate="2015-04-10T08:09:44.243" Score="8" Body="&lt;p&gt;Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at this marvel by &lt;em&gt;R. P. Mariella&lt;/em&gt; and &lt;em&gt;R. Raube&lt;/em&gt;, published in &lt;em&gt;Org. Synth.&lt;/em&gt;, &lt;strong&gt;1953&lt;/strong&gt;, &lt;em&gt;33&lt;/em&gt;, &lt;em&gt;23&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kC0JZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide &lt;em&gt;in situ&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{2Na + 2ROH -&gt; 2NaOR + H2}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr/&gt;&#xA;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example given above, no further &lt;strong&gt;direct&lt;/strong&gt; alkylation at &lt;strong&gt;that&lt;/strong&gt; position is possible, and this is exactly, what &lt;strong&gt;ron&lt;/strong&gt; pointed out in his&#xA;comment to the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He also suggested the viable solution, which I had outlined in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28467/claisen-condensation-to-ketone&quot;&gt;another question&lt;/a&gt; on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great &lt;strong&gt;synthons&lt;/strong&gt; to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using &lt;strong&gt;two&lt;/strong&gt; bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a &lt;strong&gt;strong&lt;/strong&gt; base like $\ce{LDA}$ is used. (&lt;em&gt;Exactly this is outlined in the textbook for the alkylation of an ester&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-04-11T07:04:53.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28493" PostTypeId="2" ParentId="28485" CreationDate="2015-04-10T08:19:19.743" Score="0" Body="&lt;p&gt;You were wondering about a metal filter. That's a viable idea, but not all metals are created equal. Iron indeed would require frequent replacements, it rusts rapidly when in contact with ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is a fairly noble metal (oxidizes slowly), still reasonably cheap, and it works as a catalyst to break down ozone. The oxygen radicals on the copper surface often recombine to form O2, instead of forming Cu2O/CuO. (Copper(I) oxide and copper (II) oxide). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more copper-oxygen reaction products that can form. The initial reaction forms CuO3 (basically the ozon molecule sticking to one copper atom) and CuO4 (when ozon sticks to an existing CuO molecule). Another observed reaction product is Cu2O3. With all those combinations possible, a lot of pathways for chemical reactions open up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison, a free Ozone molecule can't simply &quot;lose&quot; one of its three oxygen atoms. Creating a free oxygen atom costs energy (that's why ozone is stable), so you need two ozone molecules to react with each other. That in turn means the decomposition rate is quadratic in the ozone concentration, not linear. So with low ozone concentrations, the reaction rate is very low.&lt;/p&gt;&#xA;" OwnerUserId="7323" LastEditorUserId="7323" LastEditDate="2015-04-10T12:18:33.377" LastActivityDate="2015-04-10T12:18:33.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28494" PostTypeId="2" ParentId="28443" CreationDate="2015-04-10T08:21:38.470" Score="4" Body="&lt;p&gt;You can build molecules and calculate vibrational frequencies with &lt;a href=&quot;http://molcalc.org&quot; rel=&quot;nofollow&quot;&gt;Molcalc&lt;/a&gt;.  Molcalc does have a molecular size limit, so for bigger molecules you must download some programs.  I have written about &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/search/label/vibration&quot; rel=&quot;nofollow&quot;&gt;computing vibrations with the GAMESS program&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="30" LastEditDate="2015-05-03T13:32:16.267" LastActivityDate="2015-05-03T13:32:16.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28495" PostTypeId="1" CreationDate="2015-04-10T08:44:44.313" Score="2" ViewCount="1224" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How many grams of $\ce{CuSO4 * 5 H2O}$ are needed to prepare a $20\%$ (w/w%) solution of $\ce{CuSO4}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15487" LastEditorUserId="10185" LastEditDate="2018-01-02T13:48:13.673" LastActivityDate="2018-01-02T13:48:13.673" Title="How to find the mass of a solute given a mass fraction?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ClosedDate="2018-01-02T19:44:06.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="28496" PostTypeId="2" ParentId="28387" CreationDate="2015-04-10T11:56:41.700" Score="23" Body="&lt;p&gt;There is a bit of history behind this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the website &lt;a href=&quot;http://www.vanderkrogt.net/elements/chemical_symbols.php&quot;&gt;Development of the chemical symbols and the Periodic Table&lt;/a&gt;, it is stated that the Swedish chemist Berzelius updated Dalton's previous method.  He established a protocol for the symbols, still in use today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the class which I call metalloids, I shall employ the initial letter only, even when this letter is common to the metalloid and some metal.&lt;/li&gt;&#xA;  &lt;li&gt;In the class of metals, I shall distinguish those that have the same initials with another metal, or a metalloid, by writing the first two letters of the word.&lt;/li&gt;&#xA;  &lt;li&gt;If the first two letters be common to two metals, I shall, in that case, add to the initial letter the first consonant which they have not in common:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;for example, S = sulphur, Si = silicium, Sb = stibium (antimony), Sn =&#xA;  stannum (tin), C = carbonicum, Co = cobaltum (cobalt), Cu = cuprum&#xA;  (copper), O = oxygen, Os = osmium, &amp;amp;c&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The basis of the symbols are the Latin word for the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A notes as what Berzellius was refering to when he used the term 'metalloid', from the Wikipedia page &lt;a href=&quot;http://en.wikipedia.org/wiki/Origin_and_use_of_the_term_metalloid&quot;&gt;Origin and use of the term metalloid&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1811, Berzelius referred to nonmetallic elements as metalloids, in reference to their ability to form oxyanions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="418" LastEditDate="2015-04-10T12:05:33.120" LastActivityDate="2015-04-10T12:05:33.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28497" PostTypeId="2" ParentId="28495" CreationDate="2015-04-10T12:03:13.350" Score="3" Body="&lt;p&gt;In a $20\%$ solution of $\ce{CuSO4}$ we have $\pu{20 g}$ of $\ce{Cu}$ and $\pu{80 g}$ of $\ce{H2O}$. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in every $\pu{259.6 g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{159.6 g}$ $\ce{CuSO4}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for every $x~\pu{g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{20 g}$ $\ce{CuSO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \pu{31.87 g}$$&lt;/p&gt;&#xA;" OwnerUserId="15487" LastEditorUserId="41328" LastEditDate="2017-12-14T04:56:23.113" LastActivityDate="2017-12-14T04:56:23.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28498" PostTypeId="2" ParentId="28475" CreationDate="2015-04-10T12:03:32.170" Score="3" Body="&lt;p&gt;The uniformity of a deposition depends on more parameters, but in this case the difference is in electromigration effect. $\ce{Ag+}$ ions are atrracted to the electrode surface, but the complex $\ce{[Ag(Cl)2]-}$ is repulsed. Cyanides form  complexes as well: $\ce{[Ag(CN)2]-}$ or $\ce{[Ag(Cl)3]2-}$ and they are also  repulsed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver electrodepostion form $\ce{AgNO3}$ ($\ce{Ag+}$ in solution) is without additives nearly impossible as dendrites are created. &lt;a href=&quot;http://jes.ecsdl.org/content/136/12/3630.short&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fukunaka, Yamamoto, Kondo, Ionic Mass Transfer Associated with Dendrite Growth of Electrodeposited Silver in AgNO3 Solution, 1989 ECS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal deposition from a complex molecules is described in &lt;a href=&quot;http://download-v2.springer.com/static/pdf/792/art%253A10.1007%252FBF03214646.pdf?token2=exp=1428667646~acl=%2Fstatic%2Fpdf%2F792%2Fart%25253A10.1007%25252FBF03214646.pdf*~hmac=ccf7fe61d2a2f3c0f23901bea800b58510b102e53e8f19184252f0f01c766b60&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wilkinson, P. Understanding Gold Plating. Gold Bull 1986, 19, 75 – 81&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bKb2g.png&quot; alt=&quot;deposition form complex&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some chemicals, that are usually added to electrolyte, affect the deposition because they create a film as they are adsorbed at  inner Helmholz plane. (E.g. acidic Cu deposition with alcohol addition &lt;a href=&quot;https://www.google.cz/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CCUQFjAA&amp;amp;url=http%3A%2F%2Fwww.researchgate.net%2Fprofile%2FLiana_Muresan%2Fpublication%2F222671572_Kinetic_and_morphological_investigation_of_copper_electrodeposition_from_sulfate_electrolytes_in_the_presence_of_an_additive_based_on_ethoxyacetic_alcohol_and_triethyl-benzyl-ammonium_chloride%2Flinks%2F0912f50755620272bc000000.pdf&amp;amp;ei=QrsnVauhHImRsAHkm4ToDg&amp;amp;usg=AFQjCNE8M66i286S2pS67US7k32jV9bLVg&amp;amp;sig2=Rrbgwrp2EZxdIVVnv1AXFw&amp;amp;bvm=bv.90491159,d.ZWU&amp;amp;cad=rja&quot; rel=&quot;nofollow noreferrer&quot;&gt;Popescu et al, Materials Chemistry and Physics 72 (2001) 332–336&lt;/a&gt;.)   &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-04-10T13:03:09.010" LastActivityDate="2015-04-10T13:03:09.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28499" PostTypeId="1" AcceptedAnswerId="28504" CreationDate="2015-04-10T12:21:33.130" Score="4" ViewCount="6833" Body="&lt;p&gt;In the book that I'm reading it is mentioned higher molar mass results in stronger van der Waals force, but didn't continued why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why is that so? molar mass is actually dependable on the number of protons and neutrons, how can it be effective on the van der Waals force?&lt;/p&gt;&#xA;" OwnerUserId="15426" LastEditorUserId="7475" LastEditDate="2015-10-23T17:01:04.810" LastActivityDate="2015-10-23T17:01:04.810" Title="Relation between molar mass and van der Waals force" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28500" PostTypeId="1" AcceptedAnswerId="28523" CreationDate="2015-04-10T12:45:35.293" Score="4" ViewCount="284" Body="&lt;p&gt;How do I balance this using the half-reaction method: $$\ce{Zn(s) + HCl(aq) -&amp;gt; Zn^{2+} (aq) + H_2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first split it up in its ionic components:&#xA;$\ce{Zn(s) + H^+ + Cl^- -&amp;gt; Zn^{2+} (aq) + H_2 (g)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have first the oxidation-reaction: $\ce{Zn(s) -&amp;gt; Zn^{2+} (aq)}$. Balancing this with electrons gives $\ce{Zn(s) -&amp;gt; Zn^{2+} + 2e^-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction-reaction is $\ce{H^+ + Cl^- -&amp;gt; H_2 (g)}$. I need to balance this with $\ce{Cl}$? So that would get me $\ce{H^+ + Cl^- -&amp;gt; H_2 (g) + Cl}$. Then adding hydrogen to the left side with $\ce{H^+}$ gives $\ce{2H^+ + Cl^- -&amp;gt; H_2 (g) + Cl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I feel like I made a mistake somewhere. Any help please?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorDisplayName="user7951" LastEditDate="2015-04-11T07:19:41.200" LastActivityDate="2015-04-11T07:19:41.200" Title="Balancing Redox Equations - Half Reactions" Tags="&lt;redox&gt;&lt;aqueous-solution&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28502" PostTypeId="2" ParentId="28499" CreationDate="2015-04-10T13:11:34.793" Score="1" Body="&lt;p&gt;Somewhat related to: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28273/hydrogen-and-helium-ideal-gas-behavior-in-room-temperature/28280#28280&quot;&gt;Hydrogen and Helium ideal gas behavior in room temperature&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The induced dipole-dipole interactions operate through the polarizability of the molecules. Generally speaking heavier atoms will have more electrons and they will have some of those in the outer, higher radius shells. These higher radius elements will be more polarizable, hence stronger Van der Waals forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I can see where the statement comes from, I doubt it is a completely true statement. I would think one can have higher molar mass compounds that are less polarizable(though I cannot come up with an easy example right now)&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-04-10T13:11:34.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28503" PostTypeId="1" CreationDate="2015-04-10T13:17:54.553" Score="31" ViewCount="66718" Body="&lt;p&gt;So I came across this diagram,the other day:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r36te.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r36te.png&quot; alt=&quot;Bonds&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As  a chemistry student, I am well-versed with the dashed and wedged lines, but I was wondering what the wavy and the dotted line represent?&lt;/p&gt;&#xA;" OwnerUserId="15493" LastEditorUserId="16683" LastEditDate="2016-05-04T13:44:47.320" LastActivityDate="2020-07-25T17:13:18.857" Title="What are the meanings of dotted and wavy lines in structural formulas?" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="28504" PostTypeId="2" ParentId="28499" CreationDate="2015-04-10T13:21:46.897" Score="2" Body="&lt;p&gt;From &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Van_der_Waals_Forces&quot; rel=&quot;nofollow&quot;&gt;chemwiki.Ucdavis&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems totally legitimate. However, (as chemwiki subtly points out) the main factor in work here is the volume occupied by electrons. Polarizability increase is mainly due to a volume increase &lt;sup&gt;1&lt;/sup&gt;. So, all you need to take into consideration is whether a molecule is bigger than the other in terms of comparing the stronger London dispersion force.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Simply put, the ability of a molecule to get polarized is called polarizability. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizability&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; for more info.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastActivityDate="2015-04-10T13:21:46.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28505" PostTypeId="2" ParentId="28503" CreationDate="2015-04-10T13:23:41.840" Score="6" Body="&lt;p&gt;The wavy line usually represents either that that part of the molecule is unimportant or that the stereochemistry at that specific position is unimportant (or unknown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dotted line usually represents intermolecular interactions, like hydrogen bonds. Dashed lines refer mostly to partial bonds such as in the transition state.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-04-10T21:02:15.583" LastActivityDate="2015-04-10T21:02:15.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28506" PostTypeId="2" ParentId="28503" CreationDate="2015-04-10T13:24:32.620" Score="42" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I was wondering what the wavy and the dotted line represent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LEHOI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LEHOI.png&quot; alt=&quot;Different bonds&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A &lt;strong&gt;dashed&lt;/strong&gt; line indicates that the bond is extending &lt;strong&gt;behind the plane&lt;/strong&gt; of the drawing surface&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;bold-wedged&lt;/strong&gt; line indicates that the bond is &lt;strong&gt;protruding out&lt;/strong&gt; from the plane of the drawing surface&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;solid&lt;/strong&gt; line indicates that the bond exists &lt;strong&gt;in&lt;/strong&gt; the plane of the drawing surface.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;wavy&lt;/strong&gt; line indicates that the stereochemistry of the bond is &lt;strong&gt;unknown&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;strong&gt;dotted&lt;/strong&gt; line indicates that the bond is not a full bond, it is only a &lt;strong&gt;partial bond&lt;/strong&gt; as in a hydrogen bond or a partially formed or broken bond in a transition state. Also, as Mrigank Pawagi notes in the comments, the &lt;strong&gt;dotted&lt;/strong&gt; line can be used along with the &lt;strong&gt;solid&lt;/strong&gt; line to denote bonds that have partial double bond character due to resonance.&lt;/li&gt;&#xA;&lt;li&gt;An &lt;strong&gt;arrow&lt;/strong&gt; indicates a dative bond in which both electrons in the bond originate from one atom (not common)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-07-25T17:13:18.857" LastActivityDate="2020-07-25T17:13:18.857" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28507" PostTypeId="1" AcceptedAnswerId="28510" CreationDate="2015-04-10T13:29:33.050" Score="8" ViewCount="5546" Body="&lt;p&gt;Can all lattices be described as one of the fourteen Bravais lattices? &#xA;Is the hexagonal close packed structure also one of the fourteen Bravais lattices?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-04-10T15:03:18.953" Title="Are all lattices Bravais lattices?" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28509" PostTypeId="1" CreationDate="2015-04-10T04:34:59.203" Score="3" ViewCount="304" Body="&lt;p&gt;Based on the Wikipedia page about &lt;a href=&quot;http://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow noreferrer&quot;&gt;rust&lt;/a&gt;, there are red and green types of rust that occur on iron.  Red rust is the reaction with oxygen and water and the green with chlorine (in oxygen poor environments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the chemical reactions that forms these types of rust, but in the Wikipedia article, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Several forms of rust are distinguishable both visually and &lt;strong&gt;by spectroscopy&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(bolding mine). Visually, it'd be pretty obvious (one is red, one is green); spectroscopy-wise, I would imagine it to be a similar story - but what aspect of the types of rust cause this difference - hence, the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are the spectral reflectence features that distinguish between the red and green rust types based on what the iron reacts with (oxygen/water or chlorine), or is it based on the type of iron present (Fe&lt;sup&gt;2+&lt;/sup&gt; vs Fe&lt;sup&gt;3+&lt;/sup&gt;)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Full disclosure, I have a similar question on Open Data.SE &lt;a href=&quot;https://opendata.stackexchange.com/questions/4936/spectral-reflectance-data-for-iron-rust&quot;&gt;here&lt;/a&gt;, but are asking for an explanation in terms of spectral reflectivity.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:30.120" LastActivityDate="2015-06-30T13:07:13.607" Title="Spectral reflectance identification of types of iron rust" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28510" PostTypeId="2" ParentId="28507" CreationDate="2015-04-10T13:59:52.923" Score="9" Body="&lt;p&gt;From Ashcroft and Mermin's Solid State Physics: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fundamental concept in the description of any crystalline solid is that of the &lt;em&gt;Bravais lattice&lt;/em&gt;, which specifies the periodic array in which the repeated units of the crystal are arranged. The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, a real crystal structure is defined by the Bravais lattice, and the unit placed at each point on the Bravais lattice.  Now, it turns out that the hexagonal close packed structure (hcp) is not, in fact, a Bravais lattice.  Hcp is based on the simple hexagonal Bravais lattice, but with two atoms as the basis (one could also call it two interpenetrating simple hexagonal Bravais lattices).  The bottom line is the hcp is not itself a Bravais lattice (see further details in Ashcroft and Mermin). I'll skip any discussion of point groups and space groups, since those are just further refinements of what it means to combine the symmetry of a Bravais lattice with the symmetry of the repeating unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a hierarchy of symmetry - 7 crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups. For hcp, the point is that it can be represented as a simple hexagonal Bravais lattice with a two-atom unit, so having hcp as another &quot;Bravais lattice&quot; would totally change the definition of what a Bravais lattice is - the fundamental symmetry is simple hexagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general idea is to start with something you can repeat in a specified way to fill space.  For example, starting with a cube, it is pretty easy to see how to stack them up to fill space.  If, instead of being cubes they are stretched along one axis to get a rectangular solid, you can still fill space, but it will look different depending on what direction you are looking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we get the seven crystal systems: Cubic (with simple cubic, body centered, and face centered cubic Bravais lattices); Tetragonal, which is a cube stretched along one side (with simple tetragonal and centered tetragonal Bravais lattices); Orthorhombic, a unit with perpendicular faces but unequal lengths on each side (with 4 Bravais lattices); Monoclinic, which tilts an orthorhombic unit to make one angle non-90 degrees (2 Bravais lattices); Triclinic, which has no two faces perpendicular to each other (1 Bravais lattice); Trigonal, which is obtained by taking a cube and stretching it along a body diagonal (1 Bravais lattice), and Hexagonal, which has one Bravais lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get those 7 crystal systems and 14 Bravais lattices by being able to stack the appropriate units to fill space.  Now, while this concept is pretty fundamental, there are further symmetries imposed on the crystal by what you stack up.  Lets consider something fairly simple, simple cubic.  If you start with a big pile of white cubes, the resulting object you put together will look the same regardless of which direction you look at it, or how you rotate it around.  BUT - if you start with a pile of white cubes with one face painted red, you just changed things. To conform with the Bravais lattice, you would stack them so that all the red faces pointed in the same direction.  You would still get a big cube made of the smaller cubes.  But now it would look different depending on what direction you looked at it - you have reduced the symmetry of the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple hexagonal, one starts with a hexagonal right prism (hexagon base, some height perpendicular to the base).  Stacking those one on top of each other results in a simple hexagonal Bravais lattice.  Lets make the hexagonal prisms out of clear acrylic, but place a marble at the center of the prism.  Now when you have stacked them up you would see a simple hexagonal lattice of marbles, or heck, lets call them atoms.  Now, exactly where you put the marble in the clear prism doesn't matter - you would still see the exact same arrangement of the marbles in space.  OK, lets place the first model in the center of the base of the prism, and place a second marble halfway up one of the edges.  When you stack them up, you will now get an arrangement of marbles in space that looks like the hexagonal close packed structure.  But you stacked the prisms following the rules of the simple hexagonal Bravais lattice.  Placing that second marble in the Bravais prism generated a two atom basis.  It clearly is different from simple hexagonal, but the reason is not how you stacked the prisms, it is what the prism contains.  The underlying Bravais lattice does not change, but the symmetry of the resulting crystal is changed by the basis unit used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, use different colored marbles in the two positions, and it changes once again.  Throw three marbles in in different places in the prism and it changes yet again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the Bravais lattice occupies a fundamental level in understanding the crystal structure - it is the rules for stacking things up (and &lt;em&gt;only&lt;/em&gt; the rules for stacking things).  The objects you stack up may have their own properties, meaning that two things stacked using the same Bravais lattice may in the end look very different.  This combination of the lattice and the unit are what create the 230 space groups.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2015-04-10T15:03:18.953" LastActivityDate="2015-04-10T15:03:18.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28515" PostTypeId="1" CreationDate="2015-04-10T16:26:32.423" Score="1" ViewCount="5724" Body="&lt;p&gt;I drew a rough picture of the two cells, and marked each position with a distinct letter:&lt;img src=&quot;https://i.stack.imgur.com/jwCfH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole reason I am asking this question is that I am bit confused and unsure about some of the chemistry and physics concepts. It will be extremely helpful if you guys can tell me if what I did is correct and why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I think is correct:&lt;/strong&gt; to make this whole thing work.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ol&gt;&#xA;&lt;li&gt;Connect alligator clip &lt;strong&gt;x&lt;/strong&gt; to position &lt;strong&gt;A&lt;/strong&gt;, right on the zinc anode not the wire.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;And connect clip &lt;strong&gt;y&lt;/strong&gt; to &lt;strong&gt;B&lt;/strong&gt;, on the copper cathode plate.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From what I read, the electron is forced from Ag plate, anode, through the wire (with two alligator clips on each end), then reduces $Ag_{+}$ ions to $Ag$ metal on the spoon cathode. But from the picture, it looks more like all the electrons came from the zinc anode of the voltaic cell and all the electrons from the $Ag$ anode of the electrolytic cell go to $Cu$ cathode of the voltaic cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some physics aspect of this:&lt;/strong&gt; if &lt;strong&gt;x&lt;/strong&gt; clip was to be connected to position &lt;strong&gt;C&lt;/strong&gt; and &lt;strong&gt;y&lt;/strong&gt; clip to position &lt;strong&gt;D&lt;/strong&gt;, is it true that there will be no potential difference?(unless the wire itself has a high resistance)&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2015-04-11T05:11:23.273" Title="Electroplating a spoon (electrolytic cell) using a electrochemical voltaic cell" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28517" PostTypeId="2" ParentId="28489" CreationDate="2015-04-10T17:28:09.967" Score="5" Body="&lt;p&gt;Traditional malonic ester synthesis post-processing includes hot acidic hydrolysis, which should lead to substituted malonic acid. However, such acids are prone to decarboxylation, leading to tertiary carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you deviate from the tradition, there is no problem with getting quaternary  carbon (though general bulkiness still applies and may prevent some synthesises) &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-04-10T19:08:36.783" LastActivityDate="2015-04-10T19:08:36.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28518" PostTypeId="1" AcceptedAnswerId="31274" CreationDate="2015-04-10T17:34:11.690" Score="5" ViewCount="1911" Body="&lt;p&gt;Are rate constants for water exchange of transition metal aqua-complexes tabulated anywhere ? I haven't been able to find a resource which is anywhere near comprehensive.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="7448" LastEditDate="2015-04-10T17:35:18.093" LastActivityDate="2016-03-11T23:04:57.787" Title="Rate constants for water exchange" Tags="&lt;kinetics&gt;&lt;coordination-compounds&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28519" PostTypeId="2" ParentId="28515" CreationDate="2015-04-10T18:09:33.090" Score="1" Body="&lt;p&gt;Let's start with the last question: The points A and C have the same potential, because we  generally neglect the resistance of the wire. The same argument is valid for points B and D. But according to your scheme, you have a short circuit (as C and D are connected).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to electroplate the spoon, you want the silver to be oxidized at the silver electrode. So, silver electrode is the anode, and must be connected to pole (+) of the voltaic cell (point B):&#xA;$$\ce{Ag -&amp;gt; Ag+ + e- }$$&#xA;The spoon is the electrode where ion silver will be reduced. So, it's the cathode and must be connected to pole (-) of of the voltaic cell (point A). The electrons enter the cathode and reduced the silver ion according to the equation:&#xA;$$\ce{Ag+ + e- -&amp;gt; Ag}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you're right!&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-04-11T05:11:23.273" LastActivityDate="2015-04-11T05:11:23.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28520" PostTypeId="1" AcceptedAnswerId="28522" CreationDate="2015-04-10T18:34:17.187" Score="1" ViewCount="1038" Body="&lt;p&gt;I understand that $\ce{CdCl2}$ has a rhombohedral unit cell as shown in &lt;a href=&quot;https://books.google.co.uk/books?id=-EKCm5UQaqEC&amp;amp;pg=PA256&quot; rel=&quot;nofollow&quot;&gt;this book.&lt;/a&gt; However I often find it being referred to as a &quot;CCP analogue&quot; of $\ce{CdI2}$. Presumably it is the chlorides which form a cubic lattice; how would you draw the cubic unit cell ?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-10T19:59:42.373" Title="Unit cell of CdCl2" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28521" PostTypeId="1" CreationDate="2015-04-10T19:56:44.070" Score="2" ViewCount="2309" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the pH of CsBr?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;3.8&lt;/li&gt;&#xA;  &lt;li&gt;7&lt;/li&gt;&#xA;  &lt;li&gt;9&lt;/li&gt;&#xA;  &lt;li&gt;12&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cs+ + H2O &amp;lt;&amp;lt;=&amp;gt; CsOH + H+}$   (because $\ce{CsOH}$ is a strong base, the forward reaction should not occur, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br- + H2O &amp;lt;&amp;lt;=&amp;gt; HBr + OH-}$ (because $\ce{HBr}$ is an acid, the forward reaction should also not occur, right?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer should be $\mathrm{pH} = 7$.&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="5026" LastEditDate="2018-07-20T03:12:06.247" LastActivityDate="2018-07-20T03:12:06.247" Title="What is the pH of CsBr?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28522" PostTypeId="2" ParentId="28520" CreationDate="2015-04-10T19:59:42.373" Score="2" Body="&lt;p&gt;Cubic close packing is another (confusing) name for face-centered cubic.  So, to visualize these structures, you start with a hexagonal 2D layer of the chlorine or iodine.  Now start stacking those layers, ABAB... for hcp (iodine), or ABCABC... for fcc (chlorine).  Finally, stick the Cd in between alternate layers of the chlorine/iodine. The net result for CdCl2, is that the unit cell is indeed rhombohedral, since there is one Cd layer for every two Cl layers, so it is stretched in that direction.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-04-10T19:59:42.373" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28523" PostTypeId="2" ParentId="28500" CreationDate="2015-04-10T20:30:11.303" Score="3" Body="&lt;p&gt;You are right for the half reaction for oxidation is:&#xA;$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$&#xA;Now, you have to write properly the  half reaction for reduction:&#xA; $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, that the electrons given by zinc are taken by ion hydrogen.&#xA;So, if we add the two half reactions:&#xA;$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$&#xA;Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:&#xA;$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$&#xA;Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-04-10T20:36:59.833" LastActivityDate="2015-04-10T20:36:59.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28524" PostTypeId="2" ParentId="28521" CreationDate="2015-04-10T20:45:32.027" Score="3" Body="&lt;p&gt;The reactions you have shown could in fact occur (in the forward direction) to a slight degree in a solution of cesium bromide in water. But I would expect them to occur to a lesser degree than the autoprotolysis of water (formation of hydronium and hydroxide ions from two water molecules), so their effect on pH would be insignificant, and the pH would effectively be neutral.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-10T20:45:32.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28525" PostTypeId="1" CreationDate="2015-04-10T22:13:43.847" Score="5" ViewCount="95" Body="&lt;p&gt;Would copper wires from power transmission or telephone lines acquire a patina but remain largely intact, or oxidize into dust after a century or more without replacement or maintenance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously environmental factors would be enormous; ideally I'd like to know what the metal's state would be in ideal (underground?), typical (temperate zone), and worst-case (near salt water?) scenarios. Clearly some circumstances allow copper objects to survive largely intact for several thousand years, since such artifacts have been found by archeologists, but does this require very exceptional circumstances?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is inspired by an answer on the Worldbuilding site, which claims that a post-apocalyptic society would have a difficult time developing metallurgy as the rusted remains of human civilization would be &lt;em&gt;more&lt;/em&gt; difficult to smelt than the naturally-occurring ores which were relatively abundant during the Stone Age. Hopefully by asking about this copper wires specifically, I have an appropriate question for this site. If so, I may ask another question about iron and steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not have much luck in finding relevant tags. Any assistance there is appreciated!&lt;/p&gt;&#xA;" OwnerDisplayName="user7812" LastActivityDate="2015-05-11T06:14:12.577" Title="To what state would copper wiring revert if abandoned for centuries?" Tags="&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28526" PostTypeId="1" CreationDate="2015-04-10T22:42:54.643" Score="4" ViewCount="291" Body="&lt;p&gt;Is time a measurement of entropy? Is it because time is increasing forever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well after reading about the Second Law of Thermodynamics and working with Entropy, I began to realize that if time can be interpreted as a measure of Entropy, since it is always increasing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since almost everything decays with time, I thought that time could be&#xA;  a measure of Entropy. Since it is achieving the lowest energy state&#xA;  possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-10T22:50:23.637" LastActivityDate="2015-05-22T03:53:35.407" Title="Is time a measurement of Entropy?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28527" PostTypeId="2" ParentId="28483" CreationDate="2015-04-11T00:07:44.207" Score="1" Body="&lt;p&gt;Your issue is that you are falsely considering the $ \ce{CaSO4}$ to be part of the system. The system under consideration is the solution, which the solid is not a part of. This is the idea of the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_Product_Constant,_Ksp&quot; rel=&quot;nofollow&quot;&gt;solubility product&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reaction quotient is therefore: $Q= \ce{[Ca^{2+}]·[{SO4}^{2-}]}$&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-11T00:07:44.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28529" PostTypeId="2" ParentId="28381" CreationDate="2015-04-11T04:07:26.720" Score="4" Body="&lt;p&gt;The egg hatch toys are composed of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Superabsorbent_polymer&quot; rel=&quot;nofollow&quot;&gt;superabsorbent polymer&lt;/a&gt;. These polymers are able to absorb up to 30 to 60 time their volume in water by forming hydrogen bonds with the water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium polyacrylate ($\ce{[-CH2-CH(CO2Na){}-]}_n%edit$) is the most common superabsorbent polymer and according to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_polyacrylate&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt;, is the one used in the egg hatch toys.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="24052" LastEditDate="2016-05-27T22:18:09.753" LastActivityDate="2016-05-27T22:18:09.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28530" PostTypeId="1" AcceptedAnswerId="28532" CreationDate="2015-04-11T04:45:51.870" Score="2" ViewCount="2012" Body="&lt;p&gt;I was wondering why the melting point of $\ce{TiCl4}$ was below that of $\ce{TiCl3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The melting point of the former is  $\ce{-25 Celsius}$ whereas the latter is $\ce{425 Celsius}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-11T05:53:36.407" Title="Why is the melting point of TiCl4 lower than TiCl3" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28532" PostTypeId="2" ParentId="28530" CreationDate="2015-04-11T05:53:36.407" Score="5" Body="&lt;p&gt;-The high melting point of  $\ce{TiCl3}$&#xA;is mainly due to the short bond length $\ce{Ti-Ti}$ , which infer&#xA; a strong metallic character to these bonds between titanium ions in  $\ce{TiCl3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-While in $\ce{TiCl4}$, the main intermolecular forces are of van der Waals type.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-11T05:53:36.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28533" PostTypeId="1" AcceptedAnswerId="28555" CreationDate="2015-04-11T06:54:52.083" Score="4" ViewCount="1200" Body="&lt;p&gt;I'm interested in ab initio computational chemistry. I'm trying to figure out what I should study. I know that I should study physical chemistry, mathematics, and programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am not sure in what I should be focusing on. It seems to be a fairly new field and books on the topic are rather specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone working in computational chemistry research provide me with a list of (math, computational chemistry, physical chemistry) textbooks they studied or any additional resources to get me started?&lt;/p&gt;&#xA;" OwnerUserId="3983" LastEditorUserId="402" LastEditDate="2016-02-05T05:34:50.253" LastActivityDate="2016-02-05T05:34:50.253" Title="Computational Chemistry (Ab Initio), what should I study?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28534" PostTypeId="1" AcceptedAnswerId="28535" CreationDate="2015-04-11T10:56:19.663" Score="4" ViewCount="9054" Body="&lt;p&gt;Could someone, please, explain to me how the standard &#xA;Hydrogen Electrode works and how it is used to measure electrode potentials? Also, why does it have a potential value of zero?&lt;/p&gt;&#xA;" OwnerUserId="15507" LastEditorUserId="15235" LastEditDate="2017-07-30T00:59:13.360" LastActivityDate="2017-07-30T00:59:13.360" Title="Standard Hydrogen Electrode" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28535" PostTypeId="2" ParentId="28534" CreationDate="2015-04-11T11:44:16.903" Score="4" Body="&lt;p&gt;The &lt;strong&gt;SHE&lt;/strong&gt; (Standard Hydrogen Electrode) is a standard equipment for measuring the &lt;strong&gt;&lt;em&gt;relative electrode potential&lt;/em&gt;&lt;/strong&gt;, as the process of measuring the actual potential is quite difficult. It consists of a platinum foil dipped in 1.0M $\ce{HCl}$ in the presence of $\ce{H2}$ gas at 1 bar of pressure.&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when the SHE is connected to another half-cell, either oxidation or reduction may occur in the SHE. &#xA; &lt;p&gt;&#xA;Suppose we have to find the electrode potential of a half cell represented by &quot;A&quot; Now, on setting up the cell and connecting a voltmeter we get the standard potential(E&lt;sub&gt;standard&lt;/sub&gt;) of the cell i.e.&lt;strong&gt;&lt;br&gt;E&lt;sub&gt;standard&lt;/sub&gt; = E&lt;sub&gt;A&lt;/sub&gt; - E&lt;sub&gt;H&lt;sub&gt;2&lt;/sub&gt;&lt;/sub&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Now as our our aim is to just find E&lt;sub&gt;A&lt;/sub&gt;, we take E&lt;sub&gt;H&lt;sub&gt;2&lt;/sub&gt;&lt;/sub&gt; to be equal to 0&lt;br&gt;&#xA;Therefore,&lt;br&gt;&#xA;&lt;strong&gt;E&lt;sub&gt;standard&lt;/sub&gt; = E&lt;sub&gt;A&lt;/sub&gt; - 0&lt;/strong&gt;&lt;br&gt;&#xA;or,&lt;br&gt;&#xA;&lt;strong&gt;E&lt;sub&gt;standard&lt;/sub&gt; = E&lt;sub&gt;A&lt;/sub&gt;&lt;/strong&gt;&#xA;&lt;br&gt;&#xA;The value of E&lt;sub&gt;standard&lt;/sub&gt; is taken from the voltmeter.&lt;br&gt;&#xA;It should be noted that the electrode potential calculated using this method is &lt;strong&gt;relative&lt;/strong&gt;(to the electrode potential of hydrogen). It isn't the absolute value. &#xA;Also, the potential for the SHE is &lt;strong&gt;&lt;em&gt;taken to be 0&lt;/em&gt;&lt;/strong&gt; for the simplification of the process.&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="275" LastEditDate="2015-04-11T12:07:20.317" LastActivityDate="2015-04-11T12:07:20.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28536" PostTypeId="2" ParentId="25114" CreationDate="2015-04-11T11:51:34.043" Score="3" Body="&lt;p&gt;As annav states, the labeling H1, H2 etc is labeling of hydrogen atoms within a graphene-like hexagonal ring structure. Another way of looking at it, is in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F2yDj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://phys.org/news/2014-12-simplest-element-hydrogen-graphene.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an article published after the web article you link to, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/25369180&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aromaticity, closed-shell effects, and metallization of hydrogen.&lt;/a&gt; (Abstract only), the authors confirm that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recent theoretical and experimental studies reveal that compressed molecular hydrogen at 200-350 GPa transforms to layered structures consisting of distorted graphene sheets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This structure, according to the article, forms a stable structure at those pressures.  As per the comments, this is form graphene-like hydrogen, the abstract goes further, discussing the process that would result in hydrogen metallization:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metallization occurs upon further compression by destroying the closed shell electronic structure, which is mainly determined by the 1s electrons via lowering of the bonding bands stemming from the unoccupied atomic 2s and 2p orbitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This metallization of hydrogen is, according to the article &lt;a href=&quot;http://arxiv.org/pdf/1501.06336.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;High temperature superconductivity in sulfur and selenium hydrides at high pressure&lt;/a&gt; theorised to occur at pressures of 400 GPa or greater.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-11T11:57:44.160" LastActivityDate="2015-04-11T11:57:44.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28537" PostTypeId="1" AcceptedAnswerId="28543" CreationDate="2015-04-11T12:13:27.887" Score="2" ViewCount="82" Body="&lt;h3&gt;In case of metal deficient defects in the crystal lattice of a metal, the positive voids are filled by electrons. &lt;p&gt;&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does this result in the coloration of a substance? &lt;/li&gt;&#xA;&lt;li&gt;Does this condition arise only in the case of metals?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="15508" LastEditDate="2015-05-15T08:10:39.367" LastActivityDate="2015-05-15T08:10:39.367" Title="Coloration of substances because of F-Centers" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28539" PostTypeId="1" AcceptedAnswerId="28540" CreationDate="2015-04-11T13:51:04.917" Score="7" ViewCount="1142" Body="&lt;p&gt;Is it possible to reduce raspberry ketone (&lt;em&gt;4-(4-Hydroxyphenyl)butan-2-one&lt;/em&gt;) to ethyl methyl ketone? How would I do this in practice?&lt;/p&gt;&#xA;" OwnerUserId="15510" LastEditorUserId="10185" LastEditDate="2015-04-11T18:15:49.900" LastActivityDate="2015-04-11T18:15:49.900" Title="Synthesis of ethyl methyl ketone from 2-(4-hydroxyphenyl)ethyl methyl ketone" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28540" PostTypeId="2" ParentId="28539" CreationDate="2015-04-11T14:53:27.330" Score="5" Body="&lt;p&gt;There are a number of ways to accomplish your goal, but they all involve several synthetic steps. As mentioned in the comments, it's always easier to buy a compound then run a reaction, but if you had a small amount of some exotic derivative, you could convert it to methyl ethyl ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible route would be to convert the raspberry ketone to an ester using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baeyer-Villager&lt;/a&gt; reaction.  In this reaction the secondary (ethyl) branch migrates in preference to the primary (methyl) side.  The resulting ester can then be converted to the desired ketone by any number of methods such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Weinreb_ketone_synthesis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weinreb&lt;/a&gt; route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BN0YZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-11T15:34:16.683" LastActivityDate="2015-04-11T15:34:16.683" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28541" PostTypeId="1" CreationDate="2015-04-11T15:15:32.830" Score="1" ViewCount="273" Body="&lt;p&gt;I am interested in improving the extraction efficiency of hydrocarbons from soil using a soxhlet extractor, and I would like to know what solvents might improve this extraction efficiency, and why. Currently, I am using 1:1 hexane:acetone.&lt;/p&gt;&#xA;" OwnerUserId="11287" LastEditorUserId="11287" LastEditDate="2015-04-11T16:12:14.923" LastActivityDate="2015-04-11T16:12:14.923" Title="What is an effective extraction solvent for removing hydrocarbons from soil (using a soxhlet extractor)?" Tags="&lt;extraction&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28542" PostTypeId="1" CreationDate="2015-04-11T15:38:45.520" Score="4" ViewCount="33950" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in order to solve this, apparently all you need to do is plug it into the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = -\log[\ce{H3O+}]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-&amp;gt; H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="5026" LastEditDate="2018-02-23T02:14:21.270" LastActivityDate="2018-02-23T02:37:45.570" Title="What is the pH of a 5M solution of hydrochloric acid?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28543" PostTypeId="2" ParentId="28537" CreationDate="2015-04-11T15:52:59.667" Score="1" Body="&lt;p&gt;For a simple explanation, we could explain the coloration of the substance by the absorption of the energy of the electrons in the void, which leads to the color of the compliment color (as is suggested by &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/204/1078/406&quot; rel=&quot;nofollow&quot;&gt;http://rspa.royalsocietypublishing.org/content/204/1078/406&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a more rigorous explanation could be found here: &lt;a href=&quot;http://ptp.oxfordjournals.org/content/4/2/181.full.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ptp.oxfordjournals.org/content/4/2/181.full.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second question, it is written in Ceramic Materials: Science and Engineering that &quot;the term color center is now applied to any defect, including an impurity, that produces color in an insulator&quot;. It also notes that the original observation was made by the production of F centers, which suggests that the modern interpretation of the phenomenon applies to the defects which product color in an insulator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I answered you question exactly as you hoped, but if you look into the resources, there is a wealth of knowledge there.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-11T15:52:59.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28544" PostTypeId="2" ParentId="28542" CreationDate="2015-04-11T16:13:00.733" Score="-1" Body="&lt;p&gt;There exists seven strong acids that completely disassociate with hydrogen ( $\ce{HCl}$, $\ce{HBr}$, $\ce{H2SO4}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{HI}$, and $\ce{HNO3}$), so the only calculation you need to do is the negative log of the concentration (in molarity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &#xA;$\mathrm{pH} = -\log5.0=-0.69$. Yes, $\mathrm{pH} = −0.69$ even though the $\mathrm{pH}$ scale &quot;normally&quot; ends at $\ce{pH} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(in case of $\ce{HCl}~\mathrm{pH}$ will be negative for any solution more then $\pu{2M}$)&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="5026" LastEditDate="2018-02-23T02:22:35.853" LastActivityDate="2018-02-23T02:22:35.853" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28545" PostTypeId="1" AcceptedAnswerId="28552" CreationDate="2015-04-11T16:20:53.357" Score="3" ViewCount="6676" Body="&lt;p&gt;I have been working on this lab forever and have not been able to get a $K_\text{a}$ anywhere close to the accepted value. I’m sure I’m missing something really simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We titrated $\ce{CH3COOH}$ with $\ce{NaOH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used $10.00\ \mathrm{mL}$ of $\ce{CH3COOH}$. Volume of $\ce{NaOH}$ at the equivalence point was $10.14\ \mathrm{mL}$, and the concentration of $\ce{NaOH}$ was $0.1530\ \mathrm{mol \cdot L^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I found the concentration of $\ce{CH_3COOH}$ by using $c_\text{a} \cdot V_\text{a} = c_\text{b} \cdot V_\text{b}$, so $c_\text{a} \cdot 10.00\ \mathrm{mL} = 0.1530\ \mathrm{mol \cdot L^{-1}} \cdot 10.14\ \mathrm{mL}$&lt;br&gt;&#xA;$c_\text{a} = 0.1551\ \mathrm{mol \cdot L^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I used the equation: $\ce{CH3COOH + H2O &amp;lt;=&amp;gt; H3O+ + CH3COO-}$&#xA;and the pH (at the midpoint of $5.07\ \mathrm{mL}\ \ce{NaOH}$) which was $4.87$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{a}= \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]&#xA;[\ce{H3O+}]}=[\ce{CH3COO-}]= 10^{-4.87}= 1.36 \times 10^5&#xA;$$&#xA;$$&#xA;K_\text{a}= \frac{(1.36 \times 10^5)^2}{ 0.1551} = 1.19 \times 10^{-9}&#xA;$$&#xA;Any help on what I am doing wrong would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="15511" LastEditorDisplayName="user7951" LastEditDate="2015-04-11T18:47:28.477" LastActivityDate="2015-04-15T15:32:57.457" Title="I need to find the Ka of a weak acid in titration with a strong base" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28546" PostTypeId="2" ParentId="28542" CreationDate="2015-04-11T16:26:42.743" Score="4" Body="&lt;h2&gt;Method to calculate the pH of a solution with a strong monoacid&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|c|}\hline&#xA;&amp;amp;\text{Before}&amp;amp;&amp;amp;\text{After}&amp;amp;\\\hline&#xA;&amp;amp;\ce{AH}&amp;amp;\ce{H2O}&amp;amp;\ce{A-}&amp;amp;\ce{H3O+}\\\hline&#xA;\text{Initial}&amp;amp;n0&amp;amp;\text{excess}&amp;amp;0&amp;amp;0\\\hline&#xA;\text{Equilbrium}&amp;amp;\epsilon&amp;amp;\text{excess}&amp;amp;n0&amp;amp;n0\\\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $\mathrm{pH}=-\log\left[\ce{H3O+}\right]$ Now we have to verify if our result is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider that $\ce{H3O+}$ in the solution has exactly the same concentration as that of the acid. However they can react with $\ce{OH-}$ because of the water autoprotolysis! $K_\text{a}=10^{14}$ total reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3O^+ + HO^- -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the concentration of $\ce{H3O+}$ is approximatively constant if $\left[\ce{H3O+}\right]\gg\left[\ce{HO-}\right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if $\frac{1}{10}\left[\ce{H3O+}\right]&amp;gt;\left[\ce{HO-}\right]$ then we can find the limit for which our reasonning is false : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[\ce{HO-}\right]\times \left[\ce{H3O+}\right] &amp;lt; \frac{1}{10} \times \left[\ce{H3O+}\right] \times \left[\ce{H3O+}\right] \iff K_\text{e}&amp;lt;\frac{1}{10}\left[\ce{H3O+}\right]^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have $\mathrm{pH}&amp;lt;6.5$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if with a strong monoacid you find $\mathrm{pH}=6.8$ with this formula the result is not correct and you have to make an other approximation or reasonning! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have $\mathrm{pH}=-0.69&amp;lt;6.5$ so you find a reasonable answer. And in water $\mathrm{pH}$ is between $0$ and $14$ so your solution is at $\mathrm{pH} \approx 0$&lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="5026" LastEditDate="2018-02-23T02:21:05.593" LastActivityDate="2018-02-23T02:21:05.593" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28547" PostTypeId="1" CreationDate="2015-04-11T18:19:53.313" Score="2" ViewCount="493" Body="&lt;p&gt;I use borax to clean everything. the other day I thought it would be a good idea to mix a little borax with my shampoo while showering.  After lathering it on my scalp about 30 seconds it began to give off a really bad chemical smell, almost like a perm solution. What was the borax reacting with the product in my hair or the shampoo?  &lt;/p&gt;&#xA;" OwnerUserId="15513" LastEditorUserId="7475" LastEditDate="2015-10-30T17:12:51.183" LastActivityDate="2015-10-30T17:12:51.183" Title="Mixing borax and shampoo" Tags="&lt;everyday-chemistry&gt;&lt;surfactants&gt;&lt;smell&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-10-30T19:35:40.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="28548" PostTypeId="1" CreationDate="2015-04-11T18:33:28.167" Score="1" ViewCount="928" Body="&lt;p&gt;I understand this is a simple question (I’m in high school), but for some reason the numbers just are not working for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&#xA;$$\Delta H^\circ_\text{c} = -1301\ \mathrm{kJ}\ \text{for}\ \ce{C2H2(g)}$$&#xA;$$\Delta H^\circ_\text{f} = -394\ \mathrm{kJ}\ \text{for}\ \ce{CO2(g)}$$&#xA;$$\Delta H^\circ_\text{f} = -242\ \mathrm{kJ}\ \text{for}\ \ce{H2O(g)}$$&#xA;Calculate $\Delta H_\text{f}$ for $\ce{C2H2(g)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My process:&#xA;$$(1) : \ce{C2H2(g) + 5/2 O2(g) -&amp;gt; 2CO2(g) + H2O(g) }\quad \Delta H^\circ_\text{r} = -1301\ \mathrm{kJ}$$&#xA;$$(2) : \ce{C(s) + O2(g) -&amp;gt; CO2(g)} \quad \Delta H^\circ_\text{r} = -394\ \mathrm{kJ}$$&#xA;$$(3) : \ce{H2(g) + 1/2O2(g) -&amp;gt; H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Target:}\ \ce{2C(s) + H2(g) -&amp;gt; C2H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some experimentation I arrive at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-1\times(1) + 2\times(2) + (3) = \text{Target}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown through:&#xA;$$-1\times(1){:}\ \ce{2CO2(g) + H2O(g) -&amp;gt; C2H2(g) + 5/2 O2(g)} \quad \Delta H^\circ_\text{r} = 1301\ \mathrm{kJ}$$&#xA;$$2\times(2){:}\ \ce{2C(s) + 2O2(g) -&amp;gt; 2CO2(g)} \quad \Delta H^\circ_\text{r} = -788\ \mathrm{kJ}$$&#xA;$$(3){:}\ \ce{H2(g) + 1/2O2(g) -&amp;gt; H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$&#xA;$$\text{Sum:}\ \ce{2C(s) + H2(g) -&amp;gt; C2H2(g)} \quad \Delta H^\circ_\text{f} = 271\ \mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as far as I know, all the givens are accurate empirical values. But the empirical value for $\Delta H^\circ_\text{f}$ for $\ce{C2H2(g)}$ is $227\ \mathrm{kJ}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What’s even weirder is that this disparity $(271-227=44)$ is equal to the difference between standard enthalpies of formation for water in gas and liquid states $(-242 - -286 = 44)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I made an error somewhere, or is water formed in the liquid state during the combustion of acetylene?&lt;/p&gt;&#xA;" OwnerUserId="15512" LastEditorDisplayName="user7951" LastEditDate="2015-04-13T13:56:17.013" LastActivityDate="2015-04-13T14:31:25.380" Title="Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law)" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28549" PostTypeId="1" AcceptedAnswerId="28550" CreationDate="2015-04-11T18:59:19.457" Score="9" ViewCount="5676" Body="&lt;p&gt;I was trying to compare acidity of some organic compounds but can't understand some:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why should 2,6-dimethyl-4-nitrophenol be more acidic than 3,5-dimethyl-4-nitrophenol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to figure it out using various factors like hyperconjugation and induction effect of Me group but its not working. Suppose i take H.C effect 2,6-dimethyl-4-nitrophenol be &quot;less&quot; acidic than 3,5-dimethyl-4-nitrophenol since it destabilizes the negative charge on conjugate base. But this logic is not working.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4231" LastEditDate="2015-04-12T23:32:15.343" LastActivityDate="2019-09-25T02:46:37.040" Title="Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;resonance&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28550" PostTypeId="2" ParentId="28549" CreationDate="2015-04-11T19:36:34.040" Score="13" Body="&lt;p&gt;In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structures like &lt;strong&gt;B&lt;/strong&gt; in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  &lt;strong&gt;Resonance structure B requires that the nitro group be planar with the aromatic ring&lt;/strong&gt;.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure &lt;strong&gt;B&lt;/strong&gt; can no longer contribute and the compound is less acidic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e1fGJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-11T19:36:34.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28551" PostTypeId="1" CreationDate="2015-04-11T19:55:40.827" Score="2" ViewCount="1414" Body="&lt;p&gt;A solution is created by mixing $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{HCl}$, $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{CH3COOH}$ and $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ $\ce{CsOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $K_\mathrm{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the $\mathrm{pH}$ of this solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck between $\ce{HCl}$ and $\ce{CsOH}$ since they are both a strong acid and base (respectively) meaning that they should both have a substantial affect on the $\mathrm{pH}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when determining what the $\mathrm{pH}$ is, which species would we use in our ICE table?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="15410" LastEditDate="2015-05-13T16:15:58.180" LastActivityDate="2015-05-13T16:15:58.180" Title="After mixing, which species has the dominant affect on the pH of the solution that remains?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28552" PostTypeId="2" ParentId="28545" CreationDate="2015-04-11T20:12:27.017" Score="1" Body="&lt;p&gt;I’m not sure if I’m missing your question, but I thought that perhaps you could use the relation $\mathrm{pH}=\mathrm{p}K_\text{a}$ at the half equivalence point? The answer would be still incorrect but closer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you could probably use stoichiometry to solve for $K_\text{a}$ at the beginning or the end of the titration by constructing an ICE table or otherwise. By considering the deprotonation at the start, so the concentration of protons would be equivalent to concentration of conjugate base. Then, instead of inserting the pH at the half equivalence point, insert the pH before the titration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need help doing the algebra, I could try...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think what you did wrong was inserting the concentration at the half equivalence point, rather than before the titration.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorDisplayName="user7951" LastEditDate="2015-04-15T15:32:57.457" LastActivityDate="2015-04-15T15:32:57.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28553" PostTypeId="2" ParentId="28551" CreationDate="2015-04-11T20:15:40.117" Score="0" Body="&lt;p&gt;You need to provide the concentration of acetic acid. However, if I were to guess, I would say that the hydrochloric acid completely reacts with the cesium hydroxide, and the remaining cesium hydroxide reacts with the acetic acid. Finding the quantity of acetic acid, which probably was more than the concentration of cesium hydroxide, you could use hendersen hassebach.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-11T20:15:40.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28554" PostTypeId="1" CreationDate="2015-04-11T20:37:44.993" Score="11" ViewCount="1551" Body="&lt;p&gt;For example, which is the right way to name this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yCY5o.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yCY5o.png&quot; alt=&quot;Compound&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it 2-bromo-4-methylenehexane or 4-bromo-2-ethylpent-1-ene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the larger number of carbon in a chain take priority, as in the former, or does the carbon-carbon double bond take priority, as in the latter?&lt;/p&gt;&#xA;" OwnerUserId="15515" LastEditorUserId="16683" LastEditDate="2016-05-04T22:21:26.717" LastActivityDate="2016-05-04T22:21:26.717" Title="How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28555" PostTypeId="2" ParentId="28533" CreationDate="2015-04-11T20:52:10.070" Score="11" Body="&lt;p&gt;Computational chemistry is not a new field, but yes it can be difficult to find the right approach to it, because it is so diversified.  And that means to some degree what you need depends strongly on what you want to study.  Most of the time, you're basically solving the Schrodinger equation all kinds of different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be better if you can specify what level of knowledge you do have at this time. Assume that you've just got into college, then there's a list of things you should study:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Math.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You need linear algebra and mathematical analysis.  Any good old textbook will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you use the most everyday would be linear algebra, so you better be very very good at it, along with a good expertise in vectors, matrices and in some cases tensors.  There may also be some functional analysis involved but for most people it really just boil down to linear algebra.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematical analysis is basically just differentiation, integration, and all kinds of tricks related to it.  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Quantum mechanics.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You some of it from P-Chem II, but if you take it from the physics department it will be more helpful.  I recommend the &lt;em&gt;Cohen-Tannoudji textbook&lt;/em&gt;.  Make sure you start from the beginning, go through every chapter to the very end.  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Quantum chemistry.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With a background you can now proceed to understand the conventional methods of quantum chemistry.  For this I would recommend &lt;em&gt;Modern Quantum Chemistry by Szabo and Ostlund&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Courses to take&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to what was already mentioned above, many universities will also have quantum chemistry and computational chemistry courses. I also recommend you to take &lt;em&gt;group theory&lt;/em&gt; and &lt;em&gt;calculus of variation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Programming skills&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that prepares you for the theoretical part.  Whether you need comprehensive programming skill depends on what you want to do.  Some may make do with existing programs, some may need to write there own.  But even if you use an existing program, it requires no less amount of understanding of the quantum chemical theories you're using.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For programming skills, you will certainly benefit by taking numerical methods course, and mastery over one programming language. Take statistics and machine learning if you have time. Any one will do, it doesn't matter. For what we do, its straight forward to switch to a different language whenever you want.  Computational chemists are not usually very big on the algorithms themselves. Most of the times, you just take something right out of a numerical analysis text book or program package and adapt a little bit.  It is much more about the underlying physics, i.e., what form of wave function you choose, what approximation you make, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing about programming, I feel, is hands on experience.  If you can understand quantum mechanics, programming is not gonna be hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chemistry&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for other branches of chemistry, you don't have to be the expert, but you definitely need a good understanding because those are the problems you're trying to solve.  You need to understand very well the experimental procedure you're trying to study.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastActivityDate="2015-04-11T20:52:10.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28556" PostTypeId="2" ParentId="28554" CreationDate="2015-04-11T21:09:12.470" Score="-2" Body="&lt;p&gt;From wikipedia:&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the alkene takes priority over the alkane and &lt;strong&gt;4-bromo-2-ethylpent-1-ene&lt;/strong&gt; is the correct name.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-11T21:09:12.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28557" PostTypeId="1" AcceptedAnswerId="29310" CreationDate="2015-04-11T21:31:46.123" Score="10" ViewCount="9226" Body="&lt;p&gt;Why is arsenic more chemically similar to phosphorus than is nitrogen (to phosphorus)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that it may be because both phosphorus and arsenic have d orbitals (albeit one has one that is filled and the other has an unfilled one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility I thought was that because nitrogen held the electrons very closely, and thus the difference between the second and third period may be more pronounced than the difference between the third and fourth periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd really appreciate it if anyone could help clarify/ explain. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2018-02-19T06:38:27.947" Title="Why is arsenic more chemically similar to phosphorus than is nitrogen?" Tags="&lt;elements&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28558" PostTypeId="2" ParentId="28554" CreationDate="2015-04-11T22:02:15.800" Score="12" Body="&lt;p&gt;The IUPAC recommendations on this matter have been changed. In the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, a preferred IUPAC name is 3-methylidenehexane (not 2-ethylpent-1-ene) since the longer chain now supersedes a shorter unsaturated chain. (This example is given in the &lt;em&gt;Blue Book&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pridv.png&quot; alt=&quot;3-methylidenehexane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, the prefix ‘methylene’ is &lt;em&gt;no longer recommended&lt;/em&gt; in IUPAC nomenclature to designate the $\ce{=CH2}$ group. The preferred prefix is ‘methylidene’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the correct name for your example is &lt;strong&gt;2-bromo-4-methylidenehexane.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9rrWm.png&quot; alt=&quot;2-bromo-4-methylidenehexane&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-04-12T08:28:01.300" LastActivityDate="2015-04-12T08:28:01.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28559" PostTypeId="1" CreationDate="2015-04-12T02:10:26.177" Score="0" ViewCount="5321" Body="&lt;p&gt;Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected when compared to the expected IR for triphenylmethanol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interpreting the IR I took of my product, triphenylmethanol, to the expected IR I obtained from SDBS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My IR: &lt;img src=&quot;https://i.stack.imgur.com/52C54.png&quot; alt=&quot;my triphenylmethanol product&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SDBS: &lt;img src=&quot;https://i.stack.imgur.com/KlrsR.gif&quot; alt=&quot;SDBS triphenylmethanol IR&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We performed a Grignard Reaction in lab using bromobenzene and methyl benzoate in order to synthesize triphenylmethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ANY clues why my triphenylmethanol (O—H stretch) is not as intense? I really don't know.&lt;/p&gt;&#xA;" OwnerUserId="15518" LastActivityDate="2015-04-12T03:06:16.763" Title="Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28560" PostTypeId="2" ParentId="28559" CreationDate="2015-04-12T03:06:16.763" Score="1" Body="&lt;p&gt;A possible explanation would be the concentration of sample you used. Of course, if you are confident that you had utilized sufficient concentration then there may be another reason for the decreased intensity. However, some of the other peaks are also somewhat decreased in intensity so it is a possibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point is noted here (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Interpretation#The_Origin_of_Peak_Intensities&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Interpretation#The_Origin_of_Peak_Intensities&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-12T03:06:16.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28562" PostTypeId="1" AcceptedAnswerId="28607" CreationDate="2015-04-12T04:55:05.490" Score="0" ViewCount="288" Body="&lt;p&gt;I am designing a battery box to house lithium ion batteries in an electric vehicle. These batteries use an electrolyte which probably contains carbonate esters (eg. ethylene carbonate, dimethyl carbonate, diethyl carbonate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use polycarbonate as the battery box material, should I be worried about an interaction with the electrolyte in the event of a battery leak?&lt;/p&gt;&#xA;" OwnerUserId="15520" LastActivityDate="2015-04-13T04:44:11.103" Title="Chemical compatibility of polycarbonate and carbonate esters" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28563" PostTypeId="1" AcceptedAnswerId="28656" CreationDate="2015-04-12T07:09:35.000" Score="4" ViewCount="32576" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mole_(unit)&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Wikipedia article on the mole&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt; The term &lt;strong&gt;gram-atom&lt;/strong&gt; (abbreviated gat.) has been used for a related but distinct concept, namely a quantity of a substance that contains Avogadro's number of atoms, whether isolated or combined in molecules. Thus, for example, 1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain the difference between moles and gram atoms, and how is it that &lt;em&gt;1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="15521" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-06-28T18:23:46.647" LastActivityDate="2017-06-28T18:23:46.647" Title="What exactly does the term &quot;gram atoms&quot; mean?" Tags="&lt;stoichiometry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28564" PostTypeId="2" ParentId="28395" CreationDate="2015-04-12T07:24:50.200" Score="5" Body="&lt;p&gt;A full image similar to your diagram with chemical reactions is found at a blog &lt;a href=&quot;http://blogs.rediff.com/chemistrytutoring/2013/02/27/corrosion-of-iron/&quot; rel=&quot;noreferrer&quot;&gt;Chemistry Tutoring&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSPF8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_7%3A_Electrochemical_Corrosion&quot; rel=&quot;noreferrer&quot;&gt;Electrochemistry 7: Electrochemical Corrosion&lt;/a&gt;, the oxidation and reduction reactions occur in separate places in the metal - in the case of your diagram:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Oxidation occurs in the anode part of the iron pole submerged in water:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe_{(s)}-&amp;gt;Fe^{2+}_{(aq)} +2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Burak Ulgat correctly noted in his comment, &lt;a href=&quot;http://www.lenntech.com/periodic/water/iron/iron-and-water.htm&quot; rel=&quot;noreferrer&quot;&gt;the $\ce{Fe^{2+}_{(aq)}}$ ions are dissolved in the water&lt;/a&gt; and are outside of the pole and are transported by the water. The electrons move through the metal in the pole (as iron is a metal, these electrons conduct easily) - this part is not shown in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anodic region, for whatever reason, the iron is exposed to the seawater, may already have damage etc, so comes under 'attack'.  As $\ce{Fe^{2+}_{(aq)}}$ ions are 'liberated', pitting occurs as material is removed (dissolved) from this area of the pole. Often, these pits are deprived of oxygen, increasing the rate of pitting. The presence of seawater speeds up this process.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reduction occurs at the cathode region, oxygen from the air is reduced to form hydroxide:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2_{(g)} + H2O_{(l)} + 4e^{-}-&amp;gt;4OH^{-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then rust formation occurs as per the diagram above (summarised below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{2+}_{(aq)} + 2OH^{-}_{(aq)}-&amp;gt;Fe(OH)2_{(s)}}$$&#xA;$$\ce{4Fe(OH)2_{(s)} + O2_{(g)} + H2O_{(l)} -&amp;gt;Fe(OH)3_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the UC Davis ChemWiki, the process of corrosion continues if the presence of a &lt;em&gt;depolarizer&lt;/em&gt;, which removes electrons from the metal, depolarizers include more noble metals, acids and in the case relevant to the diagram, oxygen. So, the oxygen depolarizer is removing electrons out of the iron, resulting in more iron ions being dissolved into the water at the anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrochemical process is summarised by the HyperPhysics page about &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/corrosion.html&quot; rel=&quot;noreferrer&quot;&gt;Corrosion as an Electrochemical Process&lt;/a&gt;, in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2WTYJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-12T09:27:10.107" LastActivityDate="2015-04-12T09:27:10.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28567" PostTypeId="2" ParentId="12332" CreationDate="2015-04-12T09:17:47.140" Score="6" Body="&lt;p&gt;It depends where and how the electrodes are positioned. The UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_2%3A_Galvanic_cells_and_Electrodes&quot; rel=&quot;noreferrer&quot;&gt;Electrochemistry 2: Galvanic cells and electrodes&lt;/a&gt; describes the possible arrangements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the electrodes are placed in the same vessel, a salt bridge is not required (as shown below). Ions can pass through the porous barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8fST8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the electrodes are are placed in separate vessels (such as in the diagram below), then a salt bridge is required to facilitate ion movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IH9mD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-04-12T09:17:47.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28568" PostTypeId="1" AcceptedAnswerId="43463" CreationDate="2015-04-12T09:20:47.477" Score="25" ViewCount="22805" Body="&lt;p&gt;Caesium has a larger size, and the effective nuclear charge that the valence electron experiences will be far less compared to that of lithium's, right? But lithium is still considered the strongest reducing agent among all the alkali metals, and this is evidenced by its large and negative reduction potential. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="15517" LastEditorUserId="16683" LastEditDate="2017-04-30T22:05:06.310" LastActivityDate="2020-08-31T06:33:32.507" Title="Why is lithium the most reducing alkali metal, and not caesium?" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;reduction-potential&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28572" PostTypeId="1" AcceptedAnswerId="28573" CreationDate="2015-04-12T11:33:56.040" Score="3" ViewCount="4209" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A &lt;span class=&quot;math-container&quot;&gt;$\pu{20.0 L}$&lt;/span&gt; nickel container was charged with &lt;span class=&quot;math-container&quot;&gt;$\pu{0.500 atm}$&lt;/span&gt; of xenon gas and &lt;span class=&quot;math-container&quot;&gt;$\pu{1.50 atm}$&lt;/span&gt; of fluorine gas at &lt;span class=&quot;math-container&quot;&gt;$\pu{400 ^{\circ}C}$&lt;/span&gt;. The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming &lt;span class=&quot;math-container&quot;&gt;$100\%$&lt;/span&gt; yield?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; From the description, the reaction can be written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Xe + 2F2 -&amp;gt; XeF4}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I first calculated the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{Xe}$&lt;/span&gt; using the ideal gas law:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n(\ce{Xe}) = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Doing the same for fluorine gives:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n(\ce{F2}) = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then, for every mole of &lt;span class=&quot;math-container&quot;&gt;$\ce{Xe}$&lt;/span&gt; we need two moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{F2}$&lt;/span&gt;. Since we have more than &lt;span class=&quot;math-container&quot;&gt;$2 \cdot \pu{0.18 mol} = \pu{0.36 mol}$&lt;/span&gt; of fluorine, xenon is the limiting reactant. The molar mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{XeF4}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$207.1\,\mathrm{g/mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So now I would just do:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;n(\ce{XeF4}) M(\ce{XeF4}) &amp;amp;= m(\ce{XeF4})\\&#xA;\pu{0.18 mol}  \times \pu{207.1 g mol-1}  &amp;amp;= \pu{37.278 g}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, the answer at the back of my textbook says it should be &lt;span class=&quot;math-container&quot;&gt;$\pu{37.5 g}$&lt;/span&gt;. So did I make a mistake somewhere or is this just a roundoff-error?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorUserId="4945" LastEditDate="2021-05-11T22:32:58.703" LastActivityDate="2021-05-11T22:34:49.007" Title="Understanding gas stoichiometry for the reaction of xenon and fluorine" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28573" PostTypeId="2" ParentId="28572" CreationDate="2015-04-12T11:59:50.050" Score="5" Body="&lt;p&gt;Yes, this is a rounding error.&lt;/p&gt;&#xA;&lt;p&gt;Where you calculate:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The actual unrounded answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.181072885 mol}$&lt;/span&gt; - a little bit higher than the rounded value.&lt;/p&gt;&#xA;&lt;p&gt;So, in your final equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="4945" LastEditDate="2021-05-11T22:34:49.007" LastActivityDate="2021-05-11T22:34:49.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28576" PostTypeId="1" AcceptedAnswerId="28581" CreationDate="2015-04-12T14:06:36.783" Score="10" ViewCount="111957" Body="&lt;p&gt;Which of the following has higher boiling points? Alkanes, alkenes, or alkynes? And why? &lt;/p&gt;&#xA;" OwnerUserId="15525" LastEditorUserId="5026" LastEditDate="2018-02-25T08:11:52.173" LastActivityDate="2018-11-01T03:53:37.627" Title="Alkane, alkene, alkyne boiling point comparison" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28577" PostTypeId="1" AcceptedAnswerId="41094" CreationDate="2015-04-12T14:12:02.997" Score="11" ViewCount="14476" Body="&lt;p&gt;I'm wondering if there is a way to turn alcohol or another bacteria killing substance into a solid gel form, sort of like the gel you can use to clean a keyboard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MIBfB.jpg&quot; alt=&quot;Computer Cleaning Compound Super Clean Slimy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Source: Aliexpress.com &lt;a href=&quot;http://www.aliexpress.com/item/Hot-New-Keyboard-Cyber-Computer-Cleaning-Compound-Super-Clean-Slimy-Magic-Gel/587470728.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Computer Cleaning Compound&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if you could put either alcohol or other sanitizer into this sort of gel, without the sanitizer having a &quot;best before&quot; date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a complete beginner when it comes to chemistry so forgive me if there's an obvious reason why this wouldn't work. If it does work, or partially, could you give me a run down on what I would need or need to test to achieve my goal?&lt;/p&gt;&#xA;" OwnerUserId="15526" LastEditorUserId="402" LastEditDate="2015-10-12T03:55:12.723" LastActivityDate="2020-03-22T09:11:32.600" Title="Is there a way to turn alcohol into a solid gel?" Tags="&lt;solid-state-chemistry&gt;&lt;alcohols&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28578" PostTypeId="1" CreationDate="2015-04-12T14:27:54.390" Score="1" ViewCount="4834" Body="&lt;p&gt;&lt;em&gt;p&lt;/em&gt;-chlorophenol is more acidic than &lt;em&gt;p&lt;/em&gt;-fluorophenol due to presence of vacant d orbitals. But then is &lt;em&gt;p&lt;/em&gt;-bromophenol more acidic than &lt;em&gt;p&lt;/em&gt;-chlorophenol since it has vacant f orbitals also? But I guess that’s wrong, isn’t it? Please give your suggestions.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="19616" LastEditDate="2015-10-16T18:08:34.863" LastActivityDate="2019-02-27T02:21:10.430" Title="Is p-bromophenol more acidic than p-chlorophenol?" Tags="&lt;acid-base&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28579" PostTypeId="1" CreationDate="2015-04-12T14:43:27.240" Score="5" ViewCount="4727" Body="&lt;p&gt;A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{O3(g) + NO(g)-&amp;gt;O2(g) + NO2(g)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rate law for this reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{rate of reaction} = k\left[\ce{O3}\right]\left[\ce{NO}\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $k = 2.98 \times 10^6\ \mathrm{M^{–1}\cdot s^{–1}}$ at a certain temperature, calculate the initial reaction rate when $\left[\ce{O3}\right]$ and $\left[\ce{NO}\right]$ remain essentially constant at the values $\left[\ce{O3}\right]_0 = 5.34 \times 10^{–6}\ \mathrm{M}$ and $\left[\ce{NO}\right]_0 = 7.99 \times 10^{–5}\ \mathrm{M}$, owing to continuous production from separate sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have found that by using the rate of reaction equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{the rate of reaction} = $0.00127\ \mathrm{M/s}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I’m wondering how to calculate the amount of substance of $\ce{NO2(g)}$ produced per hour per liter of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, my idea is that you can take &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(1.27\times 10^{-3}\ \mathrm{M}\over\mathrm{s}\right)\left(60\ \mathrm{s}\over{1\ \mathrm{min}}\right)\left(60\ \mathrm{min}\over{1\ \mathrm{h}}\right) = \ce{NO2} \,\text{per hour per liter of air}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct way to go about it? I was thinking that the seconds and minutes would cancel leaving just $\mathrm{M/h}$ and then if you multiply by 1 liter it would be $\mathrm{M/(h\times L)}$. &lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="194" LastEditDate="2017-04-06T03:00:13.460" LastActivityDate="2017-04-06T03:00:13.460" Title="Calculate the number of moles produced per hour per liter of air?" Tags="&lt;kinetics&gt;&lt;numerical-analysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28580" PostTypeId="1" CreationDate="2015-04-12T15:51:12.750" Score="3" ViewCount="6607" Body="&lt;p&gt;How can we determine if the enthalpy of solution of a compound is positive or negative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to make a clear idea about what I mean, take the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3.5 g sample of $\ce{NH4NO3}$ was added to 80. mL of water in a coffee cup calorimeter of negligible heat capacity. As a result, the temperature of solution decreased from $21.6 \ ^\circ\text{C}$ to $18.1\ ^\circ\text{C}$. Calculate the enthalpy change of the solution ($\Delta H_{\text{sol}}$ ). (specific heat of the solution = $4.18 \frac{\text{J}}{\text{g}\cdot ^\circ\text{C}}$; density of water $1.00 \frac{\text{g}}{\text{mL}}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: $+28 \frac{\text{kJ}}{\text{mol}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why it is positive despite there is a decrease in temperature?&lt;/p&gt;&#xA;" OwnerUserId="15381" LastEditorUserId="275" LastEditDate="2015-04-12T17:03:21.187" LastActivityDate="2015-05-12T17:06:47.150" Title="How to determine whether the enthalpy of solution is positive or negative by calorimetry?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28581" PostTypeId="2" ParentId="28576" CreationDate="2015-04-12T16:02:41.597" Score="22" Body="&lt;p&gt;&lt;em&gt;Disclaimer&lt;/em&gt;: &lt;strong&gt;All of this &quot;jazz&quot; will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here.&lt;/strong&gt; :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;It's all about &lt;a href=&quot;http://en.wikipedia.org/wiki/Polarizability&quot; rel=&quot;noreferrer&quot;&gt;polarizability&lt;/a&gt;.&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Polarizability is the ability for a molecule to be polarized.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When determining (aka comparing) the boiling points of different &lt;em&gt;molecular&lt;/em&gt; substances, intermolecular forces known as &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=3&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCcQFjAC&amp;amp;url=https%3A%2F%2Fwww.chem.purdue.edu%2Fgchelp%2Fliquids%2Fdisperse.html&amp;amp;ei=eosqVbOoL4vaavrtgdAO&amp;amp;usg=AFQjCNEMaPTsbL6eJQpOZWNAR8iFFF1TiA&amp;amp;sig2=0RMrlg0sDiEn23odyik6Xw&amp;amp;bvm=bv.90491159,d.d2s&quot; rel=&quot;noreferrer&quot;&gt;London Dispersion Forces&lt;/a&gt; are at work here. Which means, these are the forces that are overcome when the boiling occurs. (See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/23459/why-do-simple-molecular-substances-have-weak-intermolecular-forces-and-why-do-gi/23462#23462&quot;&gt;here&lt;/a&gt; for example) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9IKF8.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://340454897494774470.weebly.com/uploads/1/6/2/0/16209380/8406765.jpg?395&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London forces get stronger with an increase in volume, and that's because the polarizability of the molecule increases. (See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28499/relation-between-molar-mass-and-van-der-waals-force/28504#28504&quot;&gt;the answer to this recent question&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alkanes vs. Alkenes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In their simplest form (where no substitution etc. has occurred) alkanes tend to have very close boiling points to alkenes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Boiling points of alkenes depends on  more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{array}{|c|c|}\hline&#xA;\text{Compound} &amp;amp; \text{Boiling point / }^\circ\mathrm{C} \\ \hline&#xA;\text{Ethene}	&amp;amp; -104 \\ \hline&#xA;\text{Propene}	&amp;amp; -47 \\ \hline&#xA;\textit{trans}\text{-2-Butene}	&amp;amp; 0.9 \\ \hline&#xA;\textit{cis}\text{-2-Butene} &amp;amp; 3.7 \\ \hline&#xA;\textit{trans}\text{-1,2-dichlorobutene} &amp;amp;	155 \\ \hline&#xA;\textit{cis}\text{-1,2-dichlorobutene} &amp;amp;	152 \\ \hline&#xA;\text{1-Pentene}	&amp;amp; 30 \\ \hline&#xA;\textit{trans}\text{-2-Pentene}	&amp;amp; 36 \\ \hline&#xA;\textit{cis}\text{-2-Pentene}	&amp;amp; 37 \\ \hline&#xA;\text{1-Heptene}	&amp;amp; 115 \\ \hline&#xA;\text{3-Octene}	&amp;amp; 122 \\ \hline&#xA;\text{3-Nonene}	&amp;amp; 147 \\ \hline&#xA;\text{5-Decene}	&amp;amp; 170 \\ \hline&#xA;\end{array}&#xA;  In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Alkanes vs. Alkynes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As explained, since there is a bigger volume to an alkane than its corresponding alkyne (i.e. with the same number of carbons) the alkane should have a higher boiling point. However, there's something else in play here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Alkynes, have a TRIPLE bond!&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently can think of two things that happen as a result of this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;London Dispersion Forces are in relation with distance. Usually, this relation is $r^{-6}$. (See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/23718/london-dispersion-forces-and-distance-relationship&quot;&gt;here&lt;/a&gt;) The triple bond allows two alkynes to get closer. The closer they are, the more the electron densities are polarised, and thus the stronger the forces are.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrons in $\pi$ bonds are more polarizable$^{10}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two factors overcome the slight difference of volume here. As a result, you have higher boiling points for alkynes than alkanes, generally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|}\hline&#xA;\text{Compound} &amp;amp; \text{Boiling point / }^\circ\mathrm{C} \\ \hline&#xA;\text{Ethyne}	&amp;amp; -84^{[1]} \\ \hline&#xA;\text{Propyne}	&amp;amp; -23.2^{[2]} \\ \hline&#xA;\text{2-Butyne}	&amp;amp; 27^{[3]} \\ \hline&#xA;\text{1,4-Dichloro-2-butyne} &amp;amp;	165.5^{[4]} \\ \hline&#xA;\text{1-Pentyne}	&amp;amp; 40.2^{[5]} \\ \hline&#xA;\text{2-Heptyne}	&amp;amp; 112\text{–}113^{[6]} \\ \hline&#xA;\text{3-Octyne}	&amp;amp; 133^{[7]} \\ \hline&#xA;\text{3-Nonyne}	&amp;amp; 157.1^{[8]} \\ \hline&#xA;\text{5-Decyne}	&amp;amp; 177\text{–}178^{[9]} \\ \hline&#xA;\end{array}&#xA;&lt;sub&gt;1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetylene&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Acetylene&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;2: &lt;a href=&quot;http://en.wikipedia.org/wiki/Propyne&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Propyne&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;3: &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Butyne&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/2-Butyne&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;4: &lt;a href=&quot;http://www.lookchem.com/1-4-Dichloro-2-butyne/&quot; rel=&quot;noreferrer&quot;&gt;http://www.lookchem.com/1-4-Dichloro-2-butyne/&lt;/a&gt; &lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;5: &lt;a href=&quot;http://en.wikipedia.org/wiki/1-Pentyne&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/1-Pentyne&lt;/a&gt; &lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;6: &lt;a href=&quot;http://www.chemsynthesis.com/base/chemical-structure-17405.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemsynthesis.com/base/chemical-structure-17405.html&lt;/a&gt; &lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;7: &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.76541.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemspider.com/Chemical-Structure.76541.html&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;8: &lt;a href=&quot;http://www.thegoodscentscompany.com/data/rw1120961.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.thegoodscentscompany.com/data/rw1120961.html&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;9: &lt;a href=&quot;http://www.chemsynthesis.com/base/chemical-structure-3310.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.chemsynthesis.com/base/chemical-structure-3310.html&lt;/a&gt;&lt;/sub&gt;&lt;br/&gt;&#xA;&lt;sub&gt;10: &lt;a href=&quot;https://chemistry.stackexchange.com/a/27531/5026&quot;&gt;https://chemistry.stackexchange.com/a/27531/5026&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;: We can't fully determine the boiling points of the whole class of alkanes, alkenes and alkynes. However, for the lighter hydrocarbons, comparing the boiling points, you get:&#xA;$$\text{Alkynes &amp;gt; Alkanes &amp;gt; Alkenes}$$&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5026" LastEditDate="2018-01-18T03:56:31.683" LastActivityDate="2018-01-18T03:56:31.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28582" PostTypeId="2" ParentId="28580" CreationDate="2015-04-12T16:24:34.787" Score="3" Body="&lt;p&gt;The sample takes in the heat from the water, so it takes in energy to &quot;form&quot; the solution and the water thereby loses it, becoming a lower temperature. You need to realize the reference point is not the water, but rather the sample in this case. The heat transferred (lost) by the water is to the sample, so it is the negative of the heat lost by water, making it positive.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-12T16:24:34.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28584" PostTypeId="1" AcceptedAnswerId="28586" CreationDate="2015-04-12T17:12:16.547" Score="5" ViewCount="881" Body="&lt;p&gt;I have been asked the following question in my problem sheet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/44S2h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only differemce between reaction &lt;strong&gt;A&lt;/strong&gt; and reaction &lt;strong&gt;B&lt;/strong&gt; is the electron withdrawing group that stabilises the anionic transition state changes from $CN$ in reaction &lt;strong&gt;A&lt;/strong&gt; to $NO_2$ in reaction &lt;strong&gt;B&lt;/strong&gt;. I can see no glaringly obvious reason why there should be a different rate. The problem is really (at least in my mind): what is the better electron withdrawing group out of $CN$ and $NO_2$. Is this a reasonable interpretation? If so, what reasoning can I bring to the answer to suggest that one is a better EWG than the other?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2015-04-12T23:31:33.770" LastActivityDate="2015-04-12T23:31:33.770" Title="Effect of different electron withdrawing/transition state stabilising groups in nucleophillic aromatic substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28585" PostTypeId="2" ParentId="28450" CreationDate="2015-04-12T17:16:34.210" Score="3" Body="&lt;p&gt;the volume seems to be already in the output:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Molar volume}=305.994~\mathrm{bohr^3/mol}\ (27.304~\mathrm{cm^3/mol})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, this is defined with the volume inside a density contour of $0.001~\mathrm{e/bohr^3}$.  That may be different from what you measure, because volume is not a quantity with well defined meaning, and what you find in an experiment depends on how you measure it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically any atom or molecule have a volume of infinity since the electron spreads to the entire space, but this is clearly not the physically meaning of volume as we usually know it.&lt;/p&gt;&#xA;" OwnerUserId="3723" LastEditorUserId="4945" LastEditDate="2015-04-13T03:59:50.057" LastActivityDate="2015-04-13T03:59:50.057" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28586" PostTypeId="2" ParentId="28584" CreationDate="2015-04-12T17:29:33.297" Score="6" Body="&lt;p&gt;Your reasoning is correct.  The relative reaction rates would probably be quite similar if we didn't have the methyl groups on both sides of the substituent X.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the substituent to stabilize the transition state through resonance, the substituent must be coplanar with the aromatic ring.  This isn't a problem for the linear $\ce{CN}$ group, but the $\ce{NO2}$ group with the two oxygens protruding out will not be able to remain planar with the aromatic ring.  The two adjacent methyl groups create steric interference with the planar nitro group; this steric interaction is minimized when the nitro group is twisted out of the aromatic plane.  When the nitro group is twisted out of plane it cannot resonance stabilize the transition state.  Hence the nitro compound reacts more slowly than the cyano compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the underlying reaction is different, the thinking behind this question is identical to that found in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28549/why-should-2-6-dimethyl-4-nitrophenol-be-more-acidic-than-3-5-dimethyl-4-nitroph/28550#28550&quot;&gt;this question from yesterday&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-04-12T17:29:33.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28589" PostTypeId="1" CreationDate="2015-04-12T19:09:19.060" Score="7" ViewCount="875" Body="&lt;p&gt;There are some seemingly contradictory things I’d like cleared up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Law states: $\mathrm dS_\text{univ} = \mathrm dS_\text{sys} + \mathrm dS_\text{surr}$, and $\mathrm dS_\text{univ}$ is always positive in a spontaneous process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, what does “spontaneous” truly mean? “It means the reaction occurs without any external energy input” – I’ve heard this one before, but it doesn’t make sense. In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs, though at a slower rate, than the spontaneous reaction. What is the most fundamental definition, intuitive and universally applicable definition of spontaneous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on, according to ChemWiki the Second Law states that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes. Who is right and why (explain in quantum terms, since I understand that in the long run of course the entropy of the universe increases by the laws of probability)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example to work through:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a living cell, large molecules are assembled from simple ones. Is this process consistent with the second law of thermodynamics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Book’s solution: To reconcile the operation of an order-producing cell with the second law of thermodynamics, we must remember that $\mathrm dS_\text{univ}$, not $\mathrm dS_\text{sys}$, must be positive for a process to be spontaneous. A process for which $\mathrm dS_\text{sys}$ is negative can be spontaneous if the associated $\mathrm dS_\text{surr}$ is both larger and positive. The operation of a cell is such a process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, are we to consider the operation of a cell in this problem spontaneous? As I recall from biology, life processes require ATP, the carrier of energy, so these would technically “require external energy input,” would they not? However, the book treats it as a spontaneous process. Again, this confuses me because the definition of spontaneity is inadequate as I now semi-understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for such a long post, but these are very important and fundamental ideas in spontaneity, entropy, and free energy, and I don’t want to read on until these ideas are well ingrained, and correctly so.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorDisplayName="user7951" LastEditDate="2015-08-14T18:53:39.253" LastActivityDate="2019-05-01T15:08:35.020" Title="Truly Understanding the Second Law of Thermodynamics and Entropy" Tags="&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28590" PostTypeId="1" AcceptedAnswerId="28593" CreationDate="2015-04-12T19:51:31.227" Score="1" ViewCount="132" Body="&lt;p&gt;&lt;strong&gt;Passage from my textbook:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Are reactions possible among any of the solution components; is so, what is their stoichiometry?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose that you are asked to calculate &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]$&lt;/span&gt; in a solution that is prepared to be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.10 M}~\ce{NaOH}$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.20 M}~\ce{NH4Cl}$&lt;/span&gt;. Before you answer that &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]=\pu{0.10M}$&lt;/span&gt;, consider whether a solution can be &lt;em&gt;simultaneously&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{0.10M}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.20M}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; . It cannot; any solution containing both &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; must also contain &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; react in a &lt;span class=&quot;math-container&quot;&gt;$1:1$&lt;/span&gt; mole ratio until &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is almost totally consumed:&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4+ +OH- -&amp;gt;NH3 + H2O}$$&lt;/span&gt;&#xA;You are now dealing with the buffer solution &lt;span class=&quot;math-container&quot;&gt;$\pu{0.10M}~\ce{NH3}-\pu{0.10M}~\ce{NH4+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I am having trouble understanding the third last sentence in the paragraph that starts with &amp;quot;2.&amp;quot; The statement reads: &amp;quot;It cannot; any solution containing both &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; must also contain &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; react in &lt;span class=&quot;math-container&quot;&gt;$1:1$&lt;/span&gt; mole ratio until &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is almost totally consumed&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Can someone please help me out with understanding this passage?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-24T12:10:27.883" Title="Understanding a passage related to salt buffer" Tags="&lt;acid-base&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28591" PostTypeId="2" ParentId="28578" CreationDate="2015-04-12T19:52:34.267" Score="2" Body="&lt;p&gt;Disclaimer: This is not necessarily the &quot;correct&quot; answer as many factors play roles and it would be difficult to describe the observation so simply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I can provide an easily &quot;correct&quot; answer, but I can try to explain some of the many factors that influence the acidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most notably, the factors we should consider are inductive effects and delocalization of charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inductive effects, based on electronegativity, would dismiss the experimental observation, so I'll look at delocalization of charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The delocalization of charge, which introduces a negative charge by the halogen is greatest by F&gt;Cl&gt;Br&gt;I. This would reduce acidity, as a more negative benzene ring would not desire to gain another negative charge. Thus, this is one factor that supports the experimental observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we may be able to explain the observation, if only partially, by the delocalization of charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I adapted the answer to a similar question on &lt;a href=&quot;http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why&quot; rel=&quot;nofollow&quot;&gt;http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why&lt;/a&gt; . I would have preferred more sources but I could not find a very good source.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-12T19:52:34.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28592" PostTypeId="1" CreationDate="2015-04-12T19:52:52.810" Score="2" ViewCount="119" Body="&lt;p&gt;This question is simpler than the previous one about intuition of entropy; I just need to know whether the $T$ term in $dS_{Surr} = \frac{-dH}{T}$ refers to the temperature of the surroundings or the system. I think it needs to be both, no? For example, for water to boil, the system must have $T \geq$ 100 degrees C. But can it be said that at high temperatures the decrease in entropy of the surroundings due to endothermicity of the boiling reaction is less significant than at low temperatures, as per division by T, so at high temperatures of the surroundings boiling occurs spontaneously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like both explain the phenomenon, especially since boiling occurs over time in both a sample of water at 110 degrees C in a 32 degrees C environment until it cools and boiling eventually occurs in a 110 degree C furnace if the temperature of the water starts at 32 degrees C.&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="275" LastEditDate="2015-04-12T20:25:43.480" LastActivityDate="2020-05-28T01:20:39.553" Title="In entropy change of surroundings, what is T?" Tags="&lt;energy&gt;&lt;entropy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28593" PostTypeId="2" ParentId="28590" CreationDate="2015-04-12T19:55:33.290" Score="5" Body="&lt;p&gt;There is no such thing as  $\ce{NH4OH}$ . They would react to form water and ammonia. Thus, to say that both exist in the solution at such high concentrations would be mistaken. They would react. Consider the reaction between ammonium and hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH4+ + OH- -&amp;gt; NH3 + H2O}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, do note that at low concentrations, they can exist according to dissociation. However that is not what your passage is saying&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*On your test, it may be good to assume the reaction goes to completion and then solve for the dissociation of ammonia. That is most likely what is expected of you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I thought of a more understandable analogy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about $\ce{OH-}$ and $\ce{H+}$ in solution at 25 celsius. If one increases, the other decreases! According to ${Kw}$. Adding a lot of acid and base together would simply neutralize, they cannot exist at any higher than ${Kw}$&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="14975" LastEditDate="2015-04-12T20:10:06.203" LastActivityDate="2015-04-12T20:10:06.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28594" PostTypeId="2" ParentId="27357" CreationDate="2015-04-12T20:02:41.570" Score="2" Body="&lt;p&gt;It's because of the activity of hydrogen ions in comparison to the metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certain metals like $\ce{Mg}$, say, will be oxidized by the hydrogen ions. Note that $\ce{Mg^2+}$ has a reduction potential of -2.37V, which is less than $\ce{H+}$'s reduction potential of 0V (not accounting for overvoltage and what not). Hence, $\ce{H2}$ gas is produced spontaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, metals like $\ce{Cu}$ (which is the example given by your Wikipedia link) and $\ce{Ag}$ have a reduction potential greater than 0V. This means that H+ will no longer be able to oxidize these metals. Instead, the $\ce{NO3-}$ ion will serve as the oxidizing agent. &lt;/p&gt;&#xA;" OwnerUserId="15531" LastEditorDisplayName="user15489" LastEditDate="2015-04-17T05:13:05.053" LastActivityDate="2015-04-17T05:13:05.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28595" PostTypeId="1" CreationDate="2015-04-12T20:16:23.410" Score="5" ViewCount="311" Body="&lt;p&gt;The problem asks to determine the &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt; in a &lt;span class=&quot;math-container&quot;&gt;$0.20~\mathrm{M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na3PO4}$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HPO4-}$&lt;/span&gt; was given as &lt;span class=&quot;math-container&quot;&gt;$4.5\times 10^{-13}$&lt;/span&gt;, which then allows one to calculate the corresponding &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{b}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$2.22\times10^{-2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As any acid-base problem, I simply set up the expression &lt;span class=&quot;math-container&quot;&gt;$\frac{x^2}{0.20~\mathrm{M}}=2.22 \times 10^{-2}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; being &lt;span class=&quot;math-container&quot;&gt;$[\ce{OH-}]$&lt;/span&gt; and solved for &lt;span class=&quot;math-container&quot;&gt;$[\ce{H+}]$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution involves an interesting iteration which I have never seen before for this type of problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y0Gjl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone shed some light on this?&lt;/p&gt;&#xA;" OwnerUserId="15531" LastEditorUserId="16683" LastEditDate="2019-08-20T13:38:05.630" LastActivityDate="2019-08-20T13:38:05.630" Title="Iterative method for calculating pH of a weak acid / base" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28597" PostTypeId="2" ParentId="27294" CreationDate="2015-04-12T20:51:43.187" Score="3" Body="&lt;p&gt;I be mistaken, but are you referring to vapor pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vapor pressure  or equilibrium vapour pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. (&lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Vapor_pressure&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a closed system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A system in which no mass is transferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, vapor pressure would not be exerted if the container is open and the liquid is constantly diffusing. It would only occur in a closed system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, liquids DO diffuse (evaporate) in air, rather than exert vapor pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can leave a bottle of water, one with a lid and one without. The one without the lid will soon be without water, but the one with a lid would remain with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Vapor pressure does relate to evaporation rate, which occurs in the open systems we normally observe in life&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-12T20:51:43.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28598" PostTypeId="2" ParentId="6489" CreationDate="2015-04-12T21:08:58.943" Score="-1" Body="&lt;p&gt;The process is quite simple. .. to determine CFSE  we need to know that the diffidence of energy levels of two sets have been arbiter taken as 10 dq. . Where t2g set is more stable then eg set octahedral and vice versa in tetrahedral complex. &#xA;-0.4x + 0.6y will help to calculate it.&#xA;X indicates electrons in t2g and y in eg. &#xA;Sum up all and find the diffidence ... &#xA;That gives cfsc&#xA;Usually tetrahedral splitting is shown in terms of octahedral as&#xA;4/9 OS&lt;/p&gt;&#xA;" OwnerUserId="15532" LastActivityDate="2015-04-12T21:08:58.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28602" PostTypeId="1" CreationDate="2015-04-12T23:54:12.697" Score="5" ViewCount="2434" Body="&lt;p&gt;Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol? I realize that NaOH generally does not deprotonate benzyl alcohol, but we performed an extraction in a Fischer esterification lab with benzyl alcohol an acetic acid to synthesize benzyl acetate. Acetic acid was added in excess, and the final ester product was purified with extraction techniques. I know that addition of NaOH removes acetic acid from the organic, diethyl ether layer, but would 1.5M NaOH also remove any remaining benzyl alcohol?&lt;/p&gt;&#xA;" OwnerUserId="15536" LastEditorUserId="12860" LastEditDate="2015-11-13T19:05:32.613" LastActivityDate="2017-08-07T02:10:39.087" Title="Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol?" Tags="&lt;acid-base&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28603" PostTypeId="2" ParentId="28589" CreationDate="2015-04-13T00:42:21.810" Score="1" Body="&lt;p&gt;Be careful in citing equilibrium as an example, as it &lt;em&gt;fully&lt;/em&gt; integrates entropy in its definition. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy#Entropy_in_chemical_thermodynamics&quot; rel=&quot;nofollow&quot; title=&quot;this&quot;&gt;this&lt;/a&gt; article for the details, but in a spontaneous reaction the entropy &lt;em&gt;is increasing&lt;/em&gt;, and part of reaching equilibrium is that the energy gain and loss among products and reactants is balanced. Also, most experimental reactions are (nearly) closed systems for short periods of time, and if no energy is gained or lost (again, for short periods), you can have an equilibrium where there is no change in entropy. If you let your experiment sit and the heat of the system changes (for example), the equilibrium will change as well. And eventually you end your experiment, clean out your flask, and thus increase the entropy of the universe again!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@MARamezani's comment pointing to the article on Biology looks like a good answer to your biology questions. Earth is most definitely &lt;em&gt;not&lt;/em&gt; a close system with that giant energy source 93 million miles away.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-04-13T00:42:21.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28604" PostTypeId="1" AcceptedAnswerId="69956" CreationDate="2015-04-13T01:03:06.617" Score="3" ViewCount="1439" Body="&lt;p&gt;I'd like to accurately calibrate a digital thermometer I'm building in the range of $50^\circ$ to $120^\circ$ Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that one can buy various chemicals that have melting points in this range, but are there any readily available household / pharmaceutical items I could use instead? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have read, it looks like the two easiest candidates would be vanillin or Sodium acetate (Thanks Curt F.), but I'm not sure how one would go about creating a pure enough calibration standard at home.&lt;/p&gt;&#xA;" OwnerUserId="1754" LastEditorUserId="39539" LastEditDate="2017-03-08T03:23:00.450" LastActivityDate="2017-03-08T03:23:00.450" Title="Household Melting Point Standards" Tags="&lt;physical-chemistry&gt;&lt;melting-point&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28605" PostTypeId="1" CreationDate="2015-04-13T02:17:57.443" Score="4" ViewCount="1277" Body="&lt;p&gt;What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, $\ce{CH3NH2}$,&#xA;with 25.00 mL of 0.10 M methylammonium chloride, $\ce{CH3NH3Cl}$? Assume that the volume of&#xA;the solutions are additive and that K&lt;sub&gt;b&lt;/sub&gt; = 3.70 × 10&lt;sup&gt;-4&lt;/sup&gt; for methylamine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) First, I multiplied 0.025 L with 0.10 M and received 0.0025 moles. I then divided 0.0025 moles with 0.050 L (The total volume of the solution) and received 0.05 M, the molar concentration of the methylamine and its conjugate acid in this mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I then created an ICE Chart (my reactant was the methylamine so that I would not need to find the K&lt;sub&gt;a&lt;/sub&gt;) and I solved for &quot;x&quot; (which would give me the concentration of $\ce{OH}$ ions at equilibrium). First, I solved for &quot;x&quot; but I assumed that is was insignificant. In the end I received 10.57 as the pH, which was the wrong answer. Then when I solved for &quot;x&quot; using the quadratic formula (assuming that &quot;x&quot; was significant), I got 0.0003646 as my &quot;x&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) When I took the negative log of 0.0003646, I got a pOH of 3.46. I subtracted this number by 14, and I still received a pH of 10.57&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I even attempted to use the Henderson Hasselbalch equation and still received 10.57 because the ratio of the base and acid is 1, meaning that the log is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what I am doing incorrect. The answer is supposed to be 10.87, can someone please assist me?&lt;/p&gt;&#xA;" OwnerUserId="15540" LastEditorUserId="13339" LastEditDate="2015-05-25T11:27:12.490" LastActivityDate="2015-05-25T11:27:12.490" Title="Finding pH of a buffer" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28607" PostTypeId="2" ParentId="28562" CreationDate="2015-04-13T04:44:11.103" Score="1" Body="&lt;p&gt;Yes, you need to verify that the housing of the battery is made of a material that is unaffected by its contents. Polycarbonate is a good material for rigidity and transparency, but may be dissolved or weakened over time by organic solvents (carbonate esters or otherwise) and also by pH extremes and high temperatures that could come into play when the battery is charged and discharged. Thermosetting resins might be a better bet for this type of application.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-13T04:44:11.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28608" PostTypeId="1" AcceptedAnswerId="28645" CreationDate="2015-04-13T05:00:03.810" Score="12" ViewCount="1836" Body="&lt;p&gt;I am curious as to why compounds with inert gases, such as $\ce{XeF4}$, $\ce{XeF2}$, and $\ce{XeO3}$ are considered powerful oxidizing agents. I would attribute the phenomenon to the highly oxidized state of the noble gas which would be unstable but I am not too sure about it. &lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="4945" LastEditDate="2015-04-13T05:15:37.277" LastActivityDate="2015-04-14T17:19:46.297" Title="Why are inert gas (especially Xenon) compounds powerful oxidizing agents?" Tags="&lt;stability&gt;&lt;periodic-trends&gt;&lt;oxidation-state&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28609" PostTypeId="1" CreationDate="2015-04-13T05:34:16.323" Score="0" ViewCount="146" Body="&lt;p&gt;Would it be feasible to jury-rig an automotive catalytic converter for use in the Ostwald Process for the production of Nitric Acid? Given reasonable safety precautions, would it be safe enough to be practical for a home lab? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Catalytic converter is a device used in cars to reduce Nitric oxide into nitrogen and Oxygen, as well as oxidizing Carbon monoxide and unburnt hydrocarbons into Co2 and H20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm proposing is feeding ammonia and oxygen into it at an appropriate temperature and pressure to produce the Nitric Oxide necessary to create Further oxidize into Nitrogen Dioxide, which is reacted with water to produce Nitric acid and Nitric Oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 NH&lt;sub&gt;3 (g)&lt;/sub&gt; + 5 O&lt;sub&gt;2 (g)&lt;/sub&gt; → 4 NO &lt;sub&gt;(g)&lt;/sub&gt; + 6 H&lt;sub&gt;2&lt;/sub&gt;O &lt;sub&gt;(g)&lt;/sub&gt; (ΔH = −905.2 kJ)&lt;br&gt;&#xA;2 NO &lt;sub&gt;(g)&lt;/sub&gt; + O&lt;sub&gt;2 (g)&lt;/sub&gt; → 2 NO&lt;sub&gt;2 (g)&lt;/sub&gt; (ΔH = −114 kJ/mol)&lt;br&gt;&#xA;3 NO&lt;sub&gt;2 (g)&lt;/sub&gt; + H&lt;sub&gt;2&lt;/sub&gt;O &lt;sub&gt;(l)&lt;/sub&gt; → 2 HNO&lt;sub&gt;3 (aq)&lt;/sub&gt; + NO &lt;sub&gt;(g)&lt;/sub&gt; (ΔH = −117 kJ/mol)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my Question is whether a Catalytic Converter salvaged from an automobile is a suitable substitute for a column packed with Platinum gauze.&lt;/p&gt;&#xA;" OwnerUserId="15543" LastEditorUserId="15543" LastEditDate="2015-04-13T17:42:26.413" LastActivityDate="2015-04-13T17:42:26.413" Title="jury-rigging an automotive catalytic converter for use in the Ostwald Process?" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28610" PostTypeId="1" CreationDate="2015-04-13T05:34:29.917" Score="4" ViewCount="6013" Body="&lt;p&gt;I have two questions regarding chemical shifts on aromatic systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what does being near an aromatic system do to protons? If we look at the spectra of 1-napthalenemethanol and then 9-anthracenemethanol, the peak of the two aliphatic protons of the benzylic methylene group get shifted way up. Can someone explain why adding in that third aromatic ring does this? &#xA;The spectra:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_1468-95-7_1HNMR.htm&quot; rel=&quot;nofollow&quot;&gt;9-anthracenemethanol&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_4780-79-4_1HNMR.htm&quot; rel=&quot;nofollow&quot;&gt;1-naphtalenemethanol&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was also under the impression that protons near electronegative elements would be deshielded and be shifted higher. However in the spectrum of Benzil, the protons para to the carbonyl group are higher shifted than the protons meta to the carbonyl. Why is this? Is it to do with conjugation of the carbonyl with the aromatic ring? I'm having trouble finding a proton nmr in the literature, so there's nothing to compare it to. &lt;/p&gt;&#xA;" OwnerUserId="15542" LastEditorUserId="4945" LastEditDate="2015-04-13T06:38:51.440" LastActivityDate="2015-11-14T16:15:08.030" Title="What happens to chemical shifts of protons close to aromatic systems in ¹H-NMR?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28613" PostTypeId="2" ParentId="24444" CreationDate="2015-04-13T07:40:09.780" Score="1" Body="&lt;p&gt;I agree with the other answers, it is definitely not HDPE, however, it could be one of the chemicals known as plasticizers that are used in the process. There is always leftover plasticizer in the bottle and these are known to leach into water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NIH sponsored multiple studies on the levels of this, one of them published here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/21400322&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/21400322&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-13T07:40:09.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28614" PostTypeId="2" ParentId="28610" CreationDate="2015-04-13T08:45:10.863" Score="3" Body="&lt;p&gt;The aromaticity found in certain ring systems (e.g. benzene, see &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel&amp;#39;s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hückel's rule&lt;/a&gt;) causes a magnetic field because of the circular motion of the electrons in the ring (see picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJ7kQ.png&quot; alt=&quot;Clayden et al. Organic Chemistry 2nd edition, p. 277&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Clayden et al. Organic Chemistry 2nd edition, p. 277)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This additional field enforces the magnetic field of the NMR machine inside the ring, while countering the field outside the ring, causing the signals of the outer hydrogens to shift downfield (i.e. to the larger shifts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your other questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, what does being near an aromatic system do to protons? If we&#xA;  look at the spectra of 1-napthalenemethanol and then&#xA;  9-anthracenemethanol, the peak of the two aliphatic protons of the&#xA;  benzylic methylene group get shifted way up. Can someone explain why&#xA;  adding in that third aromatic ring does this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lookig at the difference in shift between the benzylic methylene protons of 1-Naphthalenemethanol and 9-Anthracenemethanol we must conclude that the induced magnetic field of the second is stronger at that position than that of the former, as there is no other appreciable difference. Intuitively this seems to make sense as the field of the 9-Anthracenemethanol is more uniform at and closer to (with respect to the &quot;right&quot; side of the molecule as in the pictures) the discussed proton. Of course this could/should be confirmed by quantum mechanical calculations, but I do not have the tools or the knowledge to do that (if someone has, please feel free to expand this answer!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJUsJ.png&quot; alt=&quot;1-Naphthalenemethanol Proton NMR&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_4780-79-4_1HNMR.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-Naphthalenemethanol Proton NMR&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/L7k0O.png&quot; alt=&quot;9-Anthracenemethanol Proton NMR&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_1468-95-7_1HNMR.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;9-Anthracenemethanol Proton NMR&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was also under the impression that protons near electronegative&#xA;  elements would be deshielded and be shifted higher. However in the&#xA;  spectrum of Benzil, the protons para to the carbonyl group are higher&#xA;  shifted than the protons meta to the carbonyl. Why is this? Is it to&#xA;  do with conjugation of the carbonyl with the aromatic ring? I'm having&#xA;  trouble finding a proton nmr in the literature, so there's nothing to&#xA;  compare it to.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, conjugating effects and inductive effects can counter each other, but there is no easy general way to say which will win out in a particular situation, although usually it is the conjugating effect.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-13T08:45:10.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28615" PostTypeId="1" AcceptedAnswerId="28616" CreationDate="2015-04-13T08:54:34.653" Score="3" ViewCount="9879" Body="&lt;p&gt;In determining the empirical formula of a compound which constitutes a diatomic molecule, should we calculate the molar mass of the divalent molecule by multiplying the molar mass of the element by 2, or should we leave it as it is. Because, I happened to come across a problem, where I had to calculate the empirical formula for an oxide of iron. A similar question has been answered before, but it doesn't seem to clear my doubt - &lt;a href=&quot;https://chemistry.stackexchange.com/questions/26848/why-not-to-consider-hydrogen-and-oxygen-moles-to-determine-an-empirical-formula&quot;&gt;Why not to consider hydrogen and oxygen moles to determine an empirical formula&lt;/a&gt;. Hence, I restate it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an oxide of Iron has $69.9\%$ iron and $30.1\%$ dioxygen by mass, what is the empirical formula?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if I assume the mass of compound to be $100~\mathrm{g}$, then it contains $69.9~\mathrm{g}$ of iron and $30.1~\mathrm{g}$ of oxygen. If I calculate the no. of moles,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Moles of }\ce{Fe} = \frac{\text{given  mass}}{\text{molar mass}} = 69.9/58.5 = 1.25~\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles of oxygen - this is where the problem comes. If I take molar mass of oxygen as $32~\mathrm{g}$, as in $\ce{O2}$, I get $0.94~\mathrm{mol}$. It will not make any sense, as the ratio of iron to oxygen will not be in any reasonable quantity. However, if I take the  molar mass to be $16~\mathrm{g}$, I get $1.88~\mathrm{mol}$ of oxygen, and I can divide both the numbers by $1.25$ to get the ratio as $1.88/1.25: 1.88/1.88 = 1:1.5 = 2:3$,&#xA;hence the compound will be $\ce{Fe2O3}$, which makes perfect sense. But the problem is, why should I calculate the no. of moles from the molar mass of $\ce{O}$ rather than $\ce{O2}$? After all, its is the oxygen molecule that will exist, and no the oxygen atom right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the solution is given as $\ce{Fe2O3}$, so I need to take the molar mass of oxygen to calculate the number of moles as $16~\mathrm{g}$. The question is, Why? Why not $32~\mathrm{g}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have something to with the fact that the question states &quot;&lt;em&gt;dioxygen&lt;/em&gt;&quot; rather than &quot;oxygen&quot;? If yes, what is the difference between dioxygen being present in $30.1\%$ or oxygen present in $30.1\%$? I am in a real mess.&lt;/p&gt;&#xA;" OwnerUserId="15547" LastEditorUserId="4945" LastEditDate="2017-11-27T12:57:42.393" LastActivityDate="2021-05-28T06:19:39.273" Title="Do I have to consider the molecular mass of the oxygen atom or the diatomic oxygen molecule when determining the empirical formula of an iron oxide?" Tags="&lt;physical-chemistry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28616" PostTypeId="2" ParentId="28615" CreationDate="2015-04-13T09:36:40.510" Score="6" Body="&lt;p&gt;When determining the empirical formula of a compound, you only consider the molar mass of the element. In the compound itself, it is only the atom (or ion) that exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would assume the molar mass of dioxygen, then you would also assume, that the iron oxide is made out of iron and dioxygen molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know, and also do not understand, why the question states &quot;&lt;em&gt;dioxygen&lt;/em&gt;&quot; as that is - in my opinion - false.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-04-13T09:36:40.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28619" PostTypeId="1" AcceptedAnswerId="28628" CreationDate="2015-04-13T10:06:04.950" Score="1" ViewCount="115" Body="&lt;p&gt;I am having some trouble with understanding the NOE. The carbon NMR spectra can't be integrated, because the protons attached to the carbon atoms will make primary carbon atom relaxation faster than tertiary, &lt;strong&gt;so their peaks will have bigger intensity&lt;/strong&gt;. With a bigger acquisition time, the peaks will be more proportional. In proton NMR, nearby protons will cause the target proton relaxation to be faster. When NOE is applied, &lt;strong&gt;those nearby protons will be irradiated and the target proton peak will have a slightly bigger intensity&lt;/strong&gt;. Why will it have a bigger intensity?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-04-13T14:47:35.010" Title="The nuclear Overhauser effect" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28621" PostTypeId="1" CreationDate="2015-04-13T11:27:14.107" Score="1" ViewCount="67" Body="&lt;p&gt;How do I get an accurate surface area from a nitrogen isotherm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been doing some surface areas on MOFs and upon showing my boss the isotherms he asked me to convert them to show the &quot;range&quot;, which will give the most accurate surface area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using Micromeritics ASAP-2020 software and I believe that you measure this BET range by using points from relative pressures between 0.05 and 0.3. However, this seems to be unsuitable for some microporous compounds?&lt;/p&gt;&#xA;" OwnerUserId="15551" LastEditorUserId="15551" LastEditDate="2015-04-13T13:15:50.090" LastActivityDate="2015-04-13T13:15:50.090" Title="How to calculate &quot;accurate&quot; surface area of a microporous material using Micromeritics equipment?" Tags="&lt;nanotechnology&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28623" PostTypeId="1" CreationDate="2015-04-13T13:05:14.467" Score="8" ViewCount="2058" Body="&lt;p&gt;It is violet in trichloromethane, but brown in water. why? In the Shriver Atkins book, they tried to explain this using &lt;em&gt;Molecular Orbital Theory&lt;/em&gt; which was not clear to me&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="13184" LastEditDate="2015-07-17T07:57:14.283" LastActivityDate="2016-05-13T19:06:16.203" Title="Why does iodine have different colors in different media?" Tags="&lt;molecular-orbital-theory&gt;&lt;color&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28624" PostTypeId="1" CreationDate="2015-04-13T14:06:26.190" Score="2" ViewCount="468" Body="&lt;p&gt;Are the following processes exothermic or endothermic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) When solid $KBr$ is dissolved in water, the solution gets colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Natural gas ($CH_4$) is burned in a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) When concentrated $H_2SO_4$ is added to water, the solution gets very hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Water is boiled in a teakettle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; For a), I would say that it is exothermic, since the solution gets colder, which means that it releases contained heat to the surroundings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) is exothermic evidently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) By the same reasoning, the solution is getting very hot, so it must be endothermic since it absorbs heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Not sure about this one. I would say endothermic, because the absorbed heat makes it boil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I answered, and apparently I made atleast two mistakes. So can someone explain to me the flaws in my reasoning? &lt;/p&gt;&#xA;" OwnerUserId="15491" LastActivityDate="2015-04-13T14:58:43.803" Title="Exo- or endothermic? Clarification required regarding a few cases" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28625" PostTypeId="1" CreationDate="2015-04-13T14:13:05.917" Score="2" ViewCount="53" Body="&lt;p&gt;Many metals form compounds with s-block elements that are essentially entirely ionic. However, elements like silver, copper, lithium and so on, can form bonds that are significantly covalent in character. What are the factors that govern whether a particular metal will bond with significant covalent character? Is this linked to hardness and softness?    &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-13T14:13:05.917" Title="What makes a metal more inclined to bond covalently?" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2015-04-14T00:54:31.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="28626" PostTypeId="2" ParentId="28548" CreationDate="2015-04-13T14:31:25.380" Score="1" Body="&lt;p&gt;Like many &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28399/why-does-water-evaporate-spontaneously-at-room-temperature-despite-dg-0&quot;&gt;past chemistry.se questions&lt;/a&gt;, the answer to this one hinges on the presence of the $^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H^\circ_\text{c}$ is the $\Delta H$ value for a &quot;standard&quot; reaction where acetylene at 1 atm of pressure is combusted isothermally at 298 K with 1 atm of oxygen gas, forming 1 atm of $\ce{CO2}$ and &lt;em&gt;liquid&lt;/em&gt; water (which is the most thermodynamically stable form of water at 1 atm and 298 K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's safe to say that no combusion of acetylene, ever in the history of the world, has occured under those conditions.  But those &quot;standard&quot; conditions are the only conditions where the  $\Delta H^{\circ}_f$ is guaranteed to be -227 kJ/mol.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, in practical and engineering literatures, $\Delta H_c^{\circ}$ of a fuel is commonly called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_of_combustion#Higher_heating_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;higher heating value&lt;/em&gt;&lt;/a&gt; of the fuel.  It corresponds to $\Delta H_{rxn}^{\circ}$ of a reaction like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2H2(g) + 5/2 O2(g) -&amp;gt; 2CO2(g) + H2O(l)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_of_combustion#Lower_heating_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;lower heating value&lt;/em&gt;&lt;/a&gt;, conversely, corresponds to $\Delta H_{rxn}^{\circ}$ of reactions like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2H2(g) + 5/2 O2(g) -&amp;gt; 2CO2(g) + H2O(g)}$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-04-13T14:31:25.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28627" PostTypeId="1" AcceptedAnswerId="28633" CreationDate="2015-04-13T14:35:06.617" Score="3" ViewCount="2743" Body="&lt;p&gt;The second step of Djerassi's synthesis of 19-nor-17α-ethynyltestosterone (norethindrone), involves the protection of the ketone in position 3 via enolether formation, as in the picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6fcXI.png&quot; alt=&quot;synthesis step&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism of this reaction? I couldn't find it anywhere. I suppose it should be similar to an ether synthesis, but I'm not sure about the precise mechanism.&#xA;(On a side note, I assume that the other ketone in 17 remains intact because of controlled reaction conditions.)&lt;/p&gt;&#xA;" OwnerUserId="15553" LastEditorUserId="10185" LastEditDate="2015-04-13T16:27:45.500" LastActivityDate="2015-04-13T16:37:50.393" Title="Mechanism of enolether formation with ethyl orthoformate in estr-4-ene-3,17-dione (Djerassi's norethisterone synthesis)" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28628" PostTypeId="2" ParentId="28619" CreationDate="2015-04-13T14:47:35.010" Score="3" Body="&lt;p&gt;Consider the simplest possible NOE experiment, with two dipole-coupled protons, the &#xA;target I and nearby proton S. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha_{I}\alpha_{S}: 1 + 2\Delta $&lt;/li&gt;&#xA;&lt;li&gt;$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$&lt;/li&gt;&#xA;&lt;li&gt;$\beta_{I}\beta_{S}: 1 - 2\Delta $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $&lt;/li&gt;&#xA;&lt;li&gt;$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What happens next typically for small molecules is fast &quot;flip-flip&quot; cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\alpha_{I}\alpha_{S}: 1 + 2\Delta $&lt;/li&gt;&#xA;&lt;li&gt;$\beta_{I}\alpha_{S}: 1 + \Delta$&lt;/li&gt;&#xA;&lt;li&gt;$\alpha_{I}\beta_{S}: 1 - \Delta $&lt;/li&gt;&#xA;&lt;li&gt;$\beta_{I}\beta_{S}: 1 - 2\Delta $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-13T14:47:35.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28629" PostTypeId="1" CreationDate="2015-04-13T14:58:32.097" Score="2" ViewCount="688" Body="&lt;p&gt;Solvent leveling means that strong acids completely dissociate in that solvent so it is impossible to distinguish between two acids that completely dissociate in it. A solvent levels and acid if the $K_\text{a}$ of that acid is greater than 1. However, that’s not completely dissociated. An acid can have a $K_\text{a}$ of 2, 3 or 25 what’s the problem with being able to distinguish between them? In truth, whatever the $K_\text{a}$ it’s not actually COMPLETELY dissociated.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-04-13T17:07:47.933" LastActivityDate="2015-04-13T17:08:25.517" Title="Why does solvent leveling occur?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28630" PostTypeId="2" ParentId="28624" CreationDate="2015-04-13T14:58:43.803" Score="6" Body="&lt;p&gt;I think you are confusing terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;exothermic&lt;/em&gt; means that it releases heat, that is energy, because the energy of the products is lower than the one of the reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;endothermic&lt;/em&gt; means that it absorbs heat (energy) because the products are at a higher level of energy than the reagents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a solution becomes hot, it's because it's &lt;em&gt;releasing&lt;/em&gt; heat (energy), and likewise if it gets cold it's because the reagents inside are &lt;em&gt;absorbing&lt;/em&gt; energy from the surronding environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of boling water: does it need heat to become steam? Or does it release heat? In other words, do you have to &lt;em&gt;provide&lt;/em&gt; heat to the teakettle to make the water boil or it provides heat by itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can get to your answers by yourself now.&lt;/p&gt;&#xA;" OwnerUserId="15553" LastActivityDate="2015-04-13T14:58:43.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28633" PostTypeId="2" ParentId="28627" CreationDate="2015-04-13T16:37:50.393" Score="3" Body="&lt;p&gt;Usually a catalytic amount of acid is used in these reactions and they are run in the presence of alcohol (ethanol if you want to prepare the ethyl enol ether). Pyridine is typically added at the end of the reaction to recapture the acid catalyst.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step involves protonation of the carbonyl oxygen.  Protonation of the C-3 oxygen produces a resonance stabilized carbocation where the charge is delocalized over the carbonyl &lt;strong&gt;and&lt;/strong&gt; the adjacent double bond (protonation at the C-17 oxygen does not have an adjacent double bond,  hence protonation at the C-3 oxygen is favored).  The nucleophilic alcohol oxygen reacts at the carbonyl carbon (highest positive charge density) to produce a hemiketal.  Loss of water gives rise to the enol ether product.  The eliminated water is captured by the orthoformate and generates more alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9CGIA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HC(OEt)_3 + H2O -&amp;gt; HCO2Et + 2EtOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some argue that the reaction actually proceeds through the ketal (rather than the hemiketal) since ketals can be isolated in these reactions.  Others argue that just because the ketal is formed, doesn't prove it is actually involved in the enol ether formation reaction.  IMO, lacking definitive evidence, I opt for the simpler hemiketal mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-13T16:37:50.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28634" PostTypeId="1" AcceptedAnswerId="28749" CreationDate="2015-04-13T16:57:09.380" Score="1" ViewCount="737" Body="&lt;p&gt;When recharging a lead-acid battery, how are the terminals of the battery connected to the voltage source. I mean, do we connect the positive terminal of the battery to the negative terminal of the source and vice versa? Or does the positive go with the positive and the negative with the negative? Also, are the electrodes reversed while the battery is being recharged?&lt;/p&gt;&#xA;" OwnerUserId="15507" LastEditorUserId="15507" LastEditDate="2015-04-13T17:11:06.543" LastActivityDate="2015-04-15T11:48:10.093" Title="How is a car battery recharged?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28635" PostTypeId="1" AcceptedAnswerId="28639" CreationDate="2015-04-13T17:03:02.487" Score="5" ViewCount="5009" Body="&lt;p&gt;I'm having a doubt on the mechanism of oxymercuration Demercuration reaction. &#xA;How does the reaction, initially heading towards anti addition(after cyclic transition state formation) ,get spoiled by $\ce{NaBH4}$ and ends up being stereorandom? Can anyone point out the importance of using mercury as well? The explanation would be greatly helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="15558" LastEditorUserId="418" LastEditDate="2016-10-19T15:50:53.667" LastActivityDate="2016-10-19T15:50:53.667" Title="Why does the Oxymercuration Demercuration mechanism lead to stereorandomization?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28636" PostTypeId="2" ParentId="28629" CreationDate="2015-04-13T17:08:25.517" Score="4" Body="&lt;p&gt;Your final statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in truth, whatever the $K_a$ it's not actually COMPLETELY dissociated&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is exactly right.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong acid is very often stated to be &quot;completely dissociated&quot;, but that's just wrong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, nitric acid has a $K_a$ of 24, so 1M nitric acid about 96% dissociated. If the concentration is higher, it will be less dissociated and if the concentration is lower it will be more dissociated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think of the solvent leveling effect this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have 1 liter of 0.1M ethanol in water and you add 0.1 moles of sodium amide thinking that it will deprotonate the ethanol.  That's not going to happen because most of the protons that protonate the amide will come from the solvent (water) rather than the ethanol.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-04-13T17:08:25.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28637" PostTypeId="1" CreationDate="2015-04-13T17:42:55.087" Score="3" ViewCount="2677" Body="&lt;p&gt;Let's take a simple Daniell cell for example: &lt;img src=&quot;https://i.stack.imgur.com/DRPk1.jpg&quot; alt=&quot;Daniell cell&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the energy required to drive electrons through the external circuit and power up devices (like a light bulb) come from? Does the oxidation of zinc release that energy? And, does the reduction of copper result in more energy being released? (since we're calculating the cell potential by adding up the oxidation and reduction potentials of the two half cells). I've always wondered about this question, and I'd really appreciate a clear and concise explanation regarding this particular topic.  &lt;/p&gt;&#xA;" OwnerUserId="15507" LastActivityDate="2015-04-14T06:23:06.867" Title="What's the source of electrical energy in galvanic cells?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28638" PostTypeId="2" ParentId="28637" CreationDate="2015-04-13T18:03:38.937" Score="0" Body="&lt;p&gt;The cell is an electrochemical cell that derives electrical energy from &lt;em&gt;spontaneous redox reactions&lt;/em&gt; taking place within the cell. In other words, the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell. When metal B has a greater electronegativity than metal A, then metal B tends to steal electrons from metal A (that is, metal B tends to oxidize metal A), thus favoring one direction of the respective reaction.&lt;/p&gt;&#xA;" OwnerUserId="15429" LastActivityDate="2015-04-13T18:03:38.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28639" PostTypeId="2" ParentId="28635" CreationDate="2015-04-13T18:19:36.297" Score="9" Body="&lt;p&gt;The stereorandomization occurs in the formation of the cyclic mercurinium ion intermediate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TBTUq.png&quot; alt=&quot;formations of chiral mercurinium ion&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the mercury acetate approaches the alkene, it can do so from either the top face:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wHOz7.png&quot; alt=&quot;mercury approach from the top face&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the bottom face:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w67uA.png&quot; alt=&quot;mercury approach from the bottom face&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the initial alkene is achiral and the steric factors are the same for both approaches, we get both ions, which are enantiomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each enantiomer reacts stereospecifically with water to produce a chiral product. Note that the position where the mercury sits is still a chirality center in this example, but in many simpler cases it is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7iyt.png&quot; alt=&quot;attack of water on one enantiomer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X7g96.png&quot; alt=&quot;attach of water on other enantiomer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $\ce{NaBH4}$ is added to this compound &lt;strong&gt;as a second step&lt;/strong&gt; to reduce the $\ce{C-Hg}$ bond to a $\ce{C-H}$ bond. In most cases, if the mercury was at a chirality center, that position no longer is one because we have two $\ce{C-H}$ bonds present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/up1lV.png&quot; alt=&quot;reduction of C-Hg bond to produce racemic alcohol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do we use mercury anyway?&#xA;  It works. It works to put the OH on the more substituted position without rearrangements typical of carbocations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ejoc.201201441/abstract&quot; rel=&quot;noreferrer&quot;&gt;there is ongoing research to replace toxic mercury&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-04-13T18:19:36.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28640" PostTypeId="1" CreationDate="2015-04-13T18:23:56.280" Score="5" ViewCount="89" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Cobalt exhibits a more specific interaction with histidine tags, resulting in less nonspecific interaction than nickel. For this reason, cobalt is the preferred divalent cation for purifying His-tagged proteins when high purity is a primary concern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read this last week, and I'm wondering, why is cobalt like that? What about it causes that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source of quote: &lt;a href=&quot;https://www.lifetechnologies.com/us/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/his-tagged-proteins-production-purification.html&quot; rel=&quot;nofollow&quot;&gt;https://www.lifetechnologies.com/us/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/his-tagged-proteins-production-purification.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15559" LastEditorUserId="69830" LastEditDate="2019-11-26T08:58:38.640" LastActivityDate="2019-11-26T08:58:38.640" Title="Protein purification with Cobalt" Tags="&lt;transition-metals&gt;&lt;chromatography&gt;&lt;purification&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28643" PostTypeId="1" AcceptedAnswerId="31327" CreationDate="2015-04-13T20:10:36.597" Score="0" ViewCount="15484" Body="&lt;p&gt;I've just found read that entropy increases through these reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Changing the phase of the elements to a gas.&lt;/li&gt;&#xA;&lt;li&gt;Increasing the &lt;strong&gt;molar mass of that specific element&lt;/strong&gt; increases the entropy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I understand the first one, but I can't grasp the second one.&lt;br&gt;&#xA;Because naturally you would think that $\ce{H2}$ would have a higher entropy than $\ce{Ne}$ because one mole of it is generally lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But however it is said that Ne has a &lt;strong&gt;higher entropy because of its lower molar mass.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-05-14T00:40:53.353" LastActivityDate="2015-05-14T00:40:53.353" Title="Why is it that increasing the molar mass generally increases the Entropy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="13" FavoriteCount="2" ClosedDate="2015-05-13T17:15:40.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="28644" PostTypeId="1" AcceptedAnswerId="28651" CreationDate="2015-04-13T22:54:36.337" Score="8" ViewCount="18241" Body="&lt;p&gt;What is the difference between $\Delta G$ and $\Delta G^\circ$? I couldn't figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Delta G^\circ = -RT\ln K$&lt;/li&gt;&#xA;&lt;li&gt;$\Delta G = \Delta G^\circ + RT \ln Q$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the difference, is it that $\Delta G^\circ$ is the value predicted by the standard free energies and that $\Delta G$ is just the actual value?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2015-09-28T05:38:23.557" LastActivityDate="2019-05-01T02:52:42.597" Title="What is the difference between ΔG and ΔG &quot;with an o on top&quot;?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="5" ClosedDate="2015-12-09T09:27:34.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="28645" PostTypeId="2" ParentId="28608" CreationDate="2015-04-13T23:57:32.550" Score="7" Body="&lt;p&gt;Let's be clear here. Xenon by itself is &lt;strong&gt;not&lt;/strong&gt; an oxidizing agent. Instead, certain compounds of xenon tend to be strong oxidizing agents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's consider an analogous case - the nitro group. Molecules containing nitro groups ($\ce{-NO2}$) tend to be explosive - the nitro groups &lt;em&gt;contain&lt;/em&gt; oxygen which facilitates combustion. Remember, combustion is just a form of &lt;em&gt;oxidation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to xenon! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Xenon, being a noble gas, shouldn't like bonding too much. It already has a complete octet of outer shell (valence) electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is likely the case that compounds of xenon are only weakly held together. Compounds of xenon can therefore said to be relatively &lt;em&gt;thermodynamically&lt;/em&gt; unstable. Therefore, it is also likely that compounds of xenon tend to react in the presence of other molecules - especially ones containing atoms to which fluorine can form stronger bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We verify the above by examining bond enthalpy data: the Xe-F bond has a bond energy of about 130 kJ/mol. Compare this to a typical C-F bond with a bond energy of 485 kJ/mol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is no wonder therefore that compounds such as xenon difluoride are used in fluorinating (oxidizing) organic molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples from Wikipedia. Why the nitro group in the second photo has been reduced to an amino group - I have no clue - and I suspect it's just an error. Further updates soon ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oURpA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-04-14T17:19:46.297" LastActivityDate="2015-04-14T17:19:46.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28646" PostTypeId="1" AcceptedAnswerId="28650" CreationDate="2015-04-14T00:30:07.270" Score="0" ViewCount="23239" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Please write a dissociation-in-water equation for the compounds $\ce{BeI2}$ and $\ce{LiI}$. Make sure to add the states of matter after each compound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently, for $\ce{BeI2}$ I have the equation $$\ce{BeI2 (s) -&amp;gt; Be^2+ (s) + I2^2- (g)}.$$ I have yet to attempt the second one.&lt;/p&gt;&#xA;" OwnerUserId="15561" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2017-01-15T12:00:45.883" LastActivityDate="2017-01-15T12:03:51.560" Title="How to write equations for dissociation of ionic compounds in water?" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28647" PostTypeId="1" CreationDate="2015-04-14T00:36:44.690" Score="-3" ViewCount="3811" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Please write two equations for reacting lithium with water and barium with water. Make sure to add the states of matter after each compound.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Currently, for lithium I have the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Li(s) + 2H2O(l) -&amp;gt; 2LiOH(aq) + H2(g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not attempted the second one yet.&lt;/p&gt;&#xA;" OwnerUserId="15561" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-15T22:46:26.563" LastActivityDate="2019-06-03T07:37:09.000" Title="How to write equations for a reaction of metals with water" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28648" PostTypeId="5" CreationDate="2015-04-14T00:39:22.797" Score="0" Body="&lt;p&gt;This tag should be applied to questions regarding how chemists refer to a specific notion. Often it's about what the technical word for a concept or technical application is, or how one should technically use a scientific term.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Canonical Answers&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/88480/5354&quot;&gt;How pharmaceuticals are named&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="5354" LastEditDate="2020-02-24T02:54:44.817" LastActivityDate="2020-02-24T02:54:44.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28649" PostTypeId="4" CreationDate="2015-04-14T00:39:22.797" Score="0" Body="For questions relating to terminology and naming conventions for chemical compounds and reactions." OwnerUserId="7448" LastEditorUserId="102302" LastEditDate="2021-04-10T19:09:11.930" LastActivityDate="2021-04-10T19:09:11.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28650" PostTypeId="2" ParentId="28646" CreationDate="2015-04-14T00:49:50.743" Score="5" Body="&lt;p&gt;Your ionic charges are not correct for iodine. Looking at your attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$.  When the ions dissociate, they become aqueous or (aq) as the state of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the ionic charges need to balance, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BeI2 (s) -&amp;gt; Be^2+ (aq) + 2I- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar example (and further explanations) are provided on the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Unique_Features_of_Aqueous_Solutions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unique Features of Aqueous Solutions&lt;/a&gt; (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;determine the group, hence ionic charge of each dissociated ion&lt;/li&gt;&#xA;&lt;li&gt;balance these charges&lt;/li&gt;&#xA;&lt;li&gt;state that the dissociated ions are aqueous&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, use the process to determine the dissociation of $\ce{LiI}$&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2017-01-15T12:03:51.560" LastActivityDate="2017-01-15T12:03:51.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28651" PostTypeId="2" ParentId="28644" CreationDate="2015-04-14T01:27:22.613" Score="7" Body="&lt;p&gt;A simple explanation would be that the &quot;o on top&quot; denotes standard state and the one without the &quot;o on top&quot; denotes conditions that are not standard state. However, that may not allow you to &quot;get it&quot; so let's look at the equations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta G^{\circ} = -RT\ln K$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta G         = \Delta G^{\circ}  + RT \ln Q$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the standard state of formation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The partial pressure of any gas involved in the reaction is 0.1 MPa.&#xA;  The concentrations of all aqueous solutions are 1 M.&#xA;  Measurements are generally taken at a temperature of 25 °C (298 K).&#xA;  The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.&lt;br&gt;&#xA;  (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Free_Energy/Gibbs_Free_Energy/Gibbs_Free_Energy#Standard-State_Free_Energy_of_Formation_%28.CE.94Gfo%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or, very simply:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Partial Pressure of gases = 0.1 MPa&lt;/li&gt;&#xA;&lt;li&gt;Concentration of Aqueous solutions are = 1 M (Molar)&lt;/li&gt;&#xA;&lt;li&gt;Temperature usually 298 K&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So if it's so nice and well, why would we need the one without &quot;o on top&quot;? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="72973" LastEditDate="2019-05-01T02:52:42.597" LastActivityDate="2019-05-01T02:52:42.597" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28653" PostTypeId="2" ParentId="28647" CreationDate="2015-04-14T02:08:21.153" Score="2" Body="&lt;p&gt;With all technicalities aside, your first equation is on par. The second one should be adjusted a bit for that &lt;strong&gt;+2&lt;/strong&gt; charge. It is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{Ba(s) + 2HOH(l) -&amp;gt; Ba(OH)2(aq) + H2(g)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; If need be, you must also take into account the fact that &lt;em&gt;two&lt;/em&gt; strong bases are being generated. SO if you wish you might as well dissociate the &lt;strong&gt;Barium Hydroxide&lt;/strong&gt; and the &lt;strong&gt;Lithium Hydroxide&lt;/strong&gt; as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. Correct me if I am wrong.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-14T20:52:02.580" LastActivityDate="2015-04-14T20:52:02.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28654" PostTypeId="1" AcceptedAnswerId="28655" CreationDate="2015-04-14T03:09:04.493" Score="1" ViewCount="7559" Body="&lt;p&gt;The molar heat of fusion of ice is 6.01 kJ/mol. &#xA;Can you explain what this means? What I think it means is that it requires 6.01 kJ to melt a mole of ice. I'm not positive though..&lt;/p&gt;&#xA;" OwnerUserId="15564" LastActivityDate="2015-04-14T09:49:13.787" Title="Molar heat of fusion of ice?" Tags="&lt;kinetics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28655" PostTypeId="2" ParentId="28654" CreationDate="2015-04-14T03:22:50.857" Score="1" Body="&lt;p&gt;Your interpretation is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar heat of fusion is the amount of heat necessary to melt a mole of ice at its melting (or freezing) point. That means, to get 1 mole of water from 1 mole of ice at its melting point (0 degrees Celsius) it would require an input of $6.01\ \mathrm{kJ}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conceptual reasoning is that it requires so much energy to break the bonds and allow more motion (but not completely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, $6.01\ \mathrm{kJ}$ would be released upon the freezing of water at that point (1 mole of water anyway).&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorDisplayName="user7951" LastEditDate="2015-04-14T09:49:13.787" LastActivityDate="2015-04-14T09:49:13.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28656" PostTypeId="2" ParentId="28563" CreationDate="2015-04-14T04:45:00.920" Score="6" Body="&lt;p&gt;$\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so? Let us consider this problem in the familiar mole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were asked how many moles of the substance (molecule) were in this, we would state 1 mole, if there were $\pu{45.3 g}$ of substance. However, if we were asked how many moles of atoms were in the substance, we would accordingly state 3. Thus, in one molecule, there are 3 atoms, so if we simply consider the number of atoms, there would be the necessity of counting the number. of atoms in a molecule, as a molecule is comprised of more than a single atom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* This is purely hypothetical. One would probably not consider  $\ce{MgB2}$, as J. LS pointed out, it decomposes . However, the concept can be applied to other compounds&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="16683" LastEditDate="2017-06-28T18:07:51.487" LastActivityDate="2017-06-28T18:07:51.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28657" PostTypeId="1" CreationDate="2015-04-14T04:53:06.570" Score="2" ViewCount="1534" Body="&lt;p&gt;I have read numerous online questions similar to the one I am about to ask. But for most of them the answer has been that the reason one can add water to the unknown concentration solution during a titration is because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Water has a pH of 7&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is a volumetric analysis, so as long as one record the volume of the titrant and analyte used then there is no problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions are these: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the effect of $\ce{CO2}$ in the atmosphere mixing with the distilled water, it is obvious that the pH of distilled water is on the acidic side (which we confirmed using a pH sensor). Therefore, why does the addition of distilled water not affect the pH? If it is acidic that should mean there are $\ce{H3O+}$ molecules in the distilled water when we add it into the analyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we take into account the imbalance of the moles of $\ce{H3O+}$ molecules versus $\ce{OH-}$ in each drop of the acidic distilled water (ex. pH of 5.6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I have in mind right now is that all of these are acceptable uncertainties (such as the effect of the addition of an indicator), and that significant figures will wipe out these small errors.&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="4356" LastEditDate="2015-04-16T06:00:18.017" LastActivityDate="2015-04-16T06:00:18.017" Title="Titrating and the addition of acidic distilled water" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28658" PostTypeId="2" ParentId="28657" CreationDate="2015-04-14T05:09:26.853" Score="1" Body="&lt;p&gt;Unfortunately I could not find any sources that specifically address your question so I will be answering using logic. If you notice any fallacies or points you disagree with, please point them out.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{CO2}$ dissolves in water at such low concentrations normally, the pH would not be greatly affected. Unless you are working in extremely small concentrations that require high accuracy, it would not be a very large problem. If you were, you could try to minimize the time the distilled water is opened, and perhaps work in a vacuum really quickly or find some way to remove that factor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we were adding the distilled water to another solution, it is possible that the other solution was already in equilibrium with the partial pressure of $\ce{CO2}$ in the atmosphere and thus the concentration would remain constant ( a very small concentration ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I do agree that protons may be added, but they would be so few, especially if you are reacting the reaction with acid or some strong base that it would be insignificant. Often times, conditions are not perfect, but good enough to be estimated and be close enough. However, measures can be taken to prevent your worries if the experiment demands it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any $\ce{CO2}$ probably already dissociated and is in equilibrium, or comes to it very closely. Opening up a coke and watching it enter equilibrium may serve as a nice visual. Don't worry about it in a titration unless you are trying to do something really really unusual. The $\ce{CO2}$ is insignificant in most titrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the solubility is 1.45 g/L at 25 °C, 100 kPa max, so even a saturated solution would be not greatly affected. Furthermore, your solution is probably already in equilibrium so you have already accounted for the carbon dioxide. Think about how when you do a titration, it doesn't matter if you begin at 0 or whatever. As long as you account for it.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-14T05:09:26.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28659" PostTypeId="1" AcceptedAnswerId="28665" CreationDate="2015-04-14T05:59:36.620" Score="11" ViewCount="773" Body="&lt;p&gt;Whilst the Bohr model is incomplete and incorrect, it had limited usage in predicting spectral lines. In the same way, could it possibly with limited accuracy, be used to predict the outcome of chemical interactions to within &quot;an order of magnitude&quot; for which it was so famously known to do for many physical phenomena?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2020-10-08T17:23:37.223" Title="Usefulness of the outdated Bohr model?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28660" PostTypeId="2" ParentId="28450" CreationDate="2015-04-14T06:03:47.683" Score="2" Body="&lt;p&gt;The way to determine the volume of an atom in a molecule (Therefore, the molecular volume) is by doing a quantum calculation to obtain the wave function and, then, integrate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to obtain the wave function from GAMESS is with the AIMPAC keyword. Look at page 2-18 of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow&quot;&gt;the GAMESS manual&lt;/a&gt;. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain the volume of your desire molecule by following these steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do a quantum calculation with the desire method and a very good wave function. &lt;/li&gt;&#xA;&lt;li&gt;In the input file, you have to specify to the software of your preference to print out the wave function file (in gaussian, you can do it with output=wfn, more directions: &lt;a href=&quot;http://aim.tkgristmill.com/knownissues.html&quot; rel=&quot;nofollow&quot;&gt;Directions to print wave function from gaussian&lt;/a&gt;). &lt;/li&gt;&#xA;&lt;li&gt;After the quantum calculation is done, you have to integrate the wave function with a specialized software (see at external links of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow&quot;&gt;QTAIM software&lt;/a&gt;). &lt;/li&gt;&#xA;&lt;li&gt;After integrating the wave function, you will get the integrated values of several properties, one of them is the volume of each atom in the desire molecule. (See page 16 of &lt;a href=&quot;https://books.google.com.co/books?id=protBb8KraoC&amp;amp;pg=PA16&quot; rel=&quot;nofollow&quot;&gt;The Quantum Theory of Atoms in Molecules&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Finally, sum up all the atomic volumes to get the molecular one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-04-16T19:26:05.580" LastActivityDate="2015-04-16T19:26:05.580" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28661" PostTypeId="1" AcceptedAnswerId="29311" CreationDate="2015-04-14T06:13:42.340" Score="6" ViewCount="388" Body="&lt;p&gt;What is the most commonly used radius of an atom or molecule to describe the area that must touch another area of another molecule for the two to react. Is there one radius that always works or does it depend on the reaction? Of course the actual distance at which bonds can form or break varies but I was wondering if at some distance the atoms can somehow bring themselves together for bond 'swapping'&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2015-07-03T12:05:05.753" Title="Interaction radius of a molecule" Tags="&lt;atoms&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28662" PostTypeId="2" ParentId="28634" CreationDate="2015-04-14T06:15:00.433" Score="1" Body="&lt;p&gt;The voltage on the terminals is different by 12 V (± a few depending on the charge state), you can think of this as the positive terminal being at +12 V and the negative terminal at being 0 V. The polarity of the voltage does NOT change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the current, things are a bit different. During discharge, the current flows from the positive to the negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During charge, the current flows out of the positive terminal of the charger into the positive terminal of the battery (and out of the negative terminal of the battery into the negative terminal of the charger).&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorDisplayName="user7951" LastEditDate="2015-04-15T11:48:10.093" LastActivityDate="2015-04-15T11:48:10.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28663" PostTypeId="2" ParentId="28637" CreationDate="2015-04-14T06:23:06.867" Score="1" Body="&lt;p&gt;Judging by your above comment to beginner's answer, you seem to be confused about how reactions proceed in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this graph : &lt;a href=&quot;http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation in the beginning certainly requires energy, but at the end of the reaction, the system ends up at a lower energy state overall. This causes the reaction to be &quot;spontaneous&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-14T06:23:06.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28665" PostTypeId="2" ParentId="28659" CreationDate="2015-04-14T07:06:14.000" Score="10" Body="&lt;p&gt;I think the importance of Bohr model comes from the fact that it's the first atomic model that uses the hypothesis of quantization. More precisely Energy Quantization. It introduces the first quantum number $n$. It allows to interpret the atomic emission spectrum of hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it is an interesting pedagogic tool to teach students how the atomic models had evolved to arrive to the modern atomic theory (based on quantum physics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also a beautiful example to teach students how empirical science, like chemistry, evolved by making observation (emission spectrum of hydrogen), formulate hypotheses (atomic model based on the model of solar system but with electrostatic forces), testing the model (it works for interpreting the discrete emission spectrum), but it fails to interpret the behaviour of the atom under magnetic field (so, scientists thought about rejecting the model and used another one, the modern model of atoms). &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="5017" LastEditDate="2015-04-19T23:50:28.223" LastActivityDate="2015-04-19T23:50:28.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28666" PostTypeId="1" AcceptedAnswerId="28694" CreationDate="2015-04-14T07:09:56.360" Score="10" ViewCount="4700" Body="&lt;p&gt;I would like to understand the term stability in organic chemistry. We know empirically that isobutane (2-methylpropane) is more stable than &lt;em&gt;n&lt;/em&gt;-butane. For example, their heats of formation (taken from the NIST WebBook) are &lt;span class=&quot;math-container&quot;&gt;$\pu{-134.2 kJ/mol}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{-125.6 kJ/mol}$&lt;/span&gt; respectively, putting isobutane at &lt;em&gt;ca.&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{10 kJ/mol}$&lt;/span&gt; lower in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is this so? Is there a physical or chemical explanation for this?&lt;/p&gt;&#xA;" OwnerUserId="15567" LastEditorUserId="16683" LastEditDate="2019-04-20T23:33:09.613" LastActivityDate="2019-04-20T23:33:09.613" Title="Why is isobutane more thermodynamically stable than n-butane?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28667" PostTypeId="2" ParentId="28605" CreationDate="2015-04-14T08:27:59.200" Score="2" Body="&lt;p&gt;Your approach is correct and the result you obtain is right. Checked &lt;a href=&quot;http://ph.lattelog.com/melange&quot; rel=&quot;nofollow&quot; title=&quot;pH APP&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).&lt;/p&gt;&#xA;" OwnerUserId="3871" LastEditorDisplayName="user7951" LastEditDate="2015-04-14T20:43:57.857" LastActivityDate="2015-04-14T20:43:57.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28668" PostTypeId="1" CreationDate="2015-04-14T08:50:31.933" Score="2" ViewCount="44" Body="&lt;p&gt;I'm reading a paper about the atmospheric chemistry on Titan {1}. It quotes &quot;column photochemical and loss rates&quot; for various species in units of  $\mathrm{cm^{-2}s^{-1}}$. These figures are the overall rates at which these species are generated photochemically according to a model, and I would like to convert them into units of $\mathrm{mol/s}$ or $\mathrm{mol/s/m^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't know how to interpret the given units of $\mathrm{cm^{-2}s^{-1}}$. It feels like it should be something like CCs of gas produced per $\mathrm{cm}^2$ of the planet's surface area - but that would be $\mathrm{cm\,s^{-1}}$ rather than $\mathrm{cm^{-2}s^{-1}}$. No matter how I look at it I can't see where the $\mathrm{cm^{-2}}$ comes from - is anyone able to explain this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{1} P. P. Lavvas et al. (2008) Planetary and Space Science 56, pp. 67-99.&lt;/p&gt;&#xA;" OwnerUserId="148" LastEditorUserId="4945" LastEditDate="2015-04-15T06:13:41.693" LastActivityDate="2015-04-15T06:13:41.693" Title="How do I interpret a flux quoted in cm⁻² s⁻¹ in atmospheric chemistry?" Tags="&lt;photochemistry&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28669" PostTypeId="1" CreationDate="2015-04-14T09:01:41.627" Score="8" ViewCount="760" Body="&lt;p&gt;I read about entropy from different sources. Still don’t get why (universal) entropy change has to be greater than zero for irreversible process, i.e. $\mathrm{d}S_\mathrm{Univ}&amp;gt;0$. Is it a result of pure observation or is there any theoretical background of this phenomena?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the linked question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/24669/4945&quot;&gt;Does adsorption violate thermodynamics?&lt;/a&gt;&quot; does not answer this, as it only shows that overall entropy change is greater than zero. But doesn’t provide a reason where it comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a simple explanation in classical thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="15567" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-05-13T20:56:21.673" Title="Why does everything go in the direction of positive entropy change?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28670" PostTypeId="2" ParentId="28668" CreationDate="2015-04-14T09:02:51.617" Score="2" Body="&lt;p&gt;The &quot;strange&quot; unit describes the &lt;strong&gt;photon flux&lt;/strong&gt;, which is the number of photons (of unspecified wavelength) that hit an area (thus $\mathrm{cm^{-2}}$) in a time interval (thus $\mathrm{s^{-1}}$).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It feels like it should be something like CCs of gas produced [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really. Not every photon leads to an excited state, not every excited state triggers a chemical reaction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-14T09:02:51.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28671" PostTypeId="2" ParentId="28346" CreationDate="2015-04-14T09:30:28.753" Score="15" Body="&lt;p&gt;Since I had troubles with the notation the first time around, I did some real quantum chemistry (with a pencil and some paper) and finally was able to derive the whole RSPT ansatz. Since this is equivalent to the first approach, but yet also standalone, I decided to add it as a separate answer. In a few key steps, this approach is slightly different, so a one to one transformation will not be possible.&#xA;$$%Introducing some shortcuts&#xA;\require{cancel}&#xA;\newcommand{\op}[1]{\hat{\mathrm{#1}}}  %\op{H}&#xA;\newcommand{\order}[1]{^{(#1)}}         %E_n\order{1}&#xA;\newcommand{\overlap}[3]{\mathcal{#1}_{#2}\order{#3}} %\overlap{S}{m}{1}&#xA;\newcommand{\integral}[3]{\mathcal{#1}_{#2,#3}} %\integral{V}{i}{j}&#xA;\newcommand{\tagref}[1]{\mathrm{#1}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start again by setting up the Hamiltonian, as a reference operator with the perturbation operator.&#xA;$$\op{H} = \op{H}\order{0} +\lambda\op{V}  \tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perturbed Schrödinger equation therefore becomes&#xA;$$(\op{H}\order{0} +\lambda\op{V})|n\rangle = E_n|n\rangle, \tag2$$&#xA;and we again develop the terms of the energy and wave function into Taylor power series.&#xA;\begin{align}&#xA;E_n &amp;amp;= E_n\order{0} + \lambda E_n\order{1} + \lambda^2 E_n\order{2} + \dots &amp;amp;&#xA;    &amp;amp;= \sum_i \lambda^i E_n\order{i}  \tag3\\&#xA;|n\rangle &amp;amp;= |n\order{0}\rangle + \lambda |n\order{1}\rangle &#xA;             + \lambda^2 |n\order{2}\rangle + \dots &amp;amp;&#xA;          &amp;amp;= \sum_i \lambda^i |n\order{i}\rangle  \tag4\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write the unperturbed system, that provides us with a complete solution and therefore with a complete set of energies and wave functions, i.e. $n\in\mathbb{N}$.&#xA;$$\op{H}\order{0} |n\order{0}\rangle = E_n\order{0} |n\order{0}\rangle \tag5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also know from this, that all solutions are orthonormal, i.e.&#xA;\begin{align}&#xA;\langle n\order{0} |n\order{0}\rangle &amp;amp;= 1,&#xA;\langle n\order{0} |m\order{0}\rangle &amp;amp;= 0. \tag6&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further we choose our perturbed wave function to be (intermediate) normalised, i.e.&#xA;\begin{align}&#xA;\langle n\order0|n\rangle &amp;amp;= 1, &amp;amp;&#xA;\langle n|n\rangle &amp;amp;= 1, &amp;amp;&#xA;\langle n\order{i}|n\order{j}\rangle &amp;amp;= \delta_{ij}&#xA;   \begin{cases}&#xA;      1, &amp;amp; i=j\\&#xA;      0, &amp;amp; i\neq j\\&#xA;   \end{cases}.\tag7&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We continue with gathering the equations for all our orders.&#xA;\begin{align}&#xA;  \lambda^0 &amp;amp;:&amp;amp;&#xA;    \op{H}\order0 |n\order0\rangle &#xA;       &amp;amp;= E_n\order0 |n\order0\rangle\tag{8a}\\&#xA;  \lambda^1 &amp;amp;:&amp;amp;&#xA;    \op{H}\order0 |n\order1\rangle + \op{V} |n\order0\rangle&#xA;       &amp;amp;= E_n\order0 |n\order1\rangle + E_n\order1 |n\order0\rangle\tag{8b}\\&#xA;  \lambda^2 &amp;amp;:&amp;amp;&#xA;    \op{H}\order0 |n\order2\rangle + \op{V} |n\order1\rangle&#xA;       &amp;amp;= E_n\order0 |n\order2\rangle + E_n\order1 |n\order1\rangle&#xA;          E_n\order1 |n\order0\rangle\tag{8c}\\&#xA;  \vdots\\&#xA;  \lambda^n &amp;amp;:&amp;amp;&#xA;    \op{H}\order0 |n\order{n}\rangle + \op{V} |n\order{n-1}\rangle&#xA;       &amp;amp;= \sum_{i=0}^n E_n\order{i} |n\order{n-i}\rangle\tag{8d}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two fundamental tricks of algebra are adding zero and multiplying by one. &#xA;We use this to set up our bases, we basically expand our bases into an auxiliary base (also known as 'resolution-of-the-identity'), which is possible, since the unperturbed system provides the complete set. This is what makes this approach somewhat different. While in my other answer we are just expanding the wafer function into an ordinary boring linear combination, we are using here the complete Hilbert space to do the same thing, but a little more elegant. However, this is also the key step: Multiplying by one.&lt;br&gt;&#xA;At the same time we introduce an overlap integral $\integral{S}{m}{i}$, which makes writing a little easier further on. For any order $i\neq0$ we can therefore write:&#xA;\begin{align}&#xA;|n\order{i}\rangle &amp;amp;= &#xA;   \sum_{m} |m\order0\rangle\langle m\order0|n\order{i}\rangle\\&#xA;                   &amp;amp;=&#xA;   \sum_{m\neq n} |m\order0\rangle&#xA;    \underbrace{\langle m\order0|n\order{i}\rangle}_{\overlap{S}{m}{i}}&#xA;     + |n\order0\rangle\cancelto{0}{\langle n\order0|n\order{i}\rangle}\\&#xA;|n\order{i}\rangle &amp;amp;= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{i}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bases we need are for first and second order, therefore we write&#xA;\begin{align}&#xA;|n\order{1}\rangle &amp;amp;= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}, &amp;amp;&#xA;|n\order{2}\rangle &amp;amp;= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{2},&#xA;\end{align}&#xA;which seems rather trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is getting the expressions for the energies and corrections to the wave function.&lt;br&gt;&#xA;For $\tagref{8a}$ this is trivial and it just yields the expectation value of the unperturbed Hamiltonian. We project on (multiply from the left with) $\langle n\order0|$, hence&#xA;\begin{align}&#xA;  \langle n\order0|E_n\order0 |n\order0\rangle &amp;amp;=&#xA;  \langle n\order0|\op{H}\order0 |n\order0\rangle \\&#xA;  E_n\order0 \cancelto{1}{\langle n\order0|n\order0\rangle} &amp;amp;=&#xA;  \langle n\order0|\op{H}\order0 |n\order0\rangle \\&#xA;  E_n\order0 &amp;amp;= \langle n\order0|\op{H}\order0 |n\order0\rangle. \tag{9a}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first order correction to the energy we use the hermiticity of the Hamiltonian, i.e. we rewrite $\tagref{8a}$ in bra instead of the ket notation we used all way through.&#xA;$$\langle n\order0| \op{H}\order0 = E_n\order0 \langle n\order0|$$&#xA;We then further go about and project $\tagref{8b}$ on $\langle n\order0|$, hence&#xA;\begin{align}&#xA;\langle n\order0|\left(\op{H}\order0|n\order1\rangle&#xA;                       + \op{V}|n\order0\rangle \right) &amp;amp;=&#xA;   \langle n\order0|\left(E_n\order0|n\order1\rangle&#xA;                       + E_n\order1|n\order0\rangle \right)\\&#xA;\langle n\order0|\op{H}\order0|n\order1\rangle&#xA;     + \langle n\order0| \op{V}|n\order0\rangle  &amp;amp;=&#xA;   \langle n\order0|E_n\order0|n\order1\rangle&#xA;     + \langle n\order0| E_n\order1|n\order0\rangle\\&#xA;E_n\order0 \cancelto{0}{\langle n\order0|n\order1\rangle}&#xA;     + \langle n\order0| \op{V}|n\order0\rangle  &amp;amp;=&#xA;   E_n\order0\cancelto{0}{\langle n\order0|n\order1\rangle}&#xA;     + E_n\order1 \cancelto{1}{\langle n\order0| n\order0\rangle}\\&#xA;E_n\order1 &amp;amp;= \langle n\order0| \op{V}|n\order0\rangle  \tag{9b}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the first order correction to the waffle function. I have to admit, this was the part where I struggled most. We will project on $\langle k\order0|\neq\langle n\order0|$, therfore obtaining:&#xA;\begin{align}&#xA;\langle k\order0|\left(\op{H}\order0|n\order1\rangle&#xA;                       + \op{V}|n\order0\rangle \right) &amp;amp;=&#xA;   \langle k\order0|\left(E_n\order0|n\order1\rangle&#xA;                       + E_n\order1|n\order0\rangle \right)\\&#xA;\langle k\order0|\op{H}\order0|n\order1\rangle&#xA;     + \langle k\order0| \op{V}|n\order0\rangle  &amp;amp;=&#xA;   \langle k\order0|E_n\order0|n\order1\rangle&#xA;     + \langle k\order0| E_n\order1|n\order0\rangle\\&#xA;E_k\order0 \langle k\order0|n\order1\rangle&#xA;     + \langle k\order0| \op{V}|n\order0\rangle  &amp;amp;=&#xA;   E_n\order0\langle k\order0|n\order1\rangle&#xA;     + E_n\order1 \cancelto{0}{\langle k\order0| n\order0\rangle}\\&#xA;E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}&#xA;     + \langle k\order0| \op{V}|n\order0\rangle  &amp;amp;=&#xA;   E_n\order0\langle k\order0|\sum_{m\neq n} &#xA;                    |m\order0\rangle \overlap{S}{m}{1}\\&#xA;\end{align}&#xA;Before it gets to messy, we are going to reduce each of the terms individually and then reinsert, to obtain our expression.&#xA;\begin{align}&#xA;E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}&#xA;  &amp;amp;= \sum_{m\neq n,k}E_k\order0 &#xA;       \cancelto{0}{\langle k\order0|m\order0\rangle} \overlap{S}{m}{1}&#xA;     + E_k\order0 &#xA;       \cancelto{1}{\langle k\order0|k\order0\rangle} \overlap{S}{k}{1}\\&#xA;  &amp;amp;= E_k\order0 \overlap{S}{k}{1}\\&#xA;\end{align}&#xA;The second term stays the same, and the last one simplifies analogously.&#xA;\begin{align}&#xA;E_n\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}&#xA;  &amp;amp;= \sum_{m\neq n,k}E_n\order0 &#xA;       \cancelto{0}{\langle k\order0|m\order0\rangle} \overlap{S}{m}{1}&#xA;     + E_n\order0 &#xA;       \cancelto{1}{\langle k\order0|k\order0\rangle} \overlap{S}{k}{1}\\&#xA;  &amp;amp;= E_n\order0 \integral{S}{k}{1}\\&#xA;\end{align}&#xA;And reinsterting we can write and resubstitute for the overlap integral $\overlap{S}{k}{1}$, with $m=k$:&#xA;\begin{align}&#xA;E_k\order0 \overlap{S}{k}{1} + \langle k\order0| \op{V}|n\order0\rangle &#xA;   &amp;amp;= E_n\order0 \overlap{S}{k}{1}\\&#xA;\integral{S}{k}{1} &#xA;   &amp;amp;= \frac{\langle k\order0| \op{V}|n\order0\rangle}{E_n\order0 -E_k\order0}\\&#xA;|n\order1\rangle&#xA;   &amp;amp;= \sum_{k\neq n} |k\order0\rangle&#xA;      \frac{\langle k\order0| \op{V}&#xA;              |n\order0\rangle}{E_n\order0 -E_k\order0} \tag{10b}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make matters a little bit easier further on, we will introduce another short notation for the perturbed integrals:&#xA;$$\integral{V}{i}{j} := \langle i\order0|\op{V}|j\order0\rangle$$&#xA;And therefore we can write&#xA;\begin{align}&#xA;|n\order1\rangle&#xA;   &amp;amp;= \sum_{k\neq n} |k\order0\rangle&#xA;      \frac{\integral{V}{k}{n}}{E_n\order0 -E_k\order0} \tag{10b}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second order correction to the energy we project $\tagref{8c}$ on $\langle n\order0|$. This is again fairly straight forward, the key is to insert $\tagref{10b}$ at the end to obtain the expression.&#xA;\begin{align}&#xA;\langle n\order0|\left(\op{H}\order0|n\order2\rangle &#xA;  + \op{V}|n\order1\rangle\right)&#xA;  &amp;amp;= \langle n\order0|\left( E_n\order0|n\order2\rangle&#xA;       + E_n\order1|n\order1\rangle + E_n\order2|n\order0\rangle\right)\\&#xA;\langle n\order0|\op{H}\order0|n\order2\rangle &#xA;  + \langle n\order0|\op{V}|n\order1\rangle&#xA;    &amp;amp;=   E_n\order0 \langle n\order0|n\order2\rangle&#xA;       + E_n\order1 \langle n\order0|n\order1\rangle&#xA;       + E_n\order2 \langle n\order0|n\order0\rangle\\&#xA;\small&#xA; E_n\order0 \cancelto{0}{\langle n\order0|n\order2\rangle}&#xA;  + \langle n\order0|\op{V}|n\order1\rangle&#xA;    &amp;amp;= \small&#xA;         E_n\order0 \cancelto{0}{\langle n\order0|n\order2\rangle}&#xA;       + E_n\order1 \cancelto{0}{\langle n\order0|n\order1\rangle}&#xA;       + E_n\order2 \cancelto{1}{\langle n\order0|n\order0\rangle}\\&#xA;E_n\order2 &amp;amp;= \langle n\order0|\op{V}|n\order1\rangle\\&#xA;%           &amp;amp;= \langle n\order0|\op{V} \sum_{k\neq n} |k\order0\rangle&#xA;%      \frac{\langle k\order0| \op{V}&#xA;%              |n\order0\rangle}{E_n\order0 -E_k\order0}\\&#xA;           &amp;amp;= \sum_{k\neq n} \frac{\langle n\order0|\op{V} |k\order0\rangle&#xA;      \langle k\order0| \op{V} |n\order0\rangle}{E_n\order0 &#xA;                     -E_k\order0}\tag{9c}\\&#xA;           &amp;amp;= \sum_{k\neq n} \frac{\integral{V}{n}{k}&#xA;                \integral{V}{k}{n}}{E_n\order0 -E_k\order0} &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's get to the real fancy part: The second order correction to the basil. We project once again $\tagref{8c}$ on $\langle k\order| \neq \langle n\order 0|$. We do of course need our expression for the first order base and our expansion for the second order. But let us take one step at the time.&#xA;\begin{align}&#xA;\langle k\order0|\left( \op{H}\order0|n\order2\rangle &#xA;   + \op{V}|n\order1\rangle\right)&#xA;  &amp;amp;= \langle k\order0 \left(&#xA;      E_n\order0|n\order2\rangle&#xA;    + E_n\order1|n\order1\rangle&#xA;    + E_n\order2|n\order0\rangle\right)\\ &#xA;\small&#xA;\langle k\order0| \op{H}\order0|n\order2\rangle &#xA;   + \langle k\order0| \op{V}|n\order1\rangle&#xA;  &amp;amp;= \small &#xA;      \langle k\order0| E_n\order0|n\order2\rangle&#xA;    + \langle k\order0| E_n\order1|n\order1\rangle&#xA;    + \langle k\order0| E_n\order2|n\order0\rangle\\ &#xA;\end{align}&#xA;And we proceed again term by term, because it is pretty messy. We are starting with the right hand side, and there with the third term, as this one cancels, because of our normalisation.&#xA;\begin{align}&#xA;\langle k\order0| E_n\order2|n\order0\rangle&#xA;  &amp;amp;= E_n\order2 \cancelto{0}{\langle k\order0| n\order0\rangle} =0\\&#xA;\end{align}&#xA;For the second term on the right we need our definition of the base. Since we are projecting on $k$ we choose the base to be represented by $m$. We also need the expression for the first order correction to the energy $\tagref{9b}$. Also note, that we are using the contracted notation.&#xA;\begin{align}&#xA;\langle k\order0| E_n\order1|n\order1\rangle&#xA;  &amp;amp;= E_n\order1 \langle k\order0|n\order1\rangle\\&#xA;  &amp;amp;= E_n\order1 \langle k\order0| \sum_{m\neq n} |m\order0\rangle&#xA;       \frac{\integral{V}{m}{n}}{E_n\order0 - E_m\order0}\\&#xA;  &amp;amp;= \sum_{m\neq n,k} E_n\order1 \cancelto{0}{\langle k\order0| m\order0\rangle}&#xA;       \frac{\integral{V}{m}{n}}{E_n\order0 - E_m\order0}&#xA;     + E_n\order1 \cancelto{1}{\langle k\order0| k\order0\rangle}&#xA;       \frac{\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\&#xA;  &amp;amp;= \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first term on the right hand side we just need our expansion of the second order correction of the base.&#xA;\begin{align}&#xA;\langle k\order0| E_n\order0|n\order2\rangle&#xA;  &amp;amp;= E_n\order0 \langle k\order0|n\order2\rangle\\&#xA;  &amp;amp;= E_n\order0 \langle k\order0| \sum_{m\neq n} |m\order0\rangle&#xA;     \overlap{S}{m}{2}\\&#xA;  &amp;amp;= \sum_{m\neq n,k} E_n\order0 \cancelto{0}{\langle k\order0|m\order0\rangle}&#xA;     \overlap{S}{m}{2}&#xA;     + E_n\order0 \cancelto{1}{\langle k\order0|k\order0\rangle}&#xA;     \overlap{S}{k}{2}\\&#xA;  &amp;amp;= E_n\order0 \overlap{S}{k}{2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to use the same for the first term on the left hand side. In addition we are using the bra notation for the energy to simplify.&#xA;\begin{align}&#xA;\langle k\order0| \op{H}\order0|n\order2\rangle&#xA;  &amp;amp;= E_k\order0 \langle k\order0|n\order2\rangle\\&#xA;  &amp;amp;= E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle&#xA;     \overlap{S}{m}{2} \\&#xA;  &amp;amp;= \sum_{m\neq n,k} E_k\order0 \cancelto{0}{\langle k\order0|m\order0\rangle}&#xA;     \overlap{S}{m}{2} &#xA;     + E_k\order0 \cancelto{1}{\langle k\order0|k\order0\rangle}&#xA;     \overlap{S}{k}{2}\\&#xA;  &amp;amp;= E_k\order0 \overlap{S}{k}{2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second term on the left side we once again need the expression for the first order correction to bass, choosing $m$, as we are already projecting on $k$.&#xA;\begin{align}&#xA;\langle k\order0| \op{V}|n\order1\rangle&#xA;  &amp;amp;= \langle k\order0| \op{V} \sum_{m\neq n}|m\order0\rangle&#xA;     \frac{\integral{V}{m}{n}}{E_n\order0 -E_m\order0}\\&#xA;  &amp;amp;= \sum_{m\neq n}\frac{\integral{V}{k}{m}&#xA;     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's put all the stuffing back into the turkey. Then we rearrange to complete bird and solve for our desired expression for the base.&#xA;\begin{align}&#xA;E_k\order0 \overlap{S}{k}{2}&#xA;  + \sum_{m\neq n}\frac{\integral{V}{k}{m}&#xA;     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}&#xA;  &amp;amp;= E_n\order0 \overlap{S}{k}{2}&#xA;     + \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\&#xA;(E_n\order0 - E_k\order0)\overlap{S}{k}{2} &#xA;  &amp;amp;= \sum_{m\neq n}\frac{\integral{V}{k}{m}&#xA;     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}&#xA;    - \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\&#xA;\overlap{S}{k}{2} &#xA;  &amp;amp;= \sum_{m\neq n}\frac{\integral{V}{k}{m}&#xA;     \integral{V}{m}{n}}{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}&#xA;    - \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{(E_n\order0 - E_k\order0)^2}\\&#xA;|n\order2\rangle &#xA;  &amp;amp;= \sum_{k\neq n} \sum_{m\neq n} |k\order0\rangle&#xA;     \frac{\integral{V}{k}{m}&#xA;     \integral{V}{m}{n}}{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}\\&#xA;%split into two lines&#xA;  &amp;amp;\phantom{=~}&#xA;    - \sum_{k\neq n} |k\order0\rangle&#xA;      \frac{\integral{V}{k}{n}\integral{V}{n}{n}}{(E_n\order0 - E_k\order0)^2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last, but not least, let's remove that nasty shorthand and make everything pretty again.&#xA;\begin{align}&#xA;|n\order2\rangle &#xA;  &amp;amp;= \sum_{k\neq n} \sum_{m\neq n} |k\order0\rangle&#xA;     \frac{&#xA;       \langle k\order0|\op{V}|m\order0\rangle&#xA;       \langle m\order0|\op{V}|n\order0\rangle&#xA;          }{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}&#xA;    - \sum_{k\neq n} |k\order0\rangle&#xA;      \frac{&#xA;       \langle k\order0|\op{V}|n\order0\rangle&#xA;       \langle n\order0|\op{V}|n\order0\rangle&#xA;           }{(E_n\order0 - E_k\order0)^2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there we are. Pretty.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now you will probably have noticed, that you were one step away from the ultron goal. In your last equation you simply had to reduce to the non-zero parts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{equation}&#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n}  \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0} &#xA;\end{equation}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   &#xA; &amp;amp;= \sum_{m \neq n} &#xA;    \frac{ &#xA;          \langle k^0| \hat{V} |m^0 \rangle &#xA;          \langle m^0| \hat{V} |n^0 \rangle&#xA;         }{E_n^0 - E_m^0}\\&#xA; &amp;amp;\phantom{=~}  &#xA;  - \sum_{m \neq n,k} &#xA;      \frac{ \cancelto{0}{\langle k^0 |m^0 \rangle}&#xA;            \langle n^0 | \hat{V} | n^0 \rangle&#xA;            \langle m^0 | \hat{V} | n^0 \rangle&#xA;           }{E_n^0 - E_m^0}\\&#xA; &amp;amp;\phantom{=~}   &#xA;  -   \frac{ \cancelto{1}{\langle k^0 |k^0 \rangle}&#xA;            \langle k^0 | \hat{V} | n^0 \rangle&#xA;            \langle n^0 | \hat{V} | n^0 \rangle&#xA;           }{E_n^0 - E_k^0}\\ &#xA;( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   &#xA; &amp;amp;= \sum_{m \neq n} &#xA;    \frac{ &#xA;          \langle k^0| \hat{V} |m^0 \rangle &#xA;          \langle m^0| \hat{V} |n^0 \rangle&#xA;         }{E_n^0 - E_m^0}  &#xA;  -   \frac{ &#xA;            \langle k^0 | \hat{V} | n^0 \rangle&#xA;            \langle n^0 | \hat{V} | n^0 \rangle&#xA;           }{E_n^0 - E_k^0}\\ &#xA;\langle k^0  |n^2\rangle   &#xA; &amp;amp;= \sum_{m \neq n} &#xA;    \frac{ &#xA;          \langle k^0| \hat{V} |m^0 \rangle &#xA;          \langle m^0| \hat{V} |n^0 \rangle&#xA;         }{(E_n^0 - E_m^0)(E_n^0 - E_k^0)}  &#xA;  -   \frac{ &#xA;            \langle k^0 | \hat{V} | n^0 \rangle&#xA;            \langle n^0 | \hat{V} | n^0 \rangle&#xA;           }{\left(E_n^0 - E_k^0\right)^2}\\ &#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you just have to expand into the complete base by multiplying with $\sum_{k\neq n}|k^0\rangle$ and you get to the same expression as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as a theoretician a proof is only complete with a coffee stain, and trust me, there have been many.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Tj8C7.jpg&quot; alt=&quot;obligatory coffee stain courtesy of Roger Karlsson&quot;&gt;&lt;br&gt;&#xA;Image is courtesy of Roger Karlsson (&lt;a href=&quot;http://www.free-photo-gallery.org/photos/coffee-stain/&quot; rel=&quot;noreferrer&quot;&gt;http://www.free-photo-gallery.org/photos/coffee-stain/&lt;/a&gt;) obtained from &lt;a href=&quot;https://www.flickr.com/photos/free-photos/3375886335/&quot; rel=&quot;noreferrer&quot;&gt;flicker&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-04-14T09:30:28.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28676" PostTypeId="1" AcceptedAnswerId="28696" CreationDate="2015-04-14T10:59:21.927" Score="1" ViewCount="231" Body="&lt;p&gt;I suppose that the configurations assigned are opposite as in the left isomer of 1,2-dichlorocyclopropane, the left chlorine atom is coming out of the plane of the paper while the other is going into the plane of paper. So it must be trans. Am I going wrong wrong somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tu4RR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tu4RR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorDisplayName="user7951" LastEditDate="2016-11-24T01:03:23.190" LastActivityDate="2016-11-24T01:03:23.190" Title="Is the correct configuration assigned to the given pair of geometrical isomers of 1,2-dichlorocyclopropane?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28679" PostTypeId="2" ParentId="28669" CreationDate="2015-04-14T11:10:29.487" Score="2" Body="&lt;p&gt;I would like to precise that not every thing goes in the direction of positive entropy change. It's only true for spontaneous processes which don't   need to be driven by an outside source of energy,  such as a smell diffusing in a room, ice salt dissolving in water,  iron rusting, melting of an ice cube, aging over the years.&#xA;If you leave your room without taking care of putting every thing back in its place, it becomes messy and untidy (spontaneous process). You have to make an effort to keep it clean and tidy (non-spontaneous process). Entropy is the measurement of disorder within a system. The second law of thermodynamics states that an isolated system will become more disordered, as time increases. I think that statistical thermodynamics can explain this law,&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-14T11:10:29.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28682" PostTypeId="1" AcceptedAnswerId="34208" CreationDate="2015-04-14T12:05:17.890" Score="21" ViewCount="9188" Body="&lt;p&gt;In both the compounds the oxidation state of Chromium is +6 so why is there a difference in the colours of their aqueous solutions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{K2Cr2O7}$&lt;/span&gt; is red-yellow and &lt;span class=&quot;math-container&quot;&gt;$\ce{K2CrO4}$&lt;/span&gt; is bright yellow&lt;/p&gt;&#xA;&lt;p&gt;I and leaning towards the fact that due to the increased amount of bond pairs in &lt;span class=&quot;math-container&quot;&gt;$\ce{K2Cr2O7}$&lt;/span&gt; it becomes harder to remove the d orbital electron and that causes the light emitted to be of a higher wavelength(Lower frequency) and that's why there is colour difference.&lt;/p&gt;&#xA;&lt;h1&gt;NOTE: Even though both dichromate and chromate have the same oxidation state for Cr (i.e. +6) they show different colours. My question is how and why?&lt;/h1&gt;&#xA;" OwnerUserId="15574" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-02-10T15:35:22.460" Title="Why is there a difference in the colours of K₂Cr₂O₇ and K₂CrO₄?" Tags="&lt;transition-metals&gt;&lt;color&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28683" PostTypeId="1" AcceptedAnswerId="28708" CreationDate="2015-04-14T12:33:55.977" Score="4" ViewCount="3905" Body="&lt;p&gt;A question in a practice paper I did recently showed the aldol condensation of benzaldehyde with ethanoic acid. I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions but it is possible to protonate a carboxylic acid and then have it take part in an aldol reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, is this also possible for acyl chlorides, and if not, why?&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-15T07:25:49.707" Title="Can carboxylic acids be used in the aldol reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28684" PostTypeId="1" AcceptedAnswerId="28686" CreationDate="2015-04-14T13:00:48.383" Score="2" ViewCount="1112" Body="&lt;p&gt;Excuse me if the question is silly, I have very little knowledge of chemistry.&#xA;We can find oxygen in many places in the universe, but all the combustibles we use to light fire seems to be organic: carbon, oil, wood, alcohol... so if that is true, we could only find fire on Earth, as long as we don´t find organic materials in other planets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of asking the same question could be: can we light a fire without organic material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I mean fire as a result of a combustion. The Sun would be a non-valid example, as it works by fusion.&lt;/p&gt;&#xA;" OwnerUserId="15575" LastEditorUserId="10185" LastEditDate="2015-04-14T13:17:28.993" LastActivityDate="2016-09-02T20:39:34.047" Title="Is it possible to find fire outside the Earth?" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28685" PostTypeId="1" CreationDate="2015-04-14T13:24:12.080" Score="0" ViewCount="227" Body="&lt;p&gt;Is there any way to estimate the alcohol content of a drink with an electric test? Is alcohol in a liquid (e.g wine) dependent on the drink's conductivity?&lt;/p&gt;&#xA;" OwnerUserId="15576" LastEditorUserId="3735" LastEditDate="2015-04-14T13:40:27.283" LastActivityDate="2015-04-14T13:40:27.283" Title="How to electronically calculate alcoholic level of a liquid" Tags="&lt;alcohols&gt;&lt;conductivity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28686" PostTypeId="2" ParentId="28684" CreationDate="2015-04-14T13:29:46.227" Score="3" Body="&lt;p&gt;Many non-organic substances burn. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Reactive metals such as potassium burn when exposed to oxygen. (Thanks to @Jon Custer in his comment)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Non metal elements such as carbon, sulphur and phosphorous also burn in oxygen when heated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So yes fire is definitely possible outside earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally simple organic compounds such as methane are known to exist in large quantities on other bodies such as Titan, a moon of Saturn, which has a whole 'hydrological' cycle based around methane, and to a lesser extent ethane and heavier hydrocarbons, with a methane atmosphere and lakes of methane on the surface. (Thanks to @ron in his comment)&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-14T13:29:46.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28687" PostTypeId="1" CreationDate="2015-04-14T13:30:17.087" Score="-1" ViewCount="198" Body="&lt;p&gt;What's the byproduct of this reaction? After all, it is a &quot;lysis&quot; reaction. So ... would HCl be a byproduct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zf5Wo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-04-14T13:35:06.350" LastActivityDate="2015-04-14T13:35:43.237" Title="What is the byproduct from hydrogenolysis of an acyl chloride?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28688" PostTypeId="1" CreationDate="2015-04-14T13:34:55.867" Score="8" ViewCount="290" Body="&lt;p&gt;Zinc sulfide, copper iodide and copper bromide all adopt the zincblende structure. My lecture notes suggest that this structure is adopted preferentially over the wurtzite structure if the bonding is more covalent (why is this?); thus, it’s adopted where there is a polarizing cation and a polarizable anion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, surely a polarizing cation should be small and highly charged (i.e. not ions like $\ce{Cu+}$ surely? It has a low charge and large-ish radius). What makes this polarizing? I’ve also heard this about $\ce{Ag+}$.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-04-14T18:01:45.963" LastActivityDate="2015-06-13T19:05:28.340" Title="Compounds that adopt the zincblende structure" Tags="&lt;inorganic-chemistry&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28689" PostTypeId="2" ParentId="28687" CreationDate="2015-04-14T13:35:43.237" Score="5" Body="&lt;p&gt;Yes HCl is a byproduct of the Rosenmund reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2JNid.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.organic-chemistry.org/namedreactions/rosenmund-reduction.shtm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.organic-chemistry.org/namedreactions/rosenmund-reduction.shtm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-14T13:35:43.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28692" PostTypeId="1" AcceptedAnswerId="29120" CreationDate="2015-04-14T14:19:37.890" Score="22" ViewCount="6826" Body="&lt;p&gt;Why are the axial bond lengths greater than those of the equatorial bonds in a trigonal bi-pyramid (TBP) geometry molecule; but the opposite is true for pentagonal bi-pyramid (PBP) geometry molecules? I think that it might have something to do with the angles between the bonds, so I thought of a possible explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a TBP molecule, the equatorial bonds are spaced apart from each other by 120 degrees, resulting in lesser repulsion; compared to the axial bonds at 90 degrees. The reduced repulsion leads to greater stability of the equatorial bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the equatorial bonds in a PBP molecule are separated by 72 degrees which results in them being less stable when compared to the axial bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent is this reasoning correct? Please provide a more detailed and accurate explanation, if there is one. Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="4231" LastEditDate="2015-04-21T16:22:47.243" LastActivityDate="2020-03-28T03:44:29.240" Title="PBP vs TBP geometry?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;resonance&gt;&lt;structural-formula&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="28694" PostTypeId="2" ParentId="28666" CreationDate="2015-04-14T15:24:24.317" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to understand the term stability in organic chemistry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can speak of &lt;strong&gt;kinetic&lt;/strong&gt; stability or &lt;strong&gt;thermodynamic&lt;/strong&gt; stability when we discuss chemical compounds.  If a compound is kinetically stable, that indicates that it has a high barrier (large activation energy) to cross and is therefore less reactive than a compound that has a low barrier (small activation energy) to reaction.  In the following chart we can say that the reactants are kinetically less stable and the products are kinetically more stable since the barrier to the forward reaction is smaller than the barrier to the back reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SH3Ze.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/triple_aqa/calculating_energy_changes/energy_from_reactions/revision/5/&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one compound has a lower heat of formation than another compound, we can say that the former compound is thermodynamically more stable than the latter compound.  Again using the above chart, we can say that the product is thermodynamically more stable than the reactant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We know empirically that i-butane is more stable than n-butane. But&#xA;  why is it more stable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Steric interactions do not provide the answer (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.200902550/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use arguments based on carbon-hydrogen hyperconjugation to explain many things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1WSpV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27346/why-is-2-methylpropene-less-in-energy-than-its-alkene-counterparts/27347#27347&quot;&gt;why more highly substituted double bonds are more stable than less substituted double bonds&lt;/a&gt;.  Carbon-carbon hyperconjugation can also be used to explain various chemical phenomena.  In the case at hand, let's compare carbon-carbon hyperconjugated resonance structures for n-butane and 2-methylpropane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3qk8I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hyperconjugated resonance structure for butane involves a primary carbocation, while the hyperconjugated resonance structure for 2-methylpropane involves a more stable secondary carbocation.  This more stable carbon-carbon hyperconjugated resonance structure explains the increased thermodynamic stability of 2-methylpropane compared to the isomeric n-butane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-04-15T03:29:07.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28695" PostTypeId="1" AcceptedAnswerId="28701" CreationDate="2015-04-14T15:33:58.510" Score="5" ViewCount="925" Body="&lt;p&gt;According to Wikipedia, tin-112 is a stable nuclide, but indium-112 is radioactive (half-life only 14 minutes). What is the explanation for this? &lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorDisplayName="user7951" LastEditDate="2016-06-23T16:46:38.137" LastActivityDate="2016-07-23T17:28:04.250" Title="Why is tin-112 stable, but indium-112 radioactive?" Tags="&lt;stability&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28696" PostTypeId="2" ParentId="28676" CreationDate="2015-04-14T15:46:47.400" Score="4" Body="&lt;p&gt;Those drawings are rather confusing.  In the following drawing I've reproduced your &quot;cis&quot; drawing and then below, I've redrawn it more clearly.  We now see that your molecule is actually the &quot;trans&quot; isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y59G3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark wedge bonds should be either drawn both &quot;up&quot; or both &quot;down&quot;.  The same applies to the dashed bonds, they too are either drawn both &quot;up&quot; or both &quot;down&quot;.  Your drawing is different, it has one wedge bond &quot;up&quot; and the other &quot;down&quot;.  This is confusing and makes interpretation somewhat difficult.  In my redrawn version both dark wedge bonds are &quot;up&quot; and both dashed bonds are &quot;down&quot; making interpretation easier.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-14T15:46:47.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28698" PostTypeId="1" AcceptedAnswerId="28713" CreationDate="2015-04-14T16:29:14.827" Score="7" ViewCount="33774" Body="&lt;p&gt;What is the distance between successive tetrahedral voids in a FCC&#xA;crystal structure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between a face centered atom and a tetrahedral void is $\sqrt{\frac{3a}{4}}$ if the cube's edge length is $a$. Now If I know the angle made, between the line joining a tetrahedral void and the face centered atom, and the successive tetrahedral void and face centered atom, I can find the distance.  But how do I calculate that distance since I don't know the angle.&lt;/p&gt;&#xA;" OwnerUserId="14542" LastEditorUserId="5026" LastEditDate="2018-05-12T04:20:44.667" LastActivityDate="2020-04-25T07:27:19.970" Title="Distance between successive tetrahedral voids in FCC" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28700" PostTypeId="1" AcceptedAnswerId="28705" CreationDate="2015-04-14T16:33:50.297" Score="5" ViewCount="36370" Body="&lt;p&gt;These two molecules are both aromatic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fjZyo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fjZyo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the first molecule has a halogen substituent and is named &quot;bromobenzene&quot;. The second molecule has a chlorine and a hydroxyl group and is named &quot;4-chlorophenol&quot;. Assuming there was no alcohol group, how would we know when to use &lt;strong&gt;benzene&lt;/strong&gt; or &lt;strong&gt;phenyl&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BH2tj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BH2tj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, shouldnt the first molecule have been written as &quot;&lt;em&gt;bromophenyl&lt;/em&gt;&quot; and not as &quot;bromobenzene&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="41328" LastEditDate="2017-08-06T10:14:50.370" LastActivityDate="2017-08-06T10:14:50.370" Title="Phenyl or benzene when naming compounds" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28701" PostTypeId="2" ParentId="28695" CreationDate="2015-04-14T16:34:24.963" Score="10" Body="&lt;p&gt;For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called &lt;strong&gt;valley of β-stability&lt;/strong&gt; (see also Weizsäcker’s formula or semi-empirical mass formula).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola). Isobars with lower atomic numbers $Z$ are unstable to $\beta^-$ decay. Isobars with higher atomic numbers $Z$ are unstable to $\beta^+$ decay or electron capture. Therefore, at least one of two adjacent isobars must be radioactive (see Mattauch isobar rule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, since Sn-112 is stable, we expect In-112 to be radioactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term $(\pm d/A^{3/4})$ gives rise to three different parabola depending on whether the nuclides are even-$A$ (with even $Z$ and even $N$), even-$A$ (with odd $Z$ and odd $N$), or odd-$A$ (with even $Z$ and odd $N$, or with odd $Z$ and even $N$).&lt;br&gt;&#xA;In the last case, in which the mass number $A$ is odd, the result is a single parabola with one stable nuclide at the bottom.&lt;br&gt;&#xA;However, for even mass numbers $A$, we find two parabolas; the odd-odd-curve always lies above the even-even curve. Hence, it is possible that two or even three stable isobars exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case with two parabolas also applies to your example. The stable nuclide Sn-112 has 50 protons and 62 neutrons (even-even). The radioactive nuclide In-112 has 49 protons and 63 neutrons (odd-odd). The next neighbour, Cd-112 has 48 protons and 64 neutrons (even-even) and is stable.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-07-23T17:28:04.250" LastActivityDate="2016-07-23T17:28:04.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28702" PostTypeId="1" AcceptedAnswerId="28704" CreationDate="2015-04-14T16:37:15.153" Score="23" ViewCount="1305" Body="&lt;p&gt;We know that there are silicon analogs of alkanes, but according to Wikipedia, silicon analogs of alkenes or alkynes are virtually unknown. How come?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="16683" LastEditDate="2018-10-06T12:42:31.180" LastActivityDate="2018-10-06T12:43:25.080" Title="Why are silicon analogs of alkenes or alkynes so difficult to make?" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28704" PostTypeId="2" ParentId="28702" CreationDate="2015-04-14T17:17:36.783" Score="22" Body="&lt;p&gt;The typical Si–Si single bond length in a silane is around 2.33 Å.  This is much longer than a typical C–C single bond (~1.53 Å) and helps explain why silicon-silicon single bonds are so much weaker than carbon-carbon single bonds (bond dissociation energy: ~53 kcal/mol for silicon-silicon &lt;em&gt;vs.&lt;/em&gt; ~83 kcal/mol for carbon-carbon single bonds).  When we try to form a silicon-silicon double or triple bond, the large separation between the two silicon atoms results in even &lt;strong&gt;less effective p-orbital overlap and therefore even weaker π-bonds&lt;/strong&gt;.  The result is that while &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27839/why-does-si2h4-readily-polymerize-at-rtp-but-ethene-does-not-requires-catalyst/27854#27854&quot;&gt;disilenes&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16538/why-is-disilyne-bent/16543#16543&quot;&gt;disilynes&lt;/a&gt; are known (see the linked SE Chem posts), they are &lt;strong&gt;extremely&lt;/strong&gt; reactive.  Only when bulky substituents are placed around the double or triple bonds (to provide some &quot;steric protection&quot; against further reaction) can the molecules be isolated and characterized.          &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2018-10-06T12:43:25.080" LastActivityDate="2018-10-06T12:43:25.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28705" PostTypeId="2" ParentId="28700" CreationDate="2015-04-14T17:23:03.660" Score="13" Body="&lt;p&gt;You are confused about several overlapping but mostly unrelated issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you are &lt;strong&gt;right&lt;/strong&gt; that &quot;benzyl&quot; is the $\ce{C7H8\bond{~}}$ radical and that &quot;phenyl&quot; is the $\ce{C6H5\bond{~}}$ radical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, although &quot;phenol&quot; and &quot;phenyl&quot; sound almost the same, you would probably do well to think of them as entirely separate naming roots.  &quot;Phenol&quot; is the simplest aromatic alcohol.  Simple aromatic alcohols with other substitutions (e.g. 4-chlorophenol) can thus be named as substituted phenols.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, &quot;benzyl&quot; and &quot;benzene&quot; similarly sound the same but should be treated as distinct concepts.  Substituted aromatic rings can be named as benzene derivatives, such as chlorobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth, for shorthand and convenience, &quot;phenyl&quot; and &quot;benzyl&quot; are sometimes used to name &lt;em&gt;substituents&lt;/em&gt;.  Note that IUPAC has detailed priority rules for what is a substituent and what is the base molecule, but chemists don't always follow these.  &quot;chlorobenzene&quot; is a correct IUPAC name, and it views the molecule as a substituted benzene.  That is, benzene is the base molecule and &quot;chloro&quot; is a substitution.  Chemists sometimes find it convenient to reverse their perspective, even though the resulting names would not be IUPAC-approved.  From a reverse perspective, the chlorine atom is the &quot;base molecule&quot;, and it is substituted by a phenyl group.  Thus you could call the chlorobenzene &quot;phenyl chloride&quot; if you wanted to take this reverse perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, &quot;bromophenyl&quot; ends in &quot;-yl&quot; and so is the name of a &quot;radical&quot;, i.e. a substituent or fragment, not of a complete molecule.  Without any other context, I would think that &quot;bromophenyl&quot; was a $\ce{C6H4Br\bond{~}}$ substituent.  In contrast, &quot;bromobenzene&quot; is a complete molecule, derived from benzene by substitution.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-14T17:23:03.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28706" PostTypeId="2" ParentId="28702" CreationDate="2015-04-14T17:26:27.553" Score="11" Body="&lt;p&gt;Just to add to ron's answer with some information on the properties of the silicon analogues. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Silicon &quot;alkynes&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These exotic compounds were first reported on in 2004 in &lt;a href=&quot;http://www.sciencemag.org/content/305/5691/1755&quot;&gt;this paper.&lt;/a&gt; Typically show bond order somewhat less than 3; the HOMO has some non-bonding/lone pair character which is also manifest in a small deviation from linearity. The silicon-silicon &quot;triple&quot; bond is pretty reactive and most examples have very bulky alkyl substituents to create a steric barrier to reaction. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Silenes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These have been known since 1981; the first example being &lt;a href=&quot;http://www.sciencemag.org/content/214/4527/1343.abstract&quot;&gt;tetramesitylsilene&lt;/a&gt;. Similar to the &quot;alkynes&quot;, (trans) deviation from planarity and bond order less than 2 due to HOMO lone pair character observed. Decrease in Si-Si bond length observed but it is smaller than that typically seen for C-C bonds.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-14T17:26:27.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28707" PostTypeId="1" AcceptedAnswerId="28710" CreationDate="2015-04-14T17:28:34.597" Score="4" ViewCount="341" Body="&lt;p&gt;Is this an error? Why is the nitro group reduced as the benzene ring is oxidized? I don't see any possible reducing agents here.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the mechanism of aromatic fluorination by xenon difluoride? This appears analogous to electrophilic aromatic substitution in which benzene acts as a nucleophile and attacks an electrophile because of the expected meta-substitution pattern given the presence of the nitro group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is fluorine really electrophilic here? How so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hZj3m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2015-04-15T02:37:31.420" Title="Fluorination with xenon difluoride" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28708" PostTypeId="2" ParentId="28683" CreationDate="2015-04-14T17:34:36.577" Score="2" Body="&lt;h3&gt;Carboxylic acids form enolates&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They do, if you beat them to it:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Dissolve the carboxylic acid in a polar, aprotic solvent, such as 1,2-dimethoxyethane, 1,4-dioxane, tetrahydrofuran, etc.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add one quivalent of sodium hydride to deprotonate and form the carboxylate.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add one equivalent of a strong base, such as LDA (lithium diisopropylamide) to deprotonate the carboxylate at the &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-position.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add your aldehyde&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h3&gt;What about acyl chlorides?&lt;/h3&gt;&#xA;&lt;p&gt;Adding a base, such as &lt;em&gt;Hünig's base&lt;/em&gt; (plus catalytic amounts of a Lewis acid) to an enolizable acyl chloride under anhydrous conditions gives rise to a different pathway: a &lt;strong&gt;ketene&lt;/strong&gt; can be formed!&lt;/p&gt;&#xA;&lt;h3&gt;How do ketenes react with aldehydes?&lt;/h3&gt;&#xA;&lt;p&gt;In the presence of a Lewis acid, they form &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;-lactones!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-15T07:25:49.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28709" PostTypeId="1" CreationDate="2015-04-14T17:44:54.913" Score="3" ViewCount="1671" Body="&lt;p&gt;How could we be sure that one certain reaction in equilibrium is dynamic? Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?&lt;/p&gt;&#xA;" OwnerUserId="11549" LastActivityDate="2019-11-26T18:06:32.610" Title="The distinction between dynamic &amp; static equilibrium" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28710" PostTypeId="2" ParentId="28707" CreationDate="2015-04-14T17:53:59.793" Score="5" Body="&lt;p&gt;The reaction as pictured above is incorrect, the nitro group is not reduced.  As you suggest this is (predominantly) an electrophilic aromatic substitution reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually $\ce{HF}$ is used as a catalyst in these reactions and the following equilibrium is established.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{XeF2 + HF &amp;lt;=&amp;gt; XeF+ + HF2^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrophilic $\ce{XeF^+}$ is the reactive species forming a complex with the aromatic ring.  Xenon is expelled from the complex and the reaction proceeds to yield the &lt;em&gt;meta&lt;/em&gt;-fluoro isomer as expected for an electrophilic aromatic substitution.  Some other products, such as substituted biphenyls, are usually formed in small amounts suggesting that some minor pathway (radical-cation?) is also in play.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-15T02:37:31.420" LastActivityDate="2015-04-15T02:37:31.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28711" PostTypeId="2" ParentId="28688" CreationDate="2015-04-14T17:54:10.763" Score="4" Body="&lt;p&gt;Generally when you get a preference between two structures due to directional bonding, the first order environment of the cation is different. In this case however both are tetrahedrally coordinated. The Wurtzite structure has a marginally greater Madelung constant (1.641 versus 1.638).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the top of my head, I think the dominant factor in favour of the blende structure is stronger directional cation-cation and anion-anion interactions. In the case of Cu(I), the d-orbitals are going to be fairly diffuse due to the low charge, potentially allowing some covalent interaction between adjacent sites. A much better example would be GaAs, where both cation and anion are realtively soft/polarizable, which adopts blende, in contrast to AlN, with hard ionic interactions, which adopts wurzite. I don't think it's really as simple as this because the difference in energy between the two structures is often very small; you can see this in the relative stability of both ZnS polymorphs.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-14T17:54:10.763" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28712" PostTypeId="1" AcceptedAnswerId="30817" CreationDate="2015-04-14T18:04:38.437" Score="5" ViewCount="35210" Body="&lt;p&gt;Why does ionization energy increase as we go from left to right in a period?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my textbook, the explanation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is consistent with the idea that electrons added in the same principal quantum level do not completely shield the increasing nuclear charge caused by the added protons. Thus electrons in the same principal quantum level are generally more strongly bound as we move to the right on the periodic table, and there is a generally more increase in ionization energy values as electrons are added to a given principal quantum level.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, doesn't make any sense to me. From what I understand, in a stepwise ionization process, it is always the highest-energy electron (the one bound least tightly) that is removed first. So when we go more to the right in the periodic table, there are more electrons, and thus much more possibility for these to shield the outer electron from attraction to the nucleus. This would make it easier to remove that outer electron, and so I would say the opposite: the ionization energy should lower when we go to the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can someone explain to me why this is not so? Thanks for any clarifications.&lt;/p&gt;&#xA;" OwnerUserId="15491" LastActivityDate="2016-11-05T13:19:41.613" Title="Why does ionization energy increase as we go from left to right in a period?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;periodic-trends&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28713" PostTypeId="2" ParentId="28698" CreationDate="2015-04-14T18:13:37.683" Score="5" Body="&lt;p&gt;The distance between any two successive (adjacent) tetrahedral voids. To calculate it, draw a line joining a tetrahedral void to the corresponding vertex. We know that this distance is $\frac{\sqrt{3}a}{4}$, now take the projection of this on any of the edges. The angle between body diagonal and an edge is cos inverse of $\frac{1}{\sqrt{3}}$, so the projection of the line on an edge is $a/4$, Similarly the other tetrahedral void will also be $a/4$ away from its vertex on this edge. So the distance between these two is $a - a/4 - a/4 = a/2$&lt;/p&gt;&#xA;" OwnerUserId="3928" LastActivityDate="2015-04-14T18:13:37.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28714" PostTypeId="1" AcceptedAnswerId="28716" CreationDate="2015-04-14T18:30:02.590" Score="3" ViewCount="598" Body="&lt;p&gt;My book says that to favour a reaction by SN1 mechanism we need to use a polar protic solvent and a polar aprotic solvent in a reaction by SN2. &#xA;* My Prior Knowledge:*&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two types of polar sovents : Polar Protic Solvents contain a hydrogen bonded to a highly electronegative atom. They can solvate both positively and negatively charged species. On the other hand Polar Aprotic Solvents can only solvate positively charged species to an appreciable extent as there is no hydrogen bonding. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A question which arises now is that why should we use a Polar Aprotic Solvent in an SN2 reaction. The leaving group (which usually is negatively charged) needs to be stabilized as well. Wouldn't it be perfectly fine if I used a Polar Protic Solvent? If this is fine then why does presence of Polar Protic Solvent favor SN1 over SN2?&#xA;Also I asked this question on SE a while back. I am looking for a more satisfactory answer apart from the ones given.&lt;a href=&quot;https://chemistry.stackexchange.com/q/26775/8218&quot;&gt; here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-04-14T18:56:25.553" Title="Solvents in substitution reactions" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28716" PostTypeId="2" ParentId="28714" CreationDate="2015-04-14T18:56:25.553" Score="5" Body="&lt;p&gt;Take the example of methyl bromide reacting with chloride ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In protic solvents the chloride ion is strongly solvated by hydrogen bonding and so the energy of the reactants is lowered, compared to in aprotic solvents. For the SN2 mechanism, the transition state is less polar than the reactants because the negative charge is spread out across the chlorine and the bromine and so the transition state is less stabilised by protic solvents than the reactants, increasing the activation energy of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RSBJI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aprotic solvents, the reactants and transition state are less stabilised because of the lack of hydrogen bonding but crucially the difference between the stabilisation of the reactants and the transition state is less and so the activation energy is lower, meaning that the SN2 reaction proceeds more quickly in aprotic solvents than in protic solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAadn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse argument applies to SN1 reactions where the first transition state is more polar than the reactants and so is more stabilised by protic solvents.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-14T18:56:25.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28717" PostTypeId="1" CreationDate="2015-04-14T19:06:07.660" Score="6" ViewCount="395" Body="&lt;p&gt;I am working on a personal project that involves a balloon filled with liquefied petroleum gas (LPG) at a pressure of &lt;span class=&quot;math-container&quot;&gt;$\pu{4 - 5 bar}$&lt;/span&gt;, and I need some kind of a device/filter that would convert it to gas (instead of liquid) at a low pressure. I used an industrial vaporizer, but it gave me something stronger than I need, and &quot;not clean.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what i need is to build a very big lighter, &#xA;like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGsfW.jpg&quot; alt=&quot;as you can see, just gas, and wick that turns to a &amp;quot;clean&amp;quot; fire&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, just gas, and wick that turns to a &quot;clean&quot; fire, without all the side effects that the vaporizer creates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I am missing a lot of knowledge here, I would like to know if someone could guide me through how I should do this, where I can find the things I will need, what the names and definitions for the stuff needed, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I used the phrase &quot;home gas,&quot; which is common in my state, I meant liquefied petroleum gas (LPG).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the vaporizer I use now (it belongs to someone who knows what he is doing, I do not plan on killing myself :) ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ptUg4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I will try to re-describe my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get a &quot;clean, slow flame&quot; such as a lighter has. What I am getting, regardless of the vaporizer pressure, is an unclean flame, as though some gas is not really burning and just comes out as well, or as if something in the process makes the flame burn dirtier. &lt;/p&gt;&#xA;" OwnerUserId="15583" LastEditorUserId="23561" LastEditDate="2019-04-13T18:29:09.173" LastActivityDate="2019-04-13T18:35:30.937" Title="How can I turn LPG from liquid to gas?" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;hydrocarbons&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28718" PostTypeId="1" AcceptedAnswerId="53699" CreationDate="2015-04-14T19:47:22.563" Score="3" ViewCount="849" Body="&lt;p&gt;I'm looking for standard half-cell reduction potential $\ce{IO4-}$/$\ce{I2}$ in acid medium. Unfortunately I was unable to find the value in all comprehensible reference books. Hard to believe this values have not been measured or calculated so far.&lt;/p&gt;&#xA;" OwnerUserId="1985" LastEditorUserId="1985" LastEditDate="2015-04-14T19:56:46.910" LastActivityDate="2016-06-14T07:28:20.480" Title="Standard Half-Cell Reduction Potential for Periodate to Iodine" Tags="&lt;inorganic-chemistry&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28719" PostTypeId="1" CreationDate="2015-04-14T20:27:42.897" Score="2" ViewCount="71" Body="&lt;p&gt;Is there a porous thin material (e.g. filter paper, membrane) that will cool a stream of air passed through it to an even cooler temperature on the other side? If this material is absorbent, could there be a chemical absorbed into the material to facilitate the cooling?&lt;/p&gt;&#xA;" OwnerUserId="15585" LastEditorUserId="4356" LastEditDate="2015-04-17T05:31:06.060" LastActivityDate="2015-04-17T15:34:25.383" Title="Is there a porous thin material that will cool a stream of air passed through it to an even cooler temperature on the other side?" Tags="&lt;materials&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28720" PostTypeId="1" AcceptedAnswerId="28745" CreationDate="2015-04-14T21:03:57.453" Score="2" ViewCount="127" Body="&lt;p&gt;I keep hearing about this product known as triclosan, and that it is used in sanitizing soap. And I also heard that it is harmful, and causes some sort of muscle problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However some people aren't affected by it. Why is this, and should I&#xA; be worried about washing my hands?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="4945" LastEditDate="2015-04-15T06:16:13.440" LastActivityDate="2015-04-16T04:05:02.650" Title="Is triclosan harmful?" Tags="&lt;everyday-chemistry&gt;&lt;toxicity&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28722" PostTypeId="1" CreationDate="2015-04-14T21:28:45.910" Score="3" ViewCount="381" Body="&lt;p&gt;I am really confused about how reducing partial pressure of a gas by a certain percentage would impact the behavior of an equilibrium system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g $\ce{POCl3_{(g)} &amp;lt;=&amp;gt; POCl_{(g)} + Cl2_{(g)}}$ is at equilibrium.  How would this system behave if the partial pressure of $\ce{Cl2}$ is reduced by 75%?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the partial pressure is related to moles but having a hard time applying it here.  My guess is that the reactant would be consumed to produce more of the $\ce{Cl2}$ since its pressure and thus moles are decreasing.  However, my instructor said the system would react by consuming $\ce{POCl}$, the reason for which was not explained too well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could get an explanation, it would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15587" LastEditorDisplayName="user15489" LastEditDate="2015-04-14T21:32:55.777" LastActivityDate="2015-06-03T12:00:44.557" Title="le Chatelier's Principle" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28723" PostTypeId="2" ParentId="28261" CreationDate="2015-04-14T21:30:54.407" Score="5" Body="&lt;p&gt;This is not a definitive answer to the question but seems to at least give the insight that the premise might not be true (i.e. pKa of typical 6-membered cyclic imine is ~9). I decided to e-mailed the main author of the book that made the above claim, prof. Jonathan Clayden. He kindly replied (if you read, many thanks!) the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have been trying to find the original data for the statement that&#xA;  the cyclic imine (or rather its conjugate acid) has a pKa of around 9.&#xA;  Typically, imines have a pKa between 5 and 7, so this seems unusually&#xA;  high for a start. If this number is wrong then it means that pyridine&#xA;  may have quite a typical pKa for a cyclic imine of this type. I've not&#xA;  had time for a detailed search but I suspect this number may not be&#xA;  fully accurate. Imine pKas are notoriously hard to determine because&#xA;  of their hydrolytic instability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If any additional information comes up, I will edit this answer accordingly.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-14T21:30:54.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28724" PostTypeId="1" CreationDate="2015-04-14T21:32:34.707" Score="3" ViewCount="60211" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At &lt;span class=&quot;math-container&quot;&gt;$25~^\circ\mathrm{C}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$298~\mathrm{K}$&lt;/span&gt;, the reduction of copper(I) oxide, &lt;span class=&quot;math-container&quot;&gt;$ \Delta H = 58.1~\mathrm{kJ}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta S = 165~\mathrm{J/K}$&lt;/span&gt;, is nonspontaneous, &lt;span class=&quot;math-container&quot;&gt;$ \Delta G = 8.9~\mathrm{kJ}$&lt;/span&gt;.&lt;br&gt;&#xA;  Calculate the temperature at which the reaction becomes spontaneous.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So for this question I'm thinking I use the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta G = \Delta H - T\Delta S$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then plug in the values and solve for &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;? &lt;span class=&quot;math-container&quot;&gt;$$8.9~\mathrm{kJ} = 58.1~\mathrm{kJ} -T(0.165~\mathrm{kJ}) $$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="41328" LastEditDate="2019-04-14T23:51:41.087" LastActivityDate="2020-03-31T01:49:48.563" Title="How to calculate the temperature at which a reaction becomes spontaneous?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28725" PostTypeId="2" ParentId="28412" CreationDate="2015-04-14T21:47:30.500" Score="0" Body="&lt;p&gt;agitating the can causes the liquid / gas equilibrium to shift toward the gas phase. since the can is not a closed thermodynamic system, the required energy (latent heat of evaporation) is &quot;borrowed&quot; from the environment.&lt;/p&gt;&#xA;" OwnerUserId="15363" LastActivityDate="2015-04-14T21:47:30.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28726" PostTypeId="2" ParentId="28724" CreationDate="2015-04-14T22:19:09.420" Score="5" Body="&lt;p&gt;You have the right equation.  But the problem already gives you the temperature at which $\Delta G$ is 8.9 kJ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that $\Delta H$ and $\Delta S$ are constant and do not vary with temperature.  $\Delta G$ still will though because of the $T$ in the $\Delta G = \Delta H - T\Delta S$ equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So plug in the values for $\Delta H$ and $\Delta S$, and also the $\Delta G$ value that you think represents the transition from spontaneous to non-spontaneous.  You should be able to take it from there.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-14T22:19:09.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28727" PostTypeId="2" ParentId="28717" CreationDate="2015-04-14T22:34:35.190" Score="1" Body="&lt;p&gt;What you are asking is potentially very dangerous - but it is also what happens inside many cars in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Carburetor&quot; rel=&quot;nofollow&quot;&gt;carburetor&lt;/a&gt; - this is really an engineering / chemical engineering problem and if you are serious about doing this you might want to look up the linked wikipedia page about carburetors. Hope this is useful....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carburetor will premix fuel and air so make a potentially explosive mixture - better just to have a valve where you can regulate the flow of gas above you 'home gas' - this gas would be best burnt with a proper burner that is appropriate for the type of gas you have.&lt;/p&gt;&#xA;" OwnerUserId="11523" LastEditorUserId="16160" LastEditDate="2015-05-12T13:21:29.187" LastActivityDate="2015-05-12T13:21:29.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28728" PostTypeId="2" ParentId="28568" CreationDate="2015-04-14T23:06:37.023" Score="0" Body="&lt;p&gt;Well there might be more reasons than these two:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lithium has a &lt;strong&gt;higher reduction potential&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If you also look at the electronegativities of just &lt;strong&gt;Lithium&lt;/strong&gt; and &lt;strong&gt;Cesium&lt;/strong&gt; then you would notice that the &lt;strong&gt;shielding effect&lt;/strong&gt; is more prevalent in &lt;strong&gt;Cesium&lt;/strong&gt;, thereby reducing the &lt;em&gt;electronegativity&lt;/em&gt; and affecting the reduction potential. So Lithium however, just compared to Cesium, has a higher electronegativity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think these are the two main reasons, please correct me If I am wrong. &lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-14T23:06:37.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28729" PostTypeId="1" CreationDate="2015-04-14T23:35:59.130" Score="1" ViewCount="1016" Body="&lt;p&gt;You cannot use values but must determine from looking&#xA;The equation I have trouble with is as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{2Na(s) + 2H2O(l) -&amp;gt; 2NaOH(aq) + H2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to my teacher the entropy is positive. I also got the entropy to be positive, but now that I think of it, the moles is decreasing and one is becoming aqueous. So does turning into a gas trump everything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if NaOH was solid? Would it mean still that the entropy would be positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also finally according to the textbook the entropy values is as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{s&amp;lt;l&amp;lt;&amp;lt;g}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But where would the &lt;strong&gt;(aq)&lt;/strong&gt; be?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-15T01:15:35.307" Title="Predict Entropy based on the Balanced Equation" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28730" PostTypeId="1" AcceptedAnswerId="28733" CreationDate="2015-04-14T23:38:09.930" Score="8" ViewCount="16177" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0jMXe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0jMXe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook states that (a) has 2 gauche interactions, (b) has 3 gauche interactions and (c) has 3 interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do we determine this? I thought a gauche interaction would occur when two methyl groups are side by side, however I'm unsure where these answers are coming from.&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="16683" LastEditDate="2015-10-02T16:25:07.520" LastActivityDate="2015-10-02T16:25:07.520" Title="Number of gauche interactions in an alkane" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28732" PostTypeId="1" AcceptedAnswerId="28738" CreationDate="2015-04-15T00:01:03.037" Score="3" ViewCount="4000" Body="&lt;p&gt;The classic voltaic cell has a &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; anode and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; cathode. The reduction half reaction occurring at the cathode combines aqueous copper ions and electrons supplied by the anode to form solid &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;. So what is the purpose of the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; cathode? &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; is produced anyways from the electrons supplied by the anode. Why can't we have instead of the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt; cathode, simply the wire connecting to the anode which will supply electrons to fuel the reduction half reaction?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="61578" LastEditDate="2019-06-02T00:14:52.977" LastActivityDate="2019-06-02T03:38:49.897" Title="Why does the cathode have to be made out Cu in a Zn-Cu voltaic cell?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28733" PostTypeId="2" ParentId="28730" CreationDate="2015-04-15T00:16:07.353" Score="8" Body="&lt;p&gt;When we sight down a carbon-carbon single bond and draw a &lt;a href=&quot;http://en.wikipedia.org/wiki/Newman_projection&quot; rel=&quot;noreferrer&quot;&gt;Newman projection&lt;/a&gt; of a particular conformation, any two groups on the adjacent (vicinal) carbons that have a dihedral angle of 60° between them are said to have a gauche relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following figure, the $\ce{X}$ and $\ce{Y}$ have a gauche relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a4SSF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{X}$ and $\ce{Y}$ can be the same (they can both be methyl groups, other alkyl groups, other atoms such as halogen, &lt;em&gt;etc&lt;/em&gt;.) or they can be different from each other. In most organic classes, the gauche interaction between methyl groups or alkyl groups is frequently discussed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers you included (2,3,3) are correct.  Looking at the middle molecule (b) in your figure, there are 3 gauche interactions.  They are between&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the front methyl and the back ethyl&lt;/li&gt;&#xA;&lt;li&gt;the same front methyl and the back methyl&lt;/li&gt;&#xA;&lt;li&gt;the other front methyl and the back methyl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-15T00:16:07.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28734" PostTypeId="2" ParentId="28729" CreationDate="2015-04-15T01:15:35.307" Score="4" Body="&lt;p&gt;Aqueous salts will always have less order than solid salts. There is &lt;strong&gt;much&lt;/strong&gt; less order when ions are floating around in solution than when they are in their rigid lattice structures. This is what I think you are failing to consider. The reaction is not so much:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na_{(s)} + 2H2O_{(l)} -&amp;gt; 2NaOH_{(aq)} + H2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Na_{(s)} + 2H2O_{(l)} -&amp;gt; 2Na^{+}_{(aq)} +2OH^{-}_{(aq)} + H2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What results from the reaction of 2 atoms of a very structured metal and 2 molecules of liquid water, is 4 aqueous ions that are free to move within the solution, and one molecule of gas. Not only are there now more species present, but they are also in a relatively more free states than before.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-15T01:15:35.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28736" PostTypeId="1" CreationDate="2015-04-15T02:57:55.727" Score="13" ViewCount="1204" Body="&lt;p&gt;PROBLEM:&#xA;When I apply paint to my home’s exterior stucco (currently white), the paint changes color. The grays change to purple. The whites change to yellow. Greenish-beige becomes orange. I have purchased many samples of paint, and a color change is apparent for all but the darkest colors. The change in color starts happening immediately when I apply the paint, and it becomes more intense over subsequent days. After a few days, there is a radical difference between the paint’s intended color and the new color.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently a chemical reaction is occurring between the paint and a substance in/on the wall, but I don't know what it is, or how to remedy it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SURFACE FACTORS:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The house is stucco, so perhaps the surface is very alkaline. The house was built in 1971, and several coats of paint have been applied by the previous owners since then, so I would think the alkalinity would no longer be a factor, but perhaps it is. (Would it even be possible for improperly cured stucco to remain highly alkaline for 44 years?)&lt;/li&gt;&#xA;&lt;li&gt;The previous owner told me that, around fifteen years ago, he painted the house with &quot;pool paint.&quot; (Unfortunately, I don’t know what type of “pool paint” was applied, or why it was applied.) My guess is that this was in response to the color-changing problem rather than the cause of the problem, but I don’t know for certain.  (It is interesting that the pool paint has remained white all this time, while I’m unable to get a true white.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have painted the wood next to the stucco, and the paint on the wood appears exactly as it should and it remains constant(aside from becoming just slightly lighter as it dries). In other words, the paint only changes color on the stucco. Since the wood was never painted with pool paint, and since the wood is obviously not stucco, I still don’t know which factor (pool paint, stucco, or something else) is causing the color change. However, this does show me the paint color is truly changing on the stucco.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some chemistry background, but I'm really stumped by this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help in figuring out what’s going on would be very much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="15591" LastEditorUserId="402" LastEditDate="2016-03-01T23:48:09.720" LastActivityDate="2017-05-14T22:17:12.340" Title="Paint changes color when applied to Stucco. Chemical rxn assumed, but what?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28738" PostTypeId="2" ParentId="28732" CreationDate="2015-04-15T04:16:00.120" Score="4" Body="&lt;p&gt;You could certain have a zinc-copper cell in which copper was not the cathode material.  Graphite or silver or whatever could be the electrode material instead.   But once you close the circuit:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Current flows&lt;/li&gt;&#xA;&lt;li&gt;Cu is deposited as a metal solid on the electrode surface, whatever it is made out of.&lt;/li&gt;&#xA;&lt;li&gt;Copper metal is, as it always is, conductive.&lt;/li&gt;&#xA;&lt;li&gt;As a result of steps 2 and 3, there is now a copper metal cathode instead of a cathode made of another material.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So even if you start without a copper cathode you wind up with one.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-15T04:16:00.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28739" PostTypeId="1" AcceptedAnswerId="28776" CreationDate="2015-04-15T04:26:42.370" Score="12" ViewCount="17508" Body="&lt;p&gt;So, I'm considering an Artsy Fartsy project with hard drive platters. Being mildly aware of the world around me, I've come to the conclusion that I probably shouldn't grind up modern (or any) electronics and drink it for tea. To that end, is there anything in hard drive platters that is significantly toxic, or are they fine lying around the house? I don't intend in grinding them or anything, I'll be using them whole, so dust inhalation presumably isn't a big issue.&lt;/p&gt;&#xA;" OwnerUserId="15593" LastEditorUserId="69505" LastEditDate="2019-03-06T13:45:42.560" LastActivityDate="2019-03-06T13:45:42.560" Title="Do hard drive platters contain toxic chemicals?" Tags="&lt;toxicity&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28741" PostTypeId="2" ParentId="28114" CreationDate="2015-04-15T05:00:37.190" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;http://www.rigb.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Royal Institution of Great Britain&lt;/a&gt; has a &lt;a href=&quot;https://www.facebook.com/royalinstitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Facebook page&lt;/a&gt;, and I shared this question there, hoping they would respond here. Instead they answered &lt;a href=&quot;https://www.facebook.com/royalinstitution/posts/1007611645935356&quot; rel=&quot;nofollow noreferrer&quot;&gt;my inquiry on Facebook&lt;/a&gt; itself. Therefore I am quoting it here:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hi Martin. I'm afraid David Thompson is a little mistaken. We have multiple examples of Faraday colloids with the evidence to prove their provenance. We have over 10 samples and a further 700+ slides of his colloidal solutions which have been fixed to glass.&lt;/p&gt;&#xA;&lt;p&gt;For more information, you can look at the research of Ryan Tweney who has looked at the solutions and slides extensively: &lt;a href=&quot;http://ma.ecsdl.org/content/MA2009-02/3/202.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ma.ecsdl.org/content/MA2009-02/3/202.full.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There's also more information on our website:&#xA;&lt;a href=&quot;http://www.rigb.org/our-history/iconic-objects/iconic-objects-list/faraday-gold-colloids?hc_location=ufi&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rigb.org/our-history/iconic-objects/iconic-objects-list/faraday-gold-colloids?hc_location=ufi&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-15T05:00:37.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28742" PostTypeId="1" AcceptedAnswerId="28746" CreationDate="2015-04-15T05:10:31.807" Score="4" ViewCount="2851" Body="&lt;p&gt;I was just going through my last year's school textbooks and while I was reading this chapter on Redox, this above question crossed my mind.&#xA;Shouldn't it be true that since ALL metals lose electrons to get stable, they should also get oxidized. And since they getting oxidized themselves, they should act as reducing agents... Always?&#xA;If the answer is no, please provide an example of the exception.&lt;/p&gt;&#xA;" OwnerUserId="15594" LastActivityDate="2016-03-13T20:29:06.380" Title="If losing electrons means getting oxidized, then does it mean that ALL metals are reducing agents, since they themselves get oxidized?" Tags="&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28745" PostTypeId="2" ParentId="28720" CreationDate="2015-04-15T05:49:58.283" Score="3" Body="&lt;p&gt;Triclosan (2,4,4′-trichloro-2′-hydroxydiphenyl ether) is a nonionic, broad spectrum, antimicrobial agent that, because of its favorable safety profile, has been incorporated into a variety of many personal care products, including deodorant soaps, underarm deodorants, shower gels, and health care personnel handwashes. Triclosan exhibits a moderate degree of substantivity to the skin, and, in many products, it imparts a remnant antimicrobial effect.&lt;a href=&quot;http://www.ajicjournal.org/article/S0196-6553(96)90017-6/abstract&quot; rel=&quot;nofollow&quot;&gt;Here is the link&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;U.S. Food and Drug Administration (FDA) states that &quot;Triclosan is not currently known to be hazardous to humans.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite having significant uses, it also posses little amount of adverse effects on prolong use.It impairs muscle function and skeletal muscle contractile property.This research has only been conducted in mice and results were not good; when mice were exposed to triclosan, heart muscle function was reduced by 25 percent and grip strength was reduced by 18 %.&#xA;So result has concluded that it might cause skeletal and heart muscle failure in humans as well.&lt;a href=&quot;http://articles.mercola.com/sites/articles/archive/2012/08/29/triclosan-in-personal-care-products.aspx&quot; rel=&quot;nofollow&quot;&gt;Here is the link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exposures to triclosan can also cause dermatitis, or skin irritation, and an increase in allergic reactions, especially in children.Due to its link with dioxin, it has been suspected to cause other health problems like weakening of the immune system, decreased fertility, miscarriage, birth defects, and cancer.&lt;a href=&quot;http://www.beyondpesticides.org/antibacterial/health/&quot; rel=&quot;nofollow&quot;&gt;Here is the link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to eradicate these effects,FDA has been engaged in an ongoing scientific and regulatory review of this ingredient. FDA does not have sufficient safety evidence to recommend changing consumer use of products that contain triclosan at this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can conclude that, it cannot be considered entirely harmful for human body.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastEditorUserId="4656" LastEditDate="2015-04-16T04:05:02.650" LastActivityDate="2015-04-16T04:05:02.650" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="28746" PostTypeId="2" ParentId="28742" CreationDate="2015-04-15T05:57:42.993" Score="8" Body="&lt;p&gt;Your reasoning is sound. All (pure) metals can act as reducing agents, but they do not have to all the time. There are a few examples of metals in negative oxidation states, where they would act as oxidising agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most prominent examples are &lt;a href=&quot;http://en.wikipedia.org/wiki/Zintl_phase&quot; rel=&quot;nofollow&quot;&gt;Zintl phases&lt;/a&gt;, where alkaline or earth alkaline metals reduce post transition metals (or metalloids like antimony). One of the first reactions observed was the sodium lead reactions in liquid ammonia in 1907. Back in the day they proposed the following reaction equation:&#xA;$$\ce{Na + excess~Pb ~-&amp;gt;[\ce{NH3}][\mathrm{-50^\circ C}]~ Na+ + $“$~Pb2^{-}~$”$}$$&#xA;It was later found that the actual composition is $\ce{Na4Pb9}$. With crown ethers and/or cryptants it is possible to isolate discrete anions, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more examples like this, and a nice overview is given in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic052110i&quot; rel=&quot;nofollow&quot;&gt;John E. Ellis, &lt;em&gt;Inorg. Chem.&lt;/em&gt;, &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt; (8), 3167–3186.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="24052" LastEditDate="2016-03-13T20:29:06.380" LastActivityDate="2016-03-13T20:29:06.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28747" PostTypeId="2" ParentId="28742" CreationDate="2015-04-15T06:00:57.903" Score="6" Body="&lt;p&gt;No, not always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in the classic Galvanic cell: &lt;img src=&quot;https://i.stack.imgur.com/W9DBC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above image, you have zinc metal acting as a reducing agent at the anode, true to your line of thinking. However, it is also the case that the copper ions at the cathode are being reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can happen quite frequently in electrochemistry as the reduction reaction that occurs is largely governed by the reduction potential of that reaction. In the Galvanic cell example, the copper reduction reaction has a higher standard reduction potential (+0.34V) compared to that of the zinc reduction (-0.76V), meaning that the copper ions are more likely to get reduced, and so the copper reduction will proceed over that of the zinc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image: &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/119/ZnCu_detail.gif?revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemwiki.ucdavis.edu/@api/deki/files/119/ZnCu_detail.gif?revision=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="5234" LastEditDate="2015-04-18T21:16:42.747" LastActivityDate="2015-04-18T21:16:42.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28749" PostTypeId="2" ParentId="28634" CreationDate="2015-04-15T08:52:19.030" Score="1" Body="&lt;p&gt;The polarities are not reversed, basically current always flows from the positive to negative terminal (opposite the flow of electrons by definition). When a battery is discharged, the voltage difference across its terminals will drop. When you then connect the charger with the full voltage across the terminals of the batteries, the chemical process that lead to the discharge of the battery is reversed and the current flows &quot;backwards&quot; if you will, with the help of the charger. So you could think of the battery being discharged a little, and having only 10 volts instead of 12 volts across its terminals, connecting the 12 volts charger to those terminals  (negative to negative and positive to positive) means that there's a voltage difference of 2V from the positive terminal of the charger to the positive terminal of the battery which causes the current to flow and the battery to charge.&lt;br&gt;&#xA;You should look at redox processes if you want to understand more of the chemistry behind it! If you just want to charge your battery, remember to connect the positive terminal of the charger to the positive terminal of the battery!&lt;/p&gt;&#xA;" OwnerUserId="8038" LastActivityDate="2015-04-15T08:52:19.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28751" PostTypeId="1" AcceptedAnswerId="28752" CreationDate="2015-04-15T09:51:43.710" Score="5" ViewCount="167" Body="&lt;p&gt;In reading the article abstract &lt;a href=&quot;http://adsabs.harvard.edu/abs/2004HyInt.156..417D&quot; rel=&quot;noreferrer&quot;&gt;What Oxidation State of Iron Determines the Amethyst Colour?&lt;/a&gt;, the author states that in regards to the gemstone amethyst, from clear quartz:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The crystal was transformed into violet amethyst by gamma-irradiation. The change in colour was accompanied by changes in the Mössbauer spectrum that can be interpreted as the conversion of trivalent iron into the tetravalent state: $\ce{Fe^{3+}→Fe^{4+}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does gamma radiation facilitate the conversion of oxidation state of iron, from $\ce{Fe^{3+}}$ to $\ce{Fe^{4+}}$?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-15T11:06:54.460" LastActivityDate="2015-04-15T11:14:43.507" Title="How does radiation facilitate the formation of tetravalent iron?" Tags="&lt;electronic-configuration&gt;&lt;metal&gt;&lt;oxidation-state&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28752" PostTypeId="2" ParentId="28751" CreationDate="2015-04-15T11:14:43.507" Score="6" Body="&lt;p&gt;The ionization energy of $\ce{Fe^3+}$ to $\ce{Fe^4+}$ is 54.8 eV.&lt;br&gt;&#xA;The energy of typical gamma radiation is much higher (in the order of roughly some keV to a few MeV). Depending on the energy, the interaction of average gamma radiation with the affected electron is mainly due to Compton scattering.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-15T11:14:43.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28754" PostTypeId="1" AcceptedAnswerId="28773" CreationDate="2015-04-15T12:20:47.427" Score="10" ViewCount="5213" Body="&lt;p&gt;I am studying Syntex's naproxene syntheses and I came across a couple of electrophilic substitutions whose mechanism I can't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one is a Friedel-Crafts acylation:&#xA;the products are the ones indicated in the picture and they are separated before proceeding. No other byproducts are shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BeNJW.png&quot; alt=&quot;first reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the acetyl go in pos. 6? As far as I know, When an electrophilic substitution is made on an activated substituted naphthalene, the electrophile will go in the substituted ring, in the pos. 1 (kinetic control) or in pos. 2 (thermodynamic control), which is in &lt;em&gt;ortho&lt;/em&gt; or &lt;em&gt;para&lt;/em&gt; to the activating substituent. So I would expect these products, in theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a3Y2j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the contributing structures that represent the electron-donating effect are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YgJR9.png&quot; alt=&quot;second reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, although I would expect a 6-acetyl derivative as a byproduct (at least in theory, since (4) might allow it), I don't understand why it is the &lt;em&gt;main&lt;/em&gt; product. No clues at all about the 5-acetyl byproduct, since the 5-acetyl intermediate is not stabilized by the electron-donating effect: could it possibly be an error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reaction is a bromination on 2-naphthol:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZShQg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, I can think that bromine would first go in pos. 1, and then, since it is only slightly deactivating, another bromine can be added. But again, why in pos. 6? Why not in pos. 3, in &lt;em&gt;ortho&lt;/em&gt; to the activating substituent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the end, are these reactions done under kinetic or thermodynamic control?&lt;/p&gt;&#xA;" OwnerUserId="15553" LastEditorUserId="15553" LastEditDate="2015-04-15T13:04:37.347" LastActivityDate="2015-04-15T16:33:22.927" Title="Naproxene syntheses: electrophilic aromatic substitution on activated naphthalene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28755" PostTypeId="2" ParentId="28589" CreationDate="2015-04-15T12:39:28.787" Score="0" Body="&lt;p&gt;Some concepts in thermodynamics are unclear due to historic issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first: &quot;In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs&quot;. Thermodinamics arised as a &lt;strong&gt;macroscopic&lt;/strong&gt; subjet, in the XIX century two elemental reactions in oposite directions were not obvious. (Also, analyze your statement cautious).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should understand &quot;spontaneous&quot; term in a macroscopic sense. Of course, you may find contradictions in the literature (because of the mix that have been done with macroscopic and microscopic world ¿because of pedagogical reasons?)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first one is refering entropy changes asociated to proceses that are real. I mean, non considering results of calculations for arbitrary proceses (which are very unprobable to happend).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the example. As most theories, thermodynamics has applicability limits. Specially the restricted version treated in most text books. The example is invalid because of it is completely uncorrect to apply the thermodynamic of you book to a non-homogeneuos system (or a system composed by homogeneus parts). There is a strange tendency to apply thermodynamic to anything. (I do not know exactly what is your book but I can be sure of the kind of treatment because of your cites).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other side, there the author is trying to relate entropy with order/disorder. This is only valid for total energy constant proceses (I'm refering to the tendence to disorder). My advice is: be patient till you take a rigorous course (statistical physics or statistical thermodynamics or thermodynamics from another perspective), learn to repeat the examples for your test and forget it all as fast as you can. Or remember this taking in mind that maybe it is not exactly in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to start with a relatively simple but more serious thermodynamics book, you can check out this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471862568&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Thermodynamics-Introduction-Thermostatistics-Herbert-Callen/dp/0471862568&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14527" LastActivityDate="2015-04-15T12:39:28.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28758" PostTypeId="1" AcceptedAnswerId="28763" CreationDate="2015-04-15T13:24:02.123" Score="4" ViewCount="64405" Body="&lt;p&gt;I am trying to calculate a cost per hour use of a gas we buy in a cylinder. The details I have been working with are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boconline.co.uk/en/products-and-supply/speciality-gas/pure-gases/pure-nitrogen/pure-nitrogen.html&quot; rel=&quot;nofollow&quot;&gt;Gas used CP-grade N2 material number: 110628-L can&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gas: &lt;code&gt;N&lt;/code&gt;, density = &lt;code&gt;1.251 g/L @ 0.101 MPa&lt;/code&gt; M (molar mas) = &lt;code&gt;14.007&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gas cylinder: Pressure = &lt;code&gt;200 bar&lt;/code&gt; or &lt;code&gt;20 MPa&lt;/code&gt; Cylinder volume &lt;code&gt;9.45 m3&lt;/code&gt; or &lt;code&gt;9450 L&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Operating pressure of gas from cylinder = &lt;code&gt;4.5 bar&lt;/code&gt; or &lt;code&gt;0.45 MPa&lt;/code&gt; and flow rate = &lt;code&gt;60 cL/min&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have been applying the ideal gas law to &lt;code&gt;PV=nRT&lt;/code&gt; to figure out how many total litres of N are in the can at 4.5 bar pressure to then get a rough estimate of how many hours of flowing gas I get. However I seem to be getting unreasonable numbers, potentially from me using the equation or reasoning incorrectly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I apply the gas law under isothermal &lt;code&gt;V2=V1/(P2/P1)&lt;/code&gt; conditions to see how much the internal gas volume expands from the 20 MPa pressure, as a sanity check I wanted to see how many L at 1 atm came out. The result was 1.89 ML which makes little sense as the conversion of 1.89 ML N2 to mass = 2.364 Mg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly it equates to 420 kL @ 0.42 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the specifications mean that its 9.45m3 or 9450 L of compressed gas to 200 bar?&lt;/p&gt;&#xA;" OwnerUserId="13090" LastEditorUserId="13090" LastEditDate="2015-04-15T14:34:42.263" LastActivityDate="2019-09-28T16:31:52.970" Title="How to calculate the amount of gas in a compressed cylinder" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28759" PostTypeId="1" AcceptedAnswerId="28767" CreationDate="2015-04-15T13:57:20.507" Score="3" ViewCount="2425" Body="&lt;p&gt;A double bond/alkene reacts as a nucleophile. I have never seen a reaction where it takes part in an $S_N1$ or $S_N2$ reaction with a halogenoalkane for example. This would presumably be a very useful reaction yet I have never seen it. I assume this is because it doesn't work. Is this the case? If so why?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-04-16T01:44:33.407" Title="Alkylation of alkenes by nucleophilic substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28760" PostTypeId="2" ParentId="28644" CreationDate="2015-04-15T14:09:54.433" Score="-3" Body="&lt;p&gt;An add : the Gibbs energy is at its standard state only when the extent of reaction ( greek letter Xi ) is equal to zero. That counts also for the standard state of enthalpy. &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-04-15T14:09:54.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28761" PostTypeId="1" CreationDate="2015-04-15T14:19:01.103" Score="1" ViewCount="158" Body="&lt;p&gt;How can we say that if a charged nucleophile/base is present we can safely rule out the possibility that the reaction can take place via SN1/E1 mechanism? I have been trying to figure this out since hours. Help would be appreciated. Thank you. &lt;a href=&quot;http://www.masterorganicchemistry.com/2012/11/30/deciding-sn1sn2e1e2-2-the-nucleophilebase/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="10185" LastEditDate="2015-06-17T10:38:48.063" LastActivityDate="2015-06-17T10:38:48.063" Title="Charged nucleophile and SN1/E1/SN2/E2 reactions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28763" PostTypeId="2" ParentId="28758" CreationDate="2015-04-15T14:34:26.940" Score="3" Body="&lt;p&gt;Turns out my assumptions were wrong about the stated &lt;span class=&quot;math-container&quot;&gt;$9\,450\ \mathrm L$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; gas. The &lt;span class=&quot;math-container&quot;&gt;$9\,450\ \mathrm L$&lt;/span&gt; is the total volume of standard pressure gas compressed into the cylinder to &lt;span class=&quot;math-container&quot;&gt;$20\ \mathrm{MPa}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$V_2=V_1/(p_2/p_1)$&lt;/span&gt; having &lt;span class=&quot;math-container&quot;&gt;$p_1=0.101\ \mathrm{MPa}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_2=20\ \mathrm{MPa}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_1=9\,450\ \mathrm L$&lt;/span&gt;. The total volume of gas within the cylinder is about &lt;span class=&quot;math-container&quot;&gt;$47\ \mathrm L$&lt;/span&gt; a reasonably fitting size for the cylinder dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence to answer my own question, at the pressure of usage (&lt;span class=&quot;math-container&quot;&gt;$0.45\ \mathrm{MPa}$&lt;/span&gt;) the volume of gas (using same equation as above) is &lt;span class=&quot;math-container&quot;&gt;$2\,121\ \mathrm L$&lt;/span&gt; which at &lt;span class=&quot;math-container&quot;&gt;$60\ \mathrm{cL/min}$&lt;/span&gt; equates to &lt;span class=&quot;math-container&quot;&gt;$59\ \mathrm h$&lt;/span&gt; of flow roughly.&lt;/p&gt;&#xA;" OwnerUserId="13090" LastEditorDisplayName="user7951" LastEditDate="2019-09-28T16:31:52.970" LastActivityDate="2019-09-28T16:31:52.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28764" PostTypeId="1" AcceptedAnswerId="28765" CreationDate="2015-04-15T14:49:04.177" Score="5" ViewCount="260" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Suggest a synthetic route to amide C from the alkyl chloride B (reaction mechanisms are &lt;em&gt;not&lt;/em&gt; required). Standard reagents and solvents may also be used.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRgGC.png&quot; alt=&quot;N,N,3‐trimethylbutanamide from 1‐chloro‐2‐methylpropane&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In this question immediate reaction is to make a grignard reagent with &lt;strong&gt;B&lt;/strong&gt; by reacting it with magnesium and have it attack &lt;span class=&quot;math-container&quot;&gt;$\ce{COX(N(Me)2)}$&lt;/span&gt; to form &lt;strong&gt;C&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I'm not sure how it would be possible to make &lt;span class=&quot;math-container&quot;&gt;$\ce{COX(N(Me)2)}$&lt;/span&gt; from standard reagents.&lt;/p&gt;&#xA;&lt;p&gt;Are there any obvious synthesis ideas I'm missing in making &lt;span class=&quot;math-container&quot;&gt;$\ce{COX(N(Me)2)}$&lt;/span&gt; as surely if I started from &lt;span class=&quot;math-container&quot;&gt;$\ce{COXY}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\ce{X}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Y}$&lt;/span&gt; are Leaving groups, I would end up with &lt;span class=&quot;math-container&quot;&gt;$\ce{CO(N(Me)2)2}$&lt;/span&gt; on reaction with &lt;span class=&quot;math-container&quot;&gt;$\ce{NH(Me)2}$&lt;/span&gt; which would give an alcohol as the end product, as the amine isn't a good leaving group?&lt;/p&gt;&#xA;&lt;p&gt;Also would &lt;span class=&quot;math-container&quot;&gt;$\ce{COXY}$&lt;/span&gt; i.e. &lt;span class=&quot;math-container&quot;&gt;$\ce{COCl2}$&lt;/span&gt; even count as a standard reagent? I have no idea what that really entails.&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="-1" LastEditorDisplayName="user10259" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-26T12:42:49.360" Title="How to synthesise N,N,3‐trimethylbutanamide from 1‐chloro‐2‐methylpropane?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28765" PostTypeId="2" ParentId="28764" CreationDate="2015-04-15T15:11:27.227" Score="7" Body="&lt;p&gt;Phosgene, &lt;span class=&quot;math-container&quot;&gt;$\ce{O=CCl2}$&lt;/span&gt;, is a potent toxin and in no way would I consider it a safe or even standard laboratory reagent (at least on an undergraduate level).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allow me to lead you down the garden path of organic synthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a haloalkane. What do we know about such compounds; they have a polarised &lt;span class=&quot;math-container&quot;&gt;$\ce{C-X}$&lt;/span&gt; bond (&lt;span class=&quot;math-container&quot;&gt;$\ce{x}$&lt;/span&gt; is a halogen) and are prone to nucleophilic attack/substitution. In this case, we have a nice primary, haloalkane, hence, any nucleophilic substitution is likely to be S&lt;sub&gt;N&lt;/sub&gt;2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the target is an amide, but what do we note is different about the target versus the starting material (SM)? There's an extra carbon present (and yes there's no halogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning bells should be going off at this time. What nucleophilic, reagent do we know of that could extend a carbon chain?&#xA;I can think of one: &lt;span class=&quot;math-container&quot;&gt;$\ce{CN-}$&lt;/span&gt; (cyanide ion).&lt;br&gt;&#xA;So, first react with &lt;span class=&quot;math-container&quot;&gt;$\ce{KCN}$&lt;/span&gt;, second react with conc. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; to give a carboxylic acid, third react with &lt;span class=&quot;math-container&quot;&gt;$\ce{NH(Me)2}$&lt;/span&gt; to give the product. &#xA;Now, once you've isolated the carboxylic acid, you can react it 'neat' with &lt;span class=&quot;math-container&quot;&gt;$\ce{NH(Me)2}$&lt;/span&gt; and molecular sieves. The molecular sieves simply ensure a dry reaction and very high temperatures are needed (approx &lt;span class=&quot;math-container&quot;&gt;$\pu{160 ^\circ C}$&lt;/span&gt;). You could convert the carboxylic acid to an acyl chloride and react the amine to ensure a very high yield and 'fast' albeit more 'vigorous' reaction; in this case any ordinary organic solvent should suffice.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastEditorUserId="4945" LastEditDate="2020-02-26T12:42:49.360" LastActivityDate="2020-02-26T12:42:49.360" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28767" PostTypeId="2" ParentId="28759" CreationDate="2015-04-15T15:27:16.100" Score="9" Body="&lt;p&gt;The pi electrons in an alkene or alkyne can act as a nucleophile.  This class of reactions is often taught a bit later than the standard $\ce{S_{N}1}$ and $\ce{S_{N}2}$ reactions and comes up when &lt;a href=&quot;http://en.wikipedia.org/wiki/Neighbouring_group_participation&quot; rel=&quot;nofollow noreferrer&quot;&gt;neighboring group participation&lt;/a&gt; or anchimeric assistance is covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pi electrons in a double or triple bond are not strongly nucleophilic but when you arrange things in their favor, pi electrons can act as a nucleophile.  The reactions involving double bonds that I am familiar with involve an $\ce{S_{N}1}$ mechanism.  Here are a few examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aiE5q.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/rearrang.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In example (1) notice how the compound with the double bond solvolyzes $10^{11}$ times faster than the saturated model compound!  Note too that &lt;strong&gt;only&lt;/strong&gt; the product with acetate group anti to the double bond is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triple bonds and benzene rings can participate as well.  Often in cases involving double or triple bonds new carbon-carbon bonds are formed between the unsaturated linkage and the carbon with the leaving group [examples (2) and (3) above]. In cases involving aromatic rings aromaticity is restored so no new carbon-carbon bonds are formed between the aromatic ring and the carbon with the leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/asJoS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-16T01:44:33.407" LastActivityDate="2015-04-16T01:44:33.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28768" PostTypeId="1" AcceptedAnswerId="33267" CreationDate="2015-04-15T15:37:58.120" Score="11" ViewCount="730" Body="&lt;p&gt;I'm reading a review article on photocatalytic water splitting (using semiconductors) and I came across the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;To achieve photocatalytic water splitting using a single photocatalyst, the bandgap of the semiconductor must straddle the reduction and oxidation potentials of water, which are +0 and +1.23 V vs the normal hydrogen electrode (NHE) at pH =0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions about this statement:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What exactly does it mean when you say that the bandgap has to &quot;straddle&quot; the reduction and oxidation potentials? Does it mean that you need an electron/hole with energy at least equal to 1.23 eV for oxidation/reduction to take place. So, the bandgap has to be greater than 1.23 eV in order to ensure that. Is my understanding accurate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When using a single photocatalyst, I assume the oxygen and hydrogen evolution take place on the same material. If you had separate oxygen evolution and hydrogen evolution catalysts, what would the condition on the bandgaps of these semiconductors be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks! Sorry if these questions are too basic, I have little training in chemistry. &lt;/p&gt;&#xA;" OwnerUserId="14566" LastEditorUserId="4356" LastEditDate="2015-04-15T16:46:35.213" LastActivityDate="2015-06-25T03:23:33.417" Title="Bandgap of a semiconductor for photocatalytic water splitting" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28769" PostTypeId="1" CreationDate="2015-04-15T14:18:35.060" Score="0" ViewCount="143" Body="&lt;p&gt;Why do we find big spaces in the periods of the Periodic Table between $\ce{H}$ and $\ce{He}$, $\ce{Be}$ and $\ce{B}$, and $\ce{Mg}$ and $\ce{Al}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the logic of such organization of the periods?&lt;/p&gt;&#xA;" OwnerDisplayName="James Matthews" LastEditorUserId="1389" LastEditDate="2015-04-15T16:17:36.403" LastActivityDate="2017-05-09T12:51:47.327" Title="Organisation in Periodic table" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-04-15T17:00:15.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="28770" PostTypeId="2" ParentId="28769" CreationDate="2015-04-15T14:34:13.840" Score="1" Body="&lt;p&gt;From a historical perspective, this was done to account for the commonalities between how the various elements behaved. In other words, the Alkali metals all exhibited a strong reaction with $H_2O$, the noble gasses all exhibited inertness in reactions, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is explained in modern day science via the quantum description of electronic structure which indicates that elements within the same group tend to have the same number of valence electrons in the valence shell. Simply put, the reason why there are such gaps between $H$ and $He$; $Be$ and $B$, and otherwise is because there are simply different quantum arrangements of the electrons which are readily accessible to the element. That is, the ground state of $He$ has both accessible $l=0$ states occupied by electrons, which constitutes the valence shell and so is in an energetically favorable state to remain inert - becoming of the noble gasses. Hence, it’s moved far to the right from $H$ to the 8th group; consistent with the historical structure of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s a crude description, but about what’s happening here. I agree however that the chemists may feel more at home discussing the periodic table. I’ll stick to the chart of nuclides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the Wikipedia article is fairly informative on this topic:&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Periodic_table&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Periodic_table&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="kbh" LastEditorUserId="402" LastEditDate="2017-05-09T12:51:47.327" LastActivityDate="2017-05-09T12:51:47.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28772" PostTypeId="1" AcceptedAnswerId="28848" CreationDate="2015-04-15T16:32:47.070" Score="10" ViewCount="9216" Body="&lt;p&gt;Solomons and Fryhle has an example in which they have showed a very simple substitution reaction. The question was to synthesize methyl iodide. What they have done is taken methyl chloride and produced methyl iodide in the presence of $\ce{NaI}$ and ethanol through an $\mathrm{S_N2}$ mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did they use ethanol? I have read that $\mathrm{S_N2}$ reactions are favored under polar aprotic solvents and clearly ethanol is not that. Also, since it's a methyl and all substitution reactions on methyl go through $\mathrm{S_N2}$; so how would it affect my reaction if I would't use ethanol? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="41328" LastEditDate="2017-09-19T08:43:50.360" LastActivityDate="2017-09-19T08:43:50.360" Title="Why do we use ethanol in SN2 reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;solvents&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28773" PostTypeId="2" ParentId="28754" CreationDate="2015-04-15T16:33:22.927" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;No other byproducts are shown&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I haven't checked the original Syntex patent (if you've ever written a patent you know that sometimes important details are omitted), but it seems odd to me that those two isomers are claimed to be the most significant isomers formed. As you point out, the 2-methoxy-5-acetoxy isomer is not expected to be formed in significant amounts based on electrophilic substitution patterns in substituted naphthalenes (Note: here is a nice concise &lt;a href=&quot;http://home.cc.umanitoba.ca/~hultin/chem2220/Support/Electrophilic%20aromatic%20substitution%20in%20polycyclic%20structures.pdf&quot;&gt;reference&lt;/a&gt; that shows how to apply the same concepts used to predict which isomers will  predominate in electrophilic substitution with benzene to substituted naphthalenes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1966/J3/J39660000181#!divAbstract&quot;&gt;abstract&lt;/a&gt; the Friedel-Crafts acetylation of 2-methoxynaphthalene is studied.  By changing solvents they can shift the major product from 2-methoxy-3-acetoxy to 2-methoxy-6-acetoxynaphthalene.  These two isomers are among those expected and the experiment shows that these two isomers are close enough in energy that small factors can significantly impact the product distribution.  Two other factors to keep in mind&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the 6-acetoxy isomer corresponds to &quot;&lt;em&gt;para&lt;/em&gt;&quot; substitution&lt;/li&gt;&#xA;&lt;li&gt;When running a Friedel-Crafts with oxygenated (hydroxyl, methoxy) aromatics, the aluminum chloride will (depending to some degree upon solvent) complex with the oxygen.  In the case at hand this could make the methoxy a bulky substituent disfavoring reaction at the &lt;em&gt;ortho&lt;/em&gt; positions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But again, why in pos. 6? Why not in pos. 3, in ortho to the&#xA;  activating substituent?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Careful low temperature mono-bromination of phenol shows that &lt;em&gt;para&lt;/em&gt;-substitution is strongly favored (&gt;95%, &lt;a href=&quot;https://books.google.com/books?id=oYPRsEgL2EUC&amp;amp;pg=PA155&amp;amp;lpg=PA155&amp;amp;dq=dibromination%20of%20phenol&amp;amp;source=bl&amp;amp;ots=oJV6URTq1-&amp;amp;sig=ioYxPBybiYW8c0rvfB5jioDBzQ0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=emguVfrAB4q1sAScrIHQAw&amp;amp;ved=0CE8Q6AEwCQ#v=onepage&amp;amp;q=dibromination%20of%20phenol&amp;amp;f=false&quot;&gt;reference&lt;/a&gt;).  If additional bromine is available a second bromine will be substituted at the &lt;em&gt;ortho&lt;/em&gt; position. It seems that the same thing is happening here in your 2-naphthol example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you, I might not have predicted these outcomes, but that's why we run experiments.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;are these reactions done under kinetic or thermodynamic control?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, without more information (like product composition as a function of time, temperature, &lt;em&gt;etc&lt;/em&gt;) I can't answer that.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-15T16:33:22.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28774" PostTypeId="2" ParentId="28412" CreationDate="2015-04-15T16:38:55.180" Score="9" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Spray cans don't actually get colder when shaken.  However, shaking a can does increase heat conduction from your hand to the can, making it &lt;em&gt;feel&lt;/em&gt; colder.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Humans don't actually sense external temperature directly; our &lt;a href=&quot;//en.wikipedia.org/wiki/Thermoreceptor&quot;&gt;thermoreceptors&lt;/a&gt; are located under the skin, and thus effectively measure the rate at which body heat is lost through the skin.  This is why, for example, touching a metal surface will feel noticeably colder than a wooden or plastic surface at the same ambient temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you pick up a spray can, it is typically at room temperature, i.e. significantly below body temperature.  The can itself is a thin metal cylinder, and thus conducts heat well, but has very little heat capacity in itself.  Thus, the rate of heat loss from your hands (and thus the sensation of coolness) mainly depends on how quickly whatever's on the &lt;em&gt;inside&lt;/em&gt; of the can will absorb heat from the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a typical &lt;a href=&quot;//en.wikipedia.org/wiki/Aerosol_spray&quot;&gt;aerosol spray can&lt;/a&gt; contains a propellant (mixed with the actual payload fluid) which would be slightly above its boiling point (and thus in gaseous form) at room temperature and pressure, but is kept liquid by the elevated pressure inside the can.  As the spray valve is opened and the pressure reduced, some of the propellant fluid will boil, bringing the pressure back up until the liquid and gas phases are again at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that, in a half-empty spray can, the top half of the can will contain gaseous propellant (which, being a gas, will not absorb heat very effectively), while the bottom half will contain liquid propellant (mixed with the payload fluid) at &lt;em&gt;just&lt;/em&gt; under its boiling point.  This liquid will absorb heat &lt;em&gt;much&lt;/em&gt; better than the gas at the top, both because it's denser (and thus has higher heat conductance and capacity), and also because heating it will cause some of it to boil, converting part of the absorbed heat into its latent heat of vaporization.  (The boiling effect actually causes a half-empty spray can to act, to some extent, as a &lt;a href=&quot;//en.wikipedia.org/wiki/Heat_pipe#Diode_Heat_Pipes&quot;&gt;diode heat pipe&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, even without shaking the can, it's possible to observe that the bottom end of a half-empty spray can will feel cooler to your hand than the top end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when you hold the can and shake it, you cause the propellant liquid to slosh around inside, coating &lt;em&gt;all&lt;/em&gt; of the inner surface of the can instead of just the bottom.  The sloshing also significantly increases the heat transfer from the can to the liquid, by causing the warmer liquid near the surface of the can to mix with the colder liquid in the interior.  (This is essentially the same reason as why moving air feels cooler on your skin than still air.)  Thus, a shaken can will generally feel even cooler to touch than either end of an unshaken can.&lt;/p&gt;&#xA;" OwnerUserId="283" LastActivityDate="2015-04-15T16:38:55.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28775" PostTypeId="2" ParentId="901" CreationDate="2015-04-15T17:03:03.757" Score="0" Body="&lt;p&gt;I wrote this long answer to a question which was deemed &quot;duplicate&quot; but really asked a more specific question, why there are the gaps in the periodic table etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/28769/organisation-in-periodic-table&quot;&gt;Organisation in Periodic table&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The elements are organized in groups (the columns) because the elements in each group have very similar chemical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the noble gases are inert: today, we know that it is because all the electron shells up to a rather big gap are occupied, so there is no free electron that can create bonds with other atoms; and it is not easy for the atom of the noble gas to adopt an additional electron, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the opposite side, the group starting from hydrogen and lithium contains highly reactive elements because they have exactly one electron on top of the full shells above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one considers elements with electrons in the shell $1s$ only, there are only two elements, hydrogen and helium (the maximally reactive; or the maximally inert), and nothing in between. There is nothing in between because the shell $1s$ only contains two &quot;slots&quot; for electrons, so the number of electrons may only be $0,1,2$ (zero is not an atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as we go down the table, towards elements with a higher number of electrons (and protons in the nucleus), there are more possibilities. For example, in the second row, we have 8 elements. Why? It's because the shells $2s$ and $2p$ contain two and six states, respectively – eight in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we want to have at least one electron in the $n=2$ shell, there are eight options, from $1$ to $8$ electrons. Again, the element with $1$ electron in the $n=2$ shell is analogous to the hydrogen, and the element with the maximum number $8$ electrons, i.e. neon, is analogous to helium because the shells are fully occupied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But unless the first group which only consisted of hydrogen and helium, there are 6 new elements in between, beryllium, boron, carbon, nitrogen, and fluorine. Their occupation is something in between the minimal and maximal. Those have no counterparts among the simplest elements that only work with the $1s$, $n=1$ shell because there are simply not six extra integers in between one (hydrogen) and two (helium). That's why the gap appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we work with the more complicated shells, we always reveal some new &quot;intermediate&quot; elements that had no counterparts in the simpler periods. That's why one needs to create increasingly wide &quot;gaps&quot; in the upper periods – because the lower periods have some additional &quot;intermediate&quot; elements because there is a greater number of ways to fill more complicated, larger shells.&lt;/p&gt;&#xA;" OwnerUserId="1913" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-04-16T09:21:53.317" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28776" PostTypeId="2" ParentId="28739" CreationDate="2015-04-15T18:19:29.773" Score="11" Body="&lt;p&gt;I never opened a newer 2.5&quot; hard disk, but I've read that these may contain glass platters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've seen and touched stacks of platters from old 3.5&quot; hard disks - a friend of mine makes clocks from hard disk cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These platters have an aluminium core, on which another metallic layer, that serves as a kind of primer, and the magnetic material are applied by sputtering. The surface is covered by a protective polymer layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Platters have been drilled or cut in a lot of electronics projects, e.g. to make hdd clocks based on the persistence of vision effect. Apart from the usual safety rules for metal works (protect your eyes, don't touch the hot drill, etc.) one only has to make sure that the rotating platter is properly mounted and doesn't start flying around!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the platters aren't mechanically altered and only painted, there is even less danger. The platters are not toxic and do not even have sharp corners - it is safe to keep them around in the living room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, the magnets from these hard disks are more dangerous. These magnets are pretty strong and when playing with them, you can easily jam your fingers between them ;)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-15T18:19:29.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28777" PostTypeId="2" ParentId="28772" CreationDate="2015-04-15T18:31:31.880" Score="4" Body="&lt;p&gt;When compared to some polar, aprotic solvents, such as acetonitrile or dimethyl formamide, ethanol has a couple of advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is rather cheap&lt;/li&gt;&#xA;&lt;li&gt;it is less toxic&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;the solubility of sodium iodide is rather high&lt;/strong&gt;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-15T18:31:31.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28778" PostTypeId="1" AcceptedAnswerId="123269" CreationDate="2015-04-15T18:55:54.603" Score="9" ViewCount="5257" Body="&lt;p&gt;I would like to understand how a mixture of solvents works together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are two solvents combined, and each one alone can dissolve some material, but one acts faster.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they interact when used in combination on the same material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be paint thinner, for example a mixture of toluene and xylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason to use multiple solvents in a paint thinner is to make it dissolve more kinds of paints etc. (In German, this is called &quot;universal thinner&quot;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does the combination of multiple solvents somehow make it work better dissolving a single compound or mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.&lt;br&gt;&#xA;I do not know exactly what is in the thinner, so it does not mean much &amp;ndash; but it made me wonder about this! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in which ways interact multiple solvents in the context of the example, and are there other mechanisms in general, not covered by the example?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Addendum:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For discussing a simple example, let's just assume the thinner contains only the two solvents above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for context, I looked up the &lt;a href=&quot;http://images.raiffeisen.com/Raicom/sdb/415/Universalverduennung.pdf&quot; rel=&quot;noreferrer&quot;&gt;Sicherheitsdatenblatt (MSDS)&lt;/a&gt; for my example thinner, which shows some more solvents on the list of ingredients:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Compound                                Mass %&#xA;----------------------------------------------&#xA;Toluol                                  15- 45&#xA;Xylol                                   10- 40&#xA;Ethanol                                  2- 15 &#xA;Aceton                                   5- 20 &#xA;Ethylacetat                              5- 20 &#xA;n- Butylacetat                          10- 20&#xA;aromat. hydrocarbon mixture (C8-C10)     5- 15 &#xA;aliph. hydrocarbons (C7-C9)              2- 10 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5078" LastEditorUserId="5354" LastEditDate="2015-06-18T18:26:20.897" LastActivityDate="2019-11-02T10:30:43.250" Title="Mixing solvents, or how does paint thinner work?" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28779" PostTypeId="1" AcceptedAnswerId="28783" CreationDate="2015-04-15T19:04:43.873" Score="9" ViewCount="3129" Body="&lt;p&gt;As part of a synthesis I was considering the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OWxwK.jpg&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How selective is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clemmensen_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clemmensen reduction&lt;/a&gt;? I know that this reduction favours ketones adjacent to benzene rings but will it also reduce the other ketone, and/or affect any other parts of the molecule. Are there any other methods that would achieve this transformation more effectively.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="9961" LastEditDate="2017-01-24T19:35:11.383" LastActivityDate="2017-01-24T19:35:11.383" Title="Clemmensen reduction of 1,3 diketones" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28780" PostTypeId="2" ParentId="28778" CreationDate="2015-04-15T19:22:44.817" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some of these universal thinners do contain a lot more than just toluene and the xylenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out about the composition of these mixtures, a &lt;em&gt;material safety data sheet&lt;/em&gt; (&lt;strong&gt;MSDS&lt;/strong&gt;, in German: &lt;em&gt;Sicherheitsdatenblatt&lt;/em&gt;), provided by the manufacturer gives some information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, take this &lt;a href=&quot;http://www.colores.ch/images/pdf/EU_Sidabla/Universal_Nitroverduenner.pdf&quot; rel=&quot;nofollow&quot;&gt;MSDS&lt;/a&gt; by a swiss manufacturer, which lists the following ingredients:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lr}&#xA;\textrm{compound} &amp;amp; \textrm{amount [%]}\\&#xA;\hline&#xA;\textrm{toluene} &amp;amp; 25-50\\&#xA;\textrm{xylene} &amp;amp; 10-25\\&#xA;\textrm{acetone} &amp;amp; 10-25\\&#xA;\textrm{propan-2-ol} &amp;amp; 2.5-10\\&#xA;\textrm{4-methyl-pentan-2-one} &amp;amp; 2,5-10\\&#xA;\textrm{2-methoxy-1-methylethylacetate} &amp;amp; 2,5-10\\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly the latter two might be much better solvents for the polycaprolactone than toluene or xylene.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-15T19:22:44.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28781" PostTypeId="1" AcceptedAnswerId="28794" CreationDate="2015-04-15T19:38:45.777" Score="-1" ViewCount="344" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4uC8U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer (that is false) is A) however can someone please explain why the other options would be true and why it is in fact possible to compress a solid to produce a liquid?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastActivityDate="2015-04-26T00:25:43.703" Title="Interpreting a Phase Diagram?" Tags="&lt;intermolecular-forces&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28782" PostTypeId="2" ParentId="28717" CreationDate="2015-04-15T19:48:09.443" Score="5" Body="&lt;p&gt;You don't really need to do anything to get LPG vapour. It is a gas a room temperature and pressure so as long as you just want the clean gas not mixed with air (which would be dangerous as the mixture will be flammable and potentially explosive), all you need to do is allow the output of the canister to vent into a balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.uk/search?client=safari&amp;amp;rls=en&amp;amp;q=air%20pressure&amp;amp;ie=UTF-8&amp;amp;oe=UTF-8&amp;amp;gfe_rd=cr&amp;amp;ei=-r4uVduJHeiq8weOgIHgBg&quot; rel=&quot;nofollow noreferrer&quot;&gt;LPG&lt;/a&gt; is mostly propane and butane which liquefy under mild pressure. Butane will liquefy at about 220 kPa at normal room temperature (this is a little over twice atmospheric pressure). This is what happens in a typical cigarette lighter like the one illustrated. There is no &quot;wick&quot; involved just a valve that allows the gas to vent to the atmosphere and a flint to create a spark and ignite it (which you can avoid doing by careful handling of the lighter activating just the valve and not the flint). Large cylinders are similar and it might be possible to vent directly into a balloon just by carefully controlling the exit valve on the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that standard supplies of LPG for household use also contain stinky additives to make leaks easy to detect, but otherwise the gas should be similar to that found in a cigarette lighter.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorDisplayName="user7951" LastEditDate="2019-04-13T18:35:30.937" LastActivityDate="2019-04-13T18:35:30.937" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28783" PostTypeId="2" ParentId="28779" CreationDate="2015-04-15T19:51:22.490" Score="7" Body="&lt;p&gt;A &lt;em&gt;Clemmensen&lt;/em&gt; reduction of a 1,3-diketone probably isn't the best idea. The reduction is likely to proceed via a ketyl radical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3gd5Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a 1,3-diketone, formation of a dihydroxy cyclopropane, which undergoes opening/rearrangement cannot be excluded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction has for instance been observed for dimedone, where 2,4,4-trimethyl cyclopentanone was formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which other viable procedures exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on using the given starting material, the following &lt;strong&gt;might (!)&lt;/strong&gt; work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;regioselective (asymmetric) catalytic hydrogenation of the acetophenone-type carbonyl using a 1,2-diamine-substituted &lt;em&gt;Noyori&lt;/em&gt; catalyst&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16379/is-there-a-catalyst-that-will-reduce-an-alcohol-to-an-alkane/16619#16619&quot;&gt;Barton-McCombie&lt;/a&gt; reaction of the resulting alcohol&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are rather interested in the product, you might want to rethink the strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intended product is an &lt;strong&gt;arylbutanone&lt;/strong&gt;, related to the raspberry ketone. A possible retrosynthesis would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bdu6I.png&quot; alt=&quot;Retrosynthesis of an arylbutan-2-one&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get your hands on the necessary benzaldehyde, an aldol reaction with acetone, followed by catalytic hydrogenation might give the right product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5-tert.-butyl-2-hydroxybenzaldehyde is commerically available, but not really cheap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 4-tert.-butylphenol is cheap as dirt. In case you need significant quantities of your final product, you might want to consider the synthesis of the benzaldehyde. The substitution pattern cries for a &lt;em&gt;Reimer-Tiemann&lt;/em&gt; reaction.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-04-17T08:49:11.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28784" PostTypeId="1" AcceptedAnswerId="28802" CreationDate="2015-04-15T19:54:08.723" Score="4" ViewCount="2683" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What product is formed when curium-226 undergoes alpha decay?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Curium has atomic number 96. Since alpha decay decreases the number of protons and neutrons by 2, the product formed has atomic number 94 and atomic mass 222, thus the product is plutonium-222. However this was not an answer choice. The only answer choices were isotopes of radon, radium, and thorium. Either I don't know how to do a very simple question, or there are alternative representations of radioactive isotopes that I don't know about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to determine, if the question itself is wrong or if I am making a mistake.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="5026" LastEditDate="2018-07-05T10:54:10.277" LastActivityDate="2021-05-01T17:18:33.070" Title="What is the product of alpha decay of curium-226?" Tags="&lt;erratum&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28788" PostTypeId="1" AcceptedAnswerId="29674" CreationDate="2015-04-15T21:29:16.020" Score="8" ViewCount="359" Body="&lt;p&gt;According to Wikipedia: &quot;A magnon is a quasiparticle, a collective excitation of the electrons' spin structure in a crystal lattice.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have little pictures in my mind for other quasiparticles. For example, I can picture a phonon as (quantized) vibrations in a crystal lattice, or I can picture excitons as an electron jumping between band gaps (or HOMO-LUMO, etc.), or plasmons as quantized oscillations/sloshing of an electron gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no mental image for what a spin wave looks like. Can someone help me understand this? What would a collective spin excitation look like? Do the spins just precess faster or something of the sort? Also, how would we measure a spin excitation or spin wave?&lt;/p&gt;&#xA;" OwnerUserId="3798" LastActivityDate="2015-05-01T15:08:54.213" Title="How to visualize or think about spin waves (magnons)?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28789" PostTypeId="1" CreationDate="2015-04-15T22:41:20.273" Score="10" ViewCount="369" Body="&lt;p&gt;I am looking for a liquid that illuminates when vibrations are added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not have to be strictly chemical, it might as well be something else as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a crystal in powder form mixed with some kind of oil, or even algae in a thick:ish liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a liquid exist?&lt;/p&gt;&#xA;" OwnerUserId="15614" LastActivityDate="2020-05-25T17:04:55.393" Title="Does liquids that emit light when vibrated exist?" Tags="&lt;energy&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28790" PostTypeId="1" AcceptedAnswerId="28791" CreationDate="2015-04-15T23:03:10.733" Score="2" ViewCount="12202" Body="&lt;p&gt;I'm having quite a fair bit of trouble comprehending the hybridization of cyclohexane. I understand that this is a ringed structure with single C-C bonds, therefore all the carbons should have sp3 hybridization right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently however, this is incorrect, but I am not sure why. I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could provide some insight on this topic, that would be great!&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorDisplayName="user7951" LastEditDate="2016-08-31T19:09:55.337" LastActivityDate="2016-08-31T19:09:55.337" Title="Understanding the hybridization of cyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28791" PostTypeId="2" ParentId="28790" CreationDate="2015-04-15T23:19:33.123" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm having quite a fair bit of trouble comprehending the hybridization&#xA;  of cyclohexane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Atoms can be hybridized. Not entire molecules. Just atoms. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that this is a ringed structure with single C-C bonds,&#xA;  therefore all the carbons should have sp3 hybridization right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, all the carbons in cyclohexane can be said to be sp3 hybridized. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In benzene all the carbons are said to be sp2 hybridized because benzene is flat. Carbons which are sp2 are in &lt;em&gt;the&lt;/em&gt; same plane with their ligands. Sp3 carbons are &lt;em&gt;not&lt;/em&gt; flat with respect to &lt;em&gt;all&lt;/em&gt; their ligands. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why are the sp3 carbons not planar (not flat)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You seem to be adept at identifying sp3 carbons. So I won't go over that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the framework of VSEPR theory - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Atoms are made of charged particles, namely, protons and electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Protons are at the core (nucleus); electrons are distributed around the protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Electrons repel each other (like charges repel). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we have 4 atoms arranged around a central carbon atom the lowest energy conformation or the configuration which will keep each of the 4 atoms the furthest apart is tetrahedral. The angle between the four ligands in a perfect tetrahedron is about 109.5 degrees. If we constrain ourselves to 2 dimensions and force all the atoms to exist in the same plane then the best angle of separation we can achieve is 90 degrees (imagine carbon to be at the center of a square with each corner of the square representing an atom). Clearly, 109 degrees &gt; 90 degrees so the tetrahedral configuration is favored. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-04-15T23:26:32.637" LastActivityDate="2015-04-15T23:26:32.637" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28792" PostTypeId="1" CreationDate="2015-04-16T00:25:06.520" Score="2" ViewCount="881" Body="&lt;p&gt;When I wet an aluminum pot or cookie sheet and then dry it with a towel, it creates dark gray marks on the towel. This happens more if the aluminum is a little rough. It seems that there's a fine black powder coming off the aluminum. I would guess that's aluminum oxide, but I was under the impression that aluminum oxide forms a hard coating that protects it from corrosion. So is that stuff really aluminum oxide? Why does it come off that way? Does it mean the aluminum isn't protected from corrosion?&lt;/p&gt;&#xA;" OwnerUserId="15615" LastEditorUserId="7475" LastEditDate="2015-10-30T17:13:52.047" LastActivityDate="2015-10-30T17:13:52.047" Title="Why does the black powder that comes off aluminum pans behave the way it does?" Tags="&lt;everyday-chemistry&gt;&lt;metal&gt;&lt;cleaning&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28793" PostTypeId="1" CreationDate="2015-04-16T00:40:09.763" Score="5" ViewCount="122" Body="&lt;p&gt;I have two physical chemistry textbooks, one by Atkins and one by Levine, which seem to differ in their notation used when describing Raoult's law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atkins talks about the molar fraction term in Raoult's law as if it were of the entire system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Levine says that Raoult's law has the form $P_i = x^l_i P_i^\circ$, which seems to imply that $x^l_i$ is the molar fraction in the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is it really? And why would Levine use that notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Levine also uses this notation for Henry's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="16683" LastEditDate="2015-10-23T16:52:14.180" LastActivityDate="2015-10-23T16:52:14.180" Title="Raoult's law: is the molar fraction the molar fraction in the entire system?" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;&lt;notation&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28794" PostTypeId="2" ParentId="28781" CreationDate="2015-04-16T00:41:31.637" Score="8" Body="&lt;p&gt;Let us go through this step-by-step:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;A&lt;/strong&gt;: A is undeniably false because if we pick a point on the solid side of the graph and add pressure to it, the point eventually&#xA;  becomes the liquid.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;B&lt;/strong&gt;: B is true because the &lt;em&gt;triple point temperature&lt;/em&gt; is where they all intersect, and the temperature there is less than the &lt;em&gt;critical point&#xA;  temperature&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt;: You have to understand that C is a &lt;strong&gt;&lt;em&gt;can&lt;/em&gt;&lt;/strong&gt; question; it is true because you can form a gas or a liquid.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;D&lt;/strong&gt;: As you can see, where the substance changes phase is below &lt;strong&gt;300 K&lt;/strong&gt;, so we can assume that this one is also true.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;E&lt;/strong&gt;: The liquid will boil when the  vapor pressure equals the pressure. The line from the triple point to the critical point is&#xA;  sloping up. So this is also true. &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;That leaves us with A as the only false answer.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You might want to review some concepts on phase changes and pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However just to add as a side note, this question kind of seems like a homework question, you might want to add more detail and your process, so that people &lt;strong&gt;&lt;em&gt;can&lt;/em&gt;&lt;/strong&gt; take the time to answer you back. &lt;strong&gt;Showing some effort, gives a cause for others to care.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources&lt;/strong&gt;: &lt;a href=&quot;https://chemistry.stackexchange.com/users/5677/jon-custer&quot;&gt;Jon Custer&lt;/a&gt; helped find out E.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-04-26T00:25:43.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28796" PostTypeId="1" CreationDate="2015-04-16T01:45:55.640" Score="3" ViewCount="21441" Body="&lt;p&gt;What is the net ionic equation of sodium hydroxide when it dissolves in water?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the net ionic equation I got &#xA;$$\ce{NaOH(s) + H2O(l) -&amp;gt; NaH+(aq) + OH- (aq) + H2O(l)}$$&#xA;but it was wrong. &#xA;Then I tried:&#xA;$$\ce{NaOH(aq) -&amp;gt; Na+(aq) + OH- (aq)}$$&#xA;and it was wrong as well. &#xA;Can someone please explain?&lt;/p&gt;&#xA;" OwnerUserId="15617" LastEditorUserId="5026" LastEditDate="2018-03-09T06:09:11.560" LastActivityDate="2018-03-09T06:09:11.560" Title="What is the net ionic equation of sodium hydroxide when it dissolves in water?" Tags="&lt;aqueous-solution&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28798" PostTypeId="2" ParentId="28732" CreationDate="2015-04-16T02:29:14.600" Score="1" Body="&lt;p&gt;You could have a cathode made from something other than &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;, however, this reaction will only produce &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu{(s)}}$&lt;/span&gt; as a result if the metal used for the cathode is a weaker reducing agent than &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt;, otherwise the &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; electrode would become plated in &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn{(s)}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; electrode in the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^{2+}}$&lt;/span&gt; solution, for example, would result in the &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn}$&lt;/span&gt; electrode to become plated with &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn{(s)}}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="5026" LastEditDate="2019-06-02T03:38:49.897" LastActivityDate="2019-06-02T03:38:49.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28800" PostTypeId="2" ParentId="28772" CreationDate="2015-04-16T02:36:41.213" Score="1" Body="&lt;p&gt;Where did you find the statement: &quot;all substitution reactions on methyl go through $S_N1$&quot;? It is obviously wrong. Methyl definitely favor $S_N2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the answer. The conversion from $\ce{MeCl}$ to $\ce{MeI}$ go through $S_N2$ mechanism. The best solvent for $\ce{-Cl}$ to $\ce{-I}$ conversion is usually &lt;strong&gt;acetone&lt;/strong&gt;. It is cheap, polar aprotic, good solubility for $\ce{NaI}$ and poor solubility for $\ce{NaCl}$. I think the reason to use ethanol is just because this reaction is too easy to be picky on solvent. It just works in ethanol.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="10185" LastEditDate="2015-10-02T16:30:49.770" LastActivityDate="2015-10-02T16:30:49.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28801" PostTypeId="1" CreationDate="2015-04-16T02:56:11.223" Score="0" ViewCount="96" Body="&lt;p&gt;I have no understanding of chemistry, so these to chemical formulas mean nothing to me. What I am trying to ascertain is are these compounds the same? If not, what differences are there especially if injected into the CSF [which they were]?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Known as Ethyl Iodophenylundecylate: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CH3&#xA;I   (CH2)8 COOC2 H5&#xA;CH&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Known as Ethyl-10-(Iodophenylundecanoate)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CH3&#xA;I -CH2 (CH2)6 CH2 COOC2 H5&#xA;CH&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15619" LastEditorUserId="544" LastEditDate="2018-11-28T00:08:02.980" LastActivityDate="2021-01-11T21:10:04.627" Title="Are ethyl iodophenylundecylate and Ethyl-10-(Iodophenylundecanoate) the same compound?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28802" PostTypeId="2" ParentId="28784" CreationDate="2015-04-16T03:24:02.297" Score="8" Body="&lt;p&gt;The question is flawed (possibly a misprint), your reasoning is sound (given the information provided to you).&lt;/p&gt;&#xA;&lt;p&gt;Curium-226 does not exist, according to the &lt;a href=&quot;http://www.webelements.com/curium/isotopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Web Elements&lt;/a&gt; page on the isotopes of curium, the 'smallest' isotope is curium-240, which undergoes &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; decay to become plutonium-236.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^{240}_{96}Cm -&amp;gt; ^{236}_{94}Pu + ^{4}_{2}\alpha}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The most likely candidate of what the question should have been asking about is radium-226 decaying to radon-222,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^{226}_{88}Ra -&amp;gt; ^{222}_{86}Rn + ^{4}_{2}\alpha}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which is part of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Decay_chain#Uranium_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;uranium-series decay chain&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NwqCy.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is the most likely scenario, as the smallest isotope of thorium is thorium-227 (&lt;a href=&quot;http://www.webelements.com/thorium/isotopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WebElements&lt;/a&gt;) and for radium, it is radium-223 (&lt;a href=&quot;http://www.webelements.com/radium/isotopes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WebElements&lt;/a&gt;), both have too high a mass for anything with an atomic mass of 226 to directly decay into by alpha decay (as per the question).&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="24052" LastEditorDisplayName="user15489" LastEditDate="2021-05-01T17:18:33.070" LastActivityDate="2021-05-01T17:18:33.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28804" PostTypeId="1" CreationDate="2015-04-16T03:34:14.587" Score="1" ViewCount="56" Body="&lt;p&gt;I am interested to learn through heat claculations the quantity of ice that would achieve a chicken carcass core temperature of 8 degree C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chilling tank has a volume of 15,000 liters of flowing water where chicken pass through at a rate of 3,000 carcass per hour having an average carcass weight of 1.4kg and deep muscle temperature of 32 degree C.The flowing water drains at a rate of 5 litres per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplementary ice added to the chilled water at 250kg per hour while 8 degree C chilled water is fed into the tank at 4800 liters per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much more ice is required per hour to reduce the core temperature of chicken to 8 degree C? Appreciate a detailed calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you for your time and efforts.&lt;/p&gt;&#xA;" OwnerUserId="15621" LastActivityDate="2015-04-16T03:34:14.587" Title="How much ice will reduce the chicken carcass temperature from 32 to 8 degree C" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28805" PostTypeId="2" ParentId="28801" CreationDate="2015-04-16T03:36:07.860" Score="0" Body="&lt;p&gt;The answer, in principal, is not. But, based on your info, the correct structure is not so easy to determine and the reason is, as Martin states, there is no info related the positions of the functional groups (This is to say the Iodine and the phenyl). However and taken literally the structure name you shared, the structures are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of the named compound Ethyl 1-( [# ?]-Iodophenyl)-undecylenate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rsLbP.jpg&quot; alt=&quot;First_Compound&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of the named compound Ethyl 10-Iodo-phenyl-undecanoate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwIAu.jpg&quot; alt=&quot;Second_Compound&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most probably is Ethyl 10-( [# ?]-Iodophenyl)-undecanoate (Similar as Martin presented):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z597C.jpg&quot; alt=&quot;Second_Compound&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to the second part of the question about their activity into CSF is hard to say based on the incomplete info to determine their respective influence. May be is more easy to look at any &lt;a href=&quot;https://www.google.com.co/search?q=vademecum&amp;amp;oq=vademecum&amp;amp;aqs=chrome..69i57.3688j0j7&amp;amp;sourceid=chrome&amp;amp;es_sm=122&amp;amp;ie=UTF-8#q=vademecum&amp;amp;tbm=bks&quot; rel=&quot;nofollow noreferrer&quot;&gt;vademecum&lt;/a&gt; for a more generally medicine.&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-04-16T05:24:28.440" LastActivityDate="2015-04-16T05:24:28.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28806" PostTypeId="1" CreationDate="2015-04-16T03:36:29.997" Score="4" ViewCount="2120" Body="&lt;p&gt;I've always been taught that an exothermic dissolution means that the forming of solute-solvent bonds releases more energy than the energy consumed to break solute-solute bonds and solvent-solvent bonds. However, if the attraction that occurs between solute and solvent particles are mainly based off of intermolecular forces, why would the energy released when it is formed manage to exceed the breaking of the comparably stronger intramolecular bonds between solute-solute particles?&lt;/p&gt;&#xA;" OwnerUserId="15620" LastEditorUserId="4945" LastEditDate="2015-04-16T03:56:21.510" LastActivityDate="2015-04-16T05:20:00.573" Title="Why do exothermic dissolution reactions occur?" Tags="&lt;thermodynamics&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28807" PostTypeId="2" ParentId="28801" CreationDate="2015-04-16T03:37:00.420" Score="2" Body="&lt;p&gt;The &amp;quot;structures&amp;quot; you are trying to post and their respective names do not match, as I cannot spot a phenyl group in there. (Also it is hard to tell from this diagram.) But based on what you gave us, I believe these are the same compounds, where ethyl 10-(4-iodophenyl)undecanoate is the correct IUPAC name and Ethyl iodophenylundecylate is a shortened synonym. I also found trivial names for this compound, for example Pantopaque and Iofendylate.&lt;/p&gt;&#xA;&lt;p&gt;You can access this molecules record on &lt;a href=&quot;https://www.chemspider.com/Chemical-Structure.7178.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemSpider&lt;/a&gt; and you will find a couple more names attached to it. From the same page is the following structure taken:&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CMPbG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CMPbG.png&quot; alt=&quot;ethyl 10-(4-iodophenyl)undecanoate&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://www.chemspider.com/ImagesHandler.ashx?id=7178&amp;amp;w=200&amp;amp;h=200&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemspider.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="15751" LastEditDate="2021-01-11T21:10:04.627" LastActivityDate="2021-01-11T21:10:04.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28808" PostTypeId="1" AcceptedAnswerId="28811" CreationDate="2015-04-16T05:06:07.183" Score="6" ViewCount="6121" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How much sulfuric acid &lt;span class=&quot;math-container&quot;&gt;$(\ce{H2SO4})$&lt;/span&gt; can be obtained from 5 kg of sulfur &lt;span class=&quot;math-container&quot;&gt;$\ce{S8}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;answer: 15.312 kg&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To  make 1 molecules of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; we need 1 atoms of S. &lt;br&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now coming to &lt;span class=&quot;math-container&quot;&gt;$\ce{S8}$&lt;/span&gt;. This gives us 8 atoms of S. so from this we can obtain 8 &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; molecules if we take 1 atom of &lt;span class=&quot;math-container&quot;&gt;$\ce{S8}$&lt;/span&gt; and use it to make &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; molecules. So,&lt;br&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 molecule of &lt;span class=&quot;math-container&quot;&gt;$\ce{S_8}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{-&amp;gt;}$&lt;/span&gt; 8 molecules of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;0.019 moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{S_8}$&lt;/span&gt;   &lt;span class=&quot;math-container&quot;&gt;$\ce{-&amp;gt;}$&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$8 \times 0.019=0.15$&lt;/span&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; &lt;br&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Formula mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4} = 1(2)+32+16(4) = 98$&lt;/span&gt;&lt;br&gt;&#xA;we know the formula&lt;br /&gt;&#xA;for &lt;span class=&quot;math-container&quot;&gt;$$\text{Number of moles} = \frac{\text{Mass}}{\text{Formula Mass}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So from this &lt;span class=&quot;math-container&quot;&gt;$\text{Mass}_\ce{H2SO4} = \text{Moles} \times \text{Formula Mass} = 0.15 \times 98=14.7$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now you see that my answer doesn't match to the answer. What did I do wrong?&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-28T09:17:16.280" Title="How much sulphuric acid can be obtained from 5 kg of sulfur" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28809" PostTypeId="2" ParentId="28806" CreationDate="2015-04-16T05:20:00.573" Score="4" Body="&lt;p&gt;Obviously the intramolecular bonds are not stronger than the intermolecular bonds if the majority of the solute dissociates when placed in water. There are some salts ($\ce{AgCl}$, $\ce{BaSO4}$, $\ce{ZnS}$) that have extremely low $\ce{K_{sp}}$ values, meaning the majority of the salt placed in the solution does not dissociate, and are therefore considered &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;insoluble salts&lt;/a&gt;. In these salts, the intramolecular bonds are stronger than the intermolecular bonds formed with the water, which explains why only extremely small amounts of these salts dissociate in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, on the other hand, salts that do dissociate in water. Dissolution requires energy, and this energy is supplied by the kinetic motion of the water. This energy is used in breaking apart lattice structure of the salts, thereby lowering the temperature of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between salts with a negative enthalpy of solution and a positive enthalpy of solution, as you stated, is in formation of bonds between the water and the ions of the salt. The formation of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;hydrogen bond&lt;/a&gt; is an extremely exothermic process, and accounts for the extremely negative enthalpies of solution of $\ce{HF}$, $\ce{NaOH}$, and $\ce{CsOH}$. Weaker intermolecular bonds also form in the dissolution of these salts and many others as a result of the interaction of the water with the electron clouds of the ions. Interactions between the water and the ions always occur, though to a greater extent in some salts than others.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-16T05:20:00.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28811" PostTypeId="2" ParentId="28808" CreationDate="2015-04-16T05:38:58.427" Score="10" Body="&lt;p&gt;$\ce{S6}$, $\ce{S8}$, $\ce{S12}$ – does it make a difference or is it just a trick to make the question more complicated than it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you know for sure is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1}$ &lt;/li&gt;&#xA;&lt;li&gt;$M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$155.93\,\mathrm{mol} \cdot 98.097\,\mathrm{g}\cdot \mathrm{mol}^{-1} = 15.29\,\mathrm{kg}$, which is close enough to the answer.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-04-16T08:30:55.493" LastActivityDate="2015-04-16T08:30:55.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28812" PostTypeId="2" ParentId="28808" CreationDate="2015-04-16T05:40:55.353" Score="5" Body="&lt;p&gt;I prefer simple dimensional analysis as a method of arriving at the answer. I'm sure it will simplify many of your future calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${5 \hspace{.5mm} kg \hspace{1mm} S_8}·\frac{1000 \hspace{.5mm} g}{1 \hspace{.5mm} kg}·\frac{1\hspace{.5mm}mol\hspace{1mm}S_8}{256.528\hspace{.5mm}g\hspace{1mm}S_8}·\frac{8\hspace{.5mm}mol\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}S_8}·\frac{98.07948\hspace{.5mm}g\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}H_2SO_4}·\frac{1\hspace{.5mm}kg}{1000\hspace{.5mm}g}=15.2934\hspace{.5mm}kg\hspace{1mm}H_2SO_4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The discrepancy in the answer is likely as a result of me using very precise values for the molar mass of the atoms used in the calculation, but clearly the answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not familiar with dimensional analysis, the goal is to cancel the units via dividing with fractions. You will notice that all the units in the above equation cancel to leave us with our desired answer, with the units $kg\hspace{1mm}H_2SO_4$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-16T05:40:55.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28813" PostTypeId="1" AcceptedAnswerId="34364" CreationDate="2015-04-16T05:46:04.577" Score="3" ViewCount="3651" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A voltaic cell with $\ce{Ni/Ni^2+}$ and $\ce{Co/Co^2+}$ half cells has the initial concentrations of $\ce{Ni^2+}$ $\pu{0.80 M}$ and $\ce{Co^2+}$ $\pu{0.20 M}$.&lt;br&gt;&#xA;  (&lt;strong&gt;a&lt;/strong&gt;) Find initial cell potential;&lt;br&gt;&#xA;  (&lt;strong&gt;b&lt;/strong&gt;) Find concentration of $\ce{Ni^2+}$ when $E_\mathrm{cell}$ reaches $\pu{0.03 V}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In part (&lt;strong&gt;a&lt;/strong&gt;), I correctly found the initial $E_\mathrm{cell}$ to be $\pu{0.05 V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part (&lt;strong&gt;b&lt;/strong&gt;), the book doesn't have any other examples that are explained showing how to get one of the concentrations, it's supposed to be $\ce{0.50 M}$, but I'm not sure how they get to that. From my tables, I found $E$ half cell for $\ce{Ni}$ to be $\pu{-0.25 V}$ and $E$ half cell for $\ce{Co}$ to be $\pu{-0.28 V}$, and the overall then to be $\pu{0.03 V}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried plugging in everything I have into the Nernst equation, but that still leaves me coming up short.&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-06T09:37:19.577" LastActivityDate="2017-08-06T09:37:19.577" Title="Finding concentrations in a voltaic cell" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28817" PostTypeId="1" AcceptedAnswerId="28850" CreationDate="2015-04-16T08:30:18.140" Score="7" ViewCount="4077" Body="&lt;p&gt;The addition of $\ce{HBr}$ to $\ce{CH2=CHBr}$ produces $\ce{CH3-CHBr2}$. Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to me, halogens have a stronger $-I$ effect than a $+M$ effect (for example a halogen substituent on benzene, deactivates the ring), and hence the $\ce{CH3-CHBr+}$ carbocation should be more &lt;strong&gt;unstable&lt;/strong&gt; as compared to $\ce{CH2+-CH2Br}$. Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : I have seen &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4967/markownikoffs-addition-to-halogenated-alkene&quot;&gt;this&lt;/a&gt; question but it doesn't answer my query.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2017-04-01T17:09:10.367" Title="Addition of HBr to Vinyl bromide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28818" PostTypeId="1" AcceptedAnswerId="29645" CreationDate="2015-04-16T08:52:39.003" Score="3" ViewCount="1525" Body="&lt;p&gt;We have a drinking bottle made partly from &lt;strong&gt;polypropylene&lt;/strong&gt; that wraps when washed in a &lt;strong&gt;dishwasher&lt;/strong&gt;. We would like to change the plastic type so that it can withstand in dishwasher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know for a fact that my dishwasher at home has a lining inside made from polypropylene. After googling this  I have found a few sources that suggest that dishwashers would use a filler material in the PP called &quot;&lt;a href=&quot;http://www.ktron.com/industries_served/Plastics/Calcium_Carbonate_in_Plastics_Compounding.cfm&quot; rel=&quot;nofollow&quot;&gt;calcium carbonate&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear from someone who actually with confidence can tell us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the usual plastic type used inside dishwashers? How is this plastic different from &quot;normal&quot; PP? Will we be able to use the same mold for this kind of plastic as &quot;normal&quot; PP?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="4945" LastEditDate="2015-04-16T09:28:31.620" LastActivityDate="2018-05-06T02:35:29.537" Title="Which plastic type is used inside dishwashers?" Tags="&lt;reference-request&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28819" PostTypeId="2" ParentId="28817" CreationDate="2015-04-16T09:11:27.620" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to me, halogens have a stronger −I effect than a +M effect (for example a halogen substituent on benzene, deactivates the ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The -I effect of a bromo substituent does indeed reduce the rate in a further electrophilic aromatic substitution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, the +M effect of the same substituent is &lt;em&gt;ortho&lt;/em&gt;/&lt;em&gt;para&lt;/em&gt;-directing, as to be seen in the bromination of bromobenzene.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect, namely the stabilisation of a positive charge through interaction with the lone pair of the halogen ($\ce{H3C-CH+-Br &amp;lt;-&amp;gt; H3C-CH=Br+}$) controls the addition of $\ce{HBr}$ to vinyl bromide.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-16T09:11:27.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28820" PostTypeId="1" CreationDate="2015-04-16T09:14:52.673" Score="2" ViewCount="96" Body="&lt;p&gt;Two questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, how does the gate provide a positive charge by running a current through it? I really just don't see how this positive charge could be achieved, unless you had some sort of battery in the gate itself and attached a wire to donate some of the electrons flowing through to the cathode of some other battery. Is it possible to get electrons out of the gate by some other means?&lt;/p&gt;&#xA;" OwnerUserId="7917" LastActivityDate="2015-04-17T13:16:29.797" Title="Transistors: How does the gate work?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28821" PostTypeId="1" CreationDate="2015-04-16T09:48:45.437" Score="5" ViewCount="191" Body="&lt;p&gt;I was just reading about defects and doping in solids and read about the defect in  ionic crystal ($\ce{NaCl}$). There were examples of $\ce{SrCl2}$. Could this also occur with $\ce{CaCl2}$?&lt;/p&gt;&#xA;" OwnerUserId="15629" LastEditorUserId="3735" LastEditDate="2015-04-16T15:54:34.110" LastActivityDate="2017-07-20T22:27:31.660" Title="Is it possible to dope Ca ions in an NaCl crystal to produce a vacancy defect?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28827" PostTypeId="2" ParentId="28820" CreationDate="2015-04-16T13:38:56.180" Score="1" Body="&lt;p&gt;The gate of a FET works much like the grid of a radio tube (valve). The gate is insulated from source and drain; the only current flow (if the insulator were perfect) is to charge the intrinsic gate capacitance. Even a MOSFET has &lt;em&gt;some&lt;/em&gt; leakage, but it's pico- or femtoamperes. See &lt;a href=&quot;https://electronics.stackexchange.com/questions/31594/mosfet-when-can-we-not-assume-that-the-gate-current-is-0&quot;&gt;this discussion&lt;/a&gt; on capacitance and leakage.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:07.607" LastActivityDate="2015-04-16T13:38:56.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28828" PostTypeId="1" AcceptedAnswerId="29556" CreationDate="2015-04-16T13:46:51.737" Score="6" ViewCount="3343" Body="&lt;p&gt;Transition elements are good catalysts because they have multiple oxidation states?Why multiple oxidation states matter?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastActivityDate="2015-04-29T13:21:52.497" Title="Why are multiple oxidation states useful for a catalyst?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;transition-metals&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28829" PostTypeId="1" AcceptedAnswerId="28830" CreationDate="2015-04-16T13:52:57.350" Score="14" ViewCount="447" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why do we insert a numeric multiplier after group positions when naming an isomer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take &lt;code&gt;2,3 dimethylbutane&lt;/code&gt; for example. Why do we add the numerical multiplier &lt;code&gt;di&lt;/code&gt; when we've already explicitly mentioned two locations? Why don't we just use &lt;code&gt;2,3 methylbutane&lt;/code&gt;? This just doesn't feel intuitive.&lt;/p&gt;&#xA;" OwnerUserId="14911" LastEditorUserId="14911" LastEditDate="2018-04-09T09:32:15.570" LastActivityDate="2018-04-09T09:32:15.570" Title="Why do we insert a numeric multiplier after group positions when naming an isomer?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28830" PostTypeId="2" ParentId="28829" CreationDate="2015-04-16T14:06:03.103" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;take 2,3-dimethylbutane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the case of different substituents on different or the same atom, we use a combination of the position &lt;strong&gt;and&lt;/strong&gt; the name of the substituent, such as in&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2-chloro-3-methoxy...&lt;/li&gt;&#xA;&lt;li&gt;3-cyano-4-bromo..., etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier &lt;strong&gt;di&lt;/strong&gt;, together with the positions, just shortens and simplifies the name and still is unambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multiplier &lt;strong&gt;di&lt;/strong&gt; explicitly denotes that there are two identical substituents at the given positions and not just one that bridges two centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fOVJc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example for such a situation would be 1,4-methanonaphthalene (&lt;strong&gt;1&lt;/strong&gt;). Note that even in this context, &lt;strong&gt;di&lt;/strong&gt; can be used, as in 1,4:5,8-dimethanonaphthalene (&lt;strong&gt;2&lt;/strong&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-16T14:25:19.597" LastActivityDate="2015-04-16T14:25:19.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28834" PostTypeId="2" ParentId="28829" CreationDate="2015-04-16T14:40:52.453" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Take &lt;code&gt;2,3 dimethylbutane&lt;/code&gt; for example. Why does one add the numerical multiplier &lt;code&gt;di&lt;/code&gt; when one has already explicitly mentioned two locations? Why doesn't one just use &lt;code&gt;2,3 methylbutane&lt;/code&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, it is the other way round. Multiplicative prefixes (di, tri, tetra, etc.) are always used in names to denote multiplicity of identical features in structures, whereas locants (1, 2, 3, etc.) may be omitted under certain conditions when there is no ambiguity, e.g.&lt;br&gt;&#xA;‘trichloromethane’ (instead of ‘1,1,1-trichloromethane’) or&lt;br&gt;&#xA;‘heptafluorobutanoic acid’ (instead of ‘2,2,3,3,4,4,4-heptafluorobutanoic acid’).&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-01-03T13:27:57.453" LastActivityDate="2016-01-03T13:27:57.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28836" PostTypeId="2" ParentId="27348" CreationDate="2015-04-16T15:37:42.107" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The way to do obtain the charge of a fragment in a molecule is quiet similar to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28450/how-to-calculate-molecular-volume-from-the-wavefunction/28660#28660&quot;&gt;How to calculate molecular volume from the wavefunction?&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The way to obtain the wave function from GAMESS is with the AIMPAC keyword. Look at page 2-18 of &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the GAMESS manual&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As just repost the answer given over there and just editing the correct words to this post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to determine the charge of a fragment in a molecule is by doing a quantum calculation to obtain the wave function and, then, integrate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain the charge of your desire fragment in the molecule by following these steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do a quantum calculation with the desire method and a very good wave function. &lt;/li&gt;&#xA;&lt;li&gt;In the input file, you have to specify to the software of your preference to print out the wave function file (in gaussian, you can do it with output=wfn, more directions: &lt;a href=&quot;http://aim.tkgristmill.com/knownissues.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Directions to print wave function from gaussian&lt;/a&gt;). &lt;/li&gt;&#xA;&lt;li&gt;After the quantum calculation is done, you have to integrate the wave function with a specialized software (see at external links of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atoms_in_molecules&quot; rel=&quot;nofollow noreferrer&quot;&gt;QTAIM software&lt;/a&gt;). &lt;/li&gt;&#xA;&lt;li&gt;After integrating the wave function, you will get the integrated values of several properties, one of them is the atomic charge of each atom. (See page 16 of &lt;a href=&quot;https://books.google.com.co/books?id=protBb8KraoC&amp;amp;pg=PA16&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Quantum Theory of Atoms in Molecules&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Finally, sum up all the atomic charges of the fragment you want to know.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-04-16T19:24:53.827" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="28837" PostTypeId="1" CreationDate="2015-04-16T15:39:11.293" Score="4" ViewCount="5141" Body="&lt;p&gt;I know,  The primary function of a salt bridge is to complete the circuit, so that charge can flow from one half-cell to the other. The second function is to balance the mass by allowing the anions to move to the half-cell where addition cations are being produced.But I can't seem to figure out what would happen if the half-cell compartments of a galvanic cell weren't connected?&lt;/p&gt;&#xA;" OwnerUserId="15632" LastActivityDate="2017-08-17T11:31:36.080" Title="Describe the function of a salt bridge, What would happen if the half-cell compartments of a galvanic cell were not connected?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28840" PostTypeId="2" ParentId="28837" CreationDate="2015-04-16T15:48:58.287" Score="2" Body="&lt;p&gt;My electrochemical knowledge is a wee bit rusty, but I believe the salt bridge helps balance the &lt;strong&gt;charges&lt;/strong&gt; and not mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What charge you ask? So, let imagine you start up your galvanic cell (below). Electrons from the Zn move to Cu electrode. You start to build up a negative charge on the Cu which would resist/repel further addition of electrons after a period of time, hence production of electricity ceases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By having a salt bridge, the electrolyte ions can move from one side to the other to mitigate this issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EqRAP.png&quot; alt=&quot;Galvanic cell&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2015-04-16T15:48:58.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28841" PostTypeId="1" AcceptedAnswerId="28844" CreationDate="2015-04-16T15:51:16.033" Score="4" ViewCount="1554" Body="&lt;p&gt;My chemistry book says that delocalized electrons are the reason for electrical conductivity in a metal. However it says that nonmetals like sulfur, phosphorus and silicon have &quot;low&quot; conductivity. How can they conduct electricity if they have no delocalized electrons? Silicon has a giant macromolecular structure while sulfur and phosphorus have simple molecular structure. What causes them to conduct even for a little bit?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="3735" LastEditDate="2015-04-16T16:00:37.117" LastActivityDate="2015-08-14T16:38:15.590" Title="Why is the conductivity of non-metals not equal to zero?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;elements&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28844" PostTypeId="2" ParentId="28841" CreationDate="2015-04-16T16:37:03.937" Score="7" Body="&lt;p&gt;When we think of molecules we think of their &lt;strong&gt;molecular orbitals&lt;/strong&gt;.  If we place these molecules into an ordered solid like a crystal, then the orbitals on adjacent molecules are close enough to interact and &quot;extended&quot; molecular orbitals are formed.  We call these extended molecular orbitals &quot;&lt;strong&gt;bands&lt;/strong&gt;&quot;. Typically these bands are &quot;thicker&quot; (spread over a wider range of energies) than molecular orbitals.  That is to say that because the energy of a molecular orbital in a molecule near an edge will be slightly different than the same molecular orbital in a molecule in the middle of the crystal, there is a spread of energies associated with a band.  Ultimately, because of this &quot;thickening&quot; of bands, all of the bands containing electrons overlap and form what is termed the &lt;strong&gt;valence band&lt;/strong&gt;.  Similarly all of the empty bands overlap and form the &lt;strong&gt;conduction band&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/loVuB.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://people.seas.harvard.edu/~jones/es154/lectures/lecture_2/energy_gap/energy_gap.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The valence band and the conduction band are analogous to the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) that exist in individual molecules.  The HOMO-LUMO separation is termed the &quot;&lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Band_gap&quot; rel=&quot;nofollow noreferrer&quot;&gt;band gap&lt;/a&gt;&lt;/strong&gt;&quot; in a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In metals the band gap is essentially zero with the valence and conduction bands overlapping; electrons move readily from the valence band to the conduction band. Electrons in the conduction band then move freely throughout the solid and give metals their characteristic conductivity. In insulators the band gap is very large and very few electrons have the thermal energy needed to be promoted to the  conduction band, hence there is basically no conduction.  Semiconductors lie in between these two extremes and thermal energy can promote some electrons (not as much as in a metal, but far more than in an insulator) to the conduction band, enough to have some conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVZPn.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://sti.epfl.ch/page-61514-en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-16T23:05:21.403" LastActivityDate="2015-04-16T23:05:21.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28845" PostTypeId="1" AcceptedAnswerId="28846" CreationDate="2015-04-16T17:00:22.370" Score="11" ViewCount="21348" Body="&lt;p&gt;This is the mechanism given in my lecture handout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7w5ol.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a problem with the third species along. Surely the negatively charged oxygen would protonate before the other oxygen would. Also, why does this reaction work at all? The amine group is more basic than the &quot;alcohol group&quot; that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2015-09-13T15:37:13.493" LastActivityDate="2016-03-18T06:11:52.740" Title="Mechanism for formation of amide by reaction of amine with ester" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;esters&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28846" PostTypeId="2" ParentId="28845" CreationDate="2015-04-16T17:10:13.703" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Surely the negatively charged oxygen would protonate before the other&#xA;  oxygen would.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the time, but not all of the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The amine group is more basic than the &quot;alcohol group&quot; that must leave&#xA;  to form the amide so why wouldn't that protonate first and then just&#xA;  leave (i.e no reaction)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the time, but not all of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reasoning is correct, but remember, &lt;strong&gt;all of these steps are equilibria&lt;/strong&gt;.  Although what you say will happen most of the time, some of the time the reaction will take the forward route.  Presumably the reaction is run with a large excess of ammonia to &quot;force&quot; the reaction in the forward direction.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-16T17:10:13.703" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28847" PostTypeId="1" AcceptedAnswerId="29123" CreationDate="2015-04-16T17:54:33.463" Score="16" ViewCount="526" Body="&lt;p&gt;What type of dash does one use when indicating something like F$^{-}$? I assumed it was a minus, since you indicate positive charge with a +, but there is one professor I know who insists that it should be typeset with an endash. Is there a style manual on chemical typesetting I can consult. &lt;/p&gt;&#xA;" OwnerUserId="121" LastEditorUserId="4945" LastEditDate="2015-04-21T09:29:29.403" LastActivityDate="2021-02-06T18:32:24.293" Title="Proper dash to use for negative charge" Tags="&lt;reference-request&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28848" PostTypeId="2" ParentId="28772" CreationDate="2015-04-16T18:49:50.590" Score="9" Body="&lt;p&gt;Usually, nucleophilic substitution reactions of anionic nucleophiles occur more rapidly in polar aprotic solvents. In such solvents (e.g. DMSO) the typical relative reactivity order is&lt;br&gt;&#xA;$$ \ce{SCN-} &amp;lt; \ce{I-} &amp;lt; \ce{Br-} &amp;lt; \ce{Cl-} &amp;lt; \ce{F-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In protic hydrogen-bonding solvents (e.g. ethanol), anions are solvated. Therefore, the reactivity of nucleophiles is decreased. Since hard nucleophiles are more strongly solvated than soft nucleophiles, the &lt;em&gt;relative&lt;/em&gt; reactivity of soft anions is increased in such solvents. The typical relative reactivity order is&lt;br&gt;&#xA;$$ \ce{F-} &amp;lt; \ce{Cl-} &amp;lt; \ce{Br-} &amp;lt; \ce{I-} &amp;lt; \ce{SCN-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given example, $\ce{-Cl}$ shall be replaced by $\ce{-I}$; and in alcohols, the nucleophilicity of $\ce{I-}$ is higher than the nucleophilicity of $\ce{Cl-}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-16T18:49:50.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28849" PostTypeId="2" ParentId="28847" CreationDate="2015-04-16T18:54:51.427" Score="11" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;[…] one professor I know who insists that it should be typeset with an en dash.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;I disagree!&lt;/strong&gt; A standard minus sign should be&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;on the same height as the horizontal line of a plus sign;&lt;/li&gt;&#xA;&lt;li&gt;as long as an equal sign.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The en dash is&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lower than the horizontal bar of the plus sign;&lt;/li&gt;&#xA;&lt;li&gt;slightly longer than the equal sign.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Unicode has dedicated characters for minus and superscript minus&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;standard minus: U+2212: −&lt;/li&gt;&#xA;&lt;li&gt;superscript minus: U+207B: Cl⁻&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;See also:&lt;br /&gt;&#xA;The situation, and how to realize the correct typography using &lt;span class=&quot;math-container&quot;&gt;$\mathrm\LaTeX$&lt;/span&gt; has been described in the German article &lt;em&gt;Feinheiten bei wissenschaftlichen Publikationen – Mikrotypographie-Regeln, Teil II&lt;/em&gt; by &lt;em&gt;Marion Neubauer&lt;/em&gt; in &lt;em&gt;Die TeXnische Kommödie&lt;/em&gt;, &lt;strong&gt;1997&lt;/strong&gt;, &lt;em&gt;1&lt;/em&gt;, 25-44 (&lt;a href=&quot;http://www.thamnos.de/stuff/latex/dtk97_1_neubauer_feinheiten.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="41328" LastEditDate="2021-02-06T18:32:24.293" LastActivityDate="2021-02-06T18:32:24.293" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28850" PostTypeId="2" ParentId="28817" CreationDate="2015-04-16T20:16:32.993" Score="6" Body="&lt;p&gt;The reaction of $\ce{HBr}$ with olefins is a relatively fast reaction, in other words, the activation energy required is small.  Further, formation of the carbocation intermediate is the rate determining (slow) step in the overall process.  Considering both of these facts and applying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammond&amp;#39;s_postulate&quot; rel=&quot;noreferrer&quot;&gt;Hammond postulate&lt;/a&gt; we can say that the transition state leading from the reactants to the carbocation intermediate will be an early transition state, &lt;strong&gt;it will look more like the reactants than the carbocation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that resonance &lt;strong&gt;and&lt;/strong&gt; inductive effects need to be examined with regard to the &lt;strong&gt;reactants&lt;/strong&gt;, rather than the transition state, in order to determine how the molecule will react with an electrophile (the proton).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Bggj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the above diagram illustrates, both resonance and inductive effects (in an early transition state that resembles the reactant) suggest that the &lt;em&gt;beta&lt;/em&gt; carbon will be more nucleophilic than the &lt;em&gt;alpha&lt;/em&gt; carbon.  Therefore addition of the proton (electrophile) will occur at the &lt;em&gt;beta&lt;/em&gt; carbon.  This will be followed by rapid addition of $\ce{Br^{-}}$ at the carbon already bearing a bromine atom (the &lt;em&gt;alpha&lt;/em&gt; carbon) to yield a geminal dibromide.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-16T20:25:47.967" LastActivityDate="2015-04-16T20:25:47.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28852" PostTypeId="1" CreationDate="2015-04-16T20:23:51.633" Score="1" ViewCount="158" Body="&lt;p&gt;which conformer stable of piperazine, when hydrogen of N in axial or in equatorial, or lone pair? why?&lt;/p&gt;&#xA;" OwnerUserId="15639" LastActivityDate="2015-04-16T20:23:51.633" Title="Stable conformers of piperazine" Tags="&lt;stereochemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28853" PostTypeId="1" AcceptedAnswerId="28854" CreationDate="2015-04-16T20:23:57.657" Score="11" ViewCount="25106" Body="&lt;p&gt;I think I've heard hardening is when you dunk red-hot metal into cold water, and tempering is when you take that hardened metal, heat it slightly, and then let it cool slowly. However what is the difference between their end products?&lt;/p&gt;&#xA;" OwnerUserId="15588" LastActivityDate="2015-04-29T15:11:59.293" Title="What is the difference between tempering and hardening metals and their end products?" Tags="&lt;metal&gt;&lt;heat&gt;&lt;metallurgy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28854" PostTypeId="2" ParentId="28853" CreationDate="2015-04-16T21:34:16.673" Score="5" Body="&lt;p&gt;As the names imply, hardening makes the metal more rigid but more brittle, and tempering (from &quot;temperate&quot;, moderate), forgoes some hardness for increased toughness. Iron alloys are hardened by rapid &lt;a href=&quot;http://en.wikipedia.org/wiki/Quench&quot; rel=&quot;noreferrer&quot;&gt;quenching&lt;/a&gt; and toughened by &lt;a href=&quot;http://en.wikipedia.org/wiki/Annealing_%28metallurgy%29&quot; rel=&quot;noreferrer&quot;&gt;annealing&lt;/a&gt;, but some copper alloys are made more ductile by quenching and are hardened by working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In brief, &lt;em&gt;annealing&lt;/em&gt; allows atoms to migrate to a less-strained position, whereas &lt;em&gt;quenching&lt;/em&gt; lock atoms in place; this is used to make &lt;a href=&quot;http://en.wikipedia.org/wiki/Toughened_glass&quot; rel=&quot;noreferrer&quot;&gt;tempered glass&lt;/a&gt;, as well as to harden metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extremely rapid quenching can produce a &lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_metal&quot; rel=&quot;noreferrer&quot;&gt;metallic glass&lt;/a&gt;, where precipitation of constituents is prevented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edged-tools may need the body to be ductile, so that it does not shatter, but require for the edge to be hardened so that it retains sharpness. See &lt;a href=&quot;http://www.smt.sandvik.com/en/products/strip-steel/strip-products/knife-steel/hardening-guide/purpose-of-hardening-and-tempering/&quot; rel=&quot;noreferrer&quot;&gt;http://www.smt.sandvik.com/en/products/strip-steel/strip-products/knife-steel/hardening-guide/purpose-of-hardening-and-tempering/&lt;/a&gt; or &lt;a href=&quot;http://www.stormthecastle.com/blacksmithing/blacksmithing-a-knife/stock-removal-method-of-knifemaking-part-4.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.stormthecastle.com/blacksmithing/blacksmithing-a-knife/stock-removal-method-of-knifemaking-part-4.htm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-16T21:34:16.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28855" PostTypeId="1" AcceptedAnswerId="28861" CreationDate="2015-04-16T21:34:33.893" Score="3" ViewCount="1065" Body="&lt;p&gt;Is there a way to extract the molecular wavefunction from GAMESS output?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are packages that can extract the eigenvalues or the exponents and show them as a table but that is not of much help as anyone can easily find those values in the output file by looking at the numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is the complete wavefunction equation. Is there such a package or something available. Something like this simple extractor I have written for STO-3G (this is just a very primitive one that only works for small STO-3G outputs ) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VkPji.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastActivityDate="2017-04-09T04:55:20.443" Title="How to extract wavefunction from GAMESS output?" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28856" PostTypeId="1" CreationDate="2015-04-16T22:01:49.567" Score="2" ViewCount="268" Body="&lt;p&gt;I am using a calculation software called CASTEP, within Materials Studio, and I want to make a &quot;movie&quot; of my geometry optimization, but don't know how to. CASTEP allows one to do many many different types of calculations, and one of the simpler ones it does is a geometry optimization for a unit cell. I remember hearing that it was possible to show a &quot;movie&quot; of the geometry optimization. This &quot;movie&quot; would show all of the steps from the beginning atom positions in the unit cell to the final converged one. I don't know how to make this &quot;movie&quot; and was wondering if someone else knows how. &lt;/p&gt;&#xA;" OwnerUserId="15642" LastEditorUserId="5017" LastEditDate="2015-04-27T14:47:54.730" LastActivityDate="2015-04-27T14:47:54.730" Title="CASTEP Geometry Optimization Movie" Tags="&lt;theoretical-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28857" PostTypeId="1" CreationDate="2015-04-17T00:59:13.073" Score="5" ViewCount="1643" Body="&lt;p&gt;$\ce{Na2CO3}$ is deliquescent in nature, and $\ce{NaHCO3}$ is not, for example, so $\ce{Na2CO3}$ should have been more soluble than $\ce{NaHCO3}$ in water ($\ce{NaHCO3}$ I always thought was less soluble since it precipitated easily in the Solvay process). But my book says $\ce{Na2CO3}$ is less soluble than $\ce{NaHCO3}$ in water. Isn't deliquescence though an indicator of high solubility in water?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="15410" LastEditDate="2015-04-17T01:29:20.263" LastActivityDate="2016-02-11T11:51:22.913" Title="Are all deliquescent salts highly soluble?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28861" PostTypeId="2" ParentId="28855" CreationDate="2015-04-17T02:45:19.000" Score="1" Body="&lt;p&gt;I'm not too familiar with the input structure of GAMESS. But, I found and input in which you can find how to call the AIMPAC keyword &lt;a href=&quot;https://gc3pie.googlecode.com/svn-history/r863/trunk/htpie/examples/ghessiantest/hold/Coran.hess.inp&quot; rel=&quot;nofollow noreferrer&quot;&gt;GAMESS input with AIMPAC keyword&lt;/a&gt; and this is how &lt;a href=&quot;https://github.com/theochem/horton/blob/master/data/test/he_spdfgh_orbital.wfn&quot; rel=&quot;nofollow noreferrer&quot;&gt;wfn file&lt;/a&gt; looks like. In principle, when you indicate the AIMPAC keyword, the GAMESS itself print the wave function in a file with wfn extension at the same folder of the input/output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you get the wfn file, you have to integrate it with (I recommend) 1. &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aimpac/download/download.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;AIMPAC&lt;/a&gt; or 2. &lt;a href=&quot;http://sobereva.com/multiwfn/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MULTIWFN&lt;/a&gt; (Both are free).&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="11367" LastEditDate="2017-04-09T04:55:20.443" LastActivityDate="2017-04-09T04:55:20.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28862" PostTypeId="1" AcceptedAnswerId="28868" CreationDate="2015-04-17T02:51:06.423" Score="2" ViewCount="1275" Body="&lt;p&gt;I was presented with the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided.&#xA;Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen and water away from the iron) &#xA;If the layer of Tin is worn away or scratched, the presence of Tin will actually accelerate the corrosion of the underlying iron. (&lt;a href=&quot;http://www.mikeblaber.org/oldwine/chm1046/notes/Electro/Corrode/Corrode.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.mikeblaber.org/oldwine/chm1046/notes/Electro/Corrode/Corrode.htm&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why Iron would be oxidized when the coating breaks, as an observation of the reduction (oxidation) potential would show that it is more favorable to oxidize iron, over tin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the portion I am confused on is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would corrosion be accelerated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passage I was reading suggested that it was because Tin acts as a cathode and thus would accelerate the corrosion, but that does not make a lot of sense to me. As I would see it, there would be little difference whether the Tin is there or not, and thus the reaction (corrosion) should not be accelerated, but simply as quick as it would have been without the coating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight would be greatly appreciated! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-17T05:32:51.647" Title="Accelerated Oxidation Of Iron When Coating Breaks" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28863" PostTypeId="2" ParentId="13607" CreationDate="2015-04-17T03:13:55.200" Score="28" Body="&lt;p&gt;Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy.&#xA;Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.&lt;/p&gt;&#xA;" OwnerUserId="15647" LastEditorUserId="4945" LastEditDate="2015-04-17T05:44:31.443" LastActivityDate="2015-04-17T05:44:31.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28864" PostTypeId="1" AcceptedAnswerId="32677" CreationDate="2015-04-17T03:19:19.977" Score="4" ViewCount="11705" Body="&lt;p&gt;The axial bonds of $\ce{PF5}$ are longer than those of the equatorial positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One explanation is that because the axial bonds are experiencing more repulsion &#xA;than those of the equatorial and this leads to longer bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another explanation is that the axial bonds of p/d hybrids, whereas the equatorial bonds are s/p hybrids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to comprehend and appreciate the first explanation. However, I am unable to understand how there are differences in the hybrids. I had always considered d hybrids a dubious argument at best, and I believed the hybridization would lead to degenerate orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,  the explanation provided would suggest otherwise, that there are different hybridized orbitals of different energies at different portions of a single hybridized atom (excluding lone pairs. I understand those being non-hybridized in a hybridized atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, my question is how the hybrid argument that was presented works and how would one be able  to realize the different hybridized states of an atom in a molecule (as an approximation).&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-06-09T22:33:11.923" Title="Why are the axial bonds of PF5 longer than those of the equatorial bonds? (Hybridization)" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2015-06-10T13:21:30.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="28865" PostTypeId="1" AcceptedAnswerId="28903" CreationDate="2015-04-17T03:42:54.733" Score="2" ViewCount="491" Body="&lt;p&gt;For example, $\ce{BaHPO_{4}}$ has acidity of 2 and basicity 1. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why does this so? Its basicity is 1 because it has 1 replaceable hydrogen ion. &lt;/li&gt;&#xA;&lt;li&gt;But why does this have acidity? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11331" LastEditorUserId="15429" LastEditDate="2015-04-18T03:04:40.807" LastActivityDate="2016-05-23T07:17:58.790" Title="Why does an acid salt possess acidity as well as basicity?" Tags="&lt;acid-base&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28866" PostTypeId="1" AcceptedAnswerId="28892" CreationDate="2015-04-17T03:48:25.197" Score="2" ViewCount="1565" Body="&lt;p&gt;Does stereochemistry play any role in the heating of Cyclohexane-1,2-diol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, different products are formed depending on whether the diol is cis or trans. (I think the products are cyclohexanone and cyclopentanecarbaldehyde). Please clarify this along with the reason why different products are formed depending on the stereochemistry.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-04-18T18:22:31.667" Title="Stereochemical aspect of heating of cyclohexane-1,2-diol" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28867" PostTypeId="1" CreationDate="2015-04-17T04:51:37.363" Score="7" ViewCount="485" Body="&lt;p&gt;I am trying to obtain a potential energy surface (PES) for $\ce{N2 + N}$ combination using &lt;code&gt;scan&lt;/code&gt; feature of Gaussian. I have tried different combinations of basis sets and methods but unfortunately at some point in the iteration, the convergence criteria is not met and I get an &quot;Aborted Core dump&quot; closure in the log file. The Gaussian code is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# HF/6-31G test scan &#xA;&#xA;NNN PES&#xA;&#xA;0 2&#xA;N&#xA;N 1 R1&#xA;N 2 R2 1 120.&#xA;&#xA;R1 0.5 80 0.1&#xA;R2 0.5 80 0.1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where am I going wrong. More importantly, I would be really thankful if someone could suggest how to choose a method and basis set for a type of problem. I was thinking if I can get some answer using HF/6-31G, I could use this as an input for a higher order method like cassf. What that be a good idea?&lt;/p&gt;&#xA;" OwnerUserId="15648" LastEditorUserId="4945" LastEditDate="2015-04-17T05:38:50.593" LastActivityDate="2017-11-11T15:36:15.450" Title="Gaussian Convergence Issue for N2+N PES" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28868" PostTypeId="2" ParentId="28862" CreationDate="2015-04-17T05:15:53.840" Score="1" Body="&lt;p&gt;The main factor for this phenomena is the the values of the standard electrode potentials for of the metal ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Case_Studies/Corrosion/Corrosion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Corrosion&lt;/a&gt;, uses this scenario as an example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^2+}$ (E° = −0.14 V) and $\ce{Fe^2+}$ (E° = −0.45 V) show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Tin is therefore more of a &lt;a href=&quot;http://www.corrosion-doctors.org/Definitions/galvanic-series.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;noble metal&lt;/a&gt; than iron - a galvanic corrosion chart is shown below, showing that tin is more 'noble' than iron (&lt;a href=&quot;http://www.corrosionist.com/galvanic_corrosion_chart.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EdNdX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The significance of tin being more noble than iron is that, according to the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_7%3A_Electrochemical_Corrosion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrochemical Corrosion&lt;/a&gt;, it can act as a depolariser, removing more electrons than the water/air alone - hence, accelerate the corrosion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/moboB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the copper used in the example, it would be tin in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat simplied illustration of the process is shown in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGbnD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-17T05:32:51.647" LastActivityDate="2015-04-17T05:32:51.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28869" PostTypeId="1" CreationDate="2015-04-17T05:44:35.347" Score="1" ViewCount="460" Body="&lt;p&gt;Molecules must be within a certain distance for a reaction to begin as well as satisfying a myriad of other factors. Whilst the exact distance varies from reaction to reaction, is their any distance/molecular radius which provides a fairly good picture in terms of how close the centre of the molecules have to be to begin the interaction given they are in the correct orientation with enough energy etc.?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2015-04-17T05:44:35.347" Title="What radius best represents reaction distance?" Tags="&lt;reaction-mechanism&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;&lt;atomic-radius&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28870" PostTypeId="1" AcceptedAnswerId="28876" CreationDate="2015-04-17T06:01:45.307" Score="1" ViewCount="291" Body="&lt;p&gt;Quick question:  Any one knows a resource for &lt;strong&gt;&lt;em&gt;Cohesive Energy Density&lt;/em&gt;&lt;/strong&gt; of some organic materials ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for any specific molecule, actually I want to plan my project based on available data. I have found some tables for internal cohesive pressure but could not find any thing for CED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything including tables/papers and books are fine.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="7448" LastEditDate="2015-04-17T10:42:51.573" LastActivityDate="2016-04-01T21:49:35.017" Title="Cohesive Energy Density table request" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;reference-request&gt;&lt;equation-of-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28871" PostTypeId="2" ParentId="27357" CreationDate="2015-04-17T07:05:17.903" Score="4" Body="&lt;p&gt;Actually, it usually doesn't. It produces all types of products ranging from $\ce{NO2}$ to $\ce{NH4NO3}$ with exact products depending on concentration and conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for it is high oxidation power of nitric acid, exceeding that of hydrogen cations, so nitric acid reduces first. As result, to produce some nitrogen, the oxidation power must be brought below that of hydrogen cations. I'm not aware if it is actually possible, but possibly high dilution may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Mg, it reacts with hot water, so it's unclear if the reaction of $\ce{Mg}$ with diluted nitric acid counts. Not aware about $\ce{Mn}$ though.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-04-17T07:55:01.927" LastActivityDate="2015-04-17T07:55:01.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28873" PostTypeId="2" ParentId="28867" CreationDate="2015-04-17T09:35:01.387" Score="9" Body="&lt;p&gt;You did not specify at which geometry you have convergence problems, so at first I thought it happens when all three $\ce{N}$ atoms are far away from each other. But then I run the calculation by myself and found out that it is not the case: the SCF convergence failure happens at &lt;code&gt;R1=1.2&lt;/code&gt; and &lt;code&gt;R2=1.1&lt;/code&gt;. At this geometry SCF starts to oscillate after 20 iterations back and force as shown in a graph below, so simply increasing the number of SCF iteration would not help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/byQyr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that reason I quickly checked few other well known tricks to avoid the problems with SCF convergence (see, for instance, &lt;a href=&quot;http://www.ccl.net/cca/documents/dyoung/topics-orig/converge.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) and I recommend to use a &lt;strong&gt;quadratically convergent SCF&lt;/strong&gt; (QC-SCF) by adding &lt;code&gt;scf=qc&lt;/code&gt; to the route section:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# HF/6-31G test scan scf=qc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note though that while QC-SCF is &lt;em&gt;more reliable&lt;/em&gt; than the default DIIS algorithm it is also &lt;em&gt;slower&lt;/em&gt;, so expect an increase in the computation time. For that reason, there is also an interesting option &lt;code&gt;scf=xqc&lt;/code&gt; which first tries the default algorithm and switches to a quadratically convergent one only if the default has not converged.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# HF/6-31G test scan scf=xqc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, switching to a quadratically convergent algorithm did not help, since SCF still fails to converge but at some other geometries. So, I investigated the matter more closely, and in addition to &lt;code&gt;scf=xqc&lt;/code&gt; I also recommend to improve both &lt;strong&gt;radial and angular flexibility&lt;/strong&gt; of you basis set by adding diffuse and polarization functions into it&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# HF/6-31+G(d,p) test scan scf=xqc&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After running the calculation for an hour it reached 2656 points (which is about 40% of all points) without SCF convergence problems, but in case if this variant still fails:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Consider lowering down the convergence criteria by changing &lt;code&gt;scf=xqc&lt;/code&gt; to &lt;code&gt;scf=(xqc,conver=n)&lt;/code&gt; with &lt;code&gt;n&lt;/code&gt; being, say, 6 (it is 8 by default).&lt;/li&gt;&#xA;&lt;li&gt;If failures happen when at least one $\ce{N}$ is far away from two others, consider lowering down the maximum distance for your PES scan. In my opinion 8.5Å is way to big, 5Å would probably be more than enough.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I finally succeeded with the following route section&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# HF/6-31+G(d,p) Scan SCF=(XQC,Conver=6)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here are &lt;a href=&quot;https://gist.github.com/kemiisto/de0e3a599e729d18ca34#file-n3-pes&quot; rel=&quot;noreferrer&quot;&gt;the results&lt;/a&gt; and below is the plot of the PES. :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrprA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-04-17T19:06:52.363" LastActivityDate="2015-04-17T19:06:52.363" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28875" PostTypeId="1" CreationDate="2015-04-17T11:01:09.220" Score="1" ViewCount="164" Body="&lt;p&gt;Consider the initial boundary value problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_t=a(t)u_{xx},\,\,\,0&amp;lt;x&amp;lt;1,\,\,\,t&amp;gt;0$$&#xA;$$-ku_x(0,t)=hu(0,t),\,\,\,t&amp;gt;0$$&#xA;$$u(1,t)=u_0,\,\,\,t&amp;gt;0$$&#xA;$$u(x,0)=f(x),\,\,\,t&amp;gt;0$$&#xA;where $a(t)&amp;gt;0$ is time-dependent thermal diffusivity, $k&amp;gt;0$ constant thermal conductivity and $h&amp;gt;0$ constant heat transfer coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-I wonder that is there any chemical process that can be modelled by the above equation ?&#xA;-Is it possible for a physical process that the thermal diffusivity is time dependent but thermal conductivity and heat transfer coefficient are constant ?&lt;/p&gt;&#xA;" OwnerUserId="15458" LastEditorUserId="15458" LastEditDate="2015-04-17T13:16:58.893" LastActivityDate="2015-04-17T13:16:58.893" Title="a chemical model for pde" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28876" PostTypeId="2" ParentId="28870" CreationDate="2015-04-17T11:21:11.263" Score="2" Body="&lt;p&gt;Some resources to get you started:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ma5005474&quot; rel=&quot;nofollow&quot;&gt;Free Volume, Cohesive Energy Density, and Internal Pressure as Predictors of Polymer Miscibility&lt;/a&gt; (White and Lipson, 2014) (paywalled).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where the authors use cohesive energy density as part of their research to develop&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;strong correlations between bulk behavior and theoretical pure component character, and use those correlations to motivate a discussion of what drives polymer miscibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Materials Design webpage &lt;a href=&quot;http://www.materialsdesign.com/appnote/organic-materials-properties-densities-cohesive-energies-heats-vaporization&quot; rel=&quot;nofollow&quot;&gt;Organic Materials Properties: Densities, Cohesive Energies, and Heats of Vaporization&lt;/a&gt; contains a couple of lists and information pertaining to a computer based modelling tool.  Also included is a bibliography of further resources.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Tampere University of Technology in Finland's webpage &lt;a href=&quot;https://www.tut.fi/ms/muo/vert/4_rubber_physics/structure_cohesive_application.htm&quot; rel=&quot;nofollow&quot;&gt;Exploitation of CED&lt;/a&gt; provides a list of CEDs of several organic compound, as well as an explanation and further information on the previous and next pages.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-17T12:59:40.093" LastActivityDate="2015-04-17T12:59:40.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28877" PostTypeId="2" ParentId="28867" CreationDate="2015-04-17T11:23:10.047" Score="7" Body="&lt;p&gt;Apart from the technical aspects, that Wildcat already provided, I would assume, that the problem is far too complicated to be treated with Hartree-Fock correctly. It might be even not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hartree-Fock is a method that describes no coulomb correlation at all. I would assume, that this is vital to the system in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At long distances you will have the problem, that you cannot describe the system in a doublet state any more. Nitrogen as an atom has five valence electrons, and according to Hund's rule it should be a quartet spin state. So if you have the separation $\ce{N + N2}$ this should also be a quartet state. In this case you will have (nearly) degenerate orbitals, that will cause the SCF to oscillate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you come to the situation $\ce{N + N + N}$, you are effectively treating it as a super-molecule, where you couple three quartet states. The only way of qualitatively treating this correctly would be a complete active space ansatz. Here again it is most likely, that the SCF will oscillate because of (nearly) degenerate orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are freezing the bond angle at 120.0&amp;deg;, you should also be aware, that the equilibrium geometry will not be part of your designed potential energy surface. I would assume, that it is linear, analogous to the azide ion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I would expect the wave function to be highly spin contaminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running a scan at the CASSCF level of theory is very time consuming and involves high effort computations. I would guess you need to pick at least 15 active electrons in 15 orbitals. For starters it might work only treating the &amp;pi; system with CAS, but that still leaves 9 electrons in 6 orbitals.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-04-17T11:23:10.047" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28878" PostTypeId="1" CreationDate="2015-04-17T13:04:48.973" Score="5" ViewCount="430" Body="&lt;p&gt;Apologies in advance as I am not a science person by any definition. By most, I'm probably not a person but we'll let that be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am studying the notion of &quot;change&quot; in things and focusing on chemical change. Could someone please let me know if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the change in the color of a photogray (photosensitive, photochromic) lens is a &quot;chemical&quot; change?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the change in a baby's diaper when it gets wet is a chemical change (and if it is similar in any nature to the change in #1) and if that is the same as the change in a pregnancy test or a urinalysis stick&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if the change in a thermochromic/temperature sensitive item (a beer can that changes color in the cold, a color changing thermometer or spoon or mood ring ) is the same type of chemical change&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is too much for one question, or if it can't be explained to a lay person, but I appreciate any responses. I'm too newbie on this board to post all the links as links so the questioned items are connectable through the list of urls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.allaboutvision.com/lenses/photochromic.htm&quot; rel=&quot;nofollow&quot;&gt;Lenses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.huggies.com/en-us/products/diapers/littlesnugglers&quot; rel=&quot;nofollow&quot;&gt;Diapers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.clearblueeasy.com/healthcare/clearblue-pregnancy-test-with-color-change-tip.php&quot; rel=&quot;nofollow&quot;&gt;Pregnancy test&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.csigizmos.com/products/blooddetection/hemastix.html&quot; rel=&quot;nofollow&quot;&gt;Blood detection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemistry-blog.com/2011/08/02/coors-light-cold-activated-bottles-how-does-it-work/&quot; rel=&quot;nofollow&quot;&gt;Coors Light Cold-Activated Bottles&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.alibaba.com/product-detail/color-changing-thermometer_1048898240.html&quot; rel=&quot;nofollow&quot;&gt;Color changing thermometer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.frozensolutions.com/index.html&quot; rel=&quot;nofollow&quot;&gt;Frozensolutions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mood_ring&quot; rel=&quot;nofollow&quot;&gt;Mood ring -wiki&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-----edit---------------&#xA;The question of &quot;chemical&quot; change would be, in my mind different from &quot;physical&quot; but I don't know if that distinction comes into play here, or is an issue when the change is permanent or temporary.&lt;/p&gt;&#xA;" OwnerUserId="15649" LastEditorUserId="4097" LastEditDate="2015-04-17T15:47:45.273" LastActivityDate="2015-04-17T16:16:52.270" Title="Effecting Chemical Change" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-10-30T19:34:54.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="28879" PostTypeId="1" CreationDate="2015-04-17T13:07:59.170" Score="2" ViewCount="1362" Body="&lt;h2&gt;&lt;strong&gt;Update 31.12.2015: In a closed system of water and air, how does the concentration of CO2 change as a function of temperature, in the gas phase?&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is the graph of carbon dioxide ($\ce{CO2}$) content in the air in a compartment consisting of 80 percent air and 20 percent ocean sea water as a function of temperature, in the temperature range 0 to 60 degree Celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example but not the complete answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBOEt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen from the graph above, increasing the temperature of the water will cause the water to release more carbon dioxide ($\ce{CO2}$) in to the gas phase. This happens for most gases as can be seen from this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am interested in is to see a graph of the percentage, or how to put it, of carbon dioxide ($\ce{CO2}$) in air as a function of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what does the plot of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\%\ce{CO_2} \text{ in air (temperature)} = \frac{\text{kilogram } \ce{CO_2} \cdot 100 %}{\text{kilogram  air}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;look like in the gas phase in a compartment consisting of 20 % sea water and 80 % air as a function of temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally the compartment should consist of a column with the height upwards of average atmosphere thickness on planet earth filled with air, and a depth downwards with the average depth of sea water also at planet earth filled with sea water. But this is probably not feasible. Therefore the simplifcation 80 percent air, 20 percent sea water. If sea water is too complicated I am also satisfied with an answer with pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of air without water steam in volume percentages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrogen $\ce{N_2}$ = 78.08&lt;br&gt;&#xA;Oxygen $\ce{O_2}$ = 20.95&lt;br&gt;&#xA;Argon $\ce{Ar}$  = 0.934&lt;br&gt;&#xA;Carbon dioxide $\ce{CO_2}$ = 0.036&lt;br&gt;&#xA;Neon $\ce{Ne}$ = 0.00182&lt;br&gt;&#xA;Helium $\ce{He}$ = 0.00052&lt;br&gt;&#xA;Krypton $\ce{Kr}$ = 0.00011&lt;br&gt;&#xA;Hydrogen $\ce{H_2}$ = 0.000050&lt;br&gt;&#xA;Xenon $\ce{Xe}$ = 0.000009&lt;br&gt;&#xA;Ozone $\ce{O_3}$ = 0.000001&lt;br&gt;&#xA;Radon $\ce{Rn}$ = 0.000 000 000 000 000 006  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 18.4.2015:&#xA;I will try to clarify the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does an increase in temperature cause an increase or decrease in the carbon dioxide (CO2) content in the atmosphere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Edit 22.4.2015: According to engineering toolbox dot com, the four most common gases in the atmosphere have the following solubilities in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccpDF.jpg&quot; alt=&quot;Nitrogene solubility in water as a function of temperature&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rjzJ7.jpg&quot; alt=&quot;Oxygene solubility in water as a function of temperature&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HOQIs.jpg&quot; alt=&quot;Argon solubility in water as a function of temperature&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xdxQr.jpg&quot; alt=&quot;Carbon dioxide solubility in water as a function of temperature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question: &lt;a href=&quot;https://mathematica.stackexchange.com/questions/80586/how-can-i-get-the-solubility-value-of-nitrogen-in-water-at-10-degree-celsius-wit&quot;&gt;https://mathematica.stackexchange.com/questions/80586/how-can-i-get-the-solubility-value-of-nitrogen-in-water-at-10-degree-celsius-wit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="93" LastEditorUserId="-1" LastEditDate="2017-04-13T12:56:10.427" LastActivityDate="2015-12-31T14:52:10.913" Title="What is the CO2 content in the air in a compartment of air and water, as a function of temperature?" Tags="&lt;physical-chemistry&gt;&lt;temperature&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28880" PostTypeId="2" ParentId="28820" CreationDate="2015-04-17T13:16:29.797" Score="1" Body="&lt;p&gt;It is really not charge injection, but charge separation. When a potential is applied to the gate with respect to the SD, the charges in the gate dielectric separate. This separation creates a layer of positive or negative charges just next to the SD channel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ideal cases there is no current through the gate.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-17T13:16:29.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28882" PostTypeId="1" AcceptedAnswerId="28884" CreationDate="2015-04-17T14:24:20.357" Score="7" ViewCount="2275" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following planar structures would be aromatic? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnWY4.png&quot; alt=&quot;molecules&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me determining the aromaticity of the compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) Not an aromatic compound since it only has 4 pi electrons that does not satisfy Hückel's (4n+2) rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) Not aromatic because it only has 4 pi electrons, however I'm not sure if the oxygen contributes a p-orbital, but it shouldn't matter since 5 pi electrons still does not satisfy Hückel's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) I'm very confused because I'm not sure if the ketones are considered &quot;part of the ring&quot; therefore do they also contribute p orbitals? However if we ignore the ketones, then I believe the Nitrogens contribute 2 e- since they are sp2 hybridized? Therefore the 2 N atoms and 2 C atoms contribute a total of 4-not aromatic either?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorUserId="16683" LastEditDate="2018-01-08T15:35:24.363" LastActivityDate="2018-01-08T15:35:24.363" Title="Determining aromaticity of compounds" Tags="&lt;organic-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2018-01-08T15:35:28.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="28883" PostTypeId="1" AcceptedAnswerId="31492" CreationDate="2015-04-17T15:01:14.217" Score="10" ViewCount="519" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Kxwz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rhodamine is a fluorescent dye. What features of the molecule result in it being so highly fluorescent ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the extended conjugated system with a stabilized first excited state is necessary, to absorb light in the visible region of the spectrum, and I would also think a rigid, planar structure is preferable for good Franck-Condon factors for absorption. Are there other factors involving the energy/stability of the $S_{1}$ state I should be aware of ?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="30424" LastEditDate="2016-07-20T13:24:21.630" LastActivityDate="2016-11-01T14:42:58.947" Title="Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28884" PostTypeId="2" ParentId="28882" CreationDate="2015-04-17T15:29:57.090" Score="7" Body="&lt;p&gt;(i) that's correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) is aromatic, it fit's the 4n+2 rule with n=1.  Here is a drawing of furan showing 2 lone pair electrons in a p-orbital that completes the cyclic aromatic ring. Furan's other lone pair of electrons is perpendicular to the pi system and does not play a role in furan's aromaticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d83AV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Aromatics/Properties_of_Aromatics/Aromaticity/H%C3%BCckel&amp;#39;s_Rule&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(iii) Edit to incorporate @jerepierre's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like phenol can exist in equilibrium with its &lt;strong&gt;non-aromatic&lt;/strong&gt; tautomeric keto form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ffJbR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so can derivatives of pyrimidinedione (or uracil).  However, unlike phenol were the equilibrium lies far to the aromatic phenol side, these compounds usually (it depends on the substituents) exist in the keto form.  The strength of the keto and amide bonds outweigh the weaker aromaticity of the pyrimidine skeleton and shift the equilibrium to the keto side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6CD67.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Uracil&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question remains, &lt;strong&gt;is the keto form aromatic&lt;/strong&gt;?  As jerepierre points out, you can draw the following resonance structure for the keto form.  In this double-dipolar resonance structure there are 6 pi electrons 2 in the double bond and 2 [a lone pair] on each nitrogen) suggesting that this resonance structure would be aromatic.  Since this resonance structure contributes to the overall description of the molecule, it would be fair to say that this molecule has some aromatic character; certainly not as much as benzene or furan - but some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v8UwT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-17T15:57:58.093" LastActivityDate="2015-04-17T15:57:58.093" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="28885" PostTypeId="1" AcceptedAnswerId="28902" CreationDate="2015-04-17T16:05:04.870" Score="6" ViewCount="4920" Body="&lt;p&gt;Preface: Buffer solution (acid-base buffer). I am provided with a weak base, which I will designate B. &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt;'s conjugate acid, which I will designate &lt;span class=&quot;math-container&quot;&gt;$\ce{BH}$&lt;/span&gt;, is &lt;span class=&quot;math-container&quot;&gt;$8.1$&lt;/span&gt;, and its mole weight (sic) is &lt;span class=&quot;math-container&quot;&gt;$121.1$&lt;/span&gt;. I'm assuming the latter is the molar mass, though I don't know how that helps me solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{mol/L}$&lt;/span&gt;) of &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; was dissolved in &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0 L}$&lt;/span&gt; water, then titrated to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$8.1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$7.5$&lt;/span&gt;. I am to assume no change in volume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The last statement sounds like an odd assumption to me, how do you titrate something without [significantly] altering the volume? I always assumed titration is only done by adding a liquid to another liquid, is this not correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am to find the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{BH}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$8.1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$7.5$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question looked really easy to me at the first glance, but after a few hours I realized I had no idea on how to go about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried calculating the equilibrium concentrations and going from there. This just ends up in &lt;span class=&quot;math-container&quot;&gt;$x \neq x$&lt;/span&gt; in my calculations (where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the concentration of some substance), which obviously doesn't hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the acid dissociation-, base dissociation- and Henderson–Hasselbalch equations, but nothing seems to do the trick. I might not be the brightest person to have grazed the surface of the planet, but I'm confident that I've got a firm grasp on elementary arithmetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is provided, and should be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.05 M}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 8.1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would highly appreciate somebody showing or explaining the thought process to me. &lt;/p&gt;&#xA;" OwnerUserId="15652" LastEditorUserId="4945" LastEditDate="2019-07-11T16:04:52.817" LastActivityDate="2019-07-11T16:04:52.817" Title="How to calculate the concentration of conjugate acid from given pH and pKa values?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28886" PostTypeId="1" CreationDate="2015-04-17T16:11:32.950" Score="7" ViewCount="4627" Body="&lt;p&gt;I cannot figure out why glycine hydrochloride is more acidic than chloroacetic acid. Could I have figured this out from a $\mathrm pK_\mathrm a$ table and maybe am just not seeing it? On the table I have in my book (Organic Chemistry, Seyhan Ege, 3rd edition), neither acid nor its conjugate base are there.&lt;/p&gt;&#xA;" OwnerUserId="15654" LastEditorUserId="16683" LastEditDate="2017-05-19T23:35:57.260" LastActivityDate="2017-05-19T23:58:46.567" Title="Why is glycine hydrochloride more acidic than chloroacetic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;erratum&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28887" PostTypeId="2" ParentId="28878" CreationDate="2015-04-17T16:16:52.270" Score="2" Body="&lt;p&gt;For starters, I think most chemists would agree that a chemical change is different from a physical change by involving the formation or breaking of chemical bonds, or changes to the charge, coordination, or bonding state of some of the atoms involved. A chemical change can be reversible, however, so reversibility is not a determining factor. Either a physical change or a chemical change could result in a change of color, or not--as the case may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your examples, the absorption of water by a disposable diaper, as well as the color changing of liquid crystal-based displays (like the beer temperature indicator, thermometer, and mood ring) are clearly physical changes. The liquid crystals change their optical properties due to the physical alignment of the molecules, and not because of making or breaking of bonds. On the other hand, the pregnancy test and blood detection test probably involve chemical changes to produce a color readout (sometimes initiated by a relatively loose binding between an antigen and an antibody, or formation of a metal complex). The photochromic lenses turn dark as a result of a chemical change induced by light (according to Wikipedia, anyway--this is not my area of expertise).&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-17T16:16:52.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28888" PostTypeId="1" CreationDate="2015-04-17T16:23:32.253" Score="9" ViewCount="728" Body="&lt;p&gt;I had a conversation with somebody who had somewhat confused ideas about food. He thought that &quot;all food contains sugar&quot; and when I gave a bottle of oil as the counterexample, he said the glycerol is the sugar there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's forget for a minute that glycerol is not a sugar. The point is, glycerol is indeed a &quot;building block&quot; of a triglyceride, and glucose is a &quot;building block&quot; of glucose. But still, if you are talking about a process which needs sugar, starch won't work unless you cleave it off first. Just like starch won't spontaneously conduct electricity or emit beta radiation even though it's chock full of electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we have somebody with a very basic level of knowledge, and who is already trying to wrap his head around several confusing concepts at once, what would be an intuitive way to get him to really grok that, just because a given atom or group is present in the structure of a molecule, it does not mean that a substance &quot;contains&quot; this atom or group in the same way that a wet sponge contains water? And that without going off on a long tangent of theory? &lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2015-04-18T13:26:00.723" Title="How can I easily explain to a layperson the difference between types of &quot;contains&quot;" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28889" PostTypeId="2" ParentId="28879" CreationDate="2015-04-17T16:41:33.900" Score="2" Body="&lt;p&gt;Simple, partial pressure of $\ce{CO2}$ is misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ actually reacts with sea water:&#xA;$$\ce{H2O + CO2 -&amp;gt; H2CO3}$$&#xA;which then ionizes:&#xA;$$\ce{H2CO3 -&amp;gt; H+ + HCO3-}$$&#xA;which lowers the pH of the sea water (making it more acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In previous high $\ce{CO2}$ periods, $\ce{CaCO3}$ came out of sea water &#xA;making limestone and marble rock formations geologically. $\ce{CO2}$ at high pressure is responsible for other side effects, also killing &#xA;fish, and microbes, damaging coral reefs and generally messing up ocean life.&lt;/p&gt;&#xA;" OwnerUserId="15655" LastEditorDisplayName="user7951" LastEditDate="2015-04-18T10:09:46.633" LastActivityDate="2015-04-18T10:09:46.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28890" PostTypeId="2" ParentId="28888" CreationDate="2015-04-17T16:47:30.133" Score="10" Body="&lt;p&gt;I would say something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All matter -- that is, all substances -- are made of atoms.  The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemists refer to the &quot;types&quot; of atoms as &quot;elements&quot;.  For example, vegetable oil and table sugar (sucrose) are both composed of three types of atoms only: carbon, hydrogen, and oxygen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrangement of the atoms in the two materials differs drastically.  (You could refer to structure diagrams at this point if your friend is interested.)  First, oil has much less oxygen and slightly more hydrogen per carbon atom.  Second, the way the atoms are connected -- their arrangement -- is very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemists sometimes give a name to a group of &lt;em&gt;similar&lt;/em&gt; arrangements of atoms.  For example, &quot;sugars&quot; are compounds that (a) have carbon, hydrogen, and oxygen atoms in ratios that are &lt;em&gt;approximately&lt;/em&gt; $\ce{CH2O}$, and (b) &lt;em&gt;often&lt;/em&gt; form ring structures, as seen in table sugar, glucose, and corn starch.  But the key words are &quot;approximately&quot; and &quot;often&quot; because some molecules that chemists call sugars don't satisfy these rules.  Biochemists interested in metabolism sometimes refer to molecules that can &quot;easily&quot; be converted into &quot;true&quot; sugars as sugars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naming in chemistry is complicated by the fact that chemists use words for different purposes: to indicate metabolic similarity, to indicate structural similarity, or to indicate similarity in reactivity.  Those different purposes sometimes lead situation where chemist B says that what chemist A calls a &quot;sugar&quot;  isn't really a sugar.  But that's just an argument about humans' imperfect way of naming things, not about chemistry.  There is a lot of wiggle room in the words &quot;approxiately&quot;, &quot;often&quot;, &quot;similar&quot;, and &quot;easily&quot;.  So if we focus on metabolism only, the glycerol part of the vegetable oil arrangement (i.e. structure) is a bit like a sugar.  But it is dissimilar from sugars in terms of both structure and reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, what ultimately matters is (a) the type and (b) the arrangement of the atoms in a substance.  Those two things are together necessary and sufficient for a complete description of all substances.  Whether to call particular arrangements of particular atom types &quot;sugars&quot; or not is something people can argue about.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-04-17T16:51:57.523" LastActivityDate="2015-04-17T16:51:57.523" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="28891" PostTypeId="1" AcceptedAnswerId="28920" CreationDate="2015-04-17T17:06:06.967" Score="4" ViewCount="116" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the couple &lt;span class=&quot;math-container&quot;&gt;$\ce{Hg^2+}/\ce{Hg2^2+}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$E^\circ = \pu{0.91V}$&lt;/span&gt;.&#xA;  At the border the concentration is &lt;span class=&quot;math-container&quot;&gt;$c = \pu{0.10 mol \cdot L-1}$&lt;/span&gt; for all ions. Hence build the stability diagram of mercury.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Hg^2+ +2e- &amp;lt;=&amp;gt; Hg2^2+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by the Nernst relation I have:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E = E^\circ + \pu{0.03 V} \times \log\left( \frac{\left[\ce{Hg^2+}\right]^2}{\left[\ce{Hg2^2+}\right]} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solution of the exercise they write:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the border &lt;span class=&quot;math-container&quot;&gt;$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand why it is &lt;span class=&quot;math-container&quot;&gt;$\frac{c}{2}$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="4945" LastEditDate="2020-04-28T09:31:47.317" LastActivityDate="2020-04-28T10:46:59.300" Title="Why is the concentration halved at the intersection of Hg²⁺/Hg₂²⁺?" Tags="&lt;electrochemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28892" PostTypeId="2" ParentId="28866" CreationDate="2015-04-17T17:48:31.437" Score="1" Body="&lt;p&gt;The reaction you are thinking of is likely the thermal decomposition of the cyclic sulfites prepared from the &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-diols.  Here is a &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01634a002&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to the abstract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UYFA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both reactions $\ce{SO2}$ is extruded.  Whichever group on the vicinal carbon that is lined up in an anti-periplanar fashion to the departing oxygen is the group that migrates.  This is because the antiperiplanar geometry allows for good orbital overlap during the migration; if a choice is available then hydrogen migrates in preference to an alkyl group just as with carbocations. In the &lt;em&gt;cis&lt;/em&gt;-isomer, the vicinal hydrogen is properly aligned (I darkened the affected bonds to highlight the geometric arrangement), its migration produces cyclohexanone.  In the &lt;em&gt;trans&lt;/em&gt;-isomer, only a $\ce{C-C}$ bond in the cyclohexane ring is properly aligned, its migration produces cyclopentanecarbaldehyde.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-17T17:48:31.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28895" PostTypeId="2" ParentId="9733" CreationDate="2015-04-17T18:17:14.417" Score="-2" Body="&lt;p&gt;Endothermic, exothermic. The difference is that an endothermic molecule is unstable. From the moment you remove the extra energy, it reverts back to its original structure. From original author, Park S. Nobel, 1974, it all depends on the manner you get the four protons in order to get the monomer. (five or six monomers to build a sugar). You get either four protons with two or four water molecules. Up to now, it is utilized two water molecules. So you say that the oxygen of photosynthesis is endothermic. And that can't be. (Park S. Nobel is the father of the expression Electron Chain Transport). He also proposed the possibility that the origin of the four protons is due to four molecules of water but for that one he did not propose a model, so the winning model is Two Water Molecules, only because the other model was not described. &lt;/p&gt;&#xA;" OwnerUserId="15658" LastActivityDate="2015-04-17T18:17:14.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28896" PostTypeId="2" ParentId="28888" CreationDate="2015-04-17T18:48:48.950" Score="8" Body="&lt;p&gt;In his excellent answer, &lt;strong&gt;Curt F.&lt;/strong&gt; wrote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Let me add some examples to illustrate how subtle differences lead to substantial property changes any layman can experience.&lt;/p&gt;&#xA;&lt;h3&gt;Containing the same number of atoms means nothing&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vid52.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-pinene (&lt;strong&gt;1&lt;/strong&gt;) and camphene (&lt;strong&gt;2&lt;/strong&gt;) have the same sum formula (&lt;span class=&quot;math-container&quot;&gt;$\ce{C10H16}$&lt;/span&gt;), but a very different constitution and a very different smell, just as their names suggest.&lt;/p&gt;&#xA;&lt;h3&gt;Containing the same number of atoms and having almost the same constitution&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VSXiR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thymene (&lt;strong&gt;3&lt;/strong&gt;) and carvacrol (&lt;strong&gt;4&lt;/strong&gt;) are regioisomers and only differ in the position of the &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; group. However, thymene has a thyme odour, while carvacrol smells like oregano.&lt;/p&gt;&#xA;&lt;h3&gt;Mirror, mirror&lt;/h3&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qCAS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Compounds &lt;strong&gt;5&lt;/strong&gt; and &lt;strong&gt;6&lt;/strong&gt; have the same constitution and identical boiling points. They only differ in that one group is above/below the ring plane (the world isn't flat). These are (just) mirror images: enantiomers of limonene. However the interested layman will recognize they are &lt;strong&gt;not&lt;/strong&gt; identical: &lt;strong&gt;5&lt;/strong&gt; smells like lemon, while &lt;strong&gt;6&lt;/strong&gt; smells like orange.&lt;/p&gt;&#xA;&lt;hr/&gt;&#xA;&lt;h3&gt;Containing isn't being&lt;/h3&gt;&#xA;&lt;p&gt;A layman might know that chlorine is a toxic gas and was used in the chemical warfare of WWI. Some might even know that fluorine isn't any better.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;These unhealthy properties are not inherited by compounds that &amp;quot;contain&amp;quot; these elements!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Table salt (&lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt;) &amp;quot;contains chlorine&amp;quot; (admittedly, writing it that way makes me shudder) but is isn't as toxic as the element. Neither is fluoridated tooth-paste.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QM6WX.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each molecule of diazepam (&lt;strong&gt;7&lt;/strong&gt;) contains one chlorine atom, each molecule of fluoxetine (&lt;strong&gt;8&lt;/strong&gt;) contains three (!) fluorine atoms. Even so, these compounds are not part of modern chemical warfare, but on the &lt;a href=&quot;http://en.wikipedia.org/wiki/WHO_Model_List_of_Essential_Medicines&quot; rel=&quot;noreferrer&quot;&gt;WHO List of Essential Medicines&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-17T20:20:54.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28897" PostTypeId="1" AcceptedAnswerId="28899" CreationDate="2015-04-17T19:08:54.903" Score="10" ViewCount="1988" Body="&lt;p&gt;What is a catemer and how does it differ from a dimer? What type of compounds could form catemers and how does their formation influence its crystal structure?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="418" LastEditDate="2015-04-17T20:44:51.727" LastActivityDate="2015-04-17T20:44:51.727" Title="What is a catemer?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28899" PostTypeId="2" ParentId="28897" CreationDate="2015-04-17T19:49:25.073" Score="13" Body="&lt;p&gt;In the crystal, carboxylic acids mostly form dimers through pairs of $\ce{OH\bond{~}O=C}$ hydrogen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, infinite chains of hydrogen bonds are formed instead: here, each carboxylic acid forms hydrogen bonds to &lt;strong&gt;two different neighbours&lt;/strong&gt;. These structures are known as catemers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d7ouV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Please note that the depiction of the catemer is &lt;strong&gt;very much&lt;/strong&gt; simplified! In reality, the structure could also be a helix, stabilized by additional sets of interactions&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I have never used the term &lt;em&gt;catemer&lt;/em&gt; in a crystallographic publication. It is very imprecise, almost meaningless. Instead, I have mostly used the a notation introduced by &lt;em&gt;Margret C. Etter&lt;/em&gt; to describe hydrogen bonding pattern (dimers, stands, rings) in a crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two relevant articles in this context are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Encoding and decoding hydrogen-bond patterns of organic compounds,&#xA;&lt;em&gt;Acc. Chem. Res.&lt;/em&gt;, &lt;strong&gt;1990&lt;/strong&gt;, &lt;em&gt;23&lt;/em&gt;, 120-126 (&lt;a href=&quot;http://dx.doi.org/10.1021/ar00172a005&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Graph-set analysis of hydrogen-bond patterns in organic crystals, &lt;em&gt;Acta Cryst. B&lt;/em&gt;, &lt;strong&gt;1990&lt;/strong&gt;, &lt;em&gt;46&lt;/em&gt;, 256-262 (&lt;a href=&quot;http://dx.doi.org/10.1107/S0108768189012929&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-17T20:42:44.367" LastActivityDate="2015-04-17T20:42:44.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28900" PostTypeId="2" ParentId="28886" CreationDate="2015-04-17T22:06:53.027" Score="3" Body="&lt;p&gt;Glycine hydrochloride is the amino acid glycine adsorving a molecule of Hydrochloric acid, wich is a strong acid. I think this is the main reason of the acidicity difference, since the pKa's are similar (comparing glycine and chloroacetic acid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: glycine is amphoteric: it can either act as a base or an acid.&#xA;In water the hydrochloric acid molecule of the glycine hydrochloride will dissociate: $$\ce{H2O +HCl -&amp;gt; H3O+ +Cl-}$$&#xA;Then:&#xA;$$\ce{Gly +H+ -&amp;gt; GlyH+}$$&#xA;$$\ce{GlyH+ +Cl- &amp;lt;-&amp;gt; Gly + H+ + Cl-}$$&#xA;You can conclude, therefore, that Glycine hydrochloride is more acid than glycine (and also the chloroacetic acid). Hope it helps, and sorry for the english mistakes!&lt;/p&gt;&#xA;" OwnerUserId="15662" LastEditorUserId="15662" LastEditDate="2015-04-17T22:59:54.750" LastActivityDate="2015-04-17T22:59:54.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28901" PostTypeId="2" ParentId="14771" CreationDate="2015-04-17T23:33:32.047" Score="7" Body="&lt;p&gt;What is the chemical reaction through which propylene glycol/glycerin forms formaldehyde/acetaldehyde?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Propylene glycol is NOT the same as glycerin. Propylene glycol does not decompose on heating, it simply boils. In theory, under certain conditions (pressure/encapsulation/catalysts) it could dehydrate to either acetone or Propionaldehyde, probably the former.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for glycerine (= glycerol = VG = animal glycerine), that decomposes below its boiling point by a simple dehydration mechanism, to acrolein and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H8O3$\xrightarrow{\Delta}$C3H4O +2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acrolein could further degrade by a retro aldol condensation to acetaldehyde and formaldehyde. This could be catalysed by acids/bases (both of which are present in eliquids) or even by hot metal catalysts (such as the heating coil):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H4O\xrightarrow{H^+}H3CCHO +HCHO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H4O\xrightarrow{hot~metal}H3CCHO +HCHO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are well known reactions to any organic chemist, and are far more likely to happen than any vague/spurious 'oxidation' reactions as suggested by a few.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Why does propylene glycol form more carbonylated products than glycerol?&#xA;Because it has a dual mode of dehydration.&#xA;Really, this is a 'Have you stopped beating your wife?' question: PG and glycerol can form multiple carbonyl compounds, depending on what further side reactions happen:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Formaldehyde is stable, other than the possibility of oxidation to formic acid or carbon dioxide.&#xA;Acetalydehyde can dimerize to crotonaldehyde.&#xA;Acrolein can both dimerize, polymerise and undergo addition reactions, in all cases forming a variety of carbonyl compounds&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;One of the most bizarre things about some studies (including those of Dr Fasolinos), is that while formaldehyde and acetaldehyde are observed, there is insignificant acrolein. And he makes little comment about the latter!&#xA;Don't you wonder where it's all going?&#xA;I'd suggest that the bulk of the acrolein is decomposing to those two other carbonyl compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most aldehydes undergo oxidation over time, but this is not combustion taking place.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15664" LastEditorUserId="15410" LastEditDate="2015-04-18T06:25:02.267" LastActivityDate="2015-04-18T06:25:02.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28902" PostTypeId="2" ParentId="28885" CreationDate="2015-04-18T00:14:48.053" Score="6" Body="&lt;p&gt;To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{B + H2O&amp;lt;=&amp;gt;BH +OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equilibrium expression of the reaction is given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\log K_\mathrm{a}=8.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\log K_\mathrm{a}=-8.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$14=8.1+\mathrm{p}\ce{OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pOH}=5.9$$&#xA;and&#xA;$$-\log{[\ce{OH-}]}=5.9$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the knowledge we have gained from the problem, we can now solve (finally!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=0.1-x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2x=0.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=0.05$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try using this process to solve for the $\mathrm{pH}$ of 7.5.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2017-02-21T01:34:26.437" LastActivityDate="2017-02-21T01:34:26.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="28903" PostTypeId="2" ParentId="28865" CreationDate="2015-04-18T00:44:56.873" Score="4" Body="&lt;p&gt;This is because Barium Hydrogen Phosphatev ($\ce{BaHPO4}$) is an &lt;strong&gt;amphoteric species&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemistry, an amphoteric species is a molecule or ion that can&#xA;  react as an acid as well as a base&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Amphoterism&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it acts as a base or as an acid depends on the medium the salt is dissolved. In a pure water solution, it's an acid, since the following occurs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BaHPO4 + H2O &amp;lt;=&amp;gt; Ba^2+ + PO4^3- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phosphate ion also has its reactions with water forming other species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PO4^3- + H3O+ &amp;lt;=&amp;gt; HPO4^2- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HPO4^2- + H3O+ &amp;lt;=&amp;gt; H2PO4^- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these are all weak bases, you won't &quot;consume&quot; all $\ce{H3O+}$ and the solution will be acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you add a strong acid, the scenario changes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BaHPO4 + H3O+ &amp;lt;=&amp;gt; Ba^2+ + H2PO4^- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll still have all phosphate species as above, but now Barium Hydrogen Phosphate has acted as a base, since it received some hydrogen atoms, which is a Brønsted-Lowry base definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the same kind of behaviour happens with water, even if pure, because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow&quot;&gt;self ionization&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, water acted as both the base and the acid, which is why it's an amphoteric species.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15410" LastEditDate="2016-05-23T07:17:58.790" LastActivityDate="2016-05-23T07:17:58.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28904" PostTypeId="1" CreationDate="2015-04-18T01:55:12.460" Score="2" ViewCount="18672" Body="&lt;p&gt;I'm pretty sure the sign changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I want to be sure if it becomes endothermic or exothermic.&lt;/p&gt;&#xA;" OwnerUserId="15665" LastEditorUserId="15429" LastEditDate="2015-04-18T03:41:03.187" LastActivityDate="2015-04-18T04:30:15.500" Title="Given a forward reaction with positive enthalpy, is its reverse reaction be endothermic or exothermic?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28905" PostTypeId="1" AcceptedAnswerId="33594" CreationDate="2015-04-18T03:27:40.950" Score="5" ViewCount="22027" Body="&lt;p&gt;Would isolated silicon engage in network covalent bonding as quartz does or would it engage in a different kind of bonding? Would germanium display similar qualities?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I am asking what is the bonding of silicon, not why silicon is different from carbon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am interested in isolated silicon, not silicon as a &quot;hydrosilicon&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I believe it is network covalent, but I am not sure. &lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorDisplayName="user15489" LastEditDate="2015-07-04T11:16:58.427" LastActivityDate="2017-10-19T13:07:18.467" Title="What type of bonding occurs in isolated silicon?" Tags="&lt;bond&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28906" PostTypeId="2" ParentId="28904" CreationDate="2015-04-18T03:54:34.127" Score="2" Body="&lt;p&gt;Yes, in a reverse reaction the sign changes. That is because Enthalpy is a state function, one that depends on only the current state of the function. That means the energy required to push a reaction to its products would be directly reversed to push it back to its reactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key use of this knowledge is in Hess's law, and Born Haber Cycle.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-18T03:54:34.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28907" PostTypeId="2" ParentId="28904" CreationDate="2015-04-18T04:30:15.500" Score="0" Body="&lt;p&gt;The enthalpy of a reaction can be found by calculating the amount of energy required to break the bonds of the reactants, and then subtracting from this value the amount of energy required to form the bonds of the products. You can find a &lt;a href=&quot;http://www4.ncsu.edu/~shultz/Common_Bond_enthalpies.pdf&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; of these values here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse reaction would simply be doing the opposite, meaning that the magnitude would be the same, but the sign would be opposite, as you stated.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-18T04:30:15.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28908" PostTypeId="1" AcceptedAnswerId="28909" CreationDate="2015-04-18T05:05:40.180" Score="2" ViewCount="2406" Body="&lt;p&gt;First, I should say I never have done any spectroscopies and I am just reading about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, UV-Vis spectrophotometer has 2 beam samples: &lt;strong&gt;sample&lt;/strong&gt; and &lt;strong&gt;reference&lt;/strong&gt;. The sample has very diluted test material in a solvent and the reference is only the solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading a book about some experimental values for UV-Vis spectrophotometry and I noticed this page.  My question is about that sentence saying &lt;strong&gt;&quot;The data were obtained .. against water reference&quot;&lt;/strong&gt; .   As the reference is the solvent alone, &lt;strong&gt;&lt;em&gt;does it mean the samples are dilute solutions in water&lt;/em&gt;&lt;/strong&gt; ?  if so, how they solved benzene in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a snapshot of the text:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pEvwY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorDisplayName="user15489" LastEditDate="2015-04-18T05:15:34.410" LastActivityDate="2015-04-18T05:43:05.813" Title="What is this reference material in spectrophotometry?" Tags="&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28909" PostTypeId="2" ParentId="28908" CreationDate="2015-04-18T05:22:40.060" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As the reference is the solvent alone, does it mean the samples are dilute solutions in water ? if so, how they solved benzene in water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I come to a different conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sample: &lt;strong&gt;solution&lt;/strong&gt; of benzene, acetone, acetonitrile, etc. &lt;strong&gt;in&lt;/strong&gt; water&lt;/li&gt;&#xA;&lt;li&gt;reference: water&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Benzene is soluble in water: about $1.8\,\mathrm{g\cdot L^{-1}}$ at 25 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a reference (blank), a solution of &lt;strong&gt;everything except the sample&lt;/strong&gt; is used. In organic (photo)chemistry this usually means: just the solvent. The situation is different in biochemistry, medicinal chemistry, etc. when complex aqueous buffer systems are used. Here, &lt;strong&gt;everything except the probe&lt;/strong&gt; really means that the whole buffer is used as a blank.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-18T05:43:05.813" LastActivityDate="2015-04-18T05:43:05.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28910" PostTypeId="1" CreationDate="2015-04-18T05:34:09.647" Score="1" ViewCount="8078" Body="&lt;p&gt;In all of my physical chemistry books I find the same expression for estimating the equilibrium constant of a reaction at a non-standard temperature. The derivation starts with $R \ln K_\text{eq} = -\Delta g^\circ/T = -(\Delta h^\circ/T - \Delta s^\circ)$. Then one takes the temperature derivative of each side assuming that $\Delta h^\circ$ and $\Delta s^\circ$ can both be considered constants, which eliminates the entropy term. The equation &#xA;$\mathrm{d}\ln K = (1/R)(\Delta h^\circ/T^2)\mathrm{d}T$ is then integrated between $T_1$ and $T_2$ to give&#xA;$\ln(K_2/K_1) = (\Delta h^\circ/R)(1/T_1 - 1/T_2)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the question: $\Delta h_\text{r}^\circ$ is really NOT a constant, but a function of temperature. Assuming constant heat capacities, $\Delta h_\text{r}^\circ$ at $T = \Delta h_\text{r}^\circ + \Delta c_p(T - T_0)$. Likewise $\Delta s_\text{r}^\circ$ at $T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$. Where $\Delta c_p$ is the difference in the heat capacity of the products and the reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to insert these expressions into the typical derivation but am getting wonky results. Can someone help me with the derivation or point me to a source where $\ln K$ vs. $T$ is derived with temperature-sensitive reaction enthalpies and entropies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 4/19/2015: Would it be more effective to invoke $\mathrm{d}\Delta g = \Delta v\mathrm{d}P - \Delta s\mathrm{d}T$ which at constant pressure, and in the case of free energy of reaction, gives $\mathrm{d}\Delta g_\text{r}^\circ/\mathrm{d}T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help.&lt;/p&gt;&#xA;" OwnerUserId="15666" LastEditorDisplayName="user7951" LastEditDate="2015-04-20T08:11:41.773" LastActivityDate="2015-10-26T10:12:34.400" Title="Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28911" PostTypeId="1" AcceptedAnswerId="28918" CreationDate="2015-04-18T05:39:16.283" Score="6" ViewCount="4553" Body="&lt;p&gt;What kinds of molecules will experience hydrogen bonding (inter and intra)? Why does it occur for some H atoms but not others?&lt;/p&gt;&#xA;" OwnerUserId="5402" LastEditorUserId="9961" LastEditDate="2016-11-02T22:35:23.270" LastActivityDate="2016-11-02T22:35:23.270" Title="When does hydrogen bonding happen?" Tags="&lt;molecules&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28913" PostTypeId="1" AcceptedAnswerId="28914" CreationDate="2015-04-18T05:56:29.590" Score="3" ViewCount="2106" Body="&lt;p&gt;Consider ethanedioic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6Xtum.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it form intramolecular hydrogen bonds (between the C=O and HO)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xpjibm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5402" LastActivityDate="2015-04-18T06:19:15.097" Title="Does intramolecular hydrogen bonding occur in dicarboxylic acids?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28914" PostTypeId="2" ParentId="28913" CreationDate="2015-04-18T06:19:15.097" Score="1" Body="&lt;p&gt;No, this would not happen the way your second drawing depicts because the hydrogens do not orient themselves that way. Intramolecular hydrogen bonding would be no more feasible in ethandioic acid than any other monocarboxylic acid. See picture below:&lt;img src=&quot;https://i.stack.imgur.com/FfJXx.png&quot; alt=&quot;oxalic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Oxalic-acid-3D-balls-B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://commons.wikimedia.org/wiki/File:Oxalic-acid-3D-balls-B.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-18T06:19:15.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28915" PostTypeId="2" ParentId="28911" CreationDate="2015-04-18T06:21:53.187" Score="4" Body="&lt;p&gt;Most people seems to think that hydrogen bonds are limited to $\ce{X-H\bond{~}Y}$ interactions, where $\ce{X}$ and $\ce{Y}$ are heteroatoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the crystal, a $\ce{C-H}$ of an aromatic system or a $\ce{C=C}$ double bond can serve as a donor as well and interact with heteroatoms, such as oxygen atoms in $\ce{C=O}$ or $\ce{C_{arene}-O-C_{alkyl}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These interactions can form complex networks of rings and strands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've examined and published that a while ago for a couple of different phthalimides, such as &lt;a href=&quot;http://dx.doi.org/10.1107/S1600536806046149&quot; rel=&quot;nofollow&quot;&gt;N-(4-methoxybenzyl)phthalimide&lt;/a&gt; or &lt;a href=&quot;http://dx.doi.org/10.1107/S160053680604596X&quot; rel=&quot;nofollow&quot;&gt;N-allyl­phthalimide&lt;/a&gt;, and some more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the articles are behind a paywall (they were published before &lt;em&gt;Acta Cryst. E&lt;/em&gt; went open source), the CIF files can be downloaded and played with/examined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crucial parameter for hydrogen bonds is distance, if the acceptor is too far away, there is (obviously) no hydrogen bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, $2.3\,\mathrm{\mathring{A}}$ is a distance where hydrogen bonds can be expected.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorDisplayName="user7951" LastEditDate="2015-04-18T10:19:25.110" LastActivityDate="2015-04-18T10:19:25.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28916" PostTypeId="1" AcceptedAnswerId="28966" CreationDate="2015-04-18T06:34:27.060" Score="10" ViewCount="10343" Body="&lt;p&gt;I am learning about solutions and while giving an example of a liquid (solute) and a solid (solvent), my teacher told us that mercury forms an amalgam with all the metals except iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know why this is so?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-05-16T03:36:38.597" LastActivityDate="2019-11-21T21:38:15.863" Title="Why does mercury not form amalgam with iron?" Tags="&lt;metal&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28917" PostTypeId="1" AcceptedAnswerId="28919" CreationDate="2015-04-18T07:57:16.513" Score="1" ViewCount="2619" Body="&lt;p&gt;On a proton NMR, does the hydrogen in the CHO group cause coupling, or get coupled itself? Specifically, I a considering 2-methylpropanal. I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet, and I am unsure as to whether the single proton on carbon number two would be a heptet (from the 6 adjacent protons), an octet ( from the 6 protons plus the proton in the aldehyde group), or if it would be split into some combination of a heptet of doublets, or a doublet of heptets...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice on dealing with spin spin coupling in proton NMR would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="1887" LastEditDate="2015-04-18T08:31:58.063" LastActivityDate="2015-04-19T08:19:05.110" Title="Spins spin coupling on proton NMR for an aldehyde?" Tags="&lt;analytical-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28918" PostTypeId="2" ParentId="28911" CreationDate="2015-04-18T08:18:47.523" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What kinds of molecules will experience hydrogen bonding (inter and intra)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intramolecular hydrogen bonds:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The hydrogen bond which is within the same molecules is called intramolecular hydrogen bond.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It will form when 2 functional group of a molecule can form hydrogen bonds with each other.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In order for this to happen, both a hydrogen donor an acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;EXAMPLE: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KcqlF.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;sub&gt;&lt;a href=&quot;http://www.google.co.in/imgres?imgurl=http://cdn1.askiitians.com/cms-content/common/www.askiitians.comonlineteststudymaterial_images759_Intramolecularhydrogenbonding.JPG.jpg&amp;amp;imgrefurl=http://www.askiitians.com/iit-jee-chemical-bonding/metallic-and-hydrogen-bonding.html&amp;amp;h=111&amp;amp;w=257&amp;amp;tbnid=yMfwAGVa1N-IQM:&amp;amp;zoom=1&amp;amp;docid=DN8P-moGhOrsyM&amp;amp;ei=TVoyVbWCIteTuAT49IDACQ&amp;amp;tbm=isch&amp;amp;client=ubuntu&amp;amp;ved=0CCAQMygEMAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intermolecular hydrogen bonds:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The hydrogen bond which occurs between two separate molecules is caleed intermolecular hydrogen bond.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in positions in which they can interact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;EXAMPLE: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GPSzm.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;sub&gt;&lt;a href=&quot;http://www.google.co.in/imgres?imgurl=http://everythingscience.co.za/grade-12/04-organic-molecules/pspictures/bfed1f5b7737a289350e961e48e6ff38.png&amp;amp;imgrefurl=http://socratic.org/questions/which-liquid-water-or-alcohol-has-stronger-intermolecular-forces&amp;amp;h=292&amp;amp;w=438&amp;amp;tbnid=r_BAzYw1v6zdpM:&amp;amp;zoom=1&amp;amp;docid=cMdtwCVtGqgjwM&amp;amp;ei=chIyVcuEC4vIuASG34GACg&amp;amp;tbm=isch&amp;amp;client=ubuntu&amp;amp;ved=0CCEQMygGMAY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does it occur for some H atoms but not others?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bond depends on certain factors like electronegativity, viscosity, atomic size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Viscosity:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those substances which are capable of forming hydrogen bonds tend to have a higher viscosity than those that do not. Substances which have the possibility for multiple hydrogen bonds exhibit even higher viscosities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;electronegativity:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we see molecules such as PH&lt;sub&gt;3&lt;/sub&gt;, which no not partake in hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Atomic size:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of donors and acceptors can also effect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Hydrogen_Bonding&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; for detailed explanation &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-04-18T13:23:06.180" LastActivityDate="2015-04-18T13:23:06.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28919" PostTypeId="2" ParentId="28917" CreationDate="2015-04-18T08:36:03.433" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I cannot decide whether the H in the aldehyde group would be split into a doublet or remain as a singlet [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What would be the argument &lt;strong&gt;against&lt;/strong&gt; $^3J$ coupling between these protons? The aldehyde proton will appear as a doublet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that coupling works both ways! $\ce{(CH3)2C\color\red{H}-}$ will (at least) split to a doublet too. Coupling with this proton will also  split the signal for the methyl protons to a doublet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean for the proton at C-2? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, a &lt;strong&gt;doublet&lt;/strong&gt; (from coupling with the aldehyde proton) &lt;strong&gt;of a septet&lt;/strong&gt; (from coupling with six magnetically equivalent methyl protons) is to be expected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In really, you will observe a smaller number of lines! I suggest to try this with a pen on squared paper to figure out what happens when the two coupling constants are identical or one is twice as big as the other.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-19T08:19:05.110" LastActivityDate="2015-04-19T08:19:05.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28920" PostTypeId="2" ParentId="28891" CreationDate="2015-04-18T08:36:39.887" Score="2" Body="&lt;p&gt;The total concentration of all ions is &lt;span class=&quot;math-container&quot;&gt;$c = \pu{0.10 mol L-1}$&lt;/span&gt;. The conservation equation of element mercury requires that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, it's a convention to make the concentration of ions at the border equal. On the other hand, matter conservation implies that the concentration of each ion is &lt;span class=&quot;math-container&quot;&gt;$\frac{c}{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2020-04-28T10:46:59.300" LastActivityDate="2020-04-28T10:46:59.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28921" PostTypeId="1" AcceptedAnswerId="28927" CreationDate="2015-04-18T08:57:34.297" Score="2" ViewCount="135" Body="&lt;p&gt;If electrons were spin-$\frac {3}{2}$&#xA;instead of spin-$\frac {1}{2}$ , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what would be the atomic&#xA;number $Z$ for the first noble gas ?&lt;/p&gt;&#xA;" OwnerUserId="15668" LastEditorUserId="15668" LastEditDate="2015-04-18T10:06:54.520" LastActivityDate="2015-04-18T12:45:52.667" Title="Atomic number $Z$" Tags="&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electronic-configuration&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28923" PostTypeId="1" CreationDate="2015-04-18T11:01:14.443" Score="0" ViewCount="104" Body="&lt;p&gt;$\ce{C4H6O2}$ (10.75 g) was dissolved in water (250 cm&lt;sup&gt;3&lt;/sup&gt;) and then a 20 cm&lt;sup&gt;3&lt;/sup&gt; sample was titrated against an alkali metal hydroxide solution. This required 17.35 cm&lt;sup&gt;3&lt;/sup&gt; of the alkali. What was the concentration of the alkali? &#xA;The water was evaporated from the alkali solution (100 cm&lt;sup&gt;3&lt;/sup&gt;) to leave a white solid (3.23 g). Identify the alkali metal present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working already is that I started by dividing 10.75g by 0.25dm&lt;sup&gt;3&lt;/sup&gt; to get 43g/dm&lt;sup&gt;3&lt;/sup&gt;. Then divided 43g/dm&lt;sup&gt;3&lt;/sup&gt; by the relative formula mass of 86 to get 0.5mol/dm&lt;sup&gt;3&lt;/sup&gt;. I did 0.5mol/dm&lt;sup&gt;3&lt;/sup&gt;*0.01735dm&lt;sup&gt;3&lt;/sup&gt; to get 0.008675mol then divided that by 0.02dm&lt;sup&gt;3&lt;/sup&gt; to get 0.0434 mol/dm&lt;sup&gt;3&lt;/sup&gt; as the concentration of the alkali. I'm not sure how to proceed from there or know what I have already done is correct. &lt;/p&gt;&#xA;" OwnerUserId="15671" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T14:27:07.527" LastActivityDate="2015-05-02T14:27:07.527" Title="Titration of C₄H₆O₂" Tags="&lt;homework&gt;&lt;metal&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28924" PostTypeId="2" ParentId="28888" CreationDate="2015-04-18T11:08:38.223" Score="1" Body="&lt;p&gt;For a layman you could simply mention that &lt;strong&gt;airplanes contain aluminium, as do rollerskate rails&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrangement of the aluminium is a critical factor determining which vehicle could be used for skating around the neighbourhood and which vehicle could be used for intercontinental transport.&lt;/p&gt;&#xA;" OwnerUserId="6987" LastEditorUserId="6987" LastEditDate="2015-04-18T13:26:00.723" LastActivityDate="2015-04-18T13:26:00.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28925" PostTypeId="1" AcceptedAnswerId="28926" CreationDate="2015-04-17T14:16:02.613" Score="3" ViewCount="679" Body="&lt;p&gt;I have a device in a box which has parts of metal. So can I prevent corrosion by creating a vacuum around the metallic object?&lt;/p&gt;&#xA;" OwnerUserId="15672" OwnerDisplayName="Bhavesh" LastActivityDate="2015-04-18T11:45:59.153" Title="Can we prevent metal corrosion by creating a vacuum around it?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28926" PostTypeId="2" ParentId="28925" CreationDate="2015-04-17T14:43:44.557" Score="11" Body="&lt;p&gt;The word &lt;em&gt;corrosion&lt;/em&gt; has a range of meanings, but it is generally taken to indicate &lt;a href=&quot;http://en.wikipedia.org/wiki/Corrosion&quot;&gt;the chemical reaction of a material with its environment&lt;/a&gt;. For example iron corrodes by reacting with oxygen in the air (or water if it's underwater).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put the metal in a vacuum then it doesn't have anything around it to react with, so in this case yes putting the metal in a vacuum will stop it corroding.&lt;/p&gt;&#xA;" OwnerUserId="1470" OwnerDisplayName="John Rennie" LastActivityDate="2015-04-17T14:43:44.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28927" PostTypeId="2" ParentId="28921" CreationDate="2015-04-18T12:45:52.667" Score="1" Body="&lt;p&gt;The spin multiplicity will be $2S+1=4$. This means that the first atomic orbital can hold $4$ electrons with four different quantum numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the atomic number for the first noble gas is $4$.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-18T12:45:52.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28928" PostTypeId="2" ParentId="28923" CreationDate="2015-04-18T14:14:07.657" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My working already is that I started by dividing $10.75g$ by $0.25dm^3$ to get $43g/dm^3$. Then divided $43g/dm^3$ by the relative formula mass of $86 g/mol$to get $0.5mol/dm^3$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Up to here, you have it correct. The next step is to take $20 mL$ from $250 mL$, which would decrease the amount of acid to titrate. Then you would find the concentration of the base.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-18T14:14:07.657" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="28929" PostTypeId="2" ParentId="28923" CreationDate="2015-04-18T14:22:15.417" Score="0" Body="&lt;p&gt;I would suggest to solve a titration question by equating the meq of acids and base for a neutral condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$moles \space of \space \ce{C4H6O2}=0.5=\frac{10.75*1000}{250*86}$$&#xA; $$meq \space of \space \ce{C4H6O2}= 500$$&#xA;$$meq \space of \space \ce{C4H6O2} in \space 0.02 l=500*0.02=10$$ &#xA;$$\therefore$$&#xA;$$meq \space of \space alkali= 10$$&#xA;The alkali is sodium, as it give a white ppt and the base was NaOH&#xA;$$moles \space of \space \ce{NaOH}=0.01=conc*volume$$&#xA;$$conc \space of \space \ce{NaOH}=\frac{0.01}{17.35}*1000=0.576$$&#xA;$$\therefore$$&#xA;The concentration is 0.576M&lt;/p&gt;&#xA;" OwnerUserId="15574" LastActivityDate="2015-04-18T14:22:15.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28931" PostTypeId="1" CreationDate="2015-04-18T14:32:31.077" Score="2" ViewCount="1343" Body="&lt;p&gt;As a contact lenses user, I was looking for information about why manufacturers produce soft contacts using polymers from both classes. Common knowledge would suggest that something placed in contact over the eyes for corrective vision should be inert. Ionic usually means more reactive. So why do they even make ionic contacts?&lt;/p&gt;&#xA;" OwnerUserId="15674" LastEditorUserId="34307" LastEditDate="2016-08-30T20:03:28.960" LastActivityDate="2016-08-30T20:03:28.960" Title="Contact lenses come in two polymer classes, ionic or non-ionic. Whats the significance of having both?" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2015-11-30T12:08:23.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="28934" PostTypeId="1" AcceptedAnswerId="29092" CreationDate="2015-04-18T15:02:29.880" Score="14" ViewCount="614" Body="&lt;p&gt;I don't know if this is really applicable but I want to add a custom force (say, &lt;span class=&quot;math-container&quot;&gt;$F(x,y,z)$&lt;/span&gt;) to some atoms in my geometry optimization. Is there any way to add such forces to the DFT or molecular mechanics calculations? (preferably GAMESS or Gaussian for QC or Avogadro for MM). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It can be also done as Molecular Mechanics (MM)&lt;/em&gt; as I just need an approximate geometry after applying the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give an example. Say I have molecule A and molecule B and a force that changes in space as &lt;span class=&quot;math-container&quot;&gt;$F(x,y,z)$&lt;/span&gt; but is only applied to molecule B. Now I want to calculate the new geometry of A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think this is like adding a &lt;span class=&quot;math-container&quot;&gt;$V_F$&lt;/span&gt; to the external energy in the Schrödinger equation, but I don't know how to do that in QC packages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left[-\frac{1}{2}∇^2 + V(r) +\color{red}{V_F(r)}\right]ψ(r) = Eψ(r)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;  The idea of making a unit cell and making shrinkage on it looks great (suggested by Geoff Hutchinson here), but the problem is I couldn't figure out how to install CP2K on MacOS and also I am looking for something with a simple interface like Avogadro. I even tried Maestro by Schrödinger software, but couldn't make unit cell shrinkage for that.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="41328" LastEditDate="2019-04-05T15:40:35.533" LastActivityDate="2019-04-05T15:40:35.533" Title="How to define custom force in quantum chemistry packages?" Tags="&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28935" PostTypeId="1" CreationDate="2015-04-18T15:09:42.160" Score="3" ViewCount="1178" Body="&lt;p&gt;If one took equimolar solutions of $\ce{Li}$ and $\ce{K}$, which would have higher conductivity? Is it the same for molar conductivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, $\ce{LiCl}$ and $\ce{KCl}$, of a litre each. Also, as Molar Conductivity = Conductivity / Molarity, are molar conductivity and conductivity directly proportional? Why or why not?&lt;/p&gt;&#xA;" OwnerUserId="15676" LastEditorUserId="15410" LastEditDate="2015-04-18T17:32:43.620" LastActivityDate="2015-07-18T12:48:51.490" Title="Conductivity and Molar Conductivity of Equimolar Solutions of LiCl and KCl" Tags="&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28936" PostTypeId="2" ParentId="28883" CreationDate="2015-04-18T15:16:03.850" Score="1" Body="&lt;p&gt;Hmmm, I can't think of much else. You have a conjugated, planar and rigid structure that ensures your structure doesn't change/expand much upon promotion of an electron into the antibonding orbital; hence, ensuring a good FC factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly solvent related effects that might encourage ISC over fluoresence? I know polar solvents can encourage ISC but not sure in what chemical context and it's applicability to the particular situation.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2015-04-18T15:16:03.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28937" PostTypeId="1" AcceptedAnswerId="28938" CreationDate="2015-04-18T16:00:24.447" Score="2" ViewCount="1134" Body="&lt;p&gt;Could someone please show me the MO diagram for $\ce{N_2^{2-}}$? What I did was since N has 5 valence $\ce{e^{-}}$, then $\ce{N_2}$ must have 10, and $\ce{N_2^{2-}}$ must have 12. And $\ce{N}$ has $s$ and $p$ orbital mixing. So when I draw out the MO it has all the orbitals filled. And the bond order is $\frac{1}{2}(6-6)=0$, but this is apparently wrong. Because according to the lewis structure the bond order should be $2$ I think. &lt;/p&gt;&#xA;" OwnerUserId="15045" LastEditorDisplayName="user15489" LastEditDate="2015-04-19T10:12:55.353" LastActivityDate="2015-04-19T10:12:55.353" Title="MO for $N_2^{2-}$" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecules&gt;&lt;electronic-configuration&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28938" PostTypeId="2" ParentId="28937" CreationDate="2015-04-18T16:10:57.173" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MPLYB.jpg&quot; alt=&quot;N2 MOhttp://h2g2.com/edited_entry/A1011899&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This an image of the molecular orbital of $\ce{N2}$. If we add 2 more electrons, they will enter singly into the 2 antibonding pi orbitals and since bond strength is defined as bonding electrons minus nonbonding electrons over 2, or as I prefer, the net number of bonding/ nonbonding electrons times half, we would subtract 1 from the original bond order of $\ce{N2}$ to get 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the mathematical calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2}(8 (bonding)-4(antibonding))=2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Antibonding orbitals are shown by an asterisk. Note that you will need draw in the 2 electrons on your own.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-18T16:10:57.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28939" PostTypeId="1" AcceptedAnswerId="28944" CreationDate="2015-04-18T17:03:10.683" Score="21" ViewCount="1932" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y75z8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y75z8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="5026" LastEditDate="2018-04-07T15:23:15.950" LastActivityDate="2018-04-08T14:55:40.413" Title="Ring contraction in a carbocation due to ring strain and back bonding" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="28942" PostTypeId="2" ParentId="20189" CreationDate="2015-04-18T17:34:35.473" Score="7" Body="&lt;p&gt;Having one methyl group &lt;em&gt;ortho&lt;/em&gt; to the carboxylate group &lt;strong&gt;will not&lt;/strong&gt; prevent the carboxylate group from becoming reasonably planar with the aromatic ring.  If you had a second methyl group in the other &lt;em&gt;ortho&lt;/em&gt; position, then planarity would be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A methyl group attached to a benzene ring is electron releasing in both an inductive and a resonance sense&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inductively, the electrons in an $\ce{sp^3}$ hybridized methyl group prefer to shift towards the lower energy (more s-character in the orbital) $\ce{sp^2}$ aromatic carbon.  &lt;/li&gt;&#xA;&lt;li&gt;The methyl group is also electron releasing in a resonance sense due to carbon-hydrogen hyperconjugation (see figure below).  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The resonance structure pictured below on the left is one of many that we can draw for the &lt;em&gt;o&lt;/em&gt;-toluic carboxylate anion.  Note that it places a positive charge on the methyl bearing &lt;em&gt;ortho&lt;/em&gt; carbon. Both inductively and through resonance (see the hyperconjugated resonance structure below on the right) the methyl group will stabilize this resonance structure.  &lt;strong&gt;It is this stabilization provided by the methyl group that makes &lt;em&gt;o&lt;/em&gt;-toluic acid a stronger acid (pKa=3.91) than benzoic acid (pKa=4.19).&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y5T7D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; As pointed out by both @Mithoron and @GaurangTandon, this answer does not explain why &lt;em&gt;p&lt;/em&gt;-toluic acid is a weaker acid than &lt;em&gt;o&lt;/em&gt;-toluic acid or benzoic acid. The answers in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases&quot;&gt;this earlier question&lt;/a&gt; better answer this question.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-03-09T18:13:06.617" LastActivityDate="2018-03-09T18:13:06.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28943" PostTypeId="1" AcceptedAnswerId="28986" CreationDate="2015-04-18T17:58:36.933" Score="2" ViewCount="480" Body="&lt;p&gt;According to wikipedia, the third law of thermodynamics is &quot;The entropy of a perfect crystal, at absolute zero (zero kelvins), is exactly equal to zero.&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, theoretically, could all substances, including molecular compounds, attain absolute zero? Another way of thinking of this question is whether all compounds could become perfect crystals. &lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="14975" LastEditDate="2015-04-18T18:28:40.287" LastActivityDate="2015-04-19T12:55:50.803" Title="Does the third law of thermodynamics apply to all compounds and elements?" Tags="&lt;thermodynamics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28944" PostTypeId="2" ParentId="28939" CreationDate="2015-04-18T18:13:43.390" Score="19" Body="&lt;p&gt;The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ring_strain&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).  This isn't much of a difference, certainly not enough to drive a reaction one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization.  Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-react-with-bromine/10666#10666&quot;&gt;here&lt;/a&gt;), said differently, they have a lot of p-character in the bond.  In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/INxnP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here's another representation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WBqGo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation.  In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital.  The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap!  In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation.  Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond.  Sometimes we use resonance structures such as the following to denote this stabilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cv5Nf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cv5Nf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2018-04-08T14:55:40.413" LastActivityDate="2018-04-08T14:55:40.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28945" PostTypeId="1" AcceptedAnswerId="28951" CreationDate="2015-04-18T18:21:35.007" Score="1" ViewCount="408" Body="&lt;p&gt;I need to calculate the theoretical $\mathrm{pH}$ in a $50~\mathrm{mL}$ water solution where $6.4\times10^{-4}~\mathrm{g}$ of $\ce{Ca^{2+}}$ is present. The reaction I base this on is the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Ca^{2+}_{(aq)} + 2H2O_{(l)} \rightleftarrows Ca(OH)2_{(s)} + 2H^+_{(aq)}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K = \frac{[\ce{H+}]^2}{[\ce{Ca^{2+}}]} \Rightarrow [\ce{H+}] = \sqrt{K \cdot [\ce{Ca^{2+}}]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a database of standard equilibrium constants for such Lewis acid reactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any help!&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="5334" LastEditDate="2015-04-18T20:36:28.097" LastActivityDate="2018-03-04T02:53:06.997" Title="Equilibrium constants for lewis acids such as Ca2+" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28946" PostTypeId="2" ParentId="28866" CreationDate="2015-04-18T18:22:31.667" Score="0" Body="&lt;p&gt;Compounds containing a cyclohexane-1,2-diol structure can undergo pinacol-like rearrangements. Usually, the substituent that is &lt;em&gt;anti&lt;/em&gt; to the leaving group is the one that undergoes migration. Therefore, &lt;em&gt;cis&lt;/em&gt;-cyclohexane-1,2-diol mainly yields cyclohexanone, whereas a &lt;em&gt;trans&lt;/em&gt;-cyclohexane-1,2-diol mainyl yields cyclopentanecarbaldehyde.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-18T18:22:31.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28951" PostTypeId="2" ParentId="28945" CreationDate="2015-04-18T20:29:28.893" Score="4" Body="&lt;p&gt;You're confusing terms here. $\ce{K_{a}}$ is the (Brønsted) acid dissociation constant. This suggests that $\ce{K_a}$ refers to the degree that a &lt;strong&gt;Brønsted&lt;/strong&gt; acid dissociates into its conjugate base and hydrogen protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_{a}}$ is inapplicable to Lewis acids because Lewis acids are &lt;em&gt;not&lt;/em&gt; proton donors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brønsted acids are Lewis acids in that Brønsted acids such as the hydrogen proton can accept a pair of electrons. But the converse is &lt;strong&gt;not&lt;/strong&gt; true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lewis acids are electron acceptors. How do we measure the strength of Lewis acids? There is no universally accepted measure as there is for Brønsted acids. We could look at the enthalpy of reaction (Lewis adduct formation) between a Lewis acid and a Lewis base. This is a common method. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-04T02:53:06.997" LastActivityDate="2018-03-04T02:53:06.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28952" PostTypeId="2" ParentId="28595" CreationDate="2015-04-18T20:39:59.017" Score="4" Body="&lt;p&gt;This method of repeating iterations is interesting, though I would suggest something perhaps a little more straightforward (and rigorous) than doing something until you are &quot;acceptably close.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method I learned is called a &lt;strong&gt;RICE&lt;/strong&gt; diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reaction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}&amp;lt;=&amp;gt;HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initial&lt;/strong&gt; (concentrations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{PO4^{3-}}]=0.20~~~~~[\ce{HPO4^{2-}}]=0~~~~~[\ce{OH-}]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Change&lt;/strong&gt; (in concentrations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{PO4^{3-}}]=-x~~~~~[\ce{HPO4^{2-}}]=+x~~~~~[\ce{OH-}]=+x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equilibrium&lt;/strong&gt; (concentrations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{PO4^{3-}}]=0.20-x~~~~~[\ce{HPO4^{2-}}]=x~~~~~[\ce{OH-}]=x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write the reaction expression as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{b}=\frac{[\ce{HPO4^{2-}}]\cdot[\ce{OH-}]}{[\ce{PO4^{3-}}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We substitute in the equilibrium concentrations and the $K_\mathrm{b}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solve for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2.22\times10^{-2}=\frac{x^2}{0.20-x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4.44\times10^{-3}-2.22\times10^{-2}\cdot x=x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=x^2+2.22\times10^{-2}\cdot x-4.44\times10^{-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the quadratic formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=\frac{-(2.22\times10^{-2})\pm\sqrt{(2.22\times10^{-2})^2-4(1)(-4.44\times10^{-3})}}{2(1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\approx-7.87\times10^{-2},~5.65\times10^{-2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the negative solution is not applicable to our problem, so $x=5.65\times10^{-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $x=[\ce{OH-}]$, $[\ce{OH-}]=5.65\times10^{-2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}]\cdot [\ce{OH-}]=1\times10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] \cdot [5.65\times10^{-2}]=1\times10^{-14}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] =\frac{1\times10^{-14}}{5.65\times10^{-2}}\approx1.77\times10^{-13}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to a calculator with a numerical solver function, I would recommend that instead of using the quadratic formula, but if not, this is how it is done.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-18T20:39:59.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28953" PostTypeId="2" ParentId="28945" CreationDate="2015-04-18T20:54:35.393" Score="1" Body="&lt;p&gt;You should be able to find the constant $K_\text{a}$ for the reaction&#xA;$$\ce{Ca^2+ + H2O &amp;lt;=&amp;gt; CaOH+ + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a value of $\mathrm{p}K_\text{a} = 12.6$ (at 25 °C) in “Dissociation Constants of Inorganic Acids and Bases”, in &lt;em&gt;CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010),&lt;/em&gt; David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Werner Stumm, James J. Morgan &lt;em&gt;Aquatic Chemistry&lt;/em&gt;, 3rd Edition (1996), John Wiley &amp;amp; Sons, New York, NY gives a value of $\log K = -12.78$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-18T20:54:35.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28954" PostTypeId="1" AcceptedAnswerId="28959" CreationDate="2015-04-18T21:01:53.587" Score="3" ViewCount="2559" Body="&lt;p&gt;The equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^{。}_{cell}= \frac{RT}{nF}\ln K_{eq}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know cell potential is intensive, not affected by the amount, Because: $volt=\frac{joule}{coulomb}$. Both joule and coulomb will be doubled altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as seen from the equation, cell potential is affected by the number of electrons transferred.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2015-04-18T22:53:37.780" LastActivityDate="2015-04-18T22:53:37.780" Title="Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28955" PostTypeId="2" ParentId="28595" CreationDate="2015-04-18T21:12:37.463" Score="3" Body="&lt;p&gt;The correct equation for determining the value of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, as noted by ringo, is &lt;span class=&quot;math-container&quot;&gt;$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$&lt;/span&gt; With a &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; value sufficiently low, you can ignore the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; value on the bottom of the fraction, as it is negligible compared to the initial concentration of the species present (&lt;span class=&quot;math-container&quot;&gt;$0.20$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in this case the &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; value of &lt;span class=&quot;math-container&quot;&gt;$2.22\times10^{-2}$&lt;/span&gt; is actually quite close to the initial concentration of the &lt;span class=&quot;math-container&quot;&gt;$\ce{PO4_{(aq)}^{3-}}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$0.20$&lt;/span&gt; (only around &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times less). As a result, we cannot ignore the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in the denominator, as that would cause a significant change in the ultimate value of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. This is shown by the iterations you provided, the initial (approximate) &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; value being about &lt;span class=&quot;math-container&quot;&gt;$\pu{0.01 M}$&lt;/span&gt; different from the final. The iterations then are just a method of compensating for ignoring the bottom &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; value from the very beginning. I was taught the same method that ringo outlined, and it is the more accurate method, but the iterations seem to work as well.&lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="16683" LastEditDate="2019-08-20T13:34:59.653" LastActivityDate="2019-08-20T13:34:59.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28957" PostTypeId="1" CreationDate="2015-04-18T22:13:06.017" Score="7" ViewCount="8696" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Given information:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The boiling point of benzene at atmospheric pressure is $353~\mathrm{K}$ &lt;/li&gt;&#xA;  &lt;li&gt;the enthalpy of vaporization of benzene is $30.8~\mathrm{kJ~mol^{−1}}$ at this temperature. &lt;/li&gt;&#xA;  &lt;li&gt;The molar heat capacities of the liquid and vapour are $136.1~\mathrm{J~K^{−1}~mol^{-1}}$ and $81.7~\mathrm{J~K^{−1}~mol^{-1}}$, &#xA;  respectively, and may be assumed temperature independent.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Calculate the entropy change of the system, the surroundings and hence the universe when $1~\mathrm{mol}$ of benzene vapour at $343~\mathrm{K}$ and atmospheric pressure becomes liquid benzene at $343~\mathrm{K}$. Also, will this process occur spontaneously?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that $\mathrm{d}S = \frac{\mathrm{d}H}{T}$ therefore,&lt;br&gt;&#xA;$\displaystyle\mathrm{d}S = \frac{-30.8 \times 10^3~&#xA;\mathrm{J}}{343~\mathrm{K}}=-89.8~\mathrm{J~K^{-1}}$ which is the entropy of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I continue from there, utilising the molar heat capacities given?&lt;/p&gt;&#xA;" OwnerUserId="10257" LastEditorUserId="5026" LastEditDate="2018-05-16T10:05:29.057" LastActivityDate="2018-05-16T10:05:29.057" Title="How to calculate the entropy change of benzene when it changes from vapour into liquid?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28959" PostTypeId="2" ParentId="28954" CreationDate="2015-04-18T22:47:39.763" Score="4" Body="&lt;p&gt;No, there is no violation of &quot;intensivity&quot;.  The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the electrolysis of water:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$$\ce{2H2O_{(l)} -&amp;gt; 2H2_{(g)} + O2_{(g)}}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider this reaction:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;$$\ce{4H2O_{(l)} -&amp;gt; 4H2_{(g)} + 2O2_{(g)}}$$ &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$.  If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big.  But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-18T22:47:39.763" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="28960" PostTypeId="2" ParentId="28595" CreationDate="2015-04-19T00:27:07.473" Score="4" Body="&lt;p&gt;Just to add to what everyone is saying, the reason that the &lt;strong&gt;iterations&lt;/strong&gt; are shown is to arrive at a &lt;em&gt;more accurate&lt;/em&gt; answer. Your method was also correct as long as you assume that &lt;span class=&quot;math-container&quot;&gt;$0.20 - x \approx 0.20$&lt;/span&gt;, but the problem's provided solution was just &lt;em&gt;more&lt;/em&gt; &lt;em&gt;accurate&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to remember this is that if after you find the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{x}$&lt;/span&gt; (not doing the quadratic method), divide it by the original concentration of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.20 M}$&lt;/span&gt; and calculate the percent dissociation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$x \leq 5\%$&lt;/span&gt; then your approximation is acceptable, and you don't have to worry about solving for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; using the quadratic equation and can cancel out the insignificant &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; in the denominator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want a more precise answer, it is better to use the quadratic equation to solve for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2019-08-20T13:37:28.350" LastActivityDate="2019-08-20T13:37:28.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28961" PostTypeId="1" CreationDate="2015-04-19T01:47:44.307" Score="1" ViewCount="920" Body="&lt;p&gt;If I know the amount of oxygen in such a solution, how can I know the required concentration of sodium sulfite to be used and react as oxygen scavenger?&lt;/p&gt;&#xA;" OwnerUserId="15683" LastEditorUserId="8083" LastEditDate="2015-04-19T01:49:28.417" LastActivityDate="2015-04-20T08:25:25.200" Title="How to calculate the required concentration for sodium sulfite to remove oxygen as an oxygen scavenger from a solution?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28962" PostTypeId="1" AcceptedAnswerId="28964" CreationDate="2015-04-19T02:18:37.387" Score="1" ViewCount="2443" Body="&lt;p&gt;So while calculating for the relative yield of monochlorination for 2,3-dimethylbutane, I was counting the primary hydrogens, I counted 6, but in fact it was 12. Why is it 12? How do you find equivalent hydrogens? Since the compound is symmetrical doesn't the hydrogens on both sides be equivalent to each other?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9yrfD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15045" LastEditorUserId="15045" LastEditDate="2015-04-21T13:16:05.020" LastActivityDate="2015-04-21T13:16:05.020" Title="Equivalent hydrogens in 2,3-dimethylbutane" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;hydrogen&gt;&lt;symmetry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28963" PostTypeId="1" CreationDate="2015-04-19T02:21:12.247" Score="3" ViewCount="10318" Body="&lt;p&gt;What percent of a solution needs to be acetic acid for a particular $\mathrm{pH}$?  I have 100% acetic acid, and want $1~\mathrm{L}$ of solution with a $\mathrm{pH}$ of 5. I've done some research, but I still do not understand how to solve this problem.&lt;/p&gt;&#xA;" OwnerUserId="15684" LastEditorUserId="15410" LastEditDate="2015-04-19T05:31:35.413" LastActivityDate="2015-04-19T15:06:09.510" Title="Diluting acetic acid to obtain a solution of pH 5" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-04-19T12:54:20.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="28964" PostTypeId="2" ParentId="28962" CreationDate="2015-04-19T02:24:55.987" Score="1" Body="&lt;p&gt;The hydrogen atoms in a methyl group are equivalent to each other. There are 4 methyl groups and therefore there are 12 equivalent hydrogens. 2,3-dimethylbutane is symmetric as you stated, and so on one side you have 6 equivalent hydrogens, this implies that there are 6 more equivalent hydrogens on the opposite side. &lt;/p&gt;&#xA;" OwnerUserId="3735" LastActivityDate="2015-04-19T02:24:55.987" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28965" PostTypeId="2" ParentId="28963" CreationDate="2015-04-19T03:14:11.520" Score="3" Body="&lt;p&gt;This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is necessary first to find the molarity of the acetic acid. On &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/acetic_acid&quot; rel=&quot;nofollow&quot;&gt;PubChem&lt;/a&gt;, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have 1 L of acetic acid, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{pH} = -\log{\ce{[H3O+]}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our &lt;strong&gt;RICE&lt;/strong&gt; table,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HC2H3O2 + H2O -&amp;gt; C2H3O2- + H3O+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array} {|c|c|c|c|c|}&#xA;\hline&#xA;\text{Initial conc.} &amp;amp; x \ \text{mol} &amp;amp; - &amp;amp; \text{0 mol} &amp;amp; \text{0 mol}\\&#xA;\hline&#xA;\text{Change conc.} &amp;amp; -1 \cdot 10^{-5}\ \text{M} &amp;amp; - &amp;amp; 1 \cdot 10^{-5}\ \text{M} &amp;amp; 1 \cdot 10^{-5}\ \text{M}\\&#xA;\hline&#xA;\text{End conc.} &amp;amp; x - 1 \cdot 10^{-5}\ \text{M} &amp;amp; - &amp;amp;  1 \cdot 10^{-5}\ \text{M} &amp;amp; 1 \cdot 10^{-5}\ \text{M}\\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$&#xA;$$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$ &#xA;$$2.75 \cdot 10^{-10} = 0.0000175x$$ &#xA;$$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two assumptions,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration. &lt;/li&gt;&#xA;&lt;li&gt;Autoprotolysis of water does not significantly contribute to pH.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2015-04-19T15:06:09.510" LastActivityDate="2015-04-19T15:06:09.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28966" PostTypeId="2" ParentId="28916" CreationDate="2015-04-19T03:36:01.360" Score="7" Body="&lt;p&gt;Iron does not form an amalgam &lt;em&gt;readily&lt;/em&gt;, hence the use of iron electrodes in mercury switches.  Apparently, iron &lt;em&gt;does&lt;/em&gt; form alloys with mercury under specific conditions; see &lt;a href=&quot;http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, mercury spilled on gold (e.g. gold-alloy jewelry) soaks in immediately, discoloring the gold, and migrates further into the bulk metal in a few hours, thus restoring the color.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-19T03:36:01.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28967" PostTypeId="1" CreationDate="2015-04-19T03:45:38.203" Score="3" ViewCount="9918" Body="&lt;p&gt;&lt;img src=&quot;https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcSa8CBVhZpwdu2ByoAj6N0etmsauMwW4-bNKiVT8MWZa6ybqnAeeg&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it says that the hydrocarbon tail is non-polar. I understand that the $\ce{H}$ on top of the $\ce{C}$ and on the bottom have dipoles that cancel, but at the very end, with the $\ce{CH3}$ group, there is no opposing vector to cancel with that dipole. Why then is the hydrocarbon tail still non-polar?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="3735" LastEditDate="2015-04-19T03:54:02.673" LastActivityDate="2015-04-19T15:03:46.313" Title="Why is a hydrocarbon tail non-polar?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28968" PostTypeId="2" ParentId="28967" CreationDate="2015-04-19T04:11:12.037" Score="2" Body="&lt;p&gt;The hydrocarbon tail is considered nonpolar mainly due to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Pauling electronegativity of Carbon is 2.55 and Hydrogen is 2.2. Due to the small difference, .35, and the definition of a polar bond, C-H bonds are considered nonpolar.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Indeed, the molecule methane, $\ce{CH4}$ has a dipole moment of 0 D (debyes). I think part of your confusion arises from the picture you provided. Skeletal drawings are 2-d in nature and do not convey information about 3-d. For instance, each carbon has it's hydrogens arranged in a tetrahedral structure, with bond angles 109.5 degrees. The symmetry of this cancels all the vectors contributing to polarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/ae/Methane-2D-small.png/230px-Methane-2D-small.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the picture of methane above, and consider the top most hydrogen as the next carbon in the hydrocarbon tail, you can see that in reality, the bottom 3 hydrogens do experience a net dipole moment towards the carbon, and this increased by having another carbon further up in the tail. Relatively speaking, this distance is rather large from carbon 1 to carbon 2. We can say that the vector that arises from this is negligible, and indeed, there is a net dipole moment anyways towards the oxygen at the polar head.&lt;/p&gt;&#xA;" OwnerUserId="3735" LastEditorUserId="3735" LastEditDate="2015-04-19T15:03:46.313" LastActivityDate="2015-04-19T15:03:46.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28969" PostTypeId="1" AcceptedAnswerId="28971" CreationDate="2015-04-19T04:19:50.877" Score="5" ViewCount="139" Body="&lt;p&gt;I notice when using SMILES or a IUPAC name for many molecules and generating them in Chembio3D, and then running MM2 &gt; Minimize Energy, the structures produced are most of the time not reflections of any conformation observed physiologically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there's a way I can import protein and other macromolecules into Chembio3D, perhaps from a database somewhere, where I can view their precise arrangement either as found in crystallography or in vivo in their usual forms at physiological pH. Most importantly, the proteins would need to be accurate (something I find lacking when generating structures by name in chemdraw/chembio3d).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate that the proteins and other molecules are always changing conformation and interacting with other molecules, but at least a reasonable prediction of some realistic conformations to view them in 3D would be nice, instead of conformations that might rarely/never exist in a cell. Is this possible? And if so, how would I go about this? What would be the specific steps I need to follow to import these proteins into Chembio3D?&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="5017" LastEditDate="2015-04-20T00:12:33.260" LastActivityDate="2015-04-20T00:12:33.260" Title="Viewing a protein or bio structure in Chembio3D in its natural conformation" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28970" PostTypeId="2" ParentId="28967" CreationDate="2015-04-19T04:22:04.790" Score="1" Body="&lt;p&gt;The hydrocarbon that you have depicted (decanoic acid) does indeed have a polar head and a non-polar head, though it is not apparent in the form it is presented to you in. When viewing decanoic acid as a 3-dimensional, ball-and-stick model, it becomes more obvious:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l6sby.png&quot; alt=&quot;decanoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen on C-10 of this chain that sticks out in the plane of the rest of the carbon chain is balanced by C-9. Most importantly, it should be noted that there are no unbonded electron pairs in this region. Because the electron density along the chain remains consistent, the carbon chain is non-polar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the two oxygens in the carboxylic acid groups have two electron clouds that give these regions around the molecule a negative charge. The space around the hydrogen bonded to oxygen, and the space around C-1 that is oriented away from the oxygens are both more positive. These opposing charges result in the polarity of the carboxylic acid group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/9/98/Decanoic-acid-3D-balls.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://upload.wikimedia.org/wikipedia/commons/9/98/Decanoic-acid-3D-balls.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-19T04:22:04.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28971" PostTypeId="2" ParentId="28969" CreationDate="2015-04-19T04:56:24.830" Score="3" Body="&lt;p&gt;The site you are looking for is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Protein_Data_Bank&quot; rel=&quot;nofollow&quot;&gt;Protein Data Bank&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get to it with one of these links.&lt;br&gt;&#xA;www.wwpdb.org&lt;br&gt;&#xA;www.pdbe.org&lt;br&gt;&#xA;www.rcsb.org/pdb&lt;br&gt;&#xA;www.pdbj.org&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-04-19T13:51:54.060" LastActivityDate="2015-04-19T13:51:54.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28972" PostTypeId="1" AcceptedAnswerId="28980" CreationDate="2015-04-19T05:27:21.803" Score="0" ViewCount="121" Body="&lt;p&gt;My understanding says &lt;strong&gt;organo-metallic&lt;/strong&gt; = organic + metallic compunds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One already known to me is &lt;strong&gt;Grignard Reagent&lt;/strong&gt; ($\ce{RMgX}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But would you consider $\ce{NaOMe}$ or $\ce{NaCCH}$ as an &lt;strong&gt;organometallic compound&lt;/strong&gt;? Would you consider $\ce{LiOAc}$, $\ce{LiOMe}$, $\ce{LiN(Me)2}$ or $\ce{LiMe}$?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2015-04-19T09:54:45.533" LastActivityDate="2015-04-19T09:54:45.533" Title="Are these organo-metallic compounds?" Tags="&lt;nomenclature&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28975" PostTypeId="1" CreationDate="2015-04-19T05:39:34.817" Score="-2" ViewCount="675" Body="&lt;p&gt;I was asked which geometrical isomer of $\ce{[M(NH3)2Cl2]}$, viz. &lt;em&gt;cis&lt;/em&gt; or &lt;em&gt;trans&lt;/em&gt; would react with silver oxalate. The possible reaction could be formation of silver chloride but since the coordination entity does not break up, any reaction thus, in my opinion can not happen. But the asker says that the answer is not &lt;strong&gt;none&lt;/strong&gt;, but actually one of them.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user15489" LastEditDate="2015-05-17T08:03:53.580" LastActivityDate="2015-05-17T08:03:53.580" Title="Reaction with silver oxalate?" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28976" PostTypeId="1" AcceptedAnswerId="28991" CreationDate="2015-04-19T06:02:52.690" Score="2" ViewCount="4230" Body="&lt;p&gt;I need to find CFSE for these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{[Ti(H2O)6]^3+}.$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Ti}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(4s)^2(3d)^2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ti^3+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{(4s)^0(3d)^1}.$&lt;/span&gt; Afterwards it becomes &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d^4sp^2}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}^5e_g^4}$&lt;/span&gt; so CFSE is &lt;span class=&quot;math-container&quot;&gt;$\frac25\Delta_0$&lt;/span&gt;. And I'm given &lt;span class=&quot;math-container&quot;&gt;$\bar\nu_\mathrm{max} = \pu{20300 cm-1}.$&lt;/span&gt; So, we can use &lt;span class=&quot;math-container&quot;&gt;$E=hc\bar\nu.$&lt;/span&gt; And that doesn't seem to come close to answer, I think some conceptual mistake has occurred.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{[CoF6]^3-}.$&lt;/span&gt; It's &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3d^2}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\mathrm{t_{2g}^4e_g^2}.$&lt;/span&gt; Here CFSE would be &lt;span class=&quot;math-container&quot;&gt;$4\frac25\Delta - 2\frac35\Delta = \frac25\Delta = \pu{4 Dq}$&lt;/span&gt;. This one's correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="41328" LastEditDate="2019-12-17T16:11:41.167" LastActivityDate="2019-12-17T16:15:59.493" Title="Calculation of CFSE" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;crystal-field-theory&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28977" PostTypeId="1" AcceptedAnswerId="28979" CreationDate="2015-04-19T06:08:00.383" Score="4" ViewCount="3120" Body="&lt;p&gt;I have a solution containing $0.2\:\mathrm{mol}$ of $\ce{Br^-}$ ions and $\ce{SO_4^2-}$ ions each.&lt;br&gt;&#xA;Now if I add excess  $\ce{AgNO3}$ in this solution,&lt;br&gt;&#xA;what will precipitate?&lt;br&gt;&#xA; $\ce{Ag2SO4}$ or $\ce{AgBr}$ or both?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I add just $0.2\:\mathrm{mol}$ of $\ce{AgNO3}$, what will precipitate?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't want to calculate anything. Just wanted to know qualitatively that which compound precipitates.)&lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="15410" LastEditDate="2015-05-20T00:35:58.430" LastActivityDate="2015-05-20T00:35:58.430" Title="How to determine which salt will precipitate from a solution containing multiple ions?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28978" PostTypeId="2" ParentId="28963" CreationDate="2015-04-19T06:18:10.353" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What percent of a solution needs to be acetic acid for a particular pH? I have 100% acetic acid, and want 1 L of solution with a pH of 5. I've done some research, but I still do not understand how to solve this problem.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need to know that if you add water moles remian same so $\rm M_1V_1=M_2V_2$. You also need to know $\rm pH=-\log[H^+]$ form where you get concentration of $\rm H^+$. Now for weak acids the cubic equation used to determine pH gets some terms negligible hence we use $\rm [H^+]=\sqrt{K_aM_2}$ where $\rm K_a$ is very common and I remember it as $1.7\times10^{-5}\sim10^{-4.7}$.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-19T06:18:10.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28979" PostTypeId="2" ParentId="28977" CreationDate="2015-04-19T06:23:52.717" Score="2" Body="&lt;p&gt;You need to know that $K_\mathrm{sp}$ is used to determine the solubilities of sparingly soluble salts usually. You can get the values of $\ce{AgSO4,AgBr}$ and as $K_\mathrm{sp}=[\ce{Ag+}]^2[\ce{SO4^2-}]$ or $[\ce{Ag+}][\ce{Br-}]$. Now concentration of silver ion is constant so try calculating the other two anions and try comparing them.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2015-04-19T09:56:47.217" LastActivityDate="2015-04-19T09:56:47.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28980" PostTypeId="2" ParentId="28972" CreationDate="2015-04-19T07:14:21.850" Score="4" Body="&lt;p&gt;To my understanding, only $\ce{NaC#CH}$ and $\ce{LiCH3}$ are organometallic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://www.iupac.org/nc/home/publications/provisional-recommendations/published/published-container/nomenclature-of-organometallic-compounds-of-the-transition-elements.html?sword_list%5B%5D=organometallic&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;IUPAC&lt;/strong&gt; definition of organometallic compounds&lt;/a&gt;, these must contain&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;at least one &lt;strong&gt;metal-carbon bond&lt;/strong&gt; between an organic molecule, ion, or radical and a metal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-19T07:14:21.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28981" PostTypeId="1" AcceptedAnswerId="29047" CreationDate="2015-04-19T09:57:29.277" Score="6" ViewCount="1288" Body="&lt;p&gt;As you can see the atom consists not only of the nucleus and electron but also of &quot;empty space is the space empty or is their something else?&lt;img src=&quot;https://i.stack.imgur.com/6oOwF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14594" LastActivityDate="2015-04-20T13:55:43.240" Title="What is there in an atom between the nucleus and electrons?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28984" PostTypeId="2" ParentId="28935" CreationDate="2015-04-19T10:29:35.290" Score="1" Body="&lt;p&gt;Since Li is highly solvated due to its small size, it's mobility is reduced and hence conductivity decreased.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-19T10:29:35.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28985" PostTypeId="2" ParentId="28943" CreationDate="2015-04-19T10:36:05.487" Score="1" Body="&lt;p&gt;I strongly suggest you read &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_zero&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_zero&quot; rel=&quot;nofollow&quot;&gt;there&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The laws of thermodynamics dictate that absolute zero cannot be reached using only thermodynamic means, as the temperature of the substance being cooled approaches the temperature of the cooling agent asymptotically. A system at absolute zero still possesses quantum mechanical zero-point energy, the energy of its ground state. The kinetic energy of the ground state cannot be removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An even stronger assertion is that It is impossible by any procedure to reduce the temperature of a system to zero in a finite number of operations. (≈ Guggenheim, p. 157)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get &quot;frozen in&quot; at low temperatures, so transitions to more stable states do not occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="10256" LastEditDate="2015-04-19T12:55:50.803" LastActivityDate="2015-04-19T12:55:50.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28986" PostTypeId="2" ParentId="28943" CreationDate="2015-04-19T10:51:54.387" Score="0" Body="&lt;p&gt;No, because it is possible that perfect crystals cannot be formed for statistical reasons. A simple example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute Boltzmann entropy is defined as $S = k_\text{B} \ln \Omega$ per molecule, where $\Omega$ is the number of possible microstates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a crystal of carbon monoxide at absolute zero. Now consider exchanging the position of all C atoms with those of all O atoms. This structure will be equivalent in energy so $\Omega = 2$. Thus $S = k_\text{B} \ln 2 &amp;gt; 0$.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T12:08:17.553" LastActivityDate="2015-04-19T12:08:17.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="28987" PostTypeId="1" CreationDate="2015-04-19T11:09:14.323" Score="12" ViewCount="9105" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The intermolecular interaction that is dependent on the inverse cube of distance between molecules is:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;hydrogen bond&lt;/li&gt;&#xA;  &lt;li&gt;ion-ion interaction&lt;/li&gt;&#xA;  &lt;li&gt;ion-dipole interaction&lt;/li&gt;&#xA;  &lt;li&gt;London force&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inverse cube of distance I think means an ion-dipole interaction, since we know force due to a dipole on a charge is inversely proportional to the cube of distance. But I don't understand why the answer given is hydrogen bond. What is the distance dependence of the hydrogen bond? Shouldn't the answer be (3)?&lt;/p&gt;&#xA;" OwnerUserId="15687" LastEditorUserId="194" LastEditDate="2018-05-14T20:42:07.093" LastActivityDate="2018-05-14T20:42:56.890" Title="Which intermolecular interaction is dependent on the inverse cube of distance?" Tags="&lt;intermolecular-forces&gt;&lt;hydrogen-bond&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28988" PostTypeId="1" CreationDate="2015-04-19T13:05:15.970" Score="0" ViewCount="251" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;what are the answer for CONFIRMATION OF REGISTRATION when register in dalton forum ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;install got error, how to install dalton 2015&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&#xA; wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ uname -r&#xA;    3.8.0-29-generic&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ sudo apt-get install openmpi-bin openmpi-dev&#xA;    [sudo] password for wonder: &#xA;    Reading package lists... Done&#xA;    Building dependency tree       &#xA;    Reading state information... Done&#xA;    Note, selecting 'libopenmpi-dev' instead of 'openmpi-dev'&#xA;    libopenmpi-dev is already the newest version.&#xA;    libopenmpi-dev set to manually installed.&#xA;    The following packages were automatically installed and are no longer required:&#xA;      libghc-regex-posix-dev libghc-http-dev libghc-stm-dev libghc-quickcheck2-dev&#xA;      libxmu-headers mesa-common-dev libghc-hunit-dev libghc-cgi-dev alex&#xA;      libghc-monadcatchio-mtl-dev libghc-syb-dev ghc-haddock happy&#xA;      libglu1-mesa-dev libdrm-dev freeglut3-dev libghc-html-dev libxt-dev&#xA;      libxmu-dev libghc-regex-base-dev libghc-fgl-dev gap-libs cabal-install&#xA;      libghc-opengl-dev libgl1-mesa-dev libghc-random-dev libkms1&#xA;      libghc-haskell-src-dev libghc-regex-compat-dev libghc-xhtml-dev freeglut3&#xA;      libghc-glut-dev libghc-parallel-dev&#xA;    Use 'apt-get autoremove' to remove them.&#xA;    The following extra packages will be installed:&#xA;      blcr-util libcr0 openmpi-checkpoint&#xA;    Suggested packages:&#xA;      blcr-dkms gfortran&#xA;    The following NEW packages will be installed:&#xA;      blcr-util libcr0 openmpi-bin openmpi-checkpoint&#xA;    0 upgraded, 4 newly installed, 0 to remove and 404 not upgraded.&#xA;    Need to get 239 kB of archives.&#xA;    After this operation, 729 kB of additional disk space will be used.&#xA;    Do you want to continue [Y/n]? y&#xA;    Get:1 http://us.archive.ubuntu.com/ubuntu/ precise-updates/universe libcr0 i386 0.8.2-15ubuntu2.1 [21.3 kB]&#xA;    Get:2 http://us.archive.ubuntu.com/ubuntu/ precise-updates/universe blcr-util i386 0.8.2-15ubuntu2.1 [27.5 kB]&#xA;    Get:3 http://us.archive.ubuntu.com/ubuntu/ precise/universe openmpi-bin i386 1.4.3-2.1ubuntu3 [152 kB]&#xA;    Get:4 http://us.archive.ubuntu.com/ubuntu/ precise/universe openmpi-checkpoint i386 1.4.3-2.1ubuntu3 [38.8 kB]&#xA;    Fetched 239 kB in 1s (159 kB/s)                 &#xA;    Selecting previously unselected package libcr0.&#xA;    (Reading database ... 192321 files and directories currently installed.)&#xA;    Unpacking libcr0 (from .../libcr0_0.8.2-15ubuntu2.1_i386.deb) ...&#xA;    Selecting previously unselected package blcr-util.&#xA;    Unpacking blcr-util (from .../blcr-util_0.8.2-15ubuntu2.1_i386.deb) ...&#xA;    Selecting previously unselected package openmpi-bin.&#xA;    Unpacking openmpi-bin (from .../openmpi-bin_1.4.3-2.1ubuntu3_i386.deb) ...&#xA;    Selecting previously unselected package openmpi-checkpoint.&#xA;    Unpacking openmpi-checkpoint (from .../openmpi-checkpoint_1.4.3-2.1ubuntu3_i386.deb) ...&#xA;    Processing triggers for man-db ...&#xA;    Setting up libcr0 (0.8.2-15ubuntu2.1) ...&#xA;    Setting up blcr-util (0.8.2-15ubuntu2.1) ...&#xA;    Setting up openmpi-bin (1.4.3-2.1ubuntu3) ...&#xA;    update-alternatives: using /usr/bin/mpirun.openmpi to provide /usr/bin/mpirun (mpirun) in auto mode.&#xA;    Setting up openmpi-checkpoint (1.4.3-2.1ubuntu3) ...&#xA;    Processing triggers for libc-bin ...&#xA;    ldconfig deferred processing now taking place&#xA;    wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ ./setup --mpi&#xA;Configuring system: Ubuntu&#xA; FC=mpif90 CC=mpicc CXX=mpicxx cmake -DENABLE_MPI=ON &#xA; -DENABLE_SGI_MPT=OFF -DENABLE_OMP=OFF &#xA; -DENABLE_64BIT_INTEGERS=OFF &#xA; -DENABLE_OPENACC=OFF -DENABLE_COLLAPSE=OFF&#xA; -DENABLE_CSR=OFF -DENABLE_SCALASCA=OFF &#xA; -DENABLE_VAMPIRTRACE=OFFDENABLE_TIMINGS=OFF &#xA; -DENABLE_XCFUN=OFF -DENABLE_INTEREST=OFF -DENABLE_ICHOR=OFF &#xA; -DENABLE_STATIC_LINKING=OFF -DENABLE_SCALAPACK=OFF &#xA; -DCMAKE_BUILD_TYPE=release /home/wonder/dalton/DALTON-Source&#xA;&#xA;CMake Error at /usr/share/cmake-2.8/Modules/CMakeTestFortranCompiler.cmake:40 (MESSAGE):&#xA;  The Fortran compiler &quot;/usr/bin/mpif90&quot; is not able to compile a simple test&#xA;  program.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It fails with the following output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   Change Dir: /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp&#xA;&#xA;&#xA;&#xA;  Run Build Command:/usr/bin/make &quot;cmTryCompileExec/fast&quot;&#xA;&#xA;  /usr/bin/make -f CMakeFiles/cmTryCompileExec.dir/build.make&#xA;  CMakeFiles/cmTryCompileExec.dir/build&#xA;&#xA;  make[1]: Entering directory&#xA;  `/home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp'&#xA;&#xA;  /usr/bin/cmake -E cmake_progress_report&#xA;  /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp/CMakeFiles 1&#xA;&#xA;  Building Fortran object&#xA;  CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o&#xA;&#xA;  /usr/bin/mpif90 -c&#xA;  /home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp/testFortranCompiler.f&#xA;  -o CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o&#xA;&#xA;  --------------------------------------------------------------------------&#xA;&#xA;  The Open MPI wrapper compiler was unable to find the specified compiler&#xA;&#xA;  gfortran in your PATH.&#xA;&#xA;&#xA;&#xA;  Note that this compiler was either specified at configure time or in&#xA;&#xA;  one of several possible environment variables.&#xA;&#xA;  --------------------------------------------------------------------------&#xA;&#xA;  make[1]: Leaving directory&#xA;  `/home/wonder/dalton/DALTON-Source/build/CMakeFiles/CMakeTmp'&#xA;&#xA;  make[1]: *** [CMakeFiles/cmTryCompileExec.dir/testFortranCompiler.f.o]&#xA;  Error 1&#xA;&#xA;  make: *** [cmTryCompileExec/fast] Error 2&#xA;&#xA;&#xA;&#xA;&#xA;&#xA;  CMake will not be able to correctly generate this project.&#xA;Call Stack (most recent call first):&#xA;  CMakeLists.txt:18 (project)&#xA;&#xA;&#xA;-- The Fortran compiler identification is unknown&#xA;-- The C compiler identification is GNU&#xA;-- The CXX compiler identification is GNU&#xA;-- Check for working Fortran compiler: /usr/bin/mpif90&#xA;-- Check for working Fortran compiler: /usr/bin/mpif90  -- broken&#xA;-- Configuring incomplete, errors occurred!&#xA;&#xA;wonder@wonder-VirtualBox:~/dalton/DALTON-Source$ ls -l /usr/bin/mpif90&#xA;lrwxrwxrwx 1 root root 24 Dec 12 22:51 /usr/bin/mpif90 -&amp;gt; /etc/alternatives/mpif90&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15689" LastEditorUserId="418" LastEditDate="2015-04-19T14:36:30.413" LastActivityDate="2015-04-19T14:43:13.440" Title="installation error of dalton 2015 in ubuntu 12" Tags="&lt;quantum-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-04-19T15:46:52.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="28990" PostTypeId="1" CreationDate="2015-04-19T13:19:14.617" Score="6" ViewCount="6250" Body="&lt;p&gt;One of my exam questions is about lipstick and calculating the amount of lithium in the lipstick by AAS. The determination took place by making standard solutions with known concentrations of lithium. For the stock solution $653.3~\mathrm{mg}$ lithium chloride, $\ce{LiCl}$, was dissolved in $100~\mathrm{ml}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(\ce{Li}) = 6.49~\mathrm{g/mol}$&lt;br&gt;&#xA;$M(\ce{LiCl}) = 42.4~\mathrm{g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the stock solution the following volumes were pipetted into $100~\mathrm{ml}$ flasks: $3$, $5$, $8$ and $10~\mathrm{ml}$. Also $1~\mathrm{ml}$ $\ce{KCl}$ solution, $1.0~\mathrm{M}$, was added to prevent ionisation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the concentration Lithium in the stock as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of Lithium in the weighed $\ce{LiCl}$ is equal to the molar mass of lithium divided by the molar mass of lithium chloride multiplied by the mass that was weighed. So in the form of an equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(Li)=\frac{M(\ce{Li})}{M(\ce{LiCl})}\cdot m(\text{total})=99.9~\mathrm{mg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the concentration of the stock solution is equal to the mass of lithium divided by the volume of the flask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(\text{Li, stock}) = \frac{m(\ce{Li})}{V(\ce{Li})}=\frac{99.9}{0.1}=999~\mathrm{mg/l}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of the standard solutions therefore would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(\text{standard}) = C(\text{stock}) \times \frac{V(\text{stock})}{V(\text{flask volume})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the concentration of the standard in $\mathrm{mg/L}$ and the volumes of both stock and flash volume in $\mathrm{mL}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would give concentrations of $29.9$; $49.9$; $79.9$ and $99.9~\mathrm{mg/L}$ to the standard solutions where $3$, $5$, $8$ and $10~\mathrm{ml}$ were pipetted into. The answer sheet however provides us with $300$, $500$, $800$ and $1000~\mathrm{\mu g/L}$. I suspect I've made a mistake in unit conversion somewhere, due to the factor 100 mistake in the answer, although the units don't correspond either. &lt;/p&gt;&#xA;" OwnerUserId="15690" LastEditorDisplayName="user7951" LastEditDate="2018-03-03T14:37:20.057" LastActivityDate="2021-02-18T01:05:05.480" Title="How to calculate the concentration based on dilution of a stock solution?" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28991" PostTypeId="2" ParentId="28976" CreationDate="2015-04-19T14:29:51.613" Score="2" Body="&lt;p&gt;Calculation for the first part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\bar\nu_\mathrm{max} = \pu{20300 cm^-1},$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$\Delta = \pu{4.03E-19 J}$&lt;/span&gt; per molecule or &lt;span class=&quot;math-container&quot;&gt;$\pu{242.7 kJ mol-1}.$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\mathrm{CFSE} = \pu{97.1 kJ}.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{CFSE} = \pu{6.6E-34 J s} × \pu{3E8 m s-1} × \pu{20300 cm-1} × \pu{6.02E23 mol-1} × \pu{E-3 J kJ-1} × \frac25 = \pu{97.1 kJ mol-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="41328" LastEditDate="2019-12-17T16:15:59.493" LastActivityDate="2019-12-17T16:15:59.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28992" PostTypeId="1" AcceptedAnswerId="28996" CreationDate="2015-04-19T14:31:43.920" Score="17" ViewCount="4710" Body="&lt;p&gt;March's Advanced Organic Chemistry has the following paragraph&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The other effect operates not&#xA;  through bonds, but directly through space or solvent molecules, and is called the field&#xA;  effect. It is often very difficult to separate the two kinds of effect, but it has been&#xA;  done in a number of cases, generally by taking advantage of the fact that the field effect&#xA;  depends on the geometry of the molecule but the inductive effect depends only on the&#xA;  nature of the bonds. For example, in isomers 1 and 2 the inductive effect of the chlorine&#xA;  atoms on the position of the electrons in the COOH group (and hence on the&#xA;  acidity, see Chapter 8) should be the same since the same bonds intervene; but the field&#xA;  effect is different because the chlorines are closer in space to the COOH in 1 than they&#xA;  are in 2. Thus a comparison of the acidity of 1 and 2 should reveal whether a field&#xA;  effect is truly operating. The evidence obtained from such experiments is overwhelming&#xA;  that field effects are much more important than inductive effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The diagram: &lt;img src=&quot;https://i.stack.imgur.com/OccCm.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simply do not understand how 2 is more acidic than 1. Since the field effect is greater in 1 , should not 1 be more acidic? &lt;/p&gt;&#xA;" OwnerUserId="11587" LastEditorUserId="7448" LastEditDate="2015-12-04T22:03:05.667" LastActivityDate="2015-12-04T22:03:05.667" Title="Experimental verification of the field effect" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="28993" PostTypeId="2" ParentId="28957" CreationDate="2015-04-19T14:38:42.927" Score="1" Body="&lt;p&gt;You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this:&#xA;$$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T11:39:45.553" LastActivityDate="2015-04-30T11:39:45.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28994" PostTypeId="2" ParentId="28988" CreationDate="2015-04-19T14:43:13.440" Score="2" Body="&lt;p&gt;As pointed out by Wildcat have a look at these messages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;CMake Error at /usr/share/cmake-2.8/Modules/CMakeTestFortranCompiler.cmake:40 (MESSAGE): The Fortran compiler &quot;/usr/bin/mpif90&quot; is not able to compile a simple test program.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;The Open MPI wrapper compiler was unable to find the specified compiler gfortran in your PATH.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;mpif90&lt;/code&gt; is a environment variable set by (older versions of) MPICH and is actually deprecated in recent versions of &lt;a href=&quot;https://www.open-mpi.org/doc/v1.8/man1/mpif90.1.php&quot; rel=&quot;nofollow&quot;&gt;Open MPI&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it you didn't forget to install &lt;code&gt;gfortran&lt;/code&gt;, what is the value of &lt;code&gt;mpif90&lt;/code&gt; on your system?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-19T14:43:13.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28995" PostTypeId="1" CreationDate="2015-04-19T14:56:54.173" Score="5" ViewCount="142" Body="&lt;p&gt;I mean by &quot;combining&quot; is to make a new half-reaction equation and not an overall equation for a reaction in whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I was trying to arrive at the following half-reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO2(s) + 4H+(aq) + 2e- -&amp;gt; Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by combining&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -&amp;gt; Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ=  1.51 \,\mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -&amp;gt; MnO2(s) + 2H_2O(l)}\quad\quad E^\circ=&#xA;1.7\,\mathrm{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using Hess's Law. My calculation yielded +0.19 V, and &lt;em&gt;not&lt;/em&gt; +1.23 V as I would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why caused this?&lt;/strong&gt; I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="865" LastEditDate="2015-08-03T00:02:35.447" LastActivityDate="2015-08-03T10:15:17.573" Title="Why can some redox half-reactions be combined and some cannot?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="4" ClosedDate="2017-04-10T22:27:54.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="28996" PostTypeId="2" ParentId="28992" CreationDate="2015-04-19T16:24:01.123" Score="14" Body="&lt;p&gt;The &lt;strong&gt;inductive effect&lt;/strong&gt; is transmitted &lt;strong&gt;through bonds&lt;/strong&gt;; the &lt;strong&gt;field effect&lt;/strong&gt; is transmitted &lt;strong&gt;through space&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{C-Cl}$ bond is polarized due to the electronegativity difference between the two atoms (carbon 2.55; chlorine 3.16), the chlorine end being more negative than the carbon end. We could note this excess of electron density (partial negative charge) around the chlorines in your structures by placing a ${\delta^{-}}$ by the chlorine and a ${\delta^{+}}$ by the carbon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when the acid ionizes there is also a negative charge associated with the carboxylate group.  When the chlorines are on the same side of the bridge as the carboxylate (syn isomer) the (through space) field effect acts to destabilize the situation because we have placed the partial negative charge around chlorine closer to the negative charge on the carboxylate (placing the negative charges closer together is destabilizing since like charges repel). The destabilization results in a higher activation energy being needed to reach the transition state for ionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the anti isomer with the chlorines further from the carboxylate will be less destabilized because the negative charges are further apart.  This means that the anti isomer will have a lower activation energy for ionization compared to the syn isomer.  Therefore, &lt;strong&gt;the anti isomer will be more acidic (display a lower pKa) than the syn isomer due to the field effect&lt;/strong&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-11T00:47:15.280" LastActivityDate="2015-06-11T00:47:15.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28997" PostTypeId="1" AcceptedAnswerId="29001" CreationDate="2015-04-19T16:26:59.433" Score="15" ViewCount="1552" Body="&lt;p&gt;Does anyone know of any good software that would allow me to view the structure of a compound while also allowing me to change the bond angles in the structure (and give me numeric values for those angles)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for something freely available, but that's not a major requirement. Thank you so much!&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2015-11-17T03:25:12.857" LastActivityDate="2020-02-01T17:19:26.457" Title="Software for viewing a structure and altering bond angles" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="28998" PostTypeId="2" ParentId="28987" CreationDate="2015-04-19T16:28:42.240" Score="1" Body="&lt;p&gt;Simple Google search led me to the content at &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The energy of a Keesom interaction depends on the inverse sixth power of the distance, unlike &lt;strong&gt;the interaction energy of two spatially fixed dipoles, which depends on the inverse third power of the distance.&lt;/strong&gt; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-19T16:28:42.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="28999" PostTypeId="1" CreationDate="2015-04-19T17:24:28.737" Score="2" ViewCount="50" Body="&lt;p&gt;In a container with volume 4 litres we have 2 moles $\ce{COCl2}$ and 4 moles $\ce{Cl2}$. Equilibrium is achieved when 60 % of the reactant $\ce{COCl2}$ is wasted 60 %. What is the equilibrium constant $\ce{CO + Cl2 &amp;lt;=&amp;gt; COCl2}$?&lt;br&gt;&#xA;In my book it is explained the solution. I understand it when it says that $\ce{PCl5}$ is the substance that is wasted, but after finding the concentration of each substance to find the constant it uses the formula: $k=I$ understand that we get the concentration of $\ce{Cl2}$, $\ce{COCl2}$, but then when he finds the constant it divides $[\ce{CO}]\cdot[\ce{Cl2}]$ with $[\ce{COCl2}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think that they should divide $[\ce{COCl2}]$ with $[\ce{CO}]\cdot[\ce{Cl2}]$.&lt;/p&gt;&#xA;" OwnerUserId="9878" LastEditorDisplayName="user7951" LastEditDate="2015-04-19T18:51:39.090" LastActivityDate="2015-04-19T18:51:39.090" Title="Find the equilibrium constant" Tags="&lt;kinetics&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29000" PostTypeId="1" CreationDate="2015-04-19T17:24:48.163" Score="27" ViewCount="16284" Body="&lt;p&gt;$\ce{C2H5+ vs \ C3H7+}$&#xA;There are two conflicting trends here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductive effect of ethyl will be higher than that of methyl so carbocation attached to ethyl (3 carbons in total) should be more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand 3 hyperconjugation structures can be drawn for the carbocation attached to methyl (2 carbons in total) and only 2 hyperconjugation structure for the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which effect wins? I searched a lot of books but didn't find anything.&lt;/p&gt;&#xA;" OwnerUserId="11587" LastEditorUserId="4231" LastEditDate="2015-05-20T02:22:14.733" LastActivityDate="2020-09-06T08:22:33.830" Title="Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29001" PostTypeId="2" ParentId="28997" CreationDate="2015-04-19T17:49:27.900" Score="11" Body="&lt;p&gt;Have a look at &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;! &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;build your own structures or import files in various formats.&lt;/li&gt;&#xA;&lt;li&gt;optimize the geometry using different force fields&lt;/li&gt;&#xA;&lt;li&gt;change and measure bond length, bond angles and dihedral angles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you've used it, you don't want to miss it again. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Geoff Hutchison&lt;/em&gt;, who cannot be praised enough for his child &lt;em&gt;Avogadro&lt;/em&gt;, has given the right directions in his comment. I'm just adding some images to illustrate how easy angle adjustment is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw a molecule in Avogadro&lt;/li&gt;&#xA;&lt;li&gt;From the main menu, select Build -&gt; Properties -&gt; Angles or whatever the localized menu of your system shows (on my German machine, it is Ansicht -&gt; Eigenschaften -&gt; Winkeleigenschaften)&lt;/li&gt;&#xA;&lt;li&gt;In the table that pops ups, select the angle you want to change. Note that the atoms of the selected angle are highlighted&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xPO3y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Change the value for the angle. Note that the geometry of the molecule is immediately updated! &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSduB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again, I think that Avogadro is one of the best tools available and it is free and open source! Use it in education, use it in research and whenever you do, cite the Avogadro paper:&lt;br&gt;&#xA;Hanwell, M. D.; Curtis, D. E.; Lonie, D. C.; Vandermeersch, T.; Zurek, E.; Hutchison, G. R.; Avogadro: an advanced semantic chemical editor, visualization, and analysis platform. &lt;em&gt;J. Cheminf.&lt;/em&gt; &lt;strong&gt;2012,&lt;/strong&gt; &lt;em&gt;4&lt;/em&gt; (1), No. 17. &lt;a href=&quot;https://doi.org/10.1186/1758-2946-4-17&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1186/1758-2946-4-17&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2020-02-01T17:19:26.457" LastActivityDate="2020-02-01T17:19:26.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29002" PostTypeId="1" AcceptedAnswerId="29006" CreationDate="2015-04-19T18:01:50.973" Score="4" ViewCount="761" Body="&lt;p&gt;$$\ce{C2O4^2- + MnO2 -&amp;gt; Mn^2+ + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the half reactions are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2O4^2- -&amp;gt; CO2}$$&#xA;$$\ce{MnO2 -&amp;gt; Mn^2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to balance these by adding water, $\ce{H+}$ atoms and by adding $\ce{e-}$’s, but I’m just not sure on the method to do this as we’ve covered it extremely quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I found the oxidation numbers for the overall equation, and I think that $\ce{C2O4^2-}$ is the reducing agent because $\ce{C}$ is losing charge from +3 to +4, I just don’t know how to use that to balance this. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="15252" LastEditDate="2015-04-19T18:51:43.097" LastActivityDate="2016-03-11T23:50:57.687" Title="How can I balance the following equation atomically and electrically?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29003" PostTypeId="1" CreationDate="2015-04-19T18:23:26.143" Score="3" ViewCount="2599" Body="&lt;p&gt;A picture for reference:&lt;img src=&quot;https://i.stack.imgur.com/jfU4V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Difficulty 1:&lt;/strong&gt; I don't get why a salt bridge is needed. People say it's to maintain the overall neutrality. But in both cases where  salt bridge is a tube of solution and a membrane, two half cells can be seen as one. So as long as no thing leaves either of the two half cells, the whole is always neutral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Difficulty 2:&lt;/strong&gt; as in the picture, cations go to the right cell, and anions go to the left. I can understand that anions go the left because the number of zinc ions increase as the reaction proceed. But in the right half cell, the electrons are being absorbed, its charge cancelled by that of the copper ion; I don't see any need for sodium ions to move to the right.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastActivityDate="2017-11-20T01:13:37.467" Title="Difficulties understanding how salt bridge works" Tags="&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29004" PostTypeId="2" ParentId="29002" CreationDate="2015-04-19T18:59:35.403" Score="1" Body="&lt;p&gt;It might be a good idea to balance your half reactions first and then sum up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align*}&#xA;\ce{C2O4^{2-} &amp;amp;-&amp;gt; 2CO2 +2e-\\&#xA;MnO2 + 4H+ + 2e- &amp;amp;-&amp;gt; Mn^2+ + 2H2O}\\&#xA;\hline&#xA;\ce{MnO2 + 4H+ + C2O4^{2-} &amp;amp;-&amp;gt; Mn^2+ + 2CO2 + 2H2O}\\&#xA;\end{align*}&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="24052" LastEditDate="2016-03-11T23:50:57.687" LastActivityDate="2016-03-11T23:50:57.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29005" PostTypeId="2" ParentId="29003" CreationDate="2015-04-19T19:00:23.153" Score="3" Body="&lt;p&gt;&lt;strong&gt;Regarding Difficulty 1)&lt;/strong&gt;  When the zinc bar (unconnected) is first immersed into the solution, some zinc ions immediately dissolve away.  Now the zinc bar has extra electrons and the solution has extra positive zinc cations.  This is entropically acceptable, however, because cations in the solution (both zinc cations and hydronium cations) can align close to the surface of the zinc metal (creating what's called the 'Helmholtz layer').  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might imagine that as soon as the zinc bar was connected, the extra electrons in the zinc would drain to the other side of the cell.  But they don't because the electrostatic pull of the zinc and hydronium cations in the left-beaker solution is holding them back.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an analogous situation in the right cell:  the copper metal can't accept any electrons because of the electrostatic repulsion of the sulfate and hydroxide ions in right-beaker solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding Difficulty 2)&lt;/strong&gt; When the cell is running, there is an electrostatic gradient across the membrane where the salt bridge meets the right-side solution (among other places, for more info see note 1).  Cations (such as sodium) are driven out of the salt bridge;  anions (such as sulfate) are drawn into the salt bridge.  Sodium cations are smaller and diffusion more quickly than the larger, slower sulfate anions. Overall, many more sodium ions will leave the salt bridge compared to the number of sulfate anions that will enter the salt bridge.  There is an analogous situation at the membrane where the salt bridge meets the left-beaker solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note 1, Electrostatic Gradients)&lt;/strong&gt; When the cell is running, the zinc bar has a slightly negative charge and the copper has a slightly positive charge.  If we let ourselves ignore the metal wire for the moment, this creates an electrostatic field throughout  the cell very much like the classic &lt;a href=&quot;http://en.wikipedia.org/wiki/Electric_field&quot; rel=&quot;nofollow&quot;&gt;&quot;electrostatic field around adjacent positive and negative charges in a uniform media&quot;&lt;/a&gt;.  Notice that in the diagram of that situation, there are electrostatic field lines pointing from the positive charge straight-across to the negative charge.  Another way to describe this situation would be to say that there is an &lt;em&gt;electrostatic gradient&lt;/em&gt; between the two charges.  If you drop a much smaller, and otherwise free-floating positive charge any where in the neighborhood around those two charges, it will move down electrostatic gradient to the negative charge:  the same way rain-water flows downhill, (in that case due to a gravitational gradient).  Now imagine there was an imaginary plane exactly halfway between the two charges:  you could now also say that there is an electrostatic gradient &lt;em&gt;across&lt;/em&gt; that plane, from the one side to the other. If you dropped a positive charge right beside the plane, but on the same side as the bigger positive charge, the electrostatic gradient would drive the positive charge across the plane.  Similarly, if you dropped a negative charge next to the plane,  on the same side as the bigger negative charge, then again the electrostatic gradient would drive the negative charge across the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the electrochemical cell problem, the electrostatic field lines are going to be more complicated because there are two materials (air and water / salt-bridge) with different &lt;a href=&quot;http://en.wikipedia.org/wiki/Permittivity&quot; rel=&quot;nofollow&quot;&gt;permittivity&lt;/a&gt;.  Long story short: the electrostatic fields will actually be much stronger in the water than in the air.  The strongest part of the electrostatic field will go from the copper electrode, across the right-side solution, through the salt bridge, and across the left-side solution to the zinc.   So, among other places, when the cell is running, there will be an electrostatic gradient across the membrane where the salt bridge meets the right-side solution.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="1040" LastEditDate="2015-05-03T03:07:44.610" LastActivityDate="2015-05-03T03:07:44.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29006" PostTypeId="2" ParentId="29002" CreationDate="2015-04-19T19:04:46.320" Score="3" Body="&lt;p&gt;Deal with the two half equations separately and then combine them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the oxalate equation: $$\ce{C2O4^{2-} -&amp;gt; CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balance the atoms: $$\ce{C2O4^{2-} -&amp;gt; 2CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now balance the charge by adding electrons: $$\ce{C2O4^{2-} -&amp;gt; CO2 + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the manganese dioxide reduction: $$\ce{MnO2 -&amp;gt; Mn^{2+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can balance the half equations by adding water, hydrogen ions or electrons. Since water is the only one which contains oxygen we should add this first: $$\ce{MnO2 -&amp;gt; Mn^{2+} + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add hydrogen ions to balance the hydrogens: $$\ce{MnO2 +4H+ -&amp;gt; Mn^{2+} + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally add electrons the balance the charge: $$\ce{MnO2 +4H+ + 2e- -&amp;gt; Mn^{2+} + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now combine the equations to cancel all the electrons. In this case there are two on both sides so we don't need to multiply any of the equations by anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO2 +4H+ + C2O4^{2-} -&amp;gt; Mn^{2+} + 2CO2 + 2H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-19T19:04:46.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29009" PostTypeId="1" AcceptedAnswerId="29015" CreationDate="2015-04-19T20:46:22.030" Score="2" ViewCount="383" Body="&lt;p&gt;I am working on a project for my General Chemistry class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to compare the EtOH production (and the rate of EtOH production) between seven types of yeast strains by fermenting the YPD media, apple juice, or both, if necessary. I can describe the difference between the yeast strains if asked, but I don't think it is necessary right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I have only tested plain YPD media, which is 1% yeast extract, 2% peptone and 2% d-glucose mixture. All strains yield around 8-9 mg/mL of EtOH, and the fermentation stops after ~24 hours. My next step is to test YPD + apple juice media (because the project asks for &quot;fun&quot; element), and I was wondering if I should use the same method for analyzing this YPD + apple juice fermentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I used for YPD media is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Grow each strain (total of 7 strains) in ~5 mL YPD media at 30 degrees shaker ~250 rpm overnight.&lt;/li&gt;&#xA;&lt;li&gt;Subculture each strain into new fresh YPD media so that the new media is 50 mL and has 0.5 OD at 600 nm. Use the &quot;special&quot; flask designed for anaerobic growth. This flask is a 250 mL flask with a special cap. The cap is plastic at the outer radius and rubber at the inside. This allows me to grow yeast anaerobically and poke a needled syringe and retrieve small amount of sample without letting the air in.&lt;/li&gt;&#xA;&lt;li&gt;Flush all flasks with nitrogen gas for four minutes.&lt;/li&gt;&#xA;&lt;li&gt;After 6 hour, 12 hour, and 24 hour growths, use syringe to retrieve ~1.5 mL of sample.&lt;/li&gt;&#xA;&lt;li&gt;Measure OD600 of each sample. (This is used to get the number of yeast cells in the media)&lt;/li&gt;&#xA;&lt;li&gt;Centrifuge samples for 8 minutes in 14,000 rpm.&lt;/li&gt;&#xA;&lt;li&gt;Pipette 1 mL of the supernatant into a GC vial.&lt;/li&gt;&#xA;&lt;li&gt;Use HPLC to obtain concentration of ethanol. I am using refractive index detector, ion exclusion column(?) (I forgot the type of column used at this moment), and sulfuric acid as mobile phase.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each &quot;special flask&quot;. Will this work? Will this be good for the same HPLC setting?&lt;/p&gt;&#xA;" OwnerUserId="15695" LastActivityDate="2015-04-20T00:23:27.763" Title="Yeast Fermentation with YPD + Apple Juice" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29010" PostTypeId="2" ParentId="29000" CreationDate="2015-04-19T21:04:31.553" Score="3" Body="&lt;p&gt;You never defined which carbon is the carbocation in the case of the propylium carbocation. I will go through a comparison of all three possibilities: the case of the ethylium carbocation, the linear propylium carbocation with a terminal carbocation, and the linear propylium carbocation with an internal carbocation.&lt;/p&gt;&#xA;&lt;p&gt;Thermodynamically: Ethylium &amp;lt; terminal propylium &amp;lt; internal propylium&lt;/p&gt;&#xA;&lt;p&gt;Kinetically: Terminal propylium &amp;lt; ethylium &amp;lt; internal propylium&lt;/p&gt;&#xA;&lt;p&gt;The difference in the kinetic and thermodynamic stability arises from the fact that the terminal propylium will readily undergo a 1,2-hydride shift to create the internal propylium ion.&lt;/p&gt;&#xA;&lt;p&gt;As for the thermodynamic order, it gets a bit more complex. Remember that for alkanes, it is ALWAYS the case that the more substituted carbocation is more stable and in cases of ties, more substituted substituents make for more stable carbocations. As for why this is true, well it comes down to the fact that carbons are electron donating to carbocations (and other electron deficient carbons). There is a role from hyperconjugation but you are forgetting that you can have hyperconjugation from bonds other than &lt;span class=&quot;math-container&quot;&gt;$\ce{C-H}$&lt;/span&gt;. You also seem to think that more hyperconjugatory structures is better when really that is not the case. Hyperconjugation is NOT an example of resonance where the molecule is in a superposition of multiple states so there is no enthalpic benefit of more methods of hyperconjugation. There is an entropic effect from the additional degrees of freedom but this would be a very small difference and if I remember right, would actually benefit both propylium carbocations, not the ethylium carbocation.&lt;/p&gt;&#xA;" OwnerUserId="15697" LastEditorUserId="61578" LastEditDate="2020-09-06T08:22:33.830" LastActivityDate="2020-09-06T08:22:33.830" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29012" PostTypeId="1" CreationDate="2015-04-19T21:21:50.107" Score="2" ViewCount="8949" Body="&lt;p&gt;I'm an IGCSE Chemistry-extended student, and my question is: in electrolysis, does it make any difference if the electrolyte is in a molten, aqueous or dilute aqueous solution form?&#xA;An example mentioned in my book:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Explain  the difference in the electrolysis products of&lt;/p&gt;&#xA;&lt;p&gt;a) molten sodium chloride;&lt;br /&gt;&#xA;b) dilute aqueous solution of sodium chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15698" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-07-03T11:31:11.597" Title="In electrolysis, does it make any difference if the electrolyte is an molten, aqueous or dilute aqueous solution?" Tags="&lt;physical-chemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29013" PostTypeId="2" ParentId="29012" CreationDate="2015-04-19T21:31:51.640" Score="7" Body="&lt;p&gt;In molten sodium chloride, the only available ions are $\ce{Na^+}$ and $\ce{Cl^-}$. With electrolysis, the products will be sodium metal and chlorine gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an aqueous solution, the sodium chloride still moves the charges from one electrode to the other, but the potential for electrolysis of water is less than $\ce{Na^+}$ or $\ce{Cl^-}$. So the products of a solution of sodium chloride is hydrogen gas and oxygen gas.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorDisplayName="user15489" LastEditDate="2015-04-20T09:33:18.007" LastActivityDate="2015-04-20T09:33:18.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29014" PostTypeId="2" ParentId="28934" CreationDate="2015-04-19T23:45:10.687" Score="7" Body="&lt;p&gt;This is certainly possible in most MM packages and hence it's possible via QM/MM as well. I had to do a bit of digging to see how you might do this in each program (hint, it's not easy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Gaussian, your best bet is to read the extremely long section on &lt;a href=&quot;http://gaussian.com/g_tech/g_ur/k_mm.htm&quot;&gt;molecular mechanics paramters&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For GAMESS, you'd need to install the &lt;a href=&quot;http://dasher.wustl.edu/tinker/&quot;&gt;Tinker&lt;/a&gt; program and use the QM/MM integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro and Open Babel don't currently allow custom forces, although it wouldn’t be too hard to code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the other alternative, depending on why you need the forces, is to use a geometry constraint during optimization. For example, ensuring certain distances are retained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is basically, that without care, your geometry optimization will be forced to use numeric derivatives, since the QM or MM program won't know much about the functional form of your custom force.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-04-19T23:45:10.687" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29015" PostTypeId="2" ParentId="29009" CreationDate="2015-04-20T00:23:27.763" Score="1" Body="&lt;p&gt;Thank you for an especially clear and detailed question!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each &quot;special flask&quot;. Will this work? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this would work but it seems like a better comparison to your earlier experiments to keep using 50 mL of total medium, i.e., mix 25 mL of YPD with 25 mL of apple juice.  That way the volume of the fermentation is the same in each experiment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will this be good for the same HPLC setting?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably, although it depends on the components of the apple juice you use.  Have you injected just apple juice (unfermented) into your HPLC?  Have you verified that this negative control injection contains no ethanol or other interfering peaks in the region of interest?  If not, those are things you should do to make sure the HPLC method you have developed for YPD medium will still work in apple juice-based media.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-20T00:23:27.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29016" PostTypeId="1" CreationDate="2015-04-20T00:29:54.410" Score="3" ViewCount="14541" Body="&lt;p&gt;Is it possible for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\Delta G=-nFE$&lt;/span&gt; to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.&lt;/p&gt;&#xA;" OwnerUserId="15700" LastEditorDisplayName="user7951" LastEditDate="2019-03-11T08:21:32.627" LastActivityDate="2019-03-11T08:21:32.627" Title="Is the 'n' in Delta G = -nFE, ever negative?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29017" PostTypeId="2" ParentId="28957" CreationDate="2015-04-20T01:21:28.690" Score="0" Body="&lt;p&gt;You already determined the $\Delta S$ for benzene at 353 K by dividing the $\Delta H$ at 353 K by the temperature 353 K.  Now all you need to do is determine the $\Delta H$ at 343 K using Hess' law?  Once you do that, you can get the $\Delta S$ at 343 K.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T11:41:20.987" LastActivityDate="2015-04-30T11:41:20.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29018" PostTypeId="1" CreationDate="2015-04-20T01:24:32.733" Score="0" ViewCount="465" Body="&lt;p&gt;I thought that the ratio of silicon to oxygen in a silica tetrahedron was 1:4, so if a single chain is just many of these linked together, why does the ratio become 1:3?&lt;/p&gt;&#xA;" OwnerUserId="15252" LastActivityDate="2015-04-20T12:12:42.943" Title="Why is the ratio between Silicon and Oxygen 1:3 in single chain silicates?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29020" PostTypeId="2" ParentId="29016" CreationDate="2015-04-20T01:27:59.407" Score="8" Body="&lt;p&gt;No. This result occurs from the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Free_Energy/Gibbs_Free_Energy/Gibbs_Free_Energy&quot;&gt;Nernst equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;n&quot; is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise &quot;n&quot; is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a negative number of electrons transferred would be impossible.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-04-20T01:27:59.407" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="29021" PostTypeId="2" ParentId="28910" CreationDate="2015-04-20T02:04:22.167" Score="1" Body="&lt;p&gt;For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="15095" LastEditDate="2015-10-26T10:12:34.400" LastActivityDate="2015-10-26T10:12:34.400" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29023" PostTypeId="1" CreationDate="2015-04-20T02:27:01.097" Score="11" ViewCount="1452" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wurtz reaction&lt;/a&gt;, is the intermediate formed a carbanion or a free radical?&lt;/p&gt;&#xA;&lt;p&gt;And if we can assume both mechanisms to be true, how can we decide the reactivity of a tertiary halide?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-05-10T17:08:42.007" Title="Does Wurtz reaction follow an ionic mechanism or a free radical mechanism?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29025" PostTypeId="1" CreationDate="2015-04-20T03:39:49.937" Score="6" ViewCount="1926" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would a radical reaction end up producing an alkene in the context of the Wurtz reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do see happening is an E2 elimination creating an alkene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYlxb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this raises another question - can't Grignards also create alkenes &quot;by accident&quot;? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5026" LastEditDate="2018-03-12T04:39:01.880" LastActivityDate="2018-03-12T04:39:01.880" Title="Origin of alkenes from Wurtz Reaction" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;halides&gt;&lt;hydrocarbons&gt;&lt;elimination&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29027" PostTypeId="1" AcceptedAnswerId="29038" CreationDate="2015-04-20T04:18:22.053" Score="7" ViewCount="2202" Body="&lt;p&gt;How do you get from 2,3-dimethylcyclopentanol to 1,2-dimethylcyclopent-1-ene? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the -OH group leaves as water (being protonated by acid) forming a carbocation. Then the weak nucleophile (water) attacks the beta-hydrogen to form what I think is 2,3-dimethylcyclopent-1-ene. But why is it 1,2-dimethylcyclopent-1-ene? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Zaitsev's rule wants the more substituted alkene, but forming the 1,2-dimethylcyclopent-1-ene would leave the carbocation hanging there, wouldn't it (the carbocation would still be unsatisfied)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ODEvv.png&quot; alt=&quot;acid-catalyzed dehydration of 2,3-dimethylcyclopentanol&quot;&gt;  &lt;img src=&quot;https://i.stack.imgur.com/fZhoh.png&quot; alt=&quot;2,3-dimethylcyclopent-1-ene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;img src=&quot;https://i.stack.imgur.com/yTi69.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15045" LastEditorUserId="15045" LastEditDate="2015-04-21T03:38:16.027" LastActivityDate="2015-04-22T05:43:20.887" Title="Mechanism for acid-catalyzed dehydration of 2,3-dimethylcyclopentanol" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29029" PostTypeId="1" CreationDate="2015-04-20T04:55:18.497" Score="1" ViewCount="40" Body="&lt;p&gt;It is stated in my book &quot;All gases condense to form liquids or solids before -273.15 degree Celsius  is reached?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;could you explain this statement generally and my question in the title specifically?&lt;/p&gt;&#xA;" OwnerUserId="15381" LastActivityDate="2015-04-20T20:19:42.733" Title="How can cooling gases do not change their physical states?" Tags="&lt;temperature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29030" PostTypeId="2" ParentId="29025" CreationDate="2015-04-20T05:39:56.653" Score="7" Body="&lt;p&gt;The best explanation I was able to find is given by &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01262a003&quot; rel=&quot;noreferrer&quot;&gt;Avery A. Morton , John B. Davidson , Barton L. Hakan, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1942&lt;/strong&gt;, &lt;em&gt;64&lt;/em&gt; (10), 2242–2247.&lt;/a&gt; If I understand it correctly, this boils down exactly to that you already propose.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Disproportionation has been so universally accepted as the criterion of a&#xA;  free radical that there has been no attempt to show a derivation from other molecular species. Nevertheless, a mechanism for this phenomenon&#xA;  by way of an organometallic intermediate appears exceptionally reasonable. If an alkylsodium meets an alkyl halide in such manner that the two alkyl chains are adjacent to each other while sodium halide is being formed, the two alkyl residues will have unlike charges and a proton will be drawn to the alkyl radical having the two unsaturated electrons. The reaction is merely the conventional prototropic change taking place between adjacent portions of two molecules whose inorganic components are effecting, or have just&#xA;  completed, a union.&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/iYf3e.png&quot; alt=&quot;disproportionation mechanism according to Morton&quot;&gt;&lt;br&gt;&#xA;  If this view be accepted, the olefin produced must come from the alkyl halide rather than at random from both participant alkyl radicals. Though conclusions are to some extent obscured by a metal-halogen interchange (see later), all available facts unquestionably point to the correctness of this deduction. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, there might be another reason, which I cannot back up with literature. This is probably due to the persistence of the radical species; higher concentration of radicals increases the possibility that they react with each other. In other words, one radical abstracts a hydrogen from another radical, hence creating an alkane and an alkene, in a purely radical reaction. &#xA;$$\ce{2 R-CH2-CH2. -&amp;gt; R-CH2-CH3 + R-CH=CH2}.$$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IM7Ui.jpg&quot; alt=&quot;reaction mechanism of Wurtz side reaction&quot;&gt;&lt;br&gt;&#xA;But then again radical mechanisms are never simple and it is quite possible that this does not happen at all, yet it would appear to be at least &lt;em&gt;an&lt;/em&gt; explanation for the Wikipedia statement.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-04-20T05:39:56.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29032" PostTypeId="1" CreationDate="2015-04-20T07:25:12.277" Score="4" ViewCount="2505" Body="&lt;p&gt;I'm consulting one of higher classes books to study chemistry. I had been studying Graham's Law and question related to rate of diffusion. I came across a question&lt;sup&gt;[1]&lt;/sup&gt; which says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the ratio of diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; under constant temperature and pressure.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question mentioned the mass of elements as following:   &lt;span class=&quot;math-container&quot;&gt;$\ce{S}=32$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{O}=16$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{C}=12$&lt;/span&gt;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution of the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}=32+32=64$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}=12+32=44$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\textrm{Rate of diff. of}\ \ce{SO2}}{\textrm{Rate of diff. of}\  \ce{CO2}} = \sqrt{\frac{44}{64}} = \frac{\sqrt{11}}{4} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book has a different answer at the back, where it says &lt;span class=&quot;math-container&quot;&gt;$1:2$&lt;/span&gt;. I rechecked and performed all the maths again, but still cannot find this answer. Is the book incorrect, or have I made a mistake? If the latter, please help me find it.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;D. N. Singh: &lt;em&gt;Basic Concepts of Inorganic Chemistry, 2nd Edition.&lt;/em&gt;&#xA;Pearson India, December 2011. ISBN: 9788131768617&#xA;&lt;a href=&quot;https://www.oreilly.com/library/view/basic-concepts-of/9788131768617&quot; rel=&quot;nofollow noreferrer&quot;&gt;URL of publisher&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5466" LastEditorUserId="4945" LastEditDate="2019-03-18T19:44:55.467" LastActivityDate="2019-03-18T19:44:55.467" Title="Ratio of rate of diffusion of two gases under constant temperature and pressure" Tags="&lt;inorganic-chemistry&gt;&lt;erratum&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29033" PostTypeId="2" ParentId="28961" CreationDate="2015-04-20T08:25:25.200" Score="1" Body="&lt;p&gt;From the equation&#xA;\[\ce{2SO3^{2-} + O2 -&gt; 2SO4^{2-} } \]&#xA;you could conclude that you need (at least) 2 moles of sodium sulfite $(\ce{Na2SO3})$ for each mole of oxygen in solution.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-20T08:25:25.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29034" PostTypeId="1" AcceptedAnswerId="29035" CreationDate="2015-04-20T08:29:20.743" Score="4" ViewCount="190" Body="&lt;p&gt;Is peroxide bleaching ($\ce{H2O2}$) suitable for treating bast fibres? I read that to be successful, it needs to be combined with chlorine bleaching. Because of the end usage, I don't want to use this kind of bleaching.&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="10185" LastEditDate="2015-04-20T08:56:47.333" LastActivityDate="2015-04-20T08:56:47.333" Title="Peroxide bleaching for bast fibres?" Tags="&lt;everyday-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29035" PostTypeId="2" ParentId="29034" CreationDate="2015-04-20T08:44:55.713" Score="5" Body="&lt;p&gt;Old patents of the 1930s, particularly from I.G. Farben often describe the use of hypochlorite/chlorine in the bleaching of bast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from newer literature, such as this application manual for hydrogen peroxide bleaching from &lt;a href=&quot;http://h2o2.evonik.com/sites/dc/Downloadcenter/Evonik/Product/H2O2/applications---hydrogen-peroxide-for-textile-industry.pdf&quot;&gt;Evonik Industries&lt;/a&gt;, I would conclude that bleaching of bast can be performed &lt;strong&gt;without&lt;/strong&gt; chlorine, supposed that the peroxide bleaching is performed &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in alkaline medium&lt;/li&gt;&#xA;&lt;li&gt;at elevated temperatures $(85-90\,°\mathrm{C})$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-20T08:44:55.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29036" PostTypeId="2" ParentId="29003" CreationDate="2015-04-20T09:39:44.997" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Difficulty 1:&lt;/strong&gt; I don't get why a salt bridge is needed. People say it's to maintain the overall neutrality. But in both cases where salt bridge is a tube of solution and a membrane, two half cells can be seen as one. So as long as no thing leaves either of the two half cells, the whole is always neutral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Salt bridge is to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;maintain neutrality&lt;/li&gt;&#xA;&lt;li&gt;complete the circuit&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your problem is that through wires electrons move from one compartment to another, but if there's no salt bridge and charge cannot build up, current simply does not flow.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Difficulty 2&lt;/strong&gt;: as in the picture, cations go to the right cell, and anions go to the left. I can understand that anions go the left because the number of zinc ions increase as the reaction proceed. But in the right half cell, the electrons are being absorbed, its charge cancelled by that of the copper ion; I don't see any need for sodium ions to move to the right.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes you are right, since charge is cancelled, we need to get back the positive charge back in right compartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly in left as zinc is increasing, positive charge is increasing, so to balance sulphate moves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the total charge in a vessel must remain constant (not neutral).&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="9961" LastEditDate="2017-11-20T01:13:37.467" LastActivityDate="2017-11-20T01:13:37.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29038" PostTypeId="2" ParentId="29027" CreationDate="2015-04-20T09:49:36.190" Score="3" Body="&lt;p&gt;See this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n2dSh.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5531" LastEditDate="2015-04-22T05:43:20.887" LastActivityDate="2015-04-22T05:43:20.887" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29039" PostTypeId="2" ParentId="29018" CreationDate="2015-04-20T09:51:50.227" Score="2" Body="&lt;p&gt;Silicon has 2 complete oxygen atoms ($2\times1=2$) and 2 equally shared oxygen atoms ($2\times1/2=1$), total 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgKNj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="5017" LastEditDate="2015-04-20T12:12:42.943" LastActivityDate="2015-04-20T12:12:42.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29043" PostTypeId="1" CreationDate="2015-04-20T10:10:53.480" Score="4" ViewCount="1014" Body="&lt;p&gt;I'm modelling a small set of proline analogues - extra functional groups, sulfur in the ring. Using Gaussian / DMACRYS.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--functional B3LYP EmpiricalDispersion=GD3BJ --basis_set 6-311G**&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-fun B3LYP EmpiricalDispersion=GD3BJ -bas 6-311G**&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Would EmpiricalDispersion=GD3BJ go under -fun, or would it go under &lt;code&gt;--additional_args&lt;/code&gt;? (If so, please show how, I can't get this to run correctly, unsure of the syntax)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To model this more accurately, PBEPBE has been suggested, pure DFT rather than hybrid functional, so there is increased accuracy. Is there a sufficient increase in accuracy for the increase in computational time, or is it better to apply dispersion corrections to B3LYP? (Or PBEPBE and corrections?) Is there another level of theory you would suggest trying?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The electrostatic interactions seem to be over-modelled using B3LYP / 6-311G**, is there a basis set / functional combination (without or with corrections) that would not over-estimate the energy of the + / - charge interactions? (I've tried CAM-B3LYP, but doesn't work on my system.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15708" LastEditorUserId="402" LastEditDate="2017-01-14T08:51:04.820" LastActivityDate="2017-01-14T08:51:04.820" Title="Most efficient level of theory for sulfur-containing proline analogues (one that will work)" Tags="&lt;computational-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29044" PostTypeId="1" CreationDate="2015-04-20T11:25:47.967" Score="8" ViewCount="6031" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A buffer was made by mixing $\pu{500 mL}$ of $\pu{1 M}$ acetic acid and $\pu{500 mL}$ of $\pu{0.5 M}$ calcium acetate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What are the resulting concentrations of acetic acid (henceforth called $\ce{AcH}$), $\ce{Ca^2+}$ and acetate (henceforth called $\ce{Ac-}$, and what is the resulting $\mathrm{pH}$? $K_\mathrm{a}$(acetic acid) = $1.75\times10^{-5}.$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempted solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem gives me two equilibrium reactions to take into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{AcH + H2O &amp;amp;&amp;lt;=&amp;gt; Ac- + H3O+}\tag{1}\\&#xA;\ce{Ac- + H2O &amp;amp;&amp;lt;=&amp;gt; AcH + OH- }\tag{2}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (2), I assume that calcium acetate completely dissociates in water. I also assume that the concentration of $\ce{OH- }$ from the auto ionization of water is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I calculate the concentrations of $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ in (1), ignoring what I assume to be minuscule amounts of $\ce{H+}$ from the auto ionization of water. To do this, I use the equation for the acid dissociation constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{a}=\frac{[\ce{Ac- }][\ce{H3O+}]}{[\ce{AcH}]}\tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial values for $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ are set to 1, 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$ M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into (3), I get $x = 4.17\times10^{-3}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have now found (or can at least easily calculate) the concentrations of the compounds in (1) at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to (2), I know that $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$, and I have the values for $K_\mathrm{a}$ and $K_\mathrm{w}$, which gives me the value $5.71\times10^{-10}$ for $K_\mathrm{b}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial concentrations in (2) for $\ce{Ac- }$, $\ce{AcH}$ and $\ce{OH- }$ are 1 (because 0.5 moles of calcium acetate gives $2\times0.5=1$ moles of acetate), 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$. Plugging this into (3) (though using the equation for the base dissociation constant instead), I get $x = 2.39\times10^{-5}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have now found all the values I need to calculate the final concentrations and pH when adding the two solutions to each other, and up till this point I am fairly sure that all my calculations are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do I write the chemical equation for the mixing of the two solutions? I tried the following; $\ce{AcH + Ac- &amp;lt;=&amp;gt; Ac- + AcH},$ but this doesn’t seem to help at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From my younger days, I seem to remember a much simpler and quicker method of doing these calculations than what I've shown here; does such a method exist? Please note that if it does exist I’d like to learn about it, but I would also like answers based on the method I've used.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let’s say you mix one weak acid with one weak base, and they aren't conjugate pairs. How would the method differ from the one in this problem? I am specifically wondering about whether both the dissociation constants must be provided to be able to solve, and also how to write the chemical equation, but other observations are more than welcome as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15709" LastEditorUserId="5026" LastEditDate="2018-02-25T07:51:10.530" LastActivityDate="2018-02-25T07:51:10.530" Title="How to calculate the pH of an acetic acid/ calcium acetate buffer?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29045" PostTypeId="2" ParentId="29043" CreationDate="2015-04-20T12:38:24.970" Score="5" Body="&lt;p&gt;You seem to be running Gaussian through another program or script. Consequently, I have no idea where the various arguments should go. Conventionally, you make a Gaussian input file with keywords, such as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;%n B3LYP/6-311G** EmpiricalDispersion=GD3BJ Opt&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[... molecule goes here ..]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For dispersion-correction (i.e., GD3BJ) you would need an appropriately recent version of Gaussian and use the separate term as indicated. It is not part of the functional designation. (Although some functionals implicitly have dispersion correction, B3LYP does not.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know &lt;em&gt;a priori&lt;/em&gt; that PBEPBE would give more accurate results than B3LYP or any other functional. Your best bet is to either consider results from papers in the field (e.g., molecular crystal prediction or runs with DMACRYS) or testing a few functionals and finding out for yourself. Consider reading &lt;a href=&quot;https://chemistry.stackexchange.com/a/27418/5017&quot;&gt;this question and answers&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Intermolecular interactions are not likely to be handled well by B3LYP. This is the whole reason for range-separated functionals and dispersion corrections. As to a specific functional suggestion, I highly recommend trying a few and looking at &lt;a href=&quot;https://chemistry.stackexchange.com/a/27418/5017&quot;&gt;how to choose a functional&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-04-20T12:38:24.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29047" PostTypeId="2" ParentId="28981" CreationDate="2015-04-20T13:55:43.240" Score="4" Body="&lt;p&gt;It is not necessarily empty space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the flux of &lt;a href=&quot;http://users.phys.psu.edu/~cteq/schools/summer09/talks/morfinCTEQ09.pdf&quot; rel=&quot;nofollow&quot;&gt;neutrinos&lt;/a&gt; on Earth is &lt;a href=&quot;http://theta13.lbl.gov/diablo_canyon/diablo_canyon_05.html&quot; rel=&quot;nofollow&quot;&gt;about $3 \times 10^{12}/cm^2/s$&lt;/a&gt;.  At a given moment, a neutrino may or may not be within the atom or even within the nucleus of the atom. Similarly, other particles may be present within an atom at a given time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an electron may be inside the nucleus in accordance with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermi_contact_interaction&quot; rel=&quot;nofollow&quot;&gt;Fermi Contact Interaction&lt;/a&gt;.  So if you have a hydrogen atom, having only one electron, and the electron is inside the nucleus at a given moment, there will be no &quot;between&quot; for there to be space, empty or not.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-04-20T13:55:43.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29049" PostTypeId="2" ParentId="28340" CreationDate="2015-04-20T14:32:14.767" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;'UV light + water + carbon dioxide + methane + ammonia + a couple of weeks time = amino-acids'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answers by ron and Klaus are very good, but I'd like to add one additional point which I think is very important.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key &quot;reactant&quot; in the above equation is, to my mind &quot;UV light&quot;.  UV light comes from the sun.  The sun is very hot.  The surface of the sun is at ~5800 Kelvins, or about ~5500 &amp;deg;C, and the light it emits is effectively the equilibrium mixture of photons for that temperature (this equilibrium for photons is called black body radiation).  The high temperature for the sun means that the equilibrium mixture of light frequencies it emits has much more UV light than &quot;equilibrium&quot; light at Earth's temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view the formation of complex molecules from simple molecules via UV light as a partial, frustrated attempt for the simple molecules to &lt;em&gt;equilibrate&lt;/em&gt; with the blackbody radiation of the sun, i.e., for the simple molecules to equilibrate at temperatures of 5800 K.  Those high temperatures favor high entropies or &quot;disorders&quot; among the molecules.  You can view a &quot;soup&quot; that contains small amounts of many different amino acids and other molecules (in addition to large amounts of unreacted simple molecules) as having more disorder or entropy than a mix where the simple molecules are present in 100% purity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy is the frying of an egg.  An egg in the refrigerator is ordered and &quot;simple&quot;:.  When you try to equilibrate the egg with blackbody radiation at a higher temperature (i.e. put it on the stove), the simple mix changes considerably and the result is a complex, solid-like gel of egg white and a pasty solid goo of yolk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that your question mainly seems to be about formation of the primordial soup, and not actually formation of the replicators themselves.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-20T14:32:14.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29050" PostTypeId="1" AcceptedAnswerId="29053" CreationDate="2015-04-20T15:06:19.010" Score="11" ViewCount="7079" Body="&lt;p&gt;Today, at my chemistry class, we were learning how to convert Celsius to Kelvin with exact values. I said we add 273.15 K to given Celsius but my colleague said we add 273.16 K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I did check on internet, we use 273.15 K for conversion, but I also found some relation with 273.16 K and I don't understand the concept of that. It's something to do with thermodynamics and triple point of water. &lt;/p&gt;&#xA;" OwnerUserId="15711" LastEditorUserId="46306" LastEditDate="2018-02-20T10:49:02.853" LastActivityDate="2020-02-04T13:30:24.503" Title="What is the significance of 273.16 K?" Tags="&lt;physical-chemistry&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29051" PostTypeId="1" AcceptedAnswerId="29055" CreationDate="2015-04-20T15:08:11.110" Score="5" ViewCount="1299" Body="&lt;p&gt;Can pseudochirality be considered when three or four groups on the carbon atom are constitutionally the same, but configurationally different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when the C atom is bonded to one H, and then three more structurally identical groups with 2 chiral C each - in configurations R,R and S,S and R,S respectively - then by assigning each of the 4 groups a priority value (since they are all different - though 3 have the same structure) we can assign a configuration to this C atom.&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="5026" LastEditDate="2018-03-01T11:33:54.547" LastActivityDate="2020-01-12T23:34:47.173" Title="Pseudochirality with more than two groups on one carbon atom" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29052" PostTypeId="1" CreationDate="2015-04-20T15:28:52.613" Score="3" ViewCount="460" Body="&lt;p&gt;I'm familiar with the sea model of metals. What I'm curious about is: exactly how many electrons are lost by the iron, to be absorbed into the sea of electrons? For a general metal, how many electrons are lost to the sea of electrons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7917" LastActivityDate="2015-04-21T19:27:53.457" Title="In solid Iron, what is the charge of the positive atom?" Tags="&lt;metal&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-04-21T19:51:45.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="29053" PostTypeId="2" ParentId="29050" CreationDate="2015-04-20T15:39:48.607" Score="17" Body="&lt;p&gt;The Celcius scale was originally based on the freezing and boiling points of water, so 0 °C was chosen as the freezing point until 1954. But now, the size of one degree on both Celcius and Kelvin scales is defined as 1/273.16 of the difference between absolute zero and the triple point of &lt;a href=&quot;http://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water&quot; rel=&quot;noreferrer&quot;&gt;VSMOW&lt;/a&gt; (for various reasons relating to the actual measurement), but the 0 point of the Celcius scale was left as the freezing point of water which means that even though both scales are defined at 0 K and 273.16 K (the triple point), the Celcius scale still starts at the freezing point and 0.01 °C was defined as the triple point of VSMOW, which means 0 °C = 273.15 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absolute zero is the coldest possible thermodynamic temperature, 0 K. The triple point is the temperature and pressure at which solid, liquid, and gas phases can exist in a stable equilibrium. If you look at the phase diagram below, it's the point at which the solid, liquid, and gas regions all meet. For water, the temperature of the triple point (273.16 K) is remarkably close to its freezing point at atmospheric pressure (273.15 K). Because 0 °C was originally defined as the freezing point of water, instead of changing the zero point when they redefined the scale based on absolute zero and the triple point, they decided to define the triple point as 0.01 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We add 273.15 to °C to get kelvins because that's the difference between 0 K and 0 °C. We don't add 273.16 because the triple point is not 0 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WDOqM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Phase_diagram_of_water.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2015-04-21T17:23:16.283" LastActivityDate="2015-04-21T17:23:16.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29055" PostTypeId="2" ParentId="29051" CreationDate="2015-04-20T16:06:46.097" Score="10" Body="&lt;p&gt;Ah, now I think I understand what you're asking.  You're saying that we have a central carbon to which is attached a hydrogen and 3 groups with the same structure. Each of these 3 groups contains 2 chiral carbons.  One group is R,R one is R,S and one is S,S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the chirality of the central carbon (it is chiral, it is not meso or prochiral - there is no plane of symmetry bisecting the central carbon) can be assigned using the Cahn–Ingold–Prelog priority rules where &quot;R&quot; has higher priority than &quot;S&quot;.  Therefore the priority ranking around the central carbon in your molecule would be R,R &gt; R,S &gt; S,S &gt; H.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-20T16:06:46.097" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29056" PostTypeId="1" AcceptedAnswerId="29058" CreationDate="2015-04-20T16:09:25.433" Score="7" ViewCount="9302" Body="&lt;p&gt;I have a polymethine molecule with let's say 9 &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H-CH=CH-CH=CH-CH=CH-CH=CH-CH3}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the length of the whole molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need this number to compute the wavelengths the molecule will absorb (model of linear potential well). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are the average bond lengths and bond angles in this molecule and how are they calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Another example for polymethine (just the red part of it):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ib3A.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13004" LastEditorUserId="15751" LastEditorDisplayName="user15489" LastEditDate="2019-01-05T19:53:51.597" LastActivityDate="2019-01-05T19:53:51.597" Title="How to determine bond length and angle in a conjugated pi-system (polymethine)?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;color&gt;&lt;molecular-structure&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29057" PostTypeId="2" ParentId="22233" CreationDate="2015-04-20T16:11:19.030" Score="9" Body="&lt;p&gt;There are various designs and materials for control rods, which are more technically correctly called &lt;em&gt;control assemblies.&lt;/em&gt; The choice depends on the type and purpose of the reactor, the size of the core, the availability of other means for power control (e.g. boric acid in the reactor coolant, burnable poison rods, $\ce{UO2-Gd2O3}$ fuels, recirculation pumps, or thermal-hydraulic feedback), patent rights, and tradition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control assemblies of typical &lt;strong&gt;boiling water reactors (BWRs)&lt;/strong&gt; are inserted from the bottom because, in a BWR, the neutron flux is larger in the lower part of the core and the space above the core is occupied by the water separator and the steam dryer. Therefore, the design should consider saving weight. BWRs usually use cross-shaped control assemblies. The four blades of one cruciform control assembly are inserted between four fuel assemblies. The blades are made of stainless steel and contain tubes that are filled with boron carbide ($\ce{B4C}$) as neutron-absorbing material. In addition, hafnium ($\ce{Hf}$) absorber may be placed in selected locations of selected control assemblies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control assemblies of typical &lt;strong&gt;pressurized water reactors (PWRs)&lt;/strong&gt; are inserted from above. Each control assembly has multiple rod fingers, which enter openings at the upper end of a fuel assembly. These control assemblies are very heavy because, when the reactor is tripped, they have to quickly drop into the core against the pressure of the reactor coolant. The neutron absorbing material mainly is an $\ce{Ag-In-Cd}$ alloy (usually 80 % $\ce{Ag}$, 15 % $\ce{In}$, and 5 % $\ce{Cd}$) encapsulated in an austenitic steel cladding. These three neutron absorbers together effectively block the entire energy spectrum of fast, epithermal, and thermal neutrons. In other PWR core designs, the control rods are divided into two groups: so-called &lt;em&gt;black assemblies&lt;/em&gt; made up of rods consisting of an $\ce{Ag-In-Cd}$ alloy in the lower section and of $\ce{B4C}$ in the upper part, and so-called &lt;em&gt;grey assemblies&lt;/em&gt; containing absorber rods made of $\ce{Ag-In-Cd}$ alloy and steel.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-20T16:11:19.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29058" PostTypeId="2" ParentId="29056" CreationDate="2015-04-20T16:27:22.060" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So what are the average bond lengths and bond angles in this molecule&#xA;  and how are they calculated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;1,3-butadiene should serve as a reasonable model for your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/utKSW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the bond lengths in the literature.  The carbon-carbon double bond length is 1.338 &amp;angst;, typical for a double bond (&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/16759136&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).  The carbon-carbon single bond length is 1.454 &amp;angst;, shorter than expected due to resonance.  All of the carbons are $\ce{sp^2}$ hybridized, which means that all of the bond angles should be ~120°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this information and vector addition you can determine the length of any polymethine.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-20T16:54:40.203" LastActivityDate="2015-04-20T16:54:40.203" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="29059" PostTypeId="1" AcceptedAnswerId="29063" CreationDate="2015-04-20T16:28:21.957" Score="3" ViewCount="119" Body="&lt;p&gt;Whenever I brush teeth, there are some kind of bubbles sticking inside the cavity and that's uncomfortable, it should be SLS as found on the internet. Does anyone know how do I get rid of those bubbles things easily?&#xA;Eat/Drink sth? Tell me if anything is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition: SLS represents sodium lauryl sulfate, chemical exist in our cleaning product: toothpaste, shampoo,   etc. I don't know what it is for but seems it produce bubbles as long as the product is used. &lt;/p&gt;&#xA;" OwnerUserId="15713" LastEditorUserId="7475" LastEditDate="2015-10-30T17:11:36.373" LastActivityDate="2015-10-30T17:11:36.373" Title="What non-poisonous chemicals can get rid of SLS effectively?" Tags="&lt;everyday-chemistry&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29060" PostTypeId="1" AcceptedAnswerId="29076" CreationDate="2015-04-20T16:46:11.147" Score="4" ViewCount="156" Body="&lt;p&gt;I was wondering if there was a known algorithm which could calculate simple reaction products - not anything organic, but things like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaBr + LiF -&amp;gt; NaF + LiBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these are quite easily to be worked out mentally, I cannot think of an easy way in which these reaction products could be predicted. Is there some software which does this (most I have found simply uses databases of known reactions), or even better an algorithm which does it?&lt;/p&gt;&#xA;" OwnerUserId="15714" LastEditorUserId="5017" LastEditDate="2015-04-20T19:06:13.717" LastActivityDate="2015-04-26T15:14:18.833" Title="Calculating reaction products programmatically?" Tags="&lt;inorganic-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29061" PostTypeId="1" AcceptedAnswerId="29062" CreationDate="2015-04-20T16:56:06.580" Score="2" ViewCount="543" Body="&lt;p&gt;&lt;strong&gt;Setup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found in &lt;a href=&quot;http://store.doverpublications.com/0486691861.html&quot; rel=&quot;nofollow&quot;&gt;Szabo, Quantum chemistry&lt;/a&gt; page 88, these two ways to calculate the energy of the ground state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_0= \sum_a^N \langle a|h|a\rangle + \frac{1}{2} \sum_a^N \sum_b^N \langle ab||ab\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_0= \sum_a^N \langle a|h|a\rangle + \sum_a^N \sum_{b&amp;gt;a}^N \langle ab||ab\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-04-20T17:17:43.853" LastActivityDate="2015-04-20T18:54:39.107" Title="Why are 2 ways to calculate the energy of the ground state?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29062" PostTypeId="2" ParentId="29061" CreationDate="2015-04-20T17:04:47.693" Score="9" Body="&lt;p&gt;Both formulas have exactly the &lt;em&gt;same meaning&lt;/em&gt;. They are just two different ways of &lt;em&gt;not double counting&lt;/em&gt; the Coulomb and the exchange interactions. In principle, the goal can be achieved in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;double counting in the first place and then dividing by two (first formula);&lt;/li&gt;&#xA;&lt;li&gt;counting wisely so that there is no double counting in the first place (second formula).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously, for any practical purposes when you one does not want to double count one would almost always choose the second procedure, but the first representation in some cases provides more apparent picture when dealing with the formulas analytically.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where does all this come from?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole derivation of the expression for the ground state Hartree-Fock energy basically starts from writing down the expectation value of the ground state electronic energy for the ground state Slater determinant $\Phi_{0}$&#xA;$$&#xA;\langle \Phi_{0} \mid \hat{H}_{\mathrm{e}} \mid \Phi_{0} \rangle \, ,&#xA;$$&#xA;in terms of the individual spin-orbitals of which $\Phi_{0}$ is constructed.&#xA;And in doing so we use the so-called &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%E2%80%93Condon_rules&quot; rel=&quot;nofollow&quot;&gt;Slater rules&lt;/a&gt;&lt;/em&gt; and it happens so that the electronic Hamiltonian $\hat{H}_{\mathrm{e}}$ contains only two different kinds of operators:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one-electron operators of the form &#xA;$$&#xA;    \hat{F} = \sum\limits_{i=1}^{N}\ \hat{f}(i)\,,&#xA;$$&lt;/li&gt;&#xA;&lt;li&gt;and two-electron operators of the form&#xA;$$&#xA;    \hat{G} = \frac 12 \sum\limits_{i=1}^{N} \sum\limits_{j=1\atop{j\neq i}}^{N}\ \hat{g}(i,j)\,.&#xA;$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Already at this stage an expression for any two-electron operator $\hat{G}$ could as well be written as&#xA;$$&#xA;    \hat{G} = \sum\limits_{i=1}^{N} \sum\limits_{j&amp;gt;i}^{N}\ \hat{g}(i,j)\,.&#xA;$$&#xA;The mathematical meaning is exactly the same: we do not want to double count pairwise interactions, so that if, say $\hat{g}(1,2)$ is already counted, $\hat{g}(2,1)$ has to be excluded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note carefully that we also have to exclude physically meaningless &lt;em&gt;self-interactions&lt;/em&gt;, such as, say, $\hat{g}(1,1)$, which is done in the first case by an additional restriction on summation $j\neq i$ and in the second case by the strict inequality in the lower limit of summation $j&amp;gt;i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the exclusion of this non-physical self-interactions note now that they it is missing from the HF energy expression being written in the first form. But that is perfectly fine since each and every Coulomb self-interaction is perfectly canceled by the corresponding exchange self-interaction. Thus, you do not necessarily need the above mentioned additional restriction of the form $j\neq i$ ($b \neq a$ this time), although you could include it if you would like to,&#xA;$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b=1\atop{b \neq a}}^N \langle ab||ab\rangle \, ,$$&#xA;or adopting a rather usual convention for such summations, as follows,&#xA;$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b \neq a}^N \langle ab||ab\rangle \, .$$&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-04-20T18:54:39.107" LastActivityDate="2015-04-20T18:54:39.107" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29063" PostTypeId="2" ParentId="29059" CreationDate="2015-04-20T17:12:05.043" Score="5" Body="&lt;p&gt;SLS is &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/Sodium_dodecyl_sulfate#section=Chemical-and-Physical-Properties&quot;&gt;highly soluble in water&lt;/a&gt; (100g/L)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in the specific case of your teeth having lots of bubbles around them after brushing, you should simply &lt;a href=&quot;http://en.wikipedia.org/wiki/Gargling&quot;&gt;gargle&lt;/a&gt; your mouth with water to remove the excess SLS. &lt;/p&gt;&#xA;" OwnerUserId="11303" LastActivityDate="2015-04-20T17:12:05.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29064" PostTypeId="2" ParentId="23940" CreationDate="2015-04-20T17:31:46.230" Score="2" Body="&lt;p&gt;It could also be the case (impossible to tell, as you haven't told us what molecules you are using) that the two minima are equivalent by symmetry.&lt;/p&gt;&#xA;" OwnerUserId="15715" LastActivityDate="2015-04-20T17:31:46.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29067" PostTypeId="2" ParentId="29056" CreationDate="2015-04-20T19:40:04.730" Score="3" Body="&lt;p&gt;Think in a chain of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AUMuh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the distance between C1 and C2 (= $a$), the distance between C2 and C3 (= $b$), and the bond angle $\gamma$ are known, the distance between C1 and C3 is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = \sqrt{a^2 + b^2 -2ab\cos\gamma}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(cosine rule)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-20T19:40:04.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29070" PostTypeId="2" ParentId="29029" CreationDate="2015-04-20T20:19:42.733" Score="1" Body="&lt;p&gt;Well since the lowest temperature that can exist is -273.15 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All gases condense to liquid or solids almost all to solids because at this temperature molecular activity ceases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the title of your question seems contradicting with what you have written. Probably you might have mistaken interpreting &quot;before&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also only ideal gas can remain gas at this temperature as they have no force of attraction.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Absolute_zero&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Absolute_zero&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5424" LastActivityDate="2015-04-20T20:19:42.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29071" PostTypeId="1" CreationDate="2015-04-20T20:24:48.023" Score="5" ViewCount="132" Body="&lt;p&gt;In a paper I am reading regarding instability and decay of DNA, the author says: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The chemical price paid for the greatly increased resistance of the nucleic acid phosphodiester bond (gained by removal of the sugar 2'-OH group) is the labile N-glycosyl bond&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand the link between the two chemical bonds. Are they dependent on each other (e.g. one causes the other?)&lt;/p&gt;&#xA;" OwnerUserId="15716" LastEditorUserId="15410" LastEditDate="2015-04-23T08:20:22.110" LastActivityDate="2015-04-23T08:20:22.110" Title="Nucleic Acids Bonds" Tags="&lt;bond&gt;&lt;reactivity&gt;&lt;dna-rna&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29072" PostTypeId="2" ParentId="28542" CreationDate="2015-04-20T20:32:51.610" Score="9" Body="&lt;p&gt;As explained in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed083p752&quot; rel=&quot;nofollow noreferrer&quot;&gt;pH Paradoxes: Demonstrating That It Is Not True&#xA;That pH ≡ -log[H+]&lt;/a&gt; Journal of Chemical Education Vol. 83 pages 752-757.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH = -log (hydrogen ion activity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &quot;pH of 7.6 M HCl is about -1.85 (not -0.88)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how far off is -log (5) = -0.69 from the real answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.kayelaby.npl.co.uk/chemistry/3_9/3_9_6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table of activity coefficients&lt;/a&gt; 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the table is in terms of molality rather than molarity.  5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2018-02-23T02:37:45.570" LastActivityDate="2018-02-23T02:37:45.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29073" PostTypeId="2" ParentId="28542" CreationDate="2015-04-20T21:03:26.727" Score="-4" Body="&lt;p&gt;Just to give a quick and logical answer, if you look at the following acids:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{HCl}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HBr}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HI}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2SO4}$ - &lt;strong&gt;&lt;em&gt;first dissociation only&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HClO4}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HClO3}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HNO3}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These acids above are &lt;strong&gt;Strong Acids&lt;/strong&gt;, meaning that they completely dissociate, so for that purpose there is &lt;strong&gt;&lt;em&gt;no need of a certain K value&lt;/em&gt;&lt;/strong&gt;, since &lt;strong&gt;&lt;em&gt;none of this is an equilibrium reaction&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just a &lt;strong&gt;forward&lt;/strong&gt; reaction. &lt;strong&gt;$\ce{-&amp;gt;}$&lt;/strong&gt; not $\ce{&amp;lt;-&amp;gt;}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;NOTE: This is usually for general purposes. Not to be super accurate.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-21T19:29:41.280" LastActivityDate="2015-04-21T19:29:41.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29074" PostTypeId="2" ParentId="4797" CreationDate="2015-04-20T21:18:19.787" Score="1" Body="&lt;p&gt;To sum it up, we know that $\ce{H2CO3}$ is an acid, so for $\ce{HCO3-}$, just look at the $\ce{pK_{a}}$ and the $\ce{pK_{b}}$ values. Pick the greatest one and do it from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{pK_{a}}$ = ${7.6}$&#xA;$\ce{pK_{b}}$ = ${6.4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we know that the entity gains an $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-20T21:18:19.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29075" PostTypeId="2" ParentId="29003" CreationDate="2015-04-20T21:43:29.313" Score="2" Body="&lt;p&gt;I'll put it in a different way for you, so essentially what we are doing is that one side is &lt;strong&gt;taking&lt;/strong&gt; &lt;strong&gt;electrons&lt;/strong&gt;, so it becomes &lt;strong&gt;more negative&lt;/strong&gt;, and the other side &lt;strong&gt;becomes positive&lt;/strong&gt;. And the salt bridge is to maintain &lt;strong&gt;&lt;em&gt;this&lt;/em&gt;&lt;/strong&gt; &lt;strong&gt;&lt;em&gt;electroneutrality&lt;/em&gt;&lt;/strong&gt; or in other words, to maintain one side from becoming too negative or positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is exactly why we need a salt bridge with the $\ce{Na+}$ and $\ce{SO4^{-2}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So at the end of the day, the positive cations go to the cathode and the negative anions move to the anode.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-05-03T13:42:30.677" LastActivityDate="2015-05-03T13:42:30.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29076" PostTypeId="2" ParentId="29060" CreationDate="2015-04-20T21:52:40.217" Score="2" Body="&lt;p&gt;You are looking at the most simple form of &lt;em&gt;double displacement&lt;/em&gt; reactions of salts with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{M^1X^1 + M^2X^2 -&gt; M^1X^2 + M^2X^1}\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dissolving these salts leads to complete dissociation $\ce{MX -&amp;gt; M+ + X-}$. From the solution, pairs of anions and cations recombine.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using a lookup table with elements, you could identify the cation from each string that represents a starting material (=salt). The remaining string is the anion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Build three lists:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a list of cations &lt;code&gt;cations = ['Na', 'Li']&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list of anions   &lt;code&gt;anions = ['Br', 'F']&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;a list of the starting materials&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With Python, &lt;code&gt;itertools.product(cations, anions)&lt;/code&gt; will give a generator object that yields tuples of all possible cation-anion combinations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;These tuples can be joined to strings for a possible salts&lt;/li&gt;&#xA;&lt;li&gt;Put these strings in a list and subtract the list of the starting materials to obtain the possible products&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import itertools as it&#xA;reactants = ['NaBr', 'LiF']&#xA;# splitting the string via lookup table is something I leave up to you&#xA;cations = ['Na', 'Li']&#xA;anions = ['Br', 'F']&#xA;&#xA;pr = it.product(cations, anions)&#xA;pairs = list(pr)&#xA;# yields list of tuples [('Na', 'F'), ('Na', Br'), ...]&#xA;&#xA;salts = [''.join(pair) for pair in pairs]&#xA;# yields a list of strings by joining the tuples:&#xA;# ['NaBr', 'NaF', 'LiBr', 'LiF']&#xA;&#xA;products = [salt for salt in salts if not salt in reactants]&#xA;# only gives those salts that are not in the list of reactants&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-20T22:01:26.760" LastActivityDate="2015-04-20T22:01:26.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29077" PostTypeId="2" ParentId="29060" CreationDate="2015-04-20T22:14:55.470" Score="2" Body="&lt;p&gt;Well I do not have a software but can offer some info on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a &lt;strong&gt;double replacement&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A liquid, solid, or gas must form.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If it is a &lt;strong&gt;single replacement&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Element replacing other element must be higher on the activity series.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you really want to make sure if the reaction happens, just&#xA;  calculate the $\ce{\Delta G}$ for it. If it is &lt;strong&gt;(-)&lt;/strong&gt; then it is&#xA;  spontaneous and the reaction is thermodynamically favorable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is how I usually remember it. Works for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But otherwise Klaus got you covered :)&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-26T15:14:18.833" LastActivityDate="2015-04-26T15:14:18.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29078" PostTypeId="2" ParentId="5122" CreationDate="2015-04-20T22:21:05.840" Score="1" Body="&lt;p&gt;After you calculate everything and balance the $\ce{e-}$ too, then you will know that one side loses a certain number of $\ce{e-}$ and the other gains this certain number of $\ce{e-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't any other way apart from balancing the equation and figuring it out from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://www.dummies.com/how-to/content/how-to-balance-redox-equations.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.khanacademy.org/test-prep/mcat/physical-processes/redox-reactions/v/balance-and-redox-reactions1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for more info.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-20T22:21:05.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29081" PostTypeId="1" AcceptedAnswerId="29084" CreationDate="2015-04-20T23:28:28.887" Score="3" ViewCount="533" Body="&lt;p&gt;I'm working often with spectral data of various gases, and the bulk of the data analysis is assigning the peaks with maximum signal (y-axis) to frequency (x-axis). Often we have several thousand data points, and it can be quite difficult to manually find peaks. I've tried getting around this in Excel with clever IF/AND statements, but the major obstacle is signal noise: I simply cannot devise a clever Excel statement to help me pick peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, using an =IF(AND(greater signal than adjacent points),&quot;max&quot;,&quot;&quot;) statement gives tons of extraneous hits where the noise causes a slight fluctuation in the data. Similarly, creating a column of slopes and doing an =IF(slope goes from + to -),&quot;max&quot;,&quot;&quot;) statement also gives bad data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most elegant solution I have is to pick an arbitrary value # that most of my peaks are above and use an =IF(AND(slope goes from + to -,B&gt;#),&quot;max&quot;,&quot;&quot;) statement. That seems to work for stronger peaks (maybe 70% of the peaks on a good day), but I still have to go in and manually look at the weaker peaks (and often when doing these spectra many energetic transitions that need labeling are statistically unlikely--&gt;low signal). I'm wondering if anyone here has a better solution able to detect local maxima not resulting from noise with a better success rate.&lt;/p&gt;&#xA;" OwnerUserId="15719" LastActivityDate="2015-04-21T00:55:42.753" Title="Spectral Peak-Picking in MS Excel" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29084" PostTypeId="2" ParentId="29081" CreationDate="2015-04-21T00:55:42.753" Score="4" Body="&lt;p&gt;One thing you can do before searching for peaks is to smooth the data with a 9 point (or 15 point) moving average. I'm not sure how the ends are handled, but the average of 9 points replaces the data point so noise fluctuations are reduced.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-21T00:55:42.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29086" PostTypeId="1" CreationDate="2015-04-21T02:22:37.523" Score="5" ViewCount="1365" Body="&lt;p&gt;I broke a CFL in my home (so elemental mercury is present in a small amount) and wiped up some of the mess with a clorox wipe.&#xA;The wipe in question contains dimethyl benzyl ammonium chloride and dimethyl ethylbenzyl ammonium chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not chemical-phobic, but can someone please tell me, I didn't accidentally expose myself to dimethylmercury? I realize that it is something you don't find outside of very specific lab environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks everyone for the quick responses. It's been a long time since I took a chem class! Can you help me to better understand why this will not methylate the mercury?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I had been cleaning my carpet at the same time with resolve (water, hydrogen peroxide, sodium lauryl sulfate, probably other ingredients too) and threw out paper towels soaked with it to clean the rug into the same bag as the CFL. Am I correct to assume that despite the methyl group on the SDS, it is also stable like the BZK and thus unlikely to methylate the mercury?&lt;/p&gt;&#xA;" OwnerUserId="15722" LastEditorUserId="15725" LastEditDate="2015-04-22T18:29:15.677" LastActivityDate="2015-04-22T18:29:15.677" Title="Can dimethylmercury originate from cleaning up a broken thermometer with disposable wet wipes?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29087" PostTypeId="2" ParentId="29086" CreationDate="2015-04-21T02:25:39.480" Score="0" Body="&lt;p&gt;No, because you don't have a strong oxidizing agent to oxidize the mercury to $\ce{Hg^{2+}}$.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-04-21T02:25:39.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29089" PostTypeId="1" AcceptedAnswerId="29091" CreationDate="2015-04-21T03:22:26.453" Score="5" ViewCount="1728" Body="&lt;p&gt;How would you draw the resonance structure for the compound below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hraqmt.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of the resonance forms I think... but I'm sure there is another one too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mR9R4t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15045" LastEditorUserId="4945" LastEditDate="2015-04-21T04:54:36.510" LastActivityDate="2015-04-21T05:23:36.993" Title="What are the resonance structures of 3-aminocyclohex-2-en-1-ylium?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29090" PostTypeId="2" ParentId="29086" CreationDate="2015-04-21T03:24:13.393" Score="7" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www2.epa.gov/cfl&quot;&gt;US EPA&lt;/a&gt;, disposable wet wipes are a recommended tool for &lt;a href=&quot;http://www2.epa.gov/cfl/cleaning-broken-cfl-0&quot;&gt;cleaning up a broken CFL&lt;/a&gt;. Although having a number of adverse health hazards at high concentration &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzalkonium_chloride&quot;&gt;BZK&lt;/a&gt; is a common ingredient in many sanitizers, wet wipes and disinfecting wipes, where it occurs at very low concentrations; typically much less than 1%. These ingredients are not likely to react adversely with the released mercury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For future reference to cleaning up broken CFLs, be sure to &lt;a href=&quot;http://www2.epa.gov/cfl/cleaning-broken-cfl-0&quot;&gt;read this guide&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2015-04-21T03:24:13.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29091" PostTypeId="2" ParentId="29089" CreationDate="2015-04-21T04:49:52.870" Score="10" Body="&lt;p&gt;The resonance structures and how to construct them for 3-aminocyclohex-2-en-1-ylium are shown below. The second structure in your question is none of them, as it is a different molecule: cyclohexa-1,3-dien-1-amine. For this molecule it is also possible to construct a resonance structure, as also shown below.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xg3vJ.jpg&quot; alt=&quot;resonance structures&quot;&gt;&lt;br&gt;&#xA;However, the two structures can be interconverted via an acid-base reaction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And for the stability of 3-aminocyclohex-2-en-1-ylium, the last structure would be most stable (with the positive charge on the EN nitrogen)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as a most stable &lt;a href=&quot;http://goldbook.iupac.org/R05326.html&quot; rel=&quot;noreferrer&quot;&gt;resonance&lt;/a&gt; structure. Resonance is only a concept that breaks down a highly complicated electronic structure into imaginable contributions of Lewis structures. All of these contributions are present all of the time. However, there are usually a few, that would describe the overall character best. Sometimes there is even only one predominant one.&lt;br&gt;&#xA;The concept of resonance is necessary to describe delocalisation effects, hence the Lewis-like representation below is probably much closer to the actual structure than any of the resonance forms depicted above.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/knNkc.jpg&quot; alt=&quot;delocalised structure&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-04-21T05:23:36.993" LastActivityDate="2015-04-21T05:23:36.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29092" PostTypeId="2" ParentId="28934" CreationDate="2015-04-21T05:50:31.690" Score="9" Body="&lt;p&gt;The short answer is not. You cannot define a constant force for an atom or set of atoms in a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still want to modify force constants, this is the way to do it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gaussian, you can modify through the keyword GEOM(IHarmonic=&lt;em&gt;n&lt;/em&gt;), where &lt;em&gt;n&lt;/em&gt; is part of: $\frac{n}{10^6}$ which indicates the respective force applied to the system. Its units are $\frac{Hartree}{Bohr^2}$. To more info, &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_geom.htm&quot; rel=&quot;noreferrer&quot;&gt;please refer to this link&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If you want to introduce a specific force to an atom or set of atoms in MM, the best way to do it is by parameterizing your own system into your desire environment. The most important part is that you have to have in mind that the right part of the principal ec. of any MM describes the parameters of the specific force field of your desire system or set of systems under specific conditions (i.e. aminoacids in water like &lt;a href=&quot;https://en.wikipedia.org/wiki/OPLS&quot; rel=&quot;noreferrer&quot;&gt;OPLS&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\vec{F}=\frac{\partial V}{\partial x}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Moreover, when you talk about forces in QM you are talking about the forces acting on electrons. Not on atoms.&lt;/li&gt;&#xA;&lt;li&gt;But, if you still want to know a way to know how the forces of electrons (From DFT) and, at the same time, nuclei are, I recommend you Car-Parrinello. &lt;a href=&quot;http://www.fisica.uniud.it/~giannozz/QE-Tutorial/tutorial_cp1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Please, refer to this link&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;If you have a particularly interest on simulating systems in QM with specific forces acting on electrons, the best way to do it (As far as I know) is to develop your own density functional.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I Gaussian input file was requested to figure out how the keyword and input is used: Geom(iharmonic=&lt;em&gt;n&lt;/em&gt;). I did not find the respective file, instead, I give a try twice by crating two input files with different values of n. The used Gaussian input files were:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;chk=Test.chk&#xA;#p opt b3lyp/6-31g geom(iharmonic=n)&#xA;&#xA;Test&#xA;&#xA;0 1&#xA;&#xA;H   &#xA;H 1 0.9 &#xA;O 2 1.2 1 120.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the used values of &lt;strong&gt;n&lt;/strong&gt; were 2 (On the left) and 20 (On the right). The respective Energy and RMS profile of both simulations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6LAaU.jpg&quot; alt=&quot;Energy and RMS profile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in Gamess, you can use these keywords:  &lt;strong&gt;Statpt&lt;/strong&gt;, &lt;strong&gt;Force&lt;/strong&gt; and its related keywords like &lt;strong&gt;Zmt&lt;/strong&gt;. More info, &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf&quot; rel=&quot;noreferrer&quot;&gt;Please see this link&lt;/a&gt; at page 2-121 and 2-131.&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-04-21T23:32:06.393" LastActivityDate="2015-04-21T23:32:06.393" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29094" PostTypeId="1" CreationDate="2015-04-21T06:16:49.147" Score="2" ViewCount="7542" Body="&lt;p&gt;I am familiar with the idea that rate law must be determined experimentally. I am also familiar with the idea that there is no formulated way for coming up with the rate law besides just experimentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why isn't the rate law of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ aA + bB \rightarrow \mbox{something}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;just:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = k[A]^a[B]^b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, the equilibrium constant expression involves something to that effect, and the equilibrium constant expression should be precisely talking about when the rates for each direction are equal. So if the rate law is not the above, then why does the equilibrium have that expression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: A bit off topic, but how is it possible for something to be zeroth order?&lt;/p&gt;&#xA;" OwnerUserId="7917" LastEditorUserId="5017" LastEditDate="2015-04-21T13:57:31.250" LastActivityDate="2015-04-21T14:28:09.697" Title="Why must the rate law be experimentally determined?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29095" PostTypeId="2" ParentId="24285" CreationDate="2015-04-21T06:52:21.707" Score="2" Body="&lt;p&gt;Nano Zinc Oxide ($\ce{ZnO}$) is a nanomaterial and a nanomaterial is described as&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;materials of which a single unit&#xA;  is sized (in at least one dimension) between 1 and 1000 nanometers&#xA;  ($10^{-9} m$) but is usually between 1—100 nm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nano zinc oxide  can be synthesized using different approaches that included's both chemical (precipitation, reduction etc) and physical (depositon) methods . You can read more about nano zinc oxide and its application in this paper &lt;a href=&quot;http://iopscience.iop.org/0953-8984/16/25/R01/pdf/0953-8984_16_25_R01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of Physics: Condensed Matter Volume 16 Number 25&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JM0Kc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an image of Nano Zinc oxide grown as a flower (nano-flower) for &lt;a href=&quot;http://www.newscientist.com/article/dn10001-nanoflowers-show-promise-for-alcohol-detection.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sensor applications&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorDisplayName="user15489" LastEditDate="2015-04-21T07:12:34.210" LastActivityDate="2015-04-21T07:12:34.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29098" PostTypeId="1" CreationDate="2015-04-21T07:06:00.860" Score="5" ViewCount="578" Body="&lt;p&gt;I want to know if there is any free open source tool available for Gasteiger-Hückel charge calculation? I know it is available on Sylbyl but is there any other(free) way to calculate these charges for small molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was actually trying this paper, if any one is interested, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/23244516&quot; rel=&quot;nofollow&quot;&gt;How to improve docking accuracy of AutoDock4.2: a case study using different electrostatic potentials.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hou X1, Du J, Zhang J, Du L, Fang H, Li M.X et al.&lt;/p&gt;&#xA;" OwnerUserId="7120" LastEditorUserId="7120" LastEditDate="2015-04-21T10:12:44.500" LastActivityDate="2015-04-21T14:22:51.803" Title="Any free tool for Calculating Gasteiger-Hückel charges for ligands?" Tags="&lt;software&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29099" PostTypeId="2" ParentId="16367" CreationDate="2015-04-21T07:23:30.413" Score="5" Body="&lt;p&gt;I would like to add another shade to the already quite excellent answer of ron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use donor-acceptor complexes, like Lewis acid base adducts, as case in point. Probably one of the simplest and most popular is the ammonia and borane adduct. I consider the structures below to be the most accepted ones, with probably the widest use.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/bSAB7.jpg&quot; alt=&quot;resonance structures of the borane ammonia adduct&quot;&gt;&lt;br&gt;&#xA;These are all structures (not all of the are Lewis approved) that try to cover the concept of a rather difficult bonding situation. In this special case, how to describe a dative bond in an easy model. This also gives us a lot to think about.&lt;br&gt;&#xA;The first structure completely lacks the feature of a bond, therefore it cannot possibly be used to describe this molecule alone. But since we are talking about donor acceptor interactions, a comparatively weaker bond, the separated structure also has to be considered to contribute to bonding.&lt;br&gt;&#xA;The second structure is not Lewis approved (there are not arrows in the traditional representation). It is much like the first structure, but the bonding here is at least indicated. In my opinion this is the only structure that comes somewhat close to the actual bonding situation. However, it treats the donor acceptor bond between nitrogen and boron as a second class covalent bond - is that really true?&lt;br&gt;&#xA;The third structure is often used to indicate, that there is a real covalent bond between boron and nitrogen. To be perfectly honest, what is the difference between polar covalent and dative covalent anyway; it's shared electrons in both cases (see above). In my opinion this structure is one of the best examples why resonance structures are all too often misleading. Since we are in the Lewis picture, we have to obey the rule of two (octet rule), that means we have to assign formal charges. This is suddenly somewhat suggesting (to the untrained mind), that boron will carry a negative charge, which is of course wrong. Nitrogen still is more electronegative and will obviously have the highest negative charge, the electron sharing will hardly change that.&lt;br&gt;&#xA;The last structure is a truly awful hybrid, that I came across. It's neither fish nor flesh, nor good red herring. Don't use it, I just included is for the sake of completeness. It makes all the errors of the other structures combined.&lt;br&gt;&#xA;So all of the structures have their flaws and have to be seen in conjunction. Does any of these structures become invalid through the flaws it represents? No. If anyone argues this, then the concept of resonance - as an approximation to the true bonding situation - is not understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the validity of ionic resonance structures basically becomes a question of semantics, within the framework of the model, they work very well. Do they represent something real, then the obvious answer is no. Model systems often help us understand complicated points, but they must not be mistaken for the (sometimes unknown) truth and they should also not be confused with more elaborate models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I previously stated in the comments, that in the framework of Valence Bond Theory, for each resonance structure it is possible to construct a wave function (= configuration). It is then also possible to obtain the contributions of each configuration to the total wave function. This at least offers a quantum mechanical backbone to the whole model of resonance, and it often is described as the quantitative theory (within it's limitations) of it. The biggest problem we have with this nice theory is, that it is very expensive to calculate.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-04-21T13:18:40.607" LastActivityDate="2015-04-21T13:18:40.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29100" PostTypeId="1" AcceptedAnswerId="29104" CreationDate="2015-04-21T07:35:33.117" Score="2" ViewCount="137" Body="&lt;p&gt;I will remove the pectin from hemp fibres with Bioprep 3000L pectinase. How can I measure the % of pectin before and after for loose fibres? I only found a way to measure it for fabric, but I will perform this test on just loose hemp fibres. &lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="418" LastEditDate="2015-04-21T11:51:18.350" LastActivityDate="2015-04-21T11:51:18.350" Title="Measuring the % of pectin in a fibre after enzymatic treatment with pectinase" Tags="&lt;spectrophotometry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29101" PostTypeId="1" AcceptedAnswerId="29256" CreationDate="2015-04-21T07:40:00.613" Score="14" ViewCount="27175" Body="&lt;p&gt;$\ce{SO3}$ molecule has three double bonded oxygen to the central sulfur atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur has $\ce{sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen.&lt;br&gt;&#xA;So shouldn't the bond order be 2?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I0QVj.png&quot; width=&quot;150&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="15549" LastEditorUserId="15549" LastEditDate="2015-07-11T19:32:33.943" LastActivityDate="2015-07-12T22:23:11.023" Title="Why is the bond order in the SO₃ molecule 1.33 and not 2?" Tags="&lt;bond&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29102" PostTypeId="2" ParentId="29094" CreationDate="2015-04-21T07:49:56.167" Score="1" Body="&lt;p&gt;The order of reaction with respect to each of the reactants, sometimes called to partial order is not equal to the stoichiometric coefficients of the reaction. They are instead dependent on how the reaction proceeds, this is why it must be determined experimentally. So basically it depends on the reaction mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant has nothing to do with time as it is measured once the equilibrium has been reached using the concentrations of the reactants and products at that time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your rate could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rate=k[A]^x[B]^y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $x$ and $y$ are the orders with respect to A and B respectively rather than the stoichiometric coefficients.&lt;/p&gt;&#xA;" OwnerUserId="15191" LastEditorUserId="15191" LastEditDate="2015-04-21T07:55:50.123" LastActivityDate="2015-04-21T07:55:50.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29104" PostTypeId="2" ParentId="29100" CreationDate="2015-04-21T09:31:39.627" Score="1" Body="&lt;p&gt;&lt;em&gt;Note that I do not have professional experience in the textile industry!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposed that the stems have been (mechanically) treated, and the &quot;bark&quot; is removed, the &lt;em&gt;ruthenium red method&lt;/em&gt; should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Crystal-Structure of Ruthenium Red and Stereochemistry of its Pectic Stain&lt;/em&gt; by &lt;em&gt;Clarence Sterling&lt;/em&gt;, &lt;em&gt;Am. J. Bot.&lt;/em&gt; &lt;strong&gt;1970&lt;/strong&gt;, &lt;em&gt;57&lt;/em&gt;, 172-175, the specific binding of Ruthenium Red (&lt;strong&gt;RR&lt;/strong&gt;, CAS-RN: 11103-72-3) along a pectin chain, that is, between a $\ce{COOH}$ of one galacturonic acid unit and the $\ce{OH}$ of the next, has been described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more recent publication, &lt;em&gt;Qualitative distinction of carboxyl group distributions in pectins with ruthenium red&lt;/em&gt;, published in &lt;em&gt;Bot. Bull. Acad. Sin.&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;40&lt;/em&gt;, 115-119 (&lt;a href=&quot;http://ejournal.sinica.edu.tw/bbas/content/1999/2/bot402-02.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;), the method was applied to orange peel pectines of different origin. Here, the dye-pectin adducts were precipitated by addition of $\ce{CaCl2}$ and centrifugated. Then, the absorbtion of the supernatant solution and a stock solution of the dye at $\lambda = 534\,\mathrm{nm}$ were compared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, it should be possible to take fibre samples, suspend them and detemine whether they decrease the absorption of &lt;strong&gt;RR&lt;/strong&gt; (= still contain pectin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you planning to bleach these fibres in a way recently discussed in another question? It might be interesting to figure out whether these conditions (alkaline hydroperoxide) are sufficent to (partly) degrade the pectin.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-21T09:41:09.490" LastActivityDate="2015-04-21T09:41:09.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29105" PostTypeId="1" CreationDate="2015-04-21T06:14:11.063" Score="2" ViewCount="1471" Body="&lt;p&gt;I was wondering about this. What is the physics underlying the &quot;stringy&quot; consistency of melted cheese? That is, when the melted cheese is hot, and it is pulled apart, it forms strings of cheese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behavior also occurs with a number of other fluid substances. It seems to happen with polymer substances (like cheese which is made of proteins like casein, also plastics when melted), gums, etc. What is the technical term for this kind of behavior and what is the general mechanism by which it happens?&lt;/p&gt;&#xA;" OwnerUserId="53075" OwnerDisplayName="mike4ty4" LastActivityDate="2015-04-21T10:08:09.160" Title="What causes &quot;stringy&quot; melted cheese (and similar behavior in other materials)?" Tags="&lt;physical-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29106" PostTypeId="2" ParentId="29105" CreationDate="2015-04-21T06:50:25.090" Score="4" Body="&lt;p&gt;When cheese is heated to 140 degrees fahrenheit, the proteins line up in rows.  Soaking the cheese in a salt solution before pulling it helps to stiffen it and keep it pliable, as the salt neutralizes acids which could cause the long lines of protein to break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polymers, similar to cheese, can be stringy because their molecules are very large (macromolecules), and tend to entangle with one another.  @IanF1 is correct in saying this may be a chemistry question.  Here is a good explanation of polymerization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://courses.chem.psu.edu/chem112/materials/polymers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://courses.chem.psu.edu/chem112/materials/polymers.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Ernie" LastActivityDate="2015-04-21T06:50:25.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29107" PostTypeId="2" ParentId="29101" CreationDate="2015-04-21T10:10:10.570" Score="3" Body="&lt;p&gt;Great question, I had to do a little reading to piece together an answer as I haven't worked with sulfur oxides in quite some time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wikipedia: Sulfur Trioxide&lt;/li&gt;&#xA;&lt;li&gt;Shriver &amp;amp; Atkins, &lt;em&gt;Inorganic Chemistry&lt;/em&gt;, 4th ed., p388&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20120918160908AAKY8lZ&quot; rel=&quot;nofollow&quot;&gt;Some Yahoo Answer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Greenwood &amp;amp; Earnshaw, &lt;em&gt;Chemistry of the Elements&lt;/em&gt;, 2nd ed., p703&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a simplistic explanation, the above sources state that the lewis structure of $\ce{SO_3}$ contains a 2$^+$ charge on the central sulfur and negative charges on two of the three bonded oxygen atoms. In that case, $\ce{SO_3}$ contains one double bond and two single bonds, which is why people tend to list the overall bond-order as 1.33. The actual bonding structure of $\ce{SO_3}$ is a little more complicated than that, as J. LS points out, so you might need to brush up on molecular-orbital theory to get into the nitty-gritty of its bonding structure.&lt;/p&gt;&#xA;" OwnerUserId="2338" LastEditorUserId="2338" LastEditDate="2015-04-21T10:43:50.600" LastActivityDate="2015-04-21T10:43:50.600" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="29108" PostTypeId="1" CreationDate="2015-04-21T10:24:19.557" Score="3" ViewCount="1282" Body="&lt;p&gt;Why does a more electronegative heteroatom (i.e. $\ce{Cl}$ or $\ce{O}$) increase the stretching frequency of a carbonyl? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suspect it's that they donate their lone pairs and we end up with a negative charge on the oxygen in one of our resonance structures which makes the electrostatic attraction between the carbon and oxygen larger? &lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="10185" LastEditorDisplayName="user15489" LastEditDate="2016-05-25T08:12:24.807" LastActivityDate="2020-06-30T13:48:10.643" Title="Why do electronegative heteroatoms increase C=O stretching frequency (bond strength)" Tags="&lt;spectroscopy&gt;&lt;carbonyl-compounds&gt;&lt;ir-spectroscopy&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29109" PostTypeId="1" AcceptedAnswerId="29110" CreationDate="2015-04-21T10:52:11.347" Score="5" ViewCount="147" Body="&lt;p&gt;I have a substituent called TAPD on a molecule I'm working with. I have a hard time locating information about its properties. The bold square in the picture below is supposed to be the rest of the system on which the TAPD is attached. I am a physicist and thus not very good at chemical naming, so I would have good use of some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gog08.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13341" LastActivityDate="2015-04-21T11:02:20.937" Title="What is the the full name of TAPD? Structure provided" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29110" PostTypeId="2" ParentId="29109" CreationDate="2015-04-21T11:02:20.937" Score="5" Body="&lt;p&gt;&lt;strong&gt;TAPD = Tetraalkyl-​p-​phenylenediamine&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because you have a phenyl (benzenoid) ring, with two amino groups (diamine) opposite (&lt;strong&gt;p&lt;/strong&gt;ara) to each other, and each of those amino groups has two alkyl substituents, giving four (tetra) in total. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-04-21T11:02:20.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29111" PostTypeId="2" ParentId="29098" CreationDate="2015-04-21T14:22:51.803" Score="2" Body="&lt;p&gt;The method is not &lt;a href=&quot;http://www.ccl.net/chemistry/resources/messages/2002/01/16.005-dir/&quot; rel=&quot;nofollow&quot;&gt;fully published&lt;/a&gt; according to Gasteiger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;http://www2.chemie.uni-erlangen.de/software/petra/&quot; rel=&quot;nofollow&quot;&gt;online server&lt;/a&gt; for PETRA run by the Gasteiger group, and in principal, it's available from &lt;a href=&quot;https://www.molecular-networks.com/products&quot; rel=&quot;nofollow&quot;&gt;Molecular Networks&lt;/a&gt;. I note that PETRA is no longer listed on the products page, but it's probably part of some other product now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The open source &lt;a href=&quot;http://cdk.github.io/cdk/1.5/docs/api/org/openscience/cdk/charges/GasteigerPEPEPartialCharges.html&quot; rel=&quot;nofollow&quot;&gt;CDK&lt;/a&gt; package has an implementation of the PEPE method, although I don't know if it has been validated against Gasteiger's code.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-04-21T14:22:51.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29112" PostTypeId="2" ParentId="29094" CreationDate="2015-04-21T14:28:09.697" Score="3" Body="&lt;p&gt;Since rate of a reaction depends actually upon the way reactants react, there is not surety that the quoted stoichiometric coefficients may be the same for rate of a reaction, the rate depends on the way or mechanism of the reaction, and since all reactions have a bit different mechanism no strict formula could be quoted. It is only experimentally derived and can be theoretically approached if we know that it is an elementary (simple step) reaction or it falls under a already classified or studied reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero Order, what if a reactant is in very large excess, now the rate is not limited or dependent on its concentration, also it's concentration is almost constant, so it is also merged into the constant and hence we have zero order, some adsorption reactions are maybe zero order too, it is because it actually depends on the surface area of the substance and it's constant. &lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-21T14:28:09.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29113" PostTypeId="2" ParentId="26370" CreationDate="2015-04-21T14:32:36.477" Score="0" Body="&lt;p&gt;Yes in calcination, with limited supply carbonates are reduced to oxides and to remove organic matter or to remove water of crystallization or evaporation of volatile components such as arsenic or antimony vapors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the oxygen component is more it would actually be then a roasting process where sulfides are converted to oxides since they are easier to reduce.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-21T14:32:36.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29114" PostTypeId="2" ParentId="28934" CreationDate="2015-04-21T14:34:28.853" Score="7" Body="&lt;p&gt;If you are wanting the capability to apply forces to nuclei, to alter the equilibrium geometry, I do not know of any software with &quot;a simple user interface like Avogadro&quot; that offers built-in functionality to do what you seek. You are probably facing either a programming/scripting project or a great deal of manual iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation would be that you investigate &lt;a href=&quot;https://orcaforum.cec.mpg.de/&quot; rel=&quot;nofollow&quot; title=&quot;ORCA&quot;&gt;ORCA&lt;/a&gt;. ORCA incorporates the ability to perform geometry optimizations on systems where energy and gradient information is computed externally (&lt;code&gt;ExtOpt&lt;/code&gt; mode; see Section 6.13.5 of the ORCA v3.0.3 Manual, downloadable from the forum linked above).  I am actually planning to develop something rather like this, though with functionality not nearly as general as what you describe [arbitrary $F(x,y,z)$].  I plan to write a Python application that calls ORCA in this &lt;code&gt;ExtOpt&lt;/code&gt; mode, and then within each &lt;code&gt;ExtOpt&lt;/code&gt; cycle:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Makes a normal call to ORCA to compute the single-point energy and gradient in the absence of the external force field&lt;/li&gt;&#xA;&lt;li&gt;Calculates the contributions of the external force field to the energy and gradient&lt;/li&gt;&#xA;&lt;li&gt;Merges the QM energy and gradient computed by ORCA with the force field energy and gradient&lt;/li&gt;&#xA;&lt;li&gt;Passes the composite energy/gradient back to &lt;code&gt;ExtOpt&lt;/code&gt; ORCA&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;ORCA supports semi-empirical methods as well as full quantum methods, so even large systems should be tractable in a reasonable time frame, once you have built the execution framework to carry it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For your application, where you're trying to simulate increased pressure, the 'constrained volume' approach would probably work pretty well using a piecewise potential function, where $V = 0$ within the molecular volume calculated for a particular pressure, and then ramps to a very high value at the borders of the box.  The task then becomes appropriate definition of the box boundary, a problem which I suspect is similar to that involved in defining the cavity in implicit solvation models (&lt;a href=&quot;http://en.wikipedia.org/wiki/COSMO_Solvation_Model&quot; rel=&quot;nofollow&quot;&gt;COSMO&lt;/a&gt;$^1$, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^1$ Klamt and Schüürmann. &lt;em&gt;Perkins Transactions 2&lt;/em&gt; &lt;strong&gt;2&lt;/strong&gt;: 799-805 (1993). doi: &lt;a href=&quot;http://dx.doi.org/10.1039/P29930000799&quot; rel=&quot;nofollow&quot;&gt;10.1039/P29930000799&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-04-24T15:39:57.517" LastActivityDate="2015-04-24T15:39:57.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29115" PostTypeId="2" ParentId="27742" CreationDate="2015-04-21T14:36:09.053" Score="-1" Body="&lt;p&gt;Since the already present sodium chloride highly ionic, it's dissociation can be considered 100% and thus the contribution of chloride form silver chloride would be thus negligible (in comparison to rock salt's) or suppressed. Thus we can treat the final chloride concentration as $0.55\rm M$. Now the solubility product is given as:&#xA;$$K_{\rm sp,AgCl}=\ce{[Ag+][Cl- ]}$$&#xA;Now we can find out $\ce {[Ag+]=\frac{K_{\rm sp,AgCl}}{[Cl- ]}}$. This would actually be all from silver chloride and hence the solubility of it.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-21T14:36:09.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29116" PostTypeId="2" ParentId="29044" CreationDate="2015-04-21T14:44:43.720" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem: A buffer was made by mixing 500 mL of 1 M (mol/L) acetic acid and 500 mL of 0.5 M calcium acetate.&#xA;  What are the resulting concentrations of acetic acid, Ca2+ and acetate  and what is the resulting pH? Ka(acetic acid) = 1.75×10−5&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Final volume = $500+500$ mL = 1L&lt;br&gt;&#xA;Concentration of Acetic Acid = $\frac{\rm moles}{\rm volume}=\frac{0.5*1}{1}=0.5$ M&lt;br&gt;&#xA;Concentration of Calcium Acetate = $\frac{\rm moles}{\rm volume}=\frac{0.5*0.5}{1}=0.25$ M&lt;br&gt;&#xA;So using Henderson equation:&#xA;$$pH=pKa+\log\frac{0.25*2}{0.5}=4.76$$&#xA;Rest is easy.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-21T14:44:43.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29117" PostTypeId="1" AcceptedAnswerId="29118" CreationDate="2015-04-21T15:03:28.843" Score="3" ViewCount="5334" Body="&lt;p&gt;According to Henry's law &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The solubility of gas in a liquid is directly&#xA;  proportional to the partial pressure i.e S=Kp&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid. &#xA;I don't understand how. I know I am overseeing something really simple over here as all the books that I have and online websites seem to conclude the same from the given equation. A little help maybe. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="8218" LastActivityDate="2018-09-17T03:52:15.933" Title="Henry's constant and solubility" Tags="&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29118" PostTypeId="2" ParentId="29117" CreationDate="2015-04-21T15:17:11.690" Score="1" Body="&lt;p&gt;Actually solubility $x$ is given as a function of pressure $p$ and henry's constant $K_H$:&#xA;$$p=K_Hx$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid. I don't understand how.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since at a given pressure, $p$ is constant, if you increase $K_H$, logically solubility $x$ would decrease.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-21T15:17:11.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29119" PostTypeId="1" AcceptedAnswerId="29159" CreationDate="2015-04-21T15:53:03.067" Score="6" ViewCount="2266" Body="&lt;p&gt;Can a liquid which is saturated with a single solute, dissolve a different solute, or is saturation a universal thing? I ask because I’ve seen that different solutes have different points (amounts) at which they become saturated in water, for example, which caused me to think that maybe each solution was perhaps independent, even though that would produce some interesting extreme implications. Something like, take a water &amp;amp; sucrose mix that is saturated; can this mixture then dissolve salt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific implication I am asking about is this: If I dissolve xylitol (a sugar alcohol) in water until it is saturated (200 g xylitol to 100 g water at 25 °C / 77 °F) can I then dissolve erythritol (a different sugar alcohol) into the solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, we could heat the liquid and enter super-saturation, but that’s not what I’m asking. I’m asking if this can be done without super-saturating (i.e. risking precipitation/crystallization) once cooled.&lt;/p&gt;&#xA;" OwnerUserId="15734" LastEditorDisplayName="user7951" LastEditDate="2015-06-12T19:26:57.220" LastActivityDate="2015-06-12T19:26:57.220" Title="Saturation of multiple solutes" Tags="&lt;aqueous-solution&gt;&lt;alcohols&gt;&lt;precipitation&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29120" PostTypeId="2" ParentId="28692" CreationDate="2015-04-21T16:11:16.353" Score="19" Body="&lt;p&gt;&lt;strong&gt;General Rule #1:&lt;/strong&gt; Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, &lt;em&gt;etc&lt;/em&gt;. orbitals in bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Rule #2:&lt;/strong&gt; The more s-character in a bond the shorter the bond (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bent&amp;#39;s_rule&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).  For example &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a $\ce{C(sp^3)-C(sp^3)}$ single bond length is ~ 1.54 $\mathrm{\mathring{A}}$&lt;/li&gt;&#xA;&lt;li&gt;a $\ce{C(sp^2)-C(sp^3)}$ single bond length is ~ 1.50 $\mathrm{\mathring{A}}$&lt;/li&gt;&#xA;&lt;li&gt;a $\ce{C(sp)-C(sp^3)}$ single bond length is ~ 1.46 $\mathrm{\mathring{A}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In an &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;earlier SE Chem post&lt;/a&gt;, the structure and bonding in the &lt;strong&gt;trigonal bipyramid&lt;/strong&gt; molecule $\ce{PCl_5}$ was discussed.  The molecular hybridization is pictured below.  The axial bonds are &lt;strong&gt;hypervalent&lt;/strong&gt; (this concept is explained in the earlier post) and they are constructed from p orbitals.  The equatorial bonds are constructed from $\ce{sp^2}$ orbitals.  General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlPkL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the following structural diagram shows, this is indeed the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UtR02.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's turn our attention to the &lt;strong&gt;pentagonal bipyramid&lt;/strong&gt; structure as found in $\ce{IF_7}$ for example.  Here is the structure for this molecule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WzQJI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted in your question, the equatorial bonds are now longer than the axial bonds.  Based on what we've discussed thus far, we might make an initial &quot;guess&quot; that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds.  As it turns out, this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00057a044&quot; rel=&quot;noreferrer&quot;&gt;link to an article&lt;/a&gt; discussing the structure and bonding in $\ce{IF_7}$.  Only the abstract is freely available.  For completeness, I'll mention that there was an &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6785/how-can-the-bonding-in-if7-be-explained-using-lcao-method&quot;&gt;earlier SE Chem question&lt;/a&gt; related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation.  Returning to the published abstract, the key part of the abstract states,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;These features can be explained, however, by a bonding scheme&#xA;  involving a planar, delocalized pxy hybrid on the central atom for the&#xA;  formation of five equatorial, semi-ionic, 6-center 10-electron bonds&#xA;  and an sp hybrid for the formation of two mainly covalent axial&#xA;  bonds.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_{z}}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one $\ce{p_{z}}$ orbital and one s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result).  &lt;strong&gt;That is the answer to the question&lt;/strong&gt;.  As we guessed, there is more s-character in the axial bonds than the equatorial bonds.  Hence, the axial bonds are now shorter than the equatorial bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For completeness&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives?  First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypervalent bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment.  But note that the abstract also mentioned that the equatorial bonds are semi-ionic.  This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor.  We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds.  This means that mixed in with our hypervalent p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds.  Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypervalent p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-04-21T22:08:36.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29122" PostTypeId="1" CreationDate="2015-04-21T17:51:47.040" Score="6" ViewCount="1723" Body="&lt;p&gt;I have measured the element composition of a sample using ICP-OES, and reported my results in wt%. A study which has analyzed the same sample types, but with an electron microprobe, has reported their element concentrations in at%, which in addition are normalized to 100 %. I would like to compare my results to those in the article, but I am not sure how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am able to convert the atomic percentages to weight percentages, but I am not sure how to tackle the normalization of the data. When I add the wt% of my sample, I only get to about 50%. This loss is attributed to that I have not analyzed oxygen and carbon, which I suspect would make up a significant portion of the sample weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the study I want to cite has indeed analyzed both oxygen and carbon, and all the other elements I have measured. I do not suspect that they have a very large portion of at% unaccounted for. Therefore I believe their un-normalized data is quite similar to their normalized data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how should I proceed with comparing my data to theirs? Should I just argue, like I have done here, that I think their data is &quot;comparable enough&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any insight on the subject!&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorUserId="4945" LastEditDate="2015-04-24T07:34:32.080" LastActivityDate="2015-05-08T18:35:34.243" Title="Comparing wt% to at% which is normalized to 100%" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29123" PostTypeId="2" ParentId="28847" CreationDate="2015-04-21T18:05:51.990" Score="10" Body="&lt;p&gt;On this matter, the &lt;em&gt;Green Book&lt;/em&gt; (‘Quantities, Units and Symbols in Physical Chemistry’) refers to the &lt;em&gt;Red Book&lt;/em&gt; (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’), which clearly states that plus and minus signs are used to indicate the charge on an ion in a formula or name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this statement does not clarify which particular character shall be used as minus sign. Nevertheless, the &lt;em&gt;Red Book&lt;/em&gt; also explicitly mentions hyphens and ‘em’ dashes for other purposes but no ‘en’ dashes. Hence, I suppose, the &lt;em&gt;Red Book&lt;/em&gt; would have mentioned ‘en’ dashes if this character shall be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, signs and symbols for quantities and units etc are standardized in the international standard series ISO/IEC 80000, which has been adopted by many national standards. This standard is intended mainly for use in the natural sciences and technology. ISO 80000 Part 2 (‘Mathematical signs and symbols to be used in the natural sciences and technology’) clarifies precisely that the symbol that shall be used as minus sign really is the MINUS SIGN of ISO/IEC 10646 and Unicode (they are identical). The hexadecimal code assigned to the minus sign by both ISO/IEC 10646 and Unicode is &lt;strong&gt;2212&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-21T18:05:51.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29124" PostTypeId="1" CreationDate="2015-04-21T18:06:52.860" Score="3" ViewCount="44" Body="&lt;p&gt;From basic coordination chemistry and acid-base equilibria we understand that a metal hexa-aqua ion can be deprotonated until it's positive charge is balanced by the negative charge of the hydroxides rendering a neutral complex. We then say that this complex precipitates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently in my inorganic class we studied hydrolytic polymerisation where complexes with hydroxyl groups condense into $\ce{OH}$ bridged complexes, which can form polynuclear structures. These structures then release the hydroxide when the solubility limits are reached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are these two ideas the same, and is hydrolytic polymerisation the mechanism by which the precipitate forms from the neutral complex? Another way of phrasing it would be, are these complex precipitates extended structures or are they bound via intermolecular forces? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="7448" LastEditDate="2015-04-21T18:31:39.747" LastActivityDate="2015-04-21T18:31:39.747" Title="Is hydrolytic polymerisation the mechanism by which the precipitate forms from the neutral complex?" Tags="&lt;coordination-compounds&gt;&lt;solid-state-chemistry&gt;&lt;transition-metals&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29125" PostTypeId="1" AcceptedAnswerId="65026" CreationDate="2015-04-21T18:15:04.787" Score="14" ViewCount="3214" Body="&lt;p&gt;With highly oxidized metal centers, we expect that the hydroxido ligand's proton becomes acidic and will be lost to the surrounding media, creating an oxido ligand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxo ligand is a $\pi$ basic ligand and we expect both $\sigma $ and $\pi$ donation into the metal center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal's $\mathrm{t_{2g}}$ orbitals are lowered in energy by the presence of the oxido ligand's filled p orbital donating into the $\mathrm{t_{2g}}$. My question is regarding the symmetry of the ligand group orbitals. A p orbital does not have $\mathrm{t_{2g}}$ symmetry and as we know non-symmetry interactions are forbidden. Therefore I am struggling to understand the details of the oxido ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My only explanation is that the complex's symmetry is reduced with the presence of an oxo ligand maybe from $O_\mathrm{h}$ to $C_4$ or similar. This would then change the symmetry of the $\mathrm{t_{2g}}$ orbitals on the metal center, but I really need some help explaining this!&lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="7475" LastEditDate="2016-12-23T13:50:04.233" LastActivityDate="2017-09-22T12:48:09.570" Title="Molecular orbital diagram of a complex including an oxido ligand" Tags="&lt;molecular-orbital-theory&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29126" PostTypeId="1" CreationDate="2015-04-21T18:21:11.517" Score="20" ViewCount="6630" Body="&lt;p&gt;When studying ligand substitution (at UK year 13 level), the following example has been given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{[Cu(H2O)_6]^2+ + 2NH3 &amp;amp;&amp;lt;=&amp;gt; [Cu(OH)_2(H2O)_6] + 2NH4^+}\\&#xA;\ce{[Cu(OH)_2(H2O)_4] + 4NH3 &amp;amp;&amp;lt;=&amp;gt; [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been no explanation given of why the substitution is only partial, whereas the substitution of, say, $\ce{[Co(H2O)_6]^2+}$ is complete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? Moreover my textbook states ammonia is a better ligand than water, so why would $\ce{NH3}$ not replace those final two aqua ligands? &lt;/p&gt;&#xA;" OwnerUserId="14863" LastEditorUserId="4945" LastEditDate="2015-04-23T14:29:54.847" LastActivityDate="2019-01-07T19:17:47.520" Title="Why is ligand substitution only partial with copper(II) ions and ammonia?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29127" PostTypeId="1" CreationDate="2015-04-21T18:34:56.113" Score="2" ViewCount="539" Body="&lt;p&gt;Why are the atomic and the mass numbers of an element flipped in the textbooks relative to the periodic table?&lt;/p&gt;&#xA;" OwnerUserId="15738" LastEditorUserId="418" LastEditDate="2015-04-21T19:46:16.690" LastActivityDate="2015-04-21T19:52:36.600" Title="Placing the atomic mass and number of an element..." Tags="&lt;atoms&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-04-21T20:49:44.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="29128" PostTypeId="2" ParentId="28886" CreationDate="2015-04-21T18:51:54.417" Score="3" Body="&lt;p&gt;Glycine hydrochloride is a salt of the amino acid glycine. This means that HCl (hydrochloric acid) has been used to protonate the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand better which part of the molecule is protonated we need to think about chemical structure of glycine and it's acidity in water.&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/qNC8B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water glycine is amphoteric which means that it can react acid as well as base. The hydrochloride salt can be easily prepared by adding HCl to solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UD5ea.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we evaporate solvent we will get glycine hydrochloride salt, which has a pKa is 2.3 for first deprotonation and a pKa of 9.6 for the second deprotonation. Chloroacetic acid's pKa is 2.87, which means that glycine hydrochloride is more acidic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? It's conjugate base is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In common language conjugate base is happy with its charge and doesn't want to get protonated. Let's take look at conjugate bases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oup9n.jpg&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chloroacetic acid's conjugate base, the nearby electron withdrawing chlorine group stabilizes the negative charge in the conjugate base, which lowers the pKa of chloroacetic acid compared to acetic acid (pKa = 4.75) ~ 2 units (remember pKa is on a logarithmic scale). In glycine hydrochloride's conjugate base, the nearby positive charge stabilizes the negative charge of conjugate base, which lowers the pKa even more to 2.3.   &lt;/p&gt;&#xA;" OwnerUserId="15733" LastEditorUserId="4325" LastEditDate="2017-05-19T23:58:46.567" LastActivityDate="2017-05-19T23:58:46.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29129" PostTypeId="2" ParentId="29127" CreationDate="2015-04-21T19:28:18.943" Score="1" Body="&lt;p&gt;In the &lt;a href=&quot;http://commons.wikimedia.org/w/index.php?title=Periodic_Table_of_Elements&quot; rel=&quot;nofollow&quot;&gt;periodic table&lt;/a&gt;, the largest number is indeed the atomic number. This is the number of protons. Since this defines the element, it is considered the &lt;strong&gt;most important&lt;/strong&gt; number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you probably mean is &#xA;$^A_Z\mathrm{Element}$. Here, &lt;strong&gt;by convention&lt;/strong&gt; the lower $Z$ (from the German word &lt;em&gt;Zahl&lt;/em&gt; is the atomic number, while $A$ is the sum of protons and neutrons. This defines a particular isotope. This notation is often used in the context of NMR spectroscopy ($^{13}\ce{C}$ NMR), or when radioactive decays are described:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[\ce{^{14}_6C -&gt; ^{14}_7N} + e^- + \bar\nu_e\]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I think this is just a set of different conventions which show that in differents contexts either the element itself or a particular isotope and its properties are considered more important.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-21T19:52:36.600" LastActivityDate="2015-04-21T19:52:36.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29131" PostTypeId="1" CreationDate="2015-04-21T20:22:51.943" Score="3" ViewCount="836" Body="&lt;p&gt;The Lewis structure of nitric acid is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zsRdd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, since Pauling's rule is $\mathrm{p}K_\mathrm{a}=8-5p$, where $p$ is the number of oxo groups, how is the rule applied in this instance? $p=1.5$? $p=2$?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="14" LastEditDate="2015-06-17T15:49:23.560" LastActivityDate="2015-06-17T15:49:23.560" Title="How are Pauling's rules applied to Nitric acid?" Tags="&lt;acid-base&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29132" PostTypeId="1" CreationDate="2015-04-21T20:48:14.187" Score="2" ViewCount="818" Body="&lt;p&gt;Why is the pH of a buffer solution measured before measuring the pH of an unknown solution?&lt;br&gt;&#xA;I do not have any idea about it. I had read a chapter about ionic equilibrium and electrochemistry, but have not been able to find out an answer.&lt;/p&gt;&#xA;" OwnerUserId="15736" LastEditorDisplayName="user15489" LastEditDate="2015-04-22T11:31:32.910" LastActivityDate="2015-04-22T11:31:32.910" Title="Potentiometric titration using a pH meter" Tags="&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29133" PostTypeId="1" AcceptedAnswerId="29134" CreationDate="2015-04-21T20:59:00.433" Score="8" ViewCount="3188" Body="&lt;p&gt;I have been asked to identify suitable reagents for the following transformation:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9FdXg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies it's back to front - the question is a map of reactions and I took a snapshot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really have no idea about this one. Don't even know where to begin, sorry. For the purposes of the question I only need reagents but I would appreciate a mechanism if you know it. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="10185" LastEditDate="2016-05-30T21:42:08.103" LastActivityDate="2017-09-19T10:37:22.533" Title="Aromatic ether to phenol reagents" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29134" PostTypeId="2" ParentId="29133" CreationDate="2015-04-21T21:24:16.437" Score="13" Body="&lt;p&gt;The reaction is demethylation of methoxybenzene. You can do it by using an acid which has a nucleophilic counter ion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yt046.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not actually obvious why the iodine nucleophile attacks the methyl group and not inside benzene ring. The other option would be:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0Hrfd.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction doesn't happen because otherwise the iodine ion would appear inside the benzene ring and resulting product would be in a very strange conformation (high energy).&#xA;More information:&lt;br&gt;&#xA; Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. p. 434, 589.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it was correctly stated in the other answer there are also other reagents which can be used to cleave the methyl ether. $\ce{BBr3}$ is a Lewis acid which also can be used in this reaction. A relatively new &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ejoc.201301647/full&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; describes a computationally calculated bimolecular mechanism for this reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qmnaJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15733" LastEditorUserId="10185" LastEditDate="2016-05-30T21:44:11.753" LastActivityDate="2016-05-30T21:44:11.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29135" PostTypeId="2" ParentId="29133" CreationDate="2015-04-21T21:27:27.683" Score="8" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boron_tribromide&quot;&gt;Boron tribromide&lt;/a&gt; is a standard reagent for cleaving aryl methyl ethers to the corresponding phenol. It is a general method and can be used in aromatic methyl ethers with sensitive substituents where acid or base cannot be tolerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above Wiki link also shows the following mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) complex or adduct formation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ArOCH3 + BBr3 -&amp;gt; ArO+(^{−}BBr3)CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) bimolecular reaction of 2 adducts&#xA;$$\ce{2 ArO^{+}(^{−}BBr3)CH3 -&amp;gt; 2 ArO(BBr2) + 2CH3Br }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) hydrolysis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ArOBBr2 + 3H2O → ArOH + B(OH)3 + 2HBr}$$&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-21T21:27:27.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29136" PostTypeId="2" ParentId="29132" CreationDate="2015-04-21T21:32:33.823" Score="3" Body="&lt;p&gt;In principle, you &lt;em&gt;should&lt;/em&gt; know the pH of the buffer solution. Measuring it before an unknown sample is a straightforward way of calibrating the detector (pH detectors can get a bit funny after a while). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put the detector into a pH 5.5 solution and it reads 7, you're going to have to tweak the detector (if possible) or find another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring unknown samples with a detector that incorrectly measures a known buffer solution is a bad idea but an even worse idea would be to measure unknown samples without knowing your detector was incorrect in the first place.&lt;/p&gt;&#xA;" OwnerUserId="2338" LastActivityDate="2015-04-21T21:32:33.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29137" PostTypeId="1" AcceptedAnswerId="29140" CreationDate="2015-04-21T22:12:18.370" Score="1" ViewCount="9595" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;" OwnerUserId="15742" LastEditorUserId="5026" LastEditDate="2018-03-08T10:26:52.493" LastActivityDate="2018-03-08T10:26:52.493" Title="How to calculate the pH of a buffer after dilution" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29139" PostTypeId="1" CreationDate="2015-04-21T22:42:21.703" Score="3" ViewCount="366" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You have discovered a new strain of bacteria called &lt;em&gt;Tankeromyces subtilis&lt;/em&gt; that has a special outer membrane that is capable of withstanding high levels of osmotic pressure. These bacteria can survive osmotic pressure as high &lt;span class=&quot;math-container&quot;&gt;$\pu{1.3 kJ L-1}$&lt;/span&gt;, but not higher. The &lt;span class=&quot;math-container&quot;&gt;$(iC)_\mathrm{total}$&lt;/span&gt; value for the inside of these bacteria is &lt;span class=&quot;math-container&quot;&gt;$\pu{630 mM}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What is the lowest concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaCl}$&lt;/span&gt; (in otherwise pure water) in which these bacteria can survive without bursting due to osmotic pressure?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R = \pu{8.3153 J mol-1 K-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T = \pu{296 K}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.10 M}$&lt;/span&gt;, but I have no idea how to drive at that answer. However, I do know that it involves this equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p = iCRT$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15742" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-12-14T06:02:06.290" Title="Osmolarity and osmotic pressure" Tags="&lt;biochemistry&gt;&lt;concentration&gt;&lt;pressure&gt;&lt;osmosis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29140" PostTypeId="2" ParentId="29137" CreationDate="2015-04-21T23:02:13.017" Score="1" Body="&lt;p&gt;Adding water means diluting the buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-21T23:02:13.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29141" PostTypeId="2" ParentId="29056" CreationDate="2015-04-22T00:31:42.617" Score="1" Body="&lt;p&gt;If you have a site license (or a torrent) to use ChemDraw you can generate a 3D model of such a structure and optimize it to find the most stable configuration.  It will give you the bond angles on each  to a high degree of accuracy.  You can use the MM2 function to optimize and find bond lengths as well.  ChemDraw will give you a simple list.  It can also generate bond rotations, but you can bet that the conjugated pi system will remain planar, as the energy barrier to rotation about the sp2 carbon bonds is very high.&lt;/p&gt;&#xA;&lt;p&gt;If you utilize a simple one-dimensional particle-in-a-box model, the length L of the &amp;quot;box&amp;quot; would be the length of the conjugated system.  This would be the path along which electrons are conjugated.  It's not exactly the jagged line between all the sp2 carbons, but it's pretty close, so you can say L=(number of bonds in conjugated system)x(average length of these bonds).  Note that this is ABSOLUTELY NOT the distance between the C1-C3-C5-etc. carbons the other guy mentioned--I don't think he understands exactly what you're trying to calculate.&lt;/p&gt;&#xA;&lt;p&gt;Using this model, you can calculate the wavelength of maximum absorption from the quantum numbers of HOMO-LUMO transitions.  Look at the equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1mtfr.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this equation nf and ni are the electronic quantum numbers of the final and initial states of a transitioning electron, h is Planck's constant, m is the mass of an electron, and L is as described earlier.  The quantity in parentheses simplifies to N+1, as ni=N/2 and nf=N/2+1, where N is the number of pi electrons in the conjugated system.  To understand this conceptually, the ground state of a molecule will be populated such that the N/2 lowest energy levels will be filled (since electrons fill them in pairs), and all higher energy levels will be empty.  When it absorbs light, one of its electrons jumps from the highest filled energy level (HOMO, with ni=N/2) to the lowest unfilled level (LUMO, with nf=N/2+1).  It's important to understand that if an electron is promoted, it can't simply skip an energy level, so if you know the number of pi electrons you also know what the HOMO-LUMO transition will be.  If you can count the number of pi electrons in the conjugated system (ex. 1,6-diphenyl-1,3,5 hexatriene has 3 double bonds in its box, which means 6 pi electrons), then you can use this equation to find the desired wavelength of maximum absorption.  Surely you've seen the classical equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d94lz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;where c is the speed of light.  If you substitute this into the first equation you should be able to solve for the wavelength of maximum absorption.  Mind your units!&lt;/p&gt;&#xA;" OwnerUserId="15719" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-22T01:13:36.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29142" PostTypeId="1" AcceptedAnswerId="29461" CreationDate="2015-04-22T01:24:15.133" Score="34" ViewCount="3152" Body="&lt;p&gt;I realize that the octet rule is more a suggestion than a rule, and that it applies mainly to non-transition metal compounds. Still, compounds that don't have an octet, like $\ce{BH3}$, tend to dimerize or disproportionate in order to achieve an octet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In recent SE Chem posts discussing the bonding in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;phosphorous pentachloride&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28692/pbp-vs-tbp-geometry&quot;&gt;iodine heptafluoride&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypervalent bonding&lt;/a&gt; was used to describe the bonding in these molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UtR02.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WzQJI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count the electrons around phosphorous in $\ce{PCl5}$ there are 6 in the three equatorial bonds (2x3) and 4 in the hypervalent axial bonds; a total of 10. However 2 of the electrons in the hypervalent bond are in a non-bonding orbital, leaving 8 bonding electrons, an octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we count the electrons around iodine in $\ce{IF7}$ there are 4 in the two axial bonds (2x2) and 10 in the hypervalent equatorial bonds; a total of 14. However, 4 electrons are in a hypervalent non-bonding orbital and 2 more electrons can be viewed as being associated with a fluoride anion, leaving 8 bonding electrons, again, an octet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these compounds considered to be exceptions to the octet &quot;rule&quot;? Do all compounds involving hypervalent bonding violate the octet &quot;rule&quot;? Or do these compounds comply with the octet &quot;rule&quot;? &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-04-27T23:28:49.047" Title="Hypervalency and the octet rule" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="29143" PostTypeId="1" AcceptedAnswerId="29676" CreationDate="2015-04-22T01:58:01.113" Score="18" ViewCount="641" Body="&lt;p&gt;Initial harmonic is measured as Hartree/Bohr^2 ( like IHarmonic=n in Gaussian Software ). As I am from physics background I am used to look at features from their &lt;strong&gt;dimension&lt;/strong&gt; point of view:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FUnq3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Actually &lt;strong&gt;surface tension&lt;/strong&gt; is the only physical feature that I know is compatible with Force / Distance. I was wondering if there is any relation between the surface tension and initial harmonic.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Another physical feature is spring constant which is not irrelevant also !&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-03T22:13:11.730" Title="What is the relation between surface tension and initial molecular harmonic?" Tags="&lt;thermodynamics&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29144" PostTypeId="1" AcceptedAnswerId="29146" CreationDate="2015-04-22T02:38:34.007" Score="5" ViewCount="366" Body="&lt;p&gt;Can acid mine drainage form acid rain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since normal acid rain is formed via combustion of fossil fuels, the sulphur dioxide and nitrogen dioxide are all in the gas phase, whereas in acid mines the sulphuric acid is in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the reaction for acid rock drainage if it helps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{FeS2 + 3.75 O2 + 0.5 H2O → Fe^3+ + 2 SO4^2- + H+}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15045" LastEditorUserId="41328" LastEditDate="2019-03-13T11:50:05.640" LastActivityDate="2019-03-13T11:50:05.640" Title="Can acid mine drainage form acid rain?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;solutions&gt;&lt;minerals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29146" PostTypeId="2" ParentId="29144" CreationDate="2015-04-22T03:30:21.613" Score="5" Body="&lt;p&gt;I don't think acid &lt;em&gt;rain&lt;/em&gt; formation from liquid acid mine drainage is very likely.  The reason is that in acid rock drainage (a.k.a. acid mine drainage or AMD), the acidity is caused by sulfuric acid.  Sulfuric acid, $\ce{H2SO4}$ is much less volatile than water.  It simply won't appreciably evaporate from bulk liquids at normal conditions on the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one condition where acid mine drainage could possibly contribute to the formation of acid rain: if there were lots of aerosols formed from acid mine drainage waters.  If there were an AMD-laden stream or creek that flowed over a waterfall or was otherwise agitated enough to form very small droplets and splashes, some sulfuric acid might travel to the atmosphere in the small aerosol droplets.  That could contribute to acid rain, although a more likely effect would be a much more local (tens to hundreds of feet) acidification of the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even then, I suspect that sulfur from coal combustion would be the dominant source of airborne sulfuric acid in most regions.  In the United States, coal consumption is roughly &lt;a href=&quot;http://www.eia.gov/cfapps/ipdbproject/IEDIndex3.cfm?tid=1&amp;amp;pid=1&amp;amp;aid=2&quot; rel=&quot;nofollow&quot;&gt;one gigaton per year&lt;/a&gt;, leading to about &lt;a href=&quot;http://www.sourcewatch.org/index.php/Sulfur_dioxide_and_coal&quot; rel=&quot;nofollow&quot;&gt;20 megatons&lt;/a&gt; of sulfur emissions to the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-04-22T15:13:25.303" LastActivityDate="2015-04-22T15:13:25.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29149" PostTypeId="2" ParentId="29025" CreationDate="2015-04-22T05:45:14.890" Score="1" Body="&lt;p&gt;Yes it's possible this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kG5b6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, this raises another question - can't Grignards also create alkenes &quot;by accident&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's very common. Grignard reagents will take away any acidic proton available. That's why we use Grignard reagents in anhydrous conditions, else it would take up water's proton.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-22T05:45:14.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29150" PostTypeId="1" AcceptedAnswerId="29157" CreationDate="2015-04-22T08:01:20.850" Score="10" ViewCount="8170" Body="&lt;p&gt;Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$? It seems that 2,5-dinitropyrrole is not formed when using $\ce{Ac2O/HNO3}$ (ref 1). Is this just because of the mild conditions used, or does this reflect the pyrrole ring being more deactivated by the first nitration than a benzene ring would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Cooksey, A. R.; Morgan, K. J.; Morrey, D. P. Nitropyrroles—II : The nitration of pyrrole. &lt;em&gt;Tetrahedron&lt;/em&gt; &lt;strong&gt;1970,&lt;/strong&gt; &lt;em&gt;26&lt;/em&gt; (21), 5101–5111. &lt;a href=&quot;http://doi.org/10.1016/S0040-4020(01)93163-1&quot; rel=&quot;noreferrer&quot;&gt;DOI:10.1016/S0040-4020(01)93163-1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="16683" LastEditDate="2017-03-05T20:09:19.807" LastActivityDate="2021-02-14T01:08:29.477" Title="Nitration of pyrrole with sulfuric and nitric acids" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;&lt;nitro-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29152" PostTypeId="1" AcceptedAnswerId="29154" CreationDate="2015-04-22T08:29:44.220" Score="5" ViewCount="3383" Body="&lt;p&gt;We know that isodiaphers have same difference in neutrons and protons, but I encountered another definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Atoms which have same isotopic axis are called isodiaphers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know what it means by &quot;isotopic axis&quot;. Please try to explain in simple words.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="16683" LastEditDate="2018-11-16T22:38:16.570" LastActivityDate="2018-11-16T22:38:16.570" Title="What are isodiaphers?" Tags="&lt;inorganic-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29154" PostTypeId="2" ParentId="29152" CreationDate="2015-04-22T08:52:39.560" Score="18" Body="&lt;p&gt;These are basic terms of radiochemistry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Isotopes&lt;/em&gt; are nuclides with a common number of protons $Z$.&lt;br&gt;&#xA;&lt;em&gt;Isobars&lt;/em&gt; are nuclides with a common number of nucleons $A$.&lt;br&gt;&#xA;&lt;em&gt;Isotones&lt;/em&gt; are nuclides with a common number of neutrons $N$.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Isodiaphers&lt;/em&gt;&lt;/strong&gt; are nuclides with a common neutron excess $N-Z=A-2Z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the chart of the nuclides, isodiaphers are found in diagonals from bottom left to top right. For example, the mother and daughter nuclides of alpha decay are isodiaphers.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-22T08:52:39.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29155" PostTypeId="1" CreationDate="2015-04-22T10:36:22.437" Score="3" ViewCount="388" Body="&lt;p&gt;Are there any limits to rotating a compound in order to prove that its chirality or achirality? (assuming that you're rotating about a single bond)&lt;/p&gt;&#xA;" OwnerUserId="15517" LastActivityDate="2016-11-19T21:13:26.517" Title="Can a meso compund be rotated in ANY sort of way in order to show its achirality?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29156" PostTypeId="1" AcceptedAnswerId="29158" CreationDate="2015-04-22T11:13:41.190" Score="4" ViewCount="1939" Body="&lt;p&gt;In &lt;a href=&quot;https://chemistry.stackexchange.com/questions/20190/why-does-hydrogen-bonding-in-salicylic-acid-make-it-more-acidic&quot;&gt;this&lt;/a&gt; question, it is explained why salicylic acid is a stronger acid than benzoic acid. So, why acetylsalicylic acid (aspirin), whose conjugate base cannot hydrogen bond since there is no hydrogen, is still more acid than benzoic acid?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;aspirin: $\mathrm{p}K_\text{a}$ = 3.5&lt;/li&gt;&#xA;&lt;li&gt;benzoic acid: $\mathrm{p}K_\text{a}$ = 4.2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Acetoxyl is a moderate activating group for the aromatic electrophilic substitution, so it should mean that the resonance-driven electron-donating effect prevails over the inductive electron-withdrawing effect, therefore it should raise the $\mathrm{p}K_\text{a}$ of the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it different between electrophilic substitution reactivity and acidity? Does the inductive effect prevail in this case?&lt;/p&gt;&#xA;" OwnerUserId="15553" LastEditorUserId="41890" LastEditDate="2017-11-27T18:01:57.800" LastActivityDate="2018-03-04T07:43:55.787" Title="Why is acetylsalicylic acid still more acid than benzoic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29157" PostTypeId="2" ParentId="29150" CreationDate="2015-04-22T11:27:05.567" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I doubt it and would rather expect a lot of tar to be formed here. Pyrrole tends to undergo acid-catalyzed polymerization!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For decades, acetyl nitrate, formed by addition of fuming nitric acid to acetic anhydride, is the reagent of choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mononitration of pyrrole predominantly gives 2-nitropyrrole, accompanied by minor amounts of 3-nitropyrrole. (K.J. Morgan and D.P. Morrey, &lt;em&gt;Tetrahedron&lt;/em&gt;, &lt;strong&gt;1966&lt;/strong&gt;, &lt;em&gt;22&lt;/em&gt;, 57–62. &lt;a href=&quot;http://dx.doi.org/10.1016/0040-4020(66)80101-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of 1-methylpyrrole, the yield of the 3-nitropyrole is somewhat larger (H.J. Anderson, &lt;em&gt;Canad. J. Chem&lt;/em&gt;, &lt;strong&gt;1957&lt;/strong&gt;, &lt;em&gt;35&lt;/em&gt;, 23–29). In you want even more of the 3-nitro isomer, a more bulky alkyl substituent (&lt;em&gt;t-Bu&lt;/em&gt;) or N-silylation with $\ce{((CH3)CH)2SiCl}$ is a viable approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitration of 2-nitropyrrole at −15 °C with nitric acid in acetic anhydride  furnishes a 4:1 mixture of 2,4- and 2,5-dinitropyrrole with a total yield of 61%. (R. Alan Jones, Gerritt P. Bean, &lt;em&gt;The Chemistry of Pyrroles: Organic Chemistry: A Series of Monographs&lt;/em&gt;, Vol. 34, Academic Press, p. 124)&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2017-03-05T20:11:10.937" LastActivityDate="2017-03-05T20:11:10.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29158" PostTypeId="2" ParentId="29156" CreationDate="2015-04-22T11:43:13.170" Score="3" Body="&lt;p&gt;The acetyl moiety in acetylsalicylic acid has a $(-I)$-effect, i.e. compared to a proton attached to the aromatic ring (as in the case of benzoic acid), it withdraws electron density out of the aromatic ring.  This provides an &lt;em&gt;additional&lt;/em&gt; stabilization of the conjugated base, that is equivalent with rendering acetylsalicylic acid a stronger Brønsted acid than benzoic acid.  Hence, the $pK_a$ of acetylsalicylic acid is lower than the one of benzoic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the second paragraph, please notice $pK_a$ is a thermodynamic property -- where time does not play a role.  In contrast, the substitution reactivity you likely refer to (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hammett_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammett parameters&lt;/a&gt;) refers to reaction kinetics and rate constants.  Here, time does play a role.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="5026" LastEditDate="2018-03-04T07:43:55.787" LastActivityDate="2018-03-04T07:43:55.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29159" PostTypeId="2" ParentId="29119" CreationDate="2015-04-22T13:31:25.413" Score="4" Body="&lt;p&gt;Different solutes in the same solution indeed can affect each others' solubility. One example where this effect can be used to advantage is the &quot;salting out&quot; of a protein from water solution. Adding a salt (ammonium sulfate is especially good for this purpose) reduces the solubility of the protein in water, causing it to precipitate or crystallize out. Salting out can also be used to more fully separate an organic solvent from water. I imagine that in your example with the sugar alcohols, adding one component to water will reduce the solubility of the second component by a similar degree. But there may be cases where one solute increases the solubility of another.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-22T13:31:25.413" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29160" PostTypeId="2" ParentId="29142" CreationDate="2015-04-22T15:07:23.557" Score="8" Body="&lt;p&gt;In general, the octet rule gets less useful as you move down the periodic table (one early teacher of mine cautioned against using it for elements other than $\ce{H}$, $\ce{C}$, $\ce{N}$, $\ce{O}$ and $\ce{F}$). You can &quot;force compliance&quot; with the octet rule for molecules like $\ce{IF7}$ in the manner you describe (semi-ionic character etc.) but you quickly get to the point where the rule loses its predictive power, because more subtle theories of bonding are required to justify the octet rule's conclusions.&lt;/p&gt;&#xA;" OwnerUserId="15747" LastEditorDisplayName="user15489" LastEditDate="2015-04-22T20:15:02.763" LastActivityDate="2015-04-22T20:15:02.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29161" PostTypeId="2" ParentId="29016" CreationDate="2015-04-22T15:15:00.943" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; is the maximum non expansion work and hence should be equal to external force under consideration times the distance for a reaction to be at equilibrium, hence we need to find that external force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that from electrostatics that for a charge in a uniform electric field the force is &lt;span class=&quot;math-container&quot;&gt;$\bar F=q\bar E$&lt;/span&gt;. for &lt;span class=&quot;math-container&quot;&gt;$q=-1.6\times10^{-19}\ \mathrm C$&lt;/span&gt;. When we take our system to be &lt;span class=&quot;math-container&quot;&gt;$1\ \mathrm{mol}$&lt;/span&gt; of electrons, we have a total of &lt;span class=&quot;math-container&quot;&gt;$-1.6\times10^{-19}\ \mathrm C\times6.02\times10^{23}\approx -96500\ \mathrm C$&lt;/span&gt; and we have that &lt;span class=&quot;math-container&quot;&gt;$1\ \mathrm F=96500\ \mathrm C$&lt;/span&gt;. So total force is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=W_\text{max,add}=\bar F_\text{ext}\cdot \bar d$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G=\bar F_\text{ext}\cdot \bar d=q_\text{total}\cdot E\cdot 1\ \mathrm m=(-n) _\mathrm F\cdot E _\mathrm V=-nFE_\mathrm J$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think, can the number of moles of electrons transferred be negative?&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorDisplayName="user7951" LastEditDate="2019-03-11T08:20:50.657" LastActivityDate="2019-03-11T08:20:50.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29162" PostTypeId="1" CreationDate="2015-04-22T15:39:42.590" Score="8" ViewCount="5069" Body="&lt;p&gt;Is it possible to synthesize an ester from an ether in a single step? For example if I had an ether $\ce{(R-CH2-OR')}$, can I oxidize it so it will be an ester $\ce{(R-CO-OR')}$.&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="10185" LastEditorDisplayName="user15489" LastEditDate="2015-09-20T18:06:44.750" LastActivityDate="2017-02-13T19:46:52.733" Title="Is it possible to oxidize ether to ester?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29163" PostTypeId="2" ParentId="29162" CreationDate="2015-04-22T15:54:36.277" Score="11" Body="&lt;p&gt;Yes, it is possible.  One reagent to do so is acidic bromate salts.  &lt;a href=&quot;https://www.thevespiary.org/rhodium/Rhodium/pdf/thf2gbl.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; discusses a number of optimizations and characterizations of the bromate-powered ether oxidation reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-CH2-O-CH2-R + BrO3- -&amp;gt;[H+][H2O; {r.t.}] R-CO-O-CH2-R + Br^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrolysis of the formed ester leads to an undesired byproduct: the free acid $\ce{RCOOH}$.  In addition to that byproduct, overoxidation of the substrate can also happen, leading to anhydrides or even $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may well be plenty of other oxidants and catalysts than can achieve the same reaction, but I thought this one was notable because it works in water at room temperature.  Yield depends strongly on reaction conditions and the vagaries of the particular substrates you use, though.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2017-02-13T19:46:52.733" LastActivityDate="2017-02-13T19:46:52.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29164" PostTypeId="2" ParentId="29071" CreationDate="2015-04-22T16:19:15.723" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am trying to understand the link between the two chemical bonds. Are they dependent on each other (e.g. one causes the other?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One does not &quot;cause&quot; the other.  Instead, the author is only trying to say that removing the 2' hydroxyl group from nucleotides does two things: (a) vastly &lt;em&gt;increases&lt;/em&gt; stability of 3' bond to phosphate, and (b) &lt;em&gt;decreases&lt;/em&gt; the stability of the N-glycosyl bond.  There's no way to get just one effect without the other, since you can either have a 2'-OH group or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source to look for more information on N-glycosidic bond stability is &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/bi701480f&quot; rel=&quot;nofollow&quot;&gt;the work of Richard Wolfenden&lt;/a&gt;.  The linked paper shows that in 2'-deoxynucleosides, the N-glycosidic bond has a half-life at 37 &amp;deg;C of between 12 and 43 years.  (That is the water-catalyzed half-life.  Of course hydrolysis in strong base, strong acid, or by enzymes accelerates this rate, but the &quot;un&quot;catalyzed or merely water-catalyzed rate is of most interest as an indicator of bond stability.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0045206810000519&quot; rel=&quot;nofollow&quot;&gt;later paper also by Wolfenden et al.&lt;/a&gt; directly compares the stability of the N-glycosidic bond of adenosine to 2'-deoxyadenosine and finds large differences.  At 25 &amp;deg;C, the half-life of adenosine's N-glycosidic bond was 6000 years, while that for deoxyadenosine was only 180 years.  (Note the difference in temperature between this paragraph and the preceeding.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those papers help indicate the instability of N-glycosidic bond in 2'-deoxysugars.  &lt;a href=&quot;http://www.pnas.org/content/103/11/4052.short&quot; rel=&quot;nofollow&quot;&gt;Yet another Wolfenden paper&lt;/a&gt; examines the stability of phosphodiester bonds, although because of experimental difficulties there was no direct comparison between say A-A and dA-dA hydrolysis rates.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-22T16:19:15.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29165" PostTypeId="2" ParentId="29155" CreationDate="2015-04-22T16:26:20.453" Score="1" Body="&lt;p&gt;The properties of a molecule don't change upon rotation about a single bond and such rotations are very fast and frequent at room temperature, so you can consider the rotated form and decied the chirality of the molecule.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-22T16:26:20.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29167" PostTypeId="2" ParentId="29139" CreationDate="2015-04-22T17:00:14.440" Score="1" Body="&lt;p&gt;The highest pressure they can handle is $\pu{1.3 kJ L-1}$, or $\pu{13000 J L-1}$. &lt;a href=&quot;http://www.chem.purdue.edu/gchelp/howtosolveit/Solutions/osmoticpressure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;In the formula&lt;/a&gt; you use the ideal gas constant $0.08206$ and the temperature given to find the concentration the maximum pressure it can handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes out at $\pu{0.53 M}$ or $\pu{530 mM}$ and that's the maximum concentration it can handle. To find the minimum, you subtract $530$ from $630$ giving you $\pu{100 mM}$ or $\pu{0.1 M}$, which is the lowest concentration it can handle.&lt;/p&gt;&#xA;" OwnerUserId="15749" LastEditorUserId="41328" LastEditDate="2017-12-14T06:02:06.290" LastActivityDate="2017-12-14T06:02:06.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29168" PostTypeId="1" CreationDate="2015-04-22T17:18:52.367" Score="17" ViewCount="13160" Body="&lt;p&gt;My professor sent out a review guide. One of the questions on this guide is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The carboxylic acid carbonyl oxygen is:&lt;br&gt;&#xA;  1. $\ce{sp^2}$&lt;br&gt;&#xA;  2. $\ce{sp^3}$&lt;br&gt;&#xA;  3. $\ce{H}$-bond donor&lt;br&gt;&#xA;  4. rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$&lt;br&gt;&#xA;  5. 1 and 3&lt;br&gt;&#xA;  6. 1 and 4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I chose (4) because of the resonance that can be drawn with a negative oxygen and positive carbon. However apparently the answer is (6). I don't understand how something can be equilibrating between two states and still be one of those states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="15750" LastEditorUserId="4945" LastEditDate="2018-02-19T11:06:39.503" LastActivityDate="2018-02-19T11:23:09.353" Title="What is the hybridization of the carbonyl oxygen in a carboxylic acid?" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29169" PostTypeId="1" AcceptedAnswerId="29170" CreationDate="2015-04-22T17:35:58.393" Score="5" ViewCount="218" Body="&lt;p&gt;I have been trying to solve this NMR question for about four hours and I feel like I'm always really close but never manage to find the perfect combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway here's the info in the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formula is $\ce{C11H14O2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.triplet at 2.4 ppm (2H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.singlet at 3.3 ppm (3H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.triplet at 3.5 ppm (2H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.doublet at 7.4 ppm (2H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.doublet at 7.5 ppm (2H)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.singlet at 2.0 ppm integrates for 3H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried a number of structures and the one I feel i got very close to was this one, each number on the side corresponds to the signal that I think it is. The X is just the area where I know i am missing a bond, I just don't know how to seal it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHjV6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: After the advice given by you all, I came up with this structure which seems to be pretty sound&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SA5Go.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="402" LastEditDate="2016-11-20T01:28:38.500" LastActivityDate="2016-11-20T01:28:38.500" Title="Determination of structure with HNMR" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29170" PostTypeId="2" ParentId="29169" CreationDate="2015-04-22T17:55:51.580" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;Calculate your &lt;strong&gt;DBEs&lt;/strong&gt; (double bond equivalents):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;\[\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5\] &#xA;Note that a benzene has 4 &lt;strong&gt;DBEs&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?&lt;/li&gt;&#xA;&lt;li&gt;Largest couplings between benzene protons are typically &lt;em&gt;ortho&lt;/em&gt; couplings. How large are they in your case?&lt;/li&gt;&#xA;&lt;li&gt;Two sets of doublets at 7.5 and 7.4 ppm might indicate a 1,4-disubstituted benzene. You might want to play with some increment rules here. &lt;/li&gt;&#xA;&lt;li&gt;You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.&lt;/li&gt;&#xA;&lt;li&gt;R in (5) might be $\ce{CH3}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-22T18:19:37.057" LastActivityDate="2015-04-22T18:19:37.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29171" PostTypeId="2" ParentId="29126" CreationDate="2015-04-22T18:46:56.383" Score="13" Body="&lt;p&gt;Metal ion complexes have stepwise stability constants:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{[Cu(H2O)6]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_1$] [Cu(NH3)(H2O)5]^2+ + H2O}\\&#xA;\ce{[Cu(NH3)(H2O)5]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_2$] [Cu(NH3)2(H2O)4]^2+ + H2O}\\&#xA;\ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_3$] [Cu(NH3)3(H2O)3]^2+ + H2O}\\&#xA;\ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_4$] [Cu(NH3)4(H2O)2]^2+ + H2O}\\&#xA;\ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_5$] [Cu(NH3)5(H2O)]^2+ + H2O}\\&#xA;\ce{[Cu(NH3)5(H2O)]^2+ + NH3 &amp;amp;&amp;lt;=&amp;gt;[$K_6$] [Cu(NH3)6]^2+ + H2O}\\&#xA;\end{align}&lt;/span&gt;&#xA;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;K_1 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;K_2 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;K_3 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;K_4 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;K_5 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;K_6 &amp;amp;= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall stability constant is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In particular, note the exponent 6.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, usually &lt;span class=&quot;math-container&quot;&gt;$K_i/K_{i+1}&amp;gt;1$&lt;/span&gt;; thus, &lt;span class=&quot;math-container&quot;&gt;$K_1 &amp;gt; K_2 &amp;gt; K_3 &amp;gt; K_4 &amp;gt; K_5 &amp;gt; K_6$&lt;/span&gt;. Consequently, it gets more and more difficult to exchange further ligands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the predominating complex strongly depends on the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;.&#xA;And you probably work with dilute aqueous solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply speaking, the missing &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cu(NH3)6]^2+}$&lt;/span&gt; in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="16683" LastEditDate="2019-01-07T19:17:47.520" LastActivityDate="2019-01-07T19:17:47.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29172" PostTypeId="2" ParentId="29126" CreationDate="2015-04-22T19:58:59.230" Score="6" Body="&lt;p&gt;In this complex you're going to see an arrangement with 4 equatorial and 2 axial ligand positions.  In such compounds, ligands that are stronger σ-donors and weaker π-acceptors preferentially occupy the axial positions.  On the other hand, weaker σ-donors and stronger π-acceptors preferentially occupy equatorial positions.  A quick look at the spectrochemical series (which ranks ligands by increasing field splitting effects which have been measured experimentally, which is the same as ranking them by increasing π-acceptor strength) shows that $\ce{NH3}$ is a stronger π-acceptor than both $\ce{H2O}$ and $\ce{HO}$, which means that it will preferentially occupy equatorial positions while the $\ce{HO/H2O}$ preferentially occupies the axial positions.  This is because the equatorial plane is electron-rich, which is a more stable environment for a stronger π-acceptor.  Conversely, the axial positions are electron-deficient, and thus they attract the stronger σ-donors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you only see 4 substitutions is because these substitutions are reversible.  It is energetically favorable for a stronger π-acceptor to displace equatorial ligands, but it's not energetically favorable for a weaker σ-donor to displace an axially-substituted ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can sort of see this in the structures of the ligands.  $\ce{NH3}$ has one lone pair, $\ce{H2O}$ has 2 lone pairs, and $\ce{HO-}$ has 3 lone pairs.  σ-bonding interactions involve free lone pairs on the ligands, so in this case you can see that more lone pairs means a greater ability to σ-bond.  Now look up how the spectrochemical series ranks these three ligands in terms of increasing σ-donor/decreasing π-acceptor strength.&lt;/p&gt;&#xA;" OwnerUserId="15719" LastEditorUserId="15719" LastEditDate="2015-04-22T20:03:05.720" LastActivityDate="2015-04-22T20:03:05.720" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29173" PostTypeId="1" CreationDate="2015-04-22T20:59:21.387" Score="3" ViewCount="219" Body="&lt;p&gt;I've stumbled across an electrochemistry problem that I need help on:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Fe~|~Fe^{+2}~||~Cl^{-},Cl2~|~Pt}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I understood that &lt;strong&gt;iron&lt;/strong&gt; is at the anode and is being oxidized, however I do not know what was going on with &lt;strong&gt;chlorine&lt;/strong&gt;, it looks like it is being oxidized while it is supposed to be reduced.&lt;/p&gt;&#xA;&lt;p&gt;Maybe this is a non-spontaneous reaction? Can this be possible, can chlorine be used as an oxidation reaction? And if so how can we calculate the &lt;strong&gt;voltage&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="69830" LastEditDate="2021-05-17T15:43:37.437" LastActivityDate="2021-05-17T16:07:38.937" Title="Electrochemistry - Redox &amp; Electroplating" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29175" PostTypeId="2" ParentId="28990" CreationDate="2015-04-22T21:30:07.223" Score="0" Body="&lt;p&gt;Well, I'll start by saying I redid your math, and you have rounding errors.  Using correct sig figs, your stock solution has 100. mg of Li in it, giving a concentration of 1.00E3 mg/L.  Those dilutions therefore should be 30.0 mg/L, 50.0 mg/L, 80.0 mg/L, and 100. mg/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct aside from the rounding errors, and the answer key is wrong.  The question as you've written can't have those answers, as 300 μg/L is 0.3 mg/L, which is not the answer.&lt;/p&gt;&#xA;" OwnerUserId="15719" LastActivityDate="2015-04-22T21:30:07.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29176" PostTypeId="2" ParentId="29155" CreationDate="2015-04-22T21:36:35.593" Score="2" Body="&lt;p&gt;Rotating about a single bond would be a conformational change and not a stereochemical one. Meso compounds are not defined as having rotational symmetry, either way. For example, 1,2-diphenylethane-1,2-diol does not have rotational symmetry since one carbon is R and the other S.  However, it's still a meso compound because its &lt;em&gt;mirror image&lt;/em&gt; can be rotated to show that it is identical to the original compound.&lt;/p&gt;&#xA;" OwnerUserId="15719" LastEditorUserId="9961" LastEditDate="2016-11-19T21:13:26.517" LastActivityDate="2016-11-19T21:13:26.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29177" PostTypeId="1" CreationDate="2015-04-22T22:19:54.150" Score="15" ViewCount="489" Body="&lt;p&gt;Suppose we hold temperature, pressure, and gas composition (say, a standard atmosphere) constant.  Then we know that if we allow the water content of the air to vary it can range between 0 and some saturation limit &lt;em&gt;S&lt;/em&gt; g/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that saturation limit vary with the composition of the air?  For example, does it depend on the molecular weight and proportions of the other constituents of the air?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the water (or whatever substance is in question) does not chemically react with any other constituents at that temperature and pressure are there other constituent chemical relationships that can affect the saturation limit?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="16683" LastEditDate="2016-09-30T09:44:23.520" LastActivityDate="2017-01-23T14:34:40.473" Title="How does composition of air affect vapor saturation limit?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29178" PostTypeId="1" CreationDate="2015-04-22T23:04:43.400" Score="1" ViewCount="267" Body="&lt;p&gt;I need to calculate the molar mass and density of the shape memory alloy $\ce{Co49Ni21Ga30}$&lt;br&gt;&#xA;(I suppose the numbers indicate the mass percentage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the molar mass, is it correct to use the formula below?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mol_mass_compound = Mol_mass_Co*0.49 + Mol_mass_Ni*0.21 + Mol_mass_Ga*0.30 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Regarding the density, how do you suggest to proceed? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15754" LastEditorUserId="15754" LastEditDate="2015-04-23T06:42:00.273" LastActivityDate="2015-04-23T06:42:00.273" Title="Co49Ni21Ga30 - molar mass and density" Tags="&lt;concentration&gt;&lt;mole&gt;&lt;alloy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29179" PostTypeId="1" AcceptedAnswerId="29195" CreationDate="2015-04-22T23:21:25.637" Score="2" ViewCount="13789" Body="&lt;p&gt;I just want to know why, if we dilute a cell with water, it decreases the voltage. Assume it is a galvanic cell.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="73" LastEditDate="2015-04-23T03:02:08.407" LastActivityDate="2019-05-12T01:30:15.377" Title="How come diluting a cell decreases the voltage?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29180" PostTypeId="1" CreationDate="2015-04-23T00:51:42.153" Score="8" ViewCount="511" Body="&lt;p&gt;I'd like clarification of the meaning of $W_\text{max}$ in the following equations (and thus in the combined result in the third equation):&#xA;\begin{align}&#xA;W_\text{max} &amp;amp;= \Delta G\\&#xA;W_\text{max} &amp;amp;= -nF\varepsilon\\&#xA;\Delta G &amp;amp;= -nF\varepsilon\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first equation, we somehow come to the conclusion that the maximum useful work available from a process is that process' change in Gibbs Free Energy. I ask now what work that means - does it mean work done on the surroundings by release of energy from the system? Does it mean the maximum work done on the system via a chemical process (this makes more sense because it doesn't make sense that an increase in Gibbs Free Energy of a system would do work on anything but the system)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, in the electrochemical definition of $W_\text{max}$, we have the point of view of the system; that is, positive work is done on the system and negative by the system. Therefore, we add the negative sign because when cell potential is positive the work done on the system is negative (it does work on the surroundings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only logical conclusion I can come to is that the two $W_\text{max}$ definitions must be both with respect to the system's point of view, which means that $W_\text{max} = \Delta G$ refers to the maximum amount of work done on a system during a chemical process (the intuitive interpretation that I posed). However, it still makes me uncomfortable to define &quot;useful work&quot; as work done on the system. How can we harness work from putting energy into something?&lt;/p&gt;&#xA;" OwnerUserId="4642" LastEditorUserId="5026" LastEditDate="2018-05-05T07:19:29.750" LastActivityDate="2018-05-05T07:19:29.750" Title="What is the precise definition of maximum work in the context of electrochemistry and thermodynamics?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29182" PostTypeId="1" AcceptedAnswerId="29220" CreationDate="2015-04-23T00:56:45.943" Score="9" ViewCount="757" Body="&lt;p&gt;Given that the half-lives differ between isotopes of a particular radioactive element, wouldn't the natural abundance change (fairly rapidly)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we isolated a bunch of thorium, wouldn't the relative amount of the various isotopes of thorium in the sample differ over time, with the more slowly decaying isotope eventually making up the largest proportion?&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2016-10-09T18:20:48.203" LastActivityDate="2017-04-02T19:55:10.353" Title="Does the natural abundance of radioactive elements change?" Tags="&lt;isotope&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29183" PostTypeId="2" ParentId="29182" CreationDate="2015-04-23T01:15:23.760" Score="6" Body="&lt;p&gt;Yes, you are correct; over time the proportion of the more stable isotopes (those with longer half-lives) increases. There can be alterations to this tendency if a less stable isotope happens to be continually replenished by the decay of a heavier element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some very interesting physical proof of this. Consider uranium, whose dominant naturally-occurring isotopes are $\ce{^{238}U}$ and $\ce{^{235}U}$, with respective half-lives of $\mathrm{4.47\times 10^9\ yr}$ and $\mathrm{7.8\times 10^8\ yr}$. Currently, the uranium on Earth is composed of approximately 99.3% $\ce{^{238}U}$ and 0.7% $\ce{^{235}U}$. Given that $\ce{^{235}U}$ decays faster, we expect that in the distant past there was a higher proportion of $\ce{^{235}U}$ than we find now. Indeed that is the case - billions of years ago, the fraction of $\ce{^{235}U}$ in natural uranium was as high as 3%. It turns out that, due to its fissile nature, this concentration of $\ce{^{235}U}$ was high enough to sustain &lt;a href=&quot;http://en.wikipedia.org/wiki/Natural_nuclear_fission_reactor&quot; rel=&quot;nofollow noreferrer&quot;&gt;natural nuclear fission reactors&lt;/a&gt;. Nowadays, natural uranium contains too little of the fissile isotope to spontaneously form underground nuclear reactors, and so enrichment techniques must be used to obtain fissile uranium fuel, containing a higher proportion of $\ce{^{235}U}$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="402" LastEditDate="2017-04-02T19:55:10.353" LastActivityDate="2017-04-02T19:55:10.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29184" PostTypeId="1" AcceptedAnswerId="31045" CreationDate="2015-04-23T01:16:23.437" Score="6" ViewCount="1341" Body="&lt;p&gt;I am trying to learn to work with CP2K and now need to make its input file. CP2K has kind of &lt;a href=&quot;http://cp2k.org/tools:avogadro2&quot; rel=&quot;nofollow&quot;&gt;adaptor&lt;/a&gt; for &lt;strong&gt;Avogadro2&lt;/strong&gt; but not Avogadro ( I don't know what is the relation between them !).  I also tried that adaptor plugin with Avogadro2 but it does not work for me ( tried on Mac OS X Yosemite and Windows 7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched a lot but seems there is no way to make CP2K inputs with Avogadro . Is there any ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="5017" LastEditDate="2015-05-08T14:35:18.103" LastActivityDate="2015-11-27T22:52:43.297" Title="How to make a CP2K input file with Avogadro?" Tags="&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29187" PostTypeId="1" AcceptedAnswerId="29244" CreationDate="2015-04-23T02:52:04.297" Score="12" ViewCount="9102" Body="&lt;p&gt;About 2 years ago I purchased 2 liters of 98% concentrated sulfuric acid. It's in a standard poly bottle with leakproof lids, and originally was clear. Recently I noticed the acids have turned a dark gray; almost black color. They have both never been opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there was something in the lid that was eroded by the acid, but is there any other reason sulfuric acid would darken over time?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="15410" LastEditDate="2016-11-10T05:12:12.417" LastActivityDate="2017-07-19T11:59:09.103" Title="Why would sulfuric acid turn gray/black?" Tags="&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;safety&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29189" PostTypeId="1" CreationDate="2015-04-22T17:29:02.220" Score="3" ViewCount="1015" Body="&lt;p&gt;If I mix water in a cup with sugar at some stage the solution will become saturated and no more sugar will dissolve into the water. If I now add salt to the solution will the salt dissolve into the water or is the solution already saturated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not tried an experiment as I cannot figure out how to measure the dissolved component of the 2, intuitively I think that some salt will dissolve, but not as much as in a pure salt water mixture.&lt;/p&gt;&#xA;" OwnerDisplayName="Gobabis" LastEditorUserId="4945" LastEditDate="2015-04-23T06:37:19.987" LastActivityDate="2015-04-23T06:37:19.987" Title="What would happen if I mix salt and sugar in a cup of water?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-04-23T13:04:32.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="29190" PostTypeId="1" CreationDate="2015-04-23T05:08:39.893" Score="2" ViewCount="224" Body="&lt;p&gt;Why something jelly forms in the water if it is kept in in a ultrasonication bath for long periods?&#xA; When water is exposed to ultrasonication for say a month, some jelly substances form in the ultrasonicator bath. is this something related to some bioorganism growth?&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2015-04-23T05:56:22.490" LastActivityDate="2015-04-23T11:05:40.370" Title="Water in the ultrasonication bath" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;biochemistry&gt;&lt;solutions&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29191" PostTypeId="1" CreationDate="2015-04-23T05:14:26.557" Score="2" ViewCount="80" Body="&lt;p&gt;How can we record a IR spectrum of a substance coated on a Over Head Projector (OHP) sheet?&#xA;instrument has both transmission and ATR modes. IR spectrum is to be recorded for both the OHP sheet as well as the material that is coated on it&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2015-04-23T08:57:09.880" LastActivityDate="2015-04-23T09:35:46.007" Title="Infrared spectroscopy of substances on a OHP film" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;spectroscopy&gt;&lt;surface-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29192" PostTypeId="1" CreationDate="2015-04-23T05:24:17.527" Score="8" ViewCount="2127" Body="&lt;p&gt;Is it possible to reduce an acyl group to an alkane, but still keep a nitro group intact without its reduction to an amino group? A Clemmensen reduction would reduce the nitro group, and it seems like other reducing agents will also reduce the nitro group to an amino group. Would some other mechanism be better suited for this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, how can I selectively reduce 1-(3-nitrophenyl)propan-1-one to 1-nitro-3-propylbenzene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x2PMC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x2PMC.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="16683" LastEditDate="2018-08-14T09:09:04.317" LastActivityDate="2018-08-14T09:09:04.317" Title="How can an aromatic ketone be reduced in the presence of a nitro group?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;nitro-compounds&gt;&lt;organic-reduction&gt;&lt;chemoselectivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29193" PostTypeId="2" ParentId="29192" CreationDate="2015-04-23T05:34:02.777" Score="12" Body="&lt;p&gt;You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (&lt;a href=&quot;http://dx.doi.org/10.1021/jo00278a035&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A previous study (&lt;em&gt;Bull. Chem. Soc. Jpn.&lt;/em&gt;, &lt;strong&gt;1974&lt;/strong&gt;, &lt;em&gt;47&lt;/em&gt;, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCNBH3}$ is an even milder hydride transfer agent, that has been widely used in the reduction of imines and iminium salts. A review by &lt;em&gt;C.F.Lane&lt;/em&gt;, &lt;em&gt;Sodium Cyanoborohydride - A Highly Selective Reducing Agent for Organic Functional Groups&lt;/em&gt; in &lt;em&gt;Synthesis&lt;/em&gt;, &lt;strong&gt;1975&lt;/strong&gt;, &lt;em&gt;3&lt;/em&gt;, 135-146 gives some background information on the versatility of this reagent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitro group is definitely safe with $\ce{NaCNBH3}$. However, considering the high toxicity and the price for this reagent, you might want to run a small scale reduction of your tosylhydrazone with the much cheaper $\ce{NaBH4}$ first.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="11367" LastEditDate="2016-12-03T02:59:52.020" LastActivityDate="2016-12-03T02:59:52.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29194" PostTypeId="1" CreationDate="2015-04-23T05:43:56.217" Score="1" ViewCount="86" Body="&lt;p&gt;What is the basis set in the computational software &quot;Gaussian-03&quot; that can be used to study the interaction of $\ce{H}$-atom with $\ce{Pt-CNT}$ dropcasted electrode? How can we study the energy differences that arises due to the interaction computationally?&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2015-04-23T09:02:36.983" LastActivityDate="2015-04-23T09:02:36.983" Title="Basis set in the computational software &quot;Gaussian-03&quot; for studying adsorption of a hydrogen atom on a carbon paste electrode dropcasted with Pt-CNT" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;computational-chemistry&gt;&lt;surface-chemistry&gt;&lt;catalysis&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29195" PostTypeId="2" ParentId="29179" CreationDate="2015-04-23T07:20:38.930" Score="6" Body="&lt;p&gt;Your assertion the voltage in a cell is decreased by diluting the cell  is not necessarily true. According to the Nernst Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{cell}=E^0-\frac{RT}{nF}\ln(Q)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q$ is the reaction quotient relating $\frac{[\mathrm{metal~ion~formed~by~oxidation}]}{[\mathrm{metal~ion~reduced}]}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $Q$ can be increased by: increasing the concentration of the metal ion formed by oxidation, or by decreasing the concentration of the metal ion that is reduced. By increasing the value of $Q$, the magnitude of the term $-\frac{RT}{nF}\ln(Q)$ is increased. This means that a larger number is subtracted from the $E^0$ of the cell, and that the cell voltage decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $Q$ can be decreased by: decreasing the concentration of the metal ion formed by oxidation, or by increasing the concentration of the metal ion that is reduced. By decreasing the value of $Q$ to a fraction, the value of the term $-\frac{RT}{nF}\ln(Q)$ is made positive. This means that this term is added to the $E^0$ of the cell, and that the cell voltage increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is consistent with the Le Châtelier's Theorem. Essentially, increasing the relative concentration of your reactants drives the reaction forward (higher $\Delta V$), and increasing the relative concentration of your products drives the reaction backwards (lower $\Delta V$).&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-23T07:20:38.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29196" PostTypeId="1" CreationDate="2015-04-23T07:27:54.777" Score="19" ViewCount="16319" Body="&lt;p&gt;Is lead (Pb) radioactive? I have heard about it being radioactive but have also heard it referred to as the heaviest stable element. Radioactivity refers to release of alpha, beta, gamma rays.&lt;/p&gt;&#xA;" OwnerUserId="15758" LastEditorUserId="23561" LastEditDate="2019-03-14T00:15:21.823" LastActivityDate="2021-05-01T17:19:18.993" Title="Is lead radioactive?" Tags="&lt;radioactivity&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" ClosedDate="2017-02-13T04:41:36.590" ContentLicense="CC BY-SA 4.0" />
  <row Id="29197" PostTypeId="2" ParentId="29196" CreationDate="2015-04-23T07:48:39.507" Score="14" Body="&lt;p&gt;One radioactive isotope of lead is &lt;span class=&quot;math-container&quot;&gt;$\ce{^{210}_{82}Pb}$&lt;/span&gt;, which has a half-life of 22.20 years. The only reaction that &lt;span class=&quot;math-container&quot;&gt;$\ce{^{210}_{82}Pb}$&lt;/span&gt; undergoes is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^{210}_{82}Pb -&amp;gt; ^{210}_{83}Bi +^{0}_{-1}e}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\ce{^{0}_{-1}e}$&lt;/span&gt; released in the reaction is a high speed electron, otherwise known as a &lt;span class=&quot;math-container&quot;&gt;$\beta^-$&lt;/span&gt; particle. The fact a &lt;span class=&quot;math-container&quot;&gt;$\beta^-$&lt;/span&gt; particle is released by the &lt;span class=&quot;math-container&quot;&gt;$\ce{^{210}_{82}Pb}$&lt;/span&gt; nucleus would suggest that it is an excess number of neutrons causing the instability.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="24052" LastEditDate="2021-05-01T17:19:18.993" LastActivityDate="2021-05-01T17:19:18.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29198" PostTypeId="2" ParentId="29196" CreationDate="2015-04-23T07:51:36.163" Score="5" Body="&lt;p&gt;Before explaining why lead is radioactive, I would like to briefly explain what radioactivity is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radioactive_decay&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Radioactive decay, also known as nuclear decay or radioactivity, is&#xA;   the process by which a nucleus of an unstable atom loses energy by&#xA;   emitting ionizing radiation. A material that spontaneously emits this&#xA;   kind of radiation — which includes the emission of alpha particles,&#xA;   beta particles, gamma rays and conversion electrons — is considered&#xA;   radioactive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, radioactive decay occurs because an unstable atom &quot;prefers&quot; to release extra energy to become more stable. If you consider lead, &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotopes_of_lead&quot; rel=&quot;nofollow&quot;&gt;it has different isotopes&lt;/a&gt;. A large percentage (&gt;98%) of lead ordinarily found in nature is of stable isotopes whereas the rest of its isotopes are unstable. It is the unstable isotopes of lead that are radioactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon-12&quot; rel=&quot;nofollow&quot;&gt;Carbon-12&lt;/a&gt; with six neutrons is a stable isotope of carbon which has no radioactivity and &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon-14&quot; rel=&quot;nofollow&quot;&gt;carbon-14&lt;/a&gt; with eight neutrons is not stable and shows radioactivity.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="470" LastEditDate="2015-04-23T22:49:09.853" LastActivityDate="2015-04-23T22:49:09.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29200" PostTypeId="1" CreationDate="2015-04-23T09:11:32.107" Score="3" ViewCount="3144" Body="&lt;p&gt;What does the Mesh in the mesh size of silica gel that is used for various chromatography techniques mean?&lt;/p&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2015-04-23T09:17:28.270" Title="Mesh size of the silica gel" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29202" PostTypeId="2" ParentId="29200" CreationDate="2015-04-23T09:17:28.270" Score="2" Body="&lt;p&gt;Mesh is a unit representing the grain-size distribution of a material. So it represents either the opening of the mesh, or the number of grain per surface unit depending on what you are considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find &lt;a href=&quot;http://en.wikipedia.org/wiki/Mesh_%28scale%29&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; the conversion in grain size and surface unit related to the &quot;Mesh&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10119" LastActivityDate="2015-04-23T09:17:28.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29203" PostTypeId="1" AcceptedAnswerId="29207" CreationDate="2015-04-23T09:27:20.393" Score="4" ViewCount="845" Body="&lt;p&gt;I was wondering why when acetaldehyde is treated with $\ce{OH-}$, the $\ce{OH-}$ will attack the methyl group instead of the carbonyl carbon, forming an enolate anion. I would have though that the $\ce{OH-}$ would have attacked the carbonyl carbon due to it having a  relative positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, my professor said during lectures that form x is favored over form Y. Is this because a negative oxygen is able to delocalize the negative charge (and be protonated more easily) more efficently than a $\ce{CH2-}$ group? If so,  is the x form so dramatically favored that it would not make sense to draw any mechanisms involving the y form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AtqTk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2015-10-09T15:37:18.393" LastActivityDate="2015-10-09T15:37:18.393" Title="Enolate Ion formation" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29205" PostTypeId="2" ParentId="29191" CreationDate="2015-04-23T09:35:46.007" Score="1" Body="&lt;p&gt;Your are going to analyze both, the OHP foil alone, and the part of the OHP foil that is covered by the coating and calculate the difference of the later to get the data of your coating material.&#xA;If you do want to do the classical way (transmission IR), just cut-off a little of the sample (a hole puncher as in the office may do the job, too), attempt to grind it with KBr and press the pellet as normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to stick to ATR, place the foil on the dish without prior sample preparation.  This is quicker, and to some degree allows a space-resolved analysis, too -- the window (center dish of the table) of a single reflection geometry instrument (like the one below, Nicolet iS10 with smart-ATR-IR module) is of about 3 mm in diameter).  The window material of the ATR module may be Germanium (as here), yet there are instruments with a window of Diamond, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u5YDW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-23T09:35:46.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29206" PostTypeId="2" ParentId="29196" CreationDate="2015-04-23T10:34:06.060" Score="47" Body="&lt;p&gt;Important naturally-occurring radioactive lead nuclides are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ from $\ce{^{232}Th}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a geological time scale, their half-lives are very short. Nevertheless, these lead nuclides are members of the $\ce{^{238}U}$, $\ce{^{235}U}$, and $\ce{^{232}Th}$ decay series. Thus, they are constantly reproduced by decay of their respective mother nuclides. Therefore, all environmental samples (including soil, water, air, plants, and animals) naturally contain significant amounts of radioactive lead nuclides. However, in various materials (e.g. residues from mining or oil and gas production, fertilizers, building materials), the concentration of natural lead nuclides can be strongly enhanced by technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When lead is freshly refined (chemically purified), most of the mother nuclides are removed and the short-lived lead nuclides quickly decay. Nevertheless, the lead still contains significant amounts of $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$. The presence of $\ce{^{210}Pb}$ is a source of background mainly through the lead X-rays and bremsstrahlung caused by high-energy beta radiation emitted by its decay product $\ce{^{210}Bi}$. This background is a real problem when lead is used as a shielding material for low-background detectors. Therefore, some shields are made from selected lead with certified low $\ce{^{210}Pb}$ content. Preferably, very old lead is used. Since $\ce{^{210}Pb}$ decays with a half-life of 22.3 years, samples of lead that are many decades old are relatively free of this activity. In one extreme example, 2000-year-old lead was salvaged from a sunken Roman ship.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-10-09T15:52:48.577" LastActivityDate="2015-10-09T15:52:48.577" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="29207" PostTypeId="2" ParentId="29203" CreationDate="2015-04-23T10:50:20.900" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Enolate vs hydrate formation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both the enolate and the hydrate form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aldehydes (and ketones) react with water in the presence of acid or base to form &lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Organic_Chemistry/Aldehydes_and_Ketones/Reactions_of_Aldehydes_%26_Ketones/Addition_of_Water_to_form_Hydrates_%28Gem-Diols%29&quot;&gt;hydrates&lt;/a&gt;. However, for more aldehydes in aqueous solution (even dilute solution), the hydrate is only a minor component of the mixture (~1%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other 99% of the aldehyde molecules are perfectly able to form the enolate, but only a small number of them do because the difference in pKa between most alpha hydrogens on aldehydes and ketones (~21) and water (15.7). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in actuality most of the aldehyde molecules at any given time are neither enolates nor hydrates. Now, if a reaction should occur to consume enolate, then Le Châtelier will come out and nudge the system to make more enolate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Favored&quot; resonance contibutors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The two resonance contributors $x$ and $y$ are not separate compounds. It is not possible for one structure or another to be &quot;favored&quot;, though we often use that language because we can relate to comparing the structures of two distinct compounds, like the cis and trans isomers of an alkene. What we really mean when we say &quot;favored&quot; is that the resonance contributor is more important for describing the structure and behavior of the ion than the other. The true structure of the enolate ion is some linear combination of resonance contributors $x$ and $y$ (as well as some other minor things). Thus, the true structure might be like a weighted average over the contributors with $x$ having higher weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$enolate = \frac{2}{3}x + \frac{1}{3}y$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-04-23T10:50:20.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29208" PostTypeId="1" CreationDate="2015-04-23T11:02:31.783" Score="0" ViewCount="80" Body="&lt;p&gt;When filling an orbital diagram of an atom, usually the up-spin is filled first and then the down-spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for Nitrogen, we have : &lt;img src=&quot;https://i.stack.imgur.com/9gbLQ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in a model in which up-spin electron is preferred over electron down-spin, there would be many problems.&#xA;One such problem would be that bond formation due to orbital overlapping requires the electrons in a shared pair to be of opposite spins, so if all unpaired electrons are up-spin in an Nitrogen atom, then Dinitrogen shouldn't exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as per the title, Is one electron spin preferred over the other in filling of orbitals?&lt;/p&gt;&#xA;" OwnerUserId="1821" LastActivityDate="2015-04-23T11:02:31.783" Title="Is one electron spin preferred over the other in filling of orbitals?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;spin&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-04-23T13:05:18.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="29209" PostTypeId="2" ParentId="29190" CreationDate="2015-04-23T11:05:40.370" Score="5" Body="&lt;p&gt;To see these jelly flakes in an ultrasound bath is a normal thing -- referring to ``it happens regardless if the water was tap-water or deionized water, a just a matter of time''.  Perhaps this occurs faster in the former, than in the later, especially if the bath is used to clean glassware.  But it equally happens with glassware judged as clean (by eye) for example if the bath is deployed to initiate / support a reaction inside a reaction flask.  Ultrasound-assisted generation of Grignard reagents is one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultrasound bath works by sonication, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation&quot;&gt;cavitation&lt;/a&gt; simply detaches the oily/fatty/grease surface film covering quickly all what was touched once by naked hand (e.g. sweat), and breaks this into even smaller pieces.  Over time -- when the ultrasound bath is switched off --, this thermodynamically unstable emulsion decomposes, splitting into a water soluble phase, and a hydrophobic phase that then tends to agglomerate.  (I did not check the presence of organisms living in or on them since discharging the bath is a chore regularly done, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note:  For me, these residues appear to be more a liquid, then a solid -- hence I used the term emulsion.  If they were a solid, the thermodynamically unstable mixture would be an suspension.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-23T11:05:40.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29210" PostTypeId="2" ParentId="29126" CreationDate="2015-04-23T13:12:27.773" Score="9" Body="&lt;p&gt;To my understanding, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu(II)}$&lt;/span&gt; actually prefers four-fold coordination with ligands like water and ammonia, in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Square_planar_molecular_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;square planar&lt;/a&gt; configuration. Gas-phase DFT / MD simulations bear this out&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt;, at least for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;, and the pioneering work&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; of Jannik Bjerrum indicates, apropos Loong's answer, that the first four step-wise stability constants for the substitution of ammonia ligands for water are significantly higher than the fifth (IIRC the 6th was un-measurable at that time).  I'm pretty sure this is because the &lt;span class=&quot;math-container&quot;&gt;$\ce{[Ar]}3\mathrm d^9$&lt;/span&gt; ground state configuration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu(II)}$&lt;/span&gt; just doesn't have enough un-occupied &lt;em&gt;d&lt;/em&gt;-orbitals to support 'proper' six-coordination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're absolutely correct that &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; is a better ligand than water -- this explains why the first four molecules substitute readily into the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu(II)}$&lt;/span&gt; coordination environment.  However, if the remaining &quot;axial coordination sites&quot; are really not coordination sites at all, but instead are part of a non- or weakly-coordinating solvation shell, then it just becomes a statistical competition between excess &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; solutes and the &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; solvent, in which case the water will win out until the ammonia concentration is quite high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, the comparatively &lt;em&gt;d&lt;/em&gt;-electron-poor &lt;span class=&quot;math-container&quot;&gt;$\ce{Co(II)}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$(\ce{[Ar]}3\mathrm d^7)$&lt;/span&gt; has plenty of room in the &lt;span class=&quot;math-container&quot;&gt;$3\mathrm d$&lt;/span&gt; subshell to support six-coordination.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; Bérces et al. &lt;em&gt;J Phys Chem A&lt;/em&gt; &lt;strong&gt;103&lt;/strong&gt;: 9693-9701 (1999), doi: &lt;a href=&quot;http://dx.doi.org/10.1021/jp992612f&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1021/jp992612f&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; Bjerrum, J. &quot;Metal Ammine Formation in Aqueous Solution.&quot; Haase and son, 1941. (&lt;a href=&quot;https://books.google.com/books?id=1HTYcQAACAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="16683" LastEditDate="2019-01-07T19:07:10.853" LastActivityDate="2019-01-07T19:07:10.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29211" PostTypeId="1" AcceptedAnswerId="29215" CreationDate="2015-04-23T13:24:40.477" Score="7" ViewCount="75796" Body="&lt;p&gt;I know that relative atomic mass of $\ce{^{12}C}$ is $12~\mathrm{u}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore mass of $1~\mathrm{mol}~\ce{C} = 12~\mathrm{g}$&#xA;\begin{align}&#xA; \text{mass of }6.022\cdot 10^{23} \text{ C atoms} &amp;amp;= 12~\mathrm{g}\\ &#xA; \text{mass of }1 \text{ C atom} &amp;amp;= \frac{12~\mathrm{g}}{6.022\cdot10^{23}}\\&#xA;&amp;amp;=\boxed{1.99\cdot10^{-23}~\mathrm{g}},\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA; 1~\mathrm{u} &amp;amp;= 1.66\cdot10^{-24}~\mathrm{g}\\&#xA;\implies \text{Mass of }1~\ce{^{12}C}\text{ atom} &amp;amp;= 1.66\cdot 12 \cdot 10^{-24}~\mathrm{g}\\&#xA;&amp;amp;= 19.92\cdot 10^{-24}~\mathrm{g}\\&#xA;&amp;amp;= 1.992\cdot 10^{-23}~\mathrm{g}.\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what is dimension formula of relative atomic mass, molar mass? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is unit of molar mass $\dfrac{\text{gram}}{\text{mole}}$ and not just $\text{gram}$ Since when has this $\text{mol}$ become a unit? It's just a number.&lt;/p&gt;&#xA;" OwnerUserId="15763" LastEditorUserId="46334" LastEditDate="2018-01-24T15:30:17.303" LastActivityDate="2018-01-24T15:30:17.303" Title="What is the mass of 1 atom of carbon?" Tags="&lt;physical-chemistry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29212" PostTypeId="1" AcceptedAnswerId="29214" CreationDate="2015-04-23T13:46:23.750" Score="11" ViewCount="20149" Body="&lt;p&gt;Why does one mole of all gases occupy 22.4 L at standard temperature and pressure? According to my thinking, it may be due to equal diffusion of all particles when at a particular temperature and  pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So going by this, one mole of all gases should occupy the same $x~\mathrm{L}$ at some other temperature-pressure conditions. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="15763" LastEditorUserId="15763" LastEditDate="2015-04-23T16:58:46.560" LastActivityDate="2015-04-25T03:15:04.460" Title="Do all gases occupy same volume at equal temperature and pressure conditions?" Tags="&lt;physical-chemistry&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29213" PostTypeId="2" ParentId="29155" CreationDate="2015-04-23T13:52:06.897" Score="4" Body="&lt;p&gt;I think what you may be asking is: could there could be a nominally &quot;meso&quot; compound that has restricted rotation about its central single bond, causing it to exist in the form of two non-interconvertible stereoisomers. Example- start with meso-tartaric acid, and make the di-tert butyl ether from the OH groups and di-tert-butyl ester from the COOH groups. Maybe these groups would be bulky enough to restrict rotation. You could mentally rotate around the central C-C bond until the molecule has a plane of symmetry, so it is still a nominally &quot;meso&quot; compound, but physically you could separate the two stereoisomers.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-23T13:52:06.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29214" PostTypeId="2" ParentId="29212" CreationDate="2015-04-23T14:00:09.037" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do all gases occupy 22.4 L [&lt;strong&gt;per mol&lt;/strong&gt;] at STP? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question is based on a false premise. Only &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas&quot;&gt;ideal gases&lt;/a&gt; are guaranteed to occupy 22.4 L/mol at STP.  There are many gases that are not ideal.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So going by this, all gases should occupy same $x$ L at some other temperature-pressure conditions. Is this true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, again this would be true only for ideal gases, not all gases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At STP, many commonly encountered gases on Earth behave nearly ideally, but certainly not all.  For example, vapor of acetic acid and other simple carboxylic acids show strong deviations from ideality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia is a good place to read about the assumptions of the ideal gas law and the kinetic theory of gases.  The reason that &lt;em&gt;ideal&lt;/em&gt; gases share a molar volume of 22.4 L at STP is not related to diffusion, but rather to the assumption of: (a) non-interacting gas particles that are (b) effectively zero-sized relative to the mean distance between them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read about equations of state for non-ideal gases on Wikipedia as well.  Popular equations to describe non-ideal gases include the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_equation&quot;&gt;van der Waals equation&lt;/a&gt;, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Redlich%E2%80%93Kwong_equation_of_state&quot;&gt;Redlich-Kwong equation&lt;/a&gt;, and a whole slew of others listed at Wikipedia's &lt;a href=&quot;http://en.wikipedia.org/wiki/Real_gas&quot;&gt;real gas&lt;/a&gt; page.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-04-23T15:35:07.247" LastActivityDate="2015-04-23T15:35:07.247" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29215" PostTypeId="2" ParentId="29211" CreationDate="2015-04-23T14:10:10.377" Score="9" Body="&lt;p&gt;Both approaches are correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?na%7Csearch_for=physchem_in!&quot;&gt;Avogadro's number is&lt;/a&gt; $6.02214129\times 10^{23}$ and represents the number of carbon-12 atoms in 12 grams of unbound carbon-12 in the ground electronic state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12$grams$/6.02214129\times 10^{23} = 1.9926467\times 10^{-23}$grams&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?tukg&quot;&gt;unified atomic mass unit (u) is&lt;/a&gt; $1.660538921 \times 10^{-24}$ grams&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 \times 1.660538921 \times 10^{-24}$ grams $ = 1.9926467\times 10^{-23}$grams&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-04-23T14:34:01.180" LastActivityDate="2015-04-23T14:34:01.180" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29217" PostTypeId="1" AcceptedAnswerId="29225" CreationDate="2015-04-23T15:42:46.797" Score="4" ViewCount="92" Body="&lt;p&gt;We need to calculate the heat produced by a set of $\mathrm{N_R}$ reactions in our reactor model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathrm{Q_r = \sum_{j=1}^{Nr} Rr_j*\Delta H_{R,j}}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Rr_j}$ is the reaction rate of reaction $\mathrm{j}$, in $\mathrm{\frac{kmol}{kg_{cat} s}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{\Delta H_{R,j}}$, in  $\mathrm{\frac{kJ}{kmol}}$, is constant and calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathrm{\Delta H_{R,j} = \sum_{i=1}^{Nc}\nu_{i,j}*\Delta h_{i,j}^f}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider these:&#xA;\begin{align}&#xA;\mathrm{R1:}\ \ce{H2 + 1/2O2 \rightarrow H_2O}&#xA;\end{align}&#xA;\begin{align}&#xA;\mathrm{R2:}\ \ce{2H2 + O2 \rightarrow 2H_2O}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are undoubtedly the same reaction, but the R2 gives a  $\mathrm{\Delta H_R}$ that is twice the $\mathrm{\Delta H_R}$ in R1. It is naturally so, but greatly affects my energy balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I know which is the &quot;correct&quot; way to write the reaction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean that the reaction rate should be specific to R1 or R2?&lt;/p&gt;&#xA;" OwnerUserId="15764" LastEditorUserId="4356" LastEditDate="2015-04-24T01:36:33.673" LastActivityDate="2015-04-24T01:36:33.673" Title="How do I know which is the &quot;correct&quot; way to write the reaction?" Tags="&lt;thermodynamics&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29218" PostTypeId="1" AcceptedAnswerId="29223" CreationDate="2015-04-23T15:54:25.183" Score="3" ViewCount="304" Body="&lt;p&gt;So for the reaction of $\ce{H_2O}$ with an alkyl ethene, in the presence of $\ce{D^+}$ ions, forms two possible carbocations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ R-CHD-CH2^+}$ and $\ce{R-CH^+-CH2D}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major product formed is the secondary carbocation as it is stabler than the primary carbocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the two carbocations in an equilibrium or are they formed in&#xA;specific, fixed quantities?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If they are in equilibrium, i.e., the primary and secondary&#xA;carbocation, why and how would the secondary carbocation revert back&#xA;to being a less stable primary carbocation?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, is the hydride shift that converts the primary to a secondary&#xA;carbocation the only shift that can occur; even if after the shift,&#xA;a more stable carbocation can be formed (depending on the attached&#xA;alkyl group)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="4231" LastEditDate="2015-04-23T17:22:07.100" LastActivityDate="2015-04-23T17:27:46.890" Title="Are carbocation intermediates in equilibrium?" Tags="&lt;organic-chemistry&gt;&lt;equilibrium&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29219" PostTypeId="1" AcceptedAnswerId="29261" CreationDate="2015-04-23T15:57:35.020" Score="5" ViewCount="2635" Body="&lt;p&gt;When I freeze water with some impurities (1% NaCl by mass), the mixture seems to have a much lower enthalpy of fusion that pure water. The impurities cause the &lt;a href=&quot;http://en.wikipedia.org/wiki/Colligative_properties#Freezing_point_depression_.28cryoscopy.29&quot; rel=&quot;nofollow&quot;&gt;freezing point to depress&lt;/a&gt; to 0.6°C. I only have a simple calorimeter available but it looks like the enthalpy of fusion is dropping by 25%+ compared to pure water (tested by melting the ice into water and measuring the temperature drop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the ice does not have a nice crystal structure. It is &quot;mushy&quot; and easily deformed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This effect is surprising to me since colligative theory tells us that most properties changes in a pretty small linear way at low concentrations (like the freezing point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I observing a real phenomenon or is there some problem with my testing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its a real phenomenon, is there a way to calculate/estimate the effect?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorUserId="4945" LastEditDate="2015-04-24T07:47:46.493" LastActivityDate="2020-11-19T16:44:22.417" Title="Effect of impurities on enthalpy of fusion of ice" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29220" PostTypeId="2" ParentId="29182" CreationDate="2015-04-23T16:27:42.323" Score="7" Body="&lt;p&gt;Yes, you are right. The abundances of (radioactive and stable) nuclides change due to radioactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding this, most environmental radionuclides can be divided into four categories:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cosmogenic radionuclides&lt;/strong&gt; are constantly produced by cosmic irradiation of the atmosphere. Therefore, their average concentrations in the atmosphere stay roughly constant even if their half-lives are short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Primordial radionuclides&lt;/strong&gt; have lifetimes comparable to the age of the earth. These nuclides have been formed at the time of the formation of the earth, the solar system, or even before. Their abundance is slowly decreasing due to radioactive decay. Therefore, the radiation exposure was higher some billion years ago when life evolved on earth than today. Furthermore, due to different half-lives, the &lt;em&gt;relative&lt;/em&gt; abundance of these nuclides (e.g. the natural $\ce{^{235}U/^{238}U}$ isotopic ratio) also changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radionuclides that are part of the natural &lt;strong&gt;decay chains&lt;/strong&gt; of $\ce{^{238}U}$, $\ce{^{235}U}$ and $\ce{^{232}Th}$ are constantly reproduced by the radioactive decay of their mother nuclides. Therefore, their abundance is only slowly decreasing with the long half-life of $\ce{^{238}U}$, $\ce{^{235}U}$, or $\ce{^{232}Th}$ even if their own half-lives are short (this situation is called &lt;em&gt;secular equilibrium&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anthropogenic radionuclides&lt;/strong&gt; are produced by modern technology. Since the discovery of nuclear fission, the activity concentrations of various anthropogenic radionuclides in the environment were increasing until the limited nuclear-test-ban treaty banned atmospheric tests of nuclear weapons. After that, the activity concentrations of most anthropogenic radionuclides decreased again. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you isolate and chemically purify a radioactive sample, you remove any other mother or daughter nuclides and thus you disturb any decay chain or radioactive equilibrium. Thereby you start a &lt;em&gt;radioactive clock.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a solid rock sample may contain $\ce{^{226}Ra}$ $\left(t_{1/2}=1600\ \mathrm{a}\right)$ and its daughter nuclide $\ce{^{222}Rn}$ $\left(t_{1/2}=3.83\ \mathrm{d}\right)$ in a secular equilibrium. If you dissolve the sample and release the $\ce{^{222}Rn}$ (radon is a noble gas), the released $\ce{^{222}Rn}$ decays with its half-life of 3.83 days. If you purify the $\ce{^{226}Ra}$ and put it in a closed container, initially the container does not contain any $\ce{^{222}Rn}$, but new $\ce{^{222}Rn}$ is produced by radioactive decay of $\ce{^{226}Ra}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar thing occurs when natural minerals solidify. Radionuclides become trapped and decay. As they decay, decay products accumulate in the closed mineral.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By measuring the amount of mother and daughter nuclide, it is possible to calculate how long this sample has existed. This is the basis for &lt;em&gt;nuclear dating&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you isolate a sample that contains multiple radionuclides of the same chemical element, you can observe changing isotopic ratios – as you have expected in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example, freshly purified thorium from a natural sample, mainly contains six different thorium nuclides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{^{234}Th}$ $\left(t_{1/2}=24.10\ \mathrm{d}\right)$&lt;br&gt;&#xA;$\ce{^{232}Th}$ $\left(t_{1/2}=1.405\cdot10^{10}\ \mathrm{a}\right)$&lt;br&gt;&#xA;$\ce{^{231}Th}$ $\left(t_{1/2}=25.5\ \mathrm{h}\right)$&lt;br&gt;&#xA;$\ce{^{230}Th}$ $\left(t_{1/2}=7.54\cdot10^{4}\ \mathrm{a}\right)$&lt;br&gt;&#xA;$\ce{^{228}Th}$ $\left(t_{1/2}=1.91\ \mathrm{a}\right)$&lt;br&gt;&#xA;$\ce{^{227}Th}$ $\left(t_{1/2}=18.7\ \mathrm{d}\right)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual half-lives differ significantly; thus, the different thorium nuclides decay at different rates. The isotopic ratios change accordingly.&lt;br&gt;&#xA;However, the calculation is slightly complicated since $\ce{^{234}Th}$ indirectly decays to $\ce{^{230}Th}$, $\ce{^{231}Th}$ indirectly decays to $\ce{^{227}Th}$, and $\ce{^{232}Th}$ indirectly decays to $\ce{^{228}Th}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="402" LastEditDate="2017-04-02T19:54:55.620" LastActivityDate="2017-04-02T19:54:55.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29223" PostTypeId="2" ParentId="29218" CreationDate="2015-04-23T17:19:01.683" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-CH=CH2 + H^+ &amp;lt;=&amp;gt; R-CH2-CH2^+ + RCH^+-CH3 &amp;lt;=&amp;gt; alcohol~products}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes, the two carbocations formed are in equilibrium&lt;/strong&gt;. Initially the two carbocations will be formed in a ratio dependent upon the difference in activation energies leading to them (kinetic control).  However, since all of the steps in this reaction involve equilibria, after a while when equilibrium is established several things can happen&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;many carbocations will revert back to starting material which can then be reprotonated and reform the carbocations&lt;/li&gt;&#xA;&lt;li&gt;or alcoholic product can be reprotonated, eliminate water and reform the carbocations&lt;/li&gt;&#xA;&lt;li&gt;or the carbocations themselves can equilibrate directly by hydrogen shift&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once the equilibrium is established, the ratio of carbocations formed will be dependent upon the energy difference between the carbocations themselves (thermodynamic control).  Often, but not always, the relative stabilities of the transition states leading to the carbocations matches the relative stabilities of the carbocation intermediates themselves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back&#xA;  to being a less stable primary carbocation?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;why - that's what happens in equibria; even though phenol is much, much more stable than its enol counterpart, there will be an exceedingly small amount of the enol present at equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4XSZZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how - by any of the 3 routes mentioned in the bullets above&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a&#xA;  more stable carbocation can be formed (depending on the attached alkyl&#xA;  group)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other groups like alkyl or phenyl can shift too, but hydrogen shift is usually faster than the others.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-23T17:19:01.683" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29224" PostTypeId="2" ParentId="29218" CreationDate="2015-04-23T17:27:46.890" Score="0" Body="&lt;p&gt;Actually four possible intermediates are there, two from autolysis of water and two from added deutrium.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RXneF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $\rm 1&amp;gt;2\gg3&amp;gt;4$&lt;br&gt; Since autolysis of water is very minute and a secondary carbocation is always stable than a primary carbocation since it has more hyperconjugations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) They would be in equilibrium since every reaction is reversible, though it's rate might be minute for a specific direction. There could also be one in a gazillion hydride shift case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) This depends on the migratory aptitude which is generalised as $\rm H&amp;gt;Ph&amp;gt;1^\circ&amp;gt;2^\circ&amp;gt;3^\circ$&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-23T17:27:46.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29225" PostTypeId="2" ParentId="29217" CreationDate="2015-04-23T17:41:06.840" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I know which is the &quot;correct&quot; way to write the reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no single correct way.  Picking a different way to write the reaction affects both $\Delta H_{R_j}$ and $Rr_j$ in opposite ways that cancel out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it mean that the reaction rate should be specific to R1 or R2?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.  The unit of your reaction rate $Rr_j$ is $\frac{\textrm{kmol}}{\textrm{kg}_{cat}}$.  But the key question is &lt;strong&gt;kmol of what&lt;/strong&gt;?  A good way to think about the answer is &quot;&lt;em&gt;kmol of things that have a stoichiometric coefficient of 1 in the reaction&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the reaction is R2, then the rate of the reaction will be the rate of disappearance of $\ce{O2}$.  The rate of formation of water will be twice the reaction rate $Rr_2$.  But if the reaction is R1, then the reaction rate is equal to the rate of formation of water (or the rate of depletion of hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you don't directly measure &quot;reaction&quot; rates, you can only measure the formation or disappearance of products or reactants.  So say you measured the formation rate of water.  If you decide the reaction is R2, then $Rr_2 = \frac{1}{2}R_{\ce{H2O}}$.  If you decide the reaction is R1, then $Rr_1 = R_{\ce{H2O}}$.  The factor of $\frac{1}{2}$ will cancel out the factor of two in $\Delta H_R$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end it shouldn't matter what choice you make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the exact same idea in a totally different chemical context, see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28954/does-the-relationship-equation-between-standard-cell-potential-and-equilibrium-c/28959#28959&quot;&gt;this se.chem question on electrochemistry&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-04-23T17:47:35.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29227" PostTypeId="2" ParentId="29180" CreationDate="2015-04-23T17:47:19.757" Score="2" Body="&lt;p&gt;&quot;Maximum Additional Work that can be extracted at constant pressure and temperature&quot; to be precise, which may include electrical work, etc.  It means the work besides the expansion work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative sign signifies the sign of the charge of the electron, since electrical work is proportional to the charge.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-23T17:47:19.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29228" PostTypeId="1" CreationDate="2015-04-23T18:52:26.430" Score="5" ViewCount="102" Body="&lt;p&gt;When doing TLCs to monitor reactions, it is considered good common practice to not only spot the reactants and the reaction mixture (likely after some mini-workup), but also to include a ‘cross-spot’: every other substance (mixture) spotted somewhere on the plate is spotted together on one spot, commonly labelled something like &lt;em&gt;X&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When asking why during my bachelor's thesis, the PhD student supervising me said something that roughly translates to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You can't be sure that the components don't run differently in the mixture than by themselves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But in a perfect world, they should. Aside from the obvious case of having a polar reaction solvent that didn't entirely evaporate and skewes the elution mixture on parts of the TLC, are there other reasons why the components of a mixture should have different R&lt;sub&gt;f&lt;/sub&gt; values when spotted separately?&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-04-24T23:40:09.080" LastActivityDate="2015-04-24T23:40:09.080" Title="What mechanisms might make a substance travel different distances on the same TLC plate?" Tags="&lt;experimental-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29229" PostTypeId="1" CreationDate="2015-04-23T19:00:10.183" Score="3" ViewCount="177" Body="&lt;p&gt;I am taking the instrumental methods of analysis course. Our job is to find the unknown  mass of $\ce{Fe}$ which is added, by measuring the absorbance of different solutions with ortho-fenantroline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I can know if my calculated mass of $\ce{Fe}$ is correct or not?   Is there any way I can prove it, since I know the absorbances? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="10256" LastEditDate="2015-04-23T20:11:09.217" LastActivityDate="2015-04-23T20:11:09.217" Title="Instrumental methods of analysis" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29230" PostTypeId="2" ParentId="29228" CreationDate="2015-04-23T19:09:17.983" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;are there other reasons why the components of a mixture should have different Rf values when spotted separately?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One might speculate that two polar reactants form a less polar &quot;adduct&quot;. This should however result in an additional spot, rather than in altered $\mathrm{R_f}$ values for the individual compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is speculation. Admittedly, I have never used the &lt;em&gt;cross&lt;/em&gt; spot method. Neither have I ever observed that spots of the reactants had different $\mathrm{R_f}$ values in the reaction mixture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note: What always worked for me was:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Don't apply spots but bands and overlap them.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Sharpen&quot; the bands on the base line by developping the TLC in methanol or another polar solvent first. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Together, these two tricks allow to destinguish between two compounds with very similar  $\mathrm{R_f}$ values.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-23T19:09:17.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29231" PostTypeId="2" ParentId="29229" CreationDate="2015-04-23T19:16:22.820" Score="2" Body="&lt;p&gt;Am I right that you are planning to simply calculate the $\ce{Fe}$ concentration using the Lambert-Beer equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious proof would be to prepare $\ce{Fe}$ solutions with known concentration(s) to compare the absorbances. Using these solutions, you can also make sure that the assumption of linearity over the concentration range is actually fulfilled.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-23T19:16:22.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29232" PostTypeId="1" CreationDate="2015-04-23T20:11:00.790" Score="5" ViewCount="96" Body="&lt;p&gt;Can someone tell me if quantum yield of a reaction factors in the number of electrons needed to create the particular species? If not, is there a metric that does?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a 4 electron reduction which means that even if I have 100% efficiency, my quantum yield is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{# molecules reduced}}{\text{# photons absorbed}} = 0.25$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better metric I can use so that 1.0 will be 100% efficiency?&lt;/p&gt;&#xA;" OwnerUserId="7614" LastEditorUserId="41328" LastEditDate="2017-09-19T04:39:00.147" LastActivityDate="2017-09-19T04:39:00.147" Title="Does quantum yield of reaction factor in number of electrons needed?" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29234" PostTypeId="2" ParentId="29232" CreationDate="2015-04-23T20:51:44.347" Score="5" Body="&lt;p&gt;I'm not aware of a definition of quantum yield that factors the electrons in. The normative reference was/is Silvia Braslavsky's &lt;em&gt;Glossary of terms used in photochemistry&lt;/em&gt; (&lt;a href=&quot;http://www.degruyter.com/view/j/pac.2007.79.issue-3/pac200779030293/pac200779030293.xml&quot; rel=&quot;noreferrer&quot;&gt;SRC&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I misunderstood, you do have an overall four electron reduction, but this happens in a sequence of four excitation events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a formal point of view, the quantum yield(s) should be given for each of these processes. I reckon that this isn't possible, so an overall quantum yield of 0.25 is all you can get. In the end, I'd simply explain that, rather that looking for some &quot;tricks&quot; to come up with a more impressive number. &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-23T20:51:44.347" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29235" PostTypeId="2" ParentId="18489" CreationDate="2015-04-23T20:58:42.993" Score="7" Body="&lt;p&gt;Your professor does have a case in that many carbocations can be thought of as resonance stabilised, as was depicted by jerepierre in &lt;a href=&quot;https://chemistry.stackexchange.com/a/18519/7475&quot;&gt;his answer&lt;/a&gt;. I once had a professor who would prefer any resonance structure that has complete octets on all p-group elements over any resonance structure that doesn't have that (although he didn't go as far as downgrading for sextet cations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are also certainly reactions that &lt;strong&gt;cannot&lt;/strong&gt; be thought of as concerted, and also &lt;strong&gt;cannot&lt;/strong&gt; be resonance-stabilised to give an all octet structure: See for example the electrophilic aromatic substitution as depicted by this mechanism (sheepishly stealing the image from &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrophilic_aromatic_substitution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDPuH.png&quot; alt=&quot;basic SEAr-mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism as shown here is for unsubstituted benzene. I did it in an organic practical course with nitrobenzene. There is no way in which nitrobenzene will stabilise a carbocation (indeed, the nitro group is &lt;em&gt;meta&lt;/em&gt; directing due to that). And the mechanism can — to the best of my knowledge — cannot be thought of as concerted.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-04-23T20:58:42.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29236" PostTypeId="1" CreationDate="2015-04-23T21:14:42.553" Score="4" ViewCount="128" Body="&lt;p&gt;I had broken a CFL partially on carpet last week. It was a spent bulb, and the outer glass shell broke but not the fluorescent tube - apparently the tube did have a small hole in it which found out after all my other steps when I saw it in the garbage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I cleaned it up first, I used a mild spray cleaner which contained methyl esters and various surfactants - I found this out now, not then, or I wouldn't hzve used it. Wiped it all up with a paper towel, then vacuumed in case I missed small pieces. I did ventilated at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About two hours later, my cat made a mess on the rug. At that point, I sprayed down the carpet with resolve (which says it contains hydrogen peroxide). Wiped that all down with a towel, then revacuumed...and realized I hadn't cleaned out the dust cup. I think I then emptied that outside and cleaned out the HEPA filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern now is what I may have exposed myself to in that botched cleanup. Any ideas on what may have formed? Do I need to be concerned?&lt;/p&gt;&#xA;" OwnerUserId="15725" LastEditorUserId="4356" LastEditDate="2015-04-24T05:27:09.507" LastActivityDate="2015-04-24T05:27:09.507" Title="Can organomercury compounds synthesize accidentally from household cleaning products?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-04-24T07:16:10.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="29237" PostTypeId="2" ParentId="28231" CreationDate="2015-04-23T21:20:41.427" Score="5" Body="&lt;p&gt;There are a number of types of rapid chemical equilibria that basically cannot be turned off. Tautomers are probably the best known examples, although &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluxional_molecule&quot; rel=&quot;noreferrer&quot;&gt;fluxional&lt;/a&gt; behavior is observed in many inorganic and organometallic systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin for most tautomers is that protons are really quantum-mechanical. They can tunnel just like electrons can delocalize. Obviously different tautomers will have different stability, much like some electric resonance structures are clearly more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comments to your question above indicate that &lt;em&gt;most&lt;/em&gt; chemists would just call it &quot;acetone&quot; but of course there's some prop-1-en-2-ol present too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the reactivities might be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a partly a matter of the balance of the equilibrium (i.e., that for many tautomers, only one form predominates). There are exceptions, e.g. &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism#Phenols&quot; rel=&quot;noreferrer&quot;&gt;phenols (from Wikipedia)&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In certain aromatic compounds such as phenol, the enol is important due to the aromatic character of the enol but not the keto form. Melting the naphthalene derivative naphthalene-1,4-diol 2 at 200 °C results in a 2:1 mixture with the keto form.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's also a matter of the time-scale. For example, at room temperature if I take an NMR, do I see two sets of peaks? What about IR or other characterization techniques? If I can't tell the difference because there are one set of peaks, I think many chemists will consider there to be &quot;one compound&quot; even if multiple tautomers are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example of this (from the cheminformatics perspective), Marc Nicklaus of the NIH/National Cancer Institute gave a great talk at the 247th ACS national meeting &lt;a href=&quot;http://bulletin.acscinf.org/PDFs/247nm56.pdf&quot; rel=&quot;noreferrer&quot;&gt;&quot;The NCI/CADD Group's InChI Usage and Analysis of Tautomerism for InChI V2&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The talk focuses on tautomers and what software is trying to do to derive simple rules (as opposed to lots of quantum calculations or experiments) to determine one standard tautomer. But they also measured several tautomers by NMR:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These would obviously be chemically quite different, but one form clearly predominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5eilh.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tAyGp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-04-23T21:20:41.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29238" PostTypeId="1" CreationDate="2015-04-23T21:31:27.923" Score="7" ViewCount="223" Body="&lt;p&gt;I hope this is the best place to pose this unique question.  Please forgive me if it's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went into a closet to get a prescription medication that was filled about a year ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The label on the pill bottle has visible signs of &lt;strong&gt;&lt;em&gt;something&lt;/em&gt;&lt;/strong&gt;.   Identifying that something is the purpose of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;something&quot; can be described as light brownish orange circles of discoloration.  They look a lot like mold, but I'm hoping that's not the case.  I tried to photograph them, but they don't show up well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a look at a few other prescription bottles from that time period, and the labels on them also have this appearance.  Labels on &lt;em&gt;newer&lt;/em&gt; bottles from the same pharmacy do not have this appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one bottle, there are also discolorations that are yellow, similar to what a yellow highlighter would produce (but not quite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is mold, but I'm hoping it's not.  I don't see other signs of visible mold in the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closet is not near a bathroom.  The room is air conditioned and heated (depending on the season) and is kept at approximately 76 degrees Fahrenheit, and never outside the range of 61-89 degrees.  There is a MERV 16 filter on the HVAC system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pharmacy is very clean and is fairly new (probably less than 10 years old).  The entire pharmacy chain is probably less than 15-20 years old.  They compound medications within the confines of a special compounding enclosure; none of my medications were compounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to bias the answers, but I am hoping that the brownish orange circles can be explained by a chemical reaction.  I'm thinking that perhaps pharmacist hands are so covered with different chemicals that it could cause a reaction on the label over time.  I don't think the labels had the discolorations when I picked up the medications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The areas on the labels that show the light brownish orange circles could possibly be areas where the pharmacist pushed down the label.  Areas in which people make contact with items have increased probability of growing mold; but given that a pharmacist was the one who touched these labels, there is also the increased possibility that their fingers were coating with small amounts of medications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above facts and descriptions, are there any likely chemical reactions that would cause the observed phenomena?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please note that I'm not a chemist, so please write answers accordingly!&lt;/em&gt;  :-)&lt;/p&gt;&#xA;" OwnerUserId="15770" LastActivityDate="2015-04-23T21:31:27.923" Title="Solve this chemical or biological mystery" Tags="&lt;reaction-mechanism&gt;&lt;biochemistry&gt;&lt;reactivity&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29240" PostTypeId="2" ParentId="29236" CreationDate="2015-04-23T21:50:27.293" Score="2" Body="&lt;p&gt;Unfortunately, you seem to have already broken some of the protocols of cleaning up a broken CFL according to the &lt;a href=&quot;http://www2.epa.gov/cfl/cleaning-broken-cfl&quot; rel=&quot;nofollow&quot;&gt;EPA's website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Vacuuming is not recommended unless broken glass remains after all other cleanup steps have been taken.  Vacuuming could spread mercury-containing powder or mercury vapor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also according to data from a study of the effect of the breakage of CFLs performed by the &lt;a href=&quot;http://www.maine.gov/dep/homeowner/cflreport.html&quot; rel=&quot;nofollow&quot;&gt;Maine Department of Environmental Protection&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mercury concentration in the study room air often exceeds the Maine Ambient Air Guideline (MAAG) of 300 nanograms per cubic meter (ng/m3) for some period of time, with short excursions over 25,000 ng/m3, sometimes over 50,000 ng/m3, and possibly over 100,000 ng/m3 from the breakage of a single compact fluorescent lamp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Vacuuming a carpet where a lamp has broken and been visibly cleaned up, even weeks after the cleanup, can elevate the mercury readings over the MAAG in an un-vented room.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the end of the report, it states that although mercury vapor is extremely toxic:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...it is unclear what the exact health risks are from exposure to low levels of elemental mercury...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems as though the best course of action you could take now would be to ventilate the room and if possible your house extremely well for several hours, and to avoid vacuuming any more.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-04-23T21:58:12.737" LastActivityDate="2015-04-23T21:58:12.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29244" PostTypeId="2" ParentId="29187" CreationDate="2015-04-24T00:51:12.920" Score="10" Body="&lt;p&gt;The only reason I can think of that would cause the solutions would blacken over time is if the solutions reacted with organic substances. Sulfuric acid is an extremely strong oxidizing agent, and dehydrates many organic compounds to produce carbon in the form of graphite, and water. For example, the reaction of sulfuric acid with paper (cellulose) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\ce{C6H10O5})_{n}\ce{(s)} \xrightarrow{\ce{H2SO4}} {6n~\ce{C(s)} + 5n~\ce{H2O(l)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and can be seen in the following photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XzLco.jpg&quot; alt=&quot;Graphite Formation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a considerable quantity of organic material was reacted in your solutions, the $[\ce{H+}]$ in your solutions would actually decrease, so I would recommend titrating one of your solutions to see if this has changed since you bought them.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2016-10-11T00:16:30.230" LastActivityDate="2016-10-11T00:16:30.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29246" PostTypeId="1" CreationDate="2015-04-24T01:32:11.363" Score="5" ViewCount="6819" Body="&lt;p&gt;When I place a straw in carbontated drink and blow (making bubbling noises), more gas bubbles rise to the surface. The gas bubbles which rise to the surface are from the carbon dioxide gas dissolved in the drink, I'm not talking about the air bubbles I made with the straw? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="16447" LastEditDate="2015-05-23T02:33:20.393" LastActivityDate="2019-03-21T03:25:24.963" Title="When I place a straw in carbontated drink and blow, more gas bubbles rise to the surface?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29247" PostTypeId="2" ParentId="29246" CreationDate="2015-04-24T01:39:57.523" Score="6" Body="&lt;p&gt;When you blow into a solution, your warm breath (~98.6°F) warms the solution up. As a solution warms up, less $\ce{CO2}$ is soluble.  Therefore, blow into a solution, warm it up, less $\ce{CO2}$ is soluble, $\ce{CO2}$ escapes (bubbles out) from the warmer solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KSqCn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.rocketscientistsjournal.com/2006/10/co2_acquittal.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-24T01:39:57.523" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="29248" PostTypeId="2" ParentId="29196" CreationDate="2015-04-24T02:14:51.473" Score="10" Body="&lt;p&gt;As noted in the comments, the premise is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually: &lt;strong&gt;Lead is the heaviest stable element.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are heavier elements unstable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic nucleus consists of neutrons and protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without going into detail, these are held together by short range nuclear forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protons also repel each other due to their positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we add more protons, the charge repulsion continues to build. Each pair of protons generates a repulsive force, so this grows in total as the square of the number of protons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short range attractive force does not grow quadratically in the same way, as 'distant' nucleons do not feel as much of an effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect is that above lead the repulsion is strong enough that the overall energy of a lower number and the radiated particle is less than that of the heavy nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in total energy drives the radioactive decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the elements get heavier this energy difference increases resulting in faster decays. This is complicated somewhat by the nuclear structure (the same concept as the electronic structure) in that completed nuclear shells create stability.&lt;/p&gt;&#xA;" OwnerUserId="15781" LastActivityDate="2015-04-24T02:14:51.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29249" PostTypeId="1" CreationDate="2015-04-24T03:28:59.617" Score="6" ViewCount="1110" Body="&lt;p&gt;If you were to apply the Huckel method to some molecules which belonged to, say, $C_{2v}, D_{2h}, $ and $ D_{4h}$, which ones would you expect to have degeneracies? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is mixing up math jargon with chemistry jargon here, but I know the Mulliken symbol E is doubly degenerate and T is triply degenerate. So would the $D_{4h}$ molecule have degeneracies because it has the irreducible representations $E_g$ and $E_u$?&lt;/p&gt;&#xA;" OwnerUserId="10207" LastEditorUserId="10207" LastEditDate="2015-04-24T03:34:57.593" LastActivityDate="2015-06-09T20:00:31.180" Title="How could you predict degenerate pi molecular orbitals just knowing what symmetry point group a molecule belongs to?" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29250" PostTypeId="1" AcceptedAnswerId="31430" CreationDate="2015-04-24T03:32:32.203" Score="19" ViewCount="1845" Body="&lt;p&gt;In my inorganic class, we carried out the synthesis of chlorotribenzyltin, &lt;span class=&quot;math-container&quot;&gt;$\ce{SnCl(CH2C6H5)3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zdqx3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The synthesis begins with a reflux I'll describe below:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a hood, place 2.0 g (17 mmol) of 325 mesh tin powder, 4.0 mL of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; and 6.0 mL (6.6 g, 52 mmol) of benzyl chloride in a 50 mL round-bottom flask. After the reaction mixture has been heated to reflux for 3 hours*, remove the heat and cool the flask in ice. The lower liquid phase should solidify to form a white solid.&lt;/p&gt;&#xA;&lt;p&gt;*The white solids that appeared shortly after refluxing commenced should have largely disappeared, although small amounts of white or gray solids on the sides of the flask are usually still present. The reaction mixture at this point should consist of two liquid phases, plus a small amount of unreacted tin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So the question the textbook asks is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I have a number of ideas as to what this could be:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The chloride bonded to the tin forming some kind of salt like&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SnCl2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reaction did some kind of partial reaction creating something like&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SnCl2R2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The reaction went too far and created&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SnR4}$$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$$\ce{SnCl4}$$&lt;/span&gt; and keeping them in reflux redistributed the ligands  based on stoichiometric amounts like&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SnCl4 +3SnR4-&amp;gt;4SnClR3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This last one seems most likely based on supporting evidence from the text:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{SnCl4}$&lt;/span&gt; reactions with Grignard reagents to give organotin compounds:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SnCl4 +4RMgX-&amp;gt;SnR4 +4MgXCl}$$&lt;/span&gt; ...&#xA;(it goes on to say)...the product, chlorotribenzyltin &lt;span class=&quot;math-container&quot;&gt;$\ce{SnCl(CH2C6H5)3}$&lt;/span&gt;, is a colorless solid that is stable in air for several days. In some respects, the reaction is analogous to the preparation of Grignard reagents, &lt;span class=&quot;math-container&quot;&gt;$\ce{RMgX}$&lt;/span&gt; by treatment of magnesium metal with organic halides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is my reasoning for choice 3 solid? Is there something I'm not thinking of?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-08-30T13:39:46.453" Title="Propose a chemical formula for the white solid that forms during the initial stages of the reaction of Sn with benzyl chloride" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29251" PostTypeId="1" AcceptedAnswerId="29252" CreationDate="2015-04-24T04:35:07.843" Score="23" ViewCount="18264" Body="&lt;p&gt;Why is it so that azo coupling with phenol needs basic pH (9–10) and with aniline it needs acidic pH (4–5). How does pH participate in the mechanism? I have searched the internet for quite a while nowhere does it explain the actual ‘role’ of the pH. Any light on this matter would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorDisplayName="user7951" LastEditDate="2015-04-24T09:43:06.187" LastActivityDate="2016-07-30T15:22:54.480" Title="What is the role of pH in azo coupling reaction of diazonium with phenol and aniline?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29252" PostTypeId="2" ParentId="29251" CreationDate="2015-04-24T06:17:06.807" Score="18" Body="&lt;p&gt;Deprotonation of the phenol and protonation of aniline result in species that easily react with a diazonium cation in the intended manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at the species involved. On the one hand, there is the &lt;strong&gt;diazonium cation&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhbMx.png&quot; alt=&quot;diazonium cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deprotonation of phenol yields phenolate, for which a resonance structure with a negative charge in &lt;em&gt;para&lt;/em&gt; position to the substituent can be written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FrlEw.png&quot; alt=&quot;phenolate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude: &lt;strong&gt;Deprotonation of the phenol means activation&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qCLuD.png&quot; alt=&quot;phenolate reaction with diazonium cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of aniline, the lone pair on the nitrogen atom is the preferred centre for the reaction with the diazonium cation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHGR4.png&quot; alt=&quot;aniline reaction with diazonium&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't what we want! Protonation of aniline avoids this side reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We conclude: &lt;strong&gt;Protonation of aniline means &quot;masking&quot; the wrong centre!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-24T06:17:06.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29253" PostTypeId="2" ParentId="28318" CreationDate="2015-04-24T06:21:10.397" Score="2" Body="&lt;p&gt;Short answer (After some trials): &lt;strong&gt;It does not matter which dissociation reaction you use&lt;/strong&gt;, The system of equation will be solved anyway since the reactions are not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using the first reaction as a dissociation mechanism would generate a Kp equation that's a lot easier to solve than the others.&lt;/p&gt;&#xA;" OwnerUserId="15414" LastEditorUserId="15414" LastEditDate="2015-04-24T06:57:59.333" LastActivityDate="2015-04-24T06:57:59.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29255" PostTypeId="2" ParentId="29168" CreationDate="2015-04-24T09:13:57.633" Score="17" Body="&lt;p&gt;I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hybridisation is a model, that can be used to describe a bonding situation.&lt;/strong&gt; It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking it is not even necessary to be invoked, but it can be quite handy. Quoting from &lt;a href=&quot;https://chemistry.stackexchange.com/a/2771/4945&quot;&gt;Aditya's answer&lt;/a&gt; - I could not have said it better:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\ce{sp^2}$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning.&lt;br&gt;&#xA;  &lt;em&gt;Note: Only orbitals can be hybridised, never compounds.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Having understood that, it is easy to see, that atoms in a molecule could use many different hybridisations schemes. There are some concepts, based on quantum chemical calculations, which would find the most appropriate fit between a Lewis structure and the molecular orbital approach. See for example this fine question: &lt;a href=&quot;https://chemistry.stackexchange.com/q/285/4945&quot;&gt;What is natural bond orbital theory used for?&lt;/a&gt; There is also a more quantitative approach, which is called Valence Bond theory, but that is beyond the scope of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is just a mathematical concept, an equilibration between $\ce{sp^2}$ and $\ce{sp^3}$ would not change a thing. We can decide which way we want to look at it. Ron consequently raised the question in the &lt;a href=&quot;https://chemistry.stackexchange.com/users/4231/ron&quot;&gt;comments&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't understand how &quot;6&quot; can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) &lt;strong&gt;and also be&lt;/strong&gt; rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I pointed out, (4) should not even be an option. The premise of the question is clearly flawed. Please do not take hybridisation too seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I come to my next point. Are the oxygen's orbitals in carbon acids really $\ce{sp^2}$ (or even $\ce{sp^3}$) hybridised? Well.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is often appropriate to describe oxygen's orbitals as $\ce{sp^2}$, when it is bonded to two other elements. Two $\ce{sp^2}$ lobes are needed for the bonds, the remaining $\ce{sp^2}$ orbital hosts one lone pair, the remaining $\ce{p}$ orbital hosts the other lone pair.  &lt;/li&gt;&#xA;&lt;li&gt;More often the case is that oxygen has only one bonding partner. In these cases it is more appropriate to describe the oxygen with $\ce{sp}$ orbitals due to symmetry reasons. Consider the local symmetry to have an infinite rotational axis $C_\infty$, hence there have to be two degenerate, perpendicular orbitals. In most of these cases you will also have (partial) &amp;pi;-bonding and/or hyperconjugation.  &lt;/li&gt;&#xA;&lt;li&gt;There are very few exceptions, like $\ce{H3O+}$, where it is appropriate to describe the oxygen with $\ce{sp^3}$ orbitals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all of this it should be considered that hybridisation can be fractional. See the last note &lt;a href=&quot;https://chemistry.stackexchange.com/a/26561/4945&quot;&gt;here&lt;/a&gt; and more consequently &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/15620/4945&quot;&gt;What is Bent&amp;#39;s rule?&lt;/a&gt;&quot; and &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/15671/4945&quot;&gt;Utility of Bent&amp;#39;s Rule - What can Bent&amp;#39;s rule explain that other qualitative considerations cannot?&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important point to remember is, that when there is a &amp;pi;-bond, the hybridisation can be $\ce{sp^2}$ at maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that answer 6 is really about &lt;strong&gt;a wrong interpretation&lt;/strong&gt; of resonance stabilisation of the molecule. (Please see &lt;a href=&quot;https://chemistry.stackexchange.com/q/24509/4945&quot;&gt;Why is there a need for resonance?&lt;/a&gt; for more information)&#xA;Carbon acids can be described in the resonance scheme below. Now it appears, that the oxygen employs $\ce{sp^2}$ orbitals in the first structure (A) and $\ce{sp^3}$ in the others (B, C). It is important to recognise, that the bonding situation in this molecule is a superposition of all (possible) resonance structures. Theses structures are &lt;strong&gt;not&lt;/strong&gt; in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aemew.jpg&quot; alt=&quot;resonance carbon acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion there are several issues with this question and you might want to address this with your teacher.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2018-02-19T11:23:09.353" LastActivityDate="2018-02-19T11:23:09.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29256" PostTypeId="2" ParentId="29101" CreationDate="2015-04-24T10:27:09.500" Score="12" Body="&lt;p&gt;The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: &lt;a href=&quot;https://chemistry.stackexchange.com/q/29142/4945&quot;&gt;Hypervalency and the octet rule&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stated in &lt;a href=&quot;https://chemistry.stackexchange.com/q/13483/4945&quot;&gt;Is the bond enthalpy of S-O stronger in SO₃ or SO₃&amp;#178;⁻?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a couple of molecules that belong roughly to the same class. I call them &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromaticity#Atypical_aromatic_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Y aromatic systems&lt;/a&gt;: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see &lt;a href=&quot;https://chemistry.stackexchange.com/q/24509/4945&quot;&gt;Why is there a need for resonance?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7ez8.jpg&quot; alt=&quot;resonance of SO3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have explained the &lt;a href=&quot;https://chemistry.stackexchange.com/q/21856/4945&quot;&gt;delocalisation in the nitrate ion&lt;/a&gt;, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare &lt;a href=&quot;https://chemistry.stackexchange.com/a/29255/4945&quot;&gt;here&lt;/a&gt;). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme &lt;a href=&quot;https://i.stack.imgur.com/rR7zr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vqc2x.jpg&quot; alt=&quot;pi mo of so3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-04-24T10:27:09.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29257" PostTypeId="2" ParentId="180" CreationDate="2015-04-24T10:27:56.437" Score="0" Body="&lt;p&gt;Because boiling point of different materials depend on the intermolecular forces present between the atoms. The stronger the intermolecular forces the higher the boiling point and the weaker the intermolecular forces the lower the boiling point.&lt;/p&gt;&#xA;" OwnerUserId="15786" LastEditorUserId="7448" LastEditDate="2015-04-24T11:50:05.667" LastActivityDate="2015-04-24T11:50:05.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29259" PostTypeId="2" ParentId="15702" CreationDate="2015-04-24T11:59:13.087" Score="9" Body="&lt;p&gt;Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Philip already pointed out in the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15702/which-d-orbitals-of-sulphur-take-part-in-the-pi-bonds-of-so3#comment24198_15702&quot;&gt;comments&lt;/a&gt;: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_four-electron_bond#Hypervalent_description_with_d_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/29101/4945&quot;&gt;Why is the bond order in the SO₃ molecule 1.33 and not 2?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/13483/4945&quot;&gt;Is the bond enthalpy of S-O stronger in SO₃ or SO₃&amp;#178;⁻?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Atomic contributions to Alpha molecular orbitals:&#xA; Alpha occ 1 OE=-88.394 is S1-s=1.00&#xA; Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17&#xA; Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50&#xA; Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33&#xA; Alpha occ 5 OE=-7.922 is S1-s=1.00&#xA; Alpha occ 6 OE=-5.945 is S1-p=1.00&#xA; Alpha occ 7 OE=-5.941 is S1-p=1.00&#xA; Alpha occ 8 OE=-5.941 is S1-p=1.00&#xA; Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17&#xA; Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18&#xA; Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12&#xA; Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11&#xA; Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14&#xA; Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12&#xA; Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20&#xA; Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13&#xA; Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21&#xA; Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15&#xA; Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45&#xA; Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32&#xA; Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the &lt;code&gt;O2-p&lt;/code&gt; orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, &lt;code&gt;0.15; 0.15; 0.10; 0.10&lt;/code&gt;. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-04-24T11:59:13.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29261" PostTypeId="2" ParentId="29219" CreationDate="2015-04-24T12:24:31.333" Score="7" Body="&lt;p&gt;According to the article &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp106050r&quot; rel=&quot;noreferrer&quot;&gt;An Empirical Correlation between the Enthalpy of Solution of Aqueous Salts and Their Ability to Form Hydrates&lt;/a&gt; (Pandelov et al. 2010), your observation of a lower enthalpy is correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We have found a correlation between the enthalpy of solution of the salt and its ability to form a hydrate, namely, that the salt’s enthalpy of solution is lower than the standard enthalpy of fusion of ice (6 kJ/mol)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the drop in enthalpy you observed of 25%+ seems excessive, as can be seen in the graph below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i9Ew1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://www1.lsbu.ac.uk/water/colligative_properties.html&quot; rel=&quot;noreferrer&quot;&gt;Colligative properties of water&lt;/a&gt; cited from the original article &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0140700708001497&quot; rel=&quot;noreferrer&quot;&gt;Formulation of the latent heat of fusion of ice in aqueous solution&lt;/a&gt; (Kumano et al. 2009)&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;KUmano et al. 2009's methodology centred around the use of &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Differential scanning calorimetry analysis was performed and the measurements and calculated values compared with respect to the specific enthalpy of the ice/solution mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for the 'mushy ice' that you observed, it is likely to be a mixture of ice and brine, as per the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FLO7R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_chloride&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-04-24T12:24:31.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29262" PostTypeId="1" AcceptedAnswerId="29422" CreationDate="2015-04-24T13:06:35.810" Score="5" ViewCount="190" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/Triple_point#Triple_point_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;Triple point of water&lt;/a&gt;, the triple point of water occurs at&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at exactly 273.16 K (0.01 °C; 32.02 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shown in the illustration below (from the Wikipedia article):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4q5mz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crucially, to put this into a context, the 611.73 pascals quoted in the article corresponds to the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;minimum pressure at which liquid water can exist. At pressures below the triple point (as in outer space), solid ice when heated at constant pressure is converted directly into water vapor in a process known as sublimation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potentially useful information to determine if an exoplanet can possibly have water (if all other conditions are suitable).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user7951" LastEditDate="2016-11-26T18:59:36.227" LastActivityDate="2016-11-26T18:59:36.227" Title="What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?" Tags="&lt;water&gt;&lt;spectroscopy&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29263" PostTypeId="1" CreationDate="2015-04-24T13:53:25.203" Score="2" ViewCount="250" Body="&lt;p&gt;When a compound gains an electron, why are they more likely to give them up (i.e. become stronger reducing agents)?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastActivityDate="2015-04-24T16:14:39.070" Title="Why does a reduced compound become a better electron donor?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29264" PostTypeId="2" ParentId="10991" CreationDate="2015-04-24T14:37:13.133" Score="5" Body="&lt;p&gt;Really interesting question!&lt;br&gt;&#xA;Mechanism of acid catalyzed hydrolysis of purine nucleosides is known to proceed through unimolecular departure of base moiety(DOI: 10.1021/ja00709a055).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/kM29I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that someone else has also asked the same question: &quot;Why reaction rate with pyrimidines is so much slower?&quot; -and published paper of the reaction mechanism (DOI:10.1039/P29940000309). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper it is suggested that pyrimidines follow different reaction mechanims. This would explain huge differences in reaction rates. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words,&#xA;  the structure-reactivity correlations lend additional support to&#xA;  the previous suggestion, according to which the acid-catalysed&#xA;  hydrolysis of 4-oxopyrimidine nucleosides follows an openchain&#xA;  mechanism      &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Open chain mechanims: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZHEzE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15733" LastActivityDate="2015-04-24T14:37:13.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29265" PostTypeId="1" AcceptedAnswerId="29269" CreationDate="2015-04-24T15:23:10.487" Score="1" ViewCount="11501" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Air contains 20 % &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; by volume. How many cm³ of air will be required&#xA;for oxidation of 100 cm³ of acetylene?&lt;/p&gt;&#xA;&lt;p&gt;(a) 1064 cm³&lt;br /&gt;&#xA;(b) 212.8 cm³&lt;br /&gt;&#xA;(c) 500 cm³&lt;br /&gt;&#xA;(d) 1250 cm³&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer: (d)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;My solution:&lt;/p&gt;&#xA;&lt;p&gt;Equation: &lt;span class=&quot;math-container&quot;&gt;$\ce{2C2H2  +5O2 -&amp;gt; 4CO2 +2H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;From stoichiometric calculations:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with 5 moles of oxygen&lt;/li&gt;&#xA;&lt;li&gt;1 &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H5} \rightarrow $&lt;/span&gt; 5/2 moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{22.4L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2} \rightarrow 5/2\times \pu{22.4L}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;1 volume of air &lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; 0.2 (20/100) volume of oxygen&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="5368" LastEditorUserId="69830" LastEditDate="2021-05-22T13:19:55.383" LastActivityDate="2021-05-22T13:21:17.983" Title="Air contains 20 % O2 by volume. How many cm³ of air will be required for oxidation of 100 cm³ of acetylene?" Tags="&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29266" PostTypeId="1" CreationDate="2015-04-24T15:28:18.023" Score="1" ViewCount="101" Body="&lt;p&gt;My textbook tells me that when carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose. It goes on to say that because, during photosynthesis, carbon becomes bonded to less electronegative atoms like hydrogen, it becomes more capable of forming larger compounds. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="7475" LastEditDate="2015-10-09T16:54:14.670" LastActivityDate="2015-10-09T16:54:14.670" Title="Oxygen Preventing the Formation of Large Organic Compounds?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29267" PostTypeId="1" AcceptedAnswerId="29279" CreationDate="2015-04-24T15:51:04.180" Score="2" ViewCount="5644" Body="&lt;p&gt;The zinc ion in its +2 state can show no more pairing, since all the electrons are already paired - so in any complex it forms it will have the same magnetic moment as it does right now. $\ce{Ti}$ in its +3 state will show both outer and inner orbital complexes without any electrons pairing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while calling these complexes high spin or low spin complexes is not useful, since there is no possibility of pairing involved in either - but all the same, won't they be called high spin complexes, since no lowering of magnetic moment occurs? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher said, however, that since $\ce{Zn}$ in +2 state forms only outer orbital complexes, its complexes will be called high spin - and $\ce{Ti}$ in its +2 state similarly when it forms inner orbital complexes (which it does without pairing), will be called low spin because the complex is inner orbital. But shouldn't such a complex of $\ce{Ti}$ be a high spin complex? (I'm saying this in reference to VBT).&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorDisplayName="user15489" LastEditDate="2015-04-24T20:30:09.210" LastActivityDate="2015-04-24T20:58:03.187" Title="High or low spin complex of Zn2+ and Ti3+" Tags="&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29268" PostTypeId="2" ParentId="29263" CreationDate="2015-04-24T16:14:39.070" Score="2" Body="&lt;p&gt;Simple electrostatic repulsion can be used to explain this. For example, if you have $\ce{O^2-}$, there are 8 protons in the nucleus, but 10 electrons, so there is a net negative charge of 2. Like charges repel, pushing the electrons away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, &lt;em&gt;compounds&lt;/em&gt; rarely &quot;gain an electron&quot;, the reactive species are usually &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radical_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;radicals&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-24T16:14:39.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29269" PostTypeId="2" ParentId="29265" CreationDate="2015-04-24T16:16:31.300" Score="4" Body="&lt;p&gt;Let's begin with your answer:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2\ \mathrm{mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with &lt;span class=&quot;math-container&quot;&gt;$5\ \mathrm{mol}$&lt;/span&gt; of oxygen.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$1\ \mathrm{mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with &lt;span class=&quot;math-container&quot;&gt;$5/2\ \mathrm{mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If we suppose that the gases are under standard conditions, then&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$22.4\ \mathrm{L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with  &lt;span class=&quot;math-container&quot;&gt;$\frac{5}{2} \times 22.4\ \mathrm{L  } ~ \ce{O2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This means that:&#xA;&lt;span class=&quot;math-container&quot;&gt;$0.1\ \mathrm{L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with &lt;span class=&quot;math-container&quot;&gt;$\frac{5}{2} \times 0.1\ \mathrm{L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$0.1\ \mathrm{L}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{C2H2}$&lt;/span&gt; will react with &lt;span class=&quot;math-container&quot;&gt;$\frac{5}{2} \times 0.1\times 5\ \mathrm{L}$&lt;/span&gt; of air.&lt;br /&gt;&#xA;So, the volume of air is &lt;span class=&quot;math-container&quot;&gt;$1.25\ \mathrm{L}= 1250\ \mathrm{cm^3}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="69830" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T13:21:17.983" LastActivityDate="2021-05-22T13:21:17.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29270" PostTypeId="2" ParentId="29266" CreationDate="2015-04-24T16:48:13.177" Score="3" Body="&lt;p&gt;If we take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;nofollow noreferrer&quot;&gt;glucose&lt;/a&gt; we can see there are oxygen atoms in it, just with extra presence of the hydrogen.&lt;/p&gt;&#xA;&lt;p&gt;The assumption :&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;is wrong.&lt;/p&gt;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxocarbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polymeric carbon oxides&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio. The polymer is believed to be a linear chain of fused six-membered lactone rings, with a continuous carbon backbone of alternating single and double bonds. Physical measurements indicate that the mean number of units per molecule is about 5–6, depending on the formation temperature.&lt;/p&gt;&#xA;&lt;p&gt;Carbon monoxide compressed to 5 GPa in a diamond anvil cell yields a somewhat similar reddish polymer with a slightly higher oxygen content, which is metastable at room conditions. It is believed that &lt;span class=&quot;math-container&quot;&gt;$CO$&lt;/span&gt; disproportionates in the cell to a mixture of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C3O2}$&lt;/span&gt; ; the latter forms a polymer similar to the one described above (but with a more irregular structure), that traps some of the &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; in its matrix.&lt;/p&gt;&#xA;&lt;p&gt;Another carbon-oxygen polymer, with C:O ratio 5:1 or higher, is the classical graphite oxide and its single-sheet version &lt;a href=&quot;http://en.wikipedia.org/wiki/Graphite_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;graphene oxide&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-24T16:48:13.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29273" PostTypeId="1" AcceptedAnswerId="29281" CreationDate="2015-04-24T17:31:18.080" Score="5" ViewCount="794" Body="&lt;p&gt;I understand that complex ions are coloured due to d-orbital splitting which results in electrons being able to absorb wavelengths of visible light and become excited to the higher energy state meaning the transmitted light is coloured. However I don't see why the electron would not simply de-excite and emit the same wavelength of light that it absorbed which would result in no visible difference. Is there some reason why the same wavelength is not emitted or perhaps the electron does not de-excite at all?&lt;/p&gt;&#xA;" OwnerUserId="15790" LastActivityDate="2015-04-25T03:16:07.797" Title="Why do complex ions not emit light from de-excitations of electrons?" Tags="&lt;physical-chemistry&gt;&lt;transition-metals&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29275" PostTypeId="1" CreationDate="2015-04-24T18:25:28.453" Score="-2" ViewCount="316" Body="&lt;p&gt;I have no idea how to find a picture for these enzyems. My teacher did not explain the chemical reaction and I am really lost. Please help!&lt;/p&gt;&#xA;" OwnerUserId="15792" LastEditorUserId="7448" LastEditDate="2015-04-24T18:40:45.653" LastActivityDate="2015-04-24T19:22:06.210" Title="For protease, lipase, and amylase, include an image of the chemical, and the chemical that it operates" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-24T23:35:20.010" ContentLicense="CC BY-SA 3.0" />
  <row Id="29276" PostTypeId="2" ParentId="29275" CreationDate="2015-04-24T19:16:12.240" Score="1" Body="&lt;p&gt;Your question is too broad; protease is a family of enzymes which  performs protein digestion aka proteolysis; same goes to lipase and amylase for lipids and sugars respectively. These enzymes/proteins have a complex &lt;a href=&quot;http://en.wikipedia.org/wiki/Protein_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical structure&lt;/a&gt;. If you want to see the structure of proteins or enzymes I would recommend &lt;a href=&quot;http://www.rcsb.org/pdb/home/home.do&quot; rel=&quot;nofollow noreferrer&quot;&gt;RCSB PDB database&lt;/a&gt;. You can search for protein structure by name,PDB ID, sequence etc. in this database. It also gives detailed information about that protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a search for &lt;code&gt;Serine proteases&lt;/code&gt; ( proteases that have serine bonded at the active site) will give you &lt;a href=&quot;http://www.rcsb.org/pdb/results/results.do?qrid=DB727361&amp;amp;tabtoshow=Current&quot; rel=&quot;nofollow noreferrer&quot;&gt;458 strutural hits&lt;/a&gt;. There are different parameters such as organism, taxonomy etc available to refine your search&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An image of HIV-1 Protease&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rcsb.org/pdb/explore/explore.do?structureId=4Q5M&quot; rel=&quot;nofollow noreferrer&quot;&gt;D30N tethered HIV-1 protease dimer/saquinavir complex&lt;/a&gt; (PDB ID=4Q5M)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mqjbg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditDate="2015-04-24T19:22:06.210" LastActivityDate="2015-04-24T19:22:06.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29277" PostTypeId="1" AcceptedAnswerId="29321" CreationDate="2015-04-24T19:17:44.160" Score="8" ViewCount="17889" Body="&lt;p&gt;I am having a problem with this equation for redox potentials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = -nFE_\mathrm{cell}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this equation I never am totally sure about what the value of $n$ should be, for example for the reaction shown below, would the $n$ be 2 electrons or 1 electron? Personally, I think it should be 2 electrons because that is the number of mol of electron under the simplest whole number ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{cyt $c$(Fe^3+) + e- &amp;amp;-&amp;gt; cyt $c$(Fe^2+)}\\&#xA;\ce{NADH &amp;amp;-&amp;gt; NAD+ + H+ + 2e- }\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a different matter than in the already answered question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/q/28954/4945&quot;&gt;Does the relationship equation between standard cell potential and equilibrium constant violate potential&amp;#39;s intensive properties?&lt;/a&gt;&quot; Because in that Nernst equation the change in $n$ would be balanced by the change in the equilibrium constant. I assume that you are insinuating that there is one of the variables here which will correct for the increase in $n$. I don't see which would, as $E_\mathrm{cell}$ should be independent of quantity and $F$ is a constant.&lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="16683" LastEditDate="2017-07-31T07:45:35.647" LastActivityDate="2017-07-31T10:12:35.717" Title="How to determine the value for n in the Gibbs free energy and redox potential equation?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29279" PostTypeId="2" ParentId="29267" CreationDate="2015-04-24T20:13:34.030" Score="2" Body="&lt;p&gt;Are you familiar with ligand field stabilisation energy? This approach answers your question. Interaction of the central metal ion with the surrounding ligands causes splitting of the $d$ orbitals into an essentially upper and lower energy level, the form of the splitting is rather complex and depends directly upon the symmetry point group that the ligand complex belongs to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, high spin complexes only form if the reduction in inter electron repulsion energy is greater than the energy splitting of the orbitals, with the definition of high spin being a none aufbau filling of the split orbitals. It does not make sense to describe the electronic structure of a $\ce{Zn^2+}$ ion in a complex as high or low spin, there is only one configuration available as there are 10 $d$ electrons present, and the 5 $d$ orbitals split to form 5 new orbitals of varying degeneracies (dependent upon the point group of the complex and the symmetry of the coordination environment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ti^2+}$ has 2 valence d electrons available to be involved in bonding with the ligand environment, however there is no coordination environment that I am aware of (The splitting would have to be extremely small and other thermodynamic factors such as the ionic contribution to the Gibbs energy of complex formation would mean that the complex would never be observed anyway - such a low splitting would only be the result of very poor overlap which would be caused by large separation between the central ion and the ligands, resulting in a lower ionic term) that causes splitting where the magnitude of the splitting is less than the repulsion term invoked when both electrons occupy the same molecular orbital, be it degenerate or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{Ti^3+}$ ion has only 1 $d$ electron, it will definitely reside in the lowest available molecular orbital. However, the first excited states, ignoring hyperfine splitting, of both titanium ions will be high spin - the magnitude of the splitting usually corresponds to wavelengths of light in the visible region of the spectrum, hence many transition metal complexes are coloured.&lt;/p&gt;&#xA;" OwnerUserId="15791" LastEditorDisplayName="user15489" LastEditDate="2015-04-24T20:58:03.187" LastActivityDate="2015-04-24T20:58:03.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29280" PostTypeId="1" CreationDate="2015-04-24T20:20:21.520" Score="3" ViewCount="254" Body="&lt;p&gt;For solid-liquid mixtures, cooling crystallization process can be used to separate the solute substances from each other. Assume that we have two salts that have very close solubility in a certain solvent which we are going to use. First, we dissolve the salts in the solvent while heating. After a point, we let the mixture cool down and observe the crystallization. My question is, how can we do this experiment accurately if they have really close crystallization point that they would nearly crystallize at the same time?&lt;/p&gt;&#xA;" OwnerUserId="11549" LastActivityDate="2017-02-18T22:06:44.753" Title="Crystallization issue" Tags="&lt;solubility&gt;&lt;recrystallization&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29281" PostTypeId="2" ParentId="29273" CreationDate="2015-04-24T20:45:02.143" Score="4" Body="&lt;p&gt;The fate of electronically excited states is not immediately obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all the rate at which the electron emits a photon and falls down a level is dependent upon the energy difference between the two levels, it is only an expectation value of a macroscopic observable, on an individual atomic scale it is impossible to predict when the electron in the upper level will emit a photon, however multiple instances of the same measurement will tend towards the expectation value. The direct result of this is that there is a path difference between the emitted photon and the 'rest' of the photons that were not absorbed by the complex and the result is a different interference between photons in the transmitted light compared to the incident light and hence a new colour is observed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another fate of excited electrons, the Franck-Condon principle invokes the very plausible assumption that the nuclei do not respond immediately (at least on the time scale of electronic transitions) to the new equilibrium bond position that is determined by the new electronic state of the complex. This means the excited electronic state will most likely not be in the lowest vibrational state of the new equilibrium bond configuration, the splitting of vibrational energy levels is orders of magnitude lower than those of electronic transitions and hence the same law that governs electronic emission applies, the electron is much more likely to fall down these vibrational energy levels first, without emitting photons of visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron will then be expected to, by the time it is occupying the lowest vibrational state of the excited electronic state, invoke a transition down to the ground state energy level. The energy of the excited state is described in a basic form by &#xA;$$H=E^*+v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H$ is the energy of the absorbed photon, $E^*$ is the energy of the excited state in the lowest vibrational mode and $v$ is the contribution from the initial vibrational state of the excited complex.&lt;/p&gt;&#xA;" OwnerUserId="15791" LastEditorDisplayName="user15489" LastEditDate="2015-04-25T03:16:07.797" LastActivityDate="2015-04-25T03:16:07.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29284" PostTypeId="2" ParentId="29212" CreationDate="2015-04-24T23:56:03.223" Score="5" Body="&lt;p&gt;IUPAC has changed the definition of STP from [273 K, 1 atm (101.3 kPa)] to [273 K, 1 bar (100.0 kPa)]. Hence their molar volume at STP has changed from 22.4 L/mol to 22.7 L/mol.&lt;/p&gt;&#xA;" OwnerUserId="15794" LastEditorDisplayName="user15489" LastEditDate="2015-04-25T03:15:04.460" LastActivityDate="2015-04-25T03:15:04.460" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29285" PostTypeId="1" AcceptedAnswerId="29393" CreationDate="2015-04-25T03:28:17.340" Score="2" ViewCount="240" Body="&lt;p&gt;In the awwa notebook for $\ce{Na2SiF6}$ determination, why it is required to wash the product sample in an alcoholised $\ce{KCl}$ solution before titration?&lt;/p&gt;&#xA;" OwnerUserId="15795" LastEditorUserId="139" LastEditDate="2015-04-25T03:31:24.810" LastActivityDate="2015-04-26T20:47:44.237" Title="Why wash a sample before titration?" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29286" PostTypeId="1" CreationDate="2015-04-25T03:38:23.140" Score="6" ViewCount="1284" Body="&lt;p&gt;Why is it that energy is released from electrons when they move from one atom to a more electronegative one?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="402" LastEditDate="2015-04-25T05:23:03.770" LastActivityDate="2015-06-20T18:20:30.267" Title="Why is energy released as electrons move?" Tags="&lt;energy&gt;&lt;electrons&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29287" PostTypeId="1" CreationDate="2015-04-25T04:11:14.230" Score="4" ViewCount="105" Body="&lt;p&gt;I'm calculating the multipole moments on several fragments of a molecule. What are some procedures to ensure that the charges etc are representative of the complete molecule? How do I ensure I'm not getting an overall charge in a neutral molecule?&lt;/p&gt;&#xA;" OwnerUserId="15796" LastEditorUserId="15796" LastEditDate="2015-04-26T00:14:11.010" LastActivityDate="2015-05-08T14:32:31.937" Title="Charge consistency in fragment qm/mm methods" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29288" PostTypeId="1" AcceptedAnswerId="29314" CreationDate="2015-04-25T05:31:22.973" Score="7" ViewCount="8320" Body="&lt;p&gt;In my book it is given that earlier cylinder were filled with 2% oxygen and 98% helium. Nowadays it is filled with 11.7% helium, 56.2% nitrogen and 32.1% oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are few questions which come up in my mind&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Earlier 2% oxygen is changed to 32%. So was 2% oxygen enough for huaman?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;98% helium is reduced to 11.7% why it is so? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't get reason why helium and nitrogen are filled?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-04-25T16:43:38.007" Title="Proportion of gases filled in cylinder of sea-divers" Tags="&lt;solutions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29292" PostTypeId="1" CreationDate="2015-04-25T09:37:41.517" Score="6" ViewCount="3052" Body="&lt;p&gt;My guess is that they are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think hydrated oxides would have a lattice with the cation and oxide anions as lattice points and molecules of water trapped in interstitial spaces, while hydroxides would have the cation and hydroxide anions as lattice points with no water molecules in interstitial spaces.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2017-10-21T05:07:04.177" Title="Is there a difference between hydrated oxides and hydroxides?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29294" PostTypeId="1" CreationDate="2015-04-25T11:07:09.137" Score="13" ViewCount="15211" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the products of the reaction of 2-butanol with silver nitrate?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the moment, I am a bit stumped as I only have $\ce{AgNO3}$ as an oxidizing agent to form a carboxylic acid from an aldehyde. Additionally, the lack of $\ce{H2O}$ indicates that this is not the reaction I should be looking for. The only reaction that seems feasible to me is an S&lt;sub&gt;N&lt;/sub&gt;2 mechanism where the nitrate anion acts a nucleophile and the hydroxide ion as the leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my proposed mechanism much in the style of Tollen's reaction with an aldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xEaxa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="5026" LastEditDate="2018-03-16T05:28:53.553" LastActivityDate="2018-03-16T05:28:53.553" Title="Reaction of secondary alcohol with ammoniacal silver nitrate (Tollens' reagent)" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29295" PostTypeId="2" ParentId="29286" CreationDate="2015-04-25T11:24:58.943" Score="0" Body="&lt;p&gt;Due to difference in electronegativities the atoms have different binding energies and this difference is radiated in some form. If you're talking about a transition in a single atom the difference is the binding energies of different orbital/orbits/sheels.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-25T11:24:58.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29296" PostTypeId="1" CreationDate="2015-04-25T11:29:40.713" Score="1" ViewCount="10357" Body="&lt;p&gt;In 1955, it was proposed that hollow cathode lamps are good emission sources for monochromatic radiation, and thus for atomic absorption spectroscopy (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;AAS&lt;/a&gt;).  In looking into the topic, I've read that &quot;continuum radiation is not preferred in AAS.&quot; Why not? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="11367" LastEditDate="2015-04-25T12:53:19.070" LastActivityDate="2015-04-25T13:25:17.507" Title="Why are hollow cathode lamps used for atomic absorption spectroscopy?" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29297" PostTypeId="1" CreationDate="2015-04-25T11:36:43.853" Score="2" ViewCount="499" Body="&lt;p&gt;When a liquid (in my case I used butane) under high pressure is released into the air, it becomes gaseous, and the container seems to get colder. Why is it that turning a liquid to a gas requires energy when it was simply pressure that was keeping it a liquid? Where does it get this energy? Would the gas formed be colder, or no?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastActivityDate="2015-04-25T13:42:08.823" Title="Endothermic Conversion of Liquid to Gas" Tags="&lt;energy&gt;&lt;gas-laws&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29298" PostTypeId="2" ParentId="29297" CreationDate="2015-04-25T12:04:45.653" Score="3" Body="&lt;p&gt;This is probably a bit of a simplistic explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the butane being released from the from the container, the pressure in the container decreases, which in turn, decreases the temperature of the container according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gay-Lussac%27s_law#Pressure-temperature_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pressure-Temperature Law&lt;/a&gt; (also known as the Admonton Law), that states for an ideal gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P  \propto  T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiling point of butane is normally −1 °C at normal pressures – so in typical use at room temperature, it should have already vapourised. But, due to it being under pressure, it remains a liquid, undergoing evaporation in the container. So, butane has a vapour pressure that increases with temperature, as seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YdhRI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Butane_%28data_page%29#Vapor_pressure_of_liquid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, who sourced it from the &lt;em&gt;CRC Handbook of Chemistry and Physics 44th ed&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though butane (like all gases) is not an ideal gas, we can still surmise that if you drop the pressure (by releasing the gas), then the temperature of the gas also drops, especially considering the combined pressure due to that from the container and vapour pressure is rapidly released = rapid depressurisation = rapid loss of temperature, which is felt on the outside of the container.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user7951" LastEditDate="2015-04-25T13:42:08.823" LastActivityDate="2015-04-25T13:42:08.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29299" PostTypeId="1" CreationDate="2015-04-25T12:07:52.680" Score="3" ViewCount="244" Body="&lt;p&gt;In my opinion, it should be both, because it will affect the way in which a molecule will interact with plane polarized light (physical), but also affect chemical interactions such as receptor binding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider $\ce{CHDBrCH3}$, which is chiral, as there 4 distinct groups attached to the central carbon. This would insinuate that chirality is a purely physical property, as despite deuterium having the same chemical properties as hydrogen, we are able to form a chiral molecule.&lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="194" LastEditDate="2018-01-23T17:00:20.903" LastActivityDate="2018-01-23T17:00:20.903" Title="Is chirality a physical or a chemical property?" Tags="&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29300" PostTypeId="2" ParentId="29296" CreationDate="2015-04-25T12:14:23.293" Score="1" Body="&lt;p&gt;One of the strengths of atomic absorption and other ionizing elemental analysis techniques using optical detection (&lt;a href=&quot;https://en.wikipedia.org/wiki/Inductively_coupled_plasma_atomic_emission_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;ICP-OES&lt;/a&gt; is the main other one I know of) is that the optical absorption and emission properties at particular wavelengths are &lt;em&gt;highly&lt;/em&gt; element-specific.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this, the reasons for desiring monochromatic, element-specific AAS light sources are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Most detectors used in AAS / OES are broad-spectrum: they report just about any light that strikes them within their respective ranges of sensitivity, which usually span several hundred nanometers.&lt;/li&gt;&#xA;&lt;li&gt;While the &quot;fingerprint&quot; wavelengths of each element are highly specific, they often sit at wavelengths close to those of other elements.&lt;/li&gt;&#xA;&lt;li&gt;In AAS, any light that is passed through the sample has the potential to be absorbed by anything present in the sample.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, the use of a continuum light source in AAS would obliterate the analytical specificity afforded by the technique: anything present in the sample that absorbs light in the region of overlap of (a) the emission wavelength range of the light source and (b) the wavelength range of sensitivity of the detector would be recorded in a given measurement.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-04-25T12:14:23.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29301" PostTypeId="1" CreationDate="2015-04-25T12:30:42.520" Score="6" ViewCount="8201" Body="&lt;p&gt;Why is it that the potential energy of a system increases when the bonds present in it are broken? (i.e., when a solid turns into a liquid.) I thought the potential energy of the system was mostly contained in these bonds? If these bonds are broken, then why does the system’s potential energy increase?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="30894" LastEditDate="2017-02-05T16:57:52.833" LastActivityDate="2017-03-07T17:25:05.013" Title="Increase in Potential Energy When Bonds Are Broken" Tags="&lt;bond&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29302" PostTypeId="1" AcceptedAnswerId="29304" CreationDate="2015-04-25T12:37:06.917" Score="5" ViewCount="11594" Body="&lt;p&gt;When study biology in class, I don't know how my teacher get to the topics on pH value. So, she says that pH stands for percentage hydrogen. Then, I thought there is something strange about it cause acid has a higher percentage of hydrogen then base. But why does an acid have a low pH value? Is there any mistake on the pH definition or my understanding?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="16683" LastEditDate="2020-04-06T10:20:58.050" LastActivityDate="2020-04-06T10:20:58.050" Title="Why does an acid have a low pH value?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29303" PostTypeId="1" AcceptedAnswerId="29305" CreationDate="2015-04-25T12:47:55.517" Score="2" ViewCount="3119" Body="&lt;p&gt;I am currently studying about the vespr theory and I'm stuck on an ion which is dichromate (VI) ion $\ce{Cr2O7^2-}$. So this is what I am thinking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If I make the oxygen atom at the middle the centre atom, it is a bent.&lt;br/&gt;&#xA;2) If I make the chromium atom thecentre atom, it is a tetrahedral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my problem is that whether dichromate (VI) ion bent or tetrahedral?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="7448" LastEditDate="2015-04-25T13:16:37.793" LastActivityDate="2015-04-25T13:16:37.793" Title="Dichromate ion: Tetrahedral or bent?" Tags="&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29304" PostTypeId="2" ParentId="29302" CreationDate="2015-04-25T12:52:45.413" Score="11" Body="&lt;p&gt;Actually in pH the p is for potenz or power of hydrogen. The concentration of proton $(\rm H^+\equiv H_3O^+)$ in water is often $10^{-3},10^{-5},10^{-14}$ which is very tedious to write, so pH is defined as $3,5,14$ for these solutions. Or:&#xA;$$\rm [H^+]=10^{-pH}\iff \log_{10}[H^+]=-pH\iff pH=-\log_{10}[H^+]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that pH scale and proton concentration works in reverse direction, one increases so the other decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH Scale at normal room temperature is from 0 to 14 (debatable) and 7 is neutral (no excess proton or no deficiency of proton) and 0-7 is acidic (excess proton) wheras 7-14 is basic (deficiency of proton). This is defined because purest form of water at this temperature has pH 7 and pH more than this (and hence concentration of proton less than this) is caused by bases which take up proton and and pH less than this (and hence concentration of proton more than this) is caused by acids which release proton.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="15133" LastEditDate="2015-04-25T17:40:59.077" LastActivityDate="2015-04-25T17:40:59.077" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29305" PostTypeId="2" ParentId="29303" CreationDate="2015-04-25T13:00:25.037" Score="1" Body="&lt;p&gt;Both, Bent about Oxygen and tetrahedral about chromium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2.5in$&lt;img src=&quot;https://i.stack.imgur.com/4PAMv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-25T13:00:25.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29306" PostTypeId="2" ParentId="29299" CreationDate="2015-04-25T13:23:52.640" Score="5" Body="&lt;p&gt;I would not consider chirality as a property--either physical or chemical--it is part of the definition of a molecule's structure, which in turn gives rise to the properties it exhibits (rotation of polarized light, melting point, reactivity with X, etc.). In your example, the fact that hydrogen and deuterium are almost (but not entirely) chemically equivalent will make the substance behave almost the same as its dihydrogen analog, and the two mirror image isomers nearly indistinguishable by physical or chemical means.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-25T13:23:52.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29307" PostTypeId="2" ParentId="29296" CreationDate="2015-04-25T13:25:17.507" Score="2" Body="&lt;p&gt;Atomic absorption lines are very narrow.&lt;br&gt;&#xA;For the &lt;a href=&quot;http://dx.doi.org/10.1351/goldbook.B00626&quot; rel=&quot;nofollow&quot;&gt;Beer-Lambert law&lt;/a&gt; to be applicable, the bandwidth of the source should be narrow in comparison with the width of the absorption peak. Otherwise, the signal-to-noise ratio and the slope of the calibration curve would be low; the resulting sensitivity would be poor.&lt;br&gt;&#xA;Continuous radiation sources use monochromators; however, even good monochromators have effective bandwidths that are significantly greater than the width of atomic absorption lines.&lt;br&gt;&#xA;Therefore, the use of line sources with bandwidths even narrower than the width of absorption peaks is preferred.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-25T13:25:17.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29308" PostTypeId="2" ParentId="28661" CreationDate="2015-04-25T13:42:07.543" Score="0" Body="&lt;p&gt;According to Valence Bonding Theory, proposed by Heitler and London, bonding between two atoms A and B happens when the energy of the system is the lowest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With experimentation was found that this minimum of energy, requires the distance between nucleuses to be $\mathbf{74~pm}$ and the energy $\mathrm{-4.72~ eV}$. Calculations say that the true distance should be $\mathbf{87~pm}$ and the energy $\mathrm{-3.14~eV}$, which means we should consider other factors which indicate the interaction between nucleuses and electrons.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="7448" LastEditDate="2015-07-03T12:05:05.753" LastActivityDate="2015-07-03T12:05:05.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29310" PostTypeId="2" ParentId="28557" CreationDate="2015-04-25T13:55:09.363" Score="5" Body="&lt;p&gt;Every element which is in the first place of the group, is pretty much different from the other elements of the same group. The reason is because the first element has ALWAYS the smallest radius, which leads to some changings also in group properties. Due to the small radius they are more stable than the others and they require more energy for any specific reaction. With some research you can see that also other  first elements of other groups such as &lt;strong&gt;Li,O,C&lt;/strong&gt; etc.  differ from the other elements of that specific group.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-04-25T13:55:09.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29311" PostTypeId="2" ParentId="28661" CreationDate="2015-04-25T15:20:53.793" Score="6" Body="&lt;p&gt;The short answer: no, there is no such a general distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study these collisions is called reaction dynamics, and you can read e.g. Polanyi's &lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1986/polanyi-lecture.pdf&quot;&gt;Nobel lecture&lt;/a&gt;.You should note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is a quantum event, so technically speaking it is not deterministic.&lt;/li&gt;&#xA;&lt;li&gt;The orientation of the collision and the geometry of the potential energy surface along the reaction can be favourable, and rotation, vibration etc can help the reaction, or can prevent it.  &lt;/li&gt;&#xA;&lt;li&gt;The details are also influenced by the medium (is it a reaction in gas or condensed phase, on surface?) and what kind of orbitals are involved in the reaction. About frontier orbitals and their roles you also can find lot of materials in textbooks, eg. any text that discusses the work of Fukui and Hoffmann.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-04-25T15:20:53.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29312" PostTypeId="2" ParentId="29288" CreationDate="2015-04-25T15:41:15.657" Score="3" Body="&lt;p&gt;&lt;strong&gt;Hint:&lt;/strong&gt;&#xA;Try to use Henry's Law. According to which solubility of gases increases with increase in pressure. If we go deeper downwards, pressure increases and when we come to the surface, pressure decreases, so there might be that soem buubles are created in the blood which may be fatal. So air cylinders are made keeping this in mind.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-25T15:41:15.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29314" PostTypeId="2" ParentId="29288" CreationDate="2015-04-25T16:43:38.007" Score="2" Body="&lt;p&gt;From this &lt;a href=&quot;http://www.pa.msu.edu/sciencet/ask_st/082593.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Helium is used in scuba tanks to dilute oxygen.  Why helium?  First, it is chemically inert, that is to say, it does not participate in chemical reactions.  Second, helium is much less soluble in water than many other gases, such as nitrogen.  The low solubility means it does not enter the blood stream, even under pressures commonly experienced by deep sea divers.&#xA;  Under water, a scuba diver is subjected to added pressure.  In shallow water the pressure difference is not so great, and one can simply use compressed air to breathe.  However, at a depth of 300 feet, a diver experiences nearly 10 times normal atmospheric pressure.  Under these conditions, breathing compressed air can be harmful.  The increased pressure requires an inert gas to be used to dilute oxygen.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Oxygen is essential for breathing, above and beneath the surface.  However, one can get too much of it.  Under high pressures, greater amounts of oxygen enter into the bloodstream.  If the concentration becomes too high, &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_toxicity&quot; rel=&quot;nofollow&quot;&gt;oxygen poisoning&lt;/a&gt; can result, with symptoms that include confusion, impaired vision and nausea.  So, under high pressures, oxygen must be diluted with some other gas.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Nitrogen is very abundant and relatively inert, but it cannot be used to dilute oxygen in scuba tanks.  At the elevated pressures experienced by divers, larger amounts of nitrogen will dissolve in the blood causing &lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_narcosis&quot; rel=&quot;nofollow&quot;&gt;nitrogen narcosis&lt;/a&gt;.  Dissolved nitrogen also can cause a painful condition, called &quot;the bends&quot; if a diver makes too rapid an ascent.  As the pressure decreases, the dissolved gas can form bubbles that can stop circulation in capillaries and damage the nervous system. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Breathing_gas#Oxygen&quot; rel=&quot;nofollow&quot;&gt;breathing gas&lt;/a&gt; says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Helium (He) is an inert gas that is less narcotic than nitrogen at equivalent pressure (in fact there is no evidence for any narcosis from helium at all), so it is more suitable for deeper dives than nitrogen. Helium is equally able to cause &lt;a href=&quot;http://en.wikipedia.org/wiki/Decompression_sickness&quot; rel=&quot;nofollow&quot;&gt;decompression sickness&lt;/a&gt;. At high pressures, helium also causes &lt;a href=&quot;http://en.wikipedia.org/wiki/High-pressure_nervous_syndrome&quot; rel=&quot;nofollow&quot;&gt;high-pressure nervous syndrome&lt;/a&gt;, which is a CNS irritation syndrome which is in some ways opposite to narcosis.&#xA;  Helium fills typically cost &lt;strong&gt;ten&lt;/strong&gt; times more than an equivalent air fill.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalent_air_depth&quot; rel=&quot;nofollow&quot;&gt;Equivalent air depth&lt;/a&gt; is used to estimate the decompression requirements of a nitrox (oxygen/nitrogen) mixture. &lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalent_narcotic_depth&quot; rel=&quot;nofollow&quot;&gt;Equivalent narcotic depth&lt;/a&gt; is used to estimate the narcotic potency of &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimix_(breathing_gas)&quot; rel=&quot;nofollow&quot;&gt;trimix&lt;/a&gt; (oxygen/helium/nitrogen mixture). Many divers find that the level of narcosis caused by a 30 m (100 ft) dive, whilst breathing air, is a comfortable maximum&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect earlier dives were not so deep , as the suits, storage and general technology were probably lacking. So a 2% was probably enough for the shallow dives.&#xA;Helium is a good diluting gas , but i think the new mixture was adopted to reduce costs as helium is very expensive.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-25T16:43:38.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29316" PostTypeId="1" CreationDate="2015-04-25T18:18:41.853" Score="4" ViewCount="639" Body="&lt;p&gt;Heyo, I'm sure you all know of the experiment where you take a mixture of ice and water, plop some salt in there and the temperature of the solution progressively drops. The salt increases the entropy of the liquid state, shifting somewhat the ice/water equilibrium towards water. As ice is converted to liquid water, the endothermicity of this process leads to a decrease in the temperature of the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I was wondering if the reverse would ever be possible, say you were to have a thermos flask which would be sealed. It would contain Hot Tea and have a slot where one could pour in salt grains without releasing steam. By the logic in the previous question, won't some of the water vapor move to the liquid state (exothermic), resulting in the heatin up of your already hot tea. If possible would this change even be measurable or useful in any sort of way? &lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="7448" LastEditDate="2015-04-25T18:25:18.137" LastActivityDate="2015-04-25T21:16:01.963" Title="Thermodynamics and Coffee" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29317" PostTypeId="2" ParentId="29316" CreationDate="2015-04-25T18:51:31.420" Score="3" Body="&lt;p&gt;You're exactly right: the phenomenon is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Boiling-point_elevation&quot; rel=&quot;nofollow&quot;&gt;boiling-point elevation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. [...] It is an effect of the dilution of the solvent in the presence of a solute. It is a phenomenon that happens for all solutes in all solutions, even in ideal solutions, and does not depend on any specific solute-solvent interactions. The boiling point elevation happens both when the solute is an electrolyte, such as various salts, and a nonelectrolyte. In thermodynamic terms, the origin of the boiling point elevation is entropy and can be explained in terms of the vapor pressure or chemical potential of the solvent. In both cases, the explanation depends on the fact that many solutes are only present in the liquid phase and do not enter into the gas phase (except at extremely high temperatures).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The empirical method for computing boiling point elevation for a given solute is described &lt;a href=&quot;https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/solutions-12/colligative-properties-of-nonelectrolyte-solutions-95/boiling-point-elevation-408-3643/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. In short:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\Delta T_b = \sum_i{n_iK_{b,i}m_i}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, $\Delta T_b$ is the observed increase in boiling point, and $K_{b,i}$ and $m_i$ are the boiling point elevation constant and the molality (moles of solute per kilogram of &lt;em&gt;solvent&lt;/em&gt;), respectively, of solute $i$.  The $n_i$ factor is the number of ions into which solute $i$ dissociates upon dissolution (e.g., $n_{sucrose} = 1$ and $n_{\ce{NaCl}} = 2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet for a full thermodynamic description is a college or graduate level textbook on the topic. I don't know offhand of any that specifically address boiling-point elevation, though.  The Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Colligative_properties#Boiling_point_elevation_.28ebullioscopy.29&quot; rel=&quot;nofollow&quot;&gt;colligative properties&lt;/a&gt; actually has a bit of thermodynamic detail, providing the expression for computing the boiling point elevation constant from the enthalpy of vaporization $(\Delta H_{\mathrm{vap}})$ and other physical parameters:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$K_{b,i} = \frac{RM_iT_b^2}{\Delta H_{\mathrm{vap}}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $R$, $M$, and $T_b$ are the ideal gas constant, the formula weight of solute $i$, and the boiling point of pure solvent, respectively.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-04-25T21:16:01.963" LastActivityDate="2015-04-25T21:16:01.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29318" PostTypeId="1" AcceptedAnswerId="29319" CreationDate="2015-04-25T20:05:56.200" Score="3" ViewCount="77" Body="&lt;p&gt;My favorite groups of shapes are &lt;a href=&quot;https://en.wikipedia.org/wiki/Cupola_(geometry)&quot; rel=&quot;nofollow&quot;&gt;cupolas&lt;/a&gt;. Are there hydrocarbons in the shape of these similar to how cubane is to a cube? If so how stable are they, and could you please give some general information about each one.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastActivityDate="2015-05-06T19:46:10.783" Title="cupola based hydrocarbons" Tags="&lt;organic-chemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29319" PostTypeId="2" ParentId="29318" CreationDate="2015-04-25T20:23:30.917" Score="3" Body="&lt;p&gt;They would be highly unstable due to the inverted tetrahedral geometry of the carbons in the top face. Generally, a polyhedron can form the basis of a stable hydrocarbon if it has at most 3 edges for each corner. A big caveat is that the hydrocarbon will distort the shape if the angle strain is too great. I found that out in my computational research on prismanes.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2015-04-25T20:23:30.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29321" PostTypeId="2" ParentId="29277" CreationDate="2015-04-25T21:13:29.837" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Personally, I think it should be 2 electrons because that is the&#xA;  number of mol of electron under the simplest whole number ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is $\ce{2e-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the two reactions provided above are half-reactions, the $n$ has to be $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_\mathrm{cell}$ is independent of quantity because it is an &lt;em&gt;intensive property&lt;/em&gt;. It does not depend on the amount of matter present. The $n$ accounts for the number of moles of $\ce{e-}$, this is used to calculate $\Delta G$, &lt;em&gt;which in turn depends on the amount of matter present&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, one must understand that $E_\mathrm{cell}$ is just the &lt;em&gt;difference in cell potentials of two half-reactions&lt;/em&gt;. &lt;em&gt;Increasing&lt;/em&gt; or &lt;em&gt;decreasing&lt;/em&gt; the amount of both does not matter.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="16683" LastEditDate="2017-07-31T07:46:41.083" LastActivityDate="2017-07-31T07:46:41.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29322" PostTypeId="1" CreationDate="2015-04-25T21:17:53.450" Score="-1" ViewCount="5861" Body="&lt;p&gt;I learnt that basic character is directly proportional to the positive inductive effect and inversely proportional to steric hindrance &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the above mentioned properties how is $\ce{(CH3)2NH}$ is &lt;em&gt;more&#xA;  basic&lt;/em&gt; than $\ce{CH3NH2}$ while $\ce{(CH3)3N}$ is &lt;em&gt;less basic&lt;/em&gt; than the&#xA;  secondary amine $\ce{(CH3)2NH}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15803" LastEditorDisplayName="user7951" LastEditDate="2018-08-06T18:54:03.967" LastActivityDate="2018-08-06T18:54:03.967" Title="Basic character of amines" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-04-26T04:22:18.397" ContentLicense="CC BY-SA 4.0" />
  <row Id="29323" PostTypeId="1" CreationDate="2015-04-25T21:32:55.283" Score="1" ViewCount="124" Body="&lt;p&gt;I'm not a chemist but something of a tinkerer and adult boy scout.  Currently if you want to make soap, you need fat and lye.  Fat I'm good with, it's easy to get and you can even get it from plants. Lye usually requires wood ash though and that I would rather avoid, if possible. If not possible and burning something is mandatory, then I want the biggest bang for my buck. That means I want to get the most lye I possibly can with the least pollutant release, especially greenhouse gasses and we shouldn't be thinking of just wood. Anything that will get us some low impact soap in the end. We may even want to replace soap as an idea with something else that you can make at home and wash humans, clothing and dishes with. &lt;/p&gt;&#xA;" OwnerUserId="15804" LastEditorUserId="19616" LastEditDate="2015-10-16T17:58:23.033" LastActivityDate="2015-10-16T17:58:23.033" Title="Least evil way to make soap?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29324" PostTypeId="2" ParentId="29323" CreationDate="2015-04-25T22:33:32.480" Score="4" Body="&lt;p&gt;By far the largest source of lye nowadays is the electrolysis of concentrated solutions of $\ce{NaCl}$ or $\ce{KCl}$ in water, the famous and incredibly industrially important &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloralkali_process&quot; rel=&quot;nofollow&quot;&gt;chloralkali process&lt;/a&gt;, producing millions of tons of $\ce{NaOH}$ and $\ce{KOH}$ every year. While electrolysis is an energy-intensive process, presumably the electricity could be supplied by a low-carbon energy source such as solar or wind power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Replacing soap&quot;, in the sense of halting the usage of surfactants, is not a good idea and likely not even possible. Surfactants are too useful a class of substances. In fact, soaps are likely the most environmentally friendly surfactants, given that they're based on a simple hydrolysis and acid-base reaction with common biomolecules, fatty acids and triglycerides. Other surfactants such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_dodecylbenzenesulfonate&quot; rel=&quot;nofollow&quot;&gt;alkylbenzenesulphonates&lt;/a&gt; (dishwashing liquid), &lt;a href=&quot;http://en.wikipedia.org/wiki/Cetrimonium_bromide&quot; rel=&quot;nofollow&quot;&gt;tetraalkylammonium salts&lt;/a&gt; (hair conditioner) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Perfluorooctanoic_acid&quot; rel=&quot;nofollow&quot;&gt;perfluoroalcoxylates&lt;/a&gt;, among others, can be toxic to aquatic lifeforms and take longer to degrade in the environment. The excessive usage and improper disposal of surfactants is a recognized environmental problem, but tackling soaps won't really get you far in solving the issue.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-04-25T22:33:32.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29329" PostTypeId="2" ParentId="27112" CreationDate="2015-04-25T23:36:30.083" Score="3" Body="&lt;p&gt;Here is a harder version of the answer above.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sodium has 3 electron shells while lithium has only 2 electron shells&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sodium can donate its electrons more easily because the valence electron is further from the nucleus and the force of attraction between both is weaker.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Whereas lithium donates its electron less easily because the valence electron is closer to the nucleus and the force of attraction between both is stronger.&lt;/li&gt;&#xA;&lt;li&gt;This is why sodium will &quot;react more&quot; with oxygen compare to lithium&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="5026" LastEditDate="2018-07-20T14:59:44.833" LastActivityDate="2018-07-20T14:59:44.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29330" PostTypeId="1" CreationDate="2015-04-26T00:11:29.103" Score="31" ViewCount="67577" Body="&lt;p&gt;Pyrite ($\ce{FeS2}$) forms cubic crystals, like these ones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nTLCkm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the crystal structure is primitive cubic, but I don’t see how cubic structure on the molecular level translates to cubic structure on much larger scales. If you are given some number of cubes, there are many different ways to assemble the cubes face to face, and only one of these ways is to create a large cube. So why is it that the pyrite in this rock forms nearly perfect cubes?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="5026" LastEditDate="2018-07-26T17:58:08.753" LastActivityDate="2019-06-23T21:57:24.400" Title="Why does pyrite form cubic crystals?" Tags="&lt;crystal-structure&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29331" PostTypeId="2" ParentId="29330" CreationDate="2015-04-26T00:39:44.457" Score="15" Body="&lt;p&gt;Pyrite can actually come in a range of crystalline morphologies, as seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z21W8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: Modelling nanoscale $\ce{FeS2}$ formation in sulfur rich conditions&lt;/em&gt; (&lt;a href=&quot;https://doi.org/10.1039/B819214F&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;J. Mater. Chem.&lt;/em&gt; &lt;strong&gt;2009,&lt;/strong&gt; &lt;em&gt;19&lt;/em&gt; (21), 3389&lt;/a&gt;), who state that the shape is dependent on the concentration of sulphur during its formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modelling of cubic pyrite crystals, as in the picture in your question, are explained in the article Modeling the Shape of Ions in Pyrite-Type Crystals (&lt;a href=&quot;https://doi.org/10.3390/cryst4030390&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Crystals&lt;/em&gt; &lt;strong&gt;2014,&lt;/strong&gt; &lt;em&gt;4&lt;/em&gt; (3), 390–403.&lt;/a&gt;), modelling the following structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/djoIfm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the caption, the blue spheres represent $\ce{Fe}$ and the ellipsoids represent $\ce{S}$. The shapes of the constituent chemicals used in this diagram is based on their modelling, specifically one of the conclusions the author makes is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It turns out for $p$ valence shell ions that an isotropic ionic radius only occurs on cubic lattice sites. For all other site symmetries, however, two or three radial parameters will apply. Appropriate geometric shapes for $p$ valence shell ions are thus given by ellipsoids instead of spheres.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The $p$ valence ion in this case being sulphide.  With this model in mind, the explanation provided in the caption for this diagram states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mesh inscribed into the sulfur ions now reveals the ellipsoidal compression along &amp;lt;111&gt; directions. In this model, the number of contact points of $\ce{S}$ and $\ce{Fe}$ ions become four and six, respectively, and the ellipsoidal deformation is concluded to enable a stable packing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The molecular shape does not necessarily mean that the pyrite crystal has a cubic shape, however if the conditions are right (pressure and temperature), and the composition is 'correct' - if the formation is in a iron rich environment (as per the first reference), then this molecular shape could express as a cubic crystal.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-07-26T18:01:13.793" LastActivityDate="2018-07-26T18:01:13.793" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29333" PostTypeId="1" CreationDate="2015-04-26T01:15:09.667" Score="-1" ViewCount="6952" Body="&lt;p&gt;What prevents weak acids from dissociating completely in water?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastActivityDate="2015-04-26T12:50:49.540" Title="What prevents weak acids from dissociating completely?" Tags="&lt;acid-base&gt;" AnswerCount="4" CommentCount="2" ClosedDate="2015-04-26T08:42:29.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="29334" PostTypeId="2" ParentId="27018" CreationDate="2015-04-26T01:30:55.663" Score="-1" Body="&lt;p&gt;In chemistry dU = dQ + dW ( not dQ = dU + dW) and dW = -Pdv.The general expression for enthalpy is H = U + PV . Enthalpy is the amount of heat content used or released in a system at constant pressure.&#xA;If there is no non-expansion work on the system and the pressure is still constant, then the change in enthalpy will equal the heat consumed or released by the system (q) i.e. (dH = Q).&#xA;But,you can calculate change in enthalpy every case throuugh equation (H = U + PV) i.e. dH = dU + d(PV).&lt;/p&gt;&#xA;" OwnerUserId="15736" LastActivityDate="2015-04-26T01:30:55.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29335" PostTypeId="2" ParentId="29333" CreationDate="2015-04-26T01:47:26.850" Score="0" Body="&lt;p&gt;according to ostwald dilution law, there is relationship between dissociation constant and degree of dissociation of weak electrolyte. Since, weak acids have very small dissociation constant ('k' value ) , so they do not donate all of its hydrogen ion (H+).That's why , they do not dissociate completely.&lt;/p&gt;&#xA;" OwnerUserId="15736" LastEditorUserId="15736" LastEditDate="2015-04-26T01:57:47.060" LastActivityDate="2015-04-26T01:57:47.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29336" PostTypeId="2" ParentId="29333" CreationDate="2015-04-26T02:13:08.857" Score="2" Body="&lt;p&gt;To give a straight answer: &lt;strong&gt;intramolecular forces.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take two examples, $\ce{HCl}$ and $\ce{HF}$. We all know that $\ce{HCl}$ is a &lt;strong&gt;strong&lt;/strong&gt; acid and it dissociates completely, while $\ce{HF}$ doesn't. This is because of the strength of &lt;strong&gt;Fluorine&lt;/strong&gt; and &lt;strong&gt;Chlorine&lt;/strong&gt; differs. &lt;em&gt;Fluorine is strongly bonded to Hydrogen so less of it dissociates.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is position which is mostly found in oxoacids. &lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-26T12:50:49.540" LastActivityDate="2015-04-26T12:50:49.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29337" PostTypeId="2" ParentId="29322" CreationDate="2015-04-26T02:37:07.877" Score="6" Body="&lt;p&gt;what we expect here would be the tertiary amine to have the greatest basicity and the least would be the primary amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But through experiments it has proven that secondary amines to have the greatest basicity. the order of basicity is as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NH3 &amp;lt; primary amine &amp;lt; secondary amine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basicity of the tertiary amine is approx. equal to that of primary amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons behind this is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Steric hindrance &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solvation of ions&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;3 + inductive effect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the alkyl group attached increases the basicity reduces. It is because the size of the alkyl group is so large that it prevents an attack from Hydrogen ion. So according to this theory decreasing order would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primary amine &gt; Secondary amine &gt; Tertiary amine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's another thing you've got to consider. That is the solvation of ions. Amines dissolve in water to form protonated amines thus increasing the number of  possibilities of hydrogen bonding. More the H bondings, greater the dissolution in water. Higher the H bonding, greater is their stability and greater is their basicity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments have indicated that primary amines are the ones who have more H bonds and tertiary amines are the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by combining all those reasons scientists have shown that basicity of amines vary as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NH3 &amp;lt; primary amine ~ tertiary amine &amp;lt; secondary amine&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="9961" LastEditDate="2016-08-18T15:17:24.897" LastActivityDate="2016-08-18T15:17:24.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29338" PostTypeId="2" ParentId="29333" CreationDate="2015-04-26T03:37:11.080" Score="1" Body="&lt;p&gt;The Acid (HA), which calls as weak, will produce an ion (A-) by dissociation. According to Bronsted-Lowry acid-base theory, the produced ion (A-), which is called as Conjugate base of acid (HA), is unstable when its respective acid (HA) is weaker one. &#xA;Because of the instability, the conjugate base (A-) involves in backward directive reaction to form its parental acid (HA) which is in non-dissociated form. In other words, there could be equilibrium in between non-dissociated acid (HA) (reactant in forward direction because of the ability to donate H+ ion) and conjugate base (A-) (reactant in backward direction because of the instability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uRJBg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reason there is no complete dissociation occurs in weak acids.&#xA;    Apart from this, there could be the situation in which the weaker one is comparative term. I mean, in the presence of strong acid (H2SO4) the weak acid (CH3COOH) is in the state of non-dissociation form that is because of ‘&lt;a href=&quot;http://en.wikipedia.org/wiki/Common-ion_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Common-ion effect&lt;/a&gt;’. In this case also there is no complete dissociation from weak acid.&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorUserId="15800" LastEditDate="2015-04-26T03:43:37.773" LastActivityDate="2015-04-26T03:43:37.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29339" PostTypeId="1" CreationDate="2015-04-26T03:57:49.477" Score="6" ViewCount="12437" Body="&lt;p&gt;An opened bottle of wine developed into a sour taste and smell of vinegar. Why?&lt;/p&gt;&#xA;" OwnerUserId="15810" LastEditorUserId="7448" LastEditDate="2015-07-10T15:05:51.003" LastActivityDate="2019-09-17T10:39:22.270" Title="Why does opened wine develop the sour taste and smell of vinegar?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29340" PostTypeId="2" ParentId="29301" CreationDate="2015-04-26T04:12:04.647" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is &lt;strong&gt;not&lt;/strong&gt; true that &quot;potential energy of a system increases when its bonds are broken.&quot; For example, there is &lt;em&gt;less&lt;/em&gt; potential (chemical) energy when the &lt;a href=&quot;https://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/08_2_PHILADELPHIA_04-64_0137.pdf&quot; rel=&quot;nofollow&quot;&gt;bonds of acetylene are broken&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In an exothermic reaction, energy is released (often as heat) as bonds form. For example, heating ammonium chloride (requiring &lt;strong&gt;input&lt;/strong&gt; of thermal energy), it decomposes to ammonia an hydrogen chloride; when these reactants recombine, they &lt;strong&gt;release&lt;/strong&gt; that energy.&#xA;$\ce{NH4Cl &amp;lt;=&amp;gt; NH3 + HCl}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-26T04:12:04.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29341" PostTypeId="2" ParentId="29333" CreationDate="2015-04-26T04:35:35.323" Score="2" Body="&lt;p&gt;It seems that your fundamental misunderstanding is that some molecules will dissociate and some will not.  Who is to say that they do not dissociate?  In fact, virtually every single molecule will dissociate in solution, BUT here's the catch: the same holds for protonation of the conjugate base.  In reality, these molecules are all constantly dissociating and reprotonating, and the relative stability of the acid and conjugate base, represented by equilibrium constants, determines the rate of the forward and backward reactions, which determines where the forward and backward reactions proceed at the same rate.  This is the fundamental concept of equilibrium in a reversible process, that the system is constantly in flux, but there is a point where the overall concentrations remain constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we term &quot;strong acids&quot; are simply acids with pKa values so low that they completely dissociate and any time a conjugate base is reprotonated it almost immediately dissociates again.  There are also classes of acids that are &quot;near strong acids&quot; and &quot;weak acids&quot; with increasing pKa values.&lt;/p&gt;&#xA;" OwnerUserId="15719" LastEditorUserId="15719" LastEditDate="2015-04-26T04:41:58.817" LastActivityDate="2015-04-26T04:41:58.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29342" PostTypeId="2" ParentId="29339" CreationDate="2015-04-26T04:37:37.003" Score="3" Body="&lt;p&gt;The ethanol (alcohol content) of the wine, is oxidised, by the oxygen in the air into ethanoic acid. Vinegar is dilute acetic acid which is where it gets its smell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction happens in to stages,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ethanol is partially oxidised to ethanal&lt;/li&gt;&#xA;&lt;li&gt;The ethanal is further oxidised to ethanoic acid&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15809" LastActivityDate="2015-04-26T04:37:37.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29343" PostTypeId="1" CreationDate="2015-04-26T05:03:56.857" Score="8" ViewCount="2058" Body="&lt;p&gt;I completely don't understand how this method works. My book says that this method works by the principle of &lt;strong&gt;self-absorption&lt;/strong&gt; which occurs by the emitted radiation of hollow cathode lamp &lt;strong&gt;when it works on high intensity of electricity&lt;/strong&gt;, due to which we have more atoms who lead to an extension of emission lines. This is how self-absorption occurs.  Does anyone have any proper explanation for this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this self-absorption actually mean? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that lead to a correction? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorDisplayName="user15489" LastEditDate="2015-05-08T20:46:19.130" LastActivityDate="2019-05-30T22:07:10.610" Title="Smith Hieftje correction method" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29344" PostTypeId="1" CreationDate="2015-04-26T05:21:25.763" Score="3" ViewCount="183" Body="&lt;p&gt;If you add Alum - $\ce{Al2(SO4)3}$, to a clay water sample which contains $\ce{Ca(HCO3)2}$ - due to the water's alkalinity, the pH of this water sample would drop as Alum drops the pH levels. How would you counteract this imbalance. As in, what could you add to the sample to keep the pH the same as it was before the addition of Alum.&lt;/p&gt;&#xA;" OwnerUserId="15811" LastEditorUserId="73" LastEditDate="2015-04-26T06:14:33.343" LastActivityDate="2016-02-20T17:25:48.690" Title="Alkalinity in Water" Tags="&lt;precipitation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29347" PostTypeId="1" CreationDate="2015-04-26T06:11:57.753" Score="5" ViewCount="220" Body="&lt;p&gt;I would like to ask if the hydrogen next to carbonyl group is acidic,and organometallics are strong base, when will it act as a base and extracts the acidic proton and when will it attack the carbonyl carbon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I would also like to ask, for bases when will it attack the acidic proton to form a enolate and when will it attack the carbonyl carbon directly to form an hemiacetal/hemiketals?&lt;/p&gt;&#xA;" OwnerUserId="15813" LastEditorUserId="5456" LastEditDate="2015-04-26T06:52:25.877" LastActivityDate="2015-04-26T12:03:07.247" Title="When will organometallics act as base or nucleophile?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29348" PostTypeId="1" CreationDate="2015-04-26T06:20:59.130" Score="5" ViewCount="2645" Body="&lt;p&gt;I need help understanding the molecule FHF-'s molecular orbital diagram: &lt;img src=&quot;https://i.stack.imgur.com/bNzio.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;p&gt;So I know that &quot;a&lt;sub&gt;g&lt;/sub&gt;&quot; refers to the 2p&lt;sub&gt;z&lt;/sub&gt; and 2s orbitals of fluorine atoms...but I am not sure why &quot;a&lt;sub&gt;g&lt;/sub&gt;&quot; is labeled for the topmost orbital level? I don't really know what that is...but it's interacting with both hydrogen's 1s orbital and fluorine's 2p orbital, so does it imply that it's either antibonding or bonding orbital? And the one right below the topmost orbital--would that be the antibonding sigma 2p&lt;sub&gt;z&lt;/sub&gt; orbital? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, I'd appreciate the help!  &lt;/p&gt;&#xA;" OwnerUserId="9625" LastActivityDate="2015-04-26T11:43:32.923" Title="Understanding FHF- molecular orbital diagram?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29349" PostTypeId="1" CreationDate="2015-04-26T07:11:40.167" Score="3" ViewCount="62" Body="&lt;p&gt;I am currently researching as too which is better for purifying a water sample. I have obviously tried the standard Google search for 'Aluminium Sulphate or Ferric Chloride' and have come up with lots of results. The only problem, is that most of these results are based more around peoples opinions or psuedo-scientific observations. So, to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What chemical reasons are there for choosing Aluminium Sulphate over Ferric Chloride to refine a water sample?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15811" LastActivityDate="2015-04-26T13:45:39.350" Title="Chemical based Water Purifier" Tags="&lt;water&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29350" PostTypeId="1" AcceptedAnswerId="29369" CreationDate="2015-04-26T07:26:04.733" Score="8" ViewCount="2328" Body="&lt;p&gt;Is the peroxide effect observed only with addition of hydrogen bromide and chloroform? If so, why not with other reagents in electrophilic addition on alkenes?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="7448" LastEditDate="2015-10-02T16:58:03.963" LastActivityDate="2015-10-02T16:58:03.963" Title="Peroxide effect in reagents apart from HBr and CHCl₃" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29351" PostTypeId="2" ParentId="29349" CreationDate="2015-04-26T07:31:33.610" Score="1" Body="&lt;p&gt;On addition of inorganic coagulants like aluminium sulfate(alum) or Iron salts causes several chemical and physical interaction take place among particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inorganic coagulants neutralize the negative charge on particles. Also metal hydroxide precipitates of the aluminum and iron salts ions begin to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This precipitates combine to form larger particles under natural processes such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Brownian motion&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum hydroxides are formed within a fairly narrow pH range, typically: 5.5 to about 7.7. Iron (III) hydroxides can form over a larger pH range including pH levels lower than are effective for alum, typically: 5.0 to 8.5&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_purification#Coagulation_and_flocculation&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-04-26T13:45:39.350" LastActivityDate="2015-04-26T13:45:39.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29352" PostTypeId="2" ParentId="27995" CreationDate="2015-04-26T08:28:49.807" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the mechanism of the reduction of organic acids with borane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/75NWz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;br&gt;&#xA;Henry Rzepa. Chemistry with a twist. Mechanism of the reduction of a carboxylic acid by borane: revisited and revised. October 16th, 2011. &lt;a href=&quot;http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114&quot; rel=&quot;noreferrer&quot;&gt;http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114&lt;/a&gt;&lt;br&gt;&#xA;DOI: &lt;a href=&quot;https://dx.doi.org/10.15200/winn.142795.57029&quot; rel=&quot;noreferrer&quot;&gt;10.15200/winn.142795.57029&lt;/a&gt; provided by The Winnower, a DIY scholarly publishing platform.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2015-04-26T12:04:18.160" LastActivityDate="2015-04-26T12:04:18.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29354" PostTypeId="1" CreationDate="2015-04-26T09:14:20.733" Score="0" ViewCount="231" Body="&lt;p&gt;Below is a model example of how to interpret gathered data. Temperature on y-axis, time on x-axis.&#xA;We are supposed to read &lt;img src=&quot;https://i.stack.imgur.com/kgtJf.gif&quot; alt=&quot;enter image description here&quot;&gt; (temperature change by neutralization) and &lt;img src=&quot;https://i.stack.imgur.com/W4SJQ.gif&quot; alt=&quot;enter image description here&quot;&gt; (temperature change by resistive heating - to determine heat capacity of calorimeter) by overlaying the selected area with straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is the way the straight lines are put the way they are. Maybe the bottom one: we expect the temperature drop caused by heating the calorimeter will continue by this straight fashion so we must take it into account in subsequent temp change reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the upper line? Why read the upper value above the graph and not on its local maximum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mDa5X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15669" LastEditorUserId="5026" LastEditDate="2018-02-23T02:10:51.267" LastActivityDate="2018-02-23T02:10:51.267" Title="Determining enthalpy of neutralization and heat capacity of calorimeter by overlying with straight lines or &quot;eye-o-metric method&quot;" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29355" PostTypeId="2" ParentId="29354" CreationDate="2015-04-26T09:57:28.430" Score="1" Body="&lt;p&gt;Ah, I understand now. It's of analogous reason to the first case. By adding heat during a time interval we expect that some of it will be absorbed by the calorimeter. &lt;/p&gt;&#xA;" OwnerUserId="15669" LastActivityDate="2015-04-26T09:57:28.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29356" PostTypeId="1" AcceptedAnswerId="29358" CreationDate="2015-04-26T10:26:38.410" Score="2" ViewCount="204" Body="&lt;p&gt;I bought an electronic meter which measures the concentration of oxygen in air. But I think the sensor has failed because it is too old, as it had a crusty substance coming from it.but I can no longer find spare sensors for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might this sensor have worked? I presume the property of some substance was measured; this property changes with the concentration of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one of these: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Its resistance changed as air was passed through some sort of chemical, and the resistance measured to determine the oxygen concentration. &lt;/li&gt;&#xA;&lt;li&gt;Or perhaps its opacity to light measured with an LED and an appropriate optic sensor. &lt;/li&gt;&#xA;&lt;li&gt;Or the volume of the substance changed, or some other property, which ultimately varies the capacitance.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What substance might this have been?&lt;/strong&gt; I want to make my own device - I don't intend to fix this one, so I don't need to identify the substance used in this particular sensor. &lt;/p&gt;&#xA;" OwnerUserId="371" LastEditorDisplayName="user15489" LastEditDate="2015-04-27T08:17:42.637" LastActivityDate="2015-04-27T08:17:42.637" Title="Electronic oxygen concentration sensor" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29357" PostTypeId="2" ParentId="7638" CreationDate="2015-04-26T11:08:38.720" Score="0" Body="&lt;p&gt;As we know:&#xA;$$K=Ae^{-\Delta H/RT}$$&#xA;Now as this is a monotonously increasing function in the variable T for the positive value of $\Delta H$. So you need to increase temperature to increase the equilibrium constant.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-26T11:08:38.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29358" PostTypeId="2" ParentId="29356" CreationDate="2015-04-26T11:39:24.120" Score="4" Body="&lt;p&gt;I cannot say as to your specific sensor, but most oxygen sensors are electrochemical sensors built around concentration cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the following diagram of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_sensor#Zirconia_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple zirconia-based oxygen probe&lt;/a&gt; from Wikipedia. This type of sensor is very like the one in automobiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z7PuK.png&quot; alt=&quot;diagram of zirconia-based sensor probe for oxygen&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the middle of the sensor is a gas permeable zirconia $(\ce{ZrO2})$ membrane between two gas permeable electrodes. The membrane and electrodes form the electrochemical cell. In many cases, an electrochemical cell will use a reference electrode with a standard reaction going on for comparison. In this sensor, the standard reaction is the same one that is going on at the other electrode. The difference is that at the standard electrode, the partial pressure (&quot;concentration&quot;) of oxygen is known and at the other electrode it is unknown. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But wait, how does that work? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's take a generic electrochemical process, where $\ce{M}$ in the following equation represents something generic that we can oxidize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{M(s) + O2(g) -&amp;gt; MO2(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cell has a standard potential $E^\circ$. The actual measured potential, however, will depend on the partial pressure of oxygen $P_{\ce{O2}}$ because the standard potential is defined at $P_{\ce{O2}} = 1\ \text{atm}$. The relationship that gives us the actual electrode potential at some other oxygen level is the Nernst equation;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=E^\circ - \frac{RT}{nF}\ln{Q}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R$ is the ideal gas constant&lt;/li&gt;&#xA;&lt;li&gt;$T$ is the temperature &lt;/li&gt;&#xA;&lt;li&gt;$n$ is the number of electrons transferred (4 in my example)&lt;/li&gt;&#xA;&lt;li&gt;$F$ is the Faraday constant&lt;/li&gt;&#xA;&lt;li&gt;$Q$ is the reaction quotient, in this case $Q=\frac{1}{P_{\ce{O2}}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If $P_{\ce{O2}}$ is large, then $E$ is small and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we have two electrodes, our unknown and our reference, each with their own potential:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{ref}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},ref}}}$$&#xA;$$E_{unk}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},unk}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltmeter in the device measures the difference between these two potentials and does some maths to calculate back $P_{\ce{O2},unk}$ and then convert it to concentration or percent. Note that it does not even matter what $\ce{M}$ and $\ce{MO2}$ were, since they are not in the equation and even $E^\circ$ cancels out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{unk}-E_{ref} = \frac{RT}{nF}\ln{\frac{P_{\ce{O2},ref}}{P_{\ce{O2},unk}}}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What substance might this have been? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your sensor was this kind of sensor, the crusty mess is probably zirconia.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-04-26T11:39:24.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29359" PostTypeId="2" ParentId="29348" CreationDate="2015-04-26T11:43:32.923" Score="3" Body="&lt;p&gt;So we are treating the two fluorines as ligand groups on the right hand side and we introduce a Hydrogen on the left. We observe that each fluorine has a 2S, 2px, 2py and 2pz. If we take the 2pz as being along the internuclear axis then we assign it with sigma symmetry, while the x and y p orbitals have pi symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the 2pz that can interact with the Hydrogens 1S because it has the correct symmetry. It turns out that the symmetry of that orbital is ag. This creates as you say the bonding and anti bonding pair with the H's 1S. It is important to note that the bonding and anti bonding pair are of the same symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen only has the 1S orbital (symmetry ag) and no others, so every other orbital of a symmetry that is not ag is non-bonding since it has no respective partner in the FHF environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to the assignation of symmetry to an orbital you must have an understanding of SALCs (symmetry adapted linear combinations of molecular orbitals), if you perform a projection operator method then you will generate the symmetrys ... however this is a bit advanced and not something expected at undergraduate level (to my knowledge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2015-04-26T11:43:32.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29362" PostTypeId="2" ParentId="29344" CreationDate="2015-04-26T11:52:28.933" Score="1" Body="&lt;p&gt;One thing I can suggest are buffers which maintain the pH of a solution within a certain limit when either acid or bases are added. These often are based around the concept of weak acids or bases. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Buffer_solution&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-26T11:52:28.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29363" PostTypeId="2" ParentId="29347" CreationDate="2015-04-26T12:03:07.247" Score="1" Body="&lt;p&gt;Grignard is a base as well as an nucleophile. When there's a strong electron deficient group it'll attack there, when there' a very acidic group it'll attack there. For other bases it depends upon their relative basicity and nucleophilicity. BAsicity is thermodynamically related while nucleophiicity is kinetically decide. They're silent about each other. You need to practice and gather examples and an experience too :D&lt;/p&gt;&#xA;" OwnerUserId="5531" LastActivityDate="2015-04-26T12:03:07.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29364" PostTypeId="2" ParentId="27018" CreationDate="2015-04-26T12:13:40.333" Score="1" Body="&lt;p&gt;The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 1:  Reactants at temperature $T_0$ and pressure $P_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 2:  Products at temperature $T_0$ and pressure $P_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first law, it then follows that, for this irreversible path:&#xA;$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-04-26T12:13:40.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29365" PostTypeId="1" CreationDate="2015-04-26T13:06:34.783" Score="1" ViewCount="108" Body="&lt;p&gt;This might sound funny but my friend asked me why most - if not every - faeces are of a brown or brownish color?&#xA;Why not grey or slightly bluish or green for example?&#xA;Is there any simple scientific explanation for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="15818" LastActivityDate="2015-04-26T13:06:34.783" Title="Why most excreta are brown?" Tags="&lt;color&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29366" PostTypeId="1" AcceptedAnswerId="29377" CreationDate="2015-04-26T13:37:49.507" Score="1" ViewCount="53" Body="&lt;p&gt;I always wonder that though Both $\ce{NaCl}$ and $\ce{CsCl}$ are alkali halides, why they adopt different crystal forms, FCC and BCC respectively? and I have a doubt about $\ce{CsCl}$ structure that Why the $\ce{Cs+}$ or $\ce{Cl-}$ ions are not located at voids/lattice points as like as $\ce{NaCl}$ structure?&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorUserId="5456" LastEditDate="2015-04-26T13:51:20.783" LastActivityDate="2015-04-26T16:23:44.827" Title="Why so differance in alkali halide's crystal forms?" Tags="&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29367" PostTypeId="1" AcceptedAnswerId="29410" CreationDate="2015-04-26T14:51:23.510" Score="6" ViewCount="851" Body="&lt;p&gt;I'm a freshman and I recently learned how to do TLC and column chromatography. I also know a little about other ways to separate things by chromatography, such as affinity chromatography, gel chromatography, ion exchange chromatography, and gas chromatography, but without many details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am facing problems in solving a question requiring us to judge how to separate these combinations of substances &lt;em&gt;by chromatography&lt;/em&gt; (if there was no such confinement, this will be much easier):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;N-phenylethanamide, N-phenylpropionamide&lt;/li&gt;&#xA;  &lt;li&gt;methoxymethane, ethoxyethane&lt;/li&gt;&#xA;  &lt;li&gt;Aβ36，Aβ42 (Aβ is a kind of peptide which is related to Alzheimer's disease, while the number, 36 and 42, refer to the amount of amino acids polymerized in the peptide)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are the specific issues I've met：&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I've no idea how to deal with the first combination since their difference seems to not be significant, no matter in polarity, acidity (basicity), or molecular size. Is there any other chromatographic way, or did I get something wrong?&lt;/li&gt;&#xA;&lt;li&gt;I think I can deal with the second combination by gas chromatography, since they are both volatile. But I'm not sure if it is a good idea. Is there a better way or not?&lt;/li&gt;&#xA;&lt;li&gt;I've found some description for the separation of different Aβ, suggesting that the following ways are usually used: use antibodies (e.g. ELISA) or electrophoretic separation (e.g. PAGE). But seems that neither ways are related to chromatography. It is also said that reverse phase liquid chromatography after CNBr cleavage works, but this way is complex and time consuming. I'm wondering (a) if I can use antibodies or something else to deal with it by affinity chromatogrphy and (b) how CNBr works? (Now I know CNBr is used to cut proteins/peptides at where methionines are, but still not knowing how reverse phase liquid chromatography works.) Also I want to ask if there is a better way as well.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15820" LastEditorUserId="5026" LastEditDate="2018-06-24T11:58:15.193" LastActivityDate="2018-06-24T11:58:57.373" Title="How to choose a proper chromatography way to separate substrates?" Tags="&lt;organic-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29368" PostTypeId="1" CreationDate="2015-04-26T14:51:34.503" Score="5" ViewCount="386" Body="&lt;p&gt;In tandem mass spectrometry of proteins one will usually end up with a mix of $b$ and $y$ fragments but also $a$ fragments and immonium fragments. My problem is that wouldn't an $a_1$ ion be the exact same as an immonium ion of the first amino acid in the peptide?  &lt;/p&gt;&#xA;" OwnerUserId="15748" LastEditorUserId="13184" LastEditDate="2015-09-27T18:34:37.330" LastActivityDate="2015-09-27T18:34:37.330" Title="Tandem mass spectrometry: a1 and immonium ions" Tags="&lt;biochemistry&gt;&lt;mass-spectrometry&gt;&lt;proteins&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29369" PostTypeId="2" ParentId="29350" CreationDate="2015-04-26T15:03:36.180" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the peroxide effect observed only with addition of hydrogen bromide&#xA;  and chloroform?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the &quot;peroxide effect&quot; or &quot;anti-Markovnikov&quot; addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another  example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine the radical addition of $\ce{HX}$ to an olefin.  I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two propagation steps that are key to our analysis:&#xA;$$\ce{X* + H2C=CH2 -&amp;gt; X-H2C-CH2. ~~(1)}$$&#xA;$$\ce{HX + X-H2C-CH2. -&amp;gt; X-H2C-CH2-H ~~(2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this  step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|} \hline&#xA;\ce{C-X} &amp;amp; \text{Bond Strength} &amp;amp; \text{Pi Bond Strength} &amp;amp; \text{Overall}  \\ &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)}\\ \hline&#xA;\ce{C-F} &amp;amp; 116 &amp;amp; 66 &amp;amp; -50\\ \hline&#xA;\ce{C-Cl} &amp;amp; 81 &amp;amp; 66 &amp;amp; -15\\ \hline&#xA;\ce{C-Br} &amp;amp; 68 &amp;amp; 66 &amp;amp; -2\\ \hline&#xA;\ce{C-I} &amp;amp; 51 &amp;amp; 66 &amp;amp; 15\\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that this step is exothermic for the first 3 hydrogen halides, but is &lt;strong&gt;endothermic for the $\ce{HI}$ case&lt;/strong&gt;.  Already we see that hydrogen iodide will not add to olefins &lt;em&gt;via&lt;/em&gt; a radical pathway at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond.  Let's perform a similar thermodynamic analysis of this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|} \hline&#xA;\ce{H-X} &amp;amp; \text{Bond Strength} &amp;amp; \ce{C-H}~ \text{Bond Strength} &amp;amp; \text{Overall}  \\ &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)}\\ \hline&#xA;\ce{H-F} &amp;amp; 135 &amp;amp; 99 &amp;amp; 36\\ \hline&#xA;\ce{H-Cl} &amp;amp; 103 &amp;amp; 99 &amp;amp; 4\\ \hline&#xA;\ce{H-Br} &amp;amp; 88 &amp;amp; 99 &amp;amp; -11\\ \hline&#xA;\ce{H-I} &amp;amp; 71 &amp;amp; 99 &amp;amp; -28\\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot; rel=&quot;nofollow&quot;&gt;source for bond strengths&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both propagation steps (1) and (2) must occur for the reaction to proceed.  &lt;strong&gt;Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps.  For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-29T13:46:51.763" LastActivityDate="2015-04-29T13:46:51.763" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="29370" PostTypeId="5" CreationDate="2015-04-26T15:06:45.850" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-26T15:06:45.850" LastActivityDate="2015-04-26T15:06:45.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29371" PostTypeId="4" CreationDate="2015-04-26T15:06:45.850" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-26T15:06:45.850" LastActivityDate="2015-04-26T15:06:45.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29372" PostTypeId="1" CreationDate="2015-04-26T15:32:31.200" Score="3" ViewCount="387" Body="&lt;p&gt;I have some problems with water and metals in binding site when using Autodock Vina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to make these flexible? They unfortunately remain where they are what leads to wrong interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My other idea would be to delete it before docking, reinsert them and minimize everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a solution for metalloenzymes? (e.g. PDB:3NJY)&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2019-11-01T10:00:34.373" Title="Autodock Vina flexible metals and water molecules" Tags="&lt;computational-chemistry&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29373" PostTypeId="1" AcceptedAnswerId="29375" CreationDate="2015-04-26T15:41:49.947" Score="-5" ViewCount="39571" Body="&lt;p&gt;what is the range of strong acid and base in pH scale.I have much confusion.&lt;/p&gt;&#xA;" OwnerUserId="15822" LastEditorUserId="-1" LastEditDate="2016-07-27T17:00:26.647" LastActivityDate="2016-07-27T17:00:26.647" Title="Is there any acid or base which has a pH value under 0 and over 14?" Tags="&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-11-13T21:43:29.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="29374" PostTypeId="1" CreationDate="2015-04-26T15:43:19.697" Score="5" ViewCount="104" Body="&lt;p&gt;If I use flexible residues in the binding pocket, they are cut from the file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protein is broken and the amino acids are unbound in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get my results and the protein together again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hvVlz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5614" LastActivityDate="2019-11-01T11:01:40.257" Title="Merge rigid parts and flexible results after docking (Vina)" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29375" PostTypeId="2" ParentId="29373" CreationDate="2015-04-26T15:47:32.687" Score="-1" Body="&lt;p&gt;The solution with &lt;strong&gt;lowest pH value&lt;/strong&gt; (to the left of pH scale) will be the &lt;em&gt;strongest acid&lt;/em&gt; while the solution with &lt;strong&gt;highest pH value&lt;/strong&gt; (to the right of pH scale) will be the &lt;em&gt;strongest base&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In technical words, the acid which dissociates completely into H+ ions in water will have the lowest pH value, while the base dissociates completely into OH- ions in water will have the highest pH value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; Strongest acid has pH of 1 while strongest base has pH of 14.&#xA;(0 - 6.99 are acids. 7 is neutral. 7.1 - 14 are bases.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="15711" LastEditorUserId="15711" LastEditDate="2015-04-26T15:50:58.963" LastActivityDate="2015-04-26T15:50:58.963" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29376" PostTypeId="2" ParentId="29373" CreationDate="2015-04-26T15:51:53.107" Score="3" Body="&lt;p&gt;Usually the range of a $pH$ scale is $0-14$. A higher number on the $pH$ scale means that the substance is &lt;strong&gt;Basic&lt;/strong&gt;, a lower number means it is &lt;strong&gt;Acidic&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought to keep in mind is that there can be a $pH$ less than $0$ and greater than $14$. But let's not get into that right now. The pH scale was made for our convenience and cannot be a totally accurate representation of the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Have you done some research on this? Be more specific in what you do not understand.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible duplicates:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/4081/what-is-the-difference-between-the-titration-of-a-strong-acid-with-a-strong-base?rq=1&quot;&gt;What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/10650/why-do-strong-acids-have-weak-conjugate-bases?rq=1&quot;&gt;Why do strong acids have weak conjugate bases?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/24193/what-is-a-strong-base?rq=1&quot;&gt;What is a strong base?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ed083p1465&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; out for more information.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-26T15:58:29.747" LastActivityDate="2015-04-26T15:58:29.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29377" PostTypeId="2" ParentId="29366" CreationDate="2015-04-26T16:23:44.827" Score="2" Body="&lt;p&gt;Perhaps you have heard of the following terminology: &quot;rock salt&quot;, &quot;cesium chloride&quot;, &quot;zinc blende&quot; structures? If so, you may recall that in this case, the main factor dictating structure is &quot;size&quot; or atomic radius ratio, rather than group. For example, in the case of the rock salt structure, it is most likely that there is &quot;a cation/anion radius ratio of 0.414 to 0.732&quot;. As such, you could consider the radius of cesium in comparison to sodium: significantly larger. Thus, it adopts the cesium chloride configuration, which is preferred when &quot;ions are of roughly the same size &quot;. &lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-26T16:23:44.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29379" PostTypeId="1" CreationDate="2015-04-26T17:00:03.380" Score="3" ViewCount="153" Body="&lt;p&gt;This is a topic that I cannot find much information about. I would like to know what compounds are in the products formed from pyrolysis of canola oil. I found out by accident that when trying to ignite some used canola oil in an aluminum can without a cover, the oil had become a mix of black solids and black liquid. I have not tried heating the oil in a container that has no exposure to oxygen. I could probably get better results by limiting all exposure to oxygen. I recently tried this experiment again and the oil is slowly turning black and looking like crude oil. I did find one article on this subject that says there are alkenes in the oil. Does anyone have more information on this?&lt;/p&gt;&#xA;" OwnerUserId="15823" LastEditorUserId="30836" LastEditDate="2019-04-14T17:08:21.973" LastActivityDate="2019-04-14T17:08:21.973" Title="Pyrolysis of canola oil" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;&lt;pyrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29380" PostTypeId="1" AcceptedAnswerId="29394" CreationDate="2015-04-26T17:30:39.220" Score="4" ViewCount="90" Body="&lt;p&gt;My book doesn't seem to properly explain the importance of derivating the Gaus spectral line, neither the meaning of it, so my questions are : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt;.What do the spectral lines of 1st, 2nd, 3rd and 4th derivates of the basic Gaus specter tell us exactly ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt;.Why are the second and the fourth derivated spectral lines more used in practice ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt;.How does the amplitude of those spectral lines indicate in our results ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the lack of image.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="7448" LastEditDate="2015-04-26T17:46:37.857" LastActivityDate="2015-04-26T20:49:42.900" Title="Derivation spectrum" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29381" PostTypeId="2" ParentId="29379" CreationDate="2015-04-26T17:41:26.743" Score="2" Body="&lt;p&gt;Basically canola oil is obtained from a cultivated form of Rapeseed. Canola oil is just what &lt;a href=&quot;http://en.wikipedia.org/wiki/Rapeseed&quot; rel=&quot;nofollow noreferrer&quot;&gt;rapeseed oil&lt;/a&gt; is called in North America.  From a marketing perspective it was thought that having the term &quot;rape&quot; in the product name would not help sales.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look for articles on rapeseed oil pyrolysis you'll find a lot of information since rapeseed (canola) oil is used in biodiesel production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/18511268&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; states that the major products produced from pyrolysis in a reactor include &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gases: $\ce{CO2, CO, CH4}$ and $\ce{H2S}$&lt;/li&gt;&#xA;&lt;li&gt;oils: oleic acid, 1H-indole, 2,3,5-trimethoxy toluene, toluene, (Z)-9-octadecanamide, psoralene, phenol and phenol derivatives&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I didn't know what &quot;psoralene&quot; was, so here is a picture and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Psoralen&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1s1m6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-26T17:41:26.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29382" PostTypeId="1" AcceptedAnswerId="29383" CreationDate="2015-04-26T17:50:04.527" Score="2" ViewCount="3044" Body="&lt;p&gt;I saw a test question today that basically boiled down to comparing the ionization energies of $\ce{C}$ and $\ce{Cl}$. I know that in the periodic table, ionization energy generally increases as we move right across a row (because nuclear charge increases), and decreases as we move down a column (because atomic radius increases, so valence electrons feel less attraction to the nucleus). Note that $\ce{Cl}$ is in the row below $\ce{C}$, but is also to the right of $\ce{C}$, so it is not obvious how to compare the two. So, without looking up the ionization energies and using only the periodic table, how can I answer this question?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="9972" LastEditDate="2015-04-26T19:08:43.397" LastActivityDate="2015-04-26T19:53:45.173" Title="Compare the ionization energies of C and Cl?" Tags="&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29383" PostTypeId="2" ParentId="29382" CreationDate="2015-04-26T17:59:48.190" Score="4" Body="&lt;p&gt;It's the other way around,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;moving left to right across a row &lt;strong&gt;increases&lt;/strong&gt; the IP due to increasing nuclear charge, and &lt;/li&gt;&#xA;&lt;li&gt;moving down a column &lt;strong&gt;decreases&lt;/strong&gt; the IP, due to increasing shielding of the nuclear charge by core electrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Chlorine is only &lt;strong&gt;one&lt;/strong&gt; row below carbon, but it is &lt;strong&gt;three&lt;/strong&gt; columns to the right.  As the following chart illustrates, moving to the right by one position generally has a similar or larger effect than moving down one position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rbB8K.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Noble_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in this case moving further to the right by 3 positions (and increasing the IP) &lt;strong&gt;strongly outweighs&lt;/strong&gt; moving down by only one position (and decreasing the IP).  Hence, in this case the IP of chlorine would be predicted to be greater than the IP of carbon.  Indeed, the IP of chlorine is reported to be 13.0 eV, while that of carbon is only 11.3 eV.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-26T19:53:45.173" LastActivityDate="2015-04-26T19:53:45.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29384" PostTypeId="2" ParentId="29294" CreationDate="2015-04-26T18:12:39.653" Score="5" Body="&lt;p&gt;This question refers to the use of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tollens%27_reagent&quot; rel=&quot;nofollow&quot;&gt;Tollens' Reagent&lt;/a&gt; (ammoniacal silver nitrate). The oxidising species here is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tollens' reagent is usually used as a test for aldehydes, with which it gives a characteristic silver mirror: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://c1.staticflickr.com/3/2053/1507342827_94b449ea9a_m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often said that Tollens' reagent does not react with alcohols. However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate. Again, in my experience, secondary alcohols (such as in the question) do not react with Tollens' reagent and so I would predict that no reaction will take place here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Your proposed mechanism certainly looks plausible but whether it will take place in reality I can't say for certain (my experience of Tollens' reagent says no).&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-04-26T19:29:05.087" LastActivityDate="2015-04-26T19:29:05.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29385" PostTypeId="1" CreationDate="2015-04-26T18:22:06.753" Score="3" ViewCount="3979" Body="&lt;p&gt;If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?&lt;/p&gt;&#xA;" OwnerUserId="10201" LastEditorUserId="13184" LastEditDate="2015-04-26T18:46:04.590" LastActivityDate="2016-04-23T20:32:25.043" Title="Mass Spectrum and Molecular Ion peaks" Tags="&lt;mass-spectrometry&gt;&lt;isotope&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ClosedDate="2017-08-20T00:28:58.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="29386" PostTypeId="2" ParentId="29385" CreationDate="2015-04-26T18:29:34.170" Score="6" Body="&lt;p&gt;The major isotopes of chlorine are 35 and 37.  If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;35, 35, 35, 35&lt;/li&gt;&#xA;&lt;li&gt;35, 35, 35, 37&lt;/li&gt;&#xA;&lt;li&gt;35, 35, 37, 37&lt;/li&gt;&#xA;&lt;li&gt;35, 37, 37, 37&lt;/li&gt;&#xA;&lt;li&gt;37, 37, 37, 37&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So there will be &lt;strong&gt;5&lt;/strong&gt; molecular ion peaks.  You could calculate their relative heights by factoring in the relative isotopic proportions, along with the number of permutations of each of the above combinations.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-04-26T18:29:34.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29387" PostTypeId="1" CreationDate="2015-04-26T19:11:42.020" Score="2" ViewCount="102" Body="&lt;p&gt;Little open information can be found on the requirements of the &lt;em&gt;manufacture, development and testing&lt;/em&gt; of &lt;strong&gt;gel and printing inks&lt;/strong&gt;. &#xA;It seems that most ink manufacturers have their own specific lab configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you know of a recource, or could you list me a possible lab setup for such an endeveavor? While I have found books on the topic, few of them are of very practical nature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of specific interest to me are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;analytic methods&lt;/li&gt;&#xA;&lt;li&gt;synthetic apparatus&lt;/li&gt;&#xA;&lt;li&gt;testing procedures and protocols&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you very much! &lt;/p&gt;&#xA;" OwnerUserId="15825" LastEditorUserId="7448" LastEditDate="2015-04-26T19:34:20.120" LastActivityDate="2015-04-26T19:34:20.120" Title="Ink / formulation chemistry: What lab requirements" Tags="&lt;experimental-chemistry&gt;&lt;solutions&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29388" PostTypeId="1" CreationDate="2015-04-26T19:14:08.563" Score="3" ViewCount="686" Body="&lt;p&gt;I’m doing an electrochemistry lab that requires me to construct various galvanic cells in an attempt to solve for $\Delta H$, $\Delta G$, and $\Delta S$ values of an unknown. During one part of the experiment, I had to heat a cell to a temperature of 70 °C and record its voltage in 15 degree increments until the temperature reached 10 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cell I am using to perform this step on is represented by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn + Sn^2+ -&amp;gt; Zn^2+ + Sn}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After performing the procedure, I’m getting concerned that my data may be inaccurate. Theoretically, I was using 0.1 M concentrations of each solution in the cells. According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because $\ln(0.1/0.1) = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to my professor, I should be seeing a linear relationship when I plot $\Delta G$, calculated from the voltage measurements I took, vs the temperature in K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, theoretically, &lt;strong&gt;what should happen to the voltage as I heat the cell&lt;/strong&gt;, assuming that the concentrations of the solutions were off? I’ve calculated the $\Delta S$ value based on the &lt;a href=&quot;http://www.chemistry-reference.com/Standard%20Thermodynamic%20Values.pdf&quot; rel=&quot;nofollow&quot;&gt;standard values&lt;/a&gt; listed for these entities and I’m getting a value of −85.18 J/(K mol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculation: $\Delta S = (-112.1 + 51.55) - (41.63 + -17.0) = -85.18\ \mathrm{J/(mol\ K)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based off of this result, it would appear that the slope of this plot must come out to be a positive value because $m = - \Delta S$. Does this mean that I should be seeing an increase in voltage as temperature decreases? Regarding the $\Delta H$ of this reaction, which is represented by the Y-intercept (b) of the equation of the best fit line for my data, it appears I must also have a negative value of −145090 (in units of J) based upon the calculation I did for this, also using standard values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculation: $\Delta H = (-153.89 + 0) - (0 + -8.8) = -145.09\ \mathrm{kJ/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve tried to be as detailed as possible in describing by question, and hopefully it makes sense. I’m just getting confused on what the data should look like. Unfortunately, its not possible for me to redo the experiment. Please note that this is for my own understanding, I will not submit the assignment with fixed data. I appreciate any assistance that anyone can offer me.&lt;/p&gt;&#xA;" OwnerUserId="15824" LastEditorDisplayName="user7951" LastEditDate="2015-04-26T20:05:21.310" LastActivityDate="2015-04-26T20:46:20.103" Title="Plotting ΔG vs Temp for Electrochemical Reaction" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;enthalpy&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29389" PostTypeId="2" ParentId="29385" CreationDate="2015-04-26T20:05:35.253" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Limitations of existing answers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Ron's answer is good for molecules that consist solely of chlorine, i.e., a hypothetical $\ce{Cl4}$ molecule.  But if there are other atoms [&lt;em&gt;of elements that have more than one stable isotope&lt;/em&gt;] as well, then there will be more peaks.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An example using difluorotetrachlorobenzene&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose that the molecule is an asymmetric difluorotetrachlorobenzene, $\ce{C6Cl4F2}$.  Then all of the fluorine atoms will be $\ce{^{19}F}$, but every one of the six carbon atoms could be either $\ce{^{13}C}$ &lt;em&gt;or&lt;/em&gt; $\ce{^{12}C}$, and every one of the four chlorine atoms could be $\ce{^{35}Cl}$ or $\ce{^{37}Cl}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A complete list of all isotopologues&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since there are six carbon atoms, there are 7 possible values for the total mass of carbon in each molecule (0 $\ce{^{13}C}$ atoms, 1 $\ce{^{13}C}$ atom, ... up to 6 $\ce{^{13}C}$ atoms).  Similarly there are 5 possible values for chlorine, so there are $7\times 5=~~$ &lt;strong&gt;35 different molecular ions&lt;/strong&gt;!  Here is a complete list:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       mass log10(abund) 12C 13C 35Cl 37Cl 19F&#xA;1  249.8722   -0.5102721   6   0    4    0   2&#xA;2  251.8693   -0.4031194   6   0    3    1   2&#xA;3  253.8663   -0.7219355   6   0    2    2   2&#xA;4  255.8634   -1.3929341   6   0    1    3   2&#xA;5  257.8604   -2.4899015   6   0    0    4   2&#xA;6  250.8756   -1.6980651   5   1    4    0   2&#xA;7  252.8726   -1.5909124   5   1    3    1   2&#xA;8  254.8697   -1.9097285   5   1    2    2   2&#xA;9  256.8667   -2.5807271   5   1    1    3   2&#xA;10 258.8638   -3.6776945   5   1    0    4   2&#xA;11 251.8789   -3.2660693   4   2    4    0   2&#xA;12 253.8760   -3.1589166   4   2    3    1   2&#xA;13 255.8730   -3.4777327   4   2    2    2   2&#xA;14 257.8701   -4.1487313   4   2    1    3   2&#xA;15 259.8671   -5.2456987   4   2    0    4   2&#xA;16 252.8823   -5.1070748   3   3    4    0   2&#xA;17 254.8793   -4.9999221   3   3    3    1   2&#xA;18 256.8764   -5.3187382   3   3    2    2   2&#xA;19 258.8734   -5.9897368   3   3    1    3   2&#xA;20 260.8705   -7.0867042   3   3    0    4   2&#xA;21 253.8856   -7.1979577   2   4    4    0   2&#xA;22 255.8827   -7.0908051   2   4    3    1   2&#xA;23 257.8797   -7.4096212   2   4    2    2   2&#xA;24 259.8768   -8.0806198   2   4    1    3   2&#xA;25 261.8738   -9.1775871   2   4    0    4   2&#xA;26 254.8890   -9.5618420   1   5    4    0   2&#xA;27 256.8860   -9.4546893   1   5    3    1   2&#xA;28 258.8831   -9.7735054   1   5    2    2   2&#xA;29 260.8801  -10.4445040   1   5    1    3   2&#xA;30 262.8772  -11.5414714   1   5    0    4   2&#xA;31 255.8923  -12.3059456   0   6    4    0   2&#xA;32 257.8894  -12.1987929   0   6    3    1   2&#xA;33 259.8864  -12.5176090   0   6    2    2   2&#xA;34 261.8835  -13.1886076   0   6    1    3   2&#xA;35 263.8805  -14.2855750   0   6    0    4   2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h3&gt;Why the complete list isn't always observed&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Resolution limitations.  Many types of mass spectrometers won't be able to resolve the very small mass differences between some of these ions.  For example, $m/z=251.8693$ for $\ce{^{12}C6~^{37}Cl1~^{35}Cl3~F2}$ is only 0.0096 Daltons away in mass from $m/z=251.8789$ for $\ce{^{12}C4~^{13}C2~^{35}Cl4~F2}$.  In many instruments such peaks would not be well-resolved.  &lt;/li&gt;&#xA;&lt;li&gt;Sensitivity limitations.  Additionally, many of the 35 distinct isotopologues are very rare in abundance, well below practical limits of detection.  The abundance of $\ce{^13C6~^37Cl4~F2}$ is about &lt;em&gt;fourteen orders of magnitude&lt;/em&gt; lower than the most abundant mass isotopologue, $\ce{^{12}C6~^{35}Cl3~^{37}Cl1~F2}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;A theoretical spectrum at unit-mass resolution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The end result is that in most mass spectrometers, &lt;strong&gt;&lt;em&gt;more than 5&lt;/em&gt;&lt;/strong&gt; but many &lt;strong&gt;&lt;em&gt;fewer than 35&lt;/em&gt;&lt;/strong&gt; peaks will be observed for our example molecule of difluorotetrachlorobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are approximate results for an instrument with ~0.5 Da mass resolution, such as most quadrupole detectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fgNsw.png&quot; alt=&quot;Plot of peaks for difluorotetrachlorobenzene at 0.5 Da resolution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;How I have been calculating these results&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am a big fan of an R package called &lt;a href=&quot;http://cran.r-project.org/web/packages/ecipex/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;ecipex&lt;/code&gt;&lt;/a&gt;.  With that package these calculations are easy.  Here is my code to generate the list above.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;require(ecipex)&#xA;df &amp;lt;- ecipex('C6Cl4F2', id=T, limit=1e-16)[[1]]&#xA;dfordered &amp;lt;- df[order(df[,'12C', decreasing=True]), ]&#xA;dfordered$abundance &amp;lt;- log10(dfordered$abundance)&#xA;names(dfordered)[2] &amp;lt;- 'log10(abund)'&#xA;dfordered&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And to generate the graph:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;require(ggplot2)&#xA;ggplot(df, aes(x=mass, weights=abundance)) + &#xA;  geom_bar(stat='bin', binwidth=0.5, color='black', fill='gray') + &#xA;  theme_bw() + &#xA;  xlab('m/z, Da') + &#xA;  ylab('abundance') + &#xA;  scale_x_continuous(breaks = 248:258, limits = c(248,258))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2016-03-13T13:53:59.993" LastActivityDate="2016-03-13T13:53:59.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29390" PostTypeId="2" ParentId="29330" CreationDate="2015-04-26T20:18:30.537" Score="9" Body="&lt;p&gt;Coincidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While pyrite crystallises in spacegroup $\mathrm{Pa\bar{3}}$, i.e., a cubic spacegroup, there is no guarantee that the unit cell shape is resembled in  the macroscopic crystal shape. The faces of the crystal do however always represent a plane that can be described by some Miller indices of the crystal lattice. And indeed, there are multiple alternative crystal habits for pyrite (eg dodecahedral, see Wikipedia article and santiago's answer), while retaining the same internal structure. &#xA;Which habit forms in practice, is mainly determined by the crystallisation conditions. These have to be determined empirically, as there is no way to predict them.&lt;/p&gt;&#xA;" OwnerUserId="10050" LastEditorUserId="4945" LastEditDate="2018-07-26T08:56:12.260" LastActivityDate="2018-07-26T08:56:12.260" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29392" PostTypeId="2" ParentId="29388" CreationDate="2015-04-26T20:46:20.103" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is tough to answer because I don't know what you mean by &quot;off&quot;.  So let me ask (and answer) a few related questions that I think will help you.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What should happen to voltage as a function of temperature in an electrochemical cell&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The electrical potential of electrochemical cells &lt;em&gt;varies with temperature&lt;/em&gt;, &lt;em&gt;even when the cell is at standard conditions&lt;/em&gt;.  &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd355.pdf&quot; rel=&quot;nofollow&quot;&gt;A 1985 NIST document&lt;/a&gt; explains the situation well.  Only the standard hydrogen electrode has a standard potential that does not vary with temperature.  (The SHE potential is defined as 0 volts at every temperature.)  For other cells, usually the assumption of a linear variation with temperature is usually very good.  That is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\cal E}_T^\circ = {\cal E}_{298}^\circ  + (T-298.15)\cdot \frac{d{\cal E}_{298}^\circ}{dT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this variation is completely unrelated to the Nernst equation.  The same reference gives values for $\frac{d{\cal E}_{298}^\circ}{dT}$ for zinc(II)/zinc metal of 0.119 milliVolts per Kelvin and for tin(II)/tin metal -0.32 milliVolts per Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Don't be confused by the $T$ in the Nernst equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Nernst_equation&quot; rel=&quot;nofollow&quot;&gt;Nernst equation&lt;/a&gt; is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {\cal E}= {\cal E}^\circ + \frac{RT}{nF}\ln Q$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does contain a $T$ in there, but don't be mislead into thinking that the $T$ you see is the sole means by which a cell potential ${\cal E}$ can vary with temperature.  ${\cal E}^\circ$ also varies with temperature (see above).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because ln(0.1/0.1)=0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, hopefully my answer makes clear that this statement is wrong, because ${\cal E}^\circ$ varies with temperature.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-26T20:46:20.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29393" PostTypeId="2" ParentId="29285" CreationDate="2015-04-26T20:47:44.237" Score="1" Body="&lt;p&gt;Your protocol gives you the answer: to make sure your sediment is neutral. After all, you titrate with NaOH afterwards, so any wrong starting point in terms of pH would skew your result.&lt;/p&gt;&#xA;" OwnerUserId="10050" LastActivityDate="2015-04-26T20:47:44.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29394" PostTypeId="2" ParentId="29380" CreationDate="2015-04-26T20:49:42.900" Score="1" Body="&lt;p&gt;&lt;em&gt;Derivative spectroscopy&lt;/em&gt; you refer to is a senior technique deployed often in UV-Vis spectroscopy, yet there are examples in the field of infra-red and RAMAN spectroscopy and cyclic dichorism.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The theme in common is that at the resolution of the typical&#xA;experiment, while changing the wavelength of observation the&#xA;measured signal alters gradually and continuously and you observe&#xA;for example absorption &lt;em&gt;bands&lt;/em&gt; -- and not, like in atom absorption&#xA;spectroscopy, for example, discrete absorption &lt;em&gt;lines&lt;/em&gt;.  The following example sticks to UV-Vis absorption spectroscopy.&lt;/li&gt;&#xA;&lt;li&gt;Often, absorption bands observed consist of several contributions, i.e. there are more than one individual absorptions differing slightly only in terms of their corresponding energy of absorption -- and instead of a pure Gaussian shape of your absorption band, the enveloping absorption band observed features shoulders, for example.&lt;/li&gt;&#xA;&lt;li&gt;To determine the derivative of the spectrum recorded helps to determine the centres of the individual peaks, to ease the discrimination of the individual components of one absorption band.  This may lead to spectral deconvolution. A pictural introduction including examples and limits of application is provided &lt;a href=&quot;http://www.youngin.com/application/AN-0608-0115EN.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-26T20:49:42.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29395" PostTypeId="2" ParentId="29385" CreationDate="2015-04-26T21:28:38.417" Score="2" Body="&lt;p&gt;This answer simply aims to add to the one by Curt F. that there is software available to do these calculations, too.  Beside the ones with the mass spectrometers, and assuming Ali Naqvi participates in initial lab classes, the Chemical Calulator of the &lt;a href=&quot;http://gchemutils.nongnu.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gnome Chemistry utilities&lt;/a&gt; may be worth a trial to get familiar with typical isotope patterns of organic molecules.  As a training, I found this beneficial to recognize them in a routine analysis with ease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To return to the initial question, a $\ce{Cl4}$ would appear as this -- note there are five peaks.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HlLOb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On contrast, Ron's example of C6Cl4F2 -- nine peaks -- were this one:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/beM2c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-26T21:28:38.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29397" PostTypeId="2" ParentId="27844" CreationDate="2015-04-27T00:42:51.230" Score="0" Body="&lt;p&gt;You must understand the extra information has nothing to do with any of your calculations. It is just there to throw you off although it is satirical. You must understand that if you feel that you have arrived at the right answer, and it makes sense, then there is no reason for you to second guess yourself, especially in Chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. :)&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-27T00:42:51.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29399" PostTypeId="1" CreationDate="2015-04-27T00:53:48.400" Score="0" ViewCount="639" Body="&lt;p&gt;Calculate the $\ce{pH}$ of a $100$ $\ce{mL}$ solution containing $0$$.0375$ moles of the weak base, Sodium Benzoate, $\ce{C6H5COONa}$. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{C6H5COO + H2O -&amp;gt; C6H6COO- + OH-}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.0375 ~\mathrm{mol}}{.100~ \mathrm{L}} = 0.375~M$$&#xA;$$\ce{pOH} = 0.43$$&#xA;$$\ce{pH} = 13.57$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this right? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="15831" LastEditorUserId="14" LastEditDate="2015-06-14T15:04:21.177" LastActivityDate="2015-06-14T15:04:21.177" Title="pH of a weak base calculation" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29400" PostTypeId="1" AcceptedAnswerId="29408" CreationDate="2015-04-27T01:00:48.537" Score="6" ViewCount="16323" Body="&lt;p&gt;What is the pH of a solution of $3.4 \times 10^{-10}$ moles of $\ce{HCl}$ in a volume of $\pu{150.7L}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is dilute so I expect pH to be high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{3.4 \times 10^{-10}\ \mathrm{mol}}{150.7\ \mathrm L} = 2.26 \times 10^{-12} \frac{\mathrm{mol}}{\mathrm L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then pH is 11.65. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="15831" LastEditorUserId="5026" LastEditDate="2018-03-01T02:41:07.740" LastActivityDate="2018-03-01T02:41:07.740" Title="pH of dilute HCl solution" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2020-06-10T18:58:54.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="29401" PostTypeId="1" CreationDate="2015-04-27T01:07:12.100" Score="1" ViewCount="6861" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;One kg of each substance below is dissolved in a bathtub full of water. Will it be acidic, basic, or neutral?&lt;br&gt;&#xA;  $\ce{KNO2}$&lt;br&gt;&#xA;  $\ce{HF}$&lt;br&gt;&#xA;  $\ce{(CH3)3NHBr}$&lt;br&gt;&#xA;  $\ce{HO-CH2-CH2-O-CH2-Cl}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{KNO2}$, $\ce{K+}$ $\ce{NO2-}$ $\ce{K+}$ is inert and $\ce{NO2-}$ is the base of strong acid, so it should be basic overall.&#xA;$\ce{HF}$, $\ce{H+}$ $\ce{F-}$ $\ce{H+}$ is strong acid and $\ce{F-}$ is strong base so it should be neutral overall.&#xA;$\ce{(CH3)3NHBr}$, Molecule acts like a $\ce{H3NHBr}$ or $\ce{NH4+}$ $\ce{Br-}$ so $\ce{NH4+}$ is acidic $\ce{Br-}$ is neutral, so overall solution is acidic.&lt;br&gt;&#xA;$\ce{HO-CH2-CH2-O-CH2-Cl}$, No idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain them, if they're wrong? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="15831" LastEditorUserId="865" LastEditDate="2021-05-22T20:58:01.587" LastActivityDate="2021-05-22T20:58:01.587" Title="How to determine the acidity or basicity of dissolved salts?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29402" PostTypeId="2" ParentId="29399" CreationDate="2015-04-27T01:22:10.880" Score="4" Body="&lt;p&gt;&lt;strong&gt;Welcome to Chemistry SE!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this specific problem you need to use the $K_{b}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac{x^{2}}{.375-x}}=K_{b}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the $K_{b}$ and solving for x will give you the concentration of $\ce{OH-}$. Remember to take the $-log$ of that and solve for the $pH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also next time please show some work and take your time to &lt;a href=&quot;https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here&quot;&gt;format&lt;/a&gt; these questions so people can try to reply back to you.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="-1" LastEditDate="2017-03-16T15:50:39.383" LastActivityDate="2015-04-27T01:24:52.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29404" PostTypeId="2" ParentId="29400" CreationDate="2015-04-27T01:49:34.227" Score="4" Body="&lt;p&gt;As you may have noticed, your answer of 11.65 is counter-intuitive as it is higher than 7, meaning that dilute hydrochloric acid is somehow more BASIC than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is accounting for the $1.0 \times 10^{-7} M$ H+ ions, or $1.507 \times 10^{-5}$ moles of H+ ions for 150.7L, already present in the water due to self-ionization. An addition of $3.4 \times 10^{-10}$ more moles of H+ ions would be practically insignificant (5 orders of magnitude difference). Hence, the pH would be near 7. &lt;/p&gt;&#xA;" OwnerUserId="15531" LastEditorUserId="5456" LastEditDate="2015-04-27T05:58:17.977" LastActivityDate="2015-04-27T05:58:17.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29405" PostTypeId="1" CreationDate="2015-04-27T02:10:18.283" Score="5" ViewCount="13370" Body="&lt;p&gt;I read this today:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you mix equal amounts of $1 M$ $\ce{H_3PO_4}$ and $1 M$ $\ce{NaH_2PO_4}$ you get a buffer at $pH&amp;lt;6$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason given is that a weak acid plus a salt containing the conjugate base forms a acidic buffer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you mix equal amounts of $1 M$ $\ce{NH_3}$ and $1 M$ $\ce{NH_4Cl}$, you get a buffer at $pH&amp;gt;8$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason given is that a weak base plus a salt containing the conjugate acid forms a basic buffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this is true. Let's take the first statement for example. If you have equal amounts of a weak acid and its conjugate base, then by definition you have a buffer. Whether the buffer is acidic or basic shouldn't depend on whether the salt contains the conjugate acid or the conjugate base, right? &lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="4945" LastEditDate="2015-04-27T07:31:45.827" LastActivityDate="2019-01-23T08:26:17.837" Title="Why does a weak acid and a salt containing its conjugate base form an acidic buffer?" Tags="&lt;acid-base&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29406" PostTypeId="2" ParentId="29401" CreationDate="2015-04-27T02:25:20.833" Score="1" Body="&lt;p&gt;The $\ce{KNO2}$ is basic, the HF is acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acid too, the last one is an alcohol, I would consider it neutral for your purpose. But you are missing one step to get your answer quite right.&#xA;Last part, in the pH world, everything turn around two ions H+ (or H3O+) and OH- and the equilibrium between those two. A substance is considered acid when it either give a H+ as disolving or takes an OH-. On the other hand a substance is considered basic when it give an OH- or takes a H+. Like HF is an acid (H+, F-) and NaOH is basic (Na+, OH-). For KNO2, it can take an H+ to go back to HNO2, so it is basic. For organic compounds, alcool groups are usually considered inerte as they don't ionate in normal conditions.&#xA;Now, maybe I'm wrong but your question sound like if you put all those in a bathtub what'S the final result? Is it right or not?&lt;/p&gt;&#xA;" OwnerUserId="15795" LastEditorUserId="15795" LastEditDate="2015-04-28T01:41:13.467" LastActivityDate="2015-04-28T01:41:13.467" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29408" PostTypeId="2" ParentId="29400" CreationDate="2015-04-27T02:36:34.067" Score="6" Body="&lt;p&gt;Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&#xA;$$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and also the dissociation of the acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + HCl -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&#xA;$$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $ x = \ce{[H3O+]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 - c_0x - Kw = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be easily solved. Taking only the positive value of $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting the values of $x$ and $K_w$, the result is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is not surprising at all, but is the calculated answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-04-27T02:41:44.857" LastActivityDate="2015-04-27T02:41:44.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29410" PostTypeId="2" ParentId="29367" CreationDate="2015-04-27T03:23:06.950" Score="3" Body="&lt;p&gt;As you already mentioned, there are numerous types of chromatography, and they vary in several different ways. I can't tell you what would be best ways to separate each of these pair of substances, but I can give you some insights. Using chromatography always requires optimization of experimental procedures, and the first step is usually searching on papers and other references for information on how the specific separation is usually done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N-phenylethanamide/N-phenylpropionamide and methoxymethane/ethoxyethane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite being very similar, the presence of one or two extra $\ce{CH2}$ group in each of the second molecules is enough to give the molecules different properties, specially considering they aren't big. For simple detection purposes, I'm sure it would be possible to find the appropriate solvent mixture to separate them using TLC or column chromatography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more advanced technique, gas chromatrography is very powerful and could also be used to separate them, as long as the appropriate setup was used (injection and oven temperatures w or w/o gradients, column type and length, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aβ36/Aβ42&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These would probably decompose in elevated temperatures, so gas chromatography is not a good idea. Peptides are, nowadays, mostly separated using liquid chromatography, usually in HPLC or UPLC systems using reversed phase columns. This type of chromatography will separate the peptides mostly by their polarity and size, and once again, reference search and experimental trials are essential to optimize conditions. HPLC and UPLC are used to separate mixtures of thousands of peptides (with varying efficiency, of course), but that should be a sign that two wouldn't be a problem, even if very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the objective of the separation (identification and/or recovery), using CNBr or any other digestion step could be useful or not, but I don't think this would be the case for only two peptides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PAGE and the variations of electrophoresis were for a long time the goto separation method for peptides and proteins, but recent developments in chromatography (and mass spectrometry) have made them fall off preference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with affinity chromatography (or the whole antibody thing, to be honest), so I can't say much about that, but, what I do know is that you won't always find or be able to manufacture the chromatographic column with the desired specificity. &lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="5026" LastEditDate="2018-06-24T11:58:57.373" LastActivityDate="2018-06-24T11:58:57.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29411" PostTypeId="1" CreationDate="2015-04-27T03:24:31.563" Score="1" ViewCount="39769" Body="&lt;p&gt;Here is what seems to be a straightforward question: you dissolve some ammonium chloride in water at $25\ \mathrm{^\circ C}$, lowering the temperature of the solution. Is the change in enthalpy for the dissolving process positive or negative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning: if the temperature of the solution decreases, it means heat was released into the surroundings, so the enthalpy change is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the answer is the enthalpy change is positive. What am I doing wrong? Whatever it is, it must be stupid, but please forgive me for its late here.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorDisplayName="user7951" LastEditDate="2015-04-27T21:08:30.300" LastActivityDate="2015-04-27T21:08:30.300" Title="If the temperature of a solution decreases, what is the sign of the enthalpy change?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29412" PostTypeId="2" ParentId="28856" CreationDate="2015-04-27T04:07:16.600" Score="3" Body="&lt;p&gt;The geometry optimization routine does not appear to have a method to output frames to an &lt;code&gt;.xtd&lt;/code&gt; file for CASTEP in Materials Studio. Usually a movie makes more sense for a molecular dynamics simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you construct a script that performs a sequence if geometry optimizations with &lt;code&gt;MaxIterations = 3&lt;/code&gt; (must be &lt;code&gt;&amp;gt;2&lt;/code&gt;), you can output each result to &lt;code&gt;.std&lt;/code&gt; (study table) or &lt;code&gt;.xtd&lt;/code&gt;. You would do this by outputting a geometry optimization after three iterations, then repeat the calculation on the result, until convergence. Then you take the frames and make a video&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The help file in Materials Studio is rather a good resource, be sure to dig around in there, especially for the scripting API&lt;/p&gt;&#xA;" OwnerUserId="15065" LastEditorUserId="15065" LastEditDate="2015-04-27T13:08:30.260" LastActivityDate="2015-04-27T13:08:30.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29417" PostTypeId="2" ParentId="27341" CreationDate="2015-04-27T05:48:38.637" Score="3" Body="&lt;p&gt;This topic came up the other day but in regards to the evaporation of water.  You might find &lt;a href=&quot;https://chemistry.stackexchange.com/a/28432/13184&quot;&gt;my answer&lt;/a&gt; to that question helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I would solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Explicitly note standard state used and assumptions.&lt;/strong&gt;  I will assume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Methanol vapor is an ideal gas&lt;/li&gt;&#xA;&lt;li&gt;Standard states for all species are 1 bar of partial pressure 25 &amp;deg;C.  (This assumption matters a lot.  The meaning of the $^\circ$ symbol in $\Delta G^\circ$ can't be understood without it.  &lt;/li&gt;&#xA;&lt;li&gt;The free energy of methanol &lt;em&gt;liquid&lt;/em&gt; does not vary with pressure, i.e. $G_{liq} = G_{liq}^\circ$ at all pressures.  I'm fairly confident that this assumption is equivalent to assuming &lt;em&gt;methanol liquid is an incompressible fluid&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Note the equilibrium condition.&lt;/strong&gt;  At equilibrium, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = 0 = G_{vap} - G_{liq}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Substitute the pressure-dependent free energy of an ideal gas&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ 0 = G_{vap} - G_{liq} = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Use the assumption of incompressibility of liquid methanol that we made above&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$0 = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Rearrange and substitute&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$0 = \left [G_{vap}^\circ - G_{liq}^\circ \right ] + RT \ln{\frac{p}{p^\circ}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = \Delta G^\circ  + RT \ln{\frac{p}{p^\circ}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = 4.3 \frac{\textrm{kJ}}{\textrm{mol}} + RT \ln \frac{p}{p^\circ}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Solve for $p$&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ln{\frac{p}{p^\circ}} = \frac{-4.3 \frac{\textrm{kJ}}{\textrm{mol}}}{RT} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = p^\circ e^{\frac{-4300 \textrm{ J/mol}}{RT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = p^\circ e^{\frac{-4300}{8.314\cdot 298}}=  p^\circ e^{-1.73} = 0.176~p^\circ  $$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Figure out what $p^\circ$ is&lt;/strong&gt;.  Just like it does for $\Delta G$, adding a $^\circ$ indicates that we are referring to a standard state.  Thus $p^\circ$ refers to the partial pressure of methanol vapor at its standard state.  This is almost always defined as one atmosphere (or in more modern times, one bar, which is nearly but not exactly the same thing) of pressure. Assuming that the standard state is the usual one, then $p^\circ$ is one bar, which is equal to 750 mmHg.  This is what we have assumed in step 1.  If that is the case, then $p = 0.176~ p^\circ=0.176\cdot750~\textrm{mmHg}=134~\textrm{mmHg}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How would it know that this unitless quantity is atm and not pascal or mmHg?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hopefully my answer above helps you answer your question.  It has to do with the common use of $\circ$ to indicate a standard state of one bar of partial pressure for all gases.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2016-11-26T18:48:37.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29420" PostTypeId="2" ParentId="29401" CreationDate="2015-04-27T10:13:39.353" Score="0" Body="&lt;p&gt;Assuming your question is how to determine the acidity or basicity of salts.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Acidic Salt:&lt;/strong&gt; A salt formed between a strong acid and a weak base is an acid salt.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{HCl + NH4OH -&amp;gt; NH4Cl + HOH}$$&#xA;  The $\ce{NH4Cl}$ is an acidic salt and if it dissociates it will be acidic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Basic Salt:&lt;/strong&gt; A basic salt is formed between a weak acid and a strong base.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{NaOH + HAc -&amp;gt; NaAc + HOH}$$&#xA;  The &lt;strong&gt;Sodium Acetate&lt;/strong&gt; the basic salt that dissociated into acetate ions and takes an H+ in water. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Applying this logic to your question gives us: $\ce{KNO2}$ is a basic salt, the $\ce{HF}$ is an acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acidic too, the last one is an alcohol which has no effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-04-27T10:13:39.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29421" PostTypeId="1" AcceptedAnswerId="30830" CreationDate="2015-04-27T10:13:56.723" Score="13" ViewCount="11179" Body="&lt;p&gt;In our chemistry books, it is said that concentrated sulphuric acid is used to dry $\ce{HCl}$ gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, concentrated sulphuric acid is a strong oxidising agent so, why isn't $\ce{HCl}$ gas oxidised to $\ce{H2O}$ and chlorine gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why dry $\ce{CaCl2}$ (Calcium Chloride) isn't used as drying agent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Note: $\ce{CaCl2}$ have no reaction with $\ce{HCl}$ as they both have chloride ion in common]&lt;/p&gt;&#xA;" OwnerUserId="15837" LastEditorUserId="16683" LastEditDate="2016-11-13T17:35:25.293" LastActivityDate="2017-07-09T06:40:00.650" Title="Why isn't dry CaCl₂ used to dry HCl gas instead of concentrated sulfuric acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29422" PostTypeId="2" ParentId="29262" CreationDate="2015-04-27T10:32:45.133" Score="3" Body="&lt;p&gt;The triple point of water is point of dynamic equilibrium between three phases of water. Any small change in Pressure or Temperature will lead to one phase regions or two phase lines on the phase diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triple point is not a state of matter rather a combination/mixture of three phases . So i very much doubt we can get any unique spectroscopic signatures . We would get a distorted result as this point has three different phases. &#xA;Also Emission and Absorption spectra depends on  &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF00868550&quot; rel=&quot;nofollow&quot;&gt;temperature&lt;/a&gt; and &lt;a href=&quot;http://www.hindawi.com/journals/ijs/2012/617528/&quot; rel=&quot;nofollow&quot;&gt;pressure&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-27T10:32:45.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29426" PostTypeId="2" ParentId="29411" CreationDate="2015-04-27T12:00:29.710" Score="0" Body="&lt;p&gt;It might help to think about it with an example. $\ce{NH_4Cl}$ is the 'active' ingredient in cold packs (ice packs). The system in this case, consists of the individual ions that separate when placed in solution ($\ce{NH_4^{+}}$ and $\ce{Cl^{-}}$). The solution makes up the surroundings - since the temperature of the solution decreases, the surroundings lose heat, and the system gains heat. Enthalpy is defined by a loss or gain of the system's heat. Since the system gains heat, the enthalpic value is positive (endothermic). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for dissolving salts in solution, this is the case. &lt;/p&gt;&#xA;" OwnerUserId="15831" LastEditorDisplayName="user15489" LastEditDate="2015-04-27T20:14:22.560" LastActivityDate="2015-04-27T20:14:22.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29427" PostTypeId="1" CreationDate="2015-04-27T12:09:40.520" Score="6" ViewCount="117" Body="&lt;p&gt;In Huheey's &lt;em&gt;Inorganic Chemistry&lt;/em&gt; it is written that hexacyanidocobaltate(II) is uncommon but pentacyanidoaquacobaltate(II) is common. Something is mentioned about Jahn-Teller distortion, but I don't understand how it applies here.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="7475" LastEditDate="2016-11-05T18:07:41.800" LastActivityDate="2016-11-06T02:02:43.050" Title="Why would hexacyanidocobaltate(II) be uncommon while pentacyanidoaquacobaltate(II) is common?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29428" PostTypeId="1" AcceptedAnswerId="29432" CreationDate="2015-04-27T12:14:49.053" Score="17" ViewCount="9757" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following cyclic compounds are more acidic?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F5gBj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5gBj.png&quot; alt=&quot;Cyclopenta-1,3,-diene and cyclohepta-1,3,5-triene&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The most acidic proton is positioned on the carbon that is at the top of the above drawings (the methylene hydrogens) on each of the two species, as deprotonation allows resonance.&lt;/p&gt;&#xA;&lt;p&gt;I am aware of Hückel's rule, which states that an aromatic species has &lt;span class=&quot;math-container&quot;&gt;$4n+2$&lt;/span&gt; π-electrons. Only the five membered ring would fulfil this requirement. However, I can draw resonance structures where the seven membered conjugate base has the double bond at each position on the ring. I ask &lt;strong&gt;why is it not aromatic&lt;/strong&gt;? Which conjugate base is more stable? Essentially it's a case of aromaticity vs number of resonance structures.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-01-08T15:33:03.977" Title="Is cyclopentadiene or cycloheptatriene more acidic?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromaticity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29431" PostTypeId="1" AcceptedAnswerId="29442" CreationDate="2015-04-27T12:47:19.057" Score="10" ViewCount="1385" Body="&lt;p&gt;Why (according to provided answers) $\ce{Pu(IV)}$ disproportionates  into adjacent species whilst $\ce{Pu(V)}$ does not in acid aqueous solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\underset{(+6)}{Pu}O2^2+ -&amp;gt;[\pu{+1.02 V}] \underset{(+5)}{Pu}O2+ -&amp;gt;[\pu{+1.04 V}]\underset{(+4)}{Pu}^4+ -&amp;gt;[\pu{+1.01V}] \underset{(+3)}{Pu}^3+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it is the other way - the more positive potential of $\pu{+1.04V}$ on the right of $\ce{PuO2+}$ species will drive the disproportionation while in $\ce{Pu^4+}$ the higher value of $\pu{+1.04V}$ to the left renders the species by itself thermodynamically stable. &lt;/p&gt;&#xA;" OwnerUserId="15669" LastEditorUserId="5026" LastEditDate="2018-02-24T12:02:53.040" LastActivityDate="2018-02-24T12:02:53.040" Title="Disproportionation of plutonium species - Latimer's diagram" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29432" PostTypeId="2" ParentId="29428" CreationDate="2015-04-27T12:53:40.163" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Essentially it's a case of aromaticity vs number of resonance&#xA;  structures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well put!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromaticity is a very strong driving force so aromaticity wins out; Huckel's rule is more important than the number of resonance structures.  Therefore cyclopentadiene is more acidic than cycloheptatriene.  Even though we can draw 7 resonance structures (one with a negative charge on each carbon atom) for the tropylium anion and only 5 for the cyclopentadienyl anion, Huckel's rule tells us that the cyclopentadiene anion ($4n+2$ &amp;#x3c0; electrons, $n=1$) is aromatic while the tropylium anion with 8 &amp;#x3c0; electrons (doesn't fit $4n+2$) is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is cyclooctatetraene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wF313s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like benzene, we could draw resonance structures by shifting the double bonds in this molecule too.  But this molecule is not aromatic, it does not fit the $4n+2$ rule.  In fact, Huckel says with 8 &amp;#x3c0; electrons it is antiaromatic.  To avoid this destabilization cyclooctatetraene adopts a tub-like conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BstJw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://de.wikipedia.org/wiki/Cyclooctatetraen&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again aromaticity trumps resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Huckel's Rule: Aromaticity - Antiaromaticity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the energetic positioning of the molecular orbitals (MO's) in a cyclic, conjugated polyene, we can quickly understand the basis for Huckel's rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use &lt;a href=&quot;http://www.uea.ac.uk/~c286/aromaticnotes.htm&quot; rel=&quot;noreferrer&quot;&gt;Frost diagrams&lt;/a&gt; to determine the energetic positioning of these MO's.  Here is the diagram for benzene, we see that all of the pi electrons are in bonding MO's, the molecule is a stable aromatic that fits the $4n+2$ rule ($n=1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3U1B5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the diagram for cyclooctatetraene, and we see that not all of the electrons are in bonding MO's, two &amp;#x3c0; electrons are in non-bonding MO's.  They don't contribute to bonding or stabilization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5PRL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a $4n$ &amp;#x3c0; electron compound ($n=2$) and accordingly would be classified as antiaromatic by Huckel's rule.  As mentioned above, the compound adopts a non-planar conformation to avoid this destabilization.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-05-06T17:53:33.057" LastActivityDate="2016-05-06T17:53:33.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29433" PostTypeId="1" AcceptedAnswerId="29439" CreationDate="2015-04-27T13:49:38.860" Score="2" ViewCount="12116" Body="&lt;p&gt;Recently, I am playing with redox reaction and I come over a compound acidified potassium manganate (VII) ($\ce{KMnO4/H^+}$). I get confused that why they put $\ce{H^+}$ beside $\ce{KMnO4}$. Doesn't the hydrogen ions flow freely in the solution or something else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so, when we are writing a chemical reaction with occur $\ce{KMnO4}$, we sometimes exclude the $\ce{H^+}$. Why that so?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-04-27T19:54:40.210" Title="The hydrogen ion in acidified potassium manganate (VII)" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29437" PostTypeId="2" ParentId="29405" CreationDate="2015-04-27T13:56:05.907" Score="3" Body="&lt;p&gt;The statement &quot;a weak acid plus a salt containing the conjugate base forms an acidic buffer.&quot; is not correct. More properly, a mixture of equal amounts of a weak acid (but not too weak: with a pKa &amp;lt; 7) with its conjugate base (salt) does indeed give an acidic buffer. To get a basic buffer you would use equal amounts of an &lt;em&gt;even weaker acid&lt;/em&gt; (with a pKa &gt; 7) and its conjugate base (salt). Similar considerations apply to formation of buffers from a weak base plus a conjugate acid salt.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-27T13:56:05.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29439" PostTypeId="2" ParentId="29433" CreationDate="2015-04-27T15:51:15.123" Score="3" Body="&lt;p&gt;I think you have known that the permanganate(VII) ion is a strong oxidizing agent because manganese is in the highest +7 oxidation state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in different pH conditions, $\ce{MnO_4^-}$ Will be reduced to different product:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In a &lt;strong&gt;acidic&lt;/strong&gt; solution, it is reduced to the colourless manganese(II) ion, $\ce{Mn^{2+}}$.&#xA;$\ce{8H^+ + MnO_4^{−} + 5e^{−} → Mn^{2+} + 4H_2O}$&lt;/li&gt;&#xA;&lt;li&gt;In a &lt;strong&gt;strongly basic&lt;/strong&gt; solution, it is reduced to the green manganate ion, $\ce{MnO_4^{2−}}$.&#xA;$\ce{MnO_4^{−} + e^{−} → MnO_4^{2−}}$&lt;/li&gt;&#xA;&lt;li&gt;In a &lt;strong&gt;neutral&lt;/strong&gt; solution, it is reduced to the brown manganese dioxide $\ce{MnO_2}$.&#xA;$\ce{2H_2O + MnO_4^{−} + 3e^{−} → MnO_2 + 4OH^{−}}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But the more understandable way to say, I think, is that $\ce{MnO_4^-}$ can be reduced to different oxidation states; however, it is an idea about chemical balance. If there is $\ce{H^+}$, the first reaction dominates (since we put in more reactant). While if there is $\ce{OH^-}$, neither reaction 1 nor 3 happens (1: lack of reactant; 3: too many product), so the third dominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition: in those different reactions, there are also different tendency to take place and make $\ce{MnO_4^-}$ reduced. In fact, the first reaction (happening with acidic solution) have a much larger tendency to take place. Thus, if $\ce{H^+}$ presents, $\ce{MnO_4^-}$ will have a larger tendency to oxidate other compounds. Thus some reactions require $\ce{MnO_4^-}$ with $\ce{H^+}$ instead of only $\ce{MnO_4^-}$.&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorDisplayName="user15489" LastEditDate="2015-04-27T19:54:40.210" LastActivityDate="2015-04-27T19:54:40.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29441" PostTypeId="1" CreationDate="2015-04-27T16:22:33.780" Score="1" ViewCount="906" Body="&lt;p&gt;At $25~\mathrm{^\circ C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated it but not have the same answer as this.&#xA;I tried it this way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\mathrm{pH}$ is $8.49$, then $\mathrm{pOH}$ is $5.51$. I use this formula: $\mathrm{pOH} = -\log[\ce{OH}]$&lt;/li&gt;&#xA;&lt;li&gt;The $[\ce{OH}]$ is $10^{-5.51}$&lt;/li&gt;&#xA;&lt;li&gt;Find molarity $\ce{Cr}$: $(10.24\cdot(1/1000)\cdot(1/86))/(0.125)=9.525\times 10^{-4}$&lt;/li&gt;&#xA;&lt;li&gt;Substitute into $K_\text{sp} = [\ce{Cr}][\ce{OH}]^2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But after I substitute value of $[\ce{Cr}] = 9.525\times 10^{-4}$ and $[\ce{OH}]$ of $(2\times 10^{-5.51})^2$, I get answer of $3.6\times 10^{-14}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where have I gone wrong? Any Suggestions?&lt;/p&gt;&#xA;" OwnerUserId="15293" LastEditorUserId="865" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T19:03:20.483" LastActivityDate="2021-05-22T19:03:20.483" Title="Ksp calculation of Cr(OH)2" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29442" PostTypeId="2" ParentId="29431" CreationDate="2015-04-27T16:56:48.630" Score="5" Body="&lt;p&gt;In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar.&#xA;Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states:&#xA;$$\ce{2Pu^4+ + 2H2O &amp;lt;=&amp;gt; Pu^3+ + PuO2+ + 4H+}$$&#xA;$$\ce{Pu^4+ + PuO2+ &amp;lt;=&amp;gt; Pu^3+ + PuO2^2+}$$&#xA;Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:&lt;br&gt;&#xA;$$\ce{3Pu^4+ + 2H2O &amp;lt;=&amp;gt; 2Pu^3+ + PuO2^2+ + 4H+}$$&#xA;Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture.&#xA;Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states:&#xA;$$\ce{2PuO2+ + 4H+ &amp;lt;=&amp;gt; Pu^4+ + PuO2^2+ + 2H2O}$$&#xA;$$\ce{Pu^4+ + PuO2+ &amp;lt;=&amp;gt; Pu^3+ + PuO2^2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="5026" LastEditDate="2018-02-22T12:52:36.777" LastActivityDate="2018-02-22T12:52:36.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29443" PostTypeId="1" AcceptedAnswerId="29450" CreationDate="2015-04-27T17:04:20.900" Score="7" ViewCount="365" Body="&lt;p&gt;I am looking for a complete reference of UV/Vis spectroscopic data for organic materials ( mostly interested in λmax ) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the CRC handbook but it only has data about some organic solvents in water and very limited data on dyes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest a comprehensive reference that includes the λmax of a wide variety of molecules in different solvents ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="4231" LastEditDate="2015-04-27T17:13:14.780" LastActivityDate="2015-04-27T20:10:09.270" Title="Reference for UV/Vis Spectroscopic Data request" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;reference-request&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29445" PostTypeId="1" AcceptedAnswerId="29448" CreationDate="2015-04-27T17:55:17.167" Score="24" ViewCount="23102" Body="&lt;p&gt;If you leave something outside, its colors seem to inevitably fade or bleach due to exposure. Is this due to UV absorption? What sort of mechanism causes this - is it that man-made dyes deform on a molecular level? Are there notable materials that are exceptions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a full undergraduate background in physics but not so much training in chemistry. A technical explanation would be cool, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To turn this string of questions into a focused query, let me refer back to the title:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does sunlight cause colors to fade?&lt;/p&gt;&#xA;" OwnerUserId="15827" LastActivityDate="2017-11-27T23:54:16.847" Title="Why does sunlight cause colors to fade?" Tags="&lt;physical-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29448" PostTypeId="2" ParentId="29445" CreationDate="2015-04-27T19:12:34.983" Score="21" Body="&lt;p&gt;Dyes and pigments work by absorbing certain wavelengths of light and reflecting or transmitting the rest. When a dye molecule absorbs a photon, an electron is excited to a higher energy state. Most of the time (neglecting fluorescence), the molecule de-excites by giving off heat and returns to the ground state intact. However, because the excited state is a high energy state, it has the potential to undergo a chemical reaction, breaking a covalent bond or otherwise irreversibly reacting with another molecule. This changes the electronic structure of the molecule which changes its absorption properties: e.g. many dyes that absorb visible light have large systems of conjugated double bonds and if these are broken, the absorbance can shift to much shorter wavelengths. How likely this kind of destructive chemistry is depends on the nature of the dye. Organic dyes tend to be more susceptible to photobleaching than things like quantum dots and inorganic pigments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Touching on Nicolau's comment, red paint can often degrade faster because it absorbs higher energy (shorter wavelength) light. The more energy that is available, the more likely it is that reactions will occur. This is more pronounced as we enter UV wavelengths, which is why UV-B can cause thymine dimers in DNA, which can lead to skin cancer. For outdoor signs and such, UV-blocking coatings are often applied to extend the life of the dyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also not necessary for the light to interact directly with the dye molecules to cause bleaching: depending on what other molecules are present with the dye (solvents, etc), these other molecules may produce reactive species that then react with the dye.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-04-27T19:12:34.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29449" PostTypeId="1" CreationDate="2015-04-27T19:19:34.340" Score="1" ViewCount="236" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What methods are there to produce this compound? What are its properties as a component of automobile petroleum? How does it impact on detonation resistance of automobile petroleum?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that the title compound might be produced from catalytic cracking of petroleum. But I don't have a clue of what methods are there to produce exactly this specific compound. I also think that it helps to increase the detonation resistance of automobile petroleum but I can't explain why. So I'm seeking for the explanation. Surprisingly, there's nothing on the Internet, so this site seems to be the only place where I can possibly find some information.&lt;/p&gt;&#xA;" OwnerUserId="15847" LastEditorDisplayName="user7951" LastEditDate="2016-11-26T16:33:44.377" LastActivityDate="2018-01-26T19:00:03.190" Title="How to synthesize 6-methylhept-1-ene" Tags="&lt;isomers&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29450" PostTypeId="2" ParentId="29443" CreationDate="2015-04-27T20:01:59.750" Score="6" Body="&lt;p&gt;To know the chemical name, molecular structure, or best: the unique registry number provided by the American Chemical Society (&lt;a href=&quot;https://en.wikipedia.org/wiki/CAS_Registry_Number&quot; rel=&quot;noreferrer&quot;&gt;CAS-number&lt;/a&gt;) is beneficial in your search as trade names may vary depending on the supplier of the material you are interested in.  The index number of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Colour_Index_International&quot; rel=&quot;noreferrer&quot;&gt;Colour Index&lt;/a&gt; (like C.I. 73000 for indigo) frequently used in the business of dyes potentially is of lesser use here.  In decrease of data coverage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are commercial sources like &lt;a href=&quot;https://en.wikipedia.org/wiki/Reaxys&quot; rel=&quot;noreferrer&quot;&gt;Reaxys&lt;/a&gt; (by Elsevier) allowing to search for absorption data.  As this (in part) emerged from a much senior database about chemical reactions, &lt;a href=&quot;https://en.wikipedia.org/wiki/Beilstein_database&quot; rel=&quot;noreferrer&quot;&gt;Beilstein&lt;/a&gt; (back to 1771) this is a typical database used in Departments of Chemistry at a University or Industry. &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_Abstracts_Service&quot; rel=&quot;noreferrer&quot;&gt;SciFinder&lt;/a&gt;, based on the data indexed by the American Chemical Society equally allows you to find such data -- for both it might be helpful to request assistance / initial training to fully benefit from these sources. &lt;br&gt; Note: there is so much published data, none of both huge data bases may index / point back to all of the original journal articles, patents, conference proceedings, etc. pp.  Depending on your search strategy, typically you find the data you look for in both data bases--at different places.  Sometimes &quot;The Two&quot; are complementary to each other, sometimes an other (third) search engine finds the data on the net that neither ACS nor Reaxys indexed (yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, because of the design &quot;how the numerical data are managed&quot; I would suggest Beilstein as the better choice here: if you find your compound, the data base will list in a table wavelength and molecular absorption coefficient altogther with the bibliographic reference -- often including the solvent of the analysis.  And the thought occurs to me, it is easier to query with Boolean logic, like &quot;find a compound of molecular mass in the range of xyz to x'y'z with an $\epsilon$ greater than 20000&quot;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less in coverage -- yet worth to point to are the compilations Sadtlers &lt;em&gt;Handbook of ultraviolet spectra&lt;/em&gt; (today part in Bio-Rad's &quot;KnowItAll&quot; database) and the &lt;em&gt;handbook of stains, dyes, and indicators&lt;/em&gt; by Sigma-Aldrich, too mention some of the printed references usually found in the section of spectroscopy/chemistry in university libraries.  Here you find not only one $\epsilon$ at $\lambda_{\mbox{max}}$, but a complete print of the spectrum.  (Be aware, the elder the sources sometimes use wavenumbers or kilo-Kaysers instead of nanometers to reference the wavelength.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account your user profile related to quantum chemistry, perhaps you want to train your quantum chemical model with known experimental data, to evolve from a &lt;em&gt;descriptive&lt;/em&gt; to a &lt;em&gt;predictive&lt;/em&gt; -- and small molecules may serve you well, too.  If so, take a look at NIST's &lt;a href=&quot;http://webbook.nist.gov/&quot; rel=&quot;noreferrer&quot;&gt;webbook&lt;/a&gt; based on &lt;a href=&quot;http://webbook.nist.gov/chemistry/uv-vis/&quot; rel=&quot;noreferrer&quot;&gt;these&lt;/a&gt; references.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Last but not least: Take these data with a grain of salt, often there is some limitation: the same compound / material often exhibits a slightly different UV-Vis absorption spectrum if analyzed once in solvent A, and subsequently in solvent B.  Bathochromic / hypsochromic shifts &lt;a href=&quot;https://en.wikipedia.org/wiki/Solvatochromism&quot; rel=&quot;noreferrer&quot;&gt;(solvatochromism)&lt;/a&gt; may still occur.  Solvent polarity is among the  parameters in this play.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditDate="2015-04-27T20:10:09.270" LastActivityDate="2015-04-27T20:10:09.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29451" PostTypeId="2" ParentId="29411" CreationDate="2015-04-27T20:36:07.703" Score="2" Body="&lt;p&gt;When I was doing enthalpy, this was one of the tricky concepts that I still sometimes get wrong. Here is how you are supposed to remember it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Identify the System&lt;/strong&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The system is where the reaction happens. You cannot physically monitor the change in heat here. The change you always measure happens in the surroundings.&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Identify the Surroundings&lt;/strong&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;This is the surroundings where usually the change is expressed as.&lt;/li&gt;&#xA;  &lt;/ul&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that we got the basics cleared away, let us look at your problem. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My reasoning: if the temperature of the solution decreases, it means&#xA;  heat was released into the surroundings, so the enthalpy change is&#xA;  negative.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is what I usually think of, but now you must understand that as I've said before the &lt;em&gt;change always happens in the surroundings and is associated with the surroundings.&lt;/em&gt; &lt;strong&gt;So the temperature of the solution decreases, which means that the surroundings temperature decreased. The solution is part of the surroundings.&lt;/strong&gt; This means that the temperature of the system increases. Thus making the $\Delta H$ &lt;strong&gt;positive&lt;/strong&gt; and the reaction &lt;strong&gt;&lt;em&gt;endothermic&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Good Question, always remember that you cannot, in most situations, physically measure the change in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-27T20:56:13.710" LastActivityDate="2015-04-27T20:56:13.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29452" PostTypeId="1" CreationDate="2015-04-27T20:50:14.830" Score="2" ViewCount="4132" Body="&lt;p&gt;How to make hydrogen peroxide into a gel or just simply really thick and it has to be able to be in your mouth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing i have came up with is (Gelatins). &#xA;What should i use?&lt;/p&gt;&#xA;" OwnerUserId="15849" LastEditorUserId="9961" LastEditDate="2015-10-25T22:00:38.087" LastActivityDate="2015-10-25T22:00:38.087" Title="How to make hydrogen peroxide into gel" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29453" PostTypeId="1" CreationDate="2015-04-27T20:51:12.733" Score="4" ViewCount="4875" Body="&lt;p&gt;Functional groups like $\ce{OH}$, $\ce{NH2}$, attached to unsaturated compounds (chromophores) are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Auxochrome&quot; rel=&quot;nofollow noreferrer&quot;&gt;auxochromes&lt;/a&gt;. These groups tend to shift the absorption wavelength to the infrared region, which is called a &lt;em&gt;bathochromic shift&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why do these groups shift the light absorption wavelength?&lt;/li&gt;&#xA;&lt;li&gt;Why exactly to the infrared region; that is, why is the absorption red-shifted or lowered in energy?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="194" LastEditDate="2017-05-26T22:22:48.677" LastActivityDate="2017-05-26T22:22:48.677" Title="How do auxochrome groups work?" Tags="&lt;organic-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29454" PostTypeId="1" CreationDate="2015-04-27T21:03:41.350" Score="3" ViewCount="1615" Body="&lt;p&gt;Will sodium metal, if mixed well with iron oxide, displace the iron in a chemical reaction? I know thermite involves the mixture of aluminium powder and iron oxide, and the single displacement reaction results in molten iron and aluminium trichloride. The other day I melted some sodium metal to the point of melting but reacting with the air. I wonder if either mashing up sodium with iron oxide and then igniting, or adding melted sodium to iron oxide will act like a thermite reaction. &lt;/p&gt;&#xA;" OwnerUserId="15851" LastActivityDate="2015-04-30T12:46:36.883" Title="Single displacement reaction" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29455" PostTypeId="1" CreationDate="2015-04-27T21:18:58.003" Score="1" ViewCount="1952" Body="&lt;p&gt;How can I turn mercury metal into mercury oxide? I know that many metals will react with oxygen if enough heat is supplied in the presence of oxygen. If I heated up mercury enough and also supplied oxygen from my oxygen tank, would that result in mercury oxide? Also how about lead? Thank you &lt;/p&gt;&#xA;" OwnerUserId="15851" LastActivityDate="2015-04-28T12:44:18.567" Title="Mercury into mercury oxide" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29456" PostTypeId="1" CreationDate="2015-04-27T21:21:19.593" Score="2" ViewCount="1020" Body="&lt;p&gt;In my Physical Science class, we are learning about chemical formulas and how to balance them. I balanced this equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{Fe2O3 + Al -&amp;gt; Fe + Al2O3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the other students asked me if this chemical formula releases or absorbs heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone help me give a great explanation for this question? &lt;/p&gt;&#xA;" OwnerUserId="15852" LastEditorUserId="194" LastEditorDisplayName="user15489" LastEditDate="2017-04-01T03:15:56.800" LastActivityDate="2017-04-01T03:17:41.583" Title="Chemical formula and heat" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29457" PostTypeId="2" ParentId="29453" CreationDate="2015-04-27T21:57:35.813" Score="4" Body="&lt;p&gt;Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV.  If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the catch:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Bathchromic&lt;/em&gt; auxochromic groups (like the mentioned $\ce{-OH}$ and $\ce{-NH2}$) increase the electron density in the $\pi$-system. This is typically caused by a strong $+M$-effect. As they lower the excited $\pi^*$-level, a longer wavelength is needed for the excitation along $\pi{} \rightarrow {} \pi^*$, hence the redshift.&lt;/li&gt;&#xA;&lt;li&gt;In contrast to the former, substitution equally may alter the $n \rightarrow{} \pi^*$ transition, inducing the opposite effect, moving the absorption towards higher absorption (blue) and a &lt;em&gt;hypsochromic&lt;/em&gt; shift.  One example is the to compare the absorption of &lt;a href=&quot;https://en.wikipedia.org/wiki/Indigo_dye&quot; rel=&quot;nofollow&quot;&gt;indigo&lt;/a&gt; yielding &quot;jeans blue&quot; with 6,6'-dibromoindigo, main constitutent of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tyrian_purple&quot; rel=&quot;nofollow&quot;&gt;tyrian purple&lt;/a&gt;. A second, that heteroatoms next to a $\ce{C=O}$ carbonyl group induce a hypsochromic shift like in the case of acetaldehyde &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C75070&amp;amp;Units=SI&amp;amp;Mask=400#UV-Vis-Spec&quot; rel=&quot;nofollow&quot;&gt;example here&lt;/a&gt; in comparison to acetic acid &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C64197&amp;amp;Units=SI&amp;amp;Mask=400#UV-Vis-Spec&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note the use of the terms of &quot;bathochromic shift&quot; and &quot;hypsochromic shift&quot; along &lt;a href=&quot;https://en.wikipedia.org/wiki/Solvatochromism&quot; rel=&quot;nofollow&quot;&gt;solvatochromism&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditDate="2015-04-28T19:45:53.537" LastActivityDate="2015-04-28T19:45:53.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29458" PostTypeId="1" AcceptedAnswerId="29459" CreationDate="2015-04-27T22:11:43.690" Score="10" ViewCount="3928" Body="&lt;p&gt;I've recently encountered a problem when using an analytical (0.1 mg accuracy) scale to determine the mass of a sample of ferrocene powder. The reading in the balance seems to drift either upwards or downwards when I drop some of the powder in an erlenmeyer flask. The reading stabilizes when I add water to form a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that static electricity might cause this kinds of problems when measuring powders so I suspect this is the case here, because the problem only appears when I add the ferrocene powder. I've try reaserching some ways to get rid of the problem, a static gun seems to work in some cases (no idea if it would work with a powder though) and I've heard that air ionisers also work, but don't have access to either of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this is the first time I've encountered this situation, I've heard its pretty common, so what do people usually do to get rid of the problem in a cheap manner? Is there an easy solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using gloves to operate the scale, which I've been told to do in order to keep the flasks clean of oil from my fingers. Can the gloves be worsening the problem? What can I do about that?&lt;/p&gt;&#xA;" OwnerUserId="15854" LastActivityDate="2016-08-25T12:29:00.630" Title="How is drift due to static electricity eliminated from an analytical balance?" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;electrostatic-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29459" PostTypeId="2" ParentId="29458" CreationDate="2015-04-27T22:49:56.067" Score="4" Body="&lt;p&gt;Static charge can affect the measurements made by an analytical balance, as can magnetic materials because the mechanism in modern analytical balances involves a feedback loop between the load that is measured and an electromagnet. Insulating weighing vessels (plastic or glass) tend to accumulate static charge, especially when handled with another insulator (gloves) by the triboelectric effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should first make sure that the body of the balance is grounded properly according to the manual for your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a short-term solution to static charged glass, you might try arranging the sample so that it is father away from the balance. One way might be to place the sample on top of an empty sample bottle on the balance. As you mentioned, a more robust solution for discharging static build-up is to have a commercially available air ionizer nearby. The device will either produce an equal amount of positive and negative ions in the air at high voltage AC or DC electrodes, or with an ionizing radioactive material such as is found in anti-static brushes used for cleaning vinyl audio records.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are frequently measuring magnetic materials, many analytical balances can be configured in a bottom-loading arrangement. In this setup a platform is hooked beneath the balance, and the sample is supported sufficiently far away from the force-restoring motor so that it does not influence the measurement.&lt;/p&gt;&#xA;" OwnerUserId="15065" LastActivityDate="2015-04-27T22:49:56.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29460" PostTypeId="2" ParentId="29458" CreationDate="2015-04-27T23:01:36.413" Score="4" Body="&lt;p&gt;Wearing gloves &lt;em&gt;may contribute&lt;/em&gt; to the static charges you observe, especially if you have to work with gloves not only to protect yourself, but &lt;em&gt;per se&lt;/em&gt; you can't do otherwise (glove box) and a special, palm size ionizer is indeed a remedy.&#xA;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/C6OIM.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I do not pretend &quot;it works every time&quot;, these are my &lt;em&gt;suggestions&lt;/em&gt;, that sometimes work alone, sometimes in combination:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electric discharge of yourself.  Sometimes over the day I become &quot;charged&quot;, too, and a simple tip of my hand to the (air dry) water pipeline or heating pipe / radiator removes some of the charges that can't reach earth while walking in firm (but nonconducting) lab shoes over polymer-coated surfaces.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electric Discharge the clean spatula.  Prior to transferring material to the balance, tip the metallic spatula against something that is both conductive to the earth and clean.  Clean and air dry water pipeline, heating pipe (&lt;em&gt;vide supra&lt;/em&gt;) are handy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Attempt the mass transfer with a spatula made of a different material rather than metal.  Beside glass, different plastics &lt;em&gt;may&lt;/em&gt; be represent a surprising  alternative.  Krischenbaum &lt;em&gt;et al.&lt;/em&gt; reported recently how to deploy drinking straws as disposable spatulas in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed5006322&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of Chemical Education&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="1782" LastEditDate="2015-04-27T23:15:24.587" LastActivityDate="2015-04-27T23:15:24.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29461" PostTypeId="2" ParentId="29142" CreationDate="2015-04-27T23:28:49.047" Score="17" Body="&lt;p&gt;Out there in the real world, university students and school pupils alike favour strict rules that are true as often as possible (or at least have clear, easily remembered exceptions), while their professors and teachers on the other hand try to tell them that these rules are merely to be considered ‘guidelines’ rather than actual rules (and attempt to welcome them aboard the Black Pearl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from this point on is where the question turns philosophical very quickly. One side will say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ah, so the rule is still valid in this surrounding; I just need to remember to do &lt;em&gt;A&lt;/em&gt; and &lt;em&gt;B&lt;/em&gt; when applying it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;while the other side will reply&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ah, but this shows us that the so-called ‘rule’ can no longer be strictly applied because you need to consider data &lt;em&gt;C&lt;/em&gt; and &lt;em&gt;D&lt;/em&gt; that point towards the real truth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thing is, both sides are somehow correct while parts of their descriptions remain wrong. Maybe we can think of these as two mesomeric descriptions of the same principle?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As we all (should) know, all our concepts of describing &lt;em&gt;bonding&lt;/em&gt; and &lt;em&gt;electron location&lt;/em&gt; are mere approximations of the actual underlying truth. Hence, all the rules that were ever laid out — like &lt;em&gt;a bond consists of two electrons between two atoms&lt;/em&gt;, oxidation states or the octet rule — are again just mere approximations. They cannot be correct nor can they be properly ‘valid’. Thus, what isn't properly true can't be fulfilled or invalidated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that is the most &lt;strong&gt;theoretical&lt;/strong&gt; answer to the question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From a &lt;strong&gt;practical&lt;/strong&gt; point of view, it mostly depends on how strict you postulated the octet rule, and how quickly you would consider it violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, the octet rule wasn't valued highly for elements that weren't carbon, nitrogen, oxygen or fluorine. As such, anything that violated the octet rule in its strictest sense was considered a violation (and proof for its invalidity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, my observations imply that the octet rule is gaining support again, and some of the simpler cases that were hither considered &lt;em&gt;violations&lt;/em&gt; are now something more like &lt;em&gt;special cases&lt;/em&gt; or maybe just &lt;em&gt;octet rule with additional mesomery&lt;/em&gt;. In the eyes of those supporting this side, these compounds comply with the octet rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, there is the large grey space between the black and the white.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I for my part tend to favour rules like the octet rule very much — although I tend to use a not-as-standard defintion for it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;s&lt;/em&gt;- or &lt;em&gt;p&lt;/em&gt;-block elements will always favour surroundings where the number of valence electrons is 8 &lt;strong&gt;or less.&lt;/strong&gt; For the two elements of the first period, this reduces itself to 2 or less.&lt;br&gt;&#xA;  Mesomeric structures that allow an all-octet are to be favoured over those that do not.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under this definition, which relies heavily on mesomery for cases such as $\ce{IF7}$ or $\ce{PCl5}$ while it allows for a single, unambiguous description of $\ce{SO4^{2-}}$, these cases comply to the octet rule and actually further prove its validity. Others will surely disagree.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-04-27T23:28:49.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29462" PostTypeId="1" CreationDate="2015-04-27T23:35:24.323" Score="3" ViewCount="37349" Body="&lt;p&gt;How many hydrogen atoms are contained in 8.66 moles of $\ce{NH3}$? &#xA;The answer is given after the log has been taken, so the final answer is 25.19.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have been doing is multiplying $8.66\text{ by }6.02\times10^{23}=5.21332\times10^{24}$ and when I take the log of that I get 24.71. I'm not sure what to do to get the correct answer.   &lt;/p&gt;&#xA;" OwnerUserId="15855" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-06-14T17:34:20.617" LastActivityDate="2018-01-24T20:17:15.327" Title="How many hydrogen atoms are in a sample of NH3?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29463" PostTypeId="2" ParentId="29456" CreationDate="2015-04-28T00:36:42.203" Score="3" Body="&lt;p&gt;The best way to solve this problem is to look at the $\Delta H$ charts found online. Using the simple equation below will give you the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\sum \Delta H_{\text{Products}} - \sum \Delta H_\text{Reactants} = \Delta H_{\text{Reaction}}$$&#xA;  If the $\Delta H$ is positive in sign then the reaction is &lt;strong&gt;endothermic&lt;/strong&gt; and if it is negative in sign then it is &lt;strong&gt;exothermic&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="194" LastEditDate="2017-04-01T03:17:41.583" LastActivityDate="2017-04-01T03:17:41.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29464" PostTypeId="1" AcceptedAnswerId="31073" CreationDate="2015-04-28T00:39:00.707" Score="7" ViewCount="626" Body="&lt;p&gt;There are mainly fundamental transitions happening during excitation and UV/Vis spectroscopy (as n- &gt; σ* ,π-&gt;π* ,n- &gt; π ,..).  I am looking for a good reference that shows these transitions for molecules (for example shows n- &gt; σ* for acetone , π- &gt; π* for benzene,... ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually I am talking about tables. I can do guesswork for most of the molecules based on their structures but definitely I need references for publication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Mostly interested in these molecules: Acetone, Acetaldehyde , Benzene , Naphthalene, Anthracene , Methylene Blue, Phenanthrene , 1-Butanol&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="4945" LastEditDate="2015-05-09T09:15:46.987" LastActivityDate="2015-05-11T20:37:22.563" Title="UV/Vis spectroscopy transition reference" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;reference-request&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29465" PostTypeId="1" AcceptedAnswerId="29517" CreationDate="2015-04-28T00:42:43.340" Score="7" ViewCount="2085" Body="&lt;p&gt;I want to know what an aqueous solution is. For example, let's consider a $1~\mathrm{M}$ aqueous solution of $\ce{MgCl2}$. Does it also contain $\ce{MgOH2}$? Though it is a simple question, I feel it is important to understand something as vital to chemistry as this.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="9961" LastEditDate="2015-04-28T22:53:44.863" LastActivityDate="2019-01-25T16:17:37.360" Title="What an aqueous solution really contains?" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;hydrolysis&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29466" PostTypeId="2" ParentId="29452" CreationDate="2015-04-28T00:44:31.020" Score="2" Body="&lt;p&gt;There are a number of vegetable gums, such as carageenan, pectin and gum arabic, and man-made &lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_cellulose&quot; rel=&quot;nofollow&quot;&gt;methyl cellulose&lt;/a&gt; which &lt;em&gt;might&lt;/em&gt; be compatible with low concentrations of $\ce{H2O2}$, which you can determine easily. Any impurities, particularly transition metals, may make $\ce{H2O2}$ decompose, so it might be best to keep the gelling agent and peroxide separate until just before use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for oral use, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/8247815&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; suggests 0.75% to 1.5% solutions, one quarter and one half the strength of the common antiseptic &lt;strong&gt;N.B. $\ce{H2O2}$ bleach is much stronger and is &lt;em&gt;not&lt;/em&gt; for oral use.&lt;/strong&gt; Note that there &lt;em&gt;are&lt;/em&gt; &lt;a href=&quot;http://www.intelligentdental.com/2010/07/09/so-can-hydrogen-peroxide-be-used-as-mouthwash-safely/&quot; rel=&quot;nofollow&quot;&gt;dangers&lt;/a&gt; with use of peroxide as a mouthwash.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-28T00:44:31.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29467" PostTypeId="2" ParentId="29465" CreationDate="2015-04-28T01:04:40.630" Score="0" Body="&lt;p&gt;Aqueous means your solvent for the solution is water. It's a term to differentiate from other solution like alcoholic or etheric, etc.&#xA;On the sol chem part of your question, a molecule can take many different form in solution and their amount can usually be calculated with a constant and equilibrium equation, but for $\ce{MgCl2}$, it is considered that most of the salt that dissolve turn into $\ce{Mg^{2+}}$ and $\ce{2 Cl-}$.&lt;/p&gt;&#xA;" OwnerUserId="15795" LastEditorDisplayName="user15489" LastEditDate="2015-04-28T03:29:57.853" LastActivityDate="2015-04-28T03:29:57.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29468" PostTypeId="1" CreationDate="2015-04-28T01:06:42.930" Score="6" ViewCount="7999" Body="&lt;p&gt;I am looking for an electrolyte to use in water to perform electrolysis. It can't be salt, as salt ($\ce{NaCl}$) would cause the chemical reaction of &#xA;$$\ce{2H2O + 2NaCl -&amp;gt; Cl2 + H2 + 2NaOH}$$ &#xA;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + NaCl -&amp;gt; HCl + NaOH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which both make products with poisonous chemicals (ex: $\ce{NaOH}$ = Sodium Hydroxide (aka lye) which is poisonous). I am looking for a chemical I can use to create the pure forms of hydrogen and oxygen from the electrolysis process. I understand how electrolysis works, and have tried it before, but am looking for something that could be used as a non-poisonous electrolyte. I read that baking soda could be used, would that produce what I am looking for?&lt;/p&gt;&#xA;" OwnerUserId="15857" LastEditorUserId="15410" LastEditDate="2015-04-28T05:55:40.627" LastActivityDate="2018-04-24T19:03:29.440" Title="What can be used as an electrolyte in water to perform electrolysis other than salt?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;elements&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29469" PostTypeId="2" ParentId="29465" CreationDate="2015-04-28T01:09:09.530" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An aqueous solution is a solution in which the solvent is water. It is usually shown in chemical equations by appending (aq) to the relevant chemical formula. For example, a solution of table salt, or sodium chloride (NaCl), in water would be represented as NaCl(aq).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://en.wikipedia.org/wiki/Aqueous_solution&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15857" LastEditorUserId="5456" LastEditDate="2015-04-28T16:20:49.990" LastActivityDate="2015-04-28T16:20:49.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29470" PostTypeId="1" AcceptedAnswerId="29483" CreationDate="2015-04-28T01:41:25.500" Score="11" ViewCount="11415" Body="&lt;p&gt;I was cleaning my blinds today, and wanted to know what the primary components of dust are. I know that it is made of microorganisms and other particles, but I do not want to guess that it will be $\ce{N}$ or $\ce{Si}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess that other materials outside our house might have more than one component for dust. But what about inside the house?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="5026" LastEditDate="2018-08-16T10:07:06.963" LastActivityDate="2018-08-16T12:49:08.177" Title="What is dust made of?" Tags="&lt;everyday-chemistry&gt;&lt;elements&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29471" PostTypeId="2" ParentId="29465" CreationDate="2015-04-28T02:57:59.533" Score="2" Body="&lt;p&gt;Aqueous means that the specific compounds has dissolved in $\ce{H2O}$ or in other words, it dissociated into it's ions. When it is dissolved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Ionic Compounds&lt;/strong&gt;: $\ce{NaCl}$ for example dissociates in water to $\ce{Na+}$ and $\ce{Cl-}$ ions. &lt;strong&gt;Intramolecular&lt;/strong&gt; breaking.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;&lt;strong&gt;Covalent Compounds&lt;/strong&gt;: Covalent compounds do not break apart but they do break their &lt;strong&gt;intermolecular&lt;/strong&gt; force.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And one more thing, &lt;strong&gt;SOLIDS&lt;/strong&gt;, &lt;strong&gt;LIQUIDS&lt;/strong&gt;, &lt;strong&gt;GASSES&lt;/strong&gt; aren't &lt;em&gt;usually&lt;/em&gt; aqueous. I'm saying usually because there is always some exception to this and someone will point that out.&#xA;Hopefully this helps, comment for more information.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-04-28T19:24:54.933" LastActivityDate="2015-04-28T19:24:54.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29472" PostTypeId="1" CreationDate="2015-04-28T03:33:21.903" Score="3" ViewCount="106" Body="&lt;p&gt;From Googlebooks First Outlines of a Dictionary of Solubilities of Chemical Substances  By Frank Humphreys Storer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I find that lead tartrate is dissolved by tartrate of ammonia. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the final product?&lt;/p&gt;&#xA;" OwnerUserId="15861" LastEditorUserId="19383" LastEditDate="2015-10-30T18:30:20.527" LastActivityDate="2015-10-30T18:30:20.527" Title="ammonium tartrate + lead tartrate=?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29474" PostTypeId="1" CreationDate="2015-04-28T04:33:11.737" Score="2" ViewCount="2485" Body="&lt;p&gt;$\ce{AgNO3}$ is not suppose to cause a precipitation when mixed with $\ce{FeSO4}$, but the acidified solution turns slightly cloudy when $\ce{AgNO3}$ is added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this suppose to happen? solutions are all prepared fresh. Any reason as to why? This also happens when silver nitrate is added to magnesium sulphate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture is as attached, the left one has the $\ce{AgNO3}$ added (labelled Ag) while the one on the right has $\ce{Ba(NO3)2}$ added to show the comparison &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ujiNq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujiNqm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15862" LastEditorUserId="4945" LastEditDate="2015-04-28T05:53:39.943" LastActivityDate="2015-04-28T05:56:10.603" Title="Is an iron (II) sulphate solution supposed to turn cloudy when silver nitrate is added?" Tags="&lt;inorganic-chemistry&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29475" PostTypeId="2" ParentId="29468" CreationDate="2015-04-28T04:44:59.177" Score="0" Body="&lt;p&gt;I assure you that your worries are without basis. For the hydrolysis of water, the half reactions are:&#xA;$$\ce{2H2O_{(l)} -&amp;gt;O2_{(g)} +4H+_{(aq)} +4e-}~~~~\varepsilon^0=-1.23~\mathrm{V}$$&#xA;$$\ce{4H+_{(aq)} +4e- -&amp;gt;2H2_{(g)}}~~~~\varepsilon^0=0~\mathrm{V}~\mathrm{(by~definitoin)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\varepsilon^0$ of the reaction would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O_{(aq)} -&amp;gt;2H2_{(g)} +O2_{(g)}}~~~~\varepsilon^0=-1.23~\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the proposed half reactions your are suggesting might take place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O_{(l)} +2e- -&amp;gt;H2_{(g)} +2OH-_{(aq)}}~~~~\varepsilon^0=-0.83~\mathrm{V}$$&#xA;$$\ce{2Cl-_{(aq)} -&amp;gt;Cl2_{(g)} + 2e-}~~~~\varepsilon^0=-1.36~\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\varepsilon^0$ of the reaction would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} -&amp;gt; Cl2_{(g)} + H2_{(g)} + 2OH-_{(aq)}}~~~~\varepsilon^0=-2.19~\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the $\ce{Na+}$ ions are present in the solution, though it is not listed in the reaction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the hydrolysis of water requires less energy than the production of $\ce{Cl2}$, $\ce{H2}$, and $\ce{NaOH}$ from the ions present in the solution, the second reaction will not occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction $\ce{H2O_{(l)} + Cl^{-}_{(l)} -&amp;gt; HCl_{(g)} + OH-_{(aq)}}$ would not occur. $\ce{HCl}$ could not evolve from the solution, as there would be no mechanism forcing the $\ce{H+}$ ions onto the electronically stable $\ce{Cl-}$ ions. Rather, it would be much more likely that the $\ce{H+}$ ions would re-react with the $\ce{OH-}$ ions to reform water, meaning no net reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, go right ahead using $\ce{NaCl}$ in your hydrolysis reactions. It's actually probably one of the safest salts you could use!&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-28T04:44:59.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29476" PostTypeId="1" CreationDate="2015-04-28T05:05:02.663" Score="1" ViewCount="285" Body="&lt;p&gt;Mix equal volumes of $0.1~\mathrm{M}$ $\ce{H3PO4}$ and $0.2~\mathrm{M}$ $\ce{KOH}$. What ion remains in the largest concentration at equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: suppose we mix $1~\mathrm{L}$ of each solution. For $\ce{H3PO4}$, $K_{a_1}=7.5\cdot10^{-3}$ and $K_{a_2}=6.2\cdot10^{-8}$, so in terms of the dissociation of $\ce{H3PO4}$, the only ions produced in non-neglible quantities are $\ce{H^+}$ and $\ce{H2PO4-}$. Furthermore the quantities of $\ce{H^+}$ and of $\ce{H2PO4-}$ are both less than one mole, since there is less than 1 mole $\ce{H3PO4}$. We have 2 moles $\ce{OH^-}$ from the $\ce{KOH}$, but since of the $\ce{OH^-}$ neutralizes the $\ce{H^+}$, in particular less than a mole $\ce{OH^-}$ is used in this neutralization reaction. Thus at least one mole $\ce{OH^-}$ remains, at most 1 mole $\ce{H2PO4-}$ remains, and all other ions remain in negligible quantities, so the answer is $\ce{OH^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, this is wrong - the answer is $\ce{HPO4^2-}$. I don't see how this is possible because we already established that the amount of $\ce{HPO4^2-}$ formed is negligible from the $K_a$ values, and since we are adding a strong base to a weak acid we know the resulting solution will be basic, so we should expect from the beginning that $\ce{OH^-}$ appears in large concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what's wrong with my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="12860" LastEditDate="2015-11-13T19:21:55.907" LastActivityDate="2015-11-13T19:21:55.907" Title="What ion remains in the largest concentration at equilibrium: 0.1 M H₃PO₄ and 0.2 M KOH?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29478" PostTypeId="1" CreationDate="2015-04-28T05:08:36.617" Score="4" ViewCount="2711" Body="&lt;p&gt;I'm having trouble understanding why an ester is a meta directing/deactivating group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not talking about an ester with its oxygen bound to the ring, like an acylated phenol. I mean the R group is bound to the ring and the alkoxy group is bound only to the carbonyl carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me like the pi electrons in the carbonyl pi bond could be pushed down and delocalized into the ring. Why is this not possible?&lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorUserId="10185" LastEditDate="2015-04-28T19:12:36.567" LastActivityDate="2015-04-28T19:12:36.567" Title="Why are esters meta directors? Can't the carbonyl donate electrons into the ring?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29479" PostTypeId="1" CreationDate="2015-04-28T05:10:32.443" Score="0" ViewCount="995" Body="&lt;p&gt;Which of these compounds does not liberate heat when water is added to it: $\ce{KNO_3}$, $\ce{NaOH}$, $\ce{CaO}$, $\ce{H_2SO_4}$, $\ce{Na}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from experience that dropping sodium tablets into water has explosive results, but I am clueless about the method to solve this problem. I know the answer is $\ce{KNO_3}$, but I don't know why. If someone can provide the method, I think I can do the problem.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorDisplayName="user15489" LastEditDate="2015-04-28T05:55:56.063" LastActivityDate="2015-04-28T06:25:50.343" Title="Which of these compounds does not liberate heat when water is added to it?" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29480" PostTypeId="2" ParentId="29474" CreationDate="2015-04-28T05:50:23.003" Score="3" Body="&lt;p&gt;According to this &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;resource&lt;/a&gt; for solubility rules, $\ce{BaSO4}$ is insoluble and $\ce{Ag2SO4}$ is slightly insoluble, which would explain the the cloudiness of each solution, as well as why the $\ce{BaSO4}$ solution is cloudier than the $\ce{Ag2SO4}$ solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;https://users.stlcc.edu/gkrishnan/ksptable.html&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt;, $\ce{BaSO4}$ has a $K_{\mathrm{sp}}$ of $1.1 \times 10^{-10}$. Comparably, $\ce{Ag2SO4}$ has a $K_\mathrm{sp}$ of $1.2 \times 10^{-5}$, which is considerably higher than $\ce{BaSO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not aware the $K_{\mathrm{sp}}$ is a mathematical measurement describing the solubility of a salt at a certain temperature (usually $25^{\mathrm{o}} \mathrm{C}$), and is found by multiplying the concentration of the salt's dissolved ions, each raised to the power of their respective stoichiometric coefficients. The higher the $K_{\mathrm{sp}}$, the more soluble a salt is. For example, the $K_{\mathrm{sp}}$ for $\ce{Ag2SO4}$ would be given by the expression: $K_{\mathrm{sp}}=[\ce{Ag+}]^2\cdot[\ce{SO4^{2-}}]$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-04-28T05:56:10.603" LastActivityDate="2015-04-28T05:56:10.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29481" PostTypeId="2" ParentId="29476" CreationDate="2015-04-28T06:03:11.543" Score="1" Body="&lt;p&gt;Actually, the ion with the largest concentration is $\ce {K+}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the exact concentrations if you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;br&gt;&#xA;Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$&lt;br&gt;&#xA;Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;Calculations:&lt;br&gt;&#xA;Find $\ce {[H_2PO_4^{-}]}$:&lt;br&gt;&#xA;$$\ce {H+ + HPO4^{2-} &amp;lt;=&amp;gt; H_2PO_4^{-}}$$&lt;br&gt;&#xA;$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.&lt;br&gt;&#xA;$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$&lt;br&gt;&#xA;$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$&lt;br&gt;&#xA;$[H_2PO_4^{-}]=3.75\times 10^{-10}$M  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;Find $\ce {[PO_4^{3-}]}$:&lt;br&gt;&#xA;$$\ce {HPO4^{2-} &amp;lt;=&amp;gt; H^+ + PO_4^{3-}}$$&lt;br&gt;&#xA;$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.&lt;br&gt;&#xA;$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$&lt;br&gt;&#xA;$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$&lt;br&gt;&#xA;$[PO_4^{3-}]=2.25\times 10^{-7}$  &lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-04-29T01:39:23.547" LastActivityDate="2015-04-29T01:39:23.547" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29482" PostTypeId="2" ParentId="29479" CreationDate="2015-04-28T06:25:50.343" Score="4" Body="&lt;p&gt;It is important to note that if a molecule or formula unit is capable of undergoing hydrogen bonding ($\ce{NaOH}$, $\ce{H2SO4}$), then the formation of this hydrogen bond will be extremely exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if a substance can react to form a substance that will undergo hydrogen bonding ($\ce{CaO + H2O -&amp;gt; Ca(OH)2}$), this species will subsequently form these hydrogen bonds and liberate heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, as you have stated you know from personal experience, the reaction between a alkali metal ($\ce{Na}$) and water is extremely exothermic— so much so that many of these metals will melt from the heat liberated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, there is no fail-proof method for predicting the enthalpy of solution for any given substance. Generally, the enthalpies of solution are determined experimentally, so unless you are expected to perform all these reactions in a calorimeter yourself, I would recommend something like this: &lt;a href=&quot;http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/enthalpies%20of%20solution%20of%20electrolytes.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-04-28T06:25:50.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29483" PostTypeId="2" ParentId="29470" CreationDate="2015-04-28T06:29:42.307" Score="5" Body="&lt;p&gt;The dust in the house mostly contains sand, dirt and dead skin cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's start with the easiest, &lt;strong&gt;sand&lt;/strong&gt;. The chemical formulae of sand is $\ce{SiO2}$, so it contains silicon and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, dirt has rock, sand, and clay. Sand is made of silicon and oxygen. Rock is mostly made up of graphite (carbon) and some minerals like phosphorus and silicon. Clay is made up of silicon and oxygen and some minor minerals like aluminium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, dead skin, like other organic compounds, contains carbon, hydrogen and oxygen, maybe some nitrogen as it may have protein in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, dust is made up of carbon, hydrogen, oxygen, nitrogen, silicon, phosphorus and some minor minerals.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="7499" LastEditDate="2015-04-28T22:06:02.423" LastActivityDate="2015-04-28T22:06:02.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29484" PostTypeId="1" CreationDate="2015-04-28T06:40:35.180" Score="1" ViewCount="55" Body="&lt;p&gt;How can one electrodeposit Nickel on a substrate using a liquid crystal template?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substrate is Indium Tin Oxide coated plate and the template is cetyl trimethylammonium bromide(CTAB)+WATER+ETHANOL&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2015-04-28T08:37:57.980" LastActivityDate="2015-04-30T07:55:40.630" Title="Electrodeposition of Ni" Tags="&lt;electrochemistry&gt;&lt;liquid-crystals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29485" PostTypeId="2" ParentId="29478" CreationDate="2015-04-28T06:51:42.357" Score="4" Body="&lt;p&gt;As your concept oxygen gets positive charge and that leads to least stability which is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ua0RX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15800" LastActivityDate="2015-04-28T06:51:42.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29486" PostTypeId="1" CreationDate="2015-04-28T07:41:17.893" Score="7" ViewCount="971" Body="&lt;p&gt;Is there any way that i can detect lead in gasoline, not by sending of to a lab but to be able to do this by myself?&lt;/p&gt;&#xA;" OwnerUserId="15865" LastEditorUserId="418" LastEditDate="2015-05-13T03:56:39.927" LastActivityDate="2015-05-13T04:20:08.243" Title="Way to detect lead in gasoline" Tags="&lt;home-experiment&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29487" PostTypeId="1" AcceptedAnswerId="29509" CreationDate="2015-04-28T07:58:01.493" Score="2" ViewCount="7099" Body="&lt;p&gt;Given the concentration of bleach ($6\%\,\mathrm{m/m}$), density of bleach ($1.07\,\mathrm{g\,mL^{-1}}$), and molar mass of bleach ($74.44\,\mathrm{g\,mol^{-1}}$) how many moles of bleach are in a $4.0\,\mathrm{mL}$ sample? I don't understand where the concentration factors in, but the problem says to use it.&lt;/p&gt;&#xA;" OwnerUserId="15866" LastEditorUserId="15410" LastEditDate="2015-05-13T02:38:48.987" LastActivityDate="2015-05-13T04:53:06.187" Title="Given the concentration and density of a sample, calculate moles of bleach?" Tags="&lt;solutions&gt;&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29488" PostTypeId="1" AcceptedAnswerId="37710" CreationDate="2015-04-28T08:01:29.550" Score="4" ViewCount="4538" Body="&lt;p&gt;Ethers on hydrolysis with dilute $\ce{H2SO4}$ under pressure produce alcohols.&#xA;This is basically a type of acidic cleavage of ethers by the S&lt;sub&gt;N&lt;/sub&gt;2 mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFTPu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.masterorganicchemistry.com/reaction-guide/acidic-cleavage-of-ethers-sn2-reaction/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is inversion of configuration.&#xA;Now if $\ce{Nu}$ is $\ce{OH-}$ its called hydrolysis of ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right, that for S&lt;sub&gt;N&lt;/sub&gt;2 the leaving group must be more stable than the approaching group? In case of ether hydrolysis the leaving alkoxy group is less stable than the entering $\ce{OH-}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does the reaction succeed? Can someone please explain?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="5026" LastEditDate="2018-03-15T09:43:01.877" LastActivityDate="2018-03-15T09:43:01.877" Title="How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrolysis&gt;&lt;ethers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29489" PostTypeId="2" ParentId="29486" CreationDate="2015-04-28T08:05:24.580" Score="0" Body="&lt;p&gt;Try to use aqueous sodium chloride.&#xA;The lead in the gasoline will with the chloride ion to form lead chloride with is unsoluble white percipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb^{2+} + 2Cl^- -&amp;gt; PbCl_2}$&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-04-28T08:05:24.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29490" PostTypeId="1" AcceptedAnswerId="29528" CreationDate="2015-04-28T08:18:10.513" Score="1" ViewCount="1859" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UrwWS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the mechanism of trialkylborane. In this mechanism I can't understand why should the -OH group leave and the -R group enter in its place to form borate ester. Can someone please explain me that step?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="418" LastEditDate="2016-10-19T15:18:41.510" LastActivityDate="2016-10-19T15:18:41.510" Title="What is the mechanism of trialkylborane oxidation with hydroperoxide?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29491" PostTypeId="1" AcceptedAnswerId="29515" CreationDate="2015-04-28T08:55:07.933" Score="1" ViewCount="765" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2xWNU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During acid catalyzed hydrolysis of ester why does the carbonyl oxygen attack the $\ce{H^{+}}$ ? Why doesn't the alkoxy oxygen which is more electron rich and less electronegative not attack the $\ce{H^{+}}$ with its lone pair? Any ideas? &lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="10185" LastEditorDisplayName="user15489" LastEditDate="2015-10-02T16:59:59.203" LastActivityDate="2015-10-02T16:59:59.203" Title="Acid catalyzed hydrolysis of ester" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29493" PostTypeId="2" ParentId="29455" CreationDate="2015-04-28T10:06:57.967" Score="1" Body="&lt;p&gt;If you give enough oxygen to let it react to mercury, it still won't work just like copper and sliver. Because they are unreactive metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it will form mercury oxide when mercury and oxygen react at 350°C as the chemical equation below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Hg + O_2 -&amp;gt; 2HgO}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: The mercury oxide will decompose back to mercury and oxygen at 400°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lead is a rather more reactive subatances. So, try burning it. If it don't work, means you need a higher temperature.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2015-04-28T12:44:18.567" LastActivityDate="2015-04-28T12:44:18.567" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29494" PostTypeId="2" ParentId="29441" CreationDate="2015-04-28T10:08:35.670" Score="2" Body="&lt;p&gt;Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cr(OH)2 &amp;lt;=&amp;gt; Cr^2+ + 2 OH- }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, you know &#xA;$[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-28T10:08:35.670" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="29495" PostTypeId="2" ParentId="29405" CreationDate="2015-04-28T10:33:11.823" Score="1" Body="&lt;p&gt;I am a pretty elementary level student myself, so please excuse and ignore if you find me saying something wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition a buffer solution &lt;em&gt;resists&lt;/em&gt; the sudden change in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; on addition of a small amount of strong acid or strong base or due to dilution.&#xA;One possible way for this to occur (in case of acid buffer) is for the strong acid introduced into the system (buffer) to be converted to a weaker acid (that is why we require both a weak acid and its salt), same for a strong base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider ethanoic acid and sodium ethanoate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + CH3COONa -&amp;gt; CH3COOH}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; – proton from strong acid; &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3COOH}$&lt;/span&gt; – a weaker acid than a one introduced, hence no considerable change in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + OH- -&amp;gt; CH3COO- + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is from introduced base (mildly reversible reaction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acetate here lacks a hydroxyl group as opposed to the species you introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a basic buffer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4OH + H+ -&amp;gt; NH4+ + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; is from acid; &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4+}$&lt;/span&gt; is weakly acidic; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4Cl + OH- -&amp;gt; NH4OH + Cl-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is from base; &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4OH}$&lt;/span&gt; – weak base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, both basic and acidic buffers serve the same purpose and in nearly the same way.. they just have different compositions and consequently different &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pOH}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15868" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2019-01-23T08:26:17.837" LastActivityDate="2019-01-23T08:26:17.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29496" PostTypeId="1" AcceptedAnswerId="29506" CreationDate="2015-04-28T10:53:20.050" Score="1" ViewCount="368" Body="&lt;p&gt;You pour $50~\mathrm{mL}$ of a $0.0200~\mathrm{M}$ $\ce{HCOOH}$-solution to $150~\mathrm{mL}$ of a $0.0500~\mathrm{M}$ $\ce{HCOONa}$-solution. The $K_\mathrm{a}$ of $\ce{HCOOH}$ is $1.8 \times 10^{-4}$ and $\mathrm{p}K_\mathrm{a}$ is $3.74$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Compute the concentrations of $\ce{HCOOH}$ and its salt after dilution to this $200~\mathrm{mL}$ buffer solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Calculate the $\mathrm{pH}$ of this buffer solution with the buffer formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; Not sure how to do part a). For part b) I used the following method. We have $50 \times 10^{-3}~\mathrm{L}  \cdot \frac{0.0200~\mathrm{mol}}{1~\mathrm{L}} = 0.001~\mathrm{mol}$ of $\ce{HCOOH}$. Doing this again for $\ce{HCOONa}$ gives $0.0075~\mathrm{mol}$. Hence: \begin{align*} \mathrm{pH} = 3.74 + \log(\frac{0.0075}{0.001}) = 4.61 \end{align*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how can I solve a)? I'm having trouble with writing the reaction down. I'm not sure what's happening. Which chemical is neutralizing which? And what are the ions?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorUserId="15410" LastEditDate="2015-05-13T02:24:44.787" LastActivityDate="2015-05-13T02:24:44.787" Title="Computing pH of a buffer, and concentrations" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29497" PostTypeId="2" ParentId="29472" CreationDate="2015-04-28T11:26:16.203" Score="2" Body="&lt;p&gt;The addition of ammonium tartrate turns lead tartrate into a water-soluble tartrato complex: &#xA;\[\ce{Pb(tart) + tart^{2-} &amp;lt;=&gt;&gt; [Pb(tart)2]^2-}\]&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-04-28T11:26:16.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29499" PostTypeId="2" ParentId="29496" CreationDate="2015-04-28T11:46:37.933" Score="0" Body="&lt;p&gt;In a buffer solution, reactions primarily occur between the externally introduced species and the buffer constituents, HCOONa buffers the incoming H(+) proton from a strong acid and HCOOH buffers OH(-) from a base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is, as i believe, no other reactions involved here&lt;/p&gt;&#xA;" OwnerUserId="15868" LastActivityDate="2015-04-28T11:46:37.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29500" PostTypeId="1" CreationDate="2015-04-28T12:00:54.587" Score="5" ViewCount="35862" Body="&lt;p&gt;So here's my question: we know that melting of ice is an spontaneous reaction but is it spontaneous? We have to supply energy in the form heat for the reaction to take place.&lt;/p&gt;&#xA;" OwnerUserId="15872" LastEditorUserId="7448" LastEditDate="2015-04-28T12:16:41.920" LastActivityDate="2016-11-07T06:17:44.717" Title="Why is the melting of ice a spontaneous reaction?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;free-energy&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29502" PostTypeId="2" ParentId="29500" CreationDate="2015-04-28T12:38:07.970" Score="6" Body="&lt;p&gt;When &lt;em&gt;T&lt;/em&gt; &gt; 0 °C, ice melts spontaneously; the reverse process, liquid water turning into ice, is not spontaneous at these temperatures. However, when &lt;em&gt;T&lt;/em&gt; &amp;lt; 0 °C, the opposite is true. Liquid water converts into ice spontaneously, and the conversion of ice into water is not spontaneous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At &lt;em&gt;T&lt;/em&gt; = 0 °C the solid and liquid phases are in equilibrium. At this particular temperature the two phases are interconverting at the same rate, and there is no preferred direction for the process: Both the forward and reverse processes occur with equal preference, and the process is not spontaneously favored in one direction over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Δ&lt;em&gt;G&lt;/em&gt; is negative, a process or chemical reaction proceeds spontaneously in the forward direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also from the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Spontaneous_process&quot;&gt;Spontaneous reaction&lt;/a&gt; says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every reactant in a spontaneous process has a tendency to form the corresponding product. This tendency is related to stability. Stability is gained by a substance if it is in a minimum energy state or is in  &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot;&gt;maximum randomness&lt;/a&gt;. Only one of these can be applied at a time. e.g. Water converting to ice is a spontaneous process because ice is more stable since it is of lower energy. However, the formation of water is also a spontaneous process as water is the more random state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11310" LastEditorDisplayName="user7951" LastEditDate="2015-04-28T15:50:18.907" LastActivityDate="2015-04-28T15:50:18.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29503" PostTypeId="2" ParentId="29500" CreationDate="2015-04-28T13:17:45.433" Score="8" Body="&lt;p&gt;First of all, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what exactly is spontaneous reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let us consider melting of ice example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process took place without any intervention from your side. Heat absorbed by the ice from surrounding is due to set of conditions not due to your intervention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we can consider melting of ice as a spontaneous process(or reaction). &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-04-28T13:17:45.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29504" PostTypeId="2" ParentId="29500" CreationDate="2015-04-28T13:23:54.837" Score="6" Body="&lt;p&gt;I'm not sure if you have learnt this, so I just tell the story from the beginning. (I'm not sure, but if this is hard to understand, please tell me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a universal law usually called as &lt;strong&gt;the second law of thermodynamics&lt;/strong&gt;. It states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;entropy&lt;/strong&gt; of the universe tends to a maximum. If a process is reversible, the entropy of the universe will keep unchanged; if it is irreversible, the entropy increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, it's not the original statement. Historically, from the original statement of the law, a thermodynamic function is considered useful; that's &quot;entropy&quot; I mentioned here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is entropy? It's quite a mystery because at the first set out people do not defined what will it be; while, they just defined how it changes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An infinitesimal increment in the entropy of a system results from an infinitesimal transfer of heat is $\mathrm{d}S=\delta Q/T$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ur, I think you will not like this. Hopefully, later, Boltzmann state a hypothesis, he assumed the occupation of any microstate (means the distribution of all the particles we concerned in the space we concerned) is equally probable, then&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In most cases, the entropy is proportional to the logarithm of the number of possible microstate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You propably still think this is hard to understand... so we can say all the above in a easier way (but, I think, a less accurate way):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;the second law of thermodynamics&lt;/strong&gt; teaches us, the &lt;strong&gt;randomness&lt;/strong&gt; of the universe tend to be maximum. If a process is reversible, the randomness of the universe will keep unchanged; if it is irreversible, the randomness increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But...we still can't know how to judge whether a process is spontaneous or not yet. However, by the second law of thermodynamics we are easy to know the randomness of universe increases in a spontaneous process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we get $\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surronding} &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While we also know if the heat transferred to the system from the surroundings is $Q$, then $\Delta S_\text{surronding} = -Q/T$, where $T$ is the temperature of the surrounding. (Since the surrounding is large, we may assume the temperature is unchanged.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we get $\Delta S_\text{system} - Q/T &amp;gt; 0$, i.e. $T\Delta S_\text{system} - Q &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the process is now assumed to be &lt;strong&gt;isobaric&lt;/strong&gt; (means that the barometric pressure of system is unchanged), then $Q = \Delta H$ and we get $T\Delta S_\text{system} - \Delta H &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can define a new thermodynamic function:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;strong&gt;Gibbs free energy&lt;/strong&gt; of a system is $G = H - TS$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then we can say that &lt;strong&gt;in a spontaneous process&lt;/strong&gt;, $\Delta G &amp;lt; 0$. (If $\Delta G = 0$, it is in equilibrium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you have known it all... uh... so let's solve the question: &lt;strong&gt;Why is the melting of ice a spontaneous reaction&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know in liquid particles will arrange in a messer way then in a solid, so the entropy, or the randonness of the &quot;reaction system&quot; is increased, i.e. $\Delta S_\text{system} &amp;gt; 0$. However, when ice melt, $\Delta H &amp;gt; 0$, so for  $\Delta G &amp;lt; 0$ you need to keep the temperature higher, or supply energy to it. And simliarly, you will find if temperature is low enough, the water will spontaenously freeze. This seems to be the case you are familiar with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noticed that sometimes $\Delta G &amp;lt; 0$, however the speed of reaction is too low, so we supply energy for accerlation. For example, the diamond will burn:&#xA;$\ce{C_{(s, diamond)} + O2_{(g)} -&amp;gt; CO2_{(g)}},\ \Delta G^\circ= -397\ \mathrm{kJ}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you never see this happen in common situation: because it's too slow.&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorDisplayName="user7951" LastEditDate="2015-04-28T15:48:39.013" LastActivityDate="2015-04-28T15:48:39.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29505" PostTypeId="2" ParentId="29468" CreationDate="2015-04-28T13:28:51.543" Score="1" Body="&lt;p&gt;You can actually just use salt. It is because that the chemical reaction on the top is for one condition which is that your halide solution ($\ce{NaCl}$) has a very high concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you halide solution has lower concentration,the hydrogen ion and hydroxide ion in the water will discharged and the two half equation will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4H^+ + 4e^- -&amp;gt; 2H_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4OH^- -&amp;gt; 2H_2O + O_2 + 4e^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the overall equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{4H^+ + 4OH^- -&amp;gt; 2H_2 + 2H_2O + O_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, is okay to use salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Besides Salt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use hydrochloric acid besides salt. Just remember that don't use a high concentration hydrochloric acid as the anode will just discharge the chloride ion instead of hydroxide ion.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-04-28T13:28:51.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29506" PostTypeId="2" ParentId="29496" CreationDate="2015-04-28T13:29:14.597" Score="1" Body="&lt;p&gt;We have to compute the new concentrations after dilution:&#xA;$$\ce{[HCOOH]}=\frac{0.02\times 50}{200}=0.005\,\, \mathrm{mol/L}$$&#xA;$$\ce{[HCOO^- ]}=\frac{0.05\times 150}{200}=0.0375\,\, \mathrm{mol/L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can compute the $\mathrm{pH}$ of the buffer solution:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-04-28T13:29:14.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29507" PostTypeId="2" ParentId="26914" CreationDate="2015-04-28T13:30:58.510" Score="3" Body="&lt;p&gt;Your calculation is correct. However, there needs to be more bicarbonate in the blood because the majority of all metabolic processes tend to produce more acidic than basic products. There are also other ways to help the body maintain a pH of 7.4, not only the carbonic acid/bicarbonate buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both you and the book are right. If the calculations are done assuming there is just carbonic acid/bicarbonate acting as a way to maintian the pH in the blood, the ratio should be about 10 to one. Since there is a lot more going on though, the ratio is different than that. &lt;/p&gt;&#xA;" OwnerUserId="15874" LastActivityDate="2015-04-28T13:30:58.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29509" PostTypeId="2" ParentId="29487" CreationDate="2015-04-28T14:16:43.420" Score="4" Body="&lt;p&gt;The key issue to address this kind of problems is to determine the mass of the pure compound in the given volume of the solution. So, we'll use first the density to determine the mass of the solution. Then, the mass concentration to determine the mass of the pure compound. After that, we can readily compute the number of moles of the pure compound using the molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of $1\,\mathrm{mL}$ of bleach solution is $1.07\,\mathrm{g}$.&#xA;So, the mass of $4~\mathrm{mL}$ of bleach solution is $4 \cdot 1.07\,\mathrm{g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, each $100\,\mathrm{g}$ of bleach solution has $6\,\mathrm{g}$ of pure bleach.&#xA;So, the mass of  $4 \times 1.07\mathrm{g}$ of bleach solution has $\frac{4\cdot 6 \cdot 1.07}{100}\,\mathrm{g}$ of pure bleach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of moles in  $4\mathrm{mL}$ of bleach solution is: $$\frac{4\cdot 6 \cdot 1.07}{100\cdot 74.44}= 0.0034\,\mathrm{mol}$$ &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="15410" LastEditDate="2015-05-13T02:36:30.693" LastActivityDate="2015-05-13T02:36:30.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29511" PostTypeId="1" AcceptedAnswerId="29518" CreationDate="2015-04-28T15:00:57.470" Score="12" ViewCount="524" Body="&lt;p&gt;$\ce{[BrF6]+}$ is an octahedral species, and should have only one fluorine environment due to the symmetry. $\ce{Br}$ has nuclear spin $I=3/2$ and therefore is unlikely to cause splitting due to rapid interconversion between its spin up and spin down states averaging to a zero magnetic effect. Therefore I would expect one singlet in the $\ce{^19F}$ NMR spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However... the $\ce{^19F}$ NMR is actually two $1:1:1:1$ quartets (at low temperatures). Can someone help me find the flaws in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="16683" LastEditDate="2017-08-26T13:44:31.937" LastActivityDate="2017-08-26T13:45:37.770" Title="19F NMR of hexafluorobromine(VII) cation" Tags="&lt;inorganic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29512" PostTypeId="1" CreationDate="2015-04-28T15:08:32.653" Score="4" ViewCount="215" Body="&lt;p&gt;Suppose we have a chemical compound like ethyl 4-oxo-2-cyclopentene-1-carboxylate&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/73B6e.png&quot; alt=&quot;Ethyl 4-oxo-2-cyclopentene-1-carboxylate&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It is &lt;span class=&quot;math-container&quot;&gt;$\ce{-COOC2H5}$&lt;/span&gt; bonded to a five-membered carbon ring with one double bond and the ring has another carbonyl group.&lt;/p&gt;&#xA;&lt;p&gt;Suppose this compound is made to react with these &lt;em&gt;reducing agents&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{BH3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{THF}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2/Pt}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2/Pd-C}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Then what will be the product in each cases?&lt;/p&gt;&#xA;&lt;p&gt;I know some of the reducing agents and I want to know their reactions with a compound which is unsaturated and has an ester functionality &lt;span class=&quot;math-container&quot;&gt;$\ce{-COOR}$&lt;/span&gt; and a carbonyl group &lt;span class=&quot;math-container&quot;&gt;$\ce{R2C=O}$&lt;/span&gt; bonded to it.&lt;br /&gt;&#xA;I cannot try to understand it due to my limited knowledge in chemical mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="7434" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-10-22T13:48:02.073" Title="How do popular reducing agents react on compounds that have ester, carbonyl, and unsaturated functionalities?" Tags="&lt;organic-chemistry&gt;&lt;organic-reduction&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29515" PostTypeId="2" ParentId="29491" CreationDate="2015-04-28T15:25:33.057" Score="1" Body="&lt;p&gt;The difference between the carbonyl oxygen (red in colour) and alkoxy oxygen (blue in colour) is that the highly distorted pi-electron density (green shaded) towards to the carbonyl oxygen, due to more electro-negativity of oxygen that is the feature which is more on this oxygen than alkoxy oxygen . That feature provide the cavity for the H+ ion from acid, and the distortion of sigma- electron density at alkoxy oxygen is comparatively less than the distortion of  pi-electron density at carbonyl oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vg7kf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15800" LastActivityDate="2015-04-28T15:25:33.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29516" PostTypeId="1" CreationDate="2015-04-28T16:00:33.587" Score="3" ViewCount="97" Body="&lt;p&gt;Would the electrolysis of alcohol, specifically spirits, result in a potable end product? I'm looking for some novel cocktail ingredients.&lt;/p&gt;&#xA;" OwnerUserId="15878" LastEditorUserId="15410" LastEditDate="2015-05-13T02:15:20.960" LastActivityDate="2015-05-13T02:15:20.960" Title="Catering grade electrolysis of alcohol" Tags="&lt;electrolysis&gt;&lt;alcohols&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29517" PostTypeId="2" ParentId="29465" CreationDate="2015-04-28T16:00:49.107" Score="9" Body="&lt;p&gt;Uh... I think the insight you need is about chemical equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: &lt;strong&gt;what's the component of this mixure&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;p&gt;To answer this is not easy.&lt;/p&gt;&#xA;&lt;p&gt;In a glass of pure water, there are still a little &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; (to say presicely, &lt;span class=&quot;math-container&quot;&gt;$\ce {H3O+}$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\ce {OH-}$&lt;/span&gt;. This is because sometimes proton (&lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt;) will &amp;quot;jump&amp;quot; from one &lt;span class=&quot;math-container&quot;&gt;$\ce {H2O}$&lt;/span&gt; molecular to another one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;(Since we usually do not use the expression &lt;span class=&quot;math-container&quot;&gt;$\ce {H3O+}$&lt;/span&gt;, this process can be rewrited as &lt;span class=&quot;math-container&quot;&gt;$\ce {H2O &amp;lt;=&amp;gt; H+ + OH-}$&lt;/span&gt; to avoid disruption.)&lt;/p&gt;&#xA;&lt;p&gt;This phenomenon (called the &lt;strong&gt;autoionization&lt;/strong&gt; of the water) is very weak. The concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; in &amp;quot;pure water&amp;quot; is &lt;span class=&quot;math-container&quot;&gt;$\ce{[H+]=[OH^{-}]=}1.0 \times 10^{-7} M$&lt;/span&gt; when the temperature is 25°C (298K).&lt;/p&gt;&#xA;&lt;p&gt;Then you put something in this glass of water. As other people mentioned, most of &lt;strong&gt;covalent compounds&lt;/strong&gt;, such as glucose, will not break apart in water since the intermolecular force (covalent bond) do not break. Water moleculars may surround those covalent compounds like a cage to make them dissolved. So there won't be other things.&lt;/p&gt;&#xA;&lt;p&gt;But sometimes it breaks: when &lt;strong&gt;acid&lt;/strong&gt; compounds are dissolved into water, like &lt;span class=&quot;math-container&quot;&gt;$\ce {HCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {H2SO4}$&lt;/span&gt;, they will give their hydrogen atoms as proton to water:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {HCl + H2O -&amp;gt; H3O+ + Cl-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {H2SO4 + H2O -&amp;gt; H3O+ + HSO4-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is because they are more acidic than &lt;span class=&quot;math-container&quot;&gt;$\ce {H3O+}$&lt;/span&gt;, here &amp;quot;acid&amp;quot; means &amp;quot;who gives proton(s)&amp;quot;, while &amp;quot;base&amp;quot; means &amp;quot;who accepts proton(s)&amp;quot; (This kind of definion is called as &lt;strong&gt;Brønsted-Lowry theory&lt;/strong&gt;, and as you see, in autoionization of water, one of the &lt;span class=&quot;math-container&quot;&gt;$\ce {H2O}$&lt;/span&gt; is acid and another is base).&lt;/p&gt;&#xA;&lt;p&gt;Noticed that acid like &lt;span class=&quot;math-container&quot;&gt;$\ce {HCl}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {H2SO4}$&lt;/span&gt; will nearly definitely give their &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; to water compounds (they are too “willing” to do it), we call those acid as &lt;strong&gt;strong acid&lt;/strong&gt;. But some acid are not: protons sometimes jump back.&lt;/p&gt;&#xA;&lt;p&gt;For example, &lt;span class=&quot;math-container&quot;&gt;$\ce{HSO4-}$&lt;/span&gt; will meet an equilibrium:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HSO4- + H2O &amp;lt;=&amp;gt; H3O+ + SO4^2-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This kind of acid will be called as &lt;strong&gt;weak acid&lt;/strong&gt;. In an aqueous solution of a weak acid, most of the weak acid compounds keep unchanged.&lt;/p&gt;&#xA;&lt;p&gt;Yet we haven't mentioned what you concerned: how about aqueous solution of &lt;strong&gt;ionic compounds&lt;/strong&gt;? They usually break their intermolecular forces (ionic bond) via &lt;strong&gt;hydration&lt;/strong&gt;, i.e. combination with water compound. Water compound have some charges on them, so they will attract ions and break them apart. Especially, we will see bases we are familiar with -- those who will ionized as cations and &lt;span class=&quot;math-container&quot;&gt;$\ce {OH-}$&lt;/span&gt; -- first.&lt;/p&gt;&#xA;&lt;p&gt;In fact, the structure of hydrated cation will be complex (&amp;quot;&lt;span class=&quot;math-container&quot;&gt;$\ce {H3O+}$&lt;/span&gt;&amp;quot; is also not the actual situation), so we usually just write them as unhydrated form, such as &lt;span class=&quot;math-container&quot;&gt;$\ce {Fe^2+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce {Na+}$&lt;/span&gt;. As you know, there are weak bases and strong bases.&lt;/p&gt;&#xA;&lt;p&gt;So bases like &lt;span class=&quot;math-container&quot;&gt;$\ce {NaOH}$&lt;/span&gt; are quite &amp;quot;willing&amp;quot; to accept protons:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {NaOH -&amp;gt; Na+ + OH-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But &lt;span class=&quot;math-container&quot;&gt;$\ce {Fe(OH)3}$&lt;/span&gt; may not, and it also meet an equilibrium:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {Fe(OH)3 &amp;lt;=&amp;gt; Fe^3+ + 3OH-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again, in an aqueous solution of a weak acid, most of the weak base sturcture keep unchanged. And for this reason, weak bases are usually insoluable in water, since water &amp;quot;have no good idea&amp;quot; to deal with the whole relatively large ion sturcture. But in fact, the word &amp;quot;insoluable&amp;quot; means that there is another equilibrium like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {Fe(OH)3_{(aq)} &amp;lt;=&amp;gt; Fe(OH)3_{(s)}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and the solid form dominates.&lt;/p&gt;&#xA;&lt;p&gt;So here we meet the hardest part: what about adding some &lt;strong&gt;salts&lt;/strong&gt; into water, like &lt;span class=&quot;math-container&quot;&gt;$\ce {MgCl2}$&lt;/span&gt; you mentioned?&lt;/p&gt;&#xA;&lt;p&gt;So far you have learned about ionization of the salt, and you know most of salt will completely ionized. But how about those ions?&lt;/p&gt;&#xA;&lt;p&gt;A balance like &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg(OH)2 &amp;lt;=&amp;gt; Mg^2+ + 2OH-}$&lt;/span&gt; shows that &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg(OH)2}$&lt;/span&gt; is a  weak base. But if we see it as &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg^2+ + 2OH- &amp;lt;=&amp;gt; Mg(OH)2}$&lt;/span&gt; it means &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg^2+}$&lt;/span&gt; is a weak acid. And if you remember I have said in autoionization of water, the water compound can be both base and acid (we say this kind of substances are &lt;strong&gt;amphoteric&lt;/strong&gt;). Then we can know that there will be a new equilibrium like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce {Mg^2+ + 2H2O &amp;lt;=&amp;gt; Mg(OH)2 + 2H+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(We usually call this as &lt;strong&gt;hydrolyzation&lt;/strong&gt; of salt.)&lt;/p&gt;&#xA;&lt;p&gt;And because &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg^2+}$&lt;/span&gt; is a weak base, you may expect that there will be not a lot &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg(OH)2}$&lt;/span&gt; and it's still soluable. Also there will be not a lot &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; so the solution is just a bit more acidic then water.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Addition: You can also think this comes from these two balances:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O &amp;lt;=&amp;gt; H+ + OH-}$&lt;/span&gt; (autoionization).&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Mg(OH)2 &amp;lt;=&amp;gt; Mg^2+ + 2OH-}$&lt;/span&gt;. in further calculation you may meet in future,  this kind of idea can make you avoid some mistakes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So you may expect in a &lt;span class=&quot;math-container&quot;&gt;$1M$&lt;/span&gt; aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce {MgCl2}$&lt;/span&gt;, there will be a lot of &lt;span class=&quot;math-container&quot;&gt;$\ce {Cl-}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$2M$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg^2+}$&lt;/span&gt; (a bit less than &lt;span class=&quot;math-container&quot;&gt;$2M$&lt;/span&gt;), a little &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; and a little &lt;span class=&quot;math-container&quot;&gt;$\ce {Mg(OH)2}$&lt;/span&gt;, and there will be &lt;span class=&quot;math-container&quot;&gt;$\ce {OH-}$&lt;/span&gt;  which is negligible in the most case. (Of course there is very much water!)&lt;/p&gt;&#xA;&lt;p&gt;While you can see in the case of &lt;span class=&quot;math-container&quot;&gt;$\ce {NaCl}$&lt;/span&gt; there won't be &lt;span class=&quot;math-container&quot;&gt;$\ce {NaOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {HCl}$&lt;/span&gt; compounds in the solution: because &lt;span class=&quot;math-container&quot;&gt;$\ce {NaOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {HCl}$&lt;/span&gt; are &amp;quot;strong&amp;quot;, there won't be a hydrolyzation equilibrium. But yes, you can say there are &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {OH-}$&lt;/span&gt;, as you have known, there are &lt;span class=&quot;math-container&quot;&gt;$1.0 \times 10^{-7} M$&lt;/span&gt; of them. However, we usually not concern about this. And as you see, all those &lt;span class=&quot;math-container&quot;&gt;$\ce {H+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce {OH-}$&lt;/span&gt; in an aqueous solution, in effect, come from water compounds. (This is why there will be &amp;quot;strong&amp;quot; acid and base: because they are all giving/receiving protons from water, even if they actually have different ability, compared to water, they are behaving undistinguishably. This is called as &lt;strong&gt;leveling effect of solvent&lt;/strong&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;There are a lot of things I haven't mentioned here, because I think you are not able to calculate something about equilibrium, and maybe you are not familiar with thermodynamic idea yet. But here I just hope what I said is interesting enough for you.&lt;/p&gt;&#xA;&lt;p&gt;Well, a simple &lt;strong&gt;conclusion&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To find out what will be there in an aqueous solution, you may consider:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the solute ionize: Is it a covalent compound? Is it acid?&lt;/li&gt;&#xA;&lt;li&gt;If yes, whether the ions are weak acid or base (in Brønsted-Lowry theory)? Remember, a cation which can be formed from a weak base will be weak acid; while an anion which can be formed from a weak acid will be weak base. (Or you may use an awful word: they are conjugate acid base pair!)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I think these two step are enough for you for the time being.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15820" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-28T20:18:11.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29518" PostTypeId="2" ParentId="29511" CreationDate="2015-04-28T16:24:15.193" Score="13" Body="&lt;p&gt;$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent.  As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin &gt; 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the other nucleus into a 1:1:1:1 quartet.  In the case of bromine we don't normally see this coupling because bromine typically undergoes rapid quadrupolar relaxation (some quadrupolar nuclei do, some don't) which washes the coupling out (&lt;em&gt;e.g&lt;/em&gt;. rapid quadrupolar relaxation effectively decouples the nuclei, that's why the protons in $\ce{CH3Br}$ are a singlet, for example).  However, when the quadrupolar nucleus exists in a highly symmetric environment (octahedral, tetrahedral geometry) the electric filed gradient at the nucleus is near zero and quadrupolar relaxation becomes slow (on the nmr timescale) and coupling can often be observed in such cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, bromine occurs naturally as a mixture of two isotopes, $\ce{^{79}Br}$ (50.7%) and $\ce{^{81}Br}$ (49.3%).  Both isotopes are quadrupolar with spin = 3/2.  Because of the slight difference in electronegativity of the two bromine isotopes, they will produce a slight difference in chemical shift for the fluorines bonded to them. Hence, this mixture of $\ce{^{79}BrF6^+}$ and $\ce{^{81}BrF6^+}$ will produce &lt;strong&gt;two 1:1:1:1 quartets&lt;/strong&gt; in the $\ce{^{19}F}$ nmr spectrum. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-08-26T13:45:37.770" LastActivityDate="2017-08-26T13:45:37.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29519" PostTypeId="2" ParentId="29500" CreationDate="2015-04-28T16:37:43.097" Score="6" Body="&lt;p&gt;I wouldn't call it a reaction, it's a phase change. The term reaction refers to a change in covalent bonding between components of a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase change occurs in a seemingly spontaneous manner because the environment is capable of donating enough free energy (most likely from enthalpy, could happen otherwise) to overcome the energy needed to activate the process of reorganization of the system (increase in entropy). The other responses have done a nice job in terms of symbolic rigor, so I won't repeat them here.&lt;/p&gt;&#xA;" OwnerUserId="14523" LastActivityDate="2015-04-28T16:37:43.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29520" PostTypeId="1" AcceptedAnswerId="29537" CreationDate="2015-04-28T16:42:00.623" Score="3" ViewCount="657" Body="&lt;p&gt;I am testing electronics with many potentially unstable contacts, and I wish to monitor if they rise above 50 to 90 degrees Celsius, i.e. using carbon paper that would darken progressively from 50 to 90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The substance should stay a specific warning color if it goes too high and not change hues again same as thermochromic pigment, so that I can detect high temps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you know a chemistry trick that could help? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="14577" LastActivityDate="2015-04-28T19:53:47.347" Title="50-90'C color changing chemical to use as warning marker for hot electronics?" Tags="&lt;thermodynamics&gt;&lt;home-experiment&gt;&lt;temperature&gt;&lt;color&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29521" PostTypeId="1" CreationDate="2015-04-28T16:50:17.267" Score="2" ViewCount="164" Body="&lt;p&gt;Given a weak base such as $\ce{C2H5NH2}$, I usually see the expression for $K_\text{b}$ as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{OH-}]}{[\ce{C2H5NH2}]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the expressions &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_\text{b} = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$,&#xA;$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;equivalent? Do they represent the same process (if so, why do the latter two not work with calculations?) or fundamentally different ways of dissociation? Which one represents what's happening molecularly most accurately? Given the Brønsted-Lowry definition of a base as a proton acceptor I'd think that the second one is most accurate, but I rarely see it used. &lt;/p&gt;&#xA;" OwnerUserId="11324" LastEditorUserId="5026" LastEditDate="2018-03-04T07:43:12.283" LastActivityDate="2018-03-04T12:45:05.920" Title="Are the following expressions for base dissociation constant equivalent?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29522" PostTypeId="1" AcceptedAnswerId="29585" CreationDate="2015-04-28T16:56:38.347" Score="6" ViewCount="4910" Body="&lt;p&gt;How much energy is produced by burning a 1 ct. diamond?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out how much energy to build the diamonds in a biological process.  See here: &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/15314/are-diamond-berries-possible&quot;&gt;https://worldbuilding.stackexchange.com/questions/15314/are-diamond-berries-possible&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15880" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:35.613" LastActivityDate="2015-04-29T22:41:59.230" Title="Burning a diamond produces how much energy?" Tags="&lt;energy&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29523" PostTypeId="1" CreationDate="2015-04-28T17:01:58.943" Score="3" ViewCount="7801" Body="&lt;p&gt;Most books I have read stop at this order : &lt;br&gt;&#xA;Benzylic &gt; Allylic &gt; Tertiary &gt; secondary &gt; primary &lt;br&gt; &#xA;However If I want to compare tertiary, secondary and primary benzylic and Allylic carbocations what will be their stability order?&lt;/p&gt;&#xA;" OwnerUserId="12902" LastEditorUserId="10185" LastEditDate="2015-04-28T19:18:51.000" LastActivityDate="2015-11-30T15:17:48.490" Title="Benzylic and allylic carbocation stability order" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-11-30T20:27:44.807" ContentLicense="CC BY-SA 3.0" />
  <row Id="29524" PostTypeId="2" ParentId="29521" CreationDate="2015-04-28T17:41:17.120" Score="1" Body="&lt;p&gt;$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$&#xA;This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$ &#xA;here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.&lt;/p&gt;&#xA;" OwnerUserId="5424" LastEditorDisplayName="user7951" LastEditDate="2018-03-04T12:45:05.920" LastActivityDate="2018-03-04T12:45:05.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29525" PostTypeId="2" ParentId="29521" CreationDate="2015-04-28T17:50:25.393" Score="1" Body="&lt;p&gt;I try to say it clearer and simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second expression respects to the equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C2H5NH2 + H+ &amp;lt;=&amp;gt; C2H5NH3+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the third&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C2H5NH2 + H3O+ &amp;lt;=&amp;gt; C2H5NH3+ + H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what you want to know is the equilibrium constant of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce {C2H5NH2 + H2O &amp;lt;=&amp;gt; C2H5NH3+ + OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former two both describe how $\ce {C2H5NH3+}$ behaving as a weak acid in aqueous solution, and the third one describes how $\ce {C2H5NH2}$ behaving as a weak base in aqueous solution (hydrolyzation). They are different because in the former two equilibrium $\ce {C2H5NH2}$ are not receiving proton &lt;strong&gt;directly&lt;/strong&gt; from $\ce {H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you may look it by this view: the last one can be seen as the result of two equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce {C2H5NH2 + H+ &amp;lt;=&amp;gt; C2H5NH3+}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce {H2O &amp;lt;=&amp;gt; H+ + OH-}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or you may write as&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce {C2H5NH2 + H3O+ &amp;lt;=&amp;gt; C2H5NH3+ + H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce {2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you can see there is obvious different. The equilibrium constant of $\ce {2H2O &amp;lt;=&amp;gt; H3O+ + OH-}$ is not 1, so the base dissociation constant $K_b$ will not be the same as what you said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there is an important convention that in an aqueous solution system, the equilibrium constant should omit [$\ce {H2O}$] because it is an insignificant constant in the most cases.&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorUserId="15820" LastEditDate="2015-04-29T10:22:03.323" LastActivityDate="2015-04-29T10:22:03.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29526" PostTypeId="1" AcceptedAnswerId="29531" CreationDate="2015-04-28T17:52:10.683" Score="5" ViewCount="3013" Body="&lt;p&gt;I have a compound like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH2=C=O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it is a functional group or a normal group. As a functional group I couldn't find out.&#xA;&lt;strong&gt;Is it a normal compound or a functional group and what is it's&#xA;IUPAC name&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15884" LastEditorUserId="4945" LastEditDate="2015-06-01T02:03:29.913" LastActivityDate="2015-06-01T02:03:29.913" Title="Carbon with two double bonds and oxygen name" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29527" PostTypeId="1" CreationDate="2015-04-28T17:52:14.983" Score="6" ViewCount="313" Body="&lt;p&gt;I am trying to dissolve aspartic acid into vinegar as use as a carbon source in saltwater aquariums.   It does not dissolve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told to increase the pH and it will work?  What is the best way to increase pH with minimal negative effects?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium carbonate seems to work at dissolving; however I may be reducing the acetic acid from 5% to something lower?  Might also be creating an acetate?  Calcium hydroxide could be a better choice for the aquarium but I can't seem to get the mixing ratio right?   &lt;/p&gt;&#xA;" OwnerUserId="15883" LastEditorUserId="5456" LastEditDate="2015-04-29T09:44:28.310" LastActivityDate="2017-02-20T00:52:19.253" Title="Dissolving Aspartic acid in vinegar?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29528" PostTypeId="2" ParentId="29490" CreationDate="2015-04-28T18:10:58.700" Score="1" Body="&lt;p&gt;One aspect is the release of the stable anion $\ce{OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B-C}$ $D_{298}^ \circ  = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B-O}$ $D_{298}^ \circ  = 809\ \mathrm{kJ\ mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-28T18:10:58.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29529" PostTypeId="2" ParentId="29526" CreationDate="2015-04-28T18:22:31.243" Score="2" Body="&lt;p&gt;The compound you have drawn is called ketene, since the central carbon is part of both a &lt;strong&gt;ket&lt;/strong&gt;one and an alk&lt;strong&gt;ene&lt;/strong&gt;. Ketene is the parent compound of the ketene functional group, where the terminal carbon is substituted with some more complicated structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ketene itself is not stable and dimerizes to diketene by a [2+2]-cycloaddition. Diketene was &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/422363?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow noreferrer&quot;&gt;commercially available&lt;/a&gt; and can be converted to ketene by thermal or photochemical conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Yhe5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2015-04-28T19:10:17.707" LastActivityDate="2015-04-28T19:10:17.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29530" PostTypeId="1" CreationDate="2015-04-28T18:25:20.613" Score="5" ViewCount="649" Body="&lt;p&gt;On a test question, my organic chemistry professor gave us a really big molecule and told us to give list the formal charges on each atom. There was a $\ce{Br}$ with 3 lone pairs and a double bond to carbon (which then had a single bond to a $&#xA;\ce{H}$ and a $\ce{N}$). On the key, he labeled it with a + charge and I was wondering how is that possible (or if it is a mistake)? Shouldn't a $\ce{Br}$ with that many electrons be highly negative or does it go onto the next orbital allowing it to accommodate for this?&lt;/p&gt;&#xA;" OwnerUserId="15885" LastEditorUserId="15410" LastEditDate="2015-05-13T02:13:16.960" LastActivityDate="2015-05-13T02:13:16.960" Title="Bromine with 10 electrons" Tags="&lt;organic-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29531" PostTypeId="2" ParentId="29526" CreationDate="2015-04-28T18:28:29.037" Score="5" Body="&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; the name of the given compound is the systematically constructed name for ketones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the preferred IUPAC name (PIN) is &lt;strong&gt;ethenone.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name &lt;strong&gt;ketene&lt;/strong&gt; is retained, but only for general nomenclature.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-04-28T18:28:29.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29532" PostTypeId="2" ParentId="29405" CreationDate="2015-04-28T18:58:00.983" Score="2" Body="&lt;p&gt;Your suspicion is correct: the buffer pH depends on the $\rm{pK_a/pK_b}$ of the acid/base in question, not whether the buffer salt contains a weak conjugate acid or conjugate base. We can test this out via the application of the Henderson-Hasselbalch equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider acetic acid ($\rm{pK_a = 4.76}$) mixed with the same concentration of sodium acetate. Via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[Ac^{-}]}{[HAc]}}=pK_a + log\ 1 = pK_a = 4.76$. Thus, the acid/conjugate base buffer is acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take hydrocyanic acid ($\rm{pK_a = 9.21}$) mixed with the same concentration of sodium cyanide. Again, via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[CN^{-}]}{[HCN]}}=pK_a + log\ 1 = pK_a = 9.21$. The acid/conjugate base buffer is now basic. A similar analysis can be made for bases mixed with their conjugate acid salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does this happen? It's quite simple. If an acid has a $\rm{pK_a &amp;lt; 7}$, then it is more acidic than its conjugate base is basic ($\rm{pK_b &amp;gt; 7}$), so when mixing an equal amount of the acid and its conjugate base, the acidic character wins and the buffer is acidic. If the acid has a $\rm{pK_a &amp;gt; 7}$, then it is less acidic than the conjugate base is basic ($\rm{pK_b &amp;lt; 7}$). Mix an equal amount of both, and as you might expect, the basic character wins, and the buffer ends up alkaline. Again, a similar analysis can be made for buffers made with bases and their conjugate acid salts.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-04-28T18:58:00.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29533" PostTypeId="2" ParentId="29530" CreationDate="2015-04-28T19:06:45.430" Score="4" Body="&lt;p&gt;I suspect that the issue here is with counting the number of lone pairs.&#xA;I think the structure you are referring to looks something like the right hand one shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this structure can be described by the two resonance forms shown. The double bond arises because the bromine 'donates' one of its lone pairs to form another bond to the carbon, shifting the positive charge onto the more electronegative bromine and hence stabilising the ione. Since this is the 'more stable' resonance contributor it is usually shown as the structure of the ion in drawings but the actual structure is more accurately represented as a resonance hybrid of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hoUpC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-04-28T19:06:45.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29534" PostTypeId="1" AcceptedAnswerId="31121" CreationDate="2015-04-28T19:29:06.773" Score="11" ViewCount="2516" Body="&lt;p&gt;In &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13918/mechanism-for-tollens-classification-test-for-aldehydes&quot;&gt;this&lt;/a&gt; previous question the mechanism for the reaction of Tollens' reagent was outlined. As I understand it the oxidising agent in Tollens' is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ but the mechanism only showed the silver ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of adding ammonia to the silver nitrate to form the diamminesilver(I) complex: why not just use silver nitrate? Is it purely to provide a source of hydroxide ions or is the structure of the complex ion actually important?&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-09T17:01:07.660" Title="Purpose of ammine in Tollens' reagent" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29535" PostTypeId="2" ParentId="29530" CreationDate="2015-04-28T19:37:30.477" Score="4" Body="&lt;p&gt;Good question! As you state, &lt;em&gt;normally bromine forms negative ions&lt;/em&gt; with a charge of -1, as all orbitals are complete except for the 4p5, able to accept one electron. However, there is a &lt;em&gt;positive &lt;a href=&quot;http://courses.chem.psu.edu/chem210/mol-gallery/bromonium/bromonium.html&quot; rel=&quot;nofollow&quot;&gt;bromonium ion&lt;/a&gt;&lt;/em&gt;, first suggested by Irving Roberts and George Kimball in 1937. The generic term for a positive halogen ion is &lt;a href=&quot;http://en.wikipedia.org/wiki/Halonium_ion&quot; rel=&quot;nofollow&quot;&gt;halonium ion&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reference has a nice electron models of this odd ion.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-28T19:37:30.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29536" PostTypeId="1" AcceptedAnswerId="29555" CreationDate="2015-04-28T19:53:41.543" Score="8" ViewCount="198" Body="&lt;p&gt;Theoreticly, the width of the spectral line in  AAS (atomic absorbtion spectroscopy) is 10^-5, but in fact there happens an extention of it and it becomes 0.002-0.005. There are some factors which lead to this extention. Most common types of extentions are :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Doppler extention&lt;/li&gt;&#xA;&lt;li&gt;Pressure extention&lt;/li&gt;&#xA;&lt;li&gt;Self-absorbance extention.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to explain in a simple way why do these 3 kinds of extentions take place? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these deviations happen all the time in these measurments or are they some deviations that can happen in case the concentration of atoms is to high?&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-04-29T09:17:40.767" LastActivityDate="2015-04-30T04:36:00.913" Title="Wavelength extention in AAS" Tags="&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29537" PostTypeId="2" ParentId="29520" CreationDate="2015-04-28T19:53:47.347" Score="1" Body="&lt;p&gt;There are &lt;a href=&quot;http://www.hallcrest.com/our-products/special-effect-ink-coating/thermochromic-permanent-change-ink&quot; rel=&quot;nofollow&quot;&gt;commercial&lt;/a&gt;  &lt;a href=&quot;http://www.hallcrest.com/&quot; rel=&quot;nofollow&quot;&gt;sources&lt;/a&gt; for thermochromic permanent-change ink, but you might also try &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermal_paper&quot; rel=&quot;nofollow&quot;&gt;thermal printer paper&lt;/a&gt;. You might need to check if there is a time-dependence on the color change, as well as temperature dependence. The printer paper is available at Amazon, Staples and elsewhere for less than US$20 / package, but you might just test a cash-register receipt, many of which are on thermal paper. According to this &lt;a href=&quot;https://www.brother-usa.com/Mobile/pdfs/white_papers/ThermalPaperWhitePaper.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;, most start to darken at 79 C.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-04-28T19:53:47.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29539" PostTypeId="1" CreationDate="2015-04-28T22:17:49.033" Score="5" ViewCount="3106" Body="&lt;p&gt;Phosphorus is listed as a single element in the periodic table; however, I see that it is also a polyatomic molecule as $\ce{P4}$. I am trying to determine the number of atoms of phosphorus in an chemical equation, so I can have a balance equation. So, when would I use the polyatomic version, or the single atomic version if the chemical equation is a word equation, and the subscripts are not listed? For example the word equation is $\ce{P + Cl -&amp;gt; PCl5}$. Is $\ce{P -&amp;gt; P4}$ or just $\ce{P}$? Also, would $\ce{Cl}$ be diatomic too ($\ce{Cl2}$) in this equation?&lt;/p&gt;&#xA;" OwnerUserId="15330" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-07-01T12:10:34.137" LastActivityDate="2018-07-01T12:10:34.137" Title="When is phosphorus a polyatomic molecule?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;pnictogen&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29540" PostTypeId="2" ParentId="29539" CreationDate="2015-04-28T23:01:58.257" Score="2" Body="&lt;p&gt;Phosphorus is most common as white phosphorus (but more stable as red phosphorus) and in this form, it exists as a tetrahedron, or $\ce{P4}$. Chlorine is most stable as a diatomic molecule, as you may have learnt it with the other common gases that exists as diatomic molecules&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-04-28T23:01:58.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29541" PostTypeId="2" ParentId="29500" CreationDate="2015-04-29T00:35:42.703" Score="1" Body="&lt;p&gt;Here is the short answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When ice melts, it's molecules (H2O), have to gain energy, that is true. However, you have to look at the Gibbs Free Energy, which is the determining factor in figuring out if a reaction is spontaneous or not. Because disorder increases, the delta S (or entropy) increases. The formula calculating Gibbs Free Energy is deltaG = deltaH - TdeltaS. If the temp is high enough for a given deltaS, the deltaG will become negative. &lt;/p&gt;&#xA;" OwnerUserId="15888" LastActivityDate="2015-04-29T00:35:42.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29543" PostTypeId="1" CreationDate="2015-04-29T00:47:49.520" Score="4" ViewCount="231" Body="&lt;p&gt;Is there any way to determine the spin of $\ce{[Fe(OH)6]^{4-}}$ without looking at the spectrochemical series?&lt;/p&gt;&#xA;" OwnerUserId="15889" LastEditorUserId="8083" LastEditDate="2015-04-29T01:29:12.743" LastActivityDate="2015-04-29T11:17:13.827" Title="Determining spin of metal complex" Tags="&lt;coordination-compounds&gt;&lt;metal&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29544" PostTypeId="1" CreationDate="2015-04-29T01:22:06.187" Score="5" ViewCount="74" Body="&lt;p&gt;If at 95% Ethanol can dissolve chemical X at 1mg/1mL then theoretically if you have 500mg of chemical X you can dissolve it in 500mL of $\ce{EtOH}$ to get 1mg/mL correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if you have 75.5% $\ce{EtOH}$ for example in a 750mL solution - how many mL will you need to create the 1 mg/mL concentration of quantity of 500mg chemical X? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just 19.5% more of the 500mL volume or do you have to figure out the molarity of each and then go from there? &lt;/p&gt;&#xA;" OwnerUserId="15892" LastEditorDisplayName="user15489" LastEditDate="2015-04-29T03:40:59.873" LastActivityDate="2015-04-29T03:40:59.873" Title="Dissolving to Create a Solution with Ethanol" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29545" PostTypeId="2" ParentId="29544" CreationDate="2015-04-29T02:53:47.047" Score="1" Body="&lt;p&gt;Knowing that the material (solute)  has a solubility of 1 mg/ml in 95% ethanol does not help you predict its solubility in 75% ethanol. It could be more soluble or less soluble, depending on the properties of the solute.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-04-29T02:53:47.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29548" PostTypeId="2" ParentId="29421" CreationDate="2015-04-29T08:26:10.980" Score="-4" Body="&lt;p&gt;This is because calcium is higher than hydrogen in the activity series of metals  due to which, the objective of the reaction will be destroyed. This is why concentrated sulphuric acid is used.&lt;/p&gt;&#xA;" OwnerUserId="15896" LastEditorDisplayName="user15489" LastEditDate="2015-04-29T09:34:08.197" LastActivityDate="2015-04-29T09:34:08.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29549" PostTypeId="1" CreationDate="2015-04-29T10:06:17.100" Score="12" ViewCount="3274" Body="&lt;p&gt;Bulk carbon is an insulator, but at nanoscale (e.g. carbon nanotubes) carbon may become a good electrical conductor. If I am correct this means  carbon nanotube band structure is metal-like. E.g. it has partly filled valence band (like in the rightmost case in the figure below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/urUKa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At nanoscale materials are influenced by quantum confinement. Because of quantum confinement semiconductor nanoparticle has higher band gap and discrete energy levels compared to bulk semiconductor (see the figure about quantum confinement below). &#xA;&lt;img src=&quot;https://i.stack.imgur.com/CD3IF.png&quot; alt=&quot;Quantum confinement phenomenon in nanoparticle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that quantum confinement phenomenon also occurs for carbon at nanoscale. Then, how certain carbon nanotubes can show good electrical conductivity?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="11277" LastEditDate="2015-04-30T07:29:00.913" LastActivityDate="2015-05-10T23:36:38.470" Title="Why carbon nanotubes can act as a good electrical conductor?" Tags="&lt;nanotechnology&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29551" PostTypeId="1" AcceptedAnswerId="29571" CreationDate="2015-04-29T10:29:29.900" Score="2" ViewCount="838" Body="&lt;p&gt;If the standard gibbs free energy for a reaction is negative then the forward reaction will take place and the reaction will not be reversible. Is it true ? I have a problem understanding what is gibbs free energy and its significance in a chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance !&lt;/p&gt;&#xA;" OwnerUserId="9424" LastActivityDate="2015-04-29T16:57:49.163" Title="If for a chemical reaction the standard gibbs free energy is negative then :" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29552" PostTypeId="2" ParentId="29549" CreationDate="2015-04-29T10:44:57.890" Score="13" Body="&lt;p&gt;I am sorry to say, but your question contains several false assumptions, most importantly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Nanotubes are good conductors&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, not all of them. Certain types have metallic/semi-metallic conductance, others are semiconductors. Single wall carbon nanotubes are generally indexed with a so-called chirality index (m,n) that tells you how the carbon sheet is wrapped up into a tube. See &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_nanotube&quot; rel=&quot;nofollow&quot;&gt;wikipage.&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Nanotubes behave exactly like 3D nanoparticles in terms of band structure&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, nanosheets (graphene) and nanotubes behave in different ways and your description of quantum confinements don't work in that way in 1D infinite (nanotubes) and 2D infinite (graphene) cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive an approximate band structure of graphene/nanotubes using tight-binding models, which in most cases predicts correctly, whether the structure is metallic or semiconducting. See &lt;a href=&quot;http://demonstrations.wolfram.com/ElectronicBandStructureOfASingleWalledCarbonNanotubeByTheZon/&quot; rel=&quot;nofollow&quot;&gt;simulation of nanotube band structure.&lt;/a&gt;  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="1418" LastEditDate="2015-05-10T23:36:38.470" LastActivityDate="2015-05-10T23:36:38.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29554" PostTypeId="2" ParentId="29543" CreationDate="2015-04-29T11:17:13.827" Score="1" Body="&lt;p&gt;Are you asking experimentally or theoretically?&#xA;Theoretically speaking for general cases you should know what is the energy of the high-spin and low-spin configuration, and in given state, what is the approximate spin and vibrational entropy so you can estimate an equilibrium constant for given temperature. Now this either needs some calculation, or some sort of database from where you can make this estimations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experience (i.e. database or whatever series) says that water is a weak ligand, from that you generally guess that water complexes are generally in high spin state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally you can determine spin state by measuring magnetic susceptibility, interpreting EPR, measuring 57Fe Mossbauer spectrum etc &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-04-29T11:17:13.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29555" PostTypeId="2" ParentId="29536" CreationDate="2015-04-29T12:45:47.930" Score="0" Body="&lt;p&gt;From this source &lt;a href=&quot;http://web.nmsu.edu/~esevosti/report.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;broadenings occur due to&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Doppler effect&#xA;  This effect arises because atoms will have different components of velocity along the line of observation.&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/WD6Xn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Lorentz effect&#xA;  This effect occurs as a result of the concentration of foreign atoms present in the environment of the emitting or absorbing atoms. The magnitude of the broadening varies with the pressure of the foreign gases and their physical properties.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Quenching effect&#xA;  In a low-pressure spectral source, quenching collision can occur in flames as the result of the presence of foreign gas molecules with vibration levels very close to the excited state of the resonance line.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;Self absorption or self-reversal effect&#xA;  The atoms of the same kind as that emitting radiation will absorb maximum radiation at the center of the line than at the wings, resulting in the change of shape of the line as well as its intensity. This effect becomes serious if the vapor, which is absorbing radiation is considerably cooler than that which is emitting radiation.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Greg pointed out in the comment, You can find more detailed explanation on the topic &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectral_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure broadening should be dependent on the concentration of atoms. The other type of broadening shouldn't be dependent on the concentration &lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="11310" LastEditDate="2015-04-30T04:36:00.913" LastActivityDate="2015-04-30T04:36:00.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29556" PostTypeId="2" ParentId="28828" CreationDate="2015-04-29T13:21:52.497" Score="1" Body="&lt;p&gt;The answer has a lot to do with &quot;what is a catalyst in the first place?&quot;&#xA;A catalyst is a compound that can affect the rate of a chemical reaction&#xA;by providing an alternative and lower energy profile or pathway. That is, &#xA;it is a substance that makes it so that A -&gt; B at a lower energy cost. But&#xA;it only has an effect on the rate of the reaction. That is, it only changes&#xA;the cost of the activation energy. It is not related to the thermodynamics&#xA;of the process and hence, the final product distribution. That is, if the &#xA;above process gives 90% B via a non-catalyzed pathway in 4 days, then a &#xA;catalyst will help you get 90% B in say 4 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so a catalyst affects the transition state and activation path. How&#xA;does it do this? Typically, by complexing one of the reagents. Complexation&#xA;by transition metals affords access to a wide variety of oxidation states&#xA;for the metal. This has the property of providing electrons or withdrawing&#xA;electrons from the transition state of the reaction. That is, if the &#xA;transition state is electron rich, then the transition metal might hold &#xA;some of that electron density and those prevent too much from building up&#xA;on the reagent. This would then facilitate the reaction. Or the transition&#xA;metal might undergo formal oxidation/reduction to achieve electron transfer&#xA;to a substrate, thereby allowing a reaction to occur. This is &quot;complexation&#xA;and electron storage&quot; taken to the extreme but is a common mechanism in&#xA;organometallic chemistry. Indeed, a variety of catalytic pathways rely on&#xA;a two electron transfer between the metal and the substrate (e.g. &#xA;hydroformylation). It is the ability of the transition metal to be in a&#xA;variety of oxidation states, to undergo facile transitions between these &#xA;oxidation states, to coordinate to a substrate, and to be a good &#xA;source/sink for electrons that makes transition metals such good catalysts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is for transition metal complexes. Most industrial used &#xA;catalysts are the transition metal in a bed, as a metal or bound structure.&#xA;The above considerations are important but also the physical properties of&#xA;absorption/adsorption and the electron band structure of the material.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-29T13:21:52.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29558" PostTypeId="1" AcceptedAnswerId="29618" CreationDate="2015-04-29T13:40:24.017" Score="2" ViewCount="970" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The percent dissociation of a base $\ce{X}$ in a $0.002\ \mathrm{M}$ solution is $4.7 \cdot 10^{-3}\ \%$. What is the pH of the solution and what is the $\mathrm{p}K_\text{b}$ of base $\ce{X}$? The autoionization of water may not be neglected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; We have the following reactions: $$ \ce{X + H2O(l) &amp;lt;=&amp;gt; XH+ + OH-} \\ \ce{H2O(l) &amp;lt;=&amp;gt; H+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the percent dissociation is $4.7 \times 10^{-3}\ \%$, this enables us to find the concentration dissociated: $$ \frac{x}{0.002\ \mathrm{M}} \cdot 100 = 4.7 \times 10^{-3},$$ so $x = 9.4 \times 10^{-8}$. Now I did: $$ K_\text{a} = \frac{x^2}{0.002 - x}$$ and I let $(0.002 - x) \approx 0.002$ since $x$ is so small. But this gives me the wrong answer, and it seems I haven't taken the autoionization of water into account yet. How should I solve this problem then?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorUserId="194" LastEditDate="2017-04-01T02:26:16.307" LastActivityDate="2017-04-03T11:09:00.803" Title="Calculating pH of an unknown base, and the pKb" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29559" PostTypeId="1" AcceptedAnswerId="29563" CreationDate="2015-04-29T14:37:48.893" Score="10" ViewCount="2070" Body="&lt;p&gt;How will the cyclobutane ring behave in the case of cyclobutylmethylium (cyclobutylmethyl cation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHCDO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHCDO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially thought there would be ring expansion to a five membered ring so that there may be less angle strain and a secondary carbocation instead of a primary one. But I have also been told that there will be ring contraction for stability purposes. I would be highly obliged if someone could explain the mechanism involved here.&lt;/p&gt;&#xA;" OwnerUserId="15868" LastEditorUserId="61578" LastEditDate="2020-10-20T14:29:24.407" LastActivityDate="2020-10-20T14:29:24.407" Title="Ring expansion from a given cyclic carbocation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stability&gt;&lt;carbocation&gt;&lt;rearrangements&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29561" PostTypeId="1" AcceptedAnswerId="29574" CreationDate="2015-04-29T14:53:10.850" Score="4" ViewCount="6951" Body="&lt;p&gt;I know that most of the double displacement reactions form some kind of precipitates as their products. But do all of the double displacement reactions form precipitates? I have searched on this site as well as on Google, I've found answers, but they are not clarifying my doubts. Maybe I need a simpler explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you answer is yes, they all lead to precipitation, please justify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's no, not all of them give precipitates, please justify with an example. &lt;/p&gt;&#xA;" OwnerUserId="15711" LastEditorUserId="83758" LastEditDate="2020-03-04T09:43:10.557" LastActivityDate="2020-03-04T09:43:10.557" Title="Do all double displacement reactions lead to precipitates?" Tags="&lt;inorganic-chemistry&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29562" PostTypeId="2" ParentId="28853" CreationDate="2015-04-29T14:54:35.610" Score="3" Body="&lt;p&gt;Though the other answer provides a good insight, I think annealing and tempering could have been confused.&lt;/p&gt;&#xA;&lt;p&gt;I will talk with respect to iron-carbon alloys or steel here.&#xA;Tempering is usually a post-quenching or post hardening treatment.&#xA;It is done to relieve internal stresses, decrease brittleness, improve ductility and toughness. The usual heating range for tempering in steel is from &lt;span class=&quot;math-container&quot;&gt;$150\ \mathrm{^\circ C}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$600\ \mathrm{^\circ C}$&lt;/span&gt; and it is below the upper critical temperature or the eutectoid line. The cooling is done in still air.&lt;/p&gt;&#xA;&lt;p&gt;Annealing can induce ductility, soften material, relieve internal stresses, refine the structure by making it homogeneous (though grain coarsening can also happen), and improve cold working properties. Cooling in case of annealing is usually furnace cooling which gives the slowest of cooling rates.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XEQFG.gif&quot; alt=&quot;Annealing&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The products for annealing will depend on the carbon % and the temperature ranges from which it is cooled. But usually it is pearlite + ferrite/cementite.&lt;/p&gt;&#xA;&lt;p&gt;The product of tempering is usually just tempered martensite or bainite.&lt;/p&gt;&#xA;&lt;p&gt;Hope this was helpful.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-29T15:11:59.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29563" PostTypeId="2" ParentId="29559" CreationDate="2015-04-29T15:08:44.243" Score="10" Body="&lt;p&gt;I think your friend is thinking of the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/28939/carbocation-rearrangement-due-to-ring-strain-and-back-bonding&quot;&gt;cyclobutyl carbocation&lt;/a&gt; which does ring contract to the cycylopropyl carbinyl carbocation (and also equilibrates with the  methallyl carbocation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, just as you thought, the cyclobutyl carbinyl carbocation does ring open to the cyclopentyl carbocation (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja9828962?journalCode=jacsat&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_1&lt;/a&gt;, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/17335243&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_2&lt;/a&gt;, &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo0701334&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref_3&lt;/a&gt;).  This rearrangement is driven by carbocation stability (primary to secondary) and relief of the steric strain present in the 4-membered ring, again, just as you suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nrr7S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this question has been asked &lt;a href=&quot;https://chemistry.stackexchange.com/questions/9581/stability-of-cyclobutyl-methyl-carbocation&quot;&gt;previously&lt;/a&gt; on SE Chem, however I believe the accepted answer is incorrect, as it seems to primarily address the cyclopropyl carbinyl case, and what is said about the cyclobutyl carbinyl carbocation (little to no ring expansion) is incorrect.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-04-29T18:00:01.787" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="29564" PostTypeId="2" ParentId="27900" CreationDate="2015-04-29T15:16:35.797" Score="4" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_metal&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gallium is corrosive to all metals except tungsten and tantalum, which have a high resistance to corrosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beta-Brass is &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_metal_embrittlement&quot; rel=&quot;nofollow&quot;&gt;embrittled&lt;/a&gt; by &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0956716X9090573Y&quot; rel=&quot;nofollow&quot;&gt;Gallium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to its highly reducing character it is highly corrosive material.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-29T15:16:35.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29565" PostTypeId="1" CreationDate="2015-04-29T15:20:01.743" Score="3" ViewCount="382" Body="&lt;p&gt;I am working on a yeast fermentation project, and I am getting final OD660 reading of 1.85 for different media: yeast extracts+cane molasses, nitrogen source+cane molasses, and glucose+cane molasses at different concentrations. The final OD reading is the same for all media. Is it normal? Are my readings correct?&lt;/p&gt;&#xA;" OwnerUserId="15900" LastEditorUserId="418" LastEditDate="2015-04-29T16:40:47.990" LastActivityDate="2015-04-29T16:46:02.950" Title="OD660 readings of yeast" Tags="&lt;spectrophotometry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29567" PostTypeId="1" CreationDate="2015-04-29T15:56:00.677" Score="15" ViewCount="5975" Body="&lt;p&gt;Various sources (such as &lt;a href=&quot;http://www.scientificamerican.com/article/why-doesnt-stainless-stee/&quot;&gt;this one&lt;/a&gt;) say that stainless steel is stainless because a passivation layer of an oxide is formed, but I haven't found details on why chromium helps create this passivation layer. I can think of a couple possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Chromium changes the lattice constant of the iron to better match the spacing of the oxide layer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chromium migrates to the surface, and with enough Cr a Cr-rich layer forms on the surface which in turn creates the passivation layer. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are either of these possibilities true, and if not, how does Cr help make the passivation layer?&lt;/p&gt;&#xA;" OwnerUserId="9557" LastActivityDate="2017-07-18T08:47:21.030" Title="How does chromium help stainless steel prevent rusting?" Tags="&lt;solid-state-chemistry&gt;&lt;surface-chemistry&gt;&lt;metallurgy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29568" PostTypeId="1" CreationDate="2015-04-29T16:35:05.143" Score="6" ViewCount="140" Body="&lt;p&gt;In close packed structures there is one octahedral hole per close packed atom. In a HCP unit cell these octahedra form a chain of face sharing polyhedra. Is this this universal in close packing or only common to HCP?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally this question is &quot;how do we know when interstitial sites (T± and Oh) share edges or a faces?&quot;&lt;/p&gt;&#xA;" OwnerUserId="9690" LastActivityDate="2015-04-29T16:35:05.143" Title="Interstitial sites: edge and face sharing?" Tags="&lt;solid-state-chemistry&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29569" PostTypeId="2" ParentId="29567" CreationDate="2015-04-29T16:36:01.203" Score="10" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; &#xA;    Note that the passive layer forms &lt;em&gt;on&lt;/em&gt; the surface, there needn't be any change to lattice constant. Chromium needn't migrate , the Cr present on the surface will form the layer to protect it. The key point is how the layer develops from say a single-atom layer of oxide to the usual/maximum width by migration of electron and oxygen in the oxide layer. The process at molecular and atomic level is more complex than you think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From wikipeddia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Passivation_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;Passivation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There has been much interest in determining the mechanisms which describe how the thickness of the oxide layer on a material increases with time. Some of the important issues include: the relative volume of the oxide compared to the parent metal, the mechanism by which oxygen diffuses through the metal oxide to the metal - oxide interface and the relative chemical potential for the oxide to form. Boundaries between micro grains, if the oxide layer is crystalline, form an important pathway for oxygen to reach the unoxidized metal below. For this reason, vitreous oxide coatings – which lack grain boundaries – can retard oxidation. The conditions necessary (but not sufficient) for passivation are recorded in &lt;a href=&quot;http://en.wikipedia.org/wiki/Pourbaix_diagram&quot; rel=&quot;noreferrer&quot;&gt;Pourbaix diagrams&lt;/a&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some &lt;a href=&quot;http://en.wikipedia.org/wiki/Corrosion_inhibitor&quot; rel=&quot;noreferrer&quot;&gt;corrosion inhibitors&lt;/a&gt; help the formation of a passivation layer on the surface of the metals to which they are applied. Some compounds, dissolving in solutions (&lt;a href=&quot;http://en.wikipedia.org/wiki/Chromate_and_dichromate&quot; rel=&quot;noreferrer&quot;&gt;chromates&lt;/a&gt; , &lt;a href=&quot;http://en.wikipedia.org/wiki/Molybdic_acid&quot; rel=&quot;noreferrer&quot;&gt;molybdates&lt;/a&gt;) form non-reactive and low solubility films on metal surfaces.&#xA;   &lt;img src=&quot;https://i.stack.imgur.com/MBrwD.png&quot; alt=&quot;Pourbaix diagram&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jes.ecsdl.org/content/105/11/638&quot; rel=&quot;noreferrer&quot;&gt;The mechanism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All passivating‐type inhibitors are oxidizing agents. However, not all oxidizing agents are inhibitors. A passivating ‐type inhibitor functions by producing local‐action current which anodically polarizes a metal into the passive potential region and thereby provides the means for obtaining a noble mixed potential. This mechanism is independent of whether passivity is caused primarily by oxide or adsorbed films. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The major factors which determine whether a particular system will exhibit passivity are: &#xA;  (a) reversible potential of the redox system created by the oxidizing inhibitor;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) electrochemical parameters for reduction of the inhibitor on the metal surface (exchange current, &lt;a href=&quot;http://en.wikipedia.org/wiki/Tafel_equation&quot; rel=&quot;noreferrer&quot;&gt;Tafel slope&lt;/a&gt;, and limiting diffusion current), and &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) critical anodic current and Flade potential(The potential of a passive metal immediately preceding a final step fall from the passive to the active region.) of the metal. If oxygen is present in the system, some chemicals may function by changing its reduction kinetics. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is further shown that the amount of inhibitor found associated with the surface is not necessarily related to adsorption. Data are presented for passivation of stainless steel and titanium. The mechanism is discussed in terms of various oxidizing agents including oxygen, chromate, molybdate, and pertechnetate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some theories I could find for passivation are &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100888a028&quot; rel=&quot;noreferrer&quot;&gt;Kinetic Theory of Inhibition and Passivation in Electrochemical Reactions&lt;/a&gt; and &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/16942098&quot; rel=&quot;noreferrer&quot;&gt;Scaling theory in a model of corrosion and passivation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-29T16:36:01.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29570" PostTypeId="2" ParentId="29565" CreationDate="2015-04-29T16:46:02.950" Score="2" Body="&lt;p&gt;Many inexpensive spectrophotometers cannot measure absorbances over 1.5 or so reliably. It's possible that you're out of the measurement range of your instrument. Try diluting your samples with water first. If you dilute 1:1, then the absorbance value of your sample is double what you measure (assuming the Beer-Lambert law is valid in this situation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I imagine molasses is very absorbent at that wavelength so it might be obscuring any change from the yeast and it may not be possible to measure the yeast growth by absorbance, with such a matrix.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-04-29T16:46:02.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29571" PostTypeId="2" ParentId="29551" CreationDate="2015-04-29T16:57:49.163" Score="1" Body="&lt;p&gt;All reactions are in some sense equilibrium reactions. The Gibbs free energy is just a convenient measure of energy that we can use to understand more about our system. In particular you may be familiar with a few equations for the Gibbs energy:&#xA;\begin{equation}&#xA;\Delta G =\Delta H-T\Delta S,&#xA;\end{equation}&#xA;\begin{equation}&#xA;\Delta G=-RT\ln k&#xA;\end{equation}&#xA;\begin{equation}&#xA;\Delta G =-nFE&#xA;\end{equation}&#xA;For chemists, these three equations are the most commonly used and depend on what exactly you are investigating, but there are plenty of others too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the second of these equations we note that $k$, the equilibrium constant, appears. We can see that as the equilibrium constant increases (and since $x$ follows $\ln x$ in limits) we note that the change in Gibbs energy becomes more negative the larger the equilibrium constant becomes. In other words the more products you have relative to reactants, the more negative the change in the Gibbs energy is, we call this spontaneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could also take the third equation and note that $E$ is the reduction potential for a redox couple ($X^{n+}+e^{-}\rightarrow X^{(n-1)+}$). The more positive the potential for the reduction, the more spontaneous the reaction is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps you, oh and I'm assuming you mean reversible in the reaction sense of equilibria not in the thermodynamic sense of a reversible process?   &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2015-04-29T16:57:49.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29572" PostTypeId="2" ParentId="29567" CreationDate="2015-04-29T17:13:14.877" Score="3" Body="&lt;p&gt;I found &lt;a href=&quot;http://journals.aps.org/prb/abstract/10.1103/PhysRevB.76.220401&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which suggests that the Cr concentration at the surfaces is lower than in the bulk until ~8% Cr at. conc., at which point the concentration increases abruptly to ~20% Cr (and higher experimentally, apparently). So there is evidence that the Cr concentration at the surface may be different from the bulk and that this coincides with the corrosion resistance in Fe-Cr alloys, which means my second guess was closer to being correct. &lt;/p&gt;&#xA;" OwnerUserId="9557" LastActivityDate="2015-04-29T17:13:14.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29573" PostTypeId="1" AcceptedAnswerId="29575" CreationDate="2015-04-29T17:15:38.263" Score="4" ViewCount="142" Body="&lt;p&gt;I'm not a chemical engineer and don't have good background in it.&#xA;Just my father asked me to do something for him, as I'm a computer engineering graduate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a software for calculation of technical parameters of reactor. I haven't seen it. I've just printed output of it. Fields are like following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;5  REM**************&#xA;10 REM***********&#xA;20 REM  Name etc.&#xA;~~~~~~~~~~~~&#xA;35 REM ********** PRODUCTIVITY********&#xA;~~~~~~~~~~~~&#xA;155 REM ***** INPUTS*******&#xA;~~~~~~~~~~~~~~~~~~~~~~~~&#xA;220 DTB&#xA;240  DTS&#xA;~~~~~~~~~~~~~~~~~~~~&#xA;555 REM *********** CALCULATIONS****************&#xA;~~~~~~~~~~~~~~~~~~~~~~~~~&#xA;~&#xA;~&#xA;~&#xA;********RESULTS*************&#xA;DTO=75&#xA;G=543.2&#xA;Q=XXXX&#xA;XXXXXX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me to find what is the software for calculation. I'd be very grateful if you can help.&lt;/p&gt;&#xA;" OwnerUserId="1244" LastEditorDisplayName="user15489" LastEditDate="2015-04-29T18:28:29.527" LastActivityDate="2015-07-22T17:31:05.257" Title="Software for technical calculation of chemical reactors" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29574" PostTypeId="2" ParentId="29561" CreationDate="2015-04-29T17:23:18.560" Score="6" Body="&lt;p&gt;No, recall the definition of a precipitate &quot;insoluble ionic compound&quot;. Take this reaction into consideration $$\ce{ NaOH~(s) + HCl~(l) &amp;lt;=&amp;gt; H2O~(l) + NaCl~(aq)}$$ none of the products are insoluble. You'll learn more when you go over gas evolution reactions.  &lt;/p&gt;&#xA;" OwnerUserId="13368" LastEditorUserId="4945" LastEditDate="2015-04-30T09:23:31.800" LastActivityDate="2015-04-30T09:23:31.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29575" PostTypeId="2" ParentId="29573" CreationDate="2015-04-29T17:23:42.683" Score="3" Body="&lt;p&gt;Years ago a colleague of mine mentioned ASPEN as a typical software (bundle) used to model chemical plants.  Despite &lt;a href=&quot;http://www.aspentech.com/core/aspen-plus.aspx&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://www.chems.msu.edu/resources/tutorials/ASPEN&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; lack of examples how the print-out will look like -- and I did not work with it, either. &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-29T17:23:42.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29576" PostTypeId="2" ParentId="27034" CreationDate="2015-04-29T17:27:00.700" Score="1" Body="&lt;p&gt;From &lt;a href=&quot;http://www.eurare.eu/docs/eres2014/thirdSession/VenkatesanPrakash.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper,  it seems Electrochemical recovery of Neodymium is possible from the Magnet.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-29T17:27:00.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29577" PostTypeId="2" ParentId="17320" CreationDate="2015-04-29T17:35:39.103" Score="1" Body="&lt;p&gt;You have a misconception that tempering is hardening. It is softening done after hardening to relieve stresses, improve ductility and toughness .&lt;/p&gt;&#xA;&lt;p&gt;From this &lt;a href=&quot;http://www.asminternational.org/web/hts/materials-resources/-/journal_content/56/10192/ASMHBA0001204/PUBLICATION&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In nonferrous alloys, eutectoid transformations, which play such a prominent role in steels, are seldom encountered.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, the principles associated with chemical homogenization of cast structures are applicable to many alloys in both classes. Examination of the heat treatment used for nonferrous alloys reveals that a wide variety of processes are employed. . Annealing after cold working is a very important heat treatment for nonferrous alloys.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Basically if the non-ferrous alloy has a eutectoid region in its phase diagram , tempering can be used after hardening. But non-ferrous usually are not hardened in this way. Age hardening, Precipitation hardening, Dispersion strengthening are commonly seen in non-ferrous alloys.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-04-29T17:35:39.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29579" PostTypeId="1" CreationDate="2015-04-29T18:37:46.593" Score="7" ViewCount="1957" Body="&lt;p&gt;Is it possible to add copper nitrate solution and lead so that a single displacement reaction occurs to yield lead nitrate. I recently melted lead and added cooper nitrate to the molten lead. I then collected the copper nitrate solution I had poured and repeated the process of melting the lead and pouring the solution over it. I figured that eventually the copper will be displaced by the slightly more reactive lead. Will this method yield lead nitrate?&lt;/p&gt;&#xA;" OwnerUserId="15851" LastActivityDate="2015-04-30T06:29:07.563" Title="How to make lead nitrate from pure lead" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29580" PostTypeId="1" AcceptedAnswerId="29606" CreationDate="2015-04-29T19:49:18.863" Score="2" ViewCount="296" Body="&lt;p&gt;We are making an electrochemical research on supercapacitors that involves checking stability of our material. That requires repeated writing of cyclic voltammetry curves. Our goal is to write at least 2000 of them, but there is a problem with Nova software (version 1.8, to be specific): if we want to write even 50 cycles, Nova complains that &quot;only last 30000 point will be written&quot;. We can adjust how many points are plotted, but when it comes to 2000 cycles, it's impossible to extract any information from them (15 points per one curve is just ludicrous).&#xA;Has anyone dealt with such problem? Is there a simple solution?&lt;/p&gt;&#xA;" OwnerUserId="15902" LastEditorUserId="4356" LastEditDate="2015-07-10T18:59:16.700" LastActivityDate="2015-07-10T18:59:16.700" Title="How do I write many cyclic voltammetry curves in Nova?" Tags="&lt;electrochemistry&gt;&lt;software&gt;&lt;cyclic-voltammetry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29581" PostTypeId="1" AcceptedAnswerId="29589" CreationDate="2015-04-29T20:12:57.240" Score="5" ViewCount="6278" Body="&lt;p&gt;Suppose I have the reversible reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A +B⇌ C} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is at equilibrium with equilibrium constant $K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told that if I increase the concentration of $\ce{B}$, the rate for the forwards reaction will exceed the backwards one. Fair enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also told that $K$ will necessarily increase. Why though? Its true that Forward reaction &gt; back ward reaction until we reach a new equilibrium such that more of $\ce{C}$ is produced but I don't see why this implies in any way that the final quotient $\frac{[\ce{C}]}{\ce{[A][B]}}$ will necessarily be any greater. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly if we have a simple reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A⇌ C} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we add more of $\ce{A}$, then the equilibrium constant for the new final final state will remain as it was, ceteris paribus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I getting wrong, because my textbook suggests that $K$ will always increase, no matter what type of reaction I am dealing with (of course, as long as all reactants are in a suspended form, e.g. dissolved or gaseous and the reaction is subject to Le Châtelier's principle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, if possible, keep the answer as a level intelligible to a high-school student. I don't know much about advanced university chemistry.&lt;/p&gt;&#xA;" OwnerUserId="5114" LastEditorUserId="14" LastEditDate="2015-06-14T13:24:26.467" LastActivityDate="2015-06-14T13:24:26.467" Title="Equilibrium constant when adding more of a reactant" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29583" PostTypeId="1" AcceptedAnswerId="29584" CreationDate="2015-04-29T20:52:49.450" Score="10" ViewCount="19312" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s0KDRs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ketone has one oxygen atom in it and there are two lone pairs hanging out, both of which make it seemingly capable of hydrogen bonding. But in reality it is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did some research and found this on chemguide:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently δ+ for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the info, I am assuming that the carbon atom in carbonyl group is not electronegative enough to give a partial charge large enough for hydrogen bonding to the hydrogen atoms, say in pentanone.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="4231" LastEditDate="2015-04-29T23:13:48.903" LastActivityDate="2015-04-29T23:57:44.803" Title="Why is a ketone not capable of hydrogen bonding among themselves?" Tags="&lt;intermolecular-forces&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29584" PostTypeId="2" ParentId="29583" CreationDate="2015-04-29T21:32:12.967" Score="8" Body="&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: the ketone oxygen can participate in hydrogen bonding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound you mention, methoxymethane, is an &lt;strong&gt;ether&lt;/strong&gt; not a ketone.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mvzj0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethers don't hydrogen bond very well because the oxygen isn't very polarized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;strong&gt;the carbonyl in a ketone is polarized&lt;/strong&gt;.  We can draw resonance structures that show this polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cxrTF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_11%3A_Nucleophilic_carbonyl_addition_reactions/Section_11.1%3A_Nucleophilic_additions_to_aldehydes_and_ketones%3A_the_general_picture&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The negative polarization on the carbonyl oxygen allows it to play the role of &lt;strong&gt;hydrogen bond acceptor&lt;/strong&gt; as illustrated in the following drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VIPFa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.3%3A_Non-covalent_interactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; If you are asking if the hydrogens on the carbon adjacent to the carbonyl carbon (or any other carbon in the backbone of a simple ketone) can participate as the donor in a hydrogen bond, then the answer to this question is &quot;no&quot;. These $\ce{C-H}$ bonds are typically less  acidic and less polar than the $\ce{O-H}$ bond in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, while the oxygen in pentanone can participate in hydrogen bonding with donor hydrogens in other molecules, the hydrogens in pentanone cannot participate in hydrogen bonding.  Pentanone cannot hydrogen bond with itself.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-04-29T23:57:44.803" LastActivityDate="2015-04-29T23:57:44.803" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="29585" PostTypeId="2" ParentId="29522" CreationDate="2015-04-29T22:41:59.230" Score="5" Body="&lt;p&gt;Well it looks for me, the question was addressed by a non-Chemist.  Hence my attempt to answer this accordingly.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You have to know the reaction of the combustion, the starting materials, the product(s), and their relative quantities to balance the reaction equation. &lt;br&gt; Your case is described by &lt;span class=&quot;math-container&quot;&gt;$\ce{C} + \ce{O2} \rightarrow \ce{CO2}$&lt;/span&gt;, with &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;&amp;quot; for carbon (here: in form of diamond), &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; for oxygen and &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; to yield the product, carbon dioxide. And the reaction equation is already balanced, i.e. the number of atoms per type on the left hand side (prior the arrow) equals the number of atoms per type written on the right, too.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is a heat called &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation&quot; rel=&quot;noreferrer&quot;&gt;standard enthalpy of formation&lt;/a&gt;, describing how much heat is necessary or liberated to form a compound if one (even theoretically) were to prepare this out of elements.  Normally, you look up these tabulated data for all your products and sum this up, and subtract the sum of all these enthalpies of formation of your starting materials. &lt;br&gt; In your example, by convention elementary gases (like oxygen, &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;) were attributed a zero value.  By means of thermodynamic, diamond is &lt;em&gt;not&lt;/em&gt; the most stable form of carbon, and was attributed a value of &lt;span class=&quot;math-container&quot;&gt;${+1.88}{\,\mbox{kJ/mol}}$&lt;/span&gt; (same reference).  Carbon dioxide was attributed a value of &lt;span class=&quot;math-container&quot;&gt;${-393.5}{\,\mbox{kJ/mol}}$&lt;/span&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_dioxide&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;) and the energy balance of the reaction equation equals to &lt;span class=&quot;math-container&quot;&gt;${-391.62}{\,\mbox{kJ/mol}}$&lt;/span&gt;, i.e. if you would burn 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Mole_%28unit%29&quot; rel=&quot;noreferrer&quot;&gt;mole&lt;/a&gt; of diamond.  (This is a quantity like &lt;a href=&quot;https://en.wikipedia.org/wiki/Dozen&quot; rel=&quot;noreferrer&quot;&gt;dozen&lt;/a&gt;, used in chemistry.)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now if we express the quantity of diamond &lt;em&gt;actually&lt;/em&gt; burned (mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; of 1 &lt;a href=&quot;https://en.wikipedia.org/wiki/Carat_%28mass%29&quot; rel=&quot;noreferrer&quot;&gt;ct&lt;/a&gt;) in this above mentioned unity of mole, we need a proportional factor, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_mass&quot; rel=&quot;noreferrer&quot;&gt;molar mass&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, that happens to be &lt;span class=&quot;math-container&quot;&gt;$12{\,\mbox{g/mol}}$&lt;/span&gt; for carbon (&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Now we have the pieces.  Hence we calculate the &amp;quot;chemical quantity&amp;quot; (&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;) of carbon deployed as&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n = \frac{m}{M} = \frac{0.2\,\mbox{g}}{12\,\mbox{g/mol}} = 0.017\,\mbox{mol}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and than multiply this with the above calculated heat of the reaction, i.e&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mbox{heat} = 0.017\,\mbox{mol} \times (-391.62\,\mbox{kJ/mol}) = -6.527\,\mbox{kJ}.$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So if you burn 1 ct of diamond in oxygen at 25 degree Celsius (about 77 Fahrenheit) and would collect the heat of the product by cooling the &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; to the same 25 degrees Celsius, this offers you 6527 Joules (or about 1559 calories).&lt;/p&gt;&#xA;&lt;p&gt;Notes: The heat calculated here is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_enthalpy_of_reaction&quot; rel=&quot;noreferrer&quot;&gt;reaction enthalpy&lt;/a&gt;. The minus sign indicates energy is &amp;quot;liberated&amp;quot; in the course of this reaction. Actually, if all this heat &lt;em&gt;were completely&lt;/em&gt; converted into electrical energy, you were able to switch on a light bulb of 60 W for less than 2 minutes (only 108 seconds). &lt;br&gt; In the beginning of modern chemistry, diamonds actually were burned by concentrated solar power, for example by Lavoisier (&lt;a href=&quot;http://www.peashooter85.com/post/64922020693/antoine-lavoisiers-solar-powered-combustion&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;) although to prove their composition.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-04-29T22:41:59.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29586" PostTypeId="1" CreationDate="2015-04-30T00:21:44.353" Score="3" ViewCount="66" Body="&lt;p&gt;In all the reactions I saw online, dichlorotriazine is shown to form only one bond with cellulose (&lt;a href=&quot;https://en.wikipedia.org/wiki/Reactive_dye#Usage&quot; rel=&quot;nofollow&quot;&gt;example 1&lt;/a&gt;, &lt;a href=&quot;http://www.pburch.net/dyeing/dyeblog/C1845207367/E346098115/&quot; rel=&quot;nofollow&quot;&gt;example 2&lt;/a&gt;; &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0021979796946524&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But cellulose has &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Cellulose_strand.svg&quot; rel=&quot;nofollow&quot;&gt;many potential sites for bonding&lt;/a&gt;. Can dichlorotriazine actually give up both Cl atoms for forming two bonds with cellulose?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="16683" LastEditDate="2017-07-06T14:31:46.483" LastActivityDate="2017-07-06T14:31:46.483" Title="Can one dichlorotriazine molecule form two bonds with cellulose?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29587" PostTypeId="1" AcceptedAnswerId="32748" CreationDate="2015-04-30T00:33:51.007" Score="12" ViewCount="755" Body="&lt;p&gt;We know from basic chemistry that a reaction reaches equilibrium as described by &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot;&gt;Le Châtelier's principle&lt;/a&gt;. What's the quantum mechanical explanation to this principle? How a change in temperature/concentration causes an equilibrium to shift, from the point of view of wave functions?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2017-07-10T17:45:31.640" Title="Quantum mechanical explanation for Le Chatelier's principle?" Tags="&lt;equilibrium&gt;&lt;quantum-chemistry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29588" PostTypeId="1" CreationDate="2015-04-30T01:01:37.317" Score="14" ViewCount="540" Body="&lt;p&gt;I heard that a good way to test if a battery is dead or alive is to see if it bounces. Supposedly, dead batteries bounce higher than living ones. Can someone tell me if this is a legitimate claim, and if it is, an explanation would be nice. &lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="4945" LastEditDate="2015-04-30T08:43:56.183" LastActivityDate="2015-05-21T22:22:34.193" Title="Does the Battery Bounce Test actually work?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29589" PostTypeId="2" ParentId="29581" CreationDate="2015-04-30T01:30:29.953" Score="2" Body="&lt;p&gt;You're correct.  The equilibrium constant will not change.  The equilibrium constant is only defined &lt;strong&gt;at equilibrium&lt;/strong&gt;.  Your analysis of the situation was flawless.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorDisplayName="user15489" LastEditDate="2015-06-13T08:59:48.240" LastActivityDate="2015-06-13T08:59:48.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29590" PostTypeId="1" CreationDate="2015-04-30T02:11:46.537" Score="1" ViewCount="138" Body="&lt;p&gt;How can I figure out which two solutions will emit the largest heat, when they are mixed in same volumes?&#xA;Then  let's assume that there are five pairs of solutions given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \ce{2NaOH   + H2SO4   &amp;amp;-&amp;gt; Na2SO4    + 2H2O} \tag1\\&#xA;  \ce{Ba(OH)2 + H2SO4   &amp;amp;-&amp;gt; BaSO4     + 2H2O} \tag2\\&#xA;  \ce{2NH4OH  + H2SO4   &amp;amp;-&amp;gt; (NH4)2SO4 + 2H2O} \tag3\\&#xA;  \ce{NaOH    + H3CCOOH &amp;amp;-&amp;gt; H3CCOONa  + H2O}  \tag4\\&#xA; \ce{2NaOH    + H2C2O4  &amp;amp;-&amp;gt; Na2C2O4   + 2H2O} \tag5&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one emits the most heat, when we mix same volumes of given pairs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the balanced reaction equation, that occur when the above solutions are mixed. All of these reactions are acid base reactions. And I know when there is an acid base reaction a salt is formed. So how can I proceed with it?&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="4945" LastEditDate="2019-02-06T15:40:30.570" LastActivityDate="2019-02-06T15:46:51.807" Title="How to figure out which two solutions will emit largest heat when they are mixed in same volumes?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29591" PostTypeId="2" ParentId="29587" CreationDate="2015-04-30T02:53:59.513" Score="8" Body="&lt;p&gt;Typically speaking, QM involves single systems at zero Kelvin. So as @NicolauSakerNeto indicates, temperature and concentration would usually be reflected in statistical mechanics and thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, for large systems (e.g., nanoparticles, metals, materials), finite temperatures result in thermal broadening. That is, rather than occupying the absolute lowest energy state, there is some fractional occupation of higher states given by the amount of thermal energy available and the density of electronic (or rotational or vibrational) states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual orbitals themselves are purely mathematical and do not change with temperature - only the occupation of them.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-04-30T02:53:59.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29593" PostTypeId="1" CreationDate="2015-04-30T03:56:57.653" Score="3" ViewCount="288" Body="&lt;p&gt;An electrolyte can be electrolysed by a voltage higher than the reverse reaction’s cell potential, while conducting a current in the process. But what happens at the electrodes when the applied voltage is below this cell potential, such as below 1.23 V for water or 4.07 V for NaCl? Does the reverse reaction occur simultaneously, or is the electrolyte unable to conduct a current with such a low voltage at all?&lt;/p&gt;&#xA;" OwnerUserId="15909" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T08:27:59.363" LastActivityDate="2015-04-30T08:31:31.750" Title="How does an electrolyte react to voltages below its electrolysis potential?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29594" PostTypeId="5" CreationDate="2015-04-30T05:10:36.223" Score="0" Body="&lt;p&gt;In simple terms, catalysis is the use of a substance (a &quot;catalyst&quot;) that increases the rate of a reaction without modifying the overall standard Gibbs energy change in the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/C00876.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;catalyst&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;em&gt;Also contains definitions of:&lt;/em&gt; &lt;strong&gt;heterogeneous catalysis&lt;/strong&gt;, &lt;strong&gt;homogeneous catalysis&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A substance that increases the rate of a reaction without modifying the overall standard Gibbs energy change in the reaction; the process is called catalysis. The catalyst is both a reactant and product of the reaction. The words catalyst and catalysis should not be used when the added substance reduces the rate of reaction (see inhibitor ). Catalysis can be classified as homogeneous catalysis, in which only one phase is involved, and heterogeneous catalysis, in which the reaction occurs at or near an interface between phases. Catalysis brought about by one of the products of a reaction is called autocatalysis. Catalysis brought about by a group on a reactant molecule itself is called intramolecular catalysis. The term catalysis is also often used when the substance is consumed in the reaction (for example: base-catalysed hydrolysis of esters). Strictly, such a substance should be called an activator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several journals dedicated to catalysis: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/journal/accacs&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACS Catalysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/journal/09205861&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catalysis today&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.journals.elsevier.com/journal-of-catalysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Journal of Catalysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://link.springer.com/journal/10562&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catalysis letters&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.rsc.org/en/Journals/JournalIssues/CY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Catalysis Science &amp;amp; Technology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/journal/10.1002/(ISSN)1615-4169&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advanced Synthesis &amp;amp; Catalysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T14:43:49.617" LastActivityDate="2018-03-19T14:43:49.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29595" PostTypeId="4" CreationDate="2015-04-30T05:10:36.223" Score="0" Body="This tag is appropriate for reactions, their mechanisms, their kinetics, when catalysts of any kind are involved." OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-03-19T14:43:31.833" LastActivityDate="2018-03-19T14:43:31.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29596" PostTypeId="2" ParentId="29593" CreationDate="2015-04-30T05:33:27.280" Score="1" Body="&lt;p&gt;Electrolysing water is NOT required for it to conduct electricity. Water ionizes into $\ce{H+}$ and $\ce{OH-}$. Pure water will contain $10^{-7}\ \mathrm{M}$ of either ion at room temperature at equilibrium. This will result in $18\ \mathrm{M\Omega\ cm}$ of resistivity. If and when potentials above the electrolysis potential is applied, the local pH around the electrodes will change and more ions will be created. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl}$ dissociates as soon as it is dissolved in water. The $\ce{Na+}$ and $\ce{Cl-}$ ions created in the solution carry current and increase the conductivity. &lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T08:31:31.750" LastActivityDate="2015-04-30T08:31:31.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29597" PostTypeId="2" ParentId="29579" CreationDate="2015-04-30T06:10:28.363" Score="6" Body="&lt;p&gt;I am not sure of putting copper nitrate to lead. But the common way to produce lead nitrate is to put lead into nitric acid. It will produce lead nitrate and hydrogen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pb + 2HNO_3 -&amp;gt; Pb(NO_3)2 + H_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have lead oxide instead of lead it also can priduce lead nitrate by adding to nitric acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbO + 2HNO_3 -&amp;gt; Pb(NO_3)2 + H2O}$&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="9757" LastEditDate="2015-04-30T06:29:07.563" LastActivityDate="2015-04-30T06:29:07.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29600" PostTypeId="2" ParentId="29484" CreationDate="2015-04-30T07:55:40.630" Score="1" Body="&lt;p&gt;This seems to be a very specific case. I would say that you should start with a standard Ni electroplating solution and see the result. You could then adjust the parameters of your plating operation and the concentrations of the additives in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a bigger concern. How stable is your liquid crystal template under applied potential in an electrolyte? Before trying to deposit Ni, you should try to do simple electrochemistry find the potential window where your template stays intact.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-04-30T07:55:40.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29601" PostTypeId="1" CreationDate="2015-04-30T08:05:10.867" Score="8" ViewCount="5366" Body="&lt;p&gt;Carbohydrate like glucose undergoes periodic cleavage by $\ce{HIO4}$ and forms respective products are five moles of formic acid and one mole of formaldehyde. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question , what is mechanism and what are the products by fructose with the same? Do we have any specific rules for that?&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorDisplayName="user15489" LastEditDate="2015-04-30T08:30:19.817" LastActivityDate="2018-10-09T23:40:41.213" Title="Periodic cleavage of carbohydrates" Tags="&lt;organic-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29603" PostTypeId="1" AcceptedAnswerId="29610" CreationDate="2015-04-30T08:12:55.327" Score="21" ViewCount="3290" Body="&lt;p&gt;Do only outer electron shells take part in forming chemical bonds? Or could an inner shell create a bond under some conditions?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorUserId="7448" LastEditDate="2015-04-30T08:47:40.770" LastActivityDate="2015-05-01T05:43:12.753" Title="Do only outer electron shells take part in forming chemical bonds?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29606" PostTypeId="2" ParentId="29580" CreationDate="2015-04-30T09:22:29.460" Score="2" Body="&lt;p&gt;Try to use command &quot;repeat n-times&quot;, that will make n-data files and each can have 30 000 points. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/dBzYR.png&quot; alt=&quot;powered by P. Mazur&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-04-30T10:05:21.757" LastActivityDate="2015-04-30T10:05:21.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29609" PostTypeId="1" AcceptedAnswerId="29629" CreationDate="2015-04-30T11:26:02.760" Score="4" ViewCount="244" Body="&lt;p&gt;The solution is emerald green in colour. &#xA;When added with water, it gives a pale blue solution. &#xA;When reacted with &lt;span class=&quot;math-container&quot;&gt;$\ce{BaCl2}$&lt;/span&gt;, it gives a sky blue solution and a white precipitate. &#xA;When reacted with &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt;, it gives a sky blue solution and a white precipitate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not make any sense to me. The &lt;span class=&quot;math-container&quot;&gt;$\ce{BaCl2}$&lt;/span&gt; is used to test sulphates ions and &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; is used to test halides. How could you have two negative ions?&lt;/p&gt;&#xA;" OwnerUserId="5383" LastEditorUserId="4945" LastEditDate="2020-02-04T14:19:30.460" LastActivityDate="2020-02-04T23:41:39.627" Title="How to identify an emerald green solution with possibly two anions?" Tags="&lt;analytical-chemistry&gt;&lt;ions&gt;&lt;color&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29610" PostTypeId="2" ParentId="29603" CreationDate="2015-04-30T11:51:35.313" Score="14" Body="&lt;p&gt;Great Question. &#xA;From this &lt;a href=&quot;http://www.scientificamerican.com/article/chemical-bonds-inner-shell-electrons/&quot;&gt;article&lt;/a&gt;, it seems that extreme conditions of pressure can lead to new type of molecules,bondings and how the electrons participate in the bonding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Under very high pressures, it appears, electrons in the atom's inner shells can also take part in chemical bonds.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“It breaks our doctrine that the inner-shell electrons never react, never enter the chemistry domain,” says Mao-sheng Miao, a chemist at the University of California, Santa Barbara, and the Beijing Computational Science Research Center in China. Miao's calculations show that under extreme pressures cesium and fluorine atoms can form exotic molecules with inner-shell bonds. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Miao identified two molecules that, at high pressure, would involve cesium's inner electrons as well. To form cesium trifluoride ($\ce{CsF3}$), a cesium atom would share its single valence electron and two inner-shell electrons with three fluorine atoms. Four inner electrons would go into making cesium pentafluoride ($\ce{CsF5}$). &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;“That forms a very beautiful molecule, like a starfish,” Miao says. He reported his findings in Nature Chemistry. Both the shape of the resulting molecules and the possibility of their formation are “very surprising,” says Nobel Prize–winning chemist Ronald Hoffmann, a professor emeritus at Cornell University.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Found this in the comments about the stability of the exotic molecules at normal pressures&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Since the compound is thermodynamically unstable it would decompose to the more stable form at lower pressures. Since this means that they are going to higher states of enthalpy I would expect that this decomposition would be a endothermic reaction which means that they would get cold during the process.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="11310" LastEditDate="2015-05-01T05:43:12.753" LastActivityDate="2015-05-01T05:43:12.753" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29611" PostTypeId="1" AcceptedAnswerId="29619" CreationDate="2015-04-30T12:41:59.937" Score="5" ViewCount="23196" Body="&lt;p&gt;It is said that activation energy does not change with temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am unable to figure out why.&#xA;If we increase the temperature, the kinetic energy of the molecules will increase and they will need less extra energy and hence lesser activation energy to overcome the threshold energy barrier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I surfed the net, at most of the places it has been explained that increasing the temperature increases the no. of molecules having more energy and hence reaction rate is more but activation energy is same. This makes some sense but I still don't get why the activation energy doesn't decrease as the KE of an individual molecule should also increase on increasing the temperature.&lt;/p&gt;&#xA;" OwnerUserId="2108" LastEditorUserId="16683" LastEditDate="2017-07-06T10:03:32.827" LastActivityDate="2017-07-06T10:03:32.827" Title="Activation energy and temperature" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ClosedDate="2017-07-06T10:03:25.490" ContentLicense="CC BY-SA 3.0" />
  <row Id="29612" PostTypeId="2" ParentId="29454" CreationDate="2015-04-30T12:46:36.883" Score="1" Body="&lt;p&gt;Sodium metal will react with iron (III) oxide to produce sodium oxide and iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6Na +Fe2O3 -&amp;gt; 3Na2O + 2Fe}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also like a thermite reaction, so try to use the soam\e method as the reaction.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-04-30T12:46:36.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29613" PostTypeId="1" CreationDate="2015-04-30T12:48:36.533" Score="1" ViewCount="52" Body="&lt;p&gt;I can find in literature what the density is for untreated viscose, but not for viscose that is bleached. Does this change the density? And what is the density for bleached viscose?&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="186" LastEditDate="2015-10-02T15:42:04.693" LastActivityDate="2015-10-02T15:42:04.693" Title="Density g/cm3 of bleached viscose" Tags="&lt;physical-chemistry&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29614" PostTypeId="1" CreationDate="2015-04-30T13:06:00.550" Score="5" ViewCount="666" Body="&lt;p&gt;What is the rationalization for calling reactions with rates independent of pH but dependent on the concentration of a particular acid catalyst under &quot;general&quot; acid catalysis, and pH dependent reactions under &quot;specific&quot; acid catalysis ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem to me that the pH-independent reactions would be better considered to be under &quot;specific&quot; acid catalysis given as they are catalysed by a specific species. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="1887" LastEditDate="2015-04-30T13:25:49.433" LastActivityDate="2015-04-30T15:28:27.617" Title="Specific and general acid catalysis" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29615" PostTypeId="1" CreationDate="2015-04-30T13:14:15.557" Score="4" ViewCount="139" Body="&lt;p&gt;If you extract energy from a substrate (e.g. glucose) step-wise, a lot more free energy is released that can be used. However, if you were to do this in one single step, very little useful energy could be harnessed. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastActivityDate="2015-04-30T21:13:29.720" Title="Why is it that less useful energy can be extracted in explosive reactions?" Tags="&lt;biochemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29616" PostTypeId="1" CreationDate="2015-04-30T13:25:49.127" Score="1" ViewCount="44" Body="&lt;p&gt;I was reading an &lt;a href=&quot;http://dx.doi.org/10.1016/j.jmb.2008.05.006&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We first compared experimental (S 2 exp) and computed (S 2 calc) side-chain order parameters using the simplest approximation of side-chain conformational variability, where motions are only allowed within the rotamer well of the side-chain conformation observed in the crystal structure &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now I am wondering what side-chain order parameters are.&lt;/p&gt;&#xA;" OwnerUserId="14877" LastActivityDate="2015-04-30T13:25:49.127" Title="What are side chain order parameters?" Tags="&lt;biochemistry&gt;&lt;structural-biology&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29617" PostTypeId="1" CreationDate="2015-04-30T13:40:47.340" Score="6" ViewCount="91" Body="&lt;p&gt;Can gold recovery with acid peroxide (ap) method be done under vacuum or left in sealed chamber, as opposed to placement in open air?&lt;/p&gt;&#xA;" OwnerUserId="15913" LastEditorUserId="10185" LastEditDate="2015-04-30T19:00:22.430" LastActivityDate="2017-02-15T23:53:28.387" Title="Gold recovery under vacuum" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29618" PostTypeId="2" ParentId="29558" CreationDate="2015-04-30T13:47:27.407" Score="3" Body="&lt;p&gt;In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we still do not know what $[\ce{OH-}]$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only to consider the autoionization of water balanced, you will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is still not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer can only consider by the fact that the solution will be &lt;strong&gt;electrically neutral&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the “electrically neutral” trait will be very useful, this is a good example.&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T08:56:58.843" LastActivityDate="2015-05-01T08:56:58.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29619" PostTypeId="2" ParentId="29611" CreationDate="2015-04-30T13:51:17.750" Score="3" Body="&lt;p&gt;The two distribution plots shown here are for a lower temperature $T_{1}$ and a higher temperature $T_{2}$. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of $E_{a}$ that are shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Activation energy seems to be more of a material or reaction pertinent . Say a reaction has $10J$ as $E_{a}$ to proceed but has $20 J$ kinetic energy - I would say the the additional energy is transferred/carried on by the products after reaction is complete. So having more K.E shouldn't affect the character of a reaction but only its kinetics .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/33QCB.png&quot; alt=&quot;activation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-30T13:51:17.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29621" PostTypeId="1" CreationDate="2015-04-30T14:48:33.680" Score="4" ViewCount="116" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Dilution of the solvent adds new energy states to the liquid, but does&#xA;  not affect the solid phase.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem1.com/acad/webtext/solut/solut-3.html#SEC5&quot; rel=&quot;noreferrer&quot;&gt;The above line appeared on the following page under the section &quot;5: Colligative properties and entropy&quot; and the last paragraph&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, I can understand a little bit. According to me, while freezing the solvent freezes separately from the solute( ie If we freeze salt and water solution, the salt will be separated out from ice). Therefore, whatever the presence of the solute maybe, the solid form of solution consists of solid 'ice', which is identical in every case and hence have identical entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I going right? Is there an alternative explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not clear with the concepts of what actually happens when we freeze a solution. Any Links regarding it or any other related stuff will be welcomed.&lt;/p&gt;&#xA;" OwnerUserId="5608" LastActivityDate="2015-04-30T16:10:02.073" Title="Why the dilution of solvent doesn't affect the solid phase, in comparison to the liquid phase?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;phase&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29622" PostTypeId="1" CreationDate="2015-04-30T15:06:29.333" Score="10" ViewCount="188" Body="&lt;p&gt;An effective way to disperse rioters &amp;amp; looters would be to send an armored vehicle in to spray them with a very stinky liquid. Is there a chemical that stinks but is not  a health risk and that breaks down after a few hours, or could be neutralized with another solution after a few hours? &lt;/p&gt;&#xA;" OwnerUserId="15914" LastActivityDate="2015-05-03T00:23:31.143" Title="Is there a very stinky chemical solution that isn't a health hazard?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29624" PostTypeId="2" ParentId="29614" CreationDate="2015-04-30T15:28:27.617" Score="1" Body="&lt;p&gt;I have wondered about this too. Although I could be wrong it is reasonable to assume (or at least handy to keep as a reminder) that the term &quot;special&quot; refers to the fact that compound is protonated in a very specific state, namely before the actual reaction occurs. General acid catalysis can happen both before the reaction or at any time during the reaction (i.e. in some transition state), whatever is likely in some range around the pH of the acid. Hence it is more general version of the special case mentioned above.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-04-30T15:28:27.617" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29625" PostTypeId="2" ParentId="29621" CreationDate="2015-04-30T16:10:02.073" Score="2" Body="&lt;p&gt;Instead of &quot;making the jump&quot; from liquid to gas, the water molecules are trying to stack together into an ice crystal. The salt molecules again get in the way.&#xA;Think of making a solid as like stacking shoeboxes, but the shoeboxes are vibrating slightly. The higher the temperature, the more they vibrate. At some temperature (the melting point), the vibration gets so powerful that the whole shoebox stack collapses. Now imagine that instead of a stack of just shoeboxes (pure ice), you start putting golf balls in between some of the shoeboxes. Those are salt molecules. They get in the way of the ordered stack, they mess everything up, they make the stack collapse really easily. There has to be a lot less vibration (lower temperature) to stack the boxes when there are golf balls getting up in everyone's business. So the freezing point is reduced.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-04-30T16:10:02.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29626" PostTypeId="2" ParentId="29622" CreationDate="2015-04-30T16:21:28.707" Score="3" Body="&lt;p&gt;My vote would be for &lt;a href=&quot;http://en.wikipedia.org/wiki/Butyric_acid&quot; rel=&quot;nofollow&quot;&gt;butyric acid&lt;/a&gt; and related short-chain fatty acids with between 3 and ~8 carbons (isobutyric acid, valeric acid, isovaleric acid, caproic acid, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pure form these chemicals would be dangerous acids, but in dilute aqueous solution buffered at pH 5 or so, they are not very corrosive or toxic but are still pretty stinky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skatole could be another example.  See the Wikipedia article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Malodorant&quot; rel=&quot;nofollow&quot;&gt;malodorants&lt;/a&gt; for some more ideas.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-04-30T16:21:28.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29627" PostTypeId="1" CreationDate="2015-04-30T16:31:13.070" Score="3" ViewCount="54" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;During a chemical reaction, equal moles of two diatomic molecules $\ce{X}$ and $\ce{Y}$ are allowed to react to form one gaseous compound $\ce{X_nY_m}$. After the reaction is complete, it was observed that the density of the product formed is $1.50$ greater than the density of the reactant mixture. Specify the formula of the product $\ce{X_nY_m}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my work: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{X_m + Y_n -&amp;gt; X_nY_m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density of product: $D_\mathrm{p}$ = $(m/V)_\mathrm{p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density of reactant: $D_\mathrm{r}$ = $(m/V)_\mathrm{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;D_\mathrm{p} &amp;amp;= 1.5 \cdot D_\mathrm{r} \\&#xA;(m/V)_\mathrm{p} &amp;amp;= 1.5 \cdot (m/V)_\mathrm{r} \\&#xA;m_\mathrm{p} &amp;amp;= 1.5 \cdot m_\mathrm{r}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really dont know where to go from there. I would appreciate it if someone could give me step by step how they solved it so I could not just have the answer but also understand how they got to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, would the formula found be empirical or molecular? How do you know that it is?&lt;/p&gt;&#xA;" OwnerUserId="15917" LastEditorUserId="41328" LastEditDate="2017-12-14T05:30:28.673" LastActivityDate="2017-12-14T05:30:28.673" Title="Specify the formula of the product X_n Y_m" Tags="&lt;density&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-05-01T06:57:35.203" ContentLicense="CC BY-SA 3.0" />
  <row Id="29628" PostTypeId="1" CreationDate="2015-04-30T16:51:39.560" Score="4" ViewCount="811" Body="&lt;p&gt;for example, table salt has an a cubic structure in the atomic level, and they are often cube like in the visible level. I don't see any reason that the atomic crystal structure should reflect the crystal's shape. Can someone explain this to me. &lt;/p&gt;&#xA;" OwnerUserId="9649" LastActivityDate="2015-05-01T13:45:00.330" Title="How does the atomic structure of crystals determines its shapes?" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-01T15:06:10.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="29629" PostTypeId="2" ParentId="29609" CreationDate="2015-04-30T17:04:41.357" Score="9" Body="&lt;p&gt;You can have two different anions if your solution is a mixture of two different metal salts, a chloride and a sulfate. The color change from green to blue upon dilution with water fits well with &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt; being the corresponding cation. Concentrated solutions of &lt;span class=&quot;math-container&quot;&gt;$\ce{CuCl2}$&lt;/span&gt; are green due to the presence of chlorocuprates, like &lt;span class=&quot;math-container&quot;&gt;$\ce{[CuCl3]^{-}}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[CuCl4]^{2-}}$&lt;/span&gt;. When water is added, a stepwise ligand exchange takes place, and the pale blue color of &lt;span class=&quot;math-container&quot;&gt;$\ce{[Cu(H2O)6]^{2+}}$&lt;/span&gt; starts to appear. The image below shows copper(II) chloride solutions of different concentrations: Green when the chloride concentration is high, and blue when it is lower (&lt;a href=&quot;http://en.wikipedia.org/wiki/Copper%28II%29_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: copper chloride&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0HjYJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the observations (color change, precipitates) indicate that you have an aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{CuCl2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CuSO4}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="4945" LastEditDate="2020-02-04T14:22:27.783" LastActivityDate="2020-02-04T14:22:27.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29630" PostTypeId="1" AcceptedAnswerId="29658" CreationDate="2015-04-30T17:28:37.097" Score="7" ViewCount="1272" Body="&lt;p&gt;The thermodynamic definition of entropy is expressed as &#xA;$$dS = \frac{dq_{rev}}{T}$$&#xA;I understand that it depends on amount of heat transferred because heat transfer can be understood as the degree of dispersion of energy but what does the temperature dependence actually mean?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastActivityDate="2020-05-21T06:02:25.533" Title="Dependence of change in entropy on temperature" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29631" PostTypeId="1" AcceptedAnswerId="31811" CreationDate="2015-04-30T18:04:30.817" Score="8" ViewCount="10249" Body="&lt;p&gt;According to my textbook, vinyl has higher priority than isopropyl, because an atom which is connected by a double bond can be treated as two identical atoms (one of them is just an imaginary atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about vinyl and secondary butyl? Should the imaginary atom be treated as that nothing else is connected to it? Does secondary butyl have higher priority?&lt;/p&gt;&#xA;" OwnerUserId="15918" LastEditorUserId="194" LastEditDate="2018-03-05T03:24:13.853" LastActivityDate="2021-02-19T13:23:01.530" Title="Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29632" PostTypeId="1" CreationDate="2015-04-30T18:20:54.277" Score="3" ViewCount="36" Body="&lt;p&gt;In sulfate conjugation , is it considered as a carrier or help to make sulfonation if I got a question and I have to choose ( enzyme helps to make the reaction or is a carrier to transfer the sulfate ) I know isn't an enzyme but it does help in the reaction , that what I understand &lt;/p&gt;&#xA;" OwnerUserId="15919" LastActivityDate="2015-04-30T18:20:54.277" Title="PAPS , is it a carrier or not" Tags="&lt;medicinal-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29633" PostTypeId="5" CreationDate="2015-04-30T18:25:35.257" Score="0" Body="&lt;p&gt;For questions that deal with possible errors in a specific statement of a specified book, article, website, etc. These questions should always include a complete reference in this specified format:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Author(s), Initial(s): Year of publication, Title, Edition, Publisher, Place of Publication. (ISBN, URL if available) &lt;strong&gt;Add page number, table/figure/scheme accordingly&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Author(s), Initial(s), Journal name &lt;strong&gt;Year of publication&lt;/strong&gt; &lt;em&gt;Volume number&lt;/em&gt; (Issue or part number), first (and last page) numbers (Title, DOI, URL if available) &lt;strong&gt;Add page number (for longer publications e.g. reviews), table/ figure/ scheme accordingly&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;online resource with a permalink (e.g. Wikipedia) If the resource is fickle, consider sharing a &lt;a href=&quot;https://web.archive.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Web Archive&lt;/a&gt;d link instead.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5026" LastEditDate="2018-03-20T17:55:05.593" LastActivityDate="2018-03-20T17:55:05.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29634" PostTypeId="4" CreationDate="2015-04-30T18:25:35.257" Score="0" Body="For questions that deal with possible errors in a specific statement of a specified book, article, website, etc. These questions should always include a complete reference (see tag wiki for format)." OwnerUserId="5017" LastEditorUserId="5026" LastEditDate="2018-03-20T17:54:43.203" LastActivityDate="2018-03-20T17:54:43.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29635" PostTypeId="1" AcceptedAnswerId="29636" CreationDate="2015-04-30T18:34:14.703" Score="6" ViewCount="123" Body="&lt;p&gt;Salts can bind with some water to form hydrates. For example, we can call $\ce{CoCl_2\cdot 6H_2O}$ cobalt chloride hexahydrate. There is no reason that water should be the only solvent that can bind with salts though. Dimethyl sufloxide is pretty polar and can dissolve some salts, so I might hypothesize that we could make salt crystal that has some bound into it, say $\ce{CoCl_2\cdot 6DMSO}$. I might call that cobalt chloride hexadimethyl sulfoxate, but that is totally made up. Is there a name for that sort of thing?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T19:10:41.753" LastActivityDate="2015-04-30T19:10:41.753" Title="Name for analog of salt hydrate with non-water solvents?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29636" PostTypeId="2" ParentId="29635" CreationDate="2015-04-30T19:02:07.763" Score="4" Body="&lt;p&gt;The corresponding rule in the &lt;em&gt;Red Book&lt;/em&gt; (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’) reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaCl2.8NH3}$ calcium chloride—ammonia (1/8)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AlCl3.4EtOH}$ aluminium chloride—ethanol (1/4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BiCl3.3PCl5}$ bismuth(III) chloride—phosphorus(V) chloride (1/3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Na2CO3.3H2O2}$ sodium carbonate—hydrogen peroxide (2/3)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-04-30T19:06:30.790" LastActivityDate="2015-04-30T19:06:30.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29637" PostTypeId="5" CreationDate="2015-04-30T19:04:21.557" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-04-30T19:04:21.557" LastActivityDate="2015-04-30T19:04:21.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29638" PostTypeId="4" CreationDate="2015-04-30T19:04:21.557" Score="0" Body="For questions regarding chemicals and chemical processes that act on and interact within environmental systems." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-04-30T19:20:47.670" LastActivityDate="2015-04-30T19:20:47.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29639" PostTypeId="2" ParentId="29628" CreationDate="2015-04-30T19:27:24.433" Score="2" Body="&lt;p&gt;It’s all about minimizing the energy of the system. Imagine a 2D example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RHBJ.png&quot; alt=&quot;2D example&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine the green circles as the cations ($\ce{Na+}$) and the blue ones as anions ($\ce{Cl-}$). Where is the next chloride you add going to go?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Not a great choice – it doesn’t really balance. The nearest blue ion is already balancing its charge with two other green ions and so can’t help the new ion balance its charge as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is on the grid but still has all the same problems where the new ion doesn’t get its charge nicely balanced because the nearest blue ion is already donating part of its charge to two other ions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the best option because it can share partial charges with two other blue ions. It gets more shared charge and the whole thing ends up being lower energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This pattern will keep happening with every addition until you get a macro-scale cube. It is always lower energy to fit new ions into gaps then to branch out and make new structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ions that aren’t part of a complete unit cell will develop surface charges. These surface charges are better at attracting ions that are precipitating onto the surface from solution.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorDisplayName="user7951" LastEditDate="2015-05-01T13:45:00.330" LastActivityDate="2015-05-01T13:45:00.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29641" PostTypeId="1" AcceptedAnswerId="29644" CreationDate="2015-04-30T20:57:19.160" Score="5" ViewCount="224" Body="&lt;p&gt;I have been working on a &lt;a href=&quot;https://drive.google.com/file/d/0B1_sXH68hMWBUTYxTE13MUE3YWs/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;CIF data file&lt;/a&gt; based on experimental data &lt;a href=&quot;https://link.springer.com/article/10.1007/BF02842141&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt; (free full text isn't available, but the page 2 with the experiment details can be found in the free preview provided by Springer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know an awful lot about crystallography parameters (although I have been consulting the &lt;a href=&quot;ftp://ftp.iucr.org/pub/cifdics/cif_core_2.4.2.dic.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CIF core dictionary&lt;/a&gt;). The CIF file I linked to was created using &lt;a href=&quot;https://www.ccdc.cam.ac.uk/Community/csd-community/encifer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;EnCIFer&lt;/a&gt; &amp;amp; &lt;a href=&quot;http://jp-minerals.org/vesta/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;VESTA&lt;/a&gt; (I mostly used VESTA to get a handle for the text formatting of CIF files). The non-coordinate data mentioned in this paper includes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{ll}&#xA;\hline&#xA;Z &amp;amp; 4 \\&#xA;\text{Formula} &amp;amp; \ce{C9H7N7O2S} \\&#xA;M_r &amp;amp; 277.3 \\&#xA;\text{Coordinate system} &amp;amp; \text{monoclinic} \\&#xA;\beta &amp;amp; 105.99^\circ \\&#xA;a &amp;amp; \pu{4.488(2) Å} \\&#xA;b &amp;amp; \pu{31.886(4) Å} \\&#xA;c &amp;amp; \pu{8.067(2) Å} \\&#xA;V &amp;amp; \pu{1149.5 Å^3} \\&#xA;D_m &amp;amp; \pu{1.59 mg mm^{-3}}\\&#xA;D_c &amp;amp; \pu{1.60 mg mm^{-3}} \\&#xA;\lambda &amp;amp; \pu{0.7107 Å} \\&#xA;\text{Space group} &amp;amp; P2_1/c \\&#xA;\text{Crystal dimensions} &amp;amp; 0.7\times 0.1 \times 0.7\, \pu{mm} \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that &lt;a href=&quot;http://accelrys.com/products/collaborative-science/biovia-discovery-studio/visualization-download.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accelrys DS Visualizer&lt;/a&gt; so far renders my CIF file as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eVnri.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how some of the hydrogen atoms are disconnected from the main chunk of the azathioprine molecule. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Acharya, K. R. Proc. Indian Acad. Sci. (Chem. Sci.) 1984, 93 (2), 183–187 DOI: &lt;a href=&quot;https://link.springer.com/article/10.1007/BF02842141&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1007/BF02842141&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastEditorUserId="41328" LastEditDate="2017-09-17T07:28:06.817" LastActivityDate="2017-09-17T07:28:06.817" Title="Azathioprine CIF data file rendering problem in Accelrys" Tags="&lt;crystal-structure&gt;&lt;molecular-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29642" PostTypeId="1" AcceptedAnswerId="29655" CreationDate="2015-04-30T21:01:37.167" Score="9" ViewCount="12924" Body="&lt;p&gt;A tertiary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 18 while a primary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 16.&#xA;Shouldn't the tertiary alcohol be more acidic since its conjugate base is more stabilized due to induction?&lt;/p&gt;&#xA;" OwnerUserId="15922" LastEditorUserId="15410" LastEditDate="2015-05-01T03:03:46.687" LastActivityDate="2017-12-06T06:48:59.023" Title="What is the reason that tertiary alcohol pKa is higher than primary alcohol pKa?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29643" PostTypeId="2" ParentId="29615" CreationDate="2015-04-30T21:13:29.720" Score="1" Body="&lt;p&gt;I think there are two relevant points here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) explosive reactions can give significant useful energy - e.g. internal combustion engine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) step-wise processes may go slower and release less heat / light other energy that is not so useful. This is a bit like in thermodynamics you get maximum useful energy from slow reversible processes. &lt;/p&gt;&#xA;" OwnerUserId="11523" LastActivityDate="2015-04-30T21:13:29.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29644" PostTypeId="2" ParentId="29641" CreationDate="2015-04-30T21:27:27.663" Score="6" Body="&lt;p&gt;The coordinates of the hydrogen atoms of the methyl group in question are there (H16a, H16b, H16c). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The problem is a rendering problem of the Accelrys visualizer.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other tools don't have a problem with this CIF file. The image below is a medium-quality PovRay rendering generated through &lt;a href=&quot;http://www.cryst.chem.uu.nl/spek/xraysoft/&quot; rel=&quot;noreferrer&quot;&gt;PLATON&lt;/a&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tO2JR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-04-30T21:44:59.310" LastActivityDate="2015-04-30T21:44:59.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29645" PostTypeId="2" ParentId="28818" CreationDate="2015-04-30T21:48:46.410" Score="1" Body="&lt;p&gt;One &lt;em&gt;possible&lt;/em&gt; reason for the different behaviour of the drinking bottle, compared to the lining of the dishwasher may be indeed the different composition of the polypropylene deployed.  The part of &lt;em&gt;poly&lt;/em&gt; indicates that many hundreds and thousands smaller &lt;em&gt;propylene&lt;/em&gt; molecules are brought together to constitute the polypropylene.  Yet there are different ways this may happen, as this picture (&lt;a href=&quot;https://en.wikipedia.org/wiki/Polypropylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;source)&lt;/a&gt; depicts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQmwr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the polymer chain, there a methyl groups (marked as $\ce{CH3}$) occupying more volume than hydrogen atoms (marked by $\ce{H}$). It is imaginable that &lt;em&gt;all&lt;/em&gt; of these methyl groups are attached on one side of the chain &lt;em&gt;only&lt;/em&gt; (the upper picture), or consistently &quot;once on one, the next time on the opposite side&quot; (the lower picture).  Not shown here, there may be a randomized distribution, too, where you can't predict on which side the next methyl group will be attached (not shown here) as an intermediate situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such alterations in the molecular structure may alter how individual polymer molecules interact with each other.  It influences if the polymer becomes more crystalline, or more glass-like; and by the same token, it influences the melting point of the material.  Indeed, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Polypropylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt; states melting points of 130 degree Celsius (266 Fahrenheit), as well as 171 degree Celsius (340 Fahrenheit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think about the production of the bottle: it is done by (blow) molding.  A manufacture aims to produce a high number of bottles per unit of time, and preferentially at a low cost to spend, for example to heat the polymer that it becomes flexible enough to form it.  So it is better to select a polymer with a lower melting point, and by admixing / blending with other polymers to polypropylene, the melting point probably may be lowered further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On contrast, a manufacturer of a linings of dishwashers equally takes into account the processes while the dishwasher is used: chemicals (detergent), hot water during the washing, and heating to eventually dry the tableware.  So the selection prefers a polymer that tends to be more crystalline and hence exhibits a higher melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may encounter situations showing you that this comes with an disadvantage, too:  The more crystalline, high melting polypropylene liner often is more brisk than the more glassy, low melting polypropylene of your bottle. &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-30T21:48:46.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29646" PostTypeId="2" ParentId="19886" CreationDate="2015-04-30T22:12:42.247" Score="1" Body="&lt;p&gt;The video you reference to does not tell the whole story.  Yes, &lt;a href=&quot;https://en.wikipedia.org/wiki/Zeolite&quot; rel=&quot;nofollow noreferrer&quot;&gt;zeolites&lt;/a&gt; are minerals with open pores, similar to a filter, occurring in nature and are produced commercially, too. Equally to their structure, their composition is interesting -- as &lt;em&gt;both&lt;/em&gt; are used to tailor the reactivity and selectivity of this material.  &lt;a href=&quot;https://en.wikipedia.org/wiki/ZSM-5&quot; rel=&quot;nofollow noreferrer&quot;&gt;ZSM-5&lt;/a&gt; is one example of designed remarkably high Brønsted acidity.  Hydrocarbons in contact with these sites split into fragments and rearrange because of the intermediate generation of carbocations.  It is advantageous, as these reactions are heterogeneously catalysed by these &quot;solid state acids&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="5026" LastEditDate="2018-03-04T02:57:50.437" LastActivityDate="2018-03-04T02:57:50.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29647" PostTypeId="2" ParentId="29630" CreationDate="2015-04-30T22:21:59.037" Score="1" Body="&lt;p&gt;What this equation is saying essentially is that the level of disorder can be modeled as the number of configurations if you have a number of particles and a number of quanta of energy. It's like having three cups and three balls; all the balls can occupy one cup or one ball can occupy one cup and two can occupy another etc. Hence an increase in energy would result in an increase in entropy (more balls and more configurations). Temperature is defined as the average energy of the system. Hence if energy is concentrated (all the balls are in one cup) then the average temperature per particle will be lower and the entropy will be lower which makes sense since the system is more ordered with all the balls in one cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps answer the question somewhat&lt;/p&gt;&#xA;" OwnerUserId="15790" LastActivityDate="2015-04-30T22:21:59.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29648" PostTypeId="2" ParentId="26980" CreationDate="2015-04-30T23:07:23.880" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;The original &lt;a href=&quot;http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Rutgers' reference&quot;&lt;/a&gt; deploys $2\,\mbox{mmol}$ aldehyde in $20\,\mbox{mL}$ solvent (i.e. starting at $0.1\,\mbox{mol/L}$, or $0.1\,\mbox{M}$) and recommends a stirring of $3\,\mbox{h}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Subsequently, you mention $1\,\mbox{mmol}$ per $20\,\mbox{mL}$ solvent (i.e. starting at $0.05\,\mbox{M}$), but without an indication of reaction time.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;But if the starting concentration is lowered further ($0.01\,\mbox{mmol}$ in $5\,\mbox{mL}$ equals to $0.002\,\mbox{M}$ or $2\,\mbox{mM}$), not only the detection of the product becomes difficult. &lt;br&gt;Did you take into consideration that by diluting the reaction you over-proportionally slowed down the advancement of the intended reaction?  Compared to the initially mentioned starting concentration, your reaction starts at a dilution by a factor of 50 (fifty).  This is &lt;em&gt;more&lt;/em&gt; than a big factor for process involving a bimolecular reaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By the way, $0.01\,\mbox{mmol}$ of proline, molar weight of $115.13\,\mbox{g/mol}$ as stated &lt;a href=&quot;https://en.wikipedia.org/wiki/Proline&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; would correspond to little more than $1.1\,\mbox{mg}$.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-04-30T23:07:23.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29649" PostTypeId="1" CreationDate="2015-04-30T23:39:57.750" Score="4" ViewCount="259" Body="&lt;p&gt;I want to calculate the activity coefficients of mixed solvent salt solutions. I am seeing very strange behavior when I try calculating the activity coefficient of salts in non-polar solvents using Debye-Huckel theory though and its messing up my down-stream calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a simple example, let's compare the activity of NaCl in hexane and water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inputs: (hexane / water)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Density: 654 / 1000 kg/m^3&lt;/li&gt;&#xA;&lt;li&gt;Dielectric constant: 1.89 / 80.4&lt;/li&gt;&#xA;&lt;li&gt;Temperature: 293 K&lt;/li&gt;&#xA;&lt;li&gt;Molarity = 0.1 mol/L&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Calculations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A = consts \frac {\sqrt {\rho}}  {(\epsilon * T)^{1.5}} = $ 230 / 1&lt;/li&gt;&#xA;&lt;li&gt;$I = \frac12 \sum_{ions} cz^2 =$ 0.1 / 0.1&lt;/li&gt;&#xA;&lt;li&gt;$ln(\gamma) = -Az^2\sqrt I = $ -73 / -0.32&lt;/li&gt;&#xA;&lt;li&gt;$\gamma = 10^{-32}$ / 0.72&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In equilibrium the salt activity should be the same in each phase, $\gamma_{aq}x_{aq} = \gamma_{org} x_{org}$, which would imply that the salt should all be flying over to the organic phase. That clearly makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this give such totally wrong behavior in a limiting case?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2015-04-30T23:39:57.750" Title="Debye-Huckel theory with non-polar solvents" Tags="&lt;ions&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29650" PostTypeId="2" ParentId="27662" CreationDate="2015-04-30T23:46:56.853" Score="1" Body="&lt;p&gt;Your mixed up with the use of half orbital. When iron become to $\ce{Fe^2+}$, it needs to donate two valance electrons away which is at the highest energy (orbital $\ce{4s}$). So, that is why we write $\ce{Fe^2+}$ as $\ce{[Ar] 3d^6}$.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="25052" LastEditDate="2018-04-28T16:21:31.333" LastActivityDate="2018-04-28T16:21:31.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29651" PostTypeId="2" ParentId="27662" CreationDate="2015-05-01T01:43:22.073" Score="6" Body="&lt;p&gt;When researching the explanation for this phenomenon, I found &lt;a href=&quot;http://www.rsc.org/eic/2013/11/aufbau-electron-configuration&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is natural to question why one or two electrons are usually pushed into a higher energy orbital. The answer is because $\mathrm{3d}$ orbitals are more compact than $\mathrm{4s}$, and as a result any electrons entering $\mathrm{3d}$ orbitals will experience greater mutual repulsion. The slightly unsettling feature is that although the relevant $\mathrm{s}$ orbital can relieve such additional electron-electron repulsion, different atoms do not always make full use of this form of sheltering because the situation is more complicated than just described. One thing to consider is that nuclear charge increases as we move through the atoms, and there is a complicated set of interactions between the electrons and the nucleus as well as between the electrons themselves. This is what ultimately produces an electronic configuration and, contrary to what some educators may wish for, there is no simple qualitative rule of thumb that can cope with this complicated situation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On a semi-related note, according to one &lt;a href=&quot;http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; I found (the source sites many others as well), it is in fact the $\mathrm{3d}$ electrons that fill first, followed by the $\mathrm{4s}$ electrons. This more satisfactorily explains why the $\mathrm{s}$ electrons are first lost in the ionization of the $\mathrm{d}$-block transition metals.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-01T01:43:22.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29652" PostTypeId="5" CreationDate="2015-05-01T03:14:45.573" Score="0" Body="&lt;p&gt;Medicinal chemistry is an interdisciplinary field in which new biologically active molecules (molecules that interact with a specific biological target to produce a response) are designed and synthesised as potential therapies to treat disease and infection.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Definition:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IUPAC&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;…medicinal chemistry is a chemistry-based discipline, involving aspects of the biological, medical and pharmaceutical sciences. It is concerned with the invention, discovery, design, identification and preparation of biologically active compounds, the study of their metabolism, the interpretation of their mode of action at the molecular level and the construction of structure-activity relationships (SAR), which is the relation- ship between chemical structure and pharmacological activity for a series of compounds…&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/medicinal_chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Medicinal chemistry and pharmaceutical chemistry are disciplines at the intersection of chemistry, especially synthetic organic chemistry, and pharmacology and various other biological specialties, where they are involved with design, chemical synthesis and development for market of pharmaceutical agents, or bioactive molecules (drugs).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Applicability of the &lt;a href=&quot;/questions/tagged/medicinal-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;medicinal-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;medicinal-chemistry&lt;/a&gt; tag:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/medicinal-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;medicinal-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;medicinal-chemistry&lt;/a&gt; should be applied to any questions about the theory and practice of medicinal chemistry, in both academic and industrial situations. &lt;/li&gt;&#xA;&lt;li&gt;the &lt;a href=&quot;/questions/tagged/drugs&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;drugs&amp;#39;&quot; rel=&quot;tag&quot;&gt;drugs&lt;/a&gt; tag is likely to be relevant to many &lt;a href=&quot;/questions/tagged/medicinal-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;medicinal-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;medicinal-chemistry&lt;/a&gt; questions (those dealing with specific molecules/classes of molecules), as drugs are the product of medicinal chemistry/the pharmaceutical industry. One exception would be if a question is specifically about the practice of medicinal chemistry without any reference to specific molecules.&lt;/li&gt;&#xA;&lt;li&gt;Medicinal chemistry is an umbrella term encompassing several other fields in chemistry, meaning additional tags &lt;strong&gt;should&lt;/strong&gt; be used in addition to &lt;a href=&quot;/questions/tagged/medicinal-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;medicinal-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;medicinal-chemistry&lt;/a&gt;, for example:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;questions may deal with the synthesis/manufacture of drugs, in which case &lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/synthesis&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;synthesis&amp;#39;&quot; rel=&quot;tag&quot;&gt;synthesis&lt;/a&gt; are likely appropriate. &lt;/li&gt;&#xA;&lt;li&gt;questions may deal with the interaction of drugs with proteins/enzymes or related biological issues, so should have tags representing this. &lt;a href=&quot;/questions/tagged/biochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;biochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;biochemistry&lt;/a&gt; or &lt;a href=&quot;/questions/tagged/pharmacology&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;pharmacology&amp;#39;&quot; rel=&quot;tag&quot;&gt;pharmacology&lt;/a&gt; could be appropriate depending on the situation. (In these cases the question might also cross a line into biology warranting migration to biology.se). &lt;/li&gt;&#xA;&lt;li&gt;questions may deal with computational aspects of predicting properties such as pKa or logD, making &lt;a href=&quot;/questions/tagged/computational-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;computational-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;computational-chemistry&lt;/a&gt; a useful additional tag. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Related tags:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/medicinal-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;medicinal-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;medicinal-chemistry&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following which may be useful as additional tags on questions identified as being related to medicinal chemistry: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/drugs&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;drugs&amp;#39;&quot; rel=&quot;tag&quot;&gt;drugs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/pharmacology&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;pharmacology&amp;#39;&quot; rel=&quot;tag&quot;&gt;pharmacology&lt;/a&gt;, &lt;a href=&quot;/questions/tagged/pharmaceuticals&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;pharmaceuticals&amp;#39;&quot; rel=&quot;tag&quot;&gt;pharmaceuticals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/biochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;biochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;biochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/chemical-biology&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;chemical-biology&amp;#39;&quot; rel=&quot;tag&quot;&gt;chemical-biology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/stereochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Further reading:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Several books have been written specifically on medicinal chemistry, however, related books on organic chemistry, pharmacology, biochemistry and physiology could also be consulted for additional information (additional references can be found on the tag pages for the individual topics): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Patrick, G.; &lt;em&gt;An Introduction to Medicinal Chemistry&lt;/em&gt;; Oxford University Press: Oxford, 2013.    &lt;/li&gt;&#xA;&lt;li&gt;Abraham, D. J. (ed.); &lt;em&gt;Burger’s Medicinal Chemistry &amp;amp; Drug Discovery&lt;/em&gt;; Wiley-Interscience, Unknown. &lt;/li&gt;&#xA;&lt;li&gt;Lemke, T. L. (ed.); &lt;em&gt;Foye’s Principles of Medicinal Chemistry&lt;/em&gt;; Lippincott Williams &amp;amp; Wilkins: Baltimore, 2013&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21296" LastEditorUserId="21296" LastEditDate="2017-06-07T21:52:29.620" LastActivityDate="2017-06-07T21:52:29.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29653" PostTypeId="4" CreationDate="2015-05-01T03:14:45.573" Score="0" Body="Medicinal chemistry is the process of  design, synthesis, and testing of molecules (drugs) for the treatment of infection and disease. &#xD;&#xA;&#xD;&#xA;The medicinal-chemistry tag should be applied to all questions relating to this process, from the theoretical aspects of drug design (screening, computational studies, synthesis planning) to the practical aspects of the actual synthesis, characterisation and testing of the molecule. " OwnerDisplayName="user15489" LastEditorUserId="21296" LastEditorDisplayName="user15489" LastEditDate="2017-06-07T21:52:36.550" LastActivityDate="2017-06-07T21:52:36.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29655" PostTypeId="2" ParentId="29642" CreationDate="2015-05-01T04:01:28.890" Score="5" Body="&lt;p&gt;The short answer is that the steric hindrance of the tert butyl alcohol makes it hard to add an protonate the hydroxy group.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Based on &lt;a href=&quot;http://as.vanderbilt.edu/chemistry/Rizzo/Chem220b/Chapter_17.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lecture&lt;/a&gt; from the vanderbilt university. The answer is due to steric hindrance:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GCf3C.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The hydroxy group of the tert butyl alcohol is intramolecular stabilized by hydrogen bonds (2nd image), between the methyl hydrogens and the oxygen. In the other hand, &lt;em&gt;i.e.&lt;/em&gt; the hydroxy group of methanol is intermolecular stabilized with another &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2008/cp/b715841f#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;methanol molecules&lt;/a&gt; (&amp;quot;Behave more like a water&amp;quot;) (3th image). Therefore, the equilibrium &lt;span class=&quot;math-container&quot;&gt;$\ce{ROH}\rightleftharpoons\ce{RO^-}+\ce{H^+}$&lt;/span&gt; is more displaced to the right with the methanol than with the tert-butil alcohol.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJx8bb.png&quot; alt=&quot;enter image description here&quot; /&gt; &lt;img src=&quot;https://i.stack.imgur.com/13vy9.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The protonation of ter butyl alcohol is slower than methanol (Figure 4th and 5th), for the same reason: &lt;em&gt;steric hindrance&lt;/em&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/u0sc1.png&quot; alt=&quot;enter image description here&quot; /&gt;  &lt;img src=&quot;https://i.stack.imgur.com/VyLoK.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-01T19:11:36.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29656" PostTypeId="1" CreationDate="2015-05-01T05:36:28.287" Score="6" ViewCount="657" Body="&lt;p&gt;I'm having trouble determining if sugars are D/L in their chair forms. My textbook has generously provided this image of D-Glucose. &lt;img src=&quot;https://i.stack.imgur.com/IjNqQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know that from the fischer, when I put everything into a chair view, the things on the left point up, and the things on the right point down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what about the bottom $\ce{CH2OH}$ group. Which way does it point, and does it determine D/L in a ring? I'm imagining &quot;rotating&quot; all the groups 1 position over clockwise (does it matter if its clockwise/counterclockwise? How do you know which way to spin it?) such that the hydroxyl is now where the $\ce{CH2OH}$ was, H is on the right and $\ce{CH2OH}$ is on the left.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that sound right? I can vaguely remember this from high school, but it's become very clouded. &lt;/p&gt;&#xA;" OwnerUserId="15925" LastEditorUserId="5456" LastEditDate="2015-05-01T06:01:41.037" LastActivityDate="2015-05-01T17:09:37.357" Title="How can you tell if a sugar is D/L in its chair form?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29657" PostTypeId="1" AcceptedAnswerId="29659" CreationDate="2015-05-01T06:42:44.707" Score="18" ViewCount="65509" Body="&lt;p&gt;There was a question in one of my exam papers to draw the resonance structures for $\ce{N2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the ones I drew, but they were not awarded marks:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zNO4u.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zNO4u.png&quot; alt=&quot;putative resonance structures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones in the marking scheme are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yxBDE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yxBDE.png&quot; alt=&quot;resonance structures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please explain why mine aren't correct?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="16683" LastEditDate="2017-08-31T17:43:53.937" LastActivityDate="2018-08-18T15:58:42.847" Title="What are the correct resonance structures of nitrous oxide?" Tags="&lt;resonance&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29658" PostTypeId="2" ParentId="29630" CreationDate="2015-05-01T06:49:41.817" Score="4" Body="&lt;p&gt;An analogy that I like is one paraphrased from Peter Atkins' great book &quot;The Laws of Thermodynamics - a VSI&quot;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine two rather different places in terms of noise - a silent&#xA;  library and a busy train station. Imagine that the temperature is the amount of noise at the given place - a measure of the &quot;state of noise&quot;. Let's say that you laugh out loud,&#xA;  corresponding to supplying energy as heat. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In the library, where there is little background noise (low $T$) this would cause a large amount of disturbance and&#xA;  disorder - corresponding to a large increase in entropy. On the busy&#xA;  train station, on the other hand, you laughing would go largely&#xA;  unnoticed - there is lots of background noise already (high $T$). The (relative) amount of additional disturbance and&#xA;  disorder would be much smaller, corresponding to a smaller increase of&#xA;  entropy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a very cold system, even a tiny amount of supplied energy in terms of heat would cause a large change in entropy. On the other hand, supplying the same amount of energy to a really hot system - with lots of molecular motion going on - would cause a minuscule change in entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; Note that the analogy and explanation above are two ways of convincing yourself that the entropy change due to added heat is inversely proportional to the system temperature. They do not claim to be correct or physical in any other sense, so read them with a (large) pinch of salt. &lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2015-05-01T06:49:41.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29659" PostTypeId="2" ParentId="29657" CreationDate="2015-05-01T07:22:21.030" Score="22" Body="&lt;p&gt;Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ cannot do. The maximum number of bonds that $\ce{N}$ can participate in is 3 (covalent bonding - like $\ce{NH3}$) or 4 (coordinate bonding - like $\ce{NH4+}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you satisfy the octet rule (or at least as best as you can - here are some &lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Lewis_Theory_of_Bonding/Violations_of_the_Octet_Rule/Exceptions_to_the_Octet_Rule&quot; rel=&quot;noreferrer&quot;&gt;exceptions&lt;/a&gt;) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{FC} = \mathrm{V} - (\mathrm{N_B} + \frac{\mathrm{B}}{2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wher $\mathrm{FC}$ is the formal charge, $\mathrm{V}$ is the number of valence electrons that the atom under consideration usually has, $\mathrm{N_B}$ is the number of non-bonded electrons, and $\mathrm{B}$ is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="16683" LastEditDate="2017-08-31T17:44:41.030" LastActivityDate="2017-08-31T17:44:41.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29660" PostTypeId="2" ParentId="29657" CreationDate="2015-05-01T07:33:32.583" Score="4" Body="&lt;p&gt;Nitrogen cannot expand its octet (it cannot accommodate more than eight valence electrons). In the resonance structures that you have drawn, both the nitrogen and oxygen on the ends are stable (having no charge) but nitrogen in the middle has five bonds hence violates the above rule. So, the maximum number of bonds for nitrogen in the middle is four. That is why the middle nitrogen atom (in the correct answer)&#xA;has four bonds but with a charge.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, an expanded octet is generally observed in elements which are located below the second period. &lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="65816" LastEditDate="2018-08-18T15:58:42.847" LastActivityDate="2018-08-18T15:58:42.847" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29661" PostTypeId="1" CreationDate="2015-05-01T08:33:42.677" Score="6" ViewCount="3975" Body="&lt;p&gt;Between ethene and propene, which is more acidic and why? If they are they are the same then why so?&lt;br&gt;&#xA; Their $\mathrm{p}K_{\mathrm{a}}$ is the same $\mathrm{p}K_{\mathrm{a}}$ as ~44.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="46306" LastEditDate="2018-03-05T01:46:59.260" LastActivityDate="2018-03-05T02:09:28.463" Title="Between ethene and propene which is more acidic and why?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29662" PostTypeId="2" ParentId="27018" CreationDate="2015-05-01T08:49:48.240" Score="0" Body="&lt;p&gt;Reversible reactions are considered to be ideal, since in our everyday life we don't get to see anything broken come back together by its own. So you have to remember one thing, which I also think it's the answer youre looking for : &lt;strong&gt;every reversible reaction we talk about in thermodynamics, happens infinitezimal slowly&lt;/strong&gt;. For example :&#xA;Reversible gas compression in constant pressure, happens by removing the heat, lowing the temperature of the ambient in &lt;strong&gt;infinitezimal&lt;/strong&gt; amounts, so that the temperature of the system changes from T1 to T2. So your assumtion was right, infinitezimal is the key word in reversible reactions.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-01T18:14:28.730" LastActivityDate="2015-05-01T18:14:28.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29663" PostTypeId="2" ParentId="6251" CreationDate="2015-05-01T10:11:25.500" Score="1" Body="&lt;p&gt;By considering that mass and energy are the two faces of the same coin then the conservation is valid according to Einstein concept.&lt;/p&gt;&#xA;" OwnerUserId="15926" LastEditorUserId="7448" LastEditDate="2015-05-01T10:31:51.367" LastActivityDate="2015-05-01T10:31:51.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29664" PostTypeId="1" AcceptedAnswerId="29667" CreationDate="2015-05-01T10:59:39.337" Score="9" ViewCount="258" Body="&lt;p&gt;Smoke alarms are a necessary part of many modern day buildings, especially in laboratories and of course, the family home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the World Nuclear Association page &lt;a href=&quot;http://www.world-nuclear.org/info/non-power-nuclear-applications/radioisotopes/smoke-detectors-and-americium/&quot;&gt;Smoke Detectors and Americium&lt;/a&gt; (2014), a major part of the operation of a smoke detector is when the Americium-241 emits alpha particles collide with atmospheric oxygen and nitrogen, ionising them - which form part of an electric current between two electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is smoke (from the article) - bolding mine: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When smoke enters the space between the electrodes, &lt;strong&gt;the smoke particles attach to the charged ions, neutralizing them&lt;/strong&gt;. This causes the number of ions present – and therefore the electric current – to fall, which sets off an alarm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the chemical process that causes smoke to neutralise the charged oxygen and nitrogen ions in smoke alarms?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-01T11:04:55.303" LastActivityDate="2015-05-01T12:13:43.903" Title="Smoke alarm ion neutralisation" Tags="&lt;everyday-chemistry&gt;&lt;ions&gt;&lt;safety&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29666" PostTypeId="1" CreationDate="2015-05-01T12:00:16.427" Score="6" ViewCount="2674" Body="&lt;p&gt;I was wondering about molecular orbitals of methylene different states 1B&lt;sub&gt;1&lt;/sub&gt; 3B&lt;sub&gt;1&lt;/sub&gt; 1A&lt;sub&gt;1&lt;/sub&gt;&#xA;and how are they connected to symmetry point group lower case 1a&lt;sub&gt;1&lt;/sub&gt;, etc orbitals what those numbers and alphabets mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone can derive those orbitals from symmetry and show me which states 1B&lt;sub&gt;1&lt;/sub&gt;, 3B&lt;sub&gt;1&lt;/sub&gt; and 1A&lt;sub&gt;1&lt;/sub&gt; are referred to and what does up script, subscripts and alphabets stand for and connect them to lower case 1a&lt;sub&gt;1&lt;/sub&gt; ,etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z2HBP.gif&quot; alt=&quot;foe example in water&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15927" LastEditorUserId="5456" LastEditDate="2015-05-01T13:20:07.797" LastActivityDate="2015-06-05T18:59:47.757" Title="Molecular orbitals of CH2 an SLACs" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29667" PostTypeId="2" ParentId="29664" CreationDate="2015-05-01T12:13:43.903" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;https://www.google.co.in/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=9&amp;amp;ved=0CFAQFjAI&amp;amp;url=http%3A%2F%2Fwww.ismtii2011.org%2Farticle%2Fxml%2Fsub%2Ffile_download.kin%3Fmain_no%3D202%26mode%3Dpdf&amp;amp;ei=4mBDVZGhNY63uQS8toHYCQ&amp;amp;usg=AFQjCNH4KEifjFcA8P6pE9KxmsMyLXjbyA&amp;amp;sig2=hgnbBozls-uaDFqTWlOHew&amp;amp;bvm=bv.92189499,d.c2E&amp;amp;cad=rja&quot;&gt;this&lt;/a&gt; paper, found the mechanism for neutralisation,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Smoke particle charge interaction&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When smoke particles enter in the ionization chamber, they are&#xA;  subjected to the electric field generated by electrodes but also to the&#xA;  one generated by the charges. Close to the charges, the electric field is&#xA;  highly divergent resulting in &lt;a href=&quot;http://en.wikipedia.org/wiki/Dielectrophoresis&quot;&gt;dielectrophoretic force&lt;/a&gt; acting on smoke&#xA;  particles. The force tends to bind the smoke particles to the charges.&#xA;  When smoke particles are considered as isotropic neutral dielectric&#xA;  bodies linearly and homogeneously polarisable, dielectrophoretic&#xA;  force F can be expressed as&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke&#xA;  particle volume and $E$ is the electric field produced by the charges&#xA;  [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the&#xA;  dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $&#xA;  diameter smoke particle.&#xA;  Amplitude of electric field $E$ produced by charges is given by&#xA;  coulomb’s law:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;in which $r$ is the distance between the charge and the smoke particle.&#xA;  Due to the divergence of the field, the dielectrophoretic force&#xA;  decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time&#xA;  of interaction may be relatively long and it is necessary that the&#xA;  charge drifts sufficiently slowly in the chamber to optimize the&#xA;  interaction. Indeed if the charge passes too rapidly there will be not&#xA;  enough time for the smoke particle to bind and thus nothing would be&#xA;  detected. In fact the capture radius reduces with the electric field and&#xA;  is of the order of 10 $\mu m $ for $E= 1  V/mm$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the smoke particles attach to the free charges due to&#xA;  electrophoresis force, the charges become heavy bodies then the&#xA;  charge mobility $\mu$ decreases resulting in a drop of the electric current&#xA;  measured between the electrodes of the ionization chamber.&#xA;  Even if ionic detectors are very efficient, according to new&#xA;  standards they are no longer authorised in almost all of the developed&#xA;  countries because of the radioactive sources they include for&#xA;  generating the ions. Such radioactive sources are indeed difficult to&#xA;  collect and to recycle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-05-01T12:13:43.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29668" PostTypeId="2" ParentId="16615" CreationDate="2015-05-01T12:46:27.470" Score="1" Body="&lt;p&gt;From &lt;a href=&quot;https://books.google.co.in/books?id=o-YTN9aXAeIC&amp;amp;pg=PA334&amp;amp;lpg=PA334&amp;amp;dq=non+ferrous+alloy+showing+martensite+structure&amp;amp;source=bl&amp;amp;ots=sNzLxrp5PF&amp;amp;sig=crkye7yrKb3m65JpZkdKQHTaagg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=VHJDVZSrNdOJuwTy0IHACg&amp;amp;ved=0CEAQ6AEwCA#v=onepage&amp;amp;q=non%20ferrous%20alloy%20showing%20martensite%20structure&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; link I could find a few non-ferrous systems which show quench-induced or stress-induced martensite . Generally non-ferrous systems dont show this behaviour and they are hardened/strengthened by &lt;a href=&quot;http://en.wikipedia.org/wiki/Precipitation_hardening&quot; rel=&quot;nofollow&quot;&gt;Precipitation hardening&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_hardening&quot; rel=&quot;nofollow&quot;&gt;work hardening&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_solution_strengthening&quot; rel=&quot;nofollow&quot;&gt;Solid Solution Strengthening&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also work hardening is basically working/deforming the material plastically to strengthen it. So you can't work harden a material without contact. Induction hardening is used mainly for &lt;a href=&quot;http://en.wikipedia.org/wiki/Case_hardening&quot; rel=&quot;nofollow&quot;&gt;Case Hardening&lt;/a&gt; which is different from hardening(done to bulk of the material). So as long as the non-ferrous alloy is conductive and resistive , Induction hardening is enough for case-hardening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it is hardening , the above mentioned methods are used for non-ferrous. Only the quenching/martensitic transformation is limited to limited systems. &lt;/p&gt;&#xA;" OwnerUserId="11310" LastActivityDate="2015-05-01T12:46:27.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29670" PostTypeId="1" AcceptedAnswerId="30769" CreationDate="2015-05-01T14:18:29.500" Score="4" ViewCount="2010" Body="&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation#History&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's article about orbital hybridisation&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[...] today it is considered an effective heuristic for rationalising the&#xA;structures of organic compounds. [...] Hybrid orbitals are assumed to be mixtures of atomic orbitals, superimposed on each other in various proportions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybrid orbitals 'real'? Are they solutions of Schrodinger equation? Are they a superposition of atomic orbitals?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="61578" LastEditDate="2021-02-26T13:49:20.980" LastActivityDate="2021-02-26T13:49:20.980" Title="sp, sp2 and sp3 hybrid orbitals are actual solutions of Schrodinger's equation or only a heuristic?" Tags="&lt;hybridization&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29671" PostTypeId="2" ParentId="29590" CreationDate="2015-05-01T14:29:03.777" Score="3" Body="&lt;p&gt;You should just think about a fact: &lt;strong&gt;all chemical processes&lt;/strong&gt; will correspond to enthalpy changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic process of acid-base reaction is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + OH- -&amp;gt; H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process will have an enthalpy change of &lt;span class=&quot;math-container&quot;&gt;$\pu{-55.8 kJ mol-1}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{298 K}$&lt;/span&gt;. (I don’t know why, but in different books or articles this value will be slightly different.) However,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Precipitate formation&lt;/strong&gt; is usually exothermic, since there are some ion bonds formed.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Weak acid and base&lt;/strong&gt; usually need some energy, for they are not completely ionized, during the reaction, they will to ionize more since &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; is consumed. Ionization will be endothermic.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore the answer would be the second reaction, since &lt;span class=&quot;math-container&quot;&gt;$\ce{BaSO4}$&lt;/span&gt; is precipitated, which is exothermic, while weak acid and base reactions are endothermic.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(However, there seems to be always some exception, for example, ionization of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; will be exothermic. But to discuss this may be out of topic, and to be frankly, I’m not quite sure what its mechanism is.)&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorUserId="4945" LastEditDate="2019-02-06T15:46:51.807" LastActivityDate="2019-02-06T15:46:51.807" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29673" PostTypeId="1" AcceptedAnswerId="29680" CreationDate="2015-05-01T14:52:00.863" Score="4" ViewCount="23626" Body="&lt;p&gt;The course calendar for my university describes Chem 101 and 102 as &quot;math intensive&quot;. However, it doesn't say what kind of math is involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of math should I expect in first year chemistry?&#xA;Would it be helpful to have first year calculus under my belt first?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="10185" LastEditDate="2015-05-01T15:14:52.757" LastActivityDate="2015-07-21T18:09:54.143" Title="What kind of math should I expect in first year chemistry?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29674" PostTypeId="2" ParentId="28788" CreationDate="2015-05-01T15:08:54.213" Score="5" Body="&lt;p&gt;In short this is the way I imagine them, as slight directional changes in spins propagating through the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Heisenberg Hamiltonian for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Exchange_interaction&quot; rel=&quot;noreferrer&quot;&gt;exchange energy&lt;/a&gt; associated with magnetic coupling is a pairwise sum over an exchange integral $J_{ij}$ for two sites $i$ and $j$, with spin moments $\hat S_i$ and $\hat S_j$. &#xA;\begin{equation}&#xA;H_{ex}=-\sum _{i&amp;lt;j}J_{ij}\hat S_i\cdot \hat S_j&#xA;\end{equation}&#xA;Imagine a periodic array of lattice points each with a spin of two possible orientations $\pm\frac 12$, for a ferromagnetic ground state $(J_{ij}&amp;gt;0)$ with all the spins aligned, denoted as;&#xA;\begin{equation} |\uparrow \uparrow \uparrow \dots \uparrow\uparrow\rangle&#xA;\end{equation}&#xA;A low energy excitation would intuitively be a single flip of a spin somewhere in the lattice e.g.,&#xA;\begin{equation}&#xA;|\uparrow \downarrow\uparrow\dots\uparrow\uparrow\rangle&#xA;\end{equation}&#xA;However, &lt;strong&gt;this is not an eigenstate of the spin Hamiltonian&lt;/strong&gt;. Instead low lying excitations have a complex wave character that is akin to a fractional displacement of a spin with respect to its neighbour. Such infinitesimal variation I personally imagine as a single reversed spin distributed over all the spins in the lattice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you could view a spin wave as a superposition where all the spins contain a bit of this single inverted spin in the first excited state. It is easy to see that this is the case since we are using ladder operators to change our spins from up to down. Acting as a superposition over the lattice (&lt;strong&gt;spin wave&lt;/strong&gt;) reduces the disturbance between any given adjacent pair, whereas the naive expected eigenstate would include a really big difference in exchange energies at a localised region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may be thinking, how can I have an infinitesimally displaced spin when there are only two spin orientations? Well that is a bit of a misnomer as this oversimplified model has a discrete symmetry of 2. In reality there is a smooth symmetry associated with these spins so that the spin wave is actually a more realistic picture than the 2-fold orientation (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ising_model&quot; rel=&quot;noreferrer&quot;&gt;Ising model&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2015-05-01T15:08:54.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29675" PostTypeId="1" AcceptedAnswerId="32708" CreationDate="2015-05-01T16:01:33.080" Score="14" ViewCount="454" Body="&lt;p&gt;I have a really old sample of sodium tetraphenylborate. It should be white, but it is red. Does anyone know what was formed and how it influences precipitation of cations in water?&lt;/p&gt;&#xA;" OwnerUserId="7688" LastEditorUserId="402" LastEditDate="2017-05-12T22:48:33.130" LastActivityDate="2017-05-12T22:48:33.130" Title="Old sample of sodium tetraphenylborate—why it is red?" Tags="&lt;solubility&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29676" PostTypeId="2" ParentId="29143" CreationDate="2015-05-01T16:07:00.283" Score="15" Body="&lt;p&gt;Surface tension arises from the fewer neighbors that molecules at the surface have, relative to the number of neighbors for molecules in the bulk.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a simple model of rigid identical spheres in a &lt;a href=&quot;http://mathworld.wolfram.com/HexagonalClosePacking.html&quot;&gt;hexagonal closest packing&lt;/a&gt;.  In the bulk, there are 12 spheres in contact with a given sphere, which could be thought of as 6 spheres at vertices of a regular hexgon surrounding the given sphere, 3 spheres in a triangle below and 3 spheres in a tringle above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a surface, at least 3 of the neighboring spheres would be absent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering an interaction energy between neighboring spheres, the loss of interaction energy (due to fewer than 12 neighbors per sphere) will be proportional to the surface area of the system of spheres.  The minimum energy of the system will be when surface area of the arrangement of spheres is minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above reason, the units of surface tension are Energy/surface area, which the OP correctly points out is the same as force/distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above explanation, particularly that even for rigid spheres always in direct contact the Energy/surface area (force/distance) relationship arises, it can be understood that surface tension has no particular relationship with a force constant between molecules (or between atoms of a molecule).  Instead, surface tension is related to the energy difference between a molecule with maximum neighbors and a molecule with fewer neighbors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering a simple harmonic oscillator, energy vs distance is a parabola.  There is no horizonal asymptote.  Regardless of what the force constant is, there is always infinite energy required to remove a neighbor. In reality, intermolecular force will approach zero at infinite distance between molecules. It is the difference between energy at equilibrium seperation and infinite separation that is relavent to surface tension, and force constant does not provide this information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &quot;initial molecular harmonic&quot; and &quot;initial harmonic&quot; are not standard chemistry terms and only have meaning in the context of a particular computer program.  &quot;Initialharmonic&quot; or &quot;Iharmonic&quot; refers to a force constant that can be for linear stretching, bending about a vertex that is an atom, or dihedral bending, etc. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-05-03T22:13:11.730" LastActivityDate="2015-05-03T22:13:11.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29677" PostTypeId="2" ParentId="20218" CreationDate="2015-05-01T16:11:13.507" Score="2" Body="&lt;p&gt;For ionizing radiation, various mechanisms for interaction with matter are possible. Depending on the energy, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production. Charged particles, such as alpha and beta particles, mainly interact through coulomb forces between their charge and the negative charge of the orbital electrons of the absorber atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these processes lead to partial or complete transfer of the radiation energy to electron energy; thus, they lead to electron excitation or ionization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{H2O &amp;amp;-&amp;gt;[\gamma][\quad] H2O^{$*$} }\\&#xA;\ce{H2O &amp;amp;-&amp;gt;[\gamma][\quad] H2O+ + e- }&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interaction is very fast. The excitation or ionization event occurs within a time scale of $&amp;lt; 10^{-16}\ \mathrm{s}$. This process is too fast for immediate chemical reactions. By way of comparison, the frequencies of molecular vibrations range from less than $10^{12}\ \mathrm{s^{-1}}$ to approximately $10^{14}\ \mathrm{s^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, the first chemical reactions follow slightly later, within a time scale of $10^{-14}\ \mathrm{s}$ to $10^{-13}\ \mathrm{s}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{H2O^+ + H2O &amp;amp;-&amp;gt; *OH + H3O+ }\\&#xA;\ce{H2O^{$*$} &amp;amp;-&amp;gt; H. +\, .OH }\\&#xA;\ce{H2O^{$*$} &amp;amp;-&amp;gt; H2 + O. }&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiolysis products proceed to diffuse, and further chemical reactions occur.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="24052" LastEditDate="2016-03-12T00:09:32.123" LastActivityDate="2016-03-12T00:09:32.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29679" PostTypeId="1" AcceptedAnswerId="30795" CreationDate="2015-05-01T16:41:19.113" Score="16" ViewCount="6219" Body="&lt;p&gt;Last week, I tried synthesizing acetylsalicylic acid - the reaction is shown below - using $\ce{H2SO4}$ as a catalyst. However, as the title suggests the synthesis failed as I used too much $\ce{H2SO4}$ - approximately four times more than the prescribed volume. Needless to say, I had to redo the synthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IH3o6.png&quot; alt=&quot;Synthesis of aspirin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I try to understand why the synthesis failed - I know that catalysts are used in small amounts, but I thought that much of it would only lead to equilibrium state being reached faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that maybe the acidity of the reaction environment could've somehow explained why the synthesis didn't take place. But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="23561" LastEditDate="2019-03-19T02:39:34.417" LastActivityDate="2019-03-19T02:39:34.417" Title="Failed synthesis of aspirin due to excess sulfuric acid" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29680" PostTypeId="2" ParentId="29673" CreationDate="2015-05-01T16:55:45.420" Score="4" Body="&lt;p&gt;Let's answer three ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Specific expected math&lt;/strong&gt;  Important mathematical skills will include algebra (including understanding variables and constants conceptually), interpreting graphs, and logorithms. There is very little chance you will need to know calculus in first year chemistry. If you have specific concerns, like those described in the comments, your advisor or department assistant should be able to directly address those.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;First year chem philosophy&lt;/strong&gt;  A good goal for first year chemistry is to understand fundamental chemistry concepts so that you can apply them in future courses and work. To do this, you need to know enough math that it doesn't prohibit you from understanding the concepts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Continuing Education&lt;/strong&gt; This one is just my opinion and I'm sure many will disagree. If you want to work in chemistry, take more math and physics than the department requires. It will give you a more thorough and complete understanding of chemistry. It will give you an edge in the future in solving interesting problems. This can be done in parallel however- the department won't expect that you're doing this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bonus: As many people before me have suggested, go to 6-8 classes on syllabus day even if you're only signed up and intending to take 4. Its always better to know your options and theres no risk to going and seeing whats available. In this case, you'll learn if you've got the background the professor expects for this course and can compare it with other options.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2015-05-01T16:55:45.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29682" PostTypeId="2" ParentId="29656" CreationDate="2015-05-01T17:09:37.357" Score="3" Body="&lt;p&gt;D-glucose stays D-glucose regardless of whether it is in chair (hemiacetal) form, open chain form, or any other minor form. These different forms (if you could isolate them) would likely have different specific rotation values (interaction with plane polarized light). There are two chair forms-one with an axial OH and the other equatorial (probably favored)--but both are forms of D-glucose.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-01T17:09:37.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29685" PostTypeId="1" CreationDate="2015-05-01T17:55:38.793" Score="7" ViewCount="18786" Body="&lt;p&gt;I bought a canned pineapple and finished the content. The part of the can that is in contact with the liquid appears like galvanized steel (crystalline surface or spangles sort of appearance), while the part that are not looks pretty shiny like the outside of the can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then washed the can and added water to it, and heated the bottom of the can with a butane torch. After the water was heated up, I removed the torch and left it out for a while. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I came back later, I emptied the water but noticed that the bottom rim of can had quite a bit of rusting. After inspecting it closely, I found that there are spots of (concentrated) rust on the cylindrical wall of the can as well. There are also light patches of rust on the inside bottom of the can. The seam on the cylindrical wall that is in contact with water is also badly rusted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought galvanised steel only rusts when the layer of galvanisation wears out.&#xA;So, is the tin made with galvanised steel?&lt;br&gt;&#xA;If yes, why are there concentrated spots of rust on the wall and other rusting?&lt;br&gt;&#xA;If not, why does the inside of the can look like galvanised steel?&lt;br&gt;&#xA;And BTW, why does the emptied and washed can smell &quot;metallish&quot;?&lt;/p&gt;&#xA;" OwnerUserId="15462" LastEditorUserId="5354" LastEditDate="2015-05-04T16:52:12.467" LastActivityDate="2020-12-08T17:37:06.257" Title="Why is a tin can that appears like galvanised steel rusting?" Tags="&lt;water&gt;&lt;metal&gt;&lt;heat&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29686" PostTypeId="2" ParentId="29673" CreationDate="2015-05-01T18:00:38.273" Score="2" Body="&lt;p&gt;There is relatively little mathematics &lt;strong&gt;&lt;em&gt;required&lt;/em&gt;&lt;/strong&gt; for a typical first year chemistry course beyond what most will have studied at school. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solid understanding of algebra, trigonometry and diffrentiation/integration is necessary. More advanced topics which may be useful include partial derivatives, solving simple first and second order differential equations, solving partial differential equations, group theory, Fourier transforms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appreciation of group theory in particular is important for a good understanding of spectroscopy and quantum mechanics. A good (but difficult!) introduction to this topic with suitable examples is &lt;em&gt;Chemical Applications of Group Theory&lt;/em&gt; by F.A. Cotton. For the other mathematics you might find &lt;em&gt;Mathematics for Engineers and Scientists&lt;/em&gt; by Alan Jeffrey useful, although there are lots of other texts which cover this sort of material well. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-01T18:00:38.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29688" PostTypeId="1" AcceptedAnswerId="33106" CreationDate="2015-05-01T18:53:06.230" Score="13" ViewCount="12264" Body="&lt;p&gt;In spectroscopy we described the electric energy with the approximative separability of internal motions as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;E=E_e+E_v+E_r+E_{ns}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(energies: electronic, vibratory, rotatory; nuclear spin (neclected))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the Born Oppenheimer approximation of nuclei and electrons you get a formula which describes the degrees of freedoms for vibration motions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3N-5 for linear molecules&lt;/li&gt;&#xA;&lt;li&gt;3N-6 for non-linear molecules&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This formula can be understood by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the coordinates of 3N atoms.&lt;/li&gt;&#xA;&lt;li&gt;3 degrees of motions for the three translation of the centre-of-mass motion&lt;/li&gt;&#xA;&lt;li&gt;2 (for linear) respective 3 (for non-linear) degrees of freedoms reduced for the rotation of the whole molecule&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I guess trans buta-1,3-diene is here described as a linear molecule. [added: in the rigid rotor approximation]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I was thinking why we are so much interested on the number of freedoms for vibrations. We consider only the rotation of the whole molecule but not around bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do we not consider rotation about a bond axis, e.g. for pentane a rotation of an ethyle around the propyle group?&lt;/strong&gt; (rotation about the C(2)-C(3) axis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be because it's much another energy range as rotational energy of about 1-100 cm&lt;sup&gt;-1&lt;/sup&gt; (typical period of 0.1-10 ps).&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7201" LastEditDate="2015-05-02T16:25:01.730" LastActivityDate="2015-06-20T09:40:57.013" Title="Rotational degrees of freedom (3N-5 and 3N-6)" Tags="&lt;spectroscopy&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29690" PostTypeId="5" CreationDate="2015-05-01T19:38:21.867" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-01T19:38:21.867" LastActivityDate="2015-05-01T19:38:21.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29691" PostTypeId="4" CreationDate="2015-05-01T19:38:21.867" Score="0" Body="A colloid is a substance in which microscopically dispersed insoluble particles are suspended throughout another substance." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-01T20:43:50.100" LastActivityDate="2015-05-01T20:43:50.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29692" PostTypeId="5" CreationDate="2015-05-01T19:40:07.767" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-01T19:40:07.767" LastActivityDate="2015-05-01T19:40:07.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29693" PostTypeId="4" CreationDate="2015-05-01T19:40:07.767" Score="0" Body="Questions about the application of chemistry to interrogate and modify biological systems and processes." OwnerDisplayName="user15489" LastEditorUserId="4325" LastEditDate="2015-06-23T01:27:36.753" LastActivityDate="2015-06-23T01:27:36.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29694" PostTypeId="5" CreationDate="2015-05-01T19:49:25.073" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-01T19:49:25.073" LastActivityDate="2015-05-01T19:49:25.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29695" PostTypeId="4" CreationDate="2015-05-01T19:49:25.073" Score="0" Body="For questions about the group 15 elements — nitrogen, phosphorus, arsenic, antimony, and bismuth: their synthesis, or physical and chemical properties." OwnerDisplayName="user15489" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2019-02-20T02:28:35.853" LastActivityDate="2019-02-20T02:28:35.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29696" PostTypeId="1" CreationDate="2015-05-01T22:54:55.877" Score="1" ViewCount="144" Body="&lt;p&gt;I've been working on another CIF file, this time for (-)-&lt;em&gt;threo&lt;/em&gt;-methylphenidate based on &lt;em&gt;&lt;a href=&quot;http://link.springer.com/article/10.1023%2FA%3A1016262815984&quot; rel=&quot;nofollow&quot;&gt;Pharma Res, 1995&lt;/a&gt;&lt;/em&gt; (sadly the relevant crystallography data isn't supplied in the free preview at Springer). I have several questions relating to this file, the most pressing is that so far I have this loop:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;loop_&#xA;_atom_site_label&#xA;_atom_site_chemical_conn_number&#xA;_atom_site_fract_x&#xA;_atom_site_fract_y&#xA;_atom_site_fract_z&#xA;_atom_site_adp_type&#xA;_atom_site_B_iso_or_equiv&#xA;_atom_site_type_symbol&#xA;N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N&#xA;H1a  2    -0.9650000 0.08500000 -1.0010000             H&#xA;H1b  3    -0.8800000 -0.0800000 -1.0330000             H&#xA;C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C&#xA;H2   5    -0.7460000 -0.0800000 -0.9360000             H&#xA;C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;that is generating this error in enCIFer:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Too many or too few data values in the loop&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which I'm going to take a stab in the dark and say that it's because two fields (Biso &amp;amp; $B_{eq}$ value) aren't supplied for each atom in this loop (namely it is missing from the hydrogen atoms). The problem is that the paper I cited didn't specify these for hydrogen atoms. Should I use separate loops for hydrogen &amp;amp; other atoms (i.e., one loop for hydrogens, another for the other atoms)? I ask because I fear that this might make molecular modelling software treat the hydrogens &amp;amp; the other atoms as being part of separate molecules. &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2016-02-23T14:12:12.980" Title="CIF fields that are available for some atoms, but not others" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29697" PostTypeId="1" CreationDate="2015-05-01T22:55:14.150" Score="3" ViewCount="41830" Body="&lt;p&gt;To be honest, I just don't get it.&#xA;I had this question while reading a Chemguide article, on this topic:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/AMrse.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="16683" LastEditDate="2015-11-13T20:34:38.307" LastActivityDate="2017-04-13T21:00:01.687" Title="Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29699" PostTypeId="2" ParentId="29697" CreationDate="2015-05-01T23:21:20.077" Score="3" Body="&lt;p&gt;Look at the following equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the &lt;em&gt;half equivalence point&lt;/em&gt;, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests (&quot;half equivalence&quot;). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[WA]~ =~ [CB]}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}\mathrm{pH} &amp;amp;= \mathrm pK_{\mathrm a} + \log 1\\  &amp;amp;=\mathrm pK_{\mathrm a}\;.\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Vocabulary&lt;/strong&gt;: &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;SB : Strong Base. (eg. $\ce{NaOH}$)&lt;/li&gt;&#xA;  &lt;li&gt;CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)&lt;/li&gt;&#xA;  &lt;li&gt;WA : Weak Acid. (eg. $\ce{HAc}$)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorDisplayName="user5764" LastEditDate="2016-06-22T06:48:20.793" LastActivityDate="2016-06-22T06:48:20.793" CommentCount="9" CommunityOwnedDate="2015-05-01T23:33:12.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="29700" PostTypeId="1" CreationDate="2015-05-01T23:24:26.700" Score="1" ViewCount="2737" Body="&lt;p&gt;I'm looking for a chemical reaction that produces a large amount of a gas in a very short amount of time (like less than a second). What are my options?&lt;/p&gt;&#xA;" OwnerUserId="15934" LastActivityDate="2015-05-02T06:17:14.140" Title="What reactions create a lot of gas really quickly?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29701" PostTypeId="1" CreationDate="2015-05-01T23:58:21.920" Score="2" ViewCount="1160" Body="&lt;p&gt;Here is a problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Titrate $1 M$ sulfuric acid with $\pu{50 ml}$ of $1 M$ sodium hydroxide solution. What volume of sulfuric acid will be required for neutralization.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a simple problem. My reasoning is since both reactants are strong we will have $\pu{0.05 L}$ $\ce{OH-}$ ions and for neutralization we need an equal number of $\ce{H+}$ ions. Since $\ce{H2SO4}$ is a strong acid and ignoring the second dissociation which is weak, we get an equal number of $\ce{H+}$ as $\ce{H2SO4}$ ions. Since concentrations are equal, the volume is simply $\pu{50 mL}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently this is wrong. I think the source of confusion is what is meant by &quot;required for neutralization&quot;. If you write out the molecular equation for this reaction and do stoichiometry you get a different volume.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorUserId="5026" LastEditDate="2018-02-18T04:18:02.963" LastActivityDate="2018-02-18T04:18:02.963" Title="Calculation of volume required for titration with dibasic acid" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29702" PostTypeId="1" CreationDate="2015-05-02T00:16:45.913" Score="1" ViewCount="7930" Body="&lt;p&gt;When solving the lewis structure for the $\ce{ClO2-}$ ion, taking into consideration formal charges, the structure is represented as:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4VLc.png&quot; alt=&quot;Chlorite ion lewis structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A lewis model with &lt;strong&gt;2 double bonds&lt;/strong&gt; also fits the formal charge and lewis model requirements, however this time the negative formal charge would be on the chlorine atom. Is there a reason this form is not favoured? &lt;/p&gt;&#xA;" OwnerUserId="15935" LastEditorUserId="15410" LastEditDate="2015-05-02T01:41:05.657" LastActivityDate="2015-05-02T01:41:05.657" Title="Lewis structure of chlorite ion" Tags="&lt;lewis-structure&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-02T16:05:03.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="29703" PostTypeId="2" ParentId="29702" CreationDate="2015-05-02T00:46:11.750" Score="2" Body="&lt;p&gt;When drawing structures and taking into consideration formal charge, negative charges are assigned to the most electronegative atoms. Since the electronegativity of chlorine is less than the electronegativity of oxygen, it follows that the model with 2 double bounds is less favorable. &lt;/p&gt;&#xA;" OwnerUserId="13368" LastActivityDate="2015-05-02T00:46:11.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29704" PostTypeId="1" CreationDate="2015-05-02T00:59:05.237" Score="4" ViewCount="122" Body="&lt;p&gt;Turbidity is essentially the clarity of water. Turbidity can be measured in many ways such as a turbidimeter, secci disk or a turbidity tube (which incorporates a secci disk at the bottom). If you had a small turbid water sample - 20mL, how could you measure the turbidity without an electronic device??&lt;/p&gt;&#xA;" OwnerUserId="15811" LastEditorUserId="21964" LastEditDate="2016-07-23T01:04:07.907" LastActivityDate="2016-08-22T01:52:10.100" Title="Measurement of Water Clarity" Tags="&lt;precipitation&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29705" PostTypeId="1" AcceptedAnswerId="29709" CreationDate="2015-05-02T01:40:48.417" Score="0" ViewCount="896" Body="&lt;p&gt;I know it's probably a stupid question, but I can not figure this out. I have a liebig condenser that I usually use for distillation with a Kjeldahl bulb, which makes it easy to attach to a flask because the bulb has a stopper on both ends? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have an experiment that calls for reflux heating, where the condenser needs to be held vertically over the flask. One end of the condenser is a cork-fit opening, and the other end is a spout. The flask also has a cork-fit opening.&lt;/p&gt;&#xA;" OwnerUserId="13396" LastEditorUserId="5456" LastEditDate="2015-05-02T06:15:02.347" LastActivityDate="2016-05-28T02:43:40.987" Title="How to attach a liebig condenser to a flask for reflux heating?" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29706" PostTypeId="1" AcceptedAnswerId="29713" CreationDate="2015-05-02T01:42:50.463" Score="1" ViewCount="338" Body="&lt;p&gt;I previously asked a question about a CIF file here (namely &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29696/cif-fields-that-are-available-for-some-atoms-but-not-others&quot;&gt;CIF fields that are available for some atoms, but not others&lt;/a&gt;) I think I have the answer (namely to place dots [.] at places where the two missing terms would normally appear), although I felt I'd wait to see if anyone answers in case I am wrong. Presently I have this CIF file:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;#=====================================================================&#xA;# X-ray crystallography data from Pharma Res, 1995&#xA;#---------------------------------------------------------------------&#xA;&#xA;data_methylphenidate&#xA;&#xA;_cell_length_a                9.401(3)&#xA;_cell_length_b                7.3053(7)&#xA;_cell_length_c                11.113(3)&#xA;_cell_angle_alpha             90&#xA;_cell_angle_beta              109.260(2)&#xA;_cell_angle_gamma             90&#xA;_cell_formula_units_Z         2&#xA;_cell_volume                  720.5(3)&#xA;_cell_measurement_temperature 295.15&#xA;_cell_measurement_reflns_used 1797&#xA;&#xA;_chemical_formula_sum         'C14 H20 Cl N O2'&#xA;_chemical_formula_weight      269.77&#xA;_chemical_name_systematic     'methyl (2S)-2-phenyl-2-[(2S)-piperidin-2-yl]acetate;hydrochloride'&#xA;_chemical_identifier_inchi    'InChI=1S/C14H19NO2.ClH/c1-17-14(16)13(11-7-3-2-4-8-11)12-9-5-6-10-15-12;/h2-4,7-8,12-13,15H,5-6,9-10H2,1H3;1H/t12-,13-;/m0./s1'&#xA;_chemical_identifier_inchi_key 'JUMYIBMBTDDLNG-QNTKWALQSA-N'&#xA;&#xA;_refine_ls_R_factor_gt        0.070&#xA;&#xA;_space_group_crystal_system   monoclinic&#xA;_space_group_name_H-M_alt     'P 21'&#xA;_space_group_IT_number        4&#xA;&#xA;loop_&#xA;_symmetry_equiv_pos_as_xyz&#xA;'x, y, z'&#xA;'-x, y+1/2, -z'&#xA;&#xA;loop_&#xA;_atom_site_label&#xA;_atom_site_chemical_conn_number&#xA;_atom_site_cartn_x&#xA;_atom_site_cartn_y&#xA;_atom_site_cartn_z&#xA;_atom_site_adp_type&#xA;_atom_site_B_iso_or_equiv&#xA;_atom_site_type_symbol&#xA;N1   1  -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2)   N&#xA;H1a  2  -0.965     0.0850     -1.001     .      .      H&#xA;H1b  3  -0.880     -0.080     -1.033     .      .      H&#xA;C2   4  -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2)   C&#xA;H2   5  -0.746     -0.080     -0.936     .      .      H&#xA;C3   6  -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3)   C&#xA;H3a  7  -0.525     0.0380     -0.933     .      .      H&#xA;H3b  8  -0.603     0.2210     -0.990     .      .      H&#xA;C4   9  -0.640(1)  0.021(2)   -1.122(1)  Biso 5.0(3)   C&#xA;H4a  10 -0.639     -0.110     -1.119     .      .      H&#xA;H4b  11 -0.561     0.0640     -1.150     .      .      H&#xA;C5   12 -0.792(2)  0.086(1)   -1.213(1)  Biso 5.3(3)   C&#xA;H5a  13 -0.786     0.2150     -1.225     .      .      H&#xA;H5b  14 -0.812     0.0250     -1.291     .      .      H&#xA;C6   15 -0.915(1)  0.049(2)   -1.160(1)  Biso 4.3(2)   C&#xA;H6a  16 -1.004     0.1050     -1.214     .      .      H&#xA;H6b  17 -0.928     -0.080     -1.159     .      .      H&#xA;C7   18 -0.721(1)  0.124(1)   -0.809(1)  Biso 3.3(2)   C&#xA;H7   19 -0.737     0.2520     -0.817     .      .      H&#xA;C8   20 -0.834(1)  0.051(2)   -0.7527(9) Biso 3.8(2)   C&#xA;O9   21 -0.889(1)  0.172(1)   -0.6908(9) Biso 4.9(2)   O&#xA;C10  22 -0.998(2)  0.113(2)   -0.634(1)  Biso 5.8(4)   C&#xA;H10a 23 -1.028     0.2150     -0.594     .      .      H&#xA;H10b 24 -0.956     0.0210     -0.572     .      .      H&#xA;H10c 25 -1.085     0.0650     -0.698     .      .      H&#xA;C11  26 -0.563(1)  0.093(1)   -0.709(1)  Biso 3.9(3)   C&#xA;C12  27 -0.509(2)  -0.091(2)  -0.681(1)  Biso 5.6(3)   C&#xA;H12  28 -0.566     -0.192     -0.726     .      .      H&#xA;C13  29 -0.376(2)  -0.114(3)  -0.590(2)  Biso 7.2(4)   C&#xA;H13  30 -0.343     -0.238     -0.572     .      .      H&#xA;C14  31 -0.290(2)  0.009(6)   -0.522(2)  Biso 11.3(8)  C&#xA;H14  32 -0.1960    -0.018     -0.460     .      .      H&#xA;C15  33 -0.343(2)  0.204(4)   -0.548(2)  Biso 9.0(6)   C&#xA;H15  34 -0.284     0.3020     -0.501     .      .      H&#xA;C16  35 -0.479(2)  0.242(2)   -0.643(1)  Biso 5.9(3)   C&#xA;H16  36 -0.515     0.3640     -0.664     .      .      H&#xA;O3   37 -0.871(1)  -0.1124(9) -0.7590(9) Biso 5.7(2)   O&#xA;Cl   38 -0.8105(3) 0.5484     -1.0089(3) Biso 4.72(6)  Cl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;. The problem is that Avogadro and enCIFer are having trouble rendering it and I don't see what the problem is. e.g., this is what Avogadro is giving me at the moment:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9xzXv.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Are &lt;code&gt;_cell_measurement_theta_min&lt;/code&gt; and &lt;code&gt;_cell_measurement_theta_max&lt;/code&gt; the problem (namely that I haven't specified them)?&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-02T22:40:02.403" Title="Is there any required data missing from my CIF file?" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29707" PostTypeId="2" ParentId="29700" CreationDate="2015-05-02T02:08:33.980" Score="3" Body="&lt;p&gt;The reaction used for car airbags involves sodium azide (NaN&lt;sub&gt;3&lt;/sub&gt;). A spark initiates the reaction and it generates gaseous nitrogen really fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much more about it but you can certainly find some good informations on google!&lt;/p&gt;&#xA;" OwnerUserId="15936" LastEditorUserId="5456" LastEditDate="2015-05-02T06:17:14.140" LastActivityDate="2015-05-02T06:17:14.140" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="29708" PostTypeId="1" CreationDate="2015-05-02T02:16:15.840" Score="1" ViewCount="34" Body="&lt;p&gt;The larger purpose of this question is the engineering of liquid filled sacs that change their volume according to heat. They'd need to have diverse ranges of volume, at least a factor of ten (a 10 ml to 100ml transition).&lt;/p&gt;&#xA;" OwnerUserId="15937" LastActivityDate="2015-05-02T02:16:15.840" Title="Are there any (liquid) chemicals that experience rapid, extensive, and controllable volume changes in reaction to heat?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29709" PostTypeId="2" ParentId="29705" CreationDate="2015-05-02T02:42:43.843" Score="2" Body="&lt;p&gt;Maybe I don't understand your question well, but often in the lab when we do reflux heating, we don't really attach the liebig condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You fix the flask using a clamp. And gravity holds the liebig condenser into place. You can put a clamp around it to hold it straight but you'd better not tighten it too much because these things are fragile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it helps you :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's not clear maybe it would be easier to answer if you posted pictures of your glassware!&lt;/p&gt;&#xA;" OwnerUserId="15936" LastActivityDate="2015-05-02T02:42:43.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="29710" PostTypeId="2" ParentId="29701" CreationDate="2015-05-02T02:48:12.407" Score="1" Body="&lt;p&gt;We first observe the balanced equation. $$\ce {H2SO4 + 2NaOH &amp;lt;=&amp;gt; Na2SO4 + 2H2O}$$ we see the ratio of $\ce {NaOH}$ and $\ce {H2SO4}$ is 2:1. Given $ 0.050~\mathrm{L}$ of $\ce {1M~NaOH}$ we find moles of $\ce {NaOH}$ and use stoichiometry to find how many moles of $\ce {H2SO4}$ is needed. We find the volume to be $25~\mathrm{ml}$.   &lt;/p&gt;&#xA;" OwnerUserId="13368" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T07:22:54.227" LastActivityDate="2015-05-02T07:22:54.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29711" PostTypeId="1" AcceptedAnswerId="31096" CreationDate="2015-05-02T02:31:20.807" Score="5" ViewCount="282" Body="&lt;p&gt;I was once told on this site that it was incorrect form to use units that specify the chemical being referred to in dimensional analysis. For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$150.~\mathrm{g}~~\ce{KNO3} \cdot \frac{1~\mathrm{mol}~~\ce{KNO3}}{101.103~\mathrm{g}~~\ce{KNO3}} \cdot{} \frac{1~\mathrm{mol}~~\ce{C7H4O}}{6~\mathrm{mol}~~\ce{KNO3}} \cdot \frac{104.106~\mathrm{g}~~\ce{C7H4O}}{1~\mathrm{mol}~~\ce{C7H4O}}=25.7~\mathrm{g}~~\ce{C7H4O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the way I learned it, and I feel that this clears up confusion that might otherwise arise when from performing such a calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My overall question is: is it acceptable to do dimensional analysis as I presented it? If not, please explain why it is not correct to do it this way and show me what I should be doing instead.&lt;/p&gt;&#xA;" OwnerUserId="15410" OwnerDisplayName="ringo" LastEditorUserId="73" LastEditDate="2015-05-02T03:20:05.897" LastActivityDate="2015-05-09T12:48:00.860" Title="Dimensional Analysis in Chemistry" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29713" PostTypeId="2" ParentId="29706" CreationDate="2015-05-02T04:07:32.803" Score="2" Body="&lt;p&gt;Mamun's answer enabled me to find the fault in my CIF file. I was using cartn categories for fract data. Here's the correct file: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#=====================================================================&#xA;# X-ray crystallography data from Pharma Res, 1995&#xA;#---------------------------------------------------------------------&#xA;&#xA;data_methylphenidate&#xA;&#xA;_cell_length_a                9.401(3)&#xA;_cell_length_b                7.3053(7)&#xA;_cell_length_c                11.113(3)&#xA;_cell_angle_alpha             90&#xA;_cell_angle_beta              109.260(2)&#xA;_cell_angle_gamma             90&#xA;_cell_formula_units_Z         2&#xA;_cell_volume                  720.5(3)&#xA;_cell_measurement_temperature 295.15&#xA;_cell_measurement_reflns_used 1797&#xA;&#xA;_chemical_formula_sum         'C14 H20 Cl N O2'&#xA;_chemical_formula_weight      269.77&#xA;_chemical_name_systematic     'methyl (2S)-2-phenyl-2-[(2S)-piperidin-2-yl]acetate;hydrochloride'&#xA;_chemical_identifier_inchi    'InChI=1S/C14H19NO2.ClH/c1-17-14(16)13(11-7-3-2-4-8-11)12-9-5-6-10-15-12;/h2-4,7-8,12-13,15H,5-6,9-10H2,1H3;1H/t12-,13-;/m0./s1'&#xA;_chemical_identifier_inchi_key 'JUMYIBMBTDDLNG-QNTKWALQSA-N'&#xA;&#xA;_refine_ls_R_factor_gt        0.070&#xA;&#xA;_space_group_crystal_system   monoclinic&#xA;_space_group_name_H-M_alt     'P 21'&#xA;_space_group_IT_number        4&#xA;&#xA;loop_&#xA;_symmetry_equiv_pos_as_xyz&#xA;'x, y, z'&#xA;'-x, y+1/2, -z'&#xA;&#xA;loop_&#xA;_atom_site_label&#xA;_atom_site_occupancy&#xA;_atom_site_chemical_conn_number&#xA;_atom_site_fract_x&#xA;_atom_site_fract_y&#xA;_atom_site_fract_z&#xA;_atom_site_adp_type&#xA;_atom_site_B_iso_or_equiv&#xA;_atom_site_type_symbol&#xA;N1   1.0  1  -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2)   N&#xA;H1a  1.0  2  -0.965     0.0850     -1.001     .      .      H&#xA;H1b  1.0  3  -0.880     -0.080     -1.033     .      .      H&#xA;C2   1.0  4  -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2)   C&#xA;H2   1.0  5  -0.746     -0.080     -0.936     .      .      H&#xA;C3   1.0  6  -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3)   C&#xA;H3a  1.0  7  -0.525     0.0380     -0.933     .      .      H&#xA;H3b  1.0  8  -0.603     0.2210     -0.990     .      .      H&#xA;C4   1.0  9  -0.640(1)  0.021(2)   -1.122(1)  Biso 5.0(3)   C&#xA;H4a  1.0  10 -0.639     -0.110     -1.119     .      .      H&#xA;H4b  1.0  11 -0.561     0.0640     -1.150     .      .      H&#xA;C5   1.0  12 -0.792(2)  0.086(1)   -1.213(1)  Biso 5.3(3)   C&#xA;H5a  1.0  13 -0.786     0.2150     -1.225     .      .      H&#xA;H5b  1.0  14 -0.812     0.0250     -1.291     .      .      H&#xA;C6   1.0  15 -0.915(1)  0.049(2)   -1.160(1)  Biso 4.3(2)   C&#xA;H6a  1.0  16 -1.004     0.1050     -1.214     .      .      H&#xA;H6b  1.0  17 -0.928     -0.080     -1.159     .      .      H&#xA;C7   1.0  18 -0.721(1)  0.124(1)   -0.809(1)  Biso 3.3(2)   C&#xA;H7   1.0  19 -0.737     0.2520     -0.817     .      .      H&#xA;C8   1.0  20 -0.834(1)  0.051(2)   -0.7527(9) Biso 3.8(2)   C&#xA;O9   1.0  21 -0.889(1)  0.172(1)   -0.6908(9) Biso 4.9(2)   O&#xA;C10  1.0  22 -0.998(2)  0.113(2)   -0.634(1)  Biso 5.8(4)   C&#xA;H10a 1.0  23 -1.028     0.2150     -0.594     .      .      H&#xA;H10b 1.0  24 -0.956     0.0210     -0.572     .      .      H&#xA;H10c 1.0  25 -1.085     0.0650     -0.698     .      .      H&#xA;C11  1.0  26 -0.563(1)  0.093(1)   -0.709(1)  Biso 3.9(3)   C&#xA;C12  1.0  27 -0.509(2)  -0.091(2)  -0.681(1)  Biso 5.6(3)   C&#xA;H12  1.0  28 -0.566     -0.192     -0.726     .      .      H&#xA;C13  1.0  29 -0.376(2)  -0.114(3)  -0.590(2)  Biso 7.2(4)   C&#xA;H13  1.0  30 -0.343     -0.238     -0.572     .      .      H&#xA;C14  1.0  31 -0.290(2)  0.009(6)   -0.522(2)  Biso 11.3(8)  C&#xA;H14  1.0  32 -0.1960    -0.018     -0.460     .      .      H&#xA;C15  1.0  33 -0.343(2)  0.204(4)   -0.548(2)  Biso 9.0(6)   C&#xA;H15  1.0  34 -0.284     0.3020     -0.501     .      .      H&#xA;C16  1.0  35 -0.479(2)  0.242(2)   -0.643(1)  Biso 5.9(3)   C&#xA;H16  1.0  36 -0.515     0.3640     -0.664     .      .      H&#xA;O3   1.0  37 -0.871(1)  -0.1124(9) -0.7590(9) Biso 5.7(2)   O&#xA;Cl   1.0  38 -0.8105(3) 0.5484     -1.0089(3) Biso 4.72(6)  Cl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2015-05-02T22:40:02.403" LastActivityDate="2015-05-02T22:40:02.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29714" PostTypeId="5" CreationDate="2015-05-02T05:21:41.317" Score="0" Body="&lt;h2&gt;Wikipedia Entry:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Vapor pressure or equilibrium vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. The pressure exhibited by vapor present above a liquid surface is known as vapor pressure. As the temperature of a liquid increases, the kinetic energy of its molecules also increases. As the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How do I use this tag?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Apply this tag to questions involving clarification related to vapor pressure.&lt;/p&gt;&#xA;" OwnerUserId="42741" LastEditorUserId="42741" LastEditDate="2017-06-12T10:34:08.340" LastActivityDate="2017-06-12T10:34:08.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29715" PostTypeId="4" CreationDate="2015-05-02T05:21:41.317" Score="0" Body="This is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. Apply this tag to questions involving clarification related to vapor pressure." OwnerDisplayName="user15489" LastEditorUserId="42741" LastEditorDisplayName="user15489" LastEditDate="2017-06-12T13:05:22.263" LastActivityDate="2017-06-12T13:05:22.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29716" PostTypeId="1" AcceptedAnswerId="29725" CreationDate="2015-05-02T05:22:02.027" Score="4" ViewCount="2289" Body="&lt;p&gt;Is there an equation to figure out what the temperature of an the area will be a certain distance from the heat source? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I know there is a heat source at $0~\mathrm{m}$, and the temperature is $100~\mathrm{^\circ C}$, is there a way to find out what the temperature will be at $50~\mathrm{m}$ away from the heat source? Is there some sort of equation for this? All of this is happening in oxygen.&lt;/p&gt;&#xA;" OwnerUserId="15939" LastEditorUserId="16683" LastEditDate="2015-10-23T16:49:33.860" LastActivityDate="2015-10-23T16:49:33.860" Title="Equation for heat loss over distance?" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29717" PostTypeId="5" CreationDate="2015-05-02T05:23:28.133" Score="0" Body="&lt;p&gt;Polarity is the separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-19T04:50:23.783" LastActivityDate="2018-03-19T04:50:23.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29718" PostTypeId="4" CreationDate="2015-05-02T05:23:28.133" Score="0" Body="For questions about polar or non-polar chemical species, or their comparison. Also, for questions where polar or non-polar solvents have an effect." OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-19T04:50:32.820" LastActivityDate="2018-03-19T04:50:32.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29719" PostTypeId="5" CreationDate="2015-05-02T05:25:35.230" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-02T05:25:35.230" LastActivityDate="2015-05-02T05:25:35.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29720" PostTypeId="4" CreationDate="2015-05-02T05:25:35.230" Score="0" Body="Surface chemistry is roughly defined as the study of chemical reactions at interfaces between 2 or more phases." OwnerDisplayName="user15489" LastEditorUserId="9961" LastEditorDisplayName="user15489" LastEditDate="2017-05-10T20:08:40.610" LastActivityDate="2017-05-10T20:08:40.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29721" PostTypeId="5" CreationDate="2015-05-02T05:26:47.720" Score="0" Body="&lt;p&gt;Nanotechnology (&quot;nanotech&quot;) is manipulation of matter on an atomic, molecular, and supramolecular scale.&lt;/p&gt;&#xA;" OwnerUserId="23561" LastEditorUserId="23561" LastEditDate="2019-03-15T06:38:58.047" LastActivityDate="2019-03-15T06:38:58.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29722" PostTypeId="4" CreationDate="2015-05-02T05:26:47.720" Score="0" Body="For questions relating to the synthesis of supramolecular, or colloidal structures and their properties or manufacturing processes of characterization for nano-featured structures." OwnerDisplayName="user15489" LastEditorUserId="23561" LastEditorDisplayName="user15489" LastEditDate="2019-03-15T06:39:17.087" LastActivityDate="2019-03-15T06:39:17.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29723" PostTypeId="5" CreationDate="2015-05-02T05:28:54.933" Score="0" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Precipitation_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Precipitation is the creation of a solid from a solution. When the reaction occurs in a liquid solution, the solid formed is called the 'precipitate'. The chemical that causes the solid to form is called the 'precipitant'. Without sufficient force of gravity (settling) to bring the solid particles together, the precipitate remains in suspension&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-17T18:28:01.623" LastActivityDate="2018-03-17T18:28:01.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29724" PostTypeId="4" CreationDate="2015-05-02T05:28:54.933" Score="0" Body="For questions related to the chemical reactions and processes that result in a chemical reaching its solid phase." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T06:06:15.940" LastActivityDate="2015-05-02T06:06:15.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29725" PostTypeId="2" ParentId="29716" CreationDate="2015-05-02T05:38:26.957" Score="4" Body="&lt;p&gt;This is a heat transfer related problem. You can certainly do that. For solid medium heat transfer you need to use Fourier heat transfer equation&lt;br&gt;&#xA;\[q=-k\cdot\nabla T\]&lt;br&gt;&#xA;Here, k= material's conductivity and $\nabla T$ = 1st derivative of the Temp. along the driving force or heat&lt;br&gt;&#xA;For Liquid and gas phase you should use law for thermal convection. And for vacuum medium you should use radiation law. There are lots of ways to do this but from your problem statement i think it's a numerical grid type problem. For your better understanding, i would suggest 'Engineering heat transfer' by william s janna (I may be biased here because i like this book) but you can certainly use any other standard heat transfer book or internet.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="418" LastEditDate="2015-05-02T05:49:45.140" LastActivityDate="2015-05-02T05:49:45.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="29727" PostTypeId="1" AcceptedAnswerId="30766" CreationDate="2015-05-02T08:26:51.670" Score="5" ViewCount="1296" Body="&lt;p&gt;I always receive two type of explanation for the decreasement of atomic size for elements across the period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation according to proton number&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation is from my teacher. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The proton number of elements increases across the period.&lt;/li&gt;&#xA;&lt;li&gt;The force of attraction between the proton in the nucleus and the electrons is stronger across the period as the proton will pull the electrons towards the nucleus.&lt;/li&gt;&#xA;&lt;li&gt;So, the atomic size of the elements will decrease across the period&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation according to number of valance electron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation is from my friends&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The valance electron of elements increases across the period.&lt;/li&gt;&#xA;&lt;li&gt;The force of attraction between the valance electrons and the proton in the nucleus is stronger across the period between the valance electrons is pulled towards to the nucleus&lt;/li&gt;&#xA;&lt;li&gt;So, the atomic size of the elements will decrease across the period.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is which explanation is more suitable for the statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the atomic size of the elements decrease across the period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-05-03T05:07:13.370" Title="The explanation on atomic size" Tags="&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="29728" PostTypeId="1" CreationDate="2015-05-02T10:27:42.977" Score="1" ViewCount="3552" Body="&lt;p&gt;I know that a standard solution in volumetric analysis must have an accurately known solution. Primary standard solutions (e.g. anhydrous sodium carbonate) can be used straight-away as they follow the list of criteria (high level of purity, unaffected by air exposure, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that secondary standard solutions (hydrochloric acid) cannot be used directly like primary standard solutions as they don't follow the criteria. &lt;strong&gt;But I don't understand how standardising hydrochloric acid with a primary solution (e.g. anhydrous sodium carbonate) will make it able to be used as a standard solution.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6988" LastActivityDate="2017-11-22T13:16:16.710" Title="How does standardising a secondary standard solution make it suitable as a standard solution?" Tags="&lt;analytical-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29729" PostTypeId="1" AcceptedAnswerId="29732" CreationDate="2015-05-02T10:30:17.157" Score="20" ViewCount="6307" Body="&lt;p&gt;According to the UC Davis ChemWiki &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_18%3A_The_Noble_Gases/Chemistry_of_Helium&quot;&gt;Chemistry of Helium&lt;/a&gt;, helium has a comparatively unusual property, specifically:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Ordinary' referring to standard air pressure (1 atmosphere).  In order to solidify, there needs to be a corresponding pressure increase, with a projected density of:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$0.187 \pm 0.009~\mathrm{g~mL^{-1}}$ at $0~\mathrm{K}$ and $25~\mathrm{bar}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why can't helium be solidified at 'ordinary' pressures?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="15410" LastEditDate="2015-05-13T02:03:00.743" LastActivityDate="2016-09-29T13:55:52.333" Title="Why can't helium be solidified at 'ordinary' pressures?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29730" PostTypeId="5" CreationDate="2015-05-02T10:32:36.070" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-02T10:32:36.070" LastActivityDate="2015-05-02T10:32:36.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29731" PostTypeId="4" CreationDate="2015-05-02T10:32:36.070" Score="0" Body="Questions about compounds that lower the surface tension (or interfacial tension) between two liquids or between a liquid and a solid." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T11:53:26.977" LastActivityDate="2015-05-02T11:53:26.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29732" PostTypeId="2" ParentId="29729" CreationDate="2015-05-02T11:58:20.897" Score="21" Body="&lt;p&gt;When the temperature of helium gas is decreased to about 5.2 K, a phase transition to ordinary liquid helium ($\ce{He}$ I) occurs. The behavior of this liquid phase is normal and identical to any other liquid phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As temperature is decreased more (at moderate pressures), helium does not solidify. In fact, it undergoes a phase change to a second liquid phase known as $\ce{He}$ II at a temperature of approximately 2.17 K. As shown in the phase diagram below, the $\ce{He}$ II phase persists until absolute zero. Due to quantum effects, $\ce{He}$ II has remarkably unique properties. It is considered a superfluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pymYA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details, please &lt;a href=&quot;http://www2.esm.vt.edu/~macramer/summary/super-f.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;click here&lt;/a&gt;. Pressure is required to produce solid helium (25 atmospheres or more), because of the small mass and extremely weak forces between the helium atoms.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2016-09-29T13:55:52.333" LastActivityDate="2016-09-29T13:55:52.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="29733" PostTypeId="2" ParentId="29661" CreationDate="2015-05-02T14:41:58.337" Score="4" Body="&lt;p&gt;The difference between there acidity arises only due to the difference in inductive effect[or the electron donating capacity of the alkyl group] of the alkyl group attached to the $\ce{C=C}$ in both the cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html&quot; rel=&quot;nofollow&quot;&gt;By this site there exists a difference of total one unit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Inductive_effect&quot; rel=&quot;nofollow&quot;&gt;For more info on Inductive Effect&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative acidity of a compound is checked by removing the acidic hydrogen and then comparing the stability of the anions.The more stable the anion is the faster it is formed and hence the acid is stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anions formed are $\ce{H3C-CH=CH-}$ and  $\ce{H2C=CH-}$ . In the former one the anion or the negative charge on the anion is destabilized by the electron donating ability or the inductive effect of $\ce{CH3-}$, hence it is less stable and less rapidly formed and hence less acidic. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why $\ce{H3C-}$ has electron donating effect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an alkyl group, for example, in methyl group carbon is linked to three hydrogen atoms. Carbon is more electronegative than hydrogen. So carbon pulls bonded electrons nearer to carbon and carbon acquires a slight extra negative charge (delta-) and correspondingly hydrogen lose hold over bonded electrons and acquires a slight positive charge (indicated by delta +). Thus carbon get a little more electronic charge which the carbon will push towards the group to which it is linked.When this electron rich carbon centre (methyl group) attached with other compounds it behaves as electron donor. &#xA;And thus methyl group acts as electron donating group. &#xA;Be in mind, the electronegativity difference between carbon and hydrogen is very less.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3C-CH=CH-}$ is less stable due to presence of above factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use conjugate base method or charge holding capacity method to compare the acidity of compound to give the same result as above. &lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="5608" LastEditDate="2015-05-02T15:18:16.353" LastActivityDate="2015-05-02T15:18:16.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="29734" PostTypeId="2" ParentId="27652" CreationDate="2015-05-02T15:08:09.883" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For any steel with less than 0.02% carbon we should see only ferrite, and for steel with &gt;0.02% carbon we should see only pearlite.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have got a small misunderstanding here. The maximum solubility of carbon at $723\ \mathrm{^\circ C}$ is 0.02%. At room temperature it is very low ($0.005%$ at $0\ \mathrm{^\circ C}$). Also for carbon level between 0.02 and 0.8 you wont see only pearlite, you will see a combination of ferrite and pearlite for sure. Only at 0.8% you can observe pure pearlite(100%). Also note that the phase diagram is in such a way that if you cool from the ferrite+ austenite region at $800\ \mathrm{^\circ C}$ with carbon content &amp;lt; 0.02% we can still get a ferrite + pearlite + even austenite/martensite.&#xA;Phase diagrams give an ideal picture. In real situations, it is very tough to get very slow cooling rates to get the equilibrium structure.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorDisplayName="user7951" LastEditDate="2018-08-28T19:04:56.373" LastActivityDate="2018-08-28T19:04:56.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29736" PostTypeId="1" CreationDate="2015-05-02T18:14:25.680" Score="15" ViewCount="21104" Body="&lt;p&gt;When copper is precipitated out of a solution e.g from CuI, why does it appear pink instead of copper-coloured? Are there other situations where copper appears pink in solid form?&lt;/p&gt;&#xA;" OwnerUserId="15790" LastActivityDate="2015-05-02T19:35:39.263" Title="Why does copper precipitate out as a pink solid?" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30736" PostTypeId="2" ParentId="27622" CreationDate="2015-05-02T18:42:58.197" Score="6" Body="&lt;p&gt;The rule of thumb for hydroxide salts is actually that all hydroxides are insoluble save those of the alkali metals and the heavier alkaline earth metals (namely $\ce{Ca^{2+}}$, $\ce{Sr}^{2+}$, and $\ce{Ba}^{2+}$). There are many resources you can look up to check the solubility rules of certain ions, &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before researching this, I suspected that the solubility of the salts with a certain anion might be determined by the cation's control over its valence electrons, which is measurable through ionization energy. Sure enough, when you compare the &lt;a href=&quot;http://www.lenntech.com/periodic-chart-elements/ionization-energy.htm&quot; rel=&quot;nofollow&quot;&gt;first ionization energies&lt;/a&gt; of various metals, you find that those with the lowest first ionization energies are those that are soluble hydroxides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alkali Metals&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The salts of all alkali metals are completely soluble because they have relatively little control over the outermost electron orbiting them, in addition to this being the only electron they must lose in order to attain an octet structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alkaline Earth Metals&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the alkaline earth metals, the complete outermost $\mathrm{s}$ orbital must be lost in order to attain an octet structure. Because the orbital is complete, and among other reasons, the first ionization will require more energy than the alkali metal in the same period. In addition to this, the second electron must then also be lost. This means that overall, it requires much more energy to ionize this group than the alkali metals. This explains the insolubility of $\ce{Be(OH)2}$ and $\ce{Mg(OH)2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility of the hydroxides of the metals lower in the alkaline earth metal group can be explained by the &lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Physical_Chemistry/Quantum_Mechanics/10:_Multi-electron_Atoms/Multi-Electron_Atoms/Penetration_%26_Shielding&quot; rel=&quot;nofollow&quot;&gt;shielding effect&lt;/a&gt;, which effectively says that the ability of a positive nucleus to attract electrons decreases as the amount of orbitals surrounding the nucleus decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transition Metals&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The insolubility of the hydroxides of the transition metals can be explained both by the increasingly large ionization energies that result from successively removing electrons (or negatively charged hydroxide molecules in this case) from the metal cation, and the greater effective nuclear charge on the electrons across a period.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="16683" LastEditDate="2016-01-20T13:07:41.333" LastActivityDate="2016-01-20T13:07:41.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30737" PostTypeId="2" ParentId="29728" CreationDate="2015-05-02T18:45:54.980" Score="2" Body="&lt;p&gt;When talking about standards, there's always the dilemma: &lt;strong&gt;How is a standard standardized?&lt;/strong&gt; Well, you compare it to another standard you already have, and after you consider all the sources of errors, you have a new standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of the primary standards, you can always promptly trust the label's purity information, as well as your balance's and other measuring equipment (like glassware) accuracy, and at a much lower degree the molar masses, and so on. Having that concentration &quot;certified&quot;, anything you calculate using that as a reference will be &quot;certified&quot; as well. However...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to follow strict analytical rules, than you won't blindly trust those, which is why you cshould calculate the error (or many of the related statistical measures) by considering the accuracy of, like I said, the balance, the glassware, temperature effect on volume (density), etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common way to do this, called the standardized uncertainty, is take the square root of the sum of the squares of each instrument/measure uncertainty divided by it's quantity. You should define a minimal formula which describes the property you want to account the error for, and go for there. Let's take, for example, the concentration of 100 mL of a 0.01 M solution of potassium hydrogen phthalate (KHP), used as a primary standard to standardize sodium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of the solution is $[KHP]=\frac{n}{vol}$, but we obviously didn't measure molês, so we write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[KHP]=\frac{\frac{m}{MM}}{vol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass weighed on the balance, and $MM$ the molar mass. Now that we know what influence in your concentration, we have to use some numbers. Let's consider that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weighed mass: 0.20422 g, on a balance with uncertainty of ± 0.001 g (1 mg)&lt;/li&gt;&#xA;&lt;li&gt;Volume measured: 100 mL, in a volumetric flask with uncertainty of ± 0.1 mL&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This means that the uncertainty of the prepared solution is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{u_{[KHP]}}{0.01} = \sqrt{\left(\frac{0.001}{0.20422}\right)^2 + \left(\frac{0.1}{100}\right)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{[KHP]} \approx 5\times10^{-5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that I didn't take into account the purity of the salt (to make things simpler), and the error on the molar mass (we could assume it's much lower than the others). And what does all this have to do with your question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is, even standards have a variations in them, and when you standardize something, like a HCl or NaOH solution, you could take that into account. The concentration of NaOH, in this case, will have a relative uncertainty higher than the concentration of KHP, because all of the uncertainty of the KHP preparation must be taken into account, as well as the titration variation. The outcome is that a standard based on  a standard of another standard (...) will eventually have a very high uncertainty, which is why we try to sitck to primary and secondary standards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: The uncertainties of the equipments/glassware are usually available in the manuals or even engraved on them. This values are usually the calibration precision, and should be transformed into uncertainty using some statistical inference, which could be that the variation follows a retangular distribution, resulting in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = \frac{precision}{\sqrt3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: There's a lot more to this than what I summarized above, starting with being able to indentify what should and shouldn't be taken into account, not to mention replicates and a lot of statistics. I don't recommend blindly following this formulae to calculate uncertainties.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-05-02T18:45:54.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30738" PostTypeId="2" ParentId="29736" CreationDate="2015-05-02T19:35:39.263" Score="17" Body="&lt;p&gt;What you describe as pink is the actual colour of copper. The colour of a clean, solid surface of high-purity copper is typically salmon red. In your case, the appearance may be further affected by the particle size of the precipitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The familiar reddish-brown colour of copper is caused by copper(I) oxide, which is slowly formed on the metal surface.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-02T19:35:39.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30739" PostTypeId="1" CreationDate="2015-05-02T20:49:26.070" Score="2" ViewCount="199" Body="&lt;p&gt;Are there any known, or suspected compounds that take the crystal structure of what ever object they are touching, i.e. if it touches $\ce{NaCl}$ it would turn into a face centered cubic crystal structure.&lt;/p&gt;&#xA;" OwnerUserId="7786" LastEditorUserId="4945" LastEditDate="2015-05-07T13:11:48.340" LastActivityDate="2015-05-07T13:11:48.340" Title="Can chemicals change crystal structure by only touching?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30740" PostTypeId="5" CreationDate="2015-05-02T21:00:09.843" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-02T21:00:09.843" LastActivityDate="2015-05-02T21:00:09.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30741" PostTypeId="4" CreationDate="2015-05-02T21:00:09.843" Score="0" Body="for use with questions that pertain to theoretical compounds that do not yet or cannot exist." OwnerUserId="7786" LastEditorUserId="7786" LastEditDate="2015-05-03T01:34:19.677" LastActivityDate="2015-05-03T01:34:19.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30743" PostTypeId="1" AcceptedAnswerId="30744" CreationDate="2015-05-02T21:13:59.463" Score="1" ViewCount="1744" Body="&lt;p&gt;Objectively, what makes the solubility of $\ce{C5H11OH}$ greater than the $\ce{C5H11Cl}$ in water?&lt;/p&gt;&#xA;" OwnerUserId="15944" LastEditorDisplayName="user7951" LastEditDate="2015-05-02T21:27:04.993" LastActivityDate="2015-05-04T09:16:34.440" Title="Solubility of C5H11OH and C5H11Cl" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;alcohols&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30744" PostTypeId="2" ParentId="30743" CreationDate="2015-05-02T21:59:45.673" Score="2" Body="&lt;p&gt;I believe the $\ce{C5H11OH}$ is more soluble because it has dipole dipole, H-bonding and London Dispersion. The $\ce{C5H11Cl}$ only has dipole dipole and London Dispersion. The H-bonding is highly soluble in water&lt;/p&gt;&#xA;" OwnerUserId="12949" LastEditorDisplayName="user15489" LastEditDate="2015-05-02T22:12:14.803" LastActivityDate="2015-05-02T22:12:14.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30745" PostTypeId="1" CreationDate="2015-05-02T23:14:53.497" Score="4" ViewCount="1015" Body="&lt;p&gt;So, I have to find the $\Delta T_{\mathrm{reaction}}$ of a solution for my lab report. Earlier in the lab, it gave the equation as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta T_{\mathrm{reaction}}=T_{\mathrm{mixture}}+\frac{1}{2}(T_{\mathrm{substance~1}}+T_{\mathrm{substance~2}})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and all temperatures were at the exact time of mixing found from a linear equation of temperature points on a graph. Now, I have another mixture which is water and a substance and I am supposed to find the $\Delta T_{\mathrm{reaction}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the thing, we don't have the temperature of the substance, only the temp of the water and the solution it made, so how could I go about doing it? I thought about using room temperature since it was a solid substance but it doesn't say anything about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The substance is solid $\ce{NaOH}$.&lt;/p&gt;&#xA;" OwnerUserId="15945" LastEditorUserId="73" LastEditDate="2015-05-04T02:28:26.920" LastActivityDate="2016-01-29T05:56:31.440" Title="Is there more than one equation for Delta T(reaction)?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30746" PostTypeId="1" AcceptedAnswerId="30756" CreationDate="2015-05-02T23:48:37.437" Score="5" ViewCount="1120" Body="&lt;p&gt;I know this may look like some question easily answered with common sense, but please read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If water is the solvent wouldn't the only thing that moves be water? (Osmosis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; this is basically what I think. If you imagine the skin of whatever you are pickling as a membrane that prohibit the flow of electrolytes but allows that if water. Then salt, NaCl, cannot move anywhere. &lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="4805" LastEditDate="2015-05-03T02:30:28.163" LastActivityDate="2015-05-03T16:36:11.467" Title="Why is pickle salty? How does salt get into pickle?" Tags="&lt;osmosis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30747" PostTypeId="1" AcceptedAnswerId="30750" CreationDate="2015-05-02T23:56:57.343" Score="3" ViewCount="1154" Body="&lt;p&gt;What is the difference when following up ozonolysis with DMS (dimethyl sulfide), Zn/water, or just water? Or is there really no substantial difference at all?&#xA;In my textbook, the alkenes section uses DMS or Zn/water but then in the alkynes section it uses water with no explanation as to why that happens. My professor also uses them interchangeably, which gets confusing.&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="4231" LastEditDate="2015-05-03T01:47:23.303" LastActivityDate="2015-05-03T01:47:23.303" Title="Difference in ozonolysis reagents" Tags="&lt;organic-chemistry&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30748" PostTypeId="2" ParentId="29622" CreationDate="2015-05-03T00:23:31.143" Score="3" Body="&lt;p&gt;Already done: See, for example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Skunk_(weapon)&quot; rel=&quot;nofollow&quot;&gt;Skunk&lt;/a&gt; which has been widely used by the Israeli Defense Forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per its &lt;a href=&quot;http://www.skunk-skunk.com/&quot; rel=&quot;nofollow&quot;&gt;manufacturer&lt;/a&gt;, Skunk is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A non-lethal ... deterrent [that] consists of an extremely&#xA;  foul-smelling liquid, with the viscosity of water, that can be sprayed&#xA;  over a large area using a standard water cannon.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Skunk is not volatile, it's inflammable &amp;ndash; in fact, it can&#xA;  even be used to put out fires in a pinch &amp;ndash; and it poses no&#xA;  health hazard. In fact, despite its outrageous smell, the Skunk liquid&#xA;  is perfectly safe to consume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2015-05-03T00:23:31.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30749" PostTypeId="2" ParentId="24275" CreationDate="2015-05-03T00:42:11.713" Score="0" Body="&lt;p&gt;What I have learned at high school is that ideal gas and perfect gas are synonyms. That is, the names basically refer to the same type of gas which:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Has negligibly small forces of attraction between its molecules;&lt;/li&gt;&#xA;  &lt;li&gt;Has negligibly small distances between the particles;&lt;/li&gt;&#xA;  &lt;li&gt;Has particles randomly roaming about at high speeds;&lt;/li&gt;&#xA;  &lt;li&gt;Has molecules whose collision with walls of its container are&#xA;  perfectly elastic.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15948" LastEditorUserId="8208" LastEditDate="2015-05-03T02:06:13.937" LastActivityDate="2015-05-03T02:06:13.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30750" PostTypeId="2" ParentId="30747" CreationDate="2015-05-03T00:59:54.993" Score="1" Body="&lt;p&gt;Dimethylsulfide (DMS) is commonly added to an ozonolysis reaction after it is complete (&lt;em&gt;e.g.&lt;/em&gt; it is the first step in the reaction work-up).  I can see how zinc could serve the same purpose.  Both of these reagents can &lt;strong&gt;reduce&lt;/strong&gt; oxidized materials.  When the ozonolysis is complete the reaction contains the raw ozonide along with numerous undesirable side-products such as peroxides, hydroperoxides and tetroxanes - all of these materials are potentially &lt;strong&gt;explosive&lt;/strong&gt;.  The DMS or zinc will reduce them to carbonyls and alcohols and remove the explosion hazard.  I suppose that tap water is somewhat acidic or basic depending upon where you live, and it might, therefore, also serve to reduce these materials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is one thing to say or write &quot;add water during workup&quot; in a lecture or text, it's another thing to actually deal with explosive materials in the lab.  Personally, I wouldn't chance it. I'd use a real reducing agent such as DMS or zinc, instead of a water work-up.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-03T00:59:54.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30753" PostTypeId="1" AcceptedAnswerId="30781" CreationDate="2015-05-03T02:05:36.587" Score="4" ViewCount="980" Body="&lt;p&gt;When I learnt about polarity, I always come to the term electronegativity and always use the electronegativity chart. However, when I studied further, they have the word electroposivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been thinking why we use electronegavity more often than electropositivity? And why we have a electronegativity chart instead of electropositivity chart?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2018-02-15T03:49:10.973" Title="Why electronegativity instead of electropositivity" Tags="&lt;polarity&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30754" PostTypeId="1" AcceptedAnswerId="30764" CreationDate="2015-05-03T02:30:35.993" Score="33" ViewCount="11326" Body="&lt;p&gt;What would be the effect if someone were to drink &lt;a href=&quot;https://en.wikipedia.org/wiki/Purified_water&quot; rel=&quot;noreferrer&quot;&gt;ultra-pure water&lt;/a&gt; with an electrical resistivity of $18 \, \mathrm{M} \Omega \! \cdot \! \text{cm}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they immediately die?  Would they just need to pee more?  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the threshold for bad things to happen — a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the science here would be.&lt;/p&gt;&#xA;" OwnerUserId="15949" LastEditorUserId="7475" LastEditDate="2016-11-02T19:18:18.807" LastActivityDate="2017-01-27T06:41:12.870" Title="Effect of drinking ultra-pure water" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;biochemistry&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="30755" PostTypeId="1" CreationDate="2015-05-03T02:43:35.320" Score="1" ViewCount="1219" Body="&lt;p&gt;When I first learned this term. I was introduced to this graph: &lt;img src=&quot;https://i.stack.imgur.com/sucSu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat in this case is at 100 °C, which I had ever since believed to be the only temperature of vaporization, &lt;strong&gt;assuming vaporization = boiling.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I saw a table with all the heat of vaporization values for different compounds. Most importantly, all these values apply to reactions at 25  °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different heat of vaporization values apply to different temperature. &lt;strong&gt;but water, for ex., cannot boil, with vapor bubbles, unless at 100 °C, thus all the energy used will just be used to simply raise the temperature.&lt;/strong&gt; &lt;em&gt;wouldn't it?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="5456" LastEditDate="2015-05-03T06:28:21.733" LastActivityDate="2015-05-03T08:44:49.117" Title="What is heat of vaporization? How can it be used at temperature as low as 25 °C?" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30756" PostTypeId="2" ParentId="30746" CreationDate="2015-05-03T02:54:25.347" Score="4" Body="&lt;p&gt;Your concept of a cell is a little vague. Although the cell wall is made from a diglyceride bilayer, it is not impregnable. In fact, there are many proteins that pass through the cell wall where dissolved material can pass. One of the proteins is called the &lt;a href=&quot;https://www.youtube.com/watch?v=P-imDC1txWw&quot; rel=&quot;nofollow&quot;&gt;sodium potassium pump&lt;/a&gt;. In a living cell, the protein uses ATP to move sodium out of the cell and drag potassium into the cell. Both of these ions are against the gradient. In a non-living cell, such as the pickle, sodium ions can move across the cell membrane and make the cells salty.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-05-03T15:12:06.683" LastActivityDate="2015-05-03T15:12:06.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30758" PostTypeId="2" ParentId="30755" CreationDate="2015-05-03T03:10:53.387" Score="4" Body="&lt;p&gt;Think of vaporization as the removal of the fastest molecules. The separation of the gaseous form from the more closely-bound liquid therefore removes energy from the liquid. The heat of vaporization can apply at much lower temperatures than the BP of water &lt;em&gt;at atmospheric pressure&lt;/em&gt;. Even on a day when the temperature is well blow freezing, laundry will dry on a clothesline because ice evaporates (well, sublimes). Also, water at skin temperature soon feels cold as it loses heat through evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the boiling point of water is 100 °C only at 1 atmosphere; see this &lt;a href=&quot;https://www.youtube.com/watch?v=pOYgdQp4euc&quot; rel=&quot;nofollow&quot;&gt;video of water boiling at room temperature&lt;/a&gt;, ~25 °C.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorDisplayName="user7951" LastEditDate="2015-05-03T08:44:49.117" LastActivityDate="2015-05-03T08:44:49.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30759" PostTypeId="2" ParentId="30754" CreationDate="2015-05-03T03:14:34.070" Score="28" Body="&lt;p&gt;Pure water (i.e. 18 Megohm conductivity) is still &lt;strong&gt;just water&lt;/strong&gt;. If you don't drink so much as to cause &lt;a href=&quot;http://en.wikipedia.org/wiki/Water_intoxication&quot;&gt;water intoxication&lt;/a&gt;, it would be no different physiologically from drinking tap water or bottled water.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-03T03:14:34.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30760" PostTypeId="5" CreationDate="2015-05-03T03:14:50.483" Score="0" Body="&lt;p&gt;&lt;strong&gt;Definition from &lt;a href=&quot;https://en.wikipedia.org/wiki/Organosulfur_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Organosulfur compounds are organic compounds that contain sulfur. They are often associated with foul odors, but many of the sweetest compounds known are organosulfur derivatives, e.g., saccharin. Nature abounds with organosulfur compounds—sulfur is essential for life.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sulfur shares the chalcogen group with oxygen, selenium, and tellurium, and it is expected that organosulfur compounds have similarities with carbon–oxygen, carbon–selenium, and carbon–tellurium compounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where can I use this tag?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag is appropriate to organic functional groups that contain sulfur. &#xA;Most of these functional groups are not as common as their &lt;a href=&quot;http://meta.chemistry.stackexchange.com/q/2815/&quot;&gt;carbon analogs&lt;/a&gt;, instead of having individual tags for each this tag may be applied to all related derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are listed in the tag excerpt. For the full list, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Organosulfur_compounds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="42352" LastEditorUserId="5026" LastEditDate="2018-03-17T15:51:07.647" LastActivityDate="2018-03-17T15:51:07.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30761" PostTypeId="4" CreationDate="2015-05-03T03:14:50.483" Score="0" Body="For questions about organosulfur compounds, i.e. organic compounds that contain sulfur. These include, but are not limited to, thioesters, thioureas, xanthates, thioacetals, sulfoxides, sulfones, sulfolanes, sulfonic acids, sulfinic acids, disulfides." OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-17T14:16:19.173" LastActivityDate="2018-03-17T14:16:19.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30762" PostTypeId="5" CreationDate="2015-05-03T03:17:14.500" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-03T03:17:14.500" LastActivityDate="2015-05-03T03:17:14.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30763" PostTypeId="4" CreationDate="2015-05-03T03:17:14.500" Score="0" Body="Questions related to matter in a state that has properties between those of conventional liquid and those of solid crystal." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T10:56:33.120" LastActivityDate="2015-05-03T10:56:33.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30764" PostTypeId="2" ParentId="30754" CreationDate="2015-05-03T03:18:40.117" Score="56" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the effect if someone were to drink ultra-pure, 18 M-ohm water? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not much, although if they drank many gallons of the water it could be a problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would they immediately die? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would they just need to pee more? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices.  Same as regular water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think the effect would be that different from drinking regular water.  All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyponatremia&quot;&gt;Hyponatremia&lt;/a&gt; or possibly the far more dangerous &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypokalemia&quot;&gt;Hypokalemia&lt;/a&gt;.  The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough approximation to understanding how toxic water can be would then be to compare the concentration &lt;em&gt;difference&lt;/em&gt; of sodium (or potassium) between the ingested water and the blood.  A 25 micromolar -- i.e. 25 &amp;mu;M or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than &quot;pure&quot; 18.2 megohm water (0.055 microsiemens per cm).  According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the &quot;pure&quot; water to cause problems is (135 - 0) mM.  For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T03:19:21.020" LastActivityDate="2015-05-03T03:19:21.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30765" PostTypeId="2" ParentId="30739" CreationDate="2015-05-03T03:31:36.310" Score="2" Body="&lt;p&gt;No, there is no &quot;exact mimic&quot; crystal, but the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Crystal_habit&quot; rel=&quot;nofollow&quot;&gt;crystal habit&lt;/a&gt; can alter the crystal form&lt;/strong&gt;. Consider that the environment in the uniform inside of a substance is different from that along its boundaries; growing on a different surface can alter the shape of the crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The semiconductor industry, in particular, is often faced with difficulties in growing crystalline semiconductors with layers of varying composition (&lt;a href=&quot;http://users.wfu.edu/ucerkb/Nan242/L12-Epitaxy.pdf&quot; rel=&quot;nofollow&quot;&gt;epitaxy&lt;/a&gt;). There is often strain, leading to defects, unless an intermediate layer can smooth the transition.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-03T03:31:36.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30766" PostTypeId="2" ParentId="29727" CreationDate="2015-05-03T05:07:13.370" Score="3" Body="&lt;p&gt;Both of those explanations include part of the puzzle, but are not really complete on their own. To get a better understanding, you need to combine them, as well as understand something called Coulomb's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basics of Coulomb's law are this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The closer two oppositely charged things are, the stronger the attraction&lt;/li&gt;&#xA;&lt;li&gt;The stronger the charge on either of the things, the stronger the attraction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With those points in mind, let's combine it all. I'll keep my explanation similar to the way you phrased yours:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Each step across a row on the periodic table adds &lt;em&gt;both&lt;/em&gt; one nuclear proton &lt;em&gt;and&lt;/em&gt; one valence electron. So, that is increasing &lt;em&gt;both&lt;/em&gt; the total positive charge of the nucleus &lt;em&gt;and&lt;/em&gt; the total negative charge of the valence shell, but not the distance between them (because electrons are added to the same shell).&lt;/li&gt;&#xA;&lt;li&gt;The force of attraction between two charged objects depends on how strong each of them are (Coulomb's law). Since both the nucleus and the valence shell are stronger than in the previous step, and the distance between them has not increased (the electrons are added to the same shell) there is a stronger force of attraction between them, and the two are pulled closer together.&lt;/li&gt;&#xA;&lt;li&gt;So, the atomic size of elements decreases across a period.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15952" LastActivityDate="2015-05-03T05:07:13.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30767" PostTypeId="1" AcceptedAnswerId="30770" CreationDate="2015-05-03T09:36:18.837" Score="10" ViewCount="1228" Body="&lt;p&gt;Why do radical mechanisms favor the homolytic dissociation of a $\ce{C-H}$ bond over a $\ce{C-C}$ bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{C-H}$ bond is stronger in magnitude (and it's also slightly polarized), so why would it break more easily as compared to the $\ce{C-C}$ bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instructor explains this by claiming that the products of $\ce{C-H}$ bond dissociation are stabler than$\ce{C-C}$ bond dissociation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that radicals are unstable and highly reactive, so why does it make any difference as to which radical is formed? Wouldn't the radical easiest to form be the one formed the most?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T10:12:09.600" LastActivityDate="2015-05-03T10:44:47.283" Title="Formation of radicals from C-H and C-C bonds" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;bond&gt;&lt;stability&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30768" PostTypeId="1" AcceptedAnswerId="30772" CreationDate="2015-05-03T09:43:08.563" Score="7" ViewCount="3506" Body="&lt;p&gt;What is the effect of hybridization and double-bond formation on the rest of the atoms/molecules attached to the central atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a benzene molecule, the Carbon-Carbon bonds are intermediate between single and double bonds. But what is the effect of the presence of the double bonds between the carbons on the bonds between the carbon and hydrogen? Do the bond lengths decrease because of the sp2 hybridized bonding orbitals?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastActivityDate="2015-05-03T10:59:34.203" Title="Length of C-H bonds in hybridized bonding orbitals" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;stability&gt;&lt;hybridization&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30769" PostTypeId="2" ParentId="29670" CreationDate="2015-05-03T10:35:31.587" Score="5" Body="&lt;p&gt;As soon as you move to systems with more than one electron them all orbitals are approximations (i.e. not real). Hybridized orbitals are no more &quot;heuristic&quot; than atomic orbitals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation between hybrid and atomic orbitals can be mathematically, though not uniquely, defined (e.g. as the minimization of inter-orbital electrostatic repulsion) and the transformation does not affect the total electron density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See for example &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html&quot; rel=&quot;noreferrer&quot;&gt;http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html&lt;/a&gt; and &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html&quot; rel=&quot;noreferrer&quot;&gt;http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-03T10:35:31.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30770" PostTypeId="2" ParentId="30767" CreationDate="2015-05-03T10:44:47.283" Score="8" Body="&lt;p&gt;Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references &lt;a href=&quot;http://www.colorado.edu/chem/ellison/research/resThermo.html&quot;&gt;for the organics&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Bond_dissociation_energies..JPG&quot;&gt;here for the halogens&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Homolysis &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\ce{C-C}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-CH3 -&amp;gt; 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\ce{C-H}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3CCH2-H -&amp;gt; H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 -&amp;gt; 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chlorine radicals then react with the alkane by &lt;strong&gt;abstraction&lt;/strong&gt;, meaning that bonds are being formed as well as being broken.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Abstraction &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\ce{C-C}$&lt;/strong&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl. + H3C-CH3 -&amp;gt; Cl-CH3 +H3C.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c|} &#xA; &amp;amp; \text{Broken} &amp;amp; \text{Formed} \\ \hline&#xA;\text{bond} &amp;amp; \ce{H3C-CH3}  &amp;amp; \ce{Cl-CH3} \\ \hline&#xA;\Delta H^\circ &amp;amp; 377\ \text{kJ/mol} &amp;amp; -350 \ \text{kJ/mol} \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$\ce{C-H}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl. + H3CCH2-H -&amp;gt; Cl-H +H3CCH2.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c|c|} &#xA; &amp;amp; \text{Broken} &amp;amp; \text{Formed} \\ \hline&#xA;\text{bond} &amp;amp; \ce{H3CCH2-H}  &amp;amp; \ce{Cl-H} \\ \hline&#xA;\Delta H^\circ &amp;amp; 423\ \text{kJ/mol} &amp;amp; -432 \ \text{kJ/mol} \\ \hline&#xA;\end{array}$$&#xA;$$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-05-03T10:44:47.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30771" PostTypeId="2" ParentId="29688" CreationDate="2015-05-03T10:48:52.110" Score="4" Body="&lt;p&gt;&lt;em&gt;trans&lt;/em&gt;-buta-1,3-diene is not a linear molecule.  To be linear all atoms must be &quot;in-line&quot;, e.g. $\ce{HCCH}$ is linear but $\ce{H2CCH2}$ is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation around a bond axis is not free and is therefore described as a vibration. See for example &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/2010/06/geometry-and-molecular-motion.html&quot; rel=&quot;nofollow&quot;&gt;this blog post&lt;/a&gt;.  So, for example, for ethane the lowest of the $3N-6$ vibrational modes corresponds to a torsional motion about the $\ce{C-C}$ single bond.  The mental picture here is not 2 methyl groups that rotate freely relative to one another, but rather a rocking back and forth motion in one minimum and then an occasional rotation in to the next minimum.  The vibrationa motion is on the femto second time scale and every few nano-seconds the switch from one minimum to the next will occur. So the &quot;rotation&quot; will happen a thousand times per millisecond and look like free rotation. &lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-20T09:40:57.013" LastActivityDate="2015-06-20T09:40:57.013" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="30772" PostTypeId="2" ParentId="30768" CreationDate="2015-05-03T10:59:34.203" Score="5" Body="&lt;p&gt;Yes, the bond length decreases.  One way to think about it is that a sp3 center with 109.5 degree angles is more crowded than an sp2 center with 120 degree angles.  More crowding = more repulsion = longer bond length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is that the e-pair in an s orbital is sitting on the nucleus.  Adding a p-orbital moves the electron pair further away from the nucleus.  More p-orbital character = longer bond lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can test it for yourself using &lt;a href=&quot;http://molcalc.org/&quot; rel=&quot;noreferrer&quot;&gt;http://molcalc.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-03T10:59:34.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30775" PostTypeId="1" AcceptedAnswerId="30778" CreationDate="2015-05-03T13:12:30.910" Score="3" ViewCount="1106" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/umZXz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;technically the line between liquid and solid phase would go a long way before hitting the y-axis, but the point is it will eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question remains, can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUMxT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="9961" LastEditDate="2015-05-04T12:02:10.683" LastActivityDate="2015-09-13T18:30:22.413" Title="Can water be liquefied or solidified just by adjusting the temperature, regardless of the pressure?" Tags="&lt;water&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30776" PostTypeId="1" AcceptedAnswerId="34497" CreationDate="2015-05-03T13:30:11.063" Score="5" ViewCount="168" Body="&lt;p&gt;I am curious how, from the perspective of quantum chemistry, the hydrogen bond is modeled. In particular I am interested in using such a model to estimate the amount of energy released per $\ce{H2O}$ molecule, if say two of them approach each other e.g. in case of condensation. Any hint for precise literature would be great!&lt;/p&gt;&#xA;" OwnerUserId="15959" LastEditorUserId="10185" LastEditDate="2015-07-29T18:16:46.193" LastActivityDate="2015-08-29T06:37:24.663" Title="Theoretical model for the hydrogen bond" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30778" PostTypeId="2" ParentId="30775" CreationDate="2015-05-03T13:50:05.553" Score="7" Body="&lt;p&gt;If I actually understood your question correctly, yes there are regions above which there will be &lt;strong&gt;only&lt;/strong&gt; solid state (rather than no solid state). This is limited to the molecular region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qa2tm.png&quot; alt=&quot;Phase diagram of water&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Roman numerals indicate various ice phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above phase diagram (which can be found &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_diagram#Crystal_phase_diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) the solid-liquid equilibrium line doesn’t touch the y-axis as you suggest. But there will be only solid phase above the $10\ \mathrm{GPa}$ region(approximately). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the phase diagram you drew, if (a big if) such a PD exists for some element/molecule, then there will be no solid above the point where solid-liquid line meets y-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in the comments, when we reach high temperatures and pressures we won’t be seeing the molecular water. It will dissociate into atoms/ions as shown in the figure. &lt;img src=&quot;https://i.stack.imgur.com/BNrkK.jpg&quot; alt=&quot;water at high T and p&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorDisplayName="user7951" LastEditDate="2015-09-13T18:30:22.413" LastActivityDate="2015-09-13T18:30:22.413" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30779" PostTypeId="2" ParentId="30775" CreationDate="2015-05-03T13:50:50.177" Score="1" Body="&lt;p&gt;Theoretically, &lt;strong&gt;YES&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But why?&lt;/strong&gt;&#xA;In the real world there are a thousand different ways to accomplish a certain task, but only a few will be the most efficient. So in considering your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So the question remains, can any liquids, including water (with its unique phase diagram) be liquefied regardless of the pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, &lt;strong&gt;I do not like how you put regardless of the pressure&lt;/strong&gt;, there has to be a certain amount of pressure. Yes you can do this, but there really is an easier way to this. Use a freezer to solidify it, or leave it out in the open to liquify it. :)&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-03T13:50:50.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30781" PostTypeId="2" ParentId="30753" CreationDate="2015-05-03T14:27:01.597" Score="3" Body="&lt;p&gt;The first thing that should be highlighted here is that electropositivity is simply the opposite of electronegativity, any of the two can be used interchangeably with the necessary modifications to the sentence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true though that electronegativity is more commonly used then electropositivity. For example, the Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow&quot;&gt;article on the subject&lt;/a&gt; is electronegativity, and mentions electropositivity as its opposite. I believe there are two reasons for that, related to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First: Consistency&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very important in the scientific community, from reasons that go from facilitating students understanding to the writing of papers, that standards are used, and that everyone is talking about the same thing. It's easier if everyone is talking about the same thing, even if they are the complete opposite. It just makes us think faster. Why choose electronegativity then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second: How we explain chemistry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's unecessary to explain why and how the study of the electrons &quot;behaviour&quot; in atoms and molecules are important to chemistry. Usually, when explaining a phenomenon, we talk about where the electrons are &quot;going to&quot;. We say a reaction occurs because an electron is taken by an atom, or maybe because it's donated by one. We explain the polarity of the H-O bond by saying that the oxygen will attract the electrons of the covalent bond a lot more than hydrogen will, and not (usually) that the hydrogen doesn't attract the electrons as much as oxygen does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it's obvious that you can explain everything by means of electropositivity, but, in my experience, we usually explain things by saying which atom/group has the property that makes electrons &lt;strong&gt;go to&lt;/strong&gt; them with higher value, rather than saying which atom/group has the property that makes electrons &lt;strong&gt;leave&lt;/strong&gt; them with higher value.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastActivityDate="2015-05-03T14:27:01.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30782" PostTypeId="1" CreationDate="2015-05-03T14:49:16.897" Score="6" ViewCount="341" Body="&lt;p&gt;I know that Gaussian (or other quantum chemical packages) assumes the molecule is alone in vacuum, that is, the molecule doesn't interact with anything, when optimizing the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've found that optimization at HF or DFT levels results are almost the same as experimental data, which were from the solid state by X-ray diffraction. Everyone knows that the solid state is quite different from the vacuum, for molecules interact with each other in the solid state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why do the two generate nearly the same results? Or have I looked at so few samples that I neglect occasions when the two give quite different results? &lt;/p&gt;&#xA;" OwnerUserId="7158" LastEditorUserId="5017" LastEditDate="2015-06-17T21:59:17.987" LastActivityDate="2015-06-17T21:59:17.987" Title="Are molecular optimizations in quantum chemistry really in vacuum?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;x-ray-diffraction&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30783" PostTypeId="1" CreationDate="2015-05-03T15:02:52.647" Score="1" ViewCount="107" Body="&lt;p&gt;I have a water softener for showers and dishwasher and a reverse osmosis filter for drinking water.  For convenience of the install, the RO filter is using softened water (some calcium carbonate has been exchanged with sodium chloride -- this is my basic understanding; correct me if I'm wrong).  I don't fully understand how an RO filter works except that it is essentially just a molecule-sized physical filter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, Does the RO filter care whether the supply water is salty or hard?  Is it more effective at filtering one or the other?  Is one going to influence the life of the RO filter more than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is not a &quot;true&quot; chemistry question, but I'm looking for a bit more of an thorough explanation than I got at diy.stackexchange.&lt;/p&gt;&#xA;" OwnerUserId="15962" LastEditorUserId="15410" LastEditDate="2016-04-22T23:46:50.050" LastActivityDate="2016-04-22T23:46:50.050" Title="Water softener and RO filter" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;aqueous-solution&gt;&lt;osmosis&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30784" PostTypeId="2" ParentId="30782" CreationDate="2015-05-03T15:11:34.383" Score="4" Body="&lt;p&gt;It is possible to do crystal structure optimizations in Gaussian by adding translation vectors (DFT isn't as good with intermolecular reactions, but it will probably be a better approximation than optimizing without PBC, which is akin to a gas phase molecule).&#xA;Whether or not solid behavior and gas behavior are similar will depend on the material, and the properties you're measuring, so could you tell us more specifically what you're studying?&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-05-03T15:11:34.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30785" PostTypeId="1" CreationDate="2015-05-03T15:39:40.560" Score="5" ViewCount="3177" Body="&lt;p&gt;The atomic emission spectrum for sodium ($\ce{Na}$) is completely dominated by a line in the range of yellow, about $590~\mathrm{nm}$ (to be more precise, it's a doublet). Here is how it looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MVqSu.jpg&quot; alt=&quot;a) shows the emission spectrum and b) the absorption&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This line is due to the transition $3\mathrm{p} \to 3\mathrm{s}$.&#xA;I know the Rydberg equation which can be used to predict the transitions in hydrogen. I would like to know if there is a way to predict the transitions for other elements, such as the alkali metal $\ce{Na}$.&lt;/p&gt;&#xA;" OwnerUserId="15963" LastEditorUserId="15410" LastEditDate="2015-05-03T16:10:59.047" LastActivityDate="2015-05-03T19:05:08.783" Title="Is it possible to predict the lines in the atomic emission spectrum of Na?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30786" PostTypeId="1" AcceptedAnswerId="30790" CreationDate="2015-05-03T16:05:44.640" Score="0" ViewCount="115" Body="&lt;p&gt;Is the parent chain in the above case determined by atomic mass, alphabetical preference or any other rule?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-05-03T16:42:21.233" Title="If there are two longest chains possible in an organic compound and both have the same number of substituents, how do we decide the parent chain?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-03T17:41:24.590" ContentLicense="CC BY-SA 3.0" />
  <row Id="30787" PostTypeId="1" CreationDate="2015-05-03T16:05:53.227" Score="13" ViewCount="502" Body="&lt;p&gt;I know that one current area of research is ways to protect astronauts from ionizing radiation when they venture out of the atmosphere of Earth, but would that same ionizing radiation be a cause of concern when performing chemical reactions in space or in the atmosphere of planets with little atmosphere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that radiation like ultraviolet, which is present in sunlight here on Earth, is used in organic chemistry for the homolytic fission of halides, but I'm not sure how much higher energy radiation might affect other reactions.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="14" LastEditDate="2015-05-21T13:30:24.607" LastActivityDate="2015-12-31T14:51:07.003" Title="Radiation in space and its effects on chemical reactions?" Tags="&lt;theoretical-chemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30788" PostTypeId="2" ParentId="5513" CreationDate="2015-05-03T16:14:10.437" Score="4" Body="&lt;p&gt;Though a hydrogen atom has only one electron, it contains a large number of shells, so when this single electron jumps from one shell to another, a photon is emitted, and the energy difference of the shells causes different wavelengths to be released... hence, mono-electronic hydrogen has many spectral lines.&lt;/p&gt;&#xA;" OwnerUserId="15964" LastEditorUserId="402" LastEditDate="2017-03-30T21:54:06.870" LastActivityDate="2017-03-30T21:54:06.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30789" PostTypeId="1" CreationDate="2015-05-03T16:42:13.100" Score="6" ViewCount="1189" Body="&lt;ul&gt;&#xA;&lt;li&gt;What are the different properties that intermolecular forces and Intramolecular forces attribute to? &lt;em&gt;i.e.&lt;/em&gt; intermolecular forces contribute to BPs, vapor pressures, etc. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Specific problem: why does &lt;strong&gt;n-butylamine&lt;/strong&gt; have lower vapor pressure at room temperature than &lt;strong&gt;methanol&lt;/strong&gt;? Explain in regards to IMFs.&lt;/p&gt;&#xA;" OwnerUserId="15965" LastEditorUserId="10185" LastEditDate="2015-06-09T16:45:42.827" LastActivityDate="2017-03-02T21:56:03.313" Title="Intermolecular and Intramolecular Forces (Vapor pressure of n-butylamine and methanol)" Tags="&lt;intermolecular-forces&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30790" PostTypeId="2" ParentId="30786" CreationDate="2015-05-03T16:42:21.233" Score="0" Body="&lt;p&gt;If two longest chain are possible with same number of side chain in both of them, then we consider sum of locants of the side chain. The longest chain in which the sum of locants of the side chain is least is taken into consideration. &lt;/p&gt;&#xA;" OwnerUserId="5608" LastActivityDate="2015-05-03T16:42:21.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30791" PostTypeId="1" CreationDate="2015-05-03T17:08:14.830" Score="3" ViewCount="3571" Body="&lt;p&gt;The molecular geometry of sand is linear. Dipole moment is thus cancelled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we all know from the meniscus water makes with glass that the dipole-dipole/H-O attraction  between sand and water is stronger than the hydrogen bond between water molecule. &lt;strong&gt;maybe I am wrong&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is either that electron lone pair makes it polar or the lattice structure does.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastEditorUserId="23561" LastEditDate="2019-04-21T18:28:56.723" LastActivityDate="2019-04-21T18:28:56.723" Title="Why is sand/glass polar? How does lattice structure change polarity?" Tags="&lt;crystal-structure&gt;&lt;polarity&gt;&lt;surface-chemistry&gt;&lt;erratum&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30792" PostTypeId="1" CreationDate="2015-05-03T17:08:42.340" Score="6" ViewCount="297" Body="&lt;p&gt;I understand that the structure factor for a crystal is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{hkl} = \sum_r f_{r} e^{2{\pi}i(hx_{r}+ky_{r}+lz_{r})} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook (''Structures of crystals'', Glazier) suggests that you can calculate structure factors if you know the lattice type and motif of the crystal, simply by finding the product of the lattice and the motif structure factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this method be used generally, and if so, do you need to define the motif relative to a centre of symmetry in order to get real structure factors for centrosymmetric crystals ?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="16035" LastEditDate="2019-07-23T07:45:40.560" LastActivityDate="2020-07-14T13:23:31.110" Title="Determining structure factors for crystals from lattice and motif" Tags="&lt;crystal-structure&gt;&lt;crystallography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30794" PostTypeId="2" ParentId="30789" CreationDate="2015-05-03T17:31:59.083" Score="4" Body="&lt;p&gt;What are intermolecular molecular forces?&lt;br/&gt;&#xA;The forces(electrostatic force) between molecules of a compound. These forces are responsible for intermolecular distances and how tightly the molecules are held together. If the intermolecular forces(&lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow&quot;&gt;van der Waals force&lt;/a&gt;) are large, it means the molecules are tightly held together in a group and they are fixed in their position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is boiling point? It's a temperature at which liquid state of a compound changes into its gas state. What's the difference between gaseous and liquid phase? Gaseous phase has very large intermolecular distance when compared to that in liquid phase. If a large force exists between the molecules and they are quite fixed in their position, then large amount energy or heat will be required to move them away from each other. Large heat means greater temperature and hence greater boiling point. So the lesson is that, the stronger the intermolecular forces, the higher the boiling point.&#xA;The similar concept in applicable in melting point. the stronger the intermolecular forces, the higher will be the melting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably(&lt;a href=&quot;http://www.chem1.com/acad/webtext/states/changes.html&quot; rel=&quot;nofollow&quot;&gt;In case you may be wondering, how are they related. Click here&lt;/a&gt;). If the intermolecular forces are large, then molecule in liquid phase are not able to leave to liquid easily, hence lowering the escaping tendency and therefore lower vapour pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is a very basic explanation. &lt;/p&gt;&#xA;" OwnerUserId="5608" LastEditorUserId="7448" LastEditDate="2015-06-09T18:48:40.950" LastActivityDate="2015-06-09T18:48:40.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30795" PostTypeId="2" ParentId="29679" CreationDate="2015-05-03T17:32:24.607" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer is &quot;side reactions&quot;.  Without more information about your exact experimental conditions, or even better, data about the obtained yields of the desired product and of byproducts, it isn't possible to explain conclusively what went wrong, but here are some ideas.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Acetyl sulfate formation.  In the comments you you implied that you used anhydrous sulfuric acid.  Under those conditions, acetic anhydride can undergo solvolysis to yield acetyl sulfate and water.  $$\ce{CH3COOOCCH3 + 2H2SO4 &amp;lt;=&amp;gt; 2 CH3COOSO3H + H2O} $$  Acetyl sulfate might be a less active acetyl transfer reagent than acetic anhydride, or the water formed may hydrolyze some of the acetic anhydride or even the formed acetyl sulfate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dehydration/sulfation of the starting material instead of acetyl transfer.  If you really did use anhydrous sulfuric acid, it is a powerful dehydrating reagent.  It could be possible that you are sulfating the alcohol or even the acid group (or both?) of the substrate, forming either &lt;em&gt;o&lt;/em&gt;-sulfoxybenzoic acid (sulfation at $\ce{-OH}$ group) or sulfosalicylic acid (sulfation at $\ce{-COOH}$ group).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The heat of mixing of concentrated sulfuric acid with water is about $-74~\mathrm{kJ~mol^{-1}}$, judging from &lt;a href=&quot;http://www.nist.gov/srd/upload/jpcrd426.pdf&quot; rel=&quot;noreferrer&quot;&gt;Figure 34b of a NIST treatise on the subject&lt;/a&gt;.  The &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C108247&amp;amp;Mask=1EFF&quot; rel=&quot;noreferrer&quot;&gt;heat of reaction for hydrolysis of acetic anhydride&lt;/a&gt; is only $-41~\mathrm{kJ~mol^{-1}}$.  Thus under anhydrous conditions, I think sulfated products will be more thermodynamically favored than acetylated products.  Sulfated products will be far more soluble in water or sulfuric acid than would be acetylated products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think that using more water in the reaction would help.  The reason is that the heat of mixing of sulfuric acid is a very strong function of water concentration, but the heat of hydrolysis of acetic anhydride is not.  So raising the concentration of water a little bit will disfavor sulfation significantly by lowering the heat of mixing of sulfuric acid, but will leave the heat of hydrolysis of acetic anhydride the same.  It's worth noting that this argument is purely thermodynamic, not kinetic.  But I don't think sulfation reactions are kinetically hindered under ambient conditions, so although it isn't always true, I think thermodynamic arguments are valid in this case.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="15410" LastEditDate="2015-05-07T18:37:31.967" LastActivityDate="2015-05-07T18:37:31.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30797" PostTypeId="1" CreationDate="2015-05-03T18:06:43.753" Score="39" ViewCount="20979" Body="&lt;p&gt;Despite the fact that oxygen is much more electronegative than carbon, the bond in $\ce{CO}$ presents a weak dipole moment.&#xA;This observation can easily be explained using the concept of &quot;dative bond&quot;, that is, one bond is formed with two electrons from oxygen, producing a polarization $\ce{O\bond{-&amp;gt;}C}$ which equilibrates the expected polarization $\ce{C-&amp;gt;O}$.&#xA;I would like to know if the molecular orbital model could be used to explain this phenomenon.&#xA;Here's the diagram for $\ce{CO}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwOet.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15963" LastEditorUserId="4945" LastEditDate="2015-05-07T12:24:35.083" LastActivityDate="2018-02-06T23:12:02.423" Title="How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="30798" PostTypeId="2" ParentId="30791" CreationDate="2015-05-03T18:39:51.187" Score="6" Body="&lt;p&gt;I'm not sure where you get the idea that glass is linear. What makes a glass is the lack of long range ordering, though in fused silica, the approximate tetrahedral structure of $\ce{SiO4}$ is maintained (this is a 2D section, the fourth oxygens are not shown):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tFu90.png&quot; alt=&quot;fused silica&quot;&gt; &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Silica.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In normal glasses, there is no significant net dipole moment across the entire structure, but the reason we observe strong adhesive forces between water and glass is a surface phenomenon—in the bulk of the glass, every silicon atom is connected to four oxygen atoms and each oxygen atom to two silicon atoms, but what happens at a boundary/surface? There are oxygen atoms that only have one bond. When these are exposed to an aqueous solution, some will bear a negative charge, others will pick up a proton from water. The &lt;a href=&quot;http://physics.nyu.edu/grierlab/charge6c/&quot; rel=&quot;noreferrer&quot;&gt;theory&lt;/a&gt; behind it is pretty complicated, but you can think of it as a simple acid-base reaction—at low pH more will be protonated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Si-OH &amp;lt;=&amp;gt; Si-O- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, protonated or not, silanol groups are very happy to associate with water, accounting for the concave meniscus.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-05-03T18:39:51.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30799" PostTypeId="2" ParentId="30785" CreationDate="2015-05-03T19:05:08.783" Score="1" Body="&lt;p&gt;The origin of the Rydberg equation is the Bohr model's equation for the energy of a hydrogenic state: &#xA;$E_\text{B} = -\dfrac{R_{H}hc}{n^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is accurate for other hydrogenic atoms, with only one electron too (bearing in mind that $R_{X}$ is proportional to the atomic number squared). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get to non-hydrogenic atoms you need to model inter-electron repulsion as well. This can be done by introducing an empirical parameter $\delta$, the quantum defect: $E_\text{B} = -\dfrac{R_{X}hc}{(n-\delta_X)^2}$. For sodium the quantum defect is 1.37, allowing good approximation of the transition energies between n-states in the electronic spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a similar back-of-the-envelope method to calculate the differences in energies between multi-electron atom l-energy levels (like the 3s-3p). The theory of angular momentum tends to mathematical complexity. There are likely good computationally derived approximations though.   &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-03T19:05:08.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30800" PostTypeId="1" CreationDate="2015-05-03T19:23:58.410" Score="2" ViewCount="2337" Body="&lt;p&gt;Suppose you have the following voltaic cell: $\ce{Sn_{(s)}|Sn^{2+}_{(aq, 1.0 M)}||Cu^{2+}_{(aq, 1.0 M)}|Cu_{(s)}}$ and the salt bridge is $\ce{KNO_3}$. What I don't understand is why you need to have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell. As $\ce{Sn}$ is oxidized, electrons are supposed to travel through the wire connecting the two half cells and reduce the $\ce{Cu^{2+}}$ ions in solution, but if we have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell, the electrons might just as well reduce the $\ce{Sn^{2+}}$ ions. Why would the electrons travel all the way through the wire when there is $\ce{Sn^{2+}}$ ions right next to them that could be reduced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also don't understand why the $\ce{Cu}$ cathode has to exist. All we need is $\ce{Cu^{2+}}$ ions in solution that will be reduced, so why can't we just dip the wire directly into $\ce{Cu^{2+}}$ solutions. As the reaction proceeds, a $\ce{Cu}$ cathode will be created anyways as $\ce{Cu}$ metal precipitates on the wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am obviously misunderstanding something, because according to my understanding the following voltaic cell should work: In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire. The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution. The $\ce{Sn}$ gets oxidized and produces $\ce{Sn^{2+}}$ ions in the water. The electrons travel through the wire and reduce the $\ce{Cu^{2+}}$ ions in the cathode solution. The anode solution will become more positive and the cathode solution will become more negative, so we still need a salt bridge.&lt;/p&gt;&#xA;" OwnerUserId="9972" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T11:21:59.837" LastActivityDate="2015-06-20T11:22:37.153" Title="Why does the anode solution contain Sn2+ in a Sn-Cu voltaic cell?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30801" PostTypeId="1" AcceptedAnswerId="30803" CreationDate="2015-05-03T19:38:34.410" Score="6" ViewCount="2034" Body="&lt;p&gt;Free silicon is not found in nature (in water, it might be present in the form $\ce{Si(OH)_{4}}$. Some compounds of silicon, such as silane ($\ce{SiH_{4}}$) will react spontaneously in an aqueous environment, while the analogous methane won't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an explanation for the tendency of silicon compounds to react with water?&lt;/p&gt;&#xA;" OwnerUserId="15963" LastEditorUserId="4231" LastEditDate="2015-05-03T20:32:49.370" LastActivityDate="2015-05-03T20:32:49.370" Title="Why does free silicon react with water?" Tags="&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30803" PostTypeId="2" ParentId="30801" CreationDate="2015-05-03T20:07:49.087" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an explanation for the tendency of silicon compounds to react&#xA;  with water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reaction with water, oxygen, &lt;em&gt;etc&lt;/em&gt;., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + \frac{1}{2}O2 -&amp;gt; CH3OH}$$&#xA;$$\ce{SiH4 + \frac{1}{2}O2 -&amp;gt; SiH3OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond.  The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out.  Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Table contains the bond energies for these bonds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|c|c|} \hline&#xA;\ce{X} &amp;amp; \text{X-H Bond Strength} &amp;amp; \text{X-O Bond Strength} &amp;amp; \text{Overall Change}  \\ &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)} &amp;amp; \text{(kcal/mol)}\\ \hline&#xA;\ce{C} &amp;amp; 99 &amp;amp; 86 &amp;amp; -13\\ \hline&#xA;\ce{Si} &amp;amp; 75 &amp;amp; 110 &amp;amp; 35\\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond.  Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon.  This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it.  It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-03T20:07:49.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30804" PostTypeId="1" CreationDate="2015-05-03T20:51:07.093" Score="3" ViewCount="950" Body="&lt;p&gt;At my university there is some fields of chemistry that i don't know if i should focus on them...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How important are in biotechnology and medicinal chemistry those fields of chemistry? :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Quantum chemistry  &lt;/li&gt;&#xA;&lt;li&gt;Physical chemistry&lt;/li&gt;&#xA;&lt;li&gt;Computional chemistry&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15971" LastActivityDate="2015-05-04T06:53:27.413" Title="How important are those fields of chemistry in biotechnology and medicinal chemistry" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30805" PostTypeId="1" AcceptedAnswerId="30813" CreationDate="2015-05-03T20:56:10.720" Score="7" ViewCount="809" Body="&lt;p&gt;When neutral $\ce{Cl}$ atoms are bombarded by high energy photons, causing ejection of electrons, which subshells' ejected electrons will have highest velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose $1s$ because I thought more energy is needed to ejected those electrons so they might have higher velocity. My second choice $3s$ is right, but I can't explain why.&lt;/p&gt;&#xA;" OwnerUserId="15966" LastEditorUserId="15966" LastEditDate="2015-05-04T06:49:42.893" LastActivityDate="2017-10-12T20:39:50.903" Title="Velocity of electrons ejected from Cl atom" Tags="&lt;electrons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30807" PostTypeId="1" AcceptedAnswerId="30809" CreationDate="2015-05-03T20:58:00.497" Score="3" ViewCount="160" Body="&lt;pre&gt;&lt;code&gt;ATOM 22 N  GLN L 3 23.945 -2.849 106.974 1.00 15.00 N &#xA;ATOM 23 CA GLN L 3 23.125 -2.052 106.036 1.00 15.00 C &#xA;ATOM 24 C  GLN L 3 22.898 -0.648 106.616 1.00 15.00 C &#xA;ATOM 25 O  GLN L 3 22.613 -0.491 107.825 1.00 15.00 O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In every amino acid there are two &lt;code&gt;O&lt;/code&gt; atoms; which one is described by the (in this example) fourth entry? (I guess it's the one that's next to the next amino acid's &lt;code&gt;N&lt;/code&gt;, but then where are the coordinates of the other one or should this be obvious?)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ATOM 26 CB   GLN L 3 21.754 -2.682 105.749 1.00 15.00 C &#xA;ATOM 27 CG   GLN L 3 21.577 -3.299 104.379 1.00 15.00 C &#xA;ATOM 28 CD   GLN L 3 20.956 -4.676 104.431 1.00 15.00 C &#xA;ATOM 29 OE1  GLN L 3 20.940 -5.348 105.476 1.00 15.00 O &#xA;ATOM 30 NE2  GLN L 3 20.496 -5.137 103.279 1.00 15.00 N &#xA;ATOM 31 H    GLN L 3 24.339 -2.372 107.727 1.00 0.00  H &#xA;ATOM 32 HE21 GLN L 3 20.637 -4.559 102.496 1.00 0.00  H &#xA;ATOM 33 HE22 GLN L 3 20.049 -6.007 103.275 1.00 0.00  H &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's this entry's nr.31 supposed to describe? I thought that an entry without remoteness indicator (and branch designator) might be bound to the previous atom, but in this case the &lt;code&gt;N&lt;/code&gt;'s adjacent &lt;code&gt;H&lt;/code&gt; atoms are already dealt with with entries 32 and 33.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this doesn't quite follow the convention(?)* of increasing the branch designator by one each time a new branch emerges—is this usual when &lt;code&gt;H&lt;/code&gt; atoms' positions are given as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* like here:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ATOM 1661 CG  TRP A 214 24.705  7.905 6.526 1.00 50.03 C  &#xA;ATOM 1662 CD1 TRP A 214 23.435  7.531 6.857 1.00 49.65 C  &#xA;ATOM 1663 CD2 TRP A 214 25.087  8.914 7.475 1.00 51.52 C  &#xA;ATOM 1664 NE1 TRP A 214 22.999  8.246 7.947 1.00 51.68 N  &#xA;ATOM 1665 CE2 TRP A 214 23.993  9.098 8.350 1.00 52.47 C  &#xA;ATOM 1666 CE3 TRP A 214 26.251  9.675 7.670 1.00 52.68 C  &#xA;ATOM 1667 CZ2 TRP A 214 24.019 10.027 9.407 1.00 54.73 C  &#xA;ATOM 1668 CZ3 TRP A 214 26.279 10.600 8.724 1.00 53.86 C  &#xA;ATOM 1669 CH2 TRP A 214 25.167 10.761 9.581 1.00 54.84 C  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15084" LastActivityDate="2015-05-03T21:27:41.843" Title="About undocumented conventions(?) in PDB files" Tags="&lt;organic-chemistry&gt;&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30808" PostTypeId="2" ParentId="30800" CreationDate="2015-05-03T21:17:13.860" Score="3" Body="&lt;p&gt;A voltaic circuit needs to be a complete circuit first an foremost. So, to answer your 2 main concerns, think of a voltaic cell as a electric circuit:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Water is not a good conductor of electricity, dissolved ions are needed, such as the $\ce{Sn^{2+}}$ ions are needed to facilitate the movement of the electric current in the first place. Also, the presence of the ions allows the overall redox reaction be spontaneous to provide the electric current (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemWiki)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the ChemWiki &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voltaic Cells&lt;/a&gt;, the metals plates provide a surface for the reactions to occur on, a wire just submersed into the solution provides a very small surface area for the reaction to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram from the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Basics_of_Electrochemistry/Electrochemical_Cells&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrochemical Cells&lt;/a&gt; helps explain this (&lt;em&gt;the example uses zinc instead of tin, but the principle is the same&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQT3w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T11:22:37.153" LastActivityDate="2015-06-20T11:22:37.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30809" PostTypeId="2" ParentId="30807" CreationDate="2015-05-03T21:22:09.267" Score="4" Body="&lt;p&gt;First, your PDB data seems to have some formatting issues, in that the columns are in the wrong places.  I'm not sure if it's actually messed up, or if it simply got messed up when you posted it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In every amino acid there are two O atoms&quot;:  True for a single amino acid; not true for an amino acid residue.  I presume you're referring to the two oxygens on the carboxyl group.  However, one of these is lost when the amino acid forms an amide linkage with the next amino acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;What's this entry's nr.31 supposed to describe? &quot;:  I opened the PDB in PyMol (after I corrected the format), and it seems to be the amide hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the branch designator -- I've never heard of such a thing in a PDB file, and I don't see any mention of it in the format description.  Could you clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA:  I've tried to post the correctly formatted PDB data, but it seems to be messing up the formatting.  I'll try to figure out how to do that.&#xA;Here's an image of the residue&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pDmKY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-05-03T21:27:41.843" LastActivityDate="2015-05-03T21:27:41.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30810" PostTypeId="2" ParentId="19119" CreationDate="2015-05-03T21:36:32.740" Score="4" Body="&lt;p&gt;Not true at all.  The $\ce{C=O}$ bond is one of the most strongly IR active bonds there is (and the IR activity of $\ce{CO2}$ is the reason it's a greenhouse gas).&#xA;I suspect the person who told you this was thinking that because $\ce{CO2}$ doesn't have a static dipole, it can't be IR active.  However, IR activity is the result of dynamic dipoles (meaning the dipole changes with some type of deformation motion; in the case of $\ce{CO2}$, this occurs with bending motion and asymmetric stretching, as another answerer described), not static dipoles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some symmetric molecules, like $\ce{N2}$ or $\ce{O2}$, the only vibrational modes that can exist are stretching of the only bond, which because it's symmetric, doesn't lead to a dipole change.  Thus, those species are not IR active.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T21:38:33.840" LastActivityDate="2015-05-03T21:38:33.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30811" PostTypeId="1" AcceptedAnswerId="30812" CreationDate="2015-05-03T21:40:22.003" Score="9" ViewCount="2366" Body="&lt;p&gt;When does formal charge have more precedence over the octet rule / an atom having a full valence shell when drawing structures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you draw $\ce{BCl3}$ for example, there are no double bonds - just three single bonds. Because of this $\ce{B}$ does not end up with 8 valence electrons. The explanation I have seen for this is that a double bond makes it so that the formal charge is undesirable. When drawing structures which has a higher precedence: formal charge or ensuring the atoms have full valence shells?&lt;/p&gt;&#xA;" OwnerUserId="7612" LastEditorDisplayName="user15489" LastEditDate="2015-05-03T21:43:03.737" LastActivityDate="2015-05-03T22:10:30.167" Title="Full Valence Shell (Octet Rule) vs Formal Charge" Tags="&lt;lewis-structure&gt;&lt;molecular-structure&gt;&lt;formal-charge&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30812" PostTypeId="2" ParentId="30811" CreationDate="2015-05-03T22:00:19.870" Score="5" Body="&lt;p&gt;First, keep in mind that Lewis-structures and formal charges are huge simplifications.  In reality, at the quantum level, charges and bonds are spread out and it's somewhat unrealistic to say &quot;Atom X has a charge of Q&quot; or &quot;Atoms X and Y have an Nth order bond between them&quot;.  There's merely a cloud of electron density, and it's somewhat arbitrary to say which parts of that cloud belong to a given atom or bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's perfectly valid to draw a structure with a double bond.  And in reality, I'm sure there is hyperconjugation that leads to a small amount of double bond behavior.  However, such a resonance structure puts a negative charge on an electropositive element, and a positive charge on a very electronegative element, and thus such a structure has a smaller contribution to the overall structure.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="15960" LastEditDate="2015-05-03T22:10:30.167" LastActivityDate="2015-05-03T22:10:30.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30813" PostTypeId="2" ParentId="30805" CreationDate="2015-05-03T22:28:28.750" Score="7" Body="&lt;p&gt;It would be the $3s$ shell because less energy is required to bump it off course. Due to electron repulsion and the shielding effect, the $e-$ in the $3s$ is more prone to get dislodged. And of course because of this, it will have a higher velocity when it gets &lt;strong&gt;bumped&lt;/strong&gt; off. &lt;strong&gt;Another concept to note: $e-$ will have more energy at higher energy levels.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reason why it wouldn't be the $1s$ shell is because due to the strong pull on the $1s$ electron. Since it is on the first energy level, it is strongly &lt;strong&gt;held&lt;/strong&gt; by the nucleus. &lt;strong&gt;&lt;em&gt;Even it were to get dislodged it would move very slow because of the strong pull by the nucleus.&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-05-03T22:49:17.767" LastActivityDate="2015-05-03T22:49:17.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30815" PostTypeId="2" ParentId="28712" CreationDate="2015-05-03T23:04:21.173" Score="0" Body="&lt;p&gt;Although there're more shielding as you add more electrons, the nuclear charge also increases as you go from left to right because of the increase number of protons. The nuclear charge's pull on electrons has way more effect on the ionization energy than the gradual shielding does (shielding will have a bigger impact down the periods because of different energy levels). So the IE does increase from left to right.&lt;/p&gt;&#xA;" OwnerUserId="15966" LastActivityDate="2015-05-03T23:04:21.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30816" PostTypeId="1" AcceptedAnswerId="30819" CreationDate="2015-05-03T23:10:45.713" Score="7" ViewCount="1084" Body="&lt;p&gt;Hasok Chang, a chemist, but perhaps more a philosopher recently published the book &lt;a href=&quot;http://www.springer.com/us/book/9789400739314&quot;&gt;Is Water H2O&lt;/a&gt;? . Chang used the question and an extensive web of arguments ultimately as a platform to support a pluralistic approach to scientific endeavor. I don't know that I buy it all, but in a latter chapter he made some statements regarding 'fundamental' components of chemical reactions, and one in particular that interested me and brings me to the point of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen gas as I understand it &lt;em&gt;naturally&lt;/em&gt; occurs as a hydrogen molecule - two hydrogen atoms stuck together as a diatomic molecule. If ionized, the electron is stripped from the atoms and you have hydrogen ions or rather protons. But what about single hydrogen atoms? Are these non-existent or just a rare thing? Is there some way I can arrange to have a container of single hydrogen atoms and prevent their grouping into hydrogen molecules without ionizing them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what answers that follow - do these apply in general to the stable nature of all diatomic gases?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2016-03-27T23:08:27.547" Title="Really how rare are hydrogen atoms?" Tags="&lt;atoms&gt;&lt;ions&gt;&lt;molecules&gt;&lt;stability&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30817" PostTypeId="2" ParentId="28712" CreationDate="2015-05-03T23:24:19.510" Score="1" Body="&lt;p&gt;Crash Course on Ionization Energy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As we all know, atoms &lt;strong&gt;prefer&lt;/strong&gt; a full valence shell. So as we go right in a period, we are increasing $e-$. And also &lt;strong&gt;ADDING PROTONS&lt;/strong&gt;. Because we are adding protons, the size of the atom gets smaller because the nuclear charge will be more powerful. Adding protons in a period trumps the addition of electrons. At the end of the day, we have a small atom with many electrons in it's valence shell that does not want to let go of them. Especially the &lt;strong&gt;Noble Gasses and Halogens&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Ionization energy decreases as we move down a group because:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As we move down, a new full energy level is being added. More electrons means more repulsion.&lt;/li&gt;&#xA;&lt;li&gt;This creates the shielding effect where the addition of the shells, shields the outer electron from receiving the nucleic charge. &lt;strong&gt;NOTE&lt;/strong&gt;: Here, however the addition of another energy level trumps the addition of protons.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is just a piece of the whole picture.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-03T23:24:19.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30818" PostTypeId="1" CreationDate="2015-05-03T23:48:23.850" Score="5" ViewCount="1028" Body="&lt;p&gt;A transition metal with 7 d electrons forms a complex ion with cyanide. I know that cyanide is a high field ligand, which means the electrons will pair before moving to the next energy level, but wouldn't there still be one unpaired electron? And wouldn't that make it paramagnetic? I was told that my prediction was wrong and that it was in fact diamagnetic, but I don't understand why. &lt;/p&gt;&#xA;" OwnerUserId="15921" LastActivityDate="2015-05-04T15:10:12.827" Title="Is a transition metal with 7 d electons ever diamagnetic?" Tags="&lt;inorganic-chemistry&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30819" PostTypeId="2" ParentId="30816" CreationDate="2015-05-03T23:55:00.000" Score="10" Body="&lt;p&gt;While usually short-lived, isolated hydrogen atoms (also called hydrogen radicals) are well studied and chemically important species. Bond formation in simple species is always &lt;a href=&quot;http://en.wikipedia.org/wiki/Interatomic_potential&quot;&gt;an exothermic reaction (usually highly so) with zero activation energy&lt;/a&gt;, which means bond formation is thermodynamically very favourable and has no kinetic barrier stopping it from happening. Simply put, the isolated atoms really want to react with each other, and they'll attract each other at a distance, so if at some moment you create a container with pure monoatomic hydrogen, the hydrogen atoms will react as soon as possible to regenerate dihydrogen molecules (hydrogen gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best you can do is expose the dihydrogen molecules to energetic conditions, so that a fair amount of bonds are being constantly broken at a rate comparable to which they're formed. A simple way to do this is to strongly heat the hydrogen gas; enough thermal energy will cause the bonds to break, though the dihydrogen bond is unusually strong ($\rm{436\ kJ\ mol^{-1}}$) so even at several thousand degrees kelvin there will be only a small portion of free hydrogen atoms in the gas (see Mithoron's comment). A more interesting option is to shine a strong ultraviolet light into the gas, as the dihydrogen molecules can separate into their constituent atoms upon absorption of photons with energy around $\rm{4.52\ eV}$, which corresponds to a photon wavelength of $\rm{274\ nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any substance can have its bonds broken through the above two methods (among others), so the same applies to all other diatomic molecules. The stronger the bond between the atoms, the more uncommon is it to find the isolated atoms. For substances with very weak bonds, such as molecular dibromine or diiodine, mere exposure to visible light is enough to trigger formation of a fair amount of isolated atoms, something which is exploited in several reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping in mind that isolated atoms are very reactive, they are rather abundant in space. The average density of matter in the Universe is very, very low (about 1 atom per 5 cubic metres of space!), so isolated atoms can go a long time before finding anything to react &lt;em&gt;with&lt;/em&gt;. This makes them extremely important for understanding astrochemistry.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-05-04T00:06:39.690" LastActivityDate="2015-05-04T00:06:39.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30821" PostTypeId="2" ParentId="29685" CreationDate="2015-05-04T01:22:56.203" Score="6" Body="&lt;p&gt;First of all, any 'tin' can you are likely to find is actually made of steel. Tin is too expensive. Also, tin does not 'rust', although it oxidizes. Your rust is iron oxide. Galvanized steel is steel with a thin zinc coating, likely hot-dip galvanization. What you describe (crystalline surface or spangles sort of appearance) does sound like galvanization, although I am not aware of any steel cans that are galvanized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I thought galvanised steel only rust when the layer of galvanisation wears out.&lt;/strong&gt; This is true. If your can is indeed galvanized, then it has likely chipped off, possibly on the rims of the can where the lid it crimped or pressed together, or by cutting the can open. Strong acids will dissolve zinc  Also interesting to note: sodium chloride (salt) will dissolve zinc over time by converting the galvanization to zinc chloride, but that's not applicable here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, is the tin made with galvanised steel?&lt;/strong&gt;&#xA;  Well, that does sound like what you were describing, but galvanized steel is not fit for foodstuffs. Zinc is mildly toxic and ingesting it can make you ill. I like feetwet's suggestion; that universally the inside of metal food cans are lined, by actually an epoxy that is sprayed on. This is almost certainly to be the case. Perhaps this makes sense: if your can was indeed galvanized, then the rational by the manufacturer was that the lining will keep the zinc separated and the food safe. Sounds legit. It might be good to be aware that the type of epoxy being used here is the one that uses as its plasticizer Bisphenol A and Bisphenol S (BPA/BPS). Ill leave it up to you to decide if that should bother you or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If yes, why are there concentrated spots of rust on the wall and other rusting?&lt;/strong&gt;&#xA;Well, if there is rust, then we can deduce that water must have be coming in contact with the iron in the steel, so whatever was protecting the steel has some failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And BTW, why does the emptied and washed can smell &quot;metallish&quot;?&lt;/strong&gt;&#xA;While zinc does have a metallic smell to it, iron oxide's is much stronger. Since you visually identified rust, this is what you are smelling.&lt;/p&gt;&#xA;" OwnerUserId="1942" LastEditorUserId="1942" LastEditDate="2015-05-06T10:23:16.240" LastActivityDate="2015-05-06T10:23:16.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30822" PostTypeId="2" ParentId="30745" CreationDate="2015-05-04T02:09:09.560" Score="1" Body="&lt;p&gt;If You are stating you have your initial water temperature and the final solution of your temperature, then your change in temperature is just your final temperature minus your initial temperature. No need for linear equations. The change in temperature for the reaction is the temperature change of the water due to the heat given off or absorbed by the system to the surroundings. &lt;/p&gt;&#xA;" OwnerUserId="15977" LastActivityDate="2015-05-04T02:09:09.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30823" PostTypeId="1" AcceptedAnswerId="30824" CreationDate="2015-05-04T02:42:17.777" Score="2" ViewCount="1230" Body="&lt;p&gt;If doing a problem where you use the integrated rate law for a second order reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A_0}]}=kt$) to first find $k$, then use that to find the half-life. Does it matter seconds are used, or minutes, or whatever?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="402" LastEditDate="2016-11-12T21:14:45.510" LastActivityDate="2016-11-12T21:14:45.510" Title="Unit of time used in rate law equations" Tags="&lt;kinetics&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30824" PostTypeId="2" ParentId="30823" CreationDate="2015-05-04T03:28:36.857" Score="2" Body="&lt;p&gt;The unit of $k$ is $\frac{\mathrm{concentration}}{\mathrm{unit~time}}$. It is not necessary that you use any specific unit for concentration or time, but most often $t$ is measured in seconds, and the unit of $k$ is $\mathrm{mol~L^{-1}~s^{-1}}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-04T03:28:36.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30825" PostTypeId="1" CreationDate="2015-05-04T04:08:11.003" Score="4" ViewCount="60" Body="&lt;p&gt;Palladium hydride is an example of a hydride which contains H atoms in palladium's crystal lattice (which is fcc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if it's possible to predict if another hydride, $\ce{NbH_{x}}$ could exhibit the same behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in an ionic solid, the fraction $$\frac{r_{+}}{r_{-}}$$&#xA;Can be used to predict the geometry around a cation. &#xA;Can I use this expression for niobium hydride? In the case of niobium, the crystal structure is bcc, so it contains tetrahedral and octahedral interstial sites. &lt;/p&gt;&#xA;" OwnerUserId="15963" LastEditorUserId="10256" LastEditDate="2015-05-04T09:22:03.510" LastActivityDate="2015-05-04T09:36:21.237" Title="How to predict if hydrogen atoms could be placed in the interstials of a lattice" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30826" PostTypeId="1" AcceptedAnswerId="30833" CreationDate="2015-05-04T04:25:57.073" Score="10" ViewCount="5203" Body="&lt;p&gt;I did a &lt;a href=&quot;https://www.google.com.co/search?num=100&amp;amp;es_sm=93&amp;amp;q=exchange-correlation%20hole&amp;amp;spell=1&amp;amp;sa=X&amp;amp;ei=vvBGVZGCLsmegwS6qYHwBQ&amp;amp;ved=0CBgQvwUoAA&amp;amp;biw=769&amp;amp;bih=768&quot;&gt;google&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/search?q=exchange%20hole&quot;&gt;chemistry.exchange&lt;/a&gt; search and found several technical definitions of the exchange correlation hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions that I want to proposes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What the exchange-correlation hole means in common words?&lt;/li&gt;&#xA;&lt;li&gt;How do you explain it to a non-literate person?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastActivityDate="2017-02-21T19:10:52.377" Title="Exchange-correlation hole" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="30829" PostTypeId="1" AcceptedAnswerId="34496" CreationDate="2015-05-04T06:14:55.370" Score="3" ViewCount="637" Body="&lt;p&gt;I'm trying to calculate the energy required to dissociate a single $\ce{CO2}$ molecule into its respective elements, and I'm given that it takes $ 373.6 \times 10^{3} \, \, \mathrm{\frac{joules}{gram}} $ where  $44.01$ grams is the mass of $6.022 \times 10^{23}$ molecules of $\ce{CO2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By dimensional analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$170.43\,\, \mathrm{eV/molecule}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this derivation and the logical assumptions made herein correct?&lt;/p&gt;&#xA;" OwnerUserId="15197" LastEditorDisplayName="user7951" LastEditDate="2015-09-12T16:15:02.650" LastActivityDate="2015-09-12T16:15:19.900" Title="Dissociation Energy for CO2 per molecule" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30830" PostTypeId="2" ParentId="29421" CreationDate="2015-05-04T06:16:42.160" Score="2" Body="&lt;p&gt;Why does concentrated $\ce{H_2SO_4}$ oxidise? Due to the nascent oxygen it gives:&#xA;$$\ce{H_2SO_4}\rightleftharpoons\ce{SO_2 +H_2O +[O]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want $\ce{HCl}$ to be oxidised, the following reaction must exist:&#xA;$$\ce{HCl} +\ce{[O]}\rightarrow\ce{HClO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, you can observe that the reaction$$\ce{HClO}\rightarrow\ce{HCl} +\ce{[O]}$$&#xA;can't be backward under normal conditions (you probably have read this reaction in the bleaching action of &lt;a href=&quot;http://www.gktoday.in/bleaching-powder/&quot; rel=&quot;nofollow&quot;&gt;bleaching powder&lt;/a&gt;). So, fortunately, you can safely use $\ce{H_2SO_4}$ in this purpose...and take its advantage to $\ce{CaCl_2}$, as agha rehan abbas and Molx mentioned.&lt;/p&gt;&#xA;" OwnerUserId="15979" LastEditorUserId="15410" LastEditDate="2015-05-04T06:57:52.523" LastActivityDate="2015-05-04T06:57:52.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30831" PostTypeId="2" ParentId="30783" CreationDate="2015-05-04T06:38:28.103" Score="2" Body="&lt;p&gt;You are not wrong in your understanding of a water softener, nor are you wrong in your understanding of a reverse osmosis filtration system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the RO filter care whether the supply water is salty or hard?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to make the point that your RO filter is incapable of caring about the hardness of the water. This is more about carefully choosing your words more than anything else. I was also tripped up by this when I first began chemistry, but it was something I learned to do and I challenge you to do the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it more effective at filtering one or the other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, but only slightly. Here you can find a list of the &lt;a href=&quot;http://espwaterproducts.com/about-reverse-osmosis.htm&quot; rel=&quot;nofollow&quot;&gt;typical rejection characteristics&lt;/a&gt; of RO filters. The lower bound for the percent amount of multivalent ions ($\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$) filtered is higher than the lower bound for the percent amount of monovalent ions ($\ce{Na+}$ and $\ce{K+}$) filtered, but the upper bounds are very close in value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall you could say that RO systems are more efficient at filtering hard water, but the difference is not very large. Most systems will likely be performing considerably above the lower bounds of the typical rejection characteristics anyway, so this should not be a deciding factor in deciding to feed your RO hard or soft water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is one going to influence the life of the RO filter more than the other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes! In an ideal RO filtration system, all of the filtered materials would simply drain away, and pure water would flow to the storage tank, but nothing is ideal. If you feed hard water into your RO system, there will be scaling on the semi-permeable membrane that will eventually reduce its filtered water output. This can cause lack of flow will create a potentially damaging buildup of pressure elsewhere in the system. I would therefore recommend that you keep your setup the way it is!&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-04T06:38:28.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30832" PostTypeId="2" ParentId="30804" CreationDate="2015-05-04T06:46:57.040" Score="3" Body="&lt;p&gt;Your question is broad and these chemistry which you have pointed out in your question is interlinked (Quantum ,Physical and Computational chemistry). The best way which I can explain this question is by giving links to main research carried out using these chemistry in biological science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study of quantum mechanics in photosynthesis is one such area and this &lt;a href=&quot;http://www.nature.com/nature/journal/v446/n7137/full/nature05678.html&quot; rel=&quot;nofollow&quot;&gt;nature paper(Nature 446, 782-786 (12 April 2007)&lt;/a&gt; is about that research&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the highlights of the paper is the evidence for wavelike energy transfer through quantum coherence in photosynthetic systems&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;two-dimensional electronic spectroscopy investigations of the FMO&#xA;  bacteriochlorophyll complex, and obtain direct evidence for remarkably&#xA;  long-lived electronic quantum coherence playing an important part in&#xA;  energy transfer processes within this system. The quantum coherence&#xA;  manifests itself in characteristic, directly observable quantum&#xA;  beating signals among the excitons within the Chlorobium tepidum FMO&#xA;  complex at 77 K. This wavelike characteristic of the energy transfer&#xA;  within the photosynthetic complex can explain its extreme efficiency,&#xA;  in that it allows the complexes to sample vast areas of phase space to&#xA;  find the most efficient path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other useful links&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v431/n7006/full/431256a.html&quot; rel=&quot;nofollow&quot;&gt;Nature 431, 256-257 (16 September 2004)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nature.com/nature/journal/v446/n7137/full/446740a.html&quot; rel=&quot;nofollow&quot;&gt;Nature 446, 740-741 (12 April 2007)&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="139" LastEditDate="2015-05-04T06:53:27.413" LastActivityDate="2015-05-04T06:53:27.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30833" PostTypeId="2" ParentId="30826" CreationDate="2015-05-04T06:55:27.253" Score="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What the exchange-correlation hole means in common words?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that the probability is the common word, one could say that the &lt;em&gt;exchange-correlation hole&lt;/em&gt;, is a region of space around an electron in which the probability of finding another electron is close to zero due to &lt;em&gt;electron correlation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you explain it to a non-literate person?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long story short is that mathematically electron correlation arises as a consequence of the instantaneous Coulomb repulsive forces between electrons and also as a consequence of the fundamental principle of antisymmetry of an electronic (fermionic in general) wave function. Note that the first kind of electron correlation which is due to Coulomb repulsions is known as the &lt;em&gt;Coulomb correlation&lt;/em&gt; while the second one which is due to antisymmetry of a wave function is knows as the &lt;em&gt;exchange correlation&lt;/em&gt;. However, since the root of all ab-initio wave function methods, the HF method, already accounts for exchange correlation, the &lt;em&gt;Coulomb correlation&lt;/em&gt; is usually also referred to as just the &lt;em&gt;correlation&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to these two different kinds of electron correlation note that exchange-correlation hole does &lt;em&gt;not&lt;/em&gt; refer specifically to just the exchange correlation, rather it is the hole due to &lt;em&gt;both&lt;/em&gt; types of the electron correlation mentioned. I think that one would better call it, say, exchange-Coulomb correlation hole to avoid an ambiguity, but I'm afraid we're stuck with the exchange-correlation hole term.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long story&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\vec{r}_{1}$ - the event of finding electron-one at a point $\vec{r}_{1}$;&lt;/li&gt;&#xA;&lt;li&gt;$\vec{r}_{2}$ - the event of finding electron-two at a point $\vec{r}_{2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Probability theory reminds us that in the general case the so-called &lt;em&gt;joint probability&lt;/em&gt; of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by&#xA;\begin{equation*}&#xA;    \Pr(\vec{r}_{1} \cap \vec{r}_{2})&#xA;    = &#xA;    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \Pr(\vec{r}_{2})&#xA;    = &#xA;    \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \, ,&#xA;\end{equation*}&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\Pr(\vec{r}_{1})$ is the probability of finding electron-one at point $\vec{r}_{1}$ &lt;em&gt;irrespective&lt;/em&gt; of the position of electron-two;&lt;/li&gt;&#xA;&lt;li&gt;$\Pr(\vec{r}_{2})$ is the probability of finding electron-two at point $\vec{r}_{2}$ &lt;em&gt;irrespective&lt;/em&gt; of the position of electron-one;&lt;/li&gt;&#xA;&lt;li&gt;$\Pr(\vec{r}_{1}\,|\,\vec{r}_{2})$ is the probability of finding electron-one at point $\vec{r}_{1}$, &lt;em&gt;given&lt;/em&gt; that electron-two is at $\vec{r}_{2}$;&lt;/li&gt;&#xA;&lt;li&gt;$\Pr(\vec{r}_{2}\,|\,\vec{r}_{1})$ is the probability of finding electron-two at point $\vec{r}_{2}$, &lt;em&gt;given&lt;/em&gt; that electron-one is at $\vec{r}_{1}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two of the above probabilities are referred to as &lt;em&gt;unconditional probabilities&lt;/em&gt;, while the last two are referred to as &lt;em&gt;conditional probabilities&lt;/em&gt;, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are &lt;em&gt;independent&lt;/em&gt; of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts &#xA;\begin{equation*}&#xA;    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) = \Pr(\vec{r}_{1}) \, , \quad&#xA;    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) = \Pr(\vec{r}_{2}) \, ,&#xA;\end{equation*}&#xA;and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,&#xA;\begin{equation*}&#xA;    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) = \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .&#xA;\end{equation*}&#xA;Note that this is the picture in which electrons are sort of billiard balls, i.e. neutral macroscopic particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, electrons are, of course, not like billiard balls; they are charged microscopic particles, and consequently,&#xA;\begin{equation*}&#xA;    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \, ,&#xA;    \quad&#xA;    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \neq \Pr(\vec{r}_{2}) \, ,&#xA;\end{equation*}&#xA;and&#xA;\begin{equation}&#xA;    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .&#xA;\end{equation}&#xA;And since the spatial probabilities of finding electrons are inevitably related to their states, the inequality above means that the state of electron-one is not independent of the state of electron-two and vice versa.&#xA;This interdependency of the states of electrons is referred to as the &lt;em&gt;electron correlation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inequalities above, which are the mathematical essence of electron correlation, hold true for two reasons. First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons the conditional probability is less than half of the corresponding unconditional one&#xA;$$&#xA;    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) &amp;lt; \Pr(\vec{r}_{1}) \, ,&#xA;    \quad           &#xA;    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) &amp;lt; \Pr(\vec{r}_{2}) \, ,&#xA;$$&#xA;while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one&#xA;$$&#xA;    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) &amp;gt; \Pr(\vec{r}_{1}) \, ,&#xA;    \quad           &#xA;    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) &amp;gt; \Pr(\vec{r}_{2}) \, ,&#xA;$$&#xA;In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.&#xA;The correlation due to Coulomb repulsion is termed as &lt;em&gt;Coulomb correlation&lt;/em&gt; and one sometimes speaks about the presence of the so-called &lt;em&gt;Coulomb hole&lt;/em&gt; around each electron - a region of space around it in which the probability of finding another electron is close to zero due to Coulomb repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.&#xA;This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the later, it is still noticeable when electrons are close to each other and not just at the same location in space.&#xA;This type of electron correlation, which is a consequence of the Pauli exclusion principle, is referred to as &lt;em&gt;Fermi correlation&lt;/em&gt;, or &lt;em&gt;exchange correlation&lt;/em&gt;, and one sometimes speaks about the presence of the so-called &lt;em&gt;Fermi hole&lt;/em&gt; around each electron - a region of space around it in which the probability of finding another electron in the same spin state is close to zero.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-04T13:12:45.883" LastActivityDate="2015-05-04T13:12:45.883" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="30835" PostTypeId="1" AcceptedAnswerId="30841" CreationDate="2015-05-04T07:31:25.403" Score="4" ViewCount="1089" Body="&lt;p&gt;I'd like to print the content of property matrices (i.e., the matrices shown after selecting &lt;code&gt;View→Properties→X&lt;/code&gt; where X is one of the six options shown, including &lt;code&gt;Angle properties...&lt;/code&gt;, &lt;code&gt;Atom properties...&lt;/code&gt; and &lt;code&gt;Bond properties...&lt;/code&gt;) in Avogadro, but I can't seem to find this option at Avogadro Wiki or other forms of Avogadro documentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. this is the bond properties matrix. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/KxlWx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="876" LastEditDate="2015-05-04T08:26:15.357" LastActivityDate="2015-05-04T11:12:39.317" Title="How do I print property matrices in Avogadro?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30836" PostTypeId="1" CreationDate="2015-05-04T07:35:25.467" Score="2" ViewCount="135" Body="&lt;p&gt;For an example let's assume we have a weak acid  &lt;strong&gt;HA&lt;/strong&gt; . And we add either a strong or a weak base. Then what I want to know is whether  the weak ionization of the acid &lt;strong&gt;HA&lt;/strong&gt; taken place?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weak ionization of acid &lt;strong&gt;HA&lt;/strong&gt;. ==&gt;  &lt;img src=&quot;https://i.stack.imgur.com/YS5Vc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="9961" LastEditDate="2015-11-13T19:17:46.560" LastActivityDate="2015-11-13T19:17:46.560" Title="Does a weak acid undergo it's weak ionization upon the addition of any kind of base?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30837" PostTypeId="1" CreationDate="2015-05-04T08:04:13.307" Score="0" ViewCount="373" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOROo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marking scheme says that the molecule has both alkene and ketone.&#xA;So there is carbon to carbon double bond.ok.check there is alkene.But what about ketone.Ketone has this structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYjtI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can we technically say that there is ketone in it?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastActivityDate="2015-05-04T09:48:49.630" Title="Why is is this molecule called ketene?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-04T11:24:03.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="30838" PostTypeId="2" ParentId="30826" CreationDate="2015-05-04T08:39:44.530" Score="5" Body="&lt;p&gt;The exchange-correlation hole refers to the fact the probability of finding at electron at a certain position (&lt;strong&gt;r&lt;/strong&gt;1) given that there is an electron at another position (&lt;strong&gt;r&lt;/strong&gt;2) decreases when the distance between the two electrons decrease.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, for a two electron system, the least likely position of one electron on top of the other&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-04T08:39:44.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30839" PostTypeId="2" ParentId="30743" CreationDate="2015-05-04T09:16:34.440" Score="3" Body="&lt;p&gt;In the case of the water solubility what you want to know is how powerful the secondary interaction  of the compound that you are going to dissolve with  water?If the compound that you wish to dissolve have hydrogen bonding then it's definitely dissolve in water .It does not mean others  are  insoluble  in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But  the compounds do not have hydrogen bonding are  less soluble in water when compared to    the ones which have hydrogen bonding.Because ones which have hydrogen bonding interact with water molecules easily than others which do not have hydrogen bonding &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a look at the image   below, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kcM8T.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shows two water molecules  which posses the hydrogen bonding interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now give the eye on compounds that you have got, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The  compounds that you going to dissolve  may form  interaction force  with water as above image. &lt;strong&gt;If that interaction force is hydrogen bonding ($\ce{C5H11OH}$) then it's solubility is greater than ones which do not have hydrogen bonding($\ce{C5H11Cl}$) (dipole dipole interactions)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So because of that solubility of $\ce{C5H11Cl}$ is less than the solubility of ($\ce{C5H11OH}$) with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the bottom line is ,the solubility of a compound is vary according the  secondary interaction between the solute and solvent.In the case of water acting as  the solvent solubility is highest in  compounds which having &lt;strong&gt;H bonding&lt;/strong&gt; , than which haaving dipole-dipole forces and  weak dispersion forces.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastActivityDate="2015-05-04T09:16:34.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30840" PostTypeId="1" AcceptedAnswerId="30846" CreationDate="2015-05-04T09:17:23.557" Score="4" ViewCount="1677" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethoxylation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; page says this:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ethoxylation is an industrial process in which ethylene oxide is added to alcohols and phenols to turn it into a surfactant.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does the reaction proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/27288/&quot;&gt;Do epoxide rings react with bases?&lt;/a&gt;&quot; regarding epoxide rings reacting with bases does not include alcohols, and does not answer my question.&lt;/p&gt;&#xA;" OwnerUserId="15983" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-05-07T13:20:28.493" Title="What is the mechanism of the uncatalysed epoxide opening with alcohols?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="30841" PostTypeId="2" ParentId="30835" CreationDate="2015-05-04T09:32:34.243" Score="2" Body="&lt;p&gt;I am not aware of a menu button to print these tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is however a different way to get your hands on the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro features objects: molecules, bonds and atoms are instances of classes, whose properties can be read and processed by programming. &#xA;The easiest way is the build-in &lt;strong&gt;Python-Terminal&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;Note that I'm using Avogadro 1.1.1 on Linux in German, your menus will have different labels.&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In Avogadro, open a &lt;strong&gt;Python Terminal&lt;/strong&gt; (In German, this is done via&lt;br&gt;&#xA;&lt;em&gt;Einstellungen -&gt; Werkzeugleisten -&gt; Python-Terminal&lt;/em&gt;, which should be something like &lt;em&gt;Settings -&gt; Tool bars -&gt; Python terminal&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When run in the Python terminal of Avogradro, the following snippet writes &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;atom index&lt;/li&gt;&#xA;&lt;li&gt;atomic number&lt;/li&gt;&#xA;&lt;li&gt;x, y, z coordinates&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;for all atoms of the &lt;strong&gt;currently opened&lt;/strong&gt; molecule to a csv file into an &lt;strong&gt;existing&lt;/strong&gt; &lt;code&gt;tmp&lt;/code&gt; directory in the user's home.&#xA;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import os.path  &#xA;import csv&#xA;&#xA;home = os.path.expanduser('~')&#xA;data_dir = 'tmp'&#xA;atom_table_name = 'atoms.csv'&#xA;atom_table_path = os.path.join(home, data_dir, atom_table_name)&#xA;# generate full path for file with atom data&#xA;&#xA;with open(atom_table_path, 'wt') as fout:&#xA;    writer = csv.writer(fout)&#xA;    writer.writerow(('Index', 'Atomic number', 'X', 'Y', 'Z'))&#xA;&#xA;    mol = Avogadro.molecule&#xA;&#xA;    for atom in mol.atoms:&#xA;        index = atom.index + 1&#xA;        x, y, z = atom.pos&#xA;        writer.writerow((index, atom.atomicNumber, x, y, z))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned the Python Terminal first because I didn't want to introduce too much different things at once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have played a bit with the build-in Python Terminal, you will want to go one step beyond and try a more comfortable IDE with logging, tab-completion, build-in help, etc. My suggestion is to try &lt;a href=&quot;http://ipython.org/&quot; rel=&quot;nofollow&quot;&gt;IPython&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;em&gt;IPython&lt;/em&gt; (or any other IDE of your choice), you can make use of Avogadro through its &lt;a href=&quot;http://avogadro.cc/api/dev/index.html&quot; rel=&quot;nofollow&quot;&gt;API&lt;/a&gt; without actually running it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that your data file &lt;code&gt;propen.cml&lt;/code&gt; looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&amp;lt;molecule&amp;gt;&#xA; &amp;lt;atomArray&amp;gt;&#xA;  &amp;lt;atom id=&quot;a1&quot; elementType=&quot;C&quot; x3=&quot;-6.736896&quot; y3=&quot;1.198243&quot; z3=&quot;-0.001854&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a2&quot; elementType=&quot;C&quot; x3=&quot;-5.625957&quot; y3=&quot;2.233390&quot; z3=&quot;-0.108371&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a3&quot; elementType=&quot;H&quot; x3=&quot;-6.748313&quot; y3=&quot;0.730420&quot; z3=&quot;0.988429&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a4&quot; elementType=&quot;H&quot; x3=&quot;-6.615420&quot; y3=&quot;0.408802&quot; z3=&quot;-0.751312&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a5&quot; elementType=&quot;H&quot; x3=&quot;-7.710837&quot; y3=&quot;1.671184&quot; z3=&quot;-0.163704&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a6&quot; elementType=&quot;C&quot; x3=&quot;-4.268024&quot; y3=&quot;1.629800&quot; z3=&quot;0.106393&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a7&quot; elementType=&quot;H&quot; x3=&quot;-5.801148&quot; y3=&quot;3.017025&quot; z3=&quot;0.637857&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a8&quot; elementType=&quot;H&quot; x3=&quot;-5.668681&quot; y3=&quot;2.695849&quot; z3=&quot;-1.101309&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a9&quot; elementType=&quot;C&quot; x3=&quot;-3.137190&quot; y3=&quot;2.345874&quot; z3=&quot;0.062502&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a10&quot; elementType=&quot;H&quot; x3=&quot;-4.203709&quot; y3=&quot;0.562202&quot; z3=&quot;0.306350&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a11&quot; elementType=&quot;H&quot; x3=&quot;-3.147760&quot; y3=&quot;3.413685&quot; z3=&quot;-0.133506&quot;/&amp;gt;&#xA;  &amp;lt;atom id=&quot;a12&quot; elementType=&quot;H&quot; x3=&quot;-2.172936&quot; y3=&quot;1.873521&quot; z3=&quot;0.222912&quot;/&amp;gt;&#xA; &amp;lt;/atomArray&amp;gt;&#xA; &amp;lt;bondArray&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a1 a2&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a1 a3&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a1 a4&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a1 a5&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a2 a6&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a2 a7&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a2 a8&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a6 a9&quot; order=&quot;2&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a6 a10&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a9 a11&quot; order=&quot;1&quot;/&amp;gt;&#xA;  &amp;lt;bond atomRefs2=&quot;a9 a12&quot; order=&quot;1&quot;/&amp;gt;&#xA; &amp;lt;/bondArray&amp;gt;&#xA;&amp;lt;/molecule&amp;gt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can load the CML file and iterate over the bonds to display the index numbers of the atoms in the bond and the bond length using:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import Avogadro as av&#xA;&#xA;mol = av.MoleculeFile.readMolecule('propen.cml')&#xA;# don't forget to adjust the path for the data file!&#xA;&#xA;for bond in mol.bonds:&#xA;    first = bond.beginAtom&#xA;    last = bond.endAtom&#xA;    print first.index, last.index, bond.length&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-04T11:12:39.317" LastActivityDate="2015-05-04T11:12:39.317" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="30842" PostTypeId="2" ParentId="30825" CreationDate="2015-05-04T09:36:21.237" Score="1" Body="&lt;p&gt;Your intuition is correct; $\ce{NbH_{x}}$ retains the bcc Nb lattice, with the hydrides in the tetrahedral interstital sites, up to x = 2, when the compound switches to a fcc Nb lattice (the &lt;a href=&quot;http://www.chemtube3d.com/solidstate/_fluorite%28final%29.htm&quot; rel=&quot;nofollow&quot;&gt;Fluorite structure&lt;/a&gt;). This is pretty typical behaviour for a transition metal hydride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think radius ratio rules are going to be useful in predicting the behaviour of these systems; there is a lot of covalency going on (hydride-hydride interactions and Nb-Nb interactions), and the metal lattices will end up being distorted by the correlated interstital occupancies of the hydrides. Hydride has a radius of around 130 pm while Niobium has a radius of around 150 pm so you wouldn't predict the observed behaviour based on radius ratios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information on this particular system can be found in &lt;a href=&quot;http://journals.iucr.org/a/issues/1977/01/00/a14075/a14075.pdf&quot; rel=&quot;nofollow&quot;&gt;Hauck (1976), Ordering of Hydrogen in Niobium Hydride phases&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-04T09:36:21.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30843" PostTypeId="2" ParentId="30837" CreationDate="2015-05-04T09:48:49.630" Score="3" Body="&lt;p&gt;This molecule is a ketene because it contains the ketene functional group:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/bILkS.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Whilst the ketene appears to contain an alkene and a ketone functional group, which as you have suggested is where its name comes from, its chemistry is notably different from both of its 'constituent groups'.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-04T09:48:49.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30845" PostTypeId="2" ParentId="19495" CreationDate="2015-05-04T10:11:42.417" Score="2" Body="&lt;p&gt;As answered above water in pure form is tasteless and odourless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to various reasons there is change in taste and odour(smell) of water. Some of the different smell or taste are as following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Chlorine taste or smell:&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;this occurs when the water is treated at the water treatment plant to&#xA; disinfect it.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Metallic taste or smell:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If water is distributed using metal pipes it might get metal smell. Though nowadays plastic mostly plastic pipes are used.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Rotten egg smell:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems unusual but due to decaying of organic compounds water sometimes smells like rotten egg. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;musty smell:&lt;/strong&gt;   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This type of unpleasant smell can be caused due to mixing of pesticide with water.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;turpentine taste or smell:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This smell can be a result of MTBE contamination in your water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. As taste and smell works together it is not possible to distinguish between two. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://extoxnet.orst.edu/faqs/safedrink/colors.htm&quot; rel=&quot;nofollow&quot;&gt;extoxnet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-05-04T10:11:42.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30846" PostTypeId="2" ParentId="30840" CreationDate="2015-05-04T10:17:32.237" Score="4" Body="&lt;p&gt;Because of its special chemical struture Ethyelene oxide (EO) is highly reactive material thus easily participate in chemical reactions which includes polymerization. Mechanism of opening of EO's ring in acidic and alkaline medium is given below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8a4eH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see in both cases a molecule with alcoholic group ($\ce{-OH}$) is formed.If the epoxyethane is in excess, the reaction can continue and could lead to polymerization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is the reaction between EO and water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZC1sG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in wikipedia ethoxylation is an industrial process to make surfactant from EO. Surfactants are chemical molecule which contain both hydrophobic groups (their tails) and hydrophilic groups (their heads).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result of reaction with EO and alchol is a long chain molecule with these hydrophobic ($\ce{-R}$) and hydrophilic ($\ce{-OH}$) group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DbB0S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2015-05-04T10:17:32.237" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="30847" PostTypeId="2" ParentId="30836" CreationDate="2015-05-04T10:23:16.813" Score="1" Body="&lt;p&gt;A weak acid does not (contrary to what you show in your equation) predominantly ionize before the base is added. But keep in mind that proton transfer reactions (at least those between oxygen and/or nitrogen atoms) occur almost instantaneously in both directions, so the equilibrium you illustrated is established immediately (for a weak acid, with the big arrow pointing left). So when (strong) base is added, H3O+ is consumed immediately, then more acid is ionized to re establish the equilibrium--and this goes on as you add more base until you have consumed almost all of the weak acid. These reactions are controlled by the relative energies of the reactants and products, and not by reaction rates.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-04T10:23:16.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30849" PostTypeId="1" CreationDate="2015-05-04T12:31:47.727" Score="3" ViewCount="186" Body="&lt;p&gt;In kinetics, zero order reactions are those reactions who are independent from the concentration of the reactant or the product. So their rate is equal to the constant &lt;strong&gt;k&lt;/strong&gt;. But still we build the diagram where we find the correlation between concentration and time, where the slope is negative. &#xA;Why do we even write a diagram to find the correlation between &lt;strong&gt;concentration&lt;/strong&gt; and time, since these reactions are completely independent from concentration ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-05-04T13:06:27.207" Title="Order of reactions in kinetics" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30850" PostTypeId="1" CreationDate="2015-05-04T12:33:28.163" Score="3" ViewCount="96" Body="&lt;p&gt;In ZnTPP (Zinc tetraphenylporphyrin), there are two Q-band emission peaks. When I run steady state emission measurements at different temperatures, the ratio of their respective peak intensities decreases with increasing temperature. This seems to me like an unusual behaviour compared to many other porphyrins, where the ratio remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule has been around for a while so I would expect to find the answer in the literature, I am currently searching it and might post the answer if I find it before someone else answers.&lt;/p&gt;&#xA;" OwnerUserId="13341" LastActivityDate="2015-05-04T12:33:28.163" Title="Why might the ratio between two Q-band peaks in ZnTPP decrease with increasing temperature?" Tags="&lt;spectroscopy&gt;&lt;photochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30852" PostTypeId="2" ParentId="30849" CreationDate="2015-05-04T13:06:27.207" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So their rate is equal to the constant k.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With other words, the &lt;strong&gt;change&lt;/strong&gt; of concentration over time is linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\[ \frac{d[\ce{A}]}{dt} = k\]&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-04T13:06:27.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30853" PostTypeId="1" CreationDate="2015-05-04T13:17:47.803" Score="5" ViewCount="171" Body="&lt;p&gt;Background: Glycerin, when heated with a metal coil, decomposes below its boiling point to acrolein, which in turn appears to decompose to formaldehyde and acetaldehyde. (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/14771/how-does-propylene-glycol-glycerin-form-formaldehyde-acetaldehyde-when-vaporized&quot;&gt;see previous post&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Is there a chemical that could be added, or a process that could be applied, that would lessen the decomposition of glycerine when vaporized?&lt;/p&gt;&#xA;" OwnerUserId="15988" LastEditorUserId="72973" LastEditDate="2019-10-24T18:46:04.260" LastActivityDate="2019-11-13T10:59:29.273" Title="Reduce Decomposition Of Glycerin" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30854" PostTypeId="1" CreationDate="2015-05-04T14:42:08.383" Score="6" ViewCount="2896" Body="&lt;p&gt;Does this have something to do with the &lt;em&gt;inert pair effect&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="15993" LastEditorUserId="11367" LastEditDate="2017-01-08T12:53:14.213" LastActivityDate="2017-01-08T12:53:14.213" Title="Why does thallium form TlF and TlF3 but aluminium only forms AlF3?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;quantum-chemistry&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30855" PostTypeId="1" AcceptedAnswerId="31074" CreationDate="2015-05-04T15:03:53.897" Score="10" ViewCount="5979" Body="&lt;p&gt;I just had a question in a test asking to compare the relative acidity of picric acid (TNP) and carbonic acid, $\ce{H2CO3}$ (not phenol). It turns out that TNP is more acidic. I have no idea why? Is there some rule I'm missing or is this one of those exceptions that I just have to remember?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told that alcohols are generally less acidic than carboxyl groups. But in this case, doesn't the hydroxyl group form a chelate ring already? If the stability of the reactant is much greater than the the stability of the product, why would the reaction even take place? Sure, there is cross conjugation in carbonic acid, but shouldn't it be extremely hard to break the chelation in picric acid?&lt;/p&gt;&#xA;" OwnerUserId="15994" LastEditorUserId="5026" LastEditDate="2018-02-25T08:36:35.010" LastActivityDate="2018-02-25T09:35:03.997" Title="Why is picric acid more acidic than carbonic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;phenols&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="30856" PostTypeId="2" ParentId="30818" CreationDate="2015-05-04T15:10:12.827" Score="2" Body="&lt;p&gt;As I noted in my comment above, I think the question is whether there are 7d electrons in the &lt;strong&gt;&lt;em&gt;atom&lt;/em&gt;&lt;/strong&gt; or in the &lt;strong&gt;&lt;em&gt;ion&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all, of course, a guess given what little you posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I had a similar question in my class. Students might then assume this $\ce{M^{3+}}$ is octahedral. (Personally, I'd make it a bit more obvious in the question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the ligand is high-field, you get an initial MO diagram like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wL2Uo.png&quot; alt=&quot;d4-high spin&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;But...&lt;/em&gt;&lt;/strong&gt; we're not done. This electron configuration is subject to a weak &lt;a href=&quot;http://en.wikipedia.org/wiki/Jahn%E2%80%93Teller_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jahn-Teller distortion&lt;/a&gt;. So you'd end up with an MO diagram like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bjgcA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, of course, is diamagnetic. &lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-04T15:10:12.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30857" PostTypeId="2" ParentId="30854" CreationDate="2015-05-04T15:21:37.937" Score="7" Body="&lt;p&gt;Ahh, Thallium is a very nice element. &lt;em&gt;Because it's weird&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the inert pair effect. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_13%3A_The_Boron_Family&quot; rel=&quot;noreferrer&quot;&gt;Chemwiki.UCDavis&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As an example in group 13 the +1 oxidation state of Tl is the most stable and Tl(III) compounds are comparatively rare. &lt;a href=&quot;http://en.wikipedia.org/wiki/Inert_pair_effect&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Inert pair effect&lt;/strong&gt;, Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;So, why?&lt;/h2&gt;&#xA;&lt;p&gt;Inert pair effect was introduced to describe why the facts above are correct. However, the original explanation, provided by Sidgwick, relied on the fact that the &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; subshell has an &amp;quot;inert&amp;quot; pair of electrons. This can't be correct, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Inert_pair_effect#Description&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; uses the trend in ionization potentials of &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;&amp;quot; electrons to prove it wrong:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{In &amp;lt; Al &amp;lt; Tl &amp;lt; Ga}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A better explanation lies within the lower bond enthalpies of &lt;span class=&quot;math-container&quot;&gt;$\ce{MX}$&lt;/span&gt; where M is one of the p-block heavy elements.&lt;/p&gt;&#xA;&lt;p&gt;To reach a high oxidation number, elements will need to have a strong bond. They wanna bond in order to lose energy (i.e. become more stable) after all! If the bond is weak, this energy isn't provided. Thus, a heavy element is better off with being less oxidised.&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00045a027&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j150561a027&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;. (Sorry, they might be behind a paywall)&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-04T15:21:37.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30861" PostTypeId="2" ParentId="29685" CreationDate="2015-05-04T16:13:10.230" Score="3" Body="&lt;p&gt;Most modern steel cans for food are given a hermetic plastic coating on the interior.  Before that steel cans were tin-plated.  (I believe all-tin cans haven't been used in over a century.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how hot you got the can when testing it you could have melted or vaporized the plastic coating, exposing the unprotected steel.  Also, if the can is dented then the coating can be compromised and the steel will rust where exposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, when you open the can you are cutting through the hermetic coating, exposing the steel at least on the edges where you cut, which is why opened cans quickly develop a &quot;metallish&quot; rust smell.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2015-05-04T16:27:14.927" LastActivityDate="2015-05-04T16:27:14.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30863" PostTypeId="1" AcceptedAnswerId="31008" CreationDate="2015-05-04T16:55:40.783" Score="4" ViewCount="2125" Body="&lt;p&gt;$\ce{KBr}$ reacts with concentrated $\ce{H3PO4}$ to give $\ce{HBr}$ and ($\ce{KH2PO4}$ or $\ce{K3PO4}$) (not sure which one, if someone knows it, please tell).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't bromine gas liberated?&lt;/p&gt;&#xA;" OwnerUserId="15983" LastEditorUserId="4945" LastEditDate="2015-05-07T13:13:28.807" LastActivityDate="2016-09-15T16:14:53.247" Title="Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid?" Tags="&lt;inorganic-chemistry&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="30864" PostTypeId="1" CreationDate="2015-05-04T17:01:15.220" Score="6" ViewCount="3991" Body="&lt;p&gt;If there is a ketone and ester functional group in the same compound and I want to protonate with $\ce{HCl}$, which one will protonate first? Is it the carbonyl group in the ketone or the carbonyl group in the ester and why? Is it related to the $\mathrm{p}K_\mathrm{a}$ of the functional group?&lt;/p&gt;&#xA;" OwnerUserId="15998" LastEditorUserId="10185" LastEditDate="2015-09-20T18:08:07.983" LastActivityDate="2015-09-20T18:08:07.983" Title="Protonation of ketone and ester" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;carbonyl-compounds&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30865" PostTypeId="2" ParentId="30853" CreationDate="2015-05-04T17:03:07.810" Score="4" Body="&lt;p&gt;You could try the mechanism behind mist fountains or other &quot;cold fog&quot; machines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMISu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These use an ultrasonic oscillating diaphragm to produce a fine mist of room temp water, no heating performed. The resulting vaping machine would be larger and require more power, but wouldn't have the harmful aldehydes inherent in current vaping liquids. Nicotine itself is hygroscopic but not really water soluble at room temp, but it can form salts that are water-soluble so it could be introduced to the body this way. Various flavors currently used might be more or less water-soluble. Finally, the fog, being simple water vapor, wouldn't be visible when exhaled which is one reason the glycerol compounds are used; they produce a &quot;smoke&quot; that dissipates more quickly than tobacco smoke.&lt;/p&gt;&#xA;" OwnerUserId="367" LastEditorUserId="367" LastEditDate="2015-06-26T19:19:26.490" LastActivityDate="2015-06-26T19:19:26.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30866" PostTypeId="1" CreationDate="2015-05-04T17:09:43.203" Score="18" ViewCount="315" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m26uH.jpg&quot; alt=&quot;enter image description here&quot;&gt;Recently a Cahn TG-171 Thermogravimetric Analyzer was donated to my school. We did not get a service or user's manual with it, and some of it still needs to be put back together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to contact Thermo Fisher who bought Cahn Instruments, but they no longer support the TG-171. Any sort of information on the setup of this instrument would be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;In particular, one of the main questions I now have is &quot;what must I do to use the balance?&quot; Currently there are eccentric screws (as shown in the photo) preventing the balance beam from moving so it is unable to register mass change. Do I loosen both sets and use a counterweight of a calibrated mass? Is there a common procedure for setting up balances of this type?&lt;/p&gt;&#xA;" OwnerUserId="4813" LastEditorUserId="4813" LastEditDate="2015-05-13T19:49:49.140" LastActivityDate="2020-10-01T01:37:25.673" Title="Manuals for a Cahn TG-171 Thermogravimetric Analyzer" Tags="&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30867" PostTypeId="1" CreationDate="2015-05-04T17:30:35.320" Score="2" ViewCount="2040" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following statements is true regarding an ionic bond in $\ce{FeS}$?    &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Iron atoms gain electrons and sulfur atoms lose electrons.     &lt;/li&gt;&#xA;  &lt;li&gt;Iron and sulfur ions bond by electrostatic attraction.  &lt;/li&gt;&#xA;  &lt;li&gt;The ionic radius of an iron ion is equal to its atomic radius.  &lt;/li&gt;&#xA;  &lt;li&gt;Breaking an ionic bond between $\ce{Fe^{2+}}$ and $\ce{S^2-}$ releases energy.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know metals lose electrons. I thought the answer was iron and sulfur ions bond by electrostatic attraction, but a classmate and I have a disagreement, so now I am not sure. I don't think the ionic radius of an iron (II) ion is equal to its atomic radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is either &lt;strong&gt;2&lt;/strong&gt; or &lt;strong&gt;4&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15999" LastEditorUserId="7448" LastEditDate="2018-02-19T12:26:25.600" LastActivityDate="2018-02-19T12:35:26.770" Title="Ionic bond and ionic radii in FeS" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30868" PostTypeId="2" ParentId="30867" CreationDate="2015-05-04T18:00:55.747" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Iron and sulfur ions bond by electrostatic attraction. &lt;br/&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the right answer. I see a hard time maneuvering around it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Breaking an ionic bond between $\ce{Fe^2+}$ and $\ce{S^2-}$ releases energy.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The misconception&lt;/strong&gt;: Some students fall in the trap of believing that bonds are like boxes of energy. So, breaking these boxes of energy should release energy, right? &lt;strong&gt;&lt;em&gt;No!&lt;/em&gt;&lt;/strong&gt; In reality, ionic bonds are nothing but forces. This is like imagining that when you pull a box, you're applying or somehow putting some kinda a box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The reality&lt;/strong&gt;: Ionic bonds are just forces that are a result of attraction between two species (such as two atoms). &lt;em&gt;Why do these happen in the first place?&lt;/em&gt; Because for instance in \ce{NaCl} the sodium atom really &lt;em&gt;likes&lt;/em&gt; to lose an electron and chlorine wants to earn one. This love of giving and/or taking an electron is because of reaching the stability. Stability in this case means lesser energy. So, &lt;strong&gt;ionic bonds happen because atoms want to have lesser energy&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, what would breaking this bond result in? If the atoms bonded together to lower their energies, then what would removing the bond result in?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I don't think the ionic radius of an iron ion is equal to its atomic radius.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're right! Think of the sodium chloride example again. The &lt;em&gt;new&lt;/em&gt; sodium cation used to have 11 protons and 11 electrons. Now, after the lattice builds up, sodium &lt;em&gt;crudely&lt;/em&gt; has only 10 electrons, with the same 11 protons. The electrons are outnumbered! They can't overpower protons anymore. So, what will happen to the distance electrons were able to travel to? (The distance they could &quot;travel&quot; further from the nucleus was called the atomic radius, and it's the ionic radius now)&lt;br&gt;&#xA;So, is the atomic radius of iron equal to its ionic radius?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Rather pedantic and irrelevant, but there's no clear distinction between an ionic bond and a covalent bond. There are no &quot;pure&quot; ionic bonds.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2018-02-19T12:35:26.770" LastActivityDate="2018-02-19T12:35:26.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30869" PostTypeId="1" AcceptedAnswerId="30873" CreationDate="2015-05-04T19:04:52.067" Score="13" ViewCount="32086" Body="&lt;p&gt;Does salt vaporise? If I boil salt-water and leave it there until half of the water vaporises, will it be twice as salty as in the beginning? Or exactly the same? Or somewhere in between?&lt;/p&gt;&#xA;" OwnerUserId="16000" LastEditorUserId="-1" LastEditDate="2019-03-14T14:42:11.477" LastActivityDate="2019-03-14T16:27:56.340" Title="Does desalination work?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30870" PostTypeId="1" CreationDate="2015-05-04T19:17:41.830" Score="14" ViewCount="674" Body="&lt;p&gt;I know that the definition of molecularity of a reaction is number of species reacting in an elementary step. But considering the theory of microscopic reversibility for elementary reactions, each reaction can be assumed to be reversible. In that case, if $$\ce{A &amp;lt;=&amp;gt; B + C}$$ then what is the molecularity of the reaction? Is it 1 or 2?&lt;/p&gt;&#xA;" OwnerUserId="16001" LastEditorUserId="194" LastEditDate="2017-04-06T02:37:47.810" LastActivityDate="2017-04-06T02:37:47.810" Title="What is the molecularity of a reversible reaction?" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30871" PostTypeId="1" CreationDate="2015-05-04T19:58:02.363" Score="4" ViewCount="149" Body="&lt;p&gt;That's my question. If true, it would make my job a lot easier. I will appreciate any feedback on the matter. &lt;/p&gt;&#xA;" OwnerUserId="16002" LastEditorUserId="5456" LastEditDate="2015-05-05T03:07:03.337" LastActivityDate="2015-05-05T10:53:48.383" Title="Are glycans, such as glucose, considered alicyclic compounds?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-05-08T11:58:25.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="30872" PostTypeId="1" AcceptedAnswerId="30874" CreationDate="2015-05-04T20:28:59.197" Score="2" ViewCount="3754" Body="&lt;p&gt;Activated complex theory, tells us that due to the collision between the molecules of the reactants, they form a transition specie before the product is formed, which is called active complex. On the other hand we have consecutive reactions on which is also formed a intermediate product before forming the actual product we're interested on. My question is : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the difference between the activated complex and the intermediate product, since they are both formed &lt;strong&gt;before&lt;/strong&gt;  the actual product ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-05-04T20:55:46.773" Title="Activated complex theory vs. consecutive reactions" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-04T22:06:41.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="30873" PostTypeId="2" ParentId="30869" CreationDate="2015-05-04T20:35:35.667" Score="15" Body="&lt;p&gt;This is a classic science experiment. Where saltwater (using $\ce{NaCl}$) is heated, as it does, the water evaporates and most, if not all, of the salt remains - increasing the concentration of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporating half the water would approximately double the concentration of $\ce{NaCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, the salt concentration becomes supersaturated and crystallises. The reason is due to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the water evaporates less and less water molecules are present to keep the salt particles apart. The salt therefore recrystalises and can be collected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dynamicscience.com.au/tester/solutions1/chemistry/solutions/evprtn.htm&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is also used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_evaporation_pond&quot;&gt;salt evaporation ponds&lt;/a&gt;, where&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;seawater or brine is fed into large ponds and water is drawn out through natural evaporation which allows the salt to be subsequently harvested.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As Curt F mentioned in the comments, different salts and mixtures of salts can have different outcomes in terms of vaporisation.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-04T22:06:58.710" LastActivityDate="2015-05-04T22:06:58.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30874" PostTypeId="2" ParentId="30872" CreationDate="2015-05-04T20:44:56.277" Score="3" Body="&lt;p&gt;The key difference is that transition states occur at a &lt;strong&gt;&lt;em&gt;maximum&lt;/em&gt;&lt;/strong&gt; of the potential energy curve for the reaction whereas intermediates occur at a local &lt;strong&gt;&lt;em&gt;minimum&lt;/em&gt;&lt;/strong&gt;. Take this example of the reaction profile for an $\ce{S_{N}1}$ reaction: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e5E4R.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that the products and reactants occur at minima on the curve and the intermediate also occurs at a minimum, albeit a higher energy one. In between the minima are located the maxima where you find the transition states.  Unless the activation energy is zero (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18924/there-are-no-chemical-cliffs/18927#18927&quot;&gt;this&lt;/a&gt; question for rare examples) there will always be a transition state located between any two minima.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition states are usually represented using dashed bonds to show bonds in the process of being broken or formed as opposed to being fully formed as in intermediates or products. Additionally the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hammond%27s_postulate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammond postulate&lt;/a&gt; says that the transition state will most resemble the stable species closest to it in energy, in this case the carbocation, and this can be used to help predict the structures of transition states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition states are very short lived because they immediately 'roll downhill' on the potenetial energy curve to reach an intermediate or product. By contrast some intermediates are actually quite stable and can be isolated.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-05-04T20:55:46.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30876" PostTypeId="1" AcceptedAnswerId="30879" CreationDate="2015-05-04T21:01:11.347" Score="3" ViewCount="559" Body="&lt;p&gt;If I am given a compound which belongs to, say the cubic $\text{Ia}\bar 3\text{d}$ space group, is there a known way to calculate the lattice parameter of the compound? In general, can lattice parameters be calculate from knowing what space group the compound belongs to?&lt;/p&gt;&#xA;" OwnerUserId="2386" LastActivityDate="2015-05-06T20:43:38.113" Title="Theoretical Calculation of Lattice Parameter from Space Group" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;group-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30879" PostTypeId="2" ParentId="30876" CreationDate="2015-05-04T23:36:24.397" Score="6" Body="&lt;p&gt;No - the spacegroup only gives information about symmetry, not the lattice parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lattice parameters depend on what is in your unit cell, ie your compound - but even if you know your compound, this is not straightforward, as you don't know its arrangement. And if you do, you should already know your lattice parameters ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said: if you have additional information - e.g. the radius of a protein you crystallise - you might be able to get a rough estimate of what you are expecting as your minimal unit cell parameters. But this does not take into account any stoichiometry in the crystal, which is impossible to know in advance, so multiples thereof are just as likely.&lt;/p&gt;&#xA;" OwnerUserId="10050" LastEditorUserId="10050" LastEditDate="2015-05-06T20:43:38.113" LastActivityDate="2015-05-06T20:43:38.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30880" PostTypeId="2" ParentId="30870" CreationDate="2015-05-04T23:44:26.887" Score="9" Body="&lt;p&gt;For the reaction $$\ce {A &amp;lt;=&amp;gt; B + C}$$&#xA;its molecularity is unimolecular. If we were given $$\ce{B + C &amp;lt;=&amp;gt; A}$$ &#xA;Its molecularity would be bimolecular. Thus we see molecularity is dependent on the number of reactant particles, even if it's in equilibrium. &lt;/p&gt;&#xA;" OwnerUserId="13368" LastEditorUserId="10185" LastEditDate="2015-05-05T14:39:09.833" LastActivityDate="2015-05-05T14:39:09.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30881" PostTypeId="1" CreationDate="2015-05-05T00:05:28.973" Score="5" ViewCount="352" Body="&lt;p&gt;When using Pymol or Coot molecular visualization software to solve x-ray crystallography structures, how can I mutate a cytosine to a 5-methylcytosine?&lt;/p&gt;&#xA;" OwnerUserId="16006" LastEditorUserId="5456" LastEditDate="2015-05-05T03:45:02.333" LastActivityDate="2015-05-06T21:05:29.837" Title="Modified DNA bases in Pymol and Coot" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30882" PostTypeId="2" ParentId="30871" CreationDate="2015-05-05T00:42:12.750" Score="3" Body="&lt;p&gt;Seems I underestimated this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First passage from Wikipedia is crucial:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An alicyclic compound is an organic compound that is both aliphatic and cyclic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And second is wrong:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They contain one or more all-carbon rings(...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Alicyclic_compound&quot; rel=&quot;nofollow&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Glucose&quot; rel=&quot;nofollow&quot;&gt;Glucose&lt;/a&gt; is &lt;strong&gt;carbohydrate&lt;/strong&gt;, in its circular form which is found in &lt;a href=&quot;https://en.wikipedia.org/wiki/Glycan&quot; rel=&quot;nofollow&quot;&gt;glycans&lt;/a&gt; (also glucose isn't a glycan, but glucose residue is found in many of them) it has oxygen atom in the ring - according to second passage from Wikipedia it therefore shouldn't be alicyclic, but one can easily find mention about alicyclic ethers for ex. &lt;a href=&quot;http://www.nrcresearchpress.com/doi/pdf/10.1139/v68-161&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, with oxygen in ring. So I'm changing my answer to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes, they are.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-05-05T10:53:48.383" LastActivityDate="2015-05-05T10:53:48.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30883" PostTypeId="1" CreationDate="2015-05-05T03:22:34.050" Score="6" ViewCount="821" Body="&lt;p&gt;I'm having a bit of trouble grasping sugar stereochemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using glucose as an example, if you cyclize D and L glucose, is the stereochemistry at every carbon just inverted from the other form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like for example, on carbon 2 for D glucose, the hydroxyl is down. On carbon 2 for L glucose, the hydroxyl would be up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this trend true for every hydroxyl or $\ce{CH2OH}$ group except the one on the anomeric carbon?&lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorDisplayName="user7951" LastEditDate="2016-11-28T16:14:12.973" LastActivityDate="2021-02-24T14:02:15.233" Title="In the Haworth projections of D and L glucose, is the stereochemistry at every carbon except the anomeric carbon just inverted?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30885" PostTypeId="2" ParentId="30869" CreationDate="2015-05-05T04:18:07.483" Score="11" Body="&lt;p&gt;Note Kevin's comment on another answer. The key observation here is that it is not a question of whether $\ce{NaCl}$ vaporises but whether there is a temperature range in which water vaporises and $\ce{NaCl}$ doesn't, the answer to which question is yes. The wider the gap between the solvent and the solute, the easier and more completely you can separate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaCl}$ will certainly vaporise but it also has a liquid phase, and in this phase is used to cool nuclear reactors. Going by the 600 degrees between Kevin's figures for melting point and boiling point it doesn't tend to sublimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything vaporises at any temperature above 4K. Atmospheric agitation is knocking bits off my keyboard even as I type, but for the most part they're big heavy lumps so they fall back down and either re-bond or sit there serving as ablative armour. Molecules small and light enough to float away mostly did so fairly promptly. Think new car smell. These are called volatiles, which simply means they are small and light enough to float away with modest stimulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Why 4K? Helium liquefies. Drop the temperature even further and you get a Bose condensate which is super interesting stuff with directly observable quantum behaviour but doesn't have much to do with chemistry as we know it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where was I? Oh yes, melting point and boiling point are inflection points in a continuum. They are temperatures which bound &lt;em&gt;dominant&lt;/em&gt; behaviors.&lt;/p&gt;&#xA;" OwnerUserId="16009" LastEditorUserId="5078" LastEditDate="2018-08-22T15:47:52.587" LastActivityDate="2018-08-22T15:47:52.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30887" PostTypeId="1" CreationDate="2015-05-05T05:00:54.980" Score="7" ViewCount="835" Body="&lt;p&gt;I was reading the R.A. Millikan's oil drop experiment and the following part confused me:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In Newtonian physics, the rate of fall (or velocity) of a freely falling body does not depend on the mass of the object. So, how did Millikan measure the mass of these oil droplets? Does anyone know the equations he used?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="41328" LastEditDate="2020-07-13T07:45:04.600" LastActivityDate="2020-07-13T07:45:04.600" Title="How did Millikan measure the mass of these oil droplets?" Tags="&lt;electrons&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30888" PostTypeId="1" CreationDate="2015-05-05T06:49:24.797" Score="1" ViewCount="58" Body="&lt;p&gt;I’m really confused about this. If we consider $\ce{Ca3(PO4)2}$, it can be seen it undergoes an equilibrium. But why it’s an equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equilibrium it undergoes $$\ce{Ca3(PO4)2_{(s)}&amp;lt;=&amp;gt; 3Ca^2+_{(aq)} + 2 PO4_{(aq)}^3-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way I’m having a doubt of why some are using backward arrow for a dissociation of a particular precipitate while some are using a solid forward arrow for the same reaction. For an example let’s take $\ce{AgCl(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is a place where a backward arrow is used,&#xA;$$\ce{AgCl_{(s)}&amp;lt;=&amp;gt; Ag+_{(aq)} + Cl_{(aq)}-}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have seen in some books where there is a solid forward arrow instead of back-ward arrow for the dissociation of $\ce{AgCl(s)}$.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorDisplayName="user7951" LastEditDate="2015-05-06T11:44:34.260" LastActivityDate="2015-05-06T11:44:34.260" Title="Do all the precipitates undergo an equilibrium with its respective ions, why can't they dissociate fully?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30889" PostTypeId="1" CreationDate="2015-05-05T07:29:17.533" Score="8" ViewCount="7266" Body="&lt;p&gt;Both these processes are used in concentration of bauxite.&lt;br&gt;&#xA;They use different reagents.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is more commonly used (&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Bayer_process#History_of_the_Bayer_process&quot;&gt;wikipedia says Bayer's&lt;/a&gt;) and why?&lt;br&gt;&#xA;When is one preferred over the other?&lt;br&gt;&#xA;Which is more efficient and why?  &lt;/p&gt;&#xA;" OwnerUserId="16010" LastActivityDate="2020-04-04T07:05:50.763" Title="When is Serpeck's process preferred over Bayer's process?" Tags="&lt;metallurgy&gt;&lt;extraction&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30892" PostTypeId="1" CreationDate="2015-05-05T07:59:01.433" Score="5" ViewCount="1196" Body="&lt;p&gt;Electeothermal atomizer ( graphite furnance ) and flame atomizer, are the instruments which help us to convert the analyte into the gas state and they both achieve that. My question is, what's the difference in their way of workig ? I know the flame has its nebulisator which converts the liquid analyte into mist aerosols and then the gas helps the analyte to arrive to the flame. How does the electrothermal atomizer work ? Whats the nebulisator substitution in the graphite furnance ? &#xA;Why do we have to turn the analytie in mist aerosol in flame AAS ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-05T08:16:41.647" LastActivityDate="2015-09-19T22:19:10.737" Title="Atomizers in AAS" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30893" PostTypeId="1" AcceptedAnswerId="30894" CreationDate="2015-05-05T08:15:40.180" Score="27" ViewCount="52174" Body="&lt;p&gt;This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound? &lt;/p&gt;&#xA;" OwnerUserId="15517" LastEditorDisplayName="user15489" LastEditDate="2015-05-05T08:44:45.103" LastActivityDate="2019-09-07T09:08:30.890" Title="Why are tertiary carbocations the most reactive, if they're already stable?" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="30894" PostTypeId="2" ParentId="30893" CreationDate="2015-05-05T08:39:49.017" Score="26" Body="&lt;p&gt;This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, &lt;em&gt;transition states&lt;/em&gt; involving tertiary carbocations as opposed to primary carbocations &lt;em&gt;are more favorable&lt;/em&gt; precisely because tertiary carbocations are more stable than primary carbocations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this &lt;strong&gt;does not&lt;/strong&gt; mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the &lt;em&gt;transition&lt;/em&gt; state ... of course it's going to be reactive - of course it'll change very rapidly! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-05-06T15:50:19.897" LastActivityDate="2015-05-06T15:50:19.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30895" PostTypeId="5" CreationDate="2015-05-05T10:29:19.653" Score="0" Body="&lt;p&gt;Hydrogen bond is an interaction between hydrogen atom, bound covalently to more electronegative atom, and another atom possessing a lone pair of electrons, which it uses to provide certain covalent component to the bond.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-02-09T05:41:31.900" LastActivityDate="2016-02-09T05:41:31.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30896" PostTypeId="4" CreationDate="2015-05-05T10:29:19.653" Score="0" Body="Bond between covalently bound hydrogen atom and atom possessing a lone pair of electrons." OwnerDisplayName="user15489" LastEditorUserId="9961" LastEditorDisplayName="user15489" LastEditDate="2015-10-16T08:04:04.290" LastActivityDate="2015-10-16T08:04:04.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30897" PostTypeId="2" ParentId="30869" CreationDate="2015-05-05T11:44:32.270" Score="1" Body="&lt;p&gt;Salt does not vaporise (but does melts and vaporise at high temperature) at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sea water which is (Salt+Water) is separated by evaporation process. Salt remains and water is vaporised. You can see that in sea side salt pans.&lt;/p&gt;&#xA;" OwnerUserId="16014" LastEditorUserId="16014" LastEditDate="2015-05-05T11:51:56.623" LastActivityDate="2015-05-05T11:51:56.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30900" PostTypeId="1" CreationDate="2015-05-05T11:58:06.530" Score="9" ViewCount="1006" Body="&lt;p&gt;Why are multiple-quantum coherences (coherences with changes in magnetic quantum numbers, $|\Delta m| \neq 1$) unobservable in regular NMR experiments?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="16683" LastEditDate="2017-05-29T11:31:37.860" LastActivityDate="2017-05-30T11:45:20.340" Title="Multiple-quantum coherences in NMR" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30901" PostTypeId="2" ParentId="30888" CreationDate="2015-05-05T12:09:03.820" Score="1" Body="&lt;p&gt;In the two examples you mentioned, we have an equilibrium between the solid and the small quantity of the solid that is soluble in water and completely dissociated:&#xA;$$\ce{Ca3(PO4)2_{(s)}&amp;lt;=&amp;gt; 3Ca^2+_{(aq)}  + 2 PO4^{3-}_{(aq)} }$$ &#xA;$$\ce{AgCl_{(s)}&amp;lt;=&amp;gt; Ag^+(aq)_{(aq)}  + Cl^{-}_{(aq)} }$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's wrong to write a solid forward reaction.&#xA;It's an &lt;strong&gt;heterogeneous equilibrium&lt;/strong&gt; between the solid state (the precipitate) and the aqueous state (ions constituents of the precipitate). The constant of this equilibrium is the solubility product $K_{\mathrm{sp}}$ of the precipitate.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-05-05T14:08:14.780" LastActivityDate="2015-05-05T14:08:14.780" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="30902" PostTypeId="2" ParentId="30893" CreationDate="2015-05-05T12:54:08.713" Score="15" Body="&lt;p&gt;Very interesting question. This also popped up in my head once. And this is how I convinced myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusing term here is stability. Tertiary carbocations are stable by inductive effect and hyper conjugation, and therefore have the tendency to sustain the positive charge on the carbon atom and stay like this for long. That is why we call it stable: it can stay that way longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic theory illustrates that the reactants present in the container randomly move in a container without any energy loss. Now imagine that there is a carbocation that is not so stable (take n-butane carbocation as an example) and now if a nucleophile would want to attack the cation, how could it attack it if it is not able to hold up by itself? The cation must hold on to itself in order to let the nucleophile come to it and then attack it. Tertiary carbocations have the stability to hold up like that and allow the nucleophile to attack it. Since it can hold up, that is why the reactions involving it favors $S_N1$ mechanisms, whereas a not so stable carbocation (like n-butane) follows $S_N2$, since there is &quot;not enough time&quot; to make a carbocation and proceed the reaction. $S_N2$ is a fast process compared to $S_N1$, since 1$^O$ have to react fast in order to form the product. This could also be supported by the fact that if we want to arrange carbocations in the order of reactivity by $S_N1$ and $S_N2$ mechanisms, they are usually the opposite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise, it was stable enough to hold on to its configuration and let the attack happen.&lt;/p&gt;&#xA;" OwnerUserId="3771" LastActivityDate="2015-05-05T12:54:08.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30905" PostTypeId="2" ParentId="30870" CreationDate="2015-05-05T13:40:37.983" Score="5" Body="&lt;p&gt;Just to add onto the other answer.&lt;br&gt;&#xA;Since the equilibrium constant also changes when we reverse a reaction, so does the molecularity of the reaction.&lt;br&gt;&#xA;So many properties of reversible reactions are defined in the way we write down the reaction.  &lt;/p&gt;&#xA;" OwnerUserId="16020" LastActivityDate="2015-05-05T13:40:37.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30906" PostTypeId="1" AcceptedAnswerId="30911" CreationDate="2015-05-05T13:41:21.683" Score="14" ViewCount="14214" Body="&lt;p&gt;I wanted to ask something more in reference to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AWR3b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWR3b.jpg&quot; alt=&quot;Structures of 2- and 3-methoxyanilines&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose instead of toluidine, &lt;strong&gt;2-methoxyaniline&lt;/strong&gt; and &lt;strong&gt;3-methoxyaniline&lt;/strong&gt; are present. Now, why is &lt;strong&gt;2-methoxyaniline&lt;/strong&gt; more basic than &lt;strong&gt;3-methoxyaniline&lt;/strong&gt; (even though &lt;strong&gt;2-methoxyaniline&lt;/strong&gt; has reduction of basicity due to &lt;em&gt;ortho&lt;/em&gt;-effect but &lt;strong&gt;3-methoxyaniline&lt;/strong&gt; does not)?&lt;/p&gt;&#xA;&lt;p&gt;Also what would be the correct order of basicity if nitro group was present in place of methoxy group? Would &lt;em&gt;ortho&lt;/em&gt;-effect still take place?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="61578" LastEditDate="2021-05-26T22:48:30.193" LastActivityDate="2021-05-26T22:48:30.193" Title="Basicity Alteration Due To Ortho Effect In Methoxy Aniline System" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30907" PostTypeId="2" ParentId="30889" CreationDate="2015-05-05T14:51:36.963" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;http://www.2classnotes.com/digital_notes.asp?p=Extraction_of_Aluminium&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Baeyer's process: This method is mainly applied when ferric oxide is present as chief impurity.&lt;/p&gt;&#xA;&lt;p&gt;Serpeck's process: This process is used when silica is present in considerable amount of bauxide ore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I couldn't find any information on the efficiency , but I suspect both are efficient in their own way. We cannot compare these two processes as they are used for two different impurities.&lt;/p&gt;&#xA;" OwnerUserId="11310" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-11T16:24:55.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30908" PostTypeId="1" AcceptedAnswerId="30912" CreationDate="2015-05-05T15:03:55.717" Score="7" ViewCount="124" Body="&lt;p&gt;I would like to create protein models such as the ones in the Protein Data Bank. I have got GROMACS but I don't know if it can be used to model macromolecules.&lt;/p&gt;&#xA;" OwnerUserId="16021" LastEditorUserId="10185" LastEditDate="2015-05-05T17:26:30.083" LastActivityDate="2015-10-06T01:25:53.903" Title="What program could I use to create a protein model from scratch?" Tags="&lt;biochemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30910" PostTypeId="1" CreationDate="2015-05-05T15:38:58.633" Score="6" ViewCount="24071" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UUHqS.png&quot; alt=&quot;4-Nitrobenzoic acid&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dWTVG.png&quot; alt=&quot;4-Methoxybenzoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing the acidic strength, the nitro-substituted group is definitely more acidic. This can be reasoned by fact that a higher delta positive charge which develops on the H due to the strong &lt;b&gt;-M&lt;/b&gt; nature of the nitro group.&#xA;Another way to check the acidity is to check the stability of the conjugate base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conjugate base would be the carboxylate ion which being a &lt;b&gt;-M&lt;/b&gt; group creates a lesser electron density at the ortho and para positions. This positive charge is however stabilised by the methoxy group, which is a &lt;b&gt;+M&lt;/b&gt; group, and destabilised by the nitro group. So from this viewpoint, the methoxy derivative should have been more acidic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this not so? What is the fault in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="16022" LastEditorUserId="5026" LastEditDate="2018-03-08T09:12:33.747" LastActivityDate="2020-05-15T00:24:05.550" Title="Acidic strength of para substituted benzoic acids" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30911" PostTypeId="2" ParentId="30906" CreationDate="2015-05-05T16:07:37.380" Score="17" Body="&lt;p&gt;Take home message: &lt;strong&gt;Just because you have an &lt;em&gt;ortho&lt;/em&gt; substituent does not mean you will have an &lt;em&gt;ortho&lt;/em&gt; effect.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;To have an ortho effect &lt;strong&gt;two conditions must be met:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you need an &lt;em&gt;ortho&lt;/em&gt; substituent&lt;/li&gt;&#xA;&lt;li&gt;and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;One measure of steric bulk is the cyclohexane &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;-Value, a table of which &lt;a href=&quot;http://en.wikipedia.org/wiki/A_value&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be found here.&lt;/a&gt;  From the table we see that methoxy &lt;span class=&quot;math-container&quot;&gt;$(A=0.6)$&lt;/span&gt; and nitro &lt;span class=&quot;math-container&quot;&gt;$(A=1.1)$&lt;/span&gt; groups have a smaller steric size than that of a methyl &lt;span class=&quot;math-container&quot;&gt;$(A=1.7)$&lt;/span&gt; group.  Methoxy and nitro groups generally do not produce an &lt;em&gt;ortho&lt;/em&gt; effect.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{l|r} \hline&#xA;\text{Substituent} &amp;amp; \mathrm{p}K_\mathrm{b} \\ \hline&#xA;\ce{-H} &amp;amp; 9.13 \\ &#xA;\ce{o-methoxy} &amp;amp; 9.5 \\ &#xA;\ce{m-methoxy} &amp;amp; 9.8  \\ &#xA;\ce{p-methoxy} &amp;amp; 8.7  \\ \hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;http://web.archive.org/web/20160304032044/http://www.phs.d211.org/science/smithcw/AP%20Chemistry/Posted%20Tables/Acid%20Dissociations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;data source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;The basicity of &lt;em&gt;p&lt;/em&gt;-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant).  Drawing resonance structures we see that the &lt;em&gt;p&lt;/em&gt;-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in &lt;em&gt;p&lt;/em&gt;-methoxy aniline being more basic than aniline.&lt;/p&gt;&#xA;&lt;p&gt;The basicity of &lt;em&gt;m&lt;/em&gt;-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position).  The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.&lt;/p&gt;&#xA;&lt;p&gt;The basicity of &lt;em&gt;o&lt;/em&gt;-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions.  Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the &lt;em&gt;meta&lt;/em&gt; position had a 0.7 unit effect on the basicity, and now in the &lt;em&gt;ortho&lt;/em&gt; position it will be even stronger, while the resonance effect in the &lt;em&gt;para&lt;/em&gt; isomer only changed the basicity by 0.4 units) making &lt;em&gt;o&lt;/em&gt;-methoxy aniline less basic than aniline.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2021-05-26T22:04:35.653" LastActivityDate="2021-05-26T22:04:35.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30912" PostTypeId="2" ParentId="30908" CreationDate="2015-05-05T16:38:43.813" Score="5" Body="&lt;p&gt;You have several options to generate smaller peptides using the peptide builders in &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://avogadro.cc/wiki/Tutorials:Insert_a_Peptide&quot;&gt;Avogadro&lt;/a&gt; or &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pymolwiki.org/index.php/Peptide_Sequence&quot;&gt;PyMol&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-05T16:49:27.927" LastActivityDate="2015-05-05T16:49:27.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30913" PostTypeId="1" CreationDate="2015-05-05T17:02:49.527" Score="4" ViewCount="592" Body="&lt;p&gt;How do we number the carbon atoms in symmetrical molecules while giving R and S notation. For instance, in 2,3-Dibromobutane, there is an ambiguity over which is carbon C2 and carbon C3. Will we call it (2R,3S) or (2S,3R)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w21zN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16024" LastActivityDate="2015-05-05T18:00:37.557" Title="Stereochemistry: R and S notation" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30915" PostTypeId="2" ParentId="8387" CreationDate="2015-05-05T17:38:56.207" Score="0" Body="&lt;p&gt;Well according to my book 'General and inorganic chemistry volume 2, by Dr. Ivan Filipovic &amp;amp; Dr. Stjepan Lipanovic ', the oxidation state of superoxides is &lt;strong&gt;-1/2&lt;/strong&gt;. $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ react with oxygen and form type $\ce{MO2}$ compounds called superoxides. They are ionic crystalised compounds containing cation $\ce{M+}$ and anion $\ce{O^2-}$, and since the oxygen anion has 13 valent electrons, superoxides are colorful, paramagnetic, and &lt;strong&gt;unstable&lt;/strong&gt; compounds, and they also exist only in solid state.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorDisplayName="user15489" LastEditDate="2015-05-05T19:45:18.273" LastActivityDate="2015-05-05T19:45:18.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30916" PostTypeId="1" CreationDate="2015-05-05T17:39:12.660" Score="1" ViewCount="32" Body="&lt;p&gt;In my chemistry book writes that cooling an exothermic solution(e.g. $\ce{NaOH + H2O}$) increases it's solubility, and I can't understand why since the cooling should decrease it.&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorUserId="7448" LastEditDate="2015-05-05T17:50:21.693" LastActivityDate="2015-05-05T17:50:21.693" Title="Why cooling an exothermic solution it increases it's solubility?" Tags="&lt;solubility&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2015-05-05T18:11:13.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="30917" PostTypeId="2" ParentId="30913" CreationDate="2015-05-05T18:00:37.557" Score="4" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants&#xA;  are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(j) When there is a choice for lower locants related to the presence of stereogenic centers or stereoisomers, the lower locant is assigned to CIP stereodescriptors &lt;em&gt;Z&lt;/em&gt;, &lt;em&gt;R&lt;/em&gt;, &lt;em&gt;M&lt;/em&gt;, and &lt;em&gt;r&lt;/em&gt; (pseudoasymmetry) that are preferred to &lt;em&gt;E&lt;/em&gt;, &lt;em&gt;S&lt;/em&gt;, &lt;em&gt;P&lt;/em&gt;, and &lt;em&gt;s&lt;/em&gt;, respectively, which are preferred to the non-CIP stereodescriptors &lt;em&gt;cis&lt;/em&gt;, &lt;em&gt;trans&lt;/em&gt;, or &lt;em&gt;r&lt;/em&gt; (reference), &lt;em&gt;c&lt;/em&gt;, and &lt;em&gt;t&lt;/em&gt; (…).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the chirality center &lt;em&gt;R&lt;/em&gt; receives the lowest locant and thus a systematic name is &lt;strong&gt;(2​&lt;em&gt;R&lt;/em&gt;,3​&lt;em&gt;S&lt;/em&gt;)-2,3-dibromobutane&lt;/strong&gt; and not (2​​&lt;em&gt;S&lt;/em&gt;,3​​&lt;em&gt;R&lt;/em&gt;)‑2,3‑dibromobutane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your drawing shows (2​&lt;em&gt;R&lt;/em&gt;,3​&lt;em&gt;R&lt;/em&gt;)-2,3-dibromobutane.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-05T18:00:37.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30918" PostTypeId="1" CreationDate="2015-05-05T18:11:46.200" Score="1" ViewCount="730" Body="&lt;p&gt;Consider a sample of uranium-238 that is undergoing decay to thorium-234. During the transmutation, a gamma ray is emitted from the nucleus. However, the thorium-234 daughter particle does not decay quickly enough to be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of storage container should the uranium sample be stored in to minimise radiation exposure?&lt;/p&gt;&#xA;" OwnerUserId="16026" LastEditorUserId="16683" LastEditDate="2018-08-29T17:39:45.433" LastActivityDate="2018-08-29T22:09:08.130" Title="Container to minimise radiation exposure from U-238" Tags="&lt;safety&gt;&lt;nuclear-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30919" PostTypeId="1" AcceptedAnswerId="30920" CreationDate="2015-05-05T18:16:51.250" Score="4" ViewCount="269" Body="&lt;p&gt;Why do ketenes participate in [2+2] cycloadditions with their $\ce{C=C}$ bond and not the $\ce{C=O}$ bond? Can the reactivity be reversed somehow?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4231" LastEditDate="2015-05-05T19:17:45.863" LastActivityDate="2015-05-05T19:17:45.863" Title="Ketene cycloaddition reactivity" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30920" PostTypeId="2" ParentId="30919" CreationDate="2015-05-05T19:01:02.563" Score="7" Body="&lt;p&gt;Ketenes can participate in [2+2] cycloadditions with their $\ce{C=O}$ double bond.  Ketene dimerization is the best known example of such a cycloaddition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OcWnZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Ketene_cycloaddition&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you add substituents to the ketene $\ce{C=C}$ double bond, you begin to favor cycloadditon across the carbon-carbon double bond at the expense of cycloaddition across the carbonyl double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Diels-Alder cycloadditions also involve adding across the ketene carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tj0SC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://pubs.rsc.org/En/content/articlelanding/2006/ob/b611311g#!divAbstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With both the [2+2] and [4+2] cycloadditions often the kinetic product involves cycloaddition across the carbonyl, but on mild heating this cycloadduct can rearrange to the product from cycloaddition across the carbon-carbon double bond, the thermodynamic product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very nice review of the synthetic and mechanistic aspects of ketene cycloadditions can be found &lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/smnr_1999-2000_Janey_Jake.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-05T19:01:02.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30921" PostTypeId="2" ParentId="30918" CreationDate="2015-05-05T19:32:31.480" Score="11" Body="&lt;p&gt;Due to its very long half-life, U-238 is not particularly radioactive; the specific activity is only about 12 kBq/g. The emission probability of gamma radiation is less than 0.1 % per disintegration of U-238. Furthermore, the gamma energy is relatively low. Therefore, the direct gamma radiation from pure U-238 (i.e. disregarding Th-234 and further decay products) is usually insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the gamma dose rate in 1 m distance from 1 kg of pure U-238 (about 12 MBq) is only approximately 0.1 nSv/h – probably less due to self-absorption. By way of comparison, the dose rate from natural background radiation amounts to about 100 nSv/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly all of the alpha radiation from U-238 is absorbed within the sample. Only a small fraction originating from the surface of the sample can escape but is stopped by air within a distance of about 3 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you do not need any shielding. Just use a suitable closed container to prevent contamination or incorporation.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="16683" LastEditDate="2018-08-29T17:41:01.610" LastActivityDate="2018-08-29T17:41:01.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30922" PostTypeId="1" CreationDate="2015-05-05T20:19:31.527" Score="3" ViewCount="72" Body="&lt;p&gt;A little context: I make craft chili oils for friends and family. I use a two-step infusion process. First, I infuse the oil for flavor by simmering spices in the oil held around 250 degrees Fahrenheit. Second, I add dried chili flakes into the flavored oil around 250 degrees Fahrenheit. I have used canola and grapeseed oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions: -Is there an ideal temperature to best extract flavors/fragrances from spices (for ex, from cumin seeds or star anise) into the oil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Do the flavor compounds deteriorate with prolonged cook times even if oil temp is kept below smoke point? For example, how does the oil change when kept at 250 degrees Fahrenheit for 1 hour vs 2 hours?&lt;/p&gt;&#xA;" OwnerUserId="16028" LastEditorUserId="16028" LastEditDate="2015-05-06T20:37:59.057" LastActivityDate="2015-05-07T02:23:46.320" Title="Target temperature for infused cooking oil" Tags="&lt;experimental-chemistry&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="30923" PostTypeId="1" AcceptedAnswerId="31300" CreationDate="2015-05-05T22:50:44.497" Score="11" ViewCount="19154" Body="&lt;p&gt;I put a copper wire through a hole in the cap of a water bottle, filled the water bottle ~ half way full with vinegar, and left it over night.  The part of the wire inside the bottle that wasn't in vinegar turn black.  Really black.  Why is this?!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I wiped the black stuff off on paper, and it now looks green on the white paper.  Seems like it oxidized over night.  Why so fast?&lt;/p&gt;&#xA;" OwnerUserId="16029" LastEditorUserId="16683" LastEditDate="2017-08-21T08:54:58.970" LastActivityDate="2019-12-13T18:26:08.317" Title="What is the product of a copper and vinegar reaction?" Tags="&lt;everyday-chemistry&gt;&lt;redox&gt;&lt;metal&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30925" PostTypeId="1" CreationDate="2015-05-05T23:36:13.647" Score="7" ViewCount="1740" Body="&lt;p&gt;On &lt;a href=&quot;https://en.wikipedia.org/wiki/Nickel%28II%29_chloride#Structure_of_NiCl2_and_its_hydrates&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, it says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NiCl2·6H2O}$ consists of separated trans-$\ce{[NiCl2(H2O)4]}$ molecules&#xA;  linked more weakly to adjacent water molecules. Only four of the six&#xA;  water molecules in the formula are bound to the nickel, and the&#xA;  remaining two are water of crystallisation.[4] Cobalt(II) chloride&#xA;  hexahydrate has a similar structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So nickel chloride hexahydrate has an octahedral symmetry and the ligands are two chloride ions and 4 water molecules? Or is $\ce{NiCl2}$ a solid in water, but separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this also indicate that there are 8 electrons in the $d$ orbitals and that it will form a high-spin complex (I know that the low-spin orbitals look the same though) so that there are two unpaired electrons? I think I'm thrown off by the chloride ions because my textbook only explains complex ions like $\ce{[CoCl4]^4-}$ or $\ce{Fe(CN)6}$ which are easier to understand for me.  &lt;/p&gt;&#xA;" OwnerUserId="16031" LastEditorUserId="3846" LastEditorDisplayName="user15489" LastEditDate="2016-01-23T22:52:55.693" LastActivityDate="2016-02-24T02:36:24.237" Title="How many electrons are in the d orbitals of NiCl2·6H2O?" Tags="&lt;inorganic-chemistry&gt;&lt;orbitals&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30926" PostTypeId="1" AcceptedAnswerId="30931" CreationDate="2015-05-05T23:44:43.333" Score="5" ViewCount="557" Body="&lt;p&gt;I recently purchased a flask with a built in thermometer well, but I have discovered that simply setting a rod thermometer in the well does not produce a very accurate reading. I was wondering if there was maybe a specific method for measuring temperature in wells that I am not aware of. Here are some pictures of the flask: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pg8z4.jpg&quot; alt=&quot;Front View&quot;&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/lECyk.jpg&quot; alt=&quot;Better View of the well&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13396" LastActivityDate="2015-05-06T05:30:54.970" Title="Can I get an accurate temperature reading from a flask with a built-in thermometer well?" Tags="&lt;experimental-chemistry&gt;&lt;temperature&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30927" PostTypeId="1" CreationDate="2015-05-06T00:35:16.207" Score="27" ViewCount="3231" Body="&lt;p&gt;I know that experimentally, the rule has been verified but I am having trouble understanding why it works. I see no reason why equivalent cyclic structures can undergo resonance and be just as stable. Any clarification would be nice.&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="16683" LastEditDate="2018-01-29T22:20:55.123" LastActivityDate="2019-07-11T05:11:55.690" Title="What is the justification for Hückel's rule?" Tags="&lt;organic-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="30929" PostTypeId="1" AcceptedAnswerId="30932" CreationDate="2015-05-06T02:16:14.497" Score="7" ViewCount="1117" Body="&lt;p&gt;In Wikipedia's article for &lt;a href=&quot;http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Quasi-steady-state_approximation&quot;&gt;Michaelis-Menten kinetics&lt;/a&gt;, it titles the &quot;[ES] = constant&quot; section the &quot;Quasi-steady-state approximation&quot;. Lehninger's &lt;em&gt;Principles of Biochemistry&lt;/em&gt;, 4th Ed. pp. 203ff call this &quot;steady state kinetics&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that steady state is an assumption which is used for various kinetic models that have an intermediate, and I believe the term &quot;steady state&quot; is used in other settings as well. Is Wikipedia's use of &quot;quasi&quot; in the title for the enzyme kinetics reflective of $V_0$ only being the measurement of the &lt;em&gt;initial&lt;/em&gt; rate, and thus the further assumption that [S] is constant also adds more uncertainty to the steady state assumption? If this isn't the reason (and I'm not certain that it is), why add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?&lt;/p&gt;&#xA;" OwnerUserId="4813" LastActivityDate="2019-07-05T14:49:30.227" Title="How is the &quot;quasi-steady-state&quot; assumption of Michaelis-Menten kinetics different from the steady state assumption of ordinary kinetics?" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30930" PostTypeId="2" ParentId="30900" CreationDate="2015-05-06T03:19:02.233" Score="4" Body="&lt;p&gt;I'm not sure what you mean by 'regular' NMR experiments. NMR experiments which observe multi quantum signal are still 'regular' experiments, so I'll assume you asking why you wouldn't see DQ signal in a basic 1D 1H spectrum, for instance. I will attempt to explain without the use of product operator formalism, and scary looking maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In its simplest terms, the frequency of a double quantum coherence is about twice the normal observation frequency, and so cannot be observed without transformation back to single quantum coherence. Similarly, zero quantum coherence is typically less than a few kilohertz, and so is also not observable. If you set your receiver to the correct frequency, this transition would be observable, although significantly lower in intensity due to the lower probability of this transition path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OVL4y.png&quot; alt=&quot;Energy transitions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of an isolated, uncoupled spin, there are only two energy states, and the transition between these two states (the single quantum transition) is the only possible transition. For the case of coupled nuclei, the state of spin I is affected by the spins state of nucleus S. The diagram above shows the typical spin transitions for a spin system, IS. The most common transition is a single quantum transition, where only a single spin changes state, and this generates magnetisation in the transverse plane. When both spins change state simultaneously, you can get either zero quantum (the spins flip to opposite states) or double quantum (spins flip the same way) transitions. The frequency of the double quantum transition is roughly twice that of a single quantum transition. In the vector model, double quantum transitions have no magnetisation in the transverse plane (only transverse magnetisation can be observed), but can be transferred to single quantum transition through the process of generation of SQC(first pulse), evolution (linked to couplings) and generation of MQC (second pulse) in a multi quantum experiment. This is typical of a DQF/TDF COSY type experiment, but equally utilised in studies of dipolar couplings of small organics in liquid crystal media (where quantum orders of &gt;5 are typically observed)&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2015-05-06T03:19:02.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30931" PostTypeId="2" ParentId="30926" CreationDate="2015-05-06T05:24:21.947" Score="5" Body="&lt;p&gt;The thermal conductivity of air isn't very high, when compared to organic solvents or silicon oils used for heating baths. Consequently, one doesn't place thermometers in the vapour phase above the liquid when monitoring or controlling temperatures. (That is, of course, except for distillations, where you want to measure exactly that - the temperature of the vapour phase in the head.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, it might help to &lt;strong&gt;fill the well with the same silicon oil that you use for the heating bath&lt;/strong&gt;. In order to keep the rod thermometer in place (rather than letting it float), you might want to use a Quickfit adapter (similar to &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/z551813&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one) that fits the joint of the well.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-06T05:30:54.970" LastActivityDate="2015-05-06T05:30:54.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30932" PostTypeId="2" ParentId="30929" CreationDate="2015-05-06T05:51:33.070" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is Wikipedia's use of &quot;quasi&quot; in the title for the enzyme kinetics reflective of [...] the further assumption that &lt;span class=&quot;math-container&quot;&gt;$[S]$&lt;/span&gt; is constant[...]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No I don't think the &quot;quasi-steady state&quot; hypothesis is related to whether &lt;span class=&quot;math-container&quot;&gt;$[S]$&lt;/span&gt; varies with time.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[W]hy add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Different fields use different terms for the same concept.  For example, some folks call it the &quot;pseudo-steady state hypothesis&quot; and abbreviate it &quot;PSSH&quot;.  Some people just say &quot;steady state assumption&quot;.  It's all different vocabulary for the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there's a principled reason to prefer one term over another.  If you solve the full differential equations for kinetic models that would normally invoke the PSSH, you'll see one problem.  Before the reaction starts, the concentration of intermediate &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is zero.  But when the reaction is going on, we assume the intermediate is at steady state, i.e. that &lt;span class=&quot;math-container&quot;&gt;$\frac{dI}{dt}=0$&lt;/span&gt;.  But if &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; starts at zero and has an initial concentration of zero, then it is everywhere zero, which violates the prinicple that intermediates actually occur in real-life chemical reactions.  So there are two phases to such models: the first is the &quot;burst&quot; phase, also known as single-turnover conditions, or unsteady state kinetics.  In that phase &lt;span class=&quot;math-container&quot;&gt;$\frac{dI}{dt}$&lt;/span&gt; is not 0 but increases rapidly.  That phase ends quickly and is replaced by the pseudo-steady phase for &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;.  If you want to get precise about the strict definitions of the two phases, you'll need to invoke singular perturbation and asymptotic matching, as described in &lt;a href=&quot;https://faculty.washington.edu/hqian/amath4-523/Murray-Math-Biol-ch6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this thorough book chapter&lt;/a&gt; from &lt;em&gt;Mathematical Biology&lt;/em&gt; by James D. Murray. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could argue that the &quot;pseudo&quot; or &quot;quasi&quot; is good to have in there to remind you of these complexities.  Or you could argue that all those extra letters aren't doing a whole lot of good.  Either way, it's just a semantic issue and is unrelated to differences in enzyme kinetics and more &quot;traditional&quot; chemical reactions.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2019-07-05T14:49:30.227" LastActivityDate="2019-07-05T14:49:30.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30933" PostTypeId="1" AcceptedAnswerId="30989" CreationDate="2015-05-06T10:15:42.633" Score="9" ViewCount="1508" Body="&lt;p&gt;I would like to obtain the 3D charge density of a water molecule for plotting purposes (i.e. not necessarily super-accurate). I know this is a tough problem, but I'd like to give it a try. Since I'm currently using python, I've identified PyQuante and PySCF, and gather that I should learn how to run density functional theory (DFT) calculations. Before I do that, I'd like to find out if there is something else out there I should use instead. Multiple suggestions are welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've installed &lt;a href=&quot;http://pyquante.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyQuante&lt;/a&gt;, which according to it's README.txt may be more accessible to understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the color plots in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/1107/why-is-h%E2%82%82o-a-dipole/1110#1110&quot;&gt;this answer&lt;/a&gt; give a hint of what I'm talking about, though I'll ultimately plot the field strength contours and some electric field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to obtain a large table of values (or expressions/expansions?), so I can create and work with surfaces and field line plots myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally is it possible that there exist supplementary data in published articles. Where I might find that? Although this is about existing data, I actually do plan on putting in the time to have at least a minimal understanding of the calculation itself. If DFT is it, then I will be spending time studying that.&lt;/p&gt;&#xA;" OwnerUserId="16035" LastEditorUserId="16035" LastEditDate="2021-01-26T02:38:44.747" LastActivityDate="2021-01-26T02:38:44.747" Title="How can I calculate the charge distribution of a water molecule?" Tags="&lt;water&gt;&lt;reference-request&gt;&lt;software&gt;&lt;python&gt;" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30934" PostTypeId="2" ParentId="30933" CreationDate="2015-05-06T10:35:25.450" Score="5" Body="&lt;p&gt;If you are not looking to actually learn these calculations and just need the contour plots, you can use &lt;a href=&quot;http://www.arguslab.com/arguslab.com/ArgusLab.html&quot; rel=&quot;nofollow&quot;&gt;Arguslab&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you illustrative plots of orbitals as well as total electron density. They won't be super-accurate but they will be quick.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="5456" LastEditDate="2015-05-06T10:39:16.770" LastActivityDate="2015-05-06T10:39:16.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30935" PostTypeId="1" CreationDate="2015-05-06T10:40:09.773" Score="5" ViewCount="112" Body="&lt;p&gt;What is the specific gravity of $\ce{H2O2}$ bleached viscose? In literature, I can find that this is $1.52~\mathrm{g~cm^{-3}}$, but for bleached viscose is it not findable. Can I presume $\ce{H2O2}$ treated viscose has the same density?&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="7475" LastEditDate="2015-10-30T16:48:43.023" LastActivityDate="2015-10-30T16:48:43.023" Title="Specific gravity of H2O2 bleached viscose" Tags="&lt;everyday-chemistry&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-10-30T19:35:24.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="30936" PostTypeId="2" ParentId="30933" CreationDate="2015-05-06T11:03:39.520" Score="6" Body="&lt;p&gt;Two suggestions: &lt;a href=&quot;http://molcalc.org/&quot; rel=&quot;noreferrer&quot;&gt;MolCalc&lt;/a&gt;: Calculate Properties &gt; Polarity and Solvation gives you a 3D surface plot with the electrostatic potential superimposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is Avogadro/GAMESS/MacMolPlt. See for example &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/2009/11/electron-density.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-06T11:03:39.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30940" PostTypeId="1" AcceptedAnswerId="30953" CreationDate="2015-05-06T12:56:30.367" Score="30" ViewCount="14895" Body="&lt;p&gt;Why is the stability of (&lt;em&gt;Z&lt;/em&gt;)-cyclodecene &lt;strong&gt;2&lt;/strong&gt; greater than that of (&lt;em&gt;E&lt;/em&gt;)-cyclodecene &lt;strong&gt;1&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SKGvW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SKGvW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd think that since &lt;strong&gt;1&lt;/strong&gt; contains a &lt;em&gt;trans&lt;/em&gt; double bond instead of a &lt;em&gt;cis&lt;/em&gt; double bond, it should experience less steric repulsion and hence be more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've read that the &lt;em&gt;cis&lt;/em&gt; isomer &lt;strong&gt;2&lt;/strong&gt; is more stable than the &lt;em&gt;trans&lt;/em&gt; isomer &lt;strong&gt;1&lt;/strong&gt; if the ring size is 11 or smaller. For ring sizes larger than 11, the opposite is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so? Also, what would happen if there were more double bonds?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="16683" LastEditDate="2017-08-24T05:58:23.687" LastActivityDate="2017-08-24T05:59:47.680" Title="Relative stability of cis and trans cycloalkenes" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;isomers&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="30941" PostTypeId="2" ParentId="29666" CreationDate="2015-05-06T13:47:01.503" Score="3" Body="&lt;p&gt;The first question to always ask when you are trying to build an MO diagram is what point group does my molecule fall into?  In this case, the $\ce{CH2}$ molecule has C2V symmetry (like the water example shown above).  Each of the atomic orbitals (AO's) on carbon (left) and the combination of $1s$ orbitals on the hydrogen atoms (right) has a symmetry label that corresponds to one of the irreducible representations of the point group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can determine by symmetry label for the AO's by operating on the orbital with each of the symmetry operators in the C2V group (E, C2, σ, σ').  For example, the 2py orbital is in the plane of the nuclei and so it is symmetric with respect to E and σ' but anti-symmetric with respect to C2 and σ and has b2 symmetry from the C2V character table.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the hydrogens, you have to consider the combination of both 1s orbitals at the same time.  The a1 combination is when the phase of the wavefunction is the same and the b2 combination is when the phase is opposite (i.e. H1 is positive and H2 is negative or &lt;em&gt;vice versa&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AO's that have the same symmetry can combine to form MO's with that symmetry label.  In the above example, the 2s and 2pz orbitals on oxygen have a1 symmetry, and they combine with the a1 combination of hydrogen 1s orbitals to form three MO's with a1 symmetry.  The 2py (b2 symmetry) orbital combines with the b2 combination of the hydrogen atoms to form two MO's. Finally, the 2px orbital remains unchanged because there are no other SLAC's with b1 symmetry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MO's are qualitatively ordered based on the number of nodes that are found when you combine the SLAC's.  The numbers in front of the MO's differentiate the orbitals with the same symmetry.   &lt;/p&gt;&#xA;" OwnerUserId="16037" LastEditorDisplayName="user15489" LastEditDate="2015-06-05T18:59:47.757" LastActivityDate="2015-06-05T18:59:47.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30942" PostTypeId="1" AcceptedAnswerId="30946" CreationDate="2015-05-06T13:51:15.567" Score="2" ViewCount="2561" Body="&lt;p&gt;I need to know if a combination of bleach and the chemicals below will create poisonous gas. Will it create chloramine, and will there be any visible gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Sodium laureth sulfate&#xA;2) Sodium citrate&#xA;3) C12-14 Pareth-3&lt;/p&gt;&#xA;" OwnerUserId="5225" LastActivityDate="2016-03-11T23:04:01.883" Title="Will bleach combined with any of these chemicals create poisonous gas?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30943" PostTypeId="1" AcceptedAnswerId="30960" CreationDate="2015-05-06T13:51:15.403" Score="12" ViewCount="23161" Body="&lt;p&gt;I found a &lt;a href=&quot;http://forums.studentdoctor.net/threads/why-is-aniline-aromatic.623668/&quot; rel=&quot;nofollow noreferrer&quot;&gt;thread&lt;/a&gt; which discussed this which totally confused me. (refer to the part of the link which talks about $\ce{sp^2/sp^3}$)&lt;br&gt;&#xA;The question &quot;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/4390/&quot;&gt;How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?&lt;/a&gt;&quot; is helpful, but doesn't have aniline specifically. I understand that if the lone pair is in resonance, it becomes $\ce{sp^2}$.&lt;br&gt;&#xA;Can someone help me out?  &lt;/p&gt;&#xA;" OwnerUserId="16038" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2020-01-07T19:15:05.710" Title="Is nitrogen of aniline sp² or sp³ hybridized?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ClosedDate="2015-05-07T05:38:50.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="30944" PostTypeId="2" ParentId="30797" CreationDate="2015-05-06T14:00:05.093" Score="6" Body="&lt;p&gt;The MO diagram shows that the $\ce{CO}$ molecule forms three filled MO's with σ symmetry and two MO's with π symmetry.  Of the five filled MO's (10 electrons) formed for $\ce{CO}$, only four of them can be half-filled from carbon electrons (4 valence electrons).  So one of the filled MO's must have two electrons that originally came from the oxygen atom.  &lt;/p&gt;&#xA;" OwnerUserId="16037" LastEditorDisplayName="user15489" LastEditDate="2015-05-06T18:56:11.127" LastActivityDate="2015-05-06T18:56:11.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30945" PostTypeId="1" AcceptedAnswerId="30947" CreationDate="2015-05-06T14:07:12.513" Score="2" ViewCount="56847" Body="&lt;p&gt;I know I should divide &lt;span class=&quot;math-container&quot;&gt;$\pu{g L-1}$&lt;/span&gt; by the molar mass of the substance, but I don't seem to find the specific answer on Google. So just to be sure:&lt;/p&gt;&#xA;&lt;p&gt;If I have &lt;span class=&quot;math-container&quot;&gt;$10^{-5}~\pu{g L-1}~\ce{Cu^2+}$&lt;/span&gt; solution, do I have &lt;span class=&quot;math-container&quot;&gt;$1.57 \times 10^{-7}~\pu{mol L-1}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;I think I'm missing something.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="83758" LastEditDate="2020-06-12T16:01:45.047" LastActivityDate="2020-06-12T16:01:45.047" Title="How to convert g/L to mol/L" Tags="&lt;concentration&gt;&lt;stoichiometry&gt;&lt;units&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30946" PostTypeId="2" ParentId="30942" CreationDate="2015-05-06T14:12:53.693" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I need to know if a combination of bleach and the chemicals below will create poisonous gas.&lt;/p&gt;&#xA;&lt;p&gt;Sodium laureth sulfate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sodium laureth sulfate is &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3(CH2)11(OCH2CH2)_{\bf{3}}OSO3Na}$&lt;/span&gt;, where the boldface 3 is an approximation and can actually vary between 1 and 5 or so.  The key point is it contains no nitrogen.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Sodium citrate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Sodium citrate can refer to mono-, di-, or tri-sodium citrate, which have formulae like &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H_{7-\it{n}}O7Na_{\it{n}}}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is 1, 2, or 3, respectively.  Again there is no nitrogen.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;C12-14 Pareth-3&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;According to Google, &amp;quot;C12-14 Pareth-3&amp;quot; is a &lt;em&gt;&amp;quot;polyethylene glycol ether of a mixture of synthetic C12-14 alcohols with an average of 3 moles of ethylene oxide.&amp;quot;&lt;/em&gt;  The most representative compound would presumably be the C13 alcohol with exactly 3 moles of ethylene oxide.  That would have the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3(CH2)11CH2O-(CH2CH2-O)3H}$&lt;/span&gt;, or, if I did my math right, &lt;span class=&quot;math-container&quot;&gt;$\ce{C19H40O4}$&lt;/span&gt;.  Again the key point is there is no nitrogen.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Will it create chloramine, and will there be any visible gas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;None of the compounds have nitrogen, so &lt;a href=&quot;http://en.wikipedia.org/wiki/Chloramine&quot; rel=&quot;nofollow noreferrer&quot;&gt;chloramine&lt;/a&gt; cannot be created from bleach (chemically known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_hypochlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;sodium hypochlorite&lt;/a&gt;) and these chemicals.&lt;/p&gt;&#xA;&lt;p&gt;I think the likely products of bleach treatment would be non-volatile compounds such as inorganic sulfate plus organic compounds like tridecanoic acid, tridecanol, ethylene glycol, as well as oxidized derivatives of all of those organics, possibly including some small amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; gas.  But &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; gas is not poisonous and not visible.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-03-11T23:04:01.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30947" PostTypeId="2" ParentId="30945" CreationDate="2015-05-06T14:15:52.810" Score="8" Body="&lt;p&gt;Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$&#xA;So $$c = \frac{m}{VM}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-10-05T13:42:49.317" LastActivityDate="2015-10-05T13:42:49.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30948" PostTypeId="5" CreationDate="2015-05-06T14:21:52.303" Score="0" Body="&lt;h2&gt;Tag usage&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This tag should be used on questions about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Whether a molecule or class of molecules is chiral&lt;/li&gt;&#xA;&lt;li&gt;Group theoretical treatments of chirality&lt;/li&gt;&#xA;&lt;li&gt;Chirality in biomolecules e.g. DNA, proteins&lt;/li&gt;&#xA;&lt;li&gt;The effect of chirality on properties or the reactions of chiral compounds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Definition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Chirality&lt;/em&gt; is &lt;a href=&quot;https://goldbook.iupac.org/C01058.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;defined&lt;/a&gt; in the IUPAC Gold Book as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane, σ = &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;1&lt;/sub&gt;, a centre of inversion, &lt;em&gt;i&lt;/em&gt; = &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;2&lt;/sub&gt;, a rotation-reflection axis, &lt;em&gt;S&lt;/em&gt;&lt;sub&gt;2&lt;i&gt;n&lt;/i&gt;&lt;/sub&gt;). If the object is superposable on its mirror image the object is described as being achiral.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Further reading&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The question: &lt;a href=&quot;http://chemistry.stackexchange.com/questions/59124/what-is-the-perfect-definition-for-chirality&quot;&gt;What is the perfect definition for chirality?&lt;/a&gt; and the answers therein provide an excellent overview of chirality.&lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="16683" LastEditDate="2017-01-10T09:28:28.997" LastActivityDate="2017-01-10T09:28:28.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30949" PostTypeId="4" CreationDate="2015-05-06T14:21:52.303" Score="0" Body="Applicable to questions about the ability of a molecule (or class of molecules) to be superimposed on their mirror image(s)." OwnerUserId="10185" LastEditorUserId="16683" LastEditDate="2017-01-10T09:03:23.547" LastActivityDate="2017-01-10T09:03:23.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30950" PostTypeId="5" CreationDate="2015-05-06T14:24:11.890" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T14:24:11.890" LastActivityDate="2015-05-06T14:24:11.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30951" PostTypeId="4" CreationDate="2015-05-06T14:24:11.890" Score="0" Body="For questions relating to organic compounds which contain a bond between a carbon atom and a metal." OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-06T15:30:43.387" LastActivityDate="2015-05-06T15:30:43.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30952" PostTypeId="2" ParentId="30933" CreationDate="2015-05-06T14:34:39.370" Score="-2" Body="&lt;p&gt;If you only need the electron density distribution of a water molecule, it doesn't make much sense for you to leanr DFT method. Although I don't think there will be such data out there, it's pretty easy to run such a calculation for people who do DFT calculation regularly. It will take several minutes. My suggestion is that you find someone to help you with the calculation. The calculation could give you the data describing electron density in the 3D space. Then you can play with the data by yourself. &lt;/p&gt;&#xA;" OwnerUserId="16040" LastActivityDate="2015-05-06T14:34:39.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30953" PostTypeId="2" ParentId="30940" CreationDate="2015-05-06T14:41:25.890" Score="30" Body="&lt;p&gt;Usually &lt;em&gt;trans&lt;/em&gt;-olefins are more stable than their &lt;em&gt;cis&lt;/em&gt; isomers for steric reasons, like you suggested.  However in small and medium size rings this is not the case; here the &lt;em&gt;cis&lt;/em&gt;-cycloalkene is more stable than the corresponding  &lt;em&gt;trans&lt;/em&gt; isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;trans&lt;/em&gt;-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (&lt;em&gt;trans&lt;/em&gt;-cyclohexene and &lt;em&gt;trans&lt;/em&gt;-cycloheptene have been detected as short-lived intermediates).  In your drawing up above of &lt;em&gt;trans&lt;/em&gt;-cyclodecene (a) draw in the two hydrogens on the &lt;em&gt;trans&lt;/em&gt; double bond.  One of them is &quot;inside&quot; the ring and causes steric problems with the carbons on the other side of the ring as pictured below.  Because of this destabilizing transannular interaction &lt;em&gt;trans&lt;/em&gt;-cycloalkenes with up to around 11 carbons do not exist as planar molecules.  They exist in a conformation as pictured below for &lt;em&gt;trans&lt;/em&gt;-cyclooctene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDQsu.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can, build a model of &lt;em&gt;trans&lt;/em&gt;-cyclooctene.  Notice how you have to pull and stretch the methylene chain in order to attach that last methylene back onto the other end of the double bond - there is a lot of strain in this molecule. Now build a model of &lt;em&gt;cis&lt;/em&gt;-cyclooctene; it is easy to loop the methylene chain from one end of the double bond to the other end.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not until you get to around &lt;em&gt;trans&lt;/em&gt;-cyclododecene (12 carbons) do you have enough carbons to easily connect from one end of the &lt;em&gt;trans&lt;/em&gt; double bond back to the other end.  &lt;strong&gt;&lt;em&gt;trans&lt;/em&gt;-Cycloalkenes with less than 12 carbons have additional strain because of this difficulty in spanning the double bond&lt;/strong&gt;. Once you get to around 12 carbons there is no difficulty spanning the double bond and so once again &lt;em&gt;trans&lt;/em&gt;-cycloalkenes become more stable than their &lt;em&gt;cis&lt;/em&gt; counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since double bond carbons are $\mathrm{sp^2}$ hybridized with 120° angles, if we add more double bonds, the 120° angles make it more difficult to loop the more flexible $\mathrm{sp^3}$ methylene (with their 109.5° angles) chain around the double bonds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-08-24T05:59:47.680" LastActivityDate="2017-08-24T05:59:47.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="30954" PostTypeId="1" AcceptedAnswerId="30962" CreationDate="2015-05-06T14:44:33.100" Score="2" ViewCount="1066" Body="&lt;p&gt;I have been trying to get a relationship formula to determine the viscosity of Linseed oil. I have Absolute viscosity of the oil at 33.1 Cp (Source: &lt;a href=&quot;http://www.engineeringtoolbox.com/absolute-viscosity-liquids-d_1259.html&quot; rel=&quot;nofollow&quot;&gt;http://www.engineeringtoolbox.com/absolute-viscosity-liquids-d_1259.html&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the system where I am using the fluid has a varying temperature, and I want to be able to know what the varying viscosity is. I am aware that the viscosity reduces with increase in temperature, but wasn't able to find a specific formula to relate the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula to get the relationship between temperature and viscosity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="16041" LastActivityDate="2021-01-23T13:09:24.293" Title="Relation between Temperature and viscosity of Linseed Oil?" Tags="&lt;temperature&gt;&lt;viscosity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30955" PostTypeId="2" ParentId="30943" CreationDate="2015-05-06T14:50:39.783" Score="-2" Body="&lt;p&gt;As the non-bonding electron pair of nitrogen could partially contribute in the resonance with the other  $\pi$ electrons in the cycle. (I will  prove this point indirectly further using an experimental data), the nitrogen atom hybridization is a linear combination of sp2 and sp3.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we consider the nitrogen atom in the pyridine, it  is sp2 hybridized. Because the nitrogen atom has two neighboring atoms (2 carbon atoms) and one non-bonding electron pair, i.e. three neighboring electron pairs. But, the non-bonding electron pair can not contribute in the resonance with the other four $\pi$ electrons in the cycle. In fact, it is  perpendicular to the aromatic $\pi$ system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pKa for the pyridine systems is 5.2, for aniline it is 4.6.  So pyridine is a stronger base.  We can link this to the availability of the lone pair in pyridine, as it is not delocalized ( perpendicular to the aromatic $\pi$  system.  While in aniline the nitrogen lone pair can be delocalised using resonance with the adjacent $\pi$ system.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-05-09T13:13:21.923" LastActivityDate="2015-05-09T13:13:21.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30956" PostTypeId="5" CreationDate="2015-05-06T14:50:40.063" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T14:50:40.063" LastActivityDate="2015-05-06T14:50:40.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30957" PostTypeId="4" CreationDate="2015-05-06T14:50:40.063" Score="0" Body="Also known as cis/trans isomerism, configurational isomerism, or E/Z isomerism when referring to alkenes." OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-06T15:30:20.293" LastActivityDate="2015-05-06T15:30:20.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30958" PostTypeId="5" CreationDate="2015-05-06T14:57:00.047" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T14:57:00.047" LastActivityDate="2015-05-06T14:57:00.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30959" PostTypeId="4" CreationDate="2015-05-06T14:57:00.047" Score="0" Body="Spin is a type of angular momentum which is intrinsic to atomic and sub-atomic particles. Electrons in orbitals can be either spin paired or spin unpaired which influences the magnetic properties of the species containing these orbitals." OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-06T15:30:37.440" LastActivityDate="2015-05-06T15:30:37.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30960" PostTypeId="2" ParentId="30943" CreationDate="2015-05-06T15:08:42.867" Score="17" Body="&lt;p&gt;The nitrogen in aniline is somewhere between &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized, probably closer to the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; side.  We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized).  However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assess the nitrogen hybridization by measuring its barrier for &lt;a href=&quot;http://www.expertsmind.com/topic/properties-of-amines/pyramidal-inversion-913906.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyramidal inversion&lt;/a&gt;. If a trigonal nitrogen is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; hybridized the inversion barrier is typically around &lt;span class=&quot;math-container&quot;&gt;$\pu{4-5 kcal/mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dMxP3.jpg&quot; alt=&quot;pyramidal inversion diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(pyramidal inversion diagram)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aniline this barrier is very low, somewhere around &lt;span class=&quot;math-container&quot;&gt;$\pu{1-2 kcal/mol}$&lt;/span&gt;. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being  planar (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;) than pyramidal (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt;). &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="61578" LastEditDate="2020-01-07T19:15:05.710" LastActivityDate="2020-01-07T19:15:05.710" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30962" PostTypeId="2" ParentId="30954" CreationDate="2015-05-06T15:44:14.403" Score="1" Body="&lt;p&gt;There are many formulas!  This usually means that some of them are OK in general for estimation purposes, but none of them will be especially accurate.  Not easy to choose from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature_dependence_of_liquid_viscosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;many formulas&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most simple and popular seems to be the Wright equation, &#xA;$$\log_{10}\left(\log_{10}\left(\nu + \lambda \right)\right)=A-B\log_{10}(T),$$ &#xA;where $\nu$ is the kinematic viscosity, $\lambda$ is a parameter, usually taken to be $0.7$, and $A$ and $B$ are compound-specific parameters.  With knowledge of linseed oil viscosity at least one other temperature than the one you provided in your question, you could fit $A$ and $B$ parameters and apply the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the best answer to your question is to obtain some empirical data.  A 1977 paper[1] reports on linseed oil viscosity as a function of temperature, for example.  They found that a different equation (not the Wright) gave a reasonable fit to the data.  Specifically, they found &#xA;$$\ln(\eta) = -5.73 + \frac{2820}{T}$$&#xA;for viscosity $\eta$ in centipoises and temperature $T$ in K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to keep in mind is that linseed oil is prone to oxidation.  Oxidized linseed oil may have different properties than undergraded, &quot;pure&quot; linseed oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: T. Gallagher; &lt;em&gt;D. E. Kline; J. Am. Oil Chem. Soc.&lt;/em&gt; &lt;strong&gt;1977,&lt;/strong&gt; &lt;em&gt;54&lt;/em&gt; (2), 68-70. DOI: &lt;a href=&quot;https://doi.org/10.1007/BF02912392&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1007/BF02912392&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="4945" LastEditDate="2018-08-22T12:38:32.340" LastActivityDate="2018-08-22T12:38:32.340" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30963" PostTypeId="1" CreationDate="2015-05-06T15:51:59.640" Score="2" ViewCount="269" Body="&lt;p&gt;I was wondering why a hydrogen attached to a nitrogen or oxygen is not able to couple with p+ on adjacent carbon atoms and hence why it always appeats as a singlet in HNMR spectroscopy?&lt;/p&gt;&#xA;" OwnerUserId="15748" LastActivityDate="2016-07-14T23:32:51.747" Title="Proton attached to N or O in HNMR spectroscopy" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-06T18:51:28.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="30964" PostTypeId="1" CreationDate="2015-05-06T16:28:48.483" Score="3" ViewCount="166" Body="&lt;p&gt;I need some ethanol pure and strong enough for technical purposes (fuel, solvent, cleaning agent, dissolving rosin for soldering flux, etc.). Food quality is not required though, denaturated one is fine too. Distilling cheap vodka is possible, but even cheap vodka is expensive because of high tax on food ethanol; also now I don't have a distiller and want to try getting ethanol without building one. Pure denaturated ethanol is sold mostly in huge industrial bulks only. But we have extremely affordable 70% ethanol-based solutions in medicine stores - they are several-percent solutions of boric, formic and salicylic acids respectively. So, my question is: how can I turn these acids into precipitate or gas for easy removal, using not so exotic reagents and conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: I have realized my question may be ambiguous. Every substance I'm speaking about consists of alcohol + ONE of the three mentioned acids, not all of them together. I have a choice which acid I will be getting rid of. &lt;/p&gt;&#xA;" OwnerUserId="16044" LastEditorUserId="10185" LastEditDate="2015-05-06T17:35:57.153" LastActivityDate="2015-05-06T17:35:57.153" Title="How can I purify ethanol from formic / boric / salicylic acid?" Tags="&lt;experimental-chemistry&gt;&lt;alcohols&gt;&lt;purification&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="30965" PostTypeId="2" ParentId="30963" CreationDate="2015-05-06T16:39:39.203" Score="2" Body="&lt;p&gt;It depends on the condition of your experiment if a coupling of a proton of an amino or hydroxyl group with a proton attached to the $\alpha{}$-carbon is observable, or not.  As pointed out by the indicator set by @Mithoron, intermolecular exchange processes need to be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If your alcohol (or amine) to be analyzed is very pure, and the deuterated solvent you deploy is both very pure, and dry, and free of acids or bases &lt;em&gt;catalysing&lt;/em&gt; proton exchange, than you may observe $^3J$-couplings you expect to see.  For example a freshly opened sample of deuterated DMSO (used as solvent) may contribute to this observation.  See the &lt;em&gt;peak data&lt;/em&gt; of the 1H NMR spectrum of ethyl alcohol in DMSO &lt;a href=&quot;http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_disp.cgi?sdbsno=1300&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you aim to analyze an alcohol, not well dried, analyzed in deuterated chloroform or methanol that was frequently exposed to ambient (humid) air, then indeed instead of a triplet of an hydroxyl group  ($\ce{-OH}$)  attached to a methylene group ($\ce{-CH2-{}}$) all what may be spot is a broad singlet signal.  The same reference, 90 MHz 1H NMR in deutero chloroform provides one example.  (Note the chemical displacement of the hydroxyl protons altered, too.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="9961" LastEditDate="2016-07-14T23:32:51.747" LastActivityDate="2016-07-14T23:32:51.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30966" PostTypeId="2" ParentId="29318" CreationDate="2015-05-06T19:46:10.783" Score="1" Body="&lt;p&gt;If one forgets about the correct mathematical notation for a moment: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a cupola is like a bowl&lt;/li&gt;&#xA;&lt;li&gt;a bowl is a fragment of a spherical object&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Molecules of this shape are found in the field of fullerenes (&lt;em&gt;bucky balls&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most simple bowl-shaped molecules is &lt;em&gt;sumanene&lt;/em&gt;. If you like to play a bit with the structure, please fetch &lt;a href=&quot;http://pastebin.com/tYTjD2JL&quot; rel=&quot;nofollow noreferrer&quot;&gt;sumanene.cml&lt;/a&gt; from pastebin and open it in &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y9RAV.png&quot; alt=&quot;structure of sumanene&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape is more obvious in the following animation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3isI.gif&quot; alt=&quot;animation of sumanene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of two extended bowl-shaped molecules has been published in &#xA;&lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;133&lt;/em&gt;, 16319–16321 (&#xA;&lt;a href=&quot;http://dx.doi.org/10.1021/ja2067725&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;), and although they are not true cupolas, you might find them interesting anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've uploaded the structures &lt;a href=&quot;http://pastebin.com/MZgdHkY6&quot; rel=&quot;nofollow noreferrer&quot;&gt;bowl-01.cml&lt;/a&gt; and &lt;a href=&quot;http://pastebin.com/u1WD455w&quot; rel=&quot;nofollow noreferrer&quot;&gt;bowl-02.cml&lt;/a&gt; as CML files.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-06T19:46:10.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30967" PostTypeId="5" CreationDate="2015-05-06T20:02:53.300" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T20:02:53.300" LastActivityDate="2015-05-06T20:02:53.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30968" PostTypeId="4" CreationDate="2015-05-06T20:02:53.300" Score="0" Body="the spontaneous net movement of solvent molecules through a semi-permeable membrane into a region of higher solute concentration, in the direction that tends to equalize the solute concentrations on the two sides." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-06T21:01:08.373" LastActivityDate="2015-05-06T21:01:08.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30969" PostTypeId="5" CreationDate="2015-05-06T20:05:53.627" Score="0" Body="&lt;p&gt;Atomic radius is the measure of the size of an atom, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding &quot;cloud&quot; of electrons. Atomic radius are of three types:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ionic radius - for ionic species like $\ce{NaCl}$&lt;/li&gt;&#xA;&lt;li&gt;Covalent radius - for species like $\ce{Cl2}$&lt;/li&gt;&#xA;&lt;li&gt;van der Waal's radius - for non-bonding species like neon&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In each case, the radius is defined as half of the distance of the two atoms/ions. In the third case, it is actually half the &lt;em&gt;closest&lt;/em&gt; distance of approach, since the atoms don't bond with each other.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-17T18:27:31.690" LastActivityDate="2018-03-17T18:27:31.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30970" PostTypeId="4" CreationDate="2015-05-06T20:05:53.627" Score="0" Body="For questions about the atomic radius - a measure of the size of an atom, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding &quot;cloud&quot; of electrons." OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-17T18:26:56.033" LastActivityDate="2018-03-17T18:26:56.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30971" PostTypeId="5" CreationDate="2015-05-06T20:11:47.453" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T20:11:47.453" LastActivityDate="2015-05-06T20:11:47.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30972" PostTypeId="4" CreationDate="2015-05-06T20:11:47.453" Score="0" Body="Questions related to color as well as the reflecting and refracting properties of chemicals." OwnerDisplayName="user15489" LastEditorUserId="402" LastEditorDisplayName="user15489" LastEditDate="2017-04-06T23:30:13.923" LastActivityDate="2017-04-06T23:30:13.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30974" PostTypeId="2" ParentId="30881" CreationDate="2015-05-06T21:05:29.837" Score="2" Body="&lt;p&gt;I got the answer on a Pymol listserve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it can be done in Pymol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose the residue you want to mutate from cytosine to 5-methylcytosine.&#xA;Use &lt;strong&gt;Action&lt;/strong&gt; -  &lt;strong&gt;Hydrogens&lt;/strong&gt; - &lt;strong&gt;Add&lt;/strong&gt; function to add hydrogens.&#xA;Change mouse selection to choose atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click on the hydrogen you want to replace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click builder (there is more work to do if you're on a Mac, see below) to open builder window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're on a Mac, you won't have Builder functionality unless you change your MacPyMOL.app file name to PyMOLX11Hybrid.app&lt;/p&gt;&#xA;" OwnerUserId="16006" LastActivityDate="2015-05-06T21:05:29.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30975" PostTypeId="1" CreationDate="2015-05-06T22:40:34.263" Score="3" ViewCount="110" Body="&lt;p&gt;In my book is given the mechanism of $\ce{HBr}$ forming and I understand every step of it. We did the calculations of the rate of each reaction, which is easy depending on the sign (-,+) of compounds formed or spent, and knowing those rates we were finally able to find the rate of $\ce{HBr}$ which is well defined by the &lt;strong&gt;&lt;em&gt;empirical equation of Bodenstein and Lind&lt;/em&gt;&lt;/strong&gt;. If you know all those reactions, its not hard to find the equations you need, but my questions are : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way we can know which exactly are those reactions (mechanism)  for other examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would we know if any reaction is a branched one ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorDisplayName="user15489" LastEditDate="2015-05-06T23:20:51.570" LastActivityDate="2015-05-07T10:02:22.667" Title="Mechanisms in kinetic reactions" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30976" PostTypeId="1" AcceptedAnswerId="30977" CreationDate="2015-05-06T22:43:58.053" Score="2" ViewCount="11249" Body="&lt;p&gt;I was thinking about liquids, and I started to wonder theses related questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Besides mercury, what elements are naturally liquid at room temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What naturally found family of substances/mixtures that do not contain  $\ce{H2O}$ are naturally liquid at room temperature?&lt;/p&gt;&#xA;" OwnerUserId="16047" LastEditorUserId="16047" LastEditDate="2015-05-06T23:46:32.087" LastActivityDate="2016-05-19T11:21:52.213" Title="What elements and/or substances without water are liquid at room temperature?" Tags="&lt;solutions&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ClosedDate="2016-05-19T18:16:16.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="30977" PostTypeId="2" ParentId="30976" CreationDate="2015-05-06T22:59:55.893" Score="7" Body="&lt;p&gt;This question is a bit broad in terms of the sheer amount of chemical compounds and mixtures that are liquid at room temperature. Examples include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Compounds&lt;/strong&gt;&lt;br&gt;&#xA;Acids, bases, many hydrocarbons (e.g. hexane) and many more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mixtures&lt;/strong&gt;&lt;br&gt;&#xA;Crude oil, aqua regia and many more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of &lt;strong&gt;elements&lt;/strong&gt;, there are only two that are liquid at &lt;a href=&quot;http://en.wikipedia.org/wiki/Room_temperature&quot; rel=&quot;nofollow&quot;&gt;room temperature&lt;/a&gt; (say about 20 °C or 293 K):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury_(element)&quot; rel=&quot;nofollow&quot;&gt;Mercury&lt;/a&gt; (as you identified).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bromine&quot; rel=&quot;nofollow&quot;&gt;Bromine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/periodic/faq/liquid-elements.shtml&quot; rel=&quot;nofollow&quot;&gt;Francium, cesium, gallium and rubidium&lt;/a&gt; are close, with melting points at 300 K, 301.59 K, 303.3 K and 312.46 K respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LennTech provides a list &lt;a href=&quot;http://www.lenntech.com/periodic-chart-elements/melting-point.htm&quot; rel=&quot;nofollow&quot;&gt;The elements of the periodic table sorted by melting point&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user7951" LastEditDate="2015-06-26T13:26:55.283" LastActivityDate="2015-06-26T13:26:55.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30978" PostTypeId="1" AcceptedAnswerId="30979" CreationDate="2015-05-06T23:20:36.487" Score="1" ViewCount="288" Body="&lt;p&gt;If we take a look at the electromagnetic specter, microwave wavelength is higher than infrared, which means its &lt;em&gt;energy is very small&lt;/em&gt;. To break  chemical bonds, the energy required is 100-1000 kJ/mol, and this kind of energy can be afforded only by UV and VIS. Due to this, we can conclude that microwave radiation has only a vibrational effect on molecules of food and does &lt;em&gt;not&lt;/em&gt; break any bonds ( the food remains the same ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering those facts, why is the microwave avoided in concern for health? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-06T23:47:24.777" LastActivityDate="2015-05-07T00:02:35.297" Title="Is a microwave oven dangerous?" Tags="&lt;spectroscopy&gt;&lt;food-chemistry&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-07T01:31:44.240" ContentLicense="CC BY-SA 3.0" />
  <row Id="30979" PostTypeId="2" ParentId="30978" CreationDate="2015-05-06T23:27:43.027" Score="3" Body="&lt;p&gt;If there is no concern over the quality of the food, I'm guessing the question is more specific to whether the microwave oven is a safe piece of equipment to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fda.gov/ForConsumers/ConsumerUpdates/ucm048953.htm&quot; rel=&quot;nofollow&quot;&gt;The U.S. Food and Drug Administration&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Manufacturers must certify that their microwave ovens comply with strict FDA emission limits. The emission limits are well below the threshold for risk to public health.&#xA;  Most injuries related to microwave ovens are the result of serious thermal burns from hot containers, overheated foods, or exploding liquids.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;There have been extremely rare instances of radiation injury due to unusual circumstances or improper servicing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we're not talking about the oven itself, and rather the nature of chemicals involved with its use, then we could talk about microwaving with plastics with foods. There are sources online that raise concerns over plastics leeching dangerous chemicals into the food. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.health.harvard.edu/staying-healthy/microwaving-food-in-plastic-dangerous-or-not&quot; rel=&quot;nofollow&quot;&gt;Harvard School of Medicine&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;If you’re concerned about plastic wraps or containers in the&#xA;  microwave, transfer food to glass or ceramic containers labeled for&#xA;  use in microwave ovens.&lt;/li&gt;&#xA;  &lt;li&gt;Don’t let plastic wrap touch food during&#xA;  microwaving because it may melt. &lt;/li&gt;&#xA;  &lt;li&gt;Wax paper, kitchen parchment paper,&#xA;  white paper towels, or a domed container that fits over a plate or&#xA;  bowl are better alternatives. &lt;/li&gt;&#xA;  &lt;li&gt;Most takeout containers, water bottles,&#xA;  and plastic tubs or jars made to hold margarine, yogurt, whipped&#xA;  topping, and foods such as cream cheese, mayonnaise, and mustard are&#xA;  not microwave-safe. &lt;/li&gt;&#xA;  &lt;li&gt;Microwavable takeout dinner trays are formulated&#xA;  for one-time use only and will say so on the package. &lt;/li&gt;&#xA;  &lt;li&gt;Old, scratched,&#xA;  or cracked containers, or those that have been microwaved many times,&#xA;  may leach out more plasticizers. &lt;/li&gt;&#xA;  &lt;li&gt;Don’t microwave plastic storage bags&#xA;  or plastic bags from the grocery store. &lt;/li&gt;&#xA;  &lt;li&gt;Before microwaving food, be&#xA;  sure to vent the container: leave the lid ajar, or lift the edge of&#xA;  the cover.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-07T00:02:35.297" LastActivityDate="2015-05-07T00:02:35.297" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="30980" PostTypeId="2" ParentId="30923" CreationDate="2015-05-07T00:21:36.457" Score="6" Body="&lt;p&gt;If it looks green/blue on paper, you likely have copper (II) acetate. You likely have the anhydrous version $\ce{Cu(OAc)2}$ which is darker than the hydrous $\ce{Cu(OAc)4(H2O)2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Copper%28II%29_acetate&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia Images&lt;/a&gt; include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/si2yy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wT0W3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of formation is a property inherent to the reactants and the products, but it also depends on temperature.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-07T00:27:20.077" LastActivityDate="2015-05-07T00:27:20.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30982" PostTypeId="1" CreationDate="2015-05-07T00:41:09.950" Score="13" ViewCount="9975" Body="&lt;p&gt;A TA talked about this today in the context of ketones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stick an acetone and $\ce{NaBH4}$ together in a protic solvent, you get 2-propanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stick acetone and $\ce{NaH}$ together, you get an enolate. What's the difference? They're both hydrides.&lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorUserId="9961" LastEditDate="2015-05-09T14:01:48.430" LastActivityDate="2018-03-04T02:57:00.933" Title="Why does a hydride like NaH act as a base, but a hydride like NaBH4 act as a nucleophile?" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30983" PostTypeId="2" ParentId="30982" CreationDate="2015-05-07T01:17:04.847" Score="7" Body="&lt;p&gt;First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction. &lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorDisplayName="user15489" LastEditDate="2015-05-07T01:19:28.913" LastActivityDate="2015-05-07T01:19:28.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30985" PostTypeId="1" AcceptedAnswerId="30988" CreationDate="2015-05-07T02:37:24.780" Score="5" ViewCount="1156" Body="&lt;p&gt;In titration we usually take burette reading upto $0.01~\mathrm{mL}$, but my teacher says that we need to take reading up to $0.05~\mathrm{mL}$. What does that mean? Does it mean if I read $23.56~\mathrm{mL}$ that I should record it as $23.60~\mathrm{mL}$ and if I read $23.44~\mathrm{mL}$ I should record it as $23.40~\mathrm{mL}$?&lt;/p&gt;&#xA;" OwnerUserId="13208" LastEditorUserId="15410" LastEditDate="2015-05-07T03:18:36.340" LastActivityDate="2015-05-07T05:11:24.780" Title="What does it mean to take reading up to 0.05 mL?" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30986" PostTypeId="1" CreationDate="2015-05-07T02:52:32.137" Score="2" ViewCount="51" Body="&lt;p&gt;For example ink on paper: how can I confirm that the bonding is van der Waals and not covalent? Similarly, the same question can be asked about other interaction such as hydrogen bonding, ionic bonding, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there analytical or spectroscopic methods for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-05-07T02:52:32.137" Title="How to confirm the type of bonding taking place between two substances?" Tags="&lt;bond&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="30987" PostTypeId="2" ParentId="28359" CreationDate="2015-05-07T03:20:18.057" Score="3" Body="&lt;p&gt;Natural variables of $\Delta G$ are $p$, $T$, and $N_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ = pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ = Temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_i$ = number of particles  (or number of moles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The derivation showing this is listed on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy#Derivation&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="13184" LastEditDate="2015-05-07T15:10:56.990" LastActivityDate="2015-05-07T15:10:56.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30988" PostTypeId="2" ParentId="30985" CreationDate="2015-05-07T05:11:24.780" Score="4" Body="&lt;p&gt;I believe what your teacher means by this is that the error for your measurement is within $0.05~\mathrm{mL}$ of the estimate, since $0.05~\mathrm{mL}$ is half the value of the smallest measurable volume in a typical $50~\mathrm{mL}$ burette. This is typically how the error of an instrument reading is determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The readings you gave would then be: $23.56~\mathrm{mL} \pm 0.05~\mathrm{mL}$ and $23.44~\mathrm{mL} \pm 0.05~\mathrm{mL}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-07T05:11:24.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30989" PostTypeId="2" ParentId="30933" CreationDate="2015-05-07T06:08:38.440" Score="6" Body="&lt;p&gt;You need (some) background knowledge and the tools to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;generate an input file with the coordinates of your molecule and commands which calculations to perform&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;run the calculation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;postprocess the results&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All tools are out there! Unless you have access to the commercially available suites through your institution, I suggest to use those that are free (as in free beer &lt;strong&gt;and&lt;/strong&gt; free speech). Actually, some of these free tools are a good choice in any case!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Input files can be generated using &lt;a href=&quot;http://avogadro.cc&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt;, &lt;a href=&quot;http://www.cmbi.ru.nl/molden/&quot; rel=&quot;noreferrer&quot;&gt;Molden&lt;/a&gt;, &lt;a href=&quot;http://gabedit.sourceforge.net/&quot; rel=&quot;noreferrer&quot;&gt;Gabedit&lt;/a&gt;, or &lt;a href=&quot;https://code.google.com/p/wxmacmolplt/&quot; rel=&quot;noreferrer&quot;&gt;wxmacmolplt&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can perform your calculations using &lt;a href=&quot;http://www.nwchem-sw.org/&quot; rel=&quot;noreferrer&quot;&gt;NWChem&lt;/a&gt;, &lt;a href=&quot;http://www.mpqc.org/&quot; rel=&quot;noreferrer&quot;&gt;MPQC&lt;/a&gt; (both are in the repositories of every recent Linux distribution) or &lt;a href=&quot;http://www.msg.ameslab.gov/gamess/index.html&quot; rel=&quot;noreferrer&quot;&gt;GAMESS-US&lt;/a&gt;, which can be downloaded after registration. In addition, there are &lt;a href=&quot;https://github.com/sunqm/pyscf&quot; rel=&quot;noreferrer&quot;&gt;pyscf&lt;/a&gt; and &lt;a href=&quot;http://pyquante.sourceforge.net/&quot; rel=&quot;noreferrer&quot;&gt;PyQuante&lt;/a&gt;, which you mentioned in your question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Most of the tools used to generate the input files can be used for postprocessing too. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To quote the slogan of a shoe company:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just do it!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-07T06:08:38.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30990" PostTypeId="1" AcceptedAnswerId="31062" CreationDate="2015-05-07T06:31:48.793" Score="5" ViewCount="2243" Body="&lt;p&gt;I'm having trouble making sense of this answer. I believe I am doing all the steps right, but my main concern is that I am not arriving at a whole number. If my answer is correct, should I just round up or down?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By using the known values of $\chi$ for $\ce{HgCo(NCS)4}$ calculate the number of unpaired electrons in each of these complexes, and interpret the results in terms of the electronic structures of the complexes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;My Attempt&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HgCo(NCS)4}\ 's \   \chi = 16.44 x10^{-6}\  \mathrm{CGS (g^{-1}\cdot cm^3) \ at \ 20°C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\chi_M =16.44 \cdot 10^{-6}g^{-1}\cdot cm^3 * \frac{491.8528 g}{1 mol} =8.09\cdot 10^{-3} mol^{-1}cm^3}$$&#xA;$$\chi ' _M= \chi_M (\text{metal core electrons}) + \chi_M (\text{ligands}) + \chi_M (\text{other}) - \chi_M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamagnetic corrections include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align} \ce{Hg^{2+}}&amp;amp;=40.0\cdot 10^{-6}\\ \ce{Co^{2+}}&amp;amp;=12.8\cdot 10^{-6} \\ \ce{NCS^-} &amp;amp;=26.2\cdot 10^{-6}\end{align}$$&#xA;&lt;a href=&quot;http://wwwchem.uwimona.edu.jm/lab_manuals/DiamCorr.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi ' _M=  (40\cdot 10^{-6})+(12.8\cdot 10^{-6})+(26.2\cdot 10^{-6})(3)-(16.44\cdot 10^{-6}) =-0.02418$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then using the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\chi '_MT=\frac{1}{8}n(n+2)}$$&#xA;Assuming 298K and setting to zero&#xA;$$\mathrm{n^2 +2n-57.6722=0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;quadradic formula and i get&#xA;$$\mathrm{n=6.65}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure with all of that arithmetic it would be difficult to arrive at a whole number. So, my intuition is to round up to 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach correct? Which way should I round if so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I don't know how I would determine the electronic structure because there's two metal ions in the complex?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2018-03-08T02:26:48.900" LastActivityDate="2018-03-08T02:26:48.900" Title="By using the magnetic susceptibility, calculate the number of unpaired elections in the complex" Tags="&lt;inorganic-chemistry&gt;&lt;electrons&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30991" PostTypeId="5" CreationDate="2015-05-07T06:34:00.800" Score="0" Body="&lt;p&gt;Magnetism is a class of physical phenomena that are mediated by magnetic fields. Electric currents and the fundamental magnetic moments of elementary particles give rise to a magnetic field, which acts on other currents and magnetic moments. All materials are influenced to some extent by a magnetic field. The most familiar effect is on permanent magnets, which have persistent magnetic moments caused by ferromagnetism. Most materials do not have permanent moments. Some are attracted to a magnetic field (paramagnetism); others are repulsed by a magnetic field (diamagnetism); others have a much more complex relationship with an applied magnetic field (spin glass behavior and antiferromagnetism). Substances that are negligibly affected by magnetic fields are known as non-magnetic substances. They include copper, aluminium, gases, and plastic. Pure oxygen exhibits magnetic properties when cooled to a liquid state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic state (or phase) of a material depends on temperature (and other variables such as pressure and the applied magnetic field) so that a material may exhibit more than one form of magnetism depending on its temperature, etc.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-07T06:58:26.057" LastActivityDate="2015-05-07T06:58:26.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30992" PostTypeId="4" CreationDate="2015-05-07T06:34:00.800" Score="0" Body="For questions about determining the paramagnetism, diamagnetism, etc. of chemical species and other chemical applications of magnetism. Note that questions oriented towards physical aspects of electromagnetism are off-topic on this site." OwnerUserId="4356" LastEditorUserId="5026" LastEditDate="2018-03-15T07:10:53.330" LastActivityDate="2018-03-15T07:10:53.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30993" PostTypeId="1" CreationDate="2015-05-07T07:25:10.097" Score="6" ViewCount="958" Body="&lt;p&gt;Its known that the only method to determine mercury in atomic absorbtion spectroscopy is the &lt;em&gt;cold vapor method&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason that mercury can't be determined with other methods, such as flame, electrothermal AAS or even emission spectroscopy ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-05-14T07:26:27.147" Title="Determination of mercury in AAS" Tags="&lt;experimental-chemistry&gt;&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30994" PostTypeId="1" CreationDate="2015-05-07T08:16:01.743" Score="10" ViewCount="1456" Body="&lt;p&gt;Disclaimer: I am not a chemist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just drinking my Ice Age mineral water which is 'untouched since the Ice Age' and wondered if the bonds formed in each molecule have remained intact for all those years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a liquid that is just lying there underground I'm guessing that there is some, if not very energetic, movement of the molecules. What happens when they bump into each other? Might an atom become dislodged and bind with another molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea but the thought interested me.&lt;/p&gt;&#xA;" OwnerUserId="16053" LastActivityDate="2015-05-07T15:59:00.550" Title="How long do the bonds in a molecule hold under nominally stable conditions?" Tags="&lt;bond&gt;&lt;molecules&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30995" PostTypeId="2" ParentId="30994" CreationDate="2015-05-07T08:53:51.070" Score="10" Body="&lt;p&gt;&lt;em&gt;Water with its fluctuating hydrogen bond networks is a special case. If you feel for it, have a look at &lt;a href=&quot;http://www.pnas.org/content/98/19/10533.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water clusters: Untangling the mysteries of the liquid, one molecule at a time&lt;/a&gt;, but it doesn't answer your question ;-)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more interesting to have a look at paleochemical/geobiological findings.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KQvoW.png&quot; alt=&quot;structure of 2-alpha-methylhopane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complex organic molecule, such as &lt;strong&gt;2$\alpha$-methylhopane&lt;/strong&gt; was isolated from &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2700-million-year-old&lt;/strong&gt; shales from the Pilbara Craton, Australia&#xA;  (&lt;a href=&quot;http://dx.doi.org/10.1126/science.285.5430.1033&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Have look at the website of &lt;a href=&quot;http://summons.mit.edu/biomarkers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roger Summons at MIT&lt;/a&gt; and his freely available publications, such as &lt;a href=&quot;http://summons.mit.edu/wp-content/uploads/2014/11/SummonsPSP20FinalOPEN.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Exceptional Preservation of Interesting and Informative Biomolecules&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, bonds can last a pretty long time!&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-07T08:53:51.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30996" PostTypeId="2" ParentId="30975" CreationDate="2015-05-07T10:02:22.667" Score="3" Body="&lt;p&gt;As far as I know, the short answer is &lt;strong&gt;no&lt;/strong&gt;; in general, it is not possible to know the reaction mechanism a priori.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the approach is: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Postulate a reaction mechanism/pathway &lt;/li&gt;&#xA;&lt;li&gt;Use this to calculate the relevant reaction rate(s)&lt;/li&gt;&#xA;&lt;li&gt;Compare these rate expressions with the results of experiments&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If your proposed reaction mechanism matches the empirical results, it is likely (but you can never be sure!) that the mechanism is correct. If not, a new mechanism should be postulated and tested.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, in many examples it may be possible to make an educated guess of a plausible reaction mechanism - based on your own experience or the experience of others (i.e. from articles etc.) - and then investigate whether the proposed mechanism matches the empirical facts of relevant experiments. Often, the proposed mechanism is more or less correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, modern advances in computational chemistry may give useful information regarding likely reaction pathways based on the structure and interaction of the involved molecules. To my knowledge, it is not yet possible to use this in a standalone fashion to figure out the detailed mechanism, but it may help out in terms of making a better educated guess. &lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2015-05-07T10:02:22.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30997" PostTypeId="2" ParentId="14826" CreationDate="2015-05-07T10:22:23.283" Score="-1" Body="&lt;p&gt;Electronic configuration of $\ce{Co}$ is as follows: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Cobalt loses 2 electrons to become $\ce{Co^{2+}}$ it loses the electrons which are in $4s^2$, not the ones in $3d^7$ because the electrons in $4s^2$ have a high reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\ce{Co^{2+}}$ still contains 7 electrons in it's valence shell. The new configuration of $\ce{Co^{2+}}$ is&#xA;$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^7$.&lt;/p&gt;&#xA;" OwnerUserId="16054" LastEditorUserId="418" LastEditDate="2015-05-07T11:05:59.770" LastActivityDate="2015-05-07T11:05:59.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="30998" PostTypeId="2" ParentId="30945" CreationDate="2015-05-07T10:28:03.017" Score="0" Body="&lt;p&gt;Grams of $\ce{Cu^{+2}}$ is ill defined.  The $\ce{Cu^{+2}}$ did not get into the solution by itself. It was $\ce{CuCl2,CuSO4}$ or some other salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably the reason for the discrepancy.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-05-07T10:28:03.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="30999" PostTypeId="2" ParentId="4526" CreationDate="2015-05-07T10:55:33.727" Score="3" Body="&lt;p&gt;In short, I agree with Curt F.'s answer above. In general, you need a suitable equation of state (EoS) to handle this calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the differential form of the fundamental equation of thermodynamics (i.e. the differential form of the internal energy $U(S,V,N)$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V + \sum_{i=1}^{m} \mu_i \mathrm{d}n_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;other relevant thermodynamic potentials may be found using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Legendre_transformation&quot; rel=&quot;nofollow&quot;&gt;Legendre transform&lt;/a&gt;. Helmholtz free energy is defined as a transformation from $(S\rightarrow T)$ and Gibbs free energy by a transformation from $(S\rightarrow T)$ &lt;em&gt;and&lt;/em&gt; $(V\rightarrow p)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A(T,V,N) \triangleq U - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U - TS \\&#xA;G(T,p,N) \triangleq U - \left(\frac{\partial U}{\partial V}\right)_{S,N} - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U + pV - TS = A + pV $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an expression for the Helmholtz free energy (that is, $A(T,V,N)$) is known, along with a suitable equation of state (EoS) (e.g. $p = p(T,V,N)$), the Gibbs free energy may be calculated. Without the EoS, it is impossible to evaluate the $pV$-term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both the volume and pressure may be assumed to be constant, they will not affect the phase equilibrium calculations. An alternative may be to assume that the liquid phase volume is constant or that it follows some (simple) empirical relation. Thus, the last term may be evaluated and the Gibbs free energy may be minimized.  &lt;/p&gt;&#xA;" OwnerUserId="475" LastActivityDate="2015-05-07T10:55:33.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31000" PostTypeId="1" AcceptedAnswerId="31001" CreationDate="2015-05-07T11:05:01.960" Score="1" ViewCount="1298" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LyrlH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt; $\ce{MeO-}$ preferably gives SN2 &lt;em&gt;or&lt;/em&gt; E2, being a strong base and good nucleophile. But which reaction proceeds in this case?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="15751" LastEditDate="2017-05-04T11:56:37.787" LastActivityDate="2017-05-04T11:56:37.787" Title="What is the product of a Dieckmann condensation of dimethyl adipate?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31001" PostTypeId="2" ParentId="31000" CreationDate="2015-05-07T11:20:55.303" Score="7" Body="&lt;p&gt;&lt;em&gt;I've given a strong hint by changing the title of your question!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Methoxide as a base can abstract a proton from an acidic $\alpha$-position of an ester, that is from $\ce{-C\color\red{H}_2COOMe}$. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The resulting carbanion/enolate can add to the other ester group of your starting material.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cyclic intermediate eventually loses methoxide to form ...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;Dieckmann&lt;/em&gt; condensation is the intramolecular variant of a &lt;em&gt;Claisen&lt;/em&gt; condensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I suggest to do it on paper!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;img src=&quot;https://i.stack.imgur.com/rI785.png&quot; alt=&quot;Dieckmann condensation of methyl adipate&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-07T20:52:03.197" LastActivityDate="2015-05-07T20:52:03.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31002" PostTypeId="1" AcceptedAnswerId="31061" CreationDate="2015-05-07T12:22:01.623" Score="5" ViewCount="2243" Body="&lt;p&gt;Say I have a CIF file describing some material in terms of its symmetry space group, lattice parameters and in-cell atom positions. A simple example might be,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;data_global&#xA;_chemical_name 'Graphene'&#xA;_cell_length_a                 2.46&#xA;_cell_length_b                 2.46&#xA;_cell_length_c                 1&#xA;_cell_angle_alpha              90&#xA;_cell_angle_beta               90&#xA;_cell_angle_gamma              120&#xA;_symmetry_space_group_name_H-M 'P 3 m 1'&#xA;loop_&#xA;_atom_site_label&#xA;_atom_site_fract_x&#xA;_atom_site_fract_y&#xA;_atom_site_fract_z&#xA;C  0.00000   0.00000   0.00000&#xA;C  0.33333   0.66667   0.00000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would like to extract from this Bravais lattice vectors and atom basis vectors. For this specific example, this might look like,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 2.456   0.0     0.0&#xA;-1.228   2.126   0.0&#xA;C   &#xA; 0.0      0.0      0.0&#xA; 0.0      1.418    0.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where the first two lines give the two lattice vectors (in some basis), while the next two give the positions of the atoms in the unit cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried using &lt;a href=&quot;http://openbabel.org/wiki/Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Open Babel&lt;/a&gt; for this, converting from the CIF format to a VASP one.  This works well for the example above, but fails for a material such as tin sulfide (SnS), which has a structure similar to black phosphorus---four atoms in a unit cell, only two of which are explicitly listed in the &lt;a href=&quot;http://www.crystallography.net/cod/1011253.html&quot; rel=&quot;noreferrer&quot;&gt;CIF file&lt;/a&gt;.  The positions of the other two are implied by the symmetry group, and programs such as &lt;a href=&quot;http://www.ccdc.cam.ac.uk/Solutions/CSDSystem/Pages/Mercury.aspx&quot; rel=&quot;noreferrer&quot;&gt;Mercury&lt;/a&gt; correctly visualize the structure based on the CIF file.  However, converting the CIF into VASP only gives you two of the four unit cell atoms.  I tried all other formats in Open Babel, with no success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reliable way to convert a CIF file into a description in terms of a Bravais lattice plus atom basis vectors?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-02-25T13:07:00.410" Title="How to determine the Bravais lattice and atom basis vectors from a CIF file?" Tags="&lt;crystal-structure&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31003" PostTypeId="2" ParentId="30797" CreationDate="2015-05-07T12:28:44.247" Score="23" Body="&lt;p&gt;Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can &lt;em&gt;only&lt;/em&gt; be explained by molecular orbital theory. A complete analysis of this can be found in &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.20477/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, &lt;em&gt;J. Comp. Chem.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;28&lt;/em&gt; (1), 117-126.&lt;/a&gt; (I believe it is available free of charge.)&#xA;&lt;strong&gt;Responsible for the dipole moment is the highest occupied molecular orbital, a $\pmb{\sigma}$ orbital, which has its largest coefficient at the carbon atom.&lt;/strong&gt; In first order approximation, this orbital can be considered the lone pair of carbon. All other valence orbitals are more strongly polarised towards the oxygen. The orbital that can in first order approximation be considered as the oxygen lone pair has almost only s character and therefore contributes only little to the dipole moment.&#xA;\begin{align}&#xA;\ce{{}^{\ominus}\!:C#O:^{\oplus}} &amp;amp;&amp;amp; &#xA;\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &amp;amp;&amp;amp;&#xA;\text{Direction:}~\longleftarrow&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have reproduced the MO scheme of carbon monoxide for you below. Please note, that the blue/orange coloured orbitals are virtual (unoccupied) orbitals, which should be taken with a grain of salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c2r8J.jpg&quot; alt=&quot;MO scheme of CO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two possible decomposition schemes to explain the bonding, both of them involve donor-acceptor interactions. The term &quot;dative bonding&quot; should be avoided here, it is better to use it only for bonds, that consist purely of donor-acceptor interactions, as for example in $\ce{H3N\bond{-&amp;gt;}BH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, the two decomposition schemes are reproduced from figure 6 (b &amp;amp; c) in the linked paper. Please note, that this decomposition does not include hybridised orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xb08d.jpg&quot; alt=&quot;possible decomposition schemes for the bonding in CO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left decomposition is a better description, since it retains the $C_{\infty{}v}$ symmetry of the molecule. We can see a donor-acceptor $\sigma$ bond and two electrons sharing $\pi$ bonds.&lt;br&gt;&#xA;In the right configuration we assume an electron sharing $\sigma$ bond, an electron sharing $\pi$ bond and a donor-acceptor $\pi$ bond.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very important to understand, that the concept of a dative bond, that you are trying to employ here is &lt;em&gt;only right by coincidence&lt;/em&gt;. The reason that the dipole moment is oriented towards the carbon is only to find in the weakly bonding HOMO, the lone pair of carbon.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="42657" LastEditDate="2018-02-06T23:12:02.423" LastActivityDate="2018-02-06T23:12:02.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31004" PostTypeId="1" CreationDate="2015-05-07T12:53:55.037" Score="3" ViewCount="919" Body="&lt;p&gt;Water retention is how much water the fibre can retain. Water holding is function of hydrophilicity and construction of the fibre (fabric). What is the difference between those two? Because to retain is also to hold, in my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:10:43.600" LastActivityDate="2017-06-30T07:10:27.020" Title="What is the difference between moisture regain and water holding capacity?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31005" PostTypeId="1" CreationDate="2015-05-07T14:04:56.127" Score="14" ViewCount="903" Body="&lt;p&gt;There is another closely related post &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6366/chelation-and-its-relation-to-transition-metal-ligands&quot;&gt;here&lt;/a&gt; , and I've also read the referenced wiki article including applications of chelation, but I still don't see what's so special about chelation. I understand the structure of the bond - the 'capturing' of a metal cation by an organic agent, but I still fail to see why this chemical reaction might be so special relative to other possible organic chemical reactions with a metal cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is chelation so special?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2016-03-11T23:51:04.297" Title="What's so special about chelation?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ClosedDate="2015-05-09T21:14:03.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="31006" PostTypeId="1" AcceptedAnswerId="31800" CreationDate="2015-05-07T14:09:29.183" Score="10" ViewCount="1093" Body="&lt;p&gt;I've done a student experiment using cyclohexanol to prepare cyclohexene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We mix $10.5~\mathrm{g}$ cyclohexanol and $5~\mathrm{mL}$ $85\%$ $\ce{H3PO4}$ in a $50~\mathrm{mL}$ flask, then distill while heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know during this process what kind of side reaction happens because I notice there is something light yellow in the flask. After reaction, the liquid remaining in the flask is red and upon it there is something dark green and muddy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a guess that there is some phenol in the cyclohexanol (since cyclohexanol may be prepared from phenol), and that during reaction the phenol oxidized, producing 1,2-benzoquinone (which is red) and 1,4-benzoquinone (which is yellow). Is this idea reasonable? If it is, why does the red color appear later?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And about the dark green thing: can it be produced by some kind of carbonized organic matter? I have been told that if the $\ce{H3PO4}$ and cyclohexanol were not well mixed, there would be some cyclohexanol carbonization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Can I ask one more question?) We are required to compare the chemical properties of cyclohexene and cyclohexane. So we reacted them with $\ce{Br2}/\ce{CCl4}$ respectively, under light or covered by aluminium foil. Should it be different when &lt;strong&gt;cyclohexene&lt;/strong&gt; reacts under light or covered by aluminium foil? I notice the result is slightly different; the reactions in the light become a little opaque.&lt;/p&gt;&#xA;" OwnerUserId="15820" LastEditorUserId="15410" LastEditDate="2015-05-07T18:16:19.573" LastActivityDate="2015-05-21T19:07:25.230" Title="What happens when I use cyclohexanol to prepare cyclohexene?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31007" PostTypeId="2" ParentId="31005" CreationDate="2015-05-07T14:29:18.270" Score="19" Body="&lt;p&gt;Chelation is an entropy driven reaction, i.e. complexes with polydentate ligands are more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider what happens when you dissolve copper sulfate in water:&#xA;\begin{align}&#xA;\ce{CuSO4 + 6H2O &amp;lt;=&amp;gt; [Cu(H2O)]^2+ + SO4^2- (aq)}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This complex forms and with it comes some kind of ordering of the water molecules around it. Dissolution is often favoured because of the solvation energy. Now consider a bidentate ligand like ethylene diamine, $\ce{H2N-(CH2)2-NH2}$, short $\ce{en}$. Adding this ligand to a copper sulfate solution will result in a ligand exchange:&#xA;\begin{align}&#xA;\ce{[Cu(H2O)]^2+ + 3en &amp;lt;=&amp;gt; [Cu(en)3]^2+ + 6H2O}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this reaction you increase the number of particles from four on the left side to seven on the right side. Taking into account that you also disturb the order of the surround water molecules, this reaction is very much favoured by entropy. In an attempt of zeroth order approximation, one could assume that entropy roughly doubles. Since the Gibbs energy is inversely correlated to the entropy, this means, that the reaction will be spontaneous. This is usually referred to as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chelation#Chelate_effect&quot; rel=&quot;noreferrer&quot;&gt;chelate effect&lt;/a&gt;.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="24052" LastEditDate="2016-03-11T23:51:04.297" LastActivityDate="2016-03-11T23:51:04.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31008" PostTypeId="2" ParentId="30863" CreationDate="2015-05-07T14:35:29.043" Score="4" Body="&lt;p&gt;Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15443" LastEditDate="2015-05-07T16:48:40.070" LastActivityDate="2015-05-07T16:48:40.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31009" PostTypeId="1" CreationDate="2015-05-07T15:21:39.057" Score="7" ViewCount="2975" Body="&lt;p&gt;Every chemistry textbook I have ever seen defines intensive and extensive properties the same way: Intensive properties are those which do not depend on how much matter is present, while extensive properties do. I came across a question while looking through sample FE exam questions for which my understanding of these properties by this definition failed. The gist of my question really just revolves around pressure and volume in largely compressible substances (i.e., gases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that, when considering a gaseous substance, these definitions are not robust enough:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Volume can be shown to be independent of the amount of material present by simply compressing the gas.&lt;/li&gt;&#xA;&lt;li&gt;Density can be shown to be dependent on the amount of material present by keeping a fixed volume and simply adding more gas molecules.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While I have heard the convincing argument that we should simply consider smaller subsets of the original to determine intensiveness and extensiveness (for instance, if I consider half of the container instead, the volume halves from $V_0$ to $\frac{1}{2}V_0$ and the density remains constant at $\rho$), this doesn't help me to identify why the conventional definition remains in use with what seems like flawed implications. Can anyone help to clarify this?&lt;/p&gt;&#xA;" OwnerUserId="16055" LastEditorUserId="16683" LastEditDate="2016-09-30T09:50:34.297" LastActivityDate="2016-09-30T09:50:34.297" Title="Intensive and extensive properties of gases" Tags="&lt;physical-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31010" PostTypeId="2" ParentId="31009" CreationDate="2015-05-07T15:35:16.697" Score="5" Body="&lt;p&gt;To give you a more clear explanation of the definition of these properties, it's more easy to remember them this way :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Extensive properties are those which &lt;strong&gt;*are additive *&lt;/strong&gt; ( mass, volume ) for example 1 L + 1 L = 2 L. On the other hand, intensive properties &lt;strong&gt;&lt;em&gt;are not additive&lt;/em&gt;&lt;/strong&gt; ( temperature, pressure, concentration ) for example 2 mol/L + 3 mol/ L does not give you a 5 mol/L solution. These are properties which &lt;em&gt;don't change&lt;/em&gt; at any circumstance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other thing to be considered :&#xA;The ratio of two extensive properties, gives us an intensive property. For example density= mass/volume. Since it is a ratio it will always be the same regardless of how much of a sample is measured.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-07T16:18:30.440" LastActivityDate="2015-05-07T16:18:30.440" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31011" PostTypeId="2" ParentId="30994" CreationDate="2015-05-07T15:59:00.550" Score="7" Body="&lt;p&gt;If you are asking about the water molecules themselves (and not some other materials present such as the paleochemical structure above) the answer is that at least for liquid water, each water molecule exchanges its hydrogen atoms with other water molecules at an extremely rapid pace (millions of times per second). While frozen (a mile deep in Antarctica or Greenland?) for millions of years, this exchange process would be slower, but still I would venture that the water molecules individually are not the same ones they were way back when. Depending on how it was handled after the ice was mined from its glacier, the sample may still be collectively composed of the same atoms that were there millions of years ago.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-07T15:59:00.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31012" PostTypeId="1" AcceptedAnswerId="31026" CreationDate="2015-05-07T17:46:07.437" Score="2" ViewCount="7316" Body="&lt;p&gt;I am trying to solve the following exercise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What volume of a concentrated &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; solution, which is &lt;span class=&quot;math-container&quot;&gt;$36.0\%$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; by mass&#xA;  and has a density of &lt;span class=&quot;math-container&quot;&gt;$1.179~\mathrm{g~mL^{-1}}$&lt;/span&gt;, should be used to make &lt;span class=&quot;math-container&quot;&gt;$5.00~\mathrm{L}$&lt;/span&gt; of an &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;&#xA;  solution with a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$1.8$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I found the respective volumes that &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; contributed.&lt;/li&gt;&#xA;&lt;li&gt;I then found the molarity of the &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; aqueous solution given.&lt;/li&gt;&#xA;&lt;li&gt;I used &lt;span class=&quot;math-container&quot;&gt;$10^{-\mathrm{pH}}$&lt;/span&gt; to get the amount concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; needed, I then proceeded to find the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; by multiplying the concentration needed by &lt;span class=&quot;math-container&quot;&gt;$5.00~\mathrm{L}$&lt;/span&gt;. I then took the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; needed and multiplied by the density given to get an answer of &lt;span class=&quot;math-container&quot;&gt;$2.4~\mathrm{L}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Am I correct? &lt;/p&gt;&#xA;" OwnerUserId="6861" LastEditorUserId="4945" LastEditDate="2018-10-08T08:36:41.683" LastActivityDate="2018-10-08T08:45:06.290" Title="What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31014" PostTypeId="5" CreationDate="2015-05-07T18:35:41.697" Score="0" Body="&lt;p&gt;Conformational isomers are isomers which can be interconverted by rotation around a single (sigma) bond. These typically have a very low energy barrier to interconversion.&lt;/p&gt;&#xA;" OwnerUserId="5026" LastEditorUserId="5026" LastEditDate="2018-03-14T19:41:44.460" LastActivityDate="2018-03-14T19:41:44.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31015" PostTypeId="4" CreationDate="2015-05-07T18:35:41.697" Score="0" Body="For questions about conformational isomerism. Use this tag alongside [stereochemistry]. " OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-03-14T19:51:45.980" LastActivityDate="2018-03-14T19:51:45.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31017" PostTypeId="2" ParentId="19154" CreationDate="2015-05-07T19:53:02.347" Score="5" Body="&lt;p&gt;By convention both donor (D) and acceptor (A) potentials are listed as reduction potentials. For a nice description, see Turro's explanation of the Rehm-Weller equation in his photochemical treatise &quot;Modern Molecular Photochemistry of Organic Molecules&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, the equation that is being referenced is not actually the Rehm-Weller equation but is the &quot;Gibbs free energy of photoinduced electron transfer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to the IUPAC definition of &lt;a href=&quot;http://goldbook.iupac.org/GT07388.html&quot; rel=&quot;noreferrer&quot;&gt;Gibb's energy of photoinduced electron transfer&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs energy of photoinduced electron transfer is part of the Rehm-Weller equation. The Rehm-Weller equation is the correlation between second order rate constants and Gibbs energy of photoinduced electron transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the IUPAC definition of the &lt;a href=&quot;http://goldbook.iupac.org/RT07472.html&quot; rel=&quot;noreferrer&quot;&gt;Rehm-Weller equation&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="16059" LastActivityDate="2015-05-07T19:53:02.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31020" PostTypeId="1" AcceptedAnswerId="31022" CreationDate="2015-05-07T22:18:50.830" Score="20" ViewCount="24078" Body="&lt;p&gt;Recently I was researching $\ce{Ca(OH)_2}$, and I found this on Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|l|}\hline&#xA;{\rm \color{blue}{Solubility~in~water}} &#xA; &amp;amp; {\mathrm{0.189\,g/100\,mL}\   (0\,^\circ\mathrm{C})} \\&#xA; &amp;amp; {\mathrm{0.173\,g/100\,mL}\  (20\,^\circ\mathrm{C})} \\&#xA; &amp;amp; {\mathrm{0.066\,g/100\,mL} \ (100\,^\circ\mathrm{C})} \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain why solubility decreases when temperature increases?&lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="9961" LastEditDate="2019-01-03T19:04:56.303" LastActivityDate="2019-01-03T19:04:56.303" Title="Why does solubility of solids sometimes decrease with temperature increase?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31021" PostTypeId="1" CreationDate="2015-05-07T22:44:36.497" Score="3" ViewCount="12392" Body="&lt;p&gt;What is the difference between an esterification and a transesterification. Esterification objective is the formation of an ester by reacting an alcohol and a carboxylic acid. I am more concerned transesterification or reference to a journal article.&lt;/p&gt;&#xA;" OwnerUserId="16060" LastEditorUserId="7475" LastEditDate="2016-07-06T02:09:12.673" LastActivityDate="2017-03-22T06:15:36.943" Title="What is the difference between the esterification and transesterification?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31022" PostTypeId="2" ParentId="31020" CreationDate="2015-05-07T23:42:21.307" Score="20" Body="&lt;p&gt;In short, for this group 2 hydroxide crystal, the dissolution and solvation process is an exothermic equilibrium. Increasing the temperature drives the reaction in the reverse, favouring crystallisation! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it would be nice to rationalise some of these terms, so I will lay out a very brief overview of general solubility factors so you can understand the process in a little more detail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility is a complex thermodynamic process that takes many factors into account. It can be understood as the process of dissolving a crystal in a solvent, and then forming a solvent cage around the solute (solvation). It is a function of three interactions at heart;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;solute-solute&lt;/li&gt;&#xA;&lt;li&gt;solute-solvent &lt;/li&gt;&#xA;&lt;li&gt;solvent-solvent &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Large solute-solute interactions hinder solubility and are associated with high lattice enthalpies (of crystalline solutes), these must be overcome in order to break up the solid and dissolve it. Solvent-solvent interactions must also be overcome in order to solvate the dissolved solute, therefore for a highly interacting solvent the solubility is reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of ordered solvent structures such as clathrin cages etc orders the solvent, and therefore reduces solvent entropy significantly. Therefore solvents with high degrees of freedom hinder solubility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solute entropy favours solvation, this reflects the higher solubility of flexible molecules, this is because in the solid they are all locked up, while in solution their rotational and vibrational modes are populated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to these if the solute has multiple ionic forms then the solubility will be a function of pH too, so must be held constant in practice. The surface profile of the solid too, as well as the extent of cavitation and porosity are also key factors (since they affect the number of solvent-solute interactions possible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Importantly the type of solvent into which you are trying to dissolve your crystal in is also of fundamental importance! The rule &quot;like dissolves like&quot; is best understood by the (semi-empirical) &lt;a href=&quot;http://http:/http://www.upch.edu.pe/facien/fc/dbmbqf/zimic/cursos/modelamiento%202005/articulos/19%20Solubilidad%20de%20mol%C3%A9culas/drug%20solubility%20by%20the%20general%20solubility%20equation.pdf&quot; rel=&quot;noreferrer&quot;&gt;general solubility equation&lt;/a&gt;, &#xA;\begin{equation}&#xA;\log S =\frac 12-0.05 (T_m-T)-\log P&#xA;\end{equation}&lt;br&gt;&#xA;A large &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_coefficient&quot; rel=&quot;noreferrer&quot;&gt;partition coefficient&lt;/a&gt; (the ratio of how soluble the crystal is in organic solvent to aqueous solvent) reduces the solubility, (and mathematically since $\log x$ follows $x$ in limits). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further complicate the situation there are several definitions of solubility that we can take. In practice we choose intrinsic aqueous solubility, mainly used for high throughput and speed in drug development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility is really an experimentalists tool for parametrising the extent of Gibbs energy for the dissolving-solvation process, which in turn is usually modelled by complex thermodynamic cycles via gas phase species or super cooled liquids. Modern research into solubility focuses on finding either molecular descriptors through random forest/&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantitative_structure%E2%80%93activity_relationship&quot; rel=&quot;noreferrer&quot;&gt;QSPR&lt;/a&gt; or support vector models etc (which are all just number crunching techniques to find correlations between large sets of data) or through fundamental physics and chemistry of the process. &#xA;\begin{equation}&#xA;\log S=\frac{(-\Delta G_{sol}/RT)}{\ln 10}&#xA;\end{equation}&lt;br&gt;&#xA;So we understand solubility as an equilibrium constant for the Gibbs change of the solvation process (whatever particular cycle you use to get that data). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know something about solubility we can ask ourselves, under what conditions does increasing the temperature reduce solubility? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the temperature can effect the entropic term, but also the enthalpies of the crystal lattices, the heats given out by the solute-solvent interactions etc. It is the balance of these process that determines the solubility of something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7475" LastEditDate="2015-12-26T11:35:02.697" LastActivityDate="2015-12-26T11:35:02.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31024" PostTypeId="1" CreationDate="2015-05-08T00:43:38.010" Score="1" ViewCount="3606" Body="&lt;p&gt;My question deals with the reaction $\ce{2NO(g) + O2(g) -&amp;gt; 2NO2(g)}$ with $K_c = 1752$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have so far; $K_c$= products over reactants, so is the reverse reaction reactants over products to find the answer&#xA;Such as&lt;br&gt;&#xA;$$1752 = \frac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$&#xA;So then does the answer become&#xA;$$1752 = \frac{[x]^2}{[x]^2[x]}$$&lt;/p&gt;&#xA;" OwnerUserId="15495" LastEditorUserId="13339" LastEditDate="2015-06-26T18:36:25.923" LastActivityDate="2015-06-26T18:36:25.923" Title="What is the value for the reverse reaction?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31025" PostTypeId="1" CreationDate="2015-05-08T01:20:03.117" Score="-1" ViewCount="313" Body="&lt;p&gt;What is the freezing point of water made by dissolving $\mathrm{10.11~g}$ of magnesium chloride in $\mathrm{86.03~g}$ of water?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The freezing-point depression constant of water is $\mathrm{1.86~^\circ C~m^{-1}}$.&lt;br&gt;&#xA;To start I take the $\frac{10.11~\text{g}\ \ce{MgCl}}{86.03 \times 0.001 \times 1.86~\mathrm{^\circ C~m^{-1}})}$, which I keep getting as $\mathrm{63.18~^\circ C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="15495" LastEditorUserId="10185" LastEditDate="2015-10-02T15:54:37.507" LastActivityDate="2015-10-02T15:56:40.030" Title="Why/How am I calculating freezing point depression wrong?" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31026" PostTypeId="2" ParentId="31012" CreationDate="2015-05-08T02:22:53.983" Score="3" Body="&lt;p&gt;Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$-0.74$&lt;/span&gt;, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can solve this by making the calculations step by step. Let's start with the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt;, which will be the same as the concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, we will achieve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we're going to prepare 5 liters, this means the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To transform this into amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, we need the acid's molar mass (&lt;span class=&quot;math-container&quot;&gt;$\pu{36.46 g//mol}$&lt;/span&gt;) to convert from amount of substance to mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\require{cancel}%&#xA;g_{\ce{HCl}} &#xA;= \mathrm{\frac{36.46\ g}{\cancel{mol}}} &#xA;  \times 0.07924466\ \mathrm{\cancel{mol}} &#xA;= 2.88926 \ \mathrm{g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have the amount of substance of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, we calculate the volume, which is where the density comes in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{\mathrm{HCl}} &#xA;= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}} &#xA;= 2.450602 \ \mathrm{mL}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the last step: Our stock solution isn't pure &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;, it's diluted in water. So actually the volume we need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{\mathrm{HCl}} &#xA;= 2.450602 \times \frac{100}{36} &#xA;= \pu{6.807229 mL} &#xA;\approx \pu{6.8 mL}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might look that it is not much, but remember &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; is a very strong acid. For example, to prepare &lt;span class=&quot;math-container&quot;&gt;$\pu{1000 mL}$&lt;/span&gt; of a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.001 M}$&lt;/span&gt; solution you only need &lt;span class=&quot;math-container&quot;&gt;$\pu{0.086 mL}$&lt;/span&gt; of concentrated &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can check this result by imputing the given density, weight percentage, final volume and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; concentration (first step) into &lt;a href=&quot;http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sigma-Aldrich's Molarity Calculator&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="4945" LastEditDate="2018-10-08T08:45:06.290" LastActivityDate="2018-10-08T08:45:06.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31027" PostTypeId="2" ParentId="31024" CreationDate="2015-05-08T03:18:56.147" Score="3" Body="&lt;p&gt;${K_c}$ of reverse reaction should just be the reciprocal of forward ${K_c}$. So, ${K_c}$ of reverse reaction is $\frac{1}{1752}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am not sure, why you used 2nd equation. Suppose you have 1 unit conc of $\ce{NO}$ and $\ce{O_2}$, then after time $t$, if the conversion is $x$, then your equation should be&lt;br&gt;&#xA;$1752 = \frac{[2x]^2}{([1-x])^3}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please check your 2nd equation again?&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7777" LastEditDate="2015-05-08T06:32:32.257" LastActivityDate="2015-05-08T06:32:32.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31028" PostTypeId="2" ParentId="31025" CreationDate="2015-05-08T03:32:08.907" Score="2" Body="&lt;p&gt;The equation for freezing point depression is this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ΔTf=−K_f\cdot m \cdot i$. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where $ΔT_f$ is the freezing point depression, $K_f$ is the freezing point depression constant $m$ is the molality of the solute, and $i$ is the van 't Hoff factor. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The van 't Hoff factor is the number of moles of ions the solute dissolves into, which in the case of $\ce{MgCl2 -&amp;gt; Mg^2+ + 2Cl-}$, is 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem here is that you used the mass of $\ce{MgCl2}$ ($\ce{MgCl}$ is incorrect) instead of its molality, and did not include the van 't Hoff factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molality is calculated by dividing moles of solute by kilograms of solvent. In this case, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{0.1062 \: \mathrm{mol  \:of}  \:   \ce{MgCl2}}{0.08603  \:  \mathrm{kg \: of}  \:  \ce{H2O}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that the molality of $\ce{MgCl2}$ is $1.234 \: \mathrm{mol~kg^{-1}}$. If you use this value instead of mass to calculate $ΔT_f$, the value comes out to be $-6.888 \:\mathrm{°C}$.  &lt;/p&gt;&#xA;" OwnerUserId="15724" LastEditorUserId="10185" LastEditDate="2015-10-02T15:56:40.030" LastActivityDate="2015-10-02T15:56:40.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31029" PostTypeId="2" ParentId="31002" CreationDate="2015-05-08T03:38:16.460" Score="3" Body="&lt;p&gt;I would recommend you to use vesta (&lt;a href=&quot;http://jp-minerals.org/vesta/en/&quot; rel=&quot;nofollow&quot;&gt;http://jp-minerals.org/vesta/en/&lt;/a&gt;). It can allow you to convert almost all types of quantum chemical coordinate file to your desitred one. To get VASP format, just drag you CIF file on vesta and it will show you the structure and then you can save it as POSCAR format.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-05-08T03:38:16.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31030" PostTypeId="1" AcceptedAnswerId="31033" CreationDate="2015-05-08T04:36:26.430" Score="1" ViewCount="6866" Body="&lt;p&gt;I should probably add that I was originally given an unknown solid mixture of 2 salts. I separated the insoluble salt out by centrifuging and then decanting. I found that the insoluble salt is $\ce{MgCO_3}$ by adding $\ce{HCl}$ (bubbles) and $\ce{NaOH}$ into the solution (jelly like white ppt). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for $\ce{SO4^{2-}}$ with $\ce{BaCl_2}$ and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 soluble salts in aqueous distilled water solution. I know that one of the salts is potassium sulfate ($\ce{K_2SO_4}$) and I know that the anion of the second salt is chloride ($\ce{Cl^{-}}$). I know that the second salt is either magnesium, sodium, calcium or potassium chloride ($\ce{MgCl_2}$, $\ce{NaCl}$, $\ce{CaCl_2}$, or $\ce{KCl}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested for the anion for the second salt by adding silver nitrate ($\ce{AgNO_3}$) into the solution and observing a white precipitate. When I added sodium hydroxide ($\ce{NaOH}$) to it, I observed that the cloudy white precipitate disappeared and turned into small black precipitate which then settled to the bottom of the tube. Is there any way to get the cation of the second salt with this information, and are there any other test(s) I can do to find what is the cation for the second salt?   &lt;/p&gt;&#xA;" OwnerUserId="16063" LastEditorUserId="-1" LastEditDate="2018-07-10T13:20:32.917" LastActivityDate="2019-07-30T02:10:30.590" Title="finding cation of unknown salt in solution from precipitate" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31031" PostTypeId="1" CreationDate="2015-05-08T04:36:33.013" Score="1" ViewCount="53" Body="&lt;p&gt;Why do azimuthal quantum numbers have a value lesser than Principal Quantum number.&#xA;example-&#xA;if n=1&#xA;it will have subshell &quot;s&quot; as 0&lt;/p&gt;&#xA;" OwnerUserId="11316" LastActivityDate="2015-05-08T04:36:33.013" Title="Azimuthal Quantum Numbers" Tags="&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2015-05-29T04:07:14.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="31033" PostTypeId="2" ParentId="31030" CreationDate="2015-05-08T05:45:44.310" Score="2" Body="&lt;p&gt;Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a &lt;strong&gt;double displacement reaction&lt;/strong&gt; may occur and in one case, a precipitate will be formed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{K_2SO_4 + CaCl_2 -&amp;gt; 2KCl + CaSO_4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that the person preparing the solutions for analysis used calcium chloride, since &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this would have required to remove the precipitate by filtration or decanting off the supernatant solution&#xA;the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would do a &lt;strong&gt;flame test&lt;/strong&gt; first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$).&#xA;If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a &lt;strong&gt;titan yellow&lt;/strong&gt; solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2018-07-14T15:58:00.453" LastActivityDate="2018-07-14T15:58:00.453" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31034" PostTypeId="2" ParentId="2739" CreationDate="2015-05-08T06:12:31.100" Score="5" Body="&lt;p&gt;You are perfectly right on that you need to convert the carboxylic acid to a more reactive species to eventually obtain the amide!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing with the pH will not help: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;under acidic conditions, the amine is protonated and therefore unreactive  &lt;/li&gt;&#xA;&lt;li&gt;under alkaline conditions, the acid is deprotonated and the carboxylate won't react at all&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Schotten-Baumann&lt;/em&gt; reaction of fatty acids with $\ce{SOCl2}$ will probably work fine: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reflux 1 mol of the acid with 1.5 mol of thionyl chloride   &lt;/li&gt;&#xA;&lt;li&gt;dIstill the remaining thionyl chloride off&lt;/li&gt;&#xA;&lt;li&gt;purify the fatty acid chloride by distillation in vacuum. Note that the bp of stearoyl chloride is about 175 °C at 2 mm Hg!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is quite some work to obtain a fatty acid derivative that eventually reacts with an amine!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wouldn't it be much more convenient to form the amide in a one-pot reaction, possibly at room temperature, by activating the carboxylic acid  &lt;em&gt;in situ&lt;/em&gt;?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;In situ&lt;/em&gt; means: at the same spot, without isolating the reactive intermediate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a large number of &lt;strong&gt;coupling reagents&lt;/strong&gt; available, such as   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;carbodiimides like DCC (dicyclohexylcarbodiimide)  &lt;/li&gt;&#xA;&lt;li&gt;$1H$-benzotriazoles&lt;/li&gt;&#xA;&lt;li&gt;chloro-substituted triazines&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Eric Valeur&lt;/em&gt; and &lt;em&gt;Mark Bradley&lt;/em&gt; published a critical review titled &lt;em&gt;Amide bond formation: beyond the myth of coupling reagents&lt;/em&gt;, in &lt;em&gt;Chem. Soc. Rev.&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;38&lt;/em&gt;, 606-631 (&lt;a href=&quot;http://dx.doi.org/10.1039/B701677H&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;) on the topic.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="13184" LastEditDate="2016-12-12T17:48:09.157" LastActivityDate="2016-12-12T17:48:09.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31038" PostTypeId="1" AcceptedAnswerId="31446" CreationDate="2015-05-08T12:34:56.537" Score="3" ViewCount="691" Body="&lt;p&gt;I am trying to understand how to interpret the TDDFT results ( I know there is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18813/interpretation-of-td-dft-results&quot;&gt;another post&lt;/a&gt; for this but it does not answer my question ).   Here is part of the result from GAMESS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eDSU9.png&quot; alt=&quot;GAMESS TDDFT&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I downloaded &lt;a href=&quot;http://www.chemissian.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemissian&lt;/a&gt; and this is its result for this state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H47Y0.png&quot; alt=&quot;Chemissian Output&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two related questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-How did it calculate this 1.41 from the amplitude of this state ?&lt;br&gt;&#xA;2-Does it have anything to do with λmax ?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-05-15T23:18:49.697" Title="How to interpret TD-DFT results for finding λmax and amplitude" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;td-dft&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31039" PostTypeId="1" AcceptedAnswerId="31044" CreationDate="2015-05-08T13:11:03.190" Score="1" ViewCount="7765" Body="&lt;p&gt;Luminescence in general, is the term of emission of light from specific substances/objects. Chemiluminescence is the emission of light &lt;em&gt;due to chemical reactions&lt;/em&gt;. Bioluminescence on the other hand is the emission light from &lt;em&gt;living/biological organisms&lt;/em&gt; such as insects ( this is of course the way I understood them ), but I'm a little bit confused because what is the cause that makes insects emmit light? Doesn't that also have to do with chemical reactions but inside living organisms ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a clear explanation of the difference  between these two terms  would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="418" LastEditDate="2015-05-08T14:09:19.387" LastActivityDate="2015-05-08T14:37:28.277" Title="Chemiluminescence vs. Bioluminescence" Tags="&lt;photochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31040" PostTypeId="1" AcceptedAnswerId="31057" CreationDate="2015-05-08T13:14:06.943" Score="10" ViewCount="42157" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  One mole of ideal gas initially at a pressure of 1 atmosphere and &lt;span class=&quot;math-container&quot;&gt;$T = 298\ \mathrm K$&lt;/span&gt;, is&#xA;  expanded into a volume 50 % larger adiabatically.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this adiabatic process, no work is done on the surroundings and &lt;span class=&quot;math-container&quot;&gt;$\mathrm dQ=0$&lt;/span&gt;. And entropy is defined as &lt;span class=&quot;math-container&quot;&gt;$\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$&lt;/span&gt;. However, there must be a change in entropy surely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the definition &lt;span class=&quot;math-container&quot;&gt;$\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$&lt;/span&gt; only at a constant volume? Because I have also seen entropy defined as &lt;span class=&quot;math-container&quot;&gt;$\mathrm dS=\frac{C_V}{T}\,\mathrm dT+\frac{R}{V}\,\mathrm dV$&lt;/span&gt; and there would be a change in both temperature and volume. However, how do I find what these changes are? Would it make a difference if the expansion was reversible or irreversible (the question didn't make it explicitly clear)&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2018-12-22T20:31:08.437" LastActivityDate="2018-12-22T20:31:08.437" Title="Entropy change in an adiabatic expansion" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31041" PostTypeId="2" ParentId="31040" CreationDate="2015-05-08T13:24:23.297" Score="-1" Body="&lt;p&gt;An important thing to know is that on the adiabatic processes, the system is &lt;em&gt;isolated&lt;/em&gt;, due to that no heat can be given or taken, but that doesn't mean that the volume can't change. Because we can't take heat, we can lower/rise the temperature from T1 to T2 . We don't have any work done only when the volume is constant. Entropy will be at a constant volume or pressure only when you see the index &lt;strong&gt;v ** or **p&lt;/strong&gt; in the formula. If the process would be reversible ( which i think it is) only the change of temperature should be in infinitezimal amounts. Hope I understood the question right &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-08T13:27:04.870" LastActivityDate="2015-05-08T13:27:04.870" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31042" PostTypeId="1" AcceptedAnswerId="70899" CreationDate="2015-05-08T13:55:39.090" Score="8" ViewCount="1962" Body="&lt;p&gt;I've been asked to compare the relative acidity of $\ce{CF4}$ and $\ce{SiF4}$. As both atoms have noble gas configurations, neither can act as a lewis acid and so I assume the question is inferring that they are in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that under aqueous conditions they could react via some sort of hydrolysis to give $\ce{H+}$ and $\ce{F-}$ alongside an alcohol in the case of carbon or $\ce{Si(OH)_{(4-x)}F_{x}}$ with silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the energy match between the $\ce{Si}$ and $\ce{F}$ HOMO and LUMO is worse this means that this is likely to be a weaker bond and so $\ce{SiF4}$ is more acidic? Is there an argument that could be made about the relative stability of the hydrolysis products?&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2017-03-20T17:51:49.407" LastActivityDate="2017-03-20T18:08:36.887" Title="Comparing the Lewis acidity of carbon tetrafluoride and silicon tetrafluoride" Tags="&lt;acid-base&gt;&lt;halides&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31043" PostTypeId="2" ParentId="29287" CreationDate="2015-05-08T14:32:31.937" Score="2" Body="&lt;p&gt;&lt;strong&gt;If you're doing any sort of computational run (either quantum chemistry, solid state, or molecular dynamics), you have to set the charge on the overall system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said in your comments that you're doing a distributed multipole analysis across multiple fragments of a protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, one of the &lt;em&gt;constraints&lt;/em&gt; on the calculation is exactly that the overall charge must match that specified. The program won't arbitrarily change the charge to be an anion or a cation if you specified a neutral system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that's not to say that the individual fragments won't have fractional or integer charges. Depending on the method, it's entirely reasonable that there's some amount of partial charge transfer between fragments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just that there's absolutely no way that the sum of the charges will be different than the constraint.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-08T14:32:31.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31044" PostTypeId="2" ParentId="31039" CreationDate="2015-05-08T14:37:28.277" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Doesn't that also have to do with chemical reactions but inside living organisms?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly! That's why there is &lt;strong&gt;one&lt;/strong&gt; scientific society for &lt;strong&gt;both&lt;/strong&gt; phenomena: &lt;a href=&quot;http://www.isbc-society.org&quot; rel=&quot;nofollow&quot;&gt;ISBC - International Society for Bioluminescence and Chemiluminescence&lt;/a&gt; ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal reactions, that lead to an excited species which eventually emits a photon are quite similar. From a reactive compound, typically called a &lt;em&gt;luciferine&lt;/em&gt;, a four-membered ring, often a dioxetanone is formed. This decarboxylates and gives rise to a molecule in an electronically excited state. In reality, the decarboxylation is often not just a spontaneous process but requires an activator. W. J. Baader at IQUSP in Sao Paulo, Brazil has examined that in great detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;luciferines&lt;/em&gt; can be quite different compounds, such as  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;diaryloxalates (these are used in glow sticks, where the diferent colours come from diferent fluorphores that absorb the initial photon and emit at different wavelength)  &lt;/li&gt;&#xA;&lt;li&gt;benzothiazole derivatives (firefly)  &lt;/li&gt;&#xA;&lt;li&gt;triazine derivatives (marine organisms) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-08T14:37:28.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31045" PostTypeId="2" ParentId="29184" CreationDate="2015-05-08T14:43:57.167" Score="0" Body="&lt;p&gt;If you look at the &lt;a href=&quot;http://cp2k.org/&quot; rel=&quot;nofollow&quot;&gt;CP2K website&lt;/a&gt;, you'll see that the coordinates are only a small fraction of the input file. Most of it is text describing how a calculation should be run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key parts from a coordinates perspective look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;amp;CELL&#xA;  ABC 12.4138 12.4138 12.4138&#xA;&amp;amp;END CELL&#xA;&amp;amp;COORD&#xA;  O      12.235322       1.376642      10.869880&#xA;  H      12.415139       2.233125      11.257611&#xA;  H      11.922476       1.573799       9.986994&#xA;&amp;amp;END COORD&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As it looks, these are the Cartesian coordinates of the atoms (in regular space) and the unit cell vectors. You can either save the file in &lt;code&gt;.XYZ&lt;/code&gt; format or use an input generator like GAMESS or Gaussian to get these coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a periodic boundary calculation, you just need to estimate the bounding box. You can do this manually, or you can go to &lt;kbd&gt;Crystallography | Add Unit Cell&lt;/kbd&gt; and then tweak the volume. In Avogadro v1.1.x, you'll see the ABC lattice vectors needed for that CP2K block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it might seem, Avogadro2 will be v2.0 of Avogadro. It's still in development.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-08T14:43:57.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31046" PostTypeId="1" CreationDate="2015-05-08T14:47:41.290" Score="3" ViewCount="4940" Body="&lt;p&gt;Can we add concentrated $\ce{H2SO4}$ and concentrated $\ce{HNO3}$ to anisole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard that we can't add acids into phenol and aniline which make them blast.&lt;/p&gt;&#xA;" OwnerUserId="16070" LastEditorUserId="7475" LastEditDate="2015-10-09T15:58:23.557" LastActivityDate="2015-10-09T15:58:23.557" Title="Does anisole react with concentrated nitric or sulfuric acid" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31047" PostTypeId="1" AcceptedAnswerId="31053" CreationDate="2015-05-08T14:58:26.267" Score="10" ViewCount="6061" Body="&lt;p&gt;I am having trouble with the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the following reversible reaction in which the reaction is first order in both&#xA;  directions: $$\ce{[A] &amp;lt;=&amp;gt; [B]}$$ $k_\mathrm a$ is the rate constant for the forward reaction and $k_\mathrm b$ is the rate constant for the reverse reaction.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If only reagent $\ce{A}$ is present at $t = 0$, such that $[\ce{A}](t = 0)$ = $[\ce{A}]_0$ (and likewise for $[\ce{B}]_0$), then at all subsequent times $[\ce{A}]+[\ce{B}] = [\ce{A}]_0$. Write down a differential equation for $\frac{\mathrm d[\ce{A}]}{\mathrm dt}$ where $[\ce{A}] \equiv [\ce{A}](t)$ is the concentration of the reagent $\ce{A}$, and hence show that:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm b+k_\mathrm a \exp[-(k_\mathrm a+k_\mathrm b)t]}{k_\mathrm a+k_\mathrm b}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's my attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b[\ce{A}]\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using the identity in the question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b([\ce{A}]_0-[\ce{A}])\tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0\tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that this is a first order linear differential equation with the integrating factor: $\mathrm e^{\int (k_\mathrm a+k_\mathrm b)\,\mathrm dt}=\mathrm e^{(k_\mathrm a+k_\mathrm b)t}$ Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d}{\mathrm dt}\left([\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\right)=k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\tag{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=\int k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\,\mathrm dt\tag{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=k_\mathrm b[\ce{A}]_0\frac{\mathrm e^{(k_\mathrm a+k_\mathrm b)t}}{k_\mathrm a+k_\mathrm b}+c\tag{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't give me the final result if I put in the boundary conditions. Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4945" LastEditDate="2018-02-26T05:52:41.050" LastActivityDate="2018-12-08T18:49:20.680" Title="Chemical kinetics of a reversible reaction" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31048" PostTypeId="2" ParentId="31046" CreationDate="2015-05-08T15:05:50.610" Score="4" Body="&lt;p&gt;Anisole (methoxybenzene) is an electron-rich arene that will react with sulfuric acid or nitric acid in an &lt;strong&gt;electrophilic aromatic substitution&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I suggest that you look the reaction up. Then grab a pencil, some sheets of paper and start to draw. Draw the arenium ions (Wheland intermediates) for the addition of the electrophiles at the different positions of anisole, have a look at the possible resonance structures and enjoy chemistry&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Btw, you can do the same reaction with aniline or phenol without causing an explosion&lt;/strike&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My comment on aniline and phenol only works on paper but not in real life and &lt;strong&gt;Mithoron&lt;/strong&gt;'s comment on aniline and phenol is correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitroanilines and nitrophenols are synthesized in a different way.&#xA;In principle, nitration of aniline is possible with a mixture of nitric and nitrous acid, or after protecting the aniline as its formanilide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, 4-nitroaniline is produced by reacting 4-chloronitrobenzene with aqueous ammonia at elevated temperatures in an autoclave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details can be found in &lt;em&gt;Ullmann's Encyclopedia of Industrial Chemistry&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the alternative continuous process, molten 4-chloronitrobenzene and 40 % ammonia liquor in a molar ratio of 1 : 17 are pumped at 20 MPa through a steel reaction coil held at 237-240 °C by oil bath heating. Inflow is controlled&#xA;  to give a contact time of ca. 10 min, and outflow is directed into an expansion chamber with a tangential flow of hot water to give, in quantitative yield, a suspension of product that is readily isolated and washed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same is true for the formation of 4-nitrophenol. Again, the starting material is 4-chloronitrobenzene, which is reacted with strong aqueous $\ce{NaOH}$ to yield the phenol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two reactions are examples for &lt;strong&gt;nucleophilic aromatic substitutions&lt;/strong&gt;!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-08T17:42:27.430" LastActivityDate="2015-05-08T17:42:27.430" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31049" PostTypeId="1" CreationDate="2015-05-08T15:44:29.130" Score="15" ViewCount="3297" Body="&lt;p&gt;I was checking dipole moment orders, and came across a very peculiar result:&#xA;Methyl fluoride $(\ce{CH3F})$ has lesser dipole moment than methyl chloride ($\ce{CH3Cl}$), but hydrofluoric acid ($\ce{HF}$) has more dipole moment than hydrochloric acid ($\ce{HCl}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought it was about distance, but upon seeing the second one, it led me to think about polarizability. So my question shortens to this:&#xA;Which factor dominates while checking dipole moments and when?&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorUserId="5026" LastEditDate="2018-07-04T02:13:52.853" LastActivityDate="2018-07-04T02:20:31.557" Title="How to explain the dipole moment anomaly of methyl fluoride and methyl chloride with respect to their hydracids?" Tags="&lt;inorganic-chemistry&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31050" PostTypeId="1" CreationDate="2015-05-08T15:47:40.550" Score="9" ViewCount="18967" Body="&lt;p&gt;We talked about it in our chemistry class but we couldn't get to a conclusion, any help?&lt;/p&gt;&#xA;" OwnerUserId="16073" LastActivityDate="2021-02-25T06:21:41.320" Title="Can you melt wood?" Tags="&lt;organic-chemistry&gt;&lt;combustion&gt;&lt;melting-point&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31051" PostTypeId="2" ParentId="31047" CreationDate="2015-05-08T16:05:17.780" Score="3" Body="&lt;p&gt;We know, that at time $t = 0$, $[A] = A_0$, &#xA;utting these values in the equation you derived,&#xA;we will get $$c=\frac{K_a A_0}{K_a + K_b}. \tag{Constant of Integration}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And voilà your answer is in front of you, you just need to transpose the exponential term in the left hand side to the right hand side.&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorUserId="4945" LastEditDate="2018-02-26T05:51:49.447" LastActivityDate="2018-02-26T05:51:49.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31052" PostTypeId="2" ParentId="31050" CreationDate="2015-05-08T16:08:16.397" Score="6" Body="&lt;p&gt;NO, we can't melt wood!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From primary level we have learned that solid melts to liquid at certain temperature and on increasing the temperature further it change into gaseous. But that is not the case always. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem with melting wood revolves around what combustion is, and what temperature the combustion of wood happens at.  Combustion,&#xA;  also known as burning, is simply a chemical reaction that takes place&#xA;  where the combustible material (in this case wood) in the presence of&#xA;  an oxidizer (usually the air around the fire) changes its chemical&#xA;  composition and decomposes the material into other chemicals. The&#xA;  process is one that’s exothermic. As such, light and heat can be&#xA;  released.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Wood is mostly made up of things like cellulose, lignin, and water. As&#xA;  wood combusts, it’s broken down into products like charcoal, water,&#xA;  methanol, and carbon dioxide. Unlike water turning back into ice, if&#xA;  you cooled down the resulting products of burning wood, it obviously&#xA;  does not change back to its original composition. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All materials that combust will have a natural temperature at which&#xA;  the process will begin taking place. The higher the temperature, the&#xA;  quicker the process becomes (usually). If that temperature is lower&#xA;  than the temperature at which the material will melt, that material&#xA;  will never (naturally) melt because it just turns into other&#xA;  chemicals.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As for wood, it will begin a process known as pyrolysis at&#xA;  temperatures around 500-600 degrees Fahrenheit. Pyrolysis is also an&#xA;  exothermic reaction that tends to be self sustaining. At these&#xA;  temperatures, wood will begin giving off up to 100 chemicals,&#xA;  including methane and methanol (the same stuff they put as additives&#xA;  in gasoline), that will begin to burn. Once those chemicals begin&#xA;  burning, they will increase the temperature and the remaining char&#xA;  (the burned black bits present after the fire goes out) left behind&#xA;  will begin to further decompose, things like calcium, potassium and&#xA;  magnesium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; &lt;a href=&quot;http://www.todayifoundout.com/index.php/2013/08/why-doesnt-wood-melt/&quot; rel=&quot;noreferrer&quot;&gt;todayifoundout&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-05-08T16:08:16.397" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="31053" PostTypeId="2" ParentId="31047" CreationDate="2015-05-08T16:19:16.083" Score="7" Body="&lt;p&gt;Third equation looks right to me.  That's a &lt;a href=&quot;http://pruffle.mit.edu/3.016-2005/Lecture_20_web/node3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-homogeneous (some would say heterogeneous) first-order differential equation&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the heterogeneity is what is giving you problems.   Those equations have a general solution which is a &lt;a href=&quot;http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-NonhomgenLinEqns_Stu.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;sum of a &quot;complementary&quot; solution and a &quot;particular&quot; solution&lt;/a&gt;.  &lt;strong&gt;Don't try to use any boundary conditions until after you have combined particular and complementary solutions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;$$\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The &quot;complementary solution&quot; is the solution to the corresponding homogeneous equation $\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=0$, which is $[\ce{A}]_\mathrm c=K e^{-(k_\mathrm a+k_\mathrm b)t}$, where $K$ is an unknown constant of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;particular solution&quot; is a polynomial in $t$ of the same degree as the hetereogeneous term, which in this case is a polynomial of degree 0, $[\ce{A}]_\mathrm p=C$, where $C$ is an undetermined constant.  We solve for this constant by substitution into the original equation:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;$$0 + (k_\mathrm a+k_\mathrm b)C=k_\mathrm b[\ce{A}]_0$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This provides the value of $C$ as $C=\frac{k_\mathrm b[\ce{A}]_\mathrm 0}{k_\mathrm b+k_\mathrm a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution is thus the sum of the particular and complementary solutions:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;$$[\ce{A}]=[\ce{A}]_\mathrm c + [\ce{A}]_\mathrm p=K e^{-(k_\mathrm a+k_\mathrm b)t} + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now we can apply boundary conditions.  At time 0, $[\ce{A}]=[\ce{A}]_0$, which gives an algebraic equation we can solve for $K$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;$$[\ce{A}]_0=K + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If I did my algebra right, then the solution is $K=[\ce{A}]_\mathrm 0 \left (1-\frac{k_b}{k_\mathrm b+k_\mathrm a}  \right ) $.  Substituting this value back into the general solution from my equation 5 gives:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;$$[\ce{A}]=[\ce{A}]_0 \left (1-\frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}  \right )e^{-(k_a+k_b)t} + \frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}[\ce{A}]_0$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This solution satisfies the boundary condition and also the original equation.  It can be simplified a bit by algebraic manipulation to:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm a e^{-(k_\mathrm a+k_\mathrm b)t}+k_\mathrm b}{k_\mathrm a+k_\mathrm b}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From this expression it is easy to check that (i) the condition at $t=0$ is satisfied, that (ii) the expected exponential dependence in time is obtained, and (iii) that as $t \rightarrow \infty $, the reaction should go to equilibrium.  The last point about equilibrium can be checked by applying the two chemical definitions for the equilibrium constant (the ratio of product $B$ to reactant $A$ vs. the ratio of rate constants $k_a/k_b$), i.e. $\frac{\ce{B}_{\infty}}{\ce{A}_{\infty}}=\frac{\ce{A}_0-\ce{A}_{\infty}}{\ce{A}_{\infty}}=\frac{k_\mathrm a}{k_\mathrm b}$ and using equation 8 to check consistency between our obtained expression for $\ce{A}_{\infty}$ and the chemically intuitive definition we just made.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="5026" LastEditDate="2018-02-26T05:16:57.297" LastActivityDate="2018-02-26T05:16:57.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31054" PostTypeId="2" ParentId="31005" CreationDate="2015-05-08T16:32:43.153" Score="8" Body="&lt;p&gt;To complement Martin's answer with regard to the applications: chelation is particularly useful in the body with regard to iron transportation in siderophores and porphyrin rings and also in medicine where excess metal ions build up such as Wilkins disease ($\ce{Cu^2+}$) etc. Typically you see drugs with connected phenoxides, hydroxamates or EDTA fragments exactly for this purpose directed to iron deposits in the liver. Upon complexation you remove the chelated product easily. Without the Gibbs drive, metal plaques up. In addition catalysis and organometallic chemistry is full of chelation chemistry.&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T21:14:57.897" LastActivityDate="2015-05-09T21:14:57.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31055" PostTypeId="1" AcceptedAnswerId="31140" CreationDate="2015-05-08T16:44:58.833" Score="6" ViewCount="437" Body="&lt;p&gt;The Wikipedia article on metals describes them as elements whose polyatomic configurations* exhibit any one of a number of patterns. I could not discern any property that characterized all the patterns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge, there are two main categories of elements: metals, non-metals. There's a third category, metalloids, whose members are those that satisfy some of the criteria for inclusion in both main categories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem very odd if there were nothing that characterized all, and only, metals. In fact it would seem that metals are better conceived of as an arbitrary grouping of what is actually a number of other categories, each of which would include all, and only, the elements that exhibit a single type of polyatomic configuration*. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a characteristic that characterizes all, and only, metals?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*(&lt;em&gt;I'm using the phrase 'polyatomic configuration' in order to express 'the way that atoms of a particular element stick together when you have a number of atoms of that element connected together'. Is there a standard term that expresses that idea?&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2018-04-30T03:46:41.017" Title="Is there a characteristic that characterizes all, and only, metals?" Tags="&lt;metal&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31057" PostTypeId="2" ParentId="31040" CreationDate="2015-05-08T17:11:58.063" Score="9" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.&lt;/p&gt;&#xA;&lt;h3&gt;If &lt;strong&gt;reversible&lt;/strong&gt;:&lt;/h3&gt;&#xA;&lt;p&gt;If the expansion is done &lt;strong&gt;reversibly&lt;/strong&gt;, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic).  Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant.  In this case you can apply both of your equations.  But keep in mind that your equation for &lt;span class=&quot;math-container&quot;&gt;$dS$&lt;/span&gt; is only true for ideal gases.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$dS = \frac{\delta q_{rev}}{dT} = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is now a differential equation you can solve for temperature if you know volume.&lt;/p&gt;&#xA;&lt;p&gt;As the other answers have indicated, a &lt;em&gt;reversible&lt;/em&gt; expansion will involve the gas doing work on the surroundings.  One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings.  A loss in the potential to do work (of the universe) is an increase in entropy.  And in a reversible process that can't happen.&lt;/p&gt;&#xA;&lt;p&gt;Think of gas in a piston, where the piston is held down by a pile of sand.  The reversible expansion involves slowly removing sand, grain by grain, from the piston.  As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.&lt;/p&gt;&#xA;&lt;h3&gt;If &lt;strong&gt;irreversible&lt;/strong&gt;:&lt;/h3&gt;&#xA;&lt;p&gt;...then we still don't have enough information because there are many ways the process can be irreversible.  But one particular, extreme case of irreversibility is if the adiabatic expansion occurs &lt;strong&gt;without transfer of &lt;em&gt;any&lt;/em&gt; energy&lt;/strong&gt;, then we can solve the problem.  In this extreme scenario, neither heat nor work are not transferred.  You can thus infer that &lt;span class=&quot;math-container&quot;&gt;$dU=0$&lt;/span&gt; for the gas, so&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$dU=-TdS + p dV=0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$TdS = p dV$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using your definition for &lt;span class=&quot;math-container&quot;&gt;$dS$&lt;/span&gt; for an ideal gas,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Keep in mind that &lt;span class=&quot;math-container&quot;&gt;$\frac{RT}{V}$&lt;/span&gt; for an ideal gas is &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, so you should get that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_V dT = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which is also a differential equation that you can integrate to get the temperature change.  (The obvious solution is that the temperature &lt;strong&gt;does not change&lt;/strong&gt; in this case.)&lt;/p&gt;&#xA;&lt;p&gt;In this second case, now that we know that &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is constant, we know &lt;span class=&quot;math-container&quot;&gt;$dT$&lt;/span&gt; is zero, and thus using your equation for &lt;span class=&quot;math-container&quot;&gt;$dS$&lt;/span&gt;, we get that &lt;span class=&quot;math-container&quot;&gt;$dS = \frac{R}{V}dV &amp;gt; 0$&lt;/span&gt;, i.e., the entropy of the gas increases, which makes sense because the process was irreversible.  The potential to do work has been lost.&lt;/p&gt;&#xA;&lt;p&gt;Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum).  If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-08T17:11:58.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31058" PostTypeId="2" ParentId="20344" CreationDate="2015-05-08T17:15:15.790" Score="1" Body="&lt;p&gt;For any engine that does not have a sophisticated ECU then the addition of hydrogen into the air intake does produce significant savings on fuel and reductions in pollution.&#xA;The modern ECU will offset this effect but older engines will perform well.&#xA;The greater risk is the hydrogen that remains in the engine, but this is present at 0.001% parts per air and since I believe (though please correct me if I am wrong) hydrogen is explosive at a min of 18% even this should not be an issue&lt;/p&gt;&#xA;" OwnerUserId="16076" LastActivityDate="2015-05-08T17:15:15.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31059" PostTypeId="1" CreationDate="2015-05-08T17:21:31.883" Score="1" ViewCount="202" Body="&lt;p&gt;I would like to know if there is documented evidence of the quantity of hydrogen to air that must be present in the mixture to risk explosion or flame.&lt;/p&gt;&#xA;" OwnerUserId="16076" LastEditorUserId="5017" LastEditDate="2015-05-08T18:22:09.510" LastActivityDate="2015-05-08T18:22:09.510" Title="What fraction of hydrogen in air results in an explosive mixture?" Tags="&lt;safety&gt;&lt;hydrogen&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31060" PostTypeId="2" ParentId="31059" CreationDate="2015-05-08T17:26:50.300" Score="3" Body="&lt;p&gt;Hydrogen is explosive in mixtures in air at room temperature and ambient pressure when it is between 4% and 75% by volume.  These limits are respectively called the &quot;lower explosive limit&quot; (LEL) and the upper explosive limit (UEL). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values for hydrogen (and many other gases) are provided in many locations on the internet.  &lt;a href=&quot;http://www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html&quot; rel=&quot;nofollow&quot;&gt;Engineering toolbox&lt;/a&gt; is one such site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info on LEL and UEL is available &lt;a href=&quot;http://en.wikipedia.org/wiki/Flammability_limit&quot; rel=&quot;nofollow&quot;&gt;from Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-08T17:26:50.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31061" PostTypeId="2" ParentId="31002" CreationDate="2015-05-08T17:47:09.400" Score="5" Body="&lt;p&gt;You can use Open Babel. You just need to &quot;fill&quot; the unit cell (i.e., generate all the symmetric atoms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need the &lt;code&gt;--fillUC&lt;/code&gt; option, documented &lt;a href=&quot;https://open-babel.readthedocs.org/en/latest/Command-line_tools/babel.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a crystal structure, add atoms to fill the entire unit cell based on the unique positions, the unit cell and the spacegroup. The parameter can either be strict (the default), which only keeps atoms inside the unit cell, or keepconnect, which fills the unit cell but keeps the original connectivity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;e.g.,:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel SnS.cif -oVASP --fillUC &amp;gt;POSCAR&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Result is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Herzenbergite&#xA;1.000 &#xA;   4.330000000000000   0.000000000000000   0.000000000000000&#xA;   0.000000000000001  11.180000000000000   0.000000000000000&#xA;   0.000000000000000   0.000000000000000   3.980000000000000&#xA;Sn  S   Sn  S   &#xA;1   1   3   3   &#xA;Cartesian&#xA;     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956&#xA;     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956&#xA;     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956&#xA;     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956&#xA;     0.4979500000000001148      1.3192399999999999682      0.9949999999999999956&#xA;     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956&#xA;     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956&#xA;     2.0697400000000003573      9.5030000000000001137      0.9949999999999999956&#xA;&lt;/pre&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-08T17:47:09.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31062" PostTypeId="2" ParentId="30990" CreationDate="2015-05-08T18:07:41.770" Score="3" Body="&lt;p&gt;Let's start with &quot;what is the species?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While you write it $\ce{HgCo(NCS)4}$ it's probably better written as $\ce{Hg^{+2} [Co(NCS)4]^{2-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take the two metal ions separately as far as electronic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Hg^{+2}: d^{10}}$ So we don't expect any unpaired electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Co(NCS)4]^{2-}}$ Well, $\ce{Co}$ must be $\ce{Co^{+2}}$, so it's $\ce{d^7}$ on paper. Next you'd need to consider the shape of the complex and possibly where the ligand is in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spectrochemical_series&quot; rel=&quot;nofollow&quot;&gt;spectrochemical series&lt;/a&gt; and the likely orbital occupations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Here, $\ce{Co^{+2}}$ is tetrahedral, so it doesn't matter where NCS falls on the series. You expect a high spin complex with 3 unpaired spins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; (NCS) can be a tricky ligand, since it can coordinate with either N or S atoms, but the &lt;a href=&quot;http://journals.iucr.org/b/issues/1968/05/00/a06139/a06139.pdf&quot; rel=&quot;nofollow&quot;&gt;crystal structure&lt;/a&gt; is clearly N-coordination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mention this, because while the Hg ion is, on paper, separated from the Co complex, in the crystal structure, there's clearly some Hg-S interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's get to your question about the calculation. Yes, if you need an integer, you'd round up. But since you have many, many more than one molecule, your answer implies there are multiple magnetic species.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; This is clear from the electronic structure. Since the calculation above gives three unpaired spins in an isolated $\ce{Co^{+2}}$ complex, there &lt;em&gt;must&lt;/em&gt; be some magnetic interactions between multiple species in the solid state. &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic00054a012&quot; rel=&quot;nofollow&quot;&gt;Indeed, this is observed at different temperatures&lt;/a&gt;. I'm hardly an expert in magnetic coupling here, but it certainly implies Hg must be involved as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-05-09T15:31:15.860" LastActivityDate="2015-05-09T15:31:15.860" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31063" PostTypeId="2" ParentId="22069" CreationDate="2015-05-08T18:30:51.503" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is any one of these reaction more &quot;true&quot; (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the best answer is &quot;a little bit of everything&quot;.  Silver sulfide forms faster but requires exposure of the silver to sulfur-containing materials (like human skin, food, etc.).  Silver that isn't exposed to sulfur would still be oxidized by the air, but at slower rates.  So depending on the life history of the silver object in question, different amounts of the sulfide and oxide could be present.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-08T18:30:51.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31064" PostTypeId="2" ParentId="29122" CreationDate="2015-05-08T18:35:34.243" Score="1" Body="&lt;p&gt;This question is difficult to answer without knowing more about the data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is the issue of &quot;normalization&quot;. Do you mean that in the at% study, all measurements added together were only 96% or 90% or 80% or something, and so to get all measurements to add to 100%, the simply divided each measurement by 0.96 (or 0.9 or 0.8)?  This is what I infer from your comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second is the issue of measuring at%.  Are they measuring the raw number of total atoms in their OES experiment? Or are they just comparing their results to some calibration standard for each element measured?  Your comments make me think they are using calibration standards.  From a description of those standards, you should be able to convert the standards to wt%, and thus map from wt% to at%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you want to map to the &quot;normalized&quot; at% or the unnormalized at% is up to you.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-08T18:35:34.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31065" PostTypeId="2" ParentId="16451" CreationDate="2015-05-08T18:45:10.823" Score="2" Body="&lt;p&gt;As mentioned in the comments, a likely culprit is microbial contamination.  In a $\ce{KNO3}$ solution, you are providing plenty of water, a nitrogen source (nitrate), and even a high-energy, respiratory means for growth (nitrate respiration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Living things also need (i) food, (ii) trace amounts of sulfur, (iii) and even smaller trace amounts of phosphorus.  But if you are soaking hemp in this solution, enough leachable S and P-containing impurities are probably extracted into the solution to provide these.  (Or also low levels of S and P may already be present in your $\ce{KNO3}$, or water.)  The food could be organic compounds leached from the hemp.  Or (less likely) the microbial contaminants might be capable of photosynthesis, in which case their food is atmospheric $\ce{CO2}%edit$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:25:13.983" LastActivityDate="2016-03-11T23:25:13.983" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31067" PostTypeId="2" ParentId="19507" CreationDate="2015-05-08T19:00:33.880" Score="1" Body="&lt;p&gt;Sigma-Aldrich is a global supplier of purified chemicals for research users. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a whole page of the &lt;a href=&quot;http://www.sigmaaldrich.com/chemistry/chemistry-products.html?TablePage=16272014&quot; rel=&quot;nofollow&quot;&gt;haloalkane compounds&lt;/a&gt; they sell (in small quantities such as you desire).  You might be interested in compounds such as trichloropropane, pentachloroethane, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reason it might be difficult to get the higher-molecular weight compounds is that they are toxic and difficult to dispose of properly.  In the US, their use is subject to &lt;a href=&quot;http://www.doverchem.com/Portals/0/CP%20Regulatory%20Update%2006-13-13.pdf&quot; rel=&quot;nofollow&quot;&gt;increasing regulatory requirements&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-08T19:00:33.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31068" PostTypeId="2" ParentId="5063" CreationDate="2015-05-08T20:04:02.583" Score="3" Body="&lt;p&gt;Hydrogen bonds are three centre 4 electron interactions that lower the energy of the molecular orbitals involved. We can hypothetically imagine two scenarios, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Metal donor (M$^{\delta ^{-}}-H^{\delta ^{+}}...C$)&lt;/li&gt;&#xA;&lt;li&gt;Metal acceptor (M$^{\delta ^{-}}...H^{\delta ^{+}}-C$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(where C is just something else). Metal-Hydrogen bonds where the metal is a donor are scarce since the metal hydride bond is the opposite polarity of the metal hydrogen bond donor, however they do exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal acceptors are likely to be low oxidation state late row metals with filled $d$ orbitals to donate via $\sigma$ symmetry interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get these types of interactions all the time in crystal chemistry where we find bridging hydrogens! So in short, it depends where the metal is in the period, if its $dz^2$ is available to donate etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a specific example I would be happy to try and work it out, but it looks like you are aware of any literature I would consult!! :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2015-05-08T20:09:57.387" LastActivityDate="2015-05-08T20:09:57.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31069" PostTypeId="2" ParentId="29343" CreationDate="2015-05-08T20:04:49.413" Score="1" Body="&lt;p&gt;High intensity of electricity added to the source ( $\ce{HCL}$ ), gives us more atoms, and when we have more atoms then the chances for self-absorption are higher. One atom will get excited and then turning back will emit radiation, the other atom from the same radiation will absorb that emission. In the beginning, the instrument measures the intensities of radiation at normal intensity of electricity. After that, it makes the measurement on higher electricity. The difference between them, should give us the background absorption. &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="14968" LastEditDate="2015-05-08T21:11:51.083" LastActivityDate="2015-05-08T21:11:51.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31070" PostTypeId="1" AcceptedAnswerId="31072" CreationDate="2015-05-08T20:52:09.577" Score="6" ViewCount="985" Body="&lt;p&gt;In organic chemistry lecture we discussed the beginning of the Woodward Hoffmann Rules with a key step of the vitamin B12 synthesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The initial paper from their work can be found here:&#xA;&lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja01080a054&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/doi/pdf/10.1021/ja01080a054&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's first see the forward reaction (the electrocyclic ring closure) as Woodward and Hoffmann found (which goes by disrotatory cyclization; 6 electrons, thermal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v0zmn.png&quot; alt=&quot;Woodward and Hoffmann&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the reverse reaction, the ring opening (6 electrons, and therefore disrotatory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking now about the torquoselectivity of the ring-opening  (because in an exercise we had an example with 1-Ethyl-1-ethoxy(EDG)-cyclobut-2-ene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a rule how to see why the E (E: Electron withdrawing group) is in trans position when I start with the product?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking if they are some stereoelectronic effects here which explain the obtained selectivity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rInep.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo00244a058&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jo00244a058&lt;/a&gt;&#xA;(Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well here we have a conrotatory ring-opening. I don't know if they are still valid in our 6e case for disrotatory ring-opening of the reaction from W. and H..&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2016-11-14T23:56:30.497" LastActivityDate="2016-11-14T23:56:30.497" Title="Torquoselectivity: Disrotatory ring opening" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;molecular-orbital-theory&gt;&lt;pericyclic&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31071" PostTypeId="1" CreationDate="2015-05-08T21:32:11.877" Score="7" ViewCount="179" Body="&lt;p&gt;I have tried to find information about this case but I couldn't find anything specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point of instrumental methods of analysis, is for the student to be able to know which one of the possible methods to use for a specific determination. As for me, that is not an easy thing to do because &lt;em&gt;every&lt;/em&gt; method has its pros and cons. For the moment I am analysing AAS (Atomic Absorption Spectroscopy), and I simply cannot tell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the reason that $\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ is preferred to be determined only by the &lt;em&gt;hydrid method&lt;/em&gt;? $\ce{Pb}$ for example can be determined also with the other methods in AAS  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any tip, how to choose between methods if it's required for any element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I replied with a similar question earlier about the mercury determination.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="5026" LastEditDate="2018-03-27T13:12:12.230" LastActivityDate="2018-05-26T15:43:17.800" Title="Optional determination techniques in Atomic Absorption Spectroscopy" Tags="&lt;analytical-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31072" PostTypeId="2" ParentId="31070" CreationDate="2015-05-08T21:43:46.813" Score="3" Body="&lt;p&gt;No, there is no &lt;strong&gt;general&lt;/strong&gt; orbital symmetry rule for predicting which of the two possible disrotatory motions will be favored.  However, &lt;strong&gt;by examining steric effects in the two possible transition states we can often make a good guess as to which product will predominate&lt;/strong&gt;.  There are always a few &quot;special cases&quot; like the one you referenced in your post, where with certain substituents under certain conditions certain things will happen.  But to my knowledge there is no general rule for this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following diagram there are 3 thermal reactions controlled by orbital symmetry (6 pi electrons, therefore disrotatory motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oftGM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top line shows the symmetry allowed disrotatory closure of trans, trans-1,6-dimethylhexatriene.  There are two possible disrotatory motions - either the tops or the bottoms of the p-orbitals on carbons 1 and 6 can move towards each other in a disrotatory fashion.  One motion produces the dimethylcyclohexadiene with the two methyl groups above the plane of the screen, the other motion produces the mirror image molecule with the methyl groups below the plane of the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, cis, cis-1,6-dimethylhexatriene can undergo two disrotatory motions to produce the same two products (Note: the cis, cis isomer will react much slower and probably in lower yield; steric interactions between the methyl groups will disfavor the s-trans conformation about the two single bonds that is required for the ring closure to be able to occur).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line shows the reverse reaction, thermally converting a cyclohexadiene to a hexatriene.  As shown, two possible isomers can be formed.  The ratio of these two isomers will depend on steric interactions that develop during the two possible disrotatory motions.  In this case, we will primarily obtain the trans, trans isomer; the cis, cis isomer has steric problems in its transition state from the two methyl groups bumping into each other (build a model).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this last reaction, if we replaced the methyl groups with deuterium, we would obtain a roughly equal mixture of the two possible products, cis, cis- and trans, trans-1,6-dideuteriohexatriene.  So again, there is no orbital symmetry &quot;rule&quot; for predicting whether a group would wind up cis or trans on the double bond.  But if we examine the steric effects in the two possible disrotatory transition states, we can often make a good guess about which isomer should predominate.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-08T23:25:12.527" LastActivityDate="2015-05-08T23:25:12.527" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31073" PostTypeId="2" ParentId="29464" CreationDate="2015-05-08T22:42:47.180" Score="5" Body="&lt;p&gt;Here are a number of resources that may be of use for your research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UCLA website presents tables and data in their webpage &lt;a href=&quot;http://www.chem.ucla.edu/~bacher/UV-vis/uv_vis_tetracyclone.html.html&quot; rel=&quot;nofollow&quot;&gt;Theory of Ultraviolet-Visible (UV-Vis) Spectroscopy&lt;/a&gt;, including reference tables of common functional groups, aromatics and compounds often used in the laboratory.  (Many compounds you listed are in the tables).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference that may be of use is &lt;a href=&quot;http://www.uobabylon.edu.iq/eprints/publication_11_8282_250.pdf&quot; rel=&quot;nofollow&quot;&gt;Spectroscopy of Organic Compounds&lt;/a&gt; (Kumar, 2006) which include several tables that include information of the compounds you list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book &lt;a href=&quot;https://books.google.com.au/books?hl=en&amp;amp;lr=&amp;amp;id=QWq47QClOM0C&amp;amp;oi=fnd&amp;amp;pg=PA1&amp;amp;dq=UV-Visible%20Spectroscopy%20of%20Organic%20Compounds&amp;amp;ots=kJ2V9ub4a9&amp;amp;sig=6oqeNZYzlvxSC2eg8RaXfwlAZnE#v=onepage&amp;amp;q=UV-Visible%20Spectroscopy%20of%20Organic%20Compounds&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Spectroscopy of Organic Compounds&lt;/a&gt; (Kalsi, 2007), from page 12 provide several tables of information related to your listed organic compounds, including details of the electronic transitions. The chapter &quot;Ultraviolet (UV) and Visible Spectroscopy&quot; (starting page 9) has a subsection &quot;The Designation of Various Transitions&quot; which details the electronic transitions using examples in context (from page 11, summarised in a diagram on page 12).  Examples of solvents are discussed from page 23 onwards.  (All available online).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transitions are explored in a bit of detail in chapter 4.3 of the UC Davis ChemWiki The Dynamic Chemistry E-textbook, &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_04%3A_Structure_Determination_I/Section_4.3%3A_Ultraviolet_and_visible_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Section 4.3: Ultraviolet and visible spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-11T20:37:22.563" LastActivityDate="2015-05-11T20:37:22.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31074" PostTypeId="2" ParentId="30855" CreationDate="2015-05-08T22:48:16.913" Score="8" Body="&lt;p&gt;So you want to compare the acidity of different Brønsted–Lowry acids. This means ultimately you want to compare the equilibrium constants for reactions of the type $\ce{HX &amp;lt;=&amp;gt; H+ + X-}$. To do so, you'll have to analyse what stabilisation/destabilisation effects are present in each acid $\ce{HX}$ and each conjugate base $\ce{X-}$. This is easier said than done, especially quantitatively. There are many sorts of possible effects, often acting in a compounded fashion. Fortunately, there are some basic principles to help guide us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question specifically asks to contrast &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbonic acid&lt;/a&gt;, $\ce{H2CO3}$ and &lt;a href=&quot;http://en.wikipedia.org/wiki/Picric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;picric acid&lt;/a&gt;, more descriptively called 2,4,6-trinitrophenol. This already starts off as being a bit tricky, because they are not very similar substances; of course, the more similar the two substances are when being compared, the easier it will be to pick out the differences in stabilisation/destabilisation effects and the safer it will be to extrapolate from qualitative reasoning. With this in mind, it is best if we build up to the answer by making comparisons between several molecules, making small changes at each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let us start with &lt;a href=&quot;http://en.wikipedia.org/wiki/Alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;alcohols&lt;/a&gt;. These are substances containing hydroxyl groups ($-\ce{OH}$) linked to carbon atoms which are &lt;em&gt;saturated&lt;/em&gt; (carbon atoms with four single bonds, i.e. $\rm{sp^3}$-hybridised carbon) and whose three other bonds are only to carbon or hydrogen atoms (in other words, &lt;em&gt;linked to no heteroatoms&lt;/em&gt; other than the hydroxyl oxygen). This class represents neither of your substances, but is simple and is related to phenols, so it will be a useful benchmark. A simple alcohol is methanol, $\ce{CH3OH}$, which can act as an acid by the equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3C-OH_{(aq)} &amp;lt;=&amp;gt; H+_{(aq)} + H3C-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=3\times10^{-16}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small equilibrium constant establishes that methanol is a very weak acid in water, and exists almost completely as the unionised reactant. Values of $\mathrm pK_\mathrm a$ around 15 are typical for simple alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us take a step towards the structures we want, by examining at a hydroxyl linked to an unsaturated carbon atom, specifically a $\rm{sp^2}$-hybridised carbon atom linked to no heteroatoms other than the hydroxyl oxygen. What we have here is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Enol&quot; rel=&quot;nofollow noreferrer&quot;&gt;enol&lt;/a&gt;. Simple enols are not very stable molecules, but for simplicity we shall disregard that. For the simple molecule &lt;a href=&quot;https://i.stack.imgur.com/BMk5m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;propen-2-ol&lt;/a&gt;, the equilibrium is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2C=C(OH)-CH3_{(aq)} &amp;lt;=&amp;gt; H+_{(aq)} + H2C=C(O^{-})-CH3_{(aq)}}\ \ \ \ \ K_\mathrm a=1.3\times 10^{-11}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evidently, enols are quite a bit more acidic than alcohols. There are two new effects at play here. First, $\rm{sp^2}$-hybridised carbon atoms show an &lt;a href=&quot;http://en.wikipedia.org/w/index.php?title=Electron_withdrawing_group&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron-withdrawing effect&lt;/a&gt; compared to $\rm{sp^3}$-hybridised carbon. This ability to &quot;suck away&quot; electron density helps stabilise molecules when close to electron-rich regions such as the oxygen atom. This effect stabilises the reactant, but it stabilises the enolate product &lt;em&gt;more&lt;/em&gt; (as it has extra electron density to be pulled), which drives the equilibrium to the right. A second effect is &lt;a href=&quot;http://en.wikipedia.org/wiki/Resonance_%28chemistry%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;resonance&lt;/a&gt;. Some of the electrons from the oxygen atom are said to &lt;em&gt;conjugate&lt;/em&gt; with the carbon-carbon double bond, creating &lt;em&gt;resonance&lt;/em&gt; or &lt;em&gt;canonical structures&lt;/em&gt;, which contribute to the true description of bonding in the molecule. Once again, spreading the electrons out contributes to decrease the energy of a molecule. The resonance structures for the &lt;a href=&quot;https://i.stack.imgur.com/6xemk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;reactant&lt;/a&gt; and the &lt;a href=&quot;https://i.stack.imgur.com/EVHqe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;product&lt;/a&gt; are similar, but one of the resonance structures in the reactant contains a formal positive charge on oxygen (a very electronegative atom) and a formal negative charge on carbon (a not very electronegative atom), so this form of conjugation is not quite as favourable. In the product, though there is still a formal negative charge on carbon in one of the canonical structures, there is no formal positive charge on oxygen, making conjugation more favourable. Thus the resonance argument is also a part of the explanation for the acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next up is phenol, the parent compound of 2,4,6-trinitrophenol. The effects present here are much the same as in the enol above, except that conjugation happens in a larger extent in both the &lt;a href=&quot;https://i.stack.imgur.com/znBG6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;reactant&lt;/a&gt; and the &lt;a href=&quot;https://i.stack.imgur.com/ScgP8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;product&lt;/a&gt;. Experimentally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H5-OH_{(aq)} &amp;lt;=&amp;gt; H+_{(aq)} + C6H5-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=1.1\times10^{-10}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents only a small increase in acidity compared to propen-2-ol, which suggests &lt;a href=&quot;http://blog.chembark.com/2007/03/10/i-judge-people-by-their-grammar-and-knowledge-of-phenol/&quot; rel=&quot;nofollow noreferrer&quot;&gt;conjugation is not the main factor in phenol acidity&lt;/a&gt;, but rather the inductive effect of the $\rm{sp^2}$ carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Finally&lt;/em&gt;, we've built up to our first structure, 2,4,6-trinitrophenol. The new bit here is the addition of nitro groups ($\ce{-NO2}$), which are strongly electron-withdrawing both by induction and mesomerism (conjugation). Again, these effects decrease the energy of both the reactant and the product, but stabilize the product more because there is extra density to be spread around. As a quick interesting aside, the acidities of 2-nitrophenol and 4-nitrophenol are similar, &lt;a href=&quot;https://i.stack.imgur.com/D8RmD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;and both are somewhat larger than 3-nitrophenol&lt;/a&gt;, which suggests most of the increase in acidity comes from electron-withdrawing by mesomerism rather than induction, represented by canonical structures &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/13556/3.jpg?revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;similar to these&lt;/a&gt;. You also make a good point in mentioning the &lt;a href=&quot;https://i.stack.imgur.com/40hC0.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;intramolecular hydrogen bond&lt;/a&gt; present only in 2,4,6-trinitrophenol and not its ionised form. Indeed, this helps stabilize the reactant a slight amount, and decreases its acidity. Let's see how all these effects add up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(O2N)3C6H2-OH_{(aq)} &amp;lt;=&amp;gt; H+_{(aq)} + (O2N)3C6H2-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=0.4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acidity increase from the electron-withdrawing effects of the three nitro groups &lt;strong&gt;far&lt;/strong&gt; overwhelms the acidity decrease due to the formation of the intramolecular hydrogen bond. How could you have known this beforehand? Strictly speaking, you couldn't have deduced this from thin air (not without a computer, at any rate!), but generally you are expected to know the situation in similar cases and work out the answer based on previous knowledge. Intramolecular hydrogen bonding usually doesn't play a huge role in increasing or decreasing acidity, at least in simple molecules, as they are relatively weak interactions and so don't contribute massively to increasing/decreasing the energy of a structure. Furthermore, nitro groups are among the most electron-withdrawing there are, so if pulling electrons is going to have any role, it'll show some of its strongest effects with nitro groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove this answer comes to an end, we reach our final structure, carbonic acid. First, we notice it is related to carboxylic acids, and its acidic group is the $\ce{-COOH}$ in the structure. If you think about it, the carboxyl group looks like an enol where a $\ce{=CH2}$ group is replaced by an oxygen atom. Thus the analysis is similar to the enol. However, there is an increase electron-withdrawing effect due to induction (oxygen being more electronegative than carbon) and mesomerism (in the canonical structures, the negative formal charges are now on an oxygen atom, rather than a carbon atom as was the case for the enol). The instability of carbonic acid interferes, but &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbonic_acid#Acidity_of_carbonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;leaving that aside&lt;/a&gt;, its dissociation would be something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2CO3_{(aq)} &amp;lt;=&amp;gt; H+_{(aq)} + HCO3^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=2.5\times 10^{-4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing everything, the acidity increases in the order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple alcohols &amp;lt; Simple enols and phenols &amp;lt; Carbonic acid and simple carboxylic acids &amp;lt; Picric acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you know carbonic acid is a weaker acid than picric acid? Once again, it's not something you can deduce by pure logic; it requires some benchmarking and comparison with previous data and knowledge. The analysis above would have been helpful, but has its limits. In Chemistry, it is often far easier to explain than predict. Whether you should have been expected to know the answer to this question, I'm not sure. It would depend on how your chemistry course is conducted.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2018-02-25T09:35:03.997" LastActivityDate="2018-02-25T09:35:03.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31075" PostTypeId="1" AcceptedAnswerId="31127" CreationDate="2015-05-09T00:07:21.527" Score="19" ViewCount="4518" Body="&lt;p&gt;Say I have a molecular wavefunction as a set of molecular orbitals and want to calculate the molecule's dipole moment, but don't know how! I searched a lot but couldn't find any practical example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi _i=\sum ^N_{i=1}C_i\mathrm e^{-\alpha _ir^2}$$&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorDisplayName="user7951" LastEditDate="2015-09-03T11:49:10.853" LastActivityDate="2021-05-01T11:40:32.593" Title="How to calculate molecular dipole moment from a known wavefunction?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;dipole&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31076" PostTypeId="1" CreationDate="2015-05-09T01:13:12.720" Score="6" ViewCount="188" Body="&lt;p&gt;I've done my research on this so I've seen the questions asking about the same premise without using plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this is a thought experiment I came up with. Whether or not I perform it depends on the feasibility. This is the reaction I'm thinking of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ce{CO_2 \rightarrow C + O_2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&amp;amp;Units=SI&amp;amp;Mask=1&quot; rel=&quot;nofollow&quot;&gt;NIST&lt;/a&gt; the formation energy of carbon dioxide back into carbon and oxygen is given as -393.52 kJ/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic premise I was thinking of was that $\ce{CO_2}$ could be pressurized in a chamber with electrodes striking an arc, most likely from an HF driver source. The energy imparted to the carbon dioxide could be used to provide the dissociation energy to break the carbon-oxygen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the electrodes were designed with sufficient voltage in mind to break the dielectric constant of the gas, a few rudimentary numbers suggest that even at 5% efficiency, a sustained power input of 1kW would break the bond of 1 mol of $\ce{CO_2}$ in about 2 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concerns with the feasibility of such a process are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would it work? (i.e. would using an electric arc be a viable way to perform the bond decomposition? I'm not looking for an &lt;em&gt;industrially viable&lt;/em&gt; process, just something that could produce a small end result.)&lt;/li&gt;&#xA;&lt;li&gt;Would I have to worry about the carbon and oxygen re-forming? Obviously, this would render moot the entire experiment. I know that $\ce{CO_2}$ decomposition doesn't appear to be an equilibrium process, but it's still a point of concern.&lt;/li&gt;&#xA;&lt;li&gt;If indeed it would work, how would the carbon likely present itself? I was thinking that if it sputtered itself against the vessel, the vessel could be weighed to determine the amount of carbon formed (and therefore the reaction efficiency, once the energy input was quantified).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16013" LastEditorUserId="16013" LastEditDate="2015-05-12T01:03:54.607" LastActivityDate="2015-05-12T01:03:54.607" Title="Scission of carbon dioxide bonds using electric arc" Tags="&lt;heat&gt;" AnswerCount="0" CommentCount="8" ClosedDate="2016-05-11T06:15:54.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="31077" PostTypeId="1" CreationDate="2015-05-09T02:40:52.130" Score="1" ViewCount="64" Body="&lt;p&gt;Is a negative pH value possible in room temperature...teachers don't tend to teach about the pH of acid or base solutions which are higher than 1M.&lt;/p&gt;&#xA;" OwnerUserId="16080" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T02:47:18.097" LastActivityDate="2015-05-09T02:47:18.097" Title="Is a negative pH value possible?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-05-09T03:05:12.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="31078" PostTypeId="1" AcceptedAnswerId="31108" CreationDate="2015-05-09T03:05:47.477" Score="5" ViewCount="9563" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XB5Hr.png&quot; alt=&quot;problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very confused. According to the answer key, the circled hydrogens of 2-methylindane (2-methyl-2,3-dihydro-1⁠&lt;em&gt;H&lt;/em&gt;-indene) are diastereotopic. I thought they were enantiotopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought process is the following. They aren't homotopic because replacing one of them with another atom would create a chiral center. Due to them creating a chiral center, I thought it was enantiotopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that for it to be diastereoptopic, the carbon with the methyl group off of it would have to be chiral as well, but I don't think it is chiral. The molecule is symmetrical off both sides. Does diastereotopic rely on that carbon being chiral &lt;strong&gt;after&lt;/strong&gt; the new atom replaced the hydrogen? If so, then I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was under the assumption that the state of that carbon must be chiral before the hydrogen is even replaced.&lt;/p&gt;&#xA;" OwnerUserId="2494" LastEditorDisplayName="user7951" LastEditDate="2016-11-29T23:17:56.667" LastActivityDate="2016-11-29T23:44:26.917" Title="Are the hydrogen atoms of 2-methylindane homotopic, enantiotopic, diastereotopic or unrelated?" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31079" PostTypeId="1" AcceptedAnswerId="31122" CreationDate="2015-05-09T03:07:53.627" Score="7" ViewCount="2226" Body="&lt;p&gt;I have seen complex ionic compounds that have mixed anions and/or mixed cations. For Example I have seen this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaKCl2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also known as Sodium Potassium Chloride. The only information I can find on this is an ion cotransporter for sodium potassium and chloride ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway the structure of a single monomer of this would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{K+}$ $\ce{Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also thought of this as a possibility:&#xA;$$\ce{Na3ClO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks very similar to sodium hypochlorite in terms of its formula but has a different structure. The structure of it would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{Na+}$ $\ce{O^2-}$ $\ce{Na+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that naming mixed ionic compounds is not easy unlike naming simple ionic compounds(which itself can be hard if you don't know the name of a particular polyatomic ion). But how would I go about naming these mixed ionic compounds?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="418" LastEditDate="2015-05-09T04:54:06.303" LastActivityDate="2019-01-02T02:46:03.207" Title="Naming ionic compounds with multiple cations and anions" Tags="&lt;ions&gt;&lt;nomenclature&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31083" PostTypeId="1" AcceptedAnswerId="32287" CreationDate="2015-05-09T05:57:21.693" Score="13" ViewCount="17333" Body="&lt;p&gt;According to the Wikipedia article &lt;a href=&quot;http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry#System&quot;&gt;IUPAC nomenclature of inorganic chemistry&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;he prefix bi- is a deprecated way of indicating the presence of a single hydrogen ion&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A very common example is the commonplace 'bicarb of soda', or sodium bicarbonate (or using its correct chemical name sodium hydrogen carbonate). On the Wikipedia page for &lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_bicarbonate#cite_note-7&quot;&gt;sodium bicarbonate&lt;/a&gt; there is a note that states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The prefix &quot;bi&quot; in &quot;bicarbonate&quot; comes from an outdated naming system and is based on the observation that there is two times as much carbonate ($\ce{CO3}$) in sodium bicarbonate ($\ce{NaHCO3}$) and other bicarbonates as in sodium carbonate ($\ce{Na2CO3}$) and other carbonates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the statement above is not particularly clear, especially as carbonates are not the only compounds that used the 'bi' prefix, such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bisulfite&quot;&gt;Bisulphite&lt;/a&gt; $\ce{HSO3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sodium_bisulfate&quot;&gt;Bisulphate&lt;/a&gt; $\ce{HSO4-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why was the prefix 'bi' used to indicate the presence of a single hydrogen ion in compounds?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T06:37:28.020" LastActivityDate="2020-01-29T13:34:16.267" Title="Why was the prefix 'bi' used in compounds, such as for bicarb of soda?" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31084" PostTypeId="1" AcceptedAnswerId="31085" CreationDate="2015-05-09T05:58:53.053" Score="5" ViewCount="2207" Body="&lt;p&gt;I've been searching for the conditions necessary for the reaction of sodium (Na) metal with cycloalkenes, cyclo-di-alkenes, cyclo-tri-alkenes all over the internet but I cannot find a proper answer. Can someone explain to me the necessary conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example why does sodium metal react with cyclopentadiene but not with cycloheptatriene?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4231" LastEditDate="2015-05-09T13:38:37.537" LastActivityDate="2015-05-09T13:38:37.537" Title="When, how and why does sodium metal react with cycloalkenes, cycloalkadienes, cycloalkatrienes?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31085" PostTypeId="2" ParentId="31084" CreationDate="2015-05-09T07:14:57.887" Score="2" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5tjI.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The negatively charged particles are conjugate bases of respective compounds, which (reactants) acts due to presence of active methylene group. Depends on the stabitily of conjugate base the reaction goes forward.&lt;/p&gt;&#xA;" OwnerUserId="15800" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-09T07:14:57.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31086" PostTypeId="2" ParentId="30910" CreationDate="2015-05-09T07:39:36.143" Score="-6" Body="&lt;p&gt;In Methoxy Benzoic Acid,we have Resonance or mesomeric effect which stabilizes the carboxylate anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Nitro Benzoic Acid, we have Intermolecular Attractions (Some may call it as Hydrogen Bonding) between the Nitrogen(of Nitro group) in one molecule and the Carboxylate Anion in another molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, since the Intermolecular Attraction is more stronger than the Resonance effect, the Nitro benzoate is more stabilized than Methoxy Benzoate. So, Nitro Benzoic Acid is more acidic.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SOgDW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16079" LastEditorUserId="16079" LastEditDate="2015-05-09T10:34:44.423" LastActivityDate="2015-05-09T10:34:44.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31088" PostTypeId="1" AcceptedAnswerId="31093" CreationDate="2015-05-09T10:00:36.757" Score="2" ViewCount="794" Body="&lt;p&gt;I've done some research on google but most guides I've seen recommend adding salt to the mixture for better conductance which results in the release of chlorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to perform electrolysis on water that releases hydrogen and at the same time does not release any gases that are harmful to humans?&#xA;By that I mean can I use another electrolyte that does not produce harmful side effects (so sulfuric acid is also out, and I do realize it's next to impossible to do electrolysis on distilled water.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a way to show this experiment to kids in enclosed space without good ventilation.&lt;/p&gt;&#xA;" OwnerUserId="16084" LastEditorUserId="402" LastEditDate="2017-01-14T08:55:02.570" LastActivityDate="2017-01-14T08:55:02.570" Title="Safe way to extract hydrogen from water through electrolysis" Tags="&lt;safety&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ClosedDate="2015-05-09T13:49:02.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="31089" PostTypeId="1" CreationDate="2015-05-09T10:01:43.947" Score="2" ViewCount="12725" Body="&lt;p&gt;Here's something that makes me confuse. What will happen to the total no of moles in a system if the temp is raised. Will it increase? If so how is that possible.?&#xA;Say we have an equlibrium,&#xA;$$\ce{2P &amp;lt;=&amp;gt; Q + R}$$ &#xA;The temp is $300\:\mathrm{K}$ and total no of moles is 6 moles. &#xA;If we increase the temp upto $500\:\mathrm{K}$ will the total no of moles change to 10 moles when the volume and pressure kept constant?&lt;/p&gt;&#xA;" OwnerUserId="16085" LastEditorUserId="4945" LastEditDate="2015-05-09T10:16:57.067" LastActivityDate="2015-05-09T10:55:40.727" Title="What is the effect of the temperature on the no of moles in an equilibrium system?" Tags="&lt;equilibrium&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31091" PostTypeId="1" CreationDate="2015-05-09T10:50:02.053" Score="2" ViewCount="486" Body="&lt;p&gt;My teacher taught me the formulas 'nS' and 'nS+(n-1)d' to find the highest and lowest oxidation state of transition elements. He gave the example of $\ce{Mn}$ and showed that the numbers were +2 and +7. But I did not understand it. So can please you give me&#xA;a good explanation.&lt;/p&gt;&#xA;" OwnerUserId="16086" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T11:00:30.233" LastActivityDate="2015-05-09T11:00:30.233" Title="How to find the highest and lowest oxidation state of transitional metal" Tags="&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31092" PostTypeId="2" ParentId="31089" CreationDate="2015-05-09T10:55:40.727" Score="1" Body="&lt;p&gt;Well, your equation contains the same number of moles either side, so temperature does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the equation had different moles on either side, it should be given if the reaction is endothermic or exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eg- if a reaction is exothermic and the temperature is increased, then according to Le Chatelier's principle, the reaction should go in the backward direction (To compensate for the increase in temperature).&#xA;So, if the reactant side has more moles than product side, the number of moles will be increased. You can apply similar logic if the reaction is endothermic.&lt;/p&gt;&#xA;" OwnerUserId="16072" LastActivityDate="2015-05-09T10:55:40.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31093" PostTypeId="2" ParentId="31088" CreationDate="2015-05-09T11:27:57.107" Score="3" Body="&lt;p&gt;Some sort of advice, since you are concerned with safety and demonstrating the experiment to children. You should generally not perform an experiment in a badly ventilated space. Please be aware, that the generated hydrogen gas is highly flammable and with the addition of the right amount of oxygen, it can make very loud booms: &lt;a href=&quot;https://chemistry.stackexchange.com/q/8498/4945&quot;&gt;How explosive is hydrogen gas?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two questions, that are very closely related and already contain the answer to your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/13874/&quot;&gt;Can I perform electrolysis without electrolyte? (with distilled water)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/2806/4945&quot;&gt;How can I produce hydrogen gas efficiently and cheaply?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general any kind of salt as an electrolyte will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In low concentrations about 1-2 mol/L they are usually quite harmless. I personally would prefer diluted sulfuric acid. Any addition of group 2 salts may cause passivation of the electrodes by trapping carbon dioxide from the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diluted sodium hydroxide or potassium hydroxide are also safe to handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the safest is sodium sulfate, but it might not be the best performer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please wear safety goggles at all times during the experiment, best case: The children wear them, too.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-09T11:27:57.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31094" PostTypeId="1" AcceptedAnswerId="43364" CreationDate="2015-05-09T12:04:56.923" Score="18" ViewCount="1111" Body="&lt;p&gt;Why does the enolate of the nitrile attack the alkyl chloride from the bond shown in the picture and not from the far end pi bond?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iQrmD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2016-01-07T23:47:30.020" LastActivityDate="2016-02-03T23:53:27.687" Title="Alkylation of conjugated nitriles - regioselectivity" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31095" PostTypeId="1" AcceptedAnswerId="31097" CreationDate="2015-05-09T12:22:44.657" Score="4" ViewCount="244" Body="&lt;p&gt;I have observed this effect by rubbing together two pieces of quartz in a dark room but am unable to explain this effect. Would it be something to do with the way quartz is bonded as a covalent network?&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="402" LastEditDate="2015-05-10T05:59:37.283" LastActivityDate="2015-05-10T06:13:50.607" Title="What causes triboluminescence and what process occurs to produce light?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31096" PostTypeId="2" ParentId="29711" CreationDate="2015-05-09T12:48:00.860" Score="1" Body="&lt;p&gt;The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I think you were starting from a question like the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How much $\ce{C7H4O}$ would you have to weigh in for it to fully react&#xA;  with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by&#xA;  the following equation? $$ \ce{C7H4O + 6KNO3 -&amp;gt; Fumes^  + Tar} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is &#xA;$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the result is composed as follows:&#xA;$$\begin{align}&#xA;n_{\ce{C7H4O}} &amp;amp;= \frac{1}{6} n_{\ce{KNO3}} \\&#xA;\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &amp;amp;= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline&#xA;m_{\ce{C7H4O}} &amp;amp;= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly, I am so used to &quot;my&quot; method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-05-09T12:48:00.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31097" PostTypeId="2" ParentId="31095" CreationDate="2015-05-09T12:56:08.817" Score="3" Body="&lt;p&gt;I'm only a first year chemical engineering student, and we haven't really covered this topic in our course yet, but I will try and answer your question to the best of my knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am sure people more qualified me would provide better responses.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we apply certain stresses to a material such as a crystal, we separate charges (which is why it usually is asymmetric... the presence of anistropy allows charge separation.) These charges then can cause electric discharge, which ionizes the surrounding air and releases a flash of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist symmetric crystals which display this phenomenon as well... and it is suggested that (probably) impurities exist within such crystals which break the symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, this is my understanding of the phenomenon, and I don't claim is perfect in any way, but hopefully is helpful to you.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="402" LastEditDate="2015-05-10T06:13:50.607" LastActivityDate="2015-05-10T06:13:50.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31101" PostTypeId="2" ParentId="30982" CreationDate="2015-05-09T14:04:58.810" Score="7" Body="&lt;p&gt;To complement the answer provided by @Ian Fang and edited by @santiago, sodium hydride's action as a Brønsted base differs from other bases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Normal&quot; acid base reactions are reactions of an equilibrium, like the &lt;em&gt;partial&lt;/em&gt; deprotonation of malonic esters by 5 m% sodium carbonate in some Knoevenagel reactions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R-H} + \mbox{base} \rightleftharpoons{} \ce{R-} + \mbox{protonated base}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deploying sodium hydride as base however often intends a &lt;em&gt;quantitative&lt;/em&gt; deprotonation.  The equilibrium is driven to the right hand side, in favour of the protonated base ($\ce{H2}$) immediately leaving the reaction mixture -- so the back reaction can't take place.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="5026" LastEditDate="2018-03-04T02:57:00.933" LastActivityDate="2018-03-04T02:57:00.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31102" PostTypeId="1" CreationDate="2015-05-09T14:31:04.853" Score="12" ViewCount="2079" Body="&lt;p&gt;What is its structure? I found an image of $\ce{B6O11}$ unit, but it looks incorrect:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xN4Lg.gif&quot; alt=&quot;unit&quot;&gt; The top left corner one.   &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The oxygen can make max 2 bonds but atom $\mathrm{O8}$ has three!    &lt;/li&gt;&#xA;&lt;li&gt;There are six oxygen with only one bond, that is the unit has $-6$ net charge, but $\ce{2Ca^2+} = +4$ positive charge.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16093" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-12T19:18:49.160" LastActivityDate="2017-08-12T19:29:30.940" Title="Structure of Ca2B6O11" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;structural-formula&gt;&lt;minerals&gt;&lt;crystallography&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31103" PostTypeId="1" CreationDate="2015-05-09T14:32:48.663" Score="-3" ViewCount="266" Body="&lt;p&gt;How do I predict the spontaneity of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Metathesis_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;metathesis reaction&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a matter of summing up the differences between their ions' standard reduction potentials? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_metathesis_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;salt metathesis reaction&lt;/a&gt; between two dissolved electrolytes; sodium chloride and silver nitrate (highlighted &lt;a href=&quot;http://en.wikipedia.org/wiki/Salt_metathesis_reaction#Aqueous_metathesis_.28precipitation.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). While the &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;HSAB theory&lt;/a&gt; exists, it has not been quantified yet (as far that I know of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based my current knowledge of Chemistry, I would expect something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the symbol of each element represent its standard reduction potential. Then spontaneity would be given by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{((-Na) - (-Ag)) + (Cl - NO3)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a positive value would imply that the reaction is spontaneous, and the magnitude the degree of spontaneity.&lt;/p&gt;&#xA;" OwnerUserId="14911" LastEditorUserId="402" LastEditorDisplayName="user15489" LastEditDate="2017-04-01T00:16:48.897" LastActivityDate="2017-04-01T00:16:48.897" Title="How do I predict the spontaneity of an ionic metathesis reaction?" Tags="&lt;thermodynamics&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2017-04-07T04:52:33.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="31104" PostTypeId="2" ParentId="31102" CreationDate="2015-05-09T14:52:16.543" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The oxygen can make max 2 bonds but $\ce{O8}$ has three!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will have better idea if you see 2D structure instead of 3D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9sOuP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can clearly see no $\ce{O}$ atom have more then 3 bonds! &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T02:03:03.873" LastActivityDate="2015-06-09T02:03:03.873" CommentCount="0" CommunityOwnedDate="2015-05-09T14:52:16.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="31105" PostTypeId="1" CreationDate="2015-05-09T15:04:38.787" Score="8" ViewCount="1065" Body="&lt;p&gt;Is there some kind of database that has information about phase diagrams of different substances? I'm mostly looking for phase diagrams of elements (silicon, iron, hydrogen, etc.) because I'm trying to write a planetary generator, but I cannot find any information about the mathematics of a phase diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better yet, is there some way to simply approximate the phase diagrams of different compounds? I'm looking for simulation complexity and variation more than I want realism.&lt;/p&gt;&#xA;" OwnerUserId="16094" LastActivityDate="2015-05-21T13:44:03.433" Title="Open Database of Phase Diagrams?" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31106" PostTypeId="2" ParentId="31103" CreationDate="2015-05-09T15:13:58.123" Score="1" Body="&lt;p&gt;No, the spontaneity of the reaction does not have to do with the reaction potentials. All of the constituent parts of the salts exist as ions even in solid form, so there is no exchange of electrons occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, the insolubility of a salt is determined by the strength of the electrostatic forces between the ions. In some salts, the strength of the attractive forces that water applies is enough to break these ions apart, dissolving the salt. In other salts, the forces applied by the water might not be enough. These kinds of salts are those that are considered to be insoluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to predict whether or not a reaction between salts occurs (i.e. something precipitates out of solution), you should memorize the &lt;a href=&quot;http://www.chem.sc.edu/faculty/morgan/resources/solubility/&quot; rel=&quot;nofollow&quot;&gt;solubility rules&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-09T15:13:58.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31107" PostTypeId="1" AcceptedAnswerId="31129" CreationDate="2015-05-09T15:25:44.470" Score="9" ViewCount="1291" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wsBvV.jpg&quot; alt=&quot;Question 3b iii) &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 double bonds $\ce{C=C}$ present within each piperic acid molecule,  thus when piperic acid reacts with hot excess $\ce{KMnO4}$, it oxidised into a carboxyllic acid. Why is $\ce{CO2}$ produced?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon atom bonded directly to the benzene ring will get oxidised into a carboxylic acid and also that the carbon bonded to the $\ce{COOH}$ group will get oxidized also to a carboxylic acid to form a dicarboxylic acid. What changes will take place to the other 2 carbon atoms.&lt;/p&gt;&#xA;" OwnerUserId="14862" LastEditorUserId="46306" LastEditDate="2018-02-20T10:58:35.920" LastActivityDate="2018-02-20T10:58:35.920" Title="Piperic acid oxidation" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31108" PostTypeId="2" ParentId="31078" CreationDate="2015-05-09T15:30:19.813" Score="7" Body="&lt;p&gt;The process to determine whether two protons are enantiotopic or diastereotopic is to replace each with deuterium and look at the relationship of the resulting compounds. For the indicated hydrogens in 2-methylindane, replacement of each gives a pair of diastereomeric compounds (Figure, top). The compound also contains enantiotopic protons (Figure, center). The carbon where the replacement happens doesn't necessarily become a stereocenter in this process (Figure, bottom).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/QSRar.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-05-09T15:30:19.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31109" PostTypeId="1" CreationDate="2015-05-09T15:32:10.320" Score="6" ViewCount="216" Body="&lt;p&gt;Galvanic corrosion occurs of an active metal (i.e. iron or zinc) in contact with a passive metal (i.e. copper) in a conductive solution. A famous example is the corrosion of iron girders in contact with copper in the statue of liberty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the difference in potential can account for the protection of copper, it does not explain the girders in had a &lt;em&gt;greatly accelerated&lt;/em&gt; corrosion rate. In a zinc/copper battery there is a solution of copper sulfate that gets reduced at the copper electrode. However, if all the copper is in metallic form in the first place, no galvanic reaction can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems galvanic corrosion in real-life, does not increase total corrosion, but rather &lt;em&gt;concentrates&lt;/em&gt; it (at lest for pairs of metals with similar non-galvanic corrosion rates). Copper sheets have a large surface area that initially reacted non-galvanically to form the green color we see. Some of these ions are then reduced at the expense of the oxidizing the girders. The statue of liberty and other infamous examples of corrosion involve large amounts of inactive metals connected to small amounts of active (iron/steel) girders/fasteners/etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This logic predicts that the attaching a copper bar to an iron bar &lt;em&gt;of equal surface area&lt;/em&gt; will only double it's corrosion rate. That being said, predicting the corrosion rate is not easy so there could be some other effect I am overlooking. Is the surface area logic correct or would there a large increase in the total corrosion?&lt;/p&gt;&#xA;" OwnerUserId="1211" LastActivityDate="2015-06-09T05:37:08.153" Title="galvanic corrosion in the wild: where is the reduction?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31110" PostTypeId="2" ParentId="12332" CreationDate="2015-05-09T16:04:33.907" Score="3" Body="&lt;p&gt;You need a salt bridge / barrier to maintain electrical contact (without metal) but &lt;strong&gt;to prevent copper from plating onto the $\ce{Zn}$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A super-saturation of &lt;em&gt;metallic copper&lt;/em&gt;&lt;/strong&gt; would produce a small voltage in a homogenous electrolyte with but $\ce{Zn}$ and $\ce{Cu}$ ions. For any $\ce{Cu}$-positive voltage it is be energetically favorable for $\ce{Cu}$ to plate onto the $\ce{Zn}$. However, there is activation energy in nucleating a new phase ($\ce{Cu}$ metal). Thus we will get super-saturated: Conceptually, the &quot;concentration&quot; of $\ce{Cu}$ metal at the Zn electrode &lt;em&gt;exceeds&lt;/em&gt; the concentration of $\ce{Cu}$ metal at the copper electrode before the $\ce{Cu}$ phase appears! However, once $\ce{Cu}$ plating begins, the voltage would drop to zero since the cell is now &quot;shorted&quot;. Copper will continue to build up and zinc will erode away but no voltage will appear. If this nucleation voltage never is reached (preventing voltage-buildup requires an electrical connection between the two strips), you can get a small amount of energy from a cell in a homogenous solution, but it probably will be much less than 1.1V, and maintaining a low enough voltage would require current to keep flowing.&lt;/p&gt;&#xA;" OwnerUserId="1211" LastEditorUserId="1211" LastEditDate="2015-05-09T16:44:52.663" LastActivityDate="2015-05-09T16:44:52.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31111" PostTypeId="1" CreationDate="2015-05-09T16:12:07.307" Score="0" ViewCount="379" Body="&lt;p&gt;I am a little confused about this multiple choice question and what the requirements are to answer it. It is for a year 11 chemistry student that I am helping and I think I am going in a little too much detail to try to solve it to help them. I am sure that I am overthinking this. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The energies needed to remove electrons in succession from a neutral&#xA;  atom of a certain element &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; are given (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{kJ\cdot  mol^{-1}}$&lt;/span&gt;): &lt;span class=&quot;math-container&quot;&gt;$1010$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$2032$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$7056$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$8120$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$9300$&lt;/span&gt;.&lt;br&gt;&#xA;   The formula  of the oxide of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; is:&#xA;   &lt;span class=&quot;math-container&quot;&gt;$\ce{AO}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{A2O}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{A2O3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{A2O4}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{A3O5}$&lt;/span&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My understanding of the question and what I was trying to do to solve the problem by looking at the ratios between the energies and I was trying to apply&lt;br&gt;&#xA;1st ionization energy: &lt;span class=&quot;math-container&quot;&gt;$$\ce{X -&amp;gt; X+ + e−},$$&lt;/span&gt; &#xA;2nd ionization energy: &lt;span class=&quot;math-container&quot;&gt;$$\ce{X+ → X^2+ + e−},$$&lt;/span&gt; &#xA;and 3rd ionization energy: &lt;span class=&quot;math-container&quot;&gt;$$\ce{X^2+ -&amp;gt; X^3+ + e−}$$&lt;/span&gt; &#xA;to the energies below and find a ratio. I thought the ratio was going to be c. I understood that oxide is &lt;span class=&quot;math-container&quot;&gt;$2-$&lt;/span&gt;, however, I couldn't calculate the &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; value as I didn't know what the element was. &lt;/p&gt;&#xA;" OwnerUserId="16095" LastEditorUserId="4945" LastEditDate="2018-11-06T16:28:41.553" LastActivityDate="2018-11-06T16:28:41.553" Title="How to find the formula of an binary oxide based on the ionisation energies?" Tags="&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-09T20:19:48.873" ContentLicense="CC BY-SA 4.0" />
  <row Id="31113" PostTypeId="1" AcceptedAnswerId="31124" CreationDate="2015-05-09T16:31:12.940" Score="29" ViewCount="8962" Body="&lt;p&gt;I was studying about the periodic table recently, and was reading a topic associated with oxides of halogens, and came across the following line&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bromine oxides, $\ce{Br2O}$, $\ce{BrO2}$, $\ce{BrO3}$ are the least stable halogen oxides (&lt;strong&gt;Middle row anomaly&lt;/strong&gt;) and exist only at low temperatures. They are very powerful oxidizing agents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I went to search this on Google, and found these lines&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It refers to the instability of oxides of bromine as compared to relative stability of oxides of chlorine and Iodine at room temperature, the former being stable only at low temperatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the line above is simply restating what the book had already told. &#xA;So, is their any specific reason why this happens, or is this only due to the experimental data we have gathered?&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-24T09:33:07.823" LastActivityDate="2018-02-24T09:33:07.823" Title="&quot;Middle row anomaly&quot; of the periodic table" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31114" PostTypeId="5" CreationDate="2015-05-09T16:31:57.877" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-09T16:31:57.877" LastActivityDate="2015-05-09T16:31:57.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31115" PostTypeId="4" CreationDate="2015-05-09T16:31:57.877" Score="0" Body="For questions about the reduction potential (A measure of the tendency of a chemical species to acquire electrons and thereby be reduced) or its applications. Also see the tag [electrochemistry]." OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-20T17:54:02.840" LastActivityDate="2018-03-20T17:54:02.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31116" PostTypeId="2" ParentId="31111" CreationDate="2015-05-09T16:37:49.950" Score="1" Body="&lt;p&gt;The point of this problem is to test if you can see where there is a jump in ionization energies. The third ionization energy is considerably larger than the others. Although ionization energies will naturally increase in succession, this is the most profound increase, which would suggest that this atom will tend to lose 2 electrons before obtaining a stable octet structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean that its ion would be $\ce{A^{2+}}$. Since the oxide ion is $\ce{O^{2-}}$, the charges will be balanced if the ratio of the ions is 1:1, meaning the answer is $\ce{AO}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-09T16:37:49.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31117" PostTypeId="5" CreationDate="2015-05-09T16:40:50.730" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-09T16:40:50.730" LastActivityDate="2015-05-09T16:40:50.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31118" PostTypeId="4" CreationDate="2015-05-09T16:40:50.730" Score="0" Body="For questions related to the molecular structure of biological macromolecules (especially proteins and nucleic acids), how they acquire the structures they have, and how alterations in their structures affect their function." OwnerDisplayName="user15489" LastEditorUserId="402" LastEditorDisplayName="user15489" LastEditDate="2017-04-06T23:03:18.937" LastActivityDate="2017-04-06T23:03:18.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31119" PostTypeId="5" CreationDate="2015-05-09T16:42:53.727" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-09T16:42:53.727" LastActivityDate="2015-05-09T16:42:53.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31120" PostTypeId="4" CreationDate="2015-05-09T16:42:53.727" Score="0" Body="Questions related to abstract one-dimensional coordinates which represent progress along a reaction pathway." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T17:21:50.827" LastActivityDate="2015-05-09T17:21:50.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31121" PostTypeId="2" ParentId="29534" CreationDate="2015-05-09T17:01:07.660" Score="13" Body="&lt;p&gt;The formation of the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ due to addition of ammonia has mainly three functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt;&lt;br&gt;&#xA;The Tollens test takes place under alkaline aqueous conditions. The complexation of $\ce{Ag+}$ with $\ce{NH3}$ prevents precipitation of brown $\ce{Ag2O}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Ag+ + 2OH- &amp;lt;=&amp;gt; 2 AgOH &amp;lt;=&amp;gt; Ag2O + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;&lt;br&gt;&#xA;The Tollens test shall be selective for aldehydes. $\ce{[Ag(NH3)2]+}$ is a milder oxidising agent than $\ce{Ag+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{Ag+ + e- \;&amp;amp;&amp;lt;=&amp;gt; Ag}\quad &amp;amp;&amp;amp;E^\circ = +0.799\ \mathrm{V}\\&#xA;\ce{[Ag(NH3)2]+ + e- \;&amp;amp;&amp;lt;=&amp;gt; Ag + 2 NH3}\quad &amp;amp;&amp;amp;E^\circ = +0.373\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in redox potential can be explained using the stability constant of $\ce{[Ag(NH3)2]+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{B} = \frac{\left[\ce{[Ag(NH3)2]+}\right]}{\left[\ce{Ag+}\right]\left[\ce{NH3}\right]^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;E&amp;amp;=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Ag+}\right]\\ &#xA;&amp;amp;=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{\left[\ce{[Ag(NH3)2]+}\right]}{K_\text{B}\cdot\left[\ce{NH3}\right]^2} \\ &#xA;E_{\ce{[Ag(NH3)2]+}}^\circ&amp;amp;=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{1}{K_\text{B}} \\&#xA;&amp;amp;\approx E_{\ce{Ag+}}^\circ-0.059\ \mathrm{V}\cdot\log K_\text{B}&#xA;\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt;&lt;br&gt;&#xA;The complex formation equilibrium slows down the overall reaction. A slow, controlled reaction is important for creating the desired silver mirror. If $\ce{Ag+}$ is reduced too quickly, colloidal silver metal would appear, which would create a black cloudy liquid.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-09T17:01:07.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31122" PostTypeId="2" ParentId="31079" CreationDate="2015-05-09T17:23:43.073" Score="4" Body="&lt;p&gt;I propose you consult the recommendations of the International Union of Applied Chemistry (IUPAC), in particular &lt;a href=&quot;http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Red Book&lt;/a&gt; as reference.  Obviously, you deviated for a reason from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hill_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hill formula&lt;/a&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaKCl2}$&lt;/span&gt; (that were &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2KNa}$&lt;/span&gt;) some data bases adhere to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already presented in rules IR 4.4.2.1, and IR 4.4.2.2, the order of the elements within the formula may be arranged either by their relative electronegativity (starting by the element of least electronegativity), or by alphanumeric order, respectively.  So rules in section IR 5.4 subsequently discriminate between electropositive constituents to be named prior to electronegative ones in the &quot;overall name&quot;.  In addition (quote):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The order of citation is alphabetical within each class of constituents (multiplicative prefixes being ignored), except that hydrogen is always cited last among electropositive constituents if actually classified as&#xA;an electropositive constituent.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reference provides numerous examples of application useful to look at, too.  &lt;a href=&quot;http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf#page=88&quot; rel=&quot;nofollow noreferrer&quot;&gt;Page 76&lt;/a&gt; (of the book, not the pdf) is a good place to start.  As in nomenclature of organic compounds, there are instances you want to point the attention not only to the stoichiometric composition of the compound, rather than to different constitutional isomers.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="69490" LastEditDate="2019-01-02T02:46:03.207" LastActivityDate="2019-01-02T02:46:03.207" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31123" PostTypeId="1" CreationDate="2015-05-09T17:29:05.277" Score="3" ViewCount="175" Body="&lt;p&gt;What is the ground state conformation of dimethyl disulfide ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two which seem reasonable to me are trans to avoid methyl-methyl gauche interactions or gauche due to sulfur donation into the S-Me sigma antibonding orbital. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a discussion in &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j100547a013&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; but it doesn't seem to really answer the question definitely. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-09T17:29:05.277" Title="Ground state conformation of dimethyl disulfide" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31124" PostTypeId="2" ParentId="31113" CreationDate="2015-05-09T17:34:09.063" Score="23" Body="&lt;p&gt;This is due to the transition metal contraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-09T17:34:09.063" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="31125" PostTypeId="1" AcceptedAnswerId="31260" CreationDate="2015-05-09T17:36:27.483" Score="5" ViewCount="7317" Body="&lt;p&gt;I have a base $\ce{M(OH)_{3}}$ having 3 $\mathrm{p}K_{\mathrm{b}}$ values as,&#xA;$$&#xA;\mathrm{p}K_{\mathrm{b}_{1}} = 0.5\\&#xA;\mathrm{p}K_{\mathrm{b}_{2}} = 3.7\\&#xA;\mathrm{p}K_{\mathrm{b}_{3}} = 7.7\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a strong acid is given as the titrant and phenolphthalein and methyl orange are given as possible indicators, how can I choose an indicator using above $\mathrm{p}K_{\mathrm{b}}$ values ?&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="15410" LastEditDate="2016-04-23T01:37:43.190" LastActivityDate="2016-04-23T01:37:43.190" Title="How can I choose an indicator for determining the concentration of a base using its pKb values?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31127" PostTypeId="2" ParentId="31075" CreationDate="2015-05-09T20:58:47.753" Score="17" Body="&lt;p&gt;The dipole moment $\mu$ of a molecule is a measure of charge distribution in the molecule and the polarity formed by the nuclei and electron cloud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can perturb our system with an external electric field $\vec E$ and gauge the response of the electron cloud and nuclei by the polarisability, i.e how much the dipole moment changes. In practice the nuclei might be so heavy that their motion is not perturbed, while electrons being light are very mobile. If we imagine that the external field is caused by some other species, and that it itself is not changing, we can call this external constant electric field $\vec E$ at least over the volume of the molecule we are considering. Imagine for arbitrary book keeping that we point it down the $z$ axis. We could also investigate how the dipole moment changes with bond vibrations to discuss IR spectroscopy or if the polarisability changes during a vibration to give Raman spectroscopy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Perturbation_theory&quot; rel=&quot;noreferrer&quot;&gt;perturbation theory&lt;/a&gt; to expand the wavefunction and the molecular energy in terms of small perturbations of the field. We start by &lt;a href=&quot;http://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;noreferrer&quot;&gt;Taylor expanding&lt;/a&gt; the energy and molecular wavefunction in terms of the electric field which acts as the perturbation parameter. &#xA;\begin{equation}&#xA;E(\vec E)=E^0+\bigg(\frac{\partial E}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2E}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3E}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots &#xA;\end{equation}&#xA;\begin{equation}&#xA;\psi(\vec E)=\psi^0+\bigg(\frac{\partial \psi}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2\psi}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3\psi}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots &#xA;\end{equation}&#xA;If we use the notation that the wavefunction derivatives are given by;&#xA;\begin{equation}&#xA;\bigg(\frac {1}{i!}\bigg)\frac{\partial ^i\psi}{\partial \vec E^i}=\psi ^{(i)}&#xA;\end{equation}&#xA;The Hamiltonian for such a system under the influence of an electric field in the $z$ direction is $\hat H(\vec E)$. &#xA;\begin{equation}&#xA;\hat H(\vec E)=\hat H^0-\vec E\hat \mu _z&#xA;\end{equation}&#xA;Where $\hat \mu _z$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Dipole#Quantum_mechanical_dipole_operator&quot; rel=&quot;noreferrer&quot;&gt;dipole moment operator&lt;/a&gt; that is a summation of the charges of the nuclei and electrons in the molecule. This is the result of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hellmann%E2%80%93Feynman_theorem&quot; rel=&quot;noreferrer&quot;&gt;Hellmann-Feynman theorem&lt;/a&gt; of the energy and the electric field, e.g. $\hat H(\vec E)=\hat H^0 +\hat H^1(\vec E)$,&#xA;\begin{equation}&#xA;\frac{dE}{d\vec E}=\bigg\langle \frac{d\hat H}{d\vec E}\bigg\rangle=\bigg\langle \frac{d(-\mu _z\vec E)}{d\vec E}\bigg\rangle &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time-independent Schrödinger equation is now,&#xA;\begin{equation}&#xA;\hat H(\vec E)\psi(\vec E)=E(\vec E)\psi (\vec E)&#xA;\end{equation}&#xA;With an energy;&#xA;\begin{equation}&#xA;E(\vec E)=\big \langle \psi (\vec E)\big|\hat H(\vec E)\big|\psi (\vec E)\big \rangle=\big\langle \psi ^{(0)}+\psi ^{(1)}\vec E+\psi ^2 \vec E^2+ \dots \big|\hat H-\vec E\hat \mu _z \big|\psi ^{(0)}+\psi ^{(1)}\vec E+\psi^2\vec E^2+\dots \big \rangle&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little algebra and use of $E^{(0)}=\langle \psi ^{(0)}|\hat H^0 |\psi ^{(0)}\rangle$, as well as the Hermitian properties of the Hamiltonian. &#xA;\begin{equation}&#xA;E(\vec E)=E^{(0)}+2\vec E\big \langle \psi ^{(1)}\big|\hat H^{(0)}\big|\psi ^{(0)}\rangle -\vec E\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big \rangle +\mathcal O(\vec E^2)&#xA;\end{equation}&#xA;Using the Schrödinger equation $H\psi =E\psi$ and pulling the scalar energy out of the integral;&#xA;\begin{equation}&#xA;\big\langle \psi ^{(1)}\big|\hat H^0\big|\psi ^{(0)}\rangle =E^{(0)}\big\langle \psi ^{(1)}\big|\psi ^{(0)}\big \rangle &#xA;\end{equation}&#xA;Since $\langle \phi |\psi \rangle=0 $,&#xA;\begin{equation}&#xA;E(\vec E)=E^{(0)}-\vec E\big \langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle&#xA;\end{equation}&#xA;Therefore the expectation value of the dipole moment along the $z$ axis for a molecular state $\psi ^{(0)}$ is $\langle \mu _z \rangle$;&#xA;\begin{equation}&#xA;\langle \mu _z\rangle =\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle &#xA;\end{equation}&#xA;To understand the strength of the interaction that causes the transition between the states $\psi ^{(0)}$ and $\psi ^{(1)}$ we use the transition dipole moment which is basically exactly the same but there is a wavefunction from both of the states involved, initial and final!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to repeat this process but retain higher orders (Messy!) you would get (2nd order) the polarisability of the molecule which in essence is the susceptibility of the electron cloud to change with respect to an external electric field (so how the dipole moment changes. Third order would give the hyperpolarisability etc., etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, you could also approach this from a really different angle by interpreting the molecular orbital diagram and using computational chemistry (so &lt;a href=&quot;http://en.wikipedia.org/wiki/Variational_method_(quantum_mechanics)&quot; rel=&quot;noreferrer&quot;&gt;variational principle&lt;/a&gt; etc) to find the molecular orbital coefficients! That would give you a good idea of what is going on!&lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="9961" LastEditDate="2017-05-19T15:33:02.090" LastActivityDate="2017-05-19T15:33:02.090" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31128" PostTypeId="1" AcceptedAnswerId="31132" CreationDate="2015-05-09T21:02:38.207" Score="6" ViewCount="1624" Body="&lt;p&gt;When naming branched alkenes/alkynes, the numbers of the carbons where the branches and multiple bonds are located are supposed to be the lowest possible.  Do the branches take precedence over the multiple bonds, as for example, 2-methyl-3-pentyne?  If the counting of the carbons for this molecule is flipped so that the triple bond is lower in number, it would be 3-methyl-2-pentyne.  Is the first way correct?&lt;/p&gt;&#xA;" OwnerUserId="16099" LastEditorUserId="4231" LastEditDate="2015-05-10T16:22:22.447" LastActivityDate="2015-05-10T18:18:41.517" Title="Nomenclature Precedence Order for Unsaturation and Hydrocarbon Chain" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31129" PostTypeId="2" ParentId="31107" CreationDate="2015-05-09T22:00:17.167" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;this paper stated that $\ce{CO2}$   could be formed as well, how?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually when we consider permanganate oxidations we think of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;oxidation of alkyl benzenes to benzoic acid&lt;/li&gt;&#xA;&lt;li&gt;cleavage and oxidation of a double bond to an acid, through the aldehyde, if there is one hydrogen on the double bond&lt;/li&gt;&#xA;&lt;li&gt;cleavage and oxidation of a double bond to a ketone if there are no hydrogens on the double bond&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, there are a few &lt;strong&gt;special cases&lt;/strong&gt; such as formic acid and oxalic acid, or their corresponding aldehydes.  These compounds are oxidized by permanganate to $\ce{CO2}$ (for an idea as to why these compounds react differently than, say, acetic acid, see Scheme 1 &lt;a href=&quot;http://www.researchgate.net/profile/Laszlo_Kotai/publication/257955325_Kinetics_and_mechanism_of_oxidation_of_formic_and_oxalic_acids_by_bis(pyridine)silver_permanganate/links/0c96052c7cdb355b72000000.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information in hand we can re-examine the permanganate oxidation of piperic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9wyAd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that oxalic acid is one of the primary products produced in this reaction.  As discussed above, oxalic acid can then undergo further reaction to produce carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-09T22:00:17.167" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31132" PostTypeId="2" ParentId="31128" CreationDate="2015-05-09T23:45:11.050" Score="3" Body="&lt;p&gt;Unsaturated bonds take precedence over &quot;branches&quot;. Consider these bonds as functional groups and name the compound, it will make you number carbons correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the compound you mentioned above can't possibly exist as it has 5 bonds on $\ce{C_3}$.&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorDisplayName="user15489" LastEditDate="2015-05-09T23:45:59.307" LastActivityDate="2015-05-09T23:45:59.307" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31133" PostTypeId="1" AcceptedAnswerId="31167" CreationDate="2015-05-09T23:52:28.763" Score="9" ViewCount="3537" Body="&lt;p&gt;In objects made of plastic (and other polymers), how many molecules would one generally expect to find?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like, say, a plastic shopping bag or PET bottle couldn't be a single molecule, but on the other hand, that's one of the strengths of polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, I suppose, do items made of plastic consist of a single molecule, or many molecules fused together by way of glues or mechanical action?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Following a comment, I realise that the core of my question could be clearer, so here's an alternate phrasing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I chose an atom in a typical plastic object and gave it a colour, and then gave every atom connected to it by way of covalent bonds the same colour by flood fill, and then chose a colourless atom in the object and repeated the process until no more colourless atoms existed in the object, how many colours would a typical plastic object have?&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="5638" LastEditDate="2015-05-10T01:49:26.807" LastActivityDate="2019-06-14T01:57:23.783" Title="How many molecules are there typically in items made of plastics?" Tags="&lt;polymers&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31134" PostTypeId="1" AcceptedAnswerId="31705" CreationDate="2015-05-10T00:37:28.643" Score="15" ViewCount="715" Body="&lt;p&gt;I am a mechanical engineer, and just like everyone else I had classes on the crystal structure of metals, phase diagrams, and the various heat treatments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even after diving back into that recently and researching online, I can't make sense of how everything fits together. I know the high level, the low level, but can't link the two together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can the phase diagram of steel be linked to its physical properties? Say, given a certain carbon weight fraction, is it possible to tell from the phase diagram if the steel is going to be hard, resilient, ductile? E.g. cementite is hard and brittle, austenite is ductile and soft?&lt;/li&gt;&#xA;&lt;li&gt;How do heat treatments work? The way I see it, the phase diagram is &quot;steady state&quot;, and heat treatments are about spending a specific amount of time in a phase region to add &quot;a little bit of that&quot; etc. (or quickly cooling it to stop the reactions), but I can't link it back to the phase diagram (there is a 3rd dimension, time).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is not a very clear question, but this is exactly what I hope to find: clarification on a quite complex topic.&lt;/p&gt;&#xA;" OwnerUserId="16105" LastEditorDisplayName="user15489" LastEditDate="2015-05-10T00:38:49.927" LastActivityDate="2019-12-17T14:23:24.817" Title="What chemical characteristics give types of steel their properties?" Tags="&lt;metal&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31135" PostTypeId="1" AcceptedAnswerId="31144" CreationDate="2015-05-10T00:37:29.290" Score="3" ViewCount="112" Body="&lt;p&gt;Recently I've been looking at &lt;a href=&quot;http://www.chemteam.info/Equilibrium/Minimum-pH-required-for-precip.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;, which says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$K_{\mathrm{sp}}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(OH)3}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$6.70 \times 10^{-31}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Problem 1: What is the minimum &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; at which &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(OH)3}$&lt;/span&gt; will precipitate?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_{\mathrm{\mathrm{sp}}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{Cr^{3+}}]=s$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{OH^{-}}]=3s$$&lt;/span&gt;&#xA;Problem 2: What is the minimum &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; at which &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr(OH)3}$&lt;/span&gt; will precipitate if the solution has &lt;span class=&quot;math-container&quot;&gt;$[\ce{Cr^{3+}}] = 0.0670~\mathrm{M}$&lt;/span&gt;?&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_{\mathrm{sp}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$6.70 \times 10^{-31}=[0.0670~\mathrm{M}][s^{3}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't understand why the site wrote this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note that &lt;span class=&quot;math-container&quot;&gt;$3s$&lt;/span&gt; is not necessary. In example one, &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is not expressed in terms of some other unknown value (as it was in Problem 1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Can anyone tell me why?&lt;/p&gt;&#xA;&lt;p&gt;Also is also used to other equilibrium constant equation? What did I miss?&lt;/p&gt;&#xA;" OwnerUserId="16103" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-10T05:43:13.770" Title="Is solubility coefficient affected if ion data is given in Ksp?" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31138" PostTypeId="1" AcceptedAnswerId="31141" CreationDate="2015-05-10T02:05:43.710" Score="6" ViewCount="169" Body="&lt;p&gt;I'm trying to figure out a way to eat through concrete without damage pex tubing inside (its a radiant heated slab). Is there any way using some type of acid (hydrofluoric, muriatic ) to eat a 4 inch wide, 4 inch deep, and 71 foot long channel through the slab that could be tapped out with a blunt chisel? Pex tubing is resistant to most acids so it would not be damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this needs to be cost effective (under $1000).&lt;/p&gt;&#xA;" OwnerUserId="16108" LastEditorUserId="5591" LastEditDate="2015-11-13T19:12:48.087" LastActivityDate="2015-11-13T19:12:48.087" Title="Is it possible to create a 4 inch by 4 inch channel through concrete using acid?" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31140" PostTypeId="2" ParentId="31055" CreationDate="2015-05-10T02:50:54.840" Score="6" Body="&lt;p&gt;The Universe is full of continuous and smoothly-transitioning properties, yet which we often stubbornly try to assign a discrete yes/no answer in order to make the world easier to cope with. Metallicity is one of them. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_Arkel%E2%80%93Ketelaar_triangle&quot; rel=&quot;noreferrer&quot;&gt;Van Arkel-Ketelaar triangle&lt;/a&gt; is a diagram which attempts to show the transition between covalent, ionic and metallic types of bonding. Here is one populated by a variety of compounds:  &lt;img src=&quot;https://i.stack.imgur.com/BDPJ1.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer a substance is to one of the vertices, the more that vertex is important in describing the nature of bonding in the material. As you can see, the entire triangle is populated quite uniformly. There are certainly many more materials which would fit in the gaps between the materials shown, and even simply varying the ambient conditions (pressure, temperature, etc), a same substance can be put in different regions of the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this behaviour, it should be clear that any attempt to rigorously define a line dividing substances between metals and non-metals will be fraught with difficulty and likely not useful in the first place. No matter where the line is drawn, there will be some compounds that will look strange for &quot;barely being a metal&quot;, while others will be strange for &quot;barely being a non-metal&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-05-10T02:50:54.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31141" PostTypeId="2" ParentId="31138" CreationDate="2015-05-10T02:56:43.160" Score="5" Body="&lt;p&gt;Offhand, this seems dangerous to the structural integrity of the slab, because the acid would follow the pore structure of the concrete and weaken the wrong places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;concrete cutting saw rental&lt;/em&gt; is available; e.g. &lt;a href=&quot;http://www.unitedrentals.com/en/catalog/equipment-tools/concrete-masonry/concrete-saw-13-hp-propelled&quot; rel=&quot;noreferrer&quot;&gt;United Rentals&lt;/a&gt;&#xA;and &lt;a href=&quot;http://www6.homedepot.com/tool-truck-rental/Walk_Behind_Concrete_Saw_14/MK-CX3/index.html&quot; rel=&quot;noreferrer&quot;&gt;http://www6.homedepot.com/tool-truck-rental/Walk_Behind_Concrete_Saw_14/MK-CX3/index.html&lt;/a&gt;. Two 45-degree cuts ~6&quot; or deeper should do the job.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-10T02:56:43.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31142" PostTypeId="2" ParentId="31133" CreationDate="2015-05-10T03:07:14.513" Score="3" Body="&lt;p&gt;Some plastics have their molecular weight (m.w.) specified, e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow&quot;&gt;polyethylene&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultra-high-molecular-weight_polyethylene&quot; rel=&quot;nofollow&quot;&gt;ultra-high-molecular-weight polyethylene&lt;/a&gt;. Knowing the type of plastic, and therefore its m.w., and the mass of the object, using &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;nofollow&quot;&gt;Avogadro's constant&lt;/a&gt;, you should be able to calculate how many molecules there are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring a low-density polyethylene sandwich bag would be a convenient place to start.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-10T03:07:14.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31143" PostTypeId="2" ParentId="31083" CreationDate="2015-05-10T03:26:58.403" Score="1" Body="&lt;p&gt;The use of &quot;bi&quot; for 'two&quot; denotes that there are &lt;em&gt;two different&lt;/em&gt; cations (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+, H+}$&lt;/span&gt; in the examples), but it is confusing, because it could imply that there are two of a &lt;em&gt;single&lt;/em&gt; cation, so its use is deprecated. Similarly, in English, &lt;a href=&quot;http://www.merriam-webster.com/dictionary/biweekly&quot; rel=&quot;nofollow noreferrer&quot;&gt;biweekly&lt;/a&gt; can mean either &lt;em&gt;twice a week&lt;/em&gt; or &lt;em&gt;every two weeks&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of &quot;di&quot; for &quot;two&quot; is more consistent, i.e. it always means two of a single species, as in &lt;em&gt;carbon dioxide&lt;/em&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;. [Perhaps &quot;always&quot; is a bit strong... anyone know of contrary examples?]&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="-1" LastEditDate="2020-01-29T13:34:16.267" LastActivityDate="2020-01-29T13:34:16.267" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31144" PostTypeId="2" ParentId="31135" CreationDate="2015-05-10T03:35:40.037" Score="0" Body="&lt;p&gt;Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the site says that writing the concentration as $3t$ is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for example we had not done that and calculated the value of $t$ by using the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which you will find the $\mathrm{pH}$. So the site says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Why not calculate the value of $3t$ directly by using a variable $s$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore: $s=3t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore: $\mathrm{pOH}=-\log{s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are thinking is also correct, and the answer with that will be:&#xA;$\mathrm{pOH}=-\log{3t}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you both are doing the exact same thing here.&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorUserId="15410" LastEditDate="2015-05-10T05:43:13.770" LastActivityDate="2015-05-10T05:43:13.770" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31146" PostTypeId="1" AcceptedAnswerId="31147" CreationDate="2015-05-10T03:40:22.267" Score="3" ViewCount="1744" Body="&lt;p&gt;Just as $\ce{NaCl}$ would look like this:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cXp8z.gif&quot; alt=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/43140/IMG00003.GIF?revision=1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would a diagram of $\ce{CuSO4}$ look like?&lt;/p&gt;&#xA;" OwnerUserId="16110" LastEditorDisplayName="user15489" LastEditDate="2015-05-10T03:42:34.980" LastActivityDate="2015-05-10T04:28:52.867" Title="What would a diagram of the metal lattice structure of CuSO4 look like?" Tags="&lt;metal&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31147" PostTypeId="2" ParentId="31146" CreationDate="2015-05-10T03:51:21.387" Score="3" Body="&lt;p&gt;From the WebElements.com page on &lt;a href=&quot;http://www.webelements.com/compounds/copper/copper_sulphate.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;copper sulphate&lt;/a&gt;, the metal lattice structure looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1i7Wv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yellow pyramids are cooper and the other structures are sulphates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate view comes from the Wikipedia page of &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper(II)_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;copper (II) sulphate&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YhIJ3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pink spheres represent copper, yellow are sulphur and red are oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a 2D representation of copper sulphate, from the &lt;a href=&quot;http://www.rsc.org/learn-chemistry/wiki/Substance:Copper(II)_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Royal Society of Chemistry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEITs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question with some more insight is found in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/544/how-do-copper-sulphate-and-the-like-change-crystal-structure-on-adding-water-of&quot;&gt;How do copper sulphate and the like change crystal structure on adding water of hydration?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-10T04:28:52.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31150" PostTypeId="1" AcceptedAnswerId="33595" CreationDate="2015-05-10T05:11:42.863" Score="13" ViewCount="6832" Body="&lt;p&gt;How can I calculate the pH of a saturated solution of calcium fluoride ($\ce{CaF2}$)? I am given the following values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;K_\mathrm{sp}(\ce{CaF2}) &amp;amp;= 3.9 \cdot 10^{-11} \\&#xA;K_\mathrm{a}(\ce{HF}) &amp;amp;= 6.8 \cdot 10^{-4} \\&#xA;K_\mathrm{w} &amp;amp;= 10^{-14}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The $K_\mathrm{sp}$ and $K_\mathrm a$ values are taken from the appendices of &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0495558281&quot; rel=&quot;nofollow noreferrer&quot;&gt;Skoog et al. &lt;em&gt;Fundamentals of Analytical Chemistry&lt;/em&gt;, 9th edition&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="16103" LastEditorUserId="16683" LastEditDate="2017-09-03T09:02:09.097" LastActivityDate="2017-09-04T17:01:19.397" Title="Calculating the pH of a saturated calcium fluoride solution" Tags="&lt;solubility&gt;&lt;analytical-chemistry&gt;&lt;ph&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31151" PostTypeId="1" CreationDate="2015-05-10T05:37:15.740" Score="3" ViewCount="5162" Body="&lt;p&gt;In a given compound that contains three elements (e.g. a metal, a non-metal, and hydrogen), how can it be determined whether hydrogen is connected to the metal or the non-metal so that its oxidation number can be assigned accordingly?&lt;/p&gt;&#xA;" OwnerUserId="16113" LastEditorUserId="15410" LastEditDate="2015-05-10T06:00:37.867" LastActivityDate="2020-08-06T13:57:58.903" Title="Oxidation number of hydrogen in compounds" Tags="&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31152" PostTypeId="2" ParentId="31151" CreationDate="2015-05-10T05:54:16.183" Score="2" Body="&lt;p&gt;When hydrogen is with a metal its oxidation number is -1&#xA;When it is with a non metal element its oxidation number is +1&#xA;In this case most probably it should be -1&lt;/p&gt;&#xA;" OwnerUserId="16080" LastActivityDate="2015-05-10T05:54:16.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31154" PostTypeId="5" CreationDate="2015-05-10T06:22:43.050" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-10T06:22:43.050" LastActivityDate="2015-05-10T06:22:43.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31155" PostTypeId="4" CreationDate="2015-05-10T06:22:43.050" Score="0" Body="The description of the symmetry present in molecules. Molecular symmetry is useful for explaining or predicting of a molecule's chemical properties." OwnerUserId="402" LastEditorUserId="402" LastEditDate="2015-05-10T10:00:21.427" LastActivityDate="2015-05-10T10:00:21.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31156" PostTypeId="2" ParentId="31151" CreationDate="2015-05-10T06:23:03.873" Score="3" Body="&lt;p&gt;You can look at the electronegativity of the atom Hydrogen is bonded too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in NaH the electronegativity of H is greater than that of Na (~2.1 as opposed ~.9) so H two would have an oxidation state of -1) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In HCl, we observe that Cl is more electronegative so it will have an oxidation state of -1 and H will have an oxidation state of +1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In borane (even though B is a non-metal) each hydrogen displays an oxidation state of -1 (again electronegativity of H is 2.1 as opposed 2.0 of B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-05-10T06:23:03.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31157" PostTypeId="1" AcceptedAnswerId="31159" CreationDate="2015-05-10T06:28:48.357" Score="13" ViewCount="121555" Body="&lt;p&gt;I read my book and learned everything about enantiomers, racemization, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions, but when I'm trying to identify molecules as chiral or achiral, I get confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the two molecules below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.CH3-CH-OH&#xA;      |&#xA;      Br&#xA;&#xA;2.CH3-CH-Br&#xA;      |&#xA;      Br&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How can I tell if a compound is chiral?&lt;/p&gt;&#xA;" OwnerUserId="16114" LastEditorUserId="402" LastEditDate="2016-10-31T04:25:17.097" LastActivityDate="2020-02-15T17:00:41.543" Title="How to identify a compound as chiral or achiral" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31158" PostTypeId="2" ParentId="31157" CreationDate="2015-05-10T06:37:52.647" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw both molecules. Don't draw them flat. Alternatively, model them using &lt;a href=&quot;http://avogadro.cc&quot;&gt;Avogadro&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exchange the positions of $\ce{Br}$ and $\ce{OH}$ in the first case. Is the new molecule identical to the original one?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do the same for the dibromo-substituted compound. How is the result here?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-10T06:37:52.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31159" PostTypeId="2" ParentId="31157" CreationDate="2015-05-10T06:46:00.590" Score="8" Body="&lt;p&gt;Chirality can be determined by visualising the molecule in space. Once you do that, you can check to see if there exists a non-superimposable mirror image. If this is indeed the case, then your molecule is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look for an asymmetric carbon atom--because this often results in chirality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An asymmetric carbon (or chiral carbon) is, very simply,  a carbon atom attached to four different groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, consider your examples: 1) should thus be chiral and 2) is achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-05-10T06:46:00.590" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31160" PostTypeId="2" ParentId="31157" CreationDate="2015-05-10T06:51:36.110" Score="11" Body="&lt;p&gt;Look for carbons with four &lt;strong&gt;different&lt;/strong&gt; groups attached to identify potential chiral centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw your molecule with wedges and dashes and then draw a mirror image of the molecule. If the molecule in the mirror image is the same molecule, it is achiral. If they are different molecules, then it is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you're particularly interested in the C in the CH. However, take a look at the C in CH3. Why isn't it a chiral carbon?&lt;/p&gt;&#xA;" OwnerUserId="402" LastEditorUserId="402" LastEditDate="2017-01-28T06:30:39.647" LastActivityDate="2017-01-28T06:30:39.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31161" PostTypeId="1" AcceptedAnswerId="31206" CreationDate="2015-05-10T07:00:46.227" Score="6" ViewCount="76" Body="&lt;p&gt;Can bone mineral be obtained in pure, crystalline form outside a living organism? Had this been done ever?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="4945" LastEditDate="2015-05-11T09:42:31.833" LastActivityDate="2015-05-11T09:42:31.833" Title="Can bone mineral be obtained in pure, crystalline form?" Tags="&lt;biochemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31162" PostTypeId="1" AcceptedAnswerId="31164" CreationDate="2015-05-10T07:36:59.187" Score="1" ViewCount="252" Body="&lt;p&gt;Is it possible for an element to have a calculated relative atomic mass that is higher than the relative isotopic mass of its most massive component isotope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Relative~Atomic~Mass}=\sum[{\mathrm{(Isotopic~Mass) \cdot (Relative~Abundance)]}} \times100\%$&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="15410" LastEditDate="2015-05-12T05:25:56.480" LastActivityDate="2015-05-12T05:25:56.480" Title="Relative atomic mass" Tags="&lt;mass-spectrometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31163" PostTypeId="1" CreationDate="2015-05-10T08:12:06.733" Score="6" ViewCount="1642" Body="&lt;p&gt;Why does the current increase over time in an electrolytic cell with a $\ce{CuNO3}$ electrolyte. The current increased about $2~\mathrm{A}$, and the temperature increased during the experiment to around $55\ \mathrm{^\circ C}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not sure whether the formation of hydrogen and oxygen gas at the electrodes would decrease the internal cell resistance causing the current to increase. Or is there something else causing this? &lt;/p&gt;&#xA;" OwnerUserId="16115" LastEditorDisplayName="user7951" LastEditDate="2015-05-30T17:31:47.797" LastActivityDate="2015-05-30T17:31:47.797" Title="Change in current in an electrolytic cell over time" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31164" PostTypeId="2" ParentId="31162" CreationDate="2015-05-10T08:35:48.077" Score="3" Body="&lt;p&gt;No. Mathematically, that would not make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an imaginary element with only one known isotope: imaginarium-100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percent abundance of imaginarium-100 is $100\%$, so its relative atomic mass is: $$100 \cdot 1=100~\mathrm{amu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine a group of scientists were to discover that in fact imaginarium-99 consists of $0.0001\%$ of all imaginarium samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new relative atomic mass would be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$100 \cdot 0.999999+99 \cdot 0.000001=99.999999~\mathrm{amu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative atomic mass goes down with the discovery of a less massive isotope. So the closest that the relative atomic mass of an element can be to the mass of its most massive isotope is if the most massive isotope is also the only known isotope, but the relative atomic mass can never go over this.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-10T08:35:48.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31165" PostTypeId="2" ParentId="31163" CreationDate="2015-05-10T09:45:39.600" Score="3" Body="&lt;p&gt;The electrolyte conductivity is dependent on the temperature of the electrolyte. Higher temperature increases ions mobility, thereby increasing electrolyte conductivity. The conductivity increases around 2% per degree (assuming linear increase of conductivity). More info &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Conductivity_(electrolytic)&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Burak Ulgut mentioned, temperature increase is due to ohmic (Joule) heating. This can be estimated accordingly: $Q=V \cdot I \cdot t$&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="15410" LastEditDate="2015-05-11T07:31:36.607" LastActivityDate="2015-05-11T07:31:36.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31166" PostTypeId="1" AcceptedAnswerId="31180" CreationDate="2015-05-10T10:30:57.107" Score="9" ViewCount="11196" Body="&lt;p&gt;How do we decide the sequence of substituents in a organic compound if one of the substituents is complex. Should the complex chain must always be written first while writing the IUPAC name or are there any other rules governing it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following IUPAC names got me confused:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7-(1,2-dimethylpentyl)-5-ethyltridecane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AND&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first compound the complex substituent is written first although p(pentyl) comes after e(ethyl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second compound, the complex substituent is written later although m(methyl) comes after b(butyl).&lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2019-10-10T15:44:34.787" Title="Nomenclature of organic compounds containing complex side chains" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31167" PostTypeId="2" ParentId="31133" CreationDate="2015-05-10T11:02:10.373" Score="8" Body="&lt;p&gt;&lt;strong&gt;The short answer is a very large number and far more than you could reasonably count&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polymerization reactions typically produce very large molecules but not so large that a single molecule would make up as much material as a whole plastic bag or item of any significant size. &lt;a href=&quot;http://nzic.org.nz/ChemProcesses/polymers/10J.pdf&quot;&gt;This source&lt;/a&gt; on LDPE tells us that the typical molecular weight of the industrial product is 90,000 (some polymers are much bigger than this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-chemists molecular weights are (simplifying slightly) the weight relative to a single hydrogen atom. Carbon is 12 on this scale and ethylene (the monomer from which the plastic is made which consists of 2 carbons and 4 hydrogens) is 32. This mean that a typical molecule of the plastic is (crudely) made from a little under 3,000 ethylene molecules. The other fact you need to know is that there are about 6 x 10&lt;sup&gt;23&lt;/sup&gt; hydrogen atoms in a gramme of hydrogen (see the &lt;a href=&quot;http://en.wikipedia.org/wiki/Avogadro_constant&quot;&gt;wikipedia definition&lt;/a&gt; of Avogadro's number). Simple mathematics tells us that a gramme of LDPE will contain roughly 6.7 x 10&lt;sup&gt;18&lt;/sup&gt; molecules so that is the order of the number in small plastic item which, I expect, weighs just a few grammes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the bottom line is that even in a small plastic item consisting of a long chain polymer (LDPE is relatively short but other polymers might have chains 10-100 times more at most) there will be something like 10&lt;sup&gt;16&lt;/sup&gt; to 10&lt;sup&gt;18&lt;/sup&gt; molecules of polymer, which is a lot.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="81" LastEditDate="2015-05-10T11:07:51.930" LastActivityDate="2015-05-10T11:07:51.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31170" PostTypeId="1" CreationDate="2015-05-10T11:45:56.073" Score="9" ViewCount="4535" Body="&lt;p&gt;What is the world's finest powder substance that has the smallest granules but remains solid?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastActivityDate="2015-05-10T23:36:55.013" Title="What is the world's finest powder substance?" Tags="&lt;materials&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31171" PostTypeId="1" CreationDate="2015-05-10T11:59:57.300" Score="4" ViewCount="377" Body="&lt;p&gt;I wonder whether there is a material which consists from individual molecules (that is, fluid), which has its &lt;a href=&quot;http://en.wikipedia.org/wiki/Angle_of_repose&quot; rel=&quot;nofollow&quot;&gt;angle of repose&lt;/a&gt; thus exhibiting propertes of a powder?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="10185" LastEditDate="2015-06-09T13:10:53.703" LastActivityDate="2015-07-09T17:12:29.320" Title="Is there an otherwise liquid material that has angle of repose?" Tags="&lt;materials&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31172" PostTypeId="2" ParentId="31171" CreationDate="2015-05-10T12:32:35.843" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Liquid&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid&lt;/a&gt; by definition has no fixed shape and therefore &lt;a href=&quot;https://en.wikipedia.org/wiki/Powder_(substance)#Comparison_with_other_substances&quot; rel=&quot;nofollow noreferrer&quot;&gt;can have no angle of repose&lt;/a&gt;. I have noticed you have asked a &lt;a href=&quot;https://physics.stackexchange.com/questions/183186/is-there-a-definite-boundary-between-a-powder-and-a-fluid&quot;&gt;similar question&lt;/a&gt; on Physics.SE; the presence or absence of this kind of property can be used as a means to classify powders versus fluids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the conical &quot;fixed shape&quot; observed in powders is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Friction&quot; rel=&quot;nofollow noreferrer&quot;&gt;frictional&lt;/a&gt; forces between the particles, whereas any surface curvature in liquids from &lt;a href=&quot;https://en.wikipedia.org/wiki/Surface_tension&quot; rel=&quot;nofollow noreferrer&quot;&gt;surface tension&lt;/a&gt; (e.g., a &lt;a href=&quot;https://en.wikipedia.org/wiki/Meniscus&quot; rel=&quot;nofollow noreferrer&quot;&gt;meniscus&lt;/a&gt;) is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;intermolecular forces&lt;/a&gt; between individual molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, take into account the melting point of the substance: it's a powder if the temperature of interest is below the melting point, and a liquid if it is above.    &lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-07-09T15:38:48.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31173" PostTypeId="1" CreationDate="2015-05-10T12:56:56.537" Score="4" ViewCount="578" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/upOY4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/upOY4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The picture above is a drug called Setraline. Suggest a likely mode of degradation of sertraline hydrochloride alone in aqueous solution, giving likely structures of the degradation products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From my knowledge, degradation in aqueous solution can only mean hydrolysis and sertraline does not contain hydrolysable groups like amide or ester. What are the other forms of degradation besides hydrolysis? in aqueous solution.&lt;/p&gt;&#xA;" OwnerUserId="15253" LastEditorUserId="41890" LastEditDate="2017-07-02T13:59:22.903" LastActivityDate="2017-07-02T13:59:22.903" Title="degradation of sertraline" Tags="&lt;organic-chemistry&gt;&lt;drugs&gt;&lt;pharmacology&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31174" PostTypeId="2" ParentId="26833" CreationDate="2015-05-10T13:24:44.403" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What in the quantum nature of electrons gives rise to the octet rule?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Octet_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;octet rule&lt;/a&gt;&lt;/em&gt; is justified by the fact that in many cases an $n s^2 n p^6$ electron configuration is the most energetically favorable one. Recall that electrons are arranged in different &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_shell&quot; rel=&quot;nofollow noreferrer&quot;&gt;electron shells&lt;/a&gt;&lt;/em&gt; around the nucleus. Each shell corresponds to a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Principal_quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;principal quantum number&lt;/a&gt;&lt;/em&gt; $n$ and can hold only a certain number of electrons, namely up to $2n^2$. Each shell is composed of subshells each of which corresponds to an &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Azimuthal_quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;azimuthal quantum number&lt;/a&gt;&lt;/em&gt; $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now recall also that the binding energy of the electron decreases with both $n$ and $l$, and thus, once $n s$ and $n p$ orbitals of an $n$-th shell are filled, in many cases it is indeed energetically unfavorable to fill the subsequent orbitals: $n d$ if (there are any) or $(n+1)s$. The energy gain from binding an electron to an atom might not be enough to compensate a cost of separating it from a different atom.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How is it related to the fact that an element cannot have an oxidation&#xA;  state higher/lower than the complementary to 8?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose an atom has $x$ electrons in the outermost shell, where with the exception of transition metals $x \leq 8$. In the limiting cases the atom will either gain $8-x$ electrons to completely fill $ns$ and $np$ subshells or lose all $x$ electrons. In the first case we will end up with the atom in $-(8-x)$ oxidation state while in the second case with the atom in $+x$ oxidation state. But these are only the limiting cases, while in principle atom can gain few electrons &lt;em&gt;and&lt;/em&gt; loose few others at the same time. Besides, atoms of the same type can evenly &lt;em&gt;share&lt;/em&gt; electrons with each without satisfying the octet rule without gaining or loosing electrons. Thus, usually elements have few possible oxidation states, though, in the above mentioned limiting $[-(8-x), +x]$ range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wvdd5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excerpt from &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;'s list of oxidation states of the elements.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-10T17:58:19.687" LastActivityDate="2015-05-10T17:58:19.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31176" PostTypeId="1" AcceptedAnswerId="31265" CreationDate="2015-05-10T14:04:08.420" Score="8" ViewCount="388" Body="&lt;p&gt;I came across a unit called $\text{cmmol dm}^{-3}$ in buffer solutions. What does this unit mean?&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="16096" LastEditDate="2015-05-11T15:06:17.177" LastActivityDate="2015-05-12T09:49:29.443" Title="What is the meaning of cmmol in chemistry?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31177" PostTypeId="1" CreationDate="2015-05-10T14:23:27.360" Score="3" ViewCount="464" Body="&lt;p&gt;While the Beer's law equation is $A=a \cdot b \cdot c$, according to my book, $a$ is a coefficient which is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a characteristic property of the analyte in a specific solvent, and its values will depend on the wavelength of absorption radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is quite unclear to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have for example  $a= 1.5$, what does that exactly tell us ? &lt;em&gt;What&lt;/em&gt; do we know when this coefficient is given ? Or is this just a constant that helps us to solve the equation? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="83758" LastEditDate="2020-06-02T16:43:49.727" LastActivityDate="2020-06-02T16:43:49.727" Title="Lambert-Beer's law coefficient" Tags="&lt;physical-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31179" PostTypeId="2" ParentId="27053" CreationDate="2015-05-10T14:28:39.710" Score="0" Body="&lt;p&gt;What is your &quot;system&quot; in this case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction that you are performing--your reactants and your products--constitute your system. Since the reaction is endothermic, heat has to enter your system for the reaction to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But where does this heat come from? It is obviously drawn from the surroundings, and thus you measure a temperature that is now lower than what you started with.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-05-10T14:28:39.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31180" PostTypeId="2" ParentId="31166" CreationDate="2015-05-10T14:38:15.087" Score="11" Body="&lt;p&gt;Simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order.&lt;br /&gt;&#xA;For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.&lt;/p&gt;&#xA;&lt;p&gt;However, the name of a compound substituent is considered to begin with the first letter of its complete name.&lt;br /&gt;&#xA;For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.&lt;/p&gt;&#xA;&lt;p&gt;On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5&lt;/strong&gt; ALPHANUMERICAL ORDER&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5.1&lt;/strong&gt; Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5.2&lt;/strong&gt; The name of a prefix for a substituent is considered to begin with the first letter of its complete name.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the correct alphanumerical order corresponds to the name 7-(1,2-dimethylpentyl)-5-ethyltridecane (not 5-ethyl-7-(1,2-dimethylpentyl)tridecane) since the compound substituent name ‘dimethylpentyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘e’.&lt;br /&gt;&#xA;(Nevertheless, the preferred IUPAC name is 5-ethyl-7-(3-methylhexan-2-yl)tridecane.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cqERW.png&quot; alt=&quot;5-ethyl-7-(3-methylhexan-2-yl)tridecane&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;In your second example, the correct alphanumerical order corresponds to the name 6-(1,1-dimethylbutyl)-5,5-dimethyl-6-pentyltridecane (not 5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane) since the compound substituent name ‘dimethylbutyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘m’.&lt;br /&gt;&#xA;(Nevertheless, the preferred IUPAC name is 5,5-dimethyl-6-(2-methylpentan-2-yl)-6-pentyltridecane.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSP4p.png&quot; alt=&quot;5,5-dimethyl-6-(2-methylpentan-2-yl)-6-pentyltridecane&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-10T17:51:10.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31181" PostTypeId="2" ParentId="31176" CreationDate="2015-05-10T14:38:47.063" Score="6" Body="&lt;p&gt;Upon researching this, I found that this a measurement of charge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{cmol=10~\mathrm{meq}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1~\mathrm{eq}= \mathrm{F}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{F}$ is Faraday's constant and is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{F}=96 485.3365~\mathrm{C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-05-11T14:23:46.313" LastActivityDate="2015-05-11T14:23:46.313" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="31182" PostTypeId="2" ParentId="31157" CreationDate="2015-05-10T14:50:19.290" Score="5" Body="&lt;p&gt;The first method I used to understand chirality was to grab some different colored zip ties and make a couple small tetrahedral structures with them, assigning a color to each different radical (say blue for $\ce{CH3}$, white for $\ce{H}$, red for $\ce{OH}$ and yellow for $\ce{Br}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then look at it by rotating the first to see if I can get the second molecule, if you can't then that molecule is chiral.&lt;/p&gt;&#xA;" OwnerUserId="16121" LastEditorDisplayName="user15489" LastEditDate="2015-05-10T19:44:33.907" LastActivityDate="2015-05-10T19:44:33.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31184" PostTypeId="1" CreationDate="2015-05-10T15:10:35.397" Score="2" ViewCount="420" Body="&lt;p&gt;Whilst doing tests on two unknown solutions, I used Fehling's test. The solutions could be a mixture of compounds or just one compound dissolved in water/ another solvent - I don't know. Upon adding Fehling's solution both solutions (which were originally colourless) turned a blue colour. I then heated the solutions very gently for about 5 minutes and after this time one had completely formed a red precipitate, whereas the other still had the blue solution left in it (while forming some red precipitate). After waiting for about 20 minutes both solutions had completely formed a brick red precipitate and were the same. I assume this is to do with rate of the oxidation reaction occurring. Is there any explanation as to the rate of reaction in each solution or an explanation which is not to do with rate at all perhaps?&lt;/p&gt;&#xA;" OwnerUserId="9286" LastActivityDate="2015-05-17T14:56:16.003" Title="Fehling's Test meaning for partial formation of a precipitate?" Tags="&lt;home-experiment&gt;&lt;carbonyl-compounds&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31185" PostTypeId="1" CreationDate="2015-05-10T15:14:46.817" Score="2" ViewCount="214" Body="&lt;p&gt;There are many methods available to obtain powdered metals and alloys. &lt;a href=&quot;http://en.wikipedia.org/wiki/Powder_metallurgy#Powder_production_techniques&quot; rel=&quot;nofollow&quot;&gt;Wikipedia show some of them&lt;/a&gt;. Another form is grinding with posterior removal of wheel particles. All these methods produce metal powder with the help of specific machinery. I am asking for an alternative wet chemical method (water solution) or electrochemical method (water or ionic liquid electrolyte) to obtain metal powder (specifically iron, nickel and silicon, in pure or controlled alloy form).&lt;/p&gt;&#xA;" OwnerUserId="16122" LastEditorUserId="4356" LastEditDate="2015-05-15T21:27:33.303" LastActivityDate="2015-05-15T21:27:33.303" Title="Is there an alternative wet chemical method or electrochemical method to obtain precipitation of iron, nickel and silicon powder?" Tags="&lt;metallurgy&gt;&lt;precipitation&gt;" AnswerCount="2" CommentCount="5" ClosedDate="2015-05-20T01:36:59.430" ContentLicense="CC BY-SA 3.0" />
  <row Id="31186" PostTypeId="2" ParentId="31177" CreationDate="2015-05-10T15:36:36.477" Score="4" Body="&lt;p&gt;To start with, I recommend to reformulate Beer-Lambert's law in a way more frequently met in (analytical) chemistry, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mbox{ABS} = c \cdot{} l \cdot{} \varepsilon = \log_{10} \frac{I_0}{I}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because then it more obvious that the (dimensionless) absorption (ABS) recorded is the product of analyte concentration ($c$), optical path length ($l$) across the sample investigated, and the molecular extinction coefficient ($\varepsilon{}$).  Absorption equally is defined as the logarithm (to base of ten) of the fraction of incoming light intensity ($I_0$) and light intensity that is transmitted across the sample ($I$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to your question.  While concentration $c$ and optical path length $l$ may be changed (the later, for example, by exchanging a measurement cell of 1 cm path length to one of 2 mm or 5 cm -- depending on the setup of the spectrometer), the molecular extinction coefficient $\varepsilon$ is indeed more valuable than just assistance to solve the equation.  It relates to the ease the analyte may absorb irradiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UV-Vis spectroscopy investigates electronic transitions, prominently of $\pi{}$-systems (like of alkene double, and aromatic bonds), yet investigation of electronic transitions of $n$- (non-bonded, like of carbonyl-O electrons), $d$- (for example in transition metal complexes) and $\sigma{}$-electrons (in the short wavelength UV-range).  In principle, these transitions are discrete, i.e. only photons of this attuned energy ($h\nu{}$) that fits the energy gap between the electronic ground state and the excited state are absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analyte molecule may possess more than one allowed electronic transitions.  They may occur simultaneously, with different efficiency.  For example, many organic dyes have a strong absorption in the visible range, like indigo has one around 600 nm.  Yet if you tune the wavelength of investigation to short wavelengths, below 250 nm, for example, absorption may still occur, too, yet to a lesser extent.  (This eventually contributes to processes that indigo stained cloths fade over time, for example.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Side note:  The molecular extinction coefficient $\varepsilon$ depends on other, additional parameters than the &quot;solvent&quot; and &quot;wavelength of investigation&quot;.  Varying the pH-value, ionic strength of the solution, presence/absence of other dissolved molecules than of your analyte and solvent may alter $\varepsilon{}$, too.  Do not forget, the concentration $c$ of your analyte may influence $\varepsilon{}$, too -- pointing to (concentration dependent)  potential dimerisation / aggregation of your analyte (for example &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge-transfer_complex&quot; rel=&quot;nofollow&quot;&gt;charge transfer complexes&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-05-10T15:36:36.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31188" PostTypeId="1" CreationDate="2015-05-10T17:02:03.303" Score="-4" ViewCount="5903" Body="&lt;p&gt;Many p-block elements which are the first in their respective groups show $pπ-pπ$ multiple bonding. Why doesn't carbon do so?&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="418" LastEditDate="2015-05-10T17:34:21.543" LastActivityDate="2015-05-11T11:15:02.197" Title="Why does a diatomic carbon molecule not exist?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-05-10T19:03:55.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="31189" PostTypeId="1" CreationDate="2015-05-10T17:17:03.600" Score="17" ViewCount="245619" Body="&lt;p&gt;Recently in my chemistry classes, the teacher spoke about SPDF configuration and then said that we'll be taught about it in higher classes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm sorta curious to know that what is SPDF configuration and is there something like - obtaining electronic configuration in SPDF format?&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="402" LastEditDate="2019-05-17T06:41:29.577" LastActivityDate="2019-05-17T06:41:29.577" Title="What is SPDF configuration?" Tags="&lt;electrons&gt;&lt;atoms&gt;&lt;electronic-configuration&gt;" AnswerCount="5" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31190" PostTypeId="2" ParentId="31188" CreationDate="2015-05-10T17:19:05.447" Score="1" Body="&lt;p&gt;Okay, I believe that you mean to ask why a $C_2$ molecule does not exist?&#xA;Well, it does exist.&#xA;So, when we use molecular orbital theory to calculate the bond order of $C_2$, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B.O.=\frac{\mathrm{(No.~of~electrons~in~bonding~orbitals) - (No.~of~electrons~in~ antibonding~orbitals)}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 4 electrons in bonding and 0 in antibonding for $C_2$So putting in formula-$B.O.=(4-0)/2=2$As the bond order is 2, the molecule cannot exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when a more rigorous approach to MOT is applied, it shows that both the bonds present in the molecule must be π bonds and no sigma bonds must be present. This according to me is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Wikipedia also confirms that there is a speculation regarding the structure. It says -&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A recent paper by S. Shaik reports that a quadruple bond exists in $C_2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And hence though the actual molecule has been found, its chemistry is still being studied.&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorUserId="10185" LastEditDate="2015-05-11T11:15:02.197" LastActivityDate="2015-05-11T11:15:02.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31192" PostTypeId="1" CreationDate="2015-05-10T17:51:39.620" Score="0" ViewCount="2784" Body="&lt;p&gt;If we have  $\ce{A + T -&amp;gt; P}$ &#xA;where $\ce{A}$ is the analyte, $\ce{T}$ is the titrant, and $\ce{P}$ is the product,&#xA;photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When stable complexes formed such as $\ce{Bi}$-EDTA, they will not absorb the radiation (or in small amounts), but when unstable complexes formed such as $\ce{Cu}$-EDTA, they will absorb highly. The same goes also for the acids. Strong acids will absorb much &lt;em&gt;less&lt;/em&gt; the electromagnetic radiation than the weak acids. &#xA;The reason for this is that &lt;em&gt;strong acids are completely dissociated&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does that indicate the photometric titration ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="418" LastEditDate="2015-05-10T21:38:14.287" LastActivityDate="2015-06-09T22:06:55.813" Title="Photometric titration" Tags="&lt;analytical-chemistry&gt;&lt;titration&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31193" PostTypeId="2" ParentId="31189" CreationDate="2015-05-10T17:51:56.243" Score="20" Body="&lt;p&gt;s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms.  These orbitals have different shapes (&lt;em&gt;e.g.&lt;/em&gt; electron density distributions in space) and energies (&lt;em&gt;e.g&lt;/em&gt;. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5EcA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Electronic_Orbitals&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital&lt;/li&gt;&#xA;&lt;li&gt;a lithium atom with 3 electrons would be $\ce{1s^2~2s^1}$&lt;/li&gt;&#xA;&lt;li&gt;fluorine has 9 electrons which would be $\ce{1s^2~2s^2~2p^5}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Again, as we read from left to right the energy of the orbital increases and the superscript shows the number of electrons in the orbital. You can read more &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_configuration&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-05T18:01:32.713" LastActivityDate="2017-04-05T18:01:32.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31194" PostTypeId="2" ParentId="31128" CreationDate="2015-05-10T18:18:41.517" Score="1" Body="&lt;p&gt;With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes.&lt;/p&gt;&#xA;&lt;p&gt;On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(e) saturation/unsaturation:&lt;/p&gt;&#xA;&lt;p&gt;  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings);&lt;/p&gt;&#xA;&lt;p&gt;  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule e takes precedence over Rule f.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-31.1.1.1&lt;/strong&gt; The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.&lt;br /&gt;&#xA;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For example, a correct systematic name is &lt;strong&gt;3-methylbut-1-ene&lt;/strong&gt; (not 2-methylbut-3-ene).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzjgD.png&quot; alt=&quot;3-methylbut-1-ene&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-10T18:18:41.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31195" PostTypeId="1" AcceptedAnswerId="31197" CreationDate="2015-05-10T18:22:13.803" Score="1" ViewCount="5945" Body="&lt;p&gt;I'm teaching myself organic chemistry and am struggling with differentiating between dimethyl and ethyl substituent groups. &#xA;Example: tert-butyl (1,1-dimethylethyl)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chapter follows the one on bond-line structures, and I realize one of the rules for making a valid bond-line structure for straight-chain alkanes involves zig-zagging carbon. But I am confused on how to represent the substituent groups - is the ethyl group the V, and the methyl groups are each of the branches from the middle carbon atom of the substituent group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pkuTS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="10185" LastEditDate="2015-05-10T21:08:31.397" LastActivityDate="2015-05-10T21:08:31.397" Title="Difference between a dimethyl and ethyl substituent groups" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31196" PostTypeId="2" ParentId="31189" CreationDate="2015-05-10T18:26:16.200" Score="8" Body="&lt;p&gt;By SPDF configuration, he meant orbital configuration. Now the basic of this concept is from very fundamental quantum chemistry formulation but I don't think you need to know that now (It is usually taught in 4th year of UG or 1st year of graduate study in Engineering discipline but i am not sure about pure science discipline). Whatever I would like to think about it from a different perspective. Are you familiar with quantum numbers? There are 4 different quantum numbers:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Principal quantum number (This is like the city you live in). There are lots of electrons in an atoms. Now if we want to distinguish between those electrons than we need to name them or there should be something to distinguish. Imagine you are living in a circular states and the cities are named as the radius it has. Like sector 1 is the city which has a mean radius of 1 unit and so on. By principal quantum number, we actually mean that the probability of finding that electron is high within that particular radius. It is named as n=1,2,3...  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Azimuthal quantum number: This is like the building you live in. Now if you are living in a city that is very small and has only one building, then you don't need to specify each building of that city differently. Like for n=1, l=0 (Here l=azimuthal quantum number/building number) but if n=3, then l=0 to (n-1), that means in Sector 3 city there are 3 building named 0,1 &amp;amp; 2.    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnetic quantum number: This is like apt number on that building. m=0 to (+/-)l. So, if you are living in building no. 3, you can either live in apt -3,-2,-1,0,1,2 or 3.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Spin quantum number: Each apt has two rooms (room A and room B) (This is actually unrestricted Hartree-Fock case) but if you are living with spouse than you can have a large room by breaking the wall between those two rooms (Only room A or restricted Hartree-Fock case)   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, The building name can be rephrased as spdf orbital. If you are living in building no. 0 that means you are living in s orbital. Similarly&lt;br&gt;&#xA;building no.1= p orbital&lt;br&gt;&#xA;building no.2= d orbital&lt;br&gt;&#xA;building no.3= f orbital&lt;br&gt;&#xA;So, in your building no.0 (s orbital),&lt;br&gt;&#xA;total number of room = 1apt * 2rooms/apt = 2rooms or 2 electrons&lt;br&gt;&#xA;In building no.1 (p orbital),&lt;br&gt;&#xA;total number of electron/room = 3apt * 2rooms/apt = 6 rooms or 6 electrons  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to know more you can read:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hund's rule   &lt;/li&gt;&#xA;&lt;li&gt;Pauli's exclusion principle  &lt;/li&gt;&#xA;&lt;li&gt;Aufbau principle   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But they all are superficial theory, they can say what is going on but they can't say why. But quantum chemistry method will give you mathematical understanding of why there is 2 rooms/apt or why building 2 has 5 apt etc.  &lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="4945" LastEditDate="2015-05-18T06:24:17.650" LastActivityDate="2015-05-18T06:24:17.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31197" PostTypeId="2" ParentId="31195" CreationDate="2015-05-10T19:15:56.227" Score="2" Body="&lt;p&gt;If an organic compound has dimethyl in its name than that means that it has 2 methyl groups. If they are on the same carbon they form a V in the bond line structure. An ethyl group on the other hand  forms a 2 carbon unbranched zigzag in the bond line structure instead of the V that a dimethyl group forms in the bond line structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ethyl group it is standard on all bond line structures to represent it as a 2 carbon unbranched zigzag. For the dimethyl group it is standard to represent it as a V in all bond line structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how we know this it is because an ethyl group coming off of a carbon looks like propane if you just look at that carbon and the ethyl group. In fact some people in the early years of organic chemistry might have called propane ethylmethane before the prop prefix was used for alkanes. However if this was the case some of them would call propane dimethylmethane. This was because they were looking at different carbons as the central atom. If you look from one of the ends it looks like an ethyl group sticking off of methane. If you look at the center it looks like a dimethyl group sticking off of methane.&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2015-05-10T19:38:22.380" LastActivityDate="2015-05-10T19:38:22.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31198" PostTypeId="1" CreationDate="2015-05-10T20:48:55.613" Score="6" ViewCount="860" Body="&lt;p&gt;I have identified carvacrol and thymol via thin-layer-chromatography and via gas chromatography&#xA;On a normal silica plate the thymol has a higher Rf-value then carvacrol. The explanation for this would be the position of the hydroxyl-group and the difference size of the nearby methyl or isopropyl group. I think thymol should be more hydrophobic then carvacrol, explaining the less interaction with the silica gel resulting in a higher Rf-value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, that the analysis by gas chromatography with a non-polar column shows a different elution profile. I would expect that thymol, according to the higher hydrophobicity, should have a higher retention time then carvacrol. This isn't the case. Carvacrol (27.08 min) elutes after thymol (26.51 min).&#xA;Is there any explanation I'am not aware or is my thinking simple wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MbOxn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="1782" LastEditDate="2015-05-11T16:29:14.133" LastActivityDate="2019-10-18T19:02:30.363" Title="Different elution profiles of carvacrol and thymol" Tags="&lt;analytical-chemistry&gt;&lt;aromatic-compounds&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31199" PostTypeId="1" AcceptedAnswerId="31200" CreationDate="2015-05-10T21:02:59.387" Score="1" ViewCount="2013" Body="&lt;p&gt;Is the correct name for the following structure 3-tert-butylhexane or 3-ethyl-2,2-dimethylhexane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is 3-tert-butylhexane, but the textbook says it is 3-ethyl-2,2-dimethylhexane. Can someone please explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZeKX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="15410" LastEditDate="2015-05-10T21:40:23.290" LastActivityDate="2015-05-10T21:40:23.290" Title="Nomenclature of an Alkane" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31200" PostTypeId="2" ParentId="31199" CreationDate="2015-05-10T21:17:30.850" Score="1" Body="&lt;p&gt;According to IUPAC (&lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.6 - If chains of equal length are competing for selection as main chain in a saturated branched acyclic hydrocarbon, then the choice goes in series to:&#xA;  &lt;br&gt;(a) The chain which has the greatest number of side chains.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the correct name is 3-ethyl-2,2-dimethylhexane as opposed to 3-&lt;em&gt;tert&lt;/em&gt;-butylhexane because the former has more side chains.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-10T21:17:30.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31201" PostTypeId="2" ParentId="31192" CreationDate="2015-05-10T21:45:38.300" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Almost ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless whether you monitor a complexometric titration by eye or with the help of a spectrometer, it is a good idea to use an indicator, such as &lt;strong&gt;Eriochrome Black T&lt;/strong&gt;. Then, it doesn't matter, if or how  strong the EDTA complex of the metal absorbs!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you monitor the absorption of the indicator, that is liberated from the metal-indicator complex upon addition of EDTA.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-10T21:45:38.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31202" PostTypeId="1" AcceptedAnswerId="31205" CreationDate="2015-05-10T22:06:45.543" Score="6" ViewCount="907" Body="&lt;p&gt;My textbook tells me the following structure is 2,2,5-trimethylpentane. I try to change the parent group in different ways so as to maximize the number of substituent groups, but I think the maximum number of substituent groups for any arrangement is 3 - so my answer is 2,2,4-trimethylpentane. Am I wrong in thinking this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZbCwl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="15410" LastEditDate="2015-05-12T04:38:28.897" LastActivityDate="2015-05-12T04:38:28.897" Title="Bond-Line Structure for an Alkane" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31204" PostTypeId="2" ParentId="31170" CreationDate="2015-05-10T23:36:55.013" Score="3" Body="&lt;p&gt;Clusters of just a few atoms, such as &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2014/CP/C4CP00063C&quot; rel=&quot;nofollow&quot;&gt;gold atomic clusters&lt;/a&gt;, and &lt;a href=&quot;http://education.mrsec.wisc.edu/300.htm&quot; rel=&quot;nofollow&quot;&gt;quantum dot nanoparticles&lt;/a&gt; could be considered very fine powders, though they usually are suspended in another substance to keep the clusters from agglomerating. Particles this tiny have different properties from the bulk material, both because of quantum effects (there are few particles, so there is not a continuous valence band) and because more atoms are at the surface interface than are surrounded by similar atoms in the interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_dot&quot; rel=&quot;nofollow&quot;&gt;Quantum dot&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this fits your definition of a &quot;fine powder&quot;. You might also be interested in &lt;a href=&quot;https://www.youtube.com/watch?v=vinu8ZzNvRw&quot; rel=&quot;nofollow&quot;&gt;pyrophoric iron&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, ruby glass is made using colloidal, i.e. super-finely divided, gold.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-10T23:36:55.013" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31205" PostTypeId="2" ParentId="31202" CreationDate="2015-05-10T23:42:53.503" Score="10" Body="&lt;p&gt;There are 2 methyl groups on the second carbon of the main chain, and 1 methyl group on the fourth carbon of the main chain, so 2,2,4-trimethylpentane is correct! If there was a methyl group on the 5th carbon of a pentane, it would be a hexane, so it would not make sense to name it the way your book did.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-10T23:42:53.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31206" PostTypeId="2" ParentId="31161" CreationDate="2015-05-10T23:44:13.820" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Apatite&quot; rel=&quot;noreferrer&quot;&gt;Apatite&lt;/a&gt; is the chemical name for the mineral found in bone and teeth. It is occasionally found in the form of large crystals, such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOsP5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, just as bone is made of the mineral apatite, there are critters with skeletons of &lt;em&gt;silica&lt;/em&gt;, another common mineral, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Venus%27_flower_basket&quot; rel=&quot;noreferrer&quot;&gt;Venus' flower basket&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-10T23:44:13.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31207" PostTypeId="1" CreationDate="2015-05-11T00:26:38.637" Score="2" ViewCount="55" Body="&lt;p&gt;I have heard of an article where divalent lead $\ce{PbL2}$ ($\ce{L}$ is ligand) reacts with something (again not sure, as I dont know the article) and what it reacts with gains a ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In less confusing words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbL2 + MXn}$ ($\ce{X}$ is another ligand, not necessarily halogen) ==&gt; $\ce{PbL(n-1)X + MX(n-1)L}$&#xA;Or something similar. That is what I got from hearing about the article. I was wondering if you guys can share your thoughts as to why happens?&lt;/p&gt;&#xA;" OwnerUserId="16135" LastEditorDisplayName="user15489" LastEditDate="2015-05-11T00:41:45.500" LastActivityDate="2015-05-11T00:41:45.500" Title="Why does Pb(II) promote ligand substitution?" Tags="&lt;inorganic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31208" PostTypeId="1" CreationDate="2015-05-11T02:03:58.650" Score="4" ViewCount="179" Body="&lt;p&gt;I know some sulfur groups such as sulfones and disulfides and sulfonic acids. I also know of 1 phosphorus group. That is the phosphate ion.&#xA;I don't know of any boron groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the phosphorus equivalent of aldehydes, ketones, and carboxylic acids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphals, Phosphones, and Phosphonic acids I suppose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about the boron equivalent of aldehydes, ketones, and carboxylic acids?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="4945" LastEditDate="2015-05-11T09:37:45.560" LastActivityDate="2015-05-11T09:37:45.560" Title="Phosphorus and Boron groups" Tags="&lt;organic-chemistry&gt;&lt;pnictogen&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31209" PostTypeId="2" ParentId="24081" CreationDate="2015-05-11T03:59:09.103" Score="2" Body="&lt;p&gt;In the equation that you stated initially, water would be omitted from the equilibrium constant expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have given it some thought: in the standard case of esterification water should be included  because it is responsible for causing the reverse process ( hydrolysis of ester formed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-05-11T03:59:09.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31210" PostTypeId="1" CreationDate="2015-05-11T04:53:47.057" Score="3" ViewCount="2824" Body="&lt;p&gt;What is the structure and color of $\ce{CrO3}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to produce $\ce{CrO3}$ when dichromate is heated with sulphuric acid?&lt;/p&gt;&#xA;" OwnerUserId="16080" LastEditorUserId="15410" LastEditDate="2015-05-11T05:17:09.660" LastActivityDate="2015-05-11T05:17:09.660" Title="What is the color and structure of CrO3?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-10-02T16:54:50.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="31211" PostTypeId="2" ParentId="31210" CreationDate="2015-05-11T05:14:25.857" Score="4" Body="&lt;p&gt;According to its &lt;a href=&quot;http://en.wikipedia.org/wiki/Chromium_trioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Color&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CrO3}$ is a dark-purple solid under anhydrous conditions, that turns bright orange when wet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Production&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CrO3}$ is generated, as you said, by the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 + Na2Cr2O7 -&amp;gt; 2 CrO3 + Na2SO4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Structure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its monomeric structure is predicted to be pyramidal like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WdDEw.png&quot; alt=&quot;chromium oxide monomer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its crystalline structure is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gGmBv.png&quot; alt=&quot;chromium oxide crystal&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each chromium center shares two oxygen centers with neighbors, but two oxygen atoms are not shared, giving the overall stoichiometric ratio of 1:3.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-11T05:14:25.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31213" PostTypeId="2" ParentId="28483" CreationDate="2015-05-11T05:20:09.500" Score="9" Body="&lt;p&gt;For the reaction&#xA;$$\ce{CaSO4 (s) &amp;lt;=&amp;gt; Ca^2+ (aq) + SO4^{2-} (aq)},$$&#xA;the appropriate &lt;a href=&quot;http://goldbook.iupac.org/E02177.html&quot;&gt;equilibrium constant&lt;/a&gt; using activities can be set up as&#xA;$$K_\mathrm{eq}=\frac{a(\ce{Ca^2+})\,a(\ce{SO4^2-})}{a(\ce{CaSO4})}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The activity $a$ of a pure substances in condensed phases is approximately one, since their behaviour can be often treated ideally, i.e. the chemical potential $\mu$ at the considered conditions is approximately the same as the chemical potential at standard conditions $\mu^\circ$. This derives directly from the &lt;a href=&quot;http://goldbook.iupac.org/A00115.html&quot;&gt;definition of the relative activity&lt;/a&gt;: $$a = \exp\left\{\frac{\mu-\mu^\circ}{\mathcal{R}T}\right\}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For substances where the solubility is very low, concentrations of the ions are very low, so in a further approximation we can assume that in these solutions the &lt;a href=&quot;http://goldbook.iupac.org/A00116.html&quot;&gt;activity coefficient&lt;/a&gt; is approximately one, therefore, we can write concentrations instead of activities and obtain for the equilibrium constant the expression, that is also known a solubility product.&#xA;$$K_\mathrm{eq} \approx K_\mathrm{sp} = c(\ce{Ca^2+})\,c(\ce{SO4^2-})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is completely irrelevant to the question itself. Since the solution is saturated, it also means, that the solid is in equilibrium with the solution, which means that the kinetics of dissolution and precipitation are equal of magnitude. The Gibbs energy has a very distinct value for these equations.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;strong&gt;chemical equilibrium&lt;/strong&gt;&lt;br&gt;&#xA; Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy,  $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:&lt;br&gt;&#xA; $$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + \mathcal{R}T\,\ln\,K = 0\\\Delta G_\mathrm{r}^\circ = - \mathcal{R}T\,\ln\,K\\$$&#xA; The equilibrium constant, $K$, is given by the mass-law effect.&lt;br&gt;&#xA; Source: &lt;a href=&quot;http://goldbook.iupac.org/C01023.html&quot;&gt;IUPAC goldbook&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-11T05:20:09.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31214" PostTypeId="2" ParentId="31163" CreationDate="2015-05-11T05:37:01.667" Score="2" Body="&lt;p&gt;As the reaction proceeds, your electrodes will get rougher since the surfaces will be etched. This will in turn result in an increased electroactive surface area increasing the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature effect mentioned by Jaroslav Kotowski is there, but the growing electrode area is also a contributing factor.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-05-11T05:37:01.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31215" PostTypeId="1" CreationDate="2015-05-11T05:57:34.457" Score="2" ViewCount="1971" Body="&lt;p&gt;Consider a solution of $\ce{KOH}$ which is in equilibrium with undissolved $\ce{KOH}$ at $25\ \mathrm{^\circ C}$.&#xA;After raising the temperature to $50\ \mathrm{^\circ C}$, the mass of undissolved $\ce{KOH}$ decreases. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="16139" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T11:05:31.163" LastActivityDate="2015-05-11T11:05:31.163" Title="Why does raising the temperature affect the mass of dissolved solute?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2015-05-11T11:59:46.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="31216" PostTypeId="2" ParentId="28525" CreationDate="2015-05-11T06:14:12.577" Score="2" Body="&lt;p&gt;Let me talk first about how iron corrodes (as in rusts) and then compare that to copper. The simple answer is that when iron corrodes, the iron oxide (iron which has chemically bonded with oxygen, which it loves to do) that forms on the surface of the pure iron underneath, does not seal the surface against further intrusion of more oxygen. When copper corrodes, the copper oxide does seal the surface protecting underlying copper against further corrosion. I don't know if iron oxide is simply more porous to oxygen than copper oxide is but I do know the following. Iron can chemically combine with oxygen in multiple ratios. When you examine a highly corroded iron surface you'll notice that the first layer of rust is very red. If you brush or maybe lightly sand that red layer of rust away you'll notice a black layer underneath it. That's still rust. If you sand your way through that you'll get to the pure iron underneath which has more of a dark silver color. The red rust has more oxygen in it than the black rust but can pass some of that oxygen to the black rust turning it red. Similarly, the black rust can pass some of its oxygen on to the pure iron underneath it. That process continues until oxygen can leach all the way through the iron. Chemically, iron has more than one oxidation state. Oxidized copper atoms are much more reluctant to release their oxygen to adjacent pure copper and act as a barrier to the intrusion of more oxygen.&lt;/p&gt;&#xA;" OwnerUserId="16137" LastActivityDate="2015-05-11T06:14:12.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31217" PostTypeId="2" ParentId="31215" CreationDate="2015-05-11T06:24:55.627" Score="1" Body="&lt;p&gt;For many (though not all) solids dissolved in water, solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the intermolecular forces of the solute molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many salts and some gasses there are visual representations of solubility variation with temperature called a &lt;strong&gt;solubility curve&lt;/strong&gt;, for which the units are $\frac{\mathrm{g~solute}}{\mathrm{100~g~solvent}}$. I could not find one for $\ce{KOH}$, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; for $\ce{KOH}$ lists the solubility of $\ce{KOH}$ at various temperatures, and it is clear the trend is that it is more soluble at higher temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of Le Châtelier's principle, substances that release heat upon being dissolved tend to dissolve better in colder water. This accounts for gasses and some salts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note, however, that although $\ce{KOH}$ does evolve a large amount of heat upon being dissolved, as the concentration of $\ce{KOH}$ increases, the process becomes less exothermic, and eventually near the saturation point it becomes endothermic (it becomes less favorable to dissolve more $\ce{KOH}$), which explains why it is more soluble at higher temperatures.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-05-11T06:47:06.940" LastActivityDate="2015-05-11T06:47:06.940" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31218" PostTypeId="1" AcceptedAnswerId="31219" CreationDate="2015-05-11T07:13:01.790" Score="3" ViewCount="192" Body="&lt;p&gt;I know that this constant tells us the strength of the interaction between the metal and the ligand,  so should it also mean that the &lt;em&gt;higher the constant, the stable the complex ?&lt;/em&gt;&#xA;I need to make sure of it because in photometric titration, all that matters is the stability of complex, and if the constant is given, one should know which complex is formed earlier.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-05-11T08:05:09.080" Title="Stability constant of complexes" Tags="&lt;analytical-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31219" PostTypeId="2" ParentId="31218" CreationDate="2015-05-11T08:05:09.080" Score="2" Body="&lt;p&gt;As $K$ is a thermodynamic constant, and if you define your reaction as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A + B \rightarrow{} C + D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;leading to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = \frac{c_C \cdot{} c_D}{c_A \cdot{}c_B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then -- thermodynamically -- yes.  Especially for chelate complexes (like with EDTA), entropically favoured. This answer excludes (macroscopic) kinetics &lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-05-11T08:05:09.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31220" PostTypeId="2" ParentId="31109" CreationDate="2015-05-11T08:59:37.237" Score="2" Body="&lt;p&gt;Comparing galvanic corrosion to the zinc/copper battery is a bit unfair. A fast corrosion rate is still much much slower kinetics compared to the expected rates of the electrochemical reaction in the battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter reaction in the galvanic corrosion case is almost always reduction of water. This is why it is much worse in conditions near the sea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galvanic corrosion rate can be measured and there is an ASTM standard(G71) to make the measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your comment below, I am trying to say that it is impossible to say if it will double or triple or just increase by 10%. The factors are complicated and hard to predict a priori. That is why it is a good idea to measure using the above mentioned standard.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastEditorUserId="13355" LastEditDate="2015-06-09T05:37:08.153" LastActivityDate="2015-06-09T05:37:08.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31221" PostTypeId="1" CreationDate="2015-05-11T09:49:48.917" Score="2" ViewCount="89" Body="&lt;p&gt;Can metal ions be engineered to behave like non-metals in terms of chemical properties (by supplying or removing electrons), would the altered metal ions bond covalently? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the ions be too unstable and begin to decay? &lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="10256" LastEditDate="2015-05-11T10:08:19.083" LastActivityDate="2015-05-11T10:08:19.083" Title="Can two metals bond covalently?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2015-05-11T10:59:55.947" ContentLicense="CC BY-SA 3.0" />
  <row Id="31222" PostTypeId="1" CreationDate="2015-05-11T09:55:58.370" Score="16" ViewCount="612" Body="&lt;p&gt;On a previous answer here (&lt;a href=&quot;https://chemistry.stackexchange.com/a/30841/876&quot;&gt;https://chemistry.stackexchange.com/a/30841/876&lt;/a&gt;) I was informed in the comments that OpenBabel &amp;amp; Python terminal could be used to print the data contained in the property matrices of Avogadro (more info can be found here &lt;a href=&quot;https://chemistry.stackexchange.com/q/30835/876&quot;&gt;How do I print property matrices in Avogadro?&lt;/a&gt;). I should be clear, however, most of the files I want to be able to do this for are CIF files where the bond angles and lengths are not explicitly mentioned in the files, so these properties may need to be predicted by the software. So I felt I couldn't be the only person interested in this so I thought I'd ask this question on this site. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like any answer to be suitable for those operating on Windows machines, if possible. I have Python &amp;amp; OpenBabel installed on my 64 bit Windows 7 PC, so I believe this shouldn't be a major issue, hopefully. &lt;/p&gt;&#xA;" OwnerUserId="876" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-05-14T18:40:33.177" Title="How do I create a spreadsheet containing the coordinates and predicted bond angles, molecular properties, torsion properties, etc of a molecule?" Tags="&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31223" PostTypeId="5" CreationDate="2015-05-11T10:39:26.650" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-11T10:39:26.650" LastActivityDate="2015-05-11T10:39:26.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31224" PostTypeId="4" CreationDate="2015-05-11T10:39:26.650" Score="0" Body="Questions related to explanations of the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-11T16:24:37.833" LastActivityDate="2015-05-11T16:24:37.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31225" PostTypeId="1" CreationDate="2015-05-11T10:40:44.353" Score="6" ViewCount="86" Body="&lt;p&gt;What is the mechanism for the photolysis of $\ce{HX}$ to dihydrogen and dihalogen ? From the low quantum yield (0.3% for $\ce{HI}$ according to one paper) it seems unlikely to be a chain process but I am not sure why it would not be. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorDisplayName="user15489" LastEditDate="2015-05-11T10:42:07.493" LastActivityDate="2015-05-11T10:42:07.493" Title="Mechanism of photolysis of HX" Tags="&lt;photochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31226" PostTypeId="1" AcceptedAnswerId="31231" CreationDate="2015-05-11T12:25:18.860" Score="3" ViewCount="18784" Body="&lt;p&gt;We have a investigation in my chemistry class in which we must calculate the amount of calcium carbonate in sea shells. We can have as many sea shells as we wish, lab equipment, and hydrochloric acid of varying concentration from $0.5~\mathrm{mol~dm^{-3}}$ to $2~\mathrm{mol~dm^{-3}}$. How is it to be done?&lt;/p&gt;&#xA;" OwnerUserId="16142" LastEditorUserId="15410" LastEditDate="2015-05-11T12:38:00.853" LastActivityDate="2019-02-28T21:05:46.810" Title="How to calculate the percentage of calcium carbonate in sea shells in an experiment" Tags="&lt;homework&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2019-03-09T17:14:20.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="31227" PostTypeId="2" ParentId="28682" CreationDate="2015-05-11T12:35:51.310" Score="3" Body="&lt;p&gt;I will try to take a different approach to your question (in terms of acidity and basicity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chromate (yellow) and dichromate (orange) ion are at equilibrium in solution. The equilibrium equation can be represented by &#xA;$$\ce{Cr2O7^2- + H2O &amp;lt;=&amp;gt; 2CrO4- + 2H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Le Chatelier’s Principle, if certain conditions (concentration, temperature, pressure, volume, etc.) are changed, the amount of each ion present in solution is affected. If a shift in pH causes the solution to become more acidic (i.e. add $\ce{HCl}$), the equilibrium will shift to the left, and more dichromate ion will be present. The reaction tries to offset the increase in $\ce{H+}$ concentration and shifts to the left accordingly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if a shift in pH causes the solution to be more basic (i.e. add $\ce{NaOH}$, which will cause some $\ce{H+}$ to react with $\ce{OH-}$ and produce water), the equilibrium will shift to the right, and more chromate ion will be present. This is a means by which the reaction tries to offset the loss of $\ce{H+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of these ions is pH dependent, as indicated by the color changes when the above reactions take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a little.&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T15:01:07.633" LastActivityDate="2015-05-11T15:01:07.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31228" PostTypeId="2" ParentId="31226" CreationDate="2015-05-11T12:46:12.050" Score="2" Body="&lt;p&gt;$\ce{CaCO3}$ is basic so it will neutralize the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + 2HCl -&amp;gt; CaCl2 + H2O +CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your experiment you need to devise a way of finding out what the end point of the reaction is (using a $\mathrm{pH}$ indicator to find when the mixture is neutral). You need to be able to measure the amount of hydrochloric acid used in moles  (you should measure the volume and you know the concentration. The number of moles of water is twice the amount of moles of the calcium carbonate (from the chemical equation above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should measure the mass of the amount of shell you use at the start. You will be able to figure out the mass of the calcium carbonate in that sample of shells that you use (since you now know the number of moles of calcium carbonate). Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\frac{\mathrm{Mass_{\ce{CaCO3}}}}{\mathrm{Mass_{shells}}}) \times 100\%=\mathrm{Percent}~\ce{CaCO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to do it with just one concentration but your teacher probably wants to talk to you about reliability and experimental errors.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="15410" LastEditDate="2015-05-11T12:58:18.260" LastActivityDate="2015-05-11T12:58:18.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31229" PostTypeId="2" ParentId="28483" CreationDate="2015-05-11T13:04:05.133" Score="2" Body="&lt;p&gt;Our solution is at equilibrium so $\ce {Q} = {K}$. That is, $\ce{K_{eq}} = {K_{sp}}$.&#xA;$$\Delta G = \Delta G^\circ + RT \ln K$$&#xA;At equilibrium, $\Delta G = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $$\Delta G^\circ = - RT \ln K$$ Now, you can use the value of $\ce{K_{sp}}$ and the other known values given to determine $\Delta G^\circ$.&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="16129" LastEditDate="2015-05-11T13:13:33.293" LastActivityDate="2015-05-11T13:13:33.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31230" PostTypeId="1" AcceptedAnswerId="31254" CreationDate="2015-05-11T13:12:19.360" Score="8" ViewCount="3946" Body="&lt;p&gt;Wouldn't it be great to be able to predict the melting point of a metal based on the atomic number and not by looking in some table for the value that was measured experimentally by someone else? Or to predict the same melting point for a simple salt like NaCl or KI by applying some formula to the atomic numbers of the 2 atoms involved? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you help me understand why is it so difficult to build a theoretical model for this particular physical property? Were there attempts made in the scientific world for such a model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--EDIT--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is part of my question: &#xA;I want to understand if there is a connection between the theoretical explanations of properties like melting point and the current theoretical model of atomic structure. Can the electronic structure of $\ce{Fe 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2}$ tell us anything about it's probable melting point without measuring that point in the first place?&lt;/p&gt;&#xA;" OwnerUserId="14515" LastEditorUserId="24052" LastEditDate="2018-06-14T06:00:22.783" LastActivityDate="2018-06-14T06:00:22.783" Title="Predicting melting points of substances based on their molecular/atomic structure" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31231" PostTypeId="2" ParentId="31226" CreationDate="2015-05-11T13:20:04.970" Score="0" Body="&lt;p&gt;$\ce{CaCO3}$ reacts with $\ce{HCl}$ in the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3_{(s)} + 2HCl_{(aq)} -&amp;gt; CaCl2_{(aq)} + H2O_{(l)} +CO2_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the easiest way to measure the results of this reaction would be to collect the $\ce{CO2}$. Grind up the seashell into a powder, record the mass of this. Assemble an apparatus like this, and react the seashell with an excess of $\ce{HCl}$ in the reaction chamber to collect and measure the volume of gas evolved (&lt;a href=&quot;http://www.docbrown.info&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SedUo.png&quot; alt=&quot;apparatus&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is a gas law calculation, you will need the atmospheric pressure and ambient air temperature at the time you perform the reaction. After obtaining the final volume of the gas in the reaction, you will then need to account for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapour_pressure_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;partial pressure due to water vapor&lt;/a&gt;, and subtract this from the atmospheric pressure accordingly. Now this is simply a matter of stoichiometry, calculating how much $\ce{CaCO3}$ would be required to obtain this volume of gas at this temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then use the equation RobChem stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\frac{\mathrm{Mass_{\ce{CaCO3}}}}{\mathrm{Mass_{shells}}}) \times 100\%=\mathrm{Percent}~\ce{CaCO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to find the percent $\ce{CaCO3}$ by mass in the seashell.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-11T13:20:04.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31232" PostTypeId="2" ParentId="31230" CreationDate="2015-05-11T13:33:28.470" Score="10" Body="&lt;p&gt;With regard to organic compounds, you might be interested in group contribution methods, e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Joback_method&quot;&gt;Joback method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These methods divide a molecule into functional groups or similar small building blocks and estimate the melting point (and other thermodynamic properties) based on the contributions of the individual groups.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-11T13:33:28.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31233" PostTypeId="2" ParentId="31222" CreationDate="2015-05-11T14:20:16.057" Score="10" Body="&lt;p&gt;You want the Open Babel &quot;report&quot; format, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel benzene.cml -oreport&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;FILENAME: benzene.g03&#xA;FORMULA: C6H6&#xA;MASS: 78.1118&#xA;EXACT MASS: 78.0469502&#xA;INTERATOMIC DISTANCES&#xA;&#xA;              C   1      C   2      C   3      C   4      C   5      C   6&#xA;              ------------------------------------------------------------------&#xA;   C   1    0.0000 &#xA;   C   2    1.3975     0.0000 &#xA;   C   3    2.4207     1.3975     0.0000 &#xA;   C   4    2.7952     2.4205     1.3976     0.0000 &#xA;   C   5    2.4205     2.7948     2.4205     1.3975     0.0000 &#xA;   C   6    1.3976     2.4205     2.7952     2.4207     1.3975     0.0000 &#xA;   H   7    1.0841     2.1549     3.4030     3.8793     3.4029     2.1549 &#xA;   H   8    2.1548     1.0841     2.1548     3.4028     3.8789     3.4028 &#xA;   H   9    3.4030     2.1549     1.0841     2.1549     3.4029     3.8793 &#xA;   H  10    3.8793     3.4029     2.1549     1.0841     2.1549     3.4030 &#xA;   H  11    3.4028     3.8789     3.4028     2.1548     1.0841     2.1548 &#xA;   H  12    2.1549     3.4029     3.8793     3.4030     2.1549     1.0841 &#xA;&#xA;              H   7      H   8      H   9      H  10      H  11      H  12&#xA;              ------------------------------------------------------------------&#xA;   H   7    0.0000 &#xA;   H   8    2.4816     0.0000 &#xA;   H   9    4.2984     2.4816     0.0000 &#xA;   H  10    4.9634     4.2982     2.4817     0.0000 &#xA;   H  11    4.2982     4.9630     4.2982     2.4816     0.0000 &#xA;   H  12    2.4817     4.2982     4.9634     4.2984     2.4816     0.0000 &#xA;&#xA;&#xA;&#xA;&#xA;ATOMIC CHARGES&#xA;   C   1   -0.1860370000&#xA;   C   2   -0.1859930000&#xA;   C   3   -0.1860370000&#xA;   C   4   -0.1860370000&#xA;   C   5   -0.1859930000&#xA;   C   6   -0.1860370000&#xA;   H   7    0.1860240000&#xA;   H   8    0.1860180000&#xA;   H   9    0.1860240000&#xA;   H  10    0.1860240000&#xA;   H  11    0.1860180000&#xA;   H  12    0.1860240000&#xA;&#xA;&#xA;BOND ANGLES&#xA;   6    1    7  Car  Car    H    120.003&#xA;   2    1    6  Car  Car  Car    119.993&#xA;   2    1    7  Car  Car    H    120.004&#xA;   1    2    3  Car  Car  Car    120.014&#xA;   1    2    8  Car  Car    H    119.993&#xA;   3    2    8  Car  Car    H    119.993&#xA;   4    3    9  Car  Car    H    120.003&#xA;   2    3    4  Car  Car  Car    119.993&#xA;   2    3    9  Car  Car    H    120.004&#xA;   5    4   10  Car  Car    H    120.004&#xA;   3    4    5  Car  Car  Car    119.993&#xA;   3    4   10  Car  Car    H    120.003&#xA;   4    5   11  Car  Car    H    119.993&#xA;   6    5   11  Car  Car    H    119.993&#xA;   4    5    6  Car  Car  Car    120.014&#xA;   5    6   12  Car  Car    H    120.004&#xA;   1    6    5  Car  Car  Car    119.993&#xA;   1    6   12  Car  Car    H    120.003&#xA;&#xA;&#xA;TORSION ANGLES&#xA;  11    5    4   10     -0.000&#xA;  11    5    4    3   -180.000&#xA;   6    5    4   10   -180.000&#xA;   6    5    4    3     -0.000&#xA;  11    5    6   12      0.000&#xA;  11    5    6    1   -180.000&#xA;   4    5    6   12   -180.000&#xA;   4    5    6    1     -0.000&#xA;   5    4    3    9   -180.000&#xA;   5    4    3    2     -0.000&#xA;  10    4    3    9     -0.000&#xA;  10    4    3    2   -180.000&#xA;   5    6    1    7   -180.000&#xA;   5    6    1    2     -0.000&#xA;  12    6    1    7     -0.000&#xA;  12    6    1    2   -180.000&#xA;   6    1    2    3     -0.000&#xA;   6    1    2    8   -180.000&#xA;   7    1    2    3   -180.000&#xA;   7    1    2    8     -0.000&#xA;   4    3    2    1      0.000&#xA;   4    3    2    8   -180.000&#xA;   9    3    2    1   -180.000&#xA;   9    3    2    8     -0.000&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned wanting a CSV-parsable output, e.g., from Python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote up a fairly quick tool for this &lt;a href=&quot;https://gist.github.com/ghutchis/b692761252833ccdda5a&quot; rel=&quot;noreferrer&quot;&gt;molcsv.py&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be pretty obvious about modifying as needed in the different &lt;code&gt;for&lt;/code&gt; loops.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-05-14T18:37:51.860" LastActivityDate="2015-05-14T18:37:51.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31234" PostTypeId="1" AcceptedAnswerId="31239" CreationDate="2015-05-11T14:38:05.710" Score="23" ViewCount="5761" Body="&lt;p&gt;I know that $\ce{C-D}$ and $\ce{C-T}$ bonds are stronger than $\ce{C-H}$ bonds, but is this generally true for isotopes of all elements? (By isotopes I mean stable ones with large half lives.)&lt;/p&gt;&#xA;" OwnerUserId="16144" LastEditorUserId="10185" LastEditDate="2015-05-11T15:24:50.473" LastActivityDate="2015-05-11T23:50:16.810" Title="For a given element do heavier isotopes form stronger bonds than lighter isotopes?" Tags="&lt;bond&gt;&lt;theoretical-chemistry&gt;&lt;stability&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31236" PostTypeId="2" ParentId="31222" CreationDate="2015-05-11T14:47:25.747" Score="3" Body="&lt;p&gt;Load your CIF file in &lt;a href=&quot;http://www.cryst.chem.uu.nl/spek/platon/pl000000.html&quot; rel=&quot;nofollow&quot;&gt;PLATON&lt;/a&gt; and run &lt;code&gt;CALC-GEO&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;code&gt;*.LIS&lt;/code&gt; file generated, you'll find all the tables that you need, namely&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bond Lengths (Angstrom)&lt;/li&gt;&#xA;&lt;li&gt;Bond/Valence Angles (Degrees) &lt;/li&gt;&#xA;&lt;li&gt;Torsion/Dihedral Angles (Deg.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can still use the &lt;code&gt;regex&lt;/code&gt; module of &lt;a href=&quot;https://www.python.org/&quot; rel=&quot;nofollow&quot;&gt;Python&lt;/a&gt; to parse the data,&lt;code&gt;\s*(?P&amp;lt;atoms&amp;gt;([A-Z][a-z]?\(\d*\w\)\s*){4})\s*(?P&amp;lt;value&amp;gt;-?\d*\.\d\(\d+\))&lt;/code&gt; will split entries in the torsion table into two named groups, which you can further process and write to a CSV file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that OpenBabel is an &lt;strong&gt;excellent&lt;/strong&gt; tool! In this case, however, I suggest to use the approach above, rather than parsing the OpenBabel report file for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In a couple of cases, I noticed that the report generator of OpenBabel had some problems with CIF files that were checked both using the internal check routines of PLATON, as well as  &lt;a href=&quot;http://checkcif.iucr.org/&quot; rel=&quot;nofollow&quot;&gt;CheckCIF&lt;/a&gt;, an online tool provided by the IUCr, and were found to be valid.  &lt;/li&gt;&#xA;&lt;li&gt;In the table for the dihedral angles, the OpenBabel report does not use descriptors like &lt;code&gt;C(5), H(10A)&lt;/code&gt; to denote atoms in a dihedral, but uses the atom indices. If you would want to write the decriptors to a CSV file, you would have to look them up at another table.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you can live with the indices, the following Python snippet should suffice:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/env python3&#xA;&#xA;import csv&#xA;import argparse&#xA;&#xA;def get_torsions_from_report(report_path):&#xA;    '''Read all rows with he indices to four consecutive atoms and &#xA;       the corresponding dihedral angle from an OpenBabel report file.'''&#xA;    try:&#xA;        torsion_list = []&#xA;        with open(report_path) as infile:&#xA;            for line in infile:&#xA;                if line.startswith('TORSION ANGLES'):&#xA;                    break&#xA;&#xA;            for line in infile:&#xA;                if line.strip():&#xA;                    torsion_list.append(line.split())&#xA;&#xA;        return torsion_list&#xA;    except IOError:&#xA;        print(e)&#xA;&#xA;&#xA;def write_torsions_to_csv(csv_path, torsion_list):&#xA;    '''Write torsion data to CSV.'''&#xA;    with open(csv_path, 'wt') as fout:&#xA;        writer = csv.writer(fout)&#xA;        writer.writerow(('Atom1', 'Atom 2', 'Atom 3', 'Atom 4', 'Torsion'))&#xA;        writer.writerows(torsion_list)&#xA;&#xA;if __name__ == '__main__':&#xA;    parser = argparse.ArgumentParser(description='Read torsions &#xA;    from OpenBabel report and write them to a CSV file.')&#xA;    parser.add_argument('-i', dest=&quot;infile&quot;, required=True,&#xA;    help=&quot;OpenBabel report file&quot;)&#xA;    parser.add_argument('-o', dest=&quot;outfile&quot;, required=True, &#xA;    help=&quot;CSV file&quot;)&#xA;&#xA;    args = parser.parse_args()  &#xA;    torsions = get_torsions_from_report(args.infile)&#xA;    write_torsions_to_csv(args.outfile, torsions)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-12T06:53:54.810" LastActivityDate="2015-05-12T06:53:54.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31238" PostTypeId="2" ParentId="31234" CreationDate="2015-05-11T15:17:17.917" Score="12" Body="&lt;p&gt;When comparing isotopes, the different nuclear masses control the bond length and bond strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the $n^\text{th}$ Bohr orbit is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{n} = {n^2\hbar^2\over Zk_\text{c} e^2 m_\text{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Z$ is the atom’s atomic number, $k_\text{c}$ is Coulomb’s constant, $e$ is the electron charge, and $m_\text{e}$ is the mass of the electron. In this equation it is assumed that the electron orbits the nucleus and the nucleus remains stationary. Given the mass difference between the electron and nucleus, this is generally a reasonable assumption. However, in reality the nucleus does move too. It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\text{e}$ with the electron’s reduced mass, $\mu_\text{e}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_\text{e} = \frac{m_\text{e} \cdot m_\text{nucleus}}{m_\text{e} + m_\text{nucleus}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation for the Bohr radius becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_n = {n^2\hbar^2\over Zk_\text{c} e^2 \mu_\text{e}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_\text{heavy} &amp;lt; r_\text{light}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and consequently an electron will orbit closer to the nucleus of a heavier isotope than it will orbit the nucleus of a lighter isotope. In turn, this means that &lt;strong&gt;the heavy isotope will have shorter, stronger bonds.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-05-11T19:03:32.553" LastActivityDate="2015-05-11T19:03:32.553" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31239" PostTypeId="2" ParentId="31234" CreationDate="2015-05-11T15:30:35.290" Score="16" Body="&lt;p&gt;Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by; &#xA;\begin{equation}&#xA;\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}&#xA;\end{equation}&#xA;Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega $ with  $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Infrared_spectroscopy&quot; rel=&quot;noreferrer&quot;&gt;vibrational spectroscopy&lt;/a&gt; this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uncertainty_principle&quot; rel=&quot;noreferrer&quot;&gt;uncertainty principle&lt;/a&gt; in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum oscillator has a Hamiltonian of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harmonic_oscillator&quot; rel=&quot;noreferrer&quot;&gt;classical oscillator&lt;/a&gt; except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_mechanics&quot; rel=&quot;noreferrer&quot;&gt;classical Hamiltonian&lt;/a&gt; (or total energy) is learnt at school to be kinetic + potential,&#xA;\begin{equation}&#xA;H=\frac{p^2}{2\mu}+\frac 12 kx^2&#xA;\end{equation} &#xA;For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by &lt;a href=&quot;http://en.wikipedia.org/wiki/Momentum_operator&quot; rel=&quot;noreferrer&quot;&gt;$\hat p$&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Position_operator&quot; rel=&quot;noreferrer&quot;&gt;$\hat x$&lt;/a&gt; where,&#xA;\begin{equation}&#xA;\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x&#xA;\end{equation}&#xA;Therefore with these changes in the above classical expression we generate the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics)&quot; rel=&quot;noreferrer&quot;&gt;quantum Hamiltonian&lt;/a&gt; of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator&quot; rel=&quot;noreferrer&quot;&gt;quantum harmonic oscillator&lt;/a&gt;, &#xA;\begin{equation}&#xA;\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2 &#xA;\end{equation}&#xA;Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is another principle of quantum mechanics that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;noreferrer&quot;&gt;Schrodinger equation&lt;/a&gt; is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form. &#xA;\begin{equation}&#xA;\hat H\psi =E\psi&#xA;\end{equation}&#xA;Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;E_n=(n+\frac 12)\hbar \omega&#xA;\end{equation}&#xA;What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a zero point energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As promised here is the home-made graph of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Morse_potential&quot; rel=&quot;noreferrer&quot;&gt;molecular potential energy&lt;/a&gt;. Im sorry its not very good but honestly the best I could do with the available tools! Apologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main points to follow are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number). &lt;/li&gt;&#xA;&lt;li&gt;Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VObGy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7913" LastEditDate="2015-05-11T23:50:16.810" LastActivityDate="2015-05-11T23:50:16.810" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31240" PostTypeId="1" CreationDate="2015-05-11T15:41:22.920" Score="5" ViewCount="219" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  $\mathrm{5~mol}$ of a gas is confined in a piston (volume $\mathrm{100~cm^3}$) at $\mathrm{400~K}$. $\mathrm{10~kJ}$ of heat energy is supplied, causing a reversible and isothermal expansion of the gas. Calculate the volume of the expanded, heated, gas. You may use the equation below (assume ideal gas behaviour).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$W = -nRT(\ln V_2-\ln V_1)$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I substituted the values (after converting to SI units) into the equation and attempted to rearrange to find $V_2$, but work was unknown. I tried to use the enthalpy value (10 kJ) given but I was unsure of what to do with it.&lt;/p&gt;&#xA;" OwnerUserId="16146" LastEditorUserId="194" LastEditDate="2018-03-05T03:19:01.573" LastActivityDate="2018-03-05T03:50:49.427" Title="Calculating new volume during expansion of an ideal gas" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31241" PostTypeId="1" CreationDate="2015-05-11T15:46:49.693" Score="4" ViewCount="13793" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the contact process, the commercial process of producing $\ce{H2SO4}$, several reactions occur before the final product is produced:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{align}&#xA;\ce{S8(s) + 8O2(g) &amp;amp;-&amp;gt; 8SO2(g)} &amp;amp;\Delta H_\mathrm{f} &amp;amp;= \pu{-70.9 kcal mol^{-1}}\tag{1} \\&#xA;\ce{2SO2(g) + O2(g) &amp;amp;-&amp;gt;[V2O5] 2SO3(g)} &amp;amp;\Delta H_\mathrm{f} &amp;amp;= \pu{-94.5 kcal mol^{-1}}\tag{2} \\&#xA;\ce{SO3(g) + H2O(g) &amp;amp;-&amp;gt; H2SO4(g)} &amp;amp;\Delta H_\mathrm{f} &amp;amp;= \pu{-193.9 kcal mol^{-1}}\tag{3}&#xA;\end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Given that $\Delta H_\mathrm{f}$ for $\ce{H2O(g)}$ is $\pu{-68.3 kcal mol^{-1}}$, what is the enthalpy of formation of $\pu{1 mol}$ of $\ce{H2SO4}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I've tried to do:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.&lt;/li&gt;&#xA;&lt;li&gt;Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15525" LastEditorUserId="83758" LastEditDate="2020-02-20T15:18:57.770" LastActivityDate="2020-02-20T15:18:57.770" Title="What is the enthalpy of formation of 1 mol of sulfuric acid?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31242" PostTypeId="2" ParentId="31240" CreationDate="2015-05-11T16:36:51.317" Score="3" Body="&lt;p&gt;I dont understand where you are getting stuck up since everything is already given to you and you just have to substitute stuff using: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta U= W+Q$, the basic conservation equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ is the heat supplied. Think about the change in internal energy ($U$) during an isothermal process.&lt;/p&gt;&#xA;" OwnerUserId="16148" LastEditorUserId="5026" LastEditDate="2018-03-05T03:50:49.427" LastActivityDate="2018-03-05T03:50:49.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31243" PostTypeId="1" AcceptedAnswerId="31267" CreationDate="2015-05-11T16:52:18.540" Score="13" ViewCount="22095" Body="&lt;p&gt;What is formed when you leave iron(II) sulfate in plain air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that iron(II) is easily oxidised to iron(III), and assuming that the reactive component of air is oxygen, I solved it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{FeSO4 + O2 -&amp;gt; Fe2O3 + SO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in my textbook it's given that iron(III) sulfate, $\ce{Fe2(SO4)3}$, is formed. Is this because $\ce{Fe2O3 + SO2}$ react to form $\ce{Fe2(SO4)3}$?&lt;/p&gt;&#xA;" OwnerUserId="9963" LastEditorUserId="16683" LastEditDate="2017-09-02T18:24:16.613" LastActivityDate="2017-09-02T18:24:16.613" Title="What is formed when you leave iron(II) sulfate in plain air?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31245" PostTypeId="1" AcceptedAnswerId="31253" CreationDate="2015-05-11T17:36:36.560" Score="5" ViewCount="304" Body="&lt;p&gt;I am asked to identify the highest and lowest energy conformations for the following bond-line structure (I think it is 2,2-dimethylbutane). I drew a Newman projection (for all 6 conformations) and was able to identify the lowest energy conformation correctly but the textbook solution for the highest energy conformation does not match what I drew. It seems as though the textbook solution is missing a methyl group, and has 3 $\ce{H}$ bonds instead. Can someone please clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bMGbJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="1887" LastEditDate="2015-05-11T18:56:45.820" LastActivityDate="2015-05-11T20:16:40.370" Title="Energy of Eclipsed Interactions" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;conformers&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31248" PostTypeId="2" ParentId="31241" CreationDate="2015-05-11T18:48:15.320" Score="2" Body="&lt;p&gt;Since enthalpy is a state function, all that is needed is the $\Delta H_{\mathrm{f}}$ of each reactants that form $\ce{H2SO4}$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO3{(g)} +H2O{(g)} -&amp;gt; H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}-&amp;gt;H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -&amp;gt; SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these equations yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -&amp;gt; H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(according to the equations and numerical values provided)&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2016-10-11T00:11:48.403" LastActivityDate="2016-10-11T00:11:48.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31249" PostTypeId="1" CreationDate="2015-05-11T18:53:29.947" Score="5" ViewCount="60" Body="&lt;p&gt;I want to know if there is a way to find number of solvent molecules that are &lt;strong&gt;directly&lt;/strong&gt; embracing a solute molecule (something like a reference or table that shows &lt;strong&gt;waters of hydration&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Particularly I am interested in solvents such as water or ethanol, and the solutes acetone, acetaldehyde, 1-butanol, methylene blue, benzene, anthracene, and phenanthrene.&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="15410" LastEditDate="2016-05-05T22:42:11.123" LastActivityDate="2016-05-05T22:42:11.123" Title="Reference request for the number of solvents around a solute" Tags="&lt;solubility&gt;&lt;solutions&gt;&lt;reference-request&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="2" ClosedDate="2016-05-06T10:23:29.267" ContentLicense="CC BY-SA 3.0" />
  <row Id="31252" PostTypeId="1" AcceptedAnswerId="31255" CreationDate="2015-05-11T19:02:21.407" Score="15" ViewCount="11212" Body="&lt;p&gt;I'm wondering why maleic anhydride adds to the middle cycle of anthracene, and not the outer two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qsqZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qsqZZ.png&quot; alt=&quot;Reaction of anthracene with maleic anhydride&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have expected that a Diels–Alder with the outer ring would be better, because I expected a naphtalene part to be lower in energy than two benzene parts (more resonance stabilisation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it has to do with reactant based arguments that the atomic coefficients for the two center carbon atoms (C-9 and C-10) are higher than from the outer cycle (C-1 and C-4). I ran a calculation using &lt;a href=&quot;http://www.chem.ucalgary.ca/SHMO&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.chem.ucalgary.ca/SHMO&lt;/a&gt; and the coefficients on C-9 and C-10 were 0.44, whereas those on C-1 and C-4 were only 0.31.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2017-09-23T22:29:02.527" LastActivityDate="2017-09-23T22:29:02.527" Title="Why is it the middle ring of anthracene which reacts in a Diels–Alder?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;pericyclic&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31253" PostTypeId="2" ParentId="31245" CreationDate="2015-05-11T20:16:40.370" Score="2" Body="&lt;p&gt;You are correct in drawing this. The &quot;answer&quot; molecule depicted by your book is different from the one it first depicted, so it cannot be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You correctly depicted the high energy conformation of the molecule as a rotation of the low energy that shows the methyl groups aligned. This is high energy because of the intramolecular repulsion of the hydrogen atoms and the electron clouds.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-11T20:16:40.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31254" PostTypeId="2" ParentId="31230" CreationDate="2015-05-11T20:17:41.040" Score="10" Body="&lt;p&gt;Melting point depends not only on the structure of the molecule itself, but also on the structure of the solid (crystalline) phase and molten/liquid state of the substance. That is a much bigger computational problem than calculating energetics for a small molecule or ensemble. It is not easy to predict how a molecule (especially a complex one) will crystallize, and even if you do know that, a big computation to calculate the energy of the crystalline state.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-11T20:17:41.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31255" PostTypeId="2" ParentId="31252" CreationDate="2015-05-11T20:46:10.317" Score="11" Body="&lt;p&gt;To provide a reason for the observed &lt;em&gt;regio&lt;/em&gt;selectivity, it is helpful to draw anthracene's aromatic π-electron system in alternance of single and double bonds.  In this instance, it is more beneficial than &quot;the ring&quot; symbolizing the delocalised electron system, as this helps you to account for the precise number of π-electrons before the reaction (starting materials), during the reaction (the mechanism), and after the reaction (the product).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXmsd.png&quot; alt=&quot;Anthracene numbering&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Diels–Alder reaction, you may imagine two different pathways.  I invite you to draw the mechanisms by yourself:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A reaction that involves carbon atoms #1 and #4 (or #5 and #8).  Possible, by mechanism.  There are five double bonds remaining in conjugation, and you count one six-membered ring in the state of &quot;a benzene ring&quot; (the very left one).  In the very right six-membered ring, there is only a single double bond, too.&lt;/li&gt;&#xA;&lt;li&gt;Alternatively, a Diels–Alder reaction with carbon atoms #9 and #10.  This is more favourable then the former example, because &lt;em&gt;both&lt;/em&gt; the very left as well as the very right ring of the product are in the state of &quot;a benzene ring&quot;, each of these rings contain 6 π-electrons, the prerequisite of a Hückel aromatic compound.  It is so much favourable to the former, that this is the reaction observed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It may be helpful to add that benzene, naphthalene and anthracene are of course Hückel-aromatic compounds; with 6, 10 or 14 π-electrons they fit into the rule of $(4n + 2)$.  Yet gradually, as experimentally found, in this group of three, benzene is the most, anthracene the least aromatic compound.&lt;/p&gt;&#xA;" OwnerUserId="1782" LastEditorUserId="16683" LastEditDate="2017-06-30T09:23:50.923" LastActivityDate="2017-06-30T09:23:50.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31256" PostTypeId="1" CreationDate="2015-05-11T20:52:27.027" Score="11" ViewCount="272" Body="&lt;p&gt;I have been writing a custom MD code for some time. I have been using the Andersen thermostat, mainly because its ease of use, and the fact that I am not concerned with gathering dynamical data, but rather free energies. I have implemented the SHAKE algorithm found in LAMMPS. However, the thermostat seems to be undercooling now, when constraints are present. For instance, if I make the set point temperature 350 K, my system will equilibrate at around 330 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am calculating the number of degrees of freedom for the temperature calculation as $3N-3-$number of bond constraints. Has anyone else had a similar experience with the Andersen thermostat? Would a Langevin thermostat work better here? Why is the Andersen thermostat not playing well with the SHAKE algorithm?&lt;/p&gt;&#xA;" OwnerUserId="16152" LastEditorUserId="402" LastEditDate="2017-05-03T23:27:16.550" LastActivityDate="2017-07-02T23:55:29.573" Title="Andersen thermostat in MD with rigid bond constraints" Tags="&lt;computational-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31257" PostTypeId="1" AcceptedAnswerId="31284" CreationDate="2015-05-11T23:40:09.757" Score="12" ViewCount="673" Body="&lt;p&gt;I am trying to simulate the excitation state of acetone. I ran TDDFT for it both in gas phase and solvated state in water (both implicit and explicit water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental data say that acetone undergoes n-&gt;π* transition, which means it has a higher wavelength in gas phase ($\mathrm{\lambda_{max}}=$~$276~\mathrm{nm}$) and a shorter wavelength in the solvated state ($\mathrm{\lambda_{max}}=265~\mathrm{nm}$).  I started with gas phase and expected something a wavelength of about $276~\mathrm{nm}$ ($4.49~\mathrm{eV}$ ), but surprisingly I got $\mathrm{\lambda_{max}}=136.172~\mathrm{nm}$ ($9.326~\mathrm{eV}$)! I really cannot understand why there is such a big discrepancy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my GAMESS input file. What is wrong with this molecule ? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!   File created by the GAMESS Input Deck Generator Plugin for Avogadro&#xA; $BASIS GBASIS=N311 NGAUSS=6 $END&#xA; $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END&#xA; $CONTRL ICHARG=0 MULT=1 $END&#xA; $TDDFT NSTATE=9 $END&#xA; $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END&#xA; $SYSTEM MWORDS=1000 PARALL=.TRUE. $END&#xA; $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END&#xA;&#xA; $DATA &#xA;Title&#xA;C1&#xA;O     8.0     0.00000    -1.27900     0.00300&#xA;C     6.0    -0.00000    -0.05800     0.00100&#xA;C     6.0     1.29700     0.69100    -0.00000&#xA;C     6.0    -1.29800     0.69000    -0.00000&#xA;H     1.0     1.35900     1.32900    -0.90600&#xA;H     1.0     1.35900     1.33200     0.90300&#xA;H     1.0     2.15700    -0.01300     0.00100&#xA;H     1.0    -2.15700    -0.01400     0.00100&#xA;H     1.0    -1.35900     1.32900    -0.90600&#xA;H     1.0    -1.35900     1.33200     0.90300&#xA;&#xA; $END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="15410" LastEditDate="2015-05-12T00:49:38.257" LastActivityDate="2015-05-13T13:25:11.623" Title="Why Acetone does not behave like its computational values?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31259" PostTypeId="2" ParentId="31125" CreationDate="2015-05-12T05:03:01.503" Score="1" Body="&lt;p&gt;The first step is to determine the $\mathrm{p}K_{\mathrm{a}}$ values for the indicators and convert them to $\mathrm{p}K_{\mathrm{b}}$ values. Then determine how you would titrate the solution using the indicator(s).&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="15410" LastEditDate="2015-05-12T05:05:31.127" LastActivityDate="2015-05-12T05:05:31.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31260" PostTypeId="2" ParentId="31125" CreationDate="2015-05-12T05:13:29.887" Score="3" Body="&lt;p&gt;Since: &#xA;\begin{align}&#xA;K_\mathrm{w} &amp;amp;=K_\mathrm{a} \cdot K_\mathrm{b}\\&#xA;K_\mathrm{a} &amp;amp;=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\&#xA;\log{K_\mathrm{a}} &amp;amp;=\log K_\mathrm{w}-\log K_\mathrm{b}\\&#xA;K_\mathrm{w} &amp;amp;=1 \times 10^{-14}\\&#xA;\log{K_\mathrm{a}} &amp;amp;= -14-\log K_\mathrm{b}\\&#xA;-\log{K_\mathrm{a}} &amp;amp;=14+\log K_\mathrm{b}\\&#xA;-\log{K_\mathrm{a}} &amp;amp;=\mathrm{p}K_\mathrm{a}\\&#xA;-\log{K_\mathrm{b}} &amp;amp;=\mathrm{p}K_\mathrm{b}\\&#xA;\mathrm{p}K_\mathrm{a} &amp;amp;=14-\mathrm{p}K_\mathrm{b}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-05-12T07:31:10.107" LastActivityDate="2015-05-12T07:31:10.107" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31262" PostTypeId="1" AcceptedAnswerId="31275" CreationDate="2015-05-12T08:43:56.157" Score="4" ViewCount="919" Body="&lt;p&gt;Arranging spheres ABABAB type in 2D gives us close packing. Now, this layer if arranged in ABABAB (in 3D) gives us HCP and if arranged in ABCABC (in 3D) gives us FCC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do we get if we arrange it in AAAAA type (in 3D)? (we are talking about one type of spheres only)&lt;/p&gt;&#xA;" OwnerUserId="16157" LastEditorUserId="9961" LastEditDate="2015-10-23T17:40:17.597" LastActivityDate="2015-10-23T17:40:17.597" Title="Which unit cell do we obtain if spheres are arranged in ABABAB type in 2D and then that 2D layer is arranged in AAAAAA type in 3D?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31264" PostTypeId="1" CreationDate="2015-05-12T09:28:13.457" Score="9" ViewCount="750" Body="&lt;p&gt;If I had a container in vacuum filled with $\ce{I2}$ gas and then I bombarded it with high speed electrons using an electron gun, would  be able to get both $\ce{I+}$ and $\ce{I-}$ ions or would I only get $\ce{I2+}$ plus an ejected electron?&lt;/p&gt;&#xA;" OwnerUserId="15322" LastEditorUserId="15410" LastEditDate="2015-05-13T01:50:33.273" LastActivityDate="2016-03-11T23:50:27.920" Title="is it possible to break the bonds of diatomic elements such as fluorine or iodine and create positive and negative ions by electron bombardment?" Tags="&lt;electron-affinity&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31265" PostTypeId="2" ParentId="31176" CreationDate="2015-05-12T09:49:29.443" Score="2" Body="&lt;p&gt;The unit $cmmol\,dm^{-3}$ is the same as $mmol_{c}\,dm^{-3}$ which denotes the millimoles of charge per litre. I think the former notation is not well used anymore. On the other hand, latter is used often and a google search would return a number of references. For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://micromaintain.ucanr.edu/Prediction/Source/Groundwater/Potential_for_clogging/CP/Water_analysis_for_hazard/Levels_of_concern/&quot; rel=&quot;nofollow&quot;&gt;http://micromaintain.ucanr.edu/Prediction/Source/Groundwater/Potential_for_clogging/CP/Water_analysis_for_hazard/Levels_of_concern/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.slidefinder.net/c/converting_various_units_mmol_mmolc/32208520&quot; rel=&quot;nofollow&quot;&gt;http://www.slidefinder.net/c/converting_various_units_mmol_mmolc/32208520&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.scielo.br/scielo.php?pid=S0103-90162011000400012&amp;amp;script=sci_arttext&quot; rel=&quot;nofollow&quot;&gt;http://www.scielo.br/scielo.php?pid=S0103-90162011000400012&amp;amp;script=sci_arttext&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16096" LastActivityDate="2015-05-12T09:49:29.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31266" PostTypeId="2" ParentId="31264" CreationDate="2015-05-12T09:59:34.820" Score="7" Body="&lt;p&gt;Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms:&#xA;$$&#xA;\ce{I2 + e- -&amp;gt; I- + I} \, .&#xA;$$&#xA;But somewhere around $8.62 \pm 0.06 \, \mathrm{eV}$ according to &lt;a href=&quot;http://www.nrcresearchpress.com/doi/abs/10.1139/v60-057#.VVHO8mBmMUU&quot; rel=&quot;nofollow&quot;&gt;this study&lt;/a&gt;, you'll indeed have the ion pair formation:&#xA;$$&#xA;\ce{I2 + e- -&amp;gt; I- + I+ + e-} \, .&#xA;$$&#xA;There are some other processes which contribute to the ion pair formation. For details, read the paper quoted above.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With respect to the formation of $\ce{I2+}$ note that to form it a bombarding electron have to, formally speaking,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Either first break the $\ce{I-I}$ bond homolyticly and then remove two electrons from one of the $\ce{I}$ atom which requires it to carry about 30 eV of energy (the sum of the homolytic bond dissociation energy and of the first and the second ionization energies).&lt;/li&gt;&#xA;&lt;li&gt;Or break the $\ce{I-I}$ bond heteroliticly and then remove the second electron from $\ce{I-}$ ion which requires it to carry also about 30 eV of energy (the sum of the heterolytic bond dissociation energy and of the second ionization energy).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$%edit$&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="24052" LastEditDate="2016-03-11T23:50:27.920" LastActivityDate="2016-03-11T23:50:27.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31267" PostTypeId="2" ParentId="31243" CreationDate="2015-05-12T10:47:57.987" Score="14" Body="&lt;p&gt;Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{[Fe(H2O)6]^3+ + e- \;&amp;amp;&amp;lt;=&amp;gt; [Fe(H2O)6]^2+}\quad &amp;amp;&amp;amp;E^\circ = +0.771\ \mathrm{V}\\&#xA;\ce{O2 + 4H+ + 4e- \;&amp;amp;&amp;lt;=&amp;gt; 2H2O}\quad &amp;amp;&amp;amp;E^\circ = +1.229\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{FeO(OH) + H2O + e- \;&amp;amp;&amp;lt;=&amp;gt; Fe(OH)2 + OH-}\quad &amp;amp;&amp;amp;E^\circ = -0.69\ \mathrm{V}\\&#xA;\ce{O2 + 2H2O + 4e- \;&amp;amp;&amp;lt;=&amp;gt; 4OH-}\quad &amp;amp;&amp;amp;E^\circ = +0.401\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2FeSO4 + SO4^2- -&amp;gt; Fe2(SO4)3 + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{[Fe(H2O)6]^3+ \;&amp;amp;&amp;lt;=&amp;gt; [Fe(OH)(H2O)5]^2+ + H+}\\&#xA;\ce{[Fe(OH)(H2O)5]^2+ \;&amp;amp;&amp;lt;=&amp;gt; [Fe(OH)2(H2O)5]+ + H+}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further addition of base causes precipitation of amorphous iron(III) hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4FeSO4 + O2 + 2H2O -&amp;gt; 4Fe(OH)SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-12T10:47:57.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31268" PostTypeId="2" ParentId="31243" CreationDate="2015-05-12T10:50:27.047" Score="5" Body="&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron%28II%29_sulfate&quot;&gt;wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Upon exposure to air, it oxidizes to form a corrosive brown-yellow coating of basic ferric sulfate, which is an adduct of ferric oxide and ferric sulfate: $$\ce{12FeSO4 + 3O2 -&amp;gt; 4Fe2(SO4)3 + 2Fe2O3}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/ferrous_sulfate#section=Other-Experimental-Properties&quot;&gt;Pubchem&lt;/a&gt; page for ferrous sulphate says that this reaction occurs rapidly in moist air and that the rate of oxidation is increased by increasing the pH, temperature, or by exposure to light. However, I don't have access to the original sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Greg pointed out, oxygen appears to be acting as the oxidiser here and it forms $\ce{Fe2O3}$. There is not enough sulphate present in the starting materials to form only $\ce{Fe2(SO4)3}$ and hence the oxide is formed as well.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-12T15:46:52.747" LastActivityDate="2015-05-12T15:46:52.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31271" PostTypeId="2" ParentId="31257" CreationDate="2015-05-12T12:34:21.723" Score="2" Body="&lt;p&gt;this is your output, right?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TRANSITION DIPOLE, A.U.  OSCILLATOR&#xA;                    HARTREE          EV         X       Y       Z     STRENGTH&#xA;   0  A         -193.0290234748    0.000&#xA;   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000&#xA;   2  A         -192.7831335626    6.691     0.4051 -0.0004 -0.0000    0.027&#xA;   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002&#xA;   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042&#xA;   5  A         -192.7210047911    8.382    -0.0121 -0.0000 -0.0006    0.000&#xA;   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001&#xA;   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000&#xA;   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022&#xA;   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am confused, why do you take the 9ths excitation and assume it is the experimentally measured one? The first excitation agrees reasonably quite well with the experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the exp spectrum eg here &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022407308000307&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sciencedirect.com/science/article/pii/S0022407308000307&lt;/a&gt;&#xA;you can see that there is a 2nd absorption band below 200nm (6eV). This seems to be the 2nd excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you look at the orbitals involved in the first excitation? I would assume it shows a n-pi* transition and then your are all good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper contains some references (i cannot access the cited references therein)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/full/10.1021/jp046334i&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/doi/full/10.1021/jp046334i&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They report the following&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The n → π* electronic transition of acetone corresponds to the excitation of an electron from the nonbonding 2py orbital on the oxygen to the antibonding π* molecular orbital on the carbonyl. For an acetone molecule in its C2 or C2v geometry, the transition is&#xA;  symmetry-forbidden; as a consequence of vibronic coupling, in the gas&#xA;  phase, a weak band at ΔE = 4.4−4.5 eV is observed, with an oscillator&#xA;  strength of f ≈ 0.0004.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do we see that in the TD-DFT output? Mostly, yes! The oscillator strength is smaller than the printout (bad program), but if you use e.g. ORCA you get a non-zero oscillator strength. It is still much smaller than the reported value but that is somewhat expected from TD-DFT in this case (partially CT-state!) The small oscillator strength is troublesome, but the excitation energy is ok. It could indicate a ghost state (=purely artificial state), but these are not often seen in hybrid-functional TD-DFT calculations. Likely all states are physical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first transition is a HOMO-LUMO transition and if you bother to look at the orbital it is a nice n-pi* transition as expected. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/CnUnp.png&quot; alt=&quot;HOMO&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IHWh6.png&quot; alt=&quot;LUMO&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16161" LastEditorUserId="16161" LastEditDate="2015-05-13T13:25:11.623" LastActivityDate="2015-05-13T13:25:11.623" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="31272" PostTypeId="2" ParentId="29657" CreationDate="2015-05-12T13:03:00.793" Score="15" Body="&lt;p&gt;To append Ringo's good answer and to add some more insight into the bonding situation, I performed a calculation on the DF-BP86/def2-SVP level of theory. Since this is a linear molecule, there are symmetry restrictions. Its point group is $C_{\infty\mathrm{v}}$, which means, that there are degenerated orbitals. As we will see, these correspond to $\pi$ bonds. Here is an image of the valence orbital scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TU3tq.jpg&quot; alt=&quot;valence orbital scheme of NNO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbitals 1-3 are core orbitals (1s) at the respective elements. MO 4 corresponds to a $\ce{O-N}~\sigma$ bond. It is noteworthy, that this bond is delocalised and also describes part of the $\ce{N-N}~\sigma$ bond. Analogously this holds for MO 5, but the other way around. In MO 6 we can see the presence of the lone pair at oxygen. MOs 7 and 8 describe two delocalised $\pi$ bonds, bonding with respect to all atoms in the compound. MO 9 has mainly lone pair character and is located at the terminal nitrogen. Tho HOMOs are also $\pi$ orbitals, bonding with respect to $\ce{N-N}$ and anti-bonding with respect to $\ce{O-N}$. This somewhat implies lone pair character for these orbitals, too. Please note, that there is no orbital with lone pair character at the central nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I further analysed the wave function with the natural bond orbital theory. This method localises the orbitals (and electron density) and compares it to the ideal electron density of a given Lewis structure. High percentages mean that the configuration agrees well with the Lewis concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g5659.jpg&quot; alt=&quot;resonance configurations of NNO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I included the structures you have suggested. You can see, that while configuration &lt;em&gt;1&lt;/em&gt; is not an ideal Lewis structure, it is still agrees to a certain point with the Lewis concept. If someone would do a more rigorous approach with Valence Bond theory, I would expect this structure to have (at least a minor) contribution to the overall bonding situation. The basic flaw with this structure is the electron sextet at the terminal nitrogen, as Ringo already mentioned. The positive aspect of this structure is, that it does not need formal charges. I would not consider this structure wrong.&lt;br&gt;&#xA;The second structure, however, is not correct. This can only exist in an excited state. Nitrogen cannot have more than eight electrons binding around it.&lt;br&gt;&#xA;The third structure is the ideal structure, i.e. the one with the highest agreement. I would expect it to have the highest contribution to the overall bonding.&lt;br&gt;&#xA;Structure &lt;em&gt;3&lt;/em&gt; accounts for delocalisation effects and is one of the major contributions to the overall resonance stabilisation of the molecule. Basically you can see the character of this in the MOs 7 and 8 above.&lt;br&gt;&#xA;The last structure (&lt;em&gt;5&lt;/em&gt;) tries the same thing, but for my taste, this is a little bit far fetched. It still agrees somewhat with the Lewis-concept, but like configuration &lt;em&gt;1&lt;/em&gt; I would only expect a smaller contribution.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-12T13:03:00.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31274" PostTypeId="2" ParentId="28518" CreationDate="2015-05-12T13:53:59.940" Score="3" Body="&lt;p&gt;Yes, see Fig. 1 and tables VI-VIII of  &lt;a href=&quot;http://www.researchgate.net/profile/Lothar_Helm/publication/37429092_Solvent_exchange_on_metal_ions/links/5437daec0cf2027cbb204fc0.pdf&quot; rel=&quot;nofollow&quot;&gt;SOLVENT EXCHANGE ON METAL IONS&lt;/a&gt; Advances in Inorganic Chemistry volume 54 pages 1-69. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values vary over almost 20 orders of magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make this not a link-only answer, roughly, per second exchange rates are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ir^3+}$  $10^{-10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Rh^3+}$  $2.2 \times 10^{-9}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cr^3+}$ $10^{-6}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ru^3+}$ $10^{-5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pt^2+}$ $10^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ru^2+}$ $10^{-2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{V^2+}$, $\ce{Fe^3+}$ $10^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Pd^2+}$, $\ce{V^3+}$ $10^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ni^2+}$ $10^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ti^3+}$ $10^5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Co^2+}$ $3.2 \times 10^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe^2+}$ $4.4 \times 10^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mn^2+}$,  $2.1 \times 10^7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn^2+}$ great uncertainty $5 \times 10^7 - 10^{10}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cd^2+}$ $10^8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cr^2+}$, $\ce{Hg^2+}$ $10^9$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cu^2+}$ $5.7 \times 10^9$&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="24052" LastEditDate="2016-03-11T23:04:57.787" LastActivityDate="2016-03-11T23:04:57.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31275" PostTypeId="2" ParentId="31262" CreationDate="2015-05-12T14:14:52.257" Score="3" Body="&lt;p&gt;The question is a little confusing, so I will interpret your 'ABABAB' type in 2D to mean a simple hexagonal layer (so, the same layer as used for your hcp and fcc packing). So, a stacking of a simple hexagonal layer, each layer directly above the other, will give a structure looking like right hexagonal prisms - or right triangular prisms if you look a bit more carefully.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the resulting structure is the simple hexagonal Bravais lattice with a single atom basis. Hcp is also a simple hexagonal Bravais lattice, but with a two atom basis, one each from plane A and plane B. I do not know of any elements that crystallize in this crystal structure (but am open to corrections!).&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-05-12T14:14:52.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31276" PostTypeId="1" CreationDate="2015-05-12T15:13:02.613" Score="17" ViewCount="37304" Body="&lt;p&gt;While studying the Crystal Field Theory I was told $\mathrm{Dq}$ is a unit, related to the unit $\Delta_{\mathrm{O}}$ by the relation $\Delta_{\mathrm{O}} = 10\ \mathrm{Dq}$. But aren't $\Delta_{\mathrm{O}}$ and $\mathrm{Dq}$ variables, not units? The unit is of energy, such as $\mathrm{eV}$ or Joules, these are symbols to represent a particular value of energy, which could be anything. So what do we mean when we say $\Delta_{\mathrm{O}}= 10\ \mathrm{Dq}$? What exactly is $\mathrm{Dq}$ (I haven't been able to find this in any of my textbooks, they just either state this relation, or not at all) and are they actually units of energy?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="16683" LastEditDate="2016-03-16T15:44:36.323" LastActivityDate="2017-11-19T14:57:24.360" Title="In Crystal Field Theory what does Δo=10Dq mean?" Tags="&lt;inorganic-chemistry&gt;&lt;energy&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31277" PostTypeId="2" ParentId="31276" CreationDate="2015-05-12T15:57:02.060" Score="6" Body="&lt;p&gt;Though I don't think I can provide a full answer, I did see this in a lecture series. $\Delta_{\mathrm{O}}$ is the ligand field splitting and it is a measure for the strong or weak field that the ligands of a complex create. Depending on the ligands the energy levels of, for example octahedral iron complexes (below), are closer to each other in a low field and farther apart for a higher field. This is dependant on $\Delta_{\mathrm{O}}$. This is also an energy difference and the units are in joules or electron volts. See the picture below. You might also want to look up &lt;a href=&quot;http://en.wikipedia.org/wiki/Tanabe%E2%80%93Sugano_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tanabe-Sugano Diagrams&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told (sorry, no proper reference) that the expression $\Delta_{\mathrm{O}}= \pu{10Dq}$ comes from quantum mechanics. The reason that this term is simply stated instead of defined, lies in the origin of CFT. The theory was largely empirical and  thus $\Delta_{\mathrm{O}}$ was used. This difference in energy levels could also be described in quantum systems and is called $\pu{Dq}$ after its quantum number. The difference between the levels appeared to be $\pu{10Dq}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is not a gross oversimplification, but a rather helpful one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GL1Ct.jpg&quot; alt=&quot;Energy levels of an octahedral system&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2359" LastEditorUserId="7475" LastEditDate="2017-11-19T14:48:00.313" LastActivityDate="2017-11-19T14:48:00.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31279" PostTypeId="1" AcceptedAnswerId="31281" CreationDate="2015-05-12T17:32:05.283" Score="5" ViewCount="1619" Body="&lt;p&gt;I am asked to identify the number of stereoisomers that are possible for hepta-1,6-diene $(\ce{H2C=CHCH2CH2CH2CH=CH2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew the bond-line structure as follows, and I think that there can be &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; isomers, but my answer is wrong. The correct answer states that since neither of the double bonds exhibit stereoisomerism, this compound does not have any stereoisomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, I thought that single bonds can be considered too. But I think I my error is that by looking at the single bonds, we are considering the constitutional isomers of this compound, rather than the stereoisomers (which are determined by the 'fixed' double bonds). Since double bonds lack the free rotation that single bonds have, for any compound that may exhibit stereoisomerism, we can only consider the double bonds. Also, are we considering only one double bond at a time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this reasoning correct? Can someone clarify and expound more on this please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nysn2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorDisplayName="user7951" LastEditDate="2016-11-30T01:41:20.753" LastActivityDate="2016-11-30T01:41:20.753" Title="How many stereoisomers are possible for hepta-1,6-diene?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31280" PostTypeId="2" ParentId="31279" CreationDate="2015-05-12T17:49:09.607" Score="2" Body="&lt;p&gt;There are no stereoisomers (cis/trans) of this compound because both the double bonds have three identical substituents and one different substituent. In order to have stereoisomerism across a double bond you need to have two different groups on each side of the bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example but-2-ene has one methyl group and one hydrogen on each side of the double bond and so has two stereoisomers: &lt;em&gt;trans&lt;/em&gt;-but-2-ene and &lt;em&gt;cis&lt;/em&gt;-but-2-ene: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2lw0h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For double bonds with at least three different substituents the &lt;a href=&quot;http://en.wikipedia.org/wiki/E-Z_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;E/Z notation&lt;/a&gt; system is used. Z denotes substituents which are on the same side of a double bond and E denotes those on the opposite side. The four substituents are assigned a priority in accordance with the &lt;a href=&quot;http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cahn-Ingold-Prelog priority rules&lt;/a&gt; and then the stereochemistry is assigned based on the relative position of the two highest priority substituents. In this example, bromine has a higher priority than chlorine so the label is assigned based on the position of the bromine and the chlorine across the double bond, making this the Z isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/huJbP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-12T17:49:09.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31281" PostTypeId="2" ParentId="31279" CreationDate="2015-05-12T17:50:41.880" Score="4" Body="&lt;p&gt;&lt;strong&gt;In the case of your original drawing with the terminal double bonds, there are no &lt;em&gt;E&lt;/em&gt; or &lt;em&gt;Z&lt;/em&gt; (&lt;em&gt;trans&lt;/em&gt; or &lt;em&gt;cis&lt;/em&gt;) isomers!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpmJI.png&quot; alt=&quot;cis and trans-substituted double bonds&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;E&lt;/em&gt; and &lt;em&gt;Z&lt;/em&gt; isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;As far as the other structure is concerned, &lt;em&gt;E&lt;/em&gt; and &lt;em&gt;Z&lt;/em&gt; isomers look as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/96Q1n.png&quot; alt=&quot;ethyl-isopropyl-substituted double bond&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-12T18:42:50.187" LastActivityDate="2015-05-12T18:42:50.187" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31282" PostTypeId="2" ParentId="31279" CreationDate="2015-05-12T19:12:45.480" Score="2" Body="&lt;p&gt;There is no Cis/Trans isomerism across the double bonds because the molecule you have there is a terminal alkene which is an alkene with double bonds at the first 2 and last 2 carbons or just at the first 2 carbons if there are no higher priority groups or just at the last 2 carbons if there are higher priority groups closer to the end with all single bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had other substituents you could have R and S stereoisomers and even combinations of R and S at different chiral carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at the differences in bonding than yes you are looking at structural or constitutional isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you can use cis/trans for multiple double bonds. For example oct-2,6-diene(or 2,6-octadiene) can be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oct-(2 cis, 6 cis)-diene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oct-(2 cis, 6 trans)-diene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oct-(2 trans, 6 trans)-diene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oct-(2 trans, 6 cis)-diene&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2015-05-12T19:19:56.390" LastActivityDate="2015-05-12T19:19:56.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31283" PostTypeId="1" CreationDate="2015-05-12T19:35:11.423" Score="1" ViewCount="1556" Body="&lt;p&gt;I'm extracting DMT (N,N-dimethyltryptamine) from &lt;em&gt;Acacia confusa&lt;/em&gt; via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate.  Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used $\ce{NaCl}$ to break the emulsion, this caused a thick layer of black tar to form on top of the aqueous solution, and a glossy black tar on the bottom of the vessel.  I need to freebase the DMT acetate; however, I'm concerned that DMT-acetate migrated into one of the two polar layers.  (Correction: Either the non polar layer on top or gunk layer at the bottom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I confirm if this is so? Does $\ce{NaCl}$ cause DMT acetate  (or any acetate) to become polar in $\ce{H2O}$? Why and how? Will the DMT acetate have migrated into any of these two polar layers?&lt;/p&gt;&#xA;" OwnerUserId="16166" LastEditorUserId="16166" LastEditDate="2015-05-12T22:26:04.150" LastActivityDate="2015-05-12T22:26:04.150" Title="Does DMT-Acetate become Polar in H2O, with added NaCl?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31284" PostTypeId="2" ParentId="31257" CreationDate="2015-05-12T19:40:43.883" Score="9" Body="&lt;p&gt;It seems that the results of the calculations are more or less fine and the OP just misinterpreted the &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C67641&amp;amp;Mask=400#UV-Vis-Spec&quot; rel=&quot;noreferrer&quot;&gt;NIST data&lt;/a&gt;. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper&lt;sup&gt;[1]&lt;/sup&gt; referenced on the NIST page it is said (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two diffuse ultraviolet bands; the first at 2800 Å. is &lt;em&gt;very&#xA;  weak&lt;/em&gt; with its oscillator strength $f \sim 0.0004$ and the second at&#xA;  about 1900 Å  is moderately intense with a maximum extinction&#xA;  coefficient $\epsilon_{\mathrm{m}} \sim 1000$. McMurry has identified&#xA;  the 2800 Å. band as a &lt;em&gt;forbidden&lt;/em&gt; $\pi^* \leftarrow n$ transition&#xA;  involving excitation of a non-bonding $\ce{O}$ electron to an&#xA;  anti-bonding $\pi$ orbital between the $\ce{C}$ and $\ce{O}$ of the&#xA;  carbonyl group.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A high-resolution photoabsorption spectrum in the energy range 3.7-10.8 eV can be found, for instance, in this recent study&lt;sup&gt;[2]&lt;/sup&gt; and is more or less consistent with the result of the OP's calculations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j150521a018&quot; rel=&quot;noreferrer&quot;&gt;Noel S. Bayliss, Eion G. McRae, &lt;em&gt;J. Phys. Chem.&lt;/em&gt;, &lt;strong&gt;1954&lt;/strong&gt;, &lt;em&gt;58&lt;/em&gt; (11), 1006–1011.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2008/cp/b708580j&quot; rel=&quot;noreferrer&quot;&gt;M. Nobre, A. Fernandes, F. Ferreira da Silva, R. Antunes, D. Almeida, V. Kokhan, S. V. Hoffmann, N. J. Mason, S. Eden, P. Limão-Vieira, &lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;10&lt;/em&gt;, 550-560.&lt;/a&gt; (available at &lt;a href=&quot;https://www.researchgate.net/publication/5667732_The_VUV_electronic_spectroscopy_of_acetone_studied_by_synchrotron_radiation&quot; rel=&quot;noreferrer&quot;&gt;researchgate.net&lt;/a&gt;) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastEditorUserId="4945" LastEditDate="2015-05-13T05:31:00.537" LastActivityDate="2015-05-13T05:31:00.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31285" PostTypeId="2" ParentId="17321" CreationDate="2015-05-12T20:02:46.403" Score="1" Body="&lt;p&gt;Sodium bisulfate will also control browning. It works by lowering the pH to inhibit the polyphenol oxidase enzyme. It lowers pH without a sour taste so it does not change the flavor of the food. &lt;/p&gt;&#xA;" OwnerUserId="16167" LastActivityDate="2015-05-12T20:02:46.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31286" PostTypeId="1" AcceptedAnswerId="31288" CreationDate="2015-05-12T20:06:58.763" Score="6" ViewCount="325" Body="&lt;p&gt;Not in an atomic aspect, I learned that fluorescence is the emission of light from substances until the source of light disappears (there is no emission in the dark), on the other hand, phosphorescence, due to a blocking of electrons on their way back to normal state, is the emission of light from substances even without any source (emission in the dark). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is so, then why do we call glowing objects in the dark as &lt;em&gt;fluorescent&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="402" LastEditDate="2017-04-01T01:21:33.993" LastActivityDate="2017-04-01T01:21:33.993" Title="Fluorescence versus phosphorescence" Tags="&lt;physical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31287" PostTypeId="1" CreationDate="2015-05-12T20:11:55.617" Score="2" ViewCount="98" Body="&lt;p&gt;What are the components of and how would one go about making &lt;a href=&quot;http://www.dickblick.com/products/castincraft-clear-polyester-casting-resin/&quot; rel=&quot;nofollow&quot;&gt;polyester casting resin&lt;/a&gt;?  &lt;/p&gt;&#xA;" OwnerUserId="7677" LastActivityDate="2015-05-12T20:11:55.617" Title="How to make polyester casting resin?" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31288" PostTypeId="2" ParentId="31286" CreationDate="2015-05-12T20:16:17.623" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is so, then why do we call glowing objects in the dark as &lt;em&gt;fluoroscent&lt;/em&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is just marketing: most people have heard the term &lt;em&gt;fluorescence&lt;/em&gt;, while &lt;em&gt;phosphorescence&lt;/em&gt; or luminescence are largely unknown terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a delayed emission from a singlet state is possible, most glow-in-the-dark items do in fact show phosphorescence, i.e. emission from a tripet state.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-12T20:16:17.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31289" PostTypeId="2" ParentId="31283" CreationDate="2015-05-12T20:42:22.983" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does NaCl cause DMT acetate (or any acetate) to become polar in $\ce{H2O}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In aqueous acidic solution, that is in the presence of a sufficient amount of acetic acid, N,N-dimethyltryptamine is protonated. The acetate is the counter ion of the ammonium salt. The compound does not become more polar by adding sodium chloride!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the DMT acetate have migrated into any of these two polar layers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you having two polar layers? I doubt it. You're having a top unpolar layer, the aqueous layer and some gunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the separated aqueous layer, neutralize carefully, extract with an unpolar solvent of your choice, dry over sodium sulfate, remove the solvent in vacuum. If necessary, recrystallize and determine the yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please keep in mind that DMT is a controlled substance in many countries!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-12T20:42:22.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31290" PostTypeId="1" AcceptedAnswerId="31293" CreationDate="2015-05-12T20:50:32.540" Score="-2" ViewCount="12495" Body="&lt;p&gt;The compound below is fexofenadine and the only chirality center, as identified by the textbook is, the one circled in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why the carbonyl carbon is not a chiral center too. Isn't it attached to a &lt;em&gt;tert&lt;/em&gt;-butyl group? So isn't it attached to four different substituent groups? Or is it that the hydroxide group is attached to the carbon by the oxygen, thereby making a 'duplicate' substituent group of the carbon already double bonded to the oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A0a5R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="15410" LastEditDate="2016-11-10T04:56:07.420" LastActivityDate="2016-11-10T04:56:07.420" Title="Identifying chiral centers" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31291" PostTypeId="2" ParentId="31290" CreationDate="2015-05-12T20:58:54.777" Score="5" Body="&lt;p&gt;If in doubt, simplify, examine both centres separately and draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XqrW9.png&quot; alt=&quot;stereocentre&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you do the same for the carboxylic acid?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-12T20:58:54.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31292" PostTypeId="2" ParentId="31290" CreationDate="2015-05-12T21:04:31.407" Score="4" Body="&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the carbon circled in blue, it only has &lt;em&gt;three&lt;/em&gt; not four substituents. If you examine the three-dimensional structure around the carbon, you will find that there is a plane of reflectional symmetry through the plane of the C=O bond which means it &lt;em&gt;cannot&lt;/em&gt; be a chiral centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 different things attached to a carbon in 3D space means it will be chiral. 3 does not.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-05-12T21:04:31.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31293" PostTypeId="2" ParentId="31290" CreationDate="2015-05-12T21:06:25.883" Score="5" Body="&lt;p&gt;Take a look at your carbon circled in red, it has a hydrogen on it giving it four different groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the C circled in blue. There is no hydrogen attached to that carbon, because it already has four bonds (two to the O, one to the OH, one to the rest of the molecule.) Therefore it only has three substituent groups attached and not four.&lt;/p&gt;&#xA;" OwnerUserId="402" LastActivityDate="2015-05-12T21:06:25.883" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31294" PostTypeId="1" AcceptedAnswerId="31295" CreationDate="2015-05-12T21:14:32.047" Score="7" ViewCount="763" Body="&lt;p&gt;I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or hydrogen-2, and react it with deuterium gas. A pair of enantiomers is formed, both with the condensed formula $\ce{DCH2-CHD-CHD2}$. The answers provided say that these enantiomers are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/ivswklf.png&quot; width=&quot;200&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/sG69gLQ.png&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...where $\ce{A}$ is standing in for deuterium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't you get enantiomers where the orientation of the $\ce{C-D}$ bonds at both ends of the molecule varies? (ie. where those $\ce{C-D}$ bonds are also in varying planes, represented by different combinations of hashed and solid wedges on paper.) Here's an example of what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/RTenJOX.png&quot; width=&quot;200&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/Q4Cw2SJ.png&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't these two structures considered to be another pair of enantiomers also?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this - the orientation of bonds at the end of carbon chains counting effectively for nothing - the case with all stereoisomerism in alkanes?&lt;/p&gt;&#xA;" OwnerUserId="16169" LastEditorUserId="16169" LastEditDate="2015-05-12T21:18:54.187" LastActivityDate="2016-04-05T23:42:44.170" Title="Why does this compound only have one pair of enantiomers?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31295" PostTypeId="2" ParentId="31294" CreationDate="2015-05-12T21:34:21.110" Score="5" Body="&lt;p&gt;You should draw the molecule showing the $\ce{H}$ atoms as well as the D (or A) atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hydrogenated/deuterogenated molecule, all the $\ce{C-C}$ bonds are single bonds, meaning there is free rotation about all of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left (as you have drawn it) end of the molecule is $\ce{DH2C-~}$ and the right end is $\ce{~-CHD2}$.  Since each end-cap carbon has two identical substituents, there is no chirality.  Flipping the whole molecule about a left-to-right axis parallel to the plane of the screen, and then rotating the $\ce{ C-C}$ bond going to a particular endcap carbon, would transform the &quot;chiral&quot; representation you drew into its &quot;enantiomer&quot;.  That wouldn't / can't happen with the truly chiral central carbon atom.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorDisplayName="user15489" LastEditDate="2015-05-12T23:01:46.370" LastActivityDate="2015-05-12T23:01:46.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31296" PostTypeId="2" ParentId="31294" CreationDate="2015-05-12T22:42:24.200" Score="7" Body="&lt;p&gt;Usually it doesn't as rotation around single bonds is generally very quick in room temperature. However, sometimes it matters because rotation can be blocked by bulky substituents, or other parts of the molecule. Even double bonds usually connected with E/Z isomerism, can in certain situations cause optical activity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all examples of so called &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_chirality&quot; rel=&quot;nofollow&quot;&gt;axial chirality&lt;/a&gt;. In the first case (rotation blocked by substituents) it's called &lt;a href=&quot;https://en.wikipedia.org/wiki/Atropisomer&quot; rel=&quot;nofollow&quot;&gt;atropoisomerism&lt;/a&gt;. Second (blocking by other parts of molecule) is present for example in &lt;a href=&quot;https://en.wikipedia.org/wiki/Helicene&quot; rel=&quot;nofollow&quot;&gt;helicenes&lt;/a&gt;. Third case (cumulated double bonds) occurs in &lt;a href=&quot;https://en.wikipedia.org/wiki/Allene&quot; rel=&quot;nofollow&quot;&gt;allens&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2016-04-05T23:42:44.170" LastActivityDate="2016-04-05T23:42:44.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31297" PostTypeId="1" CreationDate="2015-05-12T22:53:37.663" Score="14" ViewCount="2020" Body="&lt;p&gt;Even chemistry textbooks less than five years old start by describing atomic structure based only on protons, neutrons and electrons. Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?&lt;/p&gt;&#xA;" OwnerUserId="16171" LastActivityDate="2015-05-14T23:39:21.720" Title="Should chemistry students worry about quarks and such?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="0" ClosedDate="2015-10-30T21:28:29.653" ContentLicense="CC BY-SA 3.0" />
  <row Id="31298" PostTypeId="2" ParentId="31297" CreationDate="2015-05-12T23:29:03.673" Score="1" Body="&lt;p&gt;I mean to be honest with you, when I was in Chem 1, we never went into the details of quarks and other sub-atomic particles except once; in nuclear chemistry. Even then, you didn't have to have an idea of these subatomic particles to understand what was going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you think about it, wouldn't they (teachers) have changed the curriculum to include, other, dozens of subatomic particles if it was &lt;strong&gt;&lt;em&gt;needed&lt;/em&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, you still have a good point. But, overall, from my definition of a chemistry student, I wouldn't think that worrying about quarks and other subatomic particles. &lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-12T23:29:03.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31299" PostTypeId="1" AcceptedAnswerId="31303" CreationDate="2015-05-12T23:45:25.980" Score="27" ViewCount="2944" Body="&lt;p&gt;I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start to think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that the plastic retains all of its chemical characteristics despite it being ripped, so I am quite confused about how to classify what is occurring.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="402" LastEditDate="2016-10-08T05:16:35.597" LastActivityDate="2019-05-12T01:35:02.177" Title="Ripping apart plastic: is this a chemical or physical change?" Tags="&lt;theoretical-chemistry&gt;&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31300" PostTypeId="2" ParentId="30923" CreationDate="2015-05-13T00:17:59.990" Score="6" Body="&lt;p&gt;I think this question is related to the Le Chatelier’s principle:&#xA;\begin{align}&#xA;\ce{2Cu + O2 &amp;amp;\;&amp;lt;=&amp;gt; 2CuO}\\&#xA;\ce{CuO + 2HOAc &amp;amp;-&amp;gt; Cu(OAc)2 + H2O} &#xA;\end{align}&#xA;You mentioned that half the copper wire was under the acetic acid, so it was not in contact with oxygen. Thus, the lower part did not get oxidized. The upper part of the wire was in contact with oxygen and reacted to form $\ce{Cu(II)}$ oxide. Because there was acetic acid vapor in the atmosphere of the closed bottle, the acid reacted further with the $\ce{Cu(II)}$ oxide, creating $\ce{Cu(II)}$ acetate and water. The removal of $\ce{CuO}$ therefore pressed the oxidation reaction to proceed forward, making more $\ce{CuO}$ and then $\ce{Cu(II)}$ acetate. As a result, the upper part of the copper wire became dark blue/green overnight due to the color of $\ce{Cu(OAc)2}$.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T08:11:54.827" LastActivityDate="2015-05-13T08:11:54.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31301" PostTypeId="2" ParentId="31299" CreationDate="2015-05-13T00:56:17.000" Score="17" Body="&lt;p&gt;Agreed: a physical action can cause a chemical reaction. For example, any attempt to manipulate $\ce{NI3}$ physically causes it to decompose (detonate). For that matter, dividing metals into tiny pieces changes chemical behavior, because bonds that would have been hidden in the interior are exposed at the surface. Finely divided iron, for example, becomes pyrophoric, yet bulk iron can be heated in air without rapid oxidation..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is one of degree: &lt;strong&gt;how much&lt;/strong&gt; of a chemical change resulted from ripping the plastic. Can you perform any chemical test that would distinguish the shreds from the whole, such as solubility? For example, bulk carbon, whether in graphite or diamond, is not soluble in most liquids at room temperature (though they dissolve in some molten metals), but as fullerene particles, carbon is &lt;a href=&quot;http://en.wikipedia.org/wiki/Fullerene#Solubility&quot;&gt;much more soluble&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be an interesting experiment: prepare some bulk samples of a plastic, and grind some others (perhaps in a chilled ball mill?) and compare some chemical characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.news-medical.net/health/Properties-of-Nanoparticles.aspx&quot;&gt;Properties of Nanoparticles&lt;/a&gt; and &lt;a href=&quot;http://web.eng.fiu.edu/wangc/size%20matters.pdf&quot;&gt;Size matters: why nanomaterials are different&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-13T00:56:17.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31302" PostTypeId="2" ParentId="31297" CreationDate="2015-05-13T01:19:58.550" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, the general chemisty books do already include &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow&quot;&gt;leptons&lt;/a&gt;, at least electrons and usually positrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than protons, neutrons, electrons and positrons (and photons), I think that when discussing &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow&quot;&gt;beta decay&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_capture&quot; rel=&quot;nofollow&quot;&gt;electron capture&lt;/a&gt;, neutrinos and antineutrinos should be included, otherwise the nuclear reactions aren't really balanced (lepton number must be conserved).  Alternatively, the general chemistry books could omit all discussion of nuclear reactions, as being physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, I think the rest of the particles can wait for physics class, rather than general chemistry class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond general chemistry, other particles are sometimes discussed in more advanced chemistry books, such as quantum chemistry texts, where examples involving muons are common.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some chemistry topics, such as &lt;a href=&quot;http://web.nmsu.edu/~snsm/classes/chem435/Lab9/&quot; rel=&quot;nofollow&quot;&gt;isotope differences in rotational-vibrational spectra&lt;/a&gt; of small molecules distinction between &lt;a href=&quot;http://en.wikipedia.org/wiki/Fermion&quot; rel=&quot;nofollow&quot;&gt;fermions&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Boson&quot; rel=&quot;nofollow&quot;&gt;bosons&lt;/a&gt; is critical.  Trying to explain the difference between fermions and bosons (and Fermi-Dirac statisitics vs Bose-Einstein statistics vs Boltzmann statistics) naturally leads to discussion of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_Model&quot; rel=&quot;nofollow&quot;&gt;standard model&lt;/a&gt; and the particles in the standard model.  So by the time a student gets to physical chemistry, around the third year of college, they should have some understanding of the standard model.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-05-14T15:03:35.000" LastActivityDate="2015-05-14T15:03:35.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31303" PostTypeId="2" ParentId="31299" CreationDate="2015-05-13T01:28:10.820" Score="24" Body="&lt;p&gt;When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material.  The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align.  To some degree this partial alignment makes continued deformation easier.  To continue tearing the polymer apart you just need to overcome the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_der_Waals_force&quot;&gt;van der Waals forces&lt;/a&gt; holding the chains together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the polymer chains being pulled apart, some chemical bonds will be broken.  I don't know how many bonds are broken per chain pulled apart, but I suspect it is low.  Low or not, some free radicals (or ions) will be generated when the chemical bonds break.  If the experiment is performed under ambient conditions, the radicals or ions will quickly react with oxygen or ions in the air.  If the experiment is performed in an inert atmosphere, then the radicals will persist (&lt;a href=&quot;http://en.wikipedia.org/wiki/Dangling_bond&quot;&gt;dangling bonds&lt;/a&gt;) and can be observed in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_paramagnetic_resonance&quot;&gt;esr spectrometer&lt;/a&gt;.  It has been demonstrated that cross-linked polymers, which are less prone to flow deformation and therefore more likely to undergo bond breaking, do produce larger esr signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most of what is going on is molecular deformation and pulling apart of the polymer chains - but some bond breaking does occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be an interesting experiment to cut two pieces of plastic and then cut one of them into finer shreds.  Place the two samples into separate tubes and then record their esr signals.  How much more intense would the signal be in the finely cut sample compared to the bulk sample?  Polycarbonate-A, straight out of the reagent bottle, produces an esr signal.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-13T01:56:02.903" LastActivityDate="2015-05-13T01:56:02.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31304" PostTypeId="2" ParentId="29486" CreationDate="2015-05-13T03:54:48.733" Score="7" Body="&lt;p&gt;As pointed out in my earlier comment, you won't find tetraalkyl lead compounds in current petrol used for cars. In most countries, it was banned decades ago and replaced by MTBE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the situation seems to be &lt;strong&gt;different&lt;/strong&gt; for fuels used in aviation,   such as &lt;em&gt;AvGas 100&lt;/em&gt; where lead concentrations up to $\mathrm{1\,g\,L^{-1}}$ can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the &lt;em&gt;low lead&lt;/em&gt; variant, AvGas 100LL contains $\mathrm{0.56\,g\,L^{-1}}$ lead, see product specifications from &lt;a href=&quot;http://www4.total.fr/germany/Flugkraftstoffe_Produktspezifikation_AVGAS-100-LL_10_2013.pdf&quot; rel=&quot;noreferrer&quot;&gt;TOTAL&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an older technical note by &lt;em&gt;R. E. Mansell&lt;/em&gt; and &lt;em&gt;T. A. Hiller&lt;/em&gt;, published in &lt;em&gt;Anal. Chem.&lt;/em&gt;, &lt;strong&gt;1973&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt;, 975-975 (&lt;a href=&quot;http://dx.doi.org/10.1021/ac60328a042&quot; rel=&quot;noreferrer&quot;&gt;DOI&lt;/a&gt;), a possible approach for the decomposition of tetraethyl lead and subsequent detection of $\ce{Pb^{2+}}$ is outlined, referring to the &lt;em&gt;ASTM D526-56&lt;/em&gt; method:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] recommends refluxing 50 ml of gasoline with 50 ml of $\ce{HCl}$ for 30 minutes to decompose the tetraethyllead.[&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the aqueous layer, $\ce{Pb^{2+}}$ can be detected by complexometric titration with EDTA, either directly, or by back titration with a solution of a $\ce{Mg^{2+}}$ salt.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-13T04:20:08.243" LastActivityDate="2015-05-13T04:20:08.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31305" PostTypeId="2" ParentId="29487" CreationDate="2015-05-13T04:53:06.187" Score="1" Body="&lt;p&gt;Dimensional analysis is a straightforward, easy, and beautiful way to solve this problem! Simply cancel the units!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4\,\mathrm{mL\,(solution)} \cdot \frac{1.07\,\mathrm{g\,(solution)}}{\mathrm{mL\,(solution)}} \cdot \frac{6\,\mathrm{g}\,\ce{NaClO}}{100\,\mathrm{g\,(solution)}} \cdot \frac{1\,\mathrm{mol}\,\ce{NaClO}}{74.44\,\mathrm{g}\,\ce{NaClO}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=0.00345\,\mathrm{mol}\,\ce{NaClO}$$&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-13T04:53:06.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31306" PostTypeId="1" AcceptedAnswerId="31319" CreationDate="2015-05-13T06:01:07.137" Score="9" ViewCount="2137" Body="&lt;p&gt;Consider the following TDDFT run with GAMESS:   &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!   File created by the GAMESS Input Deck Generator Plugin for Avogadro&#xA; $BASIS GBASIS=N311 NGAUSS=6 $END&#xA; $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END&#xA; $CONTRL ICHARG=0 MULT=1 $END&#xA; $TDDFT NSTATE=9 $END&#xA; $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END&#xA; $SYSTEM MWORDS=1000 PARALL=.TRUE. $END&#xA; $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END&#xA;&#xA; $DATA &#xA;Title&#xA;C1&#xA;O     8.0     0.00000    -1.27900     0.00300&#xA;C     6.0    -0.00000    -0.05800     0.00100&#xA;C     6.0     1.29700     0.69100    -0.00000&#xA;C     6.0    -1.29800     0.69000    -0.00000&#xA;H     1.0     1.35900     1.32900    -0.90600&#xA;H     1.0     1.35900     1.33200     0.90300&#xA;H     1.0     2.15700    -0.01300     0.00100&#xA;H     1.0    -2.15700    -0.01400     0.00100&#xA;H     1.0    -1.35900     1.32900    -0.90600&#xA;H     1.0    -1.35900     1.33200     0.90300&#xA; $END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From this we obtain the following part of the output. (The complete calculation can be found &lt;a href=&quot;http://s000.tinyupload.com/?file_id=41117960135750882611&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.) How can we interpret the orbital transitions, in terms of for example $\mathrm{n}\to\sigma^*$ or $\pi\to\pi^*$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VQwuy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="4945" LastEditDate="2015-05-13T15:16:20.460" LastActivityDate="2016-02-25T10:49:10.713" Title="How to interpret orbital transition in TDDFT?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;spectroscopy&gt;&lt;td-dft&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31307" PostTypeId="1" CreationDate="2015-05-13T07:53:20.330" Score="5" ViewCount="1905" Body="&lt;p&gt;I'm preparing for an experiment to investigate some factor affecting buffer capacity. At the moment, I'm looking at using a $\ce{NH_3}$ ($2\,\mathrm{M}$) $+$ $\ce{NH_4Cl}$ buffer, but I'm not sure which factor to test. My research tells me that only the concentrations of each will affect buffer capacity, and that the ratio of their concentrations will affect the $\mathrm{pH}$ of the buffer (by the Henderson-Hasselbalch equation). Are there other factors which I can test, because I'm trying to make it a bit more interesting? As far as I can tell, something like temperature won't affect buffer capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the actually experiment goes, what would be the best way to measure buffer capacity? I could just do a titration and record the $\mathrm{pH}$ as I go, but I'm thinking of using an indicator and measuring absorbance. I'm open to any ideas here, because I'm struggling to think of anything &quot;out of the box&quot;?&lt;/p&gt;&#xA;" OwnerUserId="14994" LastEditorUserId="15410" LastEditDate="2015-05-14T02:58:47.163" LastActivityDate="2015-05-14T02:58:47.163" Title="Investigating a factor affecting buffer capacity" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31308" PostTypeId="2" ParentId="31307" CreationDate="2015-05-13T08:51:31.197" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As far as I can tell, something like temperature won't affect buffer capacity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this particular case, I wouldn't be too sure about it. Isn't it conceivable that your buffer will lose ammonia upon heating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you already decide about the pH indicators? For the lower end, thymol blue might be a good choice. The &lt;a href=&quot;http://www.rsc.org/suppdata/tb/c4/c4tb02077d/c4tb02077d1.pdf&quot; rel=&quot;nofollow&quot;&gt;supplementary data&lt;/a&gt;&#xA;to &lt;em&gt;J. Mater. Chem. B&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;,&lt;em&gt;3&lt;/em&gt;, 2801-2807 contains a UV/VIS spectrum of this indicator.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-13T08:51:31.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31309" PostTypeId="1" AcceptedAnswerId="31311" CreationDate="2015-05-13T08:57:21.230" Score="6" ViewCount="188" Body="&lt;p&gt;I found this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0022024876902311&quot;&gt;article&lt;/a&gt;, but it was not helpful. The $\ce{In2O3}$ is probably amorphous. $\ce{HNO3}$ and $\ce{HCl}$ do not work. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="5456" LastEditDate="2015-05-13T09:00:26.447" LastActivityDate="2015-05-13T09:43:07.440" Title="How to dissolve composite In2O3 with SnO2 without affecting stainless steel?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31310" PostTypeId="2" ParentId="21696" CreationDate="2015-05-13T09:25:14.327" Score="7" Body="&lt;p&gt;Your expectations are not completely unfounded. Even without the addition of special ligands, aqueous solutions of Fe(III) often appear yellow since they contain the ions $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your experiment takes place under &lt;strong&gt;acidic conditions.&lt;/strong&gt; (I guess, sulfuric acid was added.) This is apparent because the purple permanganate is reduced to the colourless $\ce{Mn^2+}$ ion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4- + 8H+ + 5e- &amp;lt;=&amp;gt; Mn^2+ + 4H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4- + 4H+ + 3e- &amp;lt;=&amp;gt; MnO2 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the given acidic conditions, the colourless complex $\ce{[Fe(H2O)6]^3+}$ is stable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Fe(OH)2(H2O)4]+ + 2H+ &amp;lt;=&amp;gt; [Fe(OH)(H2O)5]^2+ + H+ &amp;lt;=&amp;gt; [Fe(H2O)6]^3+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the solution appears colourless.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-13T09:25:14.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31311" PostTypeId="2" ParentId="31309" CreationDate="2015-05-13T09:43:07.440" Score="4" Body="&lt;p&gt;In case your question is about the recovery of indium from ITO, the following might be helpful: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Recovery of indium from indium tin oxide by solvent extraction&lt;/em&gt;, published in &lt;em&gt;Hydrometallurgy&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;107&lt;/em&gt;, 56-61 (&lt;a href=&quot;http://dx.doi.org/10.1016/j.hydromet.2011.01.005&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;), the abstract seems to suggest that $\ce{H2SO4}$ is more suitable than $\ce{HNO3}$. Furthermore, a solvent extraction scheme for indium, using di-(2-ethylhexyl)phosphoric acid (DEHPA) is proposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;Recovery of indium from used indium–tin oxide (ITO) targets&lt;/em&gt;, published in &lt;em&gt;Hydrometallurgy&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;105&lt;/em&gt;, 207-212 (&lt;a href=&quot;http://dx.doi.org/10.1016/j.hydromet.2010.09.006&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;), the authors agree on using $\ce{H2SO4}$ (at 90 °C) and decrease the ymount of tin in solution by sulfide precipitation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-13T09:43:07.440" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31312" PostTypeId="1" CreationDate="2015-05-13T10:49:04.350" Score="6" ViewCount="6802" Body="&lt;p&gt;Top hits on google are:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt;&lt;br&gt;&#xA; Aluminum hydroxide $\ce{Al(OH)_3}$ &gt;&gt; $	1.3×10^{–33}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt;&lt;br&gt;&#xA; Aluminum hydroxide $\ce{Al(OH)_3}$ &gt;&gt; $	3×10^{–34}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is one order of magnitude different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the reason for this variation? Experimental conditions?&#xA;Where can I get accurate or reliable results/data of $K_{\mathrm{sp}}$ in general?     &lt;/p&gt;&#xA;" OwnerUserId="16177" LastEditorUserId="15410" LastEditDate="2015-05-13T15:27:04.627" LastActivityDate="2015-05-13T15:27:04.627" Title="What is the Ksp of Al(OH)3?" Tags="&lt;solubility&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31313" PostTypeId="1" CreationDate="2015-05-13T12:54:35.357" Score="6" ViewCount="694" Body="&lt;p&gt;Logically thinking, I thought that it would be high pressure and high temperature that would create high density poly(ethene)... However, that's not the case. Is it because of the catalyst? Why is the condition low pressure/temperature?&lt;/p&gt;&#xA;" OwnerUserId="16178" LastEditorUserId="16683" LastEditDate="2015-10-02T16:51:43.417" LastActivityDate="2015-10-02T16:51:43.417" Title="Why does addition polymerisation of high density poly(ethene) use low temperature and pressure?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31314" PostTypeId="2" ParentId="14244" CreationDate="2015-05-13T13:14:14.960" Score="4" Body="&lt;p&gt;For langmuir isotherm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let say we are using linear langmuir isotherm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is : $\frac{1}{q_e} = \frac{1}{q_m} + \frac{1}{q_mKL}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;find the value of $q_e$. Using the formula $q_e = \frac{(C_i-C_f)V}{\text{mass}}$ of adsorbent, and then plot the graph $\frac{1}{q_e}$ versus $\frac{1}{C_f}$. It would be easier by using excel. From the graph, we can calculate the slope. The linear graph will produce $y=mx+c$ equation. m=intercept=1/qe, hence qe=1/m. C=1/qm.KL. From the C value, we can actually find KL. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the graph generated using excel, it will shown the R2 value. This value is the important indicator to determine which isotherm model fit the system. The highest R2 value will fit the system. Let say R2 value langmuir is 0.8 and R2 for freundlich is 0.9. So, we can conclude that, Langmuir model fit your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Freundlich isotherm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot the graph log qe versus log cf. From the graph you will get linear equation, y=mx + c and the R2 value. Compare the R2 value with langmuir's R2. Highest R2 value indicate that those model fit with the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cf=final concentration&#xA;Ci= initial concentration&lt;/p&gt;&#xA;" OwnerUserId="16179" LastEditorUserId="16088" LastEditDate="2017-01-06T13:14:24.873" LastActivityDate="2017-01-06T13:14:24.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31315" PostTypeId="2" ParentId="31306" CreationDate="2015-05-13T13:28:55.660" Score="3" Body="&lt;p&gt;Form this output you dont.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find this in the output:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;             EXCITATION  DE-EXCITATION&#xA; OCC     VIR  AMPLITUDE      AMPLITUDE&#xA;  I       A     X(I-&amp;gt;A)        Y(A-&amp;gt;I)&#xA; ---     ---   --------       --------&#xA; 10      17   -0.039945       0.001926&#xA; 16      17    0.996619      -0.032612&#xA; 16      20   -0.070681       0.003443&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;it shows you the involved orbitals in the transitions. This a dominant transition from the 16ths orbital to the 17ths (HOMO to LUMO). You need to visualize the MOs and decide on the nature (sigma,pi,n) of the orbitals. Since they are MOs, it can be difficult with large molecules. You can look up &lt;em&gt;natural transition orbitals&lt;/em&gt; that aim to simplify the visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you can look at orbitals with gamess? I have no clue.&lt;/p&gt;&#xA;" OwnerUserId="16161" LastActivityDate="2015-05-13T13:28:55.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31317" PostTypeId="2" ParentId="31312" CreationDate="2015-05-13T14:57:21.113" Score="3" Body="&lt;p&gt;Let me add to the confusion: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://books.google.de/books?id=imBzjH6SHi8C&amp;amp;pg=PA303&amp;amp;lpg=PA303&amp;amp;dq=solubility+product+constant+aluminum+hydroxide&amp;amp;source=bl&amp;amp;ots=jXoPiVihnF&amp;amp;sig=eSK8saszXshTXVEKGOlThUiKVr0&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=Xl5TVcepDuj_ywOXmYDQBA&amp;amp;ved=0CG8Q6AEwCTgK#v=onepage&amp;amp;q=solubility%20product%20constant%20aluminum%20hydroxide&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;https://books.google.de/books?id=imBzjH6SHi8C&amp;amp;pg=PA303&amp;amp;lpg=PA303&amp;amp;dq=solubility+product+constant+aluminum+hydroxide&amp;amp;source=bl&amp;amp;ots=jXoPiVihnF&amp;amp;sig=eSK8saszXshTXVEKGOlThUiKVr0&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ei=Xl5TVcepDuj_ywOXmYDQBA&amp;amp;ved=0CG8Q6AEwCTgK#v=onepage&amp;amp;q=solubility%20product%20constant%20aluminum%20hydroxide&amp;amp;f=false&lt;/a&gt;&lt;br&gt;&#xA;$2\times10^{-32}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://faculty.ncc.edu/LinkClick.aspx?fileticket=smTKdcQlPY4%3D&amp;amp;tabid=1903&quot; rel=&quot;nofollow&quot;&gt;http://faculty.ncc.edu/LinkClick.aspx?fileticket=smTKdcQlPY4%3D&amp;amp;tabid=1903&lt;/a&gt;&lt;br&gt;&#xA;$1.8\times10^{-33}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.austincc.edu/dlyon/gen_chem_1_summer04/Chapter%2020%20solubiliity%20product%20f08.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.austincc.edu/dlyon/gen_chem_1_summer04/Chapter%2020%20solubiliity%20product%20f08.pdf&lt;/a&gt;&lt;br&gt;&#xA;$1.9\times 10^{-33}$  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.csudh.edu/oliver/chemdata/data-ksp.htm&lt;/a&gt;&lt;br&gt;&#xA;$4.6\times10^{-33}$  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.periodni.com/solubility_product_constants.html&quot; rel=&quot;nofollow&quot;&gt;http://www.periodni.com/solubility_product_constants.html&lt;/a&gt;&lt;br&gt;&#xA;$3\times 10^{-34}$  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://users.stlcc.edu/gkrishnan/ksptable.html&quot; rel=&quot;nofollow&quot;&gt;https://users.stlcc.edu/gkrishnan/ksptable.html&lt;/a&gt;&lt;br&gt;&#xA;$3\times 10^{-34}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Usually I'd first have a look at the CRC &lt;em&gt;Handbook of Chemistry and Physics&lt;/em&gt;. Unfortunately, I couldn't find the data for aluminium hydroxide.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-13T14:57:21.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31318" PostTypeId="2" ParentId="16402" CreationDate="2015-05-13T15:02:04.180" Score="2" Body="&lt;p&gt;At that pH, you're well outside the buffering capacity of both phosphate and boric acid, so a $\mathrm{pH}$ rise from $5.5$ to $5.9$ corresponds to a net increase in base of about $2\ \mathrm{\mu M}$.  That implies to me that it's likely something going on with one of the first five ingredients in your recipe, or the $\ce{HCl}$ added to adjust the $\mathrm{pH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Similar to Brinn's answer, perhaps a small amount of $\ce{HCl}$ has evaporated out of solution.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your $\ce{Fe-EDTA}$ was added as $\ce{Fe(II)-EDTA}$, oxidation to $\ce{Fe(III)-EDTA}$ could explain the pH rise, by reducing water to hydrogen gas and hydroxide ion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$2\ \ce{Fe(II)-EDTA} \rightarrow 2\ \ce{Fe(III)-EDTA} + 2\ e^-$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$2\ \ce{H2O} + 2\ e^- \rightarrow \ce{H2} + 2\ \ce{OH-}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't know offhand how thermodynamically or kinetically feasible #2 is.  With all of the other metals present in solution, though, it's not unreasonable to imagine that some sort of catalysis might be possible.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-05-13T15:19:05.620" LastActivityDate="2015-05-13T15:19:05.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31319" PostTypeId="2" ParentId="31306" CreationDate="2015-05-13T15:10:19.590" Score="8" Body="&lt;p&gt;To visualise the orbitals of your &lt;a href=&quot;http://s000.tinyupload.com/?file_id=41117960135750882611&quot; rel=&quot;noreferrer&quot;&gt;calculation&lt;/a&gt;, use a program of your choice. For Gamess, there are &lt;a href=&quot;http://www.msg.ameslab.gov/graphics/other.html&quot; rel=&quot;noreferrer&quot;&gt;a few options availabile&lt;/a&gt;. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2lZDS.jpg&quot; alt=&quot;orbitals of acetone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can further use the summary to identify the most interesting, strongest transitions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                          SUMMARY OF TDDFT RESULTS&#xA;&#xA;   STATE             ENERGY     EXCITATION  TRANSITION DIPOLE, A.U.  OSCILLATOR&#xA;                    HARTREE          EV         X       Y       Z     STRENGTH&#xA;   0  A         -193.0290234748    0.000&#xA;   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000&#xA;   2  A         -192.7831335626    6.691     0.4051 -0.0004  0.0000    0.027&#xA;   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002&#xA;   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042&#xA;   5  A         -192.7210047911    8.382    -0.0121  0.0000 -0.0006    0.000&#xA;   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001&#xA;   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000&#xA;   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022&#xA;   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the calculated 9 states that is the last one, with an oscillator strength of 0.290. Now skip back one section to where you find:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      -------------------&#xA;      SINGLET EXCITATIONS&#xA;      -------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Look for excited state number 9:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; STATE #   9  ENERGY =    9.326387 EV&#xA; OSCILLATOR STRENGTH =    0.290423&#xA; LAMBDA DIAGNOSTIC   =    0.561 (RYDBERG/CHARGE TRANSFER CHARACTER)&#xA; SYMMETRY OF STATE   =    A   &#xA;                 EXCITATION  DE-EXCITATION&#xA;     OCC     VIR  AMPLITUDE      AMPLITUDE&#xA;      I       A     X(I-&amp;gt;A)        Y(A-&amp;gt;I)&#xA;     ---     ---   --------       --------&#xA;      9      17    0.137515       0.018404&#xA;     15      17    0.694034      -0.078753&#xA;     13      18   -0.085768      -0.006093&#xA;     14      19   -0.059549      -0.004510&#xA;     16      19   -0.289828       0.018031&#xA;     15      20    0.051487       0.010566&#xA;     11      21    0.053380       0.009416&#xA;     13      21   -0.060120      -0.009392&#xA;     16      22   -0.607830       0.005676&#xA;     11      24    0.052546       0.012226&#xA;     13      25   -0.039117      -0.008987&#xA;     16      26    0.112895       0.016712&#xA;     16      30   -0.041659      -0.014274&#xA;     16      33    0.052880       0.019973&#xA;     16      36   -0.031783      -0.011699&#xA;     13      40   -0.030352      -0.013004&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There you find the orbital transitions. You can also have a look at the amplitudes to identify the most dominant one. In this case it is probably 15 to 17 and therefore corresponds to $\pi\to\pi^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more complicated the molecules get, the more confusing will this process be. Also the higher the level of theory, the more transitions you will need to consider. In this case Natural Transition Orbitals will certainly become very helpful, see &lt;a href=&quot;http://dx.doi.org/10.1063/1.1558471&quot; rel=&quot;noreferrer&quot;&gt;Richard L. Martin, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2003&lt;/strong&gt;, &lt;em&gt;118&lt;/em&gt;, 4775-4777&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-13T15:10:19.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31320" PostTypeId="1" CreationDate="2015-05-13T15:24:48.857" Score="5" ViewCount="229" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a greater standard electrode potential than a hydrogen ion will be reduced in its stead, and no hydrogen gas will be produced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand, that bit states that certain electrolytes will stop the production of hydrogen gas and only oxygen gas and whatever gas from the electrolyte will be produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are some electrolytes that will stop the production of hydrogen gas in water electrolysis?&lt;/li&gt;&#xA;&lt;li&gt;What kind of water will be left behind if distilled water and an aforementioned electrolyte runs through electrolysis until all of the oxygen has been exhausted from the solution?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I'm misunderstanding anything.&lt;/p&gt;&#xA;" OwnerUserId="16182" LastEditorUserId="15410" LastEditDate="2015-05-13T15:28:38.050" LastActivityDate="2015-05-13T16:13:13.010" Title="Is there an electrolyte that will halt the production of Hydrogen in water electrolysis?" Tags="&lt;water&gt;&lt;electrolysis&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31321" PostTypeId="2" ParentId="28551" CreationDate="2015-05-13T15:49:44.850" Score="1" Body="&lt;p&gt;CsOH is a strong base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.5 L X 1.5 M = 0.75 moles of hydroxide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there are 0.5 moles of monoprotic acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After mixing there will be 0.25 moles of hydroxide in 1 liter of solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words [OH-] = 0.25M&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pH is approximately 14 + log (0.25) = 13.4&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-05-13T15:49:44.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31322" PostTypeId="2" ParentId="28669" CreationDate="2015-05-13T15:50:46.367" Score="5" Body="&lt;p&gt;This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Introduction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first law of thermodynamics tells us that in any process energy is conserved, it can be transformed and converted into different forms, but the total amount is unchanged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second law imposes limits on the efficiency of energy conversion processes. For a robust understanding of the second law it is necessary first to understand Carnot cycles. As such I will divide my answer into sections. Many of the points in the following are intricate in their application and a good knowledge of the first law is required. At the time classical thermodynamics was developed the main interest of the research was into the efficiency of heat engines. Historically these are quite interesting, but that is why you need to understand a Carnot cycle first which in essence is just a really efficient engine (the most efficient). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that in any heat engine a supply of heat $Q$ can either be converted to work $W$ or internal energy $\Delta U$. Since an isothermal process means $\Delta U=0$ then the most efficient cycle will have two isotherms. However two isotherms do not constitute a cycle. The most efficient way to connect these two isotherms is with adiabatic processes (no heat exchange with the surroundings). With two adiabatic lines we finally have our cycle. Graphically to maximise the work you want to maximise the area under the cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cycle we mean that a system returns to exactly the same state as before the change or process occurred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carnot Cycles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work can be obtained from an engine of there are heat sources at different temperatures, and heat will flow from a hot body to a cold body without work being performed until thermal equilibrium is attained. Carnot realised therefore that any temperature difference either produces work, or is wasted as spontaneous flows of heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets use an ideal gas as our substance. The figure shows a Carnot cycle. It is reversible and has two reservoirs. Where $ab$ is an isotherm at temperature $T_1$ and heat $Q_1$ enters from a reservoir at $T_1$. The line $cd$ is an isotherm at a lower temperature $T_2$ where heat is given to another reservoir at $T_2$. The line $bc$ and $da$ are adiabatic processes. The work done $W$ is the area enclosed by the cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We might choose to represent the Carnot engine according to the figure below. Where $W$ is the work done by the engine. The efficiency $\eta$ is a measure of &quot;what we get out for what we put in&quot; so in this case the work output for the heat input. &#xA;\begin{equation}&#xA;\eta =\frac {W}{Q_1}&#xA;\end{equation}&#xA;If we remember the first law and that $\Delta U=0$ then we may write $W=Q_1-Q_2$ denoting the sign convention for heat into the system and work done on the system to be positive and vice versa. &#xA;\begin{equation}&#xA;\eta =1-\frac{Q_2}{Q_1}&#xA;\end{equation}&#xA;&lt;strong&gt;Kelvin-Planck statment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at a uniform temperature and the performance of an equivalent amount of work.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basically says that some heat needs to be given to a body of colder temperature otherwise by the above equation we would have 100% efficiency. If you were to violate this you could just have perpetual motion machines or cars that just absorbed heat and produced work. It may be summarised as &lt;em&gt;a process whose only effect is the complete conversion of heat into work is impossible.&lt;/em&gt; However the opposite process is allowed  where all the work is transferred to heat completely &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Clausius statement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. *&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that work must be performed if heat is to be transferred form a colder to a hotter body. We might be more familiar with this statement if we called this engine a fridge. An engine that extracts heat from a cold body and delivers it to a hot body when work is performed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kelvin-Planck statement is entirely equivalent to the Clausius statement. We will skip any proof or mathematics of why this is so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carnot's Theorem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;No engine operating between two reservoirs can be more efficient that a Carnot engine operating between those same two reservoirs.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proof is as follows. Assume that such an engine did exist $E'$ with efficiency $\eta '$ that takes heat $Q_1'$ from the hot body and does work $W'$ delivering heat $Q_2'=(Q_1'-W')$ to the cold bath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us now imagine a Carnot engine $C$ with heats $Q_1$ and $Q_2$, efficiency $\eta$ adjusted in such a way that it performs the exact amount of work $W$ as the &quot;more efficient engine $E'$ so that $W=W'$, and $Q_2=Q_1-W$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iff,&#xA;\begin{equation} &#xA;\frac{W'}{Q_1'}&amp;gt;\frac{W}{Q_1}\ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ Q_1&amp;gt;Q_1'&#xA;\end{equation}&#xA;The Carnot engine is reversible so it may be driven backwards, acting as a composite device with $E'$. It extracts positive heat $(Q_1-Q_1')$ form the cold reservoir and delivers the same heat to the hot reservoir with no external work being required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However since the reservoir is large the temperature is unchanged with the addition of this heat. This means that the Clausius statement is violated. Therefore $\eta'&amp;gt;\eta $ is not true, and $E'$ can never exist. If $\eta '=\eta$ then $Q_1'=Q_1$ and no net heat transferred for no work. Therefore we say,&#xA;\begin{equation}&#xA;\eta' \leq \eta &#xA;\end{equation} &#xA;&lt;strong&gt;Entropy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a substance that can do work under going a cycle so that at the end of its cycle its state is unchanged at $T_1$. Lets say we change the first state to infinitesimally close neighbouring state 2 at $T_2$ by adding a small amount of heat $\delta Q_1$ from . If we do this with a Carnot engine with the temperature baths of state 1 $T_1$ and some reservoir at $\bar T$ then $T_1$ supplies $\delta Q_q$ to the working substance while the Carnot engine supplies the same amount to the $T_1$ reservoir to leave it unchanged, in turn taking heat $\{\bar T/T_1\}\delta Q_1$ from a principle reservoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in short we have performed a change from states 1 to 2 with external work $\delta W_1$ and  extraction of heat $\{\bar T/T_1\}\delta Q_1$ from the principle reservoir. This is one system of a composite process of identical systems. The diagram really helps here! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composite system is assumed that $\Delta U=0$ and that the heat supplied to it is;&#xA;\begin{equation}&#xA;Q=\sum _i\frac{\bar T}{T_i}\delta Q_i&#xA;\end{equation}&lt;br&gt;&#xA;Where the sum is over all the little systems making up the composite. The external work performed is;&#xA;\begin{equation}&#xA;W=\sum _i\delta W_i&#xA;\end{equation}&#xA;The first law is then $0=Q-W$. Looking at the second diagram below for the composite system we can see that this violates the Kelvin-Planck statement. The only way that this works is if work is done &lt;em&gt;on the system&lt;/em&gt; and an equal quantity of work flows &lt;em&gt;out&lt;/em&gt;. Or it works if both $W$ and $Q$ are zero. &#xA;\begin{equation}&#xA;W=Q\leq 0&#xA;\end{equation} &#xA;So &#xA;\begin{equation}&#xA;\bar T\sum _i\frac{\delta Q_i}{T_i}\leq 0 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \sum _i \frac{\delta Q_i}{T_i}\leq 0&#xA;\end{equation}&#xA;If you take the limit of this then over one complete cycle we arrive at the Clausius inequality! &#xA;\begin{equation}&#xA;\oint \frac{dQ}{T}\leq 0&#xA;\end{equation}&#xA;It is important to know that the $T$ appearing in the integral is the temperature of the heat supplying bath!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cycle is reversible then there is no reason why we couldn't do the proof in the other direction and hence obtain the inequality sign the other way around. This only holds for reversible cycles. The only way to satisfy both inequalities physically is if for reversible cycles we equal the integral to zero exactly. This means that for reversible cycles the integral is equal to zero!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final step is to now use this Clausius inequality to prove the second law. Imagine a cycle that is completely reversible between two states $i$ and $f$ with two reversible paths $r_1$ and $r_2$ connecting them. &#xA;\begin{equation}&#xA;\oint \frac{dQ_r}{T}=_{r_1}\int^{f}_{i} \frac{dQ_r}{T}+_{r_2}\int_{f}^{i}\frac{dQ_r}{T}=0&#xA;\end{equation}&#xA;Thus,&#xA;\begin{equation}&#xA;_{r_1}\int ^{f}_{i}\frac{dQ_r}{T}=_{r_2}\int^{f}_{i}\frac{dQ_r}{T}\end{equation}&#xA;What all this means is that the integral is path independent SO THERE MUST BE A PATH INDEPENDENT FUNCTION !! We call this state function entropy. &#xA;\begin{equation}&#xA;S_f-S_i=_r\int ^f_i \frac{dQ_r}{T}&#xA;\end{equation}&#xA;For this equality to hold the heat exchanged must be defined over a reversible path! That is to say if the entire cycle is not reversible then we have to use the Clausius stamen again. Lets suppose that one off the paths is irreversible now.&lt;br&gt;&#xA;\begin{equation} &#xA; \int^f_i\frac{dQ}{T_0}+_r\int ^i_f\frac{dQ_r}{T}\leq 0&#xA;\end{equation}&#xA; With a bit of interval switching,&#xA;\begin{equation}&#xA;\int ^f_i\frac{DQ}{T_0}\leq -_r\int ^i_f\frac{dQ_r}{T}=_r\int ^f_{i}\frac{dQ_r}{T}=S_f-S_i&#xA;\end{equation}&#xA;And finally the way we normally write this is in an infinitesimal form,&#xA;\begin{equation}&#xA;\frac{dQ}{T_0}\leq dS&#xA;\end{equation}&#xA;This is more general than before since it includes the previous equation too, equality holds if the cycle is reversible. So , in an infinitesimal irreversible process between a pair of states the entropy change is $dS$, if the system is thermally isolated then $dQ=0$ and we generate the principle of increasing or &quot;maximum&quot; entropy. &#xA;\begin{equation}&#xA;S_f-S_i \geq 0&#xA;\end{equation}&#xA;Or if you prefer $dS\geq 0$. The entropy of a thermally isolated system increases in any irreversible process and is unaltered in a reversible one. &lt;/p&gt;&#xA;" OwnerUserId="7913" LastActivityDate="2015-05-13T15:50:46.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31323" PostTypeId="1" AcceptedAnswerId="31702" CreationDate="2015-05-13T16:01:47.107" Score="9" ViewCount="2083" Body="&lt;p&gt;I am trying to identify whether the following objects possess planes of symmetry or not, but my answers seem different, from the textbook solutions (Klein Organic Chemistry 2e, Page 220, Question 23). Some of them I am unsure of and would like to make sure I know the proper reasoning for. Can someone please clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) The correct response is yes. Is the plane of symmetry I have drawn acceptable?&#xA;(c) The correct response is no, but I thought it is possible to have a plane of symmetry like the one I have drawn.&#xA;(e) The correct response is yes, which I think I understand. I think this has three planes of symmetry.&#xA;(f) The correct response is no, which I agree with. However, it doesn't seem to concur with the response for the follow-up question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Follow-up: Which image has three planes of symmetry? The correct answer is apparently, (f). My response would have been (e).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F3R1U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="30836" LastEditDate="2020-04-06T08:54:27.027" LastActivityDate="2020-04-06T08:54:27.027" Title="Planes of Symmetry and Chirality" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;&lt;chirality&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31324" PostTypeId="1" AcceptedAnswerId="31328" CreationDate="2015-05-13T16:10:19.337" Score="5" ViewCount="4427" Body="&lt;p&gt;At first I thought maybe sodium sulphate in contact with water produces sulphuric acid which absorbs water but I do not think it is actually a valid reason. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="10185" LastEditDate="2015-05-13T16:15:46.207" LastActivityDate="2015-05-13T16:39:38.810" Title="Why sodium sulphate has water absorbing capability?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31325" PostTypeId="2" ParentId="31320" CreationDate="2015-05-13T16:10:23.547" Score="2" Body="&lt;p&gt;The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H+_{(aq)} + 2e^{–} -&amp;gt;H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if a species has a has a reduction potential $&amp;gt;0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -&amp;gt; H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4OH^{−}_{(aq)} -&amp;gt; O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if a species has a has a oxidation potential $&amp;gt;-.401\,\mathrm{V}$, then it will be be oxidized before the  $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="10185" LastEditDate="2015-05-13T16:13:13.010" LastActivityDate="2015-05-13T16:13:13.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31327" PostTypeId="2" ParentId="28643" CreationDate="2015-05-13T16:20:33.743" Score="4" Body="&lt;p&gt;The allowed energy levels for a particle of mass $m$ in a box of length $X$ are given by&#xA;$$E_i=\frac{h^2}{8mX^2}i^2\quad i=1,2,\ldots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Already this equation gives a hint that an ideal gas of high molar mass has more thermally accessible translational states than a gas of low molar mass. Therefore, the molar entropy of a gas of high molar mass is greater.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the partition function for translational motion in the $x$-direction is approximately&#xA;$$z=\left(\frac{2\pi mkT}{h^2}\right)^{1/2}X$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partition function for translational motion in three dimensions is&#xA;$$\begin{align}&#xA;z&amp;amp;=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\&#xA;&amp;amp;=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system of $N$ particles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}Z&amp;amp;=\frac{z^N}{N!}\\&amp;amp;=\frac{1}{N!}\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\right]^N\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statistical entropy is&#xA;$$\begin{align}&#xA;S &amp;amp;=\frac{U-U_0}{T}+k\ln Z\\&#xA;&amp;amp;=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\&#xA;&amp;amp;=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\&#xA;&amp;amp;\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\&#xA;&amp;amp;=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\&#xA;&amp;amp;=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\&#xA;&amp;amp;=\frac{U-U_0}{T}+n R \ln \left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}V}{nN_\text{A}}\right] \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal energy of a monoatomic ideal gas is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U=U_0+\tfrac{3}{2}nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to the Sackur–Tetrode equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;S &amp;amp;=\frac{U-U_0}{T}+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\&#xA;&amp;amp;=\tfrac{3}{2}nR+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\&#xA;&amp;amp;=nR\left\{\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]+\tfrac{3}{2}\right\}\\&#xA;&amp;amp;=nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}^{5/2}V}{nN_\text{A}}\right]&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation implies that the entropy $S$ of a monoatomic ideal gas depends on the mass $m$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-13T16:20:33.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31328" PostTypeId="2" ParentId="31324" CreationDate="2015-05-13T16:24:35.437" Score="5" Body="&lt;p&gt;Sodium sulphate reacts readily with water at room temperature to form hydrates up to sodium sulphate decahydrate, $\ce{NaSO4\cdot10H2O}$. $$\ce{NaSO4 + 10H2O -&amp;gt; NaSO4\cdot10H2O}$$ This means that $\ce{NaSO4}$ can absorb up to 10 mol of water for every 1 mol of salt that is used, making it one of the most effective drying agents in terms of sheer capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using data from this &lt;a href=&quot;http://www.chem.ucla.edu/~bacher/Specialtopics/Drying%20Agents.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; we can calculate the Gibbs energy change for the above reaction at different temperatures. $$\mathrm{\Delta G_{20^\circ C} = -1.33~kJ~mol^{-1}}$$ $$\mathrm{\Delta G_{30^\circ C} = 1.28~kJ~mol^{-1}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the entropy change for the reaction is negative, we can see that the cooler the solution you are trying to dry is, the more effective $\ce{NaSO4}$ will be as a drying agent.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-13T16:39:38.810" LastActivityDate="2015-05-13T16:39:38.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31329" PostTypeId="1" AcceptedAnswerId="31330" CreationDate="2015-05-13T16:26:38.943" Score="4" ViewCount="11488" Body="&lt;p&gt;The explanation I have found is that primary cells have irreversible reactions, so passing current in the opposite direction cannot recharge the cell. But as far as I understood it, no reaction is really irreversible - what one calls irreversible simply has very high values of the equilibrium constant, so for all intents and purposes proceeds to completion. Nevertheless, some small - maybe undetectable practically - amount of reactant does remain and the reaction does reach equilibrium. So even in a primary cell, when it is used up, equilibrium has been achieved, so recharging should be possible - it might take long to build up the initial value of potential since the required concentration of the electrolytes will be very different from that at equilibrium, but it should be possible. Why then is it not?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2020-08-03T21:33:02.817" Title="Why can't a primary cell be recharged?" Tags="&lt;electrochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31330" PostTypeId="2" ParentId="31329" CreationDate="2015-05-13T16:57:48.470" Score="4" Body="&lt;p&gt;In some cases, a primary cell &lt;em&gt;can&lt;/em&gt; be recharged, e.g. Mn-alkaline cells can take a partial charge. However, there are a number of reasons primary cells cannot be fully recharged, and that secondary cells have to be specifically designed to allow recharging.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Gas evolution: once most of a cell is charged (though all material has not been converted to its original state), the electrolyte is electrolyzed (apologies for the pleonastic redundancy), usually evolving hydrogen and oxygen, causing the cell to rupture. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel%E2%80%93metal_hydride_battery&quot; rel=&quot;nofollow&quot;&gt;Nickel-metal hydride cells&lt;/a&gt;, for example, get around that issue by incorporating a catalyst to recombine them, $\ce{2 H2 + O2 -&amp;gt; 2 H2O + heat}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Physical change in the reactants: the standard &quot;dry cell&quot; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Leclanch%C3%A9_cell&quot; rel=&quot;nofollow&quot;&gt;Leclanché cell&lt;/a&gt;) is made with an outer shell of zinc. As the cell discharges, parts of the shell are dissolved, eventually leaving holes. Reversing the current flow will cause zinc to redeposit, &lt;em&gt;but not necessarily filling in the weakened areas.&lt;/em&gt; Even lead-acid cells designed for storage have limited lifetime because of the buildup of large, insoluble crystals of lead sulfate (&lt;a href=&quot;http://support.rollsbattery.com/support/solutions/articles/5964-battery-sulfation&quot; rel=&quot;nofollow&quot;&gt;sulfation&lt;/a&gt;, aka sulphation in GB), which may break off from the lead plates, preventing reversal of the chemical reaction during charging.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Physical damage to the insulator: metal dendrite may grow during charge-discharge cycling, piercing the dielectric between cells, &lt;a href=&quot;http://www.mpoweruk.com/lithium_failures.htm&quot; rel=&quot;nofollow&quot;&gt;even in well-designed secondary cells&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="5456" LastEditDate="2015-05-13T17:17:40.013" LastActivityDate="2015-05-13T17:17:40.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31331" PostTypeId="2" ParentId="31329" CreationDate="2015-05-13T17:05:40.093" Score="2" Body="&lt;p&gt;&lt;em&gt;Irreversible&lt;/em&gt; in this context isn't a thermodynamic or kinetic thing, necessarily. It simply means that applying a current in reverse doesn't result in the original reagents. It may be that there is another possible reaction at a similar redox potential that leads to a different compound being formed or that the manner in which it is deposited is not suitable, e.g. the zinc in an alkaline battery being plated back onto the anode unevenly or such that it damages the ion conducting membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that recharging may make a primary cell inoperative by some other mechanism, such as gas formation in the potassium hydroxide electrolyte of an alkaline cell deforming the case or damaging the half-cell separator. Primary cells are also typically constructed cheaply as they don't have to withstand the stresses of secondary cells, e.g. the steel casing of alkaline batteries is susceptible to corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things like the various lithium battery chemistries make better secondary cells as they use organic solvents which reduce the problem of gas evolution and lithium can form intercalated compounds where the lithium can be evenly spread instead of growing dendrites of solid lithium which can damage the cell when recharged. That's the main difference between disposable and rechargeable lithium cells: disposable cells use solid lithium electrodes because the anode only has to dissolve once and the difficulty of replating the anode is not an issue, whereas a rechargeable cell needs to return the anode to its original state many times and an intercalated lithium compound, though more expensive, is far easier to replenish with lithium.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-05-13T17:05:40.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31332" PostTypeId="1" CreationDate="2015-05-13T18:44:02.690" Score="5" ViewCount="9550" Body="&lt;p&gt;When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does the change in energy level compare to the energy of the emitted photon?&lt;/p&gt;&#xA;" OwnerUserId="16184" LastEditorUserId="418" LastEditDate="2015-05-13T19:00:51.290" LastActivityDate="2015-10-22T18:41:17.540" Title="Energy of photons emitted by de-excitation of electrons" Tags="&lt;energy&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;photochemistry&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31333" PostTypeId="1" CreationDate="2015-05-13T18:45:18.780" Score="3" ViewCount="52" Body="&lt;p&gt;I am trying to understand why my nanoparticles will only electrophoretically deposit in a non-polar solvent and a non-polar/polar solvent mix but not in the polar solvent. I am hoping to use zeta potential and Hamaker's equation to understand this &lt;/p&gt;&#xA;" OwnerUserId="12821" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T18:49:32.060" LastActivityDate="2015-05-13T18:49:32.060" Title="How do I calculate the viscosity of a mixture of toluene and isopropanol?" Tags="&lt;electrochemistry&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31334" PostTypeId="2" ParentId="31332" CreationDate="2015-05-13T18:54:26.070" Score="1" Body="&lt;p&gt;When an electron de-excites from a higher energy orbital to a lower energy orbital it loses energy by emitting a photon. Since energy must be conserved, the energy of this photon is equal to the amount of energy the electron has lost, or in other words the difference in energy of the two orbitals involved. The frequency of the photon can then be calculated using $E=hf$ where $h$ is Planck's constant and has a value of $\mathrm{6.626\times10^{-34}~J~s}$.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-06-12T19:49:52.567" LastActivityDate="2015-06-12T19:49:52.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31335" PostTypeId="2" ParentId="31332" CreationDate="2015-05-13T19:20:39.583" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is one of several pathways for deexcitation. Often, &lt;em&gt;internal conversion&lt;/em&gt;, the radiationless deeexcitation happens instead.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How does the change in energy level compare to the energy of the emitted photon?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on the relaxed geometry of the excited state, as compared to the ground state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as &lt;em&gt;Stokes shift&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-13T19:20:39.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31336" PostTypeId="2" ParentId="31313" CreationDate="2015-05-13T19:22:30.100" Score="3" Body="&lt;p&gt;In fact, when polymerizing ethylene using a radical initiator at high temperature and high pressure, we favor the formation of low-density polyethylene. This is because the radical at the end of the growing chain can be transferred to any carbon atom of the chain by radical rearrangement (favored at high temperature) and due to close distances among molecules (high pressure) a branch chain can grow on the principal chain, and we get a branched polyethylene.  The chains of a branched polymer cannot align together, and eventually the density of the resultant polymer is low. We have got LDPE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, using a specific catalyst, known as Ziegler-Natta catalyst and under mild conditions of temperature and pressure, we get linear chains of polyethylene that can be aligned together and the density of the resultant polymer is high. We have got HDPE.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-05-13T19:22:30.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31337" PostTypeId="1" AcceptedAnswerId="31340" CreationDate="2015-05-13T19:36:07.627" Score="6" ViewCount="357" Body="&lt;p&gt;I found this reaction over &lt;a href=&quot;http://www.askiitians.com/iit-jee-chemistry/organic-chemistry/purification-classification-and-nomenclature-of-organic-compounds/lassaigne-method.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PgYRk.png&quot; alt=&quot;image&quot;&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which illustrates the carbonyl test but looks fuzzy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the R in the product is a typo for H, I couldn't conserve the number of Nitrogen atoms.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct reaction?&lt;br&gt;&#xA;Is this correct?&lt;br&gt;&#xA;Is there a possibility of a methyl substitution on the ring in a variant of this test?   &lt;/p&gt;&#xA;" OwnerUserId="15912" LastEditorUserId="10185" LastEditDate="2015-05-13T19:56:34.437" LastActivityDate="2015-05-13T20:07:30.340" Title="2,4-dinitrophenyl hydrazine test" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;analytical-chemistry&gt;&lt;erratum&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31338" PostTypeId="1" AcceptedAnswerId="31341" CreationDate="2015-05-13T19:51:25.350" Score="8" ViewCount="405" Body="&lt;p&gt;I know that nitrogen is versatile enough that it can form branched chains if it is all single bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2N-N(NH2)-NH2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this compound a nitrane (nitrogen alkane)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so how would I name this branched nitrogen compound?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="10185" LastEditDate="2015-05-13T20:01:09.500" LastActivityDate="2015-05-13T20:16:05.627" Title="Nitrogen chain nomenclature" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;pnictogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31339" PostTypeId="2" ParentId="31337" CreationDate="2015-05-13T19:53:48.027" Score="3" Body="&lt;p&gt;The methyl group on the product is an error. It should be a nitro group. &#xA;Also you are right that the R group should be an H (or the other way around depending on what they intended). It looks like someone stuck two different images together but forgot to check to see if they matched up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that the reaction shown is correct. The reagent used in 2,4-dinitrophenylhydrazine, commonly know as &lt;a href=&quot;http://en.wikipedia.org/wiki/2,4-Dinitrophenylhydrazine&quot; rel=&quot;nofollow&quot;&gt;DNPH&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-13T19:53:48.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31340" PostTypeId="2" ParentId="31337" CreationDate="2015-05-13T19:54:23.903" Score="4" Body="&lt;p&gt;Somebody messed up the scheme completely!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have already imagined, there is no magical $\ce{NO2}$ to $\ce{CH3}$ conversion in this classical reaction ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GHUKT.png&quot; alt=&quot;formation of a dinitrophenylhydrazone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the times before modern spectroscopy, these reactions were valuable analytical tools: the melting points of the dinitrophenylhydrazones and/or other derivatives were used to identify the carbonyl compounds.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-13T20:07:30.340" LastActivityDate="2015-05-13T20:07:30.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31341" PostTypeId="2" ParentId="31338" CreationDate="2015-05-13T20:00:46.917" Score="5" Body="&lt;p&gt;According to the &lt;a href=&quot;http://goldbook.iupac.org/A00553.html&quot;&gt;IUPAC Gold book&lt;/a&gt;, these are known as &lt;strong&gt;azanes&lt;/strong&gt;, from the information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Saturated acyclic nitrogen hydrides having the general formula $\ce{N_{n}H_{n + 2}}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the example you are looking at, according to the IUPAC document &lt;a href=&quot;http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter6-Sec67.pdf&quot;&gt;P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement&#xA;analogues as parents for naming organic compounds&lt;/a&gt; (p. 15) is named as &lt;strong&gt;2-aminotriazane&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-13T20:16:05.627" LastActivityDate="2015-05-13T20:16:05.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31342" PostTypeId="1" CreationDate="2015-05-13T20:19:50.307" Score="3" ViewCount="847" Body="&lt;p&gt;According to &lt;a href=&quot;https://physics.stackexchange.com/questions/53682/what-causes-ballpoint-pens-to-write-intermittently&quot;&gt;this question&lt;/a&gt;, ballpoint pens start writing intermittently when the ink is drying up. In other words, the ink is becoming more viscous and does not flow any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there ways to make it flow again?&lt;/p&gt;&#xA;" OwnerUserId="16187" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-05-14T06:11:52.833" Title="Reactivate dried up ink of ballpoint pens" Tags="&lt;everyday-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31343" PostTypeId="1" AcceptedAnswerId="31349" CreationDate="2015-05-13T20:41:14.983" Score="5" ViewCount="297" Body="&lt;p&gt;I'm not sure if I understand the reasoning behind this particular question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sigma and pi bonds of $\ce{C=C}$ have a combined bond dissociation energy of 632 kJ/mol. Using this information, predict whether the following reaction is exothermic or endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C2H4 + H2O -&amp;gt; CH3CH2OH}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonds being broken are the $\ce{C=C}$ pi bond and the $\ce{H-OH}$ bond. I realize there are two bonds being formed, but I am unable to correctly identify the second one. One is $\ce{CH_3CH_2-OH}$. I think the other should be $\ce{CH_2-H}$. The textbook says this bond is $\ce{CH_2H-R}$. I think R is referring to the side group, which varies (and in this case, is $\ce{CH_2-OH}$?), but the enthalpy value for breaking the $\ce{CH_2H-R}$ is listed as 410 kJ in the solutions. According to the table in my book, the value for a $\ce{CH_3CH_2-H}$ is 410 kJ/mol. I do not understand how these two bonds are equivalent. Can someone please explain this?&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="418" LastEditDate="2015-05-13T20:43:46.543" LastActivityDate="2015-05-13T23:27:44.323" Title="Enthalpy and Bond Dissociation Energies" Tags="&lt;organic-chemistry&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31344" PostTypeId="2" ParentId="28669" CreationDate="2015-05-13T20:56:21.673" Score="3" Body="&lt;p&gt;There is a very simple, non-theoretical explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short it is that there are a lot more ways for things to be disordered that there are for things to be ordered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly longer version considers the mechanism whereby the states of things change. Molecules are only at rest at absolute zero (ignoring quantum effects). At every other temperature they are moving around, banging into each other in random ways. This movement redistributes energy throughout the substance involved so it doesn't gather in one place. If you apply external heat to one end of an iron rod, the molecules or atoms there vibrate faster (that's what heat is). But they randomly knock other atoms in the rod and, in doing so, transfer energy to them. There are far more ways for this to lead to an even distribution of vibration than there are for all the hot atoms in one place to stay hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, perhaps, easier to see in a gas. There, atoms move with a range of speeds but collide a lot. The random collisions have random effects at distributing energy around. Some collisions lead to slow molecules getting slower and other getting faster. But they are far rarer than the ones where the hot molecules spread their energy out to slower molecules. The end result is a distribution of energy among all the molecules. If a corner of a vessel containing a gas is heated, the molecules near the corner will gain energy but will rapidly spread that extra energy out to the molecules in the rest of the container. This increases the entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you count the number of possible ways of distributing energy in a system you are actually measuring entropy. And the point is that there are more states available in disordered systems that in ordered ones. So if there is a way of distributing energy about, things will tend to end up in the disordered state not the ordered one. There is no mechanistic arrow forcing this to happen; it is just the implication of the statistics of large numbers and randomness in the way things happen at a molecular or atomic level.&lt;/p&gt;&#xA;" OwnerUserId="81" LastActivityDate="2015-05-13T20:56:21.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31345" PostTypeId="1" CreationDate="2015-05-13T21:19:14.977" Score="5" ViewCount="115" Body="&lt;p&gt;How do I perform this reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have prepared 4,4'-acylbiphenyl and want to make one acyl group into $\ce{-OH}$ and other into $\ce{-COOH}$ group. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="73" LastEditDate="2015-05-13T23:59:52.743" LastActivityDate="2015-05-14T09:37:31.023" Title="Transformation of acyl group to carboxyl group in biphenyl system with two acyl groups at para position" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31346" PostTypeId="2" ParentId="31342" CreationDate="2015-05-13T21:59:00.673" Score="1" Body="&lt;p&gt;As ink typically consists of non-polar dyes dissolved in a non-polar liquid, I would recommend using a straightened paper clip and a drop of acetone to clear out the dried ink clogging up the ink cartridge. This works by agitating the dried ink, as well as by replenishing the ink's solvent. Just be careful to not make too much of a mess!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method I have seen work before is heating the end of the pen under a flame, which melts the ink clogging the tip. For this method, my advice would be to not hold the pen in one place for too long, lest you risk melting the plastic.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-13T21:59:00.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31349" PostTypeId="2" ParentId="31343" CreationDate="2015-05-13T23:27:44.323" Score="2" Body="&lt;p&gt;Tabulation table such as the ones you reference, describe 'typical' bonds when using generic groups such as R.  You are right in your discomfort that the numbers should be somewhat different, but that would require a thermochemical measurement for every known molecule, which is difficult to achieve in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations such as the one you are performing are 'estimates', and while good ones, they are never as credible as an actual experiment.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-05-13T23:27:44.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31350" PostTypeId="1" CreationDate="2015-05-13T23:55:42.257" Score="7" ViewCount="220" Body="&lt;p&gt;The notion of heat capacity seemed a little odd to me. I thought that the word 'capacity' could be throwing me off. Do I understand the notion correctly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I take the word 'capacity' to express, roughly, 'the maximum amount of something that a thing can take'. My first intuition is that the maximum temperature of a system equals whatever temperature corresponds to every molecule in that system moving at the limit of the speed of light. Accordingly, if 'heat' denotes a certain species of energy transfer, then the heat capacity (i.e. the maximum amount of energy that a system can receive) should be equal to the number calculated by, for each molecule in the system, taking the difference between the speed of light and the molecule's starting speed, summing those differences, and then dividing by the number of molecules in the system. Accordingly, if everything else is equal (i.e. pressure and volume), then the heat capacity of a system is a function of the starting temperature of the system and its volume. At first blush, that conception of heat capacity seems compatible with the equation&#xA;$C_x = \frac {Q}{\Delta T_x}$.&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="15410" LastEditDate="2015-05-14T00:30:44.543" LastActivityDate="2015-05-14T10:57:30.753" Title="Do I understand why heat capacity is called 'heat *capacity*'?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31352" PostTypeId="1" AcceptedAnswerId="31381" CreationDate="2015-05-14T02:41:25.563" Score="13" ViewCount="2382" Body="&lt;p&gt;I can understand that the carbon atoms in benzene have delocalized p electrons that are shared amongst all 6 of the carbons in a cloud on either side of the ring. I also understand why this could be represented as resonance structures like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Vw7S.png&quot; alt=&quot;resonance benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also understand representing it in this form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oQz8j.jpg&quot; alt=&quot;delocalized electron cloud benzene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I cannot understand, however, is why &lt;strong&gt;only one&lt;/strong&gt; of the first form appears so often in depictions of organic molecules. For example, this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3NiVF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a &quot;resonance structure,&quot; is this not incorrect to list only one of the structures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn't sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw &lt;strong&gt;both&lt;/strong&gt; resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-05-14T02:48:19.337" LastActivityDate="2015-05-14T14:09:43.270" Title="Is the resonance depiction of benzene typically portrayed incorrectly?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31353" PostTypeId="1" AcceptedAnswerId="31354" CreationDate="2015-05-14T03:10:36.983" Score="3" ViewCount="114" Body="&lt;p&gt;For a lab, we took the temperature of 15ml of tap water, $q_{cool}$, then added 20ml of tap water warmed to 80 degrees Celsius, $q_{warm}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recorded a baseline temperature of 24.8, a temperature of 56 after adding the water, and a temperature of 42.9 100 seconds later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following equations: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$q=m × 4.184 J/(g*C) × ΔT$&lt;/li&gt;&#xA;&lt;li&gt;$q_{warm} + q_{cool} + q_{calorimeter} = 0$&lt;/li&gt;&#xA;&lt;li&gt;$q_{calorimeter} = - (q_{warm} + q_{cool} ) = C_{calorimeter} × ΔT$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the first formula $ 4.184 J/(g*C)$ is the specific heat of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge, specific heat is usually expressed in grams-Kelvin; however, the TA may have provided the info in grams-Celsius (their units are equal, so I presume they're the same thing). However, she also uses $C$ to designate heat capacity as she does in the third equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in the first equation, $C$ designates heat capacity, then I have three equations, and three unknowns: $q_{calorimeter}, q_{warm}$, and $q_{cool}$, and I'd treat it as a system of three equations and solve for them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if, in the first equation, $C$ designates 'Celsius' then I could just plug in the values for $q_{warm}$ and $q_{cool}$ and solve for $q_{calorimeter}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two related questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the first equation, what does $C$ designate. &lt;/li&gt;&#xA;&lt;li&gt;If it designates the heat capacity of the system, do I solve for $q$ by solving a system of equations?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="275" LastEditDate="2015-05-16T12:22:01.210" LastActivityDate="2015-06-03T15:09:38.993" Title="In order to solve for this equation for heat capacity, do I need to treat it as a system of equations?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31354" PostTypeId="2" ParentId="31353" CreationDate="2015-05-14T03:55:21.953" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the first equation, what does C designate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the final value is energy in joules, then it should be the the unit for temperature. She should have used °C to be clear.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it designates the heat capacity of the system, do I solve for q by solving a system of equations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, because you don't have the correct equations. You need the amount of heat released from the hot water minus the amount of heat absorbed by the cold water to give you the heat absorbed by the calorimeter.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorDisplayName="user7951" LastEditDate="2015-06-03T15:09:38.993" LastActivityDate="2015-06-03T15:09:38.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31356" PostTypeId="1" CreationDate="2015-05-14T03:57:32.577" Score="1" ViewCount="343" Body="&lt;p&gt;Do Claisen Condensation products prefer the di-keto form, or a keto/enol form due to increased conjugation and hydrogen bonding?&lt;/p&gt;&#xA;" OwnerUserId="16196" LastEditorUserId="10185" LastEditDate="2015-05-14T17:32:29.083" LastActivityDate="2015-05-14T18:38:27.190" Title="Keto Enol Equilibrium for Claisen Condensation Products" Tags="&lt;organic-chemistry&gt;&lt;equilibrium&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31357" PostTypeId="2" ParentId="31352" CreationDate="2015-05-14T04:11:11.333" Score="5" Body="&lt;p&gt;The problem of drawing all the resonance in salicylic acid (or many of the molecules with a benzene ring), is that none of them convey all the resonance involved. You can easily see that you can move the 3 double bonds so that the carbons in the ring should have equal bond lengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not evident is that the carboxylic acid is planer to the ring (the p-orbitals in the carbon and 2 oxygens are aligned with the p-orbitals in the ring) and the hydrogen is traded back and forth with the 2 oxygens. With this extra resonance, the carbon in the ring near the carboxylic acid should have a different bond length than the carbon across from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, the alcohol will also have a contribution to the resonance.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-14T04:11:11.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31358" PostTypeId="1" AcceptedAnswerId="31367" CreationDate="2015-05-14T04:23:30.010" Score="17" ViewCount="1534" Body="&lt;p&gt;The Born–Oppenheimer aproximation is said to be adiabatic.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What does &quot;adiabatic&quot; mean?&lt;/li&gt;&#xA;&lt;li&gt;What would be a non-adiabatic approximation to the Schrödinger equation?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="16683" LastEditDate="2017-05-19T22:56:40.463" LastActivityDate="2021-05-19T17:20:41.510" Title="Born–Oppenheimer adiabaticity" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="31359" PostTypeId="1" CreationDate="2015-05-14T05:15:27.760" Score="3" ViewCount="675" Body="&lt;p&gt;What will be the cost efficient and high yielding method of synthesis for 1-bromobutane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can react butanol with $\ce{HBr}$ but I am worried butanol is not present at the moment in lab. But I need to hurry and make bromobutane.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="10185" LastEditDate="2015-05-14T17:26:18.137" LastActivityDate="2016-05-18T05:56:49.770" Title="Synthesis of bromobutane" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31362" PostTypeId="2" ParentId="16300" CreationDate="2015-05-14T06:09:28.570" Score="4" Body="&lt;p&gt;What you did in your explanation is correct. First find the difference in the weight:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$20.3102-20.2367=0.0735 \ \mathrm{g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that from your density of water:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{\frac{1\ g}{1\ mL}\ \text{that }\ 0.0735 \ g = 0.0735\ mL}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the difference between the unknown and the vessel is calculated as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$20.300-20.2367 = 0.0933\ \mathrm{g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then density is expressed as mass over volume, so:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{\frac{0.0933\ g}{0.0735\ mL}=1.2693_8\ g\ mL^{-1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="61578" LastEditDate="2019-12-01T01:10:04.313" LastActivityDate="2019-12-01T01:10:04.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31363" PostTypeId="1" AcceptedAnswerId="31433" CreationDate="2015-05-14T06:30:56.150" Score="4" ViewCount="266" Body="&lt;p&gt;I am looking for a book or website such as &quot;Vogel's Textbook of Quantitative Chemical Analysis&quot; where I can find methods for analyzing/standardizing a particular species easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for an exact sub area such as water analysis or drug analysis; rather I want a resource like Vogel's which indexes generic analytical methods, both classical and modern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Actually I am looking for the method of standardizing a sodium thiosulfate solution using a potassium dichromate solution. I can't find the method in &quot;Vogel's Textbook of Quantitative Chemical Analysis 6th Edition&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="11367" LastEditDate="2015-05-14T11:28:10.947" LastActivityDate="2015-06-25T00:55:58.973" Title="What are some books, websites, etc. where I can find the detailed protocol for standardization of sodium thiosulfate with dichromate?" Tags="&lt;analytical-chemistry&gt;&lt;reference-request&gt;&lt;books&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31364" PostTypeId="2" ParentId="30993" CreationDate="2015-05-14T06:40:16.317" Score="2" Body="&lt;p&gt;Mercury is an element with poor sensitivity when analyzed by traditional flame atomic absorption. This happens because the element is extremely volatile (remember mercury is a liquid at room temperatures and has significant vapor pressure at room temperature). Cold vapor is a widely used method, that's EPA approved, and shows a significant increase in sensitivity (nearly four fold). &#xA;&lt;a href=&quot;http://www.chem.agilent.com/Library/applications/AA032.pdf&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's many different devices out there that can be used to do elemental analysis, but from a personal experience, if you're working with dry solid material (ex. soil), then X-ray fluorescence (XRF) could be worth looking at. It's relatively cheap and does multi-element analysis. With mercury however, you're looking at a LOD of about 5 ppm, which is... okay. It depends on your experiment. &lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-14T07:26:27.147" LastActivityDate="2015-05-14T07:26:27.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31365" PostTypeId="2" ParentId="31363" CreationDate="2015-05-14T06:49:52.620" Score="3" Body="&lt;p&gt;&lt;strong&gt;Book:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find one book which should have analytical methods for most of the species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://store.elsevier.com/product.jsp?isbn=9780122267703&quot; rel=&quot;nofollow noreferrer&quot;&gt;Encyclopedia of Separation Science&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is ten volume set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;websites:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are too many websites on internet but I will choose following two:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/&quot;&gt;Chemistry Stack Exchange&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wikipedia &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-05-14T06:49:52.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31366" PostTypeId="2" ParentId="19224" CreationDate="2015-05-14T07:18:19.623" Score="2" Body="&lt;p&gt;So, we know that dipotassium phtalate is a conjugate base of a diprotic acid, so the addition of it into water will remove some $\ce{[H+]}$s from $\ce{[H2O]}$ and increase the level of $\ce{[OH^{-}]}$ with means a lower pOH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We calculate the pOH by first converting to $\mathrm{K_a}$&#xA;$$\mathrm{pK_{a2}=10^{-5.408}=3.908*10^{-06}}$$&#xA;$$\mathrm{K_{b1}=\frac{K_w}{K_{a2}}= \frac{1*10^{-7}}{3.908*10^{-06}}=2.559*10^{-2}}$$&#xA;$$\ce{\frac{[HA][OH^-]}{[A^-]}=\mathrm{K_{b1}}}$$&#xA;$$\mathrm{\frac{x^2}{F-x}=2.559*10^{-2}\rightarrow 1.60*10^{-2}}$$&#xA;We did that last one assuming that x was negligibly small. &#xA;We now know&#xA;$$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$&#xA;$$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$&#xA;$$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-14T07:25:31.507" LastActivityDate="2015-05-14T07:25:31.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31367" PostTypeId="2" ParentId="31358" CreationDate="2015-05-14T07:30:12.280" Score="30" Body="&lt;p&gt;$\newcommand{\conj}[1]{\overline{#1}{}}&#xA;\newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle}&#xA;\newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle}&#xA;\newcommand{\mat}[1]{\mathbf{#1}}&#xA;\newcommand{\rel}{\vec{r}_{\mathrm{e}}}&#xA;\newcommand{\rnuc}{\vec{r}_{\mathrm{n}}}&#xA;\newcommand{\linop}[1]{\hat{#1}}&#xA;\newcommand{\Tnuc}{\linop{T}_{\mathrm{n}}}&#xA;\newcommand{\Tel}{\linop{T}_{\mathrm{e}}}&#xA;\newcommand{\Velel}{\linop{V}_{\mathrm{ee}}} &#xA;\newcommand{\Velnuc}{\linop{V}_{\mathrm{en}}}&#xA;\newcommand{\Vnucnuc}{\linop{V}_{\mathrm{nn}}}&#xA;\newcommand{\Hel}{\linop{H}_{\mathrm{e}}}&#xA;\newcommand{\psiel}{\psi_{\mathrm{e}}}&#xA;\newcommand{\psieli}{\psi_{\mathrm{e},i}}&#xA;\newcommand{\psielj}{\psi_{\mathrm{e},j}}&#xA;\newcommand{\Eel}{E_{\mathrm{e}}}&#xA;\newcommand{\Eeli}{E_{\mathrm{e},i}}&#xA;\newcommand{\Eelj}{E_{\mathrm{e},j}}&#xA;\newcommand{\psinuc}{\psi_{\mathrm{n}}}&#xA;\newcommand{\psinuci}{\psi_{\mathrm{n},i}}&#xA;\newcommand{\psinucj}{\psi_{\mathrm{n},j}}$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Setup&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In short, our goal is to separate the coordinates of all electrons $\rel = \{ \vec{r}_{i} \}_{i=1}^{n}$ from that of the nuclei $\rnuc = \{ \vec{r}_{\alpha} \}_{\alpha=1}^{\nu}$. Thus, we write the time-independent Schrödinger equation in the following form&#xA;\begin{equation}&#xA;    \linop{H} \psi(\rel, \rnuc) = E \psi(\rel, \rnuc) \, , \tag{1}&#xA;\end{equation}&#xA;where the Hamiltonian operator is given as follows&#xA;\begin{equation}&#xA;    \linop{H} = \Tnuc + \Tel + \Velnuc + \Vnucnuc + \Velel \, ,&#xA;\end{equation}&#xA;or, explicitly in atomic units,&#xA;\begin{equation}&#xA;    \linop{H}&#xA;    = - \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^2&#xA;    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^2&#xA;    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}&#xA;    + \sum\limits_{\alpha=1}^{\nu} \sum\limits_{\beta &amp;gt; \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}} &#xA;    + \sum\limits_{i=1}^{n} \sum\limits_{j &amp;gt; i}^{n} \frac{1}{r_{ij}} \, ,&#xA;\end{equation}&lt;br&gt;&#xA;where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.&#xA;Here for simplicity we ignore magnetic interactions and use the non-relativistic treatment.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Exact separation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We know that nuclei are &lt;em&gt;much heavier&lt;/em&gt; as compared to electrons, $m_{\mathrm{n}} \gg m_{\mathrm{e}}$, and assuming that the momentum of electrons and nuclei is of the same order of magnitude, $p_{\mathrm{n}} \sim p_{\mathrm{e}}$, we conclude that kinetic energy of nuclei is far smaller than kinetic energy of electrons $p_{\mathrm{n}}^2/(2 m_{\mathrm{n}}) \ll p_{\mathrm{e}}^2/(2 m_{\mathrm{e}})$, or, talking in terms of classical mechanics, nuclei are &lt;em&gt;moving considerably slower&lt;/em&gt; than electrons.&#xA;For a coupled motion of two subsystems with significantly different speeds it is often assumed that one could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (nuclei).&#xA;This assumption results in a procedure with two major steps, described below.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;First step&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;First, we solve an equation describing the motion of electrons for a number of static configurations of the nuclei.&#xA;The equation is known as the &lt;em&gt;electronic Schrödinger equation&lt;/em&gt; and it can be written as follows&#xA;\begin{equation}&#xA;    \Hel \psieli(\rel, [\rnuc]) = \Eeli([\rnuc]) \psieli(\rel, [\rnuc]) \quad \text{for} \quad i = 1, 2, \dotsc \, .&#xA;\end{equation}&#xA;where $\Hel = \Tel + \Velnuc + \Velel$ is the &lt;em&gt;electronic Hamiltonian&lt;/em&gt;, $\psieli$ is the &lt;em&gt;electronic wave function&lt;/em&gt; describing $i$-th electronic state, and $\Eeli$ is the &lt;em&gt;electronic energy&lt;/em&gt; of the $i$-th electronic state --- the total energy of a molecule with nuclei being fixed at particular coordinates less the nuclear repulsion potential energy $\Eel = U - V_{\mathrm{nn}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the electronic wave function is a function of electronic coordinates only, though &lt;em&gt;parametrically&lt;/em&gt; it still depends on nuclear coordinates. The parametric dependence of $\psieli$ and $\Eeli$ on $\rnuc$ means that for different sets of fixed nuclear coordinates $\rnuc$ we have different electronic Schrödinger equations with different solutions $\psieli$ and $\Eeli$, or, in other words, that the electronic Schrödinger equation written above is a family of equations parametrized by $\rnuc$, rather then one equation.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second step&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Once solutions of the electronic Schrödinger equation $\{ \psieli(\rel) \}_{i=1}^{\infty}$ and $\{ \Eeli \}_{i=1}^{\infty}$ are found for a number of different fixed nuclear configurations $\rnuc$, we took the second step: the task is to solve an equation of motion for nuclei which is obtained as follows.&#xA;Note first that since $\Hel$ is self-adjoint, its eigenfunctions $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ form a complete orthonormal set.&#xA;Now we use the assertion that a function of two independent variables $\psi(\rel, \rnuc)$ can be expanded over a complete set of functions of one variable $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ in the following way&#xA;\begin{equation}&#xA;    \psi(\rel, \rnuc) = \sum\limits_{i=1}^{\infty} \psieli(\rel, [\rnuc]) \psinuci(\rnuc) \, . \tag{2}&#xA;\end{equation}&#xA;Substitution of (2) in the Schrödinger equation (1), multiplication of both sides by $\conj{\psi}_{\mathrm{e},j}$, and integration over electronic coordinates $\rel$ only, leads to&#xA;\begin{equation}&#xA;    \Big( \Tnuc + \Eelj + \Vnucnuc - E \Big) \psinucj&#xA;    -&#xA;    \sum\limits_{i=1}^{\infty} \linop{\Lambda}_{ji} \psinuci&#xA;    = 0 \quad \text{for } j = 1, 2, \dotsc \, , \tag{3}&#xA;\end{equation}&#xA;where operator $\linop{\Lambda}_{ji}$ is defined as follows&#xA;\begin{equation}&#xA;    \linop{\Lambda}_{ji} = &#xA;    \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} &#xA;    \big(&#xA;        2 \bracket{ \psielj }{ \nabla_{\alpha} }{ \psieli } \nabla_{\alpha}&#xA;        + \bracket{ \psielj }{ \nabla_{\alpha}^2 }{ \psieli }&#xA;    \big) \, .&#xA;\end{equation}&#xA;and arguments $\rel$ and $\rnuc$ were dropped for brevity.&#xA;The last equation describes a system of differential equations (one equation for each value of $j$), and once the system is solved and a set of functions $\psinuci$ is obtained, one could write down the exact solution of the Schrödinger equation (1) in the form we introduced above (2).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Approximations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The task to solve the system of equations (3) is extremely difficult and the reason is that equations in the system are coupled in a sense that the solution of any $i$-th equation, $\psinuci(\rnuc)$, enters all other equations as the $\linop{\Lambda}_{ji} \psinuci$ term. One obvious way to decouple equations in the system (3) is by assuming that $\linop{\Lambda}_{ji} = 0$ either for all $i \neq j$, or for all $i$ without any exception. Operators $\linop{\Lambda}_{ji}$ may be regarded as the elements of a square matrix $\mat{\Lambda}$ and of course the fact that we equate these elements with zero is an approximation: in the first case, when we equate with zero all off-diagonal elements $\linop{\Lambda}_{ji}$, we introduce the &lt;em&gt;adiabatic approximation&lt;/em&gt;, while in the second case, when we equate with zero all elements $\linop{\Lambda}_{ji}$, we introduce the &lt;em&gt;Born–Oppenheimer approximation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both these approximations lead to a system of the so-called &lt;em&gt;nuclear Schrödinger equations&lt;/em&gt;&#xA;\begin{equation}&#xA;    \linop{H}_{\mathrm{n}} \psinucj(\rnuc) = E \psinucj(\rnuc) \quad \text{for } j = 1, 2, \dotsc \, ,&#xA;\end{equation}&#xA;where $\psinucj(\rnuc)$ is known as the &lt;em&gt;nuclear wave function&lt;/em&gt; and $\linop{H}_{\mathrm{n}}$ as the &lt;em&gt;nuclear Hamiltonian&lt;/em&gt;, with the only difference being that in the first case the nuclear Hamiltonian takes the following form&#xA;\begin{equation}&#xA;    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj} \, ,&#xA;\end{equation} &#xA;while in the second case the last term $\linop{\Lambda}_{jj}$ is absent &#xA;\begin{equation}&#xA;    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc \, .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to notice is that in both the above mentioned approximations the only terms in the nuclear Schrödinger equation for $j$-th nuclear state $\psinucj$ that depend on electronic states, $\Eelj$ and also $\linop{\Lambda}_{jj}$, are related to the same and the only electronic state $\psielj$.&#xA;This means that in both these approximations each nuclear state $\psinucj$ depends only on one electronic state $\psielj$, while in the accurate physical picture (3), electronic states are coupled through off-diagonal $\linop{\Lambda}_{ji}$ terms, known as &lt;em&gt;non-adiabatic coupling terms&lt;/em&gt; (NACT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the first term in the nuclear Hamiltonian in both adiabatic and Born–Oppenheimer approximations represents the kinetic energy of nuclear motion, and consequently, the remaining terms collectively should stand for the potential energy of nuclear motion.&#xA;It is said thus that nuclei move on the so-called &lt;em&gt;potential energy surface&lt;/em&gt; (PES) defined by the sum of these remaining terms: $\Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj}$ in the adiabatic approximation and just $\Eelj([\rnuc]) + \Vnucnuc$ in the Born–Oppenheimer one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-adiabatic coupling terms usually (but not always) can indeed be neglected, especially around the equilibrium nuclear configuration.&#xA;Besides, often the errors due to neglecting these terms are less than the errors resulting from some further approximation introduced when solving electronic and nuclear equations.&#xA;However, when two potential energy surfaces come close to each other or even cross, non-adiabatic coupling terms becomes large and are no longer negligible.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Answers&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What adiabatic means?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The term &quot;adiabatic&quot; from Ancient Greek αδιαβατος (α - &quot;not&quot;, δια - &quot;through&quot;, βατος - &quot;passable&quot;) literally means the situation when something &quot;is not passing through&quot; something else.&#xA;In thermodynamics adiabatic process is a process occurring without exchange of heat of a system with its environment, i.e. a process in which heat is not passing through system enclosure.&#xA;In quantum mechanics adiabatic refers to a process in which no abrupt transition from one state to another with respect to continuous changes of some parameters happens, and thus, energy of a system changes continuously with respect to continuous changes of these parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the reason why we call the above mentioned approximation adiabatic lies in the fact that in this approximation when solving the electronic Schrödinger equation for different fixed nuclear configurations by continuously varying the coordinates of the nuclei we indeed assume that electronic state of the system continues to be the same. In other words, we assume that we are &quot;staying&quot; at the same PES. This could be the case, of course, only when potential energy surfaces for different electronic states are well separated, so, in the regions of nuclear coordinates where potential energy surfaces come close to or even cross each other adiabatic approximation is inadequate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the case of a non adiabatic appox. to Schrodinger eq.?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Non-adiabatic solution refers to the solution of the system of coupled equations (3).&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="16683" LastEditDate="2017-05-19T22:58:12.100" LastActivityDate="2017-05-19T22:58:12.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31368" PostTypeId="2" ParentId="31359" CreationDate="2015-05-14T07:49:12.250" Score="7" Body="&lt;p&gt;Current prices are about 34 € (500 g) for 1-bromobutane and 32 € (500 mL) for 1-butanol in decent lab quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most cost efficient method is ordering 1-bromobutane from a chemical supplier of your choice!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a &lt;strong&gt;technical&lt;/strong&gt; scale, one would probably react 1-butene with anhydrous hydrogen bromide under radical conditions, using peroxides or UV light to initiate homolytic cleavage of the $\ce{H-Br}$ bond.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-14T07:49:12.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31369" PostTypeId="2" ParentId="31345" CreationDate="2015-05-14T08:16:17.083" Score="6" Body="&lt;p&gt;Is it correct that you are aiming to synthesize 4-(4-hydroxyphenyl)benzoic acid [58574-03-1]?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EgiFL.png&quot; alt=&quot;4-(4-hydroxyphenyl)benzoic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target compound is available in one step from biphenyl using potassium peroxydisulfate $(\ce{K2S2O8})$ as an oxidant, Palladium(II)-trifluoroacetate, $\ce{Pd(OCOCF3)2}$, as a catalyst, and formic acid as the carbon source for the $\ce{-COOH}$ group. The synthesis was published in&lt;br&gt;&#xA; &lt;em&gt;Org. Lett.&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;6&lt;/em&gt;, 2437-2439 (&lt;a href=&quot;http://dx.doi.org/10.1021/ol049166l&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limited resources and experience will probably rule out &lt;em&gt;Stille&lt;/em&gt;-type coupling reactions too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatives pretty much depend on which 4,4'-disubstituted biphenyl are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A methoxyarene can be converted to the corresponding phenol by $\ce{BBr3}$&lt;br&gt;&#xA;or the cheaper hydrobromic acid. The $\ce{-OCH3}$ group resists oxidation, is a &lt;strong&gt;+M&lt;/strong&gt; substituent and is not acidic. All this might be helpful (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to introduce the $\ce{-COOH}$ substituent, two approaches seem to be possible in an undergrade lab:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;oxidation of a methyl group by permanganate  &lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Grignard&lt;/em&gt; reaction of a bromoarene with carbon dioxide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3st78.png&quot; alt=&quot;retrosynthesis&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-14T09:37:31.023" LastActivityDate="2015-05-14T09:37:31.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31370" PostTypeId="2" ParentId="31350" CreationDate="2015-05-14T10:30:16.947" Score="4" Body="&lt;p&gt;Expanding on Nicolau Saker Neto's comments above, the heat capacity dictates the (reversible) heat flux required to increase the system's temperature by one unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Definitions of the heat capacity ($C_V$ and $C_p$)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it is that the heat capacity dictates the response in internal energy ($U$)/enthalpy ($H$) of the system to temperature, as described below. Note that the heat capacity at constant volume ($C_V$) is not the same as the heat capacity at constant pressure ($C_p$) - and that both may be defined in several ways, the most common being: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_V = \left(\frac{\partial U}{\partial T}\right)_{V,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{V,\mathbf{n}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_p = \left(\frac{\partial H}{\partial T}\right)_{p,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{p,\mathbf{n}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contributions to the heat capacity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In should be added that while translation - the kinetic energy of molecules in motion - are an important contribution to the heat capacity of a gas, both &lt;em&gt;vibrations&lt;/em&gt; and &lt;em&gt;rotations&lt;/em&gt; also contribute for all but the simplest molecules. For high temperatures (approx. $T \geq 3000\ \mathrm{K}) $, the excitation of outer electrons will also contribute to the heat capacity. For even higher temperatures, the substance in question will behave as &lt;a href=&quot;http://en.wikipedia.org/wiki/Plasma_(physics)&quot; rel=&quot;nofollow&quot;&gt;plasma&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many complicating factors to the precise understanding of the origin of contributions to the heat capacity - both at the high end and low end of the temperature range. In all cases, however, the heat capacity will give you the answer to the question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;How much energy do I have to add to this system at the given&#xA;  conditions in order to increase the temperature with one unit?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further reading on &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; is recommended for anyone interested. &lt;/p&gt;&#xA;" OwnerUserId="475" LastEditorDisplayName="user7951" LastEditDate="2015-05-14T10:57:30.753" LastActivityDate="2015-05-14T10:57:30.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31371" PostTypeId="2" ParentId="31356" CreationDate="2015-05-14T10:33:02.940" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Claisen_condensation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claisen reaction&lt;/a&gt; yields a product with an &lt;em&gt;acidified&lt;/em&gt; proton in $\alpha$-position of carbonyl and ester, right at the same carbon atom of R depicted in blue.&lt;img src=&quot;https://i.stack.imgur.com/vVXDo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In presence of a base, yes, $\beta$-dicarbonyl compounds undergo deprotonation at their methylene unit to yield a keto-enolate, stabilized by mesomerism with the secound carbonyl group in conjugation. Compare the $pK_a$-values of $\beta{}$-dicarbonyl compounds with ketones, for example (one &lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;), to estimate if the equilibrium is on the left, or on the right.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Vm3n4.png&quot; alt=&quot;enter image description here&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think cations &lt;em&gt;of the right size and charge&lt;/em&gt; equally favour the formation of the keto-enolate.  If you synthesize $\beta$-dicarbonyl compounds, the risk to loose product due to chelation  is lowered if initial work-up is concluded by drying the washed organic phase with sodium sulphate ($\ce{Na2SO4}$), rather than by magnesium sulphate ($\ce{MgSO4}$).&lt;/p&gt;&#xA;" OwnerUserId="1782" LastActivityDate="2015-05-14T10:33:02.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31372" PostTypeId="1" AcceptedAnswerId="35103" CreationDate="2015-05-14T10:42:43.810" Score="7" ViewCount="6617" Body="&lt;p&gt;I am slightly confused as to how I should think of changes in state (especially sublimation and deposition). Should I think of sublimation and deposition as skipping the liquid phase (fig 1) or should I think of it as a circular pattern (fig 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/05Z3Z.jpg&quot; alt=&quot;fig 1&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;fig 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFANr.gif&quot; alt=&quot;fig 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;fig 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorDisplayName="user15489" LastEditDate="2015-08-15T16:29:46.823" LastActivityDate="2015-08-17T00:38:23.693" Title="change in state (sublimation and deposition)" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31373" PostTypeId="2" ParentId="31363" CreationDate="2015-05-14T10:45:26.727" Score="4" Body="&lt;p&gt;My go-to for this is typically Kolthoff &amp;amp; Sandell's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000TCD3YE&quot; rel=&quot;nofollow&quot;&gt;Textbook of Quantitative Inorganic Analysis&lt;/a&gt;.  They unfortunately do not report a detailed protocol for dichromate standardization of thiosulfate solutions therein, but instead cite &lt;em&gt;Volumetric Analysis, Vol III&lt;/em&gt; by Kolthoff and Stenger (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00A9YOLEI&quot; rel=&quot;nofollow&quot;&gt;Amazon&lt;/a&gt; and &lt;a href=&quot;https://books.google.com/books?id=KjGNQgAACAAJ&amp;amp;dq=kolthoff%20and%20stenger&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=0p9UVcODOsWmNrSygdgB&amp;amp;ved=0CCQQ6AEwAQ&quot; rel=&quot;nofollow&quot;&gt;Google Books&lt;/a&gt;).  I don't have access to this volume; &lt;a href=&quot;http://www.worldcat.org/title/volumetric-analysis/oclc/344117&amp;amp;referer=brief_results&quot; rel=&quot;nofollow&quot;&gt;WorldCat&lt;/a&gt; shows numerous copies in university libraries in my area of the U.S.; more than likely there's a library where you are that has a copy.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-05-14T13:21:05.480" LastActivityDate="2015-05-14T13:21:05.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31374" PostTypeId="2" ParentId="31356" CreationDate="2015-05-14T11:44:30.047" Score="1" Body="&lt;p&gt;Generally, diketo compounds can react to form an enol-keto form and back again. This tautomery can &lt;strong&gt;not&lt;/strong&gt; be considered a [1,3] proton shift due to orbital chemistry and has to be thought of as a combination of deprotonation/reprotonation or protonation/deprotonation steps — ask another question if you’re interested in details (assuming they are not already posted somewhere on the site).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8cqfp.png&quot; width=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice, that in its enol-keto form, the compound can form an intramolecular hydrogen bond. If desired, the $\mathrm{p}K_{\mathrm{a}}$ of the molecule itself is enough to protonoate another one, so the process can be thought of as self-catalysed. Thus, there is a tendency of these compounds to equilibrate to a desired ratio. Only minor differences need to be considered, if $\mathrm{R} = \ce{OR''}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is that ratio going to be? It depends on the surroundings. If you dissolve it in solvents that are capable of donating hydrogen bonds, then both keto oxygens will accept H-bonds, and the diketo tautomer will be preferred. On the other hand, if the solvent cannot form any hydrogen bonds, the enol tautomer, which is capable of forming an intramolecular H-bond, is less disfavoured for that exact reason. Further parameters like other molecules dissolved, their $\mathrm{p}K_{\mathrm{a}}$’s, the dryness of the solvent, the pH-value etc. also factor in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which tautomer dominates in which solvent and what the ratio between the two is (it is extremely unlikely for one form to be the only one present) &lt;em&gt;cannot&lt;/em&gt; be predicted exactly and needs to be measured experimentally.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorDisplayName="user7951" LastEditDate="2015-05-14T18:38:27.190" LastActivityDate="2015-05-14T18:38:27.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31375" PostTypeId="1" AcceptedAnswerId="31379" CreationDate="2015-05-14T12:01:45.197" Score="34" ViewCount="4438" Body="&lt;p&gt;Suppose I have two units, Tonks, $T$, and Borks, $B$, as well as the equation $\frac{50T}{20B}=x$. May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class? For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{50T}{20B}&amp;amp;=x\\&#xA;50T&amp;amp;=20Bx\\&#xA;\frac{50T}{20}&amp;amp;=Bx\\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorDisplayName="user7951" LastEditDate="2015-09-25T23:16:29.800" LastActivityDate="2017-04-05T22:52:27.587" Title="May I treat units (e.g. joules, grams, etc.) in equations as variables?" Tags="&lt;notation&gt;&lt;units&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31376" PostTypeId="2" ParentId="31297" CreationDate="2015-05-14T12:05:44.037" Score="1" Body="&lt;p&gt;The real question is not ‘Why don’t books include &lt;em&gt;X&lt;/em&gt;?’, but &lt;strong&gt;‘What would change if books included &lt;em&gt;X&lt;/em&gt;?’&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge and understanding, what protons, electrons and neutrons are is fundamental to understanding a wide range of stuff that happens in chemistry. (Yes, neutrons play a role for deuterium exchange effects etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is hardly anything that would change if descriptions included lower-level particles like quarks. I can’t think of any &lt;em&gt;chemical reaction&lt;/em&gt; or other &lt;em&gt;chemical process&lt;/em&gt; that &lt;em&gt;cannot&lt;/em&gt; be sufficiently explained only going back to protons, neutrons and electrons — usually, even neutrons are irrelevant. So basically, you’re suggesting &lt;em&gt;increasing complexity with no added benefit to the learner&lt;/em&gt;. Why would one do that? Just for the sake of correctness? If anything, a note that protons and neutrons are made up of smaller particles suffices for any beginner’s level textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, there is quite a lot of chemistry that can be just explained with the concepts of &lt;em&gt;atoms&lt;/em&gt; and &lt;em&gt;bonds&lt;/em&gt;. The only reason why the sub-atomic particles are included in the first place, is because salts are introduced early and can be explained a lot better, if electrons and nuclei are known. (And it eases the explanation of ‘what exactly is a bond?’)&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-14T12:05:44.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31378" PostTypeId="1" CreationDate="2015-05-14T12:51:17.190" Score="4" ViewCount="293" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{25 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 \%(w/v)}$&lt;/span&gt; lactic acid (&lt;span class=&quot;math-container&quot;&gt;$\ce{C3H6O3}$&lt;/span&gt;, molecular mass &lt;span class=&quot;math-container&quot;&gt;$\pu{90.1 g mol-1}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_{\mathrm{a}} = 3.86$&lt;/span&gt; was neutralised by &lt;span class=&quot;math-container&quot;&gt;$\pu{13.1 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1010 M}$&lt;/span&gt; monobasic base. Estimate the end point &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; and calculate the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H6O3}$&lt;/span&gt; in % of the initial one.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question was part of my final paper. I can't remember the exact words used in the paper for the % part, but I think it is asking for the % purity. I am not sure how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; using the formula &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log M)$&lt;/span&gt; but I think it is wrong. How is this doen correctly?&lt;/p&gt;&#xA;" OwnerUserId="15253" LastEditorUserId="4945" LastEditDate="2020-05-30T14:01:53.003" LastActivityDate="2020-05-30T14:18:24.860" Title="How to determine the purity of lactic acid with titration?" Tags="&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31379" PostTypeId="2" ParentId="31375" CreationDate="2015-05-14T12:53:03.840" Score="39" Body="&lt;p&gt;Yes, you may. It is quite common to convert units into each other. The simplest conversion might be the prefixing of units, e.g. &#xA;$$\mathrm{\frac{km}{m}}=1000 \Longleftrightarrow \mathrm{1~km = 1000~m}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the interconversion of units of energy. In some parts of chemistry, it is still quite common to use calories, in others Joule, as a SI unit, has taken its place.&#xA;$$\mathrm{\frac{cal}{J}}=4.184 \Longleftrightarrow \mathrm{1~cal = 4.184~J}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, all units are defined to be constants, special numbers in a wider sense. You can treat them as such.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-14T12:53:03.840" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31380" PostTypeId="2" ParentId="31375" CreationDate="2015-05-14T13:18:16.717" Score="28" Body="&lt;p&gt;Treating units as if they were algebra is called “&lt;a href=&quot;http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html&quot; rel=&quot;noreferrer&quot;&gt;dimensional analysis&lt;/a&gt;”.&lt;/p&gt;&#xA;&lt;p&gt;One example given in that article is the question of how many seconds are there in two years.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2\ \mathrm{yr} \times 365\ \mathrm{day}\ \mathrm{yr}^{-1} \times 24\ \mathrm{hr}\ \mathrm{day}^{-1} \times 60\ \mathrm{min}\ \mathrm{hr}^{-1} \times 60\ \mathrm{s}\ \mathrm{min}^{-1} = 63072000\ \mathrm{s} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Another example question is,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How many atoms of hydrogen can be found in 45 g of ammonia, &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This is how I would approach this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$45\ \mathrm{g}\ \ce{NH3}\ /\ 17\ \mathrm{g}\ \mathrm{mol}^{-1} \approx 2.647\ \ce{NH3}\ \mathrm{mol}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2.647\ \ce{NH3}\ \mathrm{mol}\ \times\ 6.02 \times 10^{23}\ \mathrm{molecules}\ \mathrm{mol}^{-1} \approx 1.594 \times 10^{24}\ \ce{NH3}\ \mathrm{molecules}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$1.594 \times 10^{24}\ \ce{NH3}\ \mathrm{molecules} \times 3\ \mathrm{(H\ atoms)}\ \mathrm{(\ce{NH3}\ molecules)}^{-1} \approx 4.781 \times 10^{24}\ \ce{H}\ \mathrm{atoms}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The article works this out differently, as a single step. I myself find the multi-step approach easier to follow, but everyone’s different. The article renders this as,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$45\ \mathrm{g}\ \ce{NH3} \times \frac{1\ \mathrm{mol}\ \ce{NH3}}{17\ \mathrm{g}\ \ce{NH3}} \times \frac{6.02 \times 10^{23}\ \mathrm{molecules}\ \ce{NH3}}{1\ \mathrm{mol}\ \ce{NH3}} \times \frac{3\ \mathrm{atoms}\ \ce{H}}{1\ \mathrm{molecule}\ \ce{NH3}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ = 4.8 \times 10^{24}\ \mathrm{atoms}\ \ce{H}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(to 2 significant figures)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If the units cancel out correctly, that is further indication that you have the right formula. (And if they &lt;em&gt;don’t&lt;/em&gt; cancel out correctly, that is a certain indication that you don’t have the right formula.)&lt;/p&gt;&#xA;" OwnerUserId="177" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-16T19:17:16.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31381" PostTypeId="2" ParentId="31352" CreationDate="2015-05-14T14:09:43.270" Score="15" Body="&lt;p&gt;Starting with a short and blunt answer to the title’s question: Yes. But you likely already knew that, otherwise you wouldn’t have written the rest.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn’t sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw &lt;strong&gt;both&lt;/strong&gt; resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, there is a lot more time saved that you seem to be imagining according to your quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to suggest that there are only two correct resonance structures for benzene, which is incorrect. Even if you only consider the $\pi$ system, there are a lot more possible resonance structures which contribute to the final picture in a non-zero manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFOHu.png&quot; alt=&quot;some benzene resonances&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones I displayed in the picture are just a &lt;em&gt;very small&lt;/em&gt; subset. In fact, solely for the $\ce{H2}$ molecule, a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures and also claimed to be able to accurately calculate the energy of the molecule by adequately weighting all those structures. (‘Accurately’ meaning getting an energy comparable to one calculated using molecular orbital theory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually moves us towards another question: What actually is a bond? There is &lt;em&gt;a lot&lt;/em&gt; more to a bond than simply drawing a line between two atoms and saying that ‘they are now sharing an electron pair.’ If you calculate simple-looking molecules like formaldehyde using a program like Gaussian, you are able to see a multitude of molecular orbitals — at least twelve. Two are defined as &lt;em&gt;‘core orbitals’&lt;/em&gt;, namely they represent the 1s orbitals of carbon and oxygen, respectively; all the others &lt;em&gt;somehow contribute to bonding or antibonding interactions somewhere across the whole molecule&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what would an accurate representation be? Drawing the atoms geometrically correctly, a cloud around them and writing a number 12 inside or next to the cloud? That way, it would be made clear that there are twelve electrons delocalised across the whole of the molecule and somehow keeping it together. But what would be gained from this type of more accurate representation? Or rather: What would be lost? You would lose the possibility to easily discuss lone pairs on the oxygen, bond order between carbon and oxygen and a lot more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for benzene and all molecules that contain a six-membered ring of carbons with a $\pi$ system with six electrons (a.k.a. the benzene derivates you’re asking about): If you insisted on drawing ‘the two’ resonance structures, you would still be far from the truth. Furthermore, for most molecules that are not benzene but contain the same ring, the bonds are no longer equal and their bond orders are no longer exactly $1.5$ (true for any molecule that has lost the $\mathrm{D}_{6\mathrm{h}}$ symmetry). So drawing two sets of double bonds and claiming the structures to be equal &lt;em&gt;would be outright wrong&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, though, every chemist has learnt to get a more or less thorough understanding on the concept of a &lt;em&gt;bond&lt;/em&gt; and on the concept of &lt;em&gt;aromaticity&lt;/em&gt; that it can reasonably be assumed that what I presented is accessable somewhere at the back of the head if really required. Thus, every chemist should have a rather good understading that a line on paper is a line, a bond is not just a pair of electrons and a benzene ring can be drawn in multiple ways. When it really matters, we can either expect the chemist to instantly recognise the mattering due to a special case, or we can bring the mattering matter to the chemist’s attention by explicitly drawing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is &lt;strong&gt;a lot of time gained&lt;/strong&gt; by simplifying drawings of molecules in the way they are and also &lt;strong&gt;cases where it matters&lt;/strong&gt; can easily be &lt;strong&gt;illustrated&lt;/strong&gt; by drawing &lt;strong&gt;more elaborate drawings&lt;/strong&gt; which aren’t skipped in one’s attention because they are rarely encountered. And we also &lt;strong&gt;keep the ability&lt;/strong&gt; to easily &lt;strong&gt;recognise&lt;/strong&gt; and discuss &lt;strong&gt;simple concepts.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S: It’s not necessary for &lt;em&gt;anyone&lt;/em&gt; to interpret these concepts the same way. In fact, it’s a good thing if an artist takes the picture of benzene and interprets it completely differently. If anything, ‘anyone’ should be restricted to the chemical — or maybe the entire scientific — community, but there I can see the interpretation safely, regardless of whether aromaticity is drawn or not.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-14T14:09:43.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31382" PostTypeId="1" AcceptedAnswerId="31403" CreationDate="2015-05-14T14:21:19.270" Score="5" ViewCount="788" Body="&lt;p&gt;In our chemistry lab we performed the following experiment below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that $K_\mathrm{sp}(\ce{AgCl}) = 8.2 \times 10^{-11}$ and $K_{\mathrm{sp}}(\ce{PbCl_{2}}) = 1.7 \times 10^{-5}$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Procedure:&#xA;  $0.2\,\mathrm{mL}$ each of $\mathrm{0.05\: M \: \ce{Ag+}}$ ions and $\mathrm{ 0.05\: M \: \ce{Pb^{2+}}}$ ions were added into a centrifuge tube. The solution was diluted to $1\,\mathrm{mL}$ $6\,\mathrm{M}$ $\ce{HCl}$ was added to the solution till it completely precipitates. The quantity of $\ce{HCl}$ required was recorded. The precipitate and supernatant were separated and kept for subsequent experiments.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Observations: Precipitate formation was observed almost instantly. It took 6 drops of $\ce{HCl}$ to completely precipitate the contents. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Estimate the concentration of $\ce{HCl}$ and volume of $6\,\mathrm{M}$ $\ce{HCl}$ required to complete precipitate each of the given salts. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is my approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align} c_{1}V_{1} &amp;amp;= c_{2}V_{2} \\&#xA;\text{Let } c_{1}=0.05\:\mathrm{M}, V_{1}&amp;amp;=0.2\:\mathrm{mL}, \text{and} \ V_{2}=1\:\mathrm{mL}\\&#xA;[\ce{Pb^2+}] &amp;amp;= 0.2 \cdot 0.05\ \mathrm{M} \\&#xA;[\ce{Ag+}] &amp;amp;= 0.2 \cdot 0.05\ \mathrm{M}\\&#xA;K_{\mathrm{sp}}\:(\ce{AgCl}) &amp;amp;= [\ce{Ag+}] [\ce{Cl-}] \\&#xA;8.2 \times 10^{-11} &amp;amp;= 0.2 \cdot 0.05\ \mathrm{M} \cdot s_{1} \\&#xA;s_{1} &amp;amp;= 8.2 \times 10^{-9}\ \mathrm{M}\\&#xA;K_{\mathrm{sp}}\:(\ce{PbCl_{2}}) &amp;amp;= [\ce{Pb^{2+}}] [\ce{Cl-}]^{2}\\&#xA;1.7 \times 10^{-5} &amp;amp;= 0.2 \cdot 0.05\ \mathrm{M} \cdot (s_{2})^2\\&#xA;s_{2} &amp;amp;= 0.041\ \mathrm{M} \end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to precipitate $\ce{AgCl}$ and $\ce{PbCl_{2}}$ completely, the concentration of chloride ions in solution must exceed $8.2 \times 10^{-9}$ M and $0.041$ M respectively). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure of my computations up till this point, and I need some help on how to proceed further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; My calculations  tell me that volume of $6\,\mathrm{M}\,\ce{HCl}$ necessary is $0.00683\,\mathrm{mL}$. In the lab we have estimated the volume of one drop of our pipette as $0.037\,\mathrm{mL}$, based on this I calculated that the number of drops of hydrochloric acid necessary to complete precipitation is $0.18$ drops. But in the actual experiment I performed 6 drops of $\ce{HCl}$ were dispensed and based on the computations we did, everything should have precipitated. But after centrifugation and separation the supernatant still had lead ions which were precipitated as lead chromate. Why is this so? (This is the main reason why I suspected my calculations and turned to this website for help).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt;: I now see the flaw in my computations. what I have calculated is in fact that amount of HCl necessary to start precipitation. Thanks to everyone for their help.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="16088" LastEditDate="2015-05-16T10:43:51.773" LastActivityDate="2015-05-16T10:43:51.773" Title="Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts" Tags="&lt;analytical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31383" PostTypeId="1" AcceptedAnswerId="31391" CreationDate="2015-05-14T14:43:52.990" Score="7" ViewCount="374" Body="&lt;p&gt;I've seen before chemistry demonstrations where solutions are mixed with one another and subsequently where the resulting product goes through periodic color changes: for example blue to orange and back to blue again. The reaction seems to go on for awhile at a regular period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question - can this reaction be considered resonance in the sense that energy is flowing back and forth between two chemical states? In such reactions are there actually two chemical species between the color changes or just energy level/temperature differences?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="10185" LastEditDate="2015-05-14T18:37:43.690" LastActivityDate="2015-05-14T18:37:43.690" Title="Periodically oscillating chemical reactions?" Tags="&lt;reaction-mechanism&gt;&lt;solutions&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31384" PostTypeId="1" AcceptedAnswerId="31387" CreationDate="2015-05-14T14:44:54.693" Score="3" ViewCount="1695" Body="&lt;p&gt;Related to my previous question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31135/is-solubility-coefficient-affected-if-ion-data-is-given-in-ksp&quot;&gt;Is solubility coefficient affected if ion data is given in Ksp?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$200\,\mathrm{mL}$ solution of $0.02\,\mathrm{M}\,\ce{AgNO3}$  is added to $200\,\mathrm{mL}$ $\ce{CrO4^{2-}}$ and $\ce{PO4^3-}$ ions. Find out both of $Q_{\mathrm{sp}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Actually the question is asking &quot;Will it precipitate&quot;, but I skip it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my half approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$200\,\mathrm{mL}=2\times 10^{-1}\,\mathrm{L}$$&#xA;$$V_{\mathrm{total}}=400\,\mathrm{mL}=4\times  10^{-1}\,\mathrm{L}$$&#xA;$$[\ce{Ag+}]=[\ce{CrO4^2-}]=[\ce{PO4^3-}]=\frac{0.02\,\mathrm{M}\cdot 2\times 10^{-1}\,\mathrm{L}}{4\times 10^{-1}\,\mathrm{L}}=0.01\,\mathrm{M}$$&#xA;$$\ce{Ag2CrO4 -&amp;gt; 2Ag^{+} +CrO4^{2-}}$$&#xA;$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[\ce{Ag+}]^{2} \cdot [\ce{CrO4^2-}]$$&#xA;$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[2s]^{2}[0.01\,\mathrm{M}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that to find $K_{\mathrm{sp}}$, the $s$ needs to be multiplied (and to the power) with the coefficient from the reaction, but how about $Q_{\mathrm{sp}}$? Does the $s$ also need to be multiplied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I asked this because my teacher said that only to the power is affecting the $Q_{\mathrm{sp}}$.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16103" LastEditorUserId="5026" LastEditDate="2018-07-24T01:54:07.760" LastActivityDate="2018-07-24T01:54:07.760" Title="Is solubility in Qsp affected by coefficient?" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31385" PostTypeId="1" CreationDate="2015-05-14T14:58:09.350" Score="5" ViewCount="156" Body="&lt;p&gt;My question is based on the &lt;a href=&quot;http://icho2015.msu.az/uploads/files/Preparatory_problems_IChO_2015_May_1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Preparatory Problem Set&lt;/a&gt; for the 47th IChO Problem 26 - Holy War against Four Horsemen of the Apocalypse (it is on page 36). It is written that cooling of Y combustion products from 150 to 0 degrees Celsius does not result in any condensed phase, and addition of silver chloride to the supernatant of this combustion product provide some precipitate, so I concluded that this organic compound have either (or both) chloride and bromide, since iodine is too large for this substance. After calculating some possible answers, I've concluded that compound Y is $\ce{C3H2BrClO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if I combust compound Y ($\ce{C3H2BrClO2}$), will I get some products that aren't condensed when I cool these combustion products?&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorDisplayName="user7951" LastEditDate="2017-12-25T22:18:57.900" LastActivityDate="2017-12-25T22:18:57.900" Title="Combustion of halo-organic compound" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;combustion&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31386" PostTypeId="2" ParentId="31378" CreationDate="2015-05-14T14:59:39.337" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C' = &#xA;\frac{C_\mathrm{b}\, V_\mathrm{b,eq}}{V_\mathrm{a} + V_\mathrm{b, eq}},&#xA;$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$C_\mathrm{b}$&lt;/span&gt; is the concentration of the base, &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{b, eq}$&lt;/span&gt; is the volume of the base at the end point and &lt;span class=&quot;math-container&quot;&gt;$V_\mathrm{a}$&lt;/span&gt; is the volume of lactic acid.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C' =&#xA;\frac{\pu{0.1010 mol L-1} \times \pu{13.1 mL}}{\pu{25 mL} + \pu{13.1 mL}}&#xA; = \pu{0.03473 mol L-1}&#xA;$$&lt;/span&gt; &#xA;The &lt;span class=&quot;math-container&quot;&gt;$\ce{pH}$&lt;/span&gt; of the solution at the end point is given by the equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \mathrm{pH} &#xA;    &amp;amp;= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a} &#xA;      - \mathrm{p}C')\\&#xA;    &amp;amp;= \frac{1}{2}(14 + 3.86 + \log 0.03473) = 8.20&#xA;\end{align}&lt;/span&gt;&#xA;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To calculate the purity percentage of lactic acid, we calculate the amount of substance of pure lactic acid: At the end point &#xA;&lt;span class=&quot;math-container&quot;&gt;$$n_\mathrm{a} = n_\mathrm{b}.$$&lt;/span&gt; &#xA;This means: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;n_\mathrm{a} &#xA;  = C_\mathrm{b} V_\mathrm{b}&#xA;  = \pu{0.1010 mol L-1}\times \pu{13.1E-3 L} &#xA;  = \pu{0.001323 mol}&#xA;$$&lt;/span&gt; &#xA;The mass of pure lactic acid is: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;m &#xA;  = \pu{0.001323 mol} \times \pu{90.1 g mol-1}&#xA;  = \pu{0.1192 g}&#xA;$$&lt;/span&gt;&#xA;Or the mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 \%(w/v)}$&lt;/span&gt; lactic acid is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;m'&#xA;  = \frac{\pu{0.5 \%(w/v)} \times \pu{25 mL}}{100\%} = \pu{0.125 g}&#xA;$$&lt;/span&gt;&#xA;The purity percentage of lactic acid &lt;span class=&quot;math-container&quot;&gt;$P\%$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P\% = \frac{m\times 100\%}{m'}=95.36\%$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2020-05-30T14:18:24.860" LastActivityDate="2020-05-30T14:18:24.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31387" PostTypeId="2" ParentId="31384" CreationDate="2015-05-14T15:04:54.413" Score="3" Body="&lt;p&gt;You don't need to multiply &lt;em&gt;s&lt;/em&gt;, since $Q_\text{sp}$ is the result of multiplication of ions concentration (and you don't need to multiply with the coefficient in any $Q$). So the right way to answer $Q_\text{sp}$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO_{4}^{2-}}]$$&#xA;$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$&lt;/p&gt;&#xA;" OwnerUserId="735" LastEditorUserId="5026" LastEditDate="2018-07-24T01:53:55.687" LastActivityDate="2018-07-24T01:53:55.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31388" PostTypeId="2" ParentId="31297" CreationDate="2015-05-14T15:08:44.417" Score="2" Body="&lt;p&gt;Chemistry is governed by what electrons do, and generally... what &lt;em&gt;valence&lt;/em&gt; electrons do.  There really is no need to dive into subatomic particles.  It may help explain what keeps a nucleus held together but we generally gloss over that aspect and simply have students accept this as a fact.  Subatomic particles, of which there are many, are generally covered in physics.  We usually touch upon them in nuclear chemistry when decay processes are introduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point here is that Chemistry is essentially all about what the electrons are doing.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2015-05-14T15:08:44.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31390" PostTypeId="1" AcceptedAnswerId="31394" CreationDate="2015-05-14T15:24:12.663" Score="5" ViewCount="3460" Body="&lt;p&gt;Why does the gauche conformation of &lt;strong&gt;ethane-1,2-diol&lt;/strong&gt; show hydrogen bonding even though there is no 6-membered ring system? If we don't consider one of the $\ce{H}$ (which is what is usually done), then there is only a 5-membered ring formed. How then does hydrogen bonding take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t2zTN.gif&quot; alt=&quot;conformation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, I feel that there must be some reason apart from 6-member ring for hydrogen bonding which I'm not aware of)&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4231" LastEditDate="2015-05-14T16:18:38.533" LastActivityDate="2020-08-18T16:54:24.387" Title="Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding?" Tags="&lt;organic-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31391" PostTypeId="2" ParentId="31383" CreationDate="2015-05-14T15:24:34.807" Score="7" Body="&lt;p&gt;A couple of points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The demo is probably an oscillating chemical reaction, in which the concentration of a given species increases and decreases repeatedly over time. However, the energy of the overall system is dissipating as that is happening - not oscillating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance, on the other hand, is a (widely misunderstood) concept that applies to a single chemical species and not a bulk reaction mixture. Resonance does not consist of energy flowing back and forth between two chemical states, but rather is meant to represent a single energy state that is a hybrid between two (or more) classical Lewis structures. A double headed arrow between the two classical structures indicates &lt;em&gt;resonance stabilization&lt;/em&gt; — mostly resulting from delocalization of electrons relative to the classical structures. The double headed arrow &lt;strong&gt;does not&lt;/strong&gt; indicate oscillation between those two structures.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="7475" LastEditDate="2015-05-14T16:18:20.327" LastActivityDate="2015-05-14T16:18:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31393" PostTypeId="1" AcceptedAnswerId="31435" CreationDate="2015-05-14T15:33:21.217" Score="4" ViewCount="7399" Body="&lt;p&gt;The $K_\mathrm{D}$ (distribution coefficient) of coffee water at $25\,°\mathrm{C}$ is $7.8$ in a methylene chloride/water system. If there is $2.0\,\mathrm{g}$ of caffeine in $200\,\mathrm{mL}$ of water, calculate the grams and the % of caffeine that can be extracted with a single $100\,\mathrm{mL}$ portion of methylene chloride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{\frac{2\,\mathrm{g}}{200\,\mathrm{mL}}}=7.8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{x}{0.01\,\mathrm{g\,mL^{-1}}}=7.8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=0.078\,\mathrm{g\,mL^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x=0.078\,\mathrm{g\,mL^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.078\,\mathrm{g\,mL^{-1}}\cdot 100\,\mathrm{mL}=7.8\,\mathrm{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think I'm wrong. Am I supposed to maybe use the molar mass of caffeine ($194\,\mathrm{g\,mol^{-1}}$)?&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorDisplayName="user7951" LastEditDate="2015-05-17T08:47:36.963" LastActivityDate="2017-02-25T16:02:14.553" Title="Using Distribution Coefficient to Determine Mass and Percent of Acid Extracted" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31394" PostTypeId="2" ParentId="31390" CreationDate="2015-05-14T15:53:47.117" Score="3" Body="&lt;p&gt;Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (&lt;a href=&quot;http://ursula.chem.yale.edu/%7Echem220/chem220js/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;).  In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the &lt;span class=&quot;math-container&quot;&gt;$\ce{H-C-C-H}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C-C-C-C}$&lt;/span&gt; interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.&lt;/p&gt;&#xA;&lt;p&gt;A typical &lt;span class=&quot;math-container&quot;&gt;$\ce{O-H..H}$&lt;/span&gt; hydrogen bond can stabilize a system by ~7 kcal/mol (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;).  So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - &lt;strong&gt;a net stabilization&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$%edit$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="55119" LastEditDate="2020-08-18T16:54:24.387" LastActivityDate="2020-08-18T16:54:24.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31395" PostTypeId="1" CreationDate="2015-05-14T15:56:18.057" Score="0" ViewCount="39" Body="&lt;p&gt;If the energy of intermolecular attractions are much less than the average kinetic energy of the molecules, what is the physical state of the substance?&lt;/p&gt;&#xA;" OwnerUserId="16208" LastEditorUserId="15410" LastEditDate="2015-05-14T16:16:05.853" LastActivityDate="2015-05-14T16:16:05.853" Title="Determining phase from intermolecular interactions and kinetic energy" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31397" PostTypeId="2" ParentId="31395" CreationDate="2015-05-14T16:04:41.460" Score="1" Body="&lt;p&gt;Most likely gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the definition of an ideal gas is a gas in which its molecules have no intermolecular interactions (except elastic collisions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the strength of intermolecular attraction increases, the molecules will be brought closer together. Molecules in liquids are much more densely packed than molecules in gases.&lt;/p&gt;&#xA;" OwnerUserId="7882" LastActivityDate="2015-05-14T16:04:41.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31399" PostTypeId="1" CreationDate="2015-05-14T16:21:04.270" Score="1" ViewCount="1485" Body="&lt;p&gt;&lt;em&gt;I am having difficulty in understanding the concept of torsional strain. Can someone explain it to me in a more elaborate manner?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean why should bond restrict movement in eclipsed conformation? Doesn't that make it unstable! I tried this &lt;a href=&quot;http://en.wikipedia.org/wiki/Strain_%28chemistry%29#Torsional_strain&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; but I'm still having difficulty in understanding the concept.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="7475" LastEditDate="2015-10-16T17:26:45.827" LastActivityDate="2015-10-16T17:26:45.827" Title="What is torsional strain?" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31400" PostTypeId="1" AcceptedAnswerId="31401" CreationDate="2015-05-14T16:25:48.667" Score="1" ViewCount="1492" Body="&lt;p&gt;For my intro to chemistry class, I was asked to answer whether 'stoichiometry' refers to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To the best of my knowledge those ratios are an essential part of stoichiometry, but stoichiometry is a discipline, and not a ratio, so the answer would be false. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-05-14T16:31:19.383" Title="Does 'Stoichiometry' refer to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31401" PostTypeId="2" ParentId="31400" CreationDate="2015-05-14T16:31:19.383" Score="2" Body="&lt;p&gt;That's right -- stoichiometry is a field which uses these numbers and the relationships between reactants and products, but not the numbers itself. The mole ratio is just the mole ratio, whereas stoichiometry is the entire discipline of using mole ratios &lt;em&gt;to calculate&lt;/em&gt; relative quantities of reactants and products in reactions.&lt;/p&gt;&#xA;" OwnerUserId="13432" LastActivityDate="2015-05-14T16:31:19.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31402" PostTypeId="2" ParentId="31399" CreationDate="2015-05-14T16:31:27.453" Score="2" Body="&lt;p&gt;For the case of a single bond, consider a simple case like ethane, which can take an eclipsed conformation, a staggered conformation, or anything in between depending on the torsion angle of the C-C bond.  Each carbon has a region of especially high electron density where it's bonded to a hydrogen.  When the torsion angle is changed from staggered to eclipsed, these regions approach eachother and repel, and thus the eclipsed conformation is higher in energy than the staggered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a double bond, as in ethylene, the energy associated with torsional strain is much higher because the P orbitals involved in the pi bond have to be oriented the same way.  Thus, applying torsional strain involves breaking this pi bond, and has a high energy.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-05-14T16:31:27.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31403" PostTypeId="2" ParentId="31382" CreationDate="2015-05-14T16:35:09.577" Score="3" Body="&lt;p&gt;Your calculations are looking good. I didn’t check the numerical values, I’ll just assume you’re able to do your algebra and use a calculator correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you got are two different concentrations required to &lt;strong&gt;start to&lt;/strong&gt; precipitate all the cations in your solution. However, you are asked to &lt;em&gt;completely precipitate&lt;/em&gt; the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.&lt;br&gt;&#xA;Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-05-15T17:31:25.047" LastActivityDate="2015-05-15T17:31:25.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31405" PostTypeId="2" ParentId="19645" CreationDate="2015-05-14T16:56:37.540" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;bake starting material in Wilgerodt-Kindler reaction (it works on phenylacetylene too)&lt;/li&gt;&#xA;&lt;li&gt;hydrolyze to phenylacetic acid&lt;/li&gt;&#xA;&lt;li&gt;esterify&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16215" LastActivityDate="2015-05-14T16:56:37.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31409" PostTypeId="1" AcceptedAnswerId="31441" CreationDate="2015-05-14T18:00:21.527" Score="2" ViewCount="1483" Body="&lt;p&gt;The question asks me to draw the product for the reactions. However, I am stuck and do not understand how to draw the reactants with configurations correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) (S)-2-Chloropentane and NaSH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the reactant I drew was in S configuration. Also, I thought since it was a chirality center, inversion of configuration will occur. Can someone please explain this in detail? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CwUCr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="16129" LastEditDate="2015-05-14T18:10:30.553" LastActivityDate="2015-05-15T09:25:21.087" Title="Configurations of Reactants in Second Order Nucleophilic Substitutions" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31412" PostTypeId="1" CreationDate="2015-05-14T18:21:48.267" Score="3" ViewCount="119" Body="&lt;p&gt;This is kind of like two questions but since they're related I have put them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a monomer under a hopefully anatomically correct name 'Phenyl Carbolithia Divinylene Oxide [$\ce{C11H9LiO4}$] (For ease of understanding.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b6LRU.png&quot; alt=&quot;Skeletal Formula for molecule PCDO&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhiMVm.png&quot; alt=&quot;An image of the 3D structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible for me to find or estimate some basic properties (physical and chemical)? Or the 3D structure of the molecule (As opposed to the one I plotted, which is based on my estimation of dipole with minimal reasoning.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By separating the molecules into parts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lithonate (Lithium Carbonate)&lt;/li&gt;&#xA;&lt;li&gt;Benzene (Phenyl group)&lt;/li&gt;&#xA;&lt;li&gt;Ethylene (Vinylene group)&lt;/li&gt;&#xA;&lt;li&gt;Divinyl Ether&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Perhaps now it is possible to make some predictions based on some functional groups? (Such as ether.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though theoretical chemistry has shown that such achievement in predicting a physical structure is unlikely, I am wondering if anyone would know specific properties of this molecule and the way they have deduced it in which I haven't consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the question may lead to the solution for the first part, by synthesizing such molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My humble attempt at theorizing the chemical synthesis with my limited knowledge on the subject is based on condensation polymer. Perhaps having the separate parts of molecules mention previously dehydrated to form a chain, the molecule may be created if a similar polymer is formed except to have the carbon-oxygen backbone chain to form a ring instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have a slight inquiry on whether it is possible for the double bond to exist in this monomer; and whether my nomenclature for this substance is anatomically correct.&lt;/p&gt;&#xA;" OwnerUserId="16210" LastEditorUserId="9961" LastEditDate="2015-11-28T23:17:13.877" LastActivityDate="2020-09-04T23:55:25.040" Title="Method of deriving some properties or possible method of synthesis of a specific monomer" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31414" PostTypeId="1" CreationDate="2015-05-14T19:39:07.810" Score="4" ViewCount="133" Body="&lt;p&gt;If there are two electrons located in a shell (which contains two subshells) the molecule can either have a singlet state (if both electron are in the same subshell; because of the Pauli principle) or a triplett (if the two electrons are in the two different subshells).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign + or - means symmetric or antisymmetric with respect to reflection at a mirror plane that contains the molecular axis (for $\Sigma$ states only) of a diatomics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look how to determine if a state is of + or - symmetry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\mathbf{For~oxygen~in~the~ground~state:}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\Psi_{\mathrm{a}}^{\mathrm{R}}(q_1,1_2) = \sqrt{2}i\chi (z_1,\rho_1)\chi(z_2,\rho_2)\sin(\varphi_1-\varphi_2) \qquad \mathrm{for~the}~^3\sum~\mathrm{term}$ &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\Psi_{\mathrm{s}}^{\mathrm{R}}(q_1,1_2) = \sqrt{2}\chi (z_1,\rho_1)\chi(z_2,\rho_2)\cos(\varphi_1-\varphi_2) \qquad \mathrm{for~the}~^1\sum~\mathrm{term}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://i.stack.imgur.com/pwhCu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to determine if a state is of +/- symmetry if we know the mathematical description of the radial wave function $\Psi^R (q_1 ,q_2)$ (with the two coordinates $q_i$ of the two electrons), because exchanging $\varphi_2$ with $\varphi_1$ in $cos(\varphi_1-\varphi_2)$ part stays the same and therefore + and for the $sin(\varphi_1-\varphi_2)$ (function the sign changes and therefore we get - symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is my question now?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've some troubles to determine the +/- symmetry of a state (of a diatomics for light atoms and therefore only for s and p orbitals) if I have no mathematical description or tabular data given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it should be possible to directly derive the +/- symmetry by knowing the electronic configuration.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;A specific example of oxygen&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I read that in the two electrons in a subshell (that means the same molecular orbital shell and the same spatial orbital) of either $\lambda$ = +1 or -1 and therefore a singlet gives a state with + symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering now: Has a singlet state always + symmetry and a triplet state always - symmetry? I'mn interested to understand why this is the case. I guess this has something to do with spatial and spin part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I'm on the right track, but I think for the ground state two electrons in the same spatial orbital (=same subshell) must have opposite spins (because of Pauli principle) and I guess this means having antisymmetric spin-part. Because the product of spin and spatial part must be antisymmetric I expect the spatial part to be symmetric, what is labeled with g(from german gerade). But I guess this has nothing to do with +/-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first excited state of oxygen [electronic configuration ...$(1\Pi_u)^3(1\Pi^*_g)^3$] there exists six states: $^1{\Sigma} ^{-}_u$,$^1{\Sigma} ^{+}_u$,$^3{\Sigma} ^{-}_u$,$^3{\Sigma} ^{+}_u$,$^1{\Delta} _u$,$^3{\Delta} _u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also one case where my rule singlet gives + symmetry not work. I'm now thinking in which case we have $^1{\Sigma} ^{-}_u$ and in which case we have $^1{\Sigma} ^{+}_u$. The two electrons have opposite spin. One electron has $\lambda_1=+1$ and the other has $\lambda_2=-1$. I guess that the sign +/- depends if the electron with $\lambda_1=+1$ is in a $(1\Pi_u)$ or in a $(1\Pi^*_g)$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Symmetry properties from a table.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Instead of determining the + or - symmetry we can use tables which shows the possible terms for a specific electronic configuration:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\pi_g \otimes \pi_g = \sigma_g^- \oplus \sigma_g^+ \oplus \delta_g$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://i.stack.imgur.com/Kq94P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relation shows that two electrons both in $\pi$ orbitals results in these three different states. Sometimes this terms are written in upper case letters, sometimes they are written in lower case letters. This seems to be arbitrarily choosen.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Practical relevance of the +/- symmetry in Spectroscopy.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In Spectroscopy there are two important selection rules: The Parity selection rule and the Angular momentum selection rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I  read that a transition between two states must undergo a change of the parity from + to - and vise versa. This is the Parity selection rule. The +/- is the basis for the parity selection rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I would expect that the g/u symmetry (symmetric/antisymmetric with respect to an inversion on the centre-of-mass) is the basis of the Angular Momentum selection rule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="14" LastEditDate="2015-06-17T15:46:01.120" LastActivityDate="2015-06-17T15:46:01.120" Title="Relevance and Determination of the reflection symmetry +/- in Spectroscopy?" Tags="&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31416" PostTypeId="2" ParentId="31375" CreationDate="2015-05-14T20:16:16.573" Score="27" Body="&lt;p&gt;Not only may you, but you absolutely should. As a physics professor, watching students toss away the information contained in units is incredibly frustrating. Typically students drop the units, &quot;do the math,&quot; and then put back on whatever final units they think should be there (if they bother with the last step at all). This leads to a shockingly large number of mistakes (even multi-million dollar ones: &lt;a href=&quot;http://www.cnn.com/TECH/space/9909/30/mars.metric/&quot;&gt;http://www.cnn.com/TECH/space/9909/30/mars.metric/&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you write a physical value you should write both the number and the unit and &quot;do the math&quot; on &lt;strong&gt;both&lt;/strong&gt;, exactly as you propose. If the units don't work out to a unit that you expect then you know that you've done something wrong.&lt;/p&gt;&#xA;" OwnerUserId="16220" LastActivityDate="2015-05-14T20:16:16.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31417" PostTypeId="2" ParentId="31375" CreationDate="2015-05-14T20:17:50.390" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you can and &lt;em&gt;should&lt;/em&gt; (that is, never leave out the units when you plug values into an equation: a quantity is always the product of a numerical value and a unit, and leaving out units is &lt;a href=&quot;http://mars.jpl.nasa.gov/msp98/news/mco990930.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;asking for disaster&lt;/a&gt;). The set of rules that govern the algebraic manipulation of quantities is known as &lt;em&gt;quantity calculus&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the several references that can be found about quantity calculus and quantity equations (others can be found in &lt;a href=&quot;https://academia.stackexchange.com/a/28918/20058&quot;&gt;this answer&lt;/a&gt; on Academia.SE), I suggest you in particular &lt;a href=&quot;http://physics.nist.gov/Pubs/SP811/sec07.html#7.11&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guide, §7.11&lt;/a&gt;, and the following paper, which is geared toward chemists:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;M J ten Hoor, &quot;Quantity calculus for chemists&quot;, Chemistry in action n. 57, 1999. &lt;a href=&quot;http://www.cerg.ul.ie/Chemistry%20in%20Action/Issue%2057.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Online&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, units are not variables (we strongly believe that units should be immutable quantities: whether they are actually so is a matter of scientific investigation), and to highlight this fact it is recommended to use an upright typeface to typeset the units, instead of the italic typeface reserved to variable quantities. So your quantity equations would be better written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{50\,\mathrm{T}}{20\,\mathrm{B}} = x.$$&lt;/p&gt;&#xA;" OwnerUserId="16219" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:59.117" LastActivityDate="2015-05-14T20:17:50.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31418" PostTypeId="1" CreationDate="2015-05-14T20:29:56.210" Score="1" ViewCount="1513" Body="&lt;p&gt;What kind of solvent would you use to remove an oil based stain? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help for a chemistry lab report and I tried to look it up but nothing seems to be helping me&lt;/p&gt;&#xA;" OwnerUserId="16222" LastEditorUserId="5017" LastEditDate="2015-05-14T20:45:33.780" LastActivityDate="2017-06-28T02:34:38.853" Title="What solvents can get rid of oil stains?" Tags="&lt;everyday-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31420" PostTypeId="1" CreationDate="2015-05-14T21:19:32.907" Score="0" ViewCount="163" Body="&lt;p&gt;$\ce{C}$, $\ce{Rb}$, $\ce{Ga}$, $\ce{Ge}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused in the sense that I don't know what it means by &quot;released the smallest about of energy upon gaining?&quot; would it be $\ce{Rb}$ because the shell is so big that when it loses an electron the overall energy doesn't really matter?&lt;/p&gt;&#xA;" OwnerUserId="16227" LastEditorDisplayName="user15489" LastEditDate="2015-05-14T21:45:43.313" LastActivityDate="2015-05-14T22:46:18.020" Title="Which of the following releases the smallest amount of energy upon gaining an electron?" Tags="&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31422" PostTypeId="1" AcceptedAnswerId="31439" CreationDate="2015-05-14T21:59:35.957" Score="1" ViewCount="154" Body="&lt;p&gt;I know a little bit about unknown compounds but I am still new to the topic here. I know that you could use filtration, chromatography and distillation. You can also combine the unknown with others to form a compound. And you can measure properties and such and such to figure out what the compound is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to do this, what should I watch out for, and what would be the &lt;strong&gt;&lt;em&gt;best&lt;/em&gt;&lt;/strong&gt; way to proceed? I would appreciate it if you don't go &quot;PhD&quot; on me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question will be ambiguous for most people, but I am looking for a general idea and what you would normally do, instead of &lt;em&gt;&quot;what I would do in this situation&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-15T07:35:02.057" Title="What are some general tests that would determine an unknown compound?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31423" PostTypeId="2" ParentId="31420" CreationDate="2015-05-14T22:35:05.670" Score="2" Body="&lt;p&gt;Alright, this is similar to a question that I have answered before but I will just give you the reasoning for this phenomenon. I will &lt;strong&gt;not&lt;/strong&gt; give you the answer since this appears to be a homework question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why an atom gains energy when you take off an electron?&lt;/strong&gt;&#xA;Well for this, think of the attractions between the nucleus and the electrons, so when you must &lt;strong&gt;&lt;em&gt;strip&lt;/em&gt;&lt;/strong&gt; away an electron from this stability, you must add energy to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why an atom releases energy when you add an electron?&lt;/strong&gt;&#xA;This one might be a little hard to grasp, but at the end it is not too bad. Essentially the molecule is becoming more stable. It is getting near to having a full octet. (Ignoring noble gasses). In other words the kinetic energy added by gaining the electron is released when the atom is achieving a stable state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the general reasoning behind this, not too complicated, but I hope you will be able to understand it. Just, if you can't, remember the first one and think of the opposite for the second one. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, take a look at &lt;a href=&quot;https://chemistry.stackexchange.com/questions/30805/velocity-of-electrons-ejected-from-cl-atom/30813#30813&quot;&gt;this&lt;/a&gt; question, to help you with your question even more. You will then be able to deduce the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-05-14T22:46:18.020" LastActivityDate="2015-05-14T22:46:18.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31424" PostTypeId="1" AcceptedAnswerId="31483" CreationDate="2015-05-14T22:35:44.177" Score="4" ViewCount="1670" Body="&lt;p&gt;Consider a container that contains a mixture of many liquids. Can two of these liquids phase separate without having any significant effect on other liquids inside the container? Is there a feasible scenario where the above can actually happen? I am interested in a scenario where the two liquids only separates themselves from each other and not the rest of the mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: &#xA;Assume the container contains a mixture of liquids A and B with many other liquids (state 1). Can these two liquids separate (triggered by change in the pressure, temperature, ...) so that at the end (state 2) the liquids in the container are divided into two regions with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first region does not contain liquid B. It contains liquid A with the exact same mixture of the other liquid (same as state 1). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second region does not contain liquid A. It contains liquid B with the same mixture of other liquids (same as state 1). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Indeed we want other liquids in the container to be neutral to the phase separation of the liquids A and B.&lt;/p&gt;&#xA;" OwnerUserId="16230" LastEditorUserId="4356" LastEditDate="2015-05-15T21:21:20.510" LastActivityDate="2015-05-15T21:21:20.510" Title="Liquid-Liquid phase separation" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;phase&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31425" PostTypeId="2" ParentId="31297" CreationDate="2015-05-14T23:39:21.720" Score="0" Body="&lt;p&gt;I'm going to take a shot at this question from a different angle, aiming for a more objective analysis. For that, I wish to bring up several concepts. Bear with me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, what do we mean when asking if something is relevant or not to Chemistry? A precise and exact definition is not trivial, but let us consider a simple argument: chemical reactions are pretty important in Chemistry! In other words, we like to see atoms swapping around positions. It turns out that we have a fair idea of how quickly these events happen. We know that bonds in molecules vibrate at certain frequencies, which is how infrared spectroscopy works. The quickest bonds to vibrate do so at around $\rm{3500\ cm^{-1}}$, which implies a frequency of about $\rm{10^{14}\ Hz}$. This establishes a useful timescale - the fastest atomic displacements relevant to bonding in a molecule happen on the order of $\rm{10^{-14}\ s = 10\ fs}$. This means even the fastest chemical reactions need at least about this much of time to happen. Any event which begins and ceases within a timescale much shorter than $\rm{1\ fs}$ means that molecules don't even have time to react to the event. Even a molecule in a reaction transition state, as short-lived as it is, would be essentially frozen during time intervals much smaller than this. There are important processes which happen faster than $\rm{1\ fs}$, such as electronic transitions and nuclear decay, but they &lt;em&gt;are&lt;/em&gt; important in Chemistry because what matters here is not how long the transition takes, but that the resulting state is sustained for around $\rm{1\ fs}$ or more before it decays. As an interesting aside, &lt;em&gt;virtual electronic states&lt;/em&gt; used in Raman spectroscopy exist for times &lt;a href=&quot;https://physics.stackexchange.com/questions/89131/understanding-the-virtual-states-referenced-in-multiphoton-absorption-studies&quot;&gt;as long as $\rm{0.1\ fs}$&lt;/a&gt;, which further shows how this scale is too quick for molecules to respond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for subatomic particles. Not counting antiparticles, there are &lt;a href=&quot;http://en.wikipedia.org/wiki/Elementary_particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;at least 17 types of fundamental particles&lt;/a&gt; (though 4 of those are force carriers, and the Higgs boson is something else entirely, so we'll put those aside), and over a hundred composite subatomic particles. Yet, for some reason, almost all of Chemistry only deals with three fundamental matter particles (the electron, up quark and down quark), and two composite particles (protons and neutrons, i.e. the &lt;em&gt;up-up-down&lt;/em&gt; and &lt;em&gt;up-down-down&lt;/em&gt; triquarks, respectively). Why is that so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without getting into the nitty-gritty details of particle physics, it turns out that particles tend to decay into their lightest possible relatives while obeying certain rules, such as charge conservation. Electrons are the lightest negatively-charged particle, so they are indefinitely stable (the same is true for positrons, the lightest positively-charged particle). Protons could conceivably decay into something lighter thanks to the positron, but in this case there is another type of conservation law which strongly forbids them to do so (baryon number conservation), so if they are not indefinitely stable, they are at least very long-lived ($\rm{10^{35}}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Year#Symbol&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\rm{a}$&lt;/a&gt; or more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutrons are an interesting case. They're very slightly heavier than protons, by about 1%. This is enough to allow for decay, and indeed a free neutron has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_decay#Mean_lifetime&quot; rel=&quot;nofollow noreferrer&quot;&gt;mean lifetime&lt;/a&gt; of about $\rm{15\ min}$ before it suffers $\beta^-$ decay into a proton. If this happened to every neutron, the Universe would be filled solely with hydrogen, and life could not exist. Remarkably, the mass difference between protons and neutrons is small enough that it can be &lt;a href=&quot;http://profmattstrassler.com/articles-and-posts/particle-physics-basics/mass-energy-matter-etc/the-energy-that-holds-things-together/neutron-stability-in-atomic-nuclei/&quot; rel=&quot;nofollow noreferrer&quot;&gt;compensated by the stabilising nuclear forces&lt;/a&gt; inside the nucleus, and thus neutrons inside nuclei are stable with respect to decay for an amount of time comparable to the proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So up to now, we know that electrons, protons and neutrons are three examples of subatomic particles which are so long-lived that we can expect them to be relevant in chemical timescales. Are there any more? This is where things go awry. The next most stable subatomic particle other than the ones I just cited is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Muon&quot; rel=&quot;nofollow noreferrer&quot;&gt;muon&lt;/a&gt;, a particle identical to the electron except for its larger mass, with a mean lifetime of just $\rm{2.2\ \mu s}$. This is still many of orders of magnitude more than the femtosecond benchmark established earlier, and in fact muons can display meaning chemistry, even taking part in reactions &lt;a href=&quot;http://en.wikipedia.org/wiki/Exotic_atom#Muonic_atoms&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;, &lt;a href=&quot;http://www.sciencemag.org/content/331/6016/448&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;. However, you can imagine that processes involving muons are expected to be much, much rarer, and not have as much importance in chemistry. This is not to say that muons &lt;em&gt;are&lt;/em&gt; irrelevant to Chemistry, just that they are of interest only in a much narrower range of circumstances relative to electrons, neutrons and protons. The problem is even worse for other subatomic particles, which often have much shorter lifetimes. Looking at some lists of particles, such as &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_baryons&quot; rel=&quot;nofollow noreferrer&quot;&gt;baryons&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_mesons&quot; rel=&quot;nofollow noreferrer&quot;&gt;mesons&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Lepton#Table_of_leptons&quot; rel=&quot;nofollow noreferrer&quot;&gt;leptons&lt;/a&gt;, one sees just how short-lived they are; about half of them decay far quicker than $\rm{1\ fs}$ and could not even in principle partake in chemistry. There are still a couple dozen of particles which decay between $\rm{100\ ns}$ and $\rm{1\ fs}$, for which chemistry could exist at an even more limited scope, but again these are rarely relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all the information exposed above, should the average Chemistry student worry about these subatomic details? &lt;em&gt;No, I don't think so&lt;/em&gt;. In fact if these issues were to be mentioned as more than a passing comment in when first teaching the structure of matter, I believe it would be counter-productive. That said, I do think it's a shame that an enormous majority of chemists never seem to peer a little deeper into the structure of matter, even years after graduating. It would be nice if at least knowledge of the fundamental particles were more disseminated. The Standard Model of particle physics is quite an amazing achievement, and there's more than enough beauty to appreciate even without breaking out the massive integrals!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-05-14T23:39:21.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31426" PostTypeId="2" ParentId="31422" CreationDate="2015-05-15T00:11:48.750" Score="1" Body="&lt;p&gt;Is anything about the material known?  And how much of it do you have?&#xA;If you think the material might be inorganic, an adsorption, emission, or mass spectroscopy might be the first step, as that will tell you what elements are in it.  If the sample is large enough, and can be formed into crystals, then x-ray diffraction will likely give a thorough idea of the structure.&#xA;If it's organic, NMR is probably the most powerful technique, and only requires a minuscule amount of the sample.  Mass-spec and IR-spec are also commonly used.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-05-15T00:11:48.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31427" PostTypeId="1" AcceptedAnswerId="31453" CreationDate="2015-05-15T00:13:30.577" Score="3" ViewCount="241" Body="&lt;p&gt;I am asked to determine whether the following process is likely to involve a carbocation rearrangement or not:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4JKn1.png&quot; alt=&quot;impossible rearrangement&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think it is possible through the rearrangement given above (though the rearrangement cannot become a secondary or primary carbocation). Can someone please clarify if the carbocation rearrangement specifies a change in degree?&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="15410" LastEditDate="2015-05-19T06:12:41.390" LastActivityDate="2015-05-19T06:25:55.353" Title="Does the carbocation rearrangement in a SN1 reaction specify a change in degree?" Tags="&lt;organic-chemistry&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="31428" PostTypeId="2" ParentId="31424" CreationDate="2015-05-15T00:13:55.183" Score="3" Body="&lt;p&gt;Separation can be done in two ways: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Natural:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Natural separation term is not a good one but i'm gonna use it here. If at equilibrium (When the free energy of all the chemical constituents are same), those two specific liquid like to separate themselves out from the rest of the mixture, than you just need to give it enough time to reach equilibrium to separate them out. And this is the only method i can think of which will have no significant effect on the liquid in that container.      &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Forced:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forced separation: Now if the above mentioned equilibrium condition needs different temp. and pressure than room temp. and pressure. Then by simply changing temp. and pressure you can achieve the above separation. Or you can utilize the different in various properties to separate them, e.g. Viscosity, boiling point, melting point etc.&lt;br&gt;&#xA;Distillation: If they have separate boiling point than you can use distillation column and collecting the liquid at various height will give you separated solvent.&lt;br&gt;&#xA;Solvent extraction, Filtration, mixer settlers, centrifugal force, ion exchange, membrane filtration, using bio separation like algae etc you can also achieve liquid liquid extraction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="5456" LastEditDate="2015-05-15T04:19:43.280" LastActivityDate="2015-05-15T04:19:43.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31429" PostTypeId="2" ParentId="28483" CreationDate="2015-05-15T00:34:23.630" Score="0" Body="&lt;p&gt;You can leave this solid phase alone. Formally it's part of Nernst equation of equilibrium and for the formality's sake you can put it there, BUT its activity is ONE, so it really doesn't influence the description of the equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="16233" LastActivityDate="2015-05-15T00:34:23.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31430" PostTypeId="2" ParentId="29250" CreationDate="2015-05-15T00:39:52.317" Score="5" Body="&lt;p&gt;Turned in this report and my professor commented with his thoughts&#xA;$$\ce{Sn + 2RCl \rightarrow R2SnCl2 \rightarrow R2Sn }\text{ (white solid)} \ce{+ SnCl2 \rightarrow  R3SnCl}$$&#xA;but I didn't get the question wrong. So unless if someone contributes another answer, I'll accept this as the answer.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="16683" LastEditDate="2017-08-30T13:38:47.163" LastActivityDate="2017-08-30T13:38:47.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31431" PostTypeId="1" AcceptedAnswerId="31454" CreationDate="2015-05-15T01:29:39.830" Score="12" ViewCount="34440" Body="&lt;p&gt;If I have three aqueous ionic solutions in which I know that the cation is $\ce{Al^3+}$, $\ce{Mg^2+}$, or $\ce{Zn^2+}$, how do I find out which is which? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking to add $\ce{OH-}$ in the form of $\ce{NaOH}$, or something else to produce a precipitate, and testing these precipitates in other liquids, but I'm not sure what other liquids to try.  &lt;/p&gt;&#xA;" OwnerUserId="16234" LastEditorUserId="194" LastEditDate="2017-10-29T19:30:16.170" LastActivityDate="2017-10-29T19:30:16.170" Title="Differentiation between zinc, aluminium, and magnesium ions in solution" Tags="&lt;experimental-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;ions&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31432" PostTypeId="2" ParentId="31431" CreationDate="2015-05-15T02:09:51.780" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Al^3+}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chemical &lt;a href=&quot;http://oxfordchemserve.com/aluminon-aluminium-cations-test-kit/&quot;&gt;Aluminion&lt;/a&gt; can be added to a solution to test for the presence of $\ce{Al^3+}$ ions. This is done by first adding $\ce{NaOH}$ to the solution to precipitate out $\ce{Al(OH)3}$. In the presence of Aluminon, the $\ce{Al(OH)3}$ forms a lake dye through chemisorption, causing a pinkish red precipitate to form. Only $\ce{Be^{2+}}$ cations and $\ce{Fe^{3+}}$ cations show a similar reaction, so there is no need to worry about misidentifying in your case.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{Mg^2+}$ and $\ce{Zn^2+}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The flame test is easy enough to perform! Dip a splint into the solution of interest, place it under a bunsen burner flame, and observe the color. Different metals produce different color light when excited under flame, allowing our mystery metal to be identified. $\ce{Mg}$ produces a bright white light, while $\ce{Zn}$ produces a blue-green to pale-green color.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-15T02:09:51.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31433" PostTypeId="2" ParentId="31363" CreationDate="2015-05-15T05:52:22.617" Score="3" Body="&lt;p&gt;Finally I found the book I was looking for. The method is listed in &quot;Vogel's Textbook of Chemical Analysis 5th Edition&quot; (page 392). They have removed it from the 6th edition.&lt;/p&gt;&#xA;" OwnerUserId="16096" LastActivityDate="2015-05-15T05:52:22.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31434" PostTypeId="2" ParentId="31431" CreationDate="2015-05-15T05:58:32.067" Score="15" Body="&lt;p&gt;Apart from the methods, Ringo already described, you can do a few other tests.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Aluminium&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is loosely translated from the German &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/6/ac/versuche/kationen/_vlu/aluminium.vlu/Page/vsc/de/ch/6/ac/versuche/kationen/aluminium/nachweis.vscml.html&quot; rel=&quot;noreferrer&quot;&gt;chemgapedia.de&lt;/a&gt;. Look at the pretty pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the easiest test you can do is reacting it with &lt;a href=&quot;http://en.wikipedia.org/wiki/Morin_%28flavonol%29&quot; rel=&quot;noreferrer&quot;&gt;Morin&lt;/a&gt; in &lt;a href=&quot;http://en.wikipedia.org/wiki/Acetic_acid&quot; rel=&quot;noreferrer&quot;&gt;ethanoic acidic&lt;/a&gt; medium. It forms a yellow-green chelate complex, which has strong fluorescence under UV light. The colour and UV activity should be persistent when adding glacial acidic acid (water free), but vanish when adding concentrated hydrochloric acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oMA4e.jpg&quot; alt=&quot;Morin aluminium complexation reaction&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately sodium is also fluorescent, so keep your equipment clean and if you use a base, use potassium hydroxide instead. Also perform blind tests with all the chemicals involved, to see if they are clean enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other elements that can show fluorescence are $\ce{In^3+, Ga^3+, Sc^3+, Zr^4+, Be^2+}$, usually very pH dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my own experience in first year of university, this is a very efficient test. Another test, that can be performed from the salt, is reacting it to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobalt_blue&quot; rel=&quot;noreferrer&quot;&gt;Cobalt blue&lt;/a&gt;. If you have a solution, you have to precipitate aluminium first: $$\ce{Al^3+ + 3 {}^{-}OH -&amp;gt;C[NH3/NH4Cl] Al(OH)3 v}$$&#xA;Put the precipitate on a &lt;a href=&quot;http://shop.davinci-ls.com/INFO5054_CERAMTEC_LANG_UK.HTM&quot; rel=&quot;noreferrer&quot;&gt;magnesia groove&lt;/a&gt; and add some cobalt nitrate. Glowing the mixture will result in the characteristic cobalt blue:&#xA;\begin{align}&#xA;\ce{2 Al(OH)3 &amp;amp;-&amp;gt;[\Delta] Al2O3 + 3 H2O ^}\\&#xA;\ce{Co(NO3)2 &amp;amp;-&amp;gt;[\Delta] CoO + 2NO2 ^ + 1/2 O2 ^}\\&#xA;\ce{CoO + Al2O3 &amp;amp;-&amp;gt;[\Delta] \color{\navy}{CoAl2O4}}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an almost fool proof method for aluminium and is not influenced by other elements. And it looks quite nice.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Zinc&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Zinc is probably one of the only elements that form a white precipitate with sulfide ions. &#xA;$$\ce{Zn^2+ + H2S -&amp;gt; ZnS v + 2 H+}$$&#xA;This precipitate can be used to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobalt_green&quot; rel=&quot;noreferrer&quot;&gt;Cobalt green&lt;/a&gt;. The procedure is the same as for the above Cobalt blue.&#xA;$$\ce{ZnS + Co(NO3)2 + O2 -&amp;gt;[\Delta] \color{\green}{ZnCo2O4} + 4 NO2 ^ + SO2 ^ }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only interference with this test is, that using too much cobalt nitrate will produce the black cobalt oxide, which can make it hard to see the green colour.&#xA;$$\ce{3Co(NO3)2 -&amp;gt;[\Delta] \mathbf{Co3O4} + 6 NO2 ^ + O2 ^}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again you can have a look at the pretty pictures at &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/6/ac/versuche/kationen/_vlu/zink.vlu/Page/vsc/de/ch/6/ac/versuche/kationen/zink/nachweis.vscml.html&quot; rel=&quot;noreferrer&quot;&gt;chemgapedia.de&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Magnesium&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A common problem with magnesium is, that all tests only work well and specific if you have a pure solution. Most heavier elements react more readily with the chemicals, than magnesium itself does. Another problem is, that salts from magnesium have no flame colouration at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am again loosely translating from &lt;a href=&quot;http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/6/ac/versuche/kationen/_vlu/magnesium.vlu/Page/vsc/de/ch/6/ac/versuche/kationen/magnesium/nachweis.vscml.html&quot; rel=&quot;noreferrer&quot;&gt;chemgapedia.de&lt;/a&gt;.&lt;sup&gt;*&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can precipitate magnesium with &lt;a href=&quot;http://en.wikipedia.org/wiki/8-Hydroxyquinoline&quot; rel=&quot;noreferrer&quot;&gt;8-hydroxyquinoline&lt;/a&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uB263.jpg&quot; alt=&quot;magnesium and hydroxiquinoline&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It forms a yellow-green complex. This is only a specific test, if you know you have no heavier elements in your test tube. For example, aluminium forms the complex &lt;a href=&quot;http://en.wikipedia.org/wiki/Tris(8-hydroxyquinolinato)aluminium&quot; rel=&quot;noreferrer&quot;&gt;Tris(8-hydroxyquinolinato)aluminium&lt;/a&gt;: $$\ce{Al^3+ + 3 C9H6NOH -&amp;gt; Al(C9H6NO)3 + 3 H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is precipitation as &lt;a href=&quot;http://en.wikipedia.org/wiki/Struvite&quot; rel=&quot;noreferrer&quot;&gt;magnesium ammonium phosphate&lt;/a&gt; (struvite). And yes, you guessed it, this happens with all other alkali and earth alkali elements (except lithium). But if you are certain you only have got magnesium, you can grow nice crystals. The pH for this reaction should be buffered at around 8-9 with ammonia/ ammonium chloride. &#xA;$$\ce{Mg^2+ + NH4+ + {}^{-}OH + HPO4^2- -&amp;gt;C[H2O][NH3/NH4Cl] Mg(NH4)PO4⋅6H2O v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two more tests, that I would consider best suitable, because of the pretty colour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of magnesium hydroxide with &lt;a href=&quot;http://en.wikipedia.org/wiki/Titan_yellow&quot; rel=&quot;noreferrer&quot;&gt;titan yellow&lt;/a&gt; in presence of sodium hydroxide will produce a red precipitate. It is important to do a control/ blind experiment, as older solutions of titan yellow will form a precipitate with sodium hydroxide alone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction of magnesium hydroxide with &lt;a href=&quot;http://en.wikipedia.org/wiki/Azo_violet&quot; rel=&quot;noreferrer&quot;&gt;azo violet&lt;/a&gt; (Magneson) or &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.12625899.html&quot; rel=&quot;noreferrer&quot;&gt;Magneson II&lt;/a&gt; (&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/Magneson_II&quot; rel=&quot;noreferrer&quot;&gt;Magnezon II&lt;/a&gt;) in presence of sodium hydroxide will produce a blue precipitate. Again test with sodium hydroxide to observe the violet solution. This reaction also occurs with $\ce{Ca^2+, Al^3+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Testing for magnesium is as far as I remember is some kind of a lottery when in combination with other elements (qualitative analysis). Good luck.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; Please note that the first reaction equation has depicted &lt;a href=&quot;http://en.wikipedia.org/wiki/8-Hydroxyquinoline&quot; rel=&quot;noreferrer&quot;&gt;hydroxyquinoline&lt;/a&gt; with two circles for the aromatic system, which is wrong. In benzene one circle stands for 6 $\pi$ electrons. So that would lead to hydroxyquinoline having 12 $\pi$ electrons, two too many. &lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-15T05:58:32.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31435" PostTypeId="2" ParentId="31393" CreationDate="2015-05-15T05:58:50.980" Score="4" Body="&lt;p&gt;You got it &lt;strong&gt;almost&lt;/strong&gt; right ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution coefficient gives the ratio of &lt;strong&gt;equilibrium concentrations&lt;/strong&gt; in a biphasic mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is crucial to realize that a caffeine concentration of $2.0\,\mathrm{g}$ in $200\,\mathrm{mL}$ of water is the &lt;strong&gt;initial&lt;/strong&gt; concentration, but &lt;strong&gt;not&lt;/strong&gt; the concentration after a single extraction step! The latter will be &lt;strong&gt;lower&lt;/strong&gt;, since a part of the caffeine went into the organic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doens't matter whether you express the concentrations in $\mathrm{g\cdot L^{-1}}$ or $\mathrm{mol\cdot L^{-1}}$ since $K_D$ is a dimensionless constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_D = \frac{{[\mathrm{caffeine}]_{\ce{CH2Cl2}}}}{{[\mathrm{caffeine}]_{\ce{H2O}}}} = \frac{x}{10-x}\ \mathrm{g\cdot L^{-1}} = 7.8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = 7.8\cdot (10-x)\,\mathrm{g\cdot L^{-1}} = \mathbf{8.86} \,\mathrm{g\cdot L^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $x$ is the concentration in $\mathrm{g\cdot L^{-1}}$. In order to calculate the absolute amount of caffeine in the organic layer or the remaining amount of caffeine in the water phase, you have to consider the volumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, there's an article by G.S. Klebanov, L.N. Mednikova, and A.D. Ovcharova on this topic: &lt;em&gt;Extraction of caffeine from aqueous solutions&lt;/em&gt;, published in &#xA;&lt;em&gt;Pharmaceutical Chemistry Journal&lt;/em&gt;, &lt;strong&gt;1967&lt;/strong&gt;, &lt;em&gt;1&lt;/em&gt;, 221-223 (&lt;a href=&quot;http://dx.doi.org/10.1007/BF00770195&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-15T06:28:47.820" LastActivityDate="2015-05-15T06:28:47.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31436" PostTypeId="1" AcceptedAnswerId="31469" CreationDate="2015-05-15T06:36:17.160" Score="5" ViewCount="3437" Body="&lt;p&gt;This is a snapshot from LG Wade Organic Chemistry. I have a doubt with a particular statement in the book. It says that Grignard reagents do not displace halides as easily as acetylide ions do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But shouldn’t it be the other way round? Acetylide ions, having an sp-hybridised carbon, are much more electronegative compared to the sp³-hybridised carbon in Grignard reagents. So acetylide ions should have less tendency to displace halide ions by attacking with their lone pair compared to the alkyl group in Grignard reagent. Shouldn’t it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me sort this out! &#xA;&lt;img src=&quot;https://i.stack.imgur.com/C3FMG.png&quot; alt=&quot;snapshot from LG Wade Organic Chemistry&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4945" LastEditDate="2015-05-15T09:43:45.353" LastActivityDate="2015-05-15T15:58:18.597" Title="Why do Grignard reagents or Organolithium compounds not displace halides as easily as acetylide ions do?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31438" PostTypeId="1" AcceptedAnswerId="31451" CreationDate="2015-05-15T06:42:34.670" Score="6" ViewCount="807" Body="&lt;p&gt;I'm measuring buffer capacity by titrating $\ce{NaOH}$ into a buffer of $\ce {H3PO4 + NaH2PO4}$. I'm measuring buffer capacity by recording the amount of $\ce{NaOH}$ required to increase the $\mathrm{pH}$ of the buffer by 2 units, and then taking the ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I have no idea how much $\ce{NaOH}$ I'll need to change the $\mathrm{pH}$ by that much. If I've underestimated the strength of my buffer, I might have to reduce the drop in $\mathrm{pH}$ to just one unit. I need to have a rough idea so I can organize how much of each I'll need to prepare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, is there any way I can predict (roughly) how much $\ce{NaOH}$ I'll need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$1\:\mathrm{M}$ $\ce{H_3PO_4}$ and $1\:\mathrm{M}$ $\ce{NaH_2PO_4}$&lt;/li&gt;&#xA;&lt;li&gt;Buffer volume: $50\:\mathrm{mL}$&lt;/li&gt;&#xA;&lt;li&gt;$1\:\mathrm{M}$ $\ce{NaOH}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14994" LastEditorUserId="15410" LastEditDate="2015-05-18T03:36:28.003" LastActivityDate="2018-01-23T01:13:17.563" Title="Determining (approximately) the amount of a strong base needed to change a buffer's pH by a certain number of units" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31439" PostTypeId="2" ParentId="31422" CreationDate="2015-05-15T07:35:02.057" Score="3" Body="&lt;p&gt;As usual, it depends on what is available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass spectrometry, $^{1}\ce{H}$ and $^{13}\ce{C}$ NMR, IR, UV/VIS spectroscopy, would be obvious choices in organic chemistry nowadays. XRD of a suitable crystal would be nice to have! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in the days, (undergrad) laboratories didn't have sophisticated analytical instruments and you were happy to find a melting point apparatus, a refractometer and an IR spectrometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have used IR spectroscopy to identify funcional groups, possibly supported your hypothesis by using staining reagents for TLC, like &lt;em&gt;Dragendorff's&lt;/em&gt; reagent and then started to synthesize derivatives from which you knew that they show sharp, well-defined melting points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have converted  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;amines to sulfonamides&lt;/li&gt;&#xA;&lt;li&gt;carbohydrates to osazones  &lt;/li&gt;&#xA;&lt;li&gt;ketones and aldehydes to 2,4-dinitrophenylhydrazones or semicarbazones&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and compared the melting points of the derivatives with tabulated values. Those were the days&amp;hellip;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-15T07:35:02.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31440" PostTypeId="2" ParentId="6919" CreationDate="2015-05-15T08:56:29.373" Score="1" Body="&lt;p&gt;There are a number of resources available that provide dissociation constants for acids and bases.  The following are examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UC Davis Chemwiki pages &lt;a href=&quot;http://chemwiki.ucdavis.edu/Reference/Reference_Tables/Equilibrium_Constants/E1%3A_Acid_Dissociation_Constants_at_25%C2%B0C&quot; rel=&quot;nofollow&quot;&gt;E1: Acid Dissociation Constants at 25°C&lt;/a&gt; and &lt;a href=&quot;http://chemwiki.ucdavis.edu/Reference/Reference_Tables/Equilibrium_Constants/E2._Base_Dissociation_Constants_at_25%C2%B0C&quot; rel=&quot;nofollow&quot;&gt;E2. Base Dissociation Constants at 25°C&lt;/a&gt; provide many examples.  These are based on information taken from &lt;a href=&quot;https://www.crcpress.com/product/isbn/9780849304859&quot; rel=&quot;nofollow&quot;&gt;CRC Handbook of Chemistry and Physics, 85th Edition&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WikiBook &lt;a href=&quot;http://en.wikibooks.org/wiki/Chemical_Principles/Solution_Equilibria:_Acids_and_Bases&quot; rel=&quot;nofollow&quot;&gt;Chemical Principles/Solution Equilibria: Acids and Bases&lt;/a&gt; provides a reasonable database of dissociation constants of acids and bases (Tables 5.3 and 5.4), as well as some further information about how the values are calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The California State University webpage &lt;a href=&quot;http://www.csudh.edu/oliver/chemdata/data-ka.htm&quot; rel=&quot;nofollow&quot;&gt;p-functions for common acids and bases&lt;/a&gt;, also at 25C, of a series of acids and bases.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-15T09:52:14.097" LastActivityDate="2015-05-15T09:52:14.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31441" PostTypeId="2" ParentId="31409" CreationDate="2015-05-15T09:25:21.087" Score="7" Body="&lt;p&gt;I tried to pinpoint you in the right direction through comments, but apparantly I failed. The bottom line is that the molecule you drew and the one in the textbook answer are &lt;strong&gt;identical.&lt;/strong&gt; Here’s why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oS4T4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the top row, you can see the (&lt;em&gt;S&lt;/em&gt;)-2-chloropentane, which I have drawn as you have on the left, slowly rotating around an axis equivalent to the reaction arrows to give you &lt;em&gt;the same molecule drawn in a different orientation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you do an $\mathrm{S}_{\mathrm{N}}2$ reaction, all you need to do is replace the chlorine atom and invert the stereochemistry which I have done for both cases. You can now see that both sides are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not (and must not) matter whether you draw the 2-chloropentane with the substituted carbon to the right or to the left. Both ways are &lt;em&gt;identical.&lt;/em&gt; Usually, you will chose an orientation that is consistent with further or earlier steps of your synthesis, but if you only have one one reaction that point is moot. (It matters a lot more in total synthesis, where it helps to draw reactands in a way that show their conformation in the final molecule.) Furthermore, there is a IUPAC style guide out there somewhere, which probably says one of the two ways is to be preferred if no other reasons speak for the other … but that’s more relevant for journal publications, not for practising or for exam answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just for total consistency’s sake: Everything you did was correct; just remember that molecules can be drawn with different orientations, and learn to turn them around before your inner eye.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-15T09:25:21.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31442" PostTypeId="1" CreationDate="2015-05-15T09:36:40.543" Score="3" ViewCount="255" Body="&lt;p&gt;I am under the impression that elements from the third period onwards can promote electrons into d orbitals to form hybridised orbitals. Why would an element in the third period promote an electron to the 3d orbital when we learn that 4s orbitals are filled first because they are lower in energy (is it to do with orbital penetration?)&#xA;Why also can a 2nd period p block element not promote to 3s?&lt;/p&gt;&#xA;" OwnerUserId="16225" LastActivityDate="2015-05-15T09:51:32.383" Title="Available orbitals for promotion" Tags="&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31444" PostTypeId="2" ParentId="31442" CreationDate="2015-05-15T09:51:32.383" Score="4" Body="&lt;p&gt;There is a considerable amount of debate as to whether d-orbitals of main group elements (I assume we’re talking about them here; the discussion is rather pointless for d-block elements) are used in hybridisation or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, the octet expansion, a.k.a. using them, was postulated by Linus Pauling because he was unhappy to use charge separation to describe molecules such as sulphur trioxide. However, it is entirely possible to draw Lewis structures that &lt;em&gt;do not violate the octet rule&lt;/em&gt; and hence &lt;em&gt;do not require d-orbital hybridisation&lt;/em&gt; &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule/29461#29461&quot;&gt;for practically every case&lt;/a&gt;. There may be more mesomery required in some cases and less in others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not too good at repeating the arguments of supporters of main group elements’ d-orbital hybridisation theory, because I belong to the other group. The main argument against d-orbital participation is, as you noticed, that they are far removed on an energy scale; even behind the next s-orbital (3s, 3p, 4s, 3d) — hence the order of elements in the periodic table. I also have troubles understanding why phosphorous would only allow one d-orbital to hybridise, sulphur two and chlorine and the heavier halogens three. This makes no sense to me. (It seems to make sense for others, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion you are absolutely correct in asking that if a 3p-element can use 3d orbitals, why should it not use 4s for hybridisation, and consequently why 2p cannot use 3s. Assuming 3d orbitals to be used, I could not answer your question satisfactorily.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-15T09:51:32.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31446" PostTypeId="2" ParentId="31038" CreationDate="2015-05-15T11:05:04.913" Score="4" Body="&lt;p&gt;You are looking at the wrong numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should look for excitation  that has non-zero oscillator strength. In the text file &quot;State # 1,2 ..&quot; and in Chemissian on the left, list with columns &quot;# Energy Strength&quot; The transitions with zero or very small oscillator strength has a very low intensity in the spectrum, so you don't see them. &#xA;In your case the relevant excitation are #2 (Energy: 6.691 eV, Strength: 0.027) and #4 (Energy: 8.353 eV, Strength: 0.042), and you can see them as small peaks on the graph down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;amplitude .. 1.41&quot; has nothing to do with the oscillator strength. It tells you which orbital contributions are the most important ones for the given excitation.  &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-05-15T11:05:04.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31448" PostTypeId="1" AcceptedAnswerId="31465" CreationDate="2015-05-15T11:49:24.670" Score="12" ViewCount="6758" Body="&lt;p&gt;I learnt that halogens always form covalent bonds to becoms diatomic molecules. So why is astatine monoatomic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean they have the same properties, why shouldn't they all be diatomic?&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="5017" LastEditDate="2015-05-15T16:17:25.000" LastActivityDate="2020-01-05T14:51:17.553" Title="Why is Astatine monoatomic?" Tags="&lt;atoms&gt;&lt;elements&gt;&lt;relativistic&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31451" PostTypeId="2" ParentId="31438" CreationDate="2015-05-15T12:34:33.200" Score="4" Body="&lt;p&gt;You can calculate the quantity of sodium hydroxide, required to change the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of your system by two unites,  by calculating the concentration of sodium ion in the charge balance equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$&lt;/span&gt;&#xA;In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaH2PO4}$&lt;/span&gt;), and you calculate the final concentration of sodium ion  after adding sodium hydroxide by the above equation. So you can deduce the required quantity  of sodium hydroxide.&lt;/p&gt;&#xA;&lt;p&gt;As for the  other ions in the charge balance equation, they are calculated as follows:&#xA;At first the  &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of the buffer system &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH=pK}_a=2.15$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to change the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of your system by two unites, then the final &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$$\mathrm{pH}= 4.15\,\,\,\,(2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now you can estimate the concentrations of ion hydronium and ion hydroxide.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The use of Henderson-Hasselbalch equation gives: &lt;span class=&quot;math-container&quot;&gt;$$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The equation of matter conservation gives: &lt;span class=&quot;math-container&quot;&gt;$$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$&lt;/span&gt; As the buffer is &lt;span class=&quot;math-container&quot;&gt;$25\,\mathrm{mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$1\,\mathrm{M}$&lt;/span&gt; of each of the components, so there's a total of &lt;span class=&quot;math-container&quot;&gt;$1\,\mathrm{M}$&lt;/span&gt; in the &lt;span class=&quot;math-container&quot;&gt;$50\,\mathrm{mL}$&lt;/span&gt; solution due to dilution (not as you mentioned in your comment).&lt;/p&gt;&#xA;&lt;p&gt;Now, you can calculate the concentration of ion &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4^{-}}$&lt;/span&gt;&#xA;from equation (3) and (4):&lt;span class=&quot;math-container&quot;&gt;$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If we substitute the numerical values in equation (1):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n=\ce{  [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The volume of sodium hydroxide is &lt;span class=&quot;math-container&quot;&gt;$$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$&lt;/span&gt;&#xA;I suggest to use a  solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-16T12:32:36.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31452" PostTypeId="2" ParentId="28817" CreationDate="2015-05-15T12:36:40.280" Score="3" Body="&lt;p&gt;An additional point that I wanted to state from &lt;a href=&quot;http://www.masterorganicchemistry.com/2013/05/24/alkyne-reaction-patterns-the-carbocation-pathway/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; source.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same source has an &lt;a href=&quot;http://jamesash.wpengine.netdna-cdn.com/wp-content/uploads/2013/05/2-carbocat-alkyne.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;image&lt;/a&gt; which shows the reaction mechanism:    &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zb969.png&quot; alt=&quot;2nd equivalent&quot;&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Note the 2nd reaction in the above image.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote some relevant text:    &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that the carbocation in this case bears a chloride ion. Since carbocations are electron poor, and chlorine is quite an electronegative element, it’s interesting to point out that &lt;strong&gt;the electron releasing ability of the alkyl group [and the ability of chlorine to donate a lone pair to the carbocation] “win out” here over the electron-withdrawing character of chloride ion.&lt;/strong&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the genaralized case while in your case the R=H and it no longers qualifies as an alkyl group. But it is important to note that in the actual pathway, 3 H contribute in the hyperconjugation while in the other pathway , only 2. This effect, combined with the +M of the Cl overpower the -I effect.&lt;/p&gt;&#xA;" OwnerUserId="15912" LastActivityDate="2015-05-15T12:36:40.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31453" PostTypeId="2" ParentId="31427" CreationDate="2015-05-15T13:18:46.770" Score="5" Body="&lt;p&gt;The most common carbocation rearrangements involve [1,2] hydride or alkyl shifts. As a typical result a secondary carbocation is converted to a more stabilized tertiary carbocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the course of these reactions, the migrating group and the positive charge change place, they move to opposite directions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the case in your example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fail to see how your transformation can be described as a cationic rearrangement!&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-19T06:25:55.353" LastActivityDate="2015-05-19T06:25:55.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31454" PostTypeId="2" ParentId="31431" CreationDate="2015-05-15T14:06:22.080" Score="10" Body="&lt;p&gt;First, you add sodium hydroxide to the three different solution. (Remember to safe some of the solution, because we need it for the next test.)&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^2+}\text{ ion}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The zinc ion will react with the hydroxide ion in the sodium hydroxide solution to form zinc hydroxide which is white. So, you see white precipitate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn^2+ +2 OH^- -&amp;gt; Zn(OH)2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Adding more sodium hydroxide, the zinc hydroxide will form the zincate ion which is colourless. So, you will see a clear solution.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn(OH)2 + 2 OH^- -&amp;gt; Zn(OH)4^2-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}\text{ ion}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The aluminium ion will react with the hydroxide ion in the sodium hydroxide solution to form aluminium hydroxide which is white. So, you see white precipitate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Al^3+ +3 OH^- -&amp;gt; Al(OH)3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Adding more sodium hydroxide, the aluminium hydroxide will form aluminate ion which is colourless. So, you will see a clear solution.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Al(OH)3 + OH^- -&amp;gt; Al(OH)4^-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}\text{ ion}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The magnesium ion will react with the hydroxide ion in the sodium hydroxide solution to form magnesium hydroxide which is white. So, you see white precipitate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Mg^2+ +2 OH^- -&amp;gt; Mg(OH)2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, in this case, it won't form any else by adding more sodium hydroxide. So, you will still see the white precipitate even though you have continued to add sodium hydroxide.&lt;/p&gt;&#xA;&lt;p&gt;We can differentiate &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt; solution form the three solutions and now we are left with &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}$&lt;/span&gt;. Take another sample from the remaining two solutions and add aqueous ammonia solution (&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;) in the samples.&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^2+}\text{ ion}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The zinc ion will react with the ammonia solution to form zinc hydroxide which is white. So, you see white precipitate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn^2+ + 2NH3 + 2H2O &amp;lt;=&amp;gt; Zn(OH)2 + 2NH4^+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Adding more ammonia solution, the zinc hydroxide will form  tetraamminezinc(II) ion which is colourless. So, you will see a clear solution.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn(OH)2 + 4NH3 -&amp;gt; [Zn(NH3)4]^2+ + 2OH^-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}\text{ ion}$&lt;/span&gt;&lt;/h1&gt;&#xA;&lt;p&gt;The aluminium ion will react with the ammonia solution to form aluminium hydroxide which is white. So, you see white precipitate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Al^3+ + 3NH3 + 3H2O &amp;lt;=&amp;gt; Al(OH)3 + 3NH4^+}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, the aluminium hydroxide will not further react with ammonia. So, you will still see the white precipitate even though you have continued to add ammonia.&lt;/p&gt;&#xA;&lt;p&gt;In conclusion, first put sodium hydroxide in the three solutions (sample in test tube) and keep adding it until only one of the solutions still has white precipitate. That solution is  &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+}$&lt;/span&gt;. Next, take the remaining two solutions (sample in test tube) and  keep adding it until one of the solution still has white precipitate. That solution is &lt;span class=&quot;math-container&quot;&gt;$\ce{Al^3+}$&lt;/span&gt;. And finally the last solution is &lt;span class=&quot;math-container&quot;&gt;$\ce{Zn^2+}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-20T06:47:26.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31456" PostTypeId="2" ParentId="31185" CreationDate="2015-05-15T14:11:46.637" Score="2" Body="&lt;p&gt;Both metal powders can be preapred electrochemically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron dust is prepared using solution containing $\ce{FeCl3}$ with addition of $\ce{AlCl3}$ to make the solution more conductive. The process occurs at elevated temperature 60°C. Unfortunately exact process parameters are hard to find.  But following graph could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kcKUT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickel could be prepared electrochemically too and can be prepared by dissolving of Ni-Al alloy in hydroxide -&lt;a href=&quot;http://en.wikipedia.org/wiki/Raney_nickel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raney_nickel&lt;/a&gt;. The description of patent is &lt;a href=&quot;http://patimg2.uspto.gov/.piw?Docid=01628190&amp;amp;homeurl=http%3A%2F%2Fpatft.uspto.gov%2Fnetacgi%2Fnph-Parser%3FSect1%3DPTO2%2526Sect2%3DHITOFF%2526u%3D%25252Fnetahtml%25252FPTO%25252Fsearch-adv.htm%2526r%3D456%2526f%3DG%2526l%3D50%2526d%3DPALL%2526s1%3DMurray%2526s2%3DRaney%2526co1%3DAND%2526p%3D10%2526OS%3DMurray%252BAND%252BRaney%2526RS%3DMurray%252BAND%25&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is link to a study book &lt;a href=&quot;http://katedry.fmmi.vsb.cz/Modin_Animace/Opory/03_Materialov%C3%A9_inzenyrstvi/06_Praskova_metalurgie/Skotnicova_Praskova_metalurgie.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PRÁŠKOVÁ METALURGIE, K. Skotnicová, M. Kursa&#xA;&lt;/a&gt;, unfortunately only in czech.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-05-15T17:55:10.027" LastActivityDate="2015-05-15T17:55:10.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31457" PostTypeId="2" ParentId="31185" CreationDate="2015-05-15T14:19:51.100" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am asking for an alternative wet chemical method (water solution) or electrochemical method (water or ionic liquid electrolyte)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Water will only be a efficient electrolyte to generate metals that are more weakly reducing than hydrogen gas.  Otherwise, attempts to produce the metal will just result in mostly water hydrolysis.  The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_3%3A_Cell_potentials_and_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;chemwiki page at UC Davis&lt;/a&gt; has a good overview of this particular topic.  The standard reduction potential for iron is too low (too reducing; too negative) to be made efficiently via an aqueous electrochemical process.  &lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;: I softened the wording of this paragraph based on advice from Jaroslav Kotowski who noted that the original wording I used was wrong.  Iron dust can apparently be made via electrolysis, but efficiency is very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickel and silicon are less powerful reducing agents than iron, but are still slightly more powerful than water under standard conditions.  Some combination of pH and concentration adjustment could make it theoretically possible to generate these metals in a wet (aqueous) electrochemical process, but it would not be very efficient.  I &lt;a href=&quot;http://www.jstor.org/stable/pdf/92886.pdf&quot; rel=&quot;nofollow&quot;&gt;dug up a 1908 paper&lt;/a&gt; that reports that nickel films (not powders) can be electrodeposited from a (presumably aqueous) solution of nickel ammonium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-aqueous electrolytes such as ionic liquids do not necessarily have this problem.  You should read a review article on metal deposition from non-aqueous media &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0013468609005520&quot; rel=&quot;nofollow&quot;&gt;such as this one&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-05-15T19:36:57.540" LastActivityDate="2015-05-15T19:36:57.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31458" PostTypeId="1" AcceptedAnswerId="31502" CreationDate="2015-05-15T14:29:28.087" Score="7" ViewCount="6027" Body="&lt;p&gt;I've recently read that trinitrophenol (picric acid - relative effectiveness factor $1.20$) is more explosive than trinitrotoluene (TNT - relative effectiveness factor $1.00$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I think that the mesomeric effect of the oxygen's lone pairs in picric acid should keep the nitro groups 'happy', because of the increase in electron density on them. This should enhance the overall stability of the molecule, and reduce the relative effectiveness factor. However, this is not the case. Why is this? My guess is that the electron withdrawing nature of oxygen somehow causes this.&lt;/p&gt;&#xA;" OwnerUserId="16144" LastEditorUserId="10185" LastEditDate="2015-06-28T15:35:20.593" LastActivityDate="2015-06-28T15:35:20.593" Title="Why is picric acid more explosive than TNT?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;explosives&gt;&lt;nitro-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31459" PostTypeId="1" AcceptedAnswerId="33605" CreationDate="2015-05-15T14:34:11.083" Score="3" ViewCount="3596" Body="&lt;p&gt;Is it possible to figure out the $V_\mathrm{max}$ (or $k_\mathrm{cat}$) and/or $K_\mathrm m$ from a plot of enzyme activity expressed as specific activity over time (i.e. $\mathrm{U/mg}$ on the $Y$ and time on the $x$ axis). I've figured out how to express specific activity as turnover but I'm not sure this helps with calculating $V_\mathrm{max}$ or $k_\mathrm{cat}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction:&#xA;p38 phosphorylates MSK1 with Michaelis–Menten kinetics &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimental conditions:&#xA;$1\ \mathrm{U/mL}$ of p38 was added at time $t=0\ \mathrm s$ to a solution containing MSK1. They don't say how much MSK1 but express its activity as $\mathrm{U/mg}$ (which is analogous to $\mathrm{(\mu mol/min)/mL}$, otherwise known as specific activity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Data:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Time         MSK1 activity &#xA;(s)          ((µmol/min)/mg)&#xA;&lt;sup&gt;__________________________________________________&lt;/sup&gt;&#xA;   0               0   &#xA; 300             175    &#xA; 600             350    &#xA;1200             450&#xA;1800             475&#xA;2400             500&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Calculated turnover:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Time         MSK1 turnover &#xA;(s)          (1/(nmol s))&#xA;&lt;sup&gt;__________________________________________________&lt;/sup&gt;&#xA; 18000          300&#xA; 36000          600&#xA; 72000         1200&#xA;108000         1800&#xA;144000         2400&#xA;     0            0&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In actual fact, I though $k_{\rm cat}$ &lt;strong&gt;was&lt;/strong&gt; the turnover and that this was a constant, specific for each enzyme. This data however suggests that as the reaction progresses more MSK1 is produced the turnover number increases, which makes sense and therefore calls into question the two definitions of $k_{\rm cat}$.&lt;/p&gt;&#xA;" OwnerUserId="9434" LastEditorDisplayName="user7951" LastEditDate="2016-12-10T15:00:28.390" LastActivityDate="2016-12-10T15:00:28.390" Title="Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator" Tags="&lt;biochemistry&gt;&lt;kinetics&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31460" PostTypeId="1" CreationDate="2015-05-15T14:40:09.297" Score="2" ViewCount="4901" Body="&lt;p&gt;I would like to know if it is safe to mix organic solvents together or organic solvents with some compounds listed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a mixture with random ratios (greater than zero - all chemicals present) we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ethanol&lt;/li&gt;&#xA;&lt;li&gt;Methyl ethyl ketone (as denaturant for ethanol)&lt;/li&gt;&#xA;&lt;li&gt;Bitrex (aka Denatonium - as denaturant for ethanol)&lt;/li&gt;&#xA;&lt;li&gt;Isopropanol&lt;/li&gt;&#xA;&lt;li&gt;n-butanol (as denaturant for isopropanol)&lt;/li&gt;&#xA;&lt;li&gt;n-propanol (as denaturant for isopropanol) &lt;/li&gt;&#xA;&lt;li&gt;Ethyl acetate&lt;/li&gt;&#xA;&lt;li&gt;Acetic acid (in case that ethyl acetate hydrolyzes)&lt;/li&gt;&#xA;&lt;li&gt;Water (from ethanol and isopropanol)&lt;/li&gt;&#xA;&lt;li&gt;Acetone&lt;/li&gt;&#xA;&lt;li&gt;Methanol&lt;/li&gt;&#xA;&lt;li&gt;Naphtha (Hydrogenated)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general, we have alcohols, ketones, denatonium, ester, organic acid, water, and a liquid hydrocarbon mixture (Naphtha).Ratios can be taken as equal for all chemicals except those used for denaturation.(I think ratios for denaturants  wouldn't matter in case they would react.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could some of these chemicals react together and produce any unstable or hazardous compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What conditions should be avoided? (Light,heat,etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any reaction (slow) that could happen with atmospheric oxygen? (opening and closing a bottle containing all these chemicals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know too much about organic chemistry, just asking to be sure this is safe.I want to test something that possibly involves mixing all the above chemicals and I don't know if they could probably react together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Question unrelated to waste disposal. I'm experimenting with a DIY electronics cleaner.&lt;/p&gt;&#xA;" OwnerDisplayName="user16223" LastEditorUserId="402" LastEditDate="2017-05-25T06:35:57.087" LastActivityDate="2017-05-25T06:35:57.087" Title="Mixing solvents and other compounds - safety issues" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;&lt;solvents&gt;&lt;mixtures&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31463" PostTypeId="1" CreationDate="2015-05-15T15:07:20.450" Score="21" ViewCount="1697" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Flcnu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph of rotational heat capacity above shows a small maximum before approaching the equipartition value. What is the origin/physical explanation of this maximum?&lt;/p&gt;&#xA;" OwnerUserId="9804" LastEditorUserId="10185" LastEditDate="2015-06-03T20:56:42.387" LastActivityDate="2015-06-05T14:00:58.187" Title="What is the origin of the maximum in rotational heat capacity?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31464" PostTypeId="2" ParentId="31049" CreationDate="2015-05-15T15:13:15.850" Score="12" Body="&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A systematic examination of the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes with density functional techniques and high-level wave function theory indicates that the charge-separation within these complexes influences the relative corresponding dipole moments for these compounds.  Though fluorine is indeed more electronegative than chlorine, $\ce{CH3Cl}$ had a larger dipole moment due to the longer C-Cl bond with respect to the C-F separation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Introduction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that a dipole moment can be explained relatively simply by a separation of two charges.  The dipole moment, $\mu$, is dependent on two things: the magnitude of the charge ($q$) AND the separation between the two charges ($r$).  This can be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu = q \cdot r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from general chemistry that fluorine is the most electronegative atom on the periodic table.  Therefore, for simple molecules such as $\ce{HF}$ or $\ce{LiF}$, we would expect a rather large dipole moment than if we considered $\ce{HCl}$ or $\ce{LiCl}$.  However, a dipole moment depends on separation as well.  This can lead to some discrepancies in our general trends that we chemists love to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider chlorine.  The atomic radius for chlorine is larger than it is for fluorine.  A C-Cl bond is going to be longer than a C-F bond.  In the case of methyl fluoride ($\ce{CH3F}$) and methyl chloride ($\ce{CH3Cl}$), we will have a difference in separation of the methyl group and the halogen (larger for the latter).  Looking back at our electric dipole equation, an increase in $r$ translates into an increase in $\mu$.  Perhaps it is this larger separation between the carbon and chlorine atoms that gives rise to these relatively similar dipole moments between $\ce{CH3F}$ and $\ce{CH3Cl}$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Super Mini Research Project Time&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let us apply electronic structure theory to the problem.  Dave has already provided a link to NIST which is a superb resource for these types of simple computations. However, to be thorough, we will perform our own computations.  The experimental dipole moments for $\ce{CH3F}$ and $\ce{CH3Cl}$ (provided by DavePhD) are $\pu{-1.858 D} \text{ and } \pu{-1.897 D}$, respectively ([1] and [2]).  Dipole moments from electronic structure theory for these complexes can also be found on the NIST website (&lt;a href=&quot;http://cccbdb.nist.gov/exp2.asp?casno=593533&quot; rel=&quot;nofollow noreferrer&quot;&gt;source3&lt;/a&gt; and &lt;a href=&quot;http://cccbdb.nist.gov/exp2.asp?casno=74873#NSRDS-NBS10&quot; rel=&quot;nofollow noreferrer&quot;&gt;source4&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methodology&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full optimizations and corresponding harmonic vibrational frequency computations are performed using the B3LYP density functional technique (sorry Martin) using the Gaussian 09 software package.  The 6-31+G(d,p) basis set will be employed along with a series of Dunning's correlation consistent basis sets: heavy-aug-cc-pV&lt;em&gt;X&lt;/em&gt;Z (where &lt;em&gt;X&lt;/em&gt; = D, T, Q, 5).  This basis set denotes that cc-pV&lt;em&gt;X&lt;/em&gt;Z was employed for hydrogen atoms and the aug-cc-pV&lt;em&gt;X&lt;/em&gt;Z basis set was employed for all heavy (i.e. non-hydrogen) atoms.  A numerical integration grid consisting of 99 radial shels and 590 angular points per shell were employed for all DFT computations.  In addition, full optimizations have been performed using the CCSD(T) level of theory using the CFOUR software package.  The haDZ, haTZ, and haQZ basis sets have been employed for these computations.  All geometries were constrained to $C_{3v}$ symmetry.  Everything is converged to 'overkill'.  Frozen-core approximation applied... 5D 7F spherical basis functions used, etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Results&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dipole moments (in Debye) as well as select geometrical parameters ($R$ in Angstroms) for the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes at various levels of theory are given in the table below.  The deviation of the computed electric dipole moment from experiment is denoted as $\delta_{\mathrm{theory}}^{\mathrm{expt.}}$.  Using this deviation as our main measurement of error, we can see that for the $\ce{CH3F}$ complex, the B3LYP/ha5Z level of theory deviates from experiment by only $\pu{0.01 D}$.  A larger deviation from experiment is observed for the CH$_3$Cl complex with a magnitude of $\pu{0.1 D}$ at the B3LYP/ha5Z level of theory.  These magnitudes are qualitatively similar at the CCSD(T) level.  Therefore, with respect to the dipole moment, the B3LYP method with a triple-$\zeta$ quality basis set is necessary to describe this first-order property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/xqFcbAJ.png&quot; alt=&quot;Dipole moments (in D) and select geometrical paramaters (in Angstroms) of the CH3F and CH3Cl compounds at various levels of theory.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly $\ce{CH3Cl}$ contains a larger separation between the Cl atom and the methyl group with respect to $\ce{CH3F}$ by about 0.4 Angstroms.  If you wanted to back-calculate the charge for these complexes by assuming $R$(CX) (where X is the halogen) is equivalent to $r$ in the first equation, you will see that CH$_3$F has a larger charge (by about 0.25).  However, we still see that CH$_3$Cl has a larger dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final column in the table above ($\mu_{\mathrm{geom}}^{\ce{CH3Cl}}$) denotes the dipole moment of $\ce{CH3F}$ when using the optimized geometry of $\ce{CH3Cl}$ at the corresponding level of theory.  The dipole moment is significantly larger, by about $\pu{1.5 D}$, which is expected due to the larger C-F separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though we as chemists like to resort to using the general trends provided by the periodic table, we must take into account other factors to explain deviations from these trends.  Here we consider the separation of the halogen from the methyl group as an explanation for $\ce{CH3Cl}$ having a larger dipole moment than $\ce{CH3F}$, even though F is more electronegative.  Our computational results qualitatively agree with experiment.  In addition, B3LYP with a triple-$\zeta$ quality basis set is adequate in describing this first-order property according to CCSD(T) results.  Finally, these results indicate that it is the charge-separation in these small molecules that greatly impacts the qualitative ordering of these dipole moments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;References:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Marshall, M. D.; Muenter, J. The electric dipole moment of methyl fluoride. &lt;em&gt;J. Mol. Spectrosc.&lt;/em&gt; &lt;strong&gt;1980,&lt;/strong&gt; &lt;em&gt;83&lt;/em&gt; (2), 279–282. &lt;a href=&quot;https://doi.org/10.1016/0022-2852(80)90051-X&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/0022-2852(80)90051-X&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Wlodarczak, G.; Segard, B.; Legrand, J.; Demaison, J. The dipole moment of CH335Cl. &lt;em&gt;J. Mol. Spectrosc.&lt;/em&gt; &lt;strong&gt;1985,&lt;/strong&gt; &lt;em&gt;111&lt;/em&gt; (1), 204–206. &lt;a href=&quot;https://doi.org/10.1016/0022-2852(85)90083-9&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1016/0022-2852(85)90083-9&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14" LastEditorUserId="5026" LastEditDate="2018-07-04T02:20:31.557" LastActivityDate="2018-07-04T02:20:31.557" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31465" PostTypeId="2" ParentId="31448" CreationDate="2015-05-15T15:35:28.627" Score="14" Body="&lt;p&gt;One key problem with astatine is that it's incredibly unstable. There are no known stable isotopes, and the longest-lived has a half-life of ~8 hours. So no one has been able to (yet) prepare enough to make real measurements. Thus, we don't know for sure whether solid At is diatomic (like the other halogens) or monatomic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, high-level first-principals (quantum) calculations &lt;em&gt;have&lt;/em&gt; been performed that suggest astatine has some unusual properties. The full paper by Roald Hoffmann and Neil Ashcroft at Cornell was published in 2013 in &lt;em&gt;Phys. Rev. Lett.&lt;/em&gt;: &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevLett.111.116404&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Condensed Astatine: Monatomic and Metallic&quot;&lt;/a&gt; and also covered in multiple science news sites, including &lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/09/astatine-metallic-superconductor-predictions&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Chemistry World&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations on At:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;predict that it will be a metallic solid. That’s perhaps not so surprising given that iodine, the halogen above it in the periodic table, is itself a dark silvery solid (although not truly metallic itself) and that astatine’s melting point is 302&amp;deg;C. But the real surprise of the new results is that the solid wouldn’t be composed of diatomic molecules, like all the other halogens, but would be monatomic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would At be different than solid &lt;span class=&quot;math-container&quot;&gt;$\ce{I2}$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... The monatomic form appeared only with spin-orbit coupling included. &quot;These effects influence the atoms’ propensity to form certain bonds and crystal structures, usually by reducing molecular binding energies,&quot; says Hermann. Earlier calculations have indicated that in the gas phase astatine would still form a weakly bound diatomic molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, it's the &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;spin orbit coupling&lt;/a&gt; contribution of relativistic effects that would prevent the formation of diatomic molecules in solid At.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="17175" LastEditDate="2020-01-05T14:51:17.553" LastActivityDate="2020-01-05T14:51:17.553" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31466" PostTypeId="5" CreationDate="2015-05-15T15:45:13.513" Score="0" Body="&lt;p&gt;In some late-row elements, core electrons move with a significant fraction of the speed of light, causing &lt;a href=&quot;http://en.wikipedia.org/wiki/Relativistic_quantum_chemistry&quot; rel=&quot;nofollow&quot;&gt;relativistic effects&lt;/a&gt;. This includes properties of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gold&lt;/li&gt;&#xA;&lt;li&gt;Mercury&lt;/li&gt;&#xA;&lt;li&gt;Lead&lt;/li&gt;&#xA;&lt;li&gt;Lanthanides (e.g., the &quot;lanthanide contraction&quot;)&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The properties of these elements are often anomalies compared to other periodic trends.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-05-15T17:04:25.840" LastActivityDate="2015-05-15T17:04:25.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31467" PostTypeId="4" CreationDate="2015-05-15T15:45:13.513" Score="0" Body="Atomic and chemical properties related to the relativistic properties of electrons in heavy elements." OwnerUserId="5017" LastEditorUserId="10185" LastEditDate="2015-09-14T04:35:04.317" LastActivityDate="2015-09-14T04:35:04.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31468" PostTypeId="1" CreationDate="2015-05-15T15:45:16.657" Score="1" ViewCount="1711" Body="&lt;p&gt;I’ve got six test-tubes with $0.5\ \mathrm{ml}+0.5\ \mathrm{ml}$ of 0.005 %, 0.0025 %, 0.00125 %, 0.000625 %, 0.0003125 % and 0.0003125 % starch and 4.2 pH phosphate buffer solution (2, 4, 8, 16, and 32-fold dilution of 1 % starch solution).&#xA;0.5 ml of beta-amylase was added to each test-tube, and heated (without sixth test-tube). Then the absorbance of each specimen was measured (sixth test-tube used as reference) in 530 nm wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to find the $K_\text{M}$ of β-amylase from Lineweaver–Burk graph.&#xA;There is no problem with plotting $x$ points on graph (it’s just a reciprocal of substrate concentration).&#xA;But how can I use absorbance data to find $V$ that I need to plot on $y$-axis?&lt;/p&gt;&#xA;" OwnerUserId="16248" LastEditorUserId="865" LastEditDate="2021-05-22T20:49:09.527" LastActivityDate="2021-05-22T20:49:09.527" Title="Finding 1/V from absorbance in Lineweaver-burk plot" Tags="&lt;kinetics&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31469" PostTypeId="2" ParentId="31436" CreationDate="2015-05-15T15:58:18.597" Score="2" Body="&lt;p&gt;An acetylide ion proceeds in a nucleophilic substitution pathway (SN2 in the picture due to primary halide). The substitution involves the attack of a highly nucleophilic sp-carbon on a highly electrophilic primary carbon. The enthalpy of the C-C bond formation is relatively low due the pKa of the nucleophile being ~ 25. The ion is treated as a &quot;soft&quot; nucleophile that prefers to substitute alkyl halides instead of acting as a base and conducting an elimination reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grignard reagent is essentially a highly polarized bond between Carbon and the Bromide(Halide)/Magnesium complex. The electronegative difference between Carbon and Magnesium is 1.3 (&lt;a href=&quot;http://www.thecatalyst.org/electabl.html&quot; rel=&quot;nofollow&quot;&gt;http://www.thecatalyst.org/electabl.html&lt;/a&gt; &amp;lt;-- Source of values). The Carbon atom in a Gringnard is treated as a carbonanion. This carbonanion characterization is highly unstable in nature and treated as a &quot;hard&quot; nucleophile with pKa of ~ 50. The highly basic character of a Grignard reagent often results in an elimination reaction or no reaction at all. The transition state to substitute the alkyl halide is less stable than the Magnesium/Bromide(Halide) complex. This is due to a ligation formation between the solvent and the Magnesium atom. The solvent (ex. diethyl ether) shares non-bonding electrons with the cation and creates a stable electronic atmosphere with a lower energy than a possible C-C bond with the alkyl halide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitution isn't preferred with a hard nucleophile and therefore, the acetylide ion is more likely to preform a nucleophilic substitution reaction with an alkyl halide. Note that the electronegative difference between Carbon and Magnesium supersedes the hybridization of the sp or &quot;triple&quot; bond complex. The bond is so high in energy that the nucleophilic character of the grignard far surpasses the electronegative character of the sp-hybridization of the acetylide ion.&lt;/p&gt;&#xA;" OwnerUserId="16250" LastActivityDate="2015-05-15T15:58:18.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31470" PostTypeId="2" ParentId="31424" CreationDate="2015-05-15T16:01:03.743" Score="1" Body="&lt;p&gt;I tried to answer the above comment but couldn't due to character limitation. So, i'm posting it here as another answer:&lt;br&gt;&#xA;To answer your question we need more data. For example if we want to separate mixture of sand and sugar, we can utilize the solubility difference of these two substances. To separate water and salt, we use heat to vaporize water but that doesn't mean salt is non-volatile. Salt can be vaporized but at very high temp. (As far i can remember mp of NaCl is around 800 deg C and bp is 1500 deg C. So, if you use temp. greater than 1500 deg C to separate water and NaCl, then you will end up with a mixture of vapor of water and salt. So, in room temp. your desired separation will happen or not or do we need to exploit any physical properties, we need to know their properties. If you know the mixture components, than use CRC or Perry's handbook to find out their behavior. Even you can also find multiple component phase diagram, which will tell you everything about the mixture. Or if their is no data available, you can use software like HYSYS to exploit the QSPR.  &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-05-15T16:01:03.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31471" PostTypeId="1" CreationDate="2015-05-15T16:07:45.457" Score="39" ViewCount="4367" Body="&lt;p&gt;Fresh ice cubes are almost instantly sticky and easily cling on to fabric and other similarly rough surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few minutes later, however, the effect almost completely disappears.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the cause of this adhesion?&lt;/li&gt;&#xA;&lt;li&gt;Why does the effect vanish after some time?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="275" LastEditDate="2015-05-16T12:16:42.760" LastActivityDate="2015-06-05T07:11:29.020" Title="Why is fresh ice sticky?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;solvents&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31472" PostTypeId="2" ParentId="31471" CreationDate="2015-05-15T16:18:29.743" Score="30" Body="&lt;p&gt;Whether or not the ice cube will adhere to something is dependent on the temperature of the ice, the state of the surface of the ice, and the moisture content of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of an ice cube can range from the freezing point of water ($0\ \mathrm{^\circ C}$) to well below this temperature. Typically most materials have a small degree of moisture regardless of whether they are “wet” or not. If the ice cube is cold enough, it will freeze the water present in the material. The giant covalent network structure of the ice extends into the material, accounting for why it sticks to the ice—the moisture in the material has become part of the ice cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the surface of the ice cube is has melted, it will not be able to freeze the water in the material any more than it will be able to freeze the ice already melted on its surface. No giant covalent network solid will form, and the ice will not stick to the material.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorDisplayName="user7951" LastEditDate="2015-06-05T07:11:29.020" LastActivityDate="2015-06-05T07:11:29.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31473" PostTypeId="1" AcceptedAnswerId="31477" CreationDate="2015-05-15T16:37:16.943" Score="7" ViewCount="533" Body="&lt;p&gt;Despite the ubiquitous usefulness of electronic configurations (of elements) in explaining their behavious I have not been able to find out exactly how the configurations are determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any sure-fire technique to detect them (like x-ray diffraction for crystal structure) or is it based on guesswork, through observation of the various properties and reactions of the elements? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2015-05-15T17:48:59.670" Title="How are electronic configurations determined experimentally?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31476" PostTypeId="1" AcceptedAnswerId="32023" CreationDate="2015-05-15T17:15:45.033" Score="13" ViewCount="15005" Body="&lt;p&gt;Flame tests are a common way of identifying trace amounts of alkali metals (and to a lesser extent alkaline earth metals) in mixtures. Certain salts of $\ce{Ba, Sr}$ are also used in fireworks for their ability to impart a colour to the flame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although predominant among s block elements this behaviour is also seen in other elements like $\ce{Cu}$ and $\ce{Pb}$. My question is, what is so special about these elements that lends them this characteristic?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2019-06-24T17:43:03.257" Title="Why do s-block elements impart color to a flame?" Tags="&lt;color&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31477" PostTypeId="2" ParentId="31473" CreationDate="2015-05-15T17:23:21.770" Score="5" Body="&lt;p&gt;There are lots of techniques you could use. Most useful would be X-ray absorption techniques such as &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_absorption_near_edge_structure&quot; rel=&quot;nofollow&quot;&gt;XANES&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Extended_X-ray_absorption_fine_structure&quot; rel=&quot;nofollow&quot;&gt;EXAFS&lt;/a&gt;. X-ray emission techniques such as &lt;a href=&quot;http://en.wikipedia.org/wiki/X-ray_fluorescence&quot; rel=&quot;nofollow&quot;&gt;XPF&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy-dispersive_X-ray_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;EDS&lt;/a&gt;. You could look at electron emission spectroscopies (ESCA) such as &lt;a href=&quot;http://mmrc.caltech.edu/SS_XPS/XPS_PPT/XPS_Slides.pdf&quot; rel=&quot;nofollow&quot;&gt;XPS&lt;/a&gt; or UPS or &lt;a href=&quot;http://en.wikipedia.org/wiki/Auger_electron_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;Auger&lt;/a&gt; techniques and &lt;a href=&quot;http://en.wikipedia.org/wiki/Electron_energy_loss_spectroscopy&quot; rel=&quot;nofollow&quot;&gt;EELS&lt;/a&gt; spectroscopy, ELNES and EXELFS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point to remember is that a small perturbation to the chemical environment e.g. by a change in oxidation state or coordination number, etc. will change the signals observed, this can lead to much more than electron configurations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each has its own benefit and speciality, depending on the sample and the exact thing you are looking for but that should get you going :) &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="5017" LastEditDate="2015-05-15T17:48:59.670" LastActivityDate="2015-05-15T17:48:59.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31478" PostTypeId="2" ParentId="31476" CreationDate="2015-05-15T17:23:41.110" Score="1" Body="&lt;p&gt;When you impart heat, atom receive it as extra energy. Due to which electron in outer shell jumps to higher shell position. This electron when falls back to it original position it emits photons of specific wavelengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, different atoms have different number of valence electrons, thus different element emits photon of different wavelength.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-05-15T17:23:41.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31479" PostTypeId="2" ParentId="31463" CreationDate="2015-05-15T17:29:56.557" Score="4" Body="&lt;p&gt;OK, i tried to refresh my stat mech class materials. If you derive the formula for ${C_v}$, then you will find that at high temp. ${C_v=Nk=R}$ and at low temp. $C_v=3Nk( \frac{2\theta_r}{T})^2 \exp(\frac{-2\theta_r}{T})$. So, At low temp. limit there is a trade off between the square term and exponential term and results in the above mentioned maxima. This is the explanation from mathematical POV but from physical POV, i think the answer lies in (I am not quite sure) the separation of rotational spectra and degeneracy number.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="9804" LastEditDate="2015-05-18T15:53:09.523" LastActivityDate="2015-05-18T15:53:09.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31480" PostTypeId="2" ParentId="28328" CreationDate="2015-05-15T17:32:02.537" Score="4" Body="&lt;p&gt;The more reactive $\ce{LiAlH4}$ is required for the reduction of the amide to the corresponding amine (&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-2-2-4.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). The carbonyl carbon of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Amide&quot; rel=&quot;nofollow&quot;&gt;amide&lt;/a&gt; is less electrophilic compared to a ketone, because electron density is donated by the amide nitrogen via resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bromine and hydroxide react with an carboxylic acid amide to yield an amine with one fewer carbon atom (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hofmann_rearrangement&quot; rel=&quot;nofollow&quot;&gt;Hofmann degradation&lt;/a&gt;). $\ce{Br2}$ and $\ce{KOH}$ react to form $\ce{KOBr}$ in situ, which transforms the amide into an intermediate isocyanate ($\ce{R-N=C=O}$). Hydrolysis of the isocyanate yields a carbamic acid $\ce{R-NH-COOH}$, which decomposes to the amine $\ce{R-NH2}$ and $\ce{CO2}$. As described in &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF01172698&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, the reaction of $\ce{PCl5}$ with carboxylic acid amides produces amidochlorides $\ce{RCCl2NH2}$ as intermediates, which break down to the corresponding nitrile $\ce{RC#N}$ and $\ce{HCl}$.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2015-05-15T17:32:02.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31481" PostTypeId="2" ParentId="31468" CreationDate="2015-05-15T18:46:18.110" Score="2" Body="&lt;p&gt;First, let me tell you about the pitfalls of Lineweaver-Burke plots.  The tranformations of your data (i.e. taking the reciprocal) distort the error structure of your data and can lead to erroneous fits.  It is better to directly use nonlinear fitting techniques to find kinetic parameters.  These techniques do not require transformation of the data and thus do not distort error structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, let me answer your question.  The classical &lt;a href=&quot;http://en.wikipedia.org/wiki/Lineweaver%E2%80%93Burk_plot&quot; rel=&quot;nofollow&quot;&gt;Lineweaver-Burke plot&lt;/a&gt; puts inverse reaction rate on the $y$ axis, i.e. $1/V$.  Your question is how to find $V$ from absorbance data.  The answer is, you don't need to.  Let $A_i(t)$ be the absorbance data for tube $i$ as a function of time.  What is $V_i(t)$ for that tube?  We won't know unless you know the molar absorption coefficient of the substrate (or product -- I'm a little unclear from your question what you are hoping to detect by measuring $A_{\textrm{530 nm}}$).  Assume that it is the product that is absorbing, and that it has a molar absorption coefficient of $\epsilon$ and that you are using a photometer with path length $L$.  Then the reaction rate for the $i$th tube will be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_i(t)= \frac{1}{\epsilon L}\frac{dA_i(t)}{dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means take the slope of the absorbance measurement and divide it by $\epsilon$ and $L$ -- numbers you don't know -- to get $V$.  But the key assumption is simply that $V \propto \frac{dA}{dt}$.  Try making up any numbers you like for $\epsilon$ and $L$.  Find $K_m$.  Then make up new numbers and find $K_m$ again.  Did the result change?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-15T18:46:18.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31482" PostTypeId="2" ParentId="16010" CreationDate="2015-05-15T18:53:43.940" Score="2" Body="&lt;p&gt;Companies that sell such mixtures usually make them by dissolving everything into a solution, lyophilizing the solution, and then grinding the lyophilized residue.  This guarantees that the size of particulates in the mixture will not be related to the chemical identity of the particulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have to mix powders, I wouldn't recommend making such a large batch.  Why not make a smaller batch, enough for 10 - 20 bottles of LB, mix thoroughly by shaking (which will still not be as good as lyophilizing), and then immediately aliquot 25 g portions into a series of dry 1 L bottles?  Then you can store the dry bottles until you want to make some LB, and you don't have to worry about the powder settling etc. in each bottle once it has its own charge of powder.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-15T18:53:43.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31483" PostTypeId="2" ParentId="31424" CreationDate="2015-05-15T19:06:19.070" Score="1" Body="&lt;p&gt;I don't think what you are asking for is thermodynamically impossible, but I think it is very unlikely.  Let's say you start off with a single liquid phase containing molecules A, B, X, Y, and Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also say that when temperature changes, two phases form.  One phase is mostly A and the other phase is mostly B.  Let's call those two phases $a$ and $b$.  Your question is, is there some condition where the concentrations of molecules X, Y, and Z in phase $a$ is the same as it is in phase $b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the distribution of a solute between two phases is governed by a parameter called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Partition_coefficient&quot; rel=&quot;nofollow&quot;&gt;partition coefficient&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Distribution_constant&quot; rel=&quot;nofollow&quot;&gt;distribution constant&lt;/a&gt;.  The distribution of molecule X in phase $a$ vs phase $b$, for example, could be noted as $K_{ab, X}$.  For your condition to be satisified, the distribution coefficient of X would have to be 1.0, i.e. it would have an equal preference for both phases.  The same would be true of molecules Y and Z, i.e. $K_{ab, X}=K_{ab, Y}=K_{ab, Z}=1$ is required for your condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common type of liquid phase separation is oil and water: they don't mix.  I can't think of three molecules that each have an equal preference for water vs. oil.  These molecules would be somewhat hydrophobic and somewhat hydrophilic.  Feel free to look through a &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd367.pdf&quot; rel=&quot;nofollow&quot;&gt;giant compendium of octanol-water partition coefficients published by NIST&lt;/a&gt; to look for three molecules X, Y, and Z that all have octanol-water partition coefficients of 1.0 by yourself, though.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-15T19:06:19.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31484" PostTypeId="2" ParentId="31375" CreationDate="2015-05-15T19:06:42.517" Score="12" Body="&lt;h2&gt;Absolutely!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Having taught physics and planetary science for years now, this is how I've seen unit conversion taught most effectively. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Suppose we wish to convert  423 feet to kilometers, where we know $3280.4 \; \text{ft} = 1 \; \text{km}$, or $\frac{1 \; \text{km} }{3280.4 \; \text{ft}} =1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\require{cancel} 423 \; \text{ft} = \left( 423 \; \text{ft} \right) \times 1 = \left( 423 \; \text{ft} \right) \times \left( \frac{1 \; \text{km} }{3280.4 \; \text{ft}} \right) = \frac{423 \; \text{ft} \; \text{km}}{3280.4 \; \text{ft}} = \frac{423}{3280.4}  \frac{\cancel{\text{ft}} \; \text{km}}{\cancel{\text{ft}}} \approx 0.129 \; \text{km} $$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;There are a few catches, however...&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First, radians are a pseudo-unit.&lt;/strong&gt; Angular frequency ($\omega$) and regular frequency ($f$) are two closely related quantities: $\omega=2\pi f$, but they have different units. The units of angular frequency are radians per unit time (e.g. rad/sec), whereas the units of regular frequency are &quot;per unit time&quot; (e.g. [cycles, counts, etc.]/sec or Hz). Many scientists disregard radians as a unit and give angular and regular frequency the same units. This is acceptable, but you should be aware of the context, as an angular frequency of 20 Hz is not the same as an ordinary frequency of 20 Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, some (particularly chemists) have an unfortunate tendency to be pretty loose with units in some situations&lt;/strong&gt;—especially when it comes to calculating equilibrium constants and dimensional laws. For example, it is not permitted for the argument of a natural logarithm or exponential to have units. Many people calculate $\ln(P) = ...$ (as for the Clausius-Clapeyron equation), while it &lt;em&gt;ought&lt;/em&gt; to be something like $\ln(\frac{P}{P_0}) = ...$, where $P_0$ is a standard pressure scale used to scale all pressures in an equation. In most cases, this scale factor cancels out in the equation, but if you are keeping track of units, this is essential. Because this factor cancels out, calculations can usually be done correctly without thinking about it, and many chemists simply choose to ignore it or are even unaware of it (including some professors I've known!). Some old textbooks still do this, too, so this sloppiness is prevalent. &lt;strong&gt;The bottom line is that understanding the context is important!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third, this effectively only works for units that are directly proportional to each other.&lt;/strong&gt; One example of this not working is when converting between degrees Fahrenheit and degrees Celsius and Kelvins. The units can (and &lt;em&gt;should&lt;/em&gt;) still be treated as variables, but because these unit systems are not directly proportional to each other, special care must be taken with them. The conversion formulae must be used.&lt;/p&gt;&#xA;" OwnerUserId="16256" LastEditorUserId="16256" LastEditDate="2017-04-05T22:52:27.587" LastActivityDate="2017-04-05T22:52:27.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31485" PostTypeId="1" AcceptedAnswerId="31486" CreationDate="2015-05-15T19:13:50.577" Score="5" ViewCount="16135" Body="&lt;p&gt;Can someone explain how to calculate the number of geometrical isomers in allenes like $\ce{CH3-(CH=CH)4-CH3}$ or $\ce{CH3-(CH=CH)5-CH3}$&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="30836" LastEditDate="2020-04-05T19:07:30.657" LastActivityDate="2020-04-05T19:07:30.657" Title="How to calculate the number of geometrical isomers in allenes?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;&lt;geometrical-isomerism&gt;&lt;allenes&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31486" PostTypeId="2" ParentId="31485" CreationDate="2015-05-15T19:38:23.657" Score="13" Body="&lt;p&gt;If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of &lt;em&gt;cis-trans&lt;/em&gt; isomers is possible in compounds with a single double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhIEG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate &lt;em&gt;cis-trans&lt;/em&gt; isomers, but they can generate enantiomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JBFqO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of &lt;em&gt;cis-trans&lt;/em&gt; isomers is possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any compound with 3 or more &lt;strong&gt;cumulative&lt;/strong&gt; double bonds is a member of the &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cumulene&quot; rel=&quot;noreferrer&quot;&gt;cumulene&lt;/a&gt;&lt;/strong&gt; family. Any cumulene with an &lt;strong&gt;odd&lt;/strong&gt; number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of &lt;em&gt;cis-trans&lt;/em&gt; isomers. Cumulenes with an &lt;strong&gt;even&lt;/strong&gt; number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display &lt;em&gt;cis-trans&lt;/em&gt; isomerization, but can have enantiomers.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-15T21:51:52.050" LastActivityDate="2015-05-15T21:51:52.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31487" PostTypeId="2" ParentId="27385" CreationDate="2015-05-15T19:40:41.257" Score="5" Body="&lt;p&gt;To sum it up, the adhesive property of an epoxy is the result of &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;noreferrer&quot;&gt;inter-molecular forces&lt;/a&gt; and the formation of &lt;a href=&quot;http://en.wikipedia.org/wiki/Mechanical_bond&quot; rel=&quot;noreferrer&quot;&gt;mechanical bonding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Adhesives cure when the small resin molecules (mers) join together to form extremely large molecules known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Polymer&quot; rel=&quot;noreferrer&quot;&gt;polymers&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some adhesives are used only with specific substrates because of the cure time needed to allow for good bonding (both with the substrate and for polymer formation.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I had trouble finding information saying that all epoxies use ionic bonding, and it  seems to me that most epoxies will actually just utilize intermolecular forces such as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ion-induced dipole forces&lt;/li&gt;&#xA;&lt;li&gt;Ion-dipole forces&lt;/li&gt;&#xA;&lt;li&gt;Van der Waals forces&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen bonding&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;An important first step in applying an epoxy is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wetting&quot; rel=&quot;noreferrer&quot;&gt;wetting of the surface&lt;/a&gt; and then the polymerization of the epoxy. The ability to wet is related to &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity&quot; rel=&quot;noreferrer&quot;&gt;viscosity&lt;/a&gt; and the inter-molecular forces. If there is a low viscosity and strong attraction between the substrate and the adhesive then we can witness the formation of &lt;em&gt;mechanical bonding.&lt;/em&gt; Below is an explanation: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order for an adhesive to bond  two surfaces, there must be several types of interaction between the adhesive and both substrates. The first type of interaction is that &lt;strong&gt;the adhesive must wet the substrate&lt;/strong&gt;, meaning that the adhesive must spread itself out into a film that covers the substrate surface. In order for this to happen, the adhesive must have a low enough viscosity so that it will flow. Because viscosity is temperature dependent, the application of a cold adhesive to a substrate, or the application of an adhesive to a cold substrate, may result in poor wetting. Another factor that affects wetting is the relative strengths of cohesive forces (between like molecules, such as two adhesive molecules) and those of adhesive forces (between unlike molecules, such as an adhesive molecule and a substrate molecule). If the cohesive forces among adhesive molecules are weaker than the adhesive forces between the adhesive molecules and the substrate surface, then the adhesive molecules will spread out over the substrate and wet its surface. &lt;strong&gt;An adhesive that has a relatively low viscosity and is able to wet the substrate surface will flow into any tiny cracks or pores on the substrate surface, thus promoting what is known as mechanical bonding&lt;/strong&gt;. Mechanical bonding increases the strength of an adhesive bond and, as a result, a forced separation of the two substrate surfaces is more apt to tear the substrate surfaces.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Mechanical bonding is one of several ways that an adhesive bonds substrates. All surfaces, except those that are highly polished, have pores. If the adhesive flows into these pores and then polymerizes, a mechanical bond is formed. It is similar to placing a wick into liquid candle wax. Once the wax solidifies the wick can not be easily removed. A mechanical bond has formed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quotation and information from this answer were sourced &lt;a href=&quot;http://www.chemistryexplained.com/A-Ar/Adhesives.html&quot; rel=&quot;noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-15T19:46:11.513" LastActivityDate="2015-05-15T19:46:11.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31488" PostTypeId="1" AcceptedAnswerId="31490" CreationDate="2015-05-15T20:10:24.983" Score="4" ViewCount="915" Body="&lt;p&gt;What is the meaning of reduction and oxidation in organic chemistry? &#xA;In general chemistry I learned an &lt;strong&gt;OIL RIG&lt;/strong&gt; mnemonic, i.e. oxidation is loss and reduction is gain (of electrons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what do we say about their usage in organic chemistry? I have seen their usage a lot of times in the form of reducing agents and oxidizing agents but never really understood it. What really makes a substance a good oxidizing or a good reducing agent? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="41328" LastEditDate="2019-06-29T05:36:16.280" LastActivityDate="2019-06-29T05:41:34.323" Title="Reduction and oxidation in organic chemistry" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31489" PostTypeId="2" ParentId="31471" CreationDate="2015-05-15T21:13:46.657" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the cause of this adhesion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I propose a slightly different hypothesis from ringo. As opposed to the very cold ice freezing water in the material, I would presume the material to initially be dry, but at room temperature. This room temperature material may melt the outer layers of ice molecules, both cooling and wetting the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then as the temperature equalizes, and the material does not further conduct heat into the wet areas, the wet material becomes cold enough to freeze again, creating the sticky effect you ask about.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does the effect vanish after some time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The temperature gradient is not static, and the material will conduct some heat through it, so the sharp gradient that caused the melted ice to freeze again becomes a gradual gradient that shifts towards the center of the ice, and the water that refroze so quickly melts again, and the sides of the ice are no longer cold enough to refreeze any bits of them that happen to melt by exposure to a new contact surface.&lt;/p&gt;&#xA;" OwnerUserId="5668" LastEditorUserId="5668" LastEditDate="2015-05-16T12:05:38.040" LastActivityDate="2015-05-16T12:05:38.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31490" PostTypeId="2" ParentId="31488" CreationDate="2015-05-15T21:17:51.267" Score="6" Body="&lt;p&gt;In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction of alkenes (gain of hydrogen):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BzxH.jpg&quot; alt=&quot;Reduction of alkenes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxidation of secondary alcohols (loss of hydrogen):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oxsUN.jpg&quot; alt=&quot;Oxidation of secondary alcohols&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of an oxidising or reducing agent can be evaluated by measuring its &lt;a href=&quot;http://en.wikipedia.org/wiki/Reduction_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;reduction potential&lt;/a&gt;. Good oxidising agents have highly positive &lt;span class=&quot;math-container&quot;&gt;$E^\circ$&lt;/span&gt; values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative &lt;span class=&quot;math-container&quot;&gt;$E^\circ$&lt;/span&gt; values. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\ce{MnO4- + 8 H+ + 5 e- &amp;amp;&amp;lt;=&amp;gt; Mn^2+ + 4 H2O} &amp;amp;\quad E^\circ &amp;amp;= \pu{+1.51 V} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &amp;amp;&amp;lt;=&amp;gt; 2Cr^3+ + 7 H2O} &amp;amp;\quad E^\circ &amp;amp;= \pu{+1.33 V}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the redox potential for the &lt;span class=&quot;math-container&quot;&gt;$\ce{MnO4-}$&lt;/span&gt; system is more positive, &lt;span class=&quot;math-container&quot;&gt;$\ce{MnO4-}$&lt;/span&gt; is a stronger oxidising agent than &lt;span class=&quot;math-container&quot;&gt;$\ce{Cr2O7^2-}$&lt;/span&gt; under acidic conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction potential is affected by the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; (along with many other factors — temperature, concentration of reactants — see the Wikipedia article for more discussion). Under neutral or alkaline conditions, &lt;span class=&quot;math-container&quot;&gt;$\ce{MnO4-}$&lt;/span&gt; undergoes a different reaction with a less positive reduction potential, meaning it is a weaker oxidising agent under alkaline conditions that under acidic conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{MnO4- + 2 H2O + 3 e- &amp;lt;=&amp;gt; MnO2 + 4 OH-} \quad E^\circ = \pu{+0.59 V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactive metals, such as zinc, tend to be good reducing agents. Zinc amalgam and hydrochloric acid are used in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clemmensen_reduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clemmensen reduction&lt;/a&gt;, to reduce ketones to alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Zn^2+ + 2e- &amp;lt;=&amp;gt; Zn(Hg)} \quad E^\circ = \pu{−0.7628 V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many commonly used reducing agents in organic chemistry, such as &lt;span class=&quot;math-container&quot;&gt;$\ce{LiAlH4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4}$&lt;/span&gt;, serve as sources of the hydride ion &lt;span class=&quot;math-container&quot;&gt;$\ce{H-}$&lt;/span&gt;. As an example, in the reduction of ketones using &lt;span class=&quot;math-container&quot;&gt;$\ce{NaBH4},$&lt;/span&gt; hydrogen is added across the &lt;span class=&quot;math-container&quot;&gt;$\ce{C=O}$&lt;/span&gt; double bond though a tetrahedral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ro2hl.jpg&quot; alt=&quot;Reduction of ketones using NaBH4&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="41328" LastEditDate="2019-06-29T05:41:34.323" LastActivityDate="2019-06-29T05:41:34.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31491" PostTypeId="1" CreationDate="2015-05-15T21:18:29.150" Score="5" ViewCount="5369" Body="&lt;p&gt;What is the mechanism of formation of Grignard reagents? I couldn't find much in my book or on the Internet. &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/grignard.htm&quot;&gt;This page&lt;/a&gt; says something which I am unable to understand. Why are the $2$ electrons in the $\ce{Mg}$ separated? Shouldn't they be together? &lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="8208" LastEditDate="2015-05-16T18:34:39.863" LastActivityDate="2015-08-16T00:52:50.613" Title="What is the mechanism of formation of Grignard's Reagent?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31492" PostTypeId="2" ParentId="28883" CreationDate="2015-05-15T21:23:19.380" Score="8" Body="&lt;p&gt;Rigidity is definitely an important factor. At $295\ \mathrm{K}$, indeno[2,1-&lt;em&gt;a&lt;/em&gt;] has a fluorescence quantum yield $\Phi_F$ of 0.94, (&lt;a href=&quot;http://dx.doi.org/10.1021/ja01019a063&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;). In an organic glass at $77\ \mathrm{K}$, even &lt;em&gt;(E)&lt;/em&gt;-stilbene, which hardly shows any fluorescence at room temperature, exhibits  a $\Phi_F$ of 0.75.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eh5aX.png&quot; alt=&quot;rigid and less rigid molecules&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vibrational stretching of $\ce{C-C}$ bonds in substituents can significantly contribute to radiationless deactivation of an $S_1$ state and therefore reduce the fluorescence quantum yield $\Phi_F$. Admittedly, toluene isn't famous for its fluorescence ($\Phi_F = 0.15$), but the fluorescence of tert-butyl benzene is even lower ($\Phi_F = 0.03$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from radiationless deactivation through &lt;em&gt;internal conversion&lt;/em&gt; (&lt;strong&gt;IC&lt;/strong&gt;), an excited singlet state can be depopulated by intersystem crossing (&lt;strong&gt;ISC&lt;/strong&gt;) to a triplet state. Under certain conditions, this is even possible for compounds that usually have very low triplet quantum yields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halobenzenes added to a solution of fluorescing anthracenes in dioxane (&lt;a href=&quot;http://dx.doi.org/10.1002/bbpc.19770810605&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;) or ethyl iodide added to solutions of pyrene in alkanes (&lt;a href=&quot;http://dx.doi.org/10.1021/j100390a006&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;) have shown to quench the fluorescence normally observed in the absence of the halogenated compounds. The latter exhibit a &lt;em&gt;external heavy atom effect&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lack of $S_1 \rightarrow T_1$ intersystem crossing is probably the most important factor that makes rigid compound show large $\Phi_F$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is striking that strongly fluorescent compounds typically have an extended $\pi$-system. Here, the $S_0\rightarrow S_1$ excitation is a $(\pi,\pi^*)$ transition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side of the fence, ketones, such as acetophenone or benzophenone, hardly show fluorescence. For these compounds, $S_0\rightarrow S_1$ excitation of the $\ce{C=O}$ group is a $(n,\pi^*)$ transition. From a $^1(n,\pi^*)$ state, &lt;em&gt;spin-orbit coupling&lt;/em&gt; can both flip the spin &lt;strong&gt;and&lt;/strong&gt; move one electron to a more distant and geometrically different orbital. $S_1 \rightarrow T_1$ thus is in fact a $^1(n,\pi^*) \rightarrow {}^3(\pi,\pi^*)$ transition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compounds like rhodamin B in their $^1(\pi,\pi^*)$ states, the  corresponding transition to a $^3(\pi,\pi^*)$ state is largely spin-forbidden ans does not occur. Provided that no external effects come into play, these molecules exhibit strong fluorescence.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2016-11-01T14:42:58.947" LastActivityDate="2016-11-01T14:42:58.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31493" PostTypeId="1" CreationDate="2015-05-15T21:27:32.317" Score="4" ViewCount="418" Body="&lt;p&gt;I work at an animal shelter that uses lime sulfur dip to treat ringworm. We had an accident where someone mixed bleach instead of water with the lime sulfur dip which caused us to go through decontamination with Hazmat. Are there any other chemicals or acids that could cause this to happen?&lt;/p&gt;&#xA;" OwnerUserId="16262" LastEditorUserId="15410" LastEditDate="2015-05-16T01:30:42.353" LastActivityDate="2018-05-29T07:30:44.407" Title="What chemicals would make lime sulfur dip to produce a green colored smoke?" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31495" PostTypeId="1" CreationDate="2015-05-15T21:40:10.493" Score="3" ViewCount="233" Body="&lt;p&gt;I know that states with spin S=0 in a diatomic molecule have no spin orbit coupling, independent on the value of the projection of the total electronic angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the same is true if the absolute value of $\Lambda$ is equal to zero independent on the spin of the diatomic molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct that a diatomic molecule with a $^3\Sigma$ state has no spin orbit coupling?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="194" LastEditDate="2017-04-08T03:29:50.417" LastActivityDate="2017-04-08T06:25:21.237" Title="Does the triplet sigma state of a diatomic molecule experience spin-orbit coupling?" Tags="&lt;spectroscopy&gt;&lt;electronic-configuration&gt;&lt;spin&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31496" PostTypeId="1" AcceptedAnswerId="31500" CreationDate="2015-05-15T21:44:49.660" Score="2" ViewCount="437" Body="&lt;p&gt;I've been asked to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Descibe the ionization mechanism used in electrospray ionization&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that this is &#xA;a desorption ionization source, which uses a high electrical field as the ionizing agent. The best I've been able to put together is that there is a generation of $\ce{[M + _zH]^2+}$ ions. Surfing the web has just led me to articles saying the mechanism is still being debated. What's the mechanism, if there is a definitive one?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastActivityDate="2015-05-15T23:19:48.937" Title="What is the mechanism for electrospray ionization?" Tags="&lt;analytical-chemistry&gt;&lt;mass-spectrometry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31497" PostTypeId="1" CreationDate="2015-05-15T22:04:46.127" Score="5" ViewCount="12741" Body="&lt;p&gt;I know about s, p, d, and f-orbitals. However, my physical chemistry textbook does not mention what comes after f-orbitals. What are the next letters, and in what order are these higher orbitals filled?&lt;/p&gt;&#xA;" OwnerUserId="16265" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-09-20T13:19:50.153" LastActivityDate="2018-09-24T17:04:28.070" Title="What comes after f-orbitals, and in what order are they filled?" Tags="&lt;orbitals&gt;&lt;electronic-configuration&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31499" PostTypeId="1" AcceptedAnswerId="31503" CreationDate="2015-05-15T23:01:01.910" Score="7" ViewCount="168" Body="&lt;p&gt;This question is related to specific heat, and I thought I understood it but I've managed to confuse myself. My initial reasoning was that since in both cases, the pipe is the same, the amount of energy needed to raise the temperature will depend on the specific heat only, and since air has a lower specific heat, it will be more effective. After I started thinking about it more, I thought that the specific heat of the pipe is actually what is important in determining how much energy will be needed to raise its temperature. In this case, the amount of energy needed to unfreeze the pipe should remain constant, and since the specific heat of water is higher, it is more effective at passing on higher amounts of energy than air. Are either of these reasons correct or am I really misunderstanding something? I appreciate any help or illumination you can offer! Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16267" LastActivityDate="2015-05-16T11:23:07.390" Title="Is hot air or hot water more effective at melting a frozen pipe?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31500" PostTypeId="2" ParentId="31496" CreationDate="2015-05-15T23:19:48.937" Score="1" Body="&lt;p&gt;How electrospray works has been hotly debated for decades.  Two competing theories have been put forward.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Initial stages: theories agree on droplet collapse due to Rayleigh instability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The two theories agree on the initial stages of droplet formation.  An aerosol is formed by the extrusion of low (&quot;nano&quot;, or 25 to 100 nL / min) to high (~1 mL / min) flows of eluent through a capillary charged to ~3000 V.  The volatility of the solvent used in electrospray techniques ensure that the aerosol droplets evaporate.  Charged spheres of liquid experience a self-repulsion due to electrostatic effects.  This repulsion is counterbalanced by surface tension.  As &lt;a href=&quot;http://www.tandfonline.com/doi/pdf/10.1080/14786448208628425&quot; rel=&quot;nofollow&quot;&gt;Lord Rayleigh noted in 1887&lt;/a&gt;, as long as $T &amp;gt; \frac{Q^2}{16 \pi a_0^3}$, where $Q$ is the droplet charge, $T$ is the surface tension, and $a_0$ is the droplet radius, then the droplet is stable.  But as heat evaporates the droplet, $a_0$ shrinks, while $Q$ stays constant and $T$ nearly so.  Thus the droplet eventually becomes unstable, exploding into a mist of finer droplets, each of which has a smaller charge. What happens after that has been the subject of debate.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The two theories for what happens after droplet collapse&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;ION EVAPORATION&lt;/em&gt;.  This theory holds that as droplets shrink, eventually the field strength at the droplet surface is high enough to force remaining solvated ions into the gas phase, an effect called &quot;field desorption&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;CHARGE RESIDUE&lt;/em&gt;.  This model holds that charged analytes remain solvated in droplets through many cycles of droplet evaporation / fission.  Eventually, droplets contain on average only one analyte molecule.  Remaining solvent molecules evaporate, leaving the ion in the gas phase.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Current evidence suggests that most &lt;strong&gt;small molecules ionize primarily by ion evaporation&lt;/strong&gt;, but that &lt;strong&gt;for large molecules such as proteins, charge residue mechanisms may be important&lt;/strong&gt;.  In the ion evaporation mechanism, molecules or ions must translocate to the surface of a droplet before they can become gas phase ions.  This  means that molecules which preferentially segregate toward the surface of liquid droplets will ionize more easily.  Thus, the ion evaporation mechanism, but not the charge residue mechanism, predicts that surfactants will have higher ionization efficiencies than non-surface active agents.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An extension of the ion evaporation model&lt;/h2&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;em&gt;EQUILIBRIUM PARTITIONING MODEL&lt;/em&gt;.  This theory divides droplets into an electrically neutral core, where ion pairing maintains local electroneutrality, and a charged shell, from which small molecule ions can move into the gas phase by ion evaporation.  The amount of &quot;excess&quot; charge in the shell available for ionizing neutral molecules from the core is determined by electrospray flow parameters, in particular by the ratio of electrospray current to flow rate, or $[Q] = \frac{4.2}{F} \sqrt{\frac{K \gamma }{V_f \kappa_e}}$, where $F$ is Faraday's constant, $K$ is the conductivity of the medium, $\gamma$ is surface tension, $\kappa_e$ is the dielectric constant,  and $V_f$ is the volumetric flow rate. Conditions where excess charge is limiting could cause severe ion suppression.  For more discussion, see &lt;a href=&quot;http://pubs.acs.org/doi/ipdf/10.1021/ac303443q&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Conclusion:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This answer is very long which should be a clue that this topic is complex.  Usually, saying that things are complex is a sign that people don't understand something very well, and that is the case here.  But in general, perhaps the closest to a consensus in the community seems to be that &lt;strong&gt;small molecules ionize primarily by ion evaporation&lt;/strong&gt;, but that &lt;strong&gt;for large molecules such as proteins, charge residue mechanisms may be important&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-05-15T23:19:48.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31501" PostTypeId="2" ParentId="31493" CreationDate="2015-05-16T00:11:02.153" Score="1" Body="&lt;p&gt;Yes, mixing lime sulfur (calcium polysulfide) with an acid will react via the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{S^{2-}\:(aq) +2H+\:(aq) -&amp;gt;H2S\:(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is both extremely toxic and extremely flammable. You should always be extremely cautious and mindful when mixing any chemicals and always be sure to label everything!&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-16T00:11:02.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31502" PostTypeId="2" ParentId="31458" CreationDate="2015-05-16T00:14:35.343" Score="14" Body="&lt;p&gt;You're trying to define the stronger explosive by comparing the reagents, but in this case comparing the products is far more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trinitrotoluene is a rather oxygen-poor explosive, to the point that when detonated as a pure substance, it produces carbon soot and hydrogen gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C7H5N3O6 -&amp;gt; 3N2 + 5H2 + 12CO + 2C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The production of $\ce{CO}$, $\ce{CO2}$ and $\ce{H2O}$ is highly exothermic, so a lot of extra energy could be obtained if there were more oxygen to burn the soot and hydrogen. However, in detonations there's not nearly enough time for atmospheric oxygen to partake extensively in the reaction. This means that a fair amount of the explosive potential of TNT is wasted, decreasing its detonation yield per gram of substance. One way to compensate this yield loss in oxygen-poor explosives is to mix them with an oxidiser, usually an oxygen-rich substance such as $\ce{KClO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to picric acid. The structure and molar mass are very similar to TNT, but replacing the methyl group with a hydroxyl means the oxygen content of picric acid is higher. While still overall oxygen-poor, the detonation proceeds with much less soot and hydrogen generation, meaning a more complete burn, a higher energy release, and consequently a greater explosive yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2C6H3N3O7 -&amp;gt; 3N2 + H2 + 2H2O + 12 CO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the decomposition reactions are more complex. The balanced equations above follow a rule of thumb presented &lt;a href=&quot;http://fas.org/man/dod-101/navy/docs/es310/chemstry/chemstry.htm&quot;&gt;here&lt;/a&gt;. However, a more thorough analysis should agree with this simplified result.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-05-16T00:14:35.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31503" PostTypeId="2" ParentId="31499" CreationDate="2015-05-16T00:16:43.533" Score="5" Body="&lt;p&gt;Specific heat relates to the heat capacity; how much temperature change you get for a specific mass and energy once the heat has finished ‘moving’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you also have to consider how fast the heat is moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For melting a frozen pipe with a fluid you also need to consider the heat resistivity of the fluid which is the inverse of heat conductivity. Some fluids like air are good heat insulators - much more so than water. The resistivity will limit the rate at which energy can be transferred to the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water has a much lower resistivity than air ($\sim 1.7\ \mathrm{m\cdot K/W}$ vs $\sim 50\ \mathrm{m\cdot K/W}$) you can expect a heat transfer rate into the cold pipe about 25 times faster. This all assumes the air and water are at the same temperature.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorDisplayName="user7951" LastEditDate="2015-05-16T11:23:07.390" LastActivityDate="2015-05-16T11:23:07.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31504" PostTypeId="1" AcceptedAnswerId="31505" CreationDate="2015-05-16T01:28:09.320" Score="2" ViewCount="1042" Body="&lt;p&gt;I have seen some heating apparatus for a crucible use a retort stand with a large metal ring attached to the boss head to support the clay triangle as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/781kB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instinctively, I would have placed the clay triangle on a tripod, however, I see that this has certain advantages, such as being able to adjust the height of the crucible so that it is in the hottest part of the flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What is the name of this piece of laboratory equipment, I have seen it sometimes referred to as an iron ring, but is there a more formal name?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="5017" LastEditDate="2015-05-16T01:39:56.773" LastActivityDate="2015-05-16T01:52:23.910" Title="Crucible heating apparatus" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31505" PostTypeId="2" ParentId="31504" CreationDate="2015-05-16T01:43:54.940" Score="4" Body="&lt;p&gt;&lt;strong&gt;Iron ring&lt;/strong&gt; is a correct term often used for this piece of equipment, according to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Iron_ring_(laboratory)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;, also &lt;a href=&quot;http://www.mreisley.com/tutorials/equipment/supports.html&quot; rel=&quot;nofollow&quot;&gt;this lab equipment tutorial page&lt;/a&gt; and in a very comprehensive &lt;a href=&quot;http://avogadro.chem.iastate.edu/labmanual/documents/pictorial_guide-163.pdf&quot; rel=&quot;nofollow&quot;&gt;pictorial lab equipment guide&lt;/a&gt; from Iowa State University.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, it is also called an &lt;strong&gt;iron support ring&lt;/strong&gt; and a &lt;a href=&quot;http://www.haines.com.au/index.php/lab-equipment/retort-ring-75mm-d-with-bosshead.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;retort ring&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-16T01:52:23.910" LastActivityDate="2015-05-16T01:52:23.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31506" PostTypeId="2" ParentId="31497" CreationDate="2015-05-16T01:48:42.297" Score="7" Body="&lt;p&gt;After s, p, d, and f, the next letters follow in alphabetical sequence: g, h, i, k, l, and so on. Note that j is omitted from the sequence: see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/68251/why-is-the-letter-j-omitted-in-the-spdf-sequence&quot;&gt;Why is the letter J omitted in the spdf... sequence?&lt;/a&gt; for more information. Since g-subshells would have $l = 4$, the lowest g-subshell would be the $\mathrm{5g}$ subshell, bearing in mind $l$ must be strictly less than $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the aufbau principle, orbitals are filled in increasing order of $(n+l)$. Amongst orbitals with similar $(n+l)$, those with smaller $n$ are filled first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xA;\text{Orbital} &amp;amp; \hline (n+l) &amp;amp; n \\ \hline&#xA;\mathrm{1s} &amp;amp; 1 &amp;amp; 1 \\ \hline&#xA;\mathrm{2s} &amp;amp; 2 &amp;amp; 2 \\ \hline&#xA;\mathrm{2p} &amp;amp; 3 &amp;amp; 2 \\&#xA;\mathrm{3s} &amp;amp; 3 &amp;amp; 3 \\ \hline&#xA;\mathrm{3p} &amp;amp; 4 &amp;amp; 3 \\&#xA;\mathrm{4s} &amp;amp; 4 &amp;amp; 4 \\ \hline&#xA;\mathrm{3d} &amp;amp; 5 &amp;amp; 3 \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this principle you would theoretically have the order&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\underbrace{1s}_{\text{Period 1}} \to \underbrace{2s \to 2p}_{\text{Period 2}} \to \underbrace{3s \to 3p}_{\text{Period 3}} \to \underbrace{4s \to 3d \to 4p}_{\text{Period 4}} \to \underbrace{5s \to 4d \to 5p}_{\text{Period 5}} \to \underbrace{6s \to 4f \to 5d \to 6p}_{\text{Period 6}}} \\&#xA;\mathrm{\to \underbrace{7s \to 5f \to 6d \to 7p}_{\text{Period 7}} \to \underbrace{8s \to \mathbf{5g} \to 6f \to 7d \to 8p}_{\text{Period 8}} \to 9s \to \cdots}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at such large atomic numbers, the aufbau principle starts to fail. The subshells are much more closely spaced in energy and the exact electronic configurations are not so clear-cut: this is already evident within the d- and f-blocks where some anomalies are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this, relativistic effects (which become significant at large atomic numbers) cause subshells to split into two distinct energy levels (except for the s-subshell). See also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/58135/16683&quot;&gt;Shape of the P1/2 Orbital&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16269" LastEditorUserId="16683" LastEditDate="2018-09-20T13:45:09.997" LastActivityDate="2018-09-20T13:45:09.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31507" PostTypeId="1" AcceptedAnswerId="31510" CreationDate="2015-05-16T01:49:28.830" Score="8" ViewCount="17497" Body="&lt;p&gt;&lt;strong&gt;Do organic solvents such as pure ethanol and pure methanol conduct electricity? why is this so?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done some research and have not been able to find anything conclusive. I would think it would not conduct because I cannot see any medium for the passage of charge.&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorDisplayName="user15489" LastEditDate="2015-05-16T02:59:02.873" LastActivityDate="2020-03-19T01:47:50.200" Title="Electrical conductivity of organic solvents" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31508" PostTypeId="1" CreationDate="2015-05-16T02:06:09.137" Score="4" ViewCount="7716" Body="&lt;p&gt;Is this a combination of Boyle's and Charles' Laws or simply one of them? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused as to which applies. I don't feel I fully understand the implications of both laws individually and I'm wondering why it's easy to take a gas like butane and liquefy it under pressure at &quot;normal&quot; temperatures and why other gases need to be cooled to extremely low temperatures or very high pressures to liquefy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that temperature, pressure and volume are all related, I just don't know which rules apply and how and how they all interact. How do these laws apply to changes of state from gas to liquid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this is stuff I forgot from 10th grade but I'd love to hear an explanation again. &lt;/p&gt;&#xA;" OwnerUserId="16270" LastEditorDisplayName="user15489" LastEditDate="2015-05-16T04:05:37.687" LastActivityDate="2015-05-16T16:38:12.947" Title="Why do some gases liquefy easily under pressure and normal room temperature?" Tags="&lt;everyday-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31510" PostTypeId="2" ParentId="31507" CreationDate="2015-05-16T04:42:36.670" Score="8" Body="&lt;p&gt;Ethanol and methanol dissociate into ions, just as water does.  For water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 H2O &amp;lt;=&amp;gt; H3O+ + OH-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an alcohol &lt;span class=&quot;math-container&quot;&gt;$\ce{ROH}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2 ROH &amp;lt;=&amp;gt; ROH2+ + ^-OR}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For water the &lt;a href=&quot;http://en.wikipedia.org/wiki/Self-ionization_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;self-ionization constant&lt;/a&gt; is &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{w} \approx 10^{-14}$&lt;/span&gt;.  For ethanol or methanol, I was unable to find data on the self-ionization constant, which is essentially the same as the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of the alcohol in a solution of itself.  The &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of ethanol &lt;em&gt;in water&lt;/em&gt; is slightly higher than the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; of water in water.  Also, the lower dielectric constant in alcohols will further disfavor self-ionization relative to water.  So for methanol and ethanol the self-dissociation into ions will be weaker than in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These ions will still be present, though, and will conduct electricity even in pure ethanol and methanol solutions, but the conductivities will be &lt;em&gt;very&lt;/em&gt; low.  Very pure water has a resistivity of &lt;span class=&quot;math-container&quot;&gt;$\pu{18.2 MΩ cm}$&lt;/span&gt;.  If they self-ionize to a lesser extent than water does, the resistivities of methanol and ethanol will be even higher than water's.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="41328" LastEditDate="2020-03-18T11:26:15.063" LastActivityDate="2020-03-18T11:26:15.063" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31511" PostTypeId="2" ParentId="31508" CreationDate="2015-05-16T04:43:07.373" Score="5" Body="&lt;p&gt;The gas laws you describe, Boyle's and Charles' Laws, can be combined into one relationship known as the &lt;strong&gt;Ideal Gas Law&lt;/strong&gt;. According to this law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV=nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ is the pressure of the gas ($\ce{Pa}$), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ is the volume of the gas ($\ce{m^3}$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ is the number of moles or amount of gas ($\ce{mol}$),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ is a constant ($R\sim 8.31 \ce{\frac{J}{mol.K}}$), and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ is the temperature of the gas ($\ce{K}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how changing one property affects another property of the sample of gas, hold the other values constant and see what must happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, according to the law, an increase in Pressure will result in a decrease in Volume, if Temperature is held constant. &#xA;But if the Volume is instead held constant, then the Temperature will increase instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In effect, the ideal gas law is the combination of the other laws affecting a gas and its properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But this does not explain why some substances are liquids at room temperature while others are not and require cooling and/or a change of pressure.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe this, an understanding of what are known as &lt;em&gt;Intermolecular forces&lt;/em&gt; must be expressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between molecules of a given substance, there exist attractions. These attractions vary in strength and are what determine the pressures and temperatures needed to liquify specific chemicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (usually) weakest kind of attractions are known as &lt;strong&gt;London Dispersion Forces.&lt;/strong&gt; These forces are generated randomly due to the random movement of electrons about a chemical. Since the movement is random, there come times where the electron density is more negative on one side of a molecule than another. This results in a very slight charge, which can induce charges onto neighboring molecules and make them attract each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next class of attractions are known as &lt;strong&gt;Dipole-Dipole Interactions.&lt;/strong&gt; This is a result of a permanent charge density difference across a molecule. In these molecules, one or more atoms are much greater in their affinities for electrons and will draw the density away from the atoms with lesser affinity. This makes for a permeant slight charge on one or more sides of a molecule that are positive, while the other sides become negative. The negative ends of one molecule will attract the positive ends of neighboring molecules, resulting in a usually pretty strong attractive force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strongest by far class of interactions are known as &lt;strong&gt;Hydrogen Bonding.&lt;/strong&gt; This is when hydrogen is bonded to en extremely electronegative atom, like oxygen or fluorine. Take water for example. The Oxygen atom greatly skews the electron density, making the oxygen end much more negative in charge than the positive hydrogen ends. This makes for very strong intermolecular attractions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these forces understood, you can now see that chemicals vary in the strength of their intermolecular attractions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, &lt;strong&gt;the weaker the intermolecular attractions, the easier it is for the molecules to break their attractions and become gaseous.&lt;/strong&gt; Such chemicals with weak attractions require so little energy to become gaseous that at room temperature and standard pressure, the molecules act in the gas phase. To make these chemicals turn into liquids, one must either increase the pressure to make the formation of attractions more favorable or lower the temperature to lessen the energy of the molecules so that they don't break their attractive forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some chemicals have extremely weak attractive forces, meaning that the temperature will have to be very low or the pressure very high to force the chemical into its liquid state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other chemicals, like water, exhibit such strong forces that the temperature has to be raised or the pressure lowered to vaporize them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the gas properties of pressure, temperature, and volume do affect changes of state, but its the strength of the intermolecular attractions that are the most important and are the reasons that chemicals behave so differently under the same conditions.&lt;/p&gt;&#xA;" OwnerUserId="15236" LastEditorUserId="15236" LastEditDate="2015-05-16T16:38:12.947" LastActivityDate="2015-05-16T16:38:12.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31512" PostTypeId="1" AcceptedAnswerId="31527" CreationDate="2015-05-16T04:51:45.520" Score="12" ViewCount="566" Body="&lt;p&gt;Recently it came in the &lt;a href=&quot;https://encrypted.google.com/search?q=jahn+teller+matter&amp;amp;source=lnms&amp;amp;tbm=nws&amp;amp;sa=X&amp;amp;ei=y8pWVeziGIPs8AXgz4GAAw&amp;amp;ved=0CAgQ_AUoAg&amp;amp;biw=1366&amp;amp;bih=643#tbm=nws&amp;amp;q=jahn+teller+metal&quot;&gt;news&lt;/a&gt; that scientists have discovered a new state of matter called &lt;strong&gt;Jahn-Teller metal&lt;/strong&gt;. I tried reading up on it but couldn't understand it clearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain in simple terms why this is a new state of matter and/or what's so special about it.&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="15208" LastEditDate="2015-05-25T07:34:25.200" LastActivityDate="2015-05-25T07:34:25.200" Title="Jahn-Teller Metal : The new state of matter" Tags="&lt;metal&gt;&lt;materials&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31515" PostTypeId="1" AcceptedAnswerId="31516" CreationDate="2015-05-16T05:58:48.097" Score="4" ViewCount="2394" Body="&lt;p&gt;Recently my teacher asked us which of the following has the largest number of molecules? He gave us two options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{36 g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{54 g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2O5}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm stuck, because both of them are having two moles of each of the respective molecules. And technically, the number of molecules equals the amount of substance multiplied by Avogadro's number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So shouldn't &lt;span class=&quot;math-container&quot;&gt;$\pu{2 mol} \times N_\mathrm{A} = N(\text{molecules})$&lt;/span&gt; be the same number of molecules for both of them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yet, the teacher said that water is the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm still not able to figure out why?&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="4945" LastEditDate="2020-01-28T18:19:36.353" LastActivityDate="2020-01-28T18:24:04.307" Title="Is the largest number of molecules in 36 g of water or 54 g of dinitrogen pentaoxide?" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31516" PostTypeId="2" ParentId="31515" CreationDate="2015-05-16T06:13:17.883" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm stuck, because both of them are having two moles of each of the respective molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Are you sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;M(\ce{H2O})  &amp;amp;= \pu{18 g mol^-1} &amp;amp;&#xA;\implies \pu{36 g} &amp;amp;\mathop{\hat{=}} \pu{2 mol} \\&#xA;M(\ce{N2O5}) &amp;amp;= \pu{108 g mol^-1} &amp;amp;&#xA;\implies \pu{54 g} &amp;amp;\mathop{\hat{=}} \pu{0.5 mol} \\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2020-01-28T18:24:04.307" LastActivityDate="2020-01-28T18:24:04.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31517" PostTypeId="1" CreationDate="2015-05-16T07:23:21.440" Score="1" ViewCount="1159" Body="&lt;p&gt;From my understanding, the general rule for solubility is that polar molecules are soluble in polar liquids (e.g. water), and non polar substances are soluble in non polar liquids (organic solvents).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that when polar molecules are dissolved in polar liquids, the opposite dipoles are attracted to each other and hence the substance dissolves. For this reason, a polar liquid can't dissolve a non polar substance because there are no dipoles to attract, so it would just stay suspended in the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: &lt;strong&gt;if both the solute and the solvent are non polar, how does the solute dissolve without electrostatic attraction? would this not also just stay suspended in the liquid?&lt;/strong&gt;   &lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorDisplayName="user15489" LastEditDate="2015-05-16T07:33:14.967" LastActivityDate="2015-05-16T07:33:14.967" Title="Solubility of non polar substances in non polar solutions" Tags="&lt;solubility&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2017-09-29T20:49:57.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="31520" PostTypeId="1" AcceptedAnswerId="31557" CreationDate="2015-05-16T10:31:31.950" Score="11" ViewCount="335" Body="&lt;p&gt;Most packaged goods have the tiny &lt;strong&gt;&lt;em&gt;Silica Gel Packets&lt;/em&gt;&lt;/strong&gt; packed along with a clear warning on every single packet.&lt;p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/pDpa8.jpg&quot; alt=&quot;Silica Gel Packet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;What hazards of desiccant silica gel are they concerned about that it is necessary to post the warning &quot;THROW AWAY DO NOT EAT&quot; on the packaging?&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="23561" LastEditDate="2019-03-19T01:56:47.787" LastActivityDate="2019-03-19T01:56:47.787" Title="What would happen if someone consumed a Silica Gel Packet?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-05-19T02:09:16.247" ContentLicense="CC BY-SA 4.0" />
  <row Id="31521" PostTypeId="2" ParentId="28006" CreationDate="2015-05-16T10:46:16.710" Score="9" Body="&lt;p&gt;Your idea appears to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ &amp;lt;=&amp;gt; $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ &amp;lt;=&amp;gt; $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ &amp;lt;=&amp;gt; $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the actual complexes are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[CrCl3(H2O)3]}$ (dark green)&lt;br&gt;&#xA;$\ce{[CrCl2(H2O)4]+}$ (dark green)&lt;br&gt;&#xA;$\ce{[CrCl(H2O)5]^2+}$ (blue green)&lt;br&gt;&#xA;$\ce{[Cr(H2O)6]^3+}$ (violet)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;&amp;amp;\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&amp;amp;&amp;amp;\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\&#xA;&amp;amp;\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&amp;amp;&amp;amp;\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\&#xA;&amp;amp;\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&amp;amp;&amp;amp;\tilde\nu_3=38\,600\ \mathrm{cm^{-1}}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;&amp;amp;\ce{[Cr(H2O)6]^3+}\quad&amp;amp;&amp;amp;\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\&#xA;&amp;amp;\ce{[CrCl(H2O)5]^2+}\quad&amp;amp;&amp;amp;\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\&#xA;&amp;amp;\ce{[CrCl2(H2O)4]+}\quad&amp;amp;&amp;amp;\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\&#xA;&amp;amp;\vdots \\&#xA;&amp;amp;\ce{[CrCl6]^3-}\quad&amp;amp;&amp;amp;\tilde\nu_1=13\,200\ \mathrm{cm^{-1}}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavenumbers can be converted to the wavelength of the absorbed light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;&amp;amp;\ce{[Cr(H2O)6]^3+}\quad&amp;amp;&amp;amp;\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\&#xA;&amp;amp;\ce{[CrCl(H2O)5]^2+}\quad&amp;amp;&amp;amp;\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\&#xA;&amp;amp;\ce{[CrCl2(H2O)4]+}\quad&amp;amp;&amp;amp;\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-03-20T12:14:00.913" LastActivityDate="2016-03-20T12:14:00.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31524" PostTypeId="1" CreationDate="2015-05-16T11:29:58.133" Score="0" ViewCount="7654" Body="&lt;p&gt;According to the theory of London dispersion forces, intermolecular forces are created in non polar substances by induction of instantaneous dipole to neighbouring molecules. If that is the case then when we mix water and oil, water being polar should induce polarity in oil and they should mix. But they don't. Doesn't water induce polarity to oil? Do I have a false concept or something.&lt;/p&gt;&#xA;" OwnerUserId="15174" LastEditorUserId="15174" LastEditDate="2015-05-16T13:10:14.973" LastActivityDate="2015-05-16T14:12:04.447" Title="Mixing polar and non polar substances" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-05-16T15:52:23.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="31526" PostTypeId="1" CreationDate="2015-05-16T12:26:51.877" Score="6" ViewCount="163" Body="&lt;p&gt;$\ce{V(CO)6}$ is well known to be an unusual 17 valence electron, paramagnetic octahedral complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the corresponding tantalum and niobium species form simple dimers or more complicated structures ? &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="4945" LastEditDate="2015-05-16T16:08:56.513" LastActivityDate="2015-05-16T16:08:56.513" Title="Structure of Niobium and Tantalum carbonyls" Tags="&lt;molecular-structure&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31527" PostTypeId="2" ParentId="31512" CreationDate="2015-05-16T12:52:02.273" Score="7" Body="&lt;p&gt;The Jahn-Teller metal is not one of the 3 (or 4 including plasma) states that occur naturally on Earth; this state allows a metal to have the properties of an insulator and a superconductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYDHN.png&quot; alt=&quot;An electronic phase diagram illustrating the crossover between the Jahn-Teller metal to conventional metals.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is an intermediate state of matter confirming the Jahn-Teller distortion; by interacting with certain electrons (a more detailed description can be found in &lt;a href=&quot;http://advances.sciencemag.org/content/1/3/e1500059/&quot; rel=&quot;noreferrer&quot;&gt;this Science Advances magazine article&lt;/a&gt;, a distorted crystalline structure allows a solid depicting characteristics of an insulator to also becomes a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main breakthrough this brings is having a superconductors. Superconductors are extremely useful because they possess zero resistance, except the conditions require them to be cooled to extreme temperatures.&#xA;This state nonetheless suggests reproducing superconductors at a less extreme temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making it an economic way to make efficient electronics.&lt;/p&gt;&#xA;" OwnerUserId="16210" LastActivityDate="2015-05-16T12:52:02.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31529" PostTypeId="1" AcceptedAnswerId="31541" CreationDate="2015-05-16T13:49:29.653" Score="0" ViewCount="1234" Body="&lt;p&gt;My lecture handout proves derives an expression for entropy thus:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a perfect gas $U=U(T)$ (i.e. is not dependent on volume; only temperature). Therefore, $(\frac{\partial U}{\partial V})_T=0$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Therefore: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Equation 1) $$dU=(\frac{\partial U}{\partial V})_TdV+(\frac{\partial U}{\partial T})_VdT=C_VdT$$ Since $(\frac{\partial U}{\partial T})_V=C_V$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, for a reversible change: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Equation 2) $$dU=dq_{rev}+dw_{rev}=dq_{rev}-PdV$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Equation 3) $$dq_{rev}=PdV+dU=\frac{nRT}VdV+C_VdT$$&#xA;  Then divide by $T$ to make the equation equal to entropy rather than heat:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$dS=\frac{nR}{V}dV+C_VdT$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a couple of issues with this. Mainly stemming from the fact that $U=U(T)$ and therefore that $(\frac{\partial U}{\partial V})_T=0$. I understand it on a physical level why this is the case (no intermolecular forces). I am not asking for a proof of this; I just have an issue with the implications of this. Let me explain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equation for dU is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU=TdS-PdV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is understandable. $TdS$ accounts for heat transfer in/out of a system and $PdV$ accounts for work done by/on the system. However, surely this equation implies that $U=U(S,T)$ since both $dS$ and $dT$ appear in the expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of it is: how can $dU=C_VdT$ and $dU=TdS-PdV$. Please help me understand what's going on here.&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-05-16T17:12:53.313" Title="Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31530" PostTypeId="2" ParentId="31524" CreationDate="2015-05-16T14:12:04.447" Score="0" Body="&lt;p&gt;Water is already very 'happy' with his own guys due to strong hydrogen bonds . While van der waals force is so weak that its presence is so little for it to be seen while there is hydrogen bond. So even though there is some dipole induced it isn't enough.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the total energy is lower than when they are mixed. The system is more stable if unmixed. -Chin Yeh&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16269" LastActivityDate="2015-05-16T14:12:04.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31531" PostTypeId="1" AcceptedAnswerId="31532" CreationDate="2015-05-16T14:13:57.350" Score="14" ViewCount="33488" Body="&lt;p&gt;Does phenol undergo Friedel–Crafts reactions or does it react with Lewis acids like aniline does?&lt;/p&gt;&#xA;" OwnerUserId="14521" LastEditorUserId="10185" LastEditDate="2016-05-24T22:08:05.567" LastActivityDate="2020-04-25T11:33:25.363" Title="Friedel–Crafts reaction of phenol" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;reactivity&gt;&lt;phenols&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31532" PostTypeId="2" ParentId="31531" CreationDate="2015-05-16T14:23:00.507" Score="28" Body="&lt;p&gt;Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact most substituents with lone pair would give poor yield.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch24/ch24-3-2.html&quot; rel=&quot;noreferrer&quot;&gt;The two pathways involved in the reaction with phenol reduce the overall yeild:&lt;/a&gt; &#xA;&lt;img src=&quot;https://i.stack.imgur.com/p6ycx.gif&quot; alt=&quot;acylation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenols are examples of bidentate nucleophiles, meaning that they can react at two positions:         &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;on the aromatic ring giving an aryl ketone via C-acylation, a Friedel-Crafts reaction &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;on the phenolic oxygen giving an ester via O-acylation, an esterification&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15912" LastEditorUserId="4325" LastEditDate="2017-07-12T19:34:27.987" LastActivityDate="2017-07-12T19:34:27.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31533" PostTypeId="1" AcceptedAnswerId="31537" CreationDate="2015-05-16T15:33:04.840" Score="6" ViewCount="3796" Body="&lt;p&gt;In the complex ion $\ce{[Na(H2O)6]+}$, the sodium cation forms 6 coordinate bonds with water ligands. Typically this octahedral form is associated with $\mathrm{sp^3d^2}$ hybridization as far as I know, but in the case of sodium the $d$ orbitals aren't readily available, so how can this be described as $\mathrm{sp^3d^2}$ hybridization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that its hybridization is $\mathrm{s^2p^4}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On further thought I think the d orbitals probably are accessible and that it is indeed $\mathrm{sp^3d^2}$ but this has a high hybridisation energy and so we dont see $\ce{[Na(H2O)6]+}$ in large amounts instead we see the ion solvated by partial charges and electrostatic rather than dative covalent interactions&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="15410" LastEditDate="2016-04-20T06:55:52.047" LastActivityDate="2016-04-20T06:55:52.047" Title="Hybridization of Na in [Na(H2O)6]+" Tags="&lt;water&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31534" PostTypeId="2" ParentId="31526" CreationDate="2015-05-16T16:05:28.723" Score="3" Body="&lt;p&gt;From the abstract of &lt;em&gt;J. Chem. Theory Comput.&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &#xA;&lt;em&gt;7&lt;/em&gt;, 2112-2125 (&lt;a href=&quot;http://dx.doi.org/10.1021/ct2003513&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;) is seems that niobium forms a dinuclear $\ce{Nb2(CO)12}$ complex, in which two $\ce{Nb(CO)6}$ units are connected by a 340 pm $\ce{Nb-Nb}$ bond.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-16T16:05:28.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31535" PostTypeId="2" ParentId="31471" CreationDate="2015-05-16T16:12:45.117" Score="7" Body="&lt;p&gt;Another characteristic of fresh ice is the supercooled vapor leaving the surface (sublimation.) This process supercools the surface of the ice keeping it well below the freezing point.  When a warm object such as a towel or your skin condenses the vapor creating a thin layer of water on its surface which then freezes immediately on contact with the supercooled surface of the ice.  The surface of the ice quickly warms at ambient temperature and the effect is lost.&lt;/p&gt;&#xA;" OwnerUserId="16280" LastActivityDate="2015-05-16T16:12:45.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31536" PostTypeId="1" AcceptedAnswerId="31538" CreationDate="2015-05-16T16:17:40.140" Score="2" ViewCount="481" Body="&lt;p&gt;I am asked to explain why neither $\ce{E2}$ or $\ce{S_{N}}2$ products are formed when 3-chloro-2,2,4,4-tetramethylpentane is treated with NaOH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that since the reagent is a strong nucleophile &amp;amp; a strong base, it can only participate in either an $\ce{E2}$ reaction or in a combination of the $\ce{E2}$ and $\ce{{S_{N}}2}$ reactions, depending on the substrate. I think the substrate is secondary, so $\ce{E2}$ should predominate and $\ce{S_{N}2}$ should give a minor product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook solution says that since the substrate is tertiary, $\ce{{S_{N}}2}$ will not occur at an appreciable rate, and there are no &lt;em&gt;β&lt;/em&gt; carbons, $\ce{E2}$ cannot occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My misunderstanding seems to stem from an inability to depict the molecule as is expected, so I am hoping someone can clarify whether or not my drawing and line of reasoning are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wbjx7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16129" LastEditorUserId="16129" LastEditDate="2015-05-16T16:49:57.563" LastActivityDate="2015-05-16T16:52:08.103" Title="Predicting the Product of a Substitution/Elimination Reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31537" PostTypeId="2" ParentId="31533" CreationDate="2015-05-16T16:24:01.653" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Solvation_shell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solvation&lt;/a&gt; is the process by which a species is dissolved in a solvent. The most classic example of this is when a metal ion is dissolved in water. The electronegative oxygens in the water molecules are attracted electrostatically to the positive charge on the metal ion. A &lt;strong&gt;solvation shell&lt;/strong&gt; of water molecules result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3250073/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, which written about the hydration of alkali metal ions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In spite of many conducted studies the knowledge of the structures and bonding properties of the hydrated alkali metal ions in aqueous solution is scarce and deviating.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also according to another &lt;a href=&quot;http://www.researchgate.net/profile/Kwang_Kim3/publication/232612703_Entropy-driven_Structures_of_the_Hexa-Aqua-Sodium%28I%29/links/09e4150879f1245532000000.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;letter&lt;/a&gt;, which reports research done on the stability of the different possible forms of the ion, there is not a consensus on only one form. This letter discusses three possible forms of the ion all in the form of $n_1 +n_2$, where $n_1$ is the number of molecules in the first shell of hydration and $n_2$ is the number of molecules in the second shell of hydration. The forms discussed are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WVZ4Z.png&quot; alt=&quot;shells of hydration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;demonstrating that the water molecules do not &lt;em&gt;necessarily&lt;/em&gt; have to be coordinately bonded to the metal itself, but in fact that the second shell of hydration could include water molecules interacting with the water molecules in the first shell of hydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short the hydration of sodium (and the other alakali metals) is not well understood yet, though it is under study.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-16T16:24:01.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31538" PostTypeId="2" ParentId="31536" CreationDate="2015-05-16T16:36:39.560" Score="3" Body="&lt;p&gt;Remember that in a $S_{N}2$ reaction, the centre bearing the $\ce{Cl}$ atom would have to be attacked by $\ce{OH-}$ from the &lt;strong&gt;back side&lt;/strong&gt;. This, however, is prevented by the by two bulky tert.-butyl substituents &#xA;&lt;img src=&quot;https://i.stack.imgur.com/tV7cV.png&quot; alt=&quot;tetramethylchloropentane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the alternative $E2$ pathway is concerned: There are no $\alpha$ hydrogen atoms that could be abstracted by $\ce{OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've uploaded a &lt;a href=&quot;http://pastebin.com/13ej0bEG&quot; rel=&quot;nofollow noreferrer&quot;&gt;CML&lt;/a&gt; file of the structure to pastebin. I suggest to get yourself a copy of &lt;a href=&quot;http://avogadro.cc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt; and inspect the geometry on your own PC.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-16T16:52:08.103" LastActivityDate="2015-05-16T16:52:08.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31539" PostTypeId="1" AcceptedAnswerId="31547" CreationDate="2015-05-16T16:50:09.210" Score="9" ViewCount="1696" Body="&lt;p&gt;I've been seeing a lot of &lt;strong&gt;GAMESS&lt;/strong&gt; and &lt;strong&gt;Gaussian&lt;/strong&gt; questions around here, and most of them seem very hard to answer. I extrapolated that they might be some sort of Chemistry programs, but what do they do exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of looking it up, I feel like most of you would be more knowledgeable in this.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2017-12-06T03:35:46.633" LastActivityDate="2017-12-06T03:35:46.633" Title="What are GAMESS and Gaussian?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31540" PostTypeId="2" ParentId="4455" CreationDate="2015-05-16T17:05:33.770" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/physical/energetics/lattice.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;ChemGuide&quot;&gt;ChemGuide&lt;/a&gt; says that 'The standard atomisation enthalpy is the enthalpy change when 1 mole of gaseous atoms is &lt;strong&gt;formed&lt;/strong&gt; from the element in its standard state.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&#xA;\begin{array}{rc}&#xA;\ce{1/2 Cl2 (g) -&amp;gt; Cl (g)} &amp;amp; \Delta H_{a}^{0} = +122~\mathrm{kJ/mol} \\&#xA;\ce{1/2 Br2 (l) -&amp;gt; Br (g)} &amp;amp; \Delta H_{a}^{0} = +112~\mathrm{kJ/mol} \\&#xA;\ce{Na (s) -&amp;gt; Na (g)} &amp;amp; \Delta H_{a}^{0} = +107~\mathrm{kJ/mol}&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="16282" LastEditorUserId="194" LastEditDate="2018-03-26T19:07:06.683" LastActivityDate="2018-03-26T19:07:06.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31541" PostTypeId="2" ParentId="31529" CreationDate="2015-05-16T17:12:53.313" Score="1" Body="&lt;p&gt;Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases.  It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$.  This is because $S$ and $V$ are &lt;a href=&quot;https://physics.stackexchange.com/questions/158182/what-makes-a-thermodynamic-canonical-variable-canonical&quot;&gt;canonical variables&lt;/a&gt; for $U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's apply the ideal gas equation $dU=C_V dT$.  You can integrate this to yield $U=C_V T + K$.  Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$.  So how do we proceed?  This &lt;a href=&quot;https://physics.stackexchange.com/questions/158182/what-makes-a-thermodynamic-canonical-variable-canonical&quot;&gt;physics.se&lt;/a&gt; question has a good general answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in this specific case things are bit easier.  Let's find $dS$ from the equation you were given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C_V dT + \frac{RT}{V}dV)=C_V \frac{dT}{T}  + R \frac{dV}{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-05-16T17:12:53.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31542" PostTypeId="2" ParentId="18854" CreationDate="2015-05-16T17:13:26.423" Score="1" Body="&lt;p&gt;Oxidise it using a mild oxidizing agent like $\ce{H2O2}$ (hydrogen peroxide). When it undergoes oxidation coloured compounds get decolorized. Good luck on cleaning it.&lt;/p&gt;&#xA;" OwnerUserId="16269" LastEditorDisplayName="user7951" LastEditDate="2015-05-16T19:02:43.173" LastActivityDate="2015-05-16T19:02:43.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31543" PostTypeId="2" ParentId="28857" CreationDate="2015-05-16T17:33:34.307" Score="0" Body="&lt;p&gt;Need not be for a salt to be soluble the $\Delta_{hydration}H≥\Delta_{lattice}H$ so it must break the lattice.&lt;br&gt;&#xA;[see this][&lt;a href=&quot;http://www.britannica.com/EBchecked/topic/156605/deliquescence]&quot; rel=&quot;nofollow&quot;&gt;http://www.britannica.com/EBchecked/topic/156605/deliquescence]&lt;/a&gt;&#xA;Given there is that &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;deliquescent salts need not be soluble but all soluble salts are deliquescent&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16269" LastActivityDate="2015-05-16T17:33:34.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31545" PostTypeId="2" ParentId="6919" CreationDate="2015-05-16T18:01:13.450" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://evans.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow&quot;&gt;Evans'&lt;/a&gt; and &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/pkatable/index.htm&quot; rel=&quot;nofollow&quot;&gt;Bordwell&lt;/a&gt; $pK_a$ tables provide an excellent source of data for water and DMSO based measurements, particularly for organic compounds. &lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-16T18:01:13.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31547" PostTypeId="2" ParentId="31539" CreationDate="2015-05-16T18:26:39.997" Score="15" Body="&lt;p&gt;There exist a physical theory called &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_mechanics&quot;&gt;&lt;em&gt;quantum mechanics&lt;/em&gt;&lt;/a&gt; which describes the Universe at atomic and subatomic length scales. Chemical systems, such as, for instance, molecules, are one types of systems one encounter at this level, and as such, they can in principle be completely described by quantum mechanics. What is meant by a &quot;complete description&quot; is that all the properties of a system (say, a molecule) can be theoretically predicted by the application of the principles of quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the direct application of the principles of quantum mechanics to chemical systems is practically impossible in the absolute majority of cases due to the complexity of the resulting description. To simplify matters a series of approximation has to be introduced. But even the approximate description is usually too complex to be handled by a human being analytically (on a piece of paper), and thus, numerical methods has to be developed to shift this burden to our digital mates (computers). All these questions (approximations plus numerical treatments) are addressed by a branch of chemistry known as &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_chemistry&quot;&gt;quantum chemistry&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GAMESS&lt;sup&gt;[1]&lt;/sup&gt; and Gaussian are indeed programs; programs that implement different methods of quantum chemistry. And just for your information &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid-state_physics_software&quot;&gt;there are a lot of other such programs&lt;/a&gt;. These programs allows us to calculate the properties of molecules, to predict their behaviour in different environments, etc. And all this can be done theoretically using computers without actually synthesising a molecule in question and then doing some real experiments on it. To understand what these programs do exactly, you have to spent a lot of years by learning some advanced mathematics, quantum physics, quantum chemistry, programming, numerical methods, ... At this point I can only tell you that they are doing some magic. The quantum magic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] There are in fact 3 different programs: PC GAMESS (now Firefly), GAMESS (UK), and GAMESS (US). &lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-16T18:32:51.333" LastActivityDate="2015-05-16T18:32:51.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31548" PostTypeId="2" ParentId="31531" CreationDate="2015-05-16T18:46:21.213" Score="14" Body="&lt;p&gt;To add to @user223679's answer. Phenol can react via two pathways with acyl chlorides to give either esters, via &lt;em&gt;O&lt;/em&gt;-acylation, or hydroxyarylketones, via &lt;em&gt;C&lt;/em&gt;-acylation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, phenol esters also undergo a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fries_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fries rearrangement&lt;/a&gt; under Friedel-Crafts conditions to produce the &lt;em&gt;C&lt;/em&gt;-alkylated, hydroxyarylketones. This reaction is promoted by having an excess of catalyst present, either a Lewis acid such as &lt;span class=&quot;math-container&quot;&gt;$\ce{AlCl3}$&lt;/span&gt; or strong Brønsted acids such as &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{TfOH}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.researchgate.net/publication/49466619_Comparisons_of_O-acylation_and_FriedelCrafts_acylation_of_phenols_and_acyl_chlorides_and_Fries_rearrangement_of_phenyl_esters_in_trifluoromethanesulfonic_acid_effective_synthesis_of_optically_active_homotyrosines&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; paper reports yields of &gt;90% (up to 99%) &lt;em&gt;O&lt;/em&gt;-acylated products when phenol, and various derivatives is reacted with acyl chlorides in 1% &lt;span class=&quot;math-container&quot;&gt;$\ce{TfOH-CH3CN}$&lt;/span&gt; solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In neat &lt;span class=&quot;math-container&quot;&gt;$\ce{TfOH}$&lt;/span&gt;, yields of &gt;90% &lt;em&gt;C&lt;/em&gt;-acylated products are reported for phenol, and many &lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-substituted derivatives, but significantly lower yields of 40-50% &lt;em&gt;C&lt;/em&gt;-acylation are reported for &lt;em&gt;meta&lt;/em&gt; substituted derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, various phenol esters were heated in neat &lt;span class=&quot;math-container&quot;&gt;$\ce{TfOH}$&lt;/span&gt; and yields of &gt;90% of the &lt;em&gt;C&lt;/em&gt;-acylated Fries rearrangement products were reported. The yields were similar for &lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-substituted phenol esters but again significantly lower (30-60%) for &lt;em&gt;meta&lt;/em&gt;-substituted reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall this suggests that the ratio of &lt;em&gt;O&lt;/em&gt;- to &lt;em&gt;C&lt;/em&gt;-acylated products is strongly influenced by the concentration of the catalyst, with high concentrations favouring &lt;em&gt;C&lt;/em&gt;-acylation and low concentrations favouring &lt;em&gt;O&lt;/em&gt;-acylation.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="61578" LastEditDate="2020-04-25T11:33:25.363" LastActivityDate="2020-04-25T11:33:25.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31549" PostTypeId="2" ParentId="18980" CreationDate="2015-05-16T18:53:07.553" Score="2" Body="&lt;p&gt;The given wavelength of $\lambda=600\ \mathrm{nm}$ may be directly converted to various other quantities, namely wavenumber $\tilde\nu$, frequency $\nu$, photon energy $E_\text{p}$, and molar energy $E_\text{m}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tilde\nu=\frac{1}{\lambda}=16\,700\ \mathrm{cm^{-1}}$$&#xA;$$\nu=\tilde\nu c=\frac{c}{\lambda}=5.00\times 10^{14}\ \mathrm{s^{-1}}=500\ \mathrm{THz}$$&#xA;$$E_\text{p}=h\nu=\frac{hc}{\lambda}=3.31\times 10^{-19}\ \mathrm{J}=2.07\ \mathrm{eV}$$&#xA;$$E_\text{m}=N_\text{A}h\nu=\frac{N_\text{A}hc}{\lambda}=199\ \mathrm{kJ\ mol^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to electronic spectra of coordination compounds, usually wavenumbers $\tilde\nu$ expressed in $\mathrm{cm^{-1}}$ are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic configuration of $\ce{M^2+}$ in the given complex $\ce{[M(H2O)6]^2+}$ is unknown. Nevertheless, we may assume that the complex has octahedral geometry and that the observed absorption maximum probably corresponds to promoting one electron from the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$) to an $\mathrm{e_g}$ orbital ($\mathrm{d}_{x^2{-}y^2}$ or $\mathrm{d}_{z^2}$). Thus, it directly gives an estimate for $\Delta_\text{o}=10\ \mathrm{Dq}$ as $16\,700\ \mathrm{cm^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{NH3}$, the value of $\Delta_\text{o}=10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An empirical equation for estimating the value of $\Delta_\text{o}$ for any pair of metal and ligands is $$\Delta_\text{o}=f \cdot g$$ where $f$ is the ligand parameter and $g$ the contribution from the central atom. The parameter $f$ describes the ligand strength relative to water; i.e. the value for $\ce{H2O}$ is $f=1.00$. The empirical value for $\ce{NH3}$ is $f=1.25$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the ligand field splitting for $\ce{[M(H2O)6]^2+}$ is $\Delta_\text{o} = 16\,700\ \mathrm{cm^{-1}}$, the corresponding ligand field splitting for $\ce{[M(NH3)6]^2+}$ may be estimated as $\Delta_\text{o} = 1.25 \times 16\,700\ \mathrm{cm^{-1}}= 20\,900\ \mathrm{cm^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting wavenumber $\tilde\nu=20\,900\ \mathrm{cm^{-1}}$ may be converted to a wavelength of approximately $\lambda=480\ \mathrm{nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\lambda=\frac{1}{\tilde\nu}$, the new wavelength could have been directly calculated from the given original value of $600\ \mathrm{nm}$ using the parameter $f=1.25$ for $\ce{NH3}$:&#xA;$$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-16T18:53:07.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31550" PostTypeId="1" AcceptedAnswerId="31558" CreationDate="2015-05-16T19:01:50.223" Score="-1" ViewCount="295" Body="&lt;p&gt;Apart from the what everyone knows about, solids, liquids, gasses, plasma, BEC. What are the other states of matter and what are they used for? &#xA;I would like an explanation as well as an example of usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, which one is most used in day to day applications?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="4945" LastEditDate="2015-05-20T10:33:50.173" LastActivityDate="2015-05-20T10:33:50.173" Title="Out of the various states of matter, which one is the most used?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-08-23T18:12:33.927" ContentLicense="CC BY-SA 3.0" />
  <row Id="31551" PostTypeId="1" AcceptedAnswerId="31552" CreationDate="2015-05-16T19:11:51.067" Score="24" ViewCount="1030" Body="&lt;p&gt;Being new to chemistry, there may ought to be some properties I couldn't understand fully. One of them being cyclic compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One particular question about this is, whether these cyclic compounds, say benzene rings 'link' in-between one another like a chain. Having one molecule chain of one benzene molecule, go through the center of another, before closing back up as a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I couldn't seem to find any term describing it for further research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a stupid question or have been asked before, but since I couldn't find my answer anywhere, hopefully someone may either point out why wouldn't this exist in any case or redirect me to an answer.&lt;/p&gt;&#xA;" OwnerUserId="16210" LastEditorUserId="5026" LastEditDate="2018-03-04T07:34:34.970" LastActivityDate="2018-03-04T07:34:34.970" Title="Interlocked cyclic compounds" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;structural-formula&gt;&lt;supramolecular-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31552" PostTypeId="2" ParentId="31551" CreationDate="2015-05-16T19:44:15.800" Score="21" Body="&lt;p&gt;The compounds you are referring to are called &lt;a href=&quot;http://en.wikipedia.org/wiki/Catenane&quot; rel=&quot;nofollow noreferrer&quot;&gt;catenanes&lt;/a&gt; (see also &lt;a href=&quot;http://en.wikipedia.org/wiki/Rotaxane&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotaxanes&lt;/a&gt;, as they are a close relative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eFJas.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.depts.ttu.edu/chemistry/faculty/mayer/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;one molecule chain of one benzene molecule, go through the center of&#xA;  another, before closing back up as a ring&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While catenanes involving larger rings can be prepared, having one benzene ring threaded through a second benzene ring is not likely.  If you build a model you can see that there would be severe steric interactions, I've tried to depict this in the following drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g4Mck.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The carbon-carbon bond length in benzene is 139 pm.  Therefore, the approximate side to side distance in benzene is $\mathrm{2 \cdot 139~ pm \cdot cos(30°)= 241~pm}$.  The van der Waals radius is defined as &quot;half of the internuclear separation of two non-bonded atoms of the same element on their closest possible approach&quot; (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles/Covalent_Bond_Distance,_Radius_and_van_der_Waals_Radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;).  The &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page)#Van_der_Waals_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;van der Waals radius for carbon&lt;/a&gt; is 170 pm, meaning that the closest approach for two &lt;strong&gt;non-bonded&lt;/strong&gt; carbon atoms is around 340 pm.  This value is significantly larger than the 241 pm side-to-side distance we calculated above for benzene.  This means that a carbon atom cannot pass through the interior of a benzene ring, at least not for carbon atoms with energies typically found in an organic laboratory.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-03-04T07:34:15.297" LastActivityDate="2018-03-04T07:34:15.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31553" PostTypeId="1" AcceptedAnswerId="31590" CreationDate="2015-05-16T19:52:51.647" Score="2" ViewCount="7877" Body="&lt;p&gt;I am planning to try copper plating a piece of metal by performing electrolysis on an aqueous solution of copper sulfate. I plan run an electrical current with the metal I want to plate as the cathode and a platinum anode. I understand that the copper ions will become reduced when they plate the metal, but I am curious what will happen to the sulfate ions. Will it oxidize to form sulfur trioxide with the produced oxygen from the electrolysis?&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="194" LastEditDate="2017-04-06T01:44:07.400" LastActivityDate="2017-04-06T01:44:07.400" Title="Reaction of sulfate ion in copper sulfate electrolysis" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31557" PostTypeId="2" ParentId="31520" CreationDate="2015-05-16T21:14:52.143" Score="7" Body="&lt;p&gt;Essentially you need to understand that the silica gel packet is just &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt;. These packets are used to absorb moisture (water vapor) to protect goods from spoiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest with you here, nothing ever happens when you eat &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt;. You consume it all the time from food and water. &lt;strong&gt;But&lt;/strong&gt;, the reason these packets are so dangerous is because they might contain impurities such as &lt;span class=&quot;math-container&quot;&gt;$\ce{CoCl2}$&lt;/span&gt; (blue color) and &lt;span class=&quot;math-container&quot;&gt;$methyl~violet$&lt;/span&gt;. These colored products are used to indicate moisture. &lt;strong&gt;These products are carcinogenic and are mutatic poisons.&lt;/strong&gt; In most cases, it will not be fatal immediately, but given time, yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, if the silica gel is colored, it is highly likely that they might be very dangerous. So don't eat them. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this the silica beads tend to fracture so you have essentially consumed broken glass which will be quite painful as it tears up your digestive tract causing you internal bleeding.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Just to add a couple references: &lt;a href=&quot;http://chemistry.about.com/od/medicalhealth/f/What-Happens-If-You-Eat-Silica-Gel-Beads.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.slate.com/articles/news_and_politics/explainer/2011/07/what_happens_if_i_eat_silica_gel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.chow.com/food-news/53783/what-happens-if-you-eat-one-of-those-silica-gel-packets/&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;.&#xA;I don't know how far the the second and third ones are true but the first one is the most accurate.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="23561" LastEditDate="2019-03-19T01:54:20.770" LastActivityDate="2019-03-19T01:54:20.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31558" PostTypeId="2" ParentId="31550" CreationDate="2015-05-16T21:20:12.950" Score="3" Body="&lt;p&gt;Aside from the four fundamental states, solid, liquid, gas, and plasma. There exists more than 10 states excluding the prior four, however reproducible under extreme situation. The two main variables that make substances in different state are pressure and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wI53U.png&quot; alt=&quot;Illustration of transition between four fundamental states of matter (found from Wikipedia under creative commons)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superfluids and Supersolids are states achievable by a few substances (Like liquid helium.) cooled to a few degrees above absolute zero. Quite a few of the states were achievable by extremely low temperature. Which includes Bose-Einstein Condensate exisitng at an ultra-low temperature and within cooled medium like Photonic Matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While superfluids and supersolids are known for their property of having no resistance in flowing and moving respectively; photonic matter have been suggested to be applicable to quantum logic operation due to the properties of photons interacting with each other as though they have mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some states exist at high temperature and pressure. Supercritical fluid is an example in which the distinction between liquid and gas is nonexistent. In a phase diagram, the line that would be the boiling point under different conditions seperates liquid and gas until what is known as a critical point, adding more heat or pressure would result in this supercritical fluid. &lt;a href=&quot;http://www.supercriticalfluid.org/Applications.149.0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site of an organization lists the applications of supercritical fluid in detail&lt;/a&gt;, to my surprise, it includes dry washing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjX9D.png&quot; alt=&quot;Phase diagram depicting the critical point.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some theoretical states that scientists have minimal information on. The quark-gluon plasma is a state in which quarks and gluons move free and independent of each other or pairs. In current states, by adding energy to separate quarks only splits them into two more pairs of quarks, bonded by gluons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KA3qQ.png&quot; alt=&quot;My feeble illustration of adding energy to separate quarks at normal conditions.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jahn-Teller metal was a theoretical state, recently reproduced in lab, it possess multiple properties like an insulator-superconductor intermediate metal.&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/31512/jahn-teller-metal-the-new-state-of-matter/31527#31527&quot;&gt;It is extremely useful for making electronics energy efficient.&lt;/a&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;&lt;br/&gt;&#xA;Considering most states of matter if you exclude the four fundamental states, the most useful matter may be supercritical fluid as it is involved in a wide variety of industrial applications.&lt;br/&gt;&#xA;However, when you consider the potential of other states of matter, Photonic Matter may be used for quantum computers and Jahn-Teller metal for magnetic levitation trains to household energy efficient home appliances.&lt;/p&gt;&#xA;" OwnerUserId="16210" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-05-16T21:58:33.070" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31561" PostTypeId="5" CreationDate="2015-05-16T22:39:19.923" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-16T22:39:19.923" LastActivityDate="2015-05-16T22:39:19.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31562" PostTypeId="4" CreationDate="2015-05-16T22:39:19.923" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-16T22:39:19.923" LastActivityDate="2015-05-16T22:39:19.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31563" PostTypeId="1" CreationDate="2015-05-17T00:20:45.933" Score="9" ViewCount="1394" Body="&lt;p&gt;I performed an electrolysis of zinc, using two zinc electrodes and a zinc nitrate solution. However, after leaving the electrolysis to run for 30 min, at about 0.4 A and 6 V, I discovered that the cathode gained more mass than the electrode lost. I know that ideally the anode oxidises zinc into zinc ions, and the cathode reduces zinc ions into zinc metal, suggesting that total mass should be conserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key observation I made was that a white substance was observed on the anode, which probably led to the extra weight. Also, bubbles were observed on one of the electrodes. I am not sure which one though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My chemistry teacher suggested this could be related to zinc oxide or zinc hydroxide, but I can’t find much information on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pvcty.jpg&quot; alt=&quot;Comparison of anode and cathode&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U4VwI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16288" LastEditorDisplayName="user7951" LastEditDate="2015-05-21T11:57:33.923" LastActivityDate="2017-06-04T10:42:56.920" Title="Cathode gained mass in zinc electrolysis" Tags="&lt;redox&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31566" PostTypeId="2" ParentId="31563" CreationDate="2015-05-17T04:14:42.950" Score="2" Body="&lt;p&gt;If the differences between the two electrode masses is small its probably due to excess electrolyte left on it or a measuring error. The white substance is  dried up zinc nitrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bubbles should have been observed at both electrodes since that would have been the hydrogen, oxygen and nitrous oxide dissociating form the diluted solution of zinc nitrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie: $$\ce{3e¯ + 4H+ + NO3- -&amp;gt; NO + 2H2O}$$ $$\ce{O2(g) + 4H+(aq) + 4e -&amp;gt;  2 H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm pretty sure the bubbles didn't have anything to do with the additional mass gain. $%edit$&lt;/p&gt;&#xA;" OwnerUserId="16115" LastEditorUserId="24052" LastEditorDisplayName="user15489" LastEditDate="2016-03-13T20:29:22.050" LastActivityDate="2016-03-13T20:29:22.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31567" PostTypeId="1" AcceptedAnswerId="31579" CreationDate="2015-05-17T04:33:40.330" Score="7" ViewCount="3301" Body="&lt;p&gt;I don't really study chemistry so while my question may be very obvious, its not obvious to me. If we take an electrochemical reaction like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Fe^2+ + Au^3+ -&amp;gt; 2Fe^3+ + Au+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can find its standard potential and standard Gibb's energy by summing half reactions whose are known, and can be consulted in tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's intuitive for me that the standard Gibb's energy will be given by a sum of the corresponding reactions, and that I can multiply a reaction by minus one, or whatever, and use that to construct my complete reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I see Gibbs energy as intuitive is because you're essentially just taking &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{\mathrm{products}} \mu_{\mathrm{products}} - \sum_{\mathrm{reactants}} \mu_{\mathrm{reactants}}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this will prove the additive property of &quot;summing up reactions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't see any reason for additivity being true when considering standard potentials. I don't even see why inverting the reaction should invert my potential if my half cell is in equilibrium and is a kind of &quot;whole&quot; that is compared to the standard hydrogen electrode and should therefore have the same potential no matter how I write it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How am I screwing up here? Why can we add up potentials?&lt;/p&gt;&#xA;" OwnerUserId="8091" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-07T02:05:49.273" LastActivityDate="2017-08-07T02:05:49.273" Title="Why exactly are standard potentials additive?" Tags="&lt;electrochemistry&gt;&lt;energy&gt;&lt;free-energy&gt;&lt;reduction-potential&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31569" PostTypeId="1" AcceptedAnswerId="31574" CreationDate="2015-05-17T05:57:04.767" Score="7" ViewCount="1459" Body="&lt;p&gt;Some time ago, I was making a $\ce{NaOH}$ solution. The solution I had made was much too concentrated, and the heat released in the dissolution of the ions caused the water to begin steaming. Inadvertently, I inhaled some of this steam, and instantly my nose began to sting, but only for an instant. I can only assume I had inhaled some $\ce{NaOH}$, but how is this possible? How can ions be present in this steam, yet the process of distillation produces steam &lt;strong&gt;without&lt;/strong&gt; ions?&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-17T08:28:20.177" Title="Vaporized sodium hydroxide?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;water&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31571" PostTypeId="2" ParentId="28975" CreationDate="2015-05-17T07:21:35.640" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] but since the coordination entity does not break up, any reaction thus, IMO can not happen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your opinion contradicts experimental findings. Two chlorido ligands in metal complexes can be replaced by (other monodentate ligands or) a bidentate chelating ligand, as demonstrated for the reaction of potassium tetrachloropalladate, $\ce{K2[PdCl4]}$, with diaminosuccinic acid (and its diethylester))(&lt;a href=&quot;http://dx.doi.org/10.1016/0162-0134(94)85008-9&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the binding atoms of a bidentate ligand are close together, as in oxalate, the ligand can't just somehow wrap around the central metal. Instead specific geometric requiries have to be met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a tetra-coordinated complex, such as $\ce{[M(NH3)2Cl2]}$, two major fundamental geometries are conceivable, namely&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tetrahedral or&lt;/li&gt;&#xA;&lt;li&gt;square-planar&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Only in the latter case, &lt;em&gt;cis&lt;/em&gt;-&lt;em&gt;trans&lt;/em&gt; isomerism can be observed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the &lt;em&gt;trans&lt;/em&gt;-isomer the two square-planar complexes, the substitution of two chlorido ligands that point to opposite corners of the square is not possible: oxalate would have to bind all &lt;strong&gt;across&lt;/strong&gt; the complex. Consequently, it is &lt;strong&gt;only&lt;/strong&gt; the &lt;em&gt;cis&lt;/em&gt;-isomer which can react here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\mathit{cis}-[M(NH3)2Cl2] + ox^{2-} -&amp;gt; \mathit{cis}-[M(ox)(NH3)2] +2Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the abbreviation &lt;strong&gt;ox&lt;/strong&gt; for oxalate in coordination compounds, see the &lt;a href=&quot;http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf&quot; rel=&quot;nofollow&quot;&gt;IUPAC Red Book, Table VII, entry 118&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-17T07:21:35.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31572" PostTypeId="2" ParentId="31463" CreationDate="2015-05-17T07:32:44.420" Score="15" Body="&lt;p&gt;Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system.  At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made video about it &lt;a href=&quot;https://www.youtube.com/watch?v=Ec4rlg_EaUg&amp;amp;index=18&amp;amp;list=PLVxAq6ZYPp31uDnEw1A1HRxq-lfwFnwBs&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state.  The heat capacity can be written as&#xA;$$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$&#xA;where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively&#xA;$$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$&#xA;$$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bh5yH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first plot shows $p_0p_1$ (green), $C_V$ (red) and  $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis).  This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxEVk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis.  This shows that the shape of  $p_0p_1$ come from the shape of  $p_0$ and $p_1$, and that the peak occurs where  the ground and exited states are equally likely ($p_0 = p_1$).  The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-05T12:48:41.993" LastActivityDate="2015-06-05T12:48:41.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31574" PostTypeId="2" ParentId="31569" CreationDate="2015-05-17T08:28:20.177" Score="9" Body="&lt;p&gt;When a liquid is evaporating or even boiling, minute drops of the liquid are formed by the mechanical disintegration of the surface of the liquid (bubble bursting, splashing or foaming). Some drops are carried away from the heated liquid by the steam (by convective and vapour flow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration of non-volatile dissolved matter in the entrained drops approximately corresponds to the concentration in the bulk liquid. Therefore, if the bulk liquid contains $\ce{NaOH}$, the entrained drops also contain $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quantitative parameter that describes this transport is the &lt;em&gt;airborne release fraction&lt;/em&gt; (ARF). The ARF increases with temperature of the liquid and air velocity above the liquid. Measured values for liquids that are evaporating but not boiling range from $&amp;lt; 10^{-8}$ to $3\times10^{-5}$ (Mishima, J., L.C. Schwendiman and C.A. Radasch. November 1968. Plutonium Release Studies IV. Fractional Release From Heating Plutonium Nitrate Solutions in Flowing Air Stream, BNWL-931, Pacific Northwest Laboratory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, distillation and similar technical processes have the same problem. That is one reason why distillation shows only a limited &lt;em&gt;decontamination factor&lt;/em&gt; (DF) even for non-volatile dissolved matter.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-17T08:28:20.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31575" PostTypeId="1" CreationDate="2015-05-17T09:02:07.707" Score="2" ViewCount="35" Body="&lt;p&gt;We pipet concentrated nitric and hydrochloric acids. our current pipettor is rainin lite but its piston is metal and thus corrode easily.  We discussed this with our inhouse technician and he suggested a pipettor with a plastic piston.  So do you guys know any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We try to avoid using filtered tips as we need to leach the tips prior to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="16296" LastActivityDate="2015-05-17T09:02:07.707" Title="Know any micropipettor that is acid resistant?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31576" PostTypeId="1" AcceptedAnswerId="31578" CreationDate="2015-05-17T12:36:56.610" Score="3" ViewCount="95" Body="&lt;p&gt;In my textbook written that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that $K_\mathrm{sp}(\ce{Ag2CrO4})=9.0\times 10^{-12}$, consider the solubility of $\ce{Ag2CrO4}$ in a $0.100\:\mathrm{M}$ solution of $\ce{AgNO3}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Initial concentrations (before any $\ce{Ag2CrO4}$ dissolves) are,&#xA;  $[\ce{Ag+}]_0=0.100\:\mathrm{M}$ (from $\ce{AgNO3})$ and $[\ce{CrO4^{2-}}]_0=0\:\mathrm{M}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The system comes to equilibrium as $\ce{Ag2CrO4}$ dissolves according to the reaction:&#xA;  $$\ce{Ag2CrO4(s)\rightleftharpoons 2Ag+(aq) + CrO4^{2-}(aq)}$$&#xA;  $$K_\mathrm{sp}=[\ce{Ag+}]^{2}[\ce{CrO4^{2-}}]=9.0\times 10^{-12}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We assume that $x\:\mathrm{M}$ of $\ce{Ag2CrO4}$ dissolves to reach equilibrium, which means that:&#xA;  $$x\:\mathrm{M\:\ce{Ag2CrO4}(s)}\rightarrow 2x\:\mathrm{M\:\ce{Ag+}(aq)}+x\:\mathrm{M\:\ce{CrO4^{2-}}\:(aq)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now we can specify the equilibrium concentrations in terms of $x$:&#xA;  $$[\ce{Ag+}]=[\ce{Ag+}]_{0}+\Delta[\ce{Ag+}]=(0.100+2x)\:\mathrm{M}$$&#xA;  $$[\ce{CrO4^2-}]=[\ce{CrO4^2-}]_{0}+\Delta[\ce{CrO4^2-}]=(0+x)\:\mathrm{M}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Giving:&#xA;  $$9.0\times 10^{-12}=[0.100+2x]^{2} \cdot [x]$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, here is line I want to discuss (again, taken from the textbook):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since the $K_{\mathrm{sp}}$ value for $\ce{Ag2CrO4}$ is small (the position of the equilibrium lies far to the left), $x$ is expected to be small compared with $0.100\:\mathrm{M}$. Therefore:&#xA;  $${9.0\times 10^{-12}=[0.100+2x]^{2}[x] \approx [0.100}]^{2}[x]$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When can this approximation be made and when can it not be made?&lt;/p&gt;&#xA;" OwnerUserId="16103" LastEditorUserId="16103" LastEditDate="2015-06-07T19:11:46.940" LastActivityDate="2015-06-07T19:11:46.940" Title="Approximating the concentration of ions when a salt dissociates" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31577" PostTypeId="1" CreationDate="2015-05-17T13:01:27.027" Score="3" ViewCount="30922" Body="&lt;p&gt;Why are secondary standard solutions used in titrations? Why don't we always use primary standard solutions?&lt;/p&gt;&#xA;" OwnerUserId="16157" LastActivityDate="2017-02-15T18:58:44.180" Title="Primary and secondary standard" Tags="&lt;titration&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31578" PostTypeId="2" ParentId="31576" CreationDate="2015-05-17T13:05:50.523" Score="4" Body="&lt;p&gt;Really, your question seems to be a math question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to solve the cubic equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{9.0\cdot10^{-12}=[0.100+}2x]^{2}[x]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots&quot; rel=&quot;nofollow&quot;&gt;Cubic equations are exactly solvable&lt;/a&gt;; however, it is common to use an approximate method.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not it is appropriate to approximate 0.100 +2x as 0.100 depends upon how small x is and how precise an answer you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get an approximate solution for x, you could recalculate 0.100 +2x and use that value to re-solve the problem. By repeating this process you can get arbitrary close to the correct value by iteration.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-05-17T13:17:04.117" LastActivityDate="2015-05-17T13:17:04.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31579" PostTypeId="2" ParentId="31567" CreationDate="2015-05-17T13:36:03.030" Score="3" Body="&lt;p&gt;It helps to view this in terms of the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Δ &lt;em&gt;G°&lt;/em&gt; = -&lt;em&gt;nFE&lt;/em&gt;°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversing the reaction reverses the sign on Δ &lt;em&gt;G°&lt;/em&gt;, and therefore the sign on &lt;em&gt;E°&lt;/em&gt;. The sign on the potential determines the direction in which electricity flows. By convention, a galvanic cell (spontaneous reaction) has a positive &lt;em&gt;E°&lt;/em&gt;, while an electrolytic cell (non-spontaneous reaction) has a -&lt;em&gt;E°&lt;/em&gt;. The magnitude of the potential stays the same, but a sign is added to signify the direction of current flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as adding half cells to determine the overall potential, this can be done because of the relationship in the above equation. You can see that &lt;em&gt;E°&lt;/em&gt; depends on Δ &lt;em&gt;G°&lt;/em&gt; and &lt;em&gt;n&lt;/em&gt;. But for instance, if &lt;em&gt;n&lt;/em&gt; is multiplied by 2, Δ &lt;em&gt;G°&lt;/em&gt; will be multiplied by 2 as well (there will be twice as much Gibbs' energy because the reaction involves twice the electron transfer). This means that you can add half-cell potentials to get the overall cell potential. Be careful to note though that if you multiply a reaction by a coefficient that you don't multiply the potentials, as you would for Δ &lt;em&gt;G°&lt;/em&gt;; the potential does not depend on the stoichiometric coefficients, only on what the cathode and anode are.&lt;/p&gt;&#xA;" OwnerUserId="16290" LastActivityDate="2015-05-17T13:36:03.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31580" PostTypeId="1" CreationDate="2015-05-17T14:16:18.893" Score="5" ViewCount="1982" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Vinylic halides do not show nucleophilic substitution at all, because of their stability due to resonance. But elimination to form alkynes is possible. No nucleophile at all can displace the halide ion in a nucleophilic substitution, but a strong base suffices for elimination. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="7448" LastEditDate="2015-10-02T16:28:30.560" LastActivityDate="2015-10-02T16:28:30.560" Title="Elimination from Vinylic Halides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31581" PostTypeId="1" CreationDate="2015-05-17T14:21:32.327" Score="2" ViewCount="66" Body="&lt;p&gt;1) My textbook tells me that chromium's electronic configuration is $\ce{[Ar]}\space 3d^5 4s^1$ instead of $\ce{[Ar]} \space 3d^4 4s^2$, as fully filled and half filled orbitals have more stability. But how does that happen? Why is the fully or half filled orbital more stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Following this method as we go down group 6 we would expect tungsten's last two orbitals to be filled as $5d^5 6s^1$ to attain more stability. But the actual configuration turns out to be $5d^4 6s^2$. Isn't this supposed to be &quot;unstable&quot;?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="15410" LastEditDate="2015-05-17T17:04:26.070" LastActivityDate="2015-05-17T17:04:26.070" Title="Stability of half and fully filled orbitals" Tags="&lt;electronic-configuration&gt;&lt;orbitals&gt;&lt;stability&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-05-18T03:11:58.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="31582" PostTypeId="1" CreationDate="2015-05-17T14:24:36.017" Score="1" ViewCount="1355" Body="&lt;p&gt;We've all learnt that the reduction always accompanies oxidation and vice versa Why is this so? Why is it not possible that a species releases an electron and the electron just &quot;floats around&quot;? Is this related to the stability of a free electron?&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-05-17T16:44:38.317" Title="Why does reduction always accompany oxidation and vice versa?" Tags="&lt;redox&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31583" PostTypeId="2" ParentId="31184" CreationDate="2015-05-17T14:56:16.003" Score="1" Body="&lt;p&gt;The reason is thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For aldoses (like glucose), the copper sulfate in the reagent easily oxidizes it into the carboxylic acid forming copper oxide (a precipitate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ketoses (like fructose), the ketone cannot be oxidized, so there is no color change. Over time at elevated temperature, the α-hydroxy ketone isomerizes to a α-hydroxy aldehyde. This aldehyde is then oxidized to the carboxylic acid, producing a red precipitate.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-17T14:56:16.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31584" PostTypeId="1" CreationDate="2015-05-17T15:04:51.800" Score="8" ViewCount="35235" Body="&lt;p&gt;A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce{SF_6}$), for example, acquires an octahedral shape, which can be explained by assuming that the hybrid orbitals arrange themselves to minimize repulsion amongst themselves. An octahedron makes sense, it the only way to symmetrically arrange six hybrid orbitals on an atom. The case for an $sp^3$ hybridized atom is similar which from a similar logic can be explained to have a tetrahedral geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider the case of $sp^3d$ hybrid orbitals which arrange themselves in a trigonal bypyramidal shape ($\ce{PCl5}$). This is a gross breach of the usual adherence to symmetry. Two bonds are at the axial location and three at the equatorial location. In fact, the axial bonds are slightly longer than the equatorial ones. The case for $sp^3d^3$ hybridization is similar ($\ce{IF7}$) with five orbitals crammed into the equatorial plane and two sticking out above and below it. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V2iVg.gif&quot; alt=&quot;IF7&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="402" LastEditDate="2016-12-04T04:02:20.903" LastActivityDate="2017-01-14T08:58:52.317" Title="Why an asymmetric geometry with sp3d and sp3d3 hybridization?" Tags="&lt;hybridization&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31585" PostTypeId="1" AcceptedAnswerId="31593" CreationDate="2015-05-17T16:01:44.500" Score="3" ViewCount="2542" Body="&lt;p&gt;For example &lt;a href=&quot;https://en.wikipedia.org/wiki/PEDOT:PSS&quot; rel=&quot;nofollow&quot;&gt;PEDOT:PSS&lt;/a&gt; and P3HT:PCBM.&lt;/p&gt;&#xA;" OwnerUserId="10104" LastActivityDate="2015-05-17T19:39:38.573" Title="What is the meaning of the &quot;colon&quot; notation in chemical formulas?" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31586" PostTypeId="2" ParentId="31580" CreationDate="2015-05-17T16:03:45.683" Score="5" Body="&lt;p&gt;In the $\ce{sp^2}$ transition state for an $\mathrm{S_{N}}2$ reaction&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the attacking nucleophile, the central carbon and the leaving group prefer a linear arrangement&lt;/li&gt;&#xA;&lt;li&gt;the 3 remaining groups around carbon are arranged with 120° bond angles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rdwol.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://simple.wikipedia.org/wiki/SN2_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's look at ethylene (as a model for a vinyl halide) and let's use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;bent bond description of ethylene&lt;/a&gt; because it makes it easy to visualize what's happening (the bent bond model has been shown to be equivalent to the pi bond model in all aspects, see below for the same explanation phrased in terms of the pi bond model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ou7wN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://pirate.shu.edu/~rawncarr/Molecular%20Models/ethene.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace one of the ethylene hydrogens with halide and imagine what happens as we approach the transition state.  The bond angle in the two-membered ring wants to be small, but as we approach the transition state it needs to open up wider to 120°!  This would create a lot of strain.  It is this large strain energy that causes the $\mathrm{S_{N}}2$ reaction at $\ce{sp^2}$ hybdridized vinylic carbons to be a very high energy pathway and therefore less likely to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the elimination reaction at a vinylic carbon is quite facile.  Here as the base removes a proton we build up negative charge on an $\ce{sp^2}$ hybridized carbon.  An $\ce{sp^2}$ orbital has more s-character than an $\ce{sp^3}$ orbital; the more s-character in an orbital the lower the energy of an electron in that orbital.  Therefore, negative charge is more stable in an $\ce{sp^2}$ orbital than a $\ce{sp^3}$ orbital.  Consequently it is easier to remove a proton when the negative charge will reside on an $\ce{sp^2}$ hybridized carbon compared to an $\ce{sp^3}$ hybridized carbon (and that's why it is even easier [lower energy] to remove a proton from an $\ce{sp}$ hybridized carbon).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Explanation Based on Pi Bond Model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As expected for equivalent models, the bent bond explanation can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons are in higher energy unhybridized p orbitals and this is not offset by the energy lowering experienced by the electrons as some bonds rehybridize  from $\ce{sp^2}$ to $\ce{sp}$) thereby increasing the energy required to reach the transition state.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-18T00:13:52.033" LastActivityDate="2015-05-18T00:13:52.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31587" PostTypeId="1" AcceptedAnswerId="31592" CreationDate="2015-05-17T16:10:45.747" Score="21" ViewCount="1439" Body="&lt;p&gt;I suppose the first question supporting the main question is, has tritium water ever been synthesized in sufficient quantity to test chemical properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, and apart from the obvious radioactive nature of the molecule, and possibly self-heating nature, what unusual properties does tritium water have over protium water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, deuterium oxide is not (easily) metabolized by living things; it cannot support life in the same manner as protium oxide. Also protium water is blue and deuterium oxide is clear. What color is tritium oxide?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="4231" LastEditDate="2015-05-17T22:00:16.370" LastActivityDate="2015-07-20T16:21:37.257" Title="Are there any known chemical properties of tritium water that make it unusually different from protium water?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;optical-properties&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31588" PostTypeId="1" CreationDate="2015-05-17T16:24:44.093" Score="5" ViewCount="111" Body="&lt;p&gt;My textbook says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrostatic forces between two ions decrease by the factor $\frac{1}{d^2}$ as their separation distance, $d$, increases. Dipole–dipole forces, however, vary as  $\frac{1}{d^4}$. Because of the higher power of $d$ in the denominator, the attractive forces diminish with increasing $d$ much more rapidly than does  $\frac{1}{d^2}$. As a result, dipole forces are effective only over very short distances.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now mathematically I understand why this is true, but I don't understand why this is the case in reality. Can someone shed some light on this?&lt;/p&gt;&#xA;" OwnerUserId="16301" LastEditorUserId="16301" LastEditDate="2015-05-17T19:27:10.727" LastActivityDate="2015-05-20T11:15:47.883" Title="Why is there a greater degree of separation in a dipole-dipole interaction?" Tags="&lt;intermolecular-forces&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31589" PostTypeId="2" ParentId="31582" CreationDate="2015-05-17T16:32:43.913" Score="2" Body="&lt;p&gt;If something is to be &lt;strong&gt;reduced&lt;/strong&gt; then there must be a &lt;strong&gt;reducing agent&lt;/strong&gt; present. Also if something is &lt;strong&gt;oxidized&lt;/strong&gt; then there must be an &lt;strong&gt;oxidizing agent&lt;/strong&gt; present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must have a &lt;strong&gt;reducing agent&lt;/strong&gt; or &lt;strong&gt;oxidizing agent&lt;/strong&gt; present in order for a redox reaction to take place. An element can't just dislocate it's electrons. There must be an outside force present in order to do that.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="15410" LastEditDate="2015-05-17T16:44:38.317" LastActivityDate="2015-05-17T16:44:38.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31590" PostTypeId="2" ParentId="31553" CreationDate="2015-05-17T16:42:15.463" Score="3" Body="&lt;p&gt;A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;c(\ce{Cu^2+})&amp;amp;=0.5\ \mathrm{mol\ l^{-1}}\\&#xA;c(\ce{H+})&amp;amp;=1\ \mathrm{mol\ l^{-1}}\\&#xA;c(\ce{SO4^2-})&amp;amp;=1\ \mathrm{mol\ l^{-1}}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{Cu^2+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; Cu}\quad\quad &amp;amp;&amp;amp;E^\circ = +0.340\ \mathrm{V}\quad\quad &amp;amp;&amp;amp;&amp;amp;E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\&#xA;\ce{2H+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2}\quad\quad &amp;amp;&amp;amp;E^\circ = +0.000&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{3}&#xA;\ce{O2 + 4H+ + 4e- \;&amp;amp;&amp;lt;=&amp;gt; 2H2O}\quad\quad &amp;amp;&amp;amp;E^\circ = +1.229\ \mathrm{V}\\&#xA;\ce{S2O8^2- + 2e- \;&amp;amp;&amp;lt;=&amp;gt; 2SO4^2-}\quad\quad &amp;amp;&amp;amp;E^\circ = +2.01\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode:&#xA;$$\ce{SO4^2- -&amp;gt; SO4- + e-}$$&#xA;The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.&lt;br&gt;&#xA;Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water: &#xA;$$\ce{4SO4- + 2H2O -&amp;gt; 4SO4^2- + O2 + 4H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{4SO4^2- &amp;amp;-&amp;gt; 4SO4- + 4e-}\\&#xA;\ce{4SO4- + 2H2O &amp;amp;-&amp;gt; 4SO4^2- + O2 + 4H+}\\&#xA;\hline\ce{2H2O &amp;amp;-&amp;gt; O2 + 4H+ + 4e-}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-17T16:42:15.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31591" PostTypeId="1" AcceptedAnswerId="31594" CreationDate="2015-05-17T16:49:24.277" Score="13" ViewCount="41357" Body="&lt;p&gt;A molecule that has hydrogen bonding &lt;strong&gt;usually&lt;/strong&gt; follows these two premises.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1.) There is a hydrogen atom involved&lt;/p&gt;&#xA;&lt;p&gt;2.) Hydrogen must be bonded to a highly electronegative element which are nitrogen (&lt;span class=&quot;math-container&quot;&gt;$\ce{N}$&lt;/span&gt;), fluorine (&lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt;) and oxygen (&lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Seeing that both oxygen and chlorine have a small difference in their electronegativity (oxygen being roughly &lt;strong&gt;3.5&lt;/strong&gt; and chlorine being roughly &lt;strong&gt;3.0&lt;/strong&gt;), why does chlorine in a hydrogen chloride molecule (&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;) have a dipole-dipole interaction, while the oxygen in a water molecule (&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;) causes the water molecule to have a stronger form of dipole-dipole interaction called hydrogen bonding? I do not understand this since &lt;strong&gt;chlorine is just as electronegative as oxygen and nitrogen&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I should also added that nitrogen and chlorine have the same EN value (&lt;strong&gt;3.0&lt;/strong&gt;)&lt;/p&gt;&#xA;" OwnerUserId="16301" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2019-06-27T06:51:42.163" Title="Why is HCl not considered to have hydrogen bonding?" Tags="&lt;electronegativity&gt;&lt;dipole&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31592" PostTypeId="2" ParentId="31587" CreationDate="2015-05-17T17:06:47.157" Score="25" Body="&lt;p&gt;Yes, $\ce{T2O}$ has been prepared and is available in significant quantity.  When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil.  It must be transported in a shielded, cryogenic dewar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds.  The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest.  We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot;&gt;primary kinetic isotope effect&lt;/a&gt;. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows:&#xA;$$\frac{K_H}{K_D} \sim 6-7$$&#xA;$$\frac{K_H}{K_T} \sim 13-14$$&#xA;In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds.  The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon.  Still, as you pointed out, the effects can be disastrous in biological systems.  Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu_e=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\ce{\mu}$ is the reduced mass of the system (&lt;em&gt;e.g.&lt;/em&gt; the bond involved in the vibration that is producing the light absorption) and is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond.  We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is &lt;strong&gt;inversely&lt;/strong&gt; related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration.  Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (&lt;a href=&quot;https://books.google.com/books?id=XomQ4Qyyyy8C&amp;amp;pg=PA82&amp;amp;lpg=PA82&amp;amp;dq=d2o%20visible%20absorption&amp;amp;source=bl&amp;amp;ots=hT_IFjojPQ&amp;amp;sig=xnLwBT5BGjNHo1XcZVKEpgtuzFo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=gNVYVafQOonFsAWjtIGYDg&amp;amp;ved=0CF8Q6AEwCg#v=onepage&amp;amp;q=d2o%20visible%20absorption&amp;amp;f=false&quot;&gt;reference&lt;/a&gt;, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be &lt;strong&gt;colorless&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-07-20T16:21:37.257" LastActivityDate="2015-07-20T16:21:37.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31593" PostTypeId="2" ParentId="31585" CreationDate="2015-05-17T17:08:38.270" Score="7" Body="&lt;p&gt;All it means is that there's an intimate &lt;em&gt;physical&lt;/em&gt; mixture of the specified components in a solid or liquid phase (no chemical bonding between the components as happens in a copolymer, for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of solar cells, LEDs and related devices, they are often used as part of shorthand device architecture notation. Slashes ( / ) are used to separate different layers of the device, while colons ( : ) are used to enunciate all separate components in a single layer. Do note that every so often a slash is used incorrectly in place of a colon. Checking the experimental procedure should clarify whether two consecutively listed materials are in the same or separate layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the notation &lt;strong&gt;ITO/PEDOT:PSS/P3HT:MEH-PPV:PCBM/Ca/Al&lt;/strong&gt; for an organic polymer solar cell. There are five layers in total, deposited in sequence from left to right:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A layer of ITO (usually on top of a glass substrate which is omitted).&lt;/li&gt;&#xA;&lt;li&gt;A layer containing a binary mixture of PEDOT and PSS, usually deposited via spin casting of an aqueous solution with both materials dissolved.&lt;/li&gt;&#xA;&lt;li&gt;A layer containing a ternary mixture of P3HT, MEH-PPV and PCBM, often deposited by spin casting from a solution of all materials in a chlorinated solvent.&lt;/li&gt;&#xA;&lt;li&gt;A layer of pure calcium metal, likely deposited by thermal evaporation.&lt;/li&gt;&#xA;&lt;li&gt;A layer of pure aluminium metal, deposited on top of the calcium layer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sometimes the thickness, component proportion and concentration of the spinning solution for each layer is reported inside of parentheses, such as in ITO/PEDOT:PSS (40 nm)/P3HT:PCBM (1:0.8, 150 nm)/Ca (80 nm). Be sure to pay attention to small differences in notation resulting in &lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_solar_cell#Junction_types&quot; rel=&quot;noreferrer&quot;&gt;completely different cell architecture&lt;/a&gt;. For example, a solar cell with a ITO/PEDOT:PSS/P3HT:$\ce{C_{60}}$/Al represents a bulk heterojunction solar cell, while ITO/PEDOT:PSS/P3HT/$\ce{C_{60}}$/Al represents a bilayer (or planar heterojunction) solar cell.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-05-17T19:39:38.573" LastActivityDate="2015-05-17T19:39:38.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31594" PostTypeId="2" ParentId="31591" CreationDate="2015-05-17T17:23:30.813" Score="1" Body="&lt;p&gt;Very, very &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-why-cant-it-exist-with-atoms-other-than-f?lq=1&quot;&gt;related&lt;/a&gt;. Check it for a more in depth answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, first-of-all you need to understand that the $\ce{H}$ bonding isn't actually bonding. It is just a covalent attraction. Also since $\ce{Cl}$ is larger than $\ce{N}$, $\ce{F}$ and $\ce{O}$ it does not make a strong $\ce{H}$ bond. &lt;strong&gt;The size of the $\ce{Cl}$ makes the dipole-dipole attraction weaker&lt;/strong&gt;. However $\ce{N}$, $\ce{F}$ and $\ce{O}$ are smaller and thus have an $\ce{H}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although in reality, compared to other covalently bonded structures, $\ce{HCl}$ has a very strong covalent bond.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-17T17:23:30.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31595" PostTypeId="1" AcceptedAnswerId="31634" CreationDate="2015-05-17T17:24:10.743" Score="3" ViewCount="131" Body="&lt;p&gt;Following a question I have asked previously, what I intend to inquire this time is quite similar as well, except, with a polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So previously I asked whether such monomer PCDO is possible to synthesize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b6LRU.png&quot; alt=&quot;Phenyl Carbolithia Divinylene Oxide molecule&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That originated from this similar, but different polymer poly-PCDO. It is true that I also don't think this PCDO from my previous question is poly-PCDO after monomerization (I'm still not so sure though). But the PCDO molecule pop out when I was attempting to make this original poly-PCDO a monomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/86DgK.png&quot; alt=&quot;Poly-Phenyl-Carbolithia-Divinylene-Ether&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the synthesis of the polymer in this structure would be easier to theorize and explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can dehydrate divinyl ether with benzene (or perhaps phenol) and lithonate by condensation polymer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{n(C4H6O + C6H6O + HLiCO3) -&amp;gt; (C11H9LiO4)_{n} + nH2O + nH2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my limited knowledge in this area, I can only assume the phenyl group would most likely bond with lithonate being bonded to the ether already. Would someone explain to me how this will or will not work?&lt;/p&gt;&#xA;" OwnerUserId="16210" LastEditorUserId="3846" LastEditDate="2015-05-17T19:37:01.530" LastActivityDate="2015-05-18T11:31:25.720" Title="Possible Synthesis of a Polymer" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;synthesis&gt;&lt;theoretical-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31596" PostTypeId="1" CreationDate="2015-05-17T18:16:31.877" Score="-4" ViewCount="332" Body="&lt;p&gt;Is there any reaction between solid/ liquid elements/compounds at room temperature that produces gases??&lt;/p&gt;&#xA;" OwnerUserId="16304" LastEditorUserId="10185" LastEditDate="2015-05-17T19:12:04.297" LastActivityDate="2015-05-17T22:15:45.810" Title="Can gas get produced at room temperature?" Tags="&lt;phase&gt;" AnswerCount="2" CommentCount="2" ClosedDate="2015-05-17T22:25:36.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="31597" PostTypeId="2" ParentId="31412" CreationDate="2015-05-17T18:37:57.527" Score="2" Body="&lt;p&gt;I doubt that your &quot;monomer&quot; is very stable. Orthoesters, $\ce{R'-C(OR)3}$, are hydrolyzed under acidic conditions: $\ce{R'-C(OR)3 + 2 H+ -&amp;gt; R'-COOR + 2 ROH}$. The fate of your monomer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ke6WA.png&quot; alt=&quot;decomposition&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="9961" LastEditDate="2015-11-28T23:14:36.853" LastActivityDate="2015-11-28T23:14:36.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31599" PostTypeId="1" CreationDate="2015-05-17T18:48:31.137" Score="9" ViewCount="1582" Body="&lt;p&gt;I just read about the Pinacol rearrangement and the following reaction was presented just after it. The book (Clayden) shows path 1 being taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why is the second path not taken? I know that migratory aptitude of $\ce{Ph&amp;gt;H}$ but all we want is stability, so why can't the $\ce{H}$ migrate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfj7D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="4231" LastEditDate="2015-05-17T22:21:53.330" LastActivityDate="2015-05-18T04:45:44.107" Title="Migratory preference in a rearrangement involving carbocations" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31600" PostTypeId="2" ParentId="31596" CreationDate="2015-05-17T19:20:34.080" Score="4" Body="&lt;p&gt;Yes, there are many. One you can make in your kitchen is to mix baking soda and vinegar; it makes carbon dioxide gas.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-05-17T19:39:41.763" LastActivityDate="2015-05-17T19:39:41.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31601" PostTypeId="1" AcceptedAnswerId="31609" CreationDate="2015-05-17T19:35:13.540" Score="5" ViewCount="614" Body="&lt;p&gt;I understand the principles of electrolysis of salts in aqueous solutions, but there are two points on which i am unsure. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At the positive electrode, how can you work out which ions will preferentially donate electrons?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If so, what decides which ions are formed?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the last question, I believe the latter would be true because if the anions produced at the negative electrode react with the cations that are attracted to the electrode, the cations that are attracted to the electrode would not be available to accept electrons and hence the current would become increasingly restricted until it stops all together. Could someone negate or confirm this reasoning?&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="15410" LastEditDate="2015-05-17T19:51:47.227" LastActivityDate="2015-05-19T14:58:19.650" Title="Formation of species in electrolysis" Tags="&lt;electrochemistry&gt;&lt;electrons&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31602" PostTypeId="2" ParentId="31588" CreationDate="2015-05-17T19:37:27.197" Score="2" Body="&lt;p&gt;Like I answered my other question, this is in some ways very similar. Since electrostatic attractions are so strong, it is decreases less at a longer distances. The $+$ &amp;amp; $-$ are attracted together so strongly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dipole-dipole attractions aren't strong attractions, electrostatic attractions are stronger, since they are created by ionic bonding. Dipole-dipole occur in covalent bonding. Thus dipole-dipole attractions decrease with increasing distance faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a piece of the whole picture.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Check the refs: &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermolecular_force&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://www.chem.purdue.edu/gchelp/liquids/dipdip.html&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;. They provide more in depth detail to the entire picture.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-05-18T23:49:41.783" LastActivityDate="2015-05-18T23:49:41.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31603" PostTypeId="2" ParentId="31601" CreationDate="2015-05-17T20:11:21.897" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At the positive electrode, how can you work out which ions will preferentially donate electrons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an electrolytic cell, the positive electrode is called the anode, and the oxidation half of the reaction occurs here. The oxidation reactions which have less negative standard &lt;strong&gt;oxidation&lt;/strong&gt; potential are more likely to occur, though most resources give the standard &lt;strong&gt;reduction&lt;/strong&gt; potentials. &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; is one such resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to know which species are more likely to donate electrons, these will be the half reactions that have the less negative standard &lt;strong&gt;reduction&lt;/strong&gt; potentials.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the formation of hydroxide ions and hydrogen gas occurs at the negative electrode, can hydrogen ions and oxygen gas be formed instead?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wrote a response to this exact question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31320/is-there-an-electrolyte-that-will-halt-the-production-of-hydrogen-in-water-elect&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens to the ions formed at the negative electrode? Do they react with the cations that are attracted to the electrode or do they travel to and react with the positive electrode?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In an electrolytic cell, the negative electrode is called the cathode, and the reduction half of the reaction occurs here. When an species is reduced, typically it changes phases (goes from aqueous ion to being a gas, a liquid, or a solid). Most likely if an ion is reduced to another ion here, this ion will then also be reduced.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-05-17T20:11:21.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31604" PostTypeId="1" AcceptedAnswerId="31608" CreationDate="2015-05-17T20:12:25.950" Score="7" ViewCount="37500" Body="&lt;p&gt;Consider the Lewis dot structures of the molecules fluoroform, $\ce{CHF3}$, and tetrafluoromethane, $\ce{CF4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hq8Cz.png&quot; alt=&quot;fluoroform&quot;&gt;&amp;nbsp;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UjvVc.gif&quot; alt=&quot;tetrafluoromethane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first line of thought is that both of these molecules are symmetrical (the vectors from each individual atom naturally cancel each other out) with no lone pairs of electrons. Now I knew that just because they were symmetrical didn't automatically meant that they were non-polar. If they're symmetrical, they're usually non-polar but not always. This is one of those cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My next line of thought was that they both had to be polar and had dipole-dipole interactions. Then I later find out that the $\ce{CHF3}$ is actually the polar molecule with dipole-dipole interaction and $\ce{CF4}$ is the non-polar molecule with London-dispersion forces.&lt;br&gt;&#xA;This bothered me a little bit because both of these molecules have a very electronegative atom (fluorine) that is more electronegative than any other element in the compound. This is why I thought they were both polar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is tetrafluoromethane non-polar and fluoroform polar?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16301" LastEditorUserId="4945" LastEditDate="2018-09-07T16:46:16.120" LastActivityDate="2018-09-07T16:46:16.120" Title="Why is tetrafluoromethane non-polar and fluoroform polar?" Tags="&lt;molecules&gt;&lt;intermolecular-forces&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31605" PostTypeId="2" ParentId="31604" CreationDate="2015-05-17T20:35:53.240" Score="2" Body="&lt;p&gt;Those are not the Lewis dot structures. The fluorine in both of these molecules should have complete octet structure like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y68i7.gif&quot; alt=&quot;CF4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there are actually more electron pairs in place than you may have been considering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first line of thought is that both of these molecules are symmetrical (the vectors from each individual atom naturally canceled each other out) with no lone pairs of electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In $\ce{CF4}$, you are correct. The molecules is perfectly symmetrical, so every electron pair on each fluorine cancels out the electron pairs of every other fluorine. For this reason, this molecule is &lt;strong&gt;non-polar&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{CHF3}$, however, the hydrogen does not have 3 other electron clouds around it like the fluorine do. These means that this side of the molecule is overall more positive than the other part of the molecule, which is dense with the fluorine atoms unbonded electrons. For this reason, this molecule is &lt;strong&gt;polar&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-17T20:35:53.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31606" PostTypeId="2" ParentId="31584" CreationDate="2015-05-17T20:37:01.957" Score="5" Body="&lt;p&gt;As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently, all the cases you presented in your question can be adequately explained using &lt;em&gt;only s- and p-orbitals&lt;/em&gt;. See the links for &lt;a href=&quot;https://chemistry.stackexchange.com/a/24675/&quot;&gt;$\ce{SF6}$&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/a/18544/&quot;&gt;$\ce{PCl5}$&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/a/29120/&quot;&gt;$\ce{IF7}$&lt;/a&gt; respectively. Many thanks to ron for providing them in his comments!&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2017-01-14T08:58:52.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31607" PostTypeId="2" ParentId="31604" CreationDate="2015-05-17T20:39:51.417" Score="4" Body="&lt;p&gt;As you state, the CF4 &lt;strong&gt;is&lt;/strong&gt; symmetrical (tetrahedral, not planar), so there is no net polar moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ngltk.png&quot; alt=&quot;CF4 tetrahedral structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Replacing one F with a Cl, however, introduces an asymmetry&lt;/strong&gt; because the Cl is less &lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch01/ch1-2-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronegative&lt;/a&gt;, about 3 V vs. 4 V for F. One can imagine the electron cloud being pulled towards the 3 F atoms and away from the Cl; so the Cl is at the positive vertex of the tetrahedron, with three F atoms opposing it across the carbon.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="7448" LastEditDate="2016-08-18T22:08:09.193" LastActivityDate="2016-08-18T22:08:09.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31608" PostTypeId="2" ParentId="31604" CreationDate="2015-05-17T20:43:39.353" Score="14" Body="&lt;p&gt;Draw the structures in 3D and then you will see why one is polar and the other not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CF4}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBTIS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four $\ce{C-F}$ bonds cancel out, leaving no overall dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CHF3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/saMKV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the molecule has some symmetries, it is not perfectly symmetrical. Since the dipoles of the $\ce{C-F}$ bonds are far larger than the basically non-existent dipole of the $\ce{C-H}$ bond, the dipoles do not cancel out and you are left with a molecule which has a notable net dipole moment of 1.649 D, with the negative end over the fluorines and the positive end over the hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As general advice, when determining whether a molecule has a net dipole, always consider the molecule in 3D as opposed to just looking at individual bond dipoles.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-17T20:43:39.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31609" PostTypeId="2" ParentId="31601" CreationDate="2015-05-17T20:44:35.063" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the last question, I believe the latter would be true because if&#xA;  the anions produced at the negative electrode react with the cations&#xA;  that are attracted to the electrode, the cations that are attracted to&#xA;  the electrode would not be available to accept electrons and hence the&#xA;  current would become increasingly restricted until it stops all&#xA;  together. Could someone negate or confirm this reasoning?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19540/why-the-salts-in-a-salt-bridge&quot;&gt;Why the salts in a salt bridge?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/10203/voltaic-and-electrolytic-cells&quot;&gt;Voltaic and electrolytic cells&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/29003/difficulties-understanding-how-salt-bridge-works&quot;&gt;Difficulties understanding how salt bridge works&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/5477/why-is-it-important-to-use-a-salt-bridge-in-a-voltaic-cell-can-a-wire-be-used&quot;&gt;Why is it important to use a salt bridge in a voltaic cell? can a wire be used?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Don’t be limited to those above, keep searching until you solve your dilemma. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Now for the Hard Part&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{O2_{(g)} + 4H+ + 4e- &amp;amp;-&amp;gt; H2O_{(l)}}\quad&amp;amp;&amp;amp;1.23\ \mathrm{V}\\&#xA;\ce{F2_{(g)} + 2e- &amp;amp;-&amp;gt; 2F-}\quad&amp;amp;&amp;amp;2.87\ \mathrm{V}\\&#xA;\ce{2H2O{(l)} + 2e- &amp;amp;-&amp;gt; H2_{(g)} + 2OH-}\quad&amp;amp;&amp;amp;{-}0.83\ \mathrm{V}\\&#xA;\ce{Ni^2+ + 2e- &amp;amp;-&amp;gt; Ni_{(s)}}\quad&amp;amp;&amp;amp;{-}0.25\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can eliminate the third reaction since &lt;strong&gt;fluorine&lt;/strong&gt; already exists reduced, and it does not like being oxidized. Then you have &lt;strong&gt;nickel&lt;/strong&gt; and &lt;strong&gt;water&lt;/strong&gt;, which of them is most likely to be reduced? &lt;strong&gt;Nickel&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we must decide the compound being oxidized. &lt;strong&gt;Fluorine&lt;/strong&gt; and &lt;strong&gt;nickel&lt;/strong&gt; are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 1 (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In general …&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Eliminate which reaction cannot take place.&lt;/li&gt;&#xA;&lt;li&gt;Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T14:58:19.650" LastActivityDate="2015-05-19T14:58:19.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31610" PostTypeId="2" ParentId="31604" CreationDate="2015-05-17T21:11:01.043" Score="5" Body="&lt;p&gt;‘Symmetry’ is a rather generalised concept. It isn’t so much the existence of symmetry in a molecule, but much more &lt;em&gt;which type of symmetry&lt;/em&gt; we’re encountering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From geometry classes, you should remember that there are planes and centres of symmetry. Alongside these, rotational axes, the identity and more exotic things like improper rotation are considered &lt;em&gt;elements of symmetry&lt;/em&gt; in group theory — in a nutshell, the part of extended chemistry that discusses symmetry and its effects. From knowing some elements of symmetry in a molecule you can deduce others, but that’s not too relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point is the &lt;em&gt;effect&lt;/em&gt; that the symmetry has on your molecule. Anything that can be represented as a vector — most notably the charge separation in a dipole — needs to transform onto itself, no matter which element of symmetry you apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\ce{CF4}$, the molecule not only contains a number of planes of symmetry (not two or four, but six due to the picture bon showed). These planes also generate a &lt;em&gt;centre of symmetry&lt;/em&gt;. And the only vector that can be transformed onto itself by a centre of symmetry is the null vector, which in turn means that a dipole moment can only be zero — hence a non-polar molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{CHF3}$, we suddenly encounter the reduction of symmetry, that is already evident in your flat drawings, but only truely discussable with the 3D-model. In a nutshell, you lose three of the six planes of symmetry and also the centre of symmetry they created. The three remaining planes of symmetry share a common axis — the one which also contains the $\ce{C-H}$-bond. And a vector, that points in the same or the opposite direction as this axis will be transformed onto itself. So you have sufficient reduction in symmetry to allow a dipole momentum, creating a polar molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplified, this means that &lt;em&gt;any and all molecules&lt;/em&gt; that contain &lt;em&gt;more than one type of atom&lt;/em&gt; and &lt;em&gt;do not contain a centre of symmetry&lt;/em&gt; are polar (to some extent). However, often — especially in the case of hydrocarbons with a difference in electronegativity of only $0.4$ — the polarity is so weak that the molecules can be considered non-polar.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-17T21:11:01.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31611" PostTypeId="2" ParentId="31460" CreationDate="2015-05-17T21:23:49.910" Score="3" Body="&lt;p&gt;First, I would suggest you read the material safety data sheets (MSDS) for each of those products to be sure of any incompatibility that may arise. They are available throughout the web or on distributor web sites like Sigma or Fisher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your list of waste disposal (as it seems to be) most of your products are alcohols or water so they can be mixed without issues.&#xA;The ethyl acetate will react with water to create ethanol and acetic acid and acids may be an issue if you have great quantities. So I would suggest to react and neutralise with $\ce{NaOH}$ prior disposal.&#xA;Same thing for acetic acid: you should neutralise prior disposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also request an evaluation of price from your chemical waste service contractor, which will give you the price of each disposal category with the incompatibility management included. &lt;/p&gt;&#xA;" OwnerUserId="15795" LastEditorUserId="5354" LastEditDate="2015-06-19T15:08:56.393" LastActivityDate="2015-06-19T15:08:56.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31612" PostTypeId="2" ParentId="31599" CreationDate="2015-05-17T21:47:39.773" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why is the second path not taken?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\ce{H}$ can migrate, it is just that phenyl migrates better.  The reason that phenyl migrates better is because it is much better able to stabilize the &lt;strong&gt;developing&lt;/strong&gt; carbocation center.  The phenyl group participates by what is termed &lt;a href=&quot;http://en.wikipedia.org/wiki/Neighbouring_group_participation&quot; rel=&quot;noreferrer&quot;&gt;neighboring group participation&lt;/a&gt; (see the &quot;NGP by an aromatic ring&quot; section).  It forms a &lt;strong&gt;phenonium ion&lt;/strong&gt;.  Note the geometry of this ion, the former aromatic ring is perpendicular to the plane of the screen while the former $\ce{C-C}$ bond is contained in the plane of the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MM88p.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://rowdy.msudenver.edu/~wiederm/401chp/unit2chp/ngpchp/page12.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can draw several resonance structures to describe the phenonium ion, this leads to it being especially stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/peIxb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Neighbouring_group_participation&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such extra stabilization is not possible with a migrating hydrogen atom. Therefore the pathway involving the phenonium ion is lower energy (faster) than the pathway involving hydrogen migration.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-17T21:47:39.773" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31613" PostTypeId="2" ParentId="31577" CreationDate="2015-05-17T22:00:37.173" Score="2" Body="&lt;p&gt;As @Ridho requested, you should clarify the meaning of primary and secondary standard as those expression gets differents meaning depending on the context used. For my part I will try to answer you with this.&#xA;When you are doing a sample determination by titration, you usually use a titration compound which is easy to calculate proportion with, like NaOH or HCl for pH titration. Those two are pretty well forward, they are widely available, they are considered to be 100% dissociated when mixed in water, and 1N of NaOH/HCl is considered to neutralise 1N of anything else in the pH world. But, it comes with a catch. Those two don't give solutions that are stable in term of concentration. NaOH pellets are highly hygroscopic and HCl tend to &quot;evaporate&quot; as it is primarely a gas. So even if you weight 39.9971g of NaOH and you make a solution of 1000.000mL, you will not get a 1.00N solution. So you need to confirm or standardize your solution with another compound which is more stable, got an easy to work with pHI, but is useless for your first détermination. For NaOH, in our lab we use potassium hydrogen phthalate with phenolphthalein. &lt;/p&gt;&#xA;" OwnerUserId="15795" LastActivityDate="2015-05-17T22:00:37.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31614" PostTypeId="2" ParentId="31596" CreationDate="2015-05-17T22:15:45.810" Score="3" Body="&lt;p&gt;There is an incredible amount of such reactions. Consider the equation from thermodynamics,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ΔG = ΔH - TΔS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔG is change in Free Energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔH is change in Enthalpy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T is Temperature and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ΔS is change in Entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that a reaction is spontaneous and favorable if ΔG is negative. We therefore see that a reaction that increases entropy (ΔS &gt; 0) will be more likely to be favorable. Although not quite technically correct, Entropy can be thought of as a measure of disorder. Therefore, if a reaction results in products that are more chaotic than the reactants, it will be more likely to be favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction of solids, liquids, or both that produce a gas will increase the entropy of the system. This is because gasses are much more chaotic and unordered than solids or liquids. Therefore, there will be not just many, but an unfathomably large number of reactions that will take place at room temperature that will spontaneously occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the reaction results in a release of heat (ΔH &amp;lt; 0), then it is guaranteed to occur at any temperature if it produces gas in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples include explosives, in which much gas is produced at incredible rates. Acid-metal reactions produce Hydrogen gas as well. &lt;/p&gt;&#xA;" OwnerUserId="15236" LastActivityDate="2015-05-17T22:15:45.810" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31615" PostTypeId="1" CreationDate="2015-05-17T22:40:03.590" Score="2" ViewCount="1719" Body="&lt;p&gt;How can one read from the periodic table the number of outer shell electrons that an atom has, to predict how these atoms will make bonds with other atoms? For example to see that hydrogen ($\ce{H}$) has 1 electron free, while carbon ($\ce{C}$) has 4 and oxygen ($\ce{O}$) has 2? This allows us to infer that carbon can make four bonds with 4 different hydrogen atoms. It's not clear what in the periodic table would tell us this; it is neither in the group nor in the period of an atom, it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to this: can someone summarize concisely all the properties (like number of free outer shell electrons) that one can read off about each element from its position in the periodic table? I was having a hard time finding a concise summary of these.&lt;/p&gt;&#xA;" OwnerUserId="16310" LastEditorUserId="15410" LastEditDate="2015-05-17T23:59:04.830" LastActivityDate="2015-05-18T06:56:22.493" Title="Reading number of outer shell electrons and other properties from periodic table?" Tags="&lt;electronic-configuration&gt;&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31617" PostTypeId="1" AcceptedAnswerId="31621" CreationDate="2015-05-18T00:27:59.790" Score="5" ViewCount="160" Body="&lt;p&gt;If Omega-3 is a fatty acid with a single double bond at the third carbon atom from the omega end, then why is it regarded as a polyunsaturate and not a monounsaturate? Are there different Omega-3s with more than one double bond?&lt;/p&gt;&#xA;" OwnerUserId="16312" LastEditorUserId="15410" LastEditDate="2015-05-18T01:47:16.300" LastActivityDate="2015-05-18T01:47:16.300" Title="Why is Omega-3 a polyunsaturated and not a monounsaturated fatty acid?" Tags="&lt;biochemistry&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31618" PostTypeId="1" AcceptedAnswerId="31660" CreationDate="2015-05-18T00:36:20.503" Score="24" ViewCount="8852" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TlOsu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TlOsu.png&quot; alt=&quot;Reaction of phenylmagnesium bromide with cyclohex-2-en-1-one&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclohexenone, an α,β-unsaturated ketone, is electrophilic at both the carbonyl carbon, as well as the β-carbon. So, it can either undergo 1,2-addition to give the allylic alcohol &lt;strong&gt;1&lt;/strong&gt;, or 1,4-addition to give the ketone &lt;strong&gt;2&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the Grignard reagent add to?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="16683" LastEditDate="2017-08-31T18:20:35.507" LastActivityDate="2017-08-31T18:34:28.203" Title="Does a Grignard reagent react with enones to give the 1,2- or 1,4- product?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;grignard-reagent&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="31620" PostTypeId="1" CreationDate="2015-05-18T01:04:43.317" Score="5" ViewCount="166" Body="&lt;p&gt;I have multiple PDB files containing a single molecules of polysaccharides, with a single chain and a long list of all the atoms -they are not divided by residue. &#xA;The PDB files were created from &lt;a href=&quot;http://lipidbank.jp/cgi-bin/detail.cgi?id=CLS5224&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemDraw files&lt;/a&gt; using OpenBabel and Corina applet form molecular-networks. The workflow was: .cdx &amp;lt;-&gt; OpenBabel &amp;lt;-&gt; smiles/sdf &amp;lt;-&gt; Corina &amp;lt;-&gt; pdb.&#xA;What i end up with is this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ov8Hk.png&quot; alt=&quot;difference between my PDB and 1CAP&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do i easily break up the list into individual residues(individual monosaccharides)? I trying to something similar to &lt;a href=&quot;http://www.rcsb.org/pdb/explore.do?structureId=1cap&quot; rel=&quot;nofollow noreferrer&quot;&gt;1CAP&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what the PDB file looks like after CORINA:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HEADER    UNK                                     15-05-08    1UNK&#xA;REMARK   1 corina 3.48 0000  08.02.2010                               &#xA;HETATM    1  C1  UNK     1       6.576  -2.524   4.463  1.00 20.00&#xA;HETATM    2  H2  UNK     1       6.573  -2.048   5.444  1.00 20.00&#xA;HETATM    3  C3  UNK     1       5.876  -1.614   3.449  1.00 20.00&#xA;HETATM    4  C4  UNK     1       4.411  -1.438   3.863  1.00 20.00&#xA;HETATM    5  C5  UNK     1       3.748  -2.815   3.964  1.00 20.00&#xA;HETATM    6  H6  UNK     1       3.760  -3.296   2.986  1.00 20.00&#xA;HETATM    7  O7  UNK     1       4.467  -3.621   4.900  1.00 20.00&#xA;HETATM    8  C8  UNK     1       5.829  -3.859   4.541  1.00 20.00&#xA;HETATM    9  C9  UNK     1       6.487  -4.751   5.595  1.00 20.00&#xA;HETATM   10  O10 UNK     1       5.864  -6.037   5.587  1.00 20.00&#xA;HETATM   11  O11 UNK     1       2.398  -2.662   4.404  1.00 20.00&#xA;HETATM   12  C12 UNK     1       1.655  -3.882   4.435  1.00 20.00&#xA;HETATM   13  C13 UNK     1       0.198  -3.584   4.795  1.00 20.00&#xA;HETATM   14  C14 UNK     1      -0.566  -4.899   4.964  1.00 20.00&#xA;HETATM   15  H15 UNK     1      -0.496  -5.481   4.045  1.00 20.00&#xA;HETATM   16  C16 UNK     1      -2.036  -4.592   5.265  1.00 20.00&#xA;HETATM   17  C17 UNK     1      -2.607  -3.729   4.136  1.00 20.00&#xA;HETATM   18  C18 UNK     1      -1.764  -2.458   4.000  1.00 20.00&#xA;...&#xA;...&#xA;HETATM  286  H   UNK     1      14.974  11.280   1.314  1.00 20.00&#xA;HETATM  287  H   UNK     1      15.417  11.215  -0.409  1.00 20.00&#xA;HETATM  288  H   UNK     1      15.300  13.559   0.392  1.00 20.00&#xA;HETATM  289  H   UNK     1      14.052  13.241  -0.836  1.00 20.00&#xA;HETATM  290  H   UNK     1      13.609  13.306   0.886  1.00 20.00&#xA;HETATM  291  H   UNK     1       9.253  -4.179   6.144  1.00 20.00&#xA;HETATM  292  H   UNK     1      11.359  -2.412   4.593  1.00 20.00&#xA;CONECT    1    2    8    3  115&#xA;CONECT    2    1&#xA;CONECT    3    1    4   86  120&#xA;CONECT    4    3    5   69  121&#xA;CONECT    5    4    6    7   11&#xA;CONECT    6    5&#xA;CONECT    7    5    8&#xA;CONECT    8    7    1    9  122&#xA;...&#xA;CONECT  289  114&#xA;CONECT  290  114&#xA;CONECT  291  118&#xA;CONECT  292  119&#xA;END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="16313" LastEditorUserId="16313" LastEditDate="2015-05-18T19:09:37.553" LastActivityDate="2015-06-17T19:57:33.523" Title="How would you add saccharide residue information to PDB files?" Tags="&lt;biochemistry&gt;&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31621" PostTypeId="2" ParentId="31617" CreationDate="2015-05-18T01:21:02.260" Score="4" Body="&lt;p&gt;The term &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Omega-3_fatty_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;omega-3&lt;/a&gt;&quot; just denotes that the position of the &lt;strong&gt;first&lt;/strong&gt; double bond in the fatty acid will be at position &quot;3&quot; (counting from the omega end of the chain).  It does not mean that there is only one double bond in the fatty acid.  The most common omega-3 fatty acids contain between 3 and 6 &lt;em&gt;cis&lt;/em&gt; double bonds (therefore they are polyunsaturated).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are few examples, again note how they all have a double bond at the third carbon from the omega end of the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;α-Linolenic acid (ALA)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/tEf1B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eicosapentaenoic acid (EPA)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/o3zY1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Docosahexaenoic acid (DHA)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rXHSG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-18T01:26:41.490" LastActivityDate="2015-05-18T01:26:41.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31624" PostTypeId="2" ParentId="31599" CreationDate="2015-05-18T04:45:44.107" Score="6" Body="&lt;p&gt;In both the cases that you mentioned, the lone pair of oxygen is finally going to stabilize the carbocation by donating towards the positively charged carbon atom.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence stability of the carbocation may not be a very good criteria of comparing the mechanism. The migratory apptitude then plays a major role in deciding the faster mechanism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stability is just the thermodynamic aspect while migratory apptitude is the kinetic aspect.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both need to be considered. Since the stability brought about by the oxygen is similar in both cases, we chose the kinetic aspect as the determining criteria.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And ron's answer clearly outlines the kinetic aspect.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-05-18T04:45:44.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31625" PostTypeId="2" ParentId="31615" CreationDate="2015-05-18T05:16:05.367" Score="2" Body="&lt;p&gt;Start with a periodic table that shows the electron configurations. You might try the &lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow&quot;&gt;Dynamic Periodic Table&lt;/a&gt;; as you mouse over an element, its electron configuration is presented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then go to &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/intro2.htm&quot; rel=&quot;nofollow&quot;&gt;Electron Configurations in the Periodic Table&lt;/a&gt;, &lt;a href=&quot;http://www.mikeblaber.org/oldwine/chm1045/notes/Struct/EPeriod/Struct09.htm&quot; rel=&quot;nofollow&quot;&gt;Electronic Structure of Atoms&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Block_%28periodic_table%29&quot; rel=&quot;nofollow&quot;&gt;Block (periodic table)&lt;/a&gt; to see how shells and orbitals are added. It's pretty straightforward until the transition metals, and things get much more complicated for the lanthanides and actinides, where a new shell may start before an inner shell is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general rule is: an orbital is more &quot;stable&quot; when filled, or half-filled, so fluorine, for example, can readily accept one more electron to finish its outer &lt;strong&gt;2s2 2p5&lt;/strong&gt; shell and it will then have the configuration of neon, &lt;strong&gt;2s2 2p6&lt;/strong&gt;. Many periodic tables don't present the inner shells, using instead the shorthand of &lt;strong&gt;[Ne] 3s2 3p5&lt;/strong&gt; for chlorine, rather than the full &lt;strong&gt;1s2 2s2 2p6 3s2 3p5&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you're not alone in trying to understand the electron structure; see the current debate on the position of Lw in the table: &lt;a href=&quot;https://www.sciencenews.org/article/new-data-synthetic-element-trigger-rethink-periodic-table&quot; rel=&quot;nofollow&quot;&gt;New data on synthetic element trigger rethink of periodic table&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, note there are different ways a shell can be completed. Carbon will &lt;strong&gt;share&lt;/strong&gt; its four outer electrons with four hydrogen atoms; this is called a &lt;em&gt;covalent&lt;/em&gt; bond. Chlorine outright &quot;steals&quot; an electron from sodium; this is an ionic bond. See &lt;a href=&quot;http://www.kentchemistry.com/links/bonding/covalentlewisdot.htm&quot; rel=&quot;nofollow&quot;&gt;Covalent Lewis Dot Structures&lt;/a&gt; and &lt;a href=&quot;https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/basic-concepts-of-chemical-bonding-9/the-ionic-bond-72/ionic-bonding-and-electron-transfer-335-1484/&quot; rel=&quot;nofollow&quot;&gt;Ionic Bonding and Electron Transfer&lt;/a&gt; for more info.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="11583" LastEditDate="2015-05-18T06:56:22.493" LastActivityDate="2015-05-18T06:56:22.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31626" PostTypeId="1" CreationDate="2015-05-18T05:37:40.930" Score="0" ViewCount="203" Body="&lt;p&gt;In other words: What is the purpose of the solution that the electrodes are immersed in, in a galvanic cell, and is it possible to generate a current without any solution?&lt;/p&gt;&#xA;" OwnerUserId="16319" LastActivityDate="2015-05-18T05:37:40.930" Title="What is the purpose of the solution in a galvanic cell?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;&lt;electricity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31630" PostTypeId="1" CreationDate="2015-05-18T09:11:45.003" Score="5" ViewCount="1510" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following undergoes disproportionation with $\ce{NaOH}$?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{SiO2}$ (silicon dioxide)&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{CO2}$ (carbon dioxide)&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{SO2}$ (sulfur dioxide)&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{SO3}$ (sulfur trioxide)&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{NO2}$ (nitrogen dioxide)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the above we can conclude that $\ce{SiO2}$, $\ce{CO2}$ and $\ce{SO3}$ do not undergo disproportionation as they are all in the maximum oxidation states. $\ce{NO2}$ is in the +4 oxidation states so it disproportionates in to +3 and +5 upon addition of aqueous $\ce{NaOH}$. But what about $\ce{SO2}$? It can also disproportionates because it is in +4 oxidation no. But the answer states only $\ce{NO2}$. Can anyone please explain the reason?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="5026" LastEditDate="2018-07-18T03:06:07.783" LastActivityDate="2018-07-18T03:06:07.783" Title="Disproportionation with aqueous NaOH" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31631" PostTypeId="1" AcceptedAnswerId="31645" CreationDate="2015-05-18T09:19:02.663" Score="11" ViewCount="534" Body="&lt;p&gt;Following on from &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31587/are-there-any-known-chemical-properties-of-tritium-water-that-make-it-unusually/31592?noredirect=1#comment45832_31592&quot;&gt;this question about tritium water&lt;/a&gt;, I was wondering whether the decay of tritium to helium-3 has ever been used in organic synthesis. Such a transition would presumably result in no strong $\ce{C-He}$ interaction and thus radical or cation formation at $\ce{C}$.  &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2016-06-23T15:41:37.307" Title="Radical formation via tritium to helium decay" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;isotope&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31632" PostTypeId="2" ParentId="7190" CreationDate="2015-05-18T11:02:02.097" Score="4" Body="&lt;p&gt;According to the old IUPAC recommendations (1979 as well as 1993), compounds having the general structure $\ce{R-CH=N-OH}$ or $\ce{RR'C=N-OH}$ are called generically ‘oximes’. They are further classified as ‘aldoximes’ and ‘ketoximes’, respectively. They are named by placing the functional class name ‘oxime’ as a separate word after the name of the corresponding aldehyde $\ce{R-CH=O}$ or ketone $\ce{RR'C=O}$, respectively.&lt;br&gt;&#xA;Therefore, the systematic name of your example according to the principles of functional class nomenclature is &lt;strong&gt;butan-2-one oxime.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, functional modifiers are still acceptable for general nomenclature purposes, but the preferred IUPAC names (PIN) are substitutive names for oximes. They are named substitutively as ‘ylidene’ derivatives of hydroxylamine and not by functional class nomenclature as in previous recommendations.&lt;br&gt;&#xA;Therefore, the preferred IUPAC name of your example is &lt;strong&gt;(butan-2-ylidene)hydroxylamine.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z0cCi.png&quot; alt=&quot;(butan-2-ylidene)hydroxylamine&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-18T11:02:02.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31633" PostTypeId="1" AcceptedAnswerId="31636" CreationDate="2015-05-18T11:20:59.683" Score="1" ViewCount="241" Body="&lt;p&gt;I understand that 2-pyridone derivatives can be synthesized by oxidizing an &lt;em&gt;N&lt;/em&gt;-methyl pyridinium salt with $\ce{[Fe(CN)6]^3-}$ under basic conditions. Can a similar method be used to produce 4-pyridones ? &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="4945" LastEditDate="2015-05-18T12:01:19.807" LastActivityDate="2015-05-18T13:01:32.867" Title="Can 4-pyridones be synthesised analogously to 2-pyridones?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31634" PostTypeId="2" ParentId="31595" CreationDate="2015-05-18T11:31:25.720" Score="3" Body="&lt;p&gt;Under acidic conditions, a part of the divinylether will probably just undergo cationic polymerization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lUUlE.png&quot; alt=&quot;cationic polymerization of divinylether&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-18T11:31:25.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31636" PostTypeId="2" ParentId="31633" CreationDate="2015-05-18T12:12:14.727" Score="1" Body="&lt;p&gt;A reliable and old reference on the synthesis of 1-methyl-2-pyridone (&lt;em&gt;Org. Synth.&lt;/em&gt; &lt;strong&gt;1935&lt;/strong&gt;, &lt;em&gt;15&lt;/em&gt;, 41, &lt;a href=&quot;http://dx.doi.org/10.15227/orgsyn.015.0041&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;) does not mention the formation of 1-methyl-4-pyridone as a side product and I doubt that it and can be formed that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synthesis of 1-methyl-2-pyridone starts by reacting equimolar amounts of dry pyridine and dimethyl sulfate to form the N-methyl pyridinium salt. Then, the mixture is diluted with water and solutions of $\ce{NaOH}$ and $\ce{K3[Fe(CN)6]}$ are simultaneously added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to do the trick: the $\beta$-amino alcohol formed by adding $\ce{OH-}$ &lt;em&gt;ortho&lt;/em&gt; to the nitrogen atom is immediately oxidized, which prevents the opening of the pyridinium salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the synthesis of 1-alkyl-4-pyridones, a somewhat classical approach would be to react the corresponding 4-pyranone with a primary amine under more or less drastic conditions, as shown in &lt;em&gt;Can. J. Chem&lt;/em&gt;., &lt;strong&gt;1976&lt;/strong&gt;, &lt;em&gt;54&lt;/em&gt;, 3377. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altenatively, &lt;em&gt;Grignard&lt;/em&gt; reaction of a 2-substituted 4-benzyloxypyridine-N-oxide, followed by debenzylation/methylation with methyl triflate and subsequent treatment with $\ce{NaOH}$ will yield the desired a 1,2,5-trisubstituted 4-pyridone (&lt;a href=&quot;http://dx.doi.org/10.1016/j.tetlet.2010.06.019&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-18T13:01:32.867" LastActivityDate="2015-05-18T13:01:32.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31637" PostTypeId="5" CreationDate="2015-05-18T12:25:59.853" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-18T12:25:59.853" LastActivityDate="2015-05-18T12:25:59.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31638" PostTypeId="4" CreationDate="2015-05-18T12:25:59.853" Score="0" Body="Different forms of an element which exist in the same state but have different molecular or crystalline structures. Classic examples of allotropes include diamond and graphite (allotropes of carbon), red phosphorous and white phosphorous, and $\ce{O2}$ and $\ce{O3}$ (allotropes of oxygen)." OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-18T12:52:26.150" LastActivityDate="2015-05-18T12:52:26.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31639" PostTypeId="1" AcceptedAnswerId="31641" CreationDate="2015-05-18T12:40:16.830" Score="26" ViewCount="35993" Body="&lt;p&gt;I know how to calculate them and such stuff, but I wanted to know what they actually signify. I have a vague idea that they have something to do with an electron's position in an atom but what do all of them mean? Any help would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="16324" LastActivityDate="2016-12-09T02:54:24.007" Title="What do the quantum numbers actually signify?" Tags="&lt;quantum-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31640" PostTypeId="1" CreationDate="2015-05-18T13:03:41.577" Score="2" ViewCount="7367" Body="&lt;p&gt;If we look at the values for the atomic radii (look at the table &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_radius#Empirically_measured_atomic_radii&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), we can see that they rapidly decrease across the period initially. Looking at the second period, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lP7TL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph is pretty steep early on. But further down the period, as we enter the p-block elements, the graph levels out. This trend is repeated across the third and higher periods as well. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation of atomic radii across the d-block and f-block elements is even more gradual. Why is that again? &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorDisplayName="user84548" LastEditDate="2019-10-09T10:35:13.203" LastActivityDate="2019-10-09T10:35:13.203" Title="Variation in atomic radii of elements in different blocks?" Tags="&lt;periodic-trends&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31641" PostTypeId="2" ParentId="31639" CreationDate="2015-05-18T13:23:29.540" Score="33" Body="&lt;p&gt;Quantum numbers give information about the location of an electron or set of electrons.  A full set of quantum numbers describes a unique electron for a particular atom.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it as the mailing address to your house.  It allows one to pinpoint your exact location out of a set of $n$ locations you could possibly be in.  We can narrow the scope of this analogy even further.  Consider your daily routine.  You may begin your day at your home address but if you have an office job, you can be found at a different address during the work week.  Therefore we could say that you can be found in either of these locations depending on the time of day.  The same goes for electrons.  Electrons reside in atomic orbitals (which are very well defined 'locations').  When an atom is in the ground state, these electrons will reside in the lowest energy orbitals possible (e.g. 1$s^2$ 2$s^2$ and 2$p^2$ for carbon).  We can write out the physical 'address' of these electrons in a ground-state configuration using quantum numbers as well as the location(s) of these electrons when in some non-ground (i.e. excited) state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could describe your home location any number of ways (GPS coordinates, qualitatively describing your surroundings, etc.) but we've adapted to a particular &lt;em&gt;formalism&lt;/em&gt; in how we describe it (at least in the case of mailing addresses).  The quantum numbers have been laid out in the same way.  We could communicate with each other that an electron is &quot;located in the lowest energy, spherical atomic orbital&quot; but it is much easier to say a spin-up electron in the  1$s$ orbital instead.  The four quantum numbers allows us to communicate this information numerically without any need for a wordy description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course carbon is not always going to be in the ground state.  Given a  wavelength of light for example, one can excite carbon in any number of ways.  Where will the electron(s) go?  Regardless of what wavelength of light we use, we know that we can describe the final location(s) using the four quantum numbers.  You can do this by writing out all the possible permutations of the four quantum numbers.  Of course, with a little more effort, you could predict the exact location where the electron goes but in my example above, you know for a fact you could describe it using the quantum number formalism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantum numbers also come with a set of restrictions which inherently gives you useful information about where electrons will NOT be. For instance, you could never have the following possible quantum numbers for an atom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$=1; $l$=0; $m_l$=0; $m_s$=1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$=1; $l$=0; $m_l$=0; $m_s$=-1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$=1; $l$=0; $m_l$=0; $m_s$=1/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;em&gt;set&lt;/em&gt; of quantum numbers indicates that three electrons reside in the 1$s$ orbital which is impossible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Jan stated in his post, these quantum numbers are derived from the solutions to the Schrodinger equation for the hydrogen atom (or a 1-e$^-$ system).  There are any number of solutions to this equation that relate to the possible energy levels of they hydrogen atom.  Remember, energy is QUANTIZED (as postulated by Max Planck).  That means that an energy level may exist (arbitrarily) at 0 and 1 but NEVER in between.  There is a discrete 'jump' in energy levels and not some gradient between them.  From these solutions a &lt;em&gt;formalism&lt;/em&gt; was constructed to communicate the solutions in a very easy, numerical way just as mailing addresses are purposefully formatted in such a way that is easy that anyone can understand with minimal effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the quantum numbers not only tell you where electrons will be (ground state) and can be (excited state), &lt;em&gt;but also&lt;/em&gt; will tell you where electrons &lt;strong&gt;cannot&lt;/strong&gt; be in an atom (due to the restrictions for each quantum number).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Principle quantum number ($n$)&lt;/strong&gt; - indicates the orbital size.  Electrons in atoms reside in atomic orbitals.  These are referred to as $s,p,d,f...$ type orbitals.  A $1s$ orbital is smaller than a $2s$ orbital.  A $2p$ orbital is smaller than a $3p$ orbital.  This is because orbitals with a larger $n$ value are getting larger due to the fact that they are further away from the nucleus.  The principle quantum number is an integer value where $n$ = 1,2,3... .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Angular quantum number ($l$)&lt;/strong&gt; - indicates the shape of the orbital. Each type of orbital ($s,p,d,f..$) has a characteristic shape associated with it.  $s$-type orbitals are spherical while $p$-type orbitals have 'dumbbell' orientations.  The orbitals described by $l$=0,1,2,3... are $s,p,d,f...$ orbitals, respectively.  The angular quantum number ranges from 0 to $n$-1.  Therefore, if $n$ = 3, then the possible values of $l$ are 0, 1, 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Magnetic quantum number ($m_l$)&lt;/strong&gt; - indicates the orientation of a particular orbital in space.  Consider the $p$ orbitals.  This is a &lt;em&gt;set&lt;/em&gt; of orbitals consisting of three $p$-orbitals that have a unique orientation in space.  In Cartesian space, each orbital would like along an axis (x, y, or z) and would be centered around the origin at 0,0.  While each orbital is indeed a $p$-orbital, we can describe each orbital uniquely by assigning this third quantum number to indicate its position in space.  Therefore, for a set of $p$-orbitals, there would be three $m_l$, each uniquely describing one of these orbitals.  The magnetic quantum number can have values of $-l$ to $l$.  Therefore, in our example above (where $l$ = 0,1,2) then $m_l$ would be -2, -1, 0, 1, 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spin quantum number ($m_s$)&lt;/strong&gt; - indicates the 'spin' of the electron residing in some atomic orbital.  Thus far we have introduced three quantum numbers that localize a position to an orbital of a particular size, shape and orientation.  We now introduce the fourth quantum number that describes the type of electron that can be in that orbital.  Recall that two electrons can reside inside one atomic orbital.  We can define each one uniquely by indicating the electron's spin.  According to the Pauli-exclusion principle, no two electrons can have the exact same four quantum numbers.  This means that two electrons in one atomic orbital cannot have the same 'spin'. We generally denote 'spin-up' as $m_s$ =1/2 and spin-down as $m_s$=-1/2.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-05-29T15:03:35.743" LastActivityDate="2015-05-29T15:03:35.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31642" PostTypeId="2" ParentId="31639" CreationDate="2015-05-18T13:26:47.607" Score="17" Body="&lt;p&gt;The Schrödinger equation for most system has many solutions $\hat{H}\Psi_i=E_i\Psi_i$, where $i=1,2,3,..$. In the case of the hydrogen atom the solutions has a specific notation, which are where the quantum numbers come from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the H atom the &lt;strong&gt;principal&lt;/strong&gt; quantum number $n$ refers to solutions with different energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n&amp;gt;1$ there are several solutions with the same energy, which come in different shapes ($s$, $p$, etc with different &lt;strong&gt;angular&lt;/strong&gt; quantum numbers $l$) that can point in different directions ($p_x$, $p_y$, etc with different &lt;strong&gt;magnetic&lt;/strong&gt; quantum numbers $m$)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These quantum numbers are also applied to multi-electron atoms within the AO approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the quantum numbers are a way to count (label) the solutions to the Schrödinger equation.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-05-19T11:02:25.890" LastActivityDate="2015-05-19T11:02:25.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31643" PostTypeId="2" ParentId="31631" CreationDate="2015-05-18T13:35:32.867" Score="5" Body="&lt;p&gt;Here are two reasons why this approach would be impractical for organic synthesis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The half life of tritium is 12.3 years, so the decay is too slow to be practical as part of a synthetic method&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) While on a scale of nuclear reactions, tritium decay is relatively low in energy emitted, it still dwarfs a typical chemical reaction (18.6 keV or 429,000 kCal/mole) and may cause degradation of the molecule (or the bulk material) in unpredictable ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it could be useful for other types of chemical studies. I will defer to others in the forum to elaborate.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-18T13:35:32.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31644" PostTypeId="1" CreationDate="2015-05-18T14:25:04.987" Score="29" ViewCount="28658" Body="&lt;p&gt;If we make a resonance structure for the anilinium ion, with positive charge at either the &lt;em&gt;ortho&lt;/em&gt; or the &lt;em&gt;para&lt;/em&gt; position, we get a pentavalent nitrogen, which is not possible. So, how is the $\ce{-NH3+}$ group &lt;em&gt;meta&lt;/em&gt;-directing in electrophilic aromatic substitution reactions, e.g. in nitration of aniline under acidic conditions?&lt;/p&gt;&#xA;" OwnerUserId="16157" LastEditorUserId="16683" LastEditDate="2017-10-01T12:50:07.427" LastActivityDate="2018-12-01T07:27:38.967" Title="Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;amines&gt;&lt;regioselectivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="31645" PostTypeId="2" ParentId="31631" CreationDate="2015-05-18T14:30:57.533" Score="5" Body="&lt;p&gt;There's actually quite an interesting review article about the effects of decaying tritium atoms in organic compounds. The formation of neutral radicals is very unfavourable, and almost always neutral helium is ejected leaving a carbocation behind instead. Compare the two processes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ \ce{R-He^+ -&amp;gt; R^. + \ \ He^{+}}$$&#xA;$$\ce{R-He^+ -&amp;gt; R^+ + \ He}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter is far more plausible, as the helium cation has an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionization_energy#/media/File:First_Ionization_Energy.svg&quot; rel=&quot;nofollow&quot;&gt;extremely exothermic electron affinity&lt;/a&gt; ($\rm{2370\ kJ\ mol^{-1}}$), the largest of any atomic and likely molecular monocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the article is titled &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr00024a010&quot; rel=&quot;nofollow&quot;&gt;&quot;&lt;em&gt;Tritium for generation of carbocations&lt;/em&gt;&quot;&lt;/a&gt;. The usage of tritium for carbocation formation seems to have been first investigated in 1963. This was perhaps especially important back then because carbocation chemistry had not been yet thoroughly explored until George Olah and investigation of superacidic media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an interesting excerpt of the article, from the section &quot;&lt;em&gt;Outline of the decay technique&lt;/em&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To minimize the extent of unwanted radiolytic&#xA;  processes, the activity of the tritiated precursors used&#xA;  in the decay experiments must be kept as low as possible,&#xA;  typically below 10 mCi per mol. As a consequence, the&#xA;  number of the decay ions, hence the amount of their&#xA;  neutral decay products formed in any reasonable period&#xA;  of time, is comparatively quite small. Even in a storage&#xA;  period as long as 1 year, the decay of 10 mCi of a&#xA;  monotritiated precursor in 1 mol of an inactive gas&#xA;  (“bulk gas”) yields only about $\rm{10^{16}}$ nucleogenic ions.&#xA;  Under the favorable assumption that all ions give a&#xA;  single end product, its concentration would be of the&#xA;  order of only a few parts per billion (ppb), posing&#xA;  formidable problems even to very advanced, highly&#xA;  sophisticated analytical methodologies. Even more&#xA;  seriously, the decay products would be swamped by&#xA;  those arising from the radiolysis of the bulk gas,&#xA;  promoted by the $\beta^-$ particles of tritium, since each decay&#xA;  event produces, together with a single nucleogenic ion,&#xA;  over 200 radiolytic ions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While the reactions triggered by nucleogenic carbocations will be different from those of radiolytic ions (secondary ions formed by impacts with the beta decay electron), given that the overwhelming majority of reactions will be caused by the latter, precise application of such tritium-based carbocation production for synthetic purposes seems all but futile. It would be far simpler to just target an energetic electron beam at the sample, simulating the beta decay electron and obtaining essentially the same products.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's my previous answer, which is not quite as related but I thought interesting enough to share.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Such a transition would presumably result in no strong $\ce{C-He}$ interaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While in practice you are correct, I thought I'd point out something interesting. &lt;a href=&quot;https://books.google.com/books/about/Modeling_Marvels.html?id=IoFzgBSSCwEC&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;Some computational investigations&lt;/a&gt; have found that certain compounds such as $\ce{He^+ -C\equiv C-H}$ have a considerable bond between helium and carbon, with a bond dissociation energy of up to $\rm{92\ kJ\ mol^{-1}}$ at the B3LYP/6-31G** level of theory. Even the neutral molecule $\ce{HHeCCH}$ is expected to be bound, though with a small kinetic barrier to decomposition into $\ce{He + HCCH}$ of approximately $\rm{63\ kJ\ mol^{-1}}$. Such calculations have motivated the usage of tritiated molecules as a possible (and perhaps a bit desperate) synthetic pathway to helium compounds, which curiously are expected to be more stable than neon compounds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as has been mentioned, the slow decay of tritium (only 0.015% of a sample decays during the course of a day) precludes synthetic usage of its transmutation except perhaps for extremely specific catalytic purposes. Increasing the tritium content in a sample could quickly make costs exorbitant, and all sorts of crazy secondary reactions pathways (having their activation energies met from the abundant energy of nuclear decay) would soon turn organic reactants into a goopy radioactive mess.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-05-18T15:53:43.617" LastActivityDate="2015-05-18T15:53:43.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31646" PostTypeId="2" ParentId="31618" CreationDate="2015-05-18T14:31:45.003" Score="12" Body="&lt;p&gt;The Grignard reaction, although very well examined, there are still arguments about the mechanism and the overall process.  The reagent is usually best described via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Schlenk_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schlenk equilibrium&lt;/a&gt;, but that is not only dependent on $\ce{X}$, but also on the solvent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2RMgX &amp;lt;=&amp;gt; R2Mg + MgX2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There also have been some dimeric structures reported, i.e. the symmetrical, the asymmetrical, and the alkyl bridging dimer. According to Milton Orchin&lt;sup&gt;[1]&lt;/sup&gt; (and references therein) the latter is responsible for the formation of the above equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QyvYo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QyvYo.png&quot; alt=&quot;Grignard dimers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many mechanism possible, two of them following a non-radical pathway are depicted below. In both cases, the 1,2-addition leads to the major product. The six-membered transition state (top route) has been proposed many times and employs the alkyl bridging dimer.&lt;sup&gt;[2]&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom route considers only the monomeric Grignard reagent. Whichever pathway is taken, the result is the same: 1,2-addition, resulting in a tertiary alcohol, with the C=C double bond retained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d2jrF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2jrF.png&quot; alt=&quot;Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes and References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been correctly pointed out by ron that Grignard agents cannot coexist with acid, as my previous draft suggested. Even water poses a big threat to the stability, as pointed out by yasir. I was a little confused about this, because adding a $\ce{Cu(I)}$ co-catalyst will result in the 1,4-product,&lt;sup&gt;[3]&lt;/sup&gt; and I assumed the same would happen with an acid.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://doi.org/10.1021/ed066p586&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. Orchin, &lt;em&gt;J. Chem. Educ.&lt;/em&gt;, &lt;strong&gt;1989&lt;/strong&gt;, &lt;em&gt;66&lt;/em&gt; (7), 586.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;See for example: &lt;a href=&quot;https://doi.org/10.1351/pac198052030545&quot; rel=&quot;nofollow noreferrer&quot;&gt;E. C. Ashby. &lt;em&gt;A detailed description of the mechanism of reaction of Grignard reagents with ketones&lt;/em&gt;. International Conference on Organometallic Chemistry, ICOMC, Organometallic Chemistry, IX, University of Dijon, France, 1979-09-03–1979-09-07.&lt;/a&gt; Also available for free at &lt;a href=&quot;https://www.iupac.org/publications/pac/1980/pdf/5203x0545.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;iupac.org&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://doi.org/10.1021/ja01854a005&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. S. Kharasch, and P. O. Tawney, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1941&lt;/strong&gt;, &lt;em&gt;63&lt;/em&gt; (9), 2308–2316.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-08-31T18:34:28.203" LastActivityDate="2017-08-31T18:34:28.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31647" PostTypeId="2" ParentId="31620" CreationDate="2015-05-18T14:37:06.463" Score="0" Body="&lt;p&gt;Make sure you have different residue numbers for each residue. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;HETATM    1  C1  NAG A 600      19.340   9.045  41.827  1.00 23.83           C  &#xA;HETATM    2  C2  NAG A 600      20.265  10.267  41.992  1.00 25.36           C  &#xA;HETATM    3  N2  NAG A 600      20.722  10.713  40.685  1.00 21.83           N  &#xA;HETATM    4  C3  NAG A 600      21.468   9.913  42.881  1.00 23.92           C  &#xA;HETATM    5  O3  NAG A 600      22.240  11.098  43.157  1.00 31.96           O  &#xA;HETATM    6  C4  NAG A 600      20.974   9.305  44.194  1.00 29.02           C  &#xA;HETATM    7  O4  NAG A 600      22.087   8.931  45.028  1.00 32.10           O  &#xA;HETATM    8  C5  NAG A 600      20.112   8.085  43.874  1.00 25.92           C  &#xA;HETATM    9  O5  NAG A 600      18.973   8.502  43.103  1.00 26.24           O  &#xA;HETATM   10  C6  NAG A 600      19.585   7.353  45.095  1.00 29.15           C  &#xA;HETATM   11  O6  NAG A 600      18.639   8.141  45.801  1.00 30.55           O  &#xA;HETATM   12  C7  NAG A 600      20.897  12.007  40.404  1.00 19.02           C  &#xA;HETATM   13  O7  NAG A 600      21.415  12.386  39.350  1.00 26.03           O  &#xA;HETATM   14  C8  NAG A 600      20.437  13.046  41.410  1.00 12.16           C  &#xA;HETATM   15  C1  FUC A 601      23.465  11.198  42.483  1.00 40.56           C  &#xA;HETATM   16  C2  FUC A 601      24.142  12.529  42.836  1.00 45.83           C  &#xA;HETATM   17  O2  FUC A 601      23.243  13.605  42.609  1.00 43.55           O  &#xA;HETATM   18  C3  FUC A 601      24.578  12.521  44.304  1.00 52.54           C  &#xA;HETATM   19  O3  FUC A 601      25.277  13.719  44.606  1.00 54.93           O  &#xA;HETATM   20  C4  FUC A 601      25.476  11.312  44.578  1.00 53.83           C  &#xA;HETATM   21  O4  FUC A 601      26.684  11.438  43.843  1.00 59.66           O  &#xA;HETATM   22  C5  FUC A 601      24.755  10.026  44.164  1.00 52.07           C  &#xA;HETATM   23  O5  FUC A 601      24.344  10.110  42.778  1.00 46.32           O  &#xA;HETATM   24  C6  FUC A 601      25.630   8.795  44.302  1.00 50.48           C  &#xA;HETATM   25  C1  NAG A 602      22.004   9.334  46.354  1.00 36.94           C  &#xA;HETATM   26  C2  NAG A 602      23.011   8.547  47.198  1.00 40.27           C  &#xA;HETATM   27  N2  NAG A 602      22.674   7.137  47.173  1.00 35.92           N  &#xA;HETATM   28  C3  NAG A 602      23.018   9.060  48.643  1.00 41.77           C  &#xA;HETATM   29  O3  NAG A 602      24.056   8.420  49.372  1.00 41.95           O  &#xA;HETATM   30  C4  NAG A 602      23.233  10.575  48.670  1.00 44.23           C  &#xA;HETATM   31  O4  NAG A 602      23.108  11.054  50.004  1.00 45.34           O  &#xA;HETATM   32  C5  NAG A 602      22.198  11.260  47.766  1.00 44.11           C  &#xA;HETATM   33  O5  NAG A 602      22.295  10.741  46.420  1.00 38.50           O  &#xA;HETATM   34  C6  NAG A 602      22.376  12.768  47.690  1.00 43.86           C  &#xA;HETATM   35  O6  NAG A 602      23.507  13.121  46.907  1.00 44.99           O  &#xA;HETATM   36  C7  NAG A 602      23.592   6.247  46.811  1.00 37.09           C  &#xA;HETATM   37  O7  NAG A 602      24.596   6.552  46.167  1.00 36.50           O  &#xA;HETATM   38  C8  NAG A 602      23.367   4.802  47.226  1.00 36.20           C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The column contains the values 600/601/602 refers residue number.&lt;/p&gt;&#xA;" OwnerUserId="16164" LastActivityDate="2015-05-18T14:37:06.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31648" PostTypeId="1" AcceptedAnswerId="31671" CreationDate="2015-05-18T14:42:13.110" Score="2" ViewCount="1890" Body="&lt;p&gt;I did a lab about spectrophotometry and right now I need the molar absorptivity(ε) of thymol blue in acidic solution and also in basic solution. I have tried looking online but I haven't found the data I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the molar absorptivity of thymol blue in acidic solution and in basic solution?&#xA; &lt;strong&gt;information&lt;/strong&gt;&#xA;The specific wavelength is 540.27 nm for the acid and 601.88 nm  for the base.&lt;/p&gt;&#xA;" OwnerUserId="16328" LastEditorUserId="9961" LastEditDate="2015-11-13T19:52:00.080" LastActivityDate="2015-11-13T19:52:00.080" Title="Thymol Blue - molar absorptivity" Tags="&lt;spectrophotometry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31649" PostTypeId="1" CreationDate="2015-05-18T14:44:17.627" Score="6" ViewCount="402" Body="&lt;p&gt;There is a sentence in my book I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Every atom or ion, which has one valence electron, will have &lt;em&gt;double lines&lt;/em&gt; in the atomic emission spectrum. For example $\ce{Na}$ will produce lines at $589.0\:\mathrm{nm}$ and $589.6\:\mathrm{nm}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="11367" LastEditDate="2015-12-24T02:17:43.513" LastActivityDate="2015-12-24T02:17:43.513" Title="Atomic emission spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31650" PostTypeId="2" ParentId="31644" CreationDate="2015-05-18T15:10:45.893" Score="30" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;how is anilinium ion meta directing for electrophiles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Actually, &lt;strong&gt;anilinium is not meta directing&lt;/strong&gt; (I know it is often taught that way), but rather it inductively deactivates the entire aromatic ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain the electrophilic substitution pattern observed with any aromatic system we must consider both resonance and inductive effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resonance Effects:&lt;/strong&gt; You are correct that you can't delocalize the positive charge on the anilinium nitrogen into the benzene ring through resonance.  As you point out, resonance structures with 5 bonds to nitrogen aren't allowed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LFweR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LFweR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;strong&gt;resonance effects are not involved in explaining the reactivity of the anilinium ion towards electrophilic aromatic substitution&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inductive Effects:&lt;/strong&gt; Inductively we expect the anilinium nitrogen to strongly withdraw electron density from the &lt;em&gt;ortho&lt;/em&gt; position since it is closest to the positively charged nitrogen, this makes the &lt;em&gt;ortho&lt;/em&gt; position the most deactivated.  Inductive effects die off quickly with distance, so we would expect the &lt;em&gt;meta&lt;/em&gt; position to also be deactivated, but less so than the &lt;em&gt;ortho&lt;/em&gt; position.  Finally, we would still expect some deactivation at the distant &lt;em&gt;para&lt;/em&gt; position, but slightly less than we saw at the &lt;em&gt;meta&lt;/em&gt; position. Overall, in terms of reactivity we would expect &lt;em&gt;para&lt;/em&gt; &gt; &lt;em&gt;meta&lt;/em&gt; &gt; &lt;em&gt;ortho&lt;/em&gt;, with the difference between &lt;em&gt;para&lt;/em&gt; and &lt;em&gt;meta&lt;/em&gt; being much smaller than the difference between &lt;em&gt;meta&lt;/em&gt; and &lt;em&gt;ortho&lt;/em&gt; due to the effect distance has on inductive electron withdrawal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we examine the nitration of aniline in strong acid ($\ce{HNO3}$, $\ce{H2SO4}$), under these conditions the strong acid will completely protonate the aniline and reaction will proceed through the anilinium ion.  We see that, as expected, there is very little &lt;em&gt;ortho&lt;/em&gt; product and slightly more &lt;em&gt;para&lt;/em&gt; product than &lt;em&gt;meta&lt;/em&gt; product.  This is pretty much as we expected and is &lt;strong&gt;consistent with the electron withdrawing inductive effect of the positive anilinium nitrogen&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vWhpY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vWhpY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-09-09T07:05:19.740" LastActivityDate="2016-09-09T07:05:19.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31651" PostTypeId="1" AcceptedAnswerId="110721" CreationDate="2015-05-18T15:41:01.737" Score="6" ViewCount="12555" Body="&lt;p&gt;I am considering building a stove out of stainless steel. I want to us stainless steel because the stove will have a double outer jacket to pass water through and heat it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will a wood fire sitting directly on stainless steel corrode it? Would a higher grade of stainless steel, say 316 instead of 304, prevent this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the stainless steel be protected in some way, perhaps by a bed of fire bricks (at least) under the fire, to prolong its life? Or will stainless steel stand up to exposure to fire without corroding for a long time?&lt;/p&gt;&#xA;" OwnerUserId="16330" LastEditorUserId="865" LastEditDate="2015-08-07T15:04:53.943" LastActivityDate="2019-03-10T10:26:20.113" Title="Will fire corrode stainless steel?" Tags="&lt;everyday-chemistry&gt;&lt;materials&gt;&lt;metallurgy&gt;" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31652" PostTypeId="2" ParentId="31649" CreationDate="2015-05-18T16:08:28.963" Score="5" Body="&lt;p&gt;When an electron in any element absorbs energy it will be promoted to an excited state.  How much energy we supply determines how many \ which excited states will be created.  As the electron returns from these various excited states to the ground state light can be emitted, this is the basis of atomic emission spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Taklj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of sodium (or any other atom with just one valence electron) many excited states are possible, so many emission lines are observed.  However, a &lt;strong&gt;very likely transition&lt;/strong&gt; will be the one from the s orbital to the next highest energy orbital, the p orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic energy of an electron spin is split into energy states with different total angular momentum due to the electron spin's interaction with the internal magnetic field set up by the orbital motion of the electron, this is termed &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;spin-orbit coupling&lt;/a&gt;. In the case of a p orbital, it is split into two states (the total angular momentum j is given as j = l +/- s, where l is the angular momentum quantum number [l=1 for a p orbital] and s is the electron spin (1/2), therefore j has two values or states for an electron in a p orbital, 1/2 and 3/2). &lt;strong&gt;It is the light emission that occurs when the excited electron returns from these two p orbital energy states to the s orbital ground state that produces the doublet seen for all atoms with a ground state involving one electron in an s orbital&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eYfoo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodzee.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-19T00:05:53.933" LastActivityDate="2015-05-19T00:05:53.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31653" PostTypeId="1" CreationDate="2015-05-18T16:30:32.873" Score="4" ViewCount="437" Body="&lt;p&gt;I asked this question because the Chinese Wikipedia community is trying to decide whether it should use Latin names or IUPAC names for the first introductory sentence for its articles on chemical elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the central argument by those support Latin names is the statement that Swedish chemist Berzelius unified the names of chemical elements using the Latin language common to European countries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They cite the source in Chinese language here:&#xA;&lt;a href=&quot;http://www.cnki.com.cn/Article/CJFDTotal-ZHHJ198805015.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.cnki.com.cn/Article/CJFDTotal-ZHHJ198805015.htm&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Swedish chemist Berzelius first proposed to use Latin, a language common to European countries, to unify the naming of chemical elements, thereby changing the chaotic status of naming them.&quot; (My translation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin. If so, it begs the question why IUPAC names for Gold, Silver, etc. are not Latin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, your help is needed to answer the question as stated in the title.&lt;/p&gt;&#xA;" OwnerUserId="16331" LastEditorUserId="5017" LastEditDate="2015-05-18T18:29:41.067" LastActivityDate="2015-05-18T18:29:41.067" Title="Did Berzelius unify the names of chemical elements using the Latin language?" Tags="&lt;nomenclature&gt;&lt;elements&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31657" PostTypeId="2" ParentId="31653" CreationDate="2015-05-18T17:21:06.783" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is essentially true. Berzelius was the first to suggest using symbols abbreviated from the names of the elements and this system, with some modifications, is still used today. Most of Berzelius' symbols were abbreviated from Latin names but a large number of the elements discovered since then have had symbols drawn from their common names and some of Berzelius' original symbols have been dropped in favour of more modern versions. Dalton had earlier used a system of drawn symbols to represent the elements but this was not widely taken up as the symbols were difficult to draw and memorise. (&lt;a href=&quot;http://www.vanderkrogt.net/elements/chemical_symbols.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking though &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_chemical_element_name_etymologies&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; wikipedia article on the etymology of chemical elements, the most common language of origin appears to be Greek (argon, chlorine etc.), with Latin coming in second (mercury, rhenium etc.), including quite a large number of names which come originally from another language (often Greek) via Latin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally there are a large number of elements which are named after places (Swedish - ytterbium, yttrium, etc. ; English - americium, californium, etc. to name just a few) or people (mendelevium, einsteinium, fermium etc.) and have been given the Latinate endings &lt;em&gt;-ium&lt;/em&gt; or occasionally &lt;em&gt;-um&lt;/em&gt; (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2832/why-do-the-names-of-most-chemical-elements-end-with-um-or-ium?rq=1&quot;&gt;this&lt;/a&gt; question for more discussion). Also note that according to &lt;a href=&quot;http://www.iupac.org/publications/pac/2002/pdf/7405x0787.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC recommendations&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For linguistic consistency, the names of all new elements should end in “-ium”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The common, and IUPAC recognised, names of gold and gilver both have old Germanic origins and are recognised as official names because of their widespread usage. I have never seen the Latin names for gold, silver or lead (&lt;em&gt;aurum&lt;/em&gt;, &lt;em&gt;argentum&lt;/em&gt; and &lt;em&gt;plumbum&lt;/em&gt;) used in a serious chemical context, although of course the elements' symbols (Au, Ag and Pb) are derived from them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such I would &lt;strong&gt;not&lt;/strong&gt; recommend using the Latin names as the main recognized name in a Wikipedia page, although a side note about them might be justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Thanks to @KlausWarzecha for pointing this out. There is one exception where the latin names (aurum, etc.) come into play: in anionic complexes, the latin version of the name is officially used. $\ce{[Cu(CN)4]^{3-}}$ is tetracyanocuprate(I), $\ce{[Ag(Cl)2]^-}$ is dichloroargentate(I), $\ce{[Au(Cl)4]^- }$ is tetrachloroaurate(III). However, this does not change the viewpoint that the Latin names should not be used when commonly discussing the element.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-05-18T18:18:27.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31660" PostTypeId="2" ParentId="31618" CreationDate="2015-05-18T17:49:53.047" Score="16" Body="&lt;p&gt;The correct answer is the 1,2-addition product (i.e. the allylic alcohol &lt;strong&gt;1&lt;/strong&gt;). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This exact transformation was reported by Akai and coworkers recently (&lt;a href=&quot;https://doi.org/10.1021/ol102053a&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Org. Lett.&lt;/em&gt; &lt;strong&gt;2010,&lt;/strong&gt; &lt;em&gt;12,&lt;/em&gt; 4900&lt;/a&gt;) and they obtained compound &lt;strong&gt;1&lt;/strong&gt; in 95% yield (the supplementary information &lt;a href=&quot;http://pubs.acs.org/doi/suppl/10.1021/ol102053a/suppl_file/ol102053a_si_001.pdf&quot; rel=&quot;noreferrer&quot;&gt;(PDF)&lt;/a&gt; is free).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2017-08-31T18:24:24.633" LastActivityDate="2017-08-31T18:24:24.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31661" PostTypeId="1" AcceptedAnswerId="31662" CreationDate="2015-05-18T17:58:46.713" Score="5" ViewCount="5184" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Give the major product formed by the heating of the following ether with HI.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;        CH3&#xA;        |&#xA;CH3-CH2-CH-CH2-O-CH2-CH3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;According to me, the product should be this due to delocalization of positive charge among the neighboring carbon atoms.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;        CH3&#xA;        |    &#xA;CH3-CH2-CH-CH2-I + CH3-CH2-OH&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I checked the answer in my book and it is completely different:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;        CH3&#xA;        |&#xA;CH3-CH2-CH-CH2-OH + CH3-CH2-I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, I think they may have considered &lt;strong&gt;steric hindrance&lt;/strong&gt; to be the deciding factor of site of attack of OH but will not &lt;strong&gt;delocalization&lt;/strong&gt; be dominant over steric effect?&lt;/p&gt;&#xA;&lt;p&gt;Also, what will be the effect of hyperconjugation on the reaction?&lt;/p&gt;&#xA;" OwnerUserId="16114" LastEditorDisplayName="user7951" LastEditDate="2020-08-22T12:50:35.040" LastActivityDate="2020-09-27T10:46:52.700" Title="Heating of Ethers with HI and Cleavage of Ethers" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;ethers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31662" PostTypeId="2" ParentId="31661" CreationDate="2015-05-18T18:13:21.763" Score="7" Body="&lt;p&gt;Your ether has $\ce{CH2}$ groups at both sides of the oxygen atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would exclude a $S_N1$ reaction here! Instead, cleavage of the ether proceeds via $S_N2$. The oxonium ion resulting from addition of a proton is attacked by iodide at the least hindered carbon atom. Iodide is quite big ;-)&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-18T18:13:21.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31663" PostTypeId="2" ParentId="31620" CreationDate="2015-05-18T18:17:15.520" Score="2" Body="&lt;p&gt;&lt;strong&gt;Short answer: your workflow doesn't include residue perception, so you would need to do this manually.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't use CORINA, but I suspect the answer is the same as if you generated coordinates using Open Babel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're using ChemDraw, which as far as I know, doesn't include any sort of residue information in the file, then creating SMILES and/or SD files, which &lt;em&gt;definitely&lt;/em&gt; don't include residue information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you create 3D coordinates with CORINA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But nowhere along that chain is there a step to create / perceive / retain residue information. So I'm not surprised that you get one large unknown residue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open Babel does have code that does amino acid and nucleic acid residue perception when translating to PDB or Mol2 or other format that allows storing residue information. &lt;em&gt;It does not have support for perceiving saccharides.&lt;/em&gt; Indeed, I'm not even sure what algorithms exist for automatically perceiving sugars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Number / label the saccharides by hand (yuck)&lt;/li&gt;&#xA;&lt;li&gt;Write some sort of script in Python to label the residues - since presumably the atoms come in a particular order&lt;/li&gt;&#xA;&lt;li&gt;Use a software builder that creates the residue information &lt;em&gt;with&lt;/em&gt; the 3D coordinates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the latter case, a quick search suggested a few possibilities, although I haven't tried any of them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://glycam.org/tools/molecular-dynamics/oligosaccharide-builder/build-glycan?id=1&quot; rel=&quot;nofollow&quot;&gt;Glycam&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.yasara.org/sugarbuilder.htm&quot; rel=&quot;nofollow&quot;&gt;Yasara&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-18T18:17:15.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31664" PostTypeId="1" AcceptedAnswerId="31701" CreationDate="2015-05-18T18:19:45.473" Score="7" ViewCount="8528" Body="&lt;p&gt;Which allotrope of phosphorus is kinetically the most stable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ruled out white and yellow as being quite reactive.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red and black allotropes are both polymeric and comparitively less reactive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black looks the most stable one &lt;strong&gt;thermodynamically&lt;/strong&gt;. But kinetically, I have no idea whether it should be red or black.&lt;br&gt;&#xA;The answer provided to me is Red.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chat.stackexchange.com/rooms/3229/conversation/discussion-on-which-allotrope-of-phosphorous-is-kinetically-the-most-stable-lead&quot;&gt;Relevant chat discussion&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15912" LastEditorUserId="4945" LastEditDate="2015-05-19T06:36:47.750" LastActivityDate="2017-03-31T16:42:34.273" Title="Which allotrope of phosphorus is kinetically the most stable?" Tags="&lt;inorganic-chemistry&gt;&lt;reactivity&gt;&lt;allotropes&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31665" PostTypeId="2" ParentId="31661" CreationDate="2015-05-18T19:34:21.070" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;According to me, The product should be this due to delocalisation of positive charge among the neighbouring Carbon atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Even if you consider an &lt;span class=&quot;math-container&quot;&gt;$S_N1$&lt;/span&gt; reaction pathway for this reaction, the positive charge could be more stable on the &lt;span class=&quot;math-container&quot;&gt;$\ce{-CH2CH3}$&lt;/span&gt; branch as it has three hyperconjugation structures stabilizing the positive charge against one in the other branch. And gives the same products.&lt;/p&gt;&#xA;&lt;p&gt;Moreover, as Klaus Warzecha said, steric hindrance comes into the play while considering &lt;span class=&quot;math-container&quot;&gt;$S_N2$&lt;/span&gt; pathway which looks a more plausible pathway due to the relatively unstable primary carbocations that form in &lt;span class=&quot;math-container&quot;&gt;$S_N1$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="13339" LastEditorDisplayName="user7951" LastEditDate="2020-09-27T10:46:52.700" LastActivityDate="2020-09-27T10:46:52.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31666" PostTypeId="1" CreationDate="2015-05-18T19:41:32.860" Score="2" ViewCount="156" Body="&lt;p&gt;Does 2-Chloro-2-methylpropane have standard heats and free energies of formation and absolute entropies values in a liquid state? &#xA;&lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/thermodynamic-data&quot; rel=&quot;nofollow&quot;&gt;http://www.wiredchemist.com/chemistry/data/thermodynamic-data&lt;/a&gt; &#xA;This website only provides the values in gaseous state and I can't seem to find the values for the liquid state anywhere.   &lt;/p&gt;&#xA;" OwnerUserId="16316" LastEditorUserId="7475" LastEditDate="2015-10-16T16:45:06.303" LastActivityDate="2015-10-16T16:45:06.303" Title="Standard Heats and Free Energies of Formation and Absolute Entropies of 2-Chloro-2-methylpropane" Tags="&lt;thermodynamics&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31667" PostTypeId="2" ParentId="31666" CreationDate="2015-05-18T19:58:03.463" Score="2" Body="&lt;p&gt;$\Delta H_\mathrm{f}^\circ = -211.3~\mathrm{kJmol^{-1}}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither source provided entropy or free energy data though but this is a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf&quot; rel=&quot;nofollow&quot;&gt;http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf&lt;/a&gt;&#xA;    &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C507200&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&quot; rel=&quot;nofollow&quot;&gt;http://webbook.nist.gov/cgi/cbook.cgi?ID=C507200&amp;amp;Units=SI&amp;amp;Mask=2#Thermo-Condensed&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T15:29:55.093" LastActivityDate="2015-05-19T15:29:55.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31670" PostTypeId="1" CreationDate="2015-05-18T23:26:36.640" Score="12" ViewCount="4481" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Identify A and B.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GYhjs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GYhjs.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;My attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The first step will be the attack of &lt;span class=&quot;math-container&quot;&gt;$\ce{PhMgX}$&lt;/span&gt; on carbonyl carbon, that is, 1,2 position followed by hydrolysis, yielding an alcohol.&lt;/p&gt;&#xA;&lt;p&gt;But how does the reaction proceed with &lt;span class=&quot;math-container&quot;&gt;$\ce{HClO4}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-02T10:38:05.900" Title="Reaction of perchloric acid with cyclopropenol" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31671" PostTypeId="2" ParentId="31648" CreationDate="2015-05-18T23:41:41.490" Score="0" Body="&lt;p&gt;First of all, molar absorptivity is dependent on wavelength, so you either need to specify a wavelength or you need the entire spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When searching for absorptivity, it's often effective to search for the older name 'extinction coefficient'. Using that search term in Google, I found an absorption spectrum for thymol blue &lt;a href=&quot;http://www.phywe-es.com/index.php/fuseaction/download/lrn_file/versuchsanleitungen/P3031101/e/LEC03_11_LV.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (pdf).&lt;/p&gt;&#xA;" OwnerUserId="6907" LastActivityDate="2015-05-18T23:41:41.490" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31672" PostTypeId="1" AcceptedAnswerId="31686" CreationDate="2015-05-19T00:53:58.783" Score="1" ViewCount="2534" Body="&lt;p&gt;I have an aqueous solution of either $\ce{MgCl2}$ or $\ce{CaCl2}$ in solution with $\ce{K2SO4}$. There is a white precipitate that forms on the bottom that does not dissolve when $\ce{HCl}$ is added (maybe it needs to be in excess?). Is there any way to find out if it's $\ce{Mg^2+}$ or $\ce{Ca^2+}$? I have already tried adding $\ce{NaOH}$, and the result was a not very cloudy, white precipitate, which is not really jelly like, but when compared to stock samples of $\ce{MgCl2}$ and $\ce{CaCl2}$ with $\ce{NaOH}$, they don't match. I tried a flame test but only could see the $\ce{K+}$ (violet flame).&lt;/p&gt;&#xA;" OwnerUserId="16063" LastEditorUserId="15410" LastEditDate="2015-05-19T01:04:45.630" LastActivityDate="2018-04-15T06:03:44.417" Title="Identifying MgCl2 or CaCl2 in solution" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31673" PostTypeId="2" ParentId="31672" CreationDate="2015-05-19T01:15:16.617" Score="2" Body="&lt;p&gt;$\ce{MgSO4}$, otherwise known as epsom salt, is soluble. $\ce{CaSO4}$ ($K_\mathrm{sp}=7.1 \times 10^{-5}$), is slightly soluble. It is extremely likely that unless there are some other unknown metal ions in this solution, the white precipitate you describe in the solution is $\ce{CaSO4}$.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-19T01:15:16.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31674" PostTypeId="1" AcceptedAnswerId="31714" CreationDate="2015-05-19T02:06:36.293" Score="2" ViewCount="83" Body="&lt;p&gt;I'm trying to attach a sensor to some tubes to measure the level of liquids in them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://robotics.stackexchange.com/questions/7256/sensing-the-level-of-a-liquid-in-a-tube&quot;&gt;see here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One tube contains toluene and the other contains Recosol R55 (shellite). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of using a capacitance level sensor like &lt;a href=&quot;http://www.molex.com/molex/products/family?channel=products&amp;amp;chanName=family&amp;amp;key=capacitive_fluid_level_sensors&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;https://www.ia.omron.com/products/family/475/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am concerned about the safety of doing this. Could the capacitance sensor ignite these liquids? is this even possible? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I need to check with the suppliers of the device. However, most of datasheets for level sensors omit this information. Perhaps they are not game enough to make any claims about their sensor in this regard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand capacitance and whether it can ignite flammable liquids.  &lt;/p&gt;&#xA;" OwnerUserId="16337" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:17.997" LastActivityDate="2015-05-19T18:16:48.013" Title="Capacitance and flammable liquid" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31675" PostTypeId="2" ParentId="6530" CreationDate="2015-05-19T02:07:21.357" Score="-1" Body="&lt;p&gt;As. ortho/ para directors are positively charged so halogens, being ortho/ para directors, are positively charged. This positivity is opposed by high electronegativity of the halogens which, make it difficult to orientate negative charge into tha ring. Thus it becomes difficult to add any substituent.&lt;/p&gt;&#xA;" OwnerUserId="16339" LastEditorUserId="4945" LastEditDate="2015-05-19T02:43:51.820" LastActivityDate="2015-05-19T02:43:51.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31676" PostTypeId="1" CreationDate="2015-05-19T02:19:20.123" Score="0" ViewCount="86" Body="&lt;p&gt;Many of you might be familiar with this and I do understand this completely, however I do have a question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with this, then this is the equation and this is the question for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ClO^{-} ~(aq) + 2I^{-} ~(aq) + H2O ~(l) -&amp;gt; Cl- ~(aq) + I2 ~(aq) + 2 {}^{-}OH ~ (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the equation above I understand everything except that in the titration process my instructions say to add $\ce{HCl}$ into the solution, I do not understand how adding $\ce{HCl}$ will have any impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will provide an accurate reference later.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="4945" LastEditDate="2015-05-19T02:50:03.863" LastActivityDate="2015-05-19T02:56:10.017" Title="Which effect does adding hydrogen chloride have when determining the concentration of sodium hypochlorite in bleach via titration have?" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31677" PostTypeId="2" ParentId="31676" CreationDate="2015-05-19T02:56:10.017" Score="3" Body="&lt;p&gt;The reaction is in equilibrium and if you don't add any acid, then some of the $\ce {ClO-}$ does not react. Adding acid will change the equilibrium to the right so most of the $\ce {ClO-}$ reacts.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-19T02:56:10.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31678" PostTypeId="1" AcceptedAnswerId="31707" CreationDate="2015-05-19T03:29:19.407" Score="5" ViewCount="2045" Body="&lt;p&gt;I know that you can have 2 acids reacting with 1 base. These 4 reactions are an example of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3 HCl + H2SO4 + NaOH -&amp;gt; 3 H2O + SCl2 + O2 + NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{6 HCl +  4 H2SO4 + 2 NaOH -&amp;gt; 8 H2O + 2 S2Cl2 + 5 O2 + 2 NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H2SO4 + 3 NaOH -&amp;gt; 3 H2O + Na2S + 2 O2 + NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H2SO4 + 3 NaOH -&amp;gt; 3 H2O + Na2SO4 + NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked and all these equations are balanced. But will these reactions forming sulfur chlorides and ionic compounds with sodium and sulfur as well as $\ce{H2O}$ and for 3 of them, $\ce{O2}$, actually occur without any added heat or are they endothermic acid base reactions and thus require heat or will the stronger acid(in this case $\ce{HCl}$) be the one that reacts and the weaker acid (in this case $\ce{H2SO4}$) just make the pH of the solution lower?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="9961" LastEditDate="2015-11-13T18:58:13.483" LastActivityDate="2015-11-13T18:58:13.483" Title="Selecting between redox and neutralization reactions for a mixture of HCl, H2SO4 and NaOH" Tags="&lt;acid-base&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31679" PostTypeId="1" AcceptedAnswerId="31683" CreationDate="2015-05-19T03:40:18.357" Score="-1" ViewCount="223" Body="&lt;p&gt;With respect &lt;a href=&quot;http://www.genome.jp/dbget-bin/www_bget?rp:RP00167&quot; rel=&quot;nofollow&quot;&gt;http://www.genome.jp/dbget-bin/www_bget?rp:RP00167&lt;/a&gt; &lt;br&gt;&lt;br&gt;&#xA;can anyone tell me how the RDM pattern were obtained?? &lt;br&gt;&lt;br&gt; Information what RDM is given here: &lt;br&gt;&lt;br&gt;&lt;a href=&quot;http://www.genome.jp/kegg/reaction/&quot; rel=&quot;nofollow&quot;&gt;http://www.genome.jp/kegg/reaction/&lt;/a&gt; &lt;br&gt;&lt;br&gt;&#xA;&lt;br&gt;&lt;br&gt;Note: This may be a silly question to all chemists, but not for me, i am a computer science student , and my research field in bioinformatics and so i have to study both biology and bit of chemistry, so please just dont close the question or be harsh on me, I have very little knowledge in chemistry and also googling doesn't help much. I asked this question in Biology forum, they closed the question and asked me to ask it here&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="418" LastEditDate="2015-05-19T07:25:47.473" LastActivityDate="2015-05-19T07:25:47.473" Title="How are RDM patterns for reactions in the KEGG database constructed?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-05-24T01:25:38.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="31680" PostTypeId="1" AcceptedAnswerId="31685" CreationDate="2015-05-19T03:46:04.940" Score="0" ViewCount="61" Body="&lt;p&gt;With respect to related R class in KEGG website:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.genome.jp/tools-bin/relclass?RC00064&quot; rel=&quot;nofollow&quot;&gt;http://www.genome.jp/tools-bin/relclass?RC00064&lt;/a&gt;  &lt;br&gt;&lt;br&gt; What is the biological or statistical relevance of the score? How is it computed? And does it signify the bonding strength of how probably it is for a pair of compounds to be an R pair?&lt;br&gt;&#xA;Note: This may be a silly question to all chemists, but not for me, i am a computer science student , and my research field in bioinformatics and so i have to study both biology and bit of chemistry, so please just dont close the question or be harsh on me, I have very little knowledge in chemistry and also googling doesn't help much. I asked this question in Biology forum, they closed the question and asked me to ask it here&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="418" LastEditDate="2015-05-19T06:04:10.660" LastActivityDate="2015-05-19T06:04:10.660" Title="What is the meaning of scores when comparing reaction classes in the KEGG database" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31681" PostTypeId="1" CreationDate="2015-05-19T04:36:45.093" Score="3" ViewCount="19997" Body="&lt;p&gt;There is no heating or cooling in this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recently move to a new house and something strange is happening with the water in every tap of the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing I noticed is when you fill a bucket with the garden hose the bubbles in the water lingered around longer than they should have, ie. normal senario fill bucket with water and bubbles pop immediately, this senario the bubbles take a few good long seconds befor they pop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second thing I noticed was after waking away from the garden hose and returning after everything was dry there was white residue all over anything that the water had been on, so I wipe it down and did it again and same thing happened, so I went and tested all the taps in the house with the cold water and the same thing happened for all the taps, so all the taps are using the same water source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of white residue being left behind after heating from a source like a kettle but never heard of white residue being left behind from a non heated source.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I can find out what this is and should I be worried about this (more than I already am)?&#xA;Or alternatively does anyone here know what this is?&lt;/p&gt;&#xA;" OwnerUserId="16342" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T17:13:30.017" LastActivityDate="2017-06-13T14:23:41.473" Title="Tap water leaving white residue after drying" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31682" PostTypeId="2" ParentId="31681" CreationDate="2015-05-19T04:43:27.330" Score="6" Body="&lt;p&gt;You either have hard water or a water softener system.&lt;br&gt;&#xA;If you have hard water, then the residue is a combination of calcium / magnesium carbonate / chloride.&lt;br&gt;&#xA;If you have a water softener system, then the residue is from the salt used to remove the hard ions.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-19T04:43:27.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31683" PostTypeId="2" ParentId="31679" CreationDate="2015-05-19T05:03:55.190" Score="3" Body="&lt;p&gt;To my impression, there is no RDM &lt;strong&gt;value&lt;/strong&gt;, but a RDM &lt;strong&gt;pattern&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about mapping atoms is a reactant and a product by assigning &lt;strong&gt;R&lt;/strong&gt; (&lt;em&gt;reaction center&lt;/em&gt;), &lt;strong&gt;D&lt;/strong&gt; (&lt;em&gt;difference atoms&lt;/em&gt;) and &lt;strong&gt;M&lt;/strong&gt; (&lt;em&gt;matched atoms&lt;/em&gt;) in both compounds. Pairs for &lt;strong&gt;R&lt;/strong&gt;, &lt;strong&gt;D&lt;/strong&gt;, and &lt;strong&gt;M&lt;/strong&gt; are separated by colons. The pattern is thus given as  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;R&lt;/strong&gt;(reactant)-&lt;strong&gt;R&lt;/strong&gt;(product) : &lt;strong&gt;D&lt;/strong&gt;(reactant)-&lt;strong&gt;D&lt;/strong&gt;(product) : &lt;strong&gt;M&lt;/strong&gt;(reactant)-&lt;strong&gt;M&lt;/strong&gt;(product),   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where a an asterisk marks a void centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to figure out how these patterns are retrieved, you definitely want to have a look at &lt;em&gt;Modular Architecture of Metabolic Pathways Revealed by Conserved Sequences of Reactions&lt;/em&gt;, a freely available article by &lt;em&gt;Minoru Kanehisa&lt;/em&gt; (KEGG founder), published in &lt;em&gt;J. Chem. Inf. Model.&lt;/em&gt;, &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;53&lt;/em&gt;,613-622 (&lt;a href=&quot;http://dx.doi.org/10.1021/ci3005379&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;). Don't forget to read the &lt;a href=&quot;http://pubs.acs.org/doi/suppl/10.1021/ci3005379/suppl_file/ci3005379_si_001.pdf&quot; rel=&quot;nofollow&quot;&gt;Supporting Information&lt;/a&gt;, particularly the sections on &lt;strong&gt;KEGG atom types&lt;/strong&gt; and  &lt;strong&gt;RDM chemical transformation notation&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-19T07:22:12.450" LastActivityDate="2015-05-19T07:22:12.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31684" PostTypeId="1" AcceptedAnswerId="31689" CreationDate="2015-05-19T05:46:33.013" Score="3" ViewCount="319" Body="&lt;p&gt;I want to know why the three most simple alcohols (methanol, ethanol and propanol) are completely miscible in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what happens at molecular level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think hydrogen bonding plays an important role.&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="15410" LastEditDate="2015-05-19T06:14:23.127" LastActivityDate="2015-05-19T08:18:34.477" Title="Why are the three most simple alcohols completely miscible?" Tags="&lt;water&gt;&lt;solubility&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31685" PostTypeId="2" ParentId="31680" CreationDate="2015-05-19T05:54:17.813" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;And does [&lt;em&gt;the score&lt;/em&gt;] signify the bonding strength of how probably it is for a pair of compounds to be an R pair?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it has nothing to do with that. It is not about how probably a particular reaction is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When comparing &lt;strong&gt;RC&lt;/strong&gt;s (reaction classes), the scores in the table of related reaction classes are just a measure on how similar these reactions are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;RC00064&lt;/code&gt; apparently is an N-acylation or the formation of a peptide bond. &lt;code&gt;RC00197&lt;/code&gt; with a score of &lt;code&gt;1.0&lt;/code&gt; is a N-formylation and considered very similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the cutoff to a lower value, such as 0.4, and compare the results at the end of the table with &lt;code&gt;RC00064&lt;/code&gt;. You'll find dissimilar reaction classes like &lt;code&gt;RC02518&lt;/code&gt;, the oxidation of a selenol to a selenic acid.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-19T05:54:17.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31686" PostTypeId="2" ParentId="31672" CreationDate="2015-05-19T06:26:34.360" Score="1" Body="&lt;p&gt;First, it is no use of putting $\ce{NaOH}$ in the solution as both ion with form white percipitate and insoluble in excess of $\ce{NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try aqueous $\ce{NH3}$. If the solution contains $\ce{Mg^2+}$ ion , it will white percipitate and insoluble excess of $\ce{NH3}$. However, $\ce{Ca^2+}$ ion will not form any percipitate when $\ce{NH3}$ it added.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastActivityDate="2015-05-19T06:26:34.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31688" PostTypeId="2" ParentId="31678" CreationDate="2015-05-19T08:17:52.617" Score="2" Body="&lt;p&gt;Fortunately, sulfurdichloride and disulfurdichloride (= sulfurmonochloride), which is the decomposition product of the former,&lt;br&gt;&#xA;$$\ce{2SCl2 &amp;lt;=&amp;gt;&amp;gt; S2Cl2 + Cl2}$$ &#xA;cannot be prepared that way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest to balance the neutralisation reaction without proposing redox reactions that lead to these compounds and I would stay with your fourth equation.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-19T08:17:52.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31689" PostTypeId="2" ParentId="31684" CreationDate="2015-05-19T08:18:34.477" Score="3" Body="&lt;p&gt;This happens because the intermolecular forces between different low molecular weight alcohols are very similar; as you say, they are dominated by hydrogen bonding. You can consider mixing on a molecular level to involve exchange of molecules from the two phases at their interface; if the energy changes involved are very small, mixing will occur as it is entropically favoured (mixed phase is more disordered) and the average thermal energy is far higher than the barrier to diffusion. &lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-19T08:18:34.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31691" PostTypeId="1" AcceptedAnswerId="31703" CreationDate="2015-05-19T08:45:46.270" Score="2" ViewCount="482" Body="&lt;p&gt;It's clear that &lt;a href=&quot;http://en.wikipedia.org/wiki/Automated_pipetting_system&quot; rel=&quot;nofollow&quot;&gt;automated pipetting&lt;/a&gt; is faster than doing it manually, but how common is it for a lab to have an automated pipetting system, and what usually are the circumstances that a lab have before they purchase such a system?&lt;/p&gt;&#xA;" OwnerUserId="16344" LastEditorUserId="275" LastEditDate="2015-05-19T10:32:08.127" LastActivityDate="2015-05-19T13:01:38.333" Title="How common are automated pipetting systems" Tags="&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31692" PostTypeId="2" ParentId="31691" CreationDate="2015-05-19T08:55:16.070" Score="4" Body="&lt;p&gt;Speed isn't everything. &lt;strong&gt;Reproducibility&lt;/strong&gt; might be much more important! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you have to transfer different volumes from five different stock solutions to a 1536 well plate and adjust the total volume by adding water or the common solvent of your choice. A programmable pipetting automat might be a blessing here.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-19T08:55:16.070" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31693" PostTypeId="1" AcceptedAnswerId="43958" CreationDate="2015-05-19T09:25:58.463" Score="3" ViewCount="7930" Body="&lt;p&gt;There are a few experiments comparing the reactivity of alkali metals with cold water. Is there any purpose to use cold water? Does the temperature of water affect the reactivity of alkali metals?&lt;/p&gt;&#xA;" OwnerUserId="13415" LastActivityDate="2021-03-12T14:25:14.207" Title="How does the temperature of water affect the reactivity of alkali metals acting on it?" Tags="&lt;reactivity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31694" PostTypeId="2" ParentId="31630" CreationDate="2015-05-19T09:58:24.813" Score="4" Body="&lt;p&gt;Disproportionation is a redox reaction. Since&#xA;$$\Delta G = -zF\Delta E$$&#xA;you may &lt;strong&gt;compare the redox potentials for both half-reactions&lt;/strong&gt; in order to determine whether disproportionation occurs spontaneously ($\Delta G &amp;lt; 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$)&#xA;$$\ce{2 NO2 &amp;lt;=&amp;gt; N2O4}$$&#xA;and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{N2O4 + 2 e- \;&amp;amp;&amp;lt;=&amp;gt; 2 NO2- }\quad &amp;amp;&amp;amp;E_1^\circ = +0.867\ \mathrm{V}\\&#xA;\ce{2 NO3- + 2 H2O + 2 e- \;&amp;amp;&amp;lt;=&amp;gt; N2O4 + 4 OH-}\quad &amp;amp;&amp;amp;E_2^\circ = -0.86\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equlibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{align}&#xA;E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &amp;amp;= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\ &#xA;E_1^\circ - E_2^\circ &amp;amp;= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\ &#xA;\left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &amp;amp;= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\ &#xA;\left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &amp;amp;= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\ &#xA;{10^{58.5}} &amp;amp;= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\ &#xA;{10^{29}} &amp;amp;= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}}&#xA;\end{align} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation:&#xA;$$\ce{N2O4 + 2 OH- -&amp;gt; NO2- + NO3- + H2O}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-19T09:58:24.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31695" PostTypeId="1" AcceptedAnswerId="31698" CreationDate="2015-05-19T10:24:03.290" Score="0" ViewCount="60" Body="&lt;p&gt;if a gamma ray hits an electron and transfers energy, does it hit that electron (ionising the atom), transfer all its energy and stop or does it pass through multiple electrons, transferring a portion of its energy each time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e.) if a gamma ray has enough energy to ionise multiple atoms, will it, or will it just give all its energy to the one electron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also if is has the energy to ionise multiple atoms, about how many electrons can it remove (for a range of ray energies)? (i know it is a bit general, but lets say for the first ionisation of potassium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, why is the dose equivalent so low? is it because it can only ionise 1 atom by +1, or because its ionizations (and hence energy) are dispersed throughout many atoms (making the meaning less energy transferred per ionisation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i know this has a lot to do woth physics and so i posted it on the physics SE to, but i was wondering weather a fellow chemist could help me out, thanks)&lt;/p&gt;&#xA;" OwnerUserId="16087" LastActivityDate="2015-05-19T11:15:54.650" Title="ionising atom(s) with gamma rays" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-10-23T17:49:21.370" ContentLicense="CC BY-SA 3.0" />
  <row Id="31696" PostTypeId="1" CreationDate="2015-05-19T10:47:36.420" Score="0" ViewCount="142" Body="&lt;p&gt;Ferric chloride ($\ce{FeCl3}$) is a chemical used in water treatment. If ferric chloride is added to a water sample containing calcium bicarbonate, would the $\mathrm{pH}$ of the solution decrease or increase?&lt;/p&gt;&#xA;" OwnerUserId="15384" LastEditorUserId="15410" LastEditDate="2015-05-19T11:30:26.510" LastActivityDate="2015-05-19T11:30:26.510" Title="Purification of water with ferric chloride" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31697" PostTypeId="1" CreationDate="2015-05-19T10:58:57.430" Score="3" ViewCount="2074" Body="&lt;p&gt;Photoluminescence (PL) spectroscopy is used to find the energy levels in a compound but how does PL spectroscopy works? After exciting a material  with a particular wavelength whether PL machine looks for photons emitted with in a range of wavelenght?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The below image was taken from &lt;a href=&quot;http://www.nature.com/srep/2014/140829/srep05757/full/srep05757.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zHz9R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get this spectra the $\ce{CeO2}$ was excited at 345 nm and got peaks at 490 and 520 nm(approx) ie 2.53 and 2.38 eV respectively and its band gap is 3 eV. Whats happening here?&lt;/p&gt;&#xA;" OwnerUserId="139" LastActivityDate="2016-03-03T10:19:15.723" Title="How photoluminescence (PL) spectroscopy works" Tags="&lt;experimental-chemistry&gt;&lt;spectroscopy&gt;&lt;photochemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31698" PostTypeId="2" ParentId="31695" CreationDate="2015-05-19T11:15:54.650" Score="3" Body="&lt;p&gt;Various mechanisms for interaction of gamma radiation with matter are possible. Depending on the energy and the composition of the absorbing material, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In absorption of gamma radiation of lower energy by the &lt;strong&gt;photoelectric effect&lt;/strong&gt;, the photon is absorbed completely by the atom and an energetic electron is ejected. The energy of the emitted electron is the difference between the energy of the gamma-ray and the binding energy for that electron in the atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gamma radiation of higher energy directly interacts with one electron by the &lt;strong&gt;Compton effect&lt;/strong&gt;. Usually, this is the most probable interaction of gamma radiation. The incoming photon transfers a portion of its energy to the electron and is deflected through an angle with respects to its original direction. Since all angles of scattering are possible, the transferred energy varies from zero to a large fraction of the energy of the photon. The scattered photon may have sufficient energy to interact further by Compton effect or photoelectric effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy of the gamma radiation is very large (i.e. it exceeds $1.02\ \mathrm{MeV}$, which is twice the rest-mass energy of an electron), the process of &lt;strong&gt;pair production&lt;/strong&gt; is energetically possible. In the interaction, the photon disappears and is replaced by an electron and a positron. The process may be considered as the inverse of positron annihilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your additional question about the equivalent dose is slightly misleading. The resulting equivalent dose from gamma radiation is not particularly low. By definition, the radiation weighting factor of gamma radiation is $w_\text{R}=1$. Thus, an absorbed dose of $D=1\ \mathrm{Gy}$ leads to an equivalent dose of $H=1\ \mathrm{Sv}$. However, you may ask why some other radiation types have a higher relative biological effectiveness.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-19T11:15:54.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31699" PostTypeId="2" ParentId="31696" CreationDate="2015-05-19T11:25:50.050" Score="2" Body="&lt;p&gt;&lt;em&gt;If I'd state that the pH drops, would you know why?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you dissolve $\ce{FeCl3}$ in water, a &lt;em&gt;hexaqua&lt;/em&gt; complex is formed. These complexes are known as &lt;strong&gt;cation acids&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Fe(H2O)6]^{3+} + H2O -&amp;gt; [Fe(OH)(H2O)5]^{3+} + H3O+}$$&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-19T11:25:50.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31701" PostTypeId="2" ParentId="31664" CreationDate="2015-05-19T11:38:05.300" Score="7" Body="&lt;p&gt;I think the key to this question lies in the understanding of the &lt;a href=&quot;https://chemistry.stackexchange.com/q/2881/4945&quot;&gt;difference between thermodynamic and kinetic stability&lt;/a&gt;. In this post, I find &lt;a href=&quot;https://chemistry.stackexchange.com/a/13697/4945&quot;&gt;Thomij's answer&lt;/a&gt; the most rigorous and enlightening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you should learn, that the thermodynamically most stable allotrope corresponds to the global minimum on the potential energy surface, while a kinetically stable allotrope, i.e. hindered through an activation barrier, corresponds to local minimum. From this it can easily be seen, that the thermodynamically most stable allotrope is also the kinetically most stable allotrope. This is due to the fact, that the barriers for interconversion of the different allotropes always have to be higher for the most stable one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following two dimensional case you see that the thermodynamically most stable, i.e. the global minimum, is surrounded by two local minima. It must always be true, that $E_\mathrm{a}(1\to2)&amp;lt;E_\mathrm{a}(2\to1)$ if $1$ is a local and $2$ is a global minimum. The same applies to the other case. If this was not true, then $2$ cannot be the global minimum and hence not the thermodynamically most stable modification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gsvgs.png&quot; alt=&quot;simple 2D potential energy surface&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having established that, it is therefore true, that black phosphorus is the thermodynamically and kinetically most stable form, see &lt;a href=&quot;https://books.google.co.jp/books?id=Mtth5g59dEIC&amp;amp;dq=hollemann+wiberg&amp;amp;source=gbs_navlinks_s&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hollemann-Wiberg, p. 680&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at a different reaction, the case might not be so simple any more. This is even true for different temperatures, as the following graphic will show (quoted from the linked book).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nDh49.png&quot; alt=&quot;Hollemann Wiberg, p.680&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-05-19T11:38:05.300" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="31702" PostTypeId="2" ParentId="31323" CreationDate="2015-05-19T12:28:02.690" Score="4" Body="&lt;p&gt;Let us address the elephant in the room first and foremost.  All of these objects do not have any symmetry if you wanted to get real technical about it.  However, let us just assume certain things for this useful exercise to be relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b.) The teacup has a &lt;em&gt;vertical plane&lt;/em&gt; of symmetry splitting the teacup in half along the handle axis and intersecting with the centroid of the cup and saucer.  I would say that your guess is correct though it would be useful to have drawn in a plane rather than simply a line (for all examples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c.) At first glance the pliers have a couple planes of symmetry BUT upon further thought it may not due to the way the rotating JOINT is constructed.  See the example below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkN4w.jpg&quot; alt=&quot;Pliers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is difficult to see in your picture, below it is made very clear.  Therefore I will say that this object does not contain a plane of symmetry.  It would be POSSIBLE that it could have a plane of symmetry, one that lies in the plane of your monitor screen, cutting the object in half from tip to handle end... but it would be impossible to tell without seeing the other side. Many times pliers come with an inscription on one side of the joint, thereby breaking this symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.) The brick does indeed have multiple planes of symmetry.  One vertical plane bisects the brick along its width and one bisects the brick along its length.  In addition, one plane of symmetry lies in the plane of the brick (the plane the brick is lying on), cutting the brick in half (think of it as coming in through the side of the brick).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f.) The hand has no plane of symmetry that I can see at all.  The back of the hand doesn't look like its front... the left and right side of the hand are clearly not mirror images of each other... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brick has three planes of symmetry.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-05-19T13:29:47.037" LastActivityDate="2015-05-19T13:29:47.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31703" PostTypeId="2" ParentId="31691" CreationDate="2015-05-19T13:01:38.333" Score="3" Body="&lt;p&gt;The circumstances in a lab that would make automated liquid handling appropriate would be a need to prepare large numbers of samples by a routine method, so that it becomes more cost and time effective to install and program a machine to do the job than to do it with handheld equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is clearly the case in high throughput drug discovery work and also in analytical labs that perform identical tests on large numbers of samples; it's less common outside these fields although high throughput techniques can also be used in the development and testing of formulations, for example for personal care products or paints and coatings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard automated liquid handlers are designed to deal with volumes in the  microlitre to millilitre range and liquids that are not too volatile or viscous. Requirements outside this range, or extra constraints such as moisture or air sensitive reagents, will demand more complex and expensive equipment which may make automation less viable, although companies such as Zinsser and Chemspeed make devices for handling a wide range of liquid and solid substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a more specific answer than this you're probably going to have to narrow your question down to the particular industry and type of lab you're thinking of. If you're on LinkedIn you might get some useful responses from the &lt;a href=&quot;https://www.linkedin.com/groups/Laboratory-Robotics-Interest-Group-LRIG-705/about&quot; rel=&quot;nofollow&quot;&gt;Laboratory Robotics Interest Group&lt;/a&gt; discussion board.&lt;/p&gt;&#xA;" OwnerUserId="7439" LastActivityDate="2015-05-19T13:01:38.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31704" PostTypeId="2" ParentId="21750" CreationDate="2015-05-19T13:13:12.893" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Superabsorbent_polymer&quot; rel=&quot;nofollow&quot;&gt;superabsorbent polymer&lt;/a&gt; sounds appropriate for this application, and the wikipedia article suggests they are used for &quot;blocking water penetration in underground power or communications cable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the water be removed from the pipe again after it has intruded, and if so do you need the seal to remain impermeable? I would expect this kind of polymer to dry out and shrink with time in the absence of water and it would not necessarily then be able to swell again with the same performance.&lt;/p&gt;&#xA;" OwnerUserId="7439" LastActivityDate="2015-05-19T13:13:12.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31705" PostTypeId="2" ParentId="31134" CreationDate="2015-05-19T14:47:11.053" Score="6" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgJKm.gif&quot; alt=&quot;Steel - effect of carbon content&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: Roy Beardmore, &lt;a href=&quot;https://web.archive.org/web/20150527103902/http://www.roymech.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.roymech.co.uk/ (defunct, via the Internet Archive)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As carbon content increases, the ability to resist a sudden impact decreases, as measured by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Charpy_impact_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charpy Impact Test&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As carbon content increases, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;ultimate tensile strength&lt;/a&gt; and &lt;a href=&quot;http://mtil.illinois.edu/eq-brin.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brinell Hardness&lt;/a&gt; increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as a phase diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ieekd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;increasing &lt;span class=&quot;math-container&quot;&gt;$\ce{}$&lt;/span&gt; (cementite) content makes the steel more brittle and hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near 0.76% carbon, the steel will be &lt;a href=&quot;http://en.wikipedia.org/wiki/Pearlite&quot; rel=&quot;nofollow noreferrer&quot;&gt;pearlite&lt;/a&gt; which is ductile (good for wires).  Below 0.76% carbon, there will be increased alpha-iron (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ferrite_(iron)&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferrite&lt;/a&gt;). Above 0.76% there can be &lt;a href=&quot;http://en.wikipedia.org/wiki/Ledeburite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ledeburite&lt;/a&gt;.  Austenite only exists at high temperature.  Depending upon the cooling rate of Austenite, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bainite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bainite&lt;/a&gt; (slow cooling) or &lt;a href=&quot;http://en.wikipedia.org/wiki/Martensite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Martensite&lt;/a&gt; (rapid cooling) can form.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditDate="2019-12-17T14:23:24.817" LastActivityDate="2019-12-17T14:23:24.817" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31706" PostTypeId="1" AcceptedAnswerId="31719" CreationDate="2015-05-19T15:22:17.373" Score="5" ViewCount="4694" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is ethanoic acid most acidic in DMSO, water or methanol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that the conjugate base would be most stabilized by a polar protic solvent (meaning it's weakest in DMSO), however, water and methanol are both protic and polar. Which causes ethanoic acid to be most acidic lowest $\mathrm{p}K_\mathrm{a}$? This was asked in an exam question where the following $\mathrm{p}K_\mathrm{a}$ values were meant to be assigned to the acid in each of the solvents: $9.5$, $12.6$, $4.8$. &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="15410" LastEditDate="2015-05-19T16:19:29.593" LastActivityDate="2015-05-20T09:16:10.010" Title="Solvent effects on acidity of ethanoic acid" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31707" PostTypeId="2" ParentId="31678" CreationDate="2015-05-19T15:41:51.187" Score="7" Body="&lt;p&gt;As Klaus and I mentioned, the first reactions seem very strange from a chemical perspective, even if they are correctly balanced. This is where a chemist needs to exercise their intuition and experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balancing reactions is nothing but solving a linear system of equations, so it is entirely possible to propose reactions which balance perfectly, but which are based on questionable chemistry. For example, in all but the last reaction, the sulphur atoms in $\ce{SO_4^2-}$ anions are being reduced to lower oxidation states. While this is possible in some conditions, it takes quite strong reducing agents, and none of the reagents shown are known to possess significant reducing character. Reactions which release gaseous oxygen are also not particularly common. Furthermore, two reactions propose the formation of sulphur chlorides, which would actually hydrolyse in contact with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disregarding intuition, it should be possible to pick out which reaction is the most favourable by simply comparing the Gibbs free energy of reaction in each case; &lt;strong&gt;the most exergonic reaction, with the most negative value of $\boldsymbol{\Delta_\mathbf{r}G}$, should dominate&lt;/strong&gt; (assuming no kinetic factors bar it from happening, which is unlikely in this situation). I did a very quick and dirty calculation, out of curiosity, by taking the standard Gibbs free energy of formation $\Delta_\mathrm{f}G^\circ$ for each compound and summing them together to obtain the standard Gibbs free energy of reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta_\mathrm{r}G^\circ = \sum\limits_\text{products}\Delta_\mathrm{f}G^\circ - \sum\limits_\text{reagents}\Delta_\text{f}G^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only meant as a qualitative way to compare the reactions, as I am assuming standard state conditions for the reactions, which are not necessarily true. Furthermore, for a few species, I didn’t find the correct values in these two sources &lt;a href=&quot;http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt; &lt;a href=&quot;http://srdata.nist.gov/gateway/gateway?property=Gibbs+free+energy+of+formation&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt;. $\ce{SCl2}$ and $\ce{S2Cl2}$ probably can’t be found tabulated for the aqueous phase because of their propensity to hydrolysis, so I used the free energies of formation for the gas and solid phases, respectively. I also didn’t find a source for aqueous $\ce{Na2SO4}$, so instead I used the data for the solid decahydrate $\ce{Na2SO4 . 10 H2O (s)}$. The results I got were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \small&#xA;\begin{array}{lcc}&#xA;\hline&#xA;\text{Reaction} &amp;amp; \Delta_\mathrm{r}G^\circ / \mathrm{kJ\ mol^{-1}}\\&#xA;\hline&#xA;\text{A} &amp;amp; +440  \\&#xA;\text{B} &amp;amp; +1940  \\&#xA;\text{C} &amp;amp; +590  \\&#xA;\text{D} &amp;amp;  \mathbf{-2610}  \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, even though the calculation was very rough, the huge disparity in free energy values suggests that the first three reactions are completely overwhelmed by the last one, so that of the four possibilities listed, only the last has any chance of occurring. Strictly speaking this doesn’t mean the last reaction happens, only that it’s much more favourable than the others presented; there could be an unlisted reaction which is even more favourable than all four mentioned in the question. However, empirical data confirms that the last reaction, a simple neutralisation reaction, is the major event when mixing $\ce{HCl}$, $\ce{H2SO4}$ and $\ce{NaOH}$ in practically any conditions.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T19:40:28.750" LastActivityDate="2015-05-19T19:40:28.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31708" PostTypeId="1" AcceptedAnswerId="31770" CreationDate="2015-05-19T15:42:34.673" Score="6" ViewCount="10218" Body="&lt;p&gt;What are the reagents for the following reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6rQiF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, when thinking about ketone to alkene reactions, one thinks of the Wittig reaction. However, I don't see an appropriate form of that reaction here. This was posed as a question on my problem sheet and I suspect - after having given it some thought - that it's just a Hoffman/Saytzev question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reaction A, I think reaction with $\ce{MeMgBr}$ to make the tertiary alcohol. At this point, one has to think of a way to get the least thermodynamically stable product (the kinetic product). Perhaps sulfuric acid? It would protonate the hydroxyl so the transition state wouldn't have significant alkene character and the &quot;base&quot; is bulky. Correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reaction B, I think it should be reacted with $\ce{NaBH4}$ to make the secondary alcohol then elimination can be performed using whatever. Maybe sulfuric acid again or just tosylate it then use $^t\ce{BuOK}$. Does this sound right?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="402" LastEditDate="2017-01-14T08:41:37.553" LastActivityDate="2017-01-14T08:41:37.553" Title="Ketone to alkene regiochemistry" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31709" PostTypeId="1" AcceptedAnswerId="31716" CreationDate="2015-05-19T15:42:47.133" Score="5" ViewCount="478" Body="&lt;p&gt;Which substituent should be placed first (in case of alphabetisation)?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;isopropyl-&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;methyl-&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have come across two compounds: &lt;strong&gt;1-methyl-3-methoxy-4-isopropylbenzene&lt;/strong&gt; and &lt;strong&gt;1-isopropyl-4-methylcyclohexane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is “iso” taken care of in case of alphabetisation? And, are there any special criteria for this operation?&lt;/p&gt;&#xA;" OwnerDisplayName="user16347" LastEditorDisplayName="user7951" LastEditDate="2015-05-19T19:13:07.530" LastActivityDate="2015-05-20T13:39:16.133" Title="Nomenclature of an organic compound" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31710" PostTypeId="1" CreationDate="2015-05-19T15:45:54.510" Score="4" ViewCount="633" Body="&lt;p&gt;I'm familiar with the fact that theoretical formulae are based on already known theory and are very frequently simplified (some equations don't include interactions between molecules for example) while the purpose of empirical equations (like polynomial equations) is just to follow experimental measurements even if it doesn't make any chemical or physical sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these yield more accurate results (compared to reality)? What are the upsides and downsides of each group? Which are more universal? Which would require less compound-specific data (constants) to use it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Removed van der Waals equation as an example of theoretical equations since, as pointed out by others, it's actually an empirical equation.&lt;/p&gt;&#xA;" OwnerUserId="2253" LastEditorUserId="2253" LastEditDate="2015-05-21T15:01:44.973" LastActivityDate="2015-05-21T15:01:44.973" Title="Theoretical versus empirical equations" Tags="&lt;physical-chemistry&gt;&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31711" PostTypeId="1" AcceptedAnswerId="31712" CreationDate="2015-05-19T16:22:59.443" Score="23" ViewCount="21649" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wwTN1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand isomer is clearly the kinetically favoured product and the right hand isomer is the thermodynamically favoured product. However, I have no idea why. Can you explain?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="4231" LastEditDate="2015-05-19T18:19:53.633" LastActivityDate="2017-02-16T00:38:44.823" Title="Thermodynamic vs Kinetic Sulphonation of Naphthalene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="31712" PostTypeId="2" ParentId="31711" CreationDate="2015-05-19T16:56:51.593" Score="29" Body="&lt;p&gt;In contrast to many other electrophilic aromatic substitution reactions, &lt;strong&gt;aromatic sulfonation is reversible&lt;/strong&gt;, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate.  This makes sulfonation a nice method to 1) control where a second substituent will be introduced or 2) serve as a protecting group (block second substituent introduction) - and then when you are done, you simply de-sulfonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now to your question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the intermediates leading to the 2 isomers we see that we can draw 2 resonance structures that preserve aromaticity in one ring for the transition state leading to naphthalene-1-sulfonic acid, but only one resonance structure where the other aromatic ring remains intact for the 2-isomer (draw the resonance structures to convince yourself).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dX28h.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.thecatalyst.org/experiments/Cullen/Cullen.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the transition state leading to the 1-isomer will be lower in energy and naphthalene-1-sulfonic acid will be formed fastest - it is the &lt;strong&gt;kinetically favored product.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, naphthalene-2-sulfonic acid is a few kcal/mol more stable than its 1-isomer. This is due to an adverse 1,8 steric interaction present in the 1-isomer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDvUB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.intermediateorgchemistry.co.uk/aromatic1.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the sulfonation reaction is reversible, if we run it long enough at high temperature (favoring reversibility), eventually we will wind up with the thermodynamically preferred (more stable) product, the 2-isomer.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="36974" LastEditDate="2017-02-02T13:27:59.200" LastActivityDate="2017-02-02T13:27:59.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31714" PostTypeId="2" ParentId="31674" CreationDate="2015-05-19T18:16:48.013" Score="2" Body="&lt;p&gt;The capacitance sensor itself will certainly not present an ignition risk to your system. The sensor will be on the outside of the container, and it does not introduce any electrical potential that would cause a spark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the device works is to measure the change in the sensor's capacitance depending on the dielectric effect of the liquid, which will increase as more liquid enters the electric field of the sensor (a capacitor).&lt;/p&gt;&#xA;" OwnerUserId="15065" LastActivityDate="2015-05-19T18:16:48.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31715" PostTypeId="1" CreationDate="2015-05-19T18:53:04.293" Score="1" ViewCount="118" Body="&lt;p&gt;Is de-ionized water more corrosive to metals than tap water? I have read that it &quot;wants&quot; the removed ions back and will aggressively take them from any available source.&lt;/p&gt;&#xA;" OwnerUserId="16353" LastEditorUserId="10185" LastEditDate="2015-05-19T18:55:42.063" LastActivityDate="2015-05-19T22:06:32.210" Title="De-ionized water" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-19T23:16:30.090" ContentLicense="CC BY-SA 3.0" />
  <row Id="31716" PostTypeId="2" ParentId="31709" CreationDate="2015-05-19T19:08:55.577" Score="5" Body="&lt;p&gt;(Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)&lt;/p&gt;&#xA;&lt;p&gt;Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.&lt;br /&gt;&#xA;For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.&lt;/p&gt;&#xA;&lt;p&gt;On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5&lt;/strong&gt; ALPHANUMERICAL ORDER&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5.1&lt;/strong&gt; Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the correct alphanumerical order in your first example corresponds to the name &lt;strong&gt;&lt;em&gt;x&lt;/em&gt;-isopropyl-&lt;em&gt;y&lt;/em&gt;-methoxy-&lt;em&gt;z&lt;/em&gt;-methylbenzene.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In your second example, the correct alphanumerical order corresponds to the name &lt;strong&gt;&lt;em&gt;x&lt;/em&gt;-isopropyl-&lt;em&gt;y&lt;/em&gt;-methylcyclohexane.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(g) lowest locants for the substituent cited first as a prefix in the name;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule f takes precedence over Rule g.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, your first example is named as &lt;strong&gt;1-isopropyl-2-methoxy-4-methylbenzene&lt;/strong&gt; rather than 4-isopropyl-3-methoxy-1-methylbenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ucVu9.png&quot; alt=&quot;1-isopropyl-2-methoxy-4-methylbenzene&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your second example is named as &lt;strong&gt;1-isopropyl-4-methylcyclohexane&lt;/strong&gt; rather than 4-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ORaRl.png&quot; alt=&quot;1-isopropyl-4-methylcyclohexane&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-20T13:39:16.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31718" PostTypeId="1" CreationDate="2015-05-19T20:01:24.033" Score="6" ViewCount="335" Body="&lt;p&gt;What makes a molecule a good initiator of collisionally induced ISC? $\ce{N2}$, $\ce{O2}$ and $\ce{CH4}$ have been mentioned as examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine the most important factor is the ability to quench vibrational excitation so that the Franck-Condon factors for non-radiative decay are optimized.&lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="15410" LastEditDate="2015-05-20T00:34:06.340" LastActivityDate="2015-05-20T00:34:06.340" Title="Collisonally induced intersystem crossing" Tags="&lt;theoretical-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31719" PostTypeId="2" ParentId="31706" CreationDate="2015-05-19T20:33:46.567" Score="9" Body="&lt;p&gt;Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source &lt;a href=&quot;http://evans.harvard.edu/pdf/evans_pKa_table.pdf&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.researchgate.net/profile/Elisabeth_Bosch/publication/222627494_Dissociation_constants_of_neutral_and_charged_acids_in_methyl_alcohol._The_acid_strength_resolution/links/00463533c08df1ae05000000.pdf&quot;&gt;here&lt;/a&gt;). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) &amp;lt; pK_{a}(CH3OH) &amp;lt; pK_{a}(DMSO)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the acidity of a species in different solvents is affected by three factors: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Born_equation&quot;&gt;Born equation&lt;/a&gt; which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$&lt;/li&gt;&#xA;&lt;li&gt;The strength of solvation of the conjugate base of the acid.&lt;/li&gt;&#xA;&lt;li&gt;The strength of solvation of the conjugate acid of the solvent.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dielectric constant:&lt;/strong&gt; The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (&lt;a href=&quot;http://depts.washington.edu/eooptic/linkfiles/dielectric_chart%5B1%5D.pdf&quot;&gt;source&lt;/a&gt;). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Anion solvation:&lt;/strong&gt; As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cation solvation:&lt;/strong&gt; The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. &lt;a href=&quot;http://www.quora.com/Organic-Chemistry/Why-is-methanol-more-acidic-than-water-while-ethanol-is-a-weaker-acid-than-water&quot;&gt;This&lt;/a&gt; page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/suppmnt2.htm&quot;&gt;This&lt;/a&gt; page says some good things about the subject (some of my ideas were developed from reading this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional reading:&lt;/strong&gt; &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00961a012&quot;&gt;This&lt;/a&gt; paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-19T21:03:27.067" LastActivityDate="2015-05-19T21:03:27.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31720" PostTypeId="1" CreationDate="2015-05-19T20:51:27.157" Score="2" ViewCount="628" Body="&lt;p&gt;I made some colloidal copper with 9V battery. To make this water I had to add small amount of salt. However, the salt itself shows up in PPM count of my PPM reader. So how can I discount the salt PPM and come up with correct copper PPM in distilled water.&lt;/p&gt;&#xA;" OwnerUserId="16355" LastActivityDate="2016-02-20T09:24:42.557" Title="Measuring Colloidal copper PPM &amp; how to ignore the salt PPM in soluton" Tags="&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31722" PostTypeId="2" ParentId="31715" CreationDate="2015-05-19T22:06:32.210" Score="1" Body="&lt;p&gt;The Wikipedia article merely states that &lt;em&gt;sole consumption of purified water may deplete some electrolytes through normal excretion and lack of replacement&lt;/em&gt;,  not that it somehow &quot;wants&quot; the minerals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of fact, pure, deionized water is much &lt;strong&gt;less&lt;/strong&gt; corrosive to metals and is often used in closed systems to &lt;strong&gt;prevent&lt;/strong&gt; corrosion because it helps prevent &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_corrosion&quot; rel=&quot;nofollow&quot;&gt;galvanic (electrolytic) corrosion&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Automotive coolant antifreeze contains specific corrosion inhibitors safe for aluminum and iron parts, but if water needs to be added, it should be distilled water, to prevent galvanic corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People living in northern climates often encounter corrosion of vehicles due to road salt, and learn to wash vehicles with tap water to remove the salt (high-purity or distilled water would be better, but not cost effective).&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-19T22:06:32.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31723" PostTypeId="2" ParentId="31718" CreationDate="2015-05-19T22:10:54.043" Score="4" Body="&lt;p&gt;There are two factors that I recall that can enhance intersystem crossing (ISC, the conversion of an excited singlet state to an excited triplet state, or &lt;em&gt;vice-versa&lt;/em&gt;; $\ce{S_1 &amp;lt;=&amp;gt; T_1}$).  One is the presence of a &quot;heavy atom&quot; such as bromine or iodine in the colliding molecule (or the excited molecule itself).  The other is the presence of a paramagnetic species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both types of molecules enhance ISC by relaxing the quantum rules which forbid  singlet-triplet conversions. When a heavy atom or paramagnetic molecule collides with a molecule in its excited state it can facilitate the electron spin flip (e.g. converting paired electrons to unpaired electrons, or &lt;em&gt;vice versa&lt;/em&gt;) through spin-orbit coupling.  It is this spin-orbit coupling that relaxes the rules forbidding direct conversion between $\ce{S1}$ and $\ce{T1}$ states.  In essence spin-orbit coupling leads to a mixing of the singlet and triplet wavefunctions, thereby allowing singlet-triplet conversion.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-19T22:10:54.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31725" PostTypeId="1" CreationDate="2015-05-19T23:16:24.500" Score="4" ViewCount="184" Body="&lt;p&gt;I was wondering if there is a type of glass dish with small holes in it to use as a filter.  I've found a few but they are either fritted glass disks or a type of funnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for something similar to just a petri dish with no stems or inserts.&lt;/p&gt;&#xA;" OwnerUserId="16357" LastEditorUserId="402" LastEditDate="2015-05-20T01:37:34.583" LastActivityDate="2016-02-14T06:36:41.400" Title="Porous petri dish" Tags="&lt;equipment&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31726" PostTypeId="2" ParentId="31706" CreationDate="2015-05-19T23:28:28.763" Score="4" Body="&lt;p&gt;Thermodynamically, solvation (including (de)protonation) relies on an energy difference: energy required to make a &quot;hole&quot; in the solvent for $\ce{CH_3COOH}$, as well as the energy required to make two holes - one for $\ce{CH_3COO^-}$ and one for $\ce{H^+}$ - compared to the energy released when stabilizing interactions form (dipole-dipole interactions, charge-dipole, cation-pi interactions, etc.) between solvent and solute molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: When we say &quot;holes&quot; in the solvent, we literally mean the energy one would need to push aside the solvent molecules from each other to make a little hole big enough to put a single ethanoic acid molecule.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the energy required to make a &quot;hole&quot; in the solvent is larger than the energy released by making all the stabilizing interactions possible, the molecule will not be effectively solvated (e.g. hexanes in water: hexane cannot form lasting stabilizing interactions with water.)&lt;/li&gt;&#xA;&lt;li&gt;If the energy required to make a &quot;hole&quot; in the solvent is less than the energy released forming stabilizing interactions, the molecule will be solvated.(ethanoic acid in water)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's about the ability of a solvent to stabilize the neutral and charged species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water&lt;/strong&gt;: $\ce{CH_3COOH}$ and its acid and conjugate base are all polar molecules/ions. Water stabilizes $\ce{CH_3COOH}$ by forming excellent dipole dipole and dipole-charge interactions (and highly &lt;em&gt;symmetrically&lt;/em&gt; because water is symmetric). The $\ce{H^+}$ from the acid forms $\ce{H_3O^+}$ in water (also highly symmetric) and is also stabilized by $\ce{H_2O}$, so often you read about &quot;proton hopping&quot; in acidic aqueous solution, which also happens in methanol, but to much less extent. You can also say water's lone pairs are larger and less hindered than methanol's, or you could say that the sigma-like MO for water overlaps better (and more symmetrically) with incoming molecules than methanol's. High symmetry allows much better orbital mixing (via more degenerate MOs), lowering the energy of water and hydronium significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methanol&lt;/strong&gt;: Methanol is polar and has a $\mathrm{p}K_\mathrm{a}$ (so I've heard) around water's. Methanol $\ce{O}$ atom can accommodate a proton and when it does, it is stabilized partly by hyperconjugation (a type of orbital mixing; &lt;em&gt;see below for visual&lt;/em&gt;) but is not stabilized to the extent water is in part because methanol is not as symmetric (neutral or protonated) as water is = less MO mixing. Yet, methanol forms good dipole-dipole and dipole-charge stabilizing interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DMSO&lt;/strong&gt;: DMSO is polar, but aprotic, suggesting it does not hold on to protons very well. It can form dipole-dipole interactions, but the sulfoxide $\ce{S=O}$ part likely doesn't overlap very well in a &quot;sigma way&quot; (head-to-head), but in a &quot;pi-way&quot; resulting in weaker overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A note to illustrate hyperconjugation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If protonated methanol is stabilized, it is likely stabilized &lt;em&gt;intra&lt;/em&gt;molecularly  by hyperconjugation, which is electron overlap between the exposed (empty) $\mathrm{p}$-orbital on oxygen and the $\ce{C-H}$ sigma bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f9fIw.png&quot; alt=&quot;My attempt at hyperconjugative orbital mixing in protonated methanol&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also be stabilized &lt;em&gt;inter&lt;/em&gt;molecularly if surrounded by other polar electron-rich species (methanol, $\ce{CH_3COO^-}$ or $\ce{CH3COOH}$): These molecules will also temporarily donate electron density to the empty $\mathrm{p}$-orbital on the protonated methanol oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These effects combine to allow for methanol to harbour a proton, making it a little better base than one would think.&lt;/p&gt;&#xA;" OwnerUserId="16348" LastEditorUserId="4945" LastEditDate="2015-05-20T09:16:10.010" LastActivityDate="2015-05-20T09:16:10.010" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31728" PostTypeId="2" ParentId="31725" CreationDate="2015-05-19T23:42:09.853" Score="1" Body="&lt;p&gt;The closest I could think of are nozzle cover lids. Wouldn't such tool be ineffective at 'filter'-ing substances nonetheless?&lt;/p&gt;&#xA;" OwnerUserId="16210" LastActivityDate="2015-05-19T23:42:09.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31729" PostTypeId="2" ParentId="29000" CreationDate="2015-05-20T00:23:18.730" Score="15" Body="&lt;p&gt;The ethyl carbocation ($\ce{C2H5+}$) has a different structure than what you might have anticipated.  It is a &lt;strong&gt;non-classical ion&lt;/strong&gt; with a bridging hydrogen.&#xA;For a description of just what a non-classical carbocation is see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31732/what-is-a-non-classical-carbocation/31758#31758&quot;&gt;this answer&lt;/a&gt;. &#xA;The hallmark of a non-classical ion is 3 atoms with 2 electrons spread over them.  This is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_two-electron_bond&quot; rel=&quot;noreferrer&quot;&gt;3-center 2-electron bond&lt;/a&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypercoordinate bonding&lt;/a&gt;). In the case at hand the 3-atom 2-electron bond involves the 2 carbon atoms and the bridged hydrogen atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cU5mv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is no &quot;methyl&quot; group present in the ethyl carbocation that can inductively release electrons.  Additionally, the hyperconjugated resonance structure you mentioned has actually been replaced by a hydrogen that is partially bonded to the other carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The n-propyl carbocation would likely have a similar non-classical structure - if it existed at all.  This carbocation has not been experimentally observed, it rearranges too rapidly to the much more stable 2-propyl carbocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while we can't answer your question about whether inductive or hyperconjugative effects are more important in this series of compounds, we can assess which cation is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variety of experiments suggest that the heat of formation of the non-classical ethyl cation is around 215 kcal/mol (&lt;a href=&quot;http://www.researchgate.net/profile/Jose_Luis_Abboud/publication/237044745_Thermodynamic_Stabilities_of_Carbocations/links/00b4951af702f50e72000000.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;, see pp. 68-72). This same reference points out that the heat of formation of the classical 1-propyl cation lies approximately 20 kcal/mol above that of the 2-propyl cation.  The heat of formation of the 2-propyl cation is around 193 kcal/mol.  Adding 20 kcal/mol to that value would place the classical 1-propyl cation around 213 kcal/mol. &lt;strong&gt;But&lt;/strong&gt;, assuming the 1-propyl cation would also be a non-classical ion it would be lower in energy than the classical version, its heat of formation would therefore be less than 213 kcal/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analysis suggests that the 1-propyl cation would likely have a lower heat of formation than the ethyl cation.  In other words, the 1-propyl cation would likely be more stable than the ethyl cation.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-05-21T15:55:28.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31730" PostTypeId="1" AcceptedAnswerId="31734" CreationDate="2015-05-20T03:17:28.043" Score="13" ViewCount="1033" Body="&lt;p&gt;Rotation around the double bond of azobenzene is restricted because it would distort the P orbital overlap between the nitrogen atoms. However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why this is the case. You are taking an electron from the non-bonding orbital, and moving it into the pi-antibonding. However, the $\pi$-bonding is still present, so I see no reason why the rotation would be unrestrained. Is it something about the interaction between the electrons in the $\pi^*$ with the $\pi$ that weakens the double bond? And if so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/17vCp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5366" LastEditorUserId="418" LastEditDate="2015-05-20T05:28:01.173" LastActivityDate="2015-05-21T04:39:59.193" Title="Photoisomerization of Azobenzene" Tags="&lt;organic-chemistry&gt;&lt;orbitals&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31731" PostTypeId="1" CreationDate="2015-05-20T04:27:34.593" Score="5" ViewCount="2518" Body="&lt;p&gt;During bond formation in Carbon, one of the electrons from the &lt;strong&gt;2s&lt;/strong&gt; orbital jumps to one the empty &lt;strong&gt;&lt;em&gt;p&lt;/em&gt;&lt;/strong&gt; orbitals creating a hybrid orbital as seen in the figure below.&#xA;&lt;p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hyRiO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now clearly one electron from the &lt;strong&gt;&lt;em&gt;2s&lt;/em&gt;&lt;/strong&gt; orbital jumped to the &lt;strong&gt;&lt;em&gt;2p&lt;sub&gt;z&lt;/sub&gt;&lt;/em&gt;&lt;/strong&gt; orbital.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;But what happened to the spin?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Before hybridization, one electron had an opposite spin, but after hybridization all electrons have the same spin.&#xA;&lt;br&gt;Almost everywhere the same thing is shown. &lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;Did the spin of the promoted electron get reversed?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="15508" LastActivityDate="2015-05-21T23:30:40.000" Title="Change of electron spin in Carbon after Hybridization?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31732" PostTypeId="1" AcceptedAnswerId="31758" CreationDate="2015-05-20T05:47:07.253" Score="36" ViewCount="43575" Body="&lt;ul&gt;&#xA;&lt;li&gt;What is a non-classical carbocation? &lt;/li&gt;&#xA;&lt;li&gt;How is it different from a classical carbocation? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am confused as I have come across this term many times on Chem.SE but there seems to be nothing for my level of understanding on the Internet!&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="7448" LastEditDate="2015-05-31T20:45:41.247" LastActivityDate="2016-09-09T07:43:25.443" Title="What is a non-classical carbocation?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="31734" PostTypeId="2" ParentId="31730" CreationDate="2015-05-20T06:50:11.117" Score="8" Body="&lt;p&gt;Azobenzenes are one of the most widely used moietes in photochemically  switchable systems. Nevertheless, the pathways for the &lt;em&gt;E&lt;/em&gt;-&lt;em&gt;Z&lt;/em&gt; isomerisation of azobenzene in the $S_1$ and the $S_2$ state have been in debate for decades and I'm not quite sure whether this is finally solved.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the $S_1$ state arises from a $n \rightarrow \pi^*$ transition in a weak, broad band with $\lambda_{\mathrm{max}} = 432\,\mathrm{nm}$ and $\epsilon \approx 400\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in hexane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, &lt;em&gt;E&lt;/em&gt;-azobenzene exhibits another, much stronger absorption at $\lambda_{\mathrm{max}} = 319\,\mathrm{nm}$ with $\epsilon \approx 22800\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in the same solvent. This $\pi\rightarrow\pi^*$ transition to the $S_2$ state would indeed allow for an isomerisation through an &lt;strong&gt;one-bond-flip&lt;/strong&gt; rotational mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For the UV data, see &lt;a href=&quot;http://dx.doi.org/10.1021/ja01547a010&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1021/ja038327y&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But - &lt;em&gt;you saw that coming after this lengthy introduction&lt;/em&gt; - the situation is much more complicated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have neglected the fact that &lt;em&gt;E&lt;/em&gt;-&lt;em&gt;Z&lt;/em&gt; isomerisations can proceed by other mechanisms than just the simple one-bond-flip that is usually postulated for alkenes!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive me for not getting into all details on this, but for polyenes, to give an example, the so-called &lt;strong&gt;Bicycle Pedal&lt;/strong&gt; (&lt;strong&gt;BP&lt;/strong&gt;) and &lt;strong&gt;Hula twist&lt;/strong&gt; (&lt;strong&gt;HT&lt;/strong&gt;) pathways have been proposed and confirmed as space-conserving alternatives to the &lt;strong&gt;one-bon-flip&lt;/strong&gt; mechanism. The latter can be visualized as &lt;em&gt;&quot;let's just flip the other side of the former double bond around like a rotor&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the azobenzenes. Even from more recent works (&lt;a href=&quot;http://dx.doi.org/10.1021/ja038327y&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1021/jp031149a&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1038/ncomms6860&quot; rel=&quot;noreferrer&quot;&gt;4&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1063/1.3000008&quot; rel=&quot;noreferrer&quot;&gt;5&lt;/a&gt;), it seems that the &lt;em&gt;E&lt;/em&gt;-&lt;em&gt;Z&lt;/em&gt; isomerisation is still not fully explained, although a plain rotation around the former double bond can be excluded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two main fractions in the debate support mechanisms in which a contribution of &lt;strong&gt;inversion&lt;/strong&gt; through a change of the $\ce{C-N-N}$ angle in the excited state is assumed. Whether this is the only effect and the reaction proceeds through a transition state with a linear $\ce{C-N-N}$ arrangement or whether an &lt;strong&gt;inversion-assisted torsional pathway&lt;/strong&gt; provided the beter explanation is still unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja01547a010&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt; George Zimmerman, Lue-Yung Chow, and Un-Jin Paik, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1958&lt;/strong&gt;, &lt;em&gt;80&lt;/em&gt;, 3528-3531.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/ja038327y&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;  Alessandro Cembran, Fernando Bernardi, Marco Garavelli, Laura Gagliardi, and Giorgio Orlandi, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;126&lt;/em&gt;, 3234-3243.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/jp031149a&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt; Eric Wei-Guang Diau, &lt;em&gt;J. Phys. Chem. A&lt;/em&gt;, &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;108&lt;/em&gt;, 950-956.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1038/ncomms6860&quot; rel=&quot;noreferrer&quot;&gt;4&lt;/a&gt; Eric M. M. Tan, Saeed Amirjalayer, Szymon Smolarek, Alexander Vdovin,    Francesco Zerbetto, and Wybren Jan Buma, &lt;em&gt;Nature Communications&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;, &lt;em&gt;6&lt;/em&gt;, Article number: 5860.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.3000008&quot; rel=&quot;noreferrer&quot;&gt;5&lt;/a&gt; Junfeng Shao, Yibo Lei, Zhenyi Wen, Yusheng Dou, and Zhisong Wang, &#xA;&lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;129&lt;/em&gt;, 164111&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-21T04:39:59.193" LastActivityDate="2015-05-21T04:39:59.193" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31736" PostTypeId="2" ParentId="31588" CreationDate="2015-05-20T07:32:59.230" Score="2" Body="&lt;p&gt;Imagine the dipole as a partial positive ($q_+$) and negative ($q_-)$ separated by a small distance $\delta$.  The electrostatic potential due to the dipole is $\frac{q_+}{|d-\delta|}+\frac{q_-}{|d+\delta|}$ where $d$ is the distance to the midpoint between the two charges. (There is also an angle term which I left out for simplicity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $d$ increases $\frac{q_+}{|d-\delta|}\approx -\frac{q_-}{|d+\delta|}$, i.e. the electrostatic potential of the dipole (i.e. two very close charges of opposite charge) goes to zero faster than a single charge because at longer distances the two charges appear to be at the same position and &quot;cancel each other out&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.feynmanlectures.caltech.edu/II_06.html#Ch6-S2&quot; rel=&quot;nofollow&quot;&gt;Feynman has given as a more thorough explanation&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-05-20T11:15:47.883" LastActivityDate="2015-05-20T11:15:47.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31737" PostTypeId="2" ParentId="31710" CreationDate="2015-05-20T07:50:32.047" Score="5" Body="&lt;p&gt;As pointed out already, the van der Waals (vdW) equation is also an empirical equation.  However, I think what you are saying is that the vdW equation is a physically motivated empirical equations where the empirical parameters in principle correspond to something &quot;real&quot; (e.g. molecular size and interaction strength) in contrast to a fit to an arbitrarily chosen fitting function such as a polynomial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; you have lots of data points so that you can fit many parameters then a simple fit will be more accurate since it tends to have more fitting parameters than the physically motivated equation.  However, the simple fit will likely only be accurate for interpolation, and is still likely to be comparatively less accurate for extrapolation.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-20T07:50:32.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31739" PostTypeId="1" AcceptedAnswerId="31895" CreationDate="2015-05-20T11:45:56.940" Score="10" ViewCount="833" Body="&lt;p&gt;I have a wave number question which I have partially solved:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What transition in $\ce{He+}$ ion shall have the same wavenumber as the first line in Balmer series of $\ce{H}$ atom?&#xA;  (a) $7\to5$ (b) $5\to3$ (c) $6\to4$ (d) $4\to2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's how I solved it:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: symbol $Z$ is the atomic number&lt;br&gt;&#xA;and $1/\lambda$ is wave number (I have written frequency by mistake)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E08cT.jpg&quot; alt=&quot;Solving the problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point my teacher told me to continue using a trial and error method, is there any other way to solve this, perhaps using some other equation.&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="4945" LastEditDate="2017-11-22T07:25:06.950" LastActivityDate="2017-11-22T07:30:23.883" Title="Working with wave number equations, use trial-and-error?" Tags="&lt;physical-chemistry&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31740" PostTypeId="1" AcceptedAnswerId="31743" CreationDate="2015-05-20T12:04:23.383" Score="7" ViewCount="721" Body="&lt;p&gt;In &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_scf.htm&quot;&gt;Gaussian&lt;/a&gt; the SCF convergence criteria is set to $10^{-N}$ by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SCF=(Conver=N)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that this criteria applies to the root mean square change in density matrix between two SCF cycles, while the maximum change convergence criteria is set to $10^{-N+2}$. For instance, for the default case of $N=8$ the output says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt; Requested convergence on RMS density matrix=1.00D-08 within 128 cycles.&#xA; Requested convergence on MAX density matrix=1.00D-06.&#xA; Requested convergence on             energy=1.00D-06.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now in &lt;a href=&quot;http://myweb.liu.edu/~nmatsuna/gamess/input/SCF.html&quot;&gt;GAMESS (US)&lt;/a&gt; the SCF convergence criteria is set to $10^{-N}$ by&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$SCF CONV=1.0d-N $END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, neither manual, nor the output do not explicitly specify does this criteria refers to the root mean square change in density matrix or to the maximum change in it? Judging from the default value (1.0d-05), it is the last one, the maximum change in density matrix, otherwise it would be too small for a reliable convergence. Thus, &lt;code&gt;SCF=(Conver=N)&lt;/code&gt; in Gaussian corresponds to &lt;code&gt;$SCF CONV=1.0d-(N+2) $END&lt;/code&gt; in GAMESS (US). But I'm not sure, so can anyone confirm that?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. I tried to look at the source code, but it is an old FORTRAN mostly undocumented spaghetti mess with good-old six-lettered totally undescriptive variable names. :D&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-05-20T13:31:04.093" Title="What is the exact meaning of GAMESS (US) SCF density convergence criteria?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31742" PostTypeId="1" AcceptedAnswerId="31756" CreationDate="2015-05-20T13:28:15.357" Score="3" ViewCount="758" Body="&lt;p&gt;Why signals of carbon and hydrogen atoms (in NMR spectra) are split by coupling to $\ce{^{19}F}$? Some spectra books only states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{^{19}F}$ (natural abundance 100%) has a spin quantum number &lt;em&gt;I&lt;/em&gt; of ½. The signals of carbon and H atoms up to a distance of about four bonds are split by coupling to $\ce{^{19}F}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but there is no explanation. Could someone please explain why $\ce{^{19}F}$ can couple with H or C and split their signals.&lt;/p&gt;&#xA;" OwnerUserId="16367" LastEditorUserId="7475" LastEditDate="2015-05-20T18:56:20.120" LastActivityDate="2015-05-20T18:56:20.120" Title="Why are signals of carbon and hydrogen atoms (in NMR spectra) split by coupling to 19F?" Tags="&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31743" PostTypeId="2" ParentId="31740" CreationDate="2015-05-20T13:31:04.093" Score="7" Body="&lt;p&gt;The difference is calculated in the subroutine DDIFF (scflib.src) and corresponds to the largest absolute change in the density matrix &lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-20T13:31:04.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31744" PostTypeId="1" AcceptedAnswerId="31745" CreationDate="2015-05-20T13:31:46.243" Score="7" ViewCount="720" Body="&lt;p&gt;Formula for calculating age by using carbon-14 (source: &lt;a href=&quot;http://science.howstuffworks.com/environmental/earth/geology/carbon-142.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;HowStuffWorks&lt;/a&gt;)&#xA;$$&#xA;t = \frac{\ln (N_\mathrm f/N_\mathrm o)}{-0.693}t_{1/2}&#xA;$$&#xA;The page doesn't explain what the $-0.693$ represents. What is $-0.693$?&#xA;Secondly, it says that $N_\mathrm o$ is the original level of $\ce{C-14}$ and $N_\mathrm f$ is the measured value. How do you get the original value? Is it the same for all living organisms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says that the age of an object (e.g. a fossil) can be calculated by comparing the amount of $\ce{C-14}$ in the fossil and the amount of $\ce{C-14}$ in a sample of living tissue. The half-life of $\ce{C-14}$ ($\pu{5,730 years}$) can be used to find out how much time has passed since the death of the organism because you should be able to calculate how much time has passed by measuring how much $\ce{C-14}$ has decayed. But how can you calculate how much has decayed if you don't know how much was there in the first place?&lt;/p&gt;&#xA;" OwnerUserId="4400" LastEditorUserId="5026" LastEditDate="2018-02-25T03:08:20.417" LastActivityDate="2021-05-01T17:22:47.650" Title="Carbon dating: What to compare the new value against?" Tags="&lt;physical-chemistry&gt;&lt;radioactivity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31745" PostTypeId="2" ParentId="31744" CreationDate="2015-05-20T13:55:08.237" Score="10" Body="&lt;p&gt;Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is −0.693?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radioactive decay processes follow first-order kinetics.  A first-order reaction is one where the rate depends &lt;strong&gt;only&lt;/strong&gt; on the concentration of one of the reactants raised to the first power.  So consider the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{A} \longrightarrow \mathrm{products}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate can be expressed as the rate of change in the reactant concentration with a change in time such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simply means that [A] is being &lt;em&gt;consumed&lt;/em&gt; (i.e. converted to products) as the reaction proceeds with time.  The corresponding rate law can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{rate} = k[\mathrm{A}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is a rate constant.  We can set these two expressions equal to each other to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we must apply a bit of calculus.  Let us write the previous equation in differential form to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or} &#xA;~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate over $t=0$ to $t = t$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which results in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have our first-order integrated rate law, we can now discuss half-life.  Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to &lt;em&gt;half&lt;/em&gt; its original value (where the original value is simply [A]$_0$.  So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~&#xA;t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then simplify this equation such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~&#xA;t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2) &#xA;~~~\longrightarrow~~~&#xA;t_{\frac{1}{2}} = \dfrac{0.693}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, the 0.693 value comes from the $\ln(2)$ term.  As ringo pointed out below (and I will include here to be thorough), you can now solve your &lt;strong&gt;first-order integrated rate law&lt;/strong&gt; for $t$ and substitute in your half-life equation such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt&#xA;~~~\longrightarrow~~~&#xA;\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t&#xA;~~~\longrightarrow~~~&#xA;\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}&#xA;{&#xA;-\frac{0.693}{t_{\frac{1}{2}}}&#xA;}&#xA;=t&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or more cleanly...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$t = \dfrac{\left(t_{\frac{1}{2}}\right)&#xA;\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}&#xA;{-0.693}&#xA;$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do you get $N_0$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant.  Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!).  Therefore you could use its activity for $N_0$.  Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-05-21T13:03:02.837" LastActivityDate="2015-05-21T13:03:02.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31746" PostTypeId="1" CreationDate="2015-05-20T14:26:20.570" Score="0" ViewCount="41" Body="&lt;p&gt;When a functionalized cyclohexenyl halide undergoes $\ce{S_{N}1}$, does a racemic product form (due to ring flipping) ?&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-20T14:26:20.570" Title="Stereoselectivity solvolysis of cyclohexenyl halides" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31747" PostTypeId="1" AcceptedAnswerId="31752" CreationDate="2015-05-20T14:30:35.340" Score="6" ViewCount="136" Body="&lt;p&gt;I am currently struggling with an assignment question regarding structural determination using both $\ce{^1H}$ and $\ce{^{13}C}$ NMR data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Ib1O.png&quot; alt=&quot;screenshot data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I have done is calculate my double bond equivalences which I believe is 3. I believe the IR data corresponds to a $\ce{C=O}$ bond which is supported by the number of DBEs. As it states this is a cyclic compound I believe that the there is a ring containing a double bond and a carbon double bonded to an oxygen which accounts for the 3 DBEs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting together this information which I believe to be correct with the spectroscopic data above is where I am really struggling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite confident in saying that the 6.8 triplet is due to an aromatic carbon with two hydrogens on the adjacent carbons so I have a fragment there that I believe is correct but I cannot get anything more than that.&lt;/p&gt;&#xA;" OwnerUserId="16368" LastEditorUserId="4945" LastEditDate="2018-02-28T07:49:26.620" LastActivityDate="2018-02-28T07:49:26.620" Title="How to interpret spectroscopic data to obtain the structural formula?" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31748" PostTypeId="1" AcceptedAnswerId="31764" CreationDate="2015-05-20T14:41:49.587" Score="6" ViewCount="1521" Body="&lt;p&gt;I know that when you have $\ce{HCl}$, $\ce{H2O}$, and $\ce{NaOH}$ that you produce $\ce{H2O}$ and $\ce{NaCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However some people say water does react and so would think that this 2 step reaction is correct and the 1 step reaction is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H2O + NaOH -&amp;gt; H3O+ + Cl- + NaOH -&amp;gt; 2 H2O + NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However others say that water does not react and think that the 2 step reaction is wrong and the 1 step reaction is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + NaOH -&amp;gt; H2O + NaCl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally think that both of these happen and so water consumed = water produced and thus there is no change in volume. So does water react with the $\ce{HCl}$ via the 2 step reaction or is it just a spectator molecule waiting around until there is more water and salt?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="6875" LastEditDate="2015-09-03T20:02:16.323" LastActivityDate="2015-09-03T20:02:16.323" Title="Water reacts or doesn't react?" Tags="&lt;acid-base&gt;&lt;water&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31749" PostTypeId="1" CreationDate="2015-05-20T14:47:35.410" Score="11" ViewCount="1152" Body="&lt;p&gt;Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare for example the melting point of benzenesulfonic acid hydrate, which is lower than for anhydrous benzenesulfonic acid.&lt;/p&gt;&#xA;" OwnerUserId="16371" LastEditorUserId="3846" LastEditDate="2015-05-20T15:24:35.313" LastActivityDate="2015-10-26T16:15:21.660" Title="Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid?" Tags="&lt;melting-point&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31750" PostTypeId="2" ParentId="31744" CreationDate="2015-05-20T14:56:59.233" Score="3" Body="&lt;p&gt;Carbon-14 dating works because in a living organism, carbon-14 is constantly decaying, but is also constantly being replenished by the organism, meaning its concentration in the living tissue essentially does not change. Though the concentration of the element carbon itself is different for all organisms, the percent abundance of carbon-14 in all living organisms will be &lt;span class=&quot;math-container&quot;&gt;$\approx \pu{0.00000000001\%}$&lt;/span&gt; (one part per trillion). In a deceased organism, carbon-14 can no longer be replaced, so the supply of carbon-14 slowly diminishes over time (&lt;span class=&quot;math-container&quot;&gt;$t_{1/2}=\pu{5730 years}$&lt;/span&gt;). The concentration of carbon-14 in a deceased organism at any given time can therefore be used to find how long ago the organism stopped replenishing its supply of carbon-14.&lt;/p&gt;&#xA;&lt;p&gt;The decay of &lt;span class=&quot;math-container&quot;&gt;$\ce{^14C}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{^14N}$&lt;/span&gt; is given by the equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{^14_6C -&amp;gt; ^14_7N + e-}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The equation you see is derived from the integrated rate law. This is a first order reaction with respect to both &lt;span class=&quot;math-container&quot;&gt;$\ce{^14C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{^14N}$&lt;/span&gt;, meaning there is one of each atom in the rate determining step of the reaction. This means that the change in the concentration of each of these atoms can be used to determine the age of a sample which it decays can be modeled by the equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[\ce{^14C}]=-kt+\ln[\ce{^14C}]_0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or as you have it in your question:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[N_f]=-kt+\ln[N_0]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The rate constant for the equation, derived from the time taken for the concentration of the concentration to be half of the initial (&lt;span class=&quot;math-container&quot;&gt;$N_f=\frac{N_0}{2}$&lt;/span&gt;) is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[\frac{N_0}{2}]=-kt_{1/2}+\ln[N_0]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[\frac{N_0}{2}]-\ln[N_0]=-kt_{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[\frac{1}{2}]=-kt_{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[2]=kt_{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\ln[2]}{t_{1/2}}=k$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln(2)\approx 0.693$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k=\frac{0.693}{t_{1/2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Solving the equation for any time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[N_f]=-kt+\ln[N_0]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[N_f]-\ln[N_0]=-kt$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[N_f]-\ln[N_0]=-kt$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln[\frac{N_f}{N_0}]=-kt$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\ln[\frac{N_f}{N_0}]}{-k}=t$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Substitute in what we solved for:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\ln[\frac{N_f}{N_0}]}{-k}=t$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t=\frac{\ln[\frac{N_f}{N_0}]}{-\frac{0.693}{t_{1/2}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t=\frac{\ln[\frac{N_f}{N_0}]}{-{0.693}}t_{1/2}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="24052" LastEditDate="2021-05-01T17:22:47.650" LastActivityDate="2021-05-01T17:22:47.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31751" PostTypeId="1" AcceptedAnswerId="31761" CreationDate="2015-05-20T15:28:35.210" Score="7" ViewCount="1642" Body="&lt;p&gt;In the acid catalysed hydration of allene, acetone is obtained as the major product. This is because of electrophilic addition of $\ce{H+}$ across a double bond to give a vinyl carbocation, and subsequent addition of water and then tautomerism to yield the ketone. What I'm confused about is that why does not an allyl carbocation form, which would be resonance stabilised?(leading to  allyl alcohol?)&lt;/p&gt;&#xA;" OwnerUserId="9984" LastEditorUserId="4231" LastEditDate="2015-05-20T17:20:00.273" LastActivityDate="2015-05-20T18:33:46.963" Title="Acid catalysed hydration of allene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31752" PostTypeId="2" ParentId="31747" CreationDate="2015-05-20T15:29:24.993" Score="5" Body="&lt;p&gt;Your calculations of the double bond equivalents seems correct, and there is also another hint towards a $\ce{C=O}$ double bond: the corresponding $\ce{^{13}C}$ chemical shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while it says you have $\ce{^1H}$ decoupled $\ce{^{13}C}$ data, this is clearly wrong, because they’re showing you multiplets. The $\ce{^{13}C}$ data is $\ce{^1H}$ coupled data, so you can use the information there to deduce the number of carbons with which number of hydrogens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third hint given to you is the amoun of signals: You see signals that add up to six carbons and eight hydrogens, meaning there is no symmetry in your molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6.8 hydrogen could, by its chemical shift, be aromatic, but you do not have enough DBE’s for an aromatic ring (you would need at least 4). Do you know other types of protons that give chemical shifts in similar ranges, and are consistent with both your DBE count and the structure element(s) you have already found out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further ideas to think of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have certain integrals on your hydrogen signals, especially one that stands out (not the $6.8\,\mathrm{ppm}$). What could this integral mean, and how does it affect the rest of the structure?&lt;/li&gt;&#xA;&lt;li&gt;Make a list of the carbons you have and the amount of hydrogens attached to them.&lt;/li&gt;&#xA;&lt;li&gt;How can you turn all that into a cyclic compound?&lt;/li&gt;&#xA;&lt;li&gt;Once you have, you can check if the multiplets of the hydrogen signals can be fitted to that structure. Hint: You seem to have no H–H couplings other than across three bonds. (So no long-range couplings.) And no H–C couplings other than one bond length.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Right, in this first spoiler tag I’m just gonna state the different signals. Check the second one, if you &lt;em&gt;really&lt;/em&gt; want to know the solution.&lt;br&gt;&#xA; The carbon at 209.6 (singlet) is most likely not an aldehyde but a ketone. If it had a proton connected, it would need to be a doublet.&lt;br&gt;&#xA; The singlet at 1.8 is a methyl group, belonging to the carbon at 12.1.&lt;br&gt;&#xA; The triplet at 6.8 needs to be next to a double bond (strong low-field shift). The corresponding carbon needs to be 157.9 (doublet).&lt;br&gt;&#xA; Since it’s a triplet, it couples to two hydrogens. There is no other hydrogen on a double bond, but there is a triplet and a quartet left. One of these two must couple to the proton integral one at 6.8. Luckily they are both integral 2. A little further reasoning tells us, it must be the quartet, because the other triplet (2.4) has to couple to two protons, which have to couple back, therefore no other possibility.&lt;br&gt;&#xA; There is another carbon at a double bond (142.1) and it has no hydrogens attached to it (singlet). It needs to have the methyl group on it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; The structure I deduced is &lt;strong&gt;2-methylcyclopent-2-en-1-one&lt;/strong&gt;.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-05-20T20:07:14.190" LastActivityDate="2015-05-20T20:07:14.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31754" PostTypeId="1" AcceptedAnswerId="31759" CreationDate="2015-05-20T16:39:42.357" Score="4" ViewCount="7724" Body="&lt;p&gt;What is the correct nomenclature for $\ce{CN^-}$ and $\ce{Cl^-}$ ligands in coordination compounds?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.m.wikipedia.org/wiki/Coordination_complex#Naming_complexes&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; says the coding name(?) as cyanido and chlorido while &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Basics_of_Coordination_Chemistry/Nomenclature_of_Coordination_Complexes#Rules_for_Naming_the_Ligands&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; suggests cyano and chloro.   &lt;/p&gt;&#xA;" OwnerUserId="16377" LastActivityDate="2015-05-20T17:08:07.170" Title="Nomenclature of complexes: Cyano vs Cyanido ; Chloro vs Chlorido" Tags="&lt;nomenclature&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31755" PostTypeId="1" AcceptedAnswerId="31774" CreationDate="2015-05-20T16:42:16.110" Score="0" ViewCount="49" Body="&lt;p&gt;I have the absorption spectrum of a specific phycobiliprotein. Its maximum wavelength is 620.6 (nm),this is the only data I have. How can I know what type of phycobiliprotein it is?&lt;/p&gt;&#xA;" OwnerUserId="16328" LastEditorUserId="418" LastEditDate="2015-05-21T00:04:54.210" LastActivityDate="2015-05-21T00:11:40.327" Title="Identifying phycobiliprotein based on wavelength" Tags="&lt;spectroscopy&gt;&lt;photochemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31756" PostTypeId="2" ParentId="31742" CreationDate="2015-05-20T16:44:08.753" Score="5" Body="&lt;p&gt;You always have a coupling if you have to NMR-active nuclei connected, there is nothing unusual about fluor here. But in the most common and simplest NMR experiments on small organic molecules you usually don't encounter such heteronuclear couplings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first important aspect is that many common nuclei in organic molecules are present mostly in an NMR-inactive isotope. Carbon and nitrogen for example are only present in the NMR-active isotopes 13C and 15N in tiny amounts. So you'll be able to see small satellite peaks sometimes, but no complete splitting. If only 1% of the nuclei is NMR-active, only 1% of the signal will be affected by the coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second aspect is that some experiments like 13C 1D experiments are typically recorded with decoupling. There is a coupling between 1H and 13C, but it is intentionally suppressed as it would complicate the spectra.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-05-20T16:44:08.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31757" PostTypeId="1" AcceptedAnswerId="31814" CreationDate="2015-05-20T16:50:45.827" Score="16" ViewCount="5172" Body="&lt;p&gt;I've been taking an chemistry class online (for fun only) and one of the questions in the homework threw me a bit. Here's the wording:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Draw the Lewis structure for a nitrogen atom attached to. two methyl groups ($\ce{CH3}$) and one oxygen atom.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so I know that $\ce{NO}$ is a free radical due to the odd combined valence, i.e., the octet rule isn't satisfied. Will that carry over when the $\ce{NO}$ bonds with the two methyl groups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the sketch of what I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3BR4d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, what's the name of this compound? This is not part of the HW question; I'm just curious.&lt;/p&gt;&#xA;" OwnerUserId="16374" LastEditorUserId="16374" LastEditDate="2015-05-20T20:48:12.627" LastActivityDate="2015-05-23T04:12:04.183" Title="Lewis Structure on (CH3)2NO?" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31758" PostTypeId="2" ParentId="31732" CreationDate="2015-05-20T16:54:40.737" Score="28" Body="&lt;p&gt;Here is a picture of a &quot;classical&quot; carbocation, there is an electron deficient carbon bearing a positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IfdFC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many examples of &quot;non-classical&quot; carbocations, but the &lt;a href=&quot;http://en.wikipedia.org/wiki/2-Norbornyl_cation&quot; rel=&quot;noreferrer&quot;&gt;2-norbornyl carbocation&lt;/a&gt; is among the best known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f2mqy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f2mqy.png&quot; alt=&quot;Norbornyl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Labeling experiments have shown that the positive charge resides on more than one carbon in the 2-norbornyl ion.  Early on, the data was explained by equilibrating classical ions, but soon another possibility emerged - one involving a single non-classical ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem comes down to: are the equilibrating classical ions ground state structures with the non-classical ion serving as the transition state, or is the non-classical ion the ground state? This debate went on for a very long period of time, but now most agree that the non-classical structure is the ground state in the 2-norbornyl system.  In fact, a recent, and difficult to obtain, &lt;a href=&quot;http://www.sciencemag.org/content/341/6141/62&quot; rel=&quot;noreferrer&quot;&gt;crystal structure for the 2-norbornyl cation&lt;/a&gt; has been published proving that the ion exists with the non-classical geometry (thanks to Klaus for finding this reference, see his comment below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0MoLl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A key difference between classical and non-classical structures is the bonding&lt;/strong&gt;.  As illustrated above, a classical ion has a carbon with a sextet of electrons and 3 other bonds.  The non-classical ion, on the other hand, involves 3 carbons with 2 electrons spread over them.  This is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_two-electron_bond&quot; rel=&quot;noreferrer&quot;&gt;3-center 2-electron bond&lt;/a&gt; (&lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypercoordinate bonding&lt;/a&gt;) and is &lt;strong&gt;a clear marker for a non-classical ion&lt;/strong&gt;.  Notice if you count all of the bonds to any of these 3 carbon atoms (solid and dashed lines) you count 5!  Sounds strange, but such &quot;hypercoordinate bonding&quot; is a permitted consequence of the 3-center 2-electron bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LVZBR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LVZBR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29000/which-carbocation-is-more-stable-the-ethyl-or-1-propyl-carbocation/31729#31729&quot;&gt;link&lt;/a&gt; to a simpler non-classical carbocation discussed here yesterday.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2016-09-09T07:43:25.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31759" PostTypeId="2" ParentId="31754" CreationDate="2015-05-20T17:08:07.170" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf&quot; rel=&quot;nofollow&quot;&gt;IUPAC Red Book&lt;/a&gt; is the official reference for recommended inorganic nomenclature. The full change to the additive nomenclature system was made in the 2005 update:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some new recommendations represent breaks with tradition, in the interest of increased clarity and consistency. For example, the application of the ending ‘ido’ to all anionic ligands with ‘ide’ names in additive nomenclature (e.g. chlorido and cyanido instead of chloro and cyano, and hydrido throughout, i.e. no exception in boron nomenclature) is part of a general move to a more systematic approach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is discussed in more detail in section IR-7.1.3 on page 112. Basically, anionic ligand names used as prefixes get changed from '-ide', '-ate', and '-ite' to '-ido', '-ato', and 'ito', respectively. Neutral and cationic ligands are usually used unchanged except for things like aqua, ammine, carbonyl, etc.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-05-20T17:08:07.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31761" PostTypeId="2" ParentId="31751" CreationDate="2015-05-20T17:18:15.473" Score="11" Body="&lt;p&gt;Good question!  As the following figure illustrates, there are 2 ways we can protonate an allene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXdUS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/special1.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top line in the drawing starts by protonating at the central allene carbon.  This generates a very unstable &lt;strong&gt;primary&lt;/strong&gt; carbocation.  Initially, this primary carbocation is not stabilized by the adjacent double bond.  It is not until after this carbocation rotates 90° that its p orbital is lined up with the p orbitals in the double bond and allylic stabilization can occur.  Since there is no immediate allylic stabilization when the carbocation is formed, and since it is a primary carbocation, the pathway leading to it is very high energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, protonation at the terminal allene carbon (bottom line of figure) produces a relatively stabilized &lt;strong&gt;secondary&lt;/strong&gt; carbocation.  Formation of a secondary carbocation is a much lower energy process.  Therefore, protonation at the terminal allene carbon is lower energy and is what will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Response to OP's comment&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I always thought that a positive charge appearing on sp3 carbon (which&#xA;  then becomes sp2) is more stable than a positive charge appearing on&#xA;  sp2 carbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Generally we think of vinylic carbocations as high energy.  That is true, but what is meant by &quot;vinylic&quot; carbocation is different from what we have here.  &lt;strong&gt;Here there is an empty p orbital&lt;/strong&gt;.  &lt;strong&gt;In a vinylic carbocation the empty orbital is not a p orbital&lt;/strong&gt;, but rather something like $\ce{sp^2}$ or $\ce{sp}$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is the secondary carbocation more stable because of inductive effect,&#xA;  the rotational barrier in the primary carbocation, or both&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The secondary carbocation is more stable for the same reasons that we have the  tertiary &gt; secondary &gt; primary carbocation stability order.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inductive effects, $\ce{sp^3}$ and $\ce{sp^2}$ carbons are electron releasing towards an $\ce{sp}$ carbon due to electronegativity differences&lt;/li&gt;&#xA;&lt;li&gt;Resonance effects from hyperconjugation involving hydrogens on the carbon next to the carbocation center.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-20T18:33:46.963" LastActivityDate="2015-05-20T18:33:46.963" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31762" PostTypeId="1" CreationDate="2015-05-20T17:47:16.197" Score="2" ViewCount="76" Body="&lt;p&gt;Is there any generic method where I can know/calculate/generate the possible ionization and fragmentation of a specific chemical, for example, hydrogen chloride and water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is actually related to the mass spectrometer, where I am asked to identify the possible species/ions of a sample. Therefore I would like to know if there's any method to know (calculate) the outcome without memorising everything, for example the outcome of HCl.&lt;/p&gt;&#xA;" OwnerUserId="16247" LastEditorUserId="41328" LastEditDate="2019-05-17T20:09:07.483" LastActivityDate="2019-05-17T20:09:07.483" Title="Possible ionization and fragmentation of chemicals" Tags="&lt;analytical-chemistry&gt;&lt;ions&gt;&lt;spectroscopy&gt;&lt;oxidation-state&gt;&lt;mass-spectrometry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31763" PostTypeId="2" ParentId="8242" CreationDate="2015-05-20T18:20:17.937" Score="3" Body="&lt;p&gt;If numbering from each end of the parent chain gives the same locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name. This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(g) lowest locants for the substituent cited first as a prefix in the name;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a correct systematic name is &lt;strong&gt;4-ethyl-5-methyloctane&lt;/strong&gt; (not 5-ethyl-4-methyloctane) since ethyl is cited first as a prefix in the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fdt0F.png&quot; width=&quot;157&quot; alt=&quot;4-ethyl-5-methyloctane&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another correct systematic name is &lt;strong&gt;4-ethyl-5-propyloctane&lt;/strong&gt; (not 5-ethyl-4-propyloctane). Again, the reason is that ethyl is cited first as a prefix in the name. The size of the substituent is not relevant in this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y86jm.png&quot; width=&quot;157&quot; alt=&quot;4-ethyl-5-propyloctane&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-12-29T11:38:58.670" LastActivityDate="2016-12-29T11:38:58.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31764" PostTypeId="2" ParentId="31748" CreationDate="2015-05-20T18:30:35.600" Score="7" Body="&lt;p&gt;The answer to this depends on how the reagents are mixed. For one, you can have the reaction without water, as pointed out by @Klaus in the comments. If you blow dry HCl on dry NaOH, they will react without involving water in the reactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + NaOH -&amp;gt; H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably not the case though, and if you're dealing with solutions, then yes, water will react. If you start by adding the HCl to pure water, you'll have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This representation is prefered over the simple dissociation $\ce{HCl -&amp;gt; H+ + Cl-}$ because it shows that you don't have &quot;free&quot; protons in solution. You could actually even say that this reaction has already happened since, considering you're using hydrochloric acid it's already dissolved, and this step would be just a dilution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with the NaOH solution, you could argue that water does not react since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH -&amp;gt; Na+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but that only happens in the liquid state (usually in solution). Also, you can't ignore the ions solvation (which is one, if no the most important factor which affects the solubility of the sodium hydroxide salt) and equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O + OH- &amp;lt;=&amp;gt; OH- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what this means is that you don't have fixed hydroxil ions and water molecules, so you can't simply ignore the solvent and assume it's &lt;em&gt;just a spectator molecule waiting around&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To write the two step reaction, it's not a good idea to do it in one line, because as you can see on the first step you have NaOH on both sides, which is something that doesn't make much sense. It's better to do it like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3O+ + Cl- + NaOH -&amp;gt; 2H2O + NaCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's usually a good idea to represent salts in their ionic form when dissolved, which would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3O+ + Cl- + NaOH -&amp;gt; 2H2O + Na+_{(aq)} + Cl^{-}_{(aq})}$$&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-05-20T18:35:55.327" LastActivityDate="2015-05-20T18:35:55.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31766" PostTypeId="2" ParentId="31739" CreationDate="2015-05-20T18:58:09.380" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;[&amp;hellip;] my teacher told me to continue using a trial and error method [&amp;hellip;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But nobody told you to do it &lt;strong&gt;manually&lt;/strong&gt; once you have understood how to adjust the &lt;em&gt;Rydberg&lt;/em&gt; constant for $\ce{He+}$ and use the formula to get the energies for the given  transistions, in order to compare them with the $\ce{H_{\alpha}}$ line (656.3 nm):&lt;/p&gt;&#xA;&#xA;&lt;pre class=&quot;lang-py prettyprint-override&quot;&gt;&lt;code&gt;import scipy.constants&#xA;&#xA;transitions = ((7,5), (5,3), (6,4), (4,2))&#xA;&#xA;def get_he_line_nm(n_upper, n_lower):&#xA;    '''calculate the energy for a line in a He+ spectrum and return &#xA;       it in nanometer''' &#xA;&#xA;    energy = 4 * scipy.constants.Rydberg * (pow(n_lower, -2) - pow(n_upper, -2))&#xA;&#xA;    return 1.0/(scipy.constants.nano * energy) &#xA;    # energy in nm&#xA;&#xA;&#xA;for upper, lower  in transitions:&#xA;    wavelength = get_he_line_nm(upper, lower)&#xA;    print('{0} -&amp;gt; {1} : {2:7.2f} nm'.format(upper, lower, wavelength)) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;will give the output&#xA;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;7 -&amp;gt; 5 : 1162.81 nm&#xA;5 -&amp;gt; 3 :  320.37 nm&#xA;6 -&amp;gt; 4 :  656.11 nm&#xA;4 -&amp;gt; 2 :  121.50 nm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="418" LastEditorUserId="4945" LastEditDate="2017-11-22T07:30:23.883" LastActivityDate="2017-11-22T07:30:23.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31767" PostTypeId="2" ParentId="31640" CreationDate="2015-05-20T21:52:25.000" Score="2" Body="&lt;p&gt;Valence electrons experience an electrostatic force from the nucleus. The nucleus has its character positive charge, but due to &lt;em&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Physical_Chemistry/Quantum_Mechanics/10:_Multi-electron_Atoms/Multi-Electron_Atoms/Penetration_%26_Shielding&quot; rel=&quot;nofollow&quot;&gt;shielding&lt;/a&gt;&lt;/em&gt; by core electrons the total positive charge is not completely felt by the electrons, so the actual net positive charge felt by the valence electron has its own name and is called the effective nuclear charge $Z_{eff}$. This effective nuclear charge, largely effects the size of the atom. The general trend of the periodic table is that as you go down a group, and go from right to left, the $Z_{eff}$ decreases, and you see an increase in the size of the atom. The easy way to think of it, is that as you go from left to right, the number of valence electrons, and  protons increases but the number core electrons stays the same; so, you see an increase in $Z_{eff}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z_{eff}$ is calculated using &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;nofollow&quot;&gt;Slater's Rules&lt;/a&gt;. Where&#xA;$$Z_{eff}=Z-s$$&#xA;Where $Z$ is the atomic number and $s$ is the shielding constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules for calculating are as follows (from &lt;a href=&quot;http://en.wikipedia.org/wiki/Slater%27s_rules#Rules&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly,(1)(4) the electrons are arranged into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l, except that s- and p- orbitals are kept together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{(1s) (2s,2p) (3s,3p) (3d) (4s,4p) (4d) (4f) (5s, 5p) (5d)}$ etc.&#xA;Each group is given a different shielding constant which depends upon the number and types of electrons in those groups preceding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shielding constant for each group is formed as the sum of the following contributions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;An amount of 0.35 from each other electron within the same group except for the (1s) group, where the other electron contributes only 0.30.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the group is of the (s p) type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&#xA;If the group is of the (d) or (f), type, an amount of 1.00 for each electron &quot;closer&quot; to the atom than the group. This includes &#xA;i) electrons with a smaller principal quantum number n and &#xA;ii) electrons with an equal principal quantum number and a smaller azimuthal quantum number l.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-05-20T22:04:00.173" LastActivityDate="2015-05-20T22:04:00.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31768" PostTypeId="1" CreationDate="2015-05-20T21:59:34.647" Score="3" ViewCount="7610" Body="&lt;p&gt;I am looking to separate the small quantities of platinum and palladium in a catalytic converter. However since these are very inert metals I am unsure how to proceed about doing so. Melting the converter is an option I have considered but I am unsure whether I will be able to reach high enough temperatures in order to obtain a decent amount of metal, so I am leaning more towards performing reaction.&lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorUserId="418" LastEditDate="2015-05-22T12:42:40.877" LastActivityDate="2019-11-05T17:06:11.323" Title="Separate precious metals from catalytic converter" Tags="&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;&lt;safety&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31769" PostTypeId="2" ParentId="27200" CreationDate="2015-05-20T22:10:23.040" Score="4" Body="&lt;p&gt;Cleaning metal surfaces in industry is a very common problem.  Common enough, in fact, that an industry has grown up around the solution.  Halocarbons, either singly or in combination, are excellent at removing contaminants from metal surfaces (see the &quot;uses&quot; section in &lt;a href=&quot;http://en.wikipedia.org/wiki/Trichloroethylene&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;).  Typically a perforated wire basket is filled with the parts to be cleaned and then the basket is suspended above a bath of the halocarbon liquid.  The liquid is brought to reflux (typically less than 100 °C) and the vapors dissolve all impurities and the part comes out clean.  The dirty vapors condense, the impurities eventually precipitate out and clean vapor is ready to repeat the cleaning step.  The vapors are contained in the unit so atmospheric contamination is minimized. This approach works well for intricate parts or parts with hard to access cavities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Ultrasonic_cleaning&quot; rel=&quot;nofollow&quot;&gt;ultrasonification&lt;/a&gt;, using either an organic solvent or water containing a cleaning agent, is another possible cleaning method.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-20T22:10:23.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31770" PostTypeId="2" ParentId="31708" CreationDate="2015-05-20T22:25:53.830" Score="2" Body="&lt;p&gt;Part (a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with the previous comments that a Wittig reaction is the best way forwards especially in terms of regioselectivity of the double bond. PPh3, MeI and NaH would be an excellent combination. Perhaps draw the mechanism and you'll see how/why this work so well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested, you could use (Cp)2Ti=CH2 also known as Tebbe's Reagent. Klaus Warzecha mentioned this already (above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part (b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like your idea of reducing, tosylating and using the non-nucleophilic base tBuOK&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2015-05-20T22:25:53.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31772" PostTypeId="1" CreationDate="2015-05-20T23:08:22.990" Score="3" ViewCount="5362" Body="&lt;p&gt;Just getting interested in this stuff. People are saying I should crystallize salt, for example. What substances besides salts form a crystal structure?&lt;/p&gt;&#xA;" OwnerUserId="16029" LastEditorUserId="15410" LastEditDate="2016-05-05T22:45:33.297" LastActivityDate="2016-05-05T22:45:33.297" Title="What substances form crystal structures?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ClosedDate="2015-05-22T04:40:56.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="31773" PostTypeId="1" CreationDate="2015-05-21T00:01:59.447" Score="6" ViewCount="1159" Body="&lt;p&gt;My friend is installing copper pipes in his house, which is going to be in a humid area, and he says that copper will only be oxidized by the air and the excess water vapor in the humid air won't affect it. However, it is to my understanding that since water vapor is $\ce{H2O}$, which contains oxygen, there will be more oxidation occurring, and thus more corrosion to the pipes because of more oxygen present in the air. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="15252" LastEditorUserId="15410" LastEditDate="2015-05-21T04:02:03.260" LastActivityDate="2015-05-21T06:37:39.913" Title="Does water vapor in the air oxidize copper?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31774" PostTypeId="2" ParentId="31755" CreationDate="2015-05-21T00:11:40.327" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Its maximum wavelength is 620.6 (nm)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm afraid that this is pretty much a link-only answer, but based on   &lt;em&gt;Spectroscopic Properties of C-Phycocyanin and of Its $\alpha$ and $\beta$ Subunits&lt;/em&gt;, published in &lt;em&gt;J. Biol. Chem.&lt;/em&gt;, &lt;strong&gt;1973&lt;/strong&gt;, &lt;em&gt;248&lt;/em&gt;, 5679-5685 (&lt;a href=&quot;http://www.jbc.org/content/248/16/5679.full.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF&lt;/a&gt;) there's a good chance that the phycobiliprotein is &lt;strong&gt;C-Phycocyanin&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-21T00:11:40.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31775" PostTypeId="1" CreationDate="2015-05-21T02:08:27.330" Score="1" ViewCount="1162" Body="&lt;p&gt;I am trying to determine if the following two grades of ethanol are equivalent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ethanol absolute, 99.8% Normapur AR (Analytical Reagent)&lt;/li&gt;&#xA;&lt;li&gt;Ethanol alcohol, pure ACS reagent, ≥99.5%&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Information on the technical specifications of these grades would also be helpful for comparison.&lt;/p&gt;&#xA;" OwnerUserId="16387" LastEditorUserId="275" LastEditDate="2015-05-21T03:06:15.173" LastActivityDate="2017-05-25T21:31:26.820" Title="Comparing purity grades for ethanol" Tags="&lt;alcohols&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31776" PostTypeId="1" CreationDate="2015-05-21T05:30:40.490" Score="6" ViewCount="94" Body="&lt;p&gt;I spent a while trying to figure out which stackexchange site in which to post this question, forgive me if I got it wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://scholar.google.com/scholar?q=concrete%20leaching&amp;amp;btnG=&amp;amp;hl=en&amp;amp;as_sdt=0%2C38&amp;amp;as_vis=1&quot;&gt;number&lt;/a&gt; of scholarly articles indicate that concrete can leach calcium compounds under the right conditions. Could they ever wick upwards against gravity through dry or water-saturated soil?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be a raised bed of soil with wooden sides and a concrete floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I also wonder about plastics leaching upwards, but this is admittedly most likely a different question.&lt;/p&gt;&#xA;" OwnerUserId="16390" LastActivityDate="2021-02-27T17:04:50.417" Title="Can compounds found in concrete leach into soil against gravity?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31778" PostTypeId="1" CreationDate="2015-05-21T06:25:18.317" Score="4" ViewCount="394" Body="&lt;p&gt;I am refluxing $\ce{HBr}$ with 1,6-hexanediol for 16 hours at $80\ ^\circ\mathrm{C}$ in toluene. How can I be sure that more than $90\,\%$ of the product is brominated only at one terminal? If the amount of double brominated product is in more than $10\,\%$ of the total yield, then the product mixture will be useless for me. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorDisplayName="user7951" LastEditDate="2015-05-29T19:14:18.247" LastActivityDate="2015-05-29T19:14:18.247" Title="Preparation of 6-bromo-1-hexanol" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31779" PostTypeId="2" ParentId="31773" CreationDate="2015-05-21T06:37:39.913" Score="3" Body="&lt;p&gt;Although $\ce{H_2O}$ contains oxygen, it is there in the most reduced state $\ce{O^{2-}}$ and therefore cannot act as oxidizer. It is the oxygen in the air $\ce{O_{2}}$, which is the main and strongest oxidizing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, presence of water as a solvent can speed up some reactions. For a short overview of copper corrosion, see e.g. &lt;a href=&quot;http://www.corrosionist.com/Why_Does_Copper_Oxidize.htm&quot; rel=&quot;nofollow&quot;&gt;Why does copper oxidize&lt;/a&gt;. As the corrosion products form compact (or even protecting) layer, the corrosion itself is not a problem for copper, as compared to e.g. steel corrosion.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-05-21T06:37:39.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31780" PostTypeId="2" ParentId="9172" CreationDate="2015-05-21T06:39:24.017" Score="3" Body="&lt;p&gt;Many bases do not deprotonate (e.g. $\ce{NaOH}$), making the notion of $\mathrm{p}K_\mathrm{a}$ based on deprotonation (Brønsted-Lowry theory) extremely difficult to measure/apply. However, there are some bases which are also acids: for instance $\ce{HSO4-}$ is a base because it can accept a proton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HSO4- + H2O &amp;lt;=&amp;gt; H2SO4 + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and an acid because it can donate a proton:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HSO4- + H2O &amp;lt;=&amp;gt; SO4^2- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many bases which can act as acids and therefore those “bases” have a $\mathrm{p}K_\mathrm{a}$ for the deprotonation. In that case, the “base” would be acting as an “acid”. Measuring $\mathrm{p}K_\mathrm{a}$ for a specific substance assumes that substance undergoes acid dissociation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting the $\mathrm{p}K_\mathrm{b}$ of $\ce{HSO4-}$ from 14 &lt;em&gt;does not&lt;/em&gt; give the $\mathrm{p}K_\mathrm{a}$ of $\ce{HSO4-}$: it gives the $\mathrm{p}K_\mathrm{a}$ of the &lt;em&gt;conjugate acid&lt;/em&gt; $\ce{H2SO4}$ &lt;strong&gt;in water&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take this conjugate acid $\ce{H2SO4}$ and ask what its $\mathrm{p}K_\mathrm{a}$ is, you are assuming it is behaving as an acid by losing a proton to form $\ce{HSO4-}$. So set up the following chemical equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 + H2O &amp;lt;=&amp;gt; HSO4- + H3O+}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are asking what is the $\mathrm{p}K_\mathrm{b}$ of $\ce{H2SO4}$, you are assuming it is behaving as a base by gaining a proton to form $\ce{H3SO4+}$, a likely unstable species. Set up the following chemical equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 + H2O &amp;lt;=&amp;gt; H3SO4+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when calculating concentrations of acid/base/conjugate based on $K_\mathrm{a}$ or $K_\mathrm{b}$ values, you need to make sure you have the appropriate chemical equilibrium written out which matches either protonation or deprotonation of a specific chemical species in a specific solvent (likely water). Use this correct chemical equation to fill in the equilibrium constant expression and go from there.&lt;/p&gt;&#xA;" OwnerUserId="16348" LastEditorDisplayName="user7951" LastEditDate="2015-06-13T09:51:11.720" LastActivityDate="2015-06-13T09:51:11.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31781" PostTypeId="2" ParentId="31778" CreationDate="2015-05-21T06:44:47.250" Score="4" Body="&lt;p&gt;In an article by Katharina C. Kress, Martin Kaller, Kirill V. Axenov, Stefan Tussetschläger, and Sabine Laschat, published in &#xA;&lt;em&gt;Beilstein J. Org. Chem.&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;8&lt;/em&gt;, 371-378 (&lt;a href=&quot;http://dx.doi.org/10.3762/bjoc.8.40&quot; rel=&quot;nofollow&quot;&gt;DOI&lt;/a&gt;), this procedure is described. From the &lt;a href=&quot;http://www.beilstein-journals.org/bjoc/content/supplementary/1860-5397-8-40-S1.pdf&quot; rel=&quot;nofollow&quot;&gt;Supporting Information&lt;/a&gt;, I read that the authors obtained the product in 72% yield after &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;column chromatography on silica gel (hexanes/EtOAc, 1:1) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wouldn't be astonished if that's as good as it can get.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-21T06:44:47.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31782" PostTypeId="2" ParentId="27532" CreationDate="2015-05-21T07:11:56.157" Score="24" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt; The next in the series is called &amp;phi; bond. There is even a tiny &lt;a href=&quot;http://en.wikipedia.org/wiki/Phi_bond&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; about it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Nicolau pointed me to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_orbital#.CF.86_symmetry&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;, that had at the time a tiny section about the &amp;phi; symmetry of the bond. Ben also kindly agreed with my naming proposition. I'd like to back up just a little bit an quote one sentence of this article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The type of interaction between atomic orbitals can be further categorized by the molecular-orbital symmetry labels &amp;sigma; (sigma), &amp;pi; (pi), &amp;delta; (delta), &amp;phi; (phi), &amp;gamma; (gamma) etc. paralleling the symmetry of the atomic orbitals s, p, d, f and g. The number of nodal planes containing the internuclear axis between the atoms concerned is zero for &amp;sigma; MOs, one for &amp;pi;, two for &amp;delta;, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically this was what I was already thinking, but since the section is quite tiny overlooked it accordingly. But since this is a reference request, I'll go a little deeper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of the bond is actually quite old and I was able to track it down to at least 1984, where Bruce E. Bursten and Geoffrey A. Ozin &quot;demonstrate[d] for the first time the existance of &amp;phi; bonds between metal atoms.&quot;&lt;sup&gt;[1]&lt;/sup&gt; A little further back, 1978, N. Rösch and A. Streitwieser already performed SCF-X&amp;alpha; calculations on thorocene and uranocene, where they came across f orbital contributions in bonding. Yet, they did not coin the phrase &amp;phi; bond.&lt;sup&gt;[2]&lt;/sup&gt; Following that, there have been more theoretical studies on the diuranium molecule, with increasing level of theory.&lt;sup&gt;[3]&lt;/sup&gt; One of the more recent studies is by Laura Gagliardi and Björn O. Roos, which is available free of charge (see below).&lt;sup&gt;[4]&lt;/sup&gt; Xin Wu and Xin Lu also studied the phenomenon in endohedral matallofullerenes, i.e. diuranium in a Buckminster fullerene.&lt;sup&gt;[5]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite safe to assume, that there are more studies about this research topic or in a more general aspect the research on molecules with high bond orders. For this purpose I will include a couple of good reads.&lt;sup&gt;[6]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References and further reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic00186a039&quot; rel=&quot;noreferrer&quot;&gt;Bruce E. Bursten, Geoffrey A. Ozin, &lt;em&gt;Inorg. Chem.&lt;/em&gt;, &lt;strong&gt;1984&lt;/strong&gt;, &lt;em&gt;23&lt;/em&gt; (18), 2910–2911.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;(a) &lt;a href=&quot;http://dx.doi.org/10.1016/S0022-328X(00)91124-0&quot; rel=&quot;noreferrer&quot;&gt;Rösch, N., Streitwieser, A., Jr., &lt;em&gt;J. Organomet. Chem.&lt;/em&gt;, &lt;strong&gt;1978&lt;/strong&gt;, &lt;em&gt;115&lt;/em&gt;, 195-200.&lt;/a&gt; (b) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00363a004&quot; rel=&quot;noreferrer&quot;&gt;Rösch, N., Streitwieser, A., Jr., &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1983&lt;/strong&gt;, &lt;em&gt;105&lt;/em&gt;, 7237-40.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00177a055&quot; rel=&quot;noreferrer&quot;&gt;Melanie Pepper, Bruce E. Bursten, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1990&lt;/strong&gt;, &lt;em&gt;112&lt;/em&gt; (21),  7803–7804.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1038/nature03249&quot; rel=&quot;noreferrer&quot;&gt;Laura Gagliardi, Björn O. Roos, &lt;em&gt;Nature&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;433&lt;/em&gt;, 848-851.&lt;/a&gt; Available free at &lt;a href=&quot;http://archive-ouverte.unige.ch/unige:3652&quot; rel=&quot;noreferrer&quot;&gt;Archive ouverte UNIGE&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja067281g&quot; rel=&quot;noreferrer&quot;&gt;Xin Wu, Xin Lu, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;129&lt;/em&gt; (7), 2171–2177.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Without order: (a) &lt;a href=&quot;http://www.nature.com/nature/journal/v446/n7133/full/446276a.html&quot; rel=&quot;noreferrer&quot;&gt;Gernot Frenking, &lt;em&gt;Nature&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;446&lt;/em&gt;, 276-277.&lt;/a&gt;&#xA;(b) &lt;a href=&quot;http://www.sciencemag.org/content/310/5749/796&quot; rel=&quot;noreferrer&quot;&gt;Gernot Frenking, &lt;em&gt;Science&lt;/em&gt;, &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;310&lt;/em&gt;, 796-797.&lt;/a&gt;&#xA;(c) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja075454b&quot; rel=&quot;noreferrer&quot;&gt;Gabriel Merino, Kelling J. Donald, Jason S. D'Acchioli, Roald Hoffmann, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;129&lt;/em&gt; (49), 15295–15302.&lt;/a&gt;&#xA;(d) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic301657c&quot; rel=&quot;noreferrer&quot;&gt;Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, &lt;em&gt;Inorg. Chem.&lt;/em&gt;, &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;52&lt;/em&gt; (6), 2838–2843.&lt;/a&gt;&#xA;(e) &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic100427t&quot; rel=&quot;noreferrer&quot;&gt;Jia Zhou, Jason L. Sonnenberg, H. Bernhard Schlegel, &lt;em&gt;Inorg. Chem.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;49&lt;/em&gt; (14), 6545–6551.&lt;/a&gt;&#xA;(f) &lt;a href=&quot;http://dx.doi.org/10.1016/j.ica.2014.08.030&quot; rel=&quot;noreferrer&quot;&gt;Awal Noor, Rhett Kempe, &lt;em&gt;Inorg. Cimica Acta&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;, 424, 75-82.&lt;/a&gt;&#xA;(g) &lt;a href=&quot;http://www.nature.com/nchem/journal/v1/n7/full/nchem.359.html&quot; rel=&quot;noreferrer&quot;&gt;Frank R. Wagner, Awal Noor, Rhett Kempe, &lt;em&gt;Nature Chemistry&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;1&lt;/em&gt;, 529-536.&lt;/a&gt;&#xA;(h) &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.200603600/abstract&quot; rel=&quot;noreferrer&quot;&gt;Björn O. Roos, Antonio C. Borin, Laura Gagliardi, &lt;em&gt;Angew. Chem. Int. Ed.&lt;/em&gt;, &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;46&lt;/em&gt; (9). 1469-1472.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2016-03-03T07:09:40.390" LastActivityDate="2016-03-03T07:09:40.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31783" PostTypeId="1" AcceptedAnswerId="31787" CreationDate="2015-05-21T08:22:31.330" Score="27" ViewCount="1844" Body="&lt;p&gt;I stumbled-upon this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c3NAX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c3NAX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering how to determine the configuration of the double bond. Is it &lt;em&gt;E&lt;/em&gt; or &lt;em&gt;Z&lt;/em&gt;? How do I decide on the priorities when the substituents can only differ in the configurations on the chiral centers?&lt;/p&gt;&#xA;" OwnerUserId="16394" LastEditorUserId="16683" LastEditDate="2017-01-10T09:22:28.750" LastActivityDate="2018-04-05T12:53:04.093" Title="How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry" Tags="&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;chirality&gt;&lt;isomers&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31784" PostTypeId="2" ParentId="31772" CreationDate="2015-05-21T08:49:55.310" Score="2" Body="&lt;p&gt;Crystallization in the chemical sense, as previously noted by Nicolau Saker Neto in comment form, is when a solid forms &quot;with a regular atomic lattice&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds such as salts (e.g. NaCl, KCl) would be my first choice, but there might be others I don't know about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest method I know to do this is by supersaturating.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fill a container (such as a cup or a beaker) with water or other suitable solvent. &lt;/li&gt;&#xA;&lt;li&gt;Add salt (or other solute) while stirring until the solution becomes saturated and there is a noticeable amount of undissolved solute at the bottom of the container. &lt;/li&gt;&#xA;&lt;li&gt;Heat the solution. This increases the solubility of the solute because the particles move farther apart when at a warmer temperature. This is due to the fact that heat and particle motion are linked. This allows a greater amount of solute to dissolve.&lt;/li&gt;&#xA;&lt;li&gt;Keep stirring to make sure the solute that was previously collected at the bottom of the container is fully dissolved&lt;/li&gt;&#xA;&lt;li&gt;Allow the solution to cool. This reduces the solubility and the solute will crystallize.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Salt is an easy example. Add more salt to water until it can't dissolve anymore, heat it, wait for it to cool, and you will have your salt crystal.&lt;/p&gt;&#xA;" OwnerUserId="4400" LastEditorUserId="4400" LastEditDate="2015-05-21T08:59:14.897" LastActivityDate="2015-05-21T08:59:14.897" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31785" PostTypeId="1" CreationDate="2015-05-21T09:55:25.323" Score="8" ViewCount="268" Body="&lt;p&gt;So, I have a &lt;em&gt;big&lt;/em&gt; organic molecule consisting of $\ce{H}$, $\ce{C}$, $\ce{O}$, and $\ce{N}$ atoms and the goal is to find its equilibrium geometry and US/Vis spectrum. The molecule is so big that I was thinking about reducing the complexity of the calculations by using effective core potentials (say, Stuttgart RLC) on &lt;em&gt;light&lt;/em&gt; atoms ($\ce{C}$, $\ce{O}$, $\ce{N}$). However, I'm not sure how trustworthy the results would be comparing to using full-electron bases, thus, I'm looking for some reviews if there are any.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorDisplayName="user15489" LastEditDate="2015-05-21T10:08:59.350" LastActivityDate="2017-04-09T05:00:57.123" Title="Is it wise to use ECPs on light atoms for electronic structure calculations?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;reference-request&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31786" PostTypeId="2" ParentId="31785" CreationDate="2015-05-21T11:08:59.400" Score="2" Body="&lt;p&gt;I think you will find that the time savings are minimal. Modern integral screening will eliminate virtually all inter-atom core-core integrals anyway and if you only have a 1s core then the cost of evaluating intra-atom core-valence integrals will be comparable to evaluating the ECP-core interaction.  &lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-21T11:08:59.400" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31787" PostTypeId="2" ParentId="31783" CreationDate="2015-05-21T11:21:59.480" Score="14" Body="&lt;p&gt;In your example, the stereodescriptors &lt;em&gt;R&lt;/em&gt; and &lt;em&gt;S&lt;/em&gt; are already given. The remaining question is about the double bond which gives rise to &lt;em&gt;cis-trans&lt;/em&gt; isomerism.&lt;/p&gt;&#xA;&lt;p&gt;In the hierarchical Sequence Rules, we finally find&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-92.1.3.5&lt;/strong&gt; Sequence Rule &lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;An atom or group with descriptor ‘&lt;em&gt;R&lt;/em&gt;’, ‘&lt;em&gt;M&lt;/em&gt;’, and ‘&lt;em&gt;seqCis&lt;/em&gt;’ has priority over its enantiomorph ‘&lt;em&gt;S&lt;/em&gt;’, ‘&lt;em&gt;P&lt;/em&gt;’ or ‘&lt;em&gt;seq Trans&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source: &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the configuration would be ‘&lt;em&gt;E&lt;/em&gt;’ since ‘&lt;em&gt;R&lt;/em&gt;’ precedes ‘&lt;em&gt;S&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;p&gt;However, the ‘&lt;em&gt;E&lt;/em&gt;’ and ‘&lt;em&gt;Z&lt;/em&gt;’ stereodescriptors do not distinguish between double bonds whose descriptors are reflection invariant from the geometrically enantiomorphic double bonds whose stereodescriptors are reflection variant.&lt;/p&gt;&#xA;&lt;p&gt;Following the recommendations for enantiomorphic double bonds, the configuration is ‘&lt;em&gt;seqTrans&lt;/em&gt;’.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-21T11:21:59.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31788" PostTypeId="2" ParentId="31783" CreationDate="2015-05-21T11:26:17.267" Score="3" Body="&lt;p&gt;The authoritative article for CIP rules is &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/ange.19660780803/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spezifikation der molekularen Chiralität&lt;/a&gt;. According to the article, the double bond has the E configuration. The reason for this is that R has priority over S.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="194" LastEditDate="2018-04-05T12:53:04.093" LastActivityDate="2018-04-05T12:53:04.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31789" PostTypeId="2" ParentId="31783" CreationDate="2015-05-21T11:37:10.223" Score="9" Body="&lt;p&gt;I remember that when all the basic rules end up in a tie (including the ghost atoms), the following priority is used:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$R &amp;gt; S$ &lt;/li&gt;&#xA;&lt;li&gt;$cis &amp;gt; trans$  &lt;/li&gt;&#xA;&lt;li&gt;$Z &amp;gt; E $ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So R is given higher priority and the above mentioned molecule will be considered as $E$ and not $Z$.  &lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-05-21T11:37:10.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31790" PostTypeId="2" ParentId="31785" CreationDate="2015-05-21T11:39:16.410" Score="1" Body="&lt;p&gt;The ECP scheme will not bring substantial savings. If the molecule is really large (1000 atoms), you should better think of more efficient approximations, namely RI methods, RI-J, RI-JK or RIJCOSX. You can find their efficient implementation in &lt;a href=&quot;https://orcaforum.cec.mpg.de/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ORCA package&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I would suggest using some more recent basis sets, namely the Alrichs def2.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="11367" LastEditDate="2017-04-09T05:00:57.123" LastActivityDate="2017-04-09T05:00:57.123" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31791" PostTypeId="1" AcceptedAnswerId="31797" CreationDate="2015-05-21T11:54:54.550" Score="5" ViewCount="217" Body="&lt;p&gt;Is element 111 considered to act as an eka-aurum?&#xA;Being under the same column of group 11, which usually provides enough evidence for correlation of properties for an eka-element, would this element have the similar properties such as conductivity or inertness with gold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the understanding of the correlation between group 11 being minimal to me, is anyone able to predict perhaps the colour, ductility, or other physical properties of this element with reason?&lt;/p&gt;&#xA;" OwnerUserId="16210" LastActivityDate="2015-05-21T14:19:30.523" Title="Behaviour of element 111" Tags="&lt;theoretical-chemistry&gt;&lt;elements&gt;&lt;conductivity&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31792" PostTypeId="1" CreationDate="2015-05-21T12:35:10.853" Score="0" ViewCount="1424" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Consider the reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{CO(g) + Cl2(g) &amp;lt;=&amp;gt; COCl2(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A closed &lt;span class=&quot;math-container&quot;&gt;$\pu{5.69 L}$&lt;/span&gt; flask containing a reaction mixture at equilibrium has &lt;span class=&quot;math-container&quot;&gt;$[\ce{CO}] = \pu{1.8 \times 10^-6 M}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{Cl2}] = \pu{6.3 \times 10^-7 M}$&lt;/span&gt;. Given that the equilibrium constant &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{eq} = 2.9 \times 10^{10}$&lt;/span&gt;, calculate the mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{COCl2}$&lt;/span&gt; present in the flask.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;How can I solve the question above? I thought that equilibrium constants deal with concentrations and not masses, so I'm not sure how to use it to calculate a mass.&lt;/p&gt;&#xA;" OwnerUserId="16404" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-08-27T13:11:46.880" Title="How to calculate the mass of phosgene in the equilibrium mixture of carbon dioxide and chlorine?" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31794" PostTypeId="1" CreationDate="2015-05-21T13:15:29.670" Score="0" ViewCount="2540" Body="&lt;p&gt;Can calcium chloride be effectively used as a desiccant to absorb both heat and moisture?&lt;/p&gt;&#xA;" OwnerUserId="16405" LastEditorUserId="14" LastEditDate="2015-05-21T13:18:37.400" LastActivityDate="2015-05-21T16:45:18.067" Title="Heat and water absorption by calcium chloride" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31795" PostTypeId="2" ParentId="31105" CreationDate="2015-05-21T13:44:03.433" Score="2" Body="&lt;p&gt;I believe you can compare many of the models with a search engine.&#xA;&lt;a href=&quot;http://www1.asminternational.org/asmenterprise/apd/AdvancedSearchAPD.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www1.asminternational.org/asmenterprise/apd/AdvancedSearchAPD.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16210" LastActivityDate="2015-05-21T13:44:03.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31796" PostTypeId="1" CreationDate="2015-05-21T14:00:35.097" Score="6" ViewCount="458" Body="&lt;p&gt;The photoelectron spectrum of krypton is shown below.&lt;img src=&quot;https://i.stack.imgur.com/7KCra.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where the x-axis is in electron volts (eV))\&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both the $3p$ and $3d$ shells are more than half full, based on Hund's rules the lowest-energy level is the one with the largest J, i.e. the $2\mathrm{P}_{3/2}$ and  $2\mathrm{D}_{5/2}$ states should be lowest in energy (higher value in eV). However, based on the peak intensities, this is not the ordering observed (see labeling in diagram). Why is this the case here?&lt;/p&gt;&#xA;" OwnerUserId="9804" LastEditorUserId="402" LastEditDate="2016-11-05T17:37:21.237" LastActivityDate="2016-11-05T17:37:21.237" Title="Explaining the photoelecton spectrum of krypton" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31797" PostTypeId="2" ParentId="31791" CreationDate="2015-05-21T14:07:16.200" Score="4" Body="&lt;p&gt;As far as I understand, there has been very limited experimental investigation of the properties of roentgenium (Z=111), if at all, simply due to the fact that so few atoms have been produced and they decay so quickly. It is possible to do some chemistry with even single atoms in seconds, but as you might expect it's an extremely arduous endeavour and is riddled with massive error bars. For now, just the synthesis of the element seems to dominate experimental roentgenium research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has, however, been ample theoretical investigation of roentgenium's chemical properties, both by extrapolation of behaviour for group 11 elements and by &lt;em&gt;ab initio&lt;/em&gt; computational methods, at varying levels of approximation. One difficulty is that roentgenium displays strong relativistic effects, which make accurate calculations harder. Of course, atomic properties are easier to predict than molecular properties, which are themselves much easier to predict than condensed phase properties. So while estimates for ionization energy for atoms may be relatively accurate, physical properties of the pure materials such as ductility or melting point are to some extent little more than informed guesses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a list of some predicted basic properties of the period 7 $d$-block elements, including roentgenium (&lt;a href=&quot;https://books.google.com.books?id=9vPuV3A0UGUC&amp;amp;lpg=PA14&amp;amp;ots=W-9vK_LMHN&amp;amp;dq=the+chemistry+of+the+actinide+and+transactinide+elements&amp;amp;pg=PA1691&amp;amp;redir_esc=y#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;source, p. 1691&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P5BPd.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That source expands upon some of the general properties of roentgenium. Its electronic configuration is predicted to be $\mathrm{[Rn](5f)^{14} (6d)^9 (7s)^2}$, which is dissimilar to the configuration for gold $\mathrm{[Xe](4f)^{14} (5d)^{10} (6s)^1}$. This is because relativistic effects cause a strong stabilization of $s$ orbitals (now more accurately referred to as $s_{1/2}$ due to spin-orbit coupling) by pulling them closer to the nucleus, while simultaneously destabilising the $d$ orbitals (as well as splitting the $d$ subshell into two groups of degenerate orbitals, two $d_{3/2}$ orbitals and three $d_{5/2}$ orbitals) by pushing them away. This means complete population of the $s$ subshell becomes preferable as relativistic effects get stronger. For example, &lt;a href=&quot;https://i.stack.imgur.com/abSVf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;see the effect here&lt;/a&gt; (&lt;em&gt;ibid&lt;/em&gt;, p. 1667) in the triad Nb/Ta/Db in group 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roentgenium is expected to be a noble metal, like gold, and in fact from its predicted standard reduction potential, is &lt;em&gt;more&lt;/em&gt; noble than gold ($\mathrm{Au^{3+}(aq) + 3\ e^{-} \longrightarrow Au^0(s), \ \ \Delta E^0=+1.52\ V}$, for comparison). That said, once the roentgenium atom is ionized to Rg(III), it can reach higher oxidation states more easily due to less stable filled $6d_{5/2}$ orbitals, and so Rg(V) compounds are expected to be more stable. Interestingly, though in the 6th period transition metals gold (group 11) shows the &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev-physchem-032511-143755&quot; rel=&quot;nofollow noreferrer&quot;&gt;local &quot;relativistic maximum&quot;&lt;/a&gt; with respect to stabilization of the $6s$ subshell, in the 7th period the maximum stabilization of the $7s$ subshell shift to copernicium (group 12) rather than roentgenium. This means roentgenium is expected to be less noble than copernicium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to physical properties, not much is predicted about roentgenium. It will likely be a &lt;em&gt;very&lt;/em&gt; dense metal, with good electrical conductivity. The question about its colour is an interesting one. &lt;a href=&quot;http://en.wikipedia.org/wiki/Roentgenium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; mentions it is expected to be silvery, even though roentgenium displays stronger relativistic effects, which is likely the cause of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16633/&quot;&gt;the colour of gold&lt;/a&gt;. This may be related to the switch in electronic ground state between gold and roentgenium. Unfortunately the source for the claim on Wikipedia is no longer accessible.  &lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-05-21T14:19:30.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31799" PostTypeId="2" ParentId="31792" CreationDate="2015-05-21T15:36:28.843" Score="4" Body="&lt;p&gt;We have an equilibrium mixture for the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO (g) + Cl2 (g) &amp;lt;=&amp;gt; COCl2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We should recognize that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K_\mathrm{eq}$ equation out in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we are given concentrations of reactants in the &lt;em&gt;equilibrium mixture&lt;/em&gt;, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;2.9 \times 10^{10} &amp;amp;= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\&#xA;[\ce{COCl2}] &amp;amp;= \pu{0.033 M}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are asked how many &lt;em&gt;grams&lt;/em&gt; of $[\ce{COCl2}]$ is present in the reaction mixture.  Concentration and mass are related via the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us plug in all of our known values and solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;m(\ce{COCl2}) &amp;amp;= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\&#xA;&amp;amp;= \pu{18.6 g}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="4945" LastEditDate="2018-08-27T13:11:34.097" LastActivityDate="2018-08-27T13:11:34.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31800" PostTypeId="2" ParentId="31006" CreationDate="2015-05-21T15:52:08.663" Score="5" Body="&lt;p&gt;The acid-catalyzed dehydration of alcohols involves a carbocation intermediate, which can act as a Brønsted acid; the loss of a proton yields the desired cyclohexene. Furthermore, the carbocation may undergo other side reactions. In particular, the carbocation may be subject to a reaction with other molecules to give substitution products; or the carbocation may undergo carbocation rearrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coloured residue of the distillation is probably the result of &lt;strong&gt;cationic polymerization of cyclohexene&lt;/strong&gt;. This reaction in combination with rearrangements may yield various oligomers.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-05-21T19:07:25.230" LastActivityDate="2015-05-21T19:07:25.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31801" PostTypeId="2" ParentId="31772" CreationDate="2015-05-21T16:19:00.500" Score="4" Body="&lt;p&gt;Maybe to focus more on the question: what kinds of things can you make crystals from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically a pure element or an inorganic or organic compound can be made in crystalline form. It might be a fine powder or small granular crystals. If grown under controlled conditions, you could get large crystals. If it is a gas or liquid at room temperature, it could perhaps solidify to a crystalline form at lower temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, an impure, indeterminate, or polymeric material, or a mixture of compounds, very seldom will form crystals. Still, there are exceptions. Inorganic salts often crystallize from water solution as &quot;hydrates&quot; - with extra water molecules in the crystal lattice of the salt. Other solvents also sometimes co-crystallize with inorganic or organic substances. Neutral compounds can co-crystallize together as clathrates, such as methane hydrate - a crystalline structure containing both methane and water that forms at high pressures and low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a complex or high molecular weight substance, such as an enzyme or virus particle can form crystals. In the realm of nanotechnology, new materials are being engineered specifically to aggregate into crystal-type structures.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-05-21T16:19:00.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31803" PostTypeId="2" ParentId="31794" CreationDate="2015-05-21T16:45:18.067" Score="1" Body="&lt;p&gt;Anhydrous calcium chloride is an effective desiccant, but the absorption of water by anhydrous calcium chloride is exothermic; i.e. the reaction releases heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCl2 + 6 H2O -&amp;gt; CaCl2 . 6 H2O}\quad\Delta H = -97\ \mathrm{kJ/mol}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-21T16:45:18.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31804" PostTypeId="1" AcceptedAnswerId="31808" CreationDate="2015-05-21T18:19:20.587" Score="15" ViewCount="9536" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/COGS1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the major product of this elimination reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exocyclic product has more substituent groups on the alkene which should make it more stable but I have heard that exocyclic alkenes are less stable than their endocyclic counterparts. Why is this the case and is it a strong enough effect to outweigh the stability provided by a fourth alkyl substituent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, what influence will kinetics have on this reaction? I think that the endocyclic product should be kinetically favoured because there is less steric hindrance for deprotonation of the intermediate cation at the ring position than at the isopropyl position.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-05-22T10:30:28.907" LastActivityDate="2015-05-22T10:30:28.907" Title="Exocyclic vs. endocyclic double bonds in E1 elimination" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31806" PostTypeId="2" ParentId="31804" CreationDate="2015-05-21T20:48:03.843" Score="0" Body="&lt;p&gt;Please correct me if I'm mistaken, but in this case we would dehydrate to give the Zaitzev product (second listed product in the OP) as the major product. Consider also that adding heat as a reagent is common place to aid in dehydration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give a quick glance here for a refresher on alkene stability: &lt;a href=&quot;https://en.wikipedia.org/wiki/Zaitsev%27s_rule&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Zaitsev%27s_rule&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16412" LastActivityDate="2015-05-21T20:48:03.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31808" PostTypeId="2" ParentId="31804" CreationDate="2015-05-21T21:45:08.687" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What will be the major product of this elimination reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the acid-catalyzed dehydration of 1-methylcyclohexanol, both 1-methylcyclohexene and methylenecyclohexane are formed in &lt;em&gt;ca&lt;/em&gt;. an 85:15 ratio.  In your example we have two more substituents on the exocyclic double bond which will serve to further stabilize the exocyclic double bond and increase the amount of the exocyclic isomer. I don't know for sure how much it will push the equilibrium towards the exocyclic isomer, but I might guess that the product mixture would be at least 50:50, if not more of the exocyclic isomer - but I would expect a mixture.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what influence will kinetics have on this reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well you won't see the kinetics unless you go out of your way and look for it.  Factors that influence which isomer is kinetically favored include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;statistical factors - there are 4 ring hydrogens that can be eliminated &lt;em&gt;vs&lt;/em&gt;. only 1 on the isopropyl group; this favors kinetic formation of the endocyclic isomer&lt;/li&gt;&#xA;&lt;li&gt;alignment of the hydrogen with the p-orbital on the carbocation carbon - the hydrogen on the isopropyl group can better align with the p-orbital than those hydrogens in the ring, however the cyclohexane ring can easily distort; so while this isn't a major factor it favors the exocyclic isomer&lt;/li&gt;&#xA;&lt;li&gt;steric factors - I agree with your analysis, but would guess that the difference is small; still sterics would slightly favor the endocyclic isomer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems pretty close to call, but if I had to guess I'd say that the statistical factor wins out (and the steric argument further helps) and the endocyclic isomer would likely be the the kinetic product.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, in acid this is a fast equilibrium; the products will quickly equilibrate and the thermodynamically controlled product distribution will be observed.  Maybe if you run the reaction at low temperature with a strong dehydrating agent ($\ce{POCl3}$) you could capture some meaningful kinetic information.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-21T21:45:08.687" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31809" PostTypeId="2" ParentId="29588" CreationDate="2015-05-21T22:22:34.193" Score="5" Body="&lt;h1&gt;Yes.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The battery bounce test &lt;em&gt;kind of&lt;/em&gt; works. You can tell whether the battery still has a charge greater than 90%, but bounce hight levels off at about 50% and doesn't go higher. This means that it is not clear whether the battery is completely discharged when you observe a bounce.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Why?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2015/ta/c5ta01576f#!divAbstract&quot; rel=&quot;noreferrer&quot;&gt;Steingart &lt;em&gt;et al.&lt;/em&gt;&lt;/a&gt; just recently published a paper on the subject, and through scientific rigor could confirm this claim. The reason they give has  to do with how an alkaline battery is built:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IY5wH.png&quot; alt=&quot;Schematic cut through an alkaline battery.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Alkaline-battery-english.svg&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, a large part of the inside is devoted to the zinc anode. During the discharge, a redox reaction takes place to produce the current:&#xA;$$ \ce{Zn + 2MnO2 -&amp;gt; ZnO + Mn2O3} $$&#xA;The zinc (usually in some gel form at first) is slowly oxidized. As a result of this oxidization, crosslinks between the small zinc granules are formed, which make the whole center more rigid. As such, its &quot;bounce&quot; (or &lt;a href=&quot;http://en.wikipedia.org/wiki/Coefficient_of_restitution&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;coefficient of restitution&lt;/em&gt;&lt;/a&gt;) becomes greater, because the downward kinetic energy cannot be fully absorbed by the gel anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in more details, check out the &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2015/ta/c5ta01576f#!divAbstract&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; they published. If you don't have access you can read a summary given on the &lt;a href=&quot;http://www.princeton.edu/main/news/archive/S42/72/95S25/index.xml?section=topstories&quot; rel=&quot;noreferrer&quot;&gt;Princeton University News&lt;/a&gt; web page, where you also can see an animation of a series of discharged batteries bouncing to different heights.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-05-21T22:22:34.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31810" PostTypeId="2" ParentId="30935" CreationDate="2015-05-21T22:46:36.917" Score="3" Body="&lt;p&gt;You can assume that the density remains unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen peroxide simply oxidizes dyes, thereby destroying chromophores (groups that move the absorption of a molecule into the visible spectrum). The viscose structure should remain largely unscathed, and the change in density is so minuscule I would argue it is within the experimental error.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-05-21T22:46:36.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31811" PostTypeId="2" ParentId="29631" CreationDate="2015-05-21T23:16:40.747" Score="5" Body="&lt;p&gt;The assignment of CIP priorities has some intricate details when it comes to multiple bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the general procedure (&lt;a href=&quot;http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;&lt;p&gt;If there is a tie, we must consider the atoms at distance 2 from the stereocenter — as a list is made for each group of the atoms bonded&#xA;  to the one directly attached to the stereocenter. Each list is&#xA;  arranged in order of decreasing atomic number. Then the lists are&#xA;  compared atom by atom; at the earliest difference, the group&#xA;  containing the atom of higher atomic number receives higher priority.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;If there is still a tie, each atom in each of the two lists is replaced with a sub-list of the other atoms bonded to it (at distance&#xA;  3 from the stereocenter), the sub-lists are arranged in decreasing&#xA;  order of atomic number, and the entire structure is again compared&#xA;  atom by atom. This process is repeated, each time with atoms one bond&#xA;  farther from the stereocenter, until the tie is broken.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, concerning double bonds in particular:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If an atom A is double-bonded to an atom B, A is treated as being&#xA;  singly bonded to two atoms: B and a ghost atom that has the same&#xA;  atomic number as B but is not attached to anything except A. In turn,&#xA;  when B is replaced with a list of attached atoms, A itself is excluded&#xA;  in accordance with the general principle of not doubling back along a&#xA;  bond that has just been followed, but a ghost atom for A is included&#xA;  so that the double bond is properly represented from both ends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have constructed a table of these atom lists for the vinyl, isopropyl and &lt;em&gt;sec&lt;/em&gt;-butyl rests:&#xA;$$\small&#xA;\begin{array}{llll}&#xA;\hline&#xA;\text{Dist} &amp;amp; \text{Vinyl} &amp;amp; \text{Isopropyl} &amp;amp; sec\text{-Butyl} \\&#xA;\hline&#xA;1 &amp;amp; \text{C} &amp;amp; \text{C} &amp;amp; \text{C} \\&#xA;2 &amp;amp; \text{C} &amp;amp; \text{C} &amp;amp; \text{C} \\&#xA;  &amp;amp; \text{(C)} &amp;amp; \text{C} &amp;amp; \text{C} \\&#xA;  &amp;amp; \text{H} &amp;amp; \text{H} &amp;amp; \text{H} \\&#xA;3 &amp;amp; \text{(C)} &amp;amp; \text{H} &amp;amp; \text{C} \\&#xA;  &amp;amp; \text{H} &amp;amp; \text{H} &amp;amp; \text{H} \\&#xA;  &amp;amp; \text{H} &amp;amp; \text{H} &amp;amp; \text{H} \\&#xA;  &amp;amp; &amp;amp; \text{H} &amp;amp; \text{H} \\&#xA;  &amp;amp; &amp;amp; \vdots &amp;amp; \vdots \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;sec&lt;/em&gt;-butyl rest has thus the highest priority of the three, followed by vinyl and then isopropyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ghost atom at distance three is the ghost carbon at distance one, as seen from the ghost carbon that was doubly bonded to it. If the distance 4 was to be considered, no more ghost atoms would appear, as the ghost atom at position 1 does not have any more ghost atoms bonded to it (atom 1 however is bonded to the ghost atom at position 2).&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorDisplayName="user7951" LastEditDate="2015-08-18T15:50:32.157" LastActivityDate="2015-08-18T15:50:32.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31813" PostTypeId="2" ParentId="31731" CreationDate="2015-05-21T23:30:40.000" Score="7" Body="&lt;p&gt;Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2p&lt;sub&gt;&lt;em&gt;x&lt;/em&gt;&lt;/sub&gt;, 2p&lt;sub&gt;&lt;em&gt;y&lt;/em&gt;&lt;/sub&gt; and 2p&lt;sub&gt;&lt;em&gt;z&lt;/em&gt;&lt;/sub&gt;. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, &lt;em&gt;but waves, that are spread out across the entire molecule.&lt;/em&gt; You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in &lt;a href=&quot;http://www.cup.uni-muenchen.de/ac/kluefers/homepage/L_ac1.html&quot;&gt;this German lecture script, part A, section 4.8&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-21T23:30:40.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31814" PostTypeId="2" ParentId="31757" CreationDate="2015-05-22T00:26:47.893" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;Name of the compund: &lt;strong&gt;Dimethyl nitroxide&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Lewis structure:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aP20Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The properly way to determine the Lewis structure, based on this example, is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Total valence electrons: $4\cdot2 + 5 + 6 + 1\cdot6 = 25$&lt;/li&gt;&#xA;&lt;li&gt;Total electrons needed for octets/doublets: $8\cdot4+2\cdot6 = 44$&lt;/li&gt;&#xA;&lt;li&gt;Total shared/bonding electrons: $44 - 25 = 19$ (In other words, there is only nine single bonds and one electron left.)&lt;/li&gt;&#xA;&lt;li&gt;Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 25 - 19 = 6$ (In other words, the are only 3 pairs of lone electrons.)&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Furhter information on how to draw Lewis structures, please follow the &lt;a href=&quot;https://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_11/node3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Further information of related Lewis structures and how this knowledge is applied on real top science, I recommend seeing this &lt;a href=&quot;http://www.beilstein-journals.org/bjoc/single/articleFullText.htm?publicId=1860-5397-9-181&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference:&lt;/a&gt; Moreira, G.; Charles, L.; Major, M.; Vacandio, F.; Guillaneuf, Y.; Lefay, C. and Gigmes, D. &lt;em&gt;Beilstein J. Org. Chem.,&lt;/em&gt; &lt;strong&gt;2013&lt;/strong&gt;, 9, 1589 - 1600. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-05-23T04:12:04.183" LastActivityDate="2015-05-23T04:12:04.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31815" PostTypeId="1" CreationDate="2015-05-22T00:29:33.777" Score="3" ViewCount="14637" Body="&lt;p&gt;I'm attempting to write the balanced equation of the combination of lithium metal and fluorine gas. I understand that this will produce an ionic solid because it is a reaction of a metal and a nonmetal. However, I'm confused with the fluoride ion having a charge of -1 and fluorine only occurring as a diatomic molecule ($\ce{F2}$). Is $\ce{F2}$ two fluoride ions or just one?&lt;/p&gt;&#xA;" OwnerUserId="16417" LastEditorUserId="15410" LastEditDate="2015-05-22T03:07:09.723" LastActivityDate="2015-05-22T03:24:09.190" Title="Is F2 two fluoride ions or just one?" Tags="&lt;stoichiometry&gt;&lt;ionic-compounds&gt;" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31816" PostTypeId="1" CreationDate="2015-05-22T01:48:45.940" Score="6" ViewCount="78" Body="&lt;p&gt;For school, my teacher asked what the proper procedure/reaction would be to add a polyvinyl alcohol chain onto a polyethylene chain. Can anyone help? I'll provide additional information if needed.&lt;/p&gt;&#xA;" OwnerUserId="16418" LastEditorUserId="11367" LastEditDate="2018-05-23T17:20:01.650" LastActivityDate="2018-05-23T17:20:01.650" Title="Adding a polyvinyl alcohol chain onto polyethylene" Tags="&lt;polymers&gt;&lt;alcohols&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-06-17T19:09:12.073" ContentLicense="CC BY-SA 3.0" />
  <row Id="31817" PostTypeId="1" AcceptedAnswerId="31825" CreationDate="2015-05-22T02:06:16.230" Score="10" ViewCount="2479" Body="&lt;p&gt;In one of my MCQ questions, there was a statement saying all group 2 elements form covalent hydroxides. I know it is incorrect. I know that $\ce{BeH2}$ is a covalent hydride. What about the rest? Are all of them metallic hydrides? Is it true that metallic hydrides produce the hydroxide and give off $\ce{H2}$ with water?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="5026" LastEditDate="2018-02-25T08:17:57.930" LastActivityDate="2018-02-25T08:17:57.930" Title="What is the bonding type in hydrides of group 2 elements?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31818" PostTypeId="2" ParentId="31815" CreationDate="2015-05-22T03:18:09.280" Score="5" Body="&lt;p&gt;The fluorine &lt;strong&gt;atom&lt;/strong&gt; is far to reactive to exist on its own, so naturally it bonds with other atoms in its vicinity to stabilize itself. It should be noted that an atom is electrically neutral, meaning that it has an equal number of protons as electrons. A fluorine &lt;strong&gt;atom&lt;/strong&gt; can share electrons with another nonmetal to form a covalent bond. When sharing electrons in a covalent bond, the fluorine is still called an atom, because it has not gained complete control over the electron that is completing its $2p$ orbital. In the $\ce{F2}$ molecule, there are to fluorine &lt;strong&gt;atoms&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluoride &lt;strong&gt;ion&lt;/strong&gt; is a result of a fluorine &lt;strong&gt;atom&lt;/strong&gt; completely gaining control over an electron, which is usually donated by a metal. Electrons can be transferred from a metal to the fluorine &lt;strong&gt;atoms&lt;/strong&gt; to form metal ions and fluoride &lt;strong&gt;ions&lt;/strong&gt;. The reaction you are talking about would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Li^{(0)} +F2^{(0)} -&amp;gt; 2Li^{(+1)}F^{(-1)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I included oxidation numbers to show how each species has gained/lost electrons, though this would normally not be included in the formula.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-05-22T03:18:09.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31819" PostTypeId="2" ParentId="31815" CreationDate="2015-05-22T03:18:20.453" Score="6" Body="&lt;p&gt;$\ce{F2}$ will produce  $2\space \ce{F-}$ ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can write the balanced half cell reactions as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{F2 + 2e- -&amp;gt;2F-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Li -&amp;gt; Li+ + e-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the second equation by 2 to get $\ce{2Li -&amp;gt; 2Li+ + 2e-}$. This is to ensure that the final equation does not have any electrons on either side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add this equation with the fluorine one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll get $\ce{2Li + F2 -&amp;gt; 2Li+ + 2F-}$ which you can write as $\ce{2Li + F2 -&amp;gt; 2LiF}$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-05-22T03:18:20.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31820" PostTypeId="2" ParentId="31815" CreationDate="2015-05-22T03:19:41.300" Score="5" Body="&lt;p&gt;Fluorine gas in the diatomic $\ce{F2}$ is essentially the the covalent pair of fluorine atoms (as fluorine is a non-metal), as seen in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5SqCG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cliffsnotes.com/sciences/chemistry/chemistry/chemical-bonding/covalent-bonds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the reaction you describe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2Li(s) + F2(g) -&amp;gt;2LiF(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;requires the fluoride ion be in an ionic bond with the lithium ion, twice (to balance the equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MqMRL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.meta-synthesis.com/webbook/30_timeline/timeline.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-22T03:24:09.190" LastActivityDate="2015-05-22T03:24:09.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31821" PostTypeId="2" ParentId="31815" CreationDate="2015-05-22T03:21:37.990" Score="4" Body="&lt;p&gt;Fluorine in its elemental form is $\ce{F2}$, a neutral molecule. Every fluorine has seven valence electrons and after formation of the molecule, there is one covalent, i.e. electron sharing bond, and the remaining six valence electrons at each fluorine can be regarded as lone pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When elemental fluorine reacts with a metal it will form an ionic compound just as you stated. For this to happen, both elements involved in the reaction have to be ionised, i.e. they have to gain or lose electrons. This is called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Redox&quot; rel=&quot;nofollow&quot;&gt;redox reaction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, lithium will transfer one of the electrons to fluorine, resulting in the following equations:&#xA;\begin{align}&#xA;\ce{Li &amp;amp;~&amp;lt;=&amp;gt; Li+ + e-}&amp;amp;&amp;amp;|\cdot 2\\&#xA;\ce{F2 + 2e- &amp;amp;~&amp;lt;=&amp;gt; 2F- }\\\hline&#xA;\ce{2Li + F2 &amp;amp;~&amp;lt;=&amp;gt; 2LiF}\\&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-22T03:21:37.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31822" PostTypeId="2" ParentId="28526" CreationDate="2015-05-22T03:53:35.407" Score="2" Body="&lt;p&gt;Time and entropy has not much to do with each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From practical point of view: Equilibrium systems can hang around for infinite time, and their entropy does not change at all - so time would be an impractical measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From conceptual point of view: Entropy comes from statistical behavior of the system of many bodies. Time is unrelated to statistics; it exist even in single body systems.&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-05-22T03:53:35.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31824" PostTypeId="1" AcceptedAnswerId="31828" CreationDate="2015-05-22T05:54:13.760" Score="5" ViewCount="768" Body="&lt;p&gt;In Vogel's organic book, they have mentioned that in the Lund and Bjerrum method of purification of ethanol, magnesium is added, which forms ethanolate. But they said to activate the magnesium first by iodine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the activation needed and how exactly does this activation process work in the activation of magnesium?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="4945" LastEditDate="2015-05-22T08:26:15.127" LastActivityDate="2015-05-22T08:26:15.127" Title="Purification of Ethanol" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;alcohols&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31825" PostTypeId="2" ParentId="31817" CreationDate="2015-05-22T06:24:50.260" Score="5" Body="&lt;p&gt;There is a quite extensive chapter about hydrogen and its compounds in &lt;a href=&quot;https://books.google.co.jp/books?id=Mtth5g59dEIC&amp;amp;dq=holleman%20wiberg&quot; rel=&quot;noreferrer&quot;&gt;Holleman and Wibereg's book &quot;Inorganic Chemistry&quot;&lt;/a&gt;. As I only have the german version, I will briefly summarise what I read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are of course right, hydrogen forms salt-like structures with alkaline and earth alkaline elements, except for beryllium. Compounds like $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{RbH}$ and $\ce{CsH}$ crystallise in a sodium chloride structure. Every cation is coordinated by six hydride ions and vice versa. The compounds $\ce{CaH2}$, $\ce{SrH2}$ and $\ce{BaH2}$ crystallise in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluorite&quot; rel=&quot;noreferrer&quot;&gt;fluorite&lt;/a&gt; structure. The hydride ions are tetrahedrally coordinated by four earth alkaline cations. The cations are cubically coordinated by eight hydride ions. This &amp;beta;-structure is persistent at higher temperatures. Below $780~^\circ\mathrm{C}$ ($\ce{CaH2}$), $855~^\circ\mathrm{C}$ ($\ce{SrH2}$), and $598~^\circ\mathrm{C}$ ($\ce{BaH2}$) it changes to the &amp;alpha;-form, i.e. the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lead%28II%29_chloride&quot; rel=&quot;noreferrer&quot;&gt;lead dichloride structure&lt;/a&gt;. The magnesium compound exists in a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rutile&quot; rel=&quot;noreferrer&quot;&gt;rutile&lt;/a&gt; structure. Hydride ions are trigonally coordinated by magnesium ions. The cations are octahedrally coordinated by hydride ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the difference in the electronegativities, you would expect a predominant covalent character. The maximum is $1.3$ for $\ce{CaH2}$ and the minimum $1.0$ for $\ce{MgH2}$, which translates to a maximum $30\%$ and minimum $18\%$ ionic contribution. Maybe this is where the statement of the question originates from. Experimental findings rather support the more ionic nature of the compounds though, i.e. the found lattice energies were supported by calculations assuming a purely ionic structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion of this according to Holleman and Wiberg is that the elements are fused by ionic bonds with covalent contributions. The latter increases from $\ce{CsH}$ to $\ce{LiH}$ and from $\ce{BaH2}$ to $\ce{MgH2}$. The magnesium hydride is already borderline covalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question about their reactivity towards water is mainly answered in the chapters for the elements itself. Magnesium reacts readily with water and releases hydrogen. Because of this it is sometimes used as a drying agent. $$\ce{MgH2 + 2H2O -&amp;gt; Mg(OH)2 + H2 ^ }$$&#xA;The other elements react similar. The reaction is highly exothermic, the reaction of calcium hydride releases $228~\mathrm{kJ/mol}$. It is often used as a drying agent for gasses because of this feature. Strontium and Barium also react exothermic, i.e. they release $177~\mathrm{kJ/mol}$ and $172~\mathrm{kJ/mol}$, respectively.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-22T06:24:50.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31827" PostTypeId="2" ParentId="31768" CreationDate="2015-05-22T07:28:58.887" Score="6" Body="&lt;p&gt;&lt;em&gt;Let me try to talk you out of it ;-)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precious metals in a catalytic converter sit in a honeycomb ceramic that contains a lot of synthetic Cordierite. This material isn't only heat-resistant, but also quite hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial recycling of catalytic converters typically starts with&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;dismantling (decanning) the ceramic monolith using guillotine shears&lt;/li&gt;&#xA;&lt;li&gt;breaking down the ceramic to smaller pieces in hammer mills connected by a closed tubing system&lt;/li&gt;&#xA;&lt;li&gt;milling the material to dust in a ball-mill blender&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You might want to wear a filter mask during these steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial recycling almost exclusively uses &lt;strong&gt;pyrometallurgical&lt;/strong&gt; tretment (melting the metals) then. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having ruled that out, there are two other options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;hydrometallurgical&lt;/strong&gt; treatment&lt;br&gt;&#xA;In order to separate the metals from the ceramic, you can either try to  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dissolve the ceramic in hot alkaline solution in a closed reactor under pressure (to reach temperatures above 100 °C). This leaves the metals intact.  &lt;/li&gt;&#xA;&lt;li&gt;leach the metals out by using strong oxidizing acids, such as mixtures of concentrated nitric and hydrochlorid acid. You have the metals in solution now.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are your plans for personal protection during these steps and for the waste water management?&lt;/strong&gt;   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;biomobilization&lt;/strong&gt;&lt;br&gt;&#xA;There is some ongoing research to replace the unpleasant conditions of hydrothermal treatment with microorganisms that can leach the solid metals and store them. As an example, have a look at an article by Helmut Brandl, Stefan Lehmann, Mohammad A. Faramarzi, and Daniel Martinelli, published in  &lt;em&gt;Hydrometallurgy&lt;/em&gt;, &lt;strong&gt;2008&lt;/strong&gt;, &lt;em&gt;94&lt;/em&gt;, 14-17. The authors examined, whether metals could be converted to cyanido complexes, such as $\ce{[Pt(CN)4]^{2-}}$, by cyanide-producing microorganisms. A &lt;a href=&quot;https://www.zora.uzh.ch/10957/1/Brandl_etal08_V.pdf&quot;&gt;PDF&lt;/a&gt; of their article, titled &lt;em&gt;Biomobilization of silver, gold, and platinum from solid waste materials by HCN-forming microorganisms&lt;/em&gt; is available at no cost from &lt;em&gt;ZORA&lt;/em&gt; (Zurich Open Repository and Archive). In the case of platinum, however, the results were rather disappointing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-22T07:28:58.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31828" PostTypeId="2" ParentId="31824" CreationDate="2015-05-22T07:47:54.297" Score="7" Body="&lt;p&gt;The refluxing of methanol or ethanol over magnesium is indeed the method of choice to dry these solvents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the iodine is concerned, this is a classical trick, which is also used to initiate Grignard rections. Is is assumed that magnesium iodide is formed, which is soluble in the alkanols. As a result, a fresh, activated metal surface is left behind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might even work, when the magnesium metal is covered with a &lt;strong&gt;very thin&lt;/strong&gt; oxide layer.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-22T07:47:54.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31829" PostTypeId="2" ParentId="17109" CreationDate="2015-05-22T07:52:41.720" Score="13" Body="&lt;p&gt;Being a chemist, I know that paraffin would require a lipophilic solvent as pointed out by others. It occurred to me that ordinary vegetable oil is a cheap and of course innocuous lipophilic substance. I spilled about 2 oz molten candle wax on a marble surface and after it solidified removed a good portion by scraping with a plastic spoon. This was followed by adding about a tablespoon of vegetable oil and mixing with the spoon. The wax dissolved in the oil with no problem and could be mopped up with a paper towel. Treatment of the oily residue with sink soap removed what was left. Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="16424" LastActivityDate="2015-05-22T07:52:41.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31830" PostTypeId="1" CreationDate="2015-05-22T09:54:03.680" Score="8" ViewCount="41191" Body="&lt;p&gt;I completed a titration of Ammonium Acetate buffer solution and to it I added &lt;span class=&quot;math-container&quot;&gt;$\pu{2M}$&lt;/span&gt; Hydrochloric Acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measured the initial &lt;span class=&quot;math-container&quot;&gt;$\ce{pH}$&lt;/span&gt; of the buffer solution before any acid was added and I used methyl orange as an indicator which has a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; value of 3. &#xA;My independent variable was the temperature of the buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering how I would calculate the buffer capacity. Is it &lt;span class=&quot;math-container&quot;&gt;$\Delta\ce{pH}$&lt;/span&gt;/Volume?&lt;/p&gt;&#xA;" OwnerUserId="16291" LastEditorUserId="61578" LastEditDate="2019-03-30T21:51:37.923" LastActivityDate="2019-03-30T21:51:37.923" Title="Calculating Buffer Capacity" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31831" PostTypeId="2" ParentId="31830" CreationDate="2015-05-22T10:22:37.343" Score="6" Body="&lt;p&gt;As far as I can remember from my tutorials, the formula was&#xA;$$\beta =\frac{n(\ce{H+})}{\Delta\ce{pH}},$$&#xA;meaning: How many protons have to be added to produce a change in the pH of one unit. But there might be different definitions. The IUPAC does not provide an official one.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-22T10:22:37.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31832" PostTypeId="2" ParentId="31830" CreationDate="2015-05-22T10:27:57.110" Score="8" Body="&lt;p&gt;In addition to Martin's answer, there is at least an old recommendation &#xA;by E. B. Sandell and T. S. West in &lt;em&gt;Pure Appl. Chem.&lt;/em&gt;, &lt;strong&gt;1969&lt;/strong&gt;, &lt;em&gt;18&lt;/em&gt;, 427-436 (&lt;a href=&quot;http://dx.doi.org/10.1351/pac196918030427&quot;&gt;DOI&lt;/a&gt;), which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Buffer capacity&lt;/strong&gt; or &lt;strong&gt;buffer index&lt;/strong&gt;. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or strong base required to change the pH by one unit when added to one litre of the specified buffer solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-22T10:27:57.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31836" PostTypeId="1" AcceptedAnswerId="31838" CreationDate="2015-05-22T13:33:33.863" Score="8" ViewCount="1140" Body="&lt;p&gt;In one of our slides, the professor has written &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ \Delta E_{sys} = -\Delta e_{sur}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $E_{sys}$ symbolizes the energy of a given system, and $e_{sur}$ symbolizes the energy of the surroundings of that system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a few occasions, she's used both the upper, and the lower, case literal of a single letter as symbols of a single concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a notational convention that requires the use of an upper case literal of a letter in some circumstances and the use of its counterpart in other circumstances?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-05-22T16:10:52.870" Title="Why does my professor use both the upper, and the lower, case literal of a letter to symbolize the same concept?" Tags="&lt;notation&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31838" PostTypeId="2" ParentId="31836" CreationDate="2015-05-22T13:57:24.007" Score="9" Body="&lt;p&gt;Of course, there is a convention. In physics $E$ usually stands for energy, $e$ for elementary charge. You professor is inconsistent, to say the least. In my opinion using lower case $e$ for anything rather than for elementary charge is wrong by itself, and using it for energy is simply ridiculous. Besides, I don't see any reason for doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. During my university years I had a professor who randomly picked letters for physical quantities every lecture. :D Today I think of it with a smile, but at that time I hated both his lectures and him personally.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-22T14:02:48.093" LastActivityDate="2015-05-22T14:02:48.093" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31839" PostTypeId="1" AcceptedAnswerId="31858" CreationDate="2015-05-22T14:15:02.183" Score="3" ViewCount="1852" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Problem: 70 grams of pure, hydrated &lt;span class=&quot;math-container&quot;&gt;$\ce{MgSO_4}$&lt;/span&gt; is dissolved in 90&#xA;  grams of &lt;span class=&quot;math-container&quot;&gt;$\ce{H_2O}$&lt;/span&gt;. If the mole fraction of the solute is 0.039,&#xA;  find the molecular formula of the solute.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My flawed attempt at a solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the solute be &lt;span class=&quot;math-container&quot;&gt;$\ce{MgSO_4.xH_2O}$&lt;/span&gt;. In &lt;span class=&quot;math-container&quot;&gt;$70~\mathrm{g}$&lt;/span&gt; of solute, the amount of solute is &lt;span class=&quot;math-container&quot;&gt;$$\frac{70}{120+18x}~\mathrm{mol}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, in &lt;span class=&quot;math-container&quot;&gt;$\pu{90 g}$&lt;/span&gt; of solvent, we have &lt;span class=&quot;math-container&quot;&gt;$$\frac{90}{18} = 5~\mathrm{mol}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging these values into the formula for mole fraction and equating it to 0.039, I thought I would get the value of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; as 7. Instead, I seem to be getting approximately 12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody please help me solve this question? In general, when we have a hydrated solute in an aqueous solution, do we include the water of crystallisation in the total mass of available water while calculating the number of moles of water? &#xA;" OwnerUserId="16429" LastEditorUserId="16683" LastEditDate="2020-04-22T17:03:34.097" LastActivityDate="2020-04-22T17:03:34.097" Title="Solute mole fraction for hydrated magnesium sulfate dissolved in water" Tags="&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2020-04-23T10:15:47.720" ContentLicense="CC BY-SA 4.0" />
  <row Id="31840" PostTypeId="1" AcceptedAnswerId="31866" CreationDate="2015-05-22T14:20:47.250" Score="5" ViewCount="113" Body="&lt;p&gt;According to Wikipedia,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... the definition of the gram is not (as of 2011) mathematically tied to that of the atomic mass unit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why did we change that? &lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-05-23T04:29:06.093" Title="Why is the gram no longer a function of the atomic mass number?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31841" PostTypeId="1" AcceptedAnswerId="31847" CreationDate="2015-05-22T14:58:54.530" Score="1" ViewCount="77" Body="&lt;p&gt;In my notes, I've used phrases such as 'atom, particle, or molecule' more often than I'd like. For example, Wikipedia defines atomic mass as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the mass of an atomic particle, sub-atomic particle, or molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I remember concise definitions more easily, so a single word for 'molecules and molecular constituents' would be useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a term?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="10185" LastEditDate="2015-05-22T17:46:54.337" LastActivityDate="2015-05-22T17:46:54.337" Title="Is there a term that denotes 'molecules and molecular constituents'?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31843" PostTypeId="1" AcceptedAnswerId="31850" CreationDate="2015-05-22T15:21:32.690" Score="9" ViewCount="981" Body="&lt;p&gt;I am a second year undergraduate and have come across this when studying stereochemistry and conformations and have been unable to find a satisfactory answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\ce{sp^3}$ carbons prefer anti-periplanar and gauche conformations due to the overlap of filled orbitals with empty orbitals but cannot seem to find an answer for why $\ce{sp^2}$ carbons prefer an eclipsed conformation. I was possibly thinking about the need to avoid $\pi$ electrons however I am not sure. &lt;/p&gt;&#xA;" OwnerUserId="16427" LastEditorUserId="4231" LastEditDate="2015-05-22T17:09:16.893" LastActivityDate="2016-02-09T09:11:23.883" Title="Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31844" PostTypeId="1" CreationDate="2015-05-22T15:23:19.463" Score="5" ViewCount="15581" Body="&lt;p&gt;&lt;strong&gt;Reasoning&lt;/strong&gt;: My guess would have to be that it has to do with the strength of the intermolecular forces between each molecule. For example, molecular compounds that have a very low molar mass like &lt;strong&gt;$\ce{CO}$&lt;/strong&gt; and &lt;strong&gt;$\ce{CH4}$&lt;/strong&gt;. Both of theses molecules have a very small molar mass, a very low London dispersion force, and thus a very low boiling point. This in turns makes it easier to break the bonds that hold each individual atom and more than likely cause them to turn into a gas in normal conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when you look at ionic compounds like &lt;strong&gt;$\ce{MgCl2}$&lt;/strong&gt; and &lt;strong&gt;$\ce{NaCl}$&lt;/strong&gt;, it is more likely for these compounds to become solid because of the strong intermolecular force that they have (ionic bonds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct and is there something else that I should consider?&lt;/p&gt;&#xA;" OwnerUserId="16301" LastEditorUserId="402" LastEditDate="2017-03-31T19:38:00.940" LastActivityDate="2017-05-30T21:10:00.987" Title="How to determine if a compound is likely to become a solid at room temperature?" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31845" PostTypeId="1" AcceptedAnswerId="31882" CreationDate="2015-05-22T15:26:57.770" Score="8" ViewCount="4393" Body="&lt;p&gt;Based on WebElements, of the Noble Gases, &lt;a href=&quot;https://www.webelements.com/helium/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{He}$&lt;/a&gt; and &lt;a href=&quot;https://www.webelements.com/neon/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{Ne}$&lt;/a&gt; do not react with any of the halogens; however:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Argon_fluorohydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{HArF}$&lt;/a&gt; has been detected at low temperatures (thank you to @bon and @Martin in the comments)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.webelements.com/krypton/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{KrF2}$&lt;/a&gt; can be synthesised at low temperatures.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.webelements.com/xenon/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xenon&lt;/a&gt; can react with fluorine to produce $\ce{XeF2}$, $\ce{XeF4}$ and $\ce{XeF6}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.webelements.com/radon/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radon&lt;/a&gt; can react with fluorine to produce $\ce{RnF2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that fluorine is the only halogen that seems to react with the Noble Gases (partly due to the related question &lt;a href=&quot;https://chemistry.stackexchange.com/q/14547/15489&quot;&gt;Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?&lt;/a&gt;.  My question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are there more fluoride compounds formed with Xenon compared to the other Noble Gases?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2020-08-07T07:02:56.780" Title="Why are there more fluoride compounds formed with Xenon?" Tags="&lt;covalent-compounds&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31846" PostTypeId="1" AcceptedAnswerId="31849" CreationDate="2015-05-22T15:29:53.413" Score="11" ViewCount="2390" Body="&lt;p&gt;The mer-people demand cooked food. They don't want to go on land to make it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a chemical reaction they could use which would create enough heat to cook their food underwater in a safe manner?&lt;/p&gt;&#xA;" OwnerUserId="16432" LastEditorUserId="12860" LastEditDate="2015-10-30T18:28:36.310" LastActivityDate="2016-10-22T23:59:53.233" Title="Is there a chemical reaction which could give off enough heat underwater to cook something?" Tags="&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31847" PostTypeId="2" ParentId="31841" CreationDate="2015-05-22T15:58:19.043" Score="3" Body="&lt;p&gt;&lt;em&gt;This is pretty much link-only, but anyway...&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When searching for &lt;em&gt;official&lt;/em&gt; definitions, nomenclature, or recommended procedures, the organisation providing all these is the &lt;strong&gt;IUPAC&lt;/strong&gt; (International Union of Pure and Applied Chemistry)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An IUPAC book series, known as the &lt;strong&gt;Colour Books&lt;/strong&gt;, provides the normative definitions and guidelines on various topics. These are the results of what chemists in the different committees in the IUPAC divisions have eventually agreed upon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, all these &lt;strong&gt;Colour Books&lt;/strong&gt; are available online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;As kindly mentioned by &lt;em&gt;Loong&lt;/em&gt; in a comment, the online versions are not necessary the latest ones! The online version of the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/&quot; rel=&quot;nofollow&quot;&gt;IUPAC Nomenclature of Organic Chemistry&lt;/a&gt; is from 1993. In addition, a &lt;a href=&quot;http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/CompleteDraft.pdf&quot; rel=&quot;nofollow&quot;&gt;2004 draft copy&lt;/a&gt; is legally available.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Organic chemists will often refer to the &lt;strong&gt;Blue Book&lt;/strong&gt; (Nomenclature of Organic Chemistry), inorganic chemists to the &lt;strong&gt;Red Book&lt;/strong&gt; (Nomenclature of Inorganic Chemistry), etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your question is concerned, the &lt;strong&gt;Gold Book&lt;/strong&gt; (IUPAC Compendium of Chemical Terminology) provides the definitions for &lt;a href=&quot;http://goldbook.iupac.org/M04002.html&quot; rel=&quot;nofollow&quot;&gt;molecule&lt;/a&gt; and &lt;a href=&quot;http://goldbook.iupac.org/M03986.html&quot; rel=&quot;nofollow&quot;&gt;molecular entity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-22T17:39:37.953" LastActivityDate="2015-05-22T17:39:37.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31848" PostTypeId="2" ParentId="31836" CreationDate="2015-05-22T16:10:52.870" Score="5" Body="&lt;p&gt;The quantities and equations used with the International System of Units (SI) are called the International System of Quantities (ISQ). Many of the quantities as well as their recommended names and symbols are listed in the International Standard ISO/IEC 80000. &#xA;They are considered almost universally accepted for use throughout the physical sciences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this standard, the recommended symbol for the quantity ‘energy’ is an upper case $E$. (The quantity symbol is always written in italic type.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When, in a given context, different quantities have the same letter symbol or when, for one quantity, different applications or different values are of interest, a distinction can be made by use of subscripts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means, if you need more than one energy, you may for example write $E_1$ and $E_2$ or as in your example $E_\text{sys}$ and $E_\text{sur}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that only subscripts that represent a physical quantity or mathematical variable are printed in italic type. All other subscripts are printed in roman type. Thus, writing $E_{sys}$ or $E_{sur}$ would be wrong.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-22T16:10:52.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31849" PostTypeId="2" ParentId="31846" CreationDate="2015-05-22T16:19:03.060" Score="18" Body="&lt;p&gt;Interesting prompt!&lt;/p&gt;&#xA;&lt;p&gt;There are plenty of reactions which could be performed underwater to release a large amount of energy. In fact, many substances react vigorously with water itself. You may have heard about how alkali metals such as sodium are not compatible with water, and there's &lt;a href=&quot;https://www.youtube.com/watch?v=HY7mTCMvpEM&quot; rel=&quot;noreferrer&quot;&gt;a good reason for it!&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Explosive reactions aren't particularly suited to cooking food, though, and a sustained flame would be much more useful. Interestingly enough, there was legitimate research performed on making underwater flames, for a very specific purpose: the Olympic torch!&lt;/p&gt;&#xA;&lt;p&gt;For the 2000 Summer Olympics in Australia, during &lt;a href=&quot;http://en.wikipedia.org/wiki/2000_Summer_Olympics_torch_relay&quot; rel=&quot;noreferrer&quot;&gt;the classic Olympic torch relay&lt;/a&gt; before the start of the games, the torch was actually &lt;a href=&quot;http://news.bbc.co.uk/2/hi/asia-pacific/807872.stm&quot; rel=&quot;noreferrer&quot;&gt;carried ablaze underwater for three minutes at the Great Barrier Reef&lt;/a&gt;. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Olympic_flame#Torches&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the usual fuel for the burning torch is a mixture of liquefied propane/butane, which burns using oxygen gas in the air as an oxidizer (more details of a torch's construction can be found &lt;a href=&quot;http://entertainment.howstuffworks.com/olympic-torch5.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;). There is very little dissolved oxygen in ocean water, however, so the fuel/oxidizer mixture had to be changed. From the BBC article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The special flare was developed after nine months by a team of chemists and engineers at Melbourne pyrotechnics company Pains Wessex.&lt;/p&gt;&#xA;&lt;p&gt;The company's managing director Charles Tegner said designing the torch was a challenge - not only to produce a flare to burn underwater at such a depth, but to burn like the Olympic torch flame as well.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;It had to be clearly visible,&amp;quot; he said. &amp;quot;Such flares don't normally exist.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;The flare's chemical composition, pressed into a steel tube, produces sufficient oxygen and nitrogen to maintain a very hot flame.&lt;/p&gt;&#xA;&lt;p&gt;The flare burns so fiercely at more than 2,000 degrees Celsius that this creates enough pressure to keep the water from entering the tube.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Its intensity is produced by a mix of oxygen-generating chemicals as well as the combustible element magnesium in a finely powdered form&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I haven't found the detailed composition of the fuel/oxidizer mixture, but this is already a very good hint. Magnesium is an alkali earth metal which burns vigorously with a bright white flame in many conditions, be it in pure oxygen, air, pure nitrogen, &lt;a href=&quot;https://www.youtube.com/watch?v=_xCbal2YyaE&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;or even pure carbon dioxide&lt;/em&gt;&lt;/a&gt;, and it will also react with water when hot producing hydrogen gas, much like sodium (this is why reactive metal fires are so hazardous; don't even think about using a &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; or water extinguisher on them, &lt;a href=&quot;https://www.youtube.com/watch?v=SA6Yo5vb0wk&quot; rel=&quot;noreferrer&quot;&gt;that just feeds the fire&lt;/a&gt;!). In this case, the pyrotechnicians decided to mix finely pulverized magnesium metal with a solid oxidizer, very likely something such as potassium chlorate, &lt;span class=&quot;math-container&quot;&gt;$\ce{KClO3}$&lt;/span&gt;, which decomposes under heat releasing oxygen gas. The flame was lit in air until the reaction between the magnesium and solid oxidizer started going, and then it was safe to submerge the torch while the reactants lasted. The stunt has in fact &lt;a href=&quot;http://www.olympic.org/news/olympic-flame-carried-underwater-in-world-s-deepest-lake/217750&quot; rel=&quot;noreferrer&quot;&gt;been repeated since&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To add a few more options to this answer, there is a general class of reactions with water called &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrolysis&quot; rel=&quot;noreferrer&quot;&gt;hydrolysis reactions&lt;/a&gt;, some of which are quite energetic. Here, the starting material tears apart water molecules directly and uses the fragments to modify its structure to a more stable form. For example, &lt;a href=&quot;https://www.youtube.com/watch?v=ccPcfYQnmVw&quot; rel=&quot;noreferrer&quot;&gt;this is what happens&lt;/a&gt; when phosphorous pentoxide is allowed to come into contact with water. This avidity for water makes phosphorous pentoxide a useful material in a laboratory, as it is capable of scavenging trace amounts of water from a liquid or gas. However, these types of reactions already tend to release much less energy per gram of reactant, compared to combustion reactions.&lt;/p&gt;&#xA;&lt;p&gt;Lastly, I'll cite one more thing. Even without any chemical reaction, the dissolution of some solids in water is capable of releasing some heat. This happens when starting material has a weaker attraction to itself than to water molecules. This property can be exploited in &lt;a href=&quot;http://en.wikipedia.org/wiki/Heating_pad#Chemical&quot; rel=&quot;noreferrer&quot;&gt;disposable heat packs&lt;/a&gt;. These types of processes usually release even less heat per gram of starting material than combustion or hydrolysis, but at least the process is conceivably easily reversed, provided energy is given to dry the material again.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, it is possible to get a lot of heat from reactions underwater, though it is quite a more elaborate process than starting a fire in air.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Now that I saw the original question at Worldbuilding.SE, I realize part of it wondered about how cooking would be done in earlier stages of technological development. In this case, I'm hard-pressed to think of a chemical reaction which happens using materials found easily underwater and releases a lot of energy. The difficulty is great enough that it has actually been suggested previously as a general argument for why technologically advanced or even sentient life would be unlikely to develop underwater, as any hot thermal processing of food or materials would be severely hindered. Except for incredible strokes of luck, hydrothermal vents on the sea floor sound like the most reliable heat source which could be accessed by a primitive civilization.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-22T19:35:38.657" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31850" PostTypeId="2" ParentId="31843" CreationDate="2015-05-22T16:49:59.143" Score="6" Body="&lt;p&gt;I think you are asking about preferred conformations for $\ce{sp^3-sp^3}$ single bonds and $\ce{sp^3-sp^2}$ single bonds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that the staggered (or gauche) conformation is preferred in simple alkanes.  This is generally explained by noting that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;adverse steric interactions are minimized in the staggered conformation, and&lt;/li&gt;&#xA;&lt;li&gt;stabilizing &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperconjugation&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperconjugative interactions&lt;/a&gt; are maximized&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lqNig.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(&lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s28-02-isomers-of-organic-compounds.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;From: “Isomers of Organic Compounds”, section 24.2 from the book Principles of General Chemistry (v. 1.0).&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the $\ce{sp^3-sp^2}$ single bond as found in propene, toluene, acetaldehyde, etc., the eclipsed conformation is now the more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p3fHR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Wikitexts/UC_Davis/UCD_Chem_231A%3A_Methods_of_Organic_Synthesis/Content/Conformational_Analysis/Acyclic_Conformations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Di Maso, Jared Shaw, the UC Davis ChemWiki by University of California&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again both steric and electronic arguments can be advanced to explain this preference.  The steric argument says let's use the bent model to describe the double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJHl0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now replace one of those ethylene hydrogens with a methyl group and note that in the eclipsed conformation (build a model if you can) none of the methyl hydrogens eclipse the hydrogen on the adjacent olefinic carbon or any of the bent bonds.  In the staggered conformation all 3 of the methyl hydrogens are eclipsing something, the 2 bent bonds and the $\ce{C-H}$ bond on the adjacent olefinic carbon.  The basis of the electronic argument can be found &lt;a href=&quot;http://chemwiki.ucdavis.edu/Wikitexts/UC_Davis/UCD_Chem_231A%3A_Methods_of_Organic_Synthesis/Content/Conformational_Analysis/Acyclic_Conformations&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Personally, for me, the steric argument is easier to understand.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4945" LastEditDate="2016-02-09T09:11:23.883" LastActivityDate="2016-02-09T09:11:23.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31851" PostTypeId="2" ParentId="14533" CreationDate="2015-05-22T16:59:10.777" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Lewis structure:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlxjF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The properly way to determine the Lewis structure, based on this example, is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Total valence electrons: $4 + 6\cdot2 + 1\cdot2 = 18$&lt;/li&gt;&#xA;&lt;li&gt;Total electrons needed for octets/doublets: $8\cdot3+2\cdot2 = 28$&lt;/li&gt;&#xA;&lt;li&gt;Total shared/bonding electrons: $28-18 = 10$ (In other words, there are only five bonds.)&lt;/li&gt;&#xA;&lt;li&gt;Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 18 - 10 = 8$ (In other words, the are only 4 pairs of lone electrons (all for $\ce{O}$)&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Furhter information on how to draw Lewis structures, please follow the &lt;a href=&quot;https://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_11/node3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-05-23T04:13:09.627" LastActivityDate="2015-05-23T04:13:09.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31852" PostTypeId="2" ParentId="25128" CreationDate="2015-05-22T17:04:03.327" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;Lewis structure for HOCl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFlrD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The properly way to determine the Lewis structure, based on this example, is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Total valence electrons: $7 + 6 + 1 = 14$&lt;/li&gt;&#xA;&lt;li&gt;Total electrons needed for octets/doublets: $8\cdot2+2\cdot1 = 18$&lt;/li&gt;&#xA;&lt;li&gt;Total shared/bonding electrons: $18-14 = 4$ (In other words, there are only two single bonds.)&lt;/li&gt;&#xA;&lt;li&gt;Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 14 - 4 = 10$ (In other words, the are only 5 pairs of lone electrons (2 for $\ce{O}$ and three for $\ce{Cl}$))&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Furhter information on how to draw Lewis structures, please follow the &lt;a href=&quot;https://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_11/node3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-05-23T04:12:19.643" LastActivityDate="2015-05-23T04:12:19.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31854" PostTypeId="1" CreationDate="2015-05-22T17:13:12.100" Score="1" ViewCount="172" Body="&lt;p&gt;Methyl orange changes colors depending on the acidity or basicity of the solution. But I have read that it forms complexes with certain ions. Can anybody suggest any resource regarding the coordination complex-formation chemistry of methyl orange?&lt;/p&gt;&#xA;" OwnerUserId="16436" LastActivityDate="2015-05-22T17:13:12.100" Title="Methyl orange coordination compound chemistry" Tags="&lt;coordination-compounds&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31856" PostTypeId="2" ParentId="11092" CreationDate="2015-05-22T17:19:14.933" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Lewis structure:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vvONz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The properly way to determine the Lewis structure, based on this example, is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Total valence electrons: $7\cdot2 + 6\cdot2 = 26$&lt;/li&gt;&#xA;&lt;li&gt;Total electrons needed for octets/doublets: $8\cdot4 = 32$&lt;/li&gt;&#xA;&lt;li&gt;Total shared/bonding electrons: $32-26=6$ (In other words, there are only three bonds.)&lt;/li&gt;&#xA;&lt;li&gt;Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 26 - 6 = 20$ (In other words, the are only 5 pairs of lone electrons (2 pairs for $\ce{O}$, 6 pairs to $\ce{Cl}$ and the big &lt;strong&gt;BUT&lt;/strong&gt; of Lewis structure analysis: the remaining lone pairs correspond to the sulfur at bonding with oxygen.)&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Furhter information on how to draw Lewis structures, please follow the &lt;a href=&quot;https://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_11/node3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2016-03-26T23:00:20.020" LastActivityDate="2016-03-26T23:00:20.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31858" PostTypeId="2" ParentId="31839" CreationDate="2015-05-22T17:48:45.260" Score="3" Body="&lt;p&gt;Alright, let’s take this one step at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, you want to determine a molecular formula, and among your known data you have a mole fraction. This suggests it’s a good idea to work with moles to solve the problem. As such, the first step is to convert the amount of water added and magnesium sulfate hydrate used. Calculating the number of moles of water added is easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$90\ \mathrm{g}\ \text{water added} \times \frac{1\ \mathrm{mol}\ \ce{H2O}}{18\ \mathrm{g} \ \ce{H2O}} = 5\ \mathrm{mol}\ \ce{H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of moles of $\ce{MgSO4.$x\,$H2O}$ is unknown because even though you know how much mass was dissolved, you don’t know its formula completely. However, $\ce{MgSO4}$ contributes $120\ \mathrm{g\ mol^{-1}}$ to the molar mass, and each additional molecule of water in the hydrate stoichiometry adds a further $18\ \mathrm{g\ mol^{-1}}$, so that the molar mass of $\ce{MgSO4.$x\,$H2O}$ is $120\ \mathrm{g\ mol^{-1}} + x \times 18\ \mathrm{g\ mol^{-1}} = (120+18x)\ \mathrm{g\ mol^{-1}}$. With this molar mass, the number of moles of $\ce{MgSO4.$x\,$H2O}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$70\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O} \times \frac{1\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}}{(120+18x)\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O}} = \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the definition of mole fraction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{total}}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one is tempted to write the following and solve for $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}\ }{\frac{70}{120+18x}+5 \ }=0.039$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;strong&gt;this equation is incorrect&lt;/strong&gt;. The problem is that the following process was not taken into account:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MgSO4.$x\,$H2O(s) \xrightarrow{water} MgSO4(aq) + $x$\ H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when the magnesium sulfate hydrate dissolves in water, &lt;em&gt;the water molecules contained in the solid are released and become an indistinguishable part of the solvent&lt;/em&gt;; the final mixture is a simply a solution of the solute $\ce{MgSO4}$ in water. Therefore, the number of moles of solvent is not simply $5\ \mathrm{mol}$ as calculated at the start, but rather a larger value, which should include the number of moles of water released when the solid dissolves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the dissolution equation, we see that $1\ \mathrm{mol}$ of $\ce{MgSO4.$x\,$H2O}$ generates $1\ \mathrm{mol}$ of $\ce{MgSO4}$ and $x\ \mathrm{mol}$ of $\ce{H2O}$ from crystallized water. However, we know the amount of moles of $\ce{MgSO4.$x\,$H2O}$. This means the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_\text{solute}=n_{\ce{MgSO4}}=1 \times n_{\ce{MgSO4.$x\,$H2O}}= \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_\text{solvent} = n_\text{added water} + n_\text{crystallized water} = n_\text{added water} + x \times n_{\ce{MgSO4.$x\,$H2O}} = \bigg(5 + \frac{70x}{120+18x}\bigg)\ \mathrm{mol}\ \ce{H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can finally apply these values to the mole fraction equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}}{\frac{70}{120+18x} + \bigg(5 + \frac{70x}{120+18x}\bigg)}=0.039$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $x$ provides the answer &lt;a href=&quot;http://www.wolframalpha.com/input/?i=%2870%2F%28120%2B18x%29%29%2F%28%2870%2F%28120%2B18x%29%29%2B%285%2B%2870x%2F%28120%2B18x%29%29%29%29%3D0.039&quot; rel=&quot;nofollow&quot;&gt;$x=7.03$&lt;/a&gt;, which means there are seven moles of water per mole of $\ce{MgSO4}$, and thus the formula for the solid is $\ce{MgSO4.7H2O}$, representing magnesium sulfate heptahydrate.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorDisplayName="user7951" LastEditDate="2015-05-22T19:30:30.810" LastActivityDate="2015-05-22T19:30:30.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31859" PostTypeId="1" AcceptedAnswerId="31861" CreationDate="2015-05-22T18:43:09.423" Score="2" ViewCount="1804" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I remember using both 95% ethanol and absolute ethanol quite a few times. You'd need absolute ethanol for a reaction that would fail if water was present, such as a Grignard reaction. I can't remember whether the absolute ethanol was in an ordinary bottle or the type with a rubber septum (where you keep the liquid under nitrogen to avoid contact with air). If the ethanol was to be used as a reagent, a septum bottle would be more appropriate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://boards.straightdope.com/sdmb/showpost.php?p=5314683&amp;amp;postcount=10&quot; rel=&quot;nofollow&quot;&gt;http://boards.straightdope.com/sdmb/showpost.php?p=5314683&amp;amp;postcount=10&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that the poster just forgot all his chemistry. Ironic for someone posting on a message board that proclaims to be &quot;fighting ignorance since 1973.&quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Second (assuming that one could somehow run a Grignard reaction in ethanol), do you really &lt;em&gt;need&lt;/em&gt; anhydrous ethanol for Grignard reactions to work? Why not just use a slight excess of Grignard to account for the water present in 95% ethanol? Seems to me that anhydrous ethanol would be pretty expensive given that the usual distillation methods wouldn't get you from 95% ethanol to 100% ethanol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Finally, is there a reaction in which using excess Grignard to account for acidic protons wouldn't chemically work in some way?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-05-22T22:11:28.930" LastActivityDate="2015-05-22T22:11:28.930" Title="Grignards and anhydrous ethanol" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31861" PostTypeId="2" ParentId="31859" CreationDate="2015-05-22T19:01:41.453" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As well as being excellent nucleophiles, Grignard reagents are also very strong bases because they essentially function as alkyl anions. Therefore they will react with ethanol to form magnesium ethoxide, the alkane, and magnesium bromide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grignard reactions are usually carried out in ether solvents such as diethylether and tetrahydrofuran because they are unreactive towards the reagent and they actually stabilize it by coordinating to the magnesium ion to form a tetrahedral complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SqhQU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://employees.csbsju.edu/cschaller/Reactivity/carbonyl/COsemianion.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-22T19:01:41.453" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31863" PostTypeId="1" CreationDate="2015-05-22T19:15:45.450" Score="2" ViewCount="1437" Body="&lt;p&gt;If I have the pressure and temperature of a substance in its triple point, as well as its pressure and temperature in the critical point, how can I determine the boiling point of the substance at 1 atm of pressure?&lt;/p&gt;&#xA;" OwnerUserId="16438" LastEditorUserId="16683" LastEditDate="2020-12-23T00:57:30.660" LastActivityDate="2020-12-23T00:58:16.450" Title="Determining the boiling point of a substance given the triple point and critical point" Tags="&lt;theoretical-chemistry&gt;&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31866" PostTypeId="2" ParentId="31840" CreationDate="2015-05-22T21:17:46.850" Score="7" Body="&lt;p&gt;Unless I misunderstand the question, the gram was never defined that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For historical reasons the SI unit for mass is an oddity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The base unit is not the gram, but the kilogram. This is the only case where the base unit has a prefix. The gram is defined as $\frac{1}{1000}$ of a kilogram.&lt;/li&gt;&#xA;&lt;li&gt;The kilogram is the only remaining SI unit that is based on a man-made object, the &lt;a href=&quot;http://www.bipm.org/en/bipm/mass/prototype.html&quot; rel=&quot;nofollow&quot;&gt;International Prototype of the Kilogram&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At its 25th meeting in 2014, the General Conference on Weights and Measures (CGPM) has (again) declared its intention to abandon this atificial standard and link &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the definitions of the kilogram, ampere, kelvin, and mole to exact numerical values of the&#xA;  Planck constant $h$, elementary charge $e$, Boltzmann constant &lt;em&gt;k&lt;/em&gt;, and Avogadro constant $N_A$ [&amp;hellip;] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.bipm.org/en/CGPM/db/25/1/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the possible approaches for the redefinition of the kilogram would be a side-effect of the redefinition of the Avogadro constant with in the &lt;a href=&quot;http://www.ptb.de/cms/en/presseaktuelles/journals-magazines/ptb-news/ptb-news-ausgaben/ptb-news/news05-3/avogadro-project-first-isotope-pure-silicon-single-crystal.html&quot; rel=&quot;nofollow&quot;&gt;Avogadro project&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="418" LastEditDate="2015-05-23T04:29:06.093" LastActivityDate="2015-05-23T04:29:06.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31867" PostTypeId="1" CreationDate="2015-05-22T21:28:41.247" Score="16" ViewCount="2236" Body="&lt;p&gt;Is the azide ion ($\ce{N3-}$) paramagnetic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use molecular orbital theory to show that it is or isn't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew a sine wave diagram and filled up the $1\pi$ and $2\pi$ MOs, and since there are no unpaired electrons, can I conclude that the ion is not paramagnetic?&lt;/p&gt;&#xA;" OwnerUserId="16445" LastEditorUserId="402" LastEditDate="2016-03-01T23:53:18.590" LastActivityDate="2016-03-20T09:41:00.080" Title="Is azide paramagnetic?" Tags="&lt;molecular-orbital-theory&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31868" PostTypeId="1" CreationDate="2015-05-22T21:34:51.713" Score="12" ViewCount="435" Body="&lt;p&gt;Why is the rate of water exchange at Al(III) centres so slow ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://books.google.co.uk/books?id=VURZcorw_3gC&quot;&gt;this medical book&lt;/a&gt; (p.5) it is $10^{5}$ times faster at Mg(II).&lt;/p&gt;&#xA;" OwnerUserId="10256" LastActivityDate="2015-05-31T20:54:57.190" Title="Water exchange at Al (III)" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31869" PostTypeId="1" CreationDate="2015-05-23T01:35:37.137" Score="4" ViewCount="778" Body="&lt;p&gt;I'm trying to figure out what my optimal current density would be in respect to the limiting current density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you calculate concentration polarization and limiting current density in an electrolytic cell. There are two copper electrodes and a $2\:\mathrm{M}$ copper nitrate solution. The average current density was $0.375\:\mathrm{A\:cm^{-2}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the Tafel expression ($\eta a=a+b \log i$) can calculate the voltage drop but can I find the limiting current density just like that in the image? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVNZU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16115" LastEditorUserId="15410" LastEditDate="2015-05-23T01:48:55.837" LastActivityDate="2015-05-23T01:48:55.837" Title="Calculation of optimal current density" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31870" PostTypeId="1" CreationDate="2015-05-23T04:08:17.543" Score="7" ViewCount="352" Body="&lt;p&gt;As far as I know, for all substances other than helium, if you extrapolate the solid-gas line on the phase diagram, it passes through the origin. That is, no matter how low the temperature is, you can still vaporize the substance by lowering the pressure, and no matter how low the pressure is, you can still freeze the substance by lowering the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense to me on thermodynamic grounds. If we keep lowering the pressure at a given temperature, the entropy of vaporization increases without bound (although only logarithmically) and at some point the gas phase is favoured, whereas if we keep lowering the temperature at a given pressure, then the entropy term in free energy becomes negligible and the solid phase is favoured by enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase diagrams I've found online for helium-4, however, don't follow this rule. They seem to show that helium-4 is exclusively a liquid in a neighbourhood of the origin. Example here: &lt;a href=&quot;http://ltl.tkk.fi/research/theory/helium.html&quot;&gt;http://ltl.tkk.fi/research/theory/helium.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder whether this is true, and if so, how it can be possible. How can it be that no matter how low the pressure is, helium-4 will not boil? I understand that superfluidity is a quantum phenomenon but I can't see how my previous reasoning could not apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; since some people think this is a duplicate: Again, my question is whether helium-4 will not &lt;strong&gt;boil&lt;/strong&gt; at zero pressure, and if so, how that can be; it concerns the liquid-gas transition and not the solid-liquid one.&lt;/p&gt;&#xA;" OwnerUserId="5490" LastEditorUserId="5490" LastEditDate="2015-05-23T04:37:22.017" LastActivityDate="2015-05-23T04:37:22.017" Title="Is helium-4 a liquid at zero temperature and zero pressure?" Tags="&lt;thermodynamics&gt;&lt;phase&gt;&lt;entropy&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31871" PostTypeId="1" CreationDate="2015-05-23T05:48:57.663" Score="1" ViewCount="216" Body="&lt;p&gt;A greater positive charge means the ligand is strongly attracted and pulled closer, and thus it creates a stronger crystal field. The potential energy value of the barycentre should thus be higher, since repulsion experienced by the d electrons is stronger. Why is though the value of the Δo increased - the potential energy difference between the various differently oriented d orbitals? Each orbital experiences stronger repulsion if charge on metal ion increases and the ligand is pulled closer, but how can anything be said of the energy difference Δo? Both &lt;em&gt;t&lt;sub&gt;2g&lt;/sub&gt;&lt;/em&gt; and &lt;em&gt;e&lt;sub&gt;g&lt;/sub&gt;&lt;/em&gt; are at a higher level, that's all. &lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-05-23T05:48:57.663" Title="Why does the value of Δo increase with increased charge on the metal?" Tags="&lt;coordination-compounds&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31872" PostTypeId="1" AcceptedAnswerId="31874" CreationDate="2015-05-23T06:02:30.767" Score="20" ViewCount="3435" Body="&lt;p&gt;The boiling point of ammonia is −33 °C while that of $\ce{HCN}$ is 25 °C. In a recent AP (Advanced Placement) Chemistry test, a free response question asked why this is the case. Can someone shine light on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on Jan's answer:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Although the $\ce{C-H}$ bond does not usually exhibit good hydrogen bond, $\ce{H-CN}$ is a special case in which the bond is polar enough to provide better hydrogen bonding than $\ce{NH3}$. I agree with Jan on this point.  &lt;/li&gt;&#xA;&lt;li&gt;However, Jan's explanation is for the polarity of the $\ce{H-CN}$ bond is based on $\mathrm pK_\mathrm a$ (lower $\mathrm pK_\mathrm a$ suggests higher acidity and good hydrogen bonding). As one user pointed out, $\mathrm pK_\mathrm a$ is only defined in aqueous solution, so stating pK$_a$ as an evidence is not suitable for the system I have which is a pure $\ce{HCN}$ liquid. In such system, $\ce{HCN}$ does not significantly dissociate. A suggestion was made on using proton affinity as the evidence for the &lt;em&gt;acidity&lt;/em&gt; of $\ce{HCN}$.   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I perceive that the answers (reasons and evidence) are mostly targeted at the &lt;em&gt;chemical process&lt;/em&gt; of dissociation of $\ce{HCN}$. Should this question be addressed more in terms of a &lt;em&gt;physical process&lt;/em&gt;? I mean there is little dissociation of pure-liquid $\ce{HCN}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have been thinking about is a MO picture in which the cyano group withdraws electron density from the hydrogen atom, in a similar way to how acetate group withdraws electron from the hydrogen in acetic acid. What is your thought on this?&lt;/p&gt;&#xA;" OwnerUserId="14698" LastEditorUserId="16683" LastEditDate="2017-08-27T12:57:21.817" LastActivityDate="2017-08-27T12:57:57.433" Title="Why does HCN boil at a higher temperature than NH3?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;intermolecular-forces&gt;" AnswerCount="6" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31874" PostTypeId="2" ParentId="31872" CreationDate="2015-05-23T08:16:39.540" Score="14" Body="&lt;p&gt;The enthalpy of vaporization of &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C74908&amp;amp;Mask=4#Thermo-Phase&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{HCN}$&lt;/a&gt; is higher than for &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7664417&amp;amp;Units=SI&amp;amp;Mask=4#Thermo-Phase&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\ce{NH3}$&lt;/a&gt;, which suggests that $\ce{HCN}$ molecules interact more strongly than $\ce{NH3}$ molecules. $\ce{C-H}$ bonds are not usually considered good hydrogen bond donors, but $\ce{HCN}$ is unusual. For example $\ce{HCN}$ has a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_cyanide&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\mathrm pK_\mathrm a$ value of 9.2&lt;/a&gt;, indicating that the $\ce{CN}$ group is electron withdrawing and that it is a reasonably good hydrogen (bond) donor. This is likely due to the electronegativity of nitrogen, and also the high &quot;s-content&quot; of the sp-hybridized $\ce{CH}$ bond, which keeps the electron pair close to the nucleus.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="16683" LastEditDate="2017-08-27T12:57:57.433" LastActivityDate="2017-08-27T12:57:57.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31875" PostTypeId="2" ParentId="28254" CreationDate="2015-05-23T09:13:37.387" Score="0" Body="&lt;p&gt;The idea that the atom contains small indivisible quantities of electric charge was developed in the 19th century by various contributors (e.g. Faraday, Stoney, Laming, Weber, and Helmholtz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Thomson (Nobel Prize in physics 1906) discovered the electron in 1897 and estimated its mass and charge, he used the free electrons of cathode rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important experiment that showed that atoms contain electrons was the discovery of the Zeeman effect in 1896, i.e. the splitting of spectral lines in the presence of a magnetic field. Lorentz explained the phenomena with his electron theory. In 1902, Lorentz and Zeeman jointly received the Nobel Prize in physics.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-23T09:13:37.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31878" PostTypeId="2" ParentId="31872" CreationDate="2015-05-23T10:41:48.667" Score="8" Body="&lt;p&gt;Supporting Jan's answer, actually: Consider the points below from the book &lt;a href=&quot;https://books.google.com/books?id=xF8PV2GeT-AC&amp;amp;pg=PA102&amp;amp;lpg=PA102&amp;amp;dq=hydrogen+bonding+of+hcn+and+nh3&amp;amp;source=bl&amp;amp;ots=6XZhdYmZXA&amp;amp;sig=PpJ1aANkDc8Jec_G2uv7stWSbw0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=YFJgVY2CO5GZyQSv6IDIBQ&amp;amp;ved=0CCoQ6AEwAQ#v=onepage&amp;amp;q=hydrogen%20bonding%20of%20hcn%20and%20nh3&amp;amp;f=false&quot;&gt;Hydrogen Bonding: A Theoretical Perspective&lt;/a&gt;, p. 102.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Moreover, the hydrogen in HCN is acidic enough that the molecule may act as an effective proton donor... When paired with NH$_3$, HCN acts as a proton donor...&quot; &lt;br&gt;&#xA;This agrees with Jan's note, on the pK$_a$ of HCN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comment that &quot;the hydrogen is electronically deficient, leading to an excellent electron acceptor&quot; seems to be about hydrogen bonding, whereas there is evidence that HCN is more ionic in nature.&lt;/p&gt;&#xA;" OwnerUserId="16457" LastActivityDate="2015-05-23T10:41:48.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31880" PostTypeId="1" AcceptedAnswerId="31891" CreationDate="2015-05-23T11:59:12.627" Score="6" ViewCount="482" Body="&lt;p&gt;My professor gave my class a set of practice questions for the mid-term. One of those questions asks for the molar mass of a number of chemicals, such as lithium bromide and silver nitrate. We haven't studied those chemicals, so there's little reason for the professor to expect us to have memorized the elements that constitute those chemicals. In the case of lithium bromide, the term informs us of its constitution. Whereas,'silver nitrate' informs us that the chemical $ \mathrm{AgNO_3}$, includes silver and nitrogen, but I don't see anything in the term that would inform the reader that the chemical includes $ \mathrm{O_3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything in the term 'silver nitrate' that would inform a reader that the chemical includes $ \mathrm{O_3}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, is there a set of rules that relates the chemical constituents of chemical substances to the linguistic constituents of the terms that designate them?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="15929" LastEditDate="2015-05-23T12:09:27.990" LastActivityDate="2015-05-23T14:06:46.873" Title="Does the term 'silver nitrate' express all of its constituent chemical elements?" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31882" PostTypeId="2" ParentId="31845" CreationDate="2015-05-23T12:08:31.257" Score="13" Body="&lt;p&gt;The reason that we discovered more fluorides of xeon than helium, argon, and krypton is quite obvious and you might already know it. As lighter noble gases have more stable shell configuration, it is more difficult to make any kind of compound, including fluorides.&lt;/p&gt;&#xA;&lt;p&gt;Also, the higher amount of fluorine atoms in a molecule of the compounds means drawing more electron from the noble gas, which is thermodynamically not preferred. Thus, &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF6}$&lt;/span&gt; are far less stable than &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF2}$&lt;/span&gt;. The quantum chemical calculation confirmed this (Dixon 2007). The same reference said that they are stable, but have not been synthesized yet, and the synthesis would be very difficult.&lt;/p&gt;&#xA;&lt;br/&gt;&#xA;For radon it is a different problem. Briefly, there might be more fluoride compounds of radon than xenon, but the synthesis is difficult to be confirmed. &#xA;&lt;p&gt;Even though radon is less inert than xenon and its compounds were predicted and calculated to be more chemically stable than xenon compounds (Chernick 1962, Thayer 2005), it is difficult to study. Radon is is rare and radioactive (half-life of 3.8 days). Studying chemical compounds in trace amount is a difficult task. There are claims for discoveries of &lt;span class=&quot;math-container&quot;&gt;$\ce{RnF4}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{RnF6}$&lt;/span&gt; (also in the above references), but confirming that you have synthesized new compounds in trace amount is extremely difficult because the evidence is not obvious. &lt;span class=&quot;math-container&quot;&gt;$\ce{RnF8}$&lt;/span&gt; was calculated to be unstable, but should be more possible than &lt;span class=&quot;math-container&quot;&gt;$\ce{XeF8}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;br/&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1 David A. Dixon, Tsang-Hsiu Wang, Daniel J. Grant, Kirk A. Peterson, Karl O. Christe, Gary J. Schrobilgen, &amp;quot;Heats of formation of krypton fluorides and stability predictions for &lt;span class=&quot;math-container&quot;&gt;$\ce{KrF4}$&lt;/span&gt; and KrF6&lt;span class=&quot;math-container&quot;&gt;$\ce{KrF46}$&lt;/span&gt; from high level electronic structure calculations,&amp;quot; &lt;em&gt;Inorganic Chemistry&lt;/em&gt; &lt;strong&gt;2007&lt;/strong&gt;, &lt;em&gt;46(23)&lt;/em&gt;, 10016-10021 (&lt;a href=&quot;https://doi.org/10.1021/ic701313h&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1021/ic701313h&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;2 C. L. Chernick, H. H. Claassen, P. R. Fields, H. H. Hyman, J. G. Malm, W. M. Manning, M. S. Matheson, L. A. Quarterman, F. Schreiner, H. H. Selig, I. Sheft, S. Siegel, E. N. Sloth, L. Stein, M. H. Studier, J. L. Weeks, M. H. Zirin, &amp;quot;Fluorine Compounds of Xenon and Radon,&amp;quot; &lt;em&gt;Science&lt;/em&gt; &lt;strong&gt;1962&lt;/strong&gt;, &lt;em&gt;138(3537)&lt;/em&gt;, 136-138 (DOI: 10.1126/science.138.3537.136).&lt;/p&gt;&#xA;&lt;p&gt;3 John S. Thayer, &amp;quot;Relativistic Effects and the Chemistry of the Heaviest Main-Group Elements,&amp;quot; &lt;em&gt;Journal of Chemical Education&lt;/em&gt; &lt;strong&gt;2005&lt;/strong&gt;, &lt;em&gt;82(11)&lt;/em&gt;, 1721-1727 (&lt;a href=&quot;https://doi.org/10.1021/ed082p1721&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1021/ed082p1721&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16455" LastEditorUserId="61578" LastEditDate="2020-08-07T07:02:56.780" LastActivityDate="2020-08-07T07:02:56.780" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31883" PostTypeId="2" ParentId="31863" CreationDate="2015-05-23T12:22:09.750" Score="1" Body="&lt;p&gt;What you need is the &lt;a href=&quot;http://www.chemteam.info/GasLaw/Clasius-Clapeyron-Equation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clausius–Clapeyron equation&lt;/a&gt;, which is used to relate vapor pressure and temperature (or external pressure and boiling point).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln{\left(\dfrac{P_1}{P_2}\right)}=\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Step 1: Find the enthalpy of vaporizations &lt;span class=&quot;math-container&quot;&gt;$\Delta H_{\text{vap}}$&lt;/span&gt; by plugging in the two sets of temperatures and pressures that you have.&lt;/p&gt;&#xA;&lt;p&gt;Step 2: Use your &lt;span class=&quot;math-container&quot;&gt;$\Delta H_{\text{vap}}$&lt;/span&gt; and the temperature and pressure of the triple point to determine the boiling temperature at 1 atm.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="16683" LastEditDate="2020-12-23T00:58:16.450" LastActivityDate="2020-12-23T00:58:16.450" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31885" PostTypeId="2" ParentId="31872" CreationDate="2015-05-23T12:39:40.003" Score="7" Body="&lt;p&gt;Besides the existing answers, which focus on the acidity of HCN, note that HCN is also a considerably larger molecule than NH&lt;sub&gt;3&lt;/sub&gt;.  Thus, &lt;em&gt;even if the interactions between the molecules were qualitatively identical&lt;/em&gt;, one would still expect a higher boiling point for HCN on the basis of the size difference (and resulting stronger dispersion interactions; see comments) alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an illustrative example, we can look at &lt;a href=&quot;//en.wikipedia.org/wiki/Methylamine&quot; rel=&quot;nofollow&quot;&gt;methylamine&lt;/a&gt;, CH&lt;sub&gt;3&lt;/sub&gt;NH&lt;sub&gt;2&lt;/sub&gt;, which resembles ammonia in most respects, except for having one of its hydrogens replaced by a bulky methyl group, making it similar in size to HCN.  Its boiling point is &amp;minus;6.6 &amp;deg;C, well above the &amp;minus;33 &amp;deg;C for NH&lt;sub&gt;3&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining ~32 K difference between the boiling points of HCN and CH&lt;sub&gt;3&lt;/sub&gt;NH&lt;sub&gt;2&lt;/sub&gt; is then presumably explained by the stronger acidity of the HCN hydrogen, and thus the stronger hydrogen bonding between HCN molecules than for NH&lt;sub&gt;3&lt;/sub&gt; and CH&lt;sub&gt;3&lt;/sub&gt;NH&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;" OwnerUserId="283" LastEditorUserId="283" LastEditDate="2015-05-24T13:10:44.097" LastActivityDate="2015-05-24T13:10:44.097" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="31886" PostTypeId="2" ParentId="31872" CreationDate="2015-05-23T12:48:51.047" Score="2" Body="&lt;p&gt;Although &lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt; consist of hydrogen bonding &lt;span class=&quot;math-container&quot;&gt;$\ce{HCN}$&lt;/span&gt; has a very stronger dipole-dipole interaction which makes its boiling point equivalent to alcohols.&lt;/p&gt;&#xA;&lt;p&gt;See the diagram:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zRHOB.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="14521" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-23T17:36:18.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31890" PostTypeId="2" ParentId="31880" CreationDate="2015-05-23T13:11:05.693" Score="6" Body="&lt;p&gt;For $\ce{AgNO3}$, as Gokul said in the comment, ‘nitrate’ stands for $\ce{NO3-}$. Sodium nitrate is $\ce{NaNO3}$, ammonium nitrate is $\ce{NH4NO3}$, and so on. It is important to know that $\ce{AgNO3}$ is an &lt;strong&gt;ionic compound&lt;/strong&gt; with &lt;strong&gt;ionic bond&lt;/strong&gt; between $\ce{Ag+}$ and $\ce{NO3-}$, while nitrogen and oxygen in $\ce{NO3-}$ are bonded with &lt;strong&gt;covalent bond&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are conventions for names of compounds. Some convention such as IUPAC name for organic molecule is strict and not so confusing. For ionic compound, there are rules but there are also irregularities. In this case for anions with oxygen, here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nitrite $\ce{NO2-}$, nitrate: $\ce{NO3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;phosphite $\ce{PO3^3-}$, phosphate: $\ce{PO4^3-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sulfite $\ce{SO3^2-}$, sulfate: $\ce{SO4^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, ion with the highest oxidation state of central atom got -ate name, while the second highest got -ite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;However, if central atom is in group VII, the naming convention strangely change to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chlorite $\ce{ClO2-}$, chlorate $\ce{ClO3-}$, perchlorate $\ce{ClO4-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bromite $\ce{BrO2-}$, bromate $\ce{BrO3-}$, perbromate $\ce{BrO4-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;There are also confusing things such as nitride ($\ce{N-}$), nitrile ($\ce{-CN}$), and nitrite ($\ce{NO2-}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practical, it is easier to memorize the name of common ion and search the internet for the name of uncommon ion, rather than messing with confusing rules. I do this all the time when I do chemistry researches. However, this tactics might not work if you have to take a chemistry exam.&lt;/p&gt;&#xA;" OwnerUserId="16455" LastEditorDisplayName="user7951" LastEditDate="2015-05-23T14:06:46.873" LastActivityDate="2015-05-23T14:06:46.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31891" PostTypeId="2" ParentId="31880" CreationDate="2015-05-23T13:14:48.870" Score="7" Body="&lt;p&gt;You should know that nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfate is a polyatomic ion with the molecular formula $\ce{SO4^{2-}}$;&#xA;Sulfite is a polyatomic ion with the molecular formula $\ce{SO3^{2-}}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$;&#xA; Nitrite is a polyatomic ion with the molecular formula $\ce{NO2^{-}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, the nomenclature of the polyatomic negative ions seems difficult. &#xA;When the name of the ion  ends in either -ite or -ate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The -ite ending indicates a low oxidation state.&lt;/li&gt;&#xA;&lt;li&gt;The -ate ending indicates a high oxidation state.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I agree with you that at the beginning, it seems difficult. But by practicing, you find this quite easy.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-05-23T13:14:48.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31895" PostTypeId="2" ParentId="31739" CreationDate="2015-05-23T13:42:38.553" Score="7" Body="&lt;p&gt;Yes, there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;hydrogen: $$E=\frac{1}{hc}R\left(\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;helium ion: $$E=\frac{1}{hc}R(2)^2\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the energy is equal, we get: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = 4\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with simple rearragement: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = \frac{1}{\left(\frac{x_1}{2}\right)^2}-\frac{1}{\left(\frac{x_2}{2}\right)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;br/&gt;&#xA;Here, one (obviously) possible answer of the equation is:&#xA;$$n_1 = \frac{x_1}{2}, n_2 = \frac{x_2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;First balmer line is $n_1=3, n_2=2$. One possible answer is $x_1=6, x_2=4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, I know that with single linear equation and two variables, we can get multiple answers in the case of real numbers (actually, infinite). However, quantum number n can only be positive integer. I don't think there are other answers here.&lt;/p&gt;&#xA;" OwnerUserId="16455" LastEditorUserId="7475" LastEditDate="2016-11-13T18:00:48.090" LastActivityDate="2016-11-13T18:00:48.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31897" PostTypeId="1" CreationDate="2015-05-23T13:49:04.380" Score="6" ViewCount="198" Body="&lt;p&gt;Often crystal structures found online from past published crystallography experiments are of the &lt;strong&gt;salt&lt;/strong&gt; of a molecule of biologic interest, as opposed to the freebase form of the molecule itself. For example, I have a CIF file for mescaline hydrobromide (&lt;a href=&quot;https://summary.ccdc.cam.ac.uk/structure-summary?pid=ccdc:1211316&quot;&gt;CCDC 1211316&lt;/a&gt;) at the moment that I would like to create a 3D model of mescaline from. Would removing the extra hydrogen atom on the amine group (i.e., changing the $\ce{NH_{3}^+}$ to $\ce{NH_2}$) and the free bromide ion suffice or will I also have to optimize the structure's geometry? If I do need to optimize it, or otherwise modify it beyond the procedure I just pointed out, please tell me how to do it with free software (Avogadro, for instance). &lt;/p&gt;&#xA;" OwnerUserId="876" LastActivityDate="2015-05-28T04:30:02.277" Title="How would one construct a 3D model of a molecule based on the crystal structure of its salts?" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31898" PostTypeId="1" CreationDate="2015-05-23T13:51:19.590" Score="1" ViewCount="107" Body="&lt;p&gt;Fish oil, is susposed   to suffer from rancidity(i.e. oxidation) when stored improperly , for example in hot temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in buying fish oil from abroad(usa) it will be shipped by air and won't be kept in a cold chain,i live in a mediternean nation(climate wise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should i choose my fish oil , such that when i recieve it ,it will be OK and not rancid ? is it even possible ? &lt;/p&gt;&#xA;" OwnerUserId="2576" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T16:51:27.073" LastActivityDate="2015-10-30T16:51:27.073" Title="Choosing fish-oil that won't oxidize in long travel?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31901" PostTypeId="2" ParentId="4649" CreationDate="2015-05-23T14:11:06.553" Score="4" Body="&lt;p&gt;Our teacher told us this trick to tell if complex is going to be square planar.&lt;br&gt;&#xA;We are considering the fact that the coordination no. is 4.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the metal has a $\ce{d^7}$, $\ce{d^8}$ or $\ce{d^9}$ configuration along with a &lt;strong&gt;strong field ligand&lt;/strong&gt; or $\ce{d^4}$ with &lt;strong&gt;weak field ligand&lt;/strong&gt; then complex will be square planar otherwise tetrahedral.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One exception i have come across for this is $\ce{[PtCl4]^2-}$ which is a square planar. You can read the explanation for it &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20080308120955AAjfZWx&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5674" LastActivityDate="2015-05-23T14:11:06.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31903" PostTypeId="2" ParentId="4649" CreationDate="2015-05-23T14:30:29.950" Score="1" Body="&lt;p&gt;A method I use requires the memorization of the spectrochemical series, which arranges the ligands in increasing order of ligand strength. A strong ligand will almost always be able to cause pairing of electrons, producing a low spin, inner orbital complex, which in case of the coordination number 4, may correspond to the $dsp^2$ or $d^3s$ hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance,&lt;br&gt;&#xA;$\ce{Ni^{+2}}$ produces a square planar complex with $\ce{CN-}$, as cyanide is a strong ligand. Nickel carbonyl, on the other hand, exists as a tetrahedral complex because nickel's oxidation is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You shall be able to use this method most of the time, but in certain cases, there are some exceptions. For instance, Cobaltic ions behave differently than other $3d$ metal ions, and form a low spin complex with $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorUserId="15410" LastEditDate="2015-05-23T17:41:07.677" LastActivityDate="2015-05-23T17:41:07.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31904" PostTypeId="1" AcceptedAnswerId="32833" CreationDate="2015-05-23T15:02:26.563" Score="11" ViewCount="347" Body="&lt;p&gt;Could reduction of an organic compound by reacting it with sodium dissolved in $\ce{NH3}$ still proceed if a metal such as aluminium were used instead of sodium? I'm not sure about the aluminium amine or amide structure that would be formed.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="865" LastEditDate="2015-10-16T17:51:11.770" LastActivityDate="2015-10-16T17:51:11.770" Title="Dissolving metal reductions" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;experimental-chemistry&gt;&lt;amines&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31905" PostTypeId="2" ParentId="31749" CreationDate="2015-05-23T15:12:42.223" Score="3" Body="&lt;p&gt;The hydrated forms might be an azeotrope which means there's a content of water present that is difficult to remove via distillation techniques which may be how the P-toluene sulfonic acid was prepared. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anhydrous form would require further processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moisture present in the hydrate forms may affect the melting point if it were acting as an azeotrope with the P-toluene sulfonic acid, also it wouldn't be as high purity as the anhydrous form which would display the true melting point.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="16111" LastEditDate="2015-10-26T16:15:21.660" LastActivityDate="2015-10-26T16:15:21.660" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31908" PostTypeId="2" ParentId="31749" CreationDate="2015-05-23T15:37:31.580" Score="1" Body="&lt;p&gt;The solubility of p-toluenesulfonic acid is $670 \mathrm{g}.\mathrm{L}^{-1}$ so the solution which contains this acid and water is an &lt;a href=&quot;http://en.wikipedia.org/wiki/Azeotrope&quot; rel=&quot;nofollow&quot;&gt;azeotrop&lt;/a&gt;. &#xA;Then the melting temperature depend if you keep pressure or temperature constant in the laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take a Handbook I have not here, I think you'll find a lot of interesting things about that.&lt;/p&gt;&#xA;" OwnerUserId="15235" LastActivityDate="2015-05-23T15:37:31.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31909" PostTypeId="1" AcceptedAnswerId="31916" CreationDate="2015-05-23T16:17:33.110" Score="4" ViewCount="12666" Body="&lt;p&gt;On the table published here (&lt;a href=&quot;http://www.ptable.com/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;) each cell has one to five numerals arranged vertically in its top right corner. The sum of those numbers is always the relative atomic mass of the element. I suspect that, in most cases, two of those numerals represent the number of protons and the number of neutrons, respectively. However, I have no idea what the other numerals represent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do the numerals  on the top right corner of the cells in the periodic table represent?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-05-23T18:47:48.923" Title="What do the numerals on the top right corner of the cells in the periodic table represent?" Tags="&lt;periodic-table&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31910" PostTypeId="2" ParentId="31909" CreationDate="2015-05-23T16:20:04.313" Score="0" Body="&lt;p&gt;The numbers represent which shell the electrons occupy in the ground configuration.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-23T16:20:04.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31911" PostTypeId="1" AcceptedAnswerId="31914" CreationDate="2015-05-23T17:22:59.207" Score="11" ViewCount="3034" Body="&lt;p&gt;In particular, how do a methylene more (in the case of the pentane) and a methylene less (in the case of the octane) make the difference in harmfulness to the environment in comparison with, respectively, the butane and the nonane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw that butane and nonane are not classified as dangerous, but I have just done a research and it turns out nonane is harmful to fish, just like octane, but the difference is that the nonane becomes aqueous more difficult because of its low solubility. But I read that its soluble part is still dangerous for aquatic life. Then the question is: Shouldn't it be classified as dangerous anyway?&lt;br&gt;&#xA;The propane, as well, appears to be harmful to the fish, while I can find nothing but &quot;not classified as dangerous&quot;, as for the butane. Is LDC3's answer the explanation?&lt;/p&gt;&#xA;" OwnerUserId="16067" LastEditorUserId="23561" LastEditDate="2019-04-20T23:06:28.963" LastActivityDate="2019-04-20T23:06:28.963" Title="What makes pentane, hexane, heptane, and octane dangerous, while butane and nonane aren't?" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;&lt;environmental-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31912" PostTypeId="2" ParentId="31911" CreationDate="2015-05-23T17:52:33.950" Score="2" Body="&lt;p&gt;Butane is a gas at STP, so it would not accumulate in one location.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-23T17:52:33.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31913" PostTypeId="2" ParentId="17259" CreationDate="2015-05-23T18:00:27.210" Score="6" Body="&lt;p&gt;Pressure is the number of collisions with the container per unit area. Imagine a particle in a box of length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. Assuming for the moment that it only moves in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-direction, whenever it collides with a wall of the container the wall will gain momentum from the particle: &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\Delta p &amp;amp;= p_{\mathrm{final}, x} - p_{\mathrm{initial}, x} \\&#xA;  &amp;amp;= p_{\mathrm{final}, x} - (- p_{\mathrm{final}, x})\\&#xA;  &amp;amp;= 2p_{\mathrm{final}, x}\\ &#xA;  &amp;amp;= 2mu_x,&#xA;\end{align}&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$u_x$&lt;/span&gt; is the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-component of velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particle hits a specific wall every &lt;span class=&quot;math-container&quot;&gt;$\Delta t = \frac{2L}{u_x}$&lt;/span&gt;, since it travels the length of the box twice. Force can be defined as momentum per unit time, therefore &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{\Delta p}{\Delta t} = \frac{mu_x^2}{L}$$&lt;/span&gt; &#xA;For &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; number of particles, the expression becomes &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{Nm\overline {u_x^2}}{L},$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$\overline{u_x^2}$&lt;/span&gt; is the mean square velocity of the &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we can extend the situation so that our particle can move in three directions. Since the box is cubic, the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-, &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;-, and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-components are equivalent: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \overline{u_x^2} = \frac{\overline{u^2} }{ 3}$$&lt;/span&gt; &#xA;Therefore, &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{Nm\overline {u^2}}{3L}.$$&lt;/span&gt; &#xA;The area of any wall is &lt;span class=&quot;math-container&quot;&gt;$L^2$&lt;/span&gt;, so &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;p &amp;amp;= \frac{F}{L^2}\\&#xA;  &amp;amp;= \frac{Nm\overline {u^2}}{3L^3}\\&#xA;  &amp;amp;= \frac{Nm\overline {u^2}}{3V}&#xA;\end{align}&lt;/span&gt; &#xA;This leads to the expression &#xA;&lt;span class=&quot;math-container&quot;&gt;$$pV = \frac{1}{3}Nm\overline{u^2}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://en.wikipedia.org/wiki/Kinetic_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16290" LastEditorUserId="4945" LastEditDate="2019-01-15T18:38:11.957" LastActivityDate="2019-01-15T18:38:11.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31914" PostTypeId="2" ParentId="31911" CreationDate="2015-05-23T18:35:43.673" Score="10" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WThBut.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WThBut.gif&quot; alt=&quot;GHS09&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pentane, hexane, heptane, and octane are labelled with the ‘environment’ hazard pictogram (GHS09) since they are considered &lt;em&gt;toxic&lt;/em&gt; or &lt;em&gt;very toxic to aquatic life with long-lasting effects&lt;/em&gt; (H411 or H410) because of their toxicity to the water flea &lt;em&gt;Daphnia magna&lt;/em&gt; in standardized tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, nonane is not completely harmless; it still &lt;em&gt;may cause long-lasting harmful effects to aquatic life&lt;/em&gt; (H413). However, a pictogram is not required for this hazard category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://dx.doi.org/10.1016/0045-6535(83)90118-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;a comparative study&lt;/a&gt;, the toxicity of these substances is mainly determined by their solubility in water and their organism–water partition coefficient.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-09-07T16:00:54.753" LastActivityDate="2015-09-07T16:00:54.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31916" PostTypeId="2" ParentId="31909" CreationDate="2015-05-23T18:47:48.923" Score="5" Body="&lt;p&gt;The numbers give you the basic ground state electron configuration of the atoms by telling you how many electrons are in each shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electron's 'position' in an atom can be specified by four quantum numbers: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Principal quantum number ($n$). This tells you which shell the electron is in and can take values $1,~2,~3,~4~...$. The maximum number of electrons in each shell is given by $2n^2$ so the first shell can hold 2 electrons, the second 8, the third 18, the fourth 32 etc. The numbers given in your periodic table are the total number of electrons in each shell in the ground state (lowest energy state) of that element.&lt;/li&gt;&#xA;&lt;li&gt;Azimuthal quantum number ($l$).&lt;/li&gt;&#xA;&lt;li&gt;Magnetic quantum number ($m_e$).&lt;/li&gt;&#xA;&lt;li&gt;Spin quantum number ($m_s$).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; wikipedia article for more detail on the meaning of quantum numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron configurations can be written in more detail using $s,p,d,f$ notation. See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31189/what-is-spdf-configuration&quot;&gt;this&lt;/a&gt; question for more detail.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-05-23T18:47:48.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31917" PostTypeId="1" CreationDate="2015-05-23T19:22:06.573" Score="0" ViewCount="2012" Body="&lt;p&gt;What is the entropy change when the temperature of 1.00 mol of ideal gas is raised reversibly from 10 °C to 50 °C at constant pressure, constant volume, and assuming the heat capacity is independent of temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all the information given. I am assuming that the formula $\Delta S = C \ln (T_2/T_1)$ is used. However, I do not know how the heat capacities are supposed to be obtained from the given information. I do not have the enthalpy or internal energy of the process. I cannot calculate the heat capacity from pressure or volume. &#xA;Please say if my reasoning is incorrect.&#xA;(These aren’t the values of the original question but this is based on Atkins &lt;em&gt;Chemical Principles&lt;/em&gt; 5th ed. Ch 8 # 5 and 6)&lt;/p&gt;&#xA;" OwnerUserId="9641" LastEditorUserId="865" LastEditDate="2021-05-22T21:08:48.770" LastActivityDate="2021-05-22T21:08:48.770" Title="Entropy increase when heating a gas" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31921" PostTypeId="1" AcceptedAnswerId="31993" CreationDate="2015-05-23T20:53:46.193" Score="1" ViewCount="621" Body="&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a cell set up between a copper metal/copper(II) ion electrode and a reference electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under standard conditions, the emf of this cell was −0.07 V. The standard electrode potential of the copper metal / copper(II) ion electrode is +0.34 V. Hence the standard electrode potential of the reference electrode is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: -0.41 V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: -0.27 V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: +0.27 V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D: +0.41 V&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;My attempt:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I was taught this equation: $\mathscr{E} _{cell} = \mathscr{E} _{a} - \mathscr{E} _{b}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathscr{E} _{a}$ is the emf of the more positive electrode potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathscr{E} _{b}$ is the emf of the less positive electrode potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But using this equation, none of the combinations of electrode potentials give -0.07V as the emf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5619" LastEditorUserId="418" LastEditDate="2016-11-08T03:55:44.573" LastActivityDate="2020-05-04T11:20:20.403" Title="What is the meaning of signs in electrode potentials?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31922" PostTypeId="2" ParentId="27698" CreationDate="2015-05-23T21:00:12.563" Score="8" Body="&lt;p&gt;This is liquid methane, which boils at a temperature of −161 °C. (Methane can be condensed in a cold trap that is cooled with liquid nitrogen.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-23T21:00:12.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31923" PostTypeId="1" CreationDate="2015-05-24T00:51:26.030" Score="1" ViewCount="88" Body="&lt;p&gt;If ferric chloride was added to a water sample which had calcium bicarbonate alkalinity present, would the reaction be a ‘stoichiometry reaction’ or a ‘limiting reagent reaction’?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if 1 g of ferric chloride was added, how much precipitate would form [$\ce{Fe(OH)3}$] and what type of reaction (mentioned above) would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2FeCl3(aq) + 3Ca(HCO3)2(aq) -&amp;gt;2Fe(OH)3(s) + 3CaCl2(aq) + 6CO2(g)}$$&lt;/p&gt;&#xA;" OwnerUserId="16472" LastEditorDisplayName="user7951" LastEditDate="2015-05-24T08:07:37.283" LastActivityDate="2015-05-24T08:07:37.283" Title="Stoichiometry reactions for water clarification" Tags="&lt;stoichiometry&gt;&lt;precipitation&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31924" PostTypeId="1" CreationDate="2015-05-24T01:22:03.963" Score="13" ViewCount="47841" Body="&lt;p&gt;The Gattermann-Koch Reaction involves the formation of an aryl aldehyde using carbon monoxide and hydrochloric acid as reactants. The reaction is catalyzed by aluminium trichloride in the presence of trace amounts of cuprous chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jTMRF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand about this reaction is that according to Wikipedia along with other online texts I've used, the Gattermann-Koch reaction is inapplicable to phenol substrates. The drawback has been left unexplained by &lt;em&gt;all&lt;/em&gt; my resources (online and textbooks) and even my professor was oblivious to it.&#xA;I am unable to understand why the reaction doesn't apply to phenol and what exact side reactions might occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, why is the explanation so obscure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] I found the explanation. The Gattermann-Koch Reaction is not applicable to phenol and phenol ethers because they &quot;could not be successfully formylated at atmospheric pressure in benzene as a solvent.&quot;&#xA;&quot;This failure to react was attributed to the insolubility of the cuprous chloride in the reaction mixture.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain this in simple terms?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="10185" LastEditDate="2016-05-30T21:49:03.573" LastActivityDate="2017-10-25T14:01:51.210" Title="Gattermann-Koch Reaction with Phenol Substrates" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;phenols&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31926" PostTypeId="1" AcceptedAnswerId="31928" CreationDate="2015-05-24T03:43:32.110" Score="3" ViewCount="1265" Body="&lt;p&gt;The delocalized electrons in a metal move about because of the electrostatic attractions between them and many different positive ion cores. But why does the same not occur with the ion cores themselves? Why are electrons, which bear the same magnitude of charge as nuclei, mobile, while the nuclei themselves are not?&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="15465" LastEditDate="2015-05-24T04:03:28.740" LastActivityDate="2015-05-24T04:25:24.583" Title="Why do ion cores not move in a metal?" Tags="&lt;bond&gt;&lt;atoms&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31928" PostTypeId="2" ParentId="31926" CreationDate="2015-05-24T04:25:24.583" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The crystal structure of the metal holds the atom in place. BTW, since electrons are &lt;em&gt;shared&lt;/em&gt; in a &quot;sea&quot; of electrons or conductivity band, no one atom is necessarily ionized. The crystal structure is also important for nuclear physics, e.g. using &lt;a href=&quot;http://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment&quot;&gt;Mössbauer spectroscopy to measure gravitational effects&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Individual electrons don't actually move around the whole circuit, even in DC applications. Instead, think of it as a large stack of electrons; as one is pushed on the stack at one end, electromagnetic forces pop one off the other end. Though electrons do move a bit, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Drift_velocity&quot;&gt;drift velocity&lt;/a&gt; is much less than the speed of light (in fact, it is somewhat slower than the last-placed snail in a race).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compare the mass of a single electron and the mass of an atom, or the mass of an &lt;em&gt;ion in a plasma&lt;/em&gt;. Note that acceleration of a particle is inversely dependent on its mass. This is important in fusion research, where confining electrons and ions is a bit tricky.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See this &lt;a href=&quot;http://www2.phy.ilstu.edu/~marx/ph355/06_Metals.ppt&quot;&gt;PPT presentation&lt;/a&gt; on metallic conduction.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-24T04:25:24.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31929" PostTypeId="2" ParentId="31924" CreationDate="2015-05-24T05:01:00.680" Score="8" Body="&lt;p&gt;You have outlined that the &lt;strong&gt;formyl chloride&lt;/strong&gt; is generated as the reactive, electrophilic intermediate under these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2017-05-03T21:19:57.587" LastActivityDate="2017-05-03T21:19:57.587" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31930" PostTypeId="1" CreationDate="2015-05-24T06:02:37.973" Score="3" ViewCount="90" Body="&lt;p&gt;I have had two types of urea fertiliser analysed, the results are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{46\%N}$ (46% nitrogen) - which is sold as a urea fertiliser, tests at 46.1% ureic nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{38\%N~ 19\%S}$ (38% nitrogen, 19% sulphur) - which I understand to contain ammonium sulphate and urea, tests at 33% nitric nitrogen and 7.4% ammoniacal nitrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is-  why does the $\mathrm{38\%N~ 19\%S}$ show no ureic nitrogen when it is supposed to contain urea - the same as the $\mathrm{46\%N}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit - the analysis company has admitted a mistake- the test  result should have read 33% ureic nitrogen and not 33% nitric nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="16475" LastEditorUserId="24052" LastEditDate="2016-03-13T20:29:33.863" LastActivityDate="2016-03-13T20:29:33.863" Title="Fertiliser analysis - Why doesn't it show ureic nitrogen when it does have urea?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ClosedDate="2016-03-14T02:39:19.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="31931" PostTypeId="2" ParentId="29044" CreationDate="2015-05-24T06:38:48.790" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;In general: $\ce{acid + H2O &amp;lt;=&amp;gt; base + H_3O^+}$&lt;/li&gt;&#xA;&lt;li&gt;It is called the Henderson-Hasselbalch equation &#xA;$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$&lt;/li&gt;&#xA;&lt;li&gt;I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="4945" LastEditDate="2015-05-26T09:43:55.827" LastActivityDate="2015-05-26T09:43:55.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31932" PostTypeId="1" AcceptedAnswerId="31935" CreationDate="2015-05-24T06:42:04.053" Score="7" ViewCount="11668" Body="&lt;p&gt;This user absolutely insists that when a chloride ion is present in water (for example, when $\ce{NaCl}$ dissolves in water) that $\ce{Cl-}$ ion is ionically bound to the hydrogen atom:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55324_34449&quot;&gt;https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55324_34449&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then insist that this would form $\ce{H2OCl}$, but that's impossible due to the lack of valence electrons of $\ce{H}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55386_34449&quot;&gt;https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55386_34449&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right?&lt;/p&gt;&#xA;" OwnerUserId="2355" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:48:14.547" LastActivityDate="2015-05-24T07:56:21.867" Title="Does water ionically bond to chloride ion?" Tags="&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31933" PostTypeId="1" CreationDate="2015-05-24T06:57:21.397" Score="3" ViewCount="843" Body="&lt;p&gt;I had a fairly simple question regarding supersaturation and crystallization. When a supersaturated solution is crystallized, is &lt;em&gt;all&lt;/em&gt; of the solute crystallized or just the excess solute (such that the solution remains saturated)? After crystallization, would you expect to find some quantity of the solute still dissolved? &lt;/p&gt;&#xA;" OwnerUserId="16476" LastActivityDate="2015-06-09T21:49:28.557" Title="Supersaturation and Crystallization" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31935" PostTypeId="2" ParentId="31932" CreationDate="2015-05-24T07:16:34.350" Score="8" Body="&lt;p&gt;The most common mechanism for when $\ce{NaCl}$ dissolves is as folows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As water is a polar molecule, as per the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_Lower%27s_Chem1/M2%3A_All_About_Water&quot; rel=&quot;nofollow noreferrer&quot;&gt;All About Water&lt;/a&gt;, shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k1ql9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical &quot;image&quot; of a water molecule.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, when $\ce{NaCl}$ salt is dissolved in water, the sodium and chloride ions are pulled apart by the water molecules with the result explained in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAQvZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grandinetti.org/solution-chemistry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll note, that there is no ionic (nor covalent) bonding present, rather an &lt;strong&gt;ion-dipole interaction&lt;/strong&gt; between the water molecules and the sodium and chloride ions (&lt;a href=&quot;http://www.science.uwaterloo.ca/~cchieh/cact/applychem/hydration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Waterloo&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UCLA page &lt;a href=&quot;http://www.chem.ucla.edu/harding/IGOC/I/ion_dipole_interaction.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Illustrated Glossary of Organic Chemistry&lt;/a&gt; has a very nice definition of the ion-dipole interaction:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A noncovalent attraction between one pole of a bond dipole and an oppositely-charged ion. The δ+ end of the bond dipole is attracted to an anion, and the δ- end of the bond dipole is attracted to a cation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a neat YouTube clip showing the process &lt;a href=&quot;https://www.youtube.com/watch?v=xdedxfhcpWo&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ion $\ce{H2OCl+}$, does actually exist according to &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=B1343&amp;amp;Mask=20&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST&lt;/a&gt;, but I could not find any references that link it to the dissolution of $\ce{NaCl}$ salt. &lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-24T07:56:21.867" LastActivityDate="2015-05-24T07:56:21.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31937" PostTypeId="1" CreationDate="2015-05-24T12:51:09.817" Score="8" ViewCount="323" Body="&lt;p&gt;Several organic mechanisms I know involve [2+2] cycloadditions and retro-[2+2] cycloadditions, namely:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Wittig reaction&lt;/li&gt;&#xA;&lt;li&gt;Hydroboration of an alkene&lt;/li&gt;&#xA;&lt;li&gt;Olefin metathesis (using Grubbs' catalyst)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How are these in accordance with the Woodward–Hoffmann rules? I think that there are two $(4q+2)_s$ components and that's not an odd number, which is the criterion for a reaction to be thermally allowed.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2017-08-17T11:30:16.763" LastActivityDate="2017-08-17T11:30:16.763" Title="Grubbs, Wittig &amp; hydroboration reactions and Woodward-Hoffmann rules" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;pericyclic&gt;" AnswerCount="0" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31938" PostTypeId="1" CreationDate="2015-05-24T14:23:06.633" Score="2" ViewCount="69" Body="&lt;p&gt;Am giving presentation over Coenzyme Q10 Wednesday for my final project, but we haven't studied any biochem this year.  So questions have arisen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CoQ10 has a Quinone head, an isoprenyl tail, and these additional oxygens on the left and three methyl groups not a part of the Quinone head or tail.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for these two oxygens and three methyl groups?  Or a category?  Like the benzene ring and two oxygens is the Quinone head, what are the additional oxygens and methyl groups called, if anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CoQ10 diagram: &lt;a href=&quot;http://www.mysupplementstore.com/media/wysiwyg/structure-of-CoQ10.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.mysupplementstore.com/media/wysiwyg/structure-of-CoQ10.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16479" LastEditorUserId="5668" LastEditDate="2015-05-24T17:33:44.250" LastActivityDate="2015-05-24T17:33:44.250" Title="Name of these parts in CoQ10?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31939" PostTypeId="1" AcceptedAnswerId="31946" CreationDate="2015-05-24T14:53:05.317" Score="4" ViewCount="240" Body="&lt;p&gt;In E2 reaction, there is more than one product. According to Zaitsev’s rule – “The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.” (Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is, how can the base effect the percentage of the minor and major products? Does a stronger base give us more major product and vice versa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular question I’m trying to solve is:&#xA;In the E2 reaction of 2-bromo-2,3-dimethyl butane, which base would give more of the minor product and which of the major product?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lkFjE.jpg&quot; alt=&quot;the four given bases&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16328" LastEditorDisplayName="user7951" LastEditDate="2015-05-24T15:47:33.000" LastActivityDate="2015-05-24T16:28:41.647" Title="E2 reaction - strong base" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31940" PostTypeId="2" ParentId="31938" CreationDate="2015-05-24T15:04:06.097" Score="1" Body="&lt;p&gt;The additional groups are referred as methyl and methoxy groups. There is no category for these groups. It appears that the methoxy groups stabilizes the &lt;a href=&quot;http://en.wikipedia.org/wiki/Ubiquinol#Molecular_aspects&quot; rel=&quot;nofollow&quot;&gt;intermediates&lt;/a&gt; of the oxidation or reduction reactions. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-24T15:04:06.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31942" PostTypeId="2" ParentId="31872" CreationDate="2015-05-24T15:40:31.720" Score="2" Body="&lt;p&gt;&quot;Acidity&quot; can't be responsible, since acidity refers to what happens in water solution. (I assume the user is referring to the $\mathrm{p}K_\mathrm{a}$ data.) I don't know how great the tendency of $\ce{HCN}$ is to ionize in the pure liquid, but I doubt it's significant. But if you have data, please cite it. Even for water, $K_\mathrm{w} = 10^{-14}$, its tendency to ionize does not contribute to its high boiling point.&lt;/p&gt;&#xA;" OwnerUserId="16480" LastEditorUserId="4945" LastEditDate="2015-05-25T07:21:27.553" LastActivityDate="2015-05-25T07:21:27.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31943" PostTypeId="1" AcceptedAnswerId="31945" CreationDate="2015-05-24T15:40:34.133" Score="1" ViewCount="2929" Body="&lt;p&gt;Will an E2 reaction that has 2 potential products go faster than one that has 3 potential products? The specific question involves both molecules below reacting with methanol. I figured it's an E2 reaction (I hope I'm right) and the question is, which one will go faster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCQ8v.png&quot; alt=&quot;cyclohexylbromides&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16328" LastEditorUserId="7475" LastEditDate="2015-10-09T16:01:24.437" LastActivityDate="2015-10-09T16:01:24.437" Title="E2 reaction — which one is faster?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31944" PostTypeId="1" CreationDate="2015-05-24T15:47:16.217" Score="1" ViewCount="14299" Body="&lt;p&gt;I've already answered this question. All I want to know is if my reasoning is correct.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Rb+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Na+} \leftarrow$&lt;/span&gt; This ion is more strongly favored&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^{2+}} \leftarrow$&lt;/span&gt; This ion is more strongly favored&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Reasoning:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;For the first case, my guess would have to be because the electrons are much closer to the center of the nucleus (the atomic radius of the atom is very small) and makes it much easier to interact the positive dipole end of the water molecule (the hydrogen molecules).&lt;/p&gt;&#xA;&lt;p&gt;For the second case, the reason why the magnesium ion is more favored is because even though both ions have roughly the same molar mass, one is more positively charged than the other and thus is more likely to attract the negative dipole end of the water molecule (the oxygen molecule).&lt;/p&gt;&#xA;&lt;p&gt;Is my reasoning justified or is there something in particular that I'm not paying attention to?&lt;/p&gt;&#xA;" OwnerUserId="16301" LastEditorUserId="69830" LastEditDate="2021-05-24T06:37:01.160" LastActivityDate="2021-05-24T06:37:01.160" Title="Explain why these ions are more strongly hydrated in an aqueous solution?" Tags="&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31945" PostTypeId="2" ParentId="31943" CreationDate="2015-05-24T16:18:23.140" Score="3" Body="&lt;p&gt;For the E1 pathway, molecule A will react faster since the intermediate is more stable than from molecule B. According to this &lt;a href=&quot;http://www.masterorganicchemistry.com/2012/10/10/comparing-the-e1-and-e2-reactions/&quot; rel=&quot;nofollow&quot;&gt;web site&lt;/a&gt;, the major product for A is 1,2-dimethylcyclohexene; for B, it is 1-methylcyclohexene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the E2 pathway, both reactions will have nearly the same rate. Depending on the orientation of the α-methyl group (cis or trans to Br), the major product is the same for cis, but for trans it will be 2,3-dimethylcyclohexene for A and 2-methylcyclohexene for B.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-05-25T13:19:08.460" LastActivityDate="2015-05-25T13:19:08.460" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31946" PostTypeId="2" ParentId="31939" CreationDate="2015-05-24T16:28:41.647" Score="5" Body="&lt;p&gt;It will be more difficult for sterically hindered bases to remove the proton of the β-carbon having the fewest hydrogen substituents, since this carbon atom has more bulky groups around it compared to the the others. So base 3, being the less hindered base, will provide a higher percentage of the major Zaitsev’s product and base 2 will provide a higher percentage of the minor product compared to the other bases.&lt;/p&gt;&#xA;" OwnerUserId="16481" LastActivityDate="2015-05-24T16:28:41.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31947" PostTypeId="1" AcceptedAnswerId="31964" CreationDate="2015-05-24T16:34:28.100" Score="7" ViewCount="4461" Body="&lt;p&gt;How do you know the number of ligands surrounding a metal ion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: the obtained complex of $\ce{Ni^2+}$ ion in $\ce{NH3}$ solution is written as $\ce{[Ni(NH3)6]^2+}$ but $\ce{Cu^2+}$ in $\ce{NH3}$ solution is $\ce{[Cu(NH3)4]^2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you find the number for different metal ions like $\ce{Ag^2+}$?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="5026" LastEditDate="2018-03-01T05:44:27.710" LastActivityDate="2018-03-01T05:44:27.710" Title="Calculating the number of ligands of complex ions in d- block" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31949" PostTypeId="1" CreationDate="2015-05-24T16:47:10.077" Score="4" ViewCount="6556" Body="&lt;p&gt;My teacher said that fluorine is more electronegative than oxygen thus it will take electrons of oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now, why won't other Group 17 elements, like chlorine, bromine, iodine, would take electrons from oxygen? They too have just 7 electrons in their valence shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, they should but yet they have lesser electronegativity and this is something that I'm not able to digest. &lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorDisplayName="user15489" LastEditDate="2015-05-25T07:53:13.043" LastActivityDate="2015-05-26T21:20:32.160" Title="Why any other element except fluorine wouldn't take electrons from oxygen?" Tags="&lt;electrons&gt;&lt;oxidation-state&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31950" PostTypeId="1" CreationDate="2015-05-24T16:56:45.923" Score="42" ViewCount="6959" Body="&lt;p&gt;Is it possible for an atom to bond with 8 other elements (same or other type)? If yes, then please give some examples. If no, then what could be the possible reason for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is not about valence shell electrons. My question is about whether an element could combine with more than 8 other &lt;em&gt;same or different&lt;/em&gt; elements? It could be rather covalent bonds or any bond. &lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="41328" LastEditDate="2017-09-27T21:24:54.350" LastActivityDate="2019-01-03T23:41:52.397" Title="Can an atom bond with more than 8 other atoms?" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;coordination-compounds&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="31951" PostTypeId="1" CreationDate="2015-05-24T17:04:56.543" Score="4" ViewCount="111" Body="&lt;p&gt;In the ene-yne metathesis how to know which product will form predominately, as the mechanisms to both are plausible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLrlr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2019-01-13T14:59:21.267" LastActivityDate="2019-01-13T14:59:21.267" Title="Ene-yne metathesis product" Tags="&lt;organic-chemistry&gt;&lt;organotransition-metal-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31952" PostTypeId="1" CreationDate="2015-05-24T17:31:56.927" Score="1" ViewCount="1253" Body="&lt;p&gt;As the question suggests I am wondering the &lt;strong&gt;difference&lt;/strong&gt; between standard electrode potentials and standard reduction potentials. I know that they are &lt;strong&gt;both a measure of the Voltage&lt;/strong&gt; from certain reactions, and can be used to calculate the &lt;strong&gt;feasibility&lt;/strong&gt; of a reaction - but when do you use which / what is the difference, or are they &lt;strong&gt;interchangeable&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10189" LastActivityDate="2015-05-24T17:31:56.927" Title="What is the Difference between difference between Standard Electrode Potentials and Standard Reduction Potentials?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;reduction-potential&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31953" PostTypeId="2" ParentId="31950" CreationDate="2015-05-24T18:05:42.147" Score="40" Body="&lt;p&gt;Yes, there are coordination complexes of large elements which have coordination numbers greater than eight. Some examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[ReH9]^2-}$ with a tricapped trigonal prismatic structure. The nine hydride ligands are small enough to fit around the relatively large rhenium atom fairly easily. This ion can be isolated as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_nonahydridorhenate&quot; rel=&quot;noreferrer&quot;&gt;potassium salt&lt;/a&gt; $\ce{K2ReH9}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dLvTB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[U(NO3)6]^2-}$ with an impressive coordination number of 12, consisting of six nitrate ligands each forming two coordinate bonds to the central uranium ion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Trve9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Trve9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="41328" LastEditDate="2017-09-28T06:26:26.690" LastActivityDate="2017-09-28T06:26:26.690" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31955" PostTypeId="1" AcceptedAnswerId="31956" CreationDate="2015-05-24T18:23:48.810" Score="1" ViewCount="5335" Body="&lt;p&gt;To the best of my knowledge, energy equals work, $\mathrm{E = W}$; work equals force multiplied by distance, $\mathrm{W = Fm}$ ; force equals mass multiplied by acceleration,  $\mathrm{F = MA}$; and acceleration equals distance per second squared, $\mathrm{A = m \setminus s^2}$. However, when I substitute those values in for the dimensions of $ \mathrm{W}$, I can't transform the result of those substitutions into  $ \mathrm{W= \frac{1}{2}Mv^2}$. Perhaps I've done the algebra incorrectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the result of those substitutions transform into $ \mathrm{W= \frac{1}{2}Mv^2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't transform into that equation, why doesn't it do so, and why do we use an equation that isn't compatible with our other equations? &lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="10185" LastEditDate="2015-05-24T19:25:43.977" LastActivityDate="2015-05-25T01:51:25.877" Title="Why is it the case that KE = 0.5mv^2" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31956" PostTypeId="2" ParentId="31955" CreationDate="2015-05-24T19:16:15.147" Score="6" Body="&lt;p&gt;The kinetic energy $E_\text{k}$ equals the work $W$ done on an object:&#xA;$$E_\text{k}=W$$&#xA;The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$:&#xA;$$W=Fs$$&#xA;The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$:&#xA;$$F=ma$$&#xA;For uniform acceleration $a$, the final velocity $v$ is&#xA;$$v=at$$&#xA;and the distance $s$ is&#xA;$$s=\frac{1}{2}vt=\frac{1}{2}at^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$\begin{align}&#xA;E_\text{k}&amp;amp;=W\\&#xA;&amp;amp;=Fs\\&#xA;&amp;amp;=mas\\&#xA;&amp;amp;=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\&#xA;&amp;amp;=\frac{1}{2}mv^2&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-24T19:16:15.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31957" PostTypeId="1" AcceptedAnswerId="31958" CreationDate="2015-05-24T19:43:18.033" Score="15" ViewCount="56000" Body="&lt;p&gt;When I heat up a balloon, does the air inside the balloon increase in pressure as well as volume?  I thought pressure and volume were inversely proportional?  Or does pressure and volume increase as temperature increases?&lt;/p&gt;&#xA;" OwnerUserId="16484" LastEditorUserId="5026" LastEditDate="2018-06-26T02:54:50.277" LastActivityDate="2018-06-26T02:56:07.880" Title="When I heat up a balloon, does the air inside increase in pressure as well as volume?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31958" PostTypeId="2" ParentId="31957" CreationDate="2015-05-24T20:03:22.103" Score="14" Body="&lt;p&gt;If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gas were &lt;em&gt;completely free&lt;/em&gt; to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force per area (gas pressure) as the force per area acting on it (ambient pressure), so that the forces cancel out and the piston stops moving. Here, a temperature increase in the gas would translate solely to a &lt;em&gt;volume&lt;/em&gt; increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gas were confined in a &lt;em&gt;perfectly rigid box&lt;/em&gt;, then an increase in gas temperature would cause the molecules inside to bump harder against the inner surfaces, but to no avail, as the walls do not budge and the box stays exactly the same size. Here, a temperature increase in the gas would translate solely to a &lt;em&gt;pressure&lt;/em&gt; increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a balloon, &lt;em&gt;the gas is free to expand, but not completely free&lt;/em&gt;. In other words, it's a situation somewhere between the two described above. The balloon skin is an elastic which pulls itself, creating a force vector pointing towards the interior of the balloon, and the more the skin is stretched, the stronger the force becomes. Now the gas inside the balloon has to create enough pressure to compensate for both the ambient pressure, &lt;em&gt;and&lt;/em&gt; the elastic force trying to pull the balloon skin inwards. This means that after heating, the gas inside a balloon will expand since the balloon is not perfectly rigid, but the equilibrium pressure of the gas inside the balloon will be higher than before because the balloon is pressing against a more tightly stretched balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be provable experimentally without much difficulty. Take a rubber balloon, preferably one with as thin of a membrane as you can get, and open its mouth to the atmosphere. Then, clamp it shut without blowing any air inside (in reality, you might have to blow some air in to unstick the rubber walls, but then let all the excess air out). The air inside will have a pressure exactly equal to ambient pressure, because the elastic is not being stretched. Now cover the balloon completely in warm water for a few minutes, and it should inflate slightly. Finally, remove the balloon from the water and quickly perforate it with a sharp object. You might be able to hear a small pop, and feel a rush of air. Both are an indication that pressure inside the heated balloon was higher than ambient pressure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Here is a very related Physics.SE question - &lt;a href=&quot;https://physics.stackexchange.com/questions/127695/why-is-the-pressure-inside-a-soap-bubble-higher-than-outside&quot;&gt;Why is the pressure inside a soap bubble higher than outside?&lt;/a&gt; which deals with bubbles rather than balloons. The diagram and equations are applicable in both cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Evidently, taking into account the elasticity of the balloon requires a more subtle treatment of the problem. Take a look at Floris' interesting input in the comments. It seems that even an idealized balloon starts off acting as a quasi-rigid container. After reaching a maximum pressure, the balloon starts expanding, and from there the walls get weaker as expansion continues. This means that for a range of temperatures higher than some critical value, &lt;strong&gt;the pressure of the gas actually decreases as the temperature increases&lt;/strong&gt;. The interior pressure of the balloon will still always be greater than ambient pressure, though.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="5026" LastEditDate="2018-06-26T02:56:07.880" LastActivityDate="2018-06-26T02:56:07.880" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31961" PostTypeId="2" ParentId="31950" CreationDate="2015-05-24T20:46:05.150" Score="31" Body="&lt;p&gt;&lt;strong&gt;14 coordination&lt;/strong&gt; is claimed in $\ce{U(BH4)4}$ (&lt;a href=&quot;https://books.google.com/books?id=STxHXRR4VKIC&amp;amp;pg=PA268&amp;amp;lpg=PA268&amp;amp;dq=U%28BH4%294%20coordination%20number&amp;amp;source=bl&amp;amp;ots=EG_sKNb0ZU&amp;amp;sig=v-F-RlY3o3UNkdHDvIGVQa3d0uI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=jyhiVcq-B8LfsAXjzoDwAw&amp;amp;ved=0CDMQ6AEwAw#v=onepage&amp;amp;q=U(BH4)4%20coordination%20number&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;ref_1&lt;/a&gt;, p. 268).  The molecule exists as a polymer in the solid state.  Six hydrogens from two of the $\ce{BH4}$ groups bond between the boron and uranium (a bridge bond).  Two hydrogens from each of the two remaining $\ce{BH4}$ groups also bridge bond to uranium; the other two hydrogens bond to an adjacent uranium atom to build the polymer.  So far this takes us to 10 hydrogens (6+4) bonding to the central uranium. Two more hydrogens in two $\ce{BH4}$ groups in two adjacent uraniums also bond to the initial uranium, just as described above, as part of the polymer extension.  These 4 additional bonds to hydrogen take the coordination number to 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coordination number 15&lt;/strong&gt; has been found in $\ce{Th(H3BNMe2BH3)4}$, and it is suggested that the molecule would have a coordination number of 16 in the gas phase (&lt;a href=&quot;http://www.unige.ch/sciences/chifi/publis/refs_pdf/ref01035.pdf&quot; rel=&quot;noreferrer&quot;&gt;ref_2&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tyVSC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image source: see ref_2 above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 7 short $\ce{B-Th}$ contacts and one long $\ce{B-Th}$ contact.  Each short contact provides 2 hydrogens that bond to thorium, while the long contact provides 1 hydrogen for bonding to thorium, making for a total of 15 $\ce{Th-H}$ bonds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-24T20:46:05.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31963" PostTypeId="2" ParentId="31944" CreationDate="2015-05-24T22:25:42.060" Score="5" Body="&lt;p&gt;I believe your reasoning is correct (more or less). Solvation is influenced by charge to size ratios. Higher the charge to size ratio, the easier it is for the species to be solvated by water molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply this idea to both your cases. (note: in the second case, I don't get your argument about molar mass. The idea is much simpler, magnesium's atomic radius is smaller than sodium's and it has a higher positive charge. What can we conclude from this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, reconsider the first case: a positive ion would not be attracted to the positive end of a dipole.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="16088" LastEditDate="2015-05-25T09:52:11.917" LastActivityDate="2015-05-25T09:52:11.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31964" PostTypeId="2" ParentId="31947" CreationDate="2015-05-24T22:30:21.320" Score="2" Body="&lt;p&gt;To my knowledge, there is no &quot;rule&quot; to predict the coordination number of a given metal ion.  There are, however, guidelines, which are nicely summarized &lt;a href=&quot;http://faculty.uml.edu/ndeluca/84.334/topics/topic6.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Larger ions can accommodate more ligands&lt;/li&gt;&#xA;&lt;li&gt;Bulky ligands will reduce the overall coordination number&lt;/li&gt;&#xA;&lt;li&gt;Highly charged ions will tend to accept a greater number of Lewis Bases&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With your example, we can make an educated guess as to the coordination number of Ag&lt;sup&gt;2+&lt;/sup&gt; using &lt;a href=&quot;http://chem230.wikia.com/wiki/Trend_4.7&quot; rel=&quot;nofollow&quot;&gt;periodic trends&lt;/a&gt;.  It is in the same group as copper and will be bigger since it is lower on the periodic table.  Based on the three points above, I would predict that Ag&lt;sup&gt;2+&lt;/sup&gt; could have a coordination number greater than 4.  It &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/zaac.200900238/abstract&quot; rel=&quot;nofollow&quot;&gt;turns out&lt;/a&gt; that (distorted) octahedral coordination of the silver dication is not all that uncommon.&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2015-05-24T22:30:21.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31965" PostTypeId="2" ParentId="31947" CreationDate="2015-05-25T00:00:02.633" Score="5" Body="&lt;p&gt;1st-row $d$-element + 2nd-row $p$-element - usually octahedron except for $\ce{Ni(II),Cu(II),Co(I)}$ (often a square or a distorted octahedron with two weakly bound ligands) and $\ce{Zn}$ (usually a tetrahedron). 1st-row $d$-element + 3+-row $p$-element - usually a tetrahedron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd/3rd row $d$-element are somewhat more complicated as they allow higher coordination number up to 9 ($\ce{ReH9^{2-}}$), but the situation is so much influenced by electron configuration, ligands and metal-metal interactions here that simple rules loose a lot of meaning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the coordination number is usually a compromise between 18-electron rule and size of ligands. For example, $\ce{V(CO)6}$ should dimerize, giving c.n. of 7 to vanadium, but can't do so because of size constrains. Somewhat similar happens with bulky alcoholats of titanium, that tend to include $\ce{Ti(IV)}$ in octahedral coordination, but lower c.n. is observable in case of tritox (tris-tert-butyl) alcoholat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it is generally not recommended to guess the c.n., it is always better to google for some reference compounds.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorDisplayName="user15489" LastEditDate="2015-05-25T07:36:48.710" LastActivityDate="2015-05-25T07:36:48.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31966" PostTypeId="2" ParentId="31957" CreationDate="2015-05-25T00:51:52.337" Score="12" Body="&lt;p&gt;You may recall the &lt;a href=&quot;//en.wikipedia.org/wiki/Ideal_gas_law&quot;&gt;ideal gas law&lt;/a&gt;: $$PV = nRT.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $P$ is pressure, $V$ is volume, $n$ is the amount of gas present (in moles), $R$ is the ideal gas constant, and $T$ is temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an enclosed system, with no gas flowing in or out, $n$ is constant (as is also, obviously, $R$).  We can rearrange the equation above to pull all the constant terms to one side, like this: $$\frac{PV}{T} = nR = \text{const}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we can see that, with $nR$ staying constant, the left-hand side of the equation, $\frac{PV}{T}$, must also stay constant.  Thus, the &lt;em&gt;three&lt;/em&gt; quantities $P$, $V$ and $\frac1T$ are inversely related to each other: if you increase one of them, at least one of the other two must decrease so that their product remains constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having noted this, it should be obvious what's happening with your balloon.  By heating it, you're increasing the temperature $T$ (and thus decreasing $\frac1T$); to keep $\frac{PV}{T}$ constant, at least one of $P$ and $V$ must therefore also increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, in a rubber balloon, &lt;em&gt;both&lt;/em&gt; the pressure and the volume will typically increase, since any increase in pressure will cause the balloon skin to expand, increasing the volume (and reducing the pressure) until the inward force exerted by the stretched balloon skin is balanced by the pressure of the gas inside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a mylar balloon, however, things would be different, since mylar is not particularly elastic.  In that case, starting with a partially inflated balloon, the pressure would remain approximately constant (and equal to ambient pressure) up to the point where the mylar skin was stretched taut; at that point, the volume would stop increasing, and the pressure would start to build up instead (potentially up to the point where the balloon would burst).&lt;/p&gt;&#xA;" OwnerUserId="283" LastActivityDate="2015-05-25T00:51:52.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31967" PostTypeId="2" ParentId="31955" CreationDate="2015-05-25T01:51:25.877" Score="5" Body="&lt;p&gt;@Loong's answer is perfect but I just want to show another method which can be used even if the acceleration is &lt;strong&gt;not&lt;/strong&gt; uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$$&#xA;Hence $$a=v\frac{dv}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is mass times acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$F=ma=mv \frac{dv}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a small distance $dx$ moved, the work done by the force is $F \times dx$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the Work-Energy Theorem, this is also the gain in kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$\Delta KE = \int_{x_1}^{x_2} Fdx = \int_{x_1}^{x_2} (mv \frac{dv}{dx}) dx = \int_{v_1}^{v_2}mvdv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $$\Delta KE = \frac{1}{2}m{v_2}^2 - \frac{1}{2}m{v_1}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the body starts at rest $v_1=0$ and we get the kinetic energy as $\frac{1}{2}mv^2$&lt;/p&gt;&#xA;" OwnerUserId="13357" LastActivityDate="2015-05-25T01:51:25.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31968" PostTypeId="2" ParentId="31950" CreationDate="2015-05-25T04:02:45.627" Score="16" Body="&lt;p&gt;There has been quite some interesting work by my former co-worker and my supervisor on metal-rich molecule with the co-ordination higher than eight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KkVpI.png&quot; alt=&quot;Chem. Eur. J., 2010, 16, 13372&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more see &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/chem.201001650/abstract&quot; rel=&quot;noreferrer&quot;&gt;Timo Bollermann, Thomas Cadenbach, Christian Gemel, Moritz von Hopffgarten, Gernot Frenking, Roland A. Fischer, &lt;em&gt;Chem. Eur. J.&lt;/em&gt;, &lt;strong&gt;2010&lt;/strong&gt;, &lt;em&gt;16&lt;/em&gt; (45), 13372-13384.&lt;/a&gt; Also available through &lt;a href=&quot;https://www.researchgate.net/publication/47396676_Molecular_alloys_experimental_and_theoretical_investigations_on_the_substitution_of_zinc_by_cadmium_and_mercury_in_the_homologous_series_Mo%28M%27R%2912_and_M%28M%27R%298_%28MPd_Pt_M%27Zn_Cd_Hg%29&quot; rel=&quot;noreferrer&quot;&gt;researchgate.com&lt;/a&gt;. For the purely theoretical bonding analysis see &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp2038762&quot; rel=&quot;noreferrer&quot;&gt;Moritz von Hopffgarten, Gernot Frenking, &lt;em&gt;J. Phys. Chem. A&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;115&lt;/em&gt; (45), 12758–12768.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-25T04:02:45.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31970" PostTypeId="1" AcceptedAnswerId="31999" CreationDate="2015-05-25T05:24:56.807" Score="4" ViewCount="19676" Body="&lt;p&gt;Why is mercury the only metal that is liquid at room temperature?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am a high school student and would appreciate an answer, if possible, suited for my level of understanding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="16488" LastActivityDate="2015-05-25T19:37:37.253" Title="Why is mercury liquid?" Tags="&lt;transition-metals&gt;" AnswerCount="3" CommentCount="3" ClosedDate="2015-05-26T00:18:52.767" ContentLicense="CC BY-SA 3.0" />
  <row Id="31971" PostTypeId="2" ParentId="31970" CreationDate="2015-05-25T06:03:14.213" Score="-1" Body="&lt;p&gt;At room temp at atmospheric pressure mercury will not readily share valence electrons with other atoms. The mercury atoms won't interact with each other to form a structure or lattice to become a solid.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastActivityDate="2015-05-25T06:03:14.213" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31972" PostTypeId="2" ParentId="26723" CreationDate="2015-05-25T06:46:26.883" Score="1" Body="&lt;p&gt;Everything you do is correct and it is only the final algebraic transformation missing:&#xA;\begin{align}&#xA;0.068 &amp;amp;= \frac{1-x}{(1.25+x)(2.00+x)}\\&#xA;0.068 &amp;amp;= \frac{1-x}{x^2 + 3.25x + 2.5} &amp;amp;&amp;amp;|\cdot (x^2 + 3.25x + 2.5)\\&#xA;1-x   &amp;amp;= 0.068(x^2 + 3.25x + 2.5) &amp;amp;&amp;amp;|-1 +x\\&#xA;0.0   &amp;amp;= 0.068x^2 + 1.221x -0.83\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can solve for $x$ according to&#xA;\begin{align}&#xA;0&amp;amp;=ax^2+bx+c &amp;amp; x_{1,2}&amp;amp;=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-25T06:46:26.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31973" PostTypeId="1" AcceptedAnswerId="31976" CreationDate="2015-05-25T07:53:10.900" Score="18" ViewCount="2687" Body="&lt;p&gt;My textbook frequently mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\mathrm{sp^3}$ hybrid orbital has 25% $\mathrm{s}$-character and 75% $\mathrm{p}$-character.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are these &quot;characters&quot;? And how do these characters influence the bond and the hybrid orbital?&lt;/p&gt;&#xA;" OwnerDisplayName="user16347" LastEditorUserId="470" LastEditDate="2015-05-25T19:55:38.747" LastActivityDate="2015-07-22T17:20:01.533" Title="What does it mean when it is said that an sp³ orbital has 25% s character?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31974" PostTypeId="1" AcceptedAnswerId="31986" CreationDate="2015-05-25T08:04:09.043" Score="7" ViewCount="807" Body="&lt;p&gt;Superheated water is achieved by heating water above 100 degree Celsius under high pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose the temperature at which the water is getting heated is around 250 Celsius and sufficient pressure is provided. If, I now make a small-hole in the container, the pressure will decrease obviously, but what will happen to the superheated water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it come out in the form of vapor with lot of energy or some explosion might take place or just about something else?&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="418" LastEditDate="2015-05-25T10:05:20.250" LastActivityDate="2016-08-18T22:26:50.380" Title="What happens if I make a hole in a container with superheated water?" Tags="&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31975" PostTypeId="1" CreationDate="2015-05-25T08:16:56.770" Score="6" ViewCount="13944" Body="&lt;p&gt;I'm working on a project (school level) and in it I'll be dealing with the pressure generated by water when superheated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could understand the concept of superheated water but superheated steam is hard to understand. Are both inter-related, i mean that when water is superheated then it changes into superheated steam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it something different? And since in my project I require high-pressure-energy to be generated from water, so which one will be better, superheated steam or superheated water? &lt;/p&gt;&#xA;" OwnerUserId="16127" LastActivityDate="2016-12-07T06:21:48.577" Title="Difference between Superheated steam and Superheated water?" Tags="&lt;water&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31976" PostTypeId="2" ParentId="31973" CreationDate="2015-05-25T08:22:19.390" Score="16" Body="&lt;p&gt;Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.&lt;br&gt;&#xA;Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow&lt;sup&gt;*&lt;/sup&gt; and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;* “Is It Time To Retire the Hybrid Atomic Orbital?” &lt;a href=&quot;http://dx.doi.org/10.1021/ed100155c&quot;&gt;Alexander Grushow, &lt;em&gt;J. Chem. Educ.&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;, &lt;em&gt;88&lt;/em&gt; (7), 860–862.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="7475" LastEditDate="2015-05-27T14:56:08.183" LastActivityDate="2015-05-27T14:56:08.183" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="31977" PostTypeId="1" AcceptedAnswerId="32011" CreationDate="2015-05-25T08:33:04.860" Score="3" ViewCount="873" Body="&lt;p&gt;Are these 3 molecular structures of ammonium bicarbonate the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BrZL.png&quot; alt=&quot;http://upload.wikimedia.org/wikipedia/commons/6/65/Ammonium_bicarbonate.PNG&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aGMA9.png&quot; alt=&quot;http://www.mpbio.com/images/product-images/molecular-structure/02150107.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ssSq.gif&quot; alt=&quot;http://www.caslab.com/Chemical-Search/Chemical-Structure/1066-33-7.gif&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9616" LastEditorDisplayName="user15489" LastEditDate="2015-05-25T10:46:11.373" LastActivityDate="2015-05-29T09:42:47.523" Title="Are these 3 molecular structures of ammonium bicarbonate the same?" Tags="&lt;molecules&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="31978" PostTypeId="2" ParentId="31868" CreationDate="2015-05-25T08:50:35.753" Score="5" Body="&lt;p&gt;$\ce{Al(III)}$ has a +3 charge while $\ce{Mg(II)}$ has a +2 charge.  So the ion-dipole interaction that binds water to the ion is stronger for $\ce{Al(III)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on transition state theory the $10^5$ increase in water exchange rate translates to a difference in activation free energy of about 30 kJ/mol. The difference in interaction energy between a 2 Debye dipole that is 3 Å from a +3 and +2 charge is about 65 kJ/mol (assuming the dipole is pointed directly at the charge).&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-05-27T07:40:34.867" LastActivityDate="2015-05-27T07:40:34.867" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="31979" PostTypeId="2" ParentId="31977" CreationDate="2015-05-25T09:14:14.323" Score="3" Body="&lt;p&gt;They are certainly not the same. A is in dissociated form. Say in aqueous solution. B refers to not dissociated form but I have never come across non dissociated structure. There may not be actual bond between N atom and O atom.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-05-25T09:14:14.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31981" PostTypeId="1" CreationDate="2015-05-25T09:19:34.863" Score="3" ViewCount="1313" Body="&lt;p&gt;I have separated the sugars glucose, galactose and mannose by thin-layer-chromatography. My question is why these sugar have different running behaviors and a separation is possible.&lt;br&gt;&#xA;Glucose have the lowest Rf-value, then galactose and mannose have the highest Rf-value in my experiments.&lt;br&gt;&#xA;I thought it might be something to do with the position of the $\ce{-OH}$ groups (axial, equatorial) but I have no firm explanation at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for every explanation concerning the topic.&lt;/p&gt;&#xA;" OwnerUserId="16131" LastEditorUserId="13339" LastEditDate="2015-05-25T11:01:48.457" LastActivityDate="2015-05-25T11:01:48.457" Title="Running behavior of glucose, galactose and mannose in TLC" Tags="&lt;polarity&gt;&lt;chromatography&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="31982" PostTypeId="2" ParentId="31975" CreationDate="2015-05-25T09:50:49.120" Score="5" Body="&lt;p&gt;Boilers and similar steam generators usually produce &lt;em&gt;wet steam&lt;/em&gt;, which is a mixture of &lt;em&gt;saturated steam&lt;/em&gt; and liquid water droplets. &lt;em&gt;Saturated steam&lt;/em&gt; is steam that is in equilibrium with liquid water. The temperature of the steam is equal to the temperature of the liquid water; it corresponds to the boiling point at the given pressure. At normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$), the temperature is $T=99.974\ \mathrm{^\circ C}$. The corresponding specific enthalpy of the steam is $h = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h = 419\ \mathrm{kJ/kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If heat is supplied to the wet steam, liquid water vaporizes. At constant pressure, the temperature remains constant until all liquid water has been vaporized, i.e. until the steam is dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If further heat is supplied to the dry steam, the temperature and the specific enthalpy of the steam increase. Such steam at a temperature higher than its boiling point is called &lt;strong&gt;&lt;em&gt;superheated steam&lt;/em&gt;&lt;/strong&gt;. For example, superheated steam at normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) and a temperature of $T=150\ \mathrm{^\circ C}$ has a specific enthalpy of $h = 2777\ \mathrm{kJ/kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Superheated water&lt;/em&gt;&lt;/strong&gt; is liquid water under increased pressure at temperatures above the boiling point at normal pressure. For example, water at a temperature of $T=150\ \mathrm{^\circ C}$ and a pressure of $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$ is liquid. The specific enthalpy of the liquid water is $h = 632\ \mathrm{kJ/kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy content of superheated steam (e.g. $h = 2777\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) is much higher than the energy content of superheated water (e.g. $h = 632\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$) because of the latent heat, i.e. the enthalpy of vaporization $\Delta H_\text{vap}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-25T09:50:49.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31983" PostTypeId="2" ParentId="31970" CreationDate="2015-05-25T10:17:30.713" Score="0" Body="&lt;p&gt;There is a theory of relativistic effect. According to it, in elements with atomic number 80 or greater, the electrostatic attractive force is very high in the valence shell. This causes the 6s valence electrons of mercury to be actually strongly bonded. Now there is also a theory 'electron sea model' where metals share their valence electrons which leads to adhesive forces among metal atoms. As mentioned earlier mercury's valence shell electrons are strongly attracted to its nucleus so mercury atoms do not benefit from the adhesive forces due to very less sharing of the valence electrons. &#xA;Hope this answers your question.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-05-25T10:17:30.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31985" PostTypeId="1" CreationDate="2015-05-25T11:01:44.093" Score="7" ViewCount="712" Body="&lt;p&gt;I am currently in the process of studying enzyme catalysis and am struggling to get to grips with this concept. As far as I understand, the propinquity effect allows substrates to bind with enzymes in the conformation which would facilitate the reaction, thus increasing the reaction rate. Could someone expand on the why and/or how enzymes are able to do this? What specific reactions benefit from this the most?&lt;/p&gt;&#xA;" OwnerUserId="16427" LastActivityDate="2016-04-24T12:32:34.227" Title="The Propinquity Effect" Tags="&lt;organic-chemistry&gt;&lt;catalysis&gt;&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31986" PostTypeId="2" ParentId="31974" CreationDate="2015-05-25T11:03:37.437" Score="8" Body="&lt;p&gt;If you have water at a temperature of $T=250\ \mathrm{^\circ C}$, you need a pressure of approximately $p=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ to keep it liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1086\ \mathrm{kJ/kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you open the container, the pressure drops to the ambient pressure of $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam} = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h_\text{liquid} = 419\ \mathrm{kJ/kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flashing process is very quick, hence we may assume that there is no significant heat exchange with the environment; i.e. we may assume the following enthalpy balance in order to estimate the vaporized fraction $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;h_0 &amp;amp;= x \cdot h_\text{steam} + (1-x) \cdot h_\text{liquid}\\&#xA;x &amp;amp;= \frac{h_0 - h_\text{liquid}}{h_\text{steam} - h_\text{liquid}}\\&#xA;x &amp;amp;= \frac{1086\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}{2676\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}\\&#xA;x &amp;amp;= 0.296&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that approximately $29.6\ \%$ of the superheated water flashes to steam when the container is opened. &lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="7448" LastEditDate="2016-08-18T22:26:50.380" LastActivityDate="2016-08-18T22:26:50.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="31987" PostTypeId="1" CreationDate="2015-05-25T11:47:37.700" Score="8" ViewCount="4422" Body="&lt;p&gt;As the electronic configuration of gold is [Kr] $4d^{10} 4f^{14} 5s^2 5p^6 5d^{10} 6s^1$, the valence shell is not completely full. Why doesn't gold  corrode?&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="13339" LastEditDate="2015-05-25T13:20:52.970" LastActivityDate="2015-05-25T21:08:20.973" Title="Why doesn't gold corrode?" Tags="&lt;electronic-configuration&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-05-25T14:30:47.003" ContentLicense="CC BY-SA 3.0" />
  <row Id="31988" PostTypeId="2" ParentId="31987" CreationDate="2015-05-25T12:28:43.430" Score="6" Body="&lt;p&gt;You can look it up on James Huheeys book, gold's few properties have been qualitatively explained by relativistic theory. Due to that and high nuclear charge acting on the electron, the single electron in $6s$ is actually attracted to the nucleus strongly, thus the electron is not available to react that easily and thus gold does not corrode.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="10185" LastEditDate="2015-05-25T21:08:20.973" LastActivityDate="2015-05-25T21:08:20.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31989" PostTypeId="2" ParentId="31949" CreationDate="2015-05-25T12:32:16.530" Score="1" Body="&lt;p&gt;You can consider thinking about charge to volume density. Fluorine is very small and has highest charge to volume ratio among other group 17 elements. For that reason only fluorine is able to snatch electrons from oxygen. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorDisplayName="user15489" LastEditDate="2015-05-26T08:26:19.440" LastActivityDate="2015-05-26T08:26:19.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31990" PostTypeId="2" ParentId="5486" CreationDate="2015-05-25T12:38:37.547" Score="10" Body="&lt;p&gt;Simply it means that the equilibrium will force the conversion of certain substances to reduce the disturbance, to maintain the equilibrium itself. So, for example, $\ce{A + B &amp;lt;=&amp;gt; C + D}$ (both directions). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the concentration of the reactants — $\ce{A}$ or $\ce{B}$, then you’re increasing the conversion of reactants into products, because the rate of reaction and overall number of product formed increases. Therefore, we need to form more $\ce{C + D}$, to decrease the concentration of $\ce{A + B}$ and balance out the rate of change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, even though the reaction occurs simultaneously as a forward and backward reaction, the movement of equilibrium to the right (to the products) means the conversion of reactants to products is further emphasised, and generally — there is more of it occurring than in a usual equilibrium, in order to balance the change. Same thing would’ve happened if you increase the concentration of products, you want the equilibrium to shift to the left (reactants) so concentrations of both reactants and products balances out, reducing the disturbance (Le Chatelier’s Principle — specifically referred to by Ashu). Simply shifting equilibrium means increased rate of conversion of substances, predicating on the change in the reaction in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, it’s quite long winded and irrelevant at times, I hope you’ve benefitted from this though. I’ve also used the idea of changing concentrations in particular from Ashu, if that’s alright.&lt;/p&gt;&#xA;" OwnerUserId="16499" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2015-10-14T11:36:00.613" LastActivityDate="2015-10-14T11:36:00.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31991" PostTypeId="1" AcceptedAnswerId="31992" CreationDate="2015-05-25T13:10:30.617" Score="5" ViewCount="78" Body="&lt;p&gt;I want to create a small wind tunnel for testing of aerodynamics for Radio controlled units (Cars, planes and helicopters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real wind-tunnels water mists are used but water isn't a good thing for the electronics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Smoke bombs, but the smoke develops too fast.&lt;/li&gt;&#xA;&lt;li&gt;Burning stuff like laves or coal, but it creates odor, and i don't want my RC stuff to smell.&lt;/li&gt;&#xA;&lt;li&gt;Incense, same as above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My chemistry level is a College equivalent degree of chemistry that has been unused for 15 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the materials to make this should be somewhat harmless, and legally available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on how to do this would be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16500" LastEditorUserId="12860" LastEditDate="2015-10-30T18:28:33.383" LastActivityDate="2015-10-30T18:28:33.383" Title="Develop slow smoke for mini windtunnel" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31992" PostTypeId="2" ParentId="31991" CreationDate="2015-05-25T14:38:47.620" Score="4" Body="&lt;p&gt;The DLR (German Aerospace Center) has developed a couple of &lt;a href=&quot;http://www.dlr.de/next/desktopdefault.aspx/tabid-7757/&quot; rel=&quot;nofollow&quot;&gt;DIY instructions for school education&lt;/a&gt;. Among these is a &lt;a href=&quot;http://www.dlr.de/next/portaldata/69/resources/downloads/9_downloads/DLR_next_Anleitungen-windkanal.pdf&quot; rel=&quot;nofollow&quot;&gt;PDF in German&lt;/a&gt; for the construction of a wind tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the smoke, they suggest to use either &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;incense (the sticks or cones that you usually find in asia shops or esoteric shops) or a   &lt;/li&gt;&#xA;&lt;li&gt;fog machine, like the ones used for entertainment (dance clubs, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the latter case, a mixture of 1,2-propanediol and water is heated and pressed through a nozzle at elevated temperature. Given that properly insulated machines, build for safe use at home, are available from DIY electronics shops for roughly 30 € up, I wouldn't start to build one myself.   &lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-25T14:38:47.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31993" PostTypeId="2" ParentId="31921" CreationDate="2015-05-25T14:54:50.997" Score="5" Body="&lt;p&gt;Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, &lt;em&gt;an intuitive way to think about cell potentials.&lt;/em&gt;  Here's the corresponding graphic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7XRq.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode.  The potentials of the two half-cells are known and marked on the line-graph.  Electrons &lt;strong&gt;always&lt;/strong&gt; flow to the more positive electrode, which in this figure is towards the right.  The difference between the two half-cells is the cell potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you know the separation (-0.07 V) and one of two points (+0.34 V).  If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to &lt;strong&gt;C&lt;/strong&gt; or &lt;strong&gt;D&lt;/strong&gt;, since each of these differ from +0.34 V by 0.07.  The question now is, which direction are the electrons flowing?  When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential.  Since the cell potential in your case is &lt;em&gt;negative&lt;/em&gt;, we must be doing the opposite.  Therefore, the reference electrode potential is +0.41 V (&lt;strong&gt;D&lt;/strong&gt;).&lt;/p&gt;&#xA;" OwnerUserId="1573" LastActivityDate="2015-05-25T14:54:50.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31994" PostTypeId="1" AcceptedAnswerId="32002" CreationDate="2015-05-25T17:53:52.960" Score="6" ViewCount="1021" Body="&lt;p&gt;I just bought some (pure) antimony stones from a local store. However, I've heard that some stores deceive their customers by selling lead instead of antimony. I was wondering if there is any simple test that I can perform at home to make sure these stones are antimony and not lead. Since I don't know much about chemistry I thought of asking here...&lt;/p&gt;&#xA;" OwnerUserId="12805" LastActivityDate="2015-06-21T17:02:57.227" Title="How is it possible to distinguish antimony from lead?" Tags="&lt;home-experiment&gt;&lt;elements&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="31995" PostTypeId="1" CreationDate="2015-05-25T19:04:51.740" Score="7" ViewCount="709" Body="&lt;p&gt;What will the $\ce{^129Xe}$ NMR spectrum of $\ce{XeOF4}$ look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that its electron pair geometry is octahedral, and the molecular pair geometry is square pyramidal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the lone pair on $\ce{Xe}$ will take the equatorial position (less sterically crowded), alongside the oxygen and 2 other equatorial fluorines in the same plane, to minimize electron pair repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there will be 2 equatorial $\ce{F}$ and 2 axial $\ce{F}$. Therefore, am I right to say that the $\ce{^129Xe}$ NMR will be a triplet of triplets?&lt;/p&gt;&#xA;" OwnerUserId="16445" LastEditorUserId="4945" LastEditDate="2015-05-26T00:26:49.373" LastActivityDate="2015-05-26T00:26:49.373" Title="What does the 129Xe NMR spectrum of XeOF4 look like?" Tags="&lt;molecules&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31997" PostTypeId="1" CreationDate="2015-05-25T19:21:07.027" Score="6" ViewCount="18362" Body="&lt;p&gt;Why is the $\ce{I-Cl}$ bond length in $\ce{ICl2-}$ (charge of -1) is longer than in $\ce{ICl2+}$ (charge of +1) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few explanations that I have in mind. Please critique.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I in $\ce{ICl2-}$ is $sp^3d$ hybridized. $\ce{I}$ in $\ce{ICl+}$ is $sp^3$ hybridized. Since additional promotion energy is required to form the $sp^3d$ hybrid compared to the $sp^3$ hybrid, the bond lengths in $\ce{ICl2-}$ is longer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $\ce{ICl2-}$, if we let $\ce{I}$ be $sp^2$ hybridized such that the 3 lone pairs lie 120 apart in a plane, and then let the 2 $\ce{Cl}$ atoms and the $\ce{I}$ atom lie in a straight line, then we can form a 3-centred, 4 electron system. This is weaker than the 2-centred, 2 electron system in the $\ce{I-Cl}$ bond in $\ce{ICl+}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As $\ce{I}$ in $\ce{ICl2+}$ has a higher effective nuclear charge, its valence orbitals are smaller and more compact, hence the $\ce{I-Cl}$ covalent bond is stronger as the electrostatic attraction between the nuclei and the overlap density is stronger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16445" LastEditorUserId="4231" LastEditDate="2015-05-25T21:15:02.730" LastActivityDate="2016-08-04T11:35:10.463" Title="Bond length in ICl2- and ICl2+" Tags="&lt;bond&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="31998" PostTypeId="2" ParentId="6293" CreationDate="2015-05-25T19:25:50.240" Score="3" Body="&lt;p&gt;$\ce{B}$ in $\ce{BF3}$ is $sp^2$ hybridized (since the molecular geometry is trigonal planar), and therefore the unhybridized $p$ orbital in $\ce{B}$ can overlap with the $p$ orbitals in $\ce{F}$. You can draw resonance structures, and from the valence bond theory perspective, the bond order of $\ce{B-F}$ in $\ce{BF3}$ will be approximately 1.33 due to the partial $\pi$ bond character in the $\ce{B-F}$ bond in $\ce{BF3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{BF4-}$, the $\ce{B}$ is $sp^3$ hybridized and hence p orbital overlap is not possible. Hence, the $\ce{B-F}$ bond is simply a conventional 2-centred, 2 electron bond with bond order 1. Comparing bond orders, it becomes clear that $\ce{BF4}$ has a longer bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to, you can draw a simplified MO diagram showing stabilization due to overlap of the $\ce{B}$ $2p$ and $\ce{F}$ $2p$ orbitals, but the valence bond theory argument suffices. &lt;/p&gt;&#xA;" OwnerUserId="16445" LastEditorDisplayName="user15489" LastEditDate="2015-05-25T20:20:29.813" LastActivityDate="2015-05-25T20:20:29.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="31999" PostTypeId="2" ParentId="31970" CreationDate="2015-05-25T19:37:37.253" Score="5" Body="&lt;p&gt;There are a couple of reasons why Mercury is a liquid at STP, I will go through each of them with an explanation that won't be too complex for your understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First of all understanding why most metals are metallic is a good place to start.&lt;/strong&gt; This happens because of the sharing of valence electrons. The sharing of these electrons make metals malleable and ductile and thus not liquid. Generally the more the electrons shared means the higher the melting point and a harder metal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OK, getting into the Mercury.&lt;/strong&gt; Like others have probably mentioned above, Mercury doesn't like to share its electrons. It hangs onto them very tightly and thus Mercury-Mercury bonding is very weak and thus the compound melts and boils at very low temperatures. In other words, Mercury &lt;strong&gt;lacks&lt;/strong&gt; the sea of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So then, why doesn't it like to share electrons?&lt;/strong&gt; The $\ce{6s}$ electrons come very close to the nucleus and swing around it at near the speed of light. Due to this, &lt;strong&gt;relativistic effects&lt;/strong&gt; occur. In other words, the $\ce{s}$ electrons behave as if they are more massive than other electrons. This lowers the energy of the electrons. To summarize, the electrons are buried deep in the atomic core and thus aren't able to participate in &lt;em&gt;chemistry&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overall,&lt;/strong&gt; since you are a high school student, I would probably just say that Mercury doesn't readily bond well and that is due to the $\ce{6s}$ electrons being very attracted to the atomic core.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-05-25T19:37:37.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32000" PostTypeId="2" ParentId="31995" CreationDate="2015-05-25T19:49:39.313" Score="8" Body="&lt;p&gt;The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, &lt;strong&gt;quintuplet&lt;/strong&gt; of peaks (&lt;a href=&quot;https://books.google.com/books?id=tUmcAQAAQBAJ&amp;amp;pg=PA445&amp;amp;lpg=PA445&amp;amp;dq=xeof4%20nmr&amp;amp;source=bl&amp;amp;ots=i19TFzAU6f&amp;amp;sig=WJEgg7V-64lsYqhPbvdcFjVQqYI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=h3VjVaeKA4mLsAW5-YCQAw&amp;amp;ved=0CEYQ6AEwBg#v=onepage&amp;amp;q=xeof4%20nmr&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;, p. 445).  As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (&lt;a href=&quot;http://en.wikipedia.org/wiki/Xenon_oxytetrafluoride&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;), but the lone pair occupies an axial position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOE4B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the 4 equatorial fluorines are equivalent and since they are spin 1/2 nuclei they will couple with the $\ce{^{129}Xe}$ nucleus (also spin 1/2) producing the observed quintuplet. $\ce{XeOF4}$ is an &lt;a href=&quot;http://chem.ch.huji.ac.il/nmr/techniques/1d/row5/xe.html#ref&quot; rel=&quot;noreferrer&quot;&gt;IUPAC suggested standard&lt;/a&gt; for $\ce{^{129}Xe}$ nmr spectroscopy (but xenon gas is more commonly used because $\ce{XeOF4}$ can decompose in the presence of moisture to produce $\ce{HF}$).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-25T22:29:58.630" LastActivityDate="2015-05-25T22:29:58.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32001" PostTypeId="2" ParentId="31997" CreationDate="2015-05-25T20:26:39.417" Score="5" Body="&lt;p&gt;The iodine atom in $\ce{ICl2}$ is what is termed &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypercoordinated&lt;/a&gt; (or hypervalent). Basically, this means that the 3 atoms combine to produce 3 molecular orbitals (the 3-center 4-electron model). There is no need to invoke d-orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgN5i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In neutral $\ce{ICl2}$ there are three electrons spread across the two $\ce{I-Cl}$ bondss. Placing those electrons in our molecular orbitals, 2 go into the bonding orbital and one into the nonbonding orbital.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{ICl2^{+}}$ we have removed the one nonbonding electron.  As a consequence there will be less electron-electron repulsion and only the bonding molecular orbital is occupied, because of this we would expect shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{ICl2^{-}}$ we have added another electron and it will go into the nonbonding orbital.  This will increase the electron-electron repulsion in the molecule without increasing bonding, because of this we would expect longer bond lengths in $\ce{ICl2^{-}}$ compared to $\ce{ICl2}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These arguments tell us that we would have shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2^{-}}$.  &lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="22766" LastEditDate="2016-08-04T11:35:10.463" LastActivityDate="2016-08-04T11:35:10.463" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32002" PostTypeId="2" ParentId="31994" CreationDate="2015-05-25T20:28:33.480" Score="6" Body="&lt;p&gt;As mentioned in the comments, you could perform &lt;a href=&quot;http://www.colorado.edu/physics/phys2010/phys2010LabMan2000/2010labhtml/Lab5/EXP5LAB99.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archimedes' Experiment&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Firstly, take the mass ($m$) (in grams) of your sample.&lt;/li&gt;&#xA;&lt;li&gt;Then, measure the volume $V$ displaced (according to the first link), in a method summarised in the image below:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EzYyq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use any kitchen graduated container. Even though these have graduations for mL (millilitres) and even though the water density changes with temperature (source:&lt;a href=&quot;https://water.usgs.gov/edu/density.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;USGS&lt;/a&gt;), for the temperatures in a home experiment, the following conversion could be used &lt;strong&gt;1 mL = 1 cm&lt;sup&gt;3&lt;/sup&gt;&lt;/strong&gt; of water displaced.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then determine the density by $\rho = \frac{m}{V}$ and compare it to the known values for &lt;a href=&quot;http://www.lenntech.com/periodic/elements/pb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;lead&lt;/a&gt; at 11.3 g.cm&lt;sup&gt;-3&lt;/sup&gt; and &lt;a href=&quot;http://www.lenntech.com/periodic/elements/sb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;antimony&lt;/a&gt; at 6.7 g.cm&lt;sup&gt;-3&lt;/sup&gt;.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another observed feature is that according to the &lt;a href=&quot;http://www.lenntech.com/periodic/elements/pb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;LennTech site for lead&lt;/a&gt; is that as opposed to antimony, lead will tarnish on exposure to air.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-25T20:47:08.020" LastActivityDate="2015-05-25T20:47:08.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32003" PostTypeId="2" ParentId="8079" CreationDate="2015-05-26T01:33:40.343" Score="3" Body="&lt;p&gt;As seen on a 25kg bag of Epsom Salts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1NlM8.jpg&quot; alt=&quot;Epsom Salts bag&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16511" LastActivityDate="2015-05-26T01:33:40.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32004" PostTypeId="2" ParentId="8079" CreationDate="2015-05-26T02:08:01.680" Score="3" Body="&lt;p&gt;Chemicals from chemical makers have labels about hazards and any kind of minor components and well as the grade of purity. I don't know Epson salt specifically in your country, but most probably there is a cosmetic or food grade version, which must be safe for you. If a salt is pure enough for bath salt, it is pure enough for washing hands (you don't drink soap, so food grade is not necessary). UNLIKE many natural, organic or whatever shaman-branded products, the &quot;artificial&quot; chemicals has a strong quality and safety control and they actually make all those info public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a note:&#xA;Home-soap makers are notoriously bad in chemistry and bad in safety. I have seen many recipes that make no sense, extremely hazardous and you produce a mediocre-to-bad quality soap. From the wording of the inserted quote I would expect he/she is a nice, but uninformed amateur at best. Be careful with those recipes. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="402" LastEditDate="2017-04-09T18:23:54.737" LastActivityDate="2017-04-09T18:23:54.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32005" PostTypeId="1" CreationDate="2015-05-26T02:12:43.450" Score="7" ViewCount="55" Body="&lt;p&gt;I'm making a solution with linear &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylenimine&quot; rel=&quot;nofollow noreferrer&quot;&gt;poly(ethyleneimine)&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{[CH2CH2NH]_{n}}$&lt;/span&gt;.  The bottle indicates that the substance is hygroscopic and must be dealt with under inert gas.  Is this really necessary?  I don't currently have access to such a space - just the PEI and a balance.&lt;/p&gt;&#xA;" OwnerUserId="16512" LastEditorUserId="402" LastEditDate="2019-05-18T22:52:26.983" LastActivityDate="2019-05-18T22:52:26.983" Title="Linear PEI Solution Making" Tags="&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32006" PostTypeId="1" CreationDate="2015-05-26T02:16:22.627" Score="6" ViewCount="790" Body="&lt;p&gt;In the multistage synthesis of 4-bromobenzenamine, why must the solution be rendered alkaline prior to the steam distillation after the hydrolysis of 4-bromo-&lt;em&gt;N&lt;/em&gt;-ethanoylbenzenamine?&lt;/p&gt;&#xA;" OwnerUserId="16513" LastEditorUserId="7475" LastEditDate="2015-11-13T19:10:28.377" LastActivityDate="2015-11-13T19:10:28.377" Title="steam distillation of 4-bromoaniline , why render it basic prior" Tags="&lt;acid-base&gt;&lt;hydrolysis&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32008" PostTypeId="1" AcceptedAnswerId="32010" CreationDate="2015-05-26T05:39:08.880" Score="16" ViewCount="10975" Body="&lt;p&gt;When I start studying electrochemistry, I learn the words “molten” and “aqueous”. I don't have a problem for “aqueous”, but I'm a little bit confused about “molten”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, “molten” means melt, which means to become liquid from solid. But why do we say &lt;em&gt;molten&lt;/em&gt; lead(II) bromide instead of &lt;em&gt;liquid&lt;/em&gt; lead(II) bromide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean both are the same, but liquid is arguably easier to understand.&lt;/p&gt;&#xA;" OwnerUserId="9757" LastEditorUserId="41328" LastEditDate="2020-02-02T08:18:40.357" LastActivityDate="2020-05-13T09:04:10.273" Title="Molten vs Liquid" Tags="&lt;phase&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32009" PostTypeId="2" ParentId="32006" CreationDate="2015-05-26T05:43:20.950" Score="5" Body="&lt;p&gt;In principle, it is difficult to distill salts (ionic compounds). Therefore you first make sure that you have a free base in the solution. This is guaranteed in alkaline solution. For the steam distillation the compounds should be immiscible, which is another reason for having a free base. Let’s compare boiling points and solubility for aniline and its salt:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aniline b.p. 184 °C, solubility in water: 36 g/L&lt;/li&gt;&#xA;&lt;li&gt;Anilinium chloride b.p. 245 °C, solubility in water: 1070 g/L&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2380" LastEditorDisplayName="user7951" LastEditDate="2015-05-26T07:40:01.647" LastActivityDate="2015-05-26T07:40:01.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32010" PostTypeId="2" ParentId="32008" CreationDate="2015-05-26T05:58:10.370" Score="25" Body="&lt;p&gt;As you said, the meaning is exactly the same. Molten reduces the ambiguity, because you emphasize that you know, that it is solid substance at laboratory conditions and you heat it to become liquid (while staying pure substance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, the &quot;electrolysis of liquid sodium chloride&quot; is in principle enough to tell you all about the process, but as we often like to dissolve the salt in water, you might be slightly unsure, whether the meaning is 100% clear. Therefore &quot;electrolysis of molten sodium chloride&quot; is better choice. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-05-26T05:58:10.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32011" PostTypeId="2" ParentId="31977" CreationDate="2015-05-26T06:01:35.000" Score="9" Body="&lt;p&gt;Like Punarbasu Roy already stated, the depicted structures are certainly not the same. To be frank, only one of them (A) is correct, one of them is not completely wrong (B - but only because it is &lt;em&gt;incomplete&lt;/em&gt;), and one is completely, utterly &lt;em&gt;wrong&lt;/em&gt; (C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lieu of an actual crystal structure of ammonium hydrogencarbonate,&lt;sup&gt;[1]&lt;/sup&gt; I'll refer to the structure of the parent compound ammonium carbonate monohydrate to make my point.&lt;sup&gt;[2]&lt;/sup&gt; It is actually quite difficult to assign one correct Lewis structure. In its solid form ammonium (bi)carbonate is a salt, in which ammonium and (bi)carbonate ions are interacting via ionic bonds in a lattice. In solution this certainly is broken up and we will find separate ions, according to the following equilibria:&#xA;\begin{align}&#xA;\ce{(H4N)HCO3~(s) &amp;amp;~-&amp;gt;C[(H2O)] NH4+ ~(aq) + HCO3- ~(aq)}\\&#xA;\ce{NH4+ ~(aq) + H2O &amp;amp;~&amp;lt;=&amp;gt; H3+O~(aq) + NH3~(aq)}\\&#xA;\ce{HCO3- ~(aq) + H2O &amp;amp;~&amp;lt;=&amp;gt; H2CO3 ~(aq) + {}^{-}OH~(aq)}\\&#xA;\ce{HCO3- ~(aq) + H2O &amp;amp;~&amp;lt;=&amp;gt; CO3^2- ~(aq) + H3+O~(aq)}\\&#xA;\end{align}&#xA;The same equilibria will establish for ammonium carbonate with different concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the crystal structure of ammonium carbonate monohydrate you can see the different molecular entities (ions) held together via hydrogen bonds, reproduced from [2]:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZBWEa.jpg&quot; alt=&quot;asymmetric unit of the crystal structure of ammonium carbonate monohydrate&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia&lt;sup&gt;[3]&lt;/sup&gt; therefore offers the best possible Lewis structure:&lt;br&gt;&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/14/Ammonium_bicarbonate.png/200px-Ammonium_bicarbonate.png&quot; alt=&quot;Lewis structure of ammonium bicarbonte&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking a Lewis structure may only contain covalent bonds, i.e. two-centre-two-electron bonds, while ionic bonds are implied by placing the (molecular) ions in close proximity to each other. However, the hydrogen bond is a well known concept in chemistry and it was already extensively discussed by Linus Pauling. In his book &quot;The Nature of the Chemical Bond&quot; he uses dashed lines to represent the hydrogen bond.&lt;sup&gt;[4]&lt;/sup&gt; Certainly another interesting read is the approach of a more unified theory about the hydrogen bond.&lt;sup&gt;[5]&lt;/sup&gt;&lt;br&gt;&#xA;Another possibility is to introduce the donor-acceptor notation with arrows, to indicate a bonding interaction. Therefore one of the following structures may also be considered correct in an extended Lewis concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJcE6.jpg&quot; alt=&quot;alternative Lewis Structures (in an extended Lewis concept)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The following article reads like this is still a work in progress: &lt;a href=&quot;http://www.nature.com/nature/journal/v166/n4219/abs/166435b0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;R. Brooks, and T. C. Alcock, &lt;em&gt;Nature&lt;/em&gt;, &lt;strong&gt;1950&lt;/strong&gt;, &lt;em&gt;166&lt;/em&gt;, 435-436.&lt;/a&gt; (It does also not come with any supporting information.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1107/S205252061402126X&quot; rel=&quot;nofollow noreferrer&quot;&gt;A. D. Fortes, I. G. Wood, D. Alfè, E. R. Hernández, M. J. Gutmann, and H. A. Sparkes, &lt;em&gt;Acta Cryst. Sec. B&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;70&lt;/em&gt;, 948-962.&lt;/a&gt; This quite interesting article is &lt;strong&gt;open access&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_bicarbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ammonium bicarbonate&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Ammonium_carbonate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ammonium carbonate&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://books.google.co.jp/books?id=L-1K9HmKmUUC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Linus Pauling: &lt;em&gt;The Nature of the Chemical Bond&lt;/em&gt;. Cornell University Press, 1960.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/S0022-2860(00)00454-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;G. Gilli, P. Gilli, &lt;em&gt;J. Mol. Struc.&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;552&lt;/em&gt; (1-3), 1-15.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.567" LastActivityDate="2015-05-29T09:42:47.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32012" PostTypeId="1" AcceptedAnswerId="32016" CreationDate="2015-05-26T07:56:36.797" Score="2" ViewCount="94" Body="&lt;p&gt;Considering the Metabolic Pathway:&#xA;&lt;a href=&quot;http://www.genome.jp/kegg-bin/show_module?M00166&quot; rel=&quot;nofollow&quot;&gt;http://www.genome.jp/kegg-bin/show_module?M00166&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the reaction R01523(the very first reaction). How is it possible to know that the compound C00199(D-Ribulose 5-phosphate) will get transformed to C01182(D-Ribulose 1,5-bisphosphate)? &lt;br&gt;&lt;br&gt;What are the factors on which this transformation depends? One factor I know is structure but how does structure effect this transformation? Does two similar structured compound means that one of the compound can get transformed to the other one?&#xA;&lt;br&gt;&lt;br&gt;Note: It ia a part of biology but a chemical reaction or transformation is basically a part of chemistry, so asked this here&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="4325" LastEditDate="2015-05-26T16:05:47.737" LastActivityDate="2015-05-26T16:05:47.737" Title="Factors controlling enzymatic phosphorylation of D-ribulose 5-phosphate" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32013" PostTypeId="1" CreationDate="2015-05-26T08:26:36.840" Score="6" ViewCount="222" Body="&lt;p&gt;Why are neutron scattering factors essentially independent of $2\theta$, whereas X-ray scattering factors drop off with increasing $2\theta$ ? &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="4945" LastEditDate="2015-05-27T07:02:26.047" LastActivityDate="2019-07-06T02:01:10.167" Title="Angular dependence of x-ray and neutron scattering factors" Tags="&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32014" PostTypeId="1" AcceptedAnswerId="32018" CreationDate="2015-05-26T10:00:18.623" Score="7" ViewCount="1059" Body="&lt;p&gt;I understand the concept of incomplete combustion and that it occurs when oxygen is not a large enough supply to allow for complete combustion. My question is: how can we model combustion properly in terms of predicting the ratio of complete to incomplete combustion? Is it possible to use basic kinetic with reaction rate constants to model the rate of each reaction and then obviously model how the changing concentration from both affects each individual rate etc.&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2015-05-26T13:33:04.210" Title="Predicting incomplete combustion?" Tags="&lt;kinetics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32016" PostTypeId="2" ParentId="32012" CreationDate="2015-05-26T10:23:18.590" Score="3" Body="&lt;p&gt;After looking at the website I found the reaction is based off of an enzyme called: &lt;a href=&quot;http://www.genome.jp/dbget-bin/www_bget?ec:2.7.1.19&quot; rel=&quot;nofollow&quot;&gt;phosphoribulokinase&lt;/a&gt;. In biological systems many reactions occur under the auspice of enzymes or proteins. These are incredibly large molecules made of an extremely long chain of amino acids which are a set of 26 or so small molecules that can 'hook' into the chain structure of a protein. Each amino acid has different properties and thus experiences different forces pulling it towards and away from other amino acids on the chain. What this means is the whole chain folds up into an extremely complex structure. For a protein to allow a reaction to occur it has a small region that the reacting molecule can hook into. There are then sets of atoms and molecules that act as scaffolds that allow atoms in the reacting molecule to be moved without the need for nearly as much energy as would be required without the scaffold. Now this is how it changes structure. Now whilst the protein can just as easily allow the molecule produced to be converted back to the original, this may be less energetically favourable which brings us to equilibrium chemistry involving dynamic forward and backward reactions. What makes this more complicated is that proteins often have separate 'hooks' for molecules to disable them. That is if a lot of substance A is around then the protein will be disabled and no longer convert B into A until more A is consumed and thus stops inhibiting the protein.&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2015-05-26T10:23:18.590" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32018" PostTypeId="2" ParentId="32014" CreationDate="2015-05-26T13:33:04.210" Score="3" Body="&lt;p&gt;Before entering the kinetics realm, it is advisable to check the equilibrium. Besides the straightforward combustion reactions, you have two more, &lt;a href=&quot;https://en.wikipedia.org/wiki/Boudouard_reaction&quot; rel=&quot;nofollow&quot;&gt;Boudouard reaction&lt;/a&gt;, which favors formation of $\ce{CO}$ from $\ce{CO_2}$ and &lt;a href=&quot;https://en.wikipedia.org/wiki/Water-gas_shift_reaction&quot; rel=&quot;nofollow&quot;&gt;Water-gas shift reaction&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual reactions which contribute to the equilibrium depend on the fuel. In case of pure carbon, you at least get rid of the Water-gas shift reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might need to solve the system iteratively, as the temperature of the equilibrium mixture is result of the combustion reactions, but also affects the position of equilibrium. &lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-05-26T13:33:04.210" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32019" PostTypeId="1" CreationDate="2015-05-26T13:39:44.423" Score="5" ViewCount="1309" Body="&lt;p&gt;We know that the electron configuration in $3d$ orbitals of $\ce{Zn^2+}$, in $4d$ orbitals of $\ce{Cd^2+}$ and in $5d$ orbitals of $\ce{Hg^2+}$ ions are completely same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why the compounds of $\ce{Hg(II)}$ and $\ce{Cd}$, such as $\ce{HgO}$, $\ce{CdS}$ radiate versatile colour where $\ce{ZnO}$, $\ce{ZnS}$ are white?&lt;/p&gt;&#xA;" OwnerUserId="16407" LastEditorDisplayName="user7951" LastEditDate="2019-02-28T08:27:11.503" LastActivityDate="2019-03-01T12:26:56.710" Title="Why are HgO and CdS compounds colourful?" Tags="&lt;ions&gt;&lt;color&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32020" PostTypeId="2" ParentId="27753" CreationDate="2015-05-26T14:26:47.933" Score="3" Body="&lt;p&gt;The reaction would make more sense, and be balanced for charge, if the $\ce{H2}$ were replaced by $\ce{2H+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + 2H2O + Pb^{+2} -&amp;gt; Pb(CO3) + Ca(OH)2 + 2 H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even that is a bit weird because it suggests that the water would be acidified by this (non-redox) process.  You could also write it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + 2OH- + Pb^{+2} -&amp;gt; Pb(CO3) + Ca(OH)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Lead%28II%29_hydroxide&quot; rel=&quot;nofollow&quot;&gt;Lead hydroxide&lt;/a&gt; is apparently a compound that has never been isolated: the hydrated lead oxides are more stable than a true hydroxide salt.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-05-26T14:35:26.027" LastActivityDate="2015-05-26T14:35:26.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32021" PostTypeId="1" CreationDate="2015-05-26T14:36:02.633" Score="13" ViewCount="256" Body="&lt;p&gt;Is there a way of comparing which salt is more soluble without knowing their $ K_{sp}$ data or performing experiments?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, comparing $\ce{Hg2Cl2}$ and $\ce{Cr2(SO4)3}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure but can I compare lattice and hydration energy relatively or the ion size and estimate?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments will give concrete results but I was wondering if there is another method just to gauge the solubility.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chat.stackexchange.com/rooms/3229/conversation/relative-solubility-without-knowing-the-ksps-discussion&quot;&gt;The relevant chat discussion&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="15912" LastEditorUserId="10185" LastEditDate="2015-05-26T14:56:43.370" LastActivityDate="2016-10-25T13:22:24.933" Title="Predicting relative solubilites of salts without Ksp values" Tags="&lt;inorganic-chemistry&gt;&lt;solubility&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32022" PostTypeId="2" ParentId="31997" CreationDate="2015-05-26T15:13:42.047" Score="5" Body="&lt;p&gt;I’ll agree with ron concerning the non-hybridisation of d-orbitals and the description of $\ce{ICl2-}$. However, I disagree with ron’s description of $\ce{ICl2+}$, which was calculated to have an $\ce{Cl-I-Cl}$ bond angle of 98°.&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also believe that the contribution of the s-orbital to any bonding, non-bonding or anti-bonding orbitals in the iodine atom is neglegible, the reasoning being that the energy difference between iodine’s s- and p-orbitals is larger than for the atoms of the second period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{ICl2+}$ ion can thus be explained by assuming one double-filled s- and one double filled p-orbital plus two half-filled p-orbitals on iodine. Each of the two half-filled orbitals then create a normal $\sigma$-bonds to a corresponding p-orbital of one of the chlorine atoms, predicting a bond angle of 90° (closer to 98° than if we assumed sp³-hybridisation which would suggest 109.5°).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In $\ce{ICl2+}$, we have to bonds with a bond order of 1 to each of the chlorines. Compared to a bond order of ½ for $\ce{ICl2-}$ this means a shorter bond length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ion’s charge, which is located mostly on the iodine (electronegativity!) means that the iodine atom is contracted and the two chlorines can move closer. This effect is not present in $\ce{ICl2-}$, where the charge is distributed evenly across the two chlorine atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the reference and on the topic of predictions, I would predict the bond length of $\ce{ICl}$ to be comparable but slightly longer than that of $\ce{ICl2+}$, while still being notably shorter than that of $\ce{ICl2-}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://dx.doi.org/10.1002/1521-3749%28200011%29626:11%3C2419::AID-ZAAC2419%3E3.0.CO;2-K&quot; rel=&quot;noreferrer&quot;&gt;F. Bailly, P. Barthen, H.-J. Frohn, M. Köckerling, &lt;em&gt;Z. anorg. allgem. Chem.&lt;/em&gt; &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;626&lt;/em&gt;, 2419.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-26T15:13:42.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32023" PostTypeId="2" ParentId="31476" CreationDate="2015-05-26T15:36:30.757" Score="12" Body="&lt;p&gt;All elements colour the flame &lt;em&gt;somehow&lt;/em&gt;. The salt that is held into the flame dissociates &lt;em&gt;homogenously&lt;/em&gt; (i.e. you’re not creating unpaired ions but rather single metal atoms and their counter-radicals), the heat energy excites the electrons and when they fall back down they emit photons at certain wavelengths, corresponding to the energy differences between the orbital an electron was excited to and the one it falls down to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The further away from the nucleus the electrons are, the easier they are to excite. Thus valence electrons are most likely to be excited. The more valence electrons you have (i.e. the further right you go in a given period) the harder it gets to excite one of them because the whole atom contracts and the energy levels of the outer shells are lowered. Therefore, it is a lot easier for electrons of elements on the left half of the periodic table to be excited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the more electrons you can excite, the more complicated the resulting spectrum will be. For only one or two electrons, you usually expect (and observe) distinct lines of definite wavelengths; the more you have, the larger the possibility of broader bands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the easiest element to recognise would be hydrogen, which only creates very few single, distinct lines. Alkaline metals, having few valence electrons, will still create easy spectra with distinct lines (albeit more). It is similar for alkaline earth metals, although the spectra tend to get more complicated (see barium for an example of something with broad bands). For many, if not all other metals, the spectra just get too difficult to interpret and consistently assign very quickly. They will all seem white-ish to some degree with hardly any possibility of distinguishing characteristic lines easily. Thus, in the typical inorganic chemistry ion lottery lab course, typically only alkalines and alkaline earths are detected &lt;em&gt;via&lt;/em&gt; flame colouring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also remember from that course that copper halides seem to provide a more characteristic green colour (that I didn’t even try recognising — I relied on other methods to detect copper). I think that there is a mulitude of combined effects in action to give this combination its characteristic colour. In any case, I don’t know what exactly is going on there.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-12-08T22:18:26.177" LastActivityDate="2015-12-08T22:18:26.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32024" PostTypeId="2" ParentId="32021" CreationDate="2015-05-26T15:47:55.003" Score="1" Body="&#xA;&#xA;&lt;p&gt;Like dissolve like. Thus, a more ionic compound would be more soluble in water and a covalent one would be more soluble in organic solvents. Speaking in context of your example, and assuming solvent to be water,&#xA;The nucleus of Cr&lt;sup&gt;3+&lt;/sup&gt; ion attracts the bond pair electrons more strongly for its greater charge &amp; smaller diameter than &lt;span class=&quot;math-container&quot;&gt;$\ce{Hg+}$&lt;/span&gt; ion.&#xA;According to Fajans' rules, Cr&lt;sub&gt;2&lt;/sub&gt;(SO&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;3&lt;/sub&gt; is of more covalent character than Hg&lt;sub&gt;2&lt;/sub&gt;Cl&lt;sub&gt;2&lt;/sub&gt;. Again the factor of anion is less significant here as oxygen &amp; chlorine are quite closer  in their electronegativity.&#xA;So, from that point of view Hg&lt;sub&gt;2&lt;/sub&gt;Cl&lt;sub&gt;2 &lt;/sub&gt;has better solubility than Cr&lt;sub&gt;2&lt;/sub&gt;(SO&lt;sub&gt;4&lt;/sub&gt;)&lt;sub&gt;3&lt;/sub&gt;.&#xA;&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="16407" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-06-13T11:41:49.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32025" PostTypeId="1" CreationDate="2015-05-26T15:58:32.543" Score="12" ViewCount="1448" Body="&lt;p&gt;I'm trying to predict the solubility of long alcohols (e.g. 1-octanol, 1-decanol or 1-dodecanol) in a mixture of water and an organic co-solvent such as DMSO, acetonitrile or ethyl acetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, decanol in water with 25% DMSO or saturated with ethyl acetate (8.3%). I of course know the solubility of alcohols in water, but not in organic solvents or in mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a server to predict the solubility of any compound in any solvent, but I'm not sure how accurate it is and it only does predictions for pure solvents, not mixtures: &lt;a href=&quot;http://showme.physics.drexel.edu/onsc/models/multisolvent.php&quot;&gt;http://showme.physics.drexel.edu/onsc/models/multisolvent.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input on this would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="16530" LastActivityDate="2015-06-06T22:13:52.910" Title="How do I predict the solubility of a compound in a mixture of water and an organic co-solvent?" Tags="&lt;organic-chemistry&gt;&lt;solubility&gt;&lt;solvents&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32027" PostTypeId="1" CreationDate="2015-05-26T18:00:36.580" Score="1" ViewCount="42" Body="&lt;p&gt;Does anyone know if I were to cool a liquid crystal with a deformed nematic phase quickly, if it will preserve the deformation in the crystal lattice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't find any information in classical books on liquid crystals.&lt;/p&gt;&#xA;" OwnerUserId="16532" LastEditorDisplayName="user15489" LastEditDate="2015-05-26T19:40:29.733" LastActivityDate="2015-05-26T19:40:29.733" Title="Deformation of the nematic phase of liquid crystals surviving in the solid state" Tags="&lt;crystal-structure&gt;&lt;phase&gt;&lt;liquid-crystals&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32028" PostTypeId="1" CreationDate="2015-05-26T18:37:28.040" Score="2" ViewCount="141" Body="&lt;p&gt;Is there a way to detect whether sugar is completely dissolved in a cup of water other than looking at the bottom of the cup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a test or something that can be carried out to determine whether sugar is fully dissolved.&lt;/p&gt;&#xA;" OwnerUserId="16535" LastEditorUserId="15410" LastEditDate="2015-05-26T19:33:31.093" LastActivityDate="2017-05-16T05:12:01.180" Title="Sugar dissolution test" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;molecules&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32029" PostTypeId="1" CreationDate="2015-05-26T19:39:25.480" Score="2" ViewCount="385" Body="&lt;p&gt;$$\ce{energy + 2KClO3 (s) &amp;lt;=&amp;gt; 2KCl(s) + 3O2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction will be more active in the reverse direction if:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{O_2}$ is added   &lt;/li&gt;&#xA;&lt;li&gt;more $\ce{KCl}$ salt is added.  &lt;/li&gt;&#xA;&lt;li&gt;More $\ce{KClO3}$ is added  &lt;/li&gt;&#xA;&lt;li&gt;Increasing the temperature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think (1) and (2) are both right. But the book states that only (1) works and I can't find reason for that searching google was useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_c = [\ce{O2}]^3$, so if we increased $\ce{KCl}$ or decreased $\ce{KClO3}$ it won't affect the equilibrium?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe(OH)(s) &amp;lt;=&amp;gt; Fe^{2+}  + 2OH^-}$$ The reaction will be more active in the reverse direction if : ... is added&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{KOH}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Na_2S}   $&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Fe(OH)_2}  $ &lt;/li&gt;&#xA;&lt;li&gt;$\ce{Fe(NO3)2}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Won't it be (1) and (4)? The book says it's (4) only.&lt;/p&gt;&#xA;" OwnerDisplayName="user16485" LastEditorUserId="5026" LastEditorDisplayName="user16485" LastEditDate="2018-02-19T04:10:50.080" LastActivityDate="2018-02-19T04:10:50.080" Title="Chemical equilibrium and Le Chatelier's principle" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="17" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32030" PostTypeId="2" ParentId="32029" CreationDate="2015-05-26T20:17:47.907" Score="1" Body="&lt;p&gt;1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.&lt;br&gt;&#xA;2) Out of system due to precipitation.&lt;br&gt;&#xA;3) Out of system due to precipitation.&lt;br&gt;&#xA;4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.  &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-05-26T20:17:47.907" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32032" PostTypeId="2" ParentId="31949" CreationDate="2015-05-26T20:42:54.753" Score="1" Body="&lt;p&gt;There are two simple ways to calculate charges on atoms which can be done by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first relies on the concept of &lt;em&gt;oxidation numbers&lt;/em&gt;. To define these, draw the Lewis structure of a target molecule/ion and &lt;em&gt;break all covalent bonds by transferring both electrons in the bond to the more electronegative atom&lt;/em&gt; (usually based on the Pauling scale of electronegativity). In this sense, by definition oxygen only presents a positive charge when either an electron is directly removed from oxygen, or when it shares a bond with fluorine. There are relatively few examples of this, as $\ce{O-F}$ bonds are weak and oxygen atoms are hard to ionize. Examples of molecules with $\ce{O-F}$ bonds and positive oxidation number on the oxygen atom are &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_difluoride&quot; rel=&quot;nofollow&quot;&gt;$\ce{OF2}$&lt;/a&gt; (+2) and &lt;a href=&quot;http://en.wikipedia.org/wiki/Dioxygen_difluoride&quot; rel=&quot;nofollow&quot;&gt;$\ce{O2F2}$&lt;/a&gt; (+1). An interesting example of oxygen with positive oxidation number happens in the dioxygenyl cation, &lt;a href=&quot;http://en.wikipedia.org/wiki/Dioxygenyl&quot; rel=&quot;nofollow&quot;&gt;$\ce{O2^{+}}$&lt;/a&gt; (+0.5 on both atoms), which is quite difficult to produce except upon reaction with highly fluorinated substances such as $\ce{PtF5}$ or $\ce{AsF5}$. These are actually avid enough for electrons that they tear one off the neutral dioxygen molecule $\ce{O2}$, producing the ionic salts $\ce{O2^+[PtF5]^{-}}$ and $\ce{O2^+[AsF5]^{-}}$, among others. The sum of all oxidation numbers in a molecule/ion must equal its net charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way to calculate charges relies on the concept of &lt;em&gt;formal charges&lt;/em&gt;. To find these, draw the Lewis structure of a target molecule/ion and &lt;em&gt;break all covalent bonds right in the middle, transferring one electron to each of the bonding atoms, regardless of electronegativity&lt;/em&gt;. This definition does not rely on electronegativity, so as one might suspect, it is much more common for oxygen to present a positive formal charge. Simple examples (all of the following with a positive formal charge of +1) include the hydronium cation $\ce{H3O^+}$, trialkyloxonium cations such as trimethyloxonium &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethyloxonium_tetrafluoroborate&quot; rel=&quot;nofollow&quot;&gt;$\ce{(CH3)3O^+}$&lt;/a&gt;, the &lt;strong&gt;central&lt;/strong&gt; oxygen atom in ozone &lt;a href=&quot;http://en.wikipedia.org/wiki/Ozone&quot; rel=&quot;nofollow&quot;&gt;$\ce{O3}$&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrylium_salt&quot; rel=&quot;nofollow&quot;&gt;pyrylium aromatic rings&lt;/a&gt;, the last of which are found in many natural compounds. The sum of all formal charges in a molecule/ion must also equal its net charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A more pictorial explanation for the ways to calculate oxidation numbers and formal charges can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/Formal_charge#Formal_charge_compared_to_oxidation_state&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, which of these two concepts is correct? In reality, &lt;em&gt;neither&lt;/em&gt;. They're both extremes of a scale, and reality often sits somewhere in the middle. Both are meant more as a bookkeeping mechanism to keep track of the total charge in an ion/molecule, and provide some qualitative idea of where electrons might be more concentrated or sparse. Attempts at determining precise atomic charges rely on complicated calculations in computational chemistry. These calculations need not agree (and most often do not agree) with either of the values given by oxidation numbers or formal charges, and the actual charges can be represented by any real number, rarely being an integer or a simple fraction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of this, let's look at your claim. You suggest that fluorine takes electrons from oxygen while other halogens don't. Let us take the two simple examples $\ce{OF2}$ and $\ce{Cl2O}$. These are both gasses at room temperature, which suggests bonding in these substances is not predominantly ionic, so no atom abstracts electrons completely from any other; &lt;em&gt;instead they share their electrons&lt;/em&gt;. Thus, they are well-described by considering the oxygen and halogen atoms are covalently bound into a molecule much like water, except with halogen atoms in place of hydrogen atoms. Applying the concept of oxidation numbers, the oxygen atom in $\ce{OF2}$ has an oxidation number of +2 as stated previously, while in $\ce{Cl2O}$ the oxygen has an oxidation number of -2. Applying the concept of formal charges, we find that the formal charge on the oxygen atom is exactly zero in both molecules. Neither concept is correct, so we expect reality to be something in between. Exactly what the charge is, we don't know without resorting to computer calculations, though it is probably quite safe to say at least that the oxygen atom is more deficient in electron density in $\ce{OF2}$ than in $\ce{Cl2O}$.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-05-26T21:20:32.160" LastActivityDate="2015-05-26T21:20:32.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32033" PostTypeId="2" ParentId="24946" CreationDate="2015-05-26T21:07:53.270" Score="2" Body="&lt;p&gt;Placing the reaction vessel in ice will lower the temperature of the reaction and thus slow down the reaction. Temperature is a measure of the average kinetic energy of the molecules, so lowering the temperature lowers their kinetic energy. In order for reactants to form products, they must collide with a certain amount of energy. Thus, at a lower temperature it will take longer for the same amount of product to be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to keep in mind that at lower temperatures the equilibrium constant will change. Since this is an exothermic reaction, at lower temperatures there will be more product formed (Le Chatelier's principle). There will be more moles of oxygen present in the iced reaction. However, you're measuring the &lt;em&gt;volume&lt;/em&gt; of oxygen, and a colder gas will have a lower volume (for the same number of moles of gas). So you've got two opposing trends: the ice will cause more oxygen to be produced (which would normally correlate with a larger volume) but the colder temperature will also cause the gas to shrink. Without more information, such as the temperature of the two reactions, I don't think you can determine whether you will wind up with a larger or smaller volume, so it would be hard to determine the shape of the graph for the colder reaction.&lt;/p&gt;&#xA;" OwnerUserId="16290" LastActivityDate="2015-05-26T21:07:53.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32034" PostTypeId="1" AcceptedAnswerId="32541" CreationDate="2015-05-26T21:19:10.040" Score="14" ViewCount="1628" Body="&lt;p&gt;&lt;em&gt;Ok so this is the proposed mechanism for the Wittig Reaction as stated by my textbook, LG Wade. I have some doubts regarding this mechanism:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Njn4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/loiVT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Why is only triphenyl phosphine used here? Why not some stronger&#xA;  nucleophile like say trimethyl phosphine?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4dmMH.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;  2) Why would the lone pair on P attack the C of the alkyl halide in the&#xA;  first place? Isn't the lone pair in resonance with the benzene ring?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3) Why is butyllithium used particularly.Can't any other reagent be&#xA;  used to abstract H ?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;4) Why is the four membered ring formed in step 2?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5) Why does the ring collapse in step 3?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: I'm just a beginner in organic chemistry (Org 2) so it would be helpful if you can keep adjust your answer to suit my level of understanding.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="16683" LastEditDate="2017-08-15T12:11:59.760" LastActivityDate="2017-08-15T12:11:59.760" Title="Clarifications about the mechanism of the Wittig reaction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;&lt;organophosphorus-compounds&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ClosedDate="2017-08-15T12:11:34.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="32036" PostTypeId="1" AcceptedAnswerId="32039" CreationDate="2015-05-26T21:34:55.617" Score="5" ViewCount="234" Body="&lt;p&gt;This is the &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Dieckmann_condensation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; image for the reaction mechanism:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBWsg.png&quot; alt=&quot;diekmann reaction&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the final protonation step, why does the enol tautomerize to the keto form?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the enol form the carbon-carbon double bond is in resonance with the ester's carbonyl double bond. Also the enol's $\ce{-OH}$ group would form a H-bond.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the enol form be more stable/predominant?&lt;/p&gt;&#xA;" OwnerUserId="14846" LastEditorUserId="4231" LastEditDate="2015-05-26T22:30:55.167" LastActivityDate="2015-06-20T13:50:36.240" Title="Dieckmann condensation's final protonation step" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32038" PostTypeId="1" AcceptedAnswerId="32051" CreationDate="2015-05-26T22:01:35.447" Score="27" ViewCount="4075" Body="&lt;p&gt;The low energy portion (the part dealing with the $\ce{2s}$ and $\ce{2p}$ electrons) of the photoelectron (PE) spectrum of methane is reproduced below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEqOz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://bilbo.chm.uri.edu/CHM401/fall2008exam1-4.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction being examined is the following one-photon process&#xA;$$\ce{CH4 + ~h\nu~ -&amp;gt; [CH4]^{+} + ~e^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that 2 peaks in roughly 1:3 intensity are observed is often used to argue against the $\ce{sp^3}$ hybridization model for methane. Simply put, how can a model with 4 &lt;strong&gt;equivalent&lt;/strong&gt; $\ce{sp^3}$ bonds produce 2 peaks in the PE spectrum.  The argument then goes on to point out that, according to MO theory, methane has one triply degenerate bonding MO and a second single bonding MO and these two nonequivalent orbitals would explain the 2 peaks in the PE spectrum with their 1:3 ratio (see p. 827 &lt;a href=&quot;https://books.google.com/books?id=gY-Sxijk_tMC&amp;amp;pg=PA827&amp;amp;lpg=PA827&amp;amp;dq=photoelectron%20spectrum%20of%20methane&amp;amp;source=bl&amp;amp;ots=euuFhfImU9&amp;amp;sig=8fFX2AOQSoo4euVm3NChqvqjEak&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=l9xkVdGoHYrzsAWd7oCQCw&amp;amp;ved=0CDUQ6AEwBDgK#v=onepage&amp;amp;q=photoelectron%20spectrum%20of%20methane&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for an example of such an MO-based explanation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PE spectrum of methane shows 2 transitions with 1:3 intensity.  How does one &lt;strong&gt;know&lt;/strong&gt; that it is the methane ground state that is responsible for producing two peaks with a 1:3 ratio in the PE spectrum?  Why couldn't one argue that it is $\ce{[CH4]^+}$, where we clearly have 3 $\ce{C-H}$ bonds containing a pair of electrons and one $\ce{C-H}$ bond with one electron - two different types of $\ce{C-H}$ bonds in a 1:3 ratio - that is responsible for the two peaks in the PE spectrum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the PE spectrum of methane really be used as evidence to argue against the hybridization model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poking around on the internet, I came across this Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Orbital_hybridisation#Photoelectron_spectra&quot; rel=&quot;noreferrer&quot;&gt;Orbital Hybridization - Photoelectron Spectroscopy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;There is a popular misconception that the concept of hybrid orbitals&#xA;  incorrectly predicts the ultraviolet photoelectron spectra of many&#xA;  molecules. While this is true if Koopmans' theorem is applied to&#xA;  localized hybrids, quantum mechanics requires that the (in this case&#xA;  ionized) wavefunction obey the symmetry of the molecule which implies&#xA;  resonance in valence bond theory. &lt;strong&gt;For example, in methane&lt;/strong&gt;, the ionized&#xA;  states (CH4+) can be constructed out of four resonance structures&#xA;  attributing the ejected electron to each of the four sp3 orbitals. &lt;strong&gt;A&#xA;  linear combination of these four structures, conserving the number of&#xA;  structures, leads to a triply degenerate T2 state and a A1 state.[16]&#xA;  The difference in energy between each ionized state and the ground&#xA;  state would be an ionization energy, which yields two values in&#xA;  agreement with experiment&lt;/strong&gt;.&quot; (emphasis mine)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bold face portion seems to support my contention (if I'm understanding it correctly) that it is the energy differences between the ground state and the excited state that is responsible for the spectrum.  Therefore, orbital splitting and orbital degeneracies in the excited state are likely an explanation for the observed spectrum.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-27T17:02:10.090" LastActivityDate="2017-09-14T18:46:59.407" Title="Are the bonding orbitals in methane equivalent - photoelectron spectrum" Tags="&lt;bond&gt;&lt;spectroscopy&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32039" PostTypeId="2" ParentId="32036" CreationDate="2015-05-26T22:19:25.410" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why does the enol tautomerizes to the keto form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In 1,3-dicarbonyl systems, like the one pictured above, we would generally expect to have a measurable amount of the enol form present&lt;/strong&gt;.  The keto-enol interconversion is an &lt;strong&gt;equilibrium&lt;/strong&gt; and ratio of keto and enol will be guided by &lt;strong&gt;thermodynamic&lt;/strong&gt; factors.  Let's take a look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the structural elements in the keto and enol forms, we see that the differences between the two tautomers are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The keto form has an extra carbonyl (178), a $\ce{C-C}$ single bond (83) and a $\ce{C-H}$ bond (99)&lt;/li&gt;&#xA;&lt;li&gt;The enol form has a $\ce{C-O}$ bond (86), a $\ce{O-H}$ bond (111) and a $\ce{C=C}$ double bond (146)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The numbers in parenthesis are &lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot; rel=&quot;nofollow&quot;&gt;standard bond energies&lt;/a&gt; in kcal/mol.  If we add up the keto values we get 360 kcal/mol; while for the enol it totals to 343 kcal/mol.  The numbers strongly favor the keto form (by 17 kcal/mol).  That is why is most simple systems the keto form strongly predominates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as you point out, both resonance and hydrogen bonding can further stabilize the enol form in the current example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen bond strength in $\ce{O..H-O}$ systems is typically &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;5-6 kcal/mol&lt;/a&gt;. In this system, with an intramolecular bond and a 6-membered transition state the hydrogen bond strength is likely closer to 9-10 kcal/mol. Resonance stabilization in 1,3-pentadiene is approximately &lt;a href=&quot;https://books.google.com/books?id=1ArUAgAAQBAJ&amp;amp;pg=PA360&amp;amp;lpg=PA360&amp;amp;dq=resonance%20energy%20for%20pentadiene&amp;amp;source=bl&amp;amp;ots=ISUhVdSam8&amp;amp;sig=YfLyXxnYFH72TWBfulSy9MfBXvg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=lAhmVdPpIIGksAWb_YDQCw&amp;amp;ved=0CD8Q6AEwBQ#v=onepage&amp;amp;q=resonance%20energy%20for%20pentadiene&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;6.2 kcal/mol&lt;/a&gt;. Due to increased polarization in the enol from the carbonyl replacing a carbon-carbon double bond, the resonance stabilization probably increases to something around 8 kcal/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 9 kcal/mol effect due to hydrogen bonding and a 8 kcal/mol push due to resonance, shifts the 17 kcal/mol stabilization of the keto form to something very close to where the keto and enol forms have comparable energy. &lt;strong&gt;Therefore, in  1,3-dicarbonyl compounds we would expect a delicate balance between keto and enol forms, where both forms may be detectable.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="14846" LastEditDate="2015-06-20T13:50:36.240" LastActivityDate="2015-06-20T13:50:36.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32040" PostTypeId="1" CreationDate="2015-05-26T22:29:00.147" Score="5" ViewCount="140" Body="&lt;p&gt;Since formaldehyde gas is flammable, is there a safe way to burn it, and can this be done to dispose of the gas or neutralize its harmful effects (tear production, respiratory problems, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a more detailed form of my earlier question, including the context of &quot;gas disposal&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16538" LastEditorUserId="402" LastEditDate="2016-11-05T17:22:15.200" LastActivityDate="2016-11-05T17:22:15.200" Title="Burning formaldehyde gas for disposal?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32043" PostTypeId="2" ParentId="32038" CreationDate="2015-05-27T01:35:35.057" Score="11" Body="&lt;p&gt;&lt;strong&gt;The short answer is that's not how photoelectron spectroscopy works. It's a one-photon spectroscopy.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a sample of $\ce{CH4}$ in this case, and you shoot different energy photons at the sample. At certain energies, the photon is absorbed and you get an electron off. Note, that's a &lt;em&gt;single&lt;/em&gt; electron from one photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For you to have any significant amount of $\ce{[CH4]+}$ around, you'd have to hit it with &lt;em&gt;two&lt;/em&gt; photons. The first would do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH4 + ~energy -&amp;gt; [CH4]^{+} + ~e^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only &lt;em&gt;then&lt;/em&gt; would you be able to hit that one single $\ce{[CH4]+}$ ion with &lt;em&gt;another&lt;/em&gt; photon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a two-photon process. Assuming the incident light intensity is small, that's extremely unlikely. It's much more likely that that second photon hits a different $\ce{CH4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, normal photoelectron spectroscopy gives a set of intensities when the species absorbs a &lt;em&gt;single&lt;/em&gt; photon with enough energy to remove a &lt;em&gt;single&lt;/em&gt; electron (i.e., orbital energies). This process happens on the femtosecond timescale, so it's essentially a &lt;strong&gt;&lt;em&gt;&quot;frozen orbital&quot;&lt;/em&gt;&lt;/strong&gt; technique. There's no time to sample the electronic structure of the resulting cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the PES of $\ce{[CH4]+}$ is &lt;a href=&quot;http://www.atto.ethz.ch/publications/Woerner09a.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;completely&lt;/strong&gt; different (seen here)&lt;/a&gt;, because of the Jahn-Teller distortion. The geometry is $C_{2v}$ and thus there are many more electronic transitions - it actually gets quite complicated.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-05-27T15:06:37.047" LastActivityDate="2015-05-27T15:06:37.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32044" PostTypeId="1" AcceptedAnswerId="53742" CreationDate="2015-05-27T01:48:53.580" Score="11" ViewCount="539" Body="&lt;p&gt;I am trying to create a water drop simulation for measuring hydrophobicity of surface. I don't know how to simulate the contact angle which is related by younges equation to the surface tensions for each of the solid-liquid-gas interfaces. I have come across some models for molecular dynamics, which for chemistry novice like me (I'm a EE) are bit advanced. I am looking for simpler method in which I could simulate the interfacial tensions by some sort of derived mathematical model which would be simple enough to act as an objective function to an evolutionary optimizer.&lt;/p&gt;&#xA;" OwnerUserId="16549" LastActivityDate="2016-06-14T22:01:34.457" Title="How to simulate surface tension?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;surface-chemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32045" PostTypeId="1" CreationDate="2015-05-27T02:08:30.477" Score="8" ViewCount="2470" Body="&lt;p&gt;Suppose I have a photochemical reaction in gas phase, such as&#xA;$$&#xA;\ce{CO2 + {$h\nu$} -&amp;gt; CO + O}.&#xA;$$&#xA;I would like to work through the thermodynamics of such a reaction and understand the meaning of every term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a non-photochemical reaction, such as $\ce{CO + O -&amp;gt; CO2}$, I have a good understanding of how the change in Gibbs energy breaks down into enthalpy, internal entropy and a log-concentration term corresponding to the entropy of mixing. If I ignore the photon in the first reaction above, I can regard it as simply the reverse of this reaction, which due to its photochemical driving force is able to reduce the Gibbs energy by the same amount, $\Delta G$. All of this is unproblematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $h\nu$ itself is an energy change, which I can calculate, assuming I know the frequency $\nu$ of the absorbed light. I would like to know how to think about the thermodynamics of the whole system, including the coupling of the chemistry to the radiation field. In particular, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does the energy $h\nu$ relate to the other terms with energy units, such as $\Delta H$ and $\Delta G^0$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I know that while a single photon has no internal entropy, a beam of light of a particular frequency can be thought of as having a temperature and an entropy. How should I think about the role of radiation entropy in the thermodynamics of photochemistry, and in particular, how can I do second-law calculations for photochemical reactions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="148" LastActivityDate="2021-01-03T15:32:38.070" Title="Thermodynamics of photochemical reactions" Tags="&lt;thermodynamics&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="32047" PostTypeId="1" CreationDate="2015-05-27T02:46:40.313" Score="3" ViewCount="981" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A galvanic cell consists of a standard hydrogen electrode and a copper electrode.&#xA;  \begin{align}&#xA;  \ce{2 H+ (aq) + 2e- &amp;amp;-&amp;gt; H2 (g)}&amp;amp;&#xA;  E^\circ &amp;amp;= 0.00\ \mathrm{V}\\&#xA;  \ce{Cu^2+ (aq) + 2 e- &amp;amp;-&amp;gt; Cu (s)}&amp;amp;&#xA;  E^\circ &amp;amp;= 0.34\ \mathrm{V}&#xA;\end{align}&#xA;  Suppose the copper electrode is placed in a solution of unknown copper (II) concentration. The cell potential is measured to be $\pu{0.195 V}$. Determine the concentration of $\ce{Cu^2+}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since there's no indication that the $\ce{Cu^2+}$ is in the standard state, I started off by using the formula for non-standard cell potential:&#xA;$$E=E^\circ-\frac{RT}{nF}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can’t seem to get the right answer. I was wondering if I set up the problem incorrectly or how I go about getting the concentration of copper.&lt;/p&gt;&#xA;" OwnerUserId="16551" LastEditorUserId="4945" LastEditDate="2018-09-06T16:54:49.987" LastActivityDate="2019-05-04T03:03:25.583" Title="Calculating the concentration of copper(II) from a non-standard cell potential" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32048" PostTypeId="1" AcceptedAnswerId="32054" CreationDate="2015-05-27T03:28:48.750" Score="3" ViewCount="9953" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/foH3j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In IUPAC naming double bonds take preference over an alkane chain of any length, right? So the above should be named as &lt;code&gt;2-ethylnon-1-ene&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I verify with chemdraw it gives the answer as &lt;code&gt;3-methylenedecane&lt;/code&gt;. I cannot seem to find any mention of double bond side chains as &lt;code&gt;alkylene&lt;/code&gt; anywhere on the internet... Is this common nomenclature, and if not why would chemdraw name it as this? &lt;/p&gt;&#xA;" OwnerUserId="2142" LastEditorDisplayName="user7951" LastEditDate="2015-05-27T07:21:06.823" LastActivityDate="2016-05-04T22:08:20.473" Title="Naming double bond sidechain" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;software&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ClosedDate="2015-05-27T10:49:39.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="32049" PostTypeId="2" ParentId="32048" CreationDate="2015-05-27T03:50:12.103" Score="2" Body="&lt;p&gt;It looks like you're correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This molecule has been given 2-ethyl-1-hexene or &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Naming_the_Alkenes&quot; rel=&quot;nofollow noreferrer&quot;&gt;2-ethylhex-1-ene&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/urxn0.gif&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another name for your molecule would be 3-methylidenedecane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added from &lt;a href=&quot;http://en.wikipedia.org/wiki/Methylene_(compound)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trivial name carbene is the preferred IUPAC name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The systematic names methylidene and dihydridocarbon are constructed according to the substitutive and additive nomenclatures, respectively.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="15429" LastEditDate="2016-05-04T22:08:20.473" LastActivityDate="2016-05-04T22:08:20.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32050" PostTypeId="1" CreationDate="2015-05-27T05:46:50.340" Score="21" ViewCount="27355" Body="&lt;p&gt;In which cases is the Clemmensen reduction is preferred over the Wolff-Kishner reduction?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing is that the Clemmensen reduction is done under acidic conditions whereas the Wolff-Kishner reaction is done under basic condition, and that I think decides somewhat which to choose depending of the nature and other groups present in the reactant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there any other points to keep in mind while choosing the best suitable reaction?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="418" LastEditDate="2015-05-27T05:59:23.640" LastActivityDate="2019-06-23T07:24:43.930" Title="Clemmensen reduction vs. Wolff-Kishner reduction" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32051" PostTypeId="2" ParentId="32038" CreationDate="2015-05-27T06:14:49.480" Score="13" Body="&lt;p&gt;This question goes along the line of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31973/what-does-it-mean-when-it-is-said-that-an-sp%C2%B3-orbital-has-25-s-character&quot;&gt;what does it mean when it is said that an sp3 orbital has 25% s-character&lt;/a&gt;. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity is 3:1. You can say, it is $3\times 2p : 1 \times 2s$ and you are done. Not at all. The most important thing to deal with is the $T_{d}$ symmetry of the molecule. That means, all four $\ce{C-H}$ bonds are equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is this symmetry which has the answer. Leaving out the $1s$ orbital of carbon, the occupied orbitals &lt;strong&gt;must&lt;/strong&gt; have the following symmetries:&#xA;$1\times A_1$ and $3\times T_2$. And this is what you see in the spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how it comes with the hybridisation? Upon simple HF/STO-3G calculation, the orbital populations are following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A_1$: 0.62 C 2s + 4x 0.18 H 1s  ... fully symmetric, great&lt;/li&gt;&#xA;&lt;li&gt;$T_2$: 0.57 C pz + 0.3 1H 1s + 0.3 2H 1s - 0.3 3H 1s - 0.3 4H 1s&lt;/li&gt;&#xA;&lt;li&gt;$T_2$: 0.57 C py + 0.3 1H 1s - 0.3 2H 1s - 0.3 3H 1s + 0.3 4H 1s&lt;/li&gt;&#xA;&lt;li&gt;$T_2$: 0.57 C px + 0.3 1H 1s - 0.3 2H 1s + 0.3 3H 1s - 0.3 4H 1s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Well, as if you would copy the symmetry table ;) The downside of this picture is that all the bonds are completely delocalised (as you would expect from canonical MOs). Although observable through PES, they do not allow you to draw line between any C and H atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do we get the hybridised orbitals? If we want to localize the orbitals, whatever scheme you choose, you just form linear combinations of the existing MOs so that you have least number of atoms involved in any bond. One of such methods is NBO and as a result you obtain:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C - H1  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)&#xA;        ( 47.45%)   0.6888* H  1 s(100.00%)&#xA;C - H2  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)&#xA;        ( 47.45%)   0.6888* H  2 s(100.00%)&#xA;C - H3  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)&#xA;        ( 47.45%)   0.6888* H  3 s(100.00%)&#xA;C - H4  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)&#xA;        ( 47.45%)   0.6888* H  4 s(100.00%)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All with 2 electron occupation and energy -0.63479 Eh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where is the truth? ... $\Psi$ is the answer and you can massage it to your will, depending what you want to see. You want localized bonds? You will get them. You want orbitals with the proper symmetry required by spectroscopy? As you wish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the hybridisation concept have sense? In localized bond picture yes. Is it observable? No, and was never meant to be.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-05-27T06:14:49.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32052" PostTypeId="1" CreationDate="2015-05-27T06:20:16.617" Score="7" ViewCount="1132" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/69cL4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/69cL4.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the ketone shown, I plan to use a haloform reaction to transform the acetyl group into a carboxylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following that, there are two alkylations to be carried out: one esterification on the carboxylic acid (using propargyl bromide) and one etherification on the alcohol (using 6-bromohexan-1-ol). In what order should these be performed?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="16683" LastEditDate="2018-08-04T14:11:09.183" LastActivityDate="2018-08-04T14:12:30.320" Title="Selective alkylation of carboxylic acid versus phenol" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;phenols&gt;&lt;chemoselectivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32054" PostTypeId="2" ParentId="32048" CreationDate="2015-05-27T06:54:59.110" Score="1" Body="&lt;p&gt;According to IUPAC rules first thing you need to see is &lt;strong&gt;longest carbon chain&lt;/strong&gt; which is 10(deca) over here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then while &lt;strong&gt;numbering the Carbon&lt;/strong&gt; double bond is given preference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So above compound is 10 carbon chain in which 3rd carbon is double bonded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;strong&gt;3-methylenedecane&lt;/strong&gt; is correct name. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Few more such examples to make it more clear:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;2,6,10-trimethyldodec-1-ene&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h9UTt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;8-ethyldec-1-ene&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJJ7t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example Source: &lt;a href=&quot;http://www.chemsink.com/index.py?q=CCCCCCCC%28%3DC%29CC&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemSink&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic IUPAC rules:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the longest chain in the molecule and name it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Name all groups attached to the chain as alkyl substituents.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Number the carbons of the longest chain beginning with the end that is closest to a substituent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Write the name of the alkane by first arranging all of the substituents in numerical order (each preceded by the number of the carbon to which it is attached and a hyphen), and then add the name of the stem.&#xA;Haloalkanes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The halogen is treated as an alkyl substituent, without any priority.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ethers are named as alkoxyalkanes. The first carbon of the alkoxy substituent is numbered as C-1 of that substituent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note that these rules are changed  as of 2008. Older HW and exams have slightly different answers.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules Source: &lt;a href=&quot;http://valhalla.chem.udel.edu/nomenclature.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://valhalla.chem.udel.edu/nomenclature.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-05-27T07:08:45.993" LastActivityDate="2015-05-27T07:08:45.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32055" PostTypeId="2" ParentId="32045" CreationDate="2015-05-27T07:13:21.270" Score="5" Body="&lt;p&gt;$h\nu$ is not a thermal energy so thermodynamic considerations do not apply. For example, you cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics.  The main reason is that the excitation is not a result of (classical) thermal energy transfer, but rather the adsorption of a photon which changes the quantum state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a constant light source &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2} + h\nu \ce{-&amp;gt; CO + C}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;will reach a steady state with constant concentrations that will be different from the thermal equilibrium concentrations.  However, you can not extract a standard free energy difference by&#xA;$$ \Delta G^\circ=-RT\ln \left( \frac{p_{\ce{CO}}p_{\ce{O}}}{p_{\ce{CO2}}} \right) $$&#xA;because the products and reactant are not in thermal equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-05-27T11:19:21.053" LastActivityDate="2015-05-27T11:19:21.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32056" PostTypeId="1" AcceptedAnswerId="32078" CreationDate="2015-05-27T07:28:03.413" Score="3" ViewCount="1005" Body="&lt;p&gt;I'm using the Lennard-Jones potential for some (very simple) protein docking simulation* and couldn't find parameters for different materials (like $\ce{C}$, $\ce{N}$, $\ce{O}$, …). Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* the potential function for each configuration of the two molecules is the sum over all pairwise energies, for which I only consider the van der Waals term by using the LJ-potential&lt;/p&gt;&#xA;" OwnerUserId="15084" LastEditorUserId="5017" LastEditDate="2015-05-27T17:17:54.583" LastActivityDate="2015-05-27T19:49:10.603" Title="Fitting parameters of Lennard-Jones potential to properties of real materials" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32058" PostTypeId="1" CreationDate="2015-05-27T08:15:37.950" Score="10" ViewCount="936" Body="&lt;p&gt;I suppose that my problem is not one of great profundity, but it is an annoying one. The problem is related to the measurement units involved in rate equations of different order. Not being a chemist myself, I have in my work encountered equations of the type: $$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^b [A]^a$$ representing the decay of some species $A$. The brackets denote concentration and $k$ is the rate constant. $B$ is some other chemical species (cooking chemical) and then there are the exponents $a$, $b$. For a first order reaction with $b = 0$ and $a = 1$ everything is still under some kind of control; if time is measured in minutes the unit of the rate constant is $\mathrm{min^{-1}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we have a pseudo first order reaction where the species $B$ is present albeit constant during an experimental run, problems arise. In my line of work the species $B$ is typically hydroxide ions and their concentration is traditionally given in molar, $\mathrm{mol/L}$. Even if the concentration of $B$ does not change, the initial concentration plays a role, hence the power of $[B]$ usually appears as a separate factor in the rate equation. Unfortunately, this complicates the unit balance in the equation as frequently the exponent $b$ is a non-integer number. One workaround would be to rescale the concentration to a dimensionless number, but the choice of scaling parameter would be rather arbitrary. Another, perhaps more elegant solution would be to work with molar fractions, but this would be discordant with most of the literature in the field, where the question of unit consistency mostly is disregarded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second order rate equation ($b = 0$, $a = 2$, or $b = 1$, $a = 1$) leads to the unit $\mathrm{L\,mol^{-1}\,min^{-1}}$ for the rate constant. I suppose this is still somehow acceptable, even if I would prefer to work with scaled concentrations (divided by the initial concentrations) with unit $\mathrm{min^{-1}}$ for the rate constant, but for fractional order rate equations there is the same problem as in the earlier described pseudo first order case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any prescribed remedy to this dilemma?&lt;/p&gt;&#xA;" OwnerUserId="16557" LastEditorDisplayName="user7951" LastEditDate="2015-05-27T09:19:41.783" LastActivityDate="2015-05-28T01:09:31.627" Title="Unit consistency in rate equations" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32059" PostTypeId="1" CreationDate="2015-05-27T08:16:55.003" Score="4" ViewCount="1765" Body="&lt;p&gt;I need to know how to go about creating synthetic glucose with generally available materials. As I am aware of both photosynthesis and glucose in nature, I will not need lessons in how plants produce it. So, how may glucose be produced by artificial means?&lt;/p&gt;&#xA;" OwnerUserId="16558" LastEditorUserId="9961" LastEditDate="2015-05-28T15:24:19.587" LastActivityDate="2015-05-28T15:24:19.587" Title="What are the steps to creating synthetic glucose?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32060" PostTypeId="1" CreationDate="2015-05-27T08:22:27.927" Score="2" ViewCount="2035" Body="&lt;p&gt;I premise, I’m a physicist and I’m working on a transfer procedure for graphene via CVD, which involves the use of PMMA as a resist layer. Studying this polymer, I still haven’t found what is the cross-link temperature (or window of temperatures) for it. Can someone give me a hand? or just a good reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE&#xA;Ok, maybe I’ve been misunderstood. My PMMA is diluted in anisole, and I apply it as a photoresist over the graphene sheet via spin coating. After this deposition NO Initiator is used, but it came across various heat treatment, depending in the recipe I use. My question is: does anybody know at what temperature the PMMA starts to seriously crosslink? or better: if I do more heat treatments, all at temperatures above 150 °C, am I going to have an highly crosslinked PMMA, so that a removal process in acetone solvent is impossible because it can’t get into the chains to districate them?&lt;/p&gt;&#xA;" OwnerUserId="16556" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T08:52:10.700" LastActivityDate="2015-06-17T08:52:10.700" Title="PMMA cross-link temperature" Tags="&lt;polymers&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32061" PostTypeId="2" ParentId="32060" CreationDate="2015-05-27T09:25:54.350" Score="2" Body="&lt;p&gt;According to this &lt;a href=&quot;http://scholar.lib.vt.edu/theses/available/etd-12202001-135708/unrestricted/ThesisChapter2Part1B.PDF&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt;, the polymerization is determined by thermal initiator that is used in PMMA mixture. The polymerization (corsslinking) starts when the initator is added to the mixture, but it is really slow at low temperature. Higher temperature lowers hardening time (speeds up the crosslinking). Since the dustsheet does not provide the information about the initiator, I would suggest to determine the cross-linking temperature experimentally. Some material related to the polymerization process:&lt;a href=&quot;http://www.chemistry2011.org/ResourceFiles/10.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.chemistry2011.org/&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.listdryprocessing.com/fileadmin/user_upload/download/publications/bulk-polymerization-of-methyl-methacrylate-in-a-kneader-reactor.pdf&quot; rel=&quot;nofollow&quot;&gt;BULK POLYMERIZATION OF METHYL METHACRYLATE IN A KNEADER&#xA;REACTOR&#xA;P.-A. Fleury&#xA;LIST AG&#xA;CH-4422 Arisdorf&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemicalpapers.com/file_access.php?file=374a555.pdf&quot; rel=&quot;nofollow&quot;&gt;High conversion polymerization of methyl methacrylate&#xA;in the presence of poly(methyl methacrylate) of different relative&#xA;molecular mass&#xA;J. BARTOŇ, V. PASTUŠÁKOVÁ, V. VAŠKOVÁ, and P. AMBROVIČ&#xA;Polymer Institute, Centre of Chemical Research, ChemicalPapers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-06-17T08:31:01.683" LastActivityDate="2015-06-17T08:31:01.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32062" PostTypeId="1" CreationDate="2015-05-27T09:54:09.600" Score="6" ViewCount="268" Body="&lt;p&gt;Why is radium oxide yet to be isolated when a wide range of other radium (II) compounds are known ? &lt;/p&gt;&#xA;" OwnerUserId="10256" LastEditorUserId="402" LastEditDate="2016-11-05T17:22:25.060" LastActivityDate="2016-11-05T17:22:25.060" Title="Isolating radium oxide" Tags="&lt;inorganic-chemistry&gt;&lt;solid-state-chemistry&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32063" PostTypeId="1" CreationDate="2015-05-27T10:12:03.507" Score="5" ViewCount="1615" Body="&lt;p&gt;I read that resonance in toluene is initiated with the help of hyper-conjugation. Also hyper-conjugation is weaker than resonance/mesomeric effect. But, in toluene, its resonance is taking place with the help of hyper-conjugation, am I right? I am confused which has greater +R strength - toluene or anisole?&lt;/p&gt;&#xA;" OwnerUserId="16560" LastEditorUserId="5026" LastEditDate="2018-02-20T07:46:45.150" LastActivityDate="2018-10-22T00:23:40.397" Title="Which has greater +R strength - toluene or anisole?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32064" PostTypeId="2" ParentId="32059" CreationDate="2015-05-27T10:31:34.593" Score="6" Body="&lt;p&gt;A relatively simple process is described by Kobayashi et. al where they use cellulose, a structural component of plants, which is ball milled, mixed with activated carbons and HCl and hydrolysed to produce glucose in high yield (88%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It requires only easily available ingredients and the ability to heat (and preferably stir) to 453K. Reference is given below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-Yielding One-Pot Synthesis of Glucose from Cellulose Using Simple Activated Carbons and Trace Hydrochloric Acid&#xA;Hirokazu Kobayashi, Mizuho Yabushita, Tasuku Komanoya, Kenji Hara, Ichiro Fujita, and Atsushi Fukuoka&#xA;ACS Catalysis 2013 3 (4), 581-587&#xA;DOI: 10.1021/cs300845f&lt;/p&gt;&#xA;" OwnerUserId="15551" LastActivityDate="2015-05-27T10:31:34.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32065" PostTypeId="2" ParentId="32052" CreationDate="2015-05-27T12:22:13.470" Score="7" Body="&lt;p&gt;Using the haloform reaction to convert the methyl ketone &lt;strong&gt;1&lt;/strong&gt; to the corresponding carboxylic acid &lt;strong&gt;2&lt;/strong&gt; is a neat idea. According to an old review by &lt;em&gt;R. C. Fuson&lt;/em&gt; and &lt;em&gt;B. A. Bull&lt;/em&gt;, published in &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;1934&lt;/strong&gt;, &lt;em&gt;15&lt;/em&gt;, 275-309 (&lt;a href=&quot;http://dx.doi.org/10.1021/cr60052a001&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;), this seems to work for 4-hydroxyacetophenone. Your starting material &lt;strong&gt;1&lt;/strong&gt; is a phenyl homolog of that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EFCil.png&quot; alt=&quot;synthesis&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;2&lt;/strong&gt;, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the &lt;em&gt;Rubber Book&lt;/em&gt;), you know that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the carboxylic acid is more acidic than the phenol&lt;/li&gt;&#xA;&lt;li&gt;the difference is acidity is quite large &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as &lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-dimethylformamide (DMF) should be sufficient to deprotonate both centres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your question on the order of events:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add &lt;strong&gt;1 equivalent&lt;/strong&gt; $\ce{K2CO3}$ of to a solution of &lt;strong&gt;2&lt;/strong&gt;, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monitoring the reaction by TLC will show whether the esterification is quantitative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, I would dare to run a &lt;strong&gt;one-pot reaction&lt;/strong&gt;, add &lt;strong&gt;another equivalent&lt;/strong&gt; of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a good idea to try this first with very small amounts, just enough to check via TLC.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="16683" LastEditDate="2018-08-04T14:12:30.320" LastActivityDate="2018-08-04T14:12:30.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32066" PostTypeId="2" ParentId="16507" CreationDate="2015-05-27T12:25:05.523" Score="4" Body="&lt;p&gt;Although you should take a look at what Klaus said, copper does in fact react with hydrochloric acid, it just takes a week until all the copper is converted into copper chloride (green) and another week or so until it forms crystals and you can dissolve them in water to form copper chloride again (but depending on the amount of chloride it has, it'll be blue or green).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HCl (aq) + Cu(s) -&amp;gt; CuCl2(aq) + H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper chloride reacts with aluminium to form copper (metal) and aluminium chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{3CuCl2(aq) + 2Al(s) -&amp;gt; 3Cu(s) + 2AlCl3(aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="16562" LastEditorUserId="15410" LastEditDate="2015-05-27T13:17:39.760" LastActivityDate="2015-05-27T13:17:39.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32067" PostTypeId="1" CreationDate="2015-05-27T13:49:30.150" Score="4" ViewCount="332" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(OH)2 (s) &amp;lt;=&amp;gt; Fe^{2+} + 2OH-}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reaction will be more active in the reverse direction if .......... is added&lt;/p&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;b) &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2S}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;c) &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(OH)2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;d) &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My thoughts:&lt;/p&gt;&#xA;&lt;p&gt;Well, adding more &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt; will increase the &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions so the reaction will go in the reverse direction to get back to equilibrium. Adding &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)2}$&lt;/span&gt; Will increase &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+}}$&lt;/span&gt; ions and the reaction will go in the reverse direction too.&lt;/p&gt;&#xA;&lt;p&gt;However, the book answers states that only (d) &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)2}$&lt;/span&gt;. I searched google but all the examples involved having only choice which works.&lt;/p&gt;&#xA;&lt;p&gt;Note: Refer me to sites if necessary. I am ready to read.&lt;/p&gt;&#xA;" OwnerUserId="16563" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-22T01:00:50.693" Title="Which of the following compounds would result in the reaction being more active in the reverse direction?" Tags="&lt;equilibrium&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32068" PostTypeId="2" ParentId="32067" CreationDate="2015-05-27T15:52:15.940" Score="7" Body="&lt;p&gt;In fact, adding more ion hydroxide won't displace the equilibrium towards the reverse reaction. This is due to the fact that ion complex &lt;span class=&quot;math-container&quot;&gt;$\ce{[Fe(OH)4]^{-2}}$&lt;/span&gt;, which is stable in presence of excess of ion hydroxide, is formed by successive addition of ion hydroxide on ferrous hydroxide.&#xA;If you want to study this more in depth, I invite you to read this &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Complex_Ion_Chemistry/The_Formation_of_Complex_Ions&quot; rel=&quot;noreferrer&quot;&gt;webpage&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It's explained at the end of the page under &amp;quot;numerical problems&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-27T15:52:15.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32070" PostTypeId="1" AcceptedAnswerId="32072" CreationDate="2015-05-27T15:56:23.950" Score="3" ViewCount="836" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the most remarkable success of the Bohr's theory is that it provides a satisfactory explanation for the line spectrum of hydrogen.&quot;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what is the problem with line spectrum of hydrogen which could not explained before the Bohr's theory was introduced and how it was explained by Bohr's theory? &lt;/p&gt;&#xA;" OwnerUserId="16564" LastActivityDate="2015-05-27T16:41:07.740" Title="Bohr's model of atom" Tags="&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32071" PostTypeId="2" ParentId="31973" CreationDate="2015-05-27T16:15:25.963" Score="3" Body="&lt;p&gt;I think a good way to understand the importance of hybridization is to start with an illustration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have two structures: $\ce{SnCl2}$ and $\ce{SnCl4}$, and you are asked to rationalize why $\ce{SnCl2}$ has a stronger bond than $\ce{SnCl4}$ (this information is given). How would you do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are two ways to approach this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can skip this if you are not interested in the argument, but it is fairly straightforward)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The oxidation state of Sn is +4 in $\ce{SnCl4}$, while it is +2 in $\ce{SnCl2}$. The higher the oxidation state of the atom, the valence orbitals will be more contracted and lower in energy as the effective nuclear charge increases without a compensating increase in electron-electron repulsion (no electrons are added obviously). Therefore, as the coordination number is higher for $\ce{SnCl4}$ and the orbitals are more compact, this will weaken the $\ce{Sn-Cl}$ bond in $\ce{SnCl4}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A simpler way though, would be to say that according to the $\ce{SnCl4}$ geometry, it has to be $sp^3$ hybridized ($\ce{Sn}$ is group IV like carbon), and in $\ce{SnCl2}$, it will be $sp^2$ hybridized (2 $sp^2$-$p$ $\sigma$ bonds, 1 lone pair of electrons and 1 empty orbital). &lt;strong&gt;As the $sp^3$ bond has 25% $s$ character but the $sp^2$ bond has 33% $s$ character, and as the $s$ orbital is lower in energy than the $p$ orbital due to penetration effects, this means that the $\ce{Sn-Cl}$ bond in $\ce{SnCl2}$ will be stronger than that of $\ce{SnCl4}$.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So that's one use of the hybridization argument. Hope this helps in some way.&lt;/p&gt;&#xA;" OwnerUserId="16445" LastEditorDisplayName="user15489" LastEditDate="2015-05-27T21:41:45.140" LastActivityDate="2015-05-27T21:41:45.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32072" PostTypeId="2" ParentId="32070" CreationDate="2015-05-27T16:23:25.283" Score="5" Body="&lt;p&gt;Since this is pretty well covered in countless number of books and other sources I will try to just emphasise the key points.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The emission spectrum of atomic hydrogen was well known at that time, the only problem was that no one was able to explain it. :D Where does all this lines come from, i.e. what physical process gives rise to such discrete spectrum?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The wavelengths of spectral lines were known to be very well predicted by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_formula&quot;&gt;Rydberg formula&lt;/a&gt;, but again no one was able to explain why this formula works, i.e. what are its physical justifications.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bohr model explained discrete spectrum as follows: spectral lines are due to electron transitions between stationary orbits. Bohr was also able to determine the actual energy spacing between levels in hydrogen atom and explain the reason for the structure of the Rydberg formula.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, Bohr failed to predict the lines in the emission spectrum even for helium atom. It happened simply because Bohr model is physically  &lt;em&gt;wrong&lt;/em&gt;. Electrons do &lt;em&gt;not&lt;/em&gt; orbit around the nucleus.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-27T16:41:07.740" LastActivityDate="2015-05-27T16:41:07.740" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32073" PostTypeId="1" CreationDate="2015-05-27T16:56:33.323" Score="5" ViewCount="116" Body="&lt;p&gt;I understand the basics of carbonic acid + 1 kcal breaking down into water and carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my professor wants me to draw an energy map showing the catalyst in human blood reducing the activation energy from 21 kcal to 12 kcal. I've never run into an energy map nor have we discussed this catalyst in class yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have a resource I can look at?&lt;/p&gt;&#xA;" OwnerUserId="16566" LastEditorUserId="10185" LastEditDate="2015-05-27T18:07:03.367" LastActivityDate="2015-05-27T18:07:03.367" Title="Catalyzation in Carbonic Acid" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;energy&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32074" PostTypeId="2" ParentId="32056" CreationDate="2015-05-27T17:17:16.880" Score="2" Body="&lt;p&gt;You could, in principal, create some simulation of &quot;Lennard-Jones-ium&quot; based on a set of tailored parameters. There's probably some benefit from an applied mathematics perspective. &lt;strong&gt;However, as discussed by multiple comments, this wouldn't be at all physically realistic.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most force fields have many atom types - not just C vs. O, but $\ce{C_\alpha}$ or $\ce{C_{O2}}$ etc., because the parameters vary considerably in terms of interatomic forces, etc. for different atom types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the bonding interactions between two carbons are very different than non-bonded interactions. In other words, if you wanted a model for a protein and docking simply from LJ parameters, you'd need separate pair-wise parameters for &lt;em&gt;bonded&lt;/em&gt; and &lt;em&gt;non-bonded&lt;/em&gt; interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, for &lt;em&gt;chemistry&lt;/em&gt;, you need more than just pairwise atomic interactions. Most chemical force fields include three-atom angle-bending terms, four-atom dihedral terms, and out-of-plane distortion terms (e.g., pyramidal N or P).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, all of this ignores electrostatic interactions, which a Lennard-Jones model would sweep under the rug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to simulate something with LJ, I might go for a sample of gas (e.g., He, Ne, etc.) or atoms in a solid (e.g., Ag).&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-27T17:17:16.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32075" PostTypeId="2" ParentId="15885" CreationDate="2015-05-27T17:27:54.273" Score="5" Body="&lt;p&gt;Hand sanitizer leaves your hands feeling cool because the particles in the gel which posses the most amount of energy are able to evaporate from the gel and off your hands. This results in only the particles with low energy levels to be left on your hand which therefore reduces the hands overall temperature. &lt;/p&gt;&#xA;" OwnerUserId="16567" LastActivityDate="2015-05-27T17:27:54.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32076" PostTypeId="1" AcceptedAnswerId="32101" CreationDate="2015-05-27T17:58:19.983" Score="4" ViewCount="8785" Body="&lt;p&gt;I have to etch a very thin film (50 nm) of Nickel ($\ce{Ni}$) on a 100 nm $\ce{SiO2}$/$\ce{Si}$ substrate. Additionally, there is a ~5 nm adhesion layer of $\ce{Cr}$ under the $\ce{Ni}$ that I'd also like to etch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I found &lt;a href=&quot;http://www.cleanroom.byu.edu/wet_etch.phtml&quot; rel=&quot;nofollow&quot;&gt;this page&lt;/a&gt; which has a fairly long list of etches for $\ce{Ni}$ and $\ce{Cr}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ni:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;1 : 1 : 1 - HNO3 : HAc : Acetone&#xA;1 : 1 - HF : HNO3&#xA;30% FeCl3&#xA;3 : 1 : 5 : 1 - HNO3 : H2SO4 : HAc : H2O 85 C 10 microns/min&#xA;3 : 7 - HNO3 : H2O&#xA;1 : 1 - HNO3 : HAc&#xA;10% g/ml Ce(NH4)2(NO3)6 : H2O&#xA;HF, concentrated – slow etchant&#xA;H3PO4 – slow etchants&#xA;HNO3 – rapid etchant&#xA;HF : HNO3 – etch rate determined by ratio, the greater the amount of HF the slower the reaction&#xA;4 : 1 - HCl : HNO3 – increase HNO3 concentration increases etch rate&#xA;30% FeCl3&#xA;5g : 1ml : 150ml - 2NH4NO3.Ce(NO3)3.4(H2O) : HNO3 : H2O – decreasing HNO3 amount increases the etch rate&#xA;3 : 3 : 1 : 1 - H3PO4 : HNO3 : CH3COOH : H2O ~15min/micron @ RT with air exposure every 15 seconds&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Cr:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2 : 3 : 12  KMnO4 : NaOH : H2O&lt;br&gt;&#xA;  3 : 1   H2O : H2O2&lt;br&gt;&#xA;  concentrated and dilute     HCl&lt;br&gt;&#xA;  3 : 1   HCl : H2O2&lt;br&gt;&#xA;  2 : 1   FeCl : HCl&lt;br&gt;&#xA;      Cyantek CR-7s (Perchloric based)    7 min/micron (24A/s new)&lt;br&gt;&#xA;  1 : 1   HCl : glycerine     12min/micron after depassivation&lt;br&gt;&#xA;  1 : 3   [50gNaOH+100mlH2O] : [30g K3Fe(CN)6+100mlH2O]   1hr/micron &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd really like to avoid HF if I can (it terrifies me!), but I have pretty much everything else at my disposal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, everything I've done has been at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I tried concentrated $\ce{FeCl3}$. No bubbling or signs of anything after ~20min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I tried weak $\ce{H3PO4}$. Nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today I tried strong (98%) $\ce{H3PO4}$, nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I tried concentrated fuming $\ce{HNO3}$. Nothing after ~1hr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I tried about 1:1 $\ce{HNO3}$:$\ce{HCl}$, nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something here? Do the solutions need to be heated up or something?&lt;/p&gt;&#xA;" OwnerUserId="3764" LastEditorUserId="4097" LastEditDate="2015-05-28T07:10:46.157" LastActivityDate="2015-05-28T07:10:46.157" Title="What's the simplest way to etch Nickel and Chromium sputtered layers?" Tags="&lt;acid-base&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32077" PostTypeId="2" ParentId="32067" CreationDate="2015-05-27T18:17:42.407" Score="1" Body="&lt;p&gt;I believe the question is attempting to get you to think about the stochiometry of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse reaction takes one mole of $\ce{Fe^{2+}}$ and &lt;em&gt;two&lt;/em&gt; moles of $\ce{OH-}$. This means that if you add the same molar amount of $\ce{Fe(NO3)2}$ and $\ce{KOH}$ you would expect (roughly) half the amount of $\ce{Fe(OH)2(s)}$ formed from the $\ce{KOH}$ versus the $\ce{Fe(NO3)2}$, because it takes two (molar) units of $\ce{OH-}$ to get one unit of reaction to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both will cause the reaction to reverse because of Le Chatelier's principle, but adding $\ce{Fe(NO3)2}$ will be &quot;more active&quot; (on a mole/mole comparison basis) because of the stochiometry of the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree the question is poorly worded, though. Although using a molar comparison basis for things like this is pretty standard, it's not the only way to do it. (E.g. adding equal masses of $\ce{Fe(NO3)2}$ or $\ce{KOH}$ will result in the $\ce{KOH}$ solution having greater reverse reactions.) For completeness, how &quot;equivalent&quot; amounts of the various substances are determined should have been specified in the question.&lt;/p&gt;&#xA;" OwnerUserId="11616" LastActivityDate="2015-05-27T18:17:42.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32078" PostTypeId="2" ParentId="32056" CreationDate="2015-05-27T18:26:23.293" Score="4" Body="&lt;p&gt;All molecular simulation force fields are essentially &quot;fitted&quot; mathematical functions. The parameters are adjusted until the simulation results match some physical properties we care about. As Geoff says, Lennard-Jones force fields are not particularly realistic, in the sense that you wouldn't expect a simulation of atoms interacting solely through Lennard-Jones potentials to behave exactly like a real system. This becomes very important for things like proteins, where conformation and charge-charge interactions are very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what might not be clear from the other answers and comments is that &lt;em&gt;all&lt;/em&gt; force fields are not 100% realistic. The trick is to intelligently choose your level of realism. It could be that a Lennard-Jones force field that ignores  charge-charge interactions works perfectly well for your application. Without knowing the details, I can't tell you how suitable it would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that for polymers all sorts of real phenomena can be modeled with very simple Lennard-Jones-like force fields. A famous example is this article by Kurt Kremer and Gary Grest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/92/8/10.1063/1.458541&quot; rel=&quot;nofollow&quot;&gt;Dynamics of entangled linear polymer melts:  A molecular‐dynamics simulation. J. Chem. Phys. 92, 5057 (1990)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as Geoff mentioned, LJ potentials work pretty well for things that can be modeled by the van der Waals equation of state: noble gases, simple diatomics, and other small molecules at high temperature and low pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_dynamics#Coarse-graining_and_reduced_representations&quot; rel=&quot;nofollow&quot;&gt;coarse-grained&lt;/a&gt; simulations use LJ potentials for each &quot;blob,&quot; and depending again on what you want to study, this can be very successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while we are on the subject of unrealistic force fields, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hard_spheres&quot; rel=&quot;nofollow&quot;&gt;hard-sphere model&lt;/a&gt; has been used to successfully model things like glass transitions (I don't have a good representative link for this, but you can search for &quot;hard sphere glass transition&quot; to find many studies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in response to your actual question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe - it depends on what the goal of your simulation is. What physical behavior are you trying to model? If you just want to see how a string of soft, slightly sticky beads sticks to another string, then you don't need to differentiate between atoms. If you want to see what happens when some of those beads are really sticky (sort of like hydrogen bonding sites on a protein), then it would make sense to use different parameters for very sticky vs. slightly sticky beads. It probably would never make sense to worry about which beads were carbon and which were oxygen, nitrogen, etc., or to try to select parameters for them that were based on published values (unless those studies were specifically looking at the same system and properties that you are). Most LJ parameters are derived from fitting simulated phase behavior to real phase behavior - which is pretty far removed from what you are doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get a better idea of what real proteins would do, then the problem becomes much more complicated. Real protein behavior has a lot to do with interactions with water, which means you would need to simulate that as well. Charge-charge interactions also become much more important (and the simulations much more computationally expensive)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice (and I think what the other responses are saying as well) is to choose your system and goals carefully so that you don't need this level of realism. Or, if you do need it, then do it right; don't try to make a simple model work for a system that it doesn't fit.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-05-27T19:49:10.603" LastActivityDate="2015-05-27T19:49:10.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32079" PostTypeId="1" CreationDate="2015-05-27T18:58:10.880" Score="16" ViewCount="4994" Body="&lt;p&gt;I'm trying to understand organometallic raegents but am stuck at halogen-metal exchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you have $\ce{PhBr + BuLi}$ you get $\ce{PhLi + BuBr}$, I don't understand why $\ce{PhBu + BrLi}$ wouldn't form. &lt;/p&gt;&#xA;" OwnerUserId="16568" LastEditorUserId="10185" LastEditDate="2015-06-19T19:18:04.877" LastActivityDate="2015-06-23T19:40:49.693" Title="Why do halogen-metal exchanges happen?" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32080" PostTypeId="2" ParentId="32058" CreationDate="2015-05-27T21:01:37.057" Score="11" Body="&lt;p&gt;It is true that rate equations typically use molar concentrations for each species, and that for theoretical rate equations you can make sure the units work out by choosing the appropriate units for the rate constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two things that are important to understand about rate equations, though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that the &quot;real&quot; rate equations use &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermodynamic_activity&quot; rel=&quot;noreferrer&quot;&gt;activity&lt;/a&gt;, not concentration. Concentration is usually a &quot;good enough&quot; approximation that this doesn't matter in practice, but when you start thinking about the units of the reaction constant for things like fractional exponents in empirical rate laws, understanding that the concentrations should really be dimensionless activities makes the problem a lot easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing to understand is that real reactions don't always follow theoretical rate equations. As a result, what you are often looking at in real life is an empirical rate law. With empirical rate laws, we don't usually worry about the physical significance of fractional exponents, and in turn, how they affect the units of the rate constant. The units of the rate constant become whatever they need to be to make the equation right. In other words, if you have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^{0.6}[A]^{1.3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as an empirically derived rate law that gives $\rm{mol_A}\rm{L}^{-1}\rm{min}^{-1}$, then $k$ would need to have units of $\rm{mol_A}^{-0.3}\rm{mol_B}^{-0.6}\rm{L}^{-1}\rm{min}^{-1}$. These physically nonsensical units are ok because the equation is entirely empirical - we are just describing what we observe on a macroscopic level, without attempting to give it physical significance beyond that.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-05-28T01:09:31.627" LastActivityDate="2015-05-28T01:09:31.627" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32081" PostTypeId="2" ParentId="23632" CreationDate="2015-05-27T21:38:22.073" Score="2" Body="&lt;p&gt;While enthalpy, internal energy, and Gibbs's energy are all types of energy, they are not all measuring the same quantities. Internal energy is the energy of the molecules in your system. Enthalpy takes into account the internal energy plus whatever work is being done by or on your system. And Gibbs's energy is the maximum amount of non-&lt;em&gt;PV&lt;/em&gt; work that you can get from your system. While these three quantities are all in units of energy, they are not just different terms for the total energy. Since enthalpy and Gibbs's energy involve work, and hence exchange of energy with the surroundings, these quantities will be different than just the internal energy of the system alone.&lt;/p&gt;&#xA;" OwnerUserId="16290" LastActivityDate="2015-05-27T21:38:22.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32084" PostTypeId="1" AcceptedAnswerId="32085" CreationDate="2015-05-27T23:04:14.687" Score="2" ViewCount="1135" Body="&lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt; According to what I've learned so far, helium is in the first row of the periodic table, so its &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; should always equal &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. However, a question in my text asks about helium 2s. What is helium 2s? If helium 2s exists, why does the periodic table list helium in the first row instead of the second?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt;&#xA;I believe that He 1s is more stable than He 2s. I've reasoned thus:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The numerals '1' and '2' in 'He 1s' and 'He 2s' express the size of the orbital. The greater number represents a larger orbital.&lt;/li&gt;&#xA;&lt;li&gt;The electrons of smaller orbitals exist closer to the nucleus than the electrons of larger orbitals do.&lt;/li&gt;&#xA;&lt;li&gt;The intensity of the attractive force between electron and protons increases the nearer the protons are to the electron&lt;/li&gt;&#xA;&lt;li&gt;Therefore, the attractive force between the protons and the electron will be greater in He 1s than in He 2s.&lt;/li&gt;&#xA;&lt;li&gt;The greater the attractive force between the protons and the electron, the greater the stability of the electron.&lt;/li&gt;&#xA;&lt;li&gt;Therefore the electrons in He 1s are more stable than the electrons in He 2s.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Note the following is the original question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For each pair of orbitals determine which is more stable and explain why:&lt;/p&gt;&#xA;&lt;p&gt;(a) He 1s and He 2s;&lt;/p&gt;&#xA;&lt;p&gt;(b) Kr 5p and Kr 5s;&lt;/p&gt;&#xA;&lt;p&gt;(c) He 2s and He+ 2s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="69830" LastEditDate="2021-05-22T14:50:41.280" LastActivityDate="2021-05-22T14:50:41.280" Title="Does helium have a 2s orbital?" Tags="&lt;electrons&gt;&lt;periodic-table&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32085" PostTypeId="2" ParentId="32084" CreationDate="2015-05-27T23:56:09.187" Score="7" Body="&lt;p&gt;Part 1: The elements on the periodic table are arranged according to their ground-state electron configurations. He 2s is an excited state of He 1s. There are even He 3s and 4s depending on whether one gives enough energy for the atom to excite to such levels. In general, higher principal quantum number (n) indicates a higher energy state (less stable)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 2: Your reasoning is correct.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastActivityDate="2015-05-27T23:56:09.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32086" PostTypeId="2" ParentId="32084" CreationDate="2015-05-28T00:16:21.217" Score="2" Body="&lt;p&gt;&lt;strong&gt;Part 1:&lt;/strong&gt; The 2s is referring to the orbitals that are occupied by electrons. In the ground state the electrons in He are both in 1s orbitals, however it is possible to excite the electrons from the 1s orbital to a higher orbital like 2s. All orbitals are accessible to the electrons of an atom through excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2:&lt;/strong&gt; The reasoning you are using is appropriate to the question you are answering.&lt;/p&gt;&#xA;" OwnerUserId="16573" LastActivityDate="2015-05-28T00:16:21.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32089" PostTypeId="1" CreationDate="2015-05-28T00:53:34.087" Score="11" ViewCount="3148" Body="&lt;p&gt;As a physics student, I hardly deal with such quantities and when I do, I blunder through.&#xA;I think it’s a good time to be given a good explanation for these since I’m in a class of Nuclear physics. I have tried googling but it seems that some sources are ‘misleading’ and most have more than 1 terminology for the same thing!&#xA;Horrifying but real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me what is the difference between molar mass, molecular mass and atomic mass and what are their associated twin terminology.&lt;/p&gt;&#xA;" OwnerUserId="16575" LastEditorDisplayName="user7951" LastEditDate="2015-05-28T09:45:05.027" LastActivityDate="2016-10-22T23:54:18.087" Title="Quick and simple explanation of molar mass, molecular mass and atomic mass" Tags="&lt;terminology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32090" PostTypeId="1" AcceptedAnswerId="32094" CreationDate="2015-05-28T01:26:29.373" Score="12" ViewCount="1792" Body="&lt;p&gt;I know that orbitals are probability distributions. Are electron shells probability distributions too?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-06-02T20:34:29.837" Title="Are electron shells probability distributions?" Tags="&lt;electrons&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32091" PostTypeId="1" CreationDate="2015-05-28T01:33:51.850" Score="7" ViewCount="103575" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$.  What is the pH of the solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl -&amp;gt; H+ + Cl-}$$&#xA;$$\ce{HI -&amp;gt; H+ + I-}$$ &#xA;According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio.  So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = -\log[\ce{H+}]$$ &#xA;$$\mathrm{pH} = -\log[0.205]$$ &#xA;$$\mathrm{pH} = 0.689$$&lt;/p&gt;&#xA;" OwnerUserId="16576" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-26T02:06:45.197" LastActivityDate="2018-08-20T08:58:34.933" Title="How to find pH by mixing two solutions of different concentrations?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32092" PostTypeId="2" ParentId="32091" CreationDate="2015-05-28T03:13:51.417" Score="10" Body="&lt;p&gt;You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. &lt;a href=&quot;http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf&quot;&gt;here&lt;/a&gt;.&#xA;Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.&#xA;\begin{align}&#xA;c_{\ce{HI}}(\ce{H+}) &amp;amp;=  c(\ce{HI}) &amp;amp;&#xA;c_{\ce{HCl}}(\ce{H+}) &amp;amp;=  c(\ce{HCl}) &amp;amp;&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to add the amount of protons in each solution and determine the combined volume.&#xA;\begin{align}&#xA;n_\mathrm{tot}(\ce{H+}) &amp;amp;= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})&amp;amp;&#xA;V_\mathrm{tot} &amp;amp;= V(\ce{HI}) + V(\ce{HCl})&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the pH of the resulting concentration, just take the negative logarithm.&#xA;$$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The approach you chose can only work, if you have equal volumes of the two solutions.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-05-28T03:13:51.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32093" PostTypeId="1" CreationDate="2015-05-28T03:16:08.233" Score="3" ViewCount="549" Body="&lt;p&gt;I came across alkaline ionized water as my aunts bought a system called Kangen Water from &lt;a href=&quot;http://www.enagic.com/&quot; rel=&quot;nofollow&quot;&gt;Enagic Inc.&lt;/a&gt; The water produced by this system is claimed to contain smaller water clusters (4-6 molecules) compared to tap water with 10-12 molecules. Besides, the ORP (Oxidation Reduction Potential) of the water is negative (-250 to -350 mV), which the &lt;a href=&quot;http://jldmkangen.webs.com/PDF/Benefits%20of%20Alkaline,%20Ionized%20Water.pdf&quot; rel=&quot;nofollow&quot;&gt;company claimed&lt;/a&gt; to battle ROS (Reactive Oxygen Species). Of course, the company also advertised the water as having other incredible properties such as &lt;a href=&quot;http://www.kangendemo.com/&quot; rel=&quot;nofollow&quot;&gt;emulsifying grease&lt;/a&gt;, thereby can be used to remove pesticides on fruits and vegetables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let alone the validity of the claim on benefits, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How can one produce negative-ORP water? Is it possible to do so with added minerals? (Just the only way I can think of, is there any other way?)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What chemical/physical process is happening in the &lt;a href=&quot;http://www.kangendemo.com/&quot; rel=&quot;nofollow&quot;&gt;emulsification demo&lt;/a&gt;? Why can't tap water do so?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please refrain from discrediting the company too quickly as I think I might have and please tell me if I formulated my question clearly enough because I acknowledge this is a fairly vague question.&lt;/p&gt;&#xA;" OwnerUserId="14698" LastEditorUserId="21964" LastEditDate="2016-08-18T16:39:45.090" LastActivityDate="2016-08-18T16:39:45.090" Title="The Health Effects of Alkaline Ionized Water?" Tags="&lt;water&gt;&lt;biochemistry&gt;&lt;aqueous-solution&gt;&lt;water-treatment&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32094" PostTypeId="2" ParentId="32090" CreationDate="2015-05-28T04:22:21.500" Score="11" Body="&lt;p&gt;In order to understand you should know the differences between shell, sub-shell and orbital.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Shell&lt;/strong&gt; represents the main energy level occupied by the electron. It is given by the principal quantum number (n). For example , if n=1, the electron is present in the first main shell, called K-shell. if n=2, the electron is present in the second main shell, called L-shell and so on..&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Sub-shell&lt;/strong&gt; represents the sub-energy level occupied by the electron (as main energy level is considered to consist of a number of energy sub-levels ). it is given by azimuthal quantum number. subshell corresponding to l=0,1,2,3 are represented by s,p,d,and respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Orbitals&lt;/strong&gt; you know it well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may understand it like this- &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;when n=1 we have only s subshell&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;when n=2 we have s and p subshell&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;for n=2 we have s,p and d subshell and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Now, s, p and d and f have certain regions, which we call orbitals. which are probability regions as you know. Here is the link for you,&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;http://en.wikipedia.org/wiki/Atomic_orbital&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Must watch the video associated with it named [atomic orbitals and periodic construction]&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Last but not the least- we certainly not have electron shells probability distribution. its only orbitals. shells are created by scientists to study the atom systematically but they have physical significance too (i mean here the subshells and orbitals lie).&#xA;it seems you are not clear with - shell, sub-shell and orbitals. &#xA;you should first check out what they actually means in order to understand.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16564" LastEditorUserId="10185" LastEditDate="2015-05-28T12:09:06.833" LastActivityDate="2015-05-28T12:09:06.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32095" PostTypeId="2" ParentId="31897" CreationDate="2015-05-28T04:30:02.277" Score="4" Body="&lt;p&gt;&lt;strong&gt;In general, the crystal structure of a molecule is a highly accurate geometry.&lt;/strong&gt; So yes, you can take a salt and modify as needed. (The caveat in the comment is useful - keep in mind that different groups may be protonated or deprotonated depending on solvent and pH.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two other caveats:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The conformation in the crystal structure is not necessarily the lowest energy conformer in solution or the gas phase. You may wish to perform a conformer search to find a low-energy minima.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you are performing further calculations (e.g., DFT for electronic structure, force fields for MD, etc.) the geometry of the crystal structure will not likely be a minima of the computational method, so geometry optimization or equilibration (for MD) is needed before further simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the crystal geometry is a good place to start for many uses. Certainly if a crystal structure is available, I'd suggest you use it rather than build a geometry by hand.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-28T04:30:02.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32096" PostTypeId="1" AcceptedAnswerId="32312" CreationDate="2015-05-28T04:30:38.073" Score="19" ViewCount="8516" Body="&lt;p&gt;EDTA complexes with all the other metal ions in the periodic table except those from the group 1. Why is this so ? What is the coordination chemistry behind this ?&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="9961" LastEditDate="2015-11-01T16:32:08.810" LastActivityDate="2015-11-01T16:32:08.810" Title="Why doesn't EDTA complex with alkali metal ions?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;hsab&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32097" PostTypeId="2" ParentId="32090" CreationDate="2015-05-28T04:35:18.870" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Electron shell is well described into the old quantum mechanics developed by Bohr, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;nofollow&quot;&gt;Atom Bohr model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Probability distribution is well described into the modern quantum chemistry, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_mechanics&quot; rel=&quot;nofollow&quot;&gt;Quantum mechanics, general description&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, both definitions are not comparable among them because each one belongs to different theory conceptions.&lt;/p&gt;&#xA;" OwnerUserId="15429" LastEditorUserId="15429" LastEditDate="2015-05-28T04:41:38.947" LastActivityDate="2015-05-28T04:41:38.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32098" PostTypeId="1" AcceptedAnswerId="32403" CreationDate="2015-05-28T05:38:04.330" Score="8" ViewCount="5717" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&#xA;  What is the product of this reaction?&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/DZLKem.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The answer given:&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/nwyoMm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first, I thought this was a typical Wolff-Kishner reduction, but then I saw the reagents didn't match up (no $\ce{OH-}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this getting formed? Shouldn't the $\ce{CHO}$ group be reduced? What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="1966" LastEditorUserId="5026" LastEditDate="2018-03-05T03:52:52.370" LastActivityDate="2021-05-12T17:43:55.773" Title="Why is in the reaction of cyclic unsaturated aldehydes with hydrazine and hydrogen peroxide the aldehyde preserved?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32100" PostTypeId="1" CreationDate="2015-05-28T06:08:59.920" Score="4" ViewCount="2702" Body="&lt;p&gt;Almost everywhere I checked, the Inductive effect is described as a permanent effect with almost no mention of what that actually means. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;What exactly is permanent about the effect?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the effect is permanent, does that mean that even if the attached &lt;em&gt;electron&#xA;withdrawing/donating&lt;/em&gt; group is removed, the induced effect still continues to&#xA;exist?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15508" LastActivityDate="2015-05-28T11:25:13.837" Title="In what sense is the Inductive effect a permanent effect?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32101" PostTypeId="2" ParentId="32076" CreationDate="2015-05-28T06:24:57.720" Score="1" Body="&lt;p&gt;Nickel can be etched using diluted (1:1 V) $HCl$. 50 nm will be dissolved within seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The etching solution for the chromium layer contains $\ce{Ce(NH4)2(NO3)6}$+$\ce{CH3COOH}$+$\ce{H2O}$&#xA;in ratio 165g : 45 ml (con.) :100 ml DI water. The chromium layer is dissolved within 5 minutes of etching.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-05-28T06:45:20.227" LastActivityDate="2015-05-28T06:45:20.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32102" PostTypeId="1" AcceptedAnswerId="32105" CreationDate="2015-05-28T08:24:51.627" Score="5" ViewCount="2358" Body="&lt;p&gt;I've been asked to draw the mechanism by which benzene with &lt;span class=&quot;math-container&quot;&gt;$\ce{Me}$&lt;/span&gt; on position 1 and &lt;span class=&quot;math-container&quot;&gt;$\ce{OMe}$&lt;/span&gt; on position 2 goes to benzene with &lt;span class=&quot;math-container&quot;&gt;$\ce{Me}$&lt;/span&gt; on 1 and &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; on 2 by reaction with concentrated &lt;span class=&quot;math-container&quot;&gt;$\ce{HBr}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really unsure as to what is going on, I think its probably most likely that &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; acts as a nucleophile and attacks the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-O}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\sigma^*$&lt;/span&gt; and then &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; attacks the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-Br}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\sigma^*$&lt;/span&gt; but it doesn't seem like a likely pathway as &lt;span class=&quot;math-container&quot;&gt;$\ce{OMe}$&lt;/span&gt; has a higher pKaH than &lt;span class=&quot;math-container&quot;&gt;$\ce{Br}$&lt;/span&gt; and substitutions don't occur at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; centres.&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="15751" LastEditDate="2019-05-31T17:53:44.400" LastActivityDate="2019-05-31T17:53:44.400" Title="Ether cleavage with conc HBr" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32103" PostTypeId="2" ParentId="32089" CreationDate="2015-05-28T09:36:47.237" Score="10" Body="&lt;p&gt;The quantity &lt;strong&gt;molar mass&lt;/strong&gt; (quantity symbol: $M$) for a pure sample is defined as&#xA;$$M = m/n$$&#xA;where $m$ is mass and $n$ is amount of substance.&lt;br&gt;&#xA;The dimension of the molar mass is&#xA;$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$&#xA;The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).&lt;br&gt;&#xA;(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity &lt;strong&gt;relative molecular mass&lt;/strong&gt; (quantity symbol: $M_\mathrm{r}$) is defined as the ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass constant:&#xA;$$M_\mathrm{r} = m_\mathrm{f}/m_\mathrm{u}$$&#xA;where $m_\mathrm{f}$ is mass of entity (molecule or formula unit) and $m_\mathrm{u}$ is the unified atomic mass constant.&lt;br&gt;&#xA;The relative molecular mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):&#xA;$$\dim M_\mathrm{r} = 1$$&#xA;The coherent SI unit for relative molecular mass is the unit one (symbol: $1$).&lt;br&gt;&#xA;For historical reasons, the IUPAC accepts the use of the special name ‘molecular weight’ for the quantity relative molecular mass. The use of this traditional name is deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atoms, the relative molecular mass $M_\mathrm{r}$ is the &lt;strong&gt;relative atomic mass&lt;/strong&gt;, and the quantity symbol $A_\mathrm{r}$ may be used. The quantity relative atomic mass is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$:&#xA;$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$&#xA;where $m_\mathrm{a}$ is the atomic mass (see below) and $m_\mathrm{u}$ is the unified atomic mass constant.&lt;br&gt;&#xA;The relative atomic mass is a quantity of dimension one:&#xA;$$\dim A_\mathrm{r} = 1$$&#xA;The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).&lt;br&gt;&#xA;For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity &lt;strong&gt;atomic mass&lt;/strong&gt; (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.&lt;br&gt;&#xA;The dimension of the atomic mass is&#xA;$$\dim m_\mathrm{a} = \mathsf{M}$$&#xA;The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).  &lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-28T09:36:47.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32105" PostTypeId="2" ParentId="32102" CreationDate="2015-05-28T10:48:34.697" Score="7" Body="&lt;p&gt;Cleavage of ethers using strong acids (classically $\ce{HI}$ but $\ce{HBr}$ should work as well) can proceed by either an $S_N1$ or $S_N2$ mechanism. In this case the cation formed would be primary or phenyl, neither of which are stable, and so the reaction proceeds via an $S_N2$ mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mtENC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen is protonated in an equilibrium reaction. The bromide ion then attacks at the least hindered carbon and the $\ce{C-O}$ bond breaks in a concerted $S_N2$ mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://en.wikipedia.org/wiki/Ether_cleavage&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-28T10:48:34.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32106" PostTypeId="2" ParentId="32100" CreationDate="2015-05-28T10:57:11.240" Score="3" Body="&lt;p&gt;The effect is permanent because the group that does the 'induction' is always there in the molecule. For example, if the induction is caused by a t-butyl group, that group is held by a strong carbon to carbon ($\ce{C-C}$) bond that will not easily be broken. One could also consider the inductive effect of the Fluorine held by a very strong $\ce{C-F}$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A temporary effect might be caused by protontation of an oxygen atom where the protonation can catalyse a reaction. The protonation is reversible. The bond between the proton and oxygen would be &lt;em&gt;much&lt;/em&gt; weaker, as $\ce{C=O^+-H}$ will be in equilibrium with $\ce{C=O  +  H^+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removal of the group that does the induction will &lt;strong&gt;remove&lt;/strong&gt; the inductive effect. &lt;/p&gt;&#xA;" OwnerUserId="11523" LastEditorUserId="11523" LastEditDate="2015-05-28T11:25:13.837" LastActivityDate="2015-05-28T11:25:13.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32108" PostTypeId="1" CreationDate="2015-05-28T12:52:09.353" Score="2" ViewCount="90" Body="&lt;p&gt;I was looking at two foam materials, where I had expected one to have closed pores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfourtunately, the microscope was showing the upper layer of cut cells the same in both cases with no indication of differences in lower levels of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By now, I suspect both are open-pored;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the difference clearly defined?   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If a block of soft material can sustand high pressure based on inner air pressure in cells, the cells are clearly closed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a visual inspection shows mostly incomplete cells, such that parts of cells behind them are visible, it's clearly open pored.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;what about holding gas only at low outer forces?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which is similar to the common kitchen sponge, keeping fluid with low pressure, releasing it at high pressure. But here, I expect the pores to be open, and capularry forces to get the water inside.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, I'd like to know how to tell them appart based on actual mechanical features, and to understand the difference in case they are not trivial.&lt;/p&gt;&#xA;" OwnerUserId="5078" LastEditorUserId="5078" LastEditDate="2015-05-28T13:47:41.797" LastActivityDate="2015-05-28T13:47:41.797" Title="How to distinguish open-pored polymer foam and foam with closed pores?" Tags="&lt;materials&gt;&lt;polymers&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32109" PostTypeId="1" AcceptedAnswerId="32138" CreationDate="2015-05-28T14:29:17.913" Score="8" ViewCount="118" Body="&lt;p&gt;Here's a couple of examples of what I mean. Firstly, in the reaction with thionyl chloride to make a carboxylic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZlEE8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, in this specific example of a hydroboration reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Zqy8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really understand why you cant attack from the hydroxide group and then proton transfer. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-05-29T11:43:56.563" Title="When a carboxylic acid acts as a nucleoplile, why does it attack through the carbonyl bond?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32110" PostTypeId="2" ParentId="32090" CreationDate="2015-05-28T15:36:29.233" Score="18" Body="&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;First&lt;/em&gt;&lt;/strong&gt;, strictly speaking, the orbitals themselves in the quantum mechanical sense &lt;em&gt;are not probability distributions&lt;/em&gt;.  They are eigenfunctions $\Psi_i$ of the Hamiltonian as defined by the time-independent Schroedinger equation $H\Psi_i=E_i\Psi_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability distribution function $p(\vec r)$ for electrons is generated by summing the products of the eigenfunctions and their complex conjugates for all occupied eigenfunctions at a given 3-D point $\vec r$ and dividing by the total number $N$ of electrons in the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;p(\vec r) = \frac{1}{N}\sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $p(\vec r)$ is simply $\frac{1}{N}$ times the electron density function $\rho(\vec r) = \sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, prior to a detailed introduction to the quantum mechanical description of electronic structure, the distinction between the electronic wavefunction $\Psi(\vec r)$ and the electron density function $\rho(\vec r)$ is typically not clearly established in chemistry instruction, so your having conflated them is quite understandable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Second&lt;/em&gt;&lt;/strong&gt;, again strictly speaking, electron shells are not probability functions.  While Pushkar Soni is correct that the principal quantum number for a given electron approximately indicates the 'electron shell' to which it is assigned, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Identical_particles&quot;&gt;indistinguishability of electrons&lt;/a&gt; (among other factors) in general prohibits unambiguous translation of quantum number information into the real-space function $\rho(\vec r)$.  The best we can do is to interpret electron shells as corresponding to certain features of the structure of $\rho(\vec r)$. Offhand, I know of at least two ways of qualitatively identifying 'electron shells' from $\rho(\vec r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;First&lt;/em&gt;, as Geoff Hutchinson notes, the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_States_of_Atoms_and_Molecules/8._The_Hydrogen_Atom/The_Wavefunctions&quot;&gt;radial distribution function&lt;/a&gt; $r^2\rho(r)$ shows local maxima that can be interpreted as electron shells.  Note that  $\rho(r)$ in the r.d.f here is a function only of $r$, the distance from an isolated atom/nucleus; the angular dependencies (on $\theta$ and $\varphi$) have been integrated out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Second&lt;/em&gt;, Bader's QTAIM describes how the &lt;a href=&quot;http://www.chemistry.mcmaster.ca/aim/aim_5.html&quot;&gt;Laplacian of the electron density&lt;/a&gt;, $\nabla^2\rho(\vec r)$, exhibits minima in 3-D space at positions corresponding to the notion of electron shells.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-02T20:34:29.837" LastActivityDate="2015-06-02T20:34:29.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32111" PostTypeId="2" ParentId="32025" CreationDate="2015-05-28T15:37:48.410" Score="5" Body="&lt;p&gt;I think this is very much an open problem. As you may see from the Drexel pages, even solubility prediction in &lt;em&gt;one&lt;/em&gt; solvent is still fairly hard, partly because of the lack of experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only software that I know capable of making predictions about solubility in mixed solvents is &lt;a href=&quot;http://www.cosmologic.de/products/cosmotherm.html&quot;&gt;COSMOtherm&lt;/a&gt;, based on the &lt;a href=&quot;http://en.wikipedia.org/wiki/COSMO_Solvation_Model&quot;&gt;COSMO solvation model&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never used it, but the capability is mentioned on the website.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-28T15:37:48.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32112" PostTypeId="1" AcceptedAnswerId="32142" CreationDate="2015-05-28T16:22:27.803" Score="6" ViewCount="2761" Body="&lt;p&gt;Any quantum number that represents the size of an orbital is a positive integer, $n$. &#xA;Any quantum number that represents the shape of an orbital is a non-negative integer, $l : l &amp;lt; n$. &#xA;Any quantum number that represents the orientation of an orbital, $m$, is an integer such that $|m| \leq l$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do $n$ and $l$ restrict the ranges of $l$ and $m$, respectively? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clearer, I ask why the quantum numbers limit each other, not what they signify, which is already explained in the related question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31639/&quot;&gt;What do the quantum numbers actually signify?&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-05-31T14:27:07.680" Title="Why does the quantum number n limit the value of l, and why does l, likewise, limit the value of m?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32113" PostTypeId="1" CreationDate="2015-05-28T17:39:20.270" Score="7" ViewCount="447" Body="&lt;p&gt;I'm wondering if it's worthwhile for me as an optical spectroscopist to read up a bit on EPR. Do EPR signals reveal where, in terms of wavelength, features in the optical spectrum can be expected to appear? I know it tells me&#xA;lifetimes of states, but that's about as far as I have dug into it. I have some EPR data on a substance and I wish to figure out where to look for it's absorption with UV/Vis/NIR means.&lt;/p&gt;&#xA;" OwnerUserId="13341" LastEditorUserId="2094" LastEditDate="2016-08-24T12:27:12.377" LastActivityDate="2016-11-01T02:46:44.830" Title="Can one figure out an optical absorption spectrum from EPR data?" Tags="&lt;spectroscopy&gt;&lt;electromagnetic-radiation&gt;&lt;epr-spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32114" PostTypeId="1" CreationDate="2015-05-28T18:34:39.247" Score="-2" ViewCount="271" Body="&lt;p&gt;are eye drops with inactive ingredients called potassium chloride or sodium chloride the same as potassium dichromate and should I stop using those eye drops?&lt;/p&gt;&#xA;" OwnerUserId="16585" LastEditorUserId="4945" LastEditDate="2017-03-10T07:57:16.117" LastActivityDate="2017-03-10T07:57:16.117" Title="I recently learned I'm allergic to potassium dicromate or chromium" Tags="&lt;safety&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-28T22:00:52.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="32115" PostTypeId="2" ParentId="32114" CreationDate="2015-05-28T19:51:33.297" Score="7" Body="&lt;p&gt;Allergic reactions against dichromates are not uncommon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, dichromates and chlorides, are two pairs of shoes. Really different ones, like stiletto heels and moon boots ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, there are no eye drops containing potassium dichromate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sodium chloride that you mentioned is nothing but table salt. Isotonic solutions of sodium chloride are very common in all sorts of medical applications, like rinsing of lenses, intravenous infusions, and so on. To my knowledge, allergic reactions on these solutions are not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it up: I am not a medical doctor, and you might want to contact your ophthalmologist, but as a chemist with some experience in medical chemistry, I can assure that &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dichromates and chlorides are not the same  &lt;/li&gt;&#xA;&lt;li&gt;allergic reactions to physiological saline are not known&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-28T19:51:33.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32116" PostTypeId="1" CreationDate="2015-05-28T19:52:13.350" Score="5" ViewCount="269" Body="&lt;p&gt;I am experimenting with IR video cameras. They have a broad sensitivity peak at 940 nm. There is a filter blocking 750 nm and shorter in front of the sensor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an IR fluorescent substance to use for markers in the scene the camera is observing. Something that can be turned on with ultraviolet illumination bursts so I can differentiate between my markers and other IR sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hope is for some affordable and harmless substance. Something that could be mixed with polyester, acrylic or epoxy potting compound wold be nice. &#xA;An already solid watherproof material is even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My searches so far have found some IR fluorescent stains for microscopy but I am afraid they are expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a physics question also but I think chemists works a lot with IR absorption and IR fluorescence.&lt;/p&gt;&#xA;" OwnerUserId="15346" LastEditorUserId="418" LastEditDate="2015-05-28T19:57:01.417" LastActivityDate="2015-05-28T20:55:51.537" Title="IR fluorescent material for outdoor use" Tags="&lt;spectroscopy&gt;&lt;color&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32117" PostTypeId="2" ParentId="32116" CreationDate="2015-05-28T20:55:51.537" Score="1" Body="&lt;p&gt;A commercial solution would be a dye like IRF940A (from QCR Solutions) with an excitation maximum at &lt;span class=&quot;math-container&quot;&gt;$\lambda_{\mathrm{exc}} = 405\,\mathrm{nm}$&lt;/span&gt; and an  emission maximum at &lt;span class=&quot;math-container&quot;&gt;$\lambda_{\mathrm{em}} = 940\,\mathrm{nm}$&lt;/span&gt;. The supplier does not provide any information on the structure of the dye.&lt;/p&gt;&#xA;&lt;h3&gt;What compounds are promising candidates?&lt;/h3&gt;&#xA;&lt;p&gt;Ytterbium(III) complexes!&lt;br /&gt;&#xA;In a recent article,published in &lt;em&gt;J. Photochem. Photobiol. A&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;, *309, 65-71 (&lt;a href=&quot;http://dx.doi.org/10.1016/j.jphotochem.2015.04.022&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI&lt;/a&gt;), &lt;em&gt;J. Legendziewicz&lt;/em&gt; and coworkers prepared a &lt;span class=&quot;math-container&quot;&gt;$\ce{Yb(III)}$&lt;/span&gt; phthalocyanine with acetate as a chelating axial ligand and examined the optical properties in solution and in PMMA polymer. In PMMA, the compound shows a Soret band with a maximum at &lt;span class=&quot;math-container&quot;&gt;$\lambda_{\mathrm{abs}} = 350\,\mathrm{nm}$&lt;/span&gt;. Excitation in the Soret band result in emission with a maximum around around &lt;span class=&quot;math-container&quot;&gt;$\lambda_{\mathrm{em}} = 700\,\mathrm{nm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Other examples for even larger wavelength differences, resulting from ligand-to-metal energy transfer are known  for &lt;span class=&quot;math-container&quot;&gt;$\ce{Yb(III)}$&lt;/span&gt; complexes bound to eosines or fluoresceines as the primary chromophor.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-05-28T20:55:51.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32118" PostTypeId="1" CreationDate="2015-05-28T21:54:31.523" Score="2" ViewCount="427" Body="&lt;p&gt;I have chlorine tanks and they will be used industrially. When I use them, they will be exposed to all sorts of varying temperatures. I was wondering if there was any way that when I use them, I could figure out at what pressure I could release the $\ce{Cl2}$, depending on the temperature, so that it remains a gas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know it only takes $\pu{6.8 atm}$ at $\pu{70^\circ C}$ to liquefy $\ce{Cl2}$ to a volume of $\pu{0.322812 L}$, and its boiling point is $\pu{-33.97 ^\circ C}$ at $\pu{1 atm}$. &lt;/p&gt;&#xA;" OwnerUserId="16587" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-06T04:39:20.760" LastActivityDate="2017-08-06T04:39:20.760" Title="How to determine at temperature and pressure to keep Cl2 a gas?" Tags="&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32119" PostTypeId="1" AcceptedAnswerId="32123" CreationDate="2015-05-28T22:21:34.720" Score="27" ViewCount="6408" Body="&lt;p&gt;I have always been led to understand that the mass of an element on the periodic table is the weighted average atomic mass over all naturally occurring isotopes. This seems to make sense with all the elements I have looked at except for uranium. Wikipedia's featured article on the element assures me that uranium's relative atomic mass is 238.03. The commonest isotope is U-238, mixed with 0.72% of U-235 and 0.005% of U-234, plus traces of a couple of isotopes with smaller mass numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then can uranium's relative atomic mass &lt;em&gt;exceed&lt;/em&gt; 238? Either I have completely misunderstood something, or there is something else at work - relativistic maybe?&lt;/p&gt;&#xA;" OwnerUserId="16586" LastEditorUserId="1499" LastEditDate="2015-05-29T00:47:10.383" LastActivityDate="2015-05-29T14:30:40.517" Title="How come uranium's relative atomic mass is 238.03 when it only contains isotopes with a mass number of 238 or less?" Tags="&lt;periodic-table&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32120" PostTypeId="1" CreationDate="2015-05-28T22:49:03.253" Score="6" ViewCount="155" Body="&lt;p&gt;In violin finishing technology alkali metal Rosinates are used in varnish synthesis. Usually this alkali metal Rosinates are soluble in Turpentine. I also know that Copper, Zinc Rosinates are soluble in Diethyl ether. However, beside Turpentine I want to know if this alkali metal Rosinates ($\ce{K,\ Na,\ NH4+}$ - Rosinates) are soluble in Acetone especially or in Diethyl Ether or in any other  VOC’s?&lt;/p&gt;&#xA;" OwnerUserId="16590" LastEditorUserId="275" LastEditDate="2015-05-28T23:00:44.273" LastActivityDate="2015-05-28T23:00:44.273" Title="Alkali Metal Rosinates Solubility in Acetone" Tags="&lt;solubility&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32123" PostTypeId="2" ParentId="32119" CreationDate="2015-05-28T23:16:57.583" Score="29" Body="&lt;p&gt;Approximately 99.3% of uranium on Earth is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium-238&quot;&gt;$\mathrm{^{238}U}$ isotope&lt;/a&gt;, and this specific isotope has an atomic mass of $\mathrm{238.05\ u}$, where $\mathrm{u}$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_mass_unit&quot;&gt;atomic mass unit&lt;/a&gt;, equivalent to 1/12 the mass of a $\mathrm{^{12}C}$ atom. Including the other isotopes to obtain the average atomic mass drags the value down a little, but it still ends up being slightly larger than $\mathrm{238\ u}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to point out why the atomic mass is slightly different from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mass_number&quot;&gt;mass number&lt;/a&gt;. The masses of a free proton or neutron are both slightly larger than $\mathrm{1\ u}$ ($\mathrm{1.0073\ u}$ and $\mathrm{1.0087\ u}$, respectively). Some of this mass is converted to energy and lost when neutrons and protons (collectively called nucleons) bind to form nuclei, which means bound nuclei will always be lighter than the sum of the masses of their component nucleons. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy&quot;&gt;Different nuclei release different amounts of energy per nucleon&lt;/a&gt; depending on how many protons and neutrons are inside the nucleus, and on their relative proportions. Very interestingly, the curve for nuclear binding energy per nucleon versus mass number starts off at zero for $\mathrm{^{1}H}$ (there can be no attractive nuclear forces between nucleons if there's only one nucleon!), then rises until it shows a maximum around nickel (specifically for the isotope $\mathrm{^{62}Ni}$) and from there on slowly decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the atomic mass unit is standardized with respect to the $\mathrm{^{12}C}$ isotope of carbon, then &lt;em&gt;in approximate terms&lt;/em&gt;, the nuclei with a lower binding energy per nucleon than $\mathrm{^{12}C}$ will have less of the initial free proton/neutron mass lost to energy (and thus an isotope will have an atomic mass slightly &lt;em&gt;higher&lt;/em&gt; than its mass number), while for nuclei with a higher binding energy per nucleon than $\mathrm{^{12}C}$, more mass will be lost to energy and the resulting isotope will have an atomic mass slightly &lt;em&gt;lower&lt;/em&gt; than its mass number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a horizontal line through the $\mathrm{^{12}C}$ isotope in &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Binding_energy_curve_-_common_isotopes.svg&quot;&gt;this graph&lt;/a&gt; and looking at its intersections with the nuclear binding energy curve will show &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_mass#Mass_defects_in_atomic_masses&quot;&gt;the regions where this flip happens&lt;/a&gt;; isotopes for atoms below carbon will tend to have higher atomic mass than mass number, isotopes for atoms between carbon and thorium tend to have a lower atomic mass than mass number, and then isotopes for atoms above thorium (such as $\mathrm{^{238}U}$) will again tend to have higher atomic masses than mass numbers. This is only an approximate analysis, as protons and neutrons don't have the same mass, so isotopes very close to the transition region don't behave perfectly smoothly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The flipping effect I mention can be seen very nicely in the list of isotopes shown in &lt;a href=&quot;http://www.sisweb.com/referenc/source/exactmas.htm&quot;&gt;this site&lt;/a&gt;. Look at the (atomic) masses for the isotopes, and see how the fractional part of the mass starts out as circa $.01$ (above the isotope's mass number), then decreases to $.001$, reaches exactly zero (by definition) for $\mathrm{^{12}C}$, then soon after flips under to $.999$ and keeps going lower all the way to $.90$ around zirconium and tin isotopes before slowly rising up to $.98$ for bismuth $\mathrm{^{209}Bi}$ and then flipping back over to $.04$ for thorium and $.05$ for uranium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's a question: I said the maximum of nuclear binding energy per nucleon happened at $\mathrm{^{62}Ni}$, so why does the fractional part of the masses in the table keep decreasing all the way to tin $\mathrm{^{116}Sn}$? It's because these masses are not normalized with respect to how many protons and neutrons there are in the nucleus. If you were to take all the isotopic masses and divide them by their respective mass numbers, you would see the minimum around $\mathrm{^{62}Ni}$, as expected (actually, &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_binding_energy#The_binding_energy_maximum_and_ways_to_approach_it_by_decay&quot;&gt;the minimum ratio of isotope mass to mass number happens at $\mathrm{^{56}Fe}$&lt;/a&gt; because it has a slightly higher proportion of protons to neutrons. As I said, the small difference between proton and neutron masses creates slight irregularities which I did not take into account for simplicity).&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-05-29T14:30:40.517" LastActivityDate="2015-05-29T14:30:40.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32124" PostTypeId="1" CreationDate="2015-05-28T23:32:21.630" Score="13" ViewCount="5103" Body="&lt;p&gt;I had conducted an experiment where I reacted a $0.3\ \mathrm{g}$ strip of magnesium with $100\ \mathrm{mL}$ of varying concentrations of $\ce{HCl}$: $1\ \mathrm{mol/L}$, $2\ \mathrm{mol/L}$ and $3\ \mathrm{mol/L}$. From the data I collected, I noticed that the reaction with $1\ \mathrm{mol/L}$ $\ce{HCl}$ produced the most energy and the reaction with $3\ \mathrm{mol/L}$ $\ce{HCl}$ the least. I just wanted to know if my data is correct or not because I hypothesized that the $3\ \mathrm{mol/L}$ reaction would produce the most heat energy.&lt;/p&gt;&#xA;" OwnerUserId="12949" LastEditorUserId="194" LastEditDate="2017-04-01T01:05:53.160" LastActivityDate="2017-04-01T01:05:53.160" Title="How different concentration of HCl on magnesium affect the enthalpy change" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;metal&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32125" PostTypeId="1" CreationDate="2015-05-28T23:48:04.360" Score="6" ViewCount="928" Body="&lt;p&gt;Why do some crystal structure files not include bonds (the atoms are just placed in the respective position in the cell without bonds between any atoms)?&#xA;Shouldn't the bonds always be there?&#xA;Also, where can I look up these bonds?&lt;/p&gt;&#xA;" OwnerUserId="16591" LastEditorUserId="275" LastEditDate="2015-05-29T00:51:46.573" LastActivityDate="2015-05-29T03:14:26.123" Title="Why do some crystal structure files not include bonds?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32128" PostTypeId="2" ParentId="32125" CreationDate="2015-05-29T01:20:57.820" Score="0" Body="&lt;p&gt;Position of the atoms are sufficient information to solve schrodinger equation or to find molecular characteristics. Now, if you want to see bonds to make you more comfortable, you can use molecular visualizer such as babel, vesta, vmd, materials studio, chemshell, spartan etc. And there are plenty of visualizer that are available freely online.&lt;br&gt;&#xA;To better explain, i think i should talk about molecular orbital. In quantum mechanics a linear combination of atomic orbital is used to construct molecular orbital. Now after solving the schrodinger equation, you can see the occupancy of each molecular orbital. And they are not perfect integer as opposed to classical idea of orbital. Some of them may have fractional occupancy. So, it's irrelevant to talk about single bond, double bond or triple bond. But to understand it better we call them single bond, double bond or triple bond. Like in my calculation if C-O bond distance is ~1.5 Angstrom i take it as single bond and if it is ~1.3 Angstrom i take it as double bond. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-05-29T01:20:57.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32129" PostTypeId="1" AcceptedAnswerId="32133" CreationDate="2015-05-29T01:48:30.527" Score="2" ViewCount="871" Body="&lt;p&gt;Recently I did a reaction in the lab (the reaction itself is not important) whose product was colloidal sulfur. As I watched the sulfur particles form I noticed something strange: At first the sulfur particles appeared white, but after a while the solution turned yellowish (probably when the size of the sulfur particles grew a little). I know that sulfur's colour is yellow by nature, because of its absortion soectrum, and I understand that the white colour is probably due to Mie scattering of light by the colloidal particles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why aren't the particles yellow? if the absortion spectrum has as a consequence a yellow colour I would expect that spectrum to remain valid at all dimensions of the particle, the particle can't just randomly decide to stop absorbing light of some wavelenght when it reaches a threshold size... can it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that Mie scattering predicts that a particle will scatter all wavelenghts equally if its size is comparable or bigger than the wavelenght of light, but I would have thought that if the particle absorbed a specific wavelenght that wouldn't change with size.&lt;/p&gt;&#xA;" OwnerUserId="15854" LastActivityDate="2015-05-29T03:28:00.080" Title="Light scattering in sulfur colloids" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;optical-properties&gt;&lt;colloids&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32132" PostTypeId="2" ParentId="32125" CreationDate="2015-05-29T03:14:26.123" Score="9" Body="&lt;p&gt;&lt;strong&gt;If you solve an x-ray crystal structure, you simply get a set of atomic positions. No more, no less.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, the atomic positions don't include hydrogen atoms, only the &quot;heavy&quot; atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, most CIF (crystallographic interchange format) files, there &lt;em&gt;are&lt;/em&gt; no defined bonds. Somewhere along the way, software &lt;em&gt;perceives&lt;/em&gt; bonding information because most chemists prefer to have a valence-bond graph of atoms and bonds (and bond orders).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;em&gt;many&lt;/em&gt; algorithms for determining a bond graph from a set of atomic positions with varying &quot;accuracy.&quot; I put the word in quotes because it's sometimes unclear what the goal should be when determining bond orders. For example, is a molecule in a keto or enol tautomer form? In some cases, the $\ce{C-O}$ bond is not clearly a single bond distance or a $\ce{C=O}$ distance. Maybe there's an ensemble average of both forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard seems to be if the CIF file looks more &quot;ionic&quot; (e.g., a salt like $\ce{NaCl}$) bonds probably will be ignored, but otherwise, bonds are interpreted from interatomic distances.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-05-29T03:14:26.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32133" PostTypeId="2" ParentId="32129" CreationDate="2015-05-29T03:28:00.080" Score="2" Body="&lt;p&gt;There are a number of different effects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering&quot; rel=&quot;nofollow&quot;&gt;Rayleigh scattering&lt;/a&gt; is the predominant effect for the tiniest particles, much smaller than the wavelength of light, i.e. before they've coalesced. It is dependent on the fourth (!) power of the frequency of the radiation, so short wavelength light is scattered toward the eye. Looking &lt;em&gt;through&lt;/em&gt; the colloidal mixture, it should look redder.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tyndall_effect&quot; rel=&quot;nofollow&quot;&gt;Tyndall scattering&lt;/a&gt; applies to larger colloidal particles, and it too is dependent on the fourth power of the frequency of the radiation and preferentially scatter bluer light, so the yellow of the sulfur would be less apparent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, particles coalesce and scattering becomes less important than reflection from the larger particles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Perhaps&lt;/em&gt; sulfur nano-particles behave a bit differently from bulk material due either to &lt;a href=&quot;http://www.nano.gov/nanotech-101/special&quot; rel=&quot;nofollow&quot;&gt;quantum effects or to surface effects&lt;/a&gt;. See &lt;a href=&quot;https://www.uvm.edu/geology/documents/GarciaJr2012.pdf&quot; rel=&quot;nofollow&quot;&gt;Elemental Sulfur Nanoparticle Coarsening...&lt;/a&gt; If someone has experience with nanoscale sulfur, please feel free to comment or edit this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-05-29T03:28:00.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32134" PostTypeId="1" CreationDate="2015-05-29T03:56:15.497" Score="3" ViewCount="576" Body="&lt;p&gt;My friend can buy polyurethane epoxy from his workplace for 3 dollars a litre, except that he doesn't know precisely the data sheet for it. His factory gets it in 20 ton deliveries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He says that the mix has to be around 99% Polyurethane, 0.2% cobalt, and 0.8% of a toxic transparent catalyst liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a generic industrial chemistry used everywhere, and can be mixed roughly by eye to achieve different drying speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know what the name of this resin system is called and where i can find precise information on it? What is the catalyst?&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorUserId="14577" LastEditDate="2015-05-29T04:04:45.430" LastActivityDate="2015-05-29T14:09:13.203" Title="Polyurethane epoxy resin chemistry advice please?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32135" PostTypeId="1" CreationDate="2015-05-29T08:07:50.343" Score="-7" ViewCount="167" Body="&lt;p&gt;I am currently doing research on antioxidant activity of essential oil from Zingiber zerumbet. The essential oil contain zerumbone. The problem is I am confuse about the functional group of zerumbone is it ketone or aldehyde or what. I want to do thin layer chromatography test to identify the purity of the essential oil but I don't know how to identify it.&lt;/p&gt;&#xA;" OwnerUserId="16598" LastEditorUserId="11367" LastEditDate="2018-06-18T15:14:45.950" LastActivityDate="2018-06-18T15:14:45.950" Title="What is the functional group of zerumbone?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ClosedDate="2016-11-08T16:17:38.687" ContentLicense="CC BY-SA 3.0" />
  <row Id="32136" PostTypeId="2" ParentId="32135" CreationDate="2015-05-29T09:35:24.813" Score="5" Body="&lt;p&gt;Zerumbone, a constituent of the rhizomes of wild ginger (&lt;em&gt;Zingiber zerumbet&lt;/em&gt;), is (2​&lt;em&gt;E&lt;/em&gt;​,6​&lt;em&gt;E&lt;/em&gt;​,10​&lt;em&gt;E&lt;/em&gt;​)-​2,6,9,9-​​tetramethylcycloundeca-​2,6,10-trienone; i.e. it is a ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQh00.png&quot; alt=&quot;Zerumbone&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-29T09:35:24.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32137" PostTypeId="1" AcceptedAnswerId="32143" CreationDate="2015-05-29T10:55:43.970" Score="14" ViewCount="29697" Body="&lt;p&gt;In IR spectroscopy, the $x$-axis is used to represent wavenumber, in $\mathrm{cm^{-1}}$. Why is wavenumber, equal to $1/\lambda$, used in place of wavelength, which is simply $\lambda$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources I’ve already found explain why it was chosen rather than energy of waves, but the conversion from wavelength to wavenumber is never explained. Below are two relations from Wikipedia, which explain how it can be used in equations, but in all of these cases, $\lambda$ seems to be a choice that’s easier to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What explanations are there, if anything other than “historical reasons”, for why $1/\lambda$ is favored over $\lambda$?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A spectroscopic wavenumber $\tilde\nu$ can be converted into energy per photon $E$ via Planck’s relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = hc\tilde\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be converted into wavelength of light via&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{1}{n\tilde\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the refractive index of the medium.&lt;/p&gt;&#xA;" OwnerUserId="13432" LastEditorUserId="7113" LastEditDate="2017-10-20T03:26:43.303" LastActivityDate="2019-06-21T11:45:08.473" Title="Why is wavenumber used in IR spectroscopy rather than wavelength?" Tags="&lt;spectroscopy&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32138" PostTypeId="2" ParentId="32109" CreationDate="2015-05-29T11:43:56.563" Score="1" Body="&lt;p&gt;For both the examples that you mentioned, the carbonyl oxygen has its lone pair localized while the other oxygen's lone pair is delocalized with the double bond. The better nucleophile is the one which has greater electron density. &lt;/p&gt;&#xA;" OwnerUserId="16601" LastActivityDate="2015-05-29T11:43:56.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32139" PostTypeId="1" AcceptedAnswerId="32144" CreationDate="2015-05-29T12:00:42.213" Score="6" ViewCount="6488" Body="&lt;p&gt;I have met contradicting evidence. The phase diagrams show that $\ce{CO2}$ solidifies at $-78.5\ ^{\circ} \mathrm{C}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HPtp0.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nqNVD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_dioxide&quot; rel=&quot;noreferrer&quot;&gt;confirms this&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 1 atmosphere (near mean sea level pressure), the gas deposits directly to a solid at temperatures below $−78.5\ \mathrm{^\circ C}$ ($−109.3\ \mathrm{^\circ F}; 194.7 ~\mathrm{K}$) and the solid sublimes directly to a gas above $−78.5\ \mathrm{^\circ C}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yet &lt;a href=&quot;http://news.discovery.com/earth/global-warming/carbon-capture-antarctica-south-pole-120831.htm&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; claims it happens at $133\ \mathrm{K}$ (otherwise $\ce{CO2}$ would showfall at Earth's poles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which claim is correct?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorDisplayName="user7951" LastEditDate="2015-05-29T12:23:09.010" LastActivityDate="2015-05-29T12:38:10.807" Title="At what temperature CO2 becomes solid at normal pressure?" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ClosedDate="2015-05-29T22:26:46.030" ContentLicense="CC BY-SA 3.0" />
  <row Id="32140" PostTypeId="2" ParentId="32112" CreationDate="2015-05-29T12:01:08.740" Score="3" Body="&lt;p&gt;The shapes of the orbitals, i.e. the solutions to the Schrödinger equation for the hydrogen atom $\hat{H}\Psi_i = E_i\Psi_i $, are primarily due to the requirement that all orbitals are orthogonal: $\langle\Psi_i|\Psi_j\rangle=0$.  Orthogonality is made possible by the fact that the orbitals have &quot;positive&quot; and &quot;negative&quot; regions separated by nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One what to think about $n$ is that $n-1$ is the number of nodes in $\Psi_i$. So there can be no $1p$ orbitals since $p$-shaped orbitals have at least one node. Similarly, there can be no $2d$ orbitals, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $s$ orbitals have a spherical shape, there is only one way to orient them, i.e. $m=0$ if $l=0$. $p$-shaped orbitals lie along one axis and there are three Cartesian axes, so there are three possible - orthogonal - orientations (I'm sacrificing a bit of accuracy here for clearness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit harder to visualize why there should be 5 and only 5 $d$-shaped orbitals, but the answer is that there are only 5 ways to construct 3-dimensional shapes with (a minimum of) 2 nodes that are orthogonal to the $s$- and $p$-shaped solution as well as the other $d$-shaped solutions. &lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-05-29T12:31:56.540" LastActivityDate="2015-05-29T12:31:56.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32141" PostTypeId="2" ParentId="32137" CreationDate="2015-05-29T12:07:48.093" Score="11" Body="&lt;p&gt;Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-05-29T12:07:48.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32142" PostTypeId="2" ParentId="32112" CreationDate="2015-05-29T12:11:43.517" Score="2" Body="&lt;p&gt;While it may seem that quantum numbers were invented out of the blue, that isn’t the case. They derive directly from solving the Schrödinger equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{H}\Psi = E \Psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the simplest Schrödinger equations is that for a particle in a box — the one most quantum chemistry or quantum mechanics courses start with and therefore found in every textbook for quantum chemistry which is why I won’t repeat it here. But in a nutshell: You need a function that you can differentiate twice and which will then reproduce itself with a constant factor &lt;em&gt;E&lt;/em&gt;. That can be achieved by sine waves. You cannot say &lt;em&gt;a priori&lt;/em&gt; which frequency the sine function will have, though: If the box has size &lt;em&gt;l&lt;/em&gt;, it works for sine waves with the wavelengths $2l,\ l,\ \frac{2l}{3},\ \frac{l}{2},\ \dots$ and the corresponding wavenumbers $\frac{1}{2l},\ \frac{1}{l},\ \frac{3}{2l},\ \frac{2}{l},\ \dots$ or $\frac{n}{2l}$ with &lt;em&gt;n&lt;/em&gt; being a natural number. &lt;em&gt;n&lt;/em&gt; is therefore the simplest case of a quatum number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you solve this equation for an electron orbiting a proton, the solutions are much more complicated (and there are intermediate steps often presented in the quantum chemistry courses to make the transition easier). However, &lt;em&gt;n&lt;/em&gt; still turns up as a variable, corresponding to quanted amounts of energy. For certain values of &lt;em&gt;n&lt;/em&gt;, there are additional variables whose values cannot be derived &lt;em&gt;a priori&lt;/em&gt; but which can only assume specific values — they correspond to the other quantum numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in itself is just a solution to a mathematical equation. The real beauty lies in the fact that these quantum numbers can be used to explain experimental observations such as hydrogen’s flame colours. &lt;em&gt;All&lt;/em&gt; quantum numbers can be used to explain some property one has deduced by experiments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So tl;dr: It’s just what maths tells us will be the case.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-29T12:11:43.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32143" PostTypeId="2" ParentId="32137" CreationDate="2015-05-29T12:37:36.017" Score="7" Body="&lt;p&gt;The choice to use wavenumbers for infrared spectroscopy (rather than wavelengths, frequencies, or energies) was probably done to provide a range that has both the appearance of width (so that the difference between two peaks is more meaningful) and spans a set of reasonable values that do not contain very large or very small numbers (which are hard to conceptualize). The goal is to be able to easily compare values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following comparison of units/values for the typical range of IR spectroscopy for organic compounds and some &quot;example&quot; values for the absorptions of common bond types:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;absorption     cm⁻¹   m        µm     Hz        THz   J          kJ/mol     meV&#xA;high end       500    2.00E-5   20     1.5E+13   15    9.94E-21   5.98       62&#xA;C-O            1100   9.09E-6   9.09   3.3E+13   33    2.19E-20   13.2       136&#xA;C=C            1660   6.02E-6   6.02   5.0E+13   50    3.30E-20   19.9       206&#xA;C=O            1720   5.81E-6   5.81   5.2E+13   52    3.42E-20   20.6       213&#xA;C-H            3000   3.33E-6   3.33   9.0E+13   90    5.96E-20   35.9       372&#xA;O-H            3500   2.86E-6   2.86   1.05E+14  105   6.96E-20   41.9       434&#xA;low end        4000   2.50E-6   2.50   1.20E+14  120   7.95E-20   47.9       496&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's compare especially the peaks for $\ce{C=C}$ and $\ce{C=O}$. These peaks are easily resolvable by all modern FTIR spectrometers, and there is room for peaks to be resolved between them. Only the values in wavenumbers give this sense of resolution intuitively. Modern spectrometers can resolve data to $1.0\ \text{cm}^{-1}$ or better.  A resolution of $1.0\ \text{cm}^{-1}$ is equivalent to resolutions of $0.040\ \mu\text{m}$, $0.030\text{ THz}$, $12\text{ kJ/mol}$, and $0.12\text{ meV}$. Which of these resolutions is most intuitive to understand?&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7475" LastEditDate="2017-10-20T03:50:27.253" LastActivityDate="2017-10-20T03:50:27.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32144" PostTypeId="2" ParentId="32139" CreationDate="2015-05-29T12:38:10.807" Score="6" Body="&lt;p&gt;The phase diagram you have shown is correct and shows the sublimation point at $1~\mathrm{atm}$ as $-78.5~\mathrm{^\circ C}$ . So if you have a pure $\ce{CO2}$ atmosphere, it will solidify at $-78.5~\mathrm{^\circ C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperatues in Antarctica are known to get below $-78.5~\mathrm{^\circ C}$ so you might wonder why whe don't find dry ice in Antartica (like we do at the Martian poles [Note: they are a lot colder]). The reason is that the partial pressure of $\ce{CO2}$ in the atmosphere is so low that at $-78.5~\mathrm{^\circ C}$, dry ice is forming but it is subliming faster than it forms because the vapor pressure of $\ce{CO2}$ in the atmosphere above it is well below the saturation vapor pressure required for ice to accumulate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The vapor pressure must reach the saturation vapor pressure for dew or frost to form. This happens at the dew point or frost point temperature, which is dependent on atmospheric pressure and the absolute amount of vapor in the air. As atmospheric temperature increases, the dew/frost point temperature increases. As atmospheric pressure increases, the vapor pressure increases. At very low temperatures, the dew/frost point temperature is very low.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wattsupwiththat.com/2009/06/13/results-lab-experiment-regarding-co2-snow-in-antarctica-at-113%C2%B0f-80-5%C2%B0c-not-possible/&quot;&gt;http://wattsupwiththat.com/2009/06/13/results-lab-experiment-regarding-co2-snow-in-antarctica-at-113%C2%B0f-80-5%C2%B0c-not-possible/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://stevengoddard.wordpress.com/2014/06/03/antarctica-gets-cold-enough-to-freeze-co2/&quot;&gt;https://stevengoddard.wordpress.com/2014/06/03/antarctica-gets-cold-enough-to-freeze-co2/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-29T12:38:10.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32146" PostTypeId="1" AcceptedAnswerId="32147" CreationDate="2015-05-29T13:49:13.883" Score="4" ViewCount="54" Body="&lt;p&gt;While studying the &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;lamba sensor&lt;/a&gt; that works on yttria-strabilised-zirconium (YSZ) I came across the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is $F$ in this equation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some more info&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;wall&quot; of the lambda sensor is as shown in the picture with the voltage $V_{ref}$ measured across:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tDzDU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation apparently is derived directly from the chemical equilibrium (chemical potentials equals to zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu_{O_2}+4\mu_e=2\mu_{O^{--}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They come to the following where they end at the final formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{ref}=\frac{RT}{4F}\ln\left(\frac{p_{O_2}^{gas}}{p_{O_2}^{ref}}\right) \implies p_{O_2}^{gas}=p_{O_2}^{ref} \cdot \exp\left(\frac{4FV_{ref}}{RT}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the partial pressures. The gas constant $R$ and temperature $T$ are also clear. If I am correct, $V_{ref}$ is the voltage across the material of the sensor wall, as in the picture, so this is also clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is the final result from which the oxygen amount in the gas can be found, since the partial pressure of the oxygen in the gas can be calculated. &lt;strong&gt;My question is: What is $F$ in the equation?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16606" LastActivityDate="2015-05-29T13:56:54.370" Title="The property F in an equation for chemical potential" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32147" PostTypeId="2" ParentId="32146" CreationDate="2015-05-29T13:56:54.370" Score="4" Body="&lt;p&gt;$F$ is the Faraday constant: $9.64853399(24)\times10^4~\mathrm{C~mol^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick dimensional analysis confirms that this makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-29T13:56:54.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32148" PostTypeId="2" ParentId="32134" CreationDate="2015-05-29T14:09:13.203" Score="3" Body="&lt;p&gt;The resin is the same variety commonly used to produce fiberglass baths, boats and many other uses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resin is unsaturated polyester, using methyl ethyl ketone peroxide (MEKP) as catalyst, cobalt solution for example cobalt octoate (Coct) as an accelerator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't often sold to consumers because it is too explosive and corrosive, the MEKP reacts explosively with raw cobalt, and it's dangerous to the eyes, it can cause blindness in 15 seconds if splashed in the eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MEKP doseage is 1-2% per volume of polyester, depending on desired drying time. The cobalt is mixed first and dyes the polyester a whiskey color, you can dose it by color, the darker whisky color the faster it will dry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industry standards are 1% and 6% cobalt solution, 1% is dosed at 0.5% to 2%, however there is no way to tell without the product data sheet if the cobalt solution 1%, 6%, 10%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;safety: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MEKP and cobalt mix = flames&#xA;MEKP in eyes = blindness&#xA;wear goggles and rinse with water and go to hospital if you get some MEKP eye contact. &#xA;wear protective clothing and avoid skin contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Cobalt octoate is the most reactive and most commonly used paint drier. Cobalt octoate catalyses the oxygen uptake and accelerates peroxide formation causing a surface film to form rapidly and hence known as oxidation catalyst. It basically a metal caroxylate used to accelerate the drying process like linseed oil. It accelerates the decomposition of the organic peroxide initiators (often called catalysts) and promotes the cross-linking of polyester resins.&#xA;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that baths are finished with a gloss product called gel coat.&lt;/p&gt;&#xA;" OwnerUserId="14577" LastActivityDate="2015-05-29T14:09:13.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32149" PostTypeId="2" ParentId="32118" CreationDate="2015-05-29T14:18:49.000" Score="1" Body="&lt;p&gt;What you need to know is the vapor pressure of $\ce{Cl2}$ as a function of temperature.  Just ensure that the release pressure of the chlorine is lower than the vapor pressure, and it will be a vapor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.resistoflex.com/chlorine_graphs.htm&quot; rel=&quot;nofollow&quot;&gt;one web site&lt;/a&gt; that gives a graph (and a literature reference) of $\ce{Cl2}$ vapor pressure as a function of temperature from -30 &amp;deg;F to 220 &amp;deg;F.  The references they give are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reprinted from Chlorine Manual, the Chlorine Institute, 3rd ed. 1959. Data for graph from R. Kapoor and J.J. Martin, Thermodynamic Properties of Chlorine, Eng. Res. Inst. Univ. of Michigan, 1957.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At -30 &amp;deg;F they report $\ce{Cl2}$ vapor pressure is &quot;0&quot; psi (which has to be an approximation; the vapor pressure of all materials is non-zero at any non-zero temperature) and at 220 &amp;deg;F it is 587 psi.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:11:20.077" LastActivityDate="2016-03-11T23:11:20.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32150" PostTypeId="1" CreationDate="2015-05-29T14:30:55.423" Score="2" ViewCount="106" Body="&lt;p&gt;I have created a firefly jar with glow sticks to sell at a school fair. The firefly jar lasts only for an hour or so. I have heard that glow sticks can be reactivated. The firefly jars contain the chemicals found in good sticks. Is there any way to make the jar glow again?&lt;/p&gt;&#xA;" OwnerUserId="16604" LastEditorUserId="418" LastEditDate="2015-05-29T14:56:19.590" LastActivityDate="2015-05-29T15:36:20.437" Title="Can I reactivate a chemoluminescent reaction?" Tags="&lt;organic-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-06-15T15:26:34.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="32151" PostTypeId="1" AcceptedAnswerId="32172" CreationDate="2015-05-29T14:35:34.917" Score="6" ViewCount="74" Body="&lt;p&gt;I have been searching google and didn't found any links that describes how  this process is performed - I mean the chemical procedure, not the process of taking the blood from the vein.&lt;/p&gt;&#xA;" OwnerUserId="4641" LastEditorUserId="4231" LastEditDate="2015-05-29T22:21:53.570" LastActivityDate="2015-05-29T22:21:53.570" Title="How are hormone levels measured?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32152" PostTypeId="1" AcceptedAnswerId="32241" CreationDate="2015-05-29T14:38:15.920" Score="4" ViewCount="64" Body="&lt;p&gt;Are there softeners used in polyester or polypropylene to make these materials more flexible? Or is this not needed for this kind of synthesis?&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="83758" LastEditDate="2020-04-25T07:05:44.037" LastActivityDate="2020-04-25T07:05:44.037" Title="Are there softeners in synthetic materials?" Tags="&lt;synthesis&gt;&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32153" PostTypeId="1" CreationDate="2015-05-29T14:57:22.543" Score="1" ViewCount="974" Body="&lt;p&gt;I just want to know what particle in plants can change $\ce{H2O}$ into $\ce{O2}$ and what $\ce{CO2}$ does in this conversion system. What if we just use $\ce{H2O}$ and light to convert it into $\ce{O2}$?  If light is what we need to convert $\ce{CO2}$ into glucose and $\ce{H2O}$ into oxygen, why can't scientists make a machine to do that?&lt;/p&gt;&#xA;" OwnerUserId="16611" LastEditorUserId="5026" LastEditDate="2018-03-02T02:06:27.517" LastActivityDate="2018-03-02T02:06:27.517" Title="What is the mechanism of photosynthesis?" Tags="&lt;organic-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32154" PostTypeId="1" AcceptedAnswerId="35602" CreationDate="2015-05-29T15:23:42.810" Score="5" ViewCount="240" Body="&lt;p&gt;&lt;strong&gt;Can a flame a start on a wet piece of paper immersed in water?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video is shown below (not the full video):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/Ey3z8z4Hxtc?t=1h27m29s&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/Ey3z8z4Hxtc?t=1h27m29s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not in English so you might not understand anything. In the video they simply say it is demons or whatever that starts the fire, which is silly to me, of course. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What kind of compounds can we use (or the fraud used in the video)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What do they do in movies, tv shows, magic shows, or even the olympics, to start flames on water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What changes the color of flame. For example it was pink in the video?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16612" LastEditorUserId="9961" LastEditDate="2015-10-30T17:15:15.940" LastActivityDate="2015-10-30T17:15:15.940" Title="Can a flame start on water (video included)?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32155" PostTypeId="1" CreationDate="2015-05-29T15:26:18.557" Score="2" ViewCount="83" Body="&lt;p&gt;$\ce{LiBr}$ $Vp$ chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mediafire.com/view/x77zc8wj11gthsj/LiBr_Vp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mediafire.com/view/x77zc8wj11gthsj/LiBr_Vp.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mediafire.com/view/u4dnocvbsf3atmu/Useful_Chart_LiBr.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mediafire.com/view/u4dnocvbsf3atmu/Useful_Chart_LiBr.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lecture in question (p.13)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mediafire.com/view/gly78yz339hv1x1/RAC_Lecture_15.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mediafire.com/view/gly78yz339hv1x1/RAC_Lecture_15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HXjFV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lecture it says that at this temperature &amp;amp; at this concentration crystallisation will occur... however how can this be ascertained using the chart? I do know how to read such Duhring plots..&lt;/p&gt;&#xA;" OwnerUserId="7583" LastEditorDisplayName="user15489" LastEditDate="2015-05-29T20:11:12.770" LastActivityDate="2015-05-29T20:11:12.770" Title="Unable to comprehend graph of aqueous lithium bromide vapour pressure" Tags="&lt;vapor-pressure&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32156" PostTypeId="2" ParentId="32150" CreationDate="2015-05-29T15:36:20.437" Score="3" Body="&lt;p&gt;Short answer: No, you can't! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer: Still no, and this is why: The reaction between the oxalate and hydrogen peroxide is &lt;strong&gt;irreversible&lt;/strong&gt;. Both compounds are consumed in the reaction and there is no way back! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only compound that survives is a fluorescent material, that absorbs in the UV, emits in the visible range and gives the colour that you finally observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;strong&gt;Ben Norris&lt;/strong&gt; pointed out in his comment, your only options are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;add fresh diphenyloxalate and hydrogen peroxide (which again will be consumed) or&lt;/li&gt;&#xA;&lt;li&gt;irradiate the fluorescent compound that has survived with a UV lamp&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-29T15:36:20.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32157" PostTypeId="1" CreationDate="2015-05-29T15:57:13.283" Score="30" ViewCount="909" Body="&lt;p&gt;I want to know the exact mechanism for how silver would start to grow during a mirroring with Tollen's reagent. I've made two pictures to illustrate how I see the nucleation/growth process of silver on silica:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/xoLoDqu.png&quot; alt=&quot;nucleation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/exk2YOZ.png&quot; alt=&quot;growth&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see, I didn't draw any chemical bonds, because I think that the silver adherence is due to Van Der Waals forces. Am I correct? If that is the case, then I believe a place with &lt;strong&gt;high&lt;/strong&gt; permanent polarisation would be better suited for the nucleation of silver.&lt;/p&gt;&#xA;&lt;p&gt;Now, let's follow the logic. We have a greater electronegativity difference between &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Si}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$3.44 - 1.90 = 1.54$&lt;/span&gt;) than between &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$3.44 - 2.2 = 1.24$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;For the upper oxygen, assuming an angle of 109 degrees, you get a total dipole moment of: &lt;span class=&quot;math-container&quot;&gt;$2(1.54)\cos\left(\frac{109}{2}\right) = 1.79$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For the lower oxygen with the bonded hydrogen, assuming an angle of 120 degrees , you would get a total dipole moment of: &lt;span class=&quot;math-container&quot;&gt;$1.54 + 1.24 = 2.78$&lt;/span&gt;&#xA;(please tell me if I'm wrong to assume a straight line!)&lt;/p&gt;&#xA;&lt;p&gt;This would mean that you have a greater accumulation of electrons at the bottom oxygen, which would favor Van Der Waals attraction (Debye interaction, I guess, since you have a permanent dipole and an induced dipole). This, in turn, would mean that it would be a better nucleation site.&lt;/p&gt;&#xA;&lt;p&gt;If you make a surface treatment using silane groups, I believe you promote the presence of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; groups at your silica surface, though I'm unsure about this. I do know that it makes better silver mirroring though. I've made another drawing:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/5cOzEFM.png&quot; alt=&quot;silanisation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Following my logic, this one is wrong in many way. First of all, I drew angles on the &lt;span class=&quot;math-container&quot;&gt;$\ce{Si-O-H}$&lt;/span&gt; bonds, which I didn't draw on my first pictures. Secondly, the silver is not even near the oxygen atoms, which I believe are the nucleation points.&lt;/p&gt;&#xA;&lt;p&gt;I'm confused, could you please point out to me where I'm right or wrong in my logic/knowledge?&lt;/p&gt;&#xA;&lt;p&gt;Mainly:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a bond forming between the silica surface and the silver? Or is there only Van Der Waals (Debye) attraction?&lt;/li&gt;&#xA;&lt;li&gt;Is my depiction of nucleation &amp;amp; growth correct (for the first two pictures)?&lt;/li&gt;&#xA;&lt;li&gt;Is my logic regarding the electronegativity correct?&lt;/li&gt;&#xA;&lt;li&gt;Is my logic of &amp;quot; higher total electronegativity difference, hence better nucleation site&amp;quot; correct? (assuming I'm right about the Van Der Waals Forces)&lt;/li&gt;&#xA;&lt;li&gt;Is my depiction of silanisation correct?&lt;/li&gt;&#xA;&lt;li&gt;Am I right to say that the higher abundance of surface &lt;span class=&quot;math-container&quot;&gt;$\ce{OH}$&lt;/span&gt; groups by silanisation implies better nucleation?&lt;/li&gt;&#xA;&lt;li&gt;Could I do the same when using a strong acid on my glass surface?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16613" LastEditorUserId="102629" LastEditorDisplayName="user15489" LastEditDate="2021-01-10T21:20:45.177" LastActivityDate="2021-01-10T21:20:45.177" Title="How does the silver nucleate on glass in Tollen's reagent mirroring?" Tags="&lt;surface-chemistry&gt;&lt;silver&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32158" PostTypeId="1" AcceptedAnswerId="32159" CreationDate="2015-05-29T16:01:56.337" Score="20" ViewCount="1920" Body="&lt;p&gt;The product of $2Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively few protons. Whereas, the product of $2.5Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise relatively many protons. The product of $((2+2.5)/2)Z$ generally approximates the value of $m_\mathrm{a}$ of atoms that comprise a moderate number of protons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What function from the $Z$ values produces a value that maximally approximates the $m_\mathrm{a}$ that corresponds to $Z$?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="5017" LastEditDate="2015-05-29T16:46:41.237" LastActivityDate="2017-01-20T22:31:18.897" Title="Is there a function to approximate atomic mass from the atomic number?" Tags="&lt;periodic-table&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32159" PostTypeId="2" ParentId="32158" CreationDate="2015-05-29T16:42:14.847" Score="28" Body="&lt;p&gt;&lt;strong&gt;The short answer is that you can find a power-law fit ($1.61Z^{1.1}$) with low average error.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd never really thought about it much, but after downloading the &lt;a href=&quot;http://www.ciaaw.org/atomic-weights.htm&quot;&gt;IUPAC Atomic Weights&lt;/a&gt;, I decided to do some curve fitting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a linear fit between atomic number and atomic mass:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fxX6Y.png&quot; alt=&quot;linear fit of atomic mass from atomic number&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, the fit isn't very good for small $Z$, but the overall fit isn't bad - the mean absolute error (MAE) is $2.821 \:\mathrm{u}$, and taken as a whole, the data is &lt;em&gt;surprisingly&lt;/em&gt; linear. (Well, surprising to me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought of a &lt;strong&gt;quadratic&lt;/strong&gt; fit, requiring the intercept to be 0,0 to ensure the best fit for small $Z$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1hNd7.png&quot; alt=&quot;quadratic fit of atomic mass from atomic number&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looks better, right? Certainly the fit is much better for first and second row elements, but the MAE only reduces to $2.749 \:\mathrm{u}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I went up to a cubic fit, again requiring 0,0 for the intercepts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/47CN8.png&quot; alt=&quot;cubic fit of atomic mass from atomic number&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aha, now we're talking! We get the subtle nonlinearities, and the MAE is down to $1.36 \:\mathrm{u}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the comment by Nicolau below, I performed a power-law fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJRzo.png&quot; alt=&quot;power-law fit&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that gives a power-law fit with MAE of $0.01 \:\mathrm{u}$ and a fairly easy-to remember function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_a \approx 1.61Z^{1.1}$&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-05-29T17:10:11.123" LastActivityDate="2015-05-29T17:10:11.123" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32161" PostTypeId="1" AcceptedAnswerId="32178" CreationDate="2015-05-29T17:18:48.743" Score="5" ViewCount="1089" Body="&lt;p&gt;Because of the  8-valence electron rule I assume in CO the left resonance structure to be favored .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~&amp;lt;-&amp;gt;~ :\!C=O\!:: ~&amp;lt;-&amp;gt;~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;https://chemistry.stackexchange.com/questions/25126/what-is-the-lewis-structure-for-co&quot;&gt;What is the Lewis Structure for CO?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular orbital scheme of CO looks like this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rwOet.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the HOMO of CO is closer to the atomic orbital of O I think that the HOMO of CO is mainly localized on the oxygen atom, whereas the LUMO of CO is mainly localized on the carbon atom. Therefore I would have expected that the coefficient of the HOMO is higher on the O than on the C atom and therefore the top right resonance structure with the negative charge on the oxygen to be favored. However, in reality the top left resonance structure is favored. What is my error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From another post, calculations have shown that the coefficient is higher on the carbon atom (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/30797/how-can-the-dipole-moment-of-carbon-monoxide-be-rationalised-by-molecular-orbita&quot;&gt;How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?&lt;/a&gt;)&#xA;I guess my mistakes comes from the assumption that I say &quot;HOMO closer to the AO of oxygen&quot; means higher coefficient on oxygen. Is this assumption wrong and if yes why?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-05-30T01:12:31.230" Title="Can I predict the preferred resonance structure of carbon monoxide from its molecular orbital scheme?" Tags="&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32162" PostTypeId="2" ParentId="32153" CreationDate="2015-05-29T17:23:53.910" Score="3" Body="&lt;p&gt;I'm definitely not going to outline light harvesting in the &lt;em&gt;light harvesting complexes (LHC)&lt;/em&gt;, the role of the quinones in electron transfer, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let me give a short answer on your last question:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecular aggregates, such as the LHCs, involved in the different steps of photosynthesis are not optimized for duration. In contrast to man-made machines, they don't have to! Remember that all these fragile molecular assemblies are part of a living system. If parts break, they will resynthesized and replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is partly possible to mimic photosynthesis in the lab, it is the self-repair mechanisms that we are unable to replicate in artificial systems.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-05-29T17:23:53.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32163" PostTypeId="1" AcceptedAnswerId="32169" CreationDate="2015-05-29T18:57:28.157" Score="8" ViewCount="1568" Body="&lt;p&gt;In a study of the creation of ionic bonds of in this case $\ce{Na}^+$ and $\ce{Cl}^-$ into $\ce{NaCl}$, I have come across this equation for the overall energy of the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=-\frac{e^2}{4\pi\epsilon_0 R}+\frac{B}{R^n}+1.4 \,\mathrm{eV}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e$: electron charge. $\epsilon_0$: vacuum permittivity. $R$: the distance between them.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The first term is the attraction energy between the two equal but oppositely charged ions $\ce{Na}^+$ and $\ce{Cl}^-$.&lt;/li&gt;&#xA;&lt;li&gt;The second term is the repulsion between the like charges within the shells. When the distance $R$ is very small, this term has large (opposite) effect.&lt;/li&gt;&#xA;&lt;li&gt;The last term is the difference in free creation energy for the two free reactions ($\ce{Na \rightarrow Na^+ + e^-}$ requires $+5.1\,\mathrm{eV}$ and $\ce{Cl + e^- \rightarrow Cl^-}$ gives $-3.7\,\mathrm{eV}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is two-fold and about the repulsion energy term: &lt;strong&gt;What is $B$? And what is $n$?&lt;/strong&gt; I have been told that the exponent is around $n \approx 10$, but is this a general constant or material specific?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorDisplayName="user15489" LastEditDate="2015-05-29T19:05:28.297" LastActivityDate="2015-05-30T10:09:51.880" Title="What are B and n in this expression for charge repulsion?" Tags="&lt;energy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32164" PostTypeId="1" CreationDate="2015-05-29T20:22:17.357" Score="18" ViewCount="9412" Body="&lt;p&gt;I wanted to know whether hydrogen bonding is stronger in phenol or methanol.&#xA;I saw on Wikipedia about the enthalpies of hydrogen bonding in various cases.&#xA;but could not find the answer to this question of mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to know whether the hydrogen bond strength depends upon the carbon atom to which oxygen is attached.&lt;/p&gt;&#xA;" OwnerUserId="16619" LastEditorUserId="10185" LastEditDate="2016-05-25T09:32:28.207" LastActivityDate="2018-03-27T17:05:23.747" Title="Strength of hydrogen bonding in phenol or methanol" Tags="&lt;organic-chemistry&gt;&lt;intermolecular-forces&gt;&lt;alcohols&gt;&lt;hydrogen-bond&gt;&lt;phenols&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32167" PostTypeId="1" CreationDate="2015-05-29T21:33:44.347" Score="13" ViewCount="665" Body="&lt;p&gt;I wonder what compounds are usually called &quot;ice&quot; and what properties do determine it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can bring some examples of ice and not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not ice:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H2S}$&lt;/li&gt;&#xA;&lt;li&gt;cellulose&lt;/li&gt;&#xA;&lt;li&gt;sugar&lt;/li&gt;&#xA;&lt;li&gt;graphite&lt;/li&gt;&#xA;&lt;li&gt;lithium&lt;/li&gt;&#xA;&lt;li&gt;solid mercury&lt;/li&gt;&#xA;&lt;li&gt;polyvinylchloride&lt;/li&gt;&#xA;&lt;li&gt;solid fats&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Known as ices:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;solid water&lt;/li&gt;&#xA;&lt;li&gt;solid $\ce{CO2}$&lt;/li&gt;&#xA;&lt;li&gt;solid nitrogen&lt;/li&gt;&#xA;&lt;li&gt;solid oxygen&lt;/li&gt;&#xA;&lt;li&gt;solid methane (?)&lt;/li&gt;&#xA;&lt;li&gt;solid ethanol (?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems they all have some common properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Melting point below room temperature&lt;/li&gt;&#xA;&lt;li&gt;Transparency&lt;/li&gt;&#xA;&lt;li&gt;White or blueish color&lt;/li&gt;&#xA;&lt;li&gt;Chystalline structure (?)&lt;/li&gt;&#xA;&lt;li&gt;Easy sublimation and high saturation pressure&lt;/li&gt;&#xA;&lt;li&gt;Low strength and other mechanic properties&lt;/li&gt;&#xA;&lt;li&gt;Low surface friction (?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have I missed anything? What is the accepted definition of ices, say in astronomy or chemistry?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="11367" LastEditDate="2017-01-14T14:47:22.590" LastActivityDate="2017-04-13T08:31:16.407" Title="What properties a compound should have to be called an &quot;ice&quot;?" Tags="&lt;physical-chemistry&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32168" PostTypeId="2" ParentId="32164" CreationDate="2015-05-29T21:38:03.727" Score="-1" Body="&lt;p&gt;I'd guess phenol (though I'm not 100% sure about that).  My reasoning is that in phenol, there are several resonance structures that put extra negative charge on the oxygen, thus allowing for a stronger interaction.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastActivityDate="2015-05-29T21:38:03.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32169" PostTypeId="2" ParentId="32163" CreationDate="2015-05-29T21:58:01.977" Score="6" Body="&lt;p&gt;There isn't an exact form for the Pauli Repulsion term, but it is often approximated as a $\frac{1}{r^n}$ term for simplicity.  For example, in the Lennard-Jones expression for Van Der Waals interactions, $$E(r) = \epsilon \times[(\frac{\sigma}{r})^6-(\frac{\sigma}{r})^{12}]$$, the $\frac{1}{r^6}$ term is the London force term (which in reality, does go as $\frac{1}{r^6}$), and the $\frac{1}{r^{12}}$ is the Pauli repulsion term (just an approximation, but in this approximation, $n=12$).&lt;br&gt;&#xA;In the OPLS force field (which uses a Lennard-Jones expression for Van Der Waals interactions), the parameters for $\ce{Na+}$ and $\ce{Cl-}$ are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na+}$: $\sigma = 3.500 A$, $\epsilon = 0.0660\ \ce{kcal/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl-}$: $\sigma = 4.180 A$, $\epsilon = 0.11779\ \ce{kcal/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OPLS is parameterized for a geometric mixing rule, meaning that for an interaction between the two, the $\epsilon$ and $\sigma$ parameters are equal to the geometric mean of the parameters for each atom.  Thus, for an $\ce{Na+ - Cl-}$ interaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma = 3.8249 A, \epsilon = 0.08817\ \ce{kcal/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the Pauli part of the Lennard Jones expression is $\epsilon \times (\frac {\sigma}{r})^{12}$, so $B$ in this case is $\epsilon \times \sigma^{12} = 864554\ \ce{kcal A^12 /mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are other force fields parameterized differently, and some of these approximate the Pauli repulsion using a different value of $n$ (and some like $MM3$ don't even assume a $\frac{1}{r^n}$ dependency).  So since $n$ is assumed (and since different force fields are parameterized differently), there are other possible values for $B$ and $n$ to use in the approximation.&lt;/p&gt;&#xA;" OwnerUserId="15960" LastEditorUserId="10185" LastEditDate="2015-05-30T10:09:51.880" LastActivityDate="2015-05-30T10:09:51.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32170" PostTypeId="1" CreationDate="2015-05-29T22:14:12.870" Score="11" ViewCount="18361" Body="&lt;p&gt;The d-block sub-shell is always one shell lower than the s-block and p-block sub-shells. Assuming only the electrons in the highest energy shells count toward the set of valence electrons (is that correct?), d-block is never in the highest energy shell, and so none of the electrons in the d subshells would ever count toward the valence electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect I'm mistaken. Am I? If so, where have gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="4945" LastEditDate="2018-09-27T13:19:30.227" LastActivityDate="2018-09-27T13:27:08.510" Title="Is the d-block ever part of the valence electrons?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32171" PostTypeId="1" CreationDate="2015-05-29T22:16:11.780" Score="3" ViewCount="27" Body="&lt;p&gt;I am trying to understand why 1,2- and 1,3-dichlorobenzenes are such good solvents, able to solvate many organic compounds known to be hard to get dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any precise answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has something to do with dipolar moment and Hansen's diffusion parameter I think.&lt;/p&gt;&#xA;" OwnerUserId="16623" LastEditorDisplayName="user15489" LastEditDate="2015-05-29T23:10:34.317" LastActivityDate="2015-05-29T23:10:34.317" Title="Solubilizing ability of dichlorobenzenes" Tags="&lt;solvents&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32172" PostTypeId="2" ParentId="32151" CreationDate="2015-05-29T22:18:12.173" Score="5" Body="&lt;p&gt;The steps for hormone level measurement generally follow a similar sequence.  For the most accurate analysis &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid_chromatography%E2%80%93mass_spectrometry&quot; rel=&quot;noreferrer&quot;&gt;liquid chromatography-mass spectroscopy&lt;/a&gt; (lc-ms) is typically used.  The liquid chromatography separates the sample into fractions, conditions are selected such that a clean, separate peak is seen for the hormone being evaluated.  The purified hormone derived from the liquid chromatography step is then introduced into the mass spectrometer for quantification.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general process involves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sample collection: a sample of saliva, blood or urine is collected &lt;/li&gt;&#xA;&lt;li&gt;addition of a known quantity of hormone to act as an internal standard.  Often a $\ce{^{13}C}$ labeled hormone is used (commercially available).  The parent ion for the labeled standard will occur one mass unit higher than the parent ion for naturally occurring $\ce{^{12}C}$-containing collected sample.  If the peak height of the sample in the mass spec is half as large as that of the standard, then the concentration of hormone in the collected sample is half that in the standard.&lt;/li&gt;&#xA;&lt;li&gt;chemically separate the hormone from any large binding proteins&lt;/li&gt;&#xA;&lt;li&gt;isolate the fraction containing the hormone using liquid-liquid extraction&lt;/li&gt;&#xA;&lt;li&gt;run a second liquid-liquid extraction to remove highly polar materials&lt;/li&gt;&#xA;&lt;li&gt;run lc-ms analysis on sample that has now been purified and enriched in hormone&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a detailed &lt;a href=&quot;http://www.cdc.gov/nchs/data/nhanes/nhanes_11_12/tst_g_met.pdf&quot; rel=&quot;noreferrer&quot;&gt;procedure for testosterone analysis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep things understandable I have simplified the description of some of the&#xA;steps in the process. For example lc-&lt;a href=&quot;http://en.wikipedia.org/wiki/Tandem_mass_spectrometry&quot; rel=&quot;noreferrer&quot;&gt;tandem mass spectrometry&lt;/a&gt; is often used to improve accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research today in hormone quantification is focused on reducing cost and turn-around time.  A significant amount of work is particularly being done on analyzing the raw collected sample without further purification. &lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-05-29T22:18:12.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32173" PostTypeId="1" CreationDate="2015-05-29T22:36:44.740" Score="4" ViewCount="4602" Body="&lt;p&gt;A potassium atom comprises 19 protons and 18 core electrons. Those electrons screen 18 protons worth of Coulomb force. Therefore, there is one proton worth of Coulomb force attracting the valence electron of potassium. Whereas, krypton comprises 36 protons and 28 core electrons. Its electrons screen 28 protons wroth of Coulomb force. Therefore, there are eight protons worth of Coulomb force attracting the valence electrons of krypton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, there is more Coulomb force attracting the valence electrons of krypton than there is attracting the valence electron of potassium; accordingly, the atomic radius of krypton is smaller than the atomic radius of potassium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-07-15T17:48:02.967" LastActivityDate="2018-07-18T16:23:23.047" Title="Why is the atomic radius of krypton smaller than that of potassium?" Tags="&lt;inorganic-chemistry&gt;&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32174" PostTypeId="1" AcceptedAnswerId="48172" CreationDate="2015-05-29T23:17:39.633" Score="10" ViewCount="6088" Body="&lt;p&gt;I have been searching for the mechanism of the Mozingo reduction on Google but I haven't found anything relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the mechanism known? If so, what is it; if not, what hypothetical mechanisms can we consider?&lt;/p&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="4231" LastEditDate="2016-03-19T20:18:50.253" LastActivityDate="2016-03-19T20:18:50.253" Title="Mozingo reduction mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32175" PostTypeId="1" CreationDate="2015-05-29T23:51:13.113" Score="2" ViewCount="133" Body="&lt;p&gt;To the best of my knowledge, the following properties affect atomic radius.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The grounded-state size of the valence electron shell.&lt;/li&gt;&#xA;&lt;li&gt;The Coulomb force of the core electrons less the Coulomb force of the protons. &lt;/li&gt;&#xA;&lt;li&gt;The number of electrons in the valence shell. &lt;/li&gt;&#xA;&lt;li&gt;Whether an electron#-proton# imbalance is sufficient to change the size of the valence electron shell. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Question 1. Are there any other factors that I should know for Chem. 101? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2. If I were to estimate the size of an atom's radius relative to the radii of the atoms described on the periodic table, I would proceed thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st: Ascertain the grounded-state size of the valence electron shell (i.e. the row of the element in the periodic table). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd: If the thing in question is an ion, then adjust the number of electrons accordingly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd: If the adjustment in the second step removed, or added, enough electrons to the atom to eliminate the valence energy shell, or to create a new valence electron shell, then adjust the electron configuration accordingly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steps 1-3 give the size of the electron shell, and so give the period that comprises the atoms most comparable by radius size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4th. For the atom/ion under consideration, and for each element listed in the relevant period, subtract the Coulomb force of the protons from the Coulomb force of the core electrons. The lesser the value of the difference the smaller the size of the atomic radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5th. If steps 1-4 produce the same value for the atomic radii of the atom/ion under consideration and for one or more of the relevant elements, then the largest atomic radius characterizes the comparate whose valence electron shell comprises the most electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="15929" LastEditDate="2015-05-30T21:33:07.813" LastActivityDate="2015-05-30T21:33:07.813" Title="Is there an algorithm for ascertaining the relative size of an atomic/ionic radius?" Tags="&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32176" PostTypeId="1" CreationDate="2015-05-30T00:31:25.560" Score="9" ViewCount="4533" Body="&lt;p&gt;Looking at this equation $\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{CB}]}{[\ce{A}]}$ makes me think that it is concentration independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I mean is, when we look at $\frac{[\ce{CB}]}{[\ce{A}]}$, the volumes cancels out, so we are only left with the molar ratio of the conjugate acid and conjugate base. So does this mean that the pH is only dependent on the molar ratios? Obviously not, because $0.003\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.003\ \mathrm{mol/L}$ of its conjugate base is going to have a different $\mathrm{pH}$ from $0.3\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.3\ \mathrm{mol/L}$ of its conjugate base right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here? This is seriously confusing me. Please help, I want to use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of a weak acid as it’s being titrated with a strong base, but this is making me reluctant about the results.&lt;/p&gt;&#xA;" OwnerUserId="16624" LastEditorUserId="22102" LastEditorDisplayName="user7951" LastEditDate="2020-11-19T08:44:30.390" LastActivityDate="2020-11-19T08:44:30.390" Title="Is the Henderson–Hasselbalch equation volume independent?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;&lt;henderson-hasselbalch-equation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32177" PostTypeId="2" ParentId="32164" CreationDate="2015-05-30T00:41:41.237" Score="9" Body="&lt;p&gt;Let me answer the second question first, because the answer is so much easier: Of course the hydrogen bond strength depends &lt;em&gt;among others&lt;/em&gt; upon the carbon oxygen is attached to. Consider phenol and methanol or methanol and formic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guess as to which hydrogen bond is stronger is — I have to admit it — my speculation. But I think that methanol should be able to undergo stronger hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenol, of course, is aromatic and capable of delocalising negative (or positive) charges across its entire ring system — the reason why its $\mathrm{p}K_{\mathrm{a}} = 10$, a good five units lower than methanol’s. That means that at the same pH value of the solution, there should be less strong a negative charge on phenol’s oxygen than on methanol’s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But hydrogen bonds are, to a certain extent, electrostatic interaction. So the system which has a stronger negative charge on the oxygen (methanol) should be able to build up stronger hydrogen bonds by attracting hydrogen more strongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could also picture it differently: Phenol gets deprotonated at a lower pH, so the $\ce{O-H}$-bond must be a lot weaker than methanol’s. But if the $\ce{O-H}$-bond is weak, the oxygen has a larger tendency to simply &lt;em&gt;loose&lt;/em&gt; the hydrogen to another electronegative partner (deprotonation/protonation) rather than &lt;em&gt;share&lt;/em&gt; it (hydrogen bond).&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-05-30T00:41:41.237" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32178" PostTypeId="2" ParentId="32161" CreationDate="2015-05-30T01:12:31.230" Score="3" Body="&lt;h2&gt;On the favoured Lewis structures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Well yes, you have correctly drawn a minus sign next to the oxygen in the rightmost structure and similarly a plus sign next to the oxygen in the leftmost structure. However, &lt;em&gt;these are not true charges!&lt;/em&gt; These are &lt;strong&gt;formal charges&lt;/strong&gt;. They can be representative of real charges (cf. $\ce{SO4^2-}$ but they do not have to be. They derive from the way Lewis structures are drawn and not from any physical significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing a good Lewis structure requires balancing a lot of effects. Main group atoms should get electron octets (hydrogen dublets) is one of the main tipping points to take care of. In the leftmost structure, all atoms have electron octets. The central structure has an octet (on oxygen) and a sextet (on carbon) — less favoured. The rightmost structure only gives cabon a quartet, &lt;strong&gt;plus&lt;/strong&gt; a positive charge. This combination is so strongly disfavoured that not even the negative charge on oxygen can save anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimising the total formal charges is a second balancing effect which would prefer the central structure over the other two; however, it is less strong than the octet idea. And finally, structures that have formal charges distributed according to electronegativity are slightly favoured over those that don’t, but as I said, the former rules are more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that Lewis structures are just &lt;strong&gt;a representation; and not a particularly good one at that!&lt;/strong&gt; You can sort-of derive things like bond orders or free electrons from them, but for most uses, they are inferior to the molecular orbital model.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;On the MO diagram&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because the HOMO of CO is closer to the atomic orbital of O&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stop right there. That’s not how MO diagrams work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general gist of these diagrams is that orbitals with a lower energy will be closer to the electronegative atom(s) while those with higher energy will have stronger contributions by the electropositive partner. This is the HOMO we are talking about in a diatomic molecule. It has to have a larger contribution by carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasoning: Carbons atomic orbitals are higher &lt;em&gt;per se&lt;/em&gt;. The higher the energy of a hybrid orbital, the higher the contribution of higher energy atomic orbitals must be. Remember that you need to distribute one entire atomic orbital across all the molecular orbitals that have the same (correct) symmetry. The contributions from each atom (squared) need to add up to one. It’s a lot easier for the atom to contribute to orbitals closer to the original energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the negative pole of the dipolar molecule is closer to oxygen. Why? All populated low-energy molecular orbitals have a stronger contribution by oxygen while the high-energy unpopulated ones have a stronger carbon contribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A side note, that was mentioned in the question’s comments: You have 5 $\sigma$ orbitals and 2 $\pi$ orbitals populated in the MO drawing. 1$\sigma$ and 2$\sigma$ can be thought of the 1s orbitals of carbon and oxygen, leaving five orbitals for bonding. One of them happens to have an antibonding structure (nodal plane through the bond axis) so you get $4 - 1 = 3$ a triple bond between carbon and oxygen. (This only works this easily because there are only two atoms.) &lt;a href=&quot;https://chemistry.stackexchange.com/a/31003/7475&quot;&gt;Please also check out the pictures in the other thread&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m not sure if I touched all the points you wanted answered concerning the molecular orbitals in here. If not, comment and I’ll be back to try to explain what I missed. There’s a lot more that you can read out of it if you want to give it a try.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-05-30T01:12:31.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32181" PostTypeId="2" ParentId="32176" CreationDate="2015-05-30T04:38:21.817" Score="-2" Body="&lt;p&gt;Try plugging in 0.3 and 0.003 for each case. Either way $\ce{pH = pK_{a}}$ because the ratio of conjugate base concentration to acid concentration is 1. Therefore, the Henderson-Hasselbalch equation is independent of volume.&lt;/p&gt;&#xA;" OwnerUserId="16290" LastEditorDisplayName="user15489" LastEditDate="2015-05-31T03:05:51.663" LastActivityDate="2015-05-31T03:05:51.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32182" PostTypeId="5" CreationDate="2015-05-30T04:40:22.013" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-30T04:40:22.013" LastActivityDate="2015-05-30T04:40:22.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32183" PostTypeId="4" CreationDate="2015-05-30T04:40:22.013" Score="0" Body="Questions about how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds when a molecule is formed. " OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T13:24:45.010" LastActivityDate="2015-05-30T13:24:45.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32185" PostTypeId="5" CreationDate="2015-05-30T04:42:08.227" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-30T04:42:08.227" LastActivityDate="2015-05-30T04:42:08.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32186" PostTypeId="4" CreationDate="2015-05-30T04:42:08.227" Score="0" Body="Questions related to the physical combination of two or more substances on which the identities are retained and are mixed in the form of solutions, suspensions, and colloids." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T13:24:33.073" LastActivityDate="2015-05-30T13:24:33.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32187" PostTypeId="5" CreationDate="2015-05-30T04:44:36.860" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-30T04:44:36.860" LastActivityDate="2015-05-30T04:44:36.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32188" PostTypeId="4" CreationDate="2015-05-30T04:44:36.860" Score="0" Body="Questions related to a thermodynamic equation describing the state of matter under a given set of physical conditions." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T13:24:21.633" LastActivityDate="2015-05-30T13:24:21.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32189" PostTypeId="1" AcceptedAnswerId="32202" CreationDate="2015-05-30T04:49:17.727" Score="2" ViewCount="10135" Body="&lt;p&gt;How do we prepare potassium cyanide from potassium ferrocyanide? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is something which I am curious about, but cannot find the answer. I am studying inorganic chemistry but this reaction cannot be found.&lt;/p&gt;&#xA;" OwnerUserId="16618" LastEditorDisplayName="user7951" LastEditDate="2015-07-16T18:26:04.513" LastActivityDate="2017-10-05T10:13:10.073" Title="How to prepare potassium cyanide from potassium ferrocyanide" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32190" PostTypeId="1" AcceptedAnswerId="32216" CreationDate="2015-05-30T05:07:27.360" Score="3" ViewCount="2167" Body="&lt;p&gt;I was wondering if it would be possible to use $\ce{HCl}$ acid to dissolve $\ce{PbS}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I can dissolve it in $\ce{HNO_3}$ since, the nitric acid oxidises the sulphide ion to elemental sulphur, and some sulphides like $\ce{ZnS}$ do dissolve in $\ce{HCl}$. But what about $\ce{PbS}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the extent of $\ce{pH}$ sensitivity of lead sulphide precipitate.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T05:10:47.440" LastActivityDate="2015-05-30T15:19:23.443" Title="Action of HCl on PbS" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32197" PostTypeId="2" ParentId="27806" CreationDate="2015-05-30T05:44:39.433" Score="5" Body="&lt;p&gt;One of the first places to look at for advice is the Materials Safety Data Sheet (MSDS, or just SDS). Make sure these are kept in a handy place and are up to date. These should be provided by the supplier (but are easy enough to get online).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pertinent and thorough guide is produced by the aptly named Methanol Institute, the document is called &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanol-Safe-Handling-Manual-English.aspx&quot; rel=&quot;nofollow&quot;&gt;Methanol Safe Handling Manual&lt;/a&gt;, the chapter &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanol-Safe-Handling-Manual-English.aspx&quot; rel=&quot;nofollow&quot;&gt;Health and Safety&lt;/a&gt;, especially the &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanol-Safe-Handling-Manual-English.aspx&quot; rel=&quot;nofollow&quot;&gt;Safety Precautions&lt;/a&gt; subchapter outlines the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Smoking must be prohibited.&lt;/li&gt;&#xA;  &lt;li&gt;Vehicle access should be strictly controlled.&lt;/li&gt;&#xA;  &lt;li&gt;Ventilation must be sufficient to cope with the maximum expected vapor levels in buildings.&lt;/li&gt;&#xA;  &lt;li&gt;Positive pressure may be required for methanol-free areas, such as control, switch, and smoking rooms.&lt;/li&gt;&#xA;  &lt;li&gt;Storage tank vents to atmosphere should be sized for fire-heated emergency vapor release.&lt;/li&gt;&#xA;  &lt;li&gt;Electrical equipment must be explosion-proof to meet national electrical code requirements.&lt;/li&gt;&#xA;  &lt;li&gt;Alcohol-resistant Aqueous Film-forming Foam (AR-AFFF) with 6% foam proportioning (with water) equipment is advised for use on methanol&#xA;  fires.&lt;/li&gt;&#xA;  &lt;li&gt;Dry chemical extinguishers should be accessible for small fires. An adequate supply of handheld and wheeled types should be available.&lt;/li&gt;&#xA;  &lt;li&gt;Hydrants should be strategically placed with adequate hoses.&lt;/li&gt;&#xA;  &lt;li&gt;Small spills should be remediated with sand, earth, or other non-combustible absorbent material, and the area then flushed with&#xA;  water. Larger spills should be diluted with water and diked for later&#xA;  disposal.&lt;/li&gt;&#xA;  &lt;li&gt;Lighting should be grounded. Tall vessels and structures should be fitted with lightning conductors that are securely grounded.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These kind of directives should be on signs that are easily read alongside hazard signage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest having a readily accessed first aid kit and make a poster with first aid instructions, also included in the handbook - &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanol-Safe-Handling-Manual-English.aspx&quot; rel=&quot;nofollow&quot;&gt;First Aid Measures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of storage, the subchapter &lt;a href=&quot;http://www.methanol.org/Health-And-Safety/Safety-Resources/Health---Safety/Methanol-Safe-Handling-Manual-English.aspx&quot; rel=&quot;nofollow&quot;&gt;Methanol Storage&lt;/a&gt;, the handbook states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, fire protection for gasoline tanks is sufficient for&#xA;  methanol tanks, provided extra precaution is made for leak detection, toxic hazard, and&#xA;  availability of alcohol-compatible fire suppression foam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In terms of proximity to the house, I would consider storing these tanks as far as feasible from the house.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T05:59:06.170" LastActivityDate="2015-05-30T05:59:06.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32198" PostTypeId="2" ParentId="32176" CreationDate="2015-05-30T05:45:21.803" Score="1" Body="&lt;p&gt;If you have an aqueous solution of a weak acid and its conjugated base, the $\ce{pH }$ of this solution is given by the Henderson Hasselbalch equation. This equation states clearly that the $\ce{pH }$ is dependent on the ratio of concentrations of the weak acid and its conjugated base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have an equimolar solution of the weak acid and its conjugated base, the $\ce{pH = pK_{a}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(One way to prepare buffer solutions is to mix equimolar solutions of a weak acid and its conjugated base).&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-05-30T05:45:21.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32199" PostTypeId="1" AcceptedAnswerId="32245" CreationDate="2015-05-30T06:44:05.083" Score="5" ViewCount="2316" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A galvanic cell consists of a standard hydrogen electrode and a copper&#xA;  electrode.  Suppose that the copper electrode is immersed in a&#xA;  solution that is $0.100\ \mathrm{M}$ in $\ce{NaOH}$ and that is saturated&#xA;  with $\ce{Cu(OH)2}$. Find cell potential&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\begin{alignat}{2}\ce{2 H+ (aq) -&amp;gt; H2 (g)}\quad E_0 = 0.00\ \mathrm{V}\\&#xA;\ce{Cu^2+ (aq) + 2 e- -&amp;gt; Cu (s)}\quad E_0 = 0.34\ \mathrm{V}\end{alignat}$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$K_\text{sp} = 1.6 \times 10^{-19}$ for $\ce{Cu(OH)2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started by doing writing off the Nernst equation and realized I do not have $\ce{Cu^2+}$ concentration, and since I’m given $K_\text{sp}$ of copper, I assumed I should use it to find copper (after doing ice table for $\ce{Cu(OH)2 -&amp;gt; Cu^2+ + 2 OH-}$). But then I stumbled on $0.100\ \mathrm{M}$ of $\ce{NaOH}$, what should I do with it? and what does saturated mean in the given context of the problem?&lt;/p&gt;&#xA;" OwnerUserId="16551" LastEditorUserId="46306" LastEditDate="2018-03-04T12:00:23.673" LastActivityDate="2018-03-04T12:00:23.673" Title="Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32202" PostTypeId="2" ParentId="32189" CreationDate="2015-05-30T09:47:14.227" Score="7" Body="&lt;p&gt;Until about 1900, potassium cyanide was produced by melting dry potassium carbonate with dry potassium hexacyanidoferrate(II):&#xA;$$\ce{K4[Fe(CN)6] + K2CO3 -&amp;gt; 6KCN + FeO + CO2}$$&#xA;However, the product contains potassium cyanate, which is approximately formed as follows:&#xA;$$\ce{K4[Fe(CN)6] + K2CO3 -&amp;gt; 5KCN + KOCN + Fe + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a small scale, dry potassium hexacyanidoferrate(II) is melted in a covered porcelain crucible. It is heated as long as nitrogen escapes.&#xA;$$\ce{K4[Fe(CN)6] -&amp;gt; 4KCN + N2 +FeC2}$$&#xA;The iron carbide sinks to the bottom and the molten potassium cyanide can be poured off.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-10-26T14:12:49.973" LastActivityDate="2015-10-26T14:12:49.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32203" PostTypeId="1" AcceptedAnswerId="32207" CreationDate="2015-05-30T10:47:15.017" Score="3" ViewCount="2386" Body="&lt;p&gt;I read that continuum in line spectra is noticed beyond a certain limit. What does it mean?&lt;/p&gt;&#xA;" OwnerUserId="11331" LastActivityDate="2016-08-29T17:20:29.583" Title="What is continuum in line spectra?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32204" PostTypeId="1" CreationDate="2015-05-30T11:40:24.313" Score="2" ViewCount="232" Body="&lt;p&gt;In my studies, I have a group project to realise: to make some $\ce{Ag}$ nanoparticles using Turkevich method (boiling $\ce{AgNO3}$ with sodium citrate in water). In most of articles authors write about using sodium citrate solution, but when I searched for this compound on Sigma-Aldrich, they offer me a wide range of compounds named &quot;sodium citrate&quot;. Mono-, di-, tri- sodium citrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for a more specific description, I found this article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1878535214003645&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/S1878535214003645&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here they wrote citrate's formula. It is tri-sodium citrate. But Sigma-Aldrich offers it in hydrated versions. Which version should I use? Is hydration even relevant when I'm going to boil it in water?&lt;/p&gt;&#xA;" OwnerUserId="16635" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T11:42:57.673" LastActivityDate="2015-05-30T13:38:49.887" Title="Which sodium citrate is used in silver nanoparticles production?" Tags="&lt;organic-chemistry&gt;&lt;aqueous-solution&gt;&lt;nanotechnology&gt;&lt;silver&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32206" PostTypeId="2" ParentId="32173" CreationDate="2015-05-30T12:56:10.573" Score="1" Body="&lt;p&gt;Your concept is close to the real answer. You should think it as no. of protons per electron over all. This is the way things have been done in literature and if you are not satisfied then consider the fact that not all electrons are shielding the nucleus to same extent. There are certain rules [slaters rule] made and which works in most cases by which you can actually predict how much of nuclear charge is acting on the electron. Even with that you will see valence electron for K is more shielded than for Kr. But the numbers you came up with don't make that much sense. It does a bit but better if you go by the way it has been discussed.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-05-30T12:56:10.573" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32207" PostTypeId="2" ParentId="32203" CreationDate="2015-05-30T12:58:13.310" Score="5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/alM6T.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The four lines in the visible spectrum (designated by α through δ) were the first observed by Balmer. Notice how the lines crowd together as they approach the ionization limit in the near-ultraviolet part of the spectrum. Once the electron has left the atom, it is in an unbound state and its energy is no longer quantized.&lt;/p&gt;&#xA;&lt;p&gt;When such electrons return to the atom, they possess random amounts of kinetic energies over and above the binding energy. This reveals itself as the radiation at the short-wavelength end of the spectrum known as the &lt;strong&gt;continuum radiation&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.chem1.com/acad/webtext/atoms/atpt-3.html&quot; rel=&quot;noreferrer&quot;&gt;Chem1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-05-30T12:58:13.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32208" PostTypeId="1" AcceptedAnswerId="32212" CreationDate="2015-05-30T13:26:00.750" Score="0" ViewCount="135" Body="&lt;p&gt;I recently read somewhere that electrons do not exist. it's just the electron density. Is it true that there are no electrons (particles) but only in the form of electron density? And does this contradict the fact of dual nature of electron?&lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="9961" LastEditDate="2015-05-30T14:42:38.220" LastActivityDate="2015-05-30T14:42:38.220" Title="Electron vs electron density" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32209" PostTypeId="2" ParentId="27806" CreationDate="2015-05-30T13:27:20.520" Score="5" Body="&lt;p&gt;When I read heading I thought it is general question for which 1 answer is okay. But when I read &lt;strong&gt;200 litres&lt;/strong&gt; I thought I will dig out a bit and write few more stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found few related question on &lt;a href=&quot;http://www.methanol.org/health-and-safety/frequently-asked-questions.aspx&quot; rel=&quot;noreferrer&quot;&gt;Methanol Institute&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What kinds of drums are recommended for methanol storage, only steel or are some plastics acceptable?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well there is no Global Standards, but according to Methanol Institute following care should be taken.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mild steel is usually selected as the construction material for methanol storage containers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If moisture and trace amounts of inorganic salts are expected to exist within the container from time to time, then one should consider upgrading from carbon steel to 316 L stainless steel, or even a titanium or molybdenum stabilized grade of 316 L stainless steel &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;plastic storage containers are more susceptible to physical damage than metal ones (e.g., if dropped or hit by a forklift).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pure anhydrous methanol is mildly corrosive to lead and aluminum alloys, and more so to magnesium, and platinum. Methanol-water solutions can be corrosive to some non-ferrous alloys depending on application and environmental circumstances, including copper alloy, galvanized steel, and aluminum alloy components, and some plastics and composites.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What types of extinguishers should be used to put out small methanol fires?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Large methanol pool fires are best extinguished with &lt;strong&gt;Alcohol Resistant Aqueous Film Forming Foam (AR-AFFF) with 6% foam-water proportioning equipment&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;accidents involving methanol spills, what is the recommended evacuation distance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 2008 edition of the Emergency Response Guidebook (ERG2008) recommends an immediate isolation distance of 150 feet (50 meters) in all directions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What precautions are needed when welding or torch cutting tanks or containers that were used to store methanol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It has the potential to catch fire when hot work such as, welding, brazing, soldering, cutting, heat treating, grinding, and using power-actuated tools, is performed near methanol sources. When mixed with air, methanol can burn in the open. Methanol spills may travel short distances (yards or meters) along the ground before reaching a point of ignition and flashing back. The hazards associated with hot work can be reduced by implementing an effective hot work program that includes prior work authorization, safe welding practices, and a fire watch.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I think you should read all the questions I have provided in the beginning of Answer&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-05-30T13:27:20.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32210" PostTypeId="2" ParentId="32204" CreationDate="2015-05-30T13:38:49.887" Score="3" Body="&lt;p&gt;You can use hydrated tri-sodium citrate, it's readily soluble in water. As for its chemical reactivity, it's identical to tri-sodium citrate.&#xA;You should take into consideration the increase in the molecular weight in the hydrated form due to the presence of water molecules.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-05-30T13:38:49.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32211" PostTypeId="1" CreationDate="2015-05-30T13:55:13.423" Score="10" ViewCount="57851" Body="&lt;p&gt;My teacher told me that lone pair and lone pair are closer together hence the repulsion is greater and they take up more space, but I do not understand? can anyone explain it to me? &lt;/p&gt;&#xA;" OwnerUserId="16638" LastEditorUserId="-1" LastEditDate="2018-05-24T15:28:19.930" LastActivityDate="2018-05-24T15:28:19.930" Title="why are lp-lp repulsions greater than bp-lp and bp-bp?" Tags="&lt;electrons&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32212" PostTypeId="2" ParentId="32208" CreationDate="2015-05-30T14:03:56.357" Score="8" Body="&lt;p&gt;Of course, electrons &lt;em&gt;do&lt;/em&gt; exist. We have a myriads of observations confirming their existence. What does not exist is &lt;em&gt;working classical picture&lt;/em&gt; of an electron. Electrons can behave in certain circumstances like waves, in some other - like particles. That is already nonsense from the classical physics point of view with its universal dichotomy: something is either a particle or a wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the trickiest part of the wave-particle duality is that despite the fact that electrons can &lt;em&gt;behave&lt;/em&gt; as waves and as particles, their are, strictly speaking, &lt;em&gt;neither&lt;/em&gt; of them. Our language developed on the basis of observations of macroscopic surroundings with its well-established classical notions of waves and particles is simply &lt;em&gt;inappropriate&lt;/em&gt; for the description of the microscopic phenomena. Thus, whenever one tries to use any language except for that of mathematics to describe microscopic world he/she quickly runs into apparent paradoxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the desire of pictorial classical description of an electron I feel like the only thing I have to do is to quote Paul Dirac:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[…] the main object of physical science is not the provision of&#xA;  pictures, but is the formulation of laws governing phenomena and the&#xA;  application of these laws to the discovery of new phenomena. If a&#xA;  picture exists, so much the better; but whether a picture exists or&#xA;  not is a matter of only secondary importance. In the case of atomic&#xA;  phenomena no picture can be expected to exist in the usual sense of&#xA;  the word 'picture', by which is meant a model functioning essentially&#xA;  on classical lines. One may, however, extend the meaning of the word&#xA;  'picture' to include any way of looking at the fundamental laws which&#xA;  makes their self-consistency obvious. With this extension, one may&#xA;  gradually acquire a picture of atomic phenomena by becoming familiar&#xA;  with the laws of the quantum theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-05-30T14:24:40.293" LastActivityDate="2015-05-30T14:24:40.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32213" PostTypeId="1" AcceptedAnswerId="81162" CreationDate="2015-05-30T14:21:58.073" Score="5" ViewCount="1267" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Separation of the Chloride ($\ce{Cl-}$); Confirmation of Chloride.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;Chloride ion forms an insoluble silver compound. Silver chloride is a white solid.&lt;/em&gt;&lt;br&gt;&#xA;  $$\ce{Cl^-(aq) + Ag+(aq) -&amp;gt; AgCl(s)}$$&#xA;  &lt;em&gt;Silver chloride dissolves in $\pu{6 M}$ ammonia, $\ce{NH3}$, forming the colorless ion $\ce{Ag(NH3)2+}$. If nitric acid, $\ce{HNO3}$, is added to a solution containing this ion, the ammonia in the complex reacts with hydrogen ions to form ammonium ions, and the silver recombines with the chloride ions that are still present in solution.&lt;/em&gt;&lt;br&gt;&#xA;  \begin{align}&#xA;\ce{AgCl(s) + 2NH3(aq) &amp;amp;-&amp;gt; Ag(NH3)2+(aq) + Cl^-(aq)} \\&#xA;\ce{Ag(NH3)2+(aq) + Cl^-(aq) + 2H+(aq) &amp;amp;-&amp;gt; AgCl(s) + 2NH4+(aq)}&#xA;\end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;a&lt;/em&gt;&lt;/strong&gt;. Place 10 drops of the original test solution (or unknown solution) in a test tube. Test to see if the solution is acidic. If it is not, add $\pu{6 M}$ acetic acid, $\ce{CH3COOH}$, dropwise with stirring until the solution is acidic.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;b&lt;/em&gt;&lt;/strong&gt;. Add 10 drops of $\pu{0.1 M}$ silver nitrate, $\ce{AgNO3}$. A precipitate of $\ce{AgCl}$ will form.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;c&lt;/em&gt;&lt;/strong&gt;. Centrifuge and pour off the supernatant liquid.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;d&lt;/em&gt;&lt;/strong&gt;. Wash the solid with $\pu{0.5 mL}$ distilled water, centrifuge and discard the wash water.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;e&lt;/em&gt;&lt;/strong&gt;. Add $\pu{0.5 mL}$ $\pu{6 M}$ ammonia, $\ce{NH3}$, to the precipitate. Stir to dissolve any $\ce{AgCl}$.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;f&lt;/em&gt;&lt;/strong&gt;. Centrifuge, and pour the supernatant liquid into another test tube to test for chloride ion.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;g&lt;/em&gt;&lt;/strong&gt;. Add $\pu{1 mL}$ $\pu{6 M}$ nitric acid, $\ce{HNO3}$, to the solution containing the dissolved silver chloride. The solution will get hot and smoke from the reaction with the excess ammonia whether or not silver chloride is present.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;h&lt;/em&gt;&lt;/strong&gt;. Test with litmus or pH paper to see if the solution is acidic. If it is not, add more $\ce{HNO3}$ until the solution is acidic. The appearance of the white precipitate of $\ce{AgCl}$ in the acidic solution confirms the presence of chloride.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the procedure above, if you all could take a loot at step &lt;strong&gt;1a&lt;/strong&gt;. It makes no sense to me. I do not understand why adding $\ce{CH3COOH}$ has any effect on the solution whatsoever. What is the point of adding it when we later add $\ce{HNO3}$?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="22102" LastEditDate="2017-08-12T03:45:33.793" LastActivityDate="2017-08-12T03:45:33.793" Title="Why should I acidify twice in the procedure for qualitative analysis of chloride anions?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;&lt;identification&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32214" PostTypeId="1" AcceptedAnswerId="32228" CreationDate="2015-05-30T14:39:19.643" Score="6" ViewCount="326" Body="&lt;p&gt;The reduction potential for $\ce{AgI}$ is $-0.15\ \mathrm{V}$ and for $\ce{AgCl}$ it’s $0.22\ \mathrm{V}$. As $nFE =-\Delta{G} = T\Delta{S} - \Delta{H}$ this must imply that $\ce{AgI}$ is a stronger bond in order for it to be negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think $\ce{AgI}$ is weaker due to having a much larger ionic radii for $\ce{I-}$ than $\ce{Cl-}$ and the lack of covalence due to a smaller $\Delta\chi$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that it must be something to do with the size of the $\ce{I-}$ mean that it causes significant ordering of the water molecules in the solution and it's this entropic effect that is actually the key idea as otherwize I can't rationalise it at all.&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="275" LastEditDate="2015-06-01T11:15:51.980" LastActivityDate="2015-06-01T11:15:51.980" Title="AgI and AgCl reduction potentials" Tags="&lt;electrochemistry&gt;&lt;aqueous-solution&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32215" PostTypeId="1" CreationDate="2015-05-30T14:44:37.293" Score="9" ViewCount="202" Body="&lt;p&gt;I'm reading the chapter on gases in Chang and Overby's book &amp;quot;&lt;em&gt;General Chemistry: The Essential Concepts&lt;/em&gt;&amp;quot; [1, p. 145], and the following passage is causing me some confusion:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). &lt;strong&gt;It follows that the volume of any given gas must be proportional to the number of moles of molecules present&lt;/strong&gt;; that is,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V \propto n$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V = k_4 n \tag{5.7}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't quite understand the emphasized sentence; in particular, I don't understand how the emphasized claim &amp;quot;follows&amp;quot; from the preceding statement. In fact, I don't actually think it does.&lt;/p&gt;&#xA;&lt;p&gt;For example, suppose instead that the volume of any gas at a given temperature and pressure was &lt;em&gt;inversely&lt;/em&gt; proportional to the number of moles of the gas present; that is, suppose the relationship between volume and amount was instead&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V \stackrel{?!}{=} \frac{k}{n},$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;for some constant &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;. Then it would certainly still be true that equal volumes of different gases contain the same amount of gas. For example, if I had &lt;span class=&quot;math-container&quot;&gt;$\pu{5 L}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;, then, according to the above &amp;quot;false law&amp;quot;, I would have &lt;span class=&quot;math-container&quot;&gt;$k/5$&lt;/span&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt;. Similarly, if I had &lt;span class=&quot;math-container&quot;&gt;$\pu{5 L}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt;, I would have &lt;span class=&quot;math-container&quot;&gt;$k/5$&lt;/span&gt; moles of &lt;span class=&quot;math-container&quot;&gt;$\ce{N2}$&lt;/span&gt; — the same amount!&lt;/p&gt;&#xA;&lt;p&gt;So the claim that the proportionality &lt;em&gt;follows&lt;/em&gt; from the fact that equal volumes of different gases contain the same number of molecules seems illogical. In fact, it seems to me that the only thing one could logically deduce from the first hypothesis is that there is &lt;em&gt;some&lt;/em&gt; relationship between volume and amount which holds for all gases (that is, the relationship doesn't depend on the gas itself).&lt;/p&gt;&#xA;&lt;p&gt;Am I missing something, or is the statement in the book an error?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Added much later:&lt;/h3&gt;&#xA;&lt;p&gt;To clarify what I'm asking here (though I thought it was pretty clear), consider the following two statements:&lt;/p&gt;&#xA;&lt;p&gt;Statement &lt;strong&gt;A&lt;/strong&gt;: at a given temperature and pressure, equal volumes of different gases contain the same number of molecules.&lt;/p&gt;&#xA;&lt;p&gt;Statement &lt;strong&gt;B&lt;/strong&gt;: at a given temperature and pressure, the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; of a gas is related to the number of molecules by &lt;span class=&quot;math-container&quot;&gt;$V = kN$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; depends on temperature and pressure).&lt;/p&gt;&#xA;&lt;p&gt;My point is simply that these two statements are &lt;em&gt;not&lt;/em&gt; logically equivalent. &lt;strong&gt;B&lt;/strong&gt; implies &lt;strong&gt;A&lt;/strong&gt;, but not the other way around.&lt;/p&gt;&#xA;&lt;p&gt;So why is it that the two statements are (seemingly quite often) presented as logically equivalent?&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Chang, R.; Overby, J. General Chemistry: The Essential Concepts, 6th ed.; &lt;em&gt;McGraw-Hill: New York, NY&lt;/em&gt;, &lt;strong&gt;2011&lt;/strong&gt;. ISBN 978-0-07-337563-2.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16639" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-03-07T00:29:31.980" Title="Perplexing claim in Chang and Overby's &quot;General Chemistry&quot;" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;mole&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32216" PostTypeId="2" ParentId="32190" CreationDate="2015-05-30T15:19:23.443" Score="3" Body="&lt;p&gt;Hydrogen sulfide is a weak acid:&#xA;$$\begin{align}&#xA;\mathrm{p}K_{\text{a}1} &amp;amp;\approx 7\\&#xA;\mathrm{p}K_{\text{a}2} &amp;amp;\approx 13\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the concentration of sulfide $c_{\ce{S^2-}}$ is very low in acidic solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{c_{\ce{H2S}}}{c_{\ce{H+}}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, hydrogen sulfide is a gas. Its solubility in water is approximately $c_{\ce{H2S}}\approx 0.1\ \mathrm{mol\ l^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$, the concentration of sulfide may be estimated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{\ce{S^2-}} = \frac{K_{\text{a}1}\cdot K_{\text{a}2}\cdot c_{\ce{H2S}}}{c_{\ce{H+}}^2}\approx 10^{-20}\frac{0.1\ \mathrm{mol\ l^{-1}}}{\left(1\ \mathrm{mol\ l^{-1}}\right)^2}=10^{-21}\ \mathrm{mol\ l^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the solubility product of lead(II) sulfide is extremely low:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\text{sp} = c_{\ce{Pb^2+}} \cdot c_{\ce{S^2-}} \approx 10^{-28}\ \mathrm{mol^2\ l^{-2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the solubility of lead(II) sulfide (i.e. the maximum concentration of $\ce{Pb^2+}$) in a solution of a strong acid with $c_{\ce{H+}}=1\ \mathrm{mol\ l^{-1}}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_{\ce{Pb^2+}} = \frac{K_\text{sp}}{c_{\ce{S^2-}}} \approx \frac{10^{-28}\ \mathrm{mol^2\ l^{-2}}}{10^{-21}\ \mathrm{mol\ l^{-1}}} = 10^{-7}\ \mathrm{mol\ l^{-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, lead(II) sulfide precipitates from acidic solutions and will not significantly dissolve in dilute hydrochloric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, lead(II) sulfide can be dissolved in hot hydrochloric acid with $c_{\ce{HCl}}=8\ \mathrm{mol\ l^{-1}}$ because of the formation of anionic complexes $\ce{[PbCl3]^-}$ and $\ce{[PbCl4]^2-}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-30T15:19:23.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32218" PostTypeId="1" CreationDate="2015-05-30T16:07:50.607" Score="0" ViewCount="1450" Body="&lt;p&gt;What is flavanoid physical and chemical properties, and how alkaline reagent test for flavonoids gives yellow coulour?&lt;/p&gt;&#xA;" OwnerUserId="16640" LastActivityDate="2015-05-30T16:55:03.253" Title="basic chemistry alkaline test for flavonoids!!! with reference please" Tags="&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-05-30T20:19:16.377" ContentLicense="CC BY-SA 3.0" />
  <row Id="32219" PostTypeId="2" ParentId="32211" CreationDate="2015-05-30T16:08:37.503" Score="14" Body="&lt;p&gt;So, from &lt;a href=&quot;http://en.wikipedia.org/wiki/VSEPR_theory&quot;&gt;Wikipedia article on VSEPR theory&lt;/a&gt; we read:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The overall geometry is further refined by distinguishing between&#xA;  bonding and nonbonding electron pairs. The bonding electron pair&#xA;  shared in a sigma bond with an adjacent atom lies further from the&#xA;  central atom than a nonbonding (lone) pair of that atom, which is held&#xA;  close to its positively charged nucleus. VSEPR theory therefore views&#xA;  repulsion by the lone pair to be greater than the repulsion by a&#xA;  bonding pair. As such, when a molecule has 2 interactions with&#xA;  different degrees of repulsion, VSEPR theory predicts the structure&#xA;  where lone pairs occupy positions that allow them to experience less&#xA;  repulsion. Lone pair-lone pair (lp-lp) repulsions are considered&#xA;  stronger than lone pair-bonding pair (lp-bp) repulsions, which in turn&#xA;  are considered stronger than bonding pair-bonding pair (bp-bp)&#xA;  repulsions, distinctions that then guide decisions about overall&#xA;  geometry when 2 or more non-equivalent positions are possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I said in my comment, the key point is given in the second sentence: the bonding pair shared in a bond &lt;em&gt;lies further&lt;/em&gt; from the atom than a nonbonding pair of that atom which is held close to its positively charged nucleus. Consequently, lone pairs are closer to each other than any other combination of pairs (lone pair-bonding pair and bonding pair-bonding pair), and thus, repell each other more strongly.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-05-30T16:08:37.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32222" PostTypeId="1" AcceptedAnswerId="32226" CreationDate="2015-05-30T16:41:22.297" Score="4" ViewCount="861" Body="&lt;p&gt;When one say that the electron goes from 1 orbit , say &lt;i&gt;&lt;b&gt;2&lt;sub&gt;S&lt;/sub&gt;&lt;/i&gt;&lt;/b&gt;, to another (higher) orbital, say &lt;i&gt;&lt;b&gt;5&lt;sub&gt;P&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt;, it mean that the electron is excited to that orbit by some mean. In doing so it also it will also cross the nodes of &lt;I&gt;&lt;B&gt;2&lt;SUB&gt;P&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt; etc. But doing so does the electron actually cross the node or what? Cause doing so the, there is probability, although very less, of finding electron in node for a very short time.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2018-08-07T16:01:21.103" Title="electron passing nodes?" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32223" PostTypeId="2" ParentId="32218" CreationDate="2015-05-30T16:55:03.253" Score="1" Body="&lt;p&gt;From this &lt;a href=&quot;https://books.google.com/books?id=TXsiFVVuCqMC&amp;amp;pg=PA203&amp;amp;lpg=PA203&amp;amp;dq=flavonoid%20base&amp;amp;source=bl&amp;amp;ots=04L0QuwFfy&amp;amp;sig=xLpJ4o9VIclIybWO9IDeigYPCK8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=SelpVay0JI7LogSfyIKgBQ&amp;amp;ved=0CDEQ6AEwAg#v=onepage&amp;amp;q=flavonoid%20base&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Treatment of a flavonoid with a strong base will result in ionization of its hydroxyl groups with resulting large bathochromic shifts of both absorption maxima.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-05-30T16:55:03.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32225" PostTypeId="2" ParentId="28190" CreationDate="2015-05-30T17:27:36.880" Score="3" Body="&lt;p&gt;Phosphorus has electronic configuration of &lt;i&gt;&lt;b&gt;[Ne]3&lt;sub&gt;S2&lt;/sub&gt; 3&lt;sub&gt;P3&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt; As it is a 3&lt;sup&gt;rd&lt;/sup&gt; period element it also have a vacant &lt;i&gt;&lt;b&gt;3&lt;sub&gt;D&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt; orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of &lt;b&gt;P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;6&lt;/sub&gt;&lt;/b&gt; only the &lt;i&gt;&lt;b&gt;3&lt;sub&gt;P3&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt; electrons take part. Resulting in formation of 3 bonds with &lt;i&gt;&lt;b&gt;O&lt;/b&gt;&lt;/i&gt;. But in case of &lt;b&gt;P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt;&lt;/b&gt; one of the electron of &lt;i&gt;&lt;b&gt;3&lt;sub&gt;S&lt;/sub&gt;&lt;/b&gt;&lt;/i&gt; is excited to vacant &lt;I&gt;&lt;B&gt;3&lt;SUB&gt;D&lt;/SUB&gt;&lt;/B&gt;&lt;/i&gt; orbital. Now &lt;B&gt;P&lt;/b&gt; has 5 unpaired electron and it can form &lt;b&gt;5&lt;/b&gt; bonds. Therefore &lt;B&gt;P&lt;/B&gt; forms 5 in &lt;B&gt;P&lt;sub&gt;4&lt;/sub&gt;O&lt;sub&gt;10&lt;/sub&gt;&lt;/b&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, about the octet. Not every compound follow the octet rule. Just to understand &lt;b&gt;The whole chemistry depends upon STABILITY&lt;/b&gt; i.e lesser energy of molecule. Therefore, if by forming more bonds (and does not obeying Octet rule) the molecule is gaining more stability, it will do so.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2015-05-30T17:27:36.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32226" PostTypeId="2" ParentId="32222" CreationDate="2015-05-30T17:50:02.957" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When one say that the electron goes from 1 orbit , say 2S, to another&#xA;  (higher) orbital, say 5P, it mean that the electron is excited to that&#xA;  orbit by some mean.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, &lt;em&gt;there are no orbits.&lt;/em&gt; Electrons do &lt;em&gt;not&lt;/em&gt; circle the nucleus. Rather they are in particular &lt;em&gt;quantum states&lt;/em&gt;, or, in other words, they &quot;occupy&quot; particular &lt;em&gt;orbitals&lt;/em&gt;. Note that when we say that an electron &quot;occupies&quot; an orbital, we just mean that it is in a particular quantum state. There is no spatial occupation in a classical sense out there: eletrons do not &quot;sit&quot; or &quot;circle&quot; anywhere, they are just in their quanatum states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, when we say that electron &quot;goes&quot; from one orbital to another, what we actually mean is that the quantum state of the electrons changes. And again, an electron transition is not a mechanical motion, thus, electrons upon transitions do not &quot;move&quot; through nodes.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-05-30T17:50:02.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32227" PostTypeId="1" CreationDate="2015-05-30T20:15:29.753" Score="4" ViewCount="7433" Body="&lt;p&gt;What is the difference between &lt;em&gt;crystal radius&lt;/em&gt; and &lt;em&gt;ionic radius&lt;/em&gt;? For $\ce{NaCl}$ for example the ions combine to a crystal, and I would have guessed they are equal. But I then found an example of data, where they are not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow&quot;&gt;ionic radius&lt;/a&gt; is the radius of an atom's ion. Quite simple. So what is &lt;em&gt;crystal radius&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorDisplayName="user15489" LastEditDate="2015-05-30T21:27:09.483" LastActivityDate="2017-02-13T22:25:09.557" Title="&quot;Crystal radius&quot; vs &quot;Ionic radius&quot;" Tags="&lt;ions&gt;&lt;atomic-radius&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32228" PostTypeId="2" ParentId="32214" CreationDate="2015-05-30T20:52:56.673" Score="4" Body="&lt;p&gt;The answer is &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot; rel=&quot;nofollow&quot;&gt;hard-soft acid/base&lt;/a&gt; theory (or HSAB theory). Essentially, some Lewis acids and bases are &lt;strong&gt;hard&lt;/strong&gt; and some are &lt;strong&gt;soft&lt;/strong&gt;, with a few &lt;strong&gt;borderline&lt;/strong&gt; cases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Hard&lt;/strong&gt; acids and bases tend to have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Small atomic/ionic radius&lt;/li&gt;&#xA;&lt;li&gt;Highly charges (or highly positive/negative oxidation state)&lt;/li&gt;&#xA;&lt;li&gt;High electronegativity for bases and low electronegativity for acids&lt;/li&gt;&#xA;&lt;li&gt;Low polarizability&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Soft&lt;/strong&gt; acids and bases tend to have&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Large atomic/ionic radius&lt;/li&gt;&#xA;&lt;li&gt;Low or zero charge / oxidation state&lt;/li&gt;&#xA;&lt;li&gt;Moderate electronegativity&lt;/li&gt;&#xA;&lt;li&gt;High polarizability&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;According to HSAB theory, matches that are &lt;strong&gt;hard-hard&lt;/strong&gt; and &lt;strong&gt;soft-soft&lt;/strong&gt; form strong interactions, where &lt;strong&gt;hard-soft&lt;/strong&gt; matches are weaker interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ag+}$ is a soft Lewis acid. $\ce{Cl-}$ is a hard Lewis base, while $\ce{I-}$ is a soft Lewis base. Therefore, we would predict that $\ce{Ag+}$ and $\ce{I-}$ will have have a stronger attraction than than $\ce{Ag+}$ and $\ce{Cl-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see this difference in attraction in the solubilities of $\ce{AgCl}$ and $\ce{AgI}$ in water. $\ce{AgCl}$ is much more soluble in water than $\ce{AgI}$ (realizing that both are nearly insoluble). See the &lt;a href=&quot;http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm&quot; rel=&quot;nofollow&quot;&gt;solubility constant data&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{AgCl}\ \ \ k_{\text{sp}}=1.8\times 10^{-10}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{AgI}\ \ \ \ \ \ k_{\text{sp}}=8.5\times 10^{-17}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-05-30T20:52:56.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32229" PostTypeId="1" CreationDate="2015-05-30T21:11:26.663" Score="1" ViewCount="1451" Body="&lt;p&gt;In this &lt;a href=&quot;https://www.khanacademy.org/science/chemistry/chemical-bonds/types-chemical-bonds/v/ionic-covalent-and-metallic-bonds&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; on Khan Academy, Sal remarks that 'all atoms envy the noble gases, which are the most stable because they have eight electrons in their outer shell'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of much of a reason that the number of electrons in the outer shell would affect the stability of the electron configuration. Additionally, I read that more electrons in the outer shell slightly increases the atomic radius. A larger atomic radius usually indicates a less stable electron configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, perhaps I've been thinking about this the wrong way. I've been wondering whether some property obtains when there are eight electrons in the outer shell that causes the configuration to be more stable. Perhaps, I ought to think about thus: atoms tend toward having eight electrons in their outer shell (i.e. eight electrons in the outer shell coincides with configuration stability, but doesn't cause it). The fact that the unscreened Coulomb force increases from left to right along a period is what explains why atoms tend toward having eight electrons in the outer shell: The atoms near the left of a period exhibit relatively little unscreened Coulomb force, and so are prone to giving up electrons and, so are prone to giving up the few electrons that constitute their valence shell. Whereas, those atoms near the right of the period exhibit much Coulomb force, relatively. Accordingly, those atoms are prone to attracting more electrons, and thereby prone to filling their valence shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the stability of the electron configuration increase when the outer electron shell is full?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I have reviewed the question &lt;em&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/444/can-an-atom-have-more-than-8-valence-electrons-if-not-why-is-8-the-limit&quot;&gt;Can an atom have more than 8 valence electrons? If not, why is 8 the limit?&lt;/a&gt;&lt;/em&gt; That question asks about the maximum number of valence electrons. This question asks why eight electrons in the valence shell are more stable than 1, 2, 3, ... 6, or 7 electrons in the valence shell. &lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2017-01-12T02:34:27.853" Title="Why are electron configurations more stable when there are 8 electrons in the outer shell?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="2" ClosedDate="2017-04-12T19:00:07.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="32231" PostTypeId="2" ParentId="32176" CreationDate="2015-05-30T22:10:51.300" Score="8" Body="&lt;p&gt;For a much shorter and clearer explanation of this problem, look &lt;a href=&quot;http://www.chembuddy.com/?left=pH-calculation&amp;amp;right=pH-buffers-henderson-hasselbalch&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach somewhere around &lt;span class=&quot;math-container&quot;&gt;$\pu{10^-5 mol L-1}$&lt;/span&gt; or lower, the true &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; will deviate significantly from the value predicted by the HH equation. However, this is not the problem most of the time.&lt;/p&gt;&#xA;&lt;p&gt;The reason the HH equation might produce poor predictions when calculating buffer &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; is really because of an oft-made assumption which has nothing to do with the HH equation itself: &lt;strong&gt;the weak acid (&lt;em&gt;or weak base&lt;/em&gt;) is assumed to be so weak that its ionization contributes almost no conjugate base (&lt;em&gt;or conjugate acid&lt;/em&gt;) in comparison to the dissolution of the buffer salt&lt;/strong&gt;. In other words, we assume the &lt;em&gt;formal concentrations&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$C_X$&lt;/span&gt; of species in the buffer are equal to their &lt;em&gt;actual concentrations&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$[X]$&lt;/span&gt;; we replace&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A^-}]}{[\ce{HA}]}&#xA;$$&lt;/span&gt;&#xA;for the worse approximation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathrm{pH} &#xA;  = \mathrm{p}K_\mathrm{a} + \log \frac{C_{\ce{A^-}}}{C_{\ce{HA}}}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This assumption is incorrect, but it is used to make the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; calculation much easier at the cost of accuracy. The resulting approximation breaks down in sufficiently dilute solution, and is already quite noticeable far before reaching water self-dissociation issues. It is also a noticeably poor approximation for weak acids/bases with relatively high ionization constants (say, &lt;span class=&quot;math-container&quot;&gt;$K &amp;gt; 10^{-2}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;So how can we visualize the effect of diluting a buffer without approximations, and where do the approximations start to break down? Let us consider the simple case of solutions of a weak monoprotic acid &lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt; with acid dissociation constant &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; where the concentration of acid and conjugate base are formally equal. This can be done by titrating half of the weak acid with a strong monoprotic base (e.g. &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt;). This problem can be solved exactly (assuming all activity coefficients are equal to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, which is generally a good approximation for solutions below about &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mol L-1}$&lt;/span&gt;), and I have previously outlined the solution &lt;a href=&quot;https://chemistry.stackexchange.com/a/8074/1499&quot;&gt;here&lt;/a&gt;. The equation relevant to this problem is the following:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{multline}&#xA;0 =&#xA;  [\ce{H^+}]^3 &#xA;    + \left(K_\mathrm{a} &#xA;      + \frac{C^\circ_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}}&#xA;    \right)[\ce{H^+}]^2 \\&#xA;    + \left(&#xA;      \frac{C^\circ_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}} K_\mathrm{a} &#xA;      - \frac{C^\circ_\ce{A} V_\ce{A}}{V_\ce{A} +  V_\ce{B}}&#xA;      K_\mathrm{a} - k_\mathrm{w}&#xA;    \right)[\ce{H^+}]&#xA;    - K_\mathrm{a}  k_\mathrm{w}&#xA;\end{multline}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V_\ce{A}$&lt;/span&gt; is the volume of the weak acid solution being titrated and &lt;span class=&quot;math-container&quot;&gt;$C^\circ_\ce{A}$&lt;/span&gt; is its formal initial concentration (before mixing and reacting with the base), while &lt;span class=&quot;math-container&quot;&gt;$V_\ce{B}$&lt;/span&gt; is a variable volume of strong base solution added with formal initial concentration &lt;span class=&quot;math-container&quot;&gt;$C^\circ_\ce{B}$&lt;/span&gt; (before mixing and reacting with the acid). The presence of the self-dissociation constant of water &lt;span class=&quot;math-container&quot;&gt;$k_\mathrm{w}$&lt;/span&gt; shows that it is being taken into consideration.&lt;/p&gt;&#xA;&lt;p&gt;We can now substitute values as desired to obtain a polynomial in &lt;span class=&quot;math-container&quot;&gt;$[\ce{H^+}]$&lt;/span&gt;. I'll set up the volumes and concentrations arbitrarily so we can reach the acid and conjugate base formal concentrations of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3 mol L-1}$&lt;/span&gt; as mentioned in the question. We can start off with &lt;span class=&quot;math-container&quot;&gt;$\pu{100 mL}$&lt;/span&gt; of weak acid solution at a concentration of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.9 mol L-1}$&lt;/span&gt; with a dissociation constant &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a} = 10^{-3}$&lt;/span&gt; (this constant can be changed at will). We can reach the target concentrations of acid and conjugate base by adding &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; of strong base at the same concentration &lt;span class=&quot;math-container&quot;&gt;$\pu{0.9 mol L-1}$&lt;/span&gt; (neutralizing half the original acid). You can check the resulting concentrations after mixing the solutions, if you wish. &lt;a href=&quot;http://www.wolframalpha.com/input/?i=x%5E3%2B%281E-3%2B0.9*0.05%2F%280.1%2B0.05%29%29x%5E2%2B%280.9*0.05%2F%280.1%2B0.05%29*1E-3-0.9*0.1%2F%280.1%2B0.05%29*1E-3-1E-14%29*x-1E-17%3D0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solving this polynomial&lt;/a&gt;, the resulting buffer solution containing formal concentrations &lt;span class=&quot;math-container&quot;&gt;$C_{\ce{HA}} = \pu{0.3 mol L-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_{\ce{A^-}} = \pu{0.3 mol L-1}$&lt;/span&gt; has &lt;span class=&quot;math-container&quot;&gt;$[\ce{H^+}] = \pu{9.93399E-4 mol L-1}$&lt;/span&gt;. This results in &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 3.00288$&lt;/span&gt;, which is very close to the value predicted by the approximations made at the start (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log 1 = \mathrm{p}K_\mathrm{a} = 3$&lt;/span&gt;). For now, they work.&lt;/p&gt;&#xA;&lt;p&gt;After looking at this first case, we shall investigate the effects of dilution by factors of 10. This can be performed simply by diluting both the initial weak acid and strong base concentrations by 10 (i.e., &lt;span class=&quot;math-container&quot;&gt;$\pu{100 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.09 mol L-1}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt; plus &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.09 mol L-1}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{KOH}$&lt;/span&gt; result in a buffer with acid/conjugate base concentrations of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.03 mol L-1}$&lt;/span&gt;, and so on). Before going into the calculations, it's quite simple to see an issue with the approximation in the second paragraph. Independent of the level of dilution, so long as the concentration of acid and conjugate base remain equal, then according to the approximation, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 3$&lt;/span&gt;. However, consider the case when starting with a solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt; with initial concentration equal to &lt;span class=&quot;math-container&quot;&gt;$\pu{9E-4 mol L-1}$&lt;/span&gt;. This is a weak monoprotic acid, so the initial solution &lt;em&gt;must&lt;/em&gt; have &lt;span class=&quot;math-container&quot;&gt;$[\ce{H^+}] &amp;lt; \pu{9E-4 mol L-1}$&lt;/span&gt; and thus &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} &amp;gt; 3$&lt;/span&gt;. Clearly it's &lt;strong&gt;impossible&lt;/strong&gt; to add &lt;em&gt;any&lt;/em&gt; amount of strong base to this acid and get a buffer solution with &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 3$&lt;/span&gt;! The approximations at the start have broken down at this point.&lt;/p&gt;&#xA;&lt;p&gt;Now, for the calculations. To get the proton concentration, I replaced every value of &lt;span class=&quot;math-container&quot;&gt;$0.9$&lt;/span&gt; in the previous Wolfram Alpha link with &lt;span class=&quot;math-container&quot;&gt;$0.09$&lt;/span&gt;, then &lt;span class=&quot;math-container&quot;&gt;$0.009$&lt;/span&gt;, and so on. With the proton concentration determined, I used equations shown in my other linked answer to calculate the actual concentrations &lt;span class=&quot;math-container&quot;&gt;$[\ce{HA}]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[\ce{A^-}]$&lt;/span&gt; for comparison with the formal concentrations &lt;span class=&quot;math-container&quot;&gt;$C_\ce{HA}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_{\ce{A^-}}$&lt;/span&gt;. Note that they are almost equal at high concentration, but the error increases quite suddenly at lower concentrations. While the ratio &lt;span class=&quot;math-container&quot;&gt;$\frac{C_\ce{A^-}}{C_\ce{HA}}$&lt;/span&gt; is always equal to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, the ratio &lt;span class=&quot;math-container&quot;&gt;$\frac{[\ce{A^-}]}{[\ce{HA}]}$&lt;/span&gt; rockets away.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \scriptsize&#xA;\begin{array}{c|ccc|cc}&#xA;\hline&#xA;C_\ce{HA} = C_\ce{A^-} / (\pu{mol L^-1})  &amp;amp;&#xA;[\ce{HA}] / (\pu{mol L^-1}) &amp;amp;&#xA;[\ce{A^-}] / (\pu{mol L^-1}) &amp;amp;&#xA;\frac{[\ce{A^-}]}{[\ce{HA}]} &amp;amp; &#xA;[\ce{H^+}] / (\pu{mol L^-1}) &amp;amp; &#xA;\mathrm{pH} \\\hline&#xA;0.3 &amp;amp; 0.29901 &amp;amp; 0.30099 &amp;amp; 1.00664 &amp;amp; 9.93399\times 10^{-4} &amp;amp; 3.00288 \\&#xA;0.03 &amp;amp; 0.029061 &amp;amp; 0.030939 &amp;amp; 1.06464 &amp;amp; 9.39282 \times 10^{-4} &amp;amp; 3.02720 \\&#xA;0.003 &amp;amp; 0.00235421 &amp;amp; 0.0036458 &amp;amp; 1.54858 &amp;amp; 6.45751 \times 10^{-4} &amp;amp; 3.18993\\&#xA;3\times 10^{-4} &amp;amp; 1\times 10^{-4} &amp;amp; 5\times 10^{-4} &amp;amp; 5 &amp;amp; 2 \times 10^{-4} &amp;amp; 3.69897 \\&#xA;3\times 10^{-5} &amp;amp; 1.6539\times 10^{-6} &amp;amp; 5.83461\times 10^{-5} &amp;amp; 35.2778&#xA; &amp;amp; 2.83464 \times 10^{-5} &amp;amp; 4.54750 \\&#xA;3\times 10^{-6} &amp;amp; 1.78596\times 10^{-8} &amp;amp; 5.98214\times 10^{-6} &amp;amp; 334.953&#xA; &amp;amp; 2.98549×10^{-6} &amp;amp; 5.52498 \\&#xA;3\times 10^{-7} &amp;amp; 1.97992\times 10^{-10} &amp;amp; 5.99802\times 10^{-7} &amp;amp; 3029.42&#xA; &amp;amp; 3.30096×10^{-7} &amp;amp; 6.48136 \\&#xA;3\times 10^{-8} &amp;amp; 6.96609\times 10^{-12} &amp;amp; 5.99939\times 10^{-8}  &amp;amp; 8612.15&#xA; &amp;amp; 1.16115×10^{-7} &amp;amp; 6.93511 \\&#xA;3\times 10^{-9} &amp;amp; 6.09004\times 10^{-13} &amp;amp; 5.99939\times 10^{-9} &amp;amp; 9851.15&#xA; &amp;amp; 1.01511×10^{-7} &amp;amp; 6.99349 \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="4945" LastEditDate="2020-06-20T14:08:16.340" LastActivityDate="2020-06-20T14:08:16.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32232" PostTypeId="1" CreationDate="2015-05-30T22:30:26.837" Score="8" ViewCount="4933" Body="&lt;p&gt;As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split $\mathrm d$-orbitals. In the case of $\ce{V^5+}$ and $\ce{Cr^6+}$ however, there are no electrons in the $\mathrm d$-orbitals, and the electronic configuration of both is similar to $\ce{Ar}$, yet $\ce{V^5+}$ and $\ce{Cr^6+}$ show color(yellow/brown and orange respectively).&lt;/p&gt;&#xA;" OwnerUserId="16643" LastEditorDisplayName="user7951" LastEditDate="2015-10-28T16:21:50.127" LastActivityDate="2015-10-28T16:21:50.127" Title="Why are compounds of V5+ and Cr6+ colored?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;color&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="4" ClosedDate="2016-09-18T04:07:23.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="32234" PostTypeId="2" ParentId="32215" CreationDate="2015-05-30T23:49:27.133" Score="7" Body="&lt;p&gt;You must be familiar with the ideal gas equation: PV = nRT. Thus, an ideal gas can be completely characterised by 3 state variables (if you believe, in the ideal gas equation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, you fix pressure and temperature (say, at STP). You get V = cn (where c is some constant, and n is no. of moles).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus it is indeed true that equal moles of gas, have equal volumes (under a given set of conditions). In fact, one calculates this volume as nearly 22.4 L (for one mole, at STP)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact Avogadro was the first person to make this observation. (and even though, I started my explanation with the ideal gas equation, it was formulated afterwards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is an idealisation. A picture that is closer to reality is given by the van der waal's equation of state, which takes into account inter-molecular interactions.&#xA; $[P + a(\frac{n}{V})^2][\frac{V}{n}-b]= RT$&#xA;But this departure from ideal behaviour is usually not significant at low pressure, high volume, dilute conditions.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-05-30T23:49:27.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32236" PostTypeId="2" ParentId="32153" CreationDate="2015-05-31T00:02:25.437" Score="1" Body="&lt;p&gt;Plants can do so because of the organelles in their cells called &quot;Chloroplasts.&quot; These chloroplasts have the pigment chlorophyll, which gives plants their green color. Chlorophyll is found in different varieties, but one of them (Chlorophyll A) has the chemical formula C&lt;sub&gt;55&lt;/sub&gt;H&lt;sub&gt;72&lt;/sub&gt;O&lt;sub&gt;5&lt;/sub&gt;N&lt;sub&gt;4&lt;/sub&gt;Mg. This chlorophyll acts like a biological solar panel, providing chemical energy to carry out the photosynthesis &quot;6CO&lt;sub&gt;2&lt;/sub&gt; + 6H&lt;sub&gt;2&lt;/sub&gt;O + Energy --&gt; C&lt;SUB&gt;6&lt;/SUB&gt;H&lt;SUB&gt;12&lt;/SUB&gt;O&lt;SUB&gt;6&lt;/SUB&gt; + 6O&lt;SUB&gt;2&lt;/SUB&gt; . Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="16648" LastActivityDate="2015-05-31T00:02:25.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32238" PostTypeId="2" ParentId="967" CreationDate="2015-05-31T02:42:56.933" Score="1" Body="&lt;p&gt;The polymer slurry has the purpose of keeping the hole stable, while it is still being worked on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While building a 50 m deep hole, I think it's common to need to change or replace the tool, when it broke, or the type of ground changed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the thing on the other end of 50 m steel rod means to pull out the rod, make the change, and let it down again.&lt;br&gt;&#xA;But even without any change, the rod needs to be pulled out before filling with concrete, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slurry does not repell water on the surface like oil would, but it repels ground water from entering the bore hole by having a higher pressure.&lt;br&gt;&#xA;The pressure comes from a higher density and same depth below ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One useful effect of the polymer could be that it does not soak into dry parts of the hole wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 50 m of hole wall, a touching rod will ceartainly loosen some sand, but id will not just drop to the end - depending on the slurry and the particle size, it will sink slowly, or even not at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;From &lt;a href=&quot;http://www.ritchiewiki.com/wiki/index.php/Burj_Dubai#Preparing_the_Foundation&quot; rel=&quot;nofollow&quot;&gt;Burj Dubai - Preparing the Foundation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, large boreholes were dug 164 feet (50 m) deep and then filled with viscous polymer slurry to maintain the stability of the hole. The slurry is denser than water, but lighter than concrete. The concrete, when poured into the hole, displaces the slurry and hardens to form a foundation pile.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5078" LastActivityDate="2015-05-31T02:42:56.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32239" PostTypeId="2" ParentId="32215" CreationDate="2015-05-31T03:07:08.197" Score="8" Body="&lt;p&gt;The  &lt;a href=&quot;https://chemistry.stackexchange.com/a/32234/5017&quot;&gt;other answer&lt;/a&gt; is, I think excellent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is basically &quot;but what if there isn't a direct proportion between number of gas molecules and volume?&quot; Instead, that was Avogadro's hypothesis in a nutshell: the volume of the gas is determined &lt;strong&gt;exactly&lt;/strong&gt; proportionate to the number of molecules in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are right in principal that you could have such a &quot;false law&quot; where the number of molecules is inversely proportionate to the volume and still have equal number of molecules for $\ce{H2}$ and $\ce{N2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's follow your course of logic. That would mean if I get 10 L of $\ce{H2}$, I would then have $k/10$ molecules. Let's imagine I add more and more volume, to get 100 L then 1000 L of gas. At some point, I must necessarily divide the molecules in fractions in your false law. Certainly a molecule must occupy &lt;em&gt;some&lt;/em&gt; indivisible volume, since Avogadro believed in the atomistic theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, the obvious solution is that the number of molecules must be a measure of the amount of &quot;stuff.&quot; So if we have more &quot;stuff,&quot; we need more volume for it to occupy. The expression &lt;strong&gt;&lt;em&gt;must&lt;/em&gt;&lt;/strong&gt; be directly proportional between the volume and the number of molecules.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-05-31T03:07:08.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32240" PostTypeId="1" CreationDate="2015-05-31T04:30:51.693" Score="24" ViewCount="1782" Body="&lt;p&gt;I'm a fairly inexperienced, but nonetheless interested chemist, currently planning a small project for the coming summer: I want to isolate an unknown chemical from a plant and determine it's chemical structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ample access to &lt;em&gt;Carex grayi&lt;/em&gt;, a sedge plant, and have chosen this plant for a few reasons. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Of the plants growing near where I live, I can definitively identify &lt;em&gt;Carex grayi&lt;/em&gt;. I am relatively inexperienced in taxonomy and plant identification, and I want to know which species I am working with. &lt;/li&gt;&#xA;&lt;li&gt;As far as I can tell, the plant is relatively unresearched, and may or may not contain alkaloids.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This brings me to my next point. I know it may seem immature or disheartening, but my interest in chemistry stemmed from my interest in neuroscience and neurotransmitters. From countless hours of reading, I have learned much about the pharmacological (and sometimes psychoactive) tendencies of alkaloids. I realize that ingesting unknown alkaloids is extremely dangerous, and I do not intend to ingest &lt;em&gt;anything&lt;/em&gt;. However, due to their numerous applications in medicine and recreational use, there is a massive amount of information on the internet regarding their extraction and isolation. I am entirely open to the idea of isolating a chemical from another group, but as I am inexperienced and my interest stems from neuroscience, I only really know about alkaloids. I am only somewhat familiar with terpenoids, and know little about any other group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure? The &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_structure&quot;&gt;Chemical structure Wikipedia page&lt;/a&gt; lists the following methods for determing chemical structure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;X-ray diffraction&lt;/li&gt;&#xA;&lt;li&gt;Proton NMR&lt;/li&gt;&#xA;&lt;li&gt;Carbon-13 NMR&lt;/li&gt;&#xA;&lt;li&gt;Mass spectrometry&lt;/li&gt;&#xA;&lt;li&gt;Infrared spectroscopy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars. &lt;strong&gt;Is it too much or too difficult for a student to determine a chemical's structure over the summer?&lt;/strong&gt; Should I do anything differently (as in, is there a certain group of chemicals that would make this project possible or easier)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there a &quot;classical&quot; chemical extraction similar to this that most fledgling chemistry students would conduct? This is obviously a ridiculous made-up example: extract dandylionine from a dandy lion flower.&lt;/p&gt;&#xA;" OwnerDisplayName="user16651" LastEditorDisplayName="user16651" LastEditDate="2015-05-31T04:39:20.877" LastActivityDate="2015-06-03T02:26:25.793" Title="How can an inexperienced chemist determine the chemical structure of a molecule?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;extraction&gt;&lt;teaching-lab&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32241" PostTypeId="2" ParentId="32152" CreationDate="2015-05-31T04:50:34.100" Score="3" Body="&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;Yes. All plastics have plasticizers in them. The particular plasticizers chosen depends on what properties are desired of the plastic. Some can make the plastic much stronger but also more brittle at the same time, so that it would be much more difficult to break but when/if it did snap, it would tend to shatter or snap violently. Others make the plastic softer and less brittle, so that the plastic form has less structural strength, but will bend, stretch and disform under heavy load instead of snapping.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;Plasticizers can also affect the  temperature resistance to deformation, volatility (temperature resistance to deflagragation, and properties of said polymer if ignited, such as, will it go out on its own or will it continue to burn the whole thing). Plasticizers can be added to affect the polymer's UV resistance, resistance to chemicals or resistance to abrasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;Once again, all plastics have plasticizers in them. If one were to magically remove all of the plasticizers from a plastic bottle (PETE or PE), the plastic would become brittle, dry, and probably reduce to tiny pieces (think shattered tempered glass, but smaller) or a powder upon touching the plastic form which no longer contained any plasticizers. In fact, you can observe this with some old plastics from the 1970s. Even though the plastic may have never been exposed to UV light, a plastic that is losing its plasticizer can take on the same cracked, opaque appearance. Most plasticizers are volatile at room temperature and will slowly evaporate away just by simply being exposed to air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;Most plastics will leach their plasticizers more rapidly after being washed with soap and water, after being microwaved, while being heated, or while at a low temperature (such as in the freezer). Even still, some plasticizers will take 20 years or more to evaporate away enough to change the properties. Some plastics will never lose their plasticizer (that we know of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Plasticizer&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia&lt;/a&gt;:&#xA;Some of the most common plasticizers and their uses are: &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Phthalate&quot; rel=&quot;nofollow&quot;&gt;Phthalate&lt;/a&gt;-based plasticizers - used in situations where good resistance to water and oils is required. Examples include Bis(2-ethylhexyl) phthalate (DEHP) and Butyl benzyl phthalate (BBzP).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Adipate&quot; rel=&quot;nofollow&quot;&gt;Adipate&lt;/a&gt;-based plasticizers - used for low-temperature performance or resistance to ultraviolet light. Some examples are Bis(2-ethylhexyl)adipate (DEHA) and Dimethyl adipate (DMAD)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Trimellitate&quot; rel=&quot;nofollow&quot;&gt;Trimellitates&lt;/a&gt; - used in automobile interiors and other applications where low volatility and resistance to high temperature is desired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other common plasticizers are &lt;a href=&quot;https://en.wikipedia.org/wiki/Polybutene&quot; rel=&quot;nofollow&quot;&gt;Polybutene&lt;/a&gt;, Benzoates, &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfonamide_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;Sulfonamides&lt;/a&gt;, Glycols/polyethers and poxidized vegetable oils&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1942" LastActivityDate="2015-05-31T04:50:34.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32242" PostTypeId="2" ParentId="32240" CreationDate="2015-05-31T05:22:46.773" Score="24" Body="&lt;p&gt;&lt;strong&gt;So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then probably no (at least not with the same accuracy you would if there weren't these constraints).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass spectrometry and nuclear magnetic resonance are the two most powerful analytical tools to identify completely unknown substances. This doesn't mean it's as trivial as putting your purified substance in the equipment and finding its structure. For reliable results you need a purified substance, an optimized experimental routine and of course the instrument properly set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, before trying to determine the structure of the substance, you must purify it, and that alone is a challenge. As I'm sure you are aware, the plant contains thousands, millions of different molecules. How are you going to extract something you haven't defined? The first step would have to be defining what is it that you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider you're looking for whatever makes plants green:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could macerate the plant, add some water, filter the mixture, check the colors of the filtrate and of the residue, and you would notice most of the green wasn't extracted. You would find different results with different solvents though, and a better option would be ethanol. Even then, it wouldn't mean you have a pure chlorophyll solution, so several other steps would probably be required to purify the substance*. Generally speaking, you would use some kind of chromatography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your project, I suggest you try to first precisely define what is it you're looking for. The substance that gives the plant its color? Its smell? Its pharmacological effects? You must use this property to analyze your experimental results. You can visually inspect the color, you can smell extracts, and perhaps test physiological effects (this is obviously dangerous and shouldn't be done unless you really know what you are dealing with!). Use the target substance properties' to decide what's the best way of extracting and purifying it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, considering you did purify &lt;em&gt;something&lt;/em&gt;, without the high-tech analytical tools mentioned, there are some experiments which can indicate what you have, but they require a reference substance and usually can't be used to say that a substance is the same as another, but rather that a substance &lt;em&gt;isn't&lt;/em&gt; the same as the reference. For example, you can compare boiling/melting points and/or thin layer chromatography retardation factors, both which can be done at home home (to some extent), and also UV/Visible light absorption/emission, which might not be trivial to do at home but require instruments that are a lot easier to find in chemistry teaching labs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For precise tips on the extraction and identifications, it would be necessary to know what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to note that there's a lot you can do besides determining a chemical structure. There are several properties of a substance you can study in order to characterize it. Some examples are solubility (in different solvents, at different conditions (temperature, pH, salinity)), density, reactivity (in face of pretty much anything you find at home) and physical properties such as texture, color, smell, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, do not think I'm saying you should give up, your wish to discover what something is made of is what made chemistry a science, but you need to have a clearer problem to be solved. Decide precisely what it is you're looking for (in terms of macroscopic properties, chemical group or something like that), and then research (and ask!) what could you do about that without a full lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*Chlorophyl is actually a group of substances which have a critical role in photosynthesis, and are, of course, green.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You already have your sample. Now choose your &lt;a href=&quot;http://en.wikipedia.org/wiki/Analyte&quot; rel=&quot;noreferrer&quot;&gt;analyte&lt;/a&gt;, the molecule(s) you want to study. &lt;/li&gt;&#xA;&lt;li&gt;Do research. Look for information on how people usually study this (group of) substances. You'll have better luck looking in education-related sources, which usually have simpler experiments.&lt;/li&gt;&#xA;&lt;li&gt;Once you've found something that fits your interests and available material, do experiments. Observe the results. Change parameters, do them again, try to understand what's happening.&lt;/li&gt;&#xA;&lt;li&gt;If you can't explain the results, do research, check your references, and of course, ask here. We'll be glad to help.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-06-03T02:26:25.793" LastActivityDate="2015-06-03T02:26:25.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32243" PostTypeId="1" AcceptedAnswerId="32254" CreationDate="2015-05-31T06:04:22.997" Score="8" ViewCount="1250" Body="&lt;p&gt;what are the systems of recording atomic masses and their units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the nucleon number is the number of protons and neutrons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the mass number on the periodic table is the masses of all the isotopes and their relative abundance (I believe this is called relative atomic mass and is equal in numerical value to the molar mass of that element)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what units relative atomic mass is measured in. some say there is no units as it is a relative scale, others say that it is measured in atomic mass units (AMU), unified mass units (u) or daltons (Da).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe (but would like it to be confirmed) that the atomic mass unit (AMU) is an obsolete unit based on the mass relative to oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also believe (but would like it to be confirmed) that the unified mass units (u) and dalton (Da) are equivalent units (u=Da=1/12 mass of a carbon-12 atom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what is the difference between isotopic mass, relative atomic mass and atomic mass and what units do they all have?  &lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorDisplayName="user7951" LastEditDate="2015-09-29T08:59:24.457" LastActivityDate="2015-09-29T08:59:24.457" Title="Units of mass on the atomic scale" Tags="&lt;atoms&gt;&lt;units&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32244" PostTypeId="2" ParentId="27631" CreationDate="2015-05-31T06:08:33.770" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;nbsp;Is there any extra force holding the two components together?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;strong&gt;Correct. Namely, hydrogen bonding.&lt;/strong&gt; As a quick refresher, hydrogen bonding is an  electrostatic attraction between an electropositive hydrogen atom on one compound being (weakly) attracted to a highly electronegative atom on another, nearby molecule, typically oxygen, although it can happen between a hydrogen and another electronegative atom such as oxygen or fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;Azeotropes form when there is a 'deviation from &lt;a href=&quot;https://en.wikipedia.org/wiki/Raoult%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Raoult's law&lt;/a&gt;':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;A solution that shows positive deviation from Raoult's law forms a &lt;strong&gt;minimum&lt;/strong&gt; boiling azeotrope. A very well-known example would be the azetrope that forms from a ethanol and water mixture when it is approximately 95% ethanol by volume. Once this composition has been achieved, the liquid and vapor have the same composition, and no further separation occurs. Methods for breaking this azetrope are typically either via chemical means, such as adding a small amount of benzene to the azetrope, or by a drying agent such as $\ce{H2SO4}$, magnesium sulfate, or calcium chloride. Interesting thing to note: You do not have to dry all the way to 100% ethanol, but simply BREAK the azetrope by getting it beyond that 95.6% concentration. After you get it beyond ~96%, you can distill the rest of the way to 99% via the normal distillation method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;A solution that shows a negative deviation from Raoult's law forms a MAXIMUM boiling azeotrope, or as it is often stated: A negative deviation from Raoult's law is where the adhesive forces between different components are stronger than the average cohesive forces between like components. In consequence each component is retained in the liquid phase by attractive forces that are stronger than in the pure liquid so that its partial vapor pressure is lower. Nitric acid and water is a classic example of this class of azeotrope which forms when the solution has an approximate composition of 68% nitric acid and 32% water by mass.&lt;/p&gt;&#xA;" OwnerUserId="1942" LastEditorUserId="7448" LastEditDate="2015-07-03T12:07:25.113" LastActivityDate="2015-07-03T12:07:25.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32245" PostTypeId="2" ParentId="32199" CreationDate="2015-05-31T06:42:00.243" Score="5" Body="&lt;p&gt;In this case, &quot;saturated&quot; means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}&amp;lt;=&amp;gt;Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The $K_\mathrm{sp}$ is really really small, so we can actually ignore&#xA;its contribution to the $[\ce{OH^-}]$, but for fun, you can do the&#xA;whole ICE table thing and end up with:&#xA;$$\begin{align}&#xA;     [\ce{OH^-}] &amp;amp;= .1 + 2x \\&#xA;     [\ce{Cu^2+}] &amp;amp;= x \\&#xA;     K_\mathrm{sp} &amp;amp;= [\ce{OH^-}]^2[\ce{Cu^2+}] \\&#xA;     1.6\times 10^{-19}  &amp;amp;= (.1 + 2x)^2(x) \\&#xA;     x = [\ce{Cu^2+}] &amp;amp;= 1.6 \times 10^{-17}\ \mathrm{M}&#xA;     \end{align}&#xA;   $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$$\begin{align}&#xA;    E &amp;amp;= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\&#xA;      &amp;amp;= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\&#xA;      &amp;amp;= -0.16\ \mathrm{V}&#xA;    \end{align}&#xA;    $$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, the whole cell potential:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First, negate the calculated reduction potential to get the oxidation potential:&#xA;$$\begin{align}&#xA;    E_\mathrm{ox} &amp;amp;= -E_\mathrm{red} \\&#xA;      &amp;amp;= 0.16\ \mathrm{V}&#xA;    \end{align}&#xA;    $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;    E_\mathrm{ox} &amp;amp;= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\&#xA;                  &amp;amp;= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\&#xA;                  &amp;amp;= 0.16\ \mathrm{V}&#xA;    \end{align}&#xA;    $$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-05-31T06:42:00.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32246" PostTypeId="1" CreationDate="2015-05-31T07:59:29.187" Score="8" ViewCount="239" Body="&lt;p&gt;I know that oxidanes are oxygen chains and rings. For example trioxidane is $\ce{HO-O-OH}$. I can see this turning into $\ce{O2}$ and $\ce{H2O}$ like hydrogen peroxide does. Since the double bonds in $\ce{O2}$ are strong do oxygen compounds without nitrogen ever explode?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="9961" LastEditorDisplayName="user15489" LastEditDate="2016-06-17T17:31:14.320" LastActivityDate="2017-06-27T08:16:23.990" Title="Oxygen-based explosives" Tags="&lt;covalent-compounds&gt;&lt;explosives&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32247" PostTypeId="2" ParentId="27047" CreationDate="2015-05-31T08:13:57.600" Score="2" Body="&lt;p&gt;I wonder if, when you say UV/Vis spectroscopy, you are also including fluorescence in that. If you are, then the soft iodide / triiodide ions are pretty good reductive quenching agents and they may be responsible for the reduction in fluorescence emission you observe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, the $\text{I}^-$ will react with the excited state of your dye $\text{D*}$ and quench it by photo-induced electron transfer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{I^{-} + D}^*\ce{\quad -&amp;gt;[\text{photo-induced electron transfer}]\quad I^{\ .} + D^{\ .\ -}\quad -&amp;gt;[\text{back electron transfer}]}\quad \ce{I- + D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this process provides a &lt;em&gt;non-radiative&lt;/em&gt; pathway for the excited state of the dye to decay back to its ground state. The process is essentially &lt;em&gt;catalyzed&lt;/em&gt; by the iodide anion, which is not consumed in the process. The more iodide you have, the more prevalent this process, the more likely it is for the excited state of your dye to go through this process rather than fluorescing, so ultimately the more quenched your dye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a process can be modeled by the Stern - Volmer equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{I_0}{I}=1+k_q\tau_0[\text{Q}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $I_0$ is the fluorescence intensity in the absence of quencher &lt;strong&gt;Q&lt;/strong&gt;; $I$ the intensity in the presence of a certain concentration [Q] of quencher; $k_q$ is the kinetic constant of the quenching process; $\tau_0$ is the intrinsic lifetime of the fluorophore, i.e. the lifetime you would measure in the same conditions, but in the absence of quencher Q.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-05-31T08:13:57.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32248" PostTypeId="2" ParentId="32211" CreationDate="2015-05-31T08:24:36.963" Score="9" Body="&lt;p&gt;Simply put: a lone-pair orbital is &quot;fatter&quot; than a bonding orbital as I have &lt;a href=&quot;http://molecularmodelingbasics.blogspot.dk/2009/07/ammonia-vsepr.html&quot;&gt;illustrated here&lt;/a&gt;.  This can be rationalized as follows: in a bonding orbital the electron pair is mainly attracted to 2 different nuclei which helps localize the electron pair.  In a lone-pair orbital the electron pair is mainly attracted to a single nucleus and can therefore spread out more without reducing this attraction.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-31T08:24:36.963" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32249" PostTypeId="2" ParentId="24936" CreationDate="2015-05-31T08:26:27.460" Score="2" Body="&lt;p&gt;It is quite common to observe secondary peaks or &quot;bumps&quot; in the instrumental response function for a laser excitation. It is not always easy to pinpoint the origin of the second peak, but it is typically &lt;em&gt;not&lt;/em&gt; associated with a reflection in the instrument. If any reflection were to be involved, it would be one within the cavity of the laser itself. It is however more often the case that these secondary peaks are just an adventitious product of the laser production process (no laser is perfect). In fact, manufacturers sometimes will suggest keeping an eye on the relative ratio of these peaks in the IRF to gauge the health status of your laser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, observe the secondary peak in the red line below from a PicoQuant FluoTime instrument with a picosecond laser:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.picoquant.com/?ACT=32&amp;amp;f=decay_analysis.jpg&amp;amp;fid=27&amp;amp;d=7309&quot; alt=&quot;An example from PicoQuant&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see an example of a much more bumpy IRF in a leaflet published by Edinburgh, the makers of your own instrument &lt;a href=&quot;http://www.edinburghphotonics.com/files/file/technical-specifications/Lifespec%202%20Flyer.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LifeSpec II spectrometer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.583" LastActivityDate="2015-05-31T08:26:27.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32250" PostTypeId="1" AcceptedAnswerId="32252" CreationDate="2015-05-31T11:53:58.017" Score="5" ViewCount="3817" Body="&lt;p&gt;My teacher told me that Neon has a larger atomic radius than Fluorine.I am of the understanding that it is merely a consequence of the way we define the atomic radius and that we use Van der walls radius for neon which is larger that the covalent radius of fluorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also said that the radius of a fluorine 1- anion is more than a neon atom by using the fact that they are iso-electronic species and that since neon has a higher nuclear charge, it will pull on the 10 electrons more than fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what definition of atomic radius am i using for neon now? Aren't we always supposed to use the Van der walls definition for noble gases? In case we do use the Van der walls radius of the neon atom and the ionic radius of the fluorine anion, how am i supposed to compare these two and come to the conclusion that the fluorine anion is larger?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastActivityDate="2015-05-31T13:40:12.277" Title="Comparison of atomic radii" Tags="&lt;ions&gt;&lt;atomic-radius&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32251" PostTypeId="1" CreationDate="2015-05-31T12:29:05.673" Score="0" ViewCount="145" Body="&lt;p&gt;I would like to know how heat up a large amount of water ( assumption : 100 liter ) that people could use to take their bath and also it would not be dangerous even if they accidentally ingested it ? . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is that to design a metal container and have the chemical in it . I'll react the chemical that is inside the container and then I'll put the container into the water , which will eventually heat up the water through thermal equilibrium . &lt;/p&gt;&#xA;" OwnerUserId="16654" LastActivityDate="2015-05-31T12:29:05.673" Title="How the heat a large amount of water that is safe to use?" Tags="&lt;equilibrium&gt;&lt;water&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32252" PostTypeId="2" ParentId="32250" CreationDate="2015-05-31T13:40:12.277" Score="5" Body="&lt;p&gt;Atomic radii cannot really be uniquely nor accurately defined.  However, there is generally a very big difference between covalent, van der Waals, and ionic radii.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon.  Similarly, the ionic radius of fluoride ion will be larger than the van der Waals radius of neon. The reason for the latter is the higher nuclear charge on neon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most sources list a slightly smaller van der Waals radius for fluorine compared to neon, but a slightly larger covalent radius for fluorine compared neon. So when comparing radii you have to carefully specify what kind of radii you are talking about.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-05-31T13:40:12.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32254" PostTypeId="2" ParentId="32243" CreationDate="2015-05-31T13:49:44.577" Score="13" Body="&lt;p&gt;The quantity &lt;strong&gt;atomic mass&lt;/strong&gt; (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.&lt;br&gt;&#xA;The dimension of the atomic mass is&#xA;$$\dim m_\mathrm{a} = \mathsf{M}$$&#xA;The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity &lt;strong&gt;relative atomic mass&lt;/strong&gt; (quantity symbol: $A_\mathrm{r}$) is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative atomic mass is the ratio of the mass of an atom to the unified atomic mass constant:&#xA;$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$&#xA;where $m_\mathrm{a}$ is the atomic mass and $m_\mathrm{u}$ is the unified atomic mass constant.&lt;br&gt;&#xA;The relative atomic mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):&#xA;$$\dim A_\mathrm{r} = 1$$&#xA;The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).&lt;br&gt;&#xA;For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the &lt;strong&gt;unified atomic mass constant&lt;/strong&gt; (symbol: $m_\mathrm{u}$) is defined as 1/12 of the mass of a neutral atom of the nuclide $\ce{^12C}$ in the ground state at rest. &#xA;$$m_\mathrm{u} = \frac{m_\mathrm{a}(\ce{^12C})}{12}$$&#xA;The recommended value is&#xA;$$m_\mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$ (&lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?u&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;br&gt;&#xA;Note that the value in SI units is obtained experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit &lt;strong&gt;dalton&lt;/strong&gt; (unit symbol: $\mathrm{Da}$) and the &lt;strong&gt;unified atomic&#xA;mass unit&lt;/strong&gt; (unit symbol: $\mathrm{u}$) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the value of the unified atomic mass constant is, by definition, equal to one dalton or one unified atomic mass unit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_\mathrm{u} = 1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old relative atomic masses (unfortunately called ‘atomic weights’) and the corresponding &lt;strong&gt;atomic mass unit (amu)&lt;/strong&gt; were originally referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-31T13:49:44.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32255" PostTypeId="1" CreationDate="2015-05-31T13:55:59.850" Score="1" ViewCount="1175" Body="&lt;p&gt;$$\ce{NH_3 + HCl -&amp;gt; NH_4Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic structure of ammonia is such so that the nitrogen atom has an outer shell with a lone pair of electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When ammonia reacts with hydrochloric acid (above), the hydrogen from the $\ce{HCl}$ moves to the ammonia. However, it leaves its electrons behind because there is already a lone pair of 2 electrons on the $\ce{N}$ atom, allowing it to retain a $\mathrm{1s^2}$ shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the hydrogen take its electron over and nitrogen doesn't then &quot;donate&quot; an electron to the chlorine?&lt;/p&gt;&#xA;" OwnerUserId="16656" LastEditorUserId="865" LastEditDate="2015-08-05T15:14:28.063" LastActivityDate="2015-10-04T17:07:38.820" Title="Electron transfer in reaction between ammonia and hydrochloric acid" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32257" PostTypeId="1" AcceptedAnswerId="32260" CreationDate="2015-05-31T14:32:38.383" Score="1" ViewCount="4318" Body="&lt;p&gt;I’d like to ask a question and clear my misunderstanding.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following 0.01 M solutions have the highest osmotic pressure?&lt;br&gt;&#xA;  -$\ce{NaCl}$&lt;br&gt;&#xA;  -$\ce{C12H22O11}$&lt;br&gt;&#xA;  -$\ce{BaCl2}$&lt;br&gt;&#xA;  -$\ce{CO(NH2)2}$&lt;br&gt;&#xA;  -$\ce{[Cr(NH3)4Cl2]Cl}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the equation for osmotic pressure is&lt;br&gt;&#xA;$$\Pi = cRTi$$&#xA;I have eliminated sucrose, sodium chloride, and urea from my answer, because sucrose and urea are non-electrolyte. And sodium chloride has smaller van ’t Hoff factor than barium chloride. I ended up in barium chloride and the complex compound. I’m not really sure with the complex compound since from what I learnt, a complex compound is easier to dissolved in water &lt;strong&gt;&lt;em&gt;(my teacher in solubility chapter taught me that $\ce{AgI + NH3 -&amp;gt; [Ag(NH3)2]I}$ have a high solubility because it is a complex compound)&lt;/em&gt;&lt;/strong&gt;. However I’m not sure if it has a high van ’t Hoff factor as well.&lt;/p&gt;&#xA;" OwnerUserId="15525" LastEditorUserId="865" LastEditDate="2021-05-22T19:11:12.587" LastActivityDate="2021-05-22T19:11:12.587" Title="Highest osmotic pressure" Tags="&lt;solutions&gt;&lt;osmosis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32258" PostTypeId="1" CreationDate="2015-05-31T15:42:06.033" Score="3" ViewCount="218" Body="&lt;p&gt;This Indian Institute of Technology Bombay  &lt;a href=&quot;http://www.ircc.iitb.ac.in/IRCC-Webpage/Homeopathy-Nanoparticle-Note.pdf&quot; rel=&quot;nofollow&quot;&gt;press release&lt;/a&gt; quotes some research papers which claims to prove that homeopathic remedies do retain original substances even when diluted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The paper reports the fascinating observation that certain high potency (highly diluted)&#xA;  homeopathic remedies made from metals still contain measurable amounts of the starting&#xA;  material, even at extreme dilutions of 1 part in 10 raised to 400 parts (200C).  It is well known that a series of 1:99 dilutions done sequentially will produce a significant dilution of the starting material in very short order. Specifically, if the starting material is at one molar concentration (6.023x10e23 molecules per liter), then at about the 12th dilution (12C) there should be no or very nearly no molecules left of the starting material. At 200th dilution (200C) the likelihood of there being even one atom of the starting material approaches zero. However, dilution does not work so simply, according to this paper. Using electron microscopy (TEM), electron&#xA;  diffraction, and atomic spectroscopy, Chikramane et al. found that, contrary to the arithmetic, there are nanogram quantities of the starting material still present in these ‘high potency’ remedies in the form of nanoparticles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This was published in a homeopathy journal and not a chemical engineering journal, although the department was chemical engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A later paper, &lt;em&gt;&lt;a href=&quot;http://www.homresearch.org/wordpress/wp-content/uploads/2013/02/langmuir.pdf&quot; rel=&quot;nofollow&quot;&gt;Why Extreme Dilutions Reach Non-zero Asymptotes: A Nanoparticulate Hypothesis Based on Froth Flotation&lt;/a&gt;&lt;/em&gt; suggests that nano-particles are distributed unevenly, floating to the top of the container, which may interfere with dilution predictions,  public in a journal from the American Chemical Society.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, physicochemical studies of these solutions have unequivocally established the presence of the starting raw materials in nanoparticulate form even in these extreme (super-Avogadro, &gt;10&lt;sup&gt;23&lt;/sup&gt;) dilutions. In this article, we propose and validate a hypothesis to explain how nanoparticles are retained even at such enormous dilution levels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another researcher in a &lt;a href=&quot;http://www.researchgate.net/profile/Salvatore_Chirumbolo/publication/256465665_Homeopathy_is_not_pharmacology/links/02e7e528251b293ebd000000.pdf&quot; rel=&quot;nofollow&quot;&gt;Letter to the Editor&lt;/a&gt; that the original element was found may be be because those original elements were stuck to the inner walls of lab glass:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Further-more, soluble compounds can be captured by nanosized structure, often nanobubbles, lining the boron-silicate glasses with which dilutions, subjected to mechanical stresses, are made. Therefore, an imprecise fraction of compound may be dispersed into the solvent from the glass wall whenever a mechanical stress is applied to the dilution, possibly giving gelsemine concentration far from the theoretical evaluation into pure water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(The &quot;mechanical stress&quot; presumably refers to the &lt;a href=&quot;http://medical-dictionary.thefreedictionary.com/succussion&quot; rel=&quot;nofollow&quot;&gt;succussion&lt;/a&gt; during homeopathic preparation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, are some of the original materials still found in 200C dilutions?&lt;/p&gt;&#xA;" OwnerUserId="16657" LastEditorUserId="9961" LastEditDate="2015-10-23T17:15:11.910" LastActivityDate="2015-10-23T17:15:11.910" Title="Do extreme homeopathic dilutions still retain quantities of the starting materials?" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32259" PostTypeId="1" AcceptedAnswerId="32262" CreationDate="2015-05-31T15:44:49.283" Score="1" ViewCount="3412" Body="&lt;p&gt;I'm getting a little confused about the definition of chelation and its implications (I should probably point out I'm only an A2 Chemistry student). The IUPAC defines chelation as the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The formation or presence of two or more separate coordinate bonds between a polydentate (multiple bonded) ligand and a single central atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does this mean that any ligand that is not monodentate (e.g. even ethanedioate) is an example of chelation? I thought chelation was only for complexes with a large coordination number (e.g. $\ce{EDTA^4-}$), it would seem odd to call a single metal ion with 1 ethanedioate ligand chelated?&lt;/p&gt;&#xA;" OwnerUserId="12743" LastEditorUserId="14" LastEditDate="2015-06-01T17:37:19.297" LastActivityDate="2015-11-06T15:47:57.233" Title="Are all complexes with a polydentate ligand examples of chelation?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32260" PostTypeId="2" ParentId="32257" CreationDate="2015-05-31T15:52:54.417" Score="3" Body="&lt;p&gt;You are correct in thinking that urea and sucrose would have van't hoff factor (i) = 1, since they are non-electrolytes and don't undergo disassociation(or ion pairing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are left with are the following: $\ce{NaCl}$ , $\ce{BaCl_2}$ and $\ce{[Cr(NH_3)_4Cl_2]Cl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us begin with the the two chloride salts: $\ce{NaCl}$ , $\ce{BaCl_2}$. we assume 100% disassociation for the two (strong electrolytes) and we have i = 2 and i = 3 respectively for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the complex would indeed disassociate but not completely (the extent would be determined $K_f$ for the complex). Thus 1 &amp;lt; i &amp;lt; 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, I believe you can order them on the basis of increasing osmotic pressure.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="16088" LastEditDate="2015-05-31T16:12:13.570" LastActivityDate="2015-05-31T16:12:13.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32261" PostTypeId="1" CreationDate="2015-05-31T16:27:59.353" Score="0" ViewCount="10610" Body="&lt;p&gt;Alum is generally used in water tanks to settle down solid impurities. Since it contains aluminium I suspect it to be toxic to humans. So are there any safe chemicals to clean large overhead tanks without removing the water or  doing much effort. I am asking about solid impurities and not disinfection .&lt;/p&gt;&#xA;" OwnerUserId="16657" LastActivityDate="2019-09-13T08:25:49.990" Title="Cleaning an overhead water tank using Alum or other chemicals?" Tags="&lt;water&gt;&lt;filtering&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32262" PostTypeId="2" ParentId="32259" CreationDate="2015-05-31T16:32:44.367" Score="6" Body="&lt;p&gt;I think you should not focus on the ligand to define chelation. You should instead focus on the complex that is formed. In fact, the very same multidentate ligand may participate in binding in more than one way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;by forming more than one bond with the same central atom, therefore forming a chelated complex;&lt;/li&gt;&#xA;&lt;li&gt;by forming only one bond with that central ion, even though it could form two; this may be the case when a vast excess of this ligand is present with respect to the metal ion;&lt;/li&gt;&#xA;&lt;li&gt;by bridging between two adjacent central ions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Only the first of these possibility is referred to as chelation. In other words, forming a chelated complex doesn't depend solely on the fact that the ligand is multidentate. Being multidentate is a necessary, but not sufficient condition for a ligand to form chelated complexes.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-05-31T16:32:44.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32263" PostTypeId="2" ParentId="32261" CreationDate="2015-05-31T16:50:59.290" Score="2" Body="&lt;p&gt;Aluminum is remarkable safe for human consumption, also considering that it is present in high concentration essentially everywhere, and it is not known to have any significant specific biological roles. Wikipedia happens to have a nice well researcher section on aluminum's possible effects on human and animal health: &lt;a href=&quot;http://en.m.wikipedia.org/wiki/Aluminium#Health_concerns&quot; rel=&quot;nofollow&quot;&gt;Aluminium: health concerns&lt;/a&gt;. Alum is also routinely added to water in low concentrations for elimination of solid particulate, as you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main concern for your application is the fact that typical methods of water purification from solid particulate, just like addition of alum, will tend to form flocs in your water tank that will then need to be removed! Depending whether this scum will float on top of the water, or sink to the bottom of the tank, the removal may be more or less easy, also depending on what kind of access you have to this tank. I doubt if any of those solutions would count as not &quot;doing much effort&quot;...&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-05-31T16:50:59.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32264" PostTypeId="1" CreationDate="2015-05-31T16:54:45.550" Score="5" ViewCount="330" Body="&lt;p&gt;Some webpages describe the contact process as this (I left out $\ce{SO2}$, $\ce{SO3}$ production):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO3 + H2SO4 -&amp;gt; H2S2O7}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, it is controllable to add water to the oleum therefore making liquid sulfuric acid. However, some other sites describe this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO3 + H2O  -&amp;gt; H2SO4}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about what really happens in the absorption tower. I've read about the wet acid, but it states that adding water causes gaseous $\ce{H2SO4}$ to form, which is condensed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In the absorption tower, trioxide sulfur enters as gas, and sulfuric acid comes down as liquid. As a result the liquid comes out as oleum (first eq.). The other contact process states different: the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), giving strengthened sulfuric acid (let's say 99.5% sulfuric acid, and 0.5% water). Still confused.&lt;/p&gt;&#xA;" OwnerUserId="16660" LastEditorUserId="16660" LastEditDate="2015-05-31T21:44:52.630" LastActivityDate="2021-04-18T16:57:41.277" Title="What is the difference between these equations for the Contact process?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;water&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32265" PostTypeId="1" CreationDate="2015-05-31T17:08:44.070" Score="21" ViewCount="6074" Body="&lt;p&gt;I'm learning how to draw Lewis diagrams. Everything I've read emphasizes the octet rule. However, to the best of my knowledge, that rule only applies to elements in the first three periods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On our exam, I expect our professor will ask us to draw some Lewis diagrams of molecules whose atoms do not comport with the octet rule. However, I haven't seen any diagrams of these; it'd be useful to get a sense of what they look like so that my gut doesn't disapprove of what will otherwise be strange looking diagrams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some relatively common molecules that comprise atoms that comprise a great number of valence electrons?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="4231" LastEditDate="2015-05-31T18:11:18.807" LastActivityDate="2017-09-14T01:03:50.983" Title="Which relatively simple molecules violate the octet rule?" Tags="&lt;molecules&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32266" PostTypeId="1" AcceptedAnswerId="32267" CreationDate="2015-05-31T17:26:40.333" Score="2" ViewCount="2911" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WbZOG.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On seeing this graph closely I noticed the there a element with &lt;b&gt;no proton&lt;/b&gt;. But with no proton there will be no electron too and don't think that an lone neutron would form an element. So is that given a half-life on neutron OR there exist a element (isotope) with no electron?&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2015-05-31T17:40:15.167" Title="Element with No Proton?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32267" PostTypeId="2" ParentId="32266" CreationDate="2015-05-31T17:40:15.167" Score="8" Body="&lt;p&gt;A free neutron is not considered to be an element and there is no element that does not have at least one proton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free neutrons could be included on the graph as they are unstable and decay with a half life of around 10 minutes ($611.0\pm1.0~\mathrm{s}$) although most versions I have seen do not usually include them as they aren't really nuclei which is what the graph is about.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-05-31T17:40:15.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32268" PostTypeId="2" ParentId="32265" CreationDate="2015-05-31T17:48:15.720" Score="19" Body="&lt;p&gt;There are &lt;strong&gt;3 types&lt;/strong&gt; of octet rule &quot;violations&quot; or exceptions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;molecules with an &lt;strong&gt;odd number&lt;/strong&gt; of electrons, such as nitric oxide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Km18m.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Km18m.png&quot; alt=&quot;NO molecule&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://file.scirp.org/Html/16-8202443_44737.htm&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;molecules with &lt;strong&gt;less than&lt;/strong&gt; 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1uNMi.gif&quot; alt=&quot;BH3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Lewis_Theory_of_Bonding/Violations_of_the_Octet_Rule&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;molecules with &lt;strong&gt;more than&lt;/strong&gt; 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yb5dE.gif&quot; alt=&quot;PCl5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the image source links for other examples as well as some practice problems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding around electron deficient atoms such as the boron in $\ce{BH3}$ is best explained in terms of 3-center 2-electron bonds (&lt;a href=&quot;http://en.wikipedia.org/wiki/Three-center_two-electron_bond&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt;; &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31732/what-is-a-non-classical-carbocation/31758#31758&quot;&gt;SE Chem example_1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29000/which-carbocation-is-more-stable-the-ethyl-or-1-propyl-carbocation/31729#31729&quot;&gt;SE Chem example_2&lt;/a&gt;)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, arguments based on d-orbital involvement are advanced to explain the bonding in atoms surrounded by more than 8 electrons.  However d-orbital involvement in non-transition metals is now generally considered to be unlikely.  Instead &lt;a href=&quot;http://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypercoordinated&lt;/a&gt; (or hypervalent) bonding in which 3-center 4-electron bonds are formed has become a more generally accepted explanation.  A number of examples have previously been discussed here on SE Chem (for example, see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2016-04-29T21:19:54.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32269" PostTypeId="1" CreationDate="2015-05-31T19:01:53.970" Score="6" ViewCount="89" Body="&lt;p&gt;In the picture below a proportionality equation is shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega \propto \nu \exp\left(\frac{-\Delta G_m}{RT}\right)\propto \dots$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While $\Delta G_m$ is the Gibbs free energy, $R$ is the gas constant, and $T$ is the temperature, what are the parameters $\omega$ and $\nu$ exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lzOtH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the context on the picture: &lt;a href=&quot;http://en.wikipedia.org/wiki/Yttria-stabilized_zirconia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yttria-stabilized-zircona&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorUserId="30" LastEditDate="2016-08-09T17:24:24.413" LastActivityDate="2016-08-09T17:36:18.920" Title="What are ω and ν in this expression?" Tags="&lt;crystal-structure&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32270" PostTypeId="2" ParentId="32264" CreationDate="2015-05-31T19:03:25.547" Score="3" Body="&lt;p&gt;In principle, your sources are correct and not contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 + SO3 &amp;lt;=&amp;gt; H2S2O7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disulfuric acid can be converted to sulfuric acid by reaction with added water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2S2O7 + H2O &amp;lt;=&amp;gt; 2 H2SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the overall net reaction is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{H2SO4 + SO3 \;&amp;amp;&amp;lt;=&amp;gt; H2S2O7}\\&#xA;\ce{H2S2O7 + H2O \;&amp;amp;&amp;lt;=&amp;gt; 2 H2SO4}\\&#xA;\hline\ce{SO3 + H2O \;&amp;amp;&amp;lt;=&amp;gt; H2SO4}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-05-31T19:03:25.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32271" PostTypeId="2" ParentId="32264" CreationDate="2015-05-31T19:06:49.767" Score="3" Body="&lt;p&gt;The answer to your question is thoroughly explained in &lt;a href=&quot;http://www.sciencedirect.com/science/book/9780750633659&quot; rel=&quot;nofollow&quot;&gt;Chemistry of the Elements (Second Edition), A. Earnshaw and Norman Greenwood, 1997, p. 708&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(...) The  $\ce{SO3}$ gas cannot be absorbed directly in water because it would first come into contact with the water-vapour above the absorbant and so produce a stable mist of fine droplets of $\ce{H2SO4}$ which would then pass right through the absorber and out into the atmosphere. Instead, absorption is effected by 98% $\ce{H2SO4}$ in ceramic packed towers and sufficient water is added to the circulating acid to maintain the required concentration. (...)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Two pages following, the process is represented as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO3 + H2O\ ($\text{in 98% }$H2SO4) -&amp;gt; H2SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;Industrial Manufacture of Sulfuric Acid&lt;/strong&gt; box (p. 708-710), the $\ce{H2S2O7}$ species is not mentioned. However, the discussion about concentrated Sulfuric Acid states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is clear that &quot;pure&quot; anhydrous sulfuric acid, far from being a single substance in the bulk liquid phase, comprises a dynamic equilibrium involving at least seven well-defined species. The concentration of the self-dissociation products in $\ce{H2SO4}$ at 25º (expressed in millimoles of solute per kg solvent) are:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;\begin{array}{lcr}&#xA;\ce{HSO4-} &amp;amp; \ce{H3SO4+} &amp;amp; \ce{H3O+} &amp;amp; \ce{HS2O7-} &amp;amp; \ce{H2S2O7} &amp;amp; \ce{H2O} &amp;amp; \text{Total}\\&#xA;\hline&#xA;15.0 &amp;amp; 11.3 &amp;amp; 8.0 &amp;amp; 4.4 &amp;amp; 3.6 &amp;amp; 0.1 &amp;amp; 42.4&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering the above, you can assume that the absorption of $\ce{SO3}$ in 98% $\ce{H2SO4}$ will lead to the formation of $\ce{H2S2O7}$ but also several other species. If you are really interested in industrial production details and properties of sulfuric acid, this reference is really valuable.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="24052" LastEditDate="2016-03-13T20:29:11.957" LastActivityDate="2016-03-13T20:29:11.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32272" PostTypeId="1" AcceptedAnswerId="53661" CreationDate="2015-05-31T19:50:12.253" Score="6" ViewCount="1617" Body="&lt;p&gt;In the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kr%C3%B6ger%E2%80%93Vink_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kröger-Vink notation&lt;/a&gt; the effective charge $c$ is marked as superscript while the site location $s$ is marked as subscript on the atom itself $\ce{A}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A^{c}_{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I have learned the effective charge is &lt;em&gt;the charge difference from the perfect lattice situation&lt;/em&gt;. The Wikipedia definition agrees:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$c$ corresponds to the electronic charge of the species relative to the site that it occupies&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While neutral charge (no difference) is usually an $x$, I have always worked with a slash $/$ or similar for one negative charge and a dot $\bullet$ for one positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have run across a notation in a teacher's lecture material that says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn^{/}_{Zn}} \quad \text{ and } \quad \ce{Zn^{\bullet}_{Zn}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, adding a $\ce{Zn}$ to it's own natural position makes no charge change, so it should be $\ce{Zn^{x}_{Zn}}$. How is the above notation then possible? Does it imply that the $\ce{Zn}$ has lost or gained one electron, respectivily, and thus has a net charge?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorUserId="24052" LastEditorDisplayName="user15489" LastEditDate="2018-01-08T10:21:51.317" LastActivityDate="2018-01-08T11:58:09.623" Title="Kröger-Vink notation for defect reactions" Tags="&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32273" PostTypeId="2" ParentId="31868" CreationDate="2015-05-31T20:04:21.333" Score="10" Body="&lt;p&gt;I actually tend to agree with the answer from @JanJensen based on electrostatic effects, rather than with the &quot;covalency&quot; argument. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, it has been mentioned in the comments that the $\ce{Al^{III}-OH2}$ is a &quot;far more covalent&quot; bond than one $\ce{Mg^{II}-OH2}$ one. This remark, however, does not actually &lt;em&gt;explain&lt;/em&gt; anything, but rather it shifts the question to a new one, i.e. &lt;em&gt;why is it, then, that the $\ce{Al-O}$ bond is more covalent?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water exchange reactions at the Mg and Al centers have actually been studied extensively, but unfortunately not to the same extent as those for transition metal ions, or for selected lanthanides. The latter in particular are a subject of continuous and very intense interest, because water exchange at e.g. $\ce{Gd^{III}}$ or $\ce{Eu^{II}}$ ions play a pivotal role in the effectiveness of complexes of these ions as MRI contrast agents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, here are a few very relevant references that will certainly offer some more insight, but probably not the final answer that you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ic980333l&quot;&gt;An $\ce{^{17}O}$-NMR Study of the Exchange of Water on $\ce{AlOH(H2O)5^{2+}(aq)}$&lt;/a&gt;, &lt;em&gt;Inorganic Chemistry&lt;/em&gt;, &lt;strong&gt;1998&lt;/strong&gt;, &lt;em&gt;37&lt;/em&gt; (19), 4760–4763;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0010854599902321&quot;&gt;Water exchange on metal ions: experiments and simulations&lt;/a&gt;, &lt;em&gt;Coordination Chemistry Reviews&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;187&lt;/em&gt; (1), 151–181.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact, when studying exchange reactions at metal ions, the authors of these and other studies tend to consider ions with equal charge as homogeneous series, because &lt;em&gt;the effect of the charge will swamp out any other effect&lt;/em&gt;, some times even the relatively large effects of electronic configuration in d-transition metals!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That notwithstanding, the papers above agree that the &lt;strong&gt;mechanism&lt;/strong&gt; for the water exchange reactions for both Al and Mg metal complexes is &lt;strong&gt;dissociative&lt;/strong&gt; in nature, which means that the rate determining step involves the loss of a $\ce{H2O}$ ligand from the inner sphere of the complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can combine this piece of information with the fact that, as mentioned in the second paper above, there is a significant increase in &lt;strong&gt;charge &lt;em&gt;density&lt;/em&gt;&lt;/strong&gt; when going from $\ce{Mg^2+}$ to $\ce{Al^3+}$. First, the aluminum cation is smaller (reported values from that reference are: 72 pm for $\ce{Mg^2+}$ and 53 pm for $\ce{Al^3+}$); secondly, it has higher charge. This rather staggering difference can again be explained using the usual electrostatic arguments (i.e. the two ions are isoelectronic, but the Al nucleus contains more protons, so electrons in the latter are held more closely, leading to smaller cations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to put this all together: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;removing a water molecule determines the rate of the &lt;em&gt;overall&lt;/em&gt; exchange process according to the experimentally determined dissociative mechanism;&lt;/li&gt;&#xA;&lt;li&gt;the smaller, more highly charged $\ce{Al^3+}$ exerts a much stronger attraction on each of its bound $\ce{H2O}$ ligands, making said removal much more costly; &lt;/li&gt;&#xA;&lt;li&gt;in macroscopic terms, the $\ce{Al^3+}$ has a much higher activation energy, which ultimately manifests itself in the much lower rates of exchange that the OP referred to.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16652" LastEditorUserId="10185" LastEditDate="2015-05-31T20:54:57.190" LastActivityDate="2015-05-31T20:54:57.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32274" PostTypeId="1" CreationDate="2015-05-31T20:52:14.840" Score="1" ViewCount="95" Body="&lt;p&gt;A paragraph in my text reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonding involves the valence orbitals almost exclusively because these orbitals have the appropriate energies to interact strongly. Examine the electron energy-level diagram of a fluorine atom. The core $1 \mathrm{s}$ orbitals on a fluorine atom are at much lower energies than any of the $n = 2$ orbitals and do not participate effectively in bonding with other atoms. On the other hand, orbitals with $n &amp;gt; 2$ lie at considerably higher energy than the $n = 2$ orbitals. These orbitals can interact, but the resulting orbitals are too unstable to form strong covalent bonds. Therefore, the only orbitals of fluorine that form chemical bonds are the valence orbitals, those with $n = 2$. Typically the highest energy occupied orbitals are the ones used to bond with other atoms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In contrast this simple paragraph from SparkNotes expresses what I had believed to be the case&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The outermost orbital shell of an atom is called its valence shell, and the electrons in the valence shell are valence electrons. Valence electrons are the highest energy electrons in an atom and are therefore the most reactive. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I had thought that the outermost shell, the valence shell, the most energetic shell, and the shell involved in molecular bonds were all the same thing. However, the quoted paragraph of my text says otherwise. Where have I gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2016-04-02T01:07:38.847" Title="My text seems to say that the high(est) energy orbitals are not valence orbitals - everything else says otherwise. What's true?" Tags="&lt;electronic-configuration&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32275" PostTypeId="1" CreationDate="2015-05-31T21:55:40.483" Score="3" ViewCount="118" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_rate&quot; rel=&quot;nofollow&quot;&gt;reaction rate equation&lt;/a&gt; can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=r_0 K \prod_n a_n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $a_n$s are the activities of each reactant in the reaction equation (in the case where we find the reaction rate &lt;strong&gt;from reactants to products&lt;/strong&gt;, of course. Else the other way around). These activities are multiplied together. (This is often written as concentrations instead, I have seen, which seems to be an idealization.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is the equillibrium constant, $K=\exp\left(\frac{-\Delta G}{RT}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious as to what exactly the $r_0$ is? In the wikipedia article it is denoted $k_2$ and called the &quot;&lt;em&gt;rate constant for the second step.&lt;/em&gt;&quot; The second step here refers to the intermediate step at the actived-complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good, but &lt;em&gt;what is&lt;/em&gt; this quantity at this intermediate step? Is it a reaction rate from the activated-complex to the final product (in that case, how can this be different from the reaction rate from the reactants to this state)?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorDisplayName="user7951" LastEditDate="2015-11-20T18:28:09.023" LastActivityDate="2015-11-21T01:29:17.940" Title="What is the &quot;rate factor&quot; or &quot;second-step rate constant&quot; in the reaction rate equation?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32276" PostTypeId="1" AcceptedAnswerId="32280" CreationDate="2015-05-31T22:08:51.040" Score="1" ViewCount="5852" Body="&lt;p&gt;My text, and the other sources I've checked, include information on the effects, and kinds, of hybridized orbitals; however, they do not explain what properties of hybridized bonds conduce greater stability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does hybridization produce a (lower energy) more stable configuration? &lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="15929" LastEditDate="2015-06-01T00:10:41.023" LastActivityDate="2015-06-01T00:10:41.023" Title="Why does hybridization produce a more stable configuration?" Tags="&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32278" PostTypeId="1" AcceptedAnswerId="32282" CreationDate="2015-05-31T22:49:01.510" Score="1" ViewCount="533" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the specific heat of a $\pu{20 g}$ substance that releases $\pu{979 J}$ of heat when changing from $\pu{70 ^\circ C}$ to $\pu{20 ^\circ C}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started out with the following equation: $c = \frac{Q}{m\Delta T}$ and $Q = 979\ \mathrm{J}$ and $m = 20\ \mathrm{g}$ and $\Delta T = 45\ \mathrm{^\circ C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final answer was $2202.75\ \mathrm{J}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I use the right equations?&lt;/p&gt;&#xA;" OwnerUserId="16665" LastEditorUserId="5026" LastEditDate="2018-02-22T02:17:02.223" LastActivityDate="2018-02-22T02:17:02.223" Title="Calculate specific heat of a substance given other information" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32279" PostTypeId="1" CreationDate="2015-05-31T22:55:16.830" Score="0" ViewCount="12329" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the $\Delta H$ of the reaction where $2.6\:\mathrm{g}$ of water, $C_\mathrm{s}= 4.184 \mathrm{\frac{J}{g\:K}}$ is heated, raising the temperature increases from $298\:\mathrm{K}$ to $303\:\mathrm{K}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used the following equation: $\Delta H=mC_\mathrm{s}\Delta T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = 2.6~\mathrm{g}$, $C_\mathrm{s} = 4.184~\mathrm{J/g}$, $\Delta T= 5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I got: $(2.6)(4.184)(5)= 54.392~\mathrm{J/g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I solve this correctly?&lt;/p&gt;&#xA;" OwnerUserId="16665" LastEditorUserId="5026" LastEditDate="2018-03-09T06:33:19.807" LastActivityDate="2018-07-08T08:08:42.940" Title="How to calculate the enthalpy change of water?" Tags="&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32280" PostTypeId="2" ParentId="32276" CreationDate="2015-05-31T23:01:11.717" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What causes orbital hybridization?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hybridization only occurs in response to bond formation, that is an atom will not just spontaneously hybridize on its own.  Further, &lt;strong&gt;an atom will only hybridize if hybridization leads to the formation of a more stable (lower energy) molecule&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature is very efficient at exploring the energy surface and finding whatever energy minimum is available.  For example, ammonia ($\ce{NH3}$) is roughly a tetrahedron with $\ce{H-N-H}$ bond angles around 109.5° (the tetrahedral angle) using (roughly) $\mathrm{sp^3}$ hybridized orbitals. For ammonia, mixing (hybridizing) it's $\mathrm{2s}$ and three $\mathrm{2p}$ orbitals to form four $\mathrm{sp^3}$ hybridized orbitals that are used to form three $\ce{N-H}$ bonds and one lone pair orbital produces a more stable molecule than if it were not hybridized.  On the other hand, the analogue phosphine ($\ce{PH3}$) chooses to remain roughly unhybridized, using p-orbitals to form its $\ce{P-H}$ bonds with a resultant $\ce{H-P-H}$ bond angle around 90°.  Bond strengths, steric factors, etc. all play into determining whether a hybridized or unhybridized structure will be the more stable.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-05-31T23:22:59.027" LastActivityDate="2015-05-31T23:22:59.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32281" PostTypeId="2" ParentId="32255" CreationDate="2015-06-01T00:48:15.013" Score="2" Body="&lt;p&gt;Chlorine is the more electronegative species. It would thus retain the bonding electrons when the hydrogen is evicted. This hydrogen then forms a coordinate bond with the lone pair on nitrogen.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-06-01T00:48:15.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32282" PostTypeId="2" ParentId="32278" CreationDate="2015-06-01T00:57:07.647" Score="4" Body="&lt;p&gt;Your equation is correct, as are most of your measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit for $c$ is $\ce{Jg^{-1}K^{-1}}$ if using grams as mass, ($\ce{Jkg^{-1}K^{-1}}$ if you convert the mass to kilograms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your formula, however, does not give the answer you gave. When calculating using a calculator, consider that a parenthesis needs to be put around the denominator, as such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c = \frac{q}{(m\Delta T)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or else the calculator takes it as being $\ce{979J / 20g \times 45K}$ which gives the value you got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It needs to be $\ce{979J / (20g \times 45K)}$ which gives an answer that is much more reasonable.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-01T01:12:49.143" LastActivityDate="2015-06-01T01:12:49.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32283" PostTypeId="1" CreationDate="2015-06-01T01:12:50.063" Score="7" ViewCount="3252" Body="&lt;p&gt;How to calculate the $\mathrm{pH}$ of a $0.325~\mathrm{M}$ solution of $\ce{C5H5NHF}$? The $K_\mathrm{b}$ for $\ce{C5H5N}$ and $\ce{C5H5NH+}$ is $1.7\times 10^{-9}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my work, but it's wrong (at least I tried):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C5H5NHF -&amp;gt; C5H5NH+ + F- }$$&#xA;$$K_\mathrm{a}(\ce{HF})= 4.0\cdot10^{-4}$$&#xA;$$K=\frac{K_\mathrm{w}}{K_\mathrm{a}}$$&#xA;$$K=\frac{1\cdot10^{-14}}{4.0\cdot10^{-4}}=2.5\cdot10^{-11}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construct an ice table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|c|}\hline&#xA;&amp;amp;\ce{F-}&amp;amp;\ce{+H2O(l)-&amp;gt;}&amp;amp;\ce{HF}&amp;amp;+\ce{OH-}\\\hline&#xA;\text{I}&amp;amp;\pu{0.325M}&amp;amp;&amp;amp;0&amp;amp;0 \\\hline&#xA;\text{C}&amp;amp;-x&amp;amp;&amp;amp;x&amp;amp;x\\\hline&#xA;\text{E}&amp;amp;\pu{0.325M}-x&amp;amp;&amp;amp;x&amp;amp;x\\\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up the equation (small x-approximation):&#xA;$$2.5\cdot10^{-11} = \frac{x^2}{0.325-x}$$&#xA;Now, cross multiply:&#xA;$$x^2 = 5.87\cdot10^{-12}$$&#xA;Take the square root of both sides:&#xA;$$[\ce{OH-}] = x = 2.42\cdot10^{-6}$$&#xA;You want to get from $[\ce{OH-}]$ to $\mathrm{pOH}$:&#xA;$$\mathrm{pOH} = -\log\left(2.42\cdot10^{-6}\right) = 5.616$$&#xA;Then, from $\mathrm{pOH}$ to $\mathrm{pH}$:&#xA;$$\mathrm{pH} = 14 - 5.616 = 8.384$$ &#xA;Which is the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? How do I even use the $K_\mathrm{b}$ value in my calculation? &lt;/p&gt;&#xA;" OwnerUserId="16666" LastEditorUserId="5026" LastEditDate="2018-02-24T03:24:15.550" LastActivityDate="2018-02-24T03:24:15.550" Title="How to calculate the pH of a 0.325 M solution of pyridine hydrofloride?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32284" PostTypeId="1" CreationDate="2015-06-01T02:10:46.983" Score="11" ViewCount="735" Body="&lt;p&gt;There are two proposed mechanisms for Oxymercuration and demercuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first mechanism is concerted (i.e. single transition state connecting reactants and products) characterized by an attack of the $\pi$-bond on the $\ce{Hg(OAc)2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does mercuric acetate act as an electrophile in the first mechanism? It is in its most common 2+ oxidation state. Does this happen because mercury has empty $6d$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second mechanism is a stepwise mechanism (i.e. contains at least one intermediate species along the reaction pathway between reactants and products) that is characterized by incorporating the first ionization of $\ce{Hg(OAc)2}$ followed by an attack of the nucleophilic $\pi$-bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Hg(OAc)2}$ dissociates in water as follows:&#xA;$$\ce{Hg(OAc)2 + H2O&amp;lt;=&amp;gt;Hg+OAc + OAc-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is the actual mechanism?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="14" LastEditDate="2015-06-01T17:43:00.370" LastActivityDate="2015-06-02T11:39:48.027" Title="Oxymercuration and demercuration mechanism" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32285" PostTypeId="2" ParentId="32279" CreationDate="2015-06-01T02:55:01.473" Score="5" Body="&lt;p&gt;Yes, but your answer should be just in joules not $\mathrm{J/g}$, and your units are off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2.6~\mathrm{\color{\red}{g}})\left(4.184~\mathrm{\frac{J}{\color{\red}{g}\ K}}\right)(5~\mathrm{K})= 50~\mathrm{J}$$&lt;/p&gt;&#xA;" OwnerUserId="16667" LastEditorDisplayName="user7951" LastEditDate="2018-07-08T08:08:42.940" LastActivityDate="2018-07-08T08:08:42.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32286" PostTypeId="1" CreationDate="2015-06-01T04:28:39.560" Score="4" ViewCount="191" Body="&lt;p&gt;Let $R$ be the orthogonal matrix corresponding to an operation in $O(3)$. If&#xA;R is a proper rotation, then both vectors $\vec{V}$ and axial vectors $\vec{A}$ are transformed in the same way&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{V} \rightarrow  \vec{V}' = R\vec{V}$$&#xA;$$ \vec{A} \rightarrow  \vec{A}' = R\vec{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $R$ is an improper rotations, then&#xA;$$ \vec{V} \rightarrow  \vec{V}' = R\vec{V}$$&#xA;$$ \vec{A} \rightarrow  \vec{A}' = -R\vec{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let $\mathcal{V}$ be the vector representation of $O(3)$ and $\mathcal{P}$ be the axial-vector representation of $O(3)$. Thus we can write,&#xA;$$\vec{V} \rightarrow  \vec{V}' = \mathcal{V}(R)\vec{V}$$&#xA;$$\vec{A} \rightarrow  \vec{A}' = \mathcal{P}(R)\vec{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, considering $\mathcal{V}$ and $\mathcal{P}$ as representations of $C_\mathrm{4v}$ group, how can I find their composition in terms of irreps ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|ccccc|cc} \hline&#xA;C_\mathrm{4v} &amp;amp; E &amp;amp; 2C_4 &amp;amp; C_2 &amp;amp; 2\sigma_\mathrm{v} &amp;amp; 2\sigma_\mathrm{d} &amp;amp; &amp;amp; \\ \hline&#xA;\mathrm{A_1} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; z &amp;amp; x^2+y^2, z^2 \\&#xA;\mathrm{A_2} &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; -1 &amp;amp; R_z &amp;amp; \\&#xA;\mathrm{B_1} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; 1 &amp;amp; -1 &amp;amp; &amp;amp; x^2-y^2 \\&#xA;\mathrm{B_2} &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; -1 &amp;amp; 1 &amp;amp; &amp;amp; xy \\ &#xA;\mathrm{E} &amp;amp; 2 &amp;amp; 0 &amp;amp; -2 &amp;amp; 0 &amp;amp; 0 &amp;amp; (x,y),(R_x,R_y) &amp;amp; (xz,yz) \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="16668" LastEditorUserId="21296" LastEditDate="2017-07-05T23:05:44.500" LastActivityDate="2017-07-05T23:05:44.500" Title="Decomposition of axial vector and vector representions of C4v group" Tags="&lt;group-theory&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32287" PostTypeId="2" ParentId="31083" CreationDate="2015-06-01T04:51:08.847" Score="6" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.etymonline.com/index.php?term=bicarbonate&quot; rel=&quot;noreferrer&quot;&gt;Online Etymology Dictionary&lt;/a&gt;, the word &lt;strong&gt;bicarbonate&lt;/strong&gt; is derived as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question then becomes, what is the meaning of the prefix &lt;strong&gt;bi-&lt;/strong&gt; here?  We know it means &lt;em&gt;two&lt;/em&gt;, but it also can mean &lt;em&gt;double&lt;/em&gt; or &lt;em&gt;twice&lt;/em&gt;.  If that is the case, then sodium bicarbonate is sodium &lt;em&gt;double&lt;/em&gt; carbonate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$.  We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide.  The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2CO3 + 2HCl -&amp;gt; 2NaCl + CO2 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NaHCO3 + HCl -&amp;gt; NaCl + CO2 + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid.  In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid.  Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization.  That is, the compound is a &lt;em&gt;double carbonate.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar explanations make sense for bisulfate and bisulfite.  The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the &lt;em&gt;bi-&lt;/em&gt; versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen phosphates don't fit this pattern as nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double.  53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job.  This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2015-06-01T04:51:08.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32288" PostTypeId="1" CreationDate="2015-06-01T06:18:56.700" Score="1" ViewCount="690" Body="&lt;p&gt;Are esters soluble in Dichloromethane? I have a ketone with methyl group at one end and on the other end a big biphenyl group with a 6 carbon long aliphatic chain at 4'. I am planning to oxidise the keto group to acid using Baeyer Villiger oxidation but I am worried about the side product ester which may cause problem while separating my product.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-06-01T06:18:56.700" Title="Solublity of esters in Dichloromethane?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32289" PostTypeId="1" CreationDate="2015-06-01T07:05:28.720" Score="2" ViewCount="90" Body="&lt;p&gt;How to perform oxidation of keto group in presence of hydroxyl group. I fear there may be halogenation at $\ce{OH}$ group. Is there a way to protect the $\ce{OH}$ group ?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="4097" LastEditDate="2015-06-01T07:59:37.733" LastActivityDate="2015-06-01T07:59:37.733" Title="How to perform oxidation of keto group in presence of hydroxyl group?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;halides&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32290" PostTypeId="1" CreationDate="2015-06-01T07:44:21.680" Score="15" ViewCount="15713" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I understand that we need to supply energy to counter the nuclear attraction when we remove electrons, and that is the reason why ionization energy is endothermic. However, why does an atom release energy when we add an electron to it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With respect to ionization energy, why is energy required to remove electrons of elements like $\ce{Be}$, $\ce{Li}$ that naturally want to lose their electrons?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My teacher told me that elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="5026" LastEditDate="2018-03-21T09:34:55.153" LastActivityDate="2021-04-22T10:17:48.167" Title="Why is electron gain generally exothermic?" Tags="&lt;electron-affinity&gt;&lt;ionization-energy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32291" PostTypeId="1" AcceptedAnswerId="32296" CreationDate="2015-06-01T08:22:49.727" Score="4" ViewCount="17186" Body="&lt;p&gt;We dissolve $10\ \mathrm{g}$ of $\ce{CuCl2}$ in $0.1\ \mathrm{L}$ of water.&lt;br&gt;&#xA;Given that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(\ce{Cl})=35.5\ \mathrm{g/mol}$&lt;br&gt;&#xA;$M(\ce{Cu})=63.5\ \mathrm{g/mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all I calculated the the quantity of matter of $\ce{CuCl2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n(\ce{CuCl2})=0.074\ \mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The the molar concentration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c=\frac{0.074\ \mathrm{mol}}{0.1\ \mathrm{L}}=0.74\ \mathrm{mol/L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I don’t know how can I calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$ because I don’t know the quantity of matter of them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here im blocked!&lt;/p&gt;&#xA;" OwnerUserId="16672" LastEditorDisplayName="user7951" LastEditDate="2015-06-01T08:44:26.890" LastActivityDate="2015-06-01T09:53:31.890" Title="CuCl2 dissolved in water" Tags="&lt;aqueous-solution&gt;&lt;ions&gt;&lt;solutions&gt;&lt;concentration&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32292" PostTypeId="2" ParentId="32290" CreationDate="2015-06-01T08:33:24.670" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Energy of electron in vacuum is zero. If it can attach (however weakly), the energy is gained and process is therefore exotermic. It is not an atom which releases energy, it is the whole system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In some cases, adding of extra electron to neutral atom is not an exothermic process. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Template:Periodic_table_%28electron_affinities%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic table of electron affinities&lt;/a&gt;. For example, a helium atom has filled shell with $n=1$. The next electron would be accepted to the shell $n=2$ which has much higher energy, and therefore is not stable. For the half-filled orbitals, similar reasoning applies, but spicing it with &lt;a href=&quot;https://en.wikipedia.org/wiki/Hund%27s_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hund's rule&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="5026" LastEditDate="2018-04-12T04:10:24.927" LastActivityDate="2018-04-12T04:10:24.927" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32294" PostTypeId="2" ParentId="32291" CreationDate="2015-06-01T09:20:05.497" Score="4" Body="&lt;p&gt;When a salt is fully dissociated, than the atoms in a solution exist separately:&#xA;$$\ce{CuCl2&amp;lt;=&amp;gt;Cu^{2+}}+2\ce{Cl-}$$&#xA;Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n=\frac{m}{M}$$&#xA;$$ n \cdot M =m$$&#xA;Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g} $$ and the rest is formed by chlorine. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$ &lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-06-01T09:53:31.890" LastActivityDate="2015-06-01T09:53:31.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32295" PostTypeId="1" AcceptedAnswerId="32298" CreationDate="2015-06-01T09:20:11.493" Score="9" ViewCount="1498" Body="&lt;p&gt;PDB protein crystal structures most of the time lack hydrogen atoms. Usually I use &lt;a href=&quot;https://pymol.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PyMOL&lt;/a&gt; (&lt;code&gt;h_add&lt;/code&gt;) but it does not always seem to be giving the correct result. What can I use for accurate hydrogen addition? &lt;/p&gt;&#xA;" OwnerUserId="14877" LastEditorUserId="41328" LastEditDate="2017-08-14T18:54:25.243" LastActivityDate="2019-02-06T20:55:57.317" Title="How to add hydrogens to a crystal structure?" Tags="&lt;biochemistry&gt;&lt;crystal-structure&gt;&lt;software&gt;&lt;hydrogen&gt;&lt;crystallography&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32296" PostTypeId="2" ParentId="32291" CreationDate="2015-06-01T09:21:48.437" Score="4" Body="&lt;p&gt;Jaroslav already gave you a good hint: Assume full dissociation, which leads to the following equation: $$\ce{CuCl2 &amp;lt;=&amp;gt; Cu^2+ + 2Cl^-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you know the initial concentration of copper chloride, $c(\ce{CuCl2})=0.74~\mathrm{mol/L}$, what can you say about the ratio of $n(\ce{CuCl2})/n(\ce{Cu^2+})$ and $n(\ce{CuCl2})/n(\ce{Cl^-})$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: In one mole of of a salt of the composition $\ce{AB}$, you will find one mole of $\ce{A+}$ and one mole of $\ce{B-}$, according to $\ce{AB &amp;lt;=&amp;gt; A+ + B-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{CuCl2})/n(\ce{Cu^2+}) = 1$, thus $c(\ce{CuCl2})=n(\ce{Cu^2+})=0.74~\mathrm{mol/L}$&lt;br&gt;&#xA; and&lt;br&gt;&#xA;$n(\ce{CuCl2})/n(\ce{Cl^-}) =\frac12$, thus $c(\ce{CuCl2})=2\cdot c(\ce{Cl^-})=1.48~\mathrm{mol/L}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-01T09:21:48.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32297" PostTypeId="1" AcceptedAnswerId="32328" CreationDate="2015-06-01T10:06:28.543" Score="2" ViewCount="2754" Body="&lt;p&gt;I want to know about the reaction mechanism of haloform reaction in acidic condition. I am following a persons doctoral thesis where he used acid medium haloform reaction but unfortunately he has not given any references. &lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="189" LastEditDate="2015-06-01T22:14:03.167" LastActivityDate="2015-06-01T22:14:03.167" Title="Haloform reaction in acidic condition" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32298" PostTypeId="2" ParentId="32295" CreationDate="2015-06-01T10:10:49.057" Score="9" Body="&lt;p&gt;I've used &lt;a href=&quot;http://kinemage.biochem.duke.edu/software/reduce.php&quot; rel=&quot;noreferrer&quot;&gt;Reduce&lt;/a&gt; before, which is excellent. I'm not sure what you mean by ' does not always seem to be giving the correct result', though? Perhaps you could file a bug report with the PyMOL team if you think there is a bug in &lt;em&gt;h_add&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4665" LastActivityDate="2015-06-01T10:10:49.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32299" PostTypeId="1" AcceptedAnswerId="32300" CreationDate="2015-06-01T10:47:16.237" Score="1" ViewCount="44" Body="&lt;p&gt;When is described: &quot;In &lt;strong&gt;normal conditions&lt;/strong&gt;, viscose absorbs 11 to 14% of water vapour. In liquid water, they swell and can absorb 80 to 120% of water.&quot; (Clothing Technology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they mean with 'normal conditions'? In open air?&lt;/p&gt;&#xA;" OwnerUserId="2657" LastEditorUserId="865" LastEditDate="2015-10-30T16:54:08.703" LastActivityDate="2015-10-30T16:54:08.703" Title="Vapour absorption in 'normal conditions'" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32300" PostTypeId="2" ParentId="32299" CreationDate="2015-06-01T10:56:26.523" Score="1" Body="&lt;p&gt;Normal or &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot; rel=&quot;nofollow&quot;&gt;standard conditions&lt;/a&gt; are usually ment pressure of 101.325 kPa and temperature of 298.15 K or 293.15 K. Please see the link for reference. &lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-06-01T10:56:26.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32301" PostTypeId="2" ParentId="32170" CreationDate="2015-06-01T11:05:01.013" Score="9" Body="&lt;p&gt;The answer is merely a matter of definition of &quot;valence electrons&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many texts define &quot;valence electrons&quot; in a way that explicitly includes d electrons of unfilled (but not filled) d subshells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: Chemistry structure and dynamics by Spencer &lt;em&gt;et al.&lt;/em&gt; at page 125:&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We can define valence electrons as electrons on an atom that are not present in the previous rare gas, ignoring &lt;strong&gt;filled&lt;/strong&gt; d or f subshells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Many books published in the last 10 years use this definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That d electrons may be valence electrons is also supported by the &lt;a href=&quot;http://en.wikipedia.org/wiki/18-Electron_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;18-Electron rule&lt;/a&gt; (at least to the extent that there is such a rule).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] James N. Spencer, George M. Bodner, Lyman H. Rickard: Chemistry: Structure and Dynamics, 5th Edition. John Wiley &amp;amp; Sons: 2010. ISBN: &lt;a href=&quot;https://www.wiley.com/en-us/Chemistry%3A+Structure+and+Dynamics%2C+5th+Edition-p-9780470587119&quot; rel=&quot;nofollow noreferrer&quot;&gt;978-0-470-58711-9&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2018-09-27T13:27:08.510" LastActivityDate="2018-09-27T13:27:08.510" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32302" PostTypeId="1" AcceptedAnswerId="32303" CreationDate="2015-06-01T11:08:28.483" Score="5" ViewCount="710" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pLPjD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed the the $\ce{TsOH}$ would dissociated to produce hydronium ions that would protonate the carbonyl. This would make the carbon more delta + and so susceptible to attack by $\ce{MeOH}$. After proton exchange I thought the $\ce{H2O}$ group would leave and we would have a carbocation left over. I can't work out how to resolve this carbocation with the changes in double bonds.&lt;/p&gt;&#xA;" OwnerUserId="11592" LastEditorUserId="4325" LastEditDate="2015-06-02T12:22:24.267" LastActivityDate="2015-06-02T12:22:24.267" Title="Mechanism of enol ether formation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32303" PostTypeId="2" ParentId="32302" CreationDate="2015-06-01T11:27:49.227" Score="7" Body="&lt;p&gt;Based on what you have in the question, you are good up until formation of the carbocation by loss of a water leaving group. It looks like you are missing one of two things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The carbocation is resonance stabilized.&lt;/li&gt;&#xA;&lt;li&gt;One of the carbon-carbon double bonds in the product is new.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The scheme below has two of the major resonance contributors to that carbocation. Perhaps one of give you some insight on how to show the formation of the second alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/axrcC.png&quot; alt=&quot;a picture whereby two resonance structures of a carbocation are shown in order to help the asker see how to get to the product in a mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-01T11:27:49.227" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32304" PostTypeId="1" AcceptedAnswerId="32345" CreationDate="2015-06-01T11:50:14.493" Score="7" ViewCount="1601" Body="&lt;p&gt;If you have the forward reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2X -&amp;gt;[K] P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which of the following systems of differential equations would model the reaction's kinetics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl} \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &amp;amp;= -2K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &amp;amp;= K[\ce{X}]^2\end{array} \tag{system 1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{rl}  \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &amp;amp;= -K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &amp;amp;= K[\ce{X}]^2\end{array} \tag{system 2}$$&lt;/p&gt;&#xA;" OwnerUserId="16674" LastEditorUserId="30836" LastEditDate="2017-10-25T16:49:43.233" LastActivityDate="2017-10-25T16:49:43.233" Title="Law of mass action" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32306" PostTypeId="1" AcceptedAnswerId="32313" CreationDate="2015-06-01T13:37:57.733" Score="38" ViewCount="10561" Body="&lt;p&gt;Every covalent bond has some ionic character and every ionic bond some covalent character. I can understand why a completely ionic bond is an ideal situation. But completely covalent bonds can exist(?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the case of $\ce{H_2}$ or any other diatomic molecule between identical atoms. How can these have any ionic character?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="10185" LastEditDate="2015-06-01T14:56:39.920" LastActivityDate="2015-06-08T15:47:26.613" Title="Can 100% covalent bonds exist?" Tags="&lt;bond&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32309" PostTypeId="2" ParentId="26617" CreationDate="2015-06-01T14:35:31.010" Score="-2" Body="&lt;p&gt;Acc. to IUPAC we have to give lowest locant no. to side chain which in this case is 2-Bromo-2,4-dimethyl.&#xA;Another reason of this ans. being correct is that we take halides as a normal side chain and it is not given priority over methyl, we just only see lowest locant in this case.&lt;/p&gt;&#xA;" OwnerUserId="16681" LastActivityDate="2015-06-01T14:35:31.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32311" PostTypeId="2" ParentId="32096" CreationDate="2015-06-01T14:41:58.493" Score="10" Body="&lt;p&gt;If you're going to be picky that a &quot;complex&quot; doesn't form without covalent coordination bonds, then that's your answer. Alkali metals will tend to form cationic salts with EDTA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I agree with the other comments. EDTA does form complexes with alkali metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd classify the disodium salt of EDTA as a complex. It has a metal and an organic ligand, and there's no direct metal-carbon bond (which would make it organometallic). As discussed in another question, it's a pretty arbitrary distinction between an ionic bond and a covalent one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/19568/5017&quot;&gt;What happens if the electronegativity difference is exactly 2.1?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This distinction is &lt;em&gt;particularly&lt;/em&gt; cloudy in the case of metal complexes, since most metals readily gain positive charge.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-06-01T14:41:58.493" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32312" PostTypeId="2" ParentId="32096" CreationDate="2015-06-01T15:03:53.330" Score="7" Body="&lt;p&gt;Alkali metal cations are very &quot;hard&quot;, in the &lt;a href=&quot;http://en.wikipedia.org/wiki/HSAB_theory&quot;&gt;HSAB theory&lt;/a&gt; of acid-base interactions (roughly corresponding to 'poorly polarizable'), and thus only coordinate in the conventional sense of the term to &quot;hard&quot; ligands such as ethers.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Crown_ether&quot;&gt;Crown ethers&lt;/a&gt;, in particular, are known for their alkali metal ligating/chelating capability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amine nitrogens and carboxyl oxygens of EDTA are sufficiently 'soft' that, as Jan noted in a comment to the question, there is sufficiently little energetic benefit derived from EDTA chelation of a 'very hard' alkali cation such that the complexes are not practically observed.  Instead, the chemistry is dominated by electrostatic interactions of the positively-charged alkali cations with the negatively-charged carboxylate groups (when deprotonated, anyways).&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-06-01T15:03:53.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32313" PostTypeId="2" ParentId="32306" CreationDate="2015-06-01T15:42:47.620" Score="33" Body="&lt;p&gt;According to Pauling's famous &lt;a href=&quot;https://books.google.com/books?id=L-1K9HmKmUUC&amp;amp;printsec=frontcover&amp;amp;dq=pauling+nature+of+the+chemical+bond&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=rXpsVfLmH-OCsQSF0oOYAg&amp;amp;ved=0CB0Q6AEwAA#v=onepage&amp;amp;q=covalent&amp;amp;f=false&quot;&gt;The Nature of the Chemical Bond&lt;/a&gt; , 3rd edition, at page 73:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the hydrogen molecule a quantum-mechanical treatment has shown that the two ionic structures $\ce{H+H-}$ and $\ce{H- H+}$ enter into resonance with the extreme covalent structure $\ce{H-H}$ only to a small extent, each ionic structure contributing only about 2 percent to the normal state of the molecule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;He goes on to state that other diatomics such as chlorine should have even less ionic character.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all molecules, even homonuclear diatoms, have some ionic character. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information on diatomic hydrogen's ionic character see section 1.5 of the book and &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/1/8/10.1063/1.1749333&quot;&gt;The Normal State of the Hydrogen Molecule&lt;/a&gt; &lt;em&gt;Journal of Chemical Physics&lt;/em&gt; vol. 1, pages 593-596.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/23/1/10.1063/1.1740522&quot;&gt;Overlaps of Trial Functions for the Hydrogen Molecule. II. Covalent and Ionic Character of H2&lt;/a&gt; J. Chem. Phys. 23, 176 &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2015-06-01T19:20:12.543" LastActivityDate="2015-06-01T19:20:12.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32314" PostTypeId="1" AcceptedAnswerId="32315" CreationDate="2015-06-01T16:52:18.727" Score="7" ViewCount="8527" Body="&lt;p&gt;In a &lt;a href=&quot;http://m.bioinformatics.oxfordjournals.org/content/24/13/i196.full&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt; I have read the term GMEC, which was not further explained.&#xA;GMEC stands for Global Minimum Energy Conformation. But what is it? &#xA;Is it just the conformation with the lowest energy?&lt;/p&gt;&#xA;" OwnerUserId="14877" LastActivityDate="2017-08-01T17:36:26.950" Title="What is GMEC (Global Minimum Energy Conformation)?" Tags="&lt;computational-chemistry&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32315" PostTypeId="2" ParentId="32314" CreationDate="2015-06-01T16:58:28.323" Score="9" Body="&lt;p&gt;Yes. It is the conformer with the lowest energy out of all possible conformers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also local minima which have a lower energy than conformations either side of the them on the energy diagram but do not have the lowest possible energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cgiin.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this diagram showing the conformers of butane we can see that all three stable conformers; the two gauche conformers and the anti conformer are at local energy minima but the anti conformer is the global minimum.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-01T16:58:28.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32317" PostTypeId="1" AcceptedAnswerId="32323" CreationDate="2015-06-01T17:54:15.280" Score="12" ViewCount="211" Body="&lt;p&gt;The context of my question is rotational spectroscopy (using microwave radiation), where there is the $J = \pm 1$ selection rule. The way I understand spectroscopy is that the sample is irradiated with a range of frequencies. Is it possible, during the irradiation of the sample (say CO), for the molecule to be excited by 2 levels by 2 successive photons of the appropriate frequencies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this does not affect the locations of the peaks per se, because the relaxation of course has to happen one level at a time. I am specifically wondering about the distribution of peaks, which is supposed to follow the Boltzmann distribution. Let us say, for example, we have a sample at absolute zero, where all the molecules are at the ground state $J = 0$. Is it possible for a molecule to be excited twice (or more), leading to the rotational spectrum showing more than one peak?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2017-03-28T15:52:07.420" LastActivityDate="2017-03-28T16:12:20.500" Title="In spectroscopy, is it possible for the sample to be excited multiple times?" Tags="&lt;spectroscopy&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32318" PostTypeId="1" AcceptedAnswerId="32492" CreationDate="2015-06-01T18:24:56.893" Score="14" ViewCount="2173" Body="&lt;p&gt;So, I've been looking into polymer chemistry to model breaking polymers for a biological model, and I've been a bit stumped by the concept of persistence length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if a polymer is significantly longer than its persistence length, it needs to be treated as though the ends are completely independent of each other thanks to thermal fluctuations bending it all over (the analogy I have is treating the polymer like a piece of cooked spaghetti). If I model a polymer like this, I need to use the words &quot;3 dimensional random walk&quot; to describe its shape mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that if a polymer is significantly shorter than its persistence length, it can be treated like a rigid rod that might be a bit bendy, like a steel rod. This is significantly easier to model: I can just treat it as a line/arc that moves through space as a single particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens if a polymer's length is about equal to its persistence length? How can the polymer be modeled then? What precisely does it mean for the polymer's movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I've been debating the merits of just treating it like a flexible rod (like the significantly shorter than persistence length example) and calling it good just because of model simplification. If I do that, how off would I be from reality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2: I've been doing further research and found something called a &quot;worm like chain&quot; model. I'm pretty fuzzy on what it means for the fiber, and I'm having trouble grasping how the fiber would move from an intuitive perspective; the visual for how I'd need to treat it isn't working, and the visual would help immensely with the actual modeling for the rest of the equations I need. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-06-05T12:21:17.923" LastActivityDate="2015-06-07T03:05:22.557" Title="What happens if a polymer's length is equal to its persistence length?" Tags="&lt;computational-chemistry&gt;&lt;materials&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32319" PostTypeId="1" AcceptedAnswerId="32326" CreationDate="2015-06-01T18:37:18.933" Score="1" ViewCount="654" Body="&lt;p&gt;In my experience, most texts that address hypervalency say that it only occurs from elements in the 3rd period and onwards. This explains the occurrence of $\ce{Cl2O7}$ or chlorine heptoxide. However, some 2nd period nonmetals like $\ce{C}$ and $\ce{O}$ show hypervalency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH5}$ - This is unlikely to occur but it does sometimes happen that carbon bonds to 5 atoms instead of 4.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H3O+}$ - Here oxygen is hypervalent.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How is it possible for carbon and oxygen to each have 9 electrons if each orbital only holds 2 electrons? Do they switch between electrons or something?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastEditorUserId="7475" LastEditDate="2016-10-18T00:00:13.473" LastActivityDate="2016-10-18T00:00:13.473" Title="Hypervalency in elements in the second period" Tags="&lt;electrons&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32322" PostTypeId="1" AcceptedAnswerId="32330" CreationDate="2015-06-01T19:30:31.107" Score="9" ViewCount="5962" Body="&lt;p&gt;Somebody I know insists that the use of a catalyst increases the atom economy. They did chemistry at school and were told that a catalyst increases the atom economy. He pointed me to several past exams where this was written as a valid answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am convinced that this is false. After all, atom economy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{atom economy} = \frac{\text{molecular mass of desired product}}{\text{molecular mass of all reactants}} \times 100\%$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel this has nothing to do with a catalyst. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="4925" LastEditorUserId="402" LastEditDate="2019-10-08T23:01:13.210" LastActivityDate="2019-10-08T23:02:23.220" Title="Does a catalyst increase atom economy?" Tags="&lt;physical-chemistry&gt;&lt;catalysis&gt;" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32323" PostTypeId="2" ParentId="32317" CreationDate="2015-06-01T20:05:40.403" Score="12" Body="&lt;p&gt;Very technically? Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically? The probability is small enough that even if it does happen, the peaks for the multiple transitions are going to be small enough that we cannot really observe them on the spectrum. The lifetime of a given excited state is so small compared to the analogous time in the ground state that it can basically be considered zero for the situation you're talking about. I'm sure it has &lt;em&gt;happened&lt;/em&gt;, but compared to the one at a time transitions, your spectrum isn't really going to show it.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-01T20:05:40.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32324" PostTypeId="1" CreationDate="2015-06-01T20:26:50.420" Score="4" ViewCount="621" Body="&lt;p&gt;I am trying to repeat an experiment conducted by FDA on Malachite Green (MG).&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://wayback.archive-it.org/7993/20180424205356/https://www.fda.gov/downloads/Food/FoodScienceResearch/ucm071564.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Prepare a series of LC calibrants by aliquoting into individual 15 mL graduated &#xA;glass or disposable polypropylene centrifuge tubes, 50 µL, 100 µL, and 200 µL&#xA;of the 0.1 µg/mL MG2 solution and 50 µL and 100 µL of the 1.0 µg/mL MG1 solution.&#xA;Dilute each to 5.0mL with (1:1 vol) acetonitrile/ammonium acetate buffer&#xA;(mobile phase A), vortex mix or stopper and invert to mix thoroughly. &#xA;Based on a 5 g sample weight, these solutions will generate standards with&#xA;concentrations of 1, 2, 4, 10 and 20 ng/g of MG. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;If I am not wrong, Malachite Green used in this paper is from the salt, Malachite Green Oxalate.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is Acetonitrile used here as a pH buffer?&lt;/li&gt;&#xA;&lt;li&gt;What is the role of Ammonium acetate?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;If I plan on using Raman Spectroscopy to detect MG in my samples, shouldn't I be using the pure forms of MG as my reference signal? If so, wouldn't the addition of these two chemicals affect my readings?&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;References&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The pdf document originally linked in this post (see edits) is no longer available.&#xA;The most similar reference identified (available &lt;a href=&quot;http://wayback.archive-it.org/7993/20180424205356/https://www.fda.gov/downloads/Food/FoodScienceResearch/ucm071564.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; via the Internet Wayback Machine) is:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Wendy C. Andersen, José E. Roybal, Sherri B. Turnipseed. Determination of Malachite Green and Leucomalachite Green in Salmon with In-Situ Oxidation and Liquid Chromatography with Visible Detection. Laboratory Information Bulletin (LIB) 4334: Malachite Green and Leucomalachite Green in Salmon. Volume 20, No. 11, November 2004.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19539" LastEditorUserId="69830" LastEditDate="2021-05-01T08:57:52.060" LastActivityDate="2021-05-01T08:57:52.060" Title="Role of Acetonitrile and Ammonium Acetate buffer in Leucomalachite Green stock solution?" Tags="&lt;analytical-chemistry&gt;&lt;raman&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32325" PostTypeId="1" CreationDate="2015-06-01T20:39:16.520" Score="12" ViewCount="24007" Body="&lt;p&gt;The shapes respectively denoted by '&lt;strong&gt;tetrahedron&lt;/strong&gt;' and '&lt;strong&gt;trigonal pyramid&lt;/strong&gt;' seem to be the same. Is there a difference between the two? If not, why are the two presented as different concepts in introductory chemistry courses?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorDisplayName="user15489" LastEditDate="2015-07-04T11:17:41.277" LastActivityDate="2015-07-04T11:17:41.277" Title="What's the difference between a tetrahedron and a trigonal pyramid?" Tags="&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32326" PostTypeId="2" ParentId="32319" CreationDate="2015-06-01T20:50:16.740" Score="7" Body="&lt;p&gt;Normally when we talk about a single covalent bond, we are referring to a 2-centre 2-electron bond, which means that there are two electrons holding two atoms together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon never forms 5 bonds. The only exception that I know of is the $\ce{CH5+}$ &lt;a href=&quot;http://en.wikipedia.org/wiki/Methanium&quot; rel=&quot;nofollow&quot;&gt;methanium cation&lt;/a&gt;, the bonding in which can be explained by a 3-centre-2-electron bond. The same kind of bond appears in &lt;a href=&quot;http://en.wikipedia.org/wiki/Diborane&quot; rel=&quot;nofollow&quot;&gt;diborane&lt;/a&gt; ($\ce{B2H6}$). In both cases, the octet rule (or duplet rule in the case of the bridging hydrogens in diborane) is not violated. It is just that those 2 electrons are shared amongst 3 different atoms, so each &quot;bond&quot; is effectively &lt;em&gt;half&lt;/em&gt; a bond (in MO theory parlance we say that the bond order is 0.5). You could think of it as three of the C-H bonds being normal 2-electron bonds, and two of the C-H bonds being &lt;em&gt;half&lt;/em&gt;-bonds (having one electron each). The total number of electrons around carbon is therefore $3 \times 2 + 1 + 1 = 8$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutral species $\ce{CH5}$ does not exist, because it has one more electron than the $\ce{CH5+}$ cation. That would mean that you either have to put 9 electrons around carbon, or put 3 electrons around hydrogen, both of which are of course not allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydronium ion $\ce{H3O+}$ is not actually hypervalent. It is similar to the ammonium ion $\ce{NH4+}$ in that a dative bond is formed from the lone pair on O to a $\ce{H+}$ ion.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-10-17T14:31:22.237" LastActivityDate="2016-10-17T14:31:22.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32327" PostTypeId="2" ParentId="32325" CreationDate="2015-06-01T21:03:13.640" Score="13" Body="&lt;p&gt;In a rigorous geometrical sense, there is no difference between &lt;em&gt;tetrahedron&lt;/em&gt; and &lt;em&gt;trigonal pyramid&lt;/em&gt;--the terms both mean the &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron&quot;&gt;same thing&lt;/a&gt;. In colloquial and chemical use, however, 'tetrahedron' typically implies the '&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahedron#Formulas_for_a_regular_tetrahedron&quot;&gt;regular tetrahedron&lt;/a&gt;', where all four faces are equilateral triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemically speaking, when referring to these two shapes as descriptors of molecular geometries, there is (usually) a central atom in addition to the four atoms at the vertices.  In the symmetrically-proper regular-tetrahedral geometry, the central atom is inside the solid volume of the tetrahedron and equidistant from all of the four vertex atoms.  In the trigonal pyramidal geometry, the central atom can be located inside the solid volume, contained within a plane of the trigonal pyramid/tetrahedron, or I believe even outside the tetrahedral volume.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-02T00:57:58.757" LastActivityDate="2015-06-02T00:57:58.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32328" PostTypeId="2" ParentId="32297" CreationDate="2015-06-01T22:12:44.057" Score="6" Body="&lt;p&gt;In the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0198503466&quot; rel=&quot;noreferrer&quot;&gt;book&lt;/a&gt; by Clayden et al. you can find a very nice description of the mechanism of the halogenation of enols and the haloform reaction on pp. 533-536. Especially on p. 536 the authors give the mechanistic details of the acid catalysed halogenation of enols and some reasons why multiple halogenation steps on the same carbon atom (as happens in the haloform reaction) occur easily under base catalysis but do usually not occur under acidic conditions. Since I lack the time to write it up myself (and since the authors already did a marvelous job at describing it) I simply post the relevant passage here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MIxpo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="189" LastActivityDate="2015-06-01T22:12:44.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32329" PostTypeId="2" ParentId="32284" CreationDate="2015-06-01T22:15:59.670" Score="2" Body="&lt;p&gt;I don't know the answer (and I thought for a while that it was not knowable), but I can at least tell you how you would find out whether $\ce{Hg(OAc)2}$, $\ce{Hg(OAc)+},$ or (more likely) some combination is the active electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{Hg(OAc)2}$ dissociates, then both possible electrophiles are in solution. However, if we can effectively remove one electrophile, and see if the reaction works, then we know something.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Easy: shift equilibrium to favor $\ce{Hg(OAc)2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can shift the equilibrium to heavily favor $\ce{Hg(OAc)2}$ by adding a large excess of $\ce{OAc-}$ ions, for example as $\ce{NaOAc}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two possible outcomes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The reaction does not work (or goes very slowly). If this is the case, then $\ce{Hg(OAc)+}$ is the active electrophile.&lt;/li&gt;&#xA;&lt;li&gt;The reaction does work. We know that $\ce{Hg(OAc)2}$ can be the electrophile, but this does not rule out $\ce{Hg(OAc)+}$ as an electrophile.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hard: Shift equilibrium to favor $\ce{Hg(OAc)+}$ &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not coming up wit ha good way to do this, but if we could, then we could test whether $\ce{Hg(OAc)+}$ was necessary.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-01T22:15:59.670" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32330" PostTypeId="2" ParentId="32322" CreationDate="2015-06-01T22:54:43.433" Score="11" Body="&lt;p&gt;There are a couple of ways to think about this. First, considering your atom economy equation. Typically a catalyst is used in place of a stoichiometric reagent. A catalyst is (strictly speaking) not a reagent, so its effect on the atom economy is neutral. The alternative stoichiometric reagent is a reactant, so it will reduce the atom economy. This leads to the conclusion that a catalytic method offers a better atom economy than an alternative, non-catalytic method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a slightly different point of view, atom economy is one aspect of measuring the amount of waste that comes out of a reaction. A more relevant quantity is really the ratio of the mass of the desired product(s) to the total mass of all of the reaction inputs (including reactants, catalysts, solvents, purification materials, etc). Since catalysts can be used in smaller amounts than stoichiometric reagents, there will physically be less mass put into the process, which increases that ratio.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="402" LastEditDate="2019-10-08T23:02:23.220" LastActivityDate="2019-10-08T23:02:23.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32331" PostTypeId="2" ParentId="32324" CreationDate="2015-06-01T22:59:46.273" Score="4" Body="&lt;p&gt;Since they're using ACN/ammonium acetate to elute from solid phase extraction cartridges and that's what they're injecting into their LC analysis, they probably want to use the same mixture to avoid problems with unmatched matrices with their standards. The ammonium acetate is the pH buffer. Malachite green has different protonation states at different pH, so it's important to set the pH properly for elution from the SPE and LC analysis. MG is fairly soluble in water, but still has a lot of organic functionality, so adding an organic solvent would help elute from the SPE column and keep the analyte from being too strongly retained by the LC stationary phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for raman analysis, water has rather weak raman scattering, so it's not usually a problem. &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C75058&amp;amp;Mask=800#ESpecVS&quot; rel=&quot;nofollow&quot;&gt;Acetonitrile&lt;/a&gt; has one very strong peak at 2942 $\mathrm{cm^{-1}}$ and a few moderate peaks, so it might be an issue. Since you're not doing LC, you can likely prepare standards in buffer only. Unfortunately, ammonium acetate is also raman-active, but if you're not doing LC, you might be able to do without as well. Fortunately, one of the great powers of IR and raman spectroscopies is that there is often enough bandwidth to find one or two non-overlapping peaks in a complex mixture and use that for quantitation, so it may still work as written. They're both also very amenable to chemometric techniques like partial least squares when measuring complex samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big problem is going to be sensitivity if you're interested in the concentrations they use in the paper.. Normal raman is not that sensitive and the smallest standard is only ~3 nM, so it's probably going to be hopeless. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0308814614011777&quot; rel=&quot;nofollow&quot;&gt;These guys&lt;/a&gt; do surface-enhaced resonance raman and are able to get to about 1 nM, and the raman setup's not actually that hard since MG has a huge absorbance at 620 nm, right where HeNe lasers emit. Unfortunately, MG doesn't fluoresce much in solution or it would be very easy to measure.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-01T22:59:46.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32332" PostTypeId="2" ParentId="32232" CreationDate="2015-06-02T00:06:53.940" Score="8" Body="&lt;p&gt;Firstly, a note about terminology. As Mithoron noted, the naked ions $\ce{V^5+}$ and $\ce{Cr^6+}$ do not exist because they require a prohibitively large amount of energy to make (just take the sum of the first 5 or 6 ionisation energies of V/Cr to see why). As such your question should really be asking about the vanadium(V) and chromium(VI) oxidation states, which can exist in acidic solution as the oxoanions $\ce{VO2+}$ and $\ce{Cr2O7^2-}$ respectively. The same idea can be extended to the manganese(VII) oxidation state, which exists as $\ce{MnO4^-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on to the question itself. As you rightly note, the central atom in these oxidation states have the same electronic configuration as argon; we usually call it a $\mathrm{d^0}$ configuration to indicate that there are no d electrons. The colours of these species therefore cannot arise from d-d transitions, which are the transfer of electrons from one metal orbital to another metal orbital. They are actually associated with electron transfer from the &lt;em&gt;ligand&lt;/em&gt; (in this case, $\ce{O^2-}$) to the central metal atom. The name for this is &lt;a href=&quot;http://en.wikipedia.org/wiki/Charge-transfer_complex&quot; rel=&quot;nofollow&quot;&gt;ligand-metal charge transfer (LMCT)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2015-10-28T15:59:06.230" LastActivityDate="2015-10-28T15:59:06.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32333" PostTypeId="1" CreationDate="2015-06-02T04:44:39.270" Score="1" ViewCount="66" Body="&lt;p&gt;What is exactly the difference between these two? I think melting point is when the compound melts down to visually liquid state but actually LC state and latter is when polymer becomes completely isotropic liquid.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-06-02T04:44:39.270" Title="Melting temperature vs Isotropisation temperature in liquid crystals" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;&lt;liquid-crystals&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32334" PostTypeId="1" CreationDate="2015-06-02T04:45:57.433" Score="2" ViewCount="425" Body="&lt;p&gt;What is the most accepted theory for explaining the &lt;em&gt;odd even effect&lt;/em&gt; in liquid crystals? Can anyone explain and refer the corresponding paper?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorUserId="4945" LastEditDate="2015-06-02T06:41:24.040" LastActivityDate="2015-06-02T06:41:24.040" Title="What is the explanation of the odd even effect in liquid crystals?" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;&lt;reference-request&gt;&lt;liquid-crystals&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32336" PostTypeId="1" AcceptedAnswerId="32394" CreationDate="2015-06-02T07:37:51.207" Score="2" ViewCount="852" Body="&lt;p&gt;I read everywhere about the explosive reactivity of perchlorate salts (e.g., $\ce{Na}$, $\ce{Ca}$, $\ce{NH4}$ [especially]) when mixed with organic materials or solvents, yet I find no information about its reactivity with methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone comment please?&lt;/p&gt;&#xA;" OwnerUserId="16699" LastEditorUserId="16699" LastEditDate="2015-06-02T08:23:22.540" LastActivityDate="2015-06-03T16:00:01.480" Title="Perchlorate salts: (explosive) reactivity with methane?" Tags="&lt;organic-chemistry&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32338" PostTypeId="2" ParentId="23889" CreationDate="2015-06-02T10:46:32.013" Score="0" Body="&lt;p&gt;have you thought about using cupro-nickle, which is used exensively in marine engineering. I am aware it is expensive, but further research should reveal its suitability. I hope that you have found this helpfull. &lt;/p&gt;&#xA;" OwnerUserId="16701" LastActivityDate="2015-06-02T10:46:32.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32339" PostTypeId="1" CreationDate="2015-06-02T10:49:39.337" Score="2" ViewCount="34" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; $\ce{N2O}$ falls apart in oxygen gas and nitrogen gas according to a first order reaction rate. If you know that the rate constant is $0.38\ \mathrm{s^{-1}}$ at $760\ \mathrm{^\circ C}$, compute then the initial rate for the decomposition of $3.00\ \mathrm{g}$ $\ce{N2O}$ in a barrel of $5.0\ \mathrm{L}$. When the initial concentration doubles, what happens with the rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; I set up the reaction as $$\ce{2 N2O -&amp;gt; O2 + 2 N2}$$ The molar mass of $\ce{N2O}$ is $44.019\ \mathrm{mol/g}$, and since we got $3.00\ \mathrm{g}$, this is $132.057\ \mathrm{mol}$. Then I computed $[\ce{N2O}] = 27.01\ \mathrm{M}$ since we had to divide by $5$ because the barrel has a volume of $5.0\ \mathrm{L}$. Then we have $$v = k [\ce{N2O}] = 10.26\ \mathrm{M/s}$$ But I’m not sure if this is the correct approach. Any help please?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorDisplayName="user7951" LastEditDate="2015-06-02T11:12:40.553" LastActivityDate="2015-06-02T11:12:40.553" Title="Computing initial reaction speed given reaction constant" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32340" PostTypeId="2" ParentId="32284" CreationDate="2015-06-02T11:39:48.027" Score="5" Body="&lt;p&gt;The mechanism you are looking for is probably not known and might well be never known. According to Brown &lt;em&gt;et. al.&lt;/em&gt; there is a rapid and reversible equilibrium between associated and dissociated reactants.&lt;sup&gt;[1]&lt;/sup&gt; They tested this for cyclohexene, norbonene and apobornylene, but I believe it is fair to generalise in this matter.&#xA;$$\ce{R2C=CR2 + Hg(OOCCF3)2 &amp;lt;=&amp;gt; (F3CCOO)R2C=CR2Hg(OOCCF3)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this, even the formation of a mercurinium ion seems to be sometimes not observed. For more information please have a look at the mechanistic studies of Pasto and Gontarz,&lt;sup&gt;[2]&lt;/sup&gt; where they focussed on the stereo selectivity of substituted cyclohexenes, and references within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever molecule you might assume acting as the electrophile, it is quite obvious, that the bond mercury oxygen bond is highly ionic, with most of the electron density located in the acetyl moieties, leaving a rather substantial positive charge at the mercury itself. The following structures were optimised at DF-BP86/def2-SVP and the charges displayed are natural charges from the Natural Bond Orbital analysis to illustrate the above point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ethGJ.jpg&quot; alt=&quot;Hg+OAc charges&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hSSTi.jpg&quot; alt=&quot;Hg(OAc)2 charges structure 1&quot;&gt;&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FsGBh.jpg&quot; alt=&quot;Hg(OAc)2 charges structure 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower structure of $\ce{Hg(OAc)2}$ is slightly more stable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00709a058&quot; rel=&quot;noreferrer&quot;&gt;Herbert Charles Brown, Min-Hon Rei, Kwang Ting Liu, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1970&lt;/strong&gt;, &lt;em&gt;92&lt;/em&gt; (6), 1760–1761.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00754a035&quot; rel=&quot;noreferrer&quot;&gt;Daniel J. Pasto, John A. Gontarz, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1971&lt;/strong&gt;, &lt;em&gt;93&lt;/em&gt; (25), 6902–6908.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-02T11:39:48.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32341" PostTypeId="1" AcceptedAnswerId="32342" CreationDate="2015-06-02T12:34:16.057" Score="16" ViewCount="15727" Body="&lt;p&gt;How does one explain the trend in bond lengths and acidity of the following hydrocarbons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{ccc}&#xA;\hline&#xA;\text{Species} &amp;amp; \ce{C-H}\text{ bond length / Å} &amp;amp; \mathrm{p}K_\mathrm{a} \\&#xA;\hline&#xA;\text{Ethane, }\ce{C2H6} &amp;amp; 1.102 &amp;amp; 50 \\&#xA;\text{Ethene, }\ce{C2H4} &amp;amp; 1.085 &amp;amp; 44 \\&#xA;\text{Ethyne, }\ce{C2H2} &amp;amp; 1.061 &amp;amp; 25 \\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the shorter bonds between $\ce{C-H}$ be easier to break than longer ones, making ethane the most acidic?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="16683" LastEditDate="2016-01-09T13:01:30.773" LastActivityDate="2021-03-08T15:08:44.730" Title="Relative acidities of alkanes, alkenes, and alkynes" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;bond&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32342" PostTypeId="2" ParentId="32341" CreationDate="2015-06-02T12:49:26.443" Score="23" Body="&lt;p&gt;You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HX -&amp;gt; H + X}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the enthalpy change of that process is literally the bond dissociation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, acidity actually corresponds to the &lt;em&gt;heterolytic&lt;/em&gt; dissociation of the $\ce{H-X}$ bond, with both electrons in the bond going to $\ce{X}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HX -&amp;gt; H+ + X-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so bond strength isn't the only factor. In the case of ethane, ethene, and ethyne, the most important factor is the type of orbital that the electrons from the $\ce{C-H}$ bond end up in after deprotonation. For ethane, the two electrons end up in a carbon $\mathrm{sp^3}$ orbital; for ethene, an $\mathrm{sp^2}$ orbital; and for ethyne, an $\mathrm{sp}$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hpvg8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hpvg8.png&quot; alt=&quot;Deprotonation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, of these three, the $\mathrm{sp}$ lone pair is the &lt;em&gt;most stable&lt;/em&gt;, since the $\mathrm{sp}$ hybrid orbital contains the most s character ($50\%$) and therefore has the lowest energy. So, of the three carbanions, the lone pair in the acetylide ion $\ce{HC2-}$ is most stabilised, making ethyne the most acidic molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acidity of terminal alkynes is in fact pretty useful from a synthetic viewpoint, since you can deprotonate them with a strong base like $\ce{NaNH2}$, forming a carbon-based nucleophile which can then be used to make new $\ce{C-C}$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As for why the electronic stabilisation outweighs the bond strength, it's worth looking at some quantitative data. The $\ce{H-O}$ and $\ce{H-S}$ bonds have mean bond enthalpies of $463$ and $338~\mathrm{kJ~mol^{-1}}$ respectively (source: &lt;em&gt;Physical Chemistry&lt;/em&gt; 9th ed., Atkins &amp;amp; de Paula, p 932), which is a difference of $\mathbf{125~kJ~mol^{-1}}$. This can, to some extent, explain the difference in acidities of $\ce{H2O}$ and $\ce{H2S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the dissociation energies of the $\ce{H-C\mathrm{(sp)}}$ and $\ce{H-C\mathrm{(sp^2)}}$ bonds are given in &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/j100285a007&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Phys. Chem.&lt;/em&gt; &lt;strong&gt;1987&lt;/strong&gt;, &lt;em&gt;91&lt;/em&gt;, 17-19&lt;/a&gt; as $132.6$ and $116~\mathrm{kcal~mol^{-1}}$ respectively. The difference is roughly $17~\mathrm{kcal~mol^{-1}}$, which is equivalent to $\mathbf{69.5~kJ~mol^{-1}}$ - a somewhat smaller value than before. You can see why bond length variations might therefore play a smaller role in the hydrocarbons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;An alternative&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another equivalent way of looking at it is that an $\mathrm{sp}$-hybridised carbon is more electronegative than an $\mathrm{sp^2}$-hybridised carbon, which is in turn more electronegative than an $\mathrm{sp^3}$-hybridised carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned earlier, an $\mathrm{sp}$ orbital has the greatest s-character, and consequently electrons in a carbon $\mathrm{sp}$ orbital experience a greater effective nuclear charge and are more tightly bound to the nucleus than electrons in $\mathrm{sp^2}$ or $\mathrm{sp^3}$ orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the $\ce{C-H}$ bond also involves a contribution from the hydrogen 1s orbital. However, the above means that the electrons in $\ce{C(sp)-H}$ bonds are more strongly attracted to the carbon nucleus than electrons in $\ce{C(sp^2)-H}$ or $\ce{C(sp^3)-H}$ bonds. This means that the $\ce{C(sp)-H}$ bond is more polarised towards carbon, leaving less electron density on hydrogen, and a greater acidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of electron density on the hydrogen can in fact be probed using $\ce{^1H}$ NMR spectroscopy. A lower electron density translates into a higher chemical shift. Even though there is a second, unrelated, factor - anisotropic shielding - that serves to decrease the chemical shift of acetylenic hydrogens, they still tend to show up at higher chemical shifts than hydrogens on saturated hydrocarbons (~2 ppm, compared to ~1 ppm). This is a clear indication that the local electron density on a proton attached to $\ce{C(sp)}$ is less than that of a proton attached to $\ce{(sp^3)}$.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-05-06T23:58:49.723" LastActivityDate="2016-05-06T23:58:49.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32344" PostTypeId="1" CreationDate="2015-06-02T13:26:24.213" Score="-3" ViewCount="173" Body="&lt;p&gt;What is a photon and how is its energy directly proportional to frequency of light (please don't tell that it is because $E=h\nu$ relation)?&lt;/p&gt;&#xA;" OwnerUserId="16703" LastEditorUserId="5016" LastEditDate="2015-06-02T15:46:13.763" LastActivityDate="2015-06-02T16:14:48.827" Title="Energy of photon" Tags="&lt;energy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-06-02T22:55:28.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="32345" PostTypeId="2" ParentId="32304" CreationDate="2015-06-02T13:31:02.857" Score="6" Body="&lt;p&gt;The first answer is correct.  If the reaction is $\ce{2X-&amp;gt;P}$, then two units of $\ce{X}$ should disappear for every unit of $\ce{P}$ formed. Only the first possibility meets this criterion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate law for the reaction could be anything.  In your example you have apparently assumed that it is a second order reaction in $\ce{X}$, but we could make any other assumption.  For example, product formation rate could be $\frac{dP}{dt}=k~X^{1.2}$ or $\frac{dP}{dt}=k~\frac{X}{K_m + X^{2.4}}$.  But if the reaction you want to model is really $\ce{2X-&amp;gt;P}$, then these rate laws would imply that $\frac{dX}{dt}=-2~k~X^{1.2}$ or $\frac{dX}{dt}=-2~k~\frac{X}{K_m + X^{2.4}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if $\frac{dP}{dt}=k X^2$, &lt;strong&gt;then the rate of X depletion must be -2 times that&lt;/strong&gt;, or $\frac{dX}{dt}=-2k X^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another way to see the problem with the rate law without the $-2$: suppose there was a reaction $\ce{X-&amp;gt;P}$, i.e. only one molecule of $\ce{X}$ was needed to form $\ce{P}$.  Suppose also that this reaction followed second-order kinetics in $\ce{X}$, i.e. $\frac{dP}{dt}\propto X^2$.  What would the rate laws for this alternate reaction be?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:04:50.550" LastActivityDate="2016-03-11T23:04:50.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32346" PostTypeId="1" CreationDate="2015-06-02T13:52:06.133" Score="0" ViewCount="1455" Body="&lt;p&gt;At high temperatures nitrogen dioxide decomposes into nitrogen oxide and oxygen gas according to $$\ce{2 NO_2 -&amp;gt; 2 NO + O_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a question now. Let's say at a certain temperature I'm given data for the formation of $\ce{O_2}$. For example:&#xA;At $t = 0$ we have $[\ce{O_2}] = 0$ mol/L, at $t = 50$, $[\ce{O_2}] = 7.10 \cdot 10^{-4}$ mol/L, at $t = 100, [\ce{O_2}] = 1.205 \cdot 10^{-3}$ and so on until a certain time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm given that the initial concentration of $\ce{NO_2}$ is $0.006M$, how can I know what the concentration of $\ce{NO_2}$ is at certain times? I'm asked to determine the order of the reaction on the basis of the given data, and also to present graphically the rate constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do that I need to know if $\ln([\ce{NO_2}])$ versus time $t$ is a straight line. But how to find $[\ce{NO_2}]$ in the first place?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastActivityDate="2015-06-03T15:02:08.687" Title="Computing the concentration of a reactant given formation of products" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32347" PostTypeId="1" CreationDate="2015-06-02T14:16:07.897" Score="1" ViewCount="1728" Body="&lt;p&gt;I performed aan acid catalysed haloform reaction in acetic acid to an acid. I need to evaporate acetic acid which I used as solvent and It is around 150 ml. What are the alternatives , I have a rotavap in lab but 150ml AA is too much for a rotavap I guess.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-06-02T14:16:07.897" Title="Evaporation of acetic acid" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32348" PostTypeId="2" ParentId="32344" CreationDate="2015-06-02T14:22:07.467" Score="1" Body="&lt;p&gt;A photon is a quantum of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain more thoroughly, everything in the universe has elements of both particles and waves. So while light is usually thought of as a wave in classical mechanics, it also consists of particles. One photon is the smallest amount of light you can have while still having light at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between the energy of a photon and its frequency (as well as how its particle nature was realized) was determined by Albert Einstein in a treatise on something called the photoelectric effect. The basic experimental set up was fairly simple. People had observed that, when shining light on a metal's surface, sometimes electrons would start popping out. The hypothesis proposed by classical mechanics was that this was proportional to the intensity of the light shining on the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the experiments were carried out, the results were a bit more surprising. The light had to be a minimum &lt;em&gt;frequency&lt;/em&gt; before electrons were ejected at all. If the intensity of the light were increased at that frequency, more electrons would come flying out, but each electron would still individually have the same kinetic energy. If the frequency of the light was below the threshold, then no electrons would come out regardless of how intense the light was. And the higher the frequency of the light (once you got above the threshold frequency), the faster the electrons flew out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Einstein realized what this meant: light had to exist as a particle as well as a wave. Each light particle (which he called a &quot;photon&quot;) could interact with one electron at a time. Higher intensity light just meant more emitted photons, so the individual photon had to have enough energy to dislodge the electron from the metal surface. According to the experiments on the photoelectric effect, this energy was proportional to the frequency of the light, as noted before. When Einstein did the calculations, the constant of proportionality was $6.626\times10^{-34}~\mathrm{J~s}$, better known as Planck's constant and represented by the symbol &quot;h.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that this exact constant appeared in something seemingly completely different (the law for how blackbodies emit light) was a major basis for more investigation and eventually yielded quantum mechanics as we know it today.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="10185" LastEditDate="2015-06-02T16:14:48.827" LastActivityDate="2015-06-02T16:14:48.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32349" PostTypeId="1" CreationDate="2015-06-02T14:35:06.547" Score="0" ViewCount="116" Body="&lt;p&gt;I think I'm making this more complicated than it needs to be, but here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 7 drug solutions, all 1 mg/mL which need to have the following concentration in urine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: 25,000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: 25,000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: 100,000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D: 100,000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E: 10,000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F: 1000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G: 1000 ng/ml&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much of each 1 mg/ml solution do I need to add to urine to get a stock with the above concentrations? My ultimate goal is to take several 75 ul aliquots of urine for testing. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="16704" LastEditorUserId="4325" LastEditDate="2015-06-02T14:44:51.593" LastActivityDate="2015-06-02T15:26:29.197" Title="solution concentrations" Tags="&lt;analytical-chemistry&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-03T10:45:38.367" ContentLicense="CC BY-SA 3.0" />
  <row Id="32350" PostTypeId="1" CreationDate="2015-06-02T14:42:03.657" Score="5" ViewCount="3457" Body="&lt;p&gt;Is $\ce{Na}$ a single electron species and can I apply Bohr's Model on it? &#xA;Please explain why not as it also contains one electron in its outermost shell.&lt;/p&gt;&#xA;" OwnerUserId="16703" LastEditorUserId="1499" LastEditDate="2015-06-02T19:51:21.757" LastActivityDate="2015-06-02T19:51:21.757" Title="Can the Bohr atomic model be applied to sodium and other single valence electron species?" Tags="&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32352" PostTypeId="2" ParentId="32349" CreationDate="2015-06-02T15:26:29.197" Score="2" Body="&lt;p&gt;Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take $\text{A}$ as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n_A \ \text{in final solution} =  25000\ \ce{ng/mL} \times V_f\ \ce{mL}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to carry out these calculations for all your compounds, then add all the corresponding quantities to a calibrated vessel, e.g. a volumetric flask, then bring this mixture up to the final volume with your solvent, i.e. urine.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-06-02T15:26:29.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32353" PostTypeId="1" AcceptedAnswerId="32358" CreationDate="2015-06-02T16:10:26.660" Score="12" ViewCount="758" Body="&lt;p&gt;&lt;sub&gt;Not a duplicate of &lt;a href=&quot;https://chemistry.stackexchange.com/questions/2759/hydrazine-synthesis&quot;&gt;Hydrazine synthesis&lt;/a&gt; - I know &lt;a href=&quot;http://fscimage.fishersci.com/msds/11040.htm&quot;&gt;what hydrazine is&lt;/a&gt; and am not going near it voluntarily!&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looked at this &lt;a href=&quot;http://www.lookchem.com/Chempedia/Chemical-Technology/Inorganic-Chemical-Technology/2878.html&quot;&gt;resource on the Internet&lt;/a&gt; which cites three major commercial-scale processes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H_2O_2}$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Peroxide_process&quot;&gt;process&lt;/a&gt; (in use at &lt;a href=&quot;https://en.wikipedia.org/wiki/Arkema&quot;&gt;Arkema&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mitsubishi_Gas_Chemical_Company&quot;&gt;Mitsubishi Gas Chemical&lt;/a&gt;, developed by Produits Chimiques Ugine Kuhlmann and patented in 1976)&lt;/li&gt;&#xA;&lt;li&gt;Bayer process (at the eponymous company)&lt;/li&gt;&#xA;&lt;li&gt;Urea process&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What processes do other major large-scale producers use and what are the driving forces behind adoption of a particular hydrazine synthesis? I can guess that &lt;a href=&quot;http://www.hydrazinechina.com/aboutus.html&quot;&gt;Chinese producers&lt;/a&gt; may use the urea process, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Background: am a &lt;a href=&quot;https://space.stackexchange.com&quot;&gt;Space Exploration.SE&lt;/a&gt; regular and am interested in learning more about worldwide production of hypergolic propellants.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="5497" LastEditorUserId="10185" LastEditDate="2015-06-03T21:05:17.817" LastActivityDate="2015-06-04T20:33:30.557" Title="Industrial synthesis of hydrazine" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32354" PostTypeId="1" AcceptedAnswerId="34727" CreationDate="2015-06-02T16:25:13.397" Score="23" ViewCount="1839" Body="&lt;p&gt;The wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Trinitromethane&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; on trinitromethane claims that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is some evidence that the anion (which obeys the $4n+2$ Hückel rule) is aromatic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the citation is behind a paywall so I can't read it but my first obvious question is how can the trinitromethane anion obey Hückel's rule? Two of the requirements for the rule are that the molecule must be planar (or nearly planar) and must be cyclic and I fail to see how either of these conditions are satisfied since the anion is roughly trigonal pyramidal around the carbon and definitely not cyclic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to about Y-aromaticity. All the papers linked are behind paywalls for me so can someone please explain what Y-aromaticity is and how it applies to the example given. I found &lt;a href=&quot;https://chemistry.stackexchange.com/questions/21856/delocalization-of-pi-electrons-in-nitrate-ion&quot;&gt;this&lt;/a&gt; previous answer which gives some discussion of Y-aromaticity in nitrate ions. If someone could provide a similar MO analysis of the trinitromethane anion this might be very interesting.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="16683" LastEditDate="2018-01-07T11:09:44.580" LastActivityDate="2018-01-07T11:09:44.580" Title="What is Y-aromaticity? Is the trinitromethanide anion aromatic?" Tags="&lt;organic-chemistry&gt;&lt;nitro-compounds&gt;&lt;aromaticity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32355" PostTypeId="1" CreationDate="2015-06-02T16:35:14.033" Score="2" ViewCount="140" Body="&lt;p&gt;Does anyone have input or experience on passivation of stainless steel, particularly in outdoor environments where submersion of components is not practical due to location or size? Nitric acid (and solutions with sodium dichromate) typically call for submersion for extended periods of time based on the concentration of the solution. Citric acid has been referenced as a more suitable alternative and is more environmentally friendly but the literature on non-submersion passivation techniques is limited. Any help or direction is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16707" LastActivityDate="2015-06-02T16:35:14.033" Title="Is passivation of stainless steel feasible in the field?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32356" PostTypeId="2" ParentId="32354" CreationDate="2015-06-02T16:36:08.960" Score="5" Body="&lt;p&gt;The Wikipedia article refers to a paper discussing so-called &lt;strong&gt;Y-aromaticity&lt;/strong&gt;, a proposed extension of the idea of aromaticity to particularly stable non-cyclic structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This concept was proposed originally to explain the uncommon stability of the guanidinium cation. You can find some references in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Aromaticity#Atypical_aromatic_compounds&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on aromaticity&lt;/a&gt; under &quot;Atypical aromatic compounds&quot;. Whether this type of aromaticity really exists, however, is still somewhat controversial: other explanations have been proposed for the extra stability of guanidinium that do not necessarily require creating a new &quot;type of aromaticity&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-06-02T16:36:08.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32357" PostTypeId="1" CreationDate="2015-06-02T16:46:15.297" Score="5" ViewCount="1763" Body="&lt;p&gt;How do delocalized electrons hold metal atoms together? What are the forces of attraction between the atoms and the electrons called? &lt;/p&gt;&#xA;" OwnerUserId="16638" LastEditorUserId="865" LastEditDate="2015-08-02T20:00:46.133" LastActivityDate="2017-08-26T18:46:56.747" Title="How does a &quot;sea of electrons&quot; hold metal atom together?" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32358" PostTypeId="2" ParentId="32353" CreationDate="2015-06-02T16:48:40.230" Score="9" Body="&lt;p&gt;The Kirk-Othmer Encyclopedia of Chemical Technology reports that: &quot;About 60% is made by the [Bayer] hypochlorite–ketazine process, 25% by the peroxide–ketazine route, and the remainder by the Raschig and urea processes.&quot; Here is a link to the original article, if you can access it (e.g. at a university library): &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/0471238961.0825041819030809.a01.pub2/full&quot;&gt;Hydrazine and Its Derivatives, by Eugene F. Rothgery&lt;/a&gt;. The data cited dates back to 2004, so it may have changed somewhat, but probably not too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this also: the above makes hydrazine &lt;em&gt;hydrate&lt;/em&gt;. Propellant-grade hydrazine must be anhydrous, so there is an extra step in which water is removed from the hydrate, typically by azeotropic distillation with aniline.&lt;/p&gt;&#xA;" OwnerUserId="16652" LastActivityDate="2015-06-02T16:48:40.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32359" PostTypeId="2" ParentId="32354" CreationDate="2015-06-02T16:53:12.523" Score="5" Body="&lt;p&gt;The conjugate base of trinitromethane is flat, symmetrical (trigonal), mesomerically stabilised and it has $4n+2$ electrons. Every nitro group has 4 π electrons in resonance (an oxygen lone pair and a double bond), and together with the anionic lone pair on carbon, makes for a total of 14 π electrons ($14 = 4\times3+2$). Therefore it may be considered Y-aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, Y-aromaticity is a controversial concept, which says that in Y-shaped molecules additional stabilisation is present, analogous to the planar, cyclic compounds with $4n+2$ delocalised π electrons). Examples of such molecules include square and Y-shaped dications with 4 carbon atoms and one double bond. It seems that it still hasn't been proven as to whether it is really a valid concept.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="16683" LastEditDate="2018-01-07T11:08:55.047" LastActivityDate="2018-01-07T11:08:55.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32360" PostTypeId="1" AcceptedAnswerId="32362" CreationDate="2015-06-02T16:58:58.293" Score="9" ViewCount="6124" Body="&lt;p&gt;Liquid soap that sits in metal pump handles turns blue after a week or so, as in this photo comparing the soap in its initial clear state and then after pumping from the dispenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0aBumm.jpg&quot; alt=&quot;Clear soap turned blue&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pumps are presumably brass with either nickel or chrome electroplating on the exterior.  Can't say for sure whether or to what degree the interior where the soap makes contact is plated.  They have been in use for years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ingredients of this particular soap are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;water&lt;/li&gt;&#xA;&lt;li&gt;sodium laureth sulfate&lt;/li&gt;&#xA;&lt;li&gt;cocamidopropyl betaine&lt;/li&gt;&#xA;&lt;li&gt;decyl glucoside&lt;/li&gt;&#xA;&lt;li&gt;sodium chloride&lt;/li&gt;&#xA;&lt;li&gt;&quot;fragrance&quot;&lt;/li&gt;&#xA;&lt;li&gt;DMDM hydantoin&lt;/li&gt;&#xA;&lt;li&gt;polyquaternium-7&lt;/li&gt;&#xA;&lt;li&gt;tetrasodium EDTA&lt;/li&gt;&#xA;&lt;li&gt;citric acid&lt;/li&gt;&#xA;&lt;li&gt;sodium sulfate&lt;/li&gt;&#xA;&lt;li&gt;PEG-120 methyl glucose dioleate&lt;/li&gt;&#xA;&lt;li&gt;PEG-7 glyceryl cocoate&lt;/li&gt;&#xA;&lt;li&gt;poloxamer 124&lt;/li&gt;&#xA;&lt;li&gt;D&amp;amp;C red #33&lt;/li&gt;&#xA;&lt;li&gt;FD&amp;amp;C blue #1&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="7448" LastEditDate="2015-06-02T18:43:44.290" LastActivityDate="2017-06-20T16:24:41.093" Title="What turns clear liquid soap blue?" Tags="&lt;everyday-chemistry&gt;&lt;color&gt;&lt;surfactants&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2020-03-11T03:41:21.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="32361" PostTypeId="2" ParentId="32357" CreationDate="2015-06-02T17:15:55.840" Score="1" Body="&lt;p&gt;The metal atoms lose electrons to form positive metal ions and a sea of delocalised electrons. The positive ions are attracted to the oppositely charged electrons and form an electrostatic attraction.&lt;/p&gt;&#xA;" OwnerUserId="16708" LastActivityDate="2015-06-02T17:15:55.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32362" PostTypeId="2" ParentId="32360" CreationDate="2015-06-02T17:39:30.300" Score="11" Body="&lt;p&gt;The color of the precipitate is strongly reminiscent of &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper(II)_hydroxide#/media/File:Cupric_hydroxide.JPG&quot;&gt;copper(II) hydroxide&lt;/a&gt;. I hypothesize:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Chloride corrosion of copper from exposed brass (&lt;a href=&quot;http://www.electrochemsci.org/papers/vol4/4050654.pdf&quot;&gt;ref 1&lt;/a&gt; || &lt;a href=&quot;http://www.electrochemsci.org/papers/vol8/80709858.pdf&quot;&gt;ref 2&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. Per ref 2 above, sulfate may also participate&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alkaline precipitation of $\ce{Cu(OH)2}$ in the neutrally-buffered soap&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Insufficient $\ce{Na_4EDTA}$ to complex the amount of $\ce{Cu}$ corroded from the brass&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's also possible the blue material is $\ce{Ni(II)}$ complexed in such a way that its color is changed from the typical &lt;a href=&quot;http://en.wikipedia.org/wiki/Nickel%28II%29_hydroxide#/media/File:DSC01973_-_Nickel_(II)_reactions.JPG&quot;&gt;green-blue&lt;/a&gt; of $\ce{Ni(OH)2}$.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-02T19:42:09.200" LastActivityDate="2015-06-02T19:42:09.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32363" PostTypeId="1" AcceptedAnswerId="32384" CreationDate="2015-06-02T17:45:16.657" Score="3" ViewCount="33943" Body="&lt;p&gt;I want to address two exceptions in the trend of ionization energy across the period, that are causing me problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the second period as an example :&lt;img src=&quot;https://i.stack.imgur.com/EnMVn.gif&quot; alt=&quot;Consider only the second period&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two exceptions from the general trend are the ionization energies of B lesser than Be and that of O less than N.&#xA;My teacher told me the reason to both was that half filled and fully filled orbitals of N and Be are more stable and hence require more energy to pull off an electron. He also said that this affects only the next element in the period for a change in trend( he never said why though).&#xA;&lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/ies.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This place&lt;/a&gt; and my textbook tells me that the reasons are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) Be,B :&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electron is to be removed from the 2p orbital in B and the 2s orbital in Be. The 2s orbital penetrates more into the nucleus and hence it is more difficult to remove an electron form Be than  from B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; In that case why doesn't this apply to C? Shouldn't it be easier to remove one of the electrons from C's 2p orbital than Be's 2s orbital just like the B's 2p orbital? And so shouldn't the ionization energy of Be must be more than C too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) N,O:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N 1s2 2s2 2px1 2py1 2pz1&lt;br&gt;&#xA;  O 1s2 2s2 2px2 2py1 2pz1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I have understood here is that there are two electrons in the 2px orbital of O and because of electron-electron repulsion, it is easier to remove an electron from here as opposed to the 2px orbital of N which had only one electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Why doesn't this apply to F? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F 1s2 2s2 2px2 2py2 2pz1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F has paired electrons too like O in the px and py subshells. Wouldn't there be repulsion here too? And hence why isn't it easier to remove electrons from F than is it to from N which doesn't have any paired electrons at all? Why is the ionization energy of F more than N instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same problems can be seen in the next period.I hope they can be sorted out with the same logic that i am missing for this period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, was what my teacher said just a simplification of everything I found in the link and the textbook? Was he trying to convey the same thing? Or are they two complete different explanations?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="9298" LastEditDate="2015-06-02T18:20:27.477" LastActivityDate="2018-11-12T01:57:27.887" Title="Anomalous trends in ionization energy" Tags="&lt;periodic-trends&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32364" PostTypeId="2" ParentId="32350" CreationDate="2015-06-02T18:32:05.673" Score="9" Body="&lt;p&gt;One of the great successes of the Bohr model was the ability to explain the emission spectrum of hydrogen (and hydrogen-like monoelectronic ions), by allowing a simple theoretical derivation of the experimentally determined and remarkably accurate &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\lambda}=RZ^2 \bigg(\frac{1}{n_f^2}-\frac{1}{n_i^2}\bigg)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda$ is the wavelength of light emitted, $R$ is a proportionality constant known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Rydberg_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg constant&lt;/a&gt; (which actually varies slightly with changes in atomic mass), $Z$ is the amount of positive charge attracting the electron (the nuclear charge in multiples of $q=1.6022\times 10^{-19}\ \mathrm{C}$), and $n_f$ and $n_i$ are positive integers (called &lt;em&gt;quantum numbers&lt;/em&gt;) which previously had no experimental interpretation, but which now arise from the postulates of the Bohr model, being attributed to the concept of &lt;em&gt;electron shells&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this great new insight into atomic physics, naturally one would like to investigate how generally applicable the model is. Given the success in describing the emission spectra of hydrogen, one of the first places to look would be the spectra for other elements. So here they are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rrVzd.png&quot; alt=&quot;Click &amp;quot;view image&amp;quot; for very high resolution version, or look at source&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.umop.net/spctelem.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; - isn't it a beautiful image? Kudos to the designer! Open in a tab for gigantic resolution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, most elements have rather elaborate spectra. After undoubtedly many frustrated attempts, it was found that the Bohr model, as accurate as it was for hydrogen, applied quite poorly to the rest of the elements, and no spectra could be reliably predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the alkali metals stood out in that a modified version of the Bohr model had some success, &lt;a href=&quot;http://www.dartmouth.edu/~chem81/pdfs/Lab1a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;producing a Rydberg-like formula&lt;/a&gt;, though salvaging it came at the cost of additional experimental parameter fitting, again with no theoretical background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the energy of a electron in a hydrogen atom according to the Bohr model is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_n=-\frac{hcR}{n^2}=-\frac{13.6\ \mathrm{eV}}{n^2},\ \ \ \ n \in \mathbb{N} \ \ (n=1,2,3,4...)  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the alkali metals, the formula had to be modified to: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_n=-\frac{hcR}{n_{eff}^2}=-\frac{hcR}{(n-\delta_l)^2},\ \ \ \ n_{eff} \in \mathbb{R_+}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the same structure, the mismatch between theory and experiment is brushed into the quantum numbers $n$, which are replaced by &lt;em&gt;effective&lt;/em&gt; quantum numbers $n_{eff}$ that are no longer positive integers. To recover integer quantum numbers, a value called the &lt;em&gt;quantum defect&lt;/em&gt; ($\delta_l$) had to be introduced. This value is different for each value of $n$, and also varies with another parameter $l$. This may seem like an arbitrary mathematical attempt to salvage the equation, but while the values of $\delta_l$ had to be tabulated, it was found that they were relatively constant with changes in $n$, and that there were a limited number of different cases ($l$ was a parameter limited to only a few values). The resulting Rydberg-like equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\lambda}=R \bigg(\frac{1}{(n_f-\delta_{l_f})^2}-\frac{1}{(n_i-\delta_{l_i})^2}\bigg)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not as tidy as the result from the Bohr model, at least the equation can be used to approximately explain the atomic emission spectra for the alkali metals. Details of an experimental procedure for chemistry undergrads to determine the emission spectrum of sodium and apply the modified Rydberg formula can be found &lt;a href=&quot;http://www.dartmouth.edu/~chem81/pdfs/Lab1a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, along with theoretical background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, we know why the alkali metals seemed to be a better fit to the Bohr model than the rest of the elements. With finer knowledge about the details of the electronic structure of atoms, borne out of the full quantum mechanical model of the atom, we know that alkali metals share an important property with hydrogen; they all possess a single electron in their outermost shell, i.e. have a valence shell configuration of $(ns)^1$. This is the only electron which is important in electronic transitions in the near-ultraviolet, visible and infrared regions of the electromagnetic spectrum, as the core electrons of the alkali metals are held much more tightly to the nucleus and require high energy photons to excite. This creates a much simpler spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, atoms with multiple electrons suffer from &lt;em&gt;electron-electron repulsions&lt;/em&gt;, which are not taken into account at all in the Bohr model, and thus are a significant source of error. However, in the alkali metals the consequences of these repulsions are simplified, as all electrons are in core shells, except the single valence electron. For all other multielectronic elements, electron-electron repulsions in the valence shell preclude a simple application of the Bohr model.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-02T18:44:53.650" LastActivityDate="2015-06-02T18:44:53.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32366" PostTypeId="2" ParentId="843" CreationDate="2015-06-02T18:57:39.540" Score="6" Body="&lt;p&gt;The redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{Cu^2+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; Cu}\quad &amp;amp;&amp;amp;E^\circ = +0.340\ \mathrm{V}\\&#xA;\ce{2H+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2}\quad &amp;amp;&amp;amp;E^\circ = +0.000\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, non-oxidizing acids such as acetic acid cannot directly oxidize copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\ce{Cu}$ can be oxidized by $\ce{O2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 4H+ + 4e- &amp;lt;=&amp;gt; 2H2O}\quad E^\circ = +1.229\ \mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, copper is slowly oxidized in acetic acid in contact with air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation can be increased by addition of oxidizing agents such as hydrogen peroxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O2 + 2H+ + 2e- &amp;lt;=&amp;gt; 2H2O}\quad E^\circ = +1.763\ \mathrm{V}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-06-02T18:57:39.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32367" PostTypeId="1" AcceptedAnswerId="32397" CreationDate="2015-06-02T20:38:57.430" Score="11" ViewCount="2414" Body="&lt;p&gt;TL;DR version is the question title. Some context and data follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was creating an assignment for my organic chemistry students in which they would need to draw as many isomers as they could from a simple formula, for example $\ce{C3H6O}$ (which fits 7 structures - 9 if you count minor enol tautomers and 11 if your count stereo). In this set (ignoring stereo and enols) there are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two alcohols&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopropanol&quot;&gt;Cyclopropanol&lt;/a&gt; - boiling point $=101-102\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Allyl_alcohol&quot;&gt;Allyl alcohol&lt;/a&gt; - boiling point $=97\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two carbonyl compounds&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propanal&quot;&gt;Propanal&lt;/a&gt; - boiling point $=46\ ^\circ\text{C}$ &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Acetone&quot;&gt;Acetone&lt;/a&gt; - boiling point $=56\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three ethers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propylene_oxide&quot;&gt;Propylene oxide&lt;/a&gt; - boiling point $=34\ ^\circ\text{C}$ &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Oxetane&quot;&gt;Oxetane&lt;/a&gt; - boiling point&#xA;$=49-50\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methyl_vinyl_ether&quot;&gt;Methyl vinyl ether&lt;/a&gt; - boiling point $=6\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The boiling point of methyl vinyl ether immediately stood out as anomalous. The fully saturated &lt;a href=&quot;http://en.wikipedia.org/wiki/Methoxyethane&quot;&gt;ethyl methyl ether&lt;/a&gt; has a boiling point of $7.4\ ^\circ\text{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some more data comparing cyclic ethers to their acyclic isomers (and some analogs). The gap appears to be closing as the number of carbons increases. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Four carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahydrofuran&quot;&gt;Tetrahydrofuran&lt;/a&gt; - boiling point $=66\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/fluka/19930?lang=en&amp;amp;region=US&quot;&gt;1-Butene oxide&lt;/a&gt; - boiling point $=61-65\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/422177?lang=en&amp;amp;region=US&quot;&gt;Ethyl vinyl ether&lt;/a&gt; - boiling point $=33\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/678201?lang=en&amp;amp;region=US&quot;&gt;Allyl methyl ether&lt;/a&gt; - boiling point $=42-43\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diethyl_ether&quot;&gt;Diethyl ether&lt;/a&gt; - boiling point $=34.6\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methoxypropane&quot;&gt;Methyl propyl ether&lt;/a&gt; - boiling point $=38.8\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahydropyran&quot;&gt;Tetrahydropyran&lt;/a&gt; - boiling point $=88\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/2-Methyltetrahydrofuran&quot;&gt;2-Methyltetrahydrofuran&lt;/a&gt; - boiling point $=80.3\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/238228?lang=en&amp;amp;region=US&quot;&gt;Allyl ethyl ether&lt;/a&gt; - boiling point $=65-66\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/235377?lang=en&amp;amp;region=US&quot;&gt;Butyl methyl ether&lt;/a&gt; - boiling point $=70-71\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-05T11:39:52.843" Title="Why do cyclic ethers have higher boiling points than their acyclic isomers?" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32369" PostTypeId="1" CreationDate="2015-06-02T22:40:34.720" Score="4" ViewCount="155" Body="&lt;p&gt;This can be very useful for people who are afraid of spontaneous human combustion! But seriously, I heard there are non-harmful chemicals used in some fire extinguishers, so can they maybe be put on our skin, clothes, etc. in advance to prevent ever catching fire?&lt;/p&gt;&#xA;" OwnerUserId="16712" LastEditorUserId="9961" LastEditDate="2015-06-03T00:55:46.170" LastActivityDate="2015-06-03T06:41:26.380" Title="Are there substances people can put on themselves to prevent catching fire?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32370" PostTypeId="2" ParentId="19001" CreationDate="2015-06-02T22:54:29.740" Score="16" Body="&lt;p&gt;&lt;strong&gt;Does the activity of a solid or liquid change over the course of a reaction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too.  It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What about gases, why don't they show the same phenomenon and have fixed activities?&lt;/strong&gt; (Asked in comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is active mass given a value of 1 then?  Is this simply a convention?  Shouldn't its value depend on the compound?&lt;/strong&gt; (Asked in comments.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions.  For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar.  The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, consider the activity of solids and liquids.  Their standard state is their density at standard conditions.  Their experimental state is their density at experimental conditions.  It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2015-06-02T22:54:29.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32372" PostTypeId="2" ParentId="32369" CreationDate="2015-06-03T00:24:37.330" Score="7" Body="&lt;p&gt;It depends &lt;em&gt;how long&lt;/em&gt; one is exposed to heat, and &lt;em&gt;how hot&lt;/em&gt; the source is, but yes, there are &quot;fire retardants&quot; for skin:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A mix of alcohol (ethanol or isopropanol) and water or &lt;a href=&quot;https://www.youtube.com/watch?v=FTLuO2b2DBg&quot;&gt;butane in soap sud can be set afire on your palm without burning you&lt;/a&gt;, if done &lt;strong&gt;very carefully&lt;/strong&gt;. The water soaks up heat energy both because of its high &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_capacity&quot;&gt;specific heat&lt;/a&gt; and through evaporation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Insulation applied to the skin or &lt;a href=&quot;https://www.youtube.com/watch?v=dcUhxIbXZSk&quot;&gt;between skin and flame&lt;/a&gt; can protect (momentarily) from a torch.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A wet hand is &lt;a href=&quot;https://www.youtube.com/watch?v=yTOCAd2QhGg&quot;&gt;&lt;em&gt;temporarily&lt;/em&gt; protected from molten lead&lt;/a&gt; by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Leidenfrost_effect&quot;&gt;leidenfrost effect&lt;/a&gt;. As water on the skin boils, it acts, briefly, barrier to the heat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NB: Do not try these experiments yourself (or on others). You get &lt;em&gt;no&lt;/em&gt; points for being a crispy critter.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-03T00:24:37.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32373" PostTypeId="1" CreationDate="2015-06-03T01:57:43.533" Score="2" ViewCount="678" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A maximum of $1.89~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $25.00~^\circ\mathrm{C}$.  A maximum of $5.26~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $50.00~^\circ\mathrm{C}$.  Use this information to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the following reaction.&#xA;  $$\ce{AgCl~(s) -&amp;gt; Ag+ (aq)~+ Cl- ~(aq)}$$&#xA;  Assume that ΔH and ΔS are independent of temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I started off by converting the given masses into moles using molar mass of $\ce{AgCl}$ and since we're given the reaction, we can assume that $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$, thus&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At $25~^\circ\mathrm{C}$, after doing conversion I got $1.32\cdot 10^{-5}~\mathrm{mol}$ of $\ce{AgCl}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At $50~^\circ\mathrm{C}$ I got $3.67\cdot 10^{-5}~\mathrm{mol}$ $\ce{AgCl}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since, $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$ and mole ratio is 1:1, I calculated the two $K_\mathrm{sp}$ values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At $25~^\circ\mathrm{C}$: $K_\mathrm{sp1} = 1.74 \cdot 10^{-10}$&lt;/li&gt;&#xA;&lt;li&gt;At $50~^\circ\mathrm{C}$: $K_\mathrm{sp2}= 1.35\cdot 10^{-9}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, we do have two temperatures and two equilibrium constants which empowers me to use this formula:&#xA;$$\ln\left(\frac{K_1}{K_2}\right)=-\frac{\Delta H}{R}\cdot \frac{T_2-T_1}{T_1T_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After solving for $H^\circ$  I get $65.6~\mathrm{kJ/mol}$ which is the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find $\Delta S$? I tried calculating it via standard $$\Delta G= \Delta H - T\Delta S$$ and averaging but it does not work.&lt;/p&gt;&#xA;" OwnerUserId="16551" LastEditorUserId="4945" LastEditDate="2015-06-03T05:28:37.903" LastActivityDate="2015-06-03T05:28:37.903" Title="How to find the standard enthalpy and entropy for the dissolution of silver chloride?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;solubility&gt;&lt;free-energy&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32375" PostTypeId="2" ParentId="32164" CreationDate="2015-06-03T06:42:28.990" Score="-1" Body="&lt;p&gt;There is a strong hydrogen bonding of methanol in water than phenol in water because the partial negativity in oxygen in phenol is decreased as a result of pi electron delocalization towards the ring. Hydrogen bonding is a result of interaction between a partial negative oxygen and partial positive hydrogen. Thus, to decrease the partial negativity of the oxygen would weaken the interaction between the hydrogen of water to the oxygen in phenol.&lt;/p&gt;&#xA;" OwnerUserId="14467" LastActivityDate="2015-06-03T06:42:28.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32376" PostTypeId="1" CreationDate="2015-06-03T08:22:29.397" Score="0" ViewCount="1216" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$\pu{50.0 mg}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{50.0 mg}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; are present  in one liter of water, what will be the final (equilibrium) concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; expressed in moles per liter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16715" LastEditorUserId="4945" LastEditDate="2018-10-25T17:19:03.450" LastActivityDate="2018-10-25T17:32:57.777" Title="How to calculate the final concentration of calcium ions from known starting masses of carbonate and calcium ions?" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-03T13:19:57.177" ContentLicense="CC BY-SA 4.0" />
  <row Id="32377" PostTypeId="1" AcceptedAnswerId="32383" CreationDate="2015-06-03T08:39:11.850" Score="3" ViewCount="585" Body="&lt;p&gt;How can we look at reversible processes in an intuitive manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible, that at any moment, it can be reversed by an infinitesimal change?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T09:53:32.183" LastActivityDate="2015-07-04T01:05:15.580" Title="Reversible process in thermodynamics" Tags="&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32378" PostTypeId="5" CreationDate="2015-06-03T09:13:41.023" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-03T09:13:41.023" LastActivityDate="2015-06-03T09:13:41.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32379" PostTypeId="4" CreationDate="2015-06-03T09:13:41.023" Score="0" Body="Questions related to the chemical and physical influences that influence the rate and probability of chemical reactions. Not to be confused with [catalysis] or [reaction-mechanism]." OwnerDisplayName="user15489" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-19T20:12:09.817" LastActivityDate="2018-03-19T20:12:09.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32380" PostTypeId="5" CreationDate="2015-06-03T09:18:07.180" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-03T09:18:07.180" LastActivityDate="2015-06-03T09:18:07.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32381" PostTypeId="4" CreationDate="2015-06-03T09:18:07.180" Score="0" Body="Questions related to Time-dependent density functional theory (TDDFT) is a quantum mechanical theory used in physics and chemistry to investigate the properties and dynamics of many-body systems in the presence of time-dependent potentials." OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T11:10:26.790" LastActivityDate="2015-06-03T11:10:26.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32383" PostTypeId="2" ParentId="32377" CreationDate="2015-06-03T10:12:16.107" Score="10" Body="&lt;p&gt;I sometimes explain it this way: imagine a staircase where each step is $h$ high.  If $h$ is small then it is quite easy to take a single step up or down, i.e. the process is relatively easily reversed. However, as $h$ increases it starts to become increasingly difficult to go up (and hazardous to go down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for small steps there is a difference between going up and down since you expend more energy going up (against gravity) than going down, so it's not truly reversible, but as $h$ decreases the difference becomes minuscule (but never 0).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there is no truly reversible thermodynamic process but the more slowly you let a process proceed (and equilibrate) the more it resembles the reversible ideal process&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-03T10:34:00.607" LastActivityDate="2015-06-03T10:34:00.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32384" PostTypeId="2" ParentId="32363" CreationDate="2015-06-03T10:47:01.820" Score="4" Body="&lt;p&gt;OK. Lets get a few things cleared up first. The main factor affecting ionizing energy (IE) is the attraction between the nucleus and the electrons. The more the attraction, the more energy is needed to remove an electron from it's orbital and hence higher the IE. The other factors modify this main factor and thus change the expected IE values to the anomalous observed values. Now coming to your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As we go to the right of a row, size of the atoms get smaller. $\ce{C}$ is smaller than $\ce{Be}$ and $\ce{B}$ and hence has higher IE.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The same thing happens in case of $\ce{F}$. The size of $\ce{F}$ is smaller compared to $\ce{N}$ and this factor over powers the repulsion between the paired $p$ orbital electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1434" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T10:49:02.207" LastActivityDate="2015-06-03T10:49:02.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32385" PostTypeId="1" AcceptedAnswerId="32390" CreationDate="2015-06-03T10:56:37.873" Score="12" ViewCount="36033" Body="&lt;p&gt;While I was reading about the usefulness of the quantity $\Delta H$, I found that it can be used to calculate the how the equilibrium constant varies with temperature. How can this be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it agree with the predictions of Le Chatelier's principle (that for an exothermic reaction, increasing the temperature disfavours product formation, and vice versa)?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="16683" LastEditDate="2016-12-08T11:03:25.260" LastActivityDate="2020-01-22T03:23:25.193" Title="Derivation of van 't Hoff equation for temperature dependence of equilibrium constant" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32386" PostTypeId="2" ParentId="32376" CreationDate="2015-06-03T11:08:32.060" Score="2" Body="&lt;p&gt;The premise of your question is weird. &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mg}$&lt;/span&gt; of calcium ion and &lt;span class=&quot;math-container&quot;&gt;$\pu{50 mg}$&lt;/span&gt; of carbonate ion are not equimolar. However, let's assume that they have appropriate counterions somewhere. This problem can be approached two ways.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Limiting reactant&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have a precipitation reaction between the calcium and the carboncate ions,  as calcium carbonate is nearly insoluble in water:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ca^2+ (aq) + CO3^2- (aq) -&amp;gt; CaCO3 (s) v}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convert your masses of ions to amount of substance and figure out which ion is in excess. If calcium ion is in excess, then use the amount of substance &lt;strong&gt;in excess&lt;/strong&gt; to determine the remaining calcium concentration.&lt;br&gt;&#xA;For example: Let's say you determined that your solution has &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1663 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1221 mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO3^2-}$&lt;/span&gt;. The maximum amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCO3}$&lt;/span&gt; that could precipitate is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1221 mol}$&lt;/span&gt;, because we would run out of carbonate ion. Then there would be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0442 mol}$&lt;/span&gt; of calcium ion left in &lt;span class=&quot;math-container&quot;&gt;$\pu{1 L}$&lt;/span&gt; of solution for a concentration of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0442 M}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Solubility product&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Calcium carbonate is not completely insoluble. It's solubility product constant is &lt;span class=&quot;math-container&quot;&gt;$4.8\times 10^{-9}$&lt;/span&gt;. Thus we can set this up as an ICE equilibrium problem. I'll use the same fake amount of substances from the previous example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{CaCO3 (s) &amp;amp;&amp;lt;=&amp;gt; Ca^2+ (aq) + CO3^2- (aq)}\\&#xA;K_\mathrm{sp}&amp;amp;=[\ce{Ca^2+}][\ce{CO3^2-}]&#xA;\end{align}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lcc}&#xA;\hline&#xA;                   &amp;amp; \text{calcium}/\pu{mol}  &amp;amp; \text{carbonate}/\pu{mol}\\&#xA;\hline&#xA;\text{Initial}     &amp;amp; 0.1663                   &amp;amp; 0.1221     \\&#xA;\text{Change}      &amp;amp; -x                       &amp;amp; -x         \\&#xA;\text{Equilibrium} &amp;amp; 0.1663 - x               &amp;amp; 0.1221 - x \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;4.8\times 10^{-9}&amp;amp;=(0.1663-x)(0.1221-x)&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; gives you &lt;span class=&quot;math-container&quot;&gt;$x=0.1221,0.1663$&lt;/span&gt; and only the first number makes sense. In this case the final calcium amount in solution would still be &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0442 mol}$&lt;/span&gt;. The carbonate amount in solution is not actually zero. It is &lt;span class=&quot;math-container&quot;&gt;$\pu{1.09E-7 M}$&lt;/span&gt;, but that is for another day.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2018-10-25T17:32:57.777" LastActivityDate="2018-10-25T17:32:57.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32387" PostTypeId="2" ParentId="28722" CreationDate="2015-06-03T12:00:44.557" Score="2" Body="&lt;p&gt;Partial pressure can be imagined to be the pressure of a gas in the absence of other gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence when the pressure of $\ce{Cl2}$ is reduced by 75% in the $\ce{POCl3_{(g)} &amp;lt;=&amp;gt; POCl_{(g)} + Cl2_{(g)}}$ system, the system naturally acts to counteract the change: by increasing the pressure of $\ce{Cl2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ideal gas law, $pv = nRT$ so pressure is directly proportionate to the number of moles of $\ce{Cl2}$ present. The position of equilibrium therefore shifts to the right, favouring the forward reaction that produces more $\ce{Cl2}$ (though not as much as to increase partial pressure of $\ce{Cl2}$ to its former value). This would mean it also consumes more $\ce{POCl3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for more $\ce{POCl}$ being consumed, if all other factors remain constant I'm not sure why this would happen.&lt;/p&gt;&#xA;" OwnerUserId="5402" LastActivityDate="2015-06-03T12:00:44.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32389" PostTypeId="1" CreationDate="2015-06-03T14:40:27.953" Score="3" ViewCount="2082" Body="&lt;p&gt;In the trans-hydroxylation of alkenes to give alcohols, there is a formation of epoxide.&lt;img src=&quot;https://i.stack.imgur.com/9jZzE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the reaction stop here if we don't add water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though if we add water, why does the product depends on which medium the reaction is carried whether acidic or basic?&#xA;My text says the ring opens differently for basic and acidic medium, but both finally gives us a trans-vicinal-diol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0y4b3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15161" LastActivityDate="2015-06-09T20:56:35.663" Title="Trans-hydroxylation of alkenes" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32390" PostTypeId="2" ParentId="32385" CreationDate="2015-06-03T14:50:25.160" Score="13" Body="&lt;p&gt;The equation linking &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation&quot; rel=&quot;noreferrer&quot;&gt;van 't Hoff equation&lt;/a&gt;. Since Philipp's comment on your question already links to a very thorough discussion of where the equation &lt;span class=&quot;math-container&quot;&gt;$\Delta G^\circ = -RT\ln{K}$&lt;/span&gt; comes from, I won't repeat it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the Gibbs free energy, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, is &lt;span class=&quot;math-container&quot;&gt;$G = H - TS$&lt;/span&gt;. Using &lt;span class=&quot;math-container&quot;&gt;$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$&lt;/span&gt; we obtain the Maxwell relation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\partial G}{\partial T}\right) = -S$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence the Gibbs–Helmholtz equation (derivation &lt;a href=&quot;https://chemistry.stackexchange.com/q/73760/16683&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\partial (G/T)}{\partial T}\right) = -\frac{H}{T^2} \quad \Leftrightarrow \quad \left(\frac{\partial (\Delta G^\circ/T)}{\partial T}\right) = -\frac{\Delta H^\circ}{T^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\ln K = -\Delta G^\circ/RT$&lt;/span&gt;, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = -\frac{1}{R}\frac{\mathrm d}{\mathrm dT}\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the differential form of the van 't Hoff equation; it's not the most useful thing to us though because it only tells you the slope of a plot of &lt;span class=&quot;math-container&quot;&gt;$\ln{K}$&lt;/span&gt; against &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; at one given point. We usually separate the variables and integrate with respect to both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int_{\ln{K_1}}^{\ln{K_2}}\!\mathrm d(\ln{K}) = \int_{T_1}^{T_2}\!\frac{\Delta H^\circ}{RT^2}\,\mathrm dT$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln{K_2} - \ln{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you know the equilibrium constant &lt;span class=&quot;math-container&quot;&gt;$K_1$&lt;/span&gt; at a certain temperature &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; and you want to find the equilibrium constant &lt;span class=&quot;math-container&quot;&gt;$K_2$&lt;/span&gt; at a different temperature &lt;span class=&quot;math-container&quot;&gt;$T_2$&lt;/span&gt;, you can just plug in your values into the equation and solve for &lt;span class=&quot;math-container&quot;&gt;$K_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this equation supports what you know of Le Chatelier's principle; if the reaction is exothermic, &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ &amp;lt; 0$&lt;/span&gt;, and if you increase the temperature from &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$T_2 &amp;gt; T_1$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$(1/T_1 - 1/T_2) &amp;gt; 0$&lt;/span&gt;. The RHS of the equation is therefore negative, and that means that &lt;span class=&quot;math-container&quot;&gt;$\ln{K_2} &amp;lt; \ln{K_1} \Rightarrow K_2 &amp;lt; K_1$&lt;/span&gt; which implies that the equilibrium position has shifted to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the last step (the integration) makes the assumption that &lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ$&lt;/span&gt; is a constant over the temperature range &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$T_2$&lt;/span&gt;. Note that this is, in general, not true but if the temperature range isn't too huge you will get pretty accurate results from the use of this equation.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2018-11-12T01:19:18.467" LastActivityDate="2018-11-12T01:19:18.467" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32391" PostTypeId="2" ParentId="32346" CreationDate="2015-06-03T15:02:08.687" Score="2" Body="&lt;p&gt;In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta n$ represents a &lt;em&gt;change&lt;/em&gt; in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-06-03T15:02:08.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32392" PostTypeId="1" CreationDate="2015-06-03T15:29:53.447" Score="2" ViewCount="36" Body="&lt;p&gt;In this &lt;a href=&quot;https://www.youtube.com/watch?v=q_nf7JmB9iI&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;, at 1:30, the presenter remarks that the charge of $\mathrm{Na}$ is $1^{+}$ and that the charge of $\mathrm{Cl}$ is $1^{-}$. Then, at 2:00, he remarks that the charge of $\mathrm{Fe}$ is $3^{+}$ and that the charge of $\mathrm{O}$ is $2^{-}$. Finally, he remarks that the result of substituting $3^{+}$ and $2^{-}$ for $q_1$ and $q_2$ in $ E_{\mathrm{latice}} =\frac{Kq_1q_2}{r}$ is six times the result of substituting $1^{+}$ and $1^{-}$ in the same equation. Hence he assumes that the value of $r$ is the same for both compounds, and concludes that the lattice energy of $\mathrm{Fe_2O_3}$ is six tiems greater than the lattice energy of $\mathrm{NaCl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that the fact that $\mathrm{Fe_2O_3}$ comprises more charged atoms than $\mathrm{NaCl}$ comprises would affect the differences between their lattice energies. Accordingly, I thought that the presenter should have multiplied the charge of $\mathrm{Fe}$ by $2$ and multiplied the charge of $\mathrm{O}$ by $3$, and then substituted those two products into $q_\mathrm{1}q_\mathrm{2}$ respectively? Evidently, I was wrong about that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In estimating the relative lattice energies, should the estimator consider the number of each kind of atom in the compounds?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-06-03T15:29:53.447" Title="Should I consider the quantity of each atom when estimating the relative lattice energies of two compounds?" Tags="&lt;energy&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32393" PostTypeId="1" AcceptedAnswerId="32395" CreationDate="2015-06-03T15:50:08.070" Score="2" ViewCount="1429" Body="&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; Compute the potential of the following galvanic cell: $$(\ce{Mg (s)} \mid \ce{Mg^2+ (aq)}\ (0.30\ \mathrm{M}) \mid \mid \ce{Sn(OH)4^2- (aq)}\ (0.10\ \mathrm{M}); \mathrm{pH} = 9 \mid \ce{Sn (s)})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; I know we have to use the formula $$E_\text{cell} = E - \frac{0.0592}{n} \log Q$$ The anode is situated on the left; this means electrons are going away there. So we have for the oxidation half-reaction: $$\ce{Mg -&amp;gt; Mg^2+ + 2e-}$$ The potential for this is $E_1 = 2.36\ \mathrm{V}$. Then I’m not sure how to write my reduction half-reaction. I just wrote: $$\ce{Sn(OH)4^2- + 2e- -&amp;gt; Sn(s)}$$ and $E_2 = -0.91\ \mathrm{V}$. But this doesn’t seem correct, because if I add both reactions to get the sum, my charges aren’t balanced anymore. Also, I wonder what happens with the $\ce{(OH)4^2-}$ ? It’s not on the right side of the arrow anymore.&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorDisplayName="user7951" LastEditDate="2015-06-03T19:54:34.413" LastActivityDate="2015-06-03T19:54:34.413" Title="Balancing reduction equations for galvanic cell" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32394" PostTypeId="2" ParentId="32336" CreationDate="2015-06-03T16:00:01.480" Score="3" Body="&lt;p&gt;In general, explosion of &lt;em&gt;alkali metal&lt;/em&gt; perchlorates (at least in small quatities) is deflagration, with the perchlorate providing oxygen for combustion of another component of a mixture, rather than detonation (due to decomposition of the perchlorate). However, perchloric acid itself and other inorganic and organic perchlorates such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Pyrotechnic_initiator&quot; rel=&quot;nofollow&quot;&gt;cis-bis-(5-nitrotetrazolato)tetraminecobalt(III) perchlorate&lt;/a&gt; can transition from deflagration to brissant &lt;strong&gt;detonation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your specific question, &lt;em&gt;gaseous&lt;/em&gt; methane would burn more readily in the presence of Group I perchlorates, but the flame-front would be limited by the small contact area and low concentration of methane at the solid/gaseous interface. On the other hand, if one were to synthesize &lt;a href=&quot;https://sites.google.com/site/energeticchemical/alkyl-perchlorate-esters&quot; rel=&quot;nofollow&quot;&gt;methyl perchlorate&lt;/a&gt; or  &lt;a href=&quot;http://pipeline.corante.com/archives/2006/05/30/things_i_wont_work_with_frisky_perchlorates.php&quot; rel=&quot;nofollow&quot;&gt;ethyl perchlorate&lt;/a&gt;, one might not be able to perform subsequent syntheses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to avoid maiming or perhaps a terminal experiment, eschew perchlorates. &lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-03T16:00:01.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32395" PostTypeId="2" ParentId="32393" CreationDate="2015-06-03T16:15:57.820" Score="3" Body="&lt;p&gt;To construct reduction equations there are a set of steps you can follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -&amp;gt; Sn(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under acidic conditions the only things we can add to balance the reaction are $\ce{H=}$, $\ce{H2O}$ and $\ce{e-}$. Under alkaline conditions we can add $\ce{OH-}$ instead of $\ce{H+}$. Usually, when dealing with alkaline reactions the best way of dealing with them is to treat them as acidic and then neutralize the acid later. However, in this instance, since the only reactant involves hydroxide ions, we can immediately balance the hydrogens and oxygens just by adding four hydroxide ions to the products: $$\ce{[Sn(OH)4]^{2-}(aq) -&amp;gt; Sn(s) + 4OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then balance the charges to get the overall equation: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -&amp;gt; Sn(s) + 4OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative method which works for any alkaline reactions, which are sometimes less obvious than the example above, is to balance the reaction under acidic conditions and then neutralize the acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we only have one option for balancing the oxygens we add $\ce{H2O}$ first: $$\ce{[Sn(OH)4]^{2-}(aq) -&amp;gt; Sn(s) + 4H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now balance the hydrogens: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) -&amp;gt; Sn(s) + 4H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the charges: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) + 2e- -&amp;gt; Sn(s) + 4H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were in acidic solution then this would be the final balanced equation. However, since we are in alkaline solution, we add $\ce{OH-}$ ions to cancel out the $\ce{H+}$, forming $\ce{H2O}$. $$\ce{[Sn(OH)4]^{2-}(aq) + 4H2O(l) + 2e- -&amp;gt; Sn(s) + 4H2O(l) + 4OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we can cancel the excess $\ce{H2O}$ to give: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -&amp;gt; Sn(s) + 4OH- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same result as we got earlier by the quicker method.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-03T16:15:57.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32396" PostTypeId="2" ParentId="28029" CreationDate="2015-06-03T16:31:01.167" Score="2" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a carbonyl compound reacts with a nucleophile we form a tetrahedral intermediate.  Since the carbonyl carbon becomes $\ce{sp^3}$ hybridized in the tetrahedral intermediate there is no significant resonance stabilization - the various tetrahedral intermediates have (approximately) the same energy.  However, strong resonance effects can occur in the starting carbonyl compound, potentially stabilizing the carbonyl compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, while a substituent attached to the carbonyl carbon can stabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we place the carbonyl compound and the tetrahedral intermediate on a reaction coordinate we can see that anything that stabilizes the carbonyl compound will increase the activation energy for reaction, slowing down the reaction rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OWtmp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are the resonance structures we can draw for our carbonyl compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M1UsZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If structures II and  III contribute significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, &lt;strong&gt;lower in energy&lt;/strong&gt;, than a carbonyl compound where structures II and III contribute less.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three of your compounds are esters, one is an amide.  We know that resonance structure III is more important in an amide than in an ester (due to electronegativity differences, positive charge on nitrogen is better than positive charge on oxygen).  Therefore the amide (B) will react the slowest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to separate the three esters.  You are right that ester C will be the most reactive.  This is because the inductively electron withdrawing chlorine substituent destabilizes resonance structure II.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In compound D we can draw stabilizing resonance structures like the following;  the &lt;em&gt;para&lt;/em&gt;-hydroxy substituted aromatic ring is very effective at moving electron density towards the electron deficient carbonyl carbon thereby stabilizing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ybnDm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methyl group in compound A also donates electron density to the carbonyl carbon, but not nearly as effectively as the aromatic ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore our overall &lt;strong&gt;order of carbonyl reactivity is C &gt; A &gt; D &gt; B&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-06-03T16:31:01.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32397" PostTypeId="2" ParentId="32367" CreationDate="2015-06-03T17:22:54.813" Score="11" Body="&lt;p&gt;The data I could find suggests that cyclic ethers have higher dipole moments than their acyclic counterparts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{Four~carbons}$$&#xA;\begin{array}{c @{} c} \hline&#xA;\text{THF} &amp;amp; \mathrm{1.63~ D ^1, 1.75~ D ^2} \\&#xA;\text{1-butene oxide} &amp;amp; \mathrm{1.89~ D^2} \\&#xA;\text{ethyl vinyl ether} &amp;amp; \mathrm{1.26~ D ^2} \\&#xA;\text{diethyl ether} &amp;amp; \mathrm{1.15 ~D ^3, 1.15 D ^2} \\&#xA;\text{methyl propyl ether} &amp;amp; \mathrm{1.11 ~D ^2} \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{Five~carbons}$$&#xA;\begin{array}{c @{} c} \hline&#xA;\text{Tetrahydropyran} &amp;amp; \mathrm{1.58~D ^2, 1.87~D ^4} \\&#xA;\text{2-Methyltetrahydrofuran} &amp;amp; \mathrm{1.38~ D ^5} \\&#xA;\text{dipropyl ether} &amp;amp; \mathrm{1.21~D^2, 1.00~D^6} \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems consistent with intuitive expectations based on conformational models where, in the cyclic ethers where the molecular geometry is constrained by the ring, the lone pairs are pointing in the opposite direction from the carbon skeleton. Whereas the lone pairs and carbon chains in the conformationally more mobile acyclic isomers are less geometrically fixed and therefore less &quot;directed&quot; in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems likely that the higher dipole moments in the cyclic compounds would lead to greater alignment / ordering in the liquid phase, which in turn would lead to higher boiling points.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&#xA;  1: &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetrahydrofuran&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Tetrahydrofuran&lt;/a&gt;&lt;br/&gt;&#xA;  2: &lt;a href=&quot;https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/&quot; rel=&quot;nofollow&quot;&gt;https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/&lt;/a&gt; &lt;br/&gt;&#xA;  3: &lt;a href=&quot;http://en.wikipedia.org/wiki/Diethyl_ether&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Diethyl_ether&lt;/a&gt;&lt;br/&gt;&#xA;  4: &lt;a href=&quot;http://www.drugfuture.com/chemdata/tetrahydropyran.html&quot; rel=&quot;nofollow&quot;&gt;http://www.drugfuture.com/chemdata/tetrahydropyran.html&lt;/a&gt;&lt;br/&gt;&#xA;  5: &lt;a href=&quot;http://www.stenutz.eu/chem/solv28.php?s=1&amp;amp;p=20&quot; rel=&quot;nofollow&quot;&gt;http://www.stenutz.eu/chem/solv28.php?s=1&amp;amp;p=20&lt;/a&gt;&lt;br/&gt;&#xA;  6: &lt;a href=&quot;https://books.google.com/books?id=G6jaBwAAQBAJ&amp;amp;pg=PA50&amp;amp;lpg=PA50&amp;amp;dq=propyl%20ethyl%20ether%20dipole%20moment&amp;amp;source=bl&amp;amp;ots=VA52gqJ0kn&amp;amp;sig=GM8tua_QpXccFW3VXpy9qiNf7rk&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=rzRvVZ__Cc32yQT9gIPACQ&amp;amp;ved=0CDsQ6AEwBQ#v=onepage&amp;amp;q=propyl%20ethyl%20ether%20dipole%20moment&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;https://books.google.com/books?id=G6jaBwAAQBAJ&amp;amp;pg=PA50&amp;amp;lpg=PA50&amp;amp;dq=propyl%20ethyl%20ether%20dipole%20moment&amp;amp;source=bl&amp;amp;ots=VA52gqJ0kn&amp;amp;sig=GM8tua_QpXccFW3VXpy9qiNf7rk&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=rzRvVZ__Cc32yQT9gIPACQ&amp;amp;ved=0CDsQ6AEwBQ#v=onepage&amp;amp;q=propyl%20ethyl%20ether%20dipole%20moment&amp;amp;f=false&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7448" LastEditDate="2015-06-03T20:56:28.170" LastActivityDate="2015-06-03T20:56:28.170" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32398" PostTypeId="1" AcceptedAnswerId="32413" CreationDate="2015-06-03T18:22:38.777" Score="3" ViewCount="7075" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{CO + H2O &amp;lt;=&amp;gt; CO2 + H2}$$&#xA;  A stoichiometric mixture of $\ce{CO(g)}$ and $\ce{H2O(g)}$ is allowed to reach equilibrium. $x$ is the mole fraction of $\ce{H2(g)}$ present.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am meant to find the result that $$K_p=\frac{4x^2}{(1-2x)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I only get: $K_p=\frac{x^2}{(1-x)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I am thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p=\frac{x_{\ce{CO_2}}x_{\ce{H_2O}}}{x_{\ce{CO}}x_{\ce{H_2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But: $x(\ce{CO})=1-\alpha=x(\ce{H_2O})$ and $x(\ce{CO_2})=\alpha=x(\ce{H_2})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting this in gives me my result. What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="6925" LastEditDate="2015-06-03T21:31:14.730" LastActivityDate="2016-03-11T23:25:02.467" Title="Equilibrium constant in terms of mole fractions" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32399" PostTypeId="1" AcceptedAnswerId="32406" CreationDate="2015-06-03T18:23:46.330" Score="25" ViewCount="29066" Body="&lt;p&gt;As pointed out in the comments to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32367/why-do-cyclic-ethers-have-higher-boiling-points-than-their-acyclic-isomers&quot;&gt;this question&lt;/a&gt;, cyclic hydrocarbons have higher boiling points than their acyclic isomers. The major attractive force for hydrocarbons should be the London forces, which scale with surface area. My intuition has me thinking the cyclic molecules have less surface area, but they might enable more productive interactions over larger parts of their surface area due to their fixed conformations. &lt;strong&gt;I would love to see a study or some data that supports my intuition.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling point data&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopropane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclopropane&lt;/a&gt; - BP $=-33\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Propene&lt;/a&gt; - BP $=-47.6\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Propane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Propane&lt;/a&gt; - BP $=-42\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Four carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclobutane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclobutane&lt;/a&gt; - BP $=12.5\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Methylcyclopropane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Methylcyclopropane&lt;/a&gt; - BP $=4\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1-Butene&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-butene&lt;/a&gt; - BP $=-6.7\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Butane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Butane&lt;/a&gt; - BP $=-1\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Five carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopentane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclopentane&lt;/a&gt; - BP $=49.2\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pentene&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-Pentene&lt;/a&gt; - BP $=30\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclopentane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pentane&lt;/a&gt; - BP $=35.9\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Six carbons&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclohexane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyclohexane&lt;/a&gt; - BP $=80.7\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/1-Hexene&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-Hexene&lt;/a&gt; - BP $=63\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hexane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hexane&lt;/a&gt; - BP $=68.5\ ^\circ\text{C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2016-01-08T15:51:27.070" Title="Why do cyclic hydrocarbons have higher boiling points than their acyclic isomers?" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;hydrocarbons&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32400" PostTypeId="2" ParentId="32353" CreationDate="2015-06-03T19:15:13.000" Score="5" Body="&lt;p&gt;I would hazard a guess that the economies of scale of where the process is located comes into play. Arkema may make bulk $\ce{H2O2}$ at industrial concentration so can just feed it in without transport and hence dilution isn't required . Bayer (or a neighbour) may operate a chloralkali process on site, so use that. Everything will come down to the input costs.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T19:40:31.300" LastActivityDate="2015-06-03T19:40:31.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32401" PostTypeId="2" ParentId="32399" CreationDate="2015-06-03T19:16:53.033" Score="4" Body="&lt;p&gt;The best that I have been able to determine is that it has to do with the rigidity of the rings. With the rings locked into certain conformations, the molecules are able to come into closer contact, as the movement of the chains in the non ring compounds can clear a larger around the molecules. Rings, on the other hands, are locked into certain conformations, so their area of contact is consistently larger. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-03T19:16:53.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32402" PostTypeId="2" ParentId="32322" CreationDate="2015-06-03T19:23:40.413" Score="5" Body="&lt;p&gt;It's a bit of a blanket assumption as processes exist where the catalyst is over-stoichiometric. This would knock the atom economy argument on the head.  Also one could argue that the catalytic reaction is no different as without a catalyst, the reaction still occurs albeit more slowly, all other things being equal.  It's a completely reaction specific consideration. Technically the catalyst mass shouldn't be included in the calculation.  Atom economy is a bit of a misleading measure, mass intensity is better. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-03T19:23:40.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32403" PostTypeId="2" ParentId="32098" CreationDate="2015-06-03T19:46:40.217" Score="7" Body="&lt;p&gt;The reaction of $\ce{H2O2}$ with hydrazine forms diazene, $\ce{HN=NH}$.  This reacts in a syn fashion to deliver both hydrogens to the same side of the alkene.  &lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorDisplayName="user15489" LastEditDate="2015-06-03T19:47:26.803" LastActivityDate="2015-06-03T19:47:26.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32405" PostTypeId="1" CreationDate="2015-06-03T20:42:40.313" Score="0" ViewCount="191" Body="&lt;p&gt;Im trying to figure out the mM concentration of capric acid in coconut oil. I know that There is 8% capric acid in coconut oil, but I would like to know how much capric acid is in coconut oil in a mM ammount, not a percent. I know the molecular weight of capric acid (decanoic acid) is 172.26 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason being: I read a study that capric acid solutions are effective against microbes and fungus at a mM concentration of 20 mM. Im wondering if coconut oil's 8% of capric acid converts to roughly around a 20mM concentration of capric acid within the oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks very much for your answers!&lt;/p&gt;&#xA;" OwnerUserId="16726" LastEditorUserId="865" LastEditDate="2015-10-30T17:05:23.080" LastActivityDate="2016-01-28T04:19:16.153" Title="Finding the mM concentration of capric acid (decanoic acid) in coconut oil" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32406" PostTypeId="2" ParentId="32399" CreationDate="2015-06-03T20:59:45.503" Score="21" Body="&lt;p&gt;Your intuition is indeed correct! Several sources provide the same answer (&lt;a href=&quot;http://people.chem.ucsb.edu/neuman/robert/orgchembyneuman.book/02%20Alkanes/02Separates/02Text.pdf&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Cycloalkane#Properties&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt;). Perhaps the simplest and most direct evidence comes from comparing the densities of the liquid unbranched alkanes and cycloalkanes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FhS7g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cycloalkanes have slightly lower molecular mass than their parent unbranched alkanes with the same number of carbon atoms, yet the cycloalkanes are consistently more dense in the liquid phase by a factor of ~20%. The only way this can happen is if each individual cycloalkane molecule effectively occupies less volume in the liquid phase than the related alkane. This means cycloalkane molecules are closer on average in the liquid, and as van der Waals forces are attractive (at intermolecular separations) and their strength varies inversely with distance, the intermolecular attractions are stronger in cycloalkanes, and so the boiling point of the cycloalkanes should be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be explained by considering that the unbranched linear chains in alkanes have many possible conformations, and thus are in more geometrically irregular contact in the liquid, leaving more space on average between the molecules. Cycloalkanes do not have as many conformations available, and so their molecules will approach each other more orderly and effectively, leaving less intermolecular space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably this effect can be extended to all substances (including the ethers mentioned in the question you linked), but it need not always be the dominant factor in determining boiling points.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-03T22:04:27.877" LastActivityDate="2015-06-03T22:04:27.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32407" PostTypeId="1" CreationDate="2015-06-03T21:24:49.073" Score="25" ViewCount="6193" Body="&lt;p&gt;A question on an exam asked why there is exactly one sigma bond in double and triple covalent bonds. I looked in my text and online after the exam, but couldn't find an anawer to the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can there not be more than one sigma bond in a set of covalent bonds?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastEditorUserId="4231" LastEditDate="2015-06-06T23:06:18.547" LastActivityDate="2015-09-13T15:17:10.220" Title="Why can there not be more than one sigma bond in a set of bonds?" Tags="&lt;bond&gt;&lt;hybridization&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32409" PostTypeId="2" ParentId="32399" CreationDate="2015-06-03T21:36:18.330" Score="13" Body="&lt;p&gt;I agree with the points made above regarding attractive forces in the liquid being diminished in the gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I also think there may be &lt;strong&gt;an entropic component&lt;/strong&gt; to all of this.  When a molecule undergoes a phase change from liquid to gas (&lt;em&gt;e.g&lt;/em&gt;. it boils) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G_\mathrm{vap} = \Delta H_\mathrm{vap} - T_\mathrm{vap} \times \Delta S_\mathrm{vap} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Delta H_\mathrm{vap}}{T_\mathrm{vap}} = \Delta S_\mathrm{vap}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a ring, &lt;strong&gt;torsional vibrations&lt;/strong&gt; (rotations about the $\ce{C-C}$ single bonds) &lt;strong&gt;are constrained&lt;/strong&gt; in both the liquid and gas phase compared to an acyclic analogue where there is free rotation about the single bonds.  Hence, we would expect &lt;strong&gt;$\Delta S_\mathrm{vap}$ to be smaller for a cyclic compound compared to a noncyclic analogue&lt;/strong&gt;. For $\Delta S_\mathrm{vap}$ to be smaller either &lt;strong&gt;$T$ must be larger (higher boiling point) and/or $\Delta H_\mathrm{vap}$ must be smaller&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see how the arguments presented in the other answers can produce an &lt;strong&gt;increase&lt;/strong&gt; in the enthalpic term for rings, but this would tend to move $\Delta S_\mathrm{vap}$ in the wrong direction to completely explain the effect (&lt;em&gt;e.g.&lt;/em&gt; $\Delta S_\mathrm{vap}$ would be larger for the rings than non-rings).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when comparing compounds like cyclopentane and pentane, I suspect that the enthalpic term for the ring is similar to or larger than that for the non-ring.  Yet since the entropic term is smaller for the ring compared to the non-ring, this requires that &lt;strong&gt;$T$ must be larger for the ring compared to the non-ring&lt;/strong&gt;.  In other words, the boiling point for the ring will be higher than the boiling point of the non-ring.  Also, this reasoning would explain why, as rings get larger and are able to rotate about their bonds more freely, the magnitude of the difference between rings and non-rings diminishes.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorDisplayName="user7951" LastEditDate="2015-06-03T23:24:36.720" LastActivityDate="2015-06-03T23:24:36.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32410" PostTypeId="1" CreationDate="2015-06-03T22:17:11.823" Score="1" ViewCount="1281" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two quantities of water ($M_\mathrm{r}=18$) each of mass $216\ \mathrm{g}$ are mixed together in a&#xA;  vessel. The temperatures of the two quantities before mixing are $303\ \mathrm{K}$ and $333\ \mathrm{K}$. The entire system is perfectly insulated and the vessel has a negligible heat capacity. Calculate the change in entropy after equilibrium has been established, and comment on your result in the light of the Second Law of Thermodynamics.&lt;br&gt;&#xA;  Note that $C_{\mathrm{m},p} = 75.5\ \mathrm{J\ K^{−1}\ mol^{−1}}$ for liquid water in this temperature range.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it correct to assume that the temperature of the water at equilibrium will be half way between the two values (i.e. $318\ \mathrm{K}$)? If so, I think the rest of the problem can be solved by using $\Delta S=\frac{C_V}T\Delta T$. Is this correct or is it more complicated?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user7951" LastEditDate="2015-06-08T20:06:41.270" LastActivityDate="2015-06-08T20:06:41.270" Title="Entropy change when liquids at two different temperatures are mixed" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32411" PostTypeId="2" ParentId="32405" CreationDate="2015-06-03T22:34:59.603" Score="2" Body="&lt;p&gt;Thank you very much for a well-specified question.  If 8% of coconut oil were &lt;em&gt;free&lt;/em&gt; capric acid, then as noted in the comments the units of &quot;%&quot;, i.e. w/w, w/v, or v/v, would affect the calculation somewhat, but since capric acid likely has a density that is very close to that of bulk coconut oil, the difference will be slight.  8% w/w, and assuming a density of 0.8 g/mL for coconut oil would yield $0.08~{\frac{\rm g~capric~acid}{\rm g~oil}}\times \frac{\rm mol~capric~acid}{172~\rm g~capric~acid}\times0.8\frac{\rm g~oil}{\rm mL}\times\frac{1000~\rm mL}{\rm L}=0.372\frac{\rm mol~capric~acid}{\rm L}=372~\rm{mM}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I suspect that the capric acid in coconut oil is not free acid.  It is part of &lt;a href=&quot;http://en.wikipedia.org/wiki/Triglyceride&quot; rel=&quot;nofollow&quot;&gt;triglycerides&lt;/a&gt;.  Instead of containing capric acid, the oil contains an ester of capric acid with glycerol.  Triglycerides are molecules where all of the available 3 hydroxyl groups of glycerol are esterified with a fatty acid.  Because triglycerides are large complex molecules, they were historically difficult for chemists to analyze directly.  The method of choice for analysis of plant oils was transesterification or ester hydrolysis, converting the &lt;strong&gt;tri&lt;/strong&gt;ester triglyceride molecules into three smaller &lt;strong&gt;mono&lt;/strong&gt;ester molecules that were easier to detect by analysis.  The 8% figure you are using is probably the result of such an analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that coconut oil that is &quot;8% capric acid&quot; actually contains very little capric acid, but contains a large amount of glycerol-esterified capric acid.  But only free capric acid is likely to be antimicrobial.  So the capric acid content of coconut oil is unlikely to make it antimicrobial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of free fatty acids in plant oils is usually a sign of degradation.  In olive oil, an acidity above 0.5% (i.e. of all the fatty acid chains in the olive oil, 1 in 200 are in the free form and 199 in 200 are in the esterified form) is a sign of lower quality.  I would imagine the for coconut oil the situation is similar, although the specific numbers used as quality thresholds may be different.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-06-03T23:05:02.687" LastActivityDate="2015-06-03T23:05:02.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32413" PostTypeId="2" ParentId="32398" CreationDate="2015-06-03T23:03:32.120" Score="2" Body="&lt;p&gt;The initial mole fraction of $\ce{CO}$ present must be $y=0.5$.  Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can figure out the rest.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:25:02.467" LastActivityDate="2016-03-11T23:25:02.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32414" PostTypeId="2" ParentId="7688" CreationDate="2015-06-03T23:22:45.420" Score="2" Body="&lt;p&gt;Why would you have to solve differential equations for this problem?  What algebra have you tried so far?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C is in excess, then there is no competition!  You should be able to model the reactions as just $\ce{A &amp;lt;=&amp;gt; AC}$ and $\ce{B &amp;lt;=&amp;gt; BC}$.  The equilibrium equation would be (for A) $K_a=\frac{ac}{A_{init}-ac}$, and you can apparently already calculate $K_a$ from the thermodynamic parameters, so that is an equation where the only unknown is $ac$, the amount of AC formed.  The equation is linear!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ac=A_{init}\frac{K_a}{1+K_a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar result holds for B of course.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you can calculate $ac$ and $bc$, you should be able to calculate the parameters you want to know very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C is not in excess, the equilibrium equations will not be linear and the balance on C will couple the two equilibria (competition happens!).  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-03T23:22:45.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32415" PostTypeId="1" AcceptedAnswerId="32570" CreationDate="2015-06-04T00:50:56.830" Score="1" ViewCount="1377" Body="&lt;p&gt;In this reaction,&lt;br&gt;&#xA;$$\ce{KIO_3 + KI -&amp;gt; I_2}$$ &#xA;According to law of chemical equivalence, shouldn't $m_{eq}$ of $\ce{KIO_3}$ = $m_{eq}$ of $\ce{KI}$ = $m_{eq}$ of $\ce{I_2}$?&lt;br&gt;&#xA;However, the book that I was reading from, said:&lt;br&gt;&#xA;Total $m_{eq}$. of $\ce{I_2}$ = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO_3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$).&lt;br&gt;&#xA;Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="16702" LastEditorDisplayName="user15489" LastEditDate="2015-06-04T01:04:00.820" LastActivityDate="2015-06-10T20:03:55.033" Title="Law of chemical equivalence" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32416" PostTypeId="1" AcceptedAnswerId="32442" CreationDate="2015-06-04T00:59:37.307" Score="7" ViewCount="1881" Body="&lt;p&gt;I’m researching how the classic “alchemy” demonstration works in which zinc is plated onto a copper penny in a solution of sodium hydroxide. Most online resources refer to the electroplating of copper with zinc, which is not what I’m interested in. The &lt;a href=&quot;http://www.rsc.org/images/alchemy_tcm18-188862.pdf&quot; rel=&quot;noreferrer&quot;&gt;best article I’ve been able to find&lt;/a&gt; on the electroless plating of zinc onto copper is from the Royal Chemistry Society. However, I have several questions. The three reactions this article describes are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn(s) + 2NaOH(aq) + 2H2O(l) -&amp;gt; Na2Zn(OH)4(aq) + H2(g)}$, which is the “formation of sodium zincate”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn(s) -&amp;gt; Zn^2+(aq) + 2e- }$, which is the “electrode reaction” at the “zinc electrode” and is “followed by the complexing of the zinc ions as $\ce{Zn(OH)4^2- (aq)}$”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn(OH)4^2- (aq) + 2e- -&amp;gt; Zn(s) + 4OH- (aq)}$, which is the “electrode reaction” at the “copper electrode”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don’t understand how these three reactions come together to describe the plating of the zinc onto the copper. Specifically, I don’t understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the role of zincate, $\ce{Zn(OH)4^2- (aq)}$, in the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) On the molecular level, how are the zinc and copper actually attached together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How is this an electrochemical cell (i.e. how is electricity involved)? Where can I learn more about electrochemical cells?&lt;/p&gt;&#xA;" OwnerUserId="16730" LastEditorDisplayName="user7951" LastEditDate="2015-06-04T12:51:22.800" LastActivityDate="2017-07-05T04:08:34.340" Title="Electroless Plating of Zinc onto Copper in NaOH Solution" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32417" PostTypeId="2" ParentId="32407" CreationDate="2015-06-04T01:02:28.293" Score="10" Body="&lt;p&gt;I’m not sure if this answering attempt is correct in the light of Mithoron’s and ron’s comments on your question, but this is the way I learnt it, so if this is wrong I will at least learn something, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know what s-, p- and d-orbitals look like, but what is the significance, and why do these orbitals preferentially form $\sigma$, $\pi$ and $\delta$ bonds, respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically spoken, orbitals are functions of the hydrogen atom that solve the Schrödinger equation. The model in question is a non-rigid rotor,* i.e. the rotor’s axis is not fixed in any spatial direction (the electron can rotate freely around the nucleus). For solving this equation, it is helpful to use polar coordinates $(r, \varphi, \theta)$, mainly because the solution can be split into a radial factor (dependent only on $r$) and angular factors (dependent on $\varphi$ and $\theta$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Psi (r, \varphi, \theta) = R(r) \cdot Y(\varphi, \theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(r)$ can be thought of giving an orbital its extension into space while $Y(\varphi, \theta)$ gives it its shape. Both functions depend heavily on quantum numbers: $R(r)$ does so for $n$ and $l$ while $Y(\varphi, \theta)$ depends on $l$ and $m_l$. For the simplest case ($l = 0; m_l = 0$, s-orbital), $Y (\varphi, \theta)$ degenerates to a simple constant, meaning that the orbital will have a totally symmetrical spherical shape. $l = 1$, (the p-orbital) while loosing spherical symmetry, still keeps total symmetry with respect to one axis, i.e. every slice you take through that orbital perpendicular to the axis of symmetry will be a circle. Higher quantum numbers lose more symmetry but it’s not always as easily visualised, so I’ll stick with these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you were talking about bonds, where do they come into play? Well, bonds also have a symmetry, but they also have an axis instead of a nucleus, so their symmetry will be reduced &lt;em&gt;per se&lt;/em&gt;. The simplest symmetry along a bond axis is total rotational symmetry around the bonds axis. I hope you see the similarity between the s-orbital (total symmetry around a central point) and a  $\sigma$ bond (total symmetry around the bond’s central &lt;em&gt;axis&lt;/em&gt;). Similarly, a $\pi$ bond will always have &lt;em&gt;one&lt;/em&gt; degree of symmetry less, which turns out to mean ‘having a plane of symmetry that includes the bond axis’. And a $\delta$ bond will have two planes of symmetry — yet another degree of symmetry less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this definition, an orbital that can take part in a $\sigma$ bond needs to have full rotational symmetry along the bond’s axis. That means, that there is only one, at most two orbitals that fulfil the criterion (but if there are two, one is going to be an unmodified s-orbital and likely not take part in bonding at all). Therefore, only one $\sigma$ bond would be possible between two atoms.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Writing this up, I remembered the ‘banana bonds’ that were introduced to us to explain the extremely small ($60°$) bond angles in $\ce{P4}$. I would need to go back, recheck and rethink what I would think of those and if I would treat them as exceptions of this ‘rule’ or simply as special cases that need additional information to be discussed. They certainly deserve consideration, as they are, &lt;em&gt;de facto&lt;/em&gt; $\sigma$ bonds from the way they look, even though they do bend.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An interesting comment was left on the question pointing to sextuple bonds. I didn’t know bonds of that order existed; my knowledge was stuck at 4. For a quadruple bond, possible between certain transition metals such as in $\ce{[Re2Cl8]^2-}$, four of the five d-orbitals form a bond to the other metal; one being $\sigma$, two being $\pi$ and a fourth one of $\delta$ type (to planes of symmetry). Extending that to a quintuple bond by adding a second $\delta$ layer with the last remaining pair of d-orbitals isn’t hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sextuple bond – eg $\ce{Mo2}$ — derives from an additional $\sigma$ bond between the s-orbitals of the higher shell. You thereby solve a problem you would otherwise have: The $4\mathrm{d}_{z^2}$ orbital can take part in $\sigma$ bonding along the $z$-axis; and the higher $5\mathrm{s}$ orbital is more diffuse, extends further into space and therefore is still able to form a contact to the neighbouring atom’s counterpart. Because it is more or less a sphere, it can only form $\sigma$ bonds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* I don’t think this is the model’s correct name. In my German quantum chemistry class, the rigid rotor was a &lt;em&gt;raumstarrer Rotator&lt;/em&gt; and thus the model here was a &lt;em&gt;raumfreier Rotator&lt;/em&gt;. Somebody who might know the proper name please comment (or edit).&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-05T23:17:26.597" LastActivityDate="2015-06-05T23:17:26.597" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32418" PostTypeId="1" CreationDate="2015-06-04T03:33:43.977" Score="3" ViewCount="617" Body="&lt;p&gt;I was recently asked if $\ce{Se}$ will bond with $\ce{V}$, creating a compound $\ce{SeV}$ . I am not sure if this is even possible but if so I'd like to find out.&lt;/p&gt;&#xA;" OwnerUserId="16734" LastEditorUserId="4945" LastEditDate="2015-06-04T03:57:27.530" LastActivityDate="2015-06-04T12:42:37.623" Title="Will selenium react or bond, ionic or covalent, with vanadium?" Tags="&lt;metal&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32419" PostTypeId="2" ParentId="31463" CreationDate="2015-06-04T04:09:23.377" Score="5" Body="&lt;p&gt;Since now this question has a bounty of 400 points, I think it now deserves some well thought answer. As a part of that endeavor, I am adding another elaborate answer.&lt;br&gt;&#xA;The overall contribution to $C_v=C_{v,\text{trans}}+C_{v,\text{rot}}+C_{v,\text{vib}}$ and translational and vibrational partition function don’t show that maxima and reaches equipartition value monotonically and overall this three contribution results in $C_v=3R$, which is an expected result.&lt;br&gt;&#xA;To get the rotational partition function, we use rigid motor approximation. For a simple diatomic molecule, the moment of inertia is $I=\mu \times r_0^2$ , where ${\mu}$ is reduced mass and is calculated as $\mu=\frac{m_1 \times m_2}{m_1+m_2}$ and ${r_0}$ is equilibrium distance between two atoms. Now, from quantum mechanics we know that energy level of rotational partition function can be written as&lt;br&gt;&#xA;$$\epsilon_l=l(l+1)\frac{\hbar^2}{2I}$$ and defining characteristic temperature as $\theta_r=\frac{\hbar^2}{2Ik}$, we can rewrite that equation as $$\epsilon_l=l(l+1)k\theta_r$$ &#xA;Also from quantum mechanics, we know degeneracy of energy levels are ${g_l=(2l+1)}$&lt;br&gt;&#xA;Now the partition function $$q_r=\sum g_l\operatorname{e}^{-\epsilon_l/kT}=\sum (2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$&#xA;Now at high temperature closely spaced energy states can be considered as continuous variable. So, now we can replace the sum by integral and to make our life easier, we can perform a variable change here. Let, $x=l(l+1)$ so $\mathrm{d}x=(2l+1)$&lt;br&gt;&#xA;So, $$q_r=\int_0^\infty\operatorname{e}^{-x\theta_r/T}\mathrm{d}x=\left[-(\theta_r/T)\operatorname{e}^{-x\theta_r/T}\right]_0^\infty = \theta_r/T$$&#xA;So, at high temperature: $$E_r=NkT^2 \frac{\delta \ln q_r}{\delta T}=NkT$$&#xA;So, $$C_v=\frac{\delta E_r}{\delta T}=Nk$$&#xA;Now at low temperature: $$q_r=\sum\limits_{l=0}^\infty(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$&#xA;So, $$q_r=1+3\operatorname{e}^{-2\theta_r/T}+5\operatorname{e}^{-6\theta_r/T}+\ldots$$&lt;br&gt;&#xA;So, $$\ln q_r=\ln\left(1+3\operatorname{e}^{-2\theta_r/T}\right)$$ Neglecting higher order terms.&lt;br&gt;&#xA;Now, $\ln(1+x)=x-0.5x^2+\ldots\approx x$&lt;br&gt;&#xA;Here, at low temperature, the second term of $\ln(1+3\operatorname{e}^{-2\theta_r/T})$ is small, so &#xA;$$\ln q_r=3\operatorname{e}^{-2\theta_r/T}$$&#xA;So, $$E_r=NkT^2(\frac{\delta \ln q_r}{\delta T})=6Nk\theta_r\operatorname{e}^{-2\theta_r/T}$$&#xA;So, $$C_v=\frac{\delta E_r}{\delta T}=3Nk(\frac{2\theta_r}{T})^2\operatorname{e}^{-2\theta_r/T}$$&#xA;So, like my previous answer the explanation is the same from mathematical point of view: It’s a trade off between square term and exponential term. Moreover I tried to relate the degeneracy of level with heat capacity.&lt;br&gt;&#xA;Probability of finding a molecule in $l$th state&lt;br&gt;&#xA;$$P_l=(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}/q_r$$&#xA;So, that means the degeneracy of the highest probable energy level is the maxima of $P_l$ with respect to $l$. So, if we solve the equation $\frac{\delta P_l}{\delta l}=0$ for $l$, we will find that $$l_{\text{max}}=(T/2\theta_r)^{1/2}-1/2\approx (T/2\theta_r)^{1/2}$$&#xA;So, that means at low temperature the degeneracy of highest probable energy level is increasing with increasing temperature (since at high temperature all energy levels are equally probable, that’s why it doesn’t matter at high temperature). So, that could have a effect on the steepness of the heat capacity curve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still trying to find the physical reason for this behavior. This is what i got if i plot my equation &lt;img src=&quot;https://i.stack.imgur.com/J9iMD.jpg&quot; alt=&quot;Scale is arbitrary&quot;&gt;&lt;br&gt;&#xA;Internal energy is increasing with temp. but the rate of internal energy change is fast at the beginning but after a certain temp. it is being slowed down.&lt;br&gt;&#xA;And that's the reason for that maxima. Now, heat capacity is actually the fluctuation of energy or it can be written as&lt;br&gt;&#xA;$C_v=&amp;lt;E^2&amp;gt;-&amp;lt;E&amp;gt;^2/kT^2$&lt;br&gt;&#xA;At high temp. since all energy levels are almost equally probable and their numerical value difference is less, that's why the apparent change or fluctuation is almost constant. But at low temp. we have a certain energy level with high occupancy, that's why the change in fluctuation is very noticeable.&lt;br&gt;&#xA;Now if possible we could do a Monte Carlo simulation to see this behavior but that would be time consuming. Still i am thinking about it and hoping that we would be able to find a satisfactory answer from physical perspective.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7777" LastEditDate="2015-06-04T17:39:53.970" LastActivityDate="2015-06-04T17:39:53.970" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32420" PostTypeId="2" ParentId="31898" CreationDate="2015-06-04T04:43:55.307" Score="3" Body="&lt;p&gt;First off, a &quot;fishy&quot; smell and rancidity are two different things.  A fishy smell is likely due to impurities, primarily small volatile amines such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Trimethylamine&quot; rel=&quot;nofollow&quot;&gt;trimethylamine&lt;/a&gt;.  These impurities are unrelated to oxidative degradation of the fish oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rancidity in fish oil is the result of oxidation.  Oxidation leads to the formation of lipid hydroperoxides and short-chain aldehydes, which can have &quot;off&quot; tastes and give rise to unpleasant odors, but in a very different way than the classic fishy smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fish oil's primary components are docosahexaenoic acid and eicosapentaenoic acid.  They are more sensitive to oxidation than many other fats because they contain many double bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best bet is to look for oil that is freshly prepared and highly purified.  Some fish oils are esterified with ethanol to form ethyl esters right &quot;on the boat&quot;.  These ethyl esters can be distilled to remove most impurities.  Optionally, the purified esters can be converted back to triacylglycerols to give &quot;molecularly distilled&quot; fish oil in the triglyceride form.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-04T04:43:55.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32423" PostTypeId="1" CreationDate="2015-06-04T05:41:43.603" Score="6" ViewCount="201" Body="&lt;p&gt;I have been trying to make a monomer reached a dead end in the middle. A B C are the products I have made so far respective order. After reaching C I wanted  to prepare 2 but could not find a way to do so. haloform had a high chance of brominating Hydroxyl group at other end I speculate. Acid catalysed Haloform on A did not result in 1 but ended up with B again [with suprisingly 2 % more yield than I got with simple base hydrolysis. Now at this point I have no option but to produce 1 and then go for 2. I need references or suggestion to make 2 from C cause that will greatly reduce my work load. I am running short of time for my summer project.&lt;img src=&quot;https://i.stack.imgur.com/6aMRA.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-07-06T12:56:52.000" Title="Synthesis Challenge" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32425" PostTypeId="1" CreationDate="2015-06-04T07:23:10.647" Score="4" ViewCount="228" Body="&lt;p&gt;I have pesticide (dimethoate) dissolved in methanol in volumetric flasks and graduated cylinders. How do I clean this glassware properly before moving on to the next dilution of a different pesticide.&lt;/p&gt;&#xA;" OwnerUserId="16688" LastEditorUserId="402" LastEditDate="2017-04-15T01:37:40.363" LastActivityDate="2017-04-15T01:37:40.363" Title="How to clean glassware containing pesticide solutions" Tags="&lt;water&gt;&lt;alcohols&gt;&lt;equipment&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32426" PostTypeId="1" AcceptedAnswerId="32436" CreationDate="2015-06-04T07:31:12.137" Score="27" ViewCount="24491" Body="&lt;p&gt;All the examples of $\ce{NO3-}$ salts are soluble in water (all that I know about).&#xA;Is it always so or there is some salt which doesn't dissolve in water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so what is the reason behind it? &lt;/p&gt;&#xA;" OwnerUserId="2326" LastEditorUserId="7448" LastEditDate="2015-06-08T16:35:49.260" LastActivityDate="2019-08-10T17:52:44.990" Title="Are all NO3- salts soluble in water? If so, why?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32427" PostTypeId="1" CreationDate="2015-06-04T07:56:58.207" Score="2" ViewCount="3421" Body="&lt;p&gt;A certain salt has a solubility of 1 mg/1 L.&#xA;I have 1 mg of this certain salt dissolving in 1 L of water, I have a concentration of 1 ppm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st Question: &#xA;If the solution is left open to evaporation, 0.5 L of water (50 %) evaporated, would I still be getting concentration of 1 ppm while 0.5 mg of this salt became solid again and left at the bottom of the solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd Question: (if it's a yes to the 1st question)&#xA;If I then fill the container back to 1 L mark with water, would the concentration be 1 ppm again after the salts are completely dissolved?&lt;/p&gt;&#xA;" OwnerUserId="16688" LastEditorUserId="7448" LastEditDate="2015-06-22T13:34:07.063" LastActivityDate="2015-06-22T13:34:07.063" Title="Solubility and Evaporation - Would the concentration still be 1 ppm?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;solubility&gt;&lt;concentration&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32428" PostTypeId="1" AcceptedAnswerId="32472" CreationDate="2015-06-04T08:05:02.080" Score="5" ViewCount="1096" Body="&lt;p&gt;I want to know how the powdered formed of piezoelectric material is converted to crystal like used in quartz watches. I know that some more chemicals are mixed with the material and then it is pressed in form of tuning fork (for clocks) at particular temperature and pressure. So what are the chemicals, temperature and pressure which are used. How they are made to vibrate at that particular frequency?&lt;/p&gt;&#xA;" OwnerUserId="16737" LastEditorDisplayName="user15489" LastEditDate="2015-06-04T08:30:09.593" LastActivityDate="2015-06-05T03:24:00.520" Title="How are piezoelectric crystals made?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;&lt;materials&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32429" PostTypeId="1" CreationDate="2015-06-04T08:17:33.003" Score="2" ViewCount="2332" Body="&lt;ul&gt;&#xA;&lt;li&gt;1 ppm is 1 mg / kg&lt;/li&gt;&#xA;&lt;li&gt;Water density: 999.97 kg/m³&lt;/li&gt;&#xA;&lt;li&gt;methanol density: 791.80 kg/m³&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;solution 1 with water as solvent: at 1 $\ce{\mu}$g / mL is considered 1 ppm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solution 2 with methanol as solvent: at 1 $\ce{\mu}$g / mL, should it be considered 1 ppm? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is adjusted by the density of methanol, it wouldn't be 1 ppm. I heard it is a common practice that at the ppm / ppb levels, the density of the solvent is often omitted by only focus on the volume. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="16688" LastEditorDisplayName="user15489" LastEditDate="2015-06-04T22:00:36.750" LastActivityDate="2015-06-05T20:21:23.960" Title="how to define proper ppm concentrations for water solvent vs alcohol solvents?" Tags="&lt;water&gt;&lt;concentration&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32432" PostTypeId="1" CreationDate="2015-06-04T09:28:25.153" Score="3" ViewCount="53" Body="&lt;p&gt;For the reaction:&#xA;$$\ce{F&#xA;+ Br_2 -&amp;gt; FBr + Br}$$&#xA;The gas concentrations as a function of time are given by:&#xA;$$kt=\ce{\frac{1}{[Br2]_0-[F]_0}}\ce{\ln(\frac{[Br2][F]_0}{[F][Br_2]_0})}$$&#xA;Given that $\ce{[F]_0=4 x 10^{-9}}$ and $\ce{[Br2]=1x 10^{-10}}$ I am meant to show that the rate constant is first order with respect to $\ce{Br2}$. How do I do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorine atoms are in excess but is it a large enough excess to assume that the isolation method is valid? I could then cancel terms but would I the have to differentiate to find the rate law?  &lt;/p&gt;&#xA;" OwnerUserId="6925" LastActivityDate="2015-06-04T22:53:46.593" Title="Conditions affecting pseudo rate orders" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32433" PostTypeId="1" AcceptedAnswerId="32440" CreationDate="2015-06-04T10:34:41.240" Score="3" ViewCount="221" Body="&lt;p&gt;Are all molecular structure symmetric either in relation to a plane within itself or in relation to other molecules? Are there any completely asymmetrical structures when looking at molecular geometry? Are there any studies into the reason for symmetry?&lt;/p&gt;&#xA;" OwnerUserId="16741" LastEditorUserId="5017" LastEditDate="2015-06-04T18:37:04.843" LastActivityDate="2015-06-04T18:37:04.843" Title="Are all molecular structures symmetric?" Tags="&lt;molecules&gt;&lt;symmetry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32434" PostTypeId="2" ParentId="32433" CreationDate="2015-06-04T10:37:41.647" Score="5" Body="&lt;p&gt;It is not quite clear what do you mean by &quot;in relation to a plane within itself or in relation to other molecules&quot;, but I think the answer is &lt;em&gt;no&lt;/em&gt;, not all molecules are symmetric. Many of them belong to what is known as $\mathrm{C}_1$ point group, or, in other words they are are totally asymmetric. For more information on molecular symmetry and point groups read &lt;a href=&quot;http://en.wikipedia.org/wiki/Molecular_symmetry&quot; rel=&quot;noreferrer&quot;&gt;this Wiki article&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-06-04T10:37:41.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32435" PostTypeId="2" ParentId="32427" CreationDate="2015-06-04T11:30:36.987" Score="3" Body="&lt;p&gt;Correct, when the water evaporates you'd have more salt than is soluble in water. The excess would then fall out of solution. Adding the same amount of water that evaporated would leave you with the same kind of solution you started out with, so the concentration of dissolved salt would be the same as when you started (although you might need to stir some to get the salt back in solution).&lt;/p&gt;&#xA;" OwnerUserId="16290" LastActivityDate="2015-06-04T11:30:36.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32436" PostTypeId="2" ParentId="32426" CreationDate="2015-06-04T11:47:32.880" Score="23" Body="&lt;p&gt;No, the correct way of putting it is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Almost~all~of~the~\mathbf{inorganic}~nitrate~salts~are~soluble~in~water. }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The families of organic nitrate salts are typically nitrates of azoles and imidazoles. Some bright examples are (&lt;i&gt;R&lt;/i&gt;) &amp;amp; (&lt;i&gt;S&lt;/i&gt;)-miconazole nitrates, isoconazole nitrate and econazol nitrate. &lt;a href=&quot;http://en.wikipedia.org/wiki/Econazole&quot; rel=&quot;nofollow noreferrer&quot;&gt;Econazol nitrate&lt;/a&gt; (Other names: &lt;a href=&quot;http://www.rxlist.com/spectazole-drug.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spectazole&lt;/a&gt;, EN) is the most common organic nitrate salt, and is believed to be very slightly soluble in water, less than &lt;span class=&quot;math-container&quot;&gt;$0.1 \mathrm{\frac{g}{100~g}}$&lt;/span&gt; (&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CB0QFjAA&amp;amp;url=https%3A%2F%2Fwww.sigmaaldrich.com%2Fcontent%2Fdam%2Fsigma-aldrich%2Fdocs%2FSigma%2FProduct_Information_Sheet%2Fe4632pis.pdf&amp;amp;ei=8kxwVdPTMMKhsAGxybWQBw&amp;amp;usg=AFQjCNF-5v2mmEgDjcgbMM_NQ-MoTzuE7Q&amp;amp;sig2=6nD4qtZc2oBwHGShWvQzIQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.chemicalbook.com/ChemicalProductProperty_EN_CB3173103.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.drugbank.ca/drugs/DB01127&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In inorganic nitrate salts, with a bit of patience, I've noticed that &lt;a href=&quot;http://en.wikipedia.org/wiki/Barium_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;barium nitrate&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Mercury(I)_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;mercury(I) nitrate&lt;/a&gt;, and &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobalt(III)_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;cobalt(III) nitrate&lt;/a&gt; are the least soluble of nitrate salts with a monoatomic cation. They are in the brink of insolubility, but I still wouldn't call them insoluble, since&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Their solubility isn't that low at STP; somewhere around &lt;span class=&quot;math-container&quot;&gt;$\frac{5~\mathrm{g}}{100~\mathrm{g}}$&lt;/span&gt;. (&lt;span class=&quot;math-container&quot;&gt;$\ce{Hg2(NO3)2.2H2O}$&lt;/span&gt; decomposes in water, so as &lt;a href=&quot;http://www.rsc.org/learn-chemistry/wiki/Substance:Mercury%28I%29_nitrate_dihydrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;RSC says&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Their solubility increases greatly with temperature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The reason, you say?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, let's think of why one compound is soluble in water while the other isn't. When an ionic compound is formed, energy is released. This is called &lt;strong&gt;Lattice enthalpy&lt;/strong&gt; in constant pressure. The process of formation of lattice from separate ions is usually highly exothermic. Therefore, when an ionic compound dissolves in water, and when this lattice &quot;breaks up&quot;, the reaction would be highly endothermic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to dissolve in water, the ions have to 'overcome' the lattice energy. How? Of course, the net resulting attractions between water molecules and the ions have to be stronger. This attraction is kind of a formation of a very very weak bond (and its weakness is the reason it's &lt;em&gt;not&lt;/em&gt; categorized as such) and thus, is exothermic. This energy release when the dissolution occurs is called &lt;strong&gt;enthalpy of hydration&lt;/strong&gt;, provided that the ions be in the gaseous state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, nitrate is a really big anion, with a single charge. The lesser concentration of negative charge results in relatively lesser lattice enthalpies. Also, the possibility of the creation of hydrogen bonds with water enhances solubility by increasing enthalpy of hydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why almost all of nitrates are soluble.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="69830" LastEditDate="2019-08-10T17:52:44.990" LastActivityDate="2019-08-10T17:52:44.990" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32437" PostTypeId="1" CreationDate="2015-06-04T11:56:17.773" Score="12" ViewCount="7980" Body="&lt;p&gt;All I have found on this is that NaHSO&lt;sub&gt;3&lt;/sub&gt; reduces partially the osmate ester, so breaks off Os and leaves a syn diol. Is it possible to show this step with arrows showing the process of electron movement? I suppose this step is not the main part of the reaction, and mayn't be very important, but I would like to know if it's possible - and if it is, how it looks. &lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-06-04T12:24:58.733" LastActivityDate="2016-11-16T00:42:50.403" Title="What is the role of NaHSO3 in syn diol formation from alkenes with OsO4?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32439" PostTypeId="2" ParentId="32418" CreationDate="2015-06-04T12:42:37.623" Score="4" Body="&lt;p&gt;The ASM Alloy Phase Diagram Database (at asminternational.org if you have access) shows that $V_{2}Se_{9}$, $V_{5}Se_{8}$, $V_{3}Se_{4}$ $V_{5}Se_{4}$, and $VSe$ are all accepted stable phases in the Se-V binary system.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-06-04T12:42:37.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32440" PostTypeId="2" ParentId="32433" CreationDate="2015-06-04T12:43:29.890" Score="3" Body="&lt;p&gt;There are a couple of notable types of structurally asymmetric molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, those that are chiral due to different ligands attached to a central atom such as the improbable $\ce{CClFBrI}$ or 'bromo-chloro-fluoro-iodo-methane'. More generally, many molecules have chiral centers; sometimes multiple ones, like &lt;a href=&quot;http://en.wikipedia.org/wiki/Inositol&quot; rel=&quot;nofollow&quot;&gt;inositol&lt;/a&gt; which has 6 in a 6-carbon molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there are rarer structures where there are no chiral centers, but the whole thing is chiral. Usually this is because of steric restrictions that force it to be one shape or another. Helicene is a very nice example. Another might be a cage hydrocarbon in the shape of &lt;a href=&quot;http://en.wikipedia.org/wiki/Frucht_graph&quot; rel=&quot;nofollow&quot;&gt;this graph&lt;/a&gt; but I don't know if any exist.&lt;/p&gt;&#xA;" OwnerUserId="4665" LastEditorUserId="10185" LastEditDate="2015-06-04T15:48:02.813" LastActivityDate="2015-06-04T15:48:02.813" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32441" PostTypeId="1" AcceptedAnswerId="32445" CreationDate="2015-06-04T12:44:16.807" Score="6" ViewCount="200" Body="&lt;p&gt;I was wondering if the tin/lead/other metals bond with the metal parts that are low-temperature soldered. I know that the flux in the solder is used to remove the oxide layer, leaving a clean metal surface ready to form metallic bonds (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, should I use a tin/silver/copper eutectic to bond a silver surface and copper wire? Will that make a difference or can I use tin/lead with the same result? For example, do we get a different connection strength? In every case the solder material is mostly made of tin, which is why I was wondering what's the real impact of the other components.&lt;/p&gt;&#xA;" OwnerUserId="16613" LastEditorUserId="16613" LastEditDate="2015-06-04T23:17:26.350" LastActivityDate="2015-06-04T23:17:26.350" Title="Different solder composition affects the connection strength?" Tags="&lt;bond&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32442" PostTypeId="2" ParentId="32416" CreationDate="2015-06-04T12:53:38.993" Score="7" Body="&lt;p&gt;An explanation is provided &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed072p386&quot;&gt;in this reference&lt;/a&gt;. The idea is that the electrode reactions are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Zn(s) + 4OH^- -&amp;gt; [Zn(OH)_4]^{2-} + 2e-}$&#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Zn(OH)_4]^{2-} + 2e- -&amp;gt; Zn(alloy) + 4OH^-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the zinc powder remaining will go into solution and the zincate is reduced on the copper surface to form a form of zinc rich brass. Brass is an alloy and the atoms are held together via metallic bonding.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-04T12:53:38.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32443" PostTypeId="1" AcceptedAnswerId="32444" CreationDate="2015-06-04T13:06:40.677" Score="11" ViewCount="2902" Body="&lt;p&gt;I found different values of Avogadro constant in different places. So what is the correct value? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\pu{6.0221367*10^{23}}$ or&lt;br&gt;&#xA;$\pu{6.02214129*10^{23}}$ or&lt;br&gt;&#xA;$\pu{6.0221415*10^{23}}$ or anything else?&lt;/p&gt;&#xA;" OwnerUserId="16743" LastEditorUserId="24052" LastEditorDisplayName="user15489" LastEditDate="2018-07-26T14:39:40.453" LastActivityDate="2020-04-24T11:11:02.240" Title="What is the correct value of the Avogadro constant? And how was it derived?" Tags="&lt;mole&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32444" PostTypeId="2" ParentId="32443" CreationDate="2015-06-04T13:32:34.030" Score="24" Body="&lt;p&gt;Whenever you're looking for accurate fundamental physical constants, &lt;a href=&quot;http://www.codata.org/committees-and-groups/fundamental-physical-constants&quot; rel=&quot;nofollow noreferrer&quot;&gt;CODATA recommended values&lt;/a&gt; are the way to go. As of May 20, 2019, Avogadro's constant is now truly an &lt;strong&gt;exact&lt;/strong&gt; value, with infinite significant digits. Behold!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unlikely this value will change within our lifetimes, which is exciting in its own way!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For historical reasons, I'll leave my previous answer below:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As of 2015, the latest data for the Avogadro constant is from &lt;a href=&quot;http://physics.nist.gov/cgi-bin/cuu/Value?na|category=abbr_in&quot; rel=&quot;nofollow noreferrer&quot;&gt;2014&lt;/a&gt;. According to CODATA, the most accurate value is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative uncertainty in the measurement is thus only 12 parts per billion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the Avogadro constant may be redefined in the near future to be an &lt;em&gt;exact&lt;/em&gt; value, that is, a constant with zero uncertainty by definition, much like &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_light#Measurement&quot; rel=&quot;nofollow noreferrer&quot;&gt;the speed of light&lt;/a&gt;. This would come as a consequence of &lt;a href=&quot;http://en.wikipedia.org/wiki/Kilogram#Avogadro_project&quot; rel=&quot;nofollow noreferrer&quot;&gt;redefining the SI kilogram&lt;/a&gt; as a function of the number of atoms in an ultrapure monoisotopic &lt;span class=&quot;math-container&quot;&gt;$\ce{^{28}Si}$&lt;/span&gt; monocrystalline sphere engineered to extreme precision. A great video on this can be found in the &lt;a href=&quot;https://www.youtube.com/watch?v=ZMByI4s-D-Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Veritasium YouTube channel&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, &lt;a href=&quot;http://physics.nist.gov/cuu/Constants/bibliography.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;CODATA released updated values for the physical constants&lt;/a&gt;, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science!&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2020-04-24T11:11:02.240" LastActivityDate="2020-04-24T11:11:02.240" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32445" PostTypeId="2" ParentId="32441" CreationDate="2015-06-04T15:16:40.320" Score="7" Body="&lt;p&gt;Good question! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer: if you are soldering &lt;em&gt;bulk&lt;/em&gt; silver to copper, 63% Sn/ 37% Pb eutectic would probably be OK, but for silver plating, you &lt;em&gt;might&lt;/em&gt; be able to make a successful join using tin/lead, if heat is quickly applied and removed, but might need to use a silver-bearing solder for thin silver plate or if the base metal under the silver is not easily soldered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, &quot;cold soldered&quot; is perhaps better expressed as &lt;em&gt;low-temperature&lt;/em&gt; soldering, though both indium and gold can be joined at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A longer explanation: reasons to use a specific solder composition.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To some extent, solder dissolves the surface of the metals to be united. This can be problematic with plated parts, e.g. &lt;a href=&quot;http://www.kester.com/kester-content/uploads/2013/06/Soldering_Gold_over_Nickel.pdf&quot;&gt;gold plated over nickel&lt;/a&gt;, where the solder can completely remove the plating, making it difficult to adhere to the base metal. For that purpose, solders containing the plating metal are made, but they have other liabilities: expense and brittleness. Similarly, &lt;a href=&quot;http://www.google.com/patents/EP1391261A1?cl=en&quot;&gt;copper-containing solders&lt;/a&gt; are used for copper-plated electronic parts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not only alloys form between soldered parts, but also &lt;a href=&quot;http://www.semlab.com/papers/intermetallicsinsolderjoints.pdf&quot;&gt;intermetallic compounds&lt;/a&gt; form, such as &lt;a href=&quot;https://app.aws.org/wj/supplement/WJ_1972_11_s536.pdf&quot;&gt;$\ce{Ag3Sn}$&lt;/a&gt;, $\ce{Cu6Sn5}$ and &lt;a href=&quot;http://www.empf.org/empfasis/june04/inter.htm&quot;&gt;$\ce{AuSn4}$&lt;/a&gt;, some of which are brittle. Changing solder composition may not be practical to prevent some intermetallic compounds from weakening a joint, so barrier coatings and specific temperature cycling and storage may be used to retard their formation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pure tin solder, as well as zinc, has the property of forming &lt;a href=&quot;http://en.wikipedia.org/wiki/Whisker_%28metallurgy%29&quot;&gt;whiskers&lt;/a&gt;, which not only short-circuit electronics, but was rumored to have caught &lt;a href=&quot;http://en.wikipedia.org/wiki/Tin_pest#Napoleon.27s_buttons&quot;&gt;Napoleon's troops with their pants down&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For these and other reasons, soldering can be considered as much a craft as a science. You may have to make some tests of the solder joints if they are critical, as in aerospace or medical applications.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-04T15:16:40.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32446" PostTypeId="2" ParentId="30927" CreationDate="2015-06-04T15:19:10.277" Score="13" Body="&lt;p&gt;The justification comes from molecular orbital theory. I'm going to analyze the MO diagrams for benzene (a molecule with $4n+2$ π electrons) and an MO diagram for cyclobutadiene (a molecule with $4n$ π electrons). These results generalize for all planar cyclic molecules with π orbitals on every atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D63PI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D63PI.png&quot; alt=&quot;Diagram of benzene pi MOs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above π MO diagram of benzene may be obtained through &lt;a href=&quot;https://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;noreferrer&quot;&gt;Hückel theory&lt;/a&gt; (see also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/76809/16683&quot;&gt;Pi molecular orbitals of polyenes&lt;/a&gt;). Here, the &quot;zero&quot; of energy (referred to as $\alpha$ in Hückel theory) refers to the energy of an isolated carbon 2p orbital that does not participate in bonding. Hence, any orbitals below this energy are bonding in nature, and orbitals above it are antibonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the benzene MO diagram above, you'll notice that all the molecular orbitals that are being filled by the π electrons are more stable than leaving the carbons unbonded. On top of that, it can be shown that the stabilisation derived from delocalised π bonding in benzene is greater than the stabilisation derived from simply forming three localised C–C π bonds, in conjugation with one another, but not delocalised over the entire ring (e.g. in hexa-1,3,5-triene). This is the condition known as aromaticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at the MO diagram for a hypothetical perfectly square cyclobutadiene, for contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HoT46.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HoT46.png&quot; alt=&quot;Diagram of cyclobutadiene pi MOs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with all molecular orbitals, the most stable orbital is filled first. However, problems start to emerge when going up to next two orbitals. First, unlike the next orbitals in sequence for benzene, these are formally nonbonding within simple Hückel theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that this situation is &lt;em&gt;less&lt;/em&gt; stable than the case for an alternative version of cyclobutadiene, where the double bonds are localised between two adjacent carbons. In practice, cyclobutadiene therefore undergoes a distortion away from a square geometry to a rectangular geometry, which has longer C–C bond lengths for the single bonds and shorter bond lengths for the now localised double bonds (see also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/71249/16683&quot;&gt;Cyclobutadiene - Jahn–Teller effect or not?&lt;/a&gt;). This is the condition known as antiaromaticity.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16683" LastEditDate="2018-01-07T02:59:31.593" LastActivityDate="2018-01-07T02:59:31.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32447" PostTypeId="1" AcceptedAnswerId="67386" CreationDate="2015-06-04T15:26:44.203" Score="16" ViewCount="504" Body="&lt;p&gt;More specifically, the expression for the average expectation value of the transverse magnetization has the following form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M^+(t)=N\gamma \hbar\,\text{Tr}(\hat{\sigma}(t)\hat{F}^+)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\hat{\sigma}(t)$ the density matrix. $\hat{F}^+$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{F}^+ = \sum ^{K} _{k=1} \hat{I}^+_{\!k} = \sum^{K}_{k=1} (\hat{I}_{\!kx} + \mathrm{i}\hat{I}_{\!ky})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really see how you intuitively obtain such an expression for the transverse magnetisation (in the syllabus it was posited, not derived). On the other hand, if you would take the complex modulus then you would find the magnitude of some 2D vector. It still does not motivate (enough) for me why it should look like this. Doesn't measurement happen in a single direction? Why is it not just &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum^{K} _{k=1} \hat{I}_{\!kx}?$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also bears resemblance to the raising operator (which would explain the $+$), perhaps you measure how many quanta are absorbed, which corresponds to measuring the amount of spins that you raise to a higher energy level (but that's not what the raising operator signifies I think)? Or is it a phasor with a $90^\circ$ angle between both components?&lt;/p&gt;&#xA;" OwnerUserId="16745" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-01-25T18:47:52.473" LastActivityDate="2017-08-03T10:19:33.310" Title="Why does the transverse magnetization in Fourier Transform NMR consist of Ix+iIy ($\hat{I}_{\!x}+\mathrm{i}\hat{I}_{\!y}$)?" Tags="&lt;quantum-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32448" PostTypeId="2" ParentId="32060" CreationDate="2015-06-04T15:48:55.113" Score="0" Body="&lt;p&gt;Ok so i got the answer to my question. For what i've understood, PMMA, as it is a thermoplastic polymer, is unable to crosslink unless it's &quot;mixed&quot; with an initiator. The removal process is going to be difficult if temperatures are going to get over 150°C due to the degradation of the polymer in species that may not be diluted in organic solvents&lt;/p&gt;&#xA;" OwnerUserId="16556" LastActivityDate="2015-06-04T15:48:55.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32450" PostTypeId="1" AcceptedAnswerId="34693" CreationDate="2015-06-04T18:05:58.723" Score="6" ViewCount="24416" Body="&lt;p&gt;The question asks us to select the compound with the smallest heat of hydrogenation per mole out of the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) 1-butene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) &lt;em&gt;trans&lt;/em&gt;-2-butene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) &lt;em&gt;cis&lt;/em&gt;-2-butene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) 1,3-butadiene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all four compounds form the same product (butane), a comparison without actual data should be possible. 1,3-butadiene is conjugated so it is the most stable compound and therefore its heat of hydrogenation should be the smallest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it requires 2 moles of hydrogen for complete conversion to butane. And the answer is actually trans-2-butene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would we consider each heat of hydrogenation &quot;per mole of hydrogen&quot; or &quot;per mole of the given compound&quot;? The question isn't clear about this and also, if we do consider it as &quot;per mole of hydrogen&quot; then is it correct to simply half the value heat of hydrogenation for 2 moles?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="4231" LastEditDate="2015-08-04T13:58:19.347" LastActivityDate="2015-08-04T17:34:50.740" Title="Comparing heat of hydrogenation per mole of butenes and 1,3-butadiene" Tags="&lt;organic-chemistry&gt;&lt;hydrogen&gt;&lt;mole&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32451" PostTypeId="1" CreationDate="2015-06-04T19:00:53.060" Score="4" ViewCount="120" Body="&lt;p&gt;I'm teaching myself how to make lotion and a lot of lotions contain esters like isopropyl myristate or isopropyl palmitate. I'd like to make an ethyl ester from a popular skin-care oil (olive, apricot kernel, or castor). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there's so much controversy about using isopropyl alcohol in skin-care, I'd like to avoid that if possible. It's very difficult to find any info on the process, and the closest I could get was how to make biodiesel. All of those use methanol though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about making an ethyl ester from these oils?&lt;/p&gt;&#xA;" OwnerUserId="16746" LastEditorDisplayName="user7951" LastEditDate="2015-08-13T20:34:00.507" LastActivityDate="2015-08-13T20:34:00.507" Title="How would I go about making an ethyl ester from vegetable oil?" Tags="&lt;organic-chemistry&gt;&lt;home-experiment&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32453" PostTypeId="2" ParentId="32451" CreationDate="2015-06-04T19:07:46.793" Score="5" Body="&lt;p&gt;The biodiesel process should work well, you will need just slightly more aggressive conditions (i.e. slightly more catalyst, slightly higher EtOH excess, and/or slightly longer time) to make ethyl esters instead of methyl esters, as ethanol is slightly less reactive than methanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For skin care, you will need higher purity than biodiesel processes usually provides.  I would imagine that the cosmetics industry uses vacuum distillation to purify their esters.  This would separate the ethyl esters from unreacted triacylglycerols, any base (or acid) catalyst left over from the reaction and soluble in the organic ester phase, organic-soluble glycerol, and organic-soluble impurities.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-04T19:07:46.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32458" PostTypeId="1" CreationDate="2015-06-04T21:25:48.780" Score="4" ViewCount="415" Body="&lt;p&gt;My question is regarding setting up this equation. The givens are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gases are helium, methane and hydrogen. (Test all three)&lt;/li&gt;&#xA;&lt;li&gt;Assume balloon is spherical.&lt;/li&gt;&#xA;&lt;li&gt;Assume temperature is $25~\mathrm{^\circ C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I need to find the diameter of the balloon needed to lift a 7lb weight. I setup my equation as follows. $$m=\frac{PVM}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I subtract my first gas ($\ce{He}$) from ambient air and convert my $m$ to equal grams in the weight. so I set it up like so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$3730.15~\mathrm{g}=\frac{PVM}{RT}-\frac{PVM}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I factor out the pressure and volume and fill in my numbers.&#xA;$$3730.15~\mathrm{g}=PV\left(\frac{(28.9~\mathrm{g~mol^{-1}})}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}-\frac{8~\mathrm{g~mol^{-1}}}{(0.0821~\mathrm{L~atm^{-1}mol^{-1}K^{-1}})(298~\mathrm{K})}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I am stuck at as I don't know what makes sense to use for pressure so I can solve for volume? Or maybe this is all setup wrong and makes no sense? &lt;/p&gt;&#xA;" OwnerUserId="16414" LastEditorUserId="17175" LastEditDate="2018-03-06T13:22:42.423" LastActivityDate="2018-03-06T13:22:42.423" Title="Determine balloon diameter needed to lift a 7 pound weight with various gases" Tags="&lt;gas-laws&gt;&lt;density&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32459" PostTypeId="1" AcceptedAnswerId="32461" CreationDate="2015-06-04T22:45:35.450" Score="5" ViewCount="367" Body="&lt;p&gt;With optimal conditions can we supercool any compound or is it that just water and a certain set of others can be supercooled? If not, why?&lt;/p&gt;&#xA;" OwnerUserId="8208" LastActivityDate="2015-06-04T23:10:05.137" Title="Can all substances be supercooled?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32460" PostTypeId="2" ParentId="32432" CreationDate="2015-06-04T22:53:46.593" Score="4" Body="&lt;p&gt;Maybe it would be best to reorganise that to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right] \frac{\ce{[Br_2]_0}\ce{[F]}}{\ce{[F]_0}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as there is 40 times more $\ce{F}$ than $\ce{Br_2}$, it is safe to assume that $\ce{[F]}\approx\ce{[F]_0}$. I.e. even if all the $\ce{Br_2}$ is gone there is still $\ce{[F]}=3.9\times10^{-9}$. This simplifies the above equation to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[Br_2]}=\exp\left[-kt(\ce{[F]_0}-\ce{[Br_2]_0})\right]\ce{[Br_2]_0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the familiar form of the expression of concentration. It is also easy to see what the pseudo first-order rate constant is. &lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-04T22:53:46.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32461" PostTypeId="2" ParentId="32459" CreationDate="2015-06-04T23:04:44.357" Score="6" Body="&lt;p&gt;Supercooling is a general phenomenon and it is not limited to any particular substance, although the degree you can supercool something will vary. The idea in thermodynamics is that at the melting point the free energy of the solid and the liquid is equal and a lower temperature will favour the formation of the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-04T23:04:44.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32462" PostTypeId="1" AcceptedAnswerId="32468" CreationDate="2015-06-04T23:04:46.340" Score="5" ViewCount="1806" Body="&lt;p&gt;I have seen youtube videos of people treating some aluminium object with liquid gallium and after letting it sit for a while, they are able to break up the aluminium into tiny little pieces with their bare hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me what is exactly going? I haven't been able to find a satisfactory explanation as of yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the gallium somehow incorporate itself into the metal lattice, disrupting the bonds? or is it forming an amalgam of sorts and actually dissolving the aluminium? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a combination of these two, or are my ideas off completely? Also, what would be the exact mechanics of a process like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: How exactly does gallium &quot;unzip&quot; the grain boundaries?&#xA; Also, besides Ga or Hg, what other metals would produce a similar effect? What mechanism/property allows them to behave in such a manner?&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="16088" LastEditDate="2015-06-05T06:19:36.823" LastActivityDate="2015-06-05T06:19:36.823" Title="Gallium-induced structural failure of aluminium" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32463" PostTypeId="2" ParentId="32459" CreationDate="2015-06-04T23:10:05.137" Score="3" Body="&lt;p&gt;So called 'drop tubes' have been used for decades to study the nucleation rate vs temperature of supercooled molten droplets of (mainly) metals in vacuum or inert environments.  An (admittedly old) example is 'A review of long drop tubes as a supplement/alternative to space experiments', R. J. Bayuzick et al., Advances in Space Research 4(5) 85-90 (1984). Nucleation studies of supercooled semiconductors (Si and Ge) was also studied back in the 1980's. Googlingh 'drop tube recalescence' will find more.  Overall, I'd say the literature is quite full of supercooling on any number of different materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to note that supercooling of a liquid is much much easier than superheating of a solid - the problem is decidedly not symmetric in the general case.  The difference between nucleating a critical volume of solid in a liquid droplet is not the same as initiating melt at the surface of a solid, where there are unsatisfied bonds because of the interface.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-06-04T23:10:05.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32465" PostTypeId="1" CreationDate="2015-06-04T23:20:12.123" Score="5" ViewCount="1315" Body="&lt;p&gt;I understand that there is a galvanic reaction between aluminum and carbon fiber composite. Structural aluminum parts in contact with carbon fiber composites should be insulated by something like fiberglass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what exactly is it that happens, chemically? I know that aluminum has a tendency to be oxidized; aluminum oxidizes readily by itself in air. I also know that carbon fiber itself is a conductor due to delocalized electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming that aluminum is oxidized then the carbon fiber composite must be reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is reduced? The carbon fiber, or the epoxy? What is the role of the epoxy in all of this? Wouldn't the epoxy be electrically insulating? How exactly does this redox reaction take place across the epoxy? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-06-04T23:25:28.040" LastActivityDate="2015-06-05T23:00:05.417" Title="Galvanic reaction between aluminum and carbon fiber composite" Tags="&lt;redox&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32466" PostTypeId="1" CreationDate="2015-06-04T23:41:00.203" Score="4" ViewCount="83" Body="&lt;p&gt;I was given the formulas,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{MnO4^2- \;&amp;amp;&amp;lt;=&amp;gt; MnO4- + e-}\quad &amp;amp;&amp;amp;E^\circ = -0.56\ \mathrm{V}\\&#xA;\ce{MnO2 + 2H2O \;&amp;amp;&amp;lt;=&amp;gt; MnO4^2- + 4H+ + 2e-}\quad &amp;amp;&amp;amp;E^\circ = -2.27\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I gathered you have to see if it’s non-spontaneous or spontaneous and compare it with the original reaction, but, when I asked my teacher on how to start, he said “multiply the first reaction by 2 and the second by −1.”&#xA;I understand the ×2 to balance electrons, but why do you multiple it by −1? I don’t understand this question, any help would be appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16751" LastEditorDisplayName="user7951" LastEditDate="2015-06-08T18:36:33.007" LastActivityDate="2015-06-08T18:36:33.007" Title="If given a half reaction, how do you determine if an element can exist in a acidic conditions?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32467" PostTypeId="1" CreationDate="2015-06-05T00:20:22.203" Score="6" ViewCount="3492" Body="&lt;p&gt;Using Lewis structures to describe chemical bonds inside a molecule is still fairly common start in undergraduate education. Lewis structure however can be misleading, or just simply wrong in many cases. Mostly I see this with aromatic systems, Diels-Alder reactions etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if there is a rule of thumb when one can safely use Lewis structure and when we should just  forget them? Or should I just check it every single time and compare them with eg. MOs? &lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-06-05T17:22:27.547" Title="Limitations of Lewis structures" Tags="&lt;molecular-orbital-theory&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32468" PostTypeId="2" ParentId="32462" CreationDate="2015-06-05T02:33:45.767" Score="5" Body="&lt;p&gt;As you state, it forms an amalgam, just as &lt;a href=&quot;https://www.youtube.com/watch?v=zbUPjHHml1E&quot;&gt;mercury does&lt;/a&gt;. @Jon Custer's explanation is correct, and furthermore, the alloy &lt;em&gt;does not form a protective oxide&lt;/em&gt;, so the aluminum oxidizes to $\ce{Al2O3}$ and quickly crumbles. This &lt;a href=&quot;https://vimeo.com/32525601&quot;&gt;happens even faster with an Hg-Ga alloy&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, gallium should be safe enough, though expensive, but I would not like to work with mercury without a hood, and without careful containment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the question; this encouraged me to view that amazing demo on Vimeo!&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-05T02:33:45.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32471" PostTypeId="2" ParentId="32450" CreationDate="2015-06-05T03:14:34.827" Score="1" Body="&lt;p&gt;To saturate one mole of 1,3-butadiene, there is a need of two mole hydrogen gas. In comparison with 1-butene, &lt;em&gt;trans&lt;/em&gt;-2-butene, and &lt;em&gt;cis&lt;/em&gt;-2-butene, these molecules only need 1 mole of hydrogen gas to be saturated to butane. Thus, 1,3-butadiene has the highest heat of hydrogenation. Stability of the remaining molecules can be used to measure qualitatively the trend of the heat of hydrogenation. Since &lt;em&gt;trans&lt;/em&gt; is more stable among them, then it has the least heat of hydrogenation.&lt;/p&gt;&#xA;" OwnerUserId="14467" LastEditorDisplayName="user7951" LastEditDate="2015-06-05T07:15:59.140" LastActivityDate="2015-06-05T07:15:59.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32472" PostTypeId="2" ParentId="32428" CreationDate="2015-06-05T03:24:00.520" Score="4" Body="&lt;p&gt;Good quality quartz crystals for use in resonators can be found in nature, such as in &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartz&quot; rel=&quot;nofollow&quot;&gt;Minas Gerais, Brazil&lt;/a&gt;. They can also be produced by crystallization in a &lt;a href=&quot;http://www.4timing.com/techquartz.htm&quot; rel=&quot;nofollow&quot;&gt;hydrothermal process&lt;/a&gt;, simulating in the autoclave the same process that occurred over geologic epochs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily grow your own &lt;a href=&quot;http://en.wikipedia.org/wiki/Potassium_sodium_tartrate&quot; rel=&quot;nofollow&quot;&gt;Rochelle salt&lt;/a&gt; (potassium sodium tartrate, $\ce{KNaC4H4O6·4H2O}$) piezoelectric crystals. Rochelle salt was commonly used in inexpensive crystal headphones and phonograph cartridges. The raw materials are household chemicals and the &lt;a href=&quot;http://rimstar.org/materials/piezo/how_to_make_rochelle_salt_piezoelectric_crystal.htm&quot; rel=&quot;nofollow&quot;&gt;directions&lt;/a&gt; are straightforward. Again, these crystals are grown, not hot pressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tuning forks watches such as the Bulova Accutron did &lt;em&gt;not&lt;/em&gt; use crystals, but rather &lt;a href=&quot;http://members.iinet.net.au/~fotoplot/accspec.htm&quot; rel=&quot;nofollow&quot;&gt;metal tuning forks, made from NiSpan C&lt;/a&gt;, with a low coefficient of thermal expansion, similar to invar. These did not rely on the piezoelectric effect, but on magnetic induction.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-05T03:24:00.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32475" PostTypeId="1" CreationDate="2015-06-05T05:48:54.737" Score="2" ViewCount="773" Body="&lt;p&gt;Specifically concerning &lt;strong&gt;organic chemistry&lt;/strong&gt;, but an explanation on its significance and applications in inorganic chemistry would be great as well.&lt;/p&gt;&#xA;" OwnerUserId="16757" LastEditorUserId="5456" LastEditDate="2015-06-05T07:16:18.603" LastActivityDate="2015-06-05T08:44:20.597" Title="What significance and applications does nanotechnology have in chemistry?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-05T20:03:06.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="32477" PostTypeId="2" ParentId="32475" CreationDate="2015-06-05T07:13:37.100" Score="1" Body="&lt;p&gt;There is branch named &lt;a href=&quot;http://www.rsc.org/Membership/Networking/InterestGroups/ChemicalNanoscience/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical Nanoscience and Nanotechnology&lt;/a&gt; Which deals with chemistry asspects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Talking about organic chemistry, nanotechnology is used for Organic synthesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With help of Modern Chemistry today we can synthesis almost all complex molecules, thanks to nanotechnology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://en.wikipedia.org/wiki/Nanotechnology#Bottom-up_approaches&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia-1&lt;/a&gt;, &#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nanotechnology#Simple_to_complex:_a_molecular_perspective&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia-2&lt;/a&gt; and &lt;a href=&quot;http://www.nano.gov/you/nanotechnology-benefits&quot; rel=&quot;nofollow noreferrer&quot;&gt;nano.gov&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For organic chemistry, &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0032386108003157&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polymer nanotechnology&lt;/a&gt; is developed notably.  Polymer matrix and nanocomposites have become a prominent area of current research and development &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eng.thesaurus.rusnano.com/wiki/article1987&quot; rel=&quot;nofollow noreferrer&quot;&gt;exfoliation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://eng.thesaurus.rusnano.com/wiki/article1987&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0NHqj.png&quot; alt=&quot;exfoliation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-06-05T08:44:20.597" LastActivityDate="2015-06-05T08:44:20.597" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32479" PostTypeId="2" ParentId="32466" CreationDate="2015-06-05T09:58:08.220" Score="3" Body="&lt;p&gt;Doing so will result in the two half equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2MnO_4^2- -&amp;gt; 2MnO_4^- + 2e^-}$&#xA;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MnO_4^2- + 4H^+ +2e^- -&amp;gt; MnO_2 + 2H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which you then combine into the total equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3MnO_4^2- + 4H+ &amp;lt;=&amp;gt; 2MnO4- + MnO2 + 2 H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells. &lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-05T09:58:08.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32480" PostTypeId="1" AcceptedAnswerId="32486" CreationDate="2015-06-05T10:54:25.843" Score="6" ViewCount="207" Body="&lt;p&gt;Why do $\ce{LiAlH(O-^{t}Bu)3}$ and similar bulky reducing agents not reduce acidic chlorine further?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3-COCl + LiAlH4 + H2O -&amp;gt; CH3-CH=O -&amp;gt; CH3-CH2-OH},$$&#xA; but&#xA;$$\ce{CH3-COCl + LiAlH(O-^{t}Bu)3 + H2O -&amp;gt; CH3-CH=O -&amp;gt; no~reaction}.$$ &lt;/p&gt;&#xA;" OwnerUserId="16759" LastEditorUserId="4945" LastEditDate="2015-06-05T11:30:45.893" LastActivityDate="2015-06-05T14:05:30.743" Title="Why does LiAlH(O-tert-Bu)3 not reduce acyl chlorides to alcohols?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32481" PostTypeId="1" CreationDate="2015-06-05T10:54:57.263" Score="0" ViewCount="1370" Body="&lt;p&gt;Why when a ring has 8 carbon's and contains a double bond the trans version is very unstable to exist, for instance a molecule like cholesterol? &lt;/p&gt;&#xA;" OwnerUserId="14862" LastEditorUserId="1499" LastEditDate="2015-06-05T13:28:26.940" LastActivityDate="2015-06-05T17:36:14.510" Title="Why can't small trans-cycloalkenes exist?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-06-05T13:40:24.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="32482" PostTypeId="2" ParentId="32367" CreationDate="2015-06-05T11:23:21.647" Score="4" Body="&lt;p&gt;There are many, sometime competing, some-time reinforcing, factors that affect the boiling point of these molecules: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the polarity (e.g. approximated by the dipole moment)&lt;/li&gt;&#xA;&lt;li&gt;dispersive interactions&lt;/li&gt;&#xA;&lt;li&gt;increase in entropy upon evaporation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some illustrative examples (data taken from the &lt;a href=&quot;http://webbook.nist.gov/chemistry/name-ser.html&quot; rel=&quot;nofollow&quot;&gt;NIST Webbook&lt;/a&gt; unless otherwise noted). The boiling point of tetrahydropyran is 361 K while that of  ethyl propyl ether is 337 K.  The corresponding standard enthalpies of vaporization ($\Delta H^\circ_{\text{vap}}$) are 35 and 31 kJ/mol.  This is consistent with the higher dipole moment of cyclic structures mentioned in @ron's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I could not find the dipole moment of ethyl propyl ether, but it is probably similar to dipropyl ether and this is backed up by &lt;a href=&quot;http://molcalc.org&quot; rel=&quot;nofollow&quot;&gt;MolCalc&lt;/a&gt; calculations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, while the dipole moment of dipropyl ether (1.0-1.2 D) is lower than tetrahydropyran (1.6-1.8 D; taken from @ron's answer) the boiling point is 363 K - slightly higher than that of tetrahydropyran, as is the $\Delta H^\circ_{\text{vap}}$ (36 kJ/mol).  So polarity cannot be the whole story in general.  Dipropyl ether is slightly larger so larger dispersion interaction is the most likely explanation for the high boiling point here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the boiling point of allyl ethyl ether is lower than tetrahydropyran (340 K) while the $\Delta H^\circ_{\text{vap}}$ is the same (35 kJ/mol).  (The dipole is smaller than for tetrahydropyran but bigger than for ethyl propyl ether - according to MolCalc - but the double bond is more polarizable and will lead to an increase in dispersion interactions compared to tetrahydropyran).  Since the enthalpy change is the same, allyl ethyl ether must have a higher $\Delta S^\circ_{\text{vap}}$. It's a more &quot;floppy&quot; molecule than tetrahydropyran so the gas phase vibrational and conformational entropy will be higher.  (According to MolCalc the vibrational entropy of ethyl propyl ether  is 54 J/molK higher than for tetrahydropyran).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the higher boiling point of tetrahydropyran compared to ethyl propyl ether is likely due to a combination of factors: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;stronger interactions between tetrahydropyran molecules due to it being more polar &lt;/li&gt;&#xA;&lt;li&gt;a lower increase in entropy upon vaporization, due to tetrahydropyran being less flexible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The fact that allyl ethyl ether has roughly the same boiling point as ethyl propyl ether but the roughly the same $\Delta H^\circ_{\text{vap}}$ as tetrahydropyran suggest that point 2 contributed the most to the difference in boiling point between tetrahydropyran and ethyl propyl ether.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-05T11:39:52.843" LastActivityDate="2015-06-05T11:39:52.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32483" PostTypeId="2" ParentId="32481" CreationDate="2015-06-05T12:03:49.583" Score="2" Body="&lt;p&gt;Because you're stuck in the ring structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trans isomer has branches that, in their natural state, would want to rest antiparallel to one another. But in double bond 6 membered ring structure (like in cholesterol), these bonds would need to be twisted a lot so that they could sit in the octagonal ring structure. Given how a double bond is rigid and will already resist rotation, this theoretical trans isomer will have such a high energy requirement for it to exist that it will simply refuse to do so.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-06-05T17:36:14.510" LastActivityDate="2015-06-05T17:36:14.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32484" PostTypeId="1" AcceptedAnswerId="32488" CreationDate="2015-06-05T12:20:28.707" Score="21" ViewCount="7291" Body="&lt;p&gt;Artificially grown bismuth crystals are arguably magnificent, as seen in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LmAPp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: &lt;a href=&quot;http://www.minerals.net/mineral/bismuth.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minerals.net&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They have hopper-like growths in pseudocubic crystals and are usually coated with chemicals to prevent tarnish, thus maintaining the silver-white color. Sometimes the coating gives a colorful effect on the bismuth. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What coating gives the colourful iridescence? and what chemical processes cause the colourful iridescence to occur?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="402" LastEditDate="2017-07-21T05:17:44.013" LastActivityDate="2017-07-21T05:17:44.013" Title="What causes the iridescent colour in laboratory grown bismuth?" Tags="&lt;metal&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32485" PostTypeId="2" ParentId="31463" CreationDate="2015-06-05T14:00:58.187" Score="12" Body="&lt;p&gt;It is not a general truth that there is a maximum before approaching the equipartition value.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for $\ce{H2}$ as it exists under ordinary conditions, there is no such maximum, because &lt;a href=&quot;http://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen&quot; rel=&quot;noreferrer&quot;&gt;orthohydrogen&lt;/a&gt; predominates (in a 3:1 statistical ratio over parahydrogen) and orthohydrogen must always rotate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1Y0z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQmEQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as seen above, the existence of the greater-than-equipartition-value maximum is tied to the existence of a state where the molecule does not rotate.  Parahydrogen has such a state, while orthohydrogen does not.  In this sense, the maximum could be thought of as due to depopulation of the non-rotating state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, it is very difficult to express the form of the heat capacity curve over the full temperature range.  As seen in other answers &lt;a href=&quot;http://solar.njit.edu/~leej/lecture/ph641/ch06_summary.pdf&quot; rel=&quot;noreferrer&quot;&gt;and page 9 here&lt;/a&gt;, in the limit of temperature approaching zero or infinity, it is not extraordinarily difficult to calculate, but at intermediate values it is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, a curve with the maximum was developed by Fritz Reiche in 1919 from Planks theory.  See the presentation &lt;a href=&quot;https://www.csbsju.edu/Documents/Natural%20Sciences/H2_MP06.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Rotational Specific Heat of Hydrogen From Old Quantum Theory&lt;/a&gt; especially around slide 42 &quot;The calculations become horrendous!&quot;.  And of course he had to do them by hand, only to see that the result was an apparent failure, in the sense that it did not correctly reproduce the heat capacity curve for hydrogen. See also the 1919  &quot;A new formula for the Temperature Variation of the Specific Heat of Hydrogen&quot; Phys. Rev. vol. 11, pages 156-158. Only later when the significance of nuclear spin was understood was the problem fully explained.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-06-05T14:00:58.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32486" PostTypeId="2" ParentId="32480" CreationDate="2015-06-05T14:05:30.743" Score="7" Body="&lt;p&gt;Without being very specific it looks that $\ce{LiAlH(O-^{t}Bu)_3}$ is a lot less reactive than $\ce{LiAlH4}$, mostly because the $\ce{^{-}O-^{t}Bu}$ is a much weaker base than $\ce{H-}$. Also its much bulkier, further reducing reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aldehyde being less reactive than the acyl chloride the reducing agent first eats up all the acyl chloride, and then starts reducing the aldehyde. If you were careful to add stoichiometric amounts, you can stop right when you run out of the acyl chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also this &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/11/11/reagent-friday-lialhocch333/&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-05T14:05:30.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32487" PostTypeId="1" CreationDate="2015-06-05T15:11:09.873" Score="21" ViewCount="16085" Body="&lt;p&gt;Group 11 metals namely $\ce{Cu, Ag, Au}$ are known for their low reactivities. They are thus rightly called coinage metals and have been historically used to make currency owing to this very property of theirs. My question is, what is the reason behind this behaviour? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of the stability of fully filled and half filled orbitals? But then the counterexample of $\ce{Cu-Zn}$  presents itself. These are neighbours in the periodic table, and the differentiating electron in each goes completes a full or half-filled orbital, yet $\ce{Zn}$ is quite reactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behaviour can be seen (sort of) between $\ce{Hg-Au}$ as well which differ in one electron and yet are completely different in terms of their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The other two questions referred to as duplicates of this one cite different properties of the elements that make them unreactive. Gold has strong $\ce{Au-Au}$ bonds (due to relativistic effects) and $\ce{Zn}$ is more reducing than copper due to factors like hydration energy and the enthalpy of atomization. But this does not answer my question because:&#xA;&#xA;&lt;li&gt;I was looking for the root cause of this behaviour. If gold is stabilized by relativistic effects, why don't these effects operate on other transition metals? If zinc has high hydration enthalpy, what is the reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Since all the coinage metals occur in a single group, I really wanted an explanation that could apply to all of them, such as a peculiarity of their electronic configuration that renders them inert to most available reagents. &lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="1888" LastEditDate="2015-06-06T01:51:50.653" LastActivityDate="2017-05-23T17:11:09.457" Title="Why are Group 11 elements unreactive?" Tags="&lt;metal&gt;&lt;transition-metals&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="32488" PostTypeId="2" ParentId="32484" CreationDate="2015-06-05T16:09:25.537" Score="22" Body="&lt;p&gt;The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of &lt;a href=&quot;http://en.wikipedia.org/wiki/Thin-film_interference#Theory&quot;&gt;thin film interference&lt;/a&gt;—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air interfaces. The interference is dependant on the wavelength and incidence angle of light as well as the film thickness, which results in different wavelengths of light being attenuated by different amounts, making the surface appear coloured and the colour changing with viewing angle.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-05T16:09:25.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32489" PostTypeId="2" ParentId="32467" CreationDate="2015-06-05T17:22:27.547" Score="5" Body="&lt;p&gt;OK, here goes. Chemists use a variety of representations of molecules to try to predict their behavior and reactivity, but none of them come close to preparing you for what happens in the laboratory. In that sense, none of them are &quot;safe&quot; to use. Lewis structures, and the curly arrow formalisms, are good for organic chemists to quickly explore possible avenues of reactivity and synthetic pathways. But they are not very good at dealing with chemistry south of the second row of the periodic table, nor at phenomena such as aromaticity, electrocyclic reactions, bulk properties, solvation, steric effects, polarity, polarizability, photochemistry, free radicals, etc. So we add in to the mix some &quot;rules&quot;, try to think of molecules as being best described as &quot;multinuclear atoms&quot;, and try to improve and promulgate our best tools for computational chemistry. For all that, we still need to approach the laboratory with a degree of humility--things that look good on paper usually do not look so good when you try them out in the real world.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-05T17:22:27.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32490" PostTypeId="2" ParentId="32429" CreationDate="2015-06-05T20:13:46.760" Score="3" Body="&lt;p&gt;I think this is why it's generally not advised to use relative units such as $\%$, $\mathrm{ppm}$ and $\mathrm{ppb}$. If they're going to be used, then it's best to indicate what's the ratio being considered, which is frequently done using $\mathrm{(v/v)}$, $\mathrm{(m/m)}$ or $\mathrm{(m/v)}$, $\mathrm{m}$ being short for mass and $\mathrm{v}$ for volume. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{1\ mL_{\ methanol}}} = \mathrm{1\ ppm\ (m/v)} = \frac{\mathrm{1\ \mu g_{\ solute}}}{\mathrm{0.7918\ mg_{\ methanol}}} = \mathrm{1.263\ ppm\ (m/m)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's &lt;em&gt;common practice&lt;/em&gt; may be heavily dependent on whom we're talking about. Since we usually prepare solutions in the lab by completing the volume up to a certain amount, one could say it's best to ignore the solvent density and use $\mathrm{(/v)}$ as a rule of thumb. The solute unity will depend mostly whether it's a solid being weighed or a liquid being measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by @PLD, however, there might be cases where it's important to weigh the solvent, probably when very accurate measurements are necessary and specially if the solvent's volume changes considerably given changes in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem of using the $\mathrm{(m/v)}$ notation is that it doesn't use SI units, and might also be confusing. A more verbose way of indicating this is saying &lt;em&gt;mass to volume&lt;/em&gt;. Doing some research for this I even found the &quot;units&quot; $\mathrm{ppm_v}$ and $\mathrm{ppm_m}$, which indicates the solvent's unity, but not the solute's, but definetly aren't as common and are also unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution here is simply to use the full unity, so instead of using $\mathrm{ppm}$ (which actually isn't accepted in many cases, for example most scientific journals), use $\mathrm{mg/mL}$ (or $\mathrm{mg\ mL^{-1}}$) or whatever was really measured. This doesn't leave room for double interpretations and indicates how the solution was prepared.&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="15133" LastEditDate="2015-06-05T20:21:23.960" LastActivityDate="2015-06-05T20:21:23.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32492" PostTypeId="2" ParentId="32318" CreationDate="2015-06-05T21:41:58.803" Score="11" Body="&lt;p&gt;The persistence length is a statistically defined property - in other words, it is defined by looking at many lengths of many polymer molecules over a long period of time. It is the length (L) at which a vector drawn tangent to the polymer is no longer correlated in space with another tangent vector at distance (L). In other words, it tells you how well the position of one section of a polymer is correlated to another section. Beyond the persistence length, you would not expect to find any correlation. Below it, you would expect correlations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case &quot;correlations&quot; are measured by the cosine of the angle between the two vectors. When the angle is 90 degrees on average, the cosine is zero, and when it is zero degrees, the cosine is 1. So, for lengths beyond the persistence length, the amount of correlation is zero, and for lengths much less than the persistence length the amount of correlation is one. For lengths close to the persistence length, the correlation is between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an idea of how this works, take a look at the following illustrations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcgFr.png&quot; alt=&quot;Tangent vectors on a polymer chain.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red lines represent tangent vectors along the backbone of a polymer chain, spaced at regular intervals. We can translate all of these vectors and put their origin at the same point, and then find the angle between them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qktwK.png&quot; alt=&quot;Tangent vectors translated to share an origin.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we find another polymer chain, or watch this one for a little while and take another &quot;snapshot&quot; and do the same thing, we might end up with something that looks like this instead:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n0zdH.png&quot; alt=&quot;Another set of tangent vectors, translated to share an origin.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can even take the same very long chain and find all of the vectors that are a distance 1 apart, 2 apart, and so on. If we do this for all possible distances over hundreds of polymer molecules, and average over many snapshots for a very long time (nano- to micro- seconds for a melt or something in solution), what we will eventually find is something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7rOc5.png&quot; alt=&quot;Ensemble average of tangent vectors, translated to share an origin.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correlations drop off quickly with length, then more slowly, until finally they are completely uncorrelated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to realize here that at any given time, for any given polymer, the tangent vectors could be almost anything. You wouldn't expect two adjacent bonds to have a very small angle between them, and in general you would expect more flexibility over longer lengths of chain compared to shorter lengths, but you would not be able to predict the shape or angles with any degree of certainty, other than to say that on average, beyond the persistence length the tangent vectors would not be correlated, and that the amount of correlation would decay exponentially with length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The examples you gave of a stiff rod vs. cooked spaghetti are meant to demonstrate this intuitively, but your question is about what happens when the persistence length and the polymer itself are the same length. The answer is that the polymer will be somewhere between a very flexible piece of cooked spaghetti and a stiff rod. Imagine taking a piece of spaghetti and bending it into a loop, but not enough that it breaks. The circumference of 1/4 of the loop would be something like the persistence length, and the response of the spaghetti in the loop would be similar to a polymer that is about the same length as its persistence length. However, keep in mind that polymers are constantly in motion, and so &quot;flexible rods&quot; and &quot;cooked spaghetti&quot; are not great analogies in the first place. It's more realistic to think of them as constantly wriggling worms, or something like that. Statistical descriptions of correlation lengths and times are the best we can get, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another concept that might help you visualize this is that of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kuhn_length&quot; rel=&quot;noreferrer&quot;&gt;Kuhn length&lt;/a&gt; - this is twice the persistence length, and is the length at which sections of polymer are uncorrelated with each other in space. Practically, it means that the motion of the polymer within the Kuhn length is irrelevant to the motion of the overall chain. On a large scale, the motion of the Kuhn &quot;blobs&quot; is what matters. In your case, the polymer is about as long as the persistence length, which means it is one-half of the Kuhn length. This means motions along the polymer are correlated, and so you can't ignore them (if they are important to your model.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, what this all means is that your model is probably going to be fairly complicated, because you can't average out or ignore terms that can be ignored in either of the extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you post some more details about your model, I could probably give you more specific information.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-06-07T03:05:22.557" LastActivityDate="2015-06-07T03:05:22.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32493" PostTypeId="1" AcceptedAnswerId="32496" CreationDate="2015-06-05T21:43:33.660" Score="5" ViewCount="31958" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{2NH3 + CO2 -&amp;gt; (NH2)2CO + H2O}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If we have $637.2\ \mathrm{g}$ of $\ce{NH3}$ and $1142.0\ \mathrm{g}$ of $\ce{CO2}$ how many grams of $\ce{(NH2)2CO}$ will we get?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the explanation I was given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;First, convert reactives to moles:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$637.2\ \mathrm{g}$ of $\ce{NH3}$ are $37.5$ moles.&lt;/li&gt;&#xA;  &lt;li&gt;$1142.0\ \mathrm{g}$ of $\ce{CO2}$ are $26$ moles.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Second, define the stoichiometric proportion between reactives and&#xA;  products...&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;From $2$ moles of $\ce{NH3}$ we get $1$ mol of $\ce{(NH2)2CO}$&lt;/li&gt;&#xA;  &lt;li&gt;From $1$ mol of $\ce{CO2}$ we get $1$ mol of $\ce{(NH2)2CO}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;Third, calculate the number of product moles we would get if each&#xA;  reactive was completely consumed:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Consuming all the $\ce{NH3}$ we would get $18.75$ moles of $\ce{(NH2)2CO}$&lt;/li&gt;&#xA;  &lt;li&gt;Consuming all the $\ce{CO2}$ we would get $26$ moles of $\ce{(NH2)2CO}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;The limiting reagent is $\ce{NH3}$, so the most we will get of&#xA;  $\ce{(NH2)2CO}$ is $18.75$ moles. Now convert that to grams, we get&#xA;  $1125.0\ \mathrm{g}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are we ignoring the existence of $\ce{H2O}$? Does it not matter?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;IF&lt;/strong&gt; the formula is &lt;strong&gt;unbalanced&lt;/strong&gt;, do I &lt;em&gt;have&lt;/em&gt; to balance it first before making this calculation? What if I can't balance it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="5026" LastEditDate="2018-02-25T02:24:02.827" LastActivityDate="2018-02-25T02:24:02.827" Title="Calculating how many grams will I get from a reaction" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32496" PostTypeId="2" ParentId="32493" CreationDate="2015-06-05T22:46:12.337" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Here, $\ce{H2O}$ is a &lt;strong&gt;product&lt;/strong&gt; and you are assuming that the reaction goes to completion, so it does not affect the yield of the reaction in any way. Maybe a suitable analogy is to think of the reaction system as a car - you give the car a certain amount of petrol (say 2 gallons) and it converts it to a certain number of miles (say 70 miles), as well as a lot of exhaust gases. Now the amount of exhaust gases that your car is putting out doesn't affect the fact that if you put in 2 gallons of petrol, you get 70 miles out of it.&#xA;If you were to treat the reaction as a reversible one, where you have an equilibrium between reactants and products, then in general, the amount of products present would also matter. But don't worry about this for now, I'm sure you will come across this in your study of chemistry at some point in time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Good question - if the reaction is not balanced, then yes, you do have to balance it by finding the appropriate &lt;strong&gt;stoichiometric coefficients&lt;/strong&gt;. If you don't know exactly how many miles per gallon (35 in this case) your car consumes, then you won't be able to work out how many miles you can drive if you feed your car 2 gallons of petrol. Not all reactions can be balanced, but if a reaction can't be balanced in one way or another, it can't happen in real life. For example, you can't balance the reaction $\ce{Pb -&amp;gt; Au}$, but as we all know, alchemy was (mostly) a load of rubbish. For this reason, any reaction that you come across in your study of chemistry, as well as in real life, can be balanced.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-06-05T22:46:12.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32497" PostTypeId="2" ParentId="32465" CreationDate="2015-06-05T23:00:05.417" Score="5" Body="&lt;p&gt;The reason is that aluminium carbide, $\ce{Al4C3}$ is formed. The reaction occurs at around 500&amp;deg;C on the interface. The problem with this substance is that it is brittle, and it also reacts with water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Al4C3 + 12H2O -&amp;gt; 4Al(OH)3 + 3CH4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info at &lt;a href=&quot;http://en.wikipedia.org/wiki/Metal_matrix_composite&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Aluminium_carbide&quot;&gt;this&lt;/a&gt; wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastActivityDate="2015-06-05T23:00:05.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32498" PostTypeId="1" AcceptedAnswerId="32503" CreationDate="2015-06-06T00:02:13.600" Score="7" ViewCount="2103" Body="&lt;p&gt;I have a molecule extracted from the PDB. When I load it into Avogadro, it has some misplaced bond types. I want to correct this, by setting the bond type (single, double) for these bonds. How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="1159" LastEditDate="2015-06-08T17:15:16.190" LastActivityDate="2018-04-24T17:48:26.380" Title="How can I change the bond type (single, double) of an existing bond in Avogadro?" Tags="&lt;bond&gt;&lt;software&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32499" PostTypeId="1" CreationDate="2015-06-06T00:25:09.910" Score="3" ViewCount="40" Body="&lt;p&gt;These huge arrows are very annoying to me. I disable them every time I open Avogadro. Is there a way to disable them permanently?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastActivityDate="2015-06-06T02:29:40.127" Title="Can I disable &quot;Display visual cues&quot; by default in Avogadro?" Tags="&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32500" PostTypeId="1" CreationDate="2015-06-06T00:26:34.227" Score="6" ViewCount="937" Body="&lt;p&gt;I understand that it is possible to cool a liquid past its freezing point, and still maintain its liquid state because crystallisation would not happen in the absence of a seed crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, was marked as a duplicate and I was linked to to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32459/can-all-substances-be-supercooled&quot;&gt;Can all substances be supercooled?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the answers over there does address my question but only in part.&#xA;Here's what it says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.&#xA;  The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question still remains. Why is the lack of a nucleus a kinetic barrier to the formation of a bulk solid? If you continue past the glass transition temperature, we form an amorphous solid anyway. How do the mechanics of vitrification and crystallisation differ in this regard?&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-10-23T17:28:04.150" Title="Super-cooled liquids" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;&lt;temperature&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32501" PostTypeId="2" ParentId="18455" CreationDate="2015-06-06T01:49:37.330" Score="1" Body="&lt;p&gt;To understand the answer to (1), you have to think about what combustion actually is. It isn't a single reaction, with a single set of reactants, and a single set of products. It's actually a whole bunch of tiny steps (thousands) that all occur together. Those reactions happen among a whole bunch of unstable radical species (hundreds). When you put a match to a pool of gasoline, all that energy from the match starts tearing apart bonds in the fuel, leading to the formation of radicals. That's what triggers the initial release of energy, and what creates the cascade of ignition. So theoretically, &lt;strong&gt;no&lt;/strong&gt;, you could &lt;strong&gt;never&lt;/strong&gt; reach combustion products, through the pathway of combustion, because you'd never have that activation energy, the big kick of energy needed to create radicals and keep them alive long enough for them to create a chain reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to (2) is, no, it doesn't matter how much fuel. Remember you're measuring a mass-independent property. The autoginition temperature isn't &quot;5 degrees per gram&quot;. It's a single temperature. What matters is the ratio of oxygen to fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to (3), it's a lot easier to operate an open flask than it is to operate a piston-cylinder, and it doesn't have a substantial impact on the air-fuel ratio if you do it correctly.&lt;/p&gt;&#xA;" OwnerUserId="2599" LastActivityDate="2015-06-06T01:49:37.330" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="32502" PostTypeId="2" ParentId="32499" CreationDate="2015-06-06T02:29:40.127" Score="5" Body="&lt;p&gt;Sure. There's a big checkbox in the Navigate Tool settings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VxO7K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you quit, the settings should be remembered for next launch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, please e-mail the &lt;a href=&quot;http://lists.sourceforge.net/mailman/listinfo/avogadro-discuss&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro&lt;/a&gt; mailing list and we can figure out what's going on.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-06-06T02:29:40.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32503" PostTypeId="2" ParentId="32498" CreationDate="2015-06-06T02:33:03.073" Score="7" Body="&lt;p&gt;As discussed in the Avogadro website about the &lt;a href=&quot;https://avogadro.cc/docs/getting-started/drawing-molecules/&quot; rel=&quot;nofollow noreferrer&quot;&gt;draw tool&lt;/a&gt; you can change the bond order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By clicking on the bond - the bond order will go from single, to double, to triple, back to single.&lt;/li&gt;&#xA;&lt;li&gt;By setting the bond order and &quot;drawing over&quot; the bond. The new setting (e.g., double) will replace the existing bond.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2018-04-24T17:48:26.380" LastActivityDate="2018-04-24T17:48:26.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32504" PostTypeId="1" AcceptedAnswerId="32505" CreationDate="2015-06-06T03:02:33.740" Score="9" ViewCount="1726" Body="&lt;p&gt;The final paragraph of Dissenter's question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15885/why-does-hand-sanitizer-leave-your-skin-feeling-cool&quot;&gt;here&lt;/a&gt; is worthy of standing alone:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[H]ow does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(It's similar to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/180/why-do-different-substances-have-different-boiling-points&quot;&gt;this&lt;/a&gt; old question by Friend of Kim, but different in that it emphasizes an apparent contradiction.)&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-06-06T22:05:31.953" Title="How is it possible for a substance to have a high heat of vaporization but a low boiling point?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;boiling-point&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32505" PostTypeId="2" ParentId="32504" CreationDate="2015-06-06T03:02:33.740" Score="10" Body="&lt;p&gt;This seeming contradiction can be reconciled by examining the thermodynamic quantities involved.  First, per &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy_of_vaporization&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the &lt;em&gt;enthalpy of vaporization&lt;/em&gt; is &quot;the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure.&quot;  Written symbolically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this situation, it is most helpful to consider $H_i$ as the enthalpy per mole (&quot;molar enthalpy&quot;) of the species of interest in phase $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formally, again per &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_point&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the &lt;em&gt;boiling point&lt;/em&gt; for a substance is &quot;the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor.&quot;  This is not so helpful for answering this question, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of defining the boiling point is the temperature at which the liquid and the vapor of the substance are in equilibrium, for a closed system (no mass can get in or out) &lt;strong&gt;&lt;em&gt;containing only the pure substance of interest&lt;/em&gt;&lt;/strong&gt; at a given pressure$^*$.  The equilibrium condition here can be expressed in terms of molar &lt;a href=&quot;http://en.wikipedia.org/wiki/Gibbs_free_energy&quot; rel=&quot;nofollow&quot;&gt;Gibbs free energies&lt;/a&gt; (free energy per mole$^{**}$) as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G_{\mathrm{liq}} = G_{\mathrm{gas}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, the molar free energy of a phase is equal to the molar &lt;a href=&quot;http://en.wikipedia.org/wiki/Enthalpy&quot; rel=&quot;nofollow&quot;&gt;enthalpy&lt;/a&gt; minus the temperature times the molar &lt;a href=&quot;http://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow&quot;&gt;entropy&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G = H - TS&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the boiling point equilibrium condition can be rewritten as ($T_b$ is the boiling point):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;H_{\mathrm{liq}} - T_bS_{\mathrm{liq}} = H_{\mathrm{gas}} - T_bS_{\mathrm{gas}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting $H_{\mathrm{liq}}$ and $-T_bS_{\mathrm{gas}}$ from both sides gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_b\Delta S_{\mathrm{vap}} = T_b\left(S_{\mathrm{gas}} - S_{\mathrm{liq}}\right) = H_{\mathrm{gas}} - H_{\mathrm{liq}} = \Delta H_{\mathrm{vap}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\Delta S_{\mathrm{vap}}$ is the &lt;em&gt;molar entropy of vaporization&lt;/em&gt;, which is analogous to $\Delta H_{\mathrm{vap}}$.  The boiling point is thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_b = \frac{\Delta H_{\mathrm{vap}}}{\Delta S_{\mathrm{vap}}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So, what does this result mean?&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualitatively, as noted in answers to the linked questions, $\Delta H_{\mathrm{vap}}$ is a measure of how much energy must be added to one mole of molecules in the liquid to free them from their associations with the other liquid molecules.  The physical meaning of $\Delta S_{\mathrm{vap}}$ is somewhat harder to describe, but corresponds roughly to the increase in the 'freedom to wiggle' afforded by the transition to the gas phase from the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to have a large $\Delta H_{\mathrm{vap}}$ (lots of energy required to drive molecules into the gas phase) but also a low $T_b$ (low temperature at which molecules &quot;like&quot; to transition into the gas phase) a chemical must have a large $\Delta S_{\mathrm{vap}}$, which means that each molecule gains a significant amount of 'freedom of movement' from entering the gas phase.  This makes sense in the case of ethanol versus water in Dissenter's question: ethanol is bulkier than water, so even though they both have similar values of $\Delta H_{\mathrm{vap}}$, ethanol gains more 'freedom to wiggle' in the gas phase and thus has a lower $T_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also helps to provide a qualitative explanation for why $T_b$ is proportional to pressure: as the pressure of the gas phase increases, the 'freedom to wiggle' gained by entry to the gas phase decreases, because higher $P$ means more collisions with gas-phase molecules.  Thus, one can expect $\Delta S_{\mathrm{vap}}$ to decrease with increasing pressure ($S_{\mathrm{liq}}$ isn't affected nearly as much by pressure changes as is $S_{\mathrm{gas}}$), which translates directly to the experimentally observed increase in $T_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details on the relationship between boiling point and pressure can be found in the Wikipedia articles on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_point#Saturation_temperature_and_pressure&quot; rel=&quot;nofollow&quot;&gt;saturation temperature&lt;/a&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation&quot; rel=&quot;nofollow&quot;&gt;Clausius-Clapeyron relation&lt;/a&gt;.  More information on equilibrium and Gibbs free energy can be found on &lt;a href=&quot;http://www.chem1.com/acad/webtext/thermeq/TE4.html&quot; rel=&quot;nofollow&quot;&gt;www.chem1.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$^*$ In this model system, the &quot;knob&quot; one has available (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Phase_rule&quot; rel=&quot;nofollow&quot;&gt;Gibbs' phase rule&lt;/a&gt;) to change the equilibrium pressure and temperature is to add to or remove from the system some number of moles of the species of interest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{**}$ As orthocresol rightly notes in a comment, the equilibrium must be expressed in terms of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_potential&quot; rel=&quot;nofollow&quot;&gt;chemical potentials&lt;/a&gt; of each species present. However, in this case there is only one species present and thus examining the total molar Gibbs free energy is sufficient.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-06T22:05:31.953" LastActivityDate="2015-06-06T22:05:31.953" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32506" PostTypeId="2" ParentId="32423" CreationDate="2015-06-06T04:03:39.877" Score="2" Body="&lt;p&gt;The route I would try to prepare 1 would be to purchase 4-Bromo-4'-hydroxybiphenyl which is commercially available (Alfa Aesar sells it - see &lt;a href=&quot;http://www.alfa.com/en/catalog/A10819&quot; rel=&quot;nofollow&quot;&gt;http://www.alfa.com/en/catalog/A10819&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Step:&#xA;1) Protect phenol moeity with benzyl group via refluxing a solution of 4-Bromo-4'-hydroxybiphenyl (1.0 eq, 1.00 g), milled potassium carbonate (4.0 eq), benzyl bromide (1.1-1.25 eq), and acetone (25-30 mL). Reaction will most likely take 10-20 hours to be complete. Once reaction is complete, cool to ambient temperature and filter reaction mixture. Concentrate the filtrate to dryness. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you dont have Benzyl bromide, you can use benzyl chloride (1.1-1.25 eq) and potassium iodide (0.1-0.5 eq).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;**Alternatively, instead of filtering the reaction mixture, you could add water (50-75 mL) to the mixture to cause the product to crystallize out (this will only work if the product is a solid at room temperature). If you dont see any solids after adding water, you can try adding an ice-water bath under the mixture to encourage crystallization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once isolated the material, allow the protected product (benzyl ether) to be dried in a desiccator or vacuum oven to remove traces of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Step:&#xA;2) Perform a Grignard reaction with protected product (above).&#xA;   The Grignard reaction can be prepared either in THF or diethyl ether using Mg metal/iodine as catalyst, or using isopropylmagnesium chloride solution in THF).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the Grignard reaction has been prepared, pour the reaction onto dry ice. This will form the Protected Compound 1 as the magnesium salt. Add water to solubilize the salt, then wash the mixture with diethyl ether (1-2 times to remove impurities) and then add concentrated $\ce{HCl}$ (37%) until pH 1-2 is acheived which will precipitate out the protected Compound 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third Step:&#xA;3) Dissolve protected Compound 1 in ethyl acetate, add small amount of Palladium on carbon, and hydrogenate the mixture under hydrogen pressure (balloon pressure will work just fine too) at room temperature overnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once reaction is complete, degas the mixture with nitrogen or argon, filter, and concentrate to provide the desired Compound 1.&lt;/p&gt;&#xA;" OwnerUserId="16766" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T07:25:51.693" LastActivityDate="2015-06-06T07:25:51.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32507" PostTypeId="2" ParentId="32493" CreationDate="2015-06-06T04:13:53.383" Score="5" Body="&lt;p&gt;The product $\ce{H_2O}$ does not matter to the question.  You have some reactants and the reaction proceeds until one is used up.  As you are asked about the mass of one product, you don't care about how much of the other product there is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you need to balance the reaction.  You need to figure out which reactant is the limiting one.  Your reaction is balanced, so you use two molecules of ammonia for each of carbon dioxide.  In the answer, under Third, they assume first that $\ce{NH_3}$ is the limiting factor and that there is plenty of $\ce{CO_2}$, finding you can produce $18,75$ moles.  Then they assume there is plenty of $\ce{NH_3}$ and you are limited by $\ce{CO_2}$, finding you can produce $26$ moles.  The first assumption is correct, so that is how much you can make.&lt;/p&gt;&#xA;" OwnerUserId="11344" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T07:23:17.643" LastActivityDate="2015-06-06T07:23:17.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32509" PostTypeId="1" AcceptedAnswerId="32520" CreationDate="2015-06-06T07:36:55.443" Score="6" ViewCount="2000" Body="&lt;p&gt;How would you go about writing the reaction between water and chromium metal under standard acid and base conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you would have to look at the vertical lines at pH 0 and 14 but I don’t know how you would know which species to look at and whether or not you write the reduction or oxidation of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, at pH 14 how would you calculate the ${\Delta G}$ value for the reduction between water and chromium metal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLQVv.png&quot; alt=&quot;Pourbaix Diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16768" LastEditorDisplayName="user7951" LastEditDate="2015-06-06T15:36:14.047" LastActivityDate="2015-06-06T15:36:18.710" Title="Pourbaix diagram – writing reactions" Tags="&lt;acid-base&gt;&lt;redox&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32510" PostTypeId="2" ParentId="32500" CreationDate="2015-06-06T08:28:26.043" Score="2" Body="&lt;p&gt;When a liquid is supercooled its structure doesn't change significantly, only the particles have less kinetic energy available. As you cool even further, it could occur that particles get caught in local minima and eventually you'll end up with a glass, which can be thought of as a frozen liquid structure. As particles are trapped it will resist shear in an elastic fashion so you have a solid. In this regard it is a barrierless transition. There is no interface whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystal formation is fundamentally different as the local structure changes first to form a nucleus. This has both energetic and entropic consequences and the aforementioned kinetic barrier is observed. I would also like to add that crystal nuclei are formed all the time but they don't reach the critical size if the temperature is too high.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastEditorUserId="16740" LastEditDate="2015-06-06T09:11:03.217" LastActivityDate="2015-06-06T09:11:03.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32511" PostTypeId="1" CreationDate="2015-06-06T08:43:45.413" Score="6" ViewCount="136" Body="&lt;p&gt;I am exploring (non-experimentally!) the methylation of a perchlorate precursor, either the acid or pre-existing salts, to produce the alkali methyl perchlorate. The methylating agent would be methane itself, and I would like to understand whether:   &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The product is possible from these (or more) reactants,   &lt;/li&gt;&#xA;&lt;li&gt;Whether the methane, as an organic combustible, would not simply deflagrate with perchlorate, leading to no other products (beyond fire or explosion),   &lt;/li&gt;&#xA;&lt;li&gt;If methyl perchlorate is deliquescent (as I believe that perchlorate esters hydrolyze rapidly with water), and   &lt;/li&gt;&#xA;&lt;li&gt;If low temperature conditions (&amp;lt; 5 °C) might permit '1' and avoid '2'. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reference to any answers would be much appreciated, my interest being in recent events in the Siberian tundra...&lt;/p&gt;&#xA;" OwnerUserId="16699" LastEditorDisplayName="user7951" LastEditDate="2015-06-06T12:57:48.007" LastActivityDate="2015-06-06T12:57:48.007" Title="Methylation of perchlorate" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;combustion&gt;&lt;explosives&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32512" PostTypeId="2" ParentId="32407" CreationDate="2015-06-06T10:28:56.887" Score="10" Body="&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main thing that determines the shape of orbitals is that they must have zero net overlap with all other orbitals (i.e. that they are orthogonal). It's quite hard to construct two orthogonal $\sigma$-bonding MOs connecting the same two atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More details&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma$-bonding MOs are typically $sp^x$ hybridized orbitals with maximum overlap along the line connecting two atomic centers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PP806.jpg&quot; alt=&quot;a typical sigma bonding MO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to construct another &lt;em&gt;bonding&lt;/em&gt; MO with maximum overlap along the same line that is orthogonal to this MO. To stay orthogonal we have introduce a node and the next $\sigma$ MO becomes anti-bonding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vu7GK.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual way to get another, orthogonal, bonding MO is a $\pi$-bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9t9i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have valence $d$-electrons available then it is possible to construct another, orthogonal, $\sigma$-orbital &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pmalz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is still rare to have two $\sigma$ bonding MOs because transition metals tend to loose the $s$-electrons needed for one of the $\sigma$-MOs very readily. So double $\sigma$-bonding tend to be only observed in special cases such as gas phase $\ce{Mo2}$.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-06T10:28:56.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32513" PostTypeId="1" CreationDate="2015-06-06T10:59:24.760" Score="7" ViewCount="5303" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What would happen when you mix liquid $\ce{NH3}$ and excess of sulphur? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer key states that a solid with $\ce{N}$ and $\ce{S}$, and a gas which gives a black precipitate with lead acetate is evolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely, $\ce{H2S}$ gas will evolve. What about the solid? Is it ammonium sulphide or ammonium sulphite?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="5026" LastEditDate="2018-06-30T05:23:01.653" LastActivityDate="2018-07-31T17:38:33.760" Title="What's the solid product of the reaction of liquid ammonia with excess sulfur?" Tags="&lt;inorganic-chemistry&gt;&lt;identification&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32514" PostTypeId="1" AcceptedAnswerId="32804" CreationDate="2015-06-06T12:03:02.523" Score="11" ViewCount="1514" Body="&lt;p&gt;I'm going to parameterize AMBER force field for the simulation of some small molecules. I have found the parameter sets provided by TINKER at &lt;a href=&quot;http://dasher.wustl.edu/tinker/distribution/params/&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck here because I couldn't find any explanations about the parameter values. They should be the parameters that appear in the &lt;a href=&quot;http://en.wikipedia.org/wiki/AMBER#Functional_form&quot;&gt;formula of AMBER force field&lt;/a&gt;. However, I have no idea which numbers mean $k_b$, $l_0$, $k_a$, $θ_0$, or the rest parameters.&lt;/p&gt;&#xA;" OwnerUserId="16771" LastEditorUserId="-1" LastEditDate="2015-06-06T13:00:37.520" LastActivityDate="2018-06-28T02:47:28.083" Title="How to read TINKER's force field parameters" Tags="&lt;computational-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32515" PostTypeId="2" ParentId="32423" CreationDate="2015-06-06T12:39:53.430" Score="1" Body="&lt;p&gt;If you have the capability,  convert the ketone to an enol ether and ozonolyse / oxidative cleave the alkene portion. You might want to cleave your acetate first, make your ether and then silylate , but you'd be silylating to make the enol ether anyway.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-06T12:39:53.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32516" PostTypeId="2" ParentId="32493" CreationDate="2015-06-06T13:08:19.623" Score="2" Body="&lt;p&gt;Did anyone explain why you convert to moles? It is because moles is a certain quantity of molecules of and to balance the equation you need to work in molecules or molecule multiples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it goes to completion you will use up all of the $\ce{NH3}$ as you say, but be left with 7.25 moles of $\ce{CO2}$, 18.75 moles of $\ce{(NH2)2CO2}$ and 18.75 moles of water. By the way, the original equation is already stoichiometricly balanced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 $\ce{H}$ on both sides, 2 $\ce{O}$, and 2 $\ce{N}$ - It's just like algebra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it wasn't you would have to balance it before calculating the rest and of course we are simplifying by assuming the reaction goes to completion.&lt;/p&gt;&#xA;" OwnerUserId="16772" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T13:13:52.417" LastActivityDate="2015-06-06T13:13:52.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32517" PostTypeId="1" AcceptedAnswerId="32550" CreationDate="2015-06-06T13:59:32.020" Score="5" ViewCount="3761" Body="&lt;p&gt;I have used iodomethane to make a phenolic $\ce{OH}$ group to $\ce{OMe}$ group in Aq+Methanol medium. I have added water to precipitate the product of mine and will evaporate the solvents but I cant use rotavap until I am sure that any residual iodomethane is not contaminating the condenser of rotavap. How shall I destroy excess iodomethane? will aq. solution of $\ce{KOH}$ work?&lt;/p&gt;&#xA;" OwnerUserId="14874" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T20:29:18.850" LastActivityDate="2015-06-07T09:10:51.987" Title="how to destroy excess iodomethane in reaction mixture?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;safety&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32519" PostTypeId="1" AcceptedAnswerId="32562" CreationDate="2015-06-06T15:24:56.660" Score="4" ViewCount="676" Body="&lt;p&gt;$K_p$ is defined as $\Pi_j(\frac{P_j}{P^{\Theta}})^{\nu j}$ where $P_j$ is the partial pressure of component j and $\nu$ is the stoichiometric coefficient. I have a couple of questions already. Firstly, $K_p$ is meant to be dimensionless so therefore, $\frac{P_j}{P^{\Theta}}$ must be dimensionless. However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, suppose I want to write the expression for $K_p$ in terms of partial pressures for the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{\frac12 O2_{(g)} + C_{(s)} &amp;lt;=&amp;gt;CO_{(g)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I assume that $\frac{P_{c}}{P^\Theta}=1$ leaving me with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_p=\frac{P_{\ce{CO}}}{(P_{\ce{O_2}})^{\frac12}}\frac{1}{P^\theta}$$ &lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T20:27:35.343" LastActivityDate="2015-07-21T01:37:09.800" Title="Equilibrium constant in terms of partial pressures" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32520" PostTypeId="2" ParentId="32509" CreationDate="2015-06-06T15:36:18.710" Score="4" Body="&lt;p&gt;The red line in your diagram corresponds to the redox potential of hydrogen, which depends on $\mathrm{pH}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{3}&#xA;\ce{2H+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2}\quad\quad &amp;amp;&amp;amp;E^\circ = +0.000\ \mathrm{V}\quad\quad &amp;amp;&amp;amp;&amp;amp;(\mathrm{pH}=0)\\&#xA;\ce{2H2O + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2 + 2OH-}\quad\quad &amp;amp;&amp;amp;E^\circ = -0.828\ \mathrm{V}\quad\quad &amp;amp;&amp;amp;&amp;amp;(\mathrm{pH}=14)&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue line in your diagram corresponds to the redox potential of oxygen, which depends on $\mathrm{pH}$, too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{3}&#xA;\ce{O2 + 4H+ + 4e- \;&amp;amp;&amp;lt;=&amp;gt; 2H2O}\quad\quad &amp;amp;&amp;amp;E^\circ = +1.229\ \mathrm{V}\quad\quad &amp;amp;&amp;amp;&amp;amp;(\mathrm{pH}=0)\\&#xA;\ce{O2 + 2H2O + 4e- \;&amp;amp;&amp;lt;=&amp;gt; 4OH-}\quad\quad &amp;amp;&amp;amp;E^\circ = +0.401\ \mathrm{V}\quad\quad &amp;amp;&amp;amp;&amp;amp;(\mathrm{pH}=14)&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, only areas between these lines are stable in aqueous solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your initial species is elemental chromium ($\ce{Cr}$). It is shown below the red line; therefore, it is unstable with respect to oxidation to $\ce{Cr^3+}$ at $\mathrm{pH}=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{2H+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2}\quad\quad &amp;amp;&amp;amp;E^\circ = +0.000\ \mathrm{V}\\&#xA;\ce{Cr^3+ + e- \;&amp;amp;&amp;lt;=&amp;gt; Cr^2+}\quad\quad &amp;amp;&amp;amp;E^\circ = -0.408\ \mathrm{V}\\&#xA;\ce{Cr^2+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; Cr}\quad\quad &amp;amp;&amp;amp;E^\circ = -0.913\ \mathrm{V}\\\\&#xA;\ce{2Cr + 6H+ &amp;amp;-&amp;gt; 2Cr^3+ + 3H2}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to $\ce{Cr(OH)3}$ at $\mathrm{pH}=14$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{2H2O + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2 + 2OH-}\quad\quad &amp;amp;&amp;amp;E^\circ = -0.828\ \mathrm{V}\\&#xA;\ce{Cr(OH)3 + 3e- \;&amp;amp;&amp;lt;=&amp;gt; Cr + 3OH-}\quad\quad &amp;amp;&amp;amp;E^\circ = -1.33\ \mathrm{V}\\\\&#xA;\ce{2Cr + 6H2O &amp;amp;-&amp;gt; 2Cr(OH)3 + 3H2}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-06-06T15:36:18.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32521" PostTypeId="1" CreationDate="2015-06-06T15:54:14.280" Score="3" ViewCount="1358" Body="&lt;p&gt;Pressure holds a great significance when it comes to reactions involving gaseous species. Changing pressure can affect the kinetics and even mechanisms of such reactions. &lt;p&gt;&#xA;However, &lt;strong&gt;&lt;em&gt;does pressure hold any significance in reactions involving solid and liquid reactants only?&lt;/em&gt;&lt;/strong&gt; &#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Examples, if any, would be appreciated.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T20:20:29.917" LastActivityDate="2015-06-07T01:53:13.427" Title="Does pressure affect non-gaseous reactions?" Tags="&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;reaction-control&gt;&lt;pressure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32522" PostTypeId="2" ParentId="32407" CreationDate="2015-06-06T16:33:06.770" Score="15" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can there not be more than one sigma bond in a set of bonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There can be, even in simple carbon compounds. &lt;a href=&quot;http://en.wikipedia.org/wiki/Bent_bond&quot; rel=&quot;noreferrer&quot;&gt;Bent bonds, tau bonds or banana bonds&lt;/a&gt;; whatever you might like to call them were proposed by Linus Pauling; Erich Hückel proposed the alternative $\sigma - \pi$ bonding formalism. Hückel's description is the one commonly seen in introductory texts, but both methods produce equivalent descriptions of the electron distribution in a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to better understand the bent bond model let's first consider its application to cyclopropane and then move to ethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cyclopropane it has been found that significant electron density lies &lt;strong&gt;off&lt;/strong&gt; the internuclear axis, rather than along the axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/adHbH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://skmclasses.weebly.com/types-of-bonds.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the $\ce{H-C-H}$ angle in cyclopropane has been measured and found to be 114 degrees.  From this, and using Coulson's Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized.  Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the &quot;s&quot; character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2~=~}$$\mathrm{3.74}$, or the $\ce{C-C}$ bond is $\ce{sp^{3.74}}$ hybridized.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictorially, the bonds look as follows.  They are bent (hence the strain in cyclopropane) and concentrate their electron density off of the internuclear axis as experimentally observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M1LsF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can apply these same concepts to the description of ethylene. Using the known $\ce{H-C-H}$ bond angle of 117 degrees, Coulson's theorem and assuming that we have one p-orbital on each carbon (Hückel's $\sigma - \pi$ formalism), we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bond are $\ce{sp^{2.2}}$ hybridized and the one carbon orbital involved in the $\ce{C-C}$ sigma bond is $\ce{sp^{1.7}}$ hybridized (plus there is the unhybridized p-orbital). This is the &quot;$\ce{sp^2}$&quot; description we see in most textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, if we only change our assumption of one pi bond and one sigma bond between the two carbon atoms to two equivalent sigma bonds (Pauling's bent bond formalism), we would find that the carbon orbitals involved in the $\ce{C-H}$ bond are again $\ce{sp^{2.2}}$ hybridized, but the two equivalent $\ce{C-C}$ sigma bonds are $\ce{sp^{4.3}}$ hybridized. We have constructed a &lt;strong&gt;two-membered ring&lt;/strong&gt; cycloalkane analogue!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJHl0.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hybridization is just a mathematical construct, another model to help us understand and describe molecular bonding.  As shown in the ethylene example above, we can mix carbon's atomic s- and p-orbitals in different ways to describe ethylene. However, the two different ways we mixed these orbitals&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one s-orbital plus two p-orbitals plus one unmixed p-orbital (Hückel's  $\sigma - \pi$), or&lt;/li&gt;&#xA;&lt;li&gt;two s-orbitals plus 2 p-orbitals (Pauling's bent bond)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;must lead to &lt;strong&gt;equivalent&lt;/strong&gt; electronic descriptions of ethylene. As noted in the Wikipedia article (first link at the beginning of this answer),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is still some debate as to which of the two representations is&#xA;  better,[10] although both models are mathematically equivalent. In a&#xA;  1996 review, Kenneth B. Wiberg concluded that &quot;although a conclusive&#xA;  statement cannot be made on the basis of the currently available&#xA;  information, it seems likely that we can continue to consider the σ/π&#xA;  and bent-bond descriptions of ethylene to be equivalent.&lt;a href=&quot;https://i.stack.imgur.com/adHbH.jpg&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt; Ian&#xA;  Fleming goes further in a 2010 textbook, noting that &quot;the overall&#xA;  distribution of electrons [...] is exactly the same&quot; in the two&#xA;  models.[11]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bent bond model has the advantage of explaining both cyclopropane (and other strained molecules) as well as olefins.  The strain in cyclopropane and ethylene (&lt;em&gt;e.g.&lt;/em&gt; heats of hydrogenation) also make intuitive sense with the bent bond model where the term &quot;bent bond&quot; conjures up an image of strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, bent bonds are a mix of both sigma and pi properties and multiple bent bonds  can be used to describe the bonding between adjacent atoms.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-09-13T15:17:10.220" LastActivityDate="2015-09-13T15:17:10.220" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32523" PostTypeId="1" AcceptedAnswerId="32531" CreationDate="2015-06-06T17:32:53.117" Score="5" ViewCount="1706" Body="&lt;p&gt;The hydrogen in $\ce{C4}$, $\ce{C6}$, $\ce{C10}$ and $\ce{C8}$ belong to same hydrogen environment.&lt;br&gt;&#xA;The hydrogen in $\ce{C3}$, $\ce{C1}$, $\ce{C11}$ and $\ce{C13}$ belong to the another hydrogen environment.&lt;br&gt;&#xA;Hydrogen in $\ce{C12}$ and $\ce{C2}$ belong to the once again, another hydrogen environment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C3}$, $\ce{C1}$ , $\ce{C11}$  $\ce{C13}$, $\ce{C12}$ and $\ce{C2}$ all have two hydrogen on the neighboring carbon atoms. &#xA;So my question is why aren't they classified as one hydrogen environment? Instead, why are the categorized under two different environments? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bGOaSb.png&quot; alt=&quot;benzenephenone&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16774" LastEditorUserId="46306" LastEditDate="2018-03-04T01:37:43.360" LastActivityDate="2018-03-04T01:37:43.360" Title="hydrogen environments of benzophenone" Tags="&lt;aromatic-compounds&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32524" PostTypeId="1" CreationDate="2015-06-06T17:48:24.327" Score="6" ViewCount="382" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the reaction-enthalpy for the synthesis of 40 g hydrazine (rocketfuel): $$\ce{4NH3(g) + Cl2 (g) -&amp;gt; N2H4(l) + 2 NH4Cl(s)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt at solution:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we have to account for the decomposition of four moles of ammonia into its elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Delta H_\mathrm f^o(\ce{NH3})$ = $-46.1 \mathrm{~kJ~mol^{-1}}$, we do: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H_1 = 4(+46.1) = 184.4 \mathrm{~kJ}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl2}$ is in its standard state so we can ignore that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\Delta H_f^o(\ce{N2H4})$ =  $50.6\mathrm{~kJ~mol^{-1}}$. Therefore, $\Delta H_2 = \pu{50.6 kJ}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $\Delta H_\mathrm f^o(\ce{NH4Cl})$ =  $-314.2\mathrm{~kJ~mol^{-1}}$ to give&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H_3 = 2(-314.4) = -628.8~\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, &#xA;\begin{align}&#xA;\Delta H_{\mathrm{rxn}} &amp;amp;= \Delta H_1 + \Delta H_2 + \Delta H_3\\&#xA;&amp;amp;= 184.4 + 50.6 + (-628.8) = -393.8~\mathrm{kJ}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my textbook says the correct answer is $-492.3~\mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me what I did wrong, please? Also, I didn't account for the $40\mathrm{~g}$. How does that affect the problem?&lt;/p&gt;&#xA;" OwnerUserId="15491" LastEditorUserId="5026" LastEditDate="2018-03-01T08:18:19.440" LastActivityDate="2018-03-01T08:18:19.440" Title="Reaction-enthalpy for synthesis of 40 g hydrazine" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32525" PostTypeId="2" ParentId="32521" CreationDate="2015-06-06T17:58:26.663" Score="4" Body="&lt;p&gt;Yes. Pressure, as a concept, is simply the force exerted by random molecular motion per unit area, and since all matter is composed of molecules, and all matter above absolute zero has some random motion, everything has molecules exerting some force over some area. (This pressure is the vapor pressure of the solid, which is typically extremely low. If the external pressure on the solid is higher than the vapor pressure of the solid, the solid molecules will not fly off and escape. Things with high vapor pressure exert a high &quot;pressure&quot; on their surroundings and thus escape easily - jet fuel or nail polish remover have very high vapor pressures.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a liquid reaction between molecules A and B, bringing the liquid to a higher pressure will increase the frequency of collisions between molecules of A and B (by the definition of pressure). Since there is a finite probability of the reaction proceeding for any given collision, increasing the collisions increases the rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, for you to see that kind of effect in a laboratory would require a very pressure-sensitive reaction, large molecules whose shape changes with pressure, or a reaction with a large change in the number of moles. Biological chemistry is rich with examples pressure-sensitive non-gaseous reactions, mainly due to pressure changing the shape and behavior of large complex molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the effect of osmotic pressure, which is a pressure induced not by random molecular motion but by the tendency of the substance to seek a lower free energy state. Osmotic pressure will affect concentrations (the system will seek to reduce the concentrations of ions), which will affect reaction rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the pressure of solids and liquids can also change their physical properties (heat capacity and thermal conductivity are two excellent examples). These physical properties are innately tied to the reaction mechanism. Since most reactions occur in sequence, if the physical properties are different (and if they change over the course of the reaction), pressure will exert an effect there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is an effect technically excluded by your question, but important nonetheless, and that is the effect of pressure on solid-gas reactions (sublimation or deposition). The manufacture of integrated circuits involves depositing materials onto chips in layers, and pressure is an important way to control that rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the more nonlinear the reaction mechanism, the more likely it is that the small effects that pressure creates will cascade into larger changes that are what you would call &quot;significant.&quot;&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="2599" LastEditDate="2015-06-07T01:53:13.427" LastActivityDate="2015-06-07T01:53:13.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32527" PostTypeId="1" CreationDate="2015-06-06T18:28:18.747" Score="7" ViewCount="18976" Body="&lt;p&gt;Here are the electron affinities of the 16th and 17th groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ycQRj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ycQRj.png&quot; alt=&quot;electron affinities of the 16th and 17th group of the periodic table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general trend for electron affinity down the group is that it decreases because of the increase in atomic radius.The exception of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}&amp;gt;\ce{F}$&lt;/span&gt;, I can understand is because fluorine has a high electron density and it is unfavourable to add more electrons as it would only increase the electron -electron repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same reasoning can be applied to &lt;span class=&quot;math-container&quot;&gt;$\ce{S}&amp;gt;\ce{O}$&lt;/span&gt;. But oxygen's electron affinity is lesser than all the elements down group 16. Why is this so? And how is it so different from group 17 or why doesn't group 17 have &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; being less than the rest?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to apply the reasoning that the electron-electron repulsion in oxygen is so much that it is less favourable for &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; than any other element in group 16 to gain an electron. But in that case &lt;span class=&quot;math-container&quot;&gt;$\ce{F}$&lt;/span&gt; is an even smaller atom hence with even more repulsion. So why are these two groups so different?&lt;/p&gt;&#xA;" OwnerUserId="9298" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2019-11-19T04:20:15.087" LastActivityDate="2020-07-26T16:33:53.180" Title="Electron affinities of the chalcogens and halogens" Tags="&lt;periodic-trends&gt;&lt;electron-affinity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32528" PostTypeId="1" AcceptedAnswerId="47051" CreationDate="2015-06-06T18:32:03.823" Score="4" ViewCount="23475" Body="&lt;p&gt;I tried to google and watch YouTube videos on how to write Lewis structures properly, but I can't seem to understand how to do this correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher, when class was in session, gave me a set of instructions to complete Lewis structures:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sum Valence electrons&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add electrons to outer atoms, while obeying octet or duet rule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then add electrons to center atom, obey octet rule, unless they're third row element or below.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Do formal charges for each atom valance charge - (unbonded electrons + 1/2(bounded electrons))&lt;/li&gt;&#xA;&lt;li&gt;Once formal charges have been determined, minimize formal charges by using unbound electrons to make double or triple bonds&lt;/li&gt;&#xA;&lt;li&gt;Redo formal charges to check if formal charges are as low as possible, for central atom.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I attempted to do this for the diazomethane, but my answer (circled in green) does not match the book's answer (in red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LvSND.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LvSND.png&quot; alt=&quot;Lewis structures of diazomethane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After further considerations, I realized that my structures didn't obey the octet rule. So then would this revised structure be also correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DRgSK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DRgSK.png&quot; alt=&quot;Lewis structure of diazomethane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16224" LastEditorUserId="4945" LastEditorDisplayName="user15489" LastEditDate="2020-02-17T13:28:35.907" LastActivityDate="2020-02-17T13:28:35.907" Title="What is the correct Lewis structure of diazomethane?" Tags="&lt;organic-chemistry&gt;&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32529" PostTypeId="1" AcceptedAnswerId="32536" CreationDate="2015-06-06T19:37:20.200" Score="21" ViewCount="9920" Body="&lt;p&gt;I was wondering that like whenever I boil milk I just have to keep standing in front of it to make sure that I don't waste any milk but then I was wondering like Why does it even overflow ? In case of water it doesn't happen. What makes it to behave like this ?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastEditorUserId="10185" LastEditDate="2015-06-06T19:58:31.860" LastActivityDate="2017-09-13T13:03:01.440" Title="Why does milk overflow when boiled?" Tags="&lt;everyday-chemistry&gt;&lt;heat&gt;&lt;food-chemistry&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32530" PostTypeId="1" AcceptedAnswerId="32696" CreationDate="2015-06-06T21:08:50.987" Score="7" ViewCount="385" Body="&lt;p&gt;I've had already been aware of the fact that metallic aluminum can be degraded by coming in contact with mercury, but also came across another question more recently &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32462/gallium-induced-structural-failure-of-aluminium&quot;&gt;here&lt;/a&gt; explains a similar reaction with liquid gallium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is more general regarding corrosive action of &lt;em&gt;any&lt;/em&gt; liquid metal on other metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the more general conditions for a liquid metal to be corrosive to another, different solid metal?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the word &lt;em&gt;corrosive&lt;/em&gt; is not the most suitable choice. Corrosion refers to a chemical reaction. If I understand it right what's happening, at least in the case of $\ce{Ga}$ and $\ce{Hg}$ on $\ce{Al}$, is a weakening of the grain boundaries by displacement and amalgamation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interesting thought occurred to me. Liquid sodium is often used as a coolant in nuclear power plants. It's pumped through metal pipe (material unknown). So here is a good example of a combination that doesn't lead to corrosion. Are there examples other than I've mentioned that lead to corrosion? So what makes the difference?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-06-12T16:02:23.793" Title="What are the conditions for liquid metal to be corrosive to another, different solid metal?" Tags="&lt;physical-chemistry&gt;&lt;metallurgy&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32531" PostTypeId="2" ParentId="32523" CreationDate="2015-06-06T21:17:56.697" Score="9" Body="&lt;p&gt;Whether environments are the same or not is based on &lt;strong&gt;symmetry&lt;/strong&gt; considerations and &lt;strong&gt;not&lt;/strong&gt; the number of hydrogens attached to the adjacent carbon atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules and atoms within a molecule can be of three types:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;identical&lt;/strong&gt; - they can be interconverted by simple rotations or translations.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;enantiotopic&lt;/strong&gt; - they can be interconverted by reflection in a mirror or about a center of symmetry&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;diastereotopic&lt;/strong&gt; - they cannot be interconverted.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Molecules or atoms within a molecule that are found to be identical are said to be in the same environment.  Molecules or atoms that are identical, or in the same environment, &lt;strong&gt;must&lt;/strong&gt; be identical with respect to &lt;strong&gt;all&lt;/strong&gt; chemical and physical properties.  These properties would include chemical reactivity, spectroscopic environment, &lt;em&gt;etc.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In benzophenone, I can rotate the benzene ring about the $\ce{C5-C7}$ bond. This rotation interconverts $\ce{C4}$ and $\ce{C6}$.  If I told you to look away and I performed this rotation (and we removed the numeric subscripts), there is no way that you could tell if the original carbon or the other carbon was now in the $\ce{C4}$ location.  These two carbons (and their attached hydrogens) are therefore said to be equivalent by the above definition.  The same applies to $\ce{C1}$ and $\ce{C3}$, they are also &lt;strong&gt;equivalent to each other, or in the same environment.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also rotate the molecule around the $\ce{C7=O}$ carbonyl axis.  If I do this I can interconvert $\ce{C4, C6, C8, C_{10}}$.  This same rotation also (separately) interconverts $\ce{C1, C3, C_{11}, C_{13}}$, and also (separately) $\ce{C2}$ and $\ce{C_{12}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, these 3 sets of carbons&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{C4, C6, C8, C_{10}}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C1, C3, C_{11}, C_{13}}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C2, C_{12}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;are different (diastereotopic) from one another (I can't interconvert $\ce{C4}$ and $\ce{C_{12}}$, for example).  So we have 3 different chemical environments, but &lt;strong&gt;within each environment&lt;/strong&gt; there are several identical atoms. The same methodology and reasoning applies to the hydrogens attached to these carbons.  They also exist in 3 different environments, and within each of these 3 environments there are several equivalent hydrogens.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-06T23:04:47.480" LastActivityDate="2015-06-06T23:04:47.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32532" PostTypeId="2" ParentId="32521" CreationDate="2015-06-06T21:45:24.213" Score="1" Body="&lt;p&gt;Very simply,  if you exert pressure on a liquid reaction, you can use temperatures in excess of the RTP/STP boiling points. So, your reaction is now no longer bounded by the particular boiling point of a solvent and new regions can be investigated.  e.g.  over 100 deg C is easily achievable for THF or DCM at 20 bar.&lt;br&gt;&#xA;In addition, you can restrict off-gassing during a reaction by applying pressure.  This may or may not be desirable. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-06T21:45:24.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32533" PostTypeId="1" CreationDate="2015-06-06T22:04:41.137" Score="19" ViewCount="5322" Body="&lt;p&gt;Ionic bonds seem to be intermolecular but are classified as chemical bonds.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions.&amp;quot; -  &lt;em&gt;Wikipedia, definition of an ionic bond&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;The physical force of attraction which holds atoms and molecules in a matter is called a physical bond. van der Waals' forces and coulombic forces are physical forces.&amp;quot; - &lt;em&gt;Chemistry Stack Exchange, definition of an intermolecular force&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei or as the result of a dipole attraction.&amp;quot; - &lt;em&gt;Wikipedia, definition of a chemical bond&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Before an ionic bond is formed, oxidation and reduction must occur. For example, let's use sodium chloride. A chlorine atom will steal the outer electron from a sodium atom. The chlorine is gaining an electron, i.e. being reduced. The sodium is losing an electron, i.e. being oxidized. I agree that this is a purely chemical process.&lt;/p&gt;&#xA;&lt;p&gt;Now that the atoms are oppositely-charged ions, they can attract and form an ionic bond:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Na+ + Cl- -&amp;gt; NaCl}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;My question is this: How is this bond defined as a chemical bond?&lt;/p&gt;&#xA;&lt;p&gt;Let's take water; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;. The oxygen forms two polar-covalent bonds with two hydrogen atoms. Each hydrogen shares two electrons with the oxygen. The electrons are being shared, so this bond an intermolecular force. The atoms stay together because they both depend on the electrons they share.&lt;/p&gt;&#xA;&lt;p&gt;Now let's take more than one water molecule. These molecules will form hydrogen bonds. The hydrogens on the water molecules do not often have an electron around them because the oxygen has a greater attraction with them. This leaves the lonely proton of the hydrogen to make that end of the water positive while the oxygen is negative with its electrons. The negative end of the water (oxygen) will become attracted to the positive end of another water molecule (hydrogen), forming our anticipated intermolecular force. Does this sound familiar?&lt;/p&gt;&#xA;&lt;p&gt;I hope it does. According to my knowledge, an ionic bond is that, but instead of a dipole-dipole interaction, it's a monopole-monopole interaction. If you look at the bond itself and not the reaction forming it, it acts just like an intermolecular force.&lt;/p&gt;&#xA;" OwnerUserId="16776" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-08-07T00:07:08.573" Title="Why is an ionic bond a chemical and not a physical bond?" Tags="&lt;bond&gt;&lt;ions&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32534" PostTypeId="1" AcceptedAnswerId="32563" CreationDate="2015-06-06T22:04:58.790" Score="11" ViewCount="16197" Body="&lt;p&gt;Why is $\ce{HNO3}$ a stronger oxidising agent than $\ce{H3PO4}$? $\ce{N}$ and $\ce{P}$ have the same oxidation number. Is it the electronegativity difference between $\ce{N}$ and $\ce{P}$? &lt;/p&gt;&#xA;" OwnerUserId="16775" LastEditorDisplayName="user15489" LastEditDate="2015-06-06T23:45:34.690" LastActivityDate="2016-11-05T15:49:03.133" Title="Why is HNO3 a stronger oxidising agent than H3PO4?" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32535" PostTypeId="2" ParentId="32025" CreationDate="2015-06-06T22:13:52.910" Score="2" Body="&lt;p&gt;I would be surprised if you found anything predicting solubility in mixtures as there is poor modelling due to non-linear effects. Taking solubilities of two solvents and trying to interpolate does not work. As little as a few hundred ppm to 4% of one solvent in another is enough to confound standard predictive equations without empirical data. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-06T22:13:52.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32536" PostTypeId="2" ParentId="32529" CreationDate="2015-06-06T22:27:05.343" Score="15" Body="&lt;p&gt;Milk contains surfactants (primarily lipids), which allow stable bubbles to form. In intense boiling, these bubbles can become mixed with liquid, supporting the liquid above them and forming what I'll call for want of a better term a &quot;wet froth.&quot; Imagine a bunch of sponges made out of liquid. They take up more space than they would if you compressed them down and forced out the air. The inclusion of the water vapor in the bubbles lowers the density of the mixture, causing it to expand and overflow the pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state is only stable because the added energy keeps creating new bubbles to replace any that have popped and keeps forcing liquid upward, while it would normally fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water does not experience this mostly because its bubbles are far less stable. In fact, pure water does not form bubbles at all. The bubbles formed in water is typically due to impurities.  If you were to add soap to a boiling pot of water, you would see the exact same effect, if not worse.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-06-06T22:27:05.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32537" PostTypeId="2" ParentId="32500" CreationDate="2015-06-06T22:54:59.210" Score="2" Body="&lt;p&gt;Supercooling (and superheating) occur mainly due to the difference between surface energy and volume energy. An intersection of two phases/compounds has higher energy than a bulk material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the interface of ice and water has higher energy because frozen liquid molecules are favored by entropy to mix with the liquid. The only factor keeping the phase boundary in existence is the lack of (or excess of) energy to break the equilibrium between the phases. For any molecule to enter the liquid phase, one must enter the solid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling and freezing come into this when you consider that the temperatures tabulated for boiling and freezing actually measure these phenomena for when a phase boundary already exists. That phase boundary lowers the energy required for phase change, thus lowering the boiling point (or raising the freezing point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding/removing energy when a phase boundary already exists simply shifts the phase equilibrium I already mentioned to the appropriate side. If no boundary exists, however, it becomes more difficult for molecules to escape. When a vessel is heated from the bottom, the hottest portion of the liquid is trapped by other liquid. Any single molecule, even if it attains the required energy, is quickly collided with other molecules and pushed back to the average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substances will spontaneously change phase if enough energy exists to favor that change, even without a phase boundary. Going even further, enough energy in the right form can cause &lt;a href=&quot;https://www.youtube.com/watch?v=yBRdBrnIlTQ&quot; rel=&quot;nofollow&quot;&gt;a breakdown in the phase boundary&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not feel I can adequately answer the later part of your question, as I do not have as much background with solid chemistry. However, I will note that solids typically can exist in &lt;a href=&quot;http://en.wikipedia.org/wiki/Microstructure&quot; rel=&quot;nofollow&quot;&gt;different configurations of the molecules&lt;/a&gt;. A common example is the difference between steel that has been quickly cooled and steel that has been slow-cooled: cooling more quickly, in this case, results in a harder, more brittle material.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-06-06T22:54:59.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32539" PostTypeId="1" CreationDate="2015-06-07T01:22:20.603" Score="-3" ViewCount="1574" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the equivalent weight of an acid if it is given that $\pu{1.321 g}$ acid reacts with magnesium to give $\pu{1.72 g}$ salt. Options:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$49$ &lt;/li&gt;&#xA;  &lt;li&gt;$36.5$ &lt;/li&gt;&#xA;  &lt;li&gt;$45$&lt;/li&gt;&#xA;  &lt;li&gt;none of above. &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second choice is given as the answer.            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve the question. I used the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{wt. of acid}}{\text{equivalent weight of hydrogen}  \text{ eq. wt. of anion say 'x'}} = \frac{\text{wt. of salt}}{\text{eq. wt. of magnesium   eq. wt. of anion}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after solving I got $x=35.41$ taking equivalent weight of hydrogen as $1$, and $x=35.38$ taking equivalent weight of H as $1.008$. The answer is different in each case. Is this correct? But the correct answer given is $36.5$&lt;/p&gt;&#xA;" OwnerUserId="16778" LastEditorUserId="5026" LastEditDate="2018-02-23T16:01:59.153" LastActivityDate="2018-02-23T16:01:59.153" Title="Finding equivalent weight of acid given its reaction to form a salt" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32541" PostTypeId="2" ParentId="32034" CreationDate="2015-06-07T01:59:12.480" Score="18" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1) Why is only triphenyl phosphine used here. Why not some stronger&#xA;  nucleophile like say trimethyl phosphine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems.  Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine.  It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem.  You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups.  This would give rise to a mixture of ylides and consequently a mixture of products.  On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following additional reasons also favor usage of triphenyl phosphine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is cheap&lt;/li&gt;&#xA;&lt;li&gt;it is &lt;strong&gt;very&lt;/strong&gt; cheap&lt;/li&gt;&#xA;&lt;li&gt;it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids&lt;/li&gt;&#xA;&lt;li&gt;triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Why would the lone pair on P attack the C of the alkyl halide in the&#xA;  first place? Isn't the lone pair in resonance with the benzene ring?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely.  Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kQcZu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Why is butyllithium used particularly. Can't any other reagent be&#xA;  used to abstract H?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them.  Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost.  When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4) Why is the four membered ring formed in step 2?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The 4-membered ring may not be as strained as you imagine.  Remember, phosphorous is in the third row of the Periodic Table.  It has longer bond lengths and can tolerate smaller bond angles.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5) Why does the ring collapse in step 3?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The bond between phosphorous and oxygen is an &lt;strong&gt;extremely&lt;/strong&gt; strong bond, in other words its formation is very exothermic.  Therefore, this last step is being driven by thermodynamics.  It is the exothermicity of this step that drives the reaction to completion.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-08T00:43:29.983" LastActivityDate="2015-06-08T00:43:29.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32542" PostTypeId="1" CreationDate="2015-06-07T02:51:38.790" Score="5" ViewCount="831" Body="&lt;p&gt;\begin{equation}\ce{CaCO3(s) -&amp;gt; CaO(s) + CO2(g)}\quad\Delta H^\circ = +556\ \mathrm{kJ/mol} \tag{endothermic}\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will the equilibrium partial pressure of  $\ce{CO2}$ be affected if the volume is decreased?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I thought of course the moles will increase when you decrease the volume hence $pV = nRT$, thus $p = \frac{nRT}{V}$. Therefore, $p$ is inversely proportional to $V$. But, the right answer is NO CHANGE, can anyone help me to understand why?&lt;/p&gt;&#xA;" OwnerUserId="16551" LastEditorUserId="39539" LastEditDate="2017-04-08T07:52:06.400" LastActivityDate="2017-04-08T07:52:06.400" Title="How will the equilibrium partial pressure of carbon dioxide be affected?" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32543" PostTypeId="2" ParentId="32533" CreationDate="2015-06-07T03:11:31.857" Score="16" Body="&lt;p&gt;&lt;strong&gt;Short version:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't call bonds &quot;physical&quot;, there are chemical bonds and other types of interactions between particles. The chemical bonds are classified this way because they make up molecules, salts, polymers and such, which are the materials chemists are interested in studying, and not particularly because of their sub-atomic/electrostatic characteristcs. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your question relates to two frequent sources of confusion: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both questions have been pretty much answered here on Chemistry SE and also on Physics SE. I'll give some links to related questions and also a shrot summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question didn't go much into covalent bonds, but anyone trying to understand anything past the surfaces of chemical bonds should know that the definition of a bond as covalent &lt;strong&gt;or&lt;/strong&gt; ionic is simply an easier way for us to classify things, because when the bonds are categorized in groups we can say that one group behaves in a way, and the other behaves in a different way. The truth is, all bonds have some ionic and some covalent character to them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good answers here on Chemistry SE about that can be found here: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist&quot;&gt;Can 100% covalent bonds exist?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19568/what-happens-if-the-electronegativity-difference-is-exactly-2-1/19573&quot;&gt;What happens if the electronegativity difference is exactly 2.1?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Physics pont of view, see: &lt;a href=&quot;https://physics.stackexchange.com/questions/27837/what-distinguishes-between-physics-and-chemistry&quot;&gt;What distinguishes between physics and chemistry?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a Chemistry point of view, see: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15607/differences-between-chemical-physics-and-physical-chemistry&quot;&gt;Differences between chemical physics and physical chemistry?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it's clear that both sciences have a lot of overlapping, and perhaps the main difference is the approach taken and how the results will be used. While chemists usually focuses on understanding the structure and interactions between substances and materials, in order to be able to improve them or develop new ones, physicists are mostly interested in understanding the exact laws of nature and representing them with mathematical equations that can be used to explain and predict phenomena. This definitions do not seem so different, and that's not supposed to be wrong: We do have the Physical Chemistry and Chemical Physics fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, but &lt;strong&gt;Why is an ionic bond a chemical and not a physical bond?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two important things were mentioned by @Mithoron in the comments. The term &quot;Physical bond&quot; is not frequently used by chemists, (or anyone I suppose), pretty much because &lt;em&gt;all interactions in whole universe are physical&lt;/em&gt;. This single quote by @Mithoron is related to the two points I highlighted above. Both &quot;types&quot; bonds &quot;follow&quot; the same laws, and there is no hard line between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that the definition you gave for Physical Bond was taken from an answer without a single upvote, which is an indicator that it isn't what the Chemistry SE community thinks (despite the fact that Google liked it very much and uses it in it's &lt;a href=&quot;https://www.google.com.br/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=physical%20bond&quot; rel=&quot;noreferrer&quot;&gt;short-answer for the query&lt;/a&gt;). Another indicator that &quot;Physical Bond&quot; isn't really a type of bond is that there aren't definitions of it out there (Yahoo answers doesn't count), not even a Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do we call them &lt;strong&gt;chemical bonds&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's pretty much because the bonds we call &lt;em&gt;chemical&lt;/em&gt; are those which make up molecules, salts and other materials which are mainly studied by chemistry. The definition of chemical bond varies a lot (as explained &lt;a href=&quot;https://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds&quot;&gt;here&lt;/a&gt;), but they are the bonds which create what chemists study. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other types of bonds which aren't called &lt;em&gt;chemical&lt;/em&gt; are the intermolecular interactions, which are obviously of interest to chemistry, but also the &lt;a href=&quot;http://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;noreferrer&quot;&gt;Strong Force&lt;/a&gt;, which &lt;em&gt;is the force that binds protons and neutrons (nucleons) together to form the nucleus of an atom.&lt;/em&gt; (I hope it's okay to call Strong Force a &lt;em&gt;bond&lt;/em&gt;, I'm not sure to be honest).&lt;/p&gt;&#xA;" OwnerUserId="15133" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-07T03:11:31.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32544" PostTypeId="2" ParentId="32542" CreationDate="2015-06-07T04:26:07.860" Score="4" Body="&lt;p&gt;Okay let's recall what the Le Châtelier's principle is all about: (in short) whenever a system in equilibrium is disturbed the system will adjust itself in such a way that the effect of the change will be nullified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at your problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaCO3(s)-&amp;gt;CaO(s) +CO2(g)}$  $\Delta H* = +556 \ce{KJ/mol}$ (endothermic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ideas are correct, if we reduce the volume we are in fact increasing the pressure. (and in a way, this displeases our system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our system, naturally, reacts by shifting to a side which exerts a lower pressure or has fewer moles of gas (the left in this case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our system nullifies the effect of the stress applied to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now because this shift occurs, we observe no change in the equilibrium partial pressure of carbon dioxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense? Maybe, someone smarter than me can offer a clearer explanation.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastActivityDate="2015-06-07T04:26:07.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32545" PostTypeId="1" AcceptedAnswerId="32577" CreationDate="2015-06-07T05:05:01.943" Score="6" ViewCount="1296" Body="&lt;p&gt;If a bangle is made out of gold and silver, connected with each-other would there be corrosion happening?&#xA;If so, can it be explained using the galvanic series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also do those metals undergo oxidation under normal conditions?&lt;/p&gt;&#xA;" OwnerUserId="16779" LastEditorDisplayName="user15489" LastEditDate="2015-06-07T05:46:09.803" LastActivityDate="2015-06-08T00:44:06.470" Title="Corrosion of a galvanic couple made with silver and gold" Tags="&lt;electrochemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32546" PostTypeId="1" AcceptedAnswerId="32547" CreationDate="2015-06-07T06:42:41.073" Score="5" ViewCount="213" Body="&lt;p&gt;Why are triflates so widely used in coupling reactions, but not tosylates, mesylates nor nosylates?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4325" LastEditDate="2015-06-07T16:49:52.077" LastActivityDate="2015-06-07T16:49:52.077" Title="Triflate use in metathesis reactions" Tags="&lt;organic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32547" PostTypeId="2" ParentId="32546" CreationDate="2015-06-07T07:38:28.107" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;As a result of the -I effect of the trifluoromethyl group, triflate is a better nucleofuge than mesylate and tosylate.   &lt;/li&gt;&#xA;&lt;li&gt;As compared to tosylate and nosylate, triflate is rather small. This might be an advantage in reactions at &quot;crowded&quot; centres.  &lt;/li&gt;&#xA;&lt;li&gt;Triflates do not undergo side reactions, such as hydrogen abstraction reactions, etc. A triflate, once introduced, can easily survive several steps in a synthesis.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-06-07T07:38:28.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32548" PostTypeId="2" ParentId="32542" CreationDate="2015-06-07T08:13:11.267" Score="6" Body="&lt;p&gt;The equilibrium constant for the reaction is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ &lt;strong&gt;does not change&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ideal gas &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{\ce{CO2}}=\frac{n_{\ce{CO2}}RT}{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-07T08:13:11.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32549" PostTypeId="1" AcceptedAnswerId="32579" CreationDate="2015-06-07T08:53:21.093" Score="16" ViewCount="8266" Body="&lt;p&gt;While studying about bonding there was one statement that &quot;pi bonds can only be formed only with sigma bonds&quot; as we know that in double bond there is 1 sigma bond and 1 pi bond but then one question arises: Why is a pi bond formed only when a sigma bond is formed? Is it possible to form a pi bond without any formation of a sigma bond?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastEditorUserId="5017" LastEditDate="2015-06-09T14:40:34.290" LastActivityDate="2020-06-06T15:52:06.503" Title="Why are pi bonds only formed when sigma bonds are formed?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32550" PostTypeId="2" ParentId="32517" CreationDate="2015-06-07T08:59:16.990" Score="5" Body="&lt;p&gt;Yes, you can destroy small amounts of methyl iodide by hydrolysis with an aqueous solution of sodium hydroxide. You can just add some sodium hydroxide solution to your reaction mixture, if your molecule is stable to basic conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is: $\ce{NaOH + MeI -&amp;gt; MeOH + NaI}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: &quot;A Comprehensive Guide to the Hazardous Properties of Chemical Substances&quot; by Pradyot Patnaik, 2007&lt;/p&gt;&#xA;" OwnerUserId="11593" LastEditorDisplayName="user15489" LastEditDate="2015-06-07T09:10:51.987" LastActivityDate="2015-06-07T09:10:51.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32551" PostTypeId="1" AcceptedAnswerId="32553" CreationDate="2015-06-07T09:29:24.733" Score="5" ViewCount="821" Body="&lt;p&gt;While doing titrations, we are instructed to use pipette and they demonstrated how to use a pipette. But still after all this, we face difficulties of how to take precise reading from pipette. &#xA;Example-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Due to the concave water meniscus, until where to take the reading ?&lt;/li&gt;&#xA;&lt;li&gt;Is it really important to take every single drop of solution from pipette ?&#xA;as while putting solution in beaker we see little bit drop left inside pipette.&lt;/li&gt;&#xA;&lt;li&gt;And what's the best way to hold pipette to avoid breaking it and easy to see the reading which we are taking ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16459" LastEditorUserId="10185" LastEditDate="2015-06-07T14:20:13.903" LastActivityDate="2015-06-07T14:20:13.903" Title="How to get a precise reading from pipette?" Tags="&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32552" PostTypeId="2" ParentId="32549" CreationDate="2015-06-07T09:30:56.813" Score="6" Body="&lt;p&gt;$\sigma$-bonding MOs tend to have lower energy than $\pi$-bonding MOs so they will be formed first.  One explanation is that the $\sigma$-bonding MOs have a lot of $s$-AO character and $s$-AOs have lower energy than $p$-AOs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I don't know of any double bond that is purely $\pi$-bonding&lt;/strong&gt;. Basic MO theory suggest that $\ce{C2}$ should have a double bond made of two $\pi$-bonding MOs because the $\sigma$-anti-bonding MO is lower in the energy than the $\pi$-bonding MOs.  However, experiment tells us that $\ce{C2}$ has two unpaired electrons so this simple picture cannot be right.  This is explained in greater detail &lt;a href=&quot;http://www.madsci.org/posts/archives/2000-10/971992131.Ch.r.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.nature.com/nchem/journal/v4/n3/full/nchem.1263.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-07T09:30:56.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32553" PostTypeId="2" ParentId="32551" CreationDate="2015-06-07T10:13:36.080" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The bottom of the meniscus (in the middle) is your measurement point and should lie on the line that you are trying to read.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is completely dependent on the type of pipette that you are using. Some pipettes are designed so that you just leave the last drop and some you are supposed to just touch the tip of the pipette to the surface of what you've dispensed already to get the last bit. TD pipettes stand for &quot;to deliver&quot; (leave the last bit). TC pipettes stand for &quot;to contain&quot; (take the last bit out).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, this is dependent on the pipette. If you have a classic bulb pipette then you shouldn't hold it by the bulb as it is the weakest part and that's how you end up with glass in your hand.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15191" LastActivityDate="2015-06-07T10:13:36.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32554" PostTypeId="1" AcceptedAnswerId="32555" CreationDate="2015-06-07T10:17:46.273" Score="6" ViewCount="3329" Body="&lt;p&gt;I was wondering if there was any way of representing a dative/coordinate covalent bond on a chemical diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single bond can easily be represented as a single line, a double bond as a double line and a triple bond as a triple line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the below structure of ethene, a compound with single and double non-coordinate covalent bonds: each bond is easily represented by a single and double line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Ethene-2D-flat.png/643px-Ethene-2D-flat.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the ammonium ion does have a coordinate bond between the lone pair of electrons on the nitrogen and the H+ ion from whichever compound donated it (e.g. hydrochloric acid). Despite this, I am unable to see how a coordinate bond could be represented, as, to me, it just looks like 4 non-dative covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way of representing a coordinate/dative covalent bond on diagrams like these?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Ammonium-2D.svg/1275px-Ammonium-2D.svg.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16656" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.597" LastActivityDate="2019-02-28T22:14:53.423" Title="Is there any way of representing a dative/coordinate covalent bond?" Tags="&lt;covalent-compounds&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32555" PostTypeId="2" ParentId="32554" CreationDate="2015-06-07T10:47:10.800" Score="13" Body="&lt;p&gt;Generally, you wouldn't describe the bonds in an ammonium ion as dative bonds. They are usually considered to be proper covalent bonds. The term &amp;quot;dative bond&amp;quot; is typically used for bonds in (transition) metal complexes between ligands and a metal center. But for those complexes Lewis structures have lots of weaknesses and are rarely a good representation of the actual bonding. Yet they are used for the sake of convenience. Often an arrow from the ligand to the metal center is used to indicate a dative bond.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/20DjL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/20DjL.gif&quot; height=&quot;150&quot;&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www2.chemie.uni-erlangen.de/projects/vsc/chemie-mediziner-neu/komplexe/bilder/aufbau1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;uni-erlangen.de&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;But just as often you see normal &amp;quot;covalent bonds&amp;quot; (straight lines) like in traditional Lewis structures being used.&#xA;Often both kinds of representations are even intermixed.&lt;/p&gt;&#xA;&lt;p&gt;For a more detailed view on the subject of bonding in coordination complexes and its representation have a look at &lt;a href=&quot;https://chemistry.stackexchange.com/a/10216/189&quot;&gt;this answer of mine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-02-28T22:14:53.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32556" PostTypeId="2" ParentId="32539" CreationDate="2015-06-07T11:16:21.610" Score="1" Body="&lt;p&gt;The trick is to recognize that the acid and the salt share an anion $\ce{A}$, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HA + Mg -&amp;gt; MgA2 + H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the equivalent weights of the acid and the salt are functions of the only one unknown, the equivalent weight of $\ce{A}$. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-07T11:16:21.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32557" PostTypeId="1" AcceptedAnswerId="33080" CreationDate="2015-06-07T13:16:03.567" Score="12" ViewCount="1508" Body="&lt;p&gt;Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{ M(OH)3 &amp;amp;~&amp;lt;=&amp;gt; M(OH)^{+}_{2} + OH^{-} } &amp;amp; &#xA;\mathrm{p}K_{\mathrm{b}1} &amp;amp;= 0.5 \\&#xA;\ce{ M(OH)^{+}_{2} &amp;amp;~&amp;lt;=&amp;gt; M(OH)^{2+} + OH^{-} } &amp;amp;&#xA; \mathrm{p}K_{\mathrm{b}2} &amp;amp;= 0.7 \\&#xA;\ce{ M(OH)^{2+} &amp;amp;~&amp;lt;=&amp;gt; M^{3+} + OH^{-} } &amp;amp;&#xA; \mathrm{p}K_{\mathrm{b}3} &amp;amp;= 9.0&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the species in above equilibria are water soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Progress so far:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{HCl}$ in the medium will first react with the base and give rise to an end point.&lt;/li&gt;&#xA;&lt;li&gt;As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{ M^{3+} + {H2O} &amp;amp;~&amp;lt;=&amp;gt; M(OH)^{2+} + H+ } &amp;amp;&#xA;\mathrm{p}K_{\mathrm{a}1} &amp;amp;= 14.0 - 9.0 = 5.0 \tag{a} \\&#xA;\ce{ M(OH)^{2+} + {H2O} &amp;amp;~&amp;lt;=&amp;gt; M(OH)^{+}_{2} + H+ } &amp;amp;&#xA;\mathrm{p}K_{\mathrm{a}2} &amp;amp;= 14.0 - 0.7 = 13.3 \tag{b}\\&#xA;\ce{ M(OH)^{+}_{2} + {H2O} &amp;amp;~&amp;lt;=&amp;gt; M(OH)_3 + H+ } &amp;amp; &#xA;\mathrm{p}K_{\mathrm{a}3} &amp;amp;= 14.0 - 0.5 = 13.5 \tag{c}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.&lt;/li&gt;&#xA;&lt;li&gt;Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now the problem is how to put all these information together and deduce a method to analyze the mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="24052" LastEditDate="2016-02-16T23:10:32.427" LastActivityDate="2016-02-16T23:10:32.427" Title="How to analyze a acid and salt mixture using titrimetry?" Tags="&lt;acid-base&gt;&lt;analytical-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32558" PostTypeId="1" AcceptedAnswerId="32559" CreationDate="2015-06-07T13:42:32.310" Score="21" ViewCount="4413" Body="&lt;p&gt;I was reading N.C.E.R.T. class XI part 2 book and chapter Hydrogen,when I came across this statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Loss of the electron from hydrogen atom&#xA;  results in nucleus ($\ce{H+}$) of ~1.510–3 pm size.&#xA;  This is extremely small as compared to normal&#xA;  atomic and ionic sizes of 50 to 200pm. As a&#xA;  consequence, H+ does not exist freely and is&#xA;  always associated with other atoms or&#xA;  molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So a question arose in my mind that why it is a consequence of small size of hydrogen nucleus that $\ce{H+}$ does not exist freely and is always associated with other atoms and molecules?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="10185" LastEditDate="2015-06-07T14:08:33.423" LastActivityDate="2015-06-07T14:25:55.510" Title="Why are hydrogen ions always associated with another molecule?" Tags="&lt;ions&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32559" PostTypeId="2" ParentId="32558" CreationDate="2015-06-07T14:19:41.070" Score="14" Body="&lt;p&gt;The way I understand it is (and my understanding is by no means perfect, or complete), as you pointed out correctly: a hydrogen ion is in fact a proton. The proton is a &quot;bare charge&quot; and as you rightly said, &quot;tiny&quot;. a&#xA;This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in an aqueous solution it forms $\ce{H_3O+}$ (hydronium ion, which are further solvated by water molecules and form a manner of aggregates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extremely interesting way of looking at it was pointed out in the comments (just putting it here, so no one misses it)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An interesting way to thinking about the reactivity of a bare proton is to calculate its charge density and compare with other monocations, such as Li+. This will make it clear why free protons will abstract electrons from just about anything. –  Nicolau Saker Neto&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do note (as far as I know, someone might correct me if I am wrong) free protons are fairly stable (by free I mean not bound to electrons or other nucleons) and usually don't decay into other particles. They exist in several naturally occurring situations, high energy plasmas come to mind.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="16088" LastEditDate="2015-06-07T14:25:55.510" LastActivityDate="2015-06-07T14:25:55.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32560" PostTypeId="1" AcceptedAnswerId="32566" CreationDate="2015-06-07T14:44:10.203" Score="8" ViewCount="6065" Body="&lt;p&gt;I am currently investigating about the interaction behavior of a few atoms in certain conditions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible to use the concept of single point energy to represent the atomic interaction energies or I have to go other way around?&lt;/li&gt;&#xA;&lt;li&gt;What is the basic difference between potential energy and single point energy?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16786" LastEditorUserId="194" LastEditorDisplayName="user15489" LastEditDate="2016-11-19T17:00:54.573" LastActivityDate="2016-11-19T17:00:54.573" Title="Significance of single point energy when calculating interaction energies" Tags="&lt;computational-chemistry&gt;&lt;energy&gt;&lt;intermolecular-forces&gt;&lt;electrostatic-energy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32561" PostTypeId="1" AcceptedAnswerId="32585" CreationDate="2015-06-07T15:03:32.690" Score="1" ViewCount="292" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8fKfm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I relate the half life to the overall rate of reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I argued that from the data, doubling the partial pressure of either reactant, keeping the other constant, will half the half life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try t1/2 = &#xA;$\frac{\ce{[1]}}{\ce{[k][PA_0][PB_0]}}$ and since pressure of A is always greater than B, so t1/2 = $\frac{\ce{[1]}}{\ce{[k'][PB_0]}}$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then wrote:&#xA;$$\frac{-\mathrm{d}[PB]}{\mathrm{d}t} = k [P_B]^m $$, where &lt;strong&gt;k&lt;/strong&gt; = k[PA]   &lt;/p&gt;&#xA;" OwnerUserId="16445" LastEditorUserId="865" LastEditorDisplayName="user15489" LastEditDate="2021-05-22T21:17:57.007" LastActivityDate="2021-05-22T21:17:57.007" Title="Reaction kinetics, relating half life to reaction rate" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32562" PostTypeId="2" ParentId="32519" CreationDate="2015-06-07T15:04:09.800" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, surely $P_j$ alone is dimensionless as it's a partial pressure defined by dividing a pressure by a pressure - leaving no units. How does that work? Or am I wrong by assuming that a partial pressure has no units?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are wrong to assume that partial pressure has no units.  Partial pressure has units of pressure, such as mmHg, atm, bar, psi, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal gas mixtures, $y_j=\frac{P_j}{P_{tot}}$ or equivalently $y_j P_{tot}=P_j$, i.e. the partial pressure $P_j$ of a gas is equal to its mole fraction $y_j$ times the total pressure $P_{tot}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression you gave for $K_p$ is dimensionless because $P^\Theta$ has the same dimension as $P_j$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do I assume that $\frac{P_C}{P^Θ}=1$....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That part sounds good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;leaving me with $K_p=\frac{P_{CO}}{(P_{O_2})^{1/2}}\frac{1}{P^\Theta}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure about that one.  How is $K_p$ dimensionless in what you wrote?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-07T15:04:09.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32563" PostTypeId="2" ParentId="32534" CreationDate="2015-06-07T15:28:56.597" Score="6" Body="&lt;p&gt;Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as  in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H3PO4 + 2H+ + 2e-&amp;gt; H3PO3 + H2O}$; $E=-0.276 \ce{V}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4656" LastEditorDisplayName="user15489" LastEditDate="2015-06-07T17:34:33.810" LastActivityDate="2015-06-07T17:34:33.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32564" PostTypeId="1" CreationDate="2015-06-07T16:29:59.807" Score="13" ViewCount="8432" Body="&lt;p&gt;I have asked a question loosely asking this, where I confused terms and did not specify what I wanted to know &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond/32543?noredirect=1#comment47887_32543&quot;&gt;here&lt;/a&gt;, so I formed a new question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the differences and similarities between ionic bonds and covalent bonds and ionic bonds and dipole-dipole interactions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be that an ionic bond is nothing but a stronger dipole-dipole interaction. The only difference is where the electrons are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's take a sodium cation and chloride anion bond. The sodium had its electron taken away from it. It is now positive. Chloride had an electron added to it and it's feeling negative. Now the two will attract because of these charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take two water molecules. The hydrogen of one will be positive because it is less electronegative and thus the electron density will be greater around the oxygen, which is more electronegative, making it negative. Now the hydrogen of one and the oxygen of another will attract due to the charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like the only difference is that if you split the sodium chloride and the water's hydrogens and oxygen, the sodium and chloride will still have a charge, but the hydrogens and oxygen will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By textbook, a dipole-dipole interaction is an intermolecular force while a covalent bond, as well as an ionic bond, is a chemical bond. It seems to me that, yes, the forming of ions is a chemical reaction, but the ionic bond is an intermolecular force.&lt;/p&gt;&#xA;" OwnerUserId="16776" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-06-08T01:21:37.680" Title="Is an ionic bond more like a covalent bond or an intermolecular force?" Tags="&lt;ions&gt;&lt;intermolecular-forces&gt;&lt;ionic-compounds&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32565" PostTypeId="2" ParentId="32560" CreationDate="2015-06-07T18:03:28.093" Score="5" Body="&lt;p&gt;Single point energy is a point on the potential energy surface.&lt;br&gt;&#xA;Suppose you have a single atom which has both potential energy (Due to interaction between electrons and protons and since there is no other atom or external force, all the potential interactions come from interaction between it's own electrons and protons.) and kinetic energy due to electronic motion. So we can write a hamiltonian (PE+KE) here. Here if you want to optimize this geometry you just need to optimize electonic hamiltonian because there is only one nucleus. It can be called single point energy as well as optimized energy. Now for two H atom, each have KE and also PE due to previously discussed interaction plus interaction between electrons and nucleus of each other. For example equlibrium H-H distance is 0.745 Angstrom (B3LYP theory). So if you put two H in a distance of 0.75 or greater, they will attract each other and if you put them in a distance less than equilibrium distance they will repel each other. And eventually they will come to equilibrium distance. So your potential energy curve will look like morse potential curve. This is optimization and the lowest point you find is optimized distance. Now, you want to find the force or you just want to know the energy when they are 0.8 Angstrom apart. What you can do is you can freeze their nucleus at those position and solve the schrodinger equation. Here since there is no nuclear degree of freedom, it will only optimize electronic hamiltonian to remove any kind of electronic overlap or to make the orbital orthogonal. This is called the single point energy. If you find all the single point energy for all distance, you can plot the potential energy surface and you can also find the lowest point on the PES, which is optimized geometry or you can find the maximum energy, which would be transition energy. &lt;img src=&quot;https://i.stack.imgur.com/bAktP.gif&quot; alt=&quot;Photo courtesy WWW.chemicool.com&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-06-07T18:03:28.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32566" PostTypeId="2" ParentId="32560" CreationDate="2015-06-07T18:09:12.750" Score="11" Body="&lt;p&gt;Single point energy arises in the framework of the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation&quot; rel=&quot;noreferrer&quot;&gt;Born–Oppenheimer approximation&lt;/a&gt;&lt;/em&gt; and corresponds to just one point on the &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Potential_energy_surface&quot; rel=&quot;noreferrer&quot;&gt;potential energy surface&lt;/a&gt;&lt;/em&gt;. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-called &lt;em&gt;clamped nuclei approximation&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, if you develop the Born–Oppenheimer approximation step-by-step you can easily see that single point energy it is the sum of the electronic energy and nuclear repulsion potential energy,&#xA;$$&#xA;U = E_{\mathrm{e}} + V_{\mathrm{nn}} \, ,&#xA;$$&#xA;where the electronic energy $E_{\mathrm{e}}$ is the solution of the electronic Schrödinger equation,&#xA;$$&#xA;    \hat{H}_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) = E_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) \, .&#xA;$$&#xA;The fact that at this point we use the symbol $U$ which (alongside with $V$) is usually used for potential energy to mean the single point energy is justified a little later. Namely, when we introduce the Born-Oppenheimer approximation which give rise to the nuclear Schrödinger equation,&#xA;$$&#xA;    \Big( \hat{T}_{\mathrm{n}} + U(\vec{r}_{\mathrm{n}}) \Big) \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) = E \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) \, ,&#xA;$$&#xA;it is easy to recognize that the values of the single point energy $U$ for all possible nuclear configurations define the potential energy for nuclear motion. So, it is in this sense that the single point energy is related to the potential energy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: it became clear that OP misunderstood the notion of the single point energy $U$. Indeed, once we do few single point calculations for different nuclear configurations, the resulting $U(\vec{r}_{\mathrm{n}})$ is the potential energy for nuclear motion. However, it is &lt;em&gt;not&lt;/em&gt; the interaction energy between some fragments, though, the interaction energy contributes to it. So if one wants to obtain the interaction energy one has to decompose the potential energy $U(\vec{r}_{\mathrm{n}})$ into its parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There different ways to perform the energy decomposition, just to name a few without any particular order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.physics.udel.edu/~szalewic/SAPT/SAPT.html&quot; rel=&quot;noreferrer&quot;&gt;SAPT&lt;/a&gt; (Symmetry-Adapted Perturbation Theory) a separate program (few of them, to be more precise) which can be interfaced with different quantum chemistry codes.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://nbo6.chem.wisc.edu/tut_neda.htm&quot; rel=&quot;noreferrer&quot;&gt;NEDA&lt;/a&gt; (Natural Energy Decomposition Analysis) which is available as a part of &lt;a href=&quot;http://nbo6.chem.wisc.edu&quot; rel=&quot;noreferrer&quot;&gt;NBO&lt;/a&gt; package.&lt;/li&gt;&#xA;&lt;li&gt;Morokuma decomposition already available in some quantum chemistry codes (&lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;noreferrer&quot;&gt;GAMESS-US&lt;/a&gt;, for instance).&lt;/li&gt;&#xA;&lt;li&gt;LMO-EDA (Localised Molecular Orbital Energy Decomposition Analysis) also available in some quantum chemistry codes (&lt;a href=&quot;http://www.msg.ameslab.gov/gamess/&quot; rel=&quot;noreferrer&quot;&gt;GAMESS-US&lt;/a&gt;, for instance).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-09T08:28:43.870" LastActivityDate="2015-06-09T08:28:43.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32567" PostTypeId="1" CreationDate="2015-06-07T18:34:43.897" Score="4" ViewCount="77" Body="&lt;p&gt;From what I understand, the standard reduction/oxidation potential table is based on the hydrogen half reaction which makes a reduction and oxidation potential of 0 respectfully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a research internship I'm in now, we are looking for more efficient batteries, they are basing potential candidates in respect with magnesium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is seeing how the 'in respect' part works and how it affects the system. Maybe it's just hard for me to relate to it in my mind so maybe a more common analogy would help?&lt;/p&gt;&#xA;" OwnerUserId="16788" LastEditorDisplayName="user15489" LastEditDate="2015-06-07T18:36:21.173" LastActivityDate="2015-06-08T00:49:13.907" Title="Respect to other chemicals in reduction/oxidation potentials" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32568" PostTypeId="2" ParentId="32564" CreationDate="2015-06-07T18:40:55.700" Score="12" Body="&lt;p&gt;An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the lowest-detail end of the spectrum, you have ionic bonds - we usually model these as simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Coulomb%27s_law&quot; rel=&quot;noreferrer&quot;&gt;Coulombic&lt;/a&gt; (a.k.a. charge-charge or electrostatic) interactions. This is good enough to describe bonding in most cases where the bond is between a metal and a non-metal. We model the bond as a complete electron transfer from one atom to another, which results in an electrostatic charge. The Coulomb force between the charges holds the ion together, and so we call it a molecule. What we really mean is &quot;these atoms are now stuck together.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we move along the periodic table and the metal atoms become less metallic and the non-metals become more metallic, at some rather arbitrary point we say &quot;the ionic approximation is no longer good enough&quot; and start seeing the bond as covalent. What we really mean is that the electrons are not fully transferred between atoms, and so charge-charge interactions don't explain bonding well enough. At that point we can no longer ignore quantum mechanics, and have to acknowledge that the electrons exist in molecular orbitals, not atomic orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the main difference between ionic and covalent bonds. As for dipole-dipole interactions, it again goes back to what the electrons are doing. In a molecule with a dipole moment, what this means is that within the molecule there are polar covalent bonds, and the geometry is such that the electron density on the surface of the molecule is lopsided - one side has more electrons than the other. The dipoles can align and will be attracted to each other via the Coulomb force, but this is not the same as how we model an ionic bond - in that case, one atom gives up one or more electrons completely. In a molecule with a dipole moment, one (or more) atoms just gets &quot;more&quot; of the electron density. Then the two separate molecules, each with their own dipole, are attracted to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call ionic bonds &quot;intramolecular&quot; forces because they are what hold molecules together - without ionic bonds, you couldn't have a salt crystal, for example. When you break apart sodium chloride, you get sodium ions and chloride ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call dipole interactions &quot;intermolecular&quot; because they are what make &lt;em&gt;separate&lt;/em&gt; molecules stick together. When you pull apart two water molecules, you still have two water molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about it this way, an easy way to distinguish between intermolecular and intramolecular forces is to ask yourself whether you still have the same substance after pulling things apart. If you do, it must have been an intermolecular force. If you don't, it must have been a bond or an intramolecular force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note - For very large molecules like proteins it gets a little tricky - you can have forces between sites on the same molecule that act like what we normally describe as intermolecular forces.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="5633" LastEditDate="2015-06-08T01:21:37.680" LastActivityDate="2015-06-08T01:21:37.680" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32570" PostTypeId="2" ParentId="32415" CreationDate="2015-06-07T19:08:01.517" Score="3" Body="&lt;p&gt;This Reaction is a Disproportionation Reaction, that is, a reaction involving both oxidation and reduction of a single species, in this case : Iodine. This is not one reaction but a combination of two reactions:&#xA;$$\ce{KIO3 -&amp;gt; I2}$$&#xA;And&#xA;$$\ce{KI -&amp;gt; I2}$$&#xA;Therefore, by applying law of chemical equivalence in both reactions and adding the equations :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_{eq}$ of $\ce{I_2}$ (from both reactions) = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$) &lt;/p&gt;&#xA;" OwnerUserId="16789" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T20:03:55.033" LastActivityDate="2015-06-10T20:03:55.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32571" PostTypeId="1" CreationDate="2015-06-07T19:19:42.660" Score="1" ViewCount="2618" Body="&lt;p&gt;According to the law of equivalence, the number of equivalents of each reactant and product in a reaction are equal. Why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="16789" LastEditorUserId="402" LastEditDate="2017-01-12T02:38:16.647" LastActivityDate="2017-01-12T02:38:16.647" Title="Explanation of the law of equivalence" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32572" PostTypeId="1" AcceptedAnswerId="32575" CreationDate="2015-06-07T20:18:15.890" Score="3" ViewCount="842" Body="&lt;p&gt;For example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-bromopropane + $\ce{NaSH -&amp;gt;}$ Propanethiol + $\ce{NaBr}$&lt;br&gt;&#xA;&lt;em&gt;vs.&lt;/em&gt;&lt;br&gt;&#xA;1-bromopropane + $\ce{H2S -&amp;gt;}$ Propanethiol + $\ce{HBr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is more valid, and why? Or are they equally valid?&lt;/p&gt;&#xA;" OwnerUserId="16791" LastEditorUserId="10185" LastEditDate="2015-06-17T14:13:33.413" LastActivityDate="2015-06-17T14:13:33.413" Title="In a nucleophilic substitution reaction, does the other atom to which the nucleophile is bonded matter?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32573" PostTypeId="1" CreationDate="2015-06-07T21:27:50.237" Score="5" ViewCount="3579" Body="&lt;p&gt;Peptide bonds are synthesized when the carboxyl group of one amino acid molecule reacts with the amino group of the other amino acid molecule, causing the release of a molecule of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water molecules consist of a &lt;strong&gt;hydroxyl from a carboxyl group&lt;/strong&gt; and a &lt;strong&gt;hydrogen from an amino group&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that seems a bit strange to me, since carboxyl is an acid and should give a hydrogen (instead of hydroxyl) and amino groups are bases and should get proton (and not give a hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for me it looks like in creating this bond, the carboxyl group behaves like a base and amino group behaves like an acid. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="16792" LastEditorUserId="865" LastEditDate="2015-08-05T16:04:40.213" LastActivityDate="2015-08-05T16:04:40.213" Title="Do carboxyl groups behave like bases and amino groups behave like acids when peptide bonds are created?" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32574" PostTypeId="1" AcceptedAnswerId="32576" CreationDate="2015-06-07T22:52:26.000" Score="4" ViewCount="507" Body="&lt;p&gt;I just had a question about why does water, methanol or any other polar substances tend to bend towards the charged object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is because of a slight dipole dipole movement between the charged object and the polar substance, but not sure if that's the reason.&lt;/p&gt;&#xA;" OwnerUserId="12949" LastEditorDisplayName="user15489" LastEditDate="2015-06-07T22:58:24.130" LastActivityDate="2015-06-08T00:25:25.297" Title="Bending of water and methanol towards a charged objects" Tags="&lt;bond&gt;&lt;intermolecular-forces&gt;&lt;electronegativity&gt;&lt;dipole&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32575" PostTypeId="2" ParentId="32572" CreationDate="2015-06-07T22:55:30.210" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When undergoing an SN rxn, does the other atom to which the&#xA;  nucleophile is ionically bonded matter?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can matter.  If the counterion shifts the covalent-ionic nature of the bond connecting it to the nucleophile, then the nucleophilicity of the nucleophile can change (increase or decrease).  However in your example, the counterions don't make a significant difference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which is more valid, and why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;On paper, they are both valid but&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2S}$ is very toxic and corrosive, so the first method using sodium hydrosulfide would likely be preferred.  However, in industry, where raw material cost is very important, both of these methods are successfully &lt;a href=&quot;http://en.wikipedia.org/wiki/Thiol#Laboratory_methods&quot; rel=&quot;nofollow noreferrer&quot;&gt;used&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However both methods suffer the same drawback - once some of the desired thiol is prepared, it can react further with another molecule of alkyl bromide to produce a thioether; usually a mixture of thiol and &lt;strong&gt;thioether&lt;/strong&gt; is obtained.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the lab, neither method is used&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the chem lab, the preferred method to prepare simple thiols is through the use of thiourea.  Note the resonance structures for thiourea (just disregard the lower half of the diagram) that place negative charge on the sulfur, making it a good nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y0KYq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.osapublishing.org/ao/abstract.cfm?uri=ao-31-24-5051&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first step the thiourea reacts with the alkyl halide &lt;em&gt;via&lt;/em&gt; an $\mathrm{S_{N}2}$ reaction to yield an isothiouronium salt, which is subsequently hydrolyzed by sodium hydroxide to generate the thiol. This synthesis can be carried as a one pot, two step reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jmo2n.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch15/ch15-6-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-07T23:01:42.060" LastActivityDate="2015-06-07T23:01:42.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32576" PostTypeId="2" ParentId="32574" CreationDate="2015-06-08T00:25:25.297" Score="4" Body="&lt;p&gt;As you state, in an electric field, a polar molecule such as methanol or water, having an uneven charge distribution, is attracted (or repelled) by an electric field. This orients the molecule, but is not primarily responsible for the attraction of the bulk substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electric field causes a force on even a &lt;em&gt;nonpolar&lt;/em&gt; molecule such as methane or sulfur, though. You can pick up powdered sulfur with a charged piece of plastic because of &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrostatic_induction&quot; rel=&quot;nofollow&quot;&gt;electrostatic induction&lt;/a&gt;, which moves electrons in the bulk material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This induced charge is similar to the effect of moon (and sun) on tides, pulling the water underneath the moon more strongly than it pulls on the more distant core of the earth, and on the core more strongly than on the ocean on the opposite side of the planet, causing uneven distribution of the water. In a similar way, an external field causes an uneven distribution of charge.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-08T00:25:25.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32577" PostTypeId="2" ParentId="32545" CreationDate="2015-06-08T00:44:06.470" Score="2" Body="&lt;p&gt;As you state, though silver is not very reactive, silver &lt;em&gt;jewelry&lt;/em&gt; would corrode more rapidly when in contact with gold and a bit of sweat as electrolyte. Though the difference in electronegativity of &lt;em&gt;pure&lt;/em&gt; gold and &lt;em&gt;pure&lt;/em&gt; silver is &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_corrosion&quot; rel=&quot;nofollow&quot;&gt;0.15V, and is considered acceptably corrosion resistant&lt;/a&gt;, commercial jewelry contains copper and other metals in the alloys. There is clearly evident corrosion in &lt;a href=&quot;http://ganoksin.com/blog/binnion/2012/11/23/copper-and-silver-rings-may-not-be-such-a-good-idea/&quot; rel=&quot;nofollow&quot;&gt;photos of copper/silver jewelry&lt;/a&gt;. According to the &lt;a href=&quot;http://www.vam.ac.uk/content/articles/c/caring-for-your-metal-objects/&quot; rel=&quot;nofollow&quot;&gt;Victoria and Albert Museum&lt;/a&gt; staff, &quot;All metals, with the exception of 24 carat gold, suffer the effects of corrosion.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a bangle, physical abrasion is likely to cause more damage and loss of material than galvanic corrosion. Wear the jewelry and enjoy it, but it will not last forever.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-08T00:44:06.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32578" PostTypeId="2" ParentId="32567" CreationDate="2015-06-08T00:49:13.907" Score="1" Body="&lt;p&gt;You cannot define ‘vacuum’ oxidation potentials, or otherwise derive them &lt;em&gt;de novo&lt;/em&gt; without having &lt;em&gt;some&lt;/em&gt; zero-point. There is, technically, no self-suggesting zero-point around — unlike for example temperature, where $0\,\mathrm{K}$ is equivalent to ‘molecular movement frozen’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the reason is, that the electron(s) lost or gained during the redox reaction must go or come from &lt;em&gt;somewhere&lt;/em&gt;. Therefore, one has chosen the hydrogen cell as a centre of the scale. Why?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Hydrogen is the easiest element with the least electrons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It’s simple to build.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It actually also comes with a good further distinction: The one between noble and non-noble metals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can actually easily recalculate the standard potentials with respect to another half-cell like magnesium: Just add/subtract that cell’s potential with respect to hydrogen to all other potentials. It’s a bit like moving the temperature scale’s zero point from $0\,\mathrm{K}$ to $273\,\mathrm{K} = 0\,\mathrm{°C}$.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-08T00:49:13.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32579" PostTypeId="2" ParentId="32549" CreationDate="2015-06-08T01:02:52.573" Score="8" Body="&lt;p&gt;A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is &lt;em&gt;ortohogonal&lt;/em&gt; to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transition metals can use d-orbitals for $\pi$ bonding. They can actually point &lt;em&gt;towards&lt;/em&gt; the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing &lt;em&gt;directly&lt;/em&gt; in the bonding direction which again will bond earlier and would give a $\sigma$ bond.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-09T09:27:57.337" LastActivityDate="2015-06-09T09:27:57.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32580" PostTypeId="1" CreationDate="2015-06-08T01:27:08.667" Score="2" ViewCount="2370" Body="&lt;p&gt;we must write empirical formula of $\ce{CH3COOH}$ as $\ce{CH2O}$ or $\ce{COH2}$ ? Alike ? Any nomenclature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somebody told me it was because of bonding, different way of writing the chemical formula mean different way or bonding, but I'm not sure if it is correct or not, &lt;/p&gt;&#xA;" OwnerUserId="16796" LastEditorDisplayName="user15489" LastEditDate="2015-06-08T01:29:22.390" LastActivityDate="2015-06-08T04:16:55.900" Title="Vinegar CH2O ? COH2" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32581" PostTypeId="1" AcceptedAnswerId="32583" CreationDate="2015-06-08T01:29:16.647" Score="16" ViewCount="420" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/32529/why-does-milk-overflow-when-boiled&quot;&gt;This question about boiling milk&lt;/a&gt; reminded me of my everyday lab work at the rotavap: Some solvents and solutes are feared for the intense bubbling they create while the solution is boiling — be it at ambient pressure like the milk or at reduced pressure in the rotavap. For most of the infamous solvents, the bubbles will even create a somewhat stable foam (that usually ends up on the wrong side of the rotavap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well known in my lab are dissolved triphenylphosphane and the solvent toluene while isohexane (‘mixture of hexane isomers including methylcyclopentane’) is known to hardly ever give bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this so? Why are bubbles often so much more stable even for just a solvent (toluene)?&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-06-08T03:47:02.853" Title="Why do some solvents bubble more than others when boiling?" Tags="&lt;experimental-chemistry&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32582" PostTypeId="1" CreationDate="2015-06-08T03:38:45.973" Score="8" ViewCount="526" Body="&lt;p&gt;Just wondering if there was a systematic way to identify an alkyl ligand as a $\pi$ acceptor or $\pi$ donor in a transition metal complex? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although various sources indicate that weak field ligands are typically $\pi$ donors and strong field ligands are $\pi$ acceptors, I'm having trouble with the more uncommon ligands not listed in the spectrochemical species, namely $\ce{CH2}$ groups directly double-bonded to metals, or $\ce{CH2-tBu}$ groups double bonded to ligands.&lt;/p&gt;&#xA;" OwnerUserId="16797" LastEditorUserId="10185" LastEditDate="2015-06-08T11:25:33.257" LastActivityDate="2015-06-08T11:25:33.257" Title="Identifying Alkyl Ligands as Pi Donors or Pi Acceptors in Transition Metal Complexes" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;coordination-compounds&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32583" PostTypeId="2" ParentId="32581" CreationDate="2015-06-08T03:47:02.853" Score="11" Body="&lt;p&gt;Boiling depends on certain properties of a solvent:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Tendency to form an emulsion of bubbles due to low surface tension and separation of solvent phases, as in &lt;a href=&quot;http://www.exploratorium.edu/ronh/bubbles/soap.html&quot;&gt;soap bubbles&lt;/a&gt;. Try boiling some dish-washing detergent in water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tendency to &lt;a href=&quot;http://en.wikipedia.org/wiki/Superheating&quot;&gt;superheat&lt;/a&gt;, i.e. to go well above the boiling point and then to suddenly form bubbles, which is more pronounced at lower pressure. This can cause violent bumping, in addition to foaming. The cleaner the apparatus, the more likely, so &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiling_chip&quot;&gt;boiling stones (chips)&lt;/a&gt; are added to nucleate more gentle boiling. A glass mug of water heated in a microwave can bump violently enough to blow open the door (and blow the fuse).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-08T03:47:02.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32584" PostTypeId="2" ParentId="32580" CreationDate="2015-06-08T04:16:55.900" Score="14" Body="&lt;p&gt;The reason that acetic acid's chemical formula is often written as $\ce{CH3COOH}$ is that the $\ce{COOH}$ portion of the formula indicates that the compound is a carboxylic acid.  You'll also see it written as $\ce{HC2H3O2}$ sometimes, again to indicate an acid, this time through the leading $\ce{H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An empirical formula is only intended to give the relative number of atoms in a compound; writing the formula to indicate the type of compound is not usually done.  I would typically follow something like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hill_system&quot;&gt;Hill System&lt;/a&gt; for this purpose: carbon, hydrogen, then other elements in alphabetical order, so $\ce{CH2O}$ here.  Empirical formulas are never named.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastActivityDate="2015-06-08T04:16:55.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32585" PostTypeId="2" ParentId="32561" CreationDate="2015-06-08T04:19:37.163" Score="4" Body="&lt;p&gt;Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $k$ and do the same for the other data sets. What is the result?&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-06-08T04:19:37.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32588" PostTypeId="1" CreationDate="2015-06-08T07:01:46.377" Score="7" ViewCount="2487" Body="&lt;p&gt;Why should the Erlenmeyer flask used to store $\ce{NaOH}$ be dry? Why should it be kept covered at all times?&lt;/p&gt;&#xA;" OwnerUserId="16802" LastEditorUserId="10185" LastEditDate="2015-06-08T14:53:31.783" LastActivityDate="2015-06-08T14:53:31.783" Title="Why must NaOH be kept dry during storage?" Tags="&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32589" PostTypeId="2" ParentId="13960" CreationDate="2015-06-08T08:58:24.637" Score="11" Body="&lt;p&gt;With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) saturation/unsaturation:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that Rule e takes precedence over Rule f.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a preferred IUPAC name is 3-bromocyclohex-1-ene (not 1-bromocyclohex-2-ene): (This example is given in the &lt;em&gt;Blue Book&lt;/em&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2rM5.png&quot; width=&quot;142&quot; height=&quot;71&quot; alt=&quot;3-bromocyclohex-1-ene&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-14.3.5&lt;/strong&gt; Lowest set of locants&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with&#xA;  other locant sets, each cited in order of increasing value, has the lowest term at the first&#xA;  point of difference; (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the structure given in the question is named as 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oozeW.png&quot; width=&quot;163&quot; height=&quot;79&quot; alt=&quot;1,6-dimethylcyclohex-1-ene&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-02-17T12:02:04.730" LastActivityDate="2016-02-17T12:02:04.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32590" PostTypeId="2" ParentId="32588" CreationDate="2015-06-08T10:32:17.180" Score="11" Body="&lt;p&gt;Without a proper closed container, sodium hydroxide readily attracts water and become a viscous solution (concentrated sodium hydroxide). This solution readily attracts carbon dioxide in air and reacts with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2(g) + H2O(l)&amp;lt;=&amp;gt;H2CO3(aq)}$$&#xA;$$\ce{2OH^{-}(aq) + H2CO3(aq) -&amp;gt; 2H2O(l) + CO3^{2-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The $\ce{OH^{-}}$ in the second equation comes from $\ce{NaOH}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction decrease the amount of $\ce{NaOH}$ and thus the titration result will contains error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, the first equation can happen even without sodium hydroxide. Carbon dioxide in air is known to spontaneously dissolved into water and become carbonic acid. This is why the distilled/deionized water which in contact with air will have pH of around 6 after some time. With the presence of sodium hydroxide in water, the second equation greatly drives the equilibrium of the first solution to the right. Concentrated sodium hydroxide is used in some experiment as a carbon dioxide absorber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;Another possible problem that might occur is that concentrated sodium hydroxide can react with glass. Althought this is a very slow reaction, it might causes some unexpected error or ruin the glassware.&lt;/p&gt;&#xA;" OwnerUserId="16455" LastActivityDate="2015-06-08T10:32:17.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32591" PostTypeId="1" AcceptedAnswerId="33009" CreationDate="2015-06-08T10:49:09.250" Score="16" ViewCount="12556" Body="&lt;p&gt;I have often wondered about diagonal relationships between elements on the periodic table, and the most often cited explanations revolve around charge-density considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But other than that, what other factors could possibly contribute to this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; What I mean by &quot;diagonal relationships&quot; is that there are certain similarities in chemical properties that have been observed in diagonally adjacent neighbors in the 2nd and 3rd period example: Li-Mg, Be-Al, and B-Si.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Diagonal_relationship&quot;&gt;Link to a wikipedia article (not much to be found here though)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation that I personally have most often encountered is that the  charge-to volume ratios (charge density), say for Li-Mg cations is roughly the same, and hence could account for some of the similarities in behaviors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is what can be some &lt;em&gt;other&lt;/em&gt; contributing factors (if any) to this phenomenon. I am not expecting very concrete answers because I believe this phenomenon is not very well understood.&lt;/p&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="14" LastEditDate="2015-06-17T14:36:50.117" LastActivityDate="2017-08-13T21:59:33.343" Title="Diagonal relationships in the Periodic Table" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32593" PostTypeId="1" CreationDate="2015-06-08T15:04:40.380" Score="8" ViewCount="470" Body="&lt;p&gt;I have some PDB structures of small molecules with missing hydrogens. Is there a software available that places missing hydrogens and assigns a charge to the molecule that is reliable? I am looking for an automatic tool that returns a PDB with added Hydrogens and the overall charge (I can do some programming if I need to combine output from multiple programs to accomplish this or something.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is enough for me if you suggest a couple of FREE tools (I don't have resources to buy a commercial license, but I am academic, in case they offer free academic licenses), with associated papers that evaluate their performance that I can cite. For example, a free tool that calculates pKa values of ionizable groups would solve my problem, if you can also suggest a tool that automatically detects ionizable groups in molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used Avogadro to add hydrogens (though I don't know how to compute the charge with it), but I've found that some hydrogens aren't well placed. Is there a better tool available for this task?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="1159" LastEditDate="2015-11-16T23:07:18.103" LastActivityDate="2015-11-16T23:07:18.103" Title="Reliable automatic addition of hydrogens and charge assignment to small molecules at given pH?" Tags="&lt;computational-chemistry&gt;&lt;hydrogen&gt;&lt;software&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32594" PostTypeId="2" ParentId="15435" CreationDate="2015-06-08T15:24:23.773" Score="8" Body="&lt;p&gt;The standard redox potential of chlorine is $E^\circ = +1.358\ \mathrm{V}$; the actual potential depends on the concentration of $\ce{Cl-}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Cl2 + 2e- &amp;lt;=&amp;gt; 2 Cl- }\quad\quad E^\circ = +1.358\ \mathrm{V}$$&#xA;$$\begin{aligned}&#xA;E&amp;amp;=E^\circ+\frac{RT}{2F}\cdot\ln\frac{1}{\left[\ce{Cl-}\right]^2}\\ &#xA;&amp;amp;=1.358\ \mathrm{V}-0.05916\ \mathrm{V}\times\log{\left[\ce{Cl-}\right]}&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard redox potential of oxygen is $E^\circ = +1.229\ \mathrm{V}$; the actual potential depends on $\mathrm{pH}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 4H+ + 4e- &amp;lt;=&amp;gt; 2H2O}\quad\quad E^\circ = +1.229\ \mathrm{V}$$&#xA;$$\begin{aligned}&#xA;E&amp;amp;=E^\circ+\frac{RT}{4F}\cdot\ln\left[\ce{H+}\right]^4\\ &#xA;&amp;amp;= 1.229\ \mathrm{V}-0.05916\ \mathrm{V}\times\mathrm{pH}&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The redox potentials $(E_{\ce{O2}} &amp;lt; E_{\ce{Cl2}})$ suggest that oxidation of $\ce{H2O}$ to $\ce{O2}$ at the anode should be preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, depending on the material and shape of the anode, the &lt;strong&gt;overpotential&lt;/strong&gt; of oxygen at the anode can be very large. Therefore, the oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ at the anode is feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, at high $\mathrm{pH}$ and at a low concentration of $\ce{Cl-}$, the oxidation of $\ce{H2O}$ to $\ce{O2}$ is still preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$. Hence, if also $\ce{H2}$ and $\ce{OH-}$ are produced during the electrolysis of aqueous solutions (this is less relevant for the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ which is considered in the question, but for example important for the chloralkali process)&#xA;$$\ce{2H2O + 2e- -&amp;gt; H2 + 2OH-}$$&#xA;the electrolysis of chloride cannot be made complete in a single step since $\mathrm{pH}$ is increased and the concentration of $\ce{Cl-}$ is decreased during the electrolysis:&#xA;$$\ce{2 H2O + 2 Cl- -&amp;gt; H2 + 2 OH- + Cl2}$$&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-05-05T13:20:38.003" LastActivityDate="2016-05-05T13:20:38.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32595" PostTypeId="2" ParentId="16507" CreationDate="2015-06-08T15:36:59.173" Score="12" Body="&lt;p&gt;In principle, non-oxidizing acids cannot directly oxidize copper since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$ or to $\ce{Cu+}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\ce{2H+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; H2}\quad &amp;amp;&amp;amp;E^\circ = +0.000\ \mathrm{V}\\&#xA;\ce{Cu^2+ + 2e- \;&amp;amp;&amp;lt;=&amp;gt; Cu}\quad &amp;amp;&amp;amp;E^\circ = +0.340\ \mathrm{V}\\&#xA;\ce{Cu+ + e- \;&amp;amp;&amp;lt;=&amp;gt; Cu}\quad &amp;amp;&amp;amp;E^\circ = +0.521\ \mathrm{V}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that copper(II) is usually more stable than copper(I) in aqueous solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the situation is slightly complicated because of the low solubility of $\ce{CuCl}$ in dilute hydrochloric acid (in concentrated hydrochloric acid, copper forms chlorido complexes such as $\ce{[CuCl2]-}$ and $\ce{[CuCl4]^2-}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{sp} = \left[\ce{Cu+}\right] \cdot \left[\ce{Cl-}\right] = 1.72 \times 10^{-7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding effective redox potential may be estimated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;E&amp;amp;=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Cu+}\right]\\ &#xA;&amp;amp;=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\frac{K_\mathrm{sp}}{\left[\ce{Cl-}\right]} \\ &#xA;E_{\ce{CuCl}}^\circ&amp;amp;=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln K_\mathrm{sp}\\&#xA;&amp;amp;= E_{\ce{Cu+}}^\circ+0.0592\ \mathrm{V}\cdot\log K_\mathrm{sp}\\&#xA;&amp;amp;=0.521\ \mathrm{V}+0.0592\ \mathrm{V}\cdot\log\left(1.72 \times 10^{-7}\right)\\&#xA;&amp;amp;=0.121\ \mathrm{V}&#xA;\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, oxidation of $\ce{Cu}$ to $\ce{Cu+}$ is favoured in dilute hydrochloric acid. Nevertheless, $\ce{H+}$ is still not strong enough to oxidize $\ce{Cu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\ce{Cu}$ can be oxidized by $\ce{O2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 4H+ + 4e- &amp;lt;=&amp;gt; 2H2O}\quad E^\circ = +1.229\ \mathrm{V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, copper is slowly oxidized in dilute hydrochloric acid in contact with air.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-06-08T15:36:59.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32597" PostTypeId="1" AcceptedAnswerId="32605" CreationDate="2015-06-08T16:46:05.120" Score="5" ViewCount="215" Body="&lt;p&gt;In medical shop I have seen that there are loads of vitamins tablets available but I was wondering like does vitamins in tablet are more efficient than the vitamins which are available in their natural source like in fruits ?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastActivityDate="2015-06-08T21:59:55.143" Title="Are vitamins in tablet are more efficient than the vitamins in their natural source?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2015-06-09T14:20:24.993" ContentLicense="CC BY-SA 3.0" />
  <row Id="32600" PostTypeId="2" ParentId="20421" CreationDate="2015-06-08T17:43:39.837" Score="1" Body="&lt;p&gt;In the synthesis of &lt;a href=&quot;http://en.wikipedia.org/wiki/Dichlorotris(triphenylphosphine)ruthenium(II)&quot; rel=&quot;nofollow&quot;&gt;dichlorotris(triphenylphosphine)ruthenium(II)&lt;/a&gt;, a similar complex, $\ce{PPh3}$ is the reducing agent which is oxidized to triphenylphosphine oxide, $\ce{Ph3PO}$ (references: &lt;a href=&quot;http://dx.doi.org/10.1016%2F0022-1902%2866%2980191-4&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://dx.doi.org/10.1002%2F9780470132432.ch40&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;). The reaction is driven by the strength of the $\ce{P=O}$ double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 RuCl3 \cdot 3H2O + 7 PPh3 ~-&amp;gt;~ 2 RuCl2(PPh3)3 + 2 HCl + 5 H2O + OPPh3}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2015-06-08T17:43:39.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32601" PostTypeId="2" ParentId="32593" CreationDate="2015-06-08T18:00:06.823" Score="5" Body="&lt;p&gt;If you're looking for comprehensive protonation and evaluation of $pK_a$, I suggest my &lt;a href=&quot;https://chemistry.stackexchange.com/a/27003/5017&quot;&gt;answer&lt;/a&gt; to a previous question. There is specialized software for evaluating protonation and $pK_a$ based on large databases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avogadro has some simple rules, mostly for amino acid residues, but is not intended for automated evaluation of thousands of molecules.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-06-08T19:40:47.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32602" PostTypeId="1" CreationDate="2015-06-08T18:02:42.907" Score="1" ViewCount="1057" Body="&lt;p&gt;What is the reason for water molecules $\ce{H_2O}$ being asymmetric while other molecules like $\ce{CO_2}$ being completely symmetric? See the pictures below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uodg6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FhAXh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both case there is an electronegative difference between the atoms in the molecule. I have read that the angle in the water molecule is always a specific value. Is this a special thing for water or is there a common mechanism I should know?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorUserId="4325" LastEditDate="2015-06-08T18:37:23.263" LastActivityDate="2015-06-08T18:37:23.263" Title="Why are H2O asymmetric while CO2 is not?" Tags="&lt;water&gt;&lt;molecules&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-06-08T19:04:53.347" ContentLicense="CC BY-SA 3.0" />
  <row Id="32603" PostTypeId="7" CreationDate="2015-06-08T18:07:43.873" Score="0" Body="&lt;p&gt;&#xA;On Stack Exchange, we believe the core moderators should come &lt;i&gt;from&lt;/i&gt; the community, and be elected by the community itself through popular vote. We hold regular elections to determine who these community moderators will be.&lt;/p&gt;&#xA;&lt;p&gt;&#xA;Community moderators are &lt;a href=&quot;http://blog.stackoverflow.com/2009/05/a-theory-of-moderation/&quot;&gt;accorded the highest level of privilege on our community&lt;/a&gt;, and should themselves be exemplars of positive behavior and leaders within the community.&lt;/p&gt;&#xA;&lt;p&gt;&#xA;Our general criteria for moderators is as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;patient and fair&lt;/li&gt;&#xA;&lt;li&gt;leads by example&lt;/li&gt;&#xA;&lt;li&gt;shows respect for their fellow community members in their actions and words&lt;/li&gt;&#xA;&lt;li&gt;open to some light but firm moderation to keep the community on track and resolve (hopefully) uncommon disputes and exceptions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&#xA;Every election has three phases:&#xA;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nomination&lt;/li&gt;&#xA;&lt;li&gt;Primary&lt;/li&gt;&#xA;&lt;li&gt;Election&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&#xA;Please participate in the moderator elections by voting, and perhaps even by nominating yourself to be a community moderator!&#xA;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chat.stackexchange.com/rooms/122622/2021-chemistry-moderator-election-chat&quot;&gt;2021 election chat room&lt;/a&gt; - discuss the election, candidates, etc. (Previous chat rooms: &lt;a href=&quot;https://chat.stackexchange.com/rooms/93015/2019-chemistry-moderator-election-chat&quot;&gt;2019&lt;/a&gt;, &lt;a href=&quot;https://chat.stackexchange.com/rooms/46261/2016-chemistry-moderator-election&quot;&gt;2016&lt;/a&gt;, &lt;a href=&quot;https://chat.stackexchange.com/rooms/24590/2015-chemistry-moderator-election&quot;&gt;2015&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;For questions about the election process itself, visit &lt;a href=&quot;https://chemistry.meta.stackexchange.com/tags/election&quot;&gt;Meta&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="16683" LastEditDate="2021-04-20T08:37:20.820" LastActivityDate="2021-04-20T08:37:20.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32605" PostTypeId="2" ParentId="32597" CreationDate="2015-06-08T19:24:37.323" Score="3" Body="&lt;p&gt;For some vitamins, such as Vitamin C, which are always found in the form of a unique chemical substance, there is no chemical difference between the vitamin in a tablet--synthetic or otherwise--and the same vitamin as found in a foodstuff. Still, how well you absorb a vitamin might be affected by what else you consume at the time - for example a given amount of Vitamin C might be absorbed more or less efficiently as an ingredient in orange juice than as a pill, even though it is the same substance in both cases. But these differences might not necessarily be significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other vitamins, such as Vitamin A or Vitamin D, are found in multiple closely related chemical forms, any of which the body can use to fulfill its need for the vitamin. In that case, the body may absorb or utilize the vitamin more efficiently when it is in one form relative to another. Fats in the diet can further influence your absorption of a fatty vitamin like Vitamin D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minerals, such as iron, are notorious for being made easier or harder to absorb depending on their chemical form and how they are bound to other components of the food. So, for example, iron in greens is known to be slower/more inefficient for the body to extract than iron in liver.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-08T19:24:37.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32606" PostTypeId="1" CreationDate="2015-06-08T20:43:50.137" Score="5" ViewCount="1928" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The hydrocarbon &lt;span class=&quot;math-container&quot;&gt;$\ce{C17H36}$&lt;/span&gt; can be cracked. Which compound is the least likely to be produced in this reaction?&lt;/p&gt;&#xA;&lt;p&gt;A: &lt;span class=&quot;math-container&quot;&gt;$\ce{C3H8}$&lt;/span&gt; &lt;br&gt;B: &lt;span class=&quot;math-container&quot;&gt;$\ce{C4H8}$&lt;/span&gt; &lt;br&gt;C: &lt;span class=&quot;math-container&quot;&gt;$\ce{C8H16}$&lt;/span&gt; &lt;br&gt;D: &lt;span class=&quot;math-container&quot;&gt;$\ce{C16H34}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Is it D because cracking's purpose is the breaking of long hydrocarbon chains into smaller ones?&lt;/p&gt;&#xA;" OwnerUserId="14862" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-06T07:09:08.517" Title="Least likely product of a hydrocarbon cracking reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32607" PostTypeId="6" CreationDate="2015-06-08T20:46:05.623" Score="0" Body="&lt;p&gt;I am a physicist, having PhD in solid-state physics. Currently, I am interested in quantum chemistry, doing research on diatomic molecules. I have reasonably moderated ago, meaning I accept when I am wrong. I could try being a moderator here.&lt;/p&gt;&#xA;" OwnerUserId="1696" LastEditorUserId="1696" LastEditDate="2015-06-08T20:46:05.623" LastActivityDate="2015-06-08T20:46:05.623" CommentCount="4" CommunityOwnedDate="2015-06-08T20:46:05.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="32608" PostTypeId="6" CreationDate="2015-06-08T21:02:06.723" Score="0" Body="&lt;p&gt;Hi, I'm Manish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been with Chem.SE from the start, and am really happy to see how far it's reached!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was at odds with myself whether or not I should stand in this election but it occurred to me that I'll never have the time to contribute content to this site; moderation is the way I think I can help best!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why I'm a good candidate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I've been a pro-tem mod for this site from the beginning. I know the community and am familiar with how moderation works here&lt;/li&gt;&#xA;&lt;li&gt;I've been a mod on &lt;a href=&quot;http://physics.stackexchange.com&quot;&gt;Physics&lt;/a&gt; for more than a year.&lt;/li&gt;&#xA;&lt;li&gt;I've been quite active on the &lt;a href=&quot;http://meta.stackexchange.com&quot;&gt;mother meta&lt;/a&gt; in the past and am familiar with the network wide policies&lt;/li&gt;&#xA;&lt;li&gt;I also &lt;a href=&quot;http://www.rust-lang.org/team.html&quot; rel=&quot;nofollow&quot;&gt;help moderate&lt;/a&gt; the online community of the Rust programming language.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why I'm not a good candidate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I'm not a chemist. I'm a physics student, and may or may not have &lt;a href=&quot;http://smbc-comics.com/index.php?id=1309&quot; rel=&quot;nofollow&quot;&gt;plans&lt;/a&gt;. This never affected my ability to moderate though.&lt;/li&gt;&#xA;&lt;li&gt;I'm quite busy. I've had to cut down on participation on Stack Exchange due to my courseload and open source participation. I still moderate, though, and I expect that to continue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22" LastEditorUserId="22" LastEditDate="2015-06-08T21:02:06.723" LastActivityDate="2015-06-08T21:02:06.723" CommentCount="10" CommunityOwnedDate="2015-06-08T21:02:06.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="32610" PostTypeId="2" ParentId="32597" CreationDate="2015-06-08T21:49:37.083" Score="-2" Body="&lt;p&gt;For most vitamins, it's much better to consume them in dissolved form (as they would be in most foods). You can test how much dissolves by putting a vitamin tablet into a cup of warm-to-hot (around body temp) vinegar and seeing how much of the pill dissolves. If it doesn't dissolve, it's exiting your body without any chance to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for most vitamins, no.&lt;/p&gt;&#xA;" OwnerUserId="14887" LastEditorUserId="14887" LastEditDate="2015-06-08T21:59:55.143" LastActivityDate="2015-06-08T21:59:55.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32612" PostTypeId="1" CreationDate="2015-06-08T22:17:38.267" Score="4" ViewCount="119" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the following:&#xA;  $$\begin{alignat}{2}&#xA;\ce{PbS(s) + 3/2O2(g) &amp;amp;-&amp;gt; PbO(s) + SO2(g)}\qquad&amp;amp;&amp;amp;\Delta H^\circ = -413.7~\mathrm{kJ} \\  &#xA;\ce{PbO(s) + C(s) &amp;amp;-&amp;gt; Pb(s) + CO(g)}\qquad&amp;amp;&amp;amp;\Delta H^\circ = +106.8~\mathrm{kJ} &#xA;\end{alignat}$$&#xA;  a. Calculate the enthalpy of the following process:&#xA;  $$\ce{PbS(s) + 3/2O2(g) + C(s) -&amp;gt; Pb(s) + SO2(g) + CO(g)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b. What is the energy needed to make $454\ \mathrm g$ of $\ce{PbS}$ into $\ce{Pb}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My efforts:&lt;/strong&gt;&lt;br&gt;&#xA;a.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbS(s) + 3/2O2(g) -&amp;gt; PbO(s) + SO2(g)}$&lt;br&gt;&#xA;$\ce{PbO(s) + C(s) -&amp;gt; Pb(s) + CO(g)}$&lt;br&gt;&#xA;$\Delta H_\mathrm r^\circ = \Delta H_{\mathrm r(1)}^\circ + \Delta H_{\mathrm r(2)}^\circ$&lt;br&gt;&#xA;$\Delta H_\mathrm r^\circ = -413.7~\mathrm{kJ~mol^{-1}} + 106.8~\mathrm{kJ~mol^{-1}} = -306.9~\mathrm{kJ~mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. How do I find the energy needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular mass of $\ce{PbS}$ is   $239.2650~\mathrm{g~mol^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have $\frac{454~\mathrm{g}}{239.2650~\mathrm{g~mol^{-1}}} = 1.897~\mathrm{mol}$ of $\ce{PbS}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So energy required is $1.897~\mathrm{mol} \times 306.9~\mathrm{kJ~mol^{-1}} = 582.2~\mathrm{kJ}$&lt;/p&gt;&#xA;" OwnerUserId="11522" LastEditorUserId="5026" LastEditDate="2018-02-23T01:47:08.613" LastActivityDate="2018-02-23T01:47:08.613" Title="Calculating enthalpy change and energy required to convert lead sulphide to lead" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32613" PostTypeId="2" ParentId="32606" CreationDate="2015-06-08T22:26:12.063" Score="3" Body="&lt;p&gt;There are two ways to look at this question.  One is simple carbon atom counts.  Another is the saturation of the products.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carbon atom counts&lt;/strong&gt;.  The question doesn't specify which of the 24894 isomers of heptadecane is to be cracked.   Suppose it is &quot;normal&quot; or &lt;em&gt;n&lt;/em&gt;-heptadecane.  Cracking usually begins with the homolytic cleavage of a carbon-carbon bond.  To make a $\ce{C16}$ product from &lt;em&gt;n&lt;/em&gt;-heptadecane by cleaving carbon-carbon bonds, there are only two potential bonds to cleave, one at one end of the molecule, another at the other end.  In either case, to obtain a $\ce{C16}$ product, only a single cleavage would be allowed, because further reactions would break more C-C bonds and change the product.  Cracking is a high temperature reaction that is entropically driven.  The system is increasing the disorder.  Imagine rolling four 6-sided dice, and the total comes up 17.  Then you take away either one or two of the dice (whether its one or two is random).  What is the probability that the remaining dice sum to 16, compared to say 8 or 4?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Saturation&lt;/strong&gt;.  Choices B and C are unsaturated hydrocarbons.  Choices A and D are saturated hydrocarbons.  Both saturated and unsaturated compounds can be formed by cracking, but most modern cracking processes tend to focus on making unsaturated hydrocarbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, primarily because of simple statistical unlikelihood, but also because of saturation considerations, &lt;strong&gt;D&lt;/strong&gt; is the correct answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's article on cracking is very helpful.  See &lt;a href=&quot;http://en.wikipedia.org/wiki/Cracking_(chemistry)&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:21:23.323" LastActivityDate="2016-03-11T23:21:23.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32614" PostTypeId="2" ParentId="8259" CreationDate="2015-06-08T23:05:05.663" Score="4" Body="&lt;p&gt;From the comments: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I used 50 grams of cellulose, then I added 55 ml $\ce{H2SO4}$ solution (99%) and it was conducted at normal room temperature and normal air pressure, and in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So unless I am missing something, in an attempt to &lt;em&gt;hydrolyze&lt;/em&gt; cellulose, no water was added to cellulose powder.  The &lt;em&gt;exo&lt;/em&gt; hydrolysis of cellulose is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(C6H10O5)_{n} + H2O -&amp;gt; C6H12O6 + (C6H10O5)_{n-1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water is a reactant, adding it is probably required to make the reaction work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are some hypotheses to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mostly anhydrous sulfuric acid and cellulose probably did not react, although perhaps some solvolysis to form sulfated glucose monomers may have happened.&lt;/li&gt;&#xA;&lt;li&gt;When NaOH was added, water was formed nearly instantly by reaction with the acid, and the liberated heat strongly heated the reaction.&lt;/li&gt;&#xA;&lt;li&gt;It doesn't sound like you added enough NaOH to neutralize the acid completely.    The reactions you observed probably happened in acidic conditions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the strong heat and water formation that happened led to simultaneous hydrolysis and dehydration of the cellulose, leading to a complex mixture of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydroxymethylfurfural&quot; rel=&quot;nofollow&quot;&gt;hydroxymethylfurfural&lt;/a&gt; and further polymerized/carbonized derivatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little surprised that the 99% sulfuric acid you used did not carbonize the cellulose eventually even without water.  It's a very powerful dehydrating agent.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T22:54:35.270" LastActivityDate="2016-03-11T22:54:35.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32616" PostTypeId="1" CreationDate="2015-06-09T00:35:37.523" Score="5" ViewCount="732" Body="&lt;p&gt;Cooking oil that jumps from a frying pan in use and over time accumulates around the frying pan on a wooden surface is sticky and hard to remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why. Why is it hard to remove? What is the chemistry behind it? Does it bond with the wood somehow? If so, how is it different from oil used to give the wood a protective coating? Why is one sticky and the other not sticky?&lt;/p&gt;&#xA;" OwnerUserId="6790" LastEditorUserId="7475" LastEditDate="2015-10-30T16:54:33.937" LastActivityDate="2015-10-30T16:54:33.937" Title="Why is grease from cooking hard to remove from wood?" Tags="&lt;everyday-chemistry&gt;&lt;absorption&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32617" PostTypeId="1" AcceptedAnswerId="32621" CreationDate="2015-06-09T00:41:37.500" Score="3" ViewCount="15625" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hmnE6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried finding the slope between two points but I keep getting wrong values, how does one find the initial rate by using such graphs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I started by trying to find a slope of two points, &lt;span class=&quot;math-container&quot;&gt;$(0,0.4)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(10,0.2)$&lt;/span&gt; after calculations I got  &lt;span class=&quot;math-container&quot;&gt;$-0.02$&lt;/span&gt;, which is the opposite of the right answer in sign. I tried to check myself by trying to find the slope for other points like: &lt;span class=&quot;math-container&quot;&gt;$(10,0.2)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(20,0.1)$&lt;/span&gt; which gives me &lt;span class=&quot;math-container&quot;&gt;$-0.01$&lt;/span&gt;, I'm confused as whether my initial idea is right or not.&lt;/p&gt;&#xA;" OwnerUserId="16551" LastEditorUserId="83758" LastEditDate="2020-02-24T15:03:11.467" LastActivityDate="2020-02-24T21:41:54.517" Title="Initial rate of reaction from the graph?" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32618" PostTypeId="1" CreationDate="2015-06-09T02:07:02.590" Score="6" ViewCount="842" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is a container that has a volume of $5~\mathrm{L}$ on one side and $3~\mathrm{L}$ on the other and is separated in the middle. There are $2~\mathrm{mol}$ of $\ce{N2}$ on one side at $0~^\circ\mathrm{C}$ and $4~\mathrm{mol}$ of $\ce{H2}$ on the other at $0~^\circ\mathrm{C}$. When the valve is turned and the gases react, what will the total pressure be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The balanced reaction is $$\ce{N2 + 3H2 &amp;lt;=&amp;gt; 2NH3}$$&#xA;My thinking was that the reaction occurs at STP so the pressure would stay at 1  atm?&lt;br&gt;&#xA;Or that you could use &#xA;\begin{align}&#xA;pV &amp;amp;= nRT\\&#xA;p(8~\mathrm{L}) &amp;amp;=6(.082)(273.15)\\&#xA;p &amp;amp;= 0.06&#xA;\end{align}&#xA;The $8~\mathrm{L}$ comes from adding the two volumes, and the 6 from adding the number moles together.&lt;/p&gt;&#xA;" OwnerDisplayName="user16813" LastEditorUserId="46306" LastEditDate="2018-03-03T14:46:40.940" LastActivityDate="2018-03-03T14:46:40.940" Title="How to calculate the equilibrium pressure when ammonia and hydrogen gasses react in a closed system?" Tags="&lt;stoichiometry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32619" PostTypeId="1" CreationDate="2015-06-09T02:24:13.713" Score="11" ViewCount="825" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ds2xY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ds2xY.png&quot; alt=&quot;Alkylation of 1-naphthol with benzyl iodide in TFE/DMSO&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to work out which product is formed in which solvent when naphthalen-1-olate is treated with benzyl iodide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DMSO is a polar aprotic solvent, as it does not possess an acidic hydrogen. So, the negative charge on oxygen will be concentrated as the anion will be poorly solvated. Hence, it should act as a hard nucleophile. Since alkyl halides such as BnI are soft electrophiles, I think that the substituent should preferentially add to the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2,2,2-trifluoroethan-1-ol (TFE), I think the reaction would proceed via an S&lt;sub&gt;N&lt;/sub&gt;1 mechanism. Now, the carbocation is a hard electrophile, and thus &lt;em&gt;O&lt;/em&gt;-alkylation might be observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But considering the better solvation of the negatively charged oxygen in TFE, it would also be more hindered, which suggests that &lt;em&gt;C&lt;/em&gt;-alkylation might also be seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which products are actually formed, and am I on the right track?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="16683" LastEditDate="2018-08-14T08:54:39.933" LastActivityDate="2018-08-14T08:54:39.933" Title="Alkylation of 1-naphthol in trifluoroethanol versus in DMSO" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;&lt;nucleophilic-substitution&gt;&lt;phenols&gt;&lt;regioselectivity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32621" PostTypeId="2" ParentId="32617" CreationDate="2015-06-09T02:50:00.477" Score="8" Body="&lt;p&gt;It seems like you have the right idea to me,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;Initial Rate,&quot; you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="16814" LastEditorUserId="16814" LastEditDate="2015-06-09T14:23:03.273" LastActivityDate="2015-06-09T14:23:03.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32625" PostTypeId="2" ParentId="31102" CreationDate="2015-06-09T06:50:21.080" Score="8" Body="&lt;p&gt;1) Oxygen CAN form 3 or 4 bonds. The most common example is $\ce{H3O+}$ ion This form of bonding is critical to understand basic nature of ammonia solutions, so for exact description of it look that part of the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)Boron also can form more than 3 bonds if at least one its partner provides an electron pair for bonding, not one electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule I prefer to stay away from the chemistry of borates and alumina-silicates as it is ruled largely by ionic interactions and non-intuitive hydration/dehydration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;===========&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In structural diagrams of polymeric salts 'unit' often refers to some recurring fragment... however, it is not nessesarily has same stochiometry than formal anion. In alumosilicates the anionic frame consists of $\ce{AlO4-}$ and $\ce{SiO4}$ tetrahedral units. However, they are usually connected by shared vertexes, so the exact formula of formal anion is different. Borates are a bit more complex system, since they may include both $\ce{BO3}$ and $\ce{BO4-}$ units, and oxygen is often connected to hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;==========&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked into open cristallographic database. Unsurprisingly, it does NOT feature a compound $\ce{Ca2B6O11}$. However, salts are often written in 'dehydrated' form, so I looked for any borate with $\ce{Ca:B} = 1:3$ stochiometry. Several with rather complex crystal structure were found, so I left only two: one with rather simple structure of typical 'hidrated' borate anion and one with rather complex continous 'dehydrated' boron-oxygen layers. Note: the structures are refined from X-Ray , so they does not reveal hydrogen positions. Isolated oxygens of water molecules and cations are omitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{B3O8}$ unit from mineral Inyoite, $\ce{CaB3O3(OH)5·4(H2O)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EVsro.png&quot; alt=&quot;B3O8&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{[B3O6]_{\infty}}$ layer from mineral Fabianite $\ce{CaB3O5(OH)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pa8eW.png&quot; alt=&quot;(B3O6)_\inf&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its main building block&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZmspY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-06-09T17:10:36.650" LastActivityDate="2015-06-09T17:10:36.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32626" PostTypeId="1" AcceptedAnswerId="77044" CreationDate="2015-06-09T08:01:16.870" Score="6" ViewCount="911" Body="&lt;p&gt;It turns out pineapple juice can dissolve a jello set, and turn a beef roast to paste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out this is caused by &lt;a href=&quot;http://en.wikipedia.org/wiki/Bromelain&quot; rel=&quot;noreferrer&quot;&gt;Bromelain&lt;/a&gt;, an enzyme sometimes used in meat tenderising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to work out how you can eat pineapple without dissolving on the spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4478" LastActivityDate="2017-06-30T04:26:12.590" Title="How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)" Tags="&lt;enzymes&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32627" PostTypeId="2" ParentId="32593" CreationDate="2015-06-09T08:28:57.367" Score="2" Body="&lt;p&gt;If the molecule is a protein you can use &lt;a href=&quot;http://nbcr-222.ucsd.edu/pdb2pqr_2.0.0/&quot; rel=&quot;nofollow&quot;&gt;the PDB2PQR webserver&lt;/a&gt;.  It uses PROPKA to predict $pK_a$ values and optimizes the hydrogen bonding network&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T08:42:46.487" LastActivityDate="2015-06-09T08:42:46.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32628" PostTypeId="2" ParentId="24671" CreationDate="2015-06-09T09:44:57.323" Score="5" Body="&lt;p&gt;There is another rather more simple answer to the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SnCl2}$ can be thought of as a salt to a much greater extent than $\ce{SnCl4}$. In fact, tin is so often thought of as a metal, that it takes some people time to understand that $\ce{SnCl4}$ is in fact not a salt but much more a molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when looking at $\ce{Sn(II)}$, it becomes clear that the unaccompanied ion would have the electron configuration $\ce{[Kr] 4d^{10} 5s^2}$ — doesn’t that look stable to you? Add two chloride ions that took the electrons originally having been in the $\ce{5p}$-orbitals and voilà.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I know that this answer simplifies a lot of things that maybe shouldn’t be simplified in all cases. But to the best of my knowledge, it explains the compound and its stability well, and should therefore be considered. Once the question turns into ‘why is $\ce{Sn(II)}$ ionic, while $\ce{Sn(IV)}$ better described as molecular?’, it of course becomes invalid.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2016-02-04T14:45:17.137" LastActivityDate="2016-02-04T14:45:17.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32629" PostTypeId="1" CreationDate="2015-06-09T10:48:06.310" Score="1" ViewCount="3358" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4IAvr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My new try:&#xA;Given data&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy given by reaction =1690 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial temperature = 0 oC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final temperature = 50 oC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C water = 4.18 J&lt;em&gt;mol-1&lt;/em&gt; K-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;?Hfus ice = 6.01 kJ*mol-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate Mass of ice = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice melts at constant temperature of 0 degree C and changes the phase from solid to liquid .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then liquid water absorbs the energy and changes from 0 degree C to 50 degree C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change in temperature of the water is ?T = (50 oC - 0 oC )= 50 oC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s set up the equation as follows to calculate the mass of ice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q= (m*?Hfus) + (m*Cwater * ?T)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where , q = energy , m = mass , ?T = (Tf-Ti)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;convert given energy from unit kJ to Joules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1690 kJ * 1000 J)/1kJ = 1690000 J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convert ?H fus from kJ&lt;em&gt;mol-1 to J&lt;/em&gt;g-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.01kJ*mol-1 &lt;em&gt;(1000 J/1kJ) = 6010 J&lt;/em&gt;mol-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6010 J&lt;em&gt;mol-1 / 18.0148 g&lt;/em&gt;mol- = 333.6 J *g-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convert C water from J per mol to J per g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.18 J&lt;em&gt;mol-1&lt;/em&gt; oC /18.0148 g = 0.232 J*g-1 oC-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets put the values in the formula and calculate the mass of ice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q= (m*?Hfus) + (m*Cwater * ?T)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1690000 J = (m X 333.6 J&lt;em&gt;g-1) + (m X 0.232 J&lt;/em&gt;g-1 *oC-1 * 50 oC )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1690000 J = m X 345.2 J*g-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1690000 J /345.2 J*g-1 = m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4895.7 g = m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me!!&lt;/p&gt;&#xA;" OwnerUserId="11522" LastEditorUserId="865" LastEditDate="2021-05-22T21:37:56.667" LastActivityDate="2021-05-22T21:37:56.667" Title="When the heat is absorbed by the ice then ice will melt to form liquid water" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32631" PostTypeId="1" CreationDate="2015-06-09T11:32:24.960" Score="4" ViewCount="337" Body="&lt;p&gt;The &lt;a href=&quot;https://www.google.es/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;ved=0CDQQFjAF&amp;amp;url=http%3A%2F%2Fwww.olimex.cl%2Fpdf%2FCO2b.pdf&amp;amp;ei=hs12VYOPEYHlUaLOgcAF&amp;amp;usg=AFQjCNEyRdKTIKGQSLkfYsuZSJgMhy0lKw&amp;amp;sig2=bBJ8VvzA2ZgxPB1CDdYTaQ&amp;amp;bvm=bv.95039771,d.d24&amp;amp;cad=rja&quot; rel=&quot;nofollow&quot;&gt;MG811&lt;/a&gt; is an electrochemical sensor that measures $\ce{CO2}$. I think is something like a redox reaction but i don't understant how it produces $\ce{e-}$ so i can read them and get the $\ce{CO2}$ concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I copy you the reaction that makes the sensor measures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the sensor exposed to $\ce{CO2}$，the following electrodes reaction occurs：&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cathodic reaction: $\ce{2Li+ + CO2 + 1/2O2 + 2e- -&amp;gt; Li2CO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anodic reaction：$\ce{2Na+ + 1/2O2 + 2e- -&amp;gt; Na2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall chemical reaction：$\ce{Li2CO3 + 2Na+ -&amp;gt; Na2O + 2Li+ + CO2}$ &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The datasheet says that it workings adopting the electrolyte cell Principle and that it's composed by the following cells:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Air，$\ce{Au}$|NASICON|| carbonate|$\ce{Au}$, air，$\ce{CO2}$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I need you to explain me how actually the sensor is working and with which material are made the electrodes.&lt;/p&gt;&#xA;" OwnerUserId="16822" LastEditorUserId="16822" LastEditDate="2015-06-09T11:55:13.180" LastActivityDate="2015-06-09T12:18:16.523" Title="How does the MG811 sensor work?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32632" PostTypeId="2" ParentId="32458" CreationDate="2015-06-09T11:46:34.883" Score="3" Body="&lt;p&gt;For a real balloon, the balloon itself will have mass, and the balloon may have higher pressure inside than atmospheric pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absent additional information, for a homework problem, you would assume mass of the balloon is zero and pressure is atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: in the OP 8g/mol is wrong. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-06-09T11:46:34.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32633" PostTypeId="1" AcceptedAnswerId="32653" CreationDate="2015-06-09T12:01:55.370" Score="10" ViewCount="6130" Body="&lt;p&gt;In the reaction, $\ce{OH-}$ group attacks a hydrogen ion in an E2 reaction, forming a transition state in which simultaneously $\ce{NCH3}$ is removed. In this reaction, the $\ce{OH-}$ should attack, like in other case of E2 so as to form the Zaitsev product, but due to &quot;steric hindrance&quot; the less substituted alkene is formed. But surely the $\ce{OH-}$'s attack is not affected by $\ce{CH3}$ groups? It can be affected by the $\ce{NCH3}$ group - but whichever way it attacks, it's going at the beta carbon - so is at the same distance from that group in all cases. How can steric hindrance affect its attack? Why is the Hofmann product (less substituted product) favoured?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-06-09T13:15:53.773" LastActivityDate="2015-06-11T21:01:50.013" Title="Why is the less substituted alkene favoured in Hofmann elimination?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32635" PostTypeId="1" CreationDate="2015-06-09T12:08:21.077" Score="11" ViewCount="12780" Body="&lt;p&gt;Why does magnesium have a lower melting point than both calcium and beryllium? It does not seem to fit into the group trend.&lt;/p&gt;&#xA;" OwnerUserId="16823" LastEditorUserId="10185" LastEditDate="2015-06-09T13:01:48.837" LastActivityDate="2020-07-27T03:45:20.037" Title="Why does magnesium have an exceptionally low melting point?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32636" PostTypeId="2" ParentId="32619" CreationDate="2015-06-09T12:17:46.187" Score="0" Body="&lt;p&gt;These solvents should give identical products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't &lt;em&gt;believe&lt;/em&gt; there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="10185" LastEditDate="2015-06-09T12:29:24.607" LastActivityDate="2015-06-09T12:29:24.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32637" PostTypeId="2" ParentId="32631" CreationDate="2015-06-09T12:18:16.523" Score="3" Body="&lt;p&gt;The link you gave in the question, actually shows the equation of how this sensor works. You don't need to produce any electrons, you just need to measure the voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to write the Nernst equation for the above overall reaction and that will show you what is going on.&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-06-09T12:18:16.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32638" PostTypeId="2" ParentId="32612" CreationDate="2015-06-09T12:37:47.840" Score="4" Body="&lt;h2&gt;Part 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's easier to imagine \begin{equation}\ce{A + B -&amp;gt; C + D} \tag{1}&#xA;\end{equation}&#xA;\begin{equation}\ce{D + E -&amp;gt; F + G} \tag{2}\end{equation}&#xA;\begin{equation}\end{equation}&#xA;Now if we wanted to write a 3&lt;sup&gt;rd&lt;/sup&gt; equation where both the reactions 1 and 2 happen, we'd get&#xA;\begin{equation}\ce{A + B + E -&amp;gt; C + F + G} \tag{3}\end{equation}&#xA;Why?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;D is both a reactant and a product. So, it's a &lt;a href=&quot;http://en.wikipedia.org/wiki/Reaction_intermediate&quot; rel=&quot;nofollow&quot;&gt;reaction intermediate&lt;/a&gt;. It's like it's produced in 1 and again consumed in 2. So, if both reactions take place, it's neither on the reactants' side nor the products.&lt;/li&gt;&#xA;&lt;li&gt;The conventional way of writing reactions doesn't show the mechanism of the reaction. You have to image the reaction happening, to completely understand why D isn't written.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, as the third reaction is reaction 1 and 2 happening at the same place, $\Delta H_{r_3}$ is the algebraic sum of $\Delta H_{r_2}$ and $\Delta H_{r_1}$. Thus, your final result is correct: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Part 2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have to take a more careful look at the unit of $\Delta H$. What does it mean? It means that for &lt;em&gt;each&lt;/em&gt; mole of the reactant ($\ce{PbS}$), one mole of product ($\ce{Pb}$) is formed. (With regards to their stoichiometric coefficients in reaction 3) So, we've got to simply 'punch' the numbers in the formula $n= \frac{m}{M}$ (Where $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass) And then multiply that number by the $\Delta H_{r_3}$. Thus, your final answer is correct too.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7475" LastEditDate="2015-10-21T17:20:33.627" LastActivityDate="2015-10-21T17:20:33.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32639" PostTypeId="2" ParentId="32619" CreationDate="2015-06-09T12:56:47.200" Score="6" Body="&lt;p&gt;The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/02b08.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Thanks to @jerepierre for pointing this out in the comments: &quot;One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed.&quot;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="4325" LastEditDate="2015-06-09T16:21:11.970" LastActivityDate="2015-06-09T16:21:11.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32640" PostTypeId="1" CreationDate="2015-06-09T13:06:15.253" Score="31" ViewCount="11038" Body="&lt;p&gt;Are there gases that are not transparent at room temperature (i.e. at temperature below the point where the substance starts to radiate visible light due to heating)?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="16683" LastEditDate="2016-09-30T09:25:06.217" LastActivityDate="2016-09-30T09:25:06.217" Title="Are there non-transparent gases?" Tags="&lt;optical-properties&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="3" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32641" PostTypeId="6" CreationDate="2015-06-09T13:33:56.510" Score="0" Body="&lt;p&gt;LordStryker here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of you may know me.  I tend to camp out in chat a lot getting to know everyone. I've been a somewhat active member of Chem:SE since &lt;a href=&quot;http://chat.stackexchange.com/transcript/3229/2012/4/25&quot;&gt;day 1&lt;/a&gt;, supporting the launch of the site in Area 51, and consequently becoming user 14 for this site.  None of these things qualify me for the position but it establishes a historical context for my involvement in this wonderful community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I would be a good candidate for this position because I have witnessed and partaken in the growth and evolution of a place many of us call home.  I bring to the table a commitment and passion that would benefit this site.  I work well with others, particularly well-established users including our pro-tem mods.  I don't harbor a long list of contributed questions and answers but I do try to submit quality content, sometimes near &lt;a href=&quot;http://chemistry.stackexchange.com/questions/16342/dipole-moment-of-cis-2-butene/16374#16374&quot;&gt;production-level&lt;/a&gt; content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is simply amazing how far we have come since the very beginning.  Elected or not, I love all of you and I don't plan on going anywhere.  See you in chat!&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-06-09T13:33:56.510" LastActivityDate="2015-06-09T13:33:56.510" CommentCount="4" CommunityOwnedDate="2015-06-09T13:33:56.510" ContentLicense="CC BY-SA 3.0" />
  <row Id="32644" PostTypeId="2" ParentId="32640" CreationDate="2015-06-09T13:51:24.067" Score="19" Body="&lt;p&gt;I'd separate &lt;strong&gt;&lt;em&gt;transparent&lt;/em&gt;&lt;/strong&gt; and &lt;em&gt;colorless&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most gases are transparent or very nearly so because the concentration is low and absorptions are often weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chlorine&quot; rel=&quot;noreferrer&quot;&gt;Chlorine&lt;/a&gt;, though is yellow-green, and has a noticeable color (from Wikipedia)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4TnJj.jpg&quot; alt=&quot;chlorine ampule&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other halogens such as bromine and iodine do have observable colors as vapor, although as mentioned in the comments, you often need slightly elevated temperature for significant vapor to form.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-06-09T14:09:35.203" LastActivityDate="2015-06-09T14:09:35.203" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32645" PostTypeId="1" AcceptedAnswerId="32650" CreationDate="2015-06-09T14:04:58.857" Score="7" ViewCount="14961" Body="&lt;p&gt;The amino acids are Zwitterions. In neutral pH, an Amino acid's amino group has a postive charge and Carboxyl group has negative charge. They cancel each others charge thanks to the $Hydrogen$ that's roaming between the two groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I am not able to understand why Amino acids have positive charges at a low pH and negative at high pH? Given amino acids are Zwitterions, how can both terminals have the same charge (either positive or negative) at low and high pH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain this from general chemistry and pH perspective (acid/base)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PS: first posted in Biology site but community recommended it to be moved here.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7475" LastEditDate="2016-09-19T12:39:53.560" LastActivityDate="2019-10-23T10:03:41.507" Title="Why amino acids (Zwitterion) become either negative or positive at low and high pH solutions?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;&lt;carbonyl-compounds&gt;&lt;amines&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32646" PostTypeId="2" ParentId="32645" CreationDate="2015-06-09T14:08:54.090" Score="3" Body="&lt;p&gt;By increasing the pH you are actually increasing the concentration of hydrogen ions present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the simplistic definition, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pH = -\log\ce{[H+]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus at a low pH you have lots of $\ce{[H+]}$ present. This can protonate your amino acid resulting in a positive charge overall. The inverse is true for high pH (low concentration of $\ce{[H+]}$)&lt;/p&gt;&#xA;" OwnerUserId="15191" LastEditorUserId="10185" LastEditDate="2015-06-09T14:11:41.700" LastActivityDate="2015-06-09T14:11:41.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32647" PostTypeId="1" AcceptedAnswerId="32648" CreationDate="2015-06-09T14:12:08.577" Score="3" ViewCount="259" Body="&lt;p&gt;I am trying to compute the NMR shift for a specific atom in the middle of the surrounding crystal unit cell (1 surrounding anion sphere and 1 cation sphere) in Gaussian09. I only need the chemical shift for the middle atom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I only compute the shift for this atom, and leave out the rest? And would it even save any computation time if I did?&lt;/p&gt;&#xA;" OwnerUserId="16827" LastEditorUserId="10185" LastEditDate="2015-06-09T14:26:47.537" LastActivityDate="2015-06-09T14:38:16.587" Title="How to compute the NMR shift in Gaussian09 for only one specific atom in a bigger unit cell?" Tags="&lt;crystal-structure&gt;&lt;nmr-spectroscopy&gt;&lt;software&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32648" PostTypeId="2" ParentId="32647" CreationDate="2015-06-09T14:37:31.803" Score="3" Body="&lt;p&gt;&lt;strong&gt;Short answer, you can't.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might save a bit of time, but probably not much. The calculation would still need to go through the normal SCF on the entire system to calculate the NMR shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principal, in evaluating the property calculation, you could save a small amount of time by only computing the one atom, but for most systems, I would think the overall SCF calculation would be &lt;em&gt;vastly&lt;/em&gt; greater than computing a few properties.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-06-09T14:37:31.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32649" PostTypeId="2" ParentId="32647" CreationDate="2015-06-09T14:38:16.587" Score="3" Body="&lt;p&gt;As far as I understand the very procedure that is used to calculate NMR chemical shifts, that should not be possible. The magnetic field is basically treated as perturbation and a set of coupled-perturbed SCF (HF or DFT) equations is solved to find the response of the density matrix with respect to the field.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-06-09T14:38:16.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32650" PostTypeId="2" ParentId="32645" CreationDate="2015-06-09T14:49:46.987" Score="10" Body="&lt;p&gt;Firstly a note about terminology. The word &quot;terminus&quot; is reserved for the &lt;a href=&quot;http://en.wikipedia.org/wiki/N-terminus&quot; rel=&quot;nofollow noreferrer&quot;&gt;N- or C-termini of a polypeptide chain&lt;/a&gt;. For a free amino acid, you should refer to the carboxyl and amino groups as the &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{COOH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{NH2}$&lt;/span&gt; groups respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the -&lt;span class=&quot;math-container&quot;&gt;$\ce{COOH}$&lt;/span&gt; group is acidic; above a certain pH, typically around 2, it can be deprotonated to form a negatively charged -&lt;span class=&quot;math-container&quot;&gt;$\ce{COO-}$&lt;/span&gt; group. On the other hand, the -&lt;span class=&quot;math-container&quot;&gt;$\ce{NH2}$&lt;/span&gt; group is basic; below a certain pH, typically around 9, it can be protonated to form a positively charged -&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3+}$&lt;/span&gt; group. I made a short table describing the protonation states of the two groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{c|c|c|c}&#xA;\textbf{pH} &amp;amp; \textbf{Carboxyl group exists as} &amp;amp; \textbf{Amino group exists as} &amp;amp; \textbf{Net charge} \\&#xA;\hline&#xA;&amp;lt;2 &amp;amp; \ce{-COOH}\text{ (neutral)} &amp;amp; \ce{-NH3+}\text{ (positively charged)} &amp;amp; +1 \\&#xA;&amp;gt;2 &amp;amp; \ce{-COO^-}\text{ (negatively charged)} &amp;amp; \ce{-NH3+}\text{ (positively charged)} &amp;amp; 0 \\&#xA;&amp;gt;9 &amp;amp; \ce{-COO^-}\text{ (negatively charged)} &amp;amp; \ce{-NH2}\text{ (neutral)} &amp;amp; -1 \\&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information you can read &lt;a href=&quot;http://en.wikipedia.org/wiki/Amino_acid#Zwitterions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; or any biochemistry textbook - the first few chapters will often be devoted to discussing chemistry like this.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="7475" LastEditDate="2019-10-23T10:03:41.507" LastActivityDate="2019-10-23T10:03:41.507" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32651" PostTypeId="1" CreationDate="2015-06-09T14:51:34.100" Score="1" ViewCount="33" Body="&lt;p&gt;I'm looking for some kind of email list, forum, or group that discusses chemistry demonstrations. I'm working on an improvement to a demo that's been used for a few decades at my institution and I'm looking for a other few people (that aren't at my institution) to try it out and give feedback. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="16814" LastActivityDate="2015-06-09T14:51:34.100" Title="Looking for a Chemistry Demo Resource Group" Tags="&lt;experimental-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32652" PostTypeId="1" CreationDate="2015-06-09T15:28:11.193" Score="4" ViewCount="158" Body="&lt;p&gt;I am visualizing some molecules with Molden. Is there a way to hide hydrogens, without deleting the hydrogen entries from the structure file?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="4945" LastEditDate="2017-11-20T14:34:25.693" LastActivityDate="2017-11-20T14:34:25.693" Title="How to temporarily hide hydrogen atoms in Molden?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32653" PostTypeId="2" ParentId="32633" CreationDate="2015-06-09T15:30:40.237" Score="10" Body="&lt;p&gt;There are &lt;strong&gt;3 main types of elimination reactions, the E1, E2 and E1cb&lt;/strong&gt;.  Here is a generalized picture of the transition state for the &lt;strong&gt;E2&lt;/strong&gt; elimination reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNC6X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking of the $\ce{C-H}$ bond and the $\ce{C-LeavingGroup}$ bond are roughly equally advanced, as is formation of the double bond. Because the 2 bonds are broken to roughly the same degree, it can be said that the E2 reaction is in the &quot;middle&quot; of the possible elimination mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we change this pattern and make the $\ce{C-H}$ bond breaking more advanced than the $\ce{C-LeavingGroup}$ bond breaking (we can do this by making the leaving group a &lt;strong&gt;poor leaving group&lt;/strong&gt; [&lt;em&gt;i.e.&lt;/em&gt; groups like ammonium, $\ce{-NR3^+}$, sulfonium, $\ce{-SR2^+}$, &lt;em&gt;etc&lt;/em&gt;.]), then we shift the mechanism towards the &lt;strong&gt;E1cb&lt;/strong&gt; side of the elimination spectrum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gq80J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;E1cb&lt;/strong&gt; case, because breaking of the $\ce{C-H}$ bond is advanced, partial &lt;strong&gt;negative&lt;/strong&gt; charge develops on the carbon where the proton is leaving from.  Now primary carbanions are more stable than secondary carbanions, which, in turn, are more stable than tertiary carbanions (electron releasing alkyl groups will destabilize a negative charge).  Therefore, when we have a choice of which proton to remove, we &lt;strong&gt;remove the proton on the least substituted carbon&lt;/strong&gt; as this carbon will better stabilize the developing negative charge.  As the LeavingGroup departs and the reaction completes, this will produce the olefin with the &lt;strong&gt;least substituted double bond&lt;/strong&gt;, the &lt;strong&gt;Hofmann elimination product&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes it is argued that base size is responsible for elimination patterns and that when a large base is used the proton on the least hindered carbon is abstracted.  This is true but the effect is relatively small.  Changing the base from methoxide to &lt;em&gt;t&lt;/em&gt;-butoxide only increases the amount of Hoffmann product by a modest 10-20% in simple haloalkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while sterics play a role, it is usually a small role with the electronic factors discussed above being more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, let me briefly mention the &lt;strong&gt;E1&lt;/strong&gt; elimination which is on the &quot;other side&quot; of the elimination spectrum of mechanisms. In this case, breaking of the $\ce{C-LeavingGroup}$ bond is advanced and partial &lt;strong&gt;positive&lt;/strong&gt; charge develops on the carbon where the LeavingGroup is departing from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3TYg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elimination of a proton from this intermediate can often proceed in two directions.  Since &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27346/why-is-2-methylpropene-less-in-energy-than-its-alkene-counterparts/27347#27347&quot;&gt;more highly substituted double bonds are more stable&lt;/a&gt; than less substituted double bonds and since the transition state for proton elimination involves partial double bond character, the transition state  leading to the more stable double bond (&lt;em&gt;e.g.&lt;/em&gt; the &lt;strong&gt;more highly substituted double bond&lt;/strong&gt;) will be lower in energy.  Hence, the product with the more highly substituted double bond (&lt;strong&gt;Zaitsev product&lt;/strong&gt;) will form in &lt;strong&gt;E1&lt;/strong&gt; reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: Response to OP's comment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I meant the pyrolytic elimination of quaternary ammonium hydroxide,&#xA;  which follows E2 mechanism&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wasn't saying that this reaction (the thermal Hofmann elimination) goes through an E1cb mechanism, rather I said &quot;we shift the mechanism &lt;strong&gt;towards&lt;/strong&gt; the E1cb side of the elimination spectrum.&quot;  In other words, it is an E2 reaction but $\ce{C-H}$ bond breaking is somewhat more advanced than $\ce{C-LeavingGroup}$ bond breaking. Therefore the primary $\ce{C-H}$ bond is more likely to be involved than the secondary or tertiary $\ce{C-H}$ bond, for the reasons outlined above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here's another approach&lt;/strong&gt;.  Instead of electronic factors, let's look at steric factors - in this reaction they both lead to the same conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E2 eliminations prefer a trans or antiperiplanar geometry.  Below are Newman projections of the 2 conformations with this geometrical arrangement of the substituents.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJu5u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conformation A is set up to eliminate towards the least substituted carbon, the methyl group (Hofmann), while conformation B depicts elimination towards the more highly substituted carbon (Zaitsev).  The trialkyammonium group is extremely large and generates an unfavorcbale steric interaction in conformation B.  This makes conformation B higher in energy, hence elimination from conformation A to produce the less-substituted double bond will be preferred.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-06-11T21:01:50.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32655" PostTypeId="2" ParentId="32640" CreationDate="2015-06-09T15:47:07.043" Score="47" Body="&lt;p&gt;First, a little bit of background. Transparency is not an absolute property of a material. &lt;em&gt;Every&lt;/em&gt; substance is opaque, so long as light has to pass through enough of it, and opacity also changes according to ambient conditions.  Some substances, such as most metals, are opaque even in $100\ \mathrm{nm}$ thin films, while many gasses will let a small amount of light through mostly unperturbed even after several kilometres. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Photon_Mean_Free_Path.png&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; is a measure of how far photons of different energies can travel on average in the pure elements before interacting once with their atoms (the graph is actually for high energy x-rays rather than visible wavelengths, but they're all photons nonetheless). It makes intuitive sense that a gas should let more light through, as the atoms in it are much more spaced than in a solid or liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two general processes that work to make materials non-transparent are &lt;em&gt;photon absorption&lt;/em&gt; and &lt;em&gt;photon scattering&lt;/em&gt; (reflection being included as an example of scattering). When we say a material is &quot;transparent&quot; &lt;em&gt;colloquially&lt;/em&gt;, what we generally mean is that the material, at the thickness observed, does not scatter much light and thus allows a coherent image to pass through the material (basically, you can see through it to the other side without a lot of distortion, that is, the object is &lt;strong&gt;&quot;see-through&quot;&lt;/strong&gt; or &lt;strong&gt;clear&lt;/strong&gt;). Interestingly, this means that &lt;em&gt;a clear material need not be colourless&lt;/em&gt;, as correctly pointed out by Geoff, because photons of a given colour may be absorbed while letting most other wavelengths pass through without scattering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gasses in general are highly transparent because they are both highly colourless (absorb little light) and very clear (scatter little light) unless you're looking through several kilometres of gas, as I mentioned previously. However, there &lt;em&gt;are&lt;/em&gt; examples of highly clear but coloured gasses, as it is possible for them to absorb a significant amount of photons of a certain energy due to low energy electronic transitions, while allowing the rest of the visible spectrum to pass through unimpeded. While some gasses are coloured and others are not, I believe all pure gasses must be very clear (small amounts of gas will scatter almost no light), and they can only scatter a fair amount of light across a short distance in conditions where they are probably not best described as gasses anymore (plasmas, supercritical fluids, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most classic example of a coloured gas is nitrogen dioxide, $\ce{NO2}$, a strongly brown-red noxious gas which is &lt;a href=&quot;https://www.youtube.com/watch?v=pJSQq494oV4&quot; rel=&quot;noreferrer&quot;&gt;easily formed by decomposition of nitric acid&lt;/a&gt;, among several other ways. The unpaired electron in its structure is somewhat uncommon, and it can be easily excited to a higher electronic state upon absorption of photons in the blue region of the visible spectrum. Since blue light is strongly absorbed even in relatively thin samples (a few centimetres of $\ce{NO2}$ gas), the rest of the white light passes through with essentially no scattering, forming a perfect image of the other side but with a strong red tinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more interestingly, $\ce{NO2}$ reacts with itself at low temperatures or high pressures and dimerizes to form &lt;a href=&quot;http://en.wikipedia.org/wiki/Dinitrogen_tetroxide&quot; rel=&quot;noreferrer&quot;&gt;dinitrogen tetroxide&lt;/a&gt;, $\ce{N2O4}$, a colourless solid/liquid/gas (depending on conditions). In other words, the two substances are in a reversible equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 NO2 (g) &amp;lt;=&amp;gt; N2O4 (g) + energy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coincidentally, this equilibrium is characterized by an equilibrium constant which is close to 1 at ambient conditions, and this constant can easily be changed by varying temperature or pressure to higher or lower values. This means that under relatively easily achieved conditions, it is possible to study the reversible change of a coloured gas into a colourless one! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRuW5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(&lt;a href=&quot;http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch15_s01&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though I focused on $\ce{NO2}$ because it is a remarkable case, there are several other examples of clear but coloured gasses. As Geoff mentioned, the halogens all form coloured gasses, though bromine is a volatile liquid and iodine is a solid at room temperature, so a small amount of heating is necessary. Ozone, $\ce{O3}$, is &lt;a href=&quot;https://i.stack.imgur.com/CfI3i.jpg&quot; rel=&quot;noreferrer&quot;&gt;somewhat blue&lt;/a&gt; (&lt;a href=&quot;http://www.masterorganicchemistry.com/2011/09/16/reagent-friday-ozone-o3/&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;), though its colour is best observed in the liquid phase. The same goes for oxygen gas, $\ce{O2}$, which is slightly blue, though it is not &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_scattering#Reason_for_the_blue_color_of_the_sky&quot; rel=&quot;noreferrer&quot;&gt;the cause for the blue sky&lt;/a&gt;. There are probably a few more examples out there.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-09T16:43:55.627" LastActivityDate="2015-06-09T16:43:55.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32656" PostTypeId="2" ParentId="32645" CreationDate="2015-06-09T16:00:24.040" Score="3" Body="&lt;p&gt;Look for the $pK_a/pK_b$ values for each amino acid. When the pH of the solution is lower than both $pK_a$s, the amino acid will be in its protonated, non-zwitterionic form. When the pH is greater than both $pK_a$s, the amino acid will be in its deprotonated, anionic form. When the pH is between the two $pK_a$s (around neutral pH), the amino acid will be in its zwitterionic form.&lt;/p&gt;&#xA;" OwnerUserId="16832" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T18:58:44.537" LastActivityDate="2015-06-09T18:58:44.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32657" PostTypeId="1" AcceptedAnswerId="32658" CreationDate="2015-06-09T16:40:22.233" Score="1" ViewCount="993" Body="&lt;p&gt;If 5ml alcohol is dissolved in 1 L of water ( the purpose is to highlight the low quantity of one of the liquids) , how is it possible for 5 ml to evenly dissolve in 1 L of water ? There is not enough of alcohol to sit between water molecules , isn't it ? So how is the mixture's proportion/composition is uniform ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there anything that I miss to understand in the term - &lt;strong&gt;composition&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16834" LastEditorUserId="10185" LastEditDate="2015-06-09T16:50:12.093" LastActivityDate="2015-06-09T17:05:13.540" Title="homogeneous mixture - Are they truly uniform in composition and proportion?" Tags="&lt;solutions&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32658" PostTypeId="2" ParentId="32657" CreationDate="2015-06-09T17:04:59.453" Score="1" Body="&lt;p&gt;Your question is one of my favorite question. The concept of microscopic homogeneity and macroscopic homogeneity can explain this. Like if you add sugar and sand together and mix them well, you can see a homogeneous mixture but if you take a random sample you will find more sugar than sand or vice versa and even if your sample size is so small that it contains only one particle than it might be either sand or sugar. So the consistency of your system depends on the sample size. It has statistical significance also. In statistics if you want to make an average of large pile of data (Whatever data you have), you have to make sure that the sample you are analyzing is a true representative of the whole pile. In statistical mechanics microscopic description is used to find out the macroscopic behavior and so this simple question has lot of potential. Now this very concept is also used to find the freezing point depression or colloidal behavior. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-06-09T17:04:59.453" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32659" PostTypeId="2" ParentId="32657" CreationDate="2015-06-09T17:05:13.540" Score="2" Body="&lt;p&gt;Maybe it is just a question of what you mean by &quot;evenly dissolved&quot;. True enough, in your example, the proportion of alcohol to water is not &quot;even&quot; - rather the solution is indeed mostly water containing only a little bit of alcohol. But for a true solution such as this, the alcohol is dispersed evenly (uniformly) throughout the bulk of the solution, so that if you repeatedly withdrew a sample and analyzed it, you would get the same result no matter which part of the liquid you sampled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would not be the case for oil and water - they would exist in separate phases and which phase you sampled would give a different analysis result.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-09T17:05:13.540" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32660" PostTypeId="1" AcceptedAnswerId="32664" CreationDate="2015-06-09T17:26:07.853" Score="24" ViewCount="1604" Body="&lt;p&gt;Why isn't the &lt;a href=&quot;http://www.rsc.org/chemistryworld/News/2010/January/29011001.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;silicon analogue of benzene&lt;/a&gt; flat? As far as I understand, silicon and carbon make similar structures and it doesn't make sense that its benzene analogue should be substantially different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKDvW.jpg&quot; alt=&quot;Resonance mechanisms&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoted from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemists in the UK have constructed a structural analogue of benzene made from silicon atoms. The molecule is not flat like benzene, but it reveals a new type of aromatic stabilisation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This 'new' type of aromatic stabilisation is said to occur only around the seat of the chair structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking that silicon's bonds are too weak to be kept in a double bond, and therefore it doesn't have enough energy to change the shape of the bonding and is thus quite unstable.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="7475" LastEditDate="2017-01-21T15:15:25.220" LastActivityDate="2017-01-21T15:15:25.220" Title="Why isn't the silicon analogue of benzene flat?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32661" PostTypeId="1" AcceptedAnswerId="32663" CreationDate="2015-06-09T17:26:35.400" Score="11" ViewCount="3043" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nceyz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have this spectrum and the molecular formula of $\ce{C10H12O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MKrXf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with the peak at ~7 which is an H-Benzene signal, with quartet splitting and a relative area of 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely this means that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The benzene would have only one Hydrogen environment &lt;/li&gt;&#xA;&lt;li&gt;with 4 Hydrogens in that environment &lt;/li&gt;&#xA;&lt;li&gt;and 3 hydrogens on the adjacent carbon atoms &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But the answer has 2 H-Benzene environments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come this goes against normal? Please can answers be not too technical as I am a pre-university student &lt;/p&gt;&#xA;" OwnerUserId="5619" LastEditorUserId="5619" LastEditDate="2015-06-09T17:34:34.973" LastActivityDate="2015-11-11T21:15:02.110" Title="Phenyl group in proton NMR" Tags="&lt;analytical-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32662" PostTypeId="2" ParentId="22355" CreationDate="2015-06-09T18:30:33.080" Score="2" Body="&lt;p&gt;With regard to numbering of locants, the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name, e.g. propanoic acid) has seniority over simple substituent groups (e.g. chloro). On this matter, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(c) principal characteristic groups and free valences (suffixes);&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule c takes precedence over Rule f.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the preferred IUPAC name for the given structure is 3,3-dichloropropanoic acid (not 1,1-dichloro-3-propanoic acid).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OdweQ.png&quot; alt=&quot;3,3-dichloropropanoic acid&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;As you have correctly assumed, additional chloro substituents increase the acidity of the carboxylic acid due to their electron-withdrawing inductive effect (−I effect):&lt;/p&gt;&#xA;&lt;p&gt;  acetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 4.756$&lt;/span&gt;&lt;br /&gt;&#xA;  2-chloroacetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 2.87$&lt;/span&gt;&lt;br /&gt;&#xA;  2,2-dichloroacetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 1.35$&lt;/span&gt;&lt;br /&gt;&#xA;  2,2,2-trichloroacetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 0.66$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The acidity of chlorocarboxylic acids is hardly directly affected by the length of the carbon chain:&lt;/p&gt;&#xA;&lt;p&gt;  2-chloroacetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 2.87$&lt;/span&gt;&lt;br /&gt;&#xA;  2-chloropropanoic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 2.83$&lt;/span&gt;&lt;br /&gt;&#xA;  2-chlorobutanoic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 2.86$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, the effect decreases as the distance between the chloro substituent and the carboxyl group increases:&lt;/p&gt;&#xA;&lt;p&gt;  2-chloroacetic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 2.87$&lt;/span&gt;&lt;br /&gt;&#xA;  3-chloropropanoic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 3.98$&lt;/span&gt;&lt;br /&gt;&#xA;  4-chlorobutanoic acid: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a} = 4.52$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; values taken from “Dissociation Constants of Organic Acids and Bases”, in &lt;em&gt;CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010),&lt;/em&gt; David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-06-09T18:30:33.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32663" PostTypeId="2" ParentId="32661" CreationDate="2015-06-09T18:49:44.340" Score="13" Body="&lt;p&gt;You asked to keep it simple, which is why I’m skipping explaining some details. Check the bottom of the answer to see the parts with superscript numbers explained.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The signal at $7\,\mathrm{ppm}$ is not a quartet. It would only be a (typical; there are exceptions, however not in typical organic chemistry) quartet if both of the following two conditions are met:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The signal ratio is $1\,:\,3\,:\,3\,:\,1$ and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The distance between all four signals is equal (they have the same coupling constant&lt;sup&gt;[1]&lt;/sup&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both reasons stem from the way quartets are created due to coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the signal is approximately $3.5\,:\,6.5\,:\,6.5\,:\,3.5$ (violating the first condition) and the distances are (enlarging the image on my screen and measuring with a ruler, so only a very rough approximation) $\mathrm{7\,mm, 12\,mm, 7\,mm}$ (violating the second condition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are in fact observing is a pair of doublets. (It could be a doublet of doublets&lt;sup&gt;[2]&lt;/sup&gt; as suggested by bon, but it is not.) You can think of it as two signals, each being a doublet, that are very close together. And that fits perfectly with the theory: Each signal would have an integral of $2$ and would correspond to one set of aromatic protons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The two on the aldehyde’s side at lower field (slightly higher shift)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The two on the ethyl group’s side at higher field (slightly lower shift).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are doublets because each one couples with one hydrogen of the other group. All that is left to explain is why one of the two signals of the doublet is higher than the other, creating a quartet-like effect. This is due to an effect called &lt;em&gt;roofing&lt;/em&gt; or &lt;em&gt;tenting.&lt;/em&gt; The bottom line about roofing is that it occurs if two coupling protons have close chemical shifts, and that once you have identified its existence, you can basically ignore it (it doesn’t give you any information that you couldn’t have derived otherwise).&lt;sup&gt;[3]&lt;/sup&gt; You can observe the same effect in the triplet at $\mathrm{1\,ppm}$ and the quartet at $\mathrm{2.5\,ppm}$: The right-hand side signals of the quartet are slightly stronger than those on the left-hand side and vice-versa for the triplet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] As you know, coupling stems from two protons being close, and connected &lt;em&gt;via&lt;/em&gt; a chain of bonds — usually three for hydrogens. The reason that nuclei couple is because they have a spin. For hydrogen (and carbon), the spin can be ‘up’ or ‘down’. Consider two protons next to each other: Their spins can point in the same or in the opposite direction. Both cases have a &lt;em&gt;slight&lt;/em&gt; difference in energy which gives rise to &lt;em&gt;slightly&lt;/em&gt; different chemical shifts — the coupling you observe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons that I won’t go into detail into here, this difference does not depend on the strength of your NMR magnet (while the difference between the signals of different protons does). The ppm-scale has been chosen to remove the differences between different proton’s signals (so $2\,\mathrm{ppm}$ will always be $2\,\mathrm{ppm}$) and therefore the coupling splits will vary with the strength of the NMR magnet, which is why they aren’t given in ppm, but rather in Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have multiple protons of the same type, the coupling value (one says: &lt;em&gt;coupling constant&lt;/em&gt;) will always be the same, so $7\,\mathrm{Hz}$ three times in the event of a true quartet. The reason why the signals then need to be $1\,:\,3\,:\,3\,:\,1$ lies within Pascal’s triangle: Starting from a central signal split it into two halfes with a certain distance from the centre: You will get two $1\,:\,1$ signals. Do the same thing with &lt;em&gt;each&lt;/em&gt; signal and the &lt;em&gt;same&lt;/em&gt; distance: You will get a $1\,:\,2\,:\,1$ signal, because the two central ones will overlap. Do the same again: Voilà, $1\,:\,3\,:\,3\,:\,1$. So &lt;strong&gt;all&lt;/strong&gt; true quartets that stem from this will obey the $1\,:\,3\,:\,3\,:\,1$ rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] If you have two &lt;em&gt;different&lt;/em&gt; protons on a neighbouring atom instead of two &lt;em&gt;identical&lt;/em&gt; ones, you will receive a doublet of doublets instead of a triplet. Think of it this way: One proton has a coupling constant of (say) $\mathrm{5\,Hz}$ with the proton of interest, and would give a doublet. The second one has a coupling constant of (say) $\mathrm{7\,Hz}$ with the same proton, and by itself would give a doublet again. Together, these two doublets would &lt;em&gt;not&lt;/em&gt; meet in the middle as in the above $1\,:\,2\,:\,1$ case, because they have different distances. Instead, you will see a $1\,:\,1\,:\,1\,:\,1$ set of four signals, where the distance between the first and the second corresponds to $\mathrm{5\,Hz}$ while the distance between the first and the third corresponds to $\mathrm{7\,Hz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn’t work here, because we are only coupling to one proton (and I would estimate a coupling constant of $\mathrm{19\,Hz}$ for the second coupling which is impossible in the structure given, and isn’t seen anywhere else in the spectra. Estimate was based on the aromatic coupling being $\mathrm{7\,Hz}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] Two spins pointing in the same direction are different in energy from two pointing in different directions I just pointed out. However, if the chemical shifts of both protons that couple are similar, the difference is also noticeable &lt;em&gt;without&lt;/em&gt; an external magnet, so that even &lt;em&gt;without&lt;/em&gt; the sample being in the NMR magnet the two energy levels would differ — hence the protons are slightly more aligned &lt;em&gt;per se&lt;/em&gt; than even distribution would suggest. This in turn means that one of the sides of the signal (corresponding to an energy difference) shows stronger absorption than the other, and because it is always the inner one, one can assume to see a roof — hence roofing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information that roofing can give you — that two protons are close by in the molecule and their shifts are similar — can be derived from other information in the spectrum as well (information that corresponds to &lt;em&gt;numbers!&lt;/em&gt; Real, mathematically physical hard numbers! Hence preferred by scientists): The proximity by the coupling constants and 2D-COSY and the similarity of the shifts … well, science doesn’t care too much.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-11-11T21:15:02.110" LastActivityDate="2015-11-11T21:15:02.110" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32664" PostTypeId="2" ParentId="32660" CreationDate="2015-06-09T18:51:11.520" Score="19" Body="&lt;p&gt;The linked article is a short summary of a &lt;em&gt;Science&lt;/em&gt; article: &lt;a href=&quot;http://dx.doi.org/10.1126/science.1181771&quot;&gt;&quot;A Tricyclic Aromatic Isomer of Hexasilabenzene&quot; by Kai Abersfelder, Andrew J. P. White, Henry S. Rzepa, and David Scheschkewitz; &lt;em&gt;Science&lt;/em&gt; &lt;strong&gt;2010&lt;/strong&gt;: 327 pp. 564-566&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They propose a new term, &quot;dismutational aromaticity&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The highly dispersed $\ce{^29Si}$ nuclear magnetic resonance shifts in solution ranging from +125 to −90 parts per million indicate an inhomogeneous electron distribution due to the dismutation of formal oxidation numbers as compared with that of benzene. Theoretical analysis reveals nonetheless the cyclic delocalization of six mobile electrons of the $\pi$-, $\sigma$- and non-bonding type across the central four-membered ring. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They point out that while $\ce{Si=Si}$ bonds were ignored for a long time, multiple compounds have been prepared, even including delocalization, like 1,2-disilabenzenes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They prepared $\ce{Si6R6}$ with &quot;Tip&quot; or 2,4,6-triisopropylphenyl groups.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The surprisingly stable 3a can be exposed to air for hours as a solid or for minutes in solution without detectable changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Long story short, there is delocalization around the central 4-center ring:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MP40u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; that there is a 1,2-shift occurring, and that 2 $\ce{Si}$ atoms have two substituents, 2 have one substituent, and 2 have no substituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper argues that a fully planar hexasilabenzene might be possible, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the choice of precursor may be important in any synthetic approach.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Presumably the choice of substituent is also relevant to prevent the rearrangement here.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-06-09T18:51:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32665" PostTypeId="2" ParentId="30864" CreationDate="2015-06-09T19:35:50.143" Score="4" Body="&lt;p&gt;As was mentioned in the comments, with reference to the &lt;a href=&quot;http://www2.onu.edu/~b-myers/organic/2511_Files/pKa%20table.pdf&quot; rel=&quot;nofollow&quot;&gt;Myers $\mathrm{p}K_{\mathrm{a}}$ table&lt;/a&gt;, the $\mathrm{p}K_{\mathrm{a}}$ values of a protonated ester and a protonated ketone are largely similar (in the region of $-7$ to $-6$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it will likely depend on external factors, which group is protonated first. Are there electron withdrawing or electron donating groups in the vicinity of either? Can the proton on either maybe intramolecularly hydrogen bond to something? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these effects will also slightly modify the $\mathrm{p}K_{\mathrm{a}}$ of the functional group you’re observing. The one with the higher $\mathrm{p}K_{\mathrm{a}}$ will be protonated first. Thus, it is impossible to give a general answer here, it depends on which molecule you have and which groups are where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side fact: $\ce{HCl}$ has almost the same $\mathrm{p}K_{\mathrm{a}}$, so the protonation will be very close to equilibrated between all three components.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T19:35:50.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32666" PostTypeId="6" CreationDate="2015-06-09T19:57:57.273" Score="0" Body="&lt;p&gt;santiago reporting in!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite new here (almost 2 thirds on my way to a Fanatic badge), but I learn fast and have a passion for Chemistry. I am the one editing a fair amount and trying to prod new members who post 'homework-dumps' to provide their attempts and, &lt;em&gt;hopefully&lt;/em&gt; get them to improve their questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that respect, I am closer to a Copy Editor badge, than Fanatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would potentially be a good moderator because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I enjoy 'meeting and greeting' new members and helping them when I can.&lt;/li&gt;&#xA;&lt;li&gt;Am happy to edit and fix the formatting of posts (even if it is an epic edit).&lt;/li&gt;&#xA;&lt;li&gt;Am not afraid to face up to anyone misbehaving, and will try to calmly guide them into behaving within the rules (I am not easily intimidated).&lt;/li&gt;&#xA;&lt;li&gt;Am perpetually in chat (making bad jokes a lot of the time).&lt;/li&gt;&#xA;&lt;li&gt;My time zone is UTC +10, so that could be an advantage.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not so good reasons would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lack of experience (timewise) and never have been a mod anywhere.&lt;/li&gt;&#xA;&lt;li&gt;My field of expertise is Instrumentation Science, not Chemistry per se.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In any case, I am happy to nominate, but on the same token, are happy who ever are voted for! In either case, I am here to stay.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T19:57:57.273" LastActivityDate="2015-06-09T19:57:57.273" CommentCount="5" CommunityOwnedDate="2015-06-09T19:57:57.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="32667" PostTypeId="2" ParentId="29249" CreationDate="2015-06-09T20:00:31.180" Score="2" Body="&lt;p&gt;Which orbital symmetries you can get in a molecule &lt;em&gt;always&lt;/em&gt; depends on the molecule’s point group. Only for those point groups which have degenerate irreducible representations can you get degenerate orbitals; so out of your examples only the $D_{4h}$ molecule could. This is entirely independent of Hückel’s method to determine energy levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just because your molecule has the irreducible representations, doesn’t mean that a $\pi$ orbital &lt;em&gt;has to&lt;/em&gt; have that degeneracy. Instead, it again depends on what you get when following symmetry rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate whether you will get $\pi$ degeneracies by drawing your molecule of interest, including all the p atomic orbitals and determining the molecule’s point group. Once you have that, for each symmetry operation, check how many p orbitals would be transformed onto themselves by said operation (and note whether with or without a phase change). Add that up. Once you’ve done that, you basically have the representation of the p-orbitals of the molecule that you can reduce to irreducible representations, finding out how many discrete and degenerate energy levels you have (albeit not including energy &lt;em&gt;values&lt;/em&gt;).&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T20:00:31.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32668" PostTypeId="2" ParentId="32635" CreationDate="2015-06-09T20:42:42.980" Score="5" Body="&lt;p&gt;Following up on my comment, even though I'm not convinced that this is a 'real' answer to the question.  Consulting the SGTE database of elemental free energies [&lt;a href=&quot;https://doi.org/10.1016/0364-5916(91)90030-N&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;] (with various updates over the years), one can obtain the melting temperatures for the hcp, fcc, and bcc phases of beryllium, magnesium, calcium, strontium, and barium (rough guesses looking at plots — consider good to approx. &lt;span class=&quot;math-container&quot;&gt;$\pm\pu{5 K}):$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{array}{lrrrcc}&#xA;\hline&#xA;\text{Element} &amp;amp; T_\mathrm{m}(\text{hcp})/\pu{K} &amp;amp; T_\mathrm{m}(\text{bcc})/\pu{K} &amp;amp; T_\mathrm{m}(\text{fcc})/\pu{K} &amp;amp; \text{Phase at RT} &amp;amp; \text{Melt phase} \\ &#xA;\hline&#xA;\ce{Be} &amp;amp; 1544 &amp;amp; 1560 &amp;amp; 1018 &amp;amp; \text{hcp} &amp;amp; \text{bcc} \\&#xA;\ce{Mg} &amp;amp; 922 &amp;amp; 755 &amp;amp; 705 &amp;amp; \text{hcp} &amp;amp; \text{hcp} \\&#xA;\ce{Ca} &amp;amp; 940 &amp;amp; 1114 &amp;amp; 1063 &amp;amp; \text{fcc} &amp;amp; \text{bcc} \\&#xA;\ce{Sr} &amp;amp; 900 &amp;amp; 1050 &amp;amp; 1030 &amp;amp; \text{fcc} &amp;amp; \text{bcc} \\&#xA;\ce{Ba} &amp;amp; 635 &amp;amp; 1000 &amp;amp; 709 &amp;amp; \text{bcc} &amp;amp; \text{bcc}\\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several things to note, amongst them the variety of which crystal structure is most stable for the different elements at different temperatures. Also, the hcp phase's melting point does almost(!) monotonically decrease going down the table.  The biggest anomaly would actually seem to be the relative &lt;em&gt;instability&lt;/em&gt; of the bcc and fcc phases of magnesium (it is the only one that remains hcp up to the melting point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, it would appear that the group 2 elements display a wide variety of preferences for crystal structure, and hence bonding energy configurations.  Is &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; unusual? Not really…&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dinsdale, A. T. SGTE Data for Pure Elements. &lt;em&gt;Calphad&lt;/em&gt; &lt;strong&gt;1991&lt;/strong&gt;, 15 (4), 317–425. DOI: &lt;a href=&quot;https://doi.org/10.1016/0364-5916(91)90030-N&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1016/0364-5916(91)90030-N&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="41328" LastEditDate="2020-04-05T10:35:40.797" LastActivityDate="2020-04-05T10:35:40.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32669" PostTypeId="2" ParentId="32573" CreationDate="2015-06-09T20:47:20.663" Score="1" Body="&lt;p&gt;Yes, an amino acid consists of a carboxyl group, commonly called &lt;em&gt;carboxylic acid&lt;/em&gt;, plus the amino group. And yes, a carboxyl group can act as an acid, i.e. release its proton to give a carboxylate. Conversely, an amino group can act as base, its free electron pair accepting a hydrogen to give an ammonium derivate. That is in fact what happens in aquaeous media or in the solid state of amino acids: the carboxylate protonates the amino group creating a &lt;em&gt;zwitterion&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these are only one type of reactions both groups are capable of performing. In the case of the amino group, as long as it is not protonated (i.e. still has an unbonded electron pair) it can attack nucleophilicly. Maybe you have heard of $\mathrm{S_N}2$ type reactions, that are rather similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nucleophilic attack of the nitrogen &lt;em&gt;could&lt;/em&gt; happen on the carboxylate’s proton — but then we wouldn’t be gaining anything, since it would just be the same acid-base reaction. (And this wouldn’t be considered a nucleophilic attack, usually.) Instead, the nitrogen needs to go to the &lt;em&gt;carbonyl group&lt;/em&gt; — which happens to be the most positive carbon in an amino acid, so should be a great target for a &lt;em&gt;nucleo&lt;/em&gt;phile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial result, as I tried to show in the scheme below, is a tetrahedral carbon with a negative charge on the ex-carbonyl oxygen and a positive charge on the nitrogen. If one of the nitrogen’s protons jumps across to the ex-hydroxy oxygen, that one gets positively charged, and then electron pairs move, we release water and gain our peptide bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/90ze7.png&quot; alt=&quot;Peptide bond formation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The picture is really only the basic mock-up and no real effort put into it. It neither exactly matches the biological peptide bond mechanism, nor does it match the mechanism that a chemist would use in the lab. But it conveys the basic picture.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while you did &lt;em&gt;technically&lt;/em&gt; use the basicity (or postivie-charge loving) of the nitrogen, you see that the acidity of the carboxyl group plays no role in the reaction itself.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T20:47:20.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32670" PostTypeId="2" ParentId="32389" CreationDate="2015-06-09T20:56:35.663" Score="2" Body="&lt;p&gt;This answer has technically been given in the comments, but it bears repeating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you can stop the reaction at the epoxide stage through careful choice of conditions. If you prevent both acidic and basic aquaeous environments, then epoxide rings are stable. They are present in a wide range of molecules and even in natural products such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Epothilone&quot; rel=&quot;nofollow&quot;&gt;epothilone&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also open epoxides with other nucleophiles to create different 1,2-substitution patterns — with predefined stereochemistry, if the epoxide was stereoselective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, there are even methods out there to turn vicinal diols into epoxides (usually via sulphonylation, e.g. to a tosylate followed by intramolecular $\mathrm{S_N}2$ reaction).&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T20:56:35.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32671" PostTypeId="1" CreationDate="2015-06-09T21:22:04.347" Score="5" ViewCount="142" Body="&lt;p&gt;Is there a compound that can melt under its own weight? That is, a block of such material remains solid when put horizontally but starts to melt when put vertically or remains solid in microgravity, but melts in normal gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is motivated by a documentary which I saw on TV some years ago where they claimed that a metal block (indium or its alloy) melts under its own weight. It is suspicious though because it was quite a crank documentary which featured alleged &quot;red mercury&quot; for instance.&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorDisplayName="user15489" LastEditDate="2015-06-09T21:28:12.057" LastActivityDate="2015-06-09T22:14:33.737" Title="Are there compounds that melt under their own weight?" Tags="&lt;metal&gt;&lt;melting-point&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32672" PostTypeId="2" ParentId="21855" CreationDate="2015-06-09T21:32:48.907" Score="3" Body="&lt;p&gt;As with every process, dissolving can be described by &lt;em&gt;the&lt;/em&gt; thermodynamical equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = \Delta H - T \Delta S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which the process being spontaneous if $\Delta G$ is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two components here, enthalpic and entropic. The enthalpic component $\Delta H$ is a constant of the salt in question. The entropic component $\Delta S$ can usually be considered positive because a highly ordered crystal lattice is broken to give dissolved, much less ordered fragments — hence why for most salts they dissolve easier in hot water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the solubility &lt;em&gt;decreases&lt;/em&gt; with higher temperatures, then $\Delta S$ must be negative. The salt will then only dissolve if $\Delta H$ is negative enough to give an overall negative result. This seems to be the case for $\ce{Na2SO4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible to predict $\Delta S$ and $\Delta H$ &lt;em&gt;a priori&lt;/em&gt; without any computer quantum calculations. One has to resort to experiments to determine whether a salt will dissolve better or worse in the heat.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While I have used the term ‘salt’ most of the time, the same is true for molecules. Take sucrose: I was asked in a physical chemistry lab to guess with my colleague whether sugar would dissolve endothermicly or exothermicly. We decided, that the degree of order be a lot higher in dissolved sugar due to a high amount of water surrounding it rather tightly, that the entropy must decrease, so that the enthalpy has to be negative and sugar has to dissolve exothermically. Hence sugar should dissolve more easily in cold water, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a comment by Nicolau Saker Neto, our reasoning was false and sucrose displays &lt;a href=&quot;http://chestofbooks.com/food/science/Experimental-Cookery/The-Solubility-Of-The-Sugars.html&quot; rel=&quot;nofollow&quot;&gt;increasing solubility at higher temperatures&lt;/a&gt;. Apparantly the testing faculty member approved of it, though.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-10T19:52:38.233" LastActivityDate="2015-06-10T19:52:38.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32673" PostTypeId="2" ParentId="31933" CreationDate="2015-06-09T21:49:28.557" Score="1" Body="&lt;p&gt;Think of it this way: A saturated solution is the equilibrium. Say you had a supersaturated solution, formed a crystal out of it and somehow managed to get your solution below saturation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But once that happened, your solution is no longer saturated and there is undissolved material on the ground (your crystal). So, always striving towards equilibrium, some of your crystal would get redissolved to re-reach the saturation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, crystallisation and equilibrium are not static but rather dynamic processes. So even if there is a crystal sitting at the bottom and it looks like nothing is happening, there is a constant flux of compounds redissolving from the crystal surface or recrystallising onto the surface out of the solution. Which is also why everything seems to strive towards the equilibrium: It is the state, where the speed of one direction matches that of the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So short answer: The solution above a crystal will be saturated, but no longer supersaturated.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T21:49:28.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32674" PostTypeId="2" ParentId="32671" CreationDate="2015-06-09T21:50:01.493" Score="10" Body="&lt;p&gt;Sure. Even discounting non-equilibrium conditions, any substance which expands upon freezing will by symmetry melt under pressure, even if the pressure is caused by supporting its own weight. There's a simple example: water! A sufficiently tall self-supporting block of ice standing on a flat surface will create enough pressure at its bottom that it could melt in temperatures between approximately &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure_melting_point&quot;&gt;$-21.9\ ^\circ \mathrm{C}$&lt;/a&gt; and $0\ ^\circ \mathrm{C}$, according to &lt;a href=&quot;http://www1.lsbu.ac.uk/water/water_phase_diagram.html&quot;&gt;the phase diagram for water&lt;/a&gt;. This has very important implications for the study of glaciers, for example. However, this process requires a very large amount of pressure, and thus a very tall block of ice. Using an approximate value for the freezing point depression under pressure of &lt;a href=&quot;http://arxiv.org/ftp/arxiv/papers/1501/1501.06981.pdf&quot;&gt;$0.0072\ ^\circ \mathrm{C\ atm^{-1}}$&lt;/a&gt;, it would take an ice block about $7.5\ \mathrm{km}$ tall for the base to melt at a temperature of $-5\ ^\circ \mathrm{C}$ or higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought indium was among the substances which expanded upon freezing, but it actually turns out to &lt;a href=&quot;http://www.indium.com/metals/indium/physical-constants/&quot;&gt;contract by 2.5% when frozen&lt;/a&gt;, so this mechanism would not work (though there could be something else to consider).  Some other materials which could theoretically melt under their own weight in some regime of temperatures and pressures are gallium, bismuth and silicon. There might be some very specific conditions under which this effect could be achieved in a column of around $20\ \mathrm{cm}$, though pressure usually only slightly alters the freezing point, so it's likely that the material would have to be kept only slightly under its melting point at standard pressure for this to happen.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-09T22:14:33.737" LastActivityDate="2015-06-09T22:14:33.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32676" PostTypeId="2" ParentId="27697" CreationDate="2015-06-09T22:08:48.283" Score="1" Body="&lt;p&gt;The hydrogens in the structure you drew definitely should not be in the same environment — so while overlapping or close signals would be okay, the way your signals look isn’t. To get two signals to be identical, there needs to be some symmetrical way of corresponding the hydrogens to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can confirm a &lt;em&gt;trans&lt;/em&gt;-configured double bond (I assume you can, too?) and two disubstituted aromatic rings. The question remains which type of substitution they have (&lt;em&gt;ortho, meta&lt;/em&gt; or &lt;em&gt;para&lt;/em&gt;). If you expect two rings with the substituents you drew, remember they must be pretty different in their chemical shifts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;em&gt;ortho&lt;/em&gt;, as you drew, one would indeed expect a set of doublet, triplet, triplet, doublet &lt;em&gt;for each ring&lt;/em&gt;. &lt;em&gt;meta&lt;/em&gt; would expect one set of doublet, triplet, doublet and a singlet somewhere. &lt;em&gt;para&lt;/em&gt;, finally, would expect two doublets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that triplets &lt;em&gt;must&lt;/em&gt; have a $1\,:\,2\,:\,1$ ratio, or they aren’t triplets. With this information in hand, go back into the aromatic region and think what you could have gotten. I have dropped enough clues, I hope.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T22:08:48.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32677" PostTypeId="2" ParentId="28864" CreationDate="2015-06-09T22:33:11.923" Score="2" Body="&lt;p&gt;Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go stealing pictures and explanations from &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%e2%82%85/18544#18544&quot;&gt;ron’s great answer&lt;/a&gt;, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%e2%82%85/18544#18544&quot;&gt;ron’s answer&lt;/a&gt;. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-09T22:33:11.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32678" PostTypeId="2" ParentId="28314" CreationDate="2015-06-09T22:42:53.180" Score="7" Body="&lt;p&gt;The first thing to note about the charges that form during the formation of a coordinate bond: They are a formality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing to note about the charges that form during the formation of a coordinate bond: &lt;strong&gt;They are a mere formality.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is little correlation, if any at all, between the formal charges assigned to two atoms that have formed a coordinative bond and the actual charges observed in the compound. Take $\ce{H3N-BH3}$: Nitrogen is so much more electronegative than boron, that the electrons are still closer to the nitrogen. Thus, nitrogen has $\delta ^-$ while boron has $\delta ^+$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where does the formalism come from? When drawing Lewis structures, after having filled in all bonds and lone pairs, check every single atom and mentally cleave all its bonds homolytically. Count the electrons, subtract the amount it should have as a neutral atom and the difference is its formal charge. Boron would have four electrons if all four bonds be cleaved homolytically, but it should only have three, so it gets a $1-$ charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is only a formalism&lt;/strong&gt; so that you can basically derive all structures the same way. It has no physical significance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are enough people who were unhappy about the ‘wrong’ charge distributions that they proposed different methods to draw these compounds: some used arrows from donor to acceptor, others just drew a simple bond, knowing that it would be different from a covalent bond. Each method has its advantages and disadvantages and all have survived in at least one niche.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T22:42:53.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32679" PostTypeId="2" ParentId="23650" CreationDate="2015-06-09T22:57:46.957" Score="11" Body="&lt;p&gt;From what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is pretty simple: Pyrrole is much like the cyclopentadienyl anion except that one carbon atom is replaced by nitrogen. If it wants to coordinate in any way, it will have to do that like a slice of bread from its side, forming a $\eta ^5$ complex. Much like ferrocene in the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/i9wjHs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pyridine, on the other hand, has no hydrogen on the nitrogen atom; instead it has a lone pair pointing outwards, but still in the plane of the ring. This allows pyridine to donate two electrons in a $\sigma$-like fashion to a metal centre, $\eta ^1$-style. Of course, pyridine could also perform a sandwich-like coordination much like benzene sometimes does. However, the direct two-electron $\sigma$ donation will be much stronger and therefore largely outweigh a sandwich-type donation in most compounds.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="4231" LastEditDate="2015-06-09T23:25:39.027" LastActivityDate="2015-06-09T23:25:39.027" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32680" PostTypeId="2" ParentId="20469" CreationDate="2015-06-09T23:08:26.117" Score="13" Body="&lt;p&gt;Starch is what plants primarily use as a glucose storage. As such, it is essential that they can break it back down into glucose otherwise it would defeat its own purpose and be removed by evolution very quickly. Think about it: Requiring a lot of energy to build up something that serves no further purpose — not exactly giving a plant an advantage in survival.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different methods how to break it down. Plants, of course, use enzymes. Humans can also break down starch with the help of enzymes (which is why potatoes give us energy), much like most other animals, fungi and many single-cell organisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first amylose-breaking enzyme is present in saliva, so it’s that early that the breakdown begins. However, distinct glucose usually isn’t liberated until the gut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemically, the easiest method would be acidic hydrolysis, i.e. use hydrochloric acid and heat. The downside is, that this method might destroy the individual glucose units, too; so using enzymes is preferred if you want glucose.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-09T23:08:26.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32681" PostTypeId="2" ParentId="19394" CreationDate="2015-06-09T23:21:22.270" Score="4" Body="&lt;p&gt;The answer depends on which teaching you follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional teaching, dating back to Linus Pauling, would prefer three double bonds to minimise charge separation, requiring an expanded octet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even older traditional teaching would favour the validity of the octet rule and take charge separation as a tolerable by-product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sides have their advantage and disadvantage:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Pauling method&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;correctly shows that all three bonds are equal without resorting to mesomery&lt;/li&gt;&#xA;&lt;li&gt;incorrectly assumes a bond order of $2$ rather than the actual $1.4$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The charge-separation method&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;predicts a bond order of $1.33$ which is much closer to $1.4$&lt;/li&gt;&#xA;&lt;li&gt;requires mesomery of three structures to explain the three equal bond lengths.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I for my part prefer the mesomeric charge-separation description, because we have no trouble drawing benzene in the same way and because I like the octet rule. Others will disagree and prefer the three double bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn’t know anybody who prefers a description with two double bonds and one single bond. It is just between the two models making it all the less correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/14232&quot;&gt;Please also see this excellent answer to a related question by Martin which includes molecular orbitals for $\ce{SO3}$&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-06-09T23:21:22.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32682" PostTypeId="6" CreationDate="2015-06-09T23:31:27.623" Score="0" Body="&lt;p&gt;Greetings All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My name is Jonathan.  I have been one of the pro-tempore mods on the site during the beta and have been a member from the beginning.  Many of you already know that, but just in the interest of any newcomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a user, I have been active on many of the science-oriented sites for a few years now, and I understand the nuances that make a science site different from one that is chiefly focused on technology.  I feel passionate about striking a healthy balance between research-level questions and well-asked layperson/student ones.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been moderating with SE for about 3.5 years now, as a pro tempore on Healthcare IT and Libraries and as an elected moderator on Web Applications (12/13 to present).  I have a passion for keeping the sites clean and helping direct the curation of the valuable information that SE is assembling for present askers and future learners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of my chemistry education took place in the last century, but I still have a passion for the material.  I believe I know enough to make decisions regarding whether questions are on-topic, and certainly know to defer to the community as needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all for your consideration!&lt;/p&gt;&#xA;" OwnerUserId="73" LastEditorUserId="73" LastEditDate="2015-06-09T23:31:27.623" LastActivityDate="2015-06-09T23:31:27.623" CommentCount="7" CommunityOwnedDate="2015-06-09T23:31:27.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="32685" PostTypeId="2" ParentId="29587" CreationDate="2015-06-10T04:29:25.800" Score="6" Body="&lt;p&gt;Quantum mechanics tells us the possible stationary states of any system with a fixed energy. It would be extremely computationally difficult to use quantum mechanics to find the stationary state of a system as large as a solution but in theory this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical mechanics tells us the probability that the system is in any one of the possible stationary states for some finite temperature. In particular, stat. mech. tells us which of the quantum mechanical stationary states are most likely to be occupied at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Le Chatelier's principle tells us how to find the new equilibrium when a system is displaced from equilibrium. In other words, Le Chatelier's principle tells us which distribution of quantum mechanical stationary states the system returns to after the system is disturbed in some way. Quantum mechanics tells us the possible states of the system, but it can tell us nothing about how those states are occupied. Only statistical mechanics can tell us how those states are occupied. Therefore, quantum mechanics alone cannot explain Le Chatelier's principle; in fact, Le Chatelier's principle holds regardless of the specific interactions between molecules and does not depend on any quantum effects at all. Le Chatelier's principle would hold even if we modeled the interactions between molecules completely classically with a Lennard-Jones or Morse potential.&lt;/p&gt;&#xA;" OwnerUserId="16830" LastActivityDate="2015-06-10T04:29:25.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32686" PostTypeId="1" CreationDate="2015-06-10T04:56:22.017" Score="8" ViewCount="1779" Body="&lt;p&gt;Already I know that hydrogen, all the halogens, nitrogen and oxygen forms diatomic molecules. But I am confused about $\ce{Na}$? So I would like to know about that.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T04:59:12.533" LastActivityDate="2019-04-28T05:49:31.947" Title="Is it possible to have a diatomic molecule of sodium in gaseous state?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32687" PostTypeId="1" CreationDate="2015-06-10T05:33:28.143" Score="-1" ViewCount="1391" Body="&lt;p&gt;I am currently learning chemical bonding, specifically Lewis structures. And these nitrogen oxides, nitrogen-oxygen ions and other atoms showing variable oxidation states are confusing me. As many examples I have understood with bonding atoms having no $d$ or $f$ orbitals, the variable oxidation states are a result of dative bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when drawing the Lewis structure for $\ce{NO3-}$, I drew:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5rIG.png&quot; alt=&quot;Nitrate ion - my structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book however has only one dative bond, and has this structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cBnnG.png&quot; alt=&quot;Nitrate ion - book&amp;#39;s structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, both of them look correct. The octets seem to be satisfied in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are both of them correct? If yes, how?&lt;/li&gt;&#xA;&lt;li&gt;If my structure is incorrect, what rule am I violating? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9363" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T06:59:11.487" LastActivityDate="2015-06-10T20:44:43.817" Title="Are more than one correct Lewis structures possible (not counting resonance)?" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="32689" PostTypeId="6" CreationDate="2015-06-10T06:21:56.570" Score="0" Body="&lt;p&gt;Hello! I am Matthew, but on Chem.SE I go by &lt;strong&gt;ringo&lt;/strong&gt;! I would be honored to moderate this site, and as such, this is my nomination. It is important to know a candidate's strengths and weaknesses, so I'll be outright with mine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strengths&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am dedicated – though I have not been answering and asking questions recently, I have been checking in daily, performing review tasks, and continuing to contribute in other ways; also, I have been studying organic chemistry in my free time so that I can contribute more knowledge&lt;/li&gt;&#xA;&lt;li&gt;I am knowledgeable – with a background in chemistry, I can understand the questions and answers on the site and how they should be approached&lt;/li&gt;&#xA;&lt;li&gt;I love order – being elected a moderator would allow me to assist in keeping this community running as smoothly as it has been, as well as to make improvements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weaknesses&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I am inexperienced as a moderator – there would most certainly be things that I would need to learn if elected, but I love learning!&lt;/li&gt;&#xA;&lt;li&gt;I am relatively new – many of the incredible people on Chem.SE have been active longer than I have and are more knowledgeable, but I am working hard to catch up!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you for considering me as a candidate!&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-06-10T06:21:56.570" LastActivityDate="2015-06-10T06:21:56.570" CommentCount="2" CommunityOwnedDate="2015-06-10T06:21:56.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="32690" PostTypeId="1" CreationDate="2015-06-10T08:33:39.710" Score="5" ViewCount="280" Body="&lt;p&gt;I want to reduce vanadyl to vanadium (III), can somebody please suggest some appropriate methods for it? I am trying to make a redox flow battery for a project but I have only vanadyl sulfate. &lt;/p&gt;&#xA;" OwnerUserId="16845" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T08:36:40.463" LastActivityDate="2015-06-10T08:36:40.463" Title="Reduction of vanadyl ion" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;experimental-chemistry&gt;&lt;electrolysis&gt;&lt;reduction-potential&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32691" PostTypeId="2" ParentId="32686" CreationDate="2015-06-10T09:39:05.977" Score="8" Body="&lt;p&gt;According to molecular orbital theory, disodium should be stable in the gas phase, with a bond order of one. The molecular orbital diagram is the same for all the alkali metals since they all have one valence electron in an $s$ orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4JRk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s13-molecular-geometry-and-covalen.html#averill_1.0-ch09_s03_s02_f02&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIST chemistry webbook has a small page on &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C25681792&amp;amp;Units=SI&amp;amp;Mask=1#Thermo-Gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;disodium&lt;/a&gt; and quotes an enthalpy change of formation of $142.07~\mathrm{kJ~mol^{-1}}$ which is moderately endothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s13-molecular-geometry-and-covalen.html#averill_1.0-ch09_s03_s02_f02&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; page has a brief discussion of alkali metal dimers and claims that they are all stable in the gas phase, in agreement with the prediction from MO theory.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-06-10T15:07:22.740" LastActivityDate="2015-06-10T15:07:22.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32692" PostTypeId="6" CreationDate="2015-06-10T11:23:15.867" Score="0" Body="&lt;p&gt;Hello,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my name is Martin and I joined this site about a year ago. Since my identity is no secret for those willing to find out, I will keep it short. I hold a doctorate degree with most of my experience in computational chemistry. Currently Japan is my home, where working as a post-doctoral fellow gives me the liberty to partake on this site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First and foremost, I love this site and made a lot of friends here. I care about this site, especially keeping the quality high and the noise low. I dedicated a fair share of time to that endeavour and I will continue doing so. Occasionally I contribute some answers - the main purpose of this site - and I plan on continuing so. As a user I have also been benefiting from great advice here, which tells me that the system works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this, I consider myself to be a regular in chat - a place where almost nothing is off-topic. If you have any questions, I welcome you to ask them there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never been in a moderation position before, so I am not quite sure what to expect, but I believe I am able to learn an grow with the job. Although no one ever really can say what the future holds, I am here to stay. Thank you for your consideration.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-06-10T11:23:15.867" LastActivityDate="2015-06-10T11:23:15.867" CommentCount="10" CommunityOwnedDate="2015-06-10T11:23:15.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="32693" PostTypeId="1" AcceptedAnswerId="32694" CreationDate="2015-06-10T11:23:51.023" Score="3" ViewCount="254" Body="&lt;p&gt;Is a racemic mixture always an equi-molar solution of two enantiomers, both in the same state? Is it necessary for both isomers to be in the liquid state for racemisation to occur, or does racemisation also occur in the gaseous and solid states?&#xA;Thirdly,can the two enantiomers still form a racemic mixture if they are initially at different temperatures?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="10185" LastEditDate="2015-06-18T16:39:17.063" LastActivityDate="2015-06-18T16:39:17.063" Title="State of a Racemic Mixture" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32694" PostTypeId="2" ParentId="32693" CreationDate="2015-06-10T12:20:31.820" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is a Racemic Mixture always an equi-molar solution of two enantiomers, both in the same state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and yes. A &lt;em&gt;racemic&lt;/em&gt; mixture is equimolar in both enantiomers. Otherwise you would call the reaction enantioselective with a so-and-so selectivity.&#xA;A racemic mixture contains both enantiomers in the same state, as energy differences between enantiomers are so minuscule that you would never get one solid and one liquid at the same time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it necessary for both the mixture to be liquid in nature, or does racemisation also occur in the gaseous or solid state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Racemisation is the process where you start out with a pure enantiomer and in the end get a mixture. For &lt;em&gt;complete&lt;/em&gt; racemisation you end up with an equimolar mixture as discussed above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is thus possible that racemisation occurs in gaseous and solid phases, not only in solution or liquid. However, the reaction times will differ significantly. I could imagine that the solid racemisation reaction will take a long time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the two enantiomers still form a Racemic mixture if they are initially at different temperatures?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Higher temperatures just mean that the racemisation process proceeds faster. In the end, the warmer enantiomer will have racemised further on than the colder one, but for $t\to\infty$ the result will be the same.&lt;/p&gt;&#xA;" OwnerUserId="3846" LastActivityDate="2015-06-10T12:20:31.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32695" PostTypeId="1" CreationDate="2015-06-10T12:27:39.310" Score="4" ViewCount="1522" Body="&lt;p&gt;Chlorine can be $sp^3d^3$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, it can form $\ce{ClH_7}$ and then chlorine, being more electronegative, will have (-7) as its oxidation number. But we know that the oxidation number of $\ce{Cl}$ lies between (-1) and (+7).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="15979" LastEditorUserId="15979" LastEditDate="2015-06-11T05:04:16.767" LastActivityDate="2015-06-11T05:04:16.767" Title="Is ClH7 possible?" Tags="&lt;inorganic-chemistry&gt;&lt;hybridization&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ClosedDate="2015-06-24T18:22:45.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="32696" PostTypeId="2" ParentId="32530" CreationDate="2015-06-10T12:43:45.430" Score="6" Body="&lt;p&gt;There are no general conditions- indeed there no general conditions for the any of the forms of corrosion. Solubility is a factor and many other factors but it just comes down the actual interactions between a given liquid metal and another metal and the service conditions such as temperature, velocity, impurities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid-metal attack has the following categories:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;simple solution,&lt;/li&gt;&#xA;&lt;li&gt;alloying between liquid metal and solid metal,&lt;/li&gt;&#xA;&lt;li&gt;intergranular penetration,&lt;/li&gt;&#xA;&lt;li&gt;impurity reactions, &lt;/li&gt;&#xA;&lt;li&gt;temperature-gradient mass transfer,&lt;/li&gt;&#xA;&lt;li&gt;concentration-gradient mass transfer, or dissimilar-metal&#xA;mass transfer. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With that many means- determining what liquid metal damages what solid metal  is very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.ornl.gov/info/reports/1956/3445603500959.pdf&quot; rel=&quot;nofollow&quot;&gt;http://web.ornl.gov/info/reports/1956/3445603500959.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2581" LastEditorUserId="5017" LastEditDate="2015-06-12T16:02:23.793" LastActivityDate="2015-06-12T16:02:23.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32697" PostTypeId="2" ParentId="32695" CreationDate="2015-06-10T13:16:11.990" Score="1" Body="&lt;p&gt;First of all, $d$ orbitals play no part in hypervalency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at existing hypervalent molecules, you'll see that the central atom is always less electronegative. As H is less electronegative than $\ce{Cl}$, $\ce{ClH7}$ will be very unstable.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastEditorDisplayName="user15489" LastEditDate="2015-06-10T18:38:37.537" LastActivityDate="2015-06-10T18:38:37.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32698" PostTypeId="1" AcceptedAnswerId="32709" CreationDate="2015-06-10T13:52:49.957" Score="6" ViewCount="2396" Body="&lt;p&gt;As a follow up to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32693/state-of-a-racemic-mixture&quot;&gt;this&lt;/a&gt; question I started thinking about another question. During racemization, enantiomers are mixed in a equal ratios to form an optically inactive mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of reactions can take place between the enantiomers and how would this affect the mixture, if at all?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2020-09-07T18:57:06.193" Title="Reactions between enantiomers in a racemic mixture" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;&lt;mixtures&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32699" PostTypeId="1" AcceptedAnswerId="32711" CreationDate="2015-06-10T14:07:00.260" Score="7" ViewCount="92691" Body="&lt;p&gt;I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.&lt;/p&gt;&#xA;" OwnerUserId="16852" LastActivityDate="2015-06-10T17:22:38.883" Title="After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this?" Tags="&lt;stoichiometry&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="0" ClosedDate="2015-06-10T19:16:36.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="32700" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T14:12:12.987" Score="10" Body="&lt;p&gt;Below is a basic example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If percent yield is based on a mass, then a &gt;100% yield can result from a product contains materials that should not be there.  For instance, a product that should be anhydrous may not be completely dry and water would add mass to the product increasing your yield.  Other present contaminants can also do this and can even appear from a product that reacted with the surrounding atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="14" LastActivityDate="2015-06-10T14:12:12.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32701" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T14:27:22.037" Score="3" Body="&lt;p&gt;Yields above 100% are obviously impossible, so either you used more educts than you thought, or you don't have as much product as you think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An error when measuring the amount of educts for the reaction is one possible reason. A miscalculation at any point of the amounts needed or the final yield is also something you might want to check again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really common reason is the presence of other material in your product, often solvents like water are still present if you weren't careful enough when drying. This is really the most likely cause, especially in student labs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should think about the analytics you performed, and check if they show the presence of anything aside from your product. And then think about what kinds of substances could be in there (solvents, educts, other reaction products) that wouldn't be visible in the kind of analytics you performed. &lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-06-10T14:27:22.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32702" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T14:32:28.187" Score="3" Body="&lt;p&gt;Percentage yield exceeds 100% if impurities are present due to inadequate purification. Such impurities could be present for many reasons, such as if the vessels used to collect the product are contaminated. Also, it is possible that the product absorbs components from air like water vapor or carbon dioxide.&lt;/p&gt;&#xA;" OwnerUserId="16804" LastEditorUserId="15410" LastEditDate="2015-06-10T14:59:35.167" LastActivityDate="2015-06-10T14:59:35.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32703" PostTypeId="2" ParentId="32554" CreationDate="2015-06-10T15:04:54.490" Score="4" Body="&lt;p&gt;According to the &lt;a href=&quot;http://dx.doi.org/10.1351/pac200880020277&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Graphical Representation Standards for Chemical Structure Diagrams&lt;/em&gt; (IUPAC Recommendations 2008)&lt;/a&gt;, an ion may be best represented by its molecular formula when exact connectivity is not important. The formula should be placed in square brackets, and the charge should be written as a superscript following the rightmost bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[CuCl4]^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is important to depict atoms and bonds explicitly, the ion should be drawn in full. That structure should be enclosed in large square brackets, and the charge should be reported as a superscript following the rightmost bracket. (For single fragments, it is permissible to depict simply a corner at the top right of the diagram, with the charge as a superscript following the corner.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonds representing coordination from one atom to a single other atom should be represented as normal plain single bonds.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xahhl.png&quot; alt=&quot;preferred&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of dashed bonds or dative bonds to represent coordination is not acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OdBqg.png&quot; alt=&quot;not acceptable&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plain bonds should be used for nonstereogenic atoms, even if the substituents are not physically coplanar. If a coordination bond is attached to an atom that does have a specified stereochemical configuration, then a hashed or wedged bond depiction should be used instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure should be drawn without concern for delocalized charges. Other structural depictions that involve localized charges should be avoided, even if that produces a diagram where some atoms appear to have non-standard valences. It is not acceptable to add charges that depict formal zwitterions.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-06-10T15:04:54.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32704" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T15:27:18.157" Score="4" Body="&lt;p&gt;Imagine a reaction where your starting material is cleaved and the product has a lower molecular weight. If the conversion is incomplete and the product contaminated with some starting material, the resulting weight may suggest a yield higher than 100%.&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-06-10T15:27:18.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32705" PostTypeId="1" CreationDate="2015-06-10T15:41:59.820" Score="14" ViewCount="5330" Body="&lt;p&gt;Which elements can be diatomic and why?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen, Nitrogen, Oxygen and the Halogens tend to be thermodynamically stable as a diatomic molecule at room temperature, and are usually labelled as diatomic nonmetals. (Although I am unsure about Astatine). These are the ones you learn in school and I am fine with their reasoning. I want to know about other elements in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the rest of the non-metals: &lt;a href=&quot;http://en.wikipedia.org/wiki/Diatomic_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbon&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Diphosphorus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phosphorus&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Disulfur&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sulfur&lt;/a&gt; have diatomic molecules in gaseous states. All with a common theme of being heated at very high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with diatomic Selenium which I could not find anything on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diatomicity doesn't seem to be just a non-metal thing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lithium forms diatomic molecules and &lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s13-molecular-geometry-and-covalen.html#averill_1.0-ch09_s03_s02_f02&quot; rel=&quot;nofollow noreferrer&quot;&gt;all diatomic alkali metals are claimed to be stable in gaseous states&lt;/a&gt;. Found from &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32686/is-it-possible-to-have-a-diatomic-molecule-of-sodium-in-gaseous-state&quot;&gt;this question&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Silicon &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cen-v086n034.p010&quot; rel=&quot;nofollow noreferrer&quot;&gt;has been reported&lt;/a&gt; to have a diatomic molecule similar in structure to carbon.&lt;/li&gt;&#xA;&lt;li&gt;Tungsten, Chromium, Uranium and Molybdenum are &lt;a href=&quot;http://en.wikipedia.org/wiki/Sextuple_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;claimed to be diatomic&lt;/a&gt; with a sextuple bond.&lt;/li&gt;&#xA;&lt;li&gt;Supposedly &lt;a href=&quot;https://chemistry.stackexchange.com/questions/518/do-noble-gasses-besides-helium-form-diatomic-molecules-at-low-temperatures&quot;&gt;Noble gases can form dimer molecules&lt;/a&gt; but if these are considered diatomic is debatable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2558" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-01-14T00:33:44.833" Title="Which elements can be diatomic?" Tags="&lt;bond&gt;&lt;atoms&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32706" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T15:53:38.487" Score="1" Body="&lt;p&gt;Like others said, contaminants may be there or unreacted chemicals can be there. It happened to one or two times while I was doing summer project and left the compound in vaccumm and a trap to get rid of solvents. You should use proper solvents to separate the product and unreacted or side product stuff.&lt;/p&gt;&#xA;" OwnerUserId="14874" LastActivityDate="2015-06-10T15:53:38.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32707" PostTypeId="2" ParentId="32698" CreationDate="2015-06-10T15:56:58.727" Score="5" Body="&lt;p&gt;This is a difficult question to answer without having specific compounds to work with, but let's say that you have a pair of enantiomers with a single stereocenter, so there's an (R)-enantiomer and an (s)-enantiomer. There are four possible reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R + R -&amp;gt; RR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{R + S -&amp;gt; RS}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S + R -&amp;gt; SR}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{S + S -&amp;gt; SS}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set of products are four stereoisomers of which there are two pairs of enantiomers: (RR)/(SS) and (RS)/(SR). [Note that I'm assuming that the chemical reaction renders the (RS)/(SR) compounds asymmetrical. If they were symmetrical, then these would be achiral, meso compounds.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where this could get more interesting is that the rates of those reactions are not necessarily the same, since we have two chiral reactants interacting. Consider the transition states leading to RR and RS. Those transition states diastereomeric, and thus require different activation energies. This shouldn't be surprising since RR and RS are diastereomers, which have different energies. The reaction with the lower activation energy is called the matched case, and the reaction with the higher activation energy is called the mismatched case. I'm not aware of any chemistry where the possible starting materials are all stereoisomers of each other, but this phenomenon can be important for setting new stereocenters in compounds that already have stereocenters. &lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-06-10T15:56:58.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32708" PostTypeId="2" ParentId="29675" CreationDate="2015-06-10T15:58:04.583" Score="8" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.osti.gov/scitech/servlets/purl/4971&quot;&gt;Decomposition of Sodium Tetraphenylborate&lt;/a&gt;, which relates to the &lt;a href=&quot;http://www.srs.gov/general/about/history1.htm&quot;&gt;Savanna River site&lt;/a&gt; for nuclear weapons material production,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stable aqueous NaTPB solutions are a transparent,&#xA;  flesh-toned color. During the initial stages of decomposition (or&#xA;  instability),the solution turns a golden color. As the quantity&#xA;  of TPB- that has decomposed increases, the solution turns dark&#xA;  red, then brown, and eventually black.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the presence of oxygen &quot;phenol and phenylboronic acid&#xA;are produced as well as benzene, biphenyl, and other benzene ring&#xA;derivatives&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-06-10T15:58:04.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32709" PostTypeId="2" ParentId="32698" CreationDate="2015-06-10T16:16:06.700" Score="6" Body="&lt;p&gt;If a chiral molecule (an enantiomer) can react with itself, dimerize for example, to form a new compound, then formation of this new compound would contaminate the racemic mixture.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tX8sv.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/enrgtop.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Perhaps more importantly&lt;/strong&gt; is interaction of the racemic mixture with itself in the solid state.  Just as the interaction between a right hand and left hand is different than the interaction between two right hands, the interaction between the R form and S form of the enantiomer will be different than the interaction between two R (or two S) forms.&lt;/p&gt;&#xA;&lt;p&gt;In liquids and gases, molecular encounters are fleeting and cannot be preserved or &amp;quot;locked in&amp;quot; for significant periods of time as they can be in a solid. In other words, any favorable, lower energy molecular encounters are quickly broken apart by the energetically larger effects of thermal motion.  However, when the racemic mixture is in a solid phase, if the interaction between R and R (or S and S) is lower in energy then the interaction between R and S, then the R and S enantiomers will tend to form separate crystals and a &lt;strong&gt;mixture&lt;/strong&gt; of pure R and pure S crystals will result.  Such mixtures are often termed &amp;quot;&lt;strong&gt;conglomerates&lt;/strong&gt;&amp;quot;.  On the other hand if the interaction between R and S is more favorable (lower in energy), then a &lt;strong&gt;racemic compound&lt;/strong&gt; or &lt;strong&gt;racemate&lt;/strong&gt; is formed where each crystal contains R-S pairs.&lt;/p&gt;&#xA;&lt;p&gt;A conglomerate, being a simple mixture, will show a melting point depression compared to the pure enantiomer.  A racemic compound may have a melting point higher or lower than the pure enantiomer.&lt;/p&gt;&#xA;&lt;p&gt;Further reading on racemic conglomerates and racemic compounds can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Racemic_mixture&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.mcc-sci.com/crystallization.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-09-07T18:57:06.193" LastActivityDate="2020-09-07T18:57:06.193" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32711" PostTypeId="2" ParentId="32699" CreationDate="2015-06-10T17:22:38.883" Score="9" Body="&lt;p&gt;Preface: The problem is that you have &lt;em&gt;overstated your percentage yield&lt;/em&gt;, and the symptom is that it is above 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, and yours is 99%. Both you and your friend are wrong, for the same reason! Don't make the mistake of stopping your analysis when percentage yield drops below 100%. Don't chastise your friend for having an &quot;impossible&quot; yield while patting yourself on the back for your &quot;99%&quot;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overstatement of percentage yield happens for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Zero)&lt;/strong&gt; &lt;em&gt;Your model of the reaction chemistry is just wrong.&lt;/em&gt; Unlikely in an undergraduate setting, but worth mentioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; &lt;em&gt;Your math is wrong.&lt;/em&gt; Worth double checking and easy to rule out as a source of error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; &lt;em&gt;You used more reactant than you thought.&lt;/em&gt; Most errors in reactant measurement cause you to use less reactant than recorded (e.g. transfer errors, impure reactants), but you could possibly have used more than you thought. Beyond simple measurement errors, perhaps your limiting reagent was provided to you in solution, and that solution was of higher concentration than reported. Put a bit of time into thinking about this, but [especially in educational experiments] protocols usually minimize the chance of this kind of error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; &lt;em&gt;You actually have less product than you think.&lt;/em&gt; This is the most likely culprit. You have to be really honest with yourself here. In most experiments, you take the mass of something at the end, and call it the mass of your product. Is it &lt;em&gt;really&lt;/em&gt; your product? Or does that mass contain impurities which cause you to misrepresent impurities as product. Likely culprits are solvents (e.g. water from improper drying), side-products, and/or unreacted reagent. Have you really done your best to filter and remove them? If not, your yield will be inflated.&lt;/p&gt;&#xA;" OwnerUserId="16857" LastActivityDate="2015-06-10T17:22:38.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32712" PostTypeId="1" AcceptedAnswerId="32716" CreationDate="2015-06-10T18:21:00.827" Score="0" ViewCount="7360" Body="&lt;p&gt;My chemistry book says, an electrophile is a reagent which is atrracted to electrons, therefore positive ions and reagents with incomplete octets act as electrophiles, Sulfur trioxide belongs to neither category, so how does it act as an electrophile?&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="7448" LastEditDate="2015-10-09T17:06:48.227" LastActivityDate="2017-12-26T23:25:13.997" Title="Why is Sulfur trioxide a neutral electrophile?" Tags="&lt;acid-base&gt;&lt;oxidation-state&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32714" PostTypeId="1" AcceptedAnswerId="32758" CreationDate="2015-06-10T18:40:47.387" Score="7" ViewCount="634" Body="&lt;p&gt;I was researching and Googling and researching again, about this very interesting molecule: Diborane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I faced a very interesting structure, something that I have never seen before. Something that turns up lesser than 50 results in Google, most of which are behind paywalls, and thus leave me in pain of not understanding this species. I'm talking about &lt;strong&gt;&lt;em&gt;diborane - hydrogen chloride&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7q2Ci.png&quot; alt=&quot;Diborane - hydrogen chloride&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lb.chemie.uni-hamburg.de/search/pic.php?166/CpYaBeFg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'd like to understand this species better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bonding certainly doesn't seem ionic. What is the attraction force between $\ce{HCl}$ and $\ce{B2H6}$? There must be one or they wouldn't have written these two compounds together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I'm specifically looking for the hydrogen chloride - diborane(6) dimer you can see in the image. And the Cl is pointing towards the H in diborane, so this isn't about dipole/induced dipole interactions. You can find more info about it in &lt;a href=&quot;http://lb.chemie.uni-hamburg.de/static/CN/2_D_dibor.php?content=166/CpYaBeFg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Interestingly, in calculating molar mass, both the $M$s of $\ce{HCl}$ and $\ce{B2H6}$ are taken into account.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="10185" LastEditDate="2015-06-10T19:28:28.237" LastActivityDate="2015-06-12T12:43:42.703" Title="What's the bonding in the diborane(6)- hydrogen chloride dimer?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32716" PostTypeId="2" ParentId="32712" CreationDate="2015-06-10T18:49:47.783" Score="4" Body="&lt;p&gt;Sulfur is bonded to three oxygens, twice each. The sulfur in the middle technically has a formal charge of zero, but the oxygens it is bonded to are extremely electronegative (they're electron hogs), and so the sulfur atom in the middle has a partial plus charge on it. Ergo, it will be willing to accept electrons to try to compensate.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-10T18:49:47.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32717" PostTypeId="1" AcceptedAnswerId="32761" CreationDate="2015-06-10T19:33:02.293" Score="10" ViewCount="699" Body="&lt;p&gt;Reading the transcript of the Royal Society of Chemistry podcast &lt;a href=&quot;http://www.rsc.org/chemistryworld/2014/02/helium-hydride-podcast&quot;&gt;Helium Hydride&lt;/a&gt;, they state that helium hydride is possibly the most ancient compound to form in the Universe.  They make the assertion that despite helium hydride being most stable as an ion, but:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The neutral molecule can briefly exist, but only as an excimer, where the excess electron has enough energy to keep it at a higher level.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Under what natural conditions could a stable neutral helium hydride molecule exist?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-06-11T17:24:02.593" Title="Stable natural helium hydride?" Tags="&lt;covalent-compounds&gt;&lt;noble-gases&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32718" PostTypeId="1" CreationDate="2015-06-10T19:34:50.167" Score="1" ViewCount="35" Body="&lt;p&gt;Why equivalents of reactants and products come out to be equal in a reaction? I am seeking logic behind the law of chemical equivalence and i have just asked it for clarifying the concept.&lt;/p&gt;&#xA;" OwnerUserId="16864" LastActivityDate="2015-06-10T19:34:50.167" Title="Equivalents Chemistry" Tags="&lt;physical-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-06-11T03:02:51.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="32720" PostTypeId="2" ParentId="32695" CreationDate="2015-06-10T20:21:30.910" Score="3" Body="&lt;p&gt;Your question is based on a skewed premise, as gsurfer04 already pointed out. Rather than invoking d-orbitals, which are usually rather far away in energy from the corresponding p-orbitals (in fact, usually higher in energy than the next shell’s s-orbital, so for chlorine 3p → 4s → 3d), these hypervalent halogen compounds are considered to derive from multi-centre bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For details in the bonding mechanism, I am going to forward you to &lt;a href=&quot;https://chemistry.stackexchange.com/a/18544/7475&quot;&gt;this excellent answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven’t seen any multi-hydrogen chloride yet. In fact, when discussing interhalogens like $\ce{ClF3}$, our professor in the introductory chemistry course went as far as saying even $\ce{H3Cl}$ is impossible. I’ve forgotten the exact reasoning (and unfortunately it’s not explicitly stated in the web material any more) but I believe it to be along the lines of hydrogen’s s-orbitals not being able to form the outside parts of a four-electron three-centre bond.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-06-10T20:21:30.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32721" PostTypeId="2" ParentId="32687" CreationDate="2015-06-10T20:44:43.817" Score="2" Body="&lt;p&gt;As $\ce{O}$ is more electronegative than $\ce{N}$ the $\ce{1-}$ charge in the $\ce{O-N}$ will be on the oxygen. Even if it were to be a dative bond, its donation in an electron pair makes it very easy for oxygen to take. Oxygen is, after all, an oxidising agent.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastActivityDate="2015-06-10T20:44:43.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32722" PostTypeId="1" CreationDate="2015-06-10T21:58:21.580" Score="5" ViewCount="82" Body="&lt;p&gt;As a project I am trying to plate mercury onto a powdered metal amalgam of $\ce{Cu}$ &amp;amp; $\ce{Ag}$. I am using solution of $\ce{HgCl2}$ in normal $\ce{HCl}$ (a reducing agent), Any thoughts for increasing the efficiency of plating process on the metal alloy powder? How can we improve the $\ce{Hg}$ plating on the metal powder? Can the concept of autocatalysis / electroless plating be helpful in this case?&lt;/p&gt;&#xA;" OwnerUserId="16866" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T19:25:45.153" LastActivityDate="2015-06-11T19:25:45.153" Title="Autocatalysis / Electroless plating of mercury" Tags="&lt;electrolysis&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32723" PostTypeId="1" AcceptedAnswerId="32729" CreationDate="2015-06-10T22:26:47.147" Score="4" ViewCount="1458" Body="&lt;p&gt;In molecular orbital theory bonding of two atoms can be seen as addition or subtraction of the wave functions of these atoms. Addition seems pretty logical and straight-forward - atoms overlap their electron clouds and probability to find an electron in it increases. Subtraction is the thing I don't get. How can the probability to find an electron be negative? Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="5017" LastEditDate="2015-06-11T12:12:07.567" LastActivityDate="2015-06-11T17:11:02.087" Title="Subtracting wave functions?" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32724" PostTypeId="1" AcceptedAnswerId="46759" CreationDate="2015-06-10T23:18:15.150" Score="10" ViewCount="120" Body="&lt;p&gt;Some combinations of two elements have very complicated binary phase diagrams across the weight% horizontal axis. Others are rather simple. Is this a function of only one or both of the elements involved? I would think both. However, surely there are some elements that tend to form more phases than others. Has this tendency to form more phases (or less) been quantified for the various elements? If no, is it quantifiable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously elements form only a given number of known allotropes. However, it is not known if the number of known allotropes is exhaustive. Therefore the number of pseudomorphs in a binary phase diagram seems like a more thorough method of investigating an element's &quot;binding flexibility.&quot; &lt;/p&gt;&#xA;" OwnerUserId="114" LastEditorUserId="114" LastEditDate="2015-06-10T23:45:01.580" LastActivityDate="2016-07-19T19:36:54.887" Title="Which elements tend to form the most phases?" Tags="&lt;phase&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32725" PostTypeId="1" CreationDate="2015-06-11T00:16:13.373" Score="4" ViewCount="96" Body="&lt;p&gt;Without getting into too much/any math, I just wanna talk about intuitions if possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If atomic increase, the attraction increase, and consequently the potential energy increase. More energy leads to higher frequency, different radiations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BIsC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the reason if this diagram is universal is that the gap between 1 and 2 is too large too that the difference in energy cannot simply be compensated by adding more protons.&lt;/p&gt;&#xA;" OwnerUserId="4811" LastActivityDate="2015-06-11T02:25:24.547" Title="Can atomic number change the type of electromagnetic radiation being emitted when a electron drop from a energy level to another?" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32726" PostTypeId="2" ParentId="32725" CreationDate="2015-06-11T01:24:45.693" Score="1" Body="&lt;p&gt;Absolutely, the differences in potential energies will be different for every element. This is because the potential energies will be different for every element. Notice that the potential energy for each energy level can be determined from the ground state. Let $E_g$ be the potential energy of the ground state (with the sign). Then for energy level $n$ the potential energy is $\frac{E_g}{n^2}$. Thus Differences between energy levels $n$ and $m$ can be written as $E_g\left|\frac{1}{n^2}-\frac{1}{m^2}\right|$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly it will be different for a different ground state energy which is precisely what the number of protons affects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the only important part of the mathematics is that all differences rely on the ground state potentials.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T02:25:24.547" LastActivityDate="2015-06-11T02:25:24.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32728" PostTypeId="1" CreationDate="2015-06-11T01:52:42.290" Score="3" ViewCount="150" Body="&lt;p&gt;A question asks for physical properties of potassium nitrate, and specifically asks for &quot;magnetic charge&quot; and &quot;strength&quot; (that is how it is phrased). I have properties such as white in color, vitreous, odorless, and melting point of $\pu{334^\circ C}$. It is relatively stable, and a strong oxidizer, so it acts as a good fuel source. However, is strength asking about bond strength? And I'm not sure how to find the magnetic charge since none of my research has found anything.&lt;/p&gt;&#xA;" OwnerUserId="16854" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-03T10:52:39.653" LastActivityDate="2018-03-03T10:52:39.653" Title="Strength and magnetic charge of KNO3" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32729" PostTypeId="2" ParentId="32723" CreationDate="2015-06-11T02:12:34.640" Score="13" Body="&lt;p&gt;I think you are confusing the wavefunction $\Psi$ with the probability density function $\Psi^2$. The former is a state function describing the behavior of an electron. The latter describes the &quot;probability of finding an electron&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wavefunctions are usually complex functions, but even the simplest sine wave functions can be negative in the algebraic sense: the value of $\Psi=f(x)&amp;lt;0$. These regions where $Psi=f(x)$ has different algebraic signs are the regions corresponding to the orbital lobes of &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomic_orbital#Shapes_of_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;different phase&lt;/a&gt;, like in a p-orbital. However, $\Psi^2$ is always $\Psi^2 =f^2(x)\ge 0$. Information about the &quot;location&quot; of the electron is gained, information about its phase is lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take as an example the two wave functions representing two atomic orbitals $AO_1$ and $AO_2$ on the graph below. $\Psi_{AO_1}=\cos{(\frac{\pi x}{2})}$. $\Psi_{AO_2}=\cos{(\frac{3\pi x}{2})}$. $\Psi_{AO_2}$ has regions where $\Psi_{AO_2}&amp;lt;0$ when $\frac{1}{3}&amp;lt;|x|&amp;lt;1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eKpIf.png&quot; alt=&quot;a graph of two sine-wave functions showing different phase&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the probability density functions $\Psi^2_{AO_1}=\cos^2 {(\frac{\pi x}{2})}$ and $\Psi^2_{AO_2}=\cos^2 {(\frac{3\pi x}{2})}$. Notice that both functions are always $\ge0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jj00v.png&quot; alt=&quot;Two probability density functions showing that they are never negative&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay... now, when atomic orbitals combine to form molecular orbitals, one of those conservation laws (angular momentum) gets involved, and the number of orbitals has to remain constant. Thus, molecular orbitals are &lt;a href=&quot;http://en.wikipedia.org/wiki/Linear_combination_of_atomic_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear combinations of atomic orbitals&lt;/a&gt;. For example, if two atomic orbitals combine, two molecular orbitals are produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go back to our two sample sine wavefunctions above and pretend they are atomic orbitals overlapping. One of the molecular orbitals could then be: $\Psi_{MO_1}=\frac{1}{2}\Psi_{AO_1}+\frac{1}{2}\Psi_{AO_2}$. This is the orbital representing constructive overlap between these two wavefunctions. To get the other orbital, we can't just add again, or we would get the same orbital. If we subtract, we get a different orbital - the one representing destructive overlap. The graphs of these two &quot;molecular orbitals&quot; and their probability density functions are below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7kUng.png&quot; alt=&quot;A graph of two molecular orbitals showing constructive and destructive overlap&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gy56p.png&quot; alt=&quot;A graph of two molecular orbital density functions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again note the difference between the MO wavefunctions and their density functions. If these were atomic nuclei at $x=\frac{1}{3}$ and $x=-\frac{1}{3}$, then $\Psi_{MO_1}$ would be bonding and $\Psi_{MO2}$ would be antibonding (or perhaps nonbonding).&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="11367" LastEditDate="2015-06-11T17:11:02.087" LastActivityDate="2015-06-11T17:11:02.087" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32732" PostTypeId="1" CreationDate="2015-06-11T03:41:23.420" Score="2" ViewCount="4660" Body="&lt;p&gt;Is it possible to make a solution of ph 7 phosphate buffer solution using phosphoric acid and $\ce{K2HPO4}$ ? How can I calculate the amount of $\ce{K2HPO4}$ needed for 1L of phosphoric acid ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I have 50 mL of 0.2M $\ce{H3PO4}$ solution. How can I convert this solution into 50 mL of pH 7 buffer solution by adding (only) $\ce{K2HPO4}$ ? How can I calculate the weight of $\ce{K2HPO4}$ considering all the equilibria present in the $\ce{H3PO4}$ solution and by the application of Henderson-Hasselbalch equation ?&lt;/p&gt;&#xA;" OwnerUserId="16096" LastEditorUserId="14" LastEditDate="2015-06-14T13:25:26.180" LastActivityDate="2015-06-17T13:21:35.937" Title="Phosphate buffer from phosphoric acid and K2HPO4?" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32735" PostTypeId="1" AcceptedAnswerId="32736" CreationDate="2015-06-11T04:22:35.670" Score="0" ViewCount="431" Body="&lt;p&gt;Since water can be either hot or cold, does its temperature affect its ability to extinguish fires?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastEditorUserId="7448" LastEditDate="2015-06-11T08:47:23.120" LastActivityDate="2015-06-11T08:47:23.120" Title="Does the temperature of water affect its ability to extinguish fires?" Tags="&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32736" PostTypeId="2" ParentId="32735" CreationDate="2015-06-11T04:35:23.223" Score="2" Body="&lt;p&gt;A fire needs 3 things as shown in the image:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8ClZU.png&quot; alt=&quot;fire triangle&quot;&gt;&lt;br&gt;&#xA;Water reduces the heat as it turns into steam; and hot water turns into steam faster. But the most important thing water does, is to prevent oxygen getting to the fuel. So as long as there is liquid water, it does a great job.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-06-11T04:35:23.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32737" PostTypeId="2" ParentId="32616" CreationDate="2015-06-11T05:13:21.193" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.woodworkersjournal.com/understanding-oil-wax-finishes/&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; explains the use of oil on wood fairly well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oil is made of molecules small enough to seep down into the wood rather than merely sit on top. As a result, oil makes wood look richer and more translucent without adding a film on the surface. There are two different types of oils that woodworkers use: drying and non-drying oils. Drying oils will change from liquid to a solid film when exposed to oxygen in the air. Nut oils (boiled linseed, tung, etc.) are drying oils, but vegetable (peanut, olive) and mineral oils are non-drying. Edible mineral oil is popular on food contact items, like cutting boards. However, non-drying oils stay wet indefinitely, and they will wash off when the board is scrubbed with soap and water. Because they do not dry to a solid film, non-drying oils are considered a wood treatment, but not a finish.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-06-11T05:13:21.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32738" PostTypeId="1" AcceptedAnswerId="32744" CreationDate="2015-06-11T05:26:51.360" Score="1" ViewCount="56" Body="&lt;p&gt;I know that there are some sites like Sigma Aldrich and AIST for $\ce{^1H}$ NMR and $\ce{^13C}$ NMR, but was wondering if there were any free sites to compare $\ce{^31P}$ NMR spectra?&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastEditorUserId="15410" LastEditDate="2015-06-11T14:32:48.247" LastActivityDate="2015-06-11T14:32:48.247" Title="Are there any free phosphorus-31 NMR spectral libraries?" Tags="&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32740" PostTypeId="1" AcceptedAnswerId="32745" CreationDate="2015-06-11T05:53:10.680" Score="1" ViewCount="60" Body="&lt;p&gt;As an example, let's assume that we have a gas $\ce{P(g)}$ inside a chamber and when the temperature inside it increases to a certain value there is an equilibrium that prevails inside the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the sake of argument, let us assume that this equilibrium is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{P(g) &amp;lt;=&amp;gt; Q(g)  + R (g)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's assume the figure below shows that chamber when equilibrium is established:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2LykFt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is, what are the possible changes that might occur if I add an inert gas $\ce{I(g)}$ to this chamber?&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="11367" LastEditDate="2015-06-11T12:03:33.670" LastActivityDate="2015-06-11T12:08:40.467" Title="What are the possible changes that might occur if I add an inert gas to a chamber containing another few different gases?" Tags="&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32741" PostTypeId="2" ParentId="19540" CreationDate="2015-06-11T06:19:46.957" Score="0" Body="&lt;p&gt;In fact, it is possible if only water and some gelatin or agar-agar is present in the salt bridge. But for $\ce{SO4^{2-}}$ Or $\ce{Zn^{2+}}$ to travel through salt bridge it will have to face a lot of resistance resulting in low voltage. Instead, it is better to have strong electrolytic salt which can maintain neutrality of solution as soon as zinc atom is oxidized or $\ce{Cu}$ atom is reduced.&lt;/p&gt;&#xA;" OwnerUserId="16874" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T06:22:36.620" LastActivityDate="2015-06-11T06:22:36.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32743" PostTypeId="1" CreationDate="2015-06-11T06:25:20.730" Score="9" ViewCount="204" Body="&lt;p&gt;I've become very curious about (electrochemical) atomic layer deposition (ALD) via surface limited redox replacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the technique allows for deposition of metals in atom thick layers it's possible to create films of nearly exact atomic thickness. However, I haven't found any reference that describes how one can remove the film from the (usually Au) substrate that it's deposited on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothetically speaking, if one used ALD to generate a film of $\ce{Cu}$ that is 5 atoms thick (arbitrary length and width), how would one go about removing the film in-tact? Is it even possible?&lt;/p&gt;&#xA;" OwnerUserId="16873" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T06:27:38.083" LastActivityDate="2015-06-12T01:39:13.583" Title="How to separate a 5 atom thick layer of Cu from Au" Tags="&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32744" PostTypeId="2" ParentId="32738" CreationDate="2015-06-11T06:40:05.593" Score="2" Body="&lt;p&gt;There are quite a few it seems, a couple of examples are below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;University of Wisconsin &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/handouts/nmr/p-data.htm&quot; rel=&quot;nofollow&quot;&gt;Phosphorus NMR data&lt;/a&gt;, starts with a shifts overview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanford's Magnetic Resonance Laboratory site &lt;a href=&quot;http://smrl.stanford.edu/pnmr/&quot; rel=&quot;nofollow&quot;&gt;P NMR at the SMRL&lt;/a&gt;, provides comparisons and a protocol for testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A general NMR site is &lt;a href=&quot;http://nmrwiki.org/wiki/index.php?title=Databases&quot; rel=&quot;nofollow&quot;&gt;NMR Wiki&lt;/a&gt; - has many links to many sources.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T06:46:03.483" LastActivityDate="2015-06-11T06:46:03.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32745" PostTypeId="2" ParentId="32740" CreationDate="2015-06-11T07:13:54.040" Score="1" Body="&lt;p&gt;Quoting from the Wikipedia page on &lt;a href=&quot;http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Le Chatelier's principle&lt;/a&gt;, we have to distinguish two cases:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The total volume is constant:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Adding an inert gas into a gas-phase equilibrium  does not result in a shift of the equilibrium in this case. This is due to the fact that adding an inert gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The total volume is not constant:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If, however, the volume is increased, the partial pressures of all gases will decrease which results in a shift of the equilibrium towards the side with the greater number of moles of gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-11T12:08:40.467" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32747" PostTypeId="1" CreationDate="2015-06-11T10:13:58.547" Score="5" ViewCount="272" Body="&lt;p&gt;An $\mathrm{E2}$ (or $\mathrm{E1cb}$) mechanism requires antiperiplanar configuration. In case of restricted rotations, like in cycles, where rotation is not possible to achieve the antiperiplanar configuration, elimination occurs to favour the Hofmann product instead of the Zaitsev, if antiperiplanar configuration is possible at the less substituted β C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/39IUh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In menthyl chloride, antiperiplanar configuration is only seen with the red $\ce{H}$, so 2 menthene is the only product, although it is not the most substituted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wWGw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the molecule above, due to restricted rotation, the leaving group $\ce{Cl}$ and the base attacking the $\ce{H}$ will be on the same side. So will elimination be possible in this case?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-06-11T12:36:51.237" LastActivityDate="2015-06-11T14:17:24.533" Title="Anti elimination from vinylic halides" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32748" PostTypeId="2" ParentId="29587" CreationDate="2015-06-11T11:10:34.313" Score="12" Body="&lt;p&gt;The main thing that classical and quantum statistical mechanics have in common is probability.  In classical stat mech the equilibrium value of an observable is given by&#xA;$$\langle A \rangle = \sum_i^N p_iA_i$$&#xA;where $A_i$ is the value of $A$ in microstate $i$ and $p_i$ is the probability of microstate $i$.  The expression for $p_i$ at equilibrium depends on the ensemble (microcanonical, canonical, etc).  One way to define equilibrium is that the probabilities are constant&#xA;$$\frac{\partial p_i(t)}{\partial t}=0$$&#xA;So if one perturbs the system so that this equation doesn't hold the $p_i$'s will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_13/node1.html&quot;&gt;quantum stat mech&lt;/a&gt; the equilibrium value of an observable is given by&#xA;$$ \langle A \rangle = \tfrac{1}{N} \sum_\alpha^N \langle \Psi^{\alpha}| \hat{A} |\Psi^{\alpha}\rangle$$&#xA;where $\Psi^{\alpha}$ is the wavefunction for microstate $\alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavefunctions are typically expanded in an orthonormal basis ($\phi$'s)&#xA;$$ \langle A \rangle = \tfrac{1}{N} \sum_{k,l} \rho_{l,k}\langle \phi_l| \hat{A} |\phi_k\rangle $$&#xA;where the density matrix is can be defined by a density operator&#xA;$$ \rho_{l,k} = \langle \phi_k| \hat{\rho} |\phi_l\rangle $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium is then defined as &#xA;$$\frac{\partial \hat{\rho}(t)}{\partial t}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if one perturbs the system so that this equation doesn't hold the $\hat{\rho}$ will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $\Psi^{\alpha}$ contributions to  $\hat{\rho}$ increase of decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time dependence of $\hat{\rho}$ can be calculated by the time-dependent Schrödinger equation.  So Le Chatelier's principle is a heuristic way of predicting roughly how the time-dependent wavefunctions will evolve at the simulation approaches equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a more physical picture one can introduce the Born-Oppenheimer approximation in which case the numerical simulation of the time-dependent Schrödinger equation becomes a classical MD simulation (with a QM energy function) from which the time dependence of the probabilities $p_i$ can be computed and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-11T11:37:34.653" LastActivityDate="2015-06-11T11:37:34.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32749" PostTypeId="1" AcceptedAnswerId="32755" CreationDate="2015-06-11T11:27:22.927" Score="2" ViewCount="104" Body="&lt;p&gt;Can anyone tell me what  the difference between these two compounds is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/907&quot; rel=&quot;nofollow&quot;&gt;N(alpha)-acetyl-ornithine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/193343&quot; rel=&quot;nofollow&quot;&gt;N(delta)-acetyl-ornithine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference I can think of is the chiral properties of the nitrogen atom, i.e. nitrogen inversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, does it make sense to differ between these to compounds, or will they be constantly interconverted?&lt;/p&gt;&#xA;" OwnerUserId="16879" LastEditorUserId="7475" LastEditDate="2015-06-11T13:09:14.187" LastActivityDate="2015-06-11T13:09:14.187" Title="N(delta)-acetyl... vs N(alpha)-acetyl" Tags="&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32750" PostTypeId="1" AcceptedAnswerId="32753" CreationDate="2015-06-11T11:29:42.177" Score="3" ViewCount="1231" Body="&lt;p&gt;For a titration of potassium permanganate into a standard acidified solution of hydrated iron (II) ammonium sulfate, the solution seems to end up brown (due to the formation of manganese dioxide) during the titration unless the titration is carried out extremely slowly. How can I minimise this? Adding sulfuric acid to the solution seemed to help, but is there a physical way to prevent manganese dioxide forming when the titrant is added too rapidly?&lt;/p&gt;&#xA;" OwnerUserId="14994" LastActivityDate="2015-06-11T12:29:28.513" Title="Minimising the formation of manganese dioxide in the titration of potassium permanganate" Tags="&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32753" PostTypeId="2" ParentId="32750" CreationDate="2015-06-11T12:29:28.513" Score="4" Body="&lt;p&gt;Adding a large excess of acid helps greatly as the reduction to $\ce{MnO2}$ only takes place under neutral conditions. $$\ce{2H2O + MnO4- + 3e- -&amp;gt; MnO2 + 4OH-}$$ Under acidic conditions the $\ce{MnO4-}$ is reduced all the way to $\ce{Mn^{2+}}$.$$\ce{8H+&#xA; + MnO4- + 5e- -&amp;gt; Mn^{2+} + 4H2O}$$ Also swirling or stirring the solution as you are titrating makes a big difference as it stops the area where the permanganate is being added from becoming temporarily depleted of acid.Even if you do get some $\ce{MnO2}$ forming then as long as you have enough acid in the flask a few swirls should be enough to reduce the $\ce{MnO2}$ down to $\ce{Mn^{2+}}$.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-11T12:29:28.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32754" PostTypeId="2" ParentId="32749" CreationDate="2015-06-11T12:37:09.900" Score="1" Body="&lt;p&gt;The difference does &lt;em&gt;not&lt;/em&gt; lie in stereochemistry. Rather, &lt;em&gt;N&lt;/em&gt;$^\alpha$-acetylornithine and &lt;em&gt;N&lt;/em&gt;$^\delta$-acetylornithine are &lt;em&gt;regio&lt;/em&gt;isomers of one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The superscript Greek letter denotes which nitrogen is substituted with an acetyl group. The lettering is relative to the carboxyl group as shown in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zY4sd.png&quot; alt=&quot;ornithine with Greek-letter labelled carbons&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The carboxyl group’s carbon does not get the $\alpha$, because it is considered part as the functional group from with the labelling begins&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The $\alpha$ carbon is asymmetric. Both L- and D-compounds (or (&lt;em&gt;S&lt;/em&gt;) and (&lt;em&gt;R&lt;/em&gt;) according to CIP nomenclature) are possible for &lt;strong&gt;both&lt;/strong&gt; &lt;em&gt;N&lt;/em&gt;$^\alpha$- and &lt;em&gt;N&lt;/em&gt;$^\delta$-acetylornithine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-11T12:37:09.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32755" PostTypeId="2" ParentId="32749" CreationDate="2015-06-11T12:38:42.820" Score="1" Body="&lt;p&gt;Short answer: &lt;strong&gt;PubChem has an error&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Differences in structures as currently listed in PubChem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One other important difference is the stereochemistry of the sp3 carbon attached to the nitrogen atom, i.e., at carbon 2 of the 5-aminopetanoic acid chain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/907#section=Identification&quot; rel=&quot;nofollow&quot;&gt;N(alpha)-acetyl-ornithine&lt;/a&gt; has a computed IUPAC name in PubChem of &lt;strong&gt;2-acetamido-5-aminopentanoic acid&lt;/strong&gt;.  The stereochemistry is not listed, which leads me to assume that it is a racemate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The molecule/PubChem page that you called &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/439232#section=Identification&quot; rel=&quot;nofollow&quot;&gt;N(delta)-acetyl-ornithine&lt;/a&gt; has a computed IUPAC name in PubChem of &lt;strong&gt;(2S)-2-acetamido-5-aminopentanoic acid&lt;/strong&gt;.  The stereochemistry at the sp3 carbon bound to the nitrogen in this compound is clearly listed as &lt;em&gt;S&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Inconsistencies between the structures and the names&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Confusingly, both compounds are called &lt;em&gt;L&lt;/em&gt;-ornithines on those pubchem pages.  But since carbon 2 is the &amp;alpha; carbon, this is the position that should determine the &lt;em&gt;L&lt;/em&gt; or &lt;em&gt;D&lt;/em&gt; stereochemistry.  Thus, I'm not sure why N(&amp;alpha;)-acetyl-ornithine isn't just called N-acetyl-DL-ornithine and what you called &quot;N(&amp;delta;)-acetyl-ornithine&quot; isn't just called N(&amp;alpha;)-acetyl-L-ornithine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Secondly, there is the matter that the page you linked for &quot;N(delta)-acetylornithine&quot; lists a &quot;main&quot; name for the molecule of &quot;N-acetylornithine&quot; and &quot;MESH descriptors&quot; of &quot;delta-N-acetylornithine&quot;.  &lt;strong&gt;This use of &quot;delta&quot; is wrong because it is inconsistent with the listed structure&lt;/strong&gt;.  The &amp;alpha;, &amp;beta;, &amp;gamma;, &amp;delta; nomenclature exists to indicate &lt;em&gt;which&lt;/em&gt; nitrogen atom is substituted.  In ornithine carbons 2 and 5 of the core pentanoic acid chain have nitrogen atoms, these correspond to N(&amp;alpha;) and N(&amp;delta;).  Thus, N(&amp;delta;)-acetyl-L-ornithine should have the acetyl group substituting a different nitrogen atom than N(&amp;alpha;)-acetyl-L-ornithine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-11T12:38:42.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32756" PostTypeId="2" ParentId="32743" CreationDate="2015-06-11T13:40:06.010" Score="-1" Body="&lt;p&gt;I'd suggest electrostatic repulsion. Charge both layers with electrons and they'd separate easily. Then use a electron microscope to manipulate the layer.  &lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2015-06-11T13:40:06.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32757" PostTypeId="2" ParentId="32747" CreationDate="2015-06-11T14:17:24.533" Score="1" Body="&lt;p&gt;An E2 reaction will not occur with this reagent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An E1 reaction, on the other hand, could do so (though quite slowly: E1 is an all around pathetic reaction mechanism, and this molecule does nothing special to stabilize it). The chlorine will leave and there will be a carbocation on the end of the alkene, whereupon the proton from the other side of the double can be extracted with a base in a second step, leaving you with an alkyne. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, if you want the alkyne, it would be infinitely infinitely easier to start with the cis isomer of the molecule you have and just run it through very strong base. E2 is far far better than E1 in terms of reaction rate and yield.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-11T14:17:24.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32758" PostTypeId="2" ParentId="32714" CreationDate="2015-06-11T14:27:58.197" Score="7" Body="&lt;p&gt;I find this question to be quite interesting as it clearly pertains to non-covalent interactions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have analyzed two conformations of the diborane-HCl dimer system, one with the chlorine on HCl orientated toward diborane (HCl-Dib) and one with the chlorine oriented away from the &lt;a href=&quot;http://dx.doi.org/10.1063/1.461070&quot; rel=&quot;nofollow noreferrer&quot;&gt;diborane&lt;/a&gt; (ClH-Dib).  For comparison, I have similarly analyzed the Water-HCl dimer with the chlorine oriented toward the two hydrogens on water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Methodology&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with &lt;em&gt;Gaussian 09&lt;/em&gt; using Grimme's &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jcc.20495/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;B97D&lt;/a&gt; density functional along with the &lt;a href=&quot;http://link.springer.com/article/10.1007/BF00533485#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;6-31+G(d',p')&lt;/a&gt; basis set.  All geometries were characterized to C$_{2V}$ symmetry.  Symmetry adapted perturbation theory (SAPT) was implemented at the SAPT2+3(CCD) level using the &lt;em&gt;&lt;a href=&quot;http://sirius.chem.vt.edu/psi4manual/4.0b5/sapt.html#sec-sapt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Psi4&lt;/a&gt;&lt;/em&gt; software package.  These computations implemented Dunning's heavy-aug-cc-pVTZ (haTZ) basis set (i.e. &lt;a href=&quot;http://dx.doi.org/10.1063/1.462569&quot; rel=&quot;nofollow noreferrer&quot;&gt;aug&lt;/a&gt;-&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/90/2/10.1063/1.456153&quot; rel=&quot;nofollow noreferrer&quot;&gt;cc-pVTZ&lt;/a&gt; for non-hydrogen atoms and &lt;a href=&quot;http://dx.doi.org/10.1063/1.456153&quot; rel=&quot;nofollow noreferrer&quot;&gt;cc-pVTZ&lt;/a&gt; for all hydrogen atoms).  Finally, natural charges were quantified using the Natural Bond Orbital (NBO) analysis routine available in &lt;em&gt;Gaussian 09&lt;/em&gt; (see &lt;a href=&quot;http://nbo6.chem.wisc.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nbo6.chem.wisc.edu/&lt;/a&gt; for more details). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Results&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both diborane-containing complexes were minima on the B97D potential energy surface.  The H$_2$O-HCl dimer is a higher order saddle point exhibiting two imaginary modes of vibration.  The interaction energy ($E_{\mathrm{int}}$) can be quantified by attractive and repulsive contributions, the former consisting of electrostatics ($E_{\mathrm{elst}}$), induction ($E_{\mathrm{ind}}$), and dispersion ($E_{\mathrm{disp}}$) and the latter consisting of exchange ($E_{\mathrm{exch}}$). The table below summarizes the SAPT results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dBDXL.png&quot; alt=&quot;Contributions to the interaction energy for various dimers computed at the SAPT2+3(CCD)/haTZ level of theory.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the dominant attractive force in each case is dispersion.  The ClH-Dib complex exhibits the strongest interaction energy and contains a dipole that points from the diborane monomer to the chlorine.  The HCl-Dib complex has an interaction energy that is about 1 kcal mol$^{-1}$ smaller in magnitude and exhibits a dipole pointing toward the diborane monomer.  The analogous H$_2$O-HCl system is qualitatively similar to the HCl-Dib complex in terms of the magnitudes of the components to the interaction energy.  Surprisingly, this dipole-dipole interaction is slightly weaker (by about 0.1 kcal mol$^{-1}$) than that seen in the dipole/induced-dipole system of HCl-Dib.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have included below figures for the HCl-Dib, ClH-Dib, and H$_2$O-HCl complexes, respectively, with corresponding natural atomic charges shown on each atom.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 1&lt;/strong&gt; - HCl-Dib&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9j4bs.jpg&quot; alt=&quot;HCl-Dib&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 2&lt;/strong&gt; - ClH-Dib&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9RM0Q.jpg&quot; alt=&quot;ClH-Dib&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure 3&lt;/strong&gt; - H$_2$O-HCl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B1Dhe.jpg&quot; alt=&quot;H$_2$O-HCl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Interestingly, when chlorine is pointed toward diborane (Figure 1), the natural atomic charges for diborane remain rather symmetrical and diborane barely exhibits an individual dipole moment.  The H$\cdots$Cl separation between the two monomers is about 3.4 Angstroms, a long-range interaction.  However, when chlorine is pointed away from diborane (Figure 2), the H$\cdots$H interaction is much smaller, exhibiting a separation of about 2.2 Angstroms.  When considering water, the separation increases once again (Figure 3) to about 3.2 Angstroms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominate attractive force to the interaction energy for these dimer systems is dispersion, though we see a competing contribution from electrostatics in ClH-Dib (Figure 2).  Surprisingly, the orientation of the halogen significantly affects the geometry and energetics of these systems.  When chlorine is oriented toward the two hydrogens, a weaker interaction and a larger inter-monomer separation is seen.  However, when hydrogen is oriented toward the two hydrogens, the interaction energy increases in magnitude and the inter-monomer separation decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q/A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it more convenient to address some of Martin's comments within the answer itself (sometimes responses can become quite long).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which compound is lower in energy? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Based on absolute energies, H$_2$O-Cl is the lowest in energy of the three dimers (by about 23 Hartrees). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How stable are they wrt decomposition? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The interaction energy will give you this information (defined as the energy difference between the interacting monomers in an optimized dimer geometry and infinitely separated monomers adopting the geometry they have in the dimer complex).  Given that the magnitude of $E_{\mathrm{int}}$ is less than 2 kcal mol$^{-1}$, I would say that these are not very 'stable'.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have you tried asymmetric states?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No.  I do not believe breaking symmetry in the diborane-containing complexes will change the energy/geometry appreciably. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I find it quite surprising, that the BH--ClH distance is much longer (0.6A) than the sum of the respective vdW radii, while the BH--HCl distance is about the sum of the vdW radii. Still, in both cases there should be (about) no contribution from the orbitals. How do the bond lengths change in the two conformations? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the diborane-containing complexes, the bond lengths deviate by about 0.01 Angstroms (so... no change).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you make a rough comparison between the B--Cl distance in the two cases? &#xA;They appear to be quite similar.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In HCl-Dib, the B$\cdots$Cl separation is about 3.79 Angstroms.  In ClH$\cdots$Dib, the B$\cdots$Cl separation is about 3.86 Angstroms.  Similar they are indeed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-06-12T12:43:42.703" LastActivityDate="2015-06-12T12:43:42.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32759" PostTypeId="2" ParentId="22367" CreationDate="2015-06-11T15:17:04.950" Score="3" Body="&lt;p&gt;For the given structure, the principal chain could be a 3-hydroxyhexanoic acid or a 3-methylhexanoic acid; i.e. the name of the structure could be 3-hydroxy-2-(pentan-2-yl)hexanoic acid or 2-(1-hydroxybutyl)-3-methylhexanoic acid.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xJpxo.png&quot; alt=&quot;2-(1-hydroxybutyl)-3-methylhexanoic acid&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The IUPAC recommendations on this matter have been changed.&lt;/p&gt;&#xA;&lt;p&gt;According to the old IUPAC recommendations (1979, retained in 1993), the principal chain was the 3-hydroxyhexanoic acid. Thus, the preferred name was 3-hydroxy-2-(pentan-2-yl)hexanoic acid.&lt;/p&gt;&#xA;&lt;p&gt;However, the relevant rule in the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-45 SELECTION OF PREFERRED IUPAC NAMES&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-45.5&lt;/strong&gt; CRITERIA RELATED TO ALPHANUMERICAL ORDER OF NAMES&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-45.5.1&lt;/strong&gt; The preferred IUPAC name is the name that is earlier in alphanumerical order (…). Alphabetic letters are considered first in the order that they appear in the name; (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the preferred name is &lt;strong&gt;2-(1-hydroxybutyl)-3-methylhexanoic acid&lt;/strong&gt; (not 3-hydroxy-2-(pentan-2-yl)hexanoic acid) since ‘hydroxybutyl…’ is earlier alphabetically than ‘hydroxy-pentan-yl…’.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-11T15:23:12.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32761" PostTypeId="2" ParentId="32717" CreationDate="2015-06-11T17:24:02.593" Score="6" Body="&lt;p&gt;As explained in &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.55.2941&quot;&gt;Emission Spectra of Bound Helium Hydride&lt;/a&gt; Physical Review Letters vol. 77 pages 2941-2944:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ground state [of neutral HeH] is repulsive, as shown by&#xA;  numerous experiments and calculations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but excited states of HeH that are bound exist (as first experimentally observed by this paper, but previously calculated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that only excited states are bound, it don't think it proper to say HeH stably exists under any conditions natural or otherwise.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper does state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the abundance of He and $\ce{H_2}$ in stellar atmospheres&#xA;  and interstellar space, HeH is expected to&#xA;  be formed there. Therefore our data may be useful to&#xA;  find the HeH lines there.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, that it exists does not mean it stably exists. &lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-06-11T17:24:02.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32762" PostTypeId="1" AcceptedAnswerId="32766" CreationDate="2015-06-11T17:33:30.490" Score="3" ViewCount="88" Body="&lt;p&gt;I recently been studying different radioactive elements. The most difficult part is find the output of these elements in the form of kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a database or chart of these elements' output? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most importantly does anyone know the kWh of $1~ kg$ of $\ce{^99Tc}$? &lt;/p&gt;&#xA;" OwnerUserId="16883" LastEditorUserId="7448" LastEditDate="2015-06-11T17:38:53.080" LastActivityDate="2015-06-11T19:13:20.130" Title="kWh of radioactive elements database?" Tags="&lt;energy&gt;&lt;reference-request&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32764" PostTypeId="2" ParentId="32487" CreationDate="2015-06-11T18:28:05.420" Score="3" Body="&lt;p&gt;I'm taking a stab at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so the issue with the group 11 metals, as far as I can see, is that while, yes, their s orbitals are not filled, they do not exist in isolation: they have neighbors who all have only 1 electron in them. So, they form very strong allotropes. In other words, gold would have been reactive, but there are other golds right there it can bond with and so share electrons with it. So the group 11 elements form large networks of allotropes, where electrons freely flow and are shared with other atoms of their kind. These bonds are very stable, and ergo, very hard to break. In fact, Au-Au bonds have been found to stabilize some very strange molecules: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja511394q&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-11T18:28:05.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32765" PostTypeId="2" ParentId="7648" CreationDate="2015-06-11T18:36:06.237" Score="4" Body="&lt;p&gt;Intuition behind this concept is thermodynamically based.&lt;br&gt;&lt;br&gt;  If you think of $K_a$ as a thermodynamically favorable constant that describes the analyte concentrations associated with acid dissociation then it is not too difficult to reconcile the increase in $\ce{H+}$ ions in solution as a necessary means to maintain a thermodynamically favorable state when water is added to solution.  &lt;/p&gt;&#xA;" OwnerUserId="16882" LastEditorUserId="10185" LastEditDate="2015-06-11T18:40:45.503" LastActivityDate="2015-06-11T18:40:45.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32766" PostTypeId="2" ParentId="32762" CreationDate="2015-06-11T19:00:39.833" Score="3" Body="&lt;p&gt;$\ce{^{99m}Tc}$ is a nuclear isomer of $\ce{^{99}Tc}$, which is excited by 142.6 keV. Therefore, when one atom decays to $\ce{^{99}Tc}$, it releases 142.6 keV, mostly as gamma radiation, but also some internal conversion occurs. Therefore, for 1 kg of $\ce{^{99m}Tc}$ (10.1 mol):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(142.6\ \mathrm{keV})(10.1\ \mathrm{mol})N_A = 8.673\times 10^{29}\ \mathrm{eV} = 139\ \mathrm{GJ} = 38.6\ \mathrm{MWh}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the total energy released by a nuclear reaction in a given mass is simply the decay energy (commonly known as the Q value) multiplied by the number of atoms. The Q value is simply the difference in mass between the starting material and final product converted to energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = (m_\mathrm{i}-m_\mathrm{f})c^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q values are easy to just look up, though. (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=technetium-99&quot; rel=&quot;nofollow&quot;&gt;Wolfram|Alpha&lt;/a&gt; is a good place)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, depending on the application, all this energy is not accessible or useful. If one were trying to make an &lt;a href=&quot;http://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator&quot; rel=&quot;nofollow&quot;&gt;RTG&lt;/a&gt;, an enormous amount of shielding would be needed to turn the gamma rays into heat, since they penetrate matter very well, where things like Pu-238 work better because they release mostly alpha particles, which are easily absorbed, delivering their kinetic energy as heat.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorDisplayName="user15489" LastEditDate="2015-06-11T19:13:20.130" LastActivityDate="2015-06-11T19:13:20.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32767" PostTypeId="1" AcceptedAnswerId="32768" CreationDate="2015-06-11T19:13:15.523" Score="7" ViewCount="18692" Body="&lt;p&gt;Earlier today I was preparing an ester (propyl ethanoate) and I noticed that instead of putting the propanol, ethanoic acid and $\ce{H2SO4}$ in the test tube and warming it (taking time and external energy) I could instead mix sulfuric acid with the ethanoic acid, which made the test tube go very hot, and then by adding the propanol and shaking it split into two layers. On pouring this into water I ended up with an immiscible layer which appeared to be an ester (it was sweet smelling and considering the reaction it would be likely to be an ester). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot think of which reaction was causing the heating up initially. The test tube was clean so the only reactants were the sulfuric acid and ethanoic acid, and it appeared that a gas was being given off (as shaking it with a rubber cap on caused it to pop out of the tube, but no effervescence could easily be seen). Can anyone help? Furthermore, would the eventual product be propyl ethanoate or something else (it was suggested that it was dehydrogenating the ethanoic acid so the actual product would be something like propyl ethenoate, but from a quick look on the internet this would seem to require an iron catalyst (or another catalyst) and a high temperature, neither of which was present)?&lt;/p&gt;&#xA;" OwnerUserId="15714" LastEditorUserId="402" LastEditDate="2017-01-14T08:55:43.920" LastActivityDate="2019-11-06T14:48:19.840" Title="H2SO4 with ethanoic acid reaction?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32768" PostTypeId="2" ParentId="32767" CreationDate="2015-06-11T19:30:27.497" Score="6" Body="&lt;p&gt;I suspect that the ethanoic acid contains a residual amount of water and that the concentrated sulphuric acid is reacting with the water. Since this is a very exothermic reaction the test tube gets heated up which evaporates some of the ethanoic acid, causing the bung to pop off. If it's just slight evaporation this would also explain why you couldn't see any effervescence.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-11T19:30:27.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32769" PostTypeId="1" AcceptedAnswerId="32781" CreationDate="2015-06-11T20:32:37.843" Score="6" ViewCount="1368" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$25.0\:\mathrm{ml}$ of $0.10\:\mathrm{M}$ $\ce{CH3COOH}$ is titrated with $0.10\:\mathrm{M}$ $\ce{NaOH}$. What is the $\mathrm{pH}$ after the addition of $30.0\:\mathrm{ml}$ $\ce{NaOH}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In broad strokes, as I understand it, the sodium hydroxide in this case would completely consume the acetic acid, leaving a solution of acetate ions, $\ce{Na+}$ and $\ce{OH-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes sense to me then is to figure that the $\mathrm{pH}$ is now dominated by the concentration of hydroxide ions, so the $\mathrm{pH}$ would be given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{pH} = 14-(-\log([\ce{OH-}])= 11.96$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $[\ce{OH-}] = \frac{[(30.0 - 25.0) \times 10^{-4}]\:\mathrm{mol}}{(30.0+25.0) \times 10^{-3}\:\mathrm{l}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., the remaining moles of hydroxide ($0.0005$) per liter of total volume ($55\:\mathrm{ml}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but is this correct? Do I need to take into account the conjugate base’s contribution to deprotonation of water? If so, how is that done? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d like to be clever and use the lack of $K_\mathrm{a}$ or related constants in the problem to indicate that I’m on the right track, but that would be giving too much credit to my professor’s design of these questions (and the prior question in this series does give $\mathrm{p}K_\mathrm{a}$ for acetic acid)&lt;/p&gt;&#xA;" OwnerUserId="7078" LastEditorUserId="5026" LastEditDate="2018-03-02T02:07:57.970" LastActivityDate="2020-04-26T07:03:14.950" Title="Calculating pH in titration of monoprotic weak acid with strong base" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32770" PostTypeId="1" CreationDate="2015-06-11T20:49:56.910" Score="5" ViewCount="7441" Body="&lt;p&gt;I understand that a combination reaction is one where two substances combine to form a third substance, like with &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Na + Cl2 -&amp;gt; 2NaCl}.$$&lt;/span&gt; &#xA;I do not understand why &lt;span class=&quot;math-container&quot;&gt;$$\ce{CaO + SO2 -&amp;gt; CaSO3}$$&lt;/span&gt; is not a combination redox reaction. After all, it is creating a third substance, so why is it not a combination redox reaction? &lt;/p&gt;&#xA;" OwnerUserId="16885" LastEditorUserId="4945" LastEditDate="2019-01-15T14:27:24.927" LastActivityDate="2019-01-15T14:31:37.703" Title="Why is the reaction of calcium oxide with sulfur dioxide to form calcium sulfite not a combination redox reaction?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32771" PostTypeId="2" ParentId="32732" CreationDate="2015-06-11T22:19:56.973" Score="2" Body="&lt;p&gt;I suggest you first consider the following reaction:&#xA;$$\ce{H3PO4 + 3K2HPO4 -&amp;gt; 2HPO4^{2-} + 2H2PO4- + 6K+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$ &#xA;requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At pH = 7.0: [HPO4(2-)] &amp;lt; [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:&#xA;7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) =&gt; x = 0,1533.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please check the calculations.&lt;/p&gt;&#xA;" OwnerUserId="16886" LastEditorUserId="16886" LastEditDate="2015-06-17T13:21:35.937" LastActivityDate="2015-06-17T13:21:35.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32772" PostTypeId="2" ParentId="32770" CreationDate="2015-06-11T23:52:03.853" Score="5" Body="&lt;p&gt;This question can be solved the easiest if we break it down to its fragments. The requirements of a combination redox reaction are that the reaction must be a redox reaction and a combination reaction at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of whether this is a combination reaction or not is pretty easy: we get one compound from the original two, so yes, it is a combination reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not a redox reaction, because no particles' oxidation number changes, so no oxidation/reduction actually happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion this is a combination, but not a redox reaction, so it doesn't qualify as a combination redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="9344" LastEditorUserId="4945" LastEditDate="2019-01-15T14:31:37.703" LastActivityDate="2019-01-15T14:31:37.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32773" PostTypeId="2" ParentId="32770" CreationDate="2015-06-12T00:09:06.120" Score="8" Body="&lt;p&gt;To determine whether a reaction is a redox reaction, you first have to find the oxidation number of each element in the reactants and products. You do this by either counting the number of electrons in each bond and assigning them to the most electronegative atom in the bond, or by memorizing some rules and applying them. When you do that for this reaction, you should get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;  \ce{   &amp;amp;&amp;amp;CaO + SO2 &amp;amp;-&amp;gt; CaSO3}\\&#xA;  \ce{Ca:&amp;amp;&amp;amp; +2 &amp;amp;-&amp;gt; +2}\\&#xA;  \ce{O: &amp;amp;&amp;amp; -2, -2 &amp;amp;-&amp;gt; -2}\\&#xA;  \ce{S: &amp;amp;&amp;amp; +4 &amp;amp;-&amp;gt; +4}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the oxidation number hasn't changed for any element, no reduction or oxidation has taken place, and this is not a redox reaction.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="4945" LastEditDate="2019-01-15T14:29:53.087" LastActivityDate="2019-01-15T14:29:53.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32774" PostTypeId="2" ParentId="32743" CreationDate="2015-06-12T01:39:13.583" Score="4" Body="&lt;p&gt;In most cases I've seen, ALD is performed on the intended substrate - that is, there's no need to separate it. In other cases, you use a sacrificial substrate that you can remove easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to separate the film from the substrate, there are various techniques, but nothing comes to mind for separating Au from Cu. I find it &lt;em&gt;incredibly&lt;/em&gt; unlikely in the specific case of a 5-atom thick layer of Cu, which would likely not survive typical techniques (vs. bending, mushing, forming a larger packing, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some typical techniques for separating two films:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adhesive tape (as mentioned in a comment) - works for many films, e.g., graphene, other layered materials, where you physically pull apart layers&lt;/li&gt;&#xA;&lt;li&gt;Swelling (e.g., separating Au from mica) - including various intercalation methods used in graphenes and graphene oxides and other layered materials, so that the layers separate more easily&lt;/li&gt;&#xA;&lt;li&gt;Dissolving, etching, or eroding the substrate (e.g., back-etching a Si wafer to expose an $\ce{SiO2}$ or thin metal film deposited on the wafer)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In my group, for example, we make &quot;template-stripped gold&quot; where Au is deposited on cleaved mica substrates. Then the thin gold layer is glued to glass substrates and the mica is separated from the Au by swelling / degredation, leaving a very smooth Au surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique can work with fairly thin layers of gold (~10-30 nm) but there's a large difference between Au and mica, making separation possible. Not true with Au/Cu.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-06-12T01:39:13.583" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32775" PostTypeId="1" AcceptedAnswerId="32790" CreationDate="2015-06-12T02:16:56.567" Score="8" ViewCount="9805" Body="&lt;p&gt;As mentioned in the title, why do we use the Kovat's Index in gas phase chromatography? It seems a little arbitrary to compare everything to the order that n-alkanes (such as n-pentane, n-hexane, and n-heptane) emerge from the column, and I am curious to know reasons why we use such an index. &lt;/p&gt;&#xA;" OwnerUserId="11585" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T02:17:48.350" LastActivityDate="2017-08-02T01:05:13.823" Title="What is the Kovat's Index used for?" Tags="&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32776" PostTypeId="2" ParentId="32487" CreationDate="2015-06-12T03:18:12.847" Score="5" Body="&lt;p&gt;The classic paper of &lt;a href=&quot;https://profiles.stanford.edu/jens-noerskov&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Norskov&lt;/a&gt; explained this phenomenon very well. (Off topic: Dr. Norskov is like the godfather of catalysis and I had a chance to meet him at a summer school and that was a great experience for me.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inertness of metal surface to adsorb depends on two factors: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Degree of filling of $d$ orbital  &lt;/li&gt;&#xA;&lt;li&gt;Degree of overlapping of orbital&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Gold and Copper both has filled $d$ orbital ($d^{10})$. That means the main contributor to the reactivity of transition metal is totally absent here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second factor is very crucial to explain the inertness. In case of hydrogen adsorption on $\ce{Au}$ and $\ce{Cu}$ the anti-bonding orbital is below Fermi level, that means anti-bonding is filled for the adsorption of $\ce{H}$ on $\ce{Au}$ and $\ce{Cu}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UkZmx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5LfJA.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first picture, we can see that for noble metals Fermi level is too high and in the second (actual SCF DOS) picture we can see that $\ce{Au}$, $\ce{Cu}$ had anti-bonding DOS which is below the Fermi level. Now, How it can explain the inertness? Fermi level is almost analogous to HOMO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fermi level is the energy level where the orbital has equal probability of being filled or unfilled (Fermi-Dirac statistics). From the left picture of the 1st picture you can see the energy profile. To better explain I tried to use some abstract numerical number to define the importance of anti-bonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you want orbitals $A$ and $B$ to form the MO $AB$. Now $A$ has energy -10 and $B$ has energy -15, so they have total -25 unit energy. Now for the stability, the system will always try to minimize energy. The bonding orbital has -30 unit energy and anti-bonding orbital has 7 unit energy. So they have total -23 unit energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is why the bonding + the anti-bonding energy &gt; $A + B$. The reason is when an atomic orbital forms a molecular orbital, there is an energy penalty associated due to the orthogonalization of the orbitals. So, in case of $\ce{H2}$ formation, there is no electron in antibonding orbital, so total energy is reduced and that's why $\ce{H_2}$ is more stable than $\ce{H}$ atom; while in the case of $\ce{He}$, both bonding and anti-bonding orbitals are filled, so it's energetically more favorable for it to be in an isolated state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, In $\ce{Cu}$ and $\ce{Au}$ the antibonding state is below the Fermi level, that means anti-bonding states are filled and that's why it is energetically unfavorable. But in case of $\ce{Ni}$ and $\ce{Pt}$ antibonding orbital is above Fermi level and hence the adsorption process is more favorable. There are some other terms to explain it from the quantum mechanics viewpoint (like the overlap matrix $V_{sd}^2$ and some other fundamental terms) but the basic idea is the DOS of bonding and anti-bonding orbitals are the main reason for this inertness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://www.nature.com/nature/journal/v376/n6537/pdf/376238a0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nature.com/nature/journal/v376/n6537/pdf/376238a0.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7448" LastEditDate="2015-06-15T20:46:58.803" LastActivityDate="2015-06-15T20:46:58.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32777" PostTypeId="2" ParentId="9145" CreationDate="2015-06-12T03:56:51.893" Score="5" Body="&lt;p&gt;WebElements page about &lt;a href=&quot;http://www.webelements.com/chromium/atoms.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;chromium&lt;/a&gt; (and a number of resources) agree with the comment by @Philipp:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ground state electron configuration of ground state gaseous neutral chromium is  $\ce{[Ar]}3d^54s^1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which in some resources is written as $\ce{[Ar]}4s^13d^5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jdBaE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the Royal Society of Chemistry article &lt;a href=&quot;http://www.rsc.org/eic/2013/11/aufbau-electron-configuration&quot; rel=&quot;nofollow noreferrer&quot;&gt;The trouble with the aufbau principle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;it appears that the most stable configuration for atoms of &lt;strong&gt;chromium&lt;/strong&gt;, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum and gold involves only moving one electron into an $s$ orbital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chromium is one of a handful of Transition elements that share this electron configuration.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-30T13:22:12.663" LastActivityDate="2015-06-30T13:22:12.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32778" PostTypeId="1" CreationDate="2015-06-12T07:27:28.863" Score="2" ViewCount="88" Body="&lt;p&gt;We want to nanoencapsulate our pesticide using the following protocol; but we don’t have AOT surfactant. So, what can I use instead of AOT in the following protocol?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sodium alginate solution in water was emulsified into dioctyl sodium sulfo-succinate (AOT) solution in methylene chloride by sonication for 1 min over ice bath. The primary emulsion was further emulsified into 150 mL of aqueous PVA solution by sonication for 1 min over ice bath to form a secondary water-in-oil-in-water emulsion. The emulsion was stirred using a magnetic stirrer, and 50 mL of aqueous calcium chloride solution was added gradually to the above emulsion. The emulsion was stirred further at ambient conditions for 18 h and then under vacuum for 1 h to evaporate methylene chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16891" LastEditorDisplayName="user7951" LastEditDate="2015-06-12T07:48:33.627" LastActivityDate="2015-06-12T07:48:33.627" Title="What can I use instead of AOT in the following protocol?" Tags="&lt;nanotechnology&gt;&lt;surfactants&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32779" PostTypeId="1" CreationDate="2015-06-12T07:40:34.027" Score="5" ViewCount="204" Body="&lt;p&gt;Are (cyclic) $\ce{C}$-chain molecules as in the picture likely to occur or do they exist at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBzPt.png&quot; alt=&quot;C-chain&quot;&gt; and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering only double bonds between $\ce{C}$ atoms. This rules out most common allotropes.&lt;/p&gt;&#xA;" OwnerUserId="16890" LastEditorUserId="16890" LastEditDate="2015-06-17T08:08:27.093" LastActivityDate="2015-06-17T08:08:27.093" Title="Only-C chain molecules" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32781" PostTypeId="2" ParentId="32769" CreationDate="2015-06-12T07:53:58.027" Score="4" Body="&lt;p&gt;Your calculation is right.  Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you calculate the equilibrium constant of the neutralization reaction &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + OH- &amp;lt;=&amp;gt; CH3COO- + H2O}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you write  of reaction of acetate ion with water:&lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COO- + H2O &amp;lt;=&amp;gt; CH3COOH + OH-}$$&lt;/span&gt;&#xA;It's the opposite of the neutralization reaction. Its  equilibrium constant  is  &lt;span class=&quot;math-container&quot;&gt;$10^{-9.25}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see clearly that the neutralization reaction is the &quot;predominant reaction&quot;. So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2020-04-26T07:03:14.950" LastActivityDate="2020-04-26T07:03:14.950" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32782" PostTypeId="1" AcceptedAnswerId="32794" CreationDate="2015-06-12T08:24:14.297" Score="18" ViewCount="4097" Body="&lt;p&gt;Europium is apparently &lt;a href=&quot;http://www.rsc.org/chemistryworld/podcast/Interactive_Periodic_Table_Transcripts/Europium.asp&quot;&gt;used&lt;/a&gt; as an anti-counterfeiting measure in Euro bank notes because of its fluorescence under UV light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why it is specifically $\ce{Eu}$ and not any of the other lanthanides that fluoresce as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because $\ce{Eu}$ is relatively rare and expensive so it's harder to fake? Is it because it has a very characteristic fluorescence? Is it because they just thought it's funny to use $\ce{Eu}$ in Euros?&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T08:27:25.333" LastActivityDate="2020-09-07T21:34:15.760" Title="Why is europium used in Euro bank notes?" Tags="&lt;everyday-chemistry&gt;&lt;rare-earth-elements&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32784" PostTypeId="1" AcceptedAnswerId="32796" CreationDate="2015-06-12T08:46:17.430" Score="37" ViewCount="30659" Body="&lt;p&gt;We usually use super glue to stick any two objects but I was wondering like why don't they stick inside their own bottle ?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastActivityDate="2019-02-16T00:23:30.017" Title="Why super glue does not stick inside their own bottle?" Tags="&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32789" PostTypeId="2" ParentId="14141" CreationDate="2015-06-12T11:33:59.237" Score="16" Body="&lt;p&gt;The current version of &lt;a href=&quot;http://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book)&lt;/em&gt;&lt;/a&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., …). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So far, the systematic name of &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2}$&lt;/span&gt; should be calcium dichloride.&lt;/p&gt;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound (…).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the customary name calcium chloride is correct.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-06-12T11:33:59.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32790" PostTypeId="2" ParentId="32775" CreationDate="2015-06-12T11:53:50.803" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do we use the Kováts index? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Kováts index is used to normalize GC data. &lt;a href=&quot;https://en.wikipedia.org/wiki/Kovats_retention_index&quot; rel=&quot;noreferrer&quot;&gt;This Wikipedia page&lt;/a&gt;(though not very long), sums up why you would want to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Retention times of the same compound on even two different versions of identical instrument with the same temperature and pressure program using the same column from the same manufacturer might not be the same! Retention time is dependent on so many things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The identity of the analyte&lt;/li&gt;&#xA;&lt;li&gt;The identity of the stationary phase&lt;/li&gt;&#xA;&lt;li&gt;The length of the column (did you know that columns sometimes get trimmed as they age to remove deposits?)&lt;/li&gt;&#xA;&lt;li&gt;The diameter of the column&lt;/li&gt;&#xA;&lt;li&gt;The thickness of the stationary phase film&lt;/li&gt;&#xA;&lt;li&gt;The flow rate of the carrier gas&lt;/li&gt;&#xA;&lt;li&gt;The temperature or temperature program of the oven &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With so many variables, gas chromatography (indeed all chromatography) can seem like a forbidding black art in which one has no hope of being able to compare results from one instrument to the next or one lab to the next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, we're smarter than that. We can always compare our analyte's retention time to some retention times of well behaved standards. If we know the elution order (will not change) of a series of well-behaved standards on several instruments, and we know when our analyte elutes in relation to those standards on one instrument, we have a good starting point for predicting the retention time of that analyte on another instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ervin Kováts took this idea a step farther and created a quantitative index relating the retention time of an unknown analyte to the retention times of known standards in a way that produced &lt;strong&gt;system-independent&lt;/strong&gt; constants describing the elution behavior of compounds. The &lt;em&gt;system-independent&lt;/em&gt; part is important, given all of the factors listed above that can influence retention time. Every analyte should have (nearly) the same Kováts index value on any GC system under any set of parameters. Thus, if we determine the Kováts index of analyte $\ce{A}$ on one GC, we can know something about how $\ce{A}$ will behave on other instruments or with a different temperature program, etc.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why the linear alkanes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The choice of the &lt;em&gt;n&lt;/em&gt;-alkanes was probably not arbitrary, but I cannot say for certain. Of the two links from the Wikipedia article that might contain a rationale, one is dead and the other is in German behind a paywall. However, any well-understood series of standards would have worked. My guess would be that Kováts chose the linear alkanes for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They form a homologous series increasing one carbon at a time. &lt;/li&gt;&#xA;&lt;li&gt;A great many of them are (were - Kováts developed this index in the 1950s) available commercially in high purity (high enough purity that they can be easily purified further). At Sigma Aldrich, you can buy all of the &lt;em&gt;n&lt;/em&gt;-alkanes at least up to eicosane $\ce{C20H42}$.&lt;/li&gt;&#xA;&lt;li&gt;The first 20 linear alkanes cover a wide range of boiling points (and thus retention times): Methane $\text{BP}=-161.5\ ^\circ \text{C}$, pentane (the first liquid at room temperature) $\text{BP}=35.9\ ^\circ \text{C}$, decane $\text{BP}=173.8\ ^\circ \text{C}$, and eicosane $\text{BP}=343.1\ ^\circ \text{C}$.&lt;/li&gt;&#xA;&lt;li&gt;There is a relatively reliable linear relationship between $\log(t_r)$ and the number of carbon atoms in the chain.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How do you determine and use the Kováts index?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Lets say we have an analyte $\ce{A}$ that has a retention time of 193 seconds on some system. First, we identify the linear alkanes that have retention times on either side of the analyte. Lets say that heptane has a retention time of 170 seconds and octane has a retention time of 200 seconds in this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the index looks like this (for non-isothermal programs):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$I$ is the Kováts index&lt;/li&gt;&#xA;&lt;li&gt;$n$ is the number of carbon atoms in the alkane with the lower retention time&lt;/li&gt;&#xA;&lt;li&gt;$N$ is the number of carbon atoms in the alkane with the higher retention time&lt;/li&gt;&#xA;&lt;li&gt;$a$ is the analyte&lt;/li&gt;&#xA;&lt;li&gt;$t_r$ is retention time, so $t_{r,a}$ is the retention time of the analyte&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For our example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=100\times \left[7+\left(8-7\right)\dfrac{193-170}{200-170} \right]$$&#xA;$$I=100\times \left[7+\dfrac{23}{30}\right]=777$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we want to analyze $\ce{A}$ on a different instrument, where the retention times of heptane and octane are 217 and 289 seconds, respectively. We can plug values back into the equation for the index to predict the retention time of $\ce{A}$ on this system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$&#xA;$$777=100\times \left[7+\left(8-7\right)\dfrac{t_{r,a}-217}{289-217} \right]$$&#xA;$$7.77=\left[7+\dfrac{t_{r,a}-217}{72} \right]$$&#xA;$$0.77=\dfrac{t_{r,a}-217}{72}$$&#xA;$$55.2={t_{r,a}-217}$$&#xA;$$t_{r,a}=272$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2015-06-12T14:46:54.913" LastActivityDate="2015-06-12T14:46:54.913" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32791" PostTypeId="2" ParentId="7395" CreationDate="2015-06-12T11:54:31.577" Score="2" Body="&lt;p&gt;Aesin's answer is almost there. In geosciences and mineralogy, the notation (A,B)X means that both elements A and B exist in the same crystal structure in one mineral. This is not a mechanical mixture of iron and nickel sulfides, nor it is an amorphous mixture of ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements are sorted according to their abundance, and whatever is in parentheses should stoichiometrically sum up to whatever subscript there is. Thus (Ni,Fe)S has more Ni than Fe, and together they sum up to 1 atom(s) per formula unit. An example of what the formula can be is Ni&lt;sub&gt;0.8&lt;/sub&gt;Fe&lt;sub&gt;0.2&lt;/sub&gt;S or Ni&lt;sub&gt;0.55&lt;/sub&gt;Fe&lt;sub&gt;0.45&lt;/sub&gt;S&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="402" LastEditDate="2017-04-08T22:21:57.750" LastActivityDate="2017-04-08T22:21:57.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32792" PostTypeId="1" CreationDate="2015-06-12T12:05:53.240" Score="2" ViewCount="531" Body="&lt;p&gt;Boiling point, as I understand, generally depends on molar mass and the intermolecular interactions in the liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the other influencing factors?&lt;/li&gt;&#xA;&lt;li&gt;What intermolecular forces are present in liquid metals?&lt;/li&gt;&#xA;&lt;li&gt;Why does the boiling point of manganese deviate?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16897" LastEditorUserId="7448" LastEditDate="2015-06-12T13:01:53.230" LastActivityDate="2015-06-12T13:01:53.230" Title="What does boiling point of a metal depend on, specifically manganese?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32794" PostTypeId="2" ParentId="32782" CreationDate="2015-06-12T12:19:34.473" Score="14" Body="&lt;p&gt;As already noted by the OP, many rare earths fluoresce under ultraviolet light &lt;strike&gt;and there seems to be no particular reason why Europium is a better choice than any other element, purely based on its fluorescent properties.&lt;/strike&gt; (See update at bottom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The European Central Bank keeps the exact nature of the compounds used a secret. However, it was discovered in 2002 by Frank Suijver and Andries Meijerink &lt;sup&gt;[1]&lt;/sup&gt; that europium, complexed with two $\beta$ diketone molecules provides the red light but the other colours were more difficult to identify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the reason for the choice of Europium is unclear. According to Hugh Aldersey-Williams&lt;sup&gt;[2]&lt;/sup&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It [the European Central Bank] wilfully misunderstands my request to know who fought for Europium and drearily asks for my 'understanding that, for security reasons, we cannot comment on the chemical components of the euro banknote security features.'&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently, in the original paper by Suijver and Meijerink they say that they were visited several years beforehand by European Bank officials who were investigating potential luminescent materials so it may be that they inadvertently gave the bank the idea for using Europium in its banknotes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Europium is one of the more expensive lanthanides and so this is probably a factor in its usage to prevent against forgeries, as the OP suggests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there is also simply an element of irony in the fact that Europium is used in Euro bank notes and it may be that it was just some bank official tasked with finding a suitable compound to use who thought that it would be funny to use Europium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the bottom line appears to be that nobody knows the reason because the European Central Bank refuses to tell anyone anything about the compounds they use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; In response to @Nicolau Saker Neto's comment I did some more research into the fluorescence of Europium. Europium(III) (red) is regularly used alongside Europium(II) (blue) and Terbium(III) (green) in trichromatic lighting systems. It may therefore be the case that Europium was chosen because it is already well studied and in common use and that they created a $\beta$ diketone complex in order to fine tune the exact wavelength of the emissions and make it unique to their banknotes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://dspace.library.uu.nl/handle/1874/25831&quot;&gt;http://dspace.library.uu.nl/handle/1874/25831&lt;/a&gt; (Paper in Dutch which I can't read. However, Williams refers to some of the content in his book and I have alluded to this above. If anyone can translate the original paper and add anything to the answer that would be very helpful.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2]Hugh Aldersey-Williams; 'Periodic Tales: The Curious Lives of the Elements'; ISBN 978-0-141-04145-2 &lt;a href=&quot;http://www.amazon.co.uk/Periodic-Tales-Curious-Lives-Elements/dp/0141041455&quot;&gt;http://www.amazon.co.uk/Periodic-Tales-Curious-Lives-Elements/dp/0141041455&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-06-12T14:39:57.037" LastActivityDate="2015-06-12T14:39:57.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32796" PostTypeId="2" ParentId="32784" CreationDate="2015-06-12T13:03:25.647" Score="38" Body="&lt;p&gt;From &lt;a href=&quot;http://www.explainthatstuff.com/adhesives.html&quot;&gt;explainthatstuff.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Why doesn't glue stick to the tube?&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Adhesives are designed to work when they leave the tube—and not&#xA;  before. Different adhesives achieve this in different ways. Some are&#xA;  dissolved in chemicals called &lt;strong&gt;solvents that keep them stable and&#xA;  non-sticky in the tube&lt;/strong&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When you &lt;em&gt;squeeze them out&lt;/em&gt;, the &lt;strong&gt;solvents&#xA;  quickly evaporate in the air or get absorbed by the surfaces you're&#xA;  sticking to&lt;/strong&gt;, freeing the adhesives themselves to do their job.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Plastic modeling glue works like this. It contains molecules of&#xA;  polystyrene in&#xA;  an acetone solvent. When you squeeze the tube, the glue spurts out and&#xA;  you can usually smell the very strong acetone as it evaporates.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Once it's gone, the polystyrene molecules lock together to make strong&#xA;  chemical bonds.&#xA;  Glue doesn't smell when it's dry because all the&#xA;  solvent has vanished into the air.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Some glues (such as synthetic,&#xA;  epoxy resins) have to be mixed together before they work. They come in&#xA;  two different tubes, one containing the synthetic resin and the other&#xA;  containing a chemical that makes the resin harden. The two chemicals&#xA;  are useless by themselves but, mixed together, form a tough, permanent&#xA;  adhesive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.sciencefocus.com/blog/how-does-super-glue-work&quot;&gt;sciencefocus.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;How super glue works&lt;/h2&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;main ingredient in super glue is a chemical called &lt;em&gt;cyanoacrylate&lt;/em&gt;&lt;/strong&gt;, an acrylic resin that forms a bond (cures) almost&#xA;  instantly;&lt;/li&gt;&#xA;  &lt;li&gt;The &lt;strong&gt;only trigger it requires are the &lt;em&gt;hydroxide ions&lt;/em&gt; found in water&lt;/strong&gt;;&lt;/li&gt;&#xA;  &lt;li&gt;Traces of water are found on almost everything and in the surrounding air, which causes the glue to become sticky;&lt;/li&gt;&#xA;  &lt;li&gt;It works best in moist conditions, although it is not advised to start gluing on a soaking wet surface.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the lack of water in the tube, or the chance of solvents evaporation, doesn't allow the glue to stick to the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: that strong smell that comes out when we open the tube's cap is due to the solvents evaporation, so once they're gone the glue sticks. If a superglue tube is left open for a long time (exposed to air, allowing the water to come in contact with the glue), it's impossible (or almost) to squeeze it out since it got stuck to the tube.&lt;/p&gt;&#xA;" OwnerUserId="16900" LastEditorUserId="16900" LastEditDate="2015-06-13T08:59:56.097" LastActivityDate="2015-06-13T08:59:56.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32797" PostTypeId="1" CreationDate="2015-06-12T13:06:00.563" Score="1" ViewCount="278" Body="&lt;p&gt;I'm on the email list of an Organic Chemistry help website, and they sent out this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://orgomadesimple.com/june-9-email-practice-problem/&quot; rel=&quot;nofollow&quot;&gt;http://orgomadesimple.com/june-9-email-practice-problem/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me thinking: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What if the nitrogen here was replaced with another oxygen atom? &lt;/li&gt;&#xA;&lt;li&gt;Would it still be $sp^2$ hybridized? &lt;/li&gt;&#xA;&lt;li&gt;Can oxygen molecules hybridize?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help!&lt;/p&gt;&#xA;" OwnerUserId="16901" LastEditorUserId="14" LastEditDate="2015-06-12T13:43:28.920" LastActivityDate="2015-06-12T13:43:28.920" Title="Hybridization of Ester Oxygen" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-06-12T15:36:18.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="32799" PostTypeId="1" CreationDate="2015-06-12T13:23:44.730" Score="1" ViewCount="172" Body="&lt;p&gt;To anaesthetise the mucous membrane 2 separate drugs were given topically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxybuprocaine hydrochloride and Lidocaine hydrochloride. The reason to use both of them seem to be beginning of action and its duration (the former starts faster but shorter, the latter starts later but longer acting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead having them in separate bottles they will be mixed in one, would it cause any interactions between them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;oxybuprocaine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1dhBR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hpfDT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4548" LastEditorUserId="10185" LastEditDate="2015-06-12T13:32:23.647" LastActivityDate="2016-01-08T23:18:40.217" Title="Can Oxybuprocaine hydrochloride react with lidocaine hydrochloride?" Tags="&lt;organic-chemistry&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32800" PostTypeId="1" AcceptedAnswerId="32801" CreationDate="2015-06-12T13:35:20.463" Score="1" ViewCount="62" Body="&lt;p&gt;What is a good rule of thumb to use to determine if equilibrium is the appropriate assumption?  I assume it is mainly to do with the time scales of the problem, meaning if the reaction time scales are very short, we could probably assume that we get to equilibrium very quickly.  Is there an accepted metric for determining if the equilibrium assumption is valid?&lt;/p&gt;&#xA;" OwnerUserId="16902" LastActivityDate="2015-06-12T19:03:01.387" Title="When is equilibrium analysis appropriate?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32801" PostTypeId="2" ParentId="32800" CreationDate="2015-06-12T14:07:54.040" Score="0" Body="&lt;p&gt;It depends on the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted threshold is &quot;when everything else is slow by comparison.&quot; For example, if you have a reaction that takes place in two parts, a Jones oxidation taking a secondary alcohol to a ketone and Diels-Alder (which is far slower), you can assume for modeling purposes that the oxidation will be in equilibrium whenever you're looking at how the Diels-Alder will interact with the oxidation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, if you're reacting something like a $\ce{CN7}$ anion (which will react very very quickly, detonating with no provocation), the Jones oxidation can't be considered in equilibrium by comparison.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T19:03:01.387" LastActivityDate="2015-06-12T19:03:01.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32802" PostTypeId="1" CreationDate="2015-06-12T14:26:00.847" Score="1" ViewCount="187" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If $3.7~\mathrm{mol}$ of $\ce{Cu}$ reacts with an excess of $\ce{Cl2}$ and the exchange is $45\,\%$, how many moles of $\ce{CuCl2}$ can at maximum be created?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I start with write the reaction formula&#xA;$$\ce{Cu + Cl2 -&amp;gt; CuCl2}$$&#xA;and then balance it.&#xA;$$\ce{2Cu + 2Cl2 -&amp;gt; 2CuCl2}$$&#xA;I know that I have $3.7\ \mathrm{mol}$ of $\ce{Cu}$, which is $235.1202\ \mathrm{g}$ of $\ce{Cu}$.&#xA;I assume that $\ce{Cu}$ is limiting reagent and $\ce{Cl2}$ is an excess reagent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I get stuck, I don’t know how I’m supposed to proceed with the only information that the exchange is $45\,\%$.&lt;/p&gt;&#xA;" OwnerUserId="16903" LastEditorUserId="5026" LastEditDate="2018-02-24T03:25:29.417" LastActivityDate="2018-02-24T03:25:29.417" Title="How to proceed when percentage &quot;exchange&quot; is given in a reaction?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32804" PostTypeId="2" ParentId="32514" CreationDate="2015-06-12T14:32:51.877" Score="8" Body="&lt;p&gt;Okay I will take a stab at this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following &lt;strong&gt;bond stretching parameters&lt;/strong&gt; in the TINKER FF file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      ##################################&#xA;      ##                              ##&#xA;      ##  Bond Stretching Parameters  ##&#xA;      ##                              ##&#xA;      ##################################&#xA;&#xA;&#xA;bond          1    1          310.00     1.5260&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This tells you that this is a &lt;code&gt;bond&lt;/code&gt; parameter between &lt;code&gt;atom 1&lt;/code&gt; and &lt;code&gt;atom 1&lt;/code&gt;.  The index of atoms is given in the FF file as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   #####################################################&#xA;   ##                                                 ##&#xA;   ##  TINKER Atom Class Numbers to Amber Atom Types  ##&#xA;   ##                                                 ##&#xA;   ##    1  CT      11  CN      21  OW      31  HO    ##&#xA;   ##    2  C       12  CK      22  OH      32  HS    ##&#xA;   ##    3  CA      13  CQ      23  OS      33  HA    ##&#xA;   ##    4  CM      14  N       24  O       34  HC    ##&#xA;   ##    5  CC      15  NA      25  O2      35  H1    ##&#xA;   ##    6  CV      16  NB      26  S       36  H2    ##&#xA;   ##    7  CW      17  NC      27  SH      37  H3    ##&#xA;   ##    8  CR      18  N*      28  P       38  HP    ##&#xA;   ##    9  CB      19  N2      29  H       39  H4    ##&#xA;   ##   10  C*      20  N3      30  HW      40  H5    ##&#xA;   ##                                                 ##&#xA;   #####################################################&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So clearly this is a bond parameter between CT and CT.  What do these atom notations stand for?  They are outlined in Table 1 of &lt;a href=&quot;http://ffamber.cnsm.csulb.edu/ffamber/pdfs/cornell_amber94_1995jacs.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;.  The value &lt;code&gt;310.00&lt;/code&gt; is $K_r$ (in kcal mol$^{-1}$)  and &lt;code&gt;1.5260&lt;/code&gt; is $r_{\mathrm{eq}}$ (in Ang).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;angle bends&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      ################################&#xA;      ##                            ##&#xA;      ##  Angle Bending Parameters  ##&#xA;      ##                            ##&#xA;      ################################&#xA;&#xA;&#xA;angle         1    1    1      40.00     109.50&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;angle&lt;/code&gt; denotes what you'd expect.  The last two parameters, &lt;code&gt;40.00&lt;/code&gt; ($K_{\theta}$ in kcal (mol radian$^2$)$^{-1}$) and &lt;code&gt;109.50&lt;/code&gt; ($\theta_{\mathrm{eq}}$ in deg.) are for CT-CT-CT (atom1-atom1-atom1).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;torsional angles&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      ############################&#xA;      ##                        ##&#xA;      ##  Torsional Parameters  ##&#xA;      ##                        ##&#xA;      ############################&#xA;&#xA;&#xA;torsion       1    1    1    1            0.156    0.0  3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is clearly a torsion between CT-CT-CT-CT.  Here is where it gets a little tricky.  If anyone finds this to be erroneous, please let me know and I will correct it.  The parameter, &lt;code&gt;0.156&lt;/code&gt; is simply determined by the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\frac{V_n}{2}}{\mathrm{no.~of~paths}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V_n/2$ is the magnitude of torsion in kcal mol$^{-1}$.  In Table 14 of the previously cited paper, for Torsional Parameters you find for &lt;code&gt;X-CT-CT-X&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;no. of paths = 9&lt;/li&gt;&#xA;&lt;li&gt;$V_n/2$ = 1.40&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, $1.40/9$ $\approx$ 0.156 which is exactly what is given in the TINKER FF file.  I've done this with a few other torsional parameters and it seems to add up. I believe footnote $f$ in Table 14 explains this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;0.0&lt;/code&gt; is simply $\gamma$ (phase offset in deg.) and &lt;code&gt;3&lt;/code&gt; is $n$ (the periodicity of the torsion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Functional Form of Amber94 FF&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be useful (and convenient) to spell out the functional form as presented in the cited paper here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{\mathrm{tot}} = \sum_{\mathrm{bonds}} K_r(r-r_{\mathrm{eq}})^2 +$$&#xA;$$~~~~~~~~~~~~\sum_{\mathrm{angles}} K_{\theta}(\theta - \theta_{\mathrm{eq}} ) ^2 + $$&#xA;$$~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{\mathrm{dihedrals}} \frac{V_n}{2}[1+\cos(n\phi - \gamma)] + $$&#xA;$$~~~~~~~~~~~~~~~~~~~~~~~\sum_{i&amp;lt;j} \left[ \frac{A_{ij}}{R_{ij}^2} - \frac{B_{ij}}{R_{ij}^6} + &#xA;\frac{q_i q_j}{\epsilon R_{ij}} \right]$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; Cornell and co-workers. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1995&lt;/strong&gt;, 117, 5179.&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="5026" LastEditDate="2018-06-28T02:47:28.083" LastActivityDate="2018-06-28T02:47:28.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32805" PostTypeId="1" AcceptedAnswerId="37477" CreationDate="2015-06-12T14:39:59.340" Score="3" ViewCount="1501" Body="&lt;p&gt;Well here's a question from my book:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A solution containing one mole per liter of each &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu(NO3)2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Hg2(NO3)2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg(NO3)2}$&lt;/span&gt; is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ag+/Ag} = +0.80$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Hg2^2+/Hg} = +0.79$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+/Cu} = +0.34$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Mg^2+/Mg} = -2.37$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The sequence of deposition of the metals on the cathode will be?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer to the above question is: &lt;span class=&quot;math-container&quot;&gt;$\ce{Ag}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Hg}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Why isn't Mg deposited on the cathode? Is it because the reduction potential is negative?&lt;br&gt;&#xA;If yes, what exactly is the reason?&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-03-28T10:47:48.460" Title="Why isn't Mg deposited on the cathode after electrolysis?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32806" PostTypeId="1" CreationDate="2015-06-12T15:01:23.757" Score="0" ViewCount="575" Body="&lt;p&gt;$$\ce{HSO3- -&amp;gt; S2O6^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxidation number of $\ce{S}$ on the left is +4 and +5 so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one? &#xA;Why is 2 placed in front of the $\ce{HSO3-}$? &lt;/p&gt;&#xA;" OwnerUserId="16905" LastEditorUserId="10185" LastEditDate="2015-06-13T09:47:54.610" LastActivityDate="2015-06-13T09:47:54.610" Title="For the half reaction HSO3- -&gt; S2O6^2- why is it that 2e- are on the right hand side when balancing and not one electron?" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32807" PostTypeId="2" ParentId="32779" CreationDate="2015-06-12T15:25:19.650" Score="3" Body="&lt;p&gt;The term you are looking for is &quot;allotropes&quot;, and carbon has a whole bunch of them, but the three most readily spotted in nature are graphite, diamond and fullerene, however, none of them are simple chains such as those postulated in the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite exists in large honeycomb flakes held together on two axes by covalent bonds, and on the third by weaker Van der Waals forces, which is why they come off readily under mechanical shear, leading to their popularity in pencil lead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond are huge, multiply interlocked complexes of crystalline carbon, although often with impurities locked into their structure, perverting the crystal structure and yielding interesting colors as a byproduct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fullerene is a molecular allotrope, consisting (usually) of 80 carbon atoms arranged in a buckyball. Initially thought to be a lab-grown species, it has since been identified as a natural result of certain combustion processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, none of these are chains like what you describe; while some of the compounds you sketch out are likely possible and present, they are also likely to be highly reactive, perhaps even to the point of decomposing spontaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While amorphous carbon chains may very well exist, without a crystalline structure to shield their electrons, their reactivity would likely be such that they would be unable to exist for very long at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplistically speaking, the long stable chains of carbon found in hydrocarbons aren't entirely characteristic of carbon, as such, but rather, characteristic of hydrogenated carbon.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastActivityDate="2015-06-12T15:25:19.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32808" PostTypeId="1" CreationDate="2015-06-12T15:26:35.660" Score="15" ViewCount="153" Body="&lt;p&gt;Gasoline’s power throughput per kg is $12\,200\:\mathrm{Wh/kg}$ (Wikipedia.org). Are there any other fuels that have a higher power output that are not considered viable solutions as they are difficult to handle or have negative externalities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what characteristics of their molecular structure give them the high energy throughput?&lt;/p&gt;&#xA;" OwnerUserId="7639" LastEditorDisplayName="user7951" LastEditDate="2015-06-12T15:52:17.970" LastActivityDate="2017-01-07T08:22:21.523" Title="Disregarding negative externalities and other barriers, what are some fuels that have a higher power density than gasoline?" Tags="&lt;energy&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32809" PostTypeId="1" AcceptedAnswerId="32811" CreationDate="2015-06-12T15:38:01.170" Score="16" ViewCount="28004" Body="&lt;p&gt;This question is a little confusing. According to my teacher, the entropy should decrease, but I think that sugar is crystalline first and later it dissolves, so the randomness increases and so the entropy should also increase. What is the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="5710" LastEditorUserId="402" LastEditDate="2017-04-18T19:04:32.850" LastActivityDate="2017-06-05T01:40:35.230" Title="On adding sugar to water, does the entropy increase or decrease?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32810" PostTypeId="2" ParentId="32809" CreationDate="2015-06-12T15:53:23.013" Score="1" Body="&lt;p&gt;You're the correct one here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy will go up when a solid becomes aqueous (or dissolved at all) or when liquids mix. This is because there are an increased number of microstates that are indistinguishable from each other when a solid is dissolved or when liquids mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(However, do note that if you dissolve a gas, entropy goes down)&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-12T15:53:23.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32811" PostTypeId="2" ParentId="32809" CreationDate="2015-06-12T15:53:31.487" Score="16" Body="&lt;p&gt;While it may &lt;em&gt;seem&lt;/em&gt; that randomness always increases when a crystal is dissolved into a liquid phase, it does not &lt;em&gt;have&lt;/em&gt; to be that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will act as both a hydrogen bond donor and an acceptor, likely to two different water molecules, creating a cage of water that is rather ordered and large again when comparing to the sodium or chloride ion’s solvation cell. This will decrease the disorder (and hence entropy) of the free water so much, that it counteracts the increase in entropy of the crystal dissolving. (The effect is more pronounced with sucrose when compared to glucose due to the molecule’s size and the number of hydroxyls.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the molecules in a sugar crystal are likely less ordered than the ions of a salt crystal. They will form extensive hydrogen-bonding networks, yes; but overall if you shift one row of molecules relative to its neighbour you won’t necessarily break the crystal as you won’t suddenly have ions of the same charge next to each other. And finally, the single salt crystal will break down into much smaller fragments than the sugar molecules.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So summed up, the effects are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;large solvation cells&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;crystal breaks down into larger (less disordered) fragments upon solvation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sugar crystals themselves are less ordered than salt crystals&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Meaning that there is a lot less entropy to be gained.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="402" LastEditDate="2017-04-18T19:04:48.067" LastActivityDate="2017-04-18T19:04:48.067" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32812" PostTypeId="1" AcceptedAnswerId="32828" CreationDate="2015-06-12T16:03:40.917" Score="4" ViewCount="168" Body="&lt;p&gt;I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{3}&#xA;\ce{X- + 2e- \;&amp;amp;&amp;lt;=&amp;gt; X^3-} \qquad &amp;amp;&amp;amp;E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &amp;amp;&amp;amp;&amp;amp;\text{(a)}\\&#xA;\ce{2H+ + MO^2+ + e- \;&amp;amp;&amp;lt;=&amp;gt; M^3+ + H2O} \qquad &amp;amp;&amp;amp;E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &amp;amp;&amp;amp;&amp;amp;\text{(b)}\\ &#xA;\ce{8H+ + MnO4- + 5e- \;&amp;amp;&amp;lt;=&amp;gt; Mn^2+ + 4H2O} \qquad &amp;amp;&amp;amp;E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &amp;amp;&amp;amp;&amp;amp;\text{(c)}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find out possible reactions, following steps are taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations $\text{(a)}$ and $\text{(c)}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp;{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\&#xA;\ce{&amp;amp;16H+ +5X^3- + 2MnO4- -&amp;gt; 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\&#xA;&amp;amp;E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\&#xA;&amp;amp;\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations $\text{(b)}$ and $\text{(c)}$;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp;{-}\text{(b)} \times 5 + \text{(c)}: \\&#xA;\ce{&amp;amp;5M^3+ + H2O + MnO4- -&amp;gt; Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ &#xA;&amp;amp;E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\&#xA;&amp;amp;\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are there any errors in above calculations? If so, please suggest corrections.&lt;/li&gt;&#xA;&lt;li&gt;According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct?&lt;/li&gt;&#xA;&lt;li&gt;According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct?&lt;/li&gt;&#xA;&lt;li&gt;Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict?&lt;/li&gt;&#xA;&lt;li&gt;If not how can these results be explained?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16096" LastEditorDisplayName="user7951" LastEditDate="2015-06-13T08:10:01.867" LastActivityDate="2015-07-19T17:58:57.523" Title="Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?" Tags="&lt;thermodynamics&gt;&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32813" PostTypeId="2" ParentId="32805" CreationDate="2015-06-12T16:25:18.147" Score="0" Body="&lt;p&gt;Reduction takes place on cathode. Thus the element having most reduction potential will get reduced and get deposited. In your question the reduction potential of $\ce{Mg}$ is the least of all the ones that you have mentioned hence it will not get deposited.&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T18:47:10.117" LastActivityDate="2015-06-12T18:47:10.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32814" PostTypeId="1" AcceptedAnswerId="32815" CreationDate="2015-06-12T17:14:58.810" Score="3" ViewCount="1930" Body="&lt;p&gt;In 2010 NASA announced the discovery of microorganisms that, after being treated in laboratory, could keep growing only using arsenic, instead of the (though very small amount of) phosphate they also had in nature.&#xA;However, this was confuted by further studies in the following two years.&#xA;Has there been success in finding &quot;alien&quot; life on Earth, somewhere else, since then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: the NASA article from back then says what gilleain sais below. I naively hoped a couple of articles from different (not scientific, but not bad) journals were not modifying the facts. Thus the question actually is if non-carbon-based life has been found.&lt;/p&gt;&#xA;" OwnerUserId="16067" LastEditorUserId="16067" LastEditDate="2015-06-12T18:20:45.957" LastActivityDate="2015-06-12T18:20:45.957" Title="Has a non-carbon-based form of life been discovered since 2010?" Tags="&lt;reference-request&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32815" PostTypeId="2" ParentId="32814" CreationDate="2015-06-12T17:40:31.503" Score="11" Body="&lt;p&gt;Just because an organism &lt;em&gt;might&lt;/em&gt; need a nutrient, whether arsenic or cobalt (which &lt;em&gt;homo sapiens&lt;/em&gt; needs for making RBC's) does &lt;em&gt;not&lt;/em&gt; mean that the organism is not carbon-based. What had been announced was that an organism used arsenic to replace phosphorus, not carbon, which has &lt;a href=&quot;http://www.nature.com/news/arsenic-life-bacterium-prefers-phosphorus-after-all-1.11520&quot;&gt;since been amended&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the question whether the GFAJ-1 microbe is a &lt;em&gt;facultative&lt;/em&gt; or &lt;em&gt;obligate&lt;/em&gt; user of arsenic, or simply excludes arsenic. Though the organism can survive high concentrations of arsenic, it appears to do so by discriminating between As and P and using the phosphorus in its metabolism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are organisms such as &lt;a href=&quot;http://health.usgs.gov/dw_contaminants/Elements.As.pdf&quot;&gt;Sulfurospirillum arsenophilum and S. barnesii&lt;/a&gt; that are at least facultative As users, deriving energy from oxidation or reduction of arsenic compounds, but the arsenic is not replacing &lt;em&gt;carbon&lt;/em&gt;, much less phosphorus, in their biochemistry.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-12T17:40:31.503" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32817" PostTypeId="1" CreationDate="2015-06-12T19:01:05.307" Score="2" ViewCount="75" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Be8rh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some trouble going about this question. My first thought is to set up an ASTM separation equation to figure out the difference, but I do not know how to incorporate the reflux ratio which is 0.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASTM: TBP 50% --&gt; 550 (heavy oil)-300(light oil)= 250 degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reflux ratio is 0.5. Do I multiply the ratio by number of plates and then take $\frac{\pu{250 ^\circ C}}{\text{reflux}\times \text{number of plates}}$ to get the answer of 10 degrees?&lt;/p&gt;&#xA;" OwnerUserId="16912" LastEditorUserId="5026" LastEditDate="2018-03-08T06:19:38.777" LastActivityDate="2018-03-08T06:19:38.777" Title="Oil refinery - How many plates are needed for a 10 degrees difference in temperature?" Tags="&lt;physical-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32818" PostTypeId="2" ParentId="32799" CreationDate="2015-06-12T19:27:27.380" Score="1" Body="&lt;p&gt;Looking at the functionality, oxybuprocaine looks more likely to react with itself rather than lidocaine. As it's a salt and also no doubt from numerous medical trials, this doesn't occur,  the anticipation would be that the two compounds would be fine.  The $pK_a$'s of the tertiary amines would also be expected to be similar, so one shouldn't neutralise the other.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorDisplayName="user15489" LastEditDate="2015-06-12T19:33:47.627" LastActivityDate="2015-06-12T19:33:47.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32820" PostTypeId="1" AcceptedAnswerId="32824" CreationDate="2015-06-12T21:59:08.197" Score="5" ViewCount="3286" Body="&lt;p&gt;I understand the purpose behind infrared spectroscopy and mass spectroscopy, but UV-VIS doesn't make much sense to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through research, I see that it has to do with electron orbitals, and it has a relation to Beer's Law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what exactly can I learn about a molecule from its absorbance/transmittance of UV/Visible light?&lt;/p&gt;&#xA;" OwnerUserId="4491" LastEditorUserId="11367" LastEditDate="2015-06-13T16:28:40.473" LastActivityDate="2016-03-10T12:09:38.840" Title="What is the purpose of UV-VIS Spectroscopy?" Tags="&lt;analytical-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32821" PostTypeId="2" ParentId="32806" CreationDate="2015-06-12T22:24:39.223" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is $2$ placed in front of the $\ce{HSO3}$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is, of course, the elemental law of the conservation of mass. You have $x$ atoms that you put into your equation, you need to get $x$ atoms out again. While hydrogen and oxygen can be taken care of with water, you need to account for the sulphur immediately. Hence you need two sulphurs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The oxidation number of S on the left is +IV and +V so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The oxidation numbers are, of course, correct, but remember that in the first step you wrote a $2$ before the sulphur. Therefore you now have &lt;em&gt;two sulphurs&lt;/em&gt; that you need to treat with the appropriate number of electrons — either by stealing or by gifting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you have two sulphurs and they each lose one electron, you need to add $\ce{2 e-}$ to the right.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: The OP already understood what went wrong through hints in the comments. Unfortunately, OP decided not to write up a self-answer.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-12T22:24:39.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32822" PostTypeId="1" CreationDate="2015-06-13T01:10:53.497" Score="8" ViewCount="3125" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zt0xS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case:&lt;br&gt;&#xA;$\ce{H+}$ adds to the $\ce{OH}$ group, giving us a carbocation. The carbocation thus formed is exceptionally stable due to back bonding. I wonder why would it go under ring expansion even though the strain is not a factor here as the ring strain in a cyclobutane ring is ~$26.3\ \mathrm{kcal/mol}$, and that in a cyclopropane ring is ~$27.5\ \mathrm{kcal/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case:&lt;br&gt;&#xA;Again the $\ce{H+}$ adds to the $\ce{OH}$ group, giving us a tertiary carbocation with seven hyper-conjugating structures. Why would it go under ring expansion to give secondary carbocation with just two hyper-conjugating structures? I believe is based on ring strain in this case, as the ring strain in a five-membered ring is ~$6.2\ \mathrm{kcal/mol}$, while the ring strain in a six-membered ring is ~$0.1\ \mathrm{kcal/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source&lt;/strong&gt;: Advanced Problems In Organic Chemistry, MS Chouhan, 11th edition; Chapter - Hydrocarbons (Alkenes); Question 180 in latest edition&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="5026" LastEditDate="2018-03-06T10:53:12.023" LastActivityDate="2018-03-06T10:55:15.403" Title="Ring expansion in cyclic compounds" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32823" PostTypeId="1" AcceptedAnswerId="32830" CreationDate="2015-06-13T02:21:35.193" Score="1" ViewCount="660" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Write down an expression for the dissociation constant, $K_{\mathrm{a}}$, of a very weak monobasic acid $\ce{HA}$ in an aqueous solution in terms of the concentrations of $\ce{H+ (aq), A- (aq) and HA (aq)}$ in the solution.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Hence, show that&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;$$&#xA;\mathrm{p}K_{\mathrm{a}} = \mathrm{p}\ce{H} - \mathrm{log}_{10} \frac{[\ce{A- (aq)}]}{[\ce{HA (aq)}]} \text{ where } \mathrm{p}K_{\mathrm{a}} = - \mathrm{log}_{10} K_{\mathrm{a}}&#xA;$$&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;At a particular temperature, $\pu{2.00e-3 mol}$ of the acid $\ce{HA}$ was dissolved in water and the solution diluted until the volume was $\pu{75.00 cm^3}$. When $\pu{25.00 cm^3}$ of a $\pu{0.004 mol dm^{-3}}\,\ce{NaOH}$ solution was added to that acid solution, the $\mathrm{p}\ce{H}$ of the resulting solution was found to be 6.0. Calculate the dissociation constant, $K_{\mathrm{a}}$, of the acid $\ce{HA}$ at that temperature.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below you can see what I have tried with my current answers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For 1, I got $$K_{\mathrm{a}} = \frac{[\ce{H+ (aq)}][\ce{A- (aq)}]}{[\ce{HA (aq)}]}.$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For 2, I got the same answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I am stuck on 3. These are the things I already know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moles of NaOH added: $(25/10000) * 0.004 = \pu{0.001 mol}$&lt;/li&gt;&#xA;&lt;li&gt;Moles of acid HA present initially: $\pu{0.002 mol}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can I proceed further?&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="194" LastEditDate="2018-01-29T13:09:09.783" LastActivityDate="2018-01-29T13:09:09.783" Title="Derivation and application of the Henderson-Hasselbalch equation" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32824" PostTypeId="2" ParentId="32820" CreationDate="2015-06-13T03:10:29.953" Score="10" Body="&lt;p&gt;The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_04%3A_Structure_Determination_I/Section_4.3%3A_Ultraviolet_and_visible_spectroscopy#4.3B:_Looking_at_UV-vis_spectra&quot; rel=&quot;nofollow&quot;&gt;broad absorption bands&lt;/a&gt; of the electronic transitions observed in UV/Vis spectroscopy/spectrophotometry, as well as the myriad types of electronic transitions that might lead to absorption at a given wavelength, indeed make the technique much less powerful for identification of unknown compounds as compared to, e.g., IR, NMR, microwave, and mass$^*$ spectroscopies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jon Custer is correct in his comment to the question that UV/Vis spectra can provide key information in identifying unknown species.  However, I know of only two approaches to obtaining predictions of a given UV/Vis spectrum:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Experimental spectra of compounds suspected to be structurally related to the unknown species&lt;/li&gt;&#xA;&lt;li&gt;Quantum chemical computation of the electronic transition spectra of candidate species&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The former approach requires the relevant experimental data to exist; the latter is &lt;a href=&quot;http://dx.doi.org/10.1016/j.ccr.2008.05.014&quot; rel=&quot;nofollow&quot;&gt;not trivial&lt;/a&gt; to perform accurately. Both suffer from significant uncertainties: How similar is a reference experimental spectrum to that of the unknown species? Can I trust that my quantum computation will give realistic band locations and strengths?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In my experience, the strength of UV/Vis lies in its facility for inexpensive, convenient, routine, non-destructive quantitation of known solutes, with known &lt;a href=&quot;http://www.chemguide.co.uk/analysis/uvvisible/theory.html&quot; rel=&quot;nofollow&quot;&gt;absorption spectra&lt;/a&gt; and/or &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_attenuation_coefficient&quot; rel=&quot;nofollow&quot;&gt;extinction coefficient(s)&lt;/a&gt; at given wavelength(s).  The following list of strengths is almost certainly incomplete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The instrumentation and consumables are relatively inexpensive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No auxiliary cooling, gas supply, etc. are required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Preparation and analysis of fixed-volume samples is trivial: pour liquid into cuvette; place cuvette into instrument; measure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous analysis of dynamic systems (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromatography&quot; rel=&quot;nofollow&quot;&gt;chromatographic&lt;/a&gt; eluates and ongoing reactions) is possible using a flow cell.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A simple example is described &lt;a href=&quot;http://dx.doi.org/10.1021/ed066p604&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, unfortunately paywalled.&lt;/li&gt;&#xA;&lt;li&gt;A wide variety of configurations is presented at a commercial vendor &lt;a href=&quot;http://www.starna.com/ukhome/d_cells/d_cells_s/flow/T0585-1_2.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;(No endorsement intended. There are numerous other suppliers out there; I chose this one because it had a good variety shown in schematic.)&lt;/em&gt;&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Interference from most solvents is minimal, particularly in the visible range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law&quot; rel=&quot;nofollow&quot;&gt;Beer's Law&lt;/a&gt; in its linear form has a relatively high dynamic range for most UV/Vis-active solutes$-$that is, the concentration where absorbance deviates from linearity is much greater than the minimum concentration detectable:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{C_{\max}}{C_{\min}} \gg 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Another implication of Beer's Law is that interference(s) from a small number of absorbing co-solutes are relatively straightforward to handle, either by measuring at &lt;a href=&quot;https://en.wikipedia.org/wiki/Isosbestic_point&quot; rel=&quot;nofollow&quot;&gt;isobestic points&lt;/a&gt; or by measuring multiple wavelengths and solving the linear absorbance equations for the unknown concentrations. For example, if two species $\ce{X}$ and $\ce{Y}$ are present and absorbances are measured at wavelengths $\lambda_1$ and $\lambda_2$, the concentrations can be found by solving the following two expressions:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A_{\lambda_1} = \epsilon_{\ce{X},\lambda_1}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_1}\cdot C_\ce{Y}&#xA;$$&#xA;$$&#xA;A_{\lambda_2} = \epsilon_{\ce{X},\lambda_2}\cdot C_\ce{X} + \epsilon_{\ce{Y},\lambda_2}\cdot C_\ce{Y}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The steady-state nature of the measurements allows for 'deep' examination of a particular sample.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The full UV/Vis spectrum for a given volume of sample can be measured, giving a rich &quot;fingerprint&quot; of the particular solution and the solutes in it&lt;/li&gt;&#xA;&lt;li&gt;The absorbance at a particular wavelength can be tracked over time, allowing for detailed &lt;a href=&quot;https://en.wikipedia.org/wiki/Reaction_progress_kinetic_analysis&quot; rel=&quot;nofollow&quot;&gt;kinetic analyses&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At least in the visible range, the quality of the substance being measured is directly perceptible to the human senses. This is actually quite viscerally satisfying, to see that a sample anticipated to contain significantly more of a solute does indeed produce a noticeably stronger coloration of the solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While IR and microwave spectroscopies share some of the above advantages, they are far less suited for analysis of solutes, as the solvent absorbance is often much more likely to confound measurements.  Further, the instrumentation and sample preparation are generally much more involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass spec is terrifically expensive and complicated, is an intrinsically destructive method, and is less well suited for direct quantitation since there is no way to know &lt;em&gt;a priori&lt;/em&gt; what fraction of the injected sample is actually making it to the detector and thus careful calibration is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NMR, though non-destructive, also requires quite expensive equipment that is complicated to operate properly, and interpretation of the spectra requires quite a bit of practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I know too little about EPR/ESR, EXAFS, XANES, and others to comment on them here. I believe the above strengths hold in comparison to those methods as well, however.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^*$ &lt;em&gt;Technically, MS is &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_spectrometry&quot; rel=&quot;nofollow&quot;&gt;mass &lt;strong&gt;spectrometry&lt;/strong&gt;&lt;/a&gt;, not spectroscopy, since the measurement is of discrete ion impacts on a detector, as opposed to absorbance of portions of the electromagnetic spectrum by a detector.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2016-03-10T12:09:38.840" LastActivityDate="2016-03-10T12:09:38.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32826" PostTypeId="1" AcceptedAnswerId="33632" CreationDate="2015-06-13T03:41:39.023" Score="10" ViewCount="3873" Body="&lt;p&gt;I've read that for an oxyacid, the oxidation number of the central atom shows the power of that acid, but here, the oxidation numbers of sulphur in sulphuric acid and pyrosulphuric acid are both the same, i.e. +6. That is confusing! How should I proceed now?&lt;/p&gt;&#xA;" OwnerUserId="16919" LastEditorUserId="9961" LastEditDate="2017-05-24T15:53:55.030" LastActivityDate="2017-05-24T15:53:55.030" Title="Why is H2S2O7 stronger than H2SO4?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32827" PostTypeId="2" ParentId="32822" CreationDate="2015-06-13T04:33:20.370" Score="5" Body="&lt;p&gt;You overlooked the strain from the sp&lt;sup&gt;2&lt;/sup&gt; carbon in a small ring. The ring strain you have listed is based on sp&lt;sup&gt;3&lt;/sup&gt; carbon. sp&lt;sup&gt;2&lt;/sup&gt; supposed to have a planar triangular geometry (120°, compared to 109° for sp&lt;sup&gt;3&lt;/sup&gt;) and will be more strained in a 3 or 4 membered ring. 6 membered ring will be best followed by 5 membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you have discussed a lot about the stability of carboncations. Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-03-06T10:55:15.403" LastActivityDate="2018-03-06T10:55:15.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32828" PostTypeId="2" ParentId="32812" CreationDate="2015-06-13T04:42:03.170" Score="5" Body="&lt;p&gt;$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any errors in above calculations? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there must be.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?&#xA;  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both of these cannot be correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it useless to use E values to predict the feasibilities of redox reactions ? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it is just as good and exactly as informative as $\Delta G$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should we always resort to ΔG ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for predictions based on E values and ΔG values to contradict ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If not how can these results be explained ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;UPDATE:&lt;/em&gt;&lt;/strong&gt; &lt;strike&gt;Arithmetic error.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of &lt;em&gt;reaction&lt;/em&gt;, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{10M^3+ + 2 H2O + 2 MnO4- -&amp;gt; 2 Mn^2+ + 10 MO^2+ +4 H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of &quot;$\textbf{mol}$&quot; of &lt;em&gt;electrons&lt;/em&gt;, not of moles of a reaction of arbitrary scale.  In reality the &quot;$\textbf{mol}$&quot; of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So contrary to what I said above, the &lt;strong&gt;sign&lt;/strong&gt; of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is &lt;strong&gt;absolutely&lt;/strong&gt; spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the &lt;em&gt;relative&lt;/em&gt; spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-06-15T16:04:38.240" LastActivityDate="2015-06-15T16:04:38.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32830" PostTypeId="2" ParentId="32823" CreationDate="2015-06-13T06:53:06.957" Score="1" Body="&lt;p&gt;The reaction taking place is the quantitative reaction:&#xA;$$\ce{AH + OH- -&amp;gt; A- + H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial concentrations are:&#xA; $$\ce{[OH- ]_0}=\frac{0.001}{0.1}=0.01  \mathrm{M}$$  $$\ce{[AH ]_0}=\frac{0.002}{0.1}=0.02 \mathrm{M}$$&#xA;The equilibrium concentrations:&#xA; $$\ce{[AH ]}=\frac{0.002-0.001}{0.1}=0.01 \mathrm{M}$$ &#xA; $$\ce{[OH- ]}=10^{-8}  \mathrm{M}$$&#xA; $$\ce{[A- ]}=0.02-0.01=0.01 \mathrm{M}$$ &#xA;Using the equation of question (ii):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{pH = p}K_a=6$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-06-13T06:53:06.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32831" PostTypeId="1" AcceptedAnswerId="32878" CreationDate="2015-06-13T07:57:34.540" Score="7" ViewCount="839" Body="&lt;p&gt;Recently I started renovating a bicycle. The frame has been painted probably over 30 years ago, and despite holding well, big patches of rust are showing. So it's time to give the bike a new paint job. And before that, I need to sand the frame and remove the current paint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly it came to my attention that some bikes around that age have lead paint. Toxic, obviously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm by no means a pro chemist. Far from that actually: if you show me a bunch of formulas, I might understand it, but I'm not skilled for manipulations neither properly equipped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided that I can scratch a bit of that paint with sandpaper, what's the easiest way to test it for presence of lead? Apparently some lead test kits exist, but they're incredibly hard to find here.&lt;/p&gt;&#xA;" OwnerUserId="16920" LastEditorUserId="16920" LastEditDate="2015-06-13T08:52:29.820" LastActivityDate="2015-06-14T04:40:16.657" Title="How to detect lead paint on a bicycle?" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32832" PostTypeId="1" AcceptedAnswerId="33014" CreationDate="2015-06-13T10:00:09.743" Score="13" ViewCount="35257" Body="&lt;p&gt;My professor told us the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaHCO3}$ will react to evolve&#xA;  $\ce{CO2}$ whenever the acidic nature of the reactant exceeds the&#xA;  acidic nature of benzoic acid. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to know why this is. And is the given statement valid? He didn't give us the reason.  And in that case should $\ce{NaHCO3}$ react with 1,2-dihydroxy&#xA;cyclobutanedione to evolve carbon dioxide?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="14" LastEditDate="2015-06-14T15:06:07.910" LastActivityDate="2020-05-01T19:35:01.713" Title="Condition For NaHCO3 to react to form CO2 with organic acids" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32833" PostTypeId="2" ParentId="31904" CreationDate="2015-06-13T10:35:01.250" Score="1" Body="&lt;p&gt;Is there a goal here? If you wanted to react aluminium with concentrated aq ammonia, you would get hydrogen but the rate would be proportional to temperature.  However, this would be quite a poor way to create a reductive environment and as someone has already alluded, aluminium and sodium hydroxide or a strong mineral acid is better. From what I've seen, if you use alkali, a catalyst is needed.&#xA;Ethylamine can be more conveniently used in place of ammonia. &#xA;Calcium can also be used in reductions.I think I've also seen Indium, but that seems a bit expensive. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-13T10:35:01.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32834" PostTypeId="1" CreationDate="2015-06-13T10:37:12.783" Score="6" ViewCount="650" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At the eutectic point, what is the effect on the melting point of adding impurities?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Increase the melting point&lt;/li&gt;&#xA;  &lt;li&gt;Decrease the melting point&lt;/li&gt;&#xA;  &lt;li&gt;No effect on melting point&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know adding impurities decreases the melting point of a compound in comparison to the pure compound. But, the eutectic point is the lowest temperature where the new compound (the compound + impurity) melts. I'm indecisive between 1 and 3.&lt;/p&gt;&#xA;" OwnerUserId="16923" LastEditorUserId="194" LastEditDate="2018-03-03T15:29:07.470" LastActivityDate="2018-03-03T15:29:07.470" Title="At the eutectic point, what is the effect on the melting point of adding impurities?" Tags="&lt;phase&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32835" PostTypeId="1" AcceptedAnswerId="32850" CreationDate="2015-06-13T10:45:41.403" Score="8" ViewCount="2338" Body="&lt;p&gt;Why is it impossible in allene to have two pi-bonds in the same orientation?&#xA;=&gt; Why isn't allene planar? (4 $p$ electrons are in two 2x2 $p$-orbitals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my chemical intuition I would guess that there are 4 electrons in a $p$-system with three orbitals, but frankly I don't know how to allocate the lone pair electrons of $\ce{CS2}$ into the $sp$ and $p$ orbitals.&#xA;=&gt; Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to rationalize this by allocating electrons to different hybrid orbitals (especially in $\ce{CS2}$ I was not sure how many electrons do we have in the 2x3 $p$-orbitals):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oShdd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7201" LastEditDate="2015-06-13T14:26:56.077" LastActivityDate="2015-06-13T16:06:08.420" Title="Why allene cannot be described with an allyl system?" Tags="&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32836" PostTypeId="2" ParentId="29581" CreationDate="2015-06-13T10:49:33.210" Score="0" Body="&lt;p&gt;In your book, it was probably mentioned that the &lt;strong&gt;Reaction Quotient (Q)&lt;/strong&gt; will change, &lt;strong&gt;K&lt;/strong&gt; changes only with temperature and there is no other way to change it. Catalysts also effect only the rate (speed) of reaction. &lt;/p&gt;&#xA;" OwnerUserId="16924" LastActivityDate="2015-06-13T10:49:33.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32838" PostTypeId="1" CreationDate="2015-06-13T11:10:16.453" Score="6" ViewCount="6084" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the order of the following compounds from more polar to less polar?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Hydrochloric acid, &lt;/li&gt;&#xA;  &lt;li&gt;Methanol, &lt;/li&gt;&#xA;  &lt;li&gt;Hexane, &lt;/li&gt;&#xA;  &lt;li&gt;Petroleum benzene  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; has the largest polarity. Hexane I guess has the less. I guess the order should be like:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; &gt; Methanol &gt; Petroleum benzene &gt; Hexane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? I'm not sure of the place of petroleum benzene.&lt;/p&gt;&#xA;" OwnerUserId="16923" LastEditorDisplayName="user7951" LastEditDate="2019-06-02T19:50:42.340" LastActivityDate="2019-06-02T19:50:42.340" Title="What is the order of the following compounds based on polarity?" Tags="&lt;organic-chemistry&gt;&lt;polarity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32841" PostTypeId="1" CreationDate="2015-06-13T11:35:28.540" Score="6" ViewCount="53543" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the effect of air pressure on the melting point of a substance?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;a&lt;/strong&gt;) No effect&lt;br /&gt;&#xA;&lt;strong&gt;b&lt;/strong&gt;) The melting point increases when the air pressure increases&lt;br /&gt;&#xA;&lt;strong&gt;c&lt;/strong&gt;) The melting point decreases when the air pressure increases&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The boiling point has inverse relation with vapor pressure of the liquid and positive relation with atmosphere (air) pressure. The melting point has positive relation with vapor pressure of the substance, but what about its relation with air pressure?&lt;/p&gt;&#xA;" OwnerUserId="16923" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-03-08T17:14:34.387" Title="The effect of air pressure on the melting point" Tags="&lt;phase&gt;&lt;melting-point&gt;&lt;vapor-pressure&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32842" PostTypeId="2" ParentId="32841" CreationDate="2015-06-13T12:08:43.243" Score="9" Body="&lt;p&gt;For most substances, higher pressure (or air pressure, in your case) will cause the melting temperature to go up. To think about it intuitively, imagine that you have a certain solid. Melting it would increase the volume of that substance because liquids take more space than solids. If you increase the pressure, it becomes harder for that transformation to occur. It's harder to change into a state that needs more volume if you are under pressure! So, you need more energy, hence temperature, to melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this relation is actually reverse in water. Because ice takes more volume than liquid water, it's actually easier to melt it at high pressure and it requires less heat. Try taking an ice cube and apply pressure on it with a knife, see how it liquifies at the contact point.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2015-06-13T12:08:43.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32843" PostTypeId="1" CreationDate="2015-06-13T12:17:54.587" Score="3" ViewCount="8406" Body="&lt;p&gt;Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO2(g) &amp;lt;=&amp;gt; SO2(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How can I compare pH of an aqueous solution of $\ce{SO2}$ with that of pure water qualitatively?&lt;/li&gt;&#xA;&lt;li&gt;What happens to the pH of an aqueous solution of $\ce{SO2}$ if I aerate it by bubbling air through it?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-08-06T11:47:44.543" LastActivityDate="2017-08-06T11:47:44.543" Title="About pH of an aqueous solution of SO2" Tags="&lt;inorganic-chemistry&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32844" PostTypeId="2" ParentId="32843" CreationDate="2015-06-13T13:04:00.057" Score="-1" Body="&lt;p&gt;When you dissolve $\ce{SO2}$ in water you get $\ce{H2SO3}$. $\ce{SO2(aq)}$ means $\ce{H2SO3}$ therefor,the pH of the solution will be slightly acidic when compared to that of pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to your other question.when you aerate the solution $\ce{SO2}$ gas right above the solution goes away.The equilibrium is then disturbed.Therefore, $\ce{SO2}$ dissolved in the solution comes out.Because of that pH of the solution increases.&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorDisplayName="user15489" LastEditDate="2015-06-13T13:26:07.780" LastActivityDate="2015-06-13T13:26:07.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="32847" PostTypeId="2" ParentId="32838" CreationDate="2015-06-13T14:55:45.073" Score="0" Body="&lt;p&gt;Note that petroleum benzene is nothing but benzene itself. The order must be HCl &gt; Ethanol &gt; hexane &gt; benzene. It is because benzene is a perfectly symmetrical molecule and dipole moment vectors cancel each other completely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But hexane is not that symmetric and hence there is a dipole moment due to difference in electronegativity of C and H atoms (but it is almost negligible).&lt;/p&gt;&#xA;" OwnerUserId="16921" LastEditorUserId="41328" LastEditDate="2017-09-19T10:14:45.540" LastActivityDate="2017-09-19T10:14:45.540" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="32848" PostTypeId="1" AcceptedAnswerId="32851" CreationDate="2015-06-13T15:27:40.590" Score="4" ViewCount="5678" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The following reaction mechanism has been proposed for the reaction of $\ce{NO}$ with $\ce{Br2}$ to form $\ce{NOBr}$&lt;br /&gt;&#xA;  $$\begin{align}&#xA;\ce{NO(g) + Br2(g)&amp;amp;&amp;lt;=&amp;gt;NOBr2(g)}\\&#xA;\ce{NOBr2(g) + NO(g) &amp;amp;-&amp;gt; 2NOBr(g)}&#xA;\end{align}$$&#xA;  If the second step is the rate determining step, what is the order of the reaction with respect to $\ce{NO}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the 2nd step is the rate determining, shouldn't the order with respect to $\ce{NO}$ be 1? According to my book though, the correct answer is 2. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="5026" LastEditDate="2018-07-16T02:32:07.207" LastActivityDate="2018-07-16T02:32:07.207" Title="What is the order of the reaction with respect to NO?" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32849" PostTypeId="1" AcceptedAnswerId="33580" CreationDate="2015-06-13T15:29:14.080" Score="18" ViewCount="3831" Body="&lt;p&gt;Its really interesting to see the process of recycling of paper (as shown in picture below) but at the same time I was thinking like for how many times can one sheet of paper can be recycled ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rUa0k.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16459" LastEditorUserId="16459" LastEditDate="2015-06-14T17:37:47.067" LastActivityDate="2015-07-03T17:39:15.970" Title="How many times can we recycle one sheet of paper?" Tags="&lt;everyday-chemistry&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32850" PostTypeId="2" ParentId="32835" CreationDate="2015-06-13T16:06:08.420" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why isn't allene planar?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's start by looking at the structure of allene and seeing what happens if we make it planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/odOTU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://osxs.ch.liv.ac.uk/OldSitesDONOTVIEW/smaher/orbitalsallene.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the above image shows, the 3 carbons in allene are linear (form a straight line) and the terminal hydrogens lie in orthogonal planes (&lt;em&gt;e.g&lt;/em&gt;. the planes containing the hydrogens are rotated 90° one from the other as shown in the bottom half of the drawing).  Allene has two orthogonal pi bonds (&lt;em&gt;e.g&lt;/em&gt;. the planes of the two pi bonds are also rotated 90° one from the other).  Each one of those pi bonds contains 2 electrons.  The terminal carbons are $\ce{sp^2}$ hybridized, while the central carbon is $\ce{sp}$ hybridized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's twist one of the terminal methylenes by 90°.  It doesn't matter which one, so let's just say we twist the one on the right, and twist it so that its red p-orbital lobe lines up with the red lobes on the other double bond.  Now our molecule still has the 3 carbons arranged in a line, but all of the atoms - including the hydrogens - now lie in the same plane; &lt;strong&gt;we've made our molecule planar.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To accomplish this we broke one pi bond.  In our planar molecule we have 3 pi electrons (2 from the original pi bond on the left and 1 from the terminal methylene p-orbital that we rotated) in the 3 aligned p orbitals that exist in the plane of the screen.  We are also left with 1 electron in the p-orbital on the central carbon that is perpendicular to the screen (the p-orbital that was part of the double bond on the right side of the molecule).  So we took a molecule with &lt;strong&gt;two stable double bonds&lt;/strong&gt;, rotated a p-orbital 90° and wind up with a planar molecule that has &lt;strong&gt;3 aligned p-orbitals with 3 pi electrons and one electron by itself in a single p-orbital&lt;/strong&gt;.  This is a biradical, an allyl-like (it's just allyl-like because the 3 carbons in our molecule are still linear, whereas the 3 carbons in the allyl system are bent) radical and a radical in a p-orbital.  Converting 2 double bonds to a biradical is &lt;strong&gt;energetically uphill&lt;/strong&gt;, it does not happen at room temperature. &lt;strong&gt;Allene prefers to keep its 2 stable double bond structure rather than be a planar biradical molecule because the former geometry is much more stable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It doesn't&lt;/strong&gt;. Carbon disulfide and carbon dioxide have the same structure and hybridization as allene.  It is just a little harder to notice because we don't have hydrogens on the ends of those molecules to &quot;tip us off&quot; that the two double bonds are orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of the orbitals in carbon dioxide (it's exactly the same in carbon disulfide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tr4iP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://myweb.astate.edu/mdraganj/GenChemRevAns.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like in allene, we have 3 atoms arranged in a linear fashion and the central carbon is $\ce{sp}$ hybridized; there are two pi bonds that are orthogonal to one another.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-06-13T16:06:08.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32851" PostTypeId="2" ParentId="32848" CreationDate="2015-06-13T16:09:09.503" Score="6" Body="&lt;p&gt;The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the overall order of the reaction is determined from the rate expression involving &lt;em&gt;only the reactants&lt;/em&gt;.  The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for $R = R_2$ can thus be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-13T16:24:57.393" LastActivityDate="2015-06-13T16:24:57.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32852" PostTypeId="2" ParentId="32848" CreationDate="2015-06-13T16:17:50.423" Score="5" Body="&lt;p&gt;Even though the 2nd elementary step is the rate determining, you cannot just consider the stoichiometry of a species in one single elementary step to find out the overall order, which is what you are asked to find out in this question. This a complex reaction and thus the order of each reactant will depend on all the elementary steps of the reaction mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the equilibrium constant method(I actually forgot what this method is called) to find out the rate law for this reaction taking the help of the mechanism that you have mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate law will be $${r}=k\ce{[NO][NOBr2]}$$&#xA;Since $\ce{NOBr2}$ is the intermediate we are going to exclude it from the final rate law. (Remember, intermediates should never appear in the final rate law.)&#xA;$$K= \frac{\ce{[NOBr2]}}{\ce{[Br2]}{[NO]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[NOBr2]}=K\ce{[Br2][NO]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$r=k\ce{[NO]}K\ce{[Br2][NO]}=kK\ce{[NO]^2[Br2]}=k'\ce{[NO]^2[Br2]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can see that the order with respect to $\ce{NO}$ is 2 and not 1 as you initially though.&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorDisplayName="user15489" LastEditDate="2015-06-13T19:14:21.600" LastActivityDate="2015-06-13T19:14:21.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32854" PostTypeId="1" AcceptedAnswerId="32931" CreationDate="2015-06-13T16:30:52.583" Score="2" ViewCount="4046" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which of the following statements is true?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{11.2 dm^3}$&lt;/span&gt; of nitrogen at STP-&lt;/p&gt;&#xA;&lt;p&gt;A: has a mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{14 g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;B: consists of &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 mol}$&lt;/span&gt; atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My instinct was that the answer is B because any gas has a volume of &lt;span class=&quot;math-container&quot;&gt;$22.4$&lt;/span&gt; liters regardless of what the gas is. If &lt;span class=&quot;math-container&quot;&gt;$\pu{0.5 mol}$&lt;/span&gt; of nitrogen has a volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{11.2 dm^3}$&lt;/span&gt; at STP, then &lt;span class=&quot;math-container&quot;&gt;$\pu{1 mol}$&lt;/span&gt; of nitrogen has a volume of &lt;span class=&quot;math-container&quot;&gt;$22.4$&lt;/span&gt; liters at STP (if I am correct).&lt;/p&gt;&#xA;&lt;p&gt;It turned out the answer was A. Why is B not the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="16930" LastEditorUserId="83758" LastEditDate="2020-06-19T16:35:01.123" LastActivityDate="2020-06-19T16:35:01.123" Title="Does 11.2 dm3 of nitrogen consist of 0.5 moles at STP?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32857" PostTypeId="2" ParentId="28796" CreationDate="2015-06-13T17:27:00.397" Score="3" Body="&lt;p&gt;When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH(s) ~-&amp;gt;~ Na+(aq) + OH^{-}(aq)}$$&lt;/p&gt;&#xA;" OwnerUserId="5016" LastActivityDate="2015-06-13T17:27:00.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32860" PostTypeId="1" CreationDate="2015-06-13T18:45:20.020" Score="4" ViewCount="1215" Body="&lt;p&gt;I studied Bohr's model of atom and then the drawbacks of it and then quantum mechanical model of atom. Now quantum model is according to uncertainty principal and dual nature of matter and it says we can not talk about fixed orbits (uncertainty principle) and can only talk of  an area of maximum probability of finding an electron known as &quot;orbital&quot;. Now when I'm studying principal quantum number, it says it is the &quot;shell number&quot;. So how come we are talking about fixed shells again and returning to what Bohr said? &lt;/p&gt;&#xA;" OwnerUserId="16934" LastEditorUserId="402" LastEditDate="2016-11-15T23:42:42.900" LastActivityDate="2016-11-15T23:42:42.900" Title="Quantum mechanical model of atom and the quantum numbers" Tags="&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32862" PostTypeId="2" ParentId="32838" CreationDate="2015-06-13T20:32:58.567" Score="8" Body="&lt;p&gt;The concept of solvent polarity is commonly used to describe solvent effects. For example, the &lt;a href=&quot;http://dx.doi.org/10.1351/pac199466051077&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994)&lt;/em&gt;&lt;/a&gt; uses the following general definition (which was originally proposed by Christian Reichardt) of the term polarity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When applied to solvents, this rather ill-defined term covers their overall solvation capability (solvation power) for solutes (i.e. in chemical equilibria: reactants and products; in reaction rates: reactants and activated complex; in light absorptions: ions or molecules in the ground and excited state), which in turn depends on the action of all possible, nonspecific and specific, intermolecular interactions between solute ions or molecules and solvent molecules, excluding such interactions leading to definite chemical alterations of the ions or molecules of the solute. Occasionally, the term solvent polarity is restricted to nonspecific solute/solvent interactions only (i.e. to van der Waals forces).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the polarity of a solvent is not a physical quantity. Therefore, it is difficult to express quantitatively. Several scales exist that involve various physical properties in order to provide a measure of a solvent’s polarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the relative permittivity &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\mathrm{r}$&lt;/span&gt; is often used as a quantitative measure of solvent polarity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hexane: &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\mathrm{r} = 1.8865$&lt;/span&gt;&lt;br&gt;&#xA;Benzene: &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\mathrm{r} = 2.2825$&lt;/span&gt;&lt;br&gt;&#xA;Methanol: &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\mathrm{r} = 33.0$&lt;/span&gt;&lt;br&gt;&#xA;Water: &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\mathrm{r} = 80.100$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(values at 20 °C taken from “Laboratory Solvents and Other Liquid Reagents”, in &lt;em&gt;CRC Handbook of Chemistry and Physics, 97th Edition (2016),&lt;/em&gt; William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of comparison, the empirical &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{T}(30)$&lt;/span&gt; scale is based on the solvatochromism of Reichardt’s dye (Betaine 30):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hexane: &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{T}(30) = 31.0\ \mathrm{kcal/mol}$&lt;/span&gt;&lt;br&gt;&#xA;Benzene: &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{T}(30) = 34.3\ \mathrm{kcal/mol}$&lt;/span&gt;&lt;br&gt;&#xA;Methanol: &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{T}(30) = 55.4\ \mathrm{kcal/mol}$&lt;/span&gt;&lt;br&gt;&#xA;Water: &lt;span class=&quot;math-container&quot;&gt;$E_\mathrm{T}(30) = 63.1\ \mathrm{kcal/mol}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(values taken from &lt;a href=&quot;https://www.uni-marburg.de/fb15/ag-reichardt/et_home&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2019-06-02T19:48:02.030" LastActivityDate="2019-06-02T19:48:02.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32863" PostTypeId="1" AcceptedAnswerId="32870" CreationDate="2015-06-13T20:51:22.953" Score="8" ViewCount="2553" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nomq9.png&quot; alt=&quot;Claisen condensation mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't two last steps superfluous? I just do not see how the fourth and last forms differ.&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="10185" LastEditDate="2015-06-14T14:15:12.250" LastActivityDate="2015-06-14T14:15:12.250" Title="Claisen condensation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32864" PostTypeId="1" AcceptedAnswerId="32867" CreationDate="2015-06-13T21:04:20.950" Score="4" ViewCount="334" Body="&lt;p&gt;I tried to draw the frontier molecular orbitals of the following biradical structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMjxX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) At the top I've drawn the HOMO and LUMO, because I know that the reaction coordinate belongs to the $\ce{C_{$2v$}}$ point group. These two orbitals of the HOMO (green) lies in the O-C-O plane. In each $p$ orbital I would locate one electron (triplett state is favoured because of Hunds rule). Oxygen has six  valence electrons, so I must bring four electrons in orbitals of each oxygen. I can place two electrons in a $p$ orbital perpendicular to the greenish depicted orbital. Then I would place the other two electrons in a $sp$ hybrid orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) In the second possibility (at the right at the bottom) I also used a $sp$ hybridization for the oxygen atom. The two radicals are placed in the $p$ orbital perpendicular to the O-C-O plane. I guess the structure is less favoured than the first possibility (with the two radical in the O-C-O plane) because there are unfavoured filled-filled interactions between pink-pink and green-green while above there is onyl a pink-pink destabilizing interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) As a third possibility I've used a $sp^2$ hybridization of the oxygen. The radicals are now perpendicular to the plane. Without exact calculation of the hybrid orbitals I would predict that 4 electrons in $sp^2$ hybrid orbitals plus one electron in a $p$ orbital is lower in energy as case (A) (3 electrons in $p$ orbital and two electrons in a $sp$ hybrid orbital). However the green lobes (both two electrons and hence repulsive interaction) here in the case (C) in the $sp^2$ hybrid orbitals seems to come closer (120°) than the two electrons in the orthogonal $p$ orbitals (also filled-filled) in case (A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these reasons why the HOMO is best described by case (A)&gt;(C)&gt;(B)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably this question has similiarties to the hybridization in a previous question (please note I'm a novice concerning hybridization)&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/32835/why-allene-cannot-be-described-with-an-allyl-system&quot;&gt;Why allene cannot be described with an allyl system?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-06-14T01:37:11.540" Title="Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?" Tags="&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32865" PostTypeId="2" ParentId="32863" CreationDate="2015-06-13T21:04:43.017" Score="8" Body="&lt;p&gt;You are correct that there is no difference between the last fourth and last molecules shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the thermodynamic driving force behind the Claisen condensation is the formation of the highly stable enolate of the $\beta$ ketoester, or in your case $\beta$ diketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mrCiJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to form this stable enolate the original ester must have at least two $\alpha$ hydrogens so that there is at least one left at the end to deprotonate. Therefore the Claisen condensation will not work with esters having only one $\alpha$ hydrogen (and obviously not with esters with no $\alpha$ hydrogens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have completed the reaction to yield the enolate product, it can be acidified as shown in your reaction scheme to yield the neutral molecule.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-13T21:04:43.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32866" PostTypeId="2" ParentId="32860" CreationDate="2015-06-13T21:31:13.017" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So how come we are talking about fixed shells again and returning to&#xA;  what Bohr said?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope, we are not returning to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Old_quantum_theory&quot; rel=&quot;nofollow&quot;&gt;old quantum theory&lt;/a&gt;. We do not talk about orbits anymore, we rather have &lt;em&gt;orbitals&lt;/em&gt;. Note that it is important to bear in mind that orbitals also &lt;em&gt;aren't real&lt;/em&gt;, i.e. this description is still approximate, but it is a much better way of thinking about atoms (and molecules) than Bohr model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to understand the concept of orbitals you have to learn the basics of quantum chemistry first, namely, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;nofollow&quot;&gt;Hartree-Fock method&lt;/a&gt;. Until then just realize that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We do use hydrogen-like one-electron wave functions (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;nofollow&quot;&gt;orbitals&lt;/a&gt;, or, spin-orbitals, to be more precise) to describe the state of each and every electron in any atom.&lt;/li&gt;&#xA;&lt;li&gt;These hydrogen-like orbitals are described by a set of &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_number&quot; rel=&quot;nofollow&quot;&gt;quantum numbers&lt;/a&gt;: the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number.&lt;/li&gt;&#xA;&lt;li&gt;The set of orbitals belonging to a given principal quantum number $n$  constitutes what we call a &lt;em&gt;shell&lt;/em&gt; in the orbital model.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-12-11T10:43:45.773" LastActivityDate="2015-12-11T10:43:45.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32867" PostTypeId="2" ParentId="32864" CreationDate="2015-06-13T22:43:36.853" Score="4" Body="&lt;p&gt;I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by using cyclopropane as a model compound.  In cyclopropane the $\ce{C-H}$ bonds are approximately  $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666&quot;&gt;here&lt;/a&gt;).  That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized.  As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the &quot;initial&quot; drawing below.  If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the &quot;rotated&quot; drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z05rA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These drawings just point out the conformational dependence of the HOMO and LUMO.  Note that &lt;strong&gt;the two lone pairs on each oxygen have not entered the discussion&lt;/strong&gt;.  They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine bringing two oxygen atoms closer to each other.  If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the &quot;initial&quot; geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the &quot;initial&quot; geometry, only the orbitals are closer together).  If we push them closer and the odd electrons are pointing parallel to each other as in the &quot;rotated&quot; geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the &quot;rotated&quot; geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them.  This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could &lt;strong&gt;refine&lt;/strong&gt; our model by noting that, according to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622&quot;&gt;Bent's rule&lt;/a&gt;, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane.  Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion.  That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so.  Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals.  However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-13T22:43:36.853" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32868" PostTypeId="1" AcceptedAnswerId="32869" CreationDate="2015-06-14T00:00:52.580" Score="4" ViewCount="456" Body="&lt;p&gt;In a nearly finished bowl of oat cereal, why do the pieces of cereal stick together? I figure that since they are composed of starches it could be hydrogen bonding. Is this true or is there some other reason?&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-06-14T12:14:56.990" Title="Chemistry of cereal in milk" Tags="&lt;bond&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32869" PostTypeId="2" ParentId="32868" CreationDate="2015-06-14T00:24:55.870" Score="7" Body="&lt;p&gt;Surface tension, not hydrogen bonding, is the primary phenomenon which causes cereal rings such as Cheerios to adhere together while floating in milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total linear perimeter of the 3-phase cereal-air-milk interface is minimized when pieces cling together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.livescience.com/9350-cereal-science-floating-objects-stick.html&quot; rel=&quot;nofollow&quot;&gt;http://www.livescience.com/9350-cereal-science-floating-objects-stick.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://phys.org/news/2010-09-well-known-effect-breakfast-cereal-physicists.html&quot; rel=&quot;nofollow&quot;&gt;http://phys.org/news/2010-09-well-known-effect-breakfast-cereal-physicists.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cheerios_effect&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Cheerios_effect&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="7448" LastEditDate="2015-06-14T12:14:56.990" LastActivityDate="2015-06-14T12:14:56.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32870" PostTypeId="2" ParentId="32863" CreationDate="2015-06-14T00:55:18.230" Score="7" Body="&lt;p&gt;The Claisen reaction occurs in a strong base $(\ce{RO-})$, which is a conjugate base of an alcohol $(\text{p}K_a \approx 16-18)$; while for the $\beta$-dicarbonyl compound, $\text{p}K_a \approx 10$. The fourth molecule cannot help but be deprotonated in the strongly basic reaction mixture to produce the fifth  molecule. Acid needs to be added to reform the desired product.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="7448" LastEditDate="2015-06-14T12:52:25.103" LastActivityDate="2015-06-14T12:52:25.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32871" PostTypeId="2" ParentId="32863" CreationDate="2015-06-14T00:56:07.383" Score="7" Body="&lt;p&gt;The 4th and the last structures are same but the last two steps are not superfluous, because the 4th structure is consumed right after it is generate and will not persist while after addition of acid, the last structure persist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please pay attention to the condition of the last step (acid). All steps except the last step happened consequently once you mixed two starting materials with base. The last step happens after addition of acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under basic condition (alkoxides), the beta-keto ester generated in step 3 will be deprotonated immediately to give the second 5th enolate structure which will be protonated after addition of proton source (acid in this case).&lt;/p&gt;&#xA;" OwnerUserId="2093" LastActivityDate="2015-06-14T00:56:07.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32872" PostTypeId="2" ParentId="32843" CreationDate="2015-06-14T01:32:34.740" Score="5" Body="&lt;p&gt;Upon adding $\ce{SO2}$ gas to water, it shows its &lt;em&gt;acidic oxide&lt;/em&gt; character by hydrolyzing to sulfurous acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO2(g) + H2O(l) &amp;lt;=&amp;gt; H2SO3(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO3}$ is a &lt;strong&gt;weak&lt;/strong&gt; diprotic acid in water, with the following acid dissociation equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO3(aq) &amp;lt;=&amp;gt; H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$&#xA;$$\ce{HSO3{}^{-}(aq) &amp;lt;=&amp;gt; H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2SO2(g) + O2(g) -&amp;gt; 2SO3(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SO3(g) + H2O(l) &amp;lt;=&amp;gt; H2SO4(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This time, $\ce{H2SO4}$ is a &lt;strong&gt;strong&lt;/strong&gt; diprotic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4(aq) &amp;lt;=&amp;gt;&amp;gt; H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$&#xA;$$\ce{HSO4{}^{-}(aq) &amp;lt;=&amp;gt; H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively &lt;em&gt;replacing a weak acid with a strong acid&lt;/em&gt;, which would cause the pH to &lt;em&gt;decrease&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a &lt;em&gt;higher&lt;/em&gt; pH.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="7475" LastEditDate="2016-08-28T14:19:59.080" LastActivityDate="2016-08-28T14:19:59.080" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32873" PostTypeId="1" CreationDate="2015-06-14T02:23:02.597" Score="10" ViewCount="3001" Body="&lt;p&gt;It has been defined as the energy available for work other than expansion work. Why can't it be used for expansion work&lt;/p&gt;&#xA;" OwnerUserId="16936" LastEditorUserId="16683" LastEditDate="2016-12-08T11:10:24.737" LastActivityDate="2018-02-02T00:37:24.097" Title="Why does the Gibbs free energy only correspond to non-expansion work?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;free-energy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32874" PostTypeId="1" CreationDate="2015-06-14T02:55:14.307" Score="7" ViewCount="263" Body="&lt;p&gt;I've done some amount of searching and can't seem to find any evidence in the literature that there exists a phosphine analogue of carboxyl amides. I'm looking for compounds of the form $ \ce{ R(C=O)PR_{2}} $; basically a moeity similar to a primary or secondy amide but with a phosphorous atom instead of a nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also wonder what the proper name for these functionalities would be. Taking a shot in the dark I might guess &quot;phosphomide&quot;, &quot;acyl phosphine&quot;, or &quot;acyl phosphinide&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to specifically disqualify Acyl-phosphine oxides or any compounds where the acyl-connected phosphorous is also connected to an oxygen.&lt;/p&gt;&#xA;" OwnerUserId="16937" LastEditorDisplayName="user15489" LastEditDate="2015-06-14T03:22:55.977" LastActivityDate="2016-03-21T21:58:25.903" Title="Have P-acyl phosphines ever been synthesized?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32875" PostTypeId="2" ParentId="32831" CreationDate="2015-06-14T04:01:28.470" Score="2" Body="&lt;p&gt;If it is made of Lead (II, IV) oxide, &quot;Red Lead&quot;, it will dissolve in dilute solutions of $\ce{H2O2}$. Lead carbonate will dissolve in a hot solution of hydrogen peroxide and acetic acid. Lead chromate will form chrome red in a solution of $\ce{NaOH}$.&lt;/p&gt;&#xA;" OwnerUserId="16940" LastEditorDisplayName="user15489" LastEditDate="2015-06-14T04:20:51.673" LastActivityDate="2015-06-14T04:20:51.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32876" PostTypeId="1" CreationDate="2015-06-14T04:17:01.410" Score="5" ViewCount="6107" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Is it true that precipitates dissolve upon addition of diluted acids&lt;/em&gt;?&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If a precipitates is dissolve , does it disappear from solution , or  any thing else  happened out there?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I  am just curious about this. Because I have heard that &lt;strong&gt;precipitates can be dissolved&lt;/strong&gt; if you  add a diluted acid. But why it would happen? And is it valid for any diluted or strong acid ?.Do all the precipitates dissolve upon addition of any acid ?.I would like to know more about this.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastActivityDate="2015-06-15T01:07:27.727" Title="Can a precipitate dissolve upon addition of a diluted acid?" Tags="&lt;solutions&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32877" PostTypeId="2" ParentId="32826" CreationDate="2015-06-14T04:38:01.720" Score="4" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H2SO4 ⇌ H2O + SO3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{SO3 + H2SO4 ⇌ H2S2O7}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The mutual electron-withdrawing effects of each sulfuric acid unit on its neighbour causes a marked increase in acidity. Disulfuric acid is strong enough to protonate &amp;quot;normal&amp;quot; sulfuric acid in the (anhydrous) sulfuric acid solvent system.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sup&gt;From the Wikipedia entry on &lt;a href=&quot;https://en.wikipedia.org/wiki/Disulfuric_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Disulfuric acid&lt;/a&gt; &lt;/sup&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16924" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-06-15T02:21:07.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32878" PostTypeId="2" ParentId="32831" CreationDate="2015-06-14T04:40:16.657" Score="3" Body="&lt;p&gt;Two chemicals are used in &lt;a href=&quot;http://www.consumerreports.org/cro/lead-test-kits/buying-guide.htm&quot; rel=&quot;nofollow&quot;&gt;lead test kits&lt;/a&gt;: sodium sulfide or a rhodizonate salt. &lt;a href=&quot;http://fire.nist.gov/bfrlpubs/build00/PDF/b00034.pdf&quot; rel=&quot;nofollow&quot;&gt;Sodium sulfide&lt;/a&gt; should be easy to obtain (or synthesize), and is useful for light-colored paint, as it produces a gray or black color change. &lt;a href=&quot;https://www.youtube.com/watch?v=BLmycZ2nrt0&quot; rel=&quot;nofollow&quot;&gt;Potassium rhodizonate&lt;/a&gt; can be synthesized from inositol, but you might be better off getting sodium rhodizonate, e.g. from &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/r1609?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;Sigma-Aldrich&lt;/a&gt; for ~US$35 for 5 grams.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-14T04:40:16.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32879" PostTypeId="2" ParentId="32876" CreationDate="2015-06-14T04:53:38.693" Score="2" Body="&lt;p&gt;I would like to answer your &lt;strong&gt;second question first&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a precipitate dissolves, it basically reacts to form some other soluble compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your &lt;strong&gt;first question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, a precipitate is soluble in a dilute acid solution. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BaSO3 + 2HCl (dilute) -&amp;gt; BaCl2 (soluble) + SO2 (evolve) + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However is not necessarily true that a precipitate will dissolve in a dilute acidic solution. For example, calcium carbonate on reacting with dilute $\ce{H2SO4}$ evolves $\ce{CO2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + H2SO4 -&amp;gt; CaSO4 (insoluble) + H2O + CO2 (evolve)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on the hydration enthalpy of the (would be) product after dissolution of a precipitate, whether the precipitate will be soluble or nor. In these examples, hydration enthalpy of $\ce{BaCl2}$ is −8.8 kJ/mol and that of $\ce{CaSO4}$ is −73 kJ/mol.&lt;/p&gt;&#xA;" OwnerUserId="16924" LastEditorDisplayName="user7951" LastEditDate="2015-06-14T08:46:00.857" LastActivityDate="2015-06-14T08:46:00.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32880" PostTypeId="1" CreationDate="2015-06-14T05:20:47.440" Score="6" ViewCount="1160" Body="&lt;p&gt;We've learnt about the trans effect but I have a question regarding how this plays a role with water in the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;trans-effect&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O &amp;lt; NH3 &amp;lt; Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, given &lt;em&gt;cis&lt;/em&gt;-$\ce{[Pt(NH3)2Cl2]}$, if you were to add water to form a dicationic complex, would a substituion reaction occur between all the substituents, and then $\ce{Cl-}$ reacts again to form the trans-complex? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water has a weaker trans effect than both $\ce{NH3}$ AND $\ce{Cl-}$, I can't imagine that a cis-complex would form as any cis-combination of $\ce{NH3}$ and $\ce{Cl-}$ has a stronger trans effect than any combination of $\ce{H2O / Cl- / NH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other kinetic / electronegativity factors that I am missing?&lt;/p&gt;&#xA;" OwnerUserId="16768" LastEditorUserId="23561" LastEditorDisplayName="user15489" LastEditDate="2015-12-15T12:15:09.353" LastActivityDate="2015-12-15T12:15:09.353" Title="Trans effect on square planar complex" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;&lt;transition-metals&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ClosedDate="2016-01-24T09:15:01.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="32881" PostTypeId="1" AcceptedAnswerId="32908" CreationDate="2015-06-14T06:00:09.410" Score="11" ViewCount="7514" Body="&lt;p&gt;I am preparing to do a vacuum distillation and am unsure exactly how to manage the vacuum. The vacuum pump I have is a two-stage rotary vane pump which is capable of 250 microns or lower. It has a triple inlet, one 1/4&quot;, 3/8&quot; and 1/2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My glassware is typical distillation glassware including a Graham condenser and a vacuum takeoff joint that connects between the condenser and the receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is that if I apply too much vacuum that it will cause a glass implosion. Is this not a concern and I can just operate the vacuum pump freely, or do I need to take special steps to limit the degree of vacuum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I watched a YouTube video of a guy doing a vacuum distillation and I noticed that he did not use a capillary bubbler. Under what circumstances can the bubbler be dispensed with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another concern is the need for a manometer. I have read that I need to use a closed-end manometer (very expensive) to measure the pressure in a vacuum distillation. Why can't I use a normal vacuum gauge? Also, if I am doing an azeotropic separation such as a water-ethanol separation, is it critical I monitor the pressure? I am assuming it is.&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorUserId="4945" LastEditDate="2015-06-22T04:11:43.207" LastActivityDate="2015-09-16T22:11:35.590" Title="Proper use of vacuum pump during vacuum distillation" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;&lt;distillation&gt;" AnswerCount="2" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32882" PostTypeId="2" ParentId="32854" CreationDate="2015-06-14T07:02:10.703" Score="1" Body="&lt;p&gt;The correct answer for this question is &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to start explaining the answer for your question, I would like to show you a mistake you have done.&#xA;You have said&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;any gas has a volume of 22.4 litres regardless of what the gas is.&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This true only if the gas exist in standard temperature and pressure or (STP). And note that per one mole of any gas at STP condition occupy a volume of 22.4 litres or $22.4\ \mathrm{dm^3}$. Simply we call this as molar volume.&#xA;Here is how you can derive it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subject $V$ in the formula $pV=nRT$, and then you can substitute values. Where $T=273.15 \ \mathrm{K}$ (value of standard temperature) and $p=101325 \ \mathrm{Pa}$ (value of standard pressure) and $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_\mathrm{m}=\frac{V}{n}=\frac{RT}{p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have asked why not the answer B is correct?&#xA;It is not correct. Here is why:&#xA;In one mole of any compound you have $6.022\times10^{23}$ atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ \mathrm{mol} \Rightarrow 6.022 \times 10^{23}$ atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6.022 \times 10^{23}$ is the Avogadro constant. (If you have no idea about it check &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.)&#xA;So here in this question you have found number of moles of nitrogen molecule, it means $\ce{N2(g)}$ as $0.5\ \mathrm{mol}$. Which means you have $0.5\times2\ \mathrm{mol}$ of atoms. So as I said before if you have one mole mean you already get $6.022 \times 10^{23}$ atoms of $\ce{N}$ atom. So you have $6.022 \times 10^{23}$ atoms or $6.022 \times 10^{23}$ nitrogen mol atoms. So now you can clearly see why answer B is incorrect. I hope my answer is helpful enough to solve your doubt.&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="4532" LastEditDate="2015-06-15T08:37:58.037" LastActivityDate="2015-06-15T08:37:58.037" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32883" PostTypeId="1" AcceptedAnswerId="32886" CreationDate="2015-06-14T07:03:50.543" Score="1" ViewCount="357" Body="&lt;p&gt;I am not expert in chemistry.I learned about combustion in reaction when added glycerine to potassium permagnet.I tried it at home, I took potassium permagnet in test tube and when I added glycerine to it test tube just CRACKED as combustion started.I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any Suggestion..&lt;/p&gt;&#xA;" OwnerUserId="16941" LastActivityDate="2017-01-14T12:15:00.887" Title="test tube cracked when added glycerin to potassium permanganate" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32884" PostTypeId="1" AcceptedAnswerId="32889" CreationDate="2015-06-14T07:52:48.057" Score="12" ViewCount="5613" Body="&lt;p&gt;Is there a maximum optical rotation of plane polarized light that can be observed using a polarimeter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that in a polarimeter, light passes through a filter that converts it into plane polarized light. This type of light only oscillates in one plane. When it is passed through an optically active solution, the plane is rotated due to interactions between the molecules. To measure this rotation, we use another a filter and try to orient it so that light passing through it appears brightest. The difference in angular orientations of the two filters is read as the observed angular rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image I've taken from Solomon and Fryhle's Organic Chemistry textbook:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4tXHz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a maximum value for this rotation? Surely after a 90 degrees rotation, the angle would be the same as another angle is we'd rotated in the opposite direction. For example, I could say that a +100 degrees rotation is the same as a -10 degrees rotation.&#xA;Could the angle be greater than 90 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question might seem simple, but it's difficult for me to take into consideration multiple rotations of the plane polarized light before it passes through the second filter. Are the multiple rotations even possible?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="1499" LastEditDate="2015-06-14T16:58:57.550" LastActivityDate="2015-10-11T13:26:10.403" Title="Maximum observed optical rotation through a polarimeter" Tags="&lt;analytical-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32885" PostTypeId="1" AcceptedAnswerId="32887" CreationDate="2015-06-14T09:24:19.787" Score="-3" ViewCount="425" Body="&lt;p&gt;I am interested in the chemical lsd or &quot;acid&quot;. I wanted to know if lsd is acid in its pure form would that mean it is water? &lt;/p&gt;&#xA;" OwnerUserId="16943" LastEditorUserId="418" LastEditDate="2015-06-14T09:48:10.043" LastActivityDate="2015-06-14T09:50:41.843" Title="Ph of acid would be one?" Tags="&lt;organic-chemistry&gt;&lt;ph&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-14T22:56:13.083" ContentLicense="CC BY-SA 3.0" />
  <row Id="32886" PostTypeId="2" ParentId="32883" CreationDate="2015-06-14T09:41:35.170" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case of doubt you should not. Anyway, I recommend to avoid doing it with any significant amount. It may and will spread around and even if not it produces manganese (IV) oxide smoke that settles down nearby and is quite hard to wash out. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Potassium permanganate is a strong oxidizer, glycerol is a reducer. Mixed together they quickly start to react. The exact nature of the products is debatable, but the mixture self-ignites if warm. The thermal shock is enough to crack glass. This is a common weakness of glass and ceramics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, in case of bad luck it may produce enough gases to spread the mix around. Consequently, safety precautions must be always employed (incombustible table, safety goggles etc). A movie how it may go &lt;a href=&quot;http://www.youtube.com/watch?v=6F4cuQkKt2U&quot; rel=&quot;nofollow&quot;&gt;http://www.youtube.com/watch?v=6F4cuQkKt2U&lt;/a&gt; . The reaction has some quirks, specifically if the glycerol is cold, it may not start immediately, so a rather well mixing is achievable. Now, once it warms up to room temperature, it suddenly bursts in flame, potentially delivering itself to nearby surfaces, including your face.   &lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-06-14T09:41:35.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32887" PostTypeId="2" ParentId="32885" CreationDate="2015-06-14T09:50:41.843" Score="3" Body="&lt;p&gt;No! Don't get fooled by the slang term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LSD is the abbreviation for lysergic acid diethylamide. However, the parent lysergic acid (see structure) is indeed a carboxylic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/K7541.png&quot; alt=&quot;lysergic acid&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="418" LastActivityDate="2015-06-14T09:50:41.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32888" PostTypeId="2" ParentId="32873" CreationDate="2015-06-14T13:21:43.963" Score="10" Body="&lt;p&gt;The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from the definition of the Gibbs free energy in the first paragraph,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\mathrm{d}G &amp;amp;= \mathrm{d}(U + pV - TS)\\&#xA;&amp;amp;= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\&#xA;&amp;amp;= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\&#xA;&amp;amp;= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="27806" LastEditDate="2018-02-02T00:37:24.097" LastActivityDate="2018-02-02T00:37:24.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32889" PostTypeId="2" ParentId="32884" CreationDate="2015-06-14T14:01:18.920" Score="10" Body="&lt;p&gt;This is something I discovered after an attempt at a more mathematically rigorous investigation of how to interpret polarimetry results. It's not that hard, but I don't think you'll ever find a similar analysis in a book because in practice the details are pretty much unnecessary for experimental purposes (though fun mathematically!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we begin, it's very important to distinguish two types of optical rotation. The first is the &lt;strong&gt;real&lt;/strong&gt; optical rotation, $\alpha_{real}$. This is the twist which a substance actually causes on the plane of polarized light as it travels through the optically active solution, and therefore is the rotation which is part of the specific rotation equation, shown further below. The numerical value of this angle can be any finite real number, in degrees or radians. &lt;strong&gt;There is no constraint on how much an optically active solution is &quot;allowed&quot; to twist a plane of polarized light&lt;/strong&gt;; the fact that you can't directly see how many complete twists the plane of polarized light made is entirely an experimental limitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This experimental limitation is what gives rise to the other type of rotation: the &lt;strong&gt;observed&lt;/strong&gt; or &lt;strong&gt;apparent&lt;/strong&gt; rotation, $\alpha_{obs}$. This is an angle which is also real number, but which &lt;strong&gt;is limited&lt;/strong&gt; between $0$ and $180$ degrees or $0$ and $\pi$ radians (or you could say between $-90$ and $90$ degrees or $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians, it makes no conceptual difference). The observed rotation is what you measure in a polarimeter, and &lt;em&gt;in principle it cannot be directly put into the specific rotation equation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\alpha_{obs}$ allowed only in the range $[0, 180]$ degrees? Because you can't observe a difference between an angle $\alpha_{obs}$ and $\alpha_{obs} + 180\times n$ (where $n$ is a positive &lt;em&gt;or a negative&lt;/em&gt; integer). Look at a figure such as &lt;a href=&quot;http://cnx.org/resources/b3f35ee9f54a28afa344cc5056804a34d762ecc1/Figure_28_08_13a.jpg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; (&lt;a href=&quot;http://cnx.org/contents/0119fe7d-4d1a-42de-b580-aa46659f8228@4/Polarization&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;) and try to convince yourself of this. If it helps, remember that two identical polarizers on top of eachother with an angle of $0^\circ$ let light through completely, but rotating one will block more and more light until they are perpendicular ($90^\circ$) and no light at all comes through, at which point increasing the angle will brighten the light until it reaches a maximum again at $180^\circ$ before once again starting to dim. This means that, in polarimetry &lt;em&gt;the observed brightening or dimming by rotations of plane-polarized light are cyclic with period $180^\circ$&lt;/em&gt;, not $90^\circ$ as you suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning the previous sentence into a simple equation, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_{real} = \alpha_{obs} + 180\times n \ \ \ , \ \ n \in \mathbb{Z} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The angles in this answer will always be treated in degrees, though for simplicity, in the equations I shall forget about the $^\circ$ unit and treat angles as pure numbers*).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we use the equation for optical rotation to find the &lt;em&gt;specific rotation&lt;/em&gt; $[\alpha]^T_\lambda$, which characterizes a substance when measured at a temperature $T$ with a monochromatic light source of wavelength $\lambda$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_{real} = [\alpha]^T_\lambda \times l \times c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;, where $l$ is the optical path and $c$ is the concentration of the (enantiomerically pure) optically active substance. Substituting that into the previous equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\alpha]^T_\lambda = \frac{\alpha_{obs} + 180 n}{lc}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for each individual measurement of $\alpha_{obs}$ made, there are infinite solutions for $[\alpha]^T_\lambda$. This comes as no surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we take several measurements, then we can find a value of $[\alpha]^T_\lambda$ as a unique solution to the equation, right? Speaking in strictly mathematical terms, &lt;strong&gt;no&lt;/strong&gt;. Any finite amount of measurements, even with perfect accuracy and precision, will still yield infinite possible values of $[\alpha]^T_\lambda$. In fact, there are infinite levorotatory and dextrorotatory solutions, so in principle you can't even determine the direction of rotation by this method! See the graph below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/w3WVH.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;: In my &lt;a href=&quot;https://i.stack.imgur.com/RcfV9.png&quot; rel=&quot;noreferrer&quot;&gt;previous graph&lt;/a&gt;, I incorrectly equated the values of $\theta$ with $\alpha_{real}$. The actual identity comes from $\mathrm{tan}\ \theta = \frac{d\alpha_{obs}}{dc} = \frac{d}{dc}(\alpha_{real}+180n)=\frac{d\alpha_{real}}{dc}=\frac{d}{dc}(lc[\alpha]^T_\lambda)=l[\alpha]^T_\lambda$, from which $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$. This however does not change the theoretical development of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since the measurements are cyclic with period $180^\circ$, &lt;em&gt;straight lines wrap around&lt;/em&gt;. The two black points are two measurements of $\alpha_{obs}$ at two different concentrations. There is an &quot;obvious&quot; red line connecting them, &lt;em&gt;but it need not be the correct one&lt;/em&gt;. There are &lt;em&gt;infinite&lt;/em&gt; lines which cut both points; I have represented two more perfectly valid solutions with the green and blue lines. Each line which cuts all points represents a possible solution, with an inclination given by the angle $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$ (see derivation above). From this, $[\alpha]^T_\lambda = \frac{\mathrm{tan}\ \theta}{l}$. Interestingly, the red and blue lines are represent dextrorotatory solutions to the problem, while the green line indicates a levorotatory solution for the same compound. There are infinite more solutions, characterized by steeper lines which wrap around more times. See the comments for information about how they come into play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;in practice&lt;/em&gt;, materials only have a finite and usually relatively small ability to perform optical rotation. Most of the more commonly encountered substances have a value of $[\alpha]^T_\lambda$ which tops out at a few hundreds. Some extreme examples are &lt;a href=&quot;https://en.wikipedia.org/wiki/Helicene&quot; rel=&quot;noreferrer&quot;&gt;helicenes&lt;/a&gt;, for which specific optical rotations can reach &lt;a href=&quot;http://www.scs.illinois.edu/denmark/presentations/2014/gm-2014-5-20.pdf&quot; rel=&quot;noreferrer&quot;&gt;as high as $\approx 10000^\circ$ in [13]helicene&lt;/a&gt;. After performing enough measurements (especially if you do them intelligently, by not measuring at different concentrations which are related to eachother by simple fractions), then you basically reach a situation where you can find an infinite family of solutions such as $[\alpha]^T_\lambda = -534 + 462785\times n$, in which case the true specific rotation is probably levorotatory with a value of just $-534^\circ$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a small additional comment, usually the filters in an optical polarimeter are rotated until the &lt;em&gt;least&lt;/em&gt; amount of light comes through as the human eye is better at detecting small differences in light levels in darker conditions, which will make for a more accurate measurement of the observed rotation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Originally I had answered this question by bringing up modular arithmetic, but while it's interesting it can possibly make the answer harder to understand at first. I have relocated my previous argument here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation shown is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_{obs} = \alpha_{real}\ \ (\mathrm{mod\  180})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{mod\  180}$ part comes from &lt;a href=&quot;https://en.wikipedia.org/wiki/Modular_arithmetic&quot; rel=&quot;noreferrer&quot;&gt;modular arithmetic&lt;/a&gt;, which may sound scary and new, but it's just normal mathematics, with the small quirk that every time a number appears out of the range $[0, 180]$, you just add or subtract $180$ as many times as necessary for the answer to appear between $0$ and $180$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* In reality some of the $\alpha_{real}$ and $\alpha_{obs}$ variables in certain equations above would have to be divided by $^\circ$, so that all equalities are dimensionally valid. Equivalently, the modulus could be expressed with the unit as $\mathrm{mod\  180^\circ}$, but I'm not sure if the modulus is allowed to have physical dimensions, even if I don't see any issue with it.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-14T20:25:58.580" LastActivityDate="2015-06-14T20:25:58.580" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="32890" PostTypeId="2" ParentId="5472" CreationDate="2015-06-14T14:12:44.483" Score="1" Body="&lt;p&gt;The temperature of the flame depends (mostly) on the material being burned. Natural gas burns with a hot blue flame, while wood burns with a cooler yellow flame.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-06-14T14:12:44.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32891" PostTypeId="1" CreationDate="2015-06-14T14:22:28.180" Score="3" ViewCount="154" Body="&lt;p&gt;We have pseudo-asymmetric carbons in optical isomerism which dictates that if 2 groups on the carbon atom are different in terms in configuration then the carbon is chiral. In alkenes can the same thing happen with respect to geometrical isomerism? For example consider &lt;img src=&quot;https://i.stack.imgur.com/dslqa.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;If both the alkenes on the left were E or both Z, then certainly there wouldn't be any chance of geometrical isomerism. But as in this case if the configurations are different ,i.e, E and Z, can the alkene be contended for geometrical isomerism ,i.e., whether the compound is a Z or E isomer in terms of the middle alkene (to which a methyl group is attached)?&lt;/p&gt;&#xA;&lt;p&gt;While giving the answer to this question, it would be extremely helpful if you cite the source of your information for me to look it up.&lt;/p&gt;&#xA;" OwnerUserId="16022" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-23T13:40:02.500" Title="Pseudo-asymmetry-like situation in alkenes" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-06-23T15:57:21.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="32892" PostTypeId="5" CreationDate="2015-06-14T14:34:48.057" Score="0" Body="&lt;p&gt;The systematic discussion of molecular symmetry is known as group theory. Group theory is a systematic way of describing the symmetry of molecules using concepts/rules borrowed from mathematics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These rules can be applied in a rational, structured way, allowing chemists to quickly determine many properties of molecules. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Definition&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Group_theory#Chemistry_and_materials_science&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemistry, groups are used to classify crystal structures, regular polyhedra, and the symmetries of molecules. The assigned point groups can then be used to determine physical properties (such as polarity and chirality), spectroscopic properties (particularly useful for Raman spectroscopy, infrared spectroscopy, circular dichroism spectroscopy, magnetic circular dichroism spectroscopy, UV/Vis spectroscopy, and fluorescence spectroscopy), and to construct molecular orbitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Applicability of the &lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt; tag&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt; should be applied to any question which concerns the theory or application of point groups, symmetry elements, or character tables (i.e. any question that considers symmetry in a systematic way).  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt; tag should &lt;em&gt;not&lt;/em&gt; be applied to all  questions relating to symmetry/geometry of molecules in a more general sense. Consider 'why does methane belong to the T&lt;sub&gt;d&lt;/sub&gt; group?' vs 'why does methane have tetrahedral geometry?'. Clearly the tetrahedral point group is involved in both questions, however in the later, it is clear that molecular geometry is being discussed rather than any actual interest in the point group itself. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt; is permissible by itself, however, many questions currently tagged with &lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt; ask about &lt;em&gt;applications&lt;/em&gt; of group theory, and as such additional tags should be used to further identify the question. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Related tags&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The following tags are related to &lt;a href=&quot;/questions/tagged/group-theory&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;group-theory&amp;#39;&quot; rel=&quot;tag&quot;&gt;group-theory&lt;/a&gt;, with multiple questions on chemistry.se already tagged with one of more of the following which may be useful as additional tags on questions identified as being related to  group theory: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/symmetry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;symmetry&amp;#39;&quot; rel=&quot;tag&quot;&gt;symmetry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/stereochemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;stereochemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;stereochemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/organic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;organic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;organic-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/inorganic-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;inorganic-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;inorganic-chemistry&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;/questions/tagged/spectroscopy&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;spectroscopy&amp;#39;&quot; rel=&quot;tag&quot;&gt;spectroscopy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Further reading/resources:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Online resources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;@Orthocresol's &lt;a href=&quot;https://chemistry.meta.stackexchange.com/questions/3435/group-theory-tables&quot;&gt;MathJax Character Tables&lt;/a&gt;, hosted on meta.chem.se.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Books:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group theory is a well developed area of chemistry, and many books have been published on the topic, ranging from simple qualitative introductions to math-heavy derivations. Some good examples are listed below, however any good physical chemistry textbook, and many inorganic textbooks will also devote a chapter to chemical group theory. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Kettle, S.; &lt;em&gt;Symmetry and Structure, a readable group theory for chemists&lt;/em&gt;, Wiley: Sussex, 2007.  An easy introduction to group theory, focusing on applications and qualitative description. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cotton, F. A.; &lt;em&gt;Chemical Applications of Group Theory&lt;/em&gt;, John Wiley &amp;amp; sons inc.: New York, 1990. An older book covering a lot of topics, less readable and more math-heavy, but a good reference for more advanced topics such as MO theory. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="21296" LastEditDate="2017-07-05T21:30:52.473" LastActivityDate="2017-07-05T21:30:52.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32893" PostTypeId="4" CreationDate="2015-06-14T14:34:48.057" Score="0" Body="The systematic discussion of molecular symmetry is known as group theory. Group theory is a systematic way of describing the symmetry of molecules using concepts/rules borrowed from mathematics.  This tag should be applied to any question which involves the theory or application of point groups, symmetry elements, or character tables. " OwnerUserId="3846" LastEditorUserId="21296" LastEditDate="2017-07-05T20:44:14.547" LastActivityDate="2017-07-05T20:44:14.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32894" PostTypeId="1" AcceptedAnswerId="32939" CreationDate="2015-06-14T14:38:33.273" Score="12" ViewCount="2907" Body="&lt;p&gt;I want to know how to typeset character tables of point groups correctly. Consider the following two extremes using the $C_{\text{2v}}$ (or is it $\text{C}_{\text{2v}}$?) point group as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, one where all the symbols (except for the symbols denoting vertical) are cursive (your default LaTeX math mode behaviour):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywSFU.png&quot; alt=&quot;C2v character table in full math mode&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, a table where all the symbols are in roman font.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBFjv.png&quot; alt=&quot;C2v character table fully roman&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is correct? Or is it a mixture?&lt;/p&gt;&#xA;" OwnerUserId="3846" LastEditorUserId="14" LastEditDate="2015-06-14T14:43:00.157" LastActivityDate="2015-06-15T16:38:14.023" Title="How are point group character tables typeset correctly?" Tags="&lt;group-theory&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32895" PostTypeId="5" CreationDate="2015-06-14T14:44:51.523" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-14T14:44:51.523" LastActivityDate="2015-06-14T14:44:51.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32896" PostTypeId="4" CreationDate="2015-06-14T14:44:51.523" Score="0" Body="Style and generally accepted formatting schemes/practices are common in chemistry." OwnerUserId="14" LastEditorUserId="14" LastEditDate="2015-06-14T16:00:03.880" LastActivityDate="2015-06-14T16:00:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32897" PostTypeId="1" CreationDate="2015-06-14T15:28:51.273" Score="6" ViewCount="12549" Body="&lt;p&gt;I already know the way to determine whether if the given reaction is exothermic or endothermic by the enthalpy values. But is it possible without enthalpy values, and just by looking at the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as far as I know there are two types of chemical reactions, namely one way reactions and reversible reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can I determine this factor in &lt;strong&gt;one way reactions&lt;/strong&gt; and &lt;strong&gt;reversible reactions&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that for reversible reactions we can use equilibrium constants such as $K_p$, $K_c$, but I am not quite sure whether it is correct. On the other hand, I feel that there is no such way to determine what I want in one way reactions. So how can I decide it?&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorDisplayName="user7951" LastEditDate="2015-06-17T09:52:54.500" LastActivityDate="2015-06-17T11:09:11.333" Title="How can I figure out whether a reaction is endothermic or exothermic without its enthalpy value?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="14" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32899" PostTypeId="1" CreationDate="2015-06-14T18:00:10.233" Score="17" ViewCount="2257" Body="&lt;p&gt;Is the following molecule aromatic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MHWj3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MHWj3.png&quot; alt=&quot;Sydnone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does seem like it will be aromatic because it is similar to furan. I am confused because of the positive charge on the $\ce{N}$.&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="16683" LastEditDate="2018-12-25T19:06:53.707" LastActivityDate="2018-12-25T19:48:14.373" Title="Is this sydnone aromatic?" Tags="&lt;organic-chemistry&gt;&lt;heterocyclic-compounds&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32902" PostTypeId="2" ParentId="32860" CreationDate="2015-06-14T21:52:48.143" Score="2" Body="&lt;p&gt;We are &lt;strong&gt;not&lt;/strong&gt; talking about fixed shells. Rather, we are observing that for any given atomic orbital you can find an orbital &lt;em&gt;of very similar shape but of a higher energy&lt;/em&gt;. Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The simplest orbital is the 1s orbital of hydrogen. We can find an orbital of higher energy that has the same shape (namely: spherical symmetry): The 2s orbital.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is an orbital called $3\mathrm{d}_{z^2}$ which is rather important in transition metal chemistry. Its ‘shape’ consists of two lobes of identical phase with a nodal plane in the middle plus a ‘ring’ of the opposite phase around the middle. Again, an orbital of similar shape but higher energy can be found ($4\mathrm{d}_{z^2}$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This fitted in so nicely with the previous models, that they were called ‘same orbitals of different shells’. They’re still orbitals, so wave functions, but we grouped them into shells.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As Ian correctly mentioned in a comment, when dealing with a one-electron system (such as the basic hydrogen atom) all orbitals of the same &lt;em&gt;shell&lt;/em&gt; will have identical energies. (That’s just what Schrödinger’s maths tells us.) Another reason to still use the word &lt;em&gt;shell&lt;/em&gt; without referring to it’s meaning of the fixed Bohr orbits.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-16T08:51:35.793" LastActivityDate="2015-06-16T08:51:35.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32905" PostTypeId="1" AcceptedAnswerId="32913" CreationDate="2015-06-14T22:13:25.363" Score="3" ViewCount="1707" Body="&lt;p&gt;I know that polyethylene's melting point is between 115–135 °C, however I would like to know if can it start dissolving between 50-60 °C? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real example, if I put warm food (straight from the oven, I think it can be max. 50-60 °C) into a re-closable polyethylene plastic bag and close it, will any chemical reaction affect the food (from the polyethylene's side)? Or is polyethylene safe at that temperature, and if the bag is warm it doesn't mean that any chemical reaction is started? I didn't experienced any alteration, but I'm not an expert and don't know much about chemistry yet, so I would appreciate some guidance.       &lt;/p&gt;&#xA;" OwnerUserId="16950" LastEditorDisplayName="user467" LastEditDate="2015-06-15T00:38:16.953" LastActivityDate="2015-06-15T10:09:16.020" Title="Can Polyethylene Dissolve on 50-60 °C" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32906" PostTypeId="1" AcceptedAnswerId="33048" CreationDate="2015-06-14T22:34:01.013" Score="13" ViewCount="3260" Body="&lt;p&gt;I couldn't find a source that explicitly say this, but given the gas constant R is used, is the Arrhenius equation only valid when all reactants are gases? Do they have to be ideal gases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the above is true, is there a similar equation for reactions where at least one reactant is in liquid/solid form?&lt;/p&gt;&#xA;" OwnerUserId="8117" LastEditorUserId="8117" LastEditDate="2015-06-14T22:39:01.827" LastActivityDate="2015-06-18T16:05:44.420" Title="Is the Arrhenius Equation only valid for reactants in the gas state?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32907" PostTypeId="1" AcceptedAnswerId="32914" CreationDate="2015-06-14T22:35:28.257" Score="8" ViewCount="380" Body="&lt;p&gt;I won this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007V5TEMW&quot;&gt;&quot;toy&quot;&lt;/a&gt; at a science fair...&#xA;Now I have to make a short presentation about it in school. &#xA;On the package description they describe its energy source as &quot;fuel cell&quot;...&#xA;It consists out of a metal plate(magnesium), a black metal plate with a black coating and a paper-cotton-like isolator which is between those plates.&#xA;Through putting salt water on the isolator, there's a voltage difference of 1.76V between the black and the magnesium plate.&#xA;After a couple of seconds on load the voltage stays constant at 1.31V.&#xA;My theory is that it's only a simple battery, because there's no continious source of fuel which should be required for naming it fuel cell (according to wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thought right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea about the black coat is, that it is a rubberd graphite or &quot;coal&quot; cathode, rubberd because coal or graphite would fall apart quickly. It is described as air-cathode in the package description. My idea is that the black coal or graphite coat is used to provide a larger amount of air to the reaction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnesium plate gets damaged through this reaction, the black thing not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which material is the black coat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and what is the full reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fs1.directupload.net/images/150615/hdu8yaav.jpg&quot;&gt;a picture of the components&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16951" LastActivityDate="2015-06-15T00:44:11.500" Title="Is this product description wrong and a simple battery and not a fuel cell?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32908" PostTypeId="2" ParentId="32881" CreationDate="2015-06-14T23:59:00.953" Score="10" Body="&lt;p&gt;This pump is &lt;em&gt;way&lt;/em&gt; overkill for a vacuum distillation. The problem is not glass implosion necessarily (round bottom jointed glassware is designed to be put under vacuum, less clear on the condenser), but rather that you may not be able to condense your product given the low vacuum. Vacuum distillation is typically done with an aspirator that gets down to 25 Torr (25,000 microns) or so. At too low a pressure, you may need a very cold receiving vessel and condenser, and depending upon the solvent dry ice slush may not even be enough. You need to determine the boiling point of your solvent at low pressures to figure this out. Your pump is for Schlenk lines and similar applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see a reason not to use a vacuum gauge (and you are not supposed to use mercury anymore in any application), though again, depending upon the setup, it is possible to harm the gauge. You should try to monitor pressure as part of the experimental process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not using the capillary bubbler seems foolish, as bumping is a major problem with vacuum distillations. A &lt;em&gt;capillary bubbler&lt;/em&gt; is a very thin tube (like when you draw glass tubing after melting it) that goes into one joint of the stillpot (drawn end in the reaction mixture). When vacuum is applied, air is forced through the tube, but because the drawn end is so narrow it only allows small bubbles into the mixture. These take the place of boiling chips which will not form bubbles as all gases are extracted from the chips via the vacuum.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-09-16T22:11:35.590" LastActivityDate="2015-09-16T22:11:35.590" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32909" PostTypeId="5" CreationDate="2015-06-15T00:07:44.617" Score="0" Body="&lt;p&gt;Distillation is the process of using disparate boiling points to separate mixtures of miscible liquids.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-06-15T00:59:45.047" LastActivityDate="2015-06-15T00:59:45.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32910" PostTypeId="4" CreationDate="2015-06-15T00:07:44.617" Score="0" Body="Distillation is the process of using disparate boiling points to separate mixtures of miscible liquids." OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-06-15T00:59:41.440" LastActivityDate="2015-06-15T00:59:41.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32913" PostTypeId="2" ParentId="32905" CreationDate="2015-06-15T00:41:52.643" Score="1" Body="&lt;p&gt;There may be better answers forthcoming, but a chemical does not need to melt (i.e. be in the liquid phase) to be reactive. For example, your food very likely has some liquid component, and that could be enough to react with some solid. Now polyethylene is used all over the place and is very nonreactive in many non-rigorous situations (like food service), so you are probably just fine. But I wanted to make the more general point that you don't have to be a liquid or gas to be reactive.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-06-15T00:41:52.643" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32914" PostTypeId="2" ParentId="32907" CreationDate="2015-06-15T00:44:11.500" Score="8" Body="&lt;p&gt;Your analysis seems correct, i.e. it is only a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuel_cell&quot;&gt;fuel cell&lt;/a&gt; if the &lt;em&gt;magnesium is being oxidized by the air&lt;/em&gt;. You can prove if that is the case, and do so as part of your classroom demonstration, by excluding any air, and therefore oxygen, from the fuel cell.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use distilled water and fairly pure $\ce{NaCl}$ (rather than table salt) to make the salt solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Boil the salt water for a minute or so to remove air bubbles (some oxygen from air dissolves into water).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Put the cell with electrolyte in a plastic bag sealed around the wires and a narrow tube such as a plastic coffee stirrer or soda straw that you'll use later to admit air, carefully squeezing out air bubbles. Seal both bag and tube with plasticine modeling clay or soft wax.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Observe the open circuit voltage of the cell.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then run it for a while with a load (e.g. a &lt;a href=&quot;http://www.allelectronics.com/make-a-store/item/lp-41/1.5v-grain-of-wheat-lamp-w/long-leads/1.html&quot;&gt;75 mA 1.5 V lamp&lt;/a&gt;) or a 100 ohm resistor, which will draw 1.3 mA), depending on the current output of the cell. You would expect &lt;em&gt;some&lt;/em&gt; voltage drop over time, either due to &lt;a href=&quot;http://www.tpub.com/neets/book1/chapter2/1b.htm&quot;&gt;cell polarization&lt;/a&gt; or due to any residual oxygen being used up.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;To prove that this is a true fuel cell&lt;/strong&gt; oxidizing the magnesium using air, bubble some air into the plastic bag through the (unplugged) tube. *If the voltage goes back up, then drops down until yet more air is added, you've shown that oxygen is being used and it's a true fuel cell.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please accept my compliments for your thoughtful question; do post your findings here!&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-15T00:44:11.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32915" PostTypeId="2" ParentId="32905" CreationDate="2015-06-15T00:54:57.657" Score="0" Body="&lt;p&gt;Though polyethylene (polythene in the UK) food bags and containers are &lt;em&gt;quite safe to use&lt;/em&gt; even with hot food, oils, fats and other materials &lt;em&gt;do&lt;/em&gt; penetrate polyethylene. For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Lycopene&quot; rel=&quot;nofollow&quot;&gt;lycopene&lt;/a&gt; in tomato sauce or paste is absorbed so that the container is stained orange in a few days at room temperature, or in minutes when microwaved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this is true solution of the lycopene and/or oil carrier into the polyethylene, or penetration of microscopic pores, I do not know. This &lt;a href=&quot;http://www.google.com/patents/WO2003091018A1?cl=en&quot; rel=&quot;nofollow&quot;&gt;patent&lt;/a&gt; claims &quot;staining may occur because the relatively non-polar nature of these commodity resin materials allows greater amounts of diffusion of non-polar substances (such as oils), which may contain dispersed pigments (e.g., lycopene), into the resin.&quot;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-15T00:54:57.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32916" PostTypeId="2" ParentId="32876" CreationDate="2015-06-15T01:07:27.727" Score="3" Body="&lt;p&gt;Another example of a precipitate dissolving in acid is the reaction $\ce{AgNO3 + HCl -&amp;gt; AgCl (v) + HNO3}$. At first, the silver chloride precipitates out as a white powder (if kept away from direct light). Adding more acid, though, &lt;a href=&quot;https://en.wikipedia.org/wiki/Silver_chloride&quot; rel=&quot;nofollow&quot;&gt;forms ligands with the additional chlorine ions&lt;/a&gt;:  $\ce{AgCl(s) + Cl- (aq) -&amp;gt; AgCl2- (aq)}$. See also &lt;a href=&quot;https://www.physicsforums.com/threads/why-does-the-precipitate-dissolve-reappear-and-then-dissolve-again.169134/&quot; rel=&quot;nofollow&quot;&gt;Physics Forums&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-15T01:07:27.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32917" PostTypeId="1" CreationDate="2015-06-15T05:26:55.380" Score="12" ViewCount="8133" Body="&lt;p&gt;Below is a transcript of &lt;a href=&quot;https://i.stack.imgur.com/F9z6n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; doubtful question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A weak monobasic organic acid $\ce{HA}$ is soluble in both water&#xA;  and $\ce{CHCl3}$.  $\pu{500.0 cm3}$ of a solution of $\ce{HA}$ in $\ce{CHCl3}$, with a concentration of &#xA;  $\pu{0.057 mol dm-3}$,&#xA;  is shaken well with $\pu{500.0 cm3}$ of water and allowed&#xA;  to attain equilibrium at $\pu{27°C}$.  An aqueous layer and&#xA;  $\ce{CHCl3}$ layer then separate out; the pH of the aqueous layer is&#xA;  found to be 3.21 under these conditions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The dissociation constant of $\ce{HA}$ in water at $\pu{27°C}$&#xA;  is $\pu{1E-5 mol dm-3}$.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Calculate the partition coefficient at $\pu{27°C}$&#xA;  for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.&lt;/li&gt;&#xA;  &lt;li&gt;In a second experiment, a further $\pu{500.0 cm3}$ portion of the&#xA;  same $\ce{HA}$ solution in $\ce{CHCl3}$ in which the concentration&#xA;  of $\ce{HA}$ is $\pu{0.057 mol dm-3}$, is shaken well with&#xA;  $\pu{500.0 cm3}$ of a $\pu{0.027 mol dm-3}$ aqueous $\ce{NaOH}$&#xA;  solution and allowed to reach equilibrium at $\pu{27°C}$.&lt;br&gt;&lt;br&gt;&#xA;  Calculate the $\mathrm{pH}$ of the aqueous layer under these conditions. &lt;/li&gt;&#xA;  &lt;li&gt;State the assumptions you make, if any, in the above calculations.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My solution for Question 1.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since pH is given, $\ce{[H^+](aq)}$ is determined by&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\mathrm{pH} &amp;amp;=-\log[\ce{H^+~(aq)}]\\&#xA;[\ce{H^+}~(\ce{aq})] &amp;amp;= 10^{-3.21}\\&#xA;    &amp;amp;= 6.16 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}\\&#xA;\end{align} &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;By appropriately applying $K_\mathrm{a}$, the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer is $0.038~\mathrm{mol\,  dm^{-3}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Next, $[\ce{HA}~(\ce{aq})]$ present in the organic layer can be determined by subtracting the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer&#xA;from the initial concentration of $[\ce{HA}~({\ce{aq}})]$, which results in $0.019~\mathrm{mol\,  dm^{-3}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, I got $K_\mathrm{D} =&#xA;   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}&#xA;   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.019~\mathrm{mol\,  dm^{-3}}} = 2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope that this is correct, but  I am not quite sure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My &lt;em&gt;incomplete&lt;/em&gt; solution for Question 2.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The reaction between $\ce{HA}$ and $\ce{NaOH}$ is&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA}~(\ce{aq}) + \ce{NaOH}~(\ce{aq})  \longrightarrow \ce{NaA}~(\ce{aq}) + \ce{H2O}~(\ce{l})$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number of moles of $\ce{NaOH}$ present is $0.0135~\mathrm{mol}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am confused with the concentration of $\ce{HA}$ that I needed to take for this calculation. Initially I have found the concentration of $\ce{HA}$ present in organic layer as $0.019~\mathrm{mol~dm^{-3}}$. Do I need to use that here or $0.057~\mathrm{mol~dm^{-3}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the assumptions that I need to assume to take out this calculation? &lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="5026" LastEditDate="2018-02-25T03:31:36.313" LastActivityDate="2018-02-25T03:31:36.313" Title="How to calculate partition coefficients?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32919" PostTypeId="1" CreationDate="2015-06-15T07:10:50.843" Score="3" ViewCount="692" Body="&lt;p&gt;Who devised the term elements? Who was the one who brought the term elements to us? Also please give a bit information on background of the man.&lt;/p&gt;&#xA;" OwnerUserId="16958" LastEditorUserId="4945" LastEditDate="2015-06-15T07:23:19.837" LastActivityDate="2015-06-16T02:10:42.437" Title="Who devised the term elements?" Tags="&lt;nomenclature&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32920" PostTypeId="1" AcceptedAnswerId="32941" CreationDate="2015-06-15T07:45:19.080" Score="2" ViewCount="38490" Body="&lt;p&gt;I'm trying to understand ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, an ion is when the atom gains or loses electrons. More electrons means it is negatively charged (anion). Less electrons means it is positively charged (cation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with the following premise:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many atoms gain/lose electrons with the hope of having the same number of electrons as the closest noble gas in the periodic table.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The book uses Potassium as an example. Looking at my table, the closest noble gas to Potassium is Argon, which has $18$ electrons. Therefore Potassium wants to lose one electron...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice, I get that. But then I noticed something odd about &lt;strong&gt;&lt;em&gt;all&lt;/em&gt;&lt;/strong&gt; the examples in my book: all the elements they picked are elements conveniently one or two steps away from a noble gas. They never picked an element in the middle of the table like, say, iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron has $26$ electrons. Its closest noble gas is Argon, which has $18$ electrons. So iron wants to lose $8$ electrons (becoming a cation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^{8+}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? My book literally avoids picking elements in the middle of the table, so I feel there's something weird about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I noticed that in every example, the book mentions the Group of the element (Potassium is group 1A, etc...). How is that even relevant? All you have to do is count the number of electrons, no?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="10185" LastEditDate="2015-06-15T09:12:53.123" LastActivityDate="2016-12-07T02:13:42.510" Title="Calculating the electrons an atom wants to gain/lose to reach a noble gas" Tags="&lt;ions&gt;&lt;elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32921" PostTypeId="1" AcceptedAnswerId="32954" CreationDate="2015-06-15T07:53:33.980" Score="4" ViewCount="62" Body="&lt;p&gt;I'm working on a project that identifies average characteristics of different elements. I'm stuck on lithium, I can't find a way to ratio lithium I and lithium II that symbolizes how it is in nature. I tried looking around on the internet but I haven't found any mention of ratios to ions only ratios to isotopes. Does anyone know of anything that might help?&lt;/p&gt;&#xA;" OwnerUserId="16959" LastEditorUserId="16959" LastEditDate="2015-06-16T00:49:34.577" LastActivityDate="2015-06-16T03:56:43.153" Title="Dispersion of Lithium Ions in Nature (Ratio Similar to Isotopes?)" Tags="&lt;ions&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32922" PostTypeId="2" ParentId="32905" CreationDate="2015-06-15T10:09:16.020" Score="0" Body="&lt;p&gt;In addition to the answeres from @trb456 and DrMoishe, please be careful if you talk about PE as a homogene material. &#xA;There a quite a lot different &quot;types&quot; of PE, mostly categorised by chain lenght and side arms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LLD-PE (linear low density Polyethylene) can start to melt at temperatures as low as 45°C. Containers for food normaly have a stamp on the bottom to indicate if they can be used for microwave ovens or hot food/liquids. Same should be there for bags sold for storing food. They should have a lable if you can use them for hot food (at least in EU and US).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a reactivity standpoint, PE is mostly unreactive due to being composed of saturated hydrocarbon chains. Therefore, you can safely eat normal PE from a chemical health perspective. As long as you can't cut or otherweise harm yourself with it, it should leave your body in a state similiar to how you've eaten it (excluded are nanosized particles, which are still in discussion). Health risks from (mostly cheap) PEs normally come from additives like colours and softeners. &lt;/p&gt;&#xA;" OwnerUserId="14917" LastActivityDate="2015-06-15T10:09:16.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32924" PostTypeId="2" ParentId="6138" CreationDate="2015-06-15T10:25:47.253" Score="-2" Body="&lt;p&gt;Sometimes, dopants can go to the interstitial site of the target material. We can say this makes some distortions happen.That actually change the crystal structure. Then we can see the certain peak position changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, misalignment can make peak position changes.So we have to operate the machine properly. calibration is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strain and stress of the materials can make XRD peak shift as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="16960" LastActivityDate="2015-06-15T10:25:47.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32926" PostTypeId="1" AcceptedAnswerId="32935" CreationDate="2015-06-15T10:50:47.720" Score="11" ViewCount="1329" Body="&lt;p&gt;I always (possible naively) thought that elements are either natural or synthetic, either one or the other.  But, according to the Wikipedia page about &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_element#List_of_synthetic_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthetic Elements&lt;/a&gt;, it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All elements with atomic numbers 1 through 98 are naturally occurring at least in trace quantities, but the following elements are usually produced through synthesis. Except for Francium, they were all discovered through synthesis before being found in nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are shown in pink in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsbhS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image attribution: &quot;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Periodic_table,_good_SVG.svg#/media/File:Periodic_table,_good_SVG.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Periodic table, good SVG&lt;/a&gt;&quot; by &lt;a href=&quot;//commons.wikimedia.org/wiki/User:Incnis_Mrsi&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;User:Incnis Mrsi&quot;&gt;Incnis Mrsi&lt;/a&gt; - Own work. Licensed under &lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Creative Commons Attribution-Share Alike 3.0&quot;&gt;CC BY-SA 3.0&lt;/a&gt; via &lt;a href=&quot;//commons.wikimedia.org/wiki/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the amounts of these elements exist naturally in trace quantities, not necessarily on Earth, for example, according to the &lt;a href=&quot;http://education.jlab.org/itselemental/ele061.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jefferson Lab page about promethium&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Promethium does not occur naturally on earth, although it has been detected in the spectrum of a star in the constellation Andromeda.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leads me to my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a natural abundance threshold of a 'synthetic' element for it to be considered natural?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-06-15T21:59:49.647" Title="Is there an abundance threshold of a 'synthetic' element for it to be considered natural?" Tags="&lt;periodic-table&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32927" PostTypeId="1" AcceptedAnswerId="32971" CreationDate="2015-06-15T11:35:04.603" Score="4" ViewCount="191" Body="&lt;p&gt;For the sake of the question, let’s assume that you are provided with two labelled rods of the pure metals &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; and two unlabelled bottles containing &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0 mol dm-3}$&lt;/span&gt; aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{A^m+}$&lt;/span&gt; and  the other bottle containing &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0 mol dm-3}$&lt;/span&gt; aqueous solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{B^n+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find out, which metal out of the given two is more reducing, and identify each solution that is given in the unlabelled bottles. What are the procedures that I should follow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I immerse &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; in a solution of  &lt;span class=&quot;math-container&quot;&gt;$\ce{A^m+}$&lt;/span&gt;, I guess  there will be no change. Because if I do so there would be an equilibrium prevail with the ions in the solution and the electrode, and I think this is true for &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know how to identify the one which gets reduced more, yet I know that the anode gets reduced. So I am confused about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that the only clue I have is immersing &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; in  &lt;span class=&quot;math-container&quot;&gt;$\ce{B^n+}$&lt;/span&gt; and vice versa for &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; also. But  really I don't know what  would happen if I do so. &lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="4945" LastEditDate="2020-02-26T13:18:53.130" LastActivityDate="2020-02-26T13:22:24.523" Title="How to identify which of two given electrodes with corresponding solutions is reducing more?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32928" PostTypeId="1" CreationDate="2015-06-15T12:12:37.483" Score="13" ViewCount="4360" Body="&lt;p&gt;I'm reading document about Solid phase protein synthesis (SPPS) from &lt;a href=&quot;https://en.wikipedia.org/wiki/Peptide_synthesis#Solid-phase_synthesis&quot;&gt;Wikipedia&lt;/a&gt;, and according to the document:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SPPS is limited by yields, and typically peptides and proteins in the&#xA;  range of 70 amino acids are pushing the limits of synthetic&#xA;  accessibility. Synthetic difficulty also is sequence dependent;&#xA;  typically amyloid peptides and proteins are difficult to make. Longer&#xA;  lengths can be accessed by using native chemical ligation to couple&#xA;  two peptides together with quantitative yields.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a bit not understanding why this synthesis method is limited to 70 amino acids in length. Anyone please help me to explain what is the issue?&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="7448" LastEditDate="2015-06-15T16:15:26.837" LastActivityDate="2015-06-16T09:03:37.657" Title="Why is solid phase peptide synthesis limited to 70 amino acids?" Tags="&lt;biochemistry&gt;&lt;synthesis&gt;&lt;solid-state-chemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32929" PostTypeId="2" ParentId="32928" CreationDate="2015-06-15T12:33:47.897" Score="15" Body="&lt;p&gt;The issue is one of yield. Each step has a certain yield percentage of yield, that the synthesizers will maximize as much as possible. However, given that it is fairly impossible to generate 100% yield all the time, the amount of desired protein you get over time will deteriorate exponentially with length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say that each step gives you 95% yield. If you have a protein that is ten amino acids long, your final step will give you .95^10 yield, which is about 60%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with the same yield per step, but a 70 amino acid chain, the amount of protein you want drops to 2%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you are at 99% yield per step (a difficult feat), your final yield will be under 50% for a 70 amino acid chain. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-15T12:33:47.897" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32930" PostTypeId="2" ParentId="32928" CreationDate="2015-06-15T12:34:08.750" Score="13" Body="&lt;p&gt;You need to perform one reaction per amino acid to couple it to the previous one, and that reaction isn't 100% efficient. You always get a small amount of unreacted peptide, or some other side product. Those small inefficencies add up for longer peptides because you're doing so many reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created a quick graph showing the yield assuming 98% efficiency of the reation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U07mQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the yield drops significantly for longer peptides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't really a hard limit on size, but you get terribly inefficient for long peptides. And at a certain point it is much easier to synthesize the peptide biologically. Proteins are routinely synthesized in large amounts using bacteria like &lt;em&gt;E. coli&lt;/em&gt;, unless you need to do certain modifications you can only do with solid-phase synthesis, going the biological route can be much cheaper than using chemical synthesis.&lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-06-15T12:34:08.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32931" PostTypeId="2" ParentId="32854" CreationDate="2015-06-15T12:43:32.350" Score="5" Body="&lt;p&gt;You are right that 0.5 mol of $\ce{N2}$ is 11.2 L, but read the question carefully. Option B says that 11.2 L of $\ce{N2}$ contains 0.5 mol of &lt;strong&gt;atoms&lt;/strong&gt;, which is wrong, it contains 11.2 L of $\ce{N2}$ contains 0.5 mol of &lt;strong&gt;molecules&lt;/strong&gt;. It seems to be a trick answer. The first option is correct as 0.5 mol of $\ce{N2}$ has a mass of 14 grams.&lt;/p&gt;&#xA;" OwnerUserId="5710" LastEditorDisplayName="user15489" LastEditDate="2015-06-16T08:30:27.527" LastActivityDate="2015-06-16T08:30:27.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32932" PostTypeId="2" ParentId="32487" CreationDate="2015-06-15T13:54:16.193" Score="9" Body="&lt;p&gt;The first step toward answering this question is understanding the extent to which the statement &quot;Group 11 metals are unreactive&quot; is or isn't true.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated in &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2634185/&quot;&gt;How do plants respond to copper deficiency?&lt;/a&gt; &lt;em&gt;Plant Signaling and Behavior&lt;/em&gt; vol. 3, pages 231-232:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The transition metal copper is essential for all organisms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not just that copper is present in every living thing.  Copper is reacting in every living thing. Copper ions cycle between the +1 and +2 oxidation states in &lt;a href=&quot;https://en.wikipedia.org/wiki/Plastocyanin&quot;&gt;Plastocyanin&lt;/a&gt;,  Copper/zinc &lt;a href=&quot;https://en.wikipedia.org/wiki/Superoxide_dismutase&quot;&gt;superoxide dismutase&lt;/a&gt; and &lt;a href=&quot;http://www.life.illinois.edu/crofts/bioph354/cyt_ox.html&quot;&gt;cytochrome oxidase&lt;/a&gt;. So as you are saying copper is unreactive, it as reacting in cytochrome oxidase in every mitochondrion of your body, and in Cu/Zn superoxide dismutase in the cytosol of cells through your body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to quantify the reactivity of metal ions in through &lt;a href=&quot;http://downloads.gphysics.net/UACh/Medicina/TablasIones.pdf&quot;&gt;reduction potential&lt;/a&gt;, here listed for reduction of the ion to the zero oxidation state metal.  (limiting to d-block metals at least as unreactive as Cu)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c @{:} c} \\&#xA;\color{blue}{\text{Ion}} &amp;amp; \color{blue}{\text{Reduction potential}} \\\hline&#xA;\mathbf{\color{red}{Cu^{2+}}} &amp;amp; 0.3419  \\\hline&#xA;\ce{Tc^2+} &amp;amp; 0.400   \\\hline&#xA;\ce{Ru^2+} &amp;amp; 0.455  \\\hline&#xA;\ce{Cu+} &amp;amp; 0.521 \\ \hline&#xA;\ce{Rh+} &amp;amp; 0.600  \\&#xA;\ce{Rh^2+} &amp;amp; 0.600  \\&#xA;\ce{Rh^3+} &amp;amp; 0.758  \\ \hline&#xA;\mathbf{\color{red}{Ag^+}} &amp;amp; 0.7996   \\\hline&#xA;\ce{Os^2+} &amp;amp; 0.85  \\\hline&#xA;\ce{Hg^2+} &amp;amp; 0.851  \\\hline&#xA;\ce{Pd^2+} &amp;amp; 0.951   \\\hline&#xA;\ce{Ir^3+} &amp;amp; 1.156  \\\hline&#xA;\ce{Pt^2+} &amp;amp; 1.18   \\\hline&#xA;\mathbf{\color{red}{Au^{3+}}} &amp;amp; 1.498  \\&#xA;\ce{Au+} &amp;amp; 1.692  \\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at this list, we see that there are a substantial number of d-block metals that are less reactive than copper.  A few are less reactive than silver, and gold is the least reactive on the basis standard reduction potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall trend in d-block metals, is down and to the right in the block is less reactive. The exception is that, for a given row, the group 12 metal is more reactive than the group 11 metal, but due to the formation of a 2+ ion, not a 1+ ion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next we should look at &lt;a href=&quot;https://dept.astro.lsa.umich.edu/~cowley/ionen.htm&quot;&gt;ionization energies&lt;/a&gt; of d-block metals to see how much of the reduction potential corresponds to properties of an individual atom, versus bulk or solvent effects.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at first-ionization energies, for each row of the d-block, the highest ionization energy is always the group 12 element (Zn, Cd, Hg). This corresponds well with the fact the Zn, Cd and Hg do not form +1 ions.  The group 11 elements have significantly lower first-ionization energies, and do form +1 ions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then looking at second-ionization energies, group 11 has the highest second-ionization energies for every row of the d-block!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the reactive/unreactivity of d-block is explained by the increase of nuclear charge to the right and down the d-block.  The complete d-shell of neutral group 11 atoms shields the valence s electron well enough that it may be removed to react.  However, to lose 2 electrons, the group 11 elements would need to lose a d-electron from a completed d-shell, which is difficult.  For group 12, two s valance electrons can be lost, and the complete d-shell maintained.   &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="7448" LastEditDate="2015-06-15T21:07:37.173" LastActivityDate="2015-06-15T21:07:37.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32933" PostTypeId="2" ParentId="32926" CreationDate="2015-06-15T14:09:46.187" Score="2" Body="&lt;p&gt;I have never heard of a formal specification for such a threshold. Qualitatively speaking, though, I believe this threshold could be effectively considered &quot;enough that we can make useful stuff with it.&quot; To my mind the key phrase in your blockquote is (emphasis added):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;... the following elements are usually &lt;strong&gt;&lt;em&gt;produced&lt;/em&gt;&lt;/strong&gt; through synthesis.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;'Production' implies making stuff. 'Measurable in trace quantities by &lt;a href=&quot;http://crustal.usgs.gov/laboratories/icpms/intro.html&quot; rel=&quot;nofollow&quot;&gt;ICP-MS&lt;/a&gt;' is not helpful for making stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worth noting: since different amounts of different materials are needed to make useful things, it's possible that a hard, quantitative abundance number might not even be useful to define.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-06-15T14:09:46.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32934" PostTypeId="1" AcceptedAnswerId="32950" CreationDate="2015-06-15T14:14:53.850" Score="-4" ViewCount="355" Body="&lt;p&gt;In the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reducing_agent&quot; rel=&quot;nofollow&quot;&gt;reducing agent&lt;/a&gt;&quot; article in Wikipedia it says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Good reducing agents are reagents that deliver $\ce{H2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wouldn't that make water unsafe to dissolve oxidizing agents?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="3846" LastEditDate="2015-06-15T16:52:04.910" LastActivityDate="2021-03-15T14:50:41.003" Title="Are aqueous solutions of oxidizing agents safe?" Tags="&lt;redox&gt;&lt;aqueous-solution&gt;&lt;safety&gt;" AnswerCount="2" CommentCount="3" ClosedDate="2017-08-30T01:11:28.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="32935" PostTypeId="2" ParentId="32926" CreationDate="2015-06-15T14:35:55.020" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a natural abundance threshold of a 'synthetic' element for it&#xA;  to be considered natural?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{&amp;gt; 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, any element that can be &lt;strong&gt;detected&lt;/strong&gt; outside of the laboratory (&lt;em&gt;e.g.&lt;/em&gt; in the natural environment) is a naturally occurring element.  This would include the 80 stable elements along with the 18 radioactive elements with very long half-lives, and the shorter-lived radioactive elements produced upon their decay.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there are &lt;strong&gt;98 naturally occurring elements; all of the others are synthetic&lt;/strong&gt;, artificial, man-made, whatever you choose to call them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://en.wikipedia.org/?title=Chemical_element#Evolving_definitions&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; provides nice background, especially the following sections:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the introductory section, especially the last paragraph therein&lt;/li&gt;&#xA;&lt;li&gt;Occurrence and origin on Earth&lt;/li&gt;&#xA;&lt;li&gt;List of the 118 known chemical elements - Occurrence&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-15T21:59:49.647" LastActivityDate="2015-06-15T21:59:49.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32936" PostTypeId="1" CreationDate="2015-06-15T15:11:20.290" Score="1" ViewCount="730" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Given that &lt;span class=&quot;math-container&quot;&gt;$K=18$&lt;/span&gt; for the equilibrium below, calculate the total pressure in a sealed vessel initially containing only &lt;span class=&quot;math-container&quot;&gt;$\ce{NH4I(s)}$&lt;/span&gt;, that reaches equilibrium.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{NH4I(s) \leftrightharpoons NH3(g) + HI(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;a) &lt;span class=&quot;math-container&quot;&gt;$4.2\mathrm{~atm}$&lt;/span&gt;&lt;br/&gt;&#xA;b) &lt;span class=&quot;math-container&quot;&gt;$8.5~\mathrm{atm}$&lt;/span&gt; &lt;br/&gt;&#xA;c) &lt;span class=&quot;math-container&quot;&gt;$2.1~\mathrm{atm}$&lt;/span&gt; &lt;br/&gt;&#xA;d) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{18~atm}$&lt;/span&gt; &lt;br/&gt;&#xA;e) &lt;span class=&quot;math-container&quot;&gt;$\mathrm{6.3~atm}$&lt;/span&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$K = \ce{[NH3][HI]} = 18$&lt;/span&gt;, I found &lt;span class=&quot;math-container&quot;&gt;$\ce{[NH3]=[HI]=}~\frac{4.24mol}{L}$&lt;/span&gt;, and therefore &lt;span class=&quot;math-container&quot;&gt;$8.48~\mathrm{mol}$&lt;/span&gt; of gas is produced, but I don't know how to find the equilibrium pressure without temperature and volume.&lt;/p&gt;&#xA;" OwnerUserId="16963" LastEditorUserId="865" LastEditDate="2021-05-22T21:36:05.077" LastActivityDate="2021-05-22T21:36:05.077" Title="calculate the pressure at equilibrium" Tags="&lt;equilibrium&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32937" PostTypeId="2" ParentId="21578" CreationDate="2015-06-15T15:18:59.220" Score="1" Body="&lt;p&gt;No, you cannot. There is no way to reset the transient pulses in the neurons that carry the essence of thought processes. The information is already gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find the neural synapses all in the discharged state. All thought process, all memory, all of what makes a person is gone.&lt;/p&gt;&#xA;" OwnerUserId="16314" LastEditorUserId="16314" LastEditDate="2015-06-15T17:10:13.467" LastActivityDate="2015-06-15T17:10:13.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32938" PostTypeId="2" ParentId="32936" CreationDate="2015-06-15T15:33:14.673" Score="2" Body="&lt;p&gt;I think what is given here is $K_p= p(\ce{NH3}). p(\ce{HI})$ and not $K_c$. So, What you have here are the partial pressures of the gases. As you have a sealed vessel, so the sum of the partial pressures of the gases gives the total pressure, 8.5 atm&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-06-15T15:33:14.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32939" PostTypeId="2" ParentId="32894" CreationDate="2015-06-15T15:46:13.830" Score="11" Body="&lt;p&gt;Notations and conventions used for the description of symmetry in rigid molecules are established in &lt;a href=&quot;http://dx.doi.org/10.1351/pac199769081641&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Notations and conventions in molecular spectroscopy: Part 2. Symmetry notation (IUPAC Recommendations 1997)&lt;/em&gt;&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symbols for symmetry operators are printed italic, with sub- and super-scripts that are upright except for the variables $k$ and $n$ which are replaced by numbers for a specific operator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;$E$ – identity operator&lt;br&gt;&#xA;$\sigma$ – reflection operator for reflection across a general plane&lt;br&gt;&#xA;$\sigma_\mathrm{h}$ – reflection operator for reflection across a horizontal plane&lt;br&gt;&#xA;$\sigma_\mathrm{v}$ – reflection operator for reflection across a vertical plane&lt;br&gt;&#xA;$C_n{}^k$ – $n$-fold rotation operator for $k$ successive rotations through an angle of $2\pi/n$ about an $n$-fold rotation axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$&lt;br&gt;&#xA;$S_n{}^k$ – $n$-fold rotation-reflection operator for $k$ successive rotation-reflections about an $n$-fold rotation-reflection axis for a rotation through an angle of $2\pi/n$ followed&#xA;by a reflection in a plane perpendicular to the axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$&lt;br&gt;&#xA;$i$ – inversion operator through the centre of symmetry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symbols for symmetry groups are printed italic with upright subscripts, following common usage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;$C_n$, $S_{2n}$, $D_n$, $D_{n\mathrm{h}}$, $D_{n\mathrm{d}}$, $C_{n\mathrm{v}}$, $C_{n\mathrm{h}}$, $T$, $T_\mathrm{h}$, $T_\mathrm{d}$, $O$, $O_\mathrm{h}$, $I_\mathrm{h}$, $C_{\infty\mathrm{v}}$, $D_{\infty\mathrm{h}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Symbols for irreducible representations of point groups, which are usually called symmetry species in spectroscopy, are printed upright.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;br&gt;&#xA;$\mathrm{A}$, $\mathrm{B}$, $\mathrm{E}$, $\mathrm{A}_1$, $\mathrm{A}_2$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The example table given in the question should thus be typeset as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9CDqp.png&quot; alt=&quot;C2v character table with correct formatting&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="3846" LastEditDate="2015-06-15T16:38:14.023" LastActivityDate="2015-06-15T16:38:14.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32941" PostTypeId="2" ParentId="32920" CreationDate="2015-06-15T16:13:41.900" Score="2" Body="&lt;p&gt;The book's explanation about a noble gas configuration is somewhat accurate, but fairly incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements on the right and on the left of the periodic table (the alkali (earth) metals, the halogens, the chalcogens (the group that starts with Oxygen) and the pnictogens (Nitrogen group)) have electron configurations that make it somewhat easier to lose/gain electrons to &quot;look&quot; like noble gasses (this is why alkali metals and halogens are so reactive: the destabilization from the charge is very minimal compared to the stability of the electron configuration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you have observed, the book goes to some effort to avoid talking about transition metals. There is a reason for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noble gas configurations are a &lt;em&gt;subset&lt;/em&gt; of the stable configurations of electrons. In reality, what's actually being aimed for is an element with no incomplete electron shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an electron shell? The actual quantum mechanical definition may be a bit more complicated than you need, but for your purposes, it suffices to say that it's how electrons will tend to be distributed around a nucleus. Note that this is &lt;em&gt;extremely&lt;/em&gt; distinct from saying that the electron &quot;orbits&quot; the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a periodic table, there are four general regions to look at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J6lol.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The labels &quot;s&quot;, &quot;p&quot;, &quot;d&quot;, and &quot;f&quot; all refer to the various electron orbitals, and the blocks all indicate what the outermost (or &quot;valence&quot;) shell is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while elements in the s and p blocks can lose a small number of electrons get complete shells (which will resemble the noble gases), transition metals are a &lt;strong&gt;bit&lt;/strong&gt; more complicated. And by a bit more complicated, I mean that there are very rough rules in place that have almost more exceptions than they have actual examples of following the rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in very general terms, a transition metal will lose electrons. If they are in the first half of the transition metals in their row, they can lose a number of electrons less than or equal to the number of d electrons plus the number of s electrons. This maximum number of electrons lost will make them resemble the noble gasses in configuration, but it's only appreciably stable when the transition metal is in the first half of transition metals in its group ($\ce{Mn^{7+}}$ is well documented in ionic compounds, but $\ce{Fe^{8+}}$ doesn't exist a lot because it has too many electrons to try to lose to comfortably fall into a noble gas configuration). But transition metals can lose electrons &lt;em&gt;up to&lt;/em&gt; this point. The issue is, most transition metals can adopt multiple charge configurations for various reasons, so generally, when writing an ionic compound with a transition metal in words, you need to specify the charge of the transition metal (actually, its oxidation state, but in an ionic compound that's mostly splitting hairs). So you can have &quot;Iron (II) Oxide&quot;, which is a black powder and &quot;Iron (III) Oxide&quot; which is rust. These compounds will have different chemistries because of the charge on the transition metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suffice to say, your textbook is avoiding transition metals because transition metals are &lt;em&gt;weird&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-06-15T17:04:42.690" LastActivityDate="2015-06-15T17:04:42.690" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="32943" PostTypeId="2" ParentId="32926" CreationDate="2015-06-15T17:04:19.273" Score="3" Body="&lt;p&gt;The radioactive elements which are regarded as occurring &quot;naturally&quot; have at least one of the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;At least one isotope has a half-life sufficiently long that if it is created in a supernova, significant quantities will persist for a long time afterward.  Many isotopes of uranium fit this pattern.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They are decay by-products of other naturally-occurring radioactive isotopes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Trans-uranic elements are considered artificial because the their short half-lives mean the probability of any remaining from a million or so years ago is essentially nil, and there are no heavier natural elements that could replenish the supply.  Radon is naturally occurring because, despite its short half-life, the supply gets replenished through the decay of heavier elements.  Technetium has a much longer half-life than radon, but since no naturally-occurring radioactive isotopes of anything else produce it when they decay, any supplies that may have existed will have dwindled away to essentially nothing by now.&lt;/p&gt;&#xA;" OwnerUserId="7274" LastActivityDate="2015-06-15T17:04:19.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32945" PostTypeId="1" AcceptedAnswerId="40758" CreationDate="2015-06-15T17:53:29.193" Score="6" ViewCount="4086" Body="&lt;p&gt;As an amateur soda maker, I have noticed an odd phenomenon: lemon juice sodas do not foam. Virtually every juice I have tried foams to some degree when mixed with carbonated water. Lemon juice does not. I have read &quot;folk&quot; explanations for this as being due to the &quot;acidity&quot; of the lemon juice, but this is not correct because other acidic juices like lime and orange juice foam. &lt;strong&gt;Why is it that lemon juice has an anti-foaming action?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further information: it should be noted that &quot;pasteurized&quot; (ie, boiled) lemon juice does foam. Only fresh-squozen lemon juice is anti-foaming. Therefore, what is doing the anti-foaming may likely be some volatile compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of anti-foaming is that it is caused primarily by localized depression of surface tension, and that liquids having heterogeneous species promote anti-foaming because of the surface tension differential between those species at the surface. Therefore, one possible theory is that the lemon juice has some volatile oil which populates the surface of the water only in spots and is causing the surface tension breakage. Even if this theory is correct, it remains to determine what this mysterious volatile oil is that is the cause.&lt;/p&gt;&#xA;" OwnerUserId="9663" LastActivityDate="2017-09-22T23:28:59.067" Title="Lemon juice, the ultimate anti-foaming agent?" Tags="&lt;food-chemistry&gt;&lt;surfactants&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32946" PostTypeId="1" AcceptedAnswerId="32949" CreationDate="2015-06-15T18:02:29.613" Score="12" ViewCount="13423" Body="&lt;p&gt;Different sources have different answers. In the reduction of an aldehyde or ketone to an alcohol by Sodium Borohydride, how many moles of aldehyde are reduced by one mole of Sodium Borohydride?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While my teacher claims it is 1, Solomons-Fryhle claims it to be 4.&lt;/p&gt;&#xA;" OwnerUserId="16967" LastEditorUserId="10185" LastEditDate="2015-06-15T18:19:16.513" LastActivityDate="2015-06-15T19:40:57.810" Title="How many hydride ions are available in NaBH4?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32949" PostTypeId="2" ParentId="32946" CreationDate="2015-06-15T19:40:57.810" Score="11" Body="&lt;p&gt;Practically speaking, a chemist will use one equivalent of sodium borohydride to reduce an aldehyde or ketone. However, theoretically only 0.25 equivalents are necessary. Borohydride donates a hydride to the carbonyl carbon, generating borane, which is a weaker reducing agent than borohydride. However, borane is a Lewis acid that can react with the alkoxide formed in the reduction. This gives a new borohydride, which can reduce a second molecule of the ketone. This process can continue until all of the hydrides are consumed from the boron center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OJrrd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that sodium borohydride reductions are commonly performed in alcohol solvents, so the exact nature of these steps could be altered. The alkoxide could be protonated by solvent, and then the solvent's conjugate base could reactivate a neutral boron species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind that while sodium borohydride isn't VERY basic, it is a little basic, so it will deprotonate alcohols slowly. It's that reaction that causes the practical requirement for more than the theoretical borohydride stoichiometry.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-06-15T19:40:57.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32950" PostTypeId="2" ParentId="32934" CreationDate="2015-06-15T20:46:40.430" Score="5" Body="&lt;p&gt;If the oxidizing agent is sufficiently strong, e.g. fluorine, yes. &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fluorine&quot; rel=&quot;nofollow noreferrer&quot;&gt;[W]ater spontaneously combust[s] under a fluorine jet.&lt;/a&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;However, there are only a few oxidizers, such as fluorine, &lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorine_trifluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;chlorine trifluoride&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Dioxygen_difluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;dioxygen difluoride&lt;/a&gt; that can displace oxygen from water. More likely you'd encounter a strong reducing agent, such as Group I metals, that would attach to the oxygen in water, liberating hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="76428" LastEditDate="2021-03-15T14:50:41.003" LastActivityDate="2021-03-15T14:50:41.003" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32951" PostTypeId="1" AcceptedAnswerId="33198" CreationDate="2015-06-15T22:36:00.083" Score="1" ViewCount="719" Body="&lt;p&gt;We had the task to solve several salts in water and measure the pH value of the solution. After that we should &quot;create&quot; the protolysis equation and to tell about Chemical equilibrium[s].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get some possible result equations, but I'm really unsure about them. The [] brackets were used to make the ionic loads easier visible. solving equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{KNO3 &amp;lt;=&amp;gt; [NO3]- + [K]+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{FeCl3 &amp;lt;=&amp;gt; [Fe]^{3+} + [Cl]-}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{NH4Cl &amp;lt;=&amp;gt; [NH4]+ + [Cl]-}$`&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{NaHSO4 &amp;lt;=&amp;gt; [HSO4]- + [Na]+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;my protolysis equations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{[NO3]- + H2O &amp;lt;=&amp;gt; HNO3 + [OH]-}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{[Fe(H2O)6]^{3+} + H2O &amp;lt;=&amp;gt; [Fe(H2O)5OH]^{2+} + [H3O]+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{[NH4]+ + H2O &amp;lt;=&amp;gt; NH3 + [H3O]+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$\ce{[HSO4]- + H2O &amp;lt;=&amp;gt; [SO4]^{2-} + [H3O]+}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would say that all equations run bi-directional, because the result of solving equations can vary dependent on temperature for example, and the protolysis equations should be bi-directional because we're having a solution, which isn't a pure acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I got the question if I can draw conclusions on the pH-change through those protolysis reactions. I would say yes, but not very well, depending on the sum of all hydrogen-containing ions like $\ce{H3O+}$&lt;/p&gt;&#xA;" OwnerUserId="16951" LastEditorUserId="865" LastEditDate="2021-05-22T20:53:31.653" LastActivityDate="2021-05-22T20:53:31.653" Title="Are my protolysis equations right?" Tags="&lt;acid-base&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32952" PostTypeId="2" ParentId="32919" CreationDate="2015-06-16T02:10:42.437" Score="3" Body="&lt;p&gt;One answer is that the ancient Greeks, specifically Plato, devised the term &quot;element&quot;.  Quoting from &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_element#Greece&quot; rel=&quot;nofollow&quot;&gt;wikipedia article on Classical elements&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Plato seems to have been the first to use the term &quot;element (στοιχεῖον, stoicheion)&quot; in reference to air, fire, earth, and water. The ancient Greek word for element, &lt;em&gt;stoicheion&lt;/em&gt; (from &lt;em&gt;stoicheo&lt;/em&gt;, &quot;to line up&quot;) meant &quot;smallest division (of a sun-dial), a syllable&quot;, as the composing unit of an alphabet it could denote a letter and the smallest unit from which a word is formed. A similar alphabetic metaphor may be the origin of the equivalent Latin word elementum (from which the English word comes), possibly based on the names of the letters 'l', 'm', and 'n', though the validity of this idea is debated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you had a more modern, chemical definition of &quot;element&quot; in mind for your question, then another possible answer is Robert Boyle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting from wikipedia again, this time from its article on &lt;a href=&quot;https://en.wikipedia.org/?title=Chemical_element#History&quot; rel=&quot;nofollow&quot;&gt;chemical elements&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1661, Robert Boyle proposed his theory of corpuscularism which favoured the analysis of matter as constituted by irreducible units of matter (atoms) and, choosing to side with neither Aristotle's view of the four elements nor Paracelsus' view of three fundamental elements, left open the question of the number of elements. The first modern list of chemical elements was given in Antoine Lavoisier's 1789 Elements of Chemistry, which contained thirty-three elements, including light and caloric.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't resist adding a reference to my favorite English-language modern chemistry text &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncleftish_Beholding&quot; rel=&quot;nofollow&quot;&gt;that omits the word &quot;element&quot; entirely&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For most of its being, mankind did not know what things are made&#xA;  of, but could only guess. With the growth of worldken, we began&#xA;  to learn, and today we have a beholding of stuff and work that&#xA;  watching bears out, both in the workstead and in daily life.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The underlying kinds of stuff are the &lt;em&gt;firststuffs&lt;/em&gt;, which link&#xA;  together in sundry ways to give rise to the rest. Formerly we&#xA;  knew of ninety-two firststuffs, from waterstuff, the lightest and&#xA;  barest, to ymirstuff, the heaviest. Now we have made more, such&#xA;  as aegirstuff and helstuff. [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the Normans had never invaded England in 1066, we might not have &quot;elements&quot; at all, but &quot;firststuffs&quot; instead.  Either way, it's just a debate about the right translation for &quot;stoicheion&quot;, a concept which Plato seems to have given us.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-16T02:10:42.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32954" PostTypeId="2" ParentId="32921" CreationDate="2015-06-16T03:56:43.153" Score="0" Body="&lt;p&gt;I believe I resolved my problem, the attributes I was looking at was spectra and combining the lines of both ions. The problem was I was comparing data given by two authors that used different units, so what I did, which I think solves my problem, was divide the sums of the given intensity by the emission probabilities for which I received the ratio of Li I to Li II as roughly 33 to 6 or 18% Lithium II. More than I or apparently anyone else thought so I may have still made a mistake, either way thanks to everyone who tried and helped.&lt;/p&gt;&#xA;" OwnerUserId="16959" LastActivityDate="2015-06-16T03:56:43.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32955" PostTypeId="1" CreationDate="2015-06-16T05:38:23.500" Score="4" ViewCount="6818" Body="&lt;p&gt;What is the quickest way to dissolve a large chunk of steel? I mean a very quick reaction, assuming that the interaction occurs in a sealed or semi-sealed environment? Reaction times of the order of a few minutes versus tens of minutes, hours or days are the objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to how quickly acids like $\ce{HNO3}$ and $\ce{H2SO4}$ can react with steel containers.&lt;/p&gt;&#xA;" OwnerUserId="16971" LastEditorUserId="4945" LastEditDate="2015-06-16T07:30:08.540" LastActivityDate="2019-03-26T11:21:20.540" Title="What is the quickest way to dissolve steel?" Tags="&lt;acid-base&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32956" PostTypeId="2" ParentId="32955" CreationDate="2015-06-16T06:44:30.407" Score="1" Body="&lt;p&gt;Two to look up as &quot;homework &quot;.  Aqua Regia which is a mixture that dissolves gold. Not good enough for engineering and some other rare earth metals. Steel? Maybe. The other is Caro's acid. There's a YouTube video of it dissolving a fly. That's organic matter so it may not be good enough for steel. Worth a look though.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-16T06:44:30.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32957" PostTypeId="1" AcceptedAnswerId="32958" CreationDate="2015-06-16T06:53:35.560" Score="3" ViewCount="3423" Body="&lt;p&gt;An alloy is a material composed of two or more metals or a metal and a nonmetal. And, they are usually formed by heating the elements to their melting points, and then cooling them, so that the components mix. Now, why doesn't this works backwards i.e. if we heat the alloy again to melting point of their constituents, and they should separate?&lt;/p&gt;&#xA;" OwnerUserId="16924" LastActivityDate="2015-06-16T07:24:09.743" Title="If alloys are homogeneous mixtures, why can't we separate their components?" Tags="&lt;inorganic-chemistry&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32958" PostTypeId="2" ParentId="32957" CreationDate="2015-06-16T07:10:46.377" Score="-1" Body="&lt;p&gt;Once the alloy has been formed the atoms from the different metals will have shared there electrons with each other and come to an equilibrium. In this state the metal atoms have formed a complex structure which has a different reactivity or properties than each individual metal did in its original form .&lt;/p&gt;&#xA;" OwnerUserId="16111" LastActivityDate="2015-06-16T07:10:46.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32959" PostTypeId="2" ParentId="32957" CreationDate="2015-06-16T07:24:09.743" Score="0" Body="&lt;p&gt;If you melt the alloy and then use electrolysis, or if you use a redox reaction, you should be able to separate the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you used a basic acid, let's say hydrochloric acid, on a zinc-copper alloy, I'd imagine the zinc would be removed from the alloy and taking the hydrogen's place, forming zinc chloride, leaving hydrogen gas and copper metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based off of simple redox reaction memories and no personal experience. However, if you're curious, try it out!&lt;/p&gt;&#xA;" OwnerUserId="16776" LastActivityDate="2015-06-16T07:24:09.743" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="32960" PostTypeId="2" ParentId="32955" CreationDate="2015-06-16T08:28:40.920" Score="7" Body="&lt;p&gt;The fastest way is dissolving the stainless steel electrochemically using $\ce{H2SO4}$ at elevated temperature where the steel block is attached as an anode. Higher current densities lead to higher removal rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively a $\ce{FeCl3}$ with minimal content of $\ce{HCl}$ (up to 3%) heated at 45°C or higher.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-06-16T08:28:40.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32961" PostTypeId="2" ParentId="15414" CreationDate="2015-06-16T08:32:23.600" Score="1" Body="&lt;p&gt;Common etchant for stainless steel is $\ce{FeCl3}$ with $\ce{HCl}$ (up to 3%) heated at 45°C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The etching solution for $\ce{Ti}$ looks fine.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-06-16T08:32:23.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32962" PostTypeId="1" AcceptedAnswerId="33079" CreationDate="2015-06-16T08:49:15.463" Score="10" ViewCount="25406" Body="&lt;p&gt;I'm now learning about nomenclature. In particular, oxyanions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, when you have an anion that is a combination of a non-metal with oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It ends with -ate for the most common oxyanions of the element. It&#xA;  ends with -ite for the oxyanions that have the same charge but with&#xA;  one less atom of oxygen.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Example:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{NO_3^- \implies}$ Nitrate&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce{NO_2^- \implies}$ Nitrite&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok, if you give me an oxyanion that ends with -ate, I know that I can simply remove one oxygen atom and I will get the -ite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what I don't understand is how does the book &lt;em&gt;know&lt;/em&gt; that $\ce{NO_3^-}$ is Nitrate: how does the book know that $\ce{NO_3^-}$ is &lt;strong&gt;&quot;the most common oxyanion for the element&quot;&lt;/strong&gt;. How does it know that a charge of $-1$ and $3$ oxygen atoms create &quot;the most common&quot; Nitrogen oxyanion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, what does it even mean by &quot;most common&quot;?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorDisplayName="user15489" LastEditDate="2015-06-16T08:53:58.877" LastActivityDate="2017-06-16T16:27:47.500" Title="When to use -ate and -ite for naming oxyanions?" Tags="&lt;nomenclature&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32963" PostTypeId="2" ParentId="32928" CreationDate="2015-06-16T09:03:37.657" Score="3" Body="&lt;p&gt;Adding to the already good answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each added amino acid in solid phase peptide synthesis actually consists of &lt;em&gt;three&lt;/em&gt; steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;coupling of the new peptide&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;capping any unreacted peptide (optional)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;deprotecting the newly-added amino group.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first step is clear. Capping is required because not all free amino groups will be coupled to the new peptide and you want to prevent hard to remove side-products in your mixture. Deprotection then liberates the amino group of the new amino acid so it can form another peptide bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This now means that when you couple 70 amino acids, you are in fact looking at &lt;em&gt;140 or 210 reactions&lt;/em&gt;. Even if the deprotection step gives you $99.9\:\%$ yield, you still lose substance every coupling step — and this is where the other answers’ calculations step in.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-16T09:03:37.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32964" PostTypeId="1" CreationDate="2015-06-16T09:09:23.900" Score="-1" ViewCount="13156" Body="&lt;p&gt;I want to know how to easily make the difference or just understand the importance of stereocenters of ephedrine ?&lt;/p&gt;&#xA;" OwnerUserId="16972" LastEditorUserId="4325" LastEditDate="2015-06-16T16:03:30.620" LastActivityDate="2015-06-17T00:02:28.043" Title="What is the difference between threo/erythro ephedrine?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-06-17T13:34:10.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="32965" PostTypeId="2" ParentId="4168" CreationDate="2015-06-16T10:21:41.450" Score="1" Body="&lt;p&gt;It is usually &lt;em&gt;assumed&lt;/em&gt; that there is no finite threshold below which carcinogens will not exert an effect. Therefore, there may be no safe exposure level below which there is no health risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, a target value that is based solely on possible health risks (considering exposure over a lifetime, possibly even demanding an adequate margin of safety) would be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, permissible limit values for carcinogens should not be set as low as possible since the exposure should not be reduced as much as possible at all costs. Instead, the potential exposure and limit values should be as low as reasonably achievable taking into account economic factors, other benefits and disadvantages, and the ability to detect and remove contaminants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, there may be similar limit values for various substances with different carcinogenic potentials. On the other hand, there may be different limit values for one substance depending on the exposure pathway (e.g. for different foodstuffs).&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-06-16T10:21:41.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32966" PostTypeId="1" CreationDate="2015-06-16T10:29:33.900" Score="5" ViewCount="1010" Body="&lt;p&gt;The resin is made up from two components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Component A: silicic acid, sodium salt (less than 10% sodium silicate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Component B: Diphenylmethane-4,4'-Diisocyanate, isomers and homologs (MDI,Polymer-MDI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been cured in place, however it became stuck to a piece of equipment made out of rubber.&lt;br&gt;&#xA;It is stuck several meters down a line so access is very limited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything that would be strong enough in small quantities to dissolve or even loosen this adhesive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on a mixture of water 90%, ammonia 8% and detergent 2%? &#xA;Would I need a large quantity of this mixture or how effective would it be?&#xA;Or a mixture of the same except using sodium carbonate instead of ammonia?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or suggestions would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="16974" LastEditorUserId="10185" LastEditDate="2015-06-16T12:52:50.833" LastActivityDate="2015-06-16T12:52:50.833" Title="How to dissolve a strong resin in difficult situation?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;polymers&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32968" PostTypeId="1" AcceptedAnswerId="32982" CreationDate="2015-06-16T12:43:15.007" Score="5" ViewCount="105" Body="&lt;p&gt;I'd like to investigate into the ring opening, the migratory reverse reaction of the electrocyclic ring-closure of Dimethyldioxirane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qAMjD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first question I'd like to ask if I have drawn the through bond interaction correctly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWvgo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following scheme is wrong, because the p orbitals are perpendicular to the O-C-O as @ron clearly pointed out in his answer in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32864/are-the-p-orbitals-of-the-biradical-dioxo-compound-in-the-homo-perpendicular-to/32867?noredirect=1#comment48932_32867&quot;&gt;Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrPyn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I tried to draw the orbital correlation diagram.&#xA;I guess for this correlation diagram I don't have to draw the through bond interaction, yes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ByGrN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the dioxo biradical product I've drawn a smaller energy gap between $\pi$ and $\pi^*$ orbitals as for $\sigma$ and $\sigma^*$. I guess this should be correct.&lt;/li&gt;&#xA;&lt;li&gt;Because there is an additional through bond interaction for the product I've drawn a small exothermicity. I see no activation energy for the disrotatory ring opening. So the disrotatory ring opening must happens very fast (without a lot of thermal energy, may be even at room temperature).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, I tried to draw the state correlation diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZL3r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the photochemical reaction there is a transition from $S_0$ to $S_2$. Because of the avoided crossings the conrotatory correlation is steeper falling down, therefore I expect most of the electrons direcetly going in the pericyclic minimum and do there Interal Conversion to the $S_1$ surface and afterwards the excited product gives off a photon going back to the $S_0$ state. I expect products in excited states are not very long living and therefore undergoes emission of a photon as Internal Conversion under fluorescence here.&lt;/li&gt;&#xA;&lt;li&gt;Another way out from the pericyclic minimum can be the directly falling to the transition state of the $S_0$ state and giving the product to a good quantum yield.&lt;/li&gt;&#xA;&lt;li&gt;As third option some part of the $S_2$ states can go through the disrotatory pathway and directly emitting light from $S_2$ to $S_0$. As written in the first point I think this way is less probable because of the less steeper falling down the energy surface.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this all correct?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="4945" LastEditDate="2020-01-28T17:44:49.313" LastActivityDate="2020-01-28T17:44:49.313" Title="Through bond interaction in biradical dioxo compound and its correlation diagram" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;photochemistry&gt;&lt;hyperconjugation&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32969" PostTypeId="2" ParentId="32964" CreationDate="2015-06-16T13:28:05.797" Score="3" Body="&lt;p&gt;&lt;strong&gt;1.Ephedrine exhibits optical isomerism and has two chiral centres, giving rise to four stereo isomers. By convention, the pair of enantiomers with the stereochemistry (1R,2S and 1S,2R) is designated ephedrine, while the pair of enantiomers with the stereochemistry (1R,2R and 1S,2S) is called pseudoephedrine.&lt;/strong&gt;&#xA;Source :&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Ephedrine&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Ephedrine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Again in order to distinguish threo from erythro.. Just memorize the word :TOES =Threo - Opposite Erythro - Same&#xA;e. g., &lt;img src=&quot;https://i.stack.imgur.com/IqtLT.jpg&quot; alt=&quot;Threo - Erythro &quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16976" LastEditorUserId="16976" LastEditDate="2015-06-17T00:02:28.043" LastActivityDate="2015-06-17T00:02:28.043" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32970" PostTypeId="1" AcceptedAnswerId="32974" CreationDate="2015-06-16T13:41:46.127" Score="10" ViewCount="5392" Body="&lt;p&gt;Is there any such thing as a twist chair conformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the boat and the chair conformations are achiral, while the twist boat conformation of cyclohexane is chiral and dissymetric.&#xA;What about the half chair conformation? Am I right in thinking that it is achiral because it has a plane of symmetry?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorDisplayName="user7951" LastEditDate="2015-08-19T18:27:46.333" LastActivityDate="2015-08-19T18:27:46.333" Title="Conformations of Cyclohexane" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;conformers&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32971" PostTypeId="2" ParentId="32927" CreationDate="2015-06-16T14:34:34.107" Score="3" Body="&lt;p&gt;The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; is the more reducing metal. We can consider what happens in each case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If we immerse &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; in a solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{A^m+}$&lt;/span&gt; nothing will happen.&lt;/li&gt;&#xA;&lt;li&gt;Similarly, if we immerse &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; in a solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{B^n+}$&lt;/span&gt; nothing will happen.&lt;/li&gt;&#xA;&lt;li&gt;If we immerse &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; in a solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{A^m+}$&lt;/span&gt; nothing will happen because the electrode potential for &lt;span class=&quot;math-container&quot;&gt;$\ce{A|A^m+}$&lt;/span&gt; is more negative (more negative means its a stronger reducing agent) than for &lt;span class=&quot;math-container&quot;&gt;$\ce{B|B^n+}$&lt;/span&gt; so the reaction potential is negative and the reaction is not feasible.&lt;/li&gt;&#xA;&lt;li&gt;However, if we immerse &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; in a solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{B^n+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; will precipitate out on the surface of &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt;. This is because &lt;span class=&quot;math-container&quot;&gt;$\ce{A|A^m+}$&lt;/span&gt; has a more negative electrode potential than &lt;span class=&quot;math-container&quot;&gt;$\ce{B|B^n+}$&lt;/span&gt; and so the reaction potential will be positive. &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; will reduce &lt;span class=&quot;math-container&quot;&gt;$\ce{B^n+}$&lt;/span&gt; ions to form &lt;span class=&quot;math-container&quot;&gt;$\ce{B}$&lt;/span&gt; metal and &lt;span class=&quot;math-container&quot;&gt;$\ce{A}$&lt;/span&gt; will be oxidised to form &lt;span class=&quot;math-container&quot;&gt;$\ce{A^m+}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A (s) + $\frac{m}{n}$B^n+ (aq) -&amp;gt; A^m+ (aq) + $\frac{m}{n}$B (s)}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, the most reducing metal is the one which causes precipitation of the other metal out of its solution.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="4945" LastEditDate="2020-02-26T13:22:24.523" LastActivityDate="2020-02-26T13:22:24.523" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32973" PostTypeId="2" ParentId="30782" CreationDate="2015-06-16T15:08:22.580" Score="9" Body="&lt;p&gt;&lt;strong&gt;You raise an excellent question and the short answer is &quot;yes, quantum chemistry calculations are intrinsically in vacuum&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate methods like semiempirical (AM1, PM6, PM7) and hybrid DFT methods (e.g., B3LYP) use experimental data, which often comes from crystal structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many years, single-crystal x-ray diffraction studies were the gold standard of &quot;what is the geometry of a molecule.&quot; Performing gas-phase measurements of bond angles and bond lengths are extremely difficult, since they typically require a molecular beam of the intended target and then very high resolution IR,  rotational spectroscopy, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_electron_diffraction&quot;&gt;gas electron diffraction&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I once heard a talk by an extremely talented theorist, who asked the question &quot;is an experimental crystal structure accurate enough?&quot; He showed that in many cases, the answer is &quot;no.&quot; (Consider that establishing the position of hydrogen atoms is difficult if not impossible with many single crystals.) The method he was proposing (geometry optimization with CCSD(T) and a basis set limit) isn't really appropriate for large molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until there's some huge database of gas-phase molecular geometry measurements and CCSD(T) calculations, most methods are compared with crystal structures.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-06-16T15:50:58.137" LastActivityDate="2015-06-16T15:50:58.137" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="32974" PostTypeId="2" ParentId="32970" CreationDate="2015-06-16T15:29:37.240" Score="12" Body="&lt;p&gt;&lt;strong&gt;Short answer, no. There is not a &quot;twist-chair.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convince yourself, it probably helps to make a physical model with a chemistry model kit. If you try to twist the chair, you can't do it without significantly moving at least one atom, in which case, the conformation is basically a &quot;skew&quot; or &quot;twist boat.&quot; Try it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boat and chair conformations are indeed symmetric and achiral. The twist-boat or skew conformation has point group $D_{2}$ and is indeed chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A normal half-chair will have an axis of symmetry and is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An envelope conformation will have a plane of symmetry and thus be achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found that some of the best depictions of six-member-ring conformations come from pyranose sugars. Because of the hydroxyl groups and oxygen atoms, there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyranose#Conformations&quot; rel=&quot;noreferrer&quot;&gt;38 distinct conformations&lt;/a&gt; (2 chairs, 6 boats, 6 skew-boats, 12 half-chairs, and 12 envelopes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyranose#Conformations&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WRLqW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, in less-substituted cyclohexane, there are fewer unique shapes, but it helps to see the different &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00839a011&quot; rel=&quot;noreferrer&quot;&gt;ring-puckering&lt;/a&gt; or pseudo-rotation shapes.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-06-16T19:26:55.263" LastActivityDate="2015-06-16T19:26:55.263" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="32976" PostTypeId="1" CreationDate="2015-06-16T17:24:34.390" Score="9" ViewCount="802" Body="&lt;p&gt;I want to make ferrofluid, which is ferromagnetic particles suspended in a carrier fluid. Three methods I have seen involve using magnetic ink (for MICR codes), cassette tapes, or ferric chloride as the iron source. The last method is the only one that makes ferrofluid with nano-scale ferromagnetic particles. AFAIK, the process boils down to using ammonia to precipitate magnetite from a ferric/ferrous chloride mixture, coating the precipitated particles with oleic acid (acting as a surfactant), and suspending the coated particles in a carrier oil such as kerosene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps (based on &lt;a href=&quot;http://chemistry.about.com/od/demonstrationsexperiments/ss/liquidmagnet.htm&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Oxidize iron with ferric chloride to form ferrous chloride: $\ce{2FeCl3 + Fe -&amp;gt; 3FeCl2}$&lt;/li&gt;&#xA;&lt;li&gt;Precipitate magnetite: $\ce{2FeCl3 + FeCl2 + 8NH4OH -&amp;gt; Fe3O4 + 8NH4Cl + 4H2O}$&lt;/li&gt;&#xA;&lt;li&gt;Add oleic acid, evaporate the ammonia, and end up with surfactant coated magnetite nanoparticles?&lt;/li&gt;&#xA;&lt;li&gt;Add kerosene to magnetite suspension =&gt; ferrofluid&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is another example &lt;a href=&quot;https://sites.google.com/site/koserlab/outreach/make-your-own-ferrofluid&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; that has slightly different steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dissolve $\ce{Fe^2+}$ and $\ce{Fe^3+}$ in $\ce{HCl}$ and mix&lt;/li&gt;&#xA;&lt;li&gt;Precipitate magnetite&lt;/li&gt;&#xA;&lt;li&gt;Repeatedly:&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Separate precipitate from liquid with a magnet&lt;/li&gt;&#xA;&lt;li&gt;Decant the liquid, leaving the precipitate&lt;/li&gt;&#xA;&lt;li&gt;Add distilled water&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Add $\ce{N(CH3)4OH}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the first process generates a 'smoother' ferrofluid. I mean to say that it looks more like a homogenous fluid, whereas the ferrofluid in the videos from the second link looks more like a sludge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions. First, &lt;strong&gt;what is actually happening in the first process when the oleic acid is added?&lt;/strong&gt; From the first article, it sounds like the oleic acid reacts to form ammonium oleate, and then adding heat decomposes the ammonium oleate, releasing ammonia gas and dissolving the oleate ions, which then coat the magnetite particles. But that sounds a bit odd. Second, &lt;strong&gt;would the following work (create a suspension of magnetite nanoparticles suspended in a carrier):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Obtain a mixture of $\ce{FeCl2}$ and $\ce{FeCl3}$ via either method.&lt;/li&gt;&#xA;&lt;li&gt;Precipitate magnetite&lt;/li&gt;&#xA;&lt;li&gt;Separate the magnetite precipitate from the ammonium chloride via decanting (as in the second process)&lt;/li&gt;&#xA;&lt;li&gt;Add oleic acid (and heat?) to coat the magnetite particles&lt;/li&gt;&#xA;&lt;li&gt;Add kerosene to magnetite suspension =&gt; ferrofluid&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems like removing the ammonium chloride via washing would be easier than via evaporation. Less heat, less time, less nasty fumes. But I don't really know what's going on when the oleic acid is added, so I'm not sure if my proposed method would actually work.&lt;/p&gt;&#xA;" OwnerUserId="3711" LastEditorUserId="10185" LastEditDate="2015-06-16T17:39:08.597" LastActivityDate="2015-06-16T17:39:08.597" Title="Can I make Ferrofluid without evaporating Ammonium?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32977" PostTypeId="2" ParentId="27649" CreationDate="2015-06-16T17:54:28.153" Score="9" Body="&lt;p&gt;This questions calls for an answer from thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PmfGW.gif&quot; alt=&quot;Phase diagram of carbon.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure provided above, lifted from &lt;a href=&quot;http://www.ch.ic.ac.uk/rzepa/mim/century/html/diamond_text.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, is what we call a phase diagram. On the abscissa is temperature in units of Kelvin and on the ordinate is the pressure given in units of Gigapascals. (For your own reference, 1 GPa is nearly 10,000 times the pressure we live under, the earth's core is estimated to be over 300 GPa!) This diagram represents the input of experimental work and extrapolation with mathematical tools provided by thermodynamics. What this figure shows very clearly is that for a given temperature and pressure which phase is thermodynamically preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line between the diamond and graphite phases represents a phase boundary, and tell us precisely what temperature and pressure are required for these two phases to be in equilibrium. The phase boundary between liquid carbon and graphite represents the melting temperature for graphite; the same can be said for the diamond/carbon line, this again is the melting temperature of diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our melting temperature therefore depends on phase and pressure. So if you only have two melting points, does that really tell you which is thermodynamically preferred at room pressure (0.0001 GPa)? No, it does not. It is more subtle than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how would we determine the relative stability of graphite and diamond at room temperature and pressure? For that we compute the standard Gibbs energy of reaction (at &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot; rel=&quot;noreferrer&quot;&gt;STP&lt;/a&gt;), this requires we use some values from a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_standard_Gibbs_free_energies_of_formation&quot; rel=&quot;noreferrer&quot;&gt;table&lt;/a&gt;: $\Delta G_f^0 (diamond) = 2.90 (kJ/mol)$ and $\Delta G_f^0 (graphite) = 0 (kJ/mol)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction we want to calculate is: $C(diamond) \rightarrow C(graphite)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know that $\Delta G_{rxn} = \Sigma_{products} \Delta G_{f} - \Sigma_{reactants} \Delta G_{f} = 0 - 2.90 = -2.90 (kJ/mol)$. So $\Delta G_{rxn} &amp;lt; 0$ tells us at 298 K and 1 atm of pressure, diamond would spontaneously form graphite if it could; however, for this transition carbon atoms would have to change their location within a lattice and the rate at which this happens in a solid tends to be very slow at low temperatures. Hence, we call diamond a metastable phase under STP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you provide a sufficient energy to get over this activation barrier, then you will be able to see what thermodynamics prefers. Personally, I find &lt;a href=&quot;https://www.youtube.com/watch?v=7L7BV3IBfFA&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; video very compelling.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2015-06-16T19:09:07.397" LastActivityDate="2015-06-16T19:09:07.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32978" PostTypeId="1" AcceptedAnswerId="32979" CreationDate="2015-06-16T17:56:05.417" Score="7" ViewCount="590" Body="&lt;p&gt;I was thinking about orbital angular momentum today and  wondering how exactly do the numbers come about that quantize them. I was doing a little research on how they come about and I  came across this image many times. &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b3/Spin_half_angular_momentum.svg/2000px-Spin_half_angular_momentum.svg.png&quot; alt=&quot;1&quot;&gt;&#xA;How exactly does a chemist/physicist measure this to get the +1/2,-1/2 number. Are there other values that are possible?&lt;/p&gt;&#xA;" OwnerUserId="9987" LastActivityDate="2015-06-16T19:17:06.817" Title="Where exactly do the orbital angular momentum get their numbers e.g. +1/2" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32979" PostTypeId="2" ParentId="32978" CreationDate="2015-06-16T18:38:37.857" Score="8" Body="&lt;p&gt;$$&#xA;\newcommand{\linop}[1]{\hat{#1}}&#xA;\newcommand{\ramuno}{\mathrm{i}}&#xA;\newcommand{\ket}[1]{\,|{#1}\rangle}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Classical Mechanics&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In classical mechanics, the &lt;em&gt;angular momentum&lt;/em&gt;, $\vec{L}$, of a particle with respect to a chosen origin is given by the vector product of its position vector relative to the origin, $\vec{r}$, and its linear momentum, $\vec{p}$,&#xA;\begin{equation*}&#xA;    \vec{L} := \vec{r} \times \vec{p} \, .&#xA;\end{equation*}&#xA;By expanding $\vec{r}$ and $\vec{p}$ over a standard basis $\{ \vec{i}, \vec{j}, \vec{k} \}$ of a Cartesian coordinate system as $\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$ and $\vec{p} = p_x \vec{i} + p_y \vec{j} + p_z \vec{k}$ and then writing the cross product as the formal determinant &lt;sup&gt;[1]&lt;/sup&gt; and expanding it one could obtain &#xA;\begin{equation}&#xA;    \vec{r} \times \vec{p}&#xA;    =&#xA;    \begin{vmatrix}&#xA;    	\vec{i} &amp;amp; \vec{j} &amp;amp; \vec{k} \\&#xA;    	      x &amp;amp;       y &amp;amp;       z \\&#xA;    	    p_x &amp;amp;     p_y &amp;amp;     p_z \\&#xA;    \end{vmatrix}&#xA;    =&#xA;    (y p_z - z p_y) \vec{i} - (x p_z - z p_x) \vec{j} + (x p_y - y p_x) \vec{k} \, ,&#xA;\end{equation}&#xA;which implies that components $(L_x, L_y, L_z)$ of the angular momentum in terms of which it can be expressed over a standard basis as $\vec{L} = L_x \vec{i} + L_y \vec{j} + L_z \vec{k}$ are given as follows&#xA;\begin{equation}&#xA;	L_x = (y p_z - z p_y) \, ,&#xA;	\quad&#xA;	L_y = (z p_x - x p_z) \, ,&#xA;	\quad&#xA;	L_z = (x p_y - y p_x) \, .&#xA;\end{equation}&#xA;The magnitude, $L$, of the angular momentum is given as usual as the square root of the scalar product of the vector with itself, $L := \sqrt{\vec{L}\vec{L}}$, or, in terms of components,&#xA;\begin{equation*}&#xA;    L := \sqrt{L_x^2 + L_y^2 + L_z^2} \, .&#xA;\end{equation*}&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Quantum Mechanics&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Operators&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The quantum mechanical analogue of classical angular momentum is known as the &lt;em&gt;orbital angular momentum&lt;/em&gt; and the operators for the components of orbital angular momentum can be obtained by simply replacing the components of position and momentum in classical mechanical expressions by the corresponding operators&#xA;\begin{equation}&#xA;    \linop{L}_{x} = (\linop{y} \linop{p}_z - \linop{z} \linop{p}_y) \, ,&#xA;    \quad&#xA;    \linop{L}_y = (\linop{z} \linop{p}_x - \linop{x} \linop{p}_z) \, ,&#xA;    \quad&#xA;    \linop{L}_{z} = (\linop{x} \linop{p}_y - \linop{y} \linop{p}_x) \, .&#xA;\end{equation}&#xA;and the square of the magnitude of the angular momentum operator can be expressed in terms of these operators as follows&#xA;\begin{equation}&#xA;    \linop{L}^2 = \linop{L}_{x}^2 + \linop{L}_y^2 + \linop{L}_{z}^2 \, .&#xA;\end{equation}&#xA;Then the following commutation relations of the angular momentum operators can be easily established:&#xA;\begin{equation}&#xA;    [ \linop{L}_{x}, \linop{L}_{y} ] = \ramuno \hbar \linop{L}_{z} \, ,&#xA;    \quad&#xA;    [ \linop{L}_{y}, \linop{L}_{z} ] = \ramuno \hbar \linop{L}_{x} \, ,&#xA;    \quad&#xA;    [ \linop{L}_{z}, \linop{L}_{x} ] = \ramuno \hbar \linop{L}_{y} \, ,&#xA;\end{equation}&#xA;\begin{equation}&#xA;    [\linop{L}^2, \linop{L}_{x}] = [\linop{L}^2, \linop{L}_y] = [\linop{L}^2, \linop{L}_{z}] = 0 \, .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, in quantum mechanics there exist a different kind of angular momentum, the &lt;em&gt;intrinsic angular momentum&lt;/em&gt;, or &lt;em&gt;spin&lt;/em&gt;, which does not have a counterpart in classical mechanics. &#xA;While the intrinsic angular momentum components operators does have a different form from that of orbital angular momentum, the commutation relations between them and the square of the intrinsic angular momentum magnitude are exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interestingly, the angular momentum can be treated in such a way that the treatment relies entirely on the commutation relations just introduced, and as such, valid for any kind of angular momentum.&#xA;Thus, for generality we can use the letter $J$ for any kind of angular momentum reserving the letter $L$ specifically for orbital one.&#xA;\begin{equation}&#xA;    [ \linop{J}_{x}, \linop{J}_{y} ] = \ramuno \hbar \linop{J}_{z} \, ,&#xA;    \quad&#xA;    [ \linop{J}_{y}, \linop{J}_{z} ] = \ramuno \hbar \linop{J}_{x} \, ,&#xA;    \quad&#xA;    [ \linop{J}_{z}, \linop{J}_{x} ] = \ramuno \hbar \linop{J}_{y} \, ,&#xA;\end{equation}&#xA;\begin{equation}&#xA;    [\linop{J}^2, \linop{J}_{x}] = [\linop{J}^2, \linop{J}_y] = [\linop{J}^2, \linop{J}_{z}] = 0 \, .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Eigenvectors and eigenvalues&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;According to the commutation relations above, $\linop{J}^2$ commutes with all three components $\linop{J}_{x}, \linop{J}_y, \linop{J}_{z}$ while the components do not commute with each other.&#xA;Thus, $\linop{J}^2$ and one (and only one) of the components operators, conventionally chosen to be $\linop{J}_{z}$, admit the set of simultaneous eigenvectors, and using the so-called &lt;em&gt;ladder operators&lt;/em&gt; technique it can be shown that,&#xA;\begin{align}&#xA;    \linop{J}^2 \ket{j(j+1),m} &amp;amp;= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\&#xA;    \linop{J}_{z} \ket{j(j+1),m} &amp;amp;= m \hbar \ket{j(j+1),m} \, , &#xA;\end{align}&#xA;where $\ket{j(j+1),m}$ is a simultaneous eigenvector of $\linop{J}^2$ and $\linop{J}_{z}$, $j$ could take non-negative integer and half-integer values,&#xA;\begin{equation*}&#xA;    j = 0, 1/2, 1, 3/2, \cdots \, ,&#xA;\end{equation*}&#xA;and $m$ ranges from $-j$ to $j$ in integer steps&#xA;\begin{equation}&#xA;    m = -j, -j+1, \cdots, j-1, j \cdots \, .&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as I already said in my comment with respect to half-integer values of $j$, for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible, but where cyclic boundary conditions are not relevant, as for the intrinsic angular momentum known as spin, the half-integer values might be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Epilogue&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'm still not sure what this question is about, but if it specifically about the spin of an electron, i.e. &quot;How do we now that its component in any direction is $\pm 1/2$ in the units of $\hbar$?&quot;, then we know it both from the experiment (&lt;a href=&quot;https://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment&quot;&gt;the Stern–Gerlach one&lt;/a&gt;) and the theory (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_field_theory&quot;&gt;quantum field theory&lt;/a&gt;). And yes, for electron only these values are possible, but for other particles there might be different possibilities. Thus, for instance, a photon have component of the spin in any direction  $\pm 1$ in the units of $\hbar$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1] Here, &quot;formal&quot; means that this notation has the form of a determinant, but does not strictly adhere to the definition; it is a mnemonic used to remember the expansion of the cross product.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-16T19:17:06.817" LastActivityDate="2015-06-16T19:17:06.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32980" PostTypeId="1" CreationDate="2015-06-16T19:05:41.420" Score="1" ViewCount="5765" Body="&lt;p&gt;I’m trying to learn about ions. There is a slide with examples about combining cations with anions. It goes like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\text{For} \ \ \ce{Al^3+} \ \ \text{ and } \ \ \ce{O^2-} \ \ \text{you get} \ \  \ce{Al2O3}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wait what? So you have an atom of aluminum (in this case a cation because it loses three electrons) and an atom of oxygen (an anion since it receives two electrons). You combine them. Why do you now have TWO atoms of aluminum and THREE atoms of oxygen? Why does combining a cation and an anion suddenly spawn new atoms? And what happened to their charges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next example is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\text{For} \ \ \ce{Ca^2+} \ \ \text{ and } \ \ \ce{Br-} \ \ \text{you get} \ \  \ce{CaBr2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alright. I see a pattern. For &lt;em&gt;some reason&lt;/em&gt;, if the charge of an atom is positive or negative, such number will define the number of atoms for the other element. So since calcium has $2{+}$, there would be two bromine. And since bromine has $-1$ there will be one calcium. I don’t get &lt;strong&gt;&lt;em&gt;why&lt;/em&gt;&lt;/strong&gt;, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a third example,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\text{For} \ \ \ce{Na+} \ \ \text{ and } \ \ \ce{CO3^2-} \ \ \text{you get} \ \  \ce{Na2CO3}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok so this is a bit weird since there are now three elements involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But according to the pattern, since sodium has charge $+1$, there should be just one $\ce{CO3}$ molecule. And since $\ce{CO3^2-}$ has a charge of $-2$, there should be two of sodium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great, I suppose it’s an useful pattern, but my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why? Why does losing/gaining an electron cause the other element to gain/lose atoms?&lt;/li&gt;&#xA;&lt;li&gt;Does this pattern apply only when you combine a cation with an anion? What if it is the other way around? (An anion with a cation, or does it make no sense at all?)&lt;/li&gt;&#xA;&lt;li&gt;After combining a cation with an anion, do both atoms lose their charge? Observe how the results no longer have $+$ or $-$ superscripts.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16632" LastEditorDisplayName="user7951" LastEditDate="2015-06-18T09:38:17.700" LastActivityDate="2015-06-18T15:14:09.080" Title="About combining a cation and an anion" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="32982" PostTypeId="2" ParentId="32968" CreationDate="2015-06-16T20:04:47.460" Score="2" Body="&lt;p&gt;I would like to help here, but I am not sure where to begin. First of all, I want to point out that the reaction you have shown is not an electrocyclic reaction, because it does not involve the central saturated carbon atom. It is rather a simple homolysis reaction. The terms &quot;conrotatory&quot; and &quot;disrotatory&quot; are misapplied, since the two oxygen atoms do not have functional groups on them, and the two modes of ring opening are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dioxirane does have an electrocyclic ring opening reaction available to it - namely cleavage of a carbon-oxygen bond to form &quot;acetone-O-oxide&quot;. Perhaps you should consider that as an alternative pathway. Does that help?&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-16T20:04:47.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32984" PostTypeId="1" AcceptedAnswerId="33026" CreationDate="2015-06-16T20:58:48.453" Score="7" ViewCount="693" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies $\ldots$ - &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32980/about-combining-a-cation-and-an-anion?noredirect=1#comment49024_32980&quot;&gt;My comment&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This makes me wonder. I remember seeing $\Delta H_{\mathrm{lattice}}$ of &quot;common&quot; ionic compounds like $\mathrm{NaCl}$ and considered them fairly positive&lt;sup&gt;a&lt;/sup&gt;:&#xA;\begin{array}{|c|c|} \hline&#xA;\mathbf{Ionic~compound} &amp;amp; \mathbf{\Delta H^\circ (kJ~mol^{-1})} \\\hline&#xA;\ce{LiF} &amp;amp; 1022^\mathrm{\,b} \\\hline&#xA;\ce{NaCl} &amp;amp; 771 \\\hline&#xA;\ce{KBr} &amp;amp; 670 \\\hline&#xA;\ce{CsI} &amp;amp; 585 \\\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yeah$\ldots$ so judging by the usual trends $\ce{CsI}$ should have the least positive lattice enthalpy in ionic compounds derived only from monoatomic ions. But that's still &lt;em&gt;some&lt;/em&gt; lattice enthalpy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, could there be negative lattice enthalpies?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notes:&lt;/em&gt;&lt;br&gt;&#xA;&lt;sup&gt;a&lt;/sup&gt;: Note that there are two major definitions of lattice enthalpy. Here, we're using the definition relevant to $\Delta H_r$ of the following endothermic reaction: $$\ce{MX(s) → M+(g) + X^- (g)}$$&#xA;While the other definition uses the enthalpy of the reverse process as the reference, only the sign might differ. Note that I had the other defintion in my mind when posting that comment.&lt;br/&gt;&#xA;&lt;sup&gt;b&lt;/sup&gt;: All data obtained at $298~\mathrm{K}$. &lt;a href=&quot;http://www.sphsgator.net/sphsteachers/domanskil1/databooknew.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-18T08:51:35.810" Title="Can negative (abnormally) lattice enthalpies during the formation of ionic compounds exist?" Tags="&lt;inorganic-chemistry&gt;&lt;enthalpy&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32985" PostTypeId="1" CreationDate="2015-06-16T21:37:47.820" Score="1" ViewCount="1951" Body="&lt;p&gt;Having a real pain with nomenclature. Look at this substance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{MnO4-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, do we have an anion of manganese and oxygen, or is that just an anion of oxygen? Anyway, my book says that $\ce{MnO4-}$ is &quot;permanganate&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it ends with &quot;-ate&quot;, it seems my book thinks $\ce{MnO4-}$ is an oxyanion (because those are the ones that end with things like &quot;-ate&quot; and &quot;-ite&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it was my understanding that an oxyanion is when you have oxygen and a non-metal. But manganese is a metal! What is going on here then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not just that. Let's assume that $\ce{MnO4-}$ is indeed an oxyanion called permanganate. From what I've learned, if you remove one oxygen atom (getting  $\ce{MnO3-}$), it should become just manganate, right? But, according to my book, manganate is $\ce{MnO2-}$. It had to lose &lt;strong&gt;two&lt;/strong&gt; oxygen atoms and not one to become manganate. Why?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2017-09-20T18:39:04.957" LastActivityDate="2017-09-20T18:39:04.957" Title="Nomenclature of MnO4^-: is it an oxyanion?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32986" PostTypeId="2" ParentId="32985" CreationDate="2015-06-16T22:33:40.000" Score="1" Body="&lt;p&gt;An oxyanion is not necessarily a non-metal and oxygen, rather, it is more simply a &lt;strong&gt;chemical element&lt;/strong&gt; and an oxygen. There are a large number of elements that form oxyanions, many of which are listed &lt;a href=&quot;http://chemistry.about.com/od/chartstables/a/Common-Anions.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote &lt;a href=&quot;https://chemistry.stackexchange.com/questions/10173/about-the-nomenclature-manganate-or-manganite&quot;&gt;this&lt;/a&gt; Chem.SE post regarding a the difference between manganate and manganite:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to $\ce{Cl(VII)}$ and downward is not valid. Drawing nomenclature parallels to titanium is poor, for $\ce{Ti(IV)}$ is the highest common oxidation state. $\ce{SrO +TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is $\ce{Ti(IV)}$ on both sides.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For someone who is new to chemistry, the only metal oxyanions I would recommend that you learn are chromate ($\ce{CrO4^2-}$), dichromate ($\ce{Cr2O7^2-}$), and permanganate ($\ce{MnO4^-}$).&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-06-16T22:33:40.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="32987" PostTypeId="2" ParentId="32985" CreationDate="2015-06-16T22:34:47.817" Score="3" Body="&lt;p&gt;As far as I know, an oxoanion just refers to an anion which has oxygen covalently bonded to the central atom; the central atom can be pretty much anything apart from the Group 1 and 2 metals, since the only oxygen compounds those form are ionic oxides/peroxides/superoxides. For the 3d transition metals, you have $\ce{VO3^-}$ (vanadate(V)), $\ce{CrO4^{2-}}$ (chromate(VI)), $\ce{MnO4-}$ (manganate(VII)), and $\ce{FeO4^{2-}}$ (ferrate(VI)), amongst others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not recommend reading too much into the nomenclature, because a lot of it is just what chemists are comfortable with after using for a long time - it's like a language that you just get more familiar with over time. But there are some general rules. The 'per...ate' name, such as perchlorate $\ce{ClO4-}$, implies a higher oxidation state than the normal '...ate' compound, which is chlorate $\ce{ClO3-}$ in this case. For chlorine, only certain oxidation states are available, namely the -1, 0, +1, +3, +5 and +7 oxidation states. Apart from 0, which occurs in diatomic chlorine $\ce{Cl2}$, you will notice that these are in increments of 2 - which means that you can generate the whole list of chlorine anions by adding one more oxygen, but keeping the charge of the anion constant: $\ce{Cl-}$, $\ce{ClO-}$, $\ce{ClO2-}$, $\ce{ClO3-}$, $\ce{ClO4-}$. This tendency to form compounds 2 oxidation states apart is very pronounced for the p-block elements, i.e. Group 15, 16 and 17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for d-block elements such as $\ce{Mn}$, this isn't the case because single d electrons are easily lost and gained. So after permanganate, $\ce{MnO4-}$, which features $\ce{Mn}$ in the +7 oxidation state, the oxoanion with the next highest oxidation state is the manganate ion $\ce{MnO4^2-}$ which has $\ce{Mn}$ in the +6 oxidation state. (As far as I know the ion $\ce{MnO2-}$ is not called &quot;manganate&quot;.)&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2015-06-16T23:38:54.240" LastActivityDate="2015-06-16T23:38:54.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32988" PostTypeId="2" ParentId="19273" CreationDate="2015-06-16T22:55:12.197" Score="1" Body="&lt;p&gt;The synthesis you are referring to is outlined &lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=CV2P0264&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Potassium Carbonate acts as a drying agent as you said as the later part of the distillation requires a specific temperature to remove the toluene and alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water comes from the concentrated sulphuric acid solution added with apidic acid. The residue is also rinsed with alcohol which is a strong indicator that water needs to be removed. &lt;/p&gt;&#xA;" OwnerUserId="2558" LastActivityDate="2015-06-16T22:55:12.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32989" PostTypeId="2" ParentId="28761" CreationDate="2015-06-16T23:20:28.620" Score="1" Body="&lt;p&gt;Strong nucleophiles tend to undergo S&lt;sub&gt;N&lt;/sub&gt;1/E1 reactions whereas weak nucleophiles &lt;em&gt;tend&lt;/em&gt; to undergo S&lt;sub&gt;N&lt;/sub&gt;1/E1 reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase in electron density will make a nucleophile stronger. This effect can be clearly seen when conjugate bases of common nucleophiles are stronger than the nuclephiles themselves. For example: $\ce{NH_2^- &amp;gt; NH_3} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In S&lt;sub&gt;N&lt;/sub&gt;1/E1 reactions, the weak nucleophile can only attack the carbocation once the leaving group has left. This is precisely because it is a weak nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IN S&lt;sub&gt;N&lt;/sub&gt;2/E2 reactions, the strong nucleophile attacks the carbon before it can even form a carbocation. This forces the leaving group to leave. This is precisely because it is a strong nucleophile.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastActivityDate="2015-06-16T23:20:28.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32990" PostTypeId="2" ParentId="32834" CreationDate="2015-06-16T23:31:25.627" Score="2" Body="&lt;p&gt;Lets state my assumption that we can represent the free energies of the mixtures in fairly regular thermodynamics of phases ways (as I will get to here quickly). I will also assume fairly limited solid solubilities (think something like the Au-Si system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, just take a system of A and B. The Gibbs free energy of a binary mixture (either solid or liquid) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B})$, with $x_{A} + x_{B} = 1$, $G_{A,B}^{sol/liq}$ being the Gibbs free energy of pure A (or B) in the solid (or liquid) at that temperature, and $H_{AB}^{sol/liq}$ being the enthalpy of mixing of A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a eutectic, the free energy of the liquid drops below the tie-line connecting the free energies of the solid phases, and it clearly does this in such a way that the eutectic point has a composition that is intermediate to the compositions defining the solid tie-line. In a perfectly symmetric and regular system, this puts the eutectic at $x_{A} = x_{B} = 0.5$, but at least keep in mind that liquid composition has reasonably large values of $x_{A}$ or $x_{B}$, certainly larger than the minority composition in either solid end-member. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we add a third element. The Gibbs free energy formula becomes longer, since it now takes in to account the binary and ternary interactions.  So, one gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{C}G_{C}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + x_{A}x_{C}H_{AC}^{sol/liq} +  x_{B}x_{C}H_{BC}^{sol/liq} + x_{A}x_{B}x_{C}H_{ABC}^{sol/liq} +  RT(x_{A}\ln x_A + x_{B} \ln x_{B} + x_{C} \ln x_{C})$, with $x_{A}+ x_{B} + x_{C} = 1$. This is pretty ugly, but we just added a little bit of C, so $x_{C}$ is small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to the assumptions on the system being considered. For the liquid, $x_{A}$ and $x_{B}$ are reasonably large ($\sim 0.5$). In solid A, $x_{A}$ is near 1, $x_{B}$ is small(ish), and $x_{C}$ is really small.  In solid B, $x_{B}$ is near 1, $x_A$ is small(ish), and $x_{C}$ is small. Really small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has really changed? Well, for $x_{C}$ being really small, you can throw out all terms linear in C. The only thing left (!) is the entropy of mixing terms, where adding a little C results in a change of  $RT x_{C} \ln x_{C}$. But, this gets applied to both solids (A and B), and also to the liquid. All the free energy curves shift down by the same amount on an absolute scale. By geometry this means that the tie line connecting the two solid phases and the liquid at the eutectic point does not shift to first order in $x_{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eutectic point will not shift.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-06-16T23:31:25.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32991" PostTypeId="2" ParentId="31491" CreationDate="2015-06-16T23:35:49.797" Score="1" Body="&lt;p&gt;Looking at the mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/giT3C.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that these are not double headed arrows, but single headed arrows. This indicates the movement of single electrons, rather than electron pairs as is common in most mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanisms that involve movements of single electrons tend to form free-radicals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lewis structure of Mg has the electrons seperated in order to make it clearer to read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In step 1, the shared electron pair in $\ce{R-X}$ has been split to form two radicals: $\ce{R}{\small{•}}$ and $\ce{X}{\small{•}}$. However the $\ce{X}{\small{•}}$ radical is further attacked by an electron, this time coming from the $\ce{Mg}$. Thus forming a free-radical magnesium halide: $\ce{X^-Mg^+}{\small{•}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, the two free-radicals undergo a termination step to form the Grignard reagent.&lt;/p&gt;&#xA;" OwnerUserId="2558" LastActivityDate="2015-06-16T23:35:49.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32992" PostTypeId="1" CreationDate="2015-06-17T00:06:52.927" Score="0" ViewCount="105" Body="&lt;p&gt;I know that the glucose is oxidized into pyruvate but I don't understand what is reduced or where the electrons go.... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation:&#xA;$\ce{Glucose + 2 ADP + 2 NAD^+ -&amp;gt; 2 Pyruvate + 2 ATP + 2 NADH}$&lt;/p&gt;&#xA;" OwnerUserId="15793" LastEditorUserId="4231" LastEditDate="2015-06-17T00:31:48.680" LastActivityDate="2017-07-25T01:52:12.973" Title="According to this equation for glycolysis, what is reduced and where do the electrons go?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32993" PostTypeId="2" ParentId="31577" CreationDate="2015-06-17T01:19:46.330" Score="5" Body="&lt;p&gt;A primary standard is critical for an analytical result. If we dissolve a known mass of a material into a known volume, we must be able to know its concentration with confidence. Such a standard must be easily made very pure (&gt; 99.95%), it should not decompose, and it must be fairly robust to say heating or vacuum. We use such standard solutions as, for example, titrants. This is surprisingly hard to come by, and so we say that secondary standards are materials that do not meet these strict criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you want to titrate a sulfuric acid solution for quantitative work, you could use a known concentration of NaOH or KOH and volumetrically determine the concentration of the sulfuric acid solution. In fact, dissolving pellets of these particular bases in a known volume of water is not sufficient to calculate concentration with any confidence in analytical work because the solids are contaminated with carbonate from atmospheric CO$_2$. In general these carbonates contaminate alkaline solutions quite readily through this reaction: $\ce{OH- + CO2 -&amp;gt; HCO3-}.$ This is why we would have to titrate the concentration of $\ce{NaOH} $ solution with a primary standard before using it as a secondary standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are acid and base primary standards available, so why would one ever need to use a secondary standard? Usually it’s a matter of specific needs and convenience. Whenever possible, use a primary standard if your results actually matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us look at a specific case where a secondary standard is useful: potassium permanganate is exceptionally useful as a titrant because it serves as its own indicator where the presence of a purple color signifies the end point, but it is unstable in light or heat due to this reaction: $\ce{4 MnO4^- +2H2O -&amp;gt; 4MnO2 + 3O2 + 4OH-}$. This is why solutions of it must be stored in a dark cool place; even still, we must standardize the concentration of such a solution and we can only trust its measured concentration for a short time thereafter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A secondary standard solution of $\ce{KMnO4}$ is usually titrated by oxalic acid, a primary standard, and then used to determine the concentration of a number of other analytes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction with oxalic acid can be summarized as $\ce{5 H2C2O4 + 2 MnO4- + 6H+ -&amp;gt; 10CO2 + 2 Mn^{2+} + 8H2O}$ where everything except $\ce{MnO4-}$ is colorless. The end point of this titration is a marked by a purple color due to unconsumed $\ce{MnO4-}$. This allows you to make a standardized solution of potassium permanganate that can be used for a short while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how would we use this? Solutions containing $\ce{Ca^2+, Pb^2+, Ag+, Mg^2+, Co^2+, Zn^2+}$ and etc. can be precipitated as a metal oxalate. You would then digest this ppt in an acidic solution and titrate the released $\ce{H2C2O4}$, where the end point is marked by a purple color, allowing you to infer the original concentration of the metal ion in the original solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{KMnO4}$ was used in mineralogical surveys of uranium prior to the Manhattan project, where better techniques were developed to quantitate.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2015-06-17T19:29:12.473" LastActivityDate="2015-06-17T19:29:12.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="32995" PostTypeId="2" ParentId="32992" CreationDate="2015-06-17T05:24:54.860" Score="1" Body="&lt;p&gt;NAD+ accepts a hydride ion (proton + 2 electrons) and is reduced to NADH. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-06-17T05:24:54.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="32996" PostTypeId="1" AcceptedAnswerId="33031" CreationDate="2015-06-17T05:58:36.393" Score="1" ViewCount="10382" Body="&lt;p&gt;Learning about Oxides. Basically when oxygen is combined with a metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{FeO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called &quot;Iron Oxide (II)&quot; according to my book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the II represents the valence. But how come? Look:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe}$ has the configuration $1s^22s^22p^63s^23p^6\color{red}{4s^2}3d^6$. Since the outermost shell has $2$ electrons, the valence is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now $\ce{O}$ has the configuration $1s^2\color{red}{2s^22p^4}$. Its outermost shell has $6$ electrons so oxygen has a valence of $6$ right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how come $\ce{FeO}$ has valence $2$? Shouldn't the valence of $\ce{FeO}$ be like, the combination of the two valences? So $6 - 2 = 4$? (I read somewhere you had to subtract it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that. I am starting to think that my definition of valence is completely messed up. I just checked in &lt;a href=&quot;http://chemistry.about.com/library/weekly/aa122002a.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and apparently Oxygen has $-2$ valence? Why? And Iron has +2, +3, (+4), (+6)? What does all that even mean?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="7448" LastEditDate="2015-06-17T06:05:02.617" LastActivityDate="2019-08-24T14:31:02.823" Title="Calculating valence of oxides" Tags="&lt;nomenclature&gt;&lt;electronic-configuration&gt;&lt;ionic-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32998" PostTypeId="1" CreationDate="2015-06-17T07:06:57.147" Score="1" ViewCount="553" Body="&lt;p&gt;I’ve made magnetite &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe3O4}$&lt;/span&gt; in the lab and now I need to calculate yield percentage. How can I do it? I have these data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{FeSO4.7H2O(aq) &amp;amp;-&amp;gt; Fe^2+(aq) + SO4^2- (aq) + 7H2O(l)}\\&#xA;\ce{KOH(aq) &amp;amp;-&amp;gt; K+(aq) + OH- (aq)}\\ &#xA;\ce{KNO3(aq) &amp;amp;-&amp;gt; K+(aq) + NO3- (aq)}\\&#xA;\ce{Fe^2+(aq) + 2OH- (aq) &amp;amp;-&amp;gt; Fe(OH)2(s)}\\ &#xA;\ce{3Fe(OH)2(s) + NO3- (aq) &amp;amp;-&amp;gt; Fe3O4(s) + NO2- (aq) + 3H2O(l)}\\&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{FeSO4.7H2O &amp;amp;-&amp;gt;} 4.20~\mathrm{g}\\  &#xA;\ce{KOH &amp;amp;-&amp;gt;} 1.90~\mathrm{g}\\  &#xA;\ce{KNO3 &amp;amp;-&amp;gt;} 0.12~\mathrm{g}\\  &#xA;\text{Magnetite}~\ce{&amp;amp;-&amp;gt;} 0.36~\mathrm{g}\\&#xA;\end{align}&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to calculate the yield for simple reactions but I don’t know how to do it when there are many reactions involved.&lt;/p&gt;&#xA;" OwnerUserId="16775" LastEditorUserId="402" LastEditDate="2019-05-20T23:53:52.237" LastActivityDate="2019-05-20T23:53:52.237" Title="Calculating yield percentage of magnetite synthesis" Tags="&lt;inorganic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32999" PostTypeId="2" ParentId="32897" CreationDate="2015-06-17T08:36:19.040" Score="0" Body="&lt;p&gt;If the enthalpies of formation for all reactants and products are known then you can compute the standard enthalpy change ($\Delta H^\circ$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are not known you can estimate $\Delta H^\circ$ using bond energies as described in &lt;a href=&quot;https://youtu.be/-bqKZi63x6U&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally if you happen to know that the reaction is spontaneous ($\Delta G^\circ &amp;lt; 0$) and if the standard entropy change is (likely) negative (e.g. if the number of particles decrease) then it must be an exothermic reaction.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-17T08:36:19.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33000" PostTypeId="1" AcceptedAnswerId="33018" CreationDate="2015-06-17T08:38:12.063" Score="3" ViewCount="452" Body="&lt;p&gt;In a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32831/how-to-detect-lead-paint-on-a-bicycle&quot;&gt;previous question&lt;/a&gt;, I learned that we can detect lead with sodium / potassium rhodizonate, giving a nice purple color when reacting with pure (elemental?) lead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the paint, lead is primarily used in the form of salts such as lead oxide, lead carbonate, lead chromate...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the lead in my paint is not pure, but a salt? &lt;strong&gt;Would sodium rhodizonate react with such salts?&lt;/strong&gt; (I insist on sodium vs potassium, since it's cheaper to get). And if not, how can I detect these salts?&lt;/p&gt;&#xA;" OwnerUserId="16920" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-06-17T17:00:00.517" Title="Pb-based pigments and rhodizonate salts" Tags="&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33002" PostTypeId="2" ParentId="32897" CreationDate="2015-06-17T09:15:47.967" Score="3" Body="&lt;p&gt;First of all, the equilibrium constant $K$ in general doesn't tell too much about the enthalpy change of the reaction as it is related to the (Gibbs) free energy change $\Delta G = \Delta H - T\Delta S$. It is true that in some simple cases (such as gas phase reactions) the  entropy change can be approximated by looking at the stoichiometry of the reaction but in general it's usually more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your distinction between one-way and reversible reactions I think you may as well assume all reactions are reversible, but some have incredibly huge $K$ so the equilibrium is shifted totally to one side of the reaction - this is the result of a very negative $\Delta G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of a &quot;reversible&quot; reaction you can always look at the change of the equilibrium constant with temperature. The constant is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = \exp\left(-\frac{\Delta G}{RT}\right)=\exp\left(-\frac{\Delta H}{RT}\right)\exp\left(\frac{\Delta S}{R}\right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so from the temperature dependence you can deduce the sign of $\Delta H$. Naturally, this is assuming that $\Delta H$ and $\Delta S$ doesn't depend on the temperature in the temperature range you're looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Yyci.png&quot; alt=&quot;equilibrium constant vs temperature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure shows the change of equilibrium constant with temperature (arbitrary units) for an exothermic (yellow line) and an endothermic (blue line) reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just by looking at the reaction I don't think you can tell the enthalpy change. Normally you have bonds breaking on the reactant side and others forming on the product side and normally there is a delicate balance between the two. People have come up with bond strength values (which could be determined from known $\Delta H$ values) and you can use those in your reaction.&lt;/p&gt;&#xA;" OwnerUserId="16740" LastEditorUserId="16740" LastEditDate="2015-06-17T11:09:11.333" LastActivityDate="2015-06-17T11:09:11.333" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="33003" PostTypeId="1" AcceptedAnswerId="33247" CreationDate="2015-06-17T09:21:54.987" Score="2" ViewCount="1036" Body="&lt;p&gt;That's a long page, so you can optionally jump to the &quot;Thunderstorm it&quot; part, but a bit of context is always neat :)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Context&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Say I'm cooking an egg in vinegar. The egg shell is made of calcium carbonate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CaCO3 + CH3COOH -&amp;gt; H2O + CO2 + Ca(CH3COO)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After eating the egg, is left in the cookware: a bit of calcium acetate, lots of water, acetic acid and other unidentified stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It'd be neat to isolate that calcium acetate, as it could be of use sometime. I've seen that we can grow cool looking crystals out of it, maybe I could try that! Anyway chemistry is cool, so let's do it :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there will also be loads of water (egg cooking). I could store it all, but that wouldn't be very practical...&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Getting rid of the water&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I see three ways to get rid of that water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Boil until water evaporates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy-hungry process. I'm a kind of environmental-crazy dude, so that's not an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Filtering&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While boring, that could be an efficient option, if at all possible. It seems calcium acetate is soluble in water. Is there a way to filter it out without expensive equipment, or even at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrolysis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy-hungry as well, but I have a spare solar panel that produces 6 or 12V and anything between 0 and 4W, so that energy comes for free (albeit quite slowly in Belgium).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Thunderstorm it!&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;No matter how feasible are the other options, I'm here to experiment, so I'd love to do some electrolysis :) I got time anyway!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constraints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have very little knowledge of chemistry (but eager to learn!), and no equipment at all, except for household items. Also, this experiment would take place in living space (probably in my room), so it should be safe to do (e.g. do not produce chlorine gas or other nasty products). And if it's not smelly, it's even better :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How I would do it&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll need:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;container: a drinking glass should suffice&lt;/li&gt;&#xA;&lt;li&gt;anode&lt;/li&gt;&#xA;&lt;li&gt;cathode&lt;/li&gt;&#xA;&lt;li&gt;some electrolyte in the water&lt;/li&gt;&#xA;&lt;li&gt;electricity: the solar panel will provide what's needed&lt;/li&gt;&#xA;&lt;li&gt;time: we all have a pool of that in our life&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I still need the anode, cathode and electrolyte. And also, I'd love to understand the process that takes place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my younger age, I used copper wire as both anode and cathode, table salt as electrolyte, and it would produce tiny bubbles on one end of the wire, and go green on the other end (do I remember properly?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this experiment though, the context is a little different:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I need to &quot;dissolve&quot; a lot of water, so just a copper wire won't be enough&lt;/li&gt;&#xA;&lt;li&gt;It seems that using NaCl as electrolyte produces chlorine gas instead of oxygen, that's pretty bad&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Choosing the electrolyte&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I cooked the egg in vinegar, I'll probably have some acetic acid left in the water. Would that be a good enough electrolyte? Alternatively, I have baking soda readily available, but that would produce other salts when reacting with the remaining acetic acid...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Choosing the anode and cathode&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm lost here. First and foremost, I remember my copper wire going green when doing electrolysis. Does that mean I'll &quot;consume&quot; my anode/cathode while doing the electrolysis? That sounds pretty expensive. Also, I'll have to separate calcium acetate from copper oxyde afterwards...&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Questions&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;is the above reflection correct at all?&lt;/li&gt;&#xA;&lt;li&gt;what's the best choice for electrolyte and anode/cathode materials&lt;/li&gt;&#xA;&lt;li&gt;what's the exact reaction taking place?&lt;/li&gt;&#xA;&lt;li&gt;any other way to isolate that calcium acetate?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16920" LastEditorDisplayName="user15489" LastEditDate="2015-06-17T09:44:14.867" LastActivityDate="2015-06-23T15:58:01.000" Title="Isolating calcium acetate from water with electrolysis" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33005" PostTypeId="2" ParentId="32996" CreationDate="2015-06-17T09:37:42.153" Score="3" Body="&lt;p&gt;Valency is the &lt;i&gt;combining power of an atom&lt;/i&gt;; so, when we talk of iron having a valency of 2, it can combine either with two atoms of valency of one; or with one atom of valency of two.  Oxygen has a valency of two and so the formula for iron (II) oxide is FeO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the history of chemistry valency pre-dates the idea of electron configuration which looks like the source of your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that is clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: you mentioned that oxygen has a valency of minus 2; that's not quite right.  It has a &lt;i&gt;valency&lt;/i&gt; of 2 but it can form an &lt;i&gt;oxidation state&lt;/i&gt; of minus 2.  You need to separate out the ideas of valency and oxidation state.  Although the ideas are related, they are distinct.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-17T09:37:42.153" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33006" PostTypeId="5" CreationDate="2015-06-17T11:18:23.137" Score="0" Body="&lt;p&gt;The nitro compounds are represented by $\ce{R-NO2}$. They can be described by using resonance structures such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKiei.jpg&quot; alt=&quot;mesomeric forms of the Nitro group&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitro groups tend to be highly explosive due to their decomposition to form  $\ce{N2}$, which is highly stable due to its very strong triple bond which results in the enthalpy of decomposition for nitro compounds tending to be high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nitro compounds are divided into two main categories; the more common aromatic nitro compounds, and the less common aliphatic nitro compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromatic nitro compounds are generally synthesized using a 'nitrating mixture', a mixture of concentrated nitric and sulphuric acids, which generates the nitronium cation, $\ce{NO2+}$, that acts as an electrophile in an electrophilic aromatic substitution. However, strongly activated aromatic rings such as phenols can be nitrated using dilute nitric acid, or even in some cases nitrate salts. Aromatic nitro compounds are useful as precursors to aromatic amines or for deactivating the ring in nucleophilic aromatic substitution reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aliphatic nitro compounds are generally produced by nucleophilic substitution of halocarbons with nitrite salts, or by oxidation of primary amines. The $\alpha$ carbon is electrophilic, but the $\alpha$ hydrogen is acidic due to the electron withdrawing nature of the nitro group and so can often be deprotonated to form a nucleophilic carbanion which can undergo reactions such as the nitroaldol reaction.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nitro_compound&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reference Wikipedia page&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-03-15T17:12:52.140" LastActivityDate="2018-03-15T17:12:52.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33007" PostTypeId="4" CreationDate="2015-06-17T11:18:23.137" Score="0" Body="For questions about the synthesis, properties, and reactions of organic compounds containing the -NO2 group i.e. nitro-compounds." OwnerUserId="10185" LastEditorUserId="5026" LastEditDate="2018-03-15T17:12:42.127" LastActivityDate="2018-03-15T17:12:42.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33008" PostTypeId="2" ParentId="31785" CreationDate="2015-06-17T11:20:10.143" Score="1" Body="&lt;p&gt;If you want to make use of symmetry and reasonable parallelization for your TDDFT, you may want to try out &lt;a href=&quot;http://www.scm.com/ADF/&quot; rel=&quot;nofollow&quot;&gt;ADF&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADF doesn't use ECPs, but you can freeze the core electrons.&lt;/p&gt;&#xA;" OwnerUserId="16993" LastEditorUserId="5017" LastEditDate="2015-06-17T13:37:28.963" LastActivityDate="2015-06-17T13:37:28.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33009" PostTypeId="2" ParentId="32591" CreationDate="2015-06-17T12:11:03.337" Score="6" Body="&lt;p&gt;What I remember from studying C-P 'diagonal relationships' (for multiple bonds e.g. carbene vs phosphinidene, alkyne vs phosphaalkyne) is that similar electronegativity also played a role, also by affecting the valence orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, perhaps it is better viewed from a different vantage point:&#xA;the main group elements behave rather similarly from 3rd row onwards, but there is a distinct difference between the 2nd and 3rd row elements and the explanation is pretty straightforward:&#xA;The 2s and 2p orbitals are roughly similar in 'size' since the 2p orbitals are the first p-orbitals and are pretty compact. Therefore 2s and 2p orbitals mix happily. &#xA;Going to the third row, the 3p orbitals need to be orthogonal to the 2p orbitals and as a consequence they are much more diffuse than the 3s orbitals, resulting in less efficient s-p hybridization. This also manifests itself in the so-called 'inert pair' effect whereby e.g. Si will resist hybridization in favor of a lone pair. There is some beautiful work by Knutzelnigg from the 80s (?) describing these effects.&#xA;This 2nd-3rd row discrepancy is somewhat counterbalanced if you step to right, increasing Z and thereby contracting the orbitals giving rise to this apparent diagonal relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar arguments make that for the 3d TMs the hybridization is more efficient for 4s-3d than for the 4d/5d TMs. Consequently, 3d TMs tend to behave differently from their heavier sisters, which in turn are closer together (although for the 5d's the relativistic effects start to kick in more heavily...).&lt;/p&gt;&#xA;" OwnerUserId="16993" LastActivityDate="2015-06-17T12:11:03.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33010" PostTypeId="1" CreationDate="2015-06-17T12:39:58.127" Score="7" ViewCount="490" Body="&lt;p&gt;Electropolished stainless steel has a high corrosion resistance due to the oxide film that is created during electropolishing.  The film mainly consist of chromium(III) oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid is used as a rust removing agent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 H3PO4 + Fe2O3 -&amp;gt; 2 FePO4 + 3 H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: will 5% $\ce{H3PO4}$ solution remove this protective layer as well? It should, but how fast will be the protective layer removed? Does anyone have personal experience?&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-06-17T13:04:17.873" LastActivityDate="2015-06-23T03:37:14.000" Title="Does phosphoric acid remove stainless steel electropolished layer?" Tags="&lt;acid-base&gt;&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;surface-chemistry&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33011" PostTypeId="2" ParentId="22023" CreationDate="2015-06-17T13:02:53.733" Score="1" Body="&lt;p&gt;Although it is some time ago already, seeing this question active today made me decide to include the answer given by Prof. Clayden himself over e-mail to me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The reaction you ask about appears in the classic text on practical&#xA;  organic chemistry by Vogel: in fact this preparation is online at&#xA;  &lt;a href=&quot;http://www.erowid.org/archive/rhodium/chemistry/nitrostyrene.vogel5.html&quot; rel=&quot;nofollow&quot;&gt;http://www.erowid.org/archive/rhodium/chemistry/nitrostyrene.vogel5.html&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It’s not clear to me why amines would be preferable bases in general,&#xA;  although you are right that they avoid the Cannizzaro reaction if the&#xA;  nitro compound is hindered. They could also be preferable if the&#xA;  alcohol were the desired product, rather than the eliminated alkene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-06-17T13:02:53.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33012" PostTypeId="1" AcceptedAnswerId="33047" CreationDate="2015-06-17T13:54:31.510" Score="9" ViewCount="539" Body="&lt;p&gt;Suppose we had the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Polymer ($n$~units)_{(aq)} -&amp;gt; Polymer($a$~units)_{(aq)} + Polymer($n-a$~ units)_{(aq)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the specific monomer unit of the polymer, and assuming the polymer is straight chain, is there anyway to determine the ΔS of the reaction? I mean, I know we can determine the sign (which will almost certainly be positive), but is there a way to make even a rough approximation of magnitude? We can assume the conformational entropy is negligible, i.e. the polymers are fairly rigid.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-06-17T18:21:39.153" LastActivityDate="2015-06-19T13:14:23.580" Title="deltaS of a decomposition reaction" Tags="&lt;polymers&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33013" PostTypeId="1" CreationDate="2015-06-17T14:02:23.750" Score="2" ViewCount="1987" Body="&lt;p&gt;I'm looking at a problem where sodium methanethiolate and sodium methoxide, both in equal amounts, are being reacted with $\ce{CH3I}$ and the solvent is ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that sodium methanethiolate is a better nucleophile than sodium methoxide when the two are reacted in equal amounts with something like methyl iodide in a polar aprotic solvent. But why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is polar protic, like ethanol? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the ethanol would reduce the nucleophilicity of both, but would do so at by different amounts, given the different electronegativity of sulfur and oxygen. If ethanol reduces the nucleophilicity of both at the same rate, why even use ethanol as a solvent in this case?&lt;/p&gt;&#xA;" OwnerUserId="15654" LastEditorUserId="10185" LastEditDate="2015-06-17T14:42:13.580" LastActivityDate="2015-06-17T16:04:10.983" Title="Why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is ethanol" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;organosulfur-compounds&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33014" PostTypeId="2" ParentId="32832" CreationDate="2015-06-17T14:20:02.810" Score="9" Body="&lt;p&gt;$\ce{NaHCO3}$ is the basic sodium salt of carbonic acid, a diprotic acid which exists only in small amounts in aqueous solution, because the equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lies mainly on the educt side (this is also the reason why $\ce{CO2}$ evolution is observed). When this fact is taken into account, a $\ce{pK_{a}}$ value of 6.3 at 25 °C is obtained for the first protolysis step (&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbonic_acid&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;). Benzoic acid has a lower $\ce{pK_{a}}$ value of 4.19 and is therefore able to liberate carbonic acid from its salts, like any other acid that is significantly stronger than $\ce{H2CO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C6H5CO2H + NaHCO3 -&amp;gt; C6H5CO2Na + CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just in case that you mean 3,4-Dihydroxycyclobut-3-ene-1,2-dione, which is also known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Squaric_acid&quot; rel=&quot;noreferrer&quot;&gt;squaric acid&lt;/a&gt;, then a reaction with $\ce{CO2}$ evolution is likely. Squaric acid is a fairly strong diprotic acid, with $\ce{pK_{a}}$ values of 1.5 and 3.4, respectively. The acidity can be attributed to the stabilization of the dianion by resonance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7NSmd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If carbon dioxide evolution with 1,2-dihydroxy cyclobutanedione is possible depends on the $\ce{pK_{a}}$ value of this compound. However, I would expect it to be less acidic than squaric acid.&lt;/p&gt;&#xA;" OwnerUserId="5016" LastEditorUserId="5016" LastEditDate="2015-06-17T14:28:40.477" LastActivityDate="2015-06-17T14:28:40.477" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33015" PostTypeId="2" ParentId="33013" CreationDate="2015-06-17T14:39:15.633" Score="0" Body="&lt;p&gt;Nucleophiles can be hard or soft (as can electrophiles); a hard nucleophile tends to react preferentially with a hard electrophile, while a soft nucleophile tends to react preferentially with a soft electrophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your system the methanethiolate anion, $\ce{CH3S-}$, is a soft nucleophile while methoxide, $\ce{CH3O-}$, is a relatively hard nucleophile.  You will also have ethoxide ($\ce{CH3CH2O-}$) in your system.  As to the electrophile, methyl iodide, $\ce{CH3I}$: it is soft and hence it preferentially reacts with the softer nucleophile in the system, methanethiolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated explanation of the hard/soft idea can be found in frontier orbital analysis.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-17T14:39:15.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33016" PostTypeId="2" ParentId="28159" CreationDate="2015-06-17T15:55:26.137" Score="2" Body="&lt;p&gt;Answering this question requires assuming that all steps are &quot;elementary&quot;, e.g. that $\ce{A + B -&amp;gt; C}$ is an irreversible bimolecular reaction that is first-order in $[\ce{A}]$ and first-order in $[\ce{B}]$.  (Obviously we are also assuming that activities can approximated by concentrations.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that assumption, the fact that $\ce{C}$ conversion is &quot;fast&quot; means that we can replace the existence of $\ce{C}$ in the scheme entirely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{A + B~(slow) &amp;amp;\rightarrow D + B} \\&#xA;\ce{D + E~(fast) &amp;amp;\rightarrow F}\\&#xA;\ce{F~(fast) &amp;amp;\rightarrow G}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise if $\ce{F}$ conversion is &quot;fast&quot;, then it can be replaced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\ce{A + B~(slow) &amp;amp;\rightarrow D + B} \\&#xA;\ce{D + E~(fast) &amp;amp;\rightarrow G}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this makes it clear why $\ce{E}$ &lt;em&gt;must&lt;/em&gt; appear in any rate law for formation of $\ce{G}$: there is no source of it in the two reactions we are left with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{D}$, I agree that the situation is a bit more complex.  It might appear in the rate equation, or it might not.  Necessary reactants at the beginning of the reaction if we are ever to see product $\ce{G}$ are $\ce{A}$, $\ce{B}$, and $\ce{E}$, so it would make sense to eliminate $\ce{D}$ from the rate law if possible.  However, that could be algebraically difficult, or perhaps $\ce{D}$ is easier to measure, etc.  There is no reason $\ce{D}$ &lt;em&gt;can't&lt;/em&gt; appear in the rate law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Huge caveat:&lt;/strong&gt; &quot;fast&quot; and &quot;slow&quot; are relative terms and as such without further context they are ambiguous.  The problem as constructed sounds a bit sloppy.  For example, is the &quot;fastness&quot; of $\ce{F -&amp;gt; G}$ faster or less fast than the fastness of $\ce{D + E -&amp;gt; F}$?  The problem doesn't say but the simplifications I did above actually depend on these sorts of details.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:23:47.367" LastActivityDate="2016-03-11T23:23:47.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33017" PostTypeId="2" ParentId="33013" CreationDate="2015-06-17T16:04:10.983" Score="2" Body="&lt;p&gt;Thiolates are generally more nucleophilic than alcoholates. So whenever the choice happens to be between a nucleophilic sulphur and a nucleophilic oxygen, the sulphur will win.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the methanolate and the methanethiolate will show reduced nucleophilicity due to partial protonation, though. And of course you have introduced ethanolate as an additional nucleophile, too (which should be the least reactive, though).&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-17T16:04:10.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33018" PostTypeId="2" ParentId="33000" CreationDate="2015-06-17T17:00:00.517" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://fire.nist.gov/bfrlpubs/build00/PDF/b00034.pdf&quot; rel=&quot;nofollow&quot;&gt;Ashley &lt;em&gt;et al&lt;/em&gt;&lt;/a&gt;, &quot;a rhodizonate spot test kit [for] &lt;em&gt;in-situ&lt;/em&gt; testing of lead in paint... can be used for screening (i.e., qualitative) purposes.&quot; The test is sensitive not only to metallic (elemental) lead, but also to lead compounds used as pigments. However, the reference cited also states that the sulfide and rhodizonate tests have a high incidence of false positives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium, potassium or even cesium salts of &lt;a href=&quot;https://en.wikipedia.org/wiki/Rhodizonic_acid&quot; rel=&quot;nofollow&quot;&gt;rhodizonic acid&lt;/a&gt; should be equally effective, since the active species is the rhodizonate &lt;em&gt;ion&lt;/em&gt;, but be aware that the sodium salt is less stable in water solution and should be used within a few days.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-17T17:00:00.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33019" PostTypeId="1" AcceptedAnswerId="33022" CreationDate="2015-06-17T17:01:54.190" Score="4" ViewCount="819" Body="&lt;p&gt;I'm trying to measure capacitance of a cell I'm creating. It's composed of two copper-sheets with nano-material coated on one side, sandwiched with a membrane and wetted with $\ce{KOH}$. All of the three electrode setups I've found on the internet are for solution based voltamettry, but this cell is not in a solution. The setup that my labmates had yesterday had the reference electrode, $\ce{Ag|AgCl}$ setting in a solution, and not connected to the cell, which doesn't make any sense to me. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What reference electrode should be used to do this kind of measurement, and what should the schematic look like for our setup? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4325" LastEditDate="2015-06-17T18:29:26.313" LastActivityDate="2015-06-17T18:53:45.483" Title="How do I setup the electrodes for cyclic voltammetry on a capacitor made on dry copper sheets?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33020" PostTypeId="2" ParentId="33019" CreationDate="2015-06-17T18:10:56.140" Score="2" Body="&lt;p&gt;Well, I'm not an electrical engineer by any means, but I do have some experience with this. Basically, the potentiostat is an operational amplifier. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Potentiostat&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page for potentiostat&lt;/a&gt; shows the schematic of a test cell as a series of impedances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N72cM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to test your capacitor, simply place your capacitor in for $Z_2$, and a resistor in for $Z_1$. Unfortunately, I can't be much help in selecting a resistance value, but it will depend on your potentiostat and the capacitance of your device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a9zAp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my tests, a resistor wasn't necessary at all, so first try connecting the counter and reference to one plate of the capacitor and the working electrode to the other plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it turns out the resistive impedance is necessary, ask an electrical engineer for help, and use &quot;op amp&quot; in your explanations, because they won't know what a potentiostat is (most likely).&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2015-06-17T18:10:56.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33022" PostTypeId="2" ParentId="33019" CreationDate="2015-06-17T18:53:45.483" Score="3" Body="&lt;p&gt;The reason reference electrodes are needed in 3 electrode solution cells is that you're trying to control the interfacial potential of the working electrode (basically from the surface to a point very near the electrode in solution). The wiki's kinda lousy, but the diagram explains essentially what the potentiostat is doing: It's controlling the counter electrode (the op-amp output), to set the potential from the working electrode (ground) to a point just into the solution and the reference electrode is used to read the potential and provide feedback to the amplifier to correct the output. This way, any extra solution resistance between the counter and reference electrodes doesn't reduce the interfacial potential (known as iR drop in electrochemical circles) because you're measuring the potential near the surface. In other words, we're interested in applying a potential across $Z_2$ precisely, but we don't care about what's happening with $Z_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z56w8.png&quot; alt=&quot;potentiostat schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since you have a capacitor, what you're trying to do is control the potential of one plate versus the other plate, so you want the point where you're getting feedback to be one plate, and the other plate is ground—one plate is the &quot;reference electrode&quot; and the other is the &quot;working electrode&quot;. Your capacitor is $Z2$ and there's no solution, so there's no $Z1$. (In some cases, it still makes sense to use a separate cable for the reference input, but it's probably fine to just short them together at the inputs) Many potentiostats will work with the counter electrode and reference electrode inputs shorted together (which is how one typically connects a 2 electrode cell), but a resistor between them can help prevent it from oscillating by increasing the time constant of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're not really using a reference electrode in this case because you don't care about measuring the applied potential versus some known potential, just in controlling the potential across the capacitor. In a cell in which we're interested in studying redox reactions and such, it's important to know how the potentials relate to one another and to compensate for the various potentials that arise when electrodes are dipped in electrolyte solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're really interested in the details of potentiostats, Bank has a &lt;a href=&quot;http://www.bank-ic.de/encms/downloads/potstae2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;good primer&lt;/a&gt; with lots of details.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-17T18:53:45.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33023" PostTypeId="1" CreationDate="2015-06-17T19:09:35.810" Score="1" ViewCount="184" Body="&lt;p&gt;How exactly is the interaction between two chiral compounds so they can tell the difference between S or R?&lt;/p&gt;&#xA;" OwnerUserId="16936" LastEditorUserId="15410" LastEditDate="2015-06-17T19:27:29.310" LastActivityDate="2015-06-17T20:02:39.587" Title="Resolution of enantiomers? How to determine chiral compounds apart from each other?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-06-18T07:55:23.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="33024" PostTypeId="1" AcceptedAnswerId="33052" CreationDate="2015-06-17T19:37:55.027" Score="6" ViewCount="1031" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In the following reaction, only one of the hydroxyl groups in &lt;strong&gt;D&lt;/strong&gt; reacts under the given conditions:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K26HI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K26HI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I was wondering which OH group reacts and why? Also I have been trying to determine the structure of E based on which hydroxyl group reacts.&lt;/p&gt;&#xA;&lt;p&gt;My thoughts are that OH in the side group can react via S&lt;sub&gt;N&lt;/sub&gt;2 with the reactants, but the OH right adjacent to the ether O can react leaving a carbocation that can be stabilised by lone pair from O. This would mean S&lt;sub&gt;N&lt;/sub&gt;1 and a completely different product.&lt;/p&gt;&#xA;&lt;p&gt;Can someone please explain the correct mechanism?&lt;/p&gt;&#xA;" OwnerUserId="10257" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-06-29T03:28:27.443" Title="How to determine the most reactive hydroxyl group of α-ᴅ-glucopyranose?" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33025" PostTypeId="2" ParentId="33023" CreationDate="2015-06-17T19:40:54.250" Score="1" Body="&lt;p&gt;The method used to determine whether a specific chiral molecule is the $S$ or $R$ enantiomer is determined according to the Cahn–Ingold–Prelog priority rules. The steps to this are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Look at the four atoms directly attached to the chirality center, and rank them according to atomic number (1-4, 1 being highest ranking, 4 being lowest ranking). Isotopes of higher mass rank higher than lower mass isotopes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If a decision can't be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Having ranked the substituent groups, draw (or imagine) the lowest ranking substituent group away from yourself. The other 3 substituent groups will then appear to radiate out towards you. If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent ($\ce{1-&amp;gt;2-&amp;gt;3}$) is clockwise, the chirality center is said to have the $R$ (for &lt;em&gt;rectus&lt;/em&gt;, meaning right) configuration. Oppositely, if the curved arrow is counterclockwise, the chirality center is said to have the $S$ (for &lt;em&gt;sinister&lt;/em&gt;, meaning left) configuration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-06-17T19:40:54.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33026" PostTypeId="2" ParentId="32984" CreationDate="2015-06-17T19:53:11.060" Score="8" Body="&lt;p&gt;&lt;strong&gt;Short answer: Not for &quot;normal&quot; compounds.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see from your table, the energies decrease with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)&lt;/li&gt;&#xA;&lt;li&gt;Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Born-Lande and &lt;a href=&quot;https://en.wikipedia.org/wiki/Kapustinskii_equation&quot;&gt;Kaputinskii&lt;/a&gt; equations give some numeric estimates, e.g. (Born-Landé)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;where:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$N_A$ = Avogadro constant;&lt;/li&gt;&#xA;  &lt;li&gt;M  = Madelung constant&lt;/li&gt;&#xA;  &lt;li&gt;$z^+$ = charge number of cation&lt;/li&gt;&#xA;  &lt;li&gt;$z^−$ = charge number of anion&lt;/li&gt;&#xA;  &lt;li&gt;e = charge on an electron (elementary charge)&lt;/li&gt;&#xA;  &lt;li&gt;$\epsilon_0$ = permittivity of free space&lt;/li&gt;&#xA;  &lt;li&gt;$r_0$ = distance to closest ion&lt;/li&gt;&#xA;  &lt;li&gt;n  = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Note that I've inverted the formula from Wikipedia's version to match your reaction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these can give you a negative lattice enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could it happen? Well, you'd need something where the &lt;em&gt;solid&lt;/em&gt; is less stable than the &lt;em&gt;gas-phase ions&lt;/em&gt;. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-06-17T19:53:11.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33027" PostTypeId="2" ParentId="33023" CreationDate="2015-06-17T20:02:39.587" Score="2" Body="&lt;p&gt;If the compounds have never been made before, then the enantiomers have to be separated prior to analysis. Then analytical and spectroscopic techniques must be used. The R and S configurations cannot even be assigned until the relative stereochemistry had been elucidated. This is usually done by specific NMR experiments. Chiral HPLC can be used to give enantiomeric purity.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-17T20:02:39.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33028" PostTypeId="1" CreationDate="2015-06-17T20:34:18.127" Score="0" ViewCount="5657" Body="&lt;p&gt;$$\ce{FeO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is called &quot;Iron(II) oxide &quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe_2O_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is called &quot;Iron(III) oxide &quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number in the parenthesis is &quot;Valence&quot; of the substance. Frankly, I don't know how to calculate such number. Can you explain to me, with this example, how come $\ce{FeO}$ has valence $2$ and the other one has valence $3$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I know how to do is calculate the electron configurations of each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Iron, it is $1s^22s^22p^63s^23p^64s^23d^6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Oxygen, it is $1s^22s^22p^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told that based on the electron configuration you can get the valence of each element. And then, you can calculate the valence of $\ce{FeO}$ and $\ce{Fe_2O_3}$. &lt;strong&gt;But how?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I made a question before &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32996/calculating-valence-of-oxides&quot;&gt;Calculating valence of oxides&lt;/a&gt;, but it was wrongly formulated (they corrected me on the definition of Valence, but turns out what I actually need was &lt;em&gt;how&lt;/em&gt; to calculate it), so I'm flagging it to be closed.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt; The question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4547/how-to-get-the-valencies-of-elements&quot;&gt;How to get the valencies of elements?&lt;/a&gt; doesn't explain how to calculate the valence of a &lt;strong&gt;combination&lt;/strong&gt; of two elements.&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-06-17T21:10:39.947" Title="Calculate the valence of Fe2O3" Tags="&lt;nomenclature&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="8" ClosedDate="2015-06-18T04:08:06.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="33029" PostTypeId="2" ParentId="33028" CreationDate="2015-06-17T21:10:39.947" Score="1" Body="&lt;p&gt;@Voldemort you learn the valencies rather than calculate them.  Valency is a fact of experiment whilst electron configuration gives an underlying explanation for the observations from experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron has two different valencies (and oxidation states) which are a consequence of the stability of the electron configuration that remains when electrons are lost upon reaction.  So, iron can loose 2 electrons to leave a stable $\ce{Fe++}$ species; in your reaction those 2 electrons are gained by oxygen to give an anion that has 8 electrons in its outer shell which is a stable configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron can also loose 3 electrons to leave a stable system and again, in your system, the electrons are picked up by the appropriate number of oxygen atoms to give oxygen anions that have 8 electrons in their outer shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the either side of the periodic table, say sodium reacting with a halogen, say chlorine (to give sodium chloride); you can see the 8 electron rule working.  But you're looking at a system where the iron is found in the middle of the periodic table and is a lot more difficult to predict when compared to the 8 electron rule.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-17T21:10:39.947" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33030" PostTypeId="2" ParentId="32984" CreationDate="2015-06-17T21:38:11.200" Score="8" Body="&lt;p&gt;Geoff's answer is enlightening because it shows several of the factors influencing lattice formation energetics and how they cannot be easily made to output endothermic ionic lattices. However, an even stronger statement can be made by forgoing any reference to lattice energy formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering thermodynamics as we chemists are most used to, we have a system which is free to evolve under constant pressure and in the absence of external potentials such as gravitational, electric or magnetic fields, and which is given infinite time to reach an equilibrium condition. As such, we can use our good old friend, the Gibbs free energy equation, in its simplest form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G = \Delta H -T\Delta S$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a process to be spontaneous, we say it is &lt;em&gt;exergonic&lt;/em&gt; ($\Delta G&amp;lt;0$). This requires one of three conditions to be met:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\Delta H&amp;lt;0$, $\Delta S&amp;gt;0$ : The process is exothermic and increases system entropy; spontaneous process at any temperature.&lt;/li&gt;&#xA;&lt;li&gt;$\Delta H&amp;lt;0$, $\Delta S&amp;lt;0$ : The process is exothermic and decreases system entropy; spontaneous process only at &lt;em&gt;sufficiently low&lt;/em&gt; temperatures.&lt;/li&gt;&#xA;&lt;li&gt;$\Delta H&amp;gt;0$, $\Delta S&amp;gt;0$ : The process is endothermic and increases disorder; spontaneous process only at &lt;em&gt;sufficiently high&lt;/em&gt; temperatures&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now consider the process where gaseous ions combine to form an ionic solid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{y\ A^{+x}(g) + x\ B^{-y}(g) -&amp;gt; A_{y}B_{x}(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question essentially asks whether this process can be &lt;em&gt;exergonic while being endothermic&lt;/em&gt;. For this to be the case, the only condition which has any chance of being met is the third. Therein lies the problem. It is hard to imagine a situation where a &lt;em&gt;crystalline solid&lt;/em&gt; with long-range ordering, as is the case with most common ionic compounds, could possibly have a higher entropy than a &lt;em&gt;gas&lt;/em&gt; of ions. This is a strong, general argument against the stability and existence of ionic solids with endothermic lattice energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thermodynamic argument, however, does not constitute ultimate proof that endothermic lattices are impossible, just unlikely. For a stable endothermic lattice to be made, one of the constraints mentioned at the start must be relaxed. For example, perhaps thermodynamics under constant pressure is not adequate, but under constant &lt;em&gt;volume&lt;/em&gt; (with the associated &lt;a href=&quot;https://en.wikipedia.org/wiki/Helmholtz_free_energy&quot; rel=&quot;nofollow&quot;&gt;Helmholtz free energy&lt;/a&gt; considerations) it could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that additional potentials can be present in the system, in which case other terms must be added to the Gibbs free energy expression. As a loosely related example, &lt;a href=&quot;http://www1.lsbu.ac.uk/water/magnetic_electric_effects.html&quot; rel=&quot;nofollow&quot;&gt;water can liquefy/freeze at unusual temperatures under the action of a very strong electric field&lt;/a&gt;, even at atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another way is if, even though formation of an endothermic lattice is not spontaneous, once formed it could somehow be kinetically stabilized against dissociation. This delicate interplay of kinetics and thermodynamics in Chemistry is certainly thoroughly exploited, but whether it can applied to this particular situation, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more opportunity for the existence of a negative energy lattice can be based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;non-equilibrium thermodynamics&lt;/a&gt;, either in systems close to (but not quite at) equilibrium, or in systems very far from equilibrium. Thermodynamics in these systems is much harder to describe mathematically, however.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-06-18T08:51:35.810" LastActivityDate="2015-06-18T08:51:35.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33031" PostTypeId="2" ParentId="32996" CreationDate="2015-06-17T23:46:51.567" Score="2" Body="&lt;p&gt;Ok, finding valencies of elements in simple compounds is very easy, but you have to know some thing to start:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are treating compound as if it would be composed of discrete molecules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have compound/neutral molecule, sums of valencies of both types of atoms should be equal, as valency is number of bonds between atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of FeO you need to know that O make usually two bonds (valence = 2) and as both valencies have to be equal Fe also has val = 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of Fe2O3, 3*2=6, 6/2=3 so Fe is trivalent.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-06-17T23:46:51.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33033" PostTypeId="1" CreationDate="2015-06-18T00:49:14.997" Score="7" ViewCount="92" Body="&lt;p&gt;What is the mechanism by which phosphorus pentoxide ($\ce{P4O10}$) and a carboxylic acid form an anhydride? I can't find it anywhere online. Does $\ce{COOH}$ attack twice (I'm assuming it's the nucleophile) because there ends up being two carbonyls in the product? What leaves?&lt;/p&gt;&#xA;" OwnerUserId="17002" LastEditorUserId="10185" LastEditDate="2015-06-18T16:23:55.940" LastActivityDate="2015-06-18T16:23:55.940" Title="Mechanism for anhydride formation" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ClosedDate="2015-08-24T17:54:05.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="33034" PostTypeId="1" AcceptedAnswerId="66594" CreationDate="2015-06-18T01:29:47.010" Score="26" ViewCount="14478" Body="&lt;p&gt;When $\ce{SeO2}$ is used to oxidise aldehydes and ketones, $\ce{SeO2}$ becomes hydrated in water, and the active reagent is selenous acid $\ce{H2SeO3}$ which oxidizes aldehydes and ketones to furnish 1,2-dicarbonyl compounds such as &lt;strong&gt;9&lt;/strong&gt; in the scheme below.&lt;sup&gt;1&lt;/sup&gt; The α-hydroxyketone &lt;strong&gt;10&lt;/strong&gt; is not observed. Why is that so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YvyKQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YvyKQ.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Sharpless, K. B.; Gordon, K. M. Selenium dioxide oxidation of ketones and aldehydes. Evidence for the intermediacy of .beta.-ketoseleninic acids. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1976,&lt;/strong&gt; &lt;em&gt;98&lt;/em&gt; (1), 300–301. &lt;a href=&quot;http://dx.doi.org/10.1021/ja00417a083&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI:10.1021/ja00417a083&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="16683" LastEditDate="2017-09-23T16:56:37.940" LastActivityDate="2020-08-27T21:30:01.687" Title="Why doesn't the SeO2 oxidation of ketones stop at the hydroxyketone stage?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33036" PostTypeId="1" CreationDate="2015-06-18T02:40:47.300" Score="6" ViewCount="1460" Body="&lt;p&gt;Could anyone please tell me where I could find the the structure of cubic ice (ice I&lt;sub&gt;c&lt;/sub&gt;) as a CIF file? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been searching the cambridge structural database using &quot;H2O&quot;, &quot;water&quot;, &quot;dihyrogen oxide&quot; and &quot;ice&quot; as search key words, but cannot find anything!&lt;/p&gt;&#xA;" OwnerUserId="16152" LastEditorUserId="30" LastEditDate="2016-08-09T07:56:10.180" LastActivityDate="2016-08-09T16:26:40.013" Title="Finding cubic ice CIF files in crystal database" Tags="&lt;crystal-structure&gt;&lt;databases&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33038" PostTypeId="1" AcceptedAnswerId="33039" CreationDate="2015-06-18T05:01:48.757" Score="4" ViewCount="472" Body="&lt;p&gt;In my lab courses, we were taught to use textbooks off the shelf as elevating props to adjust the heights of heating mantles to the right height during distillations, but I was wondering if there was a safer and better way to do so? I recently used a different type of heating mantle that was like a nest and charred the cover of a textbook as well as a heat glove that I put on the second time to prevent further damage to the book.&lt;/p&gt;&#xA;" OwnerDisplayName="user9974" LastActivityDate="2015-06-18T09:13:29.833" Title="Distillation Heating Mantle Setup" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33039" PostTypeId="2" ParentId="33038" CreationDate="2015-06-18T05:41:52.233" Score="5" Body="&lt;p&gt;You need to use something called a lab jack ... just search google to see what they look like.  They are quite cheap too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should tell whoever told you to use books about lab jacks, too.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-18T05:41:52.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33040" PostTypeId="1" AcceptedAnswerId="33041" CreationDate="2015-06-18T07:10:34.967" Score="4" ViewCount="9272" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the amount of nitrate ions in &lt;span class=&quot;math-container&quot;&gt;$20.0\:\mathrm{g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)3}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The chemical formula for the nitrate ion is &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3}$&lt;/span&gt;, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molar mass of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)3}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$242~\mathrm{g~mol^{-1}}$&lt;/span&gt; of which &lt;span class=&quot;math-container&quot;&gt;$186\ \mathrm{g\ mol^{-1}}$&lt;/span&gt; belong to &lt;span class=&quot;math-container&quot;&gt;$\ce{NO3}$&lt;/span&gt;. That means that around &lt;span class=&quot;math-container&quot;&gt;$76.85\:\%$&lt;/span&gt; of the substance is nitrate ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$20.0\ \mathrm g$&lt;/span&gt; of the substance are equivalent to &lt;span class=&quot;math-container&quot;&gt;$20.0\ \mathrm g/242\ \mathrm{g\ mol^{-1}} = 0.0826~\mathrm{mol}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since roughly &lt;span class=&quot;math-container&quot;&gt;$76.85\:\%$&lt;/span&gt; are nitrate ions, there are about &lt;span class=&quot;math-container&quot;&gt;$0.06~\mathrm{mol}$&lt;/span&gt; of nitrate ions in those &lt;span class=&quot;math-container&quot;&gt;$20\ \mathrm g$&lt;/span&gt; of substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One mole is &lt;span class=&quot;math-container&quot;&gt;$6.02\times 10^{23}$&lt;/span&gt;, so if I multiply &lt;span class=&quot;math-container&quot;&gt;$(0.06\ \mathrm{mol})(6.02\times 10^{23}\ \mathrm{mol^{-1}}) = 3.612\times10^{22}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Answer:&lt;/strong&gt; There are &lt;span class=&quot;math-container&quot;&gt;$3.612\times10^{22}$&lt;/span&gt; nitrate ions in &lt;span class=&quot;math-container&quot;&gt;$20.0\:\mathrm{g}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe(NO3)3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that is &lt;strong&gt;wrong&lt;/strong&gt;. The options in the website are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$1.49\times 10^{23}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$4.98\times 10^{21}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$60.0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$8.25\times10^{21}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And my answer is not even close to any of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I do wrong?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorDisplayName="user7951" LastEditDate="2019-09-16T13:28:56.960" LastActivityDate="2019-09-16T13:28:56.960" Title="What is the amount of nitrate ions in 20 g of Fe(NO3)3?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33041" PostTypeId="2" ParentId="33040" CreationDate="2015-06-18T07:49:30.743" Score="6" Body="&lt;p&gt;The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does that mean for the number of moles of nitrate ions?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-18T07:49:30.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33042" PostTypeId="1" CreationDate="2015-06-18T08:50:57.710" Score="2" ViewCount="896" Body="&lt;p&gt;For strong electrolytes, according to Kohlrausch, molar conductivity decreases as the concentration increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;( i don't know why I'm not allowed to attach a photo here in order to show you the diagram ), but however I'm trying to explain how the diagram looks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram we have strong electrolytes like $\ce{HCl,\, H2SO4,\, KCl,\, LaCl3,\, Na2SO4}$ whose molar conductivity decreases as the concentration increases. But, what I don't understand is that in the diagram we have also $\ce{CH3COOH}$ which is a weak electrolyte, and his molar conductivity also is decreasing and even far more rapidly, as the concentration increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody explain this ? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="4097" LastEditDate="2015-06-18T10:49:58.110" LastActivityDate="2015-12-20T14:42:59.693" Title="Molar conductivity and concentration" Tags="&lt;physical-chemistry&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33045" PostTypeId="1" AcceptedAnswerId="33064" CreationDate="2015-06-18T09:19:07.977" Score="3" ViewCount="175" Body="&lt;p&gt;As I’m reading about water cations and anions, I read that the benefits we gain from water are of course lots, but the most important of them is the neutralisation of acids ($\ce{H+}$) from the anions present in water such as:  $\ce{HCO3-}$ , $\ce{CO3^2-}$ and $\ce{OH-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we choose the bottle of water with the highest concentration of these anions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier? &lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="9961" LastEditDate="2016-07-25T17:16:09.057" LastActivityDate="2016-07-25T17:16:09.057" Title="Quality of water for drinking purposes" Tags="&lt;everyday-chemistry&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33046" PostTypeId="1" CreationDate="2015-06-18T09:29:52.470" Score="5" ViewCount="7378" Body="&lt;p&gt;I understand that we need to add $\ce{FeBr3}$ while brominating benzene in order to polarize the bromine molecule. However, we do not need to add such a Lewis Acid in the case of phenol, because it is activated by the $\ce{-OH}$ group, and so polarizes the bromine molecule on its own. Anisole is also highly activated, however we still add $\ce{FeBr3}$ in order to brominate it. Why?&#xA;(An answer which compares bromination in phenol and anisole will be highly appreciated.)&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="10185" LastEditDate="2015-06-18T16:11:54.727" LastActivityDate="2018-06-22T13:24:26.760" Title="Why do we need to add FeBr3 for bromination of anisole, but not for bromination of phenol?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33047" PostTypeId="2" ParentId="33012" CreationDate="2015-06-18T10:18:04.643" Score="3" Body="&lt;p&gt;The largest contribution to the entropy change will be the translational component, which is easy to evaluate using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation&quot; rel=&quot;nofollow&quot;&gt;Sackur-Tetrode equation&lt;/a&gt;&#xA;$$S^\circ_{\rm{trans}}=R\ln(bM^{3/2}T^{3/2})$$&#xA;where $b$ is a collection of constants with a value of 3.75 (mol/gK)$^{3/2}$ $M$ is the mass in g/mol, $T$ is temperature in Kelvin, and the standard state is 1 M.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the slowly varying logarithmic dependence on $M$ and $T$ this contribution is usually on the order of 100's of J/molK.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the translational contribution is the largest, rotational and vibrational contributions can be sizable so this will give you an order of magnitude estimation at best.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Derivation starting from the wikipedia equation&#xA;$$S^\circ=kN \left( \ln \left( \frac{V^\circ}{N} \left( \frac{4\pi mU}{3h^2N}\right)^{3/2} \right)+\frac{5}{2}\right) $$ &#xA;$V^\circ$ = 1 L = 0.001 $m^3$, $N=N_A$, $U=\tfrac{3}{2}N_AkT$, $M=mN_A$&#xA;$$S^\circ=R  \ln \left( \left[e^{5/2}\frac{V^\circ}{N_A} \left( \frac{4\pi \tfrac{3}{2}k}{3h^2N_A}\right)^{3/2}\right]M^{3/2}T^{3/2} \right) $$&#xA;The term in square brackets corresponds to $b$ and &lt;a href=&quot;http://www.wolframalpha.com/input/?i=e%5E2.5*0.001%2F6.023e23*%28%284*3.14*1.5*1.38e-23%29%2F%283*6.63e-34%5E2*6.023e23%29%29%5E1.5&quot; rel=&quot;nofollow&quot;&gt;using SI units&lt;/a&gt; $b=1.20\times10^5$ mol(kg/K)$^{3/2}$ which corresponds to 3.75 mol(g/K)$^{3/2}$&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-19T13:14:23.580" LastActivityDate="2015-06-19T13:14:23.580" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33048" PostTypeId="2" ParentId="32906" CreationDate="2015-06-18T12:32:04.227" Score="4" Body="&lt;p&gt;As Ben Norris explained, $R$ can be just more convenient so that one can deal with quantities per mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. one could write either $e^{(-E_a/RT)}$ or $e^{(-E_a/K_BT)}$, where in the first instance one would use $E_a$ in $\ce{J~mol^{-1}}$, in the second instance just in $\ce{J}$. The term in the exponent should of course be dimensionless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the numbers in $\ce{kJ~mol^{-1}}$ are much more familiar for chemists, one would often use the first one to avoid having to use very small numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux is that the Boltzmann factor in the Arrhenius expression is related to the chance for particles having enough energy at a given temperature, $T$, to surmount the activation barrier. &lt;/p&gt;&#xA;" OwnerUserId="16993" LastEditorUserId="14" LastEditDate="2015-06-18T16:05:44.420" LastActivityDate="2015-06-18T16:05:44.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33049" PostTypeId="1" AcceptedAnswerId="33051" CreationDate="2015-06-18T12:49:54.423" Score="21" ViewCount="8897" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which reacts faster in the Cannizzaro Reaction?&lt;br&gt;&#xA;  a) $\ce{OHC-C6H4-NO2}$&lt;br&gt;&#xA;  b) $\ce{OHC-C6H4-OCH3}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously, a better hydride releasing group will react faster. Therefore my answer was &lt;strong&gt;b&lt;/strong&gt;, as $\ce{-OCH3}$ shows -I (inductive) effect as well as +R (resonance) effect, as it's in the &lt;em&gt;para&lt;/em&gt; position. This increases the electron concentration in the $\ce{C}$ atom of $\ce{CHO}$ and due to the excess electrons, it is a better donor of an electron to the $\ce{H}$ attached to it. Thus, leaving a  $\ce{H-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer given is &lt;strong&gt;a&lt;/strong&gt;. The answer states that $\ce{C}$ can only give electron to $\ce{H}$ when it accepts an electron from the $\ce{O}$ attached to it. For that the $\ce{C}$ should be electorn deficient. Therefore an electron withdrawing group (in this case $\ce{NO2}$, which shows -R effect at &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions) better releases hydride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think that happens. That's because if $\ce{C}$ is electron deficient, then it wouldn't give an electron to $\ce{H}$ (less electronegative) in the first place, even if it takes an electron from oxygen (which is also tough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give a proper explanation to the whole problem?&lt;/p&gt;&#xA;" OwnerUserId="17011" LastEditorUserId="5026" LastEditDate="2018-03-16T11:42:26.387" LastActivityDate="2020-05-22T19:04:31.353" Title="Which compound reacts faster in the Cannizzaro Reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;carbonyl-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33050" PostTypeId="2" ParentId="33049" CreationDate="2015-06-18T13:21:32.177" Score="3" Body="&lt;p&gt;Looking at the reaction after the fact (always easier!): Two factors favoring compound &lt;strong&gt;a&lt;/strong&gt; are (1) the equilibrium between the aldehyde and its adduct with hydroxide ion will more strongly favor the adduct when it has an electron-withdrawing p-nitro group; (2) once that intermediate is formed, it will be faster to undergo hydride transfer to another molecule of the aldehyde, when the latter has an electron-withdrawing group (nitro). &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-18T13:21:32.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33051" PostTypeId="2" ParentId="33049" CreationDate="2015-06-18T14:45:38.693" Score="23" Body="&lt;p&gt;The mechanism of the Cannizzaro reaction is illustrated below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LAdBr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate.  Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So &lt;strong&gt;we need to look at the step leading to the tetrahedral intermediate&lt;/strong&gt; - the first step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring &lt;strong&gt;can interact&lt;/strong&gt; with the carbonyl through resonance) we see that the nitro compound places positive charge &lt;strong&gt;adjacent&lt;/strong&gt; to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive).  On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fkB8x.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-18T23:24:39.033" LastActivityDate="2015-06-18T23:24:39.033" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="33052" PostTypeId="2" ParentId="33024" CreationDate="2015-06-18T14:47:44.013" Score="5" Body="&lt;p&gt;$\ce{EtOH}$ is a rather bad nucleophile and we have the assistance of a lone pair of the neighboring oxygen atom (the positive charge can hence be delocalized over the carbon and oxygen) which greatly speeds up $\ce{S_{N}1}$ in comparison to $\ce{S_{N}2}$. So what will probably happen is protonation by $\ce{HCl}$ of the alcohol next to the ether oxygen followed by nucleophilic ($\ce{S_{N}1}$) attack of $\ce{EtOH}$ on the formed (delocalized) secondary carbocation. In comparison to the possible reactions at the other secondary centers (having the formulation of the question in mind) this primary reaction will be fast enough to justify the ignoring of them.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-06-18T14:47:44.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33053" PostTypeId="2" ParentId="32980" CreationDate="2015-06-18T15:03:17.903" Score="2" Body="&lt;p&gt;I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total charges need to balance out per &quot;unit cell&quot; (an intermediate-level chemistry concept that means the &quot;smallest repeating subunit of a crystal lattice.&quot;)&#xA;So this means that you need to &lt;strong&gt;balance the total charges&lt;/strong&gt; by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic compounds that have an overall neutral charge are written without the charge superscripts,  thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -&amp;gt; Al_2O_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -&amp;gt; CaBr_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) &lt;em&gt;molecule&lt;/em&gt; instead of an &lt;em&gt;atom&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -&amp;gt; Na_2CO_3$.&lt;/p&gt;&#xA;" OwnerUserId="16814" LastEditorUserId="16814" LastEditDate="2015-06-18T15:14:09.080" LastActivityDate="2015-06-18T15:14:09.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33054" PostTypeId="2" ParentId="33046" CreationDate="2015-06-18T15:17:19.237" Score="2" Body="&lt;p&gt;I don't think polarization is the issue. Adding $\ce{FeBr3}$ is just to reduce the $\ce{Br-Br}$ bond strength, speeding up the reaction (attack of the aromatic ring on the bromine is part of the rate determining step, which is slow in the bromination of benzene). It functions as a catalyst being recycled when the $\ce{Br-}$ grabs a $\ce{H+}$ to restore aromaticity in the ring. I'm pretty sure — at least I see no reason to doubt — that the catalyst requirements for the bromination of anisole and the bromination of phenol do not differ much, if at all. That is, no catalyst is required for the bromination of anisole.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-06-18T15:17:19.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33055" PostTypeId="1" CreationDate="2015-06-18T15:38:24.737" Score="4" ViewCount="171" Body="&lt;p&gt;I understand that elements use their valence electrons for their reactions and whatnot, and that the whole idea of the electron-dot structure is all about valence electrons. But why can't the other electrons play a part in it?&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="10185" LastEditDate="2015-06-18T16:03:12.387" LastActivityDate="2015-06-25T22:31:30.070" Title="Why does an element's chemical properties rely only on its valence electrons and not on anything else?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33056" PostTypeId="1" AcceptedAnswerId="33070" CreationDate="2015-06-18T15:42:18.413" Score="18" ViewCount="3056" Body="&lt;p&gt;Wherever you find potassium, the isotopes are present in a set percentage that exists the same everywhere in nature, but why is that? Does it have something to do with how the element is produced?&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="15588" LastEditDate="2015-06-19T19:53:45.403" LastActivityDate="2015-06-19T19:53:45.403" Title="Is the relative natural abundance of isotopes of an element the same everywhere?" Tags="&lt;elements&gt;&lt;isotope&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33058" PostTypeId="1" AcceptedAnswerId="33063" CreationDate="2015-06-18T15:55:04.550" Score="17" ViewCount="49012" Body="&lt;p&gt;Let us say we have one glass of water and after that when we add one or two spoon of salt then we notice that salt dissolves in it but when we measure the water level we found that there is no difference between initial reading and final reading. So what's happening over there ? Why water level doesn't changed when salt is added ?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastActivityDate="2017-10-25T17:10:49.777" Title="Why there is no change in water level when salt is added?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33059" PostTypeId="2" ParentId="24299" CreationDate="2015-06-18T15:55:54.550" Score="20" Body="&lt;p&gt;Osmium has a bluish-gray tint.&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Osmium#Physical_properties&quot;&gt;Well; slightly.&lt;/a&gt;&lt;br&gt;&#xA;Cesium is silvery-golden!,&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Caesium&quot;&gt;But don't wear it.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15588" LastActivityDate="2015-06-18T15:55:54.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33060" PostTypeId="1" AcceptedAnswerId="33068" CreationDate="2015-06-18T15:57:08.127" Score="3" ViewCount="96" Body="&lt;p&gt;I have a bag of titanium dioxide of pigment quality, unknown source. I am curious to know if the manufacturing process of titanium dioxide makes it likely that it contains any toxic by products / impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it for example be safe to use pigment grade titanium dioxide to manufacture sunblock?&lt;/p&gt;&#xA;" OwnerUserId="17013" LastEditorUserId="10185" LastEditDate="2015-06-18T16:28:47.680" LastActivityDate="2015-06-18T18:26:08.767" Title="Is pigment grade titanium dioxide likely to contain any impurities that are toxic?" Tags="&lt;safety&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33061" PostTypeId="2" ParentId="33055" CreationDate="2015-06-18T16:00:44.313" Score="0" Body="&lt;p&gt;As the valence electrons are the furthest from the nucleus, they are more loosely held by the nucleus. So, they can readily be lost, gained or shared to form ionic or covalent bonds. The inner electrons are tightly held by the nucleus. So, they need higher energy to participate in the formation of chemical bonds.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-06-18T20:39:35.573" LastActivityDate="2015-06-18T20:39:35.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33062" PostTypeId="1" CreationDate="2015-06-18T16:11:49.377" Score="5" ViewCount="233" Body="&lt;p&gt;Why is data about wavelengths of different colors and the visible spectrum in general so different in different sources? On Wikipedia, the numbers differ by up to $\pm\mathrm{30~nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rSlVg.png&quot; alt=&quot;English Wikipedia&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/GQcDr.png&quot; alt=&quot;German Wikipedia&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="4356" LastEditDate="2015-06-18T20:33:00.500" LastActivityDate="2015-06-18T20:33:00.500" Title="Wavelengths of the visible spectrum" Tags="&lt;color&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="33063" PostTypeId="2" ParentId="33058" CreationDate="2015-06-18T17:14:57.977" Score="17" Body="&lt;p&gt;This is an example of electrostriction - see, for example, &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02907145&quot;&gt;this paper&lt;/a&gt; for a full explanation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the solvent molecules become more ordered in the vicinity of dissolved (charged) ions and less ordered as we go further away from the dissolved (charged) ions. The increase of solvent density near the ions is offset by the decrease of bulk density in the solvent as distance from the ions increases. This results in an overall net reduction in the observed volume of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;strong&gt;Discussion&lt;/strong&gt; section of the manuscript:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The interaction of the electrostatic field of an &#xA;  ion with water tends to align the dipolar water &#xA;  molecules in the direction of the field. In this way &#xA;  the field tends to disrupt hydrogen bonded &#xA;  structures in liquid water, and to compress the &#xA;  water molecules surrounding an ion. These &#xA;  electrostatic effect give rise to a shrinkage of the &#xA;  water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-06-18T18:16:57.560" LastActivityDate="2015-06-18T18:16:57.560" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33064" PostTypeId="2" ParentId="33045" CreationDate="2015-06-18T17:25:46.353" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not necessarily. This is more a subjective question involving what tastes good to you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should we choose the bottle of water with the highest concentration of these anions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming the water is actually monitored for safety/quality, the upper limit on caustic (hydroxide) alkalinity is likely too low to do any damage. And carbonate alkalinity will be limited by pH and equilibrium with atmospheric carbon dioxide. So: If it tastes good to you, then drink it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When you boil water, you are increasing the concentration of dissolved species that do not boil off at that temperature. You can, for example, increase levels of nitrate and nitrite in water by boiling it, which is a bad idea if your water is contaminated by large amounts of nitrate/nitrite. Conversely, you can kill some if not all harmful bacteria in water by boiling it. And I don't think that boiling water results in the loss of bicarbonate in any measurable way, at least not on timescales we're interested in (like cooking).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-18T17:25:46.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33065" PostTypeId="1" AcceptedAnswerId="33084" CreationDate="2015-06-18T17:30:42.063" Score="1" ViewCount="989" Body="&lt;p&gt;Alcohol and water are miscible.  I am coming across some chemicals that are soluble in alcohol and insoluble in water.  If I know solubility in two solvents separately, is solubility in a mixture of the two always straightforward, or are there confounding variables?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., if some chemical will dissolve to 100g/L alcohol at 25&amp;deg;C, then will its solubility in a 50/50 mixture by volume of water and alcohol at 25&amp;deg;C &lt;em&gt;always&lt;/em&gt; be 50g/L?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this generalize to all miscible solvents?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2015-06-18T22:00:44.853" LastActivityDate="2015-06-18T22:47:19.487" Title="Solubility in homogenous mixtures of solvents" Tags="&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-06-19T05:51:54.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="33066" PostTypeId="1" CreationDate="2015-06-18T18:14:45.673" Score="6" ViewCount="2994" Body="&lt;p&gt;I'm a bit confused.&#xA;I've got this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007V5TEMW&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;toy&quot;&lt;/a&gt; ...&#xA;with the help of this site I was abled to find out that the reaction stops with a lack of air. Link to the old &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32907/is-this-product-description-wrong-and-a-simple-battery-and-not-a-fuel-cell&quot;&gt;question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The battery/fuel cell or however you call it consists out of following components:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;magnesium plate&lt;/li&gt;&#xA;&lt;li&gt;isolator which consists out of something like paper, the surface feels more like cotton&lt;/li&gt;&#xA;&lt;li&gt;the black plate is called &quot;black cathode&quot; in the manual, my hypothesis is that it might be rubbered graphite or a polymer, because it has a very low electrical resistance.&lt;/li&gt;&#xA;&lt;li&gt;the electrolyte is a $\ce{NaCl}$ dispension- I think it just need to lower $\ce{H2O}$'s electrical resistance&lt;/li&gt;&#xA;&lt;li&gt;plastic stuff is only the case of the &quot;cell&quot;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WviYT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I did some research about this and the results are very confusing to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the manual they call it a fuel cell.&#xA;My chemistry teacher says that in his hypothesis it is impossible to be a fuel cell and has to be a battery. -Because of the great answer to my last question I doubt it or at least I'm too confused to agree.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My research results:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;►Battery:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;news about a battery for &lt;a href=&quot;http://www.electric-vehiclenews.com/2013/01/kist-develop-magnesium-air-battery-with.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cars&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an article about the upper link and about the &lt;a href=&quot;http://www.slimlife.eu/wordpress/2014/09/magnesium-luftbatterie-5x-mehr-energie-als-lithium-ionenbatterien/&quot; rel=&quot;nofollow noreferrer&quot;&gt;toy&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;►Fuel cell:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnesium-air_fuel_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia-MAFT&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a kind of datasheet of a &lt;a href=&quot;http://www.furukawadenchi.co.jp/english/csr/pdf/2014/05_3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;battery&quot;&lt;/a&gt; which is in development&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;►My hypothesis for beeing called fuel cell&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;→ because fuel cells require according to their &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuel_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition&lt;/a&gt; a constant source of fuel, it might be called fuel cell because of the fact that the &quot;thing&quot; requires air to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;►My hypothesis for beeing called battery&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;→ oxygen as fuel?- some people might see it different and call it battery because of that (then there's no constant fuel source →battery) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;►My idea about the chemical equation is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg + H2O → Mg(OH)2 + H2}$  (but I doubt it, because I can't imagine that it is responsible to sell a toy which emitts $\ce{H2}$ - or what else happens with it?-maybe the black cathode has to do something with it?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got a better idea about the chemical equation- let me know.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;I would like to know if there's a mistake in my thoughts- and if it's actually a battery or a fuel cell.(the main problem).&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;And if you've got a better idea about the black cathode component's material- let me know please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;please don't mark it as duplicate to my older question-there's a difference because I added information about the theories and reasearched a lot about it and found the reasons/arguments I've posted here out- the last question was doing it by its components.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2017-04-07T14:53:56.957" Title="Magnesium air-fuel cell or battery?" Tags="&lt;electrochemistry&gt;&lt;fuel&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33067" PostTypeId="1" AcceptedAnswerId="33077" CreationDate="2015-06-18T18:20:31.473" Score="2" ViewCount="1184" Body="&lt;p&gt;Today I was told:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When 2-chlorobutane is reacted with 2-chloropropane in the presence of sodium and dry ether the following products are formed: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2,3-dimethylpentane&lt;/li&gt;&#xA;&lt;li&gt;3,4-dimethylhexane&lt;/li&gt;&#xA;&lt;li&gt;2,3-dimethylbutane&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3,4-dimethylhexane, being most symmetric, is the major product.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not satisfied with the reason provided. Can anyone please explain the fact in a better manner?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I'm unable to understand is that both 3,4-dimethylhexane and 2,3-dimethylbutane and equally symmetric since both have a plane of symmetry, don't they? Then how are we using symmetry to compare the yield?&lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="42741" LastEditDate="2017-05-13T07:50:10.340" LastActivityDate="2017-05-13T07:50:10.340" Title="Deciding the major product in this Wurtz reaction" Tags="&lt;organic-chemistry&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33068" PostTypeId="2" ParentId="33060" CreationDate="2015-06-18T18:26:08.767" Score="1" Body="&lt;p&gt;According to the MSDS for titanium dioxide, the substance itself is suspected of causing cancer. See, for example, &lt;a href=&quot;http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=US&amp;amp;language=en&amp;amp;productNumber=14021&amp;amp;brand=SIAL&amp;amp;PageToGoToURL=http%3A%2F%2Fwww.sigmaaldrich.com%2Fcatalog%2Fproduct%2Fsial%2F14021%3Flang%3Den&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that, in general, it's not a good idea to use reagents from unknown sources for any purpose, especially something like sunblock.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-18T18:26:08.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33069" PostTypeId="1" AcceptedAnswerId="33125" CreationDate="2015-06-18T19:09:53.940" Score="4" ViewCount="6414" Body="&lt;p&gt;I would like to understand how the polyester resin or epoxy catalyst works and to what degree it must be mixed with the base material in order to be effective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical product instructions advise to thoroughly mix the catalyst for a specified period of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How thorough does this really have to be in order to have the material cure properly? Is the mixing done just to initiate a thermosetting behavior and does that just continue to propagate heat to the rest of the material to complete the curing of the entire mass? If this is the case, wouldn't the base material cure if it were heated sufficiently without the presence of a catalyst?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is something else happening when the catalyst begins the cure? And how does mixing effect the quality of the cure?&lt;/p&gt;&#xA;" OwnerUserId="17016" LastActivityDate="2015-09-18T01:12:59.897" Title="How does the catalyst in a polyester resin or epoxy work?" Tags="&lt;polymers&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33070" PostTypeId="2" ParentId="33056" CreationDate="2015-06-18T19:16:19.590" Score="17" Body="&lt;p&gt;The relative natural abundance of isotopes is &lt;strong&gt;not&lt;/strong&gt; the same everywhere.  Depending upon what you mean by &quot;everywhere&quot;, there are two cases to consider.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Extraterrestial&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Dust from before the sun was formed (&lt;a href=&quot;https://en.wikipedia.org/wiki/Presolar_grains&quot; rel=&quot;noreferrer&quot;&gt;stardust, presolar grains&lt;/a&gt;) has a very different elemental and isotopic composition than that found on earth.  Depending where a star is in its life cycle different elements and isotopes will be created and destroyed, so it is not surprising that dust from different stars has a very different composition than what is found on earth. As &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_abundance&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; notes, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In these materials, deviations from &quot;natural abundance&quot; are sometimes&#xA;  measured in factors of 100&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Terrestial&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When our solar system was forming, elements and isotopes were &lt;strong&gt;not uniformly distributed&lt;/strong&gt; - close to uniform, but not exactly uniform.  This is due to processes like diffusion in which a mass-dependent fractionation occurs.  A heavier isotope won't travel as far as a lighter isotope in a given amount of time, so the resulting distribution of the two isotopes will not be exactly equal.  While these mass-dependent processes account for most of the isotopic variation in our solar system, &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass-independent_fractionation&quot; rel=&quot;noreferrer&quot;&gt;mass-independent processes&lt;/a&gt; can play a smaller role.  It is also suggested that catastrophic events in nearby stars may have also influenced local elemental and isotopic homogeneity.  These &quot;original&quot; inhomogeneities will be further altered by a continuation of the mass-dependent (diffusion, bond-breaking and making, etc.) and -independent processes. Natural radioactive decay processes can also lead to changes in local isotope ratios.  For example, $\ce{^238U}$ ultimately decays to $\ce{^206Pb}$, so it wouldn't be surprising to find an altered lead isotope ratio (&lt;em&gt;e.g&lt;/em&gt;. enriched in $\ce{^206Pb}$) around large uranium deposits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ciaaw.org/pubs/SNIF.pdf&quot; rel=&quot;noreferrer&quot;&gt;This paper&lt;/a&gt; presents a study of the natural isotopic variation found on earth.  Here is a table from the paper that gives an idea of the range in variation.  As expected it is much smaller than the extraterrestial variation, but certainly not insignificant.  For boron and copper, the isotopic variation is on the order of parts per hundred, but more typically isotopic variation is on the order of parts per thousand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cgP9v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-18T23:06:52.260" LastActivityDate="2015-06-18T23:06:52.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33071" PostTypeId="2" ParentId="13819" CreationDate="2015-06-18T19:22:26.987" Score="1" Body="&lt;p&gt;I'll take a shot at this, so don't insta-hate me. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way I think I can show you this through is an example.&#xA;Say we have 2 apples in one bin and 3 oranges in another bin. Assuming the $\ce{E}$ is the difference of the apples and the oranges you get $1.00$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Even if we doubled the apples and the oranges the difference would be the same.&lt;/strong&gt; In other words no matter the quantity the voltage will be equivalent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now, you might be thinking: what if we just double the apples.&lt;/strong&gt;&#xA;Well it doesn't matter because we can consider apples and oranges as reactants to a apple-orange pie. So the more apples we have: it would just be excess.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;https://www.khanacademy.org/science/chemistry/oxidation-reduction/cell-potential/v/voltage-as-an-intensive-property#!&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. Explains it beautifully.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-06-19T15:39:55.167" LastActivityDate="2015-06-19T15:39:55.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33073" PostTypeId="2" ParentId="33062" CreationDate="2015-06-18T19:41:58.957" Score="5" Body="&lt;p&gt;The comments on the question discuss this may not be about chemistry.&lt;/p&gt;&#xA;&lt;p&gt;One could argue that colors are chemistry related because&#xA;they are associated with chemical substances: like minerals, chlorophyl, rust, dyes etc.&lt;/p&gt;&#xA;&lt;p&gt;So colors seem to be properties of substances.&lt;/p&gt;&#xA;&lt;p&gt;And that's fundamentally wrong.&lt;/p&gt;&#xA;&lt;p&gt;Color is a function of the visual system of the viewer!&lt;br /&gt;&#xA;If that sounds surprising, think of color blind people, which see different colors.&lt;br /&gt;&#xA;Also, there are not just us - there are other animals, with different visual spectrum, partly including what we call UV.&lt;/p&gt;&#xA;&lt;h2&gt;Colors are not well defined&lt;/h2&gt;&#xA;&lt;p&gt;Actually, there are more than enough ways, even standards, to define which part of the rainbow is green, exactly - people working with visual media like print design can tell details.&lt;br /&gt;&#xA;And there are way more details than anyone could expect.&lt;/p&gt;&#xA;&lt;p&gt;So, without a common definition of the colors, it's not the wavelengths that vary in different sources - &lt;strong&gt;it's the colors themselves&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To illustrate that color is more complex when looking closely, below you see a diagram defining color names. It is part of a formal standard which is in heavy use:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/glolR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;br /&gt;&#xA;From &lt;a href=&quot;https://en.wikipedia.org/wiki/CIE_1931_color_space&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/CIE_1931_color_space&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5078" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-18T20:02:38.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33074" PostTypeId="2" ParentId="33056" CreationDate="2015-06-18T19:55:25.917" Score="8" Body="&lt;p&gt;&lt;strong&gt;Short answer: it's not the same everywhere on earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/33070/5017&quot;&gt;Ron's answer&lt;/a&gt; is excellent, but the main point is that isotopic ratios vary from place to place and over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use these geographic variations with &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotope_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;isotope analysis&lt;/a&gt; to track locations and dates of things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The stable isotopic ratios of drinking water are a function of location, and the geology that the water percolates through. $\ce{^87Sr}$, $\ce{^88Sr}$ and oxygen isotope variations are different all over the world. These differences in isotopic ratio are then biologically 'set' in our hair as it grows and it has therefore become possible to identify recent geographic histories by the analysis of hair strands. For example, it could be possible to identify whether a terrorist suspect had recently been to a particular location from hair analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Similarly, explosives and other compounds are tested for isotopic ratios to determine their origin (e.g., if the material at a crime scene matches the isotopes in a suspect's house).&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-18T19:55:25.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33075" PostTypeId="1" AcceptedAnswerId="33076" CreationDate="2015-06-18T19:57:23.613" Score="5" ViewCount="256" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The acceptable concentration of $\ce{CO}$ in the air is $10\:\mathrm{mg/m}^3$. In&#xA;  a room that is $19\:\mathrm{m\times4.0\:m\times25\:m}$, what is the acceptable mass in&#xA;  kilograms of $\ce{CO}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The room has a volume of $1900\:\mathrm{m}^3$. Since there are $10\:\mathrm{mg}$ of $\ce{CO}$ per $\mathrm{m}^3$, there are $19000\:\mathrm{mg}$ of $\ce{CO}$ in this room. That’s $19\:\mathrm{kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book says the answer is $1.9 \times 10^{-2}\:\mathrm{kg}$. How come?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="5026" LastEditDate="2018-02-22T02:18:18.160" LastActivityDate="2018-02-22T02:18:18.160" Title="Calculating the mass of acceptable carbon monoxide in a room" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33076" PostTypeId="2" ParentId="33075" CreationDate="2015-06-18T20:00:15.320" Score="8" Body="&lt;p&gt;You have to be careful with your conversions - you are missing a step in your conversions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have 19000 mg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As 1 mg = 10&lt;sup&gt;-3&lt;/sup&gt; g  and&#xA;1 g = 10&lt;sup&gt;-3&lt;/sup&gt; kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;19000 mg = 19 g = 1.9 x 10&lt;sup&gt;-2&lt;/sup&gt; kg&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-06-18T20:00:15.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33077" PostTypeId="2" ParentId="33067" CreationDate="2015-06-18T20:16:36.113" Score="4" Body="&lt;p&gt;I think that there should be a 'statistical product distribution', if I'm correct the products should be in the ratios 2:1:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get to this conclusion as follows, call 2-chlorobutane A and 2-chloropropane B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A can react with A to give AA, and&#xA;A can react with B to give AB, while&#xA;B can react with A to give BA, while&#xA;B can react with B to give BB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you realise that AB and BA are the same molecule you get AA:2AB:BB; that is, 1:2:1.&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-18T20:16:36.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33078" PostTypeId="2" ParentId="32962" CreationDate="2015-06-18T20:40:58.920" Score="3" Body="&lt;p&gt;Historical naming conventions dictate that the &quot;-ate&quot; suffix refers to the group containing the cation with a higher oxidation state than that in the corresponding group designated with the &quot;-ite&quot; suffix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of nitrate, nitrogen carries a +5 charge; in nitrite, nitrogen is a +3. For chlorate, chlorine is +5; in chlorite, it's +3. And so on.  &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-18T20:40:58.920" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33079" PostTypeId="2" ParentId="32962" CreationDate="2015-06-18T21:13:28.267" Score="9" Body="&lt;p&gt;&quot;Most common&quot; means the forms most frequently encountered in the entire scope of chemistry, which I'm pretty sure corresponds loosely to the most thermodynamically stable oxidation state. It is not always the case that the '-ate' ion is the most stable, however—as an example, see the links for chlorine below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A detailed answer necessitates noting that the nomenclature extends beyond just '-ate' and '-ite'.  Most p-block elements that form oxo-anions form a whole series of them, each with the central atom in an oxidation state two away from each neighbor.  For all elements except the lightest ($\ce{C}$ and $\ce{N}$) and heaviest ($\ce{As}$, $\ce{Se}$, $\ce{Te}$, etc.) , usually the series is considered to contain &lt;em&gt;four members&lt;/em&gt;, though it is not guaranteed that all are stable or able to be characterized (e.g., &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bromous_acid&quot; rel=&quot;noreferrer&quot;&gt;bromite&lt;/a&gt;&quot; and &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hyposulfite&quot; rel=&quot;noreferrer&quot;&gt;hyposulfite&lt;/a&gt;&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the comments to another answer, in addition to the '-ite' and '-ate' suffixes, there are also the 'hypo-' and 'per-' &lt;em&gt;prefixes&lt;/em&gt;, where 'hypo-' only is used with '-ite', and 'per-' is only used with '-ate'.  These four combinations are used to span the series of oxo-anions for each element:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Phosphorus: &lt;a href=&quot;http://www.endmemo.com/chem/common/perphosphate.php&quot; rel=&quot;noreferrer&quot;&gt;perphosphate&lt;/a&gt; ($\ce{PO5^{3-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphate&quot; rel=&quot;noreferrer&quot;&gt;phosphate&lt;/a&gt; ($\ce{PO4^{3-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphite&quot; rel=&quot;noreferrer&quot;&gt;phosphite&lt;/a&gt; ($\ce{PO3^{3-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypophosphite&quot; rel=&quot;noreferrer&quot;&gt;hypophosphite&lt;/a&gt; ($\ce{H2PO2^{-}}$)&lt;/li&gt;&#xA;&lt;li&gt;Sulfur: &lt;a href=&quot;https://en.wikipedia.org/wiki/Persulfate&quot; rel=&quot;noreferrer&quot;&gt;persulfate&lt;/a&gt; ($\ce{SO5^{2-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfate&quot; rel=&quot;noreferrer&quot;&gt;sulfate&lt;/a&gt; ($\ce{SO4^{2-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfite&quot; rel=&quot;noreferrer&quot;&gt;sulfite&lt;/a&gt; ($\ce{SO3^{2-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyposulfite&quot; rel=&quot;noreferrer&quot;&gt;hyposulfite&lt;/a&gt; (&quot;$\ce{SO2^{2-}}$&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Chlorine: &lt;a href=&quot;https://en.wikipedia.org/wiki/Perchlorate&quot; rel=&quot;noreferrer&quot;&gt;perchlorate&lt;/a&gt; ($\ce{ClO4^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorate&quot; rel=&quot;noreferrer&quot;&gt;chlorate&lt;/a&gt; ($\ce{ClO3^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/?title=Chlorite&quot; rel=&quot;noreferrer&quot;&gt;chlorite&lt;/a&gt; ($\ce{ClO2^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypochlorite&quot; rel=&quot;noreferrer&quot;&gt;hypochlorite&lt;/a&gt; ($\ce{ClO^{-}}$)&lt;/li&gt;&#xA;&lt;li&gt;Bromine: &lt;a href=&quot;https://en.wikipedia.org/wiki/Perbromate&quot; rel=&quot;noreferrer&quot;&gt;perbromate&lt;/a&gt; ($\ce{BrO4^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Bromate&quot; rel=&quot;noreferrer&quot;&gt;bromate&lt;/a&gt; ($\ce{BrO3^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Bromous_acid&quot; rel=&quot;noreferrer&quot;&gt;bromite&lt;/a&gt; (&quot;$\ce{BrO2^{-}}$&quot;), &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypobromite&quot; rel=&quot;noreferrer&quot;&gt;hypobromite&lt;/a&gt; ($\ce{BrO^{-}}$)&lt;/li&gt;&#xA;&lt;li&gt;Iodine: &lt;a href=&quot;https://en.wikipedia.org/wiki/Periodate&quot; rel=&quot;noreferrer&quot;&gt;periodate&lt;/a&gt; ($\ce{IO4^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Iodate&quot; rel=&quot;noreferrer&quot;&gt;iodate&lt;/a&gt; ($\ce{IO3^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Iodous_acid&quot; rel=&quot;noreferrer&quot;&gt;iodite&lt;/a&gt; ($\ce{IO2^{-}}$), &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypoiodous_acid&quot; rel=&quot;noreferrer&quot;&gt;hypoiodite&lt;/a&gt; ($\ce{IO^{-}}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For carbon, '&lt;a href=&quot;https://en.wikipedia.org/?title=Carbonate&quot; rel=&quot;noreferrer&quot;&gt;carbonate&lt;/a&gt;' was given preference, per the 'most common' rubric, as the only known oxo-anion. For nitrogen, the use of the prefixes was avoided, for what I assume was the sake of simplicity.  (I would argue that &lt;a href=&quot;https://en.wikipedia.org/?title=Nitrate&quot; rel=&quot;noreferrer&quot;&gt;nitrate&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/?title=Nitrite&quot; rel=&quot;noreferrer&quot;&gt;nitrite&lt;/a&gt; are both common enough in the natural world that without such a deciding factor there would have been stiff competition for the '-ate' suffix.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible irregularities in structure within a series include peroxo-anions at high numbers of bound oxygens (e.g., I believe &lt;a href=&quot;https://en.wikipedia.org/wiki/Persulfate&quot; rel=&quot;noreferrer&quot;&gt;persulfate&lt;/a&gt; and &quot;&lt;a href=&quot;http://www.endmemo.com/chem/common/perphosphate.php&quot; rel=&quot;noreferrer&quot;&gt;perphosphate&lt;/a&gt;&quot; are both peroxo species) and variable numbers of bound oxygens for a given oxidation state in anions of heavier elements (e.g., metaperiodate, $\ce{IO4-}$, versus orthoperiodate, $\ce{IO6^{5-}}$, both of which contain heptavalent iodine). As well, there is the possibility for other, non-oxo-anion compounds of the central atoms with oxygen such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorine_dioxide&quot; rel=&quot;noreferrer&quot;&gt;chlorine dioxide&lt;/a&gt;, $\ce{ClO2}$; &lt;a href=&quot;https://en.wikipedia.org/?title=Nitrogen_dioxide&quot; rel=&quot;noreferrer&quot;&gt;nitrogen dioxide&lt;/a&gt;, $\ce{NO2}$; sulfur &lt;a href=&quot;https://en.wikipedia.org/?title=Sulfur_dioxide&quot; rel=&quot;noreferrer&quot;&gt;dioxide&lt;/a&gt;, $\ce{SO2}$, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_trioxide&quot; rel=&quot;noreferrer&quot;&gt;trioxide&lt;/a&gt;, $\ce{SO3}$; and (of course) &lt;a href=&quot;https://en.wikipedia.org/?title=Carbon_dioxide&quot; rel=&quot;noreferrer&quot;&gt;carbon dioxide&lt;/a&gt;, $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find references to oxo-anions of &lt;a href=&quot;https://en.wikipedia.org/wiki/Arsenate&quot; rel=&quot;noreferrer&quot;&gt;arsenic&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Selenate&quot; rel=&quot;noreferrer&quot;&gt;selenium&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Antimonate&quot; rel=&quot;noreferrer&quot;&gt;antimony&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Tellurate&quot; rel=&quot;noreferrer&quot;&gt;tellurium&lt;/a&gt; also (links are to the '-ates'), but to my mind, these tend to edge more toward the behavior of oxo-anion-forming metals, which either only have a single appreciably stable oxo-anion (&lt;a href=&quot;https://en.wikipedia.org/wiki/Chromate_and_dichromate&quot; rel=&quot;noreferrer&quot;&gt;chromate&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Molybdate&quot; rel=&quot;noreferrer&quot;&gt;molybdate&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tungstate&quot; rel=&quot;noreferrer&quot;&gt;tungstate&lt;/a&gt;, etc.), or break this (hypo-)-ite/(per-)-ate paradigm pretty badly (see, e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Permanganate&quot; rel=&quot;noreferrer&quot;&gt;permanganate&lt;/a&gt;, $\ce{MnO4-}$, versus &lt;a href=&quot;https://en.wikipedia.org/?title=Manganate&quot; rel=&quot;noreferrer&quot;&gt;manganate&lt;/a&gt;, $\ce{MnO4^{2-}}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to blow your mind, take a look at the &lt;em&gt;polymeric&lt;/em&gt; oxo-anions like &lt;a href=&quot;https://en.wikipedia.org/?title=Polyphosphate&quot; rel=&quot;noreferrer&quot;&gt;polyphosphate&lt;/a&gt;; or at the at-present theoretical-only &lt;a href=&quot;https://en.wikipedia.org/wiki/Orthocarbonic_acid&quot; rel=&quot;noreferrer&quot;&gt;orthocarbonate&lt;/a&gt;; or at iron, for which apparently all three of the known oxo-anions are referred to as '&lt;a href=&quot;https://en.wikipedia.org/wiki/Ferrate(VI)#Nomenclature&quot; rel=&quot;noreferrer&quot;&gt;ferrate&lt;/a&gt;'.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2017-06-16T16:27:47.500" LastActivityDate="2017-06-16T16:27:47.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33080" PostTypeId="2" ParentId="32557" CreationDate="2015-06-18T21:48:19.647" Score="9" Body="&lt;p&gt;&lt;strong&gt;Prologue&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (&lt;strong&gt;1978&lt;/strong&gt;): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think a bit about your problem first to simplify it from scratch:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.&lt;/li&gt;&#xA;&lt;li&gt;The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and&lt;/li&gt;&#xA;&lt;li&gt;the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If we take a look at the titration curve of phosphoric acid, we can see what is meant by &quot;too high to be determined&quot;. You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3fmKC.png&quot; alt=&quot;titration curve of phosphoric acid with a sodium hydroxide solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition &lt;em&gt;(Protonen-Bedingung)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration)&#xA;$$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&amp;amp;~0\\c(HA)+c(A^-) =&amp;amp;~C(HA)\end{align} \tag{2.1}$$&#xA;needs to be solved to give&#xA;$$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done analogously for the other acid and also for the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining everything so far we get the following proton condition:&#xA;$$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as&#xA;$$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get:&#xA;$$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give&#xA;$$\begin{align}C(HA) \rightarrow&amp;amp;~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&amp;amp;~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following figure shows some examples of your mixture using equation 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CXiU9.png&quot; alt=&quot;titration of a mixture of two acids with a sodium hydroxide solution&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-02T13:49:19.253" LastActivityDate="2015-07-02T13:49:19.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33081" PostTypeId="1" CreationDate="2015-06-18T21:56:31.543" Score="3" ViewCount="698" Body="&lt;p&gt;As per one of the questions in this site $$\ce{NH3 + H2O -&amp;gt; NH4 + OH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, per another perspective in absorption refrigeration&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mixture is placed in the same container. Then when it is heated the mixture is separated and $\ce{NH3}$ changes phase, becomes a gas which moves further to cooling, while water is trapped down in the reservoir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which one is correct?&lt;/p&gt;&#xA;" OwnerUserId="17018" LastEditorUserId="7448" LastEditDate="2015-06-19T11:02:48.190" LastActivityDate="2015-06-19T11:02:48.190" Title="How does Ammonia and water mixture works in an absorption refrigeration?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33082" PostTypeId="1" CreationDate="2015-06-18T21:59:25.853" Score="9" ViewCount="605" Body="&lt;p&gt;There exists &lt;a href=&quot;http://www.chem.ucalgary.ca/SHMO/&quot; rel=&quot;noreferrer&quot;&gt;a free online orbital calculator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I draw cyclopropane it plots three molecular orbitals, but unfortunately it doesn't use the Walsh orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any free online tools which can solve such the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It would be very nice if it would also be an android app. Please let me know if you know one.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The section below is for interested chemists&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I want to do that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclopropane can, to a good approximation, be described with the following orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H6Zjy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was how the molecule orbitals behave if we add an electron withdrawing group (a $\pi$ acceptor because some people call it the &quot;CXZ model&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNUGJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the substituted case all energy levels are lying more down in energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't agree because I only see overlap with the Walsh Möbius set of $sp$ orbitals in the following sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHphM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hückel set (a simpler way to put it: the energy lowest orbital in the scheme of cyclorpropane; used above) gives no overlap with the $p$ orbitals of an electron withdrawing group, e.g. X=COOH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I see from the orbitals' interaction, I believe the solution is wrong, I don't agree that all orbitals are lowered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sorry for the mistake in the sketch, there is of course only one of the two degenerate Möbius orbitals lowered.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;I guess COOH not only has a $\pi$ acceptor effect but also a $\sigma$ acceptor effect which could lower all orbitals.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7201" LastEditDate="2015-06-19T13:17:29.190" LastActivityDate="2015-06-19T13:17:29.190" Title="Free Web-Based Orbital Solver to look into the Walsh orbitals of a perturbed cyclopropane" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33083" PostTypeId="2" ParentId="33058" CreationDate="2015-06-18T22:18:59.273" Score="16" Body="&lt;p&gt;Actually, sodium chloride added to water will decrease the volume of the solution, up to around 2% for a saturated solution. Even when fully saturated, that's not a big change, so you may not have been able to observe it without something with a narrow neck like a volumetric flask. If you start with a large volume and carefully mark the level, you should be able to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Todd's answer explains why this happens with water, but it should also be noted that deviations from the ideal can happen in both directions (increasing or decreasing the volume) and changes in volume also occur when mixing two miscible liquids together, sometimes in a rather complex way. A good example is a binary mixture of acetone and chloroform (there's a good figure for this in the Engel and Reid physical chemistry textbook, but I can't find a decent one online). Starting from a pure acetone mixture, as we add chloroform, the volume is more than the sum of the two component volumes, but once we get above 20% chloroform or so, the volume drops below the sum of the two pure component volumes. Why this happens is a bit complicated, but it has to do with different competing factors affecting how compact the local structure around the molecules is.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-18T22:18:59.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33084" PostTypeId="2" ParentId="33065" CreationDate="2015-06-18T22:47:19.487" Score="2" Body="&lt;p&gt;The complication comes from solubility of the solvents with each other whilst simultaneously having some solvation power of a solute. I have made a compound before that was so water soluble that it caused sodium chloride to precipitate out. A real example was of an organic compound that had 1%w/v solubility in ethanol but around 6.5%w/v in toluene. In a 50:50 mixture,  the compound had solubility of 13%w/v. A completely non-linear result! The solubility was even higher in 90:10 toluene-ethanol. So, the complex nature of three way interactions mean it would be very difficult to predict these solubility graphs just using two data points of the two pure solvents. &#xA;If we asked the same question of boiling points, we would point straightaway to azeotropes. So, analogously, other properties exhibit this behaviour. However,  in this instance the nature of the solute as a third component makes it too complicated to predict without having the wealth of property data available for all components. If this is known and you have the software, you may have a chance. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-06-18T22:47:19.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33086" PostTypeId="1" AcceptedAnswerId="33087" CreationDate="2015-06-19T00:30:10.587" Score="2" ViewCount="391" Body="&lt;p&gt;Suppose I have a solid compound from which I want to extract a target chemical I know is soluble in acetone.  So I crush and mix the compound in acetone, dissolving the target and suspending the detritus.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose.  Are commercial coffee filters considered &quot;lab-grade?&quot;  Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone.  Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2015-06-19T01:21:04.520" Title="How to perform solvent extraction of a solid" Tags="&lt;experimental-chemistry&gt;&lt;solvents&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33087" PostTypeId="2" ParentId="33086" CreationDate="2015-06-19T01:21:04.520" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose. Are commercial coffee filters considered &quot;lab-grade?&quot; Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The method that comes to mind for this task is a Soxhlet extraction. Commercial coffee filters are not &quot;lab-grade.&quot; Soxhlet extraction thimbles are available in &lt;a href=&quot;http://www.sigmaaldrich.com/labware/glassware-catalog/extractors-thimbles.html&quot; rel=&quot;nofollow&quot;&gt;cellulose&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone. Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes - rotary evaporation, in conjunction with Soxhlet extraction.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-19T01:21:04.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33088" PostTypeId="2" ParentId="33081" CreationDate="2015-06-19T01:25:40.637" Score="1" Body="&lt;p&gt;The difference between the absorption refrigerator and the more common refrigerator is that there is no compressor in the absorption refrigerator.  The Absorption refrigerator still needs a way to convert the evaporated refrigerant back to a liquid state.  Instead of compressing it in one step, the evaporated refrigerant is absorbed by a liquid.  The liquid is heated and the refrigerant evaporates from the liquid.  The refrigerant is cooled in a heat exchanger and condensed.  It is now a liquid and ready to go back through the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation you have written is a reaction between ammonia and water.  This reaction will occur in an absorption refrigerator that uses water as the absorber and ammonia as the refrigerant.  The equilibrium constant of the reaction is low so it is not accounted for in a discussion of how an absorption refrigerator works.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-06-19T01:25:40.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33089" PostTypeId="1" CreationDate="2015-06-19T01:37:55.707" Score="2" ViewCount="1425" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Absorption_refrigerator#Simple_salt_and_water_system&quot; rel=&quot;nofollow&quot;&gt;Lithium bromide refrigerator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is lithium bromide used in absorption refrigerators?  It is more expensive than sodium chloride.  The wikipedia article on lithium bromide says it is extremely hygroscopic.  However, it is less soluble in water than NaCl.  I am thinking that a salt capable of generating a more concentrated solution would work better in the absorption refrigerator as it would increase the propensity of water to leave the gas phase and enter the solution.  Am I wrong about that?  Is the affinity of water for lithium bromide so high that it is the most important factor?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-06-19T16:56:18.327" Title="Why is lithium bromide used in absorption refrigerators?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33093" PostTypeId="1" AcceptedAnswerId="33098" CreationDate="2015-06-19T04:56:06.920" Score="2" ViewCount="5005" Body="&lt;p&gt;Can all species that undergo aldol reactions also undergo Cannizzaro reactions? Ater all, the nucleophile can always attack the carbonyl group.&lt;/p&gt;&#xA;" OwnerUserId="16618" LastEditorUserId="9961" LastEditDate="2018-03-01T15:40:35.350" LastActivityDate="2018-03-01T15:40:35.350" Title="Can all species that undergo aldol reactions also undergo Cannizzaro reactions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33094" PostTypeId="2" ParentId="33066" CreationDate="2015-06-19T06:15:16.140" Score="5" Body="&lt;p&gt;What you have is classified as a Mg-air battery. You can read more about it here : &lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Metal-air_batteries&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Category:Metal-air_batteries&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are a class of primary(i.e. not rechargeable) batteries, where the anodic reaction is a metal being oxidised and the cathodic reaction is oxygen reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Mg → Mg^{2+} + 2 e^{-}}$ on the anode&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{O2 + 4 e^{-} + 4 H+ → 2 H2O }$ on the cathode, if the reduction goes to completion. Most of the time, it is only reduced to the peroxide or the superoxide which is considered incomplete reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For it to be classified as a fuel cell, the fuel for the anodic reaction would need be brought from the outside(source: anectodal, personal thought). &lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-06-19T06:15:16.140" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33095" PostTypeId="1" CreationDate="2015-06-19T09:28:46.727" Score="0" ViewCount="7805" Body="&lt;p&gt;How does sulfur has atomicity 8, when its valence electrons are 6 like oxygen- ie even 2 sulfur atoms can share a pair of electrons and achieve the octet rule, why should it involve 8 atoms?&lt;/p&gt;&#xA;" OwnerUserId="17025" LastEditorUserId="402" LastEditDate="2016-12-04T17:20:56.693" LastActivityDate="2016-12-04T17:20:56.693" Title="How does sulfur has atomicity 8" Tags="&lt;atoms&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2015-06-19T18:54:39.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="33096" PostTypeId="2" ParentId="33095" CreationDate="2015-06-19T10:00:26.577" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Octasulfur&quot; rel=&quot;nofollow&quot;&gt;Octasulfur&lt;/a&gt; is just one of few different &lt;a href=&quot;https://en.wikipedia.org/wiki/Allotropes_of_sulfur&quot; rel=&quot;nofollow&quot;&gt;allotropic forms&lt;/a&gt; of sulfur. It happens to be the most stable form at room temperature, but different conditions favor different forms. Thus, for instance, sulfur vapor usually contains $\ce{S, S2, S4, S6, S8}$, and the higher the temperature, the lower is the amount of $\ce{S8}$ molecules since they break up into smaller allotropes of sulfur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while two sulfur atoms can share a pair of electrons by forming a diatomic &lt;a href=&quot;https://en.wikipedia.org/wiki/Disulfur&quot; rel=&quot;nofollow&quot;&gt;disulfur&lt;/a&gt; molecule with a double bond, so can also eight atoms, but by forming a puckered ring structure with few single bonds. Then it is basically the question of thermodynamic and kinetic stability, which form would dominate others at some particular conditions.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-19T15:40:01.267" LastActivityDate="2015-06-19T15:40:01.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33098" PostTypeId="2" ParentId="33093" CreationDate="2015-06-19T13:22:39.843" Score="4" Body="&lt;p&gt;No. The Cannizzaro reaction only works for aldehydes with no $\alpha$-protons, benzaldehyde being a classic example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a start, ketones and any other species that can form stable enolates, such as esters or carboxylic acids, can undergo the aldol reaction or similar reactions like the Claisen condensation. However, the Cannizzaro reaction is limited to aldehydes because only they can disproportionate to form alcohols and carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bv8V2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the Cannizzaro reaction involves strongly basic conditions, typically using sodium or potassium hydroxide. Under these conditions an aldehyde with $\alpha$-protons will undergo deprotonation to form an enolate much more readily than nucleophilic attack on the the acyl carbon by a hydroxide ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iWYFz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-19T13:22:39.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33100" PostTypeId="1" AcceptedAnswerId="62962" CreationDate="2015-06-19T13:39:15.693" Score="8" ViewCount="1236" Body="&lt;p&gt;I am investigating potential energy surface characteristics of some electrostatic monomeric interactions e.g. $\ce{F-H}$...$\ce{F-H}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using DFT and SCS-MP2 to perform the necessary calculations for single point energies. Is it possible to compare single point energy with the electrostatic dimerization energy of the monomers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concepts or references will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16786" LastEditorUserId="194" LastEditDate="2016-11-19T17:45:08.680" LastActivityDate="2016-11-20T03:49:26.673" Title="Difference between single point energy and dimerization energy" Tags="&lt;computational-chemistry&gt;&lt;energy&gt;&lt;reference-request&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33101" PostTypeId="1" AcceptedAnswerId="33112" CreationDate="2015-06-19T13:53:25.967" Score="2" ViewCount="1836" Body="&lt;p&gt;Can the structure of an azide ion ($\ce{N3-}$) be $\ce{N#N\bond{-&amp;gt;}N^-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the actual structure but was asking whether the single bond can be shifted to a N to make it a triple bond and a coordinate bond with other Nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a duplicate of &lt;em&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/19904&quot;&gt;Hybridization in azide ion?&lt;/a&gt;&lt;/em&gt; because I am asking whether it can have a coordinate bond or not while the other question asks what the hybridization of centre nitrogen in $\ce{N3-}$ is.&lt;/p&gt;&#xA;" OwnerUserId="17027" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2016-12-26T13:34:49.520" Title="Can the structure of an azide ion contain a coordinate bond?" Tags="&lt;ions&gt;&lt;terminology&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33102" PostTypeId="2" ParentId="31917" CreationDate="2015-06-19T14:07:25.217" Score="2" Body="&lt;p&gt;With constant volume heat capacity, it is defined as, for an ideal gas, that has the internal energy equation&#xA;$$dU=TdS-\rho dV= \frac{\nu}{2} Nk_{\beta} T\ (1)$$ which is the Ideal Gas Law, obtained from the first law of thermodynamics. Using moles, (1) is equivalent to&#xA;$$ \frac{\nu}{2}nRT $$&#xA;Now, onto heat capacities. &#xA;$C_V$ is defined as,&#xA;$$C_V=(\frac{\partial U}{\partial T})_V= \frac{\partial}{\partial T}\left(\frac{\nu nRT}{2}\right)=\frac{\nu}{2} nR $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constant pressure heat capacity is defined as&#xA;$$C_{\rho}=C_V + nR=\frac{\nu+2}{2} nR$$, where $n$ is the amount of substance and $R$ is the universal gas constant. $\nu$ is the degrees of freedom of the molecule, which is easily obtained by thinking about a monoatomic ideal gas and the ways it can move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is some fun theory surrounding the heat capacity for constant volume.&#xA;Note that constant-pressure specific heat is also defined as&#xA;$$C_{\rho}=(\frac{\partial U}{\partial T})_{\rho}=dU+\rho dV+Vd\rho=&#xA;(\frac{\partial H}{\partial T})  $$&#xA;Where $H$ is the enthalpy of heating the gas, which is &#xA;$$\mathrm{d}H=\mathrm{d}U+p\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S-\rho\mathrm{d}V+\rho\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S+V\mathrm{d}\rho$$&#xA;Enthalpy is commonly defined with a constant pressure process, thus $\mathrm{d}\rho=0$.&#xA;So, for a constant pressure, we have $$\mathrm{d}H=T\mathrm{d}S$$,&#xA;exploting the definition of entropy for a reversible change,&#xA;$$\mathrm{d}S=\frac{\mathrm{d}Q}{T} \rightarrow \mathrm{d}Q=T\mathrm{d}S $$&#xA;so, we finally get $$\mathrm{d}H=\mathrm{d}Q$$&#xA;which is a standard thermodynamic identity, and can also be used to get to the answer at the beginning of the post (though it is admittedly longer and more complicated to do so.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope these hints help you!&lt;/p&gt;&#xA;" OwnerUserId="7600" LastEditorUserId="7600" LastEditDate="2015-06-27T21:01:11.530" LastActivityDate="2015-06-27T21:01:11.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33104" PostTypeId="1" CreationDate="2015-06-19T14:19:46.490" Score="3" ViewCount="101" Body="&lt;p&gt;If we have an isolated pentane molecule at room temperature (RT) the Boltzmann energy $(E=k_\mathrm{B}T)$ is approxately $0.59\ \mathrm{kcal/mol} \overset{\wedge}{=} 207\ \mathrm{cm^{−1}}$. There is not enough energy for an electronic or vibrational excitation, so the molecule is in its electronic and vibratory ground state. However usually 10–20 rotational states are populated at RT. Can we attribute mathematically (to a first approximation) this $207\ \mathrm{cm^{−1}}$ amount of energy to $x\:\%$ translational energy ($E_\mathrm{kin} = \frac{m v^2}{2}$) and to $(100-x)\:\%$ rotational energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is related to the question about the internal degrees of freedom for a molecule: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29688/rotational-degrees-of-freedom-3n-5-and-3n-6/30771?noredirect=1#comment49303_30771&quot;&gt;Rotational degrees of freedom (3N-5 and 3N-6)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-06-20T10:34:39.687" Title="How to distribute the Boltzmann energy to internal and external motion?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33105" PostTypeId="1" AcceptedAnswerId="33126" CreationDate="2015-06-19T14:43:56.997" Score="14" ViewCount="4717" Body="&lt;p&gt;Both &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Amylose2.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;amylose&lt;/a&gt; and &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Cellulose_Sessel.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;cellulose&lt;/a&gt; have the same &quot;monomer&quot; structure (glucose), so what makes them look/form so differently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BuVqL.png&quot; alt=&quot;Amylose&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AohnP.png&quot; alt=&quot;Cellulose&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorDisplayName="user7951" LastEditDate="2017-01-13T17:09:34.673" LastActivityDate="2017-01-13T17:09:34.673" Title="Why starch (amylose) and cotton (cellulose) are so different?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;polymers&gt;&lt;carbohydrates&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33106" PostTypeId="2" ParentId="29688" CreationDate="2015-06-19T14:55:07.543" Score="6" Body="&lt;p&gt;Rotations around bonds are typically termed &quot;internal rotations&quot;, and represent one of the most common problematic cases for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotational_spectroscopy#Classification_of_molecular_rotors&quot; rel=&quot;nofollow&quot;&gt;rigid-rotor&lt;/a&gt;-&lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_vibration&quot; rel=&quot;nofollow&quot;&gt;harmonic-oscillator&lt;/a&gt; (RRHO) model of internal molecular motion.  This is because RRHO assumes that any vibrational amplitudes are &quot;small,&quot; and internal rotations are most definitely &lt;em&gt;not&lt;/em&gt; small!  Such rotations still involve the internal degrees of freedom of the molecule, though, and thus (as Jan notes) are considered as part of the 'vibration' of the molecule, not its 'rotation.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal rotations typically have some energy cost involved (as in your pentane example), and so cannot be treated as &quot;free&quot; rotations. They are usually termed &quot;hindered rotations,&quot; and there is a great deal of literature studying them.  Some citations I know offhand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Classic &lt;a href=&quot;http://dx.doi.org/10.1063/1.1750058&quot; rel=&quot;nofollow&quot;&gt;Pitzer&lt;/a&gt; (&lt;em&gt;J Chem Phys&lt;/em&gt; &lt;strong&gt;5&lt;/strong&gt; 469, 1937), and &lt;a href=&quot;http://dx.doi.org/10.1063/1.1723744&quot; rel=&quot;nofollow&quot;&gt;Pitzer &amp;amp; Gwinn&lt;/a&gt; (&lt;em&gt;J Chem Phys&lt;/em&gt; &lt;strong&gt;10&lt;/strong&gt; 428, 1942)&lt;/li&gt;&#xA;&lt;li&gt;Independent-rotations approximation for treating hindered internal rotation: &lt;a href=&quot;http://dx.doi.org/10.1007/s00214-007-0376-5&quot; rel=&quot;nofollow&quot;&gt;Pfaendtner&lt;/a&gt; (&lt;em&gt;Theor Chem Acc&lt;/em&gt; &lt;strong&gt;118&lt;/strong&gt; 881, 2007)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other / more-advanced treatments of internal rotation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.1743262&quot; rel=&quot;nofollow&quot;&gt;Hecht&lt;/a&gt; (&lt;em&gt;J Chem Phys&lt;/em&gt; &lt;strong&gt;26&lt;/strong&gt; 31, 1957)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.chemphys.2008.11.003&quot; rel=&quot;nofollow&quot;&gt;Strekalov&lt;/a&gt; (&lt;em&gt;Chem Phys&lt;/em&gt; &lt;strong&gt;355&lt;/strong&gt; 62, 2009)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.2161218&quot; rel=&quot;nofollow&quot;&gt;Vansteenkiste&lt;/a&gt; (&lt;em&gt;J Chem Phys&lt;/em&gt; &lt;strong&gt;124&lt;/strong&gt; 044314, 2006)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1063/1.473664&quot; rel=&quot;nofollow&quot;&gt;McClurg&lt;/a&gt; (&lt;em&gt;J Chem Phys&lt;/em&gt; &lt;strong&gt;106&lt;/strong&gt; 6675, 1997)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One species of particular interest to me on this topic has been nitromethane, which has an extraordinarily small barrier to rotation around the $\ce{C}-\ce{N}$ bond (see Strekalov, above), such that at ambient temperature it can be considered essentially a free rotation.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-06-19T15:41:27.683" LastActivityDate="2015-06-19T15:41:27.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33110" PostTypeId="1" AcceptedAnswerId="33118" CreationDate="2015-06-19T15:27:58.833" Score="2" ViewCount="3592" Body="&lt;p&gt;The rise in pH can change the protein conformation by changing the ionic interactions and hydrogen bonds between two polypeptide chains. This is a fact that I understand very well because the change in amino-acid ionization with pH. I would like to know if by changing the pH value the hydrophobic interactions between two chains can also be changed. If yes, what is the mechanism?&#xA;&lt;a href=&quot;http://www.jbc.org/content/261/23/10891.full.pdf&quot; rel=&quot;nofollow&quot;&gt;Article: a pH Induced increase in hydrophobicity...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9475" LastEditorUserId="4665" LastEditDate="2015-06-19T16:22:54.413" LastActivityDate="2015-06-19T17:52:52.743" Title="pH effect in the hydrophobic interactions between two polypeptide chains" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33112" PostTypeId="2" ParentId="33101" CreationDate="2015-06-19T16:07:48.363" Score="-3" Body="&lt;p&gt;No. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with using your proposal is very simple: $\ce{N_2}$ sucks at being reactive. It doesn't want to give up or share its electrons with any other atom. In fact, the azide ion is usually explosive &lt;em&gt;because&lt;/em&gt; it keeps trying to form nitrogen gas. $\ce{N_2}$ has no real polarity and no particular motive to coordinate to anything, a single N included.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-19T16:07:48.363" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="33113" PostTypeId="2" ParentId="33105" CreationDate="2015-06-19T16:25:03.190" Score="6" Body="&lt;p&gt;You can also add glyocogen to that list, a glycan found in us The difference is that cellulose h&lt;a href=&quot;http://www.brooklyn.cuny.edu/bc/ahp/LAD/C4c/C4c_polysaccharides.html&quot;&gt;as a linear structure, allowing parallel chains to lie in close proximity and for hydrogen bonding to predominate, strengthening the structure.&lt;/a&gt;. Starch, on the other hand, has side branches preventing as many hydrogen bonds from forming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the site above shows very similar structures for starch and cellulose. The illustrations you've shown are using different methods to display the structure, emphasizing the cyclic nature in the first image, and the bond angles in the second. It would be interesting to see these molecules in &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_force_microscopy&quot;&gt;atomic force microscopy&lt;/a&gt;, rather than in our conventional diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, consider the different structure of carbon in diamond and graphite, where there is little bonding between layers of graphite, but diamond has three-dimensional strength.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-19T16:25:03.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33114" PostTypeId="1" AcceptedAnswerId="33115" CreationDate="2015-06-19T16:28:42.603" Score="16" ViewCount="53096" Body="&lt;p&gt;So I was &amp;quot;happily&amp;quot; doing organic chemistry homework when I came across this question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}(\mathbf{a})\ &amp;amp;\ \ce{H2\overset{+}{C}-\overset{+}{C}H-CH=CH-O-CH3} &amp;amp; (\mathbf{b})\ &amp;amp;\ \ce{H2\overset{+}{C}-CH=CH-CH=\overset{+}{O}-CH3} \\&#xA;(\mathbf{c})\ &amp;amp;\ \ce{H2\overset{+}{C}-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{C}H-O-CH3} &amp;amp; (\mathbf{d})\ &amp;amp;\ \ce{H2C=CH-\overset{+}{C}H-CH=\overset{+}{O}-CH3}\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I decided to take a look through my textbook to see what I was missing. The rules to compare resonance structure energies (and therefore determine relative contributions) were in my textbook.  It said:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The following structures are considered to contribute most:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Structures having filled octet a for second row elements (C, N, O, F) are stable.&lt;/li&gt;&#xA;&lt;li&gt;Structures having minimum number of formal charges and maximum number of bonds.&lt;/li&gt;&#xA;&lt;li&gt;Structure in which negative charge appears on the most electronegative atom (C &amp;lt; N &amp;lt; O).&lt;/li&gt;&#xA;&lt;li&gt;Structure in which there is minimal charge separation while keeping the formal charges closer together.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Instead of answering my initial question, two more questions came to mind. They are (in order):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why do the octets of only the second row elements have to be filled in for greater relative contribution?&#xA;I suppose it has something to do with the greater electronegativity values of the elements, but would still like some clarification.&lt;/li&gt;&#xA;&lt;li&gt;From option (&lt;strong&gt;a&lt;/strong&gt;) and (&lt;strong&gt;c&lt;/strong&gt;) in my question, how did the two positive charges get next to each other?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Resonance structures, as stated in my textbook, can be represented by two or more Lewis structures that differ only in the positions of the electrons.&#xA;The options clearly do not just redistribute the electrons. Or at least I don't think so.&lt;/p&gt;&#xA;&lt;p&gt;So if I had to guess, the answer would be (&lt;strong&gt;b&lt;/strong&gt;) because the terminal carbocation is worse than an internal carbocation.&lt;/p&gt;&#xA;&lt;p&gt;But it turns out, the correct answer is (&lt;strong&gt;c&lt;/strong&gt;). And I think that it shouldn't even be considered a resonating structure.&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2020-02-06T16:33:17.473" Title="How to determine the worst resonance structures out of a given set?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;resonance&gt;&lt;ethers&gt;" AnswerCount="4" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33115" PostTypeId="2" ParentId="33114" CreationDate="2015-06-19T16:48:47.297" Score="9" Body="&lt;p&gt;(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the compound in the answers isn't the same as the compound named in the question, like ron said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-06-19T16:58:43.043" LastActivityDate="2015-06-19T16:58:43.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33116" PostTypeId="2" ParentId="33089" CreationDate="2015-06-19T16:56:18.327" Score="1" Body="&lt;p&gt;Your summary is correct; &lt;a href=&quot;https://en.wikipedia.org/wiki/Solubility_table&quot; rel=&quot;nofollow&quot;&gt;$\ce{NaCl}$ is more &lt;em&gt;soluble&lt;/em&gt; than $\ce{LiCl}$ or $\ce{CaCl2}$&lt;/a&gt;, for that matter, but both $\ce{LiCl}$ or $\ce{CaCl2}$ are far  more &lt;a href=&quot;https://en.wikipedia.org/wiki/Hygroscopy&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;hygroscopic&lt;/em&gt;&lt;/a&gt;. The &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/compounds/faq/why-hygroscopic.shtml&quot; rel=&quot;nofollow&quot;&gt;explanation of hygroscopicity&lt;/a&gt; involves various mechanisms, including adsorption (surface forces) and absorption (requiring transport to the interior of the substance) due to hydrogen bonding, coordinate covalent bonding etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are distinct, though related, properties: solubility (limit on how &lt;em&gt;much&lt;/em&gt; solute the solvent holds) and hygroscopicity (how &lt;em&gt;avidly&lt;/em&gt; the solute attaches to the initial molecules of solvent).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that reason, $\ce{CaCl2}$ is used to absorb water in damp closets, though it will eventually &lt;a href=&quot;https://en.wikipedia.org/wiki/Hygroscopy#Deliquescence&quot; rel=&quot;nofollow&quot;&gt;deliquesce&lt;/a&gt; into a sodden mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, Isaac Asimov wrote of a compound so hygroscopic that it dissolved &lt;em&gt;before&lt;/em&gt; it hit water: &lt;strong&gt;T&lt;a href=&quot;http://io9.com/5887014/meet-thiotimoline-the-chemical-compound-isaac-asimov-invented-to-spoof-boring-science-writing&quot; rel=&quot;nofollow&quot;&gt;he Endochronic Properties of Resublimated Thiotimoline&lt;/a&gt;&lt;/strong&gt;, often &lt;a href=&quot;https://www.aacc.org/~/media/files/meetings-and-events/resources-from-past-events/conferences/2012/mass-spectrometry/sep-6/interactive_session_handout_sep_6_2012.pdf?la=en&quot; rel=&quot;nofollow&quot;&gt;cited&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-19T16:56:18.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33117" PostTypeId="2" ParentId="10478" CreationDate="2015-06-19T17:03:40.257" Score="0" Body="&lt;p&gt;To preserve symmetry shape lanthanides are placed below main table .actually vertical elements are similar and horizontal elements are dis similar .but lanthanides shows horizontal chemical similarities so placed in f block &lt;/p&gt;&#xA;" OwnerUserId="17035" LastActivityDate="2015-06-19T17:03:40.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33118" PostTypeId="2" ParentId="33110" CreationDate="2015-06-19T17:52:52.743" Score="3" Body="&lt;p&gt;Here is one possible mechanism:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In tightly folded, compact, &quot;globular&quot; protein structures, the most hydrophobic bits of the protein tend to be in the core of the structure, away from the surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pH changes, then by mechanisms you mention in your question, this structure can be disrupted.  If it is side chain interactions among surface-oriented, polar residues that &quot;hold together&quot; the globular structure, then a pH change can disrupt those interactions, allowing the previously isolated hydrophobic &quot;core&quot; to become  exposed.  This core can now bind more surfactant molecules, aggregate with itself, or do any number of other &quot;hydrophobic&quot; things.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-19T17:52:52.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33119" PostTypeId="1" CreationDate="2015-06-19T18:01:07.893" Score="3" ViewCount="2523" Body="&lt;p&gt;Do crystals have stronger intermolecular forces than amorphous solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the properties of amorphous solids vs. crystals explained by their structures? &lt;/p&gt;&#xA;" OwnerUserId="14481" LastEditorUserId="402" LastEditDate="2015-10-12T07:15:55.713" LastActivityDate="2016-06-09T02:35:03.837" Title="Do crystals have stronger intermolecular forces than amorphous solids?" Tags="&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33120" PostTypeId="2" ParentId="32079" CreationDate="2015-06-19T19:04:46.097" Score="8" Body="&lt;p&gt;&lt;strong&gt;How Did Wittig Come to Discover Lithium-Halogen-Exchange?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Georg Wittig, discoverer of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wittig_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wittig reaction&lt;/a&gt;, was studying the effect of ring strain on the strength of two neighboring carbons; by adding enough strain he expected to coax C-C bonds into forming diradicals:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/shpNn.png&quot; alt=&quot;diyl&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately tetraphenylbenzocyclobutane rearranges to triphenyldihydroanthrancene instead of forming a diyl. This and other “failures” did not deter Wittig; they elected to continue by replacing phenyl groups with anisyls. To their surprise, reactions with anisyls gave unexpected products. They focused their efforts on a simple derivatization with benzil to characterize the major product. They were trying to synthesize $p$-anisyldiphenylcarbinol, but instead what happened is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rnsny.png&quot; alt=&quot;lithium-hydrogen exchange&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing along they discovered phenyllithium would perform the same reaction. As part of their investigation they varied other substituents on the aryl halide and discovered an even more peculiar product which violated their chemical intuition. Wittig was the first to speculate a lithium-bromide exchange reaction in 1938, below is the reaction he and coworkers carried out in diethyl ether giving a high yield (95%), the preferred product led them to infer it went through an aryllithium intermediate:&lt;img src=&quot;https://i.stack.imgur.com/CMNaV.png&quot; alt=&quot;Wittig&amp;#39;s discovery&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same reaction occured with iodine, bromine, and chlorine. Yet, fluorine gave products via an entirely different mechanism. This led them to postulate an aryne intermediate. They then set off in a new direction, by trying to weaken carbon-hydrogen bonds with a neighboring quaternary ammonium, this led them to discover N-ylides. It was found this more readily occurred with phosphonium, where they discovered P-ylides. Now you can see the rough outline of the path required to discover the Wittig reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How Does Lithium-Halogen Exchange Work?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This initial discovery by Wittig and then Gilman encouraged further efforts because it allowed the formation of a number of previously inaccessible organometallics. In the above reaction, the Grignard reagent cannot simply replace phenyllithiun (some Mg cases have been reported for other halides, but Li is the stalwart metal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today this reversible lithium-halogen exchange reaction can be formulated as,&#xA;$$\ce{RLi + R’X &amp;lt;=&amp;gt;[{K}]  R’Li + RX}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an equilibrium process, we know that the more stable lithium species is preferred. Interestingly when $\ce{R’}$ = aryl (particularly bromides and iodides) this is best written as, $$\ce{RLi + R’X &amp;lt;=&amp;gt;&amp;gt; R’Li + RX}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\ce{R’} = $ allylic or benzylic, Wurtz coupling is favored. $\ce{R’} = $ alkyl tend to give unsatisfactory results. In general, the equilibrium favors those species which most stabilize the carbonanion intermediate. Depending on the specific halide, side-product formation is a problem: &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta-Hydride_elimination&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\beta$-elimination&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Dehydrohalogenation&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\alpha$-metalation&lt;/a&gt;, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Wurtz_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wurtz-type&lt;/a&gt; reactions can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gilman discovered that alkyllithiums, for example &lt;em&gt;t&lt;/em&gt;-butyllithium, were more adaptable in their reactivity than aryllithiums, which explains their widespread success in the modern chemical literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the reaction proceeds remains an unresolved problem, however, there is evidence to suggest the following mechanisms are at play:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) intermediates proceeding through an “ate” complex&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps unsurprising to you by now, Wittig also hypothesized the existence of an &quot;ate&quot; complex intermediate: &lt;img src=&quot;https://i.stack.imgur.com/PqQRE.png&quot; alt=&quot;&amp;quot;ate&amp;quot; complex&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in other words, a nucleophilic attack of the halide by the alkyllithium with a carbonanion leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) radical mechanisms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A one electron transfer from the nucleophile to halide thereby producing a radical intermediate that then can give rise to a number of products. There are data from EPR and radical caging experiments to suggest radicals at least form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) mixtures of (1) and (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with many systems, the truth is both mechanisms compete with each other; of course which pathway, if any, dominates will depend on the solvent, temperature, substrate, and certainly the lithium species used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;tert&lt;/em&gt;-Butyllithium requires 2 equivalents, resulting in a non-equilibrium process where: $$\ce{RX + (Me)3CLi -&amp;gt; RLi + (Me)3CX}$$ $$\ce{(Me)3CX + (Me)3CLi -&amp;gt; (Me)3CH ^ + (Me)2C=CH2 ^ + LiX}.$$&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2015-06-23T19:40:49.693" LastActivityDate="2015-06-23T19:40:49.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33121" PostTypeId="1" CreationDate="2015-06-19T19:07:50.477" Score="2" ViewCount="1222" Body="&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/31531/friedel-crafts-reaction-of-phenol&quot;&gt;Friedel–Crafts reaction of phenol&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the question at the above link states that : &quot;Like aniline, phenol too reacts to a very less extent during Friedal-Crafts reaction.&#xA;The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But isn't the case with anisole, the same?&lt;/strong&gt; However, my textbook states that anisole does undergo Friedel-Crafts acylation and alkylation.&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-06-19T19:07:50.477" Title="Why does phenol not undergo Friedel-Crafts alkylation and acylation reactions, whereas anisole does?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-10-02T16:52:10.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="33122" PostTypeId="1" AcceptedAnswerId="33133" CreationDate="2015-06-19T19:31:17.217" Score="9" ViewCount="7192" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vBAqX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, 'ethoxy' should be given the locant '1' as it is a functional group. The compound should thus be named 1-ethoxy-2,2-dimethylcyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was told that the correct name is 2-ethoxy-1,1-dimethylcyclohexane. Can anyone please explain why this is so? My teacher explained this to me in terms of the lowest locant sum rule, which I just discovered &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27095/iupac-nomenclature-smallest-sum-of-locants&quot;&gt;doesn't even exist&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="16683" LastEditDate="2018-08-14T15:18:43.147" LastActivityDate="2018-08-14T15:18:43.147" Title="Nomenclature of a cyclohexyl ether and locant position enumeration" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33124" PostTypeId="1" AcceptedAnswerId="33188" CreationDate="2015-06-19T22:17:37.593" Score="3" ViewCount="203" Body="&lt;p&gt;I've read a lot about electrochromism. This &lt;a href=&quot;http://www.academia.edu/3855995/Materials_for_Electrochromic_Devices&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; on electrochromic materials has also been additionally helpful.&#xA;&lt;br&gt;I was wondering, for these electrochromic materials, can you simply attach a small electrode to either side of the film like in setup 2, then apply a small voltage? Will the electrochromism only occur at the edges in this case?&#xA;&lt;br&gt;Does anyone know how material dependent this effect is? I was hoping to use a polymer film in particular. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/YZMJU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="10185" LastEditDate="2015-06-20T09:41:41.463" LastActivityDate="2015-06-22T05:57:22.097" Title="Electrochromic Materials, electrode setup" Tags="&lt;electrochemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33125" PostTypeId="2" ParentId="33069" CreationDate="2015-06-20T00:00:41.017" Score="3" Body="&lt;p&gt;The degree of mixing is very important in &quot;two part&quot; epoxy binder systems for coatings and adhesives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases the two liquids are quite viscous and the liquids themselves are better described as suspensions or colloidal dispersions rather than solutions. Unlike say two simple aqueous solutions when combined our two prepolymer liquids will not readily homogenize in a reasonable amount of time unless they are mechanically mixed in some way.  Even if the liquids do combine very slowly they will often gel or vitrify by the molecules at the interfaces reacting before the liquids homogenize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many epoxy systems we have, as Beerhunter stated above, some oligomer or polymer with epoxy functionality in one part and some liquid with amine (or sometimes hydroxyl) functionality in the other. Every amine needs to &quot;find&quot; an oxirane ring to attack or the curing process simply can't occur. The curing process works by nucleiphillic attack from amines to open epoxied rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The presence of leftover unreacted amine in cured epoxy materials often results in a phenomenon known as amine blushing or amine blooming. leftover amines or polyamides migrate to the surface and may in some cases react with air to form carbamates. This can be disastrous for desired materials properties like tack, adhesion (for the epoxy as a substrate), or cohesive strength of the bulk material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary you can't build a robust three dimensional polymer network without thoroughly mixing because all the reactants can't react. Always mix as thoroughly as possible to get complete conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience with polyesters is limited so I will not comment on that.&lt;/p&gt;&#xA;" OwnerUserId="16937" LastActivityDate="2015-06-20T00:00:41.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33126" PostTypeId="2" ParentId="33105" CreationDate="2015-06-20T01:00:40.667" Score="8" Body="&lt;p&gt;Your structures don't clearly show the different configurations between α- and β- linked glycopyranosides. Cellulose is a non-branching (poly) β-glycopyranoside. Amylose (a component of starch) is a non-branching (poly) α-glycopyranoside. The enzymes in your mouth and digestive system are good at breaking apart the α-glycopyranoside linkage, but not the β-glycopyranoside (it doesn't fit into the enzyme).&lt;br&gt;&#xA;Amylopectin (the other component of starch) is a branched (poly) α-glycopyranoside. It branches with a 1-6 linkage. Glycogen is another branched (poly) α-glycopyranoside.&lt;br&gt;&#xA;Cellulose is not the only (poly) β-glycopyranoside; there is also β-glucans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gKWff.png&quot; alt=&quot;Cellulose and Amylopectin&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-06-20T04:40:35.130" LastActivityDate="2015-06-20T04:40:35.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33128" PostTypeId="1" AcceptedAnswerId="33129" CreationDate="2015-06-20T05:13:32.953" Score="3" ViewCount="8721" Body="&lt;p&gt;When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group, and side chain that varies from one to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amino group of one can attach to the carboxyl group of another, making a peptide bond and water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both groups are clearly important, but what is it about the amino group that that's what goes in the name &quot;amino acid&quot;? Is the amino group in some way more important? Is it just something about the way/order in which they were discovered/identified that the name just came about? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why &lt;em&gt;amino&lt;/em&gt; acid? Not carboxyl acid, or some mix of the two, or something else?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorUserId="418" LastEditDate="2016-10-30T20:28:55.770" LastActivityDate="2018-08-28T18:36:39.740" Title="Why are amino acids named amino acids?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;amino-acids&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33129" PostTypeId="2" ParentId="33128" CreationDate="2015-06-20T05:41:18.030" Score="10" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group...&lt;/p&gt;&#xA;&lt;p&gt;Why amino acid? Not carboxyl acid, or some mix of the two, or something else?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You stated why it is called an amino acid, the name comes from the amine (amino) and the carboxylic acid (carboxyl group).&lt;/p&gt;&#xA;&lt;p&gt;Amines are weak bases, so the acid portion of the name must come from the carboxylic acid.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-20T05:41:18.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33130" PostTypeId="1" CreationDate="2015-06-20T05:50:55.670" Score="0" ViewCount="141" Body="&lt;p&gt;What is the difference between different type of rhodamine, for example rhodamine red and rhodamine green?&#xA;are these two type of rhodamine (red &amp;amp; green) have a good Raman signal?&#xA;We would like examples of Raman signals of these two types of rhodamine.&lt;/p&gt;&#xA;" OwnerUserId="17040" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T05:53:30.663" LastActivityDate="2015-07-20T06:26:44.830" Title="The difference between the different types of rhodamine" Tags="&lt;homework&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33131" PostTypeId="2" ParentId="33130" CreationDate="2015-06-20T06:15:25.440" Score="2" Body="&lt;p&gt;Rhodamine green&lt;br&gt;&#xA;CAS: 189200-71-3&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0KNMj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rhodamine Red&lt;br&gt;&#xA;CAS: --&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RFSSE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-06-20T06:15:25.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33132" PostTypeId="2" ParentId="33104" CreationDate="2015-06-20T08:42:37.333" Score="1" Body="&lt;p&gt;The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an &lt;em&gt;average&lt;/em&gt; kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$.  The average reflects a Boltzmann distribution of speeds&#xA;$$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT} $$&#xA;i.e. some molecules have very low velocity and some have very high velocities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;em&gt;addition&lt;/em&gt; one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an &lt;em&gt;average&lt;/em&gt; rotational kinetic energy.  For individual molecules at any given time, the amount of translational and rotational energy is independent.  For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change.  However, on &lt;em&gt;average&lt;/em&gt; there is an equal amount of translational and rotational energy (for non-linear molecules).&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-20T10:34:39.687" LastActivityDate="2015-06-20T10:34:39.687" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="33133" PostTypeId="2" ParentId="33122" CreationDate="2015-06-20T09:21:18.513" Score="6" Body="&lt;p&gt;The &lt;em&gt;smallest sum of locants&lt;/em&gt; rule does not exist in the IUPAC recommendations. The application of this ‘rule’ can lead to wrong results &lt;a href=&quot;https://chemistry.stackexchange.com/a/28009/7951&quot;&gt;in many instances&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The corresponding section in the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; actually reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.3.5&lt;/strong&gt; Lowest set of locants&lt;/p&gt;&#xA;&lt;p&gt;The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.&lt;br /&gt;&#xA;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With regard to numbering of locants, simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl and ethoxy) are considered together with equal seniority:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the example is named as 2-ethoxy-1,1-dimethylcyclohexane rather than 1-ethoxy-2,2-dimethylcyclohexane since the locant set ‘1,1,2’ is lower than ‘1,2,2’.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-20T09:21:18.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33134" PostTypeId="2" ParentId="33128" CreationDate="2015-06-20T10:49:41.410" Score="5" Body="&lt;p&gt;You might think that the term &quot;amino acid&quot; comes from the more systematic names of these molecules, where &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/&quot; rel=&quot;noreferrer&quot;&gt;the amine is a substituent&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Common name        systematic name&#xA;glycine            2-aminoethanoic acid&#xA;alanine            2-aminopropanoic acid&#xA;valine             2-amino-3-methylbutanoic acid&#xA;cysteine           2-amino-3-mercaptopropanoic acid&#xA;serine             2-amino-3-hydroxypropanoic acid&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In all of these names &quot;amino&quot; precedes &quot;acid&quot;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You would be wrong!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Amino_acid#History&quot; rel=&quot;noreferrer&quot;&gt;&quot;Amino acid&quot;&lt;/a&gt; apparently entered the English lexicon in 1898, long before IUPAC reared its systematic nomenclature head. It is likely then, though I was not there, that &quot;amino acid&quot; came about from two considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We were already probably using the prefix &quot;amino-&quot; to denote molecules with nitrogen groups in them.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Amino acid&quot; was probably easier to pronounce than other possible combinations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-20T10:49:41.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33135" PostTypeId="1" AcceptedAnswerId="33136" CreationDate="2015-06-20T12:00:39.300" Score="16" ViewCount="8343" Body="&lt;p&gt;I have seen two different notations for &lt;strong&gt;sodium acetate&lt;/strong&gt;. The first one is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{NaCH}_3\text{COO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{CH}_3\text{COONa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm confused, which one is the best to use?&lt;/p&gt;&#xA;" OwnerUserId="13221" LastActivityDate="2019-11-17T13:55:27.637" Title="Which one is the best notation for sodium acetate" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33136" PostTypeId="2" ParentId="33135" CreationDate="2015-06-20T13:15:11.830" Score="31" Body="&lt;p&gt;There is nothing wrong with either formula. And you can use even more:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{NaC2H3O2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{C2H3NaO2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It really depends on which point you want to bring across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaCH3COO}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaC2H3O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3COONa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one stresses a more organic-chemical point of view where it’s relevant &lt;em&gt;where&lt;/em&gt; the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H3NaO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;em&gt;Beilstein&lt;/em&gt;-type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is &lt;strong&gt;not&lt;/strong&gt; considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-21T14:11:14.500" LastActivityDate="2015-06-21T14:11:14.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33137" PostTypeId="2" ParentId="33135" CreationDate="2015-06-20T13:34:52.503" Score="0" Body="&lt;p&gt;Ok , both of the formulas are absolutely correct.It depends on what you want to signify , basically if you want to look from the point of view of organic chemistry then you take the second formula. And If you any to look from just the point of view of just ions then you can render the positive ion($\ce{Na+} $) to shifted first and then the negative ion($\ce{CH3COO-}$)&#xA;.I think the second formula looks much more better because it is the usual notation for sodium acetate and my vote goes to the second.But if you want to show people something different you can obviously use the the first one. &lt;/p&gt;&#xA;" OwnerUserId="17027" LastEditorDisplayName="user15489" LastEditDate="2015-06-20T13:36:01.127" LastActivityDate="2015-06-20T13:36:01.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33138" PostTypeId="2" ParentId="33114" CreationDate="2015-06-20T13:53:04.747" Score="8" Body="&lt;p&gt;As ron stated in the question’s comments, none of the resonance structures display 1-methoxybuta-1,3-diene. In fact, they show an extremely unstable (if generateable) dication. Choose a random $+$, replace it with a $-$ (don’t choose the one on the oxygen) to get resonances for the structure the question asks for. That should also semi-answer your question about the two positive charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the question on octets: It is the more traditional viewpoint that elements of the third and higher periods such as sulphur could expand their octets (e.g. in $\ce{SO2}$). This was generally explained with contribution by d-orbitals (and Pauling’s dislike for charge separation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, however, these structures are more generally considered as &lt;em&gt;abiding&lt;/em&gt; by the octet rule invoking charge separation where necessary and multicentred bonds where applicable. The reason for this is that d-orbitals are too far removed energetically from s and p of the same shells to actively take part in bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are sub-octet structures to consider (heptets in radicals or sextets in carbocations, carbenes and the like). These exist and are undebated. This violation of the octet rule is ‘less bad’ (by a mile) than a super-octet. Why? Well, you’re not filling electrons into high-energy contributed orbitals but rather removing them from stable ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, when considering the first of these octet paragraphs, it is a lot harder for me to think of compounds that have sub-octets for elements of higher periods — most examples I know are carbon or nitrogen centred.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;On the topic of the ugliness of (c), Breaking Bioinformatics has answered that nicely. But you could have arrived at the same conclusion by checking your book’s rules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does not apply here. No resonance has an octet on every carbon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(c) has four formal charges, all the others have two. Stop here, mark (c) and grab a piece of cake.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-22T08:37:57.230" LastActivityDate="2015-06-22T08:37:57.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33142" PostTypeId="2" ParentId="29286" CreationDate="2015-06-20T18:20:30.267" Score="3" Body="&lt;p&gt;The previous answer is correct. I just wanted to add that atoms, electrons and generally matter, tries to ''organise'' itself in order to adopt a state, where it would have the lowest energy possible.&#xA;An electronegative atom tends to attract electrons and has the ability of handling their charge better. Thus when an electron moves from a less electronegative atom to a more electronegative one, it is like moving from a state of high energy to a state of low energy. This is also described as realease of energy.&lt;/p&gt;&#xA;" OwnerDisplayName="user17037" LastActivityDate="2015-06-20T18:20:30.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33146" PostTypeId="1" CreationDate="2015-06-20T22:50:03.100" Score="0" ViewCount="1283" Body="&lt;p&gt;What is the advantage of this approach over polishing the silver with an abrasive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Silver tarnish is an outer coating of silver sulfide, $\ce{Ag2S}$, formed when silver reacts with trace quantities of air-borne hydrogen sulfide, a smelly gas produced naturally by the digestion of food in living animals.  Silver atoms in silver sulfide have lost electrons to sulfur atoms in creating ionic bonds.  You can remove this tarnish with home electrochemistry.  Convert silver atoms back to their elemental state by restoring their electrons.  Sulfur atoms won't relinquish electrons to silver, but with the proper connection, aluminum atoms will.  Add about a liter of water with several heaping tablespoons of baking soda to an aluminum pan that has been scoured clean.  If you don't have an aluminum pan, use several large strips of aluminum foil at the bottom of another pan.  Bring the water to a boil and remove it from the heat source.  Slowly immerse a tarnished piece of silver; you'll see an immediate effect as the silver and aluminum make contact.  (Add more baking soda if you don't see this.)  Also, as silver is brought back to its elemental state, hydrogen sulfide is released back into the air.  You'll smell it!  The baking soda serves to remove a thin and transparent coating of aluminum oxide, which normally coats aluminum and thus prevent direct contact.  The baking soda also serves as a conductive ionic solution to permit the passage of electrons from the aluminum to the silver.&quot; &lt;/p&gt;&#xA;" OwnerUserId="17045" LastEditorUserId="11583" LastEditDate="2015-06-21T04:31:31.587" LastActivityDate="2015-06-21T04:37:10.043" Title="Cleaning silver - which method is better?" Tags="&lt;everyday-chemistry&gt;&lt;experimental-chemistry&gt;&lt;silver&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-21T13:51:16.327" ContentLicense="CC BY-SA 3.0" />
  <row Id="33147" PostTypeId="2" ParentId="27742" CreationDate="2015-06-20T23:19:56.727" Score="4" Body="&lt;p&gt;The question is most likely badly worded and should have been&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum mass of $\ce{AgCl}$ that can be dissolved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing &lt;em&gt;less&lt;/em&gt; so you can actually perform the experiment you suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it’s just a flick of dust, and because the authors likely &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;knew that, too&lt;/li&gt;&#xA;&lt;li&gt;assumend you knew that, too&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2015-06-24T13:28:20.223" LastActivityDate="2015-06-24T13:28:20.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33149" PostTypeId="1" CreationDate="2015-06-21T04:25:29.600" Score="2" ViewCount="2467" Body="&lt;p&gt;In Paulings eqn:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \%~\ce{ionic~character} = \left[ 1−e^{−(Δχ/2)^2}\right] \times 100$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is $e$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I for the love of me can't find it anywhere&lt;/p&gt;&#xA;" OwnerUserId="17050" LastEditorUserId="14" LastEditDate="2015-06-21T13:00:15.547" LastActivityDate="2015-06-21T13:00:15.547" Title="Pauling's equation on the relationship between Electronegativity and Percentage Ionic Character" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ClosedDate="2017-11-09T19:14:28.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="33150" PostTypeId="2" ParentId="33146" CreationDate="2015-06-21T04:37:10.043" Score="1" Body="&lt;p&gt;Abrading the silver sulfide removes it (and some silver underneath the tarnish). Electrolytic cleaning actually &lt;em&gt;restores&lt;/em&gt; the silver previously part of the tarnish. In addition, it cleans crevices inaccessible to manual polishing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are disadvantages, though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the article has &lt;a href=&quot;https://en.wikipedia.org/wiki/Niello&quot; rel=&quot;nofollow&quot;&gt;niello&lt;/a&gt; inlay, it could be damaged.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If other materials, such as parchment or glue, are part of the article, they would be damaged.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the article is a valuable antique, cleaning may &lt;em&gt;reduce&lt;/em&gt; its value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-21T04:37:10.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33151" PostTypeId="2" ParentId="33149" CreationDate="2015-06-21T05:47:52.227" Score="5" Body="&lt;p&gt;'e' is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Exponential_function&quot;&gt;natural exponential function&lt;/a&gt;, so not an actual measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is correctly written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% Ionic character $= (1 − e^{−(\frac{Δχ}{2})^2})×100$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $−(\frac{Δχ}{2})^2$ term is raised as a power of the natural exponent 'e'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: The University of the West Indies page &lt;a href=&quot;http://wwwchem.uwimona.edu.jm/courses/CHEM1902/IC10K_MG_Fajans.html&quot;&gt;Lecture 3. Polarizability&lt;/a&gt;, you'll need to scroll down the page a bit to the section labelled 'Bond character based on electronegativity differences'.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-21T06:00:22.823" LastActivityDate="2015-06-21T06:00:22.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33152" PostTypeId="1" AcceptedAnswerId="48749" CreationDate="2015-06-21T05:49:51.377" Score="5" ViewCount="1670" Body="&lt;p&gt;No, it's not epoxy &lt;em&gt;for&lt;/em&gt; reinforcing steel. It's &quot;steel reinforced epoxy.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jbweld.com/products/j-b-weld-twin-tube&quot; rel=&quot;noreferrer&quot;&gt;http://www.jbweld.com/products/j-b-weld-twin-tube&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How in the world does steel reinforce epoxy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand how carbon nanotubes dispersed in epoxy can help strength it, quote from paper below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The mechanical behavior of multiwalled carbon nanotube/epoxy&#xA;  composites was studied in both tension and compression. It was found&#xA;  that the compression modulus is higher than the tensile modulus,&#xA;  indicating that load transfer to the nanotubes in the composite is&#xA;  much higher in compression. In addition, it was found that the Raman&#xA;  peak position, indicating the strain in the carbon bonds under&#xA;  loading, shifts significantly under compression but not in tension. It&#xA;  is proposed that during load transfer to multiwalled nanotubes, only&#xA;  the outer layers are stressed in tension whereas all the layers&#xA;  respond in compression.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But back to steel - as far as I know, steel is a crystalline material and it doesn't possess any property akin to CNTs. Can someone explain how this particular epoxy works? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastActivityDate="2020-12-10T19:19:58.897" Title="What is &quot;Steel Reinforced Epoxy&quot;" Tags="&lt;materials&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33153" PostTypeId="1" AcceptedAnswerId="33159" CreationDate="2015-06-21T08:01:45.360" Score="7" ViewCount="114" Body="&lt;p&gt;I have this idea of creating a large domed ceiling/roof for a building that can be made clear to see the actual sky, or made fully opaque/dark grey/black to project an artificial sky onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that things similar to this effect have been chemically achieved such as UV-activated sun glasses and remote controlled window tint, but I haven't been able to find much information that explains how this kind of thing is done chemically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What options, chemically speaking, exist for making a large glass surface transform between being clear and opaque? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for any known chemical that can do this in the form of a tint applied to a glass surface, or a liquid filling a hollow glass surface, or even a gas filling a hollow glass surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be absolutely specific, I'm describing a chemical that can be either modified or added/removed to/from/within a glass surface to systematically control nearly full opacity or nearly full clarity of a layer applied to or contained within the surface of a curved glass panel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The opacity transition should be efficiently repeatable, meaning, the effect shouldn't deteriorate &lt;em&gt;significantly&lt;/em&gt; with repetition. &lt;/p&gt;&#xA;" OwnerUserId="17051" LastEditorUserId="4231" LastEditDate="2015-06-21T22:36:18.613" LastActivityDate="2015-06-21T22:36:18.613" Title="Is there a known chemical that makes it possible to control the opacity of a large surface?" Tags="&lt;color&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33154" PostTypeId="1" CreationDate="2015-06-21T08:31:39.907" Score="3" ViewCount="2407" Body="&lt;p&gt;An example would be $\ce{[Fe(H2O)6]^2+}$ in which it has 6 water molecules associated with it. What is the process of adding these water molecules called?&lt;/p&gt;&#xA;" OwnerUserId="17027" LastEditorUserId="10185" LastEditDate="2015-06-21T15:15:40.697" LastActivityDate="2019-03-22T00:43:55.067" Title="What is that process called in which we add water molecules to an ion to form a complex?" Tags="&lt;water&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33155" PostTypeId="2" ParentId="33135" CreationDate="2015-06-21T08:31:47.277" Score="-1" Body="&lt;p&gt;Both are correct, but the second one is more commonly used. Actually, the way of writing a molecular formula signifies the structure of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain: There are compounds that have the same molecular formula but have different structures. Such molecules are called structural isomers. An example for a carbon compound:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FhJ1a.png&quot; alt=&quot;carbon compound&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, even if both have the same formula they have different structures (they are structural isomers to say it precisely). Generally formulae of many compounds or molecules are written in a way so that they can signify the structure or arrangement of atoms in molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at that image again. If I want to write chemical formulae of both compounds that can signify their structures, I would write:&#xA;$\ce{CH3CH2CH2Cl}$ for the first one, and the second one is $\ce{CH3CHClCH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess now you can understand that the way of writing a formula defines the structure of the compound, so always write the formula in a way that can define or give some hint about the structure.&lt;/p&gt;&#xA;" OwnerUserId="17052" LastEditorUserId="3846" LastEditDate="2015-06-21T09:53:48.810" LastActivityDate="2015-06-21T09:53:48.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33157" PostTypeId="1" AcceptedAnswerId="33158" CreationDate="2015-06-21T13:50:51.363" Score="3" ViewCount="287" Body="&lt;p&gt;This equation has been described as an equilibrium in an article, how would you explain it to someone with not a lot of understanding of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is:&#xA;$\ce{CH3COOH &amp;lt;-&amp;gt; CH3COO- + H+}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the description is of an equilibrium, it can be clearly seen that it is a resonance arrow. So how would you describe it?&lt;/p&gt;&#xA;" OwnerUserId="17056" LastEditorUserId="9961" LastEditDate="2015-12-21T21:01:48.633" LastActivityDate="2015-12-21T21:01:48.633" Title="Dissociation reaction notation" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;notation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33158" PostTypeId="2" ParentId="33157" CreationDate="2015-06-21T14:27:48.573" Score="7" Body="&lt;p&gt;No, no it's not a resonance arrow my friend, it has to be an arrow for equilibrium because it is a reaction equation of $\ce{CH3COOH}$ in its aqueous solution. It will dissociate feebly into its constituent ions because its a really weak acid. Thus establishing equilibrium quite a lot first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover in a resonance arrow we describe the same compound with different structures (which are all correct) and that is why we establish a resonance hybrid to denote all such possible resonating diagrams in a single diagram fulfilling the properties of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless a resonance arrow can't anyway turn into a dissociation reaction arrow. &lt;/p&gt;&#xA;" OwnerUserId="17027" LastEditorUserId="7475" LastEditDate="2015-12-21T20:08:55.117" LastActivityDate="2015-12-21T20:08:55.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33159" PostTypeId="2" ParentId="33153" CreationDate="2015-06-21T14:28:55.857" Score="5" Body="&lt;p&gt;There are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermochromism&quot; rel=&quot;nofollow&quot;&gt;thermochromic&lt;/a&gt; materials that change color in response to heat&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Photochromism&quot; rel=&quot;nofollow&quot;&gt;photochromic&lt;/a&gt; materials that change color in response to light&lt;/li&gt;&#xA;&lt;li&gt;and &lt;a href=&quot;https://en.wikipedia.org/wiki/Smart_glass#Electrochromic_devices&quot; rel=&quot;nofollow&quot;&gt;electrochromic&lt;/a&gt; materials than change color in response to an applied electric field.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these materials are sub-categories of &lt;a href=&quot;https://en.wikipedia.org/wiki/Smart_glass&quot; rel=&quot;nofollow&quot;&gt;Smart Glass&lt;/a&gt;. I suspect that electric field sensitive Smart Glass materials are what you are looking for.  There are several different types of such Smart Glass materials.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Those with &lt;strong&gt;suspended particles&lt;/strong&gt; that are normally randomly aligned, but when an electric field is applied they align an alter the light transmission properties of the material.  These materials have already found commercial application (airplane) and are moving down to lower cost applications (autos)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Electrochromic&lt;/strong&gt; devices where a molecule undergoes a chemical change between colored and non-colored states when a field is applied.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Liquid crystal&lt;/strong&gt; materials dissolved in a polymer matrix are normally randomly aligned, but, when a field is applied they align thus altering the light transmission properties of the material.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also some non-electrochromic smart glass technologies that are also discussed in the Smart Glass Wikipedia article linked above.  These are generally lower tech and therefore lower cost.  They work based on&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Having several aligned windows, but with the alignments in different directions and then manually moving the windows into place to adjust the opacity.&lt;/li&gt;&#xA;&lt;li&gt;Having two reflective windows with a space in between.  When the space is filled with a material having the same refractive index as the reflective windows, the assembly becomes transparent.&lt;/li&gt;&#xA;&lt;li&gt;Using multiple polarized panels and moving them into or out of place to adjust opacity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which method you choose will depend upon how quick you need the materials to change color, the level of manual intervention you're willing to tolerate and how much your willing to spend.  If you google &quot;Smart Glass&quot; you'll find that this field is already well advanced.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-21T14:35:26.010" LastActivityDate="2015-06-21T14:35:26.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33162" PostTypeId="1" CreationDate="2015-06-21T15:28:16.650" Score="3" ViewCount="135" Body="&lt;p&gt;Does anyone know any good references which use partial differential equations to model the alcohol concentration in the bloodstream? I'm a mathematician, so I'm searching for a rigorous mathematical model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any other suggestions of models in chemistry that uses partial differential equations, feel free to post them here.&lt;/p&gt;&#xA;" OwnerUserId="17059" LastEditorUserId="15410" LastEditDate="2015-06-21T21:57:34.523" LastActivityDate="2015-06-21T21:57:34.523" Title="Alcohol in the bloodstream modeled by differential equations" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;reference-request&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33163" PostTypeId="2" ParentId="31994" CreationDate="2015-06-21T17:02:57.227" Score="3" Body="&lt;p&gt;Another method that doesn't require calculations, but does involve sacrificing only a very small portion of your sample is to do a flame test. By heating a pure sliver of antimony in a hot flame you should observe a pale green color as opposed to a blue-white color if the sample is lead. If you are a poor judge of color then its helpful to have known pure samples of the elements to compare with your unknown sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method, as well as the gravimetric method suggested by Santiago can fail if your sample is an alloy (mix) of the two elements (and perhaps other ingredients!). But here the flame test can help by using a spectroscope for a more quantitative accounting of the constituents.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-06-21T17:02:57.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33164" PostTypeId="1" CreationDate="2015-06-21T17:07:10.770" Score="4" ViewCount="1464" Body="&lt;p&gt;Why R32 Space Group is so named? I mean, if &lt;em&gt;R&lt;/em&gt; stands for Rhombohedral and &lt;em&gt;3&lt;/em&gt; for Threefold Axis, what 2 stands for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any Twofold Axis in the cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/large/155az1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://img.chem.ucl.ac.uk/sgp/large/155az1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it means 1/2, 2/3 and 1/6 on the arrows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R32 have two cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one: &lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/large/155bz1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://img.chem.ucl.ac.uk/sgp/large/155bz1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this other: &lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/large/155az1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://img.chem.ucl.ac.uk/sgp/large/155az1.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they equivalent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have traced some lines to mark the screw-axes and the twofold-axis, but i can't see any point that follow the simmetry operator and the element of simmetry. Maybe i don't understand how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the height of the green line? And, which are the points of the screw-axis of the yellow line?&lt;img src=&quot;https://i.stack.imgur.com/mnPQo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17061" LastEditorUserId="17061" LastEditDate="2015-06-22T15:23:17.457" LastActivityDate="2015-06-22T15:23:17.457" Title="R32 Space Group (Crystallography)" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33165" PostTypeId="1" CreationDate="2015-06-21T17:21:57.423" Score="7" ViewCount="3584" Body="&lt;p&gt;They are both formed when a liquid is supercooled rapidly, no free energy, and they both have irregular structures. What defines a glass other than how it is amorphous, transparent, and has a glass transition state? &lt;/p&gt;&#xA;" OwnerUserId="14481" LastEditorUserId="23561" LastEditDate="2019-03-20T21:27:34.607" LastActivityDate="2019-03-20T21:27:34.607" Title="What is the difference between an amorphous solid and glass?" Tags="&lt;crystal-structure&gt;&lt;phase&gt;&lt;materials&gt;&lt;glass&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33166" PostTypeId="1" AcceptedAnswerId="33179" CreationDate="2015-06-21T17:23:28.473" Score="4" ViewCount="5722" Body="&lt;p&gt;In several metabolic pathways, Nature (live systems) chooses to make esterification using thioesters (CoA) and not esters. I would like to know what is the main aim of the Nature with this choice: faster reactions or more extensive reactions?&lt;/p&gt;&#xA;" OwnerUserId="9475" LastEditorUserId="4231" LastEditDate="2015-06-21T22:28:08.690" LastActivityDate="2015-06-22T13:36:43.890" Title="Advantages of thioesters over esters in fatty acid catabolism" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33167" PostTypeId="1" CreationDate="2015-06-21T17:45:36.170" Score="5" ViewCount="1149" Body="&lt;p&gt;&lt;strong&gt;Requirements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I am looking for a fluid that meets these criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrophobic&lt;/li&gt;&#xA;&lt;li&gt;Denser than water&lt;/li&gt;&#xA;&lt;li&gt;Low viscosity&lt;/li&gt;&#xA;&lt;li&gt;Food safe&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are working on adding a pool and designing a backyard recreational area.  One design element that we would like to include in multiple places is waterfalls of varying sizes and styles; I love flowing water.  As an extension of that I started wondering if it would be possible to create an underwater waterfall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I was thinking this could be accomplished by creating a regular waterfall in the pool with normal water, which could then be capped with a clear glass or acrylic watertight cover.  This certainly works as an underwater waterfall, but the cover will be visible, and will likely cloud over time, not to mention the condensation issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wandering now if some sort of fluid exists that could actually flow underwater.  If it's dense enough it will flow down a rockface even under the water, and if hydrophobic will not mix with the surrounding water.  As it will be within a pool that humans will be in for extended periods it must be externally safe, and I'd like something food-safe as kids swallow pool water and it's possible this stuff might get kicked up.  And to add to that tall order, the wrong viscosity will ruin the visual effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a fluid exist in a form that a normal person can purchase for a reasonable cost?  I live in the US (Pennsylvania).  Another question, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14200/is-there-a-readily-available-liquid-thats-more-dense-than-water-and-insoluble&quot;&gt;Is there a readily available liquid that&amp;#39;s more dense than water and insoluble?&lt;/a&gt;, seemed to be looking for something similar but the answer doesn't meet my safety requirements.&lt;/p&gt;&#xA;" OwnerUserId="7763" LastEditorUserId="69830" LastEditDate="2019-10-27T11:16:01.723" LastActivityDate="2020-03-25T12:03:00.357" Title="Is there a food-safe hydrophobic liquid with a substanially higher density than water?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33168" PostTypeId="2" ParentId="33165" CreationDate="2015-06-21T17:59:58.637" Score="5" Body="&lt;p&gt;While a glass is generally considered to be a supercooled, configurationally frozen liquid, not all amorphous solids are glasses. For example, amorphous silicon is a four-fold coordinated semiconducting solid, much like crystal silicon.  Liquid silicon is 8-12 fold coordinated with metallic bonding. Amorphous silicon has been shown to display a first-order phase transition to both the crystal and the liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, all glasses are amorphous solids, but not all amorphous solids are glasses.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-06-21T17:59:58.637" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33169" PostTypeId="1" AcceptedAnswerId="33177" CreationDate="2015-06-21T18:04:39.157" Score="4" ViewCount="2632" Body="&lt;p&gt;Which of the statements below is correct?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a) The &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions are more active than the &lt;em&gt;meta&lt;/em&gt; position.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;b) The &lt;em&gt;ortho&lt;/em&gt; and &lt;strong&gt;para&lt;/strong&gt; positions are activated while the &lt;em&gt;meta&lt;/em&gt; position is deactivated.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16967" LastEditorUserId="10185" LastEditDate="2016-05-31T13:21:15.023" LastActivityDate="2016-05-31T13:21:15.023" Title="Activity of different positions in phenol with respect to electrophilic aromatic substitution" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33170" PostTypeId="1" CreationDate="2015-06-21T18:22:52.310" Score="9" ViewCount="304" Body="&lt;p&gt;In the &lt;a href=&quot;http://physwiki.ucdavis.edu/Quantum_Mechanics/Fitzpatrick&amp;#39;s_Quantum_Mechanics/08%3A_Time-Dependent_Perturbation_Theory/8.10%3A_Electric_Dipole_Approximation&quot; rel=&quot;nofollow noreferrer&quot;&gt;dipole approximation&lt;/a&gt;, the following relation holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\hat{V} = -\hat{\mu} \cdot \vec{E}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say &quot;approximation&quot;, I guess we want to point out that we are dealing with linear spectroscopy, where we have the polarizability of the molecule proportional to the electric susceptibility of a molecule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\vec{p}(\omega) = k\chi \vec{E}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wavelength of the light is not larger than the atom size, the dipole approximation is no longer a good approximation. This is probably also true for large molecules. I'm not sure if this has to do with large molecules no longer having symmetry, so transitions might be allowed which otherwise should be forbidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="194" LastEditDate="2018-01-23T16:19:46.333" LastActivityDate="2018-01-23T16:19:46.333" Title="Why do we say &quot;approximation&quot; in the dipole approximation in spectroscopy?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;terminology&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33171" PostTypeId="1" AcceptedAnswerId="33173" CreationDate="2015-06-21T19:19:51.093" Score="4" ViewCount="1951" Body="&lt;p&gt;I see heats of combustion given in kg [dot] calories per gram molecular weight (Handbook of Chemistry and Physics, 1969-1970). I don't know how that makes sense, as I don't know what a kg cal might be.  I would have expected units of energy per mass (calories per gram) or energy per mole.  What is a kg cal, and what is wrong in my thinking?&lt;/p&gt;&#xA;" OwnerUserId="17064" LastEditorUserId="17064" LastEditDate="2015-06-21T20:36:10.913" LastActivityDate="2015-06-21T20:36:10.913" Title="What is a kg.cal per gram, as a heat of combustion?" Tags="&lt;heat&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33173" PostTypeId="2" ParentId="33171" CreationDate="2015-06-21T19:27:30.487" Score="3" Body="&lt;p&gt;$\mathrm{kg~cal}$ probably represents the kilogram calorie (especially given that it is an old book). Nowadays it is usually just known as the Calorie, usually abbreviated $\mathrm{kcal}$. It is roughly the amount of energy required to raise the temperature of $1~\mathrm{kg}$ of water by $1~\mathrm{^\circ C}$. Since this is temperature and pressure dependent, there are slight variations in the actual definition depending on the context in which it is used.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-06-21T19:27:30.487" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33174" PostTypeId="2" ParentId="33164" CreationDate="2015-06-21T20:11:40.170" Score="6" Body="&lt;p&gt;The 2-fold axes are parallel to the &quot;paper plane &quot;, i.e. perpendicular to the three-fold. They are depicted as arrows, the half-arrows indicate screw-axes. The numbers next to it show the &quot;height&quot; of the axis, i.e. where the plane parallel to the paper-plane cuts the unit cell: 1/2 would be in the middle, 1/3 would be a third from the bottom/top, etc. (obviously one only gives relative numbers here, as the absolute unit cell length varies depending on the real-life crystal you are looking at).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also have a look here for the symbols: &lt;a href=&quot;http://img.chem.ucl.ac.uk/sgp/misc/guide.htm&quot; rel=&quot;nofollow&quot;&gt;http://img.chem.ucl.ac.uk/sgp/misc/guide.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, the two links you provided are equivalent, as you can see from the pictogram in the lower lefthand corner - the first link is in the &quot;tri-/hexagonal&quot;, the second link in the &quot;rhombohedral&quot; setting. The symmetry for both is obviously the same (otherwise they would not be the same space group), it is just a different way of depicting the spacegroup by changing the point of origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here how they relate to each other: &lt;a href=&quot;https://www.uwgb.edu/dutchs/symmetry/rhombo-lattice.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.uwgb.edu/dutchs/symmetry/rhombo-lattice.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one should also be quite enlightening - but I am not quite sure whether it might be behind a paywall: &lt;a href=&quot;http://it.iucr.org/Ab/ch7o1v0001/sgtable7o1o155/&quot; rel=&quot;nofollow&quot;&gt;http://it.iucr.org/Ab/ch7o1v0001/sgtable7o1o155/&lt;/a&gt;&#xA;(Alternatively, have a look at the hardcopy of the International Tables of Crystallography, Volume A and look up spacegroup 155 - any decent crystallography lab or library should have it).&lt;/p&gt;&#xA;" OwnerUserId="10050" LastEditorUserId="10050" LastEditDate="2015-06-22T12:16:11.903" LastActivityDate="2015-06-22T12:16:11.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33175" PostTypeId="2" ParentId="33162" CreationDate="2015-06-21T21:50:18.377" Score="1" Body="&lt;p&gt;One such article that I found was published by the University of South Florida. The abstract for which reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given a set of differential equations describing blood alcohol content as a function time, we integrated the&#xA;  equations to obtain a general solution. The general solution equation depends on three free parameters: the initial&#xA;  concentration of alcohol in the stomach after ingestion, the rate of alcohol absorption into the blood stream and the rate&#xA;  at which the alcohol is metabolized by the liver. We fitted our solution to experimental data to determine the unknown&#xA;  parameters for a particular subject.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If this sounds useful to you, it can be found &lt;a href=&quot;http://scholarcommons.usf.edu/cgi/viewcontent.cgi?article=4819&amp;amp;context=ujmm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-06-21T21:50:18.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33176" PostTypeId="1" CreationDate="2015-06-21T22:01:52.913" Score="3" ViewCount="131" Body="&lt;p&gt;Here's the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta S = \frac{\Delta q}{T}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The confusion I have with this equation lies in the variable for temperature, $T$.  Why is the temperature constant in this equation? Why can't the equation include $\Delta T$ instead of just $T$ and, surely, if $\Delta q$ is negative or positive the temperature must decrease or increase, respectively.&lt;/p&gt;&#xA;" OwnerUserId="16889" LastEditorUserId="15410" LastEditDate="2015-06-21T23:40:55.660" LastActivityDate="2015-07-03T20:08:56.813" Title="Can someone explain the equation for entropy?" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33177" PostTypeId="2" ParentId="33169" CreationDate="2015-06-21T22:07:16.010" Score="4" Body="&lt;p&gt;As the figure below illustrates, substituents on aromatic rings that are &lt;strong&gt;electron donating&lt;/strong&gt; like the methyl and &lt;em&gt;t&lt;/em&gt;-butyl groups (and also the hydroxy group [&lt;em&gt;e.g&lt;/em&gt;. phenol]) increase the reaction rate at &lt;strong&gt;all&lt;/strong&gt; positions (&lt;em&gt;ortho, meta&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt;) relative to benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;strong&gt;electron withdrawing&lt;/strong&gt; substituents (the chlorine and ester carbonyl examples) decrease the reaction rate at &lt;strong&gt;all&lt;/strong&gt; positions relative to benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for phenol, or any other electron donating substituent, while the &lt;em&gt;meta&lt;/em&gt; position does react more slowly than the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; positions, it still  reacts faster than benzene.  Therefore, &lt;strong&gt;it would be incorrect to say that the &lt;em&gt;meta&lt;/em&gt; position is deactivated&lt;/strong&gt;. Only when the rate at a given position is less than the rate for a single position in benzene could one say that the position is deactivated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statement &quot;a&quot; in your question is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eHx0Y.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For future reference a note on terminology. The above figure tells us that nitration of toluene proceeds 24.5 times faster than the nitration of benzene; that is to say that the &lt;strong&gt;relative rate&lt;/strong&gt; for nitration of toluene is 24.5 times greater than benzene.  The &lt;em&gt;meta&lt;/em&gt; position in toluene reacts 3 times faster than a single position in benzene, this is referred to as the &lt;strong&gt;partial rate factor&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-10-09T18:38:31.737" LastActivityDate="2015-10-09T18:38:31.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33179" PostTypeId="2" ParentId="33166" CreationDate="2015-06-21T22:26:58.460" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the main aim of the Nature with this choice: faster reactions&#xA;  or more extensive reactions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Nature is frugal&lt;/strong&gt;, its main aim is to spend as little energy as possible in a chemical reaction.  &lt;strong&gt;Nature will always evolve so as to use the lowest energy pathway&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the resonance structures for an ester and a thioester we see that we can draw the same three resonance structures to describe each structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iors5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the resonance structure on the right, the one with the double bond between the carbonyl carbon and the ester oxygen or the thioester sulfur, is less important in the sulfur case.  Why?  Because a double bond between second and third row atoms is not nearly as strong as a double bond between two second row elements (much better electronic overlap results between atoms of similar size).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result the thioester carbonyl is not as stabilized by resonance as the ester carbonyl.  Said differently, the thioester carbonyl is more reactive (&lt;strong&gt;lower activation energy to react&lt;/strong&gt;) than the ester carbonyl.  The thioester motif provides nature with a more reactive carbonyl. This way nature doesn't have to spend as much energy to bring about reaction at this more reactive carbonyl.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-06-22T13:36:43.890" LastActivityDate="2015-06-22T13:36:43.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33180" PostTypeId="1" AcceptedAnswerId="33212" CreationDate="2015-06-21T23:48:07.213" Score="6" ViewCount="2161" Body="&lt;p&gt;Are there any type of detergent/surfactant chemicals that would be good for removing dirt and grease from fabric that also would evaporate from the clothes within, say, 24 hours leaving no residue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;volatile surfactant&quot; turned up fluorocarbons and something called Surfynol 61, which is really 3,5-Dimethylhex-1-in-3-ol as possible candidates. I don't know if these would be safe for laundry.&lt;/p&gt;&#xA;" OwnerUserId="12691" LastEditorDisplayName="user7951" LastEditDate="2015-09-30T23:33:45.703" LastActivityDate="2015-12-14T21:11:58.070" Title="Is there a type of detergent/surfactant that evaporates at STP?" Tags="&lt;water&gt;&lt;safety&gt;&lt;home-experiment&gt;&lt;surfactants&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33181" PostTypeId="2" ParentId="33170" CreationDate="2015-06-21T23:55:18.570" Score="3" Body="&lt;p&gt;As per your linked reference, it's an approximation because only the first term in the expansion (here, unity) is considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with transitions that are allowed or forbidden arises as a result of using such an approximation: &quot;Forbidden&quot; transitions are not strictly forbidden, it's just that they are weak and, in the linear approximation of the transition dipole operator, are usually not observed.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-21T23:55:18.570" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33182" PostTypeId="2" ParentId="33176" CreationDate="2015-06-22T01:25:02.363" Score="1" Body="&lt;p&gt;Your relation is not correct, so you must not understand something. If I were you I will asked the question in Physics SE but nevermind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good relation is : $$\mathrm{d}S=\frac{\delta Q}{T}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let Richard Feynman explain it to you &lt;a href=&quot;http://www.feynmanlectures.caltech.edu/I_44.html#Ch44-S6&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (when you click on the link wait few seconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it will help you, have fun, live long and do chemistry ! &lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorDisplayName="user7951" LastEditDate="2015-06-22T08:57:01.197" LastActivityDate="2015-06-22T08:57:01.197" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33183" PostTypeId="1" CreationDate="2015-06-22T01:42:45.807" Score="7" ViewCount="196" Body="&lt;p&gt;I would like to ensure that my hydrangeas are blue. I am told that adding ammonium sulfate or aluminium sulfate will cause the soil to be more acidic. I am also told that adding elemental sulfur will help. Will adding magnesium sulfate to the soil help?&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="30836" LastEditDate="2019-06-22T01:41:41.903" LastActivityDate="2019-06-22T01:41:41.903" Title="Amending soil pH to change the color of hydrangea flowers" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;ph&gt;&lt;home-experiment&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33184" PostTypeId="2" ParentId="33183" CreationDate="2015-06-22T02:25:34.263" Score="3" Body="&lt;p&gt;Although it is true that soil $\mathrm{pH}$ plays a role in the color of hydrangeas, the presence of aluminium is actually what determines the flower color. Aluminium is a naturally occurring mineral that is found in most  soils, however, just because a mineral is in the soil doesn’t mean the plant is absorbing this mineral. This is where soil $\mathrm{pH}$ comes into play – plants are able to absorb aluminium when the soil $\mathrm{pH}$ is between 4.5 and 5.8 with the best range being between 5.0 and 5.5. When the $\mathrm{pH}$ moves to 6.0 and above then the plants will not be able to absorb the aluminium despite it being present in the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding either sulfur or magnesium sulfate (epsom salt) to the soil will lower the $\mathrm{pH}$, but this will not help with the presence of aluminium in the soil, which is why many aluminium sulfate is so highly recommended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that magnesium sulfate would not be helpful in other ways, however, as magnesium is the central atom of pigment chlorophyll, which will help to improve the foliage quality of your hydrangea.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-06-22T02:25:34.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33185" PostTypeId="1" CreationDate="2015-06-22T02:37:51.150" Score="6" ViewCount="922" Body="&lt;p&gt;It seems like the typical laboratory methods for determining the carbon content of steel are to cuppellate it in a stream of oxygen and then measure the volume of the evolved $\ce{CO}$ and $\ce{CO2}$, which involves many high-temperature complexities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't the carbon content of steel be measured by dissolving it in acid and reacting out the carbon at low temperature?&lt;/p&gt;&#xA;" OwnerUserId="9663" LastEditorUserId="4097" LastEditDate="2015-06-22T08:05:30.813" LastActivityDate="2016-03-29T02:52:13.380" Title="Determination of carbon content of steel by liquid chemistry" Tags="&lt;analytical-chemistry&gt;&lt;metallurgy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33186" PostTypeId="1" CreationDate="2015-06-22T03:03:40.430" Score="3" ViewCount="246" Body="&lt;p&gt;Can anyone explain to me that how solvent factor affect the wavelength and intensity of gold nanoparticle and silver nanoparticle in UV spectroscopy? &lt;/p&gt;&#xA;" OwnerUserId="17066" LastEditorUserId="4231" LastEditDate="2015-06-22T03:25:55.997" LastActivityDate="2017-04-14T16:40:43.463" Title="Solvent effect on gold nanoparticle and silver nanoparticle in UV spectroscopy?" Tags="&lt;nanotechnology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33187" PostTypeId="1" CreationDate="2015-06-22T03:09:35.703" Score="5" ViewCount="14851" Body="&lt;p&gt;can anyone tell me the difference between absorption spectroscopy and extinction spectroscopy in terms of experiment? and how to get extinction spectroscopy, how to get absorption spectroscopy? Thank you so much.&lt;/p&gt;&#xA;" OwnerUserId="17066" LastActivityDate="2015-06-22T08:55:29.517" Title="difference between absorption spectroscopy and extinction spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-09T20:02:36.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="33188" PostTypeId="2" ParentId="33124" CreationDate="2015-06-22T05:57:22.097" Score="1" Body="&lt;p&gt;I will answer your last question first, this is very material dependent. Not all materials will be electrochromic. It is a very specific effect where the material has different oxidation states that have different absorbance profiles(colors if you will). The most common ones are conducting polymers (&lt;a href=&quot;https://en.wikipedia.org/wiki/Conductive_polymer&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Conductive_polymer&lt;/a&gt;) since these materials have different oxidation states that have vastly different electronic structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two electrode electrochemistry can get pretty hard to understand. Two compare your setups 1 and 2, the main differences are the electrode area and the electrode distance. Other than that, the two are the same. So, yes, electrochromism will happen in setup2 as well. However, it is hard to predict exactly where it will happen and to what extent. It all depends on the material you select, it's resistance and doping characteristics. In either setup, it will start at the electrode and work it's way inward, obviously, setup 1 is more advantageous with the larger electrode area and shorter inter-electrode distance. &lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-06-22T05:57:22.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33189" PostTypeId="1" AcceptedAnswerId="33209" CreationDate="2015-06-22T06:02:11.377" Score="5" ViewCount="122" Body="&lt;h2&gt;Trial #1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Fe^3+(aq) + SCN^- (aq)  -&amp;gt;   FeSCN^2+(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|r|c|c|c|}\hline &#xA;\mathrm{Initial~}(M)&amp;amp;\phantom{-}6.00 &amp;amp;\phantom{-}10.00&amp;amp;\phantom{-}0.00\\\hline&#xA;\mathrm{Change}~(M) &amp;amp; –4.00	&amp;amp; –4.00 &amp;amp;+4.00  \\\hline&#xA;\mathrm{Equilibrium}~(M)&amp;amp;\phantom{-}2.00&amp;amp;\phantom{-}6.00&amp;amp;\phantom{-}4.00\\\hline &#xA;\end{array}&#xA;$$ K = 0.333$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the reaction represented by the equation:&#xA;$$\ce{Fe^3+(aq) + SCN- (aq) \longrightarrow  [Fe(SCN)]^2+(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Trial #2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Initial: $10.0 ~M\ce{~Fe^3+(aq)}$ and $8.00~M\ce{ ~SCN- (aq)}$ (same temperature as Trial #1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium: $x~M\ce{~FeSCN^2+(aq)}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried putting $10.0$ and $8.00$ in an ice table, then when I have finished I start solving to get my equation as $$0=\frac{0.33x}{(10.0-x)(8.00-x)}$$ when I'm done I come out with $$0.33=\frac{x}{(x^2-18x+80)}$$ then I plug in for quadratic equation, &#xA;$$x=\frac{-6.94\pm\sqrt{6.94^2-4\times0.33\times26.4}}{2\times0.33}$$ &#xA;my answers came out to be $-5$ and $-16.03$.  What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="194" LastEditDate="2018-03-04T15:52:58.400" LastActivityDate="2018-03-04T15:52:58.400" Title="Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33190" PostTypeId="1" AcceptedAnswerId="33191" CreationDate="2015-06-22T07:07:40.347" Score="9" ViewCount="2432" Body="&lt;p&gt;I know when a hydrogen bond is, but - can hydrogen bonds only occur between two molecules containing hydrogen? Or, only one of the molecules should contain hydrogen and the other one may not contain hydrogen, but should have a large electronegative difference between two of its sides? &lt;/p&gt;&#xA;" OwnerUserId="17072" LastEditorUserId="4945" LastEditDate="2015-06-22T07:28:47.200" LastActivityDate="2018-02-02T06:31:43.617" Title="What are the prerequisites in hydrogen bonding?" Tags="&lt;bond&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33191" PostTypeId="2" ParentId="33190" CreationDate="2015-06-22T07:28:27.657" Score="3" Body="&lt;p&gt;Actually, theoretically speaking hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. A hydrogen bond may occur between $\ce{H-O}$, $\ce{H-N}$, $\ce{H-F}$ (first two are more common). Apart from that, these hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_bond&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article about hydrogen bonding&lt;/a&gt;&#xA;A hydrogen atom attached to a relatively electronegative atom will play the role of the hydrogen bond donor An electronegative atom such as fluorine, oxygen, or nitrogen will be the hydrogen bond acceptor, irrespective of whether it is bonded to a hydrogen atom or not. An example of a hydrogen bond acceptor that does not have a hydrogen atom bonded to it is the oxygen atom in diethyl ether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/4680/image115.png?revision=1&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/@api/deki/files/4680/image115.png?revision=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17052" LastEditorUserId="17052" LastEditDate="2015-06-22T08:06:23.743" LastActivityDate="2015-06-22T08:06:23.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33192" PostTypeId="2" ParentId="33187" CreationDate="2015-06-22T07:37:55.057" Score="5" Body="&lt;p&gt;There should not be any difference in experiment, because the extinction is something you have to calculate by measuring the transmittance/absorbance of a given frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transmittance $T$ of a material is the ratio between the received intensity of a certain frequency to the transmitted intensity of the same frequency. $$T=I/I_0$$&#xA;The absorbance $A$ is defined as the decadic logarithm of the inverse transmittance. $$A=\log_{10}\left(I_0/I\right)=\log_{10}\left(1/T\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using Beer-Lambert’s law $$I=I_0\operatorname{e}^{-\varepsilon~c~d}$$ one can calculate the extinction $\varepsilon$ from the measured transmittance/absorbance by $$\varepsilon=\frac{A}{c~d}$$ where $c$ is the concentration of the absorbing material and $d$ is the distance that the light has to travel in that given material.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorDisplayName="user7951" LastEditDate="2015-06-22T08:55:29.517" LastActivityDate="2015-06-22T08:55:29.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33193" PostTypeId="2" ParentId="33186" CreationDate="2015-06-22T08:32:56.767" Score="1" Body="&lt;p&gt;Attenuation of light related to the properties of material is governed by beer-lambert law Many molecules absorb ultraviolet or visible light. The absorbance of a solution increases as attenuation of the beam increases. Absorbance is directly proportional to the path length, b, and the concentration, c, of the absorbing species. Beer's Law states that&#xA;A = ebc, where e is a constant of proportionality, called the absorbtivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different molecules absorb radiation of different wavelengths. An absorption spectrum will show a number of absorption bands corresponding to structural groups within the molecule. For example, the absorption that is observed in the UV region for the carbonyl group in acetone is of the same wavelength as the absorption from the carbonyl group in diethyl ketone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is beer law description:&lt;a href=&quot;http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm&quot; rel=&quot;nofollow&quot;&gt;http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17052" LastActivityDate="2015-06-22T08:32:56.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33194" PostTypeId="1" CreationDate="2015-06-22T09:01:26.853" Score="6" ViewCount="1742" Body="&lt;p&gt;Is it possible to say water vapour and dew is the same? I know that the water vapour is the gaseous phase of water. But what is the dew? I just wonder how can I categorised it. Is it the liquid  or the gas phase of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure below shows dew formed in grass. It seems as a liquid. But is it actually  exhibit in liquid phase or gas phase? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ala7mm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="7448" LastEditDate="2015-06-22T17:42:45.990" LastActivityDate="2019-09-04T02:06:03.197" Title="Is dew and water vapour the same?" Tags="&lt;phase&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33195" PostTypeId="2" ParentId="33194" CreationDate="2015-06-22T09:15:49.743" Score="5" Body="&lt;p&gt;Let's have a look at the definitions of dew:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;dew&lt;/strong&gt; (&lt;a href=&quot;http://www.thefreedictionary.com/dew&quot; rel=&quot;noreferrer&quot;&gt;the free dictionary&lt;/a&gt;)&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Water droplets condensed from the air, usually at night, onto cool surfaces.&lt;/li&gt;&#xA;  &lt;li&gt;Something moist, fresh, pure, or renewing: &quot;The timely dew of sleep / ... inclines / Our eye-lids&quot; (John Milton).&lt;/li&gt;&#xA;  &lt;li&gt;Moisture, as in the form of tears or perspiration, that appears in small drops.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore dew is water in the condensed phase, i.e. liquid. Water vapour on the other hand is the gaseous phase. They are definitely not the same.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-22T09:15:49.743" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="33196" PostTypeId="1" CreationDate="2015-06-22T09:17:20.933" Score="4" ViewCount="795" Body="&lt;p&gt;In my group’s seminar last Friday, we discussed the total synthesis of 6-demethyl-6-deoxytetracycline &lt;strong&gt;1&lt;/strong&gt; by Woodward &lt;em&gt;et al.&lt;/em&gt;&lt;sup&gt;[1]&lt;/sup&gt; One intermediate of their synthesis is &lt;em&gt;meta&lt;/em&gt;-substituted anisole &lt;strong&gt;2&lt;/strong&gt; that they subsequently wish to cyclise to create the first two rings of &lt;strong&gt;1&lt;/strong&gt; by a Friedel-Crafts acylation. Since the Friedel-Crafts acylation can proceed on both the &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; carbons (when viewed from anisole’s oxygen substituent), they protected the &lt;em&gt;para&lt;/em&gt;-position using a chlorine atom (&lt;strong&gt;3&lt;/strong&gt;) — introduced by electrophilic aromatic substitution of &lt;strong&gt;2&lt;/strong&gt; with chlorine gas. They reported $97\:\%$ yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9r54G.png&quot; alt=&quot;chlorination of **2**; structure of **1**&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This confused me. I remember listening to a &lt;em&gt;Concepts in Organic Chemistry&lt;/em&gt; lecture given at my university where the electrophilic addition to an electron-rich aryl was discussed. Our professor gave us numbers that approximately equalled a $2\::\:1$ ratio of &lt;em&gt;ortho&lt;/em&gt; versus &lt;em&gt;para&lt;/em&gt; substitution. The reasoning was, that the HOMO has almost equal contributions on both &lt;em&gt;ortho&lt;/em&gt; carbons and the &lt;em&gt;para&lt;/em&gt; carbon (while having close to no &lt;em&gt;meta&lt;/em&gt; contributions), meaning that none of the three be favoured. This is approximately supported by the &lt;a href=&quot;http://www.cup.lmu.de/oc/zipse/benzyl-anion-c7h7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular orbitals of a calculated benzyl anion&lt;/a&gt; that the professor has uploaded to his site. (The benzyl anion was used as a model system because it gladly delocates its anionic electron pair into the ring much like oxygens would their unbonded electron pair.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do see that &lt;strong&gt;2&lt;/strong&gt; has a further substituent in &lt;em&gt;meta&lt;/em&gt; position. This substituent (alkyl) displays a weak $+\:\mathrm{I}$ effect meaning that it enhances substitution on the same atoms as the oxygen would. One of these is now &lt;em&gt;ortho&lt;/em&gt; to two substituents (position 2) and likely reacts less — I understand that. But I would still have expected a $1\::\:1$ ratio of &lt;em&gt;ortho&lt;/em&gt;-oxygen and &lt;em&gt;para&lt;/em&gt;-oxygen substituted products rather than $97\:\%$ yield of the &lt;em&gt;para&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we rationalise Woodward’s $97\:\%$? Is it wrong to say that the contributions of &lt;em&gt;ortho&lt;/em&gt; and &lt;em&gt;para&lt;/em&gt; carbons are approximately equal, and if so, why is the &lt;em&gt;para&lt;/em&gt; contribution so much stronger that it is almost solely chlorinated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://dx.doi.org/10.1021/ja01004a041&quot; rel=&quot;nofollow noreferrer&quot;&gt;J. J. Korst, J. D. Johnston, K. Butler, E. J. Bianco, L. H. Conover, R. B. Woodward, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1968&lt;/strong&gt;, &lt;em&gt;90&lt;/em&gt;, 439.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-22T09:17:20.933" Title="Is the para-position of anisoles more reactive than ortho in electrophilic substitutions?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;aromatic-compounds&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33197" PostTypeId="2" ParentId="33194" CreationDate="2015-06-22T09:18:18.317" Score="2" Body="&lt;p&gt;As dew is condensed water vapour &lt;a href=&quot;https://en.wikipedia.org/wiki/Dew&quot; rel=&quot;nofollow&quot;&gt;[1]&lt;/a&gt; and condensation is the change of matter from gas phase to liquid phase &lt;a href=&quot;https://en.wikipedia.org/wiki/Condensation&quot; rel=&quot;nofollow&quot;&gt;[2]&lt;/a&gt; dew is clearly liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on the one hand they are somehow the same as they both consist of water but on the other hand they are not the same as they are water in different states of matter.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-06-22T09:18:18.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33198" PostTypeId="2" ParentId="32951" CreationDate="2015-06-22T09:40:45.413" Score="0" Body="&lt;p&gt;yes, they are, all of them are bidirectional.&#xA;It is not possible to draw conclusions about the pH value, because of the different abilities to solve in water and because those reaction may cause an increased or a decreased temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: my teacher)&lt;/p&gt;&#xA;" OwnerUserId="16951" LastActivityDate="2015-06-22T09:40:45.413" CommentCount="2" CommunityOwnedDate="2015-06-22T09:40:45.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="33199" PostTypeId="1" AcceptedAnswerId="33201" CreationDate="2015-06-22T09:41:18.700" Score="7" ViewCount="3650" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When propane reacts with chlorine, two different monochloropropanes are formed.&lt;/p&gt;&#xA;&lt;p&gt;a) Name and show the full structural formulae of the two products.&lt;br /&gt;&#xA;b) What kind of isomerism is shown?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Based on my understanding, the answer for part a is &lt;span class=&quot;math-container&quot;&gt;$\ce{2C8H18 + Cl2 -&amp;gt; 2C8H17Cl +H2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Name: chloropropane, formula: &lt;span class=&quot;math-container&quot;&gt;$\ce{CH3CH2CH2CH2CH2CH2CH2CH(Cl)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;For part b, is it structural isomerism? Are my answers correct?&lt;/p&gt;&#xA;" OwnerUserId="17073" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-11-01T15:19:29.560" Title="Predicting the products of the radical chlorination of propane" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33201" PostTypeId="2" ParentId="33199" CreationDate="2015-06-22T10:45:18.030" Score="10" Body="&lt;p&gt;Let me give you two helpful hints.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Propane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products of octane, so you should feel good about being able to correctly apply the reaction type and produce a correct product.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the question asking?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A careful read of the intro part of the question &lt;code&gt;When propane reacts with chlorine, two different monochloropropanes are formed&lt;/code&gt; indicates that we should be expecting two different organic products with chlorine atoms attached to the carbon skeleton. In other words this reaction should give two products with the formula $\ce{C3H7Cl}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, part a) is not worded well to indicate that you mostly likely are to draw the structures of the two monochloropropane products. You are correct in noting that a radical halogenation reaction will have two products, though the inorganic byproduct is usually considered to be $\ce{HCl}$ and not $\ce{H2}$ or $\ce{H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H8 + Cl2 -&amp;gt; C3H7Cl + HCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let me ask your question in a clearer way: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Propane reacts with chlorine to produce two different monochlorination products with the formula $\ce{C3H7Cl}$. (a) Draw the structures of  both monochlorination products and provide their names. (b) Identify the type of isomeric relationship that exists between the two structures that you drew in (a).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-22T10:45:18.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33203" PostTypeId="2" ParentId="608" CreationDate="2015-06-22T11:05:47.553" Score="7" Body="&lt;p&gt;I can't find interesting details on the practical use of the Graham column: I mean it may be more adapted to a reflux set-up or a distillation one but it looks difficult to get information regarding this.  Intuitively, I'm prone to think that the Graham design does not fit with the requirements of reflux heating: As mentioned earlier, it can be easily clogged by low-boiling materials, inner diameter of the spiral is so small that returned-dripping and upward vapors will circulate only with great difficulties.  I guess the best use for this condenser is for downward condensation in a distillation set up. Better again, avoid this old-fashion glassware: I can't remind the presence of a single Graham condenser in a chemistry lab!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reflux boiling, there are 2 very efficient condensers: the Allinh is a good one for general applications, but the Dimroth &amp;amp; the double surface coil condenser can deal with a large amount of vapors and retain every single drop of solvent (if the appropriate length is selected wrt the volume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Friedrich is very useful in addition to a standard Liebig condenser, when placed at the end of the distillation setup its cooling surface makes it unavoidable for ethers, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: The Dimroth is the condenser of choice for soxlhet extractions, since the drip tip on the terminal loop of the spiral allows a constant dripping on longitudinal axis and straight to the extraction thimble.&lt;/p&gt;&#xA;" OwnerUserId="17076" LastEditorUserId="5354" LastEditDate="2016-02-14T00:24:55.360" LastActivityDate="2016-02-14T00:24:55.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33204" PostTypeId="1" AcceptedAnswerId="33205" CreationDate="2015-06-22T11:10:32.870" Score="1" ViewCount="331" Body="&lt;p&gt;I made a PCB (printed circuit board) for the first time, and the process included covering the board with the shape we want, exposing the board to UV light and then putting it in a developer which removes the part of the resin that was exposed to the UV radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is this developer? I think it has the commercial name of CIF (&lt;a href=&quot;http://uk.farnell.com/cif/ar46/developer-ready-for-use-1l/dp/1783557&quot; rel=&quot;nofollow&quot;&gt;http://uk.farnell.com/cif/ar46/developer-ready-for-use-1l/dp/1783557&lt;/a&gt;), but what is it really? What does it do to the board and why only to the UV-exposed part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="13202" LastActivityDate="2015-06-22T11:28:30.627" Title="What is CIF (a PCB developer) and what does it do?" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33205" PostTypeId="2" ParentId="33204" CreationDate="2015-06-22T11:28:30.627" Score="3" Body="&lt;p&gt;Developers are usually solutions of $\ce{NaOH}$ or $\ce{KOH}$ (perhaps with some additives). Negative photoresists uses organic developers/solvents. From your description, I assume, that you use a positive photoresist. Positive photoresists are more soluble after exposition to the UV light - polymer bonds are disturbed and therefore more soluble (in areas where exposed). Please see illustration. For more detailed information look for &lt;a href=&quot;https://en.wikipedia.org/?title=Photoresist&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photoresists&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Photolithography&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photolithography&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LLh0l.png&quot; alt=&quot;wiki&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-06-22T11:28:30.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33206" PostTypeId="1" AcceptedAnswerId="33207" CreationDate="2015-06-22T12:18:27.507" Score="10" ViewCount="7431" Body="&lt;p&gt;I know it has something to do with carbon dioxide, but what exactly happens and causes the characteristic sound we hear (especially after we shake the bottle)? How do we manage to dissolve a non-polar substance ($\ce{CO2}$) in a polar solvent ($\ce{H2O}$)? &lt;/p&gt;&#xA;" OwnerDisplayName="user17037" LastEditorUserId="7448" LastEditDate="2015-06-22T12:19:43.067" LastActivityDate="2017-09-29T03:09:27.383" Title="Why do soft drinks fizz?" Tags="&lt;everyday-chemistry&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33207" PostTypeId="2" ParentId="33206" CreationDate="2015-06-22T12:50:26.757" Score="15" Body="&lt;p&gt;Because $\ce{CO_2}$ isn't exactly what's being dissolved. When it encounters water, it immediately encounters the following potential reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2~(g) + H2O~(l) &amp;lt;=&amp;gt; H2CO3~(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon dioxide is non polar, yes, but carbonic acid readily dissolves (and dissociates, but I didn't care to show that) in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sound we hear is because when we open the bottle, the equilibrium above shifts to the left, creating more carbon dioxide that is being released. Why does the equilibrium shift? It has to do with the pressure in the bottle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the bottle is closed, it's under a certain set pressure. When it's closed at the same temperature and pressure for a long time, the reaction above reaches an equilibrium state, where the forward and backward versions of the reaction are happening at the same rate. However, when the bottle is opened, the pressure inside the bottle decreases, as the gas in the bottle suddenly has a lot more space to expand into. A decrease in pressure favors the side of an equilibrium that has more gaseous molecules, which in this case means the carbon dioxide. Ergo, when you open the bottle, the gas is formed, escapes from solution, and rapidly expands outward, making the infamous popping noise.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="4945" LastEditDate="2015-06-23T10:52:37.120" LastActivityDate="2015-06-23T10:52:37.120" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33208" PostTypeId="2" ParentId="33206" CreationDate="2015-06-22T12:53:56.277" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do soft drinks fizz?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key word for that is &lt;strong&gt;pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manufactures close the bottles/cans by forcing carbon dioxide and water into your soda at high pressures around &lt;strong&gt;&lt;a href=&quot;http://hypertextbook.com/facts/2000/SeemaMeraj.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;40 psi&lt;/a&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you crack open the can, you release the pressure and allow the gas bubbles to wiggle free from the liquid and rise to the surface. This requires energy because in order for the gas to break free from the liquid it has to overcome the force holding the liquid together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WhY2w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.livescience.com/32492-why-does-soda-fizz.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIVESCIENCE&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-06-24T06:44:15.553" LastActivityDate="2015-06-24T06:44:15.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33209" PostTypeId="2" ParentId="33189" CreationDate="2015-06-22T13:00:14.867" Score="4" Body="&lt;p&gt;You are just making a small sign error:&#xA;\begin{align}&#xA;ax^2 +bx +c &amp;amp;= 0 &amp;amp; x_{(1;2)} &amp;amp;= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking $K=\frac13$:&#xA;\begin{align}&#xA;\frac13x^2 - \frac{21}{3}x +\frac{80}{3} &amp;amp;= 0 &amp;amp;&#xA; x_{(1;2)} &amp;amp;= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\&#xA;&amp;amp;&amp;amp; x_{(1;2)} &amp;amp;= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\&#xA;&amp;amp;&amp;amp;           &amp;amp;= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\&#xA;&amp;amp;&amp;amp; x_{(1;2)} &amp;amp;= 16; 5&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-22T13:00:14.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33212" PostTypeId="2" ParentId="33180" CreationDate="2015-06-22T15:55:48.670" Score="7" Body="&lt;h2&gt;Volatile surfactants&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are many molecules that are both surfactants (in water) and volatile, but of the ones I can think of, &lt;strong&gt;none can be used safely in home laundry applications.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Hydrocarbon derivatives.  The examples you found by googling are in this class.  3,5-dimethyl-1-hexyn-3-ol is the principal component of Surfynol 61 (probably named Surfynol 61 because it is a surface-active (surf) alkyne alcohol (ynol) with six carbons (6) and the alkyne moiety at carbon one (1).  Other examples would be medium-chain fatty alcohols like hexanol, octanol, cyclohexanol, etc.  These molecules reduce surface tension of water and have OK-ish cleaning power (not as good as larger but non-volatile alternatives like &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_dodecyl_sulfate&quot;&gt;SDS&lt;/a&gt;), but their volatility is also their downfall: their use in home environments creates a strong fire and explosion hazard when used in automatic heated dryers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fluorocarbon derivatives.  I found the &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/la9713412&quot;&gt;same article&lt;/a&gt; you probably did about volatile fluorocarbon surfactants.  The most powerful at reducing the surface tension of water seems to be &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/331023&quot;&gt;nonafluoro-&lt;em&gt;tert&lt;/em&gt;-butyl alcohol&lt;/a&gt;, which reduces water surface tension by 78%!  However this experiment required saturating the water with the vapor of this compound.  And 1 gram costs &gt;$80.  And toxicity is a concern, despite this compound apparently being researched as a &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19800006431.pdf&quot;&gt;component of or precursor to artificial blood mixtures&lt;/a&gt; in the 1970s.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other halocarbon derivatives.  These include solvents like perchloroethylene, chlorobutanol, etc.  These are all highly toxic, and with some of them, there is a flammability/explosion risk as with hydrocarbon derivatives.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Non-surfactant use&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You might consider just using organic solvents that are not (very) surface active as a way to degrease fabric.  However, these may leach dyes out of the fabric or otherwise damage them.  Isopropyl alcohol (IPA) is accessible to home users, and is relatively safe.  Be sure to do the soaking in a well-ventilated area, and &lt;strong&gt;do not use heated dryers to remove the IPA&lt;/strong&gt;!  Instead, rinse out the IPA thoroughly with water.  Depending on local regulations, you will have to dispose of all of the dirty IPA and IPA-contaminated rinsing water as solvent waste.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The best option&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Pay to have your clothes dry cleaned!  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-22T15:55:48.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33215" PostTypeId="1" AcceptedAnswerId="33216" CreationDate="2015-06-22T16:59:44.230" Score="7" ViewCount="579" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;You are asked to prepare a 0.8500 M solution of aluminum nitrate. You find that you have only 50.00 g of the solid.&lt;/p&gt;&#xA;&lt;p&gt;What is the maximum volume of solution that you can prepare?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm confusing my molar masses with grams. The molar mass of Aluminum nitrate &lt;span class=&quot;math-container&quot;&gt;$\ce{Al(NO3)3}$&lt;/span&gt; is 213.0 g. If I have 50.00 g of the entire solid &lt;span class=&quot;math-container&quot;&gt;$\ce{Al(NO3)3}$&lt;/span&gt; does that mean I have 50.00 times the amount of 213.0 g or will I end up Multiplying 50.00 g of Aluminum nitrate by 0.8500 M, then dividing that by the MM of Aluminum Nitrate. But then I don't know how to set up the rest of my dimensional analysis to reach Liters.&lt;span class=&quot;math-container&quot;&gt;$%edit$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="17082" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-28T07:00:34.053" Title="Relation between moles, molar mass and mass" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33216" PostTypeId="2" ParentId="33215" CreationDate="2015-06-22T17:28:49.133" Score="4" Body="&lt;p&gt;If you have 50 g of a compound that has a molar mass of 213 $\frac{\rm{g}}{\rm mol}$, then you have $\frac{50~\rm{g}}{213 \frac{\rm{ g}}{\rm mol}}=\frac{50}{213} \rm{mol}=0.234~\rm{mol}$ of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prepare a 0.8500 molar solution, that means the concentration you want is $0.85~\rm{M}=0.85\frac{\rm{mol}}{\rm{L}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those two numbers, the final answer should be apparent.  (Just divide them and make sure the units work out!)&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-22T17:28:49.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33217" PostTypeId="2" ParentId="33215" CreationDate="2015-06-22T17:29:33.250" Score="4" Body="&lt;p&gt;You have $x = (50~\mathrm{g} / 213~\mathrm{g})$ moles of aluminum nitrate. You need some volume of a solution where the molarity of aluminum nitrate is $0.85$. If you had $181.05~\mathrm{g}$ of aluminum nitrate in 1 liter of solution, you'd be fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you have $x$ moles. Thus, you need to scale the volume of solution accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $0.85~\mathrm{mol / L} =\frac{ x~\mathrm{mol}}{y~\mathrm{L}}$, using the value you calculate for $x$ as above, and solve for $y$, the solution volume in liters.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="4945" LastEditDate="2015-06-23T04:51:21.887" LastActivityDate="2015-06-23T04:51:21.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33218" PostTypeId="1" CreationDate="2015-06-22T17:39:26.350" Score="1" ViewCount="701" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The equilibrium constant for the following reaction at $\pu{600^{\circ}C}$ is 4.0.  Initially, two moles of $\ce{CO}$ and one mole of $\ce{H2O}$ were mixed in a $\pu{1.0 L}$ container.  Determine the concentration of all species at equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Attempt #1:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO (g) + H2O (g) -&amp;gt; CO2 (g) + H2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|r|c|c|c|}&#xA;\hline&#xA;\text{Initial}~(M)     &amp;amp; 2.0      &amp;amp; 1.0     &amp;amp; 0  &amp;amp; 0 \\ \hline&#xA;\text{Change}~(M)      &amp;amp;  -x      &amp;amp; -x      &amp;amp; +x &amp;amp; +x \\ \hline&#xA;\text{Equilibrium}~(M) &amp;amp; 2.0 – x  &amp;amp; 1.0 - x \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Plug $x$ into the equilibrium expression and solve for $x$. $4.0  = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Determine concentrations: the equilibrium values become 2.0 – 0.85, 1.0 - 0.85, 0.85, and 0.85, giving 1.2, 0.1, 0.85, and 0.85.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where does the 0.85 come from? Could it be cross multiplied some way or is there another way this is done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I want to make sure that I've done the 3.00 correctly for $\ce{Fe(SCN)2+}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider the reaction represented by the equation: $$\ce{Fe^3+ (aq) + SCN- (aq) -&amp;gt; Fe(SCN)2+ (aq)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;Attempt #2:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{ccc}&#xA;\text{Initial}     &amp;amp; \pu{6.00 M}\ \ce{Fe^{3+} (aq)} &amp;amp; \pu{6.00 M}\ \ce{SCN^{−} (aq)} \\&#xA;\text{Equilibrium} &amp;amp; \pu{? M}\ \ce{FeSCN2^+ (aq)}, \ K = 0.33&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\frac{1}{3}x^2 - \frac{124}{25}x + \frac{297}{25} &amp;amp;= 0 &amp;amp;&#xA; x_{(1;2)} &amp;amp;= \frac{\color{red}{\pmb{-}}(-\frac{124}{25})\pm\sqrt{\left(\frac{124}{25}\right)^2 - 4\cdot\frac13\frac{297}{25}}}{\frac{2}{3}} \\&#xA;&amp;amp;&amp;amp; x_{(1;2)} &amp;amp;= \frac{124/25\pm\sqrt{\frac{15376}{(625)}-\frac{15601}{1000}}}{\frac{2}{3}} \\&#xA;&amp;amp;&amp;amp;           &amp;amp;= \left(124/25\pm{3}\right)\cdot\frac{3}{2} \\&#xA;&amp;amp;&amp;amp; x_{(1;2)} &amp;amp;= 12;3&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="194" LastEditDate="2018-02-19T05:11:41.090" LastActivityDate="2018-02-19T05:11:41.090" Title="Determining equilibrium concentrations from initial conditions and equilibrium constant" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33219" PostTypeId="1" CreationDate="2015-06-22T18:05:35.637" Score="1" ViewCount="719" Body="&lt;p&gt;We were doing a titration experiment to find out how much ascorbic acid a vitamin pill actually contains. In a flask, there is a solution of &lt;strong&gt;water&lt;/strong&gt;, &lt;strong&gt;sulfuric acid&lt;/strong&gt;, &lt;strong&gt;potassium iodide&lt;/strong&gt; and &lt;strong&gt;ascorbic acid&lt;/strong&gt;. Also starch as an indicator. When adding &lt;strong&gt;potassium iodate&lt;/strong&gt; $(\text{KIO}_3)$, the following reactions will appear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{IO}_3^- + 5 \ \text{I}^- + 6 \ \text{H}^+ \longrightarrow 3 \ \text{I}_2 + 3 \ \text{H}_2\text{O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the iodine is formed, the following reaction will appear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{C}_6\text{H}_8\text{O}_6 + \ \text{I}_2 \longrightarrow \text{C}_6\text{H}_6\text{O}_6 + 2\ \text{H}^+ + 2 \ \text{I}^-$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After using the right amount of potassium iodate, there will be no ascorbic acid anymore, causing to eliminate the second reaction. When there is no ascorbic acid, iodine will not react back to iodine-ions, which causes the starch to give a brown/blue color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why isn't this possible without the sulfuric acid? &lt;/p&gt;&#xA;" OwnerUserId="13221" LastActivityDate="2018-04-23T21:06:02.693" Title="Indirect titration with ascorbic acid" Tags="&lt;titration&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33220" PostTypeId="1" CreationDate="2015-06-22T18:11:41.160" Score="1" ViewCount="27" Body="&lt;p&gt;I know atomic mass can be (over) simplified as the average mass of all isotopes of a given element weighted by the ratio of their natural occurrence. I also know that, as far as we know, refining relative atomic mass is a matter of finding new significant figures, not finding new isotopes. Are we reasonably sure these ratios won't change as we explore the universe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an over simplified example, do we have a reasonable certainty that 70% of the Calcium on Mars won't turn out to be Ca-41?&lt;/p&gt;&#xA;" OwnerUserId="17084" LastActivityDate="2015-06-22T18:11:41.160" Title="Do we have a reasonable assurance that relative atomic mass is constant throughout the universe?" Tags="&lt;physical-chemistry&gt;&lt;atoms&gt;&lt;isotope&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-06-22T18:22:57.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="33221" PostTypeId="2" ParentId="33218" CreationDate="2015-06-22T19:38:25.153" Score="4" Body="&lt;p&gt;The text&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x] &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;appears to be an error. The proper equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4  = \frac{x^2}{(2-x)(1-x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 0.85 is one of two roots of the quadratic equation you'll solve, using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after you simplify the first expression, above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the rest of your question, please expand: What does&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the 3.00&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mean?&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-06-22T19:48:04.073" LastActivityDate="2015-06-22T19:48:04.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33222" PostTypeId="1" CreationDate="2015-06-22T19:46:07.890" Score="4" ViewCount="586" Body="&lt;p&gt;My class just did a lab today where we mixed cobalt oxalate with &lt;span class=&quot;math-container&quot;&gt;$40~\mathrm{mL}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; and then titrated it with &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; (potassium permanganate). I need the mole ratio of cobalt oxalate and &lt;span class=&quot;math-container&quot;&gt;$\ce{KMnO4}$&lt;/span&gt; and am not sure how to set up the equation. She provided us with this formula with 3 variables A,B,C - &#xA;&lt;code&gt;Co(a)(C2O4)(b) * (c)H2O&lt;/code&gt; &#xA;but I am not sure what to do with it. I am also confused as cobalt oxalate appears as &lt;span class=&quot;math-container&quot;&gt;$\ce{CoC2O4}$&lt;/span&gt; in her equation yet Google says it's &lt;span class=&quot;math-container&quot;&gt;$\ce{H2C2CoO6}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My table in my lab book is asking me for &quot;Moles Oxalate Present&quot; , &quot;Mass of Oxalate in Sample&quot; , and &quot;Percent Mass of Oxalate in Sample (%)&quot;. It is also referring to the &quot;Mass of transition metal oxalate used&quot; which we took during the actual lab. (I guess this lets me know that it solely wants oxalate and to calculate these values aside from cobalt oxalate?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very confused and if any part of this question needs to be cleared up I can certainly do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can provide any/all data that I took during the lab, but explaining conceptually will still do me some good here. I really just don't know where to start with all this data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book also gave me this formula for the reaction of oxalate ions with permanganate ions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{5C2O4^{2-} + 2MnO4- + 16H+ -&amp;gt; 10CO2 + 2Mn^{2+} + 8H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this factor at all into my calculations?&lt;/p&gt;&#xA;" OwnerUserId="17085" LastEditorUserId="4945" LastEditDate="2019-04-15T14:16:08.993" LastActivityDate="2019-04-15T14:16:08.993" Title="Cobalt Oxalate / Potassium Permanganate Titration" Tags="&lt;redox&gt;&lt;titration&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33223" PostTypeId="1" CreationDate="2015-06-22T21:10:00.463" Score="3" ViewCount="886" Body="&lt;p&gt;Does anyone know what is scattering coefficient, absorption coefficient and extinction coefficient, and how to separate them experimentally?&lt;/p&gt;&#xA;" OwnerUserId="17066" LastActivityDate="2015-06-22T22:49:06.637" Title="several coefficient differences in UV-vis Spectroscopy" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33224" PostTypeId="1" CreationDate="2015-06-22T21:23:12.717" Score="2" ViewCount="211" Body="&lt;p&gt;In UV spectroscopy, when a beam of light is shone into solution, what is scattering loss during this process? &lt;/p&gt;&#xA;" OwnerUserId="17066" LastEditorDisplayName="user15489" LastEditDate="2015-06-22T21:42:14.690" LastActivityDate="2015-06-22T21:42:14.690" Title="What are scattering losses?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33225" PostTypeId="2" ParentId="33224" CreationDate="2015-06-22T21:39:25.337" Score="4" Body="&lt;p&gt;Scattering is the redirection of the 'beam' of light, as is illustrated in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dC574.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://cimss.ssec.wisc.edu/wxwise/class/optics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Wisconsin&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description from the document &lt;a href=&quot;http://www.chem.agilent.com/library/slidepresentation/public/present.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fundamentals of modern&#xA;UV-visible spectroscopy&lt;/a&gt; (Figure 46) states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Scattering causes an apparent absorbance because less light&#xA;  reaches the detector&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They also detail two main types of scattering in UV spectroscopy (Figure 47):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;• &lt;a href=&quot;https://en.wikipedia.org/?title=Rayleigh_scattering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh scattering&lt;/a&gt;: Particles small relative to wavelength&lt;br&gt;&#xA;  • &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colloid/Tyndall_Effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tyndall scattering&lt;/a&gt;: Particles large relative to wavelength&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-06-22T21:39:25.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33226" PostTypeId="2" ParentId="33223" CreationDate="2015-06-22T22:49:06.637" Score="3" Body="&lt;p&gt;In most applications, these terms are used interchangeably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most absorbance spectrophotometers work by measuring the intensity of the light beam that is &lt;em&gt;transmitted&lt;/em&gt; through the sample.  A strong light beam goes in, and a weaker one comes out the other side.  The transmission is the fraction of the light that made it through.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main ways in which the light that goes in gets diminished in intensity before it goes out.  The first is &lt;em&gt;absorption&lt;/em&gt;.  Molecules that absorb light of a certain frequency make photons of that frequency disappear.  The energy of those photons excite electrons into higher energy states.  These states may eventually decay, releasing heat or lower-energy photons, &lt;em&gt;but not photons of the same energy that were absorbed&lt;/em&gt;.  The propensity of molecules to absorb can be called the &lt;em&gt;absorption coefficient&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scattering just deflects the incident light to a new angle.  Now, when it exits the sample, it won't come straight out on the side opposite where it went in, and it won't be detected by the detector.  Molecules can cause scattering if they are very large.  The &quot;scattering coefficient&quot; describes a molecule's propensity to scatter light of a certain frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;extinction coefficient&quot; is -- in most applications -- just equal to the scattering coefficient plus the absorption coefficient, i.e., it's proportional to a molecule's ability to disrupt incident light in some way or other, regardless of the mechanism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the usual, transmission-based spectrophotometer, only extinction coefficients can truly be measured.  To resolve these measurements into scattering contributions and absorption contributions would require measuring the intensity of the scattered light, i.e. detectors would need to be present at multiple angles.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-22T22:49:06.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33227" PostTypeId="1" CreationDate="2015-06-22T23:46:19.443" Score="0" ViewCount="327" Body="&lt;p&gt;I have Ferric Chloride and Aluminum...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have steel filings from grinding steel objects, though some Iron Oxide have be separated (I do have Oxalic Acid and Citric Acid). Though that would give me steel powder, not iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the most economical and would yield purest samples?&lt;/p&gt;&#xA;" OwnerUserId="15043" LastActivityDate="2015-06-23T03:22:09.673" Title="I need Iron powder. What's the best way to produce it?" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33229" PostTypeId="2" ParentId="33227" CreationDate="2015-06-23T00:24:25.260" Score="3" Body="&lt;p&gt;Sigma-Aldrich has a &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/267953?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;product&lt;/a&gt; that is very pure but rather expensive. I suspect you can have high purity or low cost but not both.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-23T00:24:25.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33230" PostTypeId="5" CreationDate="2015-06-23T00:36:05.967" Score="0" Body="&lt;p&gt;Distillation is the process of using disparate boiling points to separate mixtures of miscible liquids.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-06-23T03:21:47.030" LastActivityDate="2015-06-23T03:21:47.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33231" PostTypeId="4" CreationDate="2015-06-23T00:36:05.967" Score="0" Body="For questions about the process of distillation. For the distillation process used in metal refining, also use the [metallurgy] tag." OwnerDisplayName="user467" LastEditorUserId="5026" LastEditorDisplayName="user467" LastEditDate="2018-03-17T03:02:10.343" LastActivityDate="2018-03-17T03:02:10.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33232" PostTypeId="1" CreationDate="2015-06-23T01:17:42.167" Score="5" ViewCount="187" Body="&lt;p&gt;I have a 8000 gallon water tank that has always had pristine clear water.   I have a brass float valve at the top input to the tank.  Someone put an aluminum ladder in the tank (15 ft deep tank) and after who knows how long, the ladder was encrusted with a white gooey powder and the water turned milky.  what happened?  Main question is IS the water safe to drink or how do I get rid of whatever is in there.&lt;/p&gt;&#xA;" OwnerUserId="17094" LastEditorDisplayName="user15489" LastEditDate="2015-06-23T01:20:01.897" LastActivityDate="2015-06-23T01:56:13.447" Title="Aluminium and water safety" Tags="&lt;water&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33233" PostTypeId="2" ParentId="33232" CreationDate="2015-06-23T01:56:13.447" Score="3" Body="&lt;p&gt;The white stuff is aluminum hydroxide. The water is fine, filter the solids with a coffee filter if it doesn't settle to the bottom. And remove the ladder.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-23T01:56:13.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33235" PostTypeId="2" ParentId="33227" CreationDate="2015-06-23T03:22:09.673" Score="0" Body="&lt;p&gt;You could use a high grade metal file on some clean cast iron or an iron nail and coldest the metal shavings which is similar consistency to a powder. An iron nail would be galvanized though so you'd have to take the coating off first. Impurities may be present from the file though.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastActivityDate="2015-06-23T03:22:09.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33238" PostTypeId="1" CreationDate="2015-06-23T05:16:15.250" Score="0" ViewCount="34013" Body="&lt;p&gt;Is it dangerous to boil dish soap (natural, ex: walgreens brand &quot;ology&quot;) while also cooking other foods in the kitchen (ex.pasta)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to clean some pots and read online that a good way to clean them is to boil dish soap in them. I was also making dinner at the same time and I am worried that I released toxic chemicals into the air and into my food.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How hot does soap need to be heated to cause it to decompose, and what substances will be formed by the decomposition of the soap ?&lt;/p&gt;&#xA;" OwnerUserId="17097" LastEditorUserId="62812" LastEditDate="2018-06-02T10:21:59.233" LastActivityDate="2019-02-24T21:20:06.230" Title="Boiling dish soap" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33239" PostTypeId="2" ParentId="33238" CreationDate="2015-06-23T07:56:11.137" Score="3" Body="&lt;p&gt;You probably don’t have to worry about toxic chemicals in the air/food. Most dish washing liquids/gels/soaps contain chemicals that can be tolerated fairly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, boiling dish soap in a pot can potentially generate harmful substances. This may happen when your dish soap is very concentrated (when nearly all water has evaporated) and the temperature in your pot may rise to a point at which the ingredients of your dish soap could decompose or generate noxious fumes. If you make sure that there is enough water in your point at any time, your maximum temperature should not go above the boiling point of water (100 °C / 212 °F) which should be fine for most detergents.&lt;/p&gt;&#xA;" OwnerUserId="17102" LastEditorDisplayName="user7951" LastEditDate="2015-06-23T08:18:35.390" LastActivityDate="2015-06-23T08:18:35.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33242" PostTypeId="1" AcceptedAnswerId="33248" CreationDate="2015-06-23T12:35:20.460" Score="8" ViewCount="2458" Body="&lt;p&gt;Cyanide and benzaldehyde both have an almond-like smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people cannot smell cyanide (specifically, HCN). Can they also not smell benzaldehyde?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, do they both trigger the same receptors?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorDisplayName="user7951" LastEditDate="2016-10-12T09:06:51.427" LastActivityDate="2016-10-12T09:06:51.427" Title="Benzaldehyde and Cyanide - if you can smell one can you smell the other?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33244" PostTypeId="1" AcceptedAnswerId="33249" CreationDate="2015-06-23T13:21:19.513" Score="4" ViewCount="480" Body="&lt;p&gt;It looks like the human body doesn't have the ability to breakdown atmospheric $\ce{N2}$, as of now $\ldots$ So an enzyme like nitrogenase (found in cyanobacteria) fixes $\ce{N2}$ and produces $\ce{NH3}$. $\ce{NH3}$ is a weak base which qualifies as a buffer component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the human body was able to react to $\ce{N2}$ inhaled from the air, would it negatively impact the body's ability to maintain pH levels of vascular systems/cells of the human body given the large amount of $\ce{N2}$ in the atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="7450" LastEditDate="2015-06-24T14:46:43.820" LastActivityDate="2016-03-11T23:03:51.760" Title="Atmospheric Nitrogen gas and human body pH buffers" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33245" PostTypeId="1" AcceptedAnswerId="33246" CreationDate="2015-06-23T14:32:44.563" Score="2" ViewCount="383" Body="&lt;p&gt;I'm interested in the concept behind this computation and was wondering if anyone could give me some pointers or hints.&#xA;I've prepared $0.28~\mathrm{L}$ solution of $0.8500~\mathrm{M}$ of $\ce{Al(NO3)3}$.&#xA;I need a $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$ of $\ce{Al(NO3)3}$ made from my original solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my dimensional analysis what I need should be up front right?&#xA;$50~\mathrm{mL}$ of the solution first converted to liters for the bridge conversion to 0.450 moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I have the molarity of my $50~\mathrm{mL}$ set to $0.450~\mathrm{mol/L}$ I divide my molarity by the molarity of the original solution I am taking this out of in the first place which gives me 0.026 liters of solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.026~\mathrm{L}$ of solution is then what I need out of my $0.28~\mathrm{L}$ $0.8500~\mathrm{M}$ solution in order to come up with my 50 milliliters of $0.450~\mathrm{M}$ of solution? (I feel that something is missing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My computations (conceptually explained above in my best effort)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$50~\mathrm{mL} \cdot 1~\mathrm{L}/1000~\mathrm{mL} \cdot 0.450~\mathrm{M}/1~\mathrm{L} \cdot 1~\mathrm{L}/0.8500~\mathrm{M} = 0.026~\mathrm{L}$ of $0.8500~\mathrm{M}$ solution needed to prepare $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$. &lt;/p&gt;&#xA;" OwnerUserId="17082" LastEditorUserId="69830" LastEditDate="2021-05-08T18:31:27.157" LastActivityDate="2021-05-08T18:31:27.157" Title="What is the concept behind the calculations for the dilution of solutions?" Tags="&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33246" PostTypeId="2" ParentId="33245" CreationDate="2015-06-23T15:09:18.970" Score="3" Body="&lt;p&gt;I'm not sure the exact question but I suppose you want to find the volume of 1st solution needed to come up with the 2nd solution. I always feel more comfortable with moles for this types of problem. Because mole is conserved (in a way).&lt;br&gt;&#xA;So, to make the 2nd solution you need  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$50~\mathrm{ml} \times \frac{1~\mathrm{l}}{1000~\mathrm{ml}} \times 0.45~\mathrm{\frac {mol}{l}}=0.0225~\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means you need $0.0225~\mathrm{mol}$ of $\ce{Al(NO3)3}$ from the first solution. You know the molarity and mole number. So the volume is &#xA;$$\frac {0.0225~\mathrm{mol}}{0.85~\mathrm{\frac{mol}{l}}}=0.0264~\mathrm{l}=26.47~\mathrm{ml}.$$ &#xA;But you need $50~\mathrm{ml}$ of the 2nd solution, so you need to add solvent (Water or whatever you used). The volume of extra solvent required is $50-26.47=23.53~\mathrm{ml}$. Now you got the 2nd solution and to check that you have got everything right, you can find the molarity of your new solution.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="4945" LastEditDate="2015-06-24T07:52:12.447" LastActivityDate="2015-06-24T07:52:12.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33247" PostTypeId="2" ParentId="33003" CreationDate="2015-06-23T15:11:51.747" Score="2" Body="&lt;p&gt;&lt;strong&gt;How to get rid of water to collect calcium acetate&lt;/strong&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to do it in an environmentally responsible way&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Electricity from the outlet cost you money and probably used some&#xA;combination of coal, natural, gas, and or nuclear to produce it. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bunsen burner, but it likely uses a fuel that was taken from the ground, made from a crop that that competes against food production for land.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solar cells to produce electricity. It uses renewable energy but&#xA;is relatively expensive and electricity is a high value energy&#xA;compared to solar thermal. Using commercial grade solar cells only&#xA;converts about 17% of the energy in sunlight into electricity that is then converted into heat to evaporate the water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solar thermal is cheap and easy to make with household items like an aluminum or soup can and black spray paint. Further more, it has the potential to be more than twice as efficient as solar cells at providing energy you need to evaporate water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrolysis&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on electrolysis for experimental reasons, calcium acetate is a salt of acetic acid and calcium from the calcium carbonate in the egg shells. As a salt, like all salts, calcium acetate will dissolve in water forming ions that will conduct electricity. So, there is no need for any electrolyte other than water and calcium acetate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, your end products will be something other than calcium acetate and would depend in part on your anode and cathode, temperature, and applied voltage. &lt;/p&gt;&#xA;" OwnerUserId="17106" LastEditorUserId="17106" LastEditDate="2015-06-23T15:58:01.000" LastActivityDate="2015-06-23T15:58:01.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33248" PostTypeId="2" ParentId="33242" CreationDate="2015-06-23T15:18:52.187" Score="5" Body="&lt;p&gt;Answer is, they are most probably not correlated. Smell requires molecules to bind to receptors, which they do by structure and functional group. CN is a very small linear molecule that does not even share its heteroatoms with benzaldehyde, an aromatic compound several times cyanide's size. Ergo, it's improbable they share the same scent receptors.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-23T15:18:52.187" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33249" PostTypeId="2" ParentId="33244" CreationDate="2015-06-23T15:59:18.333" Score="5" Body="&lt;p&gt;One chemical equation that can be written for nitrogenase is this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{N_2 + 8 \ H^+ + 8 \ e^- + 16 \ ATP \longrightarrow 2 \ NH_3 + H_2 + 16 \ ADP + 16 P_{i}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation shows that not only does nitrogenase generate the basic molecule $\ce{NH3}$, it also consumes protons (i.e. acid) to form hydrogen as an inescapable byproduct of its mechanism.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, in theory nitrogenase could perturb the human body's ability to buffer pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the relative rates are always important.  Human requirements for nitrogen are &lt;a href=&quot;http://www.ucl.ac.uk/~ucbcdab/Nbalance/Nbalance.htm&quot; rel=&quot;nofollow&quot;&gt;on the order of 80 grams of protein per day&lt;/a&gt;.  This translates to about 12 g of (fixed) nitrogen atoms per day, which corresponds to 0.85 mol of N.  If all of that nitrogen were to come from nitrogenase, then 0.85 &amp;times; 8 &amp;div; 2 = 3.4 mol of acidity would be lost via proton consumption in addition to the 0.85 mol of base generated as $\ce{NH3}$, or about 5.2 mol of &quot;base&quot; in total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But human pH homeostasis works even though &lt;a href=&quot;https://micpohling.wordpress.com/2007/03/27/math-how-much-co2-is-emitted-by-human-on-earth-annually/&quot; rel=&quot;nofollow&quot;&gt;humans breathe out about 900 g of an acid equivalent&lt;/a&gt; ($\ce{CO2}$) per day, or about 20 mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even if nitrogenase completely replaced human nitrogen requirements in some hypothetic bioengineered human, the demands on pH homeostasis would only change by about 25%, maximum.  And considering that pH homeostasis still works even when humans are exercising vigorously, when &quot;instantaneous&quot; $\ce{CO2}$ production rates are far higher, I doubt even a 25% change would be too much to handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus in conclusion I don't think nitrogenase would wreck human pH homeostasis, unless its activity in the human body was increased to a level far beyond what is necessary to replace dietary N requirements.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:03:51.760" LastActivityDate="2016-03-11T23:03:51.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33250" PostTypeId="1" CreationDate="2015-06-23T16:03:16.343" Score="13" ViewCount="25722" Body="&lt;p&gt;Or can there be a homogeneous mixture that is not a solution? &lt;/p&gt;&#xA;" OwnerUserId="14481" LastActivityDate="2016-07-08T16:17:03.967" Title="Are all homogeneous mixtures just solutions?" Tags="&lt;solutions&gt;&lt;mixtures&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33251" PostTypeId="2" ParentId="33250" CreationDate="2015-06-23T16:08:34.817" Score="3" Body="&lt;p&gt;For the liquid phase, I think it is probably true that all homogeneous mixtures are solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the situation is more complicated for solid phases.  In a diamond crystal, if the $\ce{^{13}C}$ atoms are homogeneously distributed in the crystal, could you really say that the diamond is a &quot;solution&quot; of $\ce{^{13}C}$ in a $\ce{^{12}C}$ &quot;solvent&quot;?  I don't think so.  $n$-doped or $p$-doped silicon is another example.  In contrast, other homogeneous solids, like &lt;a href=&quot;http://chemistry.about.com/od/alloys/f/Is-Brass-A-Solution.htm&quot; rel=&quot;nofollow&quot;&gt;brass&lt;/a&gt;, are correctly regarded as solid solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one usually regards gas phases as &quot;solutions&quot;, so mixtures of gases are probably not solutions.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:22:25.593" LastActivityDate="2016-03-11T23:22:25.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33252" PostTypeId="1" CreationDate="2015-06-23T16:29:20.427" Score="7" ViewCount="146" Body="&lt;p&gt;I've worked out five different conformations of propyl propanoate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9ZlE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the bisected conformations (E) should be the most stable one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are my rationals and suggestions in my scheme correct?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="5026" LastEditDate="2018-06-24T12:03:32.750" LastActivityDate="2018-06-24T12:03:32.750" Title="Which of the five conformers of propyl propanoate is most stable?" Tags="&lt;organic-chemistry&gt;&lt;conformers&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33253" PostTypeId="2" ParentId="33250" CreationDate="2015-06-23T16:30:59.817" Score="8" Body="&lt;p&gt;According to IUPAC gold book: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution is a liquid or solid phase containing more than one substance...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the concept &quot;solution&quot; is reserved to solid and liquid phases.&#xA;In gas phase: Air is an example of homogeneous mixture of nitrogen, oxygen, carbon dioxide and other gases present in air. It's obvious that air is not a solution. &lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-06-23T16:30:59.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33254" PostTypeId="2" ParentId="5974" CreationDate="2015-06-23T17:13:58.340" Score="2" Body="&lt;p&gt;&lt;strong&gt;Electrolysis:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question &quot;Is it dangerous to inhale the steam coming from electrolysis of salt-water solution?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer short answer is:  &lt;strong&gt;It is not safe&lt;/strong&gt;.&lt;br&gt;&#xA;The long answer involves relative safety, and depends on a few things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing  to understand is that chlorine is produced at the anode and sodium hydroxide produced at the cathode. Most of the chlorine will combine with sodium hydroxide producing NaOCl, in other words bleach. But, some chlorine gas may evolve from the anode because sodium hydroxide has not yet migrated to the anode in sufficient quantity. Either way, it is not safe to breath bleach or chlorine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative danger depends on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How efficient is the electrolysis: Higher voltages though causes more current to flow, are less efficient at electrolysis because it produces more heat than actual electrolysis for each unit of energy applied. More heat, thus steam more steam per electrolysis product, is comparatively less harmful. &lt;/li&gt;&#xA;&lt;li&gt;Quantity of salt: Adding only a pinch of salt means that either very little bleach or even less chlorine gas will be produced; In such low concentrations either product is likely to remain dissolved in the water rather than evolve. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Steamers:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you mentioned adding a pinch of salt to a steamer. Conventional steamers do not perform electrolysis.  Instead, electricity is passed through a restive load with the only product being heat. Adding a pinch of salt may raise the boiling point, and thus requires more heat to make steam, but it would be completely harmless, except for the dangers of heat from the steam. &lt;/p&gt;&#xA;" OwnerUserId="17106" LastEditorUserId="17106" LastEditDate="2018-01-31T22:04:39.860" LastActivityDate="2018-01-31T22:04:39.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33255" PostTypeId="2" ParentId="21968" CreationDate="2015-06-23T17:49:40.973" Score="1" Body="&lt;p&gt;Electrolize salt (NaCl) water without a proton exchange membrane is more likely to produce NaOCl. You need the membrane to keep chlorine and sodium hydroxide from combining to form bleach. The membrane may be made from a polymer which only allows positive sodium ions to pass through it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saftey note. The whole process also creates dangerous potentially deadly chlorine gas if not vented properly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its probably safer and cheaper to buy your sodium hydroxide from a commercial source as the producer is likely able to capture revenue from selling the sodium hydroxide as well as chlorine, and hydrogen that are made during the process, while buying their electricity at lower than residential rates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit,  I am not sure what will happens to the sodium ferrocyanide. The worse case is it is also broken down, and  that at above room tempurature, cyanide gas will evolve at  the same electrode as the chlorine gas,  underscoring the need for proper ventilation.&lt;/p&gt;&#xA;" OwnerUserId="17106" LastEditorUserId="17106" LastEditDate="2018-02-02T16:24:31.163" LastActivityDate="2018-02-02T16:24:31.163" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33256" PostTypeId="1" CreationDate="2015-06-23T18:09:27.797" Score="4" ViewCount="3073" Body="&lt;p&gt;I'm looking for a type of plastic membrane that would let water through it from the outside, but wouldn't allow water to escape once it's inside the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally available in a roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a plastic membrane exist?&lt;/p&gt;&#xA;" OwnerUserId="7965" LastActivityDate="2019-11-15T17:03:54.037" Title="Plastic membrane that lets water in from the outside only?" Tags="&lt;water&gt;&lt;plastics&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33257" PostTypeId="1" CreationDate="2015-06-23T18:16:46.473" Score="0" ViewCount="279" Body="&lt;p&gt;What further information would allow you to make a positive identification of an organic compound from an infrared spectrum?&lt;/p&gt;&#xA;" OwnerUserId="17112" LastActivityDate="2015-06-24T14:20:41.700" Title="Identifying organic compounds in Infrared spectrum" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="7" ClosedDate="2015-06-24T14:19:17.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="33258" PostTypeId="2" ParentId="8494" CreationDate="2015-06-23T18:19:28.987" Score="2" Body="&lt;p&gt;&lt;strong&gt;Regarding your question:&lt;/strong&gt; &quot;Is the resistance fixed, or can I do anything to lower it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy used in electrolysis produces both product and waste heat. Naturally, you want to maintain sufficient production while limiting wasted energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A few important formulas:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Amps = volts / resistance&lt;/li&gt;&#xA;&lt;li&gt;Watts = amps * volts&lt;/li&gt;&#xA;&lt;li&gt;Joules = watts * seconds. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Things that effect resistance, and their effects on product versus waste heat and rate of production:&lt;/strong&gt; Adjusting each parameter individually:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Voltage:&lt;/strong&gt; Per unit of energy supplied, in general, a lower voltage produces more product and less waste heat from electrolysis. Below a certain threshold voltage no electrolysis will occur.  Although less power is consumed to produce the same product at a low voltage at a higher efficiency, the process proceeds slower because with a lower voltage, there is less current and power is consumed more slowly. Alternativly one can increase the voltage to drive more current and produce their product faster, but this happens at an expence to efficency. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Electrolyte type and concentration:&lt;/strong&gt; Different electrolytes have different capacities to conduct and&#xA;Its concentration above or below an ideal will negatively effect the electrolyte's capactiy to conduct electricity. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Temperature:&lt;/strong&gt; Increasing the temperature of electrolyte lowers the resistance of an electrolyte. Additionally,  some of the energy used to break up water's molecular bonds is provided by the thermal energy of molecules bouncing off each other. But, the effect really only becomes appreciable with high temperature electrolysis, like say 800c. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Surface Area:&lt;/strong&gt; Surface area and current are proportional. All other parameters being equale, if you double the surface area of your electrodes, you effectivly half the resistance and double the current. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Spacing:&lt;/strong&gt; Decreasing the spacing between electrodes decreases resistance. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17106" LastEditorUserId="17106" LastEditDate="2018-09-26T15:02:05.333" LastActivityDate="2018-09-26T15:02:05.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33259" PostTypeId="2" ParentId="33250" CreationDate="2015-06-23T18:54:00.670" Score="13" Body="&lt;p&gt;Homogeneous and heterogeneous classifications are scale-dependent. What we mean when we say a mixture is homogeneous is that there are no visible phase differences on the scale of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, milk (when homogenized) and lotions are both examples of &lt;a href=&quot;https://en.wikipedia.org/wiki/Colloid&quot;&gt;colloidal suspensions&lt;/a&gt;. There are no phase differences at the human scale, but neither one is a solution. On a more microscopic scale, you could call the suspended micelles phase boundaries, and in that case might &lt;em&gt;not&lt;/em&gt; call the mixture homogeneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Solution&quot; always implies mixing on the molecular level (single-phase). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Solutions are always homogeneous mixtures, but homogeneous mixtures are not always solutions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-06-23T18:54:00.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33260" PostTypeId="1" AcceptedAnswerId="33275" CreationDate="2015-06-23T19:51:40.630" Score="8" ViewCount="4144" Body="&lt;p&gt;CH&lt;sub&gt;3&lt;/sub&gt;-O-CH&lt;sub&gt;2&lt;/sub&gt;CH&lt;sub&gt;2&lt;/sub&gt;Cl &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will this compound be named? Both, methoxy and chloro are given equal preference, since they are prefix functional groups. The answer given in the book is 2-chloro-1-methoxyethane, but why is it not 1-chloro-2-methoxyethane, as the C of chlorine comes before M in the alphabetical order?&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorDisplayName="user7951" LastEditDate="2015-09-05T15:34:27.573" LastActivityDate="2015-09-05T15:34:27.573" Title="Nomenclature of alkoxy and halo groups" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33262" PostTypeId="1" CreationDate="2015-06-23T22:09:25.810" Score="7" ViewCount="2090" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A large well-insulated container holds a mixture of &lt;span class=&quot;math-container&quot;&gt;$75\ \mathrm g$&lt;/span&gt; of ice and &lt;span class=&quot;math-container&quot;&gt;$100\ \mathrm g$&lt;/span&gt; of water at &lt;span class=&quot;math-container&quot;&gt;$0\ \mathrm{^\circ C}$&lt;/span&gt;. Using the data given below, calculate the mass in grams of solid iron, &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;, at &lt;span class=&quot;math-container&quot;&gt;$325\ \mathrm{^\circ C}$&lt;/span&gt; that you would have to add to this mixture in order to melt &lt;em&gt;all&lt;/em&gt; of the ice and raise the temperature of the resulting &lt;span class=&quot;math-container&quot;&gt;$175\ \mathrm g$&lt;/span&gt; of water to &lt;span class=&quot;math-container&quot;&gt;$22\ \mathrm{^\circ C}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Specific heat capacity of &lt;em&gt;water&lt;/em&gt; is &lt;span class=&quot;math-container&quot;&gt;$c_p(\ce{H2O})=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$&lt;/span&gt;&lt;br /&gt;&#xA;Specific heat capacity of &lt;em&gt;iron,&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{Fe}$&lt;/span&gt;, is &lt;span class=&quot;math-container&quot;&gt;$c_p(\ce{Fe})=0.45\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$&lt;/span&gt;&lt;br /&gt;&#xA;Molar enthalpy of fusion (melting) of water is &lt;span class=&quot;math-container&quot;&gt;$\Delta_\text{fus}H_\mathrm m(\ce{H2O})=6.01\ \mathrm{kJ\ mol^{-1}}$&lt;/span&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$M(\ce{H2O})=18\ \mathrm{g\ mol^{-1}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My answer was that you needed &lt;span class=&quot;math-container&quot;&gt;$25041.67\ \mathrm J$&lt;/span&gt; to melt the ice and &lt;span class=&quot;math-container&quot;&gt;$16108.4\ \mathrm J$&lt;/span&gt; to raise the temperature to 22 degrees Celsius.&lt;/p&gt;&#xA;&lt;p&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$mc \Delta T(\ce{Fe}) = -(25041.67\ \mathrm J + 16108.4\ \mathrm J)$&lt;/span&gt; and dividing this by the mass of iron and the change in temperature results in a mass of &lt;span class=&quot;math-container&quot;&gt;$301.8\ \mathrm g$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="17116" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2017-09-26T10:09:01.793" Title="How to calculate the mass of iron with a temperature of 325 °C needed to melt ice and raise the temperature of water from 0 to 22 °C?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33263" PostTypeId="2" ParentId="33262" CreationDate="2015-06-24T00:45:46.970" Score="9" Body="&lt;p&gt;I can not see anything wrong with your calculations at all.  For posterity, the following is a rough method to confirm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Melting ice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy of fusion $\Delta H_\text{ice} = 6.01\ \mathrm{kJ\ mol^{-1}}$&lt;br&gt;&#xA;Mass of ice $m = 75\ \mathrm g$&lt;br&gt;&#xA;Molar mass of $\ce{H2O}$ $M = 18\ \mathrm{g\ mol^{-1}}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hint: usually, when you are given mass and molar mass, then the amount of the substance is required.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Enthalpy of fusion ($\Delta H_\text{ice}$) is measurement in $\mathrm{kJ\ mol^{-1}}$, and as the ice is measured in grams, and you are provided with the molar mass, you'll need to use the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{m}{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_\text{ice} = \frac{75\ \mathrm g}{18\ \mathrm{kJ\ mol^{-1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_\text{ice} = 4.1666\ \mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the enthalpy of fusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta H_\text{fus} = 6.01\ \mathrm{kJ\ mol^{-1}} \times 4.1666\ \mathrm{mol} = 25.04167\ \mathrm{kJ} = 25041.67\ \mathrm J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Heating water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific heat capacity of water $c_p = 4.184\ \mathrm{J\ g^{-1}\ ^\circ C}$&lt;br&gt;&#xA;Mass of water $m = 175\ \mathrm g$&lt;br&gt;&#xA;Temperature change $\Delta T = 22\ \mathrm{^\circ C}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = mc_p \Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = 175\ \mathrm g \times 4.184\ \mathrm{J\ g^{-1}\ ^\circ C} \times 22\ \mathrm{^\circ C} = 16108.4\ \mathrm J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Total energy used by the water/ice mixture&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_\text{total} = 25041.67\ \mathrm J + 16108.40\ \mathrm J = 41150.07\ \mathrm J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Iron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific heat capacity of iron $c_p = 0.45\ \mathrm{J\ g^{-1}\ ^\circ C}$&lt;br&gt;&#xA;Mass of water $m = X$ (&lt;em&gt;unknown&lt;/em&gt;)&lt;br&gt;&#xA;Temperature change $\Delta T = 325\ \mathrm{^\circ C} - 22\ \mathrm{^\circ C} = 303\ \mathrm{^\circ C}$ decrease  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As, according to conservation laws:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{\ce{H2O}} = Q_{\ce{Fe}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$41150.07\ \mathrm J = X \times 0.45\ \mathrm{J\ g^{-1}\ ^\circ C} \times 303\ \mathrm{^\circ C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$41150.07\ \mathrm J = 136.35\ \mathrm{J\ g^{-1}}\times X$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = \frac{41150.07\ \mathrm J}{136.35\ \mathrm{J\ g^{-1}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X = 301.8\ \mathrm g$$&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user7951" LastEditDate="2016-12-03T18:33:29.133" LastActivityDate="2016-12-03T18:33:29.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33265" PostTypeId="1" CreationDate="2015-06-24T01:49:15.063" Score="4" ViewCount="114" Body="&lt;p&gt;What is the relative permittivity and relative permeability of a palladium nanocube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I change the size of this $\ce{Pd}$ nanocube, will these values be changed or remain the same?&lt;/p&gt;&#xA;" OwnerUserId="17117" LastEditorUserId="10185" LastEditDate="2015-06-24T14:04:01.850" LastActivityDate="2015-08-23T15:57:07.460" Title="Palladium Nanocube" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;&lt;materials&gt;&lt;transition-metals&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33266" PostTypeId="2" ParentId="33257" CreationDate="2015-06-24T03:03:54.287" Score="0" Body="&lt;p&gt;You can compare your IR spactra to a database entry. Usually, the database provides examples of spectra from a high purity sample and that way you can even see how pure your sample is.&lt;/p&gt;&#xA;" OwnerUserId="17118" LastEditorUserId="10185" LastEditDate="2015-06-24T14:20:41.700" LastActivityDate="2015-06-24T14:20:41.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33267" PostTypeId="2" ParentId="28768" CreationDate="2015-06-24T03:11:31.370" Score="2" Body="&lt;p&gt;The original purported advantage of semiconductor|liquid junctions proposed by those in the field of photoelectrochemistry arose from the hypothesis that charge carriers would have a shorter distance to travel than in solid-state photovoltaic systems, thereby allowing lower quality materials to be used in an integrated solar energy system and still achieve high conversion efficiencies. In theory, one would also be able to avoid forming metallurgical junctions, say those p-n junctions made by using different dopants and creating a well-defined interface to separate charge carriers. Instead of forming these solid-state structures, you can define the electric field by selecting a suitable redox couple in solution. The specific energetics of the semiconductor and redox couple will then define the maximum power obtainable from the given system, so this is why the valence band must be below the energy of the OER and the conduction band above the energy level corresponding to the HER. By natural induction, this means that the bandgap must meet or exceed 1.23 eV, but this is only a thermodynamic constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality your system’s bandgap must actually exceed 2.0 eV, this is due to the kinetic constraints based in part on the OER (and to a lesser extent the HER ) and losses encountered within the light absorber. Of course, for a single light absorber there is a tradeoff between the percentage of utilized solar flux and the bandgap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why many efforts are focused on using dual light absorbers, with one dedicated to the anode and another for the cathode. In this tandem structure they can efficiently utilize more of the solar spectrum. If you were to stack these light absorbers and use a 1.2 eV and 1.8 eV materials (with each one carrying out one of the two half-reactions required to split water), you could in theory get twice the efficiency compared to a single absorber with a bandgap of 2.2 eV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now are these semiconductor|liquid junctions going to beat a dedicated photovoltaic solar panel/electrolyzer any time soon? Maybe, but probably not. Is water splitting really worth doing? Maybe, but probably not. Is it a beautiful idea? Definitely.&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="32" LastEditDate="2015-06-25T03:23:33.417" LastActivityDate="2015-06-25T03:23:33.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33268" PostTypeId="2" ParentId="33256" CreationDate="2015-06-24T04:16:12.710" Score="2" Body="&lt;p&gt;The comments above have gently hinted that it would be a violation of thermodynamics for a passive membrane to force a substance in a single direction. Molecules have an equal probability of moving through the membrane in either direction, &lt;em&gt;unless you apply a force to make them move in a specific direction&lt;/em&gt;, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Electro-osmosis&quot; rel=&quot;nofollow noreferrer&quot;&gt;electro-osmosis&lt;/a&gt; or &lt;a href=&quot;http://www.researchgate.net/publication/230178030_Active_Water_Transport_in_Plants&quot; rel=&quot;nofollow noreferrer&quot;&gt;active water transport in plants&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that if such a membrane as you posit existed, you could use it to pump water from a lower level to a higher, and then power a water-wheel: perpetual motion. Sadly, &lt;a href=&quot;https://en.wikipedia.org/wiki/Maxwell%27s_demon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell's demon&lt;/a&gt; is getting a bit arthritic to shuttle water molecules, and the dampness only aggravates its condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dawn4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-24T04:16:12.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33269" PostTypeId="1" AcceptedAnswerId="33272" CreationDate="2015-06-24T07:50:45.440" Score="7" ViewCount="636" Body="&lt;p&gt;Take, for example, $\ce{CO2}$.  It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I wanted to build a potential energy surface for this molecule that was valid near and far from equilibrium, it should depend on these coordinates.  But how do I rectify this with the fact that $\ce{CO2}$ has four vibrational modes? As we learn in undergraduate chemistry, it has two stretches and two degenerate bends.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I build a Hamiltonian in the three valence coordinates, could it possibly describe the ground state energy correctly?  That is, would it contain the zero-point energy for all four modes?  Would it be able to find the right low-lying excited states of the bending mode(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/users/186/wildcat&quot;&gt;Wildcat&lt;/a&gt; has brought up the point of symmetry and point groups below, and perhaps that can help get to the heart of my question.  When we say that $\ce{CO2}$ is a linear molecule, that means that the lowest energy configuration has a bond angle of $\pi$.  But in no way do I want to restrict the symmetry of the molecule.  I want to work with a potential surface that is valid for symmetric and asymmetric geometries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is where I get confused.  Let's say I did restrict it to be linear.  I would then have a two-dimensional potential, a function of the two bond lengths.  Of the seven normal modes, I only keep two: the symmetric stretch and anti-symmetric stretch.  The ground state energy would be the sum of the zero-point energy of these two modes.   So therefore the number of modes is equal to the number of coordinates I used to write my Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I further constrict the symmetry, and impose the $D_{\infty h}$ point group, then the potential is now a function of one parameter.  I will now only have one vibrational mode, the symmetric stretch, and the ground state energy is the zero-point energy of this mode.  So, again, the number of modes is equal to the number of coordinates I use to write my Hamiltonian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if the symmetry were not restricted in any way?  I can write the potential as a function three internal coordinates, be they bond lengths and angle, othogonormal rectilinear coordinates, jacobi coordinates, etc.  This potential, and the Hamiltonian I construct with it, should find the same modes described above, as well as the bending modes.  But will it find both bending modes?  I don't see why it should.  Based on the two reduced-dimensional examples above, I would expect the number of modes to equal the number of coordinates.  I would expect therefore, not to find the right ground-state energy, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am concerned about the accuracy of dynamical simulations performed using this three-dimensional potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, what about a tetratomic liner molecule like $\ce{C2H2}$? I would use three bond lengths, two bond angles, and one dihedral angle, for a total of 6 coordinates - but it has 7 vibrational modes.  Would I model the dynamics properly?&lt;/p&gt;&#xA;" OwnerUserId="17121" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-10-23T18:04:51.883" Title="How many coordinates should be used for the potential energy surface of a non-rotating linear molecule? (3N-6) or (3N-5)?" Tags="&lt;bond&gt;&lt;molecular-structure&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33270" PostTypeId="2" ParentId="33269" CreationDate="2015-06-24T08:18:03.687" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is pretty clear that in the absence of rotation, the energy of&#xA;  this system depends only on the two bond lengths and the bond angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that is true, if you do not impose any restriction on geometry (such as, for instance, the linearity), the $\ce{CO2}$ molecule will indeed have $3N-6=3*3-6=3$ internal degrees of freedom and the &lt;em&gt;same number&lt;/em&gt; of independent structural parameters (as which you can naturally choose two bond lengths and the angle between them). However, the point is that when you start to impose some &lt;em&gt;restrictions&lt;/em&gt; on the geometry (symmetry), the number of internal degrees of freedom can &lt;em&gt;increase by one&lt;/em&gt; (if you require your molecule to be linear), while the number of independent structural parameters will &lt;em&gt;stay the same&lt;/em&gt; (though, few of them will be fixed and not varying anymore).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result the following three quantities which you mixed up are usually different:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of internal degrees of freedom: $3N-6$ (for a non-linear molecule) or $3N-5$ (for a linear molecule);&lt;/li&gt;&#xA;&lt;li&gt;the number of independent structural parameters: always $3N-6$;&lt;/li&gt;&#xA;&lt;li&gt;the number of varying (i.e. not-fixed) independent structural parameters: depends on the point group.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, for instance, if you require the $\ce{CO2}$ molecule to be linear the number of internal degrees of freedom will be $3N-5=3*3-5=4$. It does not mean though that you need 4 independent structural parameters to describe it, 3 is still enough. And besides, by requiring the $\ce{CO2}$ to be linear you make it to belong to $\mathrm{C_{\infty v}}$ point group which fixes one of the parameters, the angle between bonds. So, while you have 4 internal degrees of freedom, you still need just 3 structural parameters to describe the molecule one of which is fixed in addition to that. So, if you want to scan PES, you have to vary just two of them: two bond lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can decrease the number of varying structural parameters even further: to just 1, namely, just the bond length, by making an additional (but quite reasonable) assumption that two $\ce{C-O}$ bonds in the $\ce{CO2}$ molecule have the same length, i.e. that the $\ce{CO2}$ molecule belongs to $\mathrm{D_{\infty h}}$ point group. Note that the number of internal degrees of freedom and the number of independent structural parameters will still be 4 and 3 respectively.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-24T09:20:12.580" LastActivityDate="2015-06-24T09:20:12.580" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33271" PostTypeId="1" AcceptedAnswerId="33433" CreationDate="2015-06-24T08:56:23.767" Score="5" ViewCount="287" Body="&lt;p&gt;I'm currently studying about the atomic configuration of electrostatic interactions. Concept is there that in case of interactions like in &lt;strong&gt;N-H...N-H&lt;/strong&gt; (amides), the stability depends on the position of resonating electron between &lt;strong&gt;H...N&lt;/strong&gt; nuclei (&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01360a004&quot; rel=&quot;nofollow&quot;&gt;THE NATURE OF THE CHEMICAL BOND. II. THE ONE-ELECTRON BOND AND THE THREE-ELECTRON BOND - Linus. Pauling&lt;/a&gt;).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that hydrogen bonds formed between &lt;strong&gt;N-H...N-H&lt;/strong&gt; or &lt;strong&gt;N-H...O-C&lt;/strong&gt; are partial covalent in nature so the issue of directionality comes in effect.&lt;br&gt;&#xA;I'm just trying to understand, from the view-point of the electron density shared by &lt;strong&gt;H...N&lt;/strong&gt; to form the bond, that how (if possible) atomic configurations/geometry affects the electron cloud distribution ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm basically trying to understand the hydrogen bond forming amides (currently interested in) from the electronic density distribution's point of view; to calculate the single-point-energy using dft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="16786" LastEditorUserId="16786" LastEditDate="2015-06-25T17:00:37.533" LastActivityDate="2015-06-28T22:35:11.530" Title="Electronic density distribution in hydrogen bond" Tags="&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="12" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33272" PostTypeId="2" ParentId="33269" CreationDate="2015-06-24T09:48:34.380" Score="7" Body="&lt;p&gt;The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation&#xA;$$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$&#xA;where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Hessian generated in the three internal coordinates has all the information you need.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastEditorUserId="13210" LastEditDate="2015-06-24T10:52:09.923" LastActivityDate="2015-06-24T10:52:09.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33273" PostTypeId="2" ParentId="33257" CreationDate="2015-06-24T09:59:57.253" Score="3" Body="&lt;p&gt;If you're analyzing an IR spectrum to find out what compound you have, you'll first want to look past the 1500/cm region to see what functional groups you have. For example, a 1700/cm peak will tell you you have a carbonyl, alcohols will have a peak around 3000/cm, etc. From there, pull up spectra of well known compounds with those functional groups, and compare your compound's 1500/cm to 500/cm region to them. This region is called the &quot;fingerprint region&quot; because it's practically unique for all molecules, so if the spectra match there, they're likely the same compound. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-06-24T09:59:57.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33274" PostTypeId="1" AcceptedAnswerId="33371" CreationDate="2015-06-24T10:51:40.810" Score="4" ViewCount="286" Body="&lt;p&gt;One ethane molecule can create two molecules of ozone, one ethanol molecule can create one molecule of ozone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QNbfc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following table the POCP is twice as high for ethene than for propane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s8Imz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this fact I expect four molecules of ozone from ethene, but in my own formulated reaction mechanism I get only three ozone molecules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/codqc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many ozone molecules can be created by one ethene molecule and if my proposed mechanism is wrong what is the correct one?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="46306" LastEditDate="2018-12-10T17:28:40.657" LastActivityDate="2018-12-10T17:28:40.657" Title="POCP: How many ozone molecules can be created by one ethene molecule?" Tags="&lt;reaction-mechanism&gt;&lt;atmospheric-chemistry&gt;&lt;ozone&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33275" PostTypeId="2" ParentId="33260" CreationDate="2015-06-24T12:04:14.097" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Your name, 1-chloro-2-methoxyethane, is correct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is a link to the relevant page of the 1993 edition of the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_430.htm&quot; rel=&quot;noreferrer&quot;&gt;IUPAC Blue Book - Nomenclature of Organic Chemistry&lt;/a&gt;, which is published online by ACD Labs with permission of IUPAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some example ethers and sulfides from that page, and we see a similar case to yours:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KBNTG.gif&quot; alt=&quot;example IUPAC ether nomenclature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no difference in preference or priority between &lt;code&gt;halo&lt;/code&gt; and &lt;code&gt;alkoxy&lt;/code&gt; substituents (or any of the other prefix only substituents). Thus, in a case where there are two such groups and assignment of locants generates the same combination either way, you choose to assign the lower locant number to the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_317.htm&quot; rel=&quot;noreferrer&quot;&gt;alphabetically first&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-24T12:04:14.097" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33276" PostTypeId="2" ParentId="33265" CreationDate="2015-06-24T13:12:36.730" Score="1" Body="&lt;p&gt;For part (1), you should be able to find values on the internet. For (2), relative permittivity and relative permeability are intensive properties and do not change with the size of the system. However, both quantities are functions of the applied electric and/or magnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-24T13:12:36.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33277" PostTypeId="1" AcceptedAnswerId="33280" CreationDate="2015-06-24T13:19:25.900" Score="1" ViewCount="163" Body="&lt;p&gt;Which temprature is taken, the temprature of the system or the temprature of the surrounding? I think it should be the temprature of the surrounding. This is because if we take the temprature of the system then change in Gibbs free energy would always be zero as then, change in enthalpy of the system would always be equal to change in entropy of the system multiplied by temprature of the system.&lt;/p&gt;&#xA;" OwnerUserId="16157" LastEditorUserId="4325" LastEditDate="2015-06-24T20:20:37.993" LastActivityDate="2015-06-24T20:20:37.993" Title="Which temprature is taken in Gibbs free energy equation, of the system or the surrounding?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2015-06-24T18:25:45.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="33278" PostTypeId="1" CreationDate="2015-06-24T13:46:26.093" Score="12" ViewCount="34153" Body="&lt;p&gt;When salt dissolves in water it dissociates and it is no longer in a lattice structure. The electrostatic forces are no longer there and are replaced by water-ion interactions. To me that sounds like old bonds breaking and new bonds forming. Shouldn't the reaction look something like this? $$\ce{NaCl + H2O -&amp;gt; Na+ + Cl- + H2O}$$ &lt;/p&gt;&#xA;" OwnerUserId="14481" LastEditorUserId="10185" LastEditDate="2015-06-24T14:11:55.980" LastActivityDate="2015-09-30T23:32:10.597" Title="How is dissolving a physical change and not a chemical change?" Tags="&lt;bond&gt;&lt;water&gt;&lt;solutions&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33279" PostTypeId="1" AcceptedAnswerId="33347" CreationDate="2015-06-24T13:53:59.537" Score="8" ViewCount="2207" Body="&lt;p&gt;What is Hybridization? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(Spelling: Hybridisation (British English) / Hybridization (American English))&lt;br&gt;&#xA;  The hybridization is a concept that describes atomic orbitals. In other words: Hybridization is an approach to model hybrid orbitals mathematically. The geometry of a molecule determines the hybridization and not vice versa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://goldbook.iupac.org/H02874.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC Gold Book's&lt;/a&gt; definition of hybridization:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Linear combination of atomic orbitals on an atom. Hybrid orbitals are&#xA;  often used in organic chemistry to describe the bonding molecules&#xA;  containing tetrahedral ($sp^3$), trigonal ($sp^2$) and digonal ($sp$) atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a certain molecule structure we can allocate the atomic orbitals of one atom to hybrid orbitals which are also localized on this atom and which cannot be redistributed over the molecule to over atoms, e.g. a hydrogen can only have a s orbital but no sp hybrid orbitals because it has no p atomic orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The allocation can be done using the following relations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{1s + 3p -&amp;gt;4sp^3}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{1s + 2p + p -&amp;gt;3sp^2 + p}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{1s + 1p + 2p -&amp;gt; 2sp + 2p}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We say: &quot;The atom X is best described by a $\ce{s^\text{x}p^yd^z...}$ hybridization.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining two orbitals / two hybrid orbitals / one hybrid orbital (e.g. $\ce{sp^2}$) with a orbital (e.g. $\ce{s}$) we obtain bonds. $\sigma$-bonds with $\ce{sp}$ orbitals (ethyne) of the atomic centers have lower energy than $\sigma$-bonds with $\ce{sp^2}$ orbitals (ethylene), lower in energy means: they are better acceptors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is my question now?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking if we can also describe a bond (a molecular orbital) with a hybridization, e.g. &lt;strong&gt;if we can combine a $\ce{sp^2}$ hybridized carbon with a $s$ atomic orbital of hydrogen to make two $\ce{sp}$ molecular orbitals (a bonding $\sigma$ orbital with $sp$ character and an antibonding $\sigma^\star$ orbital with $\ce{sp}$ character; just be saying for a $\ce{C-H}$ bond: $\ce{s} + \ce{sp^2} = 2 \ce{sp}$, what's probably wrong&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this question has to be answered with No, because we cannot say a bond has hybridization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question might arise why I'm thinking of the hybridization of molecular orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over several years I've heard statements like this:&#xA;&lt;strong&gt;The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character (50%) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) (33%)&lt;/strong&gt; that's why ethyne has a lower $\mathrm{p}K_\mathrm{a}$ than ethylene (25 versus 44 at standard conditions in water as solvent) because a bond with higher s character can better stabilize a negative charge, because the s orbitals are penetrating deeper into the shell as the p orbitals do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I think hybridization is only a good model for atomic orbitals (=localized on atom) not for linear combinations of atomic orbitals (=bonds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can we really say the $\sigma$ bond in ethylene has $sp^2$ character?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="5026" LastEditDate="2018-02-25T08:06:29.137" LastActivityDate="2018-02-25T08:06:29.137" Title="Is the hybridization only related to atomic orbitals but not to the molecular orbitals?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33280" PostTypeId="2" ParentId="33277" CreationDate="2015-06-24T13:58:15.017" Score="5" Body="&lt;p&gt;The Gibbs free energy is a property of a system, so you should use that system's temperature when calculating it.  The Gibbs free energy is the amount of work that could be extracted from the system in a &lt;strong&gt;isothermal, isobaric&lt;/strong&gt; process: i.e. the temperature and pressure of the system won't change.  Thus Gibbs energy has to do only with the temperature of the system, not of the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a related but distinct variable called &lt;a href=&quot;https://en.wikipedia.org/wiki/Exergy&quot; rel=&quot;noreferrer&quot;&gt;exergy&lt;/a&gt; or sometimes &quot;availability&quot;.  This variable is given the symbol $B$, and is also a thermodynamic state function.  It is defined as $\Delta B = \Delta H - T_0 \Delta S$, where $T_0$ is the temperature of the &lt;em&gt;surroundings&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exergy represents the maximum amount of work that could be extracted from a system when that system is brought into thermal equilibrium with the surroundings, for example using Carnot engines.  Thus exergy indicates work obtainable through a combined (a) isothermal change in Gibbs free energy at temperature $T \neq T_0$ and (b) a temperature change occurs when Carnot engines are used to transfer heat from the system at $T$ to the surroundings at $T_0$ until thermal equilibrium is obtained.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-06-24T13:58:15.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33281" PostTypeId="2" ParentId="33278" CreationDate="2015-06-24T14:11:09.003" Score="2" Body="&lt;p&gt;The reaction you have written is correct. The physical vs. chemical change issue has to do with the presence/absence of substances with new compositions not previously present appearing after the change, and I would argue that the dissolution reaction you present represents both a physical and chemical change: Physical due to the change in properties/state of NaCl, and chemical due to the appearance of aqueous species not present initially.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-24T14:11:09.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33282" PostTypeId="1" CreationDate="2015-06-24T14:58:26.843" Score="8" ViewCount="183" Body="&lt;p&gt;Very recently Chemical Sciences published results of the discovery of a new cobalt compound that can absorb oxygen at presumably a higher capacity than what has been achievable before. The paper is &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2014/SC/C4SC01636J#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. I don't have access to the full publication, and since I'm not a chemist I don't really follow the abstract too well. But from the popular news there are claims that 10 liters of the compound can absorb &quot;a whole room of oxygen&quot;. Not very quantitative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my main question - is this really a breakthrough in terms of absorption capacity? And to what degree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - how can the absorption/release of oxygen be controlled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I found a link to the full article &lt;a href=&quot;https://findresearcher.sdu.dk:8443/ws/files/102792831/c4sc01636j.pdf&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="4945" LastEditDate="2015-06-29T12:07:32.537" LastActivityDate="2015-06-29T12:07:32.537" Title="New finding in cobalt compound that absorbs oxygen - is it really a breakthrough?" Tags="&lt;applied-chemistry&gt;&lt;adsorption&gt;&lt;absorption&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33283" PostTypeId="1" AcceptedAnswerId="33286" CreationDate="2015-06-24T15:33:38.687" Score="2" ViewCount="780" Body="&lt;p&gt;I don't get how Taylor expansion over $ \frac{1}{\overline{V}} $ of Virial equation of state is caried out: $Z = \frac{P\overline{V}}{RT} $ which yields&#xA;$$ Z = 1 + \frac{B_{2V}(T)}{\overline{V}} + \frac{B_{3V}(T)}{\overline{V^2}} + ...&#xA;$$ I thought I understood calc2 Taylor expansion of form $$f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k $$ so maybe I was in illusion. What's is $Z$ function of? At what expansion point do we expa&lt;/p&gt;&#xA;" OwnerUserId="16785" LastEditorUserId="16785" LastEditDate="2015-06-24T16:05:33.663" LastActivityDate="2015-06-24T16:05:33.663" Title="Virial equation of state - Taylor expansion" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33284" PostTypeId="1" CreationDate="2015-06-24T15:44:27.077" Score="5" ViewCount="542" Body="&lt;p&gt;Does a free software for calculating the net charge of amino acids as a function of pH exist?&lt;/p&gt;&#xA;" OwnerUserId="17129" LastEditorDisplayName="user7951" LastEditDate="2016-12-26T01:05:17.243" LastActivityDate="2016-12-26T01:05:17.243" Title="Free software for calculating the net charge of an amino acid as a function of pH?" Tags="&lt;biochemistry&gt;&lt;software&gt;&lt;amino-acids&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33285" PostTypeId="1" CreationDate="2015-06-24T15:52:37.310" Score="2" ViewCount="724" Body="&lt;p&gt;What calculations are needed to convert  $\ce{N2O}$ gas concentrations in ppm to micrograms of nitrogen per gram dry soil basis? I measured the concentration of $\ce{N2O}$ in soil samples. I extracted a small gas sample from a container (the soil is kept in this container; the container has an head space of 80 cc). Then I injected the small gas sample in the GC and I had a measurement in ppm. I want to know how many grams of N are present in my ppm measurement, taking into account that the gas comes from 5 grams of soil. &lt;/p&gt;&#xA;" OwnerUserId="17129" LastEditorUserId="10185" LastEditDate="2015-06-26T16:26:12.423" LastActivityDate="2016-03-11T23:05:13.710" Title="to convert ppm gas (N2O) concentrations in microgram nitrogen per gram dry soil basis?" Tags="&lt;analytical-chemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33286" PostTypeId="2" ParentId="33283" CreationDate="2015-06-24T15:56:47.110" Score="1" Body="&lt;p&gt;Here, Z is explicitly a function of temperature.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At what expansion point do we expand?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you clarify what you mean? I can state that it's common to truncate this particular expansion at the second term, so we'd have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce Z = 1 + {B_{2V}(T)\over \overline{V}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The virial coefficient in this case accounts for deviation from the ideal case by accounting for attractive and repulsive forces between the particles in the system.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-24T15:56:47.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33287" PostTypeId="1" AcceptedAnswerId="33291" CreationDate="2015-06-24T15:57:34.023" Score="3" ViewCount="45" Body="&lt;p&gt;When I get the pressure using the different equations of state (e.g. van der Waals, Redlich-Kwong, etc.), will the answer still be the same?&lt;/p&gt;&#xA;" OwnerUserId="17130" LastEditorUserId="5633" LastEditDate="2015-06-25T04:30:39.753" LastActivityDate="2015-06-25T04:30:39.753" Title="Getting the pressure using the equation of states - will the answer be the same?" Tags="&lt;physical-chemistry&gt;&lt;equation-of-state&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33288" PostTypeId="1" CreationDate="2015-06-24T16:01:53.910" Score="5" ViewCount="232" Body="&lt;p&gt;Elements from the carbon group tend to form compounds with two oxygen's. Let's say I want to produce $\ce{SiO_2}$ out of $\ce{SiO}$ by mixing it with some alcohol, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SiO + CH_3-CHOH-CH_3 \longrightarrow SiO_2 + CH_3-CH_2-CH_3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this likely to occur? at least before all the alcohol evaporates...&lt;/p&gt;&#xA;" OwnerUserId="16890" LastEditorUserId="16890" LastEditDate="2015-07-06T04:14:17.153" LastActivityDate="2016-07-09T17:55:10.063" Title="Reaction of Silicon monoxide with alcohols" Tags="&lt;organic-chemistry&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33289" PostTypeId="2" ParentId="33278" CreationDate="2015-06-24T16:12:03.127" Score="0" Body="&lt;p&gt;The dissolution of NaCl in water is a physical change. Because, it's a reversible process. If water evaporates, you get back the salt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dielectric constant $\epsilon_r$ of water is 80 compared with 1 for air. So, the attractive electrostatic forces between ions are weakened by a factor of 80 when salt is added to water. We say the salt NaCl is dissociated. The molecules of water are rearranged around the dissociated ions and we get the hydrated ions, $\ce{Na^+_{aq}}$, $\ce{Cl^{-}_{aq}}$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-06-24T16:12:03.127" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33290" PostTypeId="1" CreationDate="2015-06-24T16:12:05.913" Score="1" ViewCount="76" Body="&lt;p&gt;$\ce{Cl2}$ bond energy given is as 244.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/2 $\ce{Cl2}$ will still be 244 or 244/2? My teacher said it is still 244.&lt;/p&gt;&#xA;" OwnerUserId="17131" LastEditorUserId="4231" LastEditDate="2015-06-24T16:14:57.667" LastActivityDate="2015-06-24T16:32:32.763" Title="What is the bond energy" Tags="&lt;bond&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-06-26T03:16:53.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="33291" PostTypeId="2" ParentId="33287" CreationDate="2015-06-24T16:26:04.573" Score="5" Body="&lt;p&gt;In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the van der Waals EOS gives pressure values at high temperature and low pressure for noble gases that are close to what you would obtain with the ideal gas law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is typically easy to calculate pressure from an EOS given temperature and volume, you can try this yourself and see how much of an impact the choice of EOS makes for your particular system at a particular point in phase-space.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-06-24T16:26:04.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33292" PostTypeId="2" ParentId="33290" CreationDate="2015-06-24T16:32:32.763" Score="1" Body="&lt;p&gt;The quantity you refer to has units of kJ/mol. Thus, for 1/2 mole of $\ce {\rm Cl}_{2}$, the quantity is halved to 122 kJ/mol. Your teacher might be referring to the quantity on a per-bond basis, but the number you report agrees with references online for a per-mole basis.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-06-24T16:32:32.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33293" PostTypeId="2" ParentId="33278" CreationDate="2015-06-24T16:57:20.563" Score="8" Body="&lt;p&gt;This is indeed a bit of a grey area. Consider the following: chemical or physical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na+ Cl- (aq)  + Ag+ NO3- (aq)   -&amp;gt;  Ag+Cl- (s) + Na+ NO3- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a tendency in more advanced chemistry to view the formation / breaking of ionic lattices as a physical change, because there are no differences in the electron shells of the atoms. Ions are considered as independent species. As such we could consider the sodium and nitrate irrelevant and reduce the above to simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ag+ (aq) + Cl- (aq) -&amp;gt; Ag+ Cl- (s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I have shown the charges all the way through, to indicate that there are no covalent, dative, or metallic bonds made or broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions such as these are definitely &quot;chemical&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na + \frac{1}{2}Cl2 -&amp;gt; Na+ Cl-}$$&lt;br&gt;&#xA;(metallic bonds in $\ce{Na}$ and covalent bonds in $\ce{Cl2}$ broken, electron transferred from $\ce{Na}$ to $\ce{Cl}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C2H4 + H2 -&amp;gt; C2H6}$$&#xA;(double bond of $\ce{C2H4}$ opened up, bond of $\ce{H2}$ broken, new covalent bonds formed.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{SOCl2 + 2H2O -&amp;gt; H2SO3 + 2HCl}$$ (covalent bonds broken, new bonds made.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last reaction sets up another point. It is usually done with excess water, so the products (which are themselves acids) will then dissolve in water, thereby protonating it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO3 + H2O -&amp;gt; H3O+ + SO3H-}$$ &#xA;$$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These particular dissolutions are more likely to be seen as a chemical change than the dissolution of $\ce{Na+Cl-}$, because there is a rearrangement of covalent bonds (even though it's only a $\ce{H+}$ that moves from one molecule to the other.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $\ce{H2SO3}$ has another equilibrium, again involving covalent bonds being broken and made, and therefore a chemical change:&#xA;$$\ce{H2SO3 -&amp;gt; H2O + SO2 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making or breaking of metallic, covalent or dative bonds, or the formation/destruction of new ions or molecules, is always considered a chemical change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple dissolution of an ionic salt, where none of the above mentioned bonds are made or broken, and no new ions or molecules are formed/destroyed, is frequently considered a physical change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the care taken to always write $\ce{Na+Cl-}$ (and never $\ce{NaCl}$) in this answer.   &lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="10185" LastEditDate="2015-06-24T17:20:37.363" LastActivityDate="2015-06-24T17:20:37.363" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33294" PostTypeId="2" ParentId="26416" CreationDate="2015-06-24T17:55:59.070" Score="4" Body="&lt;p&gt;In case you want a safe and reliable method, consider magnesium granules/foil and acetic acid. The reaction is quite exothermic, but the reagents are reasonably safe. Do &lt;strong&gt;not&lt;/strong&gt; use magnesium powder for the reaction, it is likely to react violently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you want a cheap and reliable method, consider aluminium granules and sodium hydroxide. The reaction is, again, quite exothermic, and again, I recommend to avoid aluminium powder. Note: sodium hydroxide is unstable on air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The military option is ferrosilicon with sodium hydroxide, but ferrosilicon is not readily available and the reaction is, again, quite exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard lab option is to use zinc granules with hydrochloric acid, but zinc is quite costly. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="5026" LastEditDate="2018-04-18T06:36:38.890" LastActivityDate="2018-04-18T06:36:38.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33295" PostTypeId="2" ParentId="26416" CreationDate="2015-06-24T18:08:04.417" Score="2" Body="&lt;p&gt;The other answers are very good, so purely in the interest of diversity I'll mention the methide (methanide?) carbides such as aluminum carbide or magnesium carbide.  With aluminum, the reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Al4C3(s) + 12 H2O(l) → 4 Al(OH)3(aq) + 3 CH4(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methane ($\ce{CH4}$) is less dense than air, but not by nearly as much as $\ce{H2}$ which is generated by reactions mentioned in other answers.  I've never tried the reaction myself but I believe it is comparatively less vigorous than hydrogen generation from alkali metals or alkali earths.  But on the up-side, the relative change in mass of the system will be higher with methane evolution since it weighs more than hydrogen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why you even need a gas less dense than air.  If you're using a closed container (n.b. this is a bad, extremely dangerous idea unless it is a balloon or something), the mass of the system won't change.  If you're using an open container, the mass of the system will change as the gas is evolved, whether or not it is less dense than air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calcium carbide is not a methanide, but an acetylide, so when mixed with water it generates acetylene gas.  Acetylene is very nearly the same density as air (its molecular weight is 26 g/mol, compared to an average of 30 g/mol for air.  The reaction of calcium carbide with water was formerly used widely in &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbide_lamp&quot; rel=&quot;nofollow&quot;&gt;carbide lamps&lt;/a&gt;, so presumably it is relatively easy to control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calcium carbide is also cheap compared to alkali metals such as lithium, and acetylene evolution will cause an even higher relative change in mass of the system than methane evolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaC2(s) + 2H2O(l) -&amp;gt; HC#CH(g) + Ca(OH)2(s)}%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:21:54.700" LastActivityDate="2016-03-11T23:21:54.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33296" PostTypeId="1" CreationDate="2015-06-24T19:34:35.620" Score="3" ViewCount="643" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rqUVM.jpg&quot; alt=&quot;olive oil making grids in a nonstick pan when swirled&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever I add olive oil to my cold nonstick pan and try to swirl it, it ends up making these 90-degree grids. Is this something to do with the oil, the pan, or the interaction between the two? I find the regular spacing between lines rather interesting.&lt;/p&gt;&#xA;" OwnerUserId="17135" LastActivityDate="2015-06-24T19:34:35.620" Title="Why does olive oil make 90 degree grids on a nonstick pan?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;&lt;fats&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33297" PostTypeId="1" AcceptedAnswerId="33310" CreationDate="2015-06-24T19:35:11.840" Score="13" ViewCount="27815" Body="&lt;p&gt;This is quoted from Jim Clark's &lt;a href=&quot;http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html&quot;&gt;Chemguide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For &lt;strong&gt;reasons&lt;/strong&gt; which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;strong&gt;reason&lt;/strong&gt; stated above? Why does in only transition series, 4s has more energy than 3d &amp;amp; not in the preceeding elements?  &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="10185" LastEditDate="2015-06-24T19:46:46.700" LastActivityDate="2019-01-06T16:48:04.140" Title="Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?" Tags="&lt;physical-chemistry&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="9" ClosedDate="2015-06-25T09:02:22.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="33298" PostTypeId="2" ParentId="33284" CreationDate="2015-06-24T19:42:41.370" Score="3" Body="&lt;p&gt;I like &lt;a href=&quot;https://www.chemaxon.com/marvin/sketch/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;MarvinSketch&lt;/a&gt;. You can run it in your browser or on your computer. Under Insert &gt; Template, you can find all the amino acids. Once you've inserted one, select Calculations &gt; Protonation &gt; Isoelectric point and click OK. You'll get a window that looks like this showing the net charge versus pH (I picked a pretty boring one, but other amino acids produce more complex curves):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lf61r.jpg&quot; alt=&quot;Isoelectric point graph&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-24T19:42:41.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33301" PostTypeId="2" ParentId="33297" CreationDate="2015-06-24T20:48:15.603" Score="8" Body="&lt;p&gt;This is a difficult question to answer. Following the Aufbau Principle and the n+l rule, the 4s orbital should fill before the 3d orbital. So why is 3d lower in energy? In short, the Aufbau Principle is not entirely correct. It is a guideline (like many things in chemistry.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, orbitals fill in order of stability. That is to say, electrons will go where they will be the most stable. It takes energy to hold electrons around the nucleus. The farther away they are, the more energy is needed to keep them. So the higher the principle quantum number, the higher the energy. I.e. 3s is higher in energy than 2s. At the same time, the principle quantum number is not the only number than needs to be considered. The quantum number l, for example, is also important. The higher value of l, the higher the energy. So 3d is higher in energy than 3p which is higher in energy than 3s. The 3d orbitals are more compactly placed around the nucleus than the 4s orbitals, so they fill first, even though this contradicts the Aufbau principle. This can be seen experimentally with the electron configurations for scandium:&#xA;Sc3+: [Ar]&#xA;Sc2+: [Ar]3d(1)&#xA;Sc+:  [Ar]3d(1)4s(1)&#xA;Sc:   [Ar]3d(1)4s(2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is important to note that the 4s level does fill before 3d is entirely full. This is due to the 3d orbital's compactness. Electron repulsion &quot;pushes&quot; electrons into higher energy levels with less repulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend reading this as it explains this in much more detail: &lt;a href=&quot;http://www.rsc.org/eic/2013/11/aufbau-electron-configuration&quot; rel=&quot;noreferrer&quot;&gt;http://www.rsc.org/eic/2013/11/aufbau-electron-configuration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helped!&lt;/p&gt;&#xA;" OwnerUserId="17137" LastActivityDate="2015-06-24T20:48:15.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33302" PostTypeId="1" AcceptedAnswerId="33303" CreationDate="2015-06-24T21:55:54.970" Score="4" ViewCount="691" Body="&lt;p&gt;Is it just for historical reasons? Because the first person who did it decided to attach the C-terminus to the solid support?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there chemical reasons why it would be impossible to proceed in the opposite direction?&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2015-06-24T22:55:55.110" Title="Why does solid phase peptide synthesis (SPPS) always proceed from the C to the N terminus?" Tags="&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33303" PostTypeId="2" ParentId="33302" CreationDate="2015-06-24T22:55:55.110" Score="2" Body="&lt;p&gt;The reasons are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The use of an acylating agent in excess (e. g. acetic anhydride or &lt;em&gt;N&lt;/em&gt;-acetyl imidazole) drives the reaction forward.&lt;/li&gt;&#xA;&lt;li&gt;UNCAs (urethane-type protected  &lt;em&gt;N&lt;/em&gt;-carboxy anhydrides (e. g. Fmoc, Z, Boc, ...)) can be used. They prevent racemization of the amino acid, that it is attached to. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But there are (despite of ribosomes themselves) also instances, where N -&gt; C directionality was applied:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/anie.199710991/abstract&quot; rel=&quot;nofollow&quot;&gt;http://onlinelibrary.wiley.com/doi/10.1002/anie.199710991/abstract&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/10891165&quot; rel=&quot;nofollow&quot;&gt;http://www.ncbi.nlm.nih.gov/pubmed/10891165&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4852" LastActivityDate="2015-06-24T22:55:55.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33306" PostTypeId="2" ParentId="18524" CreationDate="2015-06-25T00:10:23.753" Score="6" Body="&lt;p&gt;This is not a simple question at all, in fact, it is an interesting insight in chemical bonding.  You make a valid assertion with your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;are ionic bonds just extremely polar covalent bonds?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This answer is a simplistic way of looking at it.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to ChemGuide page &lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/electroneg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electronegativity&lt;/a&gt;, there is no real clear-cut 'boundary' between covalent and ionic bonding. The website describes that there is a 'spectrum of bonding' based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronegativity&lt;/a&gt; (chart shown below) of the constituent elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/szCVY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;UC Davis ChemWiki&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the greater the electronegativity, the greater the polarity or &quot;pulling power&quot; on the bonding electrons towards elements with a higher electrongativity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following diagrams (from the ChemGuide site linked before), A and B represent arbitrary elements, and the green and red dots represent bonding electrons. Starting from a situation where the polarity is zero as the electronegativity of the constituent elements are the same (e.g. diatomic molecules $\ce{O2}$, $\ce{Cl2}$ etc), as in diagram I. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the polarity increases, the bonding electrons move towards the more electronegative member, causing the bond to become increasingly polar (II), then eventually to the point where the difference is enough (III) for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;To all intents and purposes&lt;/strong&gt;, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I bolded the 'to all intents and purposes' to emphasise that this is a simplistic generalisation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NB6g5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(II)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/JAbUE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(III)&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZUGVD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the lowest electronegativity is not zero (it is Francium at 0.7), there will always be an influence on the bonding electrons. So, the reality is that A does not completely lose its influence on the bonding electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we were to look at a compound combining the most and least* electrongative elements (as appear in the diagram above): &lt;a href=&quot;https://www.webelements.com/francium/chemistry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;francium fluoride $\ce{FrF}$&lt;/a&gt;, even though fluorine has a far stronger electronegativity than francium, the electronegativity of the latter is not zero, so, the influence of the francium is very small, but not zero - indicating that purely ionic bonds don't exist.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;As noted in the comments, cesium is likely to be less electronegative than francium, however the point made above still stands.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is nicely summarised by an article from the &lt;a href=&quot;https://www.um.edu.mt/__data/assets/pdf_file/0006/85443/A2.3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Malta&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pure ionic bonding is not known to exist. All ionic compounds have a degree of&#xA;  covalent bonding. The larger the difference in electronegativity between two&#xA;  atoms, the more ionic the bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-06-30T13:15:51.543" LastActivityDate="2015-06-30T13:15:51.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33308" PostTypeId="1" AcceptedAnswerId="33311" CreationDate="2015-06-25T00:51:38.963" Score="2" ViewCount="518" Body="&lt;p&gt;I am a backer of this Kickstarter: &lt;a href=&quot;https://www.kickstarter.com/projects/903107259/scio-your-sixth-sense-a-pocket-molecular-sensor-fo&quot; rel=&quot;nofollow&quot;&gt;SCiO Near IR Spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to their technology: &lt;a href=&quot;https://www.consumerphysics.com/myscio/technology&quot; rel=&quot;nofollow&quot;&gt;SCiO Technology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How short of full IR spectroscopy do you think this device is? I have asked them whether we could program the SCiO to act as an IR spectrometer, and they have said this would require additional programming (which I am not qualified to do). But I am wondering if there would still be limitations. Conversely, Consumer Physics seems to imply that our taking more observations with the SCiO will make the device smarter and more useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously a question directed at experts in IR spectroscopy, which I am most definitely not.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorUserId="7499" LastEditDate="2015-06-25T02:01:25.260" LastActivityDate="2015-06-25T02:01:25.260" Title="Near IR spectroscopy vs. IR spectroscopy" Tags="&lt;spectroscopy&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33310" PostTypeId="2" ParentId="33297" CreationDate="2015-06-25T01:18:02.227" Score="22" Body="&lt;blockquote&gt;&#xA;  &lt;h1&gt;&lt;em&gt;Disclaimer:&lt;/em&gt; I now believe this answer to be fully incorrect.&lt;/h1&gt;&#xA;  &#xA;  &lt;h3&gt;Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22.&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver &amp;amp; Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles: &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1978,&lt;/strong&gt; &lt;em&gt;55&lt;/em&gt; (1), 2 &lt;a href=&quot;https://doi.org/10.1021/ed055p2&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed055p2&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1996,&lt;/strong&gt; &lt;em&gt;73&lt;/em&gt; (6), 498 &lt;a href=&quot;https://doi.org/10.1021/ed073p498&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed073p498&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1994,&lt;/strong&gt; &lt;em&gt;71&lt;/em&gt; (6), 469 &lt;a href=&quot;https://doi.org/10.1021/ed071p469&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed071p469&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1989,&lt;/strong&gt; &lt;em&gt;66&lt;/em&gt; (6), 481 &lt;a href=&quot;https://doi.org/10.1021/ed066p481&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed066p481&lt;/a&gt;.&lt;/li&gt;&#xA;  &lt;li&gt;&lt;a href=&quot;http://www.chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Some criticism of Atkins' books by Eric Scerri.&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver &amp;amp; Atkins 5th ed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebLSy.png&quot; alt=&quot;relative&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d &gt; 4s but for Sc onwards, 4s &amp;lt; 3d.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$hcR$&lt;/span&gt; is a collection of constants, &lt;span class=&quot;math-container&quot;&gt;$Z_\text{eff}$&lt;/span&gt; is the effective nuclear charge experienced by the electron, and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the principal quantum number. Since &lt;span class=&quot;math-container&quot;&gt;$n = 4$&lt;/span&gt; for the 4s orbital and &lt;span class=&quot;math-container&quot;&gt;$n = 3$&lt;/span&gt; for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more &lt;em&gt;penetrating&lt;/em&gt; than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as &lt;span class=&quot;math-container&quot;&gt;$R(r)^2 r^2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$R(r)$&lt;/span&gt; is the radial wavefunction obtained from the Schrodinger equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MlHpi.jpg&quot; alt=&quot;3d/4s&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron &quot;tends to spend time&quot; near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron &lt;em&gt;penetrates&lt;/em&gt; the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes &lt;span class=&quot;math-container&quot;&gt;$Z_\text{eff}$&lt;/span&gt; larger. Going from the 3d to the 4s orbital, the increase in &lt;span class=&quot;math-container&quot;&gt;$Z_\text{eff}$&lt;/span&gt; wins ever so slightly over the increase in &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, which makes the energy of the 4s orbital lower.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively &lt;em&gt;radially diffuse&lt;/em&gt; (the maximum in the radial distribution function occurs at a larger value of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy &lt;span class=&quot;math-container&quot;&gt;$U = -\frac{kq_1q_2}{r}$&lt;/span&gt;. The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, and if this were to be the case, you would expect 3d &amp;lt; 4s in terms of energies, as we said at the very start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more &quot;personal space&quot; and experience less repulsion. One way of putting it is that an &lt;em&gt;empty&lt;/em&gt; 4s orbital in Sc has a higher energy than an &lt;em&gt;empty&lt;/em&gt; 3d orbital, but a &lt;em&gt;filled&lt;/em&gt; 4s orbital has a lower energy than a &lt;em&gt;filled&lt;/em&gt; 3d orbital. The fact that 4s &gt; 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation (&lt;span class=&quot;math-container&quot;&gt;$\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$&lt;/span&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an &quot;anomalous&quot; configuration that maximises exchange energy, whereas we don't get carbon adopting a &lt;span class=&quot;math-container&quot;&gt;$(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$&lt;/span&gt; configuration in order to have &quot;stable half-filled shells&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2019-01-06T16:48:04.140" LastActivityDate="2019-01-06T16:48:04.140" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33311" PostTypeId="2" ParentId="33308" CreationDate="2015-06-25T02:00:54.563" Score="5" Body="&lt;p&gt;This is more or less speculation, since they haven't released detailed technical specifications, but given the size of the instrument and the need to avoid moving parts to keep it durable, I would wager that it uses an IR LED for illumination, a diffraction grating for dispersion (although from the picture, it doesn't look like there's room for a conventional reflective diffraction grating setup so they might be doing something else), and probably some kind of silicon CCD or InGaAs diode array for detection. If this is the case, it's likely impossible to convert for use in conventional IR spectroscopy—not as a matter of programming, but simply that the hardware is not equipped to do it. Normal mid-IR spectroscopy (below about $4000\mathrm{\ cm^{-1}}$) involves much longer wavelengths beyond what LEDs can do, which generally means some kind of incandescent lamp, optics that don't involve glass (which blocks mid-IR), and a detector that can deal with such low energy photons (there are a few semiconductor-based sensors that looks promising, but most spectrometers use thermal detectors). There is also the matter of dealing with all the water in the air—many IR spectrometers have gas purge systems in their sample chambers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been a lot of interest in standoff IR spectroscopy for things like explosives detection and advances in mid-IR optics and sensors may make this kind of small instrument possible, but we're not quite there yet. Certainly an instrument designed for NIR is not likely to have any usefulness for mid-IR, without being designed with it in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related technique that is much better suited to small handheld instruments is raman spectroscopy, since the light involved is actually usually in the NIR. I doubt this thing is sensitive enough to do it well, but it might be possible to add filters to the illumination and measurement ports to give it a shot. It's definitely possible to do raman with a handheld instrument though—I saw &lt;a href=&quot;http://oceanoptics.com/product/idraman-mini/&quot;&gt;one of these&lt;/a&gt; at a conference recently. Quite expensive, but very nifty.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-06-25T02:00:54.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33312" PostTypeId="1" CreationDate="2015-06-25T02:00:56.117" Score="7" ViewCount="361" Body="&lt;p&gt;This seems like a farce... &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quantum gas goes below absolute zero&lt;br&gt;&#xA;  Ultracold atoms pave way for negative-Kelvin materials.&lt;br&gt;&#xA;  by Zeeya Merali, Nature News 03 January 2013, &lt;a href=&quot;https://doi.org/10.1038/nature.2013.12146&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1038/nature.2013.12146&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have seen several recent articles like this one  and I am extremely skeptical. Perhaps my skepticism is based upon an ignorance of physical principles, but I am less offended by apparent violations of the speed of light than by considering that temperatures below absolute zero could exist. I have studied under scientists who were using advanced techniques and vast amounts of energy to get small samples cooled down to a few nanokelvin. The physics seems to imply that if a sample were to reach zero, even the nuclear forces cohering the particles would cease. This was hypothesized in hushed tones, in awe and reverence. Now this article (which seems akin to scientific blasphemy) is causing an existential crisis. What is the deal with negative absolute temperature?&lt;/p&gt;&#xA;" OwnerUserId="15363" LastEditorUserId="4945" LastEditDate="2017-11-01T16:21:21.007" LastActivityDate="2017-11-01T16:21:21.007" Title="Temperatures Below Absolute Zero" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ClosedDate="2015-11-23T04:53:08.513" ContentLicense="CC BY-SA 3.0" />
  <row Id="33313" PostTypeId="1" CreationDate="2015-06-25T02:05:57.350" Score="1" ViewCount="1520" Body="&lt;p&gt;I have the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{Co(C2O4)*H2O}$&lt;/span&gt; in a reaction that forms cobalt oxide &lt;span class=&quot;math-container&quot;&gt;$\ce{Co3O4}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;My data here says that my oxalate hydrate weighed &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3283 g}$&lt;/span&gt; and my product oxide weighed &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1158 g}$&lt;/span&gt; which I have calculated as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.1158 \times \frac{\pu{176.79 g}\, \mathrm{(cobalt\,oxalate)}} {\pu{240.79 g}\, \mathrm{(cobalt\,oxide)}} =\pu{0.08502 g\,\mathrm{(cobalt)}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then &lt;span class=&quot;math-container&quot;&gt;$\pu{0.3283 g (reactant)} − \pu{0.1158 g (product)}$&lt;/span&gt; gives me &lt;span class=&quot;math-container&quot;&gt;$\pu{0.2125 g}$&lt;/span&gt; water removed divided by &lt;span class=&quot;math-container&quot;&gt;$\pu{18.016 g/mol}$&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$\pu{0.01179 mol}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; in the sample.&lt;/p&gt;&#xA;&lt;p&gt;How do I use this data to determine the empirical formula? I am not sure how to determine the molar ratios from this point.&lt;/p&gt;&#xA;" OwnerUserId="17085" LastEditorUserId="95133" LastEditDate="2021-05-16T06:40:31.783" LastActivityDate="2021-05-17T06:44:24.887" Title="How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced?" Tags="&lt;experimental-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33314" PostTypeId="1" AcceptedAnswerId="33315" CreationDate="2015-06-25T03:34:23.873" Score="8" ViewCount="1146" Body="&lt;p&gt;Change in enthalpy of the system is equal to its temperature multiplied by its change in entropy (as change in enthalpy is equal to heat absorbed /given out at constant pressure and change in entropy is heat absorbed/given out divided by its temperature). So, through the Gibbs free energy equation, the change in Gibbs free energy should be zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question might seem messy, as I don't know how to write equations on Stack Exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help. The problem is in the elementary concepts, I think.&lt;/p&gt;&#xA;" OwnerUserId="16157" LastEditorUserId="15751" LastEditDate="2017-10-01T12:44:50.490" LastActivityDate="2017-10-01T12:44:50.490" Title="Why is change in gibbs energy not zero through the gibbs free energy equation?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33315" PostTypeId="2" ParentId="33314" CreationDate="2015-06-25T05:22:25.413" Score="8" Body="&lt;p&gt;The change in Gibbs free energy for a closed system is given by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G = \Delta H - T \Delta S$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What you are describing is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta H = T \Delta S$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which is only true for systems at equilibrium, or when &lt;span class=&quot;math-container&quot;&gt;$\Delta G = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Why this is true is a little bit difficult to explain. It is essentially a form of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clausius_theorem&quot; rel=&quot;noreferrer&quot;&gt;Clausius Inequality&lt;/a&gt; at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta S \ge \oint \frac{dQ}{T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.&lt;/p&gt;&#xA;&lt;p&gt;If you integrate that at constant temperature you get&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta S \ge \frac{\Delta Q}{T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At constant pressure (and in the absence of non-expansion work) &lt;span class=&quot;math-container&quot;&gt;$\Delta Q = \Delta H$&lt;/span&gt;, so we can write:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta S \ge \frac{\Delta H}{T}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;...and a little rearranging gives:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta H - T \Delta S \leq 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G = \Delta H - T \Delta S$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;to give us an easier way to talk about it. Processes will happen (on their own) if &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; is negative, they won't happen if it's positive, and when they are at equilibrium, &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; will be zero.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastEditorUserId="-1" LastEditorDisplayName="user5764" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2016-02-24T10:27:21.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33316" PostTypeId="2" ParentId="27773" CreationDate="2015-06-25T05:32:49.770" Score="3" Body="&lt;p&gt;I think you are confusing two related concepts here. Spontaneity of a reaction is determined by the net change in free energy, $\Delta G$ - notice the capital Delta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which direction you will move along a reaction coordinate is determined by $dG$ - note the little $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G$ is the overall change in free energy when comparing products to reactants. It tells you whether the reactants are &quot;uphill&quot; or &quot;downhill&quot; of the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dG$ is the slope of the free energy surface at your current position on the reaction coordinate. In the equation you gave, that's given by Q/K. K basically tells you what the equilibrium ratio between reactants and products is, and Q tells you what the current ratio is. The ratio of Q/K is an indication of how far to the left or right you are of equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture might help explain it a little better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qn41n.png&quot; alt=&quot;Difference between net change in free energy and derivative of free energy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image the free energy surface as a gravitational potential energy surface (we can get away with this because free energy is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamic potential&lt;/a&gt; for systems at constant T &amp;amp; P). The ball represents the reacting system, and the x-axis is a reaction coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dG$ (the differential) tells us the slope at a given point - it tells us where the system is going to go &lt;em&gt;right now&lt;/em&gt;. $\Delta G$ tells us the overall change - it tells us where the system is going to eventually end up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look up standard state $\Delta G$ values for a reaction, you are determining whether an overall reaction is spontaneous or not - you are figuring out which side of the equilibrium has more stuff - reactants or products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at Q, you are figuring out which direction things need to go to get back to equilibrium, based on where you currently are.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-06-25T05:32:49.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33317" PostTypeId="1" AcceptedAnswerId="37475" CreationDate="2015-06-25T06:51:24.633" Score="5" ViewCount="1188" Body="&lt;p&gt;Why are salts of strongly electropositive elements less heavily hydrated in aqueous solution than the those of less electropositive elements? Shouldn't it be the other way round as the salts of electropositive elements are mostly ionic and hence strongly attracted by water molecules?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="10185" LastEditDate="2015-06-25T12:44:46.127" LastActivityDate="2015-09-17T14:00:00.993" Title="Relation Between Degree of Hydration and Electropositivity" Tags="&lt;aqueous-solution&gt;&lt;periodic-trends&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33319" PostTypeId="2" ParentId="33313" CreationDate="2015-06-25T08:56:43.460" Score="1" Body="&lt;p&gt;I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.&lt;/p&gt;&#xA;&lt;p&gt;Some parts of your calculation are correct, you do however forget that &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; is released. This would be my approach:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This means that there was three times as much &lt;span class=&quot;math-container&quot;&gt;$\ce{CoC2O4}$&lt;/span&gt; in your sample.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;That equals a weight of &lt;span class=&quot;math-container&quot;&gt;$\pu{212.1 mg}$&lt;/span&gt;. Your original sample weighed &lt;span class=&quot;math-container&quot;&gt;$\pu{328.3 mg}$&lt;/span&gt;, so there is &lt;span class=&quot;math-container&quot;&gt;$\pu{116.2 mg}$&lt;/span&gt; left that seem to be &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\pu{18.015 g/mol}$&lt;/span&gt;), giving you.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$n_\ce{H2O} = \pu{6.45 mmol}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The ratio of cobalt oxalate to water then is &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{6.450}{1.443} = 4.5\right)$&lt;/span&gt; which implies the empirical formula is &lt;span class=&quot;math-container&quot;&gt;$\ce{2CoC2O4.9H2O}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="17102" LastEditorUserId="95133" LastEditDate="2021-05-16T08:48:08.837" LastActivityDate="2021-05-16T08:48:08.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33322" PostTypeId="1" AcceptedAnswerId="33325" CreationDate="2015-06-25T12:56:35.337" Score="8" ViewCount="33526" Body="&lt;p&gt;Even when ionic compounds are strong (e.g. rubies and sapphires), they are neither malleable nor ductile in the slightest, and if sufficient stress is put on them, they will shatter, not bend. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="30836" LastEditDate="2020-04-29T19:40:07.740" LastActivityDate="2020-11-10T10:29:38.620" Title="Why are so many ionic compounds brittle?" Tags="&lt;crystal-structure&gt;&lt;ionic-compounds&gt;&lt;materials&gt;&lt;mechanical-properties&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33325" PostTypeId="2" ParentId="33322" CreationDate="2015-06-25T13:15:12.367" Score="14" Body="&lt;p&gt;Ionic crystals are hard because of tight packing lattices, say, the positive and negative ions are strongly attached among themselves.&lt;br&gt;&#xA;So, if mechanical pressure is applied to an ionic crystal then ions of similar charges may be forced to get closer to each other.&lt;br&gt;&#xA;Now, by doing so, the electrostatic repulsion can be enough to split or disorient completely the lattice infrastructure. Thus imparting the brittle character.&lt;/p&gt;&#xA;" OwnerUserId="16786" LastActivityDate="2015-06-25T13:15:12.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33328" PostTypeId="2" ParentId="18981" CreationDate="2015-06-25T13:53:01.387" Score="5" Body="&lt;p&gt;As the curve that represents the buffer capacity does not simply rely on one fraction of the whole system but on all species of the acid-base system it is not enough to only consider the derivative of $c(\ce{HA-})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to take care of the other deprotonated species in your system. For the case of a diprotic acid this means that $c(\ce{A^2-})$ is also important albeit not much as you can see in the following figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/q5VSU.png&quot; alt=&quot;comparison of different simplifications onto the buffer capacity of a diprotic acid&quot;&gt;&#xA;&lt;strong&gt;Figure 1:&lt;/strong&gt; Buffer capacities for a diprotic acid with different amounts of assumptions taken into account – denoted by their plot labels. The lower graphs are zoomed versions of the upper graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have come to the mass balance for $c(\ce{HA-})$ I don't have to show you how to come to the mass balance for $c(\ce{A^2-})$:&#xA;$$c(\ce{A^2-}) = \frac{c_0(\ce{H2A})~k_{a1}~k_{a2}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you have to derive the following equation:&#xA;$$C_B=\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$&#xA;or if you want to include he self-ionization of water it has to be&#xA;$$C_B=\frac{k_W}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{c_0(\ce{H2A})~\left(c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}\right)}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+2~k_{a1}~k_{a2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's derive (with the help of &lt;a href=&quot;http://www.jbc.org/content/52/2/525.full.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;D. D. van Slyke, &lt;em&gt;J. Biol. Chem.&lt;/em&gt; &lt;strong&gt;1922&lt;/strong&gt;, (&lt;em&gt;52&lt;/em&gt;), 525-570&lt;/a&gt; and Mathematica (or by hand if you've got time for that))&#xA;$$\beta=-\frac{\mathrm{dC_B}}{\mathrm{dpH}}=-\ln(10)~c(\ce{H3O+})~\left(\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}\right)$$&#xA;$$\frac{\mathrm{d C_B}}{\mathrm{d}c(\ce{H3O+})}=-\frac{k_W}{x}-1-\frac{c_0(\ce{H2A})~k_{a1}~c(\ce{H3O+})~(2~c(\ce{H3O+})+k_{a1})}{(c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2})^2}+\frac{c_0(\ce{H2A})~k_{a1}}{c^2(\ce{H3O+})+c(\ce{H3O+})~k_{a1}+k_{a1}~k_{a2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this final equation you should come up with plots like mine and see that the buffer capacity (for $c_0(\ce{H2A})=0.1~\ce{mol/L}$) at $\ce{pH}=\frac{1}{2}\left(\ce{pK_{a1}}+\ce{pK_{a2}}\right)$ is not zero albeit not being very high.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;complete: $\beta=0.0014~\ce{mol/L}$&lt;/li&gt;&#xA;&lt;li&gt;complete w/o &lt;em&gt;water&lt;/em&gt;: $\beta=0.0013~\ce{mol/L}$&lt;/li&gt;&#xA;&lt;li&gt;w/o $c(\ce{A^2-})$: $\beta=0.0001~\ce{mol/L}$&lt;/li&gt;&#xA;&lt;li&gt;w/o $c(\ce{A^2-})$ and &lt;em&gt;water&lt;/em&gt;: $\beta=1.1 \cdot 10^{-10}~\ce{mol/L}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So ... &lt;strong&gt;is it correct to say that at maximum concentration of the amphiprotic substance the buffer capacity is zero?&lt;/strong&gt; No but for your to acidic constants it's nearly zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But is it always zero at $c_\text{max}(\ce{HA^-})$?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at those systems a little bit closer. Imagine a few systems and their corresponding buffer capacity curves with $pK_{a1}=4$ and $pK_{a2}$ values between 4 and 10 like in the following figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MZQ2O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you create a plot with the buffer capacity over $\frac{1}{2}\left(pK_{a1}+pK_{a2}\right)$ you would end up with something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sy3GX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can see now is that the buffer capacity at $c_\text{max}(\ce{HA-})$ depends clearly on the distance between both $pK_a$ values. If they are close together the buffer capacity at $c_\text{max}(\ce{HA-})$ is high and if $\Delta pK_a \approx 3$ the buffer capacity at $c_\text{max}(\ce{HA-})$ is nearly zero but never exactly zero.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-06-25T13:53:01.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33329" PostTypeId="1" AcceptedAnswerId="33341" CreationDate="2015-06-25T14:24:11.863" Score="3" ViewCount="219" Body="&lt;p&gt;The fundamental equation of thermodynamics, as us chemists (and chemical engineers!) are used to seeing it, is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dG = - S~dT + V~dP + \sum_{i}\mu_i~dN_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the Gibbs free energy as a function of temperature, pressure, and composition, &lt;em&gt;assuming there are no other relevant forces other than mechanical pressure&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other day I watched &lt;a href=&quot;https://www.youtube.com/watch?v=uJ_NpWnCXzM&quot; rel=&quot;nofollow&quot;&gt;a video&lt;/a&gt; on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_refrigeration&quot; rel=&quot;nofollow&quot;&gt;magnetocaloric effect&lt;/a&gt;.  Obviously, there are non-pressure magnetic forces acting in such systems.  &lt;strong&gt;What's the proper form of the fundamental equation for magnetocaloric materials?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that the magnetocaloric material used is chemically pure and non-reactive during the magnetization process.  Then we could get rid of the $\sum_{i}\mu_i~dN_i$ term.  I suppose its also reasonable to assume that pressure is constant during magnetization / demagnetization process, and that the volume of the material is unchanged by magnetization so probably we could dispense with the $V~dP$ term as well (is that true?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves us with $dG = -S~dT + \rm{MAGNETIC~STUFF}$.  The $\rm{MAGNETIC ~STUFF}$ term probably has a $B$ or $H$ or something like that in it to represent the imposed magnetic field, but what else goes in there?&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-06-25T15:02:03.273" LastActivityDate="2015-06-25T18:15:32.363" Title="Fundamental equation of thermodynamics for magnetocaloric materials" Tags="&lt;thermodynamics&gt;&lt;magnetism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33330" PostTypeId="1" CreationDate="2015-06-25T14:48:11.567" Score="2" ViewCount="4463" Body="&lt;p&gt;If you have ever grown Aloe Vera and applied it to your skin, it feels a bit slimy. However, after it evaporates it has a cool feeling on your skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why does the Aloe Vera evaporate so fast? What is it made of, apart from water that makes Aloe Vera do this? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am assuming Hand Sanitizer and Aloe Vera are not directly connected.&lt;/p&gt;&#xA;" OwnerUserId="8208" LastEditorUserId="8208" LastEditDate="2015-06-25T15:03:42.433" LastActivityDate="2015-06-27T04:48:14.817" Title="How does Aloe Vera leave a cool feeling on your skin? Why does it evaporate fast?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33331" PostTypeId="1" CreationDate="2015-06-25T14:57:09.443" Score="4" ViewCount="466" Body="&lt;p&gt;I was reading Peter Atkins' &lt;em&gt;Elements of Physical Chemistry&lt;/em&gt; where I came across this line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bonds do not form &lt;strong&gt;because&lt;/strong&gt; electrons tend to pair; bonds are &lt;strong&gt;allowed&lt;/strong&gt; to form by the electrons pairing their spins.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does this line mean? What does Atkins want to say through this line &amp;amp; why did he italize the words I bolded above? &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="9961" LastEditDate="2015-06-25T16:35:10.170" LastActivityDate="2015-06-25T16:49:26.747" Title="Bonds and pairing of spins" Tags="&lt;physical-chemistry&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33332" PostTypeId="1" CreationDate="2015-06-25T15:11:50.943" Score="9" ViewCount="22625" Body="&lt;p&gt;I was reading some organometallics and I came upon the preparation of uranocene, $\ce{U(C8H8)2}$. It has two cyclooctatetraenide anions, $\ce{C8H8^2-}$, as ligands, with a uranium atom between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was mentioned that the neutral cyclooctatetraene molecule $\ce{C8H8}$ is non aromatic and has a tub-shaped conformation, but the dianion $\ce{C8H8^2-}$ is planar but aromatic. Why is this so? Can fulfilling Hückel's rule force a cyclic molecule to become planar? &lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="16683" LastEditDate="2018-01-28T12:15:55.387" LastActivityDate="2018-01-28T12:20:38.290" Title="Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar?" Tags="&lt;organic-chemistry&gt;&lt;aromaticity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33333" PostTypeId="2" ParentId="33332" CreationDate="2015-06-25T15:17:36.187" Score="11" Body="&lt;p&gt;Fulfilling Hückel's rule means that you get to be aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason cyclooctotetraene is non-planar is because its planar arrangement would be antiaromatic. It twists to avoid it, because antiaromatic compounds are unstable with respect to localisation of the π-electrons (see here: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/30927/what-is-the-justification-for-h%C3%BCckels-rule/32446#32446&quot;&gt;What is the justification for H&amp;#252;ckel&amp;#39;s Rule?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The veritable nanosecond that a molecule has the potential to become aromatic though, it will &lt;em&gt;leap&lt;/em&gt; at the opportunity. Any angle strain from being planar is so quickly overtaken by the huge jump in stability that aromaticity brings that the molecule will twist to form an aromatic structure in a blink.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16683" LastEditDate="2018-01-28T12:20:38.290" LastActivityDate="2018-01-28T12:20:38.290" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33336" PostTypeId="1" AcceptedAnswerId="33346" CreationDate="2015-06-25T16:30:37.360" Score="3" ViewCount="1010" Body="&lt;p&gt;I wonder which fundamental forces act behind osmosis process? Sum of these forces oppose gravity force, like in this figure: &lt;img src=&quot;https://i.stack.imgur.com/0orO6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder what are these forces?&lt;/p&gt;&#xA;" OwnerUserId="16780" LastActivityDate="2015-06-25T21:38:47.970" Title="Fundamental physical forces behind osmosis" Tags="&lt;intermolecular-forces&gt;&lt;osmosis&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33337" PostTypeId="2" ParentId="33331" CreationDate="2015-06-25T16:49:26.747" Score="5" Body="&lt;p&gt;I dunno if the author meant exactly this, but it is a well-known fact that pairing of electrons actually consumes energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, free nitrogen atom in ground state has three unpaired electrons. To form triple bond in dinitrogen molecule, three pairs of electrons must be, well, paired. The process of pairing does consume energy.  However, the energy freed in the bond formation is so big, that it overwhelms pairing energy by huge margin. On the other hand, in oxygen molecule two highest-energy electrons has two orbitals to occupy, and they stay unpaired, so dioxygen is paramagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by no means it is the fact of being unpaired that drives electrons to form bonds. In fact, it prevents formation of weak bonds. However, when two atoms come closer to each other and have two half-occupied orbitals that can form a bond, the fact that electrons may be paired allows this two orbitals to combine into occupied bonding orbital, forming a bond.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-06-25T16:49:26.747" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33338" PostTypeId="1" CreationDate="2015-06-25T17:10:27.150" Score="14" ViewCount="3175" Body="&lt;p&gt;Are all the &lt;span class=&quot;math-container&quot;&gt;$\ce{C-N}$&lt;/span&gt; single bonds in 2-bromo-1,3,5-trinitrobenzene the same or different? I am of the view that the &lt;span class=&quot;math-container&quot;&gt;$\ce{C_1-N}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{C_3-N}$&lt;/span&gt; bonds are of the same length and they are longer than the &lt;span class=&quot;math-container&quot;&gt;$\ce{C_5-N}$&lt;/span&gt; bond, due to steric hindrance. Am I right or am I missing out something?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4LC92.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LC92.png&quot; alt=&quot;structure of 2-bromo 1,3,5-trinitro benzene&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://webbook.nist.gov/cgi/inchi?Struct=C4185539&amp;amp;Type=Color&quot; rel=&quot;nofollow noreferrer&quot;&gt;nist.gov&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="17156" LastEditorUserId="15751" LastEditDate="2021-01-16T06:31:57.087" LastActivityDate="2021-01-16T06:31:57.087" Title="C-N bond lengths in 2-bromo-1,3,5-trinitrobenzene" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33339" PostTypeId="2" ParentId="33338" CreationDate="2015-06-25T17:36:08.473" Score="12" Body="&lt;p&gt;By rotating the molecule 180° about the $\mathrm{C_2}$ symmetry axis that passes through the bromine and the nitro group in the 5-position, you can interconvert $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$.  Groups that can be interconverted by a rotational axis &lt;strong&gt;must&lt;/strong&gt; be equivalent.  There is &lt;strong&gt;no&lt;/strong&gt; symmetry element that interconcerts $\ce{C_1-NO2}$ (or $\ce{C_3-NO2}$) and $\ce{C_5-NO2}$, therefore these groups &lt;strong&gt;cannot&lt;/strong&gt; be equivalent. So symmetry tells us that the  $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$ bond lengths are exactly the same, and both are different from the $\ce{C_5-NO2}$ bond length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetry can only tell us if things are the same or different, it can't tell us the magnitude of any difference.  So I don't know for sure which set of carbon-nitro bonds is longer or shorter, but your argument based on steric strain seems reasonable.  It is also possible that the bulky bromine causes the adjacent nitro groups to twist out of the plane of the aromatic ring, reducing resonance overlap between nitrogen and carbon and also lengthening those adjacent nitro bonds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-06-25T17:36:08.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33340" PostTypeId="1" AcceptedAnswerId="33345" CreationDate="2015-06-25T17:39:27.910" Score="5" ViewCount="531" Body="&lt;p&gt;How can hybrid orbitals be equivalent? Take, for example, $\mathrm{sp^3}$ hybridization. The wave functions of each hybrid orbitals are given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\psi_1 &amp;amp;= \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\&#xA;\psi_2 &amp;amp;= \psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\&#xA;\psi_3 &amp;amp;= \psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\&#xA;\psi_4 &amp;amp;= \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\\&#xA;\end{align}&#xA;No one of them has same linear combination. So, how can they be equivalent? I am not getting the sense.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="4945" LastEditDate="2015-06-29T08:31:20.117" LastActivityDate="2015-06-29T08:31:20.117" Title="How can hybrid orbitals be equivalent?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33341" PostTypeId="2" ParentId="33329" CreationDate="2015-06-25T18:15:32.363" Score="3" Body="&lt;p&gt;The fundamental equation for paramagnetic system is as follows&lt;br&gt;&#xA;$$dU=TdS+BdM+\mu dN$$&lt;br&gt;&#xA;Where B is the external magnetic field intensity and M is the magnetic moment (Here I have neglected the P-V work of the system). For your chemically pure and non-reactive system the above equation simplifies to&lt;br&gt;&#xA;$$dU=TdS+BdM$$&lt;br&gt;&#xA;Using Legendre Transformations:&lt;br&gt;&#xA;$$y(0)=U(S,M)$$&#xA;$$dy(0)=dU=TdS+BdM$$&#xA;$$dy(1)=dA(T,M)=-SdT+BdM$$&#xA;$$dy(2)=dG(T,B)=-SdT-MdB$$ &#xA;Now changing the order of the equation, we can find enthalpy,&#xA;$$y(0)=U(M,S)$$&#xA;$$dy(0)=dU=BdM+TdS$$&#xA;$$dy(1)=dH(S,B)=-MdB+TdS$$&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-06-25T18:15:32.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33342" PostTypeId="2" ParentId="33340" CreationDate="2015-06-25T19:48:04.150" Score="4" Body="&lt;p&gt;The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them.  Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-06-25T19:48:04.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33344" PostTypeId="2" ParentId="33336" CreationDate="2015-06-25T21:03:56.883" Score="2" Body="&lt;p&gt;If you dribble a basketball, the force of your hand on the ball, that keeps it bouncing, is not (1) gravity; (2) the weak force; or (3) the strong force. It must be electromagnetic force--the electrons in the molecules of your skin repelling the electrons in the molecules on the surface of the ball. Also, your skin and the ball remain (largely) intact because of the electromagnetic force that hold them together, respectively. Same for osmotic &quot;force&quot;. The motions of the solute and solvent molecules are not governed meaningfully by anything other than electromagnetic force.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-06-25T21:03:56.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33345" PostTypeId="2" ParentId="33340" CreationDate="2015-06-25T21:08:22.890" Score="4" Body="&lt;p&gt;You are right in that the &lt;em&gt;sign&lt;/em&gt; of each orbital, especially the p ones, is different in every hybrid orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equations are, however, lacking the normalised contribution factors that should precede the wavefunctions. A better equation for the first orbital would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi_1 = a_s \psi_s + a_{p_x} \psi_{p_x} + a_{p_y} \psi_{p_y} + a_{p_z} \psi_{p_z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, that the contribution factors are &lt;em&gt;equal&lt;/em&gt; for every sp³ hybrid orbital — every hybrid gets ¼ contribution from every p and the s-orbital. That way, all orbitals are rendered equal because their atomic orbital contribution factors are equal, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signs before each p-orbitals contribution &lt;em&gt;do not matter for the orbital’s energy value, only for its orientation.&lt;/em&gt; So inspite of all orbitals having different combinations of signs, their energy can be equal (but they will ‘point’ in different directions).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Note that this is a simplified view as are hybrid orbitals in general. Spectroscopically, you can measure two &lt;em&gt;different&lt;/em&gt; energies of the bonding orbitals in methane — the explanation cannot be given by hybridisation theory but requires proper molecular orbital theory.)&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-25T21:08:22.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33346" PostTypeId="2" ParentId="33336" CreationDate="2015-06-25T21:21:06.263" Score="1" Body="&lt;p&gt;The water molecules move across the membrane due to thermal energy. If concentration of solute on both sides of the membranes is the same, the amount of water crossing the membrane in either direction is same but when the concentration is unequal, the side with more solute particles (higher concentration, that is) attracts the water molecules and thus stopping them from going back. This process goes on until the particles cannot stop more water molecules from going back and equilibrium is achieved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering the question, the physical force that causes osmosis is the one that is responsible for dissolution of the solute in the given solvent which is (as others pointed out) electromagnetic in nature.&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="7697" LastEditDate="2015-06-25T21:38:47.970" LastActivityDate="2015-06-25T21:38:47.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33347" PostTypeId="2" ParentId="33279" CreationDate="2015-06-25T21:54:35.263" Score="10" Body="&lt;p&gt;Hybridisation is a &lt;strong&gt;concept devised by man to ease explanation of chemical observations.&lt;/strong&gt; It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the &lt;strong&gt;molecular orbital approach.&lt;/strong&gt; It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have &lt;em&gt;a lot&lt;/em&gt; more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The &lt;em&gt;wave functions&lt;/em&gt; that one can derive from all this information correspond to &lt;em&gt;molecular orbitals&lt;/em&gt; you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the &lt;em&gt;entire&lt;/em&gt; molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The second method is the &lt;strong&gt;hybridisation approach.&lt;/strong&gt; This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital &lt;em&gt;is not&lt;/em&gt; considered to be of a new sp-hybrid type, but rather &lt;em&gt;gets a new type of name:&lt;/em&gt; It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes &lt;em&gt;along&lt;/em&gt; the bond axis (the bond axis is completely part of the nodal plane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, &lt;em&gt;only we are much, much faster,&lt;/em&gt; which is why this method is often taught in schools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Onto the question of &lt;em&gt;how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$&lt;/em&gt;:&lt;br&gt;&#xA;You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that &lt;strong&gt;you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies.&lt;/strong&gt; This is what is meant with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technically, this is short for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would &lt;em&gt;recentre a p-orbital from the carbon to the hydrogen.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-25T21:54:35.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33348" PostTypeId="2" ParentId="33055" CreationDate="2015-06-25T22:31:30.070" Score="3" Body="&lt;p&gt;Technically, other electrons could play a part in determining chemistry. However, there is a &lt;em&gt;significant&lt;/em&gt; energy barrier between the p-orbitals of one period and the s-orbital of the following period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider copper. The characteristic deep blue colour of $\ce{[Cu(NH3)4(H2O)2]^2+}$ derives from the absorption of photons whose energy corresponds to the difference of the $\mathrm{3d}_{z^2}$ and $\mathrm{3d}_{x^2 - y^2}$ orbitals. This is spin-allowed because copper(II) is a d⁹ system. The wavelength corresponds to approximately $610\:\mathrm{nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the energy between a $\ce{2p}$- and a $\ce{3s}$-orbital can be measured using x-ray transition — it corresponds to a photon of a wavelength of approximately $1\:\mathrm{nm}$. This is about six-hundred times the energy difference compared to inside the d-orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is a slightly unfair comparison. The energies between $\ce{3d}$ and $\ce{4s}$ are noticeably larger than inside of the $\ce{3d}$ orbitals, and the energy difference between $\ce{4s}$ and $\ce{4p}$ should again be well larger — but they will remain &lt;em&gt;significantly&lt;/em&gt; smaller than energy differences across shells. The former are typically visible or UV, the latter are typically x-ray.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastActivityDate="2015-06-25T22:31:30.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33349" PostTypeId="1" AcceptedAnswerId="60376" CreationDate="2015-06-25T23:07:23.647" Score="25" ViewCount="293" Body="&lt;p&gt;Old papers on chemical structures often contained &quot;3D&quot; renderings of the molecule by using 2 2D images and requiring the reader to go cross-eyed to get the 3D image. A good example I recently found was figure 2 from &lt;a href=&quot;http://www.pnas.org/content/71/2/283.abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carter and Kraut, 1974&lt;/a&gt;, which I copied below: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/NR7HV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this paper is from 1974, which feels like ancient history for 3D graphics. Since I have unrelated interests in molecular biology and 3D graphics, I was wondering how they produced these images so long ago, and if the code to produce them is floating around somewhere. I know there are programs to do it for me because I have seen modern images of proteins that do this, I want the code because I might try to build it into my &lt;a href=&quot;https://www.youtube.com/watch?v=_XYj113Le58&quot; rel=&quot;nofollow noreferrer&quot;&gt;crappy 3D rendering project&lt;/a&gt; that I'm wasting too much time on.&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="402" LastEditDate="2017-05-08T00:59:29.880" LastActivityDate="2017-05-08T00:59:54.857" Title="How were old style stereographic structures produced?" Tags="&lt;structural-formula&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33350" PostTypeId="1" AcceptedAnswerId="33355" CreationDate="2015-06-26T02:20:46.747" Score="10" ViewCount="1012" Body="&lt;p&gt;On the physics site, I asked &lt;a href=&quot;https://physics.stackexchange.com/q/190726&quot;&gt;the difference between real &amp;amp; complex orbital&lt;/a&gt;. Real orbitals are the superposition of complex orbitals having definite magnetic quantum number states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;p_z &amp;amp;= p_0 \\&#xA;p_x &amp;amp;= \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\&#xA;p_y &amp;amp;= \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\&#xA;\end{align} &#xA;where $p_0 = R_{n1} Y_{10}$, $p_1 = R_{n1} Y_{11}$, and $p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $\ell = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't the orbitals of hydrogenic atoms taken as complex orbitals instead of the real ones? They are valid states for an electron, aren't they? So, what is the problem in having complex atomic orbitals?&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-06-29T08:36:53.260" Title="Why are real orbitals used to define the orbital of hydrogenic atoms instead of complex orbitals?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33351" PostTypeId="2" ParentId="16055" CreationDate="2015-06-26T08:28:53.857" Score="-1" Body="&lt;p&gt;From my thinking helium is a stable gas and is a noble gas whereas hydrogen being the lightest gas but is flammable that's why helium is filled in weather balloons.&lt;/p&gt;&#xA;" OwnerUserId="17163" LastActivityDate="2015-06-26T08:28:53.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33352" PostTypeId="1" AcceptedAnswerId="33499" CreationDate="2015-06-26T08:44:11.033" Score="6" ViewCount="204" Body="&lt;p&gt;Carbonic acid ($\ce{H2CO3}$) is known to be &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6587/is-h2co3-carbonic-acid-soluble-or-insoluble-in-water&quot;&gt;soluble in water&lt;/a&gt;, and that this particular compound is &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16076/the-instability-of-hydrated-carbon-dioxide-or-carbonic-acid&quot;&gt;unstable&lt;/a&gt;, so much so that according to the 2014 article from Berkeley Lab: &#xA;&lt;a href=&quot;https://newscenter.lbl.gov/2014/10/22/new-insights-on-carbonic-acid-in-water/&quot; rel=&quot;nofollow noreferrer&quot;&gt;New Insights on Carbonic Acid in Water&lt;/a&gt;, it only exists for a fraction of a second (300 nanoseconds according the article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newly reported experiments and observations from the Berkeley Lab found that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“Our results support an average hydration number of 3.17 with the acid’s two protons each donating a strong hydrogen bond to solvating waters, the carbonyl oxygen accepting a strong hydrogen bond from solvating water, and the hydroxyl oxygen molecules accepting weak hydrogen bonds from the water”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is not clear is how is the hydration number calculated and how does this relate to the short existence of carbonic acid in water?&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-07-03T11:33:45.260" Title="Hydration number: Carbonic acid in water" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33353" PostTypeId="1" AcceptedAnswerId="33354" CreationDate="2015-06-26T11:09:45.857" Score="5" ViewCount="3490" Body="&lt;p&gt;I'm finding it difficult to visualise the process of hybridisation. Taking the example of formation of $\ce{CH4}$, can it viewed as the collapsing of the p orbitals of the Carbon atoms as they get &quot;heavier&quot; on excitation even before the introduction of Hydrogen atoms or is it the collision with Hydrogen atoms that causes the fusion?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="7448" LastEditDate="2015-06-26T11:15:14.393" LastActivityDate="2019-11-01T01:01:15.003" Title="What Causes Atomic Orbitals to Hybridise?" Tags="&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33354" PostTypeId="2" ParentId="33353" CreationDate="2015-06-26T12:44:22.967" Score="10" Body="&lt;p&gt;Hybridization is a mathematical concept developed to cope up the flaws of Valence Bond theory. It hinges solely on the linear combination of wavefunctions of atomic orbitals having more or less same energy. Hybridized state is a purely theoretical concept &amp;amp; cannot be detected even spectroscopically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing like &quot;collapse.&quot; The wavefunction of a hybridized orbital is &quot;&lt;strong&gt;mixing&lt;/strong&gt;&quot; of the atomic orbitals only. The number of hybrid orbitals are same as the number of atomic orbitals participating in the &quot;mixing up&quot;. The wavefunctions interfere constructively or destructively to give rise to equivalent orbitals known as &lt;em&gt;hybrid orbitals.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Methane Case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon has valence configuration: $2s^2 2p_x ^1  2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then &lt;strong&gt;promotion&lt;/strong&gt; comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ &amp;amp; $p$ orbitals are not  same. To deal with the problem, concept of &lt;strong&gt;hybridization&lt;/strong&gt; was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://wps.prenhall.com/wps/media/objects/3311/3391094/imag0905/AAAVJKM2.JPG&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.620" LastActivityDate="2015-06-26T12:44:22.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33355" PostTypeId="2" ParentId="33350" CreationDate="2015-06-26T12:55:30.000" Score="7" Body="&lt;p&gt;The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize.  For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few  advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular  and closed shell systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.&lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-26T12:55:30.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33356" PostTypeId="1" AcceptedAnswerId="33920" CreationDate="2015-06-26T13:13:29.657" Score="6" ViewCount="21204" Body="&lt;p&gt;Is there a trick or a formula that given a molecular formula, allows you to know exactly how many constitutional isomers can be formed with that many atoms? Or is it more of a trial-and-error technique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for something along the lines of $2^n$ of maximum number of possible stereo-isomers, where &lt;em&gt;n&lt;/em&gt; is the number of stereogenic centers present in a compound.&lt;/p&gt;&#xA;" OwnerUserId="17167" LastEditorUserId="7448" LastEditDate="2015-06-26T13:24:26.307" LastActivityDate="2015-07-13T13:06:48.163" Title="Is there a formula to determine the total number of constitutional isomers?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33357" PostTypeId="2" ParentId="33350" CreationDate="2015-06-26T13:25:16.397" Score="7" Body="&lt;p&gt;To add my 5 cents to the discussion, note that we also usually use &lt;em&gt;real basis functions&lt;/em&gt; (&quot;atomic orbitals&quot;) in electronic structure calculations, although, molecular orbital expansion coefficients may be complex. So, each and every MO $\phi_{k}$ is usually expressed as follows,&#xA;$$&#xA;    \phi_{k}(\vec{r}) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(\vec{r}) \, ,&#xA;$$&#xA;where $\chi_{q}(\vec{r})$ are real functions and $c_{qk}$ may in general be complex numbers. The point is that there is no additional variational freedom in using both complex basis functions and complex expansion coefficients. Besides, the use of real basis functions reduces (due to permutational symmetry) the number of two-electron integrals to be calculated for a basis of $m$ real functions from $m^4$ to $M(M+1)^2$, where $M=m(m+1)/2$ is the number of distinct pairs of products of two basis functions.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-26T13:32:03.837" LastActivityDate="2015-06-26T13:32:03.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33358" PostTypeId="2" ParentId="33356" CreationDate="2015-06-26T13:58:02.647" Score="4" Body="&lt;p&gt;There seems to be some confusion here : in order to determine the number of &lt;em&gt;stereo&lt;/em&gt;-isomers of a molecular formula (say C4H11N) you would need to first know the number of &lt;em&gt;constitutional isomers&lt;/em&gt; as you need to know how many stereogenic centers across all the structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a bunch of existing questions on constitutional isomers (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/10620/is-there-a-mathematical-generalization-for-the-number-of-isomers-structural-and&quot;&gt;a&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers&quot;&gt;b&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/20237/how-do-i-approach-drawing-all-of-the-possible-constitutional-isomers&quot;&gt;c&lt;/a&gt;, ...) but the summary is : &quot;it is tricky without software for anything more than very small formulae&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;strong&gt;are&lt;/strong&gt; mathematical formulae for determining the count of constitutional isomers, although the most powerful techniques are non-trivial (such as those based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pólya enumeration theorem&lt;/a&gt;). To efficiently list a non-redundant set of compounds from an elemental formula is possibly even harder, as it requires a mechanism to avoid duplicates in the output.&lt;/p&gt;&#xA;" OwnerUserId="4665" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-06-26T13:58:02.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33359" PostTypeId="1" AcceptedAnswerId="33361" CreationDate="2015-06-26T14:14:35.207" Score="2" ViewCount="2745" Body="&lt;p&gt;Atmospheric oxygen is in O2 and a gas. Then we inhale the air, our efficient lungs do the magic to filter out the oxygen and &lt;em&gt;push&lt;/em&gt; them into the blood stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we say hemo and globin transport the oxygen using the iron ions. In what state oxygen is transported in the blood? as a gas or a liquid or an ion? It is hard for me to conceive of the idea that oxygen would be in gaseous form in the blood. &quot;GAS in blood?&quot; e.g. &lt;em&gt;Arterial Blood Gas Test&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how does the lungs &lt;em&gt;convert&lt;/em&gt; the gas into &lt;em&gt;something&lt;/em&gt; that is compatible to be in blood?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&#xA;&lt;a href=&quot;http://courses.washington.edu/conj/resp/oxygen.htm&quot; rel=&quot;nofollow&quot;&gt;Amount of Oxygen in the Blood&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="4945" LastEditDate="2015-06-29T12:05:53.423" LastActivityDate="2015-06-30T04:31:56.827" Title="What is the state of aggregation (gas, liquid) of oxygen in blood?" Tags="&lt;biochemistry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33360" PostTypeId="2" ParentId="33359" CreationDate="2015-06-26T14:37:22.980" Score="2" Body="&lt;p&gt;We inhale a lot of other atmospheric gases with oxygen. The mixture of gases go to the sacs in our lungs which is crawling with lots of blood capillaries.&lt;br&gt;&#xA;The blood (specifically RBCs) contains a pigmented substance called haemoglobin, each of them can bind 4 oxygen molecules (if working normally) forming a substance named Oxyhaemoglobin.&lt;br&gt;&#xA;This is then carried through blood vessels to the cells in the body.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binding phenomena of oxygen with haemoglobin is reversible in nature. That is, at high oxygen concentration oxygen binds with haemoglobin to form oxyhaemoglobin, but when the oxygen concentration decreases (in cells, which are in need of oxygen) the oxygen again dissociates from haemoglobin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overview thus provided here, if you wish to be more specific about chemical reactions you may find &lt;a href=&quot;http://www.rsc.org/Education/Teachers/Resources/cfb/transport.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; useful.&lt;/p&gt;&#xA;" OwnerUserId="16786" LastEditorUserId="16786" LastEditDate="2015-06-30T04:31:56.827" LastActivityDate="2015-06-30T04:31:56.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33361" PostTypeId="2" ParentId="33359" CreationDate="2015-06-26T14:40:31.987" Score="5" Body="&lt;p&gt;Regarding the state of oxygen in blood: It is in solution in the blood plasma (which mostly consists of water), in the form of single molecules. Think of water which you leave exposed to air: carbon dioxide will be captured and dissolved (along with the other gases in air), but these molecules are not gaseous or liquid, but rather &quot;in solution&quot;, which is different from the &quot;classical&quot; states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to oxygen: As your reference already states, most of the oxygen in solution will bind to hemoglobin. The actual state of oxygen in that complex has been debated, but it is believed to be reduced by the hemoglobin iron to the superoxide anion, coordinated to Fe$^{3+}$. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Hemoglobin#Iron.27s_oxidation_state_in_oxyhemoglobin&quot;&gt;Wikipedia&lt;/a&gt; on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the lungs do not &quot;convert&quot; the atmospheric oxygen to anything, they rather allow, due to their very large surface area, the quick exchange of oxygen/carbon dioxide in solution and in the air.&lt;/p&gt;&#xA;" OwnerUserId="17102" LastActivityDate="2015-06-26T14:40:31.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33362" PostTypeId="1" AcceptedAnswerId="33363" CreationDate="2015-06-26T15:36:23.007" Score="0" ViewCount="553" Body="&lt;p&gt;Consider the following electrochemical cell:&#xA;$$\ce{Zn|Zn^{2+}}(0.1~\mathrm{M})||\ce{Zn^{2+}}(0.01~\mathrm{M})\ce{|Zn}$$ How should I write the overall cell reaction for this cell? I can identify the oxidation and reduction reactions but when they are added, all the species cancel out, leaving my with no overall equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anode: $\ce{Zn -&amp;gt; Zn^{2+} + 2e}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cathode: $\ce{Zn^{2+} + 2e -&amp;gt; Zn}$&lt;/p&gt;&#xA;" OwnerUserId="16080" LastEditorUserId="10185" LastEditDate="2015-06-26T15:42:05.853" LastActivityDate="2017-03-03T21:32:18.560" Title="How to write the last cell reaction for the following cell?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33363" PostTypeId="2" ParentId="33362" CreationDate="2015-06-26T16:24:56.157" Score="3" Body="&lt;p&gt;The species do cancel out. However, the concentrations of the species do not. It is a bit of a matter of preference as to how you express this. The way I would write it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn^{2+}_{(aq)} ($c = 0.01\text{ M}$) + Zn_{(s)} &amp;lt;=&amp;gt; Zn^{2+}_{(aq)} ($c = 0.1\text{ M}$) + Zn_{(s)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for which $E = E^0 - \frac{RT}{2F}\,\ln\!\left(\frac{0.1}{0.01}\right) = -0.030 \text{ V}$ at 298 K, since $E^0 = 0 \text{ V}$. A negative value of $E$ agrees with the reverse reaction above being spontaneous, which is of course what you would expect since $0.1 &amp;gt; 0.01$.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-06-26T16:24:56.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33364" PostTypeId="1" AcceptedAnswerId="33366" CreationDate="2015-06-26T16:30:38.450" Score="15" ViewCount="1644" Body="&lt;p&gt;It seems like glass-coated stir bars are a specialty item.  Standard stir bars are coated in PTFE.  Is PTFE resistant to all typical lab solvents?  Or is there a reference to check to ensure it is inert to the chemicals being stirred?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="10185" LastEditDate="2015-06-26T16:32:11.560" LastActivityDate="2015-06-26T17:37:32.183" Title="Are PTFE stir bars resistant to all lab solvents?" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33365" PostTypeId="2" ParentId="33364" CreationDate="2015-06-26T17:29:59.057" Score="10" Body="&lt;p&gt;This may not be the answer that you are looking for, but I remember a project, for which we had to replace PTFE stirring bars with their glass-coated counterparts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that PTFE is resistant to most of the solvents used, but it is not really hard and resistant to abrasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a study where aqueous suspensions of metal silicides were stirred, we recognized very soon that the PTFE covers didn't last very long. The stirring bars were polished down to the metal core, which spoiled the experiment. Glass-covered stirring bars turned out to be a viable alternative here.&lt;/p&gt;&#xA;" OwnerUserId="418" LastEditorUserId="10185" LastEditDate="2015-06-26T17:31:27.813" LastActivityDate="2015-06-26T17:31:27.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33366" PostTypeId="2" ParentId="33364" CreationDate="2015-06-26T17:37:32.183" Score="8" Body="&lt;p&gt;Labware manufacturers will often put out chemical compatibility charts for their products. A search on Google for the phrase &lt;code&gt;ptfe chemical resistance&lt;/code&gt; yields a large number of hits, of varying thoroughness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fairly comprehensive one is &lt;a href=&quot;https://www.thermoscientific.com/content/dam/tfs/LPG/LCD/LCD%20Documents/Product%20Manuals%20%26%20Specifications/Filtration%20Products/Filter%20Units%20and%20Bottletop%20Filters/Filter%20Units/D20480~.pdf&quot;&gt;this one&lt;/a&gt; from ThermoFisher Scientific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scanning it, PTFE has &lt;em&gt;&quot;Excellent&quot;&lt;/em&gt; resistance listed for most solvents, except for pure 3-phenyl-2-propenal, which is &quot;merely&quot; &lt;em&gt;&quot;Good&quot;&lt;/em&gt; (&lt;em&gt;&quot;Little or no damage after 30 days of constant exposure.&quot;&lt;/em&gt;) Lists from other sources have a different sets of compounds, some of which caution against conditions like &quot;dry fluorine&quot; or some rarer compounds. &lt;a href=&quot;http://www.calpaclab.com/teflon-ptfe-compatibility/&quot;&gt;This one&lt;/a&gt; lists fluorine, gold monocyanide, and diethylamine as not recommended. (Though a number of other sources - including ThermoFisher - say diethylamine is perfectly fine.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So PTFE is resistant to pretty much all lab solvents which a major manufacturer of lab equipment thinks it's relevant to mention - but if you have something that you think may be atypical or which you have particular concerns about, do a search for that particular compound or contact the manufacturer for their recommendations.&lt;/p&gt;&#xA;" OwnerUserId="11616" LastActivityDate="2015-06-26T17:37:32.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33367" PostTypeId="2" ParentId="15852" CreationDate="2015-06-26T18:23:35.413" Score="2" Body="&lt;p&gt;It is because the heat energy is used to overcome the inter-molecular force of attraction but when this inter-molecular force is broken it changes it state and the temperature starts increasing.&lt;/p&gt;&#xA;" OwnerUserId="17168" LastActivityDate="2015-06-26T18:23:35.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33368" PostTypeId="1" CreationDate="2015-06-26T18:48:46.720" Score="6" ViewCount="6558" Body="&lt;p&gt;Whilst reading about stereochemistry, I came to know that the stability and melting point of cis isomers is lower than trans isomers. By searching on net I've found the reason behind that. But one question arises - why is the internal energy of cis isomers bigger than of trans isomers?&lt;/p&gt;&#xA;" OwnerUserId="16919" LastEditorUserId="10185" LastEditDate="2015-06-26T20:46:16.873" LastActivityDate="2015-07-27T09:02:15.603" Title="Cis-trans isomers internal energy" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33369" PostTypeId="2" ParentId="7648" CreationDate="2015-06-26T19:21:56.857" Score="12" Body="&lt;p&gt;Let's say we have got a solution of an acid with a pK$_a=5$. The resulting pH dependent on the concentration of that acid would be like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$ &lt;img src=&quot;https://i.stack.imgur.com/HrDec.png&quot; alt=&quot;dependence of the pH of an acid solution with pK_a 5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; $0.1~\ce{M}$: pH$=3.0$; $0.001~\ce{M}$: pH$=4.0$; $0.00001~\ce{M}$: pH$=5.2$; $0.0000001~\ce{M}$: pH$=6.8$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the pH resulting from these example concentrations we can have a look at percentage composition of the whole acid base system. (Degree of association $\alpha'$ in blue, degree of dissociation $\alpha$ in light orange.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$ &lt;img src=&quot;https://i.stack.imgur.com/K4hwb.png&quot; alt=&quot;composition of an acid base system depending on the given pH&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples:&lt;/strong&gt; pH$=3.0$: $\alpha\approx 1.0~\%$; pH$=4.0$: $\alpha\approx 9.5~\%$; pH$=5.2$: $\alpha\approx 61.8~\%$; pH$=6.8$: $\alpha\approx 98.4~\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can see that &lt;strong&gt;low acid concentrations&lt;/strong&gt; lead to &lt;strong&gt;higher pH values&lt;/strong&gt; than higher acid concentrations which automatically result in an &lt;strong&gt;increasing degree of dissociation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for me this is no big intuition thing but rather a small sequence of actions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lets say it in math:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proton concentration for your system can be described by $$c(\ce{H3O+}) = \sqrt{c_0(\ce{HA})~k_a+k_W}$$ (adding $k_W$ for very low concentrations) and the equation for the degree of dissociation is $$\alpha = \frac{k_a}{k_a+c(\ce{H3O+})}$$ which can be combined to give $$\alpha\left(c_0(\ce{HA})\right)=\frac{k_a}{k_a+\sqrt{c_0(\ce{HA})~k_a+k_W}}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;One could expand this system a little more (using the exact pH equation including the self-ionization of water) and come up with a thorough scheme like the following.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$ &lt;img src=&quot;https://i.stack.imgur.com/cF16V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-06-26T19:21:56.857" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33370" PostTypeId="2" ParentId="33368" CreationDate="2015-06-26T20:59:35.407" Score="3" Body="&lt;p&gt;I can think of two reasons:&lt;br&gt;&#xA;1) Steric repulsion&#xA;2) Electrostatic interaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steric repulsion: &lt;em&gt;cis&lt;/em&gt; means in same side. So, sterically it is suitable to be on opposite side. It is a fundamental nature of both human and chemical entity to define a personal boundary (My personal opinion :) )&lt;img src=&quot;https://i.imgur.com/Q0Gs3tD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;In the image you can see the personal boundary of $\ce{-CH_3}$ is the furthest in the &lt;em&gt;trans&lt;/em&gt; configuration and that's wh the &lt;em&gt;trans&lt;/em&gt; configuration is more stable than the &lt;em&gt;cis&lt;/em&gt; configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrostatic interaction: Due to electrostatic interaction, molecules or functional group try to distance themselve as far as possible. This also results in lower energy in trans configuration. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="16683" LastEditDate="2015-06-27T08:46:35.790" LastActivityDate="2015-06-27T08:46:35.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33371" PostTypeId="2" ParentId="33274" CreationDate="2015-06-26T21:45:11.320" Score="1" Body="&lt;p&gt;I think this question cannot be answered so easily - your proposed reaction mechanism, however, looks good. A detailed explanation and &quot;main reaction mechanism&quot; can be found &lt;a href=&quot;https://books.google.de/books?id=EHbhCQAAQBAJ&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot; title=&quot;here&quot;&gt;here&lt;/a&gt;, page 192 (just search for &quot;ethene&quot; in the book). As you already indicate ethene reacts with a hydroxyl radical, followed by the addition of oxygen. The reaction with NO generates the HOCH$_2$CH$_2$O radical which can either dissociate or react with another oxygen molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reactions of the oxidized products occuring during this pathway can also be found in the above reference, see page 372 (unfortunately only some parts of the book are available for free).&lt;/p&gt;&#xA;" OwnerUserId="17102" LastActivityDate="2015-06-26T21:45:11.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33372" PostTypeId="2" ParentId="33338" CreationDate="2015-06-26T22:04:18.970" Score="6" Body="&lt;p&gt;As I thought this could be interesting because I would have argued the same way as both of you did I have done some geometry optimizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final niveau was B3LYP-D3/def2-QZVPP in the gas phase &lt;em&gt;(no frequency calculation has been done because it would have taken too long for me to wait so I have to believe that it is a ground state)&lt;/em&gt; and the result &lt;strong&gt;from this one calculation&lt;/strong&gt; with this &lt;strong&gt;single functional&lt;/strong&gt; is that all three $\ce{C-N}$-bonds have &lt;em&gt;de facto&lt;/em&gt; the same lengths: $1.482~\mathrm{\overset{\circ}A}$ &quot;long&quot; and $1.480~\mathrm{\overset{\circ}A}$ &quot;short&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two bromine-neighboring nitro-groups do indeed turn out of plane (dihedral angle is about 61...62°) as it was suggested by &lt;a href=&quot;https://chemistry.stackexchange.com/a/33339/16622&quot;&gt;ron&lt;/a&gt;. This seems enough to compensate the proximity to the bromine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$ &lt;img src=&quot;https://i.stack.imgur.com/rWSef.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-06-26T22:04:18.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33373" PostTypeId="1" AcceptedAnswerId="33461" CreationDate="2015-06-26T23:20:26.710" Score="3" ViewCount="156" Body="&lt;p&gt;I'm growing graphene on copper substrates and need to know the mass. The graphene is grown via chemical vapor deposition; amounts of reactants are not completely known. The best approach I've thought of is recording the mass of the substrate before and after using an analytical balance, but with the nature of the material being nano, this method may be elusive. What other method could I use to find the mass?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-06-26T23:25:51.170" LastActivityDate="2015-06-29T19:10:34.720" Title="How to calculate mass of nanomaterial grown on a substrate" Tags="&lt;analytical-chemistry&gt;&lt;materials&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33374" PostTypeId="1" CreationDate="2015-06-27T01:28:16.190" Score="0" ViewCount="703" Body="&lt;p&gt;Copper(II) oxide is black, but when we find the percentage of oxygen in air by volume using ammonia, copper, and ammonium chloride, the color of $\ce{CuO}$ turns green. What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="16080" LastEditorUserId="7475" LastEditorDisplayName="user15489" LastEditDate="2016-05-08T14:45:40.277" LastActivityDate="2016-05-08T14:45:40.277" Title="Why is copper(II) oxide green?" Tags="&lt;inorganic-chemistry&gt;&lt;color&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33376" PostTypeId="2" ParentId="33285" CreationDate="2015-06-27T04:05:41.880" Score="2" Body="&lt;p&gt;I assume the ppm value you are reporting is ppm by volume (or equivalently for ideal gases, ppm by mole).  If that is true, then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The total moles of all gas in the container can be calculated from the ideal gas law:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ N_{tot} = \frac{PV}{RT}=\frac{(101325 \rm{~Pa})(\frac{80}{1000^2} \rm{~m^3})}{{(8.314 \rm{~\frac{J}{mol\cdot K}})}{(298 \rm{~K}})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have assumed that the temperature was 298 K, or about 25 &amp;deg;C.  Note that the units work out because pascals ($\rm{Pa}$) are equivalent to joules per cubic meter, $\rm{\frac{J}{m^3}}$.  Also, I assume your experiment was done at approximately atmospheric pressure, which is 101325 Pa.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The moles of $\ce{NO2}$ in this 80 cc of gas can be calculated from the ppm measurement:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$N_{\ce{NO2}}=\frac{\rm{ppm~value}}{1,000,000}\times N_{tot}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The mass of the nitrogen, in grams, in the $\ce{NO2}$ gas can be calculated from the formula weight of a nitrogen atom, $\rm{FW}_{\ce{N}}$:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$m_{\ce{NO2}}={\rm{FW}_{\ce{N}}}\times N_{\ce{NO2}}=14\frac{\rm{g}}{\rm mol} \times N_{\ce{NO2}}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The mass of soil used was 5 grams, so the mass ratio of $\ce{NO2}$ to soil, $x_{\ce{NO2}}$  can be calculated by simply dividing the masses:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$x_{\ce{NO2}} = \frac{m_{\ce{NO2}}}{5 \rm{~grams}}=\frac{m_{\ce{NO2}}\times \frac{1,000,000 \rm{~\mu g}}{\rm{g}}}{5 \rm{~grams}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part of the equation above means that if you want the mass ratio in units of &lt;em&gt;micrograms&lt;/em&gt; (of $\ce{NO2}$) per gram of soil, you have to multiply by 1,000,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the question says you want a &quot;per gram dry soil basis&quot;, but you didn't mention if the 5 g of soil you used is &quot;dry&quot;.  If it isn't, then you will first have to figure out how many grams of dry soil you added to your reaction, if it wasn't 5.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:05:13.710" LastActivityDate="2016-03-11T23:05:13.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33377" PostTypeId="2" ParentId="33330" CreationDate="2015-06-27T04:48:14.817" Score="1" Body="&lt;p&gt;The leaf of aloe vera is composed of three layers: 1) An inner clear gel that contains 99% water and rest is made of glucomannans, amino acids, lipids, sterols and vitamins. 2) The middle layer of latex which is the bitter yellow sap and contains anthraquinones and glycosides. 3) The outer thick layer of 15–20 cells called as rind which has protective function and synthesizes carbohydrates and proteins. So overally , it consists of about 99% water and remaining are vitamins, enzymes, minerals, sugars, lignin, saponins,etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what makes inner layer gelly ?? Well, the innermost part of the leaf is a clear, soft, moist and slippery tissue that consists of large thin-walled parenchyma cells in which water is held in the form of a viscous mucilage consiting of cellulose, hemicellulos and acetylated mannans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for you question, as most of it consists of water(99%) the cool feeling might has been felt due to its evaporation but some of its remaining constituents are absorbed in the skin. Mucopolysaccharides present in it help in binding moisture into the skin.Its one of the important medical applications is that it could potentially be used as a&#xA;penetration enhancement agent for the transdermal delivery of drugs and intestinal absorption of coadministerd drug.&lt;a href=&quot;http://www.mdpi.net/molecules/papers/13081599.pdf&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saponins that are the soapy substances are found in the gel of aloe vera and have cleansing and antiseptic properties &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2763764/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; so you can say that hand sanitizer and aloe vera are somewhat connected.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2015-06-27T04:48:14.817" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33378" PostTypeId="2" ParentId="7648" CreationDate="2015-06-27T05:39:14.590" Score="4" Body="&lt;p&gt;This is one of the common problems of ionic equilibrium. Generally questions are asked to calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of a diluted weak acidic solution and if the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; calculation turns out to be &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; or more than that, we have to add the contribution of water also. I am giving you an example that will resolve your matter completely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculate the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0E-8 M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reaction&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{HCl + H2O &amp;lt;=&amp;gt; H3O+ + Cl-}\tag1\label{hcl-dissociation},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you calculate the pH as &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{pH} = -\log(10^{-8}),\tag2$$&lt;/span&gt; &#xA;then &lt;span class=&quot;math-container&quot;&gt;$\text{pH}=8$&lt;/span&gt;.&#xA;Now you can see that the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; can not be &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;, it has to be less than &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt;, therefore &lt;strong&gt;you have to include the contribution of water&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{2H2O &amp;amp;&amp;lt;=&amp;gt; H3O+ + OH-}\tag3\label{selfionisation}\\&#xA;K_\mathrm{w} &amp;amp;=\ce{[OH-][H3O+]}\\&#xA;K_\mathrm{w} &amp;amp;=10^{-14}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; concentration increases from:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ionization of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; dissolved \eqref{hcl-dissociation},&lt;/li&gt;&#xA;&lt;li&gt;Self ionization of water \eqref{selfionisation}.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in very dilute solutions like this solution we have to consider the contribution of both sources of &lt;span class=&quot;math-container&quot;&gt;$\ce{H3O+}$&lt;/span&gt;; this also explains why the concentration of hydronium ions increases upon dilution of a weak acidic solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now going back to problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\text{let } x = [\ce{OH-}] &amp;amp;= [\ce{H3O+}]\\&#xA;[\ce{H3O+}] &amp;amp;= 10^{-8} + x\\&#xA;K_\mathrm{w} &amp;amp;= (10^{-8} + x)(x) = 10^{-14}\\&#xA;0 &amp;amp;= x^2 +10^{-8}x -10^{-14}\\&#xA;[\ce{OH-}] &amp;amp;= x = \pu{9.5E-8}\\&#xA;\mathrm{pOH}&amp;amp;=7.02\\&#xA;\mathrm{pH}&amp;amp;=6.98&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can see that we get &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH} = 6.98$&lt;/span&gt; after adding the contribution of water, otherwise we would have gotten the absurd answer of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}=8$&lt;/span&gt;, which can not be true.&lt;/p&gt;&#xA;" OwnerUserId="17052" LastEditorUserId="61578" LastEditDate="2018-12-14T23:13:34.360" LastActivityDate="2018-12-14T23:13:34.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33379" PostTypeId="1" CreationDate="2015-06-22T20:03:11.397" Score="28" ViewCount="12597" Body="&lt;p&gt;I understand that &lt;a href=&quot;https://en.wikipedia.org/wiki/Dipole#Molecular_dipoles&quot; rel=&quot;nofollow noreferrer&quot;&gt;molecular dipoles&lt;/a&gt; are electric dipoles. And &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_dipole_moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;electric dipole moment vectors&lt;/a&gt; point from the negative to the positive charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In class we learned to draw these special molecular dipole arrows (with a &quot;plus&quot; at the beginning) that point from the positive to the negative partial charge. I also see these arrows all over the internet, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3b0c.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for alpha-helices, the dipole moment points from the C to the N terminus. And the detail always shows the arrow pointing from the oxygen (negative) to the hydrogen (positive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jZ38i.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I just accept that even though an alpha-helix is a molecule like water, the &quot;standard&quot; way to draw its molecular dipole arrow is in the opposite direction than what people normally draw for water and other small molecules?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Another image, from a &lt;a href=&quot;http://www.nature.com/doifinder/10.1038/nchembio.1739&quot; rel=&quot;nofollow noreferrer&quot;&gt;recent publication&lt;/a&gt;, but without macro-dipole arrow...:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/t4gXd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" OwnerDisplayName="Zalumon" LastEditorUserId="2116" LastEditDate="2019-03-20T10:43:19.200" LastActivityDate="2019-03-20T10:43:19.200" Title="Confusion about direction of dipole arrow in alpha-helices and other molecules" Tags="&lt;physical-chemistry&gt;&lt;dipole&gt;&lt;structural-biology&gt;" AnswerCount="2" CommentCount="13" FavoriteCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33380" PostTypeId="1" AcceptedAnswerId="33384" CreationDate="2015-06-27T08:46:32.333" Score="5" ViewCount="1084" Body="&lt;p&gt;My book says that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Monocyclic substituted hydrocarbons are named as derivatives of benzene or compounds like cumene, toluene, xylene, etc. However, if the substituent introduced into such a compound is identical with the substituent already present in the compound, then the substituted compound is named as tge derivatives of benzene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I guess that names like toluene have stayed due to historic reasons and are not accepted in IUPAC nomenclature. Is my book correct?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="7448" LastEditDate="2015-06-27T08:57:04.323" LastActivityDate="2015-06-27T09:43:51.910" Title="Is 3,5-diethyltoluene an accepted IUPAC name?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33381" PostTypeId="2" ParentId="33379" CreationDate="2015-06-27T09:12:37.253" Score="8" Body="&lt;p&gt;Let's take a simple case of a positive charge $q_+$ and a negative charge $q_-$ with $q_+ + q_- = 0$ placed along the $x$-axis at $x_+$ and $x_-$, respectively. The dipole at the $x=0$ is&#xA;$$\mu=q_+x_+ + q_-x_- $$&#xA;which can be rewritten as &#xA;$$\mu=q_+(x_+ -x_-) $$&#xA;which is a vector pointing from $x_-$ towards $x_+$. So this is the &quot;natural&quot; direction based on the definition of a dipole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, chemists a very concerned with the &quot;flow&quot; of electrons and electronegativity. So in the case of water we like to use the bond dipole to depict the &quot;flow&quot; of electrons from the $\ce{H}$ to the $\ce{O}$ due to the higher electronegativity of $\ce{O}$.  So we tend to invert the direction of the dipole vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason it is important to always define the direction of the dipole &quot;arrow&quot; in drawings to there is no confusion. &lt;/p&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-06-27T09:12:37.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33383" PostTypeId="1" AcceptedAnswerId="33385" CreationDate="2015-06-27T09:22:22.727" Score="3" ViewCount="3406" Body="&lt;p&gt;An ideal gas in a box has pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; and temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;. This box is kept in a vacuum, within a large container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the box is punctured, what happens to the temperature of the gas as it expands to fill the entire container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was taught that the temperature will remain constant. How is this possible? I understand that the pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; will decrease and the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; will increase, but for the temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; to remain constant, the decrease in &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; must exactly counterbalance the increase in &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;. Is this the case?&lt;/p&gt;&#xA;" OwnerUserId="17108" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2018-10-28T00:36:29.323" LastActivityDate="2021-05-11T15:45:20.330" Title="Why is temperature constant when an ideal gas expands into a vacuum?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33384" PostTypeId="2" ParentId="33380" CreationDate="2015-06-27T09:43:51.910" Score="12" Body="&lt;p&gt;&lt;strong&gt;Names like toluene are &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Retained_name&quot; rel=&quot;noreferrer&quot;&gt;retained names&lt;/a&gt;&amp;quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In chemistry, a retained name is a name for a chemical compound that is recommended for use by a system of chemical nomenclature (for example, IUPAC nomenclature), but that is not fully systematic. - &lt;a href=&quot;https://en.wikipedia.org/wiki/Retained_name&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;‘Retained names’ are traditional or common well-established names that may be used either as preferred IUPAC name or as an allowed alternatives in general nomenclature, for example, naphthalene, benzoic acid and pyridine. &lt;a href=&quot;http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/CompleteDraft.pdf&quot; rel=&quot;noreferrer&quot;&gt;IUPAC's Blue book,  P-12.3&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Later in the same draft, we read&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Toluene, xylene and mesitylene are specific parent hydrides that are composed of two components, one cyclic and the other acyclic and saturated. These names are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not substitutable; toluene is substitutable under certain conditions for general nomenclature. P-22.1.3&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, anisole and toluene &lt;strong&gt;cannot be substituted by other&#xA;methoxy or methyl groups&lt;/strong&gt;, respectively. P-46.2.3.2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Names like toluene, anisole and xylene are &lt;strong&gt;retained names&lt;/strong&gt;. They &lt;em&gt;are&lt;/em&gt; preserved for historical reasons; however, their unsubstituted forms are PINs (Preferred IUPAC names).&lt;/li&gt;&#xA;&lt;li&gt;The name &lt;em&gt;3,5-diethyl toluene&lt;/em&gt; isn't acceptable according to the systematic IUPAC nomenclature.&lt;/li&gt;&#xA;&lt;li&gt;If you want to name the compound below, it's called 1,3-diethyl-5-methylbenzene.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/HSyU5.gif&quot; alt=&quot;&quot; /&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pherobase.com/pherobase/gif/1,3-diethyl-5me-benzene.GIF&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-06-27T09:43:51.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33385" PostTypeId="2" ParentId="33383" CreationDate="2015-06-27T09:47:49.423" Score="8" Body="&lt;p&gt;This process is called an expansion into a vacuum. As you described, the volume of the gas increases and the pressure drops, such that &lt;span class=&quot;math-container&quot;&gt;$pV$&lt;/span&gt; does not change at all. Since &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is constant, &lt;span class=&quot;math-container&quot;&gt;$T = \frac{pV}{nR}$&lt;/span&gt; is also constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the fact that &lt;span class=&quot;math-container&quot;&gt;$pV$&lt;/span&gt; is constant is not very intuitive if one only uses the ideal gas law. A more formal way of explaining that &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is constant would use thermodynamics. If you assume that the large container does not allow any transfer of heat (i.e. it is thermally insulating), then the large container is an isolated system: &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}q = 0$&lt;/span&gt;. Since the gas is expanding into a vacuum, &lt;span class=&quot;math-container&quot;&gt;$p_{\text{ext}} = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}w = -p_{\mathrm{ext}}\,\mathrm{d}V = 0$&lt;/span&gt;. By the First Law, &lt;span class=&quot;math-container&quot;&gt;$\mathrm{d}U = \mathrm{d}q + \mathrm{d}w = 0$&lt;/span&gt; and temperature is therefore constant (since the temperature of an ideal gas depends fully on &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2018-10-28T00:37:50.300" LastActivityDate="2018-10-28T00:37:50.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33386" PostTypeId="2" ParentId="14846" CreationDate="2015-06-27T11:38:36.760" Score="1" Body="&lt;p&gt;We may consider the pi electrons of the cyclopropenyl cation as a 3c-2e bond.  An even better example of a 3c-2e bond producing aromaticity is the trihydrogen cation, in which three protons are bound into a triangle with two electrons.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Trihydrogen_cation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17175" LastEditorUserId="10185" LastEditDate="2015-06-27T19:56:46.643" LastActivityDate="2015-06-27T19:56:46.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33387" PostTypeId="1" CreationDate="2015-06-27T12:27:27.133" Score="3" ViewCount="12109" Body="&lt;p&gt;What calculations are needed to express the grams of an element from a compound dissolved in aqueous solution? For example, I have a $0.5\,\mathrm{M}$ glucose aqueous solution and I want to know how many grams of $\ce{O}$ are in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to calculate in this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Molecular weight of glucose = $180.159 \mathrm{\frac{g}{mol}}$&lt;/li&gt;&#xA;&lt;li&gt;Atomic weight $\ce{C} = 72$; $\ce{H} = 12.09$; $\ce{O} = 96$;&lt;/li&gt;&#xA;&lt;li&gt;I am considering $0.5\,\mathrm{M}$, so in $1\,\mathrm{L}$ I have $90.08\,\mathrm{g}$ of glucose&lt;/li&gt;&#xA;&lt;li&gt;At this point I divide the molecular weight of glucose by the atomic weight of oxygen $\frac{90.08}{96}$, but I don't think it is correct. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17129" LastEditorUserId="5026" LastEditDate="2018-02-24T03:01:20.440" LastActivityDate="2018-02-24T03:06:02.767" Title="How to calculate the mass of an element in a sample of a compound?" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33388" PostTypeId="1" AcceptedAnswerId="33389" CreationDate="2015-06-27T14:07:58.803" Score="5" ViewCount="3153" Body="&lt;p&gt;Avogadro stated that for any ideal gas, 1 mole of particles will occupy a specific volume at a specific ratio of pressure and temperature. So why doesn't this apply to a liquid or solid?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorUserId="7448" LastEditDate="2015-06-27T15:09:39.947" LastActivityDate="2015-06-28T07:31:03.600" Title="Why not an Avogadro's principle for solid matter?" Tags="&lt;gas-laws&gt;&lt;intermolecular-forces&gt;&lt;mole&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33389" PostTypeId="2" ParentId="33388" CreationDate="2015-06-27T14:24:14.680" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro&amp;#39;s_law&quot; rel=&quot;nofollow&quot;&gt;Avogadro's law&lt;/a&gt;, which can be written as $V \propto n$, where $V$ is the volume of the gas and $n$ is the amount of substance of the gas (measured in moles), can be thought of as just another manifestation of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt; rewritten as follows,&#xA;$$&#xA;  V = (RT/p) n \, .&#xA;$$&#xA;Consequently, strictly speaking, Avogadro's law is applicable only for a hypothetical &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas&quot; rel=&quot;nofollow&quot;&gt;ideal gas&lt;/a&gt;, but since real gases under usual conditions (such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot; rel=&quot;nofollow&quot;&gt;standard temperature and pressure&lt;/a&gt;) are not far away in their properties from an ideal gas, Avogadro's law usually works well enough for them also. Substances in condensed phase on the other hand are milage away from an ideal gas in their properties, and thus, Avogadro's law is inapplicable for such cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify in which sense Avogadro's law is inapplicable for substances in condensed phases, note that for solids and liquids, as well as for gases, the volume is proportional to the amount of substance. But Avogadro's law states not just the proportionality. It is also insists that the coefficient of the proportionality is the &lt;em&gt;same&lt;/em&gt; for each and every gas, and this is the thing which can not be generalised to substances in condensed phases. For real gases, the ideal gas description which depends on the following assumptions (too name just a few):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;molecules are almost point-like (effectively zero-sized) particles;&lt;/li&gt;&#xA;&lt;li&gt;molecules do not exhibit any intermolecular forcers;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;usually works as a good approximation. But for liquids and solids such a description a priori is physically horrible: molecules here are relatively bigger and more tightly packed so that their size and intermolecular interaction between them couldn't be ignored anymore.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2015-06-28T07:31:03.600" LastActivityDate="2015-06-28T07:31:03.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33391" PostTypeId="1" AcceptedAnswerId="33399" CreationDate="2015-06-27T17:33:57.260" Score="5" ViewCount="104161" Body="&lt;p&gt;I've always been puzzled by this because my teachers happen to use only the KLMN method, but what is the difference between the KLMN and SPDF methods of finding electronic configuration?&lt;/p&gt;&#xA;" OwnerUserId="17174" LastActivityDate="2017-07-15T18:25:11.073" Title="What is the difference between the KLMN and SPDF methods of finding electronic configuration?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33392" PostTypeId="1" AcceptedAnswerId="33396" CreationDate="2015-06-27T18:17:09.313" Score="5" ViewCount="532" Body="&lt;p&gt;I have a question regarding mol files.For example I have this molecule-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KZYXHm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mol file for this is-&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Sample&#xA;&#xA; 22 23  0  0  0  0  0  0  0  0999 V2000&#xA;   -2.5962    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -2.5962    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.0000    0.7512    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.0000    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    3.0047    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -4.6479    1.7042    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005    3.0047    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    4.5070    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958   -1.5023    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009   -2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.8967   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.8967    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    5.1972   -2.2535    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009   -3.7559    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -3.8967    3.0047    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -5.1972    3.7559    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -3.1455    4.3052    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.9014   -4.5070    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;  1  2  2  0  0  0  0&#xA;  2  3  1  0  0  0  0&#xA;  3  4  1  0  0  0  0&#xA;  4  5  1  0  0  0  0&#xA;  5  6  1  0  0  0  0&#xA;  1  6  1  0  0  0  0&#xA;  1 19  1  0  0  0  0&#xA;  5  8  2  0  0  0  0&#xA;  6  9  1  0  0  0  0&#xA;  3 10  2  0  0  0  0&#xA;  4 11  1  0  0  0  0&#xA; 11 12  1  0  0  0  0&#xA; 12 13  2  0  0  0  0&#xA; 13 14  1  0  0  0  0&#xA; 14 15  2  0  0  0  0&#xA; 15 16  1  0  0  0  0&#xA; 11 16  2  0  0  0  0&#xA; 14 17  1  0  0  0  0&#xA; 13 18  1  0  0  0  0&#xA; 19  7  1  0  0  0  0&#xA; 19 20  1  0  0  0  0&#xA; 19 21  1  0  0  0  0&#xA; 18 22  2  0  0  0  0&#xA;M  END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So,I have read that lines 5-26 are for atoms.first three columns represent x,y and z coordinates of atoms.So,my question is how are the coordinates of atoms decided?What is taken as the origin?Also,the atoms are large,so they may occupy more than one coordinate,so how do we decide what x,y,z value to take?Also,is H atoms not taken into account for creating mol files?&lt;/p&gt;&#xA;" OwnerUserId="17178" LastEditorUserId="5017" LastEditDate="2015-07-05T18:09:30.483" LastActivityDate="2015-07-05T18:09:30.483" Title="Generating mol files from a molecular structure image?" Tags="&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33394" PostTypeId="1" CreationDate="2015-06-27T18:57:32.310" Score="3" ViewCount="143" Body="&lt;p&gt;In his 1926 thesis Robert Cady Burts &lt;a href=&quot;http://thesis.library.caltech.edu/4369/1/Burt_rc_1926.pdf&quot; rel=&quot;nofollow&quot;&gt;describes the production of sodium by electrolysis of sodium nitrate through glass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find much else on this method. Are there any modern descriptions of the experiment? Videos maybe?&lt;/p&gt;&#xA;" OwnerUserId="17180" LastEditorUserId="7448" LastEditDate="2015-06-27T19:01:41.723" LastActivityDate="2015-08-01T04:13:04.653" Title="Sodium by electrolysis through glass" Tags="&lt;metal&gt;&lt;electrolysis&gt;&lt;reference-request&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33396" PostTypeId="2" ParentId="33392" CreationDate="2015-06-27T20:40:40.483" Score="6" Body="&lt;p&gt;The MOL format specification doesn't specify an origin, so presumably any origin can be chosen, as long as the coordinate values are within the allowed range. (The format allows a range of roughly -10,000 to +100,000, but I don't know if all programs support that large a range.) For example, in one of the examples in the specification the atom closest to the origin is over 17.6 units away. Where the origin is actually set would depend on the software used to create the file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coordinates for each atom specify the nucleus of the atom. I think you have a misconception about atomic size. The size of an atom, and thus the space it takes up, can be defined in different ways. The representation of an atom as a sphere (with the nucleus at the center) is a convention, only approximating reality, and the size of the sphere is arbitrary (but usually based on measurable properties). The nucleus is the only point in the atom that has a well-defined location, and thus the only part that meaningful coordinates can be defined for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MOL files, the presence of hydrogen atoms can be implied from valence and number of other bonds. (They can also be specified explicitly if needed.) In the long set of zeros at the end of each atom line (just after the element), the fourth zero specifies to add however many single bonds to hydrogen atoms are needed for the atom's valence. The sixth zero indicates to use the element's normal valence. For example, the carbon of the $\mathrm{CH_3}$ group has only one explicit bond, so three bonds to hydrogen are implied by carbon's normal valence of 4. How implicit hydrogens are handled depends on the software (e.g., this molecule has four implicit hydrogens that, per the usual convention for skeletal formulas, are not shown in your image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://download.accelrys.com/freeware/ctfile-formats/ctfile-formats.zip&quot;&gt;specification for the MOL format&lt;/a&gt; is available from Accelrys. The this answer is based on the specification and my knowledge of chemistry. Also, it only describes the MOL V2000 format, which is what you sample file is (based on the header); there is a newer V3000 format.&lt;/p&gt;&#xA;" OwnerUserId="13334" LastActivityDate="2015-06-27T20:40:40.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33398" PostTypeId="1" AcceptedAnswerId="33481" CreationDate="2015-06-27T21:09:47.517" Score="6" ViewCount="2270" Body="&lt;p&gt;Can someone explain to me why there is a rearrangement reaction of carbocation 1 to carbocation 2? To me the second carbocation appears to be less stable than the first because it is less substituted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CVrEn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance &lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="4325" LastEditDate="2015-06-28T04:39:45.150" LastActivityDate="2015-06-30T13:16:10.523" Title="Stability of a carbocation, in part of camphor synthesis" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33399" PostTypeId="2" ParentId="33391" CreationDate="2015-06-27T21:21:32.890" Score="11" Body="&lt;p&gt;The KLMN(OP) method is based on electron shells, with the labels KLMN(OP) being derived from an experiment in which the &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/electrons/faq/why-KLMNOP.shtml&quot; rel=&quot;noreferrer&quot;&gt;spectroscopist&lt;/a&gt; wanted to leave room for lower energy transitions in case there were any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K denotes the first shell (or energy level), L the second shell, M, the third shell, and so on. In other words, the KLMN(OP) notation only indicates the number of electrons an atom has with each principal quantum number ($n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SPDF notation subdivides each shell into its subshells. For further information about the other quantum numbers, especially $l$, which defines the subshell, see the accepted answer to this question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31639/what-do-the-quantum-numbers-actually-signify&quot;&gt;What do the quantum numbers actually signify?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The K shell can hold two electrons: $n=1,\ l=0$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $l=0$, we have an s subshell, which has one orbital $m_l=0$, with room for two electrons.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The L shell can hold 8 elections: $n=2,\ l=0,1$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $l=1$, we have a p subshell, which has three orbitals $m_l=-1,0,+1$, with room for 6 electrons. The L shell also has an s subshell. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The M shell can hold 18 electrons $n=3,\ l=0,1,2$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $l=2$, we have a d subshell, which has 5 orbitals $m_l=-2,-1,0,+1,+2$, with room for 10 electrons. The M shell also has s and p subshells.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The N shell can hold 32 electrons! $n=4,\ l=0,1,2,3$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When $l=3$, we have an f subshell, which has 7 orbitals $m_l=-3,-2,-1,0,+1,+2,+3$, with room for 14 electrons. The N shell also has s, p, and d subshells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table here summarizes the relationship between the two notations for all elements. Here is an example for scandium, $\ce{Sc}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scandium has 21 electrons ($Z=21$). Its electron configuration in KLMN notation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{cccc}&#xA;\mathrm{K} &amp;amp; \mathrm{L} &amp;amp; \mathrm{M} &amp;amp; \mathrm{N} \\&#xA;2 &amp;amp; 8 &amp;amp; 9 &amp;amp; 2&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its SPDF notation (based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Aufbau_principle&quot; rel=&quot;noreferrer&quot;&gt;aufbau principle&lt;/a&gt;) is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2016-12-19T13:51:28.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33400" PostTypeId="1" CreationDate="2015-06-27T22:23:28.223" Score="2" ViewCount="2375" Body="&lt;p&gt;I used nail polish remover on an expensive phone case to remove sticker residue (and it worked, no longer sticky). I didn't know it didn't mix well with plastic materials. Did the acetone damage it?&lt;/p&gt;&#xA;" OwnerUserId="17186" LastActivityDate="2015-07-28T04:09:29.503" Title="Does acetone eat through Thermoplastic polyurethane and Polycarbonate" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33401" PostTypeId="2" ParentId="33388" CreationDate="2015-06-28T01:50:36.297" Score="2" Body="&lt;p&gt;gas molecules are small, and even largish gas molecules behave so much like small ones (example:  atoms of helium vs. carbon dioxide vapor) that Avogadro's models hold very well across all the spectrum of matter in gaseous form.  at a given temperature and pressure the individual molecules will have a certain statistically predictable average distance from one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in liquids, though the molecules are much closer together and the inter-molecular interactions--which vary widely from substance to substance--become the predominant phenomena of significance.  this is why one mole of water molecules occupies 18 cubic centimeters at STP, but one mole of ethanol occupies 58.4 cubic centimeters at STP, and one mole of mercury in liquid form occupies 14.8 cubic centimeters at STP.  One mole of the type of large, complex hydrocarbon molecules that make up the mixture we call &quot;gasoline&quot; or &quot;petrol&quot; may occupy around 162 cubic centimeters, if we use the specific example of 2, 3, 5 trimethylpentane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in solids the molecules are closer yet, and we have the additional possibility of different allotropes or crystalline forms of a given solid substance which can exhibit very different physical characteristics, resulting in even further divergences from Avogadro's ideal gases, molecules of which maintain set average distances from one another at a given temperature and pressure regardless of their mass, charge, shape, etc., and which always collide with one another or with the walls of their container perfectly elastically.&lt;/p&gt;&#xA;" OwnerUserId="17187" LastActivityDate="2015-06-28T01:50:36.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33402" PostTypeId="2" ParentId="33400" CreationDate="2015-06-28T03:15:36.960" Score="2" Body="&lt;p&gt;If there is no visual damage, e.g. cloudiness or crazing, the case should be OK. If there's some surface cloudiness, it might be possible to buff it out with mild abrasive, such as toothpaste or &lt;a href=&quot;http://www.mandolincafe.com/forum/showthread.php?115078-How-To-Get-Scratches-off-of-Mandolin&quot; rel=&quot;nofollow&quot;&gt;milk of magnesia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future, try oil (olive, safflower, mineral etc., but not motor oil) to remove stickers. A bit slower, requiring subsequent cleanup, but safer. &lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-06-28T03:15:36.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33406" PostTypeId="1" CreationDate="2015-06-28T06:31:40.583" Score="0" ViewCount="3195" Body="&lt;p&gt;What is the difference between the terms 'solve' and 'dissolve'? Up until now I have used the two terms interchangeably, not having any sound scientific definitions for each. I have thought them both to mean to make a solid into a solution through the addition of a solvent (not sure if only for solids into solutions; is the mixing of two fluids also dissolving/solving because $\ce{CO2 (g)}$ can dissolve into water?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not been able to find any scientific definitions on the internet that differentiate the two. it would also be helpful if you could clearly define any other terms related to the mixing of substances without a chemical reaction occurring (eg. immiscible, which I understand to be the state of two &lt;em&gt;FLUIDS&lt;/em&gt; being unable to homogeneously mix)      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if both substances are liquids, which one is the solvent and which one the solute?&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="39539" LastEditDate="2017-01-30T17:37:07.920" LastActivityDate="2018-12-13T22:07:34.810" Title="difference between solving and dissolving something" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;solvents&gt;" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33408" PostTypeId="1" AcceptedAnswerId="33409" CreationDate="2015-06-28T06:45:45.580" Score="2" ViewCount="540" Body="&lt;p&gt;I was doing a homework assignment and I got the question wrong because I was supposed to multiply the empirical formula of &lt;span class=&quot;math-container&quot;&gt;$\ce{C6H10S2O}$&lt;/span&gt; by 3 to get the molecular formula.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put &lt;span class=&quot;math-container&quot;&gt;$\ce{(C6H10S2O3)3}$&lt;/span&gt; and the teacher said it should be &lt;span class=&quot;math-container&quot;&gt;$\ce{C18H30S6O3}$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: isn't what I put the same thing? &lt;/p&gt;&#xA;" OwnerUserId="17192" LastEditorUserId="865" LastEditorDisplayName="user15489" LastEditDate="2020-03-26T18:03:22.657" LastActivityDate="2020-03-26T18:03:22.657" Title="Empirical and molecular formula" Tags="&lt;inorganic-chemistry&gt;&lt;molecules&gt;&lt;chemical-formula&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33409" PostTypeId="2" ParentId="33408" CreationDate="2015-06-28T06:58:14.027" Score="4" Body="&lt;p&gt;$\ce{(C6H10S2O3)3}$ implies that there are $3 ~ \ce{C6H10S2O3}$ molecules bonded together instead of $1~ \ce{C18H30S6O3}$ molecule. Although they are the same thing, the correct notation is your teachers way because it shows the isometric formula for the molecule. Also if you were using that notation you should have put $\ce{(C6H10S2O)3}$ because $\ce{(C6H10S2O3)3}$ has 6 too many oxygens. &lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="7448" LastEditDate="2015-06-28T07:15:23.317" LastActivityDate="2015-06-28T07:15:23.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33410" PostTypeId="2" ParentId="24849" CreationDate="2015-06-28T07:39:11.277" Score="-1" Body="&lt;p&gt;For $d\pi-p\pi$ bonding we need to consider other elements of group 14 except $\ce{C}$. (i.e, $\ce{Si, Ge ,Sn}$ etc.) which have got a vacant $d$ orbital of proper symmetry and energy to come into play bonding with attached group  having lone pair of electron in p orbital of comparable energy.&lt;br&gt;&#xA;    For example $\ce{SiF4}$. $\ce{GeCl4}$ etc.&#xA;In tetrahedral geometry the orientation of vacant  d orbital of the central atom (dxy. dyz,dxz) may overlap with filled p orbital of the ligating atom  and hence $d\pi -p\pi$ bonding is possible. &lt;/p&gt;&#xA;" OwnerUserId="17193" LastEditorUserId="17193" LastEditDate="2015-06-28T14:53:32.433" LastActivityDate="2015-06-28T14:53:32.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33411" PostTypeId="1" AcceptedAnswerId="33412" CreationDate="2015-06-28T07:56:14.167" Score="12" ViewCount="3268" Body="&lt;p&gt;Does food coloring somehow evaporate and turn into nothing?   Or does the beverage itself being much more color intensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a bottle of Gatorade, colored Lemon-Lime (and flavored as such). I noticed that after leaving the bottle in the car during a hot day and returned to the car, that condensation had formed on the top parts of the bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the condensation is colorless; does this mean it is solely water that is evaporating from the drink itself as condensation occurs, leaving food coloring or the color of the beverage behind?&lt;/p&gt;&#xA;" OwnerUserId="16687" LastEditorDisplayName="user15489" LastEditDate="2015-06-28T09:24:28.003" LastActivityDate="2015-06-29T02:23:53.960" Title="Why is condensation in dyed beverages colorless?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;food-chemistry&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33412" PostTypeId="2" ParentId="33411" CreationDate="2015-06-28T09:10:53.850" Score="10" Body="&lt;p&gt;Given that the majority of beverages such as Gatorade is water, as stated in the comments, the condensate is predominantly water, as can be seen in the Patent &lt;a href=&quot;http://www.google.com/patents/EP2575507A2?cl=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrolyte blends providing reduced salty taste&lt;/a&gt; (which includes Gatorade), that water amounts to well over 99% by composition, of course this will vary. The dyes themselves don't make up much of the volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common food dye used, according to the Indiana University online experiment webpage &lt;a href=&quot;http://www.indiana.edu/~l113/labs/red_dye/01_1-4_Red_40_F99.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Much Red Dye is in My Gatorade®? &lt;/a&gt;, is the red dye, also known as '&lt;a href=&quot;http://www.chemexper.com/chemicals/supplier/cas/25956-17-6+2-Naphthalenesulfonic+acid+6-hydroxy-5-2-methoxy-5-methyl-4-sulfophenylazo-+disodium+salt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Allura Red&lt;/a&gt;'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;its properties according to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Allura_Red_AC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; are stated as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It usually comes as a sodium salt, but can also be used as both calcium and potassium salts. It is soluble in water; Its melting point is above 300 °C (572 °F).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is likely that the majority of the food dye based salts would remain in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any salts (such as the example above), transported with the evaporating/condensing water would be too small of a concentration to cause any visible effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question and answer is &lt;a href=&quot;https://chemistry.stackexchange.com/q/30869/15489&quot;&gt;Does salt vaporise?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-06-28T09:29:43.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33413" PostTypeId="1" CreationDate="2015-06-28T09:42:26.290" Score="7" ViewCount="1415" Body="&lt;p&gt;Is it possible to calculate the energy of electron in any orbital and atom in the Schrödinger wave model theory? If so, how? E.g. energy of the $3s^2$ electron of the $\ce{Na-}$ ion.&lt;/p&gt;&#xA;" OwnerUserId="17196" LastEditorUserId="4945" LastEditDate="2015-06-29T08:39:42.943" LastActivityDate="2015-07-02T00:28:46.380" Title="Is it possible to calculate energy of electron in any orbital and atom in Schrödinger wave model theory?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33415" PostTypeId="1" CreationDate="2015-06-28T09:51:21.240" Score="2" ViewCount="493" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6wbKt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6wbKt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this compound, I think the ethyl group should be in the parent chain and the numbering should start from the right-most carbon. Hence, the compound should be named 2-methylhept-1-ene-1,3,4,5-tetracarbonitrile, instead of the name given above. Am I wrong somewhere?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="16683" LastEditDate="2016-01-14T20:11:49.067" LastActivityDate="2016-01-14T20:11:49.067" Title="Nomenclature of a poly-nitrile" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33416" PostTypeId="2" ParentId="33413" CreationDate="2015-06-28T11:03:13.560" Score="0" Body="&lt;p&gt;No. In QM electrons are indistinguishable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, it is possible to associate a somewhat characteristic value, using various spectroscopic methods, i.e virtually 'moving' some electron from one orbital to another (or removing it entirely) and calculating the corresponding energy. This energy, however, is unprecise, as other electrons 'feel' the move of the 'moved' one. Still, something is better than nothing, so people use what they can.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-06-28T11:03:13.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33417" PostTypeId="2" ParentId="33413" CreationDate="2015-06-28T12:09:06.820" Score="12" Body="&lt;p&gt;It is surely &lt;em&gt;possible&lt;/em&gt;. Note though, that for many-electron systems orbital picture is an &lt;em&gt;approximate&lt;/em&gt; description, but that is an approximation on which the whole general chemistry is based on, so that is fine from that perspective. The approximation is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;noreferrer&quot;&gt;Hartree-Fock&lt;/a&gt; (HF) model, and thus, to get the energies of electrons occupying some orbitals all you need to do is to solve the Schrödinger equation in this approximation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a quick exercise one can do HF calculation in, say, Gaussian, with the following input file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#P HF/aug-cc-pVTZ Pop=Full&#xA;&#xA;Na-&#xA;&#xA;-1 1&#xA; Na                 0.00000000    0.00000000    0.00000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to get this picture of occupied orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/edzNn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HOMO with the energy of $-0.01288 \, \mathrm{Hartree} = -0.3505 \, \mathrm{eV}$ is basically the $\mathrm{3s}$ orbital you're looking for (trust me). Note, however, that this energy is approximate since the basis set (aug-cc-pVTZ) is finite. We could do better than that, but that is a different story.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to permeakra (since he does not trust his colleagues) I quote an authoritative reference in the field which explicitly talks about individual electrons occupying individual spin-orbitals and orbital energies. As a reference I choose the infamous book entitled &quot;Modern Quantum Chemistry&quot; written by Attila Szabo and Neil S. Ostlund.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quote #1 (Szabo &amp;amp; Ostlund, p. 50)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This Slater determinant has $N$ electrons occupying $N$ spin orbitals&#xA;  $(\chi_i, \chi_j, \dotsc, \chi_k)$ without specifying which electron&#xA;  is in which orbital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quote #2 (Szabo &amp;amp; Ostlund, p. 54)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$   f(i) \chi(x_i) = \varepsilon \chi(x_i) \tag{2.52} $$ [...]  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The&#xA;  solution of the Hartree-Fock eigenvalue problem (2.52) yields a set &#xA;  $\{\chi_k\}$ of orthonormal Hartree-Fock spin orbitals with orbital&#xA;  energies $\{\varepsilon_k\}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below I also quote what Szabo &amp;amp; Ostlund have to say about Koopmans' theorem, because I already know that won't trust me.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Quote #3 (Szabo &amp;amp; Ostlund, p. 110)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The first theorem [Koopmans' theorem] constitutes an interpretation of&#xA;  the Hartree-Fock orbital energies as ionization potentials and &#xA;  electron affinities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See, orbitals with their energies &amp;amp; occupancies do exist in the Hartree-Fock theory. It is the interpretation of orbital energies as ionization potentials which requires an additional theorem.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-06-29T09:02:38.027" LastActivityDate="2015-06-29T09:02:38.027" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33418" PostTypeId="1" CreationDate="2015-06-28T12:12:44.997" Score="1" ViewCount="157" Body="&lt;p&gt;Is $\ce{NaCl}$ in water aqueous? For a certain amount of $\ce{NaCl}$, it can dissolve in water but if you add more $\ce{NaCl}$ it can't dissolve any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if in a certain reaction $\ce{NaCl}$ is formed with water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl(aq) + NaOH(aq) -&amp;gt; H2O(l) + NaCl(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is does $\ce{NaCl}$ in the previous reaction totally dissolve in the water formed? Or in other words is this equation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl(aq) + NaOH(aq) -&amp;gt; H2O(l) + NaCl(aq) + NaCl(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is why do we write (l) beside water rather than (aq) since it's part of the aqueous solution?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastEditorUserId="7448" LastEditDate="2015-06-28T16:08:13.083" LastActivityDate="2015-06-28T16:11:08.233" Title="Is NaCl in water considered &quot;solution&quot; despite of NaCl's having some precipitation?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33419" PostTypeId="1" CreationDate="2015-06-28T12:19:01.570" Score="1" ViewCount="7623" Body="&lt;p&gt;What's the value of standard potential of carbon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen and Carbon are in the electrochemical series despite being non metals hydrogen has zero potential difference what about carbon?&lt;/p&gt;&#xA;" OwnerUserId="4834" LastActivityDate="2015-06-28T12:57:01.067" Title="What is the standard potential of carbon?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electronegativity&gt;&lt;reactivity&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33420" PostTypeId="2" ParentId="33419" CreationDate="2015-06-28T12:57:01.067" Score="2" Body="&lt;p&gt;To look up the position of carbon in the electrochemical series, you would have to guess what species carbon is oxidised to. This is most likely carbon dioxide; you would then have to look up the value of $E^0$ for the $\ce{CO2}/\ce{C}$ couple in a textbook or some other data source. The standard reduction potential for the $\ce{CO2}/\ce{C}$ couple is given in Shriver &amp;amp; Atkins 5th ed as +0.207 V. Loosely speaking, this means that carbon is harder to oxidise than hydrogen, or in other words, carbon is below hydrogen in the electrochemical series (if you take Li to be at the top).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a note: The standard reduction potential for the $\ce{H+} / \ce{H2}$ half-cell is 0 V solely because we decided to &lt;em&gt;define&lt;/em&gt; it as such.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-06-28T12:57:01.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33421" PostTypeId="1" AcceptedAnswerId="33429" CreationDate="2015-06-28T13:14:18.053" Score="8" ViewCount="448" Body="&lt;p&gt;The epoxide on the top of the picture, in aqueous acid hydrolyzes to the compound below. What is the mechanism of this reaction? Standard epoxide opening in acidic conditions would happen at the more substituted end, therefore the secondary -OH group would end up below the plane of the paper, and the tertiary center would racemize to a significant extent. What is happening here?&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nVgWr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7448" LastEditDate="2015-06-28T15:09:42.387" LastActivityDate="2015-06-28T16:55:08.570" Title="Stereochemistry of epoxide hydrolysis under acidic conditions" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33423" PostTypeId="1" AcceptedAnswerId="33449" CreationDate="2015-06-28T13:32:47.197" Score="4" ViewCount="120" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k2bEY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k2bEY.png&quot; alt=&quot;reaction equation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the first step is the Cannizzaro reaction which will give (c) as a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since we are also adding &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; in the second step, shouldn't an esterification reaction take place? That would give (d) as a final product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the answer given is (c).&#xA;My only guess as to why esterification wouldn't take place is the fact that the phenyl groups are in perpendicular planes, therefore esterification is difficult?&#xA;Is it necessary to have all the reactants in a plane for esterification?&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="16967" LastEditorUserId="4945" LastEditDate="2020-02-13T14:58:35.597" LastActivityDate="2020-02-13T14:58:35.597" Title="Does esterification require reactants to be present in the same plane?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33424" PostTypeId="1" AcceptedAnswerId="33437" CreationDate="2015-06-28T14:58:24.107" Score="1" ViewCount="533" Body="&lt;p&gt;While naming substituted amines and amides, should we consider the &quot;N&quot; while deciding the order of substituents or simply place the substituents attached to nitrogen in the beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g. Should the compound be named: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3-chloro-N-ethyl-4-methyl-N-propylhexanamide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;N-chloro-N-propyl-3-chloro-4-methylhexanamide.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="865" LastEditDate="2015-10-16T17:54:40.407" LastActivityDate="2015-10-16T17:54:40.407" Title="Naming of substituted amines and amides" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33426" PostTypeId="2" ParentId="33418" CreationDate="2015-06-28T15:04:33.297" Score="1" Body="&lt;p&gt;When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules  in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction.&#xA;Hence the first equation&lt;br&gt;&#xA;$$\ce{HCl(aq) + NaOH(aq) -&amp;gt; NaCl(aq) + H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;will be the correct expression.&lt;/p&gt;&#xA;" OwnerUserId="17193" LastEditorUserId="10185" LastEditDate="2015-06-28T16:11:08.233" LastActivityDate="2015-06-28T16:11:08.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33428" PostTypeId="1" CreationDate="2015-06-28T16:01:03.160" Score="2" ViewCount="96" Body="&lt;p&gt;In $\ce{CH4}$, $1 s$ and $3 p$ orbitals are hybridised into four $sp^3$ orbitals. But on homolytic bond cleavage an unpaired electron remains in an unhybridised $p$ orbital?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly in heterolytic bond cleavage of $\ce{CH4}$, the methyl carbocation has an unhybridised empty $p$ orbital?&lt;/p&gt;&#xA;" OwnerUserId="17198" LastEditorUserId="10185" LastEditDate="2015-06-28T16:06:38.093" LastActivityDate="2015-06-28T16:06:38.093" Title="How does an hybridised orbital become unhybridised after bond cleavage?" Tags="&lt;hybridization&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33429" PostTypeId="2" ParentId="33421" CreationDate="2015-06-28T16:55:08.570" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Standard epoxide opening in acidic conditions would happen at the more&#xA;  substituted end&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This premise is false. Usually $\ce{S_{N}2}$ is faster than the cation-stabilized pathway (most textbooks don't even bother to explain the subtleties involved), yielding mainly the less substituted (less sterically hindered) product. It is not clear cut which of the two mechanisms will be predominant in an acid catalyzed epoxide cleavage given the reactants and reaction conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4305/regioselectivity-of-acid-catalyzed-ring-opening-of-epoxides?rq=1&quot;&gt;Regioselectivity of acid-catalyzed ring-opening of epoxides&lt;/a&gt; (especially the second answer and then the excerpt from Clayden et al.).&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-06-28T16:55:08.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33430" PostTypeId="1" CreationDate="2015-06-28T17:06:44.867" Score="7" ViewCount="2806" Body="&lt;p&gt;Is the mechanism similar to the reaction of $\ce{SOCl2}$ with alcohols?&lt;br&gt;&#xA;If not, then how is it different? We have been taught about the latter at school but my friend quizzed me about $\ce{SO2Cl2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give the detailed reaction mechanism.&lt;/p&gt;&#xA;" OwnerUserId="17203" LastEditorUserId="7448" LastEditDate="2015-06-29T19:50:40.407" LastActivityDate="2015-06-30T06:47:13.203" Title="How does SO2Cl2 react with alcohols?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33431" PostTypeId="2" ParentId="24299" CreationDate="2015-06-28T18:28:23.263" Score="7" Body="&lt;p&gt;The color of nanoparticles in solution depends on size, so one can create the whole rainbow of colors using a single elemental composition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See e.g. &lt;a href=&quot;http://nanocomposix.com/pages/plasmonics&quot; rel=&quot;noreferrer&quot;&gt;http://nanocomposix.com/pages/plasmonics&lt;/a&gt; for details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Colloidal_gold&quot; rel=&quot;noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Colloidal_gold&lt;/a&gt; shows gold being red and purple. &lt;/p&gt;&#xA;" OwnerUserId="2530" LastActivityDate="2015-06-28T18:28:23.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33432" PostTypeId="2" ParentId="33430" CreationDate="2015-06-28T18:40:17.700" Score="1" Body="&lt;p&gt;SO&lt;sub&gt;2&lt;/sub&gt;Cl&lt;sub&gt;2&lt;/sub&gt; is source of Cl&lt;sub&gt;2&lt;/sub&gt; gas rather than Cl&lt;sup&gt;-&lt;/sup&gt;. So it doesn't react the same with alcohols. It will form HCl when in contact with water - so a lot of its chemistry with organic molecules will originate from that process. Used with a radical initiator it will convert certain functional groups to alkyl chlorides.&lt;/p&gt;&#xA;" OwnerUserId="17204" LastActivityDate="2015-06-28T18:40:17.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33433" PostTypeId="2" ParentId="33271" CreationDate="2015-06-28T19:02:06.863" Score="2" Body="&lt;p&gt;There are a few descriptors in 'conceptual'  &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_functional_theory&quot; rel=&quot;nofollow&quot;&gt;DFT&lt;/a&gt; just to characterize hydrogen bonds. E.g.: non-covalent interaction (NCI): note that hydrogen bonds are often consider as NON-covalent bonds, Density Overlap Regions Indicator (DORI). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can also look at the  deformation density contributions originating from natural orbitals for chemical valence within ETS-NOCV theory. &lt;/p&gt;&#xA;" OwnerUserId="16993" LastEditorUserId="9961" LastEditDate="2015-06-28T22:35:11.530" LastActivityDate="2015-06-28T22:35:11.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33435" PostTypeId="2" ParentId="33411" CreationDate="2015-06-28T19:50:07.597" Score="9" Body="&lt;p&gt;Yes, in a rough sense, components of the mixture evaporate separately (this is a simplification which ignores things like azeotropes) with their individual vapor pressures. This is why distillation works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your gatorade example, the condensate is indeed pure water. In fact, the same principle is used to purify water using something called a solar still. The remaining gatorade at the bottom will have a deeper color, and taste more strongly, in proportion to how much water evaporated - but you won't notice this without accurate assays because of how small the amount of evaporated water is (if you manage to evaporate say 90% of the water the remainder will be 10 times more concentrated, which you can notice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an experiment, try leaving a shot glass of gatorade in a well-ventilated area with low humidity. After days or weeks the level of fluid will be half of what it was or less. Now buy another fresh gatorade of the same kind and compare taste, color (put both in shot glasses) of both.&lt;/p&gt;&#xA;" OwnerUserId="5385" LastEditorUserId="5385" LastEditDate="2015-06-29T00:19:41.187" LastActivityDate="2015-06-29T00:19:41.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33436" PostTypeId="1" CreationDate="2015-06-29T01:05:34.427" Score="3" ViewCount="1680" Body="&lt;p&gt;Why is it important to mark solvent level on chromatography paper when you remove it from the chamber?&lt;/p&gt;&#xA;" OwnerUserId="17205" LastEditorUserId="4325" LastEditDate="2015-06-29T15:14:26.683" LastActivityDate="2015-06-29T15:14:26.683" Title="Why note the solvent level on chromatography paper when it is removed from the chamber?" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33437" PostTypeId="2" ParentId="33424" CreationDate="2015-06-29T01:24:06.233" Score="3" Body="&lt;p&gt;Substituents are listed alphabetically by the name of the substituent. The &lt;em&gt;N&lt;/em&gt; in front of the substituent is a locant, and thus is not used when alphabetizing:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3-chloro-N-ethyl-4-methyl-N-propylhexanamide is correct because the substituents are in correct alphabetic order: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;chloro, ethyl, methyl, propyl&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-29T01:24:06.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33438" PostTypeId="2" ParentId="33411" CreationDate="2015-06-29T02:23:53.960" Score="-1" Body="&lt;p&gt;I thought the answer was that condensation is on the &lt;em&gt;outside&lt;/em&gt;. As such, it's the colorless water from the &lt;em&gt;air&lt;/em&gt;, not the Gatorade, that lacks color.&lt;/p&gt;&#xA;" OwnerUserId="17206" LastActivityDate="2015-06-29T02:23:53.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33439" PostTypeId="2" ParentId="33436" CreationDate="2015-06-29T02:38:08.240" Score="14" Body="&lt;p&gt;Building on my comment, marking the distance that the solvent traveled allows us to calculate the retention factor $R_f$ (or apparently &lt;a href=&quot;https://en.wikipedia.org/wiki/Retardation_factor&quot;&gt;retardation factor&lt;/a&gt;, according to IUPAC meddlers who need to rename things with perfectly good names that everyone else uses). The retardation factor is the ratio of the distance traveled by the spot to the distance traveled by the solvent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_f=\dfrac{d_{spot}}{d_{solvent}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are only performing one thin-layer chromatography experiment (TLC) ever, then you may be okay not marking the solvent distance. Since everything is on one piece of paper or one plate, you can compare readily:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0zBH0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above TLC picture, we can easily tell that unknowns A and B are probably the same compound as the standard, but unknown C is something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we have two TLC plates/papers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLGpk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can draw the same conclusions about the relationships between standard 1 and unknowns A, B, and C. Likewise, we can make some conclusions about the relationship between standard 2 and unknowns D, E, and F. However, we cannot easily determine if unknown F is the same compound as unknown A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining the $R_F$ value lets us do that. Using the same solvent system, the distance traveled by the solvent and the distance traveled by a compound are both proportionally related to the length/height of the plate/paper. The solvent can travel twice as far on a paper that is twice as long. It turns out that the compound being analyzed also travels twice as far on a paper that is twice as long. The the ratio, $R_f$ is &lt;strong&gt;constant&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's drop in some numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MBWFk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can calculate the $R_f$ values for each spot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rWifc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculating the $R_f$ values, we can make some conclusions (assuming there are only two compounds in the entire experiment):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Unknowns A, B, D, and E are the same compound as Standard 1.&lt;/li&gt;&#xA;&lt;li&gt;Unknowns C and F are the same compound as Standard 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-29T02:38:08.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33441" PostTypeId="1" CreationDate="2015-06-29T07:15:27.827" Score="7" ViewCount="188" Body="&lt;p&gt;I am trying to determine a method to dissolve a mineral salt with a lattice energy of ≈18000 kJ/mol. I know that lattice energy is the most significant factor affecting solubility of an ionic substance in highly polar solvents, and my general understanding is that the more polar (or locally polar), the greater the ability of the solvent to dissolve materials with higher lattice energies, by contributing more hydrogen bonding and/or more efficient charge distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I have been testing solubility in an array of Deep Eutectic Solvents (DES) and doing research into solubility of this and similar compounds in other ionic liquids, and reached out to an expert in the field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found that while these solvents are typically able to dissolve compounds with lattice energies unto around 12000 kJ/mol. in great quantity, solubility quickly drops off around 15000 kJ/mol. and the professor I contacted said that their research indicated that without other agents, this material would be insoluble in the ionic liquids they worked with because of it's high lattice energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question then is&lt;/strong&gt; what other factors would be most important in increasing the solubility of this compound in ionic liquids, anything from adding highly soluble cations, chelating agents, type of ionic liquid I might use, or any other other factors that play into solubility, and could be tuned to optimal conditions?&lt;/p&gt;&#xA;" OwnerUserId="17210" LastEditorUserId="7448" LastEditDate="2015-06-29T07:28:40.777" LastActivityDate="2015-06-29T07:28:40.777" Title="What factors contribute most to solubility of ionic substances in highly polar solvents, esp. ionic liquids and DES?" Tags="&lt;solubility&gt;&lt;computational-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33442" PostTypeId="1" CreationDate="2015-06-29T07:25:32.543" Score="6" ViewCount="149" Body="&lt;p&gt;I have been working with straight chain octane and nonane. The neoprene seals in some parts of the apparatus are allegedly far more vulnerable to the deleterious effects of octane than nonane - why? [I assume it's a solvent thing]&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorUserId="10185" LastEditDate="2015-06-29T20:28:09.053" LastActivityDate="2015-06-29T20:28:09.053" Title="Dissolving ability of octane vs nonane on neoprene" Tags="&lt;organic-chemistry&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33444" PostTypeId="1" AcceptedAnswerId="33447" CreationDate="2015-06-29T10:28:12.060" Score="9" ViewCount="1158" Body="&lt;p&gt;I am trying to find some good resources to understand generation of InChIKeys from the InChI. There seems to be no resource available online. I have gone through the Wikipedia article for InChI, but it hardly contains any information as to how the InChI keys are generated. Can anyone provide me with this information as to how we can generate InChIKeys from InChI. Also, if you happen to find any good resources for the same, please do share them with me.&lt;/p&gt;&#xA;" OwnerUserId="17178" LastEditorUserId="5017" LastEditDate="2015-07-05T17:42:33.110" LastActivityDate="2015-07-05T17:42:33.110" Title="InChIKey generation from InChI" Tags="&lt;computational-chemistry&gt;&lt;software&gt;&lt;reference-request&gt;&lt;cheminformatics&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33445" PostTypeId="2" ParentId="33444" CreationDate="2015-06-29T10:55:40.640" Score="0" Body="&lt;p&gt;Lots of information in this link with references,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.inchi-trust.org/technical-faq/&quot; rel=&quot;nofollow&quot;&gt;http://www.inchi-trust.org/technical-faq/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13211" LastActivityDate="2015-06-29T10:55:40.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33446" PostTypeId="2" ParentId="33444" CreationDate="2015-06-29T10:57:11.770" Score="7" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Chemical_Identifier#InChIKey&quot;&gt;Wikipedia&lt;/a&gt;, the InChikey is generated by a hashing algorithm, like the &lt;a href=&quot;https://en.wikipedia.org/wiki/SHA-2&quot;&gt;SHA-2&lt;/a&gt;. The InChi key is &lt;a href=&quot;http://www.inchi-trust.org/faq/&quot;&gt;designed for use in search&lt;/a&gt;, and thus is not necessarily designed to be comprehensible to the end user. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for a way to generate InChi keys from InChis, there are a number of software packages that do so, including the open source program from IUPAC via the InChi Trust: &lt;a href=&quot;http://www.inchi-trust.org/downloads/&quot;&gt;http://www.inchi-trust.org/downloads/&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;InChi is published under a more permissive variation of the GPL, so in addition to the standalone executable, you can download the code as well. Then, you can sift through the code for the hashing algorithm and figure out what it is doing. InChi even has a project on &lt;a href=&quot;http://sourceforge.net/projects/inchi/&quot;&gt;SourceForge&lt;/a&gt;. The provided documentation at these locations likely includes the workings of the algorithm, but remember that &lt;a href=&quot;http://www.inchi-trust.org/technical-faq/&quot;&gt;InChi key&lt;/a&gt; was designed to produce a string of characters from a structure that a computer can match; the end user is not necessarily supposed to interact with or parse the InChi key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for existing programs that can also produce InChi keys, IUPAC provides the following list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cactus.nci.nih.gov/chemical/structure&quot;&gt;http://cactus.nci.nih.gov/chemical/structure&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chemspider.com/inchi.asmx&quot;&gt;http://www.chemspider.com/inchi.asmx&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/edit/&quot;&gt;http://pubchem.ncbi.nlm.nih.gov/edit/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.acdlabs.com/download/chemsk.html&quot;&gt;http://www.acdlabs.com/download/chemsk.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first three comfortably sit in your browser.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-29T10:57:11.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33447" PostTypeId="2" ParentId="33444" CreationDate="2015-06-29T11:19:27.653" Score="8" Body="&lt;p&gt;InChI is the acronym for the IUPAC &lt;strong&gt;In&lt;/strong&gt;ternational &lt;strong&gt;Ch&lt;/strong&gt;emical &lt;strong&gt;I&lt;/strong&gt;dentifier. It is basically as system, to catalogue molecular information. Its prime advantage is, that it is just a text string. That basically means, that it can easily be understood by machines. The other advantage is, that it is also human readable. It consists of several layers: the molecular formula, connectivity of the heavy atoms, placement of the hydrogen atoms/ stereochemical information. This shall not be confused with the IChIKey, which is a string of 27 characters derived from the hash code of the InChI string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quite thorough presentation/ documentation of the software and its algorithms (including the generation of the InChI string and the InChIKey) is available as an open access paper (currently asap): &lt;a href=&quot;http://dx.doi.org/10.1186/s13321-015-0068-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stephen R Heller, Alan McNaught, Igor Pletnev, Stephen Stein and Dmitrii Tchekhovskoi, &lt;em&gt;Journal of Cheminformatics&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;, 7:23&lt;/a&gt;. Since the abstract is not very comprehensive, I will quote a short passage from the introduction, which outlines the contents of the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We then describe the workflow of InChI generation (normalization, canonicalization, and serialization stages), as well as generation of the compact hashed code derived from InChI (InChIKey); the related algorithms and implementation details are briefly discussed. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In response to R.M.'s comment I am also including the introductory sentence of the linked paper, which describes what an IChIKey is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;InChIKey is a compact chemical identifier derived from InChI. The InChIKey is always only 27-characters long.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a more technical description of the algorithms and the scope of the InChI string and the InChIKey, the reader is pointed towards the technical manual: &lt;a href=&quot;http://www.inchitrust.org/download/104/InChI_TechMan.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stein S, Heller S, Tchekhovskoi D, Pletnev I. IUPAC International Chemical Identifier (InChI) InChI version 1, software version 1.04 (2011) Technical Manual.&lt;/a&gt; The description of the InChIKey can be found in section V, pp 53-62:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The InChIKey is a character signature based on a hash code of the InChI string. A hash code is a fixed length condensed digital representation of a variable length character string. Providing a signature derived from an InChI string is helpful for search applications, including Web searching and chemical structure database indexing; also, it may serve as a checksum for verifying InChI, for example, after transmission over a network.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please note that the InChIKey by itself does not contain enough information to restore the molecular structure. This is also mentioned in the open access paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A disadvantage of the InChIKey is that one loses the ability to algorithmically restore a structure from a textual label: InChIKey is a structure-based registry-lookup identifier [...].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But these are not the only sources. As an IUPAC member, most information relating to the developement and implementation can be accessed freely at &lt;a href=&quot;http://www.iupac.org/home/publications/e-resources/inchi.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;iupac.org&lt;/a&gt;. Since itself is an non-profit organisation, it also has its own website, &lt;a href=&quot;http://www.inchi-trust.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.inchi-trust.org&lt;/a&gt;. This includes a very extensive technical FAQ (html and &lt;a href=&quot;http://www.inchi-trust.org/download/104/inchi-faq.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf&lt;/a&gt;) and since it is open source software also the source code.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The software is still actively developed by the chemical community and the trust itself. As of the current version (1.04, Sept. 2011) there are still several shortcomings, which have bean addressed in &lt;a href=&quot;https://chemistry.stackexchange.com/q/14066/4945&quot;&gt;Structure that breaks InChI&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-06-29T16:19:18.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33448" PostTypeId="1" AcceptedAnswerId="33451" CreationDate="2015-06-29T11:30:13.620" Score="8" ViewCount="1401" Body="&lt;p&gt;While reading about nomenclature of branched alkyl groups, I read that certain common names are also approved by IUPAC. Among those names, only few prefixes are italicised (namely &lt;em&gt;sec-&lt;/em&gt; and &lt;em&gt;tert-&lt;/em&gt;) and others like iso and neo are printed normally. These names are documented in the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm#a_2__25&quot; rel=&quot;noreferrer&quot;&gt;IUPAC blue book&lt;/a&gt;. What can be the reason for italicising &lt;em&gt;sec-&lt;/em&gt; and &lt;em&gt;tert-&lt;/em&gt; and leaving out the others?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="7697" LastEditDate="2015-06-29T13:40:27.273" LastActivityDate="2015-06-29T13:40:27.273" Title="Why are 'sec-' and 'tert-' prefixes printed in italics?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33449" PostTypeId="2" ParentId="33423" CreationDate="2015-06-29T12:00:42.460" Score="3" Body="&lt;p&gt;The two commenters are collectively correct. But to clarify and augment their answers I would say that structure (d) cannot be correct because it is missing 2 carbon atoms relative to (c), and if you tried to cyclize (c) by forming esters, the result would be highly strained. Try making a molecular model of it and see how it looks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the product you would get if (c) cyclized. What is the difference between this product and (d)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xyXrd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="275" LastEditDate="2015-06-29T12:08:02.127" LastActivityDate="2015-06-29T12:08:02.127" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33450" PostTypeId="1" CreationDate="2015-06-29T12:13:28.613" Score="32" ViewCount="906" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eljJp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eljJp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was to rationalize it the ring cleavage happens concerted or non-concerted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The papers &lt;em&gt;&quot;The Mechanism  of  the Thermal Decomposition  of&#xA;1-Pyrazolines and Its Relationship  to&#xA;Cyclopropane Isomerizations&quot;&lt;/em&gt; (Crawford 1965/1966; Original papers: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01091a053&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00969a014&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) have shown the loss of $\ce{N_2}$ is supposed to be concerted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the C-C/C-D(H) bonds of the central position of the ring are involved in a hyperconjugative fashion, so its hybridization does matter.  The out-of-phase linear combination of the $\sigma^\star$ orbitals looks like a p-orbital, therefore I expect a through-space interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn the the out-of-phase combination as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fXbPo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my problem is, that I don't see why the out-of-phase combination is involved in hyperconjugation, because I think the overlap is best in the educt, less good in the transition state and has no orbital overlap to the product because the out-of-phase $p$ orbital is kind of perpendicular to the $sp$ Walsh orbitals which lies in the C-C-C plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can hyperconjugation help to understand the observed kinetic isotope effect results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How does the FMO picture (HOMO-LUMO) with all involved bonds for hyperconjugation looks like?&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can the ratio 66%:34% be explained?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What does reactant 3 make compared to 2?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;How can the ratio 1.07/1.20 be explained?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2016-09-25T15:02:49.867" LastActivityDate="2016-10-06T10:10:43.427" Title="Why does hyperconjugation help for ring cleavage?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;orbitals&gt;&lt;hyperconjugation&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33451" PostTypeId="2" ParentId="33448" CreationDate="2015-06-29T12:47:01.947" Score="7" Body="&lt;p&gt;From the ACD Labs page about &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_74.htm&quot;&gt;IUPAC italicization&lt;/a&gt;,it is stated that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Italicizing mainly serves to mark letters which are not involved in the primary stage of alphabetical ordering.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://aerosol.chem.uci.edu/intranet/files/ACS_style_guide.pdf&quot;&gt;American Chemical Society Style Guide&lt;/a&gt; (p. 235), the use of italics for&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;positional, stereochemical, configurational, and descriptive&#xA;  structural prefixes when they appear with the chemical name or formula. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(There are several prefixes in the list).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are separated from the chemical name with a hyphen and are not capitalised &lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-06-29T12:47:01.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33452" PostTypeId="1" CreationDate="2015-06-29T12:50:01.700" Score="5" ViewCount="156" Body="&lt;p&gt;Why does the Ziegler–Natta polymerization display the regioselectivity shown in the image? Internal nucleophiles (coordinated to the metal) should attack the alkene on the less substituted end, like in the Heck reaction, whereas external nucleophiles should attack the more substituted end (like in the Wacker reaction). Why is this not the case for alkene polymerisation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HJltk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HJltk.png&quot; alt=&quot;Regioselectivity in Ziegler–Natta polymerisation of propene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2018-10-14T13:17:16.497" LastActivityDate="2018-10-14T13:17:16.497" Title="Regioselectivity in Ziegler–Natta polymerisation of α-olefins" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;organometallic-compounds&gt;&lt;regioselectivity&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33453" PostTypeId="1" CreationDate="2015-06-29T13:11:58.290" Score="0" ViewCount="775" Body="&lt;p&gt;In searching online, I've noticed there are a lot of different ways to group the elements of the periodic table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take mercury in the two tables linked below, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencegeek.net/tables/PToE_basic.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencegeek.net/tables/PToE_basic.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the first table, it's explicitly outside of the transition metals. On the second table, it's included in the &quot;transition metals&quot; group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mercury is an example of this inconsistent classification, but in general, why is there no canonical grouping for all elements, across the periodic table? I see some tables mentioning &quot;metalloids,&quot; but others not doing so. Or, sometimes elements are classified in entirely different groups altogether (like zinc, cadmium and mercury in the above-linked tables). Or, the lanthanides and actinides being included sometimes in the transition metals group, and sometimes not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which classification is right? Or rather, which one is the &quot;most right&quot;? Really: which one should I focus my efforts on remembering?&lt;/p&gt;&#xA;" OwnerUserId="16920" LastEditorUserId="11367" LastEditDate="2017-01-14T14:45:50.410" LastActivityDate="2017-01-14T14:45:50.410" Title="Periodic table groups - which grouping is &quot;right&quot;?" Tags="&lt;terminology&gt;&lt;elements&gt;&lt;periodic-table&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-06-29T18:59:58.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="33454" PostTypeId="1" CreationDate="2015-06-29T13:39:01.090" Score="1" ViewCount="15" Body="&lt;p&gt;Periodically in our lab, we have a reproducibility test for the various samples that we analyze. One such sample is boiler water. Two of the parameters that we test are phosphate in boiler water and iron in boiler water. Random error aside (shaking of sample, measurement bias, etc.), why are my iron results usually lower than the original and my phosphate is usually higher? Is there any chemical basis for these trends?&lt;/p&gt;&#xA;" OwnerUserId="16436" LastActivityDate="2015-06-29T13:39:01.090" Title="Trend in results for reproducibility of iron and phosphate" Tags="&lt;water&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33456" PostTypeId="1" AcceptedAnswerId="33687" CreationDate="2015-06-29T15:56:02.810" Score="1" ViewCount="5131" Body="&lt;p&gt;I was studying pinacol pinacolone reaction when I came across this. I believe the hydrogen cation will attack the hydroxyl group of carbon attached to two ethyl groups (since this carbocation is more stable due to the inductive effect), then methyl group will migrate and hence we get 3-ethyl-3-methylpentan-2-one, but the book says the answer is 2-ethyl-2-methylpentan-3-one. Is the answer in the book wrong or I am making a conceptual mistake?&lt;/p&gt;&#xA;" OwnerUserId="16618" LastEditorUserId="5026" LastEditDate="2018-02-25T02:25:28.097" LastActivityDate="2018-02-25T02:25:28.097" Title="3-ethyl-2-methylpentane-2,3-diol with sulfuric acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33458" PostTypeId="2" ParentId="33453" CreationDate="2015-06-29T17:13:57.377" Score="2" Body="&lt;p&gt;There is no &quot;most right classification scheme&quot; for the elements of the periodic table, worth memorizing above all others.  This is because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-06-29T17:13:57.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33459" PostTypeId="2" ParentId="32067" CreationDate="2015-06-29T17:14:52.833" Score="3" Body="&lt;p&gt;There are basically two processes, that make adding more hydroxide the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one is already described in Yomen's answer, but I would like to extend that a little bit further. If you dissolve iron (II) hydroxide in water, an aqueous complex will form:&#xA;$$\ce{Fe(OH)2~(s) + 4H2O &amp;lt;=&amp;gt; [Fe(H2O)4(OH)2]}$$&#xA;Apart from this, due to the auto-ionisation of water and the relative stabilities of the complexes, there is a variety of complexes in solution, among others:&#xA;$$\ce{[Fe(H2O)5(OH)1]- &amp;lt;=&amp;gt;C[H+] [Fe(H2O)4(OH)2] &amp;lt;=&amp;gt;C[OH-] [Fe(H2O)3(OH)3]+}$$&#xA;From these complexes, only the neutral one precipitates. From this equation it is also deducible, that the stability of the precipitate is highly pH dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon adding more hydroxide ions, the equilibrium will shift more and more to the right, and even further, resulting in the dissolution of the precipitate&#xA;$$\ce{[Fe(H2O)4(OH)2] &amp;lt;=&amp;gt;C[OH-] [Fe(H2O)3(OH)3]+ &amp;lt;=&amp;gt;C[OH-] [Fe(H2O)2(OH)4]^2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other possibility is, that iron (II) will easily be oxidised in basic conditions and access to air. This process is often referred to as rusting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{4Fe(OH)2 + O2 + 2H2O &amp;lt;=&amp;gt;C[OH-] 4Fe(OH)3 \equiv 2Fe2O3(OH2)3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reactions will form the more stable, but less hydrated iron (III) oxide:&#xA;$$\ce{Fe2O3(H2O)3 &amp;lt;=&amp;gt; Fe2O3(H2O)_{$x$} + $x$ H2O} \quad (x\in[0,3])$$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="5026" LastEditDate="2018-02-21T13:58:00.757" LastActivityDate="2018-02-21T13:58:00.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33460" PostTypeId="1" CreationDate="2015-06-29T18:08:33.900" Score="7" ViewCount="588" Body="&lt;p&gt;I am in XI and I am currently studying atomic orbitals, I stumbled up and am stuck at the following step of de Broglie's hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a little search on-line and found almost same derivation everywhere, the only place I found a more complex and supposedly accurate formula was the &lt;a href=&quot;https://en.wikipedia.org/wiki/Matter_wave&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; itself, though it did not explain the derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Few such sites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength&quot; rel=&quot;nofollow&quot;&gt;http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://entrancechemistry.blogspot.in/2012/09/de-broglie-equation.html&quot; rel=&quot;nofollow&quot;&gt;http://entrancechemistry.blogspot.in/2012/09/de-broglie-equation.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the derivation of the formula $$\frac{1}{\lambda}=\frac{h}{m\color{green}c}=\frac{h}{m\color{red}v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we replace $c$ with $v$. I want to ask why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of Wikipedia it gives a better formula with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lorentz_factor&quot; rel=&quot;nofollow&quot;&gt;Lorentz factor&lt;/a&gt; $\gamma$ :- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\lambda}=\frac{h}{\color{red}\gamma m_oc}=\frac{h}{m_ov}  \sqrt{\color{red}1\color{red}-\frac{\color{red}v^\color{red}2}{\color{red}c^\color{red}2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give the derivation or a hint or a logic that helps understand why the derivation from $E=mc^2$ and $E=h\nu$ gives the De Broglie equation.&lt;/p&gt;&#xA;" OwnerUserId="13190" LastEditorUserId="9961" LastEditDate="2015-06-29T20:41:58.293" LastActivityDate="2017-02-26T10:01:01.530" Title="Why replace c with v in Broglie hypothesis" Tags="&lt;quantum-chemistry&gt;&lt;relativistic&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33461" PostTypeId="2" ParentId="33373" CreationDate="2015-06-29T18:24:49.363" Score="1" Body="&lt;p&gt;One possible option for thin film thickness measurement is by Energy Dispersive Spectroscopy (EDS) in a scanning electron microscope (SEM).  There exist commercial packages for this such as &lt;a href=&quot;http://www.oxford-instruments.com/products/microanalysis/energy-dispersive-x-ray-systems-eds-edx/eds-for-sem/layerprobe&quot; rel=&quot;nofollow&quot;&gt;LayerProbe by Oxford&lt;/a&gt; and &lt;a href=&quot;http://www.samx.com/microanalysis/products/stratagem_us.html&quot; rel=&quot;nofollow&quot;&gt;STRATAGem&lt;/a&gt; (I have no financial interest in Oxford or SAMx).  If your campus/research facility has an SEM with an EDS (many do), but not a commercial package, you can still work it out yourself.  If I were tackling this in my own lab, I would probably do something along the following lines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option #1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obtain a spectrum from a known thickness of carbon on your Cu substrate.  This can be any method such as sputtering or evaporation so long as the thickness is known by other means.  Most geology labs have the ability to do this kind of deposition since they need reproducible carbon thicknesses on their samples.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, geo labs will typically place a polished brass object next to the sample to be coated in an evaporative coater.  The change in the color of the brass corresponds to the thickness of the carbon: e.g. blue is 250 Å but bluish green is 300 Å (Kerrick, D. M., et al., 1973).  So something like 25 % accuracy is possible this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you measure that thickness in the SEM using an exact beam current, size, distance from the objective lens.  Then you measure your graphene in the SEM using the &lt;strong&gt;exact same conditions&lt;/strong&gt;.  If your sample is less than about 200 nm thick, then the number of counts/second is proportional to the number of carbon atoms present in the layer.  Using the mass for your standard layer, (assume a density of graphite), then you can compute the mass of your graphene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your thickness is more than about 200 nm, then you may need to do an absorption correction – i.e. the photons are being absorbed on their way out and your count rate is no longer proportional to the number of atoms.  You can compute attenuation coefficients using the &lt;a href=&quot;http://henke.lbl.gov/optical_constants/atten2.html&quot; rel=&quot;nofollow&quot;&gt;CXRO calculator&lt;/a&gt; (and a couple other places if you google).  Then you can do a thin film correction.  At 200 nm, your error from ignoring the thin film correction is probably about 5–10 %, which is likely a bit less than the error in the determination of the standard film using the color method.  So you can think with that for your sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option #2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the mass directly using an oscillating crystal.  Commercial systems comprise a quartz crystal with electronics set to find the resonant frequency.  You place the quartz crystal in your chamber alongside your Cu substrate.  As your reaction happens, it grows both onto your Cu and onto the quartz crystal.  The resonant frequency of the crystal changes as mass is added.  These are sold commercially, though a bit pricey.  For example, at &lt;a href=&quot;http://www.tedpella.com/cressington_html/MTM-10-20-High-Resolution-Film-Thickness-Monitor.htm&quot; rel=&quot;nofollow&quot;&gt;TedPella&lt;/a&gt; (again, no financial interest).  The resolution is typically billed as 1 Å thickness resolution (assuming you know the density of your material.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your application, you would have to be sure that the graphite grows on your monitor crystal the same as on your Cu substrate.  That may require checking in an SEM, and it may require prepping the crystal somehow.  Hence, I offer this as suggestion #2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you could also figure this out using a focused ion beam and sectioning the sample from the side.  If this is too complex, there also exist labs where one can hire them to do this kind of measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope one of these solutions works for you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kerrick, D. M., Eminhizer, L. B., &amp;amp; Villaume, J. F. (1973). The Role of Carbon Film Thickness in Electron Microprobe Analysis. American Mineralogist, 58, 920–925.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastEditorDisplayName="user7951" LastEditDate="2015-06-29T19:10:34.720" LastActivityDate="2015-06-29T19:10:34.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33464" PostTypeId="2" ParentId="33460" CreationDate="2015-06-29T18:36:48.377" Score="1" Body="&lt;p&gt;C is for radioactive rays while v is for the matter waves..here c is for a particle travelling with light speed, i.e., radioactive rays, so we replace it with v velocity of any particle for its matter wave&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=mc^2$ and $E=hv$. &#xA;Therefore,  $$mc^2=hv$$&#xA;Or $$mc^2=\frac{hc}{k}$$, where k=wavelength&#xA;$$mc=\frac{h}{k}$$&#xA;Or, $$k=\frac{h}{mc}$$&#xA;   Replacing c by v we get&#xA;$$k=\frac{h}{mv}$$&#xA;Note the difference between matter waves and electromagnetic waves. E.W. are associated with electrical and magnetic field while matter waves are not so E.w. travel at 3×10^2 m/s. While matter waves travel with different velocity. The equation of de broglie with c is valid, it can be treated for finding the wavelength of electromagnetic waves, or electrons. But the velocity of matter waves are quite smaller as compared to light speed. So he used a general v in place of c V can be equal to c when the particle is travelling at light speed&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="17217" LastEditDate="2015-08-26T06:09:37.487" LastActivityDate="2015-08-26T06:09:37.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33466" PostTypeId="2" ParentId="31150" CreationDate="2015-06-29T19:50:16.223" Score="4" Body="&lt;p&gt;In homework land you are right, but not in real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $K_\mathrm b$ of $\ce{Ca^2+}$ ($\mathrm pK_\mathrm b = 2.43$ &lt;a href=&quot;https://books.google.com/books?id=CD6kqBW5U_wC&amp;amp;pg=PA74&amp;amp;lpg=PA74&amp;amp;dq=pkb%20calcium%20hydroxide&amp;amp;source=bl&amp;amp;ots=h3F15-O0OW&amp;amp;sig=D82yx8B782Xh0kKFcCan-jPSu-Q&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=r56RVZ_bDomyyQTV46Fo&amp;amp;ved=0CB0Q6AEwADgU#v=onepage&amp;amp;q=pkb%20calcium%20hydroxide&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to some sources&lt;/a&gt;) is within 1 log unit of the $K_\mathrm a$ of $\ce{HF}$.  In other words, near neutral pH, consideration of the $\ce{Ca^2+/CaOH+}$ equilibrium is almost as important as the $\ce{F-/HF}$ equilibrium.  &lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="16683" LastEditDate="2017-09-03T08:58:48.070" LastActivityDate="2017-09-03T08:58:48.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33467" PostTypeId="1" CreationDate="2015-06-29T20:07:42.383" Score="3" ViewCount="27" Body="&lt;p&gt;Following up on &lt;a href=&quot;https://chemistry.stackexchange.com/q/33086/5354&quot;&gt;How to perform solvent extraction of a solid&lt;/a&gt; I realized that I haven't quantified the size of the contaminants.  I know they aren't soluble in acetone and the target is, but I don't know how small the contaminants are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the target is dissolved presumably I must assume that suspended contaminants are as small as their molecular size.  In this case is a filter-based extraction still possible?  I don't see lab filters like extraction thimbles sold by filtration size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a case like this is the correct approach to run the solution through a centrifuge and then suck out the middle solution with a pipette?  (Or what is the lab best practice for mechanical extraction following spinning?)&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-03-08T22:00:10.030" Title="Limits on contaminant size in filtration for extraction, and standard alternatives?" Tags="&lt;experimental-chemistry&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33469" PostTypeId="1" AcceptedAnswerId="33473" CreationDate="2015-06-30T02:09:43.057" Score="2" ViewCount="1130" Body="&lt;p&gt;Magnesium shows an oxidation state of $+2$ and bismuth shows an oxidation state of $+3$. Both do not show a negative oxidation state. Then how is magnesium bismuthide ($\ce{Mg_{3}Bi_{2}}$) formed?&lt;/p&gt;&#xA;" OwnerUserId="12785" LastEditorDisplayName="user15489" LastEditDate="2015-06-30T02:13:23.760" LastActivityDate="2015-06-30T13:26:30.677" Title="Formation of magnesium bismuthide" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33470" PostTypeId="1" AcceptedAnswerId="33504" CreationDate="2015-06-30T04:35:11.580" Score="3" ViewCount="76" Body="&lt;p&gt;I am writing up a lab report investigating titanium anodisation and producing oxide layers. After some searching, all I can find is the reaction &lt;span class=&quot;math-container&quot;&gt;$\ce{Ti + O2 \to TiO2}$&lt;/span&gt;, which I am having no trouble understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I seem to be confused/mislead by the simplicity of this. The whole unit is about electrochemistry and redox reactions but I cannot find any examples (or seem to produce any either - I have tried generalizing from the rusting of iron in water) of half reactions or anything similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come about as far as including the redox reactions for the hydrolysis of water and then jumping to the above reaction but am unsure where to go from here. &lt;/p&gt;&#xA;" OwnerUserId="17221" LastEditorUserId="83758" LastEditDate="2020-04-22T13:09:48.593" LastActivityDate="2020-04-22T13:10:19.343" Title="Reactions involved in Titanium Anodisation" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33471" PostTypeId="2" ParentId="33430" CreationDate="2015-06-30T06:47:13.203" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;http://dx.doi.org/10.1021/ja01320a064&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; a mixture of products is formed, with alkylchlorsulfonates $\ce{Cl-SO2-O-R}$ being dominant products. These molecules are resistant to abstraction of sulfur trioxide, probably because trioxide is a very strong Lewis acid, unlike dioxide.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-06-30T06:47:13.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33473" PostTypeId="2" ParentId="33469" CreationDate="2015-06-30T07:27:15.173" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bismuth&quot; rel=&quot;nofollow&quot;&gt;Bismuth&lt;/a&gt; can have any oxidation state from +1 to +5 . As it is in the same group as nitrogen and phosphorous, its most commom oxidation states with strongly oxidizing substances such as halogens and oxygen are +3 and +5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However with a strongly electopositive (reducing) element, bismuth can have an oxidation state of -3, just as nitrogen can. While it is true that the oxidizing strength reduces as we go down the group, the &quot;softness&quot; (tendency to of an element to adapt its oxidation state to the preferred oxidation states of other elements) goes up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without definite information I can't discount the possibility that $\ce{Mg3Bi2}$ is an alloy, but the fact that it has an even stoichiometric ratio of elements suggests that it is in fact a compound in which magnesium has an oxidation state of +2 and bismuth has an oxidation state of -3.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorDisplayName="user15489" LastEditDate="2015-06-30T13:26:30.677" LastActivityDate="2015-06-30T13:26:30.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33474" PostTypeId="1" CreationDate="2015-06-30T10:36:54.397" Score="11" ViewCount="672" Body="&lt;p&gt;Nucleophilic substitution through an elimination-addition mechanism (involving benzyne formation) occurs with a strong base. $\ce{NH2-}$ can show this reaction. Can Grignard reagents $\ce{RMgX}$, which form $\ce{R-}$ - an even stronger base - show this reaction?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="16683" LastEditDate="2016-04-15T23:58:25.203" LastActivityDate="2016-05-16T14:38:55.987" Title="Benzyne formation with Grignard reagents" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33475" PostTypeId="1" AcceptedAnswerId="33476" CreationDate="2015-06-30T10:40:38.100" Score="4" ViewCount="1873" Body="&lt;p&gt;Grignard's Reagent will generate a strong base and nucleophile, $\ce{R-}$. Can this not show nucleophilic substitution at the C atom attached to the O atom of the ether, causing $\ce{RO-}$ to depart as a leaving group? Since $\ce{RO-}$ is a better leaving group than $\ce{R-}$ this reaction should be feasible. Additonally, $\ce{R-}$ is a very good nucleophile.&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-06-30T10:56:30.020" LastActivityDate="2015-06-30T10:57:44.423" Title="Why can Grignard's Reagent be stored in ether?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-30T13:32:14.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="33476" PostTypeId="2" ParentId="33475" CreationDate="2015-06-30T10:56:04.600" Score="6" Body="&lt;p&gt;The free nucleophile $\ce{R-}$ never really exists in a Grignard reagent. That's why we write it $\ce{RMgX}$. Ethers are used as a solvent as the oxygen coordinates to the $\ce{Mg}$, improving solubility and stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your mentioned substitution reaction, it probably would happen if a Grignard reagent was left long enough. There's a reason you make Grignard reagents immediately before use. They don't last if left on the shelf!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Grignard reagents are usually reacted with carbonyl compounds. These are much more electrophilic than ethers, because the oxygen atom is pulling electrons away from one carbón atom instead of spreading its pull between 2, and because a double $\ce{C=O}$ bond is inherently weaker than two $\ce{C-O}$ bonds. Hence the desired reaction happens much faster than the undesired one. &lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="10185" LastEditDate="2015-06-30T10:57:44.423" LastActivityDate="2015-06-30T10:57:44.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33477" PostTypeId="1" CreationDate="2015-06-30T11:23:59.660" Score="0" ViewCount="197" Body="&lt;p&gt;Can $\ce{NH3}$ substitute an alcohol? The OH group will depart as a weaker base - some O is more electronegative - so will be a better leaving group than amine. &#xA;And what about using ammonium ions - such as in an ammonium salt? They will donate an $\ce{H+}$ to protonate the alcohol and make it a good leaving group, then show substitution followed by deprotonation to form an amine?&#xA;How is the the feasibility decided in case of nucleophilic substitution reactions? Is there any rule of thumb to know what happens, and what doesn't?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-06-30T11:41:30.210" LastActivityDate="2015-06-30T11:53:06.380" Title="Feasibility of Nucleophilic Substitutions" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-30T13:31:58.860" ContentLicense="CC BY-SA 3.0" />
  <row Id="33478" PostTypeId="2" ParentId="33456" CreationDate="2015-06-30T11:39:38.593" Score="5" Body="&lt;p&gt;&lt;strong&gt;Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick analysis of this reaction gives me the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SPkAX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation &lt;strong&gt;A&lt;/strong&gt; or &lt;strong&gt;B&lt;/strong&gt;. Either of these cations can lose water to form a carbocation&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The carbocation formed from &lt;strong&gt;A&lt;/strong&gt; is a tertiary carbocation &lt;strong&gt;C&lt;/strong&gt; that might not be likely to rearrange, but we'll consider the possibility in a bit.&lt;/li&gt;&#xA;&lt;li&gt;The carbocation formed from &lt;strong&gt;B&lt;/strong&gt; is a primary(!) carbocation &lt;strong&gt;D&lt;/strong&gt; that seems unlikely to form.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pinacol rearrangement from cation &lt;strong&gt;A&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yPRm8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pinacol rearrangement from cation &lt;strong&gt;A&lt;/strong&gt; through carbocation &lt;strong&gt;C&lt;/strong&gt; forms an aldehyde: 3-ethyl-2-methylpentanal. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pinacol rearrangement from cation &lt;strong&gt;B&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1VIn8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having established that carbocation &lt;strong&gt;D&lt;/strong&gt; is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in &lt;a href=&quot;http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch12friedelcraftsalkylation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friedel-Crafts reactions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which product is more likely the major product?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rather than tell you, I will give you a hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One rearrangement produces a thermodynamically more stable cation than the other. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="275" LastEditDate="2015-07-07T19:01:01.797" LastActivityDate="2015-07-07T19:01:01.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33479" PostTypeId="2" ParentId="33477" CreationDate="2015-06-30T11:53:06.380" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Short answer - No. See below for a workaround. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Consider your acids and bases:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Substituting an alcohol with ammonia directly&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH3 + CH3CH2OH &amp;lt;=&amp;gt; CH3CH2NH2+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were an acid-base reaction (and it is, at least in the Lewis sense), then which side of this reaction would be favored by equilibrium? Which side has the stronger acid-base pair? Hydroxide is a stronger base than ammonia, so this equilibrium would favor the reactants. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Using an ammonnium salt&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first step of this mechanism would be a proton transfer to form an activated leaving group and the $\ce{NH3}$ nucleophile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4+ +CH3CH2OH &amp;lt;=&amp;gt; NH3 + CH2CH3OH2+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, which side of this equilbrium is favored? Ammonia is definitely the stronger base compared with the alcohol. Estimates of the pKa of $\ce{ROH2+}$ are in the vicinity of $-2$, while the pKa of $\ce{NH4+}$ is $9.25$. The equilbrium constant of this reaction is on the order of $10^{-11}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Workaround&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you are needing to do the transformation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{ROH -&amp;gt; RNH2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is a workaround in converting the alcohol into a molecule like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tosyl&quot; rel=&quot;nofollow&quot;&gt;tosylate ester&lt;/a&gt;, which can then react with ammonia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2OH -&amp;gt;[\ce{TsCl,base}] CH3CH2OTs }$$&#xA;$$\ce{CH3CH2OTs -&amp;gt;[\ce{NH3}] CH3CH2NH2}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-30T11:53:06.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33480" PostTypeId="1" AcceptedAnswerId="33483" CreationDate="2015-06-30T12:41:00.510" Score="6" ViewCount="40240" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{CH3COOH-&amp;gt;CH3COO- + H+}$&lt;br&gt;&#xA;  \begin{array}{c|c c c}&#xA;  \mathbf{Initial} &amp;amp;  \mathrm{1~mol} &amp;amp; \mathrm{0~mol} &amp;amp; \mathrm{0 ~mol} \\\hline&#xA;  \mathbf{Final} &amp;amp;   \mathrm{1-\alpha ~mol}&amp;amp;\mathrm{\alpha ~mol}&amp;amp;\mathrm{\alpha~ mol}&#xA; \end{array}&#xA;  Where $\alpha$ is the degree of dissociation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, $i = 1-\alpha+\alpha+\alpha =1+\alpha$, and $\alpha=(i-1)$.&lt;br&gt;&#xA;This is what is written in my book. And I understand this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{CH3COOH-&amp;gt;CH3COO- + H+}$&#xA;  \begin{array}{c|c c c}&#xA;  \mathbf{Initial} &amp;amp;  \mathrm{c~ mol}&amp;amp;\mathrm{0~mol}&amp;amp;\mathrm{0~ mol}\\\hline&#xA;  \mathbf{Final} &amp;amp;   \mathrm{c(1-\alpha)~ mol}&amp;amp;\mathrm{c\alpha~ mol}&amp;amp;\mathrm{c\alpha~ mol} &#xA;\end{array}&#xA;  Where $\alpha$ is the degree of dissociation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, &#xA;$$i = \frac{c\alpha+c\alpha+c(1-\alpha)}{c} =1+\alpha$$&#xA; (same as in case 1), and $\alpha=(i-1)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_a=\frac{c^2\alpha^2}{c(1-\alpha)} = \frac{c\alpha^2}{(1-\alpha)}$$&#xA;This is also written in my book, and I understand this, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, my question is why can we &lt;em&gt;not&lt;/em&gt; use the former method to calculate the dissociation constant, $K_a$, which will give me: &#xA;$$K_a =\frac{ \alpha^2}{(1-\alpha)}$$&#xA;as opposed to what I got in the second (and the correct) case, which was: &#xA;$$K_a = \frac{c\alpha^2}{(1-\alpha)}$$&#xA;I can see the difference of a $c$ in the two equations, but can someone explain why do we get the difference?&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="4945" LastEditDate="2017-11-10T07:04:29.960" LastActivityDate="2017-11-10T07:04:29.960" Title="Degree of association, van't Hoff factor and dissociation constant" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33481" PostTypeId="2" ParentId="33398" CreationDate="2015-06-30T13:16:10.523" Score="6" Body="&lt;p&gt;Since mechanistic studies of the camphene/ bornyl acetate system are not available, I will focus on the parent compound norbonyl, and with it explaining the stability of the cation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For quite some time there has been active debate over the structure of the 2-norbonyl cation. A very brief description of the two sides can be found in the &lt;a href=&quot;https://en.wikipedia.org/wiki/2-Norbornyl_cation&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article. A short history of the argument can be found at the homepage of the Royal Society of Chemistry (RSC): &quot;&lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/07/norbornyl-nonclassical-cation-brown-winstein-olah&quot; rel=&quot;noreferrer&quot;&gt;The nonclassical cation: a classic case of conflict&lt;/a&gt;&quot; by Mark Peplow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2013 the case was finalized after I. Krossing's group was able to obtain a crystal structure of the compound. Another blog post at the RSC gives some more insight into the matter: &quot;&lt;a href=&quot;http://www.rsc.org/chemistryworld/2013/07/norbornyl-cation-nonclassical-structure-olah-herb-brown&quot; rel=&quot;noreferrer&quot;&gt;Crystal structure closes classic carbocation case&lt;/a&gt;&quot; by Andy Extance. Finally we can speak of the 2-norbonyl cation as a true non-classical cation. The findings are also in good agreement with high-quality calculations. &lt;a href=&quot;http://dx.doi.org/10.1126/science.1238849&quot; rel=&quot;noreferrer&quot;&gt;F. Scholz, D. Himmel, F. W. Heinemann, P. v. R. Schleyer, K. Meyer, and I. Krossing, &lt;em&gt;Science&lt;/em&gt;, &lt;strong&gt;2013&lt;/strong&gt;, &lt;em&gt;341&lt;/em&gt; (6141), 62-64.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does in principle mean, that at very low temperatures, the 2-norbonyl cation is best described by the following resonance structures (courtesy of wikipedia):&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XvEt3.png&quot; alt=&quot;resonance structures of the 2-norbornyl cation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher temperature the structure is quite disordered. In solution this means basically, that there is an ultra fast equilibrium (at the rate of molecular vibrations) between several structures. This would also explain the high reactivity of this compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since camphene is a substituted norbornene, similar reactivity can be expected. The reaction is therefore most likely determined by the small steric effects. Since this reaction is highly selective, one can also safely assume, that the thermodynamic favoured product is formed. In this case it is the isobornyl acetate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have already mentioned it in the comments, but I am including it here nonetheless. The second row of your image is contains the wrong numeration and therefore depicts the mirror image. Correct is the following depiction:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XP0khm.png&quot; alt=&quot;correction to the scheme of the camphenyl cation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-06-30T13:16:10.523" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33482" PostTypeId="1" CreationDate="2015-06-30T13:51:11.720" Score="5" ViewCount="61" Body="&lt;p&gt;Cis enolates usually produce syn aldols and trans enolates produce anti aldols. How do exemptions from this &quot;rule&quot; arise?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-06-30T14:10:36.547" LastActivityDate="2015-06-30T14:10:36.547" Title="How does the stereochemistry of the enolate affect the stereochemistry of the aldol product?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ClosedDate="2015-09-22T09:14:38.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="33483" PostTypeId="2" ParentId="33480" CreationDate="2015-06-30T15:06:14.340" Score="5" Body="&lt;p&gt;There is &lt;strong&gt;absolutely no difference&lt;/strong&gt; in your two equations but you forgot the units which makes it seem like there is a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$&#xA;since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$k_a=\frac{\alpha^2}{1-\alpha}~\ce{mol~L^{-1}}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How do I know that the &lt;em&gt;c&lt;/em&gt; is the concentration?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&amp;amp;~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&amp;amp;~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$&#xA;and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-06-30T21:03:52.170" LastActivityDate="2015-06-30T21:03:52.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33484" PostTypeId="1" AcceptedAnswerId="33486" CreationDate="2015-06-30T15:15:02.693" Score="8" ViewCount="281" Body="&lt;p&gt;When talking about &lt;a href=&quot;http://goldbook.iupac.org/F02520.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;free rotation&lt;/a&gt; it is often helpful to determine about which geometric arrangement one is referring to. For example in the case of butane we talk about several eclipsed or staggered conformations when discussing the rotational barriers. This also often helps to explain the all trans (staggered) arrangement of fatty acids and surfactants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a wider sense even the pseudo rotation of cyclohexane has its own terminology, with boat, chair and twisted conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in a way to name the two rotamers that can be thought of in &amp;alpha;-keto esters (1, 2). My initial thought was calling these &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;- configuration, in analogy of the partial double bond (blue, 3, 4), but for my taste this has a too rigid connotation. Has there been anything proposed in this area? Amides would also be comparable (5, 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O74Ez.jpg&quot; alt=&quot;rotameric structures of alpha keto esters&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="10185" LastEditDate="2015-09-18T18:13:10.897" LastActivityDate="2015-09-18T18:13:10.897" Title="How can rotamers of alpha-keto esters be named to distinguish between different geometries?" Tags="&lt;terminology&gt;&lt;carbonyl-compounds&gt;&lt;molecular-structure&gt;&lt;conformers&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33486" PostTypeId="2" ParentId="33484" CreationDate="2015-06-30T15:47:50.773" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;http://goldbook.iupac.org/S05508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUPAC recommends&lt;/a&gt; &lt;em&gt;s-cis&lt;/em&gt; and &lt;em&gt;s-trans&lt;/em&gt; for the rotamers of conjugated dienes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jLVeN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nomenclature derives from having a &quot;&lt;em&gt;cis&lt;/em&gt;-like&quot; or &quot;&lt;em&gt;trans&lt;/em&gt;-like&quot; geometry about a sigma bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like something similar should be appropriate, however that same IUPAC link recommends &lt;em&gt;E&lt;/em&gt;/&lt;em&gt;Z&lt;/em&gt; or &lt;em&gt;sp&lt;/em&gt;/&lt;em&gt;ap&lt;/em&gt; (&lt;em&gt;syn&lt;/em&gt;-periplanar/&lt;em&gt;anti&lt;/em&gt;-periplanar) for the &lt;em&gt;N&lt;/em&gt;-alkyl amides (5,6). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O74Ez.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amide 5 would be &lt;em&gt;Z&lt;/em&gt; (but &lt;em&gt;ap&lt;/em&gt; w/r/t alkyl groups) and amide 6 would be &lt;em&gt;E&lt;/em&gt; (and sp). &lt;em&gt;E&lt;/em&gt;/&lt;em&gt;Z&lt;/em&gt; is less ambiguous in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ketoester 1 is &lt;em&gt;E&lt;/em&gt;, and ketoester 2 is &lt;em&gt;Z&lt;/em&gt;. I suppose that ketoester 1 is also &lt;em&gt;ap&lt;/em&gt; and ketoester 2 is also &lt;em&gt;sp&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-06-30T15:47:50.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33489" PostTypeId="2" ParentId="9145" CreationDate="2015-06-30T16:08:24.887" Score="2" Body="&lt;p&gt;Chromium and copper are examples of elements with &quot;anomalous&quot; electron configurations, meaning that they don't follow the normal rules we use for populating the configurations of other elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commonly given reason for this is that the energy of a shell is minimized when the number of electrons with the same spin is maximized (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hund&amp;#39;s_rule_of_maximum_multiplicity&quot; rel=&quot;nofollow&quot;&gt;Hund's rule&lt;/a&gt;). As a result, when the energy levels of two successively filled sub-shells are already close together (as they are with 4s and 3d sub-shells), the slightly favored half-filled configuration can &quot;win&quot; over the energy increase needed to move an electron to an even-more-slightly higher energy level. In the case of chromium, this means that one of the 4s electrons will go to the 3d orbital, resulting in two half-filled sub-shells where all electrons within each sub-shell have the same spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of copper, a similar thing happens. The difference is that the 4s electron moves into an almost-filled 3d shell in order to completely fill it. You get a slight energy decrease when all electrons are paired within a sub-shell. This, in combination with the decrease obtained from achieving a half-filled s orbital, ends up being enough to overcome the increase in energy required to move that electron to the 3d orbital in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice if these empirical rules were consistent across the entire table, but unfortunately they are not. If you look up the actual electron configuration for other d- and f-block elements, you will see there are some patterns, and similar things happen for other elements, but because they are so dependent on the delicate balance between energy levels, it is not possible to reliably predict them with simple rules for all elements. In &quot;real life&quot; we use spectroscopy and quantum mechanical calculations to find the actual electron configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since chromium and copper are common enough and reliably predictable with simple rules, we tend to use those as classroom examples to demonstrate that the reality of electron configurations is more complex than the simple rules we give you in school.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-06-30T16:08:24.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33490" PostTypeId="1" AcceptedAnswerId="33493" CreationDate="2015-06-30T16:59:25.527" Score="9" ViewCount="110000" Body="&lt;p&gt;I am curious about the flammability of diesel fuel and its properties, as I don't deal with it on a regular basis (at all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I want to know how the fuel works when powering a diesel engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I know about gasoline engines:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Compresses gasoline and air&lt;/li&gt;&#xA;&lt;li&gt;Ignites compressed gasoline and air with spark from spark plug&lt;/li&gt;&#xA;&lt;li&gt;The ignition pushes the piston back turning the crank shaft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have heard that diesel is less flammable than gasoline and that diesel engines don't require spark plugs. How does this work?&lt;/p&gt;&#xA;" OwnerUserId="17228" LastEditorUserId="11367" LastEditDate="2015-06-30T17:32:58.007" LastActivityDate="2016-05-16T06:27:23.337" Title="What are the differences between diesel and unleaded gasoline?" Tags="&lt;combustion&gt;&lt;fuel&gt;&lt;reaction-control&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33491" PostTypeId="1" AcceptedAnswerId="33501" CreationDate="2015-06-30T17:23:04.487" Score="6" ViewCount="2313" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\vec{p}_\text{ind} = \alpha \vec{E}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;induced dipole moment&lt;/strong&gt; is the &lt;strong&gt;polarizability&lt;/strong&gt; times the &lt;strong&gt;electric field vector&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}(\omega)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;polarization&lt;/strong&gt; is proportional to the &lt;strong&gt;susceptibility&lt;/strong&gt; times the &lt;strong&gt;electric field vector&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In spectroscopy we used the dipole approximation. Unfortunately, I often mix up the different formulas. Is there a relationship between both formulas, or do both equations tell us exactly the same thing?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2021-04-16T05:00:56.667" Title="Relationship between electric dipole moment and polarization?" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33492" PostTypeId="1" CreationDate="2015-06-30T17:52:46.457" Score="0" ViewCount="1415" Body="&lt;p&gt;From the Amagat's curve we can see two types of result for real gases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If we increase pressure,the value of $PV$ decreases initially and then the value starts increasing for the gases like- carbon dioxide, nitrogen, oxygen, methane, etc..&lt;/li&gt;&#xA;&lt;li&gt;If we increase pressure,the value of $PV$ starts increasing initially for the gases like - hydrogen, helium, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now please explain what're the reasons behind this kinda behaviour of real gases?&lt;/p&gt;&#xA;" OwnerUserId="16919" LastEditorUserId="16919" LastEditDate="2015-06-30T17:57:29.173" LastActivityDate="2015-06-30T18:30:46.153" Title="Deviation of real gases from ideal behaviour" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33493" PostTypeId="2" ParentId="33490" CreationDate="2015-06-30T17:55:01.957" Score="13" Body="&lt;h2&gt;Diesel engines vs. gasoline engines&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Diesel engines don't rely on spark plugs, but they still work by igniting the fuel to generate a force that moves a cylinder in the engine.  The air/fuel mixture in a diesel engine is ignited by the compression of the same cylinder during the &quot;compression&quot; stroke.  That is, like most gasoline engines in cars, diesel engines are &lt;a href=&quot;https://en.wikipedia.org/wiki/Four-stroke_engine&quot; rel=&quot;noreferrer&quot;&gt;four-stroke engines&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compression stroke, in a gasoline engine or in a diesel engine, causes heating of the cylinder contents as the compression happens.  Diesel engines are designed to have a more severe compression of the cylinder contents during the compression stroke.  This results in much higher heating of the contents; in fact the temperature becomes so high that the ignition of the fuel and air becomes spontaneous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask, where does the energy to power the compression stroke come from?  It comes from inertia in the engine as it is running.  Another way to say it is that the previous power stroke provides the energy for the compression stroke.  (The power stroke is when the exploding air fuel mixture pushes the cylinder out, doing useful work.)  There is so much work released in the power stroke that there is enough to move the car forward as well as provide inertia for the next compression stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some more web links with helpful explanations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Diesel_engine&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The diesel engine (also known as a compression-ignition or 'CI' engine) is an internal combustion engine in which ignition of the fuel that has been injected into the combustion chamber is initiated by the high temperature which a gas achieves when greatly compressed (adiabatic compression). This contrasts with spark-ignition engines such as a petrol engine (gasoline engine) or gas engine (using a gaseous fuel as opposed to gasoline), which use a spark plug to ignite an air-fuel mixture.&#xA;  The diesel engine has the highest thermal efficiency of any standard internal or external combustion engine due to its very high compression ratio and inherent lean burn which enables heat dissipation by the excess air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://auto.howstuffworks.com/diesel1.htm&quot; rel=&quot;noreferrer&quot;&gt;How stuff works&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In theory, diesel engines and gasoline engines are quite similar. They are both internal combus­tion engines designed to convert the chemical energy available in fuel into mechanical energy. This mechanical energy moves pistons up and down inside cylinders. The pistons are connected to a crankshaft, and the up-and-down motion of the pistons, known as linear motion, creates the rotary motion needed to turn the wheels of a car forward.&#xA;  Both diesel engines and gasoline engines covert fuel into energy through a series of small explosions or combustions. The major difference between diesel and gasoline is the way these explosions happen. In a gasoline engine, fuel is mixed with air, compressed by pistons and ignited by sparks from spark plugs. In a diesel engine, however, the air is compressed first, and then the fuel is injected. Because air heats up when it's compressed, the fuel ignites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Be sure to watch the animation at How Stuff Works.  It's quite helpful!&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Diesel fuel vs. gasoline&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;In diesel engines, premature combustion can be a problem.  This happens when the fuel ignites &quot;too early&quot; in the compression phase, before maximum temperature and thus maximum efficiency can be obtained.  This means that by design, diesel fuel is a bit harder to ignite than gasoline.  It's made of slightly less volatile hydrocarbons.&lt;/s&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @steveverrill in the comments, the above paragraph was incorrect.  A better explanation (straight from his comment) is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diesel is harder to ignite at ambient temp due to its lower volatility / higher flash point (approx 52 &amp;deg;C vs -43 &amp;deg;C). This means it's difficult to form an explosive mixture which could be ignited by a spark (which is a local area at much higher than ambient temp.) But inside an engine both diesel &amp;amp; gasoline are fully evaporated. Under these conditions diesel is easier to ignite due to its lower autoignition temperature (approx 265 &amp;deg;C vs 280 &amp;deg;C). Autoignition temperature is tuned by additives to help ensure diesel can be ignited just by compression, whereas gasoline cannot ignite without a spark.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Diesel_fuel&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia again&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including $n$, $iso$, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\ce{C12H23}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gasoline&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia on gasoline&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bulk of a typical gasoline consists of hydrocarbons with between 4 and 12 carbon atoms per molecule (commonly referred to as C4-C12). It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, diesel fuel contains mostly hydrocarbon molecules that have on average 12 or so carbon atoms, while gasoline contains hydrocarbons that contain on average 8 or so carbon atoms.  The heavier, bigger molecules in diesel fuel means it evaporates more slowly at ambient conditions, requires more heating (i.e. more compression) before it spontaneously ignites.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:03:58.913" LastActivityDate="2016-03-11T23:03:58.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33494" PostTypeId="2" ParentId="33490" CreationDate="2015-06-30T17:57:19.690" Score="7" Body="&lt;p&gt;&lt;strong&gt;Diesel fuel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between gasoline and diesel fuel lies in &lt;a href=&quot;http://auto.howstuffworks.com/fuel-efficiency/alternative-fuels/question1051.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the length of carbon chains&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gasoline and diesel fuel are examples of &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrocarbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrocarbons&lt;/a&gt;, molecules made up purely of hydrogen ($\ce{H}$) and carbon ($\ce{C}$) atoms. Simple examples include $\ce{C_3H_8}$ (propane) and $\ce{C_8H_18}$ (octane).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gasoline contains compounds with carbon chains containing a certain amount of carbon atoms; their formulas are typically in the range of $\ce{C_7H_16}$ to $\ce{C_11H_24}$. Diesel fuel, on the other hand, has carbon chains somewhere in the area of $\ce{C_13}$ to $\ce{C_18}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order for some set of atoms to qualify as a chain, it must have a certain structure. You could not somehow make random hoops of carbon and call them carbon chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Diesel engines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you read &lt;a href=&quot;http://en.wikipedia.org/wiki/Diesel_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article&lt;/a&gt; on diesel engines. It explains that compression is used to ignite the fuel, rather than a traditional spark (from a spark plug).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few different reasons why diesel fuel is better than gasoline in diesel engines. One major one (&lt;a href=&quot;https://engineering.stackexchange.com/questions/1903/what-happens-when-you-put-the-wrong-type-of-fuel-in-an-internal-combustion-engin/1948#1948&quot;&gt;referring to Engineering Stack Exchange&lt;/a&gt;) is that diesel has a lower autoignition temperature - which is perfect, because diesel engines don't have spark plugs like those used in gasoline engines.&lt;/p&gt;&#xA;" OwnerUserId="13294" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.913" LastActivityDate="2015-06-30T18:06:48.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33495" PostTypeId="1" CreationDate="2015-06-30T18:06:56.723" Score="4" ViewCount="494" Body="&lt;p&gt;Are there any books about MO theory that explain the theory to a certain detail but don't rely on a lot of math background or explain the math within the book?&lt;/p&gt;&#xA;" OwnerUserId="17230" LastEditorUserId="10185" LastEditDate="2015-06-30T18:18:28.487" LastActivityDate="2015-07-08T14:15:20.477" Title="What are good books about MO theory that don't require a lot of math background?" Tags="&lt;molecular-orbital-theory&gt;&lt;reference-request&gt;&lt;books&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33496" PostTypeId="2" ParentId="33492" CreationDate="2015-06-30T18:30:46.153" Score="1" Body="&lt;p&gt;The ideal gas law can be derived using kinetic molecular theory by making two very important assumptions that are not true for real gases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no attractive force between gas molecules&lt;/li&gt;&#xA;&lt;li&gt;The gas molecules have no volume&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Think about how the PV curve would change when you remove one or both of these conditions. Then think about which types of molecules the assumption of &quot;no attractive force&quot; would be more important for.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-06-30T18:30:46.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33497" PostTypeId="1" AcceptedAnswerId="33529" CreationDate="2015-06-30T19:13:26.183" Score="20" ViewCount="7534" Body="&lt;p&gt;I would like to know if this molecule has ever been synthesized, if not why not, and what would its IUPAC name be?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLv14.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="17231" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-12-19T20:01:11.793" Title="A monocyclic 6 carbon ring with 6 double bonds" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33498" PostTypeId="2" ParentId="33497" CreationDate="2015-06-30T19:24:19.587" Score="11" Body="&lt;p&gt;This is a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclocarbon&quot;&gt;cyclocarbon&lt;/a&gt;, specifically &lt;a href=&quot;http://en.wikipedia.org/wiki/Benzotriyne&quot;&gt;Benzotriyne&lt;/a&gt;. Processes using it or synthesizing it have been described in &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1996/P1/p19960001535#!divAbstract&quot;&gt;Towards the total synthesis of cyclo[n]carbons and the generation of cyclo[6]carbon&lt;/a&gt;, among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has not yet been successfully synthesized. Wikipedia states (without citations) that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Recent investigations have concluded that benzotriyne is unlikely to exist due to the large angle strain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13294" LastActivityDate="2015-06-30T19:24:19.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33499" PostTypeId="2" ParentId="33352" CreationDate="2015-06-30T19:51:33.677" Score="5" Body="&lt;p&gt;You may wish to refer to the publication of Saykally &lt;em&gt;et al&lt;/em&gt;., &lt;a href=&quot;http://www.cchem.berkeley.edu/rjsgrp/publications/papers/2014/397_Lam.pdf&quot; rel=&quot;nofollow&quot;&gt;The hydration structure of aqueous carbonic acid from X-ray&#xA;absorption spectroscopy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, to determine hydration number, X-ray spectroscopy was used. This relates to the short existence of carbonic acid as, since it is so short lived, it is necessary to analyze it in such a time frame. As such, this was &quot;the first measurement of X-ray absorption spectra for aqueous carbonic acid, enabled by the use of rapid-flow liquid microjet technology&quot; which allowed it to be captured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by analyzing the spectra, it is possible to calculate the approximate hydration number, as &quot;the increasing H-bond strength is proportional to the hydration&#xA;number (measured for distances up to $2.5\mathrm{\overset{\circ}A}$): $\ce{Na2CO3}$ (5.55), $\ce{HCO3}$&#xA;(4.26), $\ce{H2CO3}$ (3.17)&quot;.&lt;/p&gt;&#xA;" OwnerUserId="14975" LastEditorUserId="7448" LastEditDate="2015-07-03T11:33:45.260" LastActivityDate="2015-07-03T11:33:45.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33500" PostTypeId="1" AcceptedAnswerId="33512" CreationDate="2015-06-30T19:52:59.133" Score="6" ViewCount="2104" Body="&lt;p&gt;I've been reading about MO theory and am wondering if the traditional model of carboxylic acids is correct. Usually, how it's presented is that the carbonyl carbon is doubly bonded to one oxygen and singly bonded to a hydroxyl group. This would mean that one oxygen is $sp^3$ hybridization and the other is $sp^2$ hybridized, and the hydrogen is attracted to both (one through a covalent bond and the other through a hydrogen bond). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book (Ansyln and Doughtery) says that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hybridization isn't just limited to $sp^2$ and $sp^3$ and that bonding isn't localized between only two atoms but between the molecule as a whole. (paraphrased) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does this affect carboxylic acids? Is there actually clearly defined '$sp^2$' and '$sp^3$' oxygens, or are they chemically equivalent? It seems like carboxylic acids react as alcohols (see: thionyl chloride). But does some sort of spontaneous symmetry breaking take place when a reactant approaches? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find anything detailed/reputable about this on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/29168/&quot;&gt;What is the hybridization of the carbonyl oxygen in a carboxylic acid?&lt;/a&gt; does not address my concern.  I get what the classical hybridization of the carbonyl oxygen in a carboxylic acid is. I'm just wondering if there's a better non-classical description where the oxygen atoms are equivalent until approached by another molecule or something of the sort.&lt;/p&gt;&#xA;" OwnerUserId="17232" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-07-01T08:04:14.183" Title="Difference between oxygens in a carboxylic acid?" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33501" PostTypeId="2" ParentId="33491" CreationDate="2015-06-30T21:14:43.123" Score="8" Body="&lt;p&gt;In fact, the first equation is applicable on a microscopic scale  (for an atom or a molecule):&#xA;  $\space$ $\vec{p}_\text{ind}=\alpha \vec{E}_\text{local} $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the second one is applicable on a macroscopic scale (for the bulk):&#xA;$\space$ $\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}_\text{ext}(\omega) $ &#xA;So, the polarisability is a microscopic quantity, while the susceptibility is a macroscopic quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to notice that the electric field in the first equation is the local field, while it's the external field in the second one.&#xA;For more details, please see this &lt;a href=&quot;http://www.physicspages.com/2012/11/28/relation-between-polarizability-and-susceptibility/&quot; rel=&quot;nofollow noreferrer&quot;&gt;page&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="41328" LastEditDate="2018-01-23T16:28:59.483" LastActivityDate="2018-01-23T16:28:59.483" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33504" PostTypeId="2" ParentId="33470" CreationDate="2015-06-30T22:05:10.600" Score="2" Body="&lt;p&gt;I was able to locate what appears to be a useful summary of reactions and reduction potentials that are all titanium-specific &lt;a href=&quot;http://objetoseducacionais2.mec.gov.br/bitstream/handle/mec/10496/22.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction &lt;span class=&quot;math-container&quot;&gt;$\ce{ Ti + O2 -&amp;gt; TiO2}$&lt;/span&gt; is illustrative of &lt;span class=&quot;math-container&quot;&gt;$\ce{Ti}$&lt;/span&gt; being oxidized while oxygen is reduced. I agree with you that it is simplistic-looking and might not appear as traditional an example of redox chemistry as you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might investigate the high-temperature reaction of titanium metal and water (steam), which produces &lt;span class=&quot;math-container&quot;&gt;$\ce {TiO2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; gas (titanium is oxidized, hydrogen is reduced), which might be a little clearer. This reaction (and several others) are given on the page I've linked above.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="83758" LastEditDate="2020-04-22T13:10:19.343" LastActivityDate="2020-04-22T13:10:19.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33506" PostTypeId="1" CreationDate="2015-07-01T00:31:20.930" Score="2" ViewCount="525" Body="&lt;p&gt;I found a formula for the &lt;strong&gt;degree of unsaturation (DU)&lt;/strong&gt; in &lt;strong&gt;Clayden’s Organic Chemistry&lt;/strong&gt;. It explains, simply, the unsaturation with the difference in H atoms. Then I stumbled upon the &lt;a href=&quot;https://en.wikipedia.org/wiki/Degree_of_unsaturation&quot; rel=&quot;nofollow&quot;&gt;general formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$DU=1+\frac12\sum n_i\left(v_i-2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_i$ is the number of atoms with valence $v_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain from where this equation was derived? And how come the valence term was introduced?&lt;/p&gt;&#xA;" OwnerUserId="17237" LastEditorUserId="2558" LastEditDate="2015-07-01T08:59:11.913" LastActivityDate="2015-07-01T10:47:35.567" Title="The proper explanation for the general formula of the DU?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;bond&gt;&lt;atoms&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-01T13:38:41.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="33507" PostTypeId="1" CreationDate="2015-07-01T00:46:30.313" Score="3" ViewCount="268" Body="&lt;p&gt;I wanted to know if it would be possible to combine basic silicon polymers with PTFE in order to create a stable, non reactive silicon rubber that won't react to butane or other hydrocarbons in particular. Is this doable? If so what would it take to accomplish this?&lt;/p&gt;&#xA;" OwnerUserId="17238" LastEditorUserId="4231" LastEditDate="2015-07-01T00:59:02.123" LastActivityDate="2015-07-01T00:59:02.123" Title="PTFE (Teflon) combined with Silicon Polymers" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33510" PostTypeId="1" CreationDate="2015-07-01T05:56:11.470" Score="1" ViewCount="169" Body="&lt;p&gt;Is the pH of a solution (orange juice) affected by the presence of a metal (aluminium, tin or iron for example)? I know that the metal will corrode in the presence of an acid (citric acid in this case), but will the metal have any effect on the actual acidity of the solution?&lt;/p&gt;&#xA;" OwnerUserId="17243" LastActivityDate="2015-07-01T05:56:11.470" Title="Is the pH of a solution (orange juice) affected by the presence of a metal?" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;metal&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33512" PostTypeId="2" ParentId="33500" CreationDate="2015-07-01T08:04:14.183" Score="6" Body="&lt;p&gt;First of all I would like to put my usual disclaimer about hybridisation: It is just a mathematical concept, nothing too real and it should always taken with a grain of salt. In the &lt;a href=&quot;https://chemistry.stackexchange.com/q/29168/4945&quot;&gt;linked post&lt;/a&gt; I have already discussed certain aspects of that. Furthermore it has been discussed on this site in various occasions.&lt;br&gt;&#xA;Going a bit further, especially for delocalised systems, the hybridisation picture starts to break down. In acetic acid (or any other carbon acids) this is not yet a big problem and we can still argue a lot with the involved concept. But I hope the following explanations will demonstrate some of this break down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's get back to the problem at hand. Here is the structure of acetic acid in the gas phase at 0 K, calculated with Gaussian09 at the DF-BP86/def2-SVP level of theory.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wZQnf.jpg&quot; alt=&quot;molecular structure of acetic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the carbon oxygen bonds are not equal at all. In the Lewis picture, we would say there is a single bond between carbon 2 and oxygen 1, which is a somewhat valid approximation. On the other hand we would say that between carbon 2 and oxygen 2 there is a double bond, which is also a somewhat valid approximation. In the molecular orbital picture we have three &amp;pi;-type orbitals (10, 13, 15). Therefore we have partial double bond character in the carboxyl unit due to delocalistion. Note that there is also an hyperconjugation effect involved. The main difference, however lies in the the molecular orbital 15, which is bonding with respect to C2-O2 and antibonding with respect to C2-O1. This orbital is basically the key, why the Lewis structure is a valid approximation.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/BUJgv.jpg&quot; alt=&quot;molecular orbital of acetic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can unitary transform these orbitals to localised orbitals, therefore representing the bonding structure with hybrid orbitals. I am using the Natural Bond Orbital theory for this. The following summary gives the bonds of the carboxyl subunit, the numbers in brackets refer to the orbital number in the picture. Also note, that since these orbitals are no eigenfunctions of the Hamiltonian and they do not have an energy eigenvalue, hence there is no canonical order.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-----------------------------------------------------------&#xA;#  (Occupancy)  (Type)    (orbital)  || Contributions below&#xA;-----------------------------------------------------------&#xA;1. (1.99578)    BD ( 1)   C2 - O1   [6]&#xA;            ( 30.48%) C(2) s( 26.60%)  p 2.75( 73.10%)&#xA;            ( 69.52%) O(1) s( 35.03%)  p 1.85( 64.91%)&#xA;&#xA;2. (1.99728)    BD ( 1)   C2 - O2   [5]&#xA;            ( 33.43%) C2   s( 33.28%)  p 2.00( 66.51%)&#xA;            ( 66.57%) O2   s( 42.68%)  p 1.34( 57.24%)&#xA;&#xA;3. (1.99005)    BD ( 2)   C2 - O2   [14]&#xA;            ( 31.97%) C2   s(  0.00%)  p 1.00( 99.64%)&#xA;            ( 68.03%) O2   s(  0.00%)  p 1.00( 99.87%)&#xA;&#xA;4. (1.98518)    BD ( 1)   O1 - H4   [7]&#xA;            ( 74.58%) O1   s( 19.52%)  p 4.12( 80.41%)&#xA;            ( 25.42%) H4   s( 99.60%)&#xA;&#xA;5. (1.80107)    BD ( 2)   O1 - H4   [15]&#xA;            ( 99.93%) O1   s(  0.00%)  p 1.00( 99.93%)&#xA;            (  0.07%) H4   s(  0.00%)  p 1.00(100.00%)&#xA;&#xA;6. (1.97613)    LP ( 1)   O1        [10]&#xA;                           s( 45.45%)  p 1.20( 54.52%)&#xA;&#xA;7. (1.97395)    LP ( 1)   O2        [8]&#xA;                           s( 57.37%)p 0.74( 42.62%)&#xA;&#xA;8. (1.83220)    LP ( 2)   O2        [16]&#xA;                           s(  0.00%)p 1.00( 99.91%)&#xA;------------------------------------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The total compliance with the Lewis bonding picture is 98.3%. Note the artefact in line 5, which should be a lone pair of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we can see, that O1 is best described with an sp&lt;sup&gt;1.85&lt;/sup&gt; contribution towards carbon, an sp&lt;sup&gt;4.12&lt;/sup&gt; contribution towards hydrogen,&#xA;an sp&lt;sup&gt;1.20&lt;/sup&gt; and a p lone pair.&lt;br&gt;&#xA;The O2 is best described with an sp&lt;sup&gt;2&lt;/sup&gt; contribution towards carbon, an &amp;pi; orbital towards carbon, an sp&lt;sup&gt;0.74&lt;/sup&gt; and a p lone pair.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/L3Fgl.jpg&quot; alt=&quot;molecular orbitals of acetic acid in the nbo picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the intramolecular hydrogen bond, one can analyse the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). This shows that there is no bonding interaction between H4 and O2.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2e51D.jpg&quot; alt=&quot;aim plot of acetic acid&quot;&gt;&lt;br&gt;&#xA;(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho&amp;lt;0$, dashed blue lines indicate charge accumulation $\nabla^2\rho&amp;gt;0$, red spheres are bond critical points, black lines are bond paths, dark red lines are zero flux surfaces. Hydrogens H2 and H3 omitted for clarity.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue that there is an electrostatic interaction, causing the slightly smaller (105&amp;deg;) CHO angle, but since the H4-O2 distance is only slightly shorter than the sum of the van der Waals radii (H = 1.20 A, O = 1.52 A) in vacuum this might already be a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solution there are many more factors to be considered, i.e. dimerisation, protonation/ deprotonation, intermolecular hydrogen bonds, and possibly many more.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-01T08:04:14.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33515" PostTypeId="2" ParentId="33506" CreationDate="2015-07-01T10:47:35.567" Score="4" Body="&lt;p&gt;My answer is based on &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ed060p626&quot; rel=&quot;nofollow noreferrer&quot;&gt;V. Pellegrin, &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1983&lt;/strong&gt;, &lt;em&gt;60&lt;/em&gt; (8), 626&lt;/a&gt; which might be inaccessible for you by a pay wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_characteristic#Planar_graphs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eulers characteristic for planar graphs&lt;/a&gt; wich states that adding the faces and vertices of a planar graph and substracting all the edges must be 1 $$F+V-E=1$$ which can be rewritten as $$F=1+E-V$$&#xA;Now in more &lt;em&gt;chemical words&lt;/em&gt; we can redefine this by saying that &lt;strong&gt;edges are bonds&lt;/strong&gt; and &lt;strong&gt;vertices are atoms&lt;/strong&gt; and the $F$ will be our $DU$ from now on. If we know the structural formula we could stop now, count all atoms and bonds and calculate what we want to know. Those are two examples from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6FLuL.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;But what if all we have is a sum formula from let's say mass spectrometry?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have to calculate how many bonds this molecule will have based on the knowledge of how many atoms the molecule contains and how many valence electrons each atom has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each molecule consists of $$N=\sum_i n_i$$ atoms of element $i$ whereby $n_i$ denotes the amount of each element. Each atom has to be assumed in it's lowest valency state and has $v_i$ valence electrons which would make in total $$\sum_{i=1}^N n_i~v_i$$ valence electrons. As every single bond contains two electrons, the amount of all possible bonds is half the amount of all valence electrons, namely $$\sum_{i=1}^N \frac{n_i~v_i}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try it with the mass formula $\ce{C9H7NO}$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The molecule consists of $N=18$ atoms.&lt;/li&gt;&#xA;&lt;li&gt;The atoms are grouped in $4$ groups containing the elements $i=\{C,H,N,O\}$&lt;/li&gt;&#xA;&lt;li&gt;Each element is contained $n=\{9,7,1,1\}$ times.&lt;/li&gt;&#xA;&lt;li&gt;The lowest valency of each contained atom is $v=\{4,1,3,2\}$&lt;/li&gt;&#xA;&lt;li&gt;The sum of all valence electrons is $\sum_{i=1}^N n_i v_i=9\cdot4+7\cdot1+1\cdot3+1\cdot2=48$&lt;/li&gt;&#xA;&lt;li&gt;The sum of all possible bonds is $\sum_{i=1}^N \frac{n_i~v_i}{2}=48/2=24$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now let's insert those numbers into the equation $$DU=1+E-V=1+24-18=7$$ Because I knew the moleule that I have chosen we can compare&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hZQp8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and see that the calculation was correct ($5$ for the double bonds + $2$ for the ring closures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it was correct, we can combine it and derive your equation.&#xA;$$\begin{align}&#xA;F=&amp;amp;~1+E-V\\&#xA;DU=&amp;amp;~1+\text{number of bonds}-\text{number of atoms}\\&#xA;DU=&amp;amp;~1+\sum_{i=1}^N \frac{n_i~v_i}{2}-\sum_{i=1}^N n_i\\&#xA;DU=&amp;amp;~1+\sum_{i=1}^N \left(\frac{n_i~v_i}{2}-n_i\right)\\&#xA;DU=&amp;amp;~1+\frac{1}{2}\sum_{i=1}^N (n_i~v_i-2~n_i)\\&#xA;DU=&amp;amp;~1+\frac{1}{2}\sum_{i=1}^N n_i~(v_i-2)\\&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-01T10:47:35.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33517" PostTypeId="2" ParentId="19868" CreationDate="2015-07-01T11:16:12.713" Score="8" Body="&lt;p&gt;Both names, ‘benzenecarbaldehyde’ and ‘benzaldehyde’, are unambiguous and describe the same compound. However, the preferred IUPAC name (PIN) is ‘benzaldehyde’.&lt;/p&gt;&#xA;&lt;p&gt;Concerning aldehydes, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-66.6.1&lt;/strong&gt; Systematic names of aldehydes&lt;/p&gt;&#xA;&lt;p&gt;Aldehydes are systematically named in three ways:&lt;/p&gt;&#xA;&lt;p&gt;(1) substitutively, using the suffixes ‘al’ for &lt;span class=&quot;math-container&quot;&gt;$\ce{-(C)HO}$&lt;/span&gt; and ‘carbaldehyde’ for &lt;span class=&quot;math-container&quot;&gt;$\ce{-CHO}$&lt;/span&gt;;&lt;/p&gt;&#xA;&lt;p&gt;(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;&lt;/p&gt;&#xA;&lt;p&gt;(3) by using the prefixes ‘oxo’, denoting &lt;span class=&quot;math-container&quot;&gt;$\ce{=O}$&lt;/span&gt;, or ‘formyl-’, denoting the substituent group &lt;span class=&quot;math-container&quot;&gt;$\ce{-CHO}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In particular,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-66.6.1.1.3&lt;/strong&gt; The suffix ‘carbaldehyde’ is used when the &lt;span class=&quot;math-container&quot;&gt;$\ce{-CHO}$&lt;/span&gt; group is attached to a carbon atom of a ring or ring system, or to a heteroatom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For example, ‘cyclohexanecarbaldehyde’ is a PIN.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jPkTW.png&quot; alt=&quot;cyclohexanecarbaldehyde&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, according to &lt;strong&gt;P-66.6.1.2.1,&lt;/strong&gt; the PIN for the compound given in the question is ‘benzaldehyde’. (This name is analogous to the preferred name of the corresponding acid ‘benzoic acid’, see &lt;strong&gt;P-66.6.1&lt;/strong&gt; (2) above.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zqHT0.png&quot; alt=&quot;benzaldehyde&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Nevertheless, according to &lt;strong&gt;P-66.6.1.2.2,&lt;/strong&gt; the PINs for phthalaldehyde and terephthalaldehyde are the systematic names ‘benzene-1,2-dicarbaldehyde’ and ‘benzene-1,4-dicarbaldehyde’, respectively. (These names are analogous to the preferred names of the corresponding acids ‘benzene-1,2-dicarboxylic acid’ [not ‘phthalic acid’] and ‘benzene-1,4-dicarboxylic acid’ [not ‘terephthalic acid’], respectively.)&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZP1iJ.png&quot; alt=&quot;benzene-1,2-dicarbaldehyde&quot; /&gt;   &lt;img src=&quot;https://i.stack.imgur.com/nZrm3.png&quot; alt=&quot;benzene-1,4-dicarbaldehyde&quot; /&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-01T11:16:12.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33518" PostTypeId="1" AcceptedAnswerId="33528" CreationDate="2015-07-01T12:10:25.920" Score="4" ViewCount="886" Body="&lt;p&gt;from what I know, the elastic limit of a substance is the maximum amount of stress/force/pressure (a bit of confusion on my behalf here) a substance can withstand and still return to its original shape afterwards. my question is what factor(s) determine this limit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also, I have noticed that some plastics go opaque when under stress and have found the reason for this effect in  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15084/why-do-many-transparent-plastics-turn-opaque-when-deformed&quot;&gt;this question&lt;/a&gt; however, what I do not understand is why some plastics and other synthetic materials do this before the elastic limit (and turn clear upon shrinking) and others do this after the elastic limit of the substance (and the effect is permanent). why can this process of turning opaque be reversed in materials that turn opaque before the elastic limit of the substance.&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-07-01T19:59:22.410" Title="What determines the elastic limit of a substance?" Tags="&lt;crystal-structure&gt;&lt;materials&gt;&lt;polymers&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33522" PostTypeId="1" CreationDate="2015-07-01T15:45:10.650" Score="7" ViewCount="3373" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nJzG7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've spent a few days googling functional groups and reading my text assigned on organic functional groups. I'm understanding very little and I just need a little clarification. I understand that at the top right and bottom left, the hexagon structure signifies an aromatic, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top Left&lt;/strong&gt;: My first guess was carboxylic acid, but with further googling, I now think it's an amide, because it contains nitrogen and oxygen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top Right&lt;/strong&gt;: Since the hexagon is aromatic, I know it can only be the last three choices, but I'm going with C because the $\ce{-NH2}$ is not connected to an oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom Left&lt;/strong&gt;: Because of the $\ce{Br}$, I know it's a bromide and, assuming that my thinking is correct with aromatic, the answer should be B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the &lt;strong&gt;bottom right&lt;/strong&gt;, I am not sure what exactly to look at here to even guess, besides that it is not an aromatic ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that I've tried the last one, all I can find is that an alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$ and, the aromatic being the hexagon structure, I'm unable to grasp/understand which one the last group would fall into.&lt;/p&gt;&#xA;" OwnerUserId="17247" LastEditorUserId="5026" LastEditDate="2018-02-17T15:51:13.023" LastActivityDate="2020-01-31T10:12:49.027" Title="Identifying functional groups" Tags="&lt;organic-chemistry&gt;" AnswerCount="2" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="33524" PostTypeId="1" CreationDate="2015-07-01T16:13:38.997" Score="6" ViewCount="550" Body="&lt;p&gt;What chemicals are used in so-called urinal &quot;odor eaters&quot;? Do they actually destroy odors, or do they just mask them? Also, are these chemicals subject to any environmental laws or regulations?&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="7475" LastEditDate="2015-10-30T16:42:23.390" LastActivityDate="2015-10-30T16:42:23.390" Title="Chemicals in Urinal Odor Eaters" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33525" PostTypeId="1" CreationDate="2015-07-01T16:27:58.577" Score="5" ViewCount="1429" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In finding enthalpy change $\Delta H$ during adiabatic, reversible expansion of  mixture of two ideal gases $n_1$ mol of $\ce{A}$ ($C_{\mathrm{m},v}=\frac{3}{2}R$) and $n_2$ mol of $\ce{B}$ ($C_{\mathrm{m},v}=3R$) taken in a container and during this temperature change is $T_\mathrm{i}$ to $T_\mathrm{f}$. Then what should be the value of $\Delta H$? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be $$\Delta H=\left(n_1\right)\frac{5}{2}R\left(T_\mathrm{f}-T_\mathrm{i}\right)+\left(n_2\right)4R\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ taking individully for each gas. Or let say average molar heat capacity of gaseous mixture at constant pressure be $C_{\mathrm{m},p,\mathrm{avg}}$ then it can be $$\Delta H=\left(n_1+n_2\right)C_{\mathrm{m},p,\mathrm{avg}}\left(T_\mathrm{f}-T_\mathrm{i}\right)$$ so please help which is correct one. As per suggestion both are same.&#xA;$$C_{\mathrm{m},p,\mathrm{avg}}=\frac{\left(n_1\right)\frac{5}{2}R+\left(n_2\right)4R}{n_1+n_2}$$&lt;/p&gt;&#xA;" OwnerUserId="17108" LastEditorUserId="4945" LastEditDate="2019-07-11T16:08:22.920" LastActivityDate="2019-07-11T16:09:32.233" Title="Average or individual molar heat capacity?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33526" PostTypeId="2" ParentId="33524" CreationDate="2015-07-01T16:39:52.453" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What chemicals are used in so-called urinal &quot;odor eaters&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://complyplus.grainger.com/grainger/msds.asp?sheetid=4072417&quot;&gt;1,4-dichlorobenzene-based&lt;/a&gt; and &lt;a href=&quot;http://complyplus.grainger.com/grainger/msds.asp?sheetid=4072421&quot;&gt;non-PDCB&lt;/a&gt; products. In the latter, the chemical mixture from the linked MSDS is:&#xA;SODIUM SULPHATE, SODIUM DODECYL SULPHATE, COCAMIDE MEA, DIPENTENE, CITRAL, LINALOOL,BENZYL ALCOHOL&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do they actually destroy odors, or do they just mask them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They &lt;a href=&quot;https://en.wikipedia.org/wiki/Urinal_deodorizer_block&quot;&gt;mask&lt;/a&gt; odors.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, are these chemicals subject to any environmental laws or regulations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, as detailed in the MSDS sheets linked above. Please note that my knowledge of urinal cakes and related products is not exhaustive.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-01T18:09:44.443" LastActivityDate="2015-07-01T18:09:44.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33527" PostTypeId="1" AcceptedAnswerId="33530" CreationDate="2015-07-01T19:32:20.263" Score="4" ViewCount="14788" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;: I have a journal from my childhood. I applied glue (uhu brand. &lt;a href=&quot;http://www.uhu.com/uploads/tx_uhuproducts/46015_Detail_01.png&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;) to the papers almost 10 years ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q&lt;/strong&gt;: Is there a way to dissolve the glue and make the pages readable?&lt;/p&gt;&#xA;" OwnerUserId="17252" LastEditorUserId="15416" LastEditDate="2015-07-02T06:37:20.450" LastActivityDate="2015-07-02T06:37:20.450" Title="Is there a way to dissolve glue?" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33528" PostTypeId="2" ParentId="33518" CreationDate="2015-07-01T19:58:17.133" Score="3" Body="&lt;p&gt;First let me clarify that there is no way to &lt;em&gt;calculate&lt;/em&gt; the elastic limit of a substance - it has to be measured. &lt;a href=&quot;https://en.wikipedia.org/wiki/Viscoelasticity&quot; rel=&quot;nofollow&quot;&gt;Viscoelastic theory&lt;/a&gt; is pretty well-developed for only being about 150 years old, but it's not &lt;em&gt;that&lt;/em&gt; well developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, though, lots of factors affect elasticity in general and the elastic limit in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would probably be the most helpful to think about what is happening on the molecular scale. In any viscoelastic material, there is a range of deformations (strain, or the response to stress) and timescales over which molecules will go back to pretty much the same relative position that they were in before the deformation occurred. That's elasticity. What makes this happen? Each atom or molecule sits in a little potential energy well that exists because of its position relative to other atoms in the substance. Atoms like to be close to each other, but not &lt;em&gt;too&lt;/em&gt; close. As a result, when given the opportunity they will arrange themselves so that as many atoms as possible are the optimal distance apart(resulting in the lowest energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a material is deformed, we are moving atoms relative to each other. When one atom moves, its energy increases. This also changes the potential energy surface seen by its neighbors. If it moves a little bit, then the change is not usually great enough to cause the neighbors to move by much. On the other hand, for big movements, the effect can be great enough that the other nearby atoms rearrange themselves in order to find the next lowest energy level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the relative motion of atoms due to mechanical deformation is small enough that most of them go back to their original positions once the applied stress is removed, then we say the stress/strain was within the elastic limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the atoms are permanently rearranged, the shape of the material is changed and we say that we have exceeded the elastic limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ease with which atoms can move out of position (you can think of this like the depth of the wells), and their tendency to move back to the lowest energy (which can be thought of as the &quot;steepness&quot; of the wells) are the two main factors that affect how much stress/strain is needed to permanently deform a material. There are lots of things that can affect each of these, which is why there is no comprehensive theoretical model that can predict the elastic limit of materials based on molecular structure alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can summarize a few important factors here, though:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Crystallinity&quot; rel=&quot;nofollow&quot;&gt;degree of crystallinity&lt;/a&gt; in polymeric materials&lt;/li&gt;&#xA;&lt;li&gt;type and number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Dislocation&quot; rel=&quot;nofollow&quot;&gt;dislocations&lt;/a&gt; in crystalline materials&lt;/li&gt;&#xA;&lt;li&gt;length of time that the stress/strain is applied, in conjunction with the temperature of the material (this related to a phenomenon called &lt;a href=&quot;https://en.wikipedia.org/wiki/Time%E2%80%93temperature_superposition&quot; rel=&quot;nofollow&quot;&gt;time-temperature superposition&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;structure of the molecules making up the material (e.g. linear vs. &lt;a href=&quot;https://en.wikipedia.org/wiki/Cross-link&quot; rel=&quot;nofollow&quot;&gt;crosslinked&lt;/a&gt; or branched polymers, fcc vs. bcc &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_structure&quot; rel=&quot;nofollow&quot;&gt;crystal structure&lt;/a&gt;, or type of atoms in an alloy)&lt;/li&gt;&#xA;&lt;li&gt;mesoscopic structure (e.g. crystalline grain size and shape, aggregation of nanoparticles, or long-range order in polymers)&lt;/li&gt;&#xA;&lt;li&gt;macroscopic structure (e.g. fiber orientation in &lt;a href=&quot;https://en.wikipedia.org/wiki/Composite_material&quot; rel=&quot;nofollow&quot;&gt;composite materials&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are many more! And each of the ones I listed is practically an entire field of its own in materials science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the second part of your question - the cloudiness you are referring to is called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Crazing&quot; rel=&quot;nofollow&quot;&gt;crazing&lt;/a&gt;&quot; in polymer science. It is basically caused by molecules pulling apart from each other, without covalent bonds breaking. It can be reversed when it happens before the elastic limit is reached because the molecules can still go back to their original positions. If it doesn't happen until &lt;em&gt;after&lt;/em&gt; the elastic limit, then the material is permanently deformed, and the molecules cannot return to their original position.&lt;/p&gt;&#xA;" OwnerUserId="5633" LastActivityDate="2015-07-01T19:58:17.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33529" PostTypeId="2" ParentId="33497" CreationDate="2015-07-01T21:35:24.543" Score="17" Body="&lt;p&gt;There are several monocyclic $\ce{C6}$ isomers.  Three of them are pictured below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDVxf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A - the compound you've drawn above, cyclohexahexaene&lt;/li&gt;&#xA;&lt;li&gt;B - cyclohexatriyne&lt;/li&gt;&#xA;&lt;li&gt;C - cyclohexa-2,5-diyne-bis-1,4-ylidene&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While all of the carbons in these molecules are (more or less) $\ce{sp}$ hybridized, the molecules are different from one another due to differences in bond lengths and bond angles.  For example, all of the bonds in A are exactly the same length due to symmetry, whereas the triple bonds in B will be shorter than the adjacent single bonds.  Therefore, A-C are isomers, &lt;strong&gt;not&lt;/strong&gt; resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally the atoms attached to an $\ce{sp}$ hybridized carbon form a linear arrangement (180° bond angle) with the central $\ce{sp}$ carbon, acetylene and the central carbon in allene serve as examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lWB8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemistry.tutorvista.com/organic-chemistry/alkyne.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nBxSI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.organicchemistry.com/cumulated-alkadienes/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the cyclic $\ce{C6}$ compounds A-C, the $\ce{sp}$ hybridized carbons are forced by the ring to deviate from the desired 180° arrangement to something down around 120°.  Therefore, quite a bit of angle strain is incorporated into A-C making them &lt;strong&gt;extremely&lt;/strong&gt; strained and unavailable through laboratory preparation to date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned above, A-C have different bond lengths and bond angles therefore they are isomers, not resonance structures.  A fascinating question is, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do A-C each sit in a potential well, separated from one another by an energy barrier and if so, which one is the most stable (energy minimum)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there another monocyclic species lower in energy than A-C? Perhaps a molecule that could be described by resonance structures A'-C', where the prime denotes that all of the bond lengths and angles have been adjusted and made equal in A, B and C.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-07-01T21:35:24.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33530" PostTypeId="2" ParentId="33527" CreationDate="2015-07-01T22:10:54.153" Score="6" Body="&lt;p&gt;You might try different solvents on a spot that does not have writing, to see if it removes the glue, and on glue with non-critical writing, to see if it &lt;em&gt;removes or smears the writing as well as the glue&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it described as a &lt;a href=&quot;https://en.wikipedia.org/wiki/UHU&quot;&gt;clear synthetic resin&lt;/a&gt;, you might try the following sequence, from milder solvent to stronger:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Comparatively mild non-polar organic solvents, such as paint thinner or kerosene. If that does not dissolve the glue, try&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some Polar solvents such as acetone, methyl ethyl ketone or nail-polish remover. If that does not remove the glue, try&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PVC pipe solvent (&lt;a href=&quot;https://en.wikipedia.org/wiki/Tetrahydrofuran&quot;&gt;tetrahydrofuran&lt;/a&gt;).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N.B. All solvents listed are flammable, and some are fairly toxic. Use them in &lt;em&gt;small quantities&lt;/em&gt;, closing the containers after getting a bit out, to prevent flash-back of flames, and use in well-ventilated areas.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-01T22:10:54.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33532" PostTypeId="1" AcceptedAnswerId="33539" CreationDate="2015-07-01T23:49:37.963" Score="13" ViewCount="548" Body="&lt;p&gt;I am quite familiar with Hartree Fock theory. However, writing it into program is an entirely different task, and that is where I have a couple of questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a Hartree-Fock routine in Python a couple of years ago. It involved a ultra-minimal basis: only one gaussian for two hydrogen atoms each. Kinetic energy term was pretty simple, so were the one-electron terms. Coulomb and exchange integrals were tricky, which involved evaluation of an erf. Now, my question is, if I included higher angular momentum basis functions (p,d,f,g type functions), what happens? Is it difficult to evaluate the J (Coulomb) and K (exchange) integrals because of the mathematical complexity? Or something else? Can someone give the math behind this and explain why it is difficult to evaluate?&lt;/p&gt;&#xA;" OwnerUserId="6986" LastEditorDisplayName="user23061" LastEditDate="2015-12-23T19:06:09.077" LastActivityDate="2015-12-23T19:06:09.077" Title="Complexity in computing the integrals involving higher angular momentum" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33533" PostTypeId="2" ParentId="33413" CreationDate="2015-07-02T00:28:46.380" Score="0" Body="&lt;p&gt;User permeakra's observation is right, but the inference is not. User Wildcat's answer is spot-on, but missed on one minor point, which I would explain at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a summary of Wildcat's explanation: The idea is that, we have a system comprising of indistinguishable fermions, which is why we need to use the antisymmetrized wavefunction. The LCAO-MO approximation results in a linear combination of AOs resulting in MOs, which can be allowed to occupy, given by electron densities. Variationally, the ground state would lead to the all the lowest levels being occupied, resulting in HF ground state wavefunction. The HOMO, which is $3s$ orbital, would be at a certain energy level which is definitely well defined for a given &lt;strong&gt;level of theory&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, yes, you can calculate the energy of the $3s$ orbital for $\ce{Na^{-}}$. But, what is not possible is to calculate the energy of &lt;strong&gt;the&lt;/strong&gt; $3s^{2}$ electron, because the electrons occupying the $3s$ (or any of the other) energy levels are indistinguishable. Another thing to note is that, when you talk of using a wavefunction to denote electrons, they are highly delocalized, and can occupy more than one energy level at once. Think about the case of Boron atom. Which orbital do you think the $2p$ electron occupies? The $2p_x$ or $2p_y$ or $2p_z$? They are all symmetric in nature, and hence you cannot point at one particular orbital. This turns into a multiconfigurational problem, which is beyond the scope of this discussion. Hence, the implication here is that, the electrons occupying these orbitals would be at specified energy levels, depending on which orbital they occupy.&lt;/p&gt;&#xA;" OwnerUserId="6986" LastActivityDate="2015-07-02T00:28:46.380" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33534" PostTypeId="1" CreationDate="2015-07-02T00:42:07.783" Score="3" ViewCount="316" Body="&lt;p&gt;In my research we have a method of making polymer open-cell foams with high tortuosity and pore sizes on the order of 50–200 nanometers. I can make thicknesses of 100 µm to 2 mm thick (and up to 5 inch diameter disk). I am looking for some cool things to do with this material, and one idea was to try to fill it with metal and then dissolve the polymer away to leave a nanoporous metal structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that so far I cannot get water to penetrate inside the pores, I tried up to 4 MPa pressure and it just won’t go through (small pores and hydrophobic) but acetone will get absorbed in very quickly. My question is, could there be a way to use acetone as an electrolyte and deposit metals (copper, nickel, whatever) to fill these open-cell pores?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven’t taken chemistry since intro chem back in undergrad so I really don’t know much of anything about electroplating, so I will be very grateful for any help on this experiment. Thank you very much, I hope I was clear in explaining what I am trying to do.&lt;/p&gt;&#xA;" OwnerUserId="17254" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T08:38:25.257" LastActivityDate="2015-07-02T08:38:25.257" Title="Electroplating metal inside polymer foam with acetone" Tags="&lt;electrochemistry&gt;&lt;metal&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33535" PostTypeId="2" ParentId="33534" CreationDate="2015-07-02T02:45:45.603" Score="1" Body="&lt;p&gt;Electroplating might be possible, but getting good coverage of internal pores is tricky, even with aqueous electrolytes, as deposition will likely occur preferentially at the outer edge of the workpiece, leaving the internal passages lightly- or un-coated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more effective approach is likely &lt;a href=&quot;https://en.wikipedia.org/wiki/Electroless_nickel_plating&quot; rel=&quot;nofollow&quot;&gt;electroless nickel deposition&lt;/a&gt;.  As long as you can find a suitable acetone-soluble reducing agent, it might work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also recommend testing other solvents for their ability to permeate the polymer structure -- if you could find a protic solvent that wets it (even something like 50/50 ethanol/water!), you'd have a much better chance of success at electrodeposition proper, and also at making electroless work for you, since (I think) most electroless recipes are aqueous-based.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-07-02T02:45:45.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33536" PostTypeId="2" ParentId="33394" CreationDate="2015-07-02T03:35:29.823" Score="3" Body="&lt;p&gt;See Shakashiri's chemical demonstrations. He carefully outlines the steps. As I recall it requires taking sodium nitrate into a molten state and applying a current. The elemental sodium is precipitated on the inside surface of an incandescent bulb.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-07-02T03:35:29.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33537" PostTypeId="1" AcceptedAnswerId="33572" CreationDate="2015-07-02T07:58:28.143" Score="8" ViewCount="582" Body="&lt;p&gt;Studying biochemistry I noticed that equilibrium is reached for K and Na separately (not taking into account electric potential), as would be the case I assume for any molecule or atom. This is of course entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg: My understanding is, you have two types of atoms, and a membrane separating them, with different concentrations on each side. Each atom will reach equilibrium with itself across the membrane, as if the other type of atom doesn't exist (barring the influence of charge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that, for instance, a K atom is so keen to reach equilibrium with another K atom in an area of lower concentration, and yet seemingly ignores the Na, even if the Na is in high concentration where the K is low? Both Na and K still have the same charge, and would also collide with each other I imagine? Why do identical atoms behave like this but non-identical ones with very similar properties don't?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update #2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding was incorrect regarding K and Na having separate forces. The two atoms seem to be treated the same, except for selective permeability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still not understanding why the two atoms have separate gradient potentials, if they can be treated as the same. I made the following images to illustrate my point a little clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a compartment with a left and right side, separated by a wall containing two selectively permeable channels that may open or close. Only Na can cross the green channel when it is open, only K can cross the orange channel when it is open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the assumption: The main assumption behind all this is that, in answering this question, we can treat Na and K as being the same. They are effectively two marbles with identical physical and chemical properties, just with different colours. This is what I have come to understand based upon the answer by orthocrestol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cell on the left has a -70mv potential energy compared to the right compartment, caused by invisible negative ions which will always remain on the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOVvw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that has been summarized by the discussed assumption is seen in figure 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in figure 2, we open the selective channel for Na. Since Na has a gradient potential, and there is also a potential difference, the positive Na flows into the left hand compartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure 3, an equilibrium is reached at an arbitrary value of -20mv. On average, no more Na will move to the left or the right, it's at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 4, channels closed. Now this is where my understanding ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nDHeP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently now, from what I gather, in figure 5 if we open the selective channel for orange (K+), since K+ has a gradient potential, it moves to the right. In the process, it takes its positive charge with it and restores more or less the original potential difference of -70mV on the left hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if this is what happens, this must surely invalidate the marble analogy of the two atoms being the same. Because if they're the same, then whether we open the Na channel (which is already at equilibrium) or the K channel should make no difference, because once gain, they're the same atom with a different colour (except for being separately permeable via their selective channels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference that should be apparent is that if we leave both channels open, then the colours will eventually reach an equilibrium on both sides. But if an equilibrium is already reached for the charge (which it has in figures 3 and 4), I'm failing to grasp how opening then closing a different channel suddenly causes that equilibrium to change (as in figure 5 and 6). The explanation is the K potential gradient, but this is still the same marble as Na is it not? How come the potential difference changes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please direct me where my thinking has gone wrong, or if this is what even happens. And if it is, how is this possible? Many thanks.&lt;/p&gt;&#xA;" OwnerUserId="4779" LastEditorUserId="4779" LastEditDate="2015-07-10T08:09:32.593" LastActivityDate="2016-09-15T23:28:21.390" Title="Diffusion and potential gradients for two similar atoms" Tags="&lt;biochemistry&gt;" AnswerCount="2" CommentCount="17" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33539" PostTypeId="2" ParentId="33532" CreationDate="2015-07-02T08:44:22.580" Score="8" Body="&lt;p&gt;It appears that the integrals that are difficult are the two-electron repulsion integrals. There exists a recursive relationship that can turn integrals with high-angular momentum functions into integrals with only s-type functions. Refer to &lt;a href=&quot;https://ia601001.us.archive.org/9/items/Justin_T_Fermann_and_Edward_F_Valeev__Lecture_notes_on_evaluation_of_molecular_integrals/ints.pdf&quot; rel=&quot;noreferrer&quot;&gt;this lecture&lt;/a&gt; for details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the following articles for details concerning these recurrence relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M. Head Gordon and J. Pople, &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;89&lt;/strong&gt;, 5777 (1988)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S. Obara and A. Saika, &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;84&lt;/strong&gt;, 3963 (1986)&lt;/p&gt;&#xA;" OwnerUserId="66" LastEditorDisplayName="user23061" LastEditDate="2015-12-18T15:56:55.680" LastActivityDate="2015-12-18T15:56:55.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33541" PostTypeId="1" AcceptedAnswerId="33543" CreationDate="2015-07-02T10:23:23.317" Score="3" ViewCount="2827" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nomenclature priorities are in order of the degree of oxidation of the&#xA;  carbon atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am not sure if it's advisable to just take it as it is and use it or try to understand the basis of the functional group prioritization. I guess the latter interest me. So looking at the priorities for functional groups, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alkene has higher priority than Alkyne and followed by Alkane. I started reading some materials about &lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/chapt9.htm&quot; rel=&quot;nofollow&quot;&gt;Oxidative Cleavage of Double Bonds and rest of the reactions&lt;/a&gt;, which kinda went way over my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain, how and why the carbon oxidation in Alkynes are lower than Alknenes? Or put in another word, why would Alkene carbon oxidate faster than Alknenes? &lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="275" LastEditDate="2015-07-02T11:34:19.133" LastActivityDate="2015-07-02T11:34:19.133" Title="Oxidation of Carbon atom in Alkene, Alkyne and Alkanes of functional groups?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33542" PostTypeId="2" ParentId="33522" CreationDate="2015-07-02T11:02:24.780" Score="6" Body="&lt;p&gt;Some pointers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the hexagon structure signifies an aromatic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite. The hexagon structure depicted with &lt;strong&gt;alternating carbon-carbon double and single bonds&lt;/strong&gt; is aromatic. It is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzene&quot; rel=&quot;noreferrer&quot;&gt;benzene&lt;/a&gt; and is very commonly encountered in organic chemistry. It can be represented by two resonance structures as shown below and is sometimes also represented with a circle inside a ring of singly bonded carbons to show the delocalised $\pi$ system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzFzC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the are other molecules depicted as hexagons which are not aromatic such as cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/suC3J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there are other molecules which are aromatic which are not benzene, and many which are not hexagonal but you probably won't come across these at an introductory level.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My first guess was carboxylic acid, but with further googling, I'm now thinking it's an amide, because it contains nitrogen and oxygen&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are correct that it is an amide. The amide functional group contains a nitrogen adjacent to a carbon-oxygen &lt;strong&gt;double&lt;/strong&gt; bond. The carboxylic acid group is similar but the nitrogen is replaced by an $\ce{-OH}$ group. Amides can be primary, secondary or tertiary depending on the number of carbons that are bonded to the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Ptzw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are the simplest example of alkanes, alkenes and alkynes. They are called methane, ethene and ethyne respectively. However, there are more complicated molecules involving these functional groups. In general, alkanes have only carbons-carbon &lt;strong&gt;single&lt;/strong&gt; bonds and carbon-hydrogen single bonds. Alkenes are distinguished by having a carbon-carbon &lt;strong&gt;double&lt;/strong&gt; bond denoted by two lines between the two carbons which look like an equals sign. Alkynes have a carbon-carbon &lt;strong&gt;triple&lt;/strong&gt; bond which is denoted by three lines between the carbons. Therefore, your example is an alkene.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-07-02T11:02:24.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33543" PostTypeId="2" ParentId="33541" CreationDate="2015-07-02T11:32:40.677" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Nomenclature priorities are in order of the degree of oxidation of the carbon atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This may be true for the heteroatom functional groups, but not for alkenes, alkynes, and alkanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that each $\ce{C-H}$ bond contributes $-1$ to the oxidation state and each $\ce{C-C}$ bon regardless of order contributes $0$ to the oxidation state of carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hydrocarbon   formula   functional group   oxidation number of C&#xA;ethane        CH3CH3    alkane             -3&#xA;ethene        CH2CH2    alkene             -2&#xA;ethyne        CHCH      alkyne             -1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Alkynes have a higher oxidation number than alkenes. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alkenes react better in oxidation reactions than alkynes because alkenes are less oxidized than alkynes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organic chemistry textbook that I teach from explains that alkenes have higher nomenclature priority than alkynes because alkenes are more common. I cannot find a more authoritative reason/source.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-02T11:32:40.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33546" PostTypeId="1" CreationDate="2015-07-02T13:03:52.920" Score="7" ViewCount="148" Body="&lt;p&gt;Nanotechnology and sciences are the study or the manipulation of matter on an atomic, molecular, and supramolecular scale especially at nano meter range. We know that at macro molecular level a system can be described using Newtonian/classical mechanics (eg: $F=ma$) whereas for an electronic level quantum mechanics are used. In nanotechnology and science which mechanics is seen predominantly; is it Newtonian, quantum or both and why?&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="5026" LastEditDate="2018-02-28T15:12:56.510" LastActivityDate="2018-02-28T15:12:56.510" Title="Which mechanics or law is predominately seen on the nanoscale?" Tags="&lt;quantum-chemistry&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33547" PostTypeId="1" CreationDate="2015-07-02T13:07:53.770" Score="8" ViewCount="2078" Body="&lt;p&gt;I think that the ginger ale is born this way. &#xA;Anyway, if you put the ginger in a glass and add sparkling water, small drops splash out of the glass, the water is fizzier, more effervescent. &#xA;Why? what makes the two of them react?&lt;/p&gt;&#xA;" OwnerUserId="17260" LastEditorUserId="10185" LastEditDate="2015-07-02T13:13:53.590" LastActivityDate="2017-04-11T21:52:17.833" Title="Why ginger reacts with sparkling water?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33548" PostTypeId="2" ParentId="33415" CreationDate="2015-07-02T13:43:08.103" Score="3" Body="&lt;p&gt;Your approach is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given compound includes the characteristic groups $\ce{-CN}$, which are expressed as suffix ‘carbonitrile’. Since there is only one kind of suffix, the order of seniority of suffixes is not relevant in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functional suffixes are always attached to the name of the parent structure, which is modified by ‘ene’ or ‘yne’ endings if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; stipulates various criteria for the selection of the principal chain. These criteria are applied, in the order listed, until a decision is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given compound, there is a choice between two different principal chains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctJtV.png&quot; width=&quot;151&quot; height=&quot;60&quot; alt=&quot;(1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile&quot;&gt;    and   &lt;img src=&quot;https://i.stack.imgur.com/d1O7s.png&quot; width=&quot;151&quot; height=&quot;60&quot; alt=&quot;(1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first relevant criterion in this case provides that the principal chain has the maximum number of substituents corresponding to the principal characteristic group (suffix).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.1.1&lt;/strong&gt; The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the principal chain of the given structure has all four substituents $\ce{-CN}$ corresponding to the suffix ‘carbonitrile’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dhkzo.png&quot; width=&quot;151&quot; height=&quot;60&quot; alt=&quot;(1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the order of seniority provides that low locants are given first to principal characteristic groups (here: carbonitrile) before low locants are given to multiple bonds (here: the double bond) and before low locants are given to detachable alphabetized prefixes (here: methyl).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) principal characteristic groups and free valences (suffixes);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) saturation/unsaturation:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the resulting name is (1​&lt;em&gt;E&lt;/em&gt;)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile (not (6​&lt;em&gt;E&lt;/em&gt;)-6-methylhept-6-ene-3,4,5,7-tetracarbonitrile) since the locant set ‘1,3,4,5’ for the principal characteristic groups is lower than ‘3,4,5,7’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nKEF2.png&quot; width=&quot;288&quot; height=&quot;85&quot; alt=&quot;(1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-01-14T19:50:14.537" LastActivityDate="2016-01-14T19:50:14.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33550" PostTypeId="2" ParentId="20105" CreationDate="2015-07-02T14:36:42.020" Score="-3" Body="&lt;p&gt;Pyro- is for when both phosphate are together, di- means the phosphates can occur anywhere in the molecule.&lt;/p&gt;&#xA;" OwnerUserId="17262" LastActivityDate="2015-07-02T14:36:42.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33551" PostTypeId="2" ParentId="5189" CreationDate="2015-07-02T15:30:09.013" Score="10" Body="&lt;p&gt;One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zA5On.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O &amp;lt;=&amp;gt; H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently acidic no carbonic acid can/will react to form hydrogen carbonate or even further to carbonate. But from $\ce{pH}\approx4$ the possibility that hydrogen carbonate will be formed rises. This means for your titration, that you are not only titrating your acid anymore but you are also titrating the dissolved $\ce{CO2}$. The longer you wait for equlibration near the the equlibrium point the more dissolved $\ce{CO2}$ reacts to hydrogen carbonate and distorts your ideal $\ce{CO2}$-free titration curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show you the impact more clearly I created the following figures showing both of your systems with varying carbonic acid concentrations $\left(0.00\ldots0.05~\ce{mol~L^{-1}}\right)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first your system with hydrochloric acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bgQZQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And secondly the system containing acetic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MS0Nh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two remarking differences between both titration curves.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can determine the equlibrium point of the $\ce{HCl}$-titration (though it's not around pH 7 but at pH 4-5) no matter how many hydrogen carbonate will be formed during your titration.&lt;/li&gt;&#xA;&lt;li&gt;You have to pay attention during the titration of acetic acid because the inflection points gets distorted above any detectable other inflection point. However it is not that important as it seems because titrations of acetic acid are fast and there is usually not much time to dissolve enough $\ce{CO2}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The other points are already mentioned before. Your pH electrode could not be calibrated correctly and your concentrations might not be as precise as you think they are. The titration curve of hydrochloric acid looks like your acid was less concentrated which results in a higher starting pH and in a lower ending pH. The titration of acetic acid indeed looks like the electrode was not properly calibrated because it looks quite good if you just raise the pH values per measuring point and it suffers from dissolved $\ce{CO2}$ which is why the inflection point is shifted to a higher volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mentioned increase of volume that is most times neglected during the calculation of titration curves has no big impact unless the concentration of the base is much lower than that of the acid that has to be titrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cWHpw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand the first four images show that different base concentrations around your ideal concentration don't effect the curve itself very much but only the needed volume to reach the point of inflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand you can see that there is a visible influence on the titration curve if the base is 10 times as diluted as the acid.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-02T15:30:09.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33552" PostTypeId="1" CreationDate="2015-07-02T16:28:33.873" Score="7" ViewCount="3078" Body="&lt;p&gt;The fluorescence quantum yield of a sample $\Phi$ is defined as the fraction of photons emitted per number of photons absorbed. The quantum yield is often calculated by comparing a sample with a reference substance of known quantum yield $\Phi_\mathrm{r}$ such that &#xA;$$&#xA;\Phi=\Phi_\mathrm{r}\times\frac{n^2}{n_\mathrm{r}^2}\times\frac{I}{I_\mathrm{r}}\times\frac{A_\mathrm{r}}{A},&#xA;$$&#xA;where $n$ ($n_\mathrm{r}$) is the refractive index of the sample (reference), $I$ ($I_\mathrm{r}$) is the integrated emission of the sample (reference), and $A$ ($A_\mathrm{r}$) is the absorbance of the sample (reference) at the excitation wavelength (see &lt;a href=&quot;http://www.fluortools.com/software/ae/documentation/qy&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;http://www.horiba.com/fileadmin/uploads/Scientific/Documents/Fluorescence/quantumyieldstrad.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for details).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems odd to use the integrated emission because it is not proportional to the number of photons emitted unless the emission peak is extremely narrow. In particular, the integrated emission is&#xA;$$&#xA;I=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\,I_\lambda,&#xA;$$&#xA;where $I_\lambda$ is the spectral intensity. Because the number of photons for a given energy changes with wavelength ($E=hc/\lambda$) using a quantity that is proportional to the number of photons seems more sensible. In particular, &#xA;$$&#xA;I_\lambda= n_\lambda \frac{hc}{\lambda},&#xA;$$&#xA;where $n_\lambda$ is the photon density at wavelength $\lambda$. Consider the quantity&#xA;$$&#xA;N=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\, \frac{I_\lambda \lambda}{hc}.&#xA;$$&#xA;It is directly proportional to the number of photons even if the emission peak is broad. Why is $I$ used instead of $N$?&lt;/p&gt;&#xA;" OwnerUserId="17264" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T17:35:57.900" LastActivityDate="2015-07-04T15:24:43.760" Title="Calculation of fluorescence quantum yield" Tags="&lt;quantum-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33553" PostTypeId="1" AcceptedAnswerId="33556" CreationDate="2015-07-02T16:52:19.173" Score="18" ViewCount="10380" Body="&lt;p&gt;Which substances exist that are normally liquid and that will not freeze nor boil even at relatively low/high temperatures as compared to the freezing and boiling points of water? Or how can I identify such substances online?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is relatively non-reactive, non-toxic, cheap and has high thermal conductivity that is also a bonus!&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="10185" LastEditDate="2015-07-02T16:54:38.507" LastActivityDate="2017-04-17T05:58:32.717" Title="Which substance has the highest temperature range between melting and boiling point" Tags="&lt;reference-request&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="5" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33554" PostTypeId="1" CreationDate="2015-07-02T16:54:14.657" Score="2" ViewCount="510" Body="&lt;p&gt;In an exam I had to find a substance X for the following reaction to be correct:&#xA;$$\ce{X &amp;lt;=&amp;gt;[\ce{H2SO4}] Y + H2O}$$&#xA;I picked cyclobutanol. My teacher said cyclobutanol is wrong . Why?&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorUserId="16025" LastEditDate="2015-07-02T17:19:14.850" LastActivityDate="2015-07-02T17:19:14.850" Title="Does cyclobutanol undergo acidic catalysed dehydration?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;alcohols&gt;" AnswerCount="0" CommentCount="7" FavoriteCount="1" ClosedDate="2019-02-26T06:08:17.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="33556" PostTypeId="2" ParentId="33553" CreationDate="2015-07-02T18:05:57.343" Score="12" Body="&lt;p&gt;The polar aprotic solvents (for example dimethylformamide, mp −61 °C, bp 153 °C, or hexamethylphosphoramide, mp 7 °C, bp 230–232 °C) would be a place to start. Silicone oil is often used in heating baths – one product in the Aldrich catalog is advertised as having a working range of −40 °C to +230 °C.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorDisplayName="user7951" LastEditDate="2015-07-02T19:25:26.347" LastActivityDate="2015-07-02T19:25:26.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33557" PostTypeId="2" ParentId="20105" CreationDate="2015-07-02T18:13:50.867" Score="4" Body="&lt;p&gt;Great question, but unfortunately without a great answer.  Chemistry is very inconsistent in its naming of phosphates.  In addition to the terms &lt;em&gt;diphosphate&lt;/em&gt; and &lt;em&gt;pyrophosphate&lt;/em&gt;, the term &lt;em&gt;bisphosphate&lt;/em&gt; is also used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone will probably put in an answer that cites &quot;official&quot; IUPAC or &lt;a href=&quot;http://www.iubmb.org/index.php?id=33&quot; rel=&quot;nofollow noreferrer&quot;&gt;IUBMB&lt;/a&gt; rules.  In many cases (but not all) these rules are more sensical and well thought-out than the terms in common usage in the field, but unfortunately, they are not in universal (or even wide) use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough breakdown of my sense of how these three terms are used:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Inorganic chemistry&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Pyrophosphate&lt;/em&gt; indicates a phosphate anhydride.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Diphosphate&lt;/em&gt; indicates a salt that contains more than one (ortho)-phosphate anion.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Bisphosphate&lt;/em&gt; is not widely used.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Biochemistry&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Pyrophosphate&lt;/em&gt;: used inconsistently to name an organic derivative of a phosphate anhydride.  This is what you see in geranyl pyrophosphate (GPP).  It is also used to refer to &quot;inorganic&quot; pyrophosphate anion, often abbreviated as PPi.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Diphosphate&lt;/em&gt;: an unfortunate term that is ambiguous, although usage is shifting towards synonymity with pyrophosphate.  It is often used to mean &quot;pyrophosphate&quot; and but sometimes still used to mean &quot;bisphosphate&quot;.  Biochemically, the most important phosphate anhydrides are undoubtedly ATP, ADP, etc.  These are never referred to as pyrophosphates even though that term would be more accurate.  History has chosen &quot;triphosphate&quot; and &quot;diphosphate&quot; as the terms to use for adenosine-substituted phosphate anhydrides, icky though the term may be.&lt;br&gt;&#xA;Older literature will also use &quot;diphosphate&quot; to mean &quot;bisphosphate&quot;.  The name &quot;&lt;a href=&quot;https://scholar.google.com/scholar?hl=en&amp;amp;q=fructose+diphosphate&quot; rel=&quot;nofollow noreferrer&quot;&gt;fructose diphosphate&lt;/a&gt; was formerly in wide usage for what is now known as &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fructose_1,6-bisphosphate&quot; rel=&quot;nofollow noreferrer&quot;&gt;fructose-1,6-bisphosphate&lt;/a&gt;&quot;, for example.  That compound contains no phosphate anhydrides.&lt;br&gt;&#xA;&lt;a href=&quot;https://scholar.google.com/scholar?hl=en&amp;amp;q=geranyl+diphosphate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geranyl diphosphate&lt;/a&gt; is still a fairly widely used synonym for geranyl pyrophosphate.  You see abbreviations for this molecule of both GPP and GDP.  I prefer the name GPP, but unfortunately not everyone uses it.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;bisphosphate&lt;/em&gt;: Now in wide usage as the preferred term for an organic molecule with multiple ortho-phosphate esters, such as fructose 1,6-bisphosphate.  Biochemists have done a relatively good job of adopting this usage in modern times.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A common confusion that stems from this nomenclature concerns the stability of organophosphate compounds. Phosphate esters, e.g. glucose-1-phosphate or glucose-6-phosphate or adenosine monophosphate, are much more stable and less reactive than phosphate anhydrides such as acetyl phosphate (a mixed anhydride of acetic acid and phosphoric acid) or the phosphoanhydride bonds in adenosine triphosphate (ATP).&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2018-04-07T18:28:03.673" LastActivityDate="2018-04-07T18:28:03.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33559" PostTypeId="2" ParentId="33553" CreationDate="2015-07-02T20:18:11.163" Score="17" Body="&lt;p&gt;Gallium melts at 30 &amp;deg;C but doesn't boil until 2200 &amp;deg;C.  If 30 &amp;deg;C is a bit too warm to count as &quot;room temperature&quot; or &quot;normally&quot; for you, I found an &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/je60009a029&quot;&gt;old paper&lt;/a&gt; that recommends tetralkyl silanes such as tetradodecyl silane as lubricants that are liquid over very wide temperatures.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Addendum:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://msdssearch.dow.com/PublishedLiteratureDOWCOM/dh_0030/0901b803800303cd.pdf?filepath=/heattrans/pdfs/noreg/176-01337.pdf&amp;amp;fromPage=GetDoc&quot;&gt;Dowtherm A&lt;/a&gt; is a eutectic mixture of &lt;a href=&quot;https://en.wikipedia.org/wiki/Biphenyl&quot;&gt;biphenyl&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Diphenyl_ether&quot;&gt;diphenyl ether&lt;/a&gt;.  According to its manufacturer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These compounds have practically&#xA;  the same vapor pressures, so the&#xA;  mixture can be handled as if it were&#xA;  a single compound. DOWTHERM A&#xA;  fluid may be used in systems employing&#xA;  either liquid phase or vapor phase&#xA;  heating. Its normal application range&#xA;  is 60°F to 750°F (15°C to 400°C),&#xA;  and its pressure range is from&#xA;  atmospheric to 152.5 psig (10.6 bar).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The viscosity of DOWTHERM A&#xA;  fluid is low and changes only&#xA;  slightly between the melting point&#xA;  of the product and its top operating&#xA;  temperature. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-07-02T22:26:21.867" LastActivityDate="2015-07-02T22:26:21.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33560" PostTypeId="2" ParentId="33525" CreationDate="2015-07-02T22:26:52.917" Score="1" Body="&lt;p&gt;They both are same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plug &lt;span class=&quot;math-container&quot;&gt;$C_{\mathrm{m},p,\mathrm{avg}}$&lt;/span&gt; in your second equation, you will get your first equation. The average &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is calculated as a mole fraction weighted average. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="4945" LastEditDate="2019-07-11T16:09:32.233" LastActivityDate="2019-07-11T16:09:32.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33561" PostTypeId="2" ParentId="33553" CreationDate="2015-07-02T23:06:10.847" Score="9" Body="&lt;p&gt;A good place for a general list is &lt;a href=&quot;http://www.engineeringtoolbox.com/melting-boiling-temperatures-d_390.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/melting-boiling-temperatures-d_390.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high temperatures, here is a list of alloys &lt;a href=&quot;https://en.wikipedia.org/wiki/Fusible_alloy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Fusible_alloy&lt;/a&gt; Some of these have melting points below 0 °C and boiling points as high as you will ever need. Some are pretty expensive, though. Note that in general mixtures have a lower melting point than pure substances. The mixture whose percentage composition has the lowest melting point is called a Eutectic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a pure metal, tin is a good choice as it has low toxicity and a wide range. Hence it is used in the float glass process for window glass. Even though it is not particularly reactive, there is some potential for an explosive reaction of molten tin with water/steam if used in large quantities, partly due to the liberation of hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solar energy industry has used sodium and potassium nitrate (and previously sulfuric acid) but both are corrosive. Additionally, nitrates may be difficult to get hold of outside the agricultural industry in some countries as they are used in making explosives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are the organics. Silicone oil and Dowtherm A have been mentioned. Neither is particularly good for you. Cooking oil doesn’t have a bad range and is preferable to Diesel from both toxicity and flammability points of view. Among the alcohols, ethanol and methanol are probably too volatile for your use. The glycols (diols) are used as antifreeze additives with water. Ethylene glycol is quite toxic. Propylene glycol is much less toxic. Glycerol (a triol) is also nontoxic and higher boiling than propylene glycol, but it is highly viscous which will negatively impact its heat transfer properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be remembered that non-toxic organics can decompose into toxic ones, particularly when the heat flux applied to them is too high (as can happen with electric heating elements.) &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fun fact: Propylene glycol is much less toxic than ethylene glycol because the latter is metabolized to oxalic acid, whereas the former is metabolized to lactic acid, one of the most common substances in the body.&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="402" LastEditDate="2017-01-14T08:38:33.290" LastActivityDate="2017-01-14T08:38:33.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33562" PostTypeId="2" ParentId="33553" CreationDate="2015-07-02T23:17:09.697" Score="0" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_elements_by_melting_point&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c|c|c|c}&#xA; \mathrm{Element}  &amp;amp;	  \text{MP in K}	&amp;amp;  \text{BP in K}	&amp;amp;\Delta T \text{ in K} \\\hline&#xA;\mathrm{Neptunium}	&amp;amp;  917	&amp;amp;4\,273	&amp;amp;3\,356\\&#xA;\mathrm{Uranium}   &amp;amp;	1\,405	&amp;amp;4\,404	&amp;amp;2\,999\\&#xA;\mathrm{Thorium} &amp;amp; 	2\,115&amp;amp;	5\,061&amp;amp;	2\,946\\&#xA;\mathrm{Cerium}  &amp;amp; 	1\,068&amp;amp;	3\,716&amp;amp;	2\,648\\&#xA;\mathrm{Plutonium}	&amp;amp;  913&amp;amp;	3\,501&amp;amp;	2\,589\\&#xA;\mathrm{Praseodymium}	&amp;amp;1\,208&amp;amp;	3\,793&amp;amp;	2\,585\\&#xA;\mathrm{Zirconium}&amp;amp;	2\,128&amp;amp;	4\,682&amp;amp;	2\,554\\&#xA;\mathrm{Lanthanum}&amp;amp;	1\,193&amp;amp;	3\,737&amp;amp;	2\,544\\&#xA;\mathrm{Protactinium}&amp;amp;	1\,841&amp;amp;	4\,300&amp;amp;	2\,459\\&#xA;\mathrm{Tantalum}&amp;amp; 	3\,290&amp;amp;	5\,731&amp;amp;	2\,441&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is just elements (top 10 delta). &lt;/p&gt;&#xA;" OwnerUserId="17269" LastEditorDisplayName="user7951" LastEditDate="2015-07-03T11:28:41.287" LastActivityDate="2015-07-03T11:28:41.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33563" PostTypeId="2" ParentId="33553" CreationDate="2015-07-03T00:50:00.190" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Galinstan&quot; rel=&quot;noreferrer&quot;&gt;Galinstan&lt;/a&gt; is a substance similar to mercury in its basic properties, but has a liquid range at atmospheric pressure which is much larger than that of mercury's. It melts at −19 °C and boils at 1300 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, it also has good thermal conductivity (16 W/(m K)) and is relatively &lt;a href=&quot;http://www.rgmd.com/msds/msds.pdf&quot; rel=&quot;noreferrer&quot;&gt;non-toxic&lt;/a&gt; and unreactive. &lt;/p&gt;&#xA;" OwnerUserId="11303" LastEditorDisplayName="user7951" LastEditDate="2015-07-03T11:17:22.587" LastActivityDate="2015-07-03T11:17:22.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33564" PostTypeId="2" ParentId="33537" CreationDate="2015-07-03T01:55:36.193" Score="6" Body="&lt;p&gt;Each atom does its own &lt;a href=&quot;https://brainchemist.files.wordpress.com/2011/01/diffusion.gif&quot;&gt;diffusion across the membrane&lt;/a&gt;. There is a probability that a $\ce{K+}$ or $\ce{Na+}$ ion will move in the &quot;forward&quot; or the &quot;reverse&quot; direction. The probability of getting through the membrane may differ between the &lt;em&gt;types&lt;/em&gt; of ions (e.g. if the membrane pores allow one species through more easily than another because of ionic size), but, barring any external force, eventually each type of ion should come into equilibrium, with as many going forward as backward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a box of marbles of two colors, with a partial separator between, and that the box is shaken. It is simple chance that a marble crosses the separator, and if many marbles of one color are on one side, there are fewer to migrate backward from the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See this &lt;a href=&quot;http://mw.concord.org/nextgen/#interactives/biology/diffusion-bio/diffusion-bio-semipermeable-membrane&quot;&gt;online demo of diffusion across a semipermeable membrane&lt;/a&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-03T01:55:36.193" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33565" PostTypeId="1" CreationDate="2015-07-03T02:08:59.083" Score="8" ViewCount="1797" Body="&lt;p&gt;Would I have any success putting coffee in a humidifier? Or perhaps water mixed with caffeine powder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I could wake up without a jolting alarm clock, feeling like I already had a cup of coffee?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this will have to be a carefully controlled dose if working with caffeine powder, but is it feasible?&lt;/p&gt;&#xA;" OwnerUserId="17270" LastEditorUserId="30836" LastEditDate="2019-04-19T15:24:30.040" LastActivityDate="2019-04-19T19:47:08.793" Title="Can I successfully use a humidifier to distribute caffeine to wake someone up without the usual morning drowsiness?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;drugs&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33566" PostTypeId="2" ParentId="33565" CreationDate="2015-07-03T03:21:42.420" Score="2" Body="&lt;p&gt;It looks like at least one person/organization has &lt;a href=&quot;http://www.google.com/patents/US7488469&quot; rel=&quot;nofollow&quot;&gt;thought of this&lt;/a&gt; and patented a delivery mechanism (&quot;Delivery of caffeine through an inhalation route&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you might be able to couple the system outline in the patent documentation with an alarm clock of sorts to trigger the release at a specified time. I'd certainly buy one if the price was right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The inhalation part has been investigated. As far as I can tell, the business about administering caffeine to a sleeping person and then having them awaken into a fully alert (or pseudocaffeinated) state has not.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-06T18:26:14.700" LastActivityDate="2015-07-06T18:26:14.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33567" PostTypeId="1" CreationDate="2015-07-03T03:42:55.333" Score="1" ViewCount="281" Body="&lt;p&gt;I was under the impression that differences between the various long range electrostatic algorithms, ewald, PME, PPPM, to name a few, lied solely in their treatment of the reciprocal space term. Specifically, I thought all methods looped over a particles neighbor and summed q(i)*q(j)&lt;em&gt;erfc(alpha&lt;/em&gt;r)/r for the real space calculation. However, this is my impression, which I have yet to validate with others. I became confused when I recently read &quot;Ewald summation techniques in perspective: a survey&quot; whose equation 22 seems to have suggested a different way to calculate the short range potential. Could anyone else chime in here, and either tell me where I am wrong, or I am correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16152" LastEditorUserId="4945" LastEditDate="2015-07-03T03:48:35.713" LastActivityDate="2015-07-03T14:03:08.470" Title="Real space term for PPPM vs PME" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33568" PostTypeId="1" AcceptedAnswerId="33573" CreationDate="2015-07-03T03:47:38.027" Score="0" ViewCount="1037" Body="&lt;p&gt;I know that $\ce{NO2}$ has 3 effective pairs,  so it is likely to have trigonal planar structure. In trigonal planar structure we usually see three atoms bonded to central atom at 120°. But in $\ce{NO2-}$, I observe that though it has three effective pairs,  but still, it is not having the desired structure,  with only 2 O atoms bonded to N atom. Can anyone explain the situation? &lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="7475" LastEditDate="2017-02-14T00:01:15.450" LastActivityDate="2017-02-14T00:01:45.167" Title="What is the structure of nitrite?" Tags="&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33570" PostTypeId="1" AcceptedAnswerId="33575" CreationDate="2015-07-03T05:35:19.637" Score="4" ViewCount="4077" Body="&lt;p&gt;I've noticed that water soluble fertilizers tend to be blue. What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes? I haven't seen this in organic fertilizers. I figured chemists might be able to examine the ingredients of Miracle Gro or something and tell me. I'm guessing it has something to do with copper.&lt;/p&gt;&#xA;" OwnerUserId="11385" LastEditorUserId="11385" LastEditDate="2015-07-04T08:27:00.203" LastActivityDate="2015-07-04T08:27:00.203" Title="What makes many water soluble fertilizers blue?" Tags="&lt;synthesis&gt;&lt;color&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33571" PostTypeId="2" ParentId="30933" CreationDate="2015-07-03T06:40:36.610" Score="6" Body="&lt;p&gt;In &lt;a href=&quot;https://github.com/sunqm/pyscf&quot; rel=&quot;noreferrer&quot;&gt;pyscf&lt;/a&gt;, we are routinely running molden.py to generate molden file and plot the orbital surface with Jmol.  It should not be too hard to write small script to generate density or other charge surface then plot the surface with Jmol.  The basic functions you probably needed should be very close to the functions used by dft module, such as eval_ao, eval_rho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the recent Pyscf alpha 2 release, a script pyscf/tools/cubegen.py has been added to generate the Gaussian cube file format.  In my linux box, the density looks fine in Jmol.  You can use it as an example to generate other &lt;a href=&quot;http://wiki.jmol.org/index.php/File_formats/Surfaces&quot; rel=&quot;noreferrer&quot;&gt;Jmol formats&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17274" LastActivityDate="2015-07-03T06:40:36.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33572" PostTypeId="2" ParentId="33537" CreationDate="2015-07-03T07:54:51.673" Score="9" Body="&lt;p&gt;This answer has been edited quite a bit. New additions are put in blockquotes.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: &lt;strong&gt;Any mention of &amp;quot;potential&amp;quot; in this answer refers to an electric potential, not a chemical potential.&lt;/strong&gt; The definition of electric potential can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_potential&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; ions. In the left compartment, we put 1 M of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; ions and in the right compartment, we put 1 M of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; ions (with appropriate counterions). If diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; entirely ignored the presence of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; as you say, then we would expect that after a while, we would have 1 M &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and 0.5 M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; on the LHS, and on the RHS we would have 0.5 M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This does not happen. In actual fact, there would be a &lt;strong&gt;concentration gradient&lt;/strong&gt; that favours net movement of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; from RHS-&amp;gt;LHS, and there would be a &lt;strong&gt;electric potential difference&lt;/strong&gt; that favours net movement of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; from LHS-&amp;gt;RHS. We could model the concentration at any point in time with:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LHS - 1 M &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;RHS - &lt;span class=&quot;math-container&quot;&gt;$0.5 - x$&lt;/span&gt; M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is between 0 and 0.5.&lt;/p&gt;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$x = 0$&lt;/span&gt;, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; starts diffusing from RHS-&amp;gt; LHS, causing &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; to increase. Now as &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; across the membrane. This is the point where we could say that diffusion has stopped.&lt;/p&gt;&#xA;&lt;p&gt;Now, let me change the scenario and say that the membrane is permeable to BOTH &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; ions. In that case, clearly the equilibrium concentrations will be 0.5 M &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; and 0.5 M &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is important to note that the diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; is NOT directly affected by the pre-existing concentration gradient of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;. It is indirectly affected only because the concentration gradient of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;. If we were to replace &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.&lt;/p&gt;&#xA;&lt;p&gt;The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl-}$&lt;/span&gt;. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.&lt;/p&gt;&#xA;&lt;p&gt;The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; is much higher than the intracellular concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;; the converse is true for &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;. Due to the membrane having different permeabilities to &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.&lt;/p&gt;&#xA;&lt;p&gt;Neurons react to certain stimuli by having &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; channels open, making the membrane permeable to &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;At this stage, we have to consider the two factors that influence the diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since the potential difference is still negative, it &lt;strong&gt;favours influx of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Since the extracellular concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; is larger than the intracellular concentration, the concentration gradient &lt;strong&gt;also favours influx of &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Hence more &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; channels close and the voltage-gated &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; channels open. Again, we need to consider the two factors that influence the diffusion of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The potential difference, which is now positive, &lt;strong&gt;favours efflux of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Since the intracellular concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; is higher than the extracellular concentration, the concentration gradient &lt;strong&gt;also favours efflux of &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt;.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; are not the same as they were at the very start of this discussion. They are restored by the &lt;span class=&quot;math-container&quot;&gt;$\ce{Na+}$&lt;/span&gt;/&lt;span class=&quot;math-container&quot;&gt;$\ce{K+}$&lt;/span&gt; pump.&lt;/p&gt;&#xA;&lt;p&gt;The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.&lt;/p&gt;&#xA;&lt;p&gt;Prior to my edit you might have seen that I wrote that &amp;quot;diffusion does not ignore similarly charged species&amp;quot;. This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when &lt;strong&gt;both X and Y are charged species&lt;/strong&gt;, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.&lt;/p&gt;&#xA;&lt;p&gt;I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-10T10:55:29.170" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="33573" PostTypeId="2" ParentId="33568" CreationDate="2015-07-03T09:25:56.920" Score="3" Body="&lt;p&gt;Nitrogen is actually surrounded by 3 effective pairs. But 2 of them are bonding and the third one is a lone pair. According to VSEPR theory, the structure should be  trigonal planar.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The angle is not exactly $120^\circ$, as the electrostatic  repulsion between non-bonding pair and bonding pair is higher than the repulsion between two bonding pair. So, the angle of $\ce{O-N-O}$ is slightly smaller than $120^\circ$.&lt;/li&gt;&#xA;&lt;li&gt;The distribution of the three electron effective pairs is trigonal planar. But the molecule structure is bent (like letter V).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7475" LastEditDate="2017-02-14T00:01:45.167" LastActivityDate="2017-02-14T00:01:45.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33575" PostTypeId="2" ParentId="33570" CreationDate="2015-07-03T13:56:11.320" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have some Miracle Gro on hand, the water soluble kind: The blue stuff is copper sulfate, or ${\rm CuSO}_{4}$. &lt;a href=&quot;http://eldoradochemical.com/MSDS_Sheets/EDC/Third_Party_Products/MIRACLE_GRO_ALL_PURPOSE_PLANT_FOOD_24-8-16.pdf&quot;&gt;Here&lt;/a&gt; is the MSDS for Miracle Gro Water Soluble All Purpose Plant Food, which lists the chemicals in the mixture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this necessary or for aesthetic purposes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's necessary in these formulations for purposes detailed (in part) &lt;a href=&quot;http://www.extension.umn.edu/agriculture/nutrient-management/micronutrients/copper-for-crop-production/&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://npic.orst.edu/factsheets/cuso4gen.html&quot;&gt;here&lt;/a&gt;. In short, it is included in these formulations because plants need copper for biosynthesis; and because it acts as a fungicide, presumably leading to a nicer ecosystem in one's lawn/garden.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-03T13:56:11.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33576" PostTypeId="2" ParentId="33567" CreationDate="2015-07-03T14:03:08.470" Score="1" Body="&lt;p&gt;There are a number of ways of looking at Ewald's summation, but one is that instead of trying to solve the original problem where the potential is due to a set of point charges you, instead, solve the system for a set of smeared charges and add a correction. Now there are limitations on what are good ways to do the smearing, but there are still quite a lot of possible choices. The most common is to replace the original delta function charge with a Gaussian, the width of which is determined by alpha in your question. This leads to the error function and exponentials you are used to. However it's not the only possible choice. In equation 22 in the paper you refer to a different way of smearing the charges has been chosen, a linear ramp centred on the particle. This is described in equation 21. As the smearing is different the individual terms in the Ewald summation have different forms when compared to Gaussian smearing, but it's no less valid a way of proceeding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more about this look into Fourier Transforms and the Convolution theorem. A bit of reading on that and the Ewald Summation in all its forms becomes pretty straightforward.&lt;/p&gt;&#xA;" OwnerUserId="17277" LastActivityDate="2015-07-03T14:03:08.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33577" PostTypeId="1" CreationDate="2015-07-03T15:54:47.483" Score="1" ViewCount="88" Body="&lt;p&gt;This may seem like a off topic question but I really need help on it. I am currently working on a polymerization reaction in which I'm breaking down plastic polymers into smaller molecules. However, since this is a green chemistry project, I decided, with the suggestion of other chemists, that I should a method that doesn't require a lot of electrical energy to heat it. One idea was an enzyme, but I decided I want to use solar energy to reflux the RBF. I have a few ideas and please comment on them and also provide your own suggestions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First is using a parabolic mirror below the flask to focus the heat towards it. &lt;/li&gt;&#xA;&lt;li&gt;Second idea is painting the RBF black and hoping that is sufficient to heat it to reflux.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15851" LastEditorUserId="10185" LastEditDate="2015-07-03T19:36:27.827" LastActivityDate="2015-07-03T19:36:27.827" Title="How can I use sunlight to heat a RBF and reflux a reaction?" Tags="&lt;thermodynamics&gt;&lt;experimental-chemistry&gt;&lt;heat&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33580" PostTypeId="2" ParentId="32849" CreationDate="2015-07-03T17:39:15.970" Score="5" Body="&lt;p&gt;Well apart from the fact that the answer depends on the type of the paper that is used. Secondly how much is wasted and also the process used. Below is an article that might be helpful in answering the question&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Some industry sources estimate that an ordinary sheet of paper made from cellulose fibers derived from wood can survive only four to six trips through the recycling process. The Environmental Protection Agency puts the figure at five to seven times.&lt;/p&gt;&#xA;&lt;p&gt;It is not surprising that the rigors of remanufacturing take a toll on the fibers.&lt;/p&gt;&#xA;&lt;p&gt;Ideally, paper for recycling is separated into types, because paper with long fibers, like white office paper, offers the most flexibility for recycling, while newsprint, with its shorter fibers, is usually reserved for making more newsprint and other low-quality papers.&lt;/p&gt;&#xA;&lt;p&gt;The paper is shredded and chopped, then subjected to a mixture of chemicals and water and heated as it is repulped. It is centrifuged and screened to remove impurities; de-inked with more chemicals; then sprayed onto a wire screen, drained, dried and squeezed through heated rollers.&lt;/p&gt;&#xA;&lt;p&gt;With each step, the fibers become shorter, coarser and stiffer, so that eventually, recycled fiber needs to be mixed with virgin fiber to make paper of the desired quality.&lt;/p&gt;&#xA;&lt;p&gt;According to the American Forest and Paper Association, 63.4 percent of the paper consumed in the United States was recovered for recycling in 2009. C. CLAIBORNE RAY&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;SOURCES: New York Times - &lt;a href=&quot;http://www.nytimes.com/2010/12/21/science/21qna.html?_r=0&quot; rel=&quot;noreferrer&quot;&gt;http://www.nytimes.com/2010/12/21/science/21qna.html?_r=0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17280" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-03T17:39:15.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33581" PostTypeId="1" CreationDate="2015-07-03T19:41:41.420" Score="5" ViewCount="3795" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  A linear regression line was obtained; $$\log(\ce{[NaF]})~(\mathrm{mol~L^{-1}})~ \mathrm{vs~EMF}:&#xA;y = 59.8x +600.8$$&#xA;  $0.7540~\mathrm{g}$ of toothpaste was diluted in a$50~\mathrm{mL}$ and tested to obtain a EMF value of $390$. what is the ppm of $\ce{F-}$ in $198~\mathrm{g}$ of toothpaste?   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I plugged in $390$ into the eqn to get $-3.525$, taking the antilog to obtain $0.0002985$, and times it by the volume $0.05~\mathrm{L}$ to get the moles of $\ce{NaF}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to stoichiometry, this also equals $1~\mathrm{mol}$ of $\ce{F-}$. I multiplied by $18.998~\mathrm{g~mol^{-1}}$ to obtain $0.000283527~\mathrm{mol}$ of $\ce{F-}$. I then multiplied by the factor: $\frac{198}{0.750}$ to obtain $0.074454015~\mathrm{g}$ of $\ce{F-}$ in the $198~\mathrm{g}$ of toothpaste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck here, how can I calculate the ppm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected value of fluoride in toothpaste is $1000~\mathrm{ppm}$&lt;/p&gt;&#xA;" OwnerUserId="17282" LastEditorUserId="46306" LastEditDate="2018-03-06T02:23:38.960" LastActivityDate="2018-03-06T02:23:38.960" Title="Calculating Fluoride in toothpaste" Tags="&lt;electrochemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33583" PostTypeId="1" CreationDate="2015-07-03T21:37:31.897" Score="17" ViewCount="1332" Body="&lt;p&gt;The following acetate aldol reaction gives no diasteroselectivity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lntJ1.png&quot; alt=&quot;Evans&amp;#39; acetate aldol showing little diastereoselectity&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I don't know what the basis of above statement is (I heard it in a lecture and read it on some internet web pages), I expect the two products are formed as a racemic mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the related propionate aldol we can draw two (considering R of aldehyde to always be equatorial) possible transition states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5AnoS.png&quot; alt=&quot;Transition states for Evans&amp;#39; aldol reactions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is why we don't get good selectivity if we compare the acetate aldol with the same transition states for the propionate aldol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side-note: in practice chlorine is used instead of Me to obtain good selectivity and the chlorine will be eliminated finally by use of Zn/AcOH.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="21296" LastEditDate="2017-09-23T17:59:36.537" LastActivityDate="2017-09-23T17:59:36.537" Title="Why does acetate aldol with Evans auxiliary give almost no diasteroselectivity?" Tags="&lt;organic-chemistry&gt;&lt;aldol-reaction&gt;&lt;stereoselectivity&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33585" PostTypeId="2" ParentId="24723" CreationDate="2015-07-03T23:47:59.707" Score="11" Body="&lt;p&gt;You can memorize the periodic table in one night, simply by emulating best-practice memorization techniques and doing what memory experts do. Common sense, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Memory experts and world champion memory ‘athletes’ activate the enormous natural power of their visual memory by using visualization and association mnemonic techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That’s a fancy way of saying they create mental pictures and link them together in their mind. It’s incredibly simple but amazingly fast and effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch &lt;a href=&quot;https://www.youtube.com/watch?v=0nFkQ4cQhME&quot;&gt;YouTube’s #1 “How to Memorize” video&lt;/a&gt; and you’ll probably amaze yourself with how easily you can remember and recall 15 random words in order, using one of these techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foundation technique most memory experts use is the Method of Loci (or Memory Palace or Journey Method). Think of a particular journey you take every day, and picture certain locations along the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine leaving home in the morning and travelling to work or school. You might walk out your front door, through the front gate, and get on a bus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each location you visualize an object that represents what you’re trying to remember. Because the chemical elements themselves can be difficult to visualize, you substitute them with an object that you will naturally associate or link to the element itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, ‘hydrogen’ sounds similar to ‘hydrant’, so when you visualize a hydrant sitting at your front door, you’ll be prompted to remember ‘hydrogen’. When you picture a large helium balloon tied to your front gate, you’ll remember helium. And when your bus begins talking with a ‘lithp’ (how people with a lisp pronounce ‘lisp’), you’ll be prompted to recall lithium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These established memory techniques have been proven by over 50 years of academic research in fields like cognitive psychology. Google ‘memory palace’ or ‘world memory champion’ and you’ll discover they’re the fastest and most effective methods to memorize a deck of playing cards and a lot of other geeky things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method used in the video above is called the Link and Story Method, and is based on the same principles of visualization and association. The weakness of this method compared to the Memory Palace is the amount of time it takes to create the (intentionally) bizarre and crazy story to link all the words (or chemical elements) together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all the work has already been done for you at &lt;a href=&quot;https://www.memorizeperiodictable.com/&quot;&gt;How to Memorize the Periodic Table&lt;/a&gt;. This animated video course is the fastest way to memorize the periodic table because it uses best-practice visual memory techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the mental images and association links described above have already been created, and transformed into engaging whiteboard animation videos. You just need to sit back and watch, and let the amazing natural power of your visual memory do its thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about other techniques? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most other methods people suggest to memorize the periodic table rely on verbal memory, but don’t activate the enormous power of your visual memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flashcards or equivalent apps are convenient but don’t provide an association or link between chemical element names, meaning they rely on rote memorization. Repetition by itself is not meaningful, takes an unnecessarily long time and effective retention is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acronyms and acrostics are ‘first letter mnemonics’. You could use the acronym HHeLiBeBCNOF (pronounced ‘heeliebeb kernoff’) to remember the first nine elements. It's a nonsense word, but it condenses nine names into one mental prompt or cue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the acrostic “&lt;strong&gt;H&lt;/strong&gt;ere &lt;strong&gt;He&lt;/strong&gt; &lt;strong&gt;Li&lt;/strong&gt;es &lt;strong&gt;Be&lt;/strong&gt;neath &lt;strong&gt;B&lt;/strong&gt;ed &lt;strong&gt;C&lt;/strong&gt;lothes, &lt;strong&gt;N&lt;/strong&gt;othing &lt;strong&gt;O&lt;/strong&gt;n, &lt;strong&gt;F&lt;/strong&gt;eeling &lt;strong&gt;Ne&lt;/strong&gt;rvous” would equate to H He Li Be B C N O F Ne.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acronyms chunk words together, which is good (even if they are nonsense) and acrostics use more meaning, but either way you'll only have the first letter or two to remind you of each element's full name. That's tough for 118 elements!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first letter cues don’t prompt you enough to recall the complete element name, so acronyms and acrostics can be great for the first 20 elements, but not for all 118.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Songs are also popular, whether you’re a fan of Tom Lehrer or ASAP Science. A catchy tune gives better association and meaning than acronyms and acrostics, but you still have to rely on bucketloads of repetition. They’re a great way to make repetition fun, but songs only tap into your verbal memory, not your powerful visual memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line, you should play to your natural strengths. Your brain loves pictures. And that makes visual memory techniques 10X more powerful than verbal memory techniques (like songs and acronyms) to memorize the periodic table.&lt;/p&gt;&#xA;" OwnerUserId="17283" LastActivityDate="2015-07-03T23:47:59.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33586" PostTypeId="2" ParentId="32377" CreationDate="2015-07-04T01:05:15.580" Score="2" Body="&lt;p&gt;For the process to be truly reversible, you need to be able to return both the system &lt;em&gt;and&lt;/em&gt; the surroundings to their original states, without any significant change in anything else.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it useful to think of the surroundings as having a limited tool kit for accomplishing this:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A set of very small weights that can be added or removed from a massless piston at different elevations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A set of constant temperature baths featuring a broad sequence of slightly different temperatures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are considering the reversible compression or expansion of a gas inside of a cylinder in the transition between two different equilibrium states, you can do this by adding or removing the weights at different elevations and contacting the cylinder with a sequence of baths at slightly higher- or slightly lower temperatures than the gas temperature at each stage of the transition.  Then, if you wish to reverse the process, you just do the opposite sequence with the weights and the reservoirs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-04T01:05:15.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33587" PostTypeId="1" AcceptedAnswerId="33589" CreationDate="2015-07-04T02:12:56.487" Score="3" ViewCount="2741" Body="&lt;p&gt;I’m looking at a chart in my book, and while cycloheptane has a higher boiling point than cyclohexane like I’d expect, cycloheptane has a melting point of −12 °C, while cyclohexane has a melting point of 6.6 °C. Why does it have a lower melting point only?&lt;/p&gt;&#xA;" OwnerUserId="7689" LastEditorDisplayName="user7951" LastEditDate="2015-07-04T09:43:51.863" LastActivityDate="2015-07-04T09:43:51.863" Title="Boiling/melting points of cyclohexane and cycloheptane" Tags="&lt;organic-chemistry&gt;&lt;boiling-point&gt;&lt;melting-point&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33588" PostTypeId="1" AcceptedAnswerId="33603" CreationDate="2015-07-04T03:22:23.963" Score="1" ViewCount="25276" Body="&lt;p&gt;How many electrons can have the quantum number set $n=6,\ l=3,\ m_l=-1$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please explain why. I know that n describes the number of shells in an atom but what do n, l, and ml have to do with electrons?&lt;/p&gt;&#xA;" OwnerUserId="16852" LastEditorUserId="275" LastEditDate="2015-07-04T14:30:12.457" LastActivityDate="2015-07-04T14:33:28.490" Title="How many electrons can have the quantum number set n=5, l=3, ml=-1?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-07-04T17:12:44.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="33589" PostTypeId="2" ParentId="33587" CreationDate="2015-07-04T04:07:18.563" Score="4" Body="&lt;p&gt;There are not many reputable sources on the matter (as I could find) since it seems that cycloheptane isn't the most popular chemical around (poor guy) but I will present you with the best explanation/argument/insight into the situation as I can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that it would be solely my own interpretation, and a case of explaining results that we see, rather than formulating an explanation to encompass natural phenomena without observing the results prior.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key ingredient that I noticed while looking through charts and whatnot was bonding which is connected to sterics. More on that later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=tEjWioW2dH0C&amp;amp;pg=PA135&amp;amp;lpg=PA135&amp;amp;dq=cycloalkane%20melting%20boiling%20point%20chart&amp;amp;source=bl&amp;amp;ots=LvFWQe26_m&amp;amp;sig=W8A9g_dJNJddkwDCGPZxDlucxuw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=A1SXVZRKj7nJBM7MgogE&amp;amp;ved=0CF8Q6AEwDA#v=onepage&amp;amp;q=cycloalkane%20melting%20boiling%20point%20chart&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Melting and Boiling Points of Cycloalkanes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is a chart of the melting and boiling points of cycloalkanes. As expected, boiling point increases linearly. That is lovely! Thus, we can explain boiling point with natural intuition, similar to regular alkanes. More is more! If only life was so simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's take a look at melting points. HOLY! Cyclobutane? Why you so high? (it is actually an error by the book. Nonetheless, it is higher by a little than cyclopentane and that is the point) And wow, is cyclohexane a monster, past the melting point of our dearest water. Something is wrong with this universe. It must be the martians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But no. It is not the martians. At least I do not believe so. The answer, I believe, is sterics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sterics? you ask. Are you trying to say that cyclobutane bonds more structured than cyclopentane and cyclohexane bonds more structured than cyclohexane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although you may not expect much difference from cyclobutane and cyclopentane, cyclobutane is very close to a right angle (perfect square) and this would allow tight packing. Cyclopentane, on the other hand, is normally as an &quot;envelope&quot; which allows less close packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, cyclohexane has  &lt;a href=&quot;http://www.ch.imperial.ac.uk/rzepa/blog/?p=3276&quot; rel=&quot;nofollow&quot;&gt;very close cystalline packing&lt;/a&gt; which may account for its low melting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, after its crystalline structure melts away, our intuition gives way to the boiling point expectancies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Again, this is not definite. If possible, I would like to juxtapose the solid structures of each cycloalkane, but unfortunately data seems scarce, especially for cycloheptane, which is of concern.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-07-04T04:07:18.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33590" PostTypeId="1" CreationDate="2015-07-03T02:24:46.310" Score="3" ViewCount="3825" Body="&lt;p&gt;Can anyone explain the shapes of ionic compounds?  What does it mean? What does the coordinate number mean? I know that in $\ce{NaCl}$, one $\ce{Na}$ atom is surrounded by 6 $\ce{Cl}$ atoms, and vice versa, but can it be explained with the help of coordinate numbers? &lt;/p&gt;&#xA;" OwnerUserId="17217" OwnerDisplayName="Aneek" LastEditorUserId="10185" LastEditDate="2015-07-04T17:47:00.917" LastActivityDate="2016-07-15T13:30:01.613" Title="Shapes of ionic compound" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33592" PostTypeId="1" CreationDate="2015-07-04T07:24:20.343" Score="1" ViewCount="239" Body="&lt;p&gt;Which are the elements of the Twofold Screw Axis in this image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;+&quot; and &quot;-&quot; elements that intercept the red line are the elements of the Twofold Screw Axis 2/3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, is the height of the red line 2/3 plus or 2/3 minus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why the height of the red line is 2/3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the same problem with the other colored lines. Are the elements of the green line 1/3 minus or plus?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ewnbF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17061" LastActivityDate="2015-11-04T11:07:53.967" Title="R32 Space Group" Tags="&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33594" PostTypeId="2" ParentId="28905" CreationDate="2015-07-04T09:46:32.333" Score="4" Body="&lt;p&gt;In addition to @Mithoron's comments and the Wikipedia he linked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an important consideration in photovoltaics, as silicon forms the basis of the internal structure of solar technology (and image sensors). According to the PVEducation.org page &lt;a href=&quot;http://www.pveducation.org/pvcdrom/pn-junction/semiconductor-structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Semiconductor Structure&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Each silicon atom has four valence electrons which are shared, forming &lt;strong&gt;covalent bonds&lt;/strong&gt; with the four surrounding Si atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Diagrammatically, the website includes the following schematic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c9vON.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Schematic representation of covalent bonds in a silicon crystal lattice. Each line connecting the atoms represents an electron being shared between the two. Two electrons being shared are what form the covalent bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In terms of germanium, according to the WebElements page &lt;a href=&quot;http://www.webelements.com/nexus/chemistry_page/group-14-elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Group 14 Elements&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tin and lead are very metallic although one modification of tin known as grey tin has the same diamond structure [as carbon] as does germanium and silicon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-07-04T13:40:53.937" LastActivityDate="2015-07-04T13:40:53.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33595" PostTypeId="2" ParentId="31150" CreationDate="2015-07-04T10:21:56.563" Score="1" Body="&lt;p&gt;Here is my approach. The dissociation of $\ce{CaF2}$ is described by&#xA;$$\ce{CaF2 &amp;lt;=&amp;gt; Ca^2+ +2F-} \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $s$ is the molar solubility of $\ce{CaF2}$ (in $\pu{mol dm^-3}$), then&#xA;\begin{align}&#xA;K_\mathrm{sp} &amp;amp;= [\ce{Ca^2+}] \cdot [\ce{F-}]^2 \\&#xA;&amp;amp;= s \cdot (2s)^2 \\&#xA;&amp;amp;= 4s^3 = 3.9 \times 10^{-11} \\&#xA;\implies s &amp;amp;= 2.1 \times 10^{-4} \\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluoride concentration is then equal to $[\ce{F-}] = 4.2 \times 10^{-4} ~\mathrm{M}$. The pH is then governed by the equilibrium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{F- + H2O &amp;lt;=&amp;gt; HF + OH-} \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]; \quad K_\mathrm{a} = \frac{[\ce{F-}][\ce{H3O+}]}{[\ce{HF}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we find that the equilibrium constant for reaction $(2)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{[\ce{HF}][\ce{OH-}]}{[\ce{F-}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{OH-}][\ce{HF}] = \frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume now that $[\ce{OH-}] = [\ce{HF}]$ (from the stoichiometry of reaction $(2)$), and that the decrease in $[\ce{F-}]$ due to reaction $(2)$ is negligible, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;[\ce{OH-}] &amp;amp;= \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]} \\&#xA;&amp;amp;= \sqrt{\left(\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\right) (4.2 \times 10^{-4})} \\&#xA;&amp;amp;= 7.9 \times 10^{-8}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the $1 \times 10^{-7}~\mathrm{M}$ of $\ce{OH-}$ from the autodissociation of water,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;[\ce{OH-}] &amp;amp;= 1.8 \times 10^{-7} \\&#xA;\mathrm{pOH} &amp;amp;= -\log{(1.8 \cdot 10^{-7})} = 6.7\\&#xA;\mathrm{pH} &amp;amp;= \mathbf{7.3} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="16103" LastEditorUserId="16683" LastEditDate="2017-09-03T09:07:23.627" LastActivityDate="2017-09-03T09:07:23.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33596" PostTypeId="2" ParentId="32325" CreationDate="2015-07-04T10:44:08.120" Score="4" Body="&lt;p&gt;A handy guide to visualise and explain the chemical differences between tetrahedron and trigonal pyramid geometries comes from the &lt;a href=&quot;http://herzberg.dealhot.info/1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Nobel Prize in Chemistry 1971: Gerhard Herzberg&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n7ccH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a handy explanation of the geometries in a YouTube clip referenced to Grand Valley State University: &lt;a href=&quot;https://www.youtube.com/watch?v=eOQ_FqFJi0k&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular Geometry and Polarity: Tetrahedral (trigonal pyramidal)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-07-04T10:44:08.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33599" PostTypeId="1" CreationDate="2015-07-04T13:17:33.950" Score="15" ViewCount="738" Body="&lt;p&gt;In the following question, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wine/water_mixing_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;water/wine mixing problem&lt;/a&gt; was recast in chemical terms. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Partial_molar_property&quot; rel=&quot;nofollow noreferrer&quot;&gt;partial molar volumes&lt;/a&gt; can make this a much more challenging problem than it appears on the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/33579/wine-water-mixing-problem&quot;&gt;https://chemistry.stackexchange.com/questions/33579/wine-water-mixing-problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a container of &lt;span class=&quot;math-container&quot;&gt;$\pu{10 L}$&lt;/span&gt; of ethanol and another container of &lt;span class=&quot;math-container&quot;&gt;$\pu{10 L}$&lt;/span&gt; of water. To make a point, let's transfer &lt;span class=&quot;math-container&quot;&gt;$\pu{5 L}$&lt;/span&gt; of ethanol to the water container and then transfer &lt;span class=&quot;math-container&quot;&gt;$\pu{5 L}$&lt;/span&gt; of the water/ethanol solution back to the ethanol container. What is the purity in each container at the end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to consider that mixtures of water and ethanol have less volume than the two do separately:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJn5X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, after the first transfer, the water container contains &lt;span class=&quot;math-container&quot;&gt;$\pu{10.000 kg}$&lt;/span&gt; of water and &lt;span class=&quot;math-container&quot;&gt;$\pu{3.945 kg}$&lt;/span&gt; of ethanol for a total mass of &lt;span class=&quot;math-container&quot;&gt;$\pu{13.945 kg}$&lt;/span&gt;. However, we need to compute mole fraction to determine the volume of the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{c|c|c}&#xA;\ &amp;amp; \text{water} &amp;amp; \text{ethanol}\\ \hline&#xA;\text{mass in kg} &amp;amp; 10.000 &amp;amp; 3.945\\ \hline&#xA;\text{moles} &amp;amp; 555 &amp;amp; 85.6\\ \hline&#xA;\text{mole fraction} &amp;amp; 0.866 &amp;amp; 0.134&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From our graph, it looks like we have an excess volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{-0.75 mL/mol}$&lt;/span&gt;. Since we have total moles &lt;span class=&quot;math-container&quot;&gt;$=640.6$&lt;/span&gt;, we have an excess volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{-0.75 mL/mol}\times \pu{640.6 mol}\times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$&lt;/span&gt;. Thus the actual volume is not &lt;span class=&quot;math-container&quot;&gt;$\pu{15 L}$&lt;/span&gt; but more like &lt;span class=&quot;math-container&quot;&gt;$\pu{14.5 L}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we transfer &lt;span class=&quot;math-container&quot;&gt;$\pu{5 L}$&lt;/span&gt; back, we are left with less than &lt;span class=&quot;math-container&quot;&gt;$\pu{10 L}$&lt;/span&gt; in the water container. We will also have less than &lt;span class=&quot;math-container&quot;&gt;$\pu{10 L}$&lt;/span&gt; in the ethanol container! &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What volume &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt;, as a fraction of the original volume &lt;span class=&quot;math-container&quot;&gt;$V_0$&lt;/span&gt;, needs to be transferred in order for both containers to have the same final purity? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorUserId="61578" LastEditDate="2020-02-13T17:16:08.327" LastActivityDate="2020-02-16T07:06:43.633" Title="Mixing two containers of each water and ethanol to varying degrees of purity" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33600" PostTypeId="2" ParentId="33338" CreationDate="2015-07-04T13:17:34.507" Score="-2" Body="&lt;p&gt;The bond lengths of carbon 1 &amp;amp; 3 are same and they are longer than that of carbon 5 and its attached nitrogen. This is due to resonance and in resonance we see 1 structure at a time. &lt;/p&gt;&#xA;" OwnerUserId="17291" LastEditorUserId="275" LastEditDate="2015-07-04T14:15:40.127" LastActivityDate="2015-07-04T14:15:40.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33601" PostTypeId="1" AcceptedAnswerId="33626" CreationDate="2015-07-04T13:30:03.433" Score="2" ViewCount="35863" Body="&lt;p&gt;Today I was trying to clean my laptop with a nail polish remover (ingredients: ethyl acetate, peg-40, hydrogenated castor oil, panthenol, parfum, d-limonene, geraniol, ci 26100, ci 60725). Not only it didn't get clean, but some white thing remained on the keys.&#xA;I tried boiling water and a piece of cloth to make it warm and easier to remove without any success.&lt;br&gt;&#xA;What can be done to reverse the damage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wUZrel.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="17289" LastEditorUserId="4945" LastEditDate="2015-07-05T17:31:13.500" LastActivityDate="2017-08-31T16:04:27.633" Title="How to reverse damage done by nail polish remover without acetone to a laptop keyboard?" Tags="&lt;solvents&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33602" PostTypeId="1" AcceptedAnswerId="33606" CreationDate="2015-07-04T13:31:59.913" Score="14" ViewCount="1063" Body="&lt;p&gt;How is spiro[4.4]nonatetraene antiaromatic? I know that spiro compounds exhibit homoaromaticity, but how do they exhibit antiaromaticity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AiysL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4231" LastEditDate="2015-07-04T16:04:37.847" LastActivityDate="2015-07-04T16:13:07.090" Title="How is spiro[4.4]nonatetraene antiaromatic" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33603" PostTypeId="2" ParentId="33588" CreationDate="2015-07-04T14:33:28.490" Score="2" Body="&lt;p&gt;This post answers the question of the of what the other quantum numbers mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/31639/what-do-the-quantum-numbers-actually-signify/31641#31641&quot;&gt;What do the quantum numbers actually signify?&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how many electrons can have $n=5,\ l=3,\ m_l=-1$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pauli Exclusion Principle&lt;/a&gt; states that two identical fermions (an eletron is a type of fermion) cannot occupy the same quantum state. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words, two electrons (in the same atom) cannot have identical quantum numbers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For each $n$, the values of $l=0,1,2,...,n$&lt;/li&gt;&#xA;&lt;li&gt;For each $l$, the values of $m_l=-l,...,-1,0,+1,...,+l$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For each $m_l$ there are only two allowed values of $m_s$, namely $m_s = + \frac{1}{2}$ and $m_s = -\frac{1}{2}$. Thus, each set of $\{n,l,m_l\}$ can only describe two electrons. In your case, we have two electrons with $n=5,\ l=3,\ m_l=-1$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\{n,l,m_l,m_s\}=\{5,3,-1,+\frac{1}{2}\}$&lt;/li&gt;&#xA;&lt;li&gt;$\{n,l,m_l,m_s\}=\{5,3,-1,-\frac{1}{2}\}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-07-04T14:33:28.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33604" PostTypeId="2" ParentId="33552" CreationDate="2015-07-04T14:58:08.020" Score="1" Body="&lt;p&gt;After some more reading, it turns out that the protocols suggesting to use the integrated intensity are wrong and it is necessary to correct for the variable wavelength across the emission peak. &lt;a href=&quot;http://dx.doi.org/10.1038/nprot.2013.087&quot; rel=&quot;nofollow&quot;&gt;Würth et al. (2013)&lt;/a&gt; provide a description of the correct method. First, they state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;all fluorescence spectra have to be converted to a (relative) number of photons per unit time&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second, they describe the conversion as discussed in the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the radiometric quantity has to be multiplied with $\lambda/(hc_0)$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17264" LastEditorUserId="17264" LastEditDate="2015-07-04T15:24:43.760" LastActivityDate="2015-07-04T15:24:43.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33605" PostTypeId="2" ParentId="31459" CreationDate="2015-07-04T14:58:32.163" Score="3" Body="&lt;p&gt;The enzyme catalysis reaction following MM kinetics can be depicted as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Large\ce{E + S&amp;lt;=&amp;gt;[k_f][k_r]ES-&amp;gt;[k_{cat}]E + P }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope your doubt is clear.&lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="5295" LastEditDate="2015-07-27T09:15:35.137" LastActivityDate="2015-07-27T09:15:35.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33606" PostTypeId="2" ParentId="33602" CreationDate="2015-07-04T16:02:27.263" Score="11" Body="&lt;p&gt;The orbital interaction between the 2 butadiene units in spiro[4.4]nonatetraene is referred to as &lt;strong&gt;spiroconjugation&lt;/strong&gt;.  Note that the 2 rings attached through the common or &lt;strong&gt;spiro&lt;/strong&gt; carbon are perpendicular to one another.  Each cyclopentadiene ring contains a 1,3-butadiene unit.  In the figure below, I've drawn in the molecular orbitals for the highest occupied molecular orbital ($\Psi_2$) of 1,3-butadiene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/inWr9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit easier to analyze the situation using a Newman projection.  In the Newman projection on the right I only show the 4 carbons (and their P-orbitals) that are directly attached to the central spiro carbon, since these atoms and orbitals are the only ones close enough to interact between one ring and the other.  Notice how all of the through space interactions between nearby P-orbitals is stabilizing (indicated by arrows).  That is, all of the overlaps are construcive (bonding}, either + + or - -.  This indicates that $\Psi_2$ has the correct symmetry to interact with itself.  Look at the molecular orbitals (MO) for 1,3-butadiene and convince yourself that $\Psi_1$ and $\Psi_3$ do not have the appropriate symmetry to interact with $\Psi_2$ (&lt;em&gt;e.g&lt;/em&gt;. there are no + + overlaps between the carbons attached to the spiro carbon, only + - overlaps that cancel out; and energetically, $\Psi_4$ is to far away to interact.  When the two $\Psi_2$ MOs interact, as usual, new MOs are formed; one is stabilized and one is destabilized and the destabilized orbital is always slightly more destabilized than the stabilized orbital is stabilized.  The net result in this case is that spiroconjugation produces a (small) net destabilization and the molecule is termed antiaromatic in a spiroconjugation sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider spiro[2.4]heptatriene. Now the only interaction of consequence (&lt;em&gt;e.g&lt;/em&gt;. where the interactions are of the ++ or -- type rather than +-; said differently, only $\Psi_2$ of the diene and $\ce{\pi_^{*}}$ of the ethylene have the appropriate symmetry to interact) occurs between the HOMO of the diene ($\Psi_2$) and the LUMO of the ethylene in the 3-membered ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9qeZl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://books.google.com/books?id=nyx1cCjMRdYC&amp;amp;pg=PT50&amp;amp;lpg=PT50&amp;amp;dq=spiro%20aromaticity&amp;amp;source=bl&amp;amp;ots=M0m04VeZKy&amp;amp;sig=3o8AyvKdScboNHjUf6ZQULs7AdA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=7fWXVZSRJoidNtKRsvAD&amp;amp;ved=0CF8Q6AEwCQ#v=onepage&amp;amp;q=spiro%20aromaticity&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, as the orbital mixing diagram indicates, there is a net stabilization of the pi electrons that results.  Spiro[2.4]heptatriene is said to be aromatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiroconjugated compounds containing 4n+2 pi electrons will turn out to be stabilized (aromatic), while those with 4n pi electrons will be destabilized (antiaromatic).  Remember, &lt;strong&gt;these effects are small&lt;/strong&gt;, but they can be seen in UV spectral shifts and even in solvolysis reaction rates of related compounds.  If you google spiroconjugation or spiroaromaticity you can find links to the original articles that appeared in the late 60's, as well as some more recent information on the subject.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-07-04T16:13:07.090" LastActivityDate="2015-07-04T16:13:07.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33607" PostTypeId="2" ParentId="5611" CreationDate="2015-07-04T16:39:45.957" Score="6" Body="&lt;p&gt;$EC_{50}$ is concentration of a bioactive molecule which gives half the maximum response. You should note that the response is bound by a maximum which means this dose-response curve follows a saturation kinetics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturation kinetics can be modelled in many ways and Hill and &lt;a href=&quot;https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics&quot; rel=&quot;nofollow&quot;&gt;Michaelis-Menten kinetics&lt;/a&gt; are the most commonly used models in biochemistry. Again, a point to note is that many mathematical functions/models &lt;strong&gt;do not&lt;/strong&gt; explain the underlying mechanism. They just mimic the behaviour of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume that the antibody-response follows a two-step kinetics like Michaelis-Menten. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Ab + Ag &amp;lt;=&amp;gt;[k_a][k_d] AbAg -&amp;gt;[k_{res}] Response }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call $R_{max}$ as the maximum response rate. Like in the case of MM kinetics, the response rate $R$ would be (see the wiki page for the derivation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=R_{max}\frac{[Ag]}{K_m+[Ag]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $K_m$ is the concentration of antibody which would give half the maximum response rate (half response rate times times a sampling time would give you the $EC_{50}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what $K_m$ really is depends on the approximation of the model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Equilibrium model: this assumes that the binding and unbinding reactions are much faster than the response reaction and are hence at equilibrium (wrt response dynamics). In this case $K_m$ would be same as the dissociation constant. &lt;/li&gt;&#xA;&lt;li&gt;Quasi Steady State model: this assumes that enzyme-substrate complex is at a quasi steady state (QSS) i.e. it doesn't change with time. In the case of enzyme substrate reactions, the substrate gets consumed and so the QSS would result from both the binding-unbinding reactions and conversion reaction. In this case $K_m=\dfrac{k_d+k_{res}}{k_a}$. However in this case the antibody is not consumed to give the response. So QSS would be equivalent to equilibrium approximation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would also apply for the Hill kinetics in which response can be described as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=R_{max}\frac{[Ag]^n}{{K_m}^n+[Ag]^n}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is an integer which represents co-operativity in binding i.e. one bound antibody enhances the binding of the next (total number of binding sites is the upper bound of $n$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that equilibrium model may not be applicable for drugs that can get consumed in a reaction. Also, not all drug induced reaction would follow this simplified model. There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Cooperative_binding&quot; rel=&quot;nofollow&quot;&gt;other models&lt;/a&gt; for co-operative binding too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, before concluding anything about kinetics from an empirical measure like $EC_{50}$, we need to understand (or at least assume) the underlying model. &lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="5295" LastEditDate="2015-07-21T07:25:02.740" LastActivityDate="2015-07-21T07:25:02.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33609" PostTypeId="1" AcceptedAnswerId="33649" CreationDate="2015-07-05T00:38:24.553" Score="3" ViewCount="207" Body="&lt;p&gt;I understand that there are many different catalytic mechanisms that suggest different ways of accounting for the effect of a catalyst on reaction rate. My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a termolecular elementary (it is known it is an elementary reaction) reaction of the form:&#xA;$$\ce{A + B + C &amp;lt;=&amp;gt; D + E + C}$$&#xA;Does the rate of reaction simplify down (according to collision theory) to:&#xA;$$r = [\ce{C}]\bigr(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\bigr)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: To clarify my question, I have a gas phase elementary reaction of the form:&#xA;$$\ce{A + B + C -&amp;gt; D + E + C}$$&#xA;I have another of the form:&#xA;$$\ce{D + E + C -&amp;gt; A + B + C}$$&#xA;Whether they are exactly the same mechanism is irrelevant. My question relates to how quantitatively does a collision partner's concentration in a gas mixture affect reaction rate. i.e. is forward rate defined by (assuming that I know the reaction acts as an elementary reaction in that there is no intermediate):&#xA;$$r_\mathrm{fwd} = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]$$&#xA;and reverse rate defined by (reverse is also elementary reaction - not concerned if same mechanism just also acts in the rate law as a single collision event):&#xA;$$r_\mathrm{rev} = k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$&#xA;Therefore, the total rate, not assuming they're the same reaction just two elementary reactions that go in opposite directions and we've called on forward and one reverse, the total rate forward, taking into account our other reaction, is:&#xA;$$r = r_\mathrm{fwd} - r_\mathrm{rev}$$&#xA;$$r = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}] - k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$&#xA;$$r = [C]\left(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\right)$$&#xA;Therefore quantitatively a collision partner does not change the overall rates direction but rather enhances it in both directions. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="41328" LastEditDate="2017-08-06T04:36:23.813" LastActivityDate="2017-08-06T04:36:23.813" Title="Effect of Catalysis on Reaction Rate in elementary Reaction" Tags="&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33610" PostTypeId="2" ParentId="7244" CreationDate="2015-07-05T01:15:08.540" Score="0" Body="&lt;p&gt;See my recent article on this very subject at the following link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/&quot; rel=&quot;nofollow&quot;&gt;https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion is that, mechanically at least, the difference between reversible and irreversible processes is the result of viscous dissipation of mechanical energy in the irreversible case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chet&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-05T01:15:08.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33611" PostTypeId="1" CreationDate="2015-07-05T01:20:59.547" Score="18" ViewCount="85343" Body="&lt;p&gt;Looking at the Pauling electronegativities in the Periodic Table (below, from &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity&quot; rel=&quot;noreferrer&quot;&gt;ChemWiki&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/80wJk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Asides from the overall trend of increasing electronegativity across and up the Periodic Table (towards fluorine), there seems to be a minor peak in the first 3 periods of the $d$-block elements, this peak becomes more pronounced as you go down the $d$-block, with the most noticable peaks in Periods 5 and 6 - seemingly the reverse of the overall trend. This is shown in the graph below (&lt;a href=&quot;http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s11-the-periodic-table-and-periodi.html&quot; rel=&quot;noreferrer&quot;&gt;Image source)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xitIQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ChemWiki page, there is also a reference to the Allred-Rochow Electronegativity Scale, which also has peaks occurring most prominently in Periods 4 and 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nHwui.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://de.wikipedia.org/wiki/Elektronegativit%C3%A4t#Allred-Rochow-Skala&quot; rel=&quot;noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity#Methods_of_calculation&quot; rel=&quot;noreferrer&quot;&gt;Electronegativity&lt;/a&gt; details several methods of calculation, which brings into question how well-defined the electronegativity values are; however, there still seems to be the peaks occurring, particularly in Periods 5 and 6, and to a lesser extent in Period 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the reason for these electrongativity peaks in the $d$-block?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-07-05T02:10:15.903" LastActivityDate="2017-11-02T23:18:03.937" Title="Why are there peaks in electronegativities in d-block elements?" Tags="&lt;periodic-table&gt;&lt;transition-metals&gt;&lt;periodic-trends&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33612" PostTypeId="2" ParentId="31830" CreationDate="2015-07-05T03:10:54.767" Score="4" Body="&lt;p&gt;Perhaps a derivation of the definition I was taught. Consider the Henderson-Hasselbalch equation: $$pH=pK_a+\log\bigg(\frac{[A^-]}{[HA]}\bigg). $$ Given the definition provided by Klaus above and the fact that a weak acid's $pK_a$ is unique, the only term that could possible alter the $pH$ by an amount of $\pm1$ is $\log\big(\frac{[A^-]}{[HA]}\big).$ We decide to choose the case of the $pH$ increasing by $+1$ (it does not matter which direction---either more acidic or more basic---the reaction proceeds, we simply want the magnitude of the numerical change to be 1). If the $pH$ increases by 1, then the solution becomes more basic. Consequently, the term $\log\big(\frac{[A^-]}{[HA]}\big)$ changes by $\log\big(\frac{[A^-]+\beta}{[HA]-\beta}\big)$, where $\beta$ denotes the buffer capacity. We know this term must equal $+1$ because a perfect buffer has equimolar solutions of $[HA]$ and $[A^-]$. We obtain the following equation: $$1=\log\bigg(\frac{[A^-]+\beta}{[HA]-\beta}\bigg). $$ Solving for $\beta$ yields the buffer capacity equation $$\beta=\frac{10[HA]-[A^-]}{11}. $$&lt;/p&gt;&#xA;" OwnerUserId="17294" LastActivityDate="2015-07-05T03:10:54.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33613" PostTypeId="1" AcceptedAnswerId="33988" CreationDate="2015-07-05T03:34:12.290" Score="8" ViewCount="713" Body="&lt;p&gt;The Raschig process (according to the &lt;a href=&quot;https://de.wikipedia.org/wiki/Raschig-Synthese&quot; rel=&quot;nofollow noreferrer&quot;&gt;german wikipedia&lt;/a&gt;) is divided in the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{2NaOH + Cl2 -&amp;gt; NaOCl + NaCl + H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NaOCl + H2O -&amp;gt; HOCl + NaOH}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{HClO + NH3 -&amp;gt; NH2Cl + H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NH2Cl + NH3 -&amp;gt; N2H4 + HCl}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{N2H4 + 2NH2Cl -&amp;gt; N2 + 2NH4Cl}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But what happens now if one changes $\ce{NH3}$ into $\ce{R-NH2}$, with $\ce{R}=$&lt;img src=&quot;https://i.stack.imgur.com/7cd18.png&quot; alt=&quot;enter image description here&quot;&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from step 3 of the Raschig process I came up to the reaction of the produced hydrazine derivate to give an azo compound.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{HClO + R-NH2 -&amp;gt; R-NHCl + H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{R-NHCl + R-NH2 -&amp;gt; R-NH-NH-R + HCl}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{R-NH-NH-R + x~R-NHCl -&amp;gt; R-N=N-R~+~?}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The last reaction step bothers me a little ... is it simply&#xA;$$\ce{R-NH-NH-R + 2R-NHCl -&amp;gt; R-N=N-R + 2R-NH2Cl}$$&#xA;but without the cleavage?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="10185" LastEditDate="2015-07-12T16:19:07.113" LastActivityDate="2015-07-14T19:48:33.053" Title="Search for the last step of an altered Raschig process reaction mechanism" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33614" PostTypeId="1" CreationDate="2015-07-05T07:49:40.983" Score="9" ViewCount="12680" Body="&lt;p&gt;Many periodic tables place lanthanum and actinium in the f-block of elements, for example, &lt;a href=&quot;http://periodic.lanl.gov/index.shtml&quot; rel=&quot;noreferrer&quot;&gt;this periodic table&lt;/a&gt; from Los Alamos National Laboratory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;a href=&quot;http://www.rsc.org/periodic-table&quot; rel=&quot;noreferrer&quot;&gt;this table&lt;/a&gt; from the Royal Society of Chemsitry places lanthanum and actinium in the d-block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the most appropriate placement of these two elements on the periodic table? What is the most recent scientific consensus regarding this matter?&lt;/p&gt;&#xA;" OwnerUserId="16852" LastEditorUserId="275" LastEditDate="2015-07-05T11:31:16.920" LastActivityDate="2021-03-07T04:46:45.923" Title="Are the elements lanthanum and actinium considered to be in the d block or the f block of the periodic table?" Tags="&lt;periodic-table&gt;&lt;elements&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33615" PostTypeId="2" ParentId="33609" CreationDate="2015-07-05T08:11:14.003" Score="5" Body="&lt;p&gt;If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an &lt;a href=&quot;http://goldbook.iupac.org/E02035.html&quot; rel=&quot;nofollow&quot;&gt;elementary reaction&lt;/a&gt;. For example, $$\ce{A + B -&amp;gt; C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -&amp;gt; A + B}$$ is a different elementary reaction. The rates of both reactions &#xA;\begin{align}&#xA;r&amp;amp;=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\&#xA;r&amp;amp;=k_\mathrm{rev}[\ce{C}][\ce{D}]\\&#xA;\end{align}&#xA;would determine the location of the equilibrium &#xA;\begin{align}&#xA;\ce{A + B &amp;amp;&amp;lt;=&amp;gt; C + D}&amp;amp; K&amp;amp;=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}.&#xA;\end{align}&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a true &lt;em&gt;termolecular&lt;/em&gt; reaction, i.e. three species react in one concerted transition state,&#xA;$$\ce{A + B + C -&amp;gt; D + E + F},$$&#xA;the rates are always given as&#xA;\begin{align}&#xA;r&amp;amp;=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\&#xA;r&amp;amp;=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\&#xA;\end{align}&#xA;You &lt;em&gt;cannot&lt;/em&gt; simplify a rate law of an elementary reaction.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="24052" LastEditDate="2016-03-11T23:50:50.483" LastActivityDate="2016-03-11T23:50:50.483" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33616" PostTypeId="1" AcceptedAnswerId="33623" CreationDate="2015-07-05T09:15:58.727" Score="0" ViewCount="94" Body="&lt;p&gt;I have some questions regarding generation of mol files-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;For bond block the numbering can be done at random or it has to follow some order&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also in the atom block the order in which atoms occur can it be random or again it has to follow any order&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is order of numbering done in bond block and atom block related&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any free software which converts mol files to structure&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is a sample structure with mol file-&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wRExD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  Sample&#xA;&#xA; 22 23  0  0  0  0  0  0  0  0999 V2000&#xA;   -2.5962    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -2.5962    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.0000    0.7512    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.0000    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    3.0047    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -4.6479    1.7042    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005    3.0047    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958    4.5070    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.2958   -1.5023    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.3005   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009   -2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.8967   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.8967    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    5.1972   -2.2535    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0&#xA;    2.6009   -3.7559    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -3.8967    3.0047    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -5.1972    3.7559    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -3.1455    4.3052    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    3.9014   -4.5070    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0&#xA;  1  2  2  0  0  0  0&#xA;  2  3  1  0  0  0  0&#xA;  3  4  1  0  0  0  0&#xA;  4  5  1  0  0  0  0&#xA;  5  6  1  0  0  0  0&#xA;  1  6  1  0  0  0  0&#xA;  1 19  1  0  0  0  0&#xA;  5  8  2  0  0  0  0&#xA;  6  9  1  0  0  0  0&#xA;  3 10  2  0  0  0  0&#xA;  4 11  1  0  0  0  0&#xA; 11 12  1  0  0  0  0&#xA; 12 13  2  0  0  0  0&#xA; 13 14  1  0  0  0  0&#xA; 14 15  2  0  0  0  0&#xA; 15 16  1  0  0  0  0&#xA; 11 16  2  0  0  0  0&#xA; 14 17  1  0  0  0  0&#xA; 13 18  1  0  0  0  0&#xA; 19  7  1  0  0  0  0&#xA; 19 20  1  0  0  0  0&#xA; 19 21  1  0  0  0  0&#xA; 18 22  2  0  0  0  0&#xA;M  END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17178" LastEditorUserId="5017" LastEditDate="2015-07-05T15:19:20.593" LastActivityDate="2015-07-05T15:23:43.277" Title="Questions regarding mol files" Tags="&lt;cheminformatics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33617" PostTypeId="2" ParentId="24885" CreationDate="2015-07-05T11:16:59.990" Score="4" Body="&lt;p&gt;Safety is a critical issue here, and it is very possible to simulate eruption safely, with a simple combination of household items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple that I have seen and tried with some of my younger relatives - quite surprising some of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemistry.about.com/od/chemicalvolcanoes/a/ketchupvolcano.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ketchup and Baking Soda Volcano&lt;/a&gt;, this takes a bit of trial and error to get the consistency right. Mix some detergent and water with the ketchup - then react it with baking soda. (Be sure to read the ingredients list on the ketchup first). The reason the eruption occurs is due to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The ketchup contains vinegar, which is dilute acetic acid. The acetic acid reacts with the baking soda to produce carbon dioxide gas. The gas bubbles expand and rise through the liquid, bubbling out the ketchup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQ293t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also tried adding vinegar (acetic acid) to ketchup for an eruption. Once again, trial and error is needed to get the best consistency required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic Mentos in Coca Cola is a good one as well, but can be too fast and messy (use diet coke to avoid sticky residue). This may or may not be viable due to its lack of viscosity.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="7448" LastEditDate="2015-07-05T11:19:37.220" LastActivityDate="2015-07-05T11:19:37.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33618" PostTypeId="1" AcceptedAnswerId="33625" CreationDate="2015-07-05T11:47:53.343" Score="14" ViewCount="896" Body="&lt;p&gt;Why can singlet oxygen participate in cycloaddition reactions while triplet oxygen, on the other hand, can't?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="9961" LastEditDate="2016-05-06T15:00:04.300" LastActivityDate="2016-05-06T15:00:04.300" Title="Singlet/triplet oxygen cycloadditions" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;reactivity&gt;&lt;pericyclic&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33619" PostTypeId="1" AcceptedAnswerId="33621" CreationDate="2015-07-05T12:34:27.837" Score="24" ViewCount="1282" Body="&lt;p&gt;Ugh, I drank orange juice too soon after brushing my teeth and my mouth feels awful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes that foul taste with this combination of toothpaste and orange juice? Is it a reaction between the oil of wintergreen in the toothpaste and the citric acid in the OJ? Is it something else? What's the explanation for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second part to this question, is there a way to stop this reaction from happening, aside from &lt;em&gt;not&lt;/em&gt; drinking orange juice after brushing?&lt;/p&gt;&#xA;" OwnerUserId="17299" LastEditorUserId="17299" LastEditDate="2017-04-01T10:15:46.890" LastActivityDate="2017-04-01T10:15:46.890" Title="What causes that foul taste with combination of toothpaste and orange juice?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;taste&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33620" PostTypeId="2" ParentId="33614" CreationDate="2015-07-05T12:39:38.353" Score="1" Body="&lt;p&gt;Lanthanum and Actinium are generally both accepted as f-block elements. You will notice in the Royal Society's periodic table, La and Ac are both color coded to match the other f-block elements. They do this to show the beginning of both the Lanthanide and Actinide series, while maintaining a reasonable symmetry for their periodic table (that is, without &quot;holes&quot; in their table, similar to the first link you provided).&lt;/p&gt;&#xA;" OwnerUserId="17294" LastEditorUserId="17294" LastEditDate="2015-07-05T16:21:46.773" LastActivityDate="2015-07-05T16:21:46.773" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33621" PostTypeId="2" ParentId="33619" CreationDate="2015-07-05T14:51:35.920" Score="17" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do juice taste awful?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Awful taste is due to &lt;strong&gt;sodium laureth sulfate&lt;/strong&gt;, also known as &lt;strong&gt;sodium lauryl ether sulfate (SLES)&lt;/strong&gt;, or &lt;strong&gt;sodium lauryl sulfate (SLS)&lt;/strong&gt;- depending on which toothpaste you use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SLES and SLS are &lt;strong&gt;surfactants&lt;/strong&gt;(wetting agent). Both chemicals are added in toothpaste to create foam and make the paste easier to spread around your mouth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both chemicals they suppress the receptors on our taste buds that perceive sweetness, inhibiting our ability to pick up the sweet notes of food and drink. Also break up the &lt;strong&gt;phospholipids&lt;/strong&gt; on our tongue.  These fatty molecules inhibit our receptors for bitterness and keep bitter tastes from overwhelming us, but when they're broken down by the surfactants in toothpaste, bitter tastes get enhanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, anything you eat will be less sweeter and more bitter! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://mentalfloss.com/article/20151/why-does-everything-taste-bad-after-you-brush-your-teeth&quot; rel=&quot;nofollow noreferrer&quot;&gt;mentalfloss&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to prevent awful taste after brushing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;After a bit research I found few ways to avoid awful taste:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Avoid brushing. (not recommended)&lt;/li&gt;&#xA;&lt;li&gt;Wait for about 30 minutes.&lt;/li&gt;&#xA;&lt;li&gt;Look for SLS-free toothpaste.&lt;/li&gt;&#xA;&lt;li&gt;eat a little bit of cheese first.(suggested by &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33619/what-causes-that-foul-taste-with-combination-of-toothpaste-and-orange-juice/33621#comment50613_33621&quot;&gt;repurposer&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://lifehacker.com/5924488/prevent-food-and-drinks-from-tasting-minty-and-weird-after-you-brush-your-teeth&quot; rel=&quot;nofollow noreferrer&quot;&gt;hot water&lt;/a&gt; will do the trick.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-08-26T05:59:30.057" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33623" PostTypeId="2" ParentId="33616" CreationDate="2015-07-05T15:23:43.277" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For bond block the numbering can be done at random or it has to follow some order&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bonds don't need to be in any particular order, although usually bonds are created first for atom 1, then atom 2, etc. However, the first bond could be between atom 101 and atom 100 if you wish.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also in the atom block the order in which atoms occur can it be random or again it has to follow any order&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Atoms can be in any order. Some programs use &quot;canonical ordering&quot; so that the same molecule will always have the same atomic order, but this is rarely used because it can be time-consuming.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is order of numbering done in bond block and atom block related&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As indicated above, both the atom and bond blocks could be in random order. The relation, obviously, is that for a given molecular structure, there will be bonds between bonded atom pairs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there any free software which converts mol files to structure&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are many, but you don't describe whether you want 3D (presumably) or 2D structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I suggest &lt;a href=&quot;http://avogadro.cc/&quot; rel=&quot;nofollow&quot;&gt;Avogadro&lt;/a&gt; which can create 3D coordinates for files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're generating a large &lt;em&gt;batch&lt;/em&gt; of 3D structures, you will likely want to use Open Babel or RDKit or something similar. Both are open source and free.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-05T15:23:43.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33625" PostTypeId="2" ParentId="33618" CreationDate="2015-07-05T16:25:59.967" Score="14" Body="&lt;p&gt;Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, &lt;strong&gt;thermodynamically&lt;/strong&gt; should be quite reactive, yet is &lt;strong&gt;kinetically&lt;/strong&gt; unreactive - once again, a case of kinetics &lt;em&gt;vs&lt;/em&gt;. thermodynamics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it would have to produce a triplet product. For most compounds, the triplet state is considerably higher in energy than the singlet state.  Hence reaction with ground state, triplet $\ce{O2}$ would produce a higher energy, triplet product. Since ground state, triplet dioxygen doesn't have any excess energy, the activation energy required to produce a high energy triplet product just isn't there - hence, no reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, singlet $\ce{O2}$ is a spin-paired excited state with plenty of excess energy (the first singlet lies some 95 kJ/mol above the ground state triplet).  Basically, singlet $\ce{O2}$ behaves like any other electrophilic (due to the electronegativity of oxygen), reactive (due to its excess energy), compound with a double bond. It readily undergoes the Diels-Alder, ene and even 2+2 cycloaddition reactions - again, just like any other reactive olefin.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-23T20:54:50.207" LastActivityDate="2015-08-23T20:54:50.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33626" PostTypeId="2" ParentId="33601" CreationDate="2015-07-05T17:05:31.953" Score="8" Body="&lt;p&gt;The solvent appears not only to have removed paint on the keys but also to have caused &lt;a href=&quot;http://www.multitechproducts.com/content/Procedures/Aristech-TechBulletin-stress-crazing.pdf&quot;&gt;solvent crazing&lt;/a&gt; in the plastic itself. Regrettably, there is probably not much that can be done to actually &lt;em&gt;remove&lt;/em&gt; the crazing, which may have penetrated too deeply to remove by buffing, but you might be able to hide it with a substance such as an oil that has the same index of refraction as the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the appearance is unacceptable, you might replace the keyboard, use a plug-in keyboard, or get a keyboard &lt;em&gt;cover&lt;/em&gt;, such as the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00AVBHMPY&quot;&gt;Kuzy Russian Keyboard Cover&lt;/a&gt;, specific to the Mac Pro. Individual key stickers would probably not adhere well.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-05T17:05:31.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33628" PostTypeId="1" AcceptedAnswerId="33629" CreationDate="2015-07-05T19:28:40.053" Score="3" ViewCount="255" Body="&lt;p&gt;We know that equilibrium in a chemical system is attained when forward and backward reaction rates are equal. What if the reaction mixtures involve more than one reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider three inter-reacting species A, B and C. The rate constants for the forward and backward reactions of $\ce{A -&amp;gt; B}$ are $k_1$ and $k'_1$, that of $\ce{B -&amp;gt; C}$ are $k_2$ and $k'_2$ and that of $\ce{C -&amp;gt; A}$ are $k_3$ and $k'_3$. What is the relation between the six rate constants so that the system remains in equilibrium? Assume first order kinetics for all the reactions. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="12922" LastEditorUserId="194" LastEditDate="2017-04-01T02:20:54.283" LastActivityDate="2017-04-01T02:20:54.283" Title="Equilibrium of more than two inter reacting species" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33629" PostTypeId="2" ParentId="33628" CreationDate="2015-07-05T19:56:23.037" Score="2" Body="&lt;p&gt;At equilibrium $~r_1~=r_2~=r_3~=~0$.&lt;br&gt;&#xA;So, $$k_1[A]-k_{-1}[B]=0$$&lt;br&gt;&#xA;$$k_2[B]-k_{-2}[C]=0$$&lt;br&gt;&#xA;$$k_3[C]-k_{-3}[A]=0$$&lt;br&gt;&#xA;So, The criteria for equilibrium is&lt;br&gt;&#xA;1) $$k_1[A]=k_{-1}[B]$$&lt;br&gt;&#xA;2) $$k_2[B]=k_{-2}[C]$$&lt;br&gt;&#xA;3) $$k_3[C]=k_{-3}[A]$$&lt;br&gt;&#xA;Now defining equilibrium constant as $K_i=\frac{k_i}{k_{-i}}$&lt;br&gt;&#xA;We can derive a formula&lt;br&gt;&#xA;$$K_1 \times K_2 \times K_3~=~1$$&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7777" LastEditDate="2015-07-07T03:00:43.003" LastActivityDate="2015-07-07T03:00:43.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33630" PostTypeId="1" AcceptedAnswerId="33631" CreationDate="2015-07-05T21:17:29.230" Score="5" ViewCount="804" Body="&lt;p&gt;I want to study several carbohydrates that includes but not limited to the following molecules:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Glucose&lt;/li&gt;&#xA;&lt;li&gt;Galactose&lt;/li&gt;&#xA;&lt;li&gt;Mannose&lt;/li&gt;&#xA;&lt;li&gt;Fructose&lt;/li&gt;&#xA;&lt;li&gt;Tagatose&lt;/li&gt;&#xA;&lt;li&gt;Sorbose&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know that for any carbohydrate, possible list of carbohydrates isomers are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;single chain (L and D isomer)&lt;/li&gt;&#xA;&lt;li&gt;$\alpha$ and $\beta$ pyranose or six member ring structures&lt;/li&gt;&#xA;&lt;li&gt;$\alpha$ and $\beta$ furanose or five member ring structures.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, for ring structures there are several rotamers, mainly gg, gt and tg (g=gauche, t=trans) and also there are several anomers but I am not sure about them. Is there anything I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For accurate thorough computational study, I need to identify all the isomers. Is there any single book or document that can guide me to screen the full potential energy surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your kind help.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7448" LastEditDate="2015-07-05T22:20:12.130" LastActivityDate="2015-07-06T00:54:40.133" Title="Carbohydrate configurations" Tags="&lt;organic-chemistry&gt;&lt;computational-chemistry&gt;&lt;reference-request&gt;&lt;isomers&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33631" PostTypeId="2" ParentId="33630" CreationDate="2015-07-05T23:18:08.337" Score="6" Body="&lt;p&gt;There are 32 isomers of C6H12O6, of which you have named 6. The most common one is D-glucose (dextrose). I don't think all of the 16 L-saccharides are found in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;α and β refer to the connection between rings, whether the ketal (or aldehyde) oxygen is axial or equatorial.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UPMYl.jpg&quot; alt=&quot;α- and β- linkages&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/r75Og.png&quot; alt=&quot;D-Pyranoses&quot;&gt;  Structures made with ChemDraw.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/r4Hpq.png&quot; alt=&quot;D-Furanoses&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastEditorUserId="4934" LastEditDate="2015-07-06T00:54:40.133" LastActivityDate="2015-07-06T00:54:40.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33632" PostTypeId="2" ParentId="32826" CreationDate="2015-07-05T23:21:39.237" Score="7" Body="&lt;p&gt;To answer this question at a more relatable level, typically the strength of oxoacids can be determined from the number of &quot;lone&quot; oxygen atoms in the compound--that is, the number of oxygen atoms that do not have a corresponding hydrogen. What usually happens is the central atom in an oxoacid will draw electrons from the bonded oxygen atoms, thereby making them more electronegative. This increases the polarity of the O-H bond, increasing acidity. In your example, H$_2$SO$_4$ has 2 lone oxygen atoms; while in H$_2$S$_2$O$_7$, 5 lone oxygens exist. We therefore suspect that the latter is more acidic. Although there are exceptions to this rule, for a general chemistry course it typically is valid.&lt;/p&gt;&#xA;" OwnerUserId="17294" LastEditorUserId="45386" LastEditDate="2017-05-24T15:13:41.300" LastActivityDate="2017-05-24T15:13:41.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33635" PostTypeId="1" CreationDate="2015-07-06T01:04:27.477" Score="17" ViewCount="394" Body="&lt;p&gt;When I first learned about substitution reactions, I started wondering something that I never ended up figuring out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a strong nucleophile which is also a good leaving group, such as I$^{-}$. Can such a group substitute itself using the SN1 mechanism? If so, would that mean if such a group were bonded to a chiral carbon, an enantiomerically pure solution could spontaneously racemize?&lt;/p&gt;&#xA;" OwnerUserId="17140" LastActivityDate="2016-05-10T01:29:06.230" Title="Is it possible for a group to substitute itself?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33636" PostTypeId="1" AcceptedAnswerId="33660" CreationDate="2015-07-06T01:29:02.293" Score="4" ViewCount="268" Body="&lt;p&gt;Please help me figuring out what is going on with my RP-HPLC protocol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am purifying a small peptide out of E.coli Culture, heat stable enterotoxin, its amino acid sequence is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N-T-F-Y-C-C-E-L-C-C-N-P-A-C-A-G-C-Y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am collecting culture supernatant and using amberlite XAD2 to concentrate and desalting the peptide in 80% and 99% methanol fractions. These fractions were subjected to flash evaporation. Then I am using on MCI gel CHP20P (glass column) for further concentration and removing unwanted proteins (polar and less hydrophobic ones) using stepwise elution with 20%, 40%, 60%, 80% and 100% methanol fractions. Each fraction is subjected to flash evaporation. A sample (1 ml) from each fraction (60, 80 and 100%) was then loaded to RP-HPLC using xbridge C8 preparative column (19x25 mm,10 micron, 130 angstrom) on a gradient system (solvent A: water/ solvent B: 80%methanol) initial-5 minutes 100 A, 5-10 minute 30% B, 10-65 minute 100% B, 65-80 minute 100% B at flow rate 2 ml/minute. The chromatogram showed a single large peak (3 AUF) at 25 minutes, B solvent around 45-50%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering that the retention time is lower than that was reported before, 35 minute at 55-60% B (I followed almost the same protocol 5 years ago except using MCIgel methacrylate resin and Vaydc C8 column instead of mci gel CHP20P and xbridge C8 column). It is also very frustrating that I loaded 0.5 ml of the collected peak on Waters analytic C8 coulumn, 5 micron particle size and it  elutes very early at 5-6 minutes (almost in the mobile phase A: water). My frustration comes from that I am using highly hydrophobic resin particle (MCI gel CHP20P) to capture hydrophobic molecules and loaded these fractions (60%, 80% and 100% methanol MCI gel) on C8 column and eluted so early.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sorry for this long message but I rely on your understanding. Any answer would be highly appreciated. Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="17305" LastEditorUserId="13184" LastEditDate="2015-07-06T18:06:11.987" LastActivityDate="2015-07-06T18:06:11.987" Title="Reverse Phase Chromatography" Tags="&lt;biochemistry&gt;&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33637" PostTypeId="1" AcceptedAnswerId="33676" CreationDate="2015-07-06T01:42:04.157" Score="9" ViewCount="2179" Body="&lt;p&gt;Considering how electronegative oxygen is, why is the chemical shift of alcohol protons (1H-NMR) not so high? The &lt;a href=&quot;http://www2.ups.edu/faculty/hanson/Spectroscopy/NMR/HNMRshift.htm&quot;&gt;chemical shifts of the protons on 3-propanol&lt;/a&gt; are about 1.20 for the carbon-1 hydrogens, about 2.2 for the alcohol proton, and about 4.01 for the carbon-2 hydrogens. Why on earth would the electrons on the carbon-2 hydrogens feel more pull from the oxygen than those on the alcohol proton?&lt;/p&gt;&#xA;" OwnerUserId="408" LastEditorUserId="10185" LastEditDate="2015-07-06T13:34:51.667" LastActivityDate="2015-07-07T05:05:16.870" Title="Why is an alcohol proton so shielded?" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33638" PostTypeId="1" CreationDate="2015-07-06T02:45:28.897" Score="7" ViewCount="9101" Body="&lt;p&gt;I have come across the name trihydroxy propane for glycerol, which I think should've been Propan-1,2,3-triol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it named this way? Is there any  other rule of nomenclature working?&lt;/p&gt;&#xA;" OwnerUserId="17307" LastEditorDisplayName="user7951" LastEditDate="2015-07-06T15:16:51.103" LastActivityDate="2016-10-02T14:53:16.330" Title="Why is the IUPAC name of glycerol expressed in the way that it is?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33639" PostTypeId="1" AcceptedAnswerId="33642" CreationDate="2015-07-06T02:58:37.210" Score="0" ViewCount="2152" Body="&lt;p&gt;Calculate the change in enthalpy as 1kg of nitrogen is heated from 1000K to 1500K, assuming the nitrogen is an ideal gas at a constant pressure. The tempreature dependent specific heat of nitrogen is Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3 where Cp is in kJ/kg-mol and T is in K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is 697.27 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first time i've ever encountered something like &quot;Temperature Dependent Specific Heat&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I know &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H = m*C_p*(T2-T1);&#xA;Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I try to make something like C_p1:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C_p1 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1000&#xA;C_p1 = -8.1934*10^11&#xA;&#xA;C_p2 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1500&#xA;C_p2 = -2.766466215*10^12&#xA;&#xA;Cp2 - Cp1 = -3.8585 * 10^12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So yeah... Its not looking too good. Am i missing a formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?&lt;/p&gt;&#xA;" OwnerUserId="17306" LastEditorDisplayName="user15489" LastEditDate="2015-07-06T04:15:16.863" LastActivityDate="2019-02-08T03:48:33.963" Title="Temperature Dependent Specific Heat of Nitrogen" Tags="&lt;homework&gt;&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33640" PostTypeId="2" ParentId="33638" CreationDate="2015-07-06T03:18:37.440" Score="5" Body="&lt;p&gt;Glycerol and trihydroxypropane are common names. The IUPAC name is propane-1,2,3-triol. The common name glycerol comes from the root &lt;em&gt;glyco&lt;/em&gt;- which means sweet. Glucose, glycogen, and glycerin (another name for glycerol) have the same etymology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trihydroxypropane makes sense as a common name because there are only 3 locations for the hydroxyl groups on the 3 carbon chain, so the name is basically unambiguous - multiple hydroxyl groups on the same carbon are not stable (see geminal diols).&lt;/p&gt;&#xA;" OwnerUserId="17140" LastEditorUserId="17140" LastEditDate="2015-07-06T03:24:51.550" LastActivityDate="2015-07-06T03:24:51.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33642" PostTypeId="2" ParentId="33639" CreationDate="2015-07-06T04:38:30.030" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, exactly!  The equation you should be using is the differential form of the enthalpy equation you are already using.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH = m~C_p(T)~dT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{H_1}^{H_2} dH = m \int_{T_1}^{T_2} C_p(T)~dT  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can verify, &lt;strong&gt;if&lt;/strong&gt; heat capacity is &lt;em&gt;independent&lt;/em&gt; of temperature, i.e. a constant, the equation reduces to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_2 - H_1 = m~C_p~\left (T_2 - T_1 \right ) \implies \Delta H = m~C_p\Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equation you tried to use originally.  However, since we derived this equation assuming a heat capacity that is &lt;em&gt;in&lt;/em&gt;dependent of temperature, and your question explicitly involves dependence on the temperature, we should not use the already integrated equation.  Instead you'll have to do the integration yourself by substitution of temperature-dependent equation.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-07-06T15:21:50.013" LastActivityDate="2015-07-06T15:21:50.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33643" PostTypeId="1" CreationDate="2015-07-06T05:09:24.167" Score="2" ViewCount="1572" Body="&lt;p&gt;An air with mass of $0.454\ \mathrm{kg}$ and an unknown mass of $\ce{CO2}$ occupy an 85 liters tank at $2068.44\ \mathrm{kPa}$. If the partial pressure of the $\ce{CO2}$ is $344.74\ \mathrm{kPa}$, determine its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is $0.138\ \mathrm{kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok lets setup Dalton's Law: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;p_\text{total} &amp;amp;= p_\text{air} + p_{\ce{CO2}} \\&#xA;2068.44 &amp;amp;= \left(mRT/V\right)_\text{air} + 344.74 \\&#xA;m_\text{air} &amp;amp;= 0.454\ \mathrm{kg} \\&#xA;R_\text{air} &amp;amp;= 0.287\ \mathrm{kJ/(kg\ K)} \\&#xA;T_\text{air}/V_\text{air} &amp;amp;= 18847.73126&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so I need to get something out of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;T_\text{air}/V_\text{air} &amp;amp;= 18847.73126 \\&#xA;V_\text{total} &amp;amp;= V_\text{air} + V_{\ce{CO2}} = 0.085\ \mathrm{m^3}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so im basically stuck there. only 2 equations for 3 unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint? or am I totally approaching this the wrong way?&lt;/p&gt;&#xA;" OwnerUserId="17306" LastEditorUserId="69830" LastEditorDisplayName="user7951" LastEditDate="2021-05-08T18:30:01.833" LastActivityDate="2021-05-08T18:30:01.833" Title="Unknown Mass in Dalton's Law of Partial Pressures" Tags="&lt;thermodynamics&gt;&lt;pressure&gt;&lt;ideal-gas&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33644" PostTypeId="1" CreationDate="2015-07-06T05:36:33.107" Score="5" ViewCount="1385" Body="&lt;p&gt;I've been investigating chemical kinetics and stumbled across the CHEMKIN program manuals and the JANAF tables. In the original CHEMKIN database they list 'standard enthalpy'. I've checked and this is not in fact enthalpy of formation as this is a separately listed value. What I'm confused about is that I cannot find any reference to molar enthalpy any where else except in the the &lt;a href=&quot;http://kinetics.nist.gov/janaf/pdf/JANAF-FourthEd-1998-1Vol1-Intro.pdf&quot; rel=&quot;nofollow&quot;&gt;introduction to the JANAF tables&lt;/a&gt; and the &lt;a href=&quot;http://www3.nd.edu/~powers/ame.60636/kee2000.pdf&quot; rel=&quot;nofollow&quot;&gt;CHEMKIN database&lt;/a&gt;. What's more it lists the enthalpy of $\ce{H2O}$ and $\ce{H2O2}$ as zero and other substances have negative enthalpy. Questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is specific molar enthalpy in comparison to water or some other substance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is still a relative measurement of enthalpy what advantage does it provide over heat of formation in practical use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the two substances both have the same enthalpy even if it is relative if one has an extra atom and an extra bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I completely wrong and the whole thing is something else entirely?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorDisplayName="user15489" LastEditDate="2015-07-06T05:57:41.413" LastActivityDate="2018-12-23T17:35:35.890" Title="Negative and Zero Enthalpy?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33645" PostTypeId="1" CreationDate="2015-07-06T09:36:30.237" Score="7" ViewCount="18007" Body="&lt;p&gt;The form of the p orbitals that we are familiar with are the $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AecWX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AecWX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(source: &lt;a href=&quot;http://www.chemtube3d.com/A%20Level%20orbitals-p.htm&quot; rel=&quot;noreferrer&quot;&gt;ChemTube 3D&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that the p subshells have the quantum number $l = 1$, meaning that the magnetic quantum number can take the three values $m_l = -1, 0, +1$. Which orbital has which value of $m_l$?&lt;/p&gt;&#xA;" OwnerUserId="17313" LastEditorUserId="16683" LastEditDate="2016-11-04T14:21:16.200" LastActivityDate="2019-07-18T07:18:05.200" Title="What are the magnetic quantum numbers for the three real p orbitals?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33646" PostTypeId="1" CreationDate="2015-07-06T09:51:10.323" Score="3" ViewCount="472" Body="&lt;p&gt;When we perform the hydrolysis of cellulose by acid, we already knew that it can convert cellulose into glucose. As we know, cellulose is composed of beta glucose units linked by beta 1,4-glucosidic bonds. So I would like to know, resulting glucose after hydrolysis is beta or alpha glucose? If beta glucose is present, how can it be converted to alpha?&lt;/p&gt;&#xA;" OwnerUserId="17314" LastActivityDate="2015-07-07T17:39:01.583" Title="Can beta D-glucose be present hydrolysate after hydrolysis of cellulose and how can it be converted into alpha D-glucose?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33647" PostTypeId="1" CreationDate="2015-07-06T11:16:18.750" Score="6" ViewCount="3580" Body="&lt;p&gt;Benzene shows electrophilic substitution reactions. Normally, anything that can generate a carbocation shows, for example, a Friedel-Crafts alkylation reaction. I wanted to know if this reaction is also possible: taking an aldehyde or a ketone with an acid, which will protonate the oxygen atom and generate a carbocation - followed by electrophilic attack on benzene. This reaction can even go further by protonation of the alcohol and forming another carbocation for a second substitution - as I've shown in the picture below. &#xA;I couldn't find this anywhere, though it seems a simple way to make a benzyl alcohol - why isn't it more frequently carried out like other cases like Friedel-Crafts Alkylation, or Acylation, etc.?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DsSEX.jpg&quot; alt=&quot;&quot;&gt;&#xA;Something almost identical happens in synthesis of DDT using chloral and chlorobenzene.&lt;br&gt;&#xA;www.liceoagb.es/quimiorg/petroquimica_webquest/aad/DDT.ppt&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2016-01-20T16:11:19.540" Title="Acid-catalysed Reaction of Ketone with Benzene to Make Benzyl Alcohol" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33648" PostTypeId="2" ParentId="33645" CreationDate="2015-07-06T11:55:09.517" Score="26" Body="&lt;p&gt;The answer is... it is not so simple. Some quantum mechanics follow, but the &lt;strong&gt;TL;DR&lt;/strong&gt; version is that while $m_l=0$ corresponds to $p_z$, the orbitals for $m_l=+1$ and $m_l=-1$ lie in the $xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$. Thus, it is not possible to directly correlate the values of $m_l=\pm1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;em&gt;Physical Chemistry&lt;/em&gt; by Atkins and DePaula, the three wavefunctions for $n=2$ and $l=1$ are as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;&amp;amp;\Psi_{2,1,0}&amp;amp;&amp;amp;=r\cos{\theta}f(r)\\&#xA;&amp;amp;\Psi_{2,1,+1}&amp;amp;&amp;amp;=-\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{\mathrm{i}\phi}f(r)\\&#xA;&amp;amp;\Psi_{2,1,-1}&amp;amp;&amp;amp;=\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{-\mathrm{i}\phi}f(r)\\&#xA;&amp;amp;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notation is $\Psi_{n,l,m_l}$, $r$ is the radius, $\theta$ is the angle with respect to the $z$-axis and $\phi$ is the angle with respect to the $xz$-plane. &#xA;$$f(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bohr_radius&quot;&gt;Bohr radius&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos{\theta}$, so:&#xA;$$\Psi_{2,1,0}=zf(r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is $\Psi_{2p_z}$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other two wavefunctions are unhelpfully degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. Linear combinations are allowed by the maths of quantum mechanics. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because orbitals and the wavefunctions that describe them are not real physical objects. They constitute a mathematical model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the equations below, we're going to make use of some trigonometry, notably &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%27s_formula&quot;&gt;Euler's formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$$&#xA;$$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$$&#xA;$$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're also going to use $x=\sin{\theta}\cos{\phi}$ and $y=\sin{\theta}\sin{\phi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\Psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\Psi_{2,1,+1}-\Psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\&#xA;\Psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{2,1,+1}+\Psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So, while $m_l=0$ corresponds to $\Psi_{p_z}$, $m_l=+1$ and $m_l=-1$ cannot be directly assigned to $\Psi_{p_x}$ and $\Psi_{p_y}$. Rather $m_l=\pm1$ corresponds to $\{\Psi_{p_x},\Psi_{p_y} \}$. Put another way, I suppose we could say that $m_l=+1$ might correspond to $\Psi_{p_{x+y}}$ and $m_l=-1$ might correspond to $\Psi_{p_{x-y}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastEditorDisplayName="user7951" LastEditDate="2015-07-06T15:10:19.253" LastActivityDate="2015-07-06T15:10:19.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33649" PostTypeId="2" ParentId="33609" CreationDate="2015-07-06T12:29:46.893" Score="3" Body="&lt;p&gt;You are correct. What you call a collision partner is a specific example of the general concept of catalysis. &quot;C&quot; is a catalyst in your scheme. Catalysts don't change thermodynamic equilibria, just the speed at which it is approached (i.e. kinetics). Accelerating the forward rate but not the reverse rate of an equilibrium reaction would alter the equilibrium point, and is not possible for any true catalyst.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-06T12:29:46.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33650" PostTypeId="2" ParentId="33647" CreationDate="2015-07-06T12:43:57.660" Score="0" Body="&lt;p&gt;I think that it is quite difficult to generate the carbocation in the second case as the carbon atom of benzene is $\mathrm{sp^2}$ hybridised which is more electronegative than an $\mathrm{sp^3}$ hybridised carbon atom. It makes the carbon atom that is attached to it more unstable if it carries a positive charge. So formation of a carbocation could not take place.&lt;/p&gt;&#xA;" OwnerUserId="17275" LastEditorUserId="10185" LastEditDate="2015-07-06T13:37:07.163" LastActivityDate="2015-07-06T13:37:07.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33651" PostTypeId="1" AcceptedAnswerId="33655" CreationDate="2015-07-06T13:58:11.647" Score="6" ViewCount="4081" Body="&lt;p&gt;I was reading about electronic effects in organic compounds, and I found the explanation that carboxylic acids are weakening as their number of carbon atoms increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found a table of $K_\mathrm{a}$ for different acids, and it writes that $K_\mathrm{a}$ for propanoic acid is $1.34\times 10^{-5}$ while butanoic acid's acidic constant is $1.48\times 10^{-5}$. This means that butanoic acid is stronger than propanoic acid, but isn't this a contradiction to the inductive effect?&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorDisplayName="user7951" LastEditDate="2015-07-06T15:11:50.823" LastActivityDate="2015-07-06T18:46:57.210" Title="Why is propanoic acid weaker than butanoic acid?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33653" PostTypeId="2" ParentId="33651" CreationDate="2015-07-06T15:26:19.043" Score="2" Body="&lt;p&gt;The mesomeric effect requires a conjugated system of pi bonds. We do not have that in this situation: both propanoic and butanoic acids have an &quot;isolated&quot; carboxylic acid group that bonded to a saturated carbon and is not conjugated with anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could make an argument that the difference in acidity is a result of the  solvation energy difference between propanoic acid and the propanoate ion on one hand, and between butanoic acid and butanoate ion (perhaps greater) on the other.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-07-06T15:26:19.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33654" PostTypeId="1" CreationDate="2015-07-06T15:56:46.517" Score="10" ViewCount="2158" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;… catalysis, the acceleration of chemical reactions by substances not consumed in the reactions themselves—substances known as catalysts.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://global.britannica.com/science/catalysis&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as I’ve understood, to keep a reaction going, you must keep the solution heated and sometimes add more of the catalyst. Why then, is the definition that the catalyst is not consumed? Something of the catalyst must surely be consumed.&lt;/p&gt;&#xA;" OwnerUserId="17316" LastEditorDisplayName="user7951" LastEditDate="2015-07-09T16:20:03.137" LastActivityDate="2015-07-09T16:20:03.137" Title="Why is more of a catalyst added, when it is not consumed?" Tags="&lt;reaction-mechanism&gt;&lt;kinetics&gt;&lt;catalysis&gt;" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33655" PostTypeId="2" ParentId="33651" CreationDate="2015-07-06T16:04:42.157" Score="6" Body="&lt;p&gt;As a trend is usually a linear fit through experimental values it does neither mean that there are no minor variations between the values nor that there is a constant rise between values but only that one can see a difference between earlier and later values of that measured property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And besides all the answers on the actual influence of the longer alkyl chain on the acidity of the carboxylic group there are no big differences between the acidities for those small alkanoic acids between chain lengths of two to ten carbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a small overall trend $\left(+0.02~\mathrm{pK_a}\right)$ which can be seen in the following graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlfhr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one is a little bit stronger than the other but it is not very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could start to argue why exactly those two differ &quot;so much&quot; in strength but looking at those nine values above and the &quot;noise&quot; around the trend I guess that makes no sense.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="10185" LastEditDate="2015-07-06T18:46:57.210" LastActivityDate="2015-07-06T18:46:57.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33657" PostTypeId="2" ParentId="33654" CreationDate="2015-07-06T16:30:47.483" Score="17" Body="&lt;p&gt;In short, a catalyst &lt;em&gt;does&lt;/em&gt; participate in a chemical reaction the rate of which it increases, it just do so in a somewhat tricky way so that it does &lt;em&gt;not&lt;/em&gt; appear in the &lt;em&gt;overall&lt;/em&gt; reaction equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia article on catalysis provides a &lt;a href=&quot;https://en.wikipedia.org/wiki/Catalysis#Typical_mechanism&quot;&gt;concise explanation&lt;/a&gt; of why this is the case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Catalysts generally react with one or more reactants to form&#xA;  intermediates that subsequently give the final reaction product, in&#xA;  the process regenerating the catalyst. The following is a typical&#xA;  reaction scheme, where $\ce{C}$ represents the catalyst, $\ce{X}$ and $\ce{Y}$ are&#xA;  reactants, and $\ce{Z}$ is the product of the reaction of $\ce{X}$ and $\ce{Y}$:&#xA;  $$&#xA;     \ce{X + C → XC} \tag{1}&#xA;$$&#xA;  $$&#xA;     \ce{Y + XC → XYC} \tag{2}&#xA;$$&#xA;  $$&#xA;     \ce{XYC → CZ} \tag{3}&#xA;$$&#xA;  $$&#xA;     \ce{CZ → C + Z} \tag{4}&#xA;$$&#xA;  Although the catalyst is consumed by reaction 1, it is subsequently&#xA;  produced by reaction 4, so it does not occur in the overall reaction&#xA;  equation:&#xA;  $$&#xA;    \ce{X + Y → Z} \, .&#xA;$$&#xA;  As a catalyst is regenerated in a reaction, often only small amounts&#xA;  are needed to increase the rate of the reaction. In practice, however,&#xA;  catalysts are sometimes consumed in secondary processes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-07-06T16:30:47.483" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33658" PostTypeId="1" AcceptedAnswerId="33659" CreationDate="2015-07-06T16:36:51.577" Score="7" ViewCount="17329" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In which of the following diagrams is the Aufbau principle violated?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qUohQ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Ignoring the Hund principle, the Aufbau principle states that electrons have to first fill the lowest energy levels. Now the energy level of the s-orbital has to be the lowest, and only in the option (b) the s-orbital is not filled fully. So the answer is (b).&lt;/p&gt;&#xA;&lt;p&gt;Is my reasoning correct?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-06-25T07:20:25.433" Title="Is the Aufbau principle violated in the diagram (b)?" Tags="&lt;physical-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33659" PostTypeId="2" ParentId="33658" CreationDate="2015-07-06T16:58:03.723" Score="11" Body="&lt;p&gt;Since the question specifically asks about the violation of &lt;a href=&quot;https://en.wikipedia.org/wiki/Aufbau_principle&quot;&gt;Aufbau principle&lt;/a&gt;, your reasoning is correct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had it been based on the violation of &lt;a href=&quot;https://www.chem.purdue.edu/gchelp/gloss/hundsrule.html&quot;&gt;Hund's rule&lt;/a&gt; then option (a) would be the answer.&lt;/p&gt;&#xA;" OwnerUserId="13339" LastActivityDate="2015-07-06T16:58:03.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33660" PostTypeId="2" ParentId="33636" CreationDate="2015-07-06T17:58:15.807" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am just wondering how it comes that 60-100% methanol fractions from MCI gel CHP20P eluted very early on C8 column?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The gel you are using seems to be a porous divinyl benzene / styrene resin.  The aromatic groups of the resin may interact more strongly with aromatic side chains on your molecule (e.g. Y residues) than does the aliphatic C8 groups of the reversed-phase HPLC column you are using.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hydrophobic&quot; vs. &quot;hydrophilic&quot; is a very useful axis for understanding solute interactions with stationary phases, but unfortunately not everything can be collapsed into this single-variable scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that residual solvents or very large amounts of co-solutes are interfering with the RP-HPLC method you are using, causing premature elution.  As an extreme example of what I mean, imagine if you injected a few &amp;mu;L of chloroform along with your sample in the RP-HPLC method.  The chloroform would prevent column binding and cause premature elution.  Now, it's unlikely your samples have chloroform, but they could have large amounts of methanol or other &quot;solvent-like&quot; impurities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just general points; unfortunately I can't be sure that either is the precise source of your particular problem, without more (and more clearly presented) information on your former method from five years ago, your method today, and the differences in observed vs. expected results. &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-06T17:58:15.807" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33661" PostTypeId="2" ParentId="33654" CreationDate="2015-07-06T18:22:21.680" Score="3" Body="&lt;p&gt;Think of it like an assembly line. A worker in an assembly line takes two parts and puts them together. The worker isn't expended in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly a catalyst helps molecules react. But the product is just a combination of the molecules in the reaction. The catalyst isn't affected, and can go on to catalyze another reaction.&lt;/p&gt;&#xA;" OwnerUserId="5681" LastActivityDate="2015-07-06T18:22:21.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33662" PostTypeId="1" CreationDate="2015-07-06T18:30:14.650" Score="0" ViewCount="446" Body="&lt;p&gt;I have tried to solve it algebraically, using the data I'm provided with, but still, I'm not capable of proving that in this particular case, the equilibrium constant is independent from the volume, maybe because it isn't; however, I don't know how can I possibly know the concentration of each substance in the reaction without the volume, so as to find the equilibrium constant.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When a flask of constant capacity holds 1.2 mole each of acetic acid and ethanol with a catalyst and the temperature is kept at 25&amp;nbsp;°C, 0.80 mole of ethylacetate is produced at equilibrium. Calculate the equilibrium constant of the following reaction:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\ce{CH3COOH + C2H5OH &amp;lt;=&amp;gt; CH3COOC2H5 + H2O}$$&lt;/span&gt;&#xA;  (Answer option: 0.25, 0.64, 2.0, 4.0, 6.0)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17319" LastEditorUserId="4945" LastEditDate="2018-09-27T13:44:18.013" LastActivityDate="2018-09-27T13:44:18.013" Title="How to calculate the equilibrium constant for an esterification?" Tags="&lt;equilibrium&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33663" PostTypeId="2" ParentId="33662" CreationDate="2015-07-06T18:58:39.863" Score="3" Body="&lt;p&gt;It's given that this whole process is taking place at constant volume, so you can concentrate instead on what the concentrations of the species are (no pun intended) at equilibrium. The information is there, you just need to work through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have been given initial conditions:&#xA;$[{\rm CH}_{3}{\rm COOH}] = [{\rm C}_{2}{\rm H}_{5}{\rm OH}] = 1.2\;{\rm mol}; [{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = [{\rm H}_{2}{\rm O}] = 0\;{\rm mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and final conditions, where you are told that $[{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = 0.8\;{\rm mol}$. This means that at equilibrium $[{\rm H}_{2}{\rm O}] = 0.8\;{\rm mol}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your reactant equilibrium concentrations will be $1.2\;{\rm mol} - x$, where $x$ is given to you (I'm leaving that part to you. Hint: It's in the  sentence right above this one). Plug in your equilibrium values and you'll get ${\rm K}_{\rm eq}$ (also up to you).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-06T18:58:39.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33664" PostTypeId="2" ParentId="25076" CreationDate="2015-07-06T19:34:42.190" Score="10" Body="&lt;p&gt;I do have a very similar experience! What the OP calls the mollusk smell, I call it slightly rotting &quot;seaweedy&quot; smell. It has been a very unpleasant experience for us in our new home.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to OP's experience, this seems to be more intense with glasses and ceramics right after the newly washed dishes get exposed to air when the windows are open. But I also detect it with plastic cups as noted below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to eliminate other variables such as dishwasher problems and tap water quality, I bought bottled water and disposable plastic cups. I did the following experiment.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;WITH CLOSED WINDOWS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water from bottle poured in a previously unused plastic cup, sitting for 15 minutes : No smell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water from bottle poured in a cup and couple of sips taken, sitting for for 15 minutes : No smell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WITH OPEN WINDOWS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water from bottle in a previously unused plastic cup, sitting for 15 minutes : No smell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water from bottle in a cup and couple of sips taken, sitting for 5 minutes : Seaweedy smell&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm about to conclude that something in the outside air combines with the saliva (and/or chemicals/enyzmes from the detergents) to produce the seeweedy smell! The same thing happens with the newly washed dishes out of the dishwasher, and hand washed dishes after they get exposed to air with the windows open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I know this doesn't make any sense to people that haven't experienced it before! It sounds like nonsense, but it's quite annoying every time I take sips from water cup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm almost on the verge of calling up the EPA about it. There must be something in the air, but what...&lt;/p&gt;&#xA;" OwnerUserId="17320" LastActivityDate="2015-07-06T19:34:42.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33665" PostTypeId="1" AcceptedAnswerId="33683" CreationDate="2015-07-06T19:47:23.337" Score="14" ViewCount="1184" Body="&lt;p&gt;I added a branch to the example shown in the textbook. What do you do when your longest carbon chain has a $\ce{CH2Cl}$ attached to a certain carbon? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previously the book treated it as a methyl group and a halogen but my modification changes the length of the longest carbon chain so how do I name this molecule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/67cmq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/67cmq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it is named 5-chloromethylheptan-3-ol.&lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorUserId="16683" LastEditDate="2017-11-03T20:30:55.320" LastActivityDate="2017-11-03T20:30:55.320" Title="How do I name a molecule with a compound substituent?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33666" PostTypeId="2" ParentId="33665" CreationDate="2015-07-06T20:09:10.463" Score="5" Body="&lt;p&gt;The correct name should be  &lt;em&gt;5-chloromethyl-heptan-3-ol&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because, from the longest chain rule, you will have to select the longest chain (i.e. containing seven $\ce{C}$'s). Now, you are left with $\ce{-CH2Cl}$ and $\ce{-OH}$, out of which $\ce{-OH}$ has &lt;a href=&quot;https://i.stack.imgur.com/r5y18.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;greater priority&lt;/a&gt;. Then, writing alphabetically, we get the name as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5-chloromethyl-heptan-3-ol&lt;/p&gt;&#xA;" OwnerUserId="16924" LastEditorDisplayName="user7951" LastEditDate="2017-10-28T08:11:56.510" LastActivityDate="2017-10-28T08:11:56.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33667" PostTypeId="1" CreationDate="2015-07-06T20:17:53.050" Score="3" ViewCount="265" Body="&lt;p&gt;Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?&lt;/p&gt;&#xA;" OwnerUserId="9611" LastEditorUserId="7448" LastEditDate="2015-07-06T20:21:40.727" LastActivityDate="2015-07-06T20:43:28.607" Title="Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33668" PostTypeId="2" ParentId="33667" CreationDate="2015-07-06T20:27:49.363" Score="7" Body="&lt;p&gt;No, I could not think of any relation between them. Orbital energies are solutions of the &lt;em&gt;electronic&lt;/em&gt; Schrödinger equation, and as such, they describe the electronic state of a molecular system. Vibrational frequencies are solutions of the &lt;em&gt;nuclear&lt;/em&gt; Schrödinger equation, and as such, they describe the nuclear state of a molecular system. The quantities that describe the electronic state are not related to the quantities that describe the nuclear one simply because they describe two different subsystems: the electronic and the nuclear one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that usually we treat these two subsystems as being independent of each other (the Born-Oppenheimer approximation), but even if you go beyond that, what you'll see is that changes in the nuclear subsystem (say, changes in geometry during vibrations) might affect the electronic state. Consequently, molecular vibrations in principle can alter HOMO/LUMO (as well as any other orbital) energies, but I doubt this interdependence of electronic and nuclear states is the relation you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-06T20:43:28.607" LastActivityDate="2015-07-06T20:43:28.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33669" PostTypeId="1" AcceptedAnswerId="33670" CreationDate="2015-07-06T20:57:21.873" Score="2" ViewCount="32713" Body="&lt;p&gt;&lt;strong&gt;Part A)&lt;/strong&gt; Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (a) as indicated by the manometer.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/P2lqb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part B)&lt;/strong&gt;&#xA;Given a barometric pressure of $\pu{764.0 mmHg}$, calculate the pressure of gas sample (b) as indicated by the manometer.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Ijv7m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Part A) I tried to subtract $764.0-40.0$ to get $724$ but the answer is wrong.&lt;br&gt;&#xA;For Part B) I tried to add $764.0+40.0$ to get $804$, but the answer is also wrong.&lt;/p&gt;&#xA;" OwnerUserId="17322" LastEditorUserId="5026" LastEditDate="2018-02-23T01:37:50.080" LastActivityDate="2018-10-04T03:40:40.633" Title="Calculate the pressure of the gas samples as indicated by the manometer" Tags="&lt;physical-chemistry&gt;&lt;pressure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33670" PostTypeId="2" ParentId="33669" CreationDate="2015-07-06T21:20:02.087" Score="2" Body="&lt;p&gt;In part (A), ${\rm P}_{\rm gas} &amp;gt; {\rm P}_{\rm atm}$, as evidenced by the column of mercury being &quot;pushed up&quot; in the manometer on the side of ${\rm P}_{\rm atm}$ - in this case, you'll need to add (the amount it's &quot;pushed up&quot;) to atmospheric pressure (i.e., 764 mm Hg). You state that you tried to subtract 40.0 from 764.0: Look at the graphic again (I hold it's 7 cm - or 70 mm - Hg) and try adding that number to 764 mm Hg instead of 40 mm Hg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part (B), ${\rm P}_{\rm atm} &amp;gt; {\rm P}_{\rm gas}$, so reverse the process above. I believe what you state is your answer to part (A) is actually the correct answer to part (B).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-06T21:20:02.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33671" PostTypeId="1" CreationDate="2015-07-06T23:36:27.700" Score="6" ViewCount="344" Body="&lt;p&gt;I'm looking for a wet etch of Bismuth Oxide that &lt;strong&gt;does not etch $\ce{Bi}$ itself.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are obviously chemicals that etch $\ce{Bi2O3}$, but $\ce{Bi}$ is pretty fragile, so almost everything I've found that can etch $\ce{Bi2O3}$ also etches Bi, and usually more rapidly than the $\ce{Bi2O3}$ itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, someone recently mentioned to me in passing that there was a chemical that could do this, but they didn't remember what it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching Google, I found &lt;a href=&quot;https://books.google.com/books?id=ZDyTr9uJnAsC&amp;amp;pg=PA590&amp;amp;lpg=PA590&amp;amp;dq=bismuth%20oxide%20selective%20etch&amp;amp;source=bl&amp;amp;ots=MdAp64Wx1E&amp;amp;sig=qe99pBGk4uHb983fq8oYfNNBqQM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCQQ6AEwAGoVChMItvbGv9DHxgIVgRaSCh3q0QT6#v=onepage&amp;amp;q=bismuth%20oxide%20selective%20etch&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this,&lt;/a&gt; but that's RIE -- not wet etching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also read about people using $\ce{H2}$ gas to reduce the oxide back to $\ce{Bi}$, but again, not wet etching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any idea? I know it's a specific and uncommon question.&lt;/p&gt;&#xA;" OwnerUserId="3764" LastEditorDisplayName="user7951" LastEditDate="2015-10-30T18:06:53.353" LastActivityDate="2019-10-07T02:22:53.940" Title="Is there a selective wet etch of Bi2O3 Bismuth Oxide?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33672" PostTypeId="1" CreationDate="2015-07-07T01:28:12.407" Score="-1" ViewCount="2181" Body="&lt;h3&gt;Problem&lt;/h3&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nitrogen &lt;span class=&quot;math-container&quot;&gt;$(\gamma = 1.4)$&lt;/span&gt; is expanded isentropically. Its temperature changes from &lt;span class=&quot;math-container&quot;&gt;$\pu{620 °F}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\pu{60 °F}.$&lt;/span&gt; Find the pressure ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Answer&lt;/h3&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; &lt;span class=&quot;math-container&quot;&gt;$p_1/p_2 = 12.91$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;My approach&lt;/h3&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{620 °F} = (1799/3)~\pu{K}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\pu{60 °F} = (2597/3)~\pu{K}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using isentropic relation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{T_1}{T_2} = \left(\frac{p_1}{p_2}\right)^\frac{\gamma-1}{\gamma}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1799/3}{2597/3} = \left(\frac{p_1}{p_2}\right)^{\frac{1.4-1}{1.4}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{p_1}{p_2} = \left(\frac{1799/3}{2597/3}\right)^{\frac{1.4}{1.4-1}} = 0.2766669$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What am I doing wrong? Is the problem set possibly wrong?&lt;/p&gt;&#xA;" OwnerUserId="17306" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2021-04-06T11:01:30.370" LastActivityDate="2021-04-06T11:01:30.370" Title="Pressure ratio for isentropic expansion" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33673" PostTypeId="2" ParentId="33672" CreationDate="2015-07-07T02:16:02.573" Score="4" Body="&lt;p&gt;As stated by @Todd in the comments, your temperature conversions are incorrect. To convert between °F and K, use the following conversion:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pu{K} = \frac{\pu{°F}-32}{1.8}+273.15$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you want, double check your conversion using an online conversion calculator such as &lt;a href=&quot;http://www.metric-conversions.org/temperature/fahrenheit-to-kelvin.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metric Conversion&lt;/a&gt;.&#xA;Then continue your calculations as you have done (I just did and got the correct answer).&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2021-04-06T11:01:12.580" LastActivityDate="2021-04-06T11:01:12.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33674" PostTypeId="2" ParentId="33628" CreationDate="2015-07-07T03:06:37.190" Score="1" Body="&lt;p&gt;The previous answer was not done correctly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the balances on the three species:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dA}{dt}=-(k_1+k_{-3})A+k_{-1}B+k_3C$$&#xA;$$\frac{dB}{dt}=k_1A-(k_{-1}+k_2)B+k_{-2}C$$&#xA;$$\frac{C}{dt}=k_{-3}A+k_2B-(k_3+k_{-2})C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is at equilibrium, the three time derivatives must be equal to zero.  So one must have:&#xA;$$-(k_1+k_{-3})A+k_{-1}B+k_3C=0$$&#xA;$$k_1A-(k_{-1}+k_2)B+k_{-2}C=0$$&#xA;$$k_{-3}A+k_2B-(k_3+k_{-2})C=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These relationships represent 3 simultaneous homogeneous linear algebraic equations in three unknowns, A, B, and C.  The condition for these equations to be satisfied for all values of A, B, and C is that the determinant must be equal zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chet&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-07T03:06:37.190" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33675" PostTypeId="1" AcceptedAnswerId="33689" CreationDate="2015-07-07T04:45:07.700" Score="21" ViewCount="97107" Body="&lt;p&gt;The reaction of alcohols $\ce{ ROH }$ with $\ce{ PCl5 }$ and $\ce{ PCl3 }$ yields an alkyl halide $\ce{RCl}.$ With $\ce{ PCl5 }$, the reaction is quite simple leading to the formation of $\ce{ RCl }$ and $\ce{ POCl3 }.$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with $\ce{ PCl3 }$  a problem arises. Since $\ce{ PCl3 }$ has both a lone pair and vacant $\ce{3d}$ orbitals it can act both as a Lewis base and a Lewis acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first figure, $\ce{ PCl3 }$ accepts a lone pair showing its acidic character and expelling &lt;strong&gt;$\ce{Cl-}$&lt;/strong&gt; out. Now this is where I am getting confused. On one hand we have  a Lewis base $\ce{ PCl3 }$ and on the other we have the &quot;expelled&quot; &lt;strong&gt;$\ce{Cl-}$?&lt;/strong&gt; Now in this case why does $\ce{PCl3}$ take away the proton and not $\ce{Cl-}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U1L0A.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U1L0A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the next step, why does  $\ce{Cl-}$ attacks the carbon not the  $\ce{H+}$?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="5026" LastEditDate="2018-03-29T01:31:04.340" LastActivityDate="2021-03-05T17:29:31.743" Title="Reaction of alcohols with PCl5 and PCl3" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="9" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33676" PostTypeId="2" ParentId="33637" CreationDate="2015-07-07T05:05:16.870" Score="4" Body="&lt;p&gt;Oh, I was just reading about this. This excerpt points out the factor contributing to the chemical shift observed in alcohols&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in CC14 (Figure 9-29). &lt;em&gt;The hydroxyl resonance will be seen to move upfield at the lower concentrations.&lt;/em&gt; This is the result of decreasing association by hydrogen bonding through equilibria...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.organicchemistry.com/factors-that-affects-chemical-shift-and-application-of-chemical-shift-in-structure-determination/&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the reason (or a reason) for alcohols moving upfield due to hydrogen bonding, which results in decreasing association.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at your link and there are a few notable things. They mention *variable, highlighting that it is not necessarily constant based on different conditions, which connects to hydrogen bonding. For carbon-2, we have the normal ~1.7 for methine protons + 2.5 from oxygen of alcohol, coming to be a pretty close value. So no, it's not &quot;why on earth would the electrons on the carbon-2&quot; but what on earth is going on with that alcohol?&lt;/p&gt;&#xA;" OwnerUserId="14975" LastActivityDate="2015-07-07T05:05:16.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33679" PostTypeId="2" ParentId="33592" CreationDate="2015-07-07T07:37:13.440" Score="1" Body="&lt;p&gt;I would strongly recommend to consult the references I mentioned in this answer earlier: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33164/r32-space-group-crystallography&quot;&gt;R32 Space Group (Crystallography)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular have a look at the meaning of the individual symbols and how space groups are represented in these kinds of diagrams. Again, the International Tables of Crystallography, Volume A, has plenty of explanation on this in its first part. &#xA;I would also recommend not to start out with the rhombohedral groups, as they are probably the trickiest of them all due to the slightly uncommon choice of centring. Try to start with an orthorhombic space group, for instance, at least until you do not have troubles with the symbols any longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for an answer: your question unfortunately does not make a lot of sense to me. I'll go with a hunch here: Could it be that you are mistaking the &quot;+&quot; and &quot;-&quot; signs for something they are not, like having anything to do with height (which they do not)? &#xA;They describe the orientation of the motif - think &quot;upwards-/downwardsfacing&quot;. &#xA;Hence (screw-)axes obviously do not have a &quot;sign&quot;, but change the sign of the motif they operate on.&lt;/p&gt;&#xA;" OwnerUserId="10050" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-07-07T09:09:35.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33680" PostTypeId="1" AcceptedAnswerId="33681" CreationDate="2015-07-07T07:48:20.737" Score="1" ViewCount="2896" Body="&lt;p&gt;I think the substituent of option (a) should attack the meta position as the resultant carbanion gets stabilized by an inductive effect, but no such compound is given. So where am I making a mistake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBG1m.jpg&quot; alt=&quot;A chemical equation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16618" LastEditorUserId="865" LastEditDate="2021-05-22T18:54:28.100" LastActivityDate="2021-05-22T18:54:28.100" Title="What is the major product of the reaction given?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33681" PostTypeId="2" ParentId="33680" CreationDate="2015-07-07T08:03:15.920" Score="7" Body="&lt;p&gt;Your reasoning is quite correct for an uncatalysed reaction. &#xA;I think you have forgotten about the hydrochloric acid you also add (in catalytic amounts). You will therefore protonate the double bond first, with the carbo cation being more stable in the ortho position. The alcohol can now act as a nucleophile and attack at that carbon. Therefore (c) is the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-07-07T08:03:15.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33683" PostTypeId="2" ParentId="33665" CreationDate="2015-07-07T08:40:30.357" Score="16" Body="&lt;p&gt;The given compound includes the characteristic group $\ce{-OH}$, which is expressed as suffix ‘ol’. Since there is only one suffix, the order of seniority of suffixes is not relevant in this case. Functional suffixes are always attached to the name of the parent structure. Note that alcohols are named by attaching the suffix ‘ol’ to the name of the parent structure, with elision of the final letter ‘e’ in the parent structure, if present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; stipulates various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. The longest chain in the given structure has seven skeletal atoms; i.e., it is a ‘heptane’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the characteristic group $\ce{-OH}$, the parent structure has another substituent ($\ce{-CH2-Cl}$), which is expressed as the substitutive prefix ‘chloromethyl’. Note that parentheses are used around such compound prefixes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound. In particular, the order of seniority provides that low locants are given first to the principal characteristic group (here: ol) before low locants are given to prefixes (here: chloromethyl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the resulting name is 5-(chloromethyl)heptan-3-ol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4bObe.png&quot; width=&quot;164&quot; alt=&quot;5-(chloromethyl)heptan-3-ol&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2017-08-05T14:23:43.267" LastActivityDate="2017-08-05T14:23:43.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33686" PostTypeId="1" AcceptedAnswerId="33688" CreationDate="2015-07-07T11:02:33.910" Score="23" ViewCount="5236" Body="&lt;p&gt;Recently, while cleaning a neighbour's fridge (turned off for a few weeks), I came across a cup (closed with a lid).  Inside the cup was, to my olfactory horror, congealed milk, with a steel (iron) teaspoon nicely wedged into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The milk had separated into a blue cheese-like solid, overlain with a semi-translucent liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I took the teaspoon out of the congealed mess, there was a layer of rust on the teaspoon where it was in the liquid. But, I observed almost no rust on the parts of the teaspoon in the blue-cheese-like solid and that exposed to the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The illustration below is a rough schematic of where the rust occurred (the 2 layers observed are labelled A and B, and the air in the top of the container is labelled C and the diagonal line represents the teaspoon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CB1lY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A = congealed blue-cheese-like solid = virtually no rust on this part of the teaspoon&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;B = semi translucent liquid = heavily rusted part of the teaspoon&lt;/em&gt;&lt;br&gt;&#xA;&lt;em&gt;C = air = virtually no rust on this part of the teaspoon&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How could the milk, or the liquid separation, cause rust on the teaspoon?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hunch I have (and I could be way off the mark), is it possible that the corrosion could be due to, at least in some significant way, due to the fact that milk is slightly acidic $^{(1)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{(1)}$ According to the &lt;a href=&quot;http://www.ilri.org/InfoServ/Webpub/fulldocs/ilca_manual4/Milkchemistry.htm#P59_2599&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Livestock Research Institute&lt;/a&gt;, they state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fresh milk has a pH of 6.7 and is therefore slightly acidic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Related question and answer &lt;a href=&quot;https://chemistry.stackexchange.com/q/513/15489&quot;&gt;Silver and milk (or milk products)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-07-14T05:20:37.293" Title="Could milk rust a steel teaspoon?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;metal&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33687" PostTypeId="2" ParentId="33456" CreationDate="2015-07-07T12:34:25.473" Score="3" Body="&lt;p&gt;With the correct starting compound, 3-ethyl-2-methylpentan-2,3-diol, predicting the product is tough! Here is the reaction scheme for this compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dNnAG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic. &lt;/li&gt;&#xA;&lt;li&gt;Carbocations B and E are both tertiary carbocations that seem equally likely to form.&lt;/li&gt;&#xA;&lt;li&gt;The rearrangement from B to C involves an ethyl migration, which seems less likely than the methyl migration involved in the rearrangement from E to F.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Compound F is 3-ethyl-3-methylpentan-2-one, which is what you chose as the major product. Compound C is 4,4-dimethylhexan-3-one, which is what the book says is the major product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without experimental evidence, I would have trouble choosing between C and F. If I had to choose, my first instinct would be F because it involved the methyl shift instead of the ethyl shift. Methyl shifts happen faster (methyl is smaller) than ethyl shifts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My reasoning (and probably yours) is wrong, and here is why. We can only consider methyl &gt; ethyl from the same carbocation. For example, in the following rearrangement:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b2eZy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the rearrangment step at the end of the sequence, the methyl group is more likely to move than the ethyl group. We can make this comparison here because the pathway is identical up until the rearrangement.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In your reaction, the pathways toward the generation of the carbocations are not equivalent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, if carbocation B forms preferentially over carbocation E, then it might not matter that the methyl group migrates faster. We get more ethyl group migration because we have more of that carbocation formed. If carbocation B forms preferentially, one of my two assumptions about A, B, D, and E above must be incorrect. I will restate them below for convenience:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Alkonium cations A and D both equally likely because they are both tertiary. Thus both alcohols should be equally basic. &lt;/li&gt;&#xA;&lt;li&gt;Carbocations B and E are both tertiary carbocations that seem equally likely to form.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'll leave it to you now. Which assumption is not correct? Why is that assumption incorrect? Why do cations A and B form preferentially over D and E? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'll give you a hint: Have you considered steric factors?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-07T12:34:25.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33688" PostTypeId="2" ParentId="33686" CreationDate="2015-07-07T12:54:26.437" Score="26" Body="&lt;p&gt;Technically, even simple water can cause rust, so nothing surprising here.&#xA;However, spoilage of milk most probably produced a lot of organic acids (lactic acid and similar) which can speed up any corrosion process. Lactic fermentation is a natural fermentation process in milk, when bacteria start converting the sugar content of milk to lactic acid. It is not the only possible product, but a very common one. Link: &lt;a href=&quot;https://en.wikipedia.org/wiki/Lactic_acid_fermentation&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Lactic_acid_fermentation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two important factors that help corrosion: First, acidic environment (low pH) itself generally help corrosion. Second, the organic acids produced in fermentation can often form metal complexes, therefore they help remove any protecting oxide layer from the surface of metal objects as well as shift the  redox equilibrium towards an oxidized complex form. &lt;/p&gt;&#xA;" OwnerUserId="5172" LastEditorUserId="5172" LastEditDate="2015-07-14T05:20:37.293" LastActivityDate="2015-07-14T05:20:37.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33689" PostTypeId="2" ParentId="33675" CreationDate="2015-07-07T13:19:48.767" Score="14" Body="&lt;p&gt;Remember your general chemistry. In an acid-base reaction, the equilibrium favors the side of the reaction with the weaker acid/base pair. Strong bases have weak conjugate acids. Strong acids have weak conjugate bases. Weak bases have strong conjugate acids. Weak acids have strong conjugate bases. Consider the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA + B- &amp;lt;=&amp;gt; A- + HB}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{HA}$ is a stronger acid than $\ce{HB}$, then $\ce{A-}$ must be a weaker base than $\ce{B-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example with real compounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HCl + NH3 &amp;lt;=&amp;gt; Cl- + NH4+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the right side of the reaction contains ions instead of neutral molecules, the right side is favored because $\ce{HCl}$ is a strong acid and $\ce{NH4+}$ is a weak acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we study organic chemistry, we are given some rules of thumb to help us compare relative acidity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is more electronegative than $\ce{Z}$ and they are in the same period.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is larger than $\ce{Z}$ and they are in the same group.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more resonance stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more inductive stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more $\ce{s}$-character than $\ce{Z-}$ and $\ce{Y-}$ and if $\ce{Z-}$ are otherwise similar.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{YH2+}$ is always more acidic than $\ce{YH}$, though it is hard to compare $\ce{YH2+}$ and $\ce{YZ}$ or $\ce{YH}$ and $\ce{YZ2+}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With these rules in hand, it is sometimes challenging to remember that we also have an experimental measure of acid strength, and that there are only a limited number of &quot;strong&quot; acids (those acids which are stronger than $\ce{HSolvent+}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $K_\mathrm{a}$ of $\ce{HCl}$ is not easily determinable, since it is more acidic than most protonated solvents. However, the &lt;a href=&quot;http://evans.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;noreferrer&quot;&gt;Evans pKa table&lt;/a&gt; and other sources often estimate it at -7 or -8, with only $\ce{HBr}$, $\ce{HI}$, and the various &quot;superacids&quot; including such things as $\ce{HSbF6}$ being stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Evans table lists the $\mathrm{p}K_\mathrm{a}$ of various protonated phosphines. For example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH3PH3+}$ is 2.7 in DMSO ($\ce{HCl}$ is 1.8 in DMSO).&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Et3PH+}$ is 9.1 in DMSO&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Would we expect $\ce{ROPCl2H+}$, which is the intermediate in your reaction, to be more or less acidic than the two reference phosphonium ions above? Both $\ce{RO}$ amd $\ce{Cl}$ are electron withdrawing by induction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Cl-}$ maybe could attack $\ce{H}$ and not $\ce{C}$, but the $\ce{HCl}$ that forms would be so acidic in comparison to everything else in the reaction that something else would take that proton away again, regenerating $\ce{Cl-}$. Once $\ce{RCl}$ forms, there is no other good nucleophile present that is capable of displacing the chloride group.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-29T01:32:31.030" LastActivityDate="2018-03-29T01:32:31.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33690" PostTypeId="1" CreationDate="2015-07-07T13:22:20.217" Score="8" ViewCount="89" Body="&lt;p&gt;How do the 1,3-dipolar cycloadditions follow the &lt;em&gt;endo&lt;/em&gt; rule as in the reaction below, where the &lt;em&gt;anti&lt;/em&gt; diastereoisomer is formed predominantly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWhsc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2018-07-05T11:29:06.017" LastActivityDate="2018-07-05T11:29:06.017" Title="Stereoselectivity of 1,3-dipolar cycloadditions" Tags="&lt;organic-chemistry&gt;&lt;pericyclic&gt;&lt;stereoselectivity&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33691" PostTypeId="1" AcceptedAnswerId="33692" CreationDate="2015-07-07T14:10:42.530" Score="20" ViewCount="9380" Body="&lt;p&gt;To give this question context, I work in a restaurant and sometimes I'm in charge of getting all of the gum off of the tables at work. Don't mind the work, but I'd be interested in hearing what makes gum so sticky and if there is any chemical I can use to make the cleaning a bit easier and why that chemical works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess another backwards question is: Is there any chemical I can apply to the underside of tables that will prevent the gum from sticking? Would be awesome if I could fix the problem at the source and would be super interested in the chemistry behind it. &lt;/p&gt;&#xA;" OwnerUserId="17329" LastEditorUserId="7448" LastEditDate="2015-10-30T17:29:36.303" LastActivityDate="2017-05-09T10:07:36.210" Title="Why is gum so sticky and what chemicals make for easy removal of gum?" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33692" PostTypeId="2" ParentId="33691" CreationDate="2015-07-07T14:24:01.597" Score="14" Body="&lt;p&gt;The &lt;a href=&quot;http://www.gumassociation.org/&quot; rel=&quot;noreferrer&quot;&gt;International Chewing Gum Association&lt;/a&gt; tells us:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Gum base is what gives chewing gum its “chew.” It is made of a combination of food-grade polymers, waxes, and softeners that give it the texture desired by consumers and enable it to effectively deliver sweetness, flavor, and various other benefits, including dental benefits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A more complete description of &lt;a href=&quot;https://illumin.usc.edu/118/chewing-gum/&quot; rel=&quot;noreferrer&quot;&gt;what makes chewing gum sticky&lt;/a&gt; states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Terence Cosgrove, Professor of Chemistry at the University of Bristol and Chief Scientific Officer of Revolymer, explains that the chemical bonds between the molecules in a polymer-based chewing gum make it difficult to remove from a surface. When you pull a piece of used gum off a surface, most of the energy goes into stretching the polymer bonds in the gum, rather than actually breaking the bonds between the surface and the gum. Since polymers are long chains of covalently bonded molecules that give the gum its elasticity and &amp;quot;chewiness,&amp;quot; the attractive forces between the atoms are formed by the sharing of electrons [8]. Thus, the bonds between the repeating units tend to remain intact as they lengthen and contract from an applied external force.&lt;/p&gt;&#xA;&lt;p&gt;The temperature at which the polymer base is exposed to the air also affects both the elasticity and adhesiveness of the chewing gum. When a piece of gum is heated by saliva and deformed by the grinding of human teeth, its polymer chains align in the direction of these forces. The degree of alignment is a function of the magnitude of applied stress, which explains why the gum becomes tougher and less elastic the longer or more rigorously you chew [8]. After the gum is removed from your mouth and placed in a cooler environment, the drop in temperature causes the orientation of the polymer chain to freeze, resulting in a hardened piece of used gum.&lt;/p&gt;&#xA;&lt;p&gt;Most commercial polymer gum bases are hydrophobic (water-insoluble), which is the reason they stick easily to oily surfaces and are difficult to remove, even with the help of household cleaning solutions. A hydrophobic substance tends to repel polarized molecules like water and attract non-polar compounds, such as the grease and grime on streets. A hydrophilic substance, on the other hand, behaves in the opposite manner by attracting water and repelling fats and oils [8]. A compound that exhibits both hydrophobic and hydrophilic traits is known as &amp;quot;amphiphilic&amp;quot; [9]. These substances, which attract both water and oil to some degree, can be used to create synthetic polymers that could help eliminate the sticky mess of chewing gum pollution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note: The numbered references in the above block quote can be pulled from the linked article. The article also details efforts aimed at producing a new type of chewing gum that doesn't stick to things as much as normal/currently-manufactured ones seem to. It's a good read.&lt;/p&gt;&#xA;&lt;p&gt;But to your main point, I've linked the &lt;a href=&quot;http://www.uline.com/PDF/SS-19127.pdf&quot; rel=&quot;noreferrer&quot;&gt;MSDS&lt;/a&gt; for a popular product for removing gum and related compounds (it's called Goof Off), which consists almost entirely of acetone (also commonly found in nail polish remover). Given the deleterious health effects of acetone (and its high vapor pressure), it's not a good candidate for your needs.&lt;/p&gt;&#xA;&lt;p&gt;Something non-stick and more permanent, such as a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B003UTX0R8&quot; rel=&quot;noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Teflon spray&lt;/a&gt; might be ideal for your needs.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-07T15:23:57.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33694" PostTypeId="1" AcceptedAnswerId="33695" CreationDate="2015-07-07T15:12:02.983" Score="5" ViewCount="105" Body="&lt;p&gt;I'm writing a fictional story with early &lt;a href=&quot;https://en.wikipedia.org/wiki/Industrial_Age&quot;&gt;industrial age&lt;/a&gt; technology where a chemist discovers not only that diamond and graphite are made of the same substance, but that they &lt;a href=&quot;https://en.wikipedia.org/wiki/Allotropes_of_carbon&quot;&gt;differ by crystalline structure&lt;/a&gt; (cubic vs hexagonal).  However, I'd like this to be a realistic depiction if possible and the chemist does not have access to electron tunneling microscopy or advanced spectrographic methods.  The farthest spectroscopy has gotten is &lt;a href=&quot;https://en.wikipedia.org/wiki/Joseph_von_Fraunhofer&quot;&gt;Fraunhofer&lt;/a&gt;'s spectroscope and John Herschel's evaporogram, with visible light and the beginnings of an understanding of &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared#History_of_infrared_science&quot;&gt;infrared&lt;/a&gt; and ultraviolet.  Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?&lt;/p&gt;&#xA;" OwnerUserId="15055" LastActivityDate="2015-07-07T15:53:53.307" Title="Are there any early industrial age methods capable of determining diamond or graphite's crystalline structure?" Tags="&lt;spectroscopy&gt;&lt;crystal-structure&gt;&lt;history-of-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33695" PostTypeId="2" ParentId="33694" CreationDate="2015-07-07T15:22:14.117" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Would 1850's level spectroscopy or other technology of the time be able to demonstrate the crystal structures of carbon (specifically graphite and diamond) without a 1900's understanding of atomic structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, using visible light and petrographic microscopes. The field of optical mineralogy was mature enough by the mid-1800's for one to determine the crystal structure of minerals. In particular, the work of Sir William Nicol is worth investigating (thin but useful &lt;a href=&quot;https://en.wikipedia.org/wiki/William_Nicol_(geologist)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will add that there's a pretty simple way to determine whether or not the crystal structure of a mineral is the same or different than that of another mineral - the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_habit&quot; rel=&quot;nofollow&quot;&gt;crystal habit&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, diamond exhibits &lt;a href=&quot;https://en.wikipedia.org/wiki/Conchoidal_fracture&quot; rel=&quot;nofollow&quot;&gt;conchoidal fracture&lt;/a&gt;, and graphite does not. This might be the easiest way to tell that the two minerals have different crystal structures.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-07T15:53:53.307" LastActivityDate="2015-07-07T15:53:53.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33696" PostTypeId="1" CreationDate="2015-07-07T16:20:24.730" Score="1" ViewCount="303" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lrZtY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the NMR data, I have determined that the molecule must be 1-phenoxy-2-propanone, however I may be wrong because given the proton NMR, there are only 4 unique carbon environments, however in 1-phenoxy-2-propanone, there are 5 carbon environments. In addition to that, on carbon environment &quot;B&quot; there is only one proton, which isn't indicated on the NMR, which makes me doubt my choice in molecules. I've basically tried every molecule and this was the closest I could get!Could anyone please correct me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1bakN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorDisplayName="user15489" LastEditDate="2015-07-08T15:10:12.670" LastActivityDate="2015-07-08T15:10:12.670" Title="Determining Molecule Based on Proton NMR (HNMR)" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33699" PostTypeId="1" CreationDate="2015-07-07T17:09:34.867" Score="2" ViewCount="513" Body="&lt;p&gt;I seem to be getting two different answers from my book and my teacher about PCR.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does PCR require a double stranded piece of DNA that you are tryig to replicate, or does it require a circular piece of DNA (i.e. a plasmid)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that when we were doing the experiments, we used a plasmid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, I can't seem to find a source which explains how PCR works when you are also introducing a mutation at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used different primers to introduce a mutation during the PCR, but I am still very confused as to how the primer could bind when the nucleotide sequence is not exactly matching.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="16622" LastEditDate="2015-07-07T20:48:54.650" LastActivityDate="2015-07-07T20:48:54.650" Title="Confusion about PCR" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33700" PostTypeId="1" CreationDate="2015-07-07T17:18:08.143" Score="1" ViewCount="455" Body="&lt;p&gt;How can &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/152021?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;phenoxyacetone&lt;/a&gt; be synthesized from phenol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would first need to deprotonate phenol by addition of NaH in THF and then add bromopropene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure where I would go from here. Any help would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="9414" LastEditorDisplayName="user15489" LastEditDate="2015-07-08T15:53:22.660" LastActivityDate="2016-08-01T20:43:51.953" Title="Synthesis of phenoxyacetone from phenol" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33702" PostTypeId="2" ParentId="33699" CreationDate="2015-07-07T17:26:35.887" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For PCR, does PCR require a double stranded piece of DNA that you are trying to replicate, or does it require a circular piece of DNA (i.e. a plasmid).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Circular DNA is not required.  Double-stranded DNA is, but keep in mind that if you start with a single-stranded DNA that (i) can be primed by an appropriate 'foward' primer and (ii) contains the reverse complement of a 'reverse' primer binding site (&lt;strong&gt;i.e. contains the sequence of a reverse primer, see comments&lt;/strong&gt;), then during the first cycle of PCR, a double-stranded DNA will be created from the single-stranded DNA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, circular DNA, while generally a viable substrate for PCR, is less good than linear dsDNA under some extreme conditions, especially if it highly supercoiled, because strand displacement by DNA polymerase becomes more difficult.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, I can't seem to find a source wheich explains how PCR works when you are also introducing a mutation at the same time. We used different primers to introduce a mutation during the PCR, but I am stilll very confused as to how the primer could bind when the nucleotide sequence is not exactly matching?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The thermodynamics of primer binding is a complex but fairly well-understood subject.  Each nucleotide in the primer that matches the target DNA sequence will provide favorable enthalpic (and entropic) interactions with the target, favoring binding.  These favorable interactions are even more favored if the nucleotide's neighbors in the DNA chain also match.  There is a general entropic penalty for binding because the degrees of freedom in a single dsDNA molecule are way lower than in two ssDNA molecules.  Essentially, once there are enough matching sites between a primer and a target DNA, binding will be favorable (and this transition point will depend on temperature, with higher temperature always dis-favoring binding).  The presence of a few mis-matches doesn't mean a primer can never bind, it means that the temperature at which the hybridized primer/dsDNA pair dissociates (the &quot;melting temperature&quot;) is lower than for a fully-matching primer.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-07-07T17:57:11.400" LastActivityDate="2015-07-07T17:57:11.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33703" PostTypeId="2" ParentId="33646" CreationDate="2015-07-07T17:39:01.583" Score="2" Body="&lt;p&gt;In aqueous solution, $\alpha$-D-glucose and $\beta$-D-glucose equilibrate. So glucose derived from aqueous acid hydrolysis of cellulose will eventually become an equilibrium mixture of α and β. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2527694/&quot; rel=&quot;nofollow&quot;&gt;A 2006 paper&lt;/a&gt; goes into great detail on the kinetics and thermodynamics of the interconversion (in water).  They say: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At 30 °C, the rate constants for ring closing toward α-pyranose and β-pyranose are 0.044 s&lt;sup&gt;−1&lt;/sup&gt; and 0.072 s&lt;sup&gt;−1&lt;/sup&gt;, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep in mind those figures don't include the effect of acid catalysis, and are incomplete without rate constants for the formation of the ring-open aldehyde from the two different pyranoses.  But they should give you a rough sense of how quickly the process occurs.  A later figure in the same paper shows that it takes roughly 12,000 seconds or 3.3 hrs for nearly complete (~90%) relaxation to equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also see chemistry.stackexchange.com/a/7311/13184 for more information.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-07T17:39:01.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33704" PostTypeId="1" AcceptedAnswerId="33714" CreationDate="2015-07-07T17:43:32.227" Score="13" ViewCount="460" Body="&lt;p&gt;It's well known that undiluted nitroglycerin will explode when subjected to a physical shock.  What is happening as a result of the physical shock that sets off the explosive reaction?  In particular, why doesn't it explode spontaneously at room temperature without a physical shock?&lt;/p&gt;&#xA;" OwnerUserId="5298" LastEditorUserId="11367" LastEditDate="2015-07-07T19:43:35.663" LastActivityDate="2015-07-08T12:06:19.853" Title="Mechanism of initiation in contact explosives?" Tags="&lt;reaction-mechanism&gt;&lt;explosives&gt;&lt;nitro-compounds&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33705" PostTypeId="2" ParentId="33696" CreationDate="2015-07-07T17:51:15.333" Score="2" Body="&lt;p&gt;Your assignment seems reasonable. One thing to keep in mind is that even though  phenyl protons are in different chemical environments, they may be too close to resolve and with a low-field instrument, some may appear as a &quot;blob&quot; of superimposed signals. In this case, the peak from (B) is simply unresolved from that of the (C) protons, giving a single signal that integrates to 3 protons. See &lt;a href=&quot;http://sdbs.db.aist.go.jp/sdbs/cgi-bin/img_disp.cgi?disptype=disp3&amp;amp;imgdir=hsp&amp;amp;fname=HSP03910&amp;amp;sdbsno=13602&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; with a 90 MHz instrument. That spectrum is a bit different than the one you have, but the one you've been given may be a simulated spectrum.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-07-07T17:51:15.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33706" PostTypeId="1" AcceptedAnswerId="33709" CreationDate="2015-07-07T18:05:06.437" Score="8" ViewCount="23588" Body="&lt;p&gt;According to me, $\ce{Fe^{2+}}$ should be a better reducing agent because $\ce{Fe^2+}$ - after being oxidized - will attain a stable $\ce{d^5}$ configuration, whereas $\ce{Cr^2+}$ will attain a $\ce{d^3}$ configuration. I think the half filled $\ce{d^5}$ configuration is more stable than the $\ce{d^3}$ configuration. Why is this not so? &lt;/p&gt;&#xA;" OwnerUserId="17337" LastEditorUserId="5026" LastEditDate="2018-03-31T12:22:50.487" LastActivityDate="2018-11-26T15:33:56.393" Title="Why is Cr(2+) a stronger reducing agent than Fe(2+) in water?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;electronic-configuration&gt;&lt;transition-metals&gt;&lt;reduction-potential&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33707" PostTypeId="1" CreationDate="2015-07-07T18:27:05.383" Score="4" ViewCount="1358" Body="&lt;p&gt;Looking at examples of these two molecular interactions makes me wonder what the differences or similarities are exactly.&lt;/p&gt;&#xA;" OwnerUserId="17338" LastEditorUserId="10185" LastEditDate="2015-07-08T16:58:49.067" LastActivityDate="2015-07-08T16:58:49.067" Title="Is there a relationship between London dispersion forces and hydrophobic interactions?" Tags="&lt;physical-chemistry&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-07-08T19:10:16.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="33708" PostTypeId="2" ParentId="33706" CreationDate="2015-07-07T18:32:48.230" Score="3" Body="&lt;p&gt;In water both $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ are in octahedral configurations. This means that $d$-orbitals become unequal in their energy; specifically 3 of them are lower and 2 are higher. This is the premise of 'crystal field theory'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the nearest neighborhood, the splitting may be strong enough to force electron pairing or it may not. For water it usually isn't. This means that 'first half-filled shell' here is 3 electrons - $\ce{V^{2+}}$ $\ce{Cr^{3+}}$ or $\ce{Mn^{4+}}$. The second half-filled shell in low field (water ligand) is $\ce{Mn^{2+}}$ or $\ce{Fe^{3+}}$ (5 $d$-orbitals), both surprisingly stable. In strong field (say, $\ce{CN^-}$ ligand), it is 6 electrons (3 double occupied lower orbitals), like in $\ce{[Co(NH3)_{6}]^{3+}}$ and $\ce{[Fe(CN)_{6}]^{4-}}$. The next &quot;subshell&quot; is,  8 electrons, (6 on 3 lower orbitals and 2 on higher), like in $\ce{Ni^{2+}}$. Depending on the strength of the ligands, an unusual square planar coordination may become preferable, with two higher orbitals also splitting. It is typical for $\ce{Ni}$ subgroup in +2 oxidation state and $\ce{Cu}$ subgroup in +3 oxidation state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR : invest some time into reading about &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_field_theory&quot; rel=&quot;nofollow&quot;&gt;crystal field theory&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="10185" LastEditDate="2015-07-07T18:35:25.213" LastActivityDate="2015-07-07T18:35:25.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33709" PostTypeId="2" ParentId="33706" CreationDate="2015-07-07T18:45:59.810" Score="8" Body="&lt;p&gt;The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is tricky to compare the &quot;stability&quot; of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the change in (free) energy that occurs from start to finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, it may be that $\ce{Fe^3+}$ is more stable than $\ce{Cr^3+}$ on an absolute scale, but what we really care about is how much more stable $\ce{Cr^3+}$ is to $\ce{Cr^2+}$ compared to how much more stable $\ce{Fe^3+}$ is to $\ce{Fe^2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's examine all four using electron configuration as you have done:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Fe^2+}$ is $\ce{d^6}$ or more probably $\ce{[Ar] 4s^1 3d^5}$ - two half-filled half shells!&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Fe^3+}$ is $\ce{d^5}$, which is $\ce{[Ar] 3d^5}$ - one filled half shell.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The difference between the two iron ions might not be that large.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{Cr^2+}$ is $\ce{d^4}$, which is $\ce{[Ar] 3d^4}$ or $\ce{[Ar] 4s^2 3d^2}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cr^3+}$ is $\ce{d^3}$, which is $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^2 3d^1}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The energy difference between chromium ions might be larger. Represented graphically, the reaction coordinate energy diagram for the two process might be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MC0ZC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now the number part. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can go looking for some standard thermodynamic data to help make our case. The best data are for &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard reduction potentials&lt;/a&gt;, because these data are for exactly what we want!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the reduction potential data and writing the equations in the direction we care about, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;&amp;amp;\ce{Fe^2+ -&amp;gt; Fe^3+} &amp;amp;&amp;amp;&amp;amp;E^\circ=\pu{-0.77 V}\\&#xA;&amp;amp;\ce{Cr^2+ -&amp;gt; Cr^3+} &amp;amp;&amp;amp;&amp;amp;E^\circ=\pu{+0.44 V}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spontaneous reactions produce positive potential differences, so we can see right now that $\ce{Cr^2+}$ is a better reducing agent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go a step farther to &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Electrochemistry_and_Thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;free energy&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta G^\circ =-nFE^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to deal with free energy, we need a full reaction. Technically, comparing the two half-reactions in isolation is just as bad. However, the definition of the standard electrode potential and the standard free energy come with a common zero point in terms of half-reaction: $\ce{2H+ + 2e- -&amp;gt; H2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two full reactions (net ionic equations anyway) are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;&amp;amp;\ce{2Fe^2+ + 2H+ -&amp;gt; 2Fe^3+ + H2} &amp;amp;&amp;amp;E^\circ=\pu{-0.77 V}&amp;amp;&amp;amp;&amp;amp;\Delta G^\circ=\pu{+74kJ/mol}\\&#xA;&amp;amp;\ce{2Cr^2+ + 2H+ -&amp;gt; 2Cr^3+ + H2} &amp;amp;&amp;amp;E^\circ=\pu{+0.44 V}&amp;amp;&amp;amp;&amp;amp;\Delta G^\circ = \pu{-42 kJ/mol}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the winner is chromium, by a whopping $\pu{116 kJ/mol}$. In fact, the data might show that $\ce{Fe^2+}$ is &lt;strong&gt;more&lt;/strong&gt; stable than $\ce{Fe^3+}$, while $\ce{Cr^2+}$ is less stable than $\ce{Cr^3+}$. &lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="5026" LastEditDate="2018-03-31T12:19:08.530" LastActivityDate="2018-03-31T12:19:08.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33710" PostTypeId="2" ParentId="33704" CreationDate="2015-07-07T18:53:17.193" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In particular, why doesn't it explode spontaneously at room temperature without a physical shock?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://dujs.dartmouth.edu/wp-content/uploads/2012/05/12s-explosive-chemistry.pdf&quot; rel=&quot;nofollow&quot;&gt;this review article&lt;/a&gt;, the low activation energy barrier for initiating the detonation of nitroglycerin is the culprit in re: hazards in handling the substance. It also provides its own oxygen as fuel (cf. nitrocellulose, also discussed in the article), which is a contributing factor to its instability.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for random thermal motion of air molecules at room temperature, that's an average kinetic energy described by the Boltzmann distribution and is not the energetic equivalent of imparting even a small amount of energy by jarring/shaking the system for a short time (such as hitting a flask or dropping it).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is happening as a result of the physical shock that sets off the explosive reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A rapid and exothermic evolution of gaseous species is produced: That's the explosion. The mechanism of explosion is (see &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/redox/faq/nitroglycerin.shtml&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;): &#xA;$$\ce{4C3H5(ONO2)3(l) -&amp;gt; 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorDisplayName="user7951" LastEditDate="2015-07-08T12:06:19.853" LastActivityDate="2015-07-08T12:06:19.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33711" PostTypeId="5" CreationDate="2015-07-07T19:03:05.010" Score="0" Body="&lt;p&gt;This tag should be applied to questions that are about the equipment used in labs, and techniques of keeping them, their uses, or their maintenance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions should only be about the &lt;strong&gt;equipments&lt;/strong&gt; chemists use in the &lt;em&gt;lab&lt;/em&gt;; such as glassware (beakers, flasks, bottles, pipettes, watch glasses, glass rods etc.) or other equipment; namely, tongs, holders, plastic bars, burners, spatulas etc.  Please note that &lt;a href=&quot;/questions/tagged/glassware&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;glassware&amp;#39;&quot; rel=&quot;tag&quot;&gt;glassware&lt;/a&gt; falls under this category. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions with this tag should be directly about the equipment; and any other questions which do not directly deal with them and are about the experimental methods involving the mentioned equipment should be tagged with &lt;a href=&quot;/questions/tagged/experimental-chemistry&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;experimental-chemistry&amp;#39;&quot; rel=&quot;tag&quot;&gt;experimental-chemistry&lt;/a&gt; and any other relevant tags.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-07-07T22:42:12.330" LastActivityDate="2015-07-07T22:42:12.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33712" PostTypeId="4" CreationDate="2015-07-07T19:03:05.010" Score="0" Body="Questions regarding the equipment used in labs (such as flasks, tubes etc.), their operation and maintenance." OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-07-07T22:42:16.347" LastActivityDate="2015-07-07T22:42:16.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33713" PostTypeId="2" ParentId="33700" CreationDate="2015-07-07T19:09:49.013" Score="7" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oFDBM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your suggested method seems excessively complicated. Substitution of chloroacetone with phenol under basic conditions should give O-alkylation in reasonably good yield. The phenol will be deprotonated and the nucleophilic phenoxide ion will then undergo an $S_N2$ reaction with the primary alkyl chloride.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-07-07T19:09:49.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33714" PostTypeId="2" ParentId="33704" CreationDate="2015-07-07T19:28:09.193" Score="10" Body="&lt;p&gt;My guess would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation&quot; rel=&quot;noreferrer&quot;&gt;cavitation&lt;/a&gt;, presumably 'non-inertial cavitation':&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Non-inertial cavitation is the process in which a bubble in a fluid is forced to oscillate in size or shape due to some form of energy input, such as an acoustic field. Such cavitation is often employed in ultrasonic cleaning baths and can also be observed in pumps, propellers, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The energetics of cavitation are astonishingly out-sized with respect to the tiny size of bubbles, probably leading to more than enough energy to cross the initial activation barrier of the combustion reaction.  As one example, the abstract &lt;a href=&quot;http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=5192470&amp;amp;tag=1&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; summarizes other reports' results indicating that the temperature inside collapsing cavitation bubbles may reach between 10,000 K and 100,000,000 K.&lt;/p&gt;&#xA;&lt;p&gt;I don't have access to the full text to determine whether or not they met with success in attempting to ascertain cavitation's role in nitroglycerin initiation, but &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF00913238&quot; rel=&quot;noreferrer&quot;&gt;these authors&lt;/a&gt; apparently investigated the question almost fifty years ago.  The abstract (references excised):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is known that in some circumstances liquid explosives can be initiated with unexpected ease, and in other circumstances only with great diffculty.&lt;/p&gt;&#xA;&lt;p&gt;Thus, Winning has shown that nitroglycerine (NG) free of gaseous inclusions and poured into a vessel so as to leave no wall surfaces free of liquid is not exploded even by the action of a fairly strong shock wave from a detonator immersed in the NG.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, in handling liquid explosives there have been quite a few cases in which relatively weak vibrations or impacts have led to unexpected explosions, which have sometimes had serious consequences. For example, a British report describes an unfortunate accident that resulted from dropping a polyethylene bottle containing NG. Upon hitting the ground the NG exploded.&lt;/p&gt;&#xA;&lt;p&gt;The initiation of explosion by “hot spots” resulting from the adiabatic compression of gaseous inclusions even before the arrival of the shock wave has been reliably demonstrated in numerous experiments. However, some cases of initiation of liquid explosives simply cannot be attributed to the heating of such gaseous inclusions, since in these cases the adiabatic compression temperatures of the gas are so small that it is not possible to talk of a “hot spot. “Such puzzling cases include, for example, the above-mentioned explosion of NG in a polyethylene bottle. In other experiments the role of gaseous inclusions has been completely eliminated by first subjecting the liquid explosive to a constant high pressure. This so reduced the degree of compression of the gaseous inclusions by a weak shock that strong heating of the gas in the bubbles, if any were present in the liquid explosives, was completely excluded. Nonetheless, there was no reduction in the sensitivity of the explosive to weak shocks.&lt;/p&gt;&#xA;&lt;p&gt;In attempting to explain such puzzling cases it is usually pointed out that explosion can be initiated by cavitation, which may develop in a liquid even as a result of a weak impact or vibration. So far, however, no one has offered any direct experimental evidence of the possibility of cavitational initiation of explosion in liquid explosives. The object of our research was to fill that gap.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Clearly they published their results, so I figure the odds are good that they reached &lt;em&gt;some&lt;/em&gt; sort of affirmative conclusion.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-07T19:41:30.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33715" PostTypeId="2" ParentId="33699" CreationDate="2015-07-07T19:54:40.467" Score="2" Body="&lt;p&gt;In terms of a specific answer to the questions you have asked, Curt F. has provided an excellent answer. I just want to provide you with a couple of extra resources to start you off reading about PCR mutagenesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pages 3 and 4 of this manual give an intro to QuikChange plasmid mutagenesis using PCR and primers with small mismatches: &lt;a href=&quot;http://www.chem.agilent.com/library/usermanuals/Public/200523.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.chem.agilent.com/library/usermanuals/Public/200523.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This page gives an intro on introducing restriction sites to the end of a DNA sequence being amplified by PCR: &lt;a href=&quot;https://www.addgene.org/plasmid-protocols/PCR-cloning/&quot; rel=&quot;nofollow&quot;&gt;https://www.addgene.org/plasmid-protocols/PCR-cloning/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17330" LastActivityDate="2015-07-07T19:54:40.467" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33716" PostTypeId="2" ParentId="31651" CreationDate="2015-07-07T20:01:24.813" Score="2" Body="&lt;p&gt;Get a pad of steel wool.  This is stainless steel, because it's made for use in a sink, washing dishes.  Take a 9-volt battery and briefly touch both terminals to the steel wool.  &lt;em&gt;It will catch on fire.&lt;/em&gt;  (Warning: don't try this at home, kids!  But it totally will.)  When the fire goes out, you'll be left with a pad of rusty steel wool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the answer is yes: fire can corrode stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your stainless steel is a big solid block rather than a super-high-surface-area pad of steel wool, the process will be a lot slower, but it will still happen over time.&lt;/p&gt;&#xA;" OwnerUserId="3807" LastActivityDate="2015-07-07T20:01:24.813" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33717" PostTypeId="2" ParentId="24673" CreationDate="2015-07-07T21:06:45.143" Score="0" Body="&lt;p&gt;$\ce{NH2-}$ doesn't actually exist on its own, it's only shown that way for parsing a reaction,  it is transferred at the same time the group that is giving it away gains something else, for example sodium amide is Na+NH2- so in order to be separated the NH2- needs somewhere to go as well as the Na+, always wondered why they bother teaching confusing nonexistent intermediates anyway, ions only exist in the non time they are exchanged (or electrocuted).&lt;/p&gt;&#xA;" OwnerUserId="17342" LastEditorUserId="17342" LastEditDate="2015-07-09T14:52:26.260" LastActivityDate="2015-07-09T14:52:26.260" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33718" PostTypeId="1" AcceptedAnswerId="33720" CreationDate="2015-07-07T21:09:40.050" Score="7" ViewCount="499" Body="&lt;p&gt;It is well know that the sugar water can rotate the plane of linearly polarized light, so light bends when travelling in sugar water. However, do the liquid state of sugar but not the sugar solution also have this property?&lt;/p&gt;&#xA;" OwnerDisplayName="user17341" LastEditorUserId="10185" LastEditDate="2015-07-07T21:30:56.747" LastActivityDate="2015-07-08T00:10:11.083" Title="Can molten sugar also rotate polarized light as sugar water does?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;molecules&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33719" PostTypeId="1" AcceptedAnswerId="33800" CreationDate="2015-07-07T21:33:27.143" Score="4" ViewCount="428" Body="&lt;p&gt;I have to questions regarding the conformations of carbohydrates.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 38 different conformations for an aldohexose or ketohexose, including $\alpha$ and $\beta$ anomers, namely 2 chairs, 6 boats, 6 skew-boats, 12 half-chairs and 12 envelopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these 38 conformations can have 729 different isomers ($5~\ce{OH}$ and $1~\ce{CH2OH}$) because of dihedral and tetrahedral angle ($3^6$ isomers, 3 (gg, gt, tg for each dihedral angle)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, glucose has $2~(\alpha~\text{and}~\beta~\text{anomers}) \times 38 \times 3^6 = 55404$ conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this calculation correct?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The nomenclature of the 38 different conformations is still not clear to me, which leads me to my second question.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;For chair form, the literature uses $\ce{^4C_1}$ and so on. But I'm not sure, which 4 carbon constitutes the reference plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly other used terms are $\ce{^{3,O}B}$ for boats, $\ce{^3S_5}$ for skew conformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please explain how to interpret these symbols?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="16622" LastEditDate="2015-07-07T21:57:08.857" LastActivityDate="2017-10-03T14:22:42.307" Title="Carbohydrate conformations" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;conformers&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33720" PostTypeId="2" ParentId="33718" CreationDate="2015-07-08T00:10:11.083" Score="7" Body="&lt;p&gt;Molecules in every state - liquid, solid and gaseous - rotate the plane of linearly polarized light - it doesn't matter, because this rotation arises from inherent asymmetry of molecules, which is causing their &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Optical_Activity&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical activity&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to the point, not only water solutions of sugars, but also their crystals, molten sugars and even vapor, rotate the plane of linearly polarized light. If you're asking specifically about sucrose checking this may be problematic as it's decomposing during melting, as you can read in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14562/why-do-other-sugars-melt-whereas-sucrose-decomposes&quot;&gt;this question&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Sucrose&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, but I think it still can be done, under vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also check these related questions about &lt;a href=&quot;https://chemistry.stackexchange.com/questions/19196/gas-phase-chiral-molecules&quot;&gt;chiral gases&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6271/molecular-chirality-and-optical-rotation&quot;&gt;rotation of the plane of polarization by molecules&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-07-08T00:10:11.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33723" PostTypeId="1" CreationDate="2015-07-08T02:10:09.400" Score="0" ViewCount="89" Body="&lt;p&gt;What is the branch of chemistry that deals with skin adhesives? Is there an adhesive that is not water soluble that is safe to use every day?&lt;/p&gt;&#xA;" OwnerUserId="17344" LastEditorUserId="865" LastEditDate="2015-08-07T14:00:56.463" LastActivityDate="2016-01-12T19:03:08.597" Title="Skin adhesive that can be used daily without damaging skin" Tags="&lt;medicinal-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33724" PostTypeId="1" CreationDate="2015-07-08T02:56:05.653" Score="-1" ViewCount="107" Body="&lt;p&gt;I am studying organic chemistry for quite a time now, but this is the general problem I come across. Both seems likely. I only know that heat causes elimination, but this does not appear to be the only reason always. So I want a complete reference.&lt;/p&gt;&#xA;" OwnerUserId="16618" LastEditorUserId="9961" LastEditDate="2015-07-08T11:43:03.280" LastActivityDate="2015-07-08T11:43:03.280" Title="When there is elimination and when there is nucleophilic substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2017-06-29T19:45:27.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="33726" PostTypeId="2" ParentId="31651" CreationDate="2015-07-08T09:00:37.983" Score="2" Body="&lt;p&gt;Not easily from what I can find. But remember it is Stain-less not Stain-impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually there are different type of stainless steel used for different stuff. Like for protect it from fire I feel A typical composition of &lt;strong&gt;18% chromium and 10% nickel&lt;/strong&gt;, commonly known as &lt;strong&gt;18/10 stainless&lt;/strong&gt;, is often used in cutlery and high-quality cookware will be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Stainless_steel#Types_of_stainless_steel&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you see &lt;a href=&quot;http://www.hgtvgardens.com/metal-and-steel-fire-pits-options&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; fire pits are also made up of stainless steel. &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-07-08T09:00:37.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33727" PostTypeId="1" AcceptedAnswerId="58621" CreationDate="2015-07-08T09:18:07.767" Score="20" ViewCount="1672" Body="&lt;p&gt;I may be only a physicist but I'm familiar with the superscripted dot to indicate a radical.  In a paper I'm reading the authors have also used a superscripted open circle.  I can't find any reference online or in my physical chemistry textbook (Atkins). Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcmDG.png&quot; alt=&quot;Example of superscript-open-circle notation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see they've done this a few times, and also used the dot to indicate a radical.  This is in a published paper &lt;a href=&quot;http://dx.doi.org/10.1016/S0926-3373(00)00276-9&quot; rel=&quot;noreferrer&quot;&gt;&quot;Photocatalytic degradation pathway of methylene blue in water&quot;, Houas et al., Applied Catalysis B: Environmental (2001)&lt;/a&gt;.  I'm starting to wonder (given the OH&lt;sup&gt;◦&lt;/sup&gt; in the text and OH&lt;sup&gt;•&lt;/sup&gt; in the equation) if it's a typo, but I'd be surprised if something so glaring could get through review or even proofreading by coauthors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does the open circle have a meaning of its own?&lt;/p&gt;&#xA;" OwnerUserId="17347" LastEditorUserId="402" LastEditDate="2016-03-02T00:23:37.863" LastActivityDate="2016-09-07T15:24:14.790" Title="Superscript open circle notation in formula" Tags="&lt;notation&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33728" PostTypeId="1" AcceptedAnswerId="33731" CreationDate="2015-07-08T09:28:45.453" Score="11" ViewCount="1050" Body="&lt;p&gt;Yesterday, I bought some &quot;magic&quot; rollerball pens (those with water-based ink), available in red, blue and black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They look and write like a normal pen, but have an eraser at the upper end, just like a pencil. You can erase the ink with this eraser, but the real miracle is: A few seconds in the freezer at −18 °C, and the ink becomes visible again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The eraser of the pen is translucent and feels wax-like, but is quite hard for an eraser.&lt;/li&gt;&#xA;&lt;li&gt;It does not rub off anything, you don't get that typical crumbs.&lt;/li&gt;&#xA;&lt;li&gt;Normal erasers do not have any effects on the ink, just as expected.&lt;/li&gt;&#xA;&lt;li&gt;It's the perfect ink eraser: You can erase a letter and directly write a new letter on it with the same pen, like a pencil.&lt;/li&gt;&#xA;&lt;li&gt;The process can be repeated. You can erase the ink and recover it in the freezer several times.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know those ink erasers from school or those &quot;magic pens&quot; which change the color of the ink of other pens, and also know, how it works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this pens are new to me. Has anyone an explanation of their principle of work?&lt;/p&gt;&#xA;" OwnerUserId="17349" LastEditorUserId="7448" LastEditDate="2015-10-30T17:24:06.667" LastActivityDate="2015-10-30T17:24:06.667" Title="Cold-sensitive magic pen" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33729" PostTypeId="2" ParentId="33724" CreationDate="2015-07-08T11:10:26.697" Score="6" Body="&lt;p&gt;A &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=substitution%20vs%20elimination&quot;&gt;Google search&lt;/a&gt; provides several sites that address this challenging topic in organic chemistry in detail. Additionally, most organic chemistry textbooks spend several pages on the matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a summary of the conditions that favor each mechanism:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_N 2$ reactions work best with methyl and primary substrates in polar aprotic solvents. Good nucleophiles are needed.&lt;/li&gt;&#xA;&lt;li&gt;$S_N 1$ reactions work best with tertiary substrates in polar protic solvents. Nucleophiles can be anything, but strong base or heat tends to suggest elimination.&lt;/li&gt;&#xA;&lt;li&gt;$E1$ reactions work best with tertiary substrates in polar protic solvents. Generally, no other reagent but the solvent is given (solvolysis) and the reaction is heated.&lt;/li&gt;&#xA;&lt;li&gt;$E2$ reactions work best with tertiary substrates in polar protic solvents. A strong base is required, and generally it is the conjugate base of the solvent. These reactions are often heated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or we can break it down by substrate type:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Methyl substrates can only do $S_N 2$ reactions. &lt;/li&gt;&#xA;&lt;li&gt;Primary substrates will do $S_N 2$ with strong nucleophiles in aprotic solvents and $E2$ reactions with strong bases in protic solvents. All other substitution/elimination conditions react so slowly as to be &lt;strong&gt;no reaction&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Tertiary substrates will do $S_N 1$ in protic solvents with weak bases, $E1$ in protic solvents with weak bases when heated, and $E2$ in protic solvents with strong bases. Substrates that generate resonance stabilized cations will behave like tertiary substrates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What about secondary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Secondary substrates are where you need to start looking for clues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_N 2$ - polar aprotic solvent&lt;/li&gt;&#xA;&lt;li&gt;$S_N 1$ - protic solvent weak base&lt;/li&gt;&#xA;&lt;li&gt;$E2$ - protic solvent strong base&lt;/li&gt;&#xA;&lt;li&gt;$E1$ - protic solvent and heat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alcohols in the presence of strong acid generally prefer the $S_N 1$ and $E1$ mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-08T11:10:26.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33730" PostTypeId="2" ParentId="33723" CreationDate="2015-07-08T12:42:11.897" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the branch of chemistry that deals with skin adhesives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely a combination of biomedical engineering and organic chemistry, with some materials science in there as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there an adhesive that is not water soluble that is safe to use every day?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00D634M8E&quot; rel=&quot;nofollow&quot;&gt;Mastisol&lt;/a&gt; fits your criteria.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-08T12:42:11.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33731" PostTypeId="2" ParentId="33728" CreationDate="2015-07-08T13:26:24.633" Score="11" Body="&lt;p&gt;The ink you describe is an example of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Leuco_dye&quot;&gt;leuco dye&lt;/a&gt;, which are molecules that exist in colored or non-colored states depending on what they're interacting with and under what conditions (temperature, pH, electronic environment, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ink system, you need 3 components: the electron-donating leuco dye molecules; the electron-accepting molecules (the developer); and a matrix of compounds that stabilize the reaction of the first 2 components for a range of temperatures (meaning the reaction is favored one way above a certain temperature, and another way below a certain temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simple graphical explanation from the notable pen company &lt;a href=&quot;http://www.pilotpen.eu/en/the-pilot-spirit/technology-innovation/the-excellence-of-the-inks/&quot;&gt;Pilot&lt;/a&gt;, part of which is below. Note that their system uses microcapsules that interact with other molecules in the ink in certain temperature ranges that result in either colored or transparent ink. In the diagram, A is the leuco dye, B is the developer, and C is the stuff that stabilizes the reaction of A and B over a temperature range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T2356.gif&quot; alt=&quot;Erasing adds heat -&amp;gt; transparent; cooling restores color&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/patents/US20120014740&quot;&gt;This&lt;/a&gt; detailed patent application for &quot;Thermosensitive decolorable ink composition&quot; details the physical and chemical processes involving formulations of leuco dyes. When you add heat to the ink by erasing, the non-colored form is present. When you cool it down, you are letting the system revert to the colored form and the ink reappears.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-08T13:26:24.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33732" PostTypeId="2" ParentId="33495" CreationDate="2015-07-08T14:15:20.477" Score="2" Body="&lt;p&gt;I'd suggest the seminal &lt;a href=&quot;http://www.wiley.com/WileyCDA/WileyTitle/productCd-047108039X.html&quot; rel=&quot;nofollow&quot;&gt;Orbital Interactions in Chemistry&lt;/a&gt; by Albright, Burdett and Myung-Hwan Whangbo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second edition came out in 2013 and I'm sure there are many copies of the first edition around as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They focus a lot on &lt;em&gt;pictoral&lt;/em&gt; MO theory - indicating how orbitals of different shapes and symmetry combine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To truly understand MO theory at a fundamental level, it really helps to study some quantum mechanics (and thus mathematics) but you can grasp a lot from the approach used in this book.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-08T14:15:20.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33734" PostTypeId="1" CreationDate="2015-07-08T14:40:23.253" Score="0" ViewCount="1315" Body="&lt;p&gt;Group II metals have a smaller cationic size but one more valence electron than Group I metals. Why would the higher charge density of a metal cation affect the strength of the bonding of it in its elemental state?&lt;/p&gt;&#xA;" OwnerUserId="17353" LastEditorUserId="10185" LastEditDate="2015-07-08T17:01:18.550" LastActivityDate="2015-09-26T18:51:52.163" Title="Why does cationic charge density affect metallic bonding?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33735" PostTypeId="1" CreationDate="2015-07-08T15:06:12.530" Score="3" ViewCount="1640" Body="&lt;p&gt;I was reading about side chain oxidation of alkyl side chains of benzene using alkaline $\ce{KMnO4}$ and heat. My book (Solomon's and Frhyle) says that this is only for compounds having a benzylic hydrogen and has given examples of ethyl and methyl benzene. I have searched the internet but could not find a mechanism. I tried to find something of my use in that research paper given in this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/79/mechanism-for-oxidation-of-arenes-via-permanganate/86#86&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what would happen if I take something like an ester or aldehyde/keto group, or any other organic functional group attached to a benzene ring. Will this reaction always happen with any aryl compound? Also, I came across an example of acetophenone being oxidized to benzoic acid by this method. It doesn't have a benzylic hydrogen so why does it still get oxidized?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2017-01-12T02:17:25.870" Title="Oxidation of alkyl side chain using KMnO4" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-04-26T04:28:02.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="33737" PostTypeId="2" ParentId="33707" CreationDate="2015-07-08T15:48:27.227" Score="1" Body="&lt;p&gt;To summarize the excellent but rather lengthy and detailed Stack Exchange answer referred to by Andy above, hydrophobic interactions do not require London forces between the hydrocarbon chains. They are a result of the very strong attraction of water for itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would further quote the following comment to that entry (by &lt;a href=&quot;https://chemistry.stackexchange.com/users/5633/thomij&quot;&gt;https://chemistry.stackexchange.com/users/5633/thomij&lt;/a&gt;), which I think sums it up nicely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. –  thomij Jun 23 '14 at 19:02&quot;&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-07-08T15:48:27.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33738" PostTypeId="2" ParentId="31904" CreationDate="2015-07-08T16:02:08.623" Score="0" Body="&lt;p&gt;Sodium in anhydrous ammonia liberate a solvated electron for reductions and produce sodium amide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the same in the lab with aluminium then the chance is big that nothing would happen because the oxide layer on the aluminium surface and the fact that ammonia is a toxic gas that you don't smell at high concentrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing will happen if you just put aluminium in aq. ammonia. You need to transform aluminium in another form with a strong acid to make it reactive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no.&lt;/p&gt;&#xA;" OwnerUserId="17354" LastActivityDate="2015-07-08T16:02:08.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33739" PostTypeId="1" CreationDate="2015-07-08T16:13:14.567" Score="1" ViewCount="28" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why does the 57th electron enter the &lt;span class=&quot;math-container&quot;&gt;$5d$&lt;/span&gt; subshell in lanthanum, shouldn't it should enter the &lt;span class=&quot;math-container&quot;&gt;$4f$&lt;/span&gt; subshell according to the Aufbau rule?&lt;/p&gt;&#xA;&lt;p&gt;Also in thorium, the 89th and 90th electrons enter the &lt;span class=&quot;math-container&quot;&gt;$6d$&lt;/span&gt; subshell while they should enter the &lt;span class=&quot;math-container&quot;&gt;$5f$&lt;/span&gt; subshell?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17356" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-08T16:53:51.803" Title="electronic configuration of lanthinide series" Tags="&lt;physical-chemistry&gt;&lt;electronic-configuration&gt;&lt;rare-earth-elements&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-07-08T18:30:53.427" ContentLicense="CC BY-SA 3.0" />
  <row Id="33740" PostTypeId="1" CreationDate="2015-07-08T16:20:46.833" Score="2" ViewCount="13" Body="&lt;p&gt;I would like to hear more about: Will an acrylic micro emulsion applied to polyamide 6 filaments (that have been coated with PTFE) weaken the filaments?&#xA;Also would polyester filaments be effected?&lt;/p&gt;&#xA;" OwnerUserId="17357" LastActivityDate="2015-07-08T16:20:46.833" Title="Would polyamide 6 filaments be effected by an acrylic micro emulsion?" Tags="&lt;polymers&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33742" PostTypeId="1" CreationDate="2015-07-08T17:12:06.220" Score="11" ViewCount="1554" Body="&lt;p&gt;When a carboxylic acid with low molecular weight is heated with $\ce{P2O5}$ we get the corresponding anhydride. My book says that this reaction happens with dehydrating agents such as $\ce{P2O5}$. I have also learned from my teacher that $\ce{H2SO4}$ is also a good dehydrating agent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what makes a molecule a good dehydrating agent? How can I identify one (or maybe what are the common dehydrating agents?) while solving questions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have seen the use of $\ce{P4O10}$, but do $\ce{P2O5}$ and $\ce{P4O10}$ play the same role as a reagent?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="7448" LastEditDate="2015-10-09T15:40:17.460" LastActivityDate="2019-12-25T22:21:11.937" Title="What makes a compound a dehydrating agent?" Tags="&lt;molecules&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33745" PostTypeId="1" CreationDate="2015-07-08T18:06:44.540" Score="3" ViewCount="10794" Body="&lt;p&gt;$\ce{Cs}$ and $\ce{Fr}$ are in the same group of lowest first ionization energy, but which element has the lowest ionization energy, and why?&lt;/p&gt;&#xA;" OwnerUserId="17358" LastEditorUserId="10185" LastEditDate="2015-07-08T18:31:54.967" LastActivityDate="2015-07-08T19:05:48.370" Title="Does caesium or francium have a lower ionization energy?" Tags="&lt;physical-chemistry&gt;&lt;periodic-trends&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33746" PostTypeId="2" ParentId="33745" CreationDate="2015-07-08T18:18:41.497" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;which element has the lowest ionization energy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ionization energy for Cs is lower than that of Fr.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Francium&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on Francium&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Francium has a slightly higher ionization energy than caesium, 392.811(4) kJ/mol as opposed to 375.7041(2) kJ/mol for caesium, as would be expected from relativistic effects, and this would imply that caesium is the less electronegative of the two. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is, relativistic effects are more pronounced for francium than for cesium. A greater positive charge density in the nucleus for the larger atom (and higher atomic number) means higher velocity for (here) the 7s electron, which has a higher relativistic mass and thus stays closer to the nucleus than in the case of Cs (where we're dealing with the 6s electron). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relativistic effect manifests in a greater effective electronic mass which in turn decreases the effective atomic radius, which makes it harder to pull off that electron (hence the higher ionization energy). See this &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_quantum_chemistry&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; for insights on atoms with a high atomic number and associated electrons with a low principal quantum number, my source for almost all of the answer contained in this post.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-08T19:05:48.370" LastActivityDate="2015-07-08T19:05:48.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33747" PostTypeId="1" CreationDate="2015-07-08T18:20:45.997" Score="8" ViewCount="51363" Body="&lt;p&gt;Aluminium is along the dark line of the Periodic Table and it is $p$-block metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it metal or a metalloid?&#xA;Why?&lt;/p&gt;&#xA;" OwnerUserId="17358" LastEditorDisplayName="user15489" LastEditDate="2015-07-08T18:33:32.900" LastActivityDate="2019-10-29T18:33:26.900" Title="Is aluminium a metal or metalloid?" Tags="&lt;metal&gt;&lt;periodic-table&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33748" PostTypeId="2" ParentId="33747" CreationDate="2015-07-08T18:27:28.253" Score="19" Body="&lt;p&gt;Aluminium is usually considered to be a metal, as described in the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Metalloid#Aluminium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metalloids: Aluminium&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals it has a close-packed crystalline structure, and forms a cation in aqueous solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason why, in a few cases, aluminium is referred to as a metalloid, from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has some properties that are unusual for a metal; taken together, these are sometimes used as a basis to classify aluminium as a metalloid. Its crystalline structure shows some evidence of directional bonding. Aluminium bonds covalently in most compounds. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/p-Block_Elements/Group_13%3A_The_Boron_Family/Chemistry_of_Aluminum/Aluminum_Oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;The oxide $\ce{Al2O3}$ is amphoteric, and a conditional glass-former.&lt;/a&gt; Aluminium can form anionic aluminates, such behaviour being considered nonmetallic in character&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Dividing_line_between_metals_and_nonmetals&quot; rel=&quot;nofollow noreferrer&quot;&gt;line between metals and non-metals&lt;/a&gt; is somewhat arbitrary with&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Elements to the lower left of the line generally display increasing metallic behaviour; elements to the upper right display increasing nonmetallic behaviour.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boron, carbon and silicon (all on the other side of the boundary line) are hard and brittle and are semiconductors, which is a major reason why the majority of resources list Aluminium as a metal, largely due to the metallic properties listed in the first quote.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related question is &lt;a href=&quot;https://chemistry.stackexchange.com/q/16261/15489&quot;&gt;Metal Compounds that bond covalently&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-07-08T19:27:35.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33749" PostTypeId="2" ParentId="18402" CreationDate="2015-07-08T19:02:17.220" Score="0" Body="&lt;p&gt;As a semi-literate in these matters, it seems to me that, logically at heat death, there can only be low temperature photons, as any matter would represent a local concentration of energy. &lt;/p&gt;&#xA;" OwnerUserId="17359" LastActivityDate="2015-07-08T19:02:17.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33750" PostTypeId="1" CreationDate="2015-07-08T20:17:39.183" Score="1" ViewCount="833" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the solubility of $\ce{AgCl}$ ($K_{\mathrm{sp}} = 1.6\times10^{-10}$) in $100.0~\mathrm{mL}$ of $4.00\times10^{-3}~\mathrm{M}$ calcium chloride.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The stated answer is: $2.0\times10^{-8}~\mathrm{M}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My way of solving was: &#xA;$$\begin{align}&#xA;\frac{1.6\times10^{-10}}{4.00\times10^{-3}} &amp;amp;= 4.00\times10^{-14}\\\\&#xA;\sqrt{4.00\times10^{-14}} &amp;amp;= 2.0\times10^{-7}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="5026" LastEditDate="2018-07-16T02:24:41.180" LastActivityDate="2018-07-16T02:24:41.180" Title="Solubility when multiple salts are involved" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33751" PostTypeId="2" ParentId="6250" CreationDate="2015-07-08T20:33:27.023" Score="5" Body="&lt;p&gt;I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first of all, in general the ionisation energy has to be calculated as the difference in &lt;em&gt;total energy&lt;/em&gt; and not just the electronic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, for molecules there are two types ionization energies: &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionization_energy#Vertical_and_adiabatic_ionization_energy_in_molecules&quot;&gt;&lt;em&gt;adiabatic&lt;/em&gt; and &lt;em&gt;vertical&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The adiabatic ionisation energy refers to the formation of the&#xA;  molecular ion in its ground vibrational state and the vertical&#xA;  ionization energy applies to the transition to the molecular ion&#xA;  without change in geometry. (From &lt;a href=&quot;http://goldbook.iupac.org/I03199.html&quot;&gt;IUPAC Gold Book&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that in accordance with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Franck%E2%80%93Condon_principle&quot;&gt;Franck–Condon principle&lt;/a&gt; the most probable transition is the vertical one. Consequently, usually, when we say &quot;ionisation energy&quot; we actually mean the &quot;vertical ionisation energy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if the total energy is understood in the &lt;em&gt;clamped nuclei approximation&lt;/em&gt;, i.e. as the electronic energy $E_{\mathrm{e}}$ together with the nuclear repulsion energy $V_{\mathrm{nn}}$, then for the vertical ionisation energy one could indeed use the difference in electronic energy rather than in total one simply because the nuclear repulsion energy stays the same. Mathematically, $V'_{\mathrm{nn}} = V_{\mathrm{nn}}$ since geometry is not changing, and thus,&#xA;$$&#xA;  E_{\mathrm{i}}&#xA;  =&#xA;  U' - U&#xA;  =&#xA;  (E'_{\mathrm{e}} + V'_{\mathrm{nn}})&#xA;  -&#xA;  (E_{\mathrm{e}} + V_{\mathrm{nn}})&#xA;  =&#xA;  E'_{\mathrm{e}} - E_{\mathrm{e}} \, ,&#xA;$$&#xA;where primed quantities refers to the resulting molecular specie and non-primed are that of the original molecular specie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, for the vertical ionisation energy in the clamped nuclei approximation one can use the difference in electronic energy. And one can do so regardless of what are the molecular species in question. For the specific case of the vertical ionisation of $\ce{H2+}$ the electronic energy of the resulting $\ce{H2^2+}$ specie is obviously zero, so we specifically get what OP suggested&#xA;$$&#xA;  E_{\mathrm{i}}&#xA;  =&#xA;  E'_{\mathrm{e}} - E_{\mathrm{e}}&#xA;  =&#xA;  0 - E_{\mathrm{e}}&#xA;  =&#xA;  - E_{\mathrm{e}} \, .&#xA;$$&#xA;However, this is just a trivial simplification of a general formula for the vertical ionisation energy for the case of a zero-electron resulting specie.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-08T20:45:43.070" LastActivityDate="2015-07-08T20:45:43.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33753" PostTypeId="2" ParentId="33750" CreationDate="2015-07-08T21:46:53.970" Score="6" Body="&lt;p&gt;I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2&amp;lt;=&amp;gt;Ca^2+ + 2Cl-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).&lt;/p&gt;&#xA;" OwnerUserId="5234" LastEditorUserId="5026" LastEditDate="2018-07-16T02:24:13.027" LastActivityDate="2018-07-16T02:24:13.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33754" PostTypeId="1" AcceptedAnswerId="33755" CreationDate="2015-07-08T23:27:50.897" Score="7" ViewCount="723" Body="&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1023/B:RUCB.0000030813.05424.1f&quot; rel=&quot;noreferrer&quot;&gt;Fedorov et al.&lt;/a&gt; assign the $^1$H-NMR ($\ce{CDCl3}$) chemical shifts of the $\ce{CH2}$ group in 2-(bromomethyl)phenylboronic acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gr1rj.png&quot; alt=&quot;2-(bromomethyl)phenylboronic acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to three singlets at $4.75$, $5.09$ and $5.15~\ce{ppm}$ in a ratio of $3:3:14$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain their reasoning?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="1389" LastEditDate="2015-07-09T00:22:05.007" LastActivityDate="2015-07-09T00:22:05.007" Title="Strange 1H-NMR signals/signal ratio" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33755" PostTypeId="2" ParentId="33754" CreationDate="2015-07-09T00:17:18.947" Score="10" Body="&lt;p&gt;The paper specifically addresses this issue and ascribes it to the tendency for boronic acids to exist as solvent and concentration dependent oligomers:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The dependence of the intensity ratio for these signals and the&#xA;  chemical shifts in the 1H NMR spectra on the concentration of the&#xA;  samples under study as well as on the nature of the solvents used&#xA;  indicates that anhydrides of boronic acids with different degrees of&#xA;  oligomerization can exist along with monomeric forms of these&#xA;  derivatives (ArB(OH)2) in solutions of substituted arylboronic acids.&#xA;  The conclusions about the existence of oligomeric forms of boronic&#xA;  acids and the absence of byproducts in the compounds isolated are&#xA;  supported by the results published earlier.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This type of equilibrium can be expressed like:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hQ2IU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher order oligomers are also possible.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2015-07-09T00:17:18.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33759" PostTypeId="1" AcceptedAnswerId="33803" CreationDate="2015-07-09T08:25:25.440" Score="6" ViewCount="14844" Body="&lt;p&gt;What is the structure and preparation for bleaching powder?  Why there is a &lt;strong&gt;controversy&lt;/strong&gt; in the formula of bleaching powder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have encountered many forms of the formula of bleaching powder, like &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{Ca(OCl)2}$ - wikipedia (it also mentions that bleaching powder is actually a mixture having other calcium compounds in it)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Ca(OCl)Cl}$ - my textbook&lt;/li&gt;&#xA;&lt;li&gt;$\ce{CaCl2 + Ca(OH)2 + H2O -&amp;gt; CaCl2\cdot Ca(OH)2\cdot H2O}$ - here the product is &quot;apparently&quot; bleaching powder.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is posed in other sites like:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.chemicalforums.com/index.php?topic=48844.0&quot; rel=&quot;nofollow&quot;&gt;http://www.chemicalforums.com/index.php?topic=48844.0&lt;/a&gt; -&#xA;here it reads that the composition of bleaching powder has been disputed for decades.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://chem-guide.blogspot.in/2010/04/preparation-properties-and-uses-of_6609.html&quot; rel=&quot;nofollow&quot;&gt;http://chem-guide.blogspot.in/2010/04/preparation-properties-and-uses-of_6609.html&lt;/a&gt; - many forms  of the formula are given here in this site although none of it has been proved right! Also can you explain about this Odling concept of bleaching powder mentioned in this site?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="402" LastEditDate="2019-08-21T19:04:17.607" LastActivityDate="2020-12-03T14:08:52.143" Title="What is the formula of bleaching powder?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33760" PostTypeId="1" AcceptedAnswerId="34017" CreationDate="2015-07-09T08:59:36.670" Score="6" ViewCount="3070" Body="&lt;p&gt;Why does the peak adiabatic temperature usually happen for an equivalence ratio a bit higher than 1? I think it had something to do with the probability of each fuel molecule finding a corresponding oxidizer, and the fact that the molecular weight of the fuel is generally higher than that of the oxidizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CP7qK.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17369" LastEditorUserId="46306" LastEditorDisplayName="user15489" LastEditDate="2018-02-28T17:16:36.463" LastActivityDate="2021-04-13T13:21:31.150" Title="Why is the maximum adiabatic temperature found for an equivalence ratio a bit higher than 1?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33761" PostTypeId="1" AcceptedAnswerId="35630" CreationDate="2015-07-09T10:26:36.887" Score="10" ViewCount="543" Body="&lt;p&gt;Lithium aminoborohydride is a versatile reducing agent. In the scheme below, I don't understand how does unhindered LAB afford an alcohol in the reduction of amides while a hindered LAB affords an amine. Why are two different products obtained and by what mechanisms do these reductions take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HOVdE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="865" LastEditDate="2015-10-16T18:06:33.720" LastActivityDate="2015-10-16T18:06:33.720" Title="The appliance of LAB in amide reduction" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33762" PostTypeId="1" AcceptedAnswerId="33931" CreationDate="2015-07-09T11:13:43.500" Score="3" ViewCount="301" Body="&lt;p&gt;I want to convert &lt;code&gt;mol&lt;/code&gt; files to &lt;code&gt;InChiKey&lt;/code&gt;. I am using &lt;code&gt;OpenBabel&lt;/code&gt; for the conversion but I have over &lt;code&gt;500&lt;/code&gt; mol files and converting them one by one is too time consuming. Is there some way to convert them all at once?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt; This is a sample mol file&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0187.cdx&#xA;  Sample&#xA;&#xA;  8  8  0  0  0  0  0  0  0  0999 V2000&#xA;   -0.4125    0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -0.4125   -0.7145    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.4125   -0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.8250    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;    0.4125    0.7145    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0&#xA;   -1.6500    0.0000    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0&#xA;    1.6500    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0&#xA;  1  2  2  0      &#xA;  2  3  1  0      &#xA;  3  4  2  0      &#xA;  4  5  1  0      &#xA;  5  6  2  0      &#xA;  6  1  1  0      &#xA;  2  7  1  0      &#xA;  5  8  1  0      &#xA;M  END&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also I want to know is there some other format which can combine mol files into one.I think I had heard of another format like &lt;code&gt;mol&lt;/code&gt; which can combine many mol files,but I cannot remember the name of that format&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;-  What is &lt;code&gt;SDF&lt;/code&gt; format? Can it combine many mol files into one?&lt;/p&gt;&#xA;" OwnerUserId="17178" LastEditorUserId="5017" LastEditDate="2015-07-13T15:50:23.937" LastActivityDate="2015-07-13T15:50:23.937" Title="Converting mol files to InChiKey in large batches" Tags="&lt;software&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="33763" PostTypeId="1" AcceptedAnswerId="33770" CreationDate="2015-07-09T11:18:19.813" Score="1" ViewCount="138" Body="&lt;p&gt;Suppose I have a group of organic compounds say A, B, C, D, E, F, G, H. I know their chemical formula and their names. Now there exists a reaction among these compounds but we do not know which compounds would react to give which products. It may be that A and B react to give C and D or B and D react to give F and H. If we are given these compounds, is it possible to identify a reaction that will occur among any number of these compounds to give the product which would belong to this set of compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you have not understood any part of this question, please ask me before down-voting, I will edit the question and try to re-explain. Thank you. ALso I am a computer science student, so I dont have too deep knowledge in chemistry, so do help me.&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="16341" LastEditDate="2015-07-10T03:12:39.730" LastActivityDate="2015-07-10T03:12:39.730" Title="Predicting reaction among compounds" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="12" ClosedDate="2015-07-09T15:11:52.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="33764" PostTypeId="1" AcceptedAnswerId="33766" CreationDate="2015-07-09T11:18:43.300" Score="34" ViewCount="10957" Body="&lt;p&gt;The following &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33756/types-of-molecular-interaction/33758?noredirect=1#comment50935_33758&quot;&gt;comment&lt;/a&gt; by Wildcat made me think about whether density functional theory (DFT) can be considered an &lt;em&gt;ab initio&lt;/em&gt; method.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;@Martin-マーチン, this is sort of nitpicking, but DFT (where the last &quot;T&quot; comes from &quot;Theory&quot;) can be considered as an ab-initio method since the theory itself is built from the first principles. The problem with the theory is that the exact functional is unknown, and as a result, in practice we do DFA calculations (&quot;A&quot; from &quot;Approximation&quot;) with some approximate functional. It it DFA which is not an ab-initio method then, not DFT. :) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always thought that &lt;em&gt;ab initio&lt;/em&gt; refers to wave function based methods only. In principle the wave function is not necessary for the basis of DFT, but it was later introduced by Kohn and Sham for practical reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC goldbook offers a definition of &lt;a href=&quot;http://goldbook.iupac.org/AT06983.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;ab initio&lt;/em&gt; quantum mechanical methods&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;ab initio&lt;/em&gt; quantum mechanical methods&lt;/strong&gt;&lt;br&gt;&#xA;  &lt;strong&gt;Synonym&lt;/strong&gt;: non-empirical quantum mechanical methods&lt;br&gt;&#xA;  Methods of quantum mechanical calculations independent of any experiment other than the determination of fundamental constants. The methods are based on the use of the full Schroedinger equation to treat all the electrons of a chemical system. In practice, approximations are necessary to restrict the complexity of the electronic wavefunction and to make its calculation possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to this, most density functional approximations (DFA) cannot be termed &lt;em&gt;ab initio&lt;/em&gt; since almost all involve some empirical parameters and/or fitting. DFT on the other hand is independent of any of this. What I have my problems with is the second sentence. It states, that treatment of all electrons is necessary. This is technically not the case for DFT, because here only the electron density is treated. All electrons and the wavefunction are implicitly treated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An earlier definition of &lt;em&gt;ab initio&lt;/em&gt; can be found in &lt;a href=&quot;http://dx.doi.org/10.1103/RevModPhys.32.275&quot; rel=&quot;noreferrer&quot;&gt;Leland C. Allen and Arnold M. Karo, &lt;em&gt;Rev. Mod. Phys.&lt;/em&gt;, &lt;strong&gt;1960&lt;/strong&gt;, &lt;em&gt;32&lt;/em&gt;, 275.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By &lt;em&gt;ab initio&lt;/em&gt; we imply: First, consideration of all the electrons simultaneously. Second, use of the exact nonrelativistic Hamiltonian (with fixed nuclei),&#xA;  $$\mathcal{H} = -\frac12\sum_i{\nabla_i}^2 &#xA;                - \sum_{i,a}\frac{Z_a}{\mathbf{r}_{ia}}&#xA;                + \sum_{i&amp;gt;j}\frac{1}{\mathbf{r}_{ij}}&#xA;                + \sum_{a,b}\frac{Z_aZ_b}{\mathbf{r}_{ab}}$$&#xA;  the indices $i$, $j$ and $a$, $b$ refer, respectively, to the electrons and to the nuclei with nuclear charges $Z_a$, and $Z_b$. Third, an effort should have been made to evaluate all integrals rigorously. Thus, calculations are omitted in which the Mulliken integral approximations or electrostatic models have been used exclusively. These approximate schemes are valuable for many purposes, but present experience indicates that they are not sufficiently accurate to give consistent results in &lt;em&gt;ab initio&lt;/em&gt; work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This definition obviously does not include DFT, but this is probably due to the fact it was published before the Hohenberg-Kohn theorems. But in general this definition is still largely the same as in the goldbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point which confuses me are titles like: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Potential Energy Surfaces of the Gas-Phase S&lt;sub&gt;N&lt;/sub&gt;2 Reactions $\ce{X- + CH3X ~$=$~ XCH3 + X-}$ $\ce{(X ~$=$~ F, Cl, Br, I)}$: A Comparative Study by Density Functional Theory and ab Initio Methods&quot;&lt;br&gt;&#xA;    &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00102a034&quot; rel=&quot;noreferrer&quot;&gt;Liqun Deng , Vicenc Branchadell , Tom Ziegler, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt;, &lt;strong&gt;1994&lt;/strong&gt;, &lt;em&gt;116&lt;/em&gt; (23), 10645–10656.&lt;/a&gt;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then again we have titles like: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Ab Initio Density Functional Theory Study of the Structure and Vibrational Spectra of Cyclohexanone and its Isotopomers&quot;&lt;br&gt;&#xA;    &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp983267h&quot; rel=&quot;noreferrer&quot;&gt;F. J. Devlin and P. J. Stephens, &lt;em&gt;J. Phys. Chem. A&lt;/em&gt;, &lt;strong&gt;1999&lt;/strong&gt;, &lt;em&gt;103&lt;/em&gt; (4), 527–538.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately Koch and Holthausen, who wrote the probably most concise book on DFT, &lt;a href=&quot;http://as.wiley.com/WileyCDA/WileyTitle/productCd-3527303723.html&quot; rel=&quot;noreferrer&quot;&gt;A Chemist's Guide to Density Functional Theory&lt;/a&gt;, never really refer to DFT as &lt;em&gt;ab initio&lt;/em&gt; or clearly draw the line. The closest they come is on page 18: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the context of traditional wave function based ab initio quantum chemistry a large variety of computational schemes to deal with the electron correlation problem has been devised during the years. Since we will meet some of these techniques in our forthcoming discussion on the applicability of density functional theory as compared to these conventional techniques, we now briefly mention (but do not explain) the most popular ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But that does not really answer my question. Throughout the book they use the term only in the form of &lt;em&gt;conventional ab initio theory&lt;/em&gt; or in combination of explicitly stating &lt;em&gt;wave function&lt;/em&gt; and variations thereof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my quite extensive research about &lt;a href=&quot;https://chemistry.stackexchange.com/a/27418/4945&quot;&gt;DFT selection criteria&lt;/a&gt; I never came about the term '&lt;em&gt;ab initio&lt;/em&gt; DFT'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question remains:&lt;br&gt;&#xA;&lt;strong&gt;Is density functional theory an ab initio method?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2016-03-22T08:22:19.943" Title="Is density functional theory an ab initio method?" Tags="&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;&lt;ab-initio&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="33765" PostTypeId="1" CreationDate="2015-07-09T11:22:40.783" Score="4" ViewCount="2580" Body="&lt;p&gt;In my textbook of chemistry it is written that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the Lewis description of covalent bond, the Bond order is given by the number of bonds between the two atoms  of a molecule.&#xA;   ... &#xA;  Isoelectronic molecules and ions have identical bond orders; for example, $\ce{F2}$ and $\ce{O2^{2-}}$ have bond order $1$. $\ce{N2}$, $\ce{CO}$ and $\ce{NO+}$ have bond order $3$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried some more molecules and it turns out that for two diatomic isoelectronic molecules (neutral or charged) the bond orders are always the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the bond order is always same for isoelectronic species (diatomic  molecules and ions) how can I prove it? I tried and got the following result: &#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For any diatomic molecule having combination of elements of same period( either period 2 or period 3) isoelectronic molecules and ions have the same bond order.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Proof: Assume the two atoms $A$ and $B$ ( atoms may be ions also) having valence electrons in their outer shell $x$  and $y$.&#xA;  Now if they combine to form diatomic molecule or ion having bond order $m$ then they must satisfy the below two equation( octet rule):&#xA;  1) $x+m=8$&#xA;  2) $y+m=8$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now since they are along same period therefore the total no. of inner electrons( total electrons excluding valence electrons) must be equal( 2 electrons for period 2 elements and 10 electrons for period 3 elements). If we vary the atoms(or ion) $A$ and $B$ the sum of these inner electrons will obviously be constant( equal to 4 electrons for period 2 elements and 20 electrons for period 3 elements).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Therefore, For species to be isoelectronic:&#xA;  $x+y=constant$( say $k$) &#xA;  Adding equation 1) and 2) we get:&#xA;  $x+y+2m=16$ or&#xA;  $k+2m=16$ or&#xA;  *Bond order*$=m=(16-k)/2$ which is constant.  Q.E.D.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;If the result given in my textbook is not true (since it looks like the fact is universal! compare $\ce{HCl}$ and $\ce{F2}$ which are isoelectronic having same bond order but elements in $\ce{HCl}$ involves period 1 and period 3 elements) then please state any counter examples.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14914" LastEditorUserId="10185" LastEditDate="2015-07-09T12:18:45.987" LastActivityDate="2020-09-29T02:18:20.460" Title="Are bond orders always the same in isoelectronic diatomic species?" Tags="&lt;bond&gt;&lt;electronic-configuration&gt;&lt;lewis-structure&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33766" PostTypeId="2" ParentId="33764" CreationDate="2015-07-09T11:57:02.500" Score="29" Body="&lt;p&gt;First note that the acronym DFA I used in my comment originates from Axel D. Becke paper on 50 year anniversary of DFT in chemistry:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us introduce the acronym DFA at this point for “density-functional&#xA;  &lt;em&gt;approximation&lt;/em&gt;.” If you attend DFT meetings, you will know that Mel&#xA;  Levy often needs to remind us that DFT is &lt;em&gt;exact&lt;/em&gt;. The failures we&#xA;  report at meetings and in papers are not failures of DFT, but failures&#xA;  of DFAs. &lt;a href=&quot;http://dx.doi.org/10.1063/1.4869598&quot; rel=&quot;noreferrer&quot;&gt;Axel D. Becke, &lt;em&gt;J. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;140&lt;/em&gt;, 18A301&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, there are in fact two questions which must be addressed: &quot;Is DFT ab initio?&quot; and &quot;Is DFA ab initio?&quot; And in both cases the answer depend on the actual way &lt;em&gt;ab initio&lt;/em&gt; is defined.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If by &lt;em&gt;ab initio&lt;/em&gt; one means a &lt;em&gt;wave function based&lt;/em&gt; method that do not make any further approximations than HF and do not use any empirically fitted parameters, then clearly neither DFT nor DFA are ab initio methods since there is no wave function out there.&lt;/li&gt;&#xA;&lt;li&gt;But if by &lt;em&gt;ab initio&lt;/em&gt; one means a method developed &lt;em&gt;&quot;from first principles&quot;&lt;/em&gt;, i.e. on the basis of a physical theory only without any additional input, then &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;DFT &lt;em&gt;is&lt;/em&gt; ab initio;&lt;/li&gt;&#xA;&lt;li&gt;DFA might or might not be ab initio (depending on the actual functional used).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the usual scientific meaning of &lt;em&gt;ab initio&lt;/em&gt; is in fact the second one; it just happened historically that in quantum chemistry the term &lt;em&gt;ab initio&lt;/em&gt; was originally attached exclusively to Hartree–Fock based (i.e. wave function based) methods and then stuck with them. But the main point was to distinguish methods that are based solely on theory (termed &quot;ab initio&quot;) and those that uses some empirically fitted parameters to simplify the treatment (termed &quot;semi-empirical&quot;). But this distinction was done &lt;em&gt;before&lt;/em&gt; DFT even appeared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the demarcation line between ab initio and not ab initio was drawn before DFT entered the scene, so that non-wave-function-based methods were not even considered. Consequently, there is no sense to question &quot;Is DFT/DFA ab initio?&quot; with this definition of &lt;em&gt;ab initio&lt;/em&gt; historically limited to wave-function-based methods only. Today I think it is better to use the term &lt;em&gt;ab initio&lt;/em&gt; in quantum chemistry in its more usual and more general scientific sense rather then continue to give it some special meaning which it happens to have just for historical reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we stick to the second definition of &lt;em&gt;ab initio&lt;/em&gt; then, as I already said, DFT &lt;em&gt;is&lt;/em&gt; ab initio since nothing is used to formulate it except for the same physical theory used to formulate HF and post-HF methods (quantum mechanics). DFT is developed from the quantum mechanical description without any additional input: basically, DFT just reformulates the conventional quantum mechanical wave function description of a many-electron system in terms of the electron density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the situation with DFA is indeed a bit more involved. From the same viewpoint a DFA method with a functional which uses some experimental data in its construction is &lt;em&gt;not&lt;/em&gt; ab initio. So, yes, DFA with B3LYP would not qualify as ab initio, since its parameters were fitted to a set of some experimentally measure quantities. However, a DFA method with a functional which does not involve any experimental data (except the values of fundamental constants) &lt;em&gt;can&lt;/em&gt; be considered as ab initio method. Say, a DFA using some LDA functional constructed from a homogeneous electron gas model, is ab initio. It is by no means an exact method since it is based on a physically very crude approximation, but so does HF from the family of the wave function based methods. And if the later is considered to be ab initio despite the crudeness of the underlying approximation, why can't the former be also considered ab initio?&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-09T16:18:26.020" LastActivityDate="2015-07-09T16:18:26.020" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="33768" PostTypeId="1" CreationDate="2015-07-09T13:38:50.217" Score="4" ViewCount="1531" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BwMkS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Esters of catechol undergo the &lt;strong&gt;Fries rearrangement&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one product is formed more than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is due to the electron withdrawing nature of the $\ce{OCOCH3}$ group which prefers the attack on meta position.I was able to write a reaction mechanism for the product which is &lt;strong&gt;minor&lt;/strong&gt; in this case. But I couldn't write one for the &lt;strong&gt;major&lt;/strong&gt; product. I wonder how it is formed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the minor product itself rearrange to give the major product or are both of  the products formed directly in this reaction?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="15161" LastEditDate="2015-07-10T20:58:27.437" LastActivityDate="2015-07-10T20:58:27.437" Title="Esters of catechol - Fries rearrangement" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33769" PostTypeId="1" CreationDate="2015-07-09T14:04:59.920" Score="-1" ViewCount="85" Body="&lt;p&gt;Where can we find the inventors of all chemicals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discoverers of elements are easily found.&lt;/p&gt;&#xA;" OwnerUserId="9616" LastActivityDate="2015-07-09T14:11:10.153" Title="Where can we find the inventors of all chemicals?" Tags="&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-07-09T16:21:48.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="33770" PostTypeId="2" ParentId="33763" CreationDate="2015-07-09T14:10:52.197" Score="2" Body="&lt;p&gt;I think your question relates to the field of combinatorial chemistry. This is an important area especially for pharmaceutical companies that want to try out lots of different structures to see if they have therapeutic potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets talk specifics. In the case of known chemistry - e.g. formation of esters from alcohols and carboxylic acids, you can identify the possible reactions. Start with a system where A, B, C, and D are different alcohols and E, F, G, and H are different carboxylic acids. Under the right conditions (acid catalyzed dehydration), the alcohols can combine with the acids all possible ways to give 16 different esters (AE, AF, AG, ... etc.). But A (an alcohol) will not combine with the other alcohols B, C, or D -- and acid E will not combine with acids F, G, or H. Under more harsh conditions, however, they could be induced to react within each group (to form ethers or acid anhydrides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are talking about a random selection of compounds - especially ones with more complex structures and more functional groups, and a variety of reaction conditions, the outcome would be much less predictable.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-07-09T14:10:52.197" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33771" PostTypeId="2" ParentId="33769" CreationDate="2015-07-09T14:11:10.153" Score="1" Body="&lt;p&gt;I doubt you can find all the information in one place, but I suspect that between the Beilstein, the Merck Index, PubChem, CAS, and patent office filings, you can find most of what you seek. That said, there will still be a healthy component of academic and proprietary findings that will prove to be hard/impossible/expensive to get.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-09T14:11:10.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33773" PostTypeId="1" CreationDate="2015-07-09T14:34:55.937" Score="5" ViewCount="708" Body="&lt;p&gt;I'm wondering if the addition of a base of type $\ce{NR_3}$ (R: alkyl) to a ketone (e.g. 3-pentanone or 4-heptanone) results in the cis or trans-enolate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to ask here about four cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{NMe_3}$ (small)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NMeEt_2}$ (medium)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NEt_3}$ (big)&lt;/li&gt;&#xA;&lt;li&gt;$\ce{N(iPr)_2Et}$ (very bulky)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Maybe smaller bases as 1 or 2 usually give the trans-enolate, and bulkier bases such as 3 or even 4 affords the cis-enolate. (That's what I read, but I'm not able to understand why.) Is there a good explanation that rationalizes which enolate is formed, the cis or the trans?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sterically hindered bases forms the the kinetic enolate. For deprotonation of 3-Methyl-4-heptanone I assume the enolate at the linear unbranched chain site, that's clear for me, but kinetic/thermodynamic enolate has nothing to do with cis/trans enolate or does it?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related: &#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/34115/does-the-addition-of-tert-butanolate-to-3-pentanone-give-the-cis-or-trans-enolat?noredirect=1#comment51786_34115&quot;&gt;Does the addition of base to 3-pentanone give the cis or trans enolate?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="194" LastEditDate="2018-03-03T15:26:06.390" LastActivityDate="2018-03-03T15:26:06.390" Title="Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33774" PostTypeId="2" ParentId="30910" CreationDate="2015-07-09T14:38:00.943" Score="9" Body="&lt;p&gt;You are somewhat on the right track. However, consider that acidity depends on the relative stabilization of the acid vs. the conjugate base, since the expression for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; is as follows: &lt;span class=&quot;math-container&quot;&gt;$$\mathrm{p}K_\mathrm{a} = -\log\left(\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this figure for a visualization:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zSLfI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zSLfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electron-withdrawing groups like &lt;em&gt;p&lt;/em&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; (nitro group in the para position) increase acidity by making the carboxylate ion stable relative to the protonated form. However, electron-donating groups like &lt;em&gt;p&lt;/em&gt;-methoxy decrease acidity relative to standard benzoic acid by destabilizing the (already negatively charged) carboxylate anion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL; DR: An electron-withdrawing group like &lt;em&gt;p&lt;/em&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{NO2}$&lt;/span&gt; will better stabilize the negatively charged &lt;span class=&quot;math-container&quot;&gt;$\ce{COO-}$&lt;/span&gt; group than an electron-donating group like &lt;em&gt;p&lt;/em&gt;-&lt;span class=&quot;math-container&quot;&gt;$\ce{OCH3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a neat picture from UC Davis that summarizes what I have said:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MQoaS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16837" LastEditorUserId="16837" LastEditDate="2020-05-15T00:24:05.550" LastActivityDate="2020-05-15T00:24:05.550" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33775" PostTypeId="2" ParentId="33768" CreationDate="2015-07-09T14:56:21.160" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fries_rearrangement&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Wikipedia article&lt;/a&gt; provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2g7QZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$).  Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong &lt;em&gt;ortho-para&lt;/em&gt; director, so &lt;strong&gt;both &lt;em&gt;ortho&lt;/em&gt;- and &lt;em&gt;para&lt;/em&gt;-products are produced by the same mechanism&lt;/strong&gt;. The $\ce{OCOR}$ group &lt;strong&gt;does not&lt;/strong&gt; play a role in &quot;directing&quot; the substitution pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of &lt;em&gt;ortho/para&lt;/em&gt; isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic &lt;em&gt;vs&lt;/em&gt;. thermodynamic control. Typically, the &lt;em&gt;para&lt;/em&gt; product is the kinetic product (first formed), while the &lt;em&gt;ortho&lt;/em&gt; product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the &lt;em&gt;para&lt;/em&gt;-bromo product is formed in &gt;95% yield (&lt;a href=&quot;https://books.google.com/books?id=oYPRsEgL2EUC&amp;amp;pg=PA155&amp;amp;lpg=PA155&amp;amp;dq=dibromination%20of%20phenol&amp;amp;source=bl&amp;amp;ots=oJV6URTq1-&amp;amp;sig=ioYxPBybiYW8c0rvfB5jioDBzQ0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=emguVfrAB4q1sAScrIHQAw&amp;amp;ved=0CE8Q6AEwCQ#v=onepage&amp;amp;q=dibromination%20of%20phenol&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;). So your specific reaction is likely being run under kinetically controlled conditions to produce the &lt;strong&gt;para&lt;/strong&gt; product as the major product. As the Wikipedia link points out, the &lt;em&gt;para&lt;/em&gt; product can equilibrate at higher temperatures to produce more of the &lt;em&gt;ortho&lt;/em&gt; product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, just a word to address @bon 's comment.  Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (&lt;a href=&quot;https://books.google.com/books?id=AcpPyv0HWRsC&amp;amp;pg=SL1-PA172&amp;amp;lpg=SL1-PA172&amp;amp;dq=fries%20rearrangement%20of%20catechol%20esters&amp;amp;source=bl&amp;amp;ots=CXwKMjvnIc&amp;amp;sig=_tibx21YVK4_ghXCX-mUB8x51oM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=NIieVfzgMYPl-QHVjImwBQ&amp;amp;ved=0CFwQ6AEwCQ#v=onepage&amp;amp;q=fries%20rearrangement%20of%20catechol%20esters&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;, p A-172) - apparently one of the acyl groups is lost in the process.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-07-09T14:56:21.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33776" PostTypeId="2" ParentId="33764" CreationDate="2015-07-09T15:22:39.617" Score="11" Body="&lt;p&gt;The convention used by many is that &lt;em&gt;ab initio&lt;/em&gt; refers solely to wave-function based methods of various sorts and that &lt;em&gt;first principles&lt;/em&gt; refers to either wave-function or DFT methods with little approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a citation at the moment, but I know this convention is fairly widely used in, e.g., &lt;em&gt;J. Phys. Chem.&lt;/em&gt; journals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IUPAC gold book doesn't have &quot;first principles,&quot; but Google Scholar gives over 224,000 hits for &lt;a href=&quot;https://scholar.google.com/scholar?hl=en&amp;amp;q=first+principles+DFT&quot;&gt;&quot;first principles DFT&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-09T15:22:39.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33777" PostTypeId="2" ParentId="19808" CreationDate="2015-07-09T15:27:16.977" Score="5" Body="&lt;p&gt;In addition to all other good answers, I'd like to actually &lt;em&gt;show&lt;/em&gt; you why the &quot;neutralized solution&quot; is not equal to a neutral solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever we want to describe a neutralisation reaction (that is nothing else than a titration) we first need to find out where protons come from and where they go. Let's examine a simple titration of a monoprotic acid with a monoprotic base:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(\ce{H3O+})=c(\ce{OH-})+c(\ce{A-})-c(\ce{BH+})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protons come from the autoprotolysis of water and the acid residue anions and they will react with the titration base. Extending the parts of this equation by their $\mathrm{pH}$, $c_0$ and $k_\mathrm a$ containing counterparts, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c(\ce{H3O+})=\frac{k_\mathrm w}{c(\ce{H3O+})}+\frac{c_0(\ce{HA})~k_\mathrm a}{k_\mathrm a+c(\ce{H3O+})}-\frac{c_0(\ce{B})~k_\mathrm b~c(\ce{H3O+})}{k_\mathrm w+k_\mathrm b~c(\ce{H3O+})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the point of inflection, the equivalence point, the amount of both moles need to be equal and the total volume is the addition of previous acidic volume plus the added base volume $V_\mathrm{tot}=V_{\ce{A}}+V_{\ce{B}}$. The ratio of both concentrations $c(\ce{HA})$ and $c(\ce{B})$ is therefor equal to $1$.&#xA;$$\begin{align}&#xA;n_{\ce{A}}&amp;amp;=n_{\ce{B}}\\&#xA;\Leftrightarrow c_\mathrm A~V_\mathrm a&amp;amp;=c_\mathrm B~V_\mathrm b\\&#xA;\xrightarrow{V_\mathrm{tot}=V_\mathrm A+V_\mathrm B} \frac{c_B}{c_A}&amp;amp;=1=\tau~\text{or}~\gamma\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, rearranging the second equation to $c_\mathrm A/c_\mathrm B$ gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{c_\mathrm B}{c_\mathrm A}=\left(1+\frac{k_\mathrm w}{k_\mathrm b~c(\ce{H3O+})}\right)\left(\frac{k_\mathrm w}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{k_\mathrm a}{c(\ce{H3O+})+k_\mathrm a}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we now somehow have to solve for the proton concentration after setting $\tau=1$. As this is rather annoying I did that for the titration of any more or less relevant acids for $\mathrm pK_\mathrm a$ values between $-2$ and $16$, with $\ce{NaOH}$ giving the following graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1.2in$ &lt;img src=&quot;https://i.stack.imgur.com/OU66t.png&quot; alt=&quot;EP for titration of acids with NaOH&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that strong acids will be neutralized a pH 7 and extremely weak acids at pH 14. In between there is a linear range in which one can calculate the pH at the equivalence point by the simple equation $$\mathrm pH_\mathrm{eq}=7+0.5~\mathrm pK_\mathrm a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be expanded for the neutralization of every monoprotic acid with every monoprotic base and again there is a simple equation for calculating the pH at the equivalence point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1.2in$ &lt;img src=&quot;https://i.stack.imgur.com/yAotY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm pH_\mathrm {eq}=7+0.5~\mathrm pK_\mathrm a-0.5~\mathrm pK_\mathrm b$$&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorDisplayName="user7951" LastEditDate="2016-12-07T23:28:39.663" LastActivityDate="2016-12-07T23:28:39.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33778" PostTypeId="2" ParentId="33654" CreationDate="2015-07-09T15:33:54.320" Score="1" Body="&lt;p&gt;The problem you could be experiencing could be more related to the phenomenon of &quot;mass transport&quot; than anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, for a reaction to occur the two reactants need to get close to each other. It could be that a catalyst works by forming a reactive complex with the reactant and carrying it through multiple reactions.&#xA;Essentially, the catalyst &lt;em&gt;is&lt;/em&gt; speeding up the reaction but is caught up in the reaction while it is occuring. If there isn't enough catalyst for &lt;em&gt;every single&lt;/em&gt; reactant, some of them are occuring at the catalyzed speed and others are proceeding normally; adding a little more catalyst just ensures that a higher percentage of substrates are being assisted by the catalyst at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the following scenario: A room full of people trying to cut sheets of paper in half. Everyone is using scissors, and there is one movable, industrial paper cutter being passed around. The guy who has the paper cutter is going to move much more quickly while the paper cutter is at his station. Additionally, the paper cutter has to be with him &lt;em&gt;the entire time&lt;/em&gt; he is cutting, or else he slows down. The paper cutter is the catalyst.&#xA;If you add more paper cutters, you would keep speeding up. It isn't because the paper cutter disappears or stops working; it's because not everybody has a paper cutter.&lt;/p&gt;&#xA;" OwnerUserId="17374" LastEditorUserId="10185" LastEditDate="2015-07-09T15:42:07.450" LastActivityDate="2015-07-09T15:42:07.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33779" PostTypeId="1" CreationDate="2015-07-09T15:52:50.990" Score="1" ViewCount="538" Body="&lt;p&gt;Why should the enthalpy of a reaction be defined at constant pressure. Why can't it be defined at variable pressure?&lt;/p&gt;&#xA;" OwnerUserId="17378" LastEditorUserId="10185" LastEditDate="2015-07-09T15:59:02.093" LastActivityDate="2015-07-09T16:32:42.823" Title="Is enthalpy of a reaction meaningful at variable pressure?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33780" PostTypeId="1" AcceptedAnswerId="41205" CreationDate="2015-07-09T15:56:24.377" Score="37" ViewCount="109873" Body="&lt;p&gt;Hemoglobin is an iron-containing oxygen transport metalloprotein in the red blood cells of most mammals. Simply put, it's a carrier protein. Interestingly it doesn't carry carbon dioxide in the same way it does for oxygen $\ce{O2}$. Oxygen binds to the iron atoms in the protein whereas carbon dioxide $\ce{CO2}$ is bound to the protein chains of the structure. Carbon dioxide doesn't compete with oxygen in this binding process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, carbon monoxide $\ce{CO}$ is a very aggressive molecule. It's a colourless, odourless, and tasteless gas that is lighter than air and can be fatal to life. It has a greater affinity for hemoglobin than oxygen does. It displaces oxygen and quickly binds, so very little oxygen is transported through the body cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two equilibrium reactions of binding oxygen and becoming oxygenated hemoglobin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Hb (aq) + 4O2 (g) -&amp;gt; Hb(O2)4 (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Hb (aq) + 4O2 (g) &amp;lt;- Hb(O2)4 (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And carbon monoxide binding equation at equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Hb (aq) + 4CO (g) ⇌ Hb(CO)4 (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com.sg/books?id=XOL6mgnGOHoC&amp;amp;pg=PA22&amp;amp;lpg=PA22&amp;amp;dq=why+carboxyhemoglobin+bond+is+stronger&amp;amp;source=bl&amp;amp;ots=1xabW6LWuh&amp;amp;sig=wG57G9h3bCI5THVb_ITsAJ07VYk&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=lZeeVffEEcyvuQTgv7eQDA&amp;amp;ved=0CEwQ6AEwCA#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;It is said&lt;/a&gt;&#xA; the equation is shifted towards right, generating $\ce{Hb(CO)4(aq)}$, since its bond is much stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two main related questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is the carboxyhemoglobin bond stronger relative to that of oxygenated hemoglobin?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is carbon monoxide highly attracted to hemoglobin?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Does it have anything to do with the oxidation state of oxygen in each molecule? Please show some reaction equations associating $\ce{Fe}$.&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="24052" LastEditDate="2016-03-13T20:40:28.537" LastActivityDate="2019-05-10T14:53:36.483" Title="Why does carbon monoxide have a greater affinity for hemoglobin than oxygen?" Tags="&lt;biochemistry&gt;&lt;coordination-compounds&gt;&lt;oxidation-state&gt;&lt;chemical-biology&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="33781" PostTypeId="2" ParentId="33779" CreationDate="2015-07-09T16:08:13.403" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why should the enthalpy of a reaction be defined at constant pressure? Why can't it be defined at variable pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not sure &quot;should&quot; is the right choice of words: It can be defined for a reaction either at constant volume (variable pressure) or constant pressure (variable volume). You might choose one measurement over the other, but you'll get two different quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant volume, there's no work done by the system on the surroundings, so you're just getting the change in internal energy of the system (this is typical of a bomb calorimeter lab setup); in the other case, you get the internal energy change as well as the work, and that is how enthalpy is defined (mathematically). See &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_enthalpy_of_reaction&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt; for a little more detailed discussion.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-09T16:32:42.823" LastActivityDate="2015-07-09T16:32:42.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33782" PostTypeId="2" ParentId="33780" CreationDate="2015-07-09T16:59:13.710" Score="17" Body="&lt;p&gt;The answer has to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Pi_backbonding&quot; rel=&quot;noreferrer&quot;&gt;pi-backbonding&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the CO molecule has a negative formal charge on the carbon (it's neutral because of the oxygen having a positive formal charge). However, C is quite electropositive, and would like to relieve the stress caused by the negative formal charge. To relieve the stress caused by the negative charge, the CO molecule will want to bond to the iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it does this is related to CO's molecular orbitals, which look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HzGfp.png&quot; alt=&quot;CO MO Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the 4, 5, and 6 sigma orbitals, as well as the 2 pi orbital, are all antibonding. I will denote them with * from now on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As CO stands right now, it has a bond order of three (the two lower sigmas cancel each other out, since 4* is antibonding while 3 is bonding). None of its orbitals, as they stand, can participate in bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because carbon wants to rid itself of the negative charge, CO will do an interesting trick. It will promote an electron from the 5 sigma orbital into 2* pi. This reduces the bond order of CO from three to two, and introduces a separation of an electron pair. The picture of the 2* pi orbital is shown here, on the top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qMncn.gif&quot; alt=&quot;pi star orbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how is this more stable? In isolation, it is not. However, remember, we are not operating in isolation: there's an iron nearby. Most saliently, the iron has a d orbital, which looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1esYd.jpg&quot; alt=&quot;d orbital&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the iron can put an electron into this d orbital and bond it with the pi* star orbital on the CO on the carbon end. This is hugely stable (note that we were willing to promote an electron into an unoccupied orbital and break up an electron pair to do it), especially since it relieves charge stress to do it (as opposed to molecular oxygen, which does no such thing). This backbonding is also why cyanide will stable itself onto iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't hemoglobin designed to bind oxygen as strongly as possible, then? Surely it would seem this would be an advantage, evolutionarily, if oxygen could bind tightly and never let go? But oxygen &lt;em&gt;needs&lt;/em&gt; to let go: hemoglobin is a transport protein, and it needs to release oxygen into the tissues. That's why you have plenty of things that bond hemes more tightly than molecular oxygen: we need $\ce{O_2}$ in our tissues, not locked into hemoglobin forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The hybridization of the iron in the protein is often (but incorrectly) assigned as $\ce{sp^3d^2}$. This is implied by the octahedral geometry the ligands around the iron take: the heme consists of an iron bound in four coordination sites by a porphyrin ring, one site by a histidine, and the last site by the oxygen the iron needs to bind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe(II) + CO -&amp;gt; Fe(III)CO}$                 &lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="9961" LastEditDate="2017-02-19T19:47:18.887" LastActivityDate="2017-02-19T19:47:18.887" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="33784" PostTypeId="2" ParentId="31171" CreationDate="2015-07-09T17:12:29.320" Score="2" Body="&lt;p&gt;Such materials are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Bingham_plastic&quot; rel=&quot;nofollow&quot;&gt;Bingham plastics&lt;/a&gt;.  An angle of repose means that the material must withstand a non-zero shear stress (from its own weight) while still exhibiting zero shear rate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples include toothpaste, ketchup, mayonnaise, latex paint, and concentrated suspensions of coal particles in a otherwise newtonian liquid.  I'm not sure if any of these materials meets your requirement that it consist &quot;of individual molecules&quot;.  I don't think a single-component, single-phase, non-polymeric substance will ever exhibit Bingham plastic behavior, although (I might be wrong and) I can't cite any theoretical reason why this is so.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-09T17:12:29.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33789" PostTypeId="1" AcceptedAnswerId="33796" CreationDate="2015-07-09T18:23:27.053" Score="8" ViewCount="229" Body="&lt;p&gt;Why does &lt;em&gt;N&lt;/em&gt;,&lt;em&gt;N&lt;/em&gt;-dimethyl-1-phenylmethanamine (the left molecule in the picture) undergo hydrogenolysis of the C–N bond readily while 2-methyl-1,2,3,4-tetrahydroisoquinoline (right) doesn't react at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fTO7u.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5026" LastEditDate="2018-02-26T05:11:07.090" LastActivityDate="2018-02-26T05:11:07.090" Title="Hydrogenolysis of benzylic C-N bond" Tags="&lt;organic-chemistry&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33796" PostTypeId="2" ParentId="33789" CreationDate="2015-07-10T00:18:14.740" Score="12" Body="&lt;p&gt;Hydrogenolysis typically involves a metal surface ($\ce{Pd, Pt}$, etc.) and hydrogen gas. In the reaction a weak bond, that is, a bond that is either strained or a bond that can generate a stabilized radical is broken. In your examples there is no strain, but a resonance stabilized benzylic radical can be generated in one case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the benzylic amine pictured on the left, the benzylic carbon can easily adopt a geometry such that when the p-orbital on the benzylic carbon is generated it is already aligned with the p-orbitals in the aromatic ring and is therefore stabilized while it develops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molecule on the right is different. I built a model and as the C-N bond starts to break and the p-orbital starts to form, the developing p-orbital is twisted 45-60° away from alignment with the p-orbitals in the aromatic pi system - &lt;strong&gt;it is not effectively stabilized due to conformational constraints imposed by the ring system&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore this latter molecule will not undergo hydrogenolysis of the C-N bond as readily as the molecule on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some photos of my model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the molecule - note the &lt;strong&gt;blue&lt;/strong&gt; $\ce{N-CH3}$ nitrogen as a point of reference&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpawI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;conformation A - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is axial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted about (90-30) 60° away from being aligned with the p-orbitals in the aromatic ring - &lt;strong&gt;poor overlap potential&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VJW1P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;conformation B - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is equatorial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30-45° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted at least (90-45) 45° away from being aligned with the p-orbitals in the aromatic ring - &lt;strong&gt;poor overlap potential&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhAv5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-07-10T00:52:42.113" LastActivityDate="2015-07-10T00:52:42.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33797" PostTypeId="1" AcceptedAnswerId="33802" CreationDate="2015-07-10T00:39:58.730" Score="6" ViewCount="2818" Body="&lt;p&gt;After seeing one too many TV commercial about the fantastical properties of different high-profile shampoo products all &quot;with documented effect&quot; it suddenly struck me that I never actually researched what the active ingredients of a shampoo really is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what are the active ingredients of most shampoos, and how do they work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there is any significant difference between popular brands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any significant difference between cheap and expensive brands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will they actually heal your hair?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="9961" LastEditDate="2015-10-30T16:55:36.510" LastActivityDate="2021-01-26T16:19:54.937" Title="What are the active ingredients of shampoo, and how do they work?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33798" PostTypeId="1" AcceptedAnswerId="33799" CreationDate="2015-07-10T00:43:12.727" Score="2" ViewCount="193" Body="&lt;p&gt;I actually tested submerging a bunch of old rusty knives blades in a cup of plain cola for a few days not expecting much. To my amazement the myth held true, coke is a really good rust remover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the chemistry behind it?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="865" LastEditDate="2015-10-30T16:57:08.407" LastActivityDate="2015-10-30T16:57:08.407" Title="What makes cola efficient as a rust remover?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33799" PostTypeId="2" ParentId="33798" CreationDate="2015-07-10T01:04:56.083" Score="4" Body="&lt;p&gt;Rust consists of insoluble &lt;a href=&quot;https://en.wikipedia.org/wiki/Rust&quot; rel=&quot;nofollow&quot;&gt;iron oxides and hydroxides&lt;/a&gt;, $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Acidifying these compounds protonates the oxide and hydroxide anions to for water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O^{2-} + 2H^+ \rightarrow H_2O}$$&#xA;$$\ce{OH^{-} + H^+ \rightarrow H_2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the soluble iron $\ce{Fe(III)}$, which can be washed away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many colas contain phosphoric acid, and are quite acidic (Coke has a pH of ~2.5 according to &lt;a href=&quot;http://www.selah.k12.wa.us/soar/sciproj2005/Graphs/MarshalN.pdf&quot; rel=&quot;nofollow&quot;&gt;this science fair project&lt;/a&gt;), thus they can dissolve the oxides.&lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2015-07-10T01:04:56.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33800" PostTypeId="2" ParentId="33719" CreationDate="2015-07-10T01:53:40.357" Score="3" Body="&lt;p&gt;After going through several literature, finally I am able to answer these questions. It might be important starting point for someone wanting to know carbohydrate chemistry, that's why I'm posting my answer.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, I'm right. For α- and β-glucose 27702 configuration are possible. So, in total $27702 \times 2 = 55404$ possible structures for glucose [1]. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For xylose, 81 rotamers were optimized per puckering conformation for a total of 3078 conformations. For α-glucose, β-glucose, and β-mannose, 729 rotamers were generated for each pucker, resulting in 27 702 conformations for each of these six-carbon monosaccharides. For GlcNAc, 972 conformations per ring pucker were generated for a total of 36 936 geometries. Thus, the initial set of conformations considered for all molecules in this study comprised 123 201 geometries.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Actually these nomenclature is known as Cremer-Pople (&lt;a href=&quot;https://en.wikipedia.org/wiki/John_Pople&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dr. Pople&lt;/a&gt; is the Noble prize winner chemist for his outstanding contribution to formulate DFT) nomenclature. They defined puckering parameters and generalized ring structure conformations for 3/4/5/6 or more carbon species. That's a real interesting world!&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nhcvM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;sup&gt;Courtesy: Shinya Fushinobu, University of Tokyo&lt;/sup&gt;&lt;br&gt;&#xA;It might not be clear from this picture. So I will try to explain them separately.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Chair&lt;/strong&gt;: Two chair conformation possible. (O-C2-C3-C5 forms the reference plane). If C-4 is on top then $^4C_1$ and vice versa. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/rJHjt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Boats&lt;/strong&gt;: Boats forms when two plane intersects each other. The bottom two corner atoms are used to distinguish. If two bottom corners are upside down then they are used as superscript on the left and vice versa.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/dRWOn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TfBDR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Skew-Boats&lt;/strong&gt;: Same as chair. Reference plane is formed by (O-C2-C3-C5). If C5 is on the top and C1 is in bottom then it's named as $^5S_1$ &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;Reference&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mayes, H. B.; Broadbelt, L. J.; Beckham, G. T. &lt;em&gt;Journal of the American Chemical Society&lt;/em&gt; &lt;strong&gt;2014&lt;/strong&gt;, 136 (3), 1008–1022 DOI: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja410264d&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1021/ja410264d&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="41328" LastEditDate="2017-10-03T14:22:42.307" LastActivityDate="2017-10-03T14:22:42.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33802" PostTypeId="2" ParentId="33797" CreationDate="2015-07-10T02:12:38.153" Score="9" Body="&lt;p&gt;Shampoo has a few common components: Surfactants, preservatives and a number of additives meant to give the product desirable properties.&lt;/p&gt;&#xA;&lt;p&gt;When you wash your hair, you're removing debris and buildup in two steps; the polar components (such as salts and aerosols that land in your hair) are washed away by the water, and the non-polar by the surfactants.&lt;/p&gt;&#xA;&lt;p&gt;Surfactants are molecules where one side is polar and the other is non-polar. Things with similar polarity like to group together, and the opposite happens with polar/non-polar pairs (I'll call this the 'Like attracts Like rule, it's very important to this particular problem).&lt;/p&gt;&#xA;&lt;p&gt;These surfactants can group together with  each other to form little spheres known as a micelle if there are enough around. This level is known as the critical micelle concentration.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WX0fH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WX0fH.jpg&quot; alt=&quot;micelle picture&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.ekshiksha.org.in/images_carbon_its_compund_X/figure_12.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;ekshiksha.org.in&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Because the non-polar parts of the surfactants come together, the center of the micelle is also non-polar and can hold small droplets and pieces of other non-polar substances like the sebum that builds up on your hair naturally over time.&lt;/p&gt;&#xA;&lt;p&gt;Almost all shampoos use a nearly identical set of surfactants. Common examples are sulfates (ammonium lauryl sulfate, Sodium laureth sulfate, sodium pareth sulfate, sodium lauryl sulfate) and/or polysorbates (PEG20 and PEG80).&lt;/p&gt;&#xA;&lt;p&gt;There are also additives that thicken it up, keep it from going bad and make that rich wonderful lather that we've all come to know and love.&lt;/p&gt;&#xA;&lt;p&gt;What you are paying for aside from the above is mainly botanical or synthetic additives meant to soothe or beautify. The best way is to look up these ingredients on a case by case basis and decide if they would benefit you. I would say &amp;quot;be wary&amp;quot;; a large number of ingredients provide virtually zero added benefit for your money. Some work extremely well for certain people; you never know until you try.&#xA;&lt;strong&gt;Always read your labels! If you don't know what something is, google it!&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17374" LastEditorUserId="15751" LastEditDate="2021-01-26T16:19:54.937" LastActivityDate="2021-01-26T16:19:54.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33803" PostTypeId="2" ParentId="33759" CreationDate="2015-07-10T04:18:07.323" Score="7" Body="&lt;p&gt;The bleaching agent in all commonly available bleach powders is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypochlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypochlorite&lt;/a&gt;.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium_hypochlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calcium hypochlorite&lt;/a&gt; is a pure compound with the formula &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2}$&lt;/span&gt; and is the &lt;em&gt;active&lt;/em&gt; component in most bleach powders.  (Note: &lt;em&gt;active&lt;/em&gt; does not necessarily mean the most &lt;em&gt;abundant&lt;/em&gt; component).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You presented three examples of formulations for bleaching powder that you have encountered:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2}$&lt;/span&gt; &lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)Cl}$&lt;/span&gt; &lt;/li&gt;&#xA;  &lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2 + Ca(OH)2 + H2O -&amp;gt; CaCl2\cdot Ca(OH)2\cdot H2O}$&lt;/span&gt; &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The third species, &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2\cdot Ca(OH)2\cdot H2O}$&lt;/span&gt;, cannot be a bleaching agent of any type because it does not contain hypochlorite.  So I view that as plainly incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium_hypochlorite&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bleaching powder is made with slightly moist slaked lime. It is not a simple mixture of calcium hypochlorite, calcium chloride, and calcium hydroxide. Instead, it is a mixture consisting principally of calcium hypochlorite &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2}$&lt;/span&gt;, dibasic calcium hypochlorite, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca3(OCl)2(OH)4}$&lt;/span&gt;, and dibasic calcium chloride, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca3Cl2(OH)4}$&lt;/span&gt; [7].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was able to find the source cited by Wikipedia online; it seems quite authoritative.  It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calcium hypochlorite solutions are made by suspending lime or hydrated lime in water and adding&#xA;  chlorine. However, reacting lime with water generates lots of heat, and this heat must be allowed to&#xA;  dissipate before chlorination. The maximum concentration is ∼85 g/L&#xA;   of available chlorine, above&#xA;  which dibasic calcium hypochlorite begins to precipitate.&#xA;  Chlorination can be done in a batch or continuous reactor similar to those used to make sodium&#xA;  hypochlorite, except that agitation is needed to maintain the suspension. In a batch reactor, agitation&#xA;  can be achieved using a mechanical stirrer, or, more effi ciently, by circulation. A centrifugal&#xA;  pump is used to pull the solution from the bottom of a conical tank and then reinject it tangentially&#xA;  to create a strong vortex. Alternatively, the solution passes through a packed tower before it returns&#xA;  to the tank. The packing in the tower creates shear that breaks the particles apart before the chlorine&#xA;  is injected at the bottom of the tower.57 The reaction is as follows:&#xA;  &lt;span class=&quot;math-container&quot;&gt;$\ce{2Cl2 + 2Ca(OH)2 → Ca(OCl)2 + CaCl2 + 2H2O}$&lt;/span&gt; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hemibasic calcium hypochlorite, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2\cdot 0.5Ca(OH)2}$&lt;/span&gt;, can be precipitated from a solution of calcium hypochlorite at 40–80°C.143 These solutions are commercially made by chlorinating a&#xA;  slurry of hydrated lime.5&#xA;   The precipitate is removed by filtration and dried. Although it may still be&#xA;  made in Asia, it has not been made in the United States or Germany since 1955.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...] &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dibasic calcium hypochlorite, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2 · 2Ca(OH)2}$&lt;/span&gt;, is commonly used as an intermediate in making calcium hypochlorite, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2}$&lt;/span&gt;.   It has never been successfully marketed as a product because it dissolves much too slowly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although calcium hypochlorite dihydrate, &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2 · 2H2O}$&lt;/span&gt;, can be made by chlorinating a suspension&#xA;  of lime, it is difficult to filter, partly due to the presence of calcium chloride. Calcium chloride also reduces the stability of the dry product. [...] Sodium hypochlorite is commonly added to convert calcium chloride to calcium hypochlorite&#xA;  as follows: &lt;span class=&quot;math-container&quot;&gt;$\ce{CaCl2 + 2NaOCl → Ca(OCl)2 + 2NaCl}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pages and pages follow with detailed process descriptions for the manufacture of various formulations of bleaching powder.  Suffice it to say, an ideal powder would be something like pure &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca(OCl)2 · 2H2O}$&lt;/span&gt;, but this is very difficult to prepare economically, because of contamination with various amounts of chloride salts, hydroxide salts, or (depending on the route of manufacture) sodium salts.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="402" LastEditDate="2019-08-21T19:04:48.233" LastActivityDate="2019-08-21T19:04:48.233" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33805" PostTypeId="1" AcceptedAnswerId="33806" CreationDate="2015-07-10T05:11:30.603" Score="3" ViewCount="1634" Body="&lt;p&gt;What the question says. For example, if I knew the specific heat capacities of lithium and oxygen, could I work out the specific heat capacity of lithium oxide with no further information? &lt;/p&gt;&#xA;" OwnerUserId="15033" LastEditorUserId="186" LastEditDate="2015-07-10T14:36:19.977" LastActivityDate="2015-07-10T14:36:19.977" Title="Is it possible to calculate specific heat capacity of a compound based on its elements' heat capacities?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33806" PostTypeId="2" ParentId="33805" CreationDate="2015-07-10T05:22:32.820" Score="8" Body="&lt;p&gt;There is not much connection between the heat-capacity of a compound and the heat-capacity of the elements it is made of. Heat capacity is primarily coming from the low frequency vibrations of the lattice (if it is solid), and vibration+rotation+translation if it is liquid or gas. These have no connection with the properties of the lattice etc properties of the given elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So short answer: no&lt;/p&gt;&#xA;" OwnerUserId="5172" LastActivityDate="2015-07-10T05:22:32.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33809" PostTypeId="1" AcceptedAnswerId="33811" CreationDate="2015-07-10T06:41:24.827" Score="2" ViewCount="1441" Body="&lt;p&gt;In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach:I considered hydroxy group as the stronger +M group.So my guess was para position w.r.t hydroxy group.But that's not the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is actually given as ortho w.r.t hydroxy and para w.r.t -CHO group!!&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="186" LastEditDate="2015-10-02T16:22:00.363" LastActivityDate="2015-10-02T16:22:00.363" Title="In meta hydroxy benzaldehyde which position (w.r.t hydroxy group) is more prone to electrophilic attack?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33810" PostTypeId="1" CreationDate="2015-07-10T07:20:46.293" Score="1" ViewCount="2243" Body="&lt;p&gt;How to find the central atom? What is the central atom in $\ce{PCl3}$&#xA;and  how to find the shape using VSEPR theory?&lt;/p&gt;&#xA;" OwnerUserId="17390" LastEditorUserId="4945" LastEditDate="2015-07-10T08:08:44.790" LastActivityDate="2016-04-05T20:41:10.727" Title="How to know which atom is the central atom in VSEPR theory?" Tags="&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33811" PostTypeId="2" ParentId="33809" CreationDate="2015-07-10T07:30:09.473" Score="0" Body="&lt;p&gt;This is known as the &lt;strong&gt;ortho effect.&lt;/strong&gt; According to it if an &lt;strong&gt;ortho/para&lt;/strong&gt; and &lt;strong&gt;meta&lt;/strong&gt; directing group are &lt;strong&gt;meta&lt;/strong&gt; to each other, then the incoming electrophile prefers to come on &lt;strong&gt;ortho&lt;/strong&gt; of &lt;strong&gt;meta&lt;/strong&gt; directing group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the &lt;strong&gt;reference&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8BMi3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15161" LastActivityDate="2015-07-10T07:30:09.473" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33812" PostTypeId="1" AcceptedAnswerId="33813" CreationDate="2015-07-10T07:56:19.240" Score="1" ViewCount="203" Body="&lt;p&gt;My textbook says that for a $d$-electron, the orbital angular momentum is: $$ \frac{\sqrt{6}h}{2\pi} $$&#xA;But since:&#xA;$$ mvr = \frac{nh}{2\pi}$$&#xA;$$ n \in N$$&#xA;where $N$ is the set of natural numbers, how is the value of $n$ an irrational number?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="15410" LastEditDate="2016-04-14T05:39:37.253" LastActivityDate="2016-04-14T05:39:37.253" Title="Why is the principal quantum number irrational?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33813" PostTypeId="2" ParentId="33812" CreationDate="2015-07-10T08:21:06.620" Score="14" Body="&lt;p&gt;So, first, as I said in my comment, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Both can take only integer values (positive and non-negative respectively), but for totally unrelated reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the orbital angular momentum itself is related to its quantum number $l$ by the following equation,&#xA;$$&#xA;    \hat{L}^2 \Psi = l (l + 1) \hbar^2 \Psi \, ,&#xA;$$&#xA;where $\hbar^2{l(l+1)}$ is the square of the magnitude of an electron orbital angular momentum. Thus, while the orbital angular momentum quantum number $l$ is always a non-negative integer, the magnitude of an electron orbital angular momentum $\sqrt{l(l+1)} \hbar$ can take irrational values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, for a d-electron, $l=2$, and thus, the magnitude of its orbital angular momentum is $\sqrt{2(2+1)} \hbar = \sqrt{6} \hbar = \sqrt{6} h / 2 \pi$.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-11-12T18:40:40.650" LastActivityDate="2015-11-12T18:40:40.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33815" PostTypeId="1" AcceptedAnswerId="33926" CreationDate="2015-07-10T10:02:46.687" Score="6" ViewCount="19059" Body="&lt;p&gt;We know that the atomic radii increases down the group. So, less energy is required to pull out the outermost electron as we go down the group, hence,  gradually melting point decreases down the group.  The topmost element has large melting point because of its small atomic radii and so more energy is required to pull out its outermost electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it is observed that the melting point slightly increases in case of the bottom-most element of group as compared to the previous element. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Group 13 - indium - $\pu{156.6^\circ C}$ but  thallium - $\pu{304^\circ  C}$&lt;/li&gt;&#xA;&lt;li&gt;Group 14 - tin - $\pu{231.92^\circ  C}$ but lead - $\pu{327.46^\circ C}$&lt;/li&gt;&#xA;&lt;li&gt;Group 17 - iodine - $\pu{113.5^\circ  C}$ but astatine - $\pu{302^\circ C}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="9961" LastEditDate="2018-09-08T19:03:58.830" LastActivityDate="2018-09-08T19:03:58.830" Title="Trend in the melting point down the group" Tags="&lt;inorganic-chemistry&gt;&lt;periodic-trends&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33816" PostTypeId="1" AcceptedAnswerId="33838" CreationDate="2015-07-10T11:09:03.670" Score="2" ViewCount="7252" Body="&lt;p&gt;Negative of the $\Delta G$ for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way:  $\Delta H$ is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently $-\Delta H$ energy is obtained from the system after doing expansion work. Since $\Delta S$ is the entropy created in the process, at the very least $-\Delta S$ entropy must be created in the surroundings - that is, at least $-T\Delta S$ energy must be lost as heat. This comes from the $-\Delta H$, and thus leaves $-\Delta H + T\Delta S$ to do useful work. No more work can be done than this. $-\Delta H + T\Delta S$ is $-\Delta G$, so $-\Delta G$ is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why $-\Delta G$ is the maximum possible useful work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if $|\Delta H|$ is more than $|T\Delta S|$, with both being negative, then one can think of the $W_{max}$ or $-\Delta G$ as a part of the $|\Delta H|$; since $|\Delta G|&amp;lt; |\Delta H|$. Some part of $|\Delta H|$ goes as $|T\Delta S|$ to increase the surroundings' entropy, and the other part in doing useful work. But if both are positive, with $|T\Delta S|$ being more than $|\Delta H|$, again $\Delta G$ is negative, allowing useful work to be extracted. But now it seems as though $|T\Delta S|$ heat will be extracted from the surroundings, $|\Delta H|$ used up in the process, while the rest can be converted to work - in other words, useful work is a part of $|T\Delta S|$ - with the other part used by the process as $|\Delta H|$. Is this correct? Is $|T\Delta S|$ extracted from the surroundings actually?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-07-10T16:50:23.813" Title="Maximum Useful Work from a Process" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33817" PostTypeId="2" ParentId="33810" CreationDate="2015-07-10T11:13:31.507" Score="3" Body="&lt;p&gt;You know that phosphorus has 5 valence electrons and that chlorine has 7 valence electrons ... and that you somehow have to combine them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As chlorine is more electronegative (&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity#Allred-Rochow_Electronegativity&quot; rel=&quot;nofollow&quot;&gt;Allred-Rochow&lt;/a&gt;: Cl 2.83, P 2.06) you know that each chlorine will attract one electron from the phosphorus to get the noble gas configuration with 8 electrons in the valence shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that in turn means that there will be one single bond between the phosphorus and each chlorine, which automatically makes phosphorus the center.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="186" LastEditDate="2015-07-10T11:31:57.667" LastActivityDate="2015-07-10T11:31:57.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33818" PostTypeId="1" AcceptedAnswerId="33824" CreationDate="2015-07-10T11:19:56.977" Score="13" ViewCount="21649" Body="&lt;p&gt;I noticed the fact that all the hydrides of the elements belonging to group IV has bond angle 109.5° while in group V it varies from 107.3° for $\ce{NH3}$ to 91.3° for $\ce{SbH3}$. Similarly we see in group VI that water has highest bond angle then it significantly decreases to 89.5° for $\ce{H2Te}$. Is there any specific reason for hydrides of group IV metals to have remarkably equal bond angles, and other groups to have varying angles?&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="10185" LastEditDate="2015-07-10T20:25:00.130" LastActivityDate="2016-02-05T02:39:28.210" Title="Bond angles for the hydrides" Tags="&lt;bond&gt;&lt;periodic-trends&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33819" PostTypeId="1" AcceptedAnswerId="33821" CreationDate="2015-07-10T11:57:23.320" Score="21" ViewCount="2623" Body="&lt;p&gt;What chemicals are responsible for the stench emanating from socks? If it's just sweat, why don't underwear or vests stink as much? Which gas am I inhaling when smelling socks to decide whether they need a wash?&lt;/p&gt;&#xA;" OwnerUserId="17391" LastEditorUserId="7448" LastEditDate="2015-08-03T14:30:20.367" LastActivityDate="2015-08-03T14:30:20.367" Title="Why do socks stink?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33820" PostTypeId="1" AcceptedAnswerId="33844" CreationDate="2015-07-10T11:57:26.507" Score="2" ViewCount="1185" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The ionization energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{He+}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$&lt;/span&gt;. What is the energy of the first stationary state (&lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt;) of &lt;span class=&quot;math-container&quot;&gt;$\ce{Li^{2+}}$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since the question specifically states that &lt;span class=&quot;math-container&quot;&gt;$n=1$&lt;/span&gt;, I used the formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;E &amp;amp;= -2.178 \times 10^{-18} \cdot \frac{Z^2}{n^2}.\\&#xA;E &amp;amp;= -2.178 \times 10^{-18} \cdot \frac{3^2}{1}\\&#xA;E &amp;amp;= -1.96 \times 10^{-17}~\mathrm{J}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the correct answer is supposed to be &lt;span class=&quot;math-container&quot;&gt;$-4.14 \times 10^{-17}~\mathrm{J}$&lt;/span&gt; and many people are telling me to use &lt;span class=&quot;math-container&quot;&gt;$n=\frac{9}{4}$&lt;/span&gt; which gives the right answer but I don't know why I am supposed to set &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\frac{9}{4}$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="4945" LastEditDate="2019-07-23T15:19:02.583" LastActivityDate="2019-07-23T15:28:35.713" Title="Why does the first lithium stationary state have n = 9/4?" Tags="&lt;physical-chemistry&gt;&lt;ions&gt;&lt;energy&gt;&lt;atoms&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33821" PostTypeId="2" ParentId="33819" CreationDate="2015-07-10T12:06:46.943" Score="16" Body="&lt;p&gt;According to the Wikipedia page, aptly named &lt;a href=&quot;https://en.wikipedia.org/wiki/Smelly_socks&quot;&gt;Smelly Socks&lt;/a&gt;, it states that the smell&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is a mixture of ammonia, fatty acids (in particular, &lt;a href=&quot;https://en.wikipedia.org/wiki/3-Methylbutanoic_acid&quot;&gt;isovaleric acid&lt;/a&gt;), and lactic acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What makes sock odor unique is that socks often pick up small amounts of organic material when walking around without shoes on. When wearing shoes, they are in a warm semi-enclosed environment with a considerable amount of sweat infusing into them (it can be quite &lt;a href=&quot;http://www.aaqtic.org.ar/congresos/china2009/download/2-6/2-218.pdf&quot;&gt;warm and humid in shoes&lt;/a&gt;) particularly when exercising. A lot of the smell is due to bacterial action on the bits and pieces trapped in the socks themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the Wikipedia article states that it is a topic of ongoing study, in part due to a possibility that the chemicals could be used in the battle against Malaria - &lt;a href=&quot;http://www.zdnet.com/article/using-smelly-socks-to-attract-mosquitoes-and-combat-malaria/&quot;&gt;Using smelly socks to attract mosquitoes -- and combat malaria&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="11334" LastEditDate="2015-07-10T17:41:18.303" LastActivityDate="2015-07-10T17:41:18.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33822" PostTypeId="1" AcceptedAnswerId="33826" CreationDate="2015-07-10T12:10:35.937" Score="2" ViewCount="80" Body="&lt;p&gt;Sometimes I feel the water in the pool is heavily chlorinated. However it is impossible for me to swim without ingesting some water. I am only a learner but even advanced swimmers do concede that they do ingest some water at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be potentially harmful depending on the chemical content in the pool? Are there studies which advise on the extent of permissible chemicals that can be added to a pool?&lt;/p&gt;&#xA;" OwnerUserId="17391" LastEditorUserId="865" LastEditDate="2015-10-30T16:57:27.690" LastActivityDate="2015-10-30T16:57:27.690" Title="Is ingesting water during swimming harmful?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33823" PostTypeId="1" AcceptedAnswerId="33829" CreationDate="2015-07-10T12:44:24.700" Score="76" ViewCount="3385" Body="&lt;p&gt;I have a collection of old books (all 80+ years old), recently, I received a British Chemistry text from 1903 (intro page below) - this being the oldest book of my collection (112 years old at the time of writing this question): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEFTN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I notice is that there is a distinct smell that comes from the pages of older books. (the smell is not unpleasant) - it seems that the older the book, the more distinct the smell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the chemistry behind the old pages in a book smell?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-09-02T21:19:33.320" LastActivityDate="2016-02-26T10:25:42.393" Title="What causes the old book smell?" Tags="&lt;everyday-chemistry&gt;&lt;smell&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="33824" PostTypeId="2" ParentId="33818" CreationDate="2015-07-10T13:10:54.503" Score="12" Body="&lt;p&gt;The simplest way to look at this trend is through &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory&quot;&gt;VSEPR theory&lt;/a&gt;, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to VSEPR, electron domains (i.e. bonds and lone pairs) are arranged around the central atom in geometries that minimize electrostatic repulsion between the electron domains. These geometries also happen to be solutions to the Schrödinger equation for the set of molecular orbitals as linear combinations of atomic orbitals, and in the molecular orbital case, they arise not from electrostatic repulsion, but strictly from the mathematics of the linear combinations producing a minimum energy self-consistent set of orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VSEPR is qualitatively simpler for small molecules. All of the binary hydrides in Group IV, V, and VI have steric number four around the central atom. This means that the number of lone pairs and the number of bonding domains add up to four. These species can also be described using the $\ce{AXE}$ method, where $\ce{A}$ denotes the central atom, $\ce{X}$ denotes an atom bonded to $\ce{A}$ and $\ce{E}$ denotes a lone pair of electrons.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Group IV hydrides all have four bonds and no lone pairs. We describe this arrangement as $\ce{AX4}$.&lt;/li&gt;&#xA;&lt;li&gt;The Group V hydrides all have three bonds and one lone pairs. We describe this arrangement as $\ce{AX3E}$.&lt;/li&gt;&#xA;&lt;li&gt;The Group VI hydrides all have two bonds and two lone pairs. We describe this arrangement as $\ce{AX2E2}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are several factors that affect the &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory#Degree_of_repulsion&quot;&gt;degree of repulsion&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lone pairs - Lone pair domains are more diffuse and have higher repulsion than bonding domains.&lt;/li&gt;&#xA;&lt;li&gt;The nature of $\ce{X}$ - Larger and/or more electropositive atoms have higher repulsion.&lt;/li&gt;&#xA;&lt;li&gt;Bond order - Bonds with higher bond orders have higher repulsion than bonds with lower bond orders.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{X=H}$ in all cases, we only need to worry about the first factor: lone pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arrangements $\ce{AX4}$, $\ce{AX3E}$, and $\ce{AX2E2}$ all have four electron domains, and thus the base electron domain geometry group is tetrahedral. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences in bond angle between $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2O}$ originate from the increasing number of lone pairs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{CH4}$ has no lone pairs: the bond angles are the ideal tetrahedral bond angles of $109.5^\circ$.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{NH3}$ has one lone pair: the bond angles are contracted a little because of the grater repulsion of the lone pair to $107.8^\circ$.&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H2O}$ has two lone pairs: the bond angle is contracted further due to the repulsion of two lone pairs to $104.5^\circ$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All of the Group IV hydrides will have perfect tetrahedral geometry due to having four bonds to the same atom and no lone pairs. Thus $\ce{CH4}$, $\ce{SiH4}$, $\ce{GeH4}$, $\ce{SnH4}$, and $\ce{PbH4}$ all have bond angles of $109.5^\circ$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For Group V, the bond angles are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c}&#xA;\ce{AH3}&amp;amp;\text{angle}\\ \hline&#xA;\ce{NH3}&amp;amp;107.8^\circ \\ &#xA;\ce{PH3}&amp;amp;93.5^\circ \\ &#xA;\ce{AsH3}&amp;amp;91.8^\circ \\ &#xA;\ce{SbH3}&amp;amp;91.7^\circ \\ &#xA;\ce{BiH3}&amp;amp;90.5^\circ \\ \end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that there is a large decrease from $\ce{NH3}$ to $\ce{PH3}$ and then very small changes for $\ce{AsH3}$, $\ce{SbH3}$, and $\ce{BiH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make the argument that the larger orbitals of the heavier elements will have high repulsion than the smaller orbitals of nitrogen. This is where VSEPR breaks down. VSEPR works well for some compounds because there is significant $s-p$ mixing when constructing the molecular orbitals. This mixing need not happen for the heavier elements. One &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphine#Structure_and_properties&quot;&gt;molecular orbital solution&lt;/a&gt; for $\ce{PH3}$, for example, indicates very little contribution from the $\ce{P}\ 3s$ orbital. We can describe the $\ce{P-H}$ bonds as predominantly $\ce{P}_{3p}\ce{-H}_{1s}$ bonds. Thus the geometry around $\ce{P}$ is determined mostly by the orientation of the $p$ orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the Group VI hydrides, we see the same trend: a large decrease from $\ce{H2O}$ to $\ce{H2S}$ and then very little change. The same argument can be made for primarily $\ce{A}_{p}\ce{-H}_{1s}$ bonding in the heavier Group VI hydrides as for the heavier Group V hydrides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c|c}&#xA;\ce{AH2}&amp;amp;\text{angle}\\ \hline&#xA;\ce{H2O}&amp;amp;104.5^\circ \\ &#xA;\ce{H2S}&amp;amp;92.1^\circ \\ &#xA;\ce{H2Se}&amp;amp;91.0^\circ \\ &#xA;\ce{H2Te}&amp;amp;90.0^\circ \\ &#xA; \end{array}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-10T13:10:54.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33825" PostTypeId="1" AcceptedAnswerId="33828" CreationDate="2015-07-10T13:11:11.147" Score="1" ViewCount="92" Body="&lt;p&gt;In my previous question, someones answer among other things, said that boiling the water increases the concentration of nitrates and nitrites. I searched on google and it's true, but how does that happen ? What is the reaction that leads to an increasing level of nitrates? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im very courious because nitrates are too dangerous to be high in concentration, i would appreciate your help. &lt;/p&gt;&#xA;" OwnerUserId="14968" LastActivityDate="2015-07-10T13:15:37.200" Title="Nitrates in water" Tags="&lt;water&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33826" PostTypeId="2" ParentId="33822" CreationDate="2015-07-10T13:12:10.293" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could this be potentially harmful depending on the chemical content in the pool?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not, unless you managed to drink quite a bit of pool water. Worldwide, millions (hundreds of millions?) of people have collectively ingested millions (billions?) of liters of pool water - with varying chemistries - over many decades (myself included), and nobody seems to have suffered too much. I believe this falls into the &quot;generally regarded as safe&quot; area.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there studies which advise on the extent of permissible chemicals that can be added to a pool?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Numerous ones, Google away. &lt;a href=&quot;http://www.in.gov/isdh/files/Chemical_adjustment_pool.pdf&quot; rel=&quot;nofollow&quot;&gt;These guidelines&lt;/a&gt;, from the State of Indiana, are representative of the recommended levels of various chemicals present in common swimming pools.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-10T13:12:10.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33828" PostTypeId="2" ParentId="33825" CreationDate="2015-07-10T13:15:37.200" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;boiling the water increases the concentration of nitrates and nitrites.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;how does that happen? What is the reaction that leads to an increasing level of nitrates?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It happens because you boil away the water, and the nitrates and nitrites stay in what's left of the water, thus raising the concentration of those species in the now reduced amount of water. There is no &quot;reaction,&quot; per se, just you are concentrating dissolved species by removing the water.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-10T13:15:37.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33829" PostTypeId="2" ParentId="33823" CreationDate="2015-07-10T13:16:32.923" Score="52" Body="&lt;p&gt;This is a very interesting question. Provided that the materials used in making papers aren't the same around the globe, this is a very broad case of study. However, a study has been conducted in which the main goal was to identify the compounds that are the cause of the smell; &lt;a href=&quot;https://en.wikipedia.org/wiki/Volatile_organic_compound&quot;&gt;VOCs&lt;/a&gt;:&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, quoting the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Using supervised and unsuper-vised methods of multivariate data analysis, we were able to quantitatively correlate volatile degradation products with properties important for the preservation of historic paper: rosin, lignin and carbonyl group content, degree of polymerization of cellulose, and paper acidity. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is a result of the several hundred identified volatile and semi-volatile organic compounds (VOCs) off-gassing from paper and the object in general. The particular blend of compounds is a result of a network of degradation pathways and is dependent on the original composition of the object including paper substrate, applied media, and binding. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oknKB.png&quot; alt=&quot;&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JWOYS.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The 15 most abundant VOCs present in all chromatograms were selected for further analyses: &lt;strong&gt;acetic acid, benzaldehyde, 2,3-butanedione, butanol, decanal, 2,3-dihydrofuran, 2-ethylhexanol, furfural, hexadecane, hexanal, methoxyphenyloxime, nonanal, octanal, pentanal,&lt;/strong&gt; and &lt;strong&gt;undecane&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac9016049&quot;&gt;Strlič, M., Thomas, J., Trafela, T., Cséfalvayová, L., Kralj Cigić, I., Kolar, J., Cassar, M. (2009). Material degradomics: on the smell of old books. Analytical Chemistry &lt;strong&gt;2009,&lt;/strong&gt; &lt;em&gt;81&lt;/em&gt; (20), 8617-8622.&lt;/a&gt;&#xA;The main link to the article might be behind a paywall, so for the interested, the relevant &lt;a href=&quot;http://www.researchgate.net/publication/26818853_Material_Degradomics_On_the_Smell_of_Old_Books&quot;&gt;ResearchGate article is accessible&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="4945" LastEditDate="2016-02-26T10:25:42.393" LastActivityDate="2016-02-26T10:25:42.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33830" PostTypeId="1" CreationDate="2015-07-10T13:51:18.553" Score="0" ViewCount="90" Body="&lt;p&gt;I am attempting to begin synthesizing glucose, however I will require a form of ball milled cellulose as a main ingredient. Having been said, what methods would be recommended to create the Cellulose? If we were to use grass clippings or cotton as our source for example, what steps would we take in order to gain the desired material from the aforementioned source?&lt;/p&gt;&#xA;" OwnerUserId="16558" LastActivityDate="2015-07-10T20:21:13.777" Title="How Does One Create A Cellulose Ingredient For Use In The Creation Of Glucose?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33831" PostTypeId="1" CreationDate="2015-07-10T13:54:29.517" Score="6" ViewCount="2572" Body="&lt;p&gt;I know an isotherm is a line of constant temperature.&#xA;Isobar is a constant pressure line.&#xA;Isentrope is a line of constant entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what do you call a line of constant volume? An Isochore?&#xA;And what do you call a line of constant enthalpy? An &quot;isenthalp&quot;?&lt;/p&gt;&#xA;" OwnerDisplayName="user752" LastActivityDate="2015-07-10T14:04:10.407" Title="What is the proper term for lines of constant volume and constant enthalpy on a thermodynamic diagram?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;&lt;temperature&gt;&lt;entropy&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33832" PostTypeId="2" ParentId="33831" CreationDate="2015-07-10T14:04:10.407" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But what do you call a line of constant volume? An Isochore?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And what do you call a line of constant enthalpy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process is isenthalpic. More than one engineering/thermodynamic text I found by Googling indicates they are simply referred to as &quot;isenthalpic lines.&quot; See &lt;a href=&quot;https://books.google.com/books?id=Sx7SBQAAQBAJ&amp;amp;pg=SA7-PA20&amp;amp;lpg=SA7-PA20&amp;amp;dq=isenthalpic+lines&amp;amp;source=bl&amp;amp;ots=xSF_BQVcwI&amp;amp;sig=IPZ_A05qJTbneadLLbj2IcQAhQM&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=NtGfVbyABNCQyATWgK7gDg&amp;amp;ved=0CDYQ6AEwBQ#v=onepage&amp;amp;q=isenthalpic%20lines&amp;amp;f=false&quot;&gt;this&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-10T14:04:10.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33837" PostTypeId="1" CreationDate="2015-07-10T16:35:04.480" Score="5" ViewCount="1648" Body="&lt;p&gt;I've read somewhere on the Internet that heating graphite at 4200°K at a sufficiently high pressure would turn it into diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chart seems to support this idea: &lt;img src=&quot;https://i.stack.imgur.com/kOO8B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that doable industrially?&lt;/p&gt;&#xA;" OwnerUserId="17397" LastEditorUserId="17397" LastEditDate="2015-07-10T16:49:31.797" LastActivityDate="2015-07-10T16:52:45.580" Title="Is turning graphite carbon into diamond industrially doable?" Tags="&lt;everyday-chemistry&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33838" PostTypeId="2" ParentId="33816" CreationDate="2015-07-10T16:41:36.493" Score="2" Body="&lt;p&gt;&quot;Useful work&quot; does not include expansion work. For a proof of why $w_{\text{by,add}} \leq \Delta G$, you may want to first refer to my answer to this question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/32873/regarding-gibbss-function-why-is-there-non-expansion-work/32888#32888&quot;&gt;Why does the Gibbs free energy only correspond to non-expansion work?&lt;/a&gt; then come back here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that proof, I used equalities by assuming that the process is reversible. If we drop this assumption, then by the Second Law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\mathrm{d}S &amp;amp;\geq \frac{\mathrm{d}q}{T} \\&#xA;\mathrm{d}q &amp;amp;\leq T\mathrm{d}S \\&#xA;dU &amp;amp;\leq T\mathrm{d}S - p\mathrm{d}V + \mathrm{d}w_{\text{add}} \\&#xA;dG &amp;amp;\leq dw_{\text{add}} \\&#xA;\Delta G &amp;amp;\leq w_{\text{add}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equality holds if the process is reversible, or more generally, if both the initial and final states are equilibrium states. Now this seems like an obvious contradiction of the equation in the very first paragraph. The key lies in the definition of the work - whether it is done &lt;em&gt;on&lt;/em&gt; the system, or &lt;em&gt;by&lt;/em&gt; the system. In my proof, I have been referring to the work done &lt;em&gt;on&lt;/em&gt; the system; however, here we are interested in the work done &lt;em&gt;by&lt;/em&gt; the system, since that is the work that can be &quot;extracted&quot;. The work done &lt;em&gt;by&lt;/em&gt; the system must be equal to the negative of the work done &lt;em&gt;on&lt;/em&gt; the system: $w_{\text{by,add}} = -w_{\text{add}}$. This leads to the final result $w_{\text{by,add}} \leq \Delta G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Levine's &lt;em&gt;Physical Chemistry&lt;/em&gt; 6th ed. writes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but $w_{\text{by,add}}$ is the useful work output. The quantity $-\Delta G$ equals the maximum possible nonexpansion work output $w_{\text{by,add}}$ done by a system in a constant-$T$-and-$p$ process. Hence the term &quot;free energy&quot;. (Of course, for a system with P-V work only, $\mathrm{d}w_{\text{by,add}} = 0$ and $\mathrm{d}G = 0$ for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want a direct answer as to why your explanation isn't right, it's because you can't equate $\Delta H$ with the work done. Work done is given by $w = -\int\! p\,\mathrm{d}V$. You can only equate $\Delta H$ with the heat transferred at constant $p$. Apart from the First Law ($\Delta U = q + w$), heat and work are generally unrelated.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-07-10T16:50:23.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33839" PostTypeId="2" ParentId="33837" CreationDate="2015-07-10T16:49:04.577" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_diamond&quot;&gt;It has been done since 1954&lt;/a&gt; and has been a commercial success for some time. Many products now have &lt;a href=&quot;http://www.ibccoatings.com/dlc-coating-diamond-like-carbon-coating-ceratough-d&quot;&gt;diamond-like coatings&lt;/a&gt;, produced comparatively inexpensively through vapor deposition without need for high pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, in English &quot;burning&quot; means to set on fire, oxidize; I believe you mean &quot;heating graphite&quot;, and, as your graph shows, &lt;strong&gt;pressure&lt;/strong&gt; is needed as well as heat.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-10T16:49:04.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33840" PostTypeId="2" ParentId="33837" CreationDate="2015-07-10T16:52:45.580" Score="4" Body="&lt;p&gt;You want to &lt;strong&gt;heat&lt;/strong&gt; graphite at high pressure, not burn it - but yes, high-T, high-p will work. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is that doable industrially?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a rather lengthy and informative Wikipedia article for &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_diamond&quot; rel=&quot;nofollow&quot;&gt;synthetic diamond&lt;/a&gt; that illustrates (in part) the use of this process - HPHT, for &quot;high pressure, high temperature&quot;(using a high-purity source of carbon, not necessarily graphite) in industry. Note that there are several industrial processes mentioned in addition to HPHT.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-10T16:52:45.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33841" PostTypeId="2" ParentId="33830" CreationDate="2015-07-10T16:58:29.197" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.thefreedictionary.com/cotton+wool&quot; rel=&quot;nofollow&quot;&gt;Cotton wool&lt;/a&gt;, also called &quot;purified cotton&quot;, is inexpensive and available at pharmacies (&quot;chemists&quot; in the UK). I could not find the percentage of impurities (probably lignin and minerals), but it is considered &lt;a href=&quot;https://books.google.com/books?id=nuxsAAAAMAAJ&amp;amp;pg=PA772&amp;amp;lpg=PA772&amp;amp;dq=%22cotton%20wool%22%20cellulose%20lignin%20impurities&amp;amp;source=bl&amp;amp;ots=kCLojVzDRL&amp;amp;sig=JiGO7pZN7bjNdtV3jvWYS2SYPj0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=A_mfVYy0FIymgwTyy4XwBg&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=%22cotton%20wool%22%20cellulose%20lignin%20impurities&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;one of the purest forms of cellulose commercially available&lt;/a&gt;, and should not need pretreatment to make glucose.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="10185" LastEditDate="2015-07-10T20:21:13.777" LastActivityDate="2015-07-10T20:21:13.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33842" PostTypeId="1" AcceptedAnswerId="33843" CreationDate="2015-07-10T17:38:51.263" Score="3" ViewCount="10455" Body="&lt;p&gt;Determine the equivalence ratio of the following mixture: 4g of Butane, $\ce{C4H10}$, and 75 grams of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is 0.82&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok so I try converting everything into mols&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4~\mathrm{g}~(\ce{C4H10}) \times \frac{1~\mathrm{mol}}{58~\mathrm{g}} = \frac{4}{58}~\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$75~\mathrm{g~(Air)} \times \frac{1~\mathrm{mol}}{28.966~\mathrm{g}} = \frac{75}{28.966}~\mathrm{mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air-fuel ratio = $\frac{\frac{75}{28.966}}{\frac{4}{58}} = 37.54$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a guess that the air-fuel ratio is some sort of equivalence ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw that equivalence ratio =  actual air-to-fuel ratio divided by the theoretical. I don't know how to get either actual or theoretical? Is the data in the problem sufficient enough to get both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also see a formula in the reviewer: Fuel + Air = Products of combustion and the appropriate reaction, but it only works for finding the exact amount of mols like: $\ce{C4H10 + aO2 + 3.76aN2 = bCO2 + CH2O + 3.76aN2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a mechanical engineering board exam review; unfortunately, university only tapped briefly on the chemistry side of fuels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hint on what I should be doing? Point me to the right direction?&lt;/p&gt;&#xA;" OwnerUserId="17306" LastEditorUserId="10185" LastEditDate="2015-07-10T20:14:06.027" LastActivityDate="2015-07-10T20:14:06.027" Title="Determining the Equivalence Ratio" Tags="&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33843" PostTypeId="2" ParentId="33842" CreationDate="2015-07-10T17:53:23.853" Score="1" Body="&lt;p&gt;$$\begin{align}&#xA;\text{AFR (Air to Fuel ratio)} &amp;amp;= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\&#xA;\text{Equivalence ratio} &amp;amp;= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the problem statement, if you did everything right,&#xA;$$\begin{align}&#xA;\text{AFR} &amp;amp;= 37.54 \\&#xA;\text{AFR}_{\text{stoich}} &amp;amp;= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\&#xA;\ce{C4H10 + $6.5$\,O2 &amp;amp;-&amp;gt; 4 CO2 + 5 H2O} \\&#xA;\text{AFR}_{\text{stoich}} &amp;amp;= 4.76 \times 6.5 = 30.94 \\&#xA;\text{Equivalence ratio} &amp;amp;= \frac {37.54}{30.94}=1.21&#xA;\end{align}$$&#xA;Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio: &#xA;$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="7777" LastEditDate="2015-07-10T18:27:32.533" LastActivityDate="2015-07-10T18:27:32.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33844" PostTypeId="2" ParentId="33820" CreationDate="2015-07-10T18:03:14.750" Score="2" Body="&lt;p&gt;I think your calculation is correct. There is an error in this textbook.  All you have to do is to switch  between &lt;span class=&quot;math-container&quot;&gt;$\ce{Li^{2+}}$&lt;/span&gt; and  &lt;span class=&quot;math-container&quot;&gt;$\ce{He^{+}}$&lt;/span&gt;. And 'll prove my point:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Li^{2+}}$&lt;/span&gt; and  &lt;span class=&quot;math-container&quot;&gt;$\ce{He^{+}}$&lt;/span&gt; are hydrogen-like atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these species, the ionization energy is given by the equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{i}=E_\mathrm{i}(\ce{H})\times Z^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$\ce{Li^{2+}}$&lt;/span&gt;: &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;E_\mathrm{i}(\ce{Li^{2+}})&amp;amp;=E_\mathrm{i}(\ce{H})\times Z^2\\&#xA;19.6 \times 10^{-18}&amp;amp;=E_\mathrm{i}(\ce{H})\times 3^2\\&#xA;E_\mathrm{i}(\ce{H})&amp;amp;=2.18 \times 10^{-18}  \,\mathrm {J/atom}\\&#xA;E_\mathrm{i}(\ce{H})&amp;amp;=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\&#xA;  &amp;amp;= 13.6 \,\mathrm{eV}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$13.6 \,\mathrm{eV}$&lt;/span&gt; is well-known to be the first ionization energy of hydrogen. So, my assumption is true.&#xA;To solve the problem after correction:&#xA; The ionization energy of &lt;span class=&quot;math-container&quot;&gt;$\ce{He^{+}}$&lt;/span&gt; is given by the equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;E_\mathrm{i}(\ce{He^{+}})&amp;amp;=\frac{E_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\&#xA;E_\mathrm{i}(\ce{He^{+}})&amp;amp;=8.72\times 10^{-18}\,\mathrm {J/atom}&#xA;\end{align}&lt;/span&gt;&#xA;The energy of the first stationary state:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E_\mathrm{i,1}=E_\infty-E_1$$&lt;/span&gt;&#xA;But, &lt;span class=&quot;math-container&quot;&gt;$E_\infty=0$&lt;/span&gt;, we find:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2019-07-23T15:28:35.713" LastActivityDate="2019-07-23T15:28:35.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33845" PostTypeId="1" CreationDate="2015-07-10T18:14:30.410" Score="10" ViewCount="216" Body="&lt;p&gt;Part of my research is investigating sphere packings and their properties.I really would like a large database of sphere packings that I could investigate using a program I have written that tells me certain things about the packings. After doing some research I found some online databases but a lot of them haven't been taken care of so they're riddled with dead links.  I was wondering if anyone knows of any books or websites that have a lot of sphere packings that I could investigate.  Or if there is a way to interpret crystal structures as a sphere packing, seeing how there is a lot of crystal structure databases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone is confused as to what kind of data I'm looking I just need coordinates of the centers of spheres and their corresponding radii.  These need to form a packing where the spheres are touching like in this paper where the authors have found new bi-disperse packings.  Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp206115p&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jp206115p&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17399" LastEditorUserId="10185" LastEditDate="2015-07-11T17:44:08.247" LastActivityDate="2017-05-08T05:19:34.067" Title="Finding large databases of sphere packings" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;&lt;reference-request&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33846" PostTypeId="1" AcceptedAnswerId="33858" CreationDate="2015-07-10T18:36:19.247" Score="5" ViewCount="164" Body="&lt;p&gt;Notation of the reduction and oxidation of copper typically includes the aqueous state. Illustrated below:&#xA;$$\ce{Cu(s) -&amp;gt; Cu^2+(aq) + 2e-}$$&#xA;This event usually occurs at 0.34 V. What if the copper was in an organic environment instead? I ask because I am doing cyclic voltammetry of a capacitor with copper as the current collector and need to measure the &lt;a href=&quot;http://web.yonsei.ac.kr/echemlab/public_html/data/sucap.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-faradaic capacitance&lt;/a&gt;. Our current setup is using $\ce{KOH}$ as the electrolyte and we are basically just seeing a non-reversible/quasi-reversible CV like below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sj8yE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="9961" LastEditDate="2016-04-29T15:24:56.237" LastActivityDate="2016-04-29T15:24:56.237" Title="Will oxidation/reduction of copper still occur in an electrochemical cell if it is not in an aqueous environment?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrolysis&gt;&lt;cyclic-voltammetry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33847" PostTypeId="5" CreationDate="2015-07-10T18:55:34.870" Score="0" Body="&lt;p&gt;Cyclic voltammetry or CV is a type of potentiodynamic electrochemical measurement. In a cyclic voltammetry experiment the working electrode potential is ramped linearly versus time. Unlike in linear sweep voltammetry, after the set potential is reached in a CV experiment, the working electrode's potential is ramped in the opposite direction to return to the initial potential. These cycles of ramps in potential may be repeated as many times as desired. The current at the working electrode is plotted versus the applied voltage (i.e., the working electrode's potential) to give the cyclic voltammogram trace. Cyclic voltammetry is generally used to study the electrochemical properties of an analyte in solution.&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-07-10T20:53:34.433" LastActivityDate="2015-07-10T20:53:34.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33848" PostTypeId="4" CreationDate="2015-07-10T18:55:34.870" Score="0" Body="Cyclic voltammetry or CV is a type of potentiodynamic electrochemical measurement. In a cyclic voltammetry experiment the working electrode potential is ramped linearly versus time." OwnerUserId="4356" LastEditorUserId="4356" LastEditDate="2015-07-10T22:59:32.540" LastActivityDate="2015-07-10T22:59:32.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33849" PostTypeId="5" CreationDate="2015-07-10T18:58:05.730" Score="0" Body="&lt;p&gt;Cyclic voltammetry or CV is a type of potentiodynamic electrochemical measurement. In a cyclic voltammetry experiment, the working electrode potential is ramped linearly versus time. Unlike in linear sweep voltammetry, after the set potential is reached in a CV experiment, the working electrode's potential is ramped in the opposite direction to return to the initial potential. These cycles of ramps in potential may be repeated as many times as desired. The current at the working electrode is plotted versus the applied voltage (i.e., the working electrode's potential) to give the cyclic voltammogram trace. Cyclic voltammetry is generally used to study the electrochemical properties of an analyte in solution&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="402" LastEditDate="2017-05-02T21:30:32.987" LastActivityDate="2017-05-02T21:30:32.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33850" PostTypeId="4" CreationDate="2015-07-10T18:58:05.730" Score="0" Body="Cyclic voltammetry or CV is a type of potentiodynamic electrochemical measurement. In a cyclic voltammetry experiment, the working electrode potential is ramped linearly versus time. " OwnerUserId="4356" LastEditorUserId="402" LastEditDate="2017-05-02T21:30:08.053" LastActivityDate="2017-05-02T21:30:08.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33852" PostTypeId="2" ParentId="33842" CreationDate="2015-07-10T19:07:51.437" Score="1" Body="&lt;p&gt;I followed &lt;a href=&quot;http://fluid.wme.pwr.wroc.pl/~spalanie/dydaktyka/combustion_MiBM/fund/Stoichiometry.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; procedure. Note that &quot;oxygen&quot; here refers to ${\rm O}_{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the &quot;stoichiometric ratio&quot; defined as the number of moles of oxygen needed to combust 1 mole of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing 0.126/0.153 = 0.82&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-10T19:07:51.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33854" PostTypeId="1" CreationDate="2015-07-10T19:36:29.807" Score="6" ViewCount="26940" Body="&lt;p&gt;I am asked to find the color of the precipitate formed by the passage of acetylene gas into an aqueous solution of ammoniacal silver nitrate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought ammoniacal silver nitrate was Tollen's reagent, but acetylene is neither a ketone nor an aldehyde so I don't know how to figure out the color of the precipitate.&lt;/p&gt;&#xA;" OwnerUserId="17319" LastEditorUserId="5026" LastEditDate="2018-07-03T08:53:16.023" LastActivityDate="2018-07-03T08:53:16.023" Title="Color of precipitate formed on reaction of acetylene with Tollen's Reagent" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;&lt;organic-reduction&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33855" PostTypeId="2" ParentId="16028" CreationDate="2015-07-10T19:51:59.847" Score="0" Body="&lt;p&gt;Add denatured alcohol or another soluble liquid. Even if its not completely dissolved the slurry becomes greatly insensitive. TATP gets its power when allowed to dry out; even the slightest moisture (water or solvent) will greatly hinder its power until allowed to dry out. Or wait 2-3 months for sublimation to occur. Nitroglycerin may be deflagrated in an open area.&lt;/p&gt;&#xA;" OwnerUserId="17403" LastEditorUserId="10185" LastEditDate="2015-07-10T20:03:56.700" LastActivityDate="2015-07-10T20:03:56.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33857" PostTypeId="2" ParentId="33854" CreationDate="2015-07-10T20:39:19.043" Score="5" Body="&lt;p&gt;You are correct that ammoniacal silver nitrate is Tollen's reagent. Tollen's reagent is most notable for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, notably alpha-hydroxy ketones and terminal alkynes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With terminal alkynes, the alkyne reacts as a acid to form a silver acetylide which precipitates. In the case of acetylene, both hydrogens can react: $$\ce{HC#CH + 2AgNO3 + 2NH4OH -&amp;gt; AgC#CAg + 2NH4NO3 + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The colour of the precipitate is variously described as &lt;a href=&quot;https://en.wikipedia.org/wiki/Tollens%27_reagent&quot; rel=&quot;noreferrer&quot;&gt;yellow&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Silver_acetylide&quot; rel=&quot;noreferrer&quot;&gt;gray-white&lt;/a&gt;, or &lt;a href=&quot;http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf&quot; rel=&quot;noreferrer&quot;&gt;white&lt;/a&gt;. The actual colour you observe is probably strongly dependent on the purity of the sample and the size of the precipitated particles.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-07-10T20:39:19.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33858" PostTypeId="2" ParentId="33846" CreationDate="2015-07-10T20:52:40.593" Score="5" Body="&lt;p&gt;It is certainly possible for metals, including copper, to corrode in non-aqueous environments.  The details depend on the precise nature of the solvent, electrolytes, and other potentially complexing species present.  For example, in polar protic solvents (like water) that don't have any complexing ligands for copper, the equilibrium potential for copper oxidation will likely be the same as in water.  But oxidation potentials will likely shift in solvents that are non-polar, in ways that depend strongly on the composition of the electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are using potassium hydroxide as the electrolyte, your conditions might be more aqueous than you think.  You can think of &quot;hydroxide&quot; as a type of water that likes to exist at high pHs.  And hydroxide itself generates high pHs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elecrochemical water oxidation occurs via &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja205647m&quot;&gt;this reaction&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2 H2O -&amp;gt; O2 + 4 H+ + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH-dependent equilibrium potential is $1.23~\rm{V} - 0.059\frac{\rm V}{pH}\times pH $, which at a pH of 14 comes out to 0.4 volts.  So you might also possibly be irreversibly oxidizing hydroxide to oxygen during your electrolysis.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-10T20:52:40.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33860" PostTypeId="1" AcceptedAnswerId="33866" CreationDate="2015-07-11T03:58:55.053" Score="-2" ViewCount="822" Body="&lt;p&gt;As a continuation to the question: &lt;br&gt;&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/33763/predicting-reaction-among-compounds?noredirect=1#comment51084_33763&quot;&gt;Predicting reaction among compounds&lt;/a&gt; &lt;br&gt;&lt;br&gt;&#xA;I found the list of functional group on wikipaedia:&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikibooks.org/wiki/Organic_Chemistry/Overview_of_Functional_Groups&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikibooks.org/wiki/Organic_Chemistry/Overview_of_Functional_Groups&lt;/a&gt; &lt;br&gt;&lt;br&gt;&#xA;Now in the previously mentioned question, I got a clue that if the functional groups are compatible then reaction can occur, can you please tell me which are compatible with which?&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-07-11T12:56:56.097" Title="Functional Group compatibility in organic chemistry" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-07-11T13:19:45.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="33861" PostTypeId="1" AcceptedAnswerId="33873" CreationDate="2015-07-11T07:54:53.743" Score="7" ViewCount="7862" Body="&lt;p&gt;Why does the color of the halogens deepens as we go down the group? also why there is a change in physical state of halogens down the group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c|c}&#xA;\mathbf{Gas} &amp;amp; \mathbf{Color} \\\hline&#xA;\text{fluorine} &amp;amp; \text{very pale yellow (almost invisible to naked eye)}\\&#xA;\text{chlorine} &amp;amp; \text{yellow-green} \\&#xA;\text{bromine} &amp;amp; \text{red-brown}\\&#xA;\text{iodine} &amp;amp; \text{violet-black}\end{array}&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="5026" LastEditDate="2018-07-16T01:56:38.797" LastActivityDate="2018-07-16T01:56:38.797" Title="Why does color of halogen deepens and the boiling point decreases down the group?" Tags="&lt;inorganic-chemistry&gt;&lt;halides&gt;&lt;phase&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33862" PostTypeId="1" AcceptedAnswerId="33863" CreationDate="2015-07-11T08:04:24.490" Score="0" ViewCount="356" Body="&lt;p&gt;What is the chemistry behind  the bad smell of human perspiration? also what is its composition?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="544" LastEditDate="2018-11-27T23:19:04.060" LastActivityDate="2018-11-27T23:19:04.060" Title="What is the chemistry behind smell of human perspiration(sweat)?" Tags="&lt;everyday-chemistry&gt;&lt;identification&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2018-11-28T04:58:15.737" ContentLicense="CC BY-SA 3.0" />
  <row Id="33863" PostTypeId="2" ParentId="33862" CreationDate="2015-07-11T09:06:50.717" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does sweat smell?!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sweat released from sweat glands &lt;strong&gt;odourless.&lt;/strong&gt; Bad smell of sweat is due to certain bacteria on your skin mix with sweat(That is why sweat smell bad after sometime) and also due to certain hormones released during puberty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried my best to keep this part chemistry point of view(Actually this belongs to biology).  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemical composition of sweat&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main reason people perspire is so evaporation of water can cool the body. Thus &lt;strong&gt;water&lt;/strong&gt; is main component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical composition of perspiration varies between individuals and also depends on what they have been eating and drinking, why they are sweating, how long they have been perspiring and several other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On average, the mineral composition is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sodium (0.9 gram/liter)&lt;/li&gt;&#xA;&lt;li&gt;potassium (0.2 g/l)&lt;/li&gt;&#xA;&lt;li&gt;calcium (0.015 g/l)&lt;/li&gt;&#xA;&lt;li&gt;magnesium (0.0013 g/l)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Trace metals that the body excretes in sweat include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;zinc (0.4 milligrams/liter)&lt;/li&gt;&#xA;&lt;li&gt;copper (0.3–0.8 mg/l)&lt;/li&gt;&#xA;&lt;li&gt;iron (1 mg/l)&lt;/li&gt;&#xA;&lt;li&gt;chromium (0.1 mg/l)&lt;/li&gt;&#xA;&lt;li&gt;nickel (0.05 mg/l)&lt;/li&gt;&#xA;&lt;li&gt;lead (0.05 mg/l)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/18156662&quot; rel=&quot;nofollow&quot;&gt;ncbi&lt;/a&gt; and &lt;a href=&quot;http://chemistry.about.com/od/chemicalcomposition/f/What-Is-The-Chemical-Composition-Of-Human-Sweat-Or-Perspiration.htm&quot; rel=&quot;nofollow&quot;&gt;chemistry.about&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-07-11T09:06:50.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33864" PostTypeId="1" AcceptedAnswerId="77068" CreationDate="2015-07-11T11:59:40.757" Score="13" ViewCount="2421" Body="&lt;p&gt;Often (or always?) one enantiomer of a certain drug will be effective and the other ineffective or harmful. The famous example is thalidomide, where one enantiomer caused mutilation of the unborn child and the other relieved morning sickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other examples where the other enantiomer of a drug that is used today or in history causes some adverse side effect, but has obviously been separated from the product to leave the effective drug, in addition what is(are) that side effect(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does absolute configuration at chiral/stereogenic centres affect the effectiveness of a drug molecule or perhaps cause unwanted side effects?&lt;/p&gt;&#xA;" OwnerUserId="6925" LastEditorUserId="41890" LastEditDate="2017-07-25T17:51:40.430" LastActivityDate="2019-11-05T21:53:49.463" Title="Effect of enatiomers in pharmaceuticals" Tags="&lt;stereochemistry&gt;&lt;history-of-chemistry&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;&lt;pharmacology&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ClosedDate="2017-07-02T13:08:18.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="33865" PostTypeId="1" CreationDate="2015-07-11T12:12:45.280" Score="3" ViewCount="53" Body="&lt;p&gt;In my chemistry textbook, it is written that :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gibbs energy for a reaction in which all reactants and products  are in standard state, $\Delta_\mathrm{r}G^\circ$ is related to the equilibrium constant of the reaction as follows:&#xA;  $$\Delta_\mathrm{r}G^\circ=-RT\ln K$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A solved example was given there for the above equation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate $\Delta_\mathrm{r}G^\circ$ for conversion of of oxygen to ozone, $$\ce{3/2O2(g)-&amp;gt;O3(g)} $$ at $298\ \mathrm{K}$ if $K_p$ for this conversion is $2.47\times 10^{-29}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer to above problem is simple. Just substitute the values in the gibbs equation and get the value of $\Delta_\mathrm{r}G^\circ$ but the thing that I can not understand is that in the equation they substituted the value of $K_p$ directly in the equation without converting into $K_c$ for equilibrium constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is:&#xA; In the gibbs equation what does equilibrium constant stand for? ($K_c$ or $K_p$).&lt;/p&gt;&#xA;" OwnerUserId="14914" LastEditorDisplayName="user7951" LastEditDate="2015-07-11T12:40:55.600" LastActivityDate="2015-07-11T12:40:55.600" Title="Using equilibrium constant in gibbs equation" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-07-11T19:19:11.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="33866" PostTypeId="2" ParentId="33860" CreationDate="2015-07-11T12:56:56.097" Score="5" Body="&lt;p&gt;As I mentioned in my answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4798/synthesizable-functional-groups/4809#4809&quot;&gt;this question&lt;/a&gt;, you are asking for something about which my two-semester organic chemistry course sequence just scratches the surface. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;After covering the basics of organic structure and reactivity, most introductory organic chemistry texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In some cases, we can be reasonably certain that two functional groups will always react (primary amines and acyl halides), but in many, many other cases additional reagents, conditions, catalysts, etc. are needed. Different reagents and conditions can cause different outcomes from the same combination of reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a short sample of what I mean from among the earliest reactions we tend to study in the first semester organic chemistry course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Alcohol + alkyl halide&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases, this combination of compounds will not amount to much, but there are some special cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tertiary halides in primary alcohols undergo solvolysis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)3CBr + CH3CH2OH -&amp;gt; (CH3)3COCH2CH3 + HBr}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Addition of strong base promotes elimination in any alkyl halide, especially with heat.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2CH2Cl + KOCH2CH3 -&amp;gt;[\ce{CH3CH2OH}][\Delta] CH3CH=CH2 + KCl + HOCH2CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The major regioisomeric product can be controlled by which base is present.&#xA;$$\ce{CH3CH(Cl)CH2CH3 +NaOCH2CH3 -&amp;gt;[\ce{CH3CH2OH}][\Delta] CH3CH=CHCH3 + NaCl + HOCH2CH3}$$&#xA;$$\ce{CH3CH(Cl)CH2CH3 +KOC(CH3)3-&amp;gt;[\ce{(CH3)3COH}][\Delta] CH2=CHCH2CH3 + KCl + (CH3)3COH}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we take our primary alkyl halide and strong base, but the reaction is in DMSO (dimethylsulfoxide) instead of ethanol, and the reaction is kept cold, subsitution prevails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2CH2Br + NaOCH2CH3 -&amp;gt;[\ce{DMSO}] CH3CH2CH2OCH2CH3 + NaBr}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction space in organic chemistry is &lt;strong&gt;very large&lt;/strong&gt; and, while functional group analysis is always a good place to start learning, the effects of reagents and conditions, sterics and conformation, and other factors are also important. Rather than ask a question here that has an answer that amounts to whole textbooks worth of information, maybe you should sign up for a free MOOC on organic chemistry. That will give you a sense of what types of things are and are not possible (as per M.A.Ramezani's comment), and then you can ask more focused questions here.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2015-07-11T12:56:56.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33869" PostTypeId="1" CreationDate="2015-07-11T15:39:04.533" Score="2" ViewCount="1620" Body="&lt;p&gt;My textbook says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It is appropriate to remark that the trick of rationalizing apparent violations of the octet rule by invoking the participation of d-orbitals in the bonding scheme is a matter of some controversy. Some scientists feel that the d-orbitals lie so high in energy that they should not be treated as valence orbitals, but as excited orbitals which cannot confer any appreciable stability to chemical bonds. Others feel that while d-orbitals are of high energy in free atoms, their energy decreases as other atoms approach to make bonds.&lt;/p&gt;&#xA;&lt;p&gt;This author has not been convinced of the absolute validity of either side of this argument, and interprets the controversy merely as more evidence that the description of chemical bonds in terms of a few atomic orbitals is a highly approximate procedure. The most convincing reason for invoking d-orbitals in chemical bonding is that it is a simple idea which works, if it is used carefully and not over-interpreted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since this book is quite an old edition, I wanted to know whether there are any newer developments in this direction. Which of the two views is generally acceptable?&lt;/p&gt;&#xA;" OwnerUserId="17412" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-11T20:07:22.567" Title="Can d-orbitals participate in bonding?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33870" PostTypeId="1" AcceptedAnswerId="33877" CreationDate="2015-07-11T18:11:46.577" Score="7" ViewCount="35151" Body="&lt;p&gt;After reading &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33805/is-it-possible-to-calculate-specific-heat-capacity-of-a-compound-based-on-its-el&quot;&gt;this&lt;/a&gt; recent question I was interested in how to calculate the specific heat capacity of a mixture based on the specific heat capacities of its components.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.see.ed.ac.uk/%7Ejwp/procalcs/procalcs/mixtures/ideal/slides/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; website the specific heat capacity of an ideal mixture is given by&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C_{p(\mathrm{mixture})} = \sum\limits_iC_{p(i)}x_i$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$x_i$&lt;/span&gt; is the mole fraction of each component.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where does this come from and what are the assumptions involved in treating it as an ideal mixture?&lt;/li&gt;&#xA;&lt;li&gt;What deviations from this are observed in non-ideal mixtures? Can you easily predict the deviations based on the mixture and/or predict the actual specific heat capacity?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-03-05T03:25:08.500" Title="Heat capacity of mixtures" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;heat&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33872" PostTypeId="2" ParentId="33869" CreationDate="2015-07-11T20:07:22.567" Score="4" Body="&lt;p&gt;Yes it works, according to result they have with organometallic chemistry. You may not know $e_g$ and $t_{2g}$ orbitals. Have a look at this diagram of orbitals : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its diagram for an octaedric complex, where M is the metal. Positions of $e_g$ and $t_{2g}$ depends of what L is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZblCc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15235" LastActivityDate="2015-07-11T20:07:22.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33873" PostTypeId="2" ParentId="33861" CreationDate="2015-07-11T20:24:03.953" Score="5" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As for the BP:&#xA;The attraction of covalent molecules is due 'dispersion' interaction. It is a quite complex effect, commonly described as interaction of 'instantaneous dipoles', i.e. interaction of dipoles formed by nucleus and electrons around it. Its efficiency generally increases with number of electrons (due to sheer numbers) and size of electron shell. Since outer electron shells vastly grow in size in fluorine-iodine row, the efficiency of dispersion interactions grow. In fact, close halogen-halogen contacts for heavier halogens (bromine and iodine) are very common in crystal structures of organic compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As for color. As a guess, I'd say that the active transition is lone pair orbital $\rightarrow$ $\sigma$-antibonding orbital (see any introduction to MO theory for the terminology). The energy of this transition should lower from fluorine to iodine because of great decrease in bonding energy and increase of size of atoms. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_field_theory#/media/File:RBG_color_wheel.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;color wheel&lt;/a&gt;, the absorption in violet spectrum (with highest energy of photons in visible spectrum) should give yellowish apparent color (case of fluorine and chlorine), red apparent color to adsorption in lower-energy and violet color for adsorption at even lower-energy red spectrum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="485" LastEditorUserId="9961" LastEditDate="2018-07-15T22:22:38.213" LastActivityDate="2018-07-15T22:22:38.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33874" PostTypeId="1" AcceptedAnswerId="33875" CreationDate="2015-07-11T21:08:19.867" Score="0" ViewCount="8817" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose that $\pu{28 g}$ of silver is initially at $\pu{29.0^\circ C}$. What is its temperature upon absorbing $\pu{2.25 kJ}$ of heat?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My work: $2250 = 0.240 \times 28 \times t - 29$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $t$ and I get $\pu{340 ^\circ C}$. But the correct answer is $\pu{370 ^\circ C}$ and I have no clue how to get there. &lt;/p&gt;&#xA;" OwnerUserId="17322" LastEditorUserId="5026" LastEditDate="2018-03-09T05:52:18.777" LastActivityDate="2018-03-09T05:54:25.347" Title="What is the final temperature of each substance upon absorbing 2.25 kJ of heat?" Tags="&lt;physical-chemistry&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33875" PostTypeId="2" ParentId="33874" CreationDate="2015-07-11T22:42:16.067" Score="5" Body="&lt;p&gt;You are almost correct, however, to calculate these values you need to put a bracket around he $t-29$ term (which is the $\Delta T$ term from $Q = mc \Delta T$). Hence, from your working out step:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2250 = 0.240 \times 28 \times (t - 29)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need to expand the brackets.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-07-11T22:42:16.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33876" PostTypeId="1" CreationDate="2015-07-11T23:19:03.057" Score="2" ViewCount="6769" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How can glycine act as a buffer at pH 6.00 and why? Please use&#xA;chemical equations to show how it can act as a buffer at this pH&#xA;specifically?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&lt;p&gt;At pH 6.00 glycine acts as a buffer, because at pH 6.00 the condition is acidic and glycine is an amino acid, so its zwitterion forms an equilibrium, which means there is a positive charge on the N of the amino group and a negative charge on the carboxylate. So there are no sudden pH changes and we know buffer solutions resist sudden pH changes, so at pH 6.00 glycine acts as a buffer solution.&lt;/p&gt;&#xA;&lt;p&gt;Equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/44mRy.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What I need explained:&lt;/p&gt;&#xA;&lt;p&gt;HOW does glycine act as a buffer at pH 6.00 with chemical equations, meaning to show an example of how it specifically buffers another species at pH 6.00 with the proper chemical equations.&lt;/p&gt;&#xA;" OwnerUserId="17235" LastEditorUserId="69830" LastEditorDisplayName="user15489" LastEditDate="2021-05-11T04:49:45.373" LastActivityDate="2021-05-11T04:49:45.373" Title="How can glycine act as a buffer at pH 6.00 and why?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33877" PostTypeId="2" ParentId="33870" CreationDate="2015-07-12T00:12:32.163" Score="5" Body="&lt;p&gt;Suppose you have $n_1$ mole of A and $n_2$ mole of B. If they are at temp. $T_1$ and you want to raise the temp. to $T_2$ then the amount of heat you need to add is&lt;br&gt;&#xA;$$\Delta H~= n_1 \times c_{pA} \times \Delta T + n_2 \times c_{pB} \times \Delta T$$&lt;br&gt;&#xA;$$\Delta H~= (n_1 \times c_{pA} + n_2 \times c_{pB}) \times \Delta T$$&lt;br&gt;&#xA;$$\Delta H~= (n_1 +n_2)\times \frac{(n_1 \times c_{pA} + n_2 \times c_{pB})}{(n_1+n_2)} \times \Delta T$$&lt;br&gt;&#xA;$$\Delta H~= n_{total} \times(y_1 \times c_{pA} + y_2 \times c_{pB}) \times \Delta T$$&lt;br&gt;&#xA;$$\Delta H~= n_{total} \times c_{p,avg} \times \Delta T$$&lt;br&gt;&#xA;So basically you are not losing any information in case of ideal situation where molecular interaction of two different species has no effect on each other's heat capacity. But from statistical mechanics we know,&#xA;$$C_v=\frac { &amp;lt;E^2&amp;gt;-&amp;lt;E&amp;gt;^2}{k_B \times T^2}$$&lt;br&gt;&#xA;Now if we solve the schrodinger equation for individual species, we must include an extra potential energy term to account the interaction from other species, which in turn will result in fluctuation of $C_v$ from it's previous value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly off topic: When I see the question first things that came to my mind is &lt;a href=&quot;https://en.wikipedia.org/wiki/Kopp%27s_law&quot; rel=&quot;nofollow&quot;&gt;kopp's rule&lt;/a&gt;. It's used to find heat capacity of molecule using atom's heat capacity or to find alloy's heat capacity using the constituents heat capacity.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="10185" LastEditDate="2015-07-12T15:52:14.200" LastActivityDate="2015-07-12T15:52:14.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33878" PostTypeId="2" ParentId="33876" CreationDate="2015-07-12T02:18:22.997" Score="2" Body="&lt;p&gt;Your equation is not correct for pH 6; it should be for pH 1. For pH 6, the equation should be:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GBnGE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an acid is added, then the equation becomes:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zpntC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if a base is added, then the equation is:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vPAek.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-07-12T02:18:22.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33879" PostTypeId="1" AcceptedAnswerId="34264" CreationDate="2015-07-12T05:55:02.233" Score="34" ViewCount="20828" Body="&lt;p&gt;Recently I have read about both of the concepts in my book (Physical Chemistry by Atkins, Paula). It was literally a reading; though I could understand the language and superposition of orbitals, inversion symmetry, and so on, I never could really comprehend how these two theories differ (or agree) with each other. I have googled it but couldn't gain a lucid explanation between the differences in their approaches. So, now my question is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these two theories differ from each other? What are the differences between their approaches? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Also, in VB theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why it is told that &quot;MO theory provides a global, delocalized perspective on chemical bonding&quot;?&lt;/strike&gt; &lt;em&gt;This portion of the question was split to a separate question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33951/why-is-the-resonance-concept-not-required-in-molecular-orbital-theory&quot;&gt;Why is the resonance concept not required in molecular orbital theory?&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: After viewing some of the answers, I think IMHO, I had to clear what I was really asking. By saying &quot;difference between MOT &amp;amp; VBT&quot;, I am not intending to mean that just write the different predictions the two theories can separately succeed in making. What I really mean is the difference between their approaches &amp;amp; their way of explanation. Of course, the case is not like &quot;Hey, look, one theory is outdated; just follow the other one...&quot;; they are both important and hence do want to know the difference between their pathways of analyzing the cases.&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="-1" LastEditorDisplayName="user5764" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-09-14T15:13:27.790" Title="What is actually the difference between valence bond theory and molecular orbital theory?" Tags="&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="33880" PostTypeId="1" CreationDate="2015-07-12T07:08:50.360" Score="0" ViewCount="1493" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the product of 1,1-diphenylethanol + Montmorillonite K10 Clay undergoing microwave radiation? What is the mechanism for it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My effort&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,1-diphenylethanol + Montmorillonite K10 Clay catalyst used in a microwave-assisted synthesis of pyrazoles from enaminones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Support for catalysts&lt;/li&gt;&#xA;&lt;li&gt;Preparation of clay nanopaper composites of nacre-like structure&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Catalyst for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Conversion of triose sugars with alcohols to alkyl lactates&lt;/li&gt;&#xA;&lt;li&gt;Deprotection of ketone dimethylhydrazones&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is my answer correct?&lt;/p&gt;&#xA;" OwnerUserId="17416" LastEditorUserId="418" LastEditDate="2016-12-01T10:12:19.830" LastActivityDate="2016-12-02T13:32:31.173" Title="Reaction of 1,1-diphenylethanol on Montmorillonite K10 under microwave radiation" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;acid-base&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33881" PostTypeId="1" CreationDate="2015-07-12T07:22:31.917" Score="0" ViewCount="6733" Body="&lt;p&gt;An element have many oxidation numbers, then which one should we consider in the given question? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. In the compound $\ce{HCN}$ as $\ce{C}$ {carbon} has +4,+2,-4 oxidation numbers and $\ce{N}$ {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers&lt;/p&gt;&#xA;" OwnerUserId="17417" LastEditorDisplayName="user15489" LastEditDate="2015-07-12T07:28:18.440" LastActivityDate="2015-07-12T08:37:40.900" Title="Oxidation numbers" Tags="&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-13T01:30:05.497" ContentLicense="CC BY-SA 3.0" />
  <row Id="33882" PostTypeId="2" ParentId="33881" CreationDate="2015-07-12T08:37:40.900" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;e.g. In the compound HCN as C {carbon} has +4,+2,-4 oxidation numbers and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is not correct. Each atom only has one oxidation number in a compound, although one element can have different oxidative states (i.e. sodium thiosulfate). Your statement should be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C {carbon} has +4,+2,-4 oxidation numbers, and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers. In the compound HCN, carbon has +2, hydrogen has +1, and nitrogen has -3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-07-12T08:37:40.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33884" PostTypeId="1" AcceptedAnswerId="34195" CreationDate="2015-07-12T10:55:44.663" Score="15" ViewCount="933" Body="&lt;p&gt;I have been messing around with making Indigo dye while waiting for bacteria to grow. I have been following the &lt;a href=&quot;https://en.wikipedia.org/wiki/Baeyer%E2%80%93Drewson_indigo_synthesis&quot;&gt;Baeyer-Drewson&lt;/a&gt; method. Basically, you mix nitrobenzaldehyde with an acetone water mix and slowly add NaOH. The solution turns black and you get a dark sludge. I have been separating the sludge from the liquid by centrifugation. I have noticed that the supernatent has a dark reddish brown color.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I was wondering if anyone knew what the red byproduct was? &lt;/li&gt;&#xA;&lt;li&gt;My indigo is black when dried, &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4e/Indigo_cake.jpg&quot;&gt;wikipedia&lt;/a&gt; shows nice bright blue chunks. What could be causing this color difference? Maybe the red byproduct? &lt;/li&gt;&#xA;&lt;li&gt;Does indigo dissolve in anything without needing to be reduced first? I have had no luck with the organic solvents available in our lab.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="137" LastEditorUserId="10185" LastEditDate="2015-07-12T16:20:53.143" LastActivityDate="2015-07-21T09:12:20.457" Title="Byproducts in Indigo Synthesis?" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;solubility&gt;&lt;synthesis&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33885" PostTypeId="1" CreationDate="2015-07-12T11:29:53.517" Score="3" ViewCount="988" Body="&lt;p&gt;Could someone explain to me why some alkenes containing at least one couple of different substituents at the two double-bonded C atoms (that is, some of those with at least two different substituents for both C atoms) are considered achiral? For instance, for 2-chloropropene or 1-bromo-2-chloro-2-fluoro-1-iodoethene, where (and what) in the structural formula of each of them is the plane of symmetry that makes it achiral or, in other words, what do their superposable mirror images look like? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="17424" LastEditorDisplayName="user15489" LastEditDate="2015-07-12T20:32:55.247" LastActivityDate="2015-07-13T18:13:33.160" Title="Why is 2-chloropropene achiral?" Tags="&lt;chirality&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="33887" PostTypeId="1" CreationDate="2015-07-12T12:56:27.563" Score="3" ViewCount="66" Body="&lt;p&gt;In titanium anodising, solid titanium reacts with the oxygen formed at it's surface (to form titanium(IV) oxide or &lt;span class=&quot;math-container&quot;&gt;$\ce{TiO2}$&lt;/span&gt;) through the electrolysis of water. Experimentation and research has shown that the thickness of the coating is dependent on voltage and not current. The potential difference obviously plays some part in the reaction then (besides the water electrolysis) due to the fact that merely placing titanium in an environment with more oxygen gas does not affect the thickness of the naturally occurring oxide, but I am struggling to find out how (through both thought and research) the voltage affects the depth of the coating formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the actual oxidation reaction can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{Ti &amp;amp;-&amp;gt; Ti^{4+} + 4e-}\\&#xA;\ce{O2 +4e- &amp;amp;-&amp;gt; 2O^{2-}}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which are added together to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Ti + O2 -&amp;gt; TiO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also understand that redox reactions usually include a reduction potential or &lt;span class=&quot;math-container&quot;&gt;$E^0$&lt;/span&gt; value. Now this may or may not be relevant but it is the only related value I can find with a voltage associated. I have also thought that as the coating of &lt;span class=&quot;math-container&quot;&gt;$\ce{TiO2}$&lt;/span&gt; builds up, the resistance of the circuit would increase. I just cannot understand &lt;em&gt;how&lt;/em&gt; or &lt;em&gt;why&lt;/em&gt; the voltage affects the depth (Is this the right word, as it is converting not depositing?) of the formed oxide layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that my question is more like a request for any insight/resources that could help with my understanding of the voltage-dependency of the depth of the formed oxide layer (specifically titanium, but I'm assuming it would be relatively easy to generalise from other metals such as aluminium or niobium (which I have also searched for)). &lt;/p&gt;&#xA;" OwnerUserId="17221" LastEditorUserId="4945" LastEditDate="2019-04-25T13:45:05.093" LastActivityDate="2019-04-25T13:45:05.093" Title="What is the voltage dependence of the thickness of oxide coating through anodising?" Tags="&lt;electrochemistry&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33889" PostTypeId="2" ParentId="33879" CreationDate="2015-07-12T13:11:12.040" Score="4" Body="&lt;p&gt;Here, I am referring Valence Bond Theory by VBT and Molecular Orbital Theory by MOT.&lt;/p&gt;&#xA;&lt;p&gt;1.In VBT, atomic orbitals of the combining atoms retain a large amount of their individual character. (For example, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orbital Hybridization&lt;/a&gt;).&#xA;Whereas, in MOT atomic orbitals of the combining atoms lose their individual identity in the resulting &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory#Types_of_orbitals&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molecular Orbital&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;2.Secondly, VBT fails to explain properties like para-magnetism of some compounds like O2, whereas MOT easily &lt;a href=&quot;http://www.meritnation.com/ask-answer/question/explain-the-para-magnetism-in-o2-using-mot-molecular-orbital/chemistry/6758310&quot; rel=&quot;nofollow noreferrer&quot;&gt;explains&lt;/a&gt; this.&lt;/p&gt;&#xA;&lt;p&gt;3.Also, VBT fails to explain possibility and existence of ions like &lt;a href=&quot;http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;H2&lt;span class=&quot;math-container&quot;&gt;$^+$&lt;/span&gt;&lt;/a&gt;, whereas MOT is able to do this successful.&lt;/p&gt;&#xA;&lt;p&gt;&lt;code&gt;Why it is told that &amp;quot;MO theory provides a global, delocalized perspective on chemical bonding&amp;quot;?&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;Quoting from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory#Overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; of MOT,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In MO theory, any electron in a molecule may be found anywhere in the molecule, since quantum conditions allow electrons to travel under the influence of an arbitrarily large number of nuclei, as long as they are in eigenstates permitted by certain quantum rules. Thus, when excited with the requisite amount of energy through high-frequency light or other means, electrons can transition to higher-energy molecular orbitals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;code&gt;Why isn't such concept needed in MO theory? &lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33951/why-is-the-resonance-concept-not-required-in-molecular-orbital-theory&quot;&gt;this&lt;/a&gt; question, which directly addresses this. However, I would recommend you to read &lt;a href=&quot;http://socratic.org/questions/how-does-molecular-orbital-theory-deal-with-resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;How does molecular orbital theory deal with resonance?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thank You!&lt;/p&gt;&#xA;" OwnerUserId="16924" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-21T16:40:24.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33892" PostTypeId="1" AcceptedAnswerId="33893" CreationDate="2015-07-12T14:42:38.627" Score="7" ViewCount="6066" Body="&lt;p&gt;I have been using deodorant to propel potatoes (potato gun :D), but I want to switch over to pure butane. However, in order to do this safely I have to know how much butane is typically in deodorant (percentage). If you do know, please also include a source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, for a potato gun, is it wiser to use butane or propane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;And could a spark ignite butane? Or is a real flame needed?&lt;/strike&gt; Tested this already, it does.&lt;/p&gt;&#xA;" OwnerUserId="17428" LastEditorUserId="4231" LastEditDate="2015-07-12T18:16:03.793" LastActivityDate="2017-07-23T11:56:48.957" Title="Best Fuel for Potato Gun" Tags="&lt;experimental-chemistry&gt;&lt;home-experiment&gt;&lt;safety&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33893" PostTypeId="2" ParentId="33892" CreationDate="2015-07-12T16:57:42.717" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How much butane is typically in deodorant (percentage)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pg.com/productsafety/msds/beauty_care/deodorants/Secret_Antiperspirant_Deodorant_Aerosol_Canada_(99767800).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a link&lt;/a&gt; to an MSDS for a deodorant; it claims the butane range is 30-60% (they don't state if this is by volume or weight). It might be cheaper to just go to a hardware store and buy a small butane cylinder, like the ones used for outdoor cooking, then you know exactly how much butane you are using.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a potato gun, is it wiser to use butane or propane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My first though was butane because it gives off a bit more heat per mole (or per unit volume) when it burns, than does propane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C3H8 + 5 O2 -&amp;gt; 3 CO2 + 4 H2O + \pu{2219 kJ/mol}}$$&#xA;$$\ce{C4H10 + $\mathrm{6.5}$~ O2 -&amp;gt; 4 CO2 + 5 H2O + \pu{2878 kJ/mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to &quot;&lt;a href=&quot;http://www.inpharmix.com/jps/Combustion_fuels.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Common Fuels for Combustion Spud Guns&lt;/a&gt;&quot;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The key value for comparing two fuels based on their heats of combustion is not the actual heats of combustion. Instead, the &quot;Heat per mole Oxygen&quot; should be used since the amount of energy in the combustion chamber is limited by the amount of oxygen present in the chamber. Fuel is added to match that amount of oxygen. As you can see from the table, there is relatively little difference between the various fuels based on their &quot;Heat per mole Oxygen&quot; values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Both propane and butane give off $\pu{105 kcal/mol}$ of oxygen, so they should perform equivalently in a spud gun. The article goes on to note that, of common fuels, only hydrogen ($\pu{119 kcal/mol}$ of oxygen) and acetylene ($\pu{120 kcal/mol}$ of oxygen) produce significantly more energy under these conditions than other fuels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now going back to your first question, if you know the volume of your chamber, you can back-calculate the correct amount of air (oxygen) and butane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2017-07-23T11:56:48.957" LastActivityDate="2017-07-23T11:56:48.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33894" PostTypeId="1" AcceptedAnswerId="33898" CreationDate="2015-07-12T17:06:41.753" Score="1" ViewCount="99" Body="&lt;p&gt;Electrons can be anywhere in the universe and where they are is random.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they are still discrete particles and thus it takes energy for electrons to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O^+}$ and $\ce{H2O^-}$ to form?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2015-07-12T18:25:37.890" Title="Electrons jumping from 1 water molecule to another" Tags="&lt;water&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33895" PostTypeId="1" AcceptedAnswerId="33897" CreationDate="2015-07-12T17:13:33.503" Score="4" ViewCount="583" Body="&lt;p&gt;A methylene ($\ce{=CH_2}$) group ist sterically less hindered than a methyl ($\ce{CH_3}$) group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if the carbonyl ($\ce{C=O}$) group is sterically more or less larger than a methylene ($\ce{=CH_2}$) group.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bond length: ($\ce{C=O}$) is shorter than ($\ce{C=C}$)&lt;a href=&quot;http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html&quot; rel=&quot;nofollow noreferrer&quot;&gt; [1]&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Diaxial i.a.: Me//H is larger than Me//n(O) (0.9 vs. 0.4 kcal/mol; no ref. for the 2nd value)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While the shorter distance makes the carbonyl more sterically, the lone pairs are less sterically than the Hydrogens of methylene. I think this effect is larger and hence a carbonyl to be more sterically hindered than a methylene group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The A-values gives the energetical difference of a monosubstituted cyclohexane between the conformation in which the substituent is placed axial (1,3-diaxial interactions, hence higher in energy) versus equatorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reformulate above problem of the sizes I have to set following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comparing the axial conformations of vinylcyclohexane and cyclohexancarbaldehyde: Which molecule is more strained?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why I am interested in above question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we don't assume protonation of a ketone's oxygen, I'd like to know if I have to place the carbonyl axial or equatorial when R=Me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For R=long alkyl chain it makes sense to place R equatorial, but I'm not sure if a methyl group is also bigger than a carbonyl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I think methylene and carbonyl are almost of the same size from above arguments, and I'm knowing that methyl is bigger than methylene, it sounds logical to me to place the methyl equatorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVgNH.png&quot; alt=&quot;http://evans.rc.fas.harvard.edu/pdf/evans95.pdf&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://evans.rc.fas.harvard.edu/pdf/evans95.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://evans.rc.fas.harvard.edu/pdf/evans95.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="4231" LastEditDate="2015-07-12T18:13:55.353" LastActivityDate="2015-07-12T19:36:51.657" Title="Is there an A value for the carbaldehyde and vinyl groups?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33896" PostTypeId="1" AcceptedAnswerId="33899" CreationDate="2015-07-12T17:29:24.753" Score="9" ViewCount="8691" Body="&lt;p&gt;These two terms has been frequently used in Physical Chemistry by Atkins. He describes Raoult's Law using &lt;strong&gt;ideal solution&lt;/strong&gt; while Henry's law using &lt;strong&gt;ideal-dilute solution.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought, at first they were same but Atkins in order to contrast the two laws presented graphs in  which one curve was for ideal solution &amp;amp; another was for ideal-dilute solution. Though he wrote two different sections for these solutions, I really couldn't contrast what the difference actually is between &lt;strong&gt;ideal solution&lt;/strong&gt; &amp;amp; &lt;strong&gt;ideal-dilute solution&lt;/strong&gt;. I googled it but to vain! Can anyone please help what he is trying to say by these two terms? &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="9961" LastEditDate="2015-10-23T16:55:27.937" LastActivityDate="2015-10-23T16:55:27.937" Title="Difference between ideal &amp; ideal-dilute solution" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33897" PostTypeId="2" ParentId="33895" CreationDate="2015-07-12T17:59:59.020" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Comparing the axial conformations of vinylcyclohexane and&#xA;  cyclohexancarbaldehyde: Which molecule is more strained?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you point out, &quot;&lt;strong&gt;A-values&lt;/strong&gt;&quot; can be used to get some idea of the relative &quot;size&quot; of a substituent in the cyclohexane system.  More specifically, the A-value reflects both the steric size of the substituent and the length of the bond connecting the substituent to the cyclohexane ring (the longer the bond, the further away the substituent and steric interactions are therefore reduced).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The A-value is just the free energy difference between the axial and equatorial conformers. Since destabilizing steric interactions can occur in the axial conformation, the larger the substituent, less axial and more equatorial conformer will be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jaBrm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map:_McMurray_8ed_%22Organic_Chemistry%22/Unit_04:_Organic_Compounds:_Cycloalkanes_%26_their_Stereochemistry/4.7_Conformations_of_Monosubstituted_Cyclohexanes&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, the &quot;larger&quot; the substituent the larger the A-value. The A-value for a methyl group is 1.8, using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\Delta G = -RT*ln(K)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can solve for K and find that the axial/equatorial ratio for methylcyclohexane is ~ 5/95 (as expected, the less sterically strained equatorial conformer predominates) at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zIWIm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.sparknotes.com/chemistry/organic2/carbocycles/section2/page/2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/confengy.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link&lt;/a&gt; to a nice table of A-values.  The A-value for a vinyl group ($\ce{-HC=CH2}$) is 1.6, while it is only 0.7 for the aldehyde group ($\ce{-CHO}$). Hence &lt;strong&gt;the aldehyde group is the smaller of the two.&lt;/strong&gt; Or said differently, axial vinyl-cyclohexane would be more strained than axial cyclohexane-carbaldehyde&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-07-12T19:36:51.657" LastActivityDate="2015-07-12T19:36:51.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33898" PostTypeId="2" ParentId="33894" CreationDate="2015-07-12T18:25:37.890" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Electrons can be anywhere in the universe &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kind of: There is a non-zero probability that an electron can be at any distance from some point of origin (for instance, the nucleus of an atom), but there is some distance which corresponds to the place where we're most likely to find it (the place where the probability amplitude is highest) most of the time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and where they are is random.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, sort of: They are most likely located in certain regions of space for most of the time, but since it's a probability, it's not a certainty. That said, there is a common sense element that can be applied here. Yes, an electron from a hydrogen atom in a flask in your laboratory can be out near Jupiter, but realistically, it's not. And if it's on Jupiter, it's not really that hydrogen atom's electron anymore.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O}^+$ and $\ce{H2O}^−$ to form?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it is possible (but probable? I don't know, but certainly there are conditions under which such a reaction will occur).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-12T18:25:37.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33899" PostTypeId="2" ParentId="33896" CreationDate="2015-07-12T18:41:49.660" Score="7" Body="&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Changes_In_Vapor_Pressure%2C_Raoult&amp;#39;s_Law&quot; rel=&quot;noreferrer&quot;&gt;Raoult's Law&lt;/a&gt; is for &lt;strong&gt;ideal solutions&lt;/strong&gt; (all of it: solvent and solute):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. &quot;An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components.&quot; (all quoted from above link)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chm.bris.ac.uk/~chdms/Teaching/Chemical_Interactions/page_17.htm&quot; rel=&quot;noreferrer&quot;&gt;As stated here&lt;/a&gt; concerning &lt;strong&gt;ideal-dilute solutions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Many liquid mixtures deviate from the ideal behaviour predicted by Raoult's law. Ideal-dilute solutions are those for which the solvent obeys Raoult's law and the solute Henry's law. The difference in behaviour is because the solvent molecules tend to be surrounded by other solvent molecules so that they behave in an almost ideal way. However, the solute molecules tend also to be surrounded by solvent molecules so that their environment and therefore their thermodynamic behaviour is very unlike that of the pure substance.&quot; (all quoted from above link)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In summary: Ideal solutions are those which obey Raoult's Law; ideal-dilute solutions are ones where the solvent obey's Raoult's Law and the solute obeys Henry's Law.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-12T18:41:49.660" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33900" PostTypeId="2" ParentId="4270" CreationDate="2015-07-12T21:22:44.780" Score="1" Body="&lt;p&gt;Actually this paper explains the water ethanol gasoline mixture effects very well. If the ethanol will harm the equipment is more dependent on how the machine was designed. The ethanol can deteriorate carburetor parts the gasoline ethanol comes into contact with if the manufacturer hasn't replaced them yet.  Rust is not likely from humid air. &lt;a href=&quot;http://epa.gov/OMS/regs/fuels/rfg/waterphs.pdf&quot; rel=&quot;nofollow&quot;&gt;http://epa.gov/OMS/regs/fuels/rfg/waterphs.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17433" LastActivityDate="2015-07-12T21:22:44.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33901" PostTypeId="2" ParentId="29101" CreationDate="2015-07-12T21:40:20.293" Score="-2" Body="&lt;p&gt;During the reaction of $\ce{SO2}$ and $\ce{O2}$, there's a formation of nascent oxygen which oxidises sulphur. In doing this sulphur co ordinates it's lone pair to the nascent oxygen. Since only 2 electrons take part here in this co-ordinate bond formation it is just a single bond. Now the $\ce{SO3}$ thus formed has 4 bonds shared between 3 places. Hence the bond order.&lt;/p&gt;&#xA;" OwnerUserId="17432" LastEditorUserId="4231" LastEditDate="2015-07-12T22:23:11.023" LastActivityDate="2015-07-12T22:23:11.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33903" PostTypeId="1" AcceptedAnswerId="33944" CreationDate="2015-07-12T22:55:00.240" Score="3" ViewCount="54" Body="&lt;p&gt;I am into hobby machining using an old beaten CNC mill in my garage. Being novice my confidence is rather low and I have genuine fear of breaking something (even ancient CNC machines like mine cost a fortune), so when I got to the part of the users manual that listed recommended lubricants for the different parts of the machine I felt out of my league completely (especially since I just basically submerged the whole machine in a bath of general purpose lubricant spray).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to know, how are lubricants usually classified/graded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which classification/grade matters to which kind of use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the risks of using the wrong lubricants?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastActivityDate="2015-07-14T02:04:46.443" Title="What are the fundamental parameter by which lubricants are graded?" Tags="&lt;viscosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33904" PostTypeId="1" AcceptedAnswerId="33924" CreationDate="2015-07-12T23:43:47.293" Score="14" ViewCount="5650" Body="&lt;p&gt;Graphene is cool. Can I make some? What do I need? Is it at all feasible to make it in quantity?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="10185" LastEditDate="2015-07-13T10:27:19.933" LastActivityDate="2015-07-13T13:36:28.460" Title="Can I make my own graphene at home in my kitchen?" Tags="&lt;everyday-chemistry&gt;&lt;home-experiment&gt;&lt;carbon-allotropes&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33905" PostTypeId="2" ParentId="33904" CreationDate="2015-07-13T01:06:24.330" Score="5" Body="&lt;p&gt;I Googled the phrase &quot;How to make graphene&quot; and got &lt;a href=&quot;http://www.graphene-battery.net/graphene.htm&quot; rel=&quot;nofollow&quot;&gt;this result&lt;/a&gt;, which contains a section on how to make it at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is, unequivocally, yes you can. Below are excerpts from the linked page.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) The first DIY method is to use a lead pencil to deposit a thick layer of graphite onto a paper. Then use ordinary sticky tape to peel off a layer of graphite from the paper. Use another piece of sticky tape to remove a layer of graphite from the first sticky tape. Then, use a third piece of unused sticky tape to remove a layer from the second piece of sticky tape, an so on. Eventually, the graphite layers will get thinner and thinner, and you will end up with graphene, which is single-layer graphite in the strict sense, or bi-layer or few-layer graphite (which acts almost like graphene in certain uses). Even though this way of making graphene is only a proof-of-concept, the sticky tape method works. It takes patience and time, but it's the DIY method which the Manchester group used in 2004. And remember, they actually won a Nobel prize for their work, so there's no messing with sticky tape!&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) Shear exfoliation in liquids starts by pouring powdered crystal in a liquid, and then using a shear mixer to separate (exfoliate) layers of material from the crystal. The liquid that is used in such processes is chosen so that the small graphene particles don't clump back together, and the result is a liquid suspension of graphene. The suspension can then be dried to obtain graphene nanoflakes, or it could be directly used in later technological processes, such as making of graphene coatings and others.&#xA;  So, how do you make graphene at home using this method? Well, you can replace the shear mixer with a kitchen blender (since shear mixing is a fancy word for high-tech blending), and instead of special liquids you can use water and dish detergent. The detergent is added so that the particles don't clump, and acts as a surfactant. A great source of graphite powder are graphite pencils. Simply powder a few pencil leads and add the powder to the detergent solution, then blend for a while. Unless you have access to special microscopes and other equipment, you will probably be unable to confirm the existence of graphene in your detergent solution, but there's a quick rule of thumb which you can use to estimate the particle size. Generally speaking, if the graphene particles are too big, they will sink to the bottom of the vessel, while if they are smaller, they will float to the top. Nanoparticles are so small that they will be suspended in mid-water, and this is what you're aiming for if you want to make graphene. After you make enough graphene particles, you might want to filter the suspension and leave it to dry.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(3) If you pour a non-water-soluble liquid into water it will either float on top of the water, or the water will float on top of it, depending on whether the added liquid is heavier (more dense) or lighter (less dense) than water. A common example is oil floating on top of water. One group of scientists discovered that you can make graphene by pouring two solvents which do not mix together, such as heptane and water, into a glass and adding finely ground graphite powder and putting it in a sonication bath. You can buy heptane in most chemical supply stores, water is readily available (it's a good idea to use distilled water), and you can easily obtain graphite sticks in art shops. An ultrasound bath is needed, but these are not uncommon on ebay and the like. The resulting graphene is one to four layers thick, and is chemically stable. It can be extracted and deposited on a clean glass plate, or any other substrate. The idea is to mix the finely ground graphite with water and heptane (1:1 ratio of water and heptane) and sonicate it for a while. What happens next is that graphite flakes exfoliate at the interface between water and heptane and graphene actually starts &quot;climbing&quot; up the glass walls of the vial. If a hydrophilic substrate, such as a glass slide, is introduced through the interface, a thin film of graphene will climb the walls of the slide as well. The glass slide can then be extracted and after drying, you'll be left with graphene-coated glass. If you use a polyethylene vial, the resulting graphene will not climb the walls of the vial, but you can still extract it with a slide made of glass or other hydrophilic material. Graphene is deposited as a thin film on both sides of the glass plate, and its slight light-absorbing properties can be seen by looking through it (Graphene absorbs about 2.3% of visible light). The explanation of the process is that graphene self-reassembles from the small exfoliated flakes on the interface between water and heptane because the surface tension of graphene (54.8 mN/m) is nearly right between the surface tension of water (72.9 mN/m) and heptane (20.1mN/m). The thickness of the graphene film is limited by capillary forces on the interface and the excess graphene will simply fall to the bottom of the vial. The formation of thick aggregates is suppressed by the diffusion and the energy requirement needed to form a new layer. This is possibly the easiest way to make DIY graphene at home as an arbitrarily large sheet. Other methods only create graphene flakes, but these flakes could possibly be used instead of graphite flakes in this method as a refinement stage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-13T12:54:20.333" LastActivityDate="2015-07-13T12:54:20.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33907" PostTypeId="1" CreationDate="2015-07-13T02:50:06.050" Score="0" ViewCount="85" Body="&lt;p&gt;$\ce{Pb^{2+} (aq), Cl- (aq), Ag+ (aq), (SO4)^{2-} (aq)}$.  You number the four bottles, 1,2,3,4and while numbering them, you notice that bottles 1 and 3 have the word &quot;cation&quot; on their lids while bottles 2 and 4 have &quot;anion&quot; on their lids. You decide to test them by mixing them together in pairs, but you have only three empty beakers. When the solutions are mixed in pairs. &#xA;Three combination give white precipitates; &#xA;Bottle 1 + bottle 2. bottle 1 + bottle 4, bottle 2 + bottle 3. ive done each combination and theyre all insoluble? but my chart says there has to be 1 combination thats soluble and 3 that arent ? &lt;/p&gt;&#xA;" OwnerUserId="17387" LastEditorUserId="10185" LastEditDate="2015-07-13T12:29:07.167" LastActivityDate="2015-07-13T12:29:07.167" Title="Identification of solutions by solubility when mixed with each other" Tags="&lt;homework&gt;&lt;solubility&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33908" PostTypeId="2" ParentId="33907" CreationDate="2015-07-13T03:20:37.043" Score="1" Body="&lt;p&gt;Since It is a Homework Question I would Like to add a hint instead of giving full solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might not be able to Watch the difference Carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Your information&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbCl2}$ = insoluble&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbSO4}$ = insoluble &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{AgCl}$ = insoluble &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Ag2SO4}$ = slightly soluble &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try doing it again and watch the difference carefully afterall now you know which one is soluble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps.&lt;/p&gt;&#xA;" OwnerUserId="14530" LastEditorUserId="4097" LastEditDate="2015-07-13T06:07:09.520" LastActivityDate="2015-07-13T06:07:09.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33909" PostTypeId="1" CreationDate="2015-07-13T03:31:51.867" Score="2" ViewCount="26536" Body="&lt;p&gt;The type of silicone sealant that you would buy at your local hardware store. Would it break down when exposed to 90% mixture of isopropyl alcohol?&lt;/p&gt;&#xA;" OwnerUserId="17435" LastActivityDate="2019-01-04T18:41:37.803" Title="Silicone vs Alcohol" Tags="&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33911" PostTypeId="1" CreationDate="2015-07-13T05:49:15.347" Score="1" ViewCount="137" Body="&lt;p&gt;Here's an example. Say you take a 25˚C solution of NaCl, and lower the temperature suddenly. The equilibrium constant (solubility product) is immediately decreased, so ions start to join together into ionic compounds that float freely in the solvent before settling to the bottom. (In some cases they may be visible as precipitate.) &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can we refer to these free-floating ionic compounds as &quot;dissolved&quot; solute? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does &quot;XY(aq)&quot; (the aqueous solution of an ionic compound) seem like a good description of this state?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="5456" LastEditDate="2015-07-13T05:51:10.620" LastActivityDate="2015-07-13T11:13:28.410" Title="Can ionic compounds can ever &quot;dissolve&quot; as molecular compounds do?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33913" PostTypeId="1" CreationDate="2015-07-13T06:55:20.247" Score="5" ViewCount="500" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At $10~^{\circ}\mathrm{C}$, a small piece of $\ce{Zn}$ reacts with $\mathrm{2.0~M}$ $\ce{HCl}$ to produce $\pu{15.0 mL}$ of $\ce{H2}$ gas in $\pu{25 s}$. What is the rate of the reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My effort&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the $\mathrm{2.0~M~} \ce{HCl}$ concentration needs to be used and started the following calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{2.0~M} \times \frac{1}{2} \times \frac{1}{\mathrm{25~s \times 0.015~ mL}} = \mathrm{0.00060~mol}~ \ce{H2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{mol ~\ce{H2}} = \mathrm{mol~Zn} = \mathrm{0.0006~mol}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the given answer is 0.060 mol of Zn.&lt;/p&gt;&#xA;" OwnerUserId="17437" LastEditorUserId="5026" LastEditDate="2018-02-27T02:22:56.063" LastActivityDate="2018-02-27T02:22:56.063" Title="Calculate rate of reaction of a metal and an acid" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ClosedDate="2018-06-13T01:10:06.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="33915" PostTypeId="1" CreationDate="2015-07-13T08:14:38.423" Score="3" ViewCount="1686" Body="&lt;p&gt;How is Persian blue rock salt made and what does it contain that makes it look blue? &lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="41328" LastEditDate="2019-01-05T05:42:15.593" LastActivityDate="2019-01-05T05:42:15.593" Title="How is Persian blue rock salt made?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;crystal-structure&gt;&lt;color&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="33916" PostTypeId="1" CreationDate="2015-07-13T10:10:38.167" Score="0" ViewCount="590" Body="&lt;p&gt;For example, if I didn't know the compound tin chloride, and I was asked to write out its formula, how would I know how to? Because it seems unreasonable to know the formula of every compound in existence. Many thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="17441" LastEditorUserId="16683" LastEditDate="2018-08-08T16:34:03.213" LastActivityDate="2018-08-08T17:22:18.977" Title="How to know the formula of different compounds?" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="3" CommentCount="7" ClosedDate="2018-08-08T17:22:25.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="33918" PostTypeId="2" ParentId="33911" CreationDate="2015-07-13T11:13:28.410" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we refer to these free-floating ionic compounds as &quot;dissolved&quot; solute?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Dissolved implies that the ions are solvated. If the ions are in a solid ionic lattice then they are not solvated by the solvent and so would not be described as part of the dissolved solute.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does &quot;XY(aq)&quot; (the aqueous solution of an ionic compound) seem like a good description of this state?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again no. $\ce{XY(aq)}$, in the context of ionic compounds, refers to $\ce{X^{n+}}$ and $\ce{Y^{n-}}$ ions solvated by water molecules. The dissociation is generally regarded as implicit. For example, $\ce{NaCl(aq)}$ implies an aqueous solution containing $\ce{Na+}$ and $\ce{Cl-}$ ions. A better description of the state you are thinking of would be $\ce{XY(s)}$, using state symbols, or even better you could describe it as a suspension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the IUPAC &lt;a href=&quot;http://goldbook.iupac.org/S06198.html&quot; rel=&quot;nofollow&quot;&gt;Gold Book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suspension:&lt;br&gt;&#xA;  A liquid in which solid particles are dispersed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This accurately describes your system.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-07-13T11:13:28.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33919" PostTypeId="2" ParentId="33762" CreationDate="2015-07-13T12:58:40.757" Score="1" Body="&lt;p&gt;On a system where you're working in the bash shell, in your working directory where the .mol files reside, run this on the command line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;for i in $(ls *.mol); do name=${i%.mol}; babel -imol $i -oinchi $name.inchi; done&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have successfully tested this on machines running Ubuntu 12.04 and 14.04.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-13T12:58:40.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33920" PostTypeId="2" ParentId="33356" CreationDate="2015-07-13T13:06:48.163" Score="4" Body="&lt;p&gt;In principle, the maximum number of configurational stereoisomers (not constitutional isomers) is&#xA;$$N_\text{max}=2^{\left(n+m\right)}$$&#xA;where $n$ is the number of stereocentres (&lt;em&gt;R&lt;/em&gt; or &lt;em&gt;S&lt;/em&gt;) and $m$ is the number of stereogenic double bonds (&lt;em&gt;E&lt;/em&gt; or &lt;em&gt;Z&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the actual number of different stereoisomers may be smaller than the maximum number $N_\text{max}$ if constitutional symmetry is present in the molecule (&lt;em&gt;meso&lt;/em&gt;-compounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the actual number of practically possible stereoisomers may be reduced by ring strain or other geometrical limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the maximum number may also be exceeded if hindered rotation about single bonds or other steric interactions result in additional stereoisomers.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-13T13:06:48.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33922" PostTypeId="2" ParentId="33909" CreationDate="2015-07-13T13:19:14.017" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.customadvanced.com/chemical-resistance-chart.html&quot; rel=&quot;noreferrer&quot;&gt;this rather extensive chart&lt;/a&gt;, silicone can be exposed to isopropyl alcohol with no breakdown expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zid9X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-13T13:19:14.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33924" PostTypeId="2" ParentId="33904" CreationDate="2015-07-13T13:36:28.460" Score="5" Body="&lt;p&gt;There are several methods which you can use to make graphene at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the classic sticky tape method which was first used by researchers at the University of Manchester, UK and which won them the Nobel Prize for Physics in 2010.&lt;sup&gt;[1]&lt;/sup&gt; Essentially you deposit some graphite onto a surface, such as paper, using a pencil and then use sticky tape to carefully remove a layer of graphite from the paper. Then use another piece of tape to remove a thinner layer of graphite from the first piece of tape. Repeat this until you end up with a layer of graphite only a few atoms (or hopefully only one) thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method, which was discovered more recently, is to put powdered graphite in a blender along with a surfactant to stop the graphite particles clumping together.&lt;sup&gt;[2]&lt;/sup&gt; You can do this at home by getting a high speed kitchen blender and adding water, washing-up liquid and powdered graphite from a pencil lead. You want to keep blending until you end up with a suspension of very fine particles in the liquid. Unfortunately, the method isn't particularly accurate. The amount of washing-up liquid required depends on the quality of the graphite, which is difficult to assess without expensive analytical equipment and all you are left with at the end is a liquid full of tiny flakes of (hopefully) graphene and a bunch of larger particles which you didn't manage to blend all the way to single atom thickness.&lt;sup&gt;[3]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting way to make graphene at home, which albeit requires a little more effort than some of the others, involves using water and &lt;em&gt;n&lt;/em&gt;-heptane to create a thin layer of graphene at the interface between the two solvents.&lt;sup&gt;[4]&lt;/sup&gt; You can buy heptane fairly easily from chemical suppliers and distilled water is also advised to give a better result. The most expensive part is the ultrasound bath which is required to exfoliate the graphite (try buying one second hand online). Mix the two solvents in a 1:1 ratio in a hydrophobic container, such as a polyethylene vial, and then add powdered graphite. When the mixture is placed in a ultrasound bath and sonicated for a while, the graphite exfoliates and accumulates as flakes of graphene at the heptane-water interface. If you place a hydrophilic surface like a glass slide into the interface, the graphene climbs up the slide and forms a layer attached to the slide, which can then be removed, yielding a graphene coated glass slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]&lt;a href=&quot;http://www.nobelprize.org/nobel_prizes/physics/laureates/2010/press.html&quot;&gt;http://www.nobelprize.org/nobel_prizes/physics/laureates/2010/press.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2]&lt;a href=&quot;http://www.nature.com/nmat/journal/v13/n6/full/nmat3944.html&quot;&gt;http://www.nature.com/nmat/journal/v13/n6/full/nmat3944.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3]&lt;a href=&quot;https://www.newscientist.com/article/dn25442-make-graphene-in-your-kitchen-with-soap-and-a-blender/&quot;&gt;https://www.newscientist.com/article/dn25442-make-graphene-in-your-kitchen-with-soap-and-a-blender/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4]&lt;a href=&quot;http://www.researchgate.net/publication/251567939_Conductive_Thin_Films_of_Pristine_Graphene_by_Solvent_Interface_Trapping&quot;&gt;http://www.researchgate.net/publication/251567939_Conductive_Thin_Films_of_Pristine_Graphene_by_Solvent_Interface_Trapping&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-07-13T13:36:28.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33925" PostTypeId="2" ParentId="33915" CreationDate="2015-07-13T13:47:15.517" Score="7" Body="&lt;p&gt;Persian blue salt is formed by evaporating sea water, as are most if not all salt deposits. In the geosciences, we call salt deposits like this &lt;em&gt;evaporites&lt;/em&gt;. Persian blue salt is sodium chloride (the mineral name is halite).&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B003Q2BQC0&quot; rel=&quot;noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this description&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Blue salt is one of the rarest salts known to mankind, and is extracted from a salt mine in the northern province of Semnan in Persia.&lt;/p&gt;&#xA;&lt;p&gt;The intriguing blue color occurs during the forming of the salt's crystalline structure, as intense pressure is exerted on the salt deposits. The individual crystals fracture the light in an unusual way and the resulting blue (which is caused by an optical illusion), becomes visible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-13T13:58:55.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33926" PostTypeId="2" ParentId="33815" CreationDate="2015-07-13T13:56:05.113" Score="0" Body="&lt;p&gt;There's a counter effect in play here: van der Waals forces. As you become larger and larger, induced dipole-induced dipole interactions become stronger and stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gSTWT.gif&quot; alt=&quot;periodic table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that, in halogens (where abstracting an electron is a fairly tall order), the trend is that, the bigger your atom, the higher your boiling point, regardless of where you are in the group. The boron group loses the conflict of ease of abstraction versus van der Waals forces at around the gallium/indium leap, and for the carbon group, the conflict is lost to van der Waals forces only between tin and lead.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="9961" LastEditDate="2016-10-29T17:48:32.660" LastActivityDate="2016-10-29T17:48:32.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33927" PostTypeId="1" AcceptedAnswerId="33929" CreationDate="2015-07-13T14:06:41.513" Score="13" ViewCount="2430" Body="&lt;p&gt;I have a chemical formula for the mineral actinolite which is: $$\rm{Ca}_2({Mg,Fe})_5{Si}_8{O}_{22}({OH})_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula contains a &quot;,&quot; which I understand means that the magnesium and the iron can have different combinations (up to 5 total).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, if magnesium is the dominant atom, is it possible for the amount of magnesium to be less than iron and still be actinolite, or must magnesium always have the greatest amount?&lt;/p&gt;&#xA;" OwnerUserId="17041" LastEditorUserId="17041" LastEditDate="2015-07-14T02:29:20.770" LastActivityDate="2015-07-14T02:29:20.770" Title="If a chemical formula for a mineral contains a comma, are there limits on the amounts of each element?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33928" PostTypeId="1" AcceptedAnswerId="33933" CreationDate="2015-07-13T14:24:19.367" Score="12" ViewCount="5183" Body="&lt;p&gt;The famous VSEPR theory was proposed by first Sidgwick &amp;amp; Powell. It was modified then by Nyholm &amp;amp; Gillespie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading Inorganic Chemistry by J.D.Lee where he discusses very briefly the theory of Sidgwick &amp;amp; Powell; I couldn't somewhat conceive the language. So, I googled it but to my misfortune I could nowhere get more info other than this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The idea of a correlation between molecular geometry and number of valence electrons (both shared and unshared) was originally proposed in 1939 by Ryutaro Tsuchida in Japan, and was _independently presented in a Bakerian Lecture in 1940 by Nevil Sidgwick and Herbert Powell of the University of Oxford. In 1957, Ronald Gillespie and Ronald Sydney Nyholm of University College London refined this concept into a more detailed theory, capable of choosing between various alternative geometries &lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone would give the &lt;strong&gt;historical info&lt;/strong&gt; that Sidgwick &amp;amp; Powell was the first to introduce the theory which was later modified to the present VSEPR theory. I don't want that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please help me to bring me forward the original Sidgwick-Powell theory? Also the million-euro question: What are the differences between the two theories? In another way to say: What were the modifications that Mr. Nyholm &amp;amp; Gillespie brought forward that &quot;improved the original&quot; theory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]&lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory#History&quot;&gt;https://en.wikipedia.org/wiki/VSEPR_theory#History&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="10185" LastEditDate="2015-07-13T14:33:37.540" LastActivityDate="2019-06-24T13:42:46.070" Title="What are actually the differences between Sidgwick &amp; Powell theory and Nyholm &amp; Gillespie Theory?" Tags="&lt;history-of-chemistry&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33929" PostTypeId="2" ParentId="33927" CreationDate="2015-07-13T14:26:05.817" Score="11" Body="&lt;p&gt;Actinolite is a mineral that is part of a so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Solid_solution&quot;&gt;solid-solution series&lt;/a&gt;, which means that you get different but related minerals depending on the amounts of (in this case) iron and magnesium that are present. The magnesium-rich end member is &lt;a href=&quot;https://en.wikipedia.org/wiki/Tremolite&quot;&gt;tremolite&lt;/a&gt;; the iron-rich end member is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ferro-actinolite&quot;&gt;ferro-actinolite&lt;/a&gt;; and what lies between is actinolite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Actinolite&quot;&gt;Wikipedia's article on actinolite&lt;/a&gt;, the formula is more precisely:&#xA;$$\rm Ca_{2}(Mg_{4.5-2.5}Fe^{2+}_{0.5-2.5})Si_{8}O_{22}(OH)_{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is it possible for the amount of magnesium to be less than iron and still be actinolite, or must magnesium always have the greatest amount?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the answer is no: at best, Mg and Fe can be present in equal amounts for the mineral to still be called actinolite, but note that in that particular case you are on the &quot;border&quot; between actinolite and ferro-actinolite, so perhaps what you choose to call it is up to you. And as mentioned above, if magnesium is dominant, the mineral is tremolite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where iron is the dominant atom, the mineral is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Ferro-actinolite&quot;&gt;ferro-actinolite&lt;/a&gt; with the formula:&#xA;$$\rm Ca_{2}(Mg_{2.5-0.0}Fe^{2+}_{2.5-5.0})Si_{8}O_{22}(OH)_{2}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-13T15:06:25.637" LastActivityDate="2015-07-13T15:06:25.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33931" PostTypeId="2" ParentId="33762" CreationDate="2015-07-13T15:48:44.383" Score="3" Body="&lt;p&gt;This is exactly why I wrote Open Babel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;obabel *.mol -oinchikey -m&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you're not exceeding the number of command line arguments, you're done.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-13T15:48:44.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33932" PostTypeId="1" CreationDate="2015-07-13T15:55:08.890" Score="37" ViewCount="11116" Body="&lt;p&gt;I usually use Gaussian/GaussView and don't have to bother with other programs as I think that GaussView is capable of producing nice orbital/surface pictures like this one of one of uracil's Kohn-Sham-orbitals.&lt;img src=&quot;https://i.stack.imgur.com/CxzvU.png&quot; alt=&quot;GaussView cube file visualization; uracil LUMO&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when it comes that I have to use other programs than Gaussian - such as Turbomole, ORCA or Molcas - I am confronted with molden- or wfn-files that GaussView cannot handle.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Avogadro 1.1.1 (Lib 1.1.1, OB 2.3.2, Qt 4.8.5)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quite like Avogadro because it can produce as fancy pictures as GaussView but especially for molden files at least on my computer it seems to fail hard. As you can see the orbital is reproduced perfectly from the file but Avogadro somehow cannot manage to show all atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HYwv9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that I'm not the only one around who can't manage this to work properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Geoff's comments I learned that wfn-files can be imported through the Extensions $\rightarrow$ QTAIM menu which allows AIM analysis.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;gmolden 4.x or 5.x&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I have absolutely no idea which version this is.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What seems to be easier to open molden-files with molden? I guess that's something to try out. &lt;s&gt;But for me molden stops being an alternative since I am unable to arrange my molecules in space as much as I want. Also molden produces only those stacked-layer orbitals that are not really up to date.&lt;/s&gt; Molden can produce pretty good orbitals after one gets how it works. Thank you hokru!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I0VrR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;s&gt;I am admittedly not a big fan of it and didn't try to get more out of it. I was always searching for better alternatives.&lt;/s&gt; I'm still no big fan but as those things that I want are implemented and I just have been unable to click the right buttons molden is getting an appropriate way to visualize those orbitals.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Gabedit 2.4.8&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Gabedit is fancy ... there are a lot of input- and output-files that are supported - luckily molden files are part of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/O2SEI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least I can create orbitals somehow acceptable for me and arrange my molecule in space as much as I want it. There is also the povray-output stuff but I was not very successful with this approach, yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what stops me from using Gabedit at a regular basis. It is that there is everytime this point-and-click-adventurish menu -&gt; orbitals -&gt; selection -&gt; again orbital selection -&gt; quality settings -&gt; isovalue settings -&gt; wait and look. The settings were wrong? That's bad ... do the most things again and choose better values. Is there a simple way to compare orbitals fast? Not really, because the orbital slide show is a little bit tricky, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I say something about wfn-files? I think they are not supported.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Jmol 14.2.4_2014.08.03&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jmol is another program that can open molden files and I am again unable to create nice orbital images with this program. Even rising the so called &quot;resolution&quot; up to a reasonable degree only produces hexagonal-shaped orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fkUy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wfn-files? Who needs this? I guess no one except me.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Multiwfn 3.3.7&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there is this free fancy little program from china. A big 400 pages manual/tutorial-pdf comes with it. With it one can quickly have a look on orbitals with a possibility to quickly change isovalues. It lacks unfortunately my favored free arrangement in space plus there are no proper lighting settings ... and the orbitals and structures are not really that beautiful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lgoIS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But(!) it can read molden files and even wfn-files. It is also able to convert them to cube files that I afterwards can read in with GaussView to create what I want. It can also create those cube files for nearly all other surfaces that it can calculate. I like it much for this simple ability.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well ... as I can use Multiwfn to convert the files that GaussView cannot read in by itself, there is always this curious feeling that there has to be a simpler alternative to all this. A program that can produce nice images without all the lackings and misbehaviours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you do this stuff for visualisation of molden-/wfn-files to produce nice images of your orbitals and molecular surfaces?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-15T17:38:11.380" LastActivityDate="2017-07-26T09:59:51.747" Title="Generating neat orbitals/surfaces from molden/wfn-files" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="30" ContentLicense="CC BY-SA 3.0" />
  <row Id="33933" PostTypeId="2" ParentId="33928" CreationDate="2015-07-13T16:01:54.147" Score="5" Body="&lt;p&gt;Perhaps you should read the original papers: N.V. Sidgwick and H.M. Powell, &quot;&lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/176/965/153&quot; rel=&quot;nofollow&quot;&gt;Stereochemical types and valency groups&lt;/a&gt;&quot;, Proceedings of the Royal Society of London Vol 176, No. 965, pp 153-180 (1940), and R.J. Gillespie and R.S. Nyholm, &quot;&lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/1957/QR/QR9571100339&quot; rel=&quot;nofollow&quot;&gt;Inorganic Stereochemistry&lt;/a&gt;&quot;, Quarterly Reviews, Chemical Society 11 (4) 339-380 (1957).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote out of Gillespie's paper:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sidgwick and Powell assumed that lone pairs of electrons and bonding pairs were of equal importance and that they arranged themselves symmetrically so as to minimise the repulsions between them. The ligand(or crystal)- field theory is concerned with the influence of the non-bonding electrons upon the arrangement of the ligands which are assumed to be negative ions or dipoles surrounding a central metal atom. Recently the influence of covalent binding has also been considered. These various theories are in fact closely related and an attempt has been made in this Review to integrate them. It appears that the simple theory of electron-pair repulsions receives considerable justification from quantum-mechanics and when refined to take into account differences between the repulsions between lone pairs and bonding pairs of electrons provides a very satisfactory theory of the stereochemistry of inorganic molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That should get you going...&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2015-07-13T20:58:17.027" LastActivityDate="2015-07-13T20:58:17.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33934" PostTypeId="2" ParentId="33885" CreationDate="2015-07-13T18:13:33.160" Score="1" Body="&lt;p&gt;Any molecule that cannot be superimposed on its mirror image is described as being chiral. If the molecule is superimposable on its mirror image the object is called achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple alkenes, the carbon atoms of the double bond and the four atoms bonded to them all lie in the same plane, which divides the molecule into two halves, which are mirror images of each other. This plane of symmetry ensures that the mirror image forms of the molecule are superimposable. Therefore, simple alkenes like 2-chloroprop-1-ene and 1-bromo-2-chloro-2-fluoro-1-iodoethene are achiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, owing to the barrier to rotation in alkenes, 1-bromo-2-chloro-2-fluoro-1-iodoethene exists as two (achiral) stereoisomers: (&lt;em&gt;E&lt;/em&gt;)-1-bromo-2-chloro-2-fluoro-1-iodoethene and (&lt;em&gt;Z&lt;/em&gt;)-1-bromo-2-chloro-2-fluoro-1-iodoethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5zUjs.png&quot; alt=&quot;(E)-1-bromo-2-chloro-2-fluoro-1-iodoethene&quot;&gt;  &lt;img src=&quot;https://i.stack.imgur.com/XlYf5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-13T18:13:33.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33936" PostTypeId="1" CreationDate="2015-07-13T19:41:24.837" Score="3" ViewCount="181" Body="&lt;p&gt;What will happen if I incubate a clean gold surface in un-deprotected &#xA;($\ce{oligo-C6-S-SH}$) thiolated oligonucleotide solution instead of deprotected ($\ce{oligo-C6-SH}$) solution. I have read clean gold surface can cleave disulfide bond. &lt;/p&gt;&#xA;" OwnerUserId="17447" LastEditorUserId="4231" LastEditDate="2015-07-13T19:44:38.173" LastActivityDate="2015-07-13T22:51:59.863" Title="Attachment of disulfide (protected) thiolated oligonucleotide on gold surface" Tags="&lt;surface-chemistry&gt;&lt;organosulfur-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33938" PostTypeId="2" ParentId="14692" CreationDate="2015-07-13T20:17:36.030" Score="2" Body="&lt;p&gt;A Nickle Iron or iron copper battery,is most practical. A pint jar sized one is similar to a AA in voltage and similar to a lead acid battery in usable current, it's also easy enough to seal since it already has a lid. The self discharge rate is pretty high so you would have to keep your batteries empty or on a trickle charge to keep them usable. The major advantage is the long lifespan (rechargeable for nearly 70 years), and the electrolyte which is common lye. If the battery dies or spills you can neutralize the lye with vinegar,clean the plates with water and then refill it with lye and water. This will fully recharge the battery instantly and restore your battery's ability to hold a charge.  I know there are countless articles online about the dangers of lye but people have been using it for 100's of years to make soap,and it's not particularly harmful to the environment. You still want gloves goggles and vinegar on hand, it eats skin similar to the way household bleach does. (bleach has lye in it)  &lt;/p&gt;&#xA;" OwnerUserId="17448" LastActivityDate="2015-07-13T20:17:36.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33939" PostTypeId="2" ParentId="33936" CreationDate="2015-07-13T22:51:59.863" Score="3" Body="&lt;p&gt;What actually happens in sulfur-gold bonding is not very well known. There have been many proposals and studies done, well-summarized &lt;a href=&quot;http://www.nature.com/nchem/journal/v4/n6/pdf/nchem.1352.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. There is some evidence, including &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja993622x&quot; rel=&quot;nofollow&quot;&gt;computational studies&lt;/a&gt;, that suggest that dissociation into two thiolates is energetically favourable. Bridging across protruding bits of gold (we often draw gold surfaces as straight lines, but they're seldom that flat) may also be involved. This case, where you only have a chain on one of the two sulfurs, may be less interesting and I haven't found many studies of terminal disulfides so the exact process might be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the practical question, &lt;em&gt;Can I stick a disulfide-functionalized molecule onto gold?&lt;/em&gt;, yes you can, and they will stick very well. &lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-07-13T22:51:59.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33940" PostTypeId="2" ParentId="33932" CreationDate="2015-07-13T23:10:56.150" Score="7" Body="&lt;p&gt;I use Avogadro obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Avogadro v1.1.x you can not only use wfn files, but perform QTAIM analysis. Open it under the QTAIM menu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send me the Molden file if you can't read all the atoms, that's a weird bug. It works on any Molden file I've tried.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-13T23:10:56.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33941" PostTypeId="2" ParentId="33932" CreationDate="2015-07-13T23:16:41.770" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://www.ks.uiuc.edu/Research/vmd/&quot; rel=&quot;noreferrer&quot;&gt;VMD&lt;/a&gt; is my current go-to for molecular visualization.  As with any software, it takes some getting used to, but the wealth of options, what seems to be a robust feature set, and generally flexible configurability make it the king of visualizers in my book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've only ever tried it with ORCA output, and I've only been able to make it work with &quot;gOpenMol binary&quot; files as generated by the &lt;code&gt;orca_plot&lt;/code&gt; utility (ORCA's Molden files appear not to be completely read), but I'm quite satisfied with the results.  The steps I typically follow are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Run the ORCA computation to get the &lt;code&gt;.xyz&lt;/code&gt; and &lt;code&gt;.gbw&lt;/code&gt; files&lt;/li&gt;&#xA;&lt;li&gt;Run &lt;code&gt;orca_plot basename.gbw -i&lt;/code&gt; to generate the gOpenMol &lt;code&gt;.plt&lt;/code&gt; binary file of the desired function (atomic/molecular orbital, electron/spin density, etc.)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Run VMD and load the &lt;code&gt;.xyz&lt;/code&gt; and &lt;code&gt;.plt&lt;/code&gt; files&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;File&lt;/code&gt; &gt; &lt;code&gt;New Molecule&lt;/code&gt; to reach the 'open' dialog&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Browse...&lt;/code&gt; to the desired &lt;code&gt;.xyz&lt;/code&gt; file; &lt;code&gt;Determine file type&lt;/code&gt; should automatically detect the XYZ format&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Load&lt;/code&gt; to pull the geometry into VMD&lt;/li&gt;&#xA;&lt;li&gt;Change 'Load files for' to &lt;code&gt;New Molecule&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Browse...' to the desired '.plt' file;&lt;/code&gt;Determine file type` should automatically detect the 'gOpenmol plt'&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;Load&lt;/code&gt; to pull the binary information into VMD&lt;/li&gt;&#xA;&lt;li&gt;Close the 'open' dialog&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Change the molecule display mode&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Click &lt;code&gt;Graphics&lt;/code&gt; &gt; &lt;code&gt;Representations&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Under 'Selected Molecule' choose the &lt;code&gt;.xyz&lt;/code&gt; file&lt;/li&gt;&#xA;&lt;li&gt;Under 'Drawing Method' choose the desired method (I usually use 'licorice')&lt;/li&gt;&#xA;&lt;li&gt;Run up the &lt;code&gt;Sphere Resolution&lt;/code&gt; and &lt;code&gt;Bond Resolution&lt;/code&gt;, and tweak the &lt;code&gt;Bond Radius&lt;/code&gt; as desired&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apply the desired formatting to the isosurface(s)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Again in the &lt;code&gt;Graphical Representations&lt;/code&gt; dialog (don't have to close after setting the geometry drawing method), select the &lt;code&gt;.plt&lt;/code&gt; from the &lt;code&gt;Selected Molecule&lt;/code&gt; dropdown&lt;/li&gt;&#xA;&lt;li&gt;If plotting a quantity with negative regions (orbital wavefunctions, possibly spin density, etc.), click 'Create Rep' once, to get a second 'representation'.&lt;/li&gt;&#xA;&lt;li&gt;Change 'Coloring Method' to &lt;code&gt;ColorID&lt;/code&gt;, and the number in the dropdown that appears to whatever color desired for one phase (positive or negative)&lt;/li&gt;&#xA;&lt;li&gt;I usually set &lt;code&gt;Material&lt;/code&gt;:&lt;code&gt;Transparent&lt;/code&gt; (so as to see the molecule under the isosurface) ... &lt;code&gt;Drawing Method&lt;/code&gt;:&lt;code&gt;Isosurface&lt;/code&gt; ... &lt;code&gt;Draw&lt;/code&gt;:&lt;code&gt;Solid Surface&lt;/code&gt; ... &lt;code&gt;Show&lt;/code&gt;:&lt;code&gt;Isosurface&lt;/code&gt; ... &lt;code&gt;Step&lt;/code&gt;:&lt;code&gt;1&lt;/code&gt; ... &lt;code&gt;Size&lt;/code&gt;:&lt;code&gt;1&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Set the &lt;code&gt;Isovalue&lt;/code&gt; appropriately for the signedness, positive or negative, by typing the number into the box and pressing &lt;kbd&gt;Enter&lt;/kbd&gt;; I usually start at a magnitude of &lt;code&gt;0.02&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;Repeat the above for the opposite-signed portion of the plot, if required.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A quick example is below -- the SOMO of gas-phase monoaquo-Cu(II):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YG26E.png&quot; alt=&quot;Mol image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three keyboard shortcuts I use the most are those that change the mouse behavior: &lt;kbd&gt;R&lt;/kbd&gt; to rotate, &lt;kbd&gt;S&lt;/kbd&gt; to scale (zoom), and &lt;kbd&gt;T&lt;/kbd&gt; to translate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VMD can apparently export to a number of 3-D file formats, as well as PostScript vector graphics.  The direct bitmap export is fine if you don't need to zoom in, but pixelates pretty badly if you do.  The PostScript is pretty blocky in gswin64, but might do better in, e.g., Inkscape.  I don't have IS installed at the moment, so I can't tell if a better renderer does a better job of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One further nifty feature of VMD is that it can render in stereo.  Even if you don't have 3-D glasses, you can still look at things &lt;a href=&quot;https://en.wikipedia.org/wiki/Magic_Eye&quot; rel=&quot;noreferrer&quot;&gt;Magic Eye&lt;/a&gt; style by clicking &lt;code&gt;Display&lt;/code&gt; &gt; &lt;code&gt;Stereo&lt;/code&gt; &gt; &lt;code&gt;SideBySide&lt;/code&gt;. (Turn &lt;code&gt;Display&lt;/code&gt; &gt; &lt;code&gt;Stereo Eye Swap&lt;/code&gt; on if you need to cross your eyes to see it, instead of looking into the distance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would probably have to fiddle around with the different file types generable by each software package to find the best one to interface with VMD.  I don't have a copy of either Molcas or Turbomole to test them out, unfortunately.  VMD does support Gaussian CUBE files, so worst-case you can dump a (possibly huge) grid and plot it that way.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-07-14T03:05:51.633" LastActivityDate="2015-07-14T03:05:51.633" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33942" PostTypeId="2" ParentId="33916" CreationDate="2015-07-14T00:20:57.430" Score="4" Body="&lt;p&gt;In my experience if someone says something like tin chloride they are usually insinuating you know the most common oxidation state that metal is found in when complexed with the chloride atoms. This is a bad practice of them if they haven't explained further but I digress... In this example with the metal center of the complex being tin, the most common oxidation state is +2. This means the formula in its neutral state would be $\ce{Sn^{+2}}$ with 2 chlorides with a (-1) charge making the molecule neutral at $\ce{Sn(II)Cl2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different effects cause  common oxidation states for specific elements. These are outlined further with Mendeleev's Periodic Law. It breaks down into how shielded the nucleus is by electrons, and how big the radius of the atom is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further explanation of tin's common oxidation state calls upon knowledge of the inert pair effect, where elements to the right of the second and third row transition metals such as: In, Tl, &lt;strong&gt;Sn&lt;/strong&gt;, Pb, Sb, Bi form compounds that are less reactive than one would think where the oxidation state is 2 less than the expected group. This is caused by the $\mathrm{ns^2}$ electrons being relatively inert. The cause of that nonreactive state is because of two reasons specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Larger-than-normal effective&#xA;nuclear charges in these elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lower bond energies in their compounds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A good book to use referencing this subject further is Descriptive Inorganic, Coordination, and Solid-State Chemistry by Glen E. Rodgers. This is the book I own and used for when I studied periodic trends, and inorganic chemistry as an undergraduate. It's a great book for this subject, among others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are commonplace names that are used as somewhat of a slang for certain compounds used in research frequently. These can be learned through experience, or you can just google common names... All formal names are dictated by the rules set forth by the IUPAC. Learning those rules would be good idea. Also keep in mind that in some research projects you will encounter molecules that have informal nicknames too. &lt;/p&gt;&#xA;" OwnerUserId="17207" LastEditorUserId="17207" LastEditDate="2015-07-14T01:30:15.697" LastActivityDate="2015-07-14T01:30:15.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33943" PostTypeId="2" ParentId="33916" CreationDate="2015-07-14T01:18:37.037" Score="3" Body="&lt;p&gt;Actually, the rules for naming compounds have been converging on the notion that you &lt;em&gt;should&lt;/em&gt; be able to construct the compound from the name. How you do this is simply research and practice. There are sites for &lt;a href=&quot;http://www.usca.edu/chemistry/genchem/nomen.htm&quot; rel=&quot;nofollow&quot;&gt;inorganic&lt;/a&gt; and &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm&quot; rel=&quot;nofollow&quot;&gt;organic&lt;/a&gt; naming conventions. Regarding your example, I immediately knew you were talking about an inorganic salt of tin, and I further knew than tin tends to be in the $+2$ state, resulting in the formula $\ce{SnCl2}$. But I also know lots of other possible formulas simply based upon the linked rules. You may not know everything, but you can know a lot simply by learning some basic rules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, @Ryan Lee's answer points out that &quot;tin chloride&quot; is potentially ambiguous, and modern convention has been changing in a direction to make this more clear. The preferred name for this compound is &lt;strong&gt;tin(II) chloride&lt;/strong&gt;, which emphasizes the the fact the the oxidation state of tin is $+2$. Most naming conventions have been formalized so that they encode a way to completely reconstruct the molecule. Nonetheless, tradition exists, so we often still call that solvent &lt;em&gt;acetone&lt;/em&gt; rather than &lt;em&gt;propan-2-one&lt;/em&gt;, the latter being much more structurally descriptive (i.e. ketone functional group substituted at the 2nd carbon of propane).&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-07-15T00:54:32.547" LastActivityDate="2015-07-15T00:54:32.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33944" PostTypeId="2" ParentId="33903" CreationDate="2015-07-14T01:57:22.370" Score="2" Body="&lt;p&gt;Lubricants are usually graded for a host of different properties, but ultimately what you are looking for is it's thermal stability, and a viscosity index related to the temperature your mills parts run at. These properties are optimized for different parts of the machine because they run at different speeds and as a consequence temperatures. I could delve into more, but I would need more specifics about the machine itself. If you wish to pursue the problem further on your own I would suggest &lt;a href=&quot;http://www.machinerylubrication.com/&quot; rel=&quot;nofollow&quot; title=&quot;A good reference site&quot;&gt;This website&lt;/a&gt; as a reference. Also take a look at this &lt;a href=&quot;http://www.machinerylubrication.com/Read/28956/lubricant-viscosity-index&quot; rel=&quot;nofollow&quot;&gt;cautionary article&lt;/a&gt; regarding accurate measurements on viscosity index as it can be a bit tricky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence of using the wrong lubricants varies based on how wrong you are... But ultimately the lubricant will thin out to the point that is does not decrease friction and or cool the parts it is lubricating if it degrades at the temperature the part runs at.&lt;/p&gt;&#xA;" OwnerUserId="17207" LastEditorUserId="17207" LastEditDate="2015-07-14T02:04:46.443" LastActivityDate="2015-07-14T02:04:46.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33951" PostTypeId="1" CreationDate="2015-07-14T04:32:19.200" Score="8" ViewCount="1778" Body="&lt;p&gt;In valence bond theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why is it told that &quot;&lt;strong&gt;MO theory provides a global, delocalized perspective on chemical bonding&lt;/strong&gt;&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is a follow up to &lt;a href=&quot;https://chemistry.stackexchange.com/q/33879/&quot;&gt;What is actually the difference between valence bond theory and molecular orbital theory?&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-07-14T15:59:03.767" Title="Why is the resonance concept not required in molecular orbital theory?" Tags="&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;resonance&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33952" PostTypeId="1" CreationDate="2015-07-14T04:53:27.460" Score="2" ViewCount="308" Body="&lt;p&gt;I know that energy is absorbed due to chromophores which enable electron transition which uses up energy. But I am a bit unsure about what happens after the absorption and where this absorbed energy went. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not very sure if the energy is dissipated since it would contradict with the law that energy can't be destroyed. If the energy is lost as heat, then it is a bit contradictory again according to the equation $E= hc/\lambda$. &#xA;The wavelength would be big for infra-red and thus, the energy lost must be less than the absorbed energy from UV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how exactly are these flavonoids shielding UV radiation? And also, if these flavonoids are in the excited state, it makes me feel that they might be quite reactive and may form free radicals.  &lt;/p&gt;&#xA;" OwnerUserId="17455" LastEditorUserId="5456" LastEditDate="2015-07-14T06:07:34.103" LastActivityDate="2015-08-14T03:45:54.073" Title="How do flavonoids absorb UV radiation?" Tags="&lt;organic-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33953" PostTypeId="1" AcceptedAnswerId="33963" CreationDate="2015-07-14T05:06:57.627" Score="6" ViewCount="479" Body="&lt;p&gt;I'm currently looking at Principles of Chemistry by Atkins, and I have a few questions involving entropy. These questions probably arise from my poor understanding of reversible/irreversible processes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why are reversible processes called reversible? Is it necessarily true that reversible processes can be reverted spontaneously, or is this a common misconception that people have?&lt;/li&gt;&#xA;&lt;li&gt;So my chemistry textbook gives the definition of entropy as&#xA;$$\Delta S = \frac{q_{rev}}{T}$$&#xA;where $q_{rev}$ is heat transferred reversibly. However, I get lost at how entropy could increase during a reversible process.&lt;/li&gt;&#xA;&lt;li&gt;Why is the reversibility condition necessary for the definition of entropy?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17362" LastEditorUserId="10185" LastEditDate="2015-07-14T10:17:17.333" LastActivityDate="2015-07-16T03:45:48.400" Title="Trying to Understand Entropy" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33954" PostTypeId="1" CreationDate="2015-07-14T05:16:17.973" Score="2" ViewCount="1767" Body="&lt;p&gt;How to represent/ show an ionic bond visually?&#xA;For example in NaCl&lt;/p&gt;&#xA;" OwnerUserId="17456" LastActivityDate="2015-07-14T09:23:58.243" Title="How to show/represent an ionic bond visually?" Tags="&lt;bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33955" PostTypeId="2" ParentId="33954" CreationDate="2015-07-14T06:51:16.793" Score="3" Body="&lt;p&gt;Some examples are as follows, &#xA;$$&#xA;\ce{[Na]+ [Cl]-} \, , \\&#xA;\ce{[Mg]^2+ 2[Cl]-} \, , \\&#xA;\ce{3[Mg]^2+ 2[N]^3-} \, .&#xA;$$&#xA;Hopefully you get the idea, so a general rule is:&#xA;$$&#xA;\text{#ions}[\text{element symbol}]^{\text{magnitude of cation charge}} \text{#ions}[\text{element symbol}]^{\text{magnitude of anion charge}} \, .&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="16087" LastEditorUserId="186" LastEditDate="2015-07-14T08:47:25.933" LastActivityDate="2015-07-14T08:47:25.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33956" PostTypeId="2" ParentId="14523" CreationDate="2015-07-14T06:54:29.887" Score="0" Body="&lt;p&gt;Oxidized species give up electrons to an electron pool which prior to their flow to the other half cell create a large negative charge on the electrode where oxidation takes place.&lt;/p&gt;&#xA;" OwnerUserId="17457" LastActivityDate="2015-07-14T06:54:29.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33959" PostTypeId="2" ParentId="33954" CreationDate="2015-07-14T09:23:58.243" Score="3" Body="&lt;p&gt;Concerning the depiction of ionic bonds,  the &lt;a href=&quot;http://dx.doi.org/10.1351/pac200880020277&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Graphical Representation Standards for Chemical Structure Diagrams&lt;/em&gt; (IUPAC Recommendations 2008)&lt;/a&gt; read as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Structures that are known to be ionic should be depicted as such. They should be drawn with atoms bearing explicit positive and negative charges, and there should be a space (rather than a bond) between those atoms.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N8bW4.png&quot; alt=&quot;depiction of ionic bonds&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-14T09:23:58.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33963" PostTypeId="2" ParentId="33953" CreationDate="2015-07-14T10:07:17.873" Score="8" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They're called &quot;reversible&quot; because the processes can run backwards without additional input of energy. There aren't really any truly reversible processes in the universe, though (some come very close, but the construct of &quot;reversible&quot; is an ideal, not a reality).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Entropy of a &lt;em&gt;system&lt;/em&gt; can increase just fine during a reversible process: expanding a gas &quot;slowly enough&quot; is reversible (&quot;slowly enough&quot; means in an infinite number of infinitesimal steps), but the entropy of the gas increases as it expands as it has more microstates to go into. The entropy that remains constant in a reversible process is that of the system &lt;strong&gt;plus&lt;/strong&gt; the surroundings (in other words, the universe).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The process does not need to be reversible to have entropy. However, we measure entropy based on the heat of the reversible process because entropy is a state function (it depends on the beginning and end points of your process, nothing more) while heat is a path function (it depends on the path you take to get there). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-07-14T10:07:17.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33964" PostTypeId="1" CreationDate="2015-07-14T10:28:41.500" Score="1" ViewCount="1287" Body="&lt;p&gt;I am running an MD simulation of silver with $\ce{SiO2}$. Initially, my silver atom is neutral and oxygen has a coordination of 2 with silicon having a coordination of 4. After the deposition of silver atoms, I find that the coordination of silver atoms ranges up to 12 and oxygen has coordination of 3 where a bond is formed with silver. Since $\ce{Ag-Si}$ bonds are not present, silicon's coordination number does not change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to understand why oxygen is able to have a coordination number of 3 as with 2 bonds with silicon, the valency is already satisfied. I know silver forms a fcc-lattice where the coordination number is 12 but I want to get a more in depth detail of this change of coordination number. The basic physics behind it. &lt;/p&gt;&#xA;" OwnerUserId="17464" LastEditorUserId="7475" LastEditDate="2016-12-25T02:15:08.693" LastActivityDate="2016-12-25T02:15:08.693" Title="Why can oxygen have a coordination number of 3 in a complex between silver and SiO2?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33965" PostTypeId="1" AcceptedAnswerId="34011" CreationDate="2015-07-14T10:37:22.097" Score="2" ViewCount="1517" Body="&lt;p&gt;I found a question in a book about the reaction between &lt;span class=&quot;math-container&quot;&gt;$\ce{AgNO3}$&lt;/span&gt; and aromatic/aliphatic amines, but I couldn't find any information about&#xA;it. What should be the expected reaction mechanism and products formed in this reaction?&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorUserId="83758" LastEditDate="2020-09-08T19:19:27.773" LastActivityDate="2020-09-08T19:19:27.773" Title="Is there any reaction between amines and AgNO3?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33967" PostTypeId="2" ParentId="33932" CreationDate="2015-07-14T11:15:18.240" Score="12" Body="&lt;p&gt;for completeness, &lt;strong&gt;gmolden&lt;/strong&gt; can actually produce high quality pictures like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MgML.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to click the button in the top right in density mode to get the opengl rendering. It is also fully rotatable.&#xA;While i really like molden for its quick response opening structure files, usability is not its strength. Also it is slow to generate the orbitals.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another program is &lt;strong&gt;Molekel&lt;/strong&gt;, which uses openbabel (i think) to read various file formats. It works fine with molden-type files.&#xA;Much faster than molden to make the orbitals and more flexible (color, resolution, bounding box). I never tested wfn files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFZ2x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16161" LastActivityDate="2015-07-14T11:15:18.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33968" PostTypeId="1" CreationDate="2015-07-14T12:04:47.747" Score="0" ViewCount="3245" Body="&lt;p&gt;How can I convert the number of particles per liter to gram per liter? I know the size of these particle and so, I calculated the volume and then by the density I calculate the mass. After that I don't know what to do&lt;/p&gt;&#xA;" OwnerUserId="17466" LastEditorUserId="9961" LastEditDate="2015-09-12T23:01:41.960" LastActivityDate="2015-11-11T23:16:12.723" Title="How can I convert the number of particles per liter to gram per liter" Tags="&lt;concentration&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33969" PostTypeId="2" ParentId="33968" CreationDate="2015-07-14T12:14:51.830" Score="0" Body="&lt;p&gt;I would go with the Avogadro constant -&gt; calculate how many moles per liter there are in your mixture and then, using the molecular mass, you can calculate the concetration in g/L. That is if you know your compound.&lt;/p&gt;&#xA;" OwnerUserId="17371" LastActivityDate="2015-07-14T12:14:51.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33970" PostTypeId="2" ParentId="33953" CreationDate="2015-07-14T12:27:43.207" Score="4" Body="&lt;p&gt;Second law of thermodynamics tells that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Energy tends to disperse from being localized if it is not hindered from doing so.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does a hot cup of tea become cool? Because the higher vibrational energy of the hot mug &amp;amp; tea was localized only to the tea-mug system. When it is brought in the surroundings, the localized energy disperses to the surroundings. This is what second law of thermodynamics says about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give you another example of energy dispersal approach to entropy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there is a gas enclosed by a stop-cock in of the bulbs of a two-bulb container. When you open the stop-cock, the gas expands &amp;amp; covers the two bulbs. Why did it happen so? It happened because as you open the stop-cock(removed the &lt;em&gt;hinderance&lt;/em&gt;), the localized energy of the gas molecules spread over the whole volume to disperse its energy over a larger volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change in entropy measures how much energy has been dispersed at a specific temperature. Entropy is a state property &amp;amp; how it changes doesn't matter at all. However it is far easy to take the course of the change as a reversible process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy of the universe(system + surroundings) always increases or doesn't change but never decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take a hot body at temperature $T$. Now, bring in another body which is at a temperature $T - dT$. Since, the difference is infinitesimal, the process of transfer of heat can be considered as reversible. Let $dQ$ heat energy be transferred from the hotter body to the colder body. Decrease in entropy of the hotter body after the heat transfer is $\dfrac{-dQ}{T}$ &amp;amp; increase in entropy of the colder body is $\dfrac{dQ}{T - dT}$. By simple inequality, we can show that $$\dfrac{dQ}{T - dT} &amp;gt; \dfrac{dQ}{T} \implies -\dfrac{dQ}{T - dT} &amp;lt; -\dfrac{dQ}{T} \implies \dfrac{dQ}{T - dT} - \dfrac{dQ}{T} &amp;gt; 0 \implies dS &amp;gt; 0$$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In classical thermodynamics, there is no definition of entropy actually; what is concerned here is actually the &lt;em&gt;change of entropy&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in statistical Mechanics, entropy can be defined as the number of microstates the system has at a particular state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a microstate?  each particle in a system collides &amp;amp; now &amp;amp; then they redistribute the energy among themselves without changing the total energy of the system. Microstate is one of the arrangements of the particle's energy for a particular state. It describes how each molecule is distributed at various energy levels at a certain instant. ( As you are reading Atkins, you can get the taste of this approach soon; he has described it beautifully.) It is not that the particles are at various microstates at a single time; it rather describes at a certain instant how the particles are distributed among various energy levels(Total energy is always same; it is just redistributed now &amp;amp; then due to the random collisions of the particles.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More the microstates for a given state(temperature, pressure, internal energy of the system), more is the entropy at that state. Why? Because here the total energy has more ways to get &lt;em&gt;dispersed&lt;/em&gt; to a certain microstate. Suppose, you have two systems $A$ &amp;amp; $B$ having the same energy $U$. However, $A$ has more microstates than $B$. So, the energy $U$ in $A$ owing to its having larger number of microstates has more ways to redistribute among the particles at a certain instant than in case of $B$. So, the energy $U$  &lt;strong&gt;disperses&lt;/strong&gt; more in $A$( as it has more ways to distribute its internal energy) than in $B$ where the energy $U$ disperses lesser (since it has lesser number of ways to redistribute its energy) &amp;amp; hence its energy is localized. Thus entropy of $A$ is more in $A$ than in $B$ as in the former, energy isn't localized &amp;amp; disperses to more microstates than in $B$ .&lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastActivityDate="2015-07-14T12:27:43.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33971" PostTypeId="1" AcceptedAnswerId="33974" CreationDate="2015-07-14T12:50:54.130" Score="37" ViewCount="5758" Body="&lt;p&gt;&lt;em&gt;I have searched and searched, oh how I have searched.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a 3-dimensional visualization of a &lt;strong&gt;whole&lt;/strong&gt; atom, one that that includes all the orbital geometry. A proper &quot;layered&quot; view of the orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am not looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Individual orbital geometry.&lt;/li&gt;&#xA;&lt;li&gt;A &quot;picture&quot; of an atom (i.e. what looks like television static to me).&lt;/li&gt;&#xA;&lt;li&gt;Little spherical balls.&lt;/li&gt;&#xA;&lt;li&gt;Dot-and-cross diagrams.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think I understand that this cannot be expressed in just one picture alone, so I look forward to discovering software, or a collection of pictures, that could help express this. Maybe something similar to an exploded view, for atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please. The universe is weird and I want to see.&lt;/p&gt;&#xA;" OwnerUserId="14911" LastEditorUserId="14911" LastEditDate="2016-11-09T03:55:51.413" LastActivityDate="2019-03-01T17:03:10.287" Title="How do I visualize an atom?" Tags="&lt;atoms&gt;&lt;orbitals&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="33972" PostTypeId="1" CreationDate="2015-07-14T12:57:31.937" Score="1" ViewCount="384" Body="&lt;p&gt;I am currently adsorbing proteins to an aluminum hydroxide gel.  By adding a carbomer (polymer of acrylic acid cross-linked with allyl sucrose), I achieve a somewhat synergistic formulation.  Shortly after adding the carbomer, a granular precipitate forms that adheres to glass.  Questions: 1) What is this precipitate?; 2) Will my adsorbed protein release from the aluminum hydroxide after carbomer addition?; 3) Is there possible long-term instability of this mixture?  Thank you!&lt;/p&gt;&#xA;" OwnerUserId="17469" LastEditorUserId="10185" LastEditDate="2015-07-14T13:03:50.693" LastActivityDate="2015-07-14T13:03:50.693" Title="Chemical incompatibility between aluminum hydroxide [Al(OH)3] and carbomer polymer [-CH2-CH] N-COOH]" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;chemical-biology&gt;&lt;identification&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33973" PostTypeId="1" CreationDate="2015-07-14T13:02:50.150" Score="1" ViewCount="1513" Body="&lt;p&gt;If I wash chloroform (stabilized with amylene) with sulphuric acid in a separating funnel, where will the chloroform be found (top or bottom)? I think the chloroform will be at the bottom, because it is heavier than sulphuric acid. &lt;/p&gt;&#xA;" OwnerUserId="17470" LastEditorUserId="10185" LastEditDate="2015-07-14T13:07:17.503" LastActivityDate="2019-04-10T04:33:11.937" Title="Chloroform-acid wash" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33974" PostTypeId="2" ParentId="33971" CreationDate="2015-07-14T13:33:24.797" Score="35" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have searched and searched, oh how I have searched.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do you know what I always tell my mom when she asks me to find something in the Internet she was not able to find herself? I ask her: &quot;Are you sure that the thing you are looking for even exists?&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am looking for a 3 dimensional visualization of a whole (moderately&#xA;  complex, hydrogen is just a ball) atom that includes 3 dimensional&#xA;  orbital geometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hydrogen atom is &lt;em&gt;not&lt;/em&gt; &quot;just a ball&quot;.&lt;/li&gt;&#xA;&lt;li&gt;There is &lt;em&gt;no&lt;/em&gt; orbitals. In short, except for one-electron systems (such as the hydrogen atom) orbital description is just an &lt;em&gt;approximate model&lt;/em&gt; of the reality.&lt;/li&gt;&#xA;&lt;li&gt;Usually when one asks to visualize a physical object, he/she wants to visualize its physical &quot;boundaries&quot; with respect to some medium. The notion of such &quot;boundary&quot; simply looses its meaning in the microscopic world: atoms do not have &quot;boundaries&quot; like macroscopic objects do.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Please. The universe is weird and I want to see.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, but you can &lt;em&gt;not&lt;/em&gt; do this: you can &lt;em&gt;not&lt;/em&gt; visualize some things in the Universe, you can &lt;em&gt;not&lt;/em&gt; even imagine them. Our visualization &amp;amp; imagination abilities (unlike an atom) has some boundaries, because they are based ultimately on our senses which in turn has their boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Paul Dirac said on that matter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[…] the main object of physical science is not the provision of&#xA;  pictures, but is the formulation of laws governing phenomena and the&#xA;  application of these laws to the discovery of new phenomena. If a&#xA;  picture exists, so much the better; but whether a picture exists or&#xA;  not is a matter of only secondary importance. In the case of atomic&#xA;  phenomena no picture can be expected to exist in the usual sense of&#xA;  the word 'picture', by which is meant a model functioning essentially&#xA;  on classical lines. One may, however, extend the meaning of the word&#xA;  'picture' to include any &lt;em&gt;way of looking at the fundamental laws which&#xA;  makes their self-consistency obvious&lt;/em&gt;. With this extension, one may&#xA;  gradually acquire a picture of atomic phenomena by becoming familiar&#xA;  with the laws of the quantum theory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-14T13:38:53.157" LastActivityDate="2015-07-14T13:38:53.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="33975" PostTypeId="1" AcceptedAnswerId="48688" CreationDate="2015-07-14T14:06:48.297" Score="3" ViewCount="974" Body="&lt;p&gt;I have recently seen a &lt;a href=&quot;https://www.youtube.com/watch?v=H7Ng4sOVkns&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; of ruthenium on youtube. It says that ruthenium is invulnerable to many acids, especially aqua regia, but it gets destroyed by a simple household chemical, bleach. The video did not give a proper explanation about why this happens, it only gave the reaction mechanism. What is the reason for this Achilles' heel of ruthenium?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="10185" LastEditDate="2015-07-14T15:30:12.163" LastActivityDate="2016-03-29T17:28:22.353" Title="Reaction between ruthenium and bleach" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="33976" PostTypeId="2" ParentId="33971" CreationDate="2015-07-14T14:14:18.007" Score="16" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am looking for a 3 dimensional visualization of a whole (moderately complex, hydrogen is just a ball) atom that includes 3 dimensional orbital geometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;3 dimensions is only enough to represent the probability density function of hydrogen.  Given that the origin is the location of the nucleus, each point in  3-dimensional space will have a corresponding probability density of the electron being at the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For helium, 6 dimensions would be needed to represent the probability density.  This is because for each pair of points at which the nucleus and first electron are located, every point in 3-dimensional space will be associated with a probability density for the second electron.  Understanding this, I would recommend the figures in &lt;a href=&quot;http://th.physik.uni-frankfurt.de/~engel/2pd_helium.html&quot;&gt;2-Particle density: Correlated motion of electrons&lt;/a&gt; as the next step in visualizing an atom beyond hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastEditorUserId="5160" LastEditDate="2016-03-03T17:45:44.813" LastActivityDate="2016-03-03T17:45:44.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33977" PostTypeId="2" ParentId="33951" CreationDate="2015-07-14T14:18:00.487" Score="0" Body="&lt;p&gt;In MO theory, the molecule is described by Molecular Orbitals that are approximately calculated from atomic orbitals; I believe the process is referred to as constructing &quot;Linear combinations of Atomic Orbitals&quot; (LCAO).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that you use &quot;weighting coefficients&quot; to say how much a particular atomic orbital contributes to a given Molecular Orbital.You also always get out as many MO's as AO's you put in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example, Benzene. If you start adding up the atomic orbitals it would be understandable to assign each atomic orbital equal weight in the calculation. So in the end you get 6 equivalent MO's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear someone else's perspective on this, but I'm guessing that you sort of &quot;infer&quot; a structure similar to that obtained from resonance because of this equivalency.&lt;/p&gt;&#xA;" OwnerUserId="17374" LastActivityDate="2015-07-14T14:18:00.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33978" PostTypeId="2" ParentId="33968" CreationDate="2015-07-14T15:50:52.347" Score="1" Body="&lt;p&gt;$$x \frac{particles}{L} \times \frac {1 mol}{6.023 \times 10^{23}particles}\times \frac{MW gm}{mol}$$&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2015-07-14T15:50:52.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33979" PostTypeId="2" ParentId="33819" CreationDate="2015-07-14T15:54:21.097" Score="3" Body="&lt;p&gt;In addition to the answer posted by @santiago, here is some information from the following article in the UK newspaper Daily Mail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-3160002/The-science-stinky-feet-Video-reveals-processes-cause-nasty-stench-stop-it.html&quot; rel=&quot;nofollow&quot;&gt;The science behind stinky feet: Video reveals the processes that cause a nasty stench, and how to stop it&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article has a video, produced by the American Chemical Society, that explains how the bacteria that live on your skin and in your shoes eat your sweat, producing an acid by-product that causes the smells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A snippet from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The three main bacterial waste products are propanoic acid,&#xA;  methanethiol and isovaleric acid&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Methanethiol makes your feet smell like sulphur and cheese and is&#xA;  produced from dead skin.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Isovaleric acid also has a cheesy aroma, and a rancid vinegar-like&#xA;  scent, while propanoic acid has a rancid sour smell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13112" LastEditorUserId="10185" LastEditDate="2015-07-14T15:58:39.380" LastActivityDate="2015-07-14T15:58:39.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33980" PostTypeId="2" ParentId="33951" CreationDate="2015-07-14T15:59:03.767" Score="4" Body="&lt;p&gt;Because MO is closer to reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance is a convenient fiction used to simplify the visualization of some molecules, and rationalize their actions in the frame work of Lewis structures. But every resonance structure you have ever seen in your life is a lie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, this is not exactly true. It's more precise to say that resonance structures are like blind men trying to describe an elephant: they each give a partial snapshot of truth while being wholly inadequate to describing it all. The &quot;true&quot; structure is a weighted average of all the resonance structures based on how stable they would be if they were the actual molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MO cuts to the chase and attempts to directly describe the molecule in terms of its molecular orbitals: another convenient fiction, mind, but one that is closer to the truth. The reason you don't need &quot;resonance&quot; MO structures is because molecular orbitals are inherently a probabilistic construct: they already take into account the uncertainty in the electrons' positions. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-07-14T15:59:03.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="33981" PostTypeId="1" CreationDate="2015-07-14T16:34:19.367" Score="1" ViewCount="509" Body="&lt;p&gt;I am trying to calculate the total energies for a particular organic molecule that undergoes radical decomposition. I was able to calculate the neutral molecule with the exact results of a paper I found. My trouble is obtaining the correctly calculated energies for the radical molecules. I am new to computational chemistry and am looking for a bit of guidance on this. &lt;/p&gt;&#xA;" OwnerUserId="17477" LastActivityDate="2015-07-14T21:27:35.163" Title="Radical Computational Calculations" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="10" ClosedDate="2015-07-15T14:37:39.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="33983" PostTypeId="2" ParentId="33973" CreationDate="2015-07-14T17:21:18.833" Score="4" Body="&lt;p&gt;Pure, concentrated sulfuric acid has a density of 1.84 grams per cubic centimeter.  Chloroform has a density of 1.489 grams per cubic centimeter.  Thus a chloroform phase will form &lt;strong&gt;&lt;em&gt;on top of&lt;/em&gt;&lt;/strong&gt; a pure sulfuric acid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are using dilute sulfuric acid, the density will be closer to that of water (1 gram per cubic centimeter), and chloroform may form &lt;strong&gt;&lt;em&gt;beneath&lt;/em&gt;&lt;/strong&gt; an aqueous acid phase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a concentration of water in sulfuric acid at which the density of the aqueous acid phase will be equal to that of the chloroform.  In that case, the layers will never separate (at least not vertically).&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-14T17:21:18.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33984" PostTypeId="2" ParentId="33973" CreationDate="2015-07-14T18:14:05.447" Score="5" Body="&lt;p&gt;Usually, you shake chloroform with concentrated sulfuric acid, wash with water, dry over $\ce{CaCl2}$, and distill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of chloroform is approximately $\rho = 1.48\ \mathrm{g\ cm^{-3}}$, whereas the density of concentrated sulfuric acid is approximately $\rho = 1.84\ \mathrm{g\ cm^{-3}}$.&#xA;Therefore, after you allow the layers to separate, the concentrated sulfuric acid is in the lower layer and the chloroform is in the upper layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, however, when you wash with water ($\rho = 1.00\ \mathrm{g\ cm^{-3}}$), the chloroform is in the lower layer and the water is in the upper layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any doubt about the identity of a layer, add a drop of the layer in question to a small amount of water in a test tube. If the drop does not dissolve, the layer in question is the organic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caution: The mixture may become hot when the remaining sulfuric acid is removed with water, and chloroform boils at $61.2\ \mathrm{^\circ C}$.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-14T18:14:05.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33985" PostTypeId="1" CreationDate="2015-07-14T18:18:20.387" Score="12" ViewCount="731" Body="&lt;p&gt;I have recently studied superposition of states (also the famous Schrodinger's cat), measurement problem, decoherence &amp;amp; so on. I then read the resonance concept from Atkins' Physical Chemistry where it was written as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is the superposition of &lt;strong&gt;wavefunctions&lt;/strong&gt; representing different electron distributions in the same nuclear framework. [...] A better description of a molecule of &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\psi_\ce{HCl} = \psi_{\ce{H-Cl}} + \lambda\psi_{\ce{H^+} \ce{Cl^-}} $&lt;/span&gt;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus this is the general wavefunction which is superposition of all the states. &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is a numerical coefficient (can be found out by &lt;em&gt;variation theorem&lt;/em&gt;) whose square predicts the probability of being in the ionic state when we inspect the molecule. Just like typical wavefunction-collapse after measurement! This is the original Pauli's theory of resonance. Unfortunately Atkins halts down here; after all it is physical chem-book! Nowhere have I found this approach to resonance; no organic chem book I possess: Morrison, Boyd; Solomons; R.K. Gupta ... They all say it is just the &lt;em&gt;weighted average&lt;/em&gt; of all Lewis structures. No, it is not correct! As is said &lt;a href=&quot;http://academic.reed.edu/chemistry/roco/Resonance/wavefn.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This essay contains information that I have never found &lt;strong&gt;in any organic chemistry textbook.&lt;/strong&gt; This can mean two things: the information is too difficult or you don't really need to know it. Maybe both are true .&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]  When Linus Pauling developed resonance theory, he defined &quot;resonance&quot; as the &lt;strong&gt;superposition of wave functions&lt;/strong&gt;. Unfortunately, this detail has been lost in the intervening decades, and modern chemists &lt;strong&gt;incorrectly treat &quot;resonance&quot; as the superposition of Lewis structures (resonance forms)&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]Suppose you have a molecular AB that is a resonance hybrid:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xxCYP.gif&quot; alt=&quot;enter image description here&quot;&gt;&#xA;  and you want to calculate the electron density on atom A.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The usual (wrong) prediction procedure. The usual procedure says, &quot;combine what you see in each Lewis structure.&quot; In other words, combine the electron density (ED) on A in form IV with the electron density on A in form V:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ED_\text{wrong} \text{(hybrid)} = 1/2 [ED(IV) + ED(V)]$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The much-ignored (correct) prediction procedure. You may recall that electron density (ED) is given by the square of the wave function (WF). If we want to know the electron density on A, we must look at the square of the wave function near A:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ED\text{(hybrid)} = WF\text{(hybrid)} ^2 \implies ED\text{(hybrid)} = ED_\text{wrong} \text{(hybrid)} + \boxed{WF(IV)WF(V)}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The final formula shows the discrepancy between the right answer, ED(hybrid), and the usual (wrong) answer, EDwrong. The correct formula includes WF(IV)WF(V). This term is called the &quot;interaction density&quot; or &quot;overlap density,&quot; and is a product of two different wave functions. The interaction density affects the electron density everywhere in the molecule, and its appearance tells us that the electron density obtained by the &quot;usual&quot; procedure is wrong.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Electron density is just one of many properties that interest chemists, but all property predictions -- electron density, bond type, atom charge, dipole moment, geometry, energy, etc. -- have something in common with electron density prediction: the prediction procedure must start with WF(hybrid), and their answers always include &lt;strong&gt;interaction terms&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is one of the great reasons why the modern concept of resonance is wrong; they don't take the &quot;interaction term&quot;! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does only one book (which is not organic) contain the correct interpretation? Why do people follow the wrong version? Resonance is not at all the &lt;em&gt;weighted average&lt;/em&gt; of Lewis structures; rather it is just a superposition of wavefunctions? So, why do most of the sources (people &amp;amp; book) advocate the wrong theory? Can anyone tell me which organic chem book still follows Pauling's original theory of resonance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added: In my quest to find the resources which still define Pauling's theory of resonance of wavefunctions, Feynman's Lectures, which I have just completed reading, is really worthy to be gold to be noted. Feynman completely describes the benzene molecule as the &lt;em&gt;superposition&lt;/em&gt; of &lt;strong&gt;base states&lt;/strong&gt;; it can be found in the CALTECH &lt;a href=&quot;http://www.feynmanlectures.caltech.edu/III_10.html#Ch10-S4&quot; rel=&quot;nofollow noreferrer&quot;&gt;site&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorUserId="15751" LastEditorDisplayName="user5764" LastEditDate="2019-01-07T13:36:08.200" LastActivityDate="2019-01-07T13:36:08.200" Title="Why is the original Pauling's theory of resonance that uses superposition of wavefunctions not used today?" Tags="&lt;resonance&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33988" PostTypeId="2" ParentId="33613" CreationDate="2015-07-14T19:16:32.010" Score="5" Body="&lt;p&gt;It &lt;strong&gt;&lt;em&gt;can't&lt;/em&gt;&lt;/strong&gt; be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-NH-NH-R + 2R-NHCl -&amp;gt; R-N=N-R + 2R-NH2Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that while the atoms are balanced, the electrons aren't. Going from $\ce{R-NH-NH-R}$ to $\ce{R-N=N-R}$ involves the loss of two protons and two electrons, but the $\ce{R-NHCl}$ are really only picking up two protons. (Nitrogens with four substituents are positively charged.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When thinking of mechanisms, it may help to explicitly split the final listed reaction into two steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5a. $\ce{H-NH-NH-H + NH2Cl -&amp;gt; H-N=N-H + NH4Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5b. $\ce{H-N=N-H + NH2Cl -&amp;gt; N2 + NH4Cl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, each chloramine is involved in pulling one dihydrogen (two protons and two electrons) off of the diazo. - The addition of a &quot;hydride&quot; (a proton and two electrons) to the nitrogen kicks the chlorine off as a negatively charged ion, and the remaining proton protonates the lone pair on the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous last step for the $\ce{R-NH2}$ case might be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-NH-NH-R + R-NHCl -&amp;gt; R-N=N-R + R-NH3Cl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to reducing the $\ce{N-Cl}$ covalent bond to $\ce{R-NH2 + HCl}$ followed by an acid/base reaction.&lt;/p&gt;&#xA;" OwnerUserId="11616" LastEditorUserId="275" LastEditDate="2015-07-14T19:48:33.053" LastActivityDate="2015-07-14T19:48:33.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33990" PostTypeId="1" AcceptedAnswerId="33993" CreationDate="2015-07-14T20:31:57.460" Score="10" ViewCount="305" Body="&lt;p&gt;In my experience, most foods smell like they taste. Fragrant soaps, however, taste quite awful in comparison to their fragrances. How can some things smell and taste identical while others are so vastly different? I have a feeling this may have something to do with the vapor pressures of the molecules that make up the &quot;thing,&quot; but I cannot find anything to confirm my hypothesis exactly.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="23561" LastEditDate="2018-09-07T00:50:19.830" LastActivityDate="2018-09-07T00:50:19.830" Title="Discrepancies between smell and taste" Tags="&lt;food-chemistry&gt;&lt;smell&gt;&lt;taste&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33991" PostTypeId="2" ParentId="33971" CreationDate="2015-07-14T20:48:41.873" Score="9" Body="&lt;p&gt;As a matter of fact, despite the answer you accepted, these orbitals can be visualized.  The way this is done, is that an isosurface of the probability density can be plotted.  That is, a surface of constant probability around the nucleus can be visualized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There weird thing is that there is not a unique way to visualize the orbitals on the molecule.  They way you choose to visualize the orbitals will depend on what kind of chemical insight you want to have.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend software like &lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;noreferrer&quot;&gt;Avogadro&lt;/a&gt; to visualize molecules.  &lt;/p&gt;&#xA;" OwnerUserId="5717" LastEditorUserId="5717" LastEditDate="2015-07-14T21:44:25.770" LastActivityDate="2015-07-14T21:44:25.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33992" PostTypeId="2" ParentId="33981" CreationDate="2015-07-14T21:27:35.163" Score="3" Body="&lt;p&gt;&lt;em&gt;Disclaimer: this is a preliminary answer based on the current state of the discussion with OP; it might be the subject to major changes in the near future.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;OP (original poster) is trying to reproduce some &lt;em&gt;relative&lt;/em&gt; energies previously computed at ωB97X-D/6-31+G(d,p) at a different M06-2X/6-31G(d,p) level of theory. OP has exact same results for closed-shell molecular species in question, the meaning of which is &lt;em&gt;probably&lt;/em&gt; that the resulting M06-2X/6-31G(d,p) relative energies in kcal/mol or kJ/mol are exactly equal to the previously published ωB97X-D/6-31+G(d,p).&lt;/li&gt;&#xA;&lt;li&gt;OP also mentioned that he has some troubles in reproducing the relative energies for open-shell species, by which he &lt;em&gt;probably&lt;/em&gt; means that for open-shell species M06-2X/6-31G(d,p) spits out different numbers in kcal/mol or kJ/mol than ωB97X-D/6-31+G(d,p).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Before we have some numbers let me say few things which hold in generality.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The difference in few kJ/mol in relative energies computed using two different functionals is &lt;em&gt;pretty normal&lt;/em&gt; (even at exact same geometry). One should not even expect exactly same relative energies with any two different functionals in the first place.&lt;/li&gt;&#xA;&lt;li&gt;OP presumably run geometry optimization at M06-2X/6-31G(d,p) but did not mention how different are the equilibrium geometries obtained at this level from those at ωB97X-D/6-31+G(d,p) level published previously. In fact, this would be difficult to do, because I could not find the Cartesian coordinates of the species in question in the Supporting Information for the work published previously. But at least some quick visual comparison has to be done to make sure that equilibrium geometries are at least not totally different.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;OP did not mention why M06-2X functional was chosen. Even if the ultimate goal is not just to reproduce the previously published work, it is mentioned there that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ωB97X-D, a range-separated double-hybrid functional, has been shown to describe reaction energies accurately.&lt;sup&gt;45&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But, first of all, ωB97X-D is &lt;em&gt;not&lt;/em&gt; a double-hybrid functional. And secondly, the paper quoted there is fairly well-known &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/2011/cp/c0cp02984j#!divAbstract&quot; rel=&quot;nofollow&quot;&gt;benchmark of density functional methods&lt;/a&gt; done by Lars Goerigkab and Stefan Grimme, and in fact, Goerigkab and Grimme found M062X-D3 and not ωB97X-D to be &quot;statistically the best of all hybrids&quot;. Note though that M062X-D3 differs from M062X: the former includes the empirical dispersion correction while the later does not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With respect to the empirical dispersion correction, today it is usually recommended to include it in DFT calculations, since it is available at almost no (computational) cost but can substantially improve the results. Note also that dispersion correction is already included in case of ωB97X-D functional but has to be &quot;manually&quot; added for the case of M06-2X. In fact, the absence of it might explain some differences in equilibrium geometries and relative energies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-07-14T21:27:35.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33993" PostTypeId="2" ParentId="33990" CreationDate="2015-07-14T21:51:43.700" Score="11" Body="&lt;p&gt;The food that you eat regularly is something that you are used to. You know exactly which smell combines with which taste ... but that does not imply that the chemicals you smell are the same that you taste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had been eating your whole life long fragrant soap with apple smell, you would wonder about the taste of real apples as you wonder right now about the taste of apple soaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of different substances that are responsible for the special taste of everything you eat. You can find more about &lt;a href=&quot;https://en.wikipedia.org/wiki/Flavor&quot;&gt;flavor&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Aroma_compound&quot;&gt;aroma compounds&lt;/a&gt; on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to real fruits, fragrant soaps and other industrial products with a certain smell only contain those substances that are responsible for the characteristic smell of e.g. fruits (often only the main one or two aroma substances or even only synthetic aroma substances that smell similar to the natural products).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as not all products that shall smell like certain thing also need to taste like those things you only smell the &quot;apple&quot; but you won't taste it.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-15T00:59:41.183" LastActivityDate="2015-07-15T00:59:41.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33994" PostTypeId="1" AcceptedAnswerId="34041" CreationDate="2015-07-15T00:05:25.293" Score="71" ViewCount="15196" Body="&lt;p&gt;I read that the $\ce{O2}$ molecule is &lt;a href=&quot;https://en.wikipedia.org/wiki/Paramagnetism&quot;&gt;paramagnetic&lt;/a&gt;, so I'm wondering: could a strong magnet pull the $\ce{O2}$ to one part of a room – enough to cause breathing problems for the organisms in the room? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not a professional chemist, though I took some college chemistry.)&lt;/p&gt;&#xA;" OwnerUserId="4121" LastEditorUserId="-1" LastEditDate="2015-07-15T07:18:09.947" LastActivityDate="2020-04-14T20:38:19.653" Title="Could a magnet pull oxygen out of the air?" Tags="&lt;magnetism&gt;" AnswerCount="3" CommentCount="7" FavoriteCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="33995" PostTypeId="2" ParentId="33994" CreationDate="2015-07-15T00:21:33.030" Score="42" Body="&lt;p&gt;Physics says yes. And at the US Patent Office, it certainly looks like the answer is yes, according to (at least) &lt;a href=&quot;http://www.google.com/patents/US7771509&quot;&gt;this US patent application&lt;/a&gt;. The abstract reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A process for separating O$_2$ from air, that includes the steps effecting an increase in pressure of an air stream, magnetically concentrating O$_2$ in one portion of the pressurized air stream, the one portion then being an oxygen rich stream, and there being another portion of the air stream being an oxygen lean stream, compressing the oxygen rich stream and removing water and carbon dioxide therefrom, to provide a resultant stream, and cryogenically separating said resultant stream into a concentrated oxygen stream and a waste stream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would say the yes is &lt;em&gt;conditional&lt;/em&gt; in that you'd probably need a room that is sealed from communication with the atmosphere (otherwise equilibrium will be re-established with respect to oxygen pretty quickly), is vented to the outside in order to dispose of the oxygen, houses one heck of a strong magnet, and has a strongly-locked door to prevent the unfortunate organisms from leaving.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-15T01:00:25.883" LastActivityDate="2015-07-15T01:00:25.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="33996" PostTypeId="2" ParentId="33952" CreationDate="2015-07-15T01:09:31.083" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So how exactly are these flavonoids shielding UV radiation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Due to the structure of the flavonoids, there are many transitions to absorb the UV radiation. Also, there are many transitions to release the energy. Most of the time, it is released in 2 steps (2 photons) which equal the energy of the photon that exited the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorescence emits a photon that is close to the same energy as the first photon. For quercetin, the excitation is at 440 nm and the emission is at 540 nm.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-07-15T01:09:31.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="33997" PostTypeId="1" CreationDate="2015-07-15T02:17:53.243" Score="2" ViewCount="77" Body="&lt;p&gt;I have a beaten old R/C car that I am restoring for my son. It has some nice rubber tires on it, but they are kind of old and worn. The rubber itself is almost moldy and looks matte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could just fetch some wonderful product of a shelf somewhere that promises to restore the rubber, but I would like to really know what can be done for old rubber on a low level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can I treat my rubber, and which chemical process works best (if any)?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastActivityDate="2015-07-15T02:17:53.243" Title="What chemicals could be used to repair rubber tires?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-10-30T19:36:56.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="33998" PostTypeId="2" ParentId="33916" CreationDate="2015-07-15T03:38:56.370" Score="1" Body="&lt;p&gt;The other answers here have already given good responses to how you would find the answer to the specific example you gave. I want to focus on the question and this particular quote: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...it seems unreasonable to know the formula of every compound in&#xA;  existence.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're absolutely right. In principle, we're supposed to have &lt;a href=&quot;http://www.usca.edu/chemistry/genchem/nomen.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;formal&lt;/a&gt; &lt;a href=&quot;https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;naming&lt;/a&gt; &lt;a href=&quot;http://www.chemguide.co.uk/inorganic/complexions/names.html#top&quot; rel=&quot;nofollow noreferrer&quot;&gt;systems&lt;/a&gt; (thanks trb456 for the first two links) that describe exactly which molecular formula you're dealing with. Unfortunately, chemical naming conventions are a bit of a mess, with multiple names for every molecule. So what's a chemist to do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is that instead of learning the formula for every molecule in existence, you learn to extrapolate and make some pretty good guesses off of what you know. For instance, let's say someone's talking about lead chloride. Well, I don't know what the formula for that is! It could be any number of chlorines! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, I do recall that $\ce{PbO}$ is often used in microscopes. And since oxygen can only take on a 2- charge, that means that lead has to be +2, so the formula is $\ce{PbO2}$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh wait, but I also remember something in my textbook about a $\ce{PbO2}$ compound. That means that there's two possible forms of lead chloride, $\ce{PbCl2}$ and $\ce{PbCl4}$. And now I've narrowed it down to two possible formula, which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lead_chloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia disambiguation page&lt;/a&gt; helpfully shows us are the only two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to which one is specifically being talked about (and you'll have to believe me on this), I would be inclined to believe that it's the lead(II) variant simply because you see lead(II) so much more often than you see lead(IV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another trick you can use is to look at common naming conventions and try to guess why they were named that way. While common names are a lot more unpredictable than the official IUPAC rules, there are some common themes that still get involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a simple example, the following molecules are named &lt;strong&gt;acet&lt;/strong&gt;ylene, and &lt;strong&gt;acet&lt;/strong&gt;ate, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sVdko.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sVdko.png&quot; alt=&quot;Acetylene and acetate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice a trend: these are all two-carbon molecules with some functional group coming off the end. In that vein, I can guess that &lt;strong&gt;acet&lt;/strong&gt;onitrile will be a two-carbon molecule with some sort of nitrogen group coming off the end...and I'd be correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gBzmv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gBzmv.png&quot; alt=&quot;Acetonitrile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, it's silly and somewhat pointless to try to memorize the official name, common name, and molecular formulas of every single compound you'll ever use. Instead, there are two sets of tricks you can use:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nature follows a pattern. Look at compounds you already know and try to figure out what rules compounds you haven't seen should follow&lt;/li&gt;&#xA;&lt;li&gt;Humans didn't pick names for no reasons. Look at the names and see if there's a common pattern.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's not easy, but if you work hard at it for a year or two, you should have enough knowledge to piece together information about most of the compounds you'll hear about.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="16683" LastEditDate="2018-08-08T16:36:23.007" LastActivityDate="2018-08-08T16:36:23.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33999" PostTypeId="1" AcceptedAnswerId="34003" CreationDate="2015-07-15T04:30:59.647" Score="1" ViewCount="78" Body="&lt;p&gt;Is there a reason for all the bottles being 70% ethanol, or is it convention. &lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="10185" LastEditDate="2015-07-15T10:44:07.737" LastActivityDate="2015-07-15T10:44:07.737" Title="What's with the 70% ethanol?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-07-15T13:56:37.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="34001" PostTypeId="1" AcceptedAnswerId="34002" CreationDate="2015-07-15T05:53:48.250" Score="2" ViewCount="1095" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Have a dissolution of $\ce{HNO_3}$ in water. The density is&#xA;  $1.42\ce{g/mL}$ with a concentration of $\ce{16M}$. What is the&#xA;  concentration in % mass/mass?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Alright. So the % mass/mass is expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{grams of } \ce{HNO_3}}{\text{grams of the dissolution}}\cdot100$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I need to find the grams of $\ce{HNO_3}$ first. They tell me that the Molarity is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$16 = \frac{\text{moles of } \ce{HNO_3}}{\text{liters of dissolution}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HNO_3}$ has a molar mass of $63$. So to find out how many moles of $\ce{HNO_3}$ are in the dissolution I need to divide the grams of $\ce{HNO_3}$ by $63$... But how am I supposed to find out the grams of $\ce{HNO_3}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told that the numerator is $63$. But... $63$ is the molar mass of $\ce{HNO_3}$, not the number of moles of $\ce{HNO_3}$ in the dissolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, let's suppose that it is indeed $63$. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$16 = \frac{63}{\text{liters of dissolution}} \implies \text{liters of dissolution} = 3.9$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are $3.9$ liters of dissolution? I'm not sure - I feel like I got this all wrong. How do you know there are $63$ moles of $\ce{HNO_3}$?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="69830" LastEditDate="2021-05-08T18:27:25.453" LastActivityDate="2021-05-08T18:27:25.453" Title="Find number grams of the solute in a dissolution given its density and Molarity" Tags="&lt;concentration&gt;&lt;mixtures&gt;&lt;density&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34002" PostTypeId="2" ParentId="34001" CreationDate="2015-07-15T06:22:21.543" Score="3" Body="&lt;p&gt;Using 1L of solution, you have 16M of HNO3, each mole of which weighs 63g. Therefore, in 1L you'd have 63 X 16 = 1008g. Mass of 1L is given as 1420g from the density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;%mass/mass = 1008/1420 = 70% HNO3&lt;/p&gt;&#xA;" OwnerUserId="17489" LastActivityDate="2015-07-15T06:22:21.543" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34003" PostTypeId="2" ParentId="33999" CreationDate="2015-07-15T06:26:43.620" Score="2" Body="&lt;p&gt;Primarily it is used as an anti-microbial agent. There is lots of data out there showing that the efficacy of ethanol at killing bacteria peaks at the 70% mixture. As they are making tons of it (literally) for the microbiologists, it's easier to just scale production then set up a new production facility for chemists to use at an arbitrary concentration.&lt;/p&gt;&#xA;" OwnerUserId="17489" LastActivityDate="2015-07-15T06:26:43.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34004" PostTypeId="1" AcceptedAnswerId="34029" CreationDate="2015-07-15T07:47:48.650" Score="1" ViewCount="3631" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;You have a mixture of $\ce{He, Ne, Ar}$ gases with a total pressure of $8.40$ atmospheres. $\ce{He}$ has a pressure of $1.50$ and $\ce{Ne}$ has $2.00$. What is the molar fraction of $\ce{Ar}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The molar fraction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{moles of Ar}}{\text{total moles of mixture}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to know the number of moles of any of these components, I need to know their mass in the mixture. How can I know if all I am given is their pressures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is $0.583$ and I am not sure how was this concluded. All I can tell is that $\ce{Ar}$ has a pressure of $4.9$ atmospheres by Dalton's law but that's it. By the same law I know that all gases there have the same temperature and volume, but since I don't know either value for any of them it isn't useful at all.&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="5026" LastEditDate="2018-02-21T04:51:28.850" LastActivityDate="2018-02-21T04:51:28.850" Title="Calculate the molar fraction of a mixture component given only its pressure" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34005" PostTypeId="2" ParentId="33994" CreationDate="2015-07-15T09:31:31.687" Score="22" Body="&lt;p&gt;No, to separate oxygen from air you need extremely high gradients of field strength. in this paper &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs11630-007-0079-1#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://link.springer.com/article/10.1007%2Fs11630-007-0079-1#page-1&lt;/a&gt; they used about &lt;span class=&quot;math-container&quot;&gt;$\pu{0.4T}$&lt;/span&gt; per &lt;span class=&quot;math-container&quot;&gt;$\pu{mm}$&lt;/span&gt;, but to cause breathing problems it is enough to get the &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt; concentration below 17%, so lets say &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1T/mm}$&lt;/span&gt; is needed. To sustain such gradients over a whole room (lets say 4 meters), the field strength on one side of the room needs to be around &lt;span class=&quot;math-container&quot;&gt;$\pu{400T}$&lt;/span&gt;, which is very likely to kill humans (dissolved ions begin to circle due to lorentz force instead of going where they should, moving around induces huge currents…), and has never been achieved in a continuous field. (from nationalmaglab.com:  …45-Tesla hybrid magnet, which offers scientists the strongest continuous magnetic field in the world…)&lt;/p&gt;&#xA;" OwnerUserId="17492" LastEditorUserId="61578" LastEditDate="2019-12-18T02:04:32.190" LastActivityDate="2019-12-18T02:04:32.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34006" PostTypeId="1" AcceptedAnswerId="34020" CreationDate="2015-07-15T09:54:08.580" Score="15" ViewCount="39149" Body="&lt;p&gt;It's very difficult to get accurate information concerning household bleach residues. Imagine, for instance, that you pour pure (or diluted) bleach onto fabric or synthetic foam and let it dry without rinsing (as a scientific experiment). &lt;a href=&quot;http://boards.straightdope.com/sdmb/showthread.php?t=579725&quot; rel=&quot;noreferrer&quot;&gt;Some sources&lt;/a&gt; say: it will totally &quot;vanish&quot; (with air and/or light), leaving water and salt, and some others say it will leave some toxic compounds behind, such as sodium hydroxide (which is added to the bleach to slow its decomposition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: Will the bleach &lt;strong&gt;only&lt;/strong&gt; degrade into water and salt ($\ce{NaCl}$), or will it leave other substances? If so, which ones? If the first case is true (i.e., the degradation products are water and salt only), why does the solution still smell like bleach for a long time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: If bleach (a base) does leave toxic residue(s), can we use vinegar (an acid) to neutralize it? $(~\ce{OH- + H+}$ should create $\ce{H2O}$)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;White vinegar&lt;/strong&gt;: I'm confused since there is contradictory information on &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_hypochlorite&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. Some says vinegar will neutralize harmful residues from bleach, while &lt;a href=&quot;http://www.ask.com/home-garden/vinegar-neutralize-chlorine-bleach-16bbc6421df988bf&quot; rel=&quot;noreferrer&quot;&gt;another source&lt;/a&gt; says it &quot;could result in much more caustic and dangerous chemicals, such as hypochlorous acid&quot; being produced.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will an antioxidant such as &lt;strong&gt;vitamin C&lt;/strong&gt; work as neutralizer? (as suggested by Brian)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seems that everybody agrees about the use of peroxide, bisulfite, metabisulfite or thiosulfite, and sodium thiosulfate to neutralize bleach residues, but I wonder if more common products such as vinegar or vitamin C can be used instead: What do you think about these two? Any other suggestions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I removed the washing example which was confusing since it implied rinsing + I updated my question since I thought bleach was an acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Household bleach is, in general, a solution containing 3–8% sodium&#xA;  hypochlorite and 0.01–0.05% sodium hydroxide; the sodium hydroxide is&#xA;  used to slow the decomposition of sodium hypochlorite into sodium&#xA;  chloride and sodium chlorate. &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_hypochlorite&quot; rel=&quot;noreferrer&quot;&gt;wiki&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Brian and Drmoishe's answer suggest several ways to way to neutralized bleach but &lt;strong&gt;what about the sodium Hydroxide residues?&lt;/strong&gt; (I check on a french bleach, it shows only on the composition: &quot;3.7% of sodium hypochlorite&quot;, nothing else)&lt;/p&gt;&#xA;" OwnerUserId="17491" LastEditorUserId="17491" LastEditDate="2016-08-18T07:27:48.437" LastActivityDate="2016-08-18T07:27:48.437" Title="Does bleach (without rinsing) leave residues and can vinegar, baking soda neutralize them?" Tags="&lt;acid-base&gt;&lt;home-experiment&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34007" PostTypeId="1" CreationDate="2015-07-15T10:18:37.360" Score="37" ViewCount="5637" Body="&lt;p&gt;Is there a compound that &lt;strong&gt;absorbs&lt;/strong&gt; all &lt;strong&gt;visible light&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17494" LastEditorUserId="10185" LastEditDate="2015-07-15T10:48:21.723" LastActivityDate="2017-07-03T14:45:54.090" Title="A compound that absorbs all visible light" Tags="&lt;spectroscopy&gt;&lt;optical-properties&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34008" PostTypeId="1" AcceptedAnswerId="34009" CreationDate="2015-07-15T10:29:52.330" Score="11" ViewCount="1503" Body="&lt;p&gt;How can one get 3,4-dimethylhex-3-ene from but-2-ene ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$&lt;img src=&quot;https://i.stack.imgur.com/VJEdg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\ce{Br_2}$ then two equivalents of $\ce{CH_3CH_2MgBr}$, I can get the corresponding alkane (3,4-dimethylhexane) but I don't see how to do it while preserving the double bond...&lt;/p&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="46306" LastEditDate="2018-02-20T09:06:51.823" LastActivityDate="2021-03-12T01:57:48.373" Title="Getting 3,4-dimethylhex-3-ene from but-2-ene" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;grignard-reagent&gt;&lt;c-c-addition&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34009" PostTypeId="2" ParentId="34008" CreationDate="2015-07-15T10:54:35.500" Score="9" Body="&lt;p&gt;Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This approach removes the alkene and there is no obvious way to get it back (there is a way)&lt;/li&gt;&#xA;&lt;li&gt;More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases, so elimination can compete, but more importantly, they can undergo metathesis:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\ce{R-Br + R'-MgBr &amp;lt;=&amp;gt; R-MgBr + R'Br}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming your approach worked to generate 3,4-dimethylhexane, you could carefully radically brominate it to produce 3,-bromo-3,4-dimethylhexane, and I am sure you can get back to the alkene from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better approach uses two molecules of 2-butene. Do you see how you can break your target molecule up into two 4-carbon pieces? Convert one into an electrophile and the other into a nucleophile. It might take two reactions in each case. If you choose your electrophile correctly (think about what makes a great combo with a Grignard reagents), you will form the $\ce{C-C}$ bond between the two pieces and you will still have a functional group left that you can use as a handle to direct the elimination reaction to regenerate the alkene.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="10185" LastEditDate="2015-07-15T11:15:44.927" LastActivityDate="2015-07-15T11:15:44.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34010" PostTypeId="2" ParentId="34007" CreationDate="2015-07-15T11:03:05.233" Score="13" Body="&lt;p&gt;Any &lt;em&gt;black&lt;/em&gt; compound absorbs photons in all the visible spectrum; that is why essentially it appears black to our eyes. So, for instance, &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron(II,III)_oxide&quot;&gt;iron(II,III) oxide&lt;/a&gt;, $\ce{Fe3O4}$ will do so; it is even used as a black pigment. It is very unlikely though that it will absorb each and every photon of any wavelength in the visible spectrum, but the search for such a perfect absorber is meaningless in any case since the very notion of a visible spectrum is not strict.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-15T11:12:04.563" LastActivityDate="2015-07-15T11:12:04.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34011" PostTypeId="2" ParentId="33965" CreationDate="2015-07-15T11:10:35.947" Score="3" Body="&lt;p&gt;I found this reaction between amines and silver nitrate:&lt;img src=&quot;https://i.stack.imgur.com/Q9NOH.jpg&quot; alt=&quot;reaction&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17422" LastActivityDate="2015-07-15T11:10:35.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34012" PostTypeId="2" ParentId="34004" CreationDate="2015-07-15T11:10:46.460" Score="3" Body="&lt;p&gt;Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.&#xA;&lt;hr&gt;&#xA;As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$&#xA;We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three &lt;strong&gt;ideal&lt;/strong&gt; gases, as &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro%27s_law#Molar_volume&quot; rel=&quot;nofollow&quot;&gt;Avogadro's law&lt;/a&gt; says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that not only &lt;strong&gt;R and T are constant&lt;/strong&gt; but that the &lt;strong&gt;volume will also not change&lt;/strong&gt;, which in turn simplifies everything a lot.&#xA;$$n=p~\frac{V_m}{R~T}=p~k$$&#xA;Let's insert it into the mole fraction (as I told you to do) and see what happens.&#xA;$$\begin{align}&#xA;x_\ce{Ar}=&amp;amp;~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\&#xA;=&amp;amp;~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&amp;amp;~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&amp;amp;~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the simple reason why you do not need more than your pressures.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-16T09:22:58.653" LastActivityDate="2015-07-16T09:22:58.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34013" PostTypeId="2" ParentId="34007" CreationDate="2015-07-15T11:15:13.260" Score="51" Body="&lt;p&gt;Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The record is currently held by Vantablack&lt;sup&gt;[1]&lt;/sup&gt;, a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is &lt;em&gt;really&lt;/em&gt; black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnafIm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previously, the record was held by Superblack&lt;sup&gt;[2,3]&lt;/sup&gt;, a nickel-phosphorus alloy with an absorbance of up to 99.65%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, regular black paint generally absorbs around 97.5% of visible light.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sources&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;https://en.wikipedia.org/wiki/Vantablack&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Vantablack&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black&quot; rel=&quot;noreferrer&quot;&gt;https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract&quot; rel=&quot;noreferrer&quot;&gt;http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="16683" LastEditDate="2017-07-03T14:45:54.090" LastActivityDate="2017-07-03T14:45:54.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34014" PostTypeId="2" ParentId="34007" CreationDate="2015-07-15T11:47:05.277" Score="22" Body="&lt;p&gt;All metals are capable of absorbing photons of any wavelength below hard ultraviolet, as ideally there are allowed electronic transitions of arbitrarily small energy between states in the unfilled valence band. This means metals are theoretically the materials with the widest wavelength range for photon absorption (except for plasmas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this comes to odds with the idea most people have of metals, which is a collection of shiny (highly reflective) grey solids. This is because while they &lt;em&gt;can&lt;/em&gt; absorb photons of almost arbitrary wavelength, the absorption &lt;em&gt;probability&lt;/em&gt; is not as high as in other materials. As such, a secondary effect can take place where the incoming photons interact with the free electrons in the metal and cause the light to be reflected back, rather than absorbed. One simple way to decrease reflection and hence turn pure metals into black substances is to grind them into a very fine dust or otherwise make a very rough surface on the microscopic level, which causes incident light to bounce around more times when hitting the material, increasing the odds of the light being absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, while we often think of metals as materials which are composed chiefly of atoms from the $s$/$d$/$f$/early $p$ blocks of elements from the periodic table (the &quot;metallic elements&quot;), metals are actually not defined by their composition at all, but by their electronic structure. It is even possible to have metals made out of &quot;non-metals&quot; such as pure carbon, for example (carbon nanotubes of certain chiralities). Thus metallic character is actually likely a contributing factor to the darkness of Vantablack, which bon mentions in his answer.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-07-15T11:50:40.880" LastActivityDate="2015-07-15T11:50:40.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34016" PostTypeId="2" ParentId="34008" CreationDate="2015-07-15T12:58:52.400" Score="4" Body="&lt;p&gt;Ben is right that your product is simple dimer of substrate. However, there is much easier approach to conduct dimersisation of butene - it happens easily if you use strongly acidic catalyst, like conc. sulfuric acid or &lt;a href=&quot;https://en.wikipedia.org/wiki/Ion-exchange_resin&quot; rel=&quot;nofollow&quot;&gt;ion-exchange resin&lt;/a&gt;, such reaction is used for example in production of &lt;a href=&quot;https://en.wikipedia.org/wiki/2,2,4-Trimethylpentane&quot; rel=&quot;nofollow&quot;&gt;isooctane&lt;/a&gt;. It's similar to cationic polymerization of alkenes, so parameters have to be carefully adjusted to reduce creation of heavier oligomers.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-07-15T12:58:52.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34017" PostTypeId="2" ParentId="33760" CreationDate="2015-07-15T13:27:57.517" Score="3" Body="&lt;p&gt;I actually found the answer in this &lt;a href=&quot;https://web.archive.org/web/20121229075337/http://lcg.princeton.edu:80/publications/conference-publications/law-on-the-off-stoichiometric-peaking-of-adiabatic-flame-temperature.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;2011 paper&lt;/a&gt;, here is a quote from the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The characteristic rich shifting of the maximum adiabatic flame temperature from the stoichiometric value for&#xA;  hydrocarbon/air mixtures is demonstrated to be caused by product dissociation and hence reduced amount of heat release.&#xA;  Since the extent of dissociation is greater on the lean side as a result of the stoichiometry of dissociated products, the peaking&#xA;  occurs on the rich side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17369" LastEditorUserId="5026" LastEditDate="2018-02-28T14:48:36.003" LastActivityDate="2018-02-28T14:48:36.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34018" PostTypeId="1" CreationDate="2015-07-15T13:38:24.067" Score="1" ViewCount="120" Body="&lt;p&gt;I'm having a hard time figuring out how to do this and yes, I have the molecular weights, just don't know how to calculate. Been 20 years since college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm working with Copper ores. One customer reported Mo% as 5.2, S% as 5.3. The other customer reports $\ce{MoS2}$ values (this particular one is 12.34%) I need to compare these values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT #2::  I need to report 5.2% Mo and 5.3% Sulfur as ONE value - As &quot;MoS2&quot; if that is possible.&lt;/p&gt;&#xA;" OwnerUserId="17501" LastEditorUserId="17501" LastEditDate="2015-07-15T19:15:19.237" LastActivityDate="2015-07-16T08:52:18.633" Title="I need to report MoS2 values from Mo and Sulfur numbers" Tags="&lt;inorganic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34019" PostTypeId="1" AcceptedAnswerId="34036" CreationDate="2015-07-15T14:48:44.297" Score="2" ViewCount="370" Body="&lt;p&gt;Can $\ce{Ph-CH2-Ph}$ be also be named as  as (phenyl methyl) benzene instead of diphenyl methane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came to this thinking that $\ce{CH2-Ph}$ can be considered as side chain at side chain.&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="4945" LastEditDate="2015-09-07T07:16:19.167" LastActivityDate="2015-09-07T07:16:19.167" Title="What is the correct nomenclature for diphenyl methane?" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34020" PostTypeId="2" ParentId="34006" CreationDate="2015-07-15T15:03:47.153" Score="7" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Question 1&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;No. Bleach requires something to react &lt;em&gt;with&lt;/em&gt; in order for it to be consumed. There will still be bleach residues on the material.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Question 2&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;I think this question is actually based on a misperception, likely due to the multiple senses of the English word '&lt;a href=&quot;http://www.merriam-webster.com/dictionary/neutralize&quot; rel=&quot;nofollow noreferrer&quot;&gt;neutralize&lt;/a&gt;':&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;neutralize, &lt;em&gt;verb&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;: to stop (someone or something) from being effective or harmful&lt;/p&gt;&#xA;&lt;p&gt;: to cause (a chemical) to be neither an acid nor a base&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While you are completely correct that adding acid to sodium hypochlorite would neutralize it in the second sense, you would in fact be doing the &lt;em&gt;opposite&lt;/em&gt; of neutralizing it in the first sense. Your 'other source' is correct -- adding any acid to sodium hypochlorite will shift the equilibrium toward &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypochlorous_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypochlorous acid&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{HOCl}$&lt;/span&gt;), which in turn &lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorine#Hydrolysis_of_free_chlorine_or_disproportionation_in_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;equilibrates with molecular chlorine&lt;/a&gt; (&lt;span class=&quot;math-container&quot;&gt;$\ce{Cl_2}$&lt;/span&gt;), which is a nasty character you don't want to mess with&lt;sup&gt;&lt;span class=&quot;math-container&quot;&gt;$\dagger$&lt;/span&gt;&lt;/sup&gt;.&lt;/p&gt;&#xA;&lt;p&gt;What you need in order to neutralize it in the first sense, since bleach is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxidizing_agent&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxidizer&lt;/a&gt;, is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Reducing_agent&quot; rel=&quot;nofollow noreferrer&quot;&gt;reducing agent&lt;/a&gt;.  My recommendation would be to, say, crush up a few dye-free vitamin C (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ascorbic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ascorbic acid&lt;/a&gt;) tablets, dissolve in a bucket of water, and soak the article to be treated for half an hour or so.  Note that I've not ever actually tried this, so I have no idea if it will work.  It might also work to pour such a vitamin C solution into the 'liquid fabric softener' fill area of a washing machine, if it has one.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I just came across &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4945/how-much-ascorbate-to-neutralize-one-liter-of-chlorinated-tap-water&quot;&gt;this Chem.SE question&lt;/a&gt; from two years ago, which confirms that ascorbic acid is suitable for scrubbing of chlorine/hypochlorite.&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;&lt;span class=&quot;math-container&quot;&gt;$\dagger$&lt;/span&gt;&lt;/sup&gt; &lt;em&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/em&gt;: &lt;em&gt;Technically, since &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl2}$&lt;/span&gt; is volatile, soaking the material in an acid like vinegar &lt;strong&gt;would&lt;/strong&gt; probably eventually result in all of the bleach/chlorine coming out of the material, making the actual answer to Question 1 a 'conditional yes.' However, this is a really bad idea from a health/safety perspective.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-22T02:39:23.533" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34021" PostTypeId="1" CreationDate="2015-07-15T15:42:14.727" Score="1" ViewCount="25" Body="&lt;p&gt;Since some bases are not alkaline, we can't check whether they give $\ce{OH-}$ ions or not. Ethanol also has a $\ce{-OH}$ group. Why isn't it a base?&lt;/p&gt;&#xA;" OwnerUserId="17505" LastEditorUserId="10185" LastEditDate="2015-07-15T15:59:30.743" LastActivityDate="2015-07-15T15:59:30.743" Title="Confusion between bases (not alkali) and other -OH containing compounds" Tags="&lt;acid-base&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-07-15T17:58:38.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="34022" PostTypeId="1" CreationDate="2015-07-15T15:42:48.477" Score="7" ViewCount="2301" Body="&lt;p&gt;How can we balance oxygen or any other diatomic molecule with a fractional coefficient in a reaction?  I mean, if we can balance oxygen atoms in an odd number what happens to other oxygen atom, and can this sort of reaction even happen?&lt;/p&gt;&#xA;" OwnerUserId="17506" LastEditorUserId="11367" LastEditDate="2015-07-15T15:58:40.327" LastActivityDate="2017-10-24T15:43:56.840" Title="How can we balance one-half of an oxygen molecule, or any other diatomic, in a chemical reaction?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34023" PostTypeId="2" ParentId="34022" CreationDate="2015-07-15T15:57:25.143" Score="10" Body="&lt;p&gt;Whenever a chemical reaction is written with fractional coefficients, implicitly it's meant to be interpreted as the amount of substance &lt;em&gt;in moles&lt;/em&gt; of each species that is involved.  So, it's not really one half of an oxygen &lt;em&gt;molecule&lt;/em&gt; reacting, but &lt;em&gt;0.5 moles&lt;/em&gt; of $\ce{O2}$ reacting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bon's comment is spot-on -- if you are trying to think about a reaction written this way on a molecular level, you need to multiply the reaction through by whatever constant eliminates all the fractions.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2017-10-24T15:43:56.840" LastActivityDate="2017-10-24T15:43:56.840" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="34025" PostTypeId="1" CreationDate="2015-07-15T16:25:29.217" Score="6" ViewCount="333" Body="&lt;p&gt;Consider 5-cyano-2,2-dimethylpentyl 4-methylbenzene-1-sulfonate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/gITVWwI.png&quot; alt=&quot;5-cyano-2,2-dimethylpentyl 4-methylbenzene-1-sulfonate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would it yield by a Lemieux-Johonson oxydation ($\ce{OsO4}$, $\ce{NaIO4}$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess would have been that $\ce{OsO4}$ was likely to attack any of the bonds in the cycle (since they are delocalized); but would it stop here, or attack the remaining bonds? Or would something even different happen?&lt;/p&gt;&#xA;" OwnerUserId="5591" LastEditorUserId="4945" LastEditDate="2017-12-13T08:41:30.223" LastActivityDate="2017-12-13T08:41:30.223" Title="Lemieux-Johnson oxidation on cycles" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34026" PostTypeId="1" CreationDate="2015-07-15T16:27:43.390" Score="-1" ViewCount="858" Body="&lt;p&gt;I have to use dilute salt solutions incredibly frequently, and it takes valuable time to make them from pure reagents every time I need them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are concentrated salt solutions fairly stable over time?&#xA;Does anyone know of any papers describing their ability to maintain static concentrations for long periods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To qualify, I am curious about a wide range of salts. NaCl, ammonium chloride, magnesium sulfate, calcium chloride, potassium phosphate etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also not especially concerned about analytical precision for my use, but understanding how this could be affected might be useful to others.&lt;/p&gt;&#xA;" OwnerUserId="17374" LastEditorUserId="865" LastEditDate="2015-10-30T17:00:30.740" LastActivityDate="2016-05-25T20:15:45.840" Title="Are concentrated salt solutions stable over time?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34027" PostTypeId="1" CreationDate="2015-07-15T16:31:00.567" Score="1" ViewCount="665" Body="&lt;p&gt;According to the MSDS for NADPH Tetrasodium Salt, it says to store it in a place away from heat and light. Does this include UV light or just visible light? I was told that a pure NADPH stock solution (with pH 7 phosphate buffer as solvent) could be re-used after absorbing UV light at 340 nm, but based on the MSDS it seems that shouldn't be the case. So my question is: could I reuse a sample of pure NADPH stock solution that was exposed to UV for my enzyme activity assay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;datasheets.scbt.com/sc-202725.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.interprise.com.br/produtos/download/90289.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.interprise.com.br/produtos/download/90289.pdf&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="15476" LastEditorUserId="10185" LastEditDate="2015-07-15T16:34:38.647" LastActivityDate="2015-07-15T20:09:26.650" Title="Is NADPH sensitive and unstable in UV light?" Tags="&lt;biochemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34028" PostTypeId="1" AcceptedAnswerId="34030" CreationDate="2015-07-15T16:31:54.257" Score="4" ViewCount="2668" Body="&lt;p&gt;I have come across a following line in my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{PI5}$ does not exist due to steric hindrance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I searched for the explanation of steric hindrance in Wikipedia but couldn't understand it. Can anyone explain steric hindrance in simpler language?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The existence of $\ce{PI5}$ has been claimed intermittently since the&#xA;  early 1900s. The claim is disputed: &quot;The pentaiodide does not exist&#xA;  (except perhaps as $\ce{PI3\cdot I2}$, but certainly not as&#xA;  $\ce{PI4+I-}$)&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="17368" LastEditDate="2017-06-14T14:19:47.093" LastActivityDate="2017-06-14T14:19:47.093" Title="Explanation of steric hindrance and existence of PI5" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34029" PostTypeId="2" ParentId="34004" CreationDate="2015-07-15T16:40:46.667" Score="1" Body="&lt;p&gt;The molar fraction of a gas can be calculated as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_a=\frac{n_a}{n}= \frac{V_a}{V} = \frac{p_a}{p}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, firstly you need to calculate the partial pressure of Argon. Dalton's law states that in a mixture of non-reacting gases the total pressure exerted equals the sum of the partial pressures of the individual gases.&#xA;Thus $p = p_{He} + p_{Ne} + p_{Ar} $, from which you get $ p_{Ar} $ = 4,9 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the molar fraction of Argon is &#xA;$&#xA;x_a = \frac{p_a}{p} =\frac{4.90 atm}{8.40 atm} = 0,583&#xA;$ &lt;/p&gt;&#xA;" OwnerDisplayName="user17508" LastActivityDate="2015-07-15T16:40:46.667" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34030" PostTypeId="2" ParentId="34028" CreationDate="2015-07-15T16:47:16.553" Score="4" Body="&lt;p&gt;The simplest explanation is that you can not fit five iodine atoms around a phosphorous atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphorus_pentachloride&quot; rel=&quot;nofollow&quot;&gt;pentachloride&lt;/a&gt; equivalent - this is already quite crowded. Since chlorine atoms have a smaller covalent radius than iodine, you can get five of them around the phosphorous. Of course, all this depends on the bond length of P-I and P-Cl and the size of the phosphorous atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, there seems to be some suggestion of an Uranium Pentaiodide compound, for example &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/71308293&quot; rel=&quot;nofollow&quot;&gt;this PubChem compound&lt;/a&gt;. However, there are no references on that page, so presumably this has never been synthesised either?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, steric hindrance means geometric conformations of a molecule that are impossible (or difficult to achieve) due to clashes between groups and atoms within the molecule.&lt;/p&gt;&#xA;" OwnerUserId="4665" LastEditorUserId="4665" LastEditDate="2015-07-15T16:54:31.117" LastActivityDate="2015-07-15T16:54:31.117" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34033" PostTypeId="2" ParentId="34006" CreationDate="2015-07-15T18:11:08.393" Score="8" Body="&lt;p&gt;After a thorough wash, rinse and dry cycle, little or no bleach should be left in the fabric, but there is &lt;a href=&quot;http://www.cleaningservicegotham.com/green-cleaning-articles/the-dangers-of-bleach/&quot; rel=&quot;noreferrer&quot;&gt;possible concern about the effect of chlorine bleach on organic compounds&lt;/a&gt;. However, if disinfection is needed, then the infinitesimally small risk of VOC* is outweighed by the elimination of microorganisms.&lt;/p&gt;&#xA;&lt;p&gt;If any remaining smell of chlorine bothers you, an extra water rinse should be helpful. @Brian suggests a reducing agent, and &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2S2O3}$&lt;/span&gt;, sodium thiosulfate, is inexpensive and &lt;a href=&quot;http://www.skepticalaquarist.com/chlorine-chloramine&quot; rel=&quot;noreferrer&quot;&gt;is commonly used to remove residual chlorine for aquaria&lt;/a&gt;. However, leaving garments to air-dry or airing them for a day would be as effective.&lt;/p&gt;&#xA;&lt;p&gt;(*) &amp;quot;Volatile organic compounds&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-16T08:24:30.217" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34036" PostTypeId="2" ParentId="34019" CreationDate="2015-07-15T19:56:38.573" Score="4" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.1.2.2&lt;/strong&gt; Systems composed of rings and chains (exclusive of linear phanes)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two methods are recognized to name systems composed of rings and chains&#xA;  (exclusive of linear phanes).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…) For selection of a preferred IUPAC name, see P-52.2.8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-52.2.8&lt;/strong&gt; Selection between a ring and a chain as parent hydride&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Method (1): “1,1'-methylenedibenzene” (preferred IUPAC name; ring preferred to chain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method (2): “diphenylmethane” (treats phenyl groups alike as substituents)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This example is is explicitly mentioned in the Blue Book.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/a5Kiy.png&quot; alt=&quot;1,1&amp;#39;-methylenedibenzene&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-07-15T20:04:23.863" LastActivityDate="2015-07-15T20:04:23.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34037" PostTypeId="2" ParentId="34027" CreationDate="2015-07-15T20:09:26.650" Score="1" Body="&lt;p&gt;NADPH isn't destroyed by absorbing 340 nm light; if it were, then spectrophotometry wouldn't be a good technique to monitor its concentration.  The primary stability concern with NADPH and NADH is acid-catalyzed decomposition.  A great 1986 article summarizing this phenomenon is available &lt;a href=&quot;http://www.clinchem.org/content/32/2/314.full.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some highlights from the article:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;At 30 &amp;deg;C, the pseudo-first order rate constant for NADPH degradation in aqueous solution goes from 0.5 min$^{-1}$ at pH ~3 to $10^{-5}$ min$^{-1}$ at pH 10.  At pH 7 the rate constant is $10^{-3}$ min$^{-1}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The rate is elevated slightly by higher concentrations of phosphate buffer, about 10x higher for 1 molar (!) phosphate vs. extrapolation to infinite-dilution  phosphate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A secondary pathway for NAD(P)H decomposition is oxidation by atmospheric air, but it is usually pretty minor compared to acid degradation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't think there will be a big problem with light destroying your NADPH.  You might be able to re-use it, but it depends how long you let it sit in solution, and at what temperature.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-15T20:09:26.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34039" PostTypeId="1" AcceptedAnswerId="34043" CreationDate="2015-07-15T20:32:10.440" Score="6" ViewCount="815" Body="&lt;p&gt;As a similar idea to the question &lt;a href=&quot;https://chemistry.stackexchange.com/q/34007/15489&quot;&gt;A compound that absorbs all visible light&lt;/a&gt;, I am curious if there is a compound (or compounds) that absorb UV-A radiation (shown below)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, preferable the compound would have relatively good transmission in the UV-B and Visible wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HvpgE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ccohs.ca/oshanswers/phys_agents/ultravioletradiation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-07-15T22:41:09.977" Title="A compound that only absorbs UV-A radiation" Tags="&lt;spectroscopy&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34040" PostTypeId="1" AcceptedAnswerId="34046" CreationDate="2015-07-15T20:48:54.560" Score="6" ViewCount="1394" Body="&lt;p&gt;It seems such a simple formula, that each electron shell can hold in principle two times a square number. I imagine there must be a geometrical reason for this, that's to do with the way electron orbitals &quot;fit&quot; around the nucleus. If this is so, what is it?&lt;/p&gt;&#xA;" OwnerUserId="17517" LastActivityDate="2018-07-03T17:44:50.470" Title="Each electron shell can hold up to 2n² electrons. Is there a geometric reason for this?" Tags="&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34041" PostTypeId="2" ParentId="33994" CreationDate="2015-07-15T20:50:43.077" Score="53" Body="&lt;p&gt;I'm a physicist, so apologies if the answer below is in a foreign language;  but this was too interesting of a problem to pass up.  I'm going to focus on a particular question:  If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region?  The TL;DR is that thermal effects are going to make this idea basically impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on a magnetic dipole &lt;span class=&quot;math-container&quot;&gt;$\vec{m}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\vec{B}$&lt;/span&gt; is the magnetic field.  Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point:  &lt;span class=&quot;math-container&quot;&gt;$\vec{m} = \alpha \vec{B}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is what we might call the &quot;molecular magnetic susceptibility.&quot;  Then we have &lt;span class=&quot;math-container&quot;&gt;$\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$&lt;/span&gt;.  But potential energy is given by &lt;span class=&quot;math-container&quot;&gt;$\vec{F} = - \vec{\nabla} U$&lt;/span&gt;;  which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy &lt;span class=&quot;math-container&quot;&gt;$U(\vec{r}) = - \alpha B^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we're talking about a sample of gas at a temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT} &#xA;$$&lt;/span&gt;&#xA;In the limit where &lt;span class=&quot;math-container&quot;&gt;$kT \gg \alpha B^2$&lt;/span&gt;, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample.  To get a significant difference in the density of oxygen from point to point, we have to have &lt;span class=&quot;math-container&quot;&gt;$\alpha B^2 \gtrsim kT$&lt;/span&gt;;  in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how high does this have to be?  The &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; we have defined above is approximately related to the molar magnetic susceptibility by &lt;span class=&quot;math-container&quot;&gt;$\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$&lt;/span&gt;; and so we have&lt;sup&gt;1&lt;/sup&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\chi_\text{mol} B^2 \gtrsim \mu_0 RT&#xA;$$&lt;/span&gt;&#xA;and so we must have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}.&#xA;$$&lt;/span&gt;&#xA;If you believe &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_susceptibility&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, the molar susceptibility of oxygen gas is &lt;span class=&quot;math-container&quot;&gt;$4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$&lt;/span&gt;;  and plugging in the numbers, we get a requirement for a magnetic field of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;B \gtrsim \pu{258 T}.&#xA;$$&lt;/span&gt;&#xA;This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines.  Even at &lt;span class=&quot;math-container&quot;&gt;$\pu{91 Kelvin}$&lt;/span&gt; (just above the boiling point of oxygen), you would need a magnetic field of almost &lt;span class=&quot;math-container&quot;&gt;$\pu{150 T}$&lt;/span&gt;; still well out of range.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules.  A better approximation could be found by using a magnetic analog of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation&quot; rel=&quot;noreferrer&quot;&gt;Clausius-Mossotti relation&lt;/a&gt;;  and if the gas gets sufficiently dense, then all bets are off.  &lt;/p&gt;&#xA;" OwnerUserId="17515" LastEditorUserId="61578" LastEditDate="2019-12-18T01:49:42.580" LastActivityDate="2019-12-18T01:49:42.580" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34042" PostTypeId="1" CreationDate="2015-07-15T21:56:12.060" Score="1" ViewCount="6549" Body="&lt;p&gt;(Note I've refocused a previous unanswered question into a more general question. You can find the original much more specific question &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33644/negative-and-zero-enthalpy&quot;&gt;here&lt;/a&gt;)&#xA;There seems to be two main ways of looking at the enthalpy of substances: molar enthalpy and heat of formation. What are the differences/advantages and disadvantages of each and are either one or more or less suitable for different purposes?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-08-14T23:26:33.323" Title="Molar enthalpy vs Enthalpy of formation?" Tags="&lt;thermodynamics&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34043" PostTypeId="2" ParentId="34039" CreationDate="2015-07-15T22:41:09.977" Score="4" Body="&lt;p&gt;Yes, there are many such compounds.  &quot;&lt;a href=&quot;http://www.skin-care-forum.basf.com/en/articles/sun-care/the-evolution-of-uva-protection/2013/09/04?id=5cb3fae5-1a42-49c7-96df-66d08d2465c0&amp;amp;mode=Detail&quot; rel=&quot;nofollow noreferrer&quot;&gt;The evolution of UVA protection&lt;/a&gt;&quot; is a nice article detailing the progress on UVA blockers.  I'm sure you're aware that  sun care is a big business ($8.7 billion USD worldwide in 2013), so there is no shortage of chemists working in the area. In fact, there is enough (medical and clinical) focus that UVA (315-400 nm) has been subdivided into UVA1 (340-400 nm) and UVA2 (315-340 nm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a figure from the above link showing the UV spectra of some compounds.  The top part shows UVA absorbers and the bottom half shows broad-spectrum UV absorbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBpyT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a table with the names of the compounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ke5AJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BMBM was an early UVA absorber (solid blue line), but was photochemically unstable (dashed blue line), in other words, its blocking power deteriorated with UV exposure time.  More recent UVA absorbers, like DHHB are photostable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UVA absorbers belong to many different structural classes and contain many different types of functional groups.  Here are few examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BMBM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NQAmu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;DHHB&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FohYk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TDSA&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GRHEJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-07-15T22:41:09.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34044" PostTypeId="2" ParentId="34042" CreationDate="2015-07-15T23:05:04.310" Score="-1" Body="&lt;p&gt;Molar enthalpy is the energy released when the substance undergoes combustion. It is essentially the energy required to fragment the substance into its component sub-compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat of formation is the theoretical difference between the energy each atom typically has at STP and the total energy that the compound actually has. Essentially, it is the energy that goes into formulating the bonds between the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar enthalpy is generally used a lot more in experimental chemistry because it can be measured directly, and it gives useful values on the energy that can be extracted from a substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat of formation is preferred in theoretical chemistry, as it is the theoretical maximum amount of energy able to be extracted in perfect conditions. &lt;/p&gt;&#xA;" OwnerUserId="17489" LastActivityDate="2015-07-15T23:05:04.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34045" PostTypeId="1" AcceptedAnswerId="34527" CreationDate="2015-07-16T00:57:57.017" Score="7" ViewCount="1552" Body="&lt;p&gt;The modified Arrhenius equation is used to express the rate constant in a chemical mechanism model I'm working with. The equations is as follows:&#xA;$$k_\mathrm{f} = A\times T^b\times\exp\left(-\frac{E_\mathrm{a}}{RT}\right)$$&#xA;The paper states that &quot;Units are Moles, cm3, Seconds, K, and Calories/Mole&quot; so what would be the final units of the rate constant? I'm a bit confused due to the presence of the exponential function. When I simply plug the values in as given in the paper the rate becomes huge when multiplied by the molar concentration, is this because it gives molecules per second? To give you and idea for the following reaction:&#xA;$$\ce{C2H5OH + OH &amp;lt;=&amp;gt;C2H4OH + H2O}$$&#xA;The Arrhenius constants are as follows:&#xA;$$A = 1.74E+11$$&#xA;$$b = 0.27$$&#xA;$$E_a = 600.0$$&#xA;I am yet to calculate the reverse reaction using Gibbs Free Energy, is it just equally as large and thus it all cancels out or are the final units really not in $\mathrm{mol\ s^{-1}}$?&lt;/p&gt;&#xA;" OwnerUserId="7317" LastEditorUserId="7317" LastEditDate="2015-07-31T17:25:13.727" LastActivityDate="2015-07-31T17:25:13.727" Title="Units in modified Arrhenius equation?" Tags="&lt;kinetics&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34046" PostTypeId="2" ParentId="34040" CreationDate="2015-07-16T01:23:05.343" Score="8" Body="&lt;p&gt;A good insight!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_orbital&quot;&gt;number of electrons in each orbital&lt;/a&gt; are double the odd numbers. The s orbital can hold 2, p holds 6, d holds 10 and f has 14: double 1, 3, 5 and 7. If you add the odd numbers, you get a square number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 3 = 4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 3 + 5 = 9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 3 + 5 + 7 = 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;doubling&lt;/em&gt; is due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot;&gt;Pauli exclusion principle&lt;/a&gt;; in each orbital, an electron of a specific energy can be in one of two spin states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to see spectral lines split in a magnetic field (&lt;a href=&quot;http://www.daviddarling.info/encyclopedia/Z/Zeeman_effect.html&quot;&gt;Zeeman effect&lt;/a&gt;) or an electrostatic field (&lt;a href=&quot;http://www.agkador.phy.uni-bayreuth.de/en/projects_web/Single-Molecule_Spectroscopy/index.html&quot;&gt;Stark effect&lt;/a&gt;): one line may become 3, 5 or 7 lines. It clearly shows how orbital theory was derived.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-16T01:23:05.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34048" PostTypeId="1" AcceptedAnswerId="34052" CreationDate="2015-07-16T02:47:50.473" Score="7" ViewCount="465" Body="&lt;p&gt;Molecular &lt;em&gt;composition&lt;/em&gt; can be determined in some &quot;primitive&quot; ways (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/771/how-were-elements-and-molecular-formul%C3%A6-determined-in-the-old-days&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15386/how-were-the-first-molecular-formulas-determined&quot;&gt;2&lt;/a&gt;). What about the molecular structure? Is there a way to know what's bonded to what without spectroscopy etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;modern&quot; I mean any method that emits and detects particles (photons, electrons, etc.).&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="9961" LastEditDate="2017-05-21T21:41:40.753" LastActivityDate="2017-05-21T21:41:40.753" Title="Can molecular structure be determined without modern characterization methods?" Tags="&lt;analytical-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34050" PostTypeId="2" ParentId="33953" CreationDate="2015-07-16T03:45:48.400" Score="2" Body="&lt;p&gt;If you'd like to get a better understanding of entropy and the second law of thermodynamics so that you have what you need to solve homework problems, see the following link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/&quot; rel=&quot;nofollow&quot;&gt;https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-16T03:45:48.400" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34051" PostTypeId="1" CreationDate="2015-07-16T04:28:32.960" Score="3" ViewCount="49" Body="&lt;p&gt;I've been working on Tin electroplating on the side for a while now. I've formulated a mix from my usual electroplating formula template. The mix stubbornly keeps on having the Tin ions form Tin Hydroxide with water. I've tried adding different organics and complexes in hopes of preventing this. But the Hydroxides keep on forming and so the mix does not provide enough Tin ions to allow electroplating at relatively low voltages (at least 16V DC is necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I finally relented and poured in my &quot;Liquid Tin&quot; &lt;a href=&quot;http://www.mgchemicals.com/downloads/pdf/specsheets/421.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.mgchemicals.com/downloads/pdf/specsheets/421.pdf&lt;/a&gt; into the mix. Liquid Tin has Stannous Fluoroborate and its the Fluoroborate that keeps the Tin from forming Hydroxides. The effect was really good as plating happens at less than 1V, but then again Liquid Tin is for Electroless plating. After a while it needs higher voltage to electroplate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that got me wondering, does the Fluoroborate ion get used up in Electroplating? If there's a danger with Fluorine gas escaping, please also tell me.&lt;/p&gt;&#xA;" OwnerUserId="15043" LastEditorUserId="15043" LastEditDate="2015-07-16T05:13:26.393" LastActivityDate="2015-07-16T05:13:26.393" Title="Does Fluoroborate gets used up in Electroplating" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34052" PostTypeId="2" ParentId="34048" CreationDate="2015-07-16T05:00:27.193" Score="12" Body="&lt;p&gt;Structure elucidation via degradative studies and decomposition chemistry has sadly all but passed from the chemists' toolbox, but this was the method once used to determine molecular structures up until about half a century ago. One of my very favourite examples that I use to teach structure determination by NMR is the story of strychnine. &lt;a href=&quot;https://en.wikipedia.org/wiki/Strychnine&quot; rel=&quot;noreferrer&quot;&gt;Strychnine&lt;/a&gt; (below) is by no means a trivial structure; it is hard enough to solve by 'modern' spectroscopic means. The story of how &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Robinson_%28organic_chemist%29&quot; rel=&quot;noreferrer&quot;&gt;Robinson&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Burns_Woodward&quot; rel=&quot;noreferrer&quot;&gt;Woodward&lt;/a&gt;, standing upon the shoulders of those before them, unravelled the the structure of this alkaloid over a decade and a half is an intriguing story, and one well worth &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/12068914&quot; rel=&quot;noreferrer&quot;&gt;investigating further.&lt;/a&gt; (AMBIX, Vol. 48, Part 3, November 2001)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pC5DN.jpg&quot; alt=&quot;Strychnine. Structure solved by chemical degradation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the structure is slowly pieced together through a series of &lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Degradation_reactions&quot; rel=&quot;noreferrer&quot;&gt;degradation reactions&lt;/a&gt;. It is a methodology that relies on chemical processes, reactivities and properties only. For instance, it was worked out in the late 19th century that strychnine contained an &lt;a href=&quot;https://en.wikipedia.org/wiki/Amide&quot; rel=&quot;noreferrer&quot;&gt;amide&lt;/a&gt; as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lactam&quot; rel=&quot;noreferrer&quot;&gt;lactam&lt;/a&gt; through a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrolysis&quot; rel=&quot;noreferrer&quot;&gt;hydrolysis reaction&lt;/a&gt; to give strychnic acid. Were it not a lactam, two separate products would have been formed, rather than the one. Suggestions of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Quinoline&quot; rel=&quot;noreferrer&quot;&gt;quinoline&lt;/a&gt; skeleton came from the reactivity of a dimethyl derivative in a series of chemical tests. Over the next 20 or so years, this was revised and superceded by more likely candidates. So on and so forth over a number of years, each little piece of the puzzle is fitted together, all through the lengthy process of chemical reactions to give smaller easier to identify products. A number of complicated framework structures were proposed, mulled over and discarded for a number of years, and final structure was tested through &lt;a href=&quot;https://en.wikipedia.org/wiki/Total_synthesis&quot; rel=&quot;noreferrer&quot;&gt;total synthesis&lt;/a&gt; reproduction. This whole process requires an incredibly talented chemist, and it simply makes my mind boggle at the thought of how these guys operated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And though, it can be argued that UV spectroscopy did play a role in the final jigsaw piece in 1947, the correct structure was proposed by Robinson in 1942, and total synthesis confirmed this in 1954. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just one example of determination of molecular structure through degradative or decomposition chemistry, and many complicated structures were determined this way. Thank goodness for high field NMR.&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="2462" LastEditDate="2015-07-16T22:37:01.447" LastActivityDate="2015-07-16T22:37:01.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34054" PostTypeId="1" AcceptedAnswerId="34055" CreationDate="2015-07-16T07:28:48.843" Score="0" ViewCount="8310" Body="&lt;p&gt;So apparently the Lewis structure of $\ce{SO_2}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G0Zl3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was my understanding that the central atom is the one that is more electronegative. And an atom is more electronegative the closer it is to Fluorine (top-right). Oxygen is definitely closer to fluorine than sulfur is. Then, why is sulfur the central atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, one thing that has been bugging me for a while: if there are two or more atoms for the central element, how do you make the Lewis structure?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="402" LastEditDate="2017-04-01T00:33:18.860" LastActivityDate="2017-04-01T00:33:18.860" Title="Why is sulfur the central atom in the Lewis structure for SO2?" Tags="&lt;lewis-structure&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34055" PostTypeId="2" ParentId="34054" CreationDate="2015-07-16T07:34:59.823" Score="1" Body="&lt;p&gt;I think you messed up something. Because it is exactly opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Central part of atom is normally &lt;strong&gt;least&lt;/strong&gt;(not the most) electronegative atom i.e. &lt;strong&gt;S&lt;/strong&gt; in this case. &lt;/p&gt;&#xA;" OwnerUserId="5456" LastActivityDate="2015-07-16T07:34:59.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34056" PostTypeId="1" CreationDate="2015-07-16T08:02:10.323" Score="3" ViewCount="4871" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Consider $\ce{PCl3}$, $\ce{CH2Cl2}$, $\ce{HCN}$, and $\ce{SiO2}$. Which of these molecules are polar substances?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I appear to be lacking on the concept of &quot;polar substance&quot;. I was just studying about Lewis structures and this question popped up in the book, but I'm not quite sure how is this related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that a chemical bond is polar when one atom is more electronegative than another. But my basic understanding tells me that a chemical bond is non-polar only when the two atoms are the same (like $\ce{F-F}$ or $\ce{O-O}$). But all the above options have different atoms, so the answer would be &quot;none&quot;, which is clearly wrong because that's not one of the options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I tell if a molecule is polar or not?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="16683" LastEditDate="2018-09-07T17:31:26.993" LastActivityDate="2018-11-07T14:40:36.317" Title="How can I tell if a molecule is polar or not?" Tags="&lt;molecules&gt;&lt;polarity&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34057" PostTypeId="1" AcceptedAnswerId="34059" CreationDate="2015-07-16T08:28:12.530" Score="1" ViewCount="638" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the oxidation number of $\ce{Fe}$ in $\ce{Fe2O3}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWRdV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright. So we have a metal and an non metal. This means that iron will lose electrons and give them to oxygen. So it will be a positive oxidation number for iron I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, how much? A wild guess is $2+$ because, well, each iron would is bound to two oxygens so I imagine each oxygen will take one electron each?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastActivityDate="2015-07-16T08:50:26.833" Title="Find oxidation number of a metal in a compound" Tags="&lt;homework&gt;&lt;ions&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34059" PostTypeId="2" ParentId="34057" CreationDate="2015-07-16T08:50:26.833" Score="4" Body="&lt;p&gt;Oxidation number of Fe is &lt;strong&gt;+3.&lt;/strong&gt; This is because oxygen has a charge of -2. There are 3 oxygen atoms. So net charge becomes -6. Thus to balance it, each of the 2 Fe atoms must have a charge of &lt;strong&gt;+3&lt;/strong&gt; &lt;em&gt;(-6 + 6 = 0)&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17422" LastActivityDate="2015-07-16T08:50:26.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34060" PostTypeId="2" ParentId="34018" CreationDate="2015-07-16T08:52:18.633" Score="2" Body="&lt;p&gt;As the molecular mass of molybdenum is $95.95~\mathrm{g/mol} \approx 96~\mathrm{g/mol}$ and the molecular mass of sulfur is $32.06~\mathrm{g/mol} \approx 32~\mathrm{g/mol}$ one can construct a ratio between both molecular masses that in turn can be solved easily.&#xA;$$\frac{96~a}{32~b}=\frac{5.2}{5.3}=0.98\approx1$$&#xA;Which gives us after the multiplication with 32/96 on both sides&#xA;$$\frac{a}{b}=1\cdot\frac{32}{96}=\frac{1}{3}$$&#xA;So customer 1's determined &quot;compound&quot; is not $\ce{MoS2}$ but more $\ce{MoS3}$. Now I would guess that they don't analyze the copper ore for the compound $\ce{MoS2}$ but for both elements $\ce{Mo}$ and $\ce{S}$ separately which is why they came up with $5.2~\%~\ce{Mo}$ and $5.3~\%~\ce{S}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, you should ask customer 2 to split up it's $12.34~\%~\ce{MoS2}$ into the percentages of each element and compare them with $5.2~\%$ and $5.3~\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the ratio for $\ce{MoS2}$ which is $60:40$ one would estimate those $12.34~\%$ to split up into $7.40~\%~\ce{Mo}$ and $4.94~\%~\ce{S}$ or into $6.16~\%~\ce{Mo}$ and $6.18\%~\ce{S}$ if it is more like $\ce{MoS3}$.&#xA;&lt;hr&gt;&#xA;Or to answer your question ... probably you can simply add both values to give you $10.5~\%$ which nonetheless will not be equal to $\ce{MoS2}$ based upon the data you are given!&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-16T08:52:18.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34062" PostTypeId="1" CreationDate="2015-07-16T09:34:59.837" Score="5" ViewCount="299" Body="&lt;p&gt;What factors influence the partition factors across a membrane for a chemical species? Wikipedia defines the partition coefficient as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the physical sciences, a partition-coefficient (P) or distribution-coefficient (D) is the ratio of concentrations of a compound in a mixture of two immiscible phases at equilibrium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we let these &quot;two immiscible phases&quot; be either side of the membrane, In this situation the equilibrium concentrations would be determined by the solution either side of the membrane? If there is water either side of the membrane would they have a P of 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a chemist (or biologist for that matter) so apologies for any misuse of terminology or ambiguity.&lt;/p&gt;&#xA;" OwnerUserId="1990" LastEditorUserId="7448" LastEditDate="2015-07-16T09:36:49.497" LastActivityDate="2016-03-23T10:17:38.340" Title="The partition coefficient for a chemical species across a membrane" Tags="&lt;solutions&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34065" PostTypeId="1" CreationDate="2015-07-16T11:14:42.627" Score="15" ViewCount="4181" Body="&lt;p&gt;Is it blueish, perfect white or some other color? I am interested because some celestial bodies are said to be covered with methane ice.&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="30836" LastEditDate="2019-04-13T14:26:29.983" LastActivityDate="2019-04-13T14:26:29.983" Title="What color is solid methane?" Tags="&lt;phase&gt;&lt;color&gt;&lt;hydrocarbons&gt;" AnswerCount="1" CommentCount="10" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34066" PostTypeId="1" AcceptedAnswerId="91814" CreationDate="2015-07-16T11:17:26.250" Score="24" ViewCount="13063" Body="&lt;p&gt;Is there any way through which I could selectively reduce one $\ce{NO2}$ group from a dinitro- or trinitro- compound to an amine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading Solomons and Frhyle and it says that &lt;em&gt;m&lt;/em&gt;-dinitrobenzene can be selectively reduced by using $\ce{H2S}$ in aqueous or alcoholic ammonia. It also says that when it is carried out with 2,4-dinitrotoluene it leads to reduction of the 4-nitro group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there any way to predict which nitro group will be reduced, and/or any mechanism related to this. I have been frantically searching in books and even an internet article would suffice.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="865" LastEditDate="2015-10-16T18:08:43.630" LastActivityDate="2019-05-04T03:19:45.233" Title="Selective nitro reduction of poly nitro compounds" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;amines&gt;&lt;nitro-compounds&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="34067" PostTypeId="1" CreationDate="2015-07-16T11:39:36.863" Score="4" ViewCount="634" Body="&lt;p&gt;Is there any mechanism available for reduction of alkyl azides to amines.&#xA;One of the reactions is conversion by $\ce{Na}$ in ethanol or by $\ce{LiAlH4}$.&#xA;But how do these proceed?&#xA;$\ce{LiAlH4}$ will produce a hydride ion which attacks where?&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="7448" LastEditDate="2015-10-16T18:04:37.000" LastActivityDate="2015-10-16T18:04:37.000" Title="Preparing an amine from an azide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;amines&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34068" PostTypeId="1" AcceptedAnswerId="34074" CreationDate="2015-07-16T11:50:06.783" Score="6" ViewCount="2545" Body="&lt;p&gt;I am not seing tBu listed as alcohol protection group in all of these places:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Protecting_group#Alcohol_protecting_groups&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Protecting_group#Alcohol_protecting_groups&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.synarchive.com/protecting-group/&quot; rel=&quot;nofollow&quot;&gt;http://www.synarchive.com/protecting-group/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.organic-reaction.com/organic-synthesis/protecting-groups/ether/&quot; rel=&quot;nofollow&quot;&gt;http://www.organic-reaction.com/organic-synthesis/protecting-groups/ether/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand it seems to be quite common in peptide synthesis to protect the alcohol of serine, threonine and tyrosine with tBu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is listed on the German Wikipedia page: &lt;a href=&quot;https://de.wikipedia.org/wiki/Schutzgruppe#Alkohole&quot; rel=&quot;nofollow&quot;&gt;https://de.wikipedia.org/wiki/Schutzgruppe#Alkohole&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I conclude from this that tBu is very rarely used in non-peptide syntheses? Or is it just plain missing from the English Wikipedia article?&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="4325" LastEditDate="2015-07-16T15:10:48.393" LastActivityDate="2015-07-16T15:10:48.393" Title="Why is tert-Butyl (tBu) often not listed as protecting group for alcohol?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34069" PostTypeId="1" CreationDate="2015-07-16T12:22:33.323" Score="2" ViewCount="9099" Body="&lt;p&gt;I am studying Chemistry by Zumdahl and Zumdahl (9th edition) and I have come across a question in Chapter 11 that I can't figure out the explanation for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Liquid A has vapor pressure&lt;/em&gt; x &lt;em&gt;, and liquid B has vapor pressure&lt;/em&gt; y &lt;em&gt;. What is the mole fraction of the liquid mixture if the vapor above the solution is 30% A by moles?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that Raoult's Law tells us that the total vapor pressure is χx + (1-χ)y in this case. I also assume that the total vapor must be 0.3x + 0.7y given the other constraint of the question. This would mean the answer should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.3x + 0.7y = χx + (1-χ)y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this leads to the answer χ = 0.3, whereas the text gives the answer as 0.3y/(0.7x+0.3y)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I got wrong? (Also sorry for not formatting it properly but I'm a noob here and I can't figure out the equation formatting).&lt;/p&gt;&#xA;" OwnerUserId="17531" LastEditorUserId="865" LastEditDate="2015-07-16T14:58:28.307" LastActivityDate="2015-07-16T14:58:28.307" Title="Using vapor mole fraction and pressure to determine liquid mole fraction" Tags="&lt;physical-chemistry&gt;&lt;solutions&gt;&lt;vapor-pressure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34071" PostTypeId="2" ParentId="34069" CreationDate="2015-07-16T13:56:04.920" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Raoult%27s_law&quot; rel=&quot;nofollow&quot;&gt;Raoult's law&lt;/a&gt; tells us the relationship between the composition of the liquid and the vapor pressure above it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_A=\chi_A P^\circ_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P_A$ is the partial pressure of $A$ in the vapor above the solution&lt;/li&gt;&#xA;&lt;li&gt;$P^\circ_A$ is the vapor pressure of pure $A$&lt;/li&gt;&#xA;&lt;li&gt;$\chi_A$ is the mole fraction of $A$ in the solution&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a two component solution $A$ and $B$ (I don't want to use $x$ to avoid confusion with $\chi$) we can then use Dalton's Law to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_T=P_A + P_B = \chi_A P^\circ_A +\chi_B P^\circ_B= \chi_A P^\circ_A +(1-\chi_A)P^\circ_B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from the definition of mole fraction we can derive a relation between the mole fraction in the vapor (from here on out $\phi$ to avoid confusion between vapor and solution, which is what your mistake is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi_A=\frac{n_A}{n_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $n$ is the number of moles, and since at constant temperature and volume, $n\propto P$...&#xA;$$\phi_A = \frac{n_A}{n_T}=\frac{P_A}{P_T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can rewrite $P_A = \phi_A P_T$ and $P_B=(1-\phi_A)P_T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are given $\phi_A=0.3$. Thus $\phi_B=0.7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go. Substitute $P_A=\chi_A P^\circ_A$ and $P_T=\chi_A P^\circ_A +(1-\chi_A)P^\circ_B$ into $P_A=0.3 P_T$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\chi_A P^\circ_A=0.3\left(\chi_A P^\circ_A +(1-\chi_A)P^\circ_B\right)$$&#xA;$$0.7\chi_A P^\circ_A=0.3(1-\chi_A)P^\circ_B$$&#xA;$$0.7\chi_A P^\circ_A + 0.3\chi_A P^\circ_B=0.3P^\circ_B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can get it from here.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-16T13:56:04.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34072" PostTypeId="1" AcceptedAnswerId="34077" CreationDate="2015-07-16T14:14:10.280" Score="0" ViewCount="1220" Body="&lt;p&gt;I have a question on how to determine activation energy from a plot using the Arrhenius equation. I am working as an assistant with general chemistry II students and their lab is the iodination of acetone. I am familiar with using a plot of $\ln(k)$ vs $1/T$ to get the slope of $E_\mathrm{a}/R$ to solve for $E_\mathrm{a}$. Their lab manual though tells then to plot $\ln(t)$ vs $1/T$ to determine $E_\mathrm{a}$. I was wondering if anyone could shed some light on the derivation of this equation. &lt;/p&gt;&#xA;" OwnerUserId="9987" LastEditorDisplayName="user7951" LastEditDate="2015-07-16T14:19:22.503" LastActivityDate="2015-07-16T16:19:01.230" Title="Using a modified Arrhenius plot to find activation energy" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34073" PostTypeId="1" AcceptedAnswerId="34075" CreationDate="2015-07-16T14:27:18.430" Score="-1" ViewCount="837" Body="&lt;p&gt;I have 2 questions in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the structure of the superoxides? More specifically, what is the structure of $\ce{KO_2}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the structure of and dipole moment for $\ce{XeF_6}$?&lt;/p&gt;&#xA;" OwnerUserId="17532" LastEditorUserId="4945" LastEditDate="2015-07-17T07:10:07.443" LastActivityDate="2015-07-17T07:10:07.443" Title="What are the structures of potassium superoxide and xenon hexafluoride?" Tags="&lt;crystal-structure&gt;&lt;structural-formula&gt;&lt;molecular-structure&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-07-16T17:35:38.423" ContentLicense="CC BY-SA 3.0" />
  <row Id="34074" PostTypeId="2" ParentId="34068" CreationDate="2015-07-16T15:10:35.303" Score="3" Body="&lt;p&gt;Protecting group selection is a balancing act of ease of preparation and deprotection. In common practice, use of a protecting group adds two steps to a synthetic route. Installation and removing a tert-butyl group requires stronger (acidic) conditions. Of course, it's a rock while in place and inert to nearly all conditions short of strong acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peptide synthesis is particularly amenable to use of tert-butyl protection of hydroxyl groups because (1) all of the suitably protected residues are commercially available and (2) the protecting groups have been designed to be globally deprotected with trifluoroacetic acid. These factors minimize the inherent inefficiency to protecting groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The workhorse protecting groups for alcohols are silyl ethers (e.g.trimethylsilyl, triethylsilyl, tertbutyldimethylsilyl, triisopropylsilyl, tertbutyldiphenylsilyl). The silyl ethers are easier to form and cleave than standard ethers and have excellent compatibility with reactions that require alcohol protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specialized (carbon) ethers can be useful as well, including unsubstituted or substituted benzyl ether (removed by oxidative or reductive conditions) and acetals (removed by mild acid treatment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more complete description of protecting groups (including tert-butyl), see the books by &lt;a href=&quot;http://onlinelibrary.wiley.com/book/10.1002/0470053488&quot; rel=&quot;nofollow&quot;&gt;Greene and Wuts&lt;/a&gt; or &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1588903761&quot; rel=&quot;nofollow&quot;&gt;Kocienski&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-07-16T15:10:35.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34075" PostTypeId="2" ParentId="34073" CreationDate="2015-07-16T15:17:57.747" Score="7" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Xenon_hexafluoride&quot; rel=&quot;noreferrer&quot;&gt;Xenon hexafluoride&lt;/a&gt;'s geometry is not easy to be described but it is assumed to be in $C_{3v}$-symmetry - not octahedral! - in gas phase which is supported by electron diffraction measurements and quantum chemical calculations. It crystallizes in at least six to seven different known structures. [1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is further written&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Konrad Seppelt, an authority on noble gas and fluorine chemistry, says, &quot;the structure is best described in terms of a mobile electron pair that moves over the faces and edges of the octahedron and thus distorts it in a dynamic manner.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On you question regarding the dipole moment - for the $C_{3v}$-symmetry $\ce{XeF6}$ the dipole moment is calculated  to be $0.46~\mathrm{D}$. [2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure of $\ce{KO2}$ can also be found on &lt;a href=&quot;https://en.wikipedia.org/wiki/Potassium_superoxide&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;. There you find the following picture, showing it in it's      body-centered cubic unit cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQ8rh.png&quot; alt=&quot;unit cell of potassium superoxide&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://dx.doi.org/10.1016/j.jfluchem.2006.04.014&quot; rel=&quot;noreferrer&quot;&gt;S. Hoyer, T. Emmler, K. Seppelt, &lt;em&gt;J. Fluor. Chem.&lt;/em&gt; &lt;strong&gt;2006&lt;/strong&gt;, &lt;em&gt;127 (10)&lt;/em&gt;, 1415-1422&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.crawford.chem.vt.edu/PDF/JChemPhys_102_3307.pdf&quot; rel=&quot;noreferrer&quot;&gt;T. D. Crawford, K. W. Springer, H. F. Schaefer III &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;1995&lt;/strong&gt;, &lt;em&gt;102 (8)&lt;/em&gt;, 3307-3311&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-17T07:02:02.487" LastActivityDate="2015-07-17T07:02:02.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34076" PostTypeId="1" CreationDate="2015-07-16T16:02:06.273" Score="6" ViewCount="6693" Body="&lt;p&gt;Nitrogen has no available d-orbitals. Yet, it forms compounds where its oxidation state is +5, like in nitric acid. How?&lt;/p&gt;&#xA;" OwnerUserId="17534" LastEditorUserId="46306" LastEditDate="2018-12-07T18:56:39.563" LastActivityDate="2018-12-07T18:56:39.563" Title="How can nitrogen achieve +5 oxidation state?" Tags="&lt;orbitals&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34077" PostTypeId="2" ParentId="34072" CreationDate="2015-07-16T16:19:01.230" Score="4" Body="&lt;p&gt;The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions).  For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense.  In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $t_{1/2}$ is what is meant by $t$:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Arrhenius equation:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$k = A_0 e^{\frac{-E_a}{RT}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;First-order kinetics:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$C = C_0 e^{-k t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = \frac{1}{t_{1/2}} \ln 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Combining:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;N.B.&lt;/em&gt;&lt;/strong&gt; If $t$ just means &lt;em&gt;time&lt;/em&gt;, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$. &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-16T16:19:01.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34078" PostTypeId="1" AcceptedAnswerId="34079" CreationDate="2015-07-16T18:14:33.390" Score="5" ViewCount="57381" Body="&lt;p&gt;I was having trouble drawing the Lewis structure for $\ce{NCO^-}$. I found this image online:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5c5Ou.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is structure number one preferred over structure number two? Why does carbon prefer to have a triple bond with nitrogen, rather than have a double bond with nitrogen and a double bond with oxygen?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="16632" LastEditDate="2015-07-16T18:43:42.243" LastActivityDate="2015-07-16T18:57:26.027" Title="The Lewis structure for NCO-" Tags="&lt;lewis-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34079" PostTypeId="2" ParentId="34078" CreationDate="2015-07-16T18:26:08.683" Score="7" Body="&lt;p&gt;If you count electrons and determine the formal charge on each atom, you find that in structure #1, the negative charge is on the oxygen.  Do the same exercise for structure #2 and you find that the negative charge is on nitrogen.  Since oxygen is more electronegative then nitrogen, the negative charge is more stable when its on the oxygen atom.  Therefore, structure #1 is more stable (preferred) than structure #2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structures #3-#5 are even less stable.  They either violate the octet rule (#4, #5) and / or have charges (sometimes multiple) on more than one atom (#3-#5).&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-07-16T18:57:26.027" LastActivityDate="2015-07-16T18:57:26.027" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34082" PostTypeId="1" CreationDate="2015-07-16T19:22:57.870" Score="0" ViewCount="3157" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the maximum number of double bonds a carbon atom can make?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, carbon has $4$ valence electrons so... it needs $4$ more to fulfill the octet rule right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a double bond yields $4$ electrons, does it mean that the carbon atom can make only one double bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OR&lt;/strong&gt; does it mean it can create up to TWO double bonds (because they would yield $8$ electrons which fulfill the octet rule)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about Calcium? I ask because Calcium doesn't follow the Octet rule: it only wants to have 4 electrons, so...&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="402" LastEditDate="2017-04-25T23:12:14.083" LastActivityDate="2017-04-25T23:12:14.083" Title="Maximum number of double bonds a carbon atom can make" Tags="&lt;bond&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34085" PostTypeId="1" CreationDate="2015-07-16T21:58:38.453" Score="1" ViewCount="341" Body="&lt;p&gt;By clarifying butter, e.g. by melting butter and filtering it, we rid ourselves of most of the protein in the butter, and are left with mostly fat. However, the separation of protein and fat is not complete. Some people cannot eat milk proteins due to allergic reactions, and cannot therefore use unrefined butter in their cooking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple &quot;home-kitchen&quot; procedure to test whether the filtration sufficiently removed the milk proteins? Any proteins in the ghee can be precipitated by adding acetic acid, but this is not specific to milk proteins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5334" LastEditorDisplayName="user7951" LastEditDate="2015-07-16T22:05:21.307" LastActivityDate="2015-07-16T22:47:27.537" Title="Simple procedure for showing presence of milk proteins in clarified butter (ghee)" Tags="&lt;home-experiment&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34086" PostTypeId="1" CreationDate="2015-07-16T22:18:16.243" Score="2" ViewCount="315" Body="&lt;p&gt;This question dates back to a time where I used to watch something called The Curiosity Show... Kinda like Bill Nye the Science Guy's show. I just can't recall the answer to this so I'm hoping that someone could shed some light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two glasses of equal size and volume, with equivalent volumes of orange juice and coffee, one beverage in each glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that I have a hypothetical non-stick, hydrophobic teaspoon on me, if I was to get my teaspoon and take a level teaspoon of coffee out of the coffee glass and put it in the glass of orange juice, give it a stir, and then get a level teaspoon of this orange juice plus coffee mix and put this teaspoon of liquid in the coffee glass and give it a stir... Do I now:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;have more orange juice in the coffee glass than coffee in the orange juice glass;&lt;/li&gt;&#xA;&lt;li&gt;have more coffee in the orange juice glass than orange juice in the coffee glass; or&lt;/li&gt;&#xA;&lt;li&gt;have equal amounts of coffee in the orange juice glass and orange juice in the coffee glass?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17299" LastEditorUserId="4945" LastEditDate="2015-07-17T09:21:08.293" LastActivityDate="2015-07-17T09:21:08.293" Title="Coffee and orange juice concentrations with a twist" Tags="&lt;everyday-chemistry&gt;&lt;concentration&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34087" PostTypeId="1" CreationDate="2015-07-16T22:42:40.287" Score="3" ViewCount="57" Body="&lt;p&gt;Consider the following reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{2 HF(aq) + Ca(OH)2(aq) -&amp;gt; CaF2(s) + 2 H2O(l)}$&lt;br&gt;&#xA;The reaction produced a fairly insoluble salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that this reaction proceeds entirely to completion as the acid and bases were initially in equimolar quantities with respect to their stoichiometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the resulting pH of the solution at equilibrium? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the entire neutralization reaction has taken place we have the following reactions taking place in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaF2(s) -&amp;gt; Ca^2+(aq) + 2F- (aq)}$ with some $K_\mathrm{sp}$ value&lt;br&gt;&#xA;and this reaction&lt;br&gt;&#xA;$\ce{F- (aq) + H2O(aq) -&amp;gt; HF(aq) + OH- (aq)}$ with some $K_\mathrm{b}$ value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about calculating the values of $\ce{F- (aq)}$? I ask because every time some $\ce{CaF2}$ breaks off those free ions are used in the base reaction with water which according to Le Chatelier would result in more $\ce{CaF2}$ dissolving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the chemically determined $K_\mathrm{sp}$ value in textbooks take into consideration the reaction of fluoride ions deprotonating water?&lt;/p&gt;&#xA;" OwnerUserId="17321" LastEditorDisplayName="user7951" LastEditDate="2015-07-17T10:28:01.353" LastActivityDate="2015-07-17T10:28:01.353" Title="Neutralization reaction with insoluble salt (weak acid, strong base)?" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-07-17T15:10:25.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="34088" PostTypeId="2" ParentId="34085" CreationDate="2015-07-16T22:47:27.537" Score="1" Body="&lt;p&gt;A classic test for protein is &lt;a href=&quot;https://en.wikipedia.org/wiki/Biuret_test&quot; rel=&quot;nofollow&quot;&gt;Biuret solution&lt;/a&gt;, using readily available reagents:  sodium hydroxide, copper sulfate and potassium sodium tartrate (Rochelle salt). The test is not specific for milk protein, but what &lt;em&gt;other&lt;/em&gt; protein would be in pure butter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use caution with the chemicals -- the first listed is quite caustic, and the second is poisonous. BTW, save some leftover Rochelle salt if you'd like &lt;a href=&quot;http://www.instructables.com/id/Rochelle-Salt/&quot; rel=&quot;nofollow&quot;&gt;to make a piezoelectric crystal&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-16T22:47:27.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34090" PostTypeId="2" ParentId="32881" CreationDate="2015-07-16T23:49:11.457" Score="4" Body="&lt;p&gt;If you can, use a digital vacuum guage. I never use a bubbler as it can draw air in and introduce oxygen into your material as a result.  Not ideal. You could try and use nitrogen through the capillary,  but trying to manage the flow balanced with the vacuum drawing it in is a pain. Plant vessels never use bubblers and they distil under vacuum just fine.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-07-16T23:49:11.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34094" PostTypeId="2" ParentId="28623" CreationDate="2015-07-17T05:52:47.447" Score="1" Body="&lt;p&gt;color is result of electron transfer between orbitals and anything that affect energy of these orbitals can change color of that specie, so iodine in different mediums have different interactions and because of that iodine orbitals have different energies in different medium,so iodine have different color in different mediums.   &lt;/p&gt;&#xA;" OwnerUserId="17536" LastActivityDate="2015-07-17T05:52:47.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34095" PostTypeId="1" AcceptedAnswerId="34096" CreationDate="2015-07-17T06:03:29.960" Score="4" ViewCount="4488" Body="&lt;p&gt;Let's look at question &lt;strong&gt;c&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;a&lt;/strong&gt;) Write a balanced equation for the reaction.&#xA;  $$\ce{2 NaOH + H3PO4 -&amp;gt; Na2HPO4 + 2 H2O (l)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;b&lt;/strong&gt;) When some crystals of $\ce{Na2HPO4}$ were dissolved in water, the $\mathrm{pH}$ of the resulting solution was found to be $9.5$. Calculate the hydrogen ion concentration of this solution.&#xA;  $$[\ce{H+}] = 10^\mathrm{-pH}\, ;\qquad [\ce{H+}] = 10^{-9.5} = \pu{3.2e-10 mol L-1}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;c&lt;/strong&gt;) Write an equation for the reaction of the $\ce{HPO4^2-}$ ion with water to account for the measured $\mathrm{pH}$.&#xA;  $$\ce{HPO4^2- + H2O (l) &amp;lt;=&amp;gt; H2PO4- + OH-}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I'm a bit confused with what's happening. It seems that water is acting as an acid in this reaction, and donating protons to the ion. Can someone just elaborate on this whole process please because I'm not really sure what is happening. Why is it $\ce{HPO}$ and not $\ce{H2PO}$ or $\ce{H3PO}$ (as is in the original equation?) Is it a step wise process and only one step included?&lt;/p&gt;&#xA;" OwnerUserId="15053" LastEditorUserId="41328" LastEditDate="2018-01-15T15:27:58.053" LastActivityDate="2018-01-15T15:27:58.053" Title="Why does hydrogen phosphate act as a base?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34096" PostTypeId="2" ParentId="34095" CreationDate="2015-07-17T06:48:26.420" Score="5" Body="&lt;p&gt;When hydrogen phosphate salts are dissolved in water there are two main equilibria formed. This is based on the fact, that hydrogen phosphate can act as a Brønsted–Lowry base, i.e. accept protons, or as an acid, i.e. donate protons. For water the same is true. In addition to this it can react with itself, which is known as the autoprotolysis of water:&#xA;$$\ce{H2O + H2O &amp;lt;=&amp;gt; H3+O + {}^{-}OH}\tag1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this knowledge you can write&#xA;\begin{align}&#xA;\ce{HPO4^2- + H2O &amp;amp;~&amp;lt;=&amp;gt; PO4^3- + H3+O}\tag2\\&#xA;\ce{HPO4^2- + H2O &amp;amp;~&amp;lt;=&amp;gt; H2PO4- + {}^{-}OH}\tag3\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a lesser extent there is also the following equilibrium happening:&#xA;$$\ce{HPO4^2- + 2H2O &amp;lt;=&amp;gt; H3PO4 + 2 {}^{-}OH}\tag4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the acidity constant you know that $\ce{HPO4^2-}$ will react mostly as a base, i.e. &lt;a href=&quot;https://en.wikipedia.org/wiki/Disodium_phosphate&quot; rel=&quot;noreferrer&quot;&gt;$\mathrm{p}K_\mathrm{a}(\ce{Na2HPO4})=12.35$&lt;/a&gt;, so the equilibrium $(3)$ will be predominant.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-07-17T06:48:26.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34097" PostTypeId="2" ParentId="26834" CreationDate="2015-07-17T07:38:26.933" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0022024874901080&quot; rel=&quot;nofollow&quot;&gt;This paper from 1974&lt;/a&gt; seems relevant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Optically clear single crystals of calcite up to 2 cm in length have been pulled from a Li2CO3 flux in the range of 50–75 wt% CaCO3 under a pressure of 1–15 atm of carbon dioxide. The necessary phase equilibria for this growth was experimentally verified. The best growth parameters were a pull rate of 0.25 mm/hr with a rotation rate of 28 rpm. Seed orientation and perfection are critically important.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-17T07:38:26.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34098" PostTypeId="1" AcceptedAnswerId="34102" CreationDate="2015-07-17T08:15:58.010" Score="1" ViewCount="1493" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A student forgot to add the reaction mixture to the round bottomed flask at $\pu{27 ^\circ C}$ but instead he placed the flask on the flame. After a lapse of time he realised his mistake and using a pyrometer, he found the temperature of the flask was $\pu{477 ^\circ C}$. What fraction of air would have been expelled out? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By solving $$\frac{V_1}{V_1 + V_2} = \frac{T_1}{T_2}$$ I found $$\frac{V_2}{V_1} = \frac{3}{2}.$$ But according to my book it is $\frac 35$. What exactly is this ratio?&lt;/p&gt;&#xA;" OwnerUserId="17547" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-22T02:34:40.117" LastActivityDate="2018-02-22T02:34:40.117" Title="What fraction of air gets expelled upon heating an empty flask?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34099" PostTypeId="1" AcceptedAnswerId="48330" CreationDate="2015-07-17T09:52:44.880" Score="15" ViewCount="1097" Body="&lt;p&gt;What property of the elements make them form different types of carbides like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Be}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Al}$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$\ce{Be2C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Al4C3}$&lt;/span&gt; (Methanides) contains &lt;span class=&quot;math-container&quot;&gt;$\ce{C^4-}$&lt;/span&gt; ion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Na}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca}$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$\ce{Na2C2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CaC2}$&lt;/span&gt; (acetylides) contains &lt;span class=&quot;math-container&quot;&gt;$\ce{C2^{2-}}$&lt;/span&gt; ion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Li}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg}$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$\ce{Li4C3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Mg2C3}$&lt;/span&gt; (sesquicarbides) contains &lt;span class=&quot;math-container&quot;&gt;$\ce{C3^4-}$&lt;/span&gt; ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boron - &lt;span class=&quot;math-container&quot;&gt;$\ce{B4C}$&lt;/span&gt; (covalent carbides)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titanium and tungsten - &lt;span class=&quot;math-container&quot;&gt;$\ce{TiC}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{WC}$&lt;/span&gt;  (interstitial carbides)&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="17175" LastEditDate="2019-11-12T02:50:20.653" LastActivityDate="2019-12-10T14:08:44.253" Title="Why do different elements form different types of carbides?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;ionic-compounds&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34100" PostTypeId="1" CreationDate="2015-07-17T10:20:37.387" Score="1" ViewCount="40" Body="&lt;p&gt;Why is the maximum theoretical efficiency of Ar engine higher than that of Nitrogen engine? On a molecular level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an Otto Cycle&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{Eff} \le\left(1-r^{1-\gamma}\right)$&lt;/li&gt;&#xA;&lt;li&gt;$\gamma=\dfrac{C_p}{C_v}$&lt;/li&gt;&#xA;&lt;li&gt;$r=\dfrac{v_1}{v_2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sys. Fluid &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitrogen, $\gamma=1.4,\ r=10,\ \mathrm{eff(max)}=60.2%$&lt;/li&gt;&#xA;&lt;li&gt;Argon, $\gamma=1.67,\ r=10,\ \mathrm{eff(max)}=78.5%$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14624" LastEditorUserId="275" LastEditDate="2015-07-17T13:25:47.870" LastActivityDate="2015-07-17T13:25:47.870" Title="Thermodynamic-Fuel and efficiency" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;fuel&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34102" PostTypeId="2" ParentId="34098" CreationDate="2015-07-17T13:48:38.797" Score="1" Body="&lt;p&gt;Your first equation should be:&#xA;$${V_1\over V_2} = {T_1\over T_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to know what $V_2$ is in terms of $V_1$. Here, I call $V_2$ final volume. Arranging the equation and using the values for temperature in kelvin:&#xA;$${\rm final\;volume} = {V_1 \times 750\over 300}= {5\over 2}V_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the amount that was expelled, we subtract the initial volume ($V_1$) from the final volume:&#xA;$${\rm amount\;expelled} = {5\over 2}V_1 - V_1 = {3\over 2}V_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction expelled is then:&#xA;$${{\rm amount\;expelled}\over{\rm final\;volume}} = {3\over 5} $$ &lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorDisplayName="user7951" LastEditDate="2015-07-17T14:19:03.137" LastActivityDate="2015-07-17T14:19:03.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34104" PostTypeId="1" CreationDate="2015-07-17T14:03:14.897" Score="2" ViewCount="7257" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&#xA;Given the first, second and third ionization energies of an element (7 eV, 12.5 eV and 42.5 eV respectively), find the most stable oxidation state (or charge on the resultant ion) of that element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think that the element has to ionize with oxygen to reach a certain oxidation state, since oxygen has a 6 valence electrons, the element has to lose 2 electrons to attain stability. So the second ionization energy will the give the most stable oxidation state. So my answer is +2.&lt;/p&gt;&#xA;&lt;p&gt;Now that happens to be the correct answer but can someone verify my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="83758" LastEditDate="2020-09-11T20:47:04.933" LastActivityDate="2020-09-11T22:41:32.780" Title="Determine stable oxidation state from Ionization energy" Tags="&lt;physical-chemistry&gt;&lt;oxidation-state&gt;&lt;ionization-energy&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34105" PostTypeId="1" CreationDate="2015-07-17T15:02:50.763" Score="5" ViewCount="562" Body="&lt;p&gt;Can anyone tell me the difference of aggregates and clusters from the molecular point of view? Thank you so much. &lt;/p&gt;&#xA;" OwnerUserId="17216" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T17:11:32.937" LastActivityDate="2015-10-23T17:11:32.937" Title="Difference of aggregate and cluster?" Tags="&lt;terminology&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34106" PostTypeId="1" AcceptedAnswerId="34111" CreationDate="2015-07-17T16:06:36.553" Score="19" ViewCount="6226" Body="&lt;p&gt;I do not mean at the same time, of course, but I mean it appears from an overview of the common charges formed from ionizing various elements that each element forms one or more of either positive or negative ions, but they never have the opposite charge. For example Fe may have +4 or +3, but never -anything. &#xA;&lt;strong&gt;However, I am aware that this is from a &lt;em&gt;common ion&lt;/em&gt; chart. Perhaps an atom can be a cation and an anion under specific circumstances?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="9961" LastEditDate="2017-04-24T17:54:48.617" LastActivityDate="2017-04-24T17:54:48.617" Title="Do atoms form either a positive or a negative charge, but not both?" Tags="&lt;ions&gt;&lt;oxidation-state&gt;" AnswerCount="6" CommentCount="2" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34108" PostTypeId="1" AcceptedAnswerId="34113" CreationDate="2015-07-17T16:19:24.270" Score="3" ViewCount="1161" Body="&lt;p&gt;According to my textbook, fluorine has a pretty high (compared to elements like sodium) first ionization energy. &lt;em&gt;But why does it have such a high requirement to ionize?&lt;/em&gt; After all, it actively seeks out atoms to bond with, even tearing through already existing bonds in its drive to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect that fluorine would have a &lt;strong&gt;very&lt;/strong&gt; low ionization energy, as it is &lt;em&gt;soooooooo&lt;/em&gt; close to having a full octet (just like sodium). Yet unlike sodium, it has a higher first ionization energy. Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My guess is that fluorine's protons' affinity for electrons is so great that it simply &quot;overpowers&quot; the octet rule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, &lt;em&gt;neon&lt;/em&gt; can also be ionized. Doesn't that mean it can bond with cations? But it is unusually stable, so why doesn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is an underlying law behind this that can explain these, please help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Fluorine has now been explained, but why can neon be ionized? After all, if you give it a positive charge by taking away electrons, won't it be able to bond to anions? But in reality, it doesn't. What is going on?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="15588" LastEditDate="2015-07-17T22:45:44.353" LastActivityDate="2015-07-17T22:45:44.353" Title="Does ionization energy have anything to do with how reactive an element is?" Tags="&lt;ions&gt;&lt;elements&gt;&lt;electronegativity&gt;&lt;ionization-energy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34109" PostTypeId="1" AcceptedAnswerId="34135" CreationDate="2015-07-17T16:28:18.780" Score="7" ViewCount="4611" Body="&lt;p&gt;What property of lithium makes them forms the nitride, lithium nitride $\ce{Li3N}$ whereas the other group 1 elements forms the azide, like sodium azide and potassium azide - $\ce{NaN3}$ and $\ce{KN3}$ respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does lithium azide exist? If so, what is its structure?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="17368" LastEditDate="2016-05-15T15:18:44.373" LastActivityDate="2017-01-23T05:25:29.703" Title="Why does lithium from the nitride whereas the other alkali metal forms the azide?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34110" PostTypeId="1" CreationDate="2015-07-17T16:43:30.480" Score="15" ViewCount="4899" Body="&lt;p&gt;I was researching about electronegativity when I looked up what a graph of electronegativity within the periodic table is. And, &lt;a href=&quot;http://www.mrteverett.com/chemistry/pdictable/electronegativity%20table.jpg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; appeared. I scanned it, matching up everything I knew about the elements up to that table. Yep, Group 1 is all light colored; yea, the nonmetals are all really dark blue culminating in fluorine; and yes, the noble gases are mostly zero (with the exception of the traitors &lt;a href=&quot;https://en.wikipedia.org/wiki/Xenon_hexafluoride&quot; rel=&quot;noreferrer&quot;&gt;xenon&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Noble_gas_compound#Krypton_compounds&quot; rel=&quot;noreferrer&quot;&gt;krypton&lt;/a&gt;, which coincidentally are also the only electronegative noble gases!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while I was scanning the &lt;a href=&quot;https://en.wikipedia.org/wiki/Noble_gas_compound&quot; rel=&quot;noreferrer&quot;&gt;page from Wikipedia&lt;/a&gt; on noble gas compounds, I suddenly realized that there was a compound called xenon hexafluoroplatinate, meaning that xenon could bond with...platinum? Checking the table, I was astonished to find that the &lt;strong&gt;entire noble metal group was actually more electronegative then the metals surrounding them!&lt;/strong&gt; &lt;strong&gt;Why?!&lt;/strong&gt; Aren't they extremely non-reactive? How then can they crave electrons even more then regular metals?&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="4231" LastEditDate="2015-10-10T23:18:45.800" LastActivityDate="2019-01-02T01:16:31.640" Title="Why are noble metals more electronegative then most metals?" Tags="&lt;metal&gt;&lt;transition-metals&gt;&lt;covalent-compounds&gt;&lt;electronegativity&gt;&lt;relativistic&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34111" PostTypeId="2" ParentId="34106" CreationDate="2015-07-17T16:57:04.890" Score="34" Body="&lt;p&gt;Actually, in theory almost all of the elements can be found with both positive and negative oxidation numbers: it's just a matter of finding a system with the proper reagents and conditions to force it. If you isolate chemical species which have a &lt;em&gt;very&lt;/em&gt; strong tendency of displaying some specific behaviour (accepting electrons, donating electrons, coordinating ions, releasing a leaving group, bonding to metals, releasing a proton, adopting a specific molecular geometry, or any other myriad properties), then you can often obtain strange results by clashing them with substances which also have the same tendency, but not quite as strong. This often causes the substance with the weaker behaviour to &quot;operate in reverse&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give a vivid and related example. &lt;em&gt;As we all know&lt;/em&gt;, the alkali metals (group 1 elements) are exclusively present as cations with an oxidation number of +1, except in the pure metals, where it's zero, &lt;em&gt;right&lt;/em&gt;? Well, here's something which might shatter your world: most of the alkali metals (with the exception of lithium, for now) also form &lt;a href=&quot;https://en.wikipedia.org/wiki/Alkalide&quot;&gt;&lt;strong&gt;alkalides&lt;/strong&gt;&lt;/a&gt;, that is, stable salts containing discrete, clearly observed $\ce{Na^{-}}$, $\ce{K^{-}}$, $\ce{Rb^{-}}$ or $\ce{Cs^{-}}$ anions, with alkali metals displaying oxidation number -1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this done? All you need is to find a neutral substance with a much stronger tendency to donate an electron than a neutral alkali metal atom (easier said than done). Since neutral alkali metals atoms form quite stable cations upon loss of an electron, this implies you need to search for a neutral substance capable of donating an electron and forming a cation with &lt;em&gt;exceptional&lt;/em&gt; thermodynamic and/or kinetic stability. This can be achieved, for example, by the use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Cryptand&quot;&gt;cryptands&lt;/a&gt;, which are cyclic molecules capable of coordinating very strongly to cations, strongly enough that they even coordinate alkali metal cations very well. The cryptand-coordinated cation is both thermodynamically and kinetically stable enough that alkalide anions, which would be &lt;strong&gt;extremely&lt;/strong&gt; reactive otherwise, are not reactive &lt;em&gt;enough&lt;/em&gt; in this case to immediately cause charge transfer and neutralize the negative charge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amusingly, it is actually possible to prepare a single compound which contains both alkali metal cations and anions, as exemplified by $\mathrm{[Na(2,2,2- crypt)]^{+} Na^{-}}$, which contains a sodium cation coordinated by a cryptand as a counterion to a &lt;em&gt;natride&lt;/em&gt;/&lt;em&gt;sodide&lt;/em&gt; ($\ce{Na^{-}}$) anion. One can imagine this compound being made by putting together a neutral sodium atom ($\ce{Na^{0}}$) and the neutral cryptand species $\mathrm{[Na(2,2,2- crypt)]^0}$. As Brian mentions in the comments, this latter species is actually an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electride&quot;&gt;electride&lt;/a&gt;, which can be written as $\mathrm{[Na(2,2,2- crypt)]^{+}e^{-}}$ and thought of as a salt where the anion is a lone electron (!). Both the neutral sodium atom and the electride have a strong tendency to lose an electron in chemical reactions, but this tendency is much stronger for the electride. Thus, the electride ends up having its way, forcing its very loosely bound electron onto the neutral sodium atom, causing the neutral sodium atom to &quot;operate in reverse&quot; and accept an electron rather than donate it, resulting in the $\ce{Na^{-}}$ anion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements&quot;&gt;This table on Wikipedia&lt;/a&gt; is far more complete than most &quot;common oxidation number&quot; tables out there, and it lists many negative oxidation numbers for elements, including iron! For many of the transition metals, negative metal oxidation numbers are achievable by using the carbonyl ($\ce{CO}$) ligand, which removes electron density from the metal atom via backbonding. This stabilizes negative charges on the metal atom, once again allowing the resultant species to survive with the appropriate counterion.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="7412" LastEditDate="2015-07-17T22:18:31.540" LastActivityDate="2015-07-17T22:18:31.540" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34112" PostTypeId="2" ParentId="34106" CreationDate="2015-07-17T17:21:00.600" Score="6" Body="&lt;p&gt;You are correct in saying that it is possible but not common. This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases. There are small changes in what you would expect to be a straight line trend, depending on what kind of orbital the electron &quot;resides&quot; in as well. The two concepts are shown here: &lt;a href=&quot;http://crystalmaker.com/support/tutorials/crystalmaker/atomic-radii/resources/CPK_Atomic_Radii.jpg&quot; rel=&quot;nofollow&quot;&gt;Atomic radii&lt;/a&gt; and &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/1175/effective_nuclear_charge.png?size=bestfit&amp;amp;width=614&amp;amp;height=328&amp;amp;revision=1&quot; rel=&quot;nofollow&quot;&gt;Zeff image&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From both of these concepts comes the reasoning:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Q: Why are some atoms harder to ionize to a positive charge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Too high of an ionization energy required because of a lot of nucleus interaction with the outer electron.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Q: Why are some atoms not seen ionizing to a negative charge? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Electrons shield too much of the nucleus to allow for enough interaction to hold another electron.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that this holds for normal day to day reactions, and why some elements are found in the oxidation state they are on earth. Once you get into an extreme energy environments there are more possibilities.&lt;/p&gt;&#xA;" OwnerUserId="17207" LastEditorUserId="17207" LastEditDate="2015-07-18T17:50:05.833" LastActivityDate="2015-07-18T17:50:05.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34113" PostTypeId="2" ParentId="34108" CreationDate="2015-07-17T18:05:18.030" Score="3" Body="&lt;p&gt;This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.&lt;/p&gt;&#xA;" OwnerUserId="17207" LastActivityDate="2015-07-17T18:05:18.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34114" PostTypeId="1" CreationDate="2015-07-17T18:19:34.567" Score="5" ViewCount="181" Body="&lt;p&gt;I'm currently downloading .cif files from various online databases and using openbabel to convert the fractional coordinates that are in the .cif files into .xyz files.  The typical output will look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spinel&#xA;\begin{array}{c|c c c}&#xA;\ce{Mg}      &amp;amp;   1.01054     &amp;amp;   1.01054   &amp;amp;     1.01054\\\hline&#xA;\ce{Al}      &amp;amp;   1.01054     &amp;amp; 1.01054     &amp;amp;   1.01054\\\hline&#xA;\ce{Al}      &amp;amp;   4.04217     &amp;amp;   4.04217   &amp;amp;     4.04217\\\hline&#xA;\ce{Mg}      &amp;amp;   4.04217     &amp;amp;   4.04217    &amp;amp;    4.04217\\\hline&#xA;\ce{O}       &amp;amp;   2.11390     &amp;amp;   2.11390    &amp;amp; 2.11390&#xA;\end{array}&#xA;My problem is that I would like the ionic radii of each element, instead of their name.  Does anyone know of a way to recover this information from the original CIF file or am I just stuck with looking up the radii manually on, &#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Ionic_radius&lt;/a&gt; ?  &lt;/p&gt;&#xA;" OwnerUserId="17399" LastEditorUserId="7448" LastEditDate="2015-07-17T18:24:41.183" LastActivityDate="2021-02-10T17:39:08.350" Title="Extracting &quot;Crystal Radii&quot; from a CIF file" Tags="&lt;crystal-structure&gt;&lt;software&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34115" PostTypeId="1" CreationDate="2015-07-17T18:42:28.363" Score="2" ViewCount="568" Body="&lt;p&gt;I've spent several hours searching the WWW with entries like &quot;stereoselective enolate formation&quot; and similar ones, but I find almost only entries for Li-amide bases (LDA, LMTP) and no other bases like &lt;strong&gt;&lt;em&gt;t&lt;/em&gt;-BuO, NMe3, NaH, &lt;em&gt;n&lt;/em&gt;-BuLi&lt;/strong&gt; concerning the stereoselectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if the addition of these &lt;strong&gt;four bases&lt;/strong&gt; to 3-pentanone all results with preference in the &lt;strong&gt;cis or trans enolate&lt;/strong&gt; (and if possible with an explanation why).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/33773/stereoselective-enolate-formation-with-different-bases-does-the-addition-of-a-b&quot;&gt;Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-08-16T23:11:55.490" Title="Does the addition of base to 3-pentanone give the cis or trans enolate?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34116" PostTypeId="2" ParentId="33845" CreationDate="2015-07-17T18:49:36.957" Score="1" Body="&lt;p&gt;I have found a partial answer to finding a large database of sphere packings, thanks to the comment from permeakra!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have been obtaining .cif files from the large databases of crystal structures online such as, &lt;a href=&quot;http://www.crystallography.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crystallography.net/&lt;/a&gt;.  Once I have the .cif files I use openbabel to convert those files to .xyz files.  From here I have a table that describes a crystal's structure by the x,y, and z coordinates of each particular element for example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c|c c c}&#xA;\ce{Mg}      &amp;amp;   1.01054     &amp;amp;   1.01054   &amp;amp;     1.01054\\\hline&#xA;\ce{Al}      &amp;amp;   1.01054     &amp;amp; 1.01054     &amp;amp;   1.01054\\\hline&#xA;\ce{Al}      &amp;amp;   4.04217     &amp;amp;   4.04217   &amp;amp;     4.04217\\\hline&#xA;\ce{Mg}      &amp;amp;   4.04217     &amp;amp;   4.04217    &amp;amp;    4.04217\\\hline&#xA;\ce{O}       &amp;amp;   2.11390     &amp;amp;   2.11390    &amp;amp; 2.11390&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my problem from here is that babel has failed to return the ionic radius which I suppose can be retrieved online via wikipedia or something.  I posted another question that is more specific to .cif files which expands on this problem which can be found here, &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34114/extracting-crystal-radii-from-a-cif-file%5C&quot;&gt;Extracting &amp;quot;Crystal Radii&amp;quot; from a CIF file&lt;/a&gt;.\&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limitations to this approach include that I can only find 3D sphere packings and it's a rather long and clunky procedure for obtaining lots of sphere packings.  &lt;/p&gt;&#xA;" OwnerUserId="17399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-07-17T18:49:36.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34117" PostTypeId="1" AcceptedAnswerId="34118" CreationDate="2015-07-17T19:14:42.930" Score="6" ViewCount="4094" Body="&lt;p&gt;My book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{NaCl}$ shows metal excess defect, which results into the formation of an F-center in the crystal. This F-center is responsible for the yellow color of $\ce{NaCl}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How does an F-center provide color to the solid?&lt;/p&gt;&#xA;" OwnerUserId="17007" LastEditorUserId="41328" LastEditDate="2017-08-06T23:34:45.093" LastActivityDate="2018-08-11T23:14:31.333" Title="Yellow color of NaCl" Tags="&lt;physical-chemistry&gt;&lt;solid-state-chemistry&gt;&lt;color&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34118" PostTypeId="2" ParentId="34117" CreationDate="2015-07-17T19:28:06.753" Score="10" Body="&lt;p&gt;An F-center (or any color center) is a point defect in the material that generates localized electronic states. When these states are in the band gap, they will result in strong peaks in optical absorption. An F-center in an alkali-halide material is an electron bound to a negative ion vacancy.  A brief overview may be found in &lt;a href=&quot;https://books.google.com/books/about/Solid_State_Physics.html?id=1C9HAQAAIAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ashcroft and Mermin's&lt;/a&gt; Solid State Physics, Chapter 30.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2018-04-03T13:46:45.940" LastActivityDate="2018-04-03T13:46:45.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34119" PostTypeId="1" CreationDate="2015-07-17T19:48:33.690" Score="6" ViewCount="34779" Body="&lt;p&gt;Consider this picture of an $\ce{NaCl}$ unit cell:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HBKGW.gif&quot; alt=&quot;NaCl Unit cell&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to show 14 $\ce{Cl-}$ ions and only 13 $\ce{Na+}$ ions.  Given that discrepancy, how is table salt balanced for charge?  Why isn't there a charge excess?&lt;/p&gt;&#xA;" OwnerUserId="17494" LastEditorUserId="16683" LastEditDate="2019-01-22T17:46:35.073" LastActivityDate="2019-01-23T01:26:54.133" Title="Number of atoms in NaCl unit cell" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;stoichiometry&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34120" PostTypeId="2" ParentId="34114" CreationDate="2015-07-17T20:47:47.753" Score="5" Body="&lt;p&gt;The CIF file won't contain information on atomic radii, only the positions (in fractional space).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that the atom might be ionic, covalent, etc. so you'll have to choose the atomic radius definition you feel is most appropriate. (Or run some sort of quantum calculation and use the resulting [electron density to define an atomic radius].)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, Open Babel does have a data table of covalent and van der Waals atomic radii, but it sounds like you want something else.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-17T20:47:47.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34121" PostTypeId="2" ParentId="34106" CreationDate="2015-07-17T22:03:08.130" Score="12" Body="&lt;p&gt;Nicolau Saker Neto already has an excellent answer, but he doesn't mention possibly the most common instance of an element being both a anion and a cation: Hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen usually has a positive oxidation state, but in any of the metal hydrides, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_hydride&quot;&gt;sodium hydride&lt;/a&gt;, it takes on a negative oxidation state.&lt;/p&gt;&#xA;" OwnerUserId="17556" LastActivityDate="2015-07-17T22:03:08.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34122" PostTypeId="2" ParentId="34115" CreationDate="2015-07-17T22:38:22.063" Score="2" Body="&lt;p&gt;As a prenote: My answer is somehow speculative because I have no experimental data or reference values of similar reactions used. In other words: The argumentation pathway below might be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition of LDA mostly gives the trans-enolate if a primary alkyl substituent next to the carbonyl group has been used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbonyl(O) is smaller than the methyl group [&lt;a href=&quot;https://chemistry.stackexchange.com/questions/33895/is-there-an-a-value-for-the-carbaldehyde-and-vinyl-groups/33897?noredirect=1#comment51808_33897&quot;&gt;ref.&lt;/a&gt;], therefore the allylic A13 interaction of Me//Me or even Me//i-Pr is bigger than the allylic A12 interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An allylic A13 Me//Me interaction is about 3,7 kcal/mol, an allylic A12 Me//Me is may be around 3,2 kcal/mol (this is just a guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If HMPA is added Li might coordinate to four HMPA molecules forming a cluster and inhibiting the cyclic transition state of LDA and hence no longer providing the trans-enolate, &lt;strong&gt;meaning an open transition state must be lower in activation energy for the cis-enolate formation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwxvl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to draw the conformations of 3-pentanone and then treated the base as &quot;point bulk&quot;, meaning the base is like a sphere with different sizes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard bases have a bigger repulsion to the carbonyl oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used arbitrarily numbers for the stereoselectivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my answer which base gives more cis-enolate than the other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TxzCS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-07-17T22:44:22.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34123" PostTypeId="1" AcceptedAnswerId="34125" CreationDate="2015-07-17T23:12:48.367" Score="15" ViewCount="2669" Body="&lt;p&gt;First off, I am currently confused about &lt;strong&gt;why&lt;/strong&gt; neon &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34108/does-ionization-energy-have-anything-to-do-with-how-reactive-an-element-is&quot;&gt;can even be ionized at all.&lt;/a&gt; But since it &lt;em&gt;can&lt;/em&gt; be ionized, this is the energy required to give a mole of neon a charge of +8: &lt;code&gt;207,390,000&lt;/code&gt; joules!! Or ~0.5 GJ! Giving &lt;em&gt;one-fifth of a kilogram of neon the maximum charge would be like the energy from &lt;strong&gt;50 armor-piercing rounds of the ISU-150 assault tank&lt;/strong&gt;, or approximately &lt;strong&gt;one Tour-de-France!&lt;/em&gt;&lt;/strong&gt; Charging up &lt;em&gt;one kilo of neon&lt;/em&gt; would be basically either &lt;strong&gt;throwing a lightning bolt or melting a tonne of steel.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question is, what insidious chemical or procedure could rip away all of neon's electrons like that? Or is it just simply impossible to do?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that it is possible to do (otherwise how would they be able to calculate the energy expenditure), but how is the question.&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-19T00:24:18.097" Title="Can you in reality give neon a charge of +8?" Tags="&lt;ions&gt;&lt;ionization-energy&gt;&lt;noble-gases&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34124" PostTypeId="1" CreationDate="2015-07-18T00:58:50.630" Score="1" ViewCount="334" Body="&lt;p&gt;I observed the absorbance on Phenolphthalein in acidic and basic medium, and I know that the absorbances differ due to the number of conjugated double bonds, but I am not sure exactly why this happens. I believe it has something to do with the delocalised electrons, which causes lower energy differences, but I am unable to establish a solid link between the concepts. Why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="17557" LastActivityDate="2015-07-18T01:54:23.710" Title="Relation between number of conjugated double bonds and absorbance?" Tags="&lt;spectroscopy&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-07-18T11:41:16.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="34125" PostTypeId="2" ParentId="34123" CreationDate="2015-07-18T01:06:51.127" Score="14" Body="&lt;p&gt;Of course you can take all the electrons off an atom - it is then called &quot;fully stripped&quot; in atomic physics. You don't need to do it to an entire mole, mind you. In accelerators one would send energetic neon ions through a background gas or a thin foil, and the interactions will result in various charge states coming out, up to and including fully stripped.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-07-18T01:06:51.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34126" PostTypeId="2" ParentId="34106" CreationDate="2015-07-18T01:15:57.070" Score="2" Body="&lt;p&gt;The reason some atoms are generally found in negative and some are generally found in positive oxidation states is their &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity#Electronegativities_of_the_elements&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative electronegativity&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, elements with high electronegativity (such as halogens) are likely to become negative atomic ions, and elements with low electronegativity (such as alkali metals) are likely to become positive atomic ions. Of course, elements with medium levels of electronegativity (such as &lt;a href=&quot;https://chemistry.stackexchange.com/a/34121/11303&quot;&gt;hydrogen&lt;/a&gt; and carbon) can be found in both positive and negative oxidation states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, exceptions exist in most cases. &lt;a href=&quot;https://chemistry.stackexchange.com/a/34111/11303&quot;&gt;Alkali metals can form negative ions as explained in this answer&lt;/a&gt;, and halogens can also form positive atomic ions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is this &lt;a href=&quot;http://www.chem.arizona.edu/massspec/intro_html/intro.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass spectrum of methyl bromide&lt;/a&gt;, where the peaks at m/z 79 and 81 are due to the bromine cation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDW0d.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11303" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-07-18T01:15:57.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34127" PostTypeId="1" CreationDate="2015-07-18T01:18:04.143" Score="2" ViewCount="51" Body="&lt;p&gt;There's a reaction occurring at constant $T$ and $P$, and it is in chemical equilibrium. Now by adding infinitesimal quantities of reactants, it is taken to another equilibrium state in a reversible way. During this process, if entropy change is $\Delta S$ and volume change $\Delta V$, $$\Delta U = T \Delta S - P \Delta V.$$&#xA;If the same process is carried out irreversibly, by suddenly adding a large quantity of the reactants and disturbing the chemical equilibrium, then the work still remains $P \Delta V$ ($ \Delta V$, a state function, has the same value). In that case, the heat $q=T \Delta S$, which is only true for reversible processes. Even though the process is irreversible, there has been no entropy change in the universe? Where am I going wrong? It can only be that the work is not $P \Delta V$, but since the process is at constant pressure, the work should have this value. &lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-07-18T01:18:04.143" Title="Is there no irreversible way of moving between two equilibrium states at constant T and P?" Tags="&lt;thermodynamics&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34128" PostTypeId="2" ParentId="33773" CreationDate="2015-07-18T01:25:33.483" Score="1" Body="&lt;p&gt;The keto-enol tautomerization cannot take place without at least a trace of acid or base, meaning that there is no direct shift of proton from alpha-carbon to oxygen or vice versa.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two most common enolation mechanisms are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Base-catalyzed (with organolithium reagent such as LDA (secondary amine))&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the base-catalyzed enol formation, the first step, proton extraction from the alpha-carbon, is slow. (The subsequent enolate formation is fast (due to resonance)). The base is proposed to form a chair-like transition state (TS) with the ketone, extracting the alpha-carbon proton and at the same time donating the Li to the oxygen. The resulting product is the classical lithium enolate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the structure of TS, you can then predict whether the “side chain” of the base would form a unfavorable syn-pentane interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Acid-catalyzed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the acid-catalyzed enol formnation, the first step, the protonation of oxygen and subsequent “charge re-localization” to the hydroxyl carbon, is fast. (The subsequent enol formation is slow). Since the protonation can occur from any free proton in the environment, the size of the acid plays little direct role in the stereochemistry of the resulting enolate. (However, the size of the “side chain” in the ketone reactant does).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a 3rd mechanism when the “base” is a tertiary amine (as you appear to be very interested in) but the reaction is acid-catalyzed. In this mechanism, the amine behaves like a nucleophile and a carbinolamine intermediate is detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Bruice, J. Am. Chem. Soc. 1983, 105, 4982 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja00353a023&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/ja00353a023&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also see Bruice, J. Am. Chem. Soc. 1989, 111, 962 and 1990, 112, 7361)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In drawing this carbinolamine intermediate, you can predict a bulky tertiary amine to favor a &lt;em&gt;cis&lt;/em&gt;-enolate due to steric hindrance. However, as suggested in the paper, &quot;severely sterically hindered&quot; tertiary amine can be too bulky to follow this 3rd mechanism and just catalyzes enolation via the general base-catalyzed mechanism.&lt;/p&gt;&#xA;" OwnerUserId="9933" LastEditorUserId="9933" LastEditDate="2015-07-18T01:32:51.743" LastActivityDate="2015-07-18T01:32:51.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34129" PostTypeId="2" ParentId="34124" CreationDate="2015-07-18T01:54:23.710" Score="2" Body="&lt;p&gt;When 2 p orbitals mix, a pi and a pi* orbital is formed, and the electrons occupy the lower energy pi orbital.  The UV-absorbance is the energy difference between these two orbitals such that on electron may be 'promoted' to the p* orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When 3 p orbitals mix, three pi-type orbitals are formed.  One is obviously bonding and one is obviously antibonding.  There is a third one 'in the middle', frequently called non-bonding.  Because of this third orbital, there is a smaller difference between any two states, and as such absorbing a photon to promote an electron requires less energy, redshifting the absorbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a fourth p orbital and the energy differences get smaller, but with two bonding orbitals and two anti-bonding orbitals.  Add a 5th p orbital and now you have two bonding orbitals and two anti-bonding orbitals and a non-bonding orbital, with even smaller differences.  Repeat ad nauseum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller energy differences means longer wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this picture for a graphical summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Povjp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-07-18T01:54:23.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34130" PostTypeId="1" CreationDate="2015-07-18T02:18:04.720" Score="3" ViewCount="10144" Body="&lt;p&gt;Can I create carbon monoxide gas by just mixing formic acid and sulphuric acid together? Or do they have to be heated?&lt;/p&gt;&#xA;" OwnerUserId="17558" LastEditorUserId="402" LastEditDate="2017-04-01T00:44:20.070" LastActivityDate="2017-04-01T00:44:20.070" Title="Making carbon monoxide gas from formic acid and sulfuric acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34131" PostTypeId="2" ParentId="34119" CreationDate="2015-07-18T03:41:20.927" Score="12" Body="&lt;p&gt;The picture you showed does have an unequal number of sodium cations and chloride anions.  However, the picture shows only part of a crystal.  Every atom that is on a boundary of the shown cube, whether on a &lt;em&gt;face&lt;/em&gt;, &lt;em&gt;edge&lt;/em&gt;, or &lt;em&gt;vertex&lt;/em&gt; of the cube, is shared with other &quot;cubes&quot; in the crystal that aren't shown in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the 8 corner Cl atoms in your picture is shared with 8 cubes (7 not shown). The 6 face-centered Cl atoms are shared with 2 cubes. Each of the 12 edge Na atoms is shared with 4 cubes (3 not shown). The center sodium atom is not shared. Thus there are 8/8 + 6/2 = 4 Cl atoms per unit &quot;cube&quot; in your picture, and 12/4 + 1/1 = 4 Na atoms per unit &quot;cube&quot; in your picture. 4=4, so the charge balances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be thinking that this math checks out only insofar as a crystal is actually infinitely sized.  And you may have noticed that no salt crystals are infinitely big in the real world.  These things are both true.  But even tiny specks of salt crystals are gigantic relative to atoms.  The surface of a salt crystal could involve imperfections that mean that the count of sodium atoms and chloride atoms aren't precisely equal.  But instead of 14 vs. 13, the difference is more like 100,000,000,000,000,000 vs 99,999,999,999,999,999.  And since the imperfections are at the &lt;em&gt;surface&lt;/em&gt;, on the outside of the crystal, any charge imbalance can be corrected if an opposite-charged particle from outside the crystals floats by and neutralizes the extra charge from the extra atom.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-18T03:41:20.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34132" PostTypeId="1" AcceptedAnswerId="34133" CreationDate="2015-07-18T04:36:21.797" Score="5" ViewCount="3486" Body="&lt;p&gt;If oxygen has a valency of $-2$, $4$ of it would result in $-8$. But iron doesn't have a (variable) valency of anything that goes to $-8$. Then how is $\ce{Fe3O4}$ possible?&lt;/p&gt;&#xA;" OwnerUserId="17551" LastEditorUserId="5026" LastEditorDisplayName="user15489" LastEditDate="2018-02-27T11:51:40.477" LastActivityDate="2019-10-05T13:11:04.983" Title="How is the triferric tetroxide of iron possible?" Tags="&lt;inorganic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34133" PostTypeId="2" ParentId="34132" CreationDate="2015-07-18T05:10:58.517" Score="9" Body="&lt;p&gt;Iron, like many of the Transition Metals &lt;em&gt;does&lt;/em&gt; have a variable valency, it can valencies of 2+ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ferrous&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferrous iron&lt;/a&gt;), 3+ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ferric&quot; rel=&quot;nofollow noreferrer&quot;&gt;ferric iron&lt;/a&gt;) and in some cases, 4+ (&lt;a href=&quot;https://chemistry.stackexchange.com/q/28751/15489&quot;&gt;tetravalent iron&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to $\ce{Fe3O4}$, according to the Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron(II,III)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron(II,III) oxide&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;contains both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A 2-dimensional diagram of $\ce{Fe3O4}$ is below, from the relevant &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.17215625.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ChemSpider&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rLDfB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3-dimensional model is below (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron(II,III)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xx0KO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;red indicate oxygen&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="-1" LastEditorDisplayName="user15489" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-18T05:33:01.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34134" PostTypeId="2" ParentId="34110" CreationDate="2015-07-18T05:27:58.110" Score="7" Body="&lt;p&gt;Noble metals are close to filling both $s$ and $d$ subshells, so there is a certain stability in gaining electrons. Gold atoms in complexes form bonds with each other similar in strength to hydrogen bonds and can form stable $\ce{Au^-}$ salts with cations like $\ce{Cs^+}$. Platinum similarly forms $\ce{Pt^{2-}}$. There are also relativistic effects that change the characteristics of the $d$ subshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aurophilicity&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Aurophilicity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S129325580500230X?via%3Dihub&quot; rel=&quot;nofollow&quot;&gt;http://www.sciencedirect.com/science/article/pii/S129325580500230X?via%3Dihub&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4592" LastEditorUserId="9961" LastEditDate="2015-07-23T00:06:50.840" LastActivityDate="2015-07-23T00:06:50.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34135" PostTypeId="2" ParentId="34109" CreationDate="2015-07-18T06:18:05.330" Score="11" Body="&lt;p&gt;The formation of $\ce{Li3N}$ can be explained thermodynamically. The enthalpy of formation of the $\ce{N^3-}$ ion is extremely endothermic (it involves three successive additions of electrons) and with the larger counterions $\ce{Na+}$ or $\ce{K+}$ the lattice energy (which is proportional to $\frac{q_+ q_-}{r_+ + r_-}$) is not sufficiently large to compensate for the large $\Delta_\text{f}H$ of $\ce{N^3-}$. The same argument explains why magnesium also forms a nitride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my very brief research on the Internet, lithium azide does exist but is unstable. It can be prepared via reaction of $\ce{NaN3}$ and $\ce{Li2SO4}$: see &lt;a href=&quot;http://actachemscand.org/pdf/acta_vol_11_p0581-0582.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Acta Chem. Scand.&lt;/em&gt; &lt;strong&gt;1957,&lt;/strong&gt; &lt;em&gt;11,&lt;/em&gt; 581&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-10-17T14:36:48.923" LastActivityDate="2016-10-17T14:36:48.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34138" PostTypeId="1" AcceptedAnswerId="34142" CreationDate="2015-07-18T08:40:29.453" Score="5" ViewCount="1334" Body="&lt;p&gt;In the preparation of $\ce{H2SO4}$ through the Contact process, $\ce{H2SO4}$ is added to hot sulfur trioxide to form oleum. This oleum is then diluted by adding water to form $\ce{H2SO4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4(l) + SO3(g) → H2S2O7(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2S2O7(l) + H2O(l) → 2 H2SO4(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a preparatory reaction, different reagents are used to obtain the desired product but here the product is itself a reactant and is used to make the product. Indirectly, $\ce{H2SO4}$ is being used to make $\ce{H2SO4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this process be called as a preparation process?  &lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="10185" LastEditDate="2015-07-18T10:21:47.103" LastActivityDate="2016-08-03T21:25:55.207" Title="Preparation of sulfuric acid through Contact process" Tags="&lt;inorganic-chemistry&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34139" PostTypeId="1" CreationDate="2015-07-18T09:22:13.800" Score="2" ViewCount="1784" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$60\ \mathrm{cm^3}$&lt;/span&gt; of oxygen was added to &lt;span class=&quot;math-container&quot;&gt;$10\ \mathrm{cm^3}$&lt;/span&gt; of a gaseous unsaturated hydrocarbon. After explosion and cooling to room temperature, the residual gases occupied &lt;span class=&quot;math-container&quot;&gt;$45\ \mathrm{cm^3}$&lt;/span&gt;. What volume of carbon dioxide was formed in the reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I’ve tried to make an unbalanced equation out of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{(CH2)$_n$ + O2 -&amp;gt; CO2 + H2O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem now is I don’t know how to proceed to the next step. How do I even find the moles? Is the equation correct? How do I balance this equation? &lt;/p&gt;&#xA;" OwnerUserId="17562" LastEditorUserId="16683" LastEditDate="2020-05-30T17:43:40.737" LastActivityDate="2020-05-30T17:43:40.737" Title="Determining stoichiometry of an unsaturated hydrocarbon via combustion" Tags="&lt;stoichiometry&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34140" PostTypeId="1" CreationDate="2015-07-18T09:27:19.290" Score="2" ViewCount="5032" Body="&lt;p&gt;What is the hybridization of oxygen in phenol molecule? I assume it is $sp^3$ hybridized because it forms 2 bonds - 1 with benzene, another with hydrogen, as in how oxygen is hybridized in a water molecule - $sp^3$.&#xA;Now the problem is, my A - level lecturers insist that it is the p orbital of oxygen that overlaps with the delocalised electrons of benzene ring, not $sp^3$. This means that oxygen can either be unhybridized, $sp$, or $sp^2$ hybridized. So which is true?&lt;/p&gt;&#xA;" OwnerUserId="17563" LastEditorDisplayName="user15489" LastEditDate="2015-07-18T10:13:46.433" LastActivityDate="2015-07-18T10:23:02.177" Title="Hybridization of oxygen in phenol" Tags="&lt;aromatic-compounds&gt;&lt;hybridization&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ClosedDate="2015-07-27T16:20:19.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="34141" PostTypeId="2" ParentId="34106" CreationDate="2015-07-18T10:17:14.610" Score="8" Body="&lt;p&gt;Unlike some, I wouldn't bet this is not common. Let's take a look at the biggest body in our system: the Sun. In its center, you can find for example some fluorine which is completely ionized. That's not really what we are used to with this electron lover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also look for example at Titan's upper atmosphere. Here, interactions with the magnetic field of Saturn are responsible for the formation of species like $\mathrm O^+$, or almost every ionized molecule you can imagine with $\mathrm C$, $\mathrm N$, $\mathrm O$ and $\mathrm H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples like these everywhere just in our Solar system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most chemistry classes seem to forget that our &quot;standard room conditions&quot; are far from being a standard in the whole universe. That's a shame because most of the &quot;rules&quot; we learn in high school (and some colleges) are based on the assumption that we are at those said standard conditions and stash the fact that the observed chemical behavior of elements is not necessarily even representative of what happen when you look at the whole universe.&lt;/p&gt;&#xA;" OwnerUserId="17564" LastEditorUserId="17564" LastEditDate="2015-07-18T10:50:40.860" LastActivityDate="2015-07-18T10:50:40.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34142" PostTypeId="2" ParentId="34138" CreationDate="2015-07-18T10:19:24.823" Score="4" Body="&lt;p&gt;You can call it whatever you like. Personally I would say you are &lt;em&gt;preparing&lt;/em&gt; sulphuric acid because you end up with more than you started with - 1 mole is used to form 2 moles.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2016-03-08T13:17:02.523" LastActivityDate="2016-03-08T13:17:02.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34144" PostTypeId="2" ParentId="34130" CreationDate="2015-07-18T11:18:47.207" Score="5" Body="&lt;p&gt;The formation of carbon monoxide already takes place when mixing formic acid and sulfuric acid at room temperature, but the reaction is relatively slow. Since the temperature of the mixture may significantly increase during the course of the reaction, the reaction rate may increase, which makes the reaction difficult to control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For laboratory applications, concentrated sulfuric acid (or phosphoric acid) is usually heated to about 80 °C. Formic acid is added slowly, drop by drop. The formation of gas should start upon addition of the first drop, otherwise the temperature should be increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generated gas contains various impurities, which can be removed using gas washing bottles. Usually, the gas is passed first through $\ce{KOH}$ solution; however, you should take care to prevent any backflow of $\ce{KOH}$ solution to the sulfuric acid.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-18T11:18:47.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34145" PostTypeId="1" CreationDate="2015-07-18T11:21:05.450" Score="23" ViewCount="14400" Body="&lt;p&gt;Platinum and palladium are great catalysts. At the same time, other metals of the so-called platinum group metals are not. What are the atomic level reasons for this?&lt;/p&gt;&#xA;" OwnerUserId="17565" LastEditorUserId="865" LastEditorDisplayName="user15489" LastEditDate="2016-11-18T16:38:15.943" LastActivityDate="2018-02-22T06:51:16.123" Title="Platinum group metals: Why are platinum and palladium great catalysts and not the others?" Tags="&lt;metal&gt;&lt;catalysis&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34146" PostTypeId="2" ParentId="34139" CreationDate="2015-07-18T14:24:49.457" Score="4" Body="&lt;h2&gt;First step: balance your equation.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\ce{(CH2)$_n$ + O2 -&amp;gt; CO2 + H2O}$ is not balanced, this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH2)_{\it n} + \frac{3 }{2}{\it n}~O2 -&amp;gt; {\it n}~ CO2 + {\it n}~ H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Second step: perform mole balances on relevant species.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The balanced equation has $1 + \frac{3}{2}n$ moles of gas reacting, and $2 n$ moles of gas formed.  &lt;em&gt;However&lt;/em&gt;, the question states that the products are cooled to room temperature, which would eliminate the water from the gas phase.  Thus the reaction were are really interested in is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH2)_{\it n}_{\rm(g)} + \frac{3 }{2}{\it n}~O2_{(g)} -&amp;gt; {\it n}~ CO2_{(g)} + {\it n}~ H2O}_{(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $1 + \frac{3}{2}n$ moles of gas react, and $n$ moles form for each mole of reaction $\xi$.  Thus, the reaction causes a loss in moles of gas of is $-(1 + \frac{1}{2}n)$ moles per mol $\xi$ of reaction.  If we assume the ideal gas law applies, then we could also say that the loss in volume from the reaction is $(1 + \frac{1}{2}n)~{\mathrm{{cm}^3}}$ per $\xi~{\mathrm{{cm}^3}}$ of reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a total gas balance is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm GAS~IN - GAS~OUT = GAS~REACTED $$&#xA;$$(60~\mathrm{{cm}^3} + 10~\mathrm{{cm}^3}) - (45~\mathrm{{cm}^3})=(1 + \frac{1}{2}n)~\xi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A $\ce{CO2}$ balance is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm \ce{CO2}~IN - \ce{CO2}~OUT = \ce{CO2}~REACTED $$&#xA;$$ 0 - x = - n \xi $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Third step: assess the resulting equations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These equations are not solvable for $x$ because they depend on both $\xi$ and $n$.  That is, there are two equations, and three unknowns.  Writing additional balances on oxygen or on fuel will not help because we will not know the gas remaining for either species, so each new equation would introduce new variables. Thus, we will need additional assumptions to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fourth step: determine additional assumptions.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Probably we should assume that either (a) fuel was in excess or (b) oxygen was in excess, which is equivalent to assuming that (a) oxygen was completely consumed or (b) fuel was completely consumed.  At least one of these assumptions makes sense because combustion reactions are essentially irreversible.  In my answer I'll assume oxygen was in excess, meaning that the fuel was completely consumed.  &lt;em&gt;You should make the alternate assumption and repeat the procedure.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fifth step: apply additional assumptions.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming fuel was completely consumed, a fuel balance is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rm FUEL~IN - FUEL~OUT = FUEL~REACTED $$&#xA;$$\rm (10~\mathrm{{cm}^3}) - 0=\xi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This conveniently gives us a solution for $\xi$, the extent of reaction.  It is $10~\mathrm{{cm}^3}$.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sixth step: use new information to solve existing equations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can substitute that solution for $\xi$ into the total gas balance to get that $1 + \frac{1}{2}n = 3$, which implies that $n=4.$  That and our solution for $\xi$ allows us to solve the problem: &lt;strong&gt;if oxygen is in excess&lt;/strong&gt;, then 40 $\mathrm{{cm}^3}$ of $\ce{CO2}$ were formed.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:26:05.347" LastActivityDate="2016-03-11T23:26:05.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34147" PostTypeId="1" CreationDate="2015-07-18T15:20:16.473" Score="4" ViewCount="496" Body="&lt;p&gt;What does change in entropy ($\Delta S$) represent ? How does change in entropy of a universe ($\Delta S_{univ}$) give us information about energy available to do work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the physical significance of entropy of universe and how it is related to energy energy released by system that can be used to do work (Not in terms of equation. I would want to know its physical significance).&lt;/p&gt;&#xA;" OwnerUserId="14542" LastActivityDate="2015-07-21T16:39:09.597" Title="How is change in entropy of universe related to the energy released by system that can be used to do work?" Tags="&lt;thermodynamics&gt;&lt;free-energy&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34148" PostTypeId="2" ParentId="34067" CreationDate="2015-07-18T17:51:04.023" Score="-1" Body="&lt;p&gt;I googled it for a while and this was the best thing I found (&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20100525170905AA9PDiD&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This is a tough question, that is probably not known with any certainty.&#xA;  Even Jerry March (Adv Organic Chem, 4th ed, page 1219) doesn't commit himself to a mechanism.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;You'll have to speculate.   Add a hydride on the N next to the alkyl group, then&#xA;  grab a proton on the terminal N.    Add a hydride on the terminal N, then grab a proton on the middle N.  After that, it gets a little sketchy for breaking the N-N bond.    &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Try a SN2 attack by hydride on the middle N, giving hydrazine and  RNH-, which is a lousy SN2 leaving group, but we're stuck with it.&#xA;  Protonation of the RNH- gives the amine and hydrazine.   Or, you could do a SN2 on the N next to the alkyl group, and kick out a NH2NH- anion, which is also a lousy leaving group for the SN2.&#xA;  You can see why most books conveniently leave out a mechanism for this reduction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17422" LastEditorUserId="4945" LastEditDate="2015-07-24T01:56:25.117" LastActivityDate="2015-07-24T01:56:25.117" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34149" PostTypeId="1" AcceptedAnswerId="34150" CreationDate="2015-07-18T19:20:14.083" Score="17" ViewCount="4745" Body="&lt;p&gt;I have about 2g of 99.5% Potassium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get rid of it, right now it's in a glass jar sealed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the proper protocol? I was going to combust it in water and wash it down the drain but the by-product potassium hydroxide is caustic and hazardous so I don't think that's best. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I just throw it away and it goes to a garbage facility where the vial probably breaks and it reacts with rain water and the potassium hydroxide is so small and buried that it never really has an impact?&lt;/p&gt;&#xA;" OwnerUserId="17571" LastActivityDate="2015-07-19T13:51:02.683" Title="How to dispose of pure potassium" Tags="&lt;safety&gt;" AnswerCount="3" CommentCount="6" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34150" PostTypeId="2" ParentId="34149" CreationDate="2015-07-18T19:39:31.570" Score="14" Body="&lt;p&gt;In &lt;a href=&quot;http://www.rsc.org/images/Alkalimetals_with_concHCl_tcm18-194226.pdf&quot;&gt;a suggested lab&lt;/a&gt;, the &lt;a href=&quot;http://www.rsc.org/&quot;&gt;RSC&lt;/a&gt; recommends using &lt;a href=&quot;http://en.wikipedia.org/wiki/Isopropyl_alcohol&quot;&gt;propan-2-ol&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If there is any unused potassium remaining at the end of the experiment, remove it with the tweezers and place it in a beaker containing about 100 cm3 of propan-2-ol to dissolve away – it will fizz, giving off bubbles of hydrogen. When the fizzing has stopped, dispose of the resulting alkaline solution down the sink flushing it away with plenty of water. The acidic solution remaining from the experiment may also be disposed of down the sink flushing it away with plenty of water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The same is recommended for sodium.&lt;/p&gt;&#xA;" OwnerUserId="13294" LastActivityDate="2015-07-18T19:39:31.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34151" PostTypeId="2" ParentId="34149" CreationDate="2015-07-18T19:48:39.417" Score="12" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I was going to combust it in water and wash it down the drain&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, mixing it with water is a bad idea.  Hydrogen gas will be liberated along with heat.  In the presence of oxygen and heat the hydrogen will form a combustible / explosive mixture.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Could I just throw it away and it goes to a garbage facility where the&#xA;   vial probably breaks and it reacts with rain water and the potassium&#xA;   hydroxide is so small and buried that it never really has an impact?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, that's not good either, it will just explode in the dump and start a fire that someone else will have to deal with.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the proper protocol?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These two links (&lt;a href=&quot;https://books.google.com/books?id=3UArAAAAYAAJ&amp;amp;pg=PA383&amp;amp;lpg=PA383&amp;amp;dq=dispose%20of%20potassium%20metal&amp;amp;source=bl&amp;amp;ots=4PYSmiK4Ml&amp;amp;sig=O0iee-ZpCYoWLcdTAqWKZBSaZys&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CEsQ6AEwB2oVChMIw4K03rXlxgIVy4sNCh1y2Qsy#v=onepage&amp;amp;q=dispose%20of%20potassium%20metal&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;link_1&lt;/a&gt;, &lt;a href=&quot;https://books.google.com/books?id=geK5gjEn3hkC&amp;amp;pg=PA414&amp;amp;lpg=PA414&amp;amp;dq=dispose%20of%20potassium%20metal&amp;amp;source=bl&amp;amp;ots=9k9QqD8fn_&amp;amp;sig=EfjEZ5KVPrLLzYuIm2zffa6MGu8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CDQQ6AEwA2oVChMIw4K03rXlxgIVy4sNCh1y2Qsy#v=onepage&amp;amp;q=dispose%20of%20potassium%20metal&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;link_2&lt;/a&gt;) suggest the same disposal procedure.  Link_2 provides more detail. I've reproduced the key part below along with the reactions that take place prior to the final neutralization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/g0u1F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="7448" LastEditDate="2015-07-19T13:51:02.683" LastActivityDate="2015-07-19T13:51:02.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34152" PostTypeId="1" AcceptedAnswerId="34158" CreationDate="2015-07-18T20:16:08.680" Score="3" ViewCount="1387" Body="&lt;p&gt;When I put aluminium parts into my dishwasher, they get tainted and they are sometime covered with white dots similar to salt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it is aluminium oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the chemical reactions that have an effect on aluminium but not other metals?&lt;/p&gt;&#xA;" OwnerUserId="17572" LastEditorUserId="5017" LastEditDate="2015-07-19T13:26:11.963" LastActivityDate="2015-07-19T13:26:11.963" Title="Why dishwashers are corrosive to aluminum?" Tags="&lt;everyday-chemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34153" PostTypeId="1" CreationDate="2015-07-18T20:40:34.547" Score="15" ViewCount="312" Body="&lt;p&gt;I want to enolize a ketone, for instance acetone with a $\mathrm{p}K_\mathrm{a}$ value of about 20 (in $\ce{H2O}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if there exists a rule of thumb to predict how the $\mathrm{p}K_\mathrm{a}$ of the ketone will decrease if we add one of the following Lewis acids: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $\ce{(c-Hex)2BCl}$ (chlorine anion can be eliminated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $\ce{Bu2BOTf}$ (triflate anion is a good leaving group) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $\ce{NaI}$ (sodium is bigger than lithium, and I expect the bond to oxygen is less strong meaning the decrease of the $\mathrm{p}K_\mathrm{a}$ will be less than in d))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) $\ce{LiI}$ (iodine is bigger than lithium, therefore good hard-soft-mismatch; and oxygen is hard as well)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) $\ce{HCl}$ (As @SYK suggested in his answer of my recent question&#xA;[&lt;a href=&quot;https://chemistry.stackexchange.com/questions/33773/stereoselective-enolate-formation-with-different-bases-does-the-addition-of-a-b/34128#34128&quot;&gt;1&lt;/a&gt;], the proton is the strongest Lewis acid as possible and will lower the $\mathrm{p}K_\mathrm{a}$ as much as possible, that even a very weak base, may be of $\mathrm{p}K_\mathrm{a}~9$ as $\ce{NH4Cl}$ will (or won't it?) form the &lt;em&gt;cis&lt;/em&gt;-enolate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm especially interested in the lowering values for a), b) and c). My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is my qualitative lowering of $\mathrm{p}K_\mathrm{a}$ values a &amp;lt; b &amp;lt; c &amp;lt; d &amp;lt; e correct?&lt;/li&gt;&#xA;&lt;li&gt;Do you have somes guesses for decreasing values for a, b and c ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="41328" LastEditDate="2017-12-13T09:19:30.887" LastActivityDate="2017-12-13T09:19:30.887" Title="By how many pKa units will the pKa value of a ketone be minimized by addition of a Lewis acid" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34156" PostTypeId="2" ParentId="34123" CreationDate="2015-07-19T00:24:18.097" Score="8" Body="&lt;p&gt;First - in chemistry there's technically no such thing as bare multivalent cation, second - as you think, there's no such energetic chemical reaction, third - ionisation energy is physical property (although important for chemistry) and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...&quot;is usually measured in an electric discharge tube in which a fast-moving electron generated by an electric current collides with a gaseous atom of the element, causing it to eject one of its electrons.&#xA;  - &lt;em&gt;Source &quot;&lt;a href=&quot;http://www.britannica.com/science/ionization-energy&quot;&gt;Encyclopedia Britannica&lt;/a&gt;&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9961" LastActivityDate="2015-07-19T00:24:18.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34158" PostTypeId="2" ParentId="34152" CreationDate="2015-07-19T02:33:37.260" Score="8" Body="&lt;p&gt;I think it is just alkalinity. Unlike many other metals, aluminum is amophoteric, so it dissolves in both strong base and in strong acid. Most dishwashing detergents are fairly alkaline, probably at pH 11 or 12 or so. That and the heat may be enough to corrode Al.  Most other common metals are attacked by strong acids but not by alkali.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-19T02:33:37.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34160" PostTypeId="2" ParentId="34119" CreationDate="2015-07-19T05:11:03.430" Score="3" Body="&lt;p&gt;Unit cells demonstrate alignment and relative position of atoms within a crystal but provide no overt stoichiometric information. The unit cell model isn't meant to imply that atoms group to form these individual cubes or shapes. As such, the atoms/charges won't necessarily balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of NaCl, the face-centered cubic unit cell has an odd number lattice points and thus doesn't include a whole number of NaCl molecules. This is not, however, among the three unit cell criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The unit cell is the simplest repeating unit in the crystal.&lt;/li&gt;&#xA;&lt;li&gt;Opposite faces of a unit cell are parallel.&lt;/li&gt;&#xA;&lt;li&gt;The edge of the unit cell connects equivalent points.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch13/unitcell.php#simple&quot; rel=&quot;nofollow&quot;&gt;Unit Cell Overview&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17580" LastActivityDate="2015-07-19T05:11:03.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34161" PostTypeId="1" AcceptedAnswerId="34167" CreationDate="2015-07-19T05:44:30.513" Score="6" ViewCount="5066" Body="&lt;p&gt;Sodium polyacrylate is a white powder with a characteristic odor. But when we add water to this powder, instead of forming an aqueous solution, it instantaneously  solidifies and forms more amount of powder like substance, also called &quot;hot ice&quot;.&#xA;Why and how this phenomenon occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Vais8pL0w8U&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="5026" LastEditDate="2018-05-09T08:57:36.247" LastActivityDate="2019-04-15T01:52:20.310" Title="Why does sodium polyacrylate solidify, on addition of water?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;phase&gt;&lt;liquid-crystals&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34162" PostTypeId="1" AcceptedAnswerId="34175" CreationDate="2015-07-19T05:47:05.030" Score="0" ViewCount="2939" Body="&lt;p&gt;Do you just have to remember that it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I know that substances with CHs are hydrocarbons. Does such a 'code', if you will, exist for amphiprotic substances ?  &lt;/p&gt;&#xA;" OwnerUserId="15053" LastEditorUserId="15053" LastEditDate="2015-07-19T06:07:31.637" LastActivityDate="2015-07-19T15:28:44.050" Title="How do you know that Sodium Hydrogencarbonate is amphiprotic?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34163" PostTypeId="2" ParentId="34145" CreationDate="2015-07-19T06:19:33.117" Score="15" Body="&lt;p&gt;I think some of the other transition elements have also catalytic properties. For example, rhodium is generally used in addition to platinum or palladium in catalytic converters to oxidize carbon monoxide and hydrocarbons on one hand, and reduce $\ce{NO_x}$ to nitrogen on the other hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the choice is often made on the basis of relative cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of course, &lt;strong&gt;not all transition elements are adequate for these catalytic properties&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Metals such as silver and copper have a high affinity for sulfur and form metal sulfides and sulfates. As this happens, there will be progressively less metal available to function as a catalyst. While platinum tends not to become totally or irreversibly poisoned, i.e It can be efficiently recycled.&lt;/li&gt;&#xA;&lt;li&gt;Another point is linked to &quot;Tammann temperatures&quot; of these metallic elements:  Actually, in a catalytic converter, metal is used in the form of nanoparticles dispersed on the surface of a porous support material. When  temperature increases the particles starts to becomes mobile and coalesce (i.e. the catalytic activity is lost). This process is well-known under the name &quot;sintering&quot; and it becomes tangible as the metal is close to its Tammann temperature. This temperature is often defined as the half of the metal's melting point on the absolute temperature scale. In fact, metals such as gold, silver and copper have a Tammann temperature that is well below the average temperature of exhaust-gas ($600-700^0\mathrm{C}$).&#xA;Tammann Temperature is $750^0\mathrm{C}$ for platinum, and $640^0\mathrm{C}$ for palladium. While it is only $405^0\mathrm{C}$ for copper, $345^0\mathrm{C}$ for silver and $395^0\mathrm{C}$ for gold.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To sum up, the candidate transition element for use as a catalyst in a catalytic converter: &#xA;1) has a low affinity to poisons  such as sulfur compounds;&#xA;2) can be efficiently recycled;&#xA;3)  has a relatively high melting point.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-07-21T11:42:00.843" LastActivityDate="2015-07-21T11:42:00.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34164" PostTypeId="2" ParentId="34147" CreationDate="2015-07-19T06:56:48.270" Score="1" Body="&lt;p&gt;Well a reaction which is spontaneous has negative value of Gibbs or Helmholtz free energy. This free energy is the energy which is made available to be used for work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does that relate to your question? Both Gibbs and Helmholtz free energy are derived from the argument that change in entropy of the universe is greater than zero. Physically, that is an argument that no chemical system will react in such a way that decreases the number of available microstates unless it increases the number of available microstates for the surroundings by a greater amount.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-07-19T06:56:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34165" PostTypeId="2" ParentId="34147" CreationDate="2015-07-19T09:03:43.750" Score="1" Body="&lt;p&gt;Suppose you are a magician &amp;amp; want to make a rabbit having internal energy $U$. so, in order to invoke the rabbit, you have to supply $U$. But wait! In order to make the rabbit, you not only have to provide $U$ but also the energy necessary to create space for the rabbit by asiding the atmosphere. So, the total energy you supply is $U + PV$. This is enthalpy. But again wait! You don't have to supply the entire $U + PV$ as there is an associated entropy with the formation of the rabbit i.e. $ S$ i.e. the energy supplied by you, the Great Magician is $U + PV - T S$. This is the free energy as it is the only energy you will get after vanishing the rabbit. &lt;/p&gt;&#xA;" OwnerDisplayName="user5764" LastEditorDisplayName="user5764" LastEditDate="2015-07-21T16:39:09.597" LastActivityDate="2015-07-21T16:39:09.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34166" PostTypeId="2" ParentId="32812" CreationDate="2015-07-19T09:23:32.967" Score="3" Body="&lt;p&gt;(Not too good with MathJax, so sorry about the lack of elegance. If someone could suggest a good tutorial, I'd appreciate it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your methods are all correct, but it seems you are not comparing apples to apples. The $\Delta G$ of a particular reaction is going to scale with the amount of reactants you have. For example, the $\Delta G$ of $\ce{10N2 + 10H2 -&amp;gt; 10NH3}$ is going to be 10 times that of $\ce{N2 + H2 -&amp;gt; NH3}$. The same reasoning can be applied to your equations. So, in your first calculation of $-(a)\times5+(c)\times2$ the $\Delta G$ of $a$ and $c$ are 5 and 2 times greater than what they should, respectively. This is not a fair comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is to normalize your equations such that the same numbers of electrons are being transferred. You can do this with any integer, but it's probably easiest to just normalize to 1. In your first calculation, normalizing to an electron transfer of 1 leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{8/5H+ + 1/5MnO4- + 1/2X^3- -&amp;gt; 1/2X- + 1/5Mn^2+ + 4/5H2O}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ= 1.5\ \mathrm{V} + \left(-0.9\ \mathrm{V}\right) = 0.6\ \mathrm{V}$$&#xA;$$\Delta G^{\circ} =−nE^\circ F=-1\times0.6\times F=-0.6F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same normalization to the second equation yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ= 1.5\ \mathrm{V} + \left(-0.5\ \mathrm{V}\right) = 1.0\ \mathrm{V}\\&#xA;ΔG^\circ=−nE^\circ F=-1\times1.0\times F=-1.0F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard potential and the $\Delta G$ of the second overall reaction of half reactions b and c are in fact more negative than the first overall reaction. In case you're wondering why $E$ does not scale with the size of the reaction it's because $E$ is proportional to $G/n$. As you scale the size of the reaction, both $G$ and $n$ will grow or shrink in the exact same proportion and thus $E$ will always stay the same.&lt;/p&gt;&#xA;" OwnerUserId="17224" LastEditorDisplayName="user7951" LastEditDate="2015-07-19T17:58:57.523" LastActivityDate="2015-07-19T17:58:57.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34167" PostTypeId="2" ParentId="34161" CreationDate="2015-07-19T10:38:11.680" Score="6" Body="&lt;p&gt;Sodium polyacrylate is an example of a super-absorbing polymer. It is a cross-linked (network) polymer that contains sodium atoms. It absorbs water by a process called osmosis. When the (sodium-containing) polymer is placed in contact with water, there is a tendency for the sodium to distribute equally between the network and the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means, some of the sodium atoms want to leave the network and move to the water. When these sodium atoms leave, they are replaced with water molecules. Water swells the polymer network to try to keep the sodium concentration balanced between the polymer and the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-links that connect the chains together prevent them from dissolving/breaking apart in the water. The resulting product appears very similar to fake snow, as you mentioned, a solidified powder like substance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium polyacrylate can absorb it’s own weight 300x over making it an amazing chemical to aid in water cleanups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was good.&lt;/p&gt;&#xA;" OwnerUserId="17422" LastActivityDate="2015-07-19T10:38:11.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34168" PostTypeId="1" CreationDate="2015-07-19T10:49:04.990" Score="10" ViewCount="1193" Body="&lt;p&gt;We know that both solids and liquids can exhibit metallic luster. I wonder whether there is a gas that when filling a transparent envelope (such as a glass tube) would make it appear with metallic luster?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastEditorUserId="10185" LastEditDate="2015-07-27T08:16:14.400" LastActivityDate="2015-07-27T08:16:14.400" Title="Is there a gas with metallic luster?" Tags="&lt;phase&gt;&lt;color&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34169" PostTypeId="1" CreationDate="2015-07-19T11:47:01.733" Score="2" ViewCount="210" Body="&lt;p&gt;If both nitric acid and nitrate ion are stabilised through resonance, what makes the nitrate ion more stable in aqueous medium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the same argument be applied to nitrous acid and nitrite ion in aqueous medium?&lt;/p&gt;&#xA;" OwnerUserId="17586" LastEditorUserId="5026" LastEditDate="2018-03-12T06:17:14.117" LastActivityDate="2018-03-12T06:17:14.117" Title="If both nitric acid and nitrate ion are stabilised through resonance, what makes the nitrate ion more stable in aqueous medium?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;stability&gt;&lt;resonance&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34170" PostTypeId="2" ParentId="34149" CreationDate="2015-07-19T12:29:33.037" Score="1" Body="&lt;p&gt;25% isopropanol (v/v) isopropanol in toluene is the usual means.  2 g potassium is not a large amount.  Sodium potassium alloy can also be disposed of this way.&lt;/p&gt;&#xA;" OwnerUserId="16999" LastActivityDate="2015-07-19T12:29:33.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34172" PostTypeId="1" CreationDate="2015-07-19T13:10:56.330" Score="1" ViewCount="147" Body="&lt;p&gt;I have a pair of natural rubber dishwashing gloves. These particular gloves were brand new but sitting loose out of the package on top of the supplies under the sink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pulled them out today &amp;amp; found that one of them had developed small blisters all over the hand area, covering about four square inches. The blisters are not really concentrated in one area, although there are more along the thumb-forefinger. They are about two millimeters in diameter, some go in and some go out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only chemicals under the sink are vinegar, Bon Ami (mostly crushed limestone), Goo Gone (tried it; did not cause the blisters), liquid dish soap &amp;amp; detergent. I can't think of anything hot enough to cause that blister pattern that would have been down there. Any ideas? I'm totally stumped. &lt;/p&gt;&#xA;" OwnerUserId="17588" LastEditorDisplayName="user15489" LastEditDate="2015-07-21T09:26:45.390" LastActivityDate="2015-07-21T09:26:45.390" Title="Bubbles appeared on rubber gloves" Tags="&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34173" PostTypeId="2" ParentId="18836" CreationDate="2015-07-19T14:05:00.563" Score="-1" Body="&lt;p&gt;Configuration differs from conformation by a fact that conformations are only due to the orientation of the atoms in molecule around the covalent bond considering the bond as an axis.  And changing the configuration causes the cleavage and formation of new chemical bonds.&lt;/p&gt;&#xA;" OwnerUserId="17590" LastEditorUserId="7448" LastEditDate="2015-07-19T16:43:00.933" LastActivityDate="2015-07-19T16:43:00.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34175" PostTypeId="2" ParentId="34162" CreationDate="2015-07-19T15:28:44.050" Score="3" Body="&lt;p&gt;You have to consider whether or not the species is capable of both losing &lt;strong&gt;and&lt;/strong&gt; gaining a hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When sodium bicarbonate dissociates in water, it becomes a sodium ion ($\ce{Na+}$) and a bicarbonate ion ($\ce{HCO3-}$). The bicarbonate ion itself is what is amphiprotic, and this is because it can lose the hydrogen it has to become a carbonate ion ($\ce{CO3^{2-}}$), or it can gain a hydrogen to become carbonic acid ($\ce{H2CO3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many species that contain a hydrogen and an unbonded electron pair are amphiprotic, like $\ce{NH3}$ and $\ce{H2O}$, for example.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-07-19T15:28:44.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34176" PostTypeId="1" AcceptedAnswerId="34184" CreationDate="2015-07-19T16:05:03.343" Score="1" ViewCount="2636" Body="&lt;p&gt;Suppose that I take a pressure vessel that can withstand up to 2000 PSI. Now I fill it with water (not completely, but around 90%) and start to heat it with a direct flame. What I want to know is, what will happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I don't know anything about it, here are some guesses that I think might happen:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The water will begin to convert into gas. But since the volume is limited, therefore the pressure would increase (and keep on increasing) and the vessel may burst after the internal pressure crosses 2000PSI.&lt;/li&gt;&#xA;&lt;li&gt;The water won't convert into gas and would just take all the heat energy into itself. In other words, nothing would happen.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Those are my guesses and it would be great if you can help me out.&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="865" LastEditDate="2015-07-19T16:52:11.583" LastActivityDate="2015-07-20T00:26:15.283" Title="What happens when heating water in a pressure vessel?" Tags="&lt;thermodynamics&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34177" PostTypeId="1" AcceptedAnswerId="34182" CreationDate="2015-07-19T16:34:50.147" Score="0" ViewCount="2671" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;what can I do with a degree in analytical chemistry other than working in the lab?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And what is the future of being an organic chemist? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can I be a researcher in both of them?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or it's better to take organic chemistry if I want to be a researcher?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17596" LastEditorUserId="7448" LastEditDate="2015-07-19T16:44:43.630" LastActivityDate="2015-07-19T19:31:35.707" Title="What's the future of an organic and/or analytical chemist?" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-07-20T14:41:18.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="34178" PostTypeId="1" AcceptedAnswerId="34179" CreationDate="2015-07-19T16:35:04.810" Score="5" ViewCount="739" Body="&lt;p&gt;Nitrogen triiodide, on slightest physical contact, explodes to evolve dense purple fumes of iodine. Why does this phenomenon occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;video link - &lt;a href=&quot;https://www.youtube.com/watch?v=2KlAf936E90&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=2KlAf936E90&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="4185" LastEditDate="2015-07-20T01:37:13.133" LastActivityDate="2016-05-18T10:56:45.027" Title="Contact explosion of nitrogen triiodide" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;phase&gt;&lt;explosives&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34179" PostTypeId="2" ParentId="34178" CreationDate="2015-07-19T16:56:30.777" Score="9" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Nitrogen_triiodide&quot;&gt;Wikipedia&lt;/a&gt; states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The instability of $\ce{NI3}$ and $\ce{NI3 · NH3}$ can be attributed to the large steric strain caused by the three large iodine atoms being held in close proximity to each other around the relatively tiny nitrogen atom. This results in a very low activation energy for its decomposition, a reaction made even more favorable due to the great stability of $\ce{N2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The steric strain between the atoms can be better visualized by the representations &lt;a href=&quot;http://www.acornusers.org/education/Thesis/NI3andAgar.html&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decomposition reaction is&#xA;$$\ce{2NI3\to N2 + 3I2}$$&#xA;Note that there is an increase in the number of moles (4 versus 2), which means that the reaction is favored by entropy, because there are more possible microstates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_enthalpy_of_formation&quot;&gt;enthalpy of formation&lt;/a&gt; is 154.4 kJ/mol, which is incredibly favorable for a reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://chemwiki.ucdavis.edu/Development_Details/Approaches/VVV_Demos/Additional_Demos/Nitrogen_Triiodide&quot;&gt;UC Davis ChemWiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13294" LastEditorUserId="13294" LastEditDate="2015-07-19T17:06:25.993" LastActivityDate="2015-07-19T17:06:25.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34182" PostTypeId="2" ParentId="34177" CreationDate="2015-07-19T19:31:35.707" Score="5" Body="&lt;p&gt;The answers to this question are going to be as varied as the people answering them. Let me give you the advice I give my undergraduate advisees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The future is what you make of it&lt;/strong&gt;. Don't let random people on the internet tell you which field to specialize in or which research to do. The decision need to be personal, because you need to care about what you are doing so that you are motivated to excel.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Instead of worrying about which type of chemistry to study, find the problems you want to solve.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Research in chemistry is pretty interdisciplinary nowadays, both across the old subfields of &lt;code&gt;organic&lt;/code&gt;, &lt;code&gt;analytical&lt;/code&gt;, &lt;code&gt;physical&lt;/code&gt;, etc. and between chemistry and the other sciences. Rather than become obsessed about organic vs. analytical or whatever, read about current research in chemistry. Think about the types of research problems you want to solve. Usually that will help you figure out which pathway to follow (i.e. which courses you need to take, which specializations to pick up, which lab to join, etc.). Research, both in academia and elsewhere is going to be more problem- than field-focused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an undergraduate, I was torn between organic and physical chemistry. I chose organic chemistry as a focus in graduate school, joined a research group focusing on polymer synthesis, and did a lot of analytical work as part of my research. We collaborated with a theoretical group to understand how monomer sequence would affect optoelectronic properties of conjugated oligomers and polymers, so I also had to learn a lot about the physical models of my conjugated systems. Another part of my group was working on polyesters with controllable biodegredation pathways for applications in medicine. The team on that project included grad students focusing on organic, inorganic, and analytical chemistry collaborating with engineering students in another group. &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-19T19:31:35.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34184" PostTypeId="2" ParentId="34176" CreationDate="2015-07-19T19:50:07.730" Score="4" Body="&lt;p&gt;Look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_diagram&quot; rel=&quot;nofollow&quot;&gt;phase diagram&lt;/a&gt; of &lt;a href=&quot;http://www1.lsbu.ac.uk/water/water_phase_diagram.html&quot; rel=&quot;nofollow&quot;&gt;water&lt;/a&gt; to see what happens. &lt;a href=&quot;https://en.wikipedia.org/wiki/Josiah_Willard_Gibbs&quot; rel=&quot;nofollow&quot;&gt;Josiah Willard Gibbs&lt;/a&gt; invented these diagrams o help understand what happens as a substance is heated under pressure, and transitions from solid to liquid to gas and supercritical fluid (neither quite gas nor liquid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;em&gt;sufficient&lt;/em&gt; heat, the vessel will burst. Above ~650 K, the water can no longer be liquefied by pressure, and behaves more like an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas&lt;/a&gt;, so pressure continues to increase.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-19T19:50:07.730" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34185" PostTypeId="2" ParentId="34168" CreationDate="2015-07-19T19:52:24.663" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.minsocam.org/msa/collectors_corner/faq/faqminprop.htm#LU&quot; rel=&quot;nofollow&quot;&gt;Luster&lt;/a&gt; is an optical property of minerals, and a mineral is defined as a solid inorganic substance that occurs naturally (that includes metals, of course). The Mineralogic Society of America defines metallic luster (see link) as something that occurs in minerals which have a refractive index greater than 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wasn't able to find any gases with a refractive index even close to the cutoff value. I'd say the answer is no, and that what might appear to be luster is instead scattering of incident light (not refraction).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-19T19:52:24.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34186" PostTypeId="2" ParentId="34168" CreationDate="2015-07-19T20:08:45.467" Score="5" Body="&lt;p&gt;Good question... would a &lt;em&gt;plasma&lt;/em&gt; qualify? Metallic luster is due to a sea of electrons, free to move and to reflect (well, re-radiate) incident radiation. A plasma can be created in a gas, such as the interstellar medium, or in Earth's atmosphere. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Kennelly%E2%80%93Heaviside_layer&quot;&gt;Kennelly-Heaviside layer&lt;/a&gt; is such a &quot;shiny&quot;, reflective layer, causing &lt;a href=&quot;https://en.wikipedia.org/wiki/Skywave&quot;&gt;skip signal propagation&lt;/a&gt; to extend the range of radio communication. So if you saw in radio wavelenghts, rather than the visual spectrum,  plasma is shiny!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility would be a &lt;a href=&quot;https://www.physicsforums.com/threads/what-does-a-bec-look-like.284597/&quot;&gt;Bose-Einstein condensate&lt;/a&gt;... but the act of illuminating it tends to destroy it.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-19T20:08:45.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34187" PostTypeId="1" CreationDate="2015-07-19T20:55:03.517" Score="0" ViewCount="348" Body="&lt;p&gt;I'd like to produce a synthetic version of foot odor for joke purposes. My company sells novelties.  What could I use to create something virtually identical? Would simply mixing the correct proportions of 3-Methylbutanoic acid, Propionic acid, Butyric acid &amp;amp; Methanethiol do it?  Would I need Ammonia?  (I found &lt;a href=&quot;http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0008951&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I probably don't need all the other fatty acids.  Since Methanethiol is a gas, how would I integrate that?&lt;/p&gt;&#xA;" OwnerUserId="17604" LastEditorUserId="402" LastEditDate="2017-04-25T23:11:54.447" LastActivityDate="2017-04-25T23:11:54.447" Title="Synthetic version of foot odor" Tags="&lt;smell&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34188" PostTypeId="1" CreationDate="2015-07-19T21:05:46.293" Score="20" ViewCount="13868" Body="&lt;p&gt;What specifically makes oceanic desalination so expensive?&lt;/p&gt;&#xA;" OwnerUserId="17605" LastEditorUserId="4325" LastEditDate="2015-07-23T03:40:52.717" LastActivityDate="2017-10-03T17:57:44.837" Title="What makes oceanic desalination so expensive?" Tags="&lt;water&gt;&lt;purification&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34190" PostTypeId="1" AcceptedAnswerId="34192" CreationDate="2015-07-19T22:08:25.380" Score="39" ViewCount="31183" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Polytetrafluoroethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polytetrafluoroethylene&lt;/a&gt; was discovered by accident. It now is an important material in the industry mainly because of its extremely high bonding energy, which prevents corrosion, halts reaction, and reduces friction (yeah &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon%E2%80%93fluorine_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon-fluorine bonds!&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And people would have definitely put it to the test, making it contain some of the most vicious and chemically diabolical substances ever created. There is a whole HOST of items it can contain that some chemists have gone so far as to say they were 'evil':&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dioxygen_difluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dioxygen Difluoride&lt;/a&gt; &lt;br&gt;&#xA;Known as the &lt;strong&gt;gas of Lucifer&lt;/strong&gt;, &lt;a href=&quot;http://www.lateralscience.co.uk/Fluorine/Fluorine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;there is a whole list of people blown up and killed while just trying to work with one of its components, fluorine.&lt;/a&gt; It ignites stuff at temperatures that most of the stuff that we breathe in would be in &lt;em&gt;liquid form&lt;/em&gt;. No one really knows about its atomic structure (obviously).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fluoroantimonic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fluoroantimonic Acid&lt;/a&gt; &lt;br&gt;&#xA;With a staggering pH of &lt;strong&gt;&lt;em&gt;-25,&lt;/em&gt;&lt;/strong&gt; it chews through stuff you might not even believe could be corroded; like &lt;em&gt;wax&lt;/em&gt; or glass. It can even strip hydrogen off of methane&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;...There are a lot of other chemical demons it can contain, but this is not the point. Let this suffice: &lt;a href=&quot;http://www.vp-scientific.com/Chemical_Resistance_Chart.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chemical Resistance Comparison&lt;/a&gt; (Spoiler: Fluorine is &lt;strong&gt;good&lt;/strong&gt; at this corrosion thing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this kind of hyper-resistance to about anything chemically destructive, &lt;strong&gt;&lt;em&gt;is there anything that can destroy Teflon through only chemical means?&lt;/em&gt;&lt;/strong&gt; A chemical that reacts exothermically to release heat, which melts the PTFE does not count. You get the drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am very curious as to whether there is anything more resilient than Teflon? Polytetrafluoroethylene is made of many carbon-fluorine bonds in series. &lt;em&gt;However,&lt;/em&gt; carbon-fluorine is second only to the Si-F bond. Is there an &quot;overclocked&quot; Teflon made of silicon-fluorine bonds that is even stronger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Now I know that some, but very few, solvents can make a mark on Teflon; but my question has not been answered: &lt;strong&gt;Is there any more resistant substances?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(More Teflon bragging: &lt;a href=&quot;http://www.aerocomfittings.com/images/chem1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;. Take that aqua regia)&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="402" LastEditDate="2019-08-06T04:43:52.140" LastActivityDate="2020-08-19T00:49:20.350" Title="Is there ANY chemical that can destroy PTFE, or Teflon?" Tags="&lt;polymers&gt;" AnswerCount="7" CommentCount="7" FavoriteCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="34191" PostTypeId="2" ParentId="34190" CreationDate="2015-07-19T22:30:05.267" Score="33" Body="&lt;p&gt;Corrosion Resistant Products, Ltd., with the help of Dupont, has established &lt;a href=&quot;http://www.crp.co.uk/technical.aspx?page=147&quot;&gt;this source of information&lt;/a&gt; on what can and cannot eat teflon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sodium and potassium metal - these reduce and defluorinate PTFE, which finds use in etching PTFE&lt;/li&gt;&#xA;&lt;li&gt;Finely divided metal powders, like aluminum and and magnesium, cause PTFE to combust at high temperatures&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These reactions probably reduce PTFE in a manner that starts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CF2CF2)_{n} + 2Na -&amp;gt; (CF=CF)_{n} +2NaF}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The world's most powerful oxidizers like $\ce{F2}$, $\ce{OF2}$, and $\ce{ClF3}$ can oxidize PTFE at elevated temperatures, probably by:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CF2CF2)_{n} + 2nF2 -&amp;gt; 2nCF4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar things can occur under extreme conditions (temperature and pressure) with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Boranes&lt;/li&gt;&#xA;&lt;li&gt;Nitric acid&lt;/li&gt;&#xA;&lt;li&gt;80% NaOH or KOH&lt;/li&gt;&#xA;&lt;li&gt;Aluminum chloride&lt;/li&gt;&#xA;&lt;li&gt;Ammonia, some amines, and some imines&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-19T22:30:05.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34192" PostTypeId="2" ParentId="34190" CreationDate="2015-07-19T23:25:28.570" Score="32" Body="&lt;p&gt;Just to add a bit to Ben's excellent answer...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A number of fluorinating agents also react with PTFE, &lt;span class=&quot;math-container&quot;&gt;$\ce{XeF2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{CoF3}$&lt;/span&gt; being examples&lt;/li&gt;&#xA;&lt;li&gt;Ben mentioned the reaction of magnesium metal.  Typically with metals, they must be in intimate contact with the PTFE surface, so molten metals or metals dissolved in anhydrous solvents will react.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The magnesium reaction is of special interest because it serves as the basis of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermite&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermite&lt;/a&gt; flare. A pyrotechnic device commonly used in the countermeasures aircraft use to evade heat-seeking missiles.  The reaction of metals with PTFE is given by the following equation (I think this is the general description for the reaction of metals with PTFE; I'm suspect of the reaction proposed by Ben involving the formation of poly-perfluoroacetylene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Mg + -(C2F4){-} → 2MgF2 + 2C}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formation of &lt;span class=&quot;math-container&quot;&gt;$\ce{MgF2}$&lt;/span&gt; is extremely exothermic.  The heat given off along with the carbon soot provides a new, much hotter, target for the attacking missile to lock onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether there is anything more resistant, I suspect that is unlikely.  The &lt;span class=&quot;math-container&quot;&gt;$\ce{C-F}$&lt;/span&gt; bond is shorter (135 pm) than the &lt;span class=&quot;math-container&quot;&gt;$\ce{Si-F}$&lt;/span&gt; bond (160 pm) and therefore better serves to encase and protect the carbon backbone.  While there are some other polymers that have better mechanical or thermal properties, I am not aware of any that have better chemical resistance. In &lt;a href=&quot;https://books.google.com/books?id=3QQhBQAAQBAJ&amp;amp;pg=PA361&amp;amp;lpg=PA361&amp;amp;dq=polymer%20that%20is%20more%20chemical%20resistant%20than%20teflon&amp;amp;source=bl&amp;amp;ots=tLHyK9wcuT&amp;amp;sig=3wOemSwcp2LxjzKpUte998iwEqI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CB0Q6AEwADgKahUKEwjnpLHboOjGAhUDzIAKHfpNC0Y#v=onepage&amp;amp;q=polymer%20that%20is%20more%20chemical%20resistant%20than%20teflon&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Polymers for Electronic &amp;amp; Photonic Application&lt;/a&gt; from 2013, the author states, &quot;PTFE is the most chemically resistant polymer known&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="402" LastEditDate="2019-08-06T04:43:26.540" LastActivityDate="2019-08-06T04:43:26.540" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34193" PostTypeId="1" AcceptedAnswerId="34202" CreationDate="2015-07-19T23:52:11.083" Score="57" ViewCount="15348" Body="&lt;p&gt;Common saying. Diamond possesses:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;ultra hardness,&lt;/strong&gt; (&lt;em&gt;10 on the Mohs scale; 10000 HV on Vicker's Hard Test (iron merely 30-80))&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;hyper thermal conductivity,&lt;/strong&gt; (&lt;em&gt;$2320~\mathrm{W\, m^{-1}\, K^{-1}}$, or over ten times better than the heatsink in your computer!&lt;/em&gt;) &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;extreme pressure resistance,&lt;/strong&gt; &lt;em&gt;(withstands a crushing 600 gigapascals; or around 2 times the pressure at the center of the earth, enough to snap carbon nano-tubes and graphene or create metallic oxygen or overcome copper's electron degeneracy pressure, making the maximum chamber pressure of a firing pistol seem literally like popping popcorn... I digress)&lt;/em&gt; &lt;/li&gt;&#xA;&lt;li&gt;and &lt;strong&gt;excellent luster&lt;/strong&gt; (what do you expect, it's a &lt;em&gt;diamond&lt;/em&gt;) combine to make the gemstone coveted by all.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Diamonds are the stuff of awesome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do they &lt;em&gt;really&lt;/em&gt; exist forever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia notes that, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diamond is less stable than graphite, but the conversion rate from diamond&#xA;   to graphite is negligible at standard conditions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Huh. But Wikipedia doesn't mention &lt;strong&gt;how&lt;/strong&gt; long. So how long would it take for this super-material to convert to the stuff I scribble with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you doubt the claims about diamond's seemingly unbelievable properties, check out the link on Wikipedia about diamond and &lt;a href=&quot;https://en.wikipedia.org/wiki/Orders_of_magnitude_(pressure)&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_thermal_conductivities&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/34193/are-diamonds-really-forever#comment52006_34202&quot;&gt;Great point Joe made&lt;/a&gt;, that $10^{80}$ is just forever to us puny humans. Being a geek I can't resist the urge to compare the time length $10^{80}$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Makes the entire lifespan of a red dwarf star seem like the Planck second.&lt;/li&gt;&#xA;&lt;li&gt;Enough time for you to sift through all the atoms in the entire UNIVERSE at a rate of one atom per second.&lt;/li&gt;&#xA;&lt;li&gt;Getting $67 worth of US quarters and flipping them, one per second, to get all heads-up.&lt;/li&gt;&#xA;&lt;li&gt;Chance of macroscopic quantum tunneling!! (I don't know precisely how much, but quite large)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;...and this is $10^{80}$ &lt;em&gt;seconds&lt;/em&gt; I'm talking about...&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="16683" LastEditDate="2017-09-05T14:02:51.533" LastActivityDate="2020-04-17T22:35:49.577" Title="Are diamonds really forever?" Tags="&lt;carbon-allotropes&gt;&lt;pressure&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="34194" PostTypeId="1" CreationDate="2015-07-20T00:16:30.587" Score="3" ViewCount="64" Body="&lt;p&gt;I asked this over at &lt;a href=&quot;https://physics.stackexchange.com/questions/192884/what-factors-are-influencing-a-woosh-rocket?noredirect=1#comment410904_192884&quot;&gt;Physics&lt;/a&gt;, but no one seemed to know, so I'll try here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I fill a plastic bottle with a small amount of rubbing alcohol, shake it around and hold a match to the bottle neck, I'll get a flame wooshing out of the opening. This can easily be used to propel the bottle upwards, like a rocket. I tried it already, there are lots of youtube videos about this, and NASA has instructions for the experiment as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to go for maximum speed or height of the rocket, there are a few factors influencing this experiment. Here are the ones that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some related to the bottle:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;volume, shape, aerodynamics, weight, size of opening, possibly shape of the bottle (long and thin or short and fat) When the bottle neck is smaller it creates a stronger thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Others related to the propellant:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How fast does it burn, how dense is it (especially compared to air), how heavily does it vaporize. If the vapor is denser then air it all floats out of the opening of my bottle before I can light it and I don't have as much fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might be able to see, I don't know a lot about the propellants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the factors related to the propellant, that influence the launch height and speed of the woosh rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What's the best propellant to choose for a woosh rocket, and why?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17608" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-07-20T13:47:16.493" Title="What factors are influencing a woosh rocket?" Tags="&lt;home-experiment&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34195" PostTypeId="2" ParentId="33884" CreationDate="2015-07-20T00:18:24.827" Score="3" Body="&lt;p&gt;After looking further into indigo production, I came across a related structure called &lt;a href=&quot;http://wikipedia.qwika.com/de2en/Indirubin&quot; rel=&quot;nofollow noreferrer&quot;&gt;indirubin&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Indirubin&lt;/strong&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hIndf.png&quot; alt=&quot;indirubin&quot;&gt;&lt;br&gt;&#xA;&lt;strong&gt;Indigo&lt;/strong&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sMu2K.png&quot; alt=&quot;indigo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indirubin is a byproduct of indigo synthesis and has a reddish color. It may also be more soluble than indigo, and indirubin derivatives have some &lt;a href=&quot;http://www.nature.com/ncb/journal/v1/n1/full/ncb0599_60.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;anticancer activity&lt;/a&gt;. If you trust &lt;a href=&quot;http://www.alibaba.com/product-detail/High-quality-low-price-100-purity_1698536577.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this supplier&lt;/a&gt;, you can buy indirubin for pretty cheap to compare with your lab's indigo reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJbRu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found a study done on &lt;a href=&quot;http://cool.conservation-us.org/jaic/articles/jaic31-02-007_4.html#fig8&quot; rel=&quot;nofollow noreferrer&quot;&gt;old native american dyes&lt;/a&gt; that studied indigo and indirubin in fabric fibers, among other dyes. They provide a UV-VIS spectrum of both compounds, with Ind = indigo and Inr = indirubin. We can see that indigo has an absorption peak in the red region, so blue is reflected. Indirubin has a peak at around 540 nm, which is yellow-green, allowing some blue and red to reflect, but probably more red than blue. This paper also references a 1910 paper** studying indirubin levels in commercially available indigo samples, but I can't find that article. They suggest that higher levels of indirubin are produced when more indoxyl is converted to &lt;a href=&quot;https://en.wikipedia.org/wiki/Isatin&quot; rel=&quot;nofollow noreferrer&quot;&gt;isatin&lt;/a&gt;. If you can find this 1910 article, it may provide some insight into reducing isatin and indirubin levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3g0DQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** &lt;a href=&quot;http://pubs.rsc.org/en/content/articlelanding/1910/ct/ct9109701460&quot; rel=&quot;nofollow noreferrer&quot;&gt;William Popplewell Bloxam and A. G. Perkin, &lt;em&gt;J. Chem. Soc., Trans.&lt;/em&gt;, &lt;strong&gt;1910&lt;/strong&gt;, &lt;em&gt;97&lt;/em&gt;, 1460-1475.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="137" LastEditorUserId="4945" LastEditDate="2015-07-21T09:12:20.457" LastActivityDate="2015-07-21T09:12:20.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34196" PostTypeId="2" ParentId="34176" CreationDate="2015-07-20T00:26:15.283" Score="0" Body="&lt;p&gt;Why don't you just run the calculation and see for yourself (rather than forever speculating)?  Do you know how to use the steam tables?  Suppose the water is in a 1 liter container initially at 20 C, and the liquid takes up 90% of the volume, while water vapor in equilibrium with the liquid water at 20 C occupies the remainder of the volume.  Now you raise the temperature to 50 C, and you allow the system to equilibrate.  What is the new volume of the liquid water and what is the new pressure.  Try higher temperatures and see what you get.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-20T00:26:15.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34197" PostTypeId="2" ParentId="24518" CreationDate="2015-07-20T01:02:31.117" Score="3" Body="&lt;p&gt;The question is &quot;how would we determine the change in entropy of a spontaneous process i.e irreversible process.&quot;  To do this, you first focus exclusively on the initial and final thermodynamic equilibrium states of the system, resulting from the irreversible path.  Then you need to conceive of a reversible path between the exact same initial and final equilibrium states (the reversible path may or may not necessarily bear any resemblance to the actual irreversible process).  Any reversible path between the two equilibrium states will suffice.  You then calculate the integral of dq/T for this reversible path.  This will be the change in entropy.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-20T01:02:31.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34198" PostTypeId="2" ParentId="31040" CreationDate="2015-07-20T01:17:47.733" Score="1" Body="&lt;p&gt;In calculating the change in entropy of a closed system that has undergone an irreversible process, you need to first focus exclusively on the initial and final equilibrium states.  You then need to identify a reversible process path between the exact same pair of equilibrium states.  The reversible process path does not necessarily  have to bear any resemblance to the actual irreversible process path.  Therefore, even if the irreversible process path is adiabatic, the reversible path does not have to be.  You then calculate the integral of dq/T for the reversible process path.  This is equal to the entropy change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple, huh.  I don't know why it isn't taught properly to students.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-20T01:17:47.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34199" PostTypeId="1" CreationDate="2015-07-20T01:30:22.120" Score="13" ViewCount="9200" Body="&lt;p&gt;I have been studying alkanes, alkenes &amp;amp; alkynes for a while, and I see in the examples that all the basic structural formulas start with $\ce{CH3}$, then $\ce{CH2}$, and the last carbon atom is $\ce{CH3}$, but the one that is attached to the single/double/triple bond is $\ce{CH}$. (Please correct me if I'm explaining it with the right terms.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, there's a question about drawing the structural formula for compounds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw a structural formula for hex-3-ene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What my answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CH3CH2CH=CHCH2CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought this was right, then I looked at the solutions and it was&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3CCH2CH=CHCH2CH3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now my question is, why does it start with $\ce{H3C}$ not $\ce{CH3}$ like some other examples? &lt;/p&gt;&#xA;" OwnerUserId="17609" LastEditorUserId="9961" LastEditDate="2017-10-30T13:02:51.210" LastActivityDate="2020-03-25T03:38:43.667" Title="Correctness of writing H3C as opposed to CH3 in a carbon chain" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;nomenclature&gt;&lt;notation&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34200" PostTypeId="2" ParentId="34199" CreationDate="2015-07-20T01:42:09.433" Score="19" Body="&lt;p&gt;Your answer is the same as the book's. The book wrote the first carbon as $\ce{H3C}$ to stress that the chain continues via a carbon to carbon bond, not bonded a carbon to hydrogen to carbon bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why it's not like this in other examples, technically it is not necessary to write it this way, but it's strange your book lacks that consistency.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-07-20T01:43:49.043" LastActivityDate="2015-07-20T01:43:49.043" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34201" PostTypeId="1" CreationDate="2015-07-20T02:05:47.277" Score="2" ViewCount="571" Body="&lt;p&gt;I'm simple man who doesn't know much about anything. I just enjoy my life. Lately I like chemistry a lot. I want to learn chemistry at home. How do you recommend I do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do i need to do chemical experiments to become good at chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can learn about reactions with books. For example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;O + H2 -&amp;gt; H2O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Also organic chemistry like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C2H5OH + O2 -&amp;gt; CO2 + H2O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What else do I need to learn? What do chemists do all day if all formulas can be calculated on paper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does chemist invent new materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I be theoretical chemist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good hobby choice?&lt;/p&gt;&#xA;" OwnerUserId="14878" LastActivityDate="2015-07-20T02:05:47.277" Title="How to learn chemistry at home?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-07-20T23:32:25.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="34202" PostTypeId="2" ParentId="34193" CreationDate="2015-07-20T02:27:28.410" Score="46" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;how long would it take for this super-material to convert to the stuff&#xA;  I scribble with?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, despite the fact that James Bond said &quot;&lt;a href=&quot;http://www.imdb.com/title/tt0066995/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diamonds are Forever&lt;/a&gt;&quot;, that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we will soon see, even though diamond is slightly less stable than graphite (by ~ 2.5 kJ/mol), it is kinetically protected by a large activation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a comparative representation of the structures of diamond and graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zytVR.jpg&quot; alt=&quot;structures of diamond and graphite&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image source: &lt;a href=&quot;http://jidonline.com/article.asp?issn=2229-5194;year=2011;volume=1;issue=2;spage=93;epage=100;aulast=Satyanarayana;type=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Satyanarayana T, Rai R. Nanotechnology: The future. J Interdiscip Dentistry 2011;1:93-100&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9uPI3.gif&quot; alt=&quot;environment of a carbon in diamond and graphite&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://chemistry.stackexchange.com/questions/15422/is-the-valency-of-carbon-satisfied-in-graphite&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that diamond is composed of cyclohexane rings and &lt;strong&gt;each carbon is bonded to 2 more carbons external to the cyclohexane ring&lt;/strong&gt;. On the other hand, graphite is comprised of benzene rings and &lt;strong&gt;each carbon is bonded to only 1 carbon external to the benzene ring&lt;/strong&gt;. That means we need to break 6 sigma bonds in diamond and make about 2 pi bonds (remember it's an extended array of rings, don't double count) in graphite per 6-membered ring in order to convert diamond to graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical aliphatic C–C bond strength is ~340 kJ/mol and a typical pi bond strength is ~260 kJ/mol.  So to break 6 sigma bonds and make 2 pi bonds would require ~((6*340)-(2*260)) ~ 1500 kJ/mol. If the transition state were exactly midway between diamond and carbon (with roughly equal bond breaking and bond making), then we might approximate the activation energy as being half that value or ~750 kJ/mol.  Since graphite is a bit more stable than diamond, we can refine our model and realize that the transition state will occur a bit before the mid-point. So our refined model would suggest an activation energy something less than 750 kJ/mol. Had we attempted to incorporate the effect of aromaticity in graphite our estimate would be even lower.  In any case, this is an &lt;strong&gt;extremely&lt;/strong&gt; large activation energy, so, as we anticipated the reaction would be &lt;strong&gt;very&lt;/strong&gt; slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://ses.library.usyd.edu.au/bitstream/handle/2123/3682/Chapter%207%20Theoretical%20investigation.pdf;jsessionid=E93A00598143913AB59A7C644986D929?sequence=8&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimate&lt;/a&gt; (see p. 171) of the activation energy puts the reverse reaction (graphite to diamond; but since, as noted above, the energy difference between the two is very small the activation energy for the forward reaction is almost the same) at 367 kJ/mol. So at least our rough approximation was in the right ballpark, off by about a factor of 2. However, it appears that the transition state is even further from the midpoint (closer to starting material) than we might have guessed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This activation energy tells us that at 25 °C, it would take &lt;strong&gt;well over a billion years&lt;/strong&gt; to convert one cubic centimeter of diamond to graphite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note 04/17/20:&lt;/strong&gt; As mentioned in a comment, the original &quot;estimate&quot; link became  defunct and was replaced today with a new estimate link.  However the original article and estimate can can still be seen on the &lt;a href=&quot;https://web.archive.org/web/20140617023942/http://www.geo.cornell.edu/geology/classes/geo455/Problems/455%2013ProblemSet05Solutions.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wayback Machine&lt;/a&gt; and it estimates the activation energy to be 538.45 kJ/mol, reasonably close to our estimate.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2020-04-17T22:28:29.053" LastActivityDate="2020-04-17T22:28:29.053" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34203" PostTypeId="1" AcceptedAnswerId="34261" CreationDate="2015-07-20T03:12:10.303" Score="-1" ViewCount="130" Body="&lt;p&gt;What are the most common critical thermodynamic and kinetic variables  required to convert a secondary or tertiary alcohol to an alkane.? Is bond enthalpy data relevant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the process/processes commonly used to then calculate or determine the temperature and duration it would take to complete the reaction from the thermodynamic and kinetic data?&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="16111" LastEditDate="2015-11-26T07:50:02.653" LastActivityDate="2015-11-26T07:50:02.653" Title="How to calculate the requirements for completing a reaction" Tags="&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34204" PostTypeId="1" CreationDate="2015-07-20T03:33:15.297" Score="5" ViewCount="2420" Body="&lt;p&gt;As extracting scorpion venom goes, how would one most effectively store and preserve the venom for future research?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are lyophilization, however I am not sure that would be the best of all methods in this case. Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="16558" LastEditorUserId="402" LastEditDate="2015-10-12T03:53:09.157" LastActivityDate="2021-01-05T19:15:38.090" Title="What methods could be used to preserve scorpion venom?" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34206" PostTypeId="2" ParentId="34168" CreationDate="2015-07-20T05:06:55.253" Score="2" Body="&lt;p&gt;Not an answer to your question ... but This may interest you! (Or your curiosity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metallic Gas : &lt;a href=&quot;http://www.popsci.com/science/article/2011-11/researchers-turn-hydrogen-gas-metal&quot; rel=&quot;nofollow&quot;&gt;http://www.popsci.com/science/article/2011-11/researchers-turn-hydrogen-gas-metal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnetic Gas : &lt;a href=&quot;http://www.popsci.com/scitech/article/2009-09/scientists-create-first-ever-magnetic-gas&quot; rel=&quot;nofollow&quot;&gt;http://www.popsci.com/scitech/article/2009-09/scientists-create-first-ever-magnetic-gas&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As for your question :&lt;/strong&gt; I guess not!  The colours of gases are caused by selective absorption of light so that the light coming through the gas is coloured - it is not light reflected from the gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflection occurs when light passes from an optically dense medium to an optically less dense medium, or vice versa. In other words, there has to be a change in the optical properties of a gas to cause reflection. The gas molecules individually are to small to significantly effect the photons of light. &lt;/p&gt;&#xA;" OwnerUserId="7434" LastActivityDate="2015-07-20T05:06:55.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34207" PostTypeId="1" AcceptedAnswerId="34225" CreationDate="2015-07-20T05:36:27.147" Score="4" ViewCount="2226" Body="&lt;p&gt;I was helping some students today, quite successfully, until I came across this problem:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ce{A &amp;amp;&amp;lt;=&amp;gt;[$k_+$][$k_-$] B + C} \\[5pt]&#xA;\ce{C + D &amp;amp;-&amp;gt;[$k_1$] E}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation is a fast process  with forward and reverse rate constants $k_+$ and $k_-$, respectively.&#xA;The second process is slow, with rate constant $k_1$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to write down a rate law for the resulting equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A + D -&amp;gt; B + E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique that had been working was to write down the rate for the slow process...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r = k_1[\ce{C}][\ce{D}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{C}$ is a product in the fast reaction, which is an equilibrium equation, we also have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_+[\ce{A}]=k_-[\ce{B}][\ce{C}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where things went awry, because the rate law that we seek is a function of $[\ce{A}]$ and $[\ce{D}]$ only. Considering the stoichiometry, it seemed appropriate to put $[\ce{B}]=[\ce{C}]$. This leads to the substitution $[\ce{C}]=(\frac{k_+}{k_-})^{1/2}[\ce{A}]^{1/2}$. The interface has declared the result to be invalid, so it must be an illegal maneuver. How, then, do I eliminate $[\ce{B}]$ and $[\ce{C}]$ and come up with the desired reaction rate law?&lt;/p&gt;&#xA;" OwnerUserId="15363" LastEditorUserId="16683" LastEditDate="2016-12-15T09:16:00.470" LastActivityDate="2016-12-15T14:47:09.497" Title="Determine the rate equation (intermediate reaction)" Tags="&lt;kinetics&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34208" PostTypeId="2" ParentId="28682" CreationDate="2015-07-20T07:04:42.143" Score="10" Body="&lt;p&gt;I'm really excited for this because I get to reference the almighty color wheel!! Fair warning, this answer is much more qualitative than quantitative, but that's more interesting sometimes anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the structure of the chromate and dichromate ions next to each other (see here for structures: &lt;a href=&quot;https://en.wikipedia.org/wiki/Chromate_and_dichromate&quot;&gt;https://en.wikipedia.org/wiki/Chromate_and_dichromate&lt;/a&gt;), the only major difference between the two is that the Cr-O bond joining the two chromate ions (missing an oxygen) is now a single bond. This means that bond will vibrate at a lower frequency, and because frequency and wavelength are inversely related, that bond will absorb a longer wavelength of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the color wheel! It is a general chemistry (often unexplained) fact that the color we see is the complementary color of the wavelength of a bond's vibration. Thus, in the case of the chromate ion, we see yellow, and across from yellow is the purple-ish region. That means, if one of the bonds in the chromate ion, and thus two of the bonds in the dichromate ion, were absorbing a longer wavelength like we said earlier, on average we would expect something just longer than purple-ish, like blue, to be absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complementary color of blue is red slash orange, and that is in fact the color we see in the dichromate ion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the heart of all this is the principle that the colors we see are those wavelengths of light which &lt;strong&gt;on average&lt;/strong&gt; are not absorbed by a large number (on the order of Avogadro's number) of molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An approach like this will only be reliable for very similar molecules like the two we have here.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-07-20T07:04:42.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34209" PostTypeId="1" CreationDate="2015-07-20T07:34:16.083" Score="3" ViewCount="1864" Body="&lt;p&gt;I have a macbook battery that shows as completely dead (doesn’t hold the laptop on even for a second). It now started bulging, and in last week it inflated to about 50 % extra thickness, like this – I can see the inflated battery packs filled with gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://cen.acs.org/content/dam/cen/85/51/8551sci1_inflating.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big of a danger is to keep a battery like this at home? Will the reaction stop, or will it continue and will the pressure rip through the battery packs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that happens, will the battery combust? Internet is full of videos of exploding batteries, but what’s the real likelihood? One in a million? Or one out of two if it swells like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an advice to keep the battery in a bucket of salty water, does that make sense (I remember that water + lithium = intensive fire)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are Li-ion batteries disposed of when I take them into store?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask here, because I couldn’t find a qualified answers on the internet, only scare articles about explosions and vague conflicting advices. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17614" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:45.637" LastActivityDate="2016-07-26T14:12:34.377" Title="Bulging Li-ion battery – is it an immediate danger? What’s the likelihood of explosion?" Tags="&lt;electrochemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34210" PostTypeId="1" AcceptedAnswerId="34211" CreationDate="2015-07-20T07:38:03.647" Score="4" ViewCount="160" Body="&lt;p&gt;&lt;a href=&quot;http://groups.chem.ubc.ca/chem260/Stereochemistry1KEY2013.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is the link to the page in which I don't understand the answer to the last question (Question 13). The question states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Draw all of the constitutional isomers and stereoisomers of compounds named&#xA;  tribromocyclobutane (all structures must contain a cyclobutane ring).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The following image exists as a part of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrByW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is the following image not part of the answer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Wrav.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the above two images not diastereomers (hence stereoisomers)?&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="4325" LastEditDate="2015-07-20T16:20:06.683" LastActivityDate="2015-07-20T16:20:06.683" Title="Stereoisomers of compounds named tribromocyclobutane" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34211" PostTypeId="2" ParentId="34210" CreationDate="2015-07-20T08:02:30.067" Score="4" Body="&lt;p&gt;The second configuration shown cannot exist simply because of the geometry of the orbitals at the carbon center. The angle between the bromine's (those on the same carbon) is much less than 108° even though there is no forced strain.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-07-20T08:02:30.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34212" PostTypeId="1" CreationDate="2015-07-20T09:05:07.390" Score="20" ViewCount="12685" Body="&lt;p&gt;Silver, Gold and Platinum are amongst the best conductors of electricity, but also the amongst the most unreactive. Since electrical conductivity depends on the number of delocalized electrons (along with resistivity), if an element gives up its electrons readily (making it highly reactive), then shouldn't that mean it will be a great electrical conductor? Why isn't Cesium the best conductor? &lt;/p&gt;&#xA;" OwnerUserId="17615" LastEditorUserId="17615" LastEditDate="2015-07-20T12:50:14.960" LastActivityDate="2017-11-09T22:23:00.077" Title="Why is it that the least reactive metals are the best electrical conductors?" Tags="&lt;metal&gt;&lt;transition-metals&gt;&lt;metallurgy&gt;&lt;reactivity&gt;&lt;conductivity&gt;" AnswerCount="4" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34213" PostTypeId="1" AcceptedAnswerId="34216" CreationDate="2015-07-20T09:41:30.990" Score="4" ViewCount="1447" Body="&lt;p&gt;Does deuterium peroxide exist? Wikipedia has no reference. Also, the sites for delivery of chemicals like &quot;Pubchem&quot; or &quot;Chemspider&quot; have no reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my book, a preparation of $\ce{D2O2}$ is given,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{K2S2O8 + 2D2O -&amp;gt; 2KDSO4 +D2O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\ce{D2O2}$ exist, what are its physical properties (structure and appearance etc.)?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorDisplayName="user7951" LastEditDate="2015-07-20T09:49:04.913" LastActivityDate="2015-07-20T15:56:47.573" Title="Existence of Deuterium peroxide" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;structural-formula&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34214" PostTypeId="1" CreationDate="2015-07-20T10:31:02.263" Score="6" ViewCount="1543" Body="&lt;p&gt;I'm currently studying about the hydrogen bond's Potential Energy Surface (PES) pattern.&lt;br&gt;&#xA;As per literature, single point energy obtained from Density Functional Theory (DFT) using MP2 approximation performed with Gaussian software (www.gaussian.com) showed the pattern of Lennard Jones Potential (LJ-potential). A Lennard Jones is a one-well potential &lt;a href=&quot;http://www.cmbi.ru.nl/redock/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cmbi.ru.nl/redock/&lt;/a&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/r1VUP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r1VUP.png&quot; alt=&quot;LJ-potential&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;However, in certain literature it is also shown to obey the double welled potential model (PES_vs._atomic-distance response).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, I'm basically wondering about the physical significance of a double-well potential model with respect to interacting-atoms, in case of hydrogen bonds. Any insight or reference towards deciphering the concerned fact would be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16786" LastEditorUserId="4325" LastEditDate="2015-07-23T20:37:40.790" LastActivityDate="2015-07-23T20:37:40.790" Title="Physical significance of double-well potential in quantum bonding" Tags="&lt;hydrogen-bond&gt;&lt;electrostatic-energy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34215" PostTypeId="1" AcceptedAnswerId="101729" CreationDate="2015-07-20T10:44:42.423" Score="20" ViewCount="253" Body="&lt;p&gt;Recently, there has been a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34078/the-lewis-structure-for-nco&quot;&gt;question by Voldemort&lt;/a&gt; concerning different resonance structures of &lt;span class=&quot;math-container&quot;&gt;$\ce{NCO-}$&lt;/span&gt;, requesting an explanation why one resonance structure would be more preferred than another.&lt;/p&gt;&#xA;&lt;p&gt;There has been a little discussion on that between &lt;a href=&quot;https://chemistry.stackexchange.com/users/4231/ron&quot;&gt;ron&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/users/4945/martin-%E3%83%9E%E3%83%BC%E3%83%81%E3%83%B3&quot;&gt;Martin - マーチン&lt;/a&gt; that took my interest:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;ron:&lt;/strong&gt; &amp;quot;[...] structure #1 (&lt;span class=&quot;math-container&quot;&gt;$\ce{N#C-O-}$&lt;/span&gt;) is more stable (preferred) than structure #2 (&lt;span class=&quot;math-container&quot;&gt;$\ce{^{-}N=C=O}$&lt;/span&gt;).&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Martin - マーチン:&lt;/strong&gt; When talking about resonance structures, I think stable is the wrong word. I agree with your analysis in terms of contribution. Therefore #1 should have the highest contribution of all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now while I agree with ron that structure #1 is the most preferred resonance structure (referring to electronegativities), I regret with him concerning his attitude towards the &amp;quot;stability&amp;quot; as Martin does.&lt;/p&gt;&#xA;&lt;p&gt;But I mainly do this, because I used NBO6 with the Natural Resonance Theory (NRT) keyword on the Natural Bond Orbitals (NBO) produced of an MP2/6-311+G(2d) optimized wavefunction to find the contributions of all relevant resonance structures - &lt;span class=&quot;math-container&quot;&gt;$66~\%~(\ce{N#C-O-})$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$30~\%~(\ce{^{-}N=C=O})$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3~\%~(\ce{^{2-}N-C#O+})$&lt;/span&gt; - and not because I have a superior understanding on all underlying theories.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;For me, the resonance stuff lookes a lot like MCSCF treatment of some VB wavefunction, which means that the complete molecule is described best from it's contributing resonance structures.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now on a more complete comparison, &lt;strong&gt;what are the main similarities and differences between both theories&lt;/strong&gt; and as they don't exist but are rather part of an explanation, can resonance structures indeed be described as &amp;quot;stable&amp;quot;?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-09-14T23:25:30.410" Title="Similarities and Differences between Resonance and MCSCF treatments" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34216" PostTypeId="2" ParentId="34213" CreationDate="2015-07-20T11:15:17.970" Score="11" Body="&lt;p&gt;It does exist. At least Google shows one supplier situated in &lt;a href=&quot;http://www.iconisotopes.com/code/ShowPDF.asp&quot; rel=&quot;noreferrer&quot;&gt;New Jersey&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ID 1130   Deuterium Peroxide (30% in D2O) D2O2    98  5   5.0g / 10.0g&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Though for some reason I could not find any reputable supplier like Cambridge Isotopes or Sigma-Aldrich (strange!). As far as physical properties are concerned, as pointed out by @snurden there exists a significant amount of literature. Here is one article from the &lt;a href=&quot;http://dx.doi.org/10.1139/v51-022&quot; rel=&quot;noreferrer&quot;&gt;Canadian Journal of Chemistry&lt;/a&gt; that includes study of many properties (behind a pay wall though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstract:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fTPDG.png&quot; alt=&quot;Abstract of above paper&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to answer final part of your question:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/j06R0.png&quot; alt=&quot;Result of same paper&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14960" LastEditorUserId="4325" LastEditDate="2015-07-20T15:56:47.573" LastActivityDate="2015-07-20T15:56:47.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34217" PostTypeId="2" ParentId="34190" CreationDate="2015-07-20T11:18:52.683" Score="24" Body="&lt;p&gt;I am aware that this answer does not describe a ‘chemical that can destroy PTFE’. However, since you are also asking ‘is there anything that can destroy Teflon through only chemical means?’, and in order to complete the other answers, I would like to add that PTFE can be easily destroyed by means of &lt;strong&gt;radiation chemistry&lt;/strong&gt;. (Radiation chemistry is the study of the chemical effects of radiation on matter. It is not to be confused with radiochemistry, which is the chemistry of radioactive materials.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PTFE is exceptionally sensitive to radiation. PTFE experiences significant damage at lower radiation exposure levels than other polymers. In general, PTFE is considered usable without any significant restrictions only for absorbed doses of up to $100\ \mathrm{Gy}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general radiation effects on polymers are formation of gas, cross-linking of polymer chains, and scission of polymer chains. In contrast to most other polymers, all hydrogen atoms in PTFE are substituted by fluorine atoms; therefore, the elimination of hydrogen or hydrogen fluoride and the corresponding formation of double bonds do not occur. The elimination of fluorine due to $\ce{C-F}$ bond break is possible; however, since $\ce{C-C}$ bonds are clearly weaker than $\ce{C-F}$ bonds, $\ce{C-C}$ bond break predominates. Therefore, the primary effect of radiation on PTFE is the scission of the polymer chain (breaking of the large polymer molecule into smaller parts). Due to the absence of π-electrons, the excited states are not particularly stabilized, resulting in high radiation-chemical yields. Because of the absence of unsaturated bonds, the absence of functional groups that can be easily eliminated, and the general chemical inertness, PTFE cannot be cross-linked like an elastomer. Therefore, the scission of the polymer chain is not compensated for by the formation of new bonds. With advancing chain scission, the molar weight of the polymer is dramatically reduced from an initial value of about $6\times10^6\ \mathrm{g\ mol^{-1}}$. It is divided by $4$ after $250\ \mathrm{Gy}$ exposure and by about $20$ for a dose of $1000\ \mathrm{Gy}$.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of molecular weight reduction is primarily on mechanical properties. Tensile strength is reduced by $25\ \%$ after $500\ \mathrm{Gy}$ exposure and by $50\ \%$ for a dose of about $900\ \mathrm{Gy}$. For doses above $1500\ \mathrm{Gy}$, imminent material failure has to be assumed because of a brittle behaviour with unstable crack propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, PTFE components should be avoided in high radiation environments (nuclear power plants, nuclear fuel cycle facilities, irradiation facilities, or particle accelerators).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* &lt;a href=&quot;http://dx.doi.org/10.1016/S0032-3861(03)00116-2&quot;&gt;Fayolle, B.; Audouin, L.; Verdu, J.: Radiation induced embrittlement of PTFE. In: &lt;em&gt;Polymer&lt;/em&gt; 44 (2003) 2773–2780.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-07-20T11:18:52.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34218" PostTypeId="1" AcceptedAnswerId="34221" CreationDate="2015-07-20T11:39:52.947" Score="3" ViewCount="1322" Body="&lt;p&gt;How to know if the following chemical structure of morphine is an R or S? What I basically need to know is: which chirality center do we use to make the decision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2in$&lt;img src=&quot;https://i.stack.imgur.com/7dWojt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="16622" LastEditDate="2015-07-20T21:40:26.690" LastActivityDate="2015-07-20T21:40:26.690" Title="Defining enantiomer of morphine by R/S designation" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34219" PostTypeId="1" CreationDate="2015-07-20T10:08:12.833" Score="4" ViewCount="805" Body="&lt;p&gt;Why in odd electron sharing molecules (like NO ) the bond containing 3 electrons is considired to be half covalent bond?&lt;/p&gt;&#xA;" OwnerUserId="16864" OwnerDisplayName="user83120" LastEditorUserId="9961" LastEditDate="2015-07-20T19:11:30.117" LastActivityDate="2016-03-02T15:32:52.710" Title="Covalent Bonding in odd electron sharing molecules" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34220" PostTypeId="2" ParentId="34212" CreationDate="2015-07-20T12:20:34.490" Score="1" Body="&lt;p&gt;The simplest answer I can give is that silver (e.g) is a good conductor of electricity because of its sea of delocalised electrons between the layers of metal ions. However it is not highly reactive because you would need a lot of energy to break the metallic bonds within each of those layers.  &lt;/p&gt;&#xA;" OwnerUserId="17619" LastEditorUserId="17619" LastEditDate="2015-07-20T12:40:31.620" LastActivityDate="2015-07-20T12:40:31.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34221" PostTypeId="2" ParentId="34218" CreationDate="2015-07-20T12:38:48.423" Score="7" Body="&lt;p&gt;The structure that you are showing is neither R nor S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it has five chirality centers you need to look at all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip3in$&lt;img src=&quot;https://i.stack.imgur.com/cF8ib.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see in the above figure that there are three R and two S centers in the configuration of morphine that you posted.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-20T12:38:48.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34222" PostTypeId="2" ParentId="34194" CreationDate="2015-07-20T13:47:16.493" Score="3" Body="&lt;p&gt;For home projects like this I believe one of the most important things is to ensure you have a good fuel to air (oxygen) ratio. It is surprisingly easy to use too much propellant which can take up air space within the combustion chamber. More fuel doesn't necessarily mean bigger bang. &lt;/p&gt;&#xA;" OwnerUserId="17619" LastActivityDate="2015-07-20T13:47:16.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34223" PostTypeId="1" AcceptedAnswerId="34224" CreationDate="2015-07-20T15:34:35.480" Score="3" ViewCount="3847" Body="&lt;p&gt;Can anyone tell me what the difference between catenation and allotropy is? I would like the basic explanation.&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="7448" LastEditDate="2015-07-20T23:23:42.767" LastActivityDate="2015-07-20T23:23:42.767" Title="Difference between catenation and allotropy" Tags="&lt;inorganic-chemistry&gt;&lt;elements&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34224" PostTypeId="2" ParentId="34223" CreationDate="2015-07-20T16:10:01.813" Score="4" Body="&lt;p&gt;Catenation refers to the ability of an element to form long chains. The most notable example is carbon (catenation of carbon is what gives rise to the entire field of organic chemistry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allotropy is an entirely different concept; it refers to the natural existence of an element in different forms, i.e. bonded differently. For example, solid carbon exists in the forms of diamond and graphite (and more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense, the two are related because the ability of an element to catenate often leads to the formation of many different allotropes, since the chains can be linked in different ways. Apart from carbon, another example of an element that can catenate well and forms multiple allotropes would be phosphorus. On the other hand, nitrogen and oxygen prefer to form triple and double bonds respectively, instead of single bonds; this means that catenation is unlikely and that is why these elements have few allotropes (nitrogen exists as diatomic $\ce{N2}$, oxygen as diatomic $\ce{O2}$ and ozone).&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-07-20T16:10:01.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34225" PostTypeId="2" ParentId="34207" CreationDate="2015-07-20T16:47:46.450" Score="5" Body="&lt;p&gt;You can't write $[\ce{B}] = [\ce{C}]$ because  $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be &quot;negative&quot; production of $\ce{C}$ from $\ce{A}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-12-15T09:02:32.847" LastActivityDate="2016-12-15T09:02:32.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34226" PostTypeId="2" ParentId="34207" CreationDate="2015-07-20T17:32:09.763" Score="4" Body="&lt;p&gt;The best first step in any kinetics problem is writing down materials balances on all relevant species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dA}{dt}=-k_+ A  +  k_-B C$$&#xA;$$\frac{dB}{dt}=+k_+ A  -  k_-B C$$&#xA;$$\frac{dC}{dt}=+k_+ A  -  k_-B C  -  k_1 C D$$&#xA;$$\frac{dD}{dt}=-  k_1 C D$$&#xA;$$\frac{dE}{dt}=  k_1 C D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to solve for $\frac{dE}{dt}$ by eliminating $C$.  The assumption that the first reaction is &quot;fast&quot; means we can make the equilibrium assumption on $A$ and $B$ to solve for $BC$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BC = \frac{k_+}{k_-}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can eliminate $C$ by stoichiometric considerations.  The amount of $C$ in the system at any time is just equal to the net forward progress of reaction 1 minus the forward progress of reaction 2.  So $C=A_0-A - (D_0-D)=(A_0-D_0)-(A-D)$.  orthocresol's answer does a great job of explaining why.  &lt;strong&gt;But we can also make a similar argument for B: the amount of B is just equal to the net forward progress of reaction 1, or $B=A_0-A$.&lt;/strong&gt;  So after application of stoichiometry, the equilibrium equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BC = \frac{k_+}{k_-}A \implies \left[A_0-A \right] \left[(A_0-D_0)-(A-D)\right] = \frac{k_+}{k_-}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a quadratic equation in $A$ (I won't show the math) that gives $A$ if you know the constants $A_0$, $D_0$, $k_+$, and $k_-$, and the variable $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting our expression for $C$ into the equation for $D$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dD}{dt}=  - k_1 C D = -k_1 D [(A_0-D_0)-(A-D)] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could substitute the quadratic solution for $A$ into this equation and get a single differential equation that depends only on $D$ and constants.  Since the ultimate rate we care about, $\frac{dE}{dt}$, is equal to $-\frac{dD}{dt}$, this means we have a rate law. The algebra is messy but that is the path forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some problems are harder than they look!&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2016-12-15T14:47:09.497" LastActivityDate="2016-12-15T14:47:09.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34227" PostTypeId="2" ParentId="34209" CreationDate="2015-07-20T17:43:36.930" Score="3" Body="&lt;p&gt;The battery may have already ruptured due to thermal runaway, meaning that despite it being bulged, pressure from boiling electrolyte has already escaped.  In this case it is also likely that the lithium graphite electrode has already degraded through reaction with air.  If this is true, you are lucky that the previous thermal runaway did not result in a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is impossible to say that the the battery is completely dead without more information, information that would require making measurements that are difficult or impossible for most home chemists.  Thus there may well still be appreciable amounts of flammable electrolyte and air-reactive lithium graphite remaining in the battery. If heated or allowed to react with air, any such material could re-pressurize the system, resulting in a fire or small explosion.  &lt;strong&gt;In awareness of this possibility, you should still treat the battery as dangerous.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2016-07-26T14:12:34.377" LastActivityDate="2016-07-26T14:12:34.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34229" PostTypeId="1" AcceptedAnswerId="34249" CreationDate="2015-07-20T19:13:13.577" Score="4" ViewCount="1997" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution of $\pu{5 g}$ of haemoglobin in $\pu{100 cm^3}$ of solution shows a temperature raise of $\pu{0.031 ^\circ C}$ for complete oxygenation. Each mole of haemoglobin binds $\pu{4 mol}$ of oxygen. If the heat capacity of the solution is $\pu{4.18 J K^{-1} cm^{-3}}$, calculate $\Delta H$ per g of oxygen bond.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for $\pu{100 cc}$, heat capacity is $\pu{418 J K^{-1}}$&lt;/li&gt;&#xA;&lt;li&gt;for $\pu{\frac{5}{64000}mol}$ of haemoglobin, $C\Delta T=\pu{12.952 J}$ is released&lt;/li&gt;&#xA;&lt;li&gt;hence for $\pu{1 mol}$, $\pu{165862.4 J}$ energy is released&lt;/li&gt;&#xA;&lt;li&gt;This energy is released when four bonds are formed with oxygen. Hence for 1 bond, $\pu{41.47 kJ}$ energy is released.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer given in the book is $\pu{-41.47 kJ}$.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="194" LastEditDate="2018-02-19T04:46:44.387" LastActivityDate="2018-02-19T04:46:44.387" Title="Calculating enthalpy per gram mole of oxygen bond in haemoglobin" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34231" PostTypeId="1" CreationDate="2015-07-21T00:06:33.487" Score="3" ViewCount="121" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_airship&quot; rel=&quot;nofollow&quot;&gt;According to Wikipedia,&lt;/a&gt; a &lt;strong&gt;vacuum airship&lt;/strong&gt; is a hypothetical airship that is evacuated instead of using a lighter gas. Really, once I first saw this, I thought, &quot;Why did I never think about this before?!&quot; Indeed, whenever I pictured an evacuated container, I thought it would just sit there. &lt;br&gt; But &lt;em&gt;no&lt;/em&gt;. Instead here on Earth, a theoretical vacuum airship could lift 1.28 grams per liter of evacuated space, a 14% boost over helium, owing to the density of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the air is also the destroyer of this powerful capability; it exerts 14 psi, or ~0.1 MPa!! According to Wikipedia's calculations, a hemispherical shell would have to withstand $$450,000 kg^{-1} m^5 s^{-2}$$ which is over 3 times the buckling pressure of diamond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give up? Not yet. There is a couple of interesting ideas I thought about:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why hemispherical? Wouldn't other shapes do well also? For example, a sphere, or a cylinder (maybe a cylinder with radius=height). From what I know (not much..) about spherical items, I suppose it would distribute the pressure evenly. Wouldn't that mean that diamond could easily sustain 100 kPa (withstands up to 600 GPa)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there no other way around this obstacle? Could we make use of any exotic forces or structural methods? &lt;br&gt;(One interesting suggestion: Gradually evacuate the helium-filled airship, slowly going up, so that when we reach a total vacuum the airship is basically in space? In that case there wouldn't be any powerful pressure from air...)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15588" LastActivityDate="2015-07-21T00:06:33.487" Title="Could a vacuum airship be possible?" Tags="&lt;pressure&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ClosedDate="2015-07-27T13:33:49.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="34233" PostTypeId="2" ParentId="34219" CreationDate="2015-07-21T00:48:00.890" Score="5" Body="&lt;p&gt;I will answer to your question, but first let me show you that NO has a $2.5$ bonds, and I will use what I made for NO to answer your question. Here we go ! :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof that NO has a $2.5$ bonds :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the electron configurations of these two elements are :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{{}_7 N :1s^2 2s^2 2p^3}$ which have $5$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{{}_8 O :1s^2 2s^2 2p^4}$ which have $6$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we can build the orbitals-diagram of $\ce{NO}$ and I will suppose that its not a correlated diagram, and because of electronegativity of these elements verifies $\ce{\chi(N)&amp;lt;\chi(O)}$ then we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kcNBS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the electon configuration of $\ce{NO}$ is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond order $i$ is : $$i=\frac{\text{number of electrons in bonding MOs}-\text{number of electrons in anti-bonding MOs}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have $$i=\frac{2\times 4-3}{2}=\frac{5}{2}=2.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quod Erat Demonstrandum.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So now I did it for $\ce{NO}$ I have this configuration : $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you said $3$ electrons bond considered has $0.5$ bond. As I understand your question, its the case if the configuration is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)}$ (for example but it could be an other) then you have $3$ electrons in bonding bonds and $2$ in anti-bonding bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the bond order of your molecule is : $$i=\frac{3-2}{2}=0.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which conclude my answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NB:I'm seeking common examples of this kind.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="15235" LastEditDate="2015-07-21T11:31:55.120" LastActivityDate="2015-07-21T11:31:55.120" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34234" PostTypeId="2" ParentId="34193" CreationDate="2015-07-21T01:26:20.400" Score="7" Body="&lt;p&gt;I add an other answer : it depends what kind of people you are, if you are as crazy as them, of course not ! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=1QbHRLpYc-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;2012 Christmas Lectures - Burning a Diamond&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;https://chemistry.stackexchange.com/users/4231/ron&quot;&gt;ron&lt;/a&gt;'s answer is better sure ! ;)&lt;/p&gt;&#xA;" OwnerUserId="15235" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-07-21T01:26:20.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34235" PostTypeId="2" ParentId="32519" CreationDate="2015-07-21T01:37:09.800" Score="0" Body="&lt;p&gt;The standard change in free energy, which is used to determine the equilibrium constant Kp is based on the pure reactants at 1 atm going to pure products at 1 atm.  Therefore, the partial pressure must be expressed in atm in order to be consistent with Kp.  This essentially normalizes the partial pressure by 1 atm.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-07-21T01:37:09.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34239" PostTypeId="1" CreationDate="2015-07-21T06:38:20.477" Score="2" ViewCount="10920" Body="&lt;p&gt;In a book, it is stated that one reason for the failure of Bohr's theory was the fact that it violates the uncertainty principle. Is this fact true? How so?&lt;/p&gt;&#xA;" OwnerUserId="17633" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2016-12-15T09:06:40.853" LastActivityDate="2016-12-15T09:06:40.853" Title="Does the Bohr model violate the uncertainty principle?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;atoms&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34241" PostTypeId="2" ParentId="34239" CreationDate="2015-07-21T07:46:06.200" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the Bohr model of the atom the energy (thus its momentum) of the electron and the radius of its orbit (thus its position) are precisely defined quantities. This is a direct violation of the Uncertainty Principle. Since the energy levels of the hydrogen atom had been shown to be experimentally well defined by Bohr’s model ($\Delta$p ~ 0), we are left with the conclusion that we know very little about the exact location of the electrons in space ($\Delta$x ~ infinity).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.osu.edu/~woodward/ch121/ch6_quantum.htm&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16924" LastActivityDate="2015-07-21T07:46:06.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34242" PostTypeId="2" ParentId="34239" CreationDate="2015-07-21T07:50:33.690" Score="7" Body="&lt;p&gt;The Heisenberg uncertainty principle states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x \Delta p \geq \frac{\hbar}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-12-15T09:01:46.940" LastActivityDate="2016-12-15T09:01:46.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34244" PostTypeId="1" AcceptedAnswerId="34247" CreationDate="2015-07-21T09:30:10.703" Score="6" ViewCount="4216" Body="&lt;p&gt;Zinc sulfide is used in road signs because when light from cars hit the road signs, it glows. What actually is the reason for the glowing of road signs (actually glowing of $\ce{ZnS}$)?  &lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="16683" LastEditorDisplayName="user15489" LastEditDate="2017-04-01T00:54:10.847" LastActivityDate="2017-04-01T00:54:10.847" Title="Why does zinc sulfide glow?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34245" PostTypeId="1" AcceptedAnswerId="34246" CreationDate="2015-07-21T10:10:44.793" Score="1" ViewCount="162" Body="&lt;p&gt;I line dry my cloths and every now, they become rough after drying. This mostly happens with socks. They are regular cotton socks and once I start to wear them they become soft again. Any idea why this is happening?&lt;/p&gt;&#xA;" OwnerUserId="11609" LastActivityDate="2015-07-21T10:33:35.010" Title="Why some cloths become rough after drying?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34246" PostTypeId="2" ParentId="34245" CreationDate="2015-07-21T10:33:35.010" Score="4" Body="&lt;p&gt;I believe it is down to water relaxing the fibres in the material. Upon drying in one position the fibres become temporarily fixed in that position. I have also read that the detergents used in washing can cause clothes to dry stiffer. A little experiment such as soaking your clothes in hot water should kill bacteria and then see if they dry stiff. This would give an idea as to whether or not its down to the detergent or just the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, a recent study suggests that fabric softeners reduce the wettability of materials. So if you use fabric softener on your towels, over time they will become less effective at drying (being towels).&lt;/p&gt;&#xA;" OwnerUserId="17619" LastActivityDate="2015-07-21T10:33:35.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34247" PostTypeId="2" ParentId="34244" CreationDate="2015-07-21T11:08:21.013" Score="5" Body="&lt;p&gt;In the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminous_paint#Phosphorescent_paint&quot; rel=&quot;noreferrer&quot;&gt;Phosphorescent paint&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Phosphorescent paint is commonly called &quot;glow-in-the-dark&quot; paint. It is made from phosphors such as silver-activated zinc sulfide or doped strontium aluminate, and typically glows a pale green to greenish-blue color. The mechanism for producing light is similar to that of fluorescent paint&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The process is summarised in the article &lt;a href=&quot;http://scienceprojectideasforkids.com/2011/fluorescence-vs-phosphorescence/&quot; rel=&quot;noreferrer&quot;&gt;Fluorescence vs. Phosphorescence&lt;/a&gt;, &#xA;The explanation from the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Phosphorescent materials produce light in a similar way as does fluorescence materials. A visible difference between these two types of luminescence, the ability of phosphorescence materials to glow after the excitation energy source is removed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The website has a diagram exhibiting the process (in a bit of a simplistic, but understandable way - just replace the 'UV' with 'car headlights etc'):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8rScK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, photons are emitted from every step back from the excited state to the ground state, unlike fluorescence, where the photon is emitted when the electron goes straight to the ground state.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-07-21T11:31:53.567" LastActivityDate="2015-07-21T11:31:53.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34249" PostTypeId="2" ParentId="34229" CreationDate="2015-07-21T11:58:14.833" Score="4" Body="&lt;p&gt;The heat flows from each Hb molecule during the reaction at constant p and T, so $q_{p, rxn} = \Delta H_{rxn}\cdot\Delta X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of the reaction is $\Delta H_{rxn}$ as per mole of reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction itself is Hb + 4O$_{2}\rightarrow$ Hb$\cdot$(O$_{2}$)$_{4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is exothermic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of moles of Hb present is $$n_{\rm Hb} = {5\over 64,000} = 0.078\;{\rm mmol},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;all of which reacts: $\Delta X = -\Delta\,n_{\rm Hb} = n_{\rm Hb}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That much heat raises the temperature of the water (assuming all the heat is contained in the water):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = \Delta H_{\rm water} + q_{p, rxn} = C_{p}\Delta T + q_{p, rxn}$$&#xA;from which it follows that&#xA;$$q_{p, rxn} = -C_{p}\Delta T = -100\times 4.18\times 0.031 = -13\;{\rm J}$$&#xA;and&#xA;$$\Delta H_{rxn} = {q_{p}\over\Delta X} = {-13\;{\rm J}\over 0.078\;{\rm mmol}}= -167\;{{\rm kJ}\over{\rm mol-rxn}}$$&#xA;There are 4 moles of O$_{2}$ used so &#xA;$$\nu_{\rm O_{2}} = 4\;{{\rm O_{2}}\over{\rm rxn}}$$&#xA;and on an oxygen basis&#xA;$$\Delta H_{rxn} = -42\;{\rm kJ}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-21T13:22:47.463" LastActivityDate="2015-07-21T13:22:47.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34251" PostTypeId="2" ParentId="11155" CreationDate="2015-07-21T13:21:33.587" Score="0" Body="&lt;p&gt;Lead is the main toxic factor, if you work in a fume hood and make sure that the solvents used in the process are not going to be inhaled by anyone else and that you wear gloves and wear goggles for protection you are reducing your exposure to the toxic properties of the Lead if absorbed into your skin or accidentally ingested, the reason is this is a very soluble form of Lead. Some might argue lead is used everywhere in the industrial electronics etc but it is the fact that Lead in perovskite solutions and powders is very soluble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DMF is a solvent that often papers use, quite toxic as is chloroform, need to use them in the fume hood so you don't smell them, their smells are distinct so if you do smell them at least you know that you need to clear up a problem that has been created. When you anneal the perovskite substrate the solvent evaporates and you are left with a film of perovskite crystals. The HTM as you know is also drop casted or spin coated on in most cases and needs to be a solvent that will not react or dissolve the perovskite film. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most groups use Chloroform and a mixture of others for the different additives for the Spiro HTM (Standard HTM at present) such as TbP. CuI is a HTM that is much cheaper than Spiro and while it is not as good the performance is good enough considering the cost. I just did a google and found &lt;a href=&quot;https://en.wikipedia.org/wiki/Copper(I)_iodide&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Copper(I)_iodide&lt;/a&gt; }&#xA;that it is soluble in:&#xA;soluble in ammonia and potassium solutions&#xA;insoluble in dilute acids&#xA;I don't know though if those solvents would affect the HTM film after evaporation or spin coating or react with the perovskite film. Let me know or publish if you find they don't&#xA;in any case while doing it&#xA;WEAR PROTECTIVE GLOVES AND GOGGLES&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about the solvents is something I am not sure of, just make sure that when you do the work you have the relevant safety measures taken, fume cupboard, after using beakers with solvents, wipe/rinse with Acetone and Ethanol then wash the thing that contained your solvent in the relevant area.&lt;/p&gt;&#xA;" OwnerUserId="17639" LastActivityDate="2015-07-21T13:21:33.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34252" PostTypeId="1" AcceptedAnswerId="89274" CreationDate="2015-07-21T13:45:09.913" Score="3" ViewCount="388" Body="&lt;p&gt;Concrete and sand are highly unreactive to most of the chemical attacks. But chlorine triflouride is so dangerous that it will ignite concrete, sand and other fire retarding substances. Which property does ClF3 possess to cause this ignition and why? &lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="17368" LastEditDate="2016-05-15T15:25:20.750" LastActivityDate="2018-01-22T11:17:38.360" Title="Why does chlorine trifluoride ignites concrete and sand?" Tags="&lt;inorganic-chemistry&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34253" PostTypeId="1" CreationDate="2015-07-21T14:36:18.450" Score="0" ViewCount="912" Body="&lt;p&gt;Let's pretend that in other universe we have an element with atomic number 6, like our carbon, but in this other universe the element is completely different from our carbon, like a new element with other properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a possible scenario?&lt;/p&gt;&#xA;" OwnerUserId="17640" LastEditorUserId="865" LastEditDate="2015-07-21T14:44:10.183" LastActivityDate="2021-04-22T14:06:18.220" Title="Can our chemical elements be differents on other universes?" Tags="&lt;elements&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ClosedDate="2015-07-21T18:12:53.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="34254" PostTypeId="2" ParentId="34253" CreationDate="2015-07-21T16:33:52.427" Score="5" Body="&lt;p&gt;I admit this question is far out of my depth, but it's interesting enough that I feel I should give it a shot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; - The question is so hard as to be almost unanswerable in detail. Things can be very different, though not arbitrarily different, and the exact line between the two is very difficult to define. There have been some mild investigations on the chemistry of toy universes, however, one of which I describe below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with thinking about other universes is that there are potentially a &lt;strong&gt;lot&lt;/strong&gt; of different universes which can arise. Famously, string theory currently posits that there are something like $10^{500}$ &quot;valid&quot; universes, each with different details regarding physical laws and physical constants. Furthermore, universes are &lt;strong&gt;very&lt;/strong&gt; complex objects in the sense that there is a ton of interrelation between everything, so even minor changes to details can create wildly different results. Not just that, but there apparently are many parameters to fiddle with, which would create wildly different periodic tables, often apparently breaking any possibility of life. This all leads to &lt;a href=&quot;https://en.wikipedia.org/wiki/Fine-tuned_Universe&quot; rel=&quot;nofollow noreferrer&quot;&gt;fine-tuning problems&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the attraction between two protons were about 2% larger, then every proton in the Universe would be bound in the form of a diproton nucleus, i.e. the nucleus of helium-2 ($\ce{^2He}$, which is just barely unstable in our Universe), meaning the entire periodic table would collapse to a single element. Apparently there's no very interesting chemistry in that universe! Now, what exactly would have to be tweaked in order to get a 2% stronger attraction between protons while not breaking everything else, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems thinking about chemistry in different universes is far too complicated to approach in almost any situation, but still there are some investigations. One particular case I have come across is what would happen in &lt;a href=&quot;http://journals.aps.org/prd/abstract/10.1103/PhysRevD.74.035006&quot; rel=&quot;nofollow noreferrer&quot;&gt;a universe without weak interactions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of Chemistry ultimately has to do with studying different aspects of electromagnetic interactions. However, the stability of atomic nuclei depends immensely on &lt;a href=&quot;https://en.wikipedia.org/wiki/Strong_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;strong interactions&lt;/a&gt; (aka strong nuclear force), and to a lesser extent on &lt;a href=&quot;https://en.wikipedia.org/wiki/Weak_interaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;weak interactions&lt;/a&gt; (aka weak nuclear force). The authors in the article linked above analyzed the effects of taking a universe similar to ours, but removing the weak interaction. There are several differences with respect to the evolution of the universe (big bang nucleosynthesis, stellar nucleosynthesis, etc) and the abundance of the elements, but thinking just about their stability, it seems that in such a universe, a large portion of the periodic table would still exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, given that &lt;a href=&quot;https://en.wikipedia.org/wiki/Beta_decay&quot; rel=&quot;nofollow noreferrer&quot;&gt;the weak interaction mediates the decay of neutrons in our Universe&lt;/a&gt;, its absence means neutrons are completely stable subatomic particles (in our Universe free neutrons decay with a half-life of around 8 minutes). This also means that neutron-rich nuclei are more stable to decay in the hypothetical weakless universe. In our Universe, deuterium ($\ce{^2_1H}$)is the most neutron-rich stable hydrogen isotope, while tritium ($\ce{^3_1H}$) is stable enough, with a half-life of about 11 years, to be chemically important (&lt;a href=&quot;https://chemistry.stackexchange.com/q/26611/1499&quot;&gt;heavier hydrogen isotopes can essentially be neglected&lt;/a&gt;). However, in the weakless universe, it seems that even hydrogen nuclei as rich as $\ce{^6_1H}$ are indefinitely stable, and there are possibly even more neutron-rich isotopes. See the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YeqSM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://journals.aps.org/prd/abstract/10.1103/PhysRevD.74.035006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison purposes, the most neutron-rich stable isotopes for each element in our Universe are $\ce{^2_1H}$, $\ce{^4_2He}$, $\ce{^7_3Li}$, $\ce{^9_4Be}$, $\ce{^11_5B}$, $\ce{^13_6C}$, $\ce{^15_7N}$ and $\ce{^18_8O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this added nuclear stability, the periodic table in the hypothetical weakless universe could actually have more stable elements in it. In our Universe, elements beyond about the 110th are quite unstable, being produced naturally in small amounts, if any, and they will not likely have great significance in Chemistry. However, highly neutron-rich nuclei could make ultraheavy elements more accessible, even in natural conditions such as supernovae.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-07-21T16:33:52.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34255" PostTypeId="2" ParentId="34252" CreationDate="2015-07-21T17:27:10.923" Score="0" Body="&lt;p&gt;Chlorine trifluoride is very unstable so it's energetically favourable to react with almost anything to produce more stable fluorides; releasing oxygen, chlorine and compounds of the two.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2015-07-21T17:27:10.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34257" PostTypeId="2" ParentId="14711" CreationDate="2015-07-21T17:48:17.443" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If I choose GIAO HF/STO-3G, is the default solvent $\ce{CD3Cl}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the default is gas phase as usual. But you can choose whatever solvent is programmed through the scrf=(solvent=...) command. The available solvents are listed at the bottom of the &lt;a href=&quot;http://www.gaussian.com/g_tech/g_ur/k_scrf.htm&quot; rel=&quot;noreferrer&quot;&gt;related Gaussian keyword site&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the default frequency for $^1$H−NMR $500~\mathrm{MHz}$? What about the frequency for $^{13}$C−NMR?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no default frequency as there is no frequency at all. The program calculates the absolute nuclear shielding tensors that you afterwards need to relate to some other substance's nuclear shielding tensor. As In experimental spectra one chooses TMS or the solvent signal, this must be done also by calculating the shielding tensors for TMS on your chosen combination of method and basis set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BUT&lt;/strong&gt; as there are sure default values that are chosen if one chooses the &quot;NMR&quot; command, there is no such thing as a &quot;default route&quot; for NMR calculations. I mean, you can simply click &quot;NMR&quot; in GaussView which will turn on NMR=GIAO calculations but there is more to think of, as Martin already stated.&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;NMR Shieldings&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I very much like the approach by Lodewyk, Siebert, Tantillo, Rablen and Bally called &lt;a href=&quot;http://cheshirenmr.info/index.htm&quot; rel=&quot;noreferrer&quot;&gt;CHESHIRE - Chemical Shift Repository&lt;/a&gt;, because there you can find many recommended procedures for either G03 or G09.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they have done, was to calculate the shieldings for many known small substances with various DFT functionals, basis sets and solvents in many combinations to get for each of those combinations a set of scaling factors via linear regression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, I chose to optimize my structures with B3LYP/6-31+G(d,p) and afterwards calculate the NMR shieldings with mPW1PW91/6-311+G(2d,p)/PCM and nmr=giao. Then one simply has to write out the calculated shieldings and modify them by $$\delta=\frac{\sigma-\text{intercept}}{\text{slope}}$$ according to their &lt;a href=&quot;http://cheshirenmr.info/Instructions.htm#existingfactors&quot; rel=&quot;noreferrer&quot;&gt;manual&lt;/a&gt; with the appropriate scaling factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But as you mentioned that you have 200 atoms, you might consider a smaller approach.&lt;/strong&gt; Good for you, besides their &lt;a href=&quot;http://cheshirenmr.info/Recommendations.htm&quot; rel=&quot;noreferrer&quot;&gt;recommended scaling factors&lt;/a&gt; there is also a manual on &lt;a href=&quot;http://cheshirenmr.info/CalculationFiles.htm&quot; rel=&quot;noreferrer&quot;&gt;how to calculate your own scaling factors&lt;/a&gt; and a &lt;a href=&quot;http://cheshirenmr.info/ScalingFactors.htm&quot; rel=&quot;noreferrer&quot;&gt;vast table for many solvents and methods&lt;/a&gt; with their respective scaling factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another &lt;strong&gt;really imporant&lt;/strong&gt; point for such large systems is that &lt;strong&gt;they don't exist as one single configuration!&lt;/strong&gt; What you have to do is to generate and optimize an enormous amount of possible isomers, calculate the shieldings for all those isomers and Bolzmann-weight them using their energy deviation regarding the isomer with the lowest energy.&#xA;$$\delta_i = \sum_j w_j~\delta_{i,j} = \sum_j \frac{\exp\left(\frac{-E_j}{RT}\right)}{\sum_k \exp\left(\frac{-E_k}{RT}\right)}~\delta_{i,j}$$&#xA;&lt;hr&gt;&#xA;&lt;strong&gt;$^1$H,$^1$H-Coupling Constants&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an $^1$H-NMR doesn't consist of straight peaks but mostly multiplets, there could also be the need to calculate the coupling constants. There have been a few approaches towards accurate J values but one of the easiest is also stated on the &lt;a href=&quot;http://cheshirenmr.info/Recommendations.htm#ballyccc&quot; rel=&quot;noreferrer&quot;&gt;CHESHIRE website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Optimize the geometry with B3LYP/6-31G(d)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Run an NMR single-point calculation with the following route section in GAUSSIAN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;#n B3LYP/6-31G(d,p) nmr=(fconly,readatoms) iop(3/10=1100000)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the molecule specification (separated by a blank line) read in: atoms=H&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://cheshirenmr.info/sample.htm&quot; rel=&quot;noreferrer&quot;&gt;sample input file&lt;/a&gt; for chloroethane could be like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\#n B3LYP/6-31G(d,p) nmr=(fconly,readatoms) iop(3/10=1100000)&#xA;&#xA;Chloroethylene CS&#xA;&#xA;0,1&#xA;C,0,0.,0.,0.&#xA;C,0,0.,0.,1.32727906&#xA;Cl,0,1.462671003,0.,-0.9634208566&#xA;H,0,-0.8964932057,0.,-0.6096342696&#xA;H,0,-0.9458848703,0.,1.8609603221&#xA;H,0,0.9144771425,0.,1.9110938644&#xA;&#xA;atoms=H&#xA;&lt;/code&gt;&lt;/pre&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the resulting log file, extract the desired Fermi contact J values, and scale them by a factor of 0.9117&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Example:&lt;/strong&gt; Nitrobenzene&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi&quot; rel=&quot;noreferrer&quot;&gt;SDBS, AIST&lt;/a&gt; the $^1$H resp. $^{13}$C-NMR spectrum have the following peaks:&#xA;$$\begin{array}{ccccccc}&#xA;\hline&#xA;\ce{Nuc.} &amp;amp; \ce{Calc}. &amp;amp; \ce{Exp}. &amp;amp; &amp;amp; \ce{Nuc}. &amp;amp; \ce{Calc}. &amp;amp; \ce{Exp}. \\&#xA; &amp;amp; \ce{ppm} &amp;amp; \ce{ppm} &amp;amp; &amp;amp; &amp;amp; \ce{ppm} &amp;amp; \ce{ppm} \\&#xA;\hline&#xA; &amp;amp; &amp;amp; &amp;amp; &amp;amp; \ce{C_{N}} &amp;amp; 146.47 &amp;amp; 148.30 \\&#xA;\ce{H_{A,~A'}} &amp;amp; 8.18 &amp;amp; 8.25 &amp;amp; &amp;amp; \ce{C_{A,~A'}} &amp;amp; 123.25 &amp;amp; 123.46 \\&#xA;\ce{H_{B,~B'}} &amp;amp; 7.37 &amp;amp; 7.56 &amp;amp; &amp;amp; \ce{C_{B,~B'}} &amp;amp; 127.75 &amp;amp; 129.43 \\&#xA;\ce{H_{C}} &amp;amp; 7.59 &amp;amp; 7.71 &amp;amp; &amp;amp; \ce{C_{C}} &amp;amp; 135.03 &amp;amp; 134.71 \\&#xA;\hline&#xA; &amp;amp; \ce{RMSD}  &amp;amp; 0.23 &amp;amp; &amp;amp; &amp;amp; \ce{RMSD} &amp;amp; 2.52\\&#xA;\hline&#xA;\end{array}$$&#xA;The differences are not very big. As the carbon nuclei span a much wider shielding range the average error is considerably smaller when compared to the RMSD of the proton shifts. To get an image out of this pure numbers, the overlay of the calculated versus the experimental $^1$H-NMR spectrum can be seen below. It fits quite good, even if it could be better. &lt;em&gt;Maybe the WP04 functional could have done a better job.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/49AdY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-23T01:16:58.277" LastActivityDate="2015-07-23T01:16:58.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34259" PostTypeId="2" ParentId="34253" CreationDate="2015-07-21T17:57:23.693" Score="5" Body="&lt;p&gt;@NicolauSakerNeto has nicely covered the possible effects that a change in nuclear physics would have on the periodic table.  I wanted to point out two other possibilities:  a different spin for the electron, and a massive photon.&lt;/p&gt;&#xA;&lt;h2&gt;Electron spin&lt;/h2&gt;&#xA;&lt;p&gt;In particle physics, particles come in two main types:  &amp;quot;bosons&amp;quot; and &amp;quot;fermions&amp;quot;.  The electron in our Universe is a fermion, with &amp;quot;spin 1/2&amp;quot;;  however, you can in principle also have fermions with spin 3/2, 5/2, 7/2, ...  Bosons, meanwhile, have spin 0, 1, 2, ...  In terms of familiar quantum numbers, a particle of spin &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; can have a quantum number &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; anywhere between &lt;span class=&quot;math-container&quot;&gt;$-s$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$+s$&lt;/span&gt;, in integer increments.  So the electron in our Universe can have two possible values for &lt;span class=&quot;math-container&quot;&gt;$m_s$&lt;/span&gt; (-1/2 or +1/2);  a spin-one particle can be in one of three &amp;quot;spin states&amp;quot; (-1, 0, or +1);  and so forth.&lt;/p&gt;&#xA;&lt;p&gt;Now, bosons (unlike fermions) do not obey the Pauli Exclusion Principle;  in fact, they are &lt;em&gt;more&lt;/em&gt; likely than not to be found in the same state.  (This is a consequence of something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;spin-statistics theorem&lt;/a&gt;.)  This means that a Universe in which the electron was a boson would be a very boring universe indeed;  instead of filling orbitals in a sequential way, every &amp;quot;electron&amp;quot; would be in the ground state of the atom.  (This neglects interactions between the electrons themselves, but this would be a good first-order approximation.)  You therefore wouldn't get unpaired electrons in orbitals like you do in our Universe; bonding between atoms, if it exists in this Universe, would be very different from our own.  My guess would be that you'd effectively have a Universe full of noble gas atoms, with nothing around that wasn't monatomic.&lt;/p&gt;&#xA;&lt;p&gt;A Universe with a spin-3/2 electron would be more interesting.  These &lt;em&gt;do&lt;/em&gt; obey the Pauli Exclusion Principle;  and so every electron in an atom would need to have a different set of quantum numbers.  However, you could now have four electrons in every orbital (&lt;span class=&quot;math-container&quot;&gt;$m_s = -3/2, -1/2, +1/2$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$+3/2$&lt;/span&gt;);  and so the periodic table would look very different indeed.  I wouldn't want to hazard a guess as to what it would look like in the end;  the filling order of the orbitals would be very different in this Universe, and it would be a hard question to determine (for example) how many electrons would go into the &lt;span class=&quot;math-container&quot;&gt;$2s$&lt;/span&gt; orbital before the &lt;span class=&quot;math-container&quot;&gt;$2p$&lt;/span&gt; orbitals started to fill.&lt;/p&gt;&#xA;&lt;h2&gt;Massive photon&lt;/h2&gt;&#xA;&lt;p&gt;In our Universe, the photon is a massless particle;  through the magic of quantum electrodynamics, this leads to the familiar Coulomb potential for an electron orbiting a proton:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;U(r) = - \frac{Ze^2}{r}.&#xA;$$&lt;/span&gt;&#xA;However, if the photon had a mass, then this potential would be replaced by something called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yukawa_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yukawa potential&lt;/a&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;U(r) = - \frac{Ze^2}{r} \exp{\left(-\frac{mc}{\hbar}r\right)}.&#xA;$$&lt;/span&gt;&#xA;In particular, the Yukawa potential (unlike the Coulomb potential) &lt;a href=&quot;http://arxiv.org/pdf/hep-ph/0407258v2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;does not allow for bound states with arbitrarily large angular momentum;&lt;/a&gt; there is a maximum possible angular momentum above which no bound states exist.  This would again change the structure of the periodic table drastically.  The filling order of the orbitals would certainly be changed.  It's not even clear to me that a heavy nucleus (with atomic number &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;) would necessarily always have &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt; orbitals for electrons to fill, and so you could end up with the odd effect of heavy atoms tending to exist as charged ions rather than being neutral.&lt;/p&gt;&#xA;" OwnerUserId="17515" LastEditorUserId="17515" LastEditDate="2021-04-22T14:06:18.220" LastActivityDate="2021-04-22T14:06:18.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34261" PostTypeId="2" ParentId="34203" CreationDate="2015-07-21T18:56:22.510" Score="1" Body="&lt;p&gt;&lt;strong&gt;You can't&lt;/strong&gt; - at least if the only information you have is just the bond energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem you're running up against here is the difference between &lt;strong&gt;thermodynamics&lt;/strong&gt; and &lt;strong&gt;kinetics&lt;/strong&gt;. The difference can sometimes be subtle, but (roughly) thermodynamics is normally concerned with state functions, whereas kinetics deals with the process of interconversion of states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bond enthalpy is a thermodynamic quantity. It's based off a state function. It's the free energy difference between the bond-broken and bond-formed states. It doesn't matter &lt;em&gt;how&lt;/em&gt; you go from the bond-formed to bond-broken states, the free energy difference is the same. In order to calculate it, you only need to know information about the two endpoints, not the details about the states which connect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, reaction rates are kinetic quantities. The rate of reaction is highly dependent on the path (reaction mechanism and conditions) you take to interconvert the states. You can't tell anything about the path if all you have is thermodynamic information about the two end points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you can theoretically calculate reaction rates if you have additional information about the path you're taking. Most notably you would need to know the &lt;strong&gt;activation energy&lt;/strong&gt; of the reaction you're interested in. This is different from the bond energy, and is related to the free energy of the &lt;strong&gt;transition state&lt;/strong&gt;, the hypothetical high-energy intermediate along the reaction pathway. If you have the activation energy along with some other information for a single-step reaction, you can calculate reaction rates for various temperatures with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow&quot;&gt;Arrhenius equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these rates are highly dependent on the exact mechanism of the reaction you're examining. Single step versus multi-step, radical versus electron pair transfer mechanisms, the presence of catalysts, etc. all can change the activation energies of the various steps and the reaction rates. So there is no way to give a general answer about bond breakage rates.&lt;/p&gt;&#xA;" OwnerUserId="11616" LastActivityDate="2015-07-21T18:56:22.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34262" PostTypeId="1" CreationDate="2015-07-21T19:11:21.417" Score="17" ViewCount="643" Body="&lt;p&gt;I was doing some IR calculations on different 3d transition metal carbonyl complexes during some practical courses where we were introduced to effective core potentials. While doing this we found that MP2 has some problem with the calculation of at least the asymmetrical $\nu_\text{CO}$ band (that was the only one we were interested in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We made the calculations for $\ce{[Ti(CO)6]^2-}$, $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, $\ce{[Mn(CO)6]+}$ in octahedral symmetry and $\ce{[Fe(CO)4]^2-}$, $\ce{[Co(CO)4]-}$ and $\ce{[Ni(CO)4]}$ in tetrahedral symmetry. The method was MP2, as stated in the title, and the basis set was 6-31G(d). The program we used was Gaussian$~$09 Rev. A.02 but that was not the problem, as I did calculations with other programs with similar results. All optimized structures are considered to be true minima as there either is no imaginary frequency or it is calculated to be $\ge -50~\mathrm{cm^{-1}}$ with was stated to be a problem of the insufficient DFT grid but not further questioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following graphic shows that most of the calculated values are acceptable and near their experimental values but two of them, namely $\ce{[Co(CO)4]-}$ and $\ce{[Fe(CO)4]^2-}$ differ significantly from the experimental values by nearly $300\ldots1000~\mathrm{cm^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1.4in$&lt;img src=&quot;https://i.stack.imgur.com/uQywg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this I varied some parameters of the calculation that could influence this deviation as was:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;additional diffuse functions for the better description of ions&lt;/li&gt;&#xA;&lt;li&gt;tight optimization criteria (T) as this might reduce the imaginary frequency&lt;/li&gt;&#xA;&lt;li&gt;no frozen core (F) just to try it&lt;/li&gt;&#xA;&lt;li&gt;solvents: PCM(DCM), PCM(Heptane) as experimental values aren't taken in gas phase&lt;/li&gt;&#xA;&lt;li&gt;I also optfreq'ed with B3LYP/6-31G(d)/MDF10 (B) to see if it's the method itself&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The results can be seen in the following table:&#xA;$$\begin{array}{cccccccc}&#xA;\hline&#xA;\text{Basis~Set} &amp;amp; \mathrm{Solv.} &amp;amp; \mathrm{d(MC)} &amp;amp; \mathrm{d(CO)} &amp;amp; \nu_\text{CO}(calc.) &amp;amp; \mathrm{Int.} &amp;amp; \nu_\text{CO}(exp.) &amp;amp; \Delta\nu_\text{CO}\\&#xA; &amp;amp;  &amp;amp; \mathrm{Ang.} &amp;amp; \mathrm{Ang.} &amp;amp; \mathrm{cm^{-1}} &amp;amp; \mathrm{km\, mol^{-1}} &amp;amp; \mathrm{cm^{-1}} &amp;amp; \mathrm{cm^{-1}}\\&#xA;\hline&#xA;\mathrm{E}/6{-}31\mathrm{G}^\ast &amp;amp; &amp;amp; 1.802 &amp;amp; 1.184 &amp;amp; 2830 &amp;amp; 2.0\cdot10^5 &amp;amp; 1890 &amp;amp; 940\\&#xA;\mathrm{EFT}/6{-}31\mathrm{+G}^\ast &amp;amp; &amp;amp; 1.799&amp;amp; 1.187&amp;amp; 2966&amp;amp; 2.7\cdot10^5 &amp;amp; 1890 &amp;amp; 1076\\&#xA;\mathrm{E}/6{-}31\mathrm{+G}^\ast &amp;amp; &amp;amp; 1.800 &amp;amp; 1.187 &amp;amp; 2996 &amp;amp; 2.9\cdot10^5 &amp;amp; 1890 &amp;amp; 1106\\&#xA;\mathrm{E}/6{-}31\mathrm{+G}^\ast &amp;amp; \mathrm{DCM} &amp;amp; 1.795 &amp;amp; 1.186 &amp;amp; 4671 &amp;amp; 0.0 &amp;amp; 1890 &amp;amp; 2781\\&#xA;\mathrm{E}/6{-}31\mathrm{G}^\ast &amp;amp; \mathrm{Hept.} &amp;amp; 1.799 &amp;amp; 1.184 &amp;amp; 3203 &amp;amp; 4.3\cdot10^5 &amp;amp; 1890 &amp;amp; 1223\\&#xA;\hline&#xA;\mathrm{BE}/6{-}31\mathrm{G}^\ast &amp;amp; &amp;amp; 1.767 &amp;amp; 1.169 &amp;amp; 1978 &amp;amp; 1322 &amp;amp; 1890 &amp;amp; 88\\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be seen:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Within MP2 there is no significant effect on $\ce{M-C}$- or $\ce{C=O}$-bond lengths.&lt;/li&gt;&#xA;&lt;li&gt;The diffuse functions shift the frequency to higher wave numbers and&#xA;the impact of the solvent is big to huge. As without the solvents the&#xA;deviation was calculated to be somewhere around&#xA;$+1000~\mathrm{cm^{-1}}$ it is now at $+2800~\mathrm{cm^{-1}}$!&lt;/li&gt;&#xA;&lt;li&gt;Also the intensities are horribly wrong ... they are somewhere around $2$&#xA;to $4\cdot10^5~\mathrm{km\, mol^{-1}}$ instead of somewhere around&#xA;1000 or in the special case of DCM they are completely zero.&lt;/li&gt;&#xA;&lt;li&gt;B3LYP predicts everything within common error ranges&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are your ideas why MP2 fails with this special task?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="9961" LastEditDate="2016-01-19T21:14:54.250" LastActivityDate="2016-03-02T23:47:28.660" Title="Why is MP2 failing to calculate proper IR frequencies for [Co(CO)4]- and [Fe(CO)4]2-?" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;carbonyl-complexes&gt;" AnswerCount="1" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34263" PostTypeId="1" CreationDate="2015-07-21T19:14:32.367" Score="2" ViewCount="954" Body="&lt;p&gt;I have a gas ($\ce{CO2}$ for example) and I want to infuse it into a liquid ($\ce{H2O}$ for example). I need a higher pressure of $\ce{CO2}$ to do this. So I push in water into a sealed container (at 7 psi) and add $\ce{CO2}$ (at 14 psi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? Will it back up? What if I use a check valve? Can I use a non-electrical actuator of some sort to let liquid in, and then add gas 'automatically'? What about something that functions like a carburetor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to let the new solution out after it is mixed and also allow a consistent flow if I leave the exit open. If I have a liquid at 7 psi and a gas at 14 psi and I am trying to push them both into a sealed container, how is this done? Letting it out? Doing it on the fly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you do NOT carbonate water at these pressures. Just an example with real gas/liquid.&lt;/p&gt;&#xA;" OwnerUserId="17644" LastEditorUserId="865" LastEditDate="2015-07-21T19:45:10.563" LastActivityDate="2015-07-23T19:03:17.563" Title="Can you push a liquid and a gas into a container at different pressures?" Tags="&lt;solubility&gt;&lt;pressure&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34264" PostTypeId="2" ParentId="33879" CreationDate="2015-07-21T19:18:13.150" Score="18" Body="&lt;h1&gt;TL;DR&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears little resemblance to the original atomic orbitals, or to the VB theory description.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Long Version&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The key difference between the VB and MO theories lies in their respective assumptions of the underlying mathematical/physical form of the electronic wavefunction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;VB theory&lt;/strong&gt; holds tightly to the mathematical structure of &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_orbital&quot; rel=&quot;noreferrer&quot;&gt;atomic orbitals&lt;/a&gt; (AOs), assuming that the electronic structure of a molecular system is described adequately by retaining the atomic orbitals (possibly &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;noreferrer&quot;&gt;hybridized&lt;/a&gt;) from each constituent atom, and overlapping them to form the needed description of bonding.  VB calculations are computationally much more complex: the AO basis is generally non-&lt;a href=&quot;https://en.wikipedia.org/wiki/Orthogonal_functions&quot; rel=&quot;noreferrer&quot;&gt;orthogonal&lt;/a&gt;, and the linear algebra and other techniques required to cope with this non-orthogonality are much more involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More broadly, the need to invoke &lt;a href=&quot;https://en.wikipedia.org/wiki/Resonance_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;resonance&lt;/a&gt; (superposition of multiple valence-bond structures) and other concepts highlights that, in fact, often this adherence to atomic orbitals and their valence-bond overlaps is &lt;strong&gt;not&lt;/strong&gt; adequate for a proper description, at least not without extensive use of methods analogous to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Configuration_interaction&quot; rel=&quot;noreferrer&quot;&gt;configuration interaction&lt;/a&gt; of MO theory.  VB theory does hold certain advantages, however—for example, I believe the AO basis does sometimes allow for a more direct linkage to fundamental chemical concepts such as hybridization and Lewis electon-pair theory. To note, Sason Shaik is one of the strongest modern proponents of the theory I know of—he describes some of these strengths and their applications in his recent retrospective article (&lt;a href=&quot;http://dx.doi.org/10.1016/j.comptc.2017.02.011&quot; rel=&quot;noreferrer&quot;&gt;doi:10.1016/j.comptc.2017.02.011&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MO theory&lt;/strong&gt;, at its core, casts away most assumptions regarding the mathematical structure of the electronic wavefunction, as long as properties like &lt;a href=&quot;https://en.wikipedia.org/wiki/Identical_particles&quot; rel=&quot;noreferrer&quot;&gt;indistinguishability&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pauli_exclusion_principle&quot; rel=&quot;noreferrer&quot;&gt;Pauli exclusion principle&lt;/a&gt; remain satisfied.  Calculations within MO theory can use a wide variety of mathematical structures to try to represent the wavefunction, ranging from gridded numerical quadrature approaches (see &lt;a href=&quot;http://dx.doi.org/10.1021/ct501128u&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevA.76.040503&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, and also &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital-free_density_functional_theory&quot; rel=&quot;noreferrer&quot;&gt;orbital-free DFT&lt;/a&gt;) to &lt;a href=&quot;https://en.wikipedia.org/wiki/Slater-type_orbital&quot; rel=&quot;noreferrer&quot;&gt;Slater orbital&lt;/a&gt; computations to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaussian_orbital&quot; rel=&quot;noreferrer&quot;&gt;Gaussian-orbital&lt;/a&gt; based calculations used most commonly today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One potential point of confusion is that MO theory in its most common implementations &lt;em&gt;makes direct use of atomic orbitals&lt;/em&gt; in composing the molecular wavefunction (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_combination_of_atomic_orbitals&quot; rel=&quot;noreferrer&quot;&gt;LCAO theory&lt;/a&gt;). The distinction to be made here is that instead of attributing significance to the atomic orbitals as an intrinsically good descriptor of electronic structure, they are used merely because &lt;em&gt;they afford a convenient mathematical form for efficient computation&lt;/em&gt;, in trying to construct the MOs.  Particular advantages are obtained from the use of atomic orbitals constructed from Gaussian functions, since the &lt;a href=&quot;http://www.tina-vision.net/docs/memos/2003-003.pdf&quot; rel=&quot;noreferrer&quot;&gt;product of two Gaussians is another Gaussian&lt;/a&gt; (even in multiple dimensions) and &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaussian_function#Multi-dimensional_Gaussian_function&quot; rel=&quot;noreferrer&quot;&gt;numerical integration of &lt;em&gt;n&lt;/em&gt;-dimensional Gaussian functions is straightforward&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that, for as much of an improved description it affords, MO theory still is an approximation. For a given system, the electronic wavefunction just &lt;em&gt;IS&lt;/em&gt; the highly-abstract mathematical object that satisfies the appropriate &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;noreferrer&quot;&gt;Schrodinger equation&lt;/a&gt;.  MO theory is used extensively because it provides one of the most computationally tractable paths to getting in the neighborhood of the &quot;real answer.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Practical Aspects of MO theory&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;simplest version&quot; of MO theory for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Time-independent_equation&quot; rel=&quot;noreferrer&quot;&gt;time-independent&lt;/a&gt; system under the &lt;a href=&quot;https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation&quot; rel=&quot;noreferrer&quot;&gt;Born-Oppenheimer Approximation&lt;/a&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;noreferrer&quot;&gt;Hartree-Fock method&lt;/a&gt;, which provides a remarkably good description considering the &lt;em&gt;significant&lt;/em&gt; simplification represented by its &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_field_theory&quot; rel=&quot;noreferrer&quot;&gt;mean-field approximation&lt;/a&gt; of the electron-electron interactions.  Hartree-Fock is in general not good enough for computations at the level of chemical accuracy (~$1\mathrm{\ kcal/mol}$), however, and numerous methods have been developed to refine it: the most often used techniques are &lt;a href=&quot;https://en.wikipedia.org/wiki/M%C3%B8ller%E2%80%93Plesset_perturbation_theory&quot; rel=&quot;noreferrer&quot;&gt;Moller-Plesset perturbation theory&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Configuration_interaction&quot; rel=&quot;noreferrer&quot;&gt;configuration interaction&lt;/a&gt;, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Coupled_cluster&quot; rel=&quot;noreferrer&quot;&gt;coupled cluster&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In-Depth Reading on MO Theory&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a deep dive into the underlying mathematics of MO computations, it's hard to beat the classic &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0486691861&quot; rel=&quot;noreferrer&quot;&gt;Szabo &amp;amp; Ostlund&lt;/a&gt;. As a more readable, lighter-weight introduction to MO theory and its implementations, I would recommend Frank Jensen's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470011874&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Introduction to Computational Chemistry&lt;/em&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2017-03-29T16:15:15.090" LastActivityDate="2017-03-29T16:15:15.090" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34266" PostTypeId="1" CreationDate="2015-07-21T21:24:22.597" Score="0" ViewCount="78" Body="&lt;p&gt;I'm a bit new in chemistry, so I need help with a couple of things. Although I don't mind complicated answers, please try to keep it around high school level, and not rocket science. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can someone describe exactly what makes a compound a compound? Is it just the simple definition of &quot;multiple elements combined&quot; or can someone get a more specific?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sodium chloride is an ionic compound, but it's not made out of &quot;molecules&quot;. So does that mean Sodium Chloride isn't an actual molecule (like $Na^+$ and $Cl^-$ are combined as separate ions but they can't be combined in a 1 to 1 fashion)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can someone explain to me exactly what the law of gaseous reaction is, (it doesn't show up on internet specifically, but it's a law by Gay-Lussac).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="17645" LastActivityDate="2015-07-22T21:02:10.953" Title="When Compounds are Considered Compounds and Their Types" Tags="&lt;ionic-compounds&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-07-23T01:48:01.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="34267" PostTypeId="1" CreationDate="2015-07-21T23:49:07.243" Score="-1" ViewCount="978" Body="&lt;p&gt;I'm going to get different solvents with different bonding types (polar and non polar) and put different solutes with an &quot;unknown&quot; bond type to test whether or not the solute is polar or non polar The question is, can I improve this in anyway or take it further than I've already described. It sounds too easy to me and I want to take it a step (if not a few steps) further.&lt;/p&gt;&#xA;" OwnerUserId="17649" LastEditorUserId="418" LastEditDate="2016-11-08T03:16:30.813" LastActivityDate="2016-11-08T03:16:30.813" Title="How to determine whether a solvent is polar or unpolar?" Tags="&lt;polarity&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2017-09-09T01:24:32.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="34268" PostTypeId="1" AcceptedAnswerId="34270" CreationDate="2015-07-21T23:54:33.607" Score="1" ViewCount="110" Body="&lt;p&gt;Does flipping a reaction as written change the sign of the value of E or is it a completely different value? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I have the value for the half reaction $\ce{Cl2 + 2e- -&amp;gt; 2Cl-}$ but need the value for the opposite reaction ($\ce{2Cl- -&amp;gt; Cl2 + 2e-}$), do I just put a negative sign and treat it like $\Delta H$ values? &lt;/p&gt;&#xA;" OwnerUserId="16949" LastEditorUserId="865" LastEditDate="2015-07-22T00:46:22.893" LastActivityDate="2015-07-22T00:48:18.800" Title="Electrode Potential Half Reactions" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34269" PostTypeId="1" CreationDate="2015-07-22T00:01:08.620" Score="0" ViewCount="360" Body="&lt;p&gt;I have &lt;strong&gt;15 liters of aqueous solution&lt;/strong&gt; inside an aluminum stock pot which rests upon a hot plate. I want to &lt;strong&gt;reduce the boiling point from 150°C to 100°C using reduced pressure&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the &lt;strong&gt;most economical way efficiently to dry the solute&lt;/strong&gt;? Economical means does not require specialty equipment (i.e., laboratory vacuum pumps, expensive cold traps, etc.)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I own a water aspirator pump (attaches to sink, 27.4 in Hg maximum vacuum, 6 liters/minute) and a 2-stage electric vacuum pump (15-micron, 3CFM) that I use for degassing epoxies. I don't know whether either of these will be of any use here.&lt;/p&gt;&#xA;" OwnerUserId="17650" LastEditorUserId="4231" LastEditDate="2015-07-22T00:04:17.413" LastActivityDate="2015-07-22T00:31:38.280" Title="Reduced-pressure (vacuum) evaporation/drying of 15 liters of aqueous solution" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34270" PostTypeId="2" ParentId="34268" CreationDate="2015-07-22T00:12:40.670" Score="1" Body="&lt;p&gt;Yes, you are correct - just flip the sign. As quoted from &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/The_Cell_Potential&quot; rel=&quot;nofollow&quot;&gt;the venerable UC Davis ChemWiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They give the value for your (forward) reaction as +1.358 V, so for the reaction in question, you'll use &lt;strong&gt;-1.358 V&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-22T00:48:18.800" LastActivityDate="2015-07-22T00:48:18.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34271" PostTypeId="2" ParentId="34269" CreationDate="2015-07-22T00:31:38.280" Score="2" Body="&lt;p&gt;The aspirator pump would be unlikely to be damaged by the distillate (and can be cheaply repaired or replaced), but the electric vacuum pump likely will require a cold-trap to prevent the vapors from destroying it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aspirator pump would seem the way to go, but you must factor in the cost of water in your location. This &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B003OBYNBO&quot; rel=&quot;nofollow&quot;&gt;Nalgene aspirator&lt;/a&gt; uses ~6.5 liter/minute. A &lt;a href=&quot;http://www.thomassci.com/Equipment/Pumps/Vacuum/_/A-1000S-Aspirator-Pumps&quot; rel=&quot;nofollow&quot;&gt;recirculating aspirator pump&lt;/a&gt; might be the best compromise for a production environment, where efficiency is important, and it might be feasible to save and recycle the distillate.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-22T00:31:38.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34272" PostTypeId="2" ParentId="34267" CreationDate="2015-07-22T00:37:23.590" Score="1" Body="&lt;p&gt;One thing you should definitely do is a &lt;em&gt;control&lt;/em&gt; experiment where you check some known with certainty solutes as to their polarity. This will establish a baseline hypothesis experimentally; i.e. you've been taught that &quot;like dissolves like&quot;, but you should demonstrate this to yourself experimentally. Once you do this, then your testing with unknown solutes is more meaningful because you then have similar experimental observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you may find some surprises when you do this! For example, &lt;em&gt;soaps&lt;/em&gt; are interesting compounds because they are largely nonpolar molecules with polar &quot;ends&quot;. This is why they can dissolve oil (nonpolar) in water (polar), and thus is why soap &quot;cleanses&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-07-22T00:37:23.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34273" PostTypeId="2" ParentId="34263" CreationDate="2015-07-22T00:41:30.663" Score="3" Body="&lt;p&gt;Pressure is due to the random impact of molecules on the surface, regardless of whether water or carbon dioxide or anything else. Once &lt;em&gt;equilibrium&lt;/em&gt; is reached, all components are at the same pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there's no problem using a &lt;em&gt;moving&lt;/em&gt; stream of $\ce{CO2}$ to propel and to mix with water at much lower pressure, using &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's principle&lt;/a&gt;. The automotive &lt;a href=&quot;https://en.wikipedia.org/wiki/Carburetor&quot; rel=&quot;nofollow&quot;&gt;carburetor&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomizer_nozzle&quot; rel=&quot;nofollow&quot;&gt;spray atomizer&lt;/a&gt; operate on this basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative, of course, is to add the water to the container through a &lt;em&gt;check-valve&lt;/em&gt;, &lt;strong&gt;then&lt;/strong&gt; add $\ce{CO2}$ at a higher pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice is a continuous process vs. a batch process.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-22T00:41:30.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34274" PostTypeId="1" AcceptedAnswerId="34275" CreationDate="2015-07-22T02:01:39.920" Score="6" ViewCount="1273" Body="&lt;p&gt;I am curious to know if a reverse osmosis system can clean water from radioactive elements like cesium-134 and cesium-137. If it can, how efficient can it be?&lt;/p&gt;&#xA;" OwnerUserId="1312" LastEditorUserId="1215" LastEditDate="2015-07-22T07:32:10.543" LastActivityDate="2019-10-07T03:22:56.217" Title="Can a reverse osmosis system remove radioactive elements in contaminated water?" Tags="&lt;water&gt;&lt;radioactivity&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34275" PostTypeId="2" ParentId="34274" CreationDate="2015-07-22T02:16:19.510" Score="7" Body="&lt;p&gt;I don't know about cesium in particular, but according to the EPA's &lt;a href=&quot;http://water.epa.gov/drink/contaminants/basicinformation/radionuclides.cfm&quot; rel=&quot;noreferrer&quot;&gt;web site&lt;/a&gt; (see the section called &amp;quot;How will radionuclides be removed from my drinking water?&amp;quot;), both isotopes should be covered under the beta particle section.&lt;/p&gt;&#xA;&lt;p&gt;I cannot quantify the efficiency part of your question, but presume it's acceptable to the EPA (which may or may not instill confidence to you in terms of whatever methodology they're using).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The following treatment method(s) have proven to be effective in removing radionuclides at levels below their MCLs:&lt;/p&gt;&#xA;&lt;p&gt;Beta particle and Photon Radiation: ion exchange and reverse osmosis;&lt;/p&gt;&#xA;&lt;p&gt;(Gross) Alpha Emitters: reverse osmosis;&lt;/p&gt;&#xA;&lt;p&gt;Radium 226 and Radium 228 (Combined): ion exchange, reverse osmosis, lime softening;&lt;/p&gt;&#xA;&lt;p&gt;Uranium: Ion exchange, reverse osmosis, lime softening,  coagulation/filtration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-22T02:16:19.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34279" PostTypeId="2" ParentId="34188" CreationDate="2015-07-22T04:11:17.853" Score="13" Body="&lt;p&gt;Reason is salt dissolve very easily with water, forming very strong bond. Not only salt there are also other minerals. To break this bond we need technology as well as energy and both of them are very expensive(today). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two methods by which we can break this bond:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Thermal distillation&lt;/strong&gt;:&#xA;Thermal distillation involves heat: Boiling water turns it into vapor—leaving the salt behind—that is collected and condensed back into water by cooling it down&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Membrane distillation&lt;/strong&gt;: The most common type of membrane separation is called reverse osmosis. Seawater is forced through a semipermeable membrane that separates salt from water. Because the technology typically requires less energy than thermal distillation, most new plants, now use reverse osmosis.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Today about &lt;strong&gt;10 to 13 billion gallon&lt;/strong&gt;  water are desalinated worldwide per day. That's only about &lt;strong&gt;0.2%&lt;/strong&gt; of global water consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Membrane distillation is bit more cheaper than thermal distillation. Even though  large amounts of energy are needed to generate the high pressure that forces the water through the filter. Current methods require about &lt;strong&gt;14 kilowatt-hours of energy to produce 1,000 gallons&lt;/strong&gt; of desalinated seawater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can conclude that cost of distillation is high because we need notable large amount of electricity to heat water in thermal plant and generate high pressure. Also other technology(like say burner, pumps, etc) are also expensive.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.scientificamerican.com/article/why-dont-we-get-our-drinking-water-from-the-ocean/&quot; rel=&quot;noreferrer&quot;&gt;ScientificAmerica&lt;/a&gt; and &lt;a href=&quot;http://www.livescience.com/4510-desalination-work.html&quot; rel=&quot;noreferrer&quot;&gt;LiveScience&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also note we are trying to decrease the cost of distillation and day by day number of seawater distillation  are increasing. &lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-07-23T03:35:53.067" LastActivityDate="2015-07-23T03:35:53.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34280" PostTypeId="1" CreationDate="2015-07-22T05:10:05.960" Score="2" ViewCount="1250" Body="&lt;p&gt;I’m needing a method to turn cotton into a fine powder for use as a source of cellulose. What suggestions would you give to accomplish this with materials/tools that could be obtained at local stores?&lt;/p&gt;&#xA;" OwnerUserId="16558" LastActivityDate="2015-07-22T07:15:32.557" Title="What Methods Would I Use To Powder Cotton?" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34281" PostTypeId="2" ParentId="34280" CreationDate="2015-07-22T07:15:32.557" Score="1" Body="&lt;p&gt;The content of cellulose in paper is not as high as that of cotton, however, it may be easier to use as a source if you do require a fine powder. Why do you require cellulose?&lt;/p&gt;&#xA;" OwnerUserId="17619" LastActivityDate="2015-07-22T07:15:32.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34283" PostTypeId="1" AcceptedAnswerId="34284" CreationDate="2015-07-22T09:17:15.150" Score="3" ViewCount="2755" Body="&lt;p&gt;Beryllium carbonate is reported to be unstable under normal conditions and should only be kept in a atmosphere of $\ce{CO2}$. Why and how this stability is maintained? Why does this compound decompose when kept in an atmosphere of oxygen?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="865" LastEditDate="2015-07-22T12:36:45.163" LastActivityDate="2015-07-22T12:36:45.163" Title="Stability of beryllium carbonate" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;enthalpy&gt;&lt;stability&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34284" PostTypeId="2" ParentId="34283" CreationDate="2015-07-22T10:07:14.797" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why and how is this stability maintained?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Beryllium carbonate decomposes into Beryllium oxide and Carbon dioxide.&#xA;$$\ce{BeCO3 &amp;lt;=&amp;gt; BeO + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot;&gt;Le Chatelier's principle&lt;/a&gt; one simply has to store it under a $\ce{CO2}$ atmosphere to push the equilibrium to the left.&#xA;$$\ce{BeCO3 &amp;lt;&amp;lt;=&amp;gt; BeO + CO2}~\text{in excess}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why this compound decomposes when kept in atmosphere of oxygen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When there is no $\ce{CO2}$ atmosphere the equilibrium is pushed to the right, which would be the &quot;normal&quot; case under standard conditions.&#xA;$$\ce{BeCO3 &amp;lt;=&amp;gt;&amp;gt; BeO + CO2}$$&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-22T10:07:14.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34285" PostTypeId="1" CreationDate="2015-07-22T11:13:11.857" Score="8" ViewCount="13219" Body="&lt;p&gt;In the reaction of $\ce{LiAlH4}$ with carboxylic acids, deprotonation is followed by a step in which $\ce{O-AlH2-}$ acts as a leaving group. The mechanism is given &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4940/lialh%E2%82%84-with-carboxylic-acid&quot;&gt;here&lt;/a&gt; in this answer. &#xA;$\ce{LiAlH4}$ does not, however, reduce alcohols. Why doesn't a similar reaction occur with alcohols - deprotonation, followed by substitution by a hydride ion and removal of $\ce{O-AlH2-}$?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-22T18:24:19.133" Title="Why can't LiAlH4 react with alcohols?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34286" PostTypeId="2" ParentId="34263" CreationDate="2015-07-22T11:47:52.997" Score="0" Body="&lt;p&gt;We flow gases and liquids into reactors at 80 bar quite easily.  This is not into a sealed (batch) container, but into packed tubes with a back pressure regulator(ultimately)vat the end. The gas pressure is higher than the liquid pressure, which does not lead to gas in-flow due to check valves on the liquid feed.&#xA;In a sealed container, it would be expected that the headspace pressure would eventually match the gas pressure. If the liquid pressure was lower and the liquid feed couldn't hold the higher pressure, something would fail.&#xA;A mass flow meter can control your gas delivery accurately. A tube packed with inert material is a cheap way of getting good mixing.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-07-22T11:47:52.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34287" PostTypeId="1" CreationDate="2015-07-22T13:04:57.410" Score="18" ViewCount="14818" Body="&lt;p&gt;I've always known borane ($\ce{BH3}$), as a reagent for alkene hydroboration. Recently in class we talked about its use in the reduction of carboxylic acids to alcohols as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now obviously $\ce{BH3}$ is an electrophile and I don't think it reacts by donating hydrides like sodium borohydride ($\ce{NaBH4}$) does - in any case it's not charged like $\ce{BH4-}$. Also $\ce{BH3}$ should be a pretty weak hydride donor, considering the covalent nature of the bonds and lack of any charge. It can reduce carboxylic acids, however, which $\ce{NaBH4}$ can't do! Where's the difference and what's the reason for this contradiction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what prevents borane from reducing acid chlorides (my instructor told me this)? This is something $\ce{NaBH4}$ is successful at. &lt;/p&gt;&#xA;" OwnerUserId="17654" LastEditorUserId="16683" LastEditDate="2017-09-23T22:29:38.903" LastActivityDate="2017-09-23T22:29:38.903" Title="Why do BH3 and NaBH4 have different selectivities?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;carbonyl-compounds&gt;&lt;chemoselectivity&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34288" PostTypeId="1" AcceptedAnswerId="34290" CreationDate="2015-07-22T13:11:40.437" Score="2" ViewCount="258" Body="&lt;p&gt;What is the rule for the punctuation marks within the bracketed figures for a bicyclo compound? Is it a 'dot' or 'comma'? In some books there are commas whereas dots in some others. &lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="7448" LastEditDate="2015-07-22T13:15:29.813" LastActivityDate="2015-07-22T14:33:32.627" Title="What is the rule for the punctuation marks for a bicyclo compound?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34289" PostTypeId="2" ParentId="34285" CreationDate="2015-07-22T13:47:54.060" Score="5" Body="&lt;p&gt;The main issue is that the Al needs to remove its hydride. With a carboxylic acid and/or an aldehyde, it can stick its hydride onto the carbonyl carbon without issue. But the carbon bonded to the alcohol cannot take on a hydride.&lt;/p&gt;&#xA;" OwnerUserId="16685" LastEditorUserId="16685" LastEditDate="2015-07-22T18:24:19.133" LastActivityDate="2015-07-22T18:24:19.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34290" PostTypeId="2" ParentId="34288" CreationDate="2015-07-22T14:33:32.627" Score="6" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-16.2.2&lt;/strong&gt; Full stops&lt;/p&gt;&#xA;&lt;p&gt;Full stops are used to separate numbers that denote lengths of bridges in polyalicyclic names constructed according to the von Baeyer system (…) and of chain lengths between spiro atoms in von Baeyer spiro names (…).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Examples:&lt;br /&gt;&#xA;bicyclo[3.2.1]octane&lt;br /&gt;&#xA;6-oxaspiro[4.5]decane&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-07-22T14:33:32.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34291" PostTypeId="1" CreationDate="2015-07-22T14:44:55.910" Score="6" ViewCount="315" Body="&lt;p&gt;I'm a geology student completing a project on chemical staining and was after some chemistry help on some of the reactions taking place. I am using various different stains to identify carbonate and feldspar minerals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the stains, in particular, is alizarin red S (1,2-dihydroxyanthraquinone) and is used to stain calcite ($\ce{CaCO3}$) red and Fe-calcite/Fe-dolomite blue. The stain powder is mixed with a weak (2%) HCl solution to react with the carbonate and produce the stain. However, I am just unsure about what reactions are taking place between the stain solution and the mineral that produce the precipitate on the surface of the grains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not finding literature on this reaction - probably because it has been mainly used by geologists without a solid chemistry background. Any help would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorUserId="402" LastEditDate="2017-04-01T00:15:04.487" LastActivityDate="2017-04-01T00:15:04.487" Title="Chemical reactions involved in staining" Tags="&lt;inorganic-chemistry&gt;&lt;adsorption&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34292" PostTypeId="1" AcceptedAnswerId="34296" CreationDate="2015-07-22T15:51:40.883" Score="0" ViewCount="369" Body="&lt;p&gt;Non of the standard undergraduate textbooks in analytical chemistry (such as Skoog, Harris) discusses conductometry as an analytical technique. I wonder whether this technique is obsolete. Or is it taught under some other area such as physical chemistry ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please point me to some good references on this matter ? I am particularly interested in text books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16096" LastActivityDate="2015-07-22T17:12:38.560" Title="What are some good references on conductometry as an analytical technique?" Tags="&lt;analytical-chemistry&gt;&lt;reference-request&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34293" PostTypeId="1" CreationDate="2015-07-22T16:08:46.557" Score="3" ViewCount="50" Body="&lt;p&gt;I'm looking for a machine-readable database describing industrially useful reactions in terms of inputs, outputs, and (possibly) catalyzers etc. for a large portion of, or (much more preferable) the entire organic and inorganic chemistry. A tall order, I know. Non-free, paywalled data references are acceptable, although free datasets are better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1-Aminoanthraquinone -&amp;gt; Alizarin&#xA;1-Aminoanthraquinone -&amp;gt; Mitoxantrone&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;EDIT: There's of course the Wikipedia, yet I cannot currently find out how to use Wikidata to mine the relationships between compounds. Many Wikipedia articles cite Ullmann's and other chemical cyclopaedias but I'm unaware of the latter having been converted into machine-readable form.&lt;/p&gt;&#xA;" OwnerUserId="5497" LastEditorUserId="5497" LastEditDate="2015-07-24T08:36:47.400" LastActivityDate="2015-07-24T08:36:47.400" Title="Graph of compounds/reactions used on the industrial scale" Tags="&lt;synthesis&gt;&lt;reference-request&gt;&lt;applied-chemistry&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34294" PostTypeId="1" CreationDate="2015-07-22T16:32:06.087" Score="0" ViewCount="1036" Body="&lt;p&gt;I know that Quantum Dots have high quantum yield and ranges from 2nm-10nm, and Nanoclusters have high fluorescence and ranges from 1nm-10nm. But the difference between them is that Quantum Dots exhibits red shift as the size gets larger while vice versa for Nanoclusters. My guess is Quantum Clusters is the combination of both, size is less than 2nm and exhibits red shift as size gets larger. Is this correct? Are there more things that distinguish them to each other? &lt;/p&gt;&#xA;" OwnerUserId="17656" LastEditorUserId="5017" LastEditDate="2015-07-22T17:24:52.610" LastActivityDate="2015-07-23T15:34:14.020" Title="What is the main difference between Quantum Clusters and Nanoclusters?" Tags="&lt;nomenclature&gt;&lt;nanotechnology&gt;" AnswerCount="1" CommentCount="6" ClosedDate="2015-07-27T16:21:27.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="34296" PostTypeId="2" ParentId="34292" CreationDate="2015-07-22T17:12:38.560" Score="1" Body="&lt;p&gt;I am certainly no expert in the field (in fact, I know nothing about it) but I imagine this 225+ page text will prove to be useful for you:&#xA;&lt;a href=&quot;https://books.google.com/books?id=o8X9BAAAQBAJ&amp;amp;pg=PR13&amp;amp;dq=conductometry&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCIQ6AEwAWoVChMIyp63rJzvxgIVRyqICh3jhwkZ#v=onepage&amp;amp;q=conductometry&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Oscillometry and Conductometry: International Series of Monographs in Analytical Chemistry, Volume 21&lt;/a&gt; by E. Pungor is available via Google Books. You might be able to find a used copy of this book somewhere (it was published in 1965). &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-22T17:12:38.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34305" PostTypeId="2" ParentId="34266" CreationDate="2015-07-22T21:02:10.953" Score="2" Body="&lt;p&gt;A molecule is formed when two or more atoms join together chemically e.g. Na and Cl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound is a molecule that contains at least two different elements e.g. Na and Cl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All compounds are molecules but not all molecules are compounds e.g. H2 and O2 are molecules but are not compounds because bond is between two of the same elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not sure what you are asking in question 3. Gay-Lussacs Law is generally associated with the combining volumes principle. e.g. 1 volume of H2(g) + 1 volume of O2(g) = 2 volumes of H2O(g). However, you may want to consider the combined gas law, not just Gay-Lussac law. This combines Gat-Lussacs law with Boyles' and Charles' Law. &lt;/p&gt;&#xA;" OwnerUserId="17619" LastActivityDate="2015-07-22T21:02:10.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34306" PostTypeId="1" AcceptedAnswerId="34308" CreationDate="2015-07-23T01:22:04.510" Score="3" ViewCount="1767" Body="&lt;p&gt;I just did a lab that involved heating KClO&lt;sub&gt;3&lt;/sub&gt; (with MnO&lt;sub&gt;2&lt;/sub&gt; as a catalyst) to decompose it into KCl and oxygen gas. That part is easy. What's throwing me is the % yield calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical yield of O should just be the mass percentage of O in KClO&lt;sub&gt;3&lt;/sub&gt; (48.00/122.55 = KClO&lt;sub&gt;3&lt;/sub&gt; is 39.17% O by mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual yield of O&lt;sub&gt;2&lt;/sub&gt; (measured by subtraction) was 0.840 g out of a initial mass of 2.062 g KClO&lt;sub&gt;3&lt;/sub&gt;, or 40.7%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Given that the theoretical yield was 39.17% based on moles of O (not O&lt;sub&gt;2&lt;/sub&gt;) , why is the actual yield 40.7% (well with margins of error outlined by the professor) when it's based on O&lt;sub&gt;2&lt;/sub&gt;? Shouldn't it be half as much? Or did I mess my calculations up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to edit tags, I'm out of my depth on this particular SE site in that area.&lt;/p&gt;&#xA;" OwnerUserId="17084" LastEditorUserId="13184" LastEditDate="2015-07-23T21:23:12.170" LastActivityDate="2015-07-23T21:23:12.170" Title="Why are the mass ratios of O to KClO3 and O2 to KClO3 the same?" Tags="&lt;experimental-chemistry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34308" PostTypeId="2" ParentId="34306" CreationDate="2015-07-23T01:59:11.713" Score="12" Body="&lt;p&gt;Your calculations are correct. The key is that you are measuring percentage &lt;em&gt;by mass:&lt;/em&gt; the mass of oxygen in 2.062 g of $\mathrm{K}\mathrm{Cl}\mathrm{O}_3$ is the same whether it is considered the mass of individual oxygen atoms or of $\mathrm{O}_2$ (or of any other allotrope of oxygen, for that matter). You have half as many &lt;em&gt;moles&lt;/em&gt; of $\mathrm{O}_2$, but the mass is the same (since the mass of 1 mol $\mathrm{O}_2$ is twice the mass of 1 mol O).&lt;/p&gt;&#xA;" OwnerUserId="13334" LastActivityDate="2015-07-23T01:59:11.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34309" PostTypeId="1" CreationDate="2015-07-23T02:57:05.230" Score="3" ViewCount="2054" Body="&lt;p&gt;I'm confused about $\mathrm{sp^2}$ hybridization in the formation of $\ce{NO3-}$ ion. Valence electrons of N are $\mathrm{(2s)^2}$ and $\mathrm{(2p)^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one of $\mathrm{2s}$ electrons kicked up to one of the $\mathrm{2p}$ orbitals, and then the remaining $\mathrm{2s}$ electron hybridized with the two remaining $\mathrm{2p}$ orbitals which have unpaired electrons? If so, that gives $\mathrm{sp^2}$ orbitals to bond with $\mathrm{p}$ orbitals of the oxygen atoms.  However, there is now that remaining unhybridized $\mathrm{p}$ orbital of nitrogen, which has 2 electrons, so how can it be involved in $\pi$ bonding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate someone explaining this. I'm sure I'm missing something simple, but my book doesn't have an explanation of $\mathrm{sp^2}$ hybridization for Group V elements.&lt;/p&gt;&#xA;" OwnerUserId="17663" LastEditorUserId="865" LastEditDate="2015-07-23T13:26:30.430" LastActivityDate="2015-07-26T12:58:20.207" Title="sp2 hybridization Group V elements" Tags="&lt;hybridization&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34311" PostTypeId="1" CreationDate="2015-07-23T04:05:52.037" Score="3" ViewCount="965" Body="&lt;p&gt;I recently purchased some 55 gallon blue barrels for rainwater harvesting.  Upon closer inspection of the barrels, I noticed they previously housed food grade sodium hypochlorite which was used to wash produce.  I was searching for food grade barrels but wasn't expecting bleach.  Because I will be watering plants and fruit tress with the rainwater, I don't want the water chlorinated which would kill the bacteria I'm trying to promote in the soil.  These barrels are empty but I was advised to rinse them when I initially purchased them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I assume that just rinsing with water won't completely get rid of the bleach  and that the residual bleach which may be on the walls of the barrel will not decompose or neutralize by itself? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to wash the barrels with water and Hydrogen Peroxide, will this suffice to remove the chlorine?&lt;/p&gt;&#xA;" OwnerUserId="17665" LastEditorUserId="15410" LastEditDate="2015-07-23T05:18:06.673" LastActivityDate="2015-07-24T08:34:29.360" Title="Food Grade Sodium hypochlorite decomposition/neutralization" Tags="&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34312" PostTypeId="1" CreationDate="2015-07-23T06:04:38.673" Score="1" ViewCount="75" Body="&lt;p&gt;I am trying to calculate PES for Molecular Oxygen and Atomic Nitrogen Exchange reaction. This is also known as Zeldowich reaction. I found that for  asingle reactions, there are many PES's which I believe are for different electronic configurations of the NO2 PES. I found that the multiplicity can be 2 or 4. Also I find different surface being named as A' and A&quot; in the literature. I would like to what exactly are these and how can I compute these in Gaussian?&lt;/p&gt;&#xA;" OwnerUserId="15648" LastActivityDate="2015-07-23T06:04:38.673" Title="Potential energy Surfaces" Tags="&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34313" PostTypeId="2" ParentId="34291" CreationDate="2015-07-23T06:43:00.897" Score="1" Body="&lt;p&gt;Alizarin red S can stain calcium deposits through interaction of its sulfonic acid and hydroxy groups with calcium ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a reference for you: &lt;a href=&quot;http://jhc.sagepub.com/content/17/2/110.full.pdf+html&quot; rel=&quot;nofollow&quot;&gt;http://jhc.sagepub.com/content/17/2/110.full.pdf+html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, however, this research seems to suggest using a solution of pH $\approx$ 9, stating that solutions of acetic acid (pH 5) did not work that well.&lt;/p&gt;&#xA;" OwnerUserId="16837" LastActivityDate="2015-07-23T06:43:00.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34314" PostTypeId="1" CreationDate="2015-07-23T08:18:46.010" Score="2" ViewCount="35" Body="&lt;p&gt;I got stuck while reviewing the journal article &quot;Accurate reaction paths using a Hessian based predictor–corrector integrator&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the distance weighted interpolation done in the predictor steps, instead of doing it in the correcter steps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:  &lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/120/21/10.1063/1.1724823&quot; rel=&quot;nofollow&quot;&gt;H. P. Hratchian and H. B. Schlegel, &lt;em&gt;J. Chem. Phys.&lt;/em&gt; &lt;strong&gt;2004&lt;/strong&gt;, &lt;em&gt;120&lt;/em&gt;, 9918&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17667" LastEditorUserId="402" LastEditDate="2016-11-05T18:28:52.183" LastActivityDate="2016-11-05T18:28:52.183" Title="Distance weighted interpolation in predictor steps vs corrector steps" Tags="&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34316" PostTypeId="1" AcceptedAnswerId="34318" CreationDate="2015-07-23T09:33:37.843" Score="1" ViewCount="2152" Body="&lt;p&gt;Selenium is a good conductor of electricity in presence of light. This phenomenon is called photoconductivity. How does this phenomenon occur? Are there any other elements or compounds in which this phenomenon is noted?&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorDisplayName="user15489" LastEditDate="2015-07-23T10:02:30.920" LastActivityDate="2016-05-15T15:13:05.730" Title="Photoconductivity of Selenium" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;conductivity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34317" PostTypeId="1" AcceptedAnswerId="34324" CreationDate="2015-07-23T09:37:21.620" Score="3" ViewCount="20259" Body="&lt;p&gt;I have learnt that elements on the left-hand side of the periodic table such as sodium and magnesium prefer to lose electrons to form a cation because this requires less energy to obtain a stable octet, and vice-versa for the right-hand side of the periodic table e.g. fluorine. However, using this reasoning I am not sure why all transition metals tend to lose electrons rather than gain them.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2018-04-26T10:47:12.600" Title="Why do transition metals tend to form positive ions instead of negative ions?" Tags="&lt;transition-metals&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34318" PostTypeId="2" ParentId="34316" CreationDate="2015-07-23T10:02:05.610" Score="6" Body="&lt;p&gt;A very handy diagram and explanation of photoconductivity is shown below, from the American Physical Society page &lt;a href=&quot;http://physics.aps.org/articles/v6/19&quot; rel=&quot;nofollow noreferrer&quot;&gt;Viewpoint: Holey Intrinsic Photoconductivity&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LNuvt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caption explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Excitation process leading to photoconductivity in a condensed matter system (upper panel) and its cold-atom counterpart (lower panel). In the condensed matter system, incoming laser light ($hv$) excites an electron to the valence band, leaving a positively charged hole behind. In the cold-atom analogy, the optical lattice is tuned to allow an atom to move into the excited band, also leaving a hole behind. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Photoconductivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photoconductivity&lt;/a&gt;, states that some 'classic' examples of photoconductive materials (asides from selenium) include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The polymer &lt;a href=&quot;https://en.wikipedia.org/wiki/N-Vinylcarbazole&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyvinylcarbazole&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lead(II)_sulfide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lead sulfide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some more examples exist in the paper &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr00017a020&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic photoconductive materials: recent trends and developments&lt;/a&gt; include the chemical classes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Phthalocyanines&lt;/li&gt;&#xA;&lt;li&gt;Squaraines&lt;/li&gt;&#xA;&lt;li&gt;Azo pigments&lt;/li&gt;&#xA;&lt;li&gt;Perylene pigments&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-07-23T10:09:30.380" LastActivityDate="2015-07-23T10:09:30.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34320" PostTypeId="1" AcceptedAnswerId="34336" CreationDate="2015-07-23T11:14:58.113" Score="10" ViewCount="303" Body="&lt;p&gt;Does there exist any data repository of chemical compound that gives the structure of the compounds with the help of coordinate geometry. It would be like each atom has a coordinate and the list of bonds it has with other atoms so that the 3D structure of the compound can be created from this very theoretical description?&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="41328" LastEditDate="2019-05-16T08:41:00.120" LastActivityDate="2019-05-16T08:41:00.120" Title="Data repository of organic compounds" Tags="&lt;organic-chemistry&gt;&lt;crystal-structure&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34321" PostTypeId="2" ParentId="34311" CreationDate="2015-07-23T11:25:15.690" Score="0" Body="&lt;p&gt;What are you afraid of exactly? The quantity of bleach remaining after the rinse is certainly miniscule, and it will exponentially drop each time you refill the barrels with rain water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're really concerned (which IMO you shouldn't be), you can refrain from using the water you collect the first time. Also, the bleach is designed to react with organic compounds (dirt), so rinsing the barrels with dirty water will make the remaining bleach react with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen Peroxide is an oxydizer just as the bleach, so it will not help.&lt;/p&gt;&#xA;" OwnerUserId="16744" LastActivityDate="2015-07-23T11:25:15.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34323" PostTypeId="1" AcceptedAnswerId="34327" CreationDate="2015-07-23T12:21:28.403" Score="7" ViewCount="8062" Body="&lt;p&gt;I just recalled, and still see that almost every organic chemistry book starts with bonds, empahsizing hybridization. One thing is straight forward but the rest doesn't seem to connect well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that's clear is, energy level wise $\mathrm{sp &amp;lt; sp^2 &amp;lt; sp^3}$. When $\mathrm{p}$ character is higher, the energy is larger, the electrophilicity is higher and its affinity for reaction is high up as well. This is the opposite of what electrons constantly thriving to achieve : the lower energy state and be stable. However, hybridization allows molecules to have a shape minimizing the energy though. Through this bonding it also releases the energy (dissociation) by stabilizing itself - so bond formation is the tendency. Hybrids ($\mathrm{sp, sp^2, sp^3}$) form $\sigma$ bonds while pure-&lt;em&gt;breeds&lt;/em&gt; (100% $\mathrm{s}$, 100% $\mathrm{p}$) form $\pi$ bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had the impression hybridization &lt;em&gt;regulates a standard bond&lt;/em&gt; among same species of atoms, e.g &lt;em&gt;$\ce{C-H}$ bond in all alkanes are the same&lt;/em&gt;. But that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am unable to reason is, why hybridization is so important for organic chemistry than to any other (general, inorganic..., physical)? Is it about bond strength (how fast one can bond and break) or is it about acidity or basity? Why should we (learners) or organic chemists care?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide some application example. It's also great if one could point out how each $\ce{C-H}$ bond differ from alkane to alkane or any other functional group.&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="186" LastEditDate="2015-07-23T13:05:18.353" LastActivityDate="2018-07-22T23:05:52.150" Title="Why does hybridization significantly matter for organic chemistry?" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34324" PostTypeId="2" ParentId="34317" CreationDate="2015-07-23T12:42:15.850" Score="5" Body="&lt;p&gt;See the outer configuration of some transition elements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sc:      $\mathrm{[Ar]~3d^1 4s^2}$  &lt;/li&gt;&#xA;&lt;li&gt;Cr:  $\mathrm{[Ar]~3d^5 4s^\color{red}{1}}$&lt;/li&gt;&#xA;&lt;li&gt;Co:  $\mathrm{[Ar]~3d^7 4s^2}$  &lt;/li&gt;&#xA;&lt;li&gt;Cu:  $\mathrm{[Ar]~3d^{10} 4s^\color{red}{1}}$&lt;/li&gt;&#xA;&lt;li&gt;Zn:  $\mathrm{[Ar]~3d^{10} 4s^2}$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you are wondering about configuration of Cu and Cr &lt;a href=&quot;https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations&quot;&gt;read&lt;/a&gt; why this happens!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion. Gaining any electron in 4s orbital would decrease the stability of anion thus formed. &lt;strong&gt;So, they form only cations&lt;/strong&gt; &lt;em&gt;(positive ions)&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the outer electronic configuration of transition elements is $n~\mathrm{s}^2~(n-1)\mathrm{d}^{1-10}$. Because of reasons above, the electron removes from  the $\ce{ns}$  orbital, where n = principal quantum number of atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See $\ce{Zn^{2+}}$ configuration:  $\mathrm{[Ar] (3d)^{10}}$ as an example. The electron exits from 4s orbital, leading to the formation of a positively charged ion.&lt;/p&gt;&#xA;" OwnerUserId="16924" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-07-24T12:06:36.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34327" PostTypeId="2" ParentId="34323" CreationDate="2015-07-23T12:59:06.737" Score="12" Body="&lt;p&gt;Orbital hybridization is mostly useful in the &lt;em&gt;rationalisation&lt;/em&gt; of &lt;em&gt;molecular geometry&lt;/em&gt;. In fact it was invented by Linus Pauling for this purpose: to rationalise structures of organic molecules using the language of atomic orbitals. Yes, hybridization is also used to rationalise bonding properties, as well as acidity or basicity, but its primary use is to rationalise molecular geometries.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I had the impression hybridization regulates a standard bond among&#xA;  same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But&#xA;  that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, $\ce{C-H}$ bonds, of course, differ in $\ce{CH4}$ and $\ce{CH3-CH3}$, but not too much; they are &lt;em&gt;approximately&lt;/em&gt; the same. Are you naive enough to hope that a model as simple as the orbital hybridization one would perfectly predict the molecular geometry of each and every organic molecule? Surely, it won't. It can just give a crude (but simple) approximation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What I am unable to reason is, why hybridization is so important for&#xA;  organic chemistry than to any other (general, inorganic..., physical)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where did you get that? Hybridization is a concept of general chemistry, it is used in both organic and inorganic one. Its relatively big importance for organic chemistry is due to the fact that it is the only simple model which can explain (approximately) molecular geometry of organic compounds. The original valence bond theory couldn't do it, so it was extended.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;On explanation vs rationalisation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now I would like to emphasize that hybridization provides a &lt;em&gt;rationalisation&lt;/em&gt; of a molecular geometry (as well as some other properties), but not an &lt;em&gt;explanation&lt;/em&gt;, so it can &lt;em&gt;not&lt;/em&gt; be used to reliable &lt;em&gt;predict&lt;/em&gt; the properties. What is the key difference between an explanation and a rationalisation in these context? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, an &lt;em&gt;explanation&lt;/em&gt; of some phenomenon describes it in way that we can observe and test the consequences of our claims and premises. We can then tell from the observations are our claims and premises true or not, or, in other words, does our explanation work or not. The whole point of an explanation is to distinguish what is true from what is not true, in a sense, what corresponds to the actual reality from what does not do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;em&gt;rationalisation&lt;/em&gt; things work differently. We insist on some premises being true without testing them against the reality. We do observations, but we do not use them to tell are our premises true or not, rather on the basis of the desired premises and the observable evidence we start to make the observed outcome seem consistent with the premise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Rationalization_(psychology)&quot; rel=&quot;noreferrer&quot;&gt;psychology&lt;/a&gt; rationalisation is also known as making excuses. It is a defence mechanism in which a controversial (from the society viewpoint) behaviour is justified &quot;in a &lt;em&gt;seemingly&lt;/em&gt; rational or logical manner to &lt;em&gt;avoid&lt;/em&gt; the true explanation&quot;. With respect to scientific theories, it can also be thought of as a defence mechanism: when the real explanation is too complex for beginners we avoid it to defend them. Instead we first provide some simple rationalisation and later in the course of study the explanation is introduced.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The way rationalisation works for all these general chemistry models is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We do know geometry or some other property of a particular molecule, say, from an experiment, and we are seeking for some rationalisation of why the things are the way they are.&lt;/li&gt;&#xA;&lt;li&gt;The rationalisation usually has to be as simple as possible, so that, for the case of molecular geometry, we usually start from &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;noreferrer&quot;&gt;VSEPR theory&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If the simplest possible theory can not rationalise the observed property, we choose a more sophisticated model. For molecular geometry, for instance, we could then try to use the original &lt;a href=&quot;https://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;noreferrer&quot;&gt;valence bond (VB) theory&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;But even the VB theory (in its original formulation) won't work for organic compounds, so it has to be extended by two concepts: &lt;a href=&quot;https://en.wikipedia.org/wiki/Resonance_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;resonance&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_hybridisation&quot; rel=&quot;noreferrer&quot;&gt;orbital hybridization&lt;/a&gt;. Note that this concepts are also required in inorganic chemistry as well, but not always: for some inorganic molecules the original VB theory is enough.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But however deep you go into this hierarchy of chemical models, all them are just rationalisation models: they provide simplified (and in many cases oversimplified) descriptions of the reality which is approximate (and in many cases rather crude). And besides all this process of seeking for an rationalisation is a bit like cheating:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If some model doesn't work, let us try another model.&lt;/li&gt;&#xA;&lt;li&gt;If none of them works, let us introduce a new concept with some funky name in one of them, the purpose of which will be to rationalise this particular thing which couldn't be rationalised by the original model.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a result, the predictive power of such models is quite small. They can be used to rationalise what is already known, but any predictions based on such models will be unreliable. And yeah, the concept of orbital hybridization is just one of these cheats mentioned above. &lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-26T17:31:25.123" LastActivityDate="2015-07-26T17:31:25.123" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="34328" PostTypeId="2" ParentId="34311" CreationDate="2015-07-23T13:00:17.640" Score="4" Body="&lt;p&gt;Several good tap water rinses should probably get rid of any significant remaining amounts of bleach, although it's possible some bleach residues might &lt;a href=&quot;https://en.wikipedia.org/wiki/Adsorption&quot; rel=&quot;nofollow noreferrer&quot;&gt;adsorb&lt;/a&gt; to the wall of the barrel and resist these efforts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4945/how-much-ascorbate-to-neutralize-one-liter-of-chlorinated-tap-water&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34006/does-bleach-without-rinsing-leave-residues-and-could-vinegar-or-baking-soda/&quot;&gt;here&lt;/a&gt;, if you want to be very sure the residues are removed, dissolve a handful of vitamin C (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ascorbic_acid&quot; rel=&quot;nofollow noreferrer&quot;&gt;ascorbic acid&lt;/a&gt;) tablets in water and use that to rinse the barrels until all bleach odor is gone.  The ascorbic acid will act as a reducing agent and should reduce the bleach levels to next to nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dmitry is right: as an oxidizer, $\ce{H2O2}$ will probably do little to help.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-07-23T13:00:17.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34329" PostTypeId="1" AcceptedAnswerId="34331" CreationDate="2015-07-23T13:11:30.000" Score="15" ViewCount="40615" Body="&lt;p&gt;In $\ce{[Cu(NH3)4]^2+}$, the $\ce{Cu^2+}$ ion has 9 electrons in the $\mathrm{3d}$ orbital with only one unpaired electron. How is it a square planar geometry? Where is that unpaired electron going?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="4945" LastEditDate="2016-09-08T07:19:55.693" LastActivityDate="2016-09-08T07:19:55.693" Title="Why is tetraamminecopper(II) a square planar and not a tetrahedral species?" Tags="&lt;coordination-compounds&gt;&lt;transition-metals&gt;&lt;molecular-structure&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34330" PostTypeId="1" CreationDate="2015-07-23T13:14:57.297" Score="4" ViewCount="3790" Body="&lt;p&gt;Suppose $\ce{Na^+}$ and $\ce{Cl^-}$ form an ionic bond. In doing so they will come closer by attraction. Now as they come closer their speed will increase and hence the kinetic energy increases. So basically the electrostatic potential energy will be controverted into kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;My question is why do these processes that involve attraction cause a release of energy from the system? And after this release why is the system said to be more stable?&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My understanding is there might be two concepts involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, as the ions approach each other their speed increase, but as soon as they touch each other they collide and reflect back. So the ions keep on oscillating. These oscillations are what we know as heat and the temperature of $\ce{NaCl}$ will be higher than its surroundings so it will release its energy in the form of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it might be because any accelerating charge releases electromagnetic waves, and so as long as the ions oscillate they'll keep releasing energy in the form of EM waves which may be trapped by other ions or may escape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the energy will be released from the system(NaCl ion here) our system must become more stable. Stable in the sense that NaCl crystal will not convert back into ionic form separated from other ions because for coming back to that state the system must be provided the lost energy back by some external agency. This is the way how I understand the term &lt;em&gt;stabilty&lt;/em&gt;; I don't what does this term mean in chemistry. I also don't know why &quot; &lt;em&gt;every system in nature wants to be more stable&lt;/em&gt;&quot; -- This is the phrasing that I heard in school days from chemistry teacher. This fact must be related to some thermodynamic fact of kind that every system wants to release energy, but why?&lt;/p&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="4609" LastEditDate="2015-07-23T18:30:43.760" LastActivityDate="2016-03-11T23:05:20.833" Title="Why does attraction always cause release of energy making the system more stable?" Tags="&lt;thermodynamics&gt;&lt;bond&gt;&lt;ions&gt;&lt;energy&gt;&lt;electrostatic-energy&gt;" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34331" PostTypeId="2" ParentId="34329" CreationDate="2015-07-23T13:33:46.130" Score="16" Body="&lt;p&gt;It is very convenient to use crystal field theory to discuss this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of distortion is a case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Jahn%E2%80%93Teller_effect&quot; rel=&quot;noreferrer&quot;&gt;Jahn–Teller&lt;/a&gt; distortion, very common in chemistry.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="30424" LastEditDate="2016-09-06T07:15:01.460" LastActivityDate="2016-09-06T07:15:01.460" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34332" PostTypeId="2" ParentId="34320" CreationDate="2015-07-23T13:46:11.217" Score="7" Body="&lt;p&gt;A pretty good database of experimental data compiled into one place has a geometry section. Check out &#xA;&lt;a href=&quot;http://cccbdb.nist.gov/&quot;&gt;http://cccbdb.nist.gov/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to the experimental data section, then click on summary of geometry data for one molecule. The database is fairly large, but not all molecules have a full geometric data set.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-07-23T13:46:11.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34333" PostTypeId="1" CreationDate="2015-07-23T13:51:53.600" Score="6" ViewCount="1358" Body="&lt;p&gt;I have this question for a long time that in LASER when we do the pumping by heating how does molecular collision causes the atoms(molecules) to reach in excited state. E.g. in $\ce{He-Ne}$ laser when the excited $\ce{He}$ atom collides with the $\ce{Ne}$ atom the $\ce{Ne}$ atom gets excited and $\ce{He}$ atom reaches its ground state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What happens at the molecular level that causes $\ce{Ne}$'s electron to jump into higher state? Does Ne atom absorb some kind of photon by some mechanism?&lt;/li&gt;&#xA;&lt;li&gt;Another thing is if we take $\ce{He}$ atom as the frame of reference then the high energy $\ce{Ne}$ atom is striking $\ce{He}$ so the high enrgy $\ce{Ne}$ must give its energy to $\ce{He}$ not the other way round.Where did I go wrong here?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4609" LastEditorUserId="5026" LastEditDate="2018-03-04T09:08:15.747" LastActivityDate="2018-03-04T09:08:15.747" Title="How does molecular collision excites the electron?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34334" PostTypeId="2" ParentId="34333" CreationDate="2015-07-23T14:22:02.657" Score="6" Body="&lt;p&gt;The general rule is that you do &lt;em&gt;not&lt;/em&gt; need to absorb the photon to excite an electron. What you need is to provide some portion of &lt;em&gt;energy&lt;/em&gt;. &#xA;It can be done by means of a photon which is absorbed by an atom, of course, but it is not the only way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could collide an atom with some other object and upon the collision some of the &lt;em&gt;kinetic energy is converted into internal energy&lt;/em&gt; which is then redistributed by its components. One of the components of the internal energy is the &lt;em&gt;electronic energy&lt;/em&gt;, so if enough kinetic energy is converted into this particular part of internal energy, an electron gets excited. This process is usually referred to as the &lt;em&gt;collisional excitation&lt;/em&gt;. This is the process by which an electron in $\ce{He}$ atom is excited: $\ce{He}$ atom in its ground electronic state collides with an energetic electron which causes electron excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A collision between $\ce{He}$ atom in an excited electronic state and ground state $\ce{Ne}$ atom also involves an energy transfer. But this time it is the excitation energy (possibly together with some amount of kinetic energy as well) which is transferred from $\ce{He}$ atom to $\ce{Ne}$ atom. Due to a near coincidence between the $\mathrm{2\,^3S_1}$ and $\mathrm{2\,^1S_0}$ energy levels of the $\ce{He}$ electronic states, and the $\mathrm{3s^2}$ and $\mathrm{2s^2}$ energy levels of $\ce{Ne}$, the transfer of excitation energy from $\ce{He}$ atom to $\ce{Ne}$ atom is quite efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtK9u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtK9u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image courtesy of &lt;a href=&quot;https://en.wikipedia.org/wiki/Helium%E2%80%93neon_laser&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer: I'm a professional chemist, not a physicist, so I might be wrong in what follows...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... but as far as I understand the matter at the microscopic level the energy is transferred by means of &lt;em&gt;electromagnetic radiation&lt;/em&gt; the quantum of which is indeed photon. But do &lt;em&gt;not&lt;/em&gt; think about this process as a photon being somehow &quot;produced&quot; in a form of a &quot;billiard ball&quot; by one atom, then transferred to another atom and finally magically &quot;disappearing&quot; here. This classical way of thinking is the root of confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the energy transfer we have two systems each described by its own wave function. In order for the energy to be transferred the systems should start to interact with each other, but once they do so they become &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_entanglement&quot; rel=&quot;nofollow noreferrer&quot;&gt;entangled&lt;/a&gt;&lt;/em&gt; and we do not even longer have separate systems. Rather, we have a big composite system described by its own wave function. After the interaction we again can speak about the individual systems each of which is again in own quantum state, though, the final state of each system is different from the starting one due to energy transfer. And in order to predict the final state of both systems (to be more precise, the probabilities of all possible final states) one could use the rules of &lt;em&gt;quantum mechanics&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-23T19:47:33.633" LastActivityDate="2015-07-23T19:47:33.633" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34336" PostTypeId="2" ParentId="34320" CreationDate="2015-07-23T15:20:05.623" Score="10" Body="&lt;p&gt;I'm not entirely clear on what you want - I guess a database of a bunch of 3D coordinates for many organic compounds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By far the largest such database is &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/&quot;&gt;NIH PubChem&lt;/a&gt; with over 30 million compounds - mostly organic. You can &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/search/search.cgi#&quot;&gt;search&lt;/a&gt; by structure or name or identifier. Most of these have 2D and 3D geometries you can visualize or download and many computed properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar database is the &lt;a href=&quot;http://www.chemspider.com&quot;&gt;RSC ChemSpider&lt;/a&gt; with over 34 million compounds. Much like PubChem, names, 2D and 3D geometries and other properties are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll put in a plug for the &lt;a href=&quot;http://pqr.pitt.edu/&quot;&gt;Pitt Quantum Repository&lt;/a&gt; which contains ~64,000 molecules and growing, all with geometries computed using quantum chemical methods. One benefit of PQR is that you can easily access the site using smartphones and tablets and use a QR code or DOI to link to the compound.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-23T15:20:05.623" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34337" PostTypeId="2" ParentId="34294" CreationDate="2015-07-23T15:34:14.020" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short answer: &quot;quantum clusters&quot; is not widespread and likely confusing. I'd suggest &quot;small nanoparticles&quot; or perhaps &quot;metal nano clusters.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't comment on optical shifts, but many of these properties are distinct to the particular material and the onset of quantized energy levels in that material. So I doubt you can simply combine properties of quantum dots and nanoparticles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked a colleague who synthesizes gold nanoparticles. She hasn’t heard anyone use the term &quot;quantum clusters&quot; and thinks people would find it very confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the usage:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Quantum Dots&lt;/strong&gt; refer to semiconductor nanoparticles, e.g., $\ce{CdS}$ or $\ce{PbSe}$ etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Nanoparticles&lt;/strong&gt; refer to nanometer-sized particles of basically anything (proteins, polymers, metals, etc.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Clusters&lt;/strong&gt; refer to &lt;em&gt;very&lt;/em&gt; small particles (i.e., &amp;lt;1-2 nm diameter) with well-defined numbers of atoms (i.e., a few hundred at most). There is an &lt;a href=&quot;http://goldbook.iupac.org/CT06769.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC &lt;em&gt;Gold Book&lt;/em&gt;&lt;/a&gt; definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A number of metal centres grouped close together which can have direct metal bonding interactions or interactions through a bridging ligand, but are not necessarily held together by these interactions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For reference, Wikipedia suggests &lt;a href=&quot;https://en.wikipedia.org/wiki/Nanoclusters&quot; rel=&quot;nofollow&quot;&gt;nanoclusters&lt;/a&gt; are aggregates of nanoparticles. I don't think that's in common use, but certainly it suggests &lt;em&gt;your&lt;/em&gt; suggested term would be very confusing.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-07-23T15:34:14.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34338" PostTypeId="1" CreationDate="2015-07-23T15:41:18.903" Score="0" ViewCount="961" Body="&lt;p&gt;I am trying to understand more about the period of vibrational nodes. Currently, I am interested in the wavelength of the $\ce{Si-O}$ bond stretch in silica. I think this value is roughly $1000\mathrm{~cm^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am somewhat confused on how to find the period of vibrational modes. I am new to this subject. When I look up the frequency, like I said, I get units of $\mathrm{cm^{-1}}$, but I want units of $\mathrm{s^{-1}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers on how I can get this value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the way to calculate it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = \frac{1}{\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nu = c \bar{\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $T$, $\nu$, and $\bar{\nu}$ being the period, frequency, and wavenumber respectively?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;T &amp;amp;= \frac{1}{c\bar{\nu}} \\&#xA;&amp;amp;= \frac{1}{(3 \times 10^{10}\mathrm{~cm~s^{-1}})(1000\mathrm{~cm^{-1}})} \\&#xA;&amp;amp;= 33.33\mathrm{~fs}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as temperature increases, wouldn't the period decrease? But in the literature, I find that as temperature increases, the wavenumber decreases. Thus, the period increases. Why would the period increase? I would think at a higher temperature, bond stretching would occur more quickly. Is this not the case?&lt;/p&gt;&#xA;" OwnerUserId="17679" LastEditorUserId="402" LastEditDate="2017-04-25T23:11:37.287" LastActivityDate="2017-04-25T23:11:37.287" Title="Period of Vibrational Modes" Tags="&lt;physical-chemistry&gt;&lt;spectroscopy&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34339" PostTypeId="2" ParentId="34330" CreationDate="2015-07-23T16:49:29.903" Score="4" Body="&lt;p&gt;&lt;em&gt;Attraction&lt;/em&gt; is a shorthand way of saying &quot;potential energy gets lower as the distance gets lower&quot;.  The opposite is true of &lt;em&gt;repulsion&lt;/em&gt;.  The potential energy of two like-charged particles, e.g. two $\ce{Na+}$ atoms, will increase as they are brought closer together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attraction (and its opposite, repulsion), are &lt;em&gt;forces&lt;/em&gt; with units (in SI) of &lt;em&gt;Newtons&lt;/em&gt;, or $\frac{\rm kg\cdot m}{\rm s^2}= \rm N$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy has units of &lt;em&gt;Joules&lt;/em&gt;, or $\frac{\rm kg\cdot m^2}{\rm s^2} = {\rm N\cdot m} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between a force $F$ between two particles separated by a distance $r$ and the potential energy $E$ of those particles is $F = -\frac{dE}{dr}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy gets lower ($dE$ is negative) when two particles get closer together ($dr$ is negative), then the force $F$ is attractive. If the energy gets higher ($dE$ is positive) as the distance gets lower ($dr$ is negative), then the force is repulsive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force and energy are two different mathematical ways of looking at the same thing.  That's why they are always related.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:05:20.833" LastActivityDate="2016-03-11T23:05:20.833" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34340" PostTypeId="1" CreationDate="2015-07-23T16:55:56.173" Score="1" ViewCount="307" Body="&lt;p&gt;I am layman of chemistry. Latterly, I am asked to do an mathematical optimization task for a bio-chemical process, which contains a conditioning step to neutralize the sulfuric acid using ammonia hydroxide. I am trying to use the mathematical model of the reaction (kinetic model) to optimize the reaction time/energy/efficiency (target not determined yet). However, I could not find the kinetics data. Is there a way I could find the kinetic model of the reaction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is that: Should I treated this reaction as an element reaction and use third-order reaction rate formula? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused by a former post (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/16758/finding-rate-of-acid-base-reaction&quot;&gt;Finding rate of acid - base reaction&lt;/a&gt;). In that post, the accepted answer says &quot;Typical acid - base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. &quot; Should I use the diffusion control model instead of the reaction rate model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion is welcomed. &lt;/p&gt;&#xA;" OwnerUserId="17680" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-23T16:55:56.173" Title="Is there a way to find the kinetics data for the strong acid weak base reaction?" Tags="&lt;inorganic-chemistry&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34342" PostTypeId="1" AcceptedAnswerId="34348" CreationDate="2015-07-23T18:38:08.293" Score="2" ViewCount="346" Body="&lt;p&gt;It has been a long time since I studied organic chemistry, but one thing I do remember is that when we needed cyanide salts we were told not to bother keeping any remaining solution because the salts hydrolyze very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can't hydrolyzed salts be fully recovered by simply evaporating the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., we don't worry about evaporating NaCl from a distilled water solution and finding anything other than NaCl.  So I'm wondering if the lab instructions to neutralize and dispose of cyanide salt solution was just based on expediency.  Or: can something unsafe happen during the evaporation of a hydrolyzed salt solution?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2016-03-11T23:04:53.130" Title="Does evaporation reverse salt hydrolysis?" Tags="&lt;organic-chemistry&gt;&lt;hydrolysis&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="34343" PostTypeId="1" AcceptedAnswerId="34345" CreationDate="2015-07-23T18:45:10.453" Score="3" ViewCount="1920" Body="&lt;p&gt;For 10 minutes each, from two identical holes, nitrogen and an unknown gas is leaked into a common vessel of 3L capacity at 300K. The resulting pressure is 4.15 bar. If the gaseous mixture contains 0.4 moles of nitrogen, then molecular mass of the unknown gas is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;attempt: According to Graham's law of effusion, $$\frac{r_1}{r_2}=\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$$ where $V_i$ is the volume of gas effused. Let $V_1$ and $M_1$ be volume and molecular mass of nitrogen. $$V_1=22.4\times0.4=8.96;M_1=28$$&#xA;Substituting, $${V_2}=8.96\sqrt{\frac{28}{M_2}}$$&#xA;and $V_1+V_2=3$, so $$8.96(1+\sqrt{\frac{28}{M_2}})=3$$&#xA;This is not possible since $\sqrt{x}\ge0$.&#xA;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="69830" LastEditDate="2021-05-09T06:56:00.900" LastActivityDate="2021-05-09T06:56:00.900" Title="Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34344" PostTypeId="2" ParentId="34263" CreationDate="2015-07-23T19:03:17.563" Score="0" Body="&lt;p&gt;You may want to consider delivering your desired gas in a condensed form, if you want to dissolve nitrogen (using liquid nitrogen) or carbon dioxide (using dry ice). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be able to reach high pressures in a sealed container easily by just adding the dry ice or liquid nitrogen to your water. The nice thing is that you won't need any pumps, regulators, valves, etc. A safety valve would keep from accidentally shattering the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this method doesn't lend itself to a continuous flow, and you'll have to carefully regulate temperature to avoid freezing, you may save a lot of time/money by making a series of batches quickly. &lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-07-23T19:03:17.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34345" PostTypeId="2" ParentId="34343" CreationDate="2015-07-23T19:14:43.890" Score="6" Body="&lt;p&gt;Calculate total moles using the ideal gas law and solving for $n$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Graham's law of diffusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we solve for $M_{\rm X}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and plugging in numbers we get the molecular mass of the unknown gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the units on the effusion rates have been omitted for clarity and because they cancel.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-23T19:14:43.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34346" PostTypeId="1" CreationDate="2015-07-23T19:15:00.300" Score="2" ViewCount="49" Body="&lt;p&gt;How would I go about determining whether a sample of an unknown substance is either potassium bicarbonate or sodium bicarbonate (given that the sample is one or the other)?&lt;/p&gt;&#xA;" OwnerUserId="17683" LastActivityDate="2015-07-23T23:23:58.383" Title="Potassium bicarbonate or sodium bicarbonate" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="0" CommentCount="3" ClosedDate="2015-07-24T01:58:36.980" ContentLicense="CC BY-SA 3.0" />
  <row Id="34348" PostTypeId="2" ParentId="34342" CreationDate="2015-07-23T22:25:55.927" Score="3" Body="&lt;h3&gt;Simple cyanide salts&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;In aqueous solution, simple cyanide salts more or less completely dissociate into their constituent ions.  E.g. for potassium cyanide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{{KCN}(s) + H2O(l) -&amp;gt; K+(aq) + CN- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those solutions are relatively stable.  There is an acid-base equilibrium between $\ce{CN-}$ and $\ce{HCN}$, but in unbuffered or in alkaline aqueous solution this equilibrium will be almost completely on the &lt;em&gt;left&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CN- + H2O &amp;lt;&amp;lt;=&amp;gt; HCN + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over very long times, if the solution is left in an open container, some of the tiny, minute amount of $\ce{HCN}$ formed in this way could evaporate, leading to a very slow loss of cyanide from the solution.  If stored closed this will not be a problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Mixing simple cyanide salts with an acid will result in an immediate and &lt;strong&gt;&lt;em&gt;dangerous&lt;/em&gt;&lt;/strong&gt; release of gaseous $\ce{HCN}$, so in this question I assume that no acids are involved.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for simple cyanide salt solutions stored appropriately, &lt;strong&gt;&lt;em&gt;yes&lt;/em&gt;&lt;/strong&gt; you should indeed be able to recover all of the initially added the solid salt after boiling off or otherwise removing any water.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Cyanide complexes of transition metals&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Salts such as potassium ferricyanide $\ce{K3Fe(CN)6}$ consist of potassium cations and complexed ferricyanide anions, $\ce{Fe(CN)6^{~3-}}$.  They are soluble in water as well, but unlike their simple cyanide anions, they can be hydrolyzed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrolysis is catalyzed by UV light and results in the formation of free $\ce{CN-}$ anions as well as iron hydr(oxide) solid precipitates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://books.google.com/books?id=q3PLBQAAQBAJ&amp;amp;pg=PA174&amp;amp;lpg=PA174&amp;amp;dq=aqueous+cyanide+hydrolysis+by+sunlight&amp;amp;source=bl&amp;amp;ots=JvSQTB9zLr&amp;amp;sig=RDoGbbsvXLzMqitiHA3lqPG3Si0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCsQ6AEwAmoVChMIle6H1JPyxgIVDjaICh3j4wts#v=onepage&amp;amp;q=aqueous%20cyanide%20hydrolysis%20by%20sunlight&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:04:53.130" LastActivityDate="2016-03-11T23:04:53.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34350" PostTypeId="1" AcceptedAnswerId="34362" CreationDate="2015-07-23T23:18:01.177" Score="1" ViewCount="9904" Body="&lt;p&gt;When determining the number of electrons transferred in a redox reaction is it the total in both half equations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2I- + Zn^2+ -&amp;gt; I2 + Zn}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we split it up into the two half-reactions and get the following &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2I -&amp;gt; I2 + 2e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Zn^2+ + 2e- -&amp;gt; Zn}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So would the number of electrons transferred in the redox reaction be two or four?&lt;/p&gt;&#xA;" OwnerUserId="16949" LastEditorDisplayName="user7951" LastEditDate="2015-07-23T23:57:35.687" LastActivityDate="2015-07-24T12:36:44.230" Title="Determining number of electrons transferred" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34352" PostTypeId="1" CreationDate="2015-07-23T23:33:21.013" Score="6" ViewCount="132" Body="&lt;p&gt;The title says it all.  I’m new to glass making, and I need a titanium compound that will render me $\ce{TiO}$ through decomposition, and has a melting point under 1600 °C.  Hopefully I can get one that goes under 1500 °C, because that’s the max these guys usually keep the furnace at.  The compounds I’ve found so far are typically in the 1800 °C range and we can’t do that.  Any input is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17687" LastEditorDisplayName="user7951" LastEditDate="2015-07-24T09:01:30.963" LastActivityDate="2015-07-24T09:01:30.963" Title="Seeking titanium compound, melting point &lt;1600 °C, decomposes to TiO" Tags="&lt;thermodynamics&gt;&lt;synthesis&gt;&lt;heat&gt;&lt;melting-point&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34353" PostTypeId="1" CreationDate="2015-07-24T00:35:06.463" Score="1" ViewCount="65" Body="&lt;p&gt;An experiment in my lab was about the Belousov–Zhabotinsky reaction and we measured 2 times - the induction period, which was the amount of time it took for the solution to change color and back for the first time, and the oscillation period, which was the period of color change cycles after that. The lab manual says to the inverse of both those times as the rate constant K for this reaction, but I am not sure why. Please help?&lt;/p&gt;&#xA;" OwnerUserId="17557" LastEditorUserId="186" LastEditDate="2015-07-24T08:34:13.223" LastActivityDate="2015-07-24T08:34:13.223" Title="Belousov–Zhabotinsky Reaction" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2018-05-24T13:09:45.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="34354" PostTypeId="1" CreationDate="2015-07-24T03:32:46.620" Score="0" ViewCount="747" Body="&lt;p&gt;I teach Biochemistry and know that heating water can break weak bonds like hydrogen bond or ionic interaction, but not disulfide bond (without reducing agent). The bond energy of hydrogen bond in water is about 12–30 kJ/mol, while the disulfide bond is 251 kJ/mol. I initially thought the heated water would have the energy much lower than 251 kJ/mol, but higher than 30 kJ/mol. But when I looked at the energy storage in 1 L of water at 90 degrees Celsius from &lt;a href=&quot;http://www.engineeringtoolbox.com/energy-storage-water-d_1463.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, it was calculated to 294 kJ. Now I wonder whether I can compare this energy to the bond energy. What do I miss here?&lt;/p&gt;&#xA;" OwnerUserId="17691" LastEditorUserId="4945" LastEditDate="2015-10-23T17:32:19.513" LastActivityDate="2016-06-11T13:40:09.270" Title="Why can't hot water break disulfide bonds when the energy storage in hot water exceeds the bond energy?" Tags="&lt;physical-chemistry&gt;&lt;bond&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34357" PostTypeId="1" AcceptedAnswerId="34360" CreationDate="2015-07-24T09:36:47.077" Score="1" ViewCount="673" Body="&lt;p&gt;A video on YouTube shows the thermal decomposition of mercury thiocyanate. A strange thing I've observed is that $\ce{Hg(SCN)2}$, which is the reactant, is a white powder and the product is a yellow-brown solid serpent-like structure. How is this possible? What is the chemical reaction for this decomposition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;video link - &lt;a href=&quot;https://www.youtube.com/watch?v=FaC4MqxUnFw&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=FaC4MqxUnFw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17368" LastEditorUserId="865" LastEditDate="2015-07-24T12:47:18.450" LastActivityDate="2015-07-25T08:33:08.077" Title="Thermal decomposition of mercury thiocyanate" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34358" PostTypeId="1" AcceptedAnswerId="34375" CreationDate="2015-07-24T09:45:02.067" Score="4" ViewCount="1982" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many isomers are there for thioesters with the formula $\ce{C4H8OS}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer given is four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that the following two are possible constitutional isomers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amLrm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amLrm.png&quot; alt=&quot;two attempts by OP on thioester&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the other two?&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="5026" LastEditDate="2018-02-18T04:16:47.670" LastActivityDate="2018-02-18T04:16:47.670" Title="Constitutional isomers for four carbon thioesters" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34359" PostTypeId="1" CreationDate="2015-07-24T10:04:06.400" Score="6" ViewCount="531" Body="&lt;p&gt;I'd like to compare the equatorial 1-methyl-2-phenylcyclohexane conformation of with axial conformation of 1-&lt;strong&gt;chloro&lt;/strong&gt;-3-methylcyclohexane conformation. &lt;strong&gt;In which molecule there is less steric repulsion?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: &lt;strong&gt;Do you know a guess of energy in kcal/mol for a gauche Ph//Me interaction and for an axial Ph//Cl interaction in this cyclohexane ring?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to have a &lt;strong&gt;comprehensive two-dimensional table for allylic A(1,2) strain&lt;/strong&gt;, but I didn't find any. &#xA;It might sound crazy or even unbelievable: A gauche Ph//Me interaction in 1-phenylpropane is only about 0.07 kcal/mol higher in energy than its antiperiplanar conformation (according to: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0040402001915509&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sciencedirect.com/science/article/pii/S0040402001915509&lt;/a&gt;) compared to 0.9 kcal/mol for a gauche Me//Me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a diaxial Me//H only weights about 0.9 kcal/mol compaired to 2.7 kcal/mol for a diaxial Ph//H interaction; underlining that phenyl is very often more sterically hindered than a methyl, if the phenyl group can't rotate away.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You might be interested why I'm so concernced with my question:&#xA;I read the paper &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/ja035001g&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pubs.acs.org/doi/pdf/10.1021/ja035001g&lt;/a&gt; and I was curious why in Scheme 5 (with 5a) the phenyl group is placed axial for R1=Me, R2=H.&#xA;(In other words I would rather take into account the allylic A13 Ph//H plus gauche Me//Ph than the diaxial Ph//Cl interaction) It's because the the isomerization of the trans hydrazone costs too much energy!&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorDisplayName="user7951" LastEditDate="2016-12-09T21:44:50.630" LastActivityDate="2016-12-09T21:44:50.630" Title="How big is the gauche interaction of methyl with phenyl in a disubstituted cyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;energy&gt;&lt;conformers&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34360" PostTypeId="2" ParentId="34357" CreationDate="2015-07-24T10:26:40.403" Score="4" Body="&lt;p&gt;The reaction is the decomposition of mercury(II) thiocyanate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $\ce {C3N4} $ is the compound that makes up the &quot;snakes,&quot; and grow due to the evolution of gases produced by the decomposition of the other products:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{CS2 + 3O2 → CO2 + 2SO2} $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce {2C3N4 → 3(CN)2 + N2} $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$\ce {HgS + O2 → Hg + SO2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also known as the &quot;Pharaoh's Serpent&quot; in pyrotechnics. The resulting solid can range from dark graphite grey to light tan in color with the inside generally much darker than the outside.&#xA;Click &lt;a href=&quot;https://googleweblight.com/https://googleweblight.com/?lite_url=https://en.m.wikipedia.org/wiki/Mercury(II%29_thiocyanate&amp;amp;ei=7lfeNJiV&amp;amp;lc=en-IN&amp;amp;s=1&amp;amp;m=364&amp;amp;ts=1437732826&amp;amp;sig=AKQ9UO_wzGjt_MMals3Vthkl261AdoEAxg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://googleweblight.com/https://googleweblight.com/?lite_url=https://m.reddit.com/r/askscience/comments/1ivp82/this_is_what_happens_when_a_mix_of_mercury/&amp;amp;ei=i7XcfWMV&amp;amp;lc=en-IN&amp;amp;s=1&amp;amp;m=364&amp;amp;ts=1437732451&amp;amp;sig=AKQ9UO9yF1Wwa-v3APCoGRZTFSDx84yOxQ&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for reference.&lt;/p&gt;&#xA;" OwnerUserId="17422" LastEditorUserId="17422" LastEditDate="2015-07-25T08:33:08.077" LastActivityDate="2015-07-25T08:33:08.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34361" PostTypeId="1" AcceptedAnswerId="34618" CreationDate="2015-07-24T10:37:02.247" Score="2" ViewCount="848" Body="&lt;p&gt;Several sources (&lt;a href=&quot;http://www.mpoweruk.com/lithium_failures.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium-ion_battery#Extreme_temperatures&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;) suggest that charging Li-ion batteries in low-temperature conditions result in lithium plating. This seems pretty logical since low temperature reduces ion mobility, and provided there exists a high enough charge current, lithium ions don't have time to intercalate into carbon layers of the anode and are deposited on top.&lt;/p&gt;&#xA;&lt;p&gt;What I fail to understand is why this process is irreversible (Wikipedia is careful to say &amp;quot;&lt;em&gt;may&lt;/em&gt; not be removable&amp;quot;). This is counterintuitive to me. Since the plated lithium ions don't have to overcome any van der Waals forces, they should be the first to go when the battery is supplying current. So as long as the metallic lithium does not grow dendrites, thereby damaging the internal structure of the cell, a little plating should be mostly harmless.&lt;/p&gt;&#xA;&lt;p&gt;Does my reasoning make sense? And if not, what did I get wrong?&lt;/p&gt;&#xA;" OwnerUserId="16744" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2017-11-09T14:19:59.710" Title="Why is lithium plating irreversible in Li-ion cells?" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34362" PostTypeId="2" ParentId="34350" CreationDate="2015-07-24T12:36:44.230" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When determining the number of electrons transferred in a redox reaction is it the total in both half equations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the total redox reaction &lt;em&gt;that is properly balanced&lt;/em&gt;, the number of electrons transferred is equal to the stoichiometric coefficient of the $\ce{e-}$ species in either &lt;em&gt;balanced&lt;/em&gt; half-reaction.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So would the number of electrons transferred in the redox reaction be two or four?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your half-reactions and total redox reaction happen to be balanced, to the number of electrons transferred in this case is &lt;strong&gt;two&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-24T12:36:44.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34363" PostTypeId="1" CreationDate="2015-07-24T13:41:49.950" Score="1" ViewCount="3990" Body="&lt;p&gt;We know that organic compounds are compounds which contain carbon atoms. I want to know whether all biological macromolecules (such as proteins, lipids, carbohydrates, and nucleic acids) are organic compounds, or are only a subset of them considered organic?&lt;/p&gt;&#xA;" OwnerUserId="17672" LastEditorUserId="16683" LastEditDate="2015-10-16T16:47:41.797" LastActivityDate="2016-01-20T22:58:08.743" Title="Are biological macromolecules organic or inorganic compounds?" Tags="&lt;organic-chemistry&gt;&lt;biochemistry&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34364" PostTypeId="2" ParentId="28813" CreationDate="2015-07-24T14:22:06.340" Score="1" Body="&lt;p&gt;I think you were on the right track!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You were right that you need the Nernst equation, which must be how you determined the initial $E_\text{cell}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You correctly identified that your $\ce{Co|Co^2+||Ni|Ni^2+}$ cell has the cell standard potential&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=E_\text{cathode}^\circ-E_\text{anode}^\circ=-0.25\ \mathrm V--0.28\ \mathrm V=0.03\ \mathrm V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is essentially asking what concentrations are needed for your cell to have the standard potential. Again looking at the Nernst equation, you can solve for the reaction quotient $Q$ when $E_\text{cell}=E_\text{cell}^\circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_\text{cell}-E_\text{cell}^\circ=0=\frac{RT}{nF}\ln Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln Q=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, you should be able to solve for the concentration of $\ce{Ni^2+}$ with simple algebra.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorDisplayName="user7951" LastEditDate="2016-12-09T21:55:54.560" LastActivityDate="2016-12-09T21:55:54.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34365" PostTypeId="1" CreationDate="2015-07-24T14:47:22.233" Score="5" ViewCount="999" Body="&lt;p&gt;Lowering of the freezing point on adding a solute is a colligative property of solutions, it doesn't matter what the nature of the solute is (of course we have to assume ideal solutions). So when it is required to use an antifreeze - why are glycerol, glycol, methanol, etc. commonly used? For a given decrease in freezing point, a given value of molality is required - in other words a particular number of moles for the liquid's given mass. For the same number of moles, it should be simply more economical to use any substance with lower molar mass. Solubility in the solvent will obviously play a role - but there are many more solutes which are more soluble than alcohols. Why isn't $\ce{NaCl}$ used, for instance? Its molecular mass is not very high, and it is soluble to an appreciable degree. It's also more economical (moreover it will dissociate in water and generate twice the number of moles, so twice the decrease in boiling point). &lt;/p&gt;&#xA;" OwnerUserId="17654" LastEditorUserId="9961" LastEditDate="2015-10-30T16:55:42.287" LastActivityDate="2015-10-30T16:55:42.287" Title="Why is glycol used as an antifreeze?" Tags="&lt;everyday-chemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34366" PostTypeId="2" ParentId="34365" CreationDate="2015-07-24T15:04:14.017" Score="4" Body="&lt;p&gt;The main reason ionic substances like salts are not used is because they tend to be corrosive to metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polar organic liquids tend to be the best because many are very soluble in water, and they are fairly inert under the conditions antifreeze is used.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-07-24T15:04:14.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34367" PostTypeId="1" CreationDate="2015-07-24T16:24:05.837" Score="2" ViewCount="1201" Body="&lt;p&gt;I want to draw a sketch of my apparatus for my report. I am conversant with the AutoCAD drafting software. Where can I download CAD blocks for commonly used labware items (like round bottom flask, allihn condensor, drying tube, heating mantle, retort stand, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="17697" LastEditorUserId="7448" LastEditDate="2015-07-24T16:24:47.007" LastActivityDate="2016-09-05T19:25:33.523" Title="Where can I find CAD blocks for lab glassware?" Tags="&lt;reference-request&gt;&lt;equipment&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34369" PostTypeId="1" AcceptedAnswerId="34372" CreationDate="2015-07-24T16:55:09.320" Score="0" ViewCount="91" Body="&lt;p&gt;I carry around a toiletries bag. I put my toothbrush there right after brushing in a restaurant or some public area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These strangely similar articles(&lt;a href=&quot;http://www.carydental.com/blog/air-dried-toothbrush-healthy-toothbrush-raleigh/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.whitefamilydental.com/blog/an-air-dried-toothbrush-is-a-healthy-toothbrush-elizabethtown/&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;) suggest toothbrushes have to be aired. Why exactly? What harm does the wetness on the bristles cause?&lt;/p&gt;&#xA;" OwnerUserId="16928" LastActivityDate="2015-07-24T18:07:49.867" Title="Why do toothbrushes have to be dried?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-07-25T00:51:17.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="34371" PostTypeId="1" AcceptedAnswerId="34376" CreationDate="2015-07-24T17:01:13.653" Score="3" ViewCount="5515" Body="&lt;p&gt;I'm trying to create a high school level lab in which we separate the metals in an alloy. The objective is to understand and apply standard reduction potentials, and also to do some extra redox reactions to reclaim solid metals. Plus I think it might be fun!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan was to simply submerge nickels in an acidic solution. Theoretically, as long as I don't use nitric acid, the nickel will dissolve and leave the solid copper. I tried it out in the fume hood (1 nickel, 100mL 5M HCl solution), but didn't see the telltale blue/green of nickel ions. I tried sanding the nickel, in case there was an inert oxide layer. For a few minutes, I saw a tiny bit of hydrogen evolution, then the sanded surface turned black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that magnesium is sometimes added to coins for corrosion resistance, which is probably why I'm having problems. Does anyone have suggestions to make this lab possible? Maybe some electrochemistry?&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2018-05-01T20:03:57.340" Title="Separating nickel coin into copper and nickel components" Tags="&lt;inorganic-chemistry&gt;&lt;electrochemistry&gt;&lt;redox&gt;&lt;transition-metals&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34372" PostTypeId="2" ParentId="34369" CreationDate="2015-07-24T18:07:49.867" Score="5" Body="&lt;p&gt;The good people at the &lt;a href=&quot;http://www.aapd.org/&quot;&gt;American Academy of Pediatric Dentistry&lt;/a&gt; have a number of useful resources available to shed light on questions such as yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I direct you to a manuscript entitled &lt;a href=&quot;http://www.aapd.org/assets/1/25/Filho-22-05.pdf&quot;&gt;&lt;em&gt;Microbial contamination of toothbrushes and their decontamination&lt;/em&gt;&lt;/a&gt; in which it is written:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Storage conditions of toothbrushes are an important factor&#xA;  for bacterial survival. Dayoub et al. and Meier et al. reported&#xA;  that the number of microorganisms in the toothbrushes kept&#xA;  in aerated conditions was lower than in toothbrushes stored in plastic bags. Several authors have reported that bacterial contamination can be reduced by washing toothbrushes after use, and drying in aerated conditions. Caudry et al. reported that a wet environment increases bacterial growth and cross contamination. Therefore, as time increases between one toothbrushing and another, more microorganism development can occur in the toothbrushes a wet/moisture environment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The references cited in the above passage are detailed in the linked PDF. Happy brushing!&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-24T18:07:49.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34373" PostTypeId="1" CreationDate="2015-07-24T18:16:42.487" Score="0" ViewCount="2895" Body="&lt;p&gt;It's been over a semester since my most recent chemistry class, (general chemistry 2) because I could not take organic at my community college. It is only considered a sophomore level class unless taken in-residence at the university I am transferring to (which is understandable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was curious if I could narrow down what topics I could begin reviewing before the start of the semester that would help me in organic chemistry.  &lt;/p&gt;&#xA;" OwnerUserId="17699" LastEditorUserId="402" LastEditDate="2017-05-17T20:41:07.253" LastActivityDate="2017-05-17T20:41:07.253" Title="What topics to review before starting organic chemistry?" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="7" ClosedDate="2015-10-16T22:02:52.977" ContentLicense="CC BY-SA 3.0" />
  <row Id="34374" PostTypeId="1" CreationDate="2015-07-24T18:51:06.163" Score="5" ViewCount="97" Body="&lt;p&gt;I work in a winery and we have some problems about a new process involving phlorizin. We know that phlorizin is the major dihydrochalcone found in apples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been reading that this compound is also found in a grape named Albariño, a white variety of grape. What else is known about the phlorizin content in grapes?&lt;/p&gt;&#xA;" OwnerUserId="17700" LastEditorUserId="402" LastEditDate="2015-10-12T03:07:54.880" LastActivityDate="2015-10-12T03:07:54.880" Title="Phlorizin content in grapes" Tags="&lt;biochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34375" PostTypeId="2" ParentId="34358" CreationDate="2015-07-24T19:03:20.790" Score="7" Body="&lt;p&gt;One way to break things down in by the number of carbon atoms on either &quot;side&quot; of the thioester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 4 carbon atoms to work with, of which one will be taken up by the thioester moiety.  Thus there are three left to work with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the carbonyl &quot;side&quot; of the thioester, between zero and three carbons are possible.  However, on the sulfur &quot;side&quot; of the thioester, at least one carbon atom is required; otherwise the compound is not a thioester but a thioacid.  Thus the possibilities for carbon atom distribution are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one carbon on S side, two on CO side.&lt;/li&gt;&#xA;&lt;li&gt;two carbons on S side, one on CO side.&lt;/li&gt;&#xA;&lt;li&gt;three carbons on S side, zero on CO side.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You have drawn the only structures for the first two possibilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the last possibility, there are &lt;em&gt;two&lt;/em&gt; possible ways to attach a three-carbon alkyl chain to the thioester sulfur:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;as a $n$-propyl group&lt;/li&gt;&#xA;&lt;li&gt;as an isopropyl group&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, the four possible thioesters are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;methyl thiopropanoate&lt;/li&gt;&#xA;&lt;li&gt;ethyl thioacetate&lt;/li&gt;&#xA;&lt;li&gt;n-propyl thioformate&lt;/li&gt;&#xA;&lt;li&gt;isopropyl thioformate&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-24T19:03:20.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34376" PostTypeId="2" ParentId="34371" CreationDate="2015-07-24T19:25:28.813" Score="3" Body="&lt;p&gt;I recommend to dissolve the copper-nickel alloy in nitric acid and determine the dissolved $\ce{Cu(II)}$ by iodometric titration.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Start by weighing the coin, then dissolve it by placing it in nitric acid solution (under a fume hood). Continue to boil the solution for 20 min after complete dissolution of the coin (to remove any traces of remaining $\ce{NO2}$ in soution. The volume of the solution is about $40 \,\mathrm{mL}$.&#xA;$$\ce{Cu(s) + 4HNO3(aq)-&amp;gt;Cu^{2+}(aq) + 2NO3^- (aq) + 2NO2(g) + 2H2O(l)}$$ &#xA;$$\ce{Ni(s) + 2H3O^+(aq)-&amp;gt;Ni^{2+}(aq) + H2 (g)  + 2H2O(l)}$$&lt;/li&gt;&#xA;&lt;li&gt;Prepare a fresh solution of sodium thiosulfate of precise concentration $\approx 0.04\,\mathrm{M}$&lt;/li&gt;&#xA;&lt;li&gt;Add $1.00 \,\mathrm{mL}$ of the diluted copper-containing solution to $15 \,\mathrm{mL}$ of $10\% (W/V)$ potassium iodide solution.&#xA;$$\ce{Cu^{2+}(aq) +  5I^- (aq)-&amp;gt;CuI(s) + I3^- (aq) }$$ &lt;/li&gt;&#xA;&lt;li&gt;Titrate the yellow-orange suspension with sodium thiosulfate solution until the solution color has become yellow. To detect the end-point add about $1 \,\mathrm{mL}$ of starch solution.&#xA;$$\ce{I3^- (aq)  +  2S2O3^{2-} (aq)-&amp;gt;S4O6^{2-} (aq) + 3I^- (aq) }$$&lt;/li&gt;&#xA;&lt;li&gt;Repeat the titration to precisely determine the end-point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-07-24T19:25:28.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34377" PostTypeId="1" AcceptedAnswerId="34457" CreationDate="2015-07-24T22:56:38.807" Score="8" ViewCount="666" Body="&lt;p&gt;I am working on Ag-Sn-Cu alloy phase diagram to figure out the best temperature / duration to homogenize the alloy (annealing). The %age of elements in the alloy are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ag - 40 %&lt;/li&gt;&#xA;&lt;li&gt;Cu - 27.8 %&lt;/li&gt;&#xA;&lt;li&gt;Sn - 32.2 %&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(Composition can be varied if necessary but not more than 5 %)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the main goal is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find out the temperature / time duration at which the ingot of the above said alloy should be annealed to get the maximum Ag3Sn phase in the final ingot. Ag3Sn is the most desired phase with Cu3Sn being the second desired phase. Anyhow the main goal is to make as much Ag3Sn as possible.&lt;/li&gt;&#xA;&lt;li&gt;The time duration for which the ingot should be placed at the desired temperature. (The total ingot is usually 6–7 kg)&lt;/li&gt;&#xA;&lt;li&gt;[Supplementary question] Should the temperature be gradually reduced to the desired level of maximum Ag3Sn yield (or gradually increased)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The phase diagram of Ag-Sn-Cu is as follows:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XmY1b.jpg&quot; alt=&quot;Ag-Su-Cu Phase Diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16866" LastEditorDisplayName="user7951" LastEditDate="2019-09-13T16:01:52.823" LastActivityDate="2019-09-13T16:01:52.823" Title="Ag-Sn-Cu Phase diagram to formulate homogenization temperature / duration" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;metal&gt;&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34380" PostTypeId="1" AcceptedAnswerId="37526" CreationDate="2015-07-25T02:08:22.310" Score="19" ViewCount="1656" Body="&lt;p&gt;On page 88 and 89 of &lt;a href=&quot;http://library.sciencemadness.org/library/books/ignition.pdf&quot;&gt;this document&lt;/a&gt;, the author laments that $\ce{OClF5}$ has still not been discovered. He says it would be &quot;just about the ultimate possible storable oxidizer&quot; (for rockets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: it's pages 88 and 89 as labeled by the writing on the page, not the pdf metadata.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The text seems a little dated. Has anyone ever found/produced any $\ce{OClF5}$?&lt;/strong&gt; A related compound, $\ce{OClF3}$, was made in 1970 (footnote on page 89). So this seems to indicate that one of those weird loopy pairs of electrons &lt;strong&gt;can&lt;/strong&gt; be jammed/bonded with an atom of oxygen.&lt;/p&gt;&#xA;" OwnerUserId="17242" LastEditorUserId="275" LastEditDate="2015-07-29T00:52:24.037" LastActivityDate="2015-09-18T12:42:50.223" Title="Is there any experimental evidence for OClF5?" Tags="&lt;inorganic-chemistry&gt;&lt;reference-request&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34381" PostTypeId="1" CreationDate="2015-07-25T03:33:14.293" Score="8" ViewCount="3462" Body="&lt;p&gt;The question is to compare the SN1 reactivity for the following compounds. According to me it should be like R&gt;S&gt;Q&gt;P. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the correct answer is Q&gt;P&gt;R&gt;S. How can a lone pair donating into a vacant p be more stable than allyl/benzyl carbocation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the four compounds in the order PQRS respectively:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Yc8ub.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yc8ub.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="5026" LastEditDate="2018-02-26T02:10:43.220" LastActivityDate="2018-02-26T02:10:43.220" Title="SN1 reactivity order" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34382" PostTypeId="1" CreationDate="2015-07-25T04:44:00.403" Score="15" ViewCount="19009" Body="&lt;p&gt;I completely understand how phenyl group 'releases' or 'donates' electron through resonance, but how is that a phenyl group shows a weak -I effect, despite having electrons literally smeared on itself?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="15410" LastEditDate="2015-10-26T09:56:36.590" LastActivityDate="2017-08-01T14:14:21.073" Title="Inductive effect of phenyl ring" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;resonance&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34383" PostTypeId="2" ParentId="33780" CreationDate="2015-07-25T05:45:58.413" Score="11" Body="&lt;p&gt;I appreciate above answer by BreakingBioinformatics. However I have been looking for the oxidation reactions involving Fe in this case. &lt;a href=&quot;http://chemwiki.ucdavis.edu/?title=Textbook_Maps/General_Chemistry_Textbook_Maps/Map:_Averill_%26_Eldredge_%22Chemistry%22/23:_The_d-Block_Elements/23.6:_Transition_Metals_in_Biology&quot; rel=&quot;noreferrer&quot;&gt;Found some useful material here.&lt;/a&gt; It is based on the textbook, Chemistry: Principles, Patterns, and Applications, 2007, Bruce Averill, Patricia&#xA;Eldredge.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note : Following picture only shows the binding of Myoglobin.&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YEvNi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YEvNi.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's the text which explains some relevant and interesting details about the affinity of O2 and CO, although the main focus of the text is on &lt;strong&gt;primary function of the distal histidine in preventing self-poisoning of CO&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Oxygen is not unique in its ability to bind to a ferrous heme complex;&#xA;small molecules such as CO and NO bind to deoxymyoglobin even more&#xA;tightly than does O2. The interaction of the heme iron with oxygen and&#xA;other diatomic molecules involves the transfer of electron density&#xA;from the filled t2g orbitals of the low-spin d6 Fe2+ ion to the empty&#xA;π* orbitals of the ligand. In the case of the Fe2+–O2 interaction, the&#xA;transfer of electron density is so great that the Fe–O2 unit can be&#xA;described as containing low-spin Fe3+ (d5) and O2−. We can therefore&#xA;represent the binding of O2 to deoxyhemoglobin and its release as a&#xA;reversible redox reaction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Fe^{2+} + O2 ⇌ Fe^{3+} – O^{2−}}$&lt;/span&gt; (26.8.2)&lt;/p&gt;&#xA;&lt;p&gt;As shown in Figure 26.8.4 (the figure is not loading), the Fe–O2 unit is&#xA;bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in&#xA;CO are empty and those in NO are singly occupied, these ligands&#xA;interact more strongly with Fe2+ than does O2, in which the π*&#xA;orbitals of the neutral ligand are doubly occupied.&lt;/p&gt;&#xA;&lt;p&gt;Although CO has a much greater affinity for a ferrous heme than does O2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O2, which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O2 and CO provides a plausible explanation for the difference in affinities. As shown in Figure 26.8.4 &amp;quot;Binding of O&amp;quot;, the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-07-25T05:45:58.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34384" PostTypeId="1" AcceptedAnswerId="34625" CreationDate="2015-07-25T06:40:48.127" Score="-1" ViewCount="256" Body="&lt;p&gt;Is there any gas that contains oxygen so that it doesn't require oxygen from the environment in order to burn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to do is use LPG gas, which is fed through a pipe to a burner that is placed in an environment that has no air, somewhat like a vacuum. Is there any way to light the burner inside that vacuum environment? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few &lt;em&gt;wild&lt;/em&gt; ideas that I had included finding some gas that contains oxygen in itself. I may be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT - The question doesn't end here. Please read the comments section below for any doubts that you might have. And if it isn't answered in comments section then ONLY comment.&lt;/p&gt;&#xA;" OwnerUserId="16127" LastEditorUserId="865" LastEditDate="2015-08-02T00:57:12.337" LastActivityDate="2015-08-07T08:12:39.567" Title="How can I light a fire in this case?" Tags="&lt;gas-laws&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="34389" PostTypeId="1" CreationDate="2015-07-25T10:50:06.207" Score="3" ViewCount="183" Body="&lt;p&gt;I have a model railway in which I use pure copper on the tracks and in the engine to power the train with electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any solution to prevent the copper surfaces from corrosion, and maybe improve the conductivity?&lt;/p&gt;&#xA;" OwnerUserId="17713" LastEditorDisplayName="user7576" LastEditDate="2015-07-25T11:04:55.057" LastActivityDate="2015-07-26T01:50:14.977" Title="Improving copper contact and preventing oxidation" Tags="&lt;metal&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34390" PostTypeId="1" CreationDate="2015-07-25T11:20:41.773" Score="6" ViewCount="474" Body="&lt;p&gt;So basically, the whole point of this process is to extract elemental sulfur from underground deposits. Is this process essentially correct or am I missing out certain details?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Firstly, three concentric pipes are placed into the underground sulfur deposit&lt;/li&gt;&#xA;&lt;li&gt;Then, water that has been superheated is pressurised into the sulfur deposit through the outermost pipes.&lt;/li&gt;&#xA;&lt;li&gt;The result of the above is melted sulfur stuck in the ground in an &lt;em&gt;emulsion&lt;/em&gt;. The pressure of the chamber is also increased (i don't know why?)&lt;/li&gt;&#xA;&lt;li&gt;To get it out of the ground, compressed air is injected through the innermost pipe: this forces the sulfur-water emulsion to the surface through the middle pipe&lt;/li&gt;&#xA;&lt;li&gt;When this mixture is cooled, the solid sulfur separates from water and 99.5% pure sulfur is obtained&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this basically the process? Also, are these the properties of sulfur that this process relies on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sulfur has a low density, hence the compressed air is able to lift the sulfur-water emulsion to the surface&lt;/li&gt;&#xA;&lt;li&gt;Sulfur is insoluble in water and not reactive. Hence, sulfur can be recovered more easily after the sulfur-water emulsion cools as it does not mix with the water, nor react with it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are also some problems with this process if I am right:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The hot water used in this process causes thermal pollution, and may also contain some dissolved minerals. This may disrupt the chemical composition of local ecosystems, so one must recycle it.&lt;/li&gt;&#xA;&lt;li&gt;Depending on the make up of the area, remaining crypts can lead to earth subsidence. These are also harder to refill.&lt;/li&gt;&#xA;&lt;li&gt;This process releases sulfur, which must be quickly cooled otherwise it oxidises to sulfur dioxide or may be reduced to hydrogen sulfide, both of which are serious air pollutants.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17714" LastActivityDate="2015-07-25T11:20:41.773" Title="Is my understanding of the Frasch Process correct?" Tags="&lt;environmental-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34392" PostTypeId="1" CreationDate="2015-07-25T14:19:35.677" Score="5" ViewCount="277" Body="&lt;p&gt;I mean in principle, in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we tell if some phenomena was a chemical reaction(s)? Change in physical properties cannot be since large-scale structural change can change physical properties as well, like heat treatment of steel. And what about changing diamond into graphite by heating. Hence, assuming there are only chemical and physical properties, only a change in chemical properties implies chemical reaction. But how do we know if chemical properties changed? Observing chemical reactions of the result(s). But how do we know what happens is a chemical reaction? A different instance of the same problem, which we started with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can anyone sure claim that a chemical reaction happened, or equivalently that some sample is a different substance and not just an allotrope or has larger &quot;grains&quot; for example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe mass or energy change can rule out those cases?&lt;/p&gt;&#xA;" OwnerUserId="17715" LastEditorUserId="402" LastEditDate="2017-04-01T00:20:15.050" LastActivityDate="2017-04-24T22:02:21.907" Title="How can we tell whether a phenomena was a chemical reaction or not?" Tags="&lt;experimental-chemistry&gt;&lt;history-of-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34393" PostTypeId="2" ParentId="34373" CreationDate="2015-07-25T14:25:38.507" Score="4" Body="&lt;p&gt;As @inɒzɘmɒЯ.A.M pointed out in his comment, the usual prerequisite for an organic chemistry course is that a &lt;em&gt;general chemistry&lt;/em&gt; course is already taken and successfully passed. The general chemistry is, of course, a rather large topic, so I'll just name a few subtopics which I feel are absolutely essential to start a venture into the world of organic chemistry.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Structure of matter.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Modern atomic theory (Schrödinger equation, atomic orbitals, spin,  Pauli exclusion principle, electron configurations).&lt;/li&gt;&#xA;&lt;li&gt;The periodic table (periodicity, group trends).&lt;/li&gt;&#xA;&lt;li&gt;Chemical bonding (covalent and ionic bonding, Lewis structures, resonance, &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory&quot; rel=&quot;nofollow&quot;&gt;VSEPR theory &lt;/a&gt;, introductory &lt;a href=&quot;https://en.wikipedia.org/wiki/Valence_bond_theory&quot; rel=&quot;nofollow&quot;&gt;Valence Bond Theory&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_orbital_theory&quot; rel=&quot;nofollow&quot;&gt;Molecular Orbital Theory&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nomenclature &amp;amp; stoichiometry (empirical and molecular formulas, chemical equations, mole, limiting reagents, yield).&lt;/li&gt;&#xA;&lt;li&gt;Solutions (solubility, concentration, properties of solutions).&lt;/li&gt;&#xA;&lt;li&gt;Acid-base reactions (definitions of acids and bases, pH, strong/weak acids and bases, trends).&lt;/li&gt;&#xA;&lt;li&gt;Oxidation-reduction reactions (oxidation numbers, balancing oxidation-reduction reactions).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Basic concepts (energy, work, heat, enthalpy, entropy, Gibbs free energy ).&lt;/li&gt;&#xA;&lt;li&gt;The first and the second laws of thermodynamics.&lt;/li&gt;&#xA;&lt;li&gt;Hess's law.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Kinetics.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reaction rates (temperature dependence, Arrhenius equation).&lt;/li&gt;&#xA;&lt;li&gt;Reaction orders (first and second order reactions, molecularity, rate laws).&lt;/li&gt;&#xA;&lt;li&gt;Catalysis.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chemical equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Basic principles (the notion of equilibrium, the law of mass action, equilibrium constants, the direction of a reaction).&lt;/li&gt;&#xA;&lt;li&gt;Shifting the equilibrium (changing amounts, pressure, volume, temperature).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-25T14:31:03.363" LastActivityDate="2015-07-25T14:31:03.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34396" PostTypeId="1" AcceptedAnswerId="34397" CreationDate="2015-07-25T18:14:06.237" Score="3" ViewCount="661" Body="&lt;p&gt;So I'm trying to figure out how to titrate the reducing sugars on honey without generating iodine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction so far I think I could do is do a Fehling reaction with $\ce{Cu}^{2+}$and then back-titrate the remaining with EDTA, but I'm not sure that the EDTA wouldn't solubilize the $\ce{Cu2O}$, or be able to displace the tartrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sources I've looked say that this can be done as a direct titration using methylene blue as an indicator, but they haven't given a pH range, even if the reaction is fast enough to be done as a direct titration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="17719" LastEditorUserId="865" LastEditDate="2015-07-25T18:41:38.837" LastActivityDate="2015-07-25T20:33:05.960" Title="Titration of reducing sugars on honey" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34397" PostTypeId="2" ParentId="34396" CreationDate="2015-07-25T20:33:05.960" Score="3" Body="&lt;p&gt;The usual titration of glucose with Fehling's Reagent is to place the Fehling's Reagent in a flask and titrate the boiling solution with glucose. At the end point, all the $\ce {Cu^{2+}}$ ions are converted to $\ce {Cu^{+}}$; this makes the solution colorless. That is difficult to see since the red copper(I) oxide is floating in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With methylene blue, after the copper has been reduced, the glucose can then &lt;a href=&quot;https://en.wikipedia.org/wiki/Blue_bottle_(chemical_reaction)&quot; rel=&quot;nofollow&quot;&gt;reduce the methylene blue&lt;/a&gt; (in a basic solution) to a colorless solution. This makes the end point clearer since the methylene blue is much darker than the $\ce {Cu^{2+}}$ ions. The pH of the solution does not change much in this reaction. &lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-07-25T20:33:05.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34398" PostTypeId="1" CreationDate="2015-07-25T20:49:43.823" Score="6" ViewCount="160" Body="&lt;p&gt;I was so intrigued by the reported observation that sunlight precipitates the iron from $\ce{K4Fe(CN)6}$ (&lt;a href=&quot;https://books.google.com/books?id=q3PLBQAAQBAJ&amp;amp;pg=PA174&amp;amp;lpg=PA174&amp;amp;dq=Matuschek#v=onepage&amp;amp;q=Matuschek&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;originally attributed to Matuschek, 1901&lt;/a&gt;) that I wanted to see it myself.  I put a saturated aqueous solution, uncovered, in direct sunlight for four hours at mid-day.  Absolutely nothing precipitated from the solution!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding from the given reference is that in solution the $\ce{K4Fe(CN)6}$ gives $\ce{Fe(CN)6^{4−}}$ and then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe(CN)6^{4−} + 2H2O &amp;lt;-&amp;gt; Fe(CN)5 + (H2O)^{-3} + HCN + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photolysis causes $\ce{Fe(CN)6^{4−} -&amp;gt;[h\nu] Fe(CN)6^{3−} + e-}$,  so then we also have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe(CN)6^{3−} + 2H2O &amp;lt;-&amp;gt; Fe(CN)5 + (H2O)^{-2} + HCN + OH-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to assume that &lt;em&gt;if&lt;/em&gt; a $\ce{Fe(CN)6^{4-}}$ anion absorbs a photon of adequate energy (in this case it appears to require $\lambda &amp;lt; 313nm$) then with 100% probability the anion will photolyze the Fe bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And are there models that predict (at least order of magnitude) the absorption probability of a photon by an aqueous anion (being irradiated with a given spectral flux)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, what is the reaction that would precipitate iron in this case?  When we bring the K back into the equations it looks like the HCN would react to produce KCN.  I don't know what should happen to $\ce{Fe(CN)5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, after several days in open sunlight the solution is subjectively getting darker and I do see a small amount of red sediment at the bottom of the dish that appears to redissolve on swirling.  Could be an iron oxide, but would also be consistent with $\ce{K3Fe(CN)6}$.  In either case I don't have any indication from the source what should happen to the $\ce{(CN)5}$ groups if iron is precipitating, or what else might take one of the K atoms from the initial compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ultimately I had hoped to quantify the rate of photolysis by weighing the precipitate.)&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="24052" LastEditDate="2016-03-11T23:10:02.047" LastActivityDate="2016-03-11T23:10:02.047" Title="Why can't I reproduce this photolytic reaction of ferrocyanide?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="34399" PostTypeId="1" CreationDate="2015-07-25T22:30:12.877" Score="1" ViewCount="3353" Body="&lt;p&gt;I am attempting to use sodium polyacrylate as a sodium transport mechanism in a sulfur-bromine cell. I can upset the sodium balance with sodium salts, but the contamination introduced by the non-metallic ion from that salt is unacceptable. Water is a solvent that can be ultimately coerced into a solution compatible with the acrylic polymer I wish to matrix the polyacrylate within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried (with no success) alcohols, acetone, and ketones. Are there non-aqueous solvents I should consider that suit my needs that I have not yet tried? &lt;/p&gt;&#xA;" OwnerUserId="17721" LastEditorUserId="865" LastEditDate="2015-08-05T18:38:16.613" LastActivityDate="2017-01-02T06:54:27.717" Title="What are the non-aqueous solvents for sodium polyacrylate?" Tags="&lt;solvents&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34400" PostTypeId="1" AcceptedAnswerId="34644" CreationDate="2015-07-25T22:31:51.443" Score="6" ViewCount="2750" Body="&lt;p&gt;I have been synthesizing TCPO (bis(2,4,6-trichlorophenyl) oxalate) for chemiluminescent reactions, using 2,4,6-trichlorophenol and oxalyl chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jHv7O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jHv7O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I am tired of paying for oxalyl chloride (and hazard class 3 shipping), I wish to find a synthetic route to produce my own stock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common synthesis of oxalyl chloride uses oxalic acid and phosphorus pentachloride. This would reduce the cost, but is also problematic because phosphorus chlorides are considered to be drug precursor (21 CFR 1310.02), still too expensive, and hazardous to ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone recommend a synthetic route to oxalyl chloride that does not use phosphorus? Or an alternate (safer / easier / less expensive) route to TCPO?&lt;/p&gt;&#xA;" OwnerUserId="15363" LastEditorUserId="16683" LastEditDate="2017-01-08T14:35:49.507" LastActivityDate="2018-03-07T23:40:04.557" Title="Synthesis of oxalyl chloride that doesn't involve phosphorus" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;reference-request&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34401" PostTypeId="2" ParentId="34398" CreationDate="2015-07-25T22:57:59.047" Score="2" Body="&lt;p&gt;I would put the solution under a blacklight exposure source or a equivelant such as a mercury arc lamp at close range. The referenced writing makes no specifics about the the timeframe of prolonged. This could have been 3 months of exposure to sunlight.&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-07-25T22:57:59.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34402" PostTypeId="1" CreationDate="2015-07-26T01:28:24.687" Score="9" ViewCount="364" Body="&lt;p&gt;(Clarification: this is a question about terminology, not so much about when to use different concepts.  I think I do understand which characteristics of a species to use when trying to determine something about a reaction in some circumstances.  I just don't know what word to call it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When someone on the internet (here or elsewhere) is asked the difference between &quot;basicity&quot; and &quot;nucleophilicity&quot;, they typically give one of the following two answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Basicity is about how well a species donates electrons to a proton, while nucleophilicity is about how well a species donates electrons to a more general atom or ion (especially carbon).&lt;/li&gt;&#xA;&lt;li&gt;Basicity is about thermodynamics (i.e. equilibrium constant) while nucleophilicity is about kinetics (i.e. activation energy).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While these two answers may be related, they do not seem identical!  So I am wondering which one is actually correct, or whether there is some way that they are both correct.  In particular, I'm wondering what happens when you have a situation that puts those two descriptions at odds with each other.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We could have a reversible reaction involving carbon (though I know that's not the usual situation) that does in fact have some equilibrium.  When we're talking about that reaction's equilibrium, do we still use the word &quot;basicity&quot;?&lt;/li&gt;&#xA;&lt;li&gt;In a reaction with a proton rather than a carbon, when we're talking about kinetics, do we still use the word &quot;nucleophilicity&quot;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hope I've gotten my question across OK.&lt;/p&gt;&#xA;" OwnerUserId="17723" LastEditorUserId="10185" LastEditDate="2015-07-27T08:07:15.877" LastActivityDate="2017-11-09T13:38:30.990" Title="What is the relationship between the two definitions of the difference between basicity and nucleophilicity?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34403" PostTypeId="1" CreationDate="2015-07-26T01:49:46.757" Score="0" ViewCount="96" Body="&lt;p&gt;My question is: what is the primary reason that bicarbonate ($\ce{HCO3^-}$) can act as a buffer in a solution? The two (possibly) relevant equations that I can think of are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {H2CO3 + H2O &amp;lt;=&amp;gt;HCO3- + H3O+ }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ce {HCO3- + H2O &amp;lt;=&amp;gt; CO3^{2-} + H3O+} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these two equations, I can come up with two 'reasons' as to why $\ce{HCO3-}$ can act as a buffer. The first reason is that $\ce{HCO3-}$ can create $\ce{H+}$ ions in the second equation, but can also use up $\ce{H+}$ ions in the first equation. This means that if I put an excess amount of $\ce{H+}$ ions into the solution, the $\ce{HCO3-}$ can act by Le Chatelier's principle to remove the $\ce{H+}$ ions. If I put an excess of $\ce{OH-}$ ions into solution, they will react with the $\ce{H+}$ ions, and cause a shortage. The $\ce{HCO3-}$ can then react with water to replenish the $\ce{H+}$ ions in the solution. The only requirement for this explanation to occur is that enough $\ce{HCO3-}$ exists within the solution to be able to balance out whatever changes may occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the picture I had in my head as I began to study my textbook today. However, my textbook differed, listing only the first equation as the reason $\ce{HCO3-}$ can act as a buffer. In the first equation, an excess of $\ce{H+}$ ions would drive the equation to the left, removing the excess ions from the solution. An excess of $\ce{OH-}$ ions would react with the $\ce{H+}$ ions, and drive the equation to the right, replenishing the $\ce{H+}$ ions, and restoring neutrality. This is a much simpler explanation, but it requires that there be adequate amounts of both $\ce{HCO3-}$ AND $\ce{H2CO3}$ in the solution to maintain neutrality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both 'reasons'seem to operate on the same fundamental rules, and yet they require different things to operate, and I am pretty sure my textbook is right and not me. One reason I can think of why is perhaps that the second equilibrium lies far on the left, so far left that barely any $\ce{CO3^{2-}}$ is ever present in solution, which would mean the second equation could never help with buffering the solution, or if it did it would be negligible help. Is this line of reasoning correct?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="402" LastEditDate="2017-04-25T23:11:22.917" LastActivityDate="2017-04-25T23:11:22.917" Title="The reason for bicarbonate acting as a buffer" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34404" PostTypeId="2" ParentId="34389" CreationDate="2015-07-26T01:50:14.977" Score="1" Body="&lt;p&gt;Though I have not tried them &lt;em&gt;DeoxIT&lt;/em&gt; products claim to do what you want, available as &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B003D8EA7A&quot; rel=&quot;nofollow&quot;&gt;liquid&lt;/a&gt; and &lt;a href=&quot;http://www.allelectronics.com/make-a-store/item/g5s-6/deoxit-r-gold-5-oz/1.html&quot; rel=&quot;nofollow&quot;&gt;spray&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has used these, please add a comment as to effectiveness.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-26T01:50:14.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34405" PostTypeId="2" ParentId="34363" CreationDate="2015-07-26T01:59:46.277" Score="2" Body="&lt;p&gt;The definition of &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_compound&quot; rel=&quot;nofollow&quot;&gt;organic&lt;/a&gt;&quot; is somewhat vague: it can mean &lt;em&gt;containing carbon&lt;/em&gt;, or it can mean &lt;em&gt;coming from living things&lt;/em&gt;, particularly as &lt;a href=&quot;http://www.nal.usda.gov/afsic/pubs/ofp/ofp.shtml&quot; rel=&quot;nofollow&quot;&gt;defined by the United States Department of Agriculture&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further confuse the issue, some carbon compounds, such as sodium carbonate or tungsten carbide, are sometimes defined as inorganic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So state a particular definition, and examine those compounds according to that definition.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-07-26T01:59:46.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34408" PostTypeId="1" AcceptedAnswerId="34562" CreationDate="2015-07-26T04:35:58.347" Score="2" ViewCount="1884" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gLmuG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gLmuG.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to explain why treatment of 2-methyltetrahydrofuran with hydrogen iodide leads to the formation of 5-iodopentan-2-ol. However, I am getting stuck as it seems that the less stable primary carbocation is being preferentially formed, leading to the product. Is my mechanism incorrect?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="16683" LastEditDate="2018-10-19T22:53:57.697" LastActivityDate="2018-10-19T22:58:28.567" Title="How can I predict the product of the reaction of 2-methyltetrahydrofuran with HI?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34410" PostTypeId="1" CreationDate="2015-07-26T07:35:54.323" Score="1" ViewCount="34" Body="&lt;p&gt;The question is as mentioned in the title. Take for example the interaction between $\ce{Ar^+}$ and $\ce{Ar^+}$. In general, is there an acceptable potential which can represent the interaction between $\ce{X^{m+}}$ and $\ce{X^{n+}}$, where $\ce{X}$ is a given element and $m,n=0, \cdots, Z$, where $Z$ is the atomic number of the element $\ce{X}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases the coulomb potential (e.g. $V(r) = 1/r$) suffices. However, when density increases (e.g. at liquid/solid density), the size of the positive ion might become relevant. &lt;/p&gt;&#xA;" OwnerUserId="7451" LastEditorDisplayName="user15489" LastEditDate="2015-07-26T07:53:52.783" LastActivityDate="2015-07-26T07:53:52.783" Title="Is There an Acceptable Potential Which Can Represent the Interaction between Finite-sized Ions?" Tags="&lt;physical-chemistry&gt;&lt;density-functional-theory&gt;&lt;noble-gases&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34413" PostTypeId="1" CreationDate="2015-07-26T11:39:41.237" Score="3" ViewCount="1519" Body="&lt;p&gt;This can't be unique to me and would apply to anybody who has to use equipment in other departments has had departments move etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What open source packages exist to read and potentially manipulate SEM EDX (electron dispersive x-ray elemental analysis) data after the data has been collected?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(In my case I have data from an Oxford Instruments EDX but there are many more manufacturers.)&lt;/p&gt;&#xA;" OwnerUserId="1665" LastEditorUserId="1665" LastEditDate="2015-07-26T11:49:21.753" LastActivityDate="2015-07-26T14:44:32.890" Title="Is there open source software for SEM EDX data analysis?" Tags="&lt;analytical-chemistry&gt;&lt;software&gt;&lt;elements&gt;&lt;x-ray-diffraction&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34414" PostTypeId="1" CreationDate="2015-07-26T11:47:09.387" Score="0" ViewCount="935" Body="&lt;p&gt;Collecting data is one thing, however sometimes one needs to plot or analyse data at a later date when access to the original software may no longer be feasible or possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open Source software is at times a good alternative to commercial products. In this case, please list open source and/or free software for gas chromatography.&#xA;(Having said that, the databases for mass spectrometry are not open as far as I am aware, though suggestions are also welcome.)&lt;/p&gt;&#xA;" OwnerUserId="1665" LastActivityDate="2018-10-10T10:09:14.510" Title="Free Open Source Gas Chromatography Software" Tags="&lt;analytical-chemistry&gt;&lt;chromatography&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34415" PostTypeId="2" ParentId="34414" CreationDate="2015-07-26T11:47:09.387" Score="3" Body="&lt;p&gt;One software that I have come across and used is OpenChrom &lt;a href=&quot;https://www.openchrom.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.openchrom.net/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for basic tasks it performs well as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those seeking pretty chromatograms, OpenChrom can be used in conjunction with R. Open a plot in OpenChrom, export it to a csv file and then plot the data using R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A script for plotting two overlapped chromatograms in R is below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;filename = &quot;Dodecane-Neat&quot;&#xA;filename2 = &quot;Dodecane-Stressed&quot;&#xA;PlotTitle = &quot;Dodecane Neat Chromatogram - Pre and Post Test&quot;&#xA;&#xA;infilename = paste0(&quot;./&quot;,paste(filename,&quot;csv&quot;,sep=&quot;.&quot;))&#xA;infilename2 = paste0(&quot;./&quot;,paste(filename2,&quot;csv&quot;,sep=&quot;.&quot;))&#xA;outfilename = paste(filename,&quot;pdf&quot;,sep=&quot;-rect.&quot;)&#xA;&#xA;## Read Data -&amp;gt; File is called 14mielczarek11.csv&#xA;undiluted &amp;lt;- read.table(infilename, sep=&quot;;&quot;, header=TRUE, nrows=10000)&#xA;undiluted2 &amp;lt;- read.table(infilename2, sep=&quot;;&quot;, header=TRUE, nrows=10000)&#xA;## Work out Total Peak (want the Chromatogram, not the mass spec)&#xA;undiluted$peaks &amp;lt;- rowSums( undiluted[,3:404] )&#xA;undiluted2$peaks &amp;lt;- rowSums( undiluted2[,3:404] )&#xA;&#xA;GCxx &amp;lt;- c(undiluted$RT.milliseconds./1000/60, rev(undiluted$RT.milliseconds./1000/60))&#xA;GCyy &amp;lt;- c(rep(0, nrow(undiluted)), rev(undiluted$peaks)/max(undiluted$peaks))&#xA;GCxx2 &amp;lt;- c(undiluted2$RT.milliseconds./1000/60, rev(undiluted2$RT.milliseconds./1000/60))&#xA;GCyy2 &amp;lt;- c(rep(0, nrow(undiluted2)), rev(undiluted2$peaks)/max(undiluted2$peaks))&#xA;&#xA;library(survival)&#xA;library(Hmisc)&#xA;&#xA;# A4 is 8.27 × 11.7 &#xA;pdf(outfilename, height=5, width=8)&#xA;par(mar=c(5,5,3,2))&#xA;&#xA;plot(&quot;&quot;,&#xA;tck=-0.02,lab=c(10,10,4),las=1,&#xA;xaxs=&quot;i&quot;,xlab=&quot;Time (min)&quot;,&#xA;xlim=c(0,15),&#xA;yaxs=&quot;i&quot;,ylab=&quot;Detector Count Fraction&quot;,&#xA;ylim=c(0,0.15)&#xA;)&#xA;&#xA;polygon(GCxx2+0.017, GCyy2, col= rgb(0,0,1,0.5), border=NA) #Blue - Shift the blue ever so slightly to properly overlap&#xA;polygon(GCxx, GCyy, col= rgb(1,1,1,1), border=NA) # White (so we don't get violet)&#xA;polygon(GCxx, GCyy, col= rgb(1,0,0,0.6), border=NA) # Red&#xA;&#xA;lines(undiluted2$RT.milliseconds./1000/60+0.017,undiluted2$peaks/max(undiluted2$peaks),&#xA;type=&quot;l&quot;,col=4,lwd=&quot;0.05&quot;,lty=1) #Blue&#xA;lines(undiluted$RT.milliseconds./1000/60,undiluted$peaks/max(undiluted$peaks),&#xA;type=&quot;l&quot;,col=2,lwd=&quot;0.05&quot;,lty=1) # Red&#xA;&#xA;title(PlotTitle)&#xA;minor.tick(nx=5, ny=5, tick.ratio=0.3)&#xA;&#xA;dev.off()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KqM6P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KqM6P.png&quot; alt=&quot;Dodecane pre and post test - magnified view of the &amp;quot;interesting region&amp;quot;.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1665" LastActivityDate="2015-07-26T11:47:09.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34416" PostTypeId="1" AcceptedAnswerId="55030" CreationDate="2015-07-26T12:13:49.207" Score="1" ViewCount="7608" Body="&lt;p&gt;I have tried to apply the rules and basics that I learnt so far. But I am confused about the hybridization of oxygen atoms which are making the single bond in nitrate ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following are the steps I used. The average bond order of 4/3 makes the molecule unstable. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/7A7Of.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7A7Of.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the steps, the single bonds are made of $\ce{sp^3}$ of oxygen and $\ce{sp^2}$ of nitrogen, the bonds are sigma bonds. However that doesn't seem like the case. If I look at $\ce{HNO3}$ dissociation equation in the water, lone electron pair of nitrogen is entirely shared with (co-ordinative covalent bond) one of the oxygens making a single bond. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{ HNO3(aq) -&amp;gt; H+(aq) + NO3^{-}(aq) }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But considering $\ce{NO3}$ ion individually, without thinking about nitric acid, I can't find the long pair to be in the same type of co-ordinative bond. Isn't there an unbonding sp2 electron with nitrogen (probably one from the lone pair)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue, why isn't oxygen hybridizing to be $\ce{sp^3}$ here? What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="7450" LastEditorUserId="10185" LastEditDate="2015-07-27T08:02:13.967" LastActivityDate="2016-08-15T11:22:26.827" Title="Hybridization of oxygen in Nitrate ion and the location/bond of nitrogen's lone pair" Tags="&lt;inorganic-chemistry&gt;&lt;hybridization&gt;&lt;valence-bond-theory&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34417" PostTypeId="2" ParentId="34309" CreationDate="2015-07-26T12:51:33.233" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However, there is now that remaining unhybridized p orbital of&#xA;  nitrogen, which has 2 electrons, so how can it be involved in π&#xA;  bonding?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everything is more-or-less &quot;right&quot; up to these point. The thing you're missing is that there is only one electron in the unhybridized $\mathrm{p}$ orbital of $\ce{N}$ atom since to get the following usual configuration of $\ce{NO3-}$ ion (image courtesy of &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/atdvo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/atdvo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;em&gt;two&lt;/em&gt; negatively charged $\ce{O-}$ ions one electron has to be moved from $\ce{N}$ atom to $\ce{O}$ atom first. Now note that since all $\ce{O}$ atom are equivalent to each other we need to invoke the concept of the &lt;em&gt;resonance&lt;/em&gt;, but for simplicity think first about any single of the resonance structure above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, formally we start by having:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one $\ce{N}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^3}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;two $\ce{O}$ atoms with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;and one $\ce{O-}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the first step we transfer one of the $p$-electrons from $\ce{N}$ atom to an $\ce{O}$ atom, so we end up with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one $\ce{N+}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^2}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the next step we invoke the $\mathrm{sp^2}$ hybridization of $\ce{N+}$ ion, so that we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion with $\mathrm{(1s)^2 (sp^2)^3 (2p)^1}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;&lt;/li&gt;&#xA;&lt;li&gt;and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, we form bonds as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and an $\ce{O-}$ ion in forming which we use 1 of the total 3 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;&lt;/li&gt;&#xA;&lt;li&gt;Another identical $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ and another $\ce{O-}$ ion; here we use 1 of the remaining 2 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;&lt;/li&gt;&#xA;&lt;li&gt;Yet another $\sigma$ bond, but this time between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and the $\ce{O}$ atom in forming which we use the last remaining $\mathrm{sp^2}$-electron of $\ce{N+}$ ion and 1 out of total 2 unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion;&lt;/li&gt;&#xA;&lt;li&gt;What remains now is one unhybridized $\mathrm{2p}$-electrons of $\ce{N+}$ ion and one unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion which are used to form the $\pi$ bond.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pictorially (image courtesy of Professor &lt;a href=&quot;http://faculty.concordia.ca/bird/c241/notes_ch3-shriver.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peter Bird&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P4ihg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P4ihg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-26T12:58:20.207" LastActivityDate="2015-07-26T12:58:20.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34419" PostTypeId="1" AcceptedAnswerId="34435" CreationDate="2015-07-26T13:48:37.200" Score="3" ViewCount="781" Body="&lt;p&gt;I would like to know what the name is for the structure drawn below. This looks like a pyramid with 4 atoms at each vertex. Does the name of the structure depend on whether the atoms are all the same or if they are a combination of different ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PanRP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PanRP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17731" LastEditorUserId="4945" LastEditDate="2015-11-09T10:31:24.250" LastActivityDate="2015-11-09T10:31:24.250" Title="Pyramid Structure Name" Tags="&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34420" PostTypeId="2" ParentId="34413" CreationDate="2015-07-26T14:44:32.890" Score="3" Body="&lt;p&gt;You could use &lt;a href=&quot;http://hyperspy.org&quot; rel=&quot;nofollow&quot;&gt;HyperSpy library&lt;/a&gt; for Python. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples of what you can do with it are available &lt;a href=&quot;http://hyperspy.org/hyperspy-doc/current/user_guide/eds.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="7320" LastActivityDate="2015-07-26T14:44:32.890" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34422" PostTypeId="1" AcceptedAnswerId="34425" CreationDate="2015-07-26T16:08:15.100" Score="0" ViewCount="1863" Body="&lt;p&gt;In chromium and copper, the are the configuration fill up like&#xA;$$\mathrm{3d^5  4s^1}$$&#xA;but not $\mathrm{4s}$ first as it has lower energy level. and why only these 2 elements.&lt;/p&gt;&#xA;" OwnerUserId="17735" LastEditorUserId="186" LastEditDate="2015-07-26T16:21:57.243" LastActivityDate="2015-07-26T17:04:23.813" Title="Why 4s lose electron to 3d?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ClosedDate="2015-07-27T15:02:11.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="34424" PostTypeId="1" CreationDate="2015-07-26T16:40:41.607" Score="6" ViewCount="1123" Body="&lt;p&gt;Does the sulfur also oxidize the $\ce{OH-}$ radicals after the $\ce{Na2S4}$ chains form? Something like this:&#xA;$$\ce{9S + 4NaOH -&amp;gt; 2Na2S4 + SO3 + H2O}$$&lt;/p&gt;&#xA;" OwnerUserId="17721" LastEditorUserId="7448" LastEditDate="2015-07-26T20:02:32.073" LastActivityDate="2019-02-08T20:32:27.230" Title="Sodium Polysulfide production from NaOH and Sulfur" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34425" PostTypeId="2" ParentId="34422" CreationDate="2015-07-26T17:04:23.813" Score="2" Body="&lt;p&gt;In any element after Argon [1s&lt;sup&gt;2&lt;/sup&gt;2s&lt;sup&gt;2&lt;/sup&gt;2p&lt;sup&gt;6&lt;/sup&gt;3s&lt;sup&gt;2&lt;/sup&gt;3p&lt;sup&gt;6&lt;/sup&gt;],&#xA;the repulsion force between the elctrons increase at a great extent and also the atom becomes very stable. So the elctrons tend to move in far 4s orbital first and after filling it completely , fill the 3d orbital. About chromium and copper, they are exceptions. Any half filled or fully-filled orbitals are more stable then rest of the configurations, so in Chromium one electron of 4s comes to 3d orbital and a stable 3d&lt;sup&gt;5&lt;/sup&gt; is obtained, similarly in Copper one electron from 4s comes to 3d orbital and forms stable 3d&lt;sup&gt;10&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question, &#xA;If you have any further doubts you can ask me :)&lt;/p&gt;&#xA;" OwnerUserId="17737" LastActivityDate="2015-07-26T17:04:23.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34427" PostTypeId="1" CreationDate="2015-07-27T00:19:59.520" Score="6" ViewCount="673" Body="&lt;p&gt;My background: I've read several books and articles on electrochemistry and even do photoelectrochemical research in a lab, so I'm somewhat familiar with the concepts and techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In texts on electrochemistry, the standard reduction potential and its meaning is usually axiomatically presented early on and everything follows from there. What I would like to see in a treatment of the subject is an explanation of how the chemical equations of the reduction half reactions are actually determined. For example, let's take the equation $\ce{Cu^{2+} + 2e- &amp;lt;-&amp;gt; Cu}$. How did they figure out that it is in fact two electrons reducing the $\ce{Cu^{2+}}$ ion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could do open circuit potential measurements to determine the standard reduction potential, but it would seem to me that you would have to know the ionic state of the element a priori to use this technique. One could also apply a voltage until the reaction &quot;turns on&quot; then approximate that onset voltage as the standard reduction potential, but this wouldn't be strictly true as it would be a non-equilibrium measurement and would also be subject to the kinetics of the surface. Does anyone know of a book or other treatment showing how the chemical equations are determined or better yet has experience doing this?&lt;/p&gt;&#xA;" OwnerUserId="17224" LastEditorUserId="10185" LastEditDate="2015-07-27T08:14:21.183" LastActivityDate="2016-08-03T00:22:22.667" Title="How are the chemical reactions of the table of standard reduction potentials experimentally determined?" Tags="&lt;electrochemistry&gt;&lt;experimental-chemistry&gt;&lt;reduction-potential&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34428" PostTypeId="2" ParentId="34424" CreationDate="2015-07-27T01:04:19.033" Score="5" Body="&lt;p&gt;If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Disproportionation&quot; rel=&quot;noreferrer&quot;&gt;disproportionation&lt;/a&gt;. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Oxidation numbers:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Element    Reactant     Product&#xA;S           0            -1/2 in polysulfide&#xA;S           0            +6 in sulfur trioxide&#xA;Na         +1            +1&#xA;O          -2            -2 in sulfur trioxide and water&#xA;H          +1            +1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The half reactions (unbalanced) are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Oxidation: $\ce{S -&amp;gt; SO3}$&lt;/li&gt;&#xA;&lt;li&gt;Reduction: $\ce{S -&amp;gt; S4^2-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You might get some sodium hydrogen sulfate forming through the followin;&#xA;$$\ce{NaOH + SO3 &amp;lt;=&amp;gt; NaHSO4}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-27T01:04:19.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34429" PostTypeId="2" ParentId="32934" CreationDate="2015-07-27T01:32:37.247" Score="0" Body="&lt;p&gt;The safety arises from the solution's characteristics. Some oxidizers produce highly alkali solutions which pose personnel risks from chemical burns. This nature occurs to a lesser extent when dry chemical contacts the skin or mucus membranes ext. as the agent acquires water from the body and in essence forms saturated solutions at the contact sites.&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-07-27T01:32:37.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34431" PostTypeId="2" ParentId="14052" CreationDate="2015-07-27T02:04:07.310" Score="1" Body="&lt;p&gt;Steel isn't pure iron by a long shot! The dissolved carbon in steels creates a myriad of shorted electro-chemical cells. Any naturally occurring water is not pure and contains ionic impurities which make the water electrically conductive, and when in contact with these inherent shorted cells accelerates the consumption of the iron. The high Chromium content of Stainless steels mitigates this by isolating the carbon-iron cells from the aqueous solution. If you capture the off-gasses from water corroding steel, you will find it is $\ce{H2}$.&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-07-27T02:04:07.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34434" PostTypeId="1" CreationDate="2015-07-27T05:00:02.737" Score="6" ViewCount="1447" Body="&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/DiJD7EQ.jpg&quot; alt=&quot;terminal alkene on ethyl bond of cyclopentane reacts with H+ and HOCH3 to form a cyclohexane with 1 methyl group and 1 Och3 substituent&quot;&gt;&#xA;Here is my problem. The mechanism seems pretty straightforward, but I am having trouble converting the cyclopentane to a cyclohexane. During a synthesis problem, does the starting material add the extra C onto its chain during carbocation rearrangement or a methyl/hydride shift? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the problem at hand, I believe the first step is for the pi bond to attack the $\ce{H+}$, and that will leave me with a carbocation at the more substituted carbon. From there, I know $\ce{HOCH3}$ will attack, but somehow, the molecule needs to rearrange its format in order for the $\ce{HOCH3}$ to come in. What needs to happen here? &lt;/p&gt;&#xA;" OwnerUserId="17742" LastEditorUserId="10185" LastEditDate="2015-07-27T07:50:59.023" LastActivityDate="2015-07-27T08:33:43.867" Title="Cyclopentyl to cyclohexyl carbocation rearrangement" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34435" PostTypeId="2" ParentId="34419" CreationDate="2015-07-27T05:04:39.377" Score="5" Body="&lt;p&gt;If the four faces of this shape are equilateral triangles, that would make this shape a &lt;strong&gt;tetrahedron&lt;/strong&gt; (literally meaning four faces). This is the molecular shape associated with the $\mathrm{sp^3}$ hybridization. The shape is independent of what atoms constitute the corners.&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2015-07-27T05:04:39.377" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34436" PostTypeId="1" CreationDate="2015-07-27T05:09:01.173" Score="0" ViewCount="28174" Body="&lt;p&gt;Are electric current and electric charge the same thing? I'm just getting back into relearning chemistry, so a more in-depth answer about how electricity works in general would be wonderful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thank you all for clarifying that this belongs to the Physics SE. That being said, I guess this question doesn't have much relevance to me anymore. I'm honestly not sure if I should delete the question or not, so I'll leave that up to the mods to decide. Thanks again.&lt;/p&gt;&#xA;" OwnerUserId="17743" LastEditorUserId="865" LastEditDate="2015-07-27T13:47:32.813" LastActivityDate="2015-07-27T13:47:32.813" Title="What is the difference between electric current and electric charge?" Tags="&lt;electricity&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-07-27T14:38:48.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="34437" PostTypeId="2" ParentId="34436" CreationDate="2015-07-27T06:36:58.750" Score="1" Body="&lt;p&gt;Your question is very broad. The first thing. No they are not the same. Simple explanation for these terms can be:&#xA;Charge: Its the &quot;amount&quot; of unbalanced electricity in a body.  Its usually measured in the form of electrons (either excess or deficiency). Its units is coulomb.&#xA;Current: It is the &quot;flow&quot; of electrons from higher electric potential towards the lower. Its unit is ampere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To visualize how they are not the same is by considering two situations (without going into any calculations). Two bodies (A &amp;amp; B) are connected via some copper wire. A has 1 million Coulombs charge &amp;amp; the second one has 1million &amp;amp; 1 coulombs charge. Now individually these charges are enormous, but as the difference between the two is only 1 coulomb, its a very low difference in charge, so a little current will flow between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you change the bodies' charge. A has 100 Coulombs &amp;amp; B has 10 .. much more current will flow between the bodies, because there is a way bigger potential difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. and yes its a physics questions.&lt;/p&gt;&#xA;" OwnerUserId="16866" LastActivityDate="2015-07-27T06:36:58.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34438" PostTypeId="2" ParentId="34434" CreationDate="2015-07-27T08:33:43.867" Score="4" Body="&lt;p&gt;Your first step is correct. The reactive exomethylene group attacks the proton leaving a secondary carbocation (remember Markovnikovs rule). Now the five-membered ring can rearrange into a more stable six-membered ring by releasing ring strain. This is a fast intramolecular step. Finally methanol attacks the rearranged carbocation releasing a proton which is used catalytically in the next reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember though that you can often not be completely sure about a mechanism before it is verified in practice (i.e. in the lab through an experiment). Sometimes true mechanism can be quite different from what appears to be most reasonable on paper!&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-07-27T08:33:43.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34439" PostTypeId="1" CreationDate="2015-07-27T10:10:54.757" Score="2" ViewCount="501" Body="&lt;p&gt;My roommate often uses Tiger Balm and sometimes browses through my collection of vintage Star Wars figures. Can the residue on his fingers cause harm to the PVC figures or cause the paint to crack or chip?&lt;/p&gt;&#xA;" OwnerUserId="17746" LastEditorUserId="865" LastEditDate="2015-07-27T13:09:18.213" LastActivityDate="2015-07-27T14:41:25.147" Title="Can ingredients in Tiger Balm cause any harmful effects on PVC figures?" Tags="&lt;safety&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34441" PostTypeId="1" CreationDate="2015-07-27T09:48:14.877" Score="10" ViewCount="1184" Body="&lt;p&gt;In heme synthesis, porphyrinogens are colorless but upon their oxidation (from removing 6 hydrogen atoms), they become colored porphyrins. Why does the change in color occur? Is red the only color it can turn into or are there others (if so, what does that depend on)?&lt;/p&gt;&#xA;" OwnerUserId="17749" OwnerDisplayName="Magdy Shehata" LastEditorUserId="13184" LastEditDate="2015-07-28T05:38:47.943" LastActivityDate="2015-07-28T05:38:47.943" Title="Why are porphyrins colored when porphyrinogens are colorless?" Tags="&lt;biochemistry&gt;&lt;spectroscopy&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34442" PostTypeId="1" CreationDate="2015-07-27T11:40:35.183" Score="9" ViewCount="685" Body="&lt;p&gt;In one of our lab courses, we performed a variation of the BZ reaction, the experimental procedure is outlined below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Experimental Procedure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;10 mL of 3.0 M $\ce{H2SO4}$ was added to test tube T1.&lt;/li&gt;&#xA;&lt;li&gt;Test tube T1 was placed on a stand and immersed in the water bath maintained at 35 °C&lt;/li&gt;&#xA;&lt;li&gt;5.0 mL of 0.25 M $\ce{KBrO3}$ was added to test tube T2.&lt;/li&gt;&#xA;&lt;li&gt;2.5 mL 0.02 M $\ce{Ce(NH4)4(SO4)4}$ in a 1.0 M $\ce{H2SO4}$ aqueous solution was added to test tube T3.&lt;/li&gt;&#xA;&lt;li&gt;5.0 mL 1.0 M $\ce{CH2(COOH)2}$ was added to test tube T4.&lt;/li&gt;&#xA;&lt;li&gt;Test tube T2 and T3 were added to test tube T1 and then the contents were swirled.&lt;/li&gt;&#xA;&lt;li&gt;Test tube T1 and T4 were put into the water bath for 10 minutes.&lt;/li&gt;&#xA;&lt;li&gt;While test tube T1 was still in the water bath, the contents of test tube T4 was put into test tube T1 and the timer was started.&lt;/li&gt;&#xA;&lt;li&gt;Changes were observed while the tube T1 was vigorously shaken in the water bath.&lt;/li&gt;&#xA;&lt;li&gt;The length of the induction period was recorded. &lt;/li&gt;&#xA;&lt;li&gt;The reaction was allowed to oscillated 10 times, and the length of the period of oscillation was recorded. &lt;/li&gt;&#xA;&lt;li&gt;The experiment was repeated at $\pu{40 ^\circ C},~\pu{45 ^\circ C},~ \text{and}~\pu{50 ^\circ C}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, here's the data set that I obtained: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/0vbym.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0vbym.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, based on the instructions in our lab manual I plotted the following curves: &lt;a href=&quot;https://i.stack.imgur.com/ERCwe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ERCwe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my &lt;strong&gt;questions&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The manual instructs us to calculate activation energy using the period of induction and the period of oscillation from the Arrhenius plots above, which I calculate as $\pu{69.3 kJ/mol}$ (oscillation period) and $\pu{51.4 kJ/mol}$ (induction period).&lt;br&gt;&#xA;But in doing so, we utilise the fact the $k = \frac{1}{\mathrm{period}}$ (i.e., $\ln(K)$ corresponds to $\ln(\frac{1}{\mathrm{period}})$. We were just given this information with no explanation. I can intuitively guess that for a first order reaction $k$ has the dimension $\mathrm{s}^{-1}$, but is there a stronger reason for making this selection?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would guess the rate-determining step occurs during the induction period (since it is the slowest), but then the activation energy is lower. I would think that the energy barrier of the rate-determining step should be higher. Am I seeing a relation where none exists? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I noticed that the period of the reaction changed when the reaction mix was stirred more vigorously (temperature  and starting concentration being the same). Is that possible? If that is the case, then does that mean that the period isn't a reliable measure?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16088" LastEditorUserId="5026" LastEditDate="2018-02-25T03:28:39.783" LastActivityDate="2018-02-25T03:28:59.337" Title="Belousov-Zhabotinsky reaction: questions about rate determining step, k and activation energy" Tags="&lt;experimental-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34443" PostTypeId="2" ParentId="34439" CreationDate="2015-07-27T13:19:04.727" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tiger_Balm&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia entry on Tiger Balm&lt;/a&gt;, there is camphor (among other chemicals) in the formulation of this product. &lt;a href=&quot;https://www.spilltech.com/wcsstore/SpillTechUSCatalogAssetStore/Attachment/documents/ccg/CBOOM.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; chart of chemical resistance recommendations for PVC indicates that it is not recommended that PVC should be exposed to camphor. Therefore, it is probably best to reduce or eliminate contact of Tiger Balm with the figures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the paint cracking or chipping as a result of exposure to Tiger Balm, that might depend in part on the composition of the paint, and that cannot be addressed unless you can provide that information.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-27T14:41:25.147" LastActivityDate="2015-07-27T14:41:25.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34444" PostTypeId="1" CreationDate="2015-07-27T14:07:39.340" Score="6" ViewCount="1478" Body="&lt;p&gt;From a description of my experiment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Potassium iodide crystals are place in the middle of filter paper soaked with potassium sulphate and phenolphthalein. Two electrodes are placed on the filter on two opposite sides. The electrodes are connected to a battery. The objective of this experiment is to see the movement of ions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why soak the filter paper in potassium sulphate and not just soak it in potassium iodide? &lt;/p&gt;&#xA;" OwnerUserId="17748" LastEditorUserId="13184" LastEditDate="2015-07-27T14:51:36.733" LastActivityDate="2015-07-28T13:25:42.393" Title="Why soak filter paper in K2SO4?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34445" PostTypeId="1" CreationDate="2015-07-27T15:44:03.443" Score="2" ViewCount="164" Body="&lt;p&gt;Whether dissolving acids and salts is exothermic or endothermic, depends on the lattice and solvation energy. Breaking bonds in the lattice requires energy and is endothermic - that is clear. However, I do not see how solvation would release energy. I do understand that solvation is a type of creating new bonds (and creation of bonds generally releases energy), but how is it that the, eg., Al(H2O)6 3+ is at a lower energy level than Al 3+? &lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="12697" LastEditDate="2015-07-28T08:46:21.420" LastActivityDate="2015-07-28T08:46:21.420" Title="Why does solvation release energy?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;heat&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-07-28T09:02:47.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="34446" PostTypeId="1" AcceptedAnswerId="34450" CreationDate="2015-07-27T15:44:40.533" Score="6" ViewCount="10347" Body="&lt;p&gt;I have a PVA bottle, which says that the molecular weight (Mw) of PVA is between 85000 - 124000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this means that there are repeated chains of the PVA monomer, but what are the units to the molecular weight (of PVA). Is it $\frac{\mathrm{g}}{\mathrm{mol}}$? I am having trouble grasping this concept. Or does 85000 - 124000 mean the number of monomers in the chain of this PVA?&lt;/p&gt;&#xA;" OwnerUserId="17750" LastEditorUserId="7448" LastEditDate="2015-07-27T15:57:10.747" LastActivityDate="2016-02-08T07:25:27.553" Title="What are the units of 85000 Molecular Weight of a polymer" Tags="&lt;polymers&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34447" PostTypeId="1" AcceptedAnswerId="34449" CreationDate="2015-07-27T15:53:20.580" Score="4" ViewCount="1256" Body="&lt;p&gt;I have seen two electron configurations for $\ce{_{28}Ni}$ in my chemistry textbook:&#xA;$$[\ce{Ar}]~3d^8~4s^2$$&#xA;$$[\ce{Ar}]~3d^9~4s^1$$&#xA;Which of those is &lt;strong&gt;ACTUALLY&lt;/strong&gt; the correct one (in real life, reality)? Why &lt;strong&gt;exactly&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerDisplayName="user7576" LastEditorUserId="10185" LastEditDate="2015-07-27T18:05:16.263" LastActivityDate="2015-07-27T18:05:16.263" Title="Correct Electron Configuration of Ni-28" Tags="&lt;electronic-configuration&gt;&lt;isotope&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34448" PostTypeId="1" CreationDate="2015-07-27T16:18:16.837" Score="33" ViewCount="5968" Body="&lt;p&gt;I know it's possible to capture $\ce{CO2}$ with various chemical reactions. For example NASA's space shuttle had some kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_dioxide_scrubber#Regenerative_carbon_dioxide_removal_system&quot;&gt;regenerative $\ce{CO2}$ scrubber&lt;/a&gt;. But how expensive is it? Could a huge number of these devices, or something like it, significantly reduce the atmosphere's $\ce{CO2}$ level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is probably not practical, but it's an interesting thought experiment. To make my question more specific: can anyone show some rough calculations of how much $\ce{CO2}$ you could remove if you had, say, 2 trillion USD to spend? To simplify, assume that energy is at current prices, but clean. (i.e., if the device requires electricity, assume it's coming from a nuclear power plant, solar farm, etc.). &lt;/p&gt;&#xA;" OwnerUserId="4121" LastEditorUserId="402" LastEditDate="2016-11-12T01:23:16.537" LastActivityDate="2016-11-12T01:23:16.537" Title="How much CO2 could you remove from the atmosphere with 2 trillion USD?" Tags="&lt;environmental-chemistry&gt;&lt;atmospheric-chemistry&gt;" AnswerCount="4" CommentCount="8" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34449" PostTypeId="2" ParentId="34447" CreationDate="2015-07-27T16:30:09.153" Score="5" Body="&lt;p&gt;NIST suggests &lt;a href=&quot;http://physics.nist.gov/PhysRefData/DFTdata/configuration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://physics.nist.gov/PhysRefData/ASD/ionEnergy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Nickel#Electron_configuration_dispute&quot; rel=&quot;nofollow noreferrer&quot;&gt;as Wikipedia describes&lt;/a&gt;, if you don't specify the &lt;a href=&quot;https://en.wikipedia.org/wiki/Term_symbol&quot; rel=&quot;nofollow noreferrer&quot;&gt;term symbol&lt;/a&gt; for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &lt;em&gt;why&lt;/em&gt; one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in &lt;a href=&quot;https://chemistry.stackexchange.com/q/18988/1499&quot;&gt;this previous Chem.SE question&lt;/a&gt; and the several nested questions within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-07-27T16:43:27.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34450" PostTypeId="2" ParentId="34446" CreationDate="2015-07-27T17:11:16.140" Score="6" Body="&lt;p&gt;First of all $M_w$ is not the number of monomers in a chain of PVA. The number of monomers in a chain is called the degree of polymerization.&#xA;$\overline {M_w}$ is the &lt;strong&gt;weight average molecular weight&lt;/strong&gt;. It is a way of determining the molecular mass of a polymer. As polymer molecules have different chain lengths, the average molecular mass will depend on the method of averaging. For example, we can use the &lt;strong&gt;number average molecular weight&lt;/strong&gt; which is based on different type of averaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the unit, it's as you mentioned: g/mol. In biochemistry, they use the dalton (Da) as a mass unit for biopolymers (1 $N_a.$Da= g/mol, where $N_a.$ is the Avogadro number). So, when we say a protein has a molecular weight of 64000 g·mol$^{−1}$, it has  also a mass of 64 kDa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. To be rigorous, we have to say ** molecular mass** instead of ** molecular weight**. But this was an old way of naming things.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-08-22T04:23:16.963" LastActivityDate="2015-08-22T04:23:16.963" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34452" PostTypeId="2" ParentId="34448" CreationDate="2015-07-27T18:23:41.330" Score="24" Body="&lt;p&gt;This is very hard to answer precisely, as there are many different carbon capture strategies, and economics at the scale required is quite different from our normal understanding. However, I'd love to see some attempts to at least get order of magnitude estimates, or sources with more in-depth analyses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an implementation of carbon capture and storage which is useful to consider for its simplicity rather than its real-world applicability, to give a sense of scale. One way to remove anthropogenic $\ce{CO2}$ is to do &quot;inverse combustion&quot;, more specifically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CO2(g) -&amp;gt; C(s) + O2(g)}\ \ \ \ \ \ \ \mathrm{\Delta H=+390\ kJ/mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume this process can be done with perfect efficiency, and that the only energy expense in the process is to drive the reaction forwards (that is, zero energy consumed in transportation, collection, construction, etc). According to &lt;a href=&quot;http://cdiac.ornl.gov/ftp/ndp030/global.1751_2008.ems&quot; rel=&quot;noreferrer&quot;&gt;this source&lt;/a&gt;, the amount of anthropogenic $\ce{CO2}$ emissions from 1750 to 2008 has totalled about $1250\times 10^9\ \mathrm{t_{CO_2}}$. Suppose you wish to remove all this carbon dioxide (only about half is in the atmosphere, the rest is trapped in the ocean or land) using the above process. This would require &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1250+gigatonnes+%2F+%2844+g%2Fmol%29+*+%28390+kJ%2Fmol%29&quot; rel=&quot;noreferrer&quot;&gt;about $\mathrm{10^{22}\ J}$ of energy&lt;/a&gt;, which Wolfram Alpha suggests is about 50% more energy than can be retrieved from combustion of all global proven oil reserves in 2003. Put another way, this is &lt;a href=&quot;https://en.wikipedia.org/wiki/World_energy_consumption&quot; rel=&quot;noreferrer&quot;&gt;about 20 times the world energy consumption in 2012, or 150 times world electrical energy production&lt;/a&gt;. This would put the cost of this process in the range of hundreds of trillions of US dollars, meaning two trillion USD barely makes a dent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some other process where two trillion USD is close to enough? I seriously doubt it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Several comments and answers mention biological sequestration, which is a legitimate carbon capture strategy. I did not consider it, however, because its costs are far more complicated to calculate. My intention with this answer was to find a quick and comparatively simple way to attach an energy cost to carbon sequestration. Whether the monetary cost even makes sense at this scale (how do you define &quot;monetary cost&quot; when it's larger than world GDP?), I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's another amusing comparison, which should somewhat temper hopes that biological sequestration will be a magic bullet. The amount of anthropogenic carbon released between 1750 and 2008, ~350 billion metric tons of carbon, is comparable to a significant amount of the biomass on Earth (&lt;a href=&quot;https://en.wikipedia.org/wiki/Biomass_%28ecology%29&quot; rel=&quot;noreferrer&quot;&gt;all eukaryotic life contains approximately 560 billion metric tons of carbon&lt;/a&gt;, multicellular life is a fraction of this). Thus, biological sequestration of the majority of anthropogenic carbon would be broadly equivalent to sacrificing all eukaryotic life on Earth (or ~10-30% of all living organisms by mass) in order to collect and bury carbon, then seeding the Earth back to its current biological state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over 250 years of constant stimulation to produce as much energy/goods from fossil fuels, we have released &lt;strong&gt;a lot&lt;/strong&gt; of carbon.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="1499" LastEditDate="2015-07-28T18:25:03.863" LastActivityDate="2015-07-28T18:25:03.863" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="34453" PostTypeId="2" ParentId="34444" CreationDate="2015-07-27T18:25:27.363" Score="3" Body="&lt;p&gt;For this experiment, you'll be seeing an interesting color change when Iodine is oxidized at the anode to form the brownish $\ce{I2}$ on the electrode and yellow $\ce{I3-}$ in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reduction of $\ce{H+}$ to form $\ce{H2~(g)}$ and $\ce{OH-}$ will turn the indicator pink at the cathode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cool thing about your experiment is that you will not see any redox reaction at all (at either electrode) until iodide ions diffuse to the anode! That is, unless you use a high voltage battery, $\gtrapprox2.0~\mathrm{V}$, you will not have sufficient energy to oxidize $\ce{SO4^2-}$ to $\ce{S2O8^2-}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes the experiment an exercise in patience, and then sudden color display! Measuring the time it takes can yield data that can help you calculate the diffusion coefficient of $\ce{I-}$ if you wanted to take it another step!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: Soaking the filter paper with KI would result in the electrodes producing color changes immediately. I suspect the sulfate was used in this experiment to illustrate the diffusion of dissolved ions, because the redox reaction would not take place until the iodide ions reached the anode. &lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorUserId="17643" LastEditDate="2015-07-28T13:25:42.393" LastActivityDate="2015-07-28T13:25:42.393" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34454" PostTypeId="2" ParentId="34427" CreationDate="2015-07-27T19:03:10.403" Score="2" Body="&lt;p&gt;The standard reduction potentials were mostly discovered by carefully measuring the reduction and oxidation of solid metals (or dissolved ions) in electrochemical cells, and monitoring the electrical current through the wire connecting the two electrodes. (The image below should have an ammeter instead of of a voltmeter, but you get the idea)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for example, an electrochemical cell of zinc and copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kyGm3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kyGm3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the anode, the solid zinc will dissolve to become Zn&lt;sup&gt;2+&lt;/sup&gt;. At the cathode, the Cu&lt;sup&gt;2+&lt;/sup&gt; ions will be reduced, and become solid copper. During the redox reaction, the scientist measures the current across the two electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some time, the scientist could pull the two electrodes out of solution, dry them off and mass them. They could then calculate the moles of zinc that dissolved from the anode, and the moles of copper added to the cathode. In addition, by measuring the current over the period of the experiment, they would also be able to calculate the number of electrons that passed from the zinc anode to the copper cathode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this instance, they would calculate that for every mole of zinc that dissolved, 2 moles of electrons were produced. Concurrently, 2 electrons were consumed for each mole of copper reduced! These measurements, over time, made it possible to determine the oxidation states of many metals and soluble ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(image from moodle2.rockyview.ab.ca/)&lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorUserId="17643" LastEditDate="2016-08-03T00:22:22.667" LastActivityDate="2016-08-03T00:22:22.667" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34457" PostTypeId="2" ParentId="34377" CreationDate="2015-07-27T20:34:16.033" Score="7" Body="&lt;p&gt;Much of this answer is based on a document available from NIST at &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd45.pdf&quot; rel=&quot;noreferrer&quot;&gt;Properties of Ternary Copper-Silver Systems&lt;/a&gt; [originally J. Phys. Chem. Ref. Data, 6(3) 621-673 (1977)] - this has an equivalent to the diagram in the question (which is just the surface of the liquidus). Now, since you are talking about balances of solid phases, it is also worth looking at the binary diagrams to discuss those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Ag-Sn, one gets:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uTGdd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uTGdd.png&quot; alt=&quot;Ag-Sn Binary diagram calculated from Kattner et al.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Cu-Sn there is:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ErHtd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ErHtd.png&quot; alt=&quot;Cu-Sn Binary diagram calculated from Miettinen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'll throw out a few observations:&#xA;Ag$_{3}$Sn is a line compound, which in the binary system is not stable above 480C. In the NIST report this phase is referenced as $\theta$, but called Ag$_{3}$Sn($\epsilon$), causing great confusion since the $\zeta$ Ag-Sn phase is designated $\epsilon_{2}$, and Cu$_{3}$Sn($\epsilon$) is designated $\epsilon_{1}$. So, watch it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cu$_{3}$Sn line compound converts to the fcc $\gamma$ phase at around 676C. The $\gamma$ phase is sometimes also referred to as Cu$_{3}$Sn (yet more confusion possible). The Cu$_{3}$Sn and Ag$_{3}$Sn line compounds are different crystal structures, and do not form a solid solution across the ternary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For both Ag-Sn and Cu-Sn, another thing to notice is the various other phases that are in close proximity to the Ag$_{3}$Sn and Cu$_{3}$Sn line compounds. Going into the ternary system, you get questions like how adding Cu into $\zeta$Ag-Sn impacts the free energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what now? In the NIST report, on page 664 is a pseudo-binary diagram where the Sn wt% is held at 20%, and Cu wt% is varied from 0 to 80 (balance Ag). Personally I hate weight percent, but...  The point that one can eventually pick out is that the Ag$_{3}$Sn phase, designated $\theta$ appears almost nowhere - there is just a little sliver way over near 0 wt% Cu and under 250C. It appears that with the addition of Cu, the $\zeta$Ag-Sn phase rapidly replaces the Ag$_{3}$Sn phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as far as I can tell, I'm not sure that the kinetics even matter here. You basically can't get much Ag$_{3}$Sn to form within the ternary parameters you have chosen - you will get $\zeta$Ag-Sn instead.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-07-27T20:34:16.033" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="34458" PostTypeId="1" CreationDate="2015-07-27T21:13:36.980" Score="0" ViewCount="2722" Body="&lt;p&gt;How to convert ethanal to 2-hydroxy-3 butenoic acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried crossed aldol condensation with formaldehyde followed by acid addition.After that how should I proceed ?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="10185" LastEditorDisplayName="user15489" LastEditDate="2016-05-25T12:41:23.923" LastActivityDate="2016-05-25T16:15:43.593" Title="How to convert ethanal to 2-hydroxy-3 butenoic acid?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34460" PostTypeId="2" ParentId="34448" CreationDate="2015-07-27T22:17:02.413" Score="8" Body="&lt;p&gt;It's... &lt;em&gt;feasible&lt;/em&gt;. There are a number of technologies that are being considered. Costs will be high, some number of billions or trillions: you're talking planetary engineering, here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious option is to plant trees. The obvious problem with that is that even planting them at the same speed as they are being removed is infeasible. Unfortunately trees csan store only a limited amount of carbon; once a forest's got as crowded as it can, it's maxed out its carbon sequestration capacity, and will never store another gram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An option that addresses this is to heat crop residue, storing the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Biochar&quot;&gt;biochar&lt;/a&gt;&quot;, the charcoal from the burned crops, in landfill or even in farmland to enrich the soil. The output of biochar systems is typically around 20% char, 20% usable biogas, 60% usable bio-oil, making them both a net energy source, and a carbon sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system is being considered as an additional revenue stream for the sugarcane growers of Brazil, where it would (if universally adopted) sequester about 330,000,000 tons of carbon annually.&lt;/p&gt;&#xA;" OwnerUserId="13293" LastActivityDate="2015-07-27T22:17:02.413" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34461" PostTypeId="2" ParentId="34441" CreationDate="2015-07-28T00:49:53.120" Score="7" Body="&lt;p&gt;The precursor porphyrinogen does not have a conjugated pi system that spans the macrocyclic ring. On oxidation to porphyrin, however, the macrocyclic ring becomes fully &quot;conjugated&quot;, i.e. connected by sp2 hybridized atoms. The HOMO/LUMO spacing in such systems is much narrower than in non-conjugated systems, which means the photon energy required to excite electrons into the LUMO is lowered from &quot;deep&quot; UV to the visible region, resulting in strong absorbance of light in the visual frequency range.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-28T00:49:53.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34462" PostTypeId="1" CreationDate="2015-07-28T01:13:22.120" Score="1" ViewCount="555" Body="&lt;p&gt;So say that I have a normal hydrogen atom (not an isotope, just regular normal hydrogen atom) and I want to make it into an ion $\text{H}^+$. Since to make an ion of that formula I need to take away an electron, it is just the same as giving the electron energy for it to transition from quantum energy level 1 to level $\infty$, so I use the energy level formula for the hydrogen atom $E_n = -\frac{1312}{n^2}$ and substitute $n$ for $1$ and $\infty$ to get $-1312 \text{ kJ/mol}$ and $0\text{ kJ/mol}$ respectively. So if I add 1312 $\text{kJ/mol}$ of energy to the electron it should leave the atom, right? But my textbook says it is wrong.&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="9344" LastEditDate="2015-11-14T23:48:43.203" LastActivityDate="2015-11-14T23:48:43.203" Title="How much energy is required to make an ion?" Tags="&lt;electrons&gt;&lt;energy&gt;&lt;ionization-energy&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34463" PostTypeId="1" AcceptedAnswerId="34467" CreationDate="2015-07-28T01:39:44.880" Score="-1" ViewCount="1963" Body="&lt;p&gt;What are all the stereoisomers of 1-ethyl-3-methylcyclohexane? The following image shows my attempt to answer the question. But I am not sure if I am correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cs2OX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cs2OX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am treating this cyclohexane as flat since it's confomers interconvert rapidly.&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="4945" LastEditDate="2015-07-28T07:22:05.773" LastActivityDate="2015-07-28T07:22:05.773" Title="What are all the stereoisomers of 1-ethyl-3-methylcyclohexane?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;stereochemistry&gt;&lt;cyclohexane&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34464" PostTypeId="2" ParentId="34448" CreationDate="2015-07-28T03:44:00.167" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Give me a half a tanker of iron and I will give you another ice age&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the claim of people who believe in &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron_fertilization&quot;&gt;iron seeding&lt;/a&gt; the ocean. The link claims that &quot;the addition of silicic acid or choosing the proper location could, at least theoretically, eliminate and exceed all man-made CO2&quot;, but no citation is given. As to cost?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Current estimates of the amount of iron required to restore all the lost plankton and sequester 3 gigatons/year of CO&#xA;  2 range widely, from approximately 2 hundred thousand tons/year to over 4 million tons/year. The latter scenario involves 16 supertanker loads of iron and a projected cost of approximately €20 billion ($27 billion).&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but again, citation needed!&lt;/p&gt;&#xA;" OwnerUserId="17763" LastActivityDate="2015-07-28T03:44:00.167" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34466" PostTypeId="1" AcceptedAnswerId="34471" CreationDate="2015-07-28T05:16:01.497" Score="10" ViewCount="1623" Body="&lt;p&gt;Shouldn't the N pointed to by the red arrow also be a stereocenter? I am asking this because I think it's lone pair cannot take part in nitrogen inversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dVUQy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dVUQy.png&quot; alt=&quot;ampicillin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorDisplayName="user7951" LastEditDate="2015-07-28T08:41:57.230" LastActivityDate="2015-07-29T22:09:29.560" Title="Mark all stereocenters in ampicillin" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34467" PostTypeId="2" ParentId="34463" CreationDate="2015-07-28T06:05:36.513" Score="3" Body="&lt;p&gt;For me, your answer seems right ... but you should think of other abbreviations for ethyl and methyl groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip2.5in$&lt;img src=&quot;https://i.stack.imgur.com/2yu5L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly there are eight missing hydrogens per structure that you should add for clarification: three for each methyl- and five for each ethyl-group. C or C$_2$ is not a common (and as for C$_2$ not a good*) abbreviation for such groups ... Me and Et or CH$_3$ and C$_2$H$_5$ would be more usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;* e.g., C$_2$ could be $\ce{-C#CH}$ or $\ce{-CH=CH2}$&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-07-28T06:25:04.427" LastActivityDate="2015-07-28T06:25:04.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34468" PostTypeId="1" CreationDate="2015-07-28T06:45:02.940" Score="10" ViewCount="1700" Body="&lt;p&gt;That is, is there a chemical or enzymatic reaction with capsaicin that break it down into other food-safe chemicals? Or alternatively, bind to capsaicin such that it doesn't trigger the same spicy receptors in our mouths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about how casein can be used to wash capsaicin away, but it doesn't actually change it into another form or deactivate it.&lt;/p&gt;&#xA;" OwnerUserId="17767" LastActivityDate="2018-05-10T15:02:52.093" Title="Is there any food-safe way to break down capsaicin (chilli) so food is less spicy?" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34470" PostTypeId="1" AcceptedAnswerId="34473" CreationDate="2015-07-28T09:04:10.120" Score="14" ViewCount="10813" Body="&lt;p&gt;Most of the elements have isotopes, so the atomic masses are calculated depending on the percentage of the existing isotopes. That is clear. However, what about elements that have only one isotope (monoisotopic) - like fluorine? Shouldn't the atomic mass for it be a whole number and not 18.9984?&lt;/p&gt;&#xA;" OwnerUserId="12697" LastActivityDate="2018-10-13T16:49:00.657" Title="Why atomic masses aren't integers?" Tags="&lt;isotope&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34471" PostTypeId="2" ParentId="34466" CreationDate="2015-07-28T11:08:19.767" Score="2" Body="&lt;p&gt;After reading the inputs from @Curt F., @PH13, @Martin and @Mithoron, I infer that N won't be involved in nitrogen inversion due to hindrance caused by connected rings and attached sterically demanding groups. Hence, 'N' pointed to by the red arrow will be considered as a stereocenter in ampicillin.&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="9961" LastEditDate="2015-07-29T22:09:29.560" LastActivityDate="2015-07-29T22:09:29.560" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34472" PostTypeId="1" AcceptedAnswerId="34474" CreationDate="2015-07-28T11:29:11.690" Score="2" ViewCount="3251" Body="&lt;p&gt;I think my prof made a mistake. Shouldn't the answer be 1, 4 and 5 instead of 2 and 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w0yvF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0yvF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="16622" LastEditDate="2015-07-28T12:42:34.993" LastActivityDate="2015-07-28T12:42:34.993" Title="R configuration for cysteine" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34473" PostTypeId="2" ParentId="34470" CreationDate="2015-07-28T11:55:39.047" Score="17" Body="&lt;p&gt;The short answer is &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_binding_energy&quot;&gt;nuclear binding energy&lt;/a&gt;, which is the energy needed to disassemble an atom into its subatomic parts (or in some cases the energy released when this happens). The binding energy is a consequence of the strong and weak nuclear forces that hold atoms together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does this energy come from? It comes from the mass of the nucleons! What? It is true; most atoms have less mass than the sum of their parts, and that mass defect is converted into the energy that holds them together. If you do not believe me, let us look at one atom of carbon-12, which is used as the definition of the atomic mass unit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{1\ u}=\dfrac{1}{12}\text{ of the mass of one }\ce{^12C}\text{ atom}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the atomic mass of an atom of $\ce{^12C}$ is by definition $\mathrm{12\ u}$. The atom is constructed from 6 protons, 6 neutrons, and 6 electrons, and the masses of those particles are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;particle   mass                     number    total mass&#xA;p          1.00727646681290     u   6         6.0436588008774    u&#xA;n          1.0086649160043      u   6         6.0519894960258    u&#xA;e          5.485799094622×10^−4 u   6         0.0032914794567732 u&#xA;                                     TOTAL:  12.0989397763600    u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The sum of the parts of the $\ce{^12C}$ atom are more than the mass of the atom! This mass defect, $\mathrm{0.0989397763600\ u}$ in this case, is the value of the nuclear binding energy when converted to energy using $E=mc^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This handy graph from the Wikipedia article linked above shows that the binding energy per nucleon is not the same for all nuclei. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1SOKB.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1SOKB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were, then we would have integer values for monoisotopic nominal masses. However, the consequences of this change in reality could be as drastic as the heavier elements (like iron) not having enough energy to hold themselves together. An additional consequence is that there would be no productive nuclear reactions (fission, fusion, or radioactivity). All nuclear transmutations are driven by conversion of nuclei with lower binding energy per nucleon to nuclei with higher binding energy per nucleon, resulting in a net conversion of mass to energy. Without a difference in binding energy per nucleon, we would not have nuclear bombs (and probably we would be better off as a civilization), but we would also not have nuclear power and radioisotopes for compound labeling, cancer therapy, and medical diagnostics. We would also not have a ready supply of deuterium for NMR solvents, and we might not be able to use variations in isotopic distributions of $\ce{H}$, $\ce{C}$, and $\ce{O}$ to determine all sorts of things like &lt;code&gt;did that &quot;locally grown&quot; food from that restaurant you like actually come for your geographic region?&lt;/code&gt; &lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-07-28T11:55:39.047" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34474" PostTypeId="2" ParentId="34472" CreationDate="2015-07-28T12:35:03.377" Score="5" Body="&lt;p&gt;If the programmers did not make a mistake it seems that you are right - ChemDraw is on your side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip.5in$&lt;img src=&quot;https://i.stack.imgur.com/HHWKf.png&quot; alt=&quot;Bildbeschreibung und so&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-07-28T12:35:03.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34475" PostTypeId="1" AcceptedAnswerId="34480" CreationDate="2015-07-28T12:36:54.297" Score="5" ViewCount="1436" Body="&lt;p&gt;Basically I want to know what does the question mean when it says &quot;&lt;strong&gt;&lt;em&gt;taking into consideration conformational interconversions.&lt;/em&gt;&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ttwey.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ttwey.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume it means, when I draw the mirror image of the given compound, the mirror image can be same as the compound itself (the case with option (a) and option (c) ) &lt;strong&gt;&lt;em&gt;or it (the mirror image) can be same as confomer of the compound (the case for option (a) and option (b) )&lt;/em&gt;&lt;/strong&gt;, for the compound to be &lt;strong&gt;achiral&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: option (a) shows both the features because option (a) is same as it's confomer and it is same as it's mirror image as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my assumption correct?&lt;/p&gt;&#xA;" OwnerUserId="17613" LastEditorUserId="4325" LastEditDate="2015-07-28T15:09:39.933" LastActivityDate="2015-07-28T15:26:34.573" Title="Determining chirality after considering conformational interconversion in a cyclohexane" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34476" PostTypeId="1" CreationDate="2015-07-28T13:50:32.153" Score="0" ViewCount="80" Body="&lt;p&gt;Is there a way to maximize van der Waals forces between two materials, by electrifying one material, thus creating dipoles/instantaneous dipoles and thus creating stronger van der Waals forces? has this been documented?&lt;/p&gt;&#xA;" OwnerUserId="17565" LastActivityDate="2015-08-27T21:11:16.833" Title="induction of van der Waals forces" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34479" PostTypeId="2" ParentId="34476" CreationDate="2015-07-28T15:21:48.727" Score="1" Body="&lt;p&gt;Perhaps an example of what you mean can be seen with the noble gases. As you go from helium to neon to argon, etc. you are adding inner electron shells and making the electron cloud around each atom more polarizable. That in turn increases the van der Waals attraction between atoms, resulting in a dramatic increase in boiling points down the series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to make an atom more polarizable (and increase van der Waals forces) by electronically exciting it to a higher energy level. Can anyone else elaborate on that possibility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/25fMk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/25fMk.png&quot; alt=&quot;Boiling points and densities of the noble gases, from boundless.com&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5577" LastEditDate="2015-07-28T20:13:29.697" LastActivityDate="2015-07-28T20:13:29.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34480" PostTypeId="2" ParentId="34475" CreationDate="2015-07-28T15:26:34.573" Score="5" Body="&lt;p&gt;Your analysis is correct. The stipulation &quot;taking into consideration conformational interconversions&quot; means that conformational interconversion is fast. That's important because compounds that we consider achiral can have chiral conformations. I used the example of butane in a &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units/24910#24910&quot;&gt;previous question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this problem, three compounds are given that have multiple stereocenters, compounds b, c, and d. A common way to decide whether a compound with multiple stereocenters is chiral is to look for planes or points of symmetry within the compound. Compound c has a plane of symmetry, so it is clearly achiral. Neither of compounds b or d has a plane of symmetry in the drawn conformations, which means that those conformations are chiral. However, as you suggest, chair flip of b gives the mirror image of the original drawing, which means the original and mirror image are equivalent, assuming that the two conformations can interconvert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many cases where conformations cannot interconvert, and that leads to very interesting compounds that don't even require a stereocenter to be chiral (again see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units/24910#24910&quot;&gt;earlier question&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-07-28T15:26:34.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34481" PostTypeId="1" CreationDate="2015-07-28T16:25:37.037" Score="5" ViewCount="117" Body="&lt;p&gt;After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)? I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air? When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other? These questions are in reference to an automotive application.&lt;/p&gt;&#xA;" OwnerUserId="17781" LastEditorUserId="13184" LastEditDate="2015-07-29T02:07:13.837" LastActivityDate="2015-07-29T02:07:13.837" Title="N2O production vs air distillation" Tags="&lt;inorganic-chemistry&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34482" PostTypeId="2" ParentId="34448" CreationDate="2015-07-28T16:32:21.313" Score="-1" Body="&lt;p&gt;the $64,000 Question is ... Is it economically feasible to use a 'synthetic' process over a naturally occurring one. In this case, No. Zooplankton are the prime movers of CO2 in our atmosphere. They basically turn gaseous CO2 into Calcium Carbonate a lab phrase for sea shells. Increasing the number of Zooplankton is key to the process. UV and oil slicks and the Sargasso seas of plastic are lowering those numbers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the answer to the question is what process would one employ that was the most cost effective solution to the problem, clearly A. clean up the oceans and promote natural zooplankton growth or B. mimic the metabolic pathway that results in the production of calcium carbonate, as a GM/organic solution. Either A or B is a far better and sustainable solution that does not reintroduce/produce CO2 for the energy to power an artificial non-organic/biologic method.      &lt;/p&gt;&#xA;" OwnerUserId="17782" LastEditorUserId="17782" LastEditDate="2015-07-28T19:30:04.200" LastActivityDate="2015-07-28T19:30:04.200" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34483" PostTypeId="2" ParentId="8689" CreationDate="2015-07-28T16:37:30.753" Score="0" Body="&lt;p&gt;Try an electric air compressor with extremely high performance in the PSI column. liquefy the air and separate the gases by boiling points. That is how it is done now. &lt;/p&gt;&#xA;" OwnerUserId="17782" LastActivityDate="2015-07-28T16:37:30.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34484" PostTypeId="2" ParentId="33734" CreationDate="2015-07-28T17:13:23.707" Score="2" Body="&lt;p&gt;A good way to conceptualize metallic bonding is that the valence electrons on metal ions become delocalized, leaving positive ions to be electrostatically attracted to the electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group II metals contribute 2 valence electrons to the delocalized &quot;sea of electrons&quot;, and thus have a +2 charge on the remaining cation. These combined create a stronger attractive force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that the radii of Group II metal cations are smaller than those of Group I cations (within the same row). This also affects bonding, as the delocalized electrons are closer to the nucleus in the Group II case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: the increased coulombic force between a higher number of delocalized electrons and higher charges on metallic cations create a stronger metallic bond. Transition metals tend to have even stronger metallic bonding due to the participation of electrons from d orbitals.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-07-28T17:13:23.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34485" PostTypeId="1" CreationDate="2015-07-28T17:14:03.490" Score="0" ViewCount="12104" Body="&lt;p&gt;Using a weak acid as an example:&#xA;$$\ce{CH_3COOH &amp;lt;=&amp;gt; CH_3COO^- + H^+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of this equilibrium shifts to the right with increasing dilution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found textbooks that say this, but do not provide an explanation. Why is there increasing ionization with increasing dilution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it has to do with the free energy available; since ethanoic acid is a weak acid, energy has to be provided to promote the proton to a higher free energy state (transfer from $\ce{CH_3COOH}$ to $\ce{H_2O}$), therefore, &quot;more&quot; free energy for promotion and less contact with other $\ce{CH_3COOH}$ molecules?&lt;/p&gt;&#xA;" OwnerUserId="17460" LastEditorUserId="-1" LastEditDate="2018-07-23T06:04:05.697" LastActivityDate="2018-07-23T06:04:05.697" Title="percent ionization explanation" Tags="&lt;acid-base&gt;&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-07-28T20:29:21.953" ContentLicense="CC BY-SA 4.0" />
  <row Id="34486" PostTypeId="2" ParentId="34485" CreationDate="2015-07-28T18:03:48.773" Score="2" Body="&lt;p&gt;For $\ce{HA &amp;lt;=&amp;gt; A- + H+}$  with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),&lt;br&gt;&#xA;Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Another reminder&lt;/strong&gt;: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.&lt;/p&gt;&#xA;" OwnerUserId="17765" LastEditorUserId="16622" LastEditDate="2015-07-28T18:47:00.330" LastActivityDate="2015-07-28T18:47:00.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34487" PostTypeId="1" AcceptedAnswerId="34525" CreationDate="2015-07-28T18:16:18.733" Score="11" ViewCount="1052" Body="&lt;p&gt;I'm familiar with &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared_spectroscopy&quot;&gt;infrared spectroscopy&lt;/a&gt; generally, which probes mainly the vibrational behavior of molecules under study.  However, I'm seeing references in various literature to 'infrared action spectroscopy' (e.g., &lt;a href=&quot;http://link.springer.com/article/10.1007%2Fs13361-011-0208-7&quot;&gt;here&lt;/a&gt;), which I'm having a hard time finding information about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the title question: What is the difference between infrared &lt;em&gt;action&lt;/em&gt; spectroscopy and conventional IR spectroscopy?&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-07-28T18:22:14.340" LastActivityDate="2015-07-29T19:26:15.087" Title="What is the difference between infrared action spectroscopy and conventional IR spectroscopy?" Tags="&lt;spectroscopy&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34489" PostTypeId="2" ParentId="34481" CreationDate="2015-07-28T19:32:00.680" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not clear to me that $\ce{N2O}$ is produced from $\ce{N2}$: Rather, it appears that commercial quantities of $\ce{N2O}$ are produced by thermally decomposing ammonium nitrate:&#xA;$$\ce{NH4NO3(s) -&amp;gt; N2O(g) + 2H2O(g)}$$&#xA;as described (at least) by &lt;a href=&quot;http://www.nitrousoxidegasplants.com/manufacturing-process.html&quot; rel=&quot;nofollow&quot;&gt;this manufacturer and supplier of nitrous oxide gas plants&lt;/a&gt;. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrous_oxide#Production&quot; rel=&quot;nofollow&quot;&gt;comprehensive Wikipedia entry&lt;/a&gt; for more information.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you mean here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nitrogen is liquid in the range &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrogen&quot; rel=&quot;nofollow&quot;&gt;63 to 77 K&lt;/a&gt;; for oxygen, it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxygen&quot; rel=&quot;nofollow&quot;&gt;54 to 90 K&lt;/a&gt;. Thus, the temperature range for which both are liquids is bounded by the range on nitrogen, or roughly between 63 and 77 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature at which nitrous oxide solidifies (182 K per &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrous_oxide&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) is much higher than the temperature range at which oxygen and nitrogen are in the liquid phase, and thus it would have come out of solution as a solid long before the requisite temperature was reached. Thus, we would not expect to see liquid $\ce{N2O}$ co-existing with liquid oxygen and nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note bene: I am using the pressure range of 5 to 10 bars as cited &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_separation&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; in assuming that liquid and solid phases of the above moieties do not coexist during industrial gas production at these relatively low pressures.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-28T19:32:00.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34490" PostTypeId="1" AcceptedAnswerId="34500" CreationDate="2015-07-28T20:36:31.420" Score="0" ViewCount="201" Body="&lt;p&gt;Does the secondary quantum number tell how many subshells a specific principal quantum number shell has?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., if the principal quantum number is $n$, there are ($n-1$) subshells.&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="16622" LastEditDate="2015-07-28T20:51:20.593" LastActivityDate="2020-01-02T04:27:31.880" Title="Relationship between the first and the second quantum number" Tags="&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34492" PostTypeId="1" CreationDate="2015-07-28T22:14:40.093" Score="1" ViewCount="3762" Body="&lt;p&gt;I am building a set-up to produce a controlled flow of dry hydrogen. I'm aiming for around 99.5 percent purity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in thinking a PEM cell and Deterite drying tube should be fine for this? Or will I need a condenser of some kind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips?&lt;/p&gt;&#xA;" OwnerUserId="17788" LastEditorUserId="7475" LastEditDate="2015-10-27T19:44:52.487" LastActivityDate="2015-10-27T19:50:03.293" Title="Dry Hydrogen Production" Tags="&lt;experimental-chemistry&gt;&lt;purification&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34494" PostTypeId="1" CreationDate="2015-07-28T23:45:21.067" Score="2" ViewCount="76" Body="&lt;p&gt;I am trying to develop precipitation diagrams for an Ag-Cu-Sn alloy to figure out the kinetics involved using the CompuTherm Pandat Software (Demo Version).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precipitation diagrams need the thermodynamics/mobility database and it further needs the kinetics database as well. The software I have installed comes only with Al-Ni Kinetic database.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to get Pandat to generate diagrams for other alloys?&lt;br&gt;&#xA;Their website is &lt;a href=&quot;http://www.computherm.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.computherm.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16866" LastEditorUserId="16622" LastEditDate="2015-07-29T00:16:35.110" LastActivityDate="2015-07-29T00:16:35.110" Title="Generating Precipitation Diagrams with CompuTherm Pandat" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;phase&gt;&lt;precipitation&gt;&lt;alloy&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34495" PostTypeId="1" CreationDate="2015-07-29T00:23:15.053" Score="2" ViewCount="389" Body="&lt;p&gt;I am trying to find a way to keep milk from spoiling and/or turning yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently there is a process for achieving this, and I am aware that there is a lady who creates milk beads for resin jewelry and she claims that she is able to prevent the milk from spoiling/yellowing over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard that you can add baking soda to milk to preserve it, but this lady also claims that she adds nothing to the milk and that it simply undergoes a &quot;process&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, she won't share her secrets, so &lt;strong&gt;my question is&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the different ways in which milk can be preserved while preserving the white color?&lt;/p&gt;&#xA;" OwnerUserId="17789" LastEditorUserId="865" LastEditDate="2015-07-29T00:37:11.920" LastActivityDate="2015-09-13T22:19:03.930" Title="How can I keep milk from spoiling and/or turning yellow?" Tags="&lt;organic-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34496" PostTypeId="2" ParentId="30829" CreationDate="2015-07-29T04:16:48.833" Score="1" Body="&lt;p&gt;$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$170.43\,\, \mathrm{eV/molecule}$&lt;/p&gt;&#xA;" OwnerUserId="15197" LastEditorDisplayName="user7951" LastEditDate="2015-09-12T16:15:19.900" LastActivityDate="2015-09-12T16:15:19.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34497" PostTypeId="2" ParentId="30776" CreationDate="2015-07-29T05:23:18.450" Score="4" Body="&lt;p&gt;Please go through papers by Prof. Ralf Ludwig. You will certainly gain a lucid understanding of the subject both at experimental end as well as computational end. He advocates use of low frequency vibrational modes such as Far IR along with quantum chemistry for studying hydrogen bonding and allied weak interactions in very simplistic yet the most accurate way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Hydrogen Bonding in protic ionic liquids: Reminiscent of Water.&quot; K. Fumino, A. Wulf, R. Ludwig., &lt;em&gt;Angew. Chem.&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;121&lt;/em&gt;, 3230-3233. &lt;em&gt;Angew. Chem. Int. Ed.&lt;/em&gt;, &lt;strong&gt;2009&lt;/strong&gt;, &lt;em&gt;48&lt;/em&gt;, 3184-3186.  &lt;/li&gt;&#xA;&lt;li&gt;&quot;Advancing into Water’s “No Man’s Land”: Two Liquid States?&quot; D. Paschek, R. Ludwig, &lt;em&gt;Angew. Chem.&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;126&lt;/em&gt; (44), 11888-1890. &lt;em&gt;Angew. Chem. Int. Ed.&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;53&lt;/em&gt; (44), 11699–11701.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5659" LastEditorUserId="4945" LastEditDate="2015-07-30T06:09:50.973" LastActivityDate="2015-07-30T06:09:50.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34500" PostTypeId="2" ParentId="34490" CreationDate="2015-07-29T07:56:58.497" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the secondary quantum number tell how many subshells a specific&#xA;  principal quantum number shell has?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it does not. It is the principal quantum number $n$ which does it (as you actually mentioned yourself later in the question). The second quantum number $l$ describes the shape of the orbitals in a particular subshell. Say, $l=0$ for $\mathrm{s}$-subshell of any shell and all $\mathrm{s}$-orbitals have a spherical shape. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the principal quantum number is $n$, there are $(n−1)$ subshells.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why $(n−1)$? Think about the most trivial example: how many subshells the first shell ($n=1$) have? One, not zero, so $(n−1)$ is obviously wrong formula. The situation is in fact simpler: $n$-th shell has $n$ subshells.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-07-29T07:56:58.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34501" PostTypeId="1" CreationDate="2015-07-29T09:06:14.093" Score="0" ViewCount="187" Body="&lt;p&gt;I understand that melt flow rate and density can be used to classify HDPE. Does this mean that if I have the melt flow rate and density of an HDPE (with no additives), then I can know its other material properties?&lt;/p&gt;&#xA;" OwnerUserId="17798" LastActivityDate="2015-07-29T09:54:23.387" Title="Do melt flow rate and density tell you everything about pure HDPE?" Tags="&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34502" PostTypeId="2" ParentId="34501" CreationDate="2015-07-29T09:54:23.387" Score="1" Body="&lt;p&gt;Yes, as a rule the melt flow rate and density do tell you what the other material properties will be. This is why HDPE is often specified by those two properties.&lt;/p&gt;&#xA;" OwnerUserId="17798" LastActivityDate="2015-07-29T09:54:23.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34503" PostTypeId="1" CreationDate="2015-07-29T11:50:39.883" Score="8" ViewCount="710" Body="&lt;p&gt;London Dispersion forces are forces between non-polar non-ionic molecules that the random fluctuation of their electrons makes them temporarily dipoles.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the frequency that this occurs?&lt;/li&gt;&#xA;&lt;li&gt;Every how many milliseconds does a molecule like this becomes a dipole?&lt;/li&gt;&#xA;&lt;li&gt;Are there such numbers for some specific molecules out there?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17565" LastEditorUserId="11367" LastEditDate="2015-07-29T14:38:20.687" LastActivityDate="2015-07-29T19:48:19.693" Title="At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?" Tags="&lt;intermolecular-forces&gt;&lt;dipole&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34505" PostTypeId="2" ParentId="34503" CreationDate="2015-07-29T13:24:42.047" Score="1" Body="&lt;p&gt;While dispersion forces dominate in non-polar, non-ionic systems, they are actually felt by all chemical species that contain electrons. But, for simplicity's sake, let's use the example of a non-polar, non-ionic atom or molecule to describe these forces in this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, these molecules are almost never completely &quot;non-polar&quot;. There is almost always a dipole moment, but the strength and direction are random. The electron densities around these atoms and molecules are always shifting, which almost always leaves an uneven distribution of electron density around the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The London dispersion forces are simply the sum of interactions between the constantly fluctuating dipoles of one particle, and the induced dipoles within the neighboring particles. The influence of the electrostatic force is directly proportional to the product of polarizabilities for each participating particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: Molecules or atoms with no permanent dipole are almost never in a completely &quot;non-polar&quot; state, due to the constantly fluctuating electron cloud. This negates the question of instantaneous dipole frequency, because it is never not a dipole.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorUserId="17643" LastEditDate="2015-07-29T13:34:25.710" LastActivityDate="2015-07-29T13:34:25.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34507" PostTypeId="2" ParentId="34503" CreationDate="2015-07-29T13:32:31.387" Score="1" Body="&lt;p&gt;The electron rich/ electron poor regions tend to oscillate at a period of about 1-10 femtoseconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.springer.com/us/book/9783642222269&quot; rel=&quot;nofollow&quot;&gt;Logothetidis, Stergios. &lt;em&gt;Nanostructured Materials and Their Applications&lt;/em&gt;. N.p.: Springer, 2012.&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17802" LastEditorUserId="4148" LastEditDate="2015-07-29T19:48:19.693" LastActivityDate="2015-07-29T19:48:19.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34509" PostTypeId="2" ParentId="34503" CreationDate="2015-07-29T13:38:45.723" Score="9" Body="&lt;p&gt;According to &lt;a href=&quot;http://ww2.chemistry.gatech.edu/~lw26/structure/molecular_interactions/mol_int.html#DF&quot; rel=&quot;noreferrer&quot;&gt;Prof. Loren Williams of Georgia Tech&lt;/a&gt;, the fluctuations occur on the femtosecond $\left(10^{-15}\ \mathrm{s}\right)$ time scale:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zkfvE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zkfvE.jpg&quot; alt=&quot;Figure 13 from website of Prof. Loren Williams&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Figure 13 shows how fluctuating dipoles of liquid Xenon (or Helium or Neon, etc) are coupled. Darker blue indicates higher electron density. The fluctuations are correlated and are very fast, on the femtosecond ($10^{-15}$ second) timescale. Adjacent Xenon atoms experience electrostatic attraction from the transient dipoles. Two different representations of fluctuating dipoles are shown.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This corresponds to a frequency of $1\ \mathrm{PHz}$ (&lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_prefix&quot; rel=&quot;noreferrer&quot;&gt;petahertz&lt;/a&gt;), which is about five hundred thousand times faster than most modern &lt;a href=&quot;https://en.wikipedia.org/wiki/Central_processing_unit&quot; rel=&quot;noreferrer&quot;&gt;computer processors&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another estimate of the time scale (albeit one devised by me and thus of potentially suspect validity), is the SI value of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_units&quot; rel=&quot;noreferrer&quot;&gt;atomic unit&lt;/a&gt; for time, derived by dividing the reduced Planck constant by the atomic unit for energy, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree&quot; rel=&quot;noreferrer&quot;&gt;Hartree&lt;/a&gt; (values are rounded a bit):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left\{atomic\ time\right\} = \frac{\hbar}{E_\mathrm{h}} = \frac{1.05457\times 10^{-34}\ \mathrm{J\cdot s}}{4.3597\times 10^{-18}\ \mathrm{J}} = 2.41888\times 10^{-17}\ \mathrm{s} \approx 0.024\ \mathrm{fs} = 24\ \mathrm{as}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last value is $24$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Metric_prefix&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;attoseconds&lt;/em&gt;&lt;/a&gt;, which is a well and truly short span of time. For comparison, this corresponds to a frequency of $41\ \mathrm{PHz}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever the exact value, I have to believe the fluctuations are fast enough so as to be impossible to measure directly. (I would be fascinated to learn otherwise, though!)  Thus, even if different atoms/molecules do actually have different frequencies of London dipole oscillation, we have no way of finding out these values.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-07-29T14:55:50.570" LastActivityDate="2015-07-29T14:55:50.570" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34510" PostTypeId="1" CreationDate="2015-07-29T15:31:21.340" Score="5" ViewCount="5787" Body="&lt;p&gt;Of course, there are many types of coffee, so &lt;em&gt;I would like a range of lethal dose amounts for the average human, from the mildest of lattes to the most hardcore of espressos.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pure caffeine's lethal dose is easily calculable, but what about coffee?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15588" LastActivityDate="2015-07-29T16:53:07.087" Title="What is the lethal dose ( LD50 ) of coffee?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34512" PostTypeId="1" CreationDate="2015-07-29T15:42:03.903" Score="0" ViewCount="440" Body="&lt;p&gt;Being an ionic solid, salt is very susceptible to being torn apart by the polar molecules of water. Inside the water, salt exists as sodium and chlorine anions; &lt;strong&gt;but what state of matter is it in?&lt;/strong&gt;&lt;br&gt; My instinct tells me it is in a solid state, as salt's melting point is &lt;em&gt;800°C&lt;/em&gt;, but it sure doesn't really act like a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I think I ought to rephrase, I mean what state of matter are the individual sodium and chlorine ions inside the water.&lt;/p&gt;&#xA;" OwnerUserId="15588" LastEditorUserId="15588" LastEditDate="2015-07-29T16:21:14.477" LastActivityDate="2015-07-29T16:21:14.477" Title="What phase of matter is salt when dissolved in water?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2015-07-29T19:30:35.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="34514" PostTypeId="1" CreationDate="2015-07-29T15:48:26.697" Score="2" ViewCount="240" Body="&lt;p&gt;The quarters in which I live has both aluminum and copper household wiring. This results in corrosion of the contacts and can lead to sparking, overheating, and even fire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to inhibit the corrosion (if not stopping it completely), except for changing the whole wiring?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="865" LastEditDate="2015-07-29T15:54:43.680" LastActivityDate="2015-08-02T18:08:35.977" Title="protecting contact between aluminium and copper wires" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;metal&gt;&lt;surface-chemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34516" PostTypeId="2" ParentId="34510" CreationDate="2015-07-29T16:15:10.213" Score="7" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.compoundchem.com/2014/07/27/lethaldoses/&quot;&gt;one reference&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LD$_{50}$ for coffee = 118 servings at 240 ml each; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LD$_{50}$ for espresso = 175 shots (volume not specified),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a 75 kg person.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much more comprehensive quantitative description is given &lt;a href=&quot;http://scienceblogs.com/worldsfair/2009/05/01/lethal-doses-and-substance-abu/&quot;&gt;here&lt;/a&gt; for the LD$_{50}$ of coffee for an 80 kg person (113 cups at 250 mL each, 135 mg caffeine per serving). They account for half-life of elimination of caffeine, and arrive a the conclusion that one can't actually reach the LD$_{50}$ (15,360 mg) for caffeine due to metabolism, elimination, and the limit on how much water can be processed by the kidneys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, as noted in the last reference and hinted at above, the amount of water a person would imbibe on the way to reaching the 113 cup total (about 30 L) would likely kill the person (water intoxication).&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-29T16:53:07.087" LastActivityDate="2015-07-29T16:53:07.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34517" PostTypeId="1" CreationDate="2015-07-29T16:18:06.700" Score="2" ViewCount="77" Body="&lt;p&gt;&lt;em&gt;Salt&lt;/em&gt; is a famous regularly repeating cubic crystal, but there are always natural deformities and flaws present. I have heard that graphene is an extremely uniform substance. &lt;strong&gt;With these in mind,&lt;/strong&gt;&lt;br&gt; &lt;strong&gt;what is the most regular crystal pattern, natural or man-made?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15588" LastActivityDate="2015-10-26T19:47:10.567" Title="What bond\crystal has the most regular pattern (i.e. least irregularities)?" Tags="&lt;bond&gt;&lt;crystal-structure&gt;" AnswerCount="2" CommentCount="1" ClosedDate="2015-10-27T09:52:20.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="34519" PostTypeId="1" CreationDate="2015-07-29T16:34:36.490" Score="9" ViewCount="573" Body="&lt;p&gt;I'd like to reduce enantioselectively a ketone in presence of an ester (which should NOT be reduced to an aldehyde and further to an alcohol).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Adding $\ce{LiAlH4}$ would also reduce the ester and gives no enantioselective reduction of the ketone.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the following reaction (&lt;a href=&quot;https://sites.google.com/site/anthonycrastoorganicchemistry/f&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frater-Seebach Alkylation&lt;/a&gt;) they used Bakers yeast:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/l3BVf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3BVf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Probably S-BINAP also give &lt;strong&gt;2&lt;/strong&gt;. I think there is a $\ce{RuCl2BINAP}$ transition metal complex in combination with $\ce{H2}$ used.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would first use KOH, then adding Evans auxiliary, adding LiCl for chelation of acid carbonyl to Evans auxiliar carbonyl, then adding $\ce{NaBH4}$ and to a standard Felkin Anh reduction. And as a last step I would use $\ce{LiOEt}$. Does this work or will the chelating ketone also be reduced?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;May be my own suggestion is easier to reproduce in lab than the use of asymmetric reduction using $\ce{H2}$ under high pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for further ways to get the (S)-hydroxy compound &lt;strong&gt;2&lt;/strong&gt;. Are there easier ways?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2017-10-13T22:12:21.287" LastActivityDate="2017-10-13T22:12:21.287" Title="How can a ketone be enantioselectively reduced, in the presence of an ester?" Tags="&lt;organic-chemistry&gt;&lt;redox&gt;&lt;carbonyl-compounds&gt;&lt;stereoselectivity&gt;&lt;chemoselectivity&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34520" PostTypeId="1" CreationDate="2015-07-29T17:21:31.910" Score="6" ViewCount="2396" Body="&lt;p&gt;I'd like to deprotect a benzyl group for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fr%C3%A1ter%E2%80%93Seebach_alkylation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fráter–Seebach alkylation&lt;/a&gt;. Would it be possible to do this with a Birch reduction using Li/NH&lt;sub&gt;3&lt;/sub&gt; (or Na/NH&lt;sub&gt;3&lt;/sub&gt;), without also reducing the ester carbonyl group?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vv8cW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vv8cW.png&quot; alt=&quot;Selective debenzylation of methyl (S)-3-(benzyloxy)butanoate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, hydrogenation over Pd/C seems to be a plausible alternative. How else can I remove the benzyl group on oxygen?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2018-10-06T15:43:43.070" LastActivityDate="2018-10-06T16:00:21.380" Title="Can a benzyl ether be removed in the presence of an ester?" Tags="&lt;organic-chemistry&gt;&lt;chemoselectivity&gt;&lt;protecting-groups&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34521" PostTypeId="1" CreationDate="2015-07-29T18:17:57.097" Score="3" ViewCount="85" Body="&lt;p&gt;Why is this reaction successful:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5i9V1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5i9V1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't metal-halide exchange occur easily to get Ar-MgCl and R-Br?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-07-29T18:25:06.760" LastActivityDate="2015-07-31T20:23:30.633" Title="Grignard reaction - exchange of groups" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;reactivity&gt;&lt;reaction-control&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34522" PostTypeId="2" ParentId="34514" CreationDate="2015-07-29T18:23:35.967" Score="0" Body="&lt;p&gt;You won't be able to stop the corrosion unless the two pieces of metal are galvanically separated, that is, in this case, there is no ion-containg water around (I assume it's inside walls/floors/ceilings). I don't think it's really practical in a house where the wires are already mounted inside structures very apt to contain moisture.&lt;/p&gt;&#xA;" OwnerUserId="17808" LastActivityDate="2015-07-29T18:23:35.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34523" PostTypeId="2" ParentId="34514" CreationDate="2015-07-29T18:26:29.547" Score="2" Body="&lt;p&gt;This may depend on your location and local electrical codes.  I have the same issue in my house.  In the USA, there are NEC (National Electrical Code) compliant wire nuts rated for Al to Cu connections.  These come pre-packed with a protective goop. The goop can also be purchased separately in a tube for use with ordinary wire nuts. You may have to ask at your local electrical supply store.  In my area the pre-packed wire nuts are a distinctive purple color not found on regular wire nuts (for example: &lt;a href=&quot;http://www.homedepot.com/p/Ideal-Twister-Al-Cu-Wire-Connectors-65-Purple-10-per-Package-30-1765S/202935623&quot; rel=&quot;nofollow&quot;&gt;Home Depot Al/Cu nuts&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For code compliance, use approved solutions only, even if you think you know more chemistry than the electricians. Having your house burn down from a wiring issue is no fun.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-07-29T18:26:29.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34524" PostTypeId="2" ParentId="34492" CreationDate="2015-07-29T18:27:11.203" Score="1" Body="&lt;p&gt;Using molecular sieves comes to my mind, they're easy to use and efficient. You can find some practical information about how this is done for example here &lt;a href=&quot;http://www.klmtechgroup.com/PDF/Articles/articles/Molecular-Sieve-Applications.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.klmtechgroup.com/PDF/Articles/articles/Molecular-Sieve-Applications.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17808" LastActivityDate="2015-07-29T18:27:11.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34525" PostTypeId="2" ParentId="34487" CreationDate="2015-07-29T18:36:49.157" Score="8" Body="&lt;p&gt;After 30 minutes of Googling, I'm thinking that we both (EDIT: nope, it was just me) misparsed the name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When reading the name &quot;infrared action spectroscopy,&quot; my first instinct was to mentally split it up like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{infrared action} \qquad\qquad \textrm{spectroscopy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is, spectroscopy of the &quot;infrared action&quot; flavor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after running into more dead ends than a drunk rat in a maze with no exits, I realized that the correct way to split it might be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\textrm{infrared} \qquad\qquad \textrm{action spectroscopy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &quot;action spectroscopy&quot; that happens to use infrared light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lo and behold, the first hit on google was &lt;a href=&quot;https://en.wikipedia.org/wiki/Action_spectrum&quot; rel=&quot;noreferrer&quot;&gt;a wikipedia page on action spectroscopy.&lt;/a&gt; That link seems to have more of a biological tint to it (as do most results), but there are several &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0079681614000318&quot; rel=&quot;noreferrer&quot;&gt;physical science papers out there as well&lt;/a&gt;. (See Edit)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that action spectroscopy in general refers to hitting a system with light and watching some kinetic property of the system (rate of photosynthesis, conversion rate of isomers, etc.) In the case of the paper you cited, they report the relative intensity, so my guess would be that they did something like a scan over frequencies and watched how the ions fragmented at each frequency, which is supported by a cursory glance over the text surrounding Equation (1). Unfortunately, I don't have time to read the paper in full, so let me know if I'm totally wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral of the story is that English is hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;=============&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It turns out I'm the dolt and everyone else processed this correctly. As Brian mentions in the comments, the action spectroscopy page on Wikipedia is pretty awful, so I'll try to expand on action spectroscopy below, given what little I know about the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your usual spectroscopic setup, you hit the sample with some form of EM radiation, then look at what's reflected or absorbed to learn something about the system (e.g. UV-vis, IR, NMR, rotational)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that in action spectroscopy, your goal is to use the light to cause some sort of a change in the system, and measure that instead. In the example given on Wikipedia, you measure the rate of photosynthesis as you change wavelengths--this tells you at which wavelengths photosynthesis is most effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper I linked above, they seem to mediate the conversion of cis-butene to trans-butene using light. Presumably, you could start off with low-energy photons, then gradually ramp the energy up until you see the isomerization happen. The wavelength at which the transformations first start happening would tell you the activation energy barrier for the cis-trans isomerization. (Note that I can't actually read that paper because of the paywall, so I may be wrong about this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the paper Brian linked, they seemed to use IR to study the fragmentation of an ion sample. They mention &quot;loss paths&quot; as observed through an MS, so it seems like they hit a sample with IR light, then figure out what it breaks up into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Brian said in a comment, it seems that action spectroscopy is where light is used &quot;to &lt;em&gt;alter&lt;/em&gt; the system under study, not to just probe it passively.&quot; I'd like to add that, in general, it seems that the quantity of interest is not the amount of light that comes back out, but something else that happens in the system as a result of the exposure to light (e.g. isomerization, ion fragmentation, glucose production).&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="11367" LastEditDate="2015-07-29T19:26:15.087" LastActivityDate="2015-07-29T19:26:15.087" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34527" PostTypeId="2" ParentId="34045" CreationDate="2015-07-29T21:23:48.587" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The exponential must be dimensionless.  That means if $E_a$ is in units of $\mathrm{cal\cdot mol^{-1}}$, then $R$ must be units of $\mathrm {cal\cdot mol^{-1}\cdot K^{-1}}$.  Make sure you use the right value of $R$ for these units; $R=8.314~ \mathrm{J \cdot mol^{-1}\cdot K^{-1}}$, but $R = 1.987~ \mathrm{cal \cdot mol^{-1}\cdot K^{-1}}$.  Temperature must obviously be in Kelvins.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The example reaction you gave is bimolecular, so I would think $k_f$ is meant to have units of $\mathrm{cm^3 \cdot s^{-1}\cdot mol^{-1}}$.  That way, when you multiply $k_f$ by the molar concentration of both reactants, e.g. ethanol and hydroxyl, you get a reaction rate that is in units of $\mathrm{mol \cdot s^{-1}\cdot cm^{-3}}$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The remaining terms must therefore combine to give the right units for $k_f$. Since the units of $T^b$ will have units of $\mathrm K^b$, then the units for $A$ will be the units for $k_f$ divided by $\mathrm K^b$, i.e. the units of $A$ are $\mathrm{{cm}^3 \cdot s^{-1}\cdot mol^{-1} \cdot K^{\it-b}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-07-30T00:52:09.267" LastActivityDate="2015-07-30T00:52:09.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34528" PostTypeId="1" CreationDate="2015-07-29T22:43:14.050" Score="10" ViewCount="168" Body="&lt;p&gt;The claim that &quot;the tars in cannabis are considerably more dangerous than those in tobacco&quot; is made in Patricks An Introduction to Medicinal Chemistry (4th edition, p. 169). It's just a simple one line statement and not much attention is given to it, but from what I already knew this is to say the least a very controversial statement (see for example the independent website about drugs &lt;a href=&quot;https://www.erowid.org/plants/cannabis/cannabis_health.shtml&quot; rel=&quot;nofollow&quot;&gt;Erowid&lt;/a&gt;) and I expect better for a frequently used University level book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to what degree (from a chemical point of view) there is truth (or falsehood) to this statement given current more or less objective (not like the Heath monkey studies) knowledge? Is the author in his right to boldly state this?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="1887" LastEditDate="2015-08-03T14:40:06.143" LastActivityDate="2015-09-20T16:16:39.243" Title="Are the tars in cannabis considerably more dangerous than those in tobacco?" Tags="&lt;toxicity&gt;&lt;medicinal-chemistry&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="6" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34532" PostTypeId="1" AcceptedAnswerId="34533" CreationDate="2015-07-30T04:00:04.120" Score="4" ViewCount="1473" Body="&lt;p&gt;I recently saw a video on making liquid $\ce{CO2}$:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=8AN_XMcD3yI&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=8AN_XMcD3yI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically we seal some dry ice in a container. As the dry ice sublimates the pressure inside the container increases and so at about 5 atm we reach the triple point of $\ce{CO2}$, at which point the solid $\ce{CO2}$ melts instead of sublimating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point in the video, the teacher cautions that the same experiment should not be done with a Gatorade bottle (similar to a water bottle). However, I'm not sure why this is the case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly the usual &quot;dry ice bomb&quot;, which also has warm water poured into the sealed bottle, is not a good idea. In the present scenario, once all of the solid $\ce{CO2}$ had melted and the liquid $\ce{CO2}$ then goes into the gas phase, the increase in pressure could lead to an explosion: isn't this the same principle as in the case of the &quot;dry ice bomb&quot;? Is the warning because a Gatorade bottle (or water bottle, etc.) cannot withstand 5 atm of pressure?&lt;/p&gt;&#xA;" OwnerUserId="17814" LastEditorUserId="865" LastEditDate="2015-07-30T12:49:01.120" LastActivityDate="2015-07-30T23:52:25.333" Title="Making Liquid CO2" Tags="&lt;safety&gt;&lt;phase&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34533" PostTypeId="2" ParentId="34532" CreationDate="2015-07-30T04:13:52.893" Score="3" Body="&lt;p&gt;I think &lt;a href=&quot;http://www.pressdemocrat.com/news/2185542-181/bottle-bomb-explodes-in-womans&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://www.dailybreeze.com/general-news/20140425/4-year-old-hurt-when-dry-ice-filled-plastic-bottle-explodes-in-his-hand&quot; rel=&quot;nofollow&quot;&gt;three&lt;/a&gt; &lt;a href=&quot;http://www.nbcchicago.com/news/local/High-Schooler-Loses-Eye-After-Chemistry-Explosion-130134653.html&quot; rel=&quot;nofollow&quot;&gt;links&lt;/a&gt; nicely summarize what can potentially go wrong when a plastic bottle explodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question about how much pressure the Gatorade container can hold, that's a little trickier to figure out. It looks like the &lt;a href=&quot;https://en.wikipedia.org/wiki/MythBusters_%282005_season%29#Bottle_Rocket_Blast-Off&quot; rel=&quot;nofollow&quot;&gt;Mythbusters found that soda bottles can hold up to 150psi (~11 atm)&lt;/a&gt;, but gatorade and water bottles aren't designed to hold carbonation, so I'm guessing they'd be closer to the 60psi (~4 atm) that they found for water coolers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think ultimately the answer is that the teacher is playing on the safe side: sure, you might be able to hold 10 atm of pressure in a soda bottle at home, and get some cool-looking CO2 out of it, but seeing as you could lose an eye if your bottle was even slightly damaged (and seeing that this has happened before!), is it something you'd want to let high schoolers do on their own?&lt;/p&gt;&#xA;" OwnerUserId="2087" LastActivityDate="2015-07-30T04:13:52.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34534" PostTypeId="1" CreationDate="2015-07-30T04:55:41.500" Score="2" ViewCount="1067" Body="&lt;p&gt;My textbook said Le Chatelier's principle normally applies to gases but there is no explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some Googling and it said if you increase the pressure, it will shift to the side with less moles of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this really mean? Could someone clarify Le Chatelier's principle for me?&lt;/p&gt;&#xA;" OwnerUserId="17815" LastEditorUserId="402" LastEditDate="2015-10-12T02:14:12.553" LastActivityDate="2016-04-29T06:14:14.227" Title="How does Le Chatelier's principle work in a gaseous system?" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34537" PostTypeId="1" CreationDate="2015-07-30T10:32:16.590" Score="5" ViewCount="787" Body="&lt;p&gt;I know how the Kucherov reaction works for aliphatic alkynes, but what if the alkyne has a phenyl attached? How will the aromatic ring affect the reaction of alkynes with water?&lt;/p&gt;&#xA;" OwnerUserId="16025" LastEditorUserId="865" LastEditDate="2015-07-30T12:49:55.693" LastActivityDate="2015-07-30T12:49:55.693" Title="How does the Kucherov reaction work if the alkyne is aromatic?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34539" PostTypeId="1" CreationDate="2015-07-30T11:54:09.980" Score="5" ViewCount="671" Body="&lt;p&gt;Periodic acids cleave 1,2 diols by forming an periodate ester as an intermediate. Osmium tetroxide also forms an osmate ester with alkenes and yields syn diols after a hydrolysis step. Given that their structures are so similar, why doesn't periodic acid also succeed in forming diols with alkenes? (These in another step will be further cleaved to ketones). Instead in the Lemieux-Johnson oxidation, osmium tetroxide is taken in the first step - followed by periodic acid. &#xA;Osmium tetroxide does not cleave 1,2 diols - though I thought it might be in its being not such a strong oxidising agent as periodate - it may not suffice to cleave a C-C bond. Why doesn't though periodic acid carry out diol formation as well?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-07-31T20:50:33.340" LastActivityDate="2015-08-07T09:02:59.267" Title="Why don't periodic acids convert alkenes to 1,2 diols?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34541" PostTypeId="2" ParentId="20480" CreationDate="2015-07-30T14:17:50.533" Score="6" Body="&lt;p&gt;According to &lt;a href=&quot;http://shs.umsystem.edu/manuscripts/ramsay/ramsay_newton.html&quot;&gt;The State Historical Society of Missouri&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sweetwater is the name of a little trading-point established on Sweetwater Branch just after the Civil War. It was named for the branch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sweetwater Branch was named by the pioneers for the unusually good water in the stream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The place (and branch) are both in Newton County, Missouri. The same reference linked above indicates that what was once called Sweetwater is now referred to as Boulder City (because of the boulders nearby).&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-07-30T14:17:50.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34542" PostTypeId="1" AcceptedAnswerId="34547" CreationDate="2015-07-30T15:36:20.990" Score="5" ViewCount="916" Body="&lt;p&gt;Our lab uses a sodium hydroxide solution of a (theoretically) known normality (0.8000 - 1.0000) to titrate product samples for quality control.  However, we suspect that the normality of the $\ce{NaOH}$ left in the burette/tubing connected to our container is changing if we let it sit (our tests come out wrong, and the answers seem to get better if we run a couple of burettes worth of $\ce{NaOH}$ out first).  I expect it's absorbing moisture from the air, but I don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've tried different container types and we've put a desiccant in line with the tubing to absorb stray moisture (which it seems to be doing, there's been an observable color change), but the situation remains that we need to just throw out up to 200mL of $\ce{NaOH}$ any time we need to run a test.  Does anyone on here have any recommendations for countering this normality drift?&lt;/p&gt;&#xA;" OwnerUserId="17820" LastEditorUserId="10185" LastEditDate="2015-07-30T21:58:21.970" LastActivityDate="2015-07-30T21:58:21.970" Title="NaOH Normality Slowly Changing" Tags="&lt;experimental-chemistry&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34543" PostTypeId="2" ParentId="34520" CreationDate="2015-07-30T15:40:17.777" Score="2" Body="&lt;p&gt;In Wutz's book on protecting groups,&lt;sup&gt;[1]&lt;/sup&gt; I found an example of almost instantaneous reductive debenzylation which leaves an ester untouched:&lt;sup&gt;[2]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O4Apr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O4Apr.png&quot; alt=&quot;Debenzylation example with Na/NH3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of BCl&lt;sub&gt;3&lt;/sub&gt; is also possible:&lt;sup&gt;[3]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EX5xp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EX5xp.png&quot; alt=&quot;Debenzylation with BCl3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Wuts, P. G. M. &lt;em&gt;Greene's Protective Groups in Organic Synthesis&lt;/em&gt;, 5th ed.; Wiley: Hoboken, NJ, 2014, p 131.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sulikowski, M. M.; Davies, G. E. R. E.; Smith, A. B., III. Avermectin–milbemycin synthetic studies. Part 7. An approach to the southern hemisphere of milbemycin α&lt;sub&gt;1&lt;/sub&gt;. &lt;em&gt;J. Chem. Soc., Perkin Trans. 1&lt;/em&gt; &lt;strong&gt;1992,&lt;/strong&gt; 979–989. &lt;a href=&quot;https://doi.org/10.1039/P19920000979&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1039/P19920000979&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Williams, D. R.; Brown, D. L.; Benbow, J. W. Studies of Stemona alkaloids. Total synthesis of (+)-croomine. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1989,&lt;/strong&gt; &lt;em&gt;111&lt;/em&gt; (5), 1923–1925. &lt;a href=&quot;https://doi.org/10.1021/ja00187a081&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja00187a081&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2018-10-06T16:00:21.380" LastActivityDate="2018-10-06T16:00:21.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34544" PostTypeId="1" AcceptedAnswerId="34552" CreationDate="2015-07-30T15:41:42.160" Score="6" ViewCount="9057" Body="&lt;p&gt;I have a large amount of copper sulfate available and I am interested in obtaining copper chloride from it. What is the easiest way to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered mixing an aqueous solution of copper sulfate with sodium chloride, but I think that it would be difficult to separate the resulting sodium sulfate from the copper chloride. My other idea is to first react the copper sulfate with sodium carbonate and then react the resulting copper carbonate with hydrochloric acid, which would produce copper chloride (the other products would be $\ce{CO2}$ gas and water).  I am trying to find a cost effective method for performing this reaction.&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="16106" LastEditDate="2015-07-30T16:40:37.170" LastActivityDate="2019-12-12T17:41:48.917" Title="Produce CuCl2 from CuSO4" Tags="&lt;aqueous-solution&gt;&lt;synthesis&gt;&lt;ionic-compounds&gt;" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34547" PostTypeId="2" ParentId="34542" CreationDate="2015-07-30T16:29:31.920" Score="8" Body="&lt;p&gt;I would be more worried about carbon dioxide (rather than water) being absorbed into the NaOH solution. A 1 molar NaOH solution is not going to be appreciably hygroscopic. But it will absorb CO2: two moles of hydroxide yield one mole of carbonate ion, which, depending on what you are titrating, may act as only deprotonate one mole of acid per mole base - reducing the effective normality of the NaOH solution. You could try protecting any air vents with a drying tube filled with sodium or calcium hydroxide, or use a regulated supply of chemically pure nitrogen as a blanket.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-07-30T16:29:31.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34548" PostTypeId="2" ParentId="34534" CreationDate="2015-07-30T16:57:27.827" Score="-1" Body="&lt;p&gt;Just as if it was a chemical reaction of aqueous solutions, a gaseous chemical reaction has its reagents and products and these exist in an equilibrium kind of state that can be shifted either way depending on the conditons the reaction is in and the amount of reagents and products you have (in moles). Adding pressure to the reaction is like compressing the gasses into a smaller volume and this causes more molecule collisions. Since the space between the molecules involved in the chemical reaction is decreased, you have more molecule collitions among them, this makes the molecules to react with each other more rapidly and speeds up the reaction. The reaction will go the way the equilibrium constant tells it to go, but with this addition of pressure it will do it faster. Molecule collitions also creates and destroys molecule's bonds, thus this can create or absorb heat depending on the reaction that is taking place.  &lt;/p&gt;&#xA;" OwnerUserId="17806" LastActivityDate="2015-07-30T16:57:27.827" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34549" PostTypeId="1" AcceptedAnswerId="34553" CreationDate="2015-07-30T17:04:11.290" Score="7" ViewCount="1178" Body="&lt;p&gt;The periodic table has 7 periods and they have 2,8,8,18,18... elements respectively from 1 to 7. But from what I understand, the periods each state the number of electron shells that the elements in that period has. So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2? (Since period 3 has a M electron shell, it can also have a p orbital, and in total therefore can have 18 elements.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same could be said for elements in group 4, group 5, group 6 and so on since they can have f,g, and h orbitals as well. Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastActivityDate="2017-12-07T16:04:15.840" Title="Confusion with the Periodic Table" Tags="&lt;periodic-table&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34550" PostTypeId="1" CreationDate="2015-07-30T17:15:02.923" Score="29" ViewCount="57068" Body="&lt;p&gt;It is said that covalent bonds are directional, while ionic bonds are not. Why? Is it because of the orientation/directional properties of the overlapping orbitals?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="10185" LastEditDate="2015-07-30T17:38:23.920" LastActivityDate="2019-01-26T20:40:35.077" Title="Why are covalent bonds directional?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34551" PostTypeId="2" ParentId="34550" CreationDate="2015-07-30T17:42:53.373" Score="36" Body="&lt;p&gt;The meaning of covalent bonds being &lt;em&gt;directional&lt;/em&gt; is that atoms bonded covalently prefer specific &lt;em&gt;orientations&lt;/em&gt; in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you said, by overlapping the atomic orbitals of participant atoms. And usually only few patterns of overlap are possible, consequently, only few spatial arrangements of atoms are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionic bonds are different: there is no electron sharing (or atomic orbital overlaps) and the number of anions surrounding a cation is limited by the charges of the ions, their sizes, and the efficiency of the lattice packing.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2019-01-26T20:40:35.077" LastActivityDate="2019-01-26T20:40:35.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34552" PostTypeId="2" ParentId="34544" CreationDate="2015-07-30T18:49:03.923" Score="5" Body="&lt;p&gt;You cannot obtain $\ce{CuCl2}$ mixing $\ce{CuSO4}$ and $\ce{NaCl}$ because both are very soluble, the second strategy seems good because $\ce{CuCO3}$ is insoluble in water, and you can react the new salt with $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="17823" LastEditorUserId="4231" LastEditDate="2015-07-30T18:53:50.643" LastActivityDate="2015-07-30T18:53:50.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34553" PostTypeId="2" ParentId="34549" CreationDate="2015-07-30T18:52:15.120" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So if that is the case, shouldn't period 3 have more elements, since&#xA;  it can hold up to 18 electrons, and therefore it can have up to 18&#xA;  more protons from the largest atomic number element in period 2?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed, elements of the same period have the same number of electron shells, but the &quot;problem&quot; is that in accordance with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule&quot;&gt;&lt;em&gt;Madelung/Janet/Klechkowski  rule&lt;/em&gt;&lt;/a&gt;, the $\mathrm{4s}$ orbital is occupied &lt;em&gt;before&lt;/em&gt; the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied &lt;em&gt;before&lt;/em&gt; the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-30T19:21:20.033" LastActivityDate="2015-07-30T19:21:20.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34554" PostTypeId="1" CreationDate="2015-07-30T19:22:16.253" Score="2" ViewCount="230" Body="&lt;p&gt;I am a high school student and I was reading a textbook and it gave me the equation of this hydrochloric acid which was put into water and the products formed were hydronium and chlorine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HCl + H2O -&amp;gt; H3O+ + Cl-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that $\ce{HCl}$ (Hydrochloric acid) breaks up in the water to become 2 free ions (Hydrogen &amp;amp; Chlorine) and the Hydrogen bonds to the water because of electronegativity that causes the oxygen to be partially charged and so the $\ce{H+}$ will bond to water giving us $\ce{H3O+}$. This makes sense to me however I do not quite understand why $\ce{Cl-}$ is left &quot;alone&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a lot of water with only little amount of acid added surely it should lead to the $\ce{Cl-}$ also bonding to water or hydronium? Why does it not bond to them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Hypothesis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attempted to use my knowledge of physics and chemistry to tackle this problem, I am not as well-versed in chemistry so please bare with my &quot;beginners&quot; attempt at understanding this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water ($\ce{H2O}$) is a polar molecule due to electronegativity so perhaps as the chlorine is trying to attach to the hydrogen part of the molecule the oxygen's negative field is slightly stronger than the attractive forces of the hydrogen in the molecule of water thus the chlorine cannot bond with the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next hydrogen is covalently bonded to water so their outer orbitals are filled up so there is no need for the chlorine to even bond as the hydrogen has filled its orbitals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please criticize my thinking as it will only improve my scientific curiosity and intuition.&lt;/p&gt;&#xA;" OwnerUserId="17824" LastEditorUserId="10185" LastEditDate="2015-07-30T20:15:50.670" LastActivityDate="2015-07-30T20:15:50.670" Title="Why do chloride ions not bond to water molecules/hydronium ions when hydrochloric acid dissociates?" Tags="&lt;acid-base&gt;&lt;water&gt;" AnswerCount="0" CommentCount="5" ClosedDate="2015-07-31T12:28:57.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="34555" PostTypeId="1" AcceptedAnswerId="34556" CreationDate="2015-07-30T20:37:03.587" Score="12" ViewCount="1348" Body="&lt;p&gt;$\ce{S^{2-}}$, $\ce{Cl-}$, $\ce{K+}$, and $\ce{Ca^{2+}}$ each have the same number of electrons. Their effective ionic radii are 182, 181, 138, and 100 picometers, respectively, and the number of protons for each are 16, 17, 19, and 20. My question is, why does the effective ionic radius drastically go from 181 to 138 picometers between chlorine and potassium? Is it really necessarily because they are in different periods or is there something that I am missing here?  &lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="11367" LastEditDate="2015-07-31T13:31:35.550" LastActivityDate="2017-04-09T04:56:39.553" Title="Why are there drastic changes sometimes in radius of isoelectronic ions?" Tags="&lt;ions&gt;&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34556" PostTypeId="2" ParentId="34555" CreationDate="2015-07-30T21:16:40.597" Score="8" Body="&lt;p&gt;The whole concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionic_radius&quot;&gt;ionic radius&lt;/a&gt; is &lt;em&gt;not&lt;/em&gt; sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical boundaries. So from the very beginning one should not expect a perfect trend in a not so well defined property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the &lt;em&gt;distance between two ions&lt;/em&gt;, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. But whatever way the division is done, the ionic radius of an ions is &lt;em&gt;not&lt;/em&gt; a fixed, perfectly defined property, since for the same ion, the radius can &lt;em&gt;differ&lt;/em&gt; in different crystal lattices due certain variables such as &lt;em&gt;coordination number&lt;/em&gt; and &lt;em&gt;electron spin&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, you mentioned the ionic radius of 138 pm for $\ce{K+}$. But this value is for 6-coordinate compounds only while for 8- and 12-coordinate compounds of $\ce{K+}$ the ionic radius is 151 pm and 160 pm respectively. Same story for $\ce{Ca^2+}$: 100 pm, 112 pm, and 135 pm for 6-, 8-, and 12-coordinate compounds respectively.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, for anions there is a little variation in the ionic radius depending on coordination number, so that for many of them only one value independent of coordination number is specified. To be honest I could not come up with a quick explanation why, so it feels like I have to dig into the matter deeper. But I'm quite reluctant to do so, since, as I said in the beginning, the whole concept of ionic radius looks suspicious, so I expect to find just some obscure rule arriving out of the blue which will explain this little variation in the anionic radius relatively to cationic ones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Slater's rules to the rescue!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I guess, I know which rule can be used to explain this drastic change in ionic radii. Let us examine the application of the good-ol' &lt;a href=&quot;https://en.wikipedia.org/wiki/Slater%27s_rules&quot;&gt;&lt;em&gt;Slater's rules&lt;/em&gt;&lt;/a&gt; to the matter. These rules allows us to calculate the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_nuclear_charge&quot;&gt;effective nuclear charge&lt;/a&gt;&lt;/em&gt;  $Z_{\mathrm{eff}}$, which is the net positive charge experienced by an electron in a multi-electron atom.&#xA;$$&#xA;    Z_{\mathrm{eff}} = Z - s \, ,&#xA;$$ &#xA;where $s$ is the screening constant. Slater's rules provide a way to calculate the values for the screening constants, and consequently, to obtain the values of effective nuclear charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the ions $\ce{S^2-}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Ca^2+}$ are isoelectronic: the electron configuration of them is that of $\ce{Ar}$, i.e. $\mathrm{[Ne] 3s^2 3p^6}$. Thus, we have to compare effective nuclear charges for $\mathrm{3p}$ electrons in these ions. And as a first approximation we could compare effective nuclear charges for $\mathrm{3p}$ electrons in the corresponding atoms. Now without going into the details (listing the actual Slater's rules &amp;amp; applying them) I will just leave here the final numbers, effective nuclear charges for $\mathrm{3p}$ electrons for the corresponding atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{c|cccc}&#xA;     &amp;amp; \ce{S} &amp;amp; \ce{Cl} &amp;amp; \ce{K} &amp;amp; \ce{Ca} \\&#xA;\hline&#xA;Z &amp;amp; 16 &amp;amp; 17 &amp;amp; 19 &amp;amp; 20 \\&#xA;Z_{\mathrm{eff}}(\mathrm{3p}) &amp;amp; 5.482 &amp;amp; 6.116 &amp;amp; 7.726 &amp;amp; 8.658 \\&#xA;r_{\mathrm{ion}} &amp;amp; 182 &amp;amp; 181 &amp;amp; 138 &amp;amp; 100 \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can see that despite a relatively small change in the nuclear charge going from $\ce{Cl}$ to $\ce{K}$&#xA;$$&#xA;   (19-17)/17 * 100 \% \approx 11.8 \% \, ,&#xA;$$&#xA;there is a rather drastic change in effective nuclear charge &#xA;$$&#xA;   (7.726-6.116)/6.116 * 100 \% \approx 26.3 \% \, .&#xA;$$&#xA;And the drastic change in ionic radius is exactly in line with this drastic change in effective nuclear charge &#xA;$$&#xA;   (138-181)/181 * 100 \% \approx -23.8 \% \, .&#xA;$$&#xA;To summarize: the effective nuclear charge for $\mathrm{3p}$ electrons increases by about 26.3% when going from $\ce{Cl}$ to $\ce{K}$ and the ionic radius accordingly decreases by 23.8%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we used effective nuclear charges of the atoms just as the first approximation for that of the ions. For anions an additional electron will occupy an outer orbital and there will be an increased electron-electron repulsion. This will push the electrons further apart and increase the shielding, so that $Z_{\mathrm{eff}}$ for anions will be smaller than for the corresponding atoms. For cations the effect of the removal of an electron is the opposite: when an electron is lost, electron-electron repulsion and therefore, shielding decreases, so that $Z_{\mathrm{eff}}$ for cations will be bigger than for the corresponding atoms. It would not change the above mentioned trend though; in fact, it will make it even more prominent.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Ionic radii values are taken from &lt;a href=&quot;http://www.kayelaby.npl.co.uk/chemistry/3_7/3_7_6.html&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-07-31T10:45:27.123" LastActivityDate="2015-07-31T10:45:27.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34557" PostTypeId="1" CreationDate="2015-07-30T22:14:04.687" Score="1" ViewCount="44" Body="&lt;p&gt;Is there a strong adhesive (can carry about 2 kg in the air) that can be switched on or off, for example with electricity or changing the acidity of the surrounding medium, though I prefer electricity if possible. I have found a few articles on the internet but can't verify them. &lt;a href=&quot;http://www.technologyreview.com/news/408325/glue-with-an-on-and-off-switch/&quot; rel=&quot;nofollow&quot;&gt;This is one of them&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17828" LastEditorUserId="4231" LastEditDate="2015-07-30T22:20:28.810" LastActivityDate="2015-07-30T22:20:28.810" Title="Is there an adhesive with an on-and-off switch?" Tags="&lt;bond&gt;&lt;polymers&gt;&lt;materials&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2018-10-24T20:56:59.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="34558" PostTypeId="1" AcceptedAnswerId="34560" CreationDate="2015-07-30T23:24:19.550" Score="0" ViewCount="6715" Body="&lt;p&gt;I'm working through problems in my textbook. How much heat is required to convert 2.55 g of $\ce{H2O}$ at 28 °C to steam at 100 °C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simpler problems I've used the fact that $\Delta H_\text{fus} = 6.02\ \mathrm{kJ/mol}$ for $\ce{H2O}$ at a melting point of 0 °C to do these kinds of calculations. But here the temperatures are different. How can I approach this problem? Is there some other conversion factor I should be aware of?&lt;/p&gt;&#xA;" OwnerUserId="17115" LastEditorUserId="865" LastEditDate="2015-10-02T16:07:11.150" LastActivityDate="2015-10-02T16:07:11.150" Title="Convert grams of water to steam" Tags="&lt;homework&gt;&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;numerical-analysis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34559" PostTypeId="2" ParentId="34532" CreationDate="2015-07-30T23:52:25.333" Score="2" Body="&lt;p&gt;Most likely the plastic would suffers thermal and physical shock so it would shatter- that is when it suffers rapid temperature changes and rapid physical stress- plastics have very poor thermal shock resistance due to poor crack resistance and low thermal conductivity- conductivity that would stop the heat transfer occurring in one place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the conditions aren't rapid enough to cause shock then it probably could still fail though its hard to tell with the CO2 phase diagram (quick google search finds it) so the plastic bottles go below their glass temperature and have very low strength in their glassy phase whereas a pyrex bottle would have thermal stability to maintain strength but I don't know the temperature involved and it would depend on which plastic a bit. It may even be chemical damage of the plastic.&lt;/p&gt;&#xA;" OwnerUserId="2581" LastActivityDate="2015-07-30T23:52:25.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34560" PostTypeId="2" ParentId="34558" CreationDate="2015-07-31T00:49:32.417" Score="3" Body="&lt;p&gt;You need the vaporization $\Delta H_\text{vap}$ and you need to use this equation: $$Q = m~c~\Delta T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q$ equals the energy to heat $1\ \mathrm{mol}$ of water up to $\mathrm{100~^\circ C}$. $m$ is the mass of water, $c$ is the specific heat capacity and $\Delta T$ is the difference between the initial and final temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this stage you have the energy that is used to heat the liquid water up to $\mathrm{100~^\circ C}$ but the water is still liquid and you need steam water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to calculate the energy for phase transition from liquid to steam and the equation for this is: $$\Delta H_\text{vap} \cdot n = Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ is the necessary energy for the phase transition, $n$ is the amount of substance and $\Delta H_\text{vap}$ is the energy used for phase change of $1\ \mathrm{mol}$ of the substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to convert the mass to amount of substance (or use latent steam heat instead) and to calculate the $Q$ in the two stages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the energy used for heat and change the phase.&lt;/p&gt;&#xA;" OwnerUserId="17830" LastEditorDisplayName="user7951" LastEditDate="2015-07-31T09:09:22.780" LastActivityDate="2015-07-31T09:09:22.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34561" PostTypeId="1" AcceptedAnswerId="34569" CreationDate="2015-07-31T01:45:25.627" Score="8" ViewCount="3676" Body="&lt;p&gt;I want to run computation of my system in sulphuric acid with Gaussian 09; however sulphuric acid is not among the built-in solvents in G09. I know I have to give some parameters for the new solvent I define (including static and dynamic dielectric constants). Here's an example of my input files where I try to redefine a solvent water2. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#p b3lyp/6-31G(2df,p) int(grid=ultrafine)  SCRF(Solvent=Water2,Read)&#xA;&#xA;Comment&#xA;&#xA;0 1&#xA;&#xA;...&#xA;&#xA;stoichiometry=H2O1&#xA;solventname=Water2&#xA;eps=78.3553&#xA;epsinf=1.77556&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;G09 keeps telling me that &quot;SCRF(Solvent=Water2,Read)&quot; is not correct. I want to know what the correct input will be.&lt;/p&gt;&#xA;" OwnerUserId="17831" LastEditorUserId="16622" LastEditDate="2015-07-31T07:45:27.963" LastActivityDate="2017-01-19T09:27:30.027" Title="How to define a new solvent in Gaussian09 to run SCRF?" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34562" PostTypeId="2" ParentId="34408" CreationDate="2015-07-31T02:15:09.210" Score="2" Body="&lt;p&gt;Initially, the oxygen in the ring gets protonated. However, there is no formation of a carbocation. The ring opening proceeds by an S&lt;sub&gt;N&lt;/sub&gt;2 pathway, where iodide ion attacks at the less hindered carbon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5xVHc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5xVHc.png&quot; alt=&quot;Mechanism of reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="16683" LastEditDate="2018-10-19T22:58:28.567" LastActivityDate="2018-10-19T22:58:28.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34563" PostTypeId="1" AcceptedAnswerId="34567" CreationDate="2015-07-31T02:33:09.873" Score="17" ViewCount="2082" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/L-phenylalanine&quot; rel=&quot;noreferrer&quot;&gt;PubChem compound 6140&lt;/a&gt; is L-phenylalanine in its neutral (not zwitterionic) form.  According this PubChem, this molecule has the following SMILES and InChI indentifiers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SMILES: C1=CC=C(C=C1)CC(C(=O)O)N&lt;/li&gt;&#xA;&lt;li&gt;InChI: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/6925665#section=Top&quot; rel=&quot;noreferrer&quot;&gt;PubChem compound 6925665&lt;/a&gt; is the zwitterionic form of L-phenylalanine (protonated amine and deprotonated carboxylate).  Pubchem has decided that this species should be called &quot;(2S)-2-azaniumyl-3-phenylpropanoate&quot;.  The SMILES and InChI identifiers are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SMILES: C1=CC=C(C=C1)CC(C(=O)[O-])[NH3+] &lt;/li&gt;&#xA;&lt;li&gt;Inchi: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Confusion&lt;/strong&gt;.  My confusion is why these different compounds have different PubChem entries (CIDs), and different SMILES identifiers...but the &lt;em&gt;same&lt;/em&gt; InChI structure.  The different SMILES identifiers each appear to reflect the respective structures displayed by PubChem, but the single InChI identifier given for both compounds seems to reflect only the neutral form.  I even put the InChI into &lt;a href=&quot;http://www.rdkit.org/&quot; rel=&quot;noreferrer&quot;&gt;rdkit&lt;/a&gt; and converted it back to InChI.  The result was the same and rdkit interpreted this InChI as the neutral (not zwitterionic) species.  &lt;strong&gt;What is the reason for this discrepancy between InChI duplicity yet structural uniqueness?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here are some possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;PubChem is in error.  They should change the InChI for the zwitterionic compound.  (If so , to what?)&lt;/li&gt;&#xA;&lt;li&gt;PubChem is right but is using definitions of compound and CID that are different than mine.&lt;/li&gt;&#xA;&lt;li&gt;I have some kind of fundamental misunderstanding of the purpose of InChI, which I had thought would uniquely specify a molecular structure.  But InChI is designed to handle ambiguities like zwitterion vs. neutral.&lt;/li&gt;&#xA;&lt;li&gt;rdkit interprets InChIs weird. &lt;/li&gt;&#xA;&lt;li&gt;Something else.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="33746" LastEditDate="2016-08-13T13:40:17.837" LastActivityDate="2018-05-23T16:04:31.893" Title="Pubchem, InChI, SMILES, and uniqueness" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;cheminformatics&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34566" PostTypeId="1" CreationDate="2015-07-31T06:46:57.507" Score="3" ViewCount="4228" Body="&lt;p&gt;I am currently working on an investigation where I am burning isomers of butanol to test for the heat of combustion values to see which fuel is most efficient. The fuels I am testing are 1-butanol, 2-butanol, 2-methylpropan-1-ol (isobutanol) and 2-methylpopan-2-ol (tert-butyl alcohol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have researched that the heat of combustion values for these fuels are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1-butanol: −2671 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;2-butanol: −2661.1 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;Isobutanol: −2662.6 kJ/mol&lt;/li&gt;&#xA;&lt;li&gt;Tert-butyl alcohol: −2644.8 kJ/mol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So evidently, my experimental results should also show similarity in the ordering of greatest to lowest heat of combustion values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am wondering why isomers of butanol have different heat of combustions. I know it has to do with the shape of the molecules and whatnot, but how does this affect the energy released? What about bond angles? &lt;/p&gt;&#xA;" OwnerUserId="17550" LastEditorDisplayName="user7951" LastEditDate="2015-08-14T12:04:55.067" LastActivityDate="2015-09-03T21:50:30.267" Title="What causes the difference in heat energy released by combustion for the butanol isomers?" Tags="&lt;experimental-chemistry&gt;&lt;energy&gt;&lt;heat&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34567" PostTypeId="2" ParentId="34563" CreationDate="2015-07-31T06:53:07.143" Score="15" Body="&lt;p&gt;Unfortunately Pubchem is right, the two structures have the same InChI string and key, since the protonation state is the same in the zwitterion and the neutral form. So the reason for the discrepancy is &lt;strong&gt;by design&lt;/strong&gt;.&lt;br&gt;&#xA;I also always thought, InChI was designed for distinguishing between these conformations, but it turns out to just be one of the limitations of the system. The issue is addressed in &lt;a href=&quot;http://www.inchi-trust.org/technical-faq/#13.2&quot; rel=&quot;noreferrer&quot;&gt;section 13.2 of the technical FAQ of the InChI trust&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The different protonation states of the same compound will have InChIKeys differing only by the protonation indicator (unless both states have a number of inserted/removed protons greater than 12; in this case the protonation flag will also be the same, ‘A’).&lt;br&gt;&#xA;  This is exemplified below by standard InChIKeys as well as standard InChI strings for neutral, zwitterionic, anionic and cationic states of glycine (note that neutral and zwitterionic states do not differ in the total number of protons so they have the same standard InChI/InChIKey):&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/fcgIk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fcgIk.png&quot; alt=&quot;InChI for glycin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-07-31T06:59:03.380" LastActivityDate="2015-07-31T06:59:03.380" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34568" PostTypeId="1" CreationDate="2015-07-31T06:56:32.507" Score="8" ViewCount="192" Body="&lt;p&gt;Most of the common organic solvents are regarded as noncorrosive, but  stainless steel can be attacked by formic, acetic, and propanoic acids.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac50019a044&quot; rel=&quot;noreferrer&quot;&gt;Corrosion of stainless steel by organic solvent mixtures, Alice Y. Ku , David H. Freeman, Anal. Chem., 1977, 49 (11), pp 1637–1638&#xA;DOI: 10.1021/ac50019a044&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Are there some aromatic hydrocarbon that attack stainless steel? Which organic solvents can corrode copper and aluminium?&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-02T16:01:48.720" Title="Organic solutions that corrode metals" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;metal&gt;&lt;surface-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34569" PostTypeId="2" ParentId="34561" CreationDate="2015-07-31T07:35:45.403" Score="6" Body="&lt;p&gt;The correct input should be more like (replacing your &lt;em&gt;Water2&lt;/em&gt; with &lt;em&gt;Generic&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#p b3lyp/6-31G(2df,p) int(grid=ultrafine) SCRF(Solvent=Generic,Read)&#xA;&#xA;Comment&#xA;&#xA;0 1&#xA;...&#xA;&#xA;stoichiometry=H2O1&#xA;solventname=Water2&#xA;eps=78.3553&#xA;epsinf=1.77556&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;as it is stated in the &lt;a href=&quot;https://web.archive.org/web/20160922154735/http://www.gaussian.com/g_tech/g_ur/k_scrf.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Documentation&lt;/a&gt; under &lt;em&gt;Additional Input for PCM Calculations&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note - this only gets your calculation running. Your shown current input section might be insufficient to properly define sulphuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Geoff noted, the Gaussian Documentary states an important fact short before showing some $\varepsilon$-values of common solvents that should be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We list the ε values here for convenience, but be aware it is only one of many internal parameters used to define solvents. Thus, &lt;strong&gt;simply changing the ε value will not define a new solvent properly&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="4945" LastEditDate="2017-01-19T09:27:30.027" LastActivityDate="2017-01-19T09:27:30.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34571" PostTypeId="2" ParentId="34550" CreationDate="2015-07-31T08:35:27.787" Score="1" Body="&lt;p&gt;Very well put. Another explanation is that as covalent bonds are made by 'sharing' of electron, the electron will 'spend' more time with the element which is more electronegative. e.g in HCl, H has the electronegativity of 2.1 while Cl has 3, so the 'pull' of Cl on the shared electron is greater than H, hence the direction. Hence covalent bonds show directionality because of the shape of their electron wavefunction. Its the probability of finding an electron in a certain direction which is in turn effected by the partial charges. Even Ionic bonds show directionality (because there is no 'true' ionic' bond) but its very slight.&lt;/p&gt;&#xA;" OwnerUserId="16866" LastEditorUserId="16866" LastEditDate="2015-08-01T05:27:09.567" LastActivityDate="2015-08-01T05:27:09.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34572" PostTypeId="1" AcceptedAnswerId="34575" CreationDate="2015-07-31T08:49:21.037" Score="7" ViewCount="704" Body="&lt;p&gt;I understand that there is a pressure gradient between an aqueous solution, e.g. salty water, and, say, pure water due to osmosis. My question is, is it possible to match or almost match, the osmotic pressure between two solutions of different solutes, but using a solution with a different solute. For example, would it be possible to make the osmotic pressure gradient between salty water and some other solution zero, assuming both are at the same temperature and pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemist, so purely qualitative answers are helpful.&lt;/p&gt;&#xA;" OwnerUserId="17838" LastActivityDate="2015-07-31T15:34:18.983" Title="Can I equalise osmotic pressure using two different solutes?" Tags="&lt;aqueous-solution&gt;&lt;solutions&gt;&lt;osmosis&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34574" PostTypeId="1" CreationDate="2015-07-31T14:18:56.443" Score="5" ViewCount="23175" Body="&lt;p&gt;How can we know whether a compound has a coordinate bond or not? I know that I have to draw the Lewis Structure for the compound. But what to do after that? Is there anything to do with formal charges?&lt;/p&gt;&#xA;" OwnerUserId="11316" LastEditorUserId="10185" LastEditDate="2015-07-31T14:22:09.467" LastActivityDate="2021-03-08T18:09:47.970" Title="Identifying coordinate bonds" Tags="&lt;bond&gt;&lt;lewis-structure&gt;&lt;formal-charge&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34575" PostTypeId="2" ParentId="34572" CreationDate="2015-07-31T14:34:02.410" Score="3" Body="&lt;p&gt;Yes, I think so. Based on our comments above, I believe you want to prepare two solutions that are &lt;em&gt;isotonic&lt;/em&gt; with each other (have the same concentrations of different solutes) and allow them to communicate across a membrane that is selectively-permeable to water only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tonicity is described &lt;a href=&quot;https://en.wikipedia.org/wiki/Tonicity&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tonicity is a measure of the effective osmotic pressure gradient (as defined by the water potential of the two solutions) of two solutions separated by a semipermeable membrane. In other words, tonicity is the relative concentration of solutions that determine the direction and extent of diffusion. It is commonly used when describing the response of cells immersed in an external solution. Unlike osmotic pressure, tonicity is influenced only by solutes that cannot cross the membrane, as only these exert an effective osmotic pressure. Solutes able to freely cross the membrane do not affect tonicity because they will always be in equal concentrations on both sides of the membrane.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.vivo.colostate.edu/hbooks/pathphys/misc_topics/osmosis.html&quot; rel=&quot;nofollow&quot;&gt;this nice explanation with illustrations&lt;/a&gt; and &lt;a href=&quot;http://www.vivo.colostate.edu/hbooks/pathphys/misc_topics/osmosis_eg.html&quot; rel=&quot;nofollow&quot;&gt;one example&lt;/a&gt; where it shows that you could prepare a 10 mM solution of albumin and a 10 mM solution of glycine, separated by a water-permeable membrane, and no net flow of water occurs in either direction.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-07-31T15:34:18.983" LastActivityDate="2015-07-31T15:34:18.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34576" PostTypeId="2" ParentId="34544" CreationDate="2015-07-31T14:57:39.553" Score="0" Body="&lt;p&gt;I would guess you can directly mix copper sulfate solution with concentrated $\ce{HCl}$ to yield dissolved $\ce{CuCl2}$ and $\ce{H2SO4}$. This is of course in competition with (re-)formation of $\ce{CuSO4}$ but I would be interested in seeing how excess $\ce{HCl}$ would affect yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also -- I made $\ce{CuCl2}$ by dissolving copper metal in concentrated $\ce{HCl}$, adding concentrated $\ce{H2O2}$ (slowly) and then evaporating off the water solvent. The dehydrated form is a brown powder and moderately irratiting. Easily soluble in water and has a blue/green solution color. -- So some method of reducing the copper sulfate to copper-metal (e.g. electrolysis), and then this dissolution in $\ce{HCl/H2O2}$ should do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acid fumes deem a fume hood necessary, but decomposition of $\ce{H2O2}$ just gives off water and oxygen gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the lines of your precipitating idea, using barium chloride or lead chloride to form the corresponding insoluble sulfate would potentially be a faster, one step process, although perhaps more expensive. After, you decant the remaining $\ce{CuCl2}$ and evaporate solvent.&lt;/p&gt;&#xA;" OwnerUserId="17821" LastEditorUserId="10185" LastEditDate="2015-07-31T15:12:51.980" LastActivityDate="2015-07-31T15:12:51.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34577" PostTypeId="2" ParentId="34544" CreationDate="2015-07-31T16:24:05.533" Score="5" Body="&lt;p&gt;Here's a much more direct method: just add a stoichiometric amount of &lt;strong&gt;calcium&lt;/strong&gt; chloride to a concentrated aqueous solution of copper sulfate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuSO4(aq) + CaCl2(aq) -&amp;gt; CaSO4(s) + CuCl2(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the copper sulfate you have is in the anhydrous form (white powder), then add 695.3 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4}$. If your copper sulfate is in the form of the pentahydrate ($\ce{CuSO4.5H2O}$, striking blue crystals), then add 444.5 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4.5H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low solubility of calcium sulfate (~0.2 g per 100 g of water at 20°C) is what drives the reaction forwards. This &lt;a href=&quot;https://en.wikipedia.org/wiki/Salt_metathesis_reaction&quot; rel=&quot;noreferrer&quot;&gt;double displacement reaction&lt;/a&gt; does not happen if you use sodium chloride, because sodium sulfate is highly soluble. Simply filter the solid out and obtain a relatively pure solution of copper(II) chloride, with small amounts of calcium and sulfate contamination. Drying the solution completely will net $\ce{CuCl2}$ crystals.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastActivityDate="2015-07-31T16:24:05.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34578" PostTypeId="2" ParentId="34563" CreationDate="2015-07-31T16:30:02.290" Score="7" Body="&lt;p&gt;Martin's answer led me to discover an important extension of InChI that does allow for specification of some tautomer and zwitterion identification.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;InChi identifiers that begin with &lt;code&gt;InChI=1S/...&lt;/code&gt; are &lt;em&gt;standard&lt;/em&gt; InChI.  In standard InChI, the InChI identifier &quot;must be the same for any arrangement of mobile hydrogen atoms&quot;, with the quote from &lt;a href=&quot;http://www.inchi-trust.org/technical-faq/#6&quot; rel=&quot;noreferrer&quot;&gt;Section 6 of the Technical FAQ of the Inchi Trust&lt;/a&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However, non-standard InChIs are also possible.  These start with &lt;code&gt;InChI=1/...&lt;/code&gt;.  Note the missing &lt;code&gt;S&lt;/code&gt;.  In non-standard InChI, there can be an extra layer beginning with &lt;code&gt;/f&lt;/code&gt; that is called the fixed hydrogen layer.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Playing around with rdkit (via its Python API) I was able to produce a non-standard InChI that I think corresponds to &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/6925665&quot; rel=&quot;noreferrer&quot;&gt;Pubchem compound 6925665&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&lt;code&gt;&#xA;from rdkit import Chem&#xA;zwitterion_phe_smiles = 'C1=CC=C(C=C1)CC(C(=O)[O-])[NH3+]'&#xA;zwitterion_phe_mol = Chem.MolFromSmiles(zwitterion_phe_smiles)&#xA;&#xA;# produces &quot;standard&quot; InChI so not explicitly zwitterionic&#xA;print Chem.MolToInchi(zwitterion_phe_mol)&#xA;&#xA;     InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)&#xA;&#xA;&#xA;# produces &quot;non-standard&quot; InChI with fixed-H layer so zwitterion can be IDed&#xA;print Chem.MolToInchi(zwitterion_phe_mol, options='/FixedH')&#xA;&#xA;     InChI=1/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/f/h10H&#xA;&#xA;&#xA;# going from respective InChIs to SMILES&#xA;## The standard InChI produces neutral SMILES&#xA;zwitterion_nonstandard_inchi = 'InChI=1/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/f/h10H'&#xA;standard_inchi = 'InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)'&#xA;&#xA;print Chem.MolToSmiles(Chem.MolFromInchi(zwitterion_nonstandard_inchi))&#xA;print Chem.MolToSmiles(Chem.MolFromInchi(standard_inchi))&#xA;&#xA;     [NH3+]C(Cc1ccccc1)C(=O)[O-]&#xA;     NC(Cc1ccccc1)C(=O)O&#xA;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So for zwitterions tautomeric InChI's are possible.  Worryingly, however, some kinds of tautomerism are not handled even by non-standard InChI, again quoting from section 6 of the FAW:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In its current state, InChI recognizes the most common form of H migration (for the full list, see Table 6, Section IVb of the InChI Technical Manual). However, several ways of tautomeric migration that are not supported by default may appear important for some chemists. In particular, these are keto-enol and long-range tautomerisms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why Pubchem chose to list standard InChI instead of non-standard InChI is not entirely clear to me.  I suppose it is difficult to figure out programmatically when non-standard InChI would be required.  Ideally I suppose Pubchem would have both standard and non-standard InChIs for each of their molecules, but I'm not sure when/if they will ever make that change.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-07-31T16:30:02.290" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34580" PostTypeId="1" CreationDate="2015-07-31T16:56:12.520" Score="2" ViewCount="9902" Body="&lt;p&gt;These dishwasher tablets (&lt;a href=&quot;http://www.finish.co.uk/shine-and-protect/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Finish Powerball Shine&lt;/a&gt;) seem to be covered in plastic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HExHS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HExHS.jpg&quot; alt=&quot;Film-covered dishwasher tablets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that can't possibly be true since there's no unwrapping, and they're completely soluble in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the film made of? &lt;/p&gt;&#xA;" OwnerUserId="17845" LastEditorUserId="10185" LastEditDate="2015-07-31T17:07:13.730" LastActivityDate="2019-05-22T05:36:47.050" Title="What is the transparent film that covers water-soluble dishwasher tablets?" Tags="&lt;everyday-chemistry&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34582" PostTypeId="2" ParentId="34563" CreationDate="2015-07-31T17:26:03.430" Score="11" Body="&lt;p&gt;InChI is &lt;em&gt;intended&lt;/em&gt; to ignore &lt;a href=&quot;https://chemistry.stackexchange.com/a/29237/5017&quot;&gt;tautomeric forms&lt;/a&gt;. As &lt;a href=&quot;https://chemistry.stackexchange.com/a/34567/5017&quot;&gt;Martin indicates&lt;/a&gt; this also means zwitterions are considered identical to the neutral form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike you and Martin, I'm not sure I see this as a bug, since predicting the most stable tautomer or zwitterion/neutral is a complicated issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to keep track of zwitterions, I think &lt;a href=&quot;http://opensmiles.org/opensmiles.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SMILES&lt;/a&gt; is a better format, since you can specify exactly what you want as far as explicit hydrogens and charges. You'll need to stick to a particular toolkit to create a canonical ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there's a complicated relationship between CIDs and InChI / InChI keys. &#xA;There are &lt;strong&gt;other&lt;/strong&gt; cases where PubChem will have separate records for compounds that might be &quot;the same&quot; under InChI. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/14193/5017&quot;&gt;Axial or non-traditional stereochemistry&lt;/a&gt;. For example &lt;a href=&quot;https://pubchem.ncbi.nlm.nih.gov/compound/98863&quot; rel=&quot;nofollow noreferrer&quot;&gt;hexihelicene (CID=98863)&lt;/a&gt; should have two enantiomers, but the InChI reflects no stereochemistry.&lt;/li&gt;&#xA;&lt;li&gt;&quot;Extra&quot; stereo centers. PubChem takes depositions in the SD file format, which allows 2D stereochemistry in wedge/hash notation. If you take the actual 3D geometry, you might realize the wedge/hash were used for appearance, not for indicating a stereo center (i.e., multiple CIDs generate the same InChI from the full 3D molecule).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also cases where PubChem indicates an InChI key computed from the 2D depiction in the SD file, but there are missing or undefined stereo centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd say that because of incomplete stereochemistry and/or inconsistencies in representations, PubChem CIDs will not always match up with &quot;structural uniqueness&quot; and this is by design, both for PubChem and InChI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral of the story, frankly, is that chemistry is complicated and coming up with &quot;perfect&quot; unique identifiers is incredibly hard.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-07-31T17:26:03.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34583" PostTypeId="1" AcceptedAnswerId="34643" CreationDate="2015-07-31T18:04:17.597" Score="7" ViewCount="2735" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A tangent drawn on “conc. of product vs. time” curve of a 1st order reaction makes an angle 30 degrees with the $y$-axis at 20 min. Find the rate after 20 min.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I first tried to find the equation like this:&#xA;$$k=\frac{2.303}{t}\log\frac{a}{a-x}$$&#xA;Let $\frac{k}{2.303}$ be $c$ to make calculations less messy. Here, $a$ is the initial concentration of reactant.&#xA;$$\frac{a}{a-x}=10^{ct}$$&#xA;$$x=a\left(1-10^{-ct}\right)$$ &#xA;$$\frac{\mathrm dx}{\mathrm dt}=a\left(c\cdot\ln10\cdot10^{-ct}\right)=ak10^{\frac{kt}{2.303}}$$&#xA;Now, the tangent makes an angle of 30 degrees with concentration axis (or $y$-axis in this case). So the angle made with time axis (or $x$-axis) is 60 degrees. Hence $\frac{\mathrm dx}{\mathrm dt}=\tan\theta=\sqrt{3}$. But the value of initial concentration is not given for finding rate constant. Should I go for a different approach or am I making any conceptual mistake?&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="5026" LastEditDate="2018-02-26T05:18:41.383" LastActivityDate="2018-02-26T05:18:41.383" Title="Finding rate of reaction from a concentration vs. time graph" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34584" PostTypeId="1" CreationDate="2015-07-31T19:11:41.680" Score="2" ViewCount="4959" Body="&lt;p&gt;I'm asking this because I found a &lt;a href=&quot;http://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; for imidazole-HCl buffer: , but from what I understand, a buffer is a weak acid or base with its corresponding salt. Would imidazole-HCl buffer be a salt solution, or does it have a weak acid or base as well (if it does, what is the weak acid/base)? Also, I found two different pictures for imidazole hydrochloride:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mdyV1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mdyV1.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UbWfDm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UbWfDm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;Which one is correct? Is imidazole-HCl a salt or a buffer?&lt;/p&gt;&#xA;" OwnerUserId="17847" LastEditorUserId="16683" LastEditDate="2017-04-21T22:28:21.780" LastActivityDate="2017-04-21T22:28:21.780" Title="Does imidazole and hydrochloric acid yield imidazole hydrochloride salt?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;aromatic-compounds&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34585" PostTypeId="1" CreationDate="2015-07-31T19:13:12.933" Score="5" ViewCount="68" Body="&lt;p&gt;I'm studying Chemistry at A-level and a large part of the course is a researcher paper that takes most of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductive polymers interest me. I'm struggling to find any surfeited researcher on the synthesis of polypyrrole. I not sure if I need to buy a paper or another course of action would be more suitable.&lt;/p&gt;&#xA;" OwnerUserId="17848" LastEditorUserId="10185" LastEditDate="2015-07-31T20:23:57.873" LastActivityDate="2015-07-31T20:23:57.873" Title="Where can I obtain research material on the synthesis of polypyrrole?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;polymers&gt;&lt;reference-request&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34587" PostTypeId="2" ParentId="34381" CreationDate="2015-07-31T20:33:52.327" Score="0" Body="&lt;p&gt;In carbocation stability, mesomeric effect has highest priority, then hyperconjugation and then resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mesomeric effect of NH in 'Q' is greater than O in 'P'.&#xA;'R' has resonance and in 'S' carbocation has least stability.&lt;/p&gt;&#xA;" OwnerUserId="17849" LastActivityDate="2015-07-31T20:33:52.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34588" PostTypeId="2" ParentId="34584" CreationDate="2015-07-31T20:55:09.930" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oR4IU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oR4IU.png&quot; alt=&quot;Imidazole hydrochloride&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct structure of the salt is as shown above. This structure is more stable than the one that has two hydrogens on the same nitrogen - due to electron delocalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a water solution of about 50/50 imidazole base and the hydrochloride salt, you get the maximum buffer capacity, due to the fact that both acid and base forms are present, and a small amount of a strong acid or base subsequently added to the mixture will react with these imidazole species, and therefore have little effect on the pH. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-07-31T20:55:09.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34589" PostTypeId="1" CreationDate="2015-07-31T21:54:51.783" Score="3" ViewCount="1279" Body="&lt;p&gt;It would be useful to be able to predict relative evaporation rates of liquids.  For small reduced temperatures the heat of vaporization is relatively constant, but it is not one of the more common properties cited for substances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any sort of functional relationship between enthalpy of vaporization and common properties like viscosity, melting point, and boiling point?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="16622" LastEditDate="2015-08-01T14:58:31.613" LastActivityDate="2019-01-31T01:50:25.670" Title="Is there a relationship between viscosity and enthalpy of vaporization?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34591" PostTypeId="1" AcceptedAnswerId="34593" CreationDate="2015-07-31T17:11:33.077" Score="18" ViewCount="75194" Body="&lt;p&gt;I was under the impression that chemistry almost exclusively involves valence electrons because there isn't enough energy to strip off electrons located closer to the nucleus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is true, and elements in the same period have similar properties because they have the same number of valence electrons, then why is $\ce{SiO2}$ a solid and $\ce{CO2}$ a gas? Surely a difference in mass by a factor of 2.5 can't be that big a deal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because of a van der Waals force difference, because silicon has extra electrons which result in compounds being formed from it being more symmetrical than those formed from carbon?&lt;/p&gt;&#xA;" OwnerUserId="17888" OwnerDisplayName="Dov" LastEditorUserId="402" LastEditDate="2017-08-01T19:59:44.330" LastActivityDate="2020-10-13T18:32:23.533" Title="Why is SiO2 a solid while CO2 is a gas?" Tags="&lt;physical-chemistry&gt;&lt;crystal-structure&gt;" AnswerCount="7" CommentCount="6" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34592" PostTypeId="2" ParentId="34591" CreationDate="2015-07-31T17:32:09.547" Score="7" Body="&lt;p&gt;If you look at the crystal structure of silicon dioxide ($\ce{SiO2}$), you will see it is comprised of tetrahedra (one silicon surrounded by 4 oxygen atoms) and these tetrahedra are in turn connected to each other, and subsequently form a 2-dimensional network which is large enough to make this substance a solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the factor of 2.5 in mass makes a big difference, but it is not relevant to this discussion. I would recommend reading any inorganic chemistry book where such topics are discussed in depth. Van der Vaals forces only play significant roles in long-chain carbohydrates, such as fats, and are not really observed in &quot;inorganic&quot; molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO_2}$ molecules cannot form a crystal in the way $\ce{SiO2}$ units do, and I do not think I have to explain why. Also, it has no dipole moment and no van der Waals forces between the molecules.&lt;/p&gt;&#xA;" OwnerDisplayName="EpsilonDelta" LastEditorUserId="402" LastEditDate="2017-08-01T19:58:56.303" LastActivityDate="2017-08-01T19:58:56.303" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34593" PostTypeId="2" ParentId="34591" CreationDate="2015-07-31T19:52:53.273" Score="-5" Body="&lt;p&gt;It is because of the structure of the &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;. Two of carbon's valence electrons hybridize into two &lt;span class=&quot;math-container&quot;&gt;$sp$&lt;/span&gt; hybrid orbitals. As a result, the  molecule is one dimensional with an angle of 180&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt; between bonds and completely non-polar. The &lt;span class=&quot;math-container&quot;&gt;$\ce{Si}$&lt;/span&gt;, on the other hand, does not form such bonds and the angle is far from 180°, which in conjunction with oxygens high electronegativity makes it quite polar. Thus the interaction between neighboring &lt;span class=&quot;math-container&quot;&gt;$\ce{Si}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; atoms of different &lt;span class=&quot;math-container&quot;&gt;$\ce{SiO2}$&lt;/span&gt; molecules is much higher and as a consequence you need much more energy to break the solid, giving it an increased melting point.&lt;/p&gt;&#xA;&lt;p&gt;The mass (as discussed in various comments) does not play any role here since it is a matter of interaction or forces. The gravitational pull of single atoms or molecules is ridiculously small and never finds any considerations in such calculations (as it should!).&lt;/p&gt;&#xA;" OwnerUserId="17863" OwnerDisplayName="t0xic" LastEditorUserId="17175" LastEditDate="2020-10-13T18:32:23.533" LastActivityDate="2020-10-13T18:32:23.533" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="34596" PostTypeId="1" CreationDate="2015-08-01T04:01:13.973" Score="8" ViewCount="1580" Body="&lt;p&gt;I found a question asking the above which states that sulfur has a lesser second electron affinity than oxygen. But since the inter-electronic repulsion in sulfur is lesser, shouldn't it be willing to take up another electron more easily than an oxygen ion with a -1 charge?&lt;/p&gt;&#xA;" OwnerUserId="14616" LastEditorUserId="194" LastEditDate="2017-10-09T13:56:09.493" LastActivityDate="2017-10-09T13:56:09.493" Title="Which element has a greater second electron affinity, sulfur or oxygen?" Tags="&lt;periodic-trends&gt;&lt;periodic-table&gt;&lt;electron-affinity&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34600" PostTypeId="1" CreationDate="2015-08-01T09:02:48.247" Score="5" ViewCount="177" Body="&lt;p&gt;I am an electric engineer, and I read about battery self discharge phenomenon. I was wondering whether a battery self discharge rate is the same when it is unconnected (i.e. in storage) and connected (i.e. when it is plugged into a device and being discharged &quot;normally&quot;).&lt;br&gt;&#xA;In other words, when I calculate battery life time for a product, should I take add the rated self discharge of a battery to the general current calculations or not? And also-is the answer similar to all types of battery?&lt;/p&gt;&#xA;" OwnerUserId="17854" LastEditorDisplayName="user7951" LastEditDate="2015-08-01T09:50:00.573" LastActivityDate="2015-08-02T00:28:15.117" Title="Does battery self discharge when connected?" Tags="&lt;electrochemistry&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34602" PostTypeId="2" ParentId="34519" CreationDate="2015-08-01T10:34:18.357" Score="3" Body="&lt;p&gt;I have had great results for enantioselective ketone reduction using the Noyori Ru-TsDPEN system.&#xA;This has advantages through being catalytic with low stoichiometry and doesn't require high pressure, or molecular hydrogen for that matter. The downside might be the initial cost of the catalyst / chiral ligand but you don't need very much. The typical reduction reagents are triethylamine and formic acid, making this a very mild system to work with in terms of conditions.&lt;br&gt;&#xA;It might be cheaper to purchase the RuCl2(p-cymene) and chiral TsDPEN separately. That way you can also play with the ligand-metal ratio if you need to.&#xA;This saves a lot of hassle rather than adding on and removing chiral auxiliaries &lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorUserId="16724" LastEditDate="2015-08-03T16:42:39.273" LastActivityDate="2015-08-03T16:42:39.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34603" PostTypeId="2" ParentId="34591" CreationDate="2015-08-01T10:53:24.180" Score="3" Body="&lt;p&gt;Elements of the answer are also contained in other answers here but this needs some more teasing out. It has to do with the bond energies of C-O, Si-O, C=O, and Si=O. I don't have the numbers at hand, but work it out and it will point you to carbon being the most stable in the form of $\ce{CO2}$ and silicon the most stable as a tetrahedral network (similar to diamond.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why $\ce{CO2}$ is a gas can easily be explained using intermolecular bonding principles, which you can look up easily.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorUserId="402" LastEditDate="2017-08-01T19:58:00.813" LastActivityDate="2017-08-01T19:58:00.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34604" PostTypeId="2" ParentId="34568" CreationDate="2015-08-01T11:05:15.997" Score="1" Body="&lt;p&gt;Picric acid (2,4,6-trinitrophenol) is an aromatic compound has been used as an etchant in the past. Picral is an ethanol solution of picric acid and could etch steel. Look up www.metallographic.com/etchants/etchants.htm &lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorUserId="16724" LastEditDate="2015-08-01T19:38:47.627" LastActivityDate="2015-08-01T19:38:47.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34605" PostTypeId="1" CreationDate="2015-08-01T11:32:12.700" Score="2" ViewCount="21208" Body="&lt;p&gt;I have been trying for a while to determine what products are produced when copper sulfate reacts with sodium bicarbonate at room temperature and standard pressure. However, while I was looking, I found conflicting information concerning what products are formed. Some sources claim that the products are copper hydroxide and sodium sulfate, while other sources state that the products are copper carbonate and sodium sulfate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{CuSO4 + 2NaHCO3 -&amp;gt; CuCO3 + CO2 + H2O + Na2SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="865" LastEditDate="2015-08-01T18:26:53.073" LastActivityDate="2020-04-22T15:28:19.747" Title="What are the products of a reaction between copper sulfate and sodium bicarbonate?" Tags="&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34606" PostTypeId="1" AcceptedAnswerId="34609" CreationDate="2015-08-01T13:42:11.937" Score="18" ViewCount="629" Body="&lt;p&gt;According to Wikipedia, there are not many problems for which the Schrödinger equation can be solved analytically. [&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions&quot;&gt;1&lt;/a&gt;] The $\ce{H2+}$ ion is probably the most complex molecule that can be treated this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, every diatomic homonuclear system like this can be solved, e.g., $\ce{U2+}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to this point the molecules always consisted of two equal atoms which makes me curious about the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Schrödinger equation not only solvable analytically for diatomic homonuclear systems $\ce{A2+}$ but also for diatomic heteronuclear systems like $\ce{AB+}$?&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-08-01T16:09:32.370" Title="Analytical solution of the Schrödinger Equation for AB^+ systems" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34607" PostTypeId="1" CreationDate="2015-08-01T14:08:13.053" Score="2" ViewCount="301" Body="&lt;p&gt;I'm comparing a model with chemistry at equilibrium versus a kinetic one.&lt;br&gt;&#xA;I take a network of two reactions: methane steam reforming (MSR) and water gas shift (WGS) at 700°C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated extent of reaction of both reaction, with the equilibrium hypothesis, solving a system of two non-linear equation containing the two $ K _{eq} (T) =f(\lambda_{MSR}, \lambda_{WGS}) $ equations, which give me these results:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lambda_{MSR} =  1.4167e-4 [mol/s]$&lt;br&gt;&#xA;  $\lambda_{WGS} = -6.6788e-5 [mol/s]$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I calculate rates of reaction with an homogeneous and stoichiometric coefficient expression, with Temperature dependence constant (Arrhenius) taken &lt;a href=&quot;http://dx.doi.org/10.1016/j.ijheatmasstransfer.2004.04.010&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, I get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r _{MSR} = 20.9761e-4 [mol/s/m^3]$&lt;br&gt;&#xA;  $r _{WGS} =        0.0 [mol/s/m^3]$ (I have no reactants nor products...)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to compare them I have to multiply $r$ for reactor's volume.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here start my doubt:&lt;br&gt;&#xA;taking for sake of simplicity $1[m^3]$ reactor, I get  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$r_{MSR}&amp;gt;\lambda_{MSR}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;this what does it mean?&lt;/li&gt;&#xA;&lt;li&gt;does it have a physical meaning?&lt;/li&gt;&#xA;&lt;li&gt;what is the meaning of a so large reactor?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My thoughts, my idea, is that equilibrium shows the maximum rate of reaction so kinetic should, or better must, respect it giving me a lower value of reaction rate; but, as you can easily see in my above example, this &lt;strong&gt;&lt;em&gt;statement&lt;/em&gt;&lt;/strong&gt; (that I have never read somewhere in any chemistry book, so I can take it as &lt;strong&gt;&lt;em&gt;truth&lt;/em&gt;&lt;/strong&gt;) can be superimposed, forced. How do you explain it?&lt;/p&gt;&#xA;" OwnerUserId="17857" LastEditorUserId="17857" LastEditDate="2015-08-13T13:39:51.500" LastActivityDate="2015-08-13T13:39:51.500" Title="kinetic reaction rate greater than extent of reaction for equilibrium" Tags="&lt;equilibrium&gt;&lt;kinetics&gt;&lt;theoretical-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34609" PostTypeId="2" ParentId="34606" CreationDate="2015-08-01T15:42:33.387" Score="14" Body="&lt;p&gt;First, one has to be really careful with the words: when it is said that the &lt;em&gt;Schrödinger equation&lt;/em&gt; can be solved the dihydrogen cation completely &lt;em&gt;analytically&lt;/em&gt;, several remarks has to be made.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There are few different Schrödinger equations out there, so it is better to explicitly mention that one which can be solved analytically for $\ce{H2+}$ is the &lt;em&gt;electronic Schrödinger equation&lt;/em&gt;. So, first of all, we are talking about the &lt;em&gt;time-independent Schrödinger equation&lt;/em&gt; here, and secondly, we employ the &lt;em&gt;Born-Oppenheimer approximation&lt;/em&gt; and solve &lt;em&gt;only the electronic problem&lt;/em&gt;. No claim is maid with respect to analytical solutions of the whole time-independent Schrödinger equation without any approximation introduced, as well as regarding analytical solutions of its time-dependent counterpart.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second remark is about the meaning of the word &quot;analytical&quot;, or &quot;analytic&quot;. The definitions taken from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Closed-form_expression&quot;&gt;Wikipedia&lt;/a&gt; are as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An equation is said to have an &lt;em&gt;analytic solution&lt;/em&gt; if and only if at&#xA;  least one solution can be expressed as an &lt;em&gt;analytic expression&lt;/em&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;An &lt;em&gt;analytic expression&lt;/em&gt; is a mathematical expression constructed using well-known operations that lend themselves readily to calculation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These definition seems to be rather vague for me, especially taking into account that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[..] the set of well-known functions allowed can vary according to context [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so we have to be careful relying on some intuitive notions of &quot;analytic expression&quot; and &quot;analytic solution&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having said so, and in hope that my intuitive notions of &quot;analytic expression&quot; and &quot;analytic solution&quot; are not miles away from the truth,  it is important to note that $\ce{H2+}$, for which analytic solutions are known, is (a special case of) a &lt;em&gt;three-body&lt;/em&gt; system. And since there is no claim about analytic solutions for a four-body system, a five-body one, etc., from the beginning we have to restrict ourselves for three-body systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that viewpoint the notation $\ce{AB+}$ used by OP is misleading, because the only three-body chemical system of $\ce{AB+}$ kind is $\ce{H2+}$. Already for the system composed of $\ce{H}$ and $\ce{He}$ the charge has to be +2 since we have to remove 2 electrons so that we end up with just one of them which together with 2 nuclei constitutes a three-body system, and not a four-body one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what about analytical solutions for $\ce{AB^{n+}}$ systems where $n$ is chosen appropriately so that the system is a three-body one? Well, I think, as for $\ce{H2+}$, there are analytic solutions for electronic Schrödinger equation in the Born-Oppenheimer approximation. The reason is pretty simple in fact: the clamped nuclei model &lt;em&gt;reduces the three-body  problem to a single-body problem&lt;/em&gt; of the motion of a single electron within the fixed nuclear potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why there are no related references (at least I could not find one)? It is very likely because chemist has little or no interest in three-body chemical systems of $\ce{AB^{n+}}$ type in general, and specifically, little or no interest in analytic solutions for these systems. Except for the simplest $\ce{H2+}$ system which serves as a primary model for the concept of the one-electron bond.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-08-01T16:09:32.370" LastActivityDate="2015-08-01T16:09:32.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34610" PostTypeId="1" CreationDate="2015-08-01T16:17:23.397" Score="21" ViewCount="5682" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pJMQs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJMQs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the first structure on the left more stable than the second one on the right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first one the angle between the two pair of nonbonding electrons is about 120° which is a lot less than that in the second structure. And it is also very well known that &lt;strong&gt;lone&lt;/strong&gt; &lt;strong&gt;pair&lt;/strong&gt;-&lt;strong&gt;lone&lt;/strong&gt; &lt;strong&gt;pair repulsions&lt;/strong&gt; are worse than &lt;strong&gt;bonding pair-bonding pair repulsions&lt;/strong&gt;. So why is the first structure more stable?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="418" LastEditDate="2016-10-30T22:14:08.070" LastActivityDate="2017-10-10T22:36:47.380" Title="What is the favoured geometry of chlorine trifluoride?" Tags="&lt;inorganic-chemistry&gt;&lt;stability&gt;&lt;vsepr-theory&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34611" PostTypeId="2" ParentId="34610" CreationDate="2015-08-01T17:59:39.557" Score="7" Body="&lt;p&gt;Using the usual &lt;a href=&quot;https://en.wikipedia.org/wiki/VSEPR_theory#AXE_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;AXE method&quot;&lt;/a&gt; all molecules of $\ce{AX3E2}$ type, where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{A}$ represents the central atom;&lt;/li&gt;&#xA;&lt;li&gt;$\ce{X}$ represents the number of ligands (atoms) bonded to A; &lt;/li&gt;&#xA;&lt;li&gt;$\ce{E}$ represents the number of lone electron pairs surrounding the central atom;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;are assumed to be &lt;a href=&quot;https://en.wikipedia.org/wiki/T-shaped_molecular_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;T-shaped&lt;/a&gt;. I think it does not matter what the central atom $\ce{A}$ and ligands $\ce{X}$ are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jKqBi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jKqBi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP is right in that lone pairs repulsion is smaller in the second structure, however, we have to consider repulsions between lone pairs and bonding pairs as well. Yes, generally, lone pair - lone pair (lp-lp) repulsions are greater than lone pair - bonding pair (lp-bp) and bonding pair - bonding pair (bp-bp) ones, but the later could not be just  neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first structure we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One lp-lp repulsion at 120°;&lt;/li&gt;&#xA;&lt;li&gt;Two lp-bp repulsions at 120° (one bonding pair in the same plane interacts with two lone pairs);&lt;/li&gt;&#xA;&lt;li&gt;Four lp-bp repulsions at 90° (two bonding pairs out of plane interacts with two lone pairs);&lt;/li&gt;&#xA;&lt;li&gt;Two bp-bp repulsions at 90° and one at 180°.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the second structure we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One lp-lp repulsion at 180°;&lt;/li&gt;&#xA;&lt;li&gt;Six lp-bp repulsions at 90° (three bonding pairs out of plane interacts with two lone pairs);&lt;/li&gt;&#xA;&lt;li&gt;Three bp-bp repulsions at 120°.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in the second structure the lp-lp repulsion is smaller, but the overall lp-bp repulsion is bigger while the overall bp-bp repulsion is basically the same. Well, it is difficult to say &lt;em&gt;a priori&lt;/em&gt;, will the bigger lp-bp repulsion outweigh the smaller lp-lp repulsion, but it assumed in VSEPR theory that it will, so the first structure is more favorable.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="41328" LastEditDate="2017-10-10T22:36:47.380" LastActivityDate="2017-10-10T22:36:47.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34612" PostTypeId="1" CreationDate="2015-08-01T19:09:47.997" Score="2" ViewCount="489" Body="&lt;p&gt;I am confused about a specific part of Le Chatelier's Principle. My question revolves around a homework assignment question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2SO2(g)  + O2(g) &amp;lt;=&amp;gt; 2SO3(g)} \qquad\Delta H^\circ = -198.2\:\mathrm{\frac{kJ}{mol}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Comment on the changes in the concentrations of &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ce{O2}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\ce{SO3}$&lt;/span&gt; at equilibrium if we were to increase the temperature of the system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now I understand that increasing the temperature of an exothermic process will cause the equilibrium to shift towards the left (in the direction of the endothermic reaction); however, what does the temperature do to the concentrations of the gases?  I think it does nothing, but I am not sure.&lt;/p&gt;&#xA;" OwnerUserId="17865" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-04-05T05:12:32.053" Title="What does a change in temperature do to the concentration of gases at equilibrium?" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34613" PostTypeId="2" ParentId="34610" CreationDate="2015-08-01T19:38:08.820" Score="17" Body="&lt;p&gt;&lt;strong&gt;Short Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure on the left is &quot;preferred&quot; because the structure on the right cannot exist.  You cannot put 4 electrons in a p-orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Detailed Explanation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized.  This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons  or bond to a ligand.  Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawing on the right has two lone pairs placed in one p-orbital.  &lt;strong&gt;A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question really becomes&lt;/strong&gt;, does $\ce{ClF3}$ prefer to exist with&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 axial fluorines and 1 equatorial fluorine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 axial fluorine and 2 equatorial fluorines?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can use &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622&quot;&gt;Bent's rule&lt;/a&gt; to answer this question.  Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital).  An electronegative substituent will draw electron density away from the central atom.  Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals.  &lt;strong&gt;Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A final note on the bonding in $\ce{ClF3}$&lt;/strong&gt;, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;noreferrer&quot;&gt;hypercoordinated&lt;/a&gt; or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons.  See these earlier answers for more detail on hypercoordinate bonding.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544&quot;&gt;Why does F replace an axial bond in $\ce{PCl5}$?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/19433/how-can-iodine-bond-with-5-fluorines-in-iodine-pentafluoride/19437#19437&quot;&gt;How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule&quot;&gt;Hypervalency and the octet rule&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-08-01T22:28:17.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34615" PostTypeId="2" ParentId="34612" CreationDate="2015-08-01T21:59:04.837" Score="3" Body="&lt;p&gt;You are correct in saying that the increase in temperature causes the equilibrium to shift to the left, in the direction of the endothermic reaction. As the reaction continues in the endothermic direction, the concentrations of $\ce{SO3}$ will decrease in favour of the evolution of more $\ce{SO2}$ and $\ce{O2}$, due to the additional heat available.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-01T23:00:31.633" LastActivityDate="2015-08-01T23:00:31.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34616" PostTypeId="2" ParentId="34384" CreationDate="2015-08-02T00:15:09.537" Score="3" Body="&lt;p&gt;Yes. &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxyhydrogen&quot; rel=&quot;nofollow&quot;&gt;Oxyhydrogen&lt;/a&gt; is a mixture of oxygen and hydrogen, and has been or is being used in applications ranging from lighting to welding. The oxygen is mixed with hydrogen in a ratio such that the hydrogen is burned with the oxygen supplied in the gas mixture. This should meet your requirements.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-02T00:15:09.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34617" PostTypeId="2" ParentId="34600" CreationDate="2015-08-02T00:20:23.153" Score="2" Body="&lt;p&gt;Great question! I studied battery chemistry as part of my graduate work, and I think the answer depends on the mechanism of self-discharge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that by discharging the battery through an external circuit, you will generally cut off the self-discharge mechanisms (depending on the discharge rate), so if you add the discharge rate and the self-discharge rate, you will probably underestimate the battery life. You should also keep in mind that the rate of self-discharge depends on the state-of-charge for the battery, and is generally worse at high state of charge. Thus, even the manufacturer specification for self-discharge rate isn't a constant. &lt;/p&gt;&#xA;" OwnerUserId="17868" LastEditorUserId="865" LastEditDate="2015-08-02T00:28:15.117" LastActivityDate="2015-08-02T00:28:15.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34618" PostTypeId="2" ParentId="34361" CreationDate="2015-08-02T00:57:16.013" Score="2" Body="&lt;p&gt;It's true that when the battery is re-discharged, it's thermodynamically 'easier' to oxidize lithium than lithiated graphite. However, keep in mind that the lithium will react with electrolyte to form an SEI (solid electrolyte interphase) layer. This process consumes lithium ions, so on recharge you don't get all the lithium ions back. This is why the process is considered &quot;irreversible.&quot;&lt;/p&gt;&#xA;" OwnerUserId="17868" LastActivityDate="2015-08-02T00:57:16.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34619" PostTypeId="2" ParentId="34555" CreationDate="2015-08-02T01:23:25.003" Score="6" Body="&lt;p&gt;I completely agree with Wildcat's answer.  Excellent stuff. This part, though, got me thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In particular, I started to wonder: what might quantum chemistry have to say about ionic radii of isoelectronic species, for isolated ions in vacuum away from the influence of crystal packing and counter-ions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To try to answer this, I computed $\ce{S^{2-}}$ through $\ce{Ca^{2+}}$, including $\ce{Ar^0}$, at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;RHF&lt;/a&gt;/&lt;a href=&quot;http://dx.doi.org/10.1039/b508541a&quot; rel=&quot;nofollow noreferrer&quot;&gt;def2-QZVPP&lt;/a&gt; level of theory, with the minimal diffuse functions of &lt;a href=&quot;http://dx.doi.org/10.1007/s00214-010-0846-z&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zheng et al.&lt;/a&gt; added, in &lt;a href=&quot;http://orcaforum.cec.mpg.de&quot; rel=&quot;nofollow noreferrer&quot;&gt;ORCA&lt;/a&gt; v3.0. I also computed the correlation energies with fully-correlated orbital-optimized &lt;a href=&quot;https://en.wikipedia.org/wiki/Coupled_cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;CCSD(T)&lt;/a&gt;, to confirm that the RHF orbitals were a reasonable representation of the systems&lt;sup&gt;$\dagger$&lt;/sup&gt;. As an example, the input I used for neutral argon was$^\ddagger$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;! RHF CCSD(T) ma-def2-QZVPP NOFROZENCORE&#xA;! TIGHTSCF&#xA;! PAL2&#xA;&#xA;%mdci&#xA;   DenMat OrbOpt&#xA;   end&#xA;&#xA;%output&#xA;   Print[ P_OrbPopMO_L ] 1&#xA;   end&#xA;&#xA;* xyz 0 1&#xA;  Ar 0 0 0&#xA;*&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With these data in hand, I then calculated the &lt;a href=&quot;https://www.physicsforums.com/threads/radial-distribution-function-concept.577490/&quot; rel=&quot;nofollow noreferrer&quot;&gt;radial distribution functions&lt;/a&gt; (RDFs) of the wavefunctions using &lt;a href=&quot;http://sobereva.com/multiwfn/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MultiWFN&lt;/a&gt;. For completeness, the radial distribution function $R(r)$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;R(r) = 4\pi r^2 \left|\Psi(r)\right|^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It bears units of electrons per Angstrom, and represents the number of electrons one can expect to see, on average, within a spherical shell of inner radius $r$ and differential thickness $dr$.  The integral $\int_0^\infty{R(r)dr}$ theoretically should evaluate to the number of electrons in the system (here, $18$), and serves as a quality check on the computed $R(r)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RDF Results&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RDF all-space integrals for all five of the systems each fell short of the target value of $18$ by less than $0.06$ electrons $(0.33\%)$, with those for all but $\ce{S^{2-}}$ falling short less than $0.01$ electrons $(0.056\%)$.  Seems pretty solid to me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a plot of $R(r)$ for the five systems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NfZrM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NfZrM.png&quot; alt=&quot;RDF linear plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a better view of those inner-shell peaks, the following is the same plot but with a logarithmic $x$-axis$^*$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0XFPh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0XFPh.png&quot; alt=&quot;RDF Log Plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trend of the positions of the local maxima in the RDF$^\#$ is the same for all three peaks: a gradual shifting to larger distances from the nucleus as you go from $\ce{Ca^{2+}}$ to $\ce{S^{2-}}$. Looking at the second figure, it almost seems that the shift is log-linear with $Z$, but a more refined calculation would be necessary to investigate the behavior in detail. At an initial glance, though, this doesn't appear to be a trend quite as dramatic as that noted in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more subtle feature of the curves (for me, anyways, on my first look at them) is the substantial decrease in the &lt;em&gt;height&lt;/em&gt; of each peak as $Z$ is decreased from $Z=20$ for $\ce{Ca^{2+}}$ down to $Z=16$ for $\ce{S^{2-}}$. At the same time, the 'tail' of the distribution shifts substantially outward, especially for the anions $\ce{Cl-}$ and $\ce{S^{2-}}$.  Zooming in the $y$-axis on the plot with the linear $x$-axis confirms that the electron distributions of the anions indeed are quite diffuse (I'll explain the dashed line in a minute):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/njbDs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/njbDs.png&quot; alt=&quot;RDF linear zoom&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, fine, but how to quantify the ionic radii from this data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RDFs and Ionic/van der Waals Radii&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I ended up going about it was to look at a proxy for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionic_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionic&lt;/a&gt;/&lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;van der Waals radii&lt;/a&gt; of the various ions, in the form of a cutoff value of the RDF function$^\&amp;amp;$.  In other words, what value of $R(r)$ might correspond, if roughly, to these radii?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the neutral species, argon is the most reliable reference point, since it's really hard to obtain a mass of isolated cations or anions to take measurements from.  &lt;a href=&quot;http://www.webelements.com/argon/atom_sizes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WebElements&lt;/a&gt; reports argon's vdW radius as $1.88\ \mathring{\mathrm{A}}^{**}$.  Looking at the RDF for $\ce{Ar^0}$, the value of $R(1.88\ \mathring{\mathrm{A}})$ is approximately $0.23\ \frac{e^-}{\mathring{\mathrm{A}}}$.  The radial distances at which $R(r)$ falls to this value for the five species are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{|c|c|c|}&#xA;\hline&#xA;\mathrm{\bf{Species}} &amp;amp; \mathbf{r\ @\ R=0.23\ \frac{e^-}{\mathring{\mathrm{A}}}} &amp;amp; \mathrm{\bf{Lit.\ Ionic\ Radius}} \\&#xA;\hline&#xA;\ce{Ca^{2+}} &amp;amp; 1.44\ \mathring{\mathrm{A}} &amp;amp; 1.00\ \mathring{\mathrm{A}} \\&#xA;\hline&#xA;\ce{K^{+}} &amp;amp; 1.62\ \mathring{\mathrm{A}}  &amp;amp; 1.38\ \mathring{\mathrm{A}} \\&#xA;\hline&#xA;\mathbf{\ce{Ar^0}} &amp;amp; \mathbf{1.88\ \mathring{\mathrm{A}}}  &amp;amp; \mathbf{-} \\&#xA;\hline&#xA;\ce{Cl^{-}} &amp;amp; 2.27\ \mathring{\mathrm{A}} &amp;amp; 1.81\ \mathring{\mathrm{A}}  \\&#xA;\hline&#xA;\ce{S^{2-}} &amp;amp; 3.08\ \mathring{\mathrm{A}} &amp;amp; 1.82\ \mathring{\mathrm{A}}  \\&#xA;\hline&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So... my estimates don't compare terribly well to the ionic radius values.  But, recall, my calculations were for isolated ions in vacuum, not in the context of a crystal structure. So, it makes sense that the experimental ionic radii, however determined, would be smaller than those determined &lt;em&gt;in vacuo&lt;/em&gt; due to packing effects in the crystal structure.  Further, it also makes sense that the anions would be &quot;squashed down&quot; more when packed into a crystal structure as compared to cations, because the cations are already much more compact than the anions even without external 'constraints.'  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isoelectronic anions can get bigger much faster per-proton-removed than can  isoelectronic cations.  Just how much faster they get bigger depends on the size metric you choose, and in what chemical context and by what method you measure (or compute) that size metric.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\dagger$ Aside from a slight shift to high-angular-momentum functions, the results were quite similar. T1 diagnostics were all well below the 0.02 warning threshold.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\ddagger$ &lt;code&gt;NOFROZENCORE&lt;/code&gt; correlates all electrons. &lt;code&gt;TIGHTSCF&lt;/code&gt; requests more stringent convergence criteria. &lt;code&gt;PAL2&lt;/code&gt; makes ORCA run on two cores. &lt;code&gt;%mdci DenMat OrbOpt end&lt;/code&gt; enables orbital optimization. The &lt;code&gt;P_OrbPopMO_L&lt;/code&gt; part requests a full printout of the final orbital compositions for inspection.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^*$ Note that the spacing of data points near the nucleus is quite sparse for all systems, and so the curves are only as smooth as they are because I plotted with Excel's 'smoothed line' feature. If this were an actual publication, I would have worked the data up with a much denser grid in the core region.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\#$ While it would be tempting to associate the two sets of peaks squished up against the $y$-axis with the $n=1$ and $n=2$ shells, and the broader, smaller set in the vicinity of $0.5\ \mathring{\textrm{A}}\leq r\leq 0.8\ \mathring{\textrm{A}}$ with the $n=3$ shell, in reality electrons from all three shells $1\leq n\leq 3$ contribute to the inner peaks, only those from $n=2$ and $n=3$ contribute to the middle peaks, and only those from $n=3$ contribute to the outermost peaks.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\&amp;amp;$ I initially started out looking at the radial integral function of the RDF, $I(r)=\int_0^r{R(x)dx}$, but realized the plain RDF gave better results.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^{**}$ Of course, there are multiple ways of determining/defining the vdW radius, but let's take that &lt;a href=&quot;http://dx.doi.org/10.1021/j100785a001&quot; rel=&quot;nofollow noreferrer&quot;&gt;used&lt;/a&gt; for &lt;a href=&quot;http://www.webelements.com/periodicity/van_der_waals_radius/&quot; rel=&quot;nofollow noreferrer&quot;&gt;WebElements's data&lt;/a&gt; as representative. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2017-04-09T04:56:39.553" LastActivityDate="2017-04-09T04:56:39.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34620" PostTypeId="2" ParentId="6182" CreationDate="2015-08-02T01:57:36.610" Score="3" Body="&lt;p&gt;Sodium metal reacts violently with water because an electron is transferred from sodium to water (or protons in water, depending on pH) to form H2 gas. However, Na+ has already lost its electron, and Na+ does &lt;strong&gt;not&lt;/strong&gt; transfer an additional electron to water. &lt;/p&gt;&#xA;" OwnerUserId="17868" LastActivityDate="2015-08-02T01:57:36.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34621" PostTypeId="1" CreationDate="2015-08-02T02:00:34.197" Score="6" ViewCount="19494" Body="&lt;p&gt;If you have some sort of sports drink that contains sucrose, how would you take that sucrose and break it down into its components, glucose and fructose? I know that sucrose + water will break it apart, but how can it still remain together (as sucrose) while in the bottle of sports drink?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to compare the levels of glucose in various sport drinks. So I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the colouring from it before placing in the spectroscope). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will prepare standards by adding 5 grams of glucose to 100 mL of water, 10 grams of glucose to 100 mL water, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a high school student I only have limited resources. Any information about my proposed plan or alternatives that are low-cost are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17869" LastEditorUserId="7475" LastEditDate="2016-09-15T09:55:40.803" LastActivityDate="2017-02-09T22:55:39.687" Title="How to Convert Sucrose to Glucose and Fructose" Tags="&lt;aqueous-solution&gt;&lt;home-experiment&gt;&lt;food-chemistry&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34622" PostTypeId="2" ParentId="6182" CreationDate="2015-08-02T02:22:43.240" Score="4" Body="&lt;p&gt;The answer is that &lt;strong&gt;&lt;em&gt;the universe is lazy&lt;/em&gt;&lt;/strong&gt;. And that means that the universe is always seeking the lowest energy states it can find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $\ce{Na}$ is less lazy than $\ce{Na+}$, and similarly for $\ce{Cl}$ and $\ce{Cl-}$, because both these transitions release energy and thus are lazy. $\ce{Na}$ is dangerous, and $\ce{Cl}$ is poisonous, precisely because they have not achieved their most lazy state and thus want to react more, $\ce{Na}$ donating an electron and $\ce{Cl}$ accepting one, both releasing energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;universe is lazy&quot; hypothesis is further supported by the 'most reactions are random&quot; hypothesis. Most molecules are randomly hitting each other, and those that release energy are going to be favored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, $\ce{Na+}$ and $\ce{Cl-}$ are not dangerous because they are in their lazy states and thus don't need to react with much of anything.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastActivityDate="2015-08-02T02:22:43.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34624" PostTypeId="2" ParentId="34621" CreationDate="2015-08-02T03:47:43.610" Score="5" Body="&lt;p&gt;Sounds like a great experiment! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you state, a sucrose solution in water &lt;em&gt;eventually&lt;/em&gt; hydrolyzes to fructose and glucose, but without a catalyst, the reaction rate is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sucrose#Hydrolysis&quot;&gt;extremely slow&lt;/a&gt;; beverages are stable at room temperature for years. However it is hydrolyzed in hours at room temperature by adding &lt;a href=&quot;https://en.wikipedia.org/wiki/Sucrase&quot;&gt;&lt;em&gt;sucrase&lt;/em&gt; enzyme&lt;/a&gt; (secreted in the small intestine), &lt;a href=&quot;https://en.wikipedia.org/wiki/Invertase&quot;&gt;&lt;em&gt;invertase&lt;/em&gt; enzyme&lt;/a&gt; (derived from yeast) or even a weak acid such as citric acid, tartaric acid, or even dilute hydrochloric acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may need to test whichever of these are available for compatibility with Benedict's solution, though, and you may need to test some of the other ingredients in the drinks to see if they also cause false positive otr negative effects. You might compare the results from the Benedict's test with those from a blood-sugar glucometer (the solutions may need to be diluted in both tests to stay within a measurable range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, study of yeast fermentation of sucrose is responsible for the name &lt;em&gt;enzyme&lt;/em&gt;, meaning &lt;a href=&quot;https://en.wikipedia.org/wiki/Enzyme#Etymology_and_history&quot;&gt;&lt;em&gt;leavened with yeast&lt;/em&gt;&lt;/a&gt;, so one could say &lt;em&gt;invertase&lt;/em&gt; is the epitome of an enzyme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please share your experimental results here.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-02T03:47:43.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34625" PostTypeId="2" ParentId="34384" CreationDate="2015-08-02T06:56:48.943" Score="2" Body="&lt;p&gt;The most convenient solution for your question can be the Hydrooxy gas (also sometimes called the Brown’s gas). Simply put its water split into hydrogen and oxygen. Hydrogen and oxygen can be combined back by ignition and can create a maximum temperature up to 2800 °C (around 600–700 °C hotter than burning hydrogen in air), which makes it a good fuel for metal welding and cutting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though 2:1 hydrogen and oxygen ratio is enough to produce water via combustion, but on a practical solution you will need around 3:1 to 5:1 ratio to avoid oxidizing flames. The temperature you can achieve by burning hydrogen oxygen mix varies, depending on the ratio of both gases used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen and oxygen can be obtained via simple electrolysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O + Energy -&amp;gt; 2H2 + O2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and combined back as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2 + O2 -&amp;gt; 2H2O + Energy}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be worth noting that for all practical purposes, the energy you use to split hydrogen and oxygen will always be greater than what you can get by combining them back (like what happens in every combustion engine, humans have ever created).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are planning to develop an actual application, there are many precautions that you would need to consider, the most important of which would be back-fire protection (a common problem with gas based welding), so that the flame doesn’t reach back into the gas tank, which of course will explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So goes without saying, but you might want to adopt some precautionary measures if you do use it in any practical application. Temperatures this high can seriously hurt a person even if the contact was for some milliseconds. It can also damage other materials if the flame is directed at them.&lt;/p&gt;&#xA;" OwnerUserId="16866" LastEditorDisplayName="user7951" LastEditDate="2015-08-02T08:23:20.127" LastActivityDate="2015-08-02T08:23:20.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34626" PostTypeId="1" CreationDate="2015-08-02T07:45:33.120" Score="3" ViewCount="1676" Body="&lt;p&gt;I really see no relation between the terms microcosmic and the salt's structural formula, $\ce{Na(NH4)HPO4.4H2O}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Devoting sufficient time on Google didn't help at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information about the name of this compound is what I'm searching for, thanks in advance for sharing.&lt;/p&gt;&#xA;" OwnerUserId="4751" LastEditorUserId="865" LastEditDate="2015-08-02T16:50:49.937" LastActivityDate="2015-08-02T16:50:49.937" Title="Why is microcosmic salt called so?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34627" PostTypeId="1" AcceptedAnswerId="34648" CreationDate="2015-08-02T08:36:02.910" Score="2" ViewCount="2317" Body="&lt;p&gt;I am trying to calculate molar volume of certain crystal lattices to be used for crystal growth kinetics calculations. The information I have is the crystal volume &amp;amp; crystal lattices. For example the crystal $\ce{Cu41Sn11}$ has the structure cF416 and the information taken from a certain crystal database is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atom Formula: Cu&lt;sub&gt;41.5&lt;/sub&gt;Sn&lt;sub&gt;10.5&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cell volume:    5.81258 nm&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cell density (calculated):  8.85 Mg m&lt;sup&gt;-3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have tried 2 approaches to it. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First was the &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_volume#Crystalline_solids&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; formula for molar volumes &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;V&lt;sub&gt;m&lt;/sub&gt;= N&lt;sub&gt;A&lt;/sub&gt;V&lt;sub&gt;cell&lt;/sub&gt; / Z&lt;/p&gt;&#xA;&#xA;&lt;p&gt;molar volume = (Avorgados number x crystal cell volume) / number of formula units in a crystal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the following output for the above mentioned crystal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar Volume = 6.022 x 10&lt;sup&gt;23&lt;/sup&gt; x 5.81258 x 10 &lt;sup&gt;-27&lt;/sup&gt; / 51&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar volume = 6.731 x 10 &lt;sup&gt;-5&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;The second one is i figured out the formula&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Molar volume = Molar Mass of crystal / Cell Density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar mass of crystal $\ce{Cu41Sn11}$ = 41 x 63.546 + 11 x 118.71 = 3911.2 gms (3.9112 Kg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molar Volume = 3.9112 / 8850 = 4.41944 x 10 &lt;sup&gt;-4&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two approaches yield two different results, so i wanted to make sure where am i making the mistake? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you're interested in the crystal database, here it is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kqq3l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kqq3l.jpg&quot; alt=&quot;Database&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16866" LastActivityDate="2015-08-03T05:11:36.527" Title="Calculating Molar Volume from crystal volume" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34628" PostTypeId="1" CreationDate="2015-08-02T09:49:44.757" Score="4" ViewCount="163" Body="&lt;p&gt;I'm an engineering student preparing for my graduation, in one step of my graduation project I used a rechargeable battery(1.2 volt), i was shocked when this battery is starting to recharge it self until it reach 1.04 volt (not fully) for many many cycles and immediately when the voltmeter reads 0 volt from it, when i asked my doctors they told me that maybe its chemical issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so is there any explanations please?&lt;/p&gt;&#xA;" OwnerUserId="17873" LastEditorUserId="5456" LastEditDate="2015-08-02T11:18:00.977" LastActivityDate="2015-08-03T10:50:05.227" Title="How could a rechargeable battery recharge itself?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34630" PostTypeId="2" ParentId="34626" CreationDate="2015-08-02T12:35:36.850" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Devoting sufficient time on Google didn't help at all.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Hmmm... I found &lt;a href=&quot;http://www.geo.fu-berlin.de/geol/fachrichtungen/geochemhydromin/mineralogie/pdf/quickassay.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PDF&lt;/a&gt; in just a few seconds, where we read the following.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The origin of the name microcosmic salt.&lt;/p&gt;&#xA;&lt;p&gt;Beads of sodium metaphosphate&#xA;saturated with some oxides cool to a clear glass. Small crystals grow&#xA;in the glass on repeated heating and cooling. A microscope with a 80&#xA;to 100 x magnification is needed to inspect these crystals with&#xA;various morphologies. Such beads resemble a microcosmos. Some authors&#xA;tried to use the effect for analytical purposes in the 19th century,&#xA;but these efforts failed. Borax beads sometimes show the same effect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Microcosmos here is understood (as usual) as a &amp;quot;small world&amp;quot;, a &amp;quot;world in miniature&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-02T12:35:36.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34634" PostTypeId="1" AcceptedAnswerId="34667" CreationDate="2015-08-02T16:03:52.453" Score="3" ViewCount="1297" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the average life of a radio nuclide which decays by parallel paths,&#xA;  &lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;A  &amp;amp;\rightarrow B\\&#xA;2A &amp;amp;\rightarrow B,&#xA;\end{align}&lt;/span&gt;&#xA;  where the decay constants are &lt;span class=&quot;math-container&quot;&gt;$\lambda_1 = \pu{0.018 s-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\lambda_2 = \pu{0.001 s-1}$&lt;/span&gt;, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I used the formula &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{t} = \frac{t_1 t_2}{t_1 + t_2,}$$&lt;/span&gt; &#xA;where &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; represents the mean life. The equation can be written as: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{t} = \lambda_1 + \lambda_2$$&lt;/span&gt; &#xA;But I am getting the correct answer only if I take &lt;span class=&quot;math-container&quot;&gt;$\lambda_2$&lt;/span&gt; as 2 times the given decay constant for the second reaction. Is this because of &lt;span class=&quot;math-container&quot;&gt;$\underline{2}A$&lt;/span&gt; on the reactant side instead of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="4945" LastEditDate="2020-02-26T11:28:39.357" LastActivityDate="2020-02-26T15:09:54.580" Title="How to find the average half life of radioactive nuclide which undergoes two different decays?" Tags="&lt;kinetics&gt;&lt;radioactivity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34635" PostTypeId="1" CreationDate="2015-08-02T16:05:33.510" Score="3" ViewCount="145" Body="&lt;p&gt;As a computational materials chemist studying photocatalysis, how close to realistic systems can your modeling get? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a sort of high-level question about the theoretical modeling of photocatalysis. Most of the computational work on photocatalysis involves DFT calculations looking at band gaps, free energy profiles, and reaction pathways. I have also seen a handful of papers that use reactive MD to probe how the surrounding water molecules get affected by adsorbed species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a computational chemist or materials engineer, and you want to take your work closer to a real system, what is the logical next step? Would it be device modeling, where you take inputs from DFT (such as overpotentials), and then try to calculate the I-V curves, for instance? Are there any other multi-scale modeling techniques or coarse-grain methods that can be used? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this question can be extended more generally to different classes of applications, such as $\ce{CO2}$ sensors, other electrochemical reactions, and so on. I'm not really sure what the next step would be to go beyond DFT in order to model more macroscopic length and time scales.&lt;/p&gt;&#xA;" OwnerUserId="14566" LastEditorUserId="418" LastEditDate="2016-11-06T07:37:44.940" LastActivityDate="2016-11-06T07:37:44.940" Title="Modeling of photocatalytic and photoelectric devices on a macroscopic device level" Tags="&lt;electrochemistry&gt;&lt;computational-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-18T20:48:26.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="34636" PostTypeId="1" CreationDate="2015-08-02T16:17:48.500" Score="15" ViewCount="2569" Body="&lt;p&gt;Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are some compounds hygroscopic and others are not? What is it about their structures that allows them to hold water? Is it due to the atoms being arranged in a certain structure due to bonding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you tell if a compound is more hygroscopic than another?&lt;/p&gt;&#xA;" OwnerUserId="17883" LastEditorUserId="9961" LastEditDate="2015-08-02T19:57:33.880" LastActivityDate="2015-11-04T15:49:04.023" Title="What is the cause of hygroscopy?" Tags="&lt;water&gt;&lt;absorption&gt;" AnswerCount="2" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34637" PostTypeId="1" CreationDate="2015-08-02T16:32:01.750" Score="1" ViewCount="55" Body="&lt;p&gt;I have the following reactions from a book, but it is not explained how the different substances come into being. I don't know what happens to $\ce{3H2O (H3O)}$ in the second line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZaYy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZaYy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17884" LastEditorUserId="9961" LastEditDate="2015-08-02T17:25:22.147" LastActivityDate="2015-08-02T17:25:22.147" Title="How does the following work in Alkyne?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34638" PostTypeId="1" AcceptedAnswerId="34639" CreationDate="2015-08-02T16:44:35.567" Score="10" ViewCount="8413" Body="&lt;p&gt;Do aromatic dienes undergo Diels-Alder reactions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the $\pi$ bonds in aromatic compounds are not as localised as they are in other compounds, such as aliphatic dienes. Therefore, the dienophile shouldn't be as willing to go to these compounds. Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="4870" LastEditorUserId="4231" LastEditDate="2015-08-02T18:37:43.670" LastActivityDate="2019-11-15T12:13:02.750" Title="Do aromatic dienes undergo the Diels-Alder reaction?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34639" PostTypeId="2" ParentId="34638" CreationDate="2015-08-02T17:06:16.787" Score="15" Body="&lt;p&gt;Yes, you are correct.  Aromatic compounds are stabilized by their resonance energy. This means that the activation energy for reaction of an aromatic compound will typically be greater than that for a similar reaction of a non-aromatic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, aromatics will react as the diene component in a Diels-Alder reaction if &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the dienophile is extremely reactive (&lt;em&gt;o&lt;/em&gt;-benzyne for example)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TOKFh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TOKFh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if the aromatic system is destabilized (for example by steric interactions or structural constraints) so that the resonance stabilization is decreased&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X3MaQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3MaQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;or if the aromatic system is large enough that loss of one aromatic ring does not significantly destabilize the entire system.  Anthracene is such an example where the middle ring will react &lt;em&gt;via&lt;/em&gt; a Diels-Alder reaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rOXyP.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOXyP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://organicsynthesisinternational.blogspot.com/2014/12/preparation-of-910-dihydroanthracene.html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-02T19:15:29.260" LastActivityDate="2015-08-02T19:15:29.260" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34640" PostTypeId="2" ParentId="34514" CreationDate="2015-08-02T18:08:35.977" Score="0" Body="&lt;p&gt;Wire nuts are a considerably unfavorable means of connecting the house wiring to stubs for what I presume are your receptacles. It would be far safer for you to purchase the Aluminum compatible receptacles and use the binding screws for the connections. Wire nuts provide too little contact area and offer no heat dissipation ability to deal with the heat generated by the bimetallic interface of the aluminum/copper joint.&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-08-02T18:08:35.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34641" PostTypeId="2" ParentId="4331" CreationDate="2015-08-02T18:12:56.873" Score="0" Body="&lt;p&gt;Also don't rule out thermal gradients at the points where the probes/wiring are attached to the graphite strips. You can be measuring thermal electric effects as well.&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-08-02T18:12:56.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34642" PostTypeId="2" ParentId="7854" CreationDate="2015-08-02T18:19:10.310" Score="4" Body="&lt;p&gt;As pointed out in the other answer, platinum makes the oxidation of hydrogen gas to protons occur faster at any given potential, which in general allows you to get more energy out of the system at a given power level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &lt;strong&gt;why&lt;/strong&gt; this happens, it has to do with how platinum metal binds to hydrogen atoms. Hydrogen gas (H2) adsorbs (&quot;sticks&quot;) to platinum metal, which causes the H-H bond to break as two H-Pt bonds form. This makes it easier to remove an electron from each H atom, generating two H+ from H2. The thing that makes platinum such a good catalyst for this process is that it binds hydrogen gas &lt;strong&gt;strongly enough, but not too strongly&lt;/strong&gt;. Metals that have weaker interactions with hydrogen gas will not be as effective because the first step won't happen often enough (adsorption of hydrogen gas) and metals that bind hydrogen gas too strongly will not release the products, also resulting in lower rate. &lt;/p&gt;&#xA;" OwnerUserId="17868" LastActivityDate="2015-08-02T18:19:10.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34643" PostTypeId="2" ParentId="34583" CreationDate="2015-08-02T22:43:47.573" Score="4" Body="&lt;ol&gt;&#xA;&lt;li&gt;Let the reactant be $S$ and the product be $P$.  Let $p$ be the dimenionless product concentration, $\frac{P}{S_0}$.  First order kinetics:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P}{S_0} = p = 1- e^{-k t}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Slope of first-order kinetics:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dp}{dt} = k e ^{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Slopes and angles.  One confusing thing about this problem is the definition of $\theta$ is different than what you usually see in math class.  It's the angle with respect to the $y$ axis, not the $x$ axis.  As you note, if the $y$ angle is 30 degrees, then the $x$ angle is 60 degrees, meaning the slope is $\tan~60^{\circ} = \sqrt{3}$.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Units problem.  However there is a big units problem in this question.  The slope is $\sqrt{3}$ what?  Gigamolar per nanosecond?  Zeptomolar per eon?  The &lt;em&gt;angle&lt;/em&gt; of the tangent curve will change depending on the units used to plot the data.  If you use nanoseconds as the time unit, the angle of the tangent curve will be very flat, but conversely if eons is the time unit, the the angle will be very sharp.  The question does not explicitly say what units are used in the plot.  And the expression for the derivative of a first-order reaction shows that the slope should have units of concentration per time, but $\sqrt{3}$ is dimensionless.  So there is not really enough information to solve the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dicey assumptions about units.  Let's assume that the time unit is minutes.  Let's further assume that the $y$ axis plots a dimensionless concentration, i.e., $p$. Then the derivative (slope) is $-k e^{-k t}$ and apparently has a value of $\sqrt{3}~\mathbf{{min}^{-1}}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question is asking for the rate, so you are done.  But there are a lot of weird assumptions in this problem.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-02T22:56:18.713" LastActivityDate="2015-08-02T22:56:18.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34644" PostTypeId="2" ParentId="34400" CreationDate="2015-08-03T00:21:49.943" Score="3" Body="&lt;p&gt;Most amide coupling agents are suitable for esters. Feast your eyes on: &lt;a href=&quot;http://www.chtf.stuba.sk/~szolcsanyi/education/files/Organicka%20chemia%20II/Prednaska%208_Aminokyseliny%20a%20peptidy/Doplnkove%20studijne%20materialy/Amide%20bond%20formation/Amide%20bond%20formation%20-%20beyond%20the%20myth%20of%20coupling%20reagents.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amide bond formation - beyond the myth of coupling reagents&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lists old favourites such as CDI, carbodiimides, etc. A lot of it is overkill. The water soluble carbodiimides might be preferred for ease of urea removal later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The likes of CDI or $\ce{Ac2O/DMAP}$ can react with your phenol when over-stoichiometric amounts are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatives could be: boronic acid derivatives e.g.  phenylboronic acid catalysis; scandium or yttrium triflate catalysts (all requiring water removal, but a Dean-Stark could be used if a water immiscible solvent is used); dimethyl oxalate (already suggested)  but in conjunction with Takasago's ZnTAC24 catalyst. MeOH removal by distillation is required, which is quite trivial for such a volatile component.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorUserId="16683" LastEditDate="2017-01-08T14:36:25.260" LastActivityDate="2017-01-08T14:36:25.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34645" PostTypeId="1" AcceptedAnswerId="34646" CreationDate="2015-08-03T01:03:57.317" Score="6" ViewCount="2668" Body="&lt;p&gt;I am looking to produce hydrogen gas in a small amount with supplies I currently have. Most of the methods I have seen involve reacting zinc or magnesium with $\ce{HCl}$ to obtain the gas, but I do not have zinc or magnesium at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why aluminum cannot be used just as effectively as zinc? According to the reactivity series, aluminum should be even more reactive than zinc.&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="12860" LastEditDate="2015-10-30T19:03:29.280" LastActivityDate="2018-04-12T18:14:42.700" Title="Can aluminum and hydrochloric acid be used to generate hydrogen gas?" Tags="&lt;acid-base&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34646" PostTypeId="2" ParentId="34645" CreationDate="2015-08-03T01:05:33.457" Score="6" Body="&lt;p&gt;Yes. See &lt;a href=&quot;https://www.youtube.com/watch?v=LaaORZ0ofCY&quot; rel=&quot;noreferrer&quot;&gt;this video for a demonstration&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Aluminium_chloride&quot; rel=&quot;noreferrer&quot;&gt;reaction is&lt;/a&gt;:&#xA;$$\ce{ 2Al (s) + 6 HCl (aq) -&amp;gt; 2AlCl3 (s) + 3H2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can capture the hydrogen and proceed with your experiments.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-03T01:18:01.950" LastActivityDate="2015-08-03T01:18:01.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34648" PostTypeId="2" ParentId="34627" CreationDate="2015-08-03T05:11:36.527" Score="2" Body="&lt;p&gt;well figured out the answer, I had to take the correct value of Z from the database. The difference in values was because of the incorrect value of Z (which was actually 8) . When you use Z =8 , the two solution yield 'approximately' equal results.&lt;/p&gt;&#xA;" OwnerUserId="16866" LastActivityDate="2015-08-03T05:11:36.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34649" PostTypeId="1" AcceptedAnswerId="34662" CreationDate="2015-08-03T05:55:20.607" Score="2" ViewCount="3057" Body="&lt;p&gt;I am researching this reaction for my chemistry assessment. Various sources I have dug up state different answers and opinions on what the stoichiometry is and what the products are, sometimes giving no answer at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one source, sulfuric acid is a catalyst that speeds up the reaction while in another source, sulfuric acid is one of the reactants, and is therefore part of the stochiometry. The final products also vary from source to source, and include pure carbon or potassium chloride in addition to carbon dioxide and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone clarify the subject for me?&lt;/p&gt;&#xA;" OwnerUserId="17889" LastEditorUserId="865" LastEditDate="2015-08-03T12:28:56.920" LastActivityDate="2015-08-03T20:37:49.150" Title="What is the stochiometric equation of the sucrose and potassium chlorate reaction with sulfuric acid?" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34650" PostTypeId="1" AcceptedAnswerId="34655" CreationDate="2015-08-03T06:51:05.743" Score="3" ViewCount="1051" Body="&lt;p&gt;I am currently conducting an experiment into the glucose content of certain sports drinks. However, I have hit a roadblock. A majority of the drinks that I intend to test, contain sucrose. I am thinking that by adding &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0M~ HCl}$&lt;/span&gt;, to say, &lt;span class=&quot;math-container&quot;&gt;$\pu{100mL}$&lt;/span&gt; of solution (sports drink), it will decompose the sucrose into its constituents. I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the coloring from it before placing in the spectroscope). As a high school student I only have limited resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there is 10 grams of sucrose per &lt;span class=&quot;math-container&quot;&gt;$\pu{100mL}$&lt;/span&gt; of solution, what volume of &lt;span class=&quot;math-container&quot;&gt;$\pu{1.0M~ HCl}$&lt;/span&gt; would be required to completely break apart sucrose into glucose and fructose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I have to use stoichiometry to find the amount? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I know I have &lt;span class=&quot;math-container&quot;&gt;$\ce{C12H22O11 + HCl}$&lt;/span&gt;, What happens to the &lt;span class=&quot;math-container&quot;&gt;$\ce{Cl}$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; when fructose and glucose is formed?&lt;/p&gt;&#xA;" OwnerUserId="17871" LastEditorUserId="83758" LastEditDate="2020-03-20T15:49:22.820" LastActivityDate="2020-03-20T15:49:22.820" Title="How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;stoichiometry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34651" PostTypeId="2" ParentId="28995" CreationDate="2015-08-03T10:15:17.573" Score="5" Body="&lt;p&gt;In general, you cannot simply subtract electrode potentials like this to find the potential of another half-reaction. They have to be weighted by $n$ (the stoichiometric coefficient of $\ce{e-}$). In cases where your method works, it is only because $n$ is coincidentally the same for both half-reactions you are combining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foolproof way is always to convert electrode potentials to Gibbs free energy changes:&#xA;$$\Delta G^\circ = -nFE^\circ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 8H+ + 5e- -&amp;gt; Mn^2+ + 4H2O}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-5)(96485 \text{ C mol}^{-1})(1.51 \text{ V}) = -728.4 \text{ kJ mol}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $\Delta G^\circ$ for the half-reaction $\ce{MnO4- + 4H+ + 3e- -&amp;gt; MnO2 + 2H2O}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-3)(96485 \text{ C mol}^{-1})(1.7 \text{ V}) = -492.1 \text{ kJ mol}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtracting the second equation from the first, you will find that $\Delta G^\circ$ for the half-reaction of interest ($\ce{MnO2 + 4H+ + 2e- -&amp;gt; Mn2+ + 2H2O}$) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(-728.4) - (-492.1) = -236.3 \text{ kJ mol}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E^\circ = -\frac{-236.3 \text{ kJ mol}^{-1}}{(2)(96485 \text{ C mol}^{-1})} = 1.22 \text{ V.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume some rounding errors led to the slight discrepancy between my answer and the answer you want. You may leave out the Faraday constant in the calculations once you are more familiar with them; basically that means you work with the quantity $nE^\circ$ instead of $\Delta G^\circ$. The answer should be the same since they are directly proportional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information you can refer to the redox chapters in any inorganic chemistry textbook (Shriver/Atkins, Housecroft, etc.). These calculations are dealt with under the sections discussing Latimer and Frost diagrams.&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-08-03T10:15:17.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34652" PostTypeId="2" ParentId="4331" CreationDate="2015-08-03T10:43:16.813" Score="0" Body="&lt;p&gt;What is the input impedance of your voltmeter? How much DC current does it pass to measure a potential?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, the input impedance is infinite and the DC current is zero, but in practice, this isn't the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the NaCl solution with graphite electrodes, any DC current has to pass through electrolysis or impurities. This will lead to the behavior you are seeing with the impurities making the potential fluctuate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are you really after?&lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-08-03T10:43:16.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34653" PostTypeId="2" ParentId="34628" CreationDate="2015-08-03T10:50:05.227" Score="5" Body="&lt;p&gt;As the comment indicated, it is probably equilibrating. When you are discharging the battery, you are causing the materials in the electrodes and the electrolyte to rearrange. The surface of the electrodes are where the reactions are happening, so the products of the discharge reactions tend to accumulate at or near the surfaces of electrodes. This causes the measured potential to go lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the discharge stops, the contents start to homogenize within the cell bringing the discharge products away from the electrodes and bringing the material from the charged state closer to the electrodes. This in turn increases the measured potential. &lt;/p&gt;&#xA;" OwnerUserId="13355" LastActivityDate="2015-08-03T10:50:05.227" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34655" PostTypeId="2" ParentId="34650" CreationDate="2015-08-03T11:11:08.303" Score="4" Body="&lt;p&gt;The hydrolysis of sucrose is acid-catalysed, so HCl will not be in the chemical equation (it will be listed above the arrow). In theory, this means that you could have little HCl, as it will never be used up and will keep reacting to decompose all of your sucrose. So it isn't so important how much you use, but you should still use enough that it won't take ages to decompose all the molecules. If you're interested you could look up mechanisms for acid catalysis, but you wouldn't be expected to know this at high school level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the chemical equation for the decomposition, you can add up the number of elements of fructose + glucose and compare to sucrose. You will find that two hydrogens and one oxygen is needed, so it is simply a hydrolysis reaction (adding a water molecule).&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2015-08-03T11:11:08.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34656" PostTypeId="1" CreationDate="2015-08-03T12:44:37.433" Score="6" ViewCount="4813" Body="&lt;p&gt;When you calculate the solubility product constant for a salt and one of its hydrates, you sometimes get drastically different numbers! I don't know how this can be because they both dissociate into the same ions. For example, sodium sulfate heptahydrate $\ce{Na2SO4.(H2O)7}$ is more soluble than anhydrous sodium sulfate $\ce{Na2SO4}$. If you work through the math, the discrepancy in solubility is not completely explained by the larger mass of the heptahydrate (see calculations below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can explain the differences in solubility by assuming that the water molecules within the crystal structure of sodium sulfate heptahydrate lower the enthalpy of crystallization, which makes solvation energetically more favorable than the anhydrous salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me is the logical series of events that would happen when you add the solid hydrate into solution. The hydrate will dissolve into $\ce{Na+}$ and $\ce{SO4^2-}$ions. The concentration of those ions will increase until the solution is saturated (with respect to the anhydrous salt). Anhydrous $\ce{Na2SO4}$ will begin to precipitate out, keeping the ion concentrations below $K_\mathrm{sp}(\ce{Na2SO4})$ The hydrate will dissolve completely, because the solution will never reach saturation (with respect to the heptahydrate). How, then, was the solubility of the hydrate ever experimentally determined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using $\ce{Na2SO4}$ as an example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility of anhydrous $\ce{Na2SO4}$ = $\frac{4.76\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)&lt;br&gt;&#xA;Solubility of $\ce{Na2SO4.(H2O)7}$ = $\frac{19.5\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating saturated concentration:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility of anhydrous $\ce{Na2SO4}$ = $$\frac{4.76\ \mathrm g\ \ce{Na2SO4}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4}}{142.04\ \mathrm g\ \ce{Na2SO4}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.335\ \mathrm M\ \ce{Na2SO4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solubility of $\ce{Na2SO4.(H2O)7}$ = $$\frac{19.5\ \mathrm g\ \ce{Na2SO4.(H2O)7}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4.(H2O)7}}{268.1491\ \mathrm g\ \ce{Na2SO4.(H2O)7}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.727\ \mathrm M\ \ce{Na2SO4.(H2O)7}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating $K_\mathrm{sp}$:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anhydrous $\ce{Na2SO4}$: $$\ce{Na2SO4 &amp;lt;=&amp;gt;2Na+ + SO4^2-}$$  $$K_\mathrm{sp}(\ce{Na2SO4})=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.335\times 2\right)^2\times\left(0.335\right)=0.150$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Na2SO4.(H2O)7}$: $$\ce{Na2SO4.(H2O)7 &amp;lt;=&amp;gt;2Na+ + SO4^2- +7H2O}$$  $$K_\mathrm{sp}\left(\ce{Na2SO4.(H2O)7}\right)=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.666\times 2\right)^2\times\left(0.666\right)=1.18$$  &lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorDisplayName="user7951" LastEditDate="2015-08-16T12:27:01.740" LastActivityDate="2015-08-16T12:27:01.740" Title="How are solubilities of hydrate salts determined when competing equilibria are unavoidable?" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="8" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34657" PostTypeId="1" CreationDate="2015-08-03T12:53:45.070" Score="3" ViewCount="475" Body="&lt;p&gt;How was the formula (3) for viscosity index &lt;span class=&quot;math-container&quot;&gt;$\mathrm{VI}$&lt;/span&gt; derived in section &lt;em&gt;5. Procedure A—Oils of Viscosity Index Up to and Including 100&lt;/em&gt; [&lt;a href=&quot;https://doi.org/10.1520/D2270-04&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;]?&#xA;Specifically with reference to the reference oils considered and the table for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{VI} = \frac{L - U}{L - H} \times 100$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;D02 Committee. ASTM D2270 - 04 Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100 °C; ASTM International. DOI: &lt;a href=&quot;https://doi.org/10.1520/D2270-04&quot; rel=&quot;nofollow noreferrer&quot;&gt;10.1520/D2270-04&lt;/a&gt;. (&lt;a href=&quot;http://www.ptplab.net/upfile/201402/17/112156706.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17892" LastEditorUserId="41328" LastEditDate="2020-04-05T08:23:02.057" LastActivityDate="2020-05-05T09:07:15.713" Title="ASTM D2270 Viscosity Index Formula/Table derivation" Tags="&lt;viscosity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34660" PostTypeId="1" CreationDate="2015-08-03T13:34:31.657" Score="6" ViewCount="292" Body="&lt;p&gt;How is this hydride shift reaction type called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1PIzh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1PIzh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's not one of following types:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cycloaddition (Diels Alder)&lt;/li&gt;&#xA;&lt;li&gt;electrocyclic ring opening&lt;/li&gt;&#xA;&lt;li&gt;sigmatropic rearrangement&lt;/li&gt;&#xA;&lt;li&gt;ene reaction&lt;/li&gt;&#xA;&lt;li&gt;cheleotropic reaction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore I'm thinking what else can been the name of the observed reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ring opening reaction with hydride shift has been taken from &lt;em&gt;Heterocycles&lt;/em&gt; &lt;strong&gt;2004,&lt;/strong&gt; &lt;em&gt;64&lt;/em&gt; (1), 447–461 (&lt;a href=&quot;http://doi.org/10.3987/COM-04-S(P)48&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.3987/COM-04-S(P)48&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2017-04-08T23:25:38.937" LastActivityDate="2017-04-08T23:25:38.937" Title="How to classify the hydride shift in the following diazacyclopenta[cd]pentalene system?" Tags="&lt;organic-chemistry&gt;&lt;pericyclic&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34661" PostTypeId="1" AcceptedAnswerId="34672" CreationDate="2015-08-03T13:53:31.460" Score="2" ViewCount="1223" Body="&lt;p&gt;The textbooks starts with the phenomena that double helix is an expression of the rules of chemistry. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hydrogen bonds are important in determining the formation of &lt;strong&gt;specific&lt;/strong&gt;&#xA;   base pairs in the double helix. However in single staranded DNA, the hydrogen bond donors and acceptors are exposed to solution and can form hydrogen bonds with water molecules. &#xA;  When two single strands come together, these hydrogen bonds with water are broken and new hydrogen bonds between the bases are formed. Because the number of hydrogen bonds broken is the same as the number formed, these hydrogen bonds do not contribute substantially to driving the overall process of double helix-formation. However, they contribute greately to the &lt;strong&gt;specificity&lt;/strong&gt; of binding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I see the point that some neucleotides which form stable links are the complimentary bases (purine and pyramidine). Some bases do not form stable links. I did search my question online, I am not able to get near anything that's directly related or comprehensible at my level. So following is what I do not understand. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suppose two bases that cannot form Watson-Crick base pairs are brought together. Hydrogen bonds with water must be broken as the bases come into contact. &lt;strong&gt;Because the bases are not complementary in structure&lt;/strong&gt;, not all of these bonds can be simultaneously replaced by hydrogen bonds between the bases. Thus, the formation of a double helix between noncomplementary sequences is disfavoured.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the specificity mentioned here? What does it mean by non complementary in structure? An example is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7450" LastActivityDate="2015-08-19T11:52:16.530" Title="Why is the formation of double helix between non complementary sequences disfavoured?" Tags="&lt;biochemistry&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34662" PostTypeId="2" ParentId="34649" CreationDate="2015-08-03T13:54:55.413" Score="3" Body="&lt;p&gt;In &lt;a href=&quot;http://www.researchgate.net/profile/Todd_Silverstein/publication/231264372_Sugar_Dehydration_without_Sulfuric_Acid_No_More_Choking_Fumes_in_the_Classroom!/links/0fcfd50f0a8b8c9ed2000000.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; [&lt;em&gt;Sugar Dehydration without Sulfuric Acid: No More Choking in the Classroom!&lt;/em&gt; T.P. Silverstein and Y. Zhang. J. Chem. Ed. &lt;strong&gt;75&lt;/strong&gt; (6) 748-749 (1998)], the authors outline two reactions involving sucrose and sulfuric acid. The first one does not include $\ce{KClO3}$ and utilizes a significant amount of $\ce{H2SO4}$ in order to fuel the dehydration of sucrose: this is not the reaction you're interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reaction detailed herein makes use of a very small amount of sulfuric acid to start the reaction (a drop is the commonly-cited amount): the sucrose is oxidized by $\ce{KClO3}$. This is the reaction you want to examine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors give the reaction stoichiometry as&#xA;$$\ce{C12H22O11(s) + 8KClO3(s) \xrightarrow{H2SO4} 12CO2(g) + 11H2O(l) + 8KCl(s) + heat}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They note that the fumes from the reaction contain $\ce{Cl2, ClO}$, and $\ce{ClO2}$, but the amounts are not quantified. I suspect (but don't know for sure) that the amounts are finite but negligible. And even with the scant amount of sulfuric acid used here, I think you're safe going with the reaction as presented above.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-03T20:37:49.150" LastActivityDate="2015-08-03T20:37:49.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34663" PostTypeId="5" CreationDate="2015-08-03T13:58:06.280" Score="0" Body="&lt;p&gt;This tag should be applied to &lt;strong&gt;chemistry&lt;/strong&gt; questions which involve one of the human senses - &lt;a href=&quot;https://en.wikipedia.org/wiki/Olfaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Olfaction&lt;/a&gt; - or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Olfaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;odor&lt;/a&gt; behind them. The question must be about the chemical process or the chemicals that cause the specific odor behind a natural phenomenon. Under-researched &lt;a href=&quot;/questions/tagged/smell&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;smell&amp;#39;&quot; rel=&quot;tag&quot;&gt;smell&lt;/a&gt; questions might be identified as &lt;a href=&quot;/questions/tagged/homework&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;homework&amp;#39;&quot; rel=&quot;tag&quot;&gt;homework&lt;/a&gt; questions per community consensus, and dealt with in accordance to our &lt;a href=&quot;http://meta.chemistry.stackexchange.com/questions/141/&quot;&gt;homework&lt;/a&gt; policy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please check the &lt;a href=&quot;http://chemistry.stackexchange.com/help&quot;&gt;help center&lt;/a&gt; to ensure that your question is a good fit for Chemistry.StackExchange before posting it. The existence of this tag by no means indicates that all &lt;a href=&quot;/questions/tagged/smell&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;smell&amp;#39;&quot; rel=&quot;tag&quot;&gt;smell&lt;/a&gt;-related questions are on-topic.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="402" LastEditDate="2017-05-03T00:11:33.100" LastActivityDate="2017-05-03T00:11:33.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34664" PostTypeId="4" CreationDate="2015-08-03T13:58:06.280" Score="0" Body="For questions regarding the chemical processes or chemicals involved in distinguishable smells" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-08-03T14:28:45.677" LastActivityDate="2015-08-03T14:28:45.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34665" PostTypeId="1" CreationDate="2015-08-03T14:55:59.180" Score="6" ViewCount="1792" Body="&lt;p&gt;In crystalline solids, an atom may be displaced from its original position in order to create a vacancy in the original location. This is referred to as a Frenkel defect. Why does this defect result in an increase of the dielectric constant of the solid?&lt;/p&gt;&#xA;" OwnerUserId="17894" LastEditorUserId="865" LastEditDate="2015-08-03T15:44:27.107" LastActivityDate="2018-03-19T07:22:46.177" Title="Why does the dielectric constant of a solid increase due to a Frenkel defect?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34666" PostTypeId="1" AcceptedAnswerId="34668" CreationDate="2015-08-03T15:57:54.037" Score="1" ViewCount="898" Body="&lt;p&gt;For lithium, the electronic configuration is $\ce{(1s)^2(2s)^1}$. I tried to calculate $Z_{eff}$ for the $\ce{2s}$ orbital of the neutral atom as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z_{eff} = Z-\sigma$$&#xA;$$Z_{eff} = 3-(2\times0.3)$$&#xA;$$Z_{eff} = 2.4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Wikipedia article's table of effective nuclear charges (linked below) states that $Z_{eff}$ is 1.279 for the $\ce{2s}$ lithium orbital, which is not even close to my answer (above). So do Slater's rules not apply to lithium or I have done something wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wikipedia article:&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_nuclear_charge&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Effective_nuclear_charge&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="865" LastEditDate="2015-08-03T16:07:51.763" LastActivityDate="2015-08-03T16:19:55.423" Title="Do Slater's rules apply to neutral lithium (at least approximately)?" Tags="&lt;physical-chemistry&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34667" PostTypeId="2" ParentId="34634" CreationDate="2015-08-03T16:00:45.507" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But I am getting the correct answer only if I take &lt;span class=&quot;math-container&quot;&gt;$\lambda_2$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; times the given decay constant for the second reaction. Is this because of &lt;span class=&quot;math-container&quot;&gt;$\underline{2}A$&lt;/span&gt; on the reactant side instead of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations.  Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A],&#xA;$$&lt;/span&gt;&#xA;and that of reaction 2 is&lt;sup&gt;&amp;dagger;&lt;/sup&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A].&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; in the differential equation for reaction 2 is because two &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; nuclei are involved per &quot;unit&quot; of reaction progression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total rate of loss of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the sum of the above:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{\mathrm{d}[A]}{\mathrm{d}t} &#xA;= -\lambda_1[A] -2\lambda_2[A] &#xA;= -\left(\lambda_1 +  2\lambda_2\right)[A].&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&amp;dagger;&lt;/sup&gt; While it seems quite unusual to me to have a second-order reaction (&lt;span class=&quot;math-container&quot;&gt;$2A\longrightarrow B$&lt;/span&gt;) with a first-order rate constant (units of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{s}^{-1}$&lt;/span&gt;), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, simultaneously having &lt;span class=&quot;math-container&quot;&gt;$A \longrightarrow B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2A \longrightarrow B$&lt;/span&gt; processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. &lt;span class=&quot;math-container&quot;&gt;$2A \longrightarrow C$&lt;/span&gt; would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2020-02-26T15:09:54.580" LastActivityDate="2020-02-26T15:09:54.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34668" PostTypeId="2" ParentId="34666" CreationDate="2015-08-03T16:19:55.423" Score="4" Body="&lt;p&gt;You need to use the factor of 0.85 for each of the $\ce{1s}$ electrons:&#xA;$$Z_{eff} = 3 - (2\times 0.85) = 1.3$$&#xA;as per rule #2 in &lt;a href=&quot;https://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;If the group is of the [s p] type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The $\ce{1s}$ electrons both qualify as having &quot;a principal quantum number $n$ one less than that of the group&quot; (here, the principal quantum number of the group of the $\ce{2s}$ electron is 2). &lt;a href=&quot;http://chemistry.umeche.maine.edu/~amar/spring2011/zeff.html&quot; rel=&quot;nofollow&quot;&gt;Here is a nicely worked example for further edification&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-03T16:19:55.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34669" PostTypeId="2" ParentId="29023" CreationDate="2015-08-03T16:21:08.547" Score="2" Body="&lt;p&gt;I think both mechanisms are valid. As far as I know, both are supported by some experimental evidence. Byproducts, like alkanes (other than the expected one) and alkenes, can be explained by the free radical mechanism. On the other hand, the formation of Organometallic compounds can be verified by the ionic mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The essential thing to note is that the mechanism involves a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{S_N2}$&lt;/span&gt; step. And because of that sterically hindered alkyl halides, like tertiary alkyl halides, instead of giving a substitution product yield an elimination product and you get an alkene.&lt;/p&gt;&#xA;" OwnerUserId="4870" LastEditorUserId="61578" LastEditDate="2020-05-10T17:08:42.007" LastActivityDate="2020-05-10T17:08:42.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34670" PostTypeId="1" CreationDate="2015-08-03T16:30:21.907" Score="1" ViewCount="777" Body="&lt;p&gt;Do electrons in covalent bonds always necessarily obey the octet rule? &lt;em&gt;Always&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17897" LastEditorUserId="865" LastEditDate="2015-08-03T16:52:56.240" LastActivityDate="2015-08-05T14:09:10.240" Title="Do electrons in covalent bonds always obey the octet rule?" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-08-03T19:49:34.043" ContentLicense="CC BY-SA 3.0" />
  <row Id="34672" PostTypeId="2" ParentId="34661" CreationDate="2015-08-03T16:52:51.630" Score="2" Body="&lt;p&gt;If you take a look at the base pairings, as shown below, you can see that adenine and thymine form two hydrogen bonds when they are matched, while guanine and cytosine form three hydrogen bonds when they are matched. If you try to match an adenine with cytosine or guanine with thymine, the orientation of hydrogens and lone pairs will not line up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, guanine has two hydrogens and one lone pair available for hydrogen bonding, while cytosine has one hydrogen and two lone pairs available for hydrogen bonding. They fit! However, if you try to match adenine (with one hydrogen and one lone pair available for hydrogen bonding) with cytosine, the elements will not match or line up, resulting in an energetically less-favorable interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: A base on one strand of DNA requires another, specific base to complement the exposed hydrogens and lone pairs to form the most most energetically favorable interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zzpm8.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zzpm8.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-08-03T16:52:51.630" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34673" PostTypeId="2" ParentId="34670" CreationDate="2015-08-03T17:19:46.673" Score="-1" Body="&lt;p&gt;The elements in borons row do not always follow the octet rule as in the case of boron fluoride as boron is in the center of the periodic table and will usually only bond to fill up its outer shell with 6 electrons. Additionally, elements with can form expanded octets which can share more than 8 electrons such as phosphorous pentachloride.  Of course hydrogen only has 2 electrons to bond with so it is also an exception to the octet rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9YsF6.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9YsF6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pkXfN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pkXfN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="16106" LastEditDate="2015-08-05T14:09:10.240" LastActivityDate="2015-08-05T14:09:10.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34675" PostTypeId="1" AcceptedAnswerId="34677" CreationDate="2015-08-03T18:22:01.853" Score="4" ViewCount="836" Body="&lt;p&gt;Here's a very random introduction. I heard the term, &quot;dirty Sprite&quot; (it's some rap album/slang term) on a YouTube advertisement. I Googled &quot;dirty Sprite&quot;, and it's apparently a recreational drug: codeine syrup in Sprite lemon-lime soda. People allegedly get high off of it, meaning the opiates in codeine is metabolized properly. My question is, how does harsh acidity of the Sprite (pH of 3.4-3.8 if you trust Google) not interfere with the codeine structure and mitigate its metabolism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, this turned into a meta philosophical question: how does the acidity of the stomach, which is more acidic than Sprite, not interfere with not only codeine, but other substances? &lt;/p&gt;&#xA;" OwnerUserId="17900" LastActivityDate="2015-08-04T07:16:19.217" Title="How does the acidity of soft drinks not affect certain drugs, such as opiates? What about drugs and stomach acid?" Tags="&lt;acid-base&gt;&lt;drugs&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34677" PostTypeId="2" ParentId="34675" CreationDate="2015-08-03T19:32:18.137" Score="3" Body="&lt;p&gt;Many drugs are not very soluble in water what is a common property for a lot of organic substances, including Codeine. Therefor it is very common to turn them into their hydrochlorid which is simply $\text{Drug}\ce{.HCl}$. [&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrochloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;] E.g., Crystal Meth got it's name from the crystals that the hydrochloride forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1.7in$&lt;img src=&quot;https://i.stack.imgur.com/ant3sm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Crystal_Meth.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other examples are, e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Cocaine#Physical_and_chemical_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cocaine&lt;/a&gt;, &lt;a href=&quot;http://www.drugs.com/pro/cetirizine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cetirizine&lt;/a&gt;, &lt;a href=&quot;http://www.drugs.com/pro/hydroxyzine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydroxyzine&lt;/a&gt;, and &lt;a href=&quot;https://www.google.de/?gws_rd=ssl#q=hydrochloride+drugs&quot; rel=&quot;nofollow noreferrer&quot;&gt;many more&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your substance is either the hydrochloride or would turn into the hydrochloride in stomach there should not be any destructive influence on the structure of the drug. In fact it even promotes the absorption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the better solubility of the corresponding hydrochloride is a relatively big problem for those drugs that should not get absorbed in the stomach. They need to be coated to &quot;survive&quot; the stomach acidity to reach the intestine where they should get absorbed.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-08-04T07:16:19.217" LastActivityDate="2015-08-04T07:16:19.217" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34679" PostTypeId="1" AcceptedAnswerId="47417" CreationDate="2015-08-03T21:41:23.900" Score="7" ViewCount="1024" Body="&lt;p&gt;Today, I bought some graphite for my pencil. It is rated HB on the &lt;a href=&quot;http://pencils.com/hb-graphite-grading-scale/&quot;&gt;HB graphite scale&lt;/a&gt;, a measurement of the hardness of graphite. My question is how can there be different hardnesses of the same substance? Is there a difference between 4B graphite and 6H graphite on the molecular level? Are the different hardnesses not all arranged in the large, hexagonal carbon sheets that I had thought all graphite existed?&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2016-10-04T11:47:59.657" Title="Molecular differences in the HB graphite hardness scale" Tags="&lt;carbon-allotropes&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34683" PostTypeId="1" CreationDate="2015-08-04T01:07:03.773" Score="2" ViewCount="1120" Body="&lt;p&gt;I've got a few very very basic questions about ionic bonds and their role in batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumptions (correct me if I'm wrong):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Ionic bonds occur between a metal and a non-metal (the non-metal takes&#xA;the metal's electron to fill its valence shell). Therefore the metal atoms are left with a positive charge and the non-metal a negative charge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of this, the atom tries to stabilise its ionised charge and tries to bond with the oppositely charged atom.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Batteries take advantage of this chemical bonding to produce electricity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In a lead battery there is a PbO2, the electrolyte and Pb. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The electrolyte is a non-metal and therefore it tries to fill it's valence shell with the electrons from PbO2. This creates an overall positive charge in the PbO2 and a negative charge in the solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For some reason this negative charge is then transferred over to the Pb which gives it an overall negative charge. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Pb then wants to release this negative charge and pushes excess electrons which flow through a circuit to the PbO2 that is lacking electrons. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This leaves me with a few questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the electrolyte solution choose to take electrons from PbO2 (I assume it has something to do with the O2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the electrons are taken and there is a positive charge in the PbO2, and a negative charge in the solution, why don't they just bind together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought ionic bonds occurred when an metal had it's electrons taken from it. How then does the Pb receive additional electrons from the non-metal solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Pb does want to receive the extra electrons, why would it instantly try to push them away through the circuit? I'd understand if the whole negatively charged atom was trying to be pulled to the positively charged PbO2 - but from what I've read it tries to push the excess electrons, not the atom - so why even receive the electron if it doesn't want it?&lt;/p&gt;&#xA;" OwnerUserId="17905" LastEditorUserId="17905" LastEditDate="2015-08-04T03:34:46.220" LastActivityDate="2015-08-04T03:34:46.220" Title="Few basic questions about ionic bonding and batteries (lead)" Tags="&lt;electrochemistry&gt;&lt;bond&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;ionic-compounds&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34684" PostTypeId="2" ParentId="34199" CreationDate="2015-08-04T01:45:58.810" Score="2" Body="&lt;p&gt;The &quot;H&quot; part of H3C is to the outside of the center to stress that the carbon atom is oriented towards the other carbon atoms in the center.&lt;/p&gt;&#xA;" OwnerUserId="17907" LastActivityDate="2015-08-04T01:45:58.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34685" PostTypeId="1" CreationDate="2015-08-04T02:46:29.827" Score="4" ViewCount="993" Body="&lt;p&gt;I have to determine the products of the ammonium nitrate and copper nitrate. Is the below reaction correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NH4NO3 + Cu(NO3)2 -&amp;gt; Cu(NH3)4(NO3)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ammonium nitrate + copper nitrate → tetrammine copper(ii) nitrate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what would be the products of the other two reactions below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;copper nitrate + lead nitrate &lt;/li&gt;&#xA;&lt;li&gt;lead nitrate + ammonium nitrate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically, what happens when you react any two nitrates?&lt;/p&gt;&#xA;" OwnerUserId="17906" LastEditorUserId="5026" LastEditDate="2018-07-13T17:42:35.550" LastActivityDate="2018-07-13T17:42:35.550" Title="What happens when two nitrates react with each other?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34686" PostTypeId="1" CreationDate="2015-08-04T03:30:41.300" Score="5" ViewCount="8558" Body="&lt;p&gt;It is a common observation that a concentrated solution of hydrochloric acid kept in the lab fumes. What is the chemistry behind this phenomenon? Does $\ce{HCl}$ somehow react with the moisture in the air due to its hygroscopic nature?&lt;/p&gt;&#xA;" OwnerUserId="1888" LastActivityDate="2015-08-04T17:11:38.433" Title="Why does concentrated HCl fume?" Tags="&lt;inorganic-chemistry&gt;&lt;everyday-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34687" PostTypeId="1" AcceptedAnswerId="34694" CreationDate="2015-08-04T07:52:26.793" Score="5" ViewCount="527" Body="&lt;p&gt;Consider the reaction: $$\ce{H2O2(g) &amp;lt;=&amp;gt; 2OH^.(g)}\label{a}\tag{1}$$ to be at at equilibrium ($\ce{OH^.}$ is a radical). As the pressure is increased, the reaction moves towards forming $\ce{H2O2}$ as this direction reduces the amount of substance in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is well-established that the forward rate constant of $\ref{a}$ increases with increase in pressure, meaning forming more $\ce{OH}$. This seems to be in conflict to what the Le Chatelier's principle suggests. Could someone help me get my head around this?&lt;/p&gt;&#xA;" OwnerUserId="17915" LastEditorUserId="5026" LastEditDate="2018-03-27T10:05:56.110" LastActivityDate="2018-03-27T10:05:56.110" Title="Le Chatelier's Principle and Rate constants" Tags="&lt;equilibrium&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34688" PostTypeId="1" CreationDate="2015-08-04T08:38:49.883" Score="8" ViewCount="1202" Body="&lt;p&gt;You have a regular balloon, and you decide to heat it up. The gas particles inside the balloon will gain kinetic energy, and so, the gaseous particles will collide with the internal wall of the balloon with a higher frequency and more energy. This is what we call pressure, which will, as a result, increase. This is more commonly generalised to be Gay-Lussac's law, which formally states that the pressure of a gas is directly proportional to its absolute temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as pressure has increased, meaning that the gaseous particles are colliding with the internal wall of the balloon with a higher frequency and more energy, the balloon will begin to inflate, or in other words, the volume of the balloon will increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to contradict Boyle's law, which states that the pressure of a gas is inversely proportional to its volume. In the scenario I described above, it seems clear to me that it was the increase in frequency of the gaseous particles colliding with the internal wall of the balloon, that too with more energy, which altogether is known as an increase in pressure, that caused the balloon to inflate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I perfectly understand Boyle's law when the idea of a pump is used, where you can use a slider to change the volume of a gas and clearly see its inversely proportional relationship with the pressure of the gas, however, can't seem to understand what is incorrect in my logic, from which I seemingly have contradicted Boyle's law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a high-school student who just learnt about these laws, and out of curiosity decided to ask this question in hopes of clearing my doubts and fundamentally gaining a better understanding. Any help will be greatly appreciated, thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="17917" LastActivityDate="2015-08-04T23:38:37.890" Title="Ideal Gas Equation Contradiction" Tags="&lt;gas-laws&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2015-08-04T16:33:22.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="34689" PostTypeId="2" ParentId="34688" CreationDate="2015-08-04T09:16:47.350" Score="8" Body="&lt;p&gt;Your arguments do not contradict each other. You are trying to separate a process into two processes, which are coupled of course, but in first approximation this is a good way at looking it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's state our model system again: A balloon with the pressure $p_1$, the volume $V_1$ is heated from temperature $T_1$. As a result we see an inflated balloon. Let's call those variables with $2$ as an index. Of course we treat everything as ideal as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look at the coupled process as two separate processes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Isochoric heating, i.e. $V=\mathrm{const}$:&#xA;\begin{align}&#xA;&amp;amp;&amp;amp;p_1V&amp;amp;=n\mathcal{R}T_1\\ &#xA;&amp;amp;&amp;amp;p_2V&amp;amp;=n\mathcal{R}T_2\\ &#xA;\therefore &#xA;&amp;amp;&amp;amp;\frac{p_1}{p_2}&amp;amp;=\frac{T_1}{T_2}\\ &#xA;\therefore&#xA;&amp;amp;&amp;amp;p&amp;amp;\propto T\\ &#xA;\end{align}&#xA;That is &lt;a href=&quot;https://en.wikipedia.org/wiki/Gay-Lussac%27s_law&quot; rel=&quot;nofollow&quot;&gt;Gay-Lussac's law&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Isothermal expansion, ie. $T=\mathrm{const}$&#xA;\begin{align}&#xA;&amp;amp;&amp;amp;p_1V_1&amp;amp;=n\mathcal{R}T\\ &#xA;&amp;amp;&amp;amp;p_2V_2&amp;amp;=n\mathcal{R}T\\ &#xA;\therefore &#xA;&amp;amp;&amp;amp;\frac{p_1}{p_2}&amp;amp;=\frac{V_1}{V_2}\\ &#xA;\therefore&#xA;&amp;amp;&amp;amp;p&amp;amp;\propto \frac{1}{V}\\ &#xA;\end{align}&#xA;That is &lt;a href=&quot;https://en.wikipedia.org/wiki/Boyle%27s_law&quot; rel=&quot;nofollow&quot;&gt;Boyle's law&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can also look at the whole process as the pressure stays the same while the balloon is heated. That then is an isobaric process, i.e. $p=\mathrm{const}$:&#xA;\begin{align}&#xA;&amp;amp;&amp;amp;pV_1&amp;amp;=n\mathcal{R}T_1\\ &#xA;&amp;amp;&amp;amp;pV_2&amp;amp;=n\mathcal{R}T_2\\ &#xA;\therefore &#xA;&amp;amp;&amp;amp;\frac{V_1}{V_2}&amp;amp;=\frac{T_1}{T_2}\\ &#xA;\therefore&#xA;&amp;amp;&amp;amp;V&amp;amp;\propto T\\ &#xA;\end{align}&#xA;That is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Charles%27s_law&quot; rel=&quot;nofollow&quot;&gt;Charles' law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in reality there will be a mixture happening as the rubber of the balloon will expand, but the pressure that is possible is dependent on the material of it. In combination you will arrive at:&#xA;\begin{align}&#xA;&amp;amp;&amp;amp;p_1V_1&amp;amp;=n\mathcal{R}T_1\\ &#xA;&amp;amp;&amp;amp;p_2V_2&amp;amp;=n\mathcal{R}T_2\\ &#xA;\therefore &#xA;&amp;amp;&amp;amp;\frac{p_1V_1}{T_1}&amp;amp;=\frac{p_2V_2}{T_2}\\ &#xA;\therefore&#xA;&amp;amp;&amp;amp;\frac{pV}{T}&amp;amp;\propto \mathrm{const}\\ &#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="4945" LastEditDate="2015-08-04T23:38:37.890" LastActivityDate="2015-08-04T23:38:37.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34691" PostTypeId="2" ParentId="34688" CreationDate="2015-08-04T11:02:50.023" Score="3" Body="&lt;p&gt;There's no contradiction because Gay-Lussac's law only applies if the volume is held constant and Boyle's law only applies if the temperature is held constant. In the balloon example, both the volume and temperature change so neither law applies.&lt;/p&gt;&#xA;" OwnerUserId="7412" LastActivityDate="2015-08-04T11:02:50.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34693" PostTypeId="2" ParentId="32450" CreationDate="2015-08-04T13:48:31.887" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;a comparison without actual data should be possible. 1,3-butadiene is&#xA;  conjugated so it is the most stable compound and therefore its heat of&#xA;  hydrogenation should be the smallest&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, 1,3-butadiene is not the most stable of these compounds, either for the whole molecule or on a per double bond basis (see below).  The actual numbers (see table below) are too close, we need to look at some data in order to answer the question correctly.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The question asks us to select the compound with the smallest heat of&#xA;  hydrogenation per mole out of the following:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So would we consider each heat of hydrogenation &quot;per mole of hydrogen&quot;&#xA;  or &quot;per mole of the given compound&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is some data that will help answer the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8t2oC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8t2oC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://blog.sciencenet.cn/blog-94786-626892.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a &quot;&lt;strong&gt;per mole of given compound&lt;/strong&gt;&quot;, trans-2-butene has the lowest heat of hydrogenation, 27.6 kcal/mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a &quot;&lt;strong&gt;per mole of hydrogen&lt;/strong&gt;&quot;, 1,3-butadiene has a heat of hydrogenation between 28.2 (56.5/2) and 28.6 (57.1/2) kcal/mole, so &lt;em&gt;trans&lt;/em&gt;-2-butene still has the lowest heat of hydrogenation at 27.6 kcal/mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this particular list of compounds, &lt;strong&gt;it doesn't matter whether the question is asking per mole of hydrogen or per mole of given compound, the answer is the same, &lt;em&gt;trans&lt;/em&gt;-2-butene.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought &quot;trick question&quot;, but then later thought, not really.  The questioner did his homework and wanted the student to work out the answer both ways and see that since the you get the same answer whether you answer on a per mole compound or per mole hydrogen, you don't need to know whether the question is being asked on a per mole compound or per mole hydrogen basis.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-04T17:34:50.740" LastActivityDate="2015-08-04T17:34:50.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34694" PostTypeId="2" ParentId="34687" CreationDate="2015-08-04T13:59:42.343" Score="2" Body="&lt;p&gt;It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastActivityDate="2015-08-04T13:59:42.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34695" PostTypeId="1" CreationDate="2015-08-04T14:54:42.887" Score="1" ViewCount="471" Body="&lt;p&gt;I watched the Crash Course chemistry video about the electron. In the video Hank showed an example configuration of iron as $\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6}$ and explained that due to energy constraints, electrons go to the s orbital of the 4th shell before going into the d orbital of the 3rd shell and so they are written in this order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron&quot; rel=&quot;nofollow&quot;&gt;iron in Wikipedia&lt;/a&gt; and there the configuration is given as $\mathrm{[Ar](3d)^6 (4s)^2}$. I understand how it's shortened with argon, but why in Wikipedia is the d orbital of the 3rd shell written before the s orbital of the 4th shell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is preferred?&lt;/p&gt;&#xA;" OwnerUserId="17923" LastEditorUserId="4325" LastEditDate="2015-08-04T15:16:08.367" LastActivityDate="2015-08-04T15:16:08.367" Title="Confused about writing out electron configurations" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34696" PostTypeId="2" ParentId="34695" CreationDate="2015-08-04T15:11:30.447" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For a given configuration, the order of writing the orbitals is not&#xA;  completely fixed since only the orbital occupancies have physical&#xA;  significance. For example, the electron configuration of the titanium&#xA;  ground state can be written as either $\mathrm{[Ar] \, 4s^2 3d^2}$ or $\mathrm{[Ar] \, 3d^2 4s^2}$.&#xA;  The first notation follows the order based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule&quot; rel=&quot;nofollow&quot;&gt;Madelung rule&lt;/a&gt; for&#xA;  the configurations of neutral atoms; $\mathrm{4s}$ is filled before $\mathrm{3d}$ in the&#xA;  sequence $\ce{Ar, K, Ca, Sc, Ti}$. The second notation groups all orbitals&#xA;  with the same value of $n$ together, corresponding to the&#xA;  &quot;spectroscopic&quot; order of orbital energies that is the reverse of the&#xA;  order in which electrons are removed from a given atom to form&#xA;  positive ions; $\mathrm{3d}$ is filled before $\mathrm{4s}$ in the sequence $\ce{Ti^4+, Ti^3+, Ti^2+, Ti+, Ti}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_configuration&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on electron configuration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-08-04T15:11:30.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34697" PostTypeId="1" AcceptedAnswerId="34704" CreationDate="2015-08-04T15:40:05.353" Score="1" ViewCount="76" Body="&lt;p&gt;Can someone please explain why the carbon in 8-hydroxyquinoline is numbered 8, while in 1,10-phenanthroline the nitrogens are numbered 1 and 10? I'm counting carbon atoms and just cannot see why the hydroxy is in position 8 and nitrogen atoms are in positions 1 and 10.&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="865" LastEditDate="2015-08-04T15:45:28.140" LastActivityDate="2015-08-04T16:44:02.757" Title="Numbering heteroatoms and groups in benzene derivatives" Tags="&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34698" PostTypeId="1" CreationDate="2015-08-04T15:43:45.490" Score="8" ViewCount="202" Body="&lt;p&gt;I am trying to perform nucleation simulations with molecular dynamics. To prepare the metastable fluid, I am equilibrating at temperatures about 30 K above the melting temperature, and then &quot;quenching&quot; down to a sub cooled liquid state. For those familiar with lammps, the method I am using to perform the quench is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fix 1 all nve&#xA;fix 2 all viscous 0.01&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This just adds a drag force proportional to the velocity to suck heat out of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the nucleation simulation literature, I can't think of a time I have come across a paper where the author mentions an actual quench rate, or discusses whether the quench rate can have implications for the final results. Is anyone familiar with what an appropriate quench rate to prepare a metastable liquid, or whether I can produce spurious physical artifacts depending not quench rate? Currently for my water simulation, I am quenching from 300 K to 190 K in 50 steps with a 2 fs time step.&lt;/p&gt;&#xA;" OwnerUserId="16152" LastEditorUserId="16683" LastEditDate="2015-10-23T16:45:58.330" LastActivityDate="2015-10-23T16:45:58.330" Title="Quenching rates in nucleation MD simulations" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34699" PostTypeId="1" CreationDate="2015-08-04T15:56:15.060" Score="0" ViewCount="476" Body="&lt;p&gt;Here is the picture of monobasic potassium phosphate:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rIaybm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rIaybm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the picture of dibasic potassium phosphate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b9tFBm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b9tFBm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for the phosphorous-containing anions to further break down into actual phosphates and floating protons?&lt;/p&gt;&#xA;" OwnerUserId="17924" LastEditorUserId="16622" LastEditDate="2015-08-04T16:18:53.417" LastActivityDate="2015-08-04T16:18:53.417" Title="Is it possible for monobasic and dibasic potassium phosphate to complete break down into phosphates?" Tags="&lt;acid-base&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-05T14:27:30.890" ContentLicense="CC BY-SA 3.0" />
  <row Id="34700" PostTypeId="1" AcceptedAnswerId="34711" CreationDate="2015-08-04T16:06:58.523" Score="5" ViewCount="362" Body="&lt;p&gt;I used Hückel's method along with a Linear Combination of Atomic Orbitals (LCAO) to calculate an estimate for the orbital energies of cyclobutadiene ($\ce{C4H4}$) and butadiene ($\ce{C4H6}$). For butadiene, I ended up with the result&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \alpha \pm 1.62  \beta$ and $E = \alpha \pm 0.62 \beta$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha$ and $\beta$ are both negative (and are the Coulomb integral and the exchange integral, respectively). I'm fairly confident with the method and interpretation of these integrals so far, but I don't know how to use these results to &quot;guess and draw&quot; the shapes and phases of the molecular orbitals, and I'm also not sure how to tie in what $\alpha$ and $\beta$ represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand things, in these molecules, hybridisation of the orbitals occur and $\mathrm{sp^2}$ orbitals form, taking up three electrons from each carbon atom and leaving one electron left over in the third p orbital. This p orbital is the one I assume is portrayed in the below images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please give me a hint towards the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;relating the molecular orbital shapes/phases with the Coulomb integral (overlap of one carbon's electron wavefunction with a neighbouring carbon's potential) and the exchange integral (the overlap of two electrons' wavefunctions in the field of one of the carbons&lt;/li&gt;&#xA;&lt;li&gt;how to get any sort of information about the phases of the different orbitals. I understand that $E = \alpha + 2\beta$ will have the lowest energy, as both $\alpha$ and $\beta$ are negative, but how does that correspond to the phase configurations shown below? Why do the size of the circles in the first image vary?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I've been sent here from the Physics Stack Exchange, please be nice. :) Thanks for any help! This is my first post so please also let me know if the posting etiquette passes muster.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dt4N1.gif&quot; alt=&quot;Molecular orbitals of H4C6&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/y1jJn.png&quot; alt=&quot;Molecular orbitals of H4C6&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://postimg.org/image/ej0oupbh1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;First image&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;second image&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17925" LastEditorUserId="16622" LastEditDate="2015-08-04T16:42:25.350" LastActivityDate="2015-08-04T21:00:29.940" Title="Guess the shape and phase of the MOs of butadiene using LCAO and Hückel's method?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34701" PostTypeId="2" ParentId="34699" CreationDate="2015-08-04T16:07:56.033" Score="1" Body="&lt;p&gt;At basic conditions over pH 10, those &lt;em&gt;phosphorus-containing anions&lt;/em&gt; can get further deprotonated, as can easily be seen on the corresponding &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphate#Chemical_properties&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on phosphates.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphoric acid $\rightarrow$ Dihydrogen phosphate $\rightarrow$ Hydrogen phosphate $\rightarrow$ Phosphate&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-08-04T16:07:56.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34702" PostTypeId="1" CreationDate="2015-08-04T16:19:42.797" Score="1" ViewCount="190" Body="&lt;p&gt;I searched a lot for this and couldn't find any answers.  Is water softening a purely ratio-based equilibrium process, ie, more $\ce{Na+}$ cations on resin than there are $\ce{Ca^2+}$ cations in the water, causing most $\ce{Ca^2+}$ in solution to be displaced? Or do kinetic factors come into play, where at equilibrium $\ce{Ca^2+}$ cations will end up stuck on the resin over $\ce{Na+}$ cations because they are bigger and thus more restricted from leaving the resin due to steric reasons?&lt;/p&gt;&#xA;" OwnerUserId="17926" LastEditorUserId="4945" LastEditDate="2015-10-23T17:00:35.813" LastActivityDate="2015-10-23T17:00:35.813" Title="Do ion-exchange resins soften water through a kinetic equilibrium process?" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34703" PostTypeId="2" ParentId="34686" CreationDate="2015-08-04T16:33:33.973" Score="2" Body="&lt;p&gt;Hydrogen chloride ($\ce{H-Cl}$) is a gas at room temperature (boils already at −85.05 °C). The fumes you see is HCl escaping from the reaction mixture when the equilibrium is distorted due to e.g. heat. The more concentrated your HCl is, the more the equilibrium between the aqueous phase and the ambivalent air shifts towards the ambivalent air, creating fumes. These fumes are nasty. Rather not work too much with concentrated HCl (&gt; 30%) outside of the hood!&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-08-04T16:33:33.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34704" PostTypeId="2" ParentId="34697" CreationDate="2015-08-04T16:44:02.757" Score="1" Body="&lt;p&gt;Quaternary carbon atoms i.e. Those with no hydrogen atoms attached and in these examples, the ones that are common to both or all rings, are not counted. Try it with 1,8-dihydroxynaphthalene. Both OH groups should be pointing in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your examples, as you probably know, the heteroatom is numbered first. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-08-04T16:44:02.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34705" PostTypeId="1" CreationDate="2015-08-04T16:55:03.163" Score="3" ViewCount="110" Body="&lt;p&gt;This is regarding to the newly-discovered 2-D materials that can support electron flow, such as phosphorene and transition metal dichalcogenides. How do they conduct electricity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read that their electric properties change with the number of layers. For example, the band gap of black phosphorus changes from 0 - 2 eV with the number of layers. Why is this so? &lt;/p&gt;&#xA;" OwnerUserId="17927" LastEditorUserId="10185" LastEditDate="2015-08-04T17:10:01.940" LastActivityDate="2015-08-04T17:10:01.940" Title="Electrical properties of 2D materials" Tags="&lt;materials&gt;&lt;electricity&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34706" PostTypeId="2" ParentId="34686" CreationDate="2015-08-04T17:11:38.433" Score="4" Body="&lt;p&gt;In the case of $\ce{HNO3}$, according the abstract of &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/0095852263900552&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;$^\dagger$, nitric acid vapors act as nucleation sites for ambient water, causing the mists/fumes that are observed above concentrated nitric acid solutions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The gas phase reaction [forming nitric acid from other reagents] supplies $\ce{HNO3}$ nuclei on which the water vapor readily condenses to form strong nitric acid mist droplets.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://books.google.com/books?id=NtoEz62LlAQC&amp;amp;pg=PA591&amp;amp;lpg=PA591&amp;amp;dq=hydrochloric%20acid%20vapor%20nucleation&amp;amp;source=bl&amp;amp;ots=dhwUlwsDkJ&amp;amp;sig=2y_mskNmH6A45KOy3WLVUHl3-cg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CDwQ6AEwBGoVChMI2t2J2--PxwIVBDU-Ch2LIAEn#v=onepage&amp;amp;q=hydrochloric%20acid%20vapor%20nucleation&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt;$^\ddagger$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hydrochloric acid has thermodynamic properties rather similar to those of nitric acid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I would hypothesize that the physical chemistry phenomena of evaporation and hygroscopic nucleation are responsible for the fuming of $\ce{HCl}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\dagger$ Goyer, &lt;em&gt;J Colloid Sci&lt;/em&gt; &lt;strong&gt;18&lt;/strong&gt;(7), 616 (1963)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\ddagger$ &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0080420303&quot; rel=&quot;nofollow&quot;&gt;Kulmala and Wagner, &quot;Nucleation and Atmospheric Aerosols,&quot; 1996, p591.&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-08-04T17:11:38.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34707" PostTypeId="1" CreationDate="2015-08-04T17:26:16.327" Score="4" ViewCount="3885" Body="&lt;p&gt;I am looking for the less harmful chemical solution to serve as an etching solution. After many searches, my interest went to a vinegar/salt/hydrogen peroxide solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to ask then if the residue of this solution is harmful and if it is ok to pour it in the regular water waste system.&lt;/p&gt;&#xA;" OwnerUserId="17928" LastEditorUserId="16622" LastEditDate="2015-08-04T18:01:09.010" LastActivityDate="2018-09-12T18:30:11.727" Title="Reaction between a vinegar/salt/hydrogen peroxide mixture as an etching solution and copper" Tags="&lt;inorganic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34708" PostTypeId="1" CreationDate="2015-08-04T17:47:12.213" Score="2" ViewCount="424" Body="&lt;p&gt;&lt;a href=&quot;http://www.dicholding.com/en/diamond/general-informations/diamonds-mining&quot; rel=&quot;nofollow&quot;&gt;Diamond mining&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link says that diamonds will stick to &quot;anything oleaginous&quot; and that other materials slide down the fat that is applied to an inclined plane used in the mining apparatus.  I infer that the surface of the diamond interacts via dispersion forces and that other minerals are more polar.  Is this the case?  If so, would diamonds stick to wax or plastic?&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-08-04T18:20:47.280" Title="Why is it possible to use lipids to recover diamonds in mining?" Tags="&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34709" PostTypeId="2" ParentId="34707" CreationDate="2015-08-04T17:52:07.873" Score="5" Body="&lt;p&gt;The etching solution itself is fine to dispose of down the drain. However, if you use it to etch copper, the dissolved copper ions are quite toxic, and should not be disposed of through the sewer system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can bypass the problem by collecting the etch solution, and submerge aluminum foil. $$\ce{3Cu(CH3COO)2(aq) + 2Al(s)-&amp;gt;2Al(CH3COO)3(aq) +3Cu(s)}$$React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.&lt;/p&gt;&#xA;" OwnerUserId="17643" LastEditorUserId="17643" LastEditDate="2018-09-12T18:30:11.727" LastActivityDate="2018-09-12T18:30:11.727" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34710" PostTypeId="1" CreationDate="2015-08-04T17:57:07.870" Score="6" ViewCount="9100" Body="&lt;p&gt;I need clarity on saturated vapour pressure in a closed system at equilibrium. How does saturated vapour pressure relate to vapour pressure?&lt;/p&gt;&#xA;" OwnerUserId="17929" LastEditorUserId="865" LastEditDate="2015-08-04T17:58:55.583" LastActivityDate="2015-11-03T07:22:25.703" Title="Relationship between vapour pressure and saturated vapour pressure" Tags="&lt;physical-chemistry&gt;&lt;vapor-pressure&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34711" PostTypeId="2" ParentId="34700" CreationDate="2015-08-04T17:57:59.303" Score="5" Body="&lt;p&gt;The shape of the molecular orbitals is &lt;em&gt;not&lt;/em&gt; guessed, it is determined by the LCAO-MO expansion coefficients which one needs to find out in the first place. The hint is that an eigenvalue equation is usually solved in two steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We start by writing down the eigenvalue equation in the matrix form and proceed then by writing down and solving the &lt;em&gt;characteristic equation&lt;/em&gt; in which the secular determinant is equated with zero. By solving the characteristic equation we find the &lt;em&gt;eigenvalues&lt;/em&gt; (energies).&lt;/li&gt;&#xA;&lt;li&gt;Then we put eigenvalues (energies) back into the matrix eigenvalue equation and solve it for the coefficients (LCAO-MO expansion coefficients) that determine &lt;em&gt;eigenvectors&lt;/em&gt; (molecular orbitals).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, the second step was not simply done yet. Note that any scalar multiple of an eigenvector is also an eigenvector with the&#xA;same eigenvalue, so that eigenvectors are not unique. In solving the matrix eigenvalue equation one will find not the actual values of the LCAO-MO expansion coefficients, but rather some relationship between them which could be satisfied by a number of different coefficients. In principle one is free to choose any set of coefficients satisfying the relationship, and usually there is some obvious choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simplest case in which just two coefficient are related with each other through some ratio the obvious choice is to take some coefficient to be one, then the value of the second coefficient is nothing but the value of the ratio. One could also normalize the wave function then if desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that one can use symmetry to simplify the math. OP did not mention were the symmetry adapted linear combinations (SALCs) of atomic orbitals used at the first step, so I mention just in case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. Fully worked examples could be found in different places, for instance, &lt;a href=&quot;http://brouard.chem.ox.ac.uk/teaching/vallectures78.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (PDF, 150 KB).&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-08-04T21:00:29.940" LastActivityDate="2015-08-04T21:00:29.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34712" PostTypeId="2" ParentId="34708" CreationDate="2015-08-04T18:10:01.750" Score="2" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia piece on diamonds&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diamonds are naturally lipophilic and hydrophobic, which means the diamonds' surface cannot be wet by water but can be easily wet and stuck by oil. This property can be utilized to extract diamonds using oil when making synthetic diamonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It looks like the surface of diamond has dangling bonds terminated with hydrogen, so your non-polar hypothesis is correct, as applied to terminal moieties on the surface. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Material_properties_of_diamond&quot; rel=&quot;nofollow&quot;&gt;this article about the material properties of diamond&lt;/a&gt; where it is stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diamond surfaces are hydrophobic when the surface carbon atoms terminate with a hydrogen atom and hydrophilic when the surface atoms terminate with an oxygen atom or hydroxyl radical.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen and the behavior is moderately hydrophobic. This allows for separation from other minerals at the mine using the so-called &quot;grease-belt.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, would diamonds stick to wax or plastic?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I believe the answer is yes, but certainly factors such as the transport of diamond with grit in a slurry over/through a hydrophobic material such as the two you name might come into play: I think mixing the mined slurry with goop is more effective than sheeting it across a 2-D material, and a higher extract yield results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at higher temperatures and with other reactants, the surface properties can change (as stated in the linked article immediately above):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Treatment with gases or plasmas containing the appropriate gas, at temperatures of 450 °C or higher, can change the surface property completely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-04T18:20:47.280" LastActivityDate="2015-08-04T18:20:47.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34713" PostTypeId="1" CreationDate="2015-08-04T18:13:43.073" Score="6" ViewCount="488" Body="&lt;p&gt;A vial of pyrrole (that was occasionally exposed to air) changed its color while degrading from pale yellow to dark orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism of pyrrole degradation? What are the resulting products?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="16683" LastEditDate="2017-03-22T16:36:34.773" LastActivityDate="2017-03-22T16:36:34.773" Title="What is the degradation mechanism of pyrrole?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;identification&gt;&lt;heterocyclic-compounds&gt;" AnswerCount="0" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34714" PostTypeId="1" AcceptedAnswerId="34715" CreationDate="2015-08-04T18:54:36.073" Score="12" ViewCount="396" Body="&lt;p&gt;In the example calculation from the conformer search package &lt;a href=&quot;https://github.com/adrianasupady/fafoom/blob/master/examples/parameters_ff.txt&quot; rel=&quot;noreferrer&quot;&gt;fafoom&lt;/a&gt; is a mysterious pattern that I'd like to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SMILES code of the given example structure is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CC(=O)N[C@H](C(=O)NC)C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is a first not so hard pattern for the smart_cistrans pattern identifier&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C~[$(C=O)]-[$(NC)]~[C]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which tells the program that peptide bonds behave like cis-trans isomers. That is relatively easy to see but I don't really understand it to adapt it on further problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is another identifier for torsion angles that for me is even harder to understand:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[*]~[!$(*#*)&amp;amp;!D1]-&amp;amp;!@[!$(*#*)&amp;amp;!D1]~[*]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The [*]'s at the beginning and at the end look like defining any atom ... but the cursing signs in the middle don't tell me anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear reader, if you understand how those patterns are generated, I would ask you break them down piece for piece that I can use them afterwards on my own.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-08-05T07:28:05.860" LastActivityDate="2015-08-05T07:28:05.860" Title="Trying to understand SMILES(?) patterns" Tags="&lt;computational-chemistry&gt;&lt;software&gt;&lt;cheminformatics&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34715" PostTypeId="2" ParentId="34714" CreationDate="2015-08-04T21:49:26.297" Score="10" Body="&lt;p&gt;First off this is &lt;a href=&quot;http://www.daylight.com/dayhtml/doc/theory/theory.smarts.html&quot; rel=&quot;noreferrer&quot;&gt;SMARTS&lt;/a&gt; not SMILES.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of SMARTS as like a regular expression language for molecules (e.g., SMILES).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's break down some of those characters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;code&gt;~&lt;/code&gt; is &quot;any bond&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;!&lt;/code&gt; is &quot;not&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;$&lt;/code&gt; indicates a recursive SMARTS expression.&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;@&lt;/code&gt; indicates &quot;any ring bond&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;code&gt;$(*#*)&lt;/code&gt; means &quot;two atoms with a triple bond between them&quot;. The &lt;code&gt;!D1&lt;/code&gt; means and not an atom with one explicit bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken as a whole, &lt;code&gt;[*]~[!$(*#*)&amp;amp;!D1]&lt;/code&gt; means any atom with a bond to an atom that doesn't have just one explicit bond or a triple bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the pattern is trying to find a single bond in the middle with some rotatable side-groups:&#xA;&lt;code&gt;[*]~[...]-&amp;amp;[..]~[*]&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a very nice tool called &lt;a href=&quot;http://smartsview.zbh.uni-hamburg.de/smartsview/view&quot; rel=&quot;noreferrer&quot;&gt;SMARTS Viewer&lt;/a&gt; that can help to debug and construct SMARTS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xulsC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xulsC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like there are also now more polished commercial versions called SMARTStools through &lt;a href=&quot;http://www.biosolveit.de/SMARTStools/&quot; rel=&quot;noreferrer&quot;&gt;BioSolveIT&lt;/a&gt; including an editor.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-04T21:49:26.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34716" PostTypeId="1" CreationDate="2015-08-04T23:08:43.107" Score="1" ViewCount="2140" Body="&lt;p&gt;Why do ionic compounds have higher melting and boiling points despite having smaller radii?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them (and at further distances apart, a lower melting point is observed because there is less attraction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a question of ionic compounds &lt;em&gt;vs.&lt;/em&gt; covalent compounds, which is covered in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5394/why-is-the-melting-boiling-point-of-ionic-bond-higher-than-covalent-bond&quot;&gt;this question&lt;/a&gt;. To be clear, I am interested in the comparison of ionic compounds with ionic compounds.&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2017-06-02T04:34:58.830" Title="Why do ionic compounds have higher melting and boiling points despite having smaller radii?" Tags="&lt;ions&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34717" PostTypeId="1" CreationDate="2015-08-05T01:39:45.497" Score="24" ViewCount="25151" Body="&lt;p&gt;I distinctly remember a side-by-side comparison from a book where there are two nails submerged in water, in two beakers: one nail had a layer of oil on top of the water, and that nail didn't rust; the other did. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that oiling iron/steel products is supposed to prevent them from rusting: &#xA;&lt;a href=&quot;http://www.justanswer.com/home-improvement/0n1r2-does-olive-oil-prevent-rust.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.justanswer.com/home-improvement/0n1r2-does-olive-oil-prevent-rust.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://outdoors.stackexchange.com/questions/1404/can-oiling-keep-tools-from-rusting&quot;&gt;https://outdoors.stackexchange.com/questions/1404/can-oiling-keep-tools-from-rusting&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation given, I think, is that the oil somehow prevents oxygen from reacting with the metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't help but wonder why. Oxygen is a non-polar molecule, so it should actually dissolve better in oils (which are also non-polar) than in water (which is highly polar) and therefore allow more oxygen molecules to contact the iron (metal) surface in some sort of equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the conductivity of water and its ability to form electrochemical chains between metals is irrelevant in this thought experiment (i.e., a nail submerged in water in a glass). From what I understand, the other key item that has an effect on the formation of rust in water is that the oxygen molecules have better contact with the iron surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is supported by data: the concentration of oxygen in air is &lt;em&gt;far&lt;/em&gt; higher than that in water. We know from experience, though, that wet iron rusts way quicker than dry iron does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am reading it correctly, the same data shows higher oxygen concentrations in non-polar solvents than in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the data: &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd219.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nist.gov/data/PDFfiles/jpcrd219.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am reading it wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does oiling prevent rust?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: For the sake of clarity, let's say we use deionized water in a vacuum, drop in the nail, then put some oil (let's say it's caprylic acid) on it, and expose it to standard conditions. &lt;/p&gt;&#xA;" OwnerUserId="17932" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:01.370" LastActivityDate="2017-02-22T13:06:53.883" Title="How does oil on the surface of water prevent rust?" Tags="&lt;physical-chemistry&gt;&lt;electrochemistry&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="4" CommentCount="7" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34718" PostTypeId="2" ParentId="34710" CreationDate="2015-08-05T02:30:25.447" Score="2" Body="&lt;p&gt;If you're at equilibrium, you're saturated.   If you're saturated, you're at equilibrium.  The end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If things aren't at equilibrium, the situation is different.  Consider a dish of water in the middle of the Sahara desert.  The amount of water in the atomsphere in the Sahara desert is generally nonzero, say in the range of 1 $\mathrm g~{kg}^{-1}$.  Even though the water vapor pressure (even in the Sahara!) is nonzero, it is way less than &quot;saturated&quot;.  Therefore it isn't at equilibrium, and the pan of water quickly evaporates.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-08-05T02:30:25.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34720" PostTypeId="1" AcceptedAnswerId="34725" CreationDate="2015-08-05T03:54:46.433" Score="14" ViewCount="27718" Body="&lt;p&gt;We have to select the set of elements which will have almost the same atomic radii:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\ce{O, S, Se, Te}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Li, Be, B, C}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{Fe, Co, Ni, Cu}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I googled the atomic radii of the third set and found them to be extremely close to one another: all of them are nearly $\pu{125 pm}$ (according to empirical data from Wikipedia). Is there a specific reason for this, or is it just a coincidence? &lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="5026" LastEditDate="2018-04-22T09:11:30.700" LastActivityDate="2018-04-22T09:11:30.700" Title="Why are the atomic radii of iron, cobalt, nickel, and copper almost the same?" Tags="&lt;periodic-trends&gt;&lt;atomic-radius&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34721" PostTypeId="1" AcceptedAnswerId="34722" CreationDate="2015-08-05T03:58:33.240" Score="6" ViewCount="100" Body="&lt;p&gt;I came across a &lt;a href=&quot;http://www.kegg.jp/dbget-bin/www_bget?R02189&quot; rel=&quot;noreferrer&quot;&gt;reaction&lt;/a&gt; that looks as follows: &lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZkN19.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZkN19.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;br&gt; &#xA;In this reaction the polyphosphate is common in both the sides of the reaction. Does this act as enzyme? But from the website it seems that the enzyme is ADP-specific glucokinase(in the enzyme section click 2.7.1.63). Then what is the function of polyphosphate here?&lt;/p&gt;&#xA;" OwnerUserId="16341" LastActivityDate="2015-08-05T13:27:08.957" Title="Understanding a chemical reaction from kegg" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34722" PostTypeId="2" ParentId="34721" CreationDate="2015-08-05T04:07:50.747" Score="5" Body="&lt;p&gt;The $n$ of the polyphosphate should be decreased by 1 on the reactant side relative to the product side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{(P)_{n} + Glc -&amp;gt; G 6P + (P)_{n-1} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyphosphate, like ATP, contains several phosphoanhydride bonds.  These are &quot;high energy&quot; bonds that can be used for biosynthesis.  But to do so one of these bonds needs to be broken, and that is exactly what happens when one of the end phosphates on the phosphoanhydride chain is peeled off and trasferred to glucose.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-05T13:27:08.957" LastActivityDate="2015-08-05T13:27:08.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34723" PostTypeId="2" ParentId="34717" CreationDate="2015-08-05T04:43:01.747" Score="3" Body="&lt;p&gt;Oil does two things.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Prevents oxygen molecules from colliding with the surface of the metal.&lt;/li&gt;&#xA;&lt;li&gt;More importantly, oil drives away water.  Water promotes rust by creating miniature electrochemical cells on the surface of iron objects.  A cartoon of one of these cells can be seen here:&#xA;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Case_Studies/Corrosion/Corrosion&quot; rel=&quot;nofollow&quot;&gt;Electrochemical cell promoting rust&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Another point.  While I am unaware of how well oxygen dissolves in oil, I'd be interested to know how the higher viscosity of oil reduces the mobility of the oxygen dissolved in it, and possibly reduces the rate of rusting.&lt;/p&gt;&#xA;" OwnerUserId="5601" LastActivityDate="2015-08-05T04:43:01.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34724" PostTypeId="1" AcceptedAnswerId="34737" CreationDate="2015-08-05T06:51:13.463" Score="6" ViewCount="1665" Body="&lt;p&gt;I understand thermodynamics in a physicsy way - at the level of Callen's &lt;em&gt;Thermodynamics&lt;/em&gt;. However, my chemical thermodynamics is quite rusty: concepts like fugacity and activity coefficients are only vaguely familiar from undergrad courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a &lt;em&gt;well written, rigorous and succinct&lt;/em&gt; guide to advanced chemical thermodynamics - at the level where I can read papers on ENRTL equations of state, or understand Pitzer models fully. If it is geared towards chemical engineers (like myself) that would be a bonus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I consider Callen's book a classic, and wonder if there's anything that good within chemical thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a book exist? Thanks&lt;/p&gt;&#xA;" OwnerUserId="17933" LastEditorUserId="10185" LastEditDate="2015-08-05T10:35:46.643" LastActivityDate="2015-08-05T18:21:28.053" Title="Best Guide to Chemical Thermodynamics" Tags="&lt;thermodynamics&gt;&lt;books&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34725" PostTypeId="2" ParentId="34720" CreationDate="2015-08-05T08:32:26.383" Score="3" Body="&lt;p&gt;Atomic radius is inversely proportional to the effective nuclear charge. As we move from left to right in a period the effective nuclear charge increases. This will decrease the radius of an atom. At the same time, in transition elements the number of electrons in the 3d sub-shell will increase. This will repel the already present 4s electrons. This repulsion will increase the atomic radius. In Fe, Co, Ni &amp;amp; Cu both the effects nearly balance each other keeping the atomic radius same.&lt;/p&gt;&#xA;" OwnerUserId="17934" LastActivityDate="2015-08-05T08:32:26.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34726" PostTypeId="1" AcceptedAnswerId="94294" CreationDate="2015-08-05T09:23:05.217" Score="18" ViewCount="5343" Body="&lt;p&gt;The Maxwell-Boltzmann distribution can be used to determine the fraction of particles with sufficient energy to react. I know that the curve applies to gaseous reactants and would like to know whether solids and/or liquids are also described by a similar distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?&lt;/p&gt;&#xA;" OwnerUserId="17936" LastEditorUserId="31313" LastEditDate="2017-03-31T23:14:25.780" LastActivityDate="2019-11-09T09:51:43.777" Title="Does the Maxwell-Boltzmann distribution apply to gases only?" Tags="&lt;physical-chemistry&gt;&lt;temperature&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="6" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34727" PostTypeId="2" ParentId="32354" CreationDate="2015-08-05T10:43:38.910" Score="24" Body="&lt;h2&gt;Introduction&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is fairly obvious that the statement given by Wikipedia cannot be entirely correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is some evidence that the anion (which obeys the 4​&lt;em&gt;n&lt;/em&gt;+2 Hückel rule) is aromatic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trinitromethane anion cannot obey the &lt;a href=&quot;https://doi.org/10.1351/goldbook.H02867&quot; rel=&quot;noreferrer&quot;&gt;Hückel (4​&lt;em&gt;n&lt;/em&gt;+2) rule&lt;/a&gt; because it is neither monocyclic nor planar. And assigning a number of &amp;pi; electrons is fuzzy at best, so the (4​&lt;em&gt;n&lt;/em&gt;+2) part is also questionable. Hückel's rule is a very strict set, but an &lt;a href=&quot;https://doi.org/10.1351/goldbook.A00441&quot; rel=&quot;noreferrer&quot;&gt;aromatic&lt;/a&gt; compound does not necessarily have to fulfil it. The easiest compound to understand this is &lt;a href=&quot;https://en.wikipedia.org/wiki/Naphthalene&quot; rel=&quot;noreferrer&quot;&gt;naphtalene&lt;/a&gt;, which is clearly aromatic, but does not fulfil the criteria of Hückel's rule, i.e. it is not monocyclic. In extension to this, &lt;a href=&quot;https://en.wikipedia.org/wiki/Helicene&quot; rel=&quot;noreferrer&quot;&gt;helicene&lt;/a&gt; is an aromatic compound that fails two of the given conditions. Another case is thiophene, which due to the included sulfur certainly does not entirely consist of trigonally hybridised (sp&lt;sup&gt;2&lt;/sup&gt;) atoms (see Bent's rule).&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can the anion still be regarded as aromatic? In order to explain that, we need to look at the concept of Y-aromaticity.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What is Y-aromaticity?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The term and the concept is not without controversial discussion. It is an attempt to give the &quot;traditional&quot; concept of resonance energy an new, fancy and more catchy name. It is often used to explain the extraordinary stability of the guanidinium cation, which has tree equivalent resonance structures, i.e. the electronic ground state is heavily delocalised.&lt;br&gt;&#xA;Conventional Lewis structures employ two electron two centre bonds only, which therefore cannot represent the highly delocalised electronic structures, e.g. of aromatic compounds. Hence the concept of resonance had to be introduced to account for such stabilising effects. Obviously the same holds true for any compounds, that exhibit excessively delocalised electronic structures.&lt;br&gt;&#xA;The term Y-aromaticity has been used for compounds, that are Y-shaped, with a trigonal planar coordinated (carbon) centre. In addition to that the &amp;pi; orbitals hold six electrons.&lt;sup&gt;[2]&lt;/sup&gt; These compounds have a similar orbital picture and could therefore be grouped.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The electronic structure of Y-delocalised &amp;pi; systems&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The &amp;pi; orbitals of a trigonal planar coordinated centre can be generalised by the following orbital picture, which was generated for the nitrate anion at the DF-BP86/def2-SVP level of theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vqc2x.jpg&quot; alt=&quot;Pi orbitals of carbonate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can clearly see is, that there is one totally symmetric bonding orbital. Two degenerate (orthogonal) slightly bonding (or non-bonding) orbitals and another totally symmetric anti-bonding orbital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remarkable stabilisation comes from the lowest lying &amp;pi; orbital, which is delocalised over all atoms, just like in the more obvious cases like benzene. Additionally, such compounds often retain their planarity by undergoing substitution reactions, rather than addition reactions.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Pro and contra of Y-aromaticity&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The concept itself allows the discussion of certain electronic properties in quite simple terms. It is related enough to the general concept of aromaticity so that it can easily be understood at the same level of teaching. Like aromaticity, it allows predictions - to a certain degree - about the reactivity of such compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peter Grund makes the following assessment, arguing in favour of the concept of acyclic aromaticity.&lt;sup&gt;[3]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Despite recent trends, it would appear useful to designate systems as &quot;aromatic&quot; if they represent a delocalized system exhibiting peculiar chemical stability, in order to focus attention on such systems even when they are buried in complex structures. Since guanidine and its derivatives appear to possess really exceptional stability, and since their physical and chemical properties appear to be dominated by the tendency to retain the closed-shell, Y-delocalized 6 pi-electron configuration, we suggest that it is profitable to consider such substances as possessing a special type of &quot;aromatic character,&quot; despite their acyclic nature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a general understanding of this feature it certainly does not hurt to keep it in the back of ones head.&lt;sup&gt;[4]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the applicability of this concept is limited to a quite small subset of compound (the same holds true for Hückel's rule, too). It therefore seems to be quite a superfluous concept, since it is in a more general sense already covered by resonance stabilisation.&lt;br&gt;&#xA;After all, electronic structures are often not well enough described with simple models and in some cases these simple models can lead to wrong assumptions. This is one of these cases, where one can only use this concept as long as its restrictions are well known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people might argue, that the naming itself is the problem and it may lead to confusion. I can partly agree with that. That does not change the fact, that the description in resonance, valence bond and molecular orbital theory terms will produce this result. In general I think the term aromaticity is unfortunate right in the beginnings. The concept has changed and extended quite a few times already. I often see problems when the term &quot;Hückel rule&quot; is used synonymously with aromaticity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no true theoretical and at the same time all general explanation for aromaticity. It is often determined on a case by case basis experimentally. This limitation also has to apply to any derivations of it.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Electronic structure of the trinitromethane anion (TNM&lt;sup&gt;&amp;ominus;&lt;/sup&gt;)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Trinitromethane&quot; rel=&quot;noreferrer&quot;&gt;Trinitromethane&lt;/a&gt; is a quite acidic substance, which can in part be attributed to the presence of the well delocalised &amp;pi; system. Cioslowski &lt;em&gt;et. al.&lt;/em&gt; have investigated the anion and came to this conclusion. In their paper they also use the term Y-aromaticity.&lt;sup&gt;[5]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Our calculations, however, show that at the HF/6-31G* level, the carbon atom is coplanar with the nitrogens while the oxygens of the nitro groups are rotated out of plane. The result is a structure with $D_3$ symmetry. The $\ce{C(CN)3^-}$  anion is likewise found to be planar, while the $\ce{CF3^-}$ ion exhibits a&#xA;  pyramidal geometry. In addition, we observe in $\ce{C(CN)3^-}$ a shortening of the carbon-carbon bond and a lengthening of the carbon-nitrogen bond relative to $\ce{CH(CN)3}$. We also find an analogous shortening of the carbon-nitrogen bond and lengthening of the nitrogen-oxygen bond in $\ce{C(NO2)3^-}$ relative to $\ce{CH(NO2)3}$. In contrast, the $\ce{C-F}$ bond length increases upon going from $\ce{CHF3}$ to $\ce{CF3^-}$. The planarity of the $\ce{C(NO2)3^-}$ and $\ce{C(CN)3^-}$ ions and the changes in bond lengths are obvious consequences of Y-aromaticity in these systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously the bonding situation in the trinitromethane anion is a little more complicated than in guanidine, as the nitro group itself cannot be described by a single Lewis structure. The out of plane rotation of these groups further complicate things, but since the perpendicular $C_2$ axis is retained it is still somewhat possible to compare the resulting orbitals to a planar species. Therefore when you look at the orbitals you still find a remarkable resemblance to the picture above. For this purpose I have included the same schematic drawings next to the orbitals that resemble these best.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4zMon.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zMon.jpg&quot; alt=&quot;&amp;quot;pi&amp;quot; orbitals on the trinitromethane anion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is noteworthy, that because of the propeller structure there is no strict separation between &amp;sigma; and &amp;pi; orbitals, since the latter require a horizontal mirror plane, which is absent here.&lt;br&gt;&#xA;But that is not all. Try counting the number of &amp;pi; electrons. This is far from trivial and it is in no case that makes sense six.&lt;sup&gt;[6]&lt;/sup&gt; Don't let your brain explode about that fact, we are clearly dealing with a borderline case at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we have not even started to talk about the possibility of bonds between the oxygens of the nitro group.&lt;sup&gt;[7]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Take home message for TNM&lt;sup&gt;&amp;ominus;&lt;/sup&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The trinitromethane anion has a strongly delocalised electronic system. It is extraordinarily stable due to this. If you accept the concept of Y-aromaticity as valid, then it is still a borderline case, where additional considerations have to be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Notes, References and further reading&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bent's Rule&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bent, H. A. &lt;em&gt;Chem. Rev.&lt;/em&gt; &lt;strong&gt;1961,&lt;/strong&gt; &lt;em&gt;61&lt;/em&gt; (3), 275–311. &lt;a href=&quot;https://doi.org/10.1021/cr60211a005&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/cr60211a005&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/15620/4945&quot;&gt;What is Bent&amp;#39;s rule?&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/15671/4945&quot;&gt;Utility of Bent&amp;#39;s Rule - What can Bent&amp;#39;s rule explain that other qualitative considerations cannot?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The six &amp;pi; electrons are probably the reason for the statement in the Wikipedia article.&lt;/li&gt;&#xA;&lt;li&gt;Gund, P. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;1972,&lt;/strong&gt; &lt;em&gt;49&lt;/em&gt; (2), 100. &lt;a href=&quot;https://doi.org/10.1021/ed049p100&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/ed049p100&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;I have used this concept already various times to explain certain electronic structures. From a personal point of view I find it quite helpful.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/21856/4945&quot;&gt;Delocalization of pi electrons in nitrate ion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/29101/4945&quot;&gt;Why is the bond order in the SO₃ molecule 1.33 and not 2?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/q/13483/4945&quot;&gt;Is the bond enthalpy of S-O stronger in SO₃ or SO₃&amp;#178;⁻?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Cioslowski, J.; Mixon, S. T.; Fleischmann, E. D. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1991,&lt;/strong&gt; &lt;em&gt;113&lt;/em&gt; (13), 4751–4755. &lt;a href=&quot;https://doi.org/10.1021/ja00013a007&quot; rel=&quot;noreferrer&quot;&gt;DOI: 10.1021/ja00013a007&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;There are several possibilities, that can be considered to count electrons towards the number of &amp;pi; electrons. &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The probably easiest way would be to flatten the molecule for this thought experiment, then count electrons perpendicular to the plane. This gives you two electrons for the central carbon (sp&lt;sup&gt;2&lt;/sup&gt;), no electrons from the nitrogens (sp&lt;sup&gt;2&lt;/sup&gt;) and two electrons each from the oxygens (sp). That gives you a total of 14 &amp;pi; electrons.&lt;/li&gt;&#xA;&lt;li&gt;Maybe as simple as this, but somewhat the easy way out would be to disregard all contributions from the nitro groups, leaving only 2 &amp;pi; electrons.&lt;/li&gt;&#xA;&lt;li&gt;Regard the nitro groups as &amp;pi;-backdonor. Since only one of the intrinsic &amp;pi; orbitals can effectively overlap with the central carbon, each of them would donate two electrons (maybe less, because &lt;em&gt;back&lt;/em&gt;donation). Plus two from the central carbon, makes up to eight electrons. &lt;em&gt;This certainly is a very fuzzy approach.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Count four electrons per oxygen, which is kind of cheeky, but it could work. This would give 26 &amp;pi; electrons.&lt;/li&gt;&#xA;&lt;li&gt;Write a Lewis structure and count only the double bonds (&amp;times; 2), which gives a total of eight electrons.&lt;/li&gt;&#xA;&lt;li&gt;I am almost certain, there are still more ways to count the electrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;In [5] they analysed the anion with the quantum theory of atoms in molecules. They found bond critical points between the oxygen atoms. How these can be explained and what they mean is certainly a story for another time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2018-01-07T11:04:27.913" LastActivityDate="2018-01-07T11:04:27.913" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34728" PostTypeId="2" ParentId="34720" CreationDate="2015-08-05T10:46:16.920" Score="4" Body="&lt;p&gt;I think the point of this question is for you to realise that options 1 and 2 &lt;em&gt;can't&lt;/em&gt; be the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effective nuclear charge increases across a period because the nuclear charge increases but the shielding stays roughly the same (at least until you get to transition metals). Therefore atomic radius will decrease across the group and so option 2 cannot be correct and the answer must be option 3.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-08-05T10:46:16.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34729" PostTypeId="1" CreationDate="2015-08-05T10:59:44.353" Score="4" ViewCount="854" Body="&lt;p&gt;I am an ESL teacher, and am designing an activity to teach my engineering students the differences between the words &quot;invention,&quot; &quot;discovery,&quot; &quot;innovation,&quot; and &quot;creation,&quot; words that are often confusing for non-native speakers (and as I'm finding out, native speakers.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked on the internet for usage and it seems to be all over the map: Nobel either invented dynamite, created it, or discovered it, and Teflon was called a discovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my view, physical laws are discovered, as are molecules, fundamental particles, and forces because they already exist naturally.  Teflon cannot be said to be discovered because it does not exist naturally; neither does DDT and thousands of other molecules and compounds.  Is there a plant that produces nitrous oxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the usage of &quot;creation&quot; I feel that this word more aptly describes the &quot;process&quot; of chemistry.  Teflon was created, not discovered or invented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As chemists what usage do you prefer?  Do you prefer the use of inventor, creator, or discoverer?  If there are multiple answers can you help me see why you choose that word?&lt;/p&gt;&#xA;" OwnerUserId="17937" LastEditorUserId="10185" LastEditDate="2015-08-05T11:06:04.033" LastActivityDate="2015-08-05T15:10:17.567" Title="Do Chemists &quot;discover,&quot; &quot;invent,&quot; or &quot;create&quot; things?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="11" ClosedDate="2015-08-05T19:44:43.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="34730" PostTypeId="2" ParentId="34702" CreationDate="2015-08-05T11:48:49.313" Score="1" Body="&lt;p&gt;(1) Water softeners use an ion exchange material that is reversible, so that after it has been used a while and the ion exchanger becomes loaded with calcium ions, it can be regenerated to the sodium form by flushing with salt (sodium chloride) solution. This would suggest that the affinity of sodium and calcium ions for the resin do not differ that greatly. I would doubt that either kinetic or steric factors count much here either, although a divalent ion like calcium might bind more tightly by being held electrostatically by more than one anionic group on the resin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) I think the primary reason water softeners work is that they typically use the ion exchange resin in the form of a vertical column: you put the hard water in the top and soft water exits at the bottom. When hard water is introduced at the top of the column, a local equilibrium is established between ions in solution and ions bound to the resin, with a somewhat lower concentration of calcium ions and a corresponding higher concentration (at a 2:1 ratio) of sodium ions. As the water moves down the column, this equilibrium is reestablished again, but this time, the concentration of calcium is lower still. This repeats a large number of times, until all the calcium ions are replaced with sodium ions. I have heard this effect referred to as &quot;mass action&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-08-05T11:48:49.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34731" PostTypeId="1" AcceptedAnswerId="40017" CreationDate="2015-08-05T11:52:52.580" Score="12" ViewCount="15623" Body="&lt;p&gt;Despite very similar skeletal structures, the difference in acid strength between oxalic acid (ethanedioic acid) and malonic acid (propanedioic acid) is quite significant. What is the reason for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the inductive effect plays a role, but what is the explanation for the inductive effect in oxalic acid as there is same group on opposite sides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the inductive effect plays different roles in both acids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/48/Malons%C3%A4ure.svg/187px-Malons%C3%A4ure.svg.png&quot; alt=&quot;propenedioidc acid&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1b/Oxalic_acid.png&quot; height=&quot;150&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16347" LastEditorUserId="13184" LastEditDate="2015-08-05T19:36:45.747" LastActivityDate="2021-01-03T20:22:35.080" Title="Difference in acid strength of oxalic acid and malonic acid" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34732" PostTypeId="2" ParentId="34717" CreationDate="2015-08-05T12:41:13.327" Score="16" Body="&lt;p&gt;Well, I went and searched for the solubility of oxygen in water and oil, and found this summary paper on the NIST web site: &lt;a href=&quot;http://www.nist.gov/data/PDFfiles/jpcrd219.pdf&quot;&gt;&quot;The Solubility of Oxygen and Ozone in Liquids&quot;&lt;/a&gt; by Battino, Rettich and Tominaga, &lt;em&gt;J. Phys. Chem.Ref. Data.&lt;/em&gt;, vol 12, no. 2, 1983.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently, the paper gives solubility data for oxygen in both water and olive oil.  The solubility is given both as molar fractions, and also as the Ostwald coefficient $L$, defined as the volume of gas absorbed per volume of absorbing liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the very different molar volumes of water and olive oil, I believe the latter provides a much more useful comparison here.  Comparing tables 1 (water; note the $10^2$ factor in the header) and 25 (olive oil) in the paper, we see that the Ostwald solubility coefficient for oxygen in water is about 2,000 times higher than in olive oil:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Temp. (K) | Temp. (°C) | L (water) | L (olive oil)&#xA;-----------+------------+-----------+---------------&#xA;    298.15 |         25 |     310.4 |        0.1269&#xA;    308.15 |         35 |     276.6 |        0.1326&#xA;    318.15 |         45 |     254.5 |        0.1383&#xA;    328.15 |         55 |     240.9 |        0.1441&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, yes, oxygen is a &lt;em&gt;lot&lt;/em&gt; more soluble in water than in oil.  But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if &lt;a href=&quot;http://antoine.frostburg.edu/chem/senese/101/solutions/faq/predicting-DO.shtml&quot;&gt;this web site&lt;/a&gt; is to be believed, it's because of the strong hydrogen bonding between water molecules, which causes even liquid water to maintain some degree of an ice-like lattice structure, with gaps that small gas molecules like $\ce{O2}$ can easily slip into.  Once there, the gas becomes &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j150651a018&quot;&gt;solvated&lt;/a&gt;, with the strongly polar water molecules inducing a temporary dipole in the $\ce{O2}$ molecule and thus creating an attractive &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermolecular_force#Debye_.28permanent-induced_dipoles.29_force&quot;&gt;Debye interaction&lt;/a&gt;.  The overall solvation process is exothermic (i.e. releases heat), explaining why the solubility of oxygen in water decreases with temperature.&lt;/p&gt;&#xA;" OwnerUserId="283" LastActivityDate="2015-08-05T12:41:13.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34733" PostTypeId="1" CreationDate="2015-08-05T12:48:19.613" Score="25" ViewCount="388" Body="&lt;p&gt;I performed an experiment where siloxene and cerium(IV) sulphate were mixed together: when I did so  the mixture produced an orange-yellow glow. Why does it glow? What is it about the two chemicals in particular that makes this happen?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Method for synthesising siloxene:&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Carefully measure out 50 cm&lt;sup&gt;3&lt;/sup&gt; of concentrated hydrochloric acid into a 100 cm&lt;sup&gt;3&lt;/sup&gt; measuring cylinder. Put the acid in a 1 dm&lt;sup&gt;3&lt;/sup&gt; beaker containing a magnetic stirrer bar and place the beaker on a heater/magnetic stirrer.&lt;/p&gt;&#xA;&lt;p&gt;Switch on the stirrer and slowly add 5 g of calcium silicide, one spatula measure at a time, to the stirred acid. Continue stirring the mixture until the reaction subsides.&lt;/p&gt;&#xA;&lt;p&gt;Measure out a further 25 cm&lt;sup&gt;3&lt;/sup&gt; of concentrated hydrochloric acid and carefully add this to the stirred mixture. Now turn the heater on and heat the mixture to boiling. Boil the mixture for five minutes. After boiling, slowly add 150 cm&lt;sup&gt;3&lt;/sup&gt; of distilled water then boil again for approximately five minutes.&lt;/p&gt;&#xA;&lt;p&gt;Switch off the heater and leave the mixture stirring while it cools.&lt;/p&gt;&#xA;&lt;p&gt;Set up a Buchner funnel and flask in a fume cupboard.&lt;/p&gt;&#xA;&lt;p&gt;Carefully pour the yellow siloxene that you have made onto a filter paper in a Buchner funnel (try to leave the unreacted grey/black solid calcium silicide in the beaker). Now wash your sample of siloxene – first with 10 cm&lt;sup&gt;3&lt;/sup&gt; of ethanol and then with 10 cm&lt;sup&gt;3&lt;/sup&gt; of ethoxyethane. Drawing air through the sample will help to dry the yellow, solid siloxene.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Method for producing the glow:&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Place 0.8 g of siloxene and 75 cm&lt;sup&gt;3&lt;/sup&gt; of 1M sulfuric acid in a 250 cm&lt;sup&gt;3&lt;/sup&gt; conical flask. In a second 250 cm&lt;sup&gt;3&lt;/sup&gt; conical flask, dissolve the 0.5 g of cerium(IV) sulphate in 75 cm&lt;sup&gt;3&lt;/sup&gt; of 1M sulphuric acid. Label and stopper both flasks.&lt;/p&gt;&#xA;&lt;p&gt;Go into a dark room. Turn out the light, add the cerium(IV) sulphate to the siloxene and swirl the mixture. You should see a fairly bright orange-yellow glow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17939" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2018-05-22T21:30:55.750" Title="Why does a mixture of siloxene and cerium(IV) sulfate luminesce?" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34734" PostTypeId="1" CreationDate="2015-08-05T13:34:40.910" Score="7" ViewCount="4227" Body="&lt;p&gt;Some intramolecular bonds are rotatable in the sense that the torsion angle around this bond is flexible (for example, $\psi$ and $\phi$ angles in a protein backbone). Others are essentially fixed (for example, the $\omega$ angle in the peptide bond in a protein backbone). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an operational definition, or a published study where a general definition is given, which would allow me to recognize in a generic molecule (or at least an organic molecule) whether a given bond is rotatable or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no general theory for this, what are the most common indicators of whether a bond can or cannot be rotated? Are there some rules of thumb?&lt;/p&gt;&#xA;" OwnerUserId="1159" LastEditorUserId="10185" LastEditDate="2015-08-05T16:30:51.717" LastActivityDate="2015-08-05T18:00:49.963" Title="Is there a definition of rotatable bond?" Tags="&lt;bond&gt;&lt;stereochemistry&gt;&lt;conformers&gt;" AnswerCount="3" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34735" PostTypeId="2" ParentId="34734" CreationDate="2015-08-05T14:50:50.410" Score="1" Body="&lt;p&gt;For every bond, there is an energy barrier required to rotate around the bond. Would the thermal energy (at room temperature) be enough to allow rotation, it can be considered rotatable. For the reasonable rate constant I would consider $s ^{-1}$ or faster.&#xA;This means that single $\ce{C^{sp3}-C^{sp3}}$ bonds are rotatable, double bonds rarely, in other cases use common sense or quantum chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Medicinal chemists use more straightforward definition, &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jm020017n&quot; rel=&quot;nofollow&quot;&gt;http://pubs.acs.org/doi/abs/10.1021/jm020017n&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-08-05T14:50:50.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34736" PostTypeId="2" ParentId="34729" CreationDate="2015-08-05T15:10:17.567" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/discover?s=t&quot; rel=&quot;nofollow&quot;&gt;discover&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/invent?s=t&quot; rel=&quot;nofollow&quot;&gt;invent&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/create?s=t&quot; rel=&quot;nofollow&quot;&gt;create&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dictionary.reference.com/browse/innovate?s=t&quot; rel=&quot;nofollow&quot;&gt;innovate&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Create is totally weird with chemistry, you can create item, but substance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Innovate is more about improving something existing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves us with discover and invent, which in case of chemistry seems to me almost equivalent. For &quot;invent&quot; I would imagine targeted effort towards a new substance with desired properties, whereas &quot;discover&quot; evokes more trial-and-error approach.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastActivityDate="2015-08-05T15:10:17.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34737" PostTypeId="2" ParentId="34724" CreationDate="2015-08-05T15:23:48.797" Score="3" Body="&lt;p&gt;I would suggest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071247084&quot; rel=&quot;nofollow&quot;&gt;Introduction to Chemical Engineering Thermodynamics&lt;/a&gt; by Smith, Van Ness, and Abbott;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;a href=&quot;http://books.unbeatablesale.com/chemical_biochemical_engineering_thermodynamics_by_sandler_stanley_i_1152064025.php&quot; rel=&quot;nofollow&quot;&gt;Chemical, Biochemical, and Engineering Thermodynamics&lt;/a&gt; by Sandler; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) This is more suitable for advanced reading or for a graduate student: &lt;a href=&quot;http://web.mit.edu/testerel/thermo/&quot; rel=&quot;nofollow&quot;&gt;Thermodynamics and Its Applications&lt;/a&gt; by Tester and Modell.&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorUserId="865" LastEditDate="2015-08-05T18:21:28.053" LastActivityDate="2015-08-05T18:21:28.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34738" PostTypeId="1" CreationDate="2015-08-05T15:57:21.787" Score="2" ViewCount="491" Body="&lt;p&gt;How can I measure the resistance of a gel electrolyte using a multimeter? I know the formula to measure the conductivity of a gel electrolyte is k = (1/R) * (d/a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k is the conductivity (S/cm), R is the resistance (of the electrolyte), d is the distance between the two electrodes, and a is the effective area of the electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to measure the conductivity of the gel electrolyte with just a multimeter. Since this is a gel electrolyte, I was thinking to substitute the thickness of the gel electrolyte as d and substitute the area of the gel electrolyte as a. (I know the area (a) has to be that of the electrodes, but I want to measure the conductivity of the gel electrolyte without the electrodes). So now all that remains is the resistance of the gel electrolyte. To do this can I just measure its resistance using a multimeter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I understand that you have to use two similar electrodes sandwiched between the electrolyte in order to measure the conductivity of it, but is there a way around?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="17750" LastActivityDate="2015-08-05T15:57:21.787" Title="How to measure resistance of a gel electrolyte using a multimeter" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;&lt;conductivity&gt;&lt;electricity&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-08-05T19:59:42.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="34739" PostTypeId="2" ParentId="8411" CreationDate="2015-08-05T16:45:08.437" Score="0" Body="&lt;p&gt;Just apply the bond method which is the simplest and most effective. Count the number of bond pairs of central atoms excluding pi-bonds and the number Of lone pairs and match the corresponding hybridisation.&lt;/p&gt;&#xA;" OwnerUserId="17941" LastEditorUserId="10185" LastEditDate="2015-08-05T17:29:39.413" LastActivityDate="2015-08-05T17:29:39.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34740" PostTypeId="2" ParentId="34734" CreationDate="2015-08-05T16:47:45.803" Score="4" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is there an operational definition, or a published study where a&#xA;general definition is given, which would allow me to recognize in a&#xA;generic molecule (or at least an organic molecule) whether a given&#xA;bond is rotatable or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It depends on your purpose. Why do you want to identify some bonds and call the rotatable? For instance, &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PubChem&lt;/a&gt;, one of the biggest databases of molecules, defines rotatable bond as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Rotatable bond is defined as any single non-ring bond, bounded to&#xA;nonterminal heavy (i.e., non-hydrogen) atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To some extent all this boils down to the following IUPAC definition of free rotation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;free rotation&lt;/strong&gt; (&lt;strong&gt;hindered rotation&lt;/strong&gt;, &lt;strong&gt;restricted rotation&lt;/strong&gt;)&lt;/p&gt;&#xA;&lt;p&gt;In a stereochemical context the rotation about a bond is called 'free'&#xA;when the rotational barrier is so low that different conformations are&#xA;not perceptible as different chemical species on the time scale of the&#xA;experiment. The inhibition of rotation of groups about a bond due to&#xA;the presence of a sufficiently large rotational barrier to make the&#xA;phenomenon observable on the time scale of the experiment is termed&#xA;hindered rotation or restricted rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;At usual lab conditions a rotation around a single non-ring bond, bounded to nonterminal heavy atom will be indeed almost free, thus, PubChem definition can be used as a very simple rule-of-thumb to identify such bonds. But in general it depends on conditions. And purposes.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;With respect to protein backbones being rotatable or not: I'm not a speacialist in this area, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Peptide_bond&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on peptide bond&lt;/a&gt; claims that (emphasis mine)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Significant delocalisation of the lone pair of electrons on the&#xA;nitrogen atom gives the group a &lt;em&gt;partial double bond character&lt;/em&gt;. The&#xA;partial double bond renders the amide group planar, occurring in&#xA;either the cis or trans isomers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, this partial double bond character might contribute in protein backbones being not quite rotatable. Besides, the Wikipedia article also claims that (emphasis again mine):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In the unfolded state of proteins, the peptide groups are &lt;em&gt;free to&#xA;isomerize&lt;/em&gt; and adopt both isomers; however, in the folded state, only a&#xA;single isomer is adopted at each position (with rare exceptions).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, at least in the unfolded state protein backbones are rotatable. The fact that only two configurations (cis and trans) are possible is irrelevant. The notion of rotatable bond does not imply having few specific possible configurations.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-05T18:00:49.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34741" PostTypeId="2" ParentId="34734" CreationDate="2015-08-05T17:14:59.300" Score="2" Body="&lt;p&gt;The IUPAC Gold Book states at least somthing to &lt;a href=&quot;http://goldbook.iupac.org/F02520.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;free rotation&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;free rotation (hindered rotation, restricted rotation)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In a stereochemical context the rotation about a bond is called 'free' when the rotational barrier is so low that different conformations are not perceptible as different chemical species on the time scale of the experiment. The inhibition of rotation of groups about a bond due to the presence of a sufficiently large rotational barrier to make the phenomenon observable on the time scale of the experiment is termed hindered rotation or restricted rotation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What they don't tell is what causes a &lt;em&gt;sufficiently large rotational barrier&lt;/em&gt;. Those are, e.g., ring strains or resonance stabilized bonds.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-05T17:14:59.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34742" PostTypeId="2" ParentId="8411" CreationDate="2015-08-05T17:32:12.060" Score="1" Body="&lt;p&gt;For hydrocarbons, if the carbon atom forms only &amp;sigma; bonds, the orbitals of it will be sp&lt;sup&gt;3&lt;/sup&gt; hybridised. If the carbon atom forms a &amp;pi; bond, its orbitals are sp&lt;sup&gt;2&lt;/sup&gt; hybridised. If the carbon atom forms two &amp;pi; bonds, its orbitals are sp hybridised.&lt;/p&gt;&#xA;" OwnerUserId="17934" LastEditorUserId="4945" LastEditDate="2018-10-15T09:30:59.960" LastActivityDate="2018-10-15T09:30:59.960" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34743" PostTypeId="1" CreationDate="2015-08-05T17:45:41.267" Score="2" ViewCount="67" Body="&lt;p&gt;Yesterday, my girlfriend remarked on how pleased she was with the effects of the hair moisturizer she has been using: &quot;I started using it just a month ago, and look, it's healed all the damage.&quot; I complimented her, then said &quot;...but hair is dead, so it couldn't have helped heal it.&quot; She replied, &quot;Well it fills in the cracks&quot;. I accepted her answer; lots of hard-surface sealers work that way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, hair moisturizer seems odd. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Unlike hair moisturizer, the effects of hard-surface sealers typically don't need dozens of reapplications to obtain. &lt;/li&gt;&#xA;&lt;li&gt;The exceptional sealers that do need several coats, typically do not need to cure before reapplication (although I suspect that some do). However, if a person were to apply coat after coat of hair moisturizer, they'd just get greasy hair. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suspect that the porousness of hair necessitates the use of a &quot;wetter&quot; solvent than the one used in most hard-surface sealers, which protracts the curing time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder whether the porousness of hair protracts the time it takes for the moisturizer to penetrate its fractures, and thus protracts the time the user ought to wait before reapplying: assuming the presence of the solvent along the path of penetration slows the penetration of the moisturizer then, the coats applied immediately after the first coat would penetrate the hair more slowly, and so they would cure closer to its surface, and would thus produce results that differed from the ones my girlfriend was so pleased with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there substances that exhibit the properties that hair moisturizer seems to exhibit. At the molecular level, what explains why these substances exhibit those properties?&lt;/p&gt;&#xA;" OwnerUserId="15929" LastActivityDate="2015-08-05T17:45:41.267" Title="Do any substances work like consumers of hair moisturizer believe it to work?" Tags="&lt;everyday-chemistry&gt;" AnswerCount="0" CommentCount="0" ClosedDate="2015-10-30T18:34:37.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="34744" PostTypeId="1" AcceptedAnswerId="34749" CreationDate="2015-08-05T17:51:59.413" Score="2" ViewCount="78" Body="&lt;p&gt;Electrical properties of very simple systems can be calculated using &lt;a href=&quot;http://www.physics.udel.edu/~bnikolic/QTTG/shared/reviews/nanoscale_device_modeling_datta.pdf&quot; rel=&quot;nofollow&quot;&gt;NEGF&lt;/a&gt;. Is there a way to calculate the resistance (and other electrical properties) of a conjugated polymer, as a function of degree of polymerization (DP)?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="5017" LastEditDate="2015-08-05T19:43:02.243" LastActivityDate="2015-08-05T19:43:02.243" Title="Can you predict the resistance of conjugated polymers as a function of degree of polymerization (DP)?" Tags="&lt;computational-chemistry&gt;&lt;polymers&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34746" PostTypeId="1" CreationDate="2015-08-05T18:52:41.110" Score="3" ViewCount="3169" Body="&lt;p&gt;After writing &lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/21908/how-can-i-liquefy-my-enemies/21911#21911&quot;&gt;an answer&lt;/a&gt; on worldbuilding.SE, I came here to ask a question. How long would it take to dissolve a human body using fluoroantimonic acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that since the human body isn't homogenous, it's a bit tough to get the exact calculation, but lets say we create a mannequin using the material in the human body that fluoroantimonic acid takes the longest time to dissolve - how long would it take?&lt;/p&gt;&#xA;" OwnerUserId="17942" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:35.613" LastActivityDate="2015-08-05T20:12:28.593" Title="How long would it take to dissolve someone using fluoroantimonic acid?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="0" CommentCount="8" FavoriteCount="2" ClosedDate="2015-08-14T06:59:05.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="34748" PostTypeId="2" ParentId="32050" CreationDate="2015-08-05T18:55:00.283" Score="9" Body="&lt;p&gt;A key consideration for the Wolff-Kishner reduction is thermal stability of the substrate. High temperature is required for the classical reaction using hydrazine and KOH, sometimes up to 200 °C. This can rule out plenty of compounds. &#xA;The Clemmensen reduction may be fine for the laboratory,  but fundamentally many industries wouldn't use it due to scaling up the use of mercury.&lt;br&gt;&#xA;Scalable alternatives are molecular hydrogen with a catalyst under acidic conditions or, the trifluoroacetic acid-silane combination, such as triethylsilane or tetramethyldisiloxane.&lt;/p&gt;&#xA;" OwnerUserId="16724" LastEditorDisplayName="user7951" LastEditDate="2019-06-23T07:24:43.930" LastActivityDate="2019-06-23T07:24:43.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34749" PostTypeId="2" ParentId="34744" CreationDate="2015-08-05T19:41:58.227" Score="3" Body="&lt;p&gt;&lt;strong&gt;Short answer: No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, the degree of polymerization does not &lt;em&gt;directly&lt;/em&gt; relate to the charge transport properties of conducting polymers (more below). You can find highly conductive oligomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, in my opinion, NEGF is not a suitable method to understand the charge transport in polymers. It works for nanoscale devices and molecular wires, but most polymer devices are on the ~100 nm to micron scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's not to say that different oligomer and polymer lengths don't have different transport properties. Absolutely, different molecular weight polymers have different charge mobility, etc. The effects are usually due to a number of hard to characterize factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Concentration of defects and traps&lt;/li&gt;&#xA;&lt;li&gt;Morphology of the film (e.g., higher MW might yield smoother or more ordered films)&lt;/li&gt;&#xA;&lt;li&gt;Polydispersity (e.g., broader MW distribution is often problematic)&lt;/li&gt;&#xA;&lt;li&gt;Effective conjugation length&lt;/li&gt;&#xA;&lt;li&gt;Packing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronic structure of oligomers and conjugated polymers can be simulated fairly well using first-principals methods, including DFT. This can often predict components like the energy levels, band gaps, band dispersion, etc. for crystals and parameters for amorphous films such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Marcus_theory&quot; rel=&quot;nofollow&quot;&gt;Marcus reorganization energy&lt;/a&gt; which is often related to the charge mobility in many of these materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you usually find is that as a function of chain length, charges become more delocalized and the reorganization energy drops (increasing mobility). In reality, this trend should &lt;a href=&quot;http://dx.doi.org/10.1021/ja0461421&quot; rel=&quot;nofollow&quot;&gt;saturate&lt;/a&gt;, although most DFT methods incorrectly delocalize charges completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, most devices with conjugated polymers are amorphous or polycrystalline, making device performance predictions difficult. Transport is not usually band-like, but some form of variable-range hopping between chains or sites within a chain.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-05T19:41:58.227" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34750" PostTypeId="1" AcceptedAnswerId="34774" CreationDate="2015-08-05T20:22:04.253" Score="10" ViewCount="500" Body="&lt;p&gt;For example, polyacetylene is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Conjugated_system&quot; rel=&quot;noreferrer&quot;&gt;conjugated system&lt;/a&gt;, i.e. with de-localized p-orbitals, but accroding to &lt;a href=&quot;http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.39.1098&quot; rel=&quot;noreferrer&quot;&gt;Chiang et. al. (1977)&lt;/a&gt; polyacetylene is semiconducting, and becomes conductive only when doped with Cl/Br/I/AsF&lt;sub&gt;5&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the conjugation alone enough to allow conductivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oQ6K3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oQ6K3.png&quot; alt=&quot;Electrical conductivity of trans-polyacetylene as a function of AsF5 dopant concentration&quot;&gt;&lt;/a&gt;(&lt;a href=&quot;http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.39.1098&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One reason they mention in the paper is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the idealized situation of a uniform chain, the resulting&#xA;  conduction band would give rise to metallic behavior. However, such a&#xA;  system is unstable with respect to bond alternation, which causes the&#xA;  formation of an energy gap in the electronic spectrum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But if this is true then (1) how the halogens improve the situation, and (2) why isn't there a similar phenomenon in all conjugated systems?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="10104" LastEditDate="2015-08-06T04:01:11.197" LastActivityDate="2015-08-06T15:41:05.950" Title="Does the delocalization of p-orbitals ensure conductivity?" Tags="&lt;computational-chemistry&gt;&lt;orbitals&gt;&lt;polymers&gt;&lt;conductivity&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34751" PostTypeId="2" ParentId="34717" CreationDate="2015-08-05T21:17:40.267" Score="6" Body="&lt;p&gt;Before you read this answer, please read the &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Case_Studies/Corrosion/Corrosion&quot; rel=&quot;nofollow&quot;&gt;ChemWiki article on the corrosion of iron&lt;/a&gt;.  This explains the normal (Earth-surface) process for iron rusting, which involves &lt;em&gt;both&lt;/em&gt; oxygen and water.  The water is a catalyst, but the oxygen is a reactant: the reaction consumes oxygen and will stop if it becomes unavailable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you put a nail in a beaker, add enough water to completely cover the nail, and then add enough oil to completely cover the surface of the water, there won't be enough oxygen dissolved in the water to rust the nail more than a little bit.  The only way for more oxygen to get into the water is by diffusing through the oil.  We can model this as a chain of equilibria:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{O}_2\,\text{(air)} \rightleftharpoons \text{O}_2\,\text{(oil)} \rightleftharpoons \text{O}_2\,\text{(water)} + \text{Fe}\,\text{(nail)} \rightleftharpoons \text{Fe}_2\text{O}_3\,\text{(rust)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermodynamically, the equilibrium lies all the way to the right: completely rusted nail, having soaked up as much oxygen as possible, ultimately from the air.  But, as described in &lt;a href=&quot;https://chemistry.stackexchange.com/a/34732/4921&quot;&gt;Ilmari's answer to this question&lt;/a&gt;, oxygen is much less soluble in oil than in water.  Therefore, oxygen will diffuse through the oil layer very slowly, and that will be the rate-limiting step.  The rate constants for the air⇌oil and oil⇌water steps will depend on which oil is used and how thick the oil layer is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens is, after most of the oxygen that was initially dissolved in the water has been consumed by the rust, the nail does continue to rust, but very slowly — only as fast as additional oxygen molecules can get through the oil layer.&lt;/p&gt;&#xA;" OwnerUserId="4921" LastEditorUserId="4921" LastEditDate="2015-08-06T15:32:02.443" LastActivityDate="2015-08-06T15:32:02.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34752" PostTypeId="1" AcceptedAnswerId="34761" CreationDate="2015-08-05T22:25:16.267" Score="-1" ViewCount="734" Body="&lt;p&gt;Beryllium hydride does not obey the octet rule. Isn't this impossible, unless there are more elements comprising the molecule?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-08-06T01:34:55.330" LastActivityDate="2015-08-06T08:40:55.717" Title="Beryllium hydride violates the octet rule" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-06T08:07:35.840" ContentLicense="CC BY-SA 3.0" />
  <row Id="34761" PostTypeId="2" ParentId="34752" CreationDate="2015-08-06T04:11:10.553" Score="4" Body="&lt;p&gt;Technically, it does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{BeH2}$ forms a network of $\ce{BeH4}$ tetrahedrons sharing corners. Since each hydrogen belongs to two such tetrahedron, the net formula is $\ce{BeH2}$. In this structure, each electron pair belongs to a triple of atoms $\ce{Be-H-Be}$. Such bonding is known as three-centre two-electron bonding, and is fairly common in electron-deficient compounds that would otherwise violate the octet rule. For example, this kind of bonding is essential for boron-hydrogen chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, unambiguous violations of the octet rule are not unknown. The most obvious and reasonable stable example would be nitric oxide, $\ce{NO}$. However, this example again features strong 'extra' electron delocalization with 'extra' electron shared between the oxygen and nitrogen atoms. The molecule still shows tendency to dimerize at low temperatures. This kind of molecule is more of an exception and not a rule. The exact way such particles are stabilised is highly dependent on the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, if we are talking about transition elements or lanthanides, the situation is more complex, because of the involvement of fairly strongly bonded electrons in d- or f-orbitals. Still, the extra stability of complexes following the equivalent 18-electron rule is clearly observed.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2015-08-06T08:40:55.717" LastActivityDate="2015-08-06T08:40:55.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34762" PostTypeId="1" CreationDate="2015-08-06T06:18:24.323" Score="14" ViewCount="8076" Body="&lt;p&gt;When you put $\ce{Mg}$ into water a few $\ce{H2}$ bubbles appear. But when you put $\ce{Mg}$ into a $\ce{NaCl}$ solution there is a vigorous release of $\ce{H2}$, why is this and what reactions are taking place?&lt;/p&gt;&#xA;" OwnerUserId="17948" LastEditorUserId="10185" LastEditDate="2015-08-06T12:23:41.417" LastActivityDate="2018-10-12T15:53:17.357" Title="Why does Mg react vigorously in NaCl solution and less so in water?" Tags="&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34763" PostTypeId="1" CreationDate="2015-08-06T06:58:51.373" Score="0" ViewCount="187" Body="&lt;p&gt;There is a question on my assignment about stoichiometry. I spent 2 whole days searching and review this topic but still clueless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction in lithium ion battery&#xA;$$&#xA;\ce{Li(s) + CoO2(s) → LiCoO2(s)} \, .&#xA;$$&#xA;When the battery is recharged, electrical energy will reverse the reaction. How do the relatives amounts of lithium and cobalt oxide change as the battery is recharged and how?&lt;/p&gt;&#xA;" OwnerUserId="17949" LastEditorUserId="186" LastEditDate="2015-08-06T08:12:35.293" LastActivityDate="2015-08-06T14:12:08.343" Title="Does relative amount of lithium and cobalt oxide change in recharged battery?" Tags="&lt;everyday-chemistry&gt;&lt;electrochemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34764" PostTypeId="2" ParentId="34726" CreationDate="2015-08-06T07:42:34.257" Score="7" Body="&lt;p&gt;I'm willing to put myself out there and say that no, a Maxwell-Boltzmann distribution will not be able to make any statement about the ability of liquids or solids to react. Meaning also that you could not study reaction rates with a Maxwell-Boltzmann distribution for the condensed phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I am fairly confident this is true is because the Maxwell-Boltzmann distribution makes the assumption that it is working with an ideal gas. Making that assumption is both powerful and very limiting. The ideal gas approximation is quite good in a lot of cases, but it could not even come close to describing the behavior of a condensed phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some such distribution were used to study condensed phase reaction rates, it simply wouldn't be a Maxwell-Boltzmann distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also point out that the Maxwell-Boltzmann distribution doesn't describe all gas-phase reactions all that well. This can be inferred quite easily from the fact that the M-B distribution treats all gases as ideal, so the only possible difference between two systems is the mass of the particle and the temperature. That means also that any gas phase reaction which depends strongly on the orientation of the collision would likely be assumed by M-B to happen much more frequently than it really does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that to say, M-B is deeply rooted in the gas phase, and even there it falls short for certain systems, so it probably isn't possible to apply to the condensed phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it is possible that a partition function for a liquid or a solid phase could be used to study reactions in some way, and the partition function for a generic gas particle can be used to derive the M-B distribution so . . . There's that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with that is that the partition function requires you to sum the Boltzmann Factor over all states, and for a condensed phase system . . . That's basically gonna be an infinite number of states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anytime you approach stat. mech. or the name Boltzmann, it's wise to keep this quote by David Goodstein in mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. Perhaps it will be wise to approach the subject cautiously.” &lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-08-06T07:42:34.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34766" PostTypeId="1" AcceptedAnswerId="58196" CreationDate="2015-08-06T08:02:52.650" Score="9" ViewCount="3496" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5oqiw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5oqiw.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism for the above reaction? &lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2017-06-28T13:40:49.233" LastActivityDate="2017-06-28T13:40:49.233" Title="Diazo transfer to 1,3-dicarbonyl with tosyl azide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34767" PostTypeId="1" CreationDate="2015-08-06T10:18:32.857" Score="3" ViewCount="1252" Body="&lt;p&gt;I completed an electrolysis experiment. I used copper electrodes and a copper sulphate solution, and my independant variable was the temperature of the electrolyte. I was just wondering if increasing the temperature has a positive correlation?&lt;/p&gt;&#xA;" OwnerUserId="16291" LastActivityDate="2019-02-04T09:24:55.480" Title="Does Temperature Affect Rate of Reaction of Electrolytic Cell" Tags="&lt;redox&gt;&lt;electrolysis&gt;" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34768" PostTypeId="2" ParentId="34763" CreationDate="2015-08-06T10:19:27.650" Score="1" Body="&lt;p&gt;About maximum $20~\%$ (as I remember, perhaps more now) of the $\ce{Li}$ goes (as $\ce{Li+}$) out of the structure of $\ce{LiCoO2}$, through the elctrolyte, and into the electrode on the other side of the battery. That electrode consists mainly of graphite, which intercalates the $\ce{Li}$ in it's layers in a quasi-metallic form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is $\ce{Li_{1-x}CoO2 + x~Li}$ after charging. $\ce{LiCoO2}$ has a special structure that does not break down when some $\ce{Li}$ is missing, except if you keep a charged battery at over $\mathrm{50~^\circ C}$ or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium-ion_battery#Electrochemistry&quot; rel=&quot;nofollow&quot;&gt;related wikipedia article&lt;/a&gt; tells you all this. Two days of researching? ;-)&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="16622" LastEditDate="2015-08-06T14:12:08.343" LastActivityDate="2015-08-06T14:12:08.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34769" PostTypeId="1" CreationDate="2015-08-06T10:58:48.267" Score="4" ViewCount="344" Body="&lt;p&gt;In the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hoesch_reaction&quot; rel=&quot;nofollow&quot;&gt;Hoesch reaction&lt;/a&gt;, an alkyl cyanide reacts with a polyhydric benzene ring, and forms an imine salt by attacking the activated benzene ring. Most acid derivatives (and even alkyl cyanide itself with phenol) results in attack at the alcohol's oxygen, which is more nucleophilic than the benzene ring. In this case however, the major product doesn't form by attack on the alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where lies the difference? If it were in the fact that benzene is more activated because of the many alcohol groups in the polyhydric compound, wouldn't it mean that even another acid derivative, like acid chloride, would react at the benzene instead and not at the O (thus carrying out a Friedel-Crafts acylation)?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="9961" LastEditDate="2015-08-06T21:40:16.373" LastActivityDate="2015-08-06T21:40:16.373" Title="Why does RCN attack the benzene ring in Hoesch Reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34770" PostTypeId="2" ParentId="4063" CreationDate="2015-08-06T12:01:16.330" Score="1" Body="&lt;p&gt;Water shows its acidic nature when strong bases are used. Group I elements reacts with water (like any other acid) to liberate hydrogen gas.&#xA;However, you have to remember that water is a polar compound. Oxygen has enough electronegativity to pull the bonded pair of electrons toward itself but hydrogen being a partial member of group 17, has the power (I mean electron withholding property) to resist the oxygen's effort to pull all electron's toward itself. Hence water is unable to show its acidic nature.&#xA;Moreover, water has low ionisation potential to liberate hydrogen ions. Even if it dissociate, equal number of $\ce{OH-}$ and $\ce{H+}$ are produced to cancel each other's effect and remain neutral. &lt;/p&gt;&#xA;" OwnerUserId="15837" LastEditorUserId="10185" LastEditDate="2015-08-06T17:26:30.837" LastActivityDate="2015-08-06T17:26:30.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34772" PostTypeId="1" CreationDate="2015-08-06T12:32:51.767" Score="0" ViewCount="840" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LudWU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LudWU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first example, if you add up the number of carbon which has a substituent, you will get 2+4+4 = 10. However, if you start from left to right, then you'd get 2+2+4=8. Why don't we follow the latter one when it gives us a lower number? (Lowest set of locant rule.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies for the second problem: we could've gotten 2+3+4, but we chose 2+4+5, why so?&lt;/p&gt;&#xA;" OwnerUserId="6845" LastEditorUserId="10185" LastEditDate="2015-08-07T09:07:13.093" LastActivityDate="2015-08-07T09:07:13.093" Title="Assigning locants in halo alkanes/alkenes" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34774" PostTypeId="2" ParentId="34750" CreationDate="2015-08-06T13:39:22.817" Score="10" Body="&lt;p&gt;Delocalized $\pi$ orbitals &lt;strong&gt;do&lt;/strong&gt; allow for mobile charges. The catch there is &quot;allow.&quot; Just because these systems &lt;em&gt;can&lt;/em&gt; have mobile charges does not make them &lt;em&gt;conductive&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductivity can be defined as:&#xA;$\sigma = n e \mu$&#xA;where $e$ is the charge on an electron, $\mu$ is the charge mobility, and $n$ is the number of charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $doped$ conjugated polymers like you indicate above, the halogen vapors or other dopants &lt;em&gt;usually&lt;/em&gt; oxidize the polymer and create more charges (i.e., $n$) and the conductivity increases. This &lt;strong&gt;&lt;em&gt;does&lt;/em&gt;&lt;/strong&gt; happen with pretty much every conjugated system I've seen: oxidize or reduce the neutral system and you'll have more charge carriers = greater conductivity. (NB: most conjugated organics are easier to oxidize than reduce, so you get + charges (holes) as the charge carrier, rather than electrons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the quick explanation as to why adding halogens increases the conductivity -- you get more charge carriers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, there are other ways to increase the charge carriers. Chemical or electrochemical doping usually makes a conductive system, but you can also use electrical bias (e.g., a &lt;a href=&quot;https://en.wikipedia.org/wiki/Field-effect_transistor&quot; rel=&quot;noreferrer&quot;&gt;field-effect transistor&lt;/a&gt;) or light (i.e., a &lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_solar_cell&quot; rel=&quot;noreferrer&quot;&gt;photovoltaic&lt;/a&gt;) which will also increase carrier concentrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As to why &lt;em&gt;trans&lt;/em&gt;-polyacetylene isn't a metal in the first place?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use a simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Particle_in_a_box&quot; rel=&quot;noreferrer&quot;&gt;particle-in-a-box&lt;/a&gt; model, the difference in energy levels goes to zero as the box becomes very large. This would suggest &lt;em&gt;trans&lt;/em&gt;-polyacetylene should be a metal (i.e., no band gap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also the &lt;a href=&quot;https://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;noreferrer&quot;&gt;Hückel model&lt;/a&gt; for polyacetylene. We can build up the nearly-infinite system by solving for ethylene, butadiene, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CgrgO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgrgO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the energy gap between levels becomes smaller and smaller. We can plot the HOMO-LUMO gap as a function of chain length:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f1O0C.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f1O0C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gap becomes smaller and smaller! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/H%C3%BCckel_method&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; even gives the general formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_k = \alpha + 2\beta \cos \frac{k\pi}{(n+1)}$$&#xA;$$\Delta E = -4\beta \sin \frac{\pi}{2(n+1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both particle in a box and Hückel predict no band gap when $n$ gets very large. Theory predicts a metal!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the &lt;em&gt;real&lt;/em&gt; polyacetylene behave as a metal. For any one-dimensional system, there will be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Peierls_transition&quot; rel=&quot;noreferrer&quot;&gt;Peierls distortion&lt;/a&gt; which lowers the total energy by distorting the geometry (long-short bond alternation) and increasing the band gap.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short, a truly one-dimensional metal is impossible, but doped polymers are certainly very good electrical conductors. Record films of &lt;em&gt;trans&lt;/em&gt;-polyacetlyene have conductivity close to conventional metals.&lt;/strong&gt;&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P1pB2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P1pB2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&#xA;&lt;sup&gt;1&lt;/sup&gt; The long-short bond alternation is the Peierls distortion. It means some electrons are localized (shorter double bonds) and thus lower in energy. This creates a band gap (i.e., lower E for valence band).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; The film isn't a 1D material, but the trans-polyacetylene chains are each 1D.&#xA;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="10104" LastEditDate="2015-08-06T15:41:05.950" LastActivityDate="2015-08-06T15:41:05.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34775" PostTypeId="2" ParentId="34772" CreationDate="2015-08-06T15:49:42.590" Score="1" Body="&lt;p&gt;Indeed, the first compound in your sketch is numbered incorrectly according to IUPAC. &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm&quot; rel=&quot;nofollow&quot;&gt;IUPAC rules for branched hydrocarbons&lt;/a&gt; say: &quot;When series of locants containing the same number of terms are compared term by term, that series is &quot;lowest&quot; which contains the lowest number on the occasion of the first difference. This principle is applied &lt;strong&gt;irrespective of the nature of the substituents.&lt;/strong&gt; &quot; (emphasis mine). In your top case, this means that you have two numbering choices: either 2,2,4 or 2,4,4. The 2,2,4 choice has a lower locant at the first point of difference, so it is preferred. When you construct the name, you indicate the substituents in alphabetical order, so the full name of the top compound is 4-chloro-2,2-dimethylpentane.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If unsaturations are present, then these have a higher priority than alkyl and halogen substituents. From the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_53.htm&quot; rel=&quot;nofollow&quot;&gt;rules for unsaturated hydrocarbons&lt;/a&gt;: &quot;The chain is so numbered as to give the lowest possible numbers to the double bonds.&quot; This rule is sufficient to impose the numbering you report.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As Loong correctly pointed out, the links provided above refer to the text of the now obsolete 1979 recommendations. Both the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/&quot; rel=&quot;nofollow&quot;&gt;1979 and 1993 recommendations&lt;/a&gt; are available on the same site, maintained by ACD Labs with IUPAC's permission, and are in accordance with what shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the text of the most recent (2013) recommendations is not freely accessible (that I know of). &lt;a href=&quot;http://pubs.rsc.org/en/content/chapterpdf/2013/9781849733069-00001?isbn=978-0-85404-182-4&amp;amp;sercode=bk&quot; rel=&quot;nofollow&quot;&gt;This is a link&lt;/a&gt; to the electronic version of the book, published by The Royal Society of Chemistry. The relevant section is &lt;strong&gt;P-14.3.5&lt;/strong&gt; in Chapter P-1 (General Principles, Rules, and Conventions): &quot;The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set '2,3,5,8' is lower than '3,4,6,8' and '2,4,5,7'.&quot; &lt;/p&gt;&#xA;" OwnerUserId="16652" LastEditorUserId="16652" LastEditDate="2015-08-06T18:47:10.477" LastActivityDate="2015-08-06T18:47:10.477" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34777" PostTypeId="2" ParentId="6182" CreationDate="2015-08-06T15:58:08.770" Score="2" Body="&lt;p&gt;Sodium ion Na+ should not be confused with metallic sodium Na. They are actually two different chemical species (atoms vs ions), so it is expected that they chemically behave in a different way. The same applies to chlorine Cl and chloride ion Cl-.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Na can &quot;switch&quot; to Na+ when it is oxidated; this process of oxidation requires a &lt;strong&gt;first ionization energy&lt;/strong&gt;: Na is &lt;strong&gt;supplied&lt;/strong&gt; with energy in order to become Na+, so it is an &lt;strong&gt;endoergonic process&lt;/strong&gt;; a strongly energetic one. When NaCl dissolves in water, the partially negative side of water molecule bonds weakly with Na+ (intermolecular bond, which has electrostatic character); in a qualitative view, forming a bond is an &lt;strong&gt;exoergonic process&lt;/strong&gt;; since this type of bond is weak, the energy lost by Na+ isn't enough to bring Na+ back to Na.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cl- has a lower energy than Cl has, so when Cl- forms a weak bond with the partially positive side of water molecule, its energy lowers and it's thermodinamically impossible for it to become Cl.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What you asked about poisoning power of Cl is a Biochemistry topic, but it's reasonable thinking that in a healty person's body at room conditions (e.g. &lt;strong&gt;not&lt;/strong&gt; a scuba diver underwater) the total amount of Cl (Cl• radicals or Cl2 molecules) isn't large enough to cause health issues and it won't appreciably raise by drinking salted water. On the other hand, Cl- is part of our daily consumption of mineral salts.&lt;/p&gt;&#xA;" OwnerUserId="17957" LastEditorUserId="17957" LastEditDate="2015-08-06T16:06:30.717" LastActivityDate="2015-08-06T16:06:30.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34779" PostTypeId="1" AcceptedAnswerId="34946" CreationDate="2015-08-06T16:22:39.890" Score="14" ViewCount="501" Body="&lt;p&gt;Since I calculate NMR spectra for a while by simply using one of recommendations by &lt;a href=&quot;http://cheshirenmr.info/Recommendations.htm&quot; rel=&quot;noreferrer&quot;&gt;CHEmical SHIft REpository&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mPW1PW91/6-311+G(2d,p)-SCRF//B3LYP/6-31+G(d,p)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;without thinking to much how it actually works*, there is a common misprediction that I'd like to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be that the shift/shielding of the brominated carbons is always way off the measured shift. Compared to all other &quot;normal&quot; substituted carbons (C, H, O, N...) it is also far from being predicted good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily for me, as I don't want to show my own molecules :D, the Repository gives an &lt;a href=&quot;http://cheshirenmr.info/Examples.htm&quot; rel=&quot;noreferrer&quot;&gt;example&lt;/a&gt; (Aplydactone) with a brominated carbon that shows exactly what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdDqs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall $^1$H Mean Absolute Deviation (MAD) is about 0.08 ppm and for the $^{13}$C shifts the MAD is about 1.5 ppm but &lt;strong&gt;without&lt;/strong&gt; the brominated carbon (right side of picture). The predicted shift has a failure of 12.25 ppm (measured 65.5 ppm vs. calculated 77.75 ppm) which is about ten times the MAD of all other carbon shifts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a common explanation for that? Is it simply that there were no brominated samples within the test set or is there a much deeper reason? Like Bromine must be treated with a different/super special basis set or sth similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This comes right now into my mind ...&lt;/em&gt; As there might already be some influences of the relativistic effect, could that be a reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me of your deeper insight into the calculation of NMR shieldings that could yield to this misprediction that I either understand how to correct those errors or that I at least have an explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I did no further research throughout the internet yet, as I have not much time right now. But as this question keeps in my mind, maybe someone of you has done the research and would share it with me.&lt;/em&gt;&#xA;&lt;hr&gt;&#xA;* I know that it uses a regression function which was calculated through comparing lots of experimental vs. predicted shifts for a bunch of small organic substances. What I mean is more the quantum theoretical aspect.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-08-17T08:23:47.317" Title="Calculated 13C NMR Shifts of brominated Carbons" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34781" PostTypeId="2" ParentId="34772" CreationDate="2015-08-06T16:52:22.897" Score="-1" Body="&lt;p&gt;First point of difference is applied here. Chloro comes first in alphabetical order and lowest sum rule is applied when first point of difference is not working.&#xA;In 2nd case double bond has higher priority over substituent.&lt;/p&gt;&#xA;" OwnerUserId="17918" LastActivityDate="2015-08-06T16:52:22.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34782" PostTypeId="1" CreationDate="2015-08-06T17:05:30.523" Score="3" ViewCount="1305" Body="&lt;p&gt;I have a test result that I am trying to fully grasp. I need to determine how much of a contaminant I need to reach that amount in 4,800 gallons. Making the conversion to liters is no problem, nor is the math necessary to figure how much I need by weight to get this much contaminant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one thing I am unclear on is my assumption that $44.5~\mathrm{mg~L^{-1}}$ is by weight.&#xA;In other words, if I have a result that measures $50~\mathrm{mg~L^{-1}}$ of a metal (nickel, for instance), that means if I were to be able to extract the metal and weigh it, it would weigh $50~\mathrm{mg}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help you can offer would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17960" LastEditorUserId="10185" LastEditDate="2015-08-06T17:12:16.110" LastActivityDate="2017-07-26T16:13:46.713" Title="Is 44.5 mg/L a measurement of weight or volume?" Tags="&lt;units&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34783" PostTypeId="2" ParentId="34782" CreationDate="2015-08-06T17:15:59.293" Score="8" Body="&lt;p&gt;$50~\mathrm{mg~L^{-1}}$ is an expression of concentration by mass. It means that if you took one litre of the solution and extracted all the metal, the resulting metal would weigh 50 milligrams.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other common option is to express concentration in terms of moles per litre ($\mathrm{mol~L^{-1}}$) which means that if you extracted all the solute from one litre of solution, you would have 50 moles of solute.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-08-07T08:41:05.947" LastActivityDate="2015-08-07T08:41:05.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34784" PostTypeId="2" ParentId="34782" CreationDate="2015-08-06T17:18:57.643" Score="5" Body="&lt;p&gt;The test result is mass per volume, or &lt;em&gt;concentration&lt;/em&gt;. It just says there are 44.5 mg of (whatever) in each liter of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking how much total contaminant there is in 4800 US gallons given a test result of 44.5 mg/L:&#xA;$$\left(44.5\,\mathrm{mg\over L}\right)\cdot\left({1\,\mathrm{L}\over 0.264172052 \,\mathrm{US\,gal}}\right)\cdot(4800\,\mathrm{US\,gal})\cdot\left({1\,\mathrm{g}\over 1000\,\mathrm{mg}}\right) = 808.6\,\mathrm{g}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2017-07-26T16:13:46.713" LastActivityDate="2017-07-26T16:13:46.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34786" PostTypeId="1" AcceptedAnswerId="34791" CreationDate="2015-08-06T17:25:32.513" Score="5" ViewCount="4049" Body="&lt;p&gt;Today, my chemistry teacher gave the following organic conversion as homework: benzene to 1-bromo-3-iodobenzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qcWzq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcWzq.png&quot; alt=&quot;synthesis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about it, can someone review my conversion (as I have got an exam tomorrow)?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="4945" LastEditDate="2020-02-19T13:46:51.330" LastActivityDate="2020-02-19T13:46:51.330" Title="How to convert benzene to 1-bromo-3-iodobenzene?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34791" PostTypeId="2" ParentId="34786" CreationDate="2015-08-06T17:52:13.550" Score="4" Body="&lt;p&gt;It looks pretty good overall. The order of substitution is correct to achieve the required pattern but there are two points I would pick up on.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The final substitution of the diazonium salt is best done with $\ce{CuI}$ rather than $\ce{KI}$ as the copper(I) ions catalyse the reaction. Also heating is not required; the reaction can be run at room temperature (or possible slightly above but not much).&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The diazonium salt is drawn incorrectly since it is ionic in nature. There is a triple bond between the nitrogens and a formal positive charge on the middle nitrogen, although in reality the charge is delocalised over both atoms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5XqIO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5XqIO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-08-06T17:52:13.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34792" PostTypeId="2" ParentId="34399" CreationDate="2015-08-06T18:06:57.403" Score="1" Body="&lt;p&gt;Sodium polyacrylate can locally organize solvent molecules to become soluble. Since the backbone is huge compared to water, the water aligns with its more positively charged end towards the carboxylate site while the sodium ions end up on the negative end of the water. I can't find the good papers right now, but if you search google scholar for using polyacrylic acid as a draw solute for forward osmosis you can dig way deeper into the mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means practically is that you need really polar solvents that can line up with the ions. &lt;a href=&quot;https://en.wikipedia.org/wiki/Protic_solvent&quot; rel=&quot;nofollow&quot;&gt;Wikipedia has a good page&lt;/a&gt; explaining solvent classification in more detail with a table of solvents including their dipole moment (quantifying the polarity). As suggested by Mithoron, DMSO seems to be a good option. Unfortunately, after that you start getting into much nastier solvents like acetonitrile and DMF.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2015-08-06T18:06:57.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34793" PostTypeId="1" CreationDate="2015-08-06T18:17:41.980" Score="2" ViewCount="72" Body="&lt;p&gt;The internet is full of videos like &lt;a href=&quot;https://www.youtube.com/watch?v=7NCEJqFC1uw&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; showing water bottles for drinking used to supercool water. These bottles, not unexpectedly, appear a bit finicky, since they have lots of edges to serve as nucleation sites for ice crystals. What container would work best for this sort of demonstration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the most important quality just to eliminate edges or does the material have an important effect? It's important in lab to add boiling stones as nucleation sites when boiling water in glassware. Could I use this trait for freezing too? Is the important quality of the glass the high surface energy or the smoothness of the surface? Alternatively, it seems like something like PTFE coating with super low surface energy might be effective?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorUserId="10185" LastEditDate="2015-08-06T18:19:21.203" LastActivityDate="2017-02-18T22:46:41.347" Title="Container to supercool water" Tags="&lt;physical-chemistry&gt;&lt;experimental-chemistry&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34796" PostTypeId="1" CreationDate="2015-08-06T21:11:51.403" Score="6" ViewCount="339" Body="&lt;p&gt;I'm looking for a reasonably accurate density of a mixture of two gases that act quite unlike ideal gases (generally steam and water-miscible solvents). We can get a first order estimate by taking a molar average but that can be an order of magnitude off, as in the case of water/ammonia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literature is the obvious place to start but I wasn't able to find any general sources and only a very few individual pairs in a narrow temperature/pressure range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess you should be able to derive an excess volume from fugacity but I haven't been able to find a formula that doesn't take even more difficult-to-measure inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard modeling references like Refprop seem to use an ideal mixing model which isn't right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also imagine measuring them myself or doing molecular dynamics modeling but these are both quite a bit more work than I was hoping for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any literature sources for this data that I couldn't find? Anyone have an idea of another way to calculate it?&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorUserId="16683" LastEditDate="2015-10-23T16:39:38.737" LastActivityDate="2015-10-23T16:39:38.737" Title="Density of non-ideal vapor mixtures" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34797" PostTypeId="2" ParentId="33269" CreationDate="2015-08-07T03:50:20.303" Score="2" Body="&lt;p&gt;You need to incorporate all $3N-5$ modes (both bends and both stretches) in order to accurately calculate the ZPE even if, per Jan's answer, you can neglect one of the bending modes in constructing an accurate model of the PES. Even though the bends appear identical in mind's eye, both of their (degenerate) ZPE's must be included in order for the ground state calculated by the model to be accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I think about it is, imagine a $\ce{CO2}$ molecule is oriented along the $z$-axis, with the carbon at the origin. If you only allow one bending mode, say with both oxygens translating toward $+x$ and the carbon toward $-x$, you have to hold the orientation of that motion fixed and your system will be unable to accommodate vibrational movement of any atoms in the $y$-direction. Accurate evaluation of the ZPE requires a model that allows bending along both axes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The maddening thing about vibrating linear molecules in general, though, is that most linear combinations of a certain pair of these degenerate bending motions &lt;em&gt;actually contain a rotational component&lt;/em&gt;.  Using $\ce{CO2}$ as an example and ignoring &lt;a href=&quot;https://en.wikipedia.org/wiki/Anharmonicity#Examples_in_physics&quot; rel=&quot;nofollow noreferrer&quot;&gt;anharmonicity&lt;/a&gt;, if the bending vibrations are $90^\circ$ out of phase with one another, the oxygen atoms gyrate around the $\ce{O=C=O}$ axis just like they would in a rotating molecule with a bent equilibrium geometry (imagine the orange dot is $\ce{C}$ and the green dot is a projection of the positions of the two $\ce{O}$'s):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Circle_cos_sin.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJmSq.gif&quot; alt=&quot;Unit circle animation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Circle_cos_sin.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the radius of the circle of projection of the motion of the oxygens will be &lt;em&gt;quite&lt;/em&gt; small, I believe less than a degree, but it still is a rotation about the molecular axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phase angles other than $90^\circ$ result in some combination of a bending motion and a rotation, where only in the $\theta=0^\circ$ case does the rotational aspect vanish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DvZKC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DvZKC.png&quot; alt=&quot;Rotations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;For any linear molecule with three or more atoms, the degenerate bends involving the three atoms at each terminus of the molecule are the pair that can generate rotation.  For $\ce{CO2}$ that means the entire molecule; for acetylene each vibration involves the two carbons and one of the hydrogens; etc.&lt;/strike&gt; &lt;strong&gt;EDIT:&lt;/strong&gt; The more I think about it, the more confident I am that every pair of degenerate bends in a linear molecule actually has this capability to generate rotation. It's just that any bends not centered at the central atom (if it exists) will cause the molecule to &lt;a href=&quot;https://en.wikipedia.org/wiki/Precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precess&lt;/a&gt;, as well as rotate, due to the asymmetry of the two 'arms' of the bending motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to some extent, I would actually argue your question is ill-posed: &lt;em&gt;Linear molecules are &lt;strong&gt;always rotating&lt;/strong&gt;!&lt;/em&gt; They're just rotating in a manner inconsistent with the standard &lt;a href=&quot;https://en.wikipedia.org/wiki/Rigid_rotor&quot; rel=&quot;nofollow noreferrer&quot;&gt;rigid-rotor&lt;/a&gt; assumption, and (I think?) at energies much higher than those one would expect to see in conventional rotational spectroscopy.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-08-07T04:19:58.237" LastActivityDate="2015-08-07T04:19:58.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34798" PostTypeId="1" CreationDate="2015-08-07T04:47:26.893" Score="7" ViewCount="244" Body="&lt;p&gt;If the transfer of $1~\mathrm{kJ}$ of energy to a quantity of water causes its temperature to rise $10~^\circ\mathrm{C}$, what is the mass of the water?  I know how to do the math and everything but I'm so confused with significant figures. If $1~\mathrm{kJ}$ or $100~\mathrm{J}$ only has 1 significant figure, then would I write $\pu{23.90 g}$ as $\pu{20 g}$?&lt;/p&gt;&#xA;" OwnerUserId="17969" LastEditorUserId="5026" LastEditDate="2018-02-21T13:28:22.137" LastActivityDate="2018-02-21T13:28:22.137" Title="How to recognise the significant figures in given input values for the calculated output value?" Tags="&lt;physical-chemistry&gt;&lt;significant-figures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34799" PostTypeId="2" ParentId="34798" CreationDate="2015-08-07T05:18:14.100" Score="7" Body="&lt;p&gt;You have a reasonable question: if the data supplied specified 1.00 $\mathrm{kJ}$, and the temperature change was given as $10.00~^{\circ}\mathrm{C}$, then &quot;23.90 g&quot; would be a justifiable answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the data precision was given to only one digit for energy, then the mass would be &quot;20 g&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, given enough space, I'd write an explanation, of the sort you've give, e.g. &quot;Though I calculated 23.90 g, it could be rounded to 20 g due to lack of significant digits in the data given.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, you've shown good effort in &lt;em&gt;thinking through&lt;/em&gt; the homework assignment, in my opinion, which is as important as a numerically correct answer.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="865" LastEditDate="2015-08-07T12:51:55.327" LastActivityDate="2015-08-07T12:51:55.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34801" PostTypeId="2" ParentId="34384" CreationDate="2015-08-07T08:12:39.567" Score="0" Body="&lt;p&gt;TNT(trinitrotoluene),Potassium permanganate,Potassium dichromate,hydrogen peroxide,fluorine are some of the oxidising agent which you can use light up LPG without requirement of atmospheric oxygen.&lt;/p&gt;&#xA;" OwnerUserId="15837" LastActivityDate="2015-08-07T08:12:39.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34802" PostTypeId="1" CreationDate="2015-08-07T08:25:43.503" Score="5" ViewCount="391" Body="&lt;p&gt;According to the article &lt;a href=&quot;http://www.teagasc.ie/research/reports/foodprocessing/4196/eopr-4196.pdf&quot; rel=&quot;nofollow&quot;&gt;Extending the Shelf Life of Fresh Sliced Mushrooms&lt;/a&gt;, it states that 5% (v/v) hydrogen peroxide is an effective means for the preservation of sliced mushrooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to the preservation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the USA hydrogen peroxide&#xA;  is an approved bactericide for some dairy products and is used for disinfecting fruit&#xA;  and vegetables. Its activity is due to its oxidising effects on&#xA;  bacteria and it also bleaches mushrooms during the soaking period.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does oxidising effects cause the bleaching, hence preservation of sliced mushrooms?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-07T11:18:46.180" LastActivityDate="2016-07-26T14:47:48.853" Title="Preservation and bleaching of sliced mushrooms with hydrogen peroxide" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34804" PostTypeId="2" ParentId="34539" CreationDate="2015-08-07T09:02:59.267" Score="1" Body="&lt;p&gt;Sodium periodate can be used indirectly to form diols from alkene, see:&#xA;Emmanuvel, T. M. A. Shaikh, A. Sudalai, Org. Lett., 2005, 7, 5071-5074.&#xA;The mechanism for periodate cleavage of diols also strongly suggests the periodate is an electrophile.  This may explain why in general, alkenes are not nucleophilic enough to react under normal conditions without a co-reagent. The osmium tetroxide reaction is favoured as a [3+2] cycloaddition mechanism. &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2015-08-07T09:02:59.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34805" PostTypeId="1" CreationDate="2015-08-07T09:36:03.513" Score="0" ViewCount="70" Body="&lt;p&gt;Seems so simple but is confusing me as when I take into account final weight I don't get one of the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An iron ore was analysed by dissolving a $1.1324~\mathrm{g}$ sample in concentrated $\ce{HCl}$ and converting it to give $0.5394~\mathrm{g}$ of pure $\ce{Fe2O3}$. The $\mathrm{m\%}_\ce{Fe}$ in the sample is either $18.9~\%$, $25.2~\%$, $37.7~\%$ or $50.4~\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My approach:&lt;/strong&gt;&lt;br&gt;&#xA;If I take the molar mass of iron as $55.85~\mathrm{g/mol}$ and oxygen as $15.99~\mathrm{g/mol}$. I have a $159.67~\mathrm{g/mol}$ weight for $\ce{Fe2O3}$. From this I calculated that amount of $\ce{Fe2O3}$ is$$\frac{0.5394~\mathrm{g}}{159.67~\mathrm{g/mol}}=3.378 \cdot 10^{-3}~\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is then multiplied by ($2 \times 55.85~\mathrm{g/mol}$) giving $0.3773~\mathrm{g}$ iron but this is what is confusing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't the answer I got just for $0.5394~\mathrm{g}$ so shouldn't I take into account that I started off with $1.1324~\mathrm{g}$? If I do I get $33.32~\%$ so doesn't match an answer.&lt;/p&gt;&#xA;" OwnerUserId="17973" LastEditorUserId="16622" LastEditDate="2015-08-07T12:36:36.987" LastActivityDate="2015-08-07T12:36:36.987" Title="% Fe in a sample" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="34806" PostTypeId="1" CreationDate="2015-08-07T09:58:18.067" Score="2" ViewCount="176" Body="&lt;p&gt;I want to prepare a sample of lithium metaborate ($\ce{LiBO2}$) and need to dissolve it for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solvents I found so far are water and ethanol. In ethanol, one cannot dissolve enough of the solute. Water is quite nice from the solubility point of view, but its high surface tension and tendency of wetting forbids (more or less) its use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have an idea for another solvent as an alternative for water?&lt;/p&gt;&#xA;" OwnerUserId="17974" LastEditorUserId="865" LastEditDate="2015-08-07T12:30:50.810" LastActivityDate="2015-08-07T12:30:50.810" Title="Alternative to water as solvent for lithium metaborate?" Tags="&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solvents&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34809" PostTypeId="1" CreationDate="2015-08-07T14:03:18.247" Score="6" ViewCount="4599" Body="&lt;p&gt;My 10th grade chemistry book states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. Metals like potassium and &lt;strong&gt;sodium&lt;/strong&gt; react violently with cold water. In case of sodium and potassium, the reaction is so violent and exothermic, that evolved &lt;strong&gt;hydrogen immediately catches fire&lt;/strong&gt;.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So $\ce{Na}$ should react like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) $\ce{2Na + H2O -&amp;gt; Na2O + H2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) $\ce{Na2O + H2O -&amp;gt; 2NaOH}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When hydrogen catches fire -- that is, when hydrogen reacts with $\ce{O2}$, right? -- does it form $\ce{H2O}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If water is not formed, then does the water level decrease?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is (ii) an exothermic reaction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17979" LastEditorUserId="194" LastEditDate="2017-04-01T03:22:08.580" LastActivityDate="2017-04-01T03:22:08.580" Title="Does sodium produce water when it reacts with water?" Tags="&lt;water&gt;&lt;energy&gt;&lt;metal&gt;&lt;heat&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34810" PostTypeId="2" ParentId="31651" CreationDate="2015-08-07T14:19:16.510" Score="3" Body="&lt;p&gt;Here's a device that's been around for at least several decades, as far as I know:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KERkJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KERkJ.jpg&quot; alt=&quot;water heater&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.teploterm.ru/public/files/000001.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;teploterm.ru&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;I've been using something similar for around 15 years. The idea is you put firewood into the oven you can see at the bottom, and the chimney goes through a 90 liter water tank. The bottom is something like .5 - 1 cm thick cast iron, the rest is made from stainless steel sheet metal.&lt;/p&gt;&#xA;&lt;p&gt;There are variants to the construction, but I think here are the two main points, coming back to your question:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This thing needs replacement around once a decade, being in use around 500-1000 hours a year.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the stainless steel is thick, then rusting gradually becomes irrelevant. Does make the thing heavy, though.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, there is corrosion, especially around the fiery hot parts.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Also, note that the plumbing within this thing is kind of special, as it needs to refill cold water while simultaneously mixing your hot and cold water. I remember that it was very damaging to run it empty.&lt;/p&gt;&#xA;&lt;p&gt;And yeah, this is more engineery than chemicky...&lt;/p&gt;&#xA;" OwnerUserId="17932" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-03-09T22:09:08.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34811" PostTypeId="2" ParentId="34687" CreationDate="2015-08-07T16:02:47.027" Score="-1" Body="&lt;p&gt;With increasing pressure volume of system increases and thus active mass increaeas  then to keep the constant value of Keq whole setup is disturbed in direction where no of moles of reacting substance decreases. Thus increment in rate of one reaction is preferred over other.&lt;/p&gt;&#xA;" OwnerUserId="17918" LastActivityDate="2015-08-07T16:02:47.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34815" PostTypeId="2" ParentId="34809" CreationDate="2015-08-07T18:10:24.167" Score="1" Body="&lt;p&gt;The reaction of sodium metal and water could (and usually does) indirectly produce more water, if enough heat is produced to spontaneously ignite the hydrogen gas which is produced by the initial reaction. This cyclic type of reaction is often found in explosive or extremely exothermic reactions, e.g. $\ce{K+ + I- + H2O2}$ will also react violently in a cyclic way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion of hydrogen in air is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2 + O2 -&amp;gt; 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a (well) closed system with excess oxygen, water could re-form and collect. It's worth noting that dissolution of $\ce{NaOH}$ in water ($\ce{NaOH -&amp;gt; Na+ + OH-}$) is also exothermic and adds even more heat to the system.&lt;/p&gt;&#xA;" OwnerUserId="17821" LastEditorUserId="17821" LastEditDate="2015-08-07T18:21:22.153" LastActivityDate="2015-08-07T18:21:22.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34816" PostTypeId="1" CreationDate="2015-08-07T18:57:37.567" Score="24" ViewCount="114426" Body="&lt;p&gt;I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{N2}$ the orbitals in increasing energy are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_\mathrm{1s} &amp;lt; \sigma^*_\mathrm{1s} &amp;lt; \sigma_\mathrm{2s} &amp;lt; \sigma^*_\mathrm{2s} &amp;lt; \pi_{\mathrm{2p}_x}, \pi_{\mathrm{2p}_y} &amp;lt; \sigma_{\mathrm{2p}_z} &amp;lt; \pi^*_{\mathrm{2p}_x}, \pi^*_{\mathrm{2p}_y} &amp;lt; \sigma^*_{\mathrm{2p}_z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because it has 14 electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as that of $\ce{N2}$ or not? Why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because this is a good chance to ask, why is there this difference in orbital energies anyway?&lt;/p&gt;&#xA;" OwnerUserId="17987" LastEditorUserId="16683" LastEditDate="2017-01-21T17:40:31.100" LastActivityDate="2017-12-25T10:43:30.953" Title="Molecular orbital (MO) diagram for N2 and N2^-" Tags="&lt;electronic-configuration&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34818" PostTypeId="1" AcceptedAnswerId="34829" CreationDate="2015-08-07T20:19:52.900" Score="19" ViewCount="6062" Body="&lt;p&gt;Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of $\ce{HF}$ than any other hydrogen halide, which means the positive charge on hydrogen atom is the greatest in this compound and hence a comparatively small negative charge is needed to attract it. If so, why is it classified as least acidic hydrogen halide if the $\ce{H+}$ is easy to remove?&lt;/p&gt;&#xA;" OwnerUserId="7697" LastEditorUserId="9961" LastEditDate="2020-03-03T16:20:44.620" LastActivityDate="2021-05-26T22:14:52.407" Title="Is HF the least acidic hydrogen halide?" Tags="&lt;acid-base&gt;&lt;electronegativity&gt;&lt;halides&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34820" PostTypeId="2" ParentId="34818" CreationDate="2015-08-07T20:39:44.250" Score="10" Body="&lt;p&gt;Hydrofluoric acid is the least acidic hydrogen halide because of fluorine's electronegativity. Because of the fluoride ion's small size, it cannot disperse the negative charge over a larger space and will have an extremely high affinity for an electrophile (like $\ce{H+}$), and because of this it will remain mostly as $\ce{HF}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high concentrations, through hydrogen bonding with its conjugate base, hydrofluoric acid is actually a significantly stronger acid due a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Homoassociation&quot;&gt;homoassociation&lt;/a&gt;. (credit to Wildcat)&lt;/p&gt;&#xA;" OwnerUserId="15410" LastEditorUserId="15410" LastEditDate="2015-08-07T20:54:43.543" LastActivityDate="2015-08-07T20:54:43.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34821" PostTypeId="1" CreationDate="2015-08-07T20:39:51.197" Score="2" ViewCount="61" Body="&lt;p&gt;I know that the exact (OK most of the times not the exact) energy of an electron can be calculated by solving Schrodinger's differential equation, but can we explain an electron's energy of a specific orbital using the Penetration and Shielding effect?&lt;/p&gt;&#xA;" OwnerUserId="17989" LastEditorUserId="10185" LastEditDate="2015-08-07T21:22:35.403" LastActivityDate="2015-08-07T21:22:35.403" Title="Electron's Energy In The Atom" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34823" PostTypeId="1" CreationDate="2015-08-07T20:57:54.510" Score="1" ViewCount="2947" Body="&lt;p&gt;The effective nuclear charge $Z_\mathrm{eff}$ increases from left to right and from top to bottom. Can you explain why it increases from top to bottom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also can we explain the periodic trend of electronegativity using the shielding effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;The calculation of $Z_\mathrm{eff}$ for Li would be 1 (3 protons − 2 core electrons). For Na is 1 (11 protons − 10 core electrons). So far it doesn't seem to me that $Z_\mathrm{eff}$ should increase from top to bottom. Considering the fact though that for Na there are more electrons shielding the only valence electron than Li as well as Na's electron is relatively more distant than that of Li, the first should be experiencing less $Z_\mathrm{eff}$. This is not true however. As far electronegativity is concerned I can't come up with a neat explanation. &lt;/p&gt;&#xA;" OwnerUserId="17990" LastEditorUserId="865" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T21:00:28.610" LastActivityDate="2021-05-22T21:00:28.610" Title="Periodic Trends of Zeff and Electronegativity" Tags="&lt;atoms&gt;&lt;periodic-trends&gt;&lt;electronegativity&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34825" PostTypeId="2" ParentId="34821" CreationDate="2015-08-07T21:01:42.707" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we explain an electron's energy of a specific orbital using the&#xA;  Penetration and Shielding effect?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, &lt;em&gt;to some extent&lt;/em&gt; it is possible. Look no further than &lt;a href=&quot;https://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slater's rules&lt;/a&gt; which specifies the values of the effective nuclear charge $Z^*$, and the effective principal quantum number $n^*$ that can be used to replace the nuclear charge $Z$ and the principal quantum number $n$ in the hydrogen-like atom orbital energy expression. The energy of each and every orbital can be approximated then by&#xA;$$&#xA;    E = - \left( \frac{Z^*}{n^{*}} \right)^{2}. &#xA;$$&#xA;But I would be careful calling this an explanation, since it is rather a &lt;em&gt;rationalization&lt;/em&gt;. I wrote about the difference between those two &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34323/why-does-hybridization-significantly-matter-for-organic-chemistry/34327#34327&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-08-07T21:10:57.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34827" PostTypeId="2" ParentId="34823" CreationDate="2015-08-07T22:52:52.483" Score="0" Body="&lt;p&gt;The electronegativity is the greatest at the top of the periodic table because fewer electrons are shielding the outermost electrons from the attraction of the nucleus.  As more electrons are added the electrons closer to the nucleus repel some of the outermost electrons and block the nucleus's attraction.  Within one left to right row the electronegativity increases as you move right because more protons exert an attractive pull on the electrons without adding more electrons that shield the charge of the nucleus.&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="16106" LastEditDate="2015-08-08T11:38:50.207" LastActivityDate="2015-08-08T11:38:50.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34828" PostTypeId="1" CreationDate="2015-08-07T23:33:12.187" Score="1" ViewCount="8631" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Iron%28II%29_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iron(II) sulfate&lt;/a&gt; (ferrous sulfate) is a salt that, in an aqueous solution, dissolves and yields $\ce{Fe^{2+}(aq)}$ and $\ce{SO4^{2-}(aq)}$. It then &lt;a href=&quot;https://chemistry.stackexchange.com/questions/31243/what-is-formed-when-you-leave-iron-ii-sulfate-in-plain-air&quot;&gt;decomposes to ferric sulfate&lt;/a&gt; shortly thereafter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happens to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron%28III%29_sulfate&quot; rel=&quot;nofollow noreferrer&quot;&gt;iron(III) sulfate&lt;/a&gt; (ferric sulfate) in solution? Is the $\ce{Fe_2(SO_4)_3}$ compound preserved?&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-08-08T18:11:51.470" Title="What happens to iron(III) sulfate (ferric sulfate) in aqueous solution?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;ions&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34829" PostTypeId="2" ParentId="34818" CreationDate="2015-08-08T00:38:55.427" Score="18" Body="&lt;p&gt;First of all as @chipbuster says &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; in diluted solutions in water is nearly completely dissociated and therefore &lt;strong&gt;shouldn't be called weak&lt;/strong&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrofluoric_acid#Properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; describes this nicely and cites &lt;a href=&quot;https://dx.doi.org/10.1021%2Fja00537a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;several&lt;/a&gt; &lt;a href=&quot;https://dx.doi.org/10.1021%2Fja077846o&quot; rel=&quot;nofollow noreferrer&quot;&gt;sources&lt;/a&gt; for this claim.&lt;/p&gt;&#xA;&lt;p&gt;It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Tight_ion_pair&quot; rel=&quot;nofollow noreferrer&quot;&gt;tight ion pairs&lt;/a&gt;. It prevents from detecting the true strength of &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; with methods like &lt;a href=&quot;https://en.wikipedia.org/wiki/Acid%E2%80%93base_titration&quot; rel=&quot;nofollow noreferrer&quot;&gt;acid–base titration&lt;/a&gt; - they show dissociation of ionic pair. Similar effect is present in &lt;a href=&quot;https://en.wikipedia.org/wiki/Ion-exchange_resin&quot; rel=&quot;nofollow noreferrer&quot;&gt;ion exchange resins&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Strength of this acid is revealed in more concentrated solutions, where &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; molecules replace hydronium ions, creating &lt;a href=&quot;https://en.wikipedia.org/w/index.php?title=Bifluoride&quot; rel=&quot;nofollow noreferrer&quot;&gt;bifluoride&lt;/a&gt; anions, freeing them - it's homoassotiation mentioned by Wildcat.&lt;/p&gt;&#xA;&lt;p&gt;More to the point - does it mean that &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.&lt;/p&gt;&#xA;&lt;p&gt;In terms of &lt;span class=&quot;math-container&quot;&gt;$\ce{K_a}$&lt;/span&gt; in diluted aqueous solutions &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; is probably still weaker than &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; and the rest but getting exact value is greater problem than normally (generally values of &lt;span class=&quot;math-container&quot;&gt;$\ce{K_a}$&lt;/span&gt; for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/13609/hydrogen-halide-acidity&quot;&gt;this question&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;For concentrated solutions one needs to use &lt;a href=&quot;https://en.wikipedia.org/wiki/Hammett_acidity_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hammett &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; function&lt;/a&gt; which gets lower than -10 in pure &lt;span class=&quot;math-container&quot;&gt;$\ce{HF}$&lt;/span&gt; (pure sulfuric acid has -12) - it means it's &lt;strong&gt;very strong&lt;/strong&gt;, although not &lt;a href=&quot;https://en.wikipedia.org/wiki/Superacid&quot; rel=&quot;nofollow noreferrer&quot;&gt;superacidic&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;See also:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja01571a016&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Hammett Acidity Function &lt;span class=&quot;math-container&quot;&gt;$H_0$&lt;/span&gt; for Hydrofluoric Acid Solutions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2021-05-26T22:14:52.407" LastActivityDate="2021-05-26T22:14:52.407" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34830" PostTypeId="1" CreationDate="2015-08-08T05:21:19.257" Score="4" ViewCount="4259" Body="&lt;p&gt;I know that density decreases across period 4 because the last two elements are liquids and gases which means that their densities are less but why are the densities increasing and then decreasing within the transition metals themselves?&lt;/p&gt;&#xA;" OwnerUserId="17998" LastEditorUserId="10185" LastEditDate="2015-08-08T09:21:43.957" LastActivityDate="2018-01-29T05:48:02.947" Title="Why does density decrease across the period 4 and 5 transition metals?" Tags="&lt;transition-metals&gt;&lt;periodic-trends&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34832" PostTypeId="2" ParentId="34816" CreationDate="2015-08-08T09:30:27.810" Score="3" Body="&lt;p&gt;The reason for the special order for molecules with 14 or less electrons is due to the s-p mixing that arises due to the s and p atomic orbitals being close in energy. It is not actually possible to definitely answer your question without looking at the actual orbital energies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite $\ce{N2-}$ being isoelectronic to $\ce{O2+}$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. Hence s-p mixing should occur. However there are other considerations that we need to take in account, such as the extra electron repulsion that occur due to the extra electron. Yet, as I have stated above we can't definitively say which effect is more important without looking at the actual orbital energies.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-08-08T10:58:36.147" LastActivityDate="2015-08-08T10:58:36.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34833" PostTypeId="1" CreationDate="2015-08-08T09:33:00.317" Score="0" ViewCount="2831" Body="&lt;p&gt;As in weight per volume? I mean.. you always see them specified by weight but they have to have a volume, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose different fat(for example) molecules would have different densities but then how do they universally have the same energy density? Are they converted to like a standardized type of fat? Thats the fat I'm interested in.&lt;/p&gt;&#xA;" OwnerUserId="18004" LastActivityDate="2015-08-08T13:28:03.140" Title="What is the density of carbohydrates/protein/fat?" Tags="&lt;organic-chemistry&gt;&lt;carbohydrates&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-08-08T17:46:29.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="34834" PostTypeId="2" ParentId="34816" CreationDate="2015-08-08T10:23:27.647" Score="23" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have been taught that the MO diagram is different for molecules with&#xA;  14 or less electrons than the one used for molecules with 15 or more&#xA;  electrons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, it is rationalized by a successive decrease of the $\mathrm{s}$-$\mathrm{p}$ interaction moving from $\ce{Li2}$ to $\ce{F2}$. The $\mathrm{s}$-$\mathrm{p}$ interaction is the bonding interaction between the $\mathrm{2s}$ orbital of one atom and the $\mathrm{2p_{z}}$ orbital of another atom which (among other things) increases the energy of the $\mathrm{\sigma_{2p_{z}}}$ MO relative to the hypothetical case without $\mathrm{s}$-$\mathrm{p}$ interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the difference in energy between the $\mathrm{2s}$ and $\mathrm{2p_{z}}$ AOs increases from $\ce{Li2}$ to $\ce{F2}$ due to increasing nuclear charge and poor screening of the $\mathrm{2s}$ electrons by electrons in the $\mathrm{2p}$ subshell. As a result, the rising effect of $\mathrm{s}$-$\mathrm{p}$ interaction on $\mathrm{\sigma_{2p_{z}}}$ MO is getting less and less prominent, so that eventually to the right of $\ce{N2}$ $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\pi_{2p_{x,y}}}$ MO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4KdEO.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4KdEO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image courtesy of the &lt;a href=&quot;https://web.archive.org/web/20150706110342/http://chemwiki.ucdavis.edu/Wikitexts/UC_Davis/UCD_Chem_2A/UCD_Chem_2A:_Hayashi/Unit_IV:_Electronic_Structure_and_Bonding/10:_Bonding_II:_Additional_Aspects/10.5:_Molecular_Orbital_Theory&quot; rel=&quot;noreferrer&quot;&gt;UC Davis ChemWiki&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to the question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as&#xA;  that of $\ce{N2}$ (because it is actually an ionized molecule of $\ce{N2}$) or not?&#xA;  Why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The short answer is: we could not tell it using the primitive molecular orbital theory introduced in the general chemistry courses. In exact same way we could not tell why $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ and not to the left of, say, $\ce{C2}$. All this is simply because the primitive molecular orbital theory does not explain the things, rather it rationalizes them (for differences between explanation and rationalization, please, read &lt;a href=&quot;https://chemistry.stackexchange.com/a/34327/186&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then how do we know in the first place that $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ so that we can start to rationalize it? Well, we know that from a more advanced molecular orbital theory which is based on the laws of quantum mechanics. And the same advanced theory should be applied for the case of $\ce{N2-}$ to get the MO picture of this chemical species. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now note that even in this advanced molecular orbital theory a bunch of approximations is introduced, and the answer in general depends on at which level of theory calculations are done. The simplest picture would be the so-called &lt;em&gt;frozen-orbital approximation&lt;/em&gt; (in the Hartree-Fock picture) in which MOs of an anion (as well as that of a cation) are assumed to be the same as for the corresponding neutral molecule. But this approximate picture might be far away from the reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question in fact does not make a lot of sense since $\ce{N2-}$ is known to be electronically &lt;em&gt;unstable&lt;/em&gt;, i.e. its energetically less favourable then the system of $\ce{N2}$ plus a free electron. There exist few metastable resonance states of $\ce{N2-}$ that correspond to attachment of the scattered electron to excited valence states&lt;sup&gt;1&lt;/sup&gt;, but that's a different story which is beyond the scope of the question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1) Krauss, M., Neumann, D., Valence Resonance States of N2, &lt;em&gt;JResNatlStandSecA&lt;/em&gt;, 77A, 4, 1973. &lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/jres/77A/jresv77An4p411_A1b.pdf&quot; rel=&quot;noreferrer&quot;&gt;PDF&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2017-01-22T11:25:27.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34835" PostTypeId="1" AcceptedAnswerId="38722" CreationDate="2015-08-08T11:46:51.870" Score="2" ViewCount="4984" Body="&lt;p&gt;I came across this concerning phenols:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The hydroxyl group, in phenol is directly attached to&#xA;  the $\ce{sp^2}$&#xA;   hybridised carbon of benzene ring which acts as an&#xA;  electron withdrawing group. Due to the higher electronegativity of $\ce{sp^2}$&#xA;   hybridised carbon&#xA;  of phenol to which $\ce{–OH}$ is attached, electron density decreases&#xA;  on oxygen. This increases the polarity of $\ce{O–H}$ bond and results&#xA;  in an increase in ionisation of phenols than that of alcohols.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My concepts seem to be flawed, and hence I have few questions on the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the $\ce{sp^2}$ carbon atom withdraws electrons from the oxygen atom, the oxygen's electron density decreases. How does this increase the polarity in the $\ce{O-H}$ bond? Shouldn't it be the opposite as oxygen becomes more electropositive due to the carbon atom? I'm missing something here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does the carbon atom in this case act as electron withdrawing because it is less electronegative than the oxygen atom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note that this question isn't about the comparison of acidic character between alcohols and phenols, and the relative stability of alkoxides and phenoxides.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15676" LastEditorUserId="10185" LastEditDate="2016-06-17T14:54:17.800" LastActivityDate="2016-06-17T14:54:17.800" Title="What is the reason for the acidic character in phenols?" Tags="&lt;organic-chemistry&gt;&lt;electronegativity&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34836" PostTypeId="2" ParentId="34833" CreationDate="2015-08-08T13:28:03.140" Score="3" Body="&lt;p&gt;I assume the &quot;energy density&quot; you are referring to are the widely known statistics that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fat contains 9 &quot;calories&quot; per gram (actually kilocalories)&lt;/li&gt;&#xA;&lt;li&gt;protein and carbohydrates both contain 4 &quot;calories&quot; per gram&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can all fats contain &quot;9&quot; kcal per gram but have different mass densities?  Here are a couple of reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The 9 kcal/g figure is an approximation that assumes a typical dietary blend of fats.  The figure would if considering chemically pure fat molecules.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The mass densities of most common fats doesn't vary that much.  The densest &quot;fat&quot; molecules are probably the &lt;em&gt;extremely&lt;/em&gt; short chain triglycerides like &lt;a href=&quot;https://en.wikipedia.org/wiki/Triformin&quot; rel=&quot;nofollow&quot;&gt;triformin&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Triacetin&quot; rel=&quot;nofollow&quot;&gt;triacetin&lt;/a&gt;.  They have densities of approx. $1.3~ \frac{\mathrm g}{\mathrm{cm}^3}$.  These triglycerides have chain lengths of 1 or 2 carbons per fatty acid chain, a value far lower than common in natural fats.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Triolein&quot; rel=&quot;nofollow&quot;&gt;Triolein&lt;/a&gt; is more representative of dietary fats and has a density of 0.95$~\frac{\mathrm g}{\mathrm{cm}^3}$.  So the mass density variation among different fat molecules is pretty small: a 30% increase is all that is really possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-08-08T13:28:03.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34837" PostTypeId="1" CreationDate="2015-08-08T13:44:44.913" Score="5" ViewCount="1613" Body="&lt;p&gt;I am working on a mathematical model for the dissolution of coffee in water. I need to know some precise details about the effects of dissolution on the density of the mixture of the substances (that is, the density of the solution). The system is isothermal and the pressure is constant. I just want to focus on mass, volume, density, and concentration as the relevant variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the drawing (below), there is a graphical representation of the system in question (in the bottom image, I should have mentioned &lt;em&gt;concentration&lt;/em&gt; of water and coffee instead of &lt;em&gt;density&lt;/em&gt;). We have a container of water with a single coffee solid and I've drawn in 20 molecules for both the water and the coffee. The volumes are obviously not to scale (it looks like approximately 50% of the coffee has dissolved in the second pic, not 20%!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Start&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the beginning, $t = 0$, no coffee has dissolved so far and the solution liquid is simply the water. We have the following information for the density and the volume of the coffee, water, and the solution, respectively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_c = 1500$ kg m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_w = 1000$ kg m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_s = \rho_w = 1000$ kg m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_c = 0.1$ m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_w = 0.5$ m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_s = V_w = 0.5$ m$^{-3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is just the water at the start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finish&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At, say, $t = 100$, 20% of the coffee has dissolved and has diffused out evenly into the solution and the system is in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the solution contains both water and coffee. I realise that the chemical reaction dissolution and interactions at a molecular level between the coffee and the water have an impact on the volume and density of the solution. I need information on the density of the solution to decide whether the compressible Navier-Stokes or the incompressible Navier-Stokes govern the flow in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the density  and volume of the solution?&lt;/strong&gt; If experimental measurement is required for this can you instead give me an approximate answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know how much of a change in solution density occurs from $t = 0$ to $t = 100$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yu0xK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yu0xK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18007" LastEditorUserId="865" LastEditDate="2015-08-08T15:02:13.567" LastActivityDate="2015-08-08T18:17:47.347" Title="Is the density of the solution significantly different from the density of the solvent?" Tags="&lt;everyday-chemistry&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34838" PostTypeId="1" CreationDate="2015-08-08T15:13:38.607" Score="1" ViewCount="9470" Body="&lt;p&gt;Like when an e- withdrawing group (sp2 carbon) is attached to oxygen of the OH bond, it attracts the electrons. So will the polarity of the OH bond increase or decrease?&lt;/p&gt;&#xA;" OwnerUserId="15676" LastActivityDate="2020-12-05T10:37:22.287" Title="In an O-H bond, will polarity increase/decrease if some of the electron density on the oxygen atom is reduced?" Tags="&lt;bond&gt;&lt;polarity&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34839" PostTypeId="2" ParentId="34837" CreationDate="2015-08-08T15:42:19.397" Score="1" Body="&lt;p&gt;This depends a great deal both on the solute and the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the concentrations that we normally drink coffee at, you can almost certainly ignore the difference in density of the solution. But ccording to your question you are trying to dissolve 0.1m3 coffee (I assume the minus sign is an error) in 0.5m3 water. That's 100L of coffee in 500L of water, a 100L/600L=16.667% volume/volume solution! A lot of very strong coffee! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, the density rho is given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rho = (mass component 1 + mass component 2) / (volume component 1 + volume component 2 )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes (correctly) that masses can be added. Strictly speaking, the volume of a solution is not equal to the sum of the volumes of its components, as there is a slight volume change on dissolution, but this is normally insignificant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is to know the solid density of coffee, because it cannot be readily measured, as coffee comes in powder form. If you are not using instant coffee there is the problem that the husks of the grains will not dissolve. But in your case, you are told the density of the coffee is 1500kg/m3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can easily work out the masses of water and coffee, and then the density of your enormous vat of extra strong coffee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: it's going to be more than water, but not different enough to make any difference regarding the correct navier stokes model to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: on re-reading, it seems you are assuming that only 20% of your 0.1m3 of coffee will dissolve: that is 0.02m3 in 0.5m3 of water, an approximate concentration of 4%. That's not quite as absurdly strong as before. I was a little confused because 0.1m3 is 20% of 0.5m3. However I'm not sure how you decided that you would reach equilibrium with 20% of the available coffee dissolved. &lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="5541" LastEditDate="2015-08-08T18:17:47.347" LastActivityDate="2015-08-08T18:17:47.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34840" PostTypeId="1" AcceptedAnswerId="34843" CreationDate="2015-08-08T16:30:39.697" Score="1" ViewCount="4491" Body="&lt;p&gt;In chemistry one can recognize that the four states of matter are solid, liquid, gas and plasma. The first is rigid, and has a definite shape and volume. The second doesn't have a shape, and assumes the shape of its container, but it has a fixed volume. The third doesn't have either a shape or a fixed volume and assumes the volume and shape of its container. What about the fourth one (plasma)?&lt;/p&gt;&#xA;" OwnerUserId="19009" LastEditorUserId="10185" LastEditDate="2015-08-08T17:34:51.113" LastActivityDate="2015-08-08T17:34:51.113" Title="Properties of plasmas" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34841" PostTypeId="1" CreationDate="2015-08-08T16:58:21.500" Score="8" ViewCount="419" Body="&lt;p&gt;Following a book of science experiments for children with my daughter, we did the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;put some water in a shallow dish;&lt;/li&gt;&#xA;&lt;li&gt;put some ground pepper floating on the water surface;&lt;/li&gt;&#xA;&lt;li&gt;touch the water surface with the tip of a small stick previously dipped in liquid soap;&lt;/li&gt;&#xA;&lt;li&gt;then the floating pepper escapes radially from the soap.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I was at a loss to explain why. I mumbled something about “surface tension”, but didn't convince even myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was going on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure that the pepper in itself plays no role but showing the surface motion of water. Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2015-09-09T10:54:50.520" Title="Pepper escaping soap on water" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34842" PostTypeId="1" CreationDate="2015-08-08T17:26:10.303" Score="37" ViewCount="21678" Body="&lt;p&gt;What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="16683" LastEditDate="2016-12-13T03:18:25.640" LastActivityDate="2017-04-09T04:56:07.967" Title="Why is BCl3 a monomer whereas AlCl3 exists as a dimer?" Tags="&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;periodic-trends&gt;" AnswerCount="7" CommentCount="3" FavoriteCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="34843" PostTypeId="2" ParentId="34840" CreationDate="2015-08-08T17:27:32.717" Score="2" Body="&lt;p&gt;Plasma is made up of &lt;strong&gt;ionized gas&lt;/strong&gt;, so molecules have a positive electric charge and valency electrons are totally or partially separated by their nuclei.&lt;br&gt;&lt;br&gt;&#xA;Plasma is different from a gas since it has a &lt;strong&gt;high temperature&lt;/strong&gt; and is &lt;strong&gt;radiation emitting&lt;/strong&gt;; think of the Sun and other stars, which are made up of plasma and show both these properties.&lt;br&gt;&lt;br&gt;&#xA;Plasma &lt;strong&gt;hasn't got a proper volume&lt;/strong&gt;, like gases; e.g. stars can expand or contract under the opposite effects of gravity and nuclear fusion. For example, this property is important to comprehend the formation of white dwarfs and neutron stars, process caused by high pression due to gravity.&lt;br&gt;&lt;br&gt;&#xA;Little trivia: there is also a fifth state of matter, whose name is Bose-Einstein Condensate (BEC).&lt;/p&gt;&#xA;" OwnerUserId="17957" LastActivityDate="2015-08-08T17:27:32.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34844" PostTypeId="2" ParentId="34828" CreationDate="2015-08-08T18:11:51.470" Score="5" Body="&lt;p&gt;In aqueous solution $\ce{Fe2(SO4)3}$ dissociates into $\ce{Fe^{3+} (aq)}$ and $\ce{SO4^{2-} (aq)}$ ions. The sulphate ions will be solvated by hydrogen bonding with the water and the iron ions will form the hexaaquairon(III) complex, $\ce{[Fe(H2O)6]^{3+}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Mithoron mentioned, the $\ce{Fe^{3+}}$ ions cannot easily be further oxidised so no additional reaction takes place.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-08-08T18:11:51.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34845" PostTypeId="1" AcceptedAnswerId="34846" CreationDate="2015-08-08T18:24:31.400" Score="2" ViewCount="6094" Body="&lt;p&gt;According to the lowest locant rule, 2,3-dimethyl cyclopentene should be the correct IUPAC name as the sum of the locants of the substituents(2+3=5) of 2,3-dimethyl cyclopentene is lower than the sum of the locants of the substituents(1+5=6) of 1,5-dimethyl cyclopentene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my book says that the correct name of the compound is 1,5-dimethyl cyclopentene. I guess this is due to the fact that methyl gets the 1st locant position. But I am doubtful about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct name?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-08-08T19:13:03.153" Title="Which is the correct name: 2,3-dimethyl cyclopentene or 1,5-dimethyl cyclopentene?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-08T21:47:50.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="34846" PostTypeId="2" ParentId="34845" CreationDate="2015-08-08T18:35:57.893" Score="4" Body="&lt;p&gt;The correct name is 1,5-dimethylcyclopent-1-ene. The double bond has the highest priority and so these two carbons must be numbered 1 and 2 if possible (yes in this case because the molecule is cyclic). We then have to choose between numbering the methyl substituents 1,5 or 2,3. The lowest locant sum rule that you mentioned doesn't actually exist, even though it is often taught. It is a agglomeration of several rules which works most, but not all of the time. The actual rule we are interested in is the first point of difference rule which &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm&quot; rel=&quot;nofollow&quot;&gt;states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2.2 - The longest chain is numbered from one end to the other by Arabic numerals, the direction being so chosen as to give the lowest numbers possible to the side chains. When series of locants containing the same number of terms are compared term by term, that series is &quot;lowest&quot; which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore the locant set 1,5 is lower than the locant set 2,3 since at the first point of difference, 1 is lower than 2.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-08-08T19:13:03.153" LastActivityDate="2015-08-08T19:13:03.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34847" PostTypeId="1" AcceptedAnswerId="71210" CreationDate="2015-08-08T20:19:04.767" Score="6" ViewCount="987" Body="&lt;p&gt;There are many do-it-yourself cleaner solutions out there, using a combination of vinegar, alcohol (in a &quot;cheap vodka&quot; concentration), dish soap, baking soda and essential oils. I like these recipes for two reasons, first for cost and second for health (sometimes my toddler helps me clean his bathtub, and I wouldn't want him to use too harsh stuff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many of these recipes claim to be &quot;disinfecting&quot;, and as far as my understanding goes, alcohol has that effect, as has stuff that offsets the pH-level (such as the acidity of vinegar). But given that this stuff is just sprayed onto the surface, can a mixture of 2 parts water 1 part vodka really be concentrated enough to be &quot;disinfecting&quot;? And if you mix baking soda and vinegar, won't that bring you closer to a neutral pH, i.e. offset any &quot;disinfecting powers&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some sort of guideline as to how much alcohol/other agent should be in the final solution for it to be of good use in cleaning a bathroom?&lt;/p&gt;&#xA;" OwnerUserId="19013" LastEditorUserId="4325" LastEditDate="2015-08-08T20:59:25.157" LastActivityDate="2017-03-24T20:38:33.223" Title="What alcohol concentration makes bathroom cleaner &quot;disinfecting&quot;?" Tags="&lt;acid-base&gt;&lt;everyday-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34848" PostTypeId="1" CreationDate="2015-08-08T21:29:22.450" Score="1" ViewCount="52" Body="&lt;p&gt;I am trying to figure how would I go about finding a dopant element that would fit in a copper lattice? I already have the data for the atomic dimensions and structure (available below), however how would I find a dopant element that would make it a P-Type material? (Relatively new to chemistry, so bear with me if I ask a stupid question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atomic dimensions and structure&#xA;Atomic Radius 145 pm &#xA;Covalent Radius 138 pm &#xA;Van der Waals Radius 140 pm &#xA;Crystal Structure Face Centered Cubic &#xA;Lattice Angles &#xA;π/2, π/2, π/2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lattice Constants &#xA;361.49, 361.49, 361.49 pm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space Group Name Fm_ 3m &#xA;Space Group Number 225 &lt;/p&gt;&#xA;" OwnerUserId="19015" LastActivityDate="2015-08-08T21:29:22.450" Title="How would I find an apporpriate dopant element for a lattice" Tags="&lt;atoms&gt;&lt;crystal-structure&gt;&lt;metallurgy&gt;&lt;atomic-radius&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34849" PostTypeId="1" AcceptedAnswerId="34852" CreationDate="2015-08-08T21:47:33.983" Score="0" ViewCount="2433" Body="&lt;p&gt;It is probably easiest for me to explain the question in an example. Consider the following chemical reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{MgF2(s) -&amp;gt; Mg^{2+} (aq) + 2F- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this into the formula for the solubility product constant, $K_{sp}=[M^{y+}]^{x}[A^{x-}]^{y}$, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solubility product constant for magnesium fluoride is $5.16\times10^{-11}$. To determine molar solubility, the equation to solve for magnesium and fluoride is $5.16\times10^{-11}=X\times (2X)^{2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$X$ is magnesium and $(2X)^{2}$ is fluoride. This is because there are two moles of fluoride for every one mole of $\ce{MgFl_{2}}$, and one mole of magnesium for every one mole of $\ce{MgFl_{2}}$. (Please correct me if I'm wrong anywhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here's whats confusing to me. Why is fluoride denoted as $(2X)^{2}$? I know that there are two moles of fluoride per one mole of magnesium fluoride, but why does it need to be denoted by being both multiplied to its coefficient and raised to the power of that coefficient? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf&quot; rel=&quot;nofollow&quot;&gt;Magnesium Fluoride Solubility product constant&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19014" LastEditorUserId="865" LastEditDate="2015-08-08T23:57:15.190" LastActivityDate="2019-04-26T05:37:46.907" Title="Why are the reactants in the solubility product constant multiplied by their coefficients and raised to the power of their coefficients?" Tags="&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34850" PostTypeId="1" CreationDate="2015-08-08T22:02:21.547" Score="5" ViewCount="1701" Body="&lt;p&gt;This is going to be a long shot, and I might be wrong in any and all of my assumptions, be warned (... and please correct me). I might also be trivially right, I have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, each degree of freedom has the kinetic energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{1}{2}k_\mathrm{B}T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have water at, say, $300\ \mathrm K$. Water has three atoms, so 9 degrees of freedom (3 per atom, because space is three-dimensional). One of them is for a $\ce{O-H}$ stretching. I know that usually, you would need to divide into modes of oscillations, but let's say we combine half the energy of the symmetric and antisymmetric stretchings, giving that one $\ce{O-H}$ stretching has a full $\frac{1}{2}k_\mathrm{B}T$ in it. So we have an average, roll of drums,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \frac{1}{2}(1.3806488 \times 10^{-23})(300) =  2.071\times 10^{-21}\ \mathrm J$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a $\ce{O-H}$ stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOW, let's say this stretching behaves like an harmonic oscillator, its' energy is given by the amplitude of oscillation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = \frac{1}{2}kA^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is the stiffness constant of the chemical bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's nice, but we have two unknowns, $k$ and $A$. So let's go further. If we know the absorption band's wavenumber, we can say that the amplitude corresponds to this wavelength, $2.734\ \mathrm{\mu m}$ (source = Wikipedia). This would mean that we have a stiffness constant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k = 5.5412 \times 10^{-10}\ \mathrm{N/m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that, although it was fun to blindly math around, this result is wrong. Just the fact that the stiffness constant would depend on temperature doesn't sound right. Neither does the bit about wavelength = amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it possible to know the amplitude of oscillation/stiffness constant of a $\ce{O-H}$ bond?&lt;/p&gt;&#xA;" OwnerUserId="16613" LastEditorUserId="16683" LastEditDate="2016-10-22T12:03:32.940" LastActivityDate="2016-10-22T13:43:12.357" Title="Can we determine the amplitude of vibration of O-H bonds in water?" Tags="&lt;bond&gt;&lt;water&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34851" PostTypeId="2" ParentId="32124" CreationDate="2015-08-08T23:38:34.260" Score="1" Body="&lt;p&gt;In fact all three reactions produced the same amount of heat energy and hence had the same enthalpy change. By changing the concentration of the HCl only effects the reaction rate and how much energy is produced in a certain time. However eventually, each reaction will produce the same amount of energy provided that all the Mg reacts which does happen in your experiment as HCl is in excess. The reason for this can be explained quick easily using the formula: $$\Delta H=\Delta U+pV$$ Then using the ideal gas equation, we can substitute $pV$ with $\Delta n(g)RT$. Note that the only variable that actually changes is n(g) which is the change in the number of moles of gas molecules. So the equation becomes: $$\Delta H= \Delta U + \Delta  n(g)RT$$ So the only thing that effects the enthalpy change of the reaction in the change in internal energy (U) and the number of moles of gas molecules formed which are both constant for all three reactions. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-08T23:38:34.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34852" PostTypeId="2" ParentId="34849" CreationDate="2015-08-09T01:52:29.930" Score="1" Body="&lt;p&gt;The reason why in the equilibrium constant expression, the concentration is to the power of coefficient is because the above equation can be written like this:&#xA;$$\ce{MgF_2(s) -&amp;gt; Mg^{2+} (aq) + F^- + F^- (aq)}$$&#xA;Hence when you plug this into the equilibrium constant expression, you get this:&#xA;$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}][\ce{F-}]$$ which becomes the equivalent to:$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$$ So really the equilibrium constant is proportional to the square of the concentration of fluoride. So that is the reason for the square. However, the concentration of fluoride is twice as much as the concentration of Mg, hence that is the reason for the multiplying the concentration of fluoride by 2. So that is why fluoride is denoted as $(2X)^2$.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-09T01:52:29.930" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34855" PostTypeId="1" CreationDate="2015-08-09T03:19:19.717" Score="0" ViewCount="160" Body="&lt;p&gt;Maybe I'm wrong about it but I wanted to know whether the concept of absolute temperature is real or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As every element behaves differently at a particular temperatures how could we assign a common temperature to all these elements where they possess zero point energy? If they will do so then what will happen if we find an new element which ground state energy is at $\mathrm{-10\ K}$ theoretically?&lt;/p&gt;&#xA;" OwnerUserId="17275" LastEditorUserId="10185" LastEditDate="2015-08-09T09:09:44.063" LastActivityDate="2015-08-09T09:09:44.063" Title="Is the absolute temperature concept 100% viable?" Tags="&lt;thermodynamics&gt;&lt;temperature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34856" PostTypeId="2" ParentId="34838" CreationDate="2015-08-09T03:28:08.720" Score="0" Body="&lt;p&gt;Yes, polarity can be changed. Example, between alcohol and phenol the O-H polarity is more in phenol nd less in alcohol. It is due to the extent of attraction of electron by oxygen from hydrogen which is more in phenol.&lt;/p&gt;&#xA;" OwnerUserId="17275" LastActivityDate="2015-08-09T03:28:08.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34857" PostTypeId="2" ParentId="34855" CreationDate="2015-08-09T03:46:21.970" Score="3" Body="&lt;p&gt;The concept of absolute temperature is real. Temperature can be thought of as thee measure of the concentration of heat energy in a substance and isn't effected by different elements or compounds. Absolute temperature which is −273.15 degrees Celsius (0 kelvin) is the temperature where there is no heat energy in the substance at all. That means that all the molecules are existing in their lowest possible energy state. So conventionally is thought that it is impossible to get lower than absolute zero as to reduce the temperature, you have to remove heat energy but you can't remove heat energy in a substance which doesn't have any heat energy in the first place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However theoretically, the Boltzmann equation does allow for substance to go minus Kelvin and in fact it actually lies above infinity Kelvin. Recently there have been several claims that this has been achieved. If you search this up on Google I am sure that you will be able to find more information on this. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorDisplayName="user7951" LastEditDate="2015-08-09T08:39:30.920" LastActivityDate="2015-08-09T08:39:30.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34858" PostTypeId="1" CreationDate="2015-08-09T04:06:54.863" Score="2" ViewCount="138" Body="&lt;p&gt;Can anybody give me a schematic representation of how quantum tunneling allows chemical reactions to occur at very low temperatures like in the interstellar medium.&lt;/p&gt;&#xA;" OwnerUserId="17275" LastEditorUserId="10185" LastEditDate="2015-08-09T10:49:03.310" LastActivityDate="2015-08-09T10:49:03.310" Title="Effect of quantum tunnelling on chemical reactions at very low temperature" Tags="&lt;quantum-chemistry&gt;&lt;temperature&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2015-08-10T13:41:06.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="34859" PostTypeId="2" ParentId="34838" CreationDate="2015-08-09T04:17:23.647" Score="2" Body="&lt;p&gt;The polarity on the oxygen atom will decrease. This is because of the sp2 hybridised carbon atom which is highly electronegative. This will result in a decreased polarity in the OH bond and hence it will disassociate more easily in solvents such as water. One example of this effect is the increased acidity of phenol compared to most alcohols. This is because the OH is bonded to sp2 hybridised carbon in phenol and hence will disassociate in water easily when compared to normal alcohols.  &lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-09T04:17:23.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34862" PostTypeId="1" CreationDate="2015-08-09T08:31:01.907" Score="6" ViewCount="1030" Body="&lt;p&gt;I've read an article that equilibrium exists when $G$ (Gibbs free energy) becomes minimum. If formation of products decreases Gibbs free energy, why won't the reaction proceed to completion so that Gibbs energy becomes minimum. &lt;/p&gt;&#xA;" OwnerUserId="11331" LastEditorUserId="4325" LastEditDate="2016-03-16T21:06:13.537" LastActivityDate="2020-02-18T13:32:37.023" Title="Why does equilibrium exist?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;&lt;free-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34863" PostTypeId="2" ParentId="34862" CreationDate="2015-08-09T10:05:48.037" Score="4" Body="&lt;p&gt;This is because the Gibbs free energy also depends on the concentration of the reactants and products. Hence, for a reaction, if the energy change is negative it moves forwards, forming more product, which also changes the concentrations of the species involved. The Gibbs energy change at this new concentration will be less negative than that in the initial case, but the reaction still proceeds since the reaction is accompanied by a negative change of free energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point comes, where the concentrations of the reactions and the products is such, that the Gibbs energy change now becomes zero for the particular reaction. This is the equilibrium condition (minimum free energy, maximum entropy). If the energy change is positive, the reverse reaction becomes favorable, again inching towards the zero energy change condition, i.e. equilibrium. So, the reaction is already at the minimum Gibbs energy when it is at equilibrium. Further proceeding of the reaction will be accompanied by a positive change, and hence a rise in energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For how it depends on the concentrations, see these questions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/6927&quot;&gt;Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/6838&quot;&gt;Derivation of relationship between equilibrium constant and Gibbs free energy change&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="4945" LastEditDate="2020-02-18T13:32:37.023" LastActivityDate="2020-02-18T13:32:37.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34864" PostTypeId="1" AcceptedAnswerId="34868" CreationDate="2015-08-09T10:37:51.743" Score="2" ViewCount="791" Body="&lt;p&gt;Are names of substituents like isopropyl and sec-butyl accepted by IUPAC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do we need to name them according to standard IUPAC conventions?&#xA;(Ex- 1-methylethyl for isopropyl)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My book uses common names for substituents at a lot of places but I am doubtful about this. &lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-11-24T22:42:50.740" Title="Are common names of substituents accepted by IUPAC?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34865" PostTypeId="2" ParentId="34862" CreationDate="2015-08-09T10:39:29.153" Score="4" Body="&lt;p&gt;Just because the Gibbs energy change of reaction ($\Delta _r G$) is negative, doesn't mean that it reaches a minimum when all reactants have been converted to products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the reaction: $$\ce{A &amp;lt;=&amp;gt; B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can plot the total Gibbs energy of the system against the percentage conversion from $\ce{A}$ to $\ce{B}$ and we get a graph like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PZ9j3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PZ9j3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically the curve should be a parabola but it was much easier to draw an arc in the program I used. As you can see, the minimum Gibbs energy is reached when there is a mixture of products and reactants. This is in fact true for all chemical reactions: the minimum on the Gibbs energy graph cannot be at the point where only reactants or products exist. It can be extremely close to that point, in which case we tend to call the reaction irreversible (if the minimum is close to 100% products) or we say that the reaction doesn't happen (if the minimum is close to 100% reactants).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that there is a relationship between $\Delta _r G$ and the equilibrium constant which is expressed as: $$\Delta _r G = -RT\ln K$$&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="4325" LastEditDate="2016-03-16T21:06:45.913" LastActivityDate="2016-03-16T21:06:45.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34867" PostTypeId="1" AcceptedAnswerId="34883" CreationDate="2015-08-09T10:50:28.430" Score="6" ViewCount="399" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Atkin's Physical Chemistry Book; Chapter 3; question 3.3(a):&lt;br&gt;&#xA;  &lt;strong&gt;Question&lt;/strong&gt;&lt;br&gt;&#xA;  Calculate $\Delta S$ when the state of $\mathrm{3~mol}$ of perfect gas atoms, for which $C_{p,m}$ = $\frac{5}{2}R$, is changed from $\mathrm{298~K}$ and $\mathrm{1~atm}$ to $\mathrm{398~K}$ and $\mathrm{5~atm}$. How do you rationalize the sign of $\Delta S$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is at a constant pressure, $dq = dH = C_pdT$. Then I used the equation $$dS = \frac{C_pdT}{T}$$&#xA;I integrated both sides to get the following equation: $$\Delta S = C_p   \ln\frac{398}{298}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subbing in the values I got that $\Delta S =  \mathrm{+18.3~J}$. However, the answer in the book is $\mathrm{-21~J}$. Shouldn't that be definitely wrong because the temperature of the system increased and hence the total entropy should have increased?&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="46306" LastEditDate="2018-03-04T18:45:47.727" LastActivityDate="2018-03-04T19:10:42.363" Title="Calculating Entropy" Tags="&lt;thermodynamics&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34868" PostTypeId="2" ParentId="34864" CreationDate="2015-08-09T11:10:38.507" Score="5" Body="&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; various prefixes are retained for use in &lt;em&gt;general (IUPAC) nomenclature&lt;/em&gt;. However, many prefixes are no longer recommended.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The prefix &lt;strong&gt;“isopropyl”&lt;/strong&gt; for $\ce{(CH3)2CH-{}}$ is retained only for use in &lt;em&gt;general (IUPAC) nomenclature&lt;/em&gt;.&lt;br&gt;&#xA;For the &lt;em&gt;preferred IUPAC name&lt;/em&gt; (PIN), the preferred prefix is &lt;strong&gt;“propan-2-yl”&lt;/strong&gt;.&lt;br&gt;&#xA;The prefix &lt;strong&gt;“1-methylethyl”&lt;/strong&gt; may be used in &lt;em&gt;general (IUPAC) nomenclature&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prefix &lt;strong&gt;“&lt;em&gt;sec&lt;/em&gt;-butyl”&lt;/strong&gt; for $\ce{CH3-CH2-CH(CH3)-{}}$ was still contained in the 1993 recommendations but is no longer recommended as approved prefix.&lt;br&gt;&#xA;For the &lt;em&gt;preferred IUPAC name&lt;/em&gt; (PIN), the preferred prefix is &lt;strong&gt;“butan-2-yl”&lt;/strong&gt;.&lt;br&gt;&#xA;The prefix &lt;strong&gt;“1-methylpropyl”&lt;/strong&gt; may be used in &lt;em&gt;general (IUPAC) nomenclature&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-08-09T11:10:38.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34869" PostTypeId="1" CreationDate="2015-08-09T11:22:42.617" Score="1" ViewCount="10090" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dmHqO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dmHqO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Source:&lt;/strong&gt; an IB Chemistry past paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked it up and found out that brass is an alloy of copper and zinc - however, I don't see zinc in any of the equations, and therefore any brass being separated into copper and zinc, which I had originally thought would be the method to determine the percentage of copper in brass (by looking at the coefficients of the compounds in the resulting chemical equation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how do these reactions help determine the percentage of copper in brass?&lt;/p&gt;&#xA;" OwnerUserId="17917" LastEditorUserId="5026" LastEditDate="2018-03-08T09:07:16.010" LastActivityDate="2018-03-08T09:07:16.010" Title="Determining percentage of copper in brass" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34870" PostTypeId="2" ParentId="34869" CreationDate="2015-08-09T11:33:40.073" Score="2" Body="&lt;p&gt;Well the question would had more to it. The question would have probably gone something like this. A sample of brass was dissolved with excess nitric acid. The $\ce{Cu^{2+}}$ ions are then reduced with excess iodide ions. Iodine is then titrated with thiosulfate (giving its concentration and volume used). Find the percentage of copper in brass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this question, calculate the amount of moles of thiosulfate used, using its concentration and volume. Then using equation 3, you can see the number of moles of iodine is half the number moles of thiosulfate. Then using equation 2, you can see that the number of moles of $\ce{Cu^{2+}}$ ions are twice as the number of moles of iodine and hence are equal to the number of moles of iodine. Finally, using equation 1, the number of moles of copper is equal to the number of moles of $\ce{Cu^{2+}}$ ions and hence are equal to the number of moles of thiosulfate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know the number of moles of copper, you can multiply by its molar mass to get its mass and then find the percentage of copper in brass. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="10185" LastEditDate="2015-08-09T12:38:46.150" LastActivityDate="2015-08-09T12:38:46.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34873" PostTypeId="1" AcceptedAnswerId="35474" CreationDate="2015-08-09T12:41:29.753" Score="4" ViewCount="1877" Body="&lt;p&gt;I'm not a chemist and don't remember much chemistry from school, so please forgive a naive question (I've read the Wikipedia article on sodium silicate though).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm researching this topic to make a positive photo resist developer for making PCBs.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Will dissolving $\ce{Na2SiO3}$ (the anhydrous form) in water produce essentially the same solution as dissolving $\ce{Na2SiO3*5H2O}$ but in different concentration or are these intrinsically different?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will $\ce{Na2SiO3}$, if left in contact with open air, eventually hydrate to $\ce{Na2SiO3*5H2O}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_silicate&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt; talks about the $\ce{SiO2:Na2O}$ ratio affecting the pH of the water solution. Is the information about this ratio already present in the chemical formula so if I buy $\ce{Na2SiO3*5H2O}$ it is always the same ratio or is this a different parameter that is specified in addition to the chemical formula? (I understand that different concentrations of the solution will produce different pH, but consider equal concentrations)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe I need the developer to be an alkaline solution so I need to understand what will affect the pH when dissolving sodium metasilicate in water (e.g. if I need $\ce{Na2SiO3}$ or $\ce{Na2SiO3*5H2O}$, if I need to consider other specifications besides the chemical formula, if I need to consider the shelf life of the chemical if it changes with time e.g. gets hydrated from $\ce{Na2SiO3*5H2O}$ into $\ce{Na2SiO3*9H2O}$).&#xA; Searching electronics forums on the internet I haven't found a clear answer. For some people using this sodium metasilicate as a developer works for some it doesn't develop at all, some add $\ce{NaOH}$ to it but there are places that advertise their sodium metasilicate developer as sodium hydroxide free.&lt;/p&gt;&#xA;" OwnerUserId="19023" LastEditorUserId="10185" LastEditDate="2015-08-09T17:32:18.963" LastActivityDate="2015-08-25T21:08:12.817" Title="pH of sodium metasilicate anhydrous vs pentahydrate when dissolved in water" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34874" PostTypeId="2" ParentId="34850" CreationDate="2015-08-09T14:02:57.977" Score="3" Body="&lt;p&gt;Finally, I have a day off to read that question and response. Half-way reading it I was already a bit suspicious mostly because of the half-jolly and half-formal style OP used. :) Don't get me wrong, in principle I have nothing against such style. But the thing is that usually when such style is used some nonsense is eventually introduced at some point, and this nonsense is not so easily detectable since it is buried under all these jolly informalities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is exactly what happened this time too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;we can say that the amplitude corresponds to this wavelength&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wait, what? &lt;em&gt;Amplitude has no relation whatsoever to wavelength&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is: is it possible to know the amplitude of&#xA;  oscillation/stiffness constant of a $\ce{O-H}$ bond?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you can calculate the force constants of molecular vibrations (as well as frequencies). These calculations are routinely done in the domain of quantum chemistry these days.&lt;/p&gt;&#xA;" OwnerUserId="186" LastEditorUserId="186" LastEditDate="2015-08-09T19:00:05.237" LastActivityDate="2015-08-09T19:00:05.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34875" PostTypeId="1" AcceptedAnswerId="34876" CreationDate="2015-08-09T14:56:41.037" Score="5" ViewCount="1848" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3akW7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3akW7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is cyclopentyl attached to one terminal carbon atom and cyclohexyl attached to the other terminal carbon atom and there are three carbon atoms linking them. Since the cyclohexyl ring has the greatest number of carbon atoms, it should be the parent chain. So the name should be 3-cyclopentylpropyl cyclohexane.&lt;/p&gt;&#xA;&lt;p&gt;But my book says that the middle propyl chain should be the parent chain, the cyclohexyl and cyclopentyl groups should be the substituents, and the correct answer is 1-cyclohexyl-3-cyclopently propane.&lt;/p&gt;&#xA;&lt;p&gt;What is the correct name. And if my book is correct, why is it so?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-13T11:06:11.977" Title="What is the parent chain in a hydrocarbon with two different rings?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34876" PostTypeId="2" ParentId="34875" CreationDate="2015-08-09T15:37:32.453" Score="10" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.1.2.2&lt;/strong&gt; Systems composed of rings and chains (exclusive of linear phanes)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two methods are recognized to name systems composed of rings and chains&#xA;  (exclusive of linear phanes).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far, we have three possible names:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method (1): “(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane” (ring preferred to chain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Method (2): “1-cyclohexyl-3-cyclopentylpropane” (chain preferred to ring)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(…) For selection of a preferred IUPAC name, see P-52.2.8.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-52.2.8&lt;/strong&gt; Selection between a ring and a chain as parent hydride&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This leaves us with two possible preferred IUPAC names:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“(3-cyclopentylpropyl)cyclohexane” or “(3-cyclohexylpropyl)cyclopentane”&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.1.3&lt;/strong&gt; Seniority order only for rings and ring systems. Criteria that apply only when the choice for parent structure is between two or more rings or ring systems are given in P-44.2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-44.2&lt;/strong&gt; SENIORITY ORDER ONLY FOR RINGS AND RING SYSTEMS&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;P-44.2.1&lt;/strong&gt; If application of P-44.1 does not effect a choice, general criteria for determining ring seniority given below are applied successively until there are no alternatives remaining. (…)&lt;br&gt;&#xA;  The senior ring or ring system:&lt;br&gt;&#xA;  (…)&lt;br&gt;&#xA;  (e) has the greater number of skeletal atoms;&lt;br&gt;&#xA;  (…)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, the preferred IUPAC name is &lt;strong&gt;“(3-cyclopentylpropyl)cyclohexane”&lt;/strong&gt; (not “(3-cyclohexylpropyl)cyclopentane”) since cyclohexane has more skeletal atoms than cyclopentane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uBquO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBquO.png&quot; alt=&quot;(3-cyclopentylpropyl)cyclohexane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-08-09T15:48:02.610" LastActivityDate="2015-08-09T15:48:02.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34877" PostTypeId="1" CreationDate="2015-08-09T16:26:05.230" Score="13" ViewCount="18048" Body="&lt;p&gt;I wear corrective eyeglasses, and because of this, I frequently experience problems with fogging when I wear safety goggles. How can I prevent or reduce fogging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The safety goggles I use are the standard-issue goggles supplied by the campus store, but I can use other goggles rated at least Z87+D3 (&quot;+&quot;=impact, &quot;D3&quot;=splash) if I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method of reducing the fogging I have tried is wedging a folded paper towel under the top edge of the goggles. It works, but I don't feel comfortable with it because it compromises the splash protection of the goggles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried placing desiccant packets (of the sort found in various packaging) in the goggles. This has helped somewhat but doesn't solve the problem completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution would be to get safety goggle with prescription corrective lenses built-in (making them Z87-2D3 instead of Z87D3), but I have not been able to find a supplier online for such eyewear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What can I do to reduce or eliminate fogging?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8074" LastEditorUserId="402" LastEditDate="2017-02-12T03:16:37.997" LastActivityDate="2018-03-01T10:12:40.543" Title="How to stop goggles from fogging?" Tags="&lt;safety&gt;&lt;equipment&gt;" AnswerCount="5" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34878" PostTypeId="1" CreationDate="2015-08-09T16:26:39.877" Score="3" ViewCount="11917" Body="&lt;p&gt;What kind of storage solution is required for a glass electrode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is 3 M $\ce{KCl}$ sufficient for this purpose? If so, what is the approximate pH of that solution (the solution I use presently has a pH $\approx$ 5)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can filling solutions be used as storage solutions for glass electrodes?&lt;/p&gt;&#xA;" OwnerUserId="17129" LastEditorUserId="865" LastEditDate="2015-08-09T18:31:30.863" LastActivityDate="2015-08-10T11:40:57.847" Title="Solution for storing a glass electrode" Tags="&lt;aqueous-solution&gt;&lt;ph&gt;&lt;equipment&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34879" PostTypeId="2" ParentId="34877" CreationDate="2015-08-09T17:05:11.610" Score="11" Body="&lt;p&gt;Be careful with desiccant: silica may create tiny dust particles that could scratch the cornea, and hygroscopic salts may liquefy, possibly irritating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A traditional anti-fog treatment is to rub the interior of the lens with bar soap (or a drop of mild detergent) and polish it. The soap decreases the surface tension of water, causing droplets to make a smoother, more transparent film. &lt;a href=&quot;https://en.wikipedia.org/wiki/Anti-fog&quot;&gt;Wikipedia suggests some other anti-fog treatments&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://www.grainger.com/product/CREWS-Chem-Splsh-Goggles-3RYE9?s_pp=false&amp;amp;picUrl=//static.grainger.com/rp/s/is/image/Grainger/3RYE9_AS01&quot;&gt;vented anti-fog goggles&lt;/a&gt; (the vents should have foam or other filter to make them splash resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My personal choice for lab work or mechanical work is a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B009R2TLMU&quot;&gt;full-face splash shield&lt;/a&gt;, which may offer more protection and does not fog.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-09T17:05:11.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34880" PostTypeId="2" ParentId="34877" CreationDate="2015-08-09T17:06:01.183" Score="3" Body="&lt;p&gt;There are sprayable hydrophobic coatings available such as NeverWet, though you should check compatibility with the reagents you're using.&lt;/p&gt;&#xA;" OwnerUserId="4592" LastActivityDate="2015-08-09T17:06:01.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34881" PostTypeId="2" ParentId="34878" CreationDate="2015-08-09T17:07:03.760" Score="1" Body="&lt;p&gt;Most pH electrode storage solutions are 4 M KCl with a pH about 4. The solution you have should be fine if you can change the internal solution on a monthly basis (more often if you use it continuously). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal solution is also saturated with silver chloride. I would only use it for the external storage solution if I couldn't make the external solution, but it should not be a problem for a short term.&lt;/p&gt;&#xA;" OwnerUserId="4934" LastActivityDate="2015-08-09T17:07:03.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34882" PostTypeId="1" AcceptedAnswerId="34885" CreationDate="2015-08-09T17:48:35.833" Score="4" ViewCount="270" Body="&lt;p&gt;According to Ahn.N.T &quot;Frontier Orbitals&quot; hydroboration isn't a pericyclic reaction because boron uses two AOs and not one. The same applies for cheletropic reactions, which aren't pericyclic either. But the IUPAC goldbook says different: &lt;a href=&quot;http://goldbook.iupac.org/P04491.html&quot; rel=&quot;nofollow&quot;&gt;http://goldbook.iupac.org/P04491.html&lt;/a&gt; - that cheletropic reactions are pericyclic. So is hydroboration, too?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-08-09T18:13:18.087" Title="Is hydroboration a pericyclic reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34883" PostTypeId="2" ParentId="34867" CreationDate="2015-08-09T17:58:36.960" Score="6" Body="&lt;p&gt;The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\frac{dT}{T},\frac{dP}{P}\right)$, which is the data we are given.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's the derivation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second Law of Thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dS = \frac{\delta Q}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Law of Thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dU = \delta Q + \delta W = \delta Q - PdV_m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of Enthalpy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$&#xA;$$\delta Q = dH - V_mdP$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV_m = RT$$&#xA;$$V_m = \frac{RT}{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dH = C_p dT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute to get rid of $V_m$, since we don't have volume data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dQ = C_p dT - \frac{RT}{P}dP$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute into the second law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$dS = C_p \frac{dT}{T} - R\frac{dP}{P}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate (now we account for change in pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \frac{dT}{T} - R \int_{P_1}^{P_2} \frac{dP}{P}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this method, I get the correct answer after I multiply the molar entropy change by the number of moles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that if you are given volume change instead, you can use a similar derivation to reach a similar equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta S = C_v \ln \left(\frac{T_2}{T_1}\right) + R \ln \left(\frac{V_{m,2}}{V_{m,1}} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/entropy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nasa Glenn Research Center&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="275" LastEditorUserId="41556" LastEditDate="2018-03-04T19:10:42.363" LastActivityDate="2018-03-04T19:10:42.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34884" PostTypeId="1" AcceptedAnswerId="34886" CreationDate="2015-08-09T18:10:27.090" Score="9" ViewCount="28667" Body="&lt;p&gt;I buy lactose-free milk. The sell-by date is usually as much as eight weeks from the date of purchase, which is in the neighborhood of twice that of regular milk. This seems puzzling; lactose-free milk, which still has the same mass of sugar, has twice as many sugar molecules as regular milk, since all the lactose molecules have been split in to galactose and glucose. With twice as many sugar molecules per unit volume, one would think lactose-free milk would last half as long as regular milk, not twice as long.&lt;/p&gt;&#xA;" OwnerUserId="15953" LastEditorUserId="10185" LastEditDate="2015-08-09T18:14:12.210" LastActivityDate="2018-12-06T10:46:52.557" Title="Why does lactose-free milk last so much longer than regular milk?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34885" PostTypeId="2" ParentId="34882" CreationDate="2015-08-09T18:13:18.087" Score="4" Body="&lt;p&gt;Based on the &lt;a href=&quot;http://goldbook.iupac.org/P04491.html&quot; rel=&quot;nofollow&quot;&gt;IUPAC Goldbook&lt;/a&gt; definition, the only two requirements for a pericyclic reaction are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Concerted mechanism&lt;/li&gt;&#xA;&lt;li&gt;Cyclic transition state&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hydroboration reactions (and similar reactions) are pericylic reactions by these criteria, as are all cheletropic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The language about &lt;code&gt;fully conjugated transition states&lt;/code&gt; can be used to explain some pericyclic reactions, but not all, and thus IUPAC uses the phrasing &lt;code&gt;may be viewed as...&lt;/code&gt;. Hydride shifts, (and all sigmatropic rearrangements ene reactions) involve non-$\pi$ orbitals (even so, the transition states do tend to follow the Hückel rule).&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-09T18:13:18.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34886" PostTypeId="2" ParentId="34884" CreationDate="2015-08-09T18:50:10.860" Score="10" Body="&lt;p&gt;Apparently the process used by manufacturers (ultra-pasteurization) to produce lactose-free milk is responsible for the longer shelf life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shelflifeadvice.com/content/why-does-lactose-free-milk-last-longer-regular-milk&quot; rel=&quot;noreferrer&quot;&gt;One source reports&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lactose-free milk is pasteurized at a higher temperature than regular milk. The process, known as ultra-pasteurization, is designed to remove the bacteria content entirely, giving lactose-free milk a refrigerated shelf-life of 60-90 days, compared with regular pasteurized milk, which retains some bacteria. It has a shelf life of 1-3 weeks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A producer (in the United States) of organic dairy products (Organic Valley) describes ultra-pasteurization and the extended shelf-life &lt;a href=&quot;http://organicvalley.custhelp.com/app/answers/detail/a_id/77/~/ultra-pasteurization-(up)-definition&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ultra Pasteurization (UP), or Ultra High Temperature (UHT) pasteurization, is the process of heating milk to approximately  280 °F for just 2 seconds and then chilling it back down rapidly.  The result is milk that's 99.9% free from bacteria.  This pasteurization process creates an extended shelf-life for milk products of up to three times the length as HTST pasteurization; while providing the same wholesome, quality dairy product.  This allows us to distribute UP milk products regionally as well as to other areas of the country that might not have access to our dairy products. This is referred to as 'ultra pasteurized' on the milk package.  Organic Valley offers its quarts and some of its half gallons of milk with Ultra Pasteurization (UP).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-10T20:43:04.013" LastActivityDate="2015-08-10T20:43:04.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34887" PostTypeId="1" AcceptedAnswerId="34889" CreationDate="2015-08-09T19:37:27.817" Score="3" ViewCount="412" Body="&lt;p&gt;So, at the anode of a fuel cell hydrogen is split into a positively charged proton and an electron. Then, there is the membrane which can let only protons through to the cathode, whereas electrons need to follow external circuit (effectively powering it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: why do we need any other reaction on the cathode? Isn't the fact that positively charged ions travel through the membrane to the cathode enough to create the EMF?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to solve it I assume that the reaction is there to actually apply some kind of force to move the ions though the membrane to the cathode, otherwise they'd just stay and reunite with electrons on the anode. What is that force and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="19027" LastEditorUserId="275" LastEditDate="2015-08-09T20:15:58.073" LastActivityDate="2015-08-09T20:27:45.997" Title="Why is the cathode reaction required in a fuel cell?" Tags="&lt;reaction-mechanism&gt;&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34888" PostTypeId="2" ParentId="34850" CreationDate="2015-08-09T19:57:35.540" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is: is it possible to know the amplitude of&#xA;  oscillation/stiffness constant of a O−H bond?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your approach is along the right lines &lt;strong&gt;but&lt;/strong&gt; - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Hooke's Law formula you used&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac{1}{2}kA^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is for the stretching of a spring.  Hooke's law changes if the spring has mas(ses) attached at the end(s), now it is a harmonic oscillator.  In this case Hooke's Law becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = \frac {h}{2 \pi} \sqrt{\frac{k}{\mu}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where k is the force constant you are interested in and $\mathrm{\mu}$ is the reduced mass of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=h\nu$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency, which can be experimentally measured, is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{frequency} = \frac {1}{2 \pi} \sqrt{\frac{k}{\mu}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency is independent of the amplitude, affected only by the masses and the stiffness of the spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the masses of the atoms at the ends of the spring are given as A and B, then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu =\frac{A \cdot B}{A+B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for an $\ce{O-H}$ bond $\mathrm{\mu=\frac{1 \cdot 16}{1+16}}\ \mathrm{u}$. If we measure the frequency of the $\ce{O-H}$ stretch absorption, then we can calculate the force constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a &lt;a href=&quot;http://www.napavalley.edu/people/sfawl/Documents/Chem%20241/Calculation%20of%20Vibrational%20Frequency.pdf&quot; rel=&quot;nofollow&quot;&gt;one-page example&lt;/a&gt; that shows the actual computation and explains the specific units to be used very nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask about the amplitude of oscillation.  If, by this you mean the bond length, then yes this can also be calculated from spectra (relatively easy in simple diatiomic cases, but more difficult in complex polyatomic cases.  The methodology involves assuming a rigid rotor and then using the Schrodinger equation to solve for the energy states and extracting the bond length.  A nice example using $\ce{HCl}$ can be found &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot.html#c2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, look down the page for the section entitled &quot;Bond Length of HCl&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-10-22T13:43:12.357" LastActivityDate="2016-10-22T13:43:12.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34889" PostTypeId="2" ParentId="34887" CreationDate="2015-08-09T20:27:45.997" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do we need two half reactions, one at the anode and another at the cathode? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The whole thing eventually grinds to a halt without the other reaction. The electrons generated at the anode travel through the circuit to the cathode. What happens then? Something needs to happen to the electron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, what happens to that proton? As the concentration of protons increases, what will balance the charge that is produced in the cell, given that the electron is elsewhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can imagine that we can set up a cathode in which the reverse reaction of the anode occurs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cathode: $\ce{H2 -&amp;gt; 2H+ + 2e-}$&lt;/li&gt;&#xA;&lt;li&gt;Anode: $\ce{2H+ +2e- -&amp;gt; H2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, in this case, there is overall no net reaction, which means that $\Delta G =0$ and since $\Delta G = nFE$, then the cell potential $E=0$. We need a different reaction for the cathode to consume electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{O2 + 2H2O +4e- -&amp;gt; 4OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this reaction produces hydroxide anions, then we have something for the proton to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H+ + OH- -&amp;gt; H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have the following combination of reactions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Anode: $\ce{H2 -&amp;gt; 2H+ + 2e-}$&lt;/li&gt;&#xA;&lt;li&gt;Cathode: $\ce{O2 + 2H2O +4e- -&amp;gt; 4OH-}$&lt;/li&gt;&#xA;&lt;li&gt;Neutralization $\ce{H+ + OH- -&amp;gt; H2O}$&lt;/li&gt;&#xA;&lt;li&gt;Net: $\ce{2H2 + O2 -&amp;gt; 2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This net reaction is exergonic, and thus produces a potential.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-09T20:27:45.997" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34890" PostTypeId="2" ParentId="34877" CreationDate="2015-08-09T20:50:46.110" Score="2" Body="&lt;p&gt;I used to have a brown wax crayon that I would rub onto my glasses and polish in order to prevent fogging. After a quick google search I couldn't find the same product (mind you, my experience was almost 20 years ago). I found something that might be similar, albeit with a interesting name, &lt;a href=&quot;http://ekusa.com/cat-crap&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1798" LastEditorUserId="865" LastEditDate="2015-08-09T21:06:24.113" LastActivityDate="2015-08-09T21:06:24.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34891" PostTypeId="1" CreationDate="2015-08-09T21:07:02.707" Score="3" ViewCount="70" Body="&lt;p&gt;Let's say I am seeking to find an alloy with certain properties, is there any method or process that enables one to backtrack from desired properties to actual structure and composition of a material?&lt;/p&gt;&#xA;" OwnerUserId="19015" LastActivityDate="2015-08-10T19:20:35.893" Title="Reverse-Engineer a material" Tags="&lt;experimental-chemistry&gt;&lt;materials&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34894" PostTypeId="1" CreationDate="2015-08-09T23:37:14.097" Score="2" ViewCount="171" Body="&lt;p&gt;How could one calculate the Fermi level of a copper-nickel alloy? They are both structurally face-centered cubic, however how would this structure affect the Fermi level of the individual atom or the entirety of the material?&lt;/p&gt;&#xA;" OwnerUserId="19015" LastEditorUserId="19015" LastEditDate="2015-08-10T16:41:12.780" LastActivityDate="2015-08-10T16:41:12.780" Title="Fermi Level of an alloy" Tags="&lt;crystal-structure&gt;&lt;electronic-configuration&gt;&lt;materials&gt;&lt;metallurgy&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34895" PostTypeId="2" ParentId="34877" CreationDate="2015-08-09T23:52:07.257" Score="3" Body="&lt;p&gt;NASA had this problem in the late 1960s, on the Moon.  They developed a very effective solution.  Sub-Aquatic Systems (SAS) licensed their formula and sold it to sport SCUBA divers.  (I used it on my dive mask and on my motorcycle face shield.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, there have been MANY anti-fogging solutions formulated for diving.  A former dive instructor of mine swore by no-tears baby shampoo.  (You OBVIOUSLY want something that won't irritate your eyes.)  I personally used Sea Drops (tm), when I dropped back into diving after being away for many years and learning that SAS was no longer in business.&lt;/p&gt;&#xA;" OwnerUserId="19031" LastActivityDate="2015-08-09T23:52:07.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34896" PostTypeId="1" CreationDate="2015-08-10T00:55:28.900" Score="2" ViewCount="43" Body="&lt;p&gt;Two solutions were prepared.&#xA;'Solution A' being $\ce {Na2S4}$ prepared from aqueous $\ce {NaOH}$ and elemental sulfur, the mixture heated and additional sulfur added until sulfur remained as a precipitate.&#xA;'Solution B' being a solution of  $\ce {FeSO4}$ and $\ce {Fe2(SO4)2}$ prepared from reacting steel wool with $\ce{H2SO4}$, adding steel wool until reaction ceased. The liquid portion drained off and collected for use.&#xA;Solution A was added to Solution B via squeeze pipette in intervals allowing precipitate to settle between additions. This was continued until the net solution began to turn yellow from a rising conventration of  $\ce{Na2S4}$.&#xA;The solution was washed drained and the precipitate collected, however its character is not as expected from micro particulate $\ce{FeS2}$ but rather a black oily colloidal mass was the collected product. Did I somehow manage to acquire $\ce{Fe2S4}$?&lt;/p&gt;&#xA;" OwnerUserId="17721" LastActivityDate="2015-08-10T00:55:28.900" Title="Curious precipitate from substitution reaction" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34897" PostTypeId="1" AcceptedAnswerId="34902" CreationDate="2015-08-10T01:24:37.693" Score="1" ViewCount="6009" Body="&lt;p&gt;I want to know if I can calculate the dipole moment of a molecule given the electronegativity values for the atoms comprising it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that dipole moments determine the polarity of molecules, so if one molecule has a larger dipole moment than another, the one with the larger dipole moment is more polar than the other. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the magnitude of how polar a molecule is be calculated from the difference in electronegativities among the elements comprising the molecule?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-08-10T13:09:01.737" LastActivityDate="2018-09-17T02:45:29.927" Title="Relationship between dipole moment of a molecule and electronegativities of constituent atoms" Tags="&lt;physical-chemistry&gt;&lt;electronegativity&gt;&lt;polarity&gt;&lt;dipole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34898" PostTypeId="1" CreationDate="2015-08-10T02:54:03.630" Score="0" ViewCount="78887" Body="&lt;p&gt;Baking soda (sodium bicarbonate) + citric acid react when mixed with water to form some amount of carbon dioxide gas ($\ce{CO2}$). I am trying to determine what percentage of the products is carbon dioxide: in other words, how much does reaction make? &lt;/p&gt;&#xA;" OwnerUserId="19033" LastEditorUserId="865" LastEditDate="2015-08-10T12:54:41.757" LastActivityDate="2020-12-26T18:34:51.903" Title="Baking Soda + Citric Acid + Water = CO2, how much though?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="2" ClosedDate="2017-10-02T17:02:21.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="34899" PostTypeId="2" ParentId="34656" CreationDate="2015-08-10T03:15:35.083" Score="1" Body="&lt;p&gt;Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is about differences in the two equilibria $\ce{H &amp;lt;=&amp;gt; S}$ and $\ce{A &amp;lt;=&amp;gt; S}$.  Can $H$ really have a different solubility than $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the answer is yes, but only if the equilibrium $\ce{H&amp;lt;=&amp;gt;A}$ cannot and is not established.  In the presence of an &quot;infinite&quot; kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H&amp;lt;=&amp;gt;S}$ and $\ce{H&amp;lt;=&amp;gt;A}$ and independent equilibria which might have different equilibrium constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a comment to your question, I made it seem like &quot;any&quot; precipitation from a (super)saturated solution of $S$ would be the hydrate.  On further reflection, I think that is true, &lt;em&gt;unless&lt;/em&gt; (i) the equilibrium $\ce{A&amp;lt;=&amp;gt;H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution.  In that case, the supersaturation could just result in &quot;precipitation&quot; of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How, then, was the solubility of the hydrate ever experimentally determined?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-16T06:26:09.770" LastActivityDate="2015-08-16T06:26:09.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34900" PostTypeId="2" ParentId="34830" CreationDate="2015-08-10T04:12:13.597" Score="3" Body="&lt;p&gt;Density is generally proportional to how far from a full octet (or in your case, an 18 electron set of outer-most orbitals) a material is, hence this effect, and its inverse - that earlier elements in a period are less dense. This does not apply only to transition metals, but all elements, &lt;a href=&quot;http://chemistry.bd.psu.edu/jircitano/density.html&quot; rel=&quot;nofollow&quot;&gt;as illustrated here&lt;/a&gt;. (Note: I am not sure what pressure is in that chart, so I suggest ignoring gas-phase elements, including O, F, and Cl.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is fairly simple. The further from a complete octet an element is, the more energetically favorable orbital overlap becomes. The more favorable orbital overlap becomes, the closer elements will pack at a given temperature. There is no energetic benefit to an element with a full octet (ideal gases) overlapping, so they typically do not, and thus form a gas. There is an intense energy benefit in elements with both half-full octets overlapping.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-08-10T04:12:13.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34902" PostTypeId="2" ParentId="34897" CreationDate="2015-08-10T06:25:45.603" Score="1" Body="&lt;p&gt;Yes you can. By looking at the different electronegativities of each element in a molecule, you are able to determine the dipole moment for each bond. Then by adding the dipole moments for each bond, you are able to determine whether a molecule is non-polar or polar and if it is polar, how polar it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in $\ce{CH4}$, the carbon atom is bonded to four hydrogen atoms. The electronegativity of carbon is 2.55, while the electronegativity of hydrogen is 2.2 This means that for each $\ce{C-H}$ bond, there will be a dipole moment from the hydrogen to carbon atom. However, since methane has a perfect tetrahedral structure, the four dipole moments cancel each other out so that there is no net dipole moment. Hence, methane despite having polar bonds, is actually not a polar molecule, shown below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fcnOg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fcnOg.png&quot; alt=&quot;dipole moment for methane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example that we can consider is ammonia, $$\ce{NH3}$$. Ammonia consists of 3 $\ce{N-H}$ bonds. The electronegativity of nitrogen is 3.04. That means that ammonia consists of 3 polar $\ce{N-H}$ bonds. The molecular structure of ammonia is trigonal pyrimadal and hence unlike methane, when you add up the 3 dipole moments, they don't cancel each other out. Instead, they result in a net upward dipole moment. Hence, the molecule itself is also polar, shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MitHD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MitHD.png&quot; alt=&quot;dipole moment of ammonia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="10185" LastEditDate="2015-08-10T08:17:58.783" LastActivityDate="2015-08-10T08:17:58.783" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34903" PostTypeId="1" AcceptedAnswerId="34905" CreationDate="2015-08-10T09:31:33.160" Score="1" ViewCount="564" Body="&lt;p&gt;This is a question from Atkin's Physical Chemistry Book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that 3mmol $\mathrm{N_2\,(g)}$ occupies $\mathrm{36~cm^3}$ at $\mathrm{300~K}$ and expands to $\mathrm{60~cm^3}$. Calculate $\Delta G$ for the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Attempt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to split this process up into two separate process (I am not sure if this allowed. Please correct me if this wrong). The first process is an reversible isothermal expansion of the gas to $\mathrm{60~cm^3}$. Then the second step is an adiabatic cooling of the gas. I was planning to calculate $\Delta G$ for each step then add them up to get the answer. For the first step, $\Delta U = 0$, therefore:&#xA;$$q = -w = n~R~T~\ln\frac{60}{36}$$&#xA;$$\Delta S = \frac{q}{T} = n~R~\ln\frac{60}{36}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since pressure is constant, $$\Delta H = q = n~R~T~\ln\frac{60}{36}$$&#xA;Therefore: $$\Delta G = \Delta H - T\Delta S = 0$$&#xA;I stopped her because I am sure that this is incorrect as Gibbs Free Energy shouldn't be equal to $0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please clarify if the above procedure is correct and perhaps a better way to attempt this question?&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="865" LastEditDate="2021-05-22T19:07:12.523" LastActivityDate="2021-05-22T19:07:12.523" Title="Calculating Gibbs Free Energy" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34904" PostTypeId="1" CreationDate="2015-08-10T09:58:54.517" Score="1" ViewCount="8792" Body="&lt;p&gt;I have seen these terms being used almost interchangebly, but are decomposition and dissociation reactions the same ?&lt;/p&gt;&#xA;" OwnerUserId="17698" LastEditorUserId="9961" LastEditDate="2015-08-10T18:30:24.247" LastActivityDate="2015-08-11T06:54:51.473" Title="Are decomposition and dissociation reactions the same?" Tags="&lt;terminology&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34905" PostTypeId="2" ParentId="34903" CreationDate="2015-08-10T10:23:01.557" Score="1" Body="&lt;p&gt;&lt;em&gt;I'm quite certain about the following, because it's from the related (german) woorkbook to the 4th (german) edition. The beneath numbering is from the linked online version, which should be the 9th english edition.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your way to the solution is part of subsection &lt;strong&gt;3.9 Properties of the Gibbs energy&lt;/strong&gt; and subsubsection &lt;a href=&quot;https://books.google.de/books?id=BV6cAQAAQBAJ&amp;amp;lpg=PP1&amp;amp;hl=de&amp;amp;pg=PA126#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;3.9 c) The variation of the Gibbs energy with pressure&lt;/strong&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seriously entrust you to read this subsubsection and try to solve it with the given information. If this does not help, the following &quot;hidden&quot; section reveals what you need to solve it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to insert the equation for the molar volume of an ideal gas $V_m = R~T/p$ into (3-57b) $$G_m(p_f) = G_m(p_i)+\int_{p_i}^{p_f}V_m~\mathrm{d}p \tag{3-57.b}$$ which yields one of the two equations that you need: $$G_m(p_f)=G_m(p_i)+R~T\int_{p_i}^{p_f}\frac{\mathrm{d}p}{p}=G_m(p_i)+R~T\ln\frac{p_f}{p_i}\tag{3-59}$$ The other one is the Boyle's law.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-08-10T10:23:01.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34906" PostTypeId="2" ParentId="34904" CreationDate="2015-08-10T10:50:45.873" Score="1" Body="&lt;p&gt;Decomposition and disassociation refer to similar processes, however they are different. The main difference between them is that decomposition reactions are usually irreversible while disassociation reactions are reversible and exist in equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A decomposition reaction is where a compound breaks up into smaller elements or compounds. For example: $$\ce{2H2O -&amp;gt; 2H2 + O2}$$ is an example of a decomposition reaction. Note that in this case that this is actually a reversible reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A disassociation reaction is the general process where molecules split into smaller particles which can include elements, compounds, &lt;strong&gt;ions and radicals.&lt;/strong&gt; For example: $$\ce{NaCl(s) -&amp;gt; Na+ (aq) + Cl- (aq)}$$ is a disassociation reaction. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-08-11T06:54:51.473" LastActivityDate="2015-08-11T06:54:51.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34907" PostTypeId="2" ParentId="34878" CreationDate="2015-08-10T11:40:57.847" Score="3" Body="&lt;p&gt;A glass electrode is usually stored in a 3 or 4M KCl solution, depending on whatever the manufacturer advices you to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the pH that you measure from this solution is absolutely not the pH. It's the measured potential of this concentrated KCl solution that is (in this case wrongly) converted into the pH scale.&#xA;&lt;hr&gt;&#xA;The normal pH measurements with glass electrodes cannot be done without the calibration of your device through references solutions, which are common buffer solutions with acidic, neutral and basic pH values like, e.g., pH 4, 7, 9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calibrate the pH electrode, you or actually your device saves the measured potential for each of these known buffer solutions and calculates a regression function that should give you something near: $$E = 414.12 - 59.16~pH~~~~~(\text{at}~25~^\circ C)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The time when I had to do this in practical courses gave us $E = 401.93 - 56.711~pH$ ... I have no idea what the temperature was :D )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For further informations on all this you can read Mettler-Toledo's &lt;a href=&quot;http://www.mt.com/dam/non-indexed/po/pro/Brochures/IB_pH_Theory_Guide_EN_230113.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pH guide&lt;/a&gt;.&lt;/em&gt;&#xA;&lt;hr&gt;&#xA;Now, as your KCl solution has a certain potential the device converts it through this regression equation into a pH value. But as I said, that is wrong - I guess that pH indicators like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Universal_indicator&quot; rel=&quot;nofollow noreferrer&quot;&gt;universal indicator&lt;/a&gt; would show no big difference from pH 7 (I've never done this actually, so this is seriously a guess).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your actual question should rather be about what the potential of a 3M or 4M KCl solution should be. And that, as can be found &lt;a href=&quot;http://www.ags.gov.ab.ca/publications/GEO/PDF/GEO_2002_11.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gwGvr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwGvr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastActivityDate="2015-08-10T11:40:57.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34908" PostTypeId="1" AcceptedAnswerId="34928" CreationDate="2015-08-10T11:47:55.177" Score="3" ViewCount="1678" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Here's the problem: &#xA;  &lt;img src=&quot;https://i.stack.imgur.com/ubhSL.jpg&quot; alt=&quot;Determining structures of 2 compounds via NMR spectroscopy&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I already know the &lt;span class=&quot;math-container&quot;&gt;$\pu{3600 cm-1}$&lt;/span&gt; line indicates the &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; group and the &lt;span class=&quot;math-container&quot;&gt;$\pu{1640 cm-1}$&lt;/span&gt; line indicates the &lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt;-double-bond-&lt;span class=&quot;math-container&quot;&gt;$\ce{C}$&lt;/span&gt; alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;span class=&quot;math-container&quot;&gt;$\ce{^13C}$&lt;/span&gt; NMR data: the two carbons with chemical shifts of 100+ are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; carbons and the remaining two are &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^3}$&lt;/span&gt; carbons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to work out compound D, and suggest this structure: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/tYHgL.jpg&quot; alt=&quot;Possible structure for compound D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on figuring out compound C. I suggest the structure below, but am unsure whether if it is correct.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VF3Xy.jpg&quot; alt=&quot;Possible structure for compound C&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10257" LastEditorUserId="83758" LastEditDate="2020-05-07T09:55:22.373" LastActivityDate="2020-05-07T09:55:22.373" Title="Deducing compounds from spectroscopic data for isomers of formula C4H8O" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34909" PostTypeId="1" CreationDate="2015-08-10T12:01:45.453" Score="2" ViewCount="10004" Body="&lt;p&gt;Why is it that ionic reactions (i.e., reactions involving ionic compounds) are faster than covalent reactions (reactions involving covalent compounds), even though the ionic bond is stronger than the covalent bond and hence more time will be required to break the bond?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="402" LastEditDate="2017-03-30T21:54:36.473" LastActivityDate="2017-09-23T11:00:39.290" Title="Why are rates of reaction faster for ionic compounds than for covalent compounds?" Tags="&lt;kinetics&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34911" PostTypeId="2" ParentId="34898" CreationDate="2015-08-10T12:30:59.060" Score="7" Body="&lt;p&gt;I'm not sure what you mean by 90% and 10%. If you mean 90% yield (i.e. 90% or more of the bicarbonate ions in sodium bicarbonate will become carbon dioxide), then yes, that will happen if there is enough acid. How much is enough acid? We'll get to that in a minute. If you mean something else, like 90% of the mass of the components will become carbon dioxide, then that probably will not happen.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Let's look at why 90% of the mass of products is not carbon dioxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Bicarbonate anions react with protons from acid to produce carbonic acid, which decomposes to form carbon dioxide:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H+ + HCO3- -&amp;gt; H2CO3 -&amp;gt; H2O + CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's start with a gram of sodium bicarbonate. How many bicarbonate ions are present? The molar mass of sodium bicarbonate is &lt;span class=&quot;math-container&quot;&gt;$84.0066\mathrm{\ g/mol}$&lt;/span&gt;, where a mole (mol) is a quantity equal to the number of carbon-12 atoms in 12 grams of carbon-12 or an &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro's number&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$(6.022\times 10^{23})$&lt;/span&gt; of things. Atoms and molecules are very small, and the numbers of them in macro measurements are large.&lt;/p&gt;&#xA;&lt;p&gt;So we have&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{1.00\ g \times \dfrac{mol}{84.0066\ g}= 0.0119\ mol}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{0.0119\ mol \times \dfrac{6.022\times 10^{23}}{mol}=7.17\times 10^{21}\ molecules}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's stick with the mole unit since it gives us easier to conceptualize numbers.&lt;/p&gt;&#xA;&lt;p&gt;The molar mass of carbon dioxide is &lt;span class=&quot;math-container&quot;&gt;$\mathrm{44.01\ g/mol}$&lt;/span&gt;, so if all of the bicarbonate ion in sodium bicarbonate is converted into carbon dioxide, we will get &lt;span class=&quot;math-container&quot;&gt;$\mathrm{0.0119\ mol}$&lt;/span&gt; of carbon dioxide which is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{0.0119\ mol \times \dfrac{44.01\ g}{mol}=0.524 \ g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So 1.00 gram of sodium bicarbonate will product 0.524 grams of carbon dioxide. We have not even accounted for the citric acid yet. Citric acid is &lt;span class=&quot;math-container&quot;&gt;$\ce{H3C6H5O7}$&lt;/span&gt;, and it is tempting to think that some of those carbon and oxygen atoms in that molecule might become molecules of carbon dioxide. They can, but not in a reaction with sodium bicarbonate (acid-base). If you react citric acid with oxygen (combustion), then you can produce carbon dioxide.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How much acid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Citric acid serves as a proton source - in fact it has three acidic protons. That means in theory, each molecule of citric acid can donate three protons to convert three bicarbonate anions into carbon dioxide:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H3C6H5O7 + 3HCO3- -&amp;gt; C6H5O7^3- + 3H2O + 3CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, those three acidic protons are not equal (and the reason for that would make a great follow-up question). Two of the protons are more acidic than carbonic acid, and the third one is slightly less acidic than carbonic acid. In theory, since carbonic acid decomposes into carbon dioxide, which escapes from solution, we can count on Le Châtelier's principle to pull even that third proton. However, in practice, we might not want to wait around. Let's use one equivalent of citric acid per two equivalents of sodium bicarbonate:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{H3C6H5O7 + 2HCO3- -&amp;gt; HC6H5O7^2- + 2H2O + 2CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, to figure out how much citric acid (molar mass &lt;span class=&quot;math-container&quot;&gt;$\mathrm{192.12\ g/mol}$&lt;/span&gt;) we need, we have to do some calculations. Remember that 1 gram of sodium bicarboncate is 0.0119 moles, and we need 0.5 moles of citric acid for every mole of bicarbonate:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{0.0119 \ mol\ bicarb\times \dfrac{1\ mol\ CA}{2\ mol\ bicarb}\times\dfrac{192.12\ g\ CA}{1\ mol\ CA}=1.14\ g\ citric\ acid}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="102623" LastEditDate="2020-12-26T18:34:51.903" LastActivityDate="2020-12-26T18:34:51.903" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34913" PostTypeId="1" AcceptedAnswerId="35039" CreationDate="2015-08-10T13:46:04.853" Score="17" ViewCount="1736" Body="&lt;p&gt;In the last exam we were asked to rationalize the mechanism of the cyclopropylmethylene carbene cleavage into acetylene and ethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VYpqe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VYpqe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the lecture we discussed pericyclic reactions (cycloadditions, electrocyclic reactions, sigmatropic reactions, ene reactions). I don't think it's a con-/disrotatory ring opening, because in my view only a $\sigma$-plane is conserved over the entire reaction coordinate. What type of ring opening is this; &lt;strong&gt;is it a concerted or radical based C-C bond cleavage?&lt;/strong&gt; Because there is highly probable an elongation of the bond length (as drawn in the stretched cyclopropane moiety), I don't assume it to be concerted but rather radical based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to classify the reaction in terms of symmetry along its entire reaction coordinate. Because from the task, a singlet carbene, I know it's a $sp^2$ hybrid, so I strongly assume the empty p-orbital to be in conjugation with the Walsh orbitals, more precisely with the Möbius type orbitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tcXG0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tcXG0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is very similar to cyclopropylcarbinyl cation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eAZsk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eAZsk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the exam I was asked to draw the orbital correlation diagram. For me it looks like two radicals forms the double bond of ethylene and the &quot;biradical&quot; like centers recombines with the &quot;biradicaloid&quot; (in $sp^2$ orbital). &lt;strong&gt;I call a carbene a diradicaloid or biradicaloid because the two radical electrons share the same orbital.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To correlate reactants with products I need to combine orbitals with the same symmetry. We only draw the orbitals which are involved either in bond braking or in bond making. &lt;strong&gt;How should I account for the biradicaloid singlet $sp^2$ orbital in the orbital correlation diagram?&lt;/strong&gt; (It is forming the triple bond of acetylene, so there must be a change of hybridization.) For me it looks like to have no symmetry (neither A' nor A'' because it lies in the mirror plane, because it is eclipsed with the $\alpha$-carbon-H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a radical mechanism I assume the Möbius orbital of A'' symmetry (depicted in green) to correlate with the $\pi^*$ of ethylene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume, the cleavage happens preferently with light, because in this way I can put a photon from A'' (green) into the orbital of A'' symmetry on the reactant site and therefore lowering the activation energy for the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the orbital correlation diagram to look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3QaNq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3QaNq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some points unclear to me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Has the empty p-orbital A'' symmetry ?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;For me it looks somehow ugly to correlate orbitals which don't have symmetry labels&lt;/strong&gt; (I think this procedure is not allowed and I need to build a proper set.) &lt;strong&gt;How can I build a linear combination in this particulate case, so that the orbital correlation from red to red makes sense?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;There is a &quot;$\pi^*$ orbital acetylene,2&quot; unused (not correlated). Did I miss an orbital from the reactant site? (because there is one orbital too many on the product site)&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="16683" LastEditDate="2017-09-02T13:13:14.620" LastActivityDate="2017-09-02T13:13:14.620" Title="Orbital correlation diagram for the cleavage of cyclopropylmethylene carbene" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;molecular-orbital-theory&gt;&lt;pericyclic&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34916" PostTypeId="1" CreationDate="2015-08-10T19:11:29.703" Score="5" ViewCount="211" Body="&lt;p&gt;Can mechanical agitation catalyze reactions?  &lt;a href=&quot;http://en.wikipedia.org/wiki/Mechanochemistry&quot; rel=&quot;nofollow&quot;&gt;It appears so&lt;/a&gt; (thanks to Mithoron for the pointer!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We observed the following: We were using 10-micron hBN powder to impact-plate copper bearings (e.g., by running them together in a vibratory tumbler, and other similar processes).  Even though the mechanical processes ran in a standard atmosphere and did not exceed 60C we consistently observed that ammonia was released.  As a control:  I have run the same copper with all sorts of other media through the same mechanical processes and never detected ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our conclusion is that the BN is decomposing during the mechanical processing.  But hBN is &lt;em&gt;very&lt;/em&gt; thermally and chemically stable: &lt;a href=&quot;http://en.wikipedia.org/wiki/Boron_nitride#Thermal_stability&quot; rel=&quot;nofollow&quot;&gt;hBN doesn't react with Cu below 1000C&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it plausible that mechanical abrasion at this low a temperature would break the B-N bond?  If so, what subsequent reactions (again, in a standard atmosphere) might cause the free N to form $\ce{NH3}$ in detectable amounts, instead of just forming $\ce{N2}%edit$?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="24052" LastEditDate="2016-03-11T23:23:21.177" LastActivityDate="2017-03-09T03:02:03.713" Title="How might ammonia be created in this mechanochemical reaction?" Tags="&lt;reaction-mechanism&gt;&lt;catalysis&gt;&lt;applied-chemistry&gt;&lt;mechanochemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34917" PostTypeId="1" CreationDate="2015-08-10T19:11:47.917" Score="3" ViewCount="102" Body="&lt;p&gt;I have made up an o-phthaldialdehyde (OPA) reagent according to &lt;a href=&quot;https://books.google.com/books?id=4mj_DArD5n0C&amp;amp;pg=PA715&amp;amp;lpg=PA715&amp;amp;dq=handbook%20of%20hplc%20opa%20derivatization&amp;amp;source=bl&amp;amp;ots=FSOz79VpQF&amp;amp;sig=oq4vpcMvK5SEGz3Xo3FS9wgwR_8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCQQ6AEwAGoVChMI0uqS55mfxwIVip6ACh2TPQDM#v=onepage&amp;amp;q=handbook%20of%20hplc%20opa%20derivatization&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Handbook of HPLC page 715&lt;/a&gt; that will be used to derivatize amino acids before an HPLC run. There is no information given on reagent storage, but the OPA crystals themselves must be stored in the fridge. However, storing this reagent in the fridge causes clear crystals to form that cannot be redissolved into solution. Has anyone had similar issues? Is it possible to store this reagent at room temperature without having the reagent lose strength?&lt;/p&gt;&#xA;" OwnerUserId="19045" LastEditorUserId="10185" LastEditDate="2015-08-10T21:11:51.737" LastActivityDate="2015-08-10T21:11:51.737" Title="Storage issues with o-phthaldialdehyde (OPA) reagent used to derivatize amino acids" Tags="&lt;experimental-chemistry&gt;&lt;chromatography&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34918" PostTypeId="2" ParentId="34891" CreationDate="2015-08-10T19:20:35.893" Score="2" Body="&lt;p&gt;&lt;strong&gt;In general no.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're describing is known as &lt;em&gt;inverse design,&lt;/em&gt; that is, designing something with particular properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree completely with the comment above, that you can find many things from databases. But when you mix together multiple components, into new inter-metallic compounds or alloys, the results do not necessarily match the individual components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, for example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Galinstan&quot; rel=&quot;nofollow&quot;&gt;&quot;Galinstan&quot;&lt;/a&gt;, the various alloys of gallium, indium, and tin. While each of these have low melting points, the eutectic point for the ternary alloy is $\:\mathrm{-19 ^{\circ} C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other examples, but the point remains that without careful experiments or theory for materials design, it's tricky to do inverse design.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-10T19:20:35.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34919" PostTypeId="1" CreationDate="2015-08-10T20:13:25.757" Score="5" ViewCount="736" Body="&lt;p&gt;Generally water soluble salts tend to 'fit between' the molecules of water such that the volume of the resulting solution does not increase much relative to the volume and added mass of the salt. So the result is that the solution generally has a higher density than pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are there any special cases; any salts or compounds that result in an expansion of the resulting solution volume such that its density actually decreases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium polyacrylate comes to mind, but if I'm not mistaken, the expansion doesn't take place until a significant amount of the compound is added, and then only after the liquid is changed to rather a gel or solid. My question relates to the resulting solution actually being a liquid with density &amp;lt; 1.0 g/cm³.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then if not for water, are there solutions of other solvents that might behave this way?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastEditorDisplayName="user7951" LastEditDate="2015-08-29T10:59:26.263" LastActivityDate="2015-08-29T10:59:26.263" Title="Is there a soluble salt, or compound that when dissolved in water results in a solution with density less than pure water?" Tags="&lt;physical-chemistry&gt;&lt;aqueous-solution&gt;&lt;solutions&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34920" PostTypeId="2" ParentId="34919" CreationDate="2015-08-10T21:00:43.297" Score="1" Body="&lt;p&gt;No, you can't dissolve a salt to make a lower density solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of thinking about salt ions just fitting into water molecules, think about it as if the salt ions are letting everything fit together better. The water molecules are achieving their current density because of the presence of hydrogen bonding. When you add the ions you introduce better bonds between the charged ions and the polar water molecules. These bonds allow everything to fit together tighter leading to a rise in density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This decrease in overall bond length is a key to the spontaneity of the solvation reaction. If you introduce an ion that increases the bond length you would see no solubility because they system would be more energetically stable as a bulk solid and water with hydrogen bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polyelectrolyte you bring up raises a good point, but it is not an exception to this rule. Instead, the polyelectrolyte is so large that it &quot;solvates&quot; into water to form a gel. This new material is not really a solution so much as just a new material with water in it. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrate&quot; rel=&quot;nofollow&quot;&gt;Hydrated salts&lt;/a&gt; would be another example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be the case for all solute-solvent systems. However, this is not to say what you want can't be done. You may have to looks at some reversible reaction or &lt;a href=&quot;https://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism&quot; rel=&quot;nofollow&quot;&gt;cis/trans type switching material&lt;/a&gt;. A simple solute-solvent system won't get it done. &lt;/p&gt;&#xA;" OwnerUserId="17967" LastActivityDate="2015-08-10T21:00:43.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34921" PostTypeId="1" CreationDate="2015-08-10T21:17:53.283" Score="6" ViewCount="2093" Body="&lt;p&gt;I work in a restaurant where we dry age our uncut steaks in a cooler. Dry aging is a wonderfully complex process that improves the flavor of richly marbled steaks. Our particular type of aging cooler has a wall of himalayan pink salt (NaCl with other minierals and enough iron oxide to make a pink hue). Our process is even patented by a famous chef. The patent claims that the salt improves the flavor of the beef, and that the closer the beef is to the salt wall, the more flavorful the beef becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a limited understanding of NaCl's basic properties (this is necessary in my line of work). It seems to me that no appreciable amount of salt is making its way to the beef through the air. Air simply circulates around the cooler keeping the beef dry and mitigating microbial grotwh. There is also moisture control that keeps the humidity within a certain range, somewhere in the 60% ball park (relative humidity, i think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I assert to some of my co-workers that the salt does nothing besides pull some moisture from the air (via hygroscopy), they contest. They cite the gradually diminishing volume of the salt and the smell of sea air (by the sea, not in the room) to support the claims that the salt is present in the aging room atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would counter that the salt is simply ending up on the floor as it is dissolved by the water it pulls from the air via hygroscopy (The floor is cleaned daily). And on the sea air: I know that near the ocean, salt, along with other minerals and organic material, can become suspended in the air in microscopic water droplets and these droplets can deposit these salt, minerals, etc on surfaces. However, the formation of these &quot;sea salt aerosols&quot; involves the churning of the sea water, formation of bubbles, and the action of wind, two of which are not present in the salt wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the best summary of my question would be the following: Can damp salt with an air current flowing around it transfer any amount of salt, via the air, to another surface?&lt;/p&gt;&#xA;" OwnerUserId="19047" LastEditorUserId="87609" LastEditDate="2020-11-19T20:45:57.167" LastActivityDate="2020-11-19T20:45:57.167" Title="The Salt Room, is there salt in the air?" Tags="&lt;water&gt;&lt;solubility&gt;&lt;food-chemistry&gt;&lt;atmospheric-chemistry&gt;&lt;salt&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34922" PostTypeId="2" ParentId="34921" CreationDate="2015-08-10T21:53:34.490" Score="3" Body="&lt;p&gt;In short, no. The &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7647145&amp;amp;Mask=4&amp;amp;Type=ANTOINE&amp;amp;Plot=on#ref-2&quot; rel=&quot;nofollow&quot;&gt;vapor pressure of salt&lt;/a&gt; is trivial compared to the grams of salt you would put on a steak when cooking it. I think your analysis is good- the most likely impact is that the salt modulates the humidity.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2015-08-10T21:53:34.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34923" PostTypeId="2" ParentId="34916" CreationDate="2015-08-10T21:56:48.900" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://sandblastingabrasives.com/hexagonal-boron-nitride-powder-order-page-781.html&quot; rel=&quot;nofollow&quot;&gt;Fine hexagonal boron nitride powder is hygroscopic&lt;/a&gt;, so this implies there might be some water in your system. According to &lt;a href=&quot;http://www.znaturforsch.com/s63b/s63b0742.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Reaction of Hexagonal Boron Nitride Nano-crystals under Mild&#xA;Hydrothermal Conditions&lt;/em&gt;&lt;/a&gt; &quot;The reaction between water and hBN nano-crystals starts at very low pressure and temperature, i.e. 220 C and 1.0 MPa [~10 atmospheres].&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those conditions are likely to be met in a ball-mill or vibratory tumbler, at least momentarily at the points of impact (in a similar way, cavitation can heat water to luminescence!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple test would be to &lt;em&gt;preheat&lt;/em&gt; the tumbler and its contents to 150 or 200 C for an hour or two, then seal it &lt;em&gt;airtight&lt;/em&gt; before running the tumbling cycle. If there's &lt;em&gt;no&lt;/em&gt; ammonia odor, then it indicates that adsorbed water in the hBN had been reacting with it, perhaps catalyzed by the copper bearings, as well as by impact heating and pressure.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-10T21:56:48.900" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34924" PostTypeId="1" CreationDate="2015-08-10T22:18:34.940" Score="3" ViewCount="213" Body="&lt;p&gt;The system is an aqueous solution of sulfuric acid and sulfur dioxide. I know how much sulfur there is and I know the pH.  The system is in equilibrium.  I would like to use Octave to solve the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set up the following equations, ignoring $\ce{SO3^2-}$ because the pH is around 1.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition:&#xA;$$ \mathrm{pH} = -\log [\ce{H+}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of mass:&#xA;$$[\ce{H2SO4}] + [\ce{H2SO3}] + [\ce{HSO4-}] + [\ce{SO4^2-}] + [\ce{HSO3-}] = M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Henderson-Hasselbalch equations for the system:&#xA;\begin{align}&#xA;\mathrm{pH} &amp;amp;= \mathrm{p}K_\mathrm{a}(\ce{HSO3-}) + \log \frac{[\ce{HSO3-}]}{[\ce{H2SO3}]}\\&#xA;\mathrm{pH} &amp;amp;= \mathrm{p}K_\mathrm{a}(\ce{SO4^2-}) + \log  \frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}\\&#xA;\mathrm{pH} &amp;amp;= \mathrm{p}K_\mathrm{a}(\ce{HSO4-}) + \log \frac{[\ce{HSO4-}]}{[\ce{H2SO4}]}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then defined the solution vector to have the following order:&#xA;$$ x = [ [\ce{H+}], [\ce{HSO4-}], [\ce{SO4^2-}], [\ce{HSO3-}], [\ce{H2SO4}], [\ce{H2SO3}] ]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and my code for Octave is attached.  For some reason I always get negative values for some of the results.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function y = phCalc(x)&#xA;   pH = 1.87;&#xA;   pKa1 = -5;   % this is a guess to drive H2SO4 to ionization in this system&#xA;   pKa2 = 1.987;% from a textbook&#xA;   pKa3 = 1.857;% from a text book&#xA;   M = 0.0012 / 32 * 1000 ; % total sulfur moles per volume&#xA;   y(1) = 10^(-pH) - x(1);&#xA;   y(2) = x(5) * 10^(pH-pKa1) - x(2);&#xA;   y(3) = x(2) * 10^(pH-pKa2) - x(3);&#xA;   y(4) = x(1) - x(2) - x(3) - x(4);&#xA;   y(5) = x(2) + x(3) + x(4) + x(5) + x(6) - M;&#xA;   y(6) = x(4) / 10^(pH-pKa3) - x(6);&#xA;endfunction&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="19048" LastEditorUserId="13184" LastEditDate="2015-08-11T12:36:00.487" LastActivityDate="2015-08-17T10:13:58.190" Title="How to find the concentrations of the acid-base equilibrium between sulfuric acid and sulfur dioxide for a known pH using Octave?" Tags="&lt;acid-base&gt;&lt;cheminformatics&gt;" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="34925" PostTypeId="1" CreationDate="2015-08-10T22:25:42.330" Score="3" ViewCount="279" Body="&lt;p&gt;Why are there no formal charges on the 1st compound? I am specifically referring to the nitrogen atom, which I would think should exhibit a +2 formal charge? However, in David Klein's organic chemistry textbook it is said that the 1st compound shows no formal charge anywhere. The second compound, of course, exhibits a +1 formal charge on the nitrogen atom and -1 formal charge on the oxygen atom. This would make more sense if I were to take into account the lone pair for Nitrogen in compound 1 (and the 2 lone pairs for oxygen), &lt;strong&gt;but then why would they hide the lone pairs in compound 1 and exhibit them in compound 2?&lt;/strong&gt; Please, correct my incorrect thinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z04DZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z04DZ.png&quot; alt=&quot;Compound 1 and 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17904" LastEditorUserId="10185" LastEditDate="2015-08-11T08:28:00.393" LastActivityDate="2015-08-18T00:47:33.303" Title="Identifying lone pairs when looking for formal charges" Tags="&lt;formal-charge&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34926" PostTypeId="1" AcceptedAnswerId="35080" CreationDate="2015-08-10T22:42:47.640" Score="4" ViewCount="924" Body="&lt;p&gt;When a metal cation is complexed, there is strong UV-Vis absorption due to the splitting of its $d$ orbitals, thereby allowing electronic transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that ligands contribute very little to overall observed absorption. This is evident, for example, in copper solutions. Copper(II) chloride, copper(II) sulfate, and copper(II) nitrate – all solutions of complexed ions – are all very similar in color and have similar absorption spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the simple answer is that transition metals have the capacity to allow for electronic transitions while most ligands do not, but I don't believe it is this simple. Also, it is curious that sulfate, chloride, and nitrate are all colorless when dissolved in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why do ligands have such a small effect on overall absorption of a complexed metal cation?&lt;/p&gt;&#xA;" OwnerUserId="19049" LastEditorUserId="5017" LastEditDate="2015-08-14T23:42:30.483" LastActivityDate="2016-12-06T15:46:09.383" Title="Why do ligands have such a small effect on overall absorption of a complexed ion?" Tags="&lt;inorganic-chemistry&gt;&lt;ions&gt;&lt;coordination-compounds&gt;&lt;spectrophotometry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34927" PostTypeId="1" CreationDate="2015-08-10T22:48:54.653" Score="3" ViewCount="1137" Body="&lt;p&gt;What is the meaning of overpotential in the context of photocatalytic water-splitting? That is light drives charge separation inside TiO2 and the generated electrons and holes react with water to produce hydrogen and oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many papers talk about the high overpotential associated with the oxygen evolution half of the reaction. I understand overpotentials as the potential you apply to (an electrochemical) cell in order to drive the reaction without energy barriers. However, in a photocatalytic system, there are no bias voltages. So, what is the physical significance of overpotential here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14566" LastActivityDate="2015-08-11T01:57:54.703" Title="Overpotential for photocatalytic water-splitting with TiO2" Tags="&lt;electrochemistry&gt;&lt;photochemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34928" PostTypeId="2" ParentId="34908" CreationDate="2015-08-10T23:06:44.853" Score="2" Body="&lt;p&gt;Your two proposed structures are correct, although your rationalisation of the 13C spectrum for Cpd D is not quite correct around the double bond. In a nutshell, carbon chemical shifts can be calculated as a function of their &amp;alpha;, &amp;beta; and &amp;gamma; contributions. &amp;alpha; for what is directly bonded to that carbon, &amp;beta; for substituents one carbon away, and &amp;gamma; for substituents two carbons away. Almost without exception, alkene carbons exhibit a downfield &amp;beta; shift and upfield &amp;gamma; shift for substituted alkenes. So, for both compounds C and D, the carbon closest to the substituents will be shifted furthest downfield. The &amp;beta; and &amp;gamma; contributions for the following substituents are shown (someone more tech savvy than me might tidy this up) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Substituent; &amp;beta; ;&amp;gamma;]&#xA;[-H; 0 ; 0]&#xA;[-CH2O-; +14.2 ; -8.2]&#xA;[-CH3 ; +9.0 ; -7.0]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronics contributing to &amp;beta; and &amp;gamma; shifts is not well understood, but is more complex than a consideration of just electronegativities/electron withdrawing potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are the spectra for the two compounds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aeFtY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aeFtY.png&quot; alt=&quot;Cpd C&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LYzCX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYzCX.png&quot; alt=&quot;Cpd D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your confusion with the description of the quintet label arises from how people report splitting patterns in the literature. I have discussed this on &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18627/how-to-assign-the-signals-in-the-h-nmr-of-aspirin/18634#18634&quot;&gt;another question&lt;/a&gt;, but essentially the two methods of reporting splittings are to (a) report the &lt;em&gt;observed&lt;/em&gt; splitting pattern (here a quintet) and (b) report the &lt;em&gt;calculated/expected&lt;/em&gt; splitting pattern (here a doublet of quartets). As you can see, (or least I can), reporting the observed pattern can lead to some confusion, and gives no real information about how this splitting pattern arises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, reporting couplings to the nearest Hz (here 6Hz) might be fine for some occasions where the difference in coupling for the two different partners is less or approximate to the observed linewidth, however, with a good sample and a good operator on a good magnet, what is reported as a quintet may very well appear different, especially once some apodization is applied. For example, below is a expansion of a simulated spectrum for compound C with Jbc=5.8 and Jbe=6.2 (smaller couplings from d and f ignored), and observed linewidth of 0.5Hz. The top spectrum is the normal spectrum, and the bottom spectrum is what it looks like with a gaussian linewidth function applied (gb 0.1, lb -1). They look quite different, and the bottom spectrum is very hard to rationalise as a quintet, but easily recognizable as a doublet of quartets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rae8d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rae8d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-08-11T00:09:18.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34929" PostTypeId="1" AcceptedAnswerId="34931" CreationDate="2015-08-10T23:39:17.447" Score="28" ViewCount="5695" Body="&lt;p&gt;Freezing a full bottle of water tends to shatter the glass bottle. What if you used something tougher than glass, like diamond? What would happen if you kept dropping the temperature, but restrained the liquid volume so it couldn't freeze and what sort of force would the liquid exert on the container's walls?&lt;/p&gt;&#xA;" OwnerUserId="17933" LastEditorUserId="10185" LastEditDate="2015-08-11T14:23:07.220" LastActivityDate="2015-08-11T16:20:48.557" Title="What happens if you cool water in a container too small for it to freeze?" Tags="&lt;water&gt;&lt;phase&gt;&lt;melting-point&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34930" PostTypeId="2" ParentId="34929" CreationDate="2015-08-11T00:17:57.930" Score="9" Body="&lt;p&gt;If, at atmospheric pressure, ice is less dense than liquid water (it is) then we have to use pressure to keep it from expanding during the freezing process. The amount of pressure required depends on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Properties_of_water#Compressibility&quot;&gt;compressibility of water and ice&lt;/a&gt; in a fairly complex way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass bottle in your example breaks because the pressure exceeds the tensile strength of the bottle. The details of converting pressure to stress in a material is more mechanical engineering than chemistry, but I would refer you to &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylinder_stress&quot;&gt;Wikipedia's cylinder stress page&lt;/a&gt; if you care to actually work through the calculations. If you made your bottle out of some magical material with infinite strength, the pressure would just keep increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of what happens are pretty neat. Water can actually form lots of different kinds of ice with different crystal structures. Check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase_diagram_of_water.svg&quot;&gt;Wikipedia's phase diagram&lt;/a&gt; for more info. If you follow the freezing curve on that line, you'll see that with enough pressure the freeze temperature drops to ~250K!&lt;/p&gt;&#xA;" OwnerUserId="4754" LastEditorUserId="5541" LastEditDate="2015-08-11T11:52:28.937" LastActivityDate="2015-08-11T11:52:28.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34931" PostTypeId="2" ParentId="34929" CreationDate="2015-08-11T00:33:47.970" Score="28" Body="&lt;p&gt;Good question.  Let's assume the container is infinitely strong, non-deformable, and constant in volume.  Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any &lt;a href=&quot;https://www.youtube.com/watch?v=Fot3m7kyLn4&quot; rel=&quot;noreferrer&quot;&gt;supercooling&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from.  This will raise the pressure on the rest of the water.  Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water.  According to &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0165232X03001277&quot; rel=&quot;noreferrer&quot;&gt;this paper from 2004&lt;/a&gt;, ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 4 from that same paper gives the freezing point depression of water as a function of pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jZKSG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jZKSG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed.  If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html&quot; rel=&quot;noreferrer&quot;&gt;46.4 ppm per atm&lt;/a&gt; we can come up with a very simple version of that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before freezing of a fraction $X$ of the water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{ice} = X V_{tot}$$&#xA;$$V_{water} = (1-X) V_{tot}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After freezing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_{ice} = X V_{tot} 1.11 $$&#xA;$$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining those equations, you can get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water.  If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres!  Freezing 10% of the water would mean a pressure of 260 atmospheres.  Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 &amp;deg;C or -2&amp;deg;C.  Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 &amp;deg;C, the setting of the average home residential freezer!  ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-11T15:59:58.163" LastActivityDate="2015-08-11T15:59:58.163" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34932" PostTypeId="2" ParentId="34927" CreationDate="2015-08-11T00:59:37.920" Score="2" Body="&lt;p&gt;Photon absorption by titania results in an excitation of an electron from the valence band to the conduction band.  The &lt;em&gt;difference&lt;/em&gt; in energy of an electron in the valence band and the conduction band is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Band_gap&quot; rel=&quot;nofollow&quot;&gt;band gap&lt;/a&gt;, which for titanium dioxide is about 3.3 eV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What matters for electrochemistry, though, is the absolute voltage of (i) the excited electron, and (ii) the absolute voltage of the &quot;hole&quot; left in the valence band, not the just the difference between them (gap).  That is far more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective electrochemical voltages of the excited electron (or hole) can differ from what the desired electrochemical reaction is.  So there could still be an overpotential.  For titania, &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0165232X03001277&quot; rel=&quot;nofollow&quot;&gt;one paper I found&lt;/a&gt; estimates of the absolute energies of the valence band &quot;hole&quot; and the conduction band electron of -7.8 eV and -4.5 eV, respectively (at pH 0).  The thermodynamic potential for water oxidation is -5.8 eV, 2 eV higher than the hole energy.  That energy is lost if the hole is used to power water oxidation.  You could thus view excited titania having an overpotential of 2 eV for the water oxidation reaction.  In contrast, the theoretical potential for proton reduction is -4.5 eV, close to the conduction band electron energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the energetics of bulk titania provide lots of overpotential for the water oxidation reaction but the tradeoff is that a high band gap means only high-energy photons can power the electrochemistry.  Lowering the band gap would make more and longer wavelengths of radiation available to power water photolysis, but the band gap can only be lowered in certain ways.  For titania, it would not be very desirable to lower the energy of the conduction band, because that would mean excited electrons would not be able to reduce protons to hydrogen .  Thus, the only way to lower the band gap is by raising the energy of the valence band, which would both lower the overpotential and and make more types of radiation accessible for powering water photolysis.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-11T01:57:54.703" LastActivityDate="2015-08-11T01:57:54.703" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34933" PostTypeId="2" ParentId="34924" CreationDate="2015-08-11T01:16:29.487" Score="5" Body="&lt;p&gt;Your code seems to be calculating the &lt;em&gt;differences&lt;/em&gt; between a guessed concentration vector &lt;code&gt;x&lt;/code&gt; and a calculated concentration vector.  These differences in concentrations between guessed and calculated are &lt;code&gt;y&lt;/code&gt;.  For example, the guessed pH is apparently &lt;code&gt;x(1)&lt;/code&gt;, and the calculated pH is &lt;code&gt;10^(-pH)&lt;/code&gt;.  The difference between them is &lt;code&gt;y(1)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, its no surprise that the &lt;code&gt;y&lt;/code&gt; values are sometimes negative.  To solve the problem you need to feed your function to a numerical solver such as &lt;a href=&quot;https://www.gnu.org/software/octave/doc/interpreter/Solvers.html&quot; rel=&quot;nofollow&quot;&gt;&lt;code&gt;fsolve&lt;/code&gt;&lt;/a&gt;.  That function will take your guessed solution &lt;code&gt;x&lt;/code&gt;, other parameters such as &lt;code&gt;pH&lt;/code&gt;, the various &lt;code&gt;pKa&lt;/code&gt;s, and &lt;code&gt;M&lt;/code&gt;, and your function &lt;code&gt;phCalc&lt;/code&gt; as input, and return the value for &lt;code&gt;x&lt;/code&gt; which makes all the &lt;code&gt;y&lt;/code&gt; variables zero.  That solution is the one you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the chemistry side, you mention sulfur dioxide, but I don't see any equations for the solubility of sulfur dioxide gas in water.  Additionally, sulfur dioxide is reduced relative to sulfuric acid, so there would need to be an additional reaction, and parameter (extent of reaction or degree of reduction or something) that describes the relative amounts of reduced sulfur (as sulfur dioxide, bisulfite, or sulfite) vs. oxidized sulfur (as sulfate, bisulfate, sulfuric acid, or sulfur trioxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue of charge balance: what concentration of inert cations (e.g. sodium or potassium or whatever) are you assuming?  Charge balance must hold in bulk aqueous solution.  If you don't have any sodium / potassium etc., then you should include the self ionization of water as an equation, hydroxide as a species, and &lt;code&gt;y(7) = x(1) - x(2) - 2*x(3) - x(4) - x(OH)&lt;/code&gt; as an equation.  If you do have sodium/potassium/whatever at significant concentration, you can neglect hydroxide and just go with the equation &lt;code&gt;y(7) = x(1) - x(2) - 2*x(3) - x(4) + x(Na)&lt;/code&gt; where &lt;code&gt;x(Na)&lt;/code&gt; is the cation concentration, which I have assumed is sodium.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;#&lt;/h1&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="13184" LastEditDate="2015-08-11T15:58:21.840" LastActivityDate="2015-08-11T15:58:21.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34934" PostTypeId="2" ParentId="34867" CreationDate="2015-08-11T04:08:02.200" Score="1" Body="&lt;p&gt;Since the change in entropy does not depend on path, you can do it as a two step process.  First you heat it at constant pressure to the final temperature, then you compress it at constant temperature to the final pressure.  The change in entropy for each of these steps is easy to determine.  You already got the result for the first step.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-11T04:08:02.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34935" PostTypeId="1" CreationDate="2015-08-11T06:28:54.413" Score="3" ViewCount="101" Body="&lt;p&gt;Is it possible that some chemical reactions can't be balanced through the redox method? Because it seems to me that the elements on the reactants side have the same oxidation states with the product side.&lt;/p&gt;&#xA;" OwnerUserId="19053" LastEditorDisplayName="user15489" LastEditDate="2015-08-11T07:14:54.423" LastActivityDate="2015-08-11T07:14:54.423" Title="Is the redox method applicable to all chemical reactions?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34936" PostTypeId="2" ParentId="34935" CreationDate="2015-08-11T07:07:41.293" Score="1" Body="&lt;p&gt;Sometimes you aren't able to use the redox method to balance a chemical equation. This is because the redox method is only used to balanced an equation where a element gets oxidised or reduced. However if there is a chemical equation where no element is getting oxidised or reduced, then you won't be able to use the redox method to balance the equation. However, fortunately most of the time these type of equations can be easily balanced. Consider the following unbalanced equation: $$\ce{Ba(OH)2.8H2O (s) + NH4SCN (s) -&amp;gt; Ba(SCN)2 (s) +  H2O (l) + NH3 (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here none of the elements are reduced or oxidised, hence you would not be able to balance this equation using the redox method.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-11T07:07:41.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34937" PostTypeId="2" ParentId="6181" CreationDate="2015-08-11T09:25:34.597" Score="1" Body="&lt;p&gt;The relationship of osmotic pressure and molecular weight is given by the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi}{C}=\dfrac{\mathrm{R}T}{M_r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a graph of $\frac{\pi}{C}$ against $C$ the intercept is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{intercept}=\dfrac{\mathrm{R}T}{M_r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C=$ concentration&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{R}=$ gas constant&lt;/li&gt;&#xA;&lt;li&gt;$T=$ absolute temperature&lt;/li&gt;&#xA;&lt;li&gt;$M_r=$ molecular weight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19056" LastEditorUserId="275" LastEditDate="2015-08-11T11:40:44.307" LastActivityDate="2015-08-11T11:40:44.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34938" PostTypeId="1" CreationDate="2015-08-11T09:43:04.810" Score="1" ViewCount="6017" Body="&lt;p&gt;I'm supposed to produce dry and pure oxygen using the decomposition of hydrogen peroxide with a manganese dioxide catalyst.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2H2O2 (aq) → 2H2O (l) + O2 (g)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas will be collected in a bottle by downward displacement of water. How can I safely move the oxygen gas through the drying agent without mixing in atmospheric gas?&lt;/p&gt;&#xA;" OwnerUserId="19057" LastEditorUserId="10185" LastEditDate="2015-08-11T15:51:12.930" LastActivityDate="2016-12-31T12:14:08.590" Title="How to produce dry pure oxygen" Tags="&lt;experimental-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34941" PostTypeId="1" CreationDate="2015-08-11T12:34:54.383" Score="1" ViewCount="1563" Body="&lt;p&gt;In making cast iron from pig iron - the cruder form of iron - coke and limestone are added to pig iron and scrap iron in a furnace. Cast iron is   more refined form of iron and has lesser carbon content as well. What exactly occurs in this process of making cast iron - what reactions happen in the furnace? Why is coke added as well, when the purpose is to decrease the carbon content?&lt;/p&gt;&#xA;" OwnerUserId="17654" LastActivityDate="2015-08-11T13:02:19.213" Title="Making Cast Iron from Pig Iron" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34942" PostTypeId="2" ParentId="34941" CreationDate="2015-08-11T13:02:19.213" Score="5" Body="&lt;p&gt;I have quoted the following answers from &lt;a href=&quot;http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/1b_Properties_of_Transition_Metals/Metallurgy/The_Extraction_of_Iron&quot;&gt;the UC Davis ChemWiki page on &quot;The Extraction of Iron.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is coke added as well, when the purpose is to decrease the carbon content?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coke is used primarily as a reducing agent (see below) and comes out as carbon dioxide (and thus does not increase the carbon content of the molten iron).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What exactly occurs in this process of making cast iron - what reactions happen in the furnace?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reactions are given below in the narrative of the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Blast Furnace&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common ores of iron are both iron oxides, and these can be reduced to iron by heating them with carbon in the form of coke. Coke is produced by heating coal in the absence of air. Coke is cheap and provides both the reducing agent for the reaction and also the heat source. The most commonly used iron ores are hematite, ($\ce{Fe_2O_3}$), and magnetite, ($\ce{Fe_3O_4}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air blown into the bottom of the furnace is heated using the hot waste gases from the top. Heat energy is valuable, and it is important not to waste any. The coke (essentially impure carbon) burns in the blast of hot air to form carbon dioxide - a strongly exothermic reaction. This reaction is the main source of heat in the furnace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C + O_2 \rightarrow CO_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Reduction of the Ore&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the high temperature at the bottom of the furnace, carbon dioxide reacts with carbon to produce carbon monoxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{C + CO_2 \rightarrow 2CO}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the carbon monoxide which is the main reducing agent in the furnace:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe_2O_3 +3CO \rightarrow 2Fe + 3CO_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hotter parts of the furnace, the carbon itself also acts as a reducing agent. Notice that at these temperatures, the other product of the reaction is carbon monoxide, not carbon dioxide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe_2O_3 + 3C \rightarrow 2Fe + 3CO}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the furnace is hot enough to melt the iron which trickles down to the bottom where it can be tapped off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Function of the Limestone&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron ore is not pure iron oxide - it also contains an assortment of rocky material that would not melt at the temperature of the furnace, and would eventually clog it up. The limestone is added to convert this into slag which melts and runs to the bottom. The heat of the furnace decomposes the limestone to give calcium oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaCO_3 \rightarrow CaO + CO_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an endothermic reaction, absorbing heat from the furnace. It is therefore important not to add too much limestone because it would otherwise cool the furnace. Calcium oxide is a basic oxide and reacts with acidic oxides such as silicon dioxide present in the rock. Calcium oxide reacts with silicon dioxide to give calcium silicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CaO + SiO_2 \rightarrow CaSiO_3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calcium silicate melts and runs down through the furnace to form a layer on top of the molten iron. It can be tapped off from time to time as slag. Slag is used in road making and as &quot;slag cement&quot; - a final ground slag which can be used in cement, often mixed with Portland cement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cast Iron&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The molten iron from the bottom of the furnace can be used as cast iron. Cast iron is very runny when it is molten and doesn't shrink much when it solidifies. It is therefore ideal for making castings - hence its name. However, it is very impure, containing about 4% of carbon. This carbon makes it very hard, but also very brittle. If you hit it hard, it tends to shatter rather than bend or dent. Cast iron is used for things like manhole covers, guttering and drainpipes, cylinder blocks in car engines, Aga-type cookers, and very expensive and very heavy cookware.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-11T13:02:19.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34944" PostTypeId="2" ParentId="19702" CreationDate="2015-08-11T13:16:30.180" Score="7" Body="&lt;p&gt;The correct and standardized quantity name is &lt;strong&gt;amount-of-substance concentration&lt;/strong&gt;.&lt;br&gt;&#xA;In chemistry, however, the name is generally abbreviated to the single word &lt;strong&gt;concentration&lt;/strong&gt;, it being assumed that the adjective ‘amount-of-substance’ is intended.&lt;br&gt;&#xA;The quantity symbol is a lower case $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quantity ‘amount-of-substance concentration of $\ce{B}$’ is defined as&#xA;$$c_{\ce{B}} = n_{\ce{B}}/V$$&#xA;where $n_{\ce{B}}$ is the amount of substance of $\ce{B}$ and $V$ is the volume of the solution.&lt;br&gt;&#xA;The dimension of the amount-of substance concentration is&#xA;$$\dim c = \mathsf{L}^{-3} \mathsf{N}$$&#xA;The coherent SI unit for amount-of substance concentration is ‘mole per cubic metre’ (unit symbol: $\mathrm{mol/m^3}$).&lt;br&gt;&#xA;However, the usually used unit is mole per litre ($\mathrm{mol/l}$) or any decimal multiple or submultiple rather than mole per cubic metre ($\mathrm{mol/m^3}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the older literature, this quantity was often called &lt;strong&gt;molarity&lt;/strong&gt;. According to current IUPAC recommendations, the traditional name ‘molarity’ should be avoided due to the risk of confusion with the quantity molality. Current international standards do not list ‘molarity’ as a valid special name for the quantity amount-of-substance concentration.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-08-11T13:16:30.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34945" PostTypeId="1" CreationDate="2015-08-11T13:16:41.347" Score="2" ViewCount="33" Body="&lt;p&gt;Is there a chemical property (perhaps electron affinity?) that can be used to predict where a substance would fall in a triboelectric series such as &lt;a href=&quot;https://www.trifield.com/content/tribo-electric-series/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;? Or would the cause of the charge imbalance vary on a case by case basis?&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="10185" LastEditDate="2015-08-11T14:18:29.980" LastActivityDate="2015-08-11T14:18:29.980" Title="Is there a specific chemical property or concept that explains/predicts the triboelectric series?" Tags="&lt;physical-chemistry&gt;&lt;materials&gt;&lt;electrostatic-energy&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34946" PostTypeId="2" ParentId="34779" CreationDate="2015-08-11T13:45:42.637" Score="10" Body="&lt;p&gt;The authors note in their &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/cr200106v&quot; rel=&quot;noreferrer&quot;&gt;review article&lt;/a&gt;&lt;sup&gt;*&lt;/sup&gt; that such effects are expected as &quot;heavy atom effects&quot; due to electron-correlation effects and spin-orbit coupling: (references stripped for clarity)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This being said, one way to improve the computed chemical shifts of heavy-atom-bound carbon atoms is to utilize a method that better captures electron correlation effects, such as MP2 or higher-level methods along with correlation-consistent basis sets (see section 4.3). This will likely result in significant improvement for carbon atoms attached to sulfur or phosphorus as well as to halogens, although the improvement is expected to be greater for sulfur or phosphorus. Further improvement requires the use of methods that determine relativistic spin–orbit corrections. &#xA;  [...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Another option for improving chemical shift calculations with heavy-atom errors is to utilize a linear regression method, as discussed in section 3.6. Due to the systematic nature of heavy-atom effects, they are amenable to improvement via this approach, provided that the linear regression data consists of nuclei similar to the nucleus of interest. That is, a linear correlation correction to carbon nuclei attached to a given number of bromine atoms, for example, should be derived from empirical data for other carbon nuclei attached to the same number and type of halogen atoms. In fact, this approach has been demonstrated specifically for such cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The latter section cites two relevant articles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Theoretical investigation on $\ce{^1H}$ and $\ce{^13C}$ NMR chemical shifts of small alkanes and chloroalkanes&quot; &lt;a href=&quot;http://dx.doi.org/10.1063/1.2353830&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;J. Chem. Phys&lt;/em&gt;. (&lt;strong&gt;2006&lt;/strong&gt;) 125, 144309.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;A hybrid quantum mechanical and empirical model for the prediction of isotropic $\ce{^13C}$ shielding constants of organic molecules&quot; &lt;a href=&quot;http://dx.doi.org/10.1039/B206245C&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Phys. Chem. Chem. Phys.&lt;/em&gt;, &lt;strong&gt;2002&lt;/strong&gt;, 4, 5498-5507.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both articles suggest ways to use DFT methods with empirical corrections that work fairly well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Michael W. Lodewyk, Matthew R. Siebert, and Dean J. Tantillo, &lt;em&gt;Chem. Rev.&lt;/em&gt;, &lt;strong&gt;2012&lt;/strong&gt;, &lt;em&gt;112&lt;/em&gt; (3), 1839–1862.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="4945" LastEditDate="2015-08-17T08:23:47.317" LastActivityDate="2015-08-17T08:23:47.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34948" PostTypeId="1" CreationDate="2015-08-11T15:00:14.017" Score="15" ViewCount="6750" Body="&lt;p&gt;The Sandmeyer reaction is an extremely useful reaction for the functionalization of aromatic rings through a diazonium intermediate. For conversion of a diazonium intermediate to the corresponding chloride, bromide, or cyanide, the copper(I) salt is used. Surprisingly (to me at least) for iodination, potassium iodide is the reagent of choice, as described in Maitland Jones's &lt;em&gt;Organic Chemistry&lt;/em&gt; (and elsewhere I'm sure), but no rationale is given. SciFinder searches also show that potassium iodide is most frequently used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something special about copper(I) iodide that makes it unsuitable for use in the Sandmeyer reaction?&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2019-01-10T16:13:07.390" LastActivityDate="2019-01-10T16:13:07.390" Title="Why use potassium iodide instead of copper iodide in Sandmeyer reaction?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;&lt;halides&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34949" PostTypeId="1" CreationDate="2015-08-11T15:18:58.290" Score="2" ViewCount="1191" Body="&lt;p&gt;My book doesn't classify them as positional isomers but the position of the methoxy group is different in both the molecules. So, I guess they should be positional isomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the two compounds are metamers. But can't a metamer be a positional isomer?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastActivityDate="2015-08-16T02:00:36.807" Title="Is 1-methoxy-2-methylpropane a positional isomer of 2-methoxy-2-methylpropane?" Tags="&lt;organic-chemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34950" PostTypeId="1" CreationDate="2015-08-11T15:28:32.110" Score="0" ViewCount="287" Body="&lt;p&gt;The following electronic configuration and drawing of the structure are given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (3d)^{10} (4s)^2 (4p)^4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GtDI3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GtDI3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this image correspond with the electronic configuration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My train of thoughts so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the configuration has 34 electrons, I assume the element Z is actually selenium (atomic number 34). It could also be gallium since Z has no charge and has five covalent bonds to look like krypton. This is mere speculation since I just don't know how to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure what kind of bonds are present here. The tetrahedral form, even though it consists of a total of five bonds, indicates hybridization, but I am completely unfamiliar with hybridization outside of carbon-like hydbridization (four bonds). I am guessing one double bond leads to sp$^2$ hybridization.&lt;/p&gt;&#xA;" OwnerUserId="19068" LastEditorUserId="865" LastEditDate="2015-08-11T15:41:30.700" LastActivityDate="2015-09-13T17:58:35.813" Title="How does the electronic configuration for this acid oxide correspond to its structure?" Tags="&lt;electronic-configuration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34951" PostTypeId="1" CreationDate="2015-08-11T15:38:39.830" Score="11" ViewCount="13695" Body="&lt;p&gt;The structure of DDT is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yWmDB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWmDB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the benzene molecules should be the parent chain as it has a greater number of carbon atoms than the straight chain (which has just two carbon atoms). Taking benzene as the parent chain, I think that the name of the compound should be 1-Chloro-4-((1'-(4&quot;-chlorophenyl))-2',2',2'-tricholoroethyl) benzene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the Wikipedia article states that the IUPAC name of the compound is &#xA;1,1,1-Trichloro-2,2-bis(4-chlorophenyl)ethane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the ethyl chain taken as the parent chain when it has lesser number of carbon atoms than the benzene ring?&lt;/p&gt;&#xA;" OwnerUserId="10212" LastEditorUserId="865" LastEditDate="2015-08-11T16:08:07.807" LastActivityDate="2017-05-25T07:33:59.663" Title="What is the rationale of the IUPAC name for DDT?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34952" PostTypeId="1" CreationDate="2015-08-11T16:11:20.527" Score="5" ViewCount="7504" Body="&lt;p&gt;I understand why cysteine has a $\mathrm pK_{\mathrm a}$ of 8.3. But the $\mathrm pK_{\mathrm a}$ of thiol in &lt;em&gt;N&lt;/em&gt;-acetylcysteine is 9.27. Since there is an acetyl group attached to the nitrogen, doesn't this lower the $\mathrm pK_{\mathrm a}$ &lt;em&gt;via&lt;/em&gt; the inductive effect?&lt;/p&gt;&#xA;" OwnerUserId="19069" LastEditorDisplayName="user7951" LastEditDate="2015-08-11T21:01:22.787" LastActivityDate="2015-08-12T21:36:32.860" Title="Why is the pKa of the thiol group in N-acetylcysteine higher then the pKa of the thiol group in cysteine?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34954" PostTypeId="1" CreationDate="2015-08-11T16:31:46.153" Score="7" ViewCount="733" Body="&lt;p&gt;Lithium batteries seem to be mainstream now, offering high energy density, high output current capability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepping down the periodic table there are alkali metals with greater electronegativity than Lithium: sodium and potassium, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why not sodium and potassium batteries. Can't these offer higher energy density, higher current output? Is the reactivity just so great that the chemistry can't be engineered to provide a practical battery? Or are the safety issues just too great?&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-08-11T22:59:40.633" Title="Lithium batteries a reality - why not sodium, pottasium batteries?" Tags="&lt;electrochemistry&gt;&lt;safety&gt;&lt;electronegativity&gt;&lt;applied-chemistry&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34955" PostTypeId="1" CreationDate="2015-08-11T17:34:30.877" Score="2" ViewCount="1128" Body="&lt;p&gt;I would like to know what reaction occurs when  soaking a piece of meat in apple cider vinegar. Additionally, how does it happen?&#xA;I read a topic which was about the bad reputation of apple cider vinegar, which can actually &quot;cook&quot; your skin texture. &lt;/p&gt;&#xA;" OwnerUserId="19072" LastEditorUserId="10185" LastEditDate="2015-08-11T17:40:30.963" LastActivityDate="2017-03-05T19:34:44.613" Title="What reaction occurs when soaking a piece of meat in apple cider vinegar?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34956" PostTypeId="1" AcceptedAnswerId="50890" CreationDate="2015-08-11T17:38:37.950" Score="0" ViewCount="22082" Body="&lt;p&gt;I've just learned how to predict the shapes of molecules in class today using VSEPR theory. &#xA;I would like to ask is there anyway to find the number of bond pairs and lone pairs without drawing a dot-and-cross diagram?&lt;/p&gt;&#xA;" OwnerUserId="14831" LastEditorUserId="10185" LastEditDate="2015-08-11T17:53:11.453" LastActivityDate="2016-06-08T03:10:40.250" Title="Identifying the number of bonding and lone pairs without a dot and cross diagram" Tags="&lt;molecules&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34957" PostTypeId="2" ParentId="34954" CreationDate="2015-08-11T17:53:27.810" Score="5" Body="&lt;p&gt;There is relatively little research in sodium ion batteries, and as a result, breakthroughs in sodium battery technology are lacking. Recently, that has changed, and more scientists are looking into the potential of sodium ion batteries. However, the Na-ion battery's biggest drawback is its slow charge and discharge rates. This makes them impractical for everyday use. However, due to its lower cost, they are a cost-effective alternative to Li-ion batteries for applications in energy storage for wind and solar power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;a href=&quot;http://phys.org/news/2015-03-na-ion-batteries-closer-li-ion.html&quot; rel=&quot;noreferrer&quot;&gt;more and more research&lt;/a&gt; in sodium-ion batteries, they may one day take over Li-ion batteries, and some companies such as &lt;a href=&quot;http://www.aquionenergy.com/&quot; rel=&quot;noreferrer&quot;&gt;Aquion Energy&lt;/a&gt; have already begun manufacturing consumer grade sodium-ion batteries.&lt;/p&gt;&#xA;" OwnerUserId="9281" LastActivityDate="2015-08-11T17:53:27.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34958" PostTypeId="1" AcceptedAnswerId="34962" CreationDate="2015-08-11T18:04:19.833" Score="7" ViewCount="1033" Body="&lt;p&gt;Ozone readily decomposes when impacted by UV radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfur dioxide, having the same bond structure and valence electrons in each atom as ozone, is much more stable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would someone please help me understand what makes sulfur dioxide a more stable compound?&lt;/p&gt;&#xA;" OwnerUserId="19073" LastEditorUserId="10185" LastEditDate="2015-08-11T19:26:57.560" LastActivityDate="2016-07-14T10:44:32.840" Title="Why is sulfur dioxide more stable to UV radiation than ozone?" Tags="&lt;resonance&gt;&lt;valence-bond-theory&gt;&lt;ozone&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34959" PostTypeId="1" AcceptedAnswerId="34966" CreationDate="2015-08-11T18:50:36.560" Score="2" ViewCount="224" Body="&lt;p&gt;I am trying to research potential modifications to lead-free solders (intended to be used in electronics). Base alloy is eutectic $\ce{Bi_{58}Sn_{42}}$, and I am going to add various combinations of a bit of copper, silver, gold, platinum, gallium, lithium, and indium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something there which is known to not work? I am wondering this because I see that gold, platinum, gallium, and lithium are not typically used as dopants for electronic soldering alloys. Should I expect a lot of problems attributed to intermetallic compounds? (I know there are issues due to a Sn-Cu reaction, but nothing about the rest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it theoretically possible to find a combination with significantly high copper content (&gt; 10-20%) while still having low melting temperature (&amp;lt; 250 $^{\circ}\mathrm{C}$)?&lt;/p&gt;&#xA;" OwnerUserId="9675" LastEditorUserId="9675" LastEditDate="2015-08-12T19:37:22.770" LastActivityDate="2015-08-12T19:37:22.770" Title="Alloys/metallurgy : promising elements for for Sn-based solders" Tags="&lt;metallurgy&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34960" PostTypeId="2" ParentId="34951" CreationDate="2015-08-11T19:14:45.317" Score="4" Body="&lt;h2&gt;In short&lt;/h2&gt;&#xA;&lt;p&gt;The rule for identifying the parent chain isn't just &amp;quot;longest&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;In fact, &amp;quot;longest&amp;quot; is the fallback for when no other means are appropriate. The most important decider is the number of suffix substituent groups, which in this case is (chloro-)phenyl.&lt;/p&gt;&#xA;&lt;p&gt;It's not a benzene compound, because each benzene ring only has one (n-ethyl), whereas the ethane chain has two.&lt;/p&gt;&#xA;&lt;h2&gt;In long&lt;/h2&gt;&#xA;&lt;p&gt;To understand why this should be so; let's consider the actual precedence list for naming the parent chain (from &lt;a href=&quot;https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Basic_principles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum number of multiple bonds&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum number of single bonds.&lt;/li&gt;&#xA;&lt;li&gt;It should have the maximum length.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;For DDT, we can use rule one: The ethane in the middle has two phenyls stuck to it, but each phenyl only has one ethyl; so the ethane wins.&lt;/p&gt;&#xA;&lt;p&gt;For a simpler example of the same principle at work, consider 1,4-dioctylbenzene: A benzene ring with two octyl tails coming off of it. There are again two candidates for the parent chain, one making it a benzene, one making it an octane, 1-(4'-octylphenyl)-octane.&lt;/p&gt;&#xA;&lt;p&gt;But because choosing the benzene gives a parent chain with two substituents, it becomes 1,4-dioctylbenzene.&lt;/p&gt;&#xA;&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&lt;p&gt;There seems to be some confusion as to whether the rules presented above are the currently recommended ones. A careful reading of the 2013 Blue Book would be necessary to figure that out, and the text isn't online yet.&lt;/p&gt;&#xA;&lt;p&gt;However: The priority rules above were the rules used to arrive at 1,1,1-trichloro-2,2-bis(4-chlorophenyl)etane.&lt;/p&gt;&#xA;" OwnerUserId="5638" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-12T08:27:08.623" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34961" PostTypeId="2" ParentId="34924" CreationDate="2015-08-11T19:16:47.120" Score="5" Body="&lt;p&gt;What you want to solve, is the following system:&#xA;$$&#xA;\frac{k_W}{\ce{[H+]}} - \ce{[H+]} +&#xA;\frac{\ce{[H2SO4]0}\left(\ce{[H+]}~k_{a1}+2~k_{a1}k_{a2}\right)}&#xA;     {\ce{[H+]}^2+\ce{[H+]}~k_{a1}+k_{a1}k_{a2}} +&#xA;\frac{\ce{[H2SO3]0}\left(\ce{[H+]}~k_{a3}+2~k_{a3}k_{a4}\right)}&#xA;     {\ce{[H+]}^2+\ce{[H+]}~k_{a3}+k_{a3}k_{a4}}&#xA;= 0&#xA;$$&#xA;for both concentrations of $\ce{[H2SO4]0}$ and $\ce{[H2SO3]0}$ with the constraint that both concentrations add up to $\mathrm{0.0375~mol/L}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have &lt;s&gt;no idea how to&lt;/s&gt; solved this with Octave (see below), but as I am more used to Mathematica, that was my first approach to see where the error came from. The problem that you have with those negative solutions is, that the constrain about the concentrations does not allow two positive concentrations. Have a look at the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip3.5cm$&lt;img src=&quot;https://i.stack.imgur.com/lgHbD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pH of a pure 0.0375 M sulfuric acid solution would be somewhat around 1.35. Changing the mixture by reducing the amount of sulfuric acid and increasing the amount of sulfurous acid leads to a rising pH. A pure 0.0375 M sulfurous acid solution would then have a pH of about 1.77.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for concentrations below around $\mathrm{0.025~mol/L}$ the system is solvable, what can be seen in the next image. You could obtain the pH of 1.87 with a mixture of 0.00086859742 M $\ce{H2SO4}$ and 0.024131403 M $\ce{H2SO3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip3.5cm$&lt;img src=&quot;https://i.stack.imgur.com/IMVKN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Now doing this with Octave&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I never used Octave before, so there might be a better solution. But this is my code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt; function y = Conc(x)&#xA;ka1 = 10^3; % wikipedia, sulfuric_acid&#xA;ka2 = 10^(-1.99); % wikipedia, sulfuric_acid&#xA;ka3 = 10^(-1.857); % wikipedia, sulfurous_acid&#xA;ka4 = 10^(-7.172); % wikipedia, sulfurous_acid&#xA;ctot = 0.025; % total sulfur concentration in mol/L&#xA;ph = 1.87; h = 10^-(ph);&#xA;oh = 10^(-14)/h;&#xA;xH2SO4 = (h * ka1 + 2 * ka1 * ka2)/(h^2 + h * ka1 + ka1 * ka2);&#xA;xH2SO3 = (h * ka3 + 2 * ka3 * ka4)/(h^2 + h * ka3 + ka3 * ka4);&#xA;y(1) = x(1) + x(2) - ctot;&#xA;y(2) = oh - h + x(1) * xH2SO4 + x(2) * xH2SO3;&#xA;endfunction&#xA;&amp;gt;&amp;gt; [x, fval, info] = fsolve(@Conc, [1; 2])&#xA;x =&#xA;  8.6860e-004&#xA;  2.4131e-002&#xA;fval =&#xA;  6.3704e-011  2.8976e-011&#xA;info =  1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And as you can see, it gives exactle the same solutions as Mathematica. (&lt;em&gt;fval&lt;/em&gt; is the value for y(1) and y(2) that are near zero, so everything is fine. &lt;em&gt;info = 1&lt;/em&gt; means, that everything the solution has converged.)&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-08-17T10:13:58.190" LastActivityDate="2015-08-17T10:13:58.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34962" PostTypeId="2" ParentId="34958" CreationDate="2015-08-11T20:00:41.407" Score="6" Body="&lt;p&gt;I'm going to attempt to answer this from three points of view.&lt;/p&gt;&#xA;&lt;h2&gt;Qualitative considerations and MO theory:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;First&lt;/strong&gt;, why isn't ozone linear? You can imagine &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt; like a &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt; molecule that lacks two electrons. This lack of electrons will result in less &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;-character, and less bonding &amp;quot;attitude&amp;quot;. That's why ozone is bent.&lt;/p&gt;&#xA;&lt;p&gt;Still, the repulsion in its electrons (&lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; is a small central atom, &lt;em&gt;very&lt;/em&gt; small) will want it to be free. Oxygen's van der Waals radius is &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.5 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{\overset{\circ}A}$&lt;/span&gt;&lt;/a&gt;. Thus, &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt; has a bond length of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ozone&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\approx 1.2\ \mathrm{\overset{\circ}A}$&lt;/span&gt;&lt;/a&gt;. That's really &lt;em&gt;cramped&lt;/em&gt;. Electrons aren't happy with this and will try to get free as soon as &lt;span class=&quot;math-container&quot;&gt;$E_a$&lt;/span&gt; and other conditions are provided.&lt;/p&gt;&#xA;&lt;p&gt;Now compare that with sulfur's &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_radius&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.8 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{\overset{\circ}A}$&lt;/span&gt;&lt;/a&gt; radius; which, results in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_dioxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\approx 1.43\ \mathrm{\overset{\circ}A}$&lt;/span&gt;&lt;/a&gt; bond length. Less cramped, the electrons aren't as much complaining as they do in &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt;'s case. We can't get really further as these two molecules aren't isoelectronic.&lt;/p&gt;&#xA;&lt;h1&gt;_____________________________________________&lt;/h1&gt;&#xA;&lt;h2&gt;Approach using classic &lt;em&gt;high school&lt;/em&gt; knowledge:&lt;/h2&gt;&#xA;&lt;p&gt;It's because &lt;strong&gt;oxygen has an arbitrarily positive &lt;a href=&quot;https://en.wikipedia.org/wiki/Formal_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;formal charge&lt;/a&gt;&lt;/strong&gt;. Take a look at &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt;'s electron-dot diagram:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hspace{18ex}$&lt;/span&gt; &lt;a href=&quot;https://i.stack.imgur.com/9ewCA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ewCA.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\hspace{45ex}$&lt;/span&gt; &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/a/ae/Ozone-resonance-Lewis-2D.png/400px-Ozone-resonance-Lewis-2D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\small{Source}$&lt;/span&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The oxygen atom, which has an electronegativity of &lt;span class=&quot;math-container&quot;&gt;$3.6$&lt;/span&gt; in &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronegativity#Allen_electronegativity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Allen scale&lt;/a&gt;, has formed &lt;span class=&quot;math-container&quot;&gt;$sp^2$&lt;/span&gt; orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very &amp;quot;non-bonding&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;The sulfur, on the other hand, has an electronegativity of &lt;span class=&quot;math-container&quot;&gt;$\approx 2.5$&lt;/span&gt;. It will also get a positive formal charge. But since it's less electronegative, it &lt;em&gt;bears&lt;/em&gt; it.&lt;/p&gt;&#xA;&lt;h1&gt;_____________________________________________&lt;/h1&gt;&#xA;&lt;h2&gt;Thermodynamics and kinetics:&lt;/h2&gt;&#xA;&lt;p&gt;Ozone decays into diatomic oxygen under certain conditions:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2O3 -&amp;gt; 3O2}$$&lt;/span&gt;&#xA;Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is &lt;a href=&quot;http://butane.chem.uiuc.edu/pshapley/GenChem2/A4/2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$142~ \mathrm{kJ/mol}$&lt;/span&gt;&lt;/a&gt;, while it's 0 for diatomic oxygen. So&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$&lt;/span&gt;&#xA;Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.&lt;/p&gt;&#xA;&lt;p&gt;Above that, its energy of activation is a measly &lt;a href=&quot;http://scholar.lib.vt.edu/theses/available/etd-05222005-141806/unrestricted/Chapter1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$\approx 32 \mathrm{~kJ/mol}$&lt;/span&gt;&lt;/a&gt;, which is a very small kinetic barrier.&lt;/p&gt;&#xA;&lt;p&gt;Thus, you can also conclude that &lt;span class=&quot;math-container&quot;&gt;$\ce{O3}$&lt;/span&gt; is exceptionally &amp;quot;inclined to decompose&amp;quot;. Also, note that ozone is a very strong oxidant; while &lt;span class=&quot;math-container&quot;&gt;$\ce{SO2}$&lt;/span&gt; is a weaker one.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-11T22:27:29.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34963" PostTypeId="1" CreationDate="2015-08-11T20:46:12.220" Score="1" ViewCount="1289" Body="&lt;p&gt;This is a simple question but it has been so long since I have done any chemistry. Basically I would like to keep the concentration of A constant and the concentration of B variable and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we want to make a 50 ml solution with constant concentration of A and variable concentration of B.From my knowledge I would use C1V1=C2V2 in order to dilute A to its fixed concentration lets say from 4% to 3%. But how do i go about keeping it constant and B concentration variable for a final 50 ml solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd EDIT:&#xA;Apologies, this is my first time using this site and was told to edit my question and add my comment. I didn't put values as I believed it was clear enough but I guess not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@ y chung that is what I have carried out in order to create the solutions, so I'm guessing I did it correctly then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@ bon thanks for the comment. I will tell you exactly what Im trying to accomplish. I have two solutions one is A with a concentration of 9% and one is B with a concentration of 7%. I want to create 5 solutions each 50 ml with a constant 5% concentration of A and varying concentration of B from 1-5 %.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have done is as follows:&#xA;I reduced my A concentration to the required 5% by C1V1=C2V2 for a 50 ml solution getting 28 ml. I put 28 ml in each volumetric flask of 50 ml capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then calculated how much volume of B with the 7% concentration I would need to get it to say the 15 concentration based on my remaining volume of 22 (50-28). For the 1 % concentration this yielded a volume of B at 7% of 3.14 ml. I added that to my flask contain the 28 ml of A and made up the solution to 50ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is this correct? from what I gathered from y chung ,it is.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19075" LastEditorUserId="19086" LastEditDate="2015-08-12T16:54:27.457" LastActivityDate="2015-08-12T16:54:27.457" Title="Keeping one component concentration constant and varying another in a solution" Tags="&lt;concentration&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="34964" PostTypeId="1" CreationDate="2015-08-11T21:19:51.983" Score="4" ViewCount="168" Body="&lt;p&gt;Every chemical has a decomposition temperature.  My understanding is that above that temperature molecular bonds are broken.  And if we raise the heat high enough then &lt;em&gt;all&lt;/em&gt; molecular bonds will break and we will be left with a &quot;plasma&quot; of ionized atoms, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the decomposition temperature of bonds be predicted by models, and if so on what data do those models depend?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose we heat a chemical in an inert (noble) atmosphere, or in a vacuum, to the point of total decomposition.  Now we cool the plasma back to STP.  Are there models that can predict the products we would expect to find &amp;ndash; i.e., the right hand side of the reaction equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that thermodynamics stipulates we should find the lowest-&quot;energy&quot; compounds that can be created from the constituent elements on the right hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice I imagine that achieving that result could require that cooling be done at some &quot;sufficiently low rate.&quot;  So, to increase the complexity of the question: Are there any models that can predict what products will be found as the cooling rate is increased?  I believe that in the limit (a virtually instantaneous quench) one ends up with a &quot;glass&quot; (or combination of glass, condensate, and gas) but even then one can't be left with free ions so can we predict the composition of the quenched plasma?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="194" LastEditDate="2017-04-06T03:02:07.637" LastActivityDate="2017-04-06T03:02:07.637" Title="Plasma creation and condensation" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;energy&gt;&lt;theoretical-chemistry&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34966" PostTypeId="2" ParentId="34959" CreationDate="2015-08-11T22:35:25.657" Score="3" Body="&lt;p&gt;Lithium might be tricky to alloy in an oxidizing atmosphere (e.g. air), and Pt might take some effort to form an alloy, because of its high m.p., which, however, is still less than than the boiling points of Bi or Sn. Other than that, there should be little trouble forming alloys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can search on line for &quot;phase diagram &lt;em&gt;alloy components&lt;/em&gt;&quot; to find the melting point for at least some of the alloys. E.G. an online search for&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;phase diagram &quot;copper-indium&quot; alloy&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;turned up &lt;a href=&quot;http://link.springer.com/article/10.1007%2FBF02882415#page-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; at &lt;em&gt;Springer-Verlag GmbH&lt;/em&gt;, with the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BaEnZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BaEnZ.png&quot; alt=&quot;phase diagram Cu-In alloy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I would expect the &quot;10-20 % copper&quot; requirement precludes melting point less than 300 C, even with low-melting In, Hg, Ga etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I would not use &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dopant&quot; rel=&quot;nofollow noreferrer&quot;&gt;dopant&lt;/a&gt;&quot; for a component of an alloy that will not appreciably change its electronic properties; it is just another component, be it minor, of a ternary, quaternary etc. alloy.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-11T22:35:25.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34967" PostTypeId="2" ParentId="34954" CreationDate="2015-08-11T22:59:40.633" Score="7" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium%E2%80%93sulfur_battery&quot;&gt;sodium–sulfur battery&lt;/a&gt; is already in use for power leveling and other stationary applications, such as &lt;a href=&quot;http://www.energystorageexchange.org/projects/269&quot;&gt;a 1.2 MWh installation in WV, USA&lt;/a&gt;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ford_Ecostar&quot;&gt;Ford Ecostar&lt;/a&gt; did use a Na-S battery, but it did not have sufficient energy for distant travel (and some of the experimental batteries caught fire, but so do Li-ion cells on occasion. Oops!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium-ion_battery&quot;&gt;Sodium-ion batteries&lt;/a&gt; are also in development. In the JACS, the article &lt;a href=&quot;http://phys.org/news/2015-03-na-ion-batteries-closer-li-ion.html&quot;&gt;&lt;em&gt;Na-ion batteries get closer to replacing Li-ion batteries&lt;/em&gt;&lt;/a&gt; discusses some improvements in maximum current capacity and total energy storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, for &lt;em&gt;mobile use&lt;/em&gt;, lithium has a higher energy/weight ratio than heavier metals, but for other use, sodium is less expensive and is readily available.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-11T22:59:40.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34968" PostTypeId="2" ParentId="34938" CreationDate="2015-08-11T23:13:05.313" Score="1" Body="&lt;p&gt;Commercial drying agents, such as &lt;a href=&quot;https://secure.drierite.com/catalog3/page4a.cfm&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Drierite&lt;/em&gt; (anhydrous calcium sulfate)&lt;/a&gt;, or other desiccant, should do the job for small quantities of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most inorganic desiccants should &lt;em&gt;not&lt;/em&gt; react with diatomic oxygen; as long as the oxygen is produced separately from the desiccant, there should be no issue. (Dropping 90% or stronger $\ce{H2O2}$ on many substances can cause dangerously rapid decomposition, but with 30% or lower concentration, that danger is minimal.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as preventing atmospheric gases from entering, first bleed the apparatus of residual air by discarding the initial output. Seal the later containers underwater, e.g. by sliding a glass plate across the mouth. See &lt;a href=&quot;https://www.youtube.com/watch?v=xmL2Pax4yUQ&quot; rel=&quot;nofollow&quot;&gt;Janice Crowleys' demo on YouTube&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-11T23:13:05.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34971" PostTypeId="1" CreationDate="2015-08-08T21:20:19.047" Score="9" ViewCount="5677" Body="&lt;p&gt;Silicon exists in a structure similar to diamond, with 4 silicon atoms bonded to each other in a tetrahedron. &#xA;Why do we not see it exist in a structure like graphite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even possible to get silicon to exist like this? If so could we make graphene but with silicon atoms? &#xA;Not sure why you would, since I understand carbon graphene will still be cheaper. But I would just like to know.&lt;/p&gt;&#xA;" OwnerDisplayName="Shaurya Bhave" LastEditorUserId="10185" LastEditDate="2015-08-12T10:23:21.800" LastActivityDate="2015-08-13T06:08:26.873" Title="Why do we not see silicon in a structure like graphite?" Tags="&lt;inorganic-chemistry&gt;&lt;crystal-structure&gt;&lt;allotropes&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34972" PostTypeId="2" ParentId="34971" CreationDate="2015-08-12T01:49:12.523" Score="8" Body="&lt;p&gt;It's possible, and it's already been done! The compound is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silicene&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gxPqr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gxPqr.jpg&quot; alt=&quot;An STM image of silicene, the silicon analog to graphene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, silicene generally isn't perfectly flat. (Then again, as  Karl Ratzsch points out in the comments, graphene isn't perfectly flat either). The Wikipedia article notes that it usually takes a puckered shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the reasons for the puckered shape, but I recall being quite surprised that silicene could be synthesized at all. The reason why is that I was taught in high school that silicon very rarely forms double bonds due to its large size. For a visualization, please refer to the absolutely awful image below (if anyone could provide an actual graphic I'd be grateful).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lweQu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lweQu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black atoms on the left are carbon. The $p$ orbitals overlap to a significant degree, which means that pi-bonds can easily form. The green atoms to the right are silicon. You can see that, because of the larger atomic radius, the p-orbitals don't overlap as much, meaning that double bonds are generally not favorable in silicon. This explains why compounds like &lt;a href=&quot;https://en.wikipedia.org/wiki/Silenes&quot; rel=&quot;nofollow noreferrer&quot;&gt;silenes&lt;/a&gt; are relatively unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pi systems are crucial to the chemistry of carbon--they allow carbon to form multiple bonds, flat rings (like in graphene), and generally be stable in a lot of configurations. The lack of pi-bonding in silicon is sometimes given as a rationalization for why silicon-based lifeforms can't exist, and explains why it's so hard to get silicon into the flat sheets that form silicene.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="2087" LastEditDate="2015-08-13T06:08:26.873" LastActivityDate="2015-08-13T06:08:26.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34975" PostTypeId="1" AcceptedAnswerId="35037" CreationDate="2015-08-12T06:41:18.263" Score="1" ViewCount="13005" Body="&lt;p&gt;How many significant figures are there in $5~\mathrm{ft}\ 10~\mathrm{in}$.&#xA;I know that $5~\mathrm{ft}\ 10~\mathrm{in}= 5.8\bar{3}~\mathrm{ft}=70~\mathrm{in}$, so I am not sure exactly how much significant digits are in this number. &lt;/p&gt;&#xA;" OwnerUserId="19087" LastEditorUserId="5026" LastEditDate="2018-02-22T13:17:47.623" LastActivityDate="2018-02-22T13:22:11.757" Title="How determine the number of significant figures of a feet/inch system?" Tags="&lt;significant-figures&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34976" PostTypeId="1" CreationDate="2015-08-12T07:34:56.730" Score="22" ViewCount="4937" Body="&lt;p&gt;I encountered this question which asks how many possible radicals are formed when $\ce{CH3CH2C(CH3)3}$ is monosubstituted by $\ce{Br2}$. The answer given is 3 while I think it  should be 4, reasoning that there is a &quot;chiral&quot; carbon if the radical is formed on carbon 3. While arguing with my tutor two questions came to my mind:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the geometry of the $\ce{CH3\dot{\ce{C}}HC(CH3)3}$ radical?&lt;/li&gt;&#xA;&lt;li&gt;If the radical is trigonal pyramidal, does it transition quickly between the two possible states? If so are they considered two radicals?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My findings so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/19377/19088&quot;&gt;This answer on SE&lt;/a&gt; hints that the exact shape depends on the substituents on the alkyl radical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com.sg/books?id=gY-Sxijk_tMC&amp;amp;lpg=PP1&amp;amp;pg=PA57#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modern Physical Organic Chemistry&lt;/a&gt; states that all other (non-methyl) localized radicals are not planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com.sg/books?id=bga3xjLVCo0C&amp;amp;lpg=PT257&amp;amp;ots=nTv_o3k-QC&amp;amp;dq=geometry%20of%20free%20radicals&amp;amp;pg=PT258#v=onepage&amp;amp;q=geometry%20of%20free%20radicals&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Reaction Mechanisms&lt;/a&gt; states that the geometry of free radicals is still controversial.&lt;/p&gt;&#xA;" OwnerUserId="19088" LastEditorUserId="41328" LastEditDate="2017-12-02T18:03:22.110" LastActivityDate="2017-12-02T18:03:22.110" Title="What is the geometry of an alkyl radical?" Tags="&lt;organic-chemistry&gt;&lt;molecular-structure&gt;&lt;radicals&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34977" PostTypeId="2" ParentId="34975" CreationDate="2015-08-12T07:43:14.260" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;All non-zero digits are significant(62 has 2 sig fig)&lt;/li&gt;&#xA;&lt;li&gt;In a number without a decimal point, only zeros between non-zero digits are significant(206's 0 is significant). &lt;/li&gt;&#xA;&lt;li&gt;In a number with a decimal point, all zeros to the right of the first non-zero digits are significant(eg. 15.30 the zero is significant but 015.3 the zero is not significant. 15.30 has 4 sig fig but 015.3 only has 3).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19089" LastActivityDate="2015-08-12T07:43:14.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34980" PostTypeId="2" ParentId="34963" CreationDate="2015-08-12T10:16:09.680" Score="-1" Body="&lt;p&gt;Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.&lt;/p&gt;&#xA;" OwnerUserId="17765" LastEditorUserId="17765" LastEditDate="2015-08-12T10:41:25.570" LastActivityDate="2015-08-12T10:41:25.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="34981" PostTypeId="2" ParentId="34956" CreationDate="2015-08-12T10:22:24.020" Score="0" Body="&lt;p&gt;Terminal atom must obey octet (or duplet) rule. Using this, you can determine the number of bond pair around central atom, the rest are lone pair&lt;/p&gt;&#xA;" OwnerUserId="17765" LastEditorUserId="17765" LastEditDate="2015-08-12T11:25:54.027" LastActivityDate="2015-08-12T11:25:54.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34983" PostTypeId="2" ParentId="34976" CreationDate="2015-08-12T11:47:18.750" Score="2" Body="&lt;p&gt;As your books clearly indicate, nobody seems to have been able to find a chemical proof for the geometry of C-radicals. If it's not indeed planar, it equilibrates extremely quickly between the two pyramidal shapes. Unless there is some further steric constraint that keeps it in one conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the argument with your tutor: The correct answer is &quot;two diffent primary radicals and one secondary radical&quot;. If someone asks for a number, and you have the slightest doubt, write two, three senctences explaining it. If you're lucky, the tutor might say &quot;extra point for the one guy who noticed the question was stupid&quot;. If not, he'll still happily give you 8 out of ten. And if you're at a school where this is not the case: Get the hell out of there.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastActivityDate="2015-08-12T11:47:18.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34984" PostTypeId="2" ParentId="34952" CreationDate="2015-08-12T12:48:49.463" Score="1" Body="&lt;p&gt;Assume you are titrating the substance with base. You are already at pH 7 and going up. Removing a proton from a zwitterion (cysteine) to form an anion will require less energy than removing a proton from an already anionic species (N-acetylcysteinate anion) to form a dianion.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="5577" LastEditDate="2015-08-12T17:17:04.590" LastActivityDate="2015-08-12T17:17:04.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34985" PostTypeId="1" AcceptedAnswerId="34996" CreationDate="2015-08-12T14:36:08.917" Score="4" ViewCount="9750" Body="&lt;p&gt;I read on a   &lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/science/triple_aqa/medical_applications_physics/xrays/revision/2/&quot; rel=&quot;nofollow&quot;&gt;bbc gcse bitesize webpage&lt;/a&gt; &quot;In older X-ray machines, white photographic film&quot; [&lt;strong&gt;edit&lt;/strong&gt;- I now know that bbc webpage is wrong, and it's not a white film. I'll leave the rest of the post after the next sentence, as is, but note that from the next paragraph, the rest of the post is based on the misunderstanding from that bbc link that there is white film  (And I guess it's talking of any pre digital when it says 'older' - and anyhow prior to film they used photographic plates - glass rather than plastic/film ).]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that X-ray images are black and white. You have the X-rays themselves, the white film, and an object/person in-between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white parts of the image are bones and/or things the X-rays couldn't get through, the white film remains white, and the black part is where the X-rays did pass through, and it blackens the white film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also heard that they have a cassette/casing, and in that is fluorescent material or coating and film.  The X-rays hit the fluorescent material and (I guess by absorption and reflection, or absorption and emission) converts it into regular light, and there is a phosphorescent material/coating there, too, so the light remains for a while, and the light, I guess, blackens the film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand they don't work like that anymore and now you can have other types of X-ray detection mechanisms. But I want to understand at least how the traditional ones are black and white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that X-ray machines used to use a white photographic film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the white photographic film made from that distinguishes it from a black photographic film? Can I buy white photographic film?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen old-fashioned camera film before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In black and white photography, was it a black photographic film that went white when exposed to light?&lt;/p&gt;&#xA;" OwnerUserId="2020" LastEditorUserId="2020" LastEditDate="2015-08-15T10:33:21.957" LastActivityDate="2015-08-15T10:41:10.567" Title="What makes an X-ray machine's white photographic film go black?" Tags="&lt;photochemistry&gt;" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="34986" PostTypeId="2" ParentId="28402" CreationDate="2015-08-12T14:58:49.437" Score="2" Body="&lt;p&gt;As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?&#xA;$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total amount of substance in buffer solution:&#xA;$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total volume of buffer solution:&#xA;$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molarity of buffer solution:&#xA;$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2017-05-30T13:20:25.510" LastActivityDate="2017-05-30T13:20:25.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34987" PostTypeId="1" CreationDate="2015-08-12T15:02:21.070" Score="0" ViewCount="437" Body="&lt;p&gt;My textbook says that all the bonds of $\ce{SF4} $ are not equal while in the case of $\ce{XeF_2}$, $ \ce{BF4-} $ or $\ce{SiF4} $ all the bonds are equal. No reason is stated for this observation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a general way or formula to check if all the bond lengths are equal or not for a given molecule?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="402" LastEditDate="2017-04-25T23:09:43.437" LastActivityDate="2017-04-25T23:09:43.437" Title="Comparing bond lengths of a molecule using valence bond theory?" Tags="&lt;valence-bond-theory&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="34988" PostTypeId="2" ParentId="34976" CreationDate="2015-08-12T15:06:17.090" Score="13" Body="&lt;p&gt;First off, it was very astute of you to recognize that &lt;strong&gt;if&lt;/strong&gt; the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RwJLR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RwJLR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-12T18:30:54.267" LastActivityDate="2015-08-12T18:30:54.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34989" PostTypeId="2" ParentId="34951" CreationDate="2015-08-12T16:33:21.650" Score="9" Body="&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;by ordinary &lt;strong&gt;substitutive nomenclature&lt;/strong&gt;, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes&lt;/li&gt;&#xA;&lt;li&gt;by &lt;strong&gt;multiplicative nomenclature&lt;/strong&gt;, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a &lt;strong&gt;preferred IUPAC name&lt;/strong&gt; (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-45.1&lt;/strong&gt; MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-45.1.1&lt;/strong&gt; Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).&lt;br /&gt;&#xA;(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and&lt;br /&gt;&#xA;(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and&lt;br /&gt;&#xA;(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.&lt;/p&gt;&#xA;&lt;p&gt;When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The given structure of the DDT molecule fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-15.3.1.3&lt;/strong&gt; Multiplicative name formation&lt;br /&gt;&#xA;Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.&lt;br /&gt;&#xA;When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:&lt;/p&gt;&#xA;&lt;p&gt;(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here: 1,1'&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(b) the name of the linking multiplicative substituent atom or group;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here: 2,2,2-trichloroethane-1,1-diyl&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here: bis&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here: 4-chlorobenzene&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here: 1,1'-(…)bis(4-chlorobenzene) (not 4,4'-(…)bis(1-chlorobenzene))&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the preferred IUPAC name is &lt;strong&gt;1,1'-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-12T16:33:21.650" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34991" PostTypeId="2" ParentId="34964" CreationDate="2015-08-12T17:11:21.973" Score="5" Body="&lt;p&gt;To begin with, dissociating molecules to a plasma at the atomic level will not only create ions but also neutral atoms. These will indeed reassociate on cooling, and the question is how to predict what the products will be. In the extreme case of a simple system such as pure hydrogen gas (dissociating, then recombining to H2) or, say chlorine plus bromine dissociating/recombining to give BrCl, it should be easy to predict the products formed and perhaps more about the behavior of the system as a function of temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a more complex system, including (for the sake of argument) atoms such as carbon that can form complex ring and chain structures, predicting the outcome will be impossible - since (1) the different ways to combine atoms may lead to a huge number of possible intermediates, many of whose energies/reactivity are not known; (2) even if you could predict these intermediate structures, the number of further possible reactions between them becomes astronomical, and (3) kinetic barriers (activation energies) for  reactions are generally harder to predict that ground state energies. As is often the case when theory comes up short, you may have to go into the laboratory and do some experiments. Possible approach: aim the plasma (along with a stream of inert gas such as argon) at a supercooled probe in a vacuum and spectrally characterize the species in the resulting matrix. I imagine that you could end up with some pretty amazing chemical species that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would urge those among my colleagues who are familiar with the current state of theoretical chemistry to add to this discussion. I love a question that makes you go back to basics and rethink everything from the ground up.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-08-12T17:11:21.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34992" PostTypeId="1" AcceptedAnswerId="34995" CreationDate="2015-08-12T18:26:15.357" Score="3" ViewCount="386" Body="&lt;p&gt;I was surprised to learn that &lt;a href=&quot;http://en.wikipedia.org/wiki/Fractional_freezing#Freeze_distillation&quot; rel=&quot;nofollow&quot;&gt;freeze distillation has the same inefficiencies&lt;/a&gt; as vapor distillation: Namely, that it is impossible to get two miscible fluids to fully separate in one &quot;pass&quot; of subjecting the mix to a temperature between the melting point of its constituents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand why this is the case in vapor distillation, because vaporization of a liquid still occurs below its boiling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But AFAIK the same is not generally true of freezing: assuming it has a nucleus, a single liquid will either crystallize completely at a given temperature or it won't at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the inefficiency of freeze distilling has something to do with the chemistry of miscible fluids: I.e., whatever phenomenon allows them to mix also drags molecules of the lower-melting chemical out of the mix above its melting point when the higher-melting chemical crystallizes.  If so then the efficiency of freeze distilling should have some relationship to a measurable quantity (like a miscibility coefficient?).  What theories and models of miscible fluids are applicable to the question of efficiency of freeze distillation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If there is no attraction between the two fluid chemicals then presumably &quot;single-pass freeze distillation&quot; could be perfect.  It would probably require agitation to help the more volatile of the two stay out of the crystals forming from the first chemical to freeze.  For example, agitation during freezing is how ice is formed without visible bubbles from dissolved gasses.  But if &quot;there is no attraction&quot; then presumably the fluid chemicals are not miscible, right?)&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2015-11-18T03:22:54.670" Title="Freeze distillation: Does efficiency depend on solubility?" Tags="&lt;thermodynamics&gt;&lt;mixtures&gt;&lt;distillation&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34993" PostTypeId="1" CreationDate="2015-08-12T19:23:02.220" Score="7" ViewCount="315" Body="&lt;p&gt;The $\ce{.CH3}$ radical is planar but the $\ce{.SiH3}$ radical is a trigonal pyramid. Why are they different? After all, $\ce{C}$ and $\ce{Si}$ are in the same group.&lt;/p&gt;&#xA;" OwnerUserId="17591" LastEditorUserId="10185" LastEditDate="2015-08-12T20:25:27.033" LastActivityDate="2015-08-12T20:41:04.323" Title="Geometries of methyl and silyl radicals" Tags="&lt;organic-chemistry&gt;&lt;molecular-orbital-theory&gt;&lt;hybridization&gt;&lt;molecular-structure&gt;&lt;radicals&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="34994" PostTypeId="2" ParentId="34993" CreationDate="2015-08-12T20:14:21.787" Score="6" Body="&lt;p&gt;As is currently being discussed in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34976/what-is-the-geometry-of-an-alkyl-radical&quot;&gt;this question&lt;/a&gt;, I'm not convinced that the methyl radical is planar, &lt;strong&gt;but&lt;/strong&gt; even if not planar, it is certainly very close to planarity.  It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in electronegativity between $\ce{Si}$ (1.9) and $\ce{C}$ (2.5) and hydrogen (2.1) and how these differences play into &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule/15622#15622&quot;&gt;Bent's rule&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a nutshell, Bent's rule states that, &quot;atomic s character concentrates in orbitals directed toward electropositive substituents&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Bent%27s_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since hydrogen is more electronegative than silicon it will draw more of the electron denisty in a $\ce{Si-H}$ bond towards its end of the bond (the reverse of what happens in the case of a $\ce{C-H}$ bond where the carbon is more electronegative than hydrogen) decreasing the electron density in the silicon atomic orbitals involved in the bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is less electron density in the silicon atomic orbital used in the $\ce{Si-H}$ bond, then according to Bent's rule, the silicon atom will hybridize so that less s-character is used in these atomic orbitals that now have a lower electron density, and will instead use that s-character in the orbital containing the single electron and stabilize it.  Increasing the p-character in the $\ce{Si-H}$ bonds and increasing the s-character in the orbital holding the single electron will tend to push the molecule towards an $\ce{sp^3}$-like, pyramidal geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse is true in the case of carbon.  Due to electronegativities, there will be more electron density towards the carbon side of the $\ce{C-H}$ bond (compared to the $\ce{Si-H}$ bond).  Due to this increased electron density around carbon, it is energetically less favorable to remove s-character from the carbon atomic orbitals involved in the $\ce{C-H}$ bonds (and destabilize this increased electron density) and transfer it to the orbital containing the single electron.  As a result the carbon atomic orbitals involved in the $\ce{C-H}$ bonds will have higher s-character compared to the analogous silicon case and $\ce{.CH3}$ will be less pyramidal.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-08-12T20:41:04.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34995" PostTypeId="2" ParentId="34992" CreationDate="2015-08-12T21:03:29.260" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.corrosionist.com/what_is_the_melting_point_of_steel.htm&quot; rel=&quot;nofollow&quot;&gt;Solids have solubility&lt;/a&gt;, just as liquids do. This is easily seen in metal alloys, where different amounts of different compounds can cause different crystal structure, low-level phase separation, and large-scale phase separation (if the solid solution is not stable). The changing of these nanostructures of the material can even occur with a material fully in the solid state, such as tempering of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relates to your question in that liquids freezing will behave very similarly with respect to phase composition as vapors condensing. The composition of the solid will be proportional to the composition of the liquid and affected by differing freezing points, just like you would expect from a liquid-vapor Txy diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some differences, however. Molecules in solids can, as I said above, move around within the solid. However, this diffusion is substantially slower in solids than in liquids, and in non-metal solids, slow enough to be largely insignificant within the process of freezing. This means that the freezing process will create regions of differing composition within the solid. Further and related to this phenomena, the speed of freezing will impact the concentration of the solid, as faster freezing will push it even further from equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slow diffusion in solids also means that once trapped in the crystal structure, compounds with higher freezing points will not as easily exit the solid as they might from a liquid. This means that melting can behave very differently from vaporization, and solids might become a slush in the process of melting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation as to why this all occurs could go as follows: solubility is a factor of the strength of the interaction between different types of molecules. Those interactions must be stronger than interactions within the individual types to avoid phase separation. The process of freezing tends to not fundamentally shift the strength of these interactions. It is possible for materials partially soluble in liquid to be mostly insoluble in solid, but pairs of &lt;strong&gt;miscible&lt;/strong&gt; liquids have such intense cross-compound interactions that freezing typically can only impact concentration of the separate phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really important chemistry in this case takes place at the phase boundary between the solid and liquid, which behaves almost identically to that between a liquid and a gas. Molecules with weaker interactions with the solid phase (weaker interactions with other molecules in that phase) will tend to leave the solid phase more often - but there is an equilibrium where molecules will enter their less-favored phase. Essentially, everything you would have learned about liquids and vapors applies here, except diffusion is too slow to allow good mixing within the more condensed phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are, in specific cases, more complicated explanations. One is of interstitial compounds, where a second compound might occupy spaces within a solid. While this happens to a degree in all solids, interstitials that are small enough to fit fully in a crystal structure without deforming it might exhibit unusual solid solution behavior, such as a compound with a lower freezing point showing a greater affinity for solid phase than the other compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the title question, yes, separation depends on solubility. Unlike vapor distillation, though, it can also depend on temperature difference (thus speed) and mixing.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastEditorDisplayName="user7652" LastEditDate="2015-11-18T03:22:54.670" LastActivityDate="2015-11-18T03:22:54.670" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="34996" PostTypeId="2" ParentId="34985" CreationDate="2015-08-12T21:11:31.063" Score="6" Body="&lt;p&gt;X-ray film is very similar to standard black and white negative film. It's basically a transparent plastic base coated with an emulsion of silver halide particles (a salt of $\ce{Ag+}$ with a halide, usually $\ce{Cl-}$ or $\ce{Br-}$). As shown below, when light hits one of these particles, it reduces a small amount of the halide to elemental silver (1). When this is exposed to a photographic developer (essentially a mild reducing agent), more elemental silver is produced around the initial sites (2).$$\ce{Ag^+ +e- -&amp;gt; Ag}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a fixing agent is added which solubilizes the unexposed silver halide so it can be removed from the emulsion, preventing further light exposure from fogging the film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the regions that are exposed to light appear dark when developed as the tiny silver particles ($\ce{Ag}$) block light passing through the film and the unexposed regions are transparent, taking on the colour of the plastic base (often blue or grey). X-ray films are typically viewed on an illuminated white background, making the unexposed areas appear white (or off-white, coloured by the plastic base).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal B&amp;amp;W film is actually slightly sensitive to x-rays but, the emulsion of dedicated x-ray film contains other components to facilitate the silver reduction from x-ray exposure. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/I2hS9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I2hS9.png&quot; alt=&quot;Film development process&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:PhotoDeveloptKeySteps.png&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7499" LastEditorUserId="7499" LastEditDate="2015-08-12T21:47:13.143" LastActivityDate="2015-08-12T21:47:13.143" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="34997" PostTypeId="2" ParentId="34952" CreationDate="2015-08-12T21:36:32.860" Score="5" Body="&lt;p&gt;Shown below are the relevant ionization events of cysteine and N-acetylcysteine. The question is asking about the second ionization of each, the deprotonation of the thiol groups, but we see a similar difference in pKas for the carboxylic acid groups as well. The difference in each case is an ammonium group vs. an acylated nitrogen. It's true that the acetyl group is an electron withdrawing group, but it turns out that the cationic ammonium group is even more inductively electron withdrawing. For both the first and second ionizations, the difference in electron withdrawing power is enough to lower the pKa by &gt;1 pKa unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rnTUd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rnTUd.png&quot; alt=&quot;ionization of cysteine and N-acetylcysteine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iad22agp's explanation is a good way of rationalizing this. Looking at the first ionization, cysteine is cationic while N-acetyl cysteine is neutral. Since the rest of the structures are similar, we would assume that it's easier to take a positive charge away from a positively charged compound than a neutral compound. Similarly looking at the second ionization, it's easier to take a positive charge away from a neutral compound than a compound that's already negatively charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pKa refs: &lt;a href=&quot;http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html&quot; rel=&quot;noreferrer&quot;&gt;cysteine&lt;/a&gt;, &lt;a href=&quot;http://toxnet.nlm.nih.gov/cgi-bin/sis/search/a?dbs%20hsdb:@term%20@DOCNO%203003&quot; rel=&quot;noreferrer&quot;&gt;N-acetylcysteine&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-08-12T21:36:32.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34998" PostTypeId="2" ParentId="34877" CreationDate="2015-08-12T22:24:34.480" Score="0" Body="&lt;p&gt;My experience on this is also at least 20 years old, but I used to have a small stick type anti-fogging material that worked well.  It resembled a chapstick and was rubbed onto the surface.  I don't remember the name, but some of the other answers sound similar.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Google search for &quot;anti fog for glasses&quot; or for &quot;anti fog for safety glasses&quot; yields a LOT of hits, including sprays, home made methods, and anti fog glasses.  &lt;/p&gt;&#xA;" OwnerUserId="17020" LastActivityDate="2015-08-12T22:24:34.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="34999" PostTypeId="1" CreationDate="2015-08-12T23:00:17.697" Score="1" ViewCount="80" Body="&lt;p&gt;What does the process of prilling something like, say, calcium nitrate, involve? Do you need other chemicals to prill the substance?&lt;/p&gt;&#xA;" OwnerUserId="11385" LastActivityDate="2015-09-09T01:01:42.590" Title="How are chemicals prilled?" Tags="&lt;elements&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35000" PostTypeId="2" ParentId="34921" CreationDate="2015-08-13T00:06:53.523" Score="2" Body="&lt;p&gt;Unless the salt is a very fine powder and the air is being agitated throughout the room at a very high rate, the salt simply would not be prone to becoming airbone. Based on your description, it doesn't sound like it would be anywhere near fine enough to worry about salt dust possibly contributing anything. Even were it fine, the moisture would quickly clump it and make any possibility of it entering the air in the first place pointless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it's marketing. It's a dumb idea from a science standpoint, but the power of suggestion is often more potent than scientific principles. I wouldn't question it too hard if you value your job based on this very concept - it's quite likely the chef knows it does nothing.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-08-13T00:06:53.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35002" PostTypeId="1" CreationDate="2015-08-13T00:30:43.127" Score="2" ViewCount="4554" Body="&lt;p&gt;Carbon has 4 valence electrons, so it means that it can have a total of four (covalent) bonds, correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's the case, why can't carbon share four bonds, such as with another carbon atom (to form $\ce{C2}$, for example)?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="9961" LastEditDate="2015-08-13T19:01:53.390" LastActivityDate="2015-08-13T19:01:53.390" Title="Can carbon have quadruple bonds?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-08-19T15:38:48.890" ContentLicense="CC BY-SA 3.0" />
  <row Id="35003" PostTypeId="1" AcceptedAnswerId="35021" CreationDate="2015-08-13T01:40:28.283" Score="15" ViewCount="2164" Body="&lt;p&gt;Plastic balls have been added to a reservoir in drought stricken California in order to reduce evaporation.  (&lt;a href=&quot;http://abcnews.go.com/US/los-angeles-reservoir-covered-96-million-shade-balls/story?id=33038319&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to news story here&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the balls colored black instead of white?  I would think that white would decrease heating and therefore evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Apparently the balls are there to block sunlight from converting bromide to carcinogenic bromate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=uxPdPpi5W4o&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video explaining the purpose of the balls&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5601" LastEditorUserId="5601" LastEditDate="2019-05-11T17:59:38.177" LastActivityDate="2019-05-11T17:59:38.177" Title="Why are the plastic balls used to reduce evaporation in this reservoir black instead of white?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;color&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35004" PostTypeId="2" ParentId="35002" CreationDate="2015-08-13T02:20:41.760" Score="6" Body="&lt;p&gt;s orbitals have spherical symmetry, and p orbitals have a dumb-bell shape.  When these orbitals overlap, then they form the bonding orbitals (and corresponding antibonding orbitals).  The p-orbitals are aligned along Cartesian (xyz) axes.  If all three p-orbitals overlap, then the s-orbitals can't get 'close enough' to overlap and form a fourth bond.  If we start mixing s and p orbitals, the geometry always puts electron density away from the other bonding atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concisely: there are no arrangements which allow overlap of atomic orbitals that create four bonding molecular orbitals with the proper orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C2 as a molecule can exist, it will simply only have 3 bonds plus two unpaired electrons confined to each carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more elaborate answer we have to unpack the heavy quantum mechanics, and you won't be seeing that until your junior year.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-08-13T02:20:41.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35005" PostTypeId="1" CreationDate="2015-08-13T02:37:23.977" Score="1" ViewCount="95" Body="&lt;p&gt;If I add sodium percarbonate to my natural laundry soap recipe, can I still claim &quot;No Artificial Anything&quot;?&lt;br&gt;&#xA;I know it's used in many &quot;natural&quot; products, but is it considered an &quot;artificial ingredient&quot;?&#xA;I don't want to make any false claims with my product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you!&lt;/p&gt;&#xA;" OwnerUserId="19100" LastActivityDate="2015-08-13T09:22:18.100" Title="Is sodium percarbonate artificial?" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-08-13T19:37:19.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="35006" PostTypeId="1" AcceptedAnswerId="35061" CreationDate="2015-08-13T03:33:05.627" Score="4" ViewCount="295" Body="&lt;p&gt;I've had success electroplating $\ce{Ni}$ and $\ce{Cu}$, but interestingly never $\ce{Al}$. My focus here is on plating with $\ce{Au}$, specifically onto an $\ce{Ni}$ surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general from my own work and others' suggestions, I have found applying a voltage (direct current) of 5 V through a relatively dilute solution of relevant metal, e.g. 2.53 g $\ce{Ni(NH4)2(SO4)2*6H2O}$ dissolved in 81.9 mL 0.42 % acetic acid at ~60 °C makes a very nice, lustrous plating of $\ce{Ni}$ on a clean $\ce{Cu}$ surface after ~90 s of submersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to dissolve $\ce{Au}$ in warm, concentrated $\ce{HCl}$ and minimal $\ce{HNO3}$, and have heard that you can dissolve it in $\ce{HCl}$ and $\ce{H2O2}$, which makes sense because $\ce{Cu}$ does the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for an ideal solution here, starting with pure metallic gold. I want to avoid any mistakes because I only have 1 g to work with. By default I'm going to try a comparable concentration of dissolved $\ce{Au}$ to what I mentioned with $\ce{Ni}$, but any advice is appreciated – especially a specific method for making an &quot;ideal&quot; solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found that several commercial solutions contain $\ce{(CN)4AuK}$, but I really want to avoid using cyanide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.bren.ucsb.edu/~dturney/port/papers/Modern%20Electroplating/04.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by Paul A. Kohl is immensely helpful and interesting, but I'm wondering if anyone has had any success with something simple.&#xA;It makes mention of using sulphide, thiosulfite, thallium, etc.&lt;/p&gt;&#xA;" OwnerUserId="17821" LastEditorUserId="17821" LastEditDate="2015-08-14T12:34:18.647" LastActivityDate="2015-08-14T14:20:24.377" Title="Making a non-CN dissolved gold solution ideal for electroplating" Tags="&lt;electrochemistry&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35007" PostTypeId="1" AcceptedAnswerId="35032" CreationDate="2015-08-13T05:53:22.767" Score="2" ViewCount="3099" Body="&lt;p&gt;Esters can be formed in two ways according to the syllabus I'm following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Mixing an alcohol with a carboxylic acid in the presence of sulfuric acid; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Mixing an alcohol with an acid chloride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which method produces the highest yield of esters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think method (2) does because the $\ce{Cl-}$ will attach with the $\ce{-OH}$ in alcohol. However, I'm not exactly sure of the reason. Can someone explain it clearly with full details please?&lt;/p&gt;&#xA;" OwnerUserId="15072" LastEditorUserId="865" LastEditDate="2015-08-13T13:18:19.450" LastActivityDate="2015-08-14T03:47:31.190" Title="Routes of formation of esters with highest yield" Tags="&lt;organic-chemistry&gt;&lt;esters&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35008" PostTypeId="1" CreationDate="2015-08-13T07:33:33.007" Score="2" ViewCount="1413" Body="&lt;p&gt;I was going through this article about how Louis Pasteur studied and explained the &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/19687573&quot; rel=&quot;nofollow&quot;&gt;absence of optical activity in Racemic acid.&lt;/a&gt; Here's an extract: &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;W&lt;/em&gt;hen Pasteur next examined crystals of the same&#xA;  salt of racemic acid, he found that they, too, were chiral&#xA;  but that some of the crystals were left-handed and others&#xA;  were right-handed. The crystals were related to one&#xA;  another as an object and its non-superimposable mirror&#xA;  image, and they were present in equal amounts. &lt;strong&gt;&lt;em&gt;With a&#xA;  magnifying lens and a pair of tweezers, Pasteur carefully&#xA;  separated these crystals into two piles&lt;/em&gt;&lt;/strong&gt;: the left-handed&#xA;  ones and the right-handed ones.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;My guess is that the left/right handedness refers to the &lt;em&gt;molecular&lt;/em&gt; spatial alignment. So,&lt;br&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;How exactly can you assign the property of individual molecules to an entire crystal?&lt;/li&gt;&#xA;&lt;li&gt;How can you determine if a crystal is left/right handed just by looking at it?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15508" LastActivityDate="2015-08-13T21:59:41.707" Title="How do we identify left handed and right handed crystals?" Tags="&lt;crystal-structure&gt;&lt;stereochemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35010" PostTypeId="1" CreationDate="2015-08-13T08:36:14.200" Score="0" ViewCount="3322" Body="&lt;p&gt;When wine is exposed to oxygen, acetic acid forms because of an oxidation reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wine then loses acetic acid. Why? Is it because it is oxidised itself to form carbon dioxide?&lt;/p&gt;&#xA;" OwnerUserId="19102" LastEditorUserId="865" LastEditDate="2015-08-13T13:12:48.190" LastActivityDate="2015-08-13T15:30:34.480" Title="Oxidation of acetic acid in wine?" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35012" PostTypeId="2" ParentId="35005" CreationDate="2015-08-13T09:22:18.100" Score="2" Body="&lt;p&gt;Yes, it is artificial. Sodium carbonate is a substance that occurs in nature, even though any of it is most likely manufactured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium percarbonate, on the other hand, does not occur in nature and it has to be manufactured. It is an oxidiser, and oxidisers don't occur in nature normally. That's a problem for you, because oxidisers (aka bleach) are desired in laundry applications.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastActivityDate="2015-08-13T09:22:18.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35013" PostTypeId="1" AcceptedAnswerId="35015" CreationDate="2015-08-13T10:12:52.940" Score="0" ViewCount="329" Body="&lt;p&gt;I have heard a theory from someone and just wanted to confirm if its true or not. Is it true that more bulkier an organic compound is, more likely it is to react? So does this mean that higher the molecular weight of an organic compound, more will be the probability of it to react?&lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="418" LastEditDate="2016-10-19T14:50:43.987" LastActivityDate="2016-10-19T14:50:43.987" Title="Are bulkier molecules more likely to react?" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;biochemistry&gt;&lt;covalent-compounds&gt;" AnswerCount="1" CommentCount="3" ClosedDate="2015-08-13T17:41:32.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="35014" PostTypeId="1" CreationDate="2015-08-13T10:49:28.053" Score="7" ViewCount="1324" Body="&lt;p&gt;While reading about the size of atoms and ions of the Group 1 elements in the textbook &quot;Concise Inorganic Chemistry&quot; by JD Lee, I came across this line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Li+ is much smaller than the other ions. For this reason, Li only mixes with Na and above 380 Celsius, and it is immiscible with the metals K, Rb and Cs, even when molten; nor will Li form sunstitutional alloys with them. In contrast the other metals Na, K, Rb, Cs are miscible with each other in all proportions. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand why the elements cannot form alloys with Li and what makes the other elements miscible with each other in all proportions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observed that the metallic radius of Li is 1.52 $\unicode{xC5}$ and that the difference between its radius and that of Na is about 0.34 $\unicode{xC5}$. But the difference between the radii of the other successive elements in this group is also about the same, so it doesn't help explain miscibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate some insight and clarification. &lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="865" LastEditDate="2015-08-13T13:09:57.520" LastActivityDate="2015-08-17T15:12:41.980" Title="Miscibility of Molten Metals" Tags="&lt;inorganic-chemistry&gt;&lt;metal&gt;&lt;elements&gt;&lt;mixtures&gt;&lt;alloy&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35015" PostTypeId="2" ParentId="35013" CreationDate="2015-08-13T10:51:52.670" Score="6" Body="&lt;p&gt;The relative rates are going to depend not on molecular weight so much but the structure of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific reaction is also very important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nucleophilic substitution reactions (some of the first reactions most students learn about in an organic chemistry course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us compare 1-chlorobutane and 2-chloro-2-methylpropane, both $\ce{C4H9Cl}$. However, the 1-chlorobutane is a primary alkyl halide (not very bulky) and 2-chloro-2-methylpropane is a tertiary alkyl halide (bulky). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bulkier compound reacts faster in a solvolysis reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)3CCl -&amp;gt;[\ce{H2O}] (CH3)3COH}\ \ \mathrm{fast}$$&#xA;$$\ce{CH3CH2CH2CH2Cl -&amp;gt;[\ce{H2O}] CH3CH2CH2CH2OH}\ \ \mathrm{slow\approx no\ reaction}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that the tertiary halide reacts faster in this reaction is that its mechanism involves the formation of a carbocation intermediate and the extra carbon groups at the tertiary center provide more inductive stabilization to that carbocaion than the single alkyl group attached to a primary position. In general, in reactions that generate carbocation intermediates, tertiary substrates will react faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in many reactions bulkier molecules have steric interactions that negatively effect the rate. In a nucleophilic displacement with $\ce{NaI}$ in acetone, 1-chlorobutane will react faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(CH3)2CCl -&amp;gt;[\ce{NaI}][\mathrm{acetone}](CH3)3CI}\ \ \mathrm{slow\approx no\ reaction}$$&#xA;$$\ce{CH3CH2CH2CH2Cl -&amp;gt;[\ce{NaI}][\mathrm{acetone}]CH3CH2CH2CH2I}\ \ \mathrm{fast}$$&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-13T10:51:52.670" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="35016" PostTypeId="1" AcceptedAnswerId="35017" CreationDate="2015-08-13T11:11:42.687" Score="28" ViewCount="59851" Body="&lt;p&gt;What are the general ways to treat an accident involving highly concentrated or similarly dangerous acids?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a lab instructor who recently debated with our group about treating an acid spill on the skin. According to him, one should immediately rinse off the acid with running water for about 15-20 minutes. He says that one should never apply bases to the site because an Acid-Base is highly exothermic and would only increase the damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I clearly remember learning once that one should apply baking soda (sodium bicarbonate) to a bee sting or ant bite in order to neutralize the acid and relieve the burning. &#xA;I also know that the dilution of an acid in water is done by adding the acid slowly to the water containing vessel because of the heat released in the process. And especially when preparing a dilute solution of sulfuric acid, one wears safety goggles and gently stirs the solution to prevent splashing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which is it? And does it depend on the severity of the burn or type acid?&lt;/p&gt;&#xA;" OwnerUserId="15031" LastEditorUserId="1677" LastEditDate="2017-11-06T23:58:08.043" LastActivityDate="2019-05-09T19:30:50.167" Title="Acid Accident Treatment" Tags="&lt;acid-base&gt;&lt;safety&gt;" AnswerCount="5" CommentCount="4" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35017" PostTypeId="2" ParentId="35016" CreationDate="2015-08-13T11:30:51.523" Score="42" Body="&lt;p&gt;My Lab First Aid book$^{*1}$ tells the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The local therapy mainly consists of the instant, intense &lt;strong&gt;rinsing&lt;/strong&gt; (at least 10 minutes) with tap water (ideally 15 to 20 °C). Don't try to neutralize the chemical burn, because the heat generation could lead to further tissue damages.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means, that your lab instructor is absolutely right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rinsing water should wash the cold water away faster than the heat from the acid base reaction could harm you. It's not for nothing that there are emergency showers and eye douches in every lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a friend who is a chemical lab assistant and once had to carry concentrated sulfuric acid in a big glass beaker. Suddenly the bottom of the beaker broke off and all the acid poured over her etc. She not even thought of anything else than getting out of her clothes and under the next emergency shower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;$^{*1}$ The book is: R. Rossi, &lt;em&gt;Erste hilfe bei akuten Notfällen - Begleitheft zu Jander - Blasius&lt;/em&gt;, S. Hirzel Verlag, &lt;strong&gt;2005&lt;/strong&gt;, p. 11f.. &lt;strong&gt;The text is in german and I translated it.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-08-13T11:42:38.930" LastActivityDate="2015-08-13T11:42:38.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35018" PostTypeId="2" ParentId="35016" CreationDate="2015-08-13T11:41:08.623" Score="12" Body="&lt;p&gt;To add to @PH13's answer, according to the Texas A&amp;amp;M University's page &lt;a href=&quot;http://www.bio.tamu.edu/USERS/TONNA/SAFETY.HTM&quot;&gt;Safety Guidelines&lt;/a&gt;, that rinsing is crucial, from the site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemicals may cause severe burns. Immediately flush skin with cold water. Remove contaminated clothing (including socks and shoes) and rinse at least 10x as long as the chemical was in contact with the skin. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, as all chemicals, including acids, have differences in how they react, &lt;strong&gt;it is crucial to be aware of the first-aid guidelines laid out in the Materials Safety Data Sheet (MSDS)&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Read the MSDS sheets and product labels to learn the first aid procedures, health hazards and physical hazards for the chemicals used in each exercise.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This should be done as part of the lab preparation and hazard assessment, also to have on hand in case of emergencies such as a spill on skin or in the eyes.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-13T12:33:33.580" LastActivityDate="2015-08-13T12:33:33.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35019" PostTypeId="2" ParentId="35016" CreationDate="2015-08-13T12:00:36.593" Score="17" Body="&lt;p&gt;Speaking from experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minor spills (drops) on skin of concentrated nitric or sulfuric acid are nothing to speak of. Wash them off with plenty of water for a minute or two and you are fine. I'm pretty sure that this is the case for phosphoric acid as well. Hydrochloric acid is a bit more troublesome, mosly because it fumes intensively, but still nothing spectacular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common aftereffect of small nitric acid spills is local skin yellow coloring and a quickened loss of upper layer with fresh skin below. Again, nothing to panic about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These acids cause proteins to denaturate, forming strong barriers for further diffusion into skin. As such, short contact with skin is safe as long as the acid is immediately washed off and the skin is undamaged. Diluted (0.1-0.5 M) hydrochloric acid is so tame, it is pretty safe to wash briefly glassware in it without gloves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, however, acids, that are trouble. One is glacial acetic acid. It diffuses rather readily through lipid membranes and may cause deep, troublesome chemical burns requiring special treatment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that while our skin is surprisingly resilient, our mucous membranes, and eyes in particular are not. Even one unlucky drop may require special help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, summing up.&#xA;Wash off sulfuric acid, nitric acid and hydrochloric acid, but as long as the contact is brief and small, it is nothing to care about. Longer and/or wider contacts should get medical attention. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wash longer acetic acid, formic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally avoid (and I mean it) hydrofluoric acid and derivitives (toxin, rapidly diffusing through sking) and $\ce{HClO4}$ (may form explosives on storage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BY ALL COSTS PROTECT YOUR EYES. THEY CANNOT BE REPAIRED. &lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="485" LastEditDate="2015-08-15T08:35:56.640" LastActivityDate="2015-08-15T08:35:56.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35020" PostTypeId="2" ParentId="35016" CreationDate="2015-08-13T12:12:03.903" Score="11" Body="&lt;p&gt;Skin is fairly resistant to acids. Rinse it off with water for a minute and you're good. Everything that has already penetrated the skin (nitrous gases, HCl, HF(!)) will not be removed by washing afterwards. Further cooling might keep the swelling and inflammation down, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;strong&gt;ONE&lt;/strong&gt; exception: Wipe off conc. &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; with a dry piece of cloth (or with &lt;em&gt;anything&lt;/em&gt; dry) &lt;em&gt;before&lt;/em&gt; applying water. Otherwise you'll get a severe burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bases take longer to wash off, they enter the skin more easily (by dissolving it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never try to neutralize anything, it takes much longer than jumping under the emergency shower, is totally useless and likely dangerous, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how does an &quot;inexperienced&quot; lab instructor &lt;em&gt;discuss&lt;/em&gt; this with you? This is a subject for a very serious, compulsory lecture by the responsible head of laboratory, before anyone as much as opens a lab door!&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditorDisplayName="user15489" LastEditDate="2019-05-09T19:30:50.167" LastActivityDate="2019-05-09T19:30:50.167" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35021" PostTypeId="2" ParentId="35003" CreationDate="2015-08-13T12:39:47.433" Score="11" Body="&lt;p&gt;&lt;a href=&quot;http://www.bloomberg.com/news/articles/2015-08-11/who-s-behind-the-96-million-shade-balls-they-just-rolled-into-l-a-s-reservoirs-&quot;&gt;This piece in Bloomberg News&lt;/a&gt; is about Sydney Chase, the founder of &lt;a href=&quot;http://www.xavierc.net/index.html&quot;&gt;XavierC&lt;/a&gt;, the company which makes the &quot;shade balls&quot; in question (there are two other suppliers, &lt;a href=&quot;http://www.orangeproducts.com/index.html&quot;&gt;Orange Products, Inc.&lt;/a&gt; and Artisan Screen Printing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the section of the XavierC site &lt;a href=&quot;http://www.xavierc.net/howwork.html&quot;&gt;on how they work&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Two main causes of water evaporation in reservoirs are temperature and surface area. Our hollow plastic conservation balls address these two issues effectively and inexpensively.&#xA;  The black conservation balls provide a floating cover on the reservoir taking up approximately 90% of the surface area. Simultaneously the balls prevent UV rays reaching the water and therefore reducing algae.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the above-mentioned Bloomberg piece:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Los Angeles Department of Water and Power has now dumped 96 million balls into local reservoirs to reduce evaporation and block sunlight from encouraging algae growth and toxic chemical reactions. The balls are coated with a chemical that blocks ultraviolet light and helps the spheres last as long as 25 years. Las Virgenes, north of L.A., now uses shade balls, too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://www.latimes.com/local/lanow/la-me-ln-shade-ball-water-20150812-htmlstory.html&quot;&gt;Los Angeles Times story on August 12, 2015&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are they made of?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The 4-inch-diameter balls are made from high-density polyethylene, which is the same material you would find in a one-gallon milk jug. This plastic is approved to come into contact with drinking water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are they black?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbon black was added to the plastic to stabilize the balls in UV light from the sun.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and, from &lt;a href=&quot;http://articles.latimes.com/2008/jun/10/local/me-balls10&quot;&gt;another LA Times piece about the &quot;shade balls&quot; written in 2008&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The balls, which cost 40 cents each, are made of polyethylene. The coating contains carbon. Black is the only color strong enough to deflect ultraviolet rays, said Paul Sachdev, president of Orange Products.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-13T12:39:47.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35022" PostTypeId="1" CreationDate="2015-08-13T12:45:08.603" Score="7" ViewCount="803" Body="&lt;p&gt;I noticed (lim) is sometimes added to the SN1 or SN2 notation. I suppose it means limiting, but still do not understand it completely. Can someone please explain how is SN1 and SN2 different to SN1 (lim) and SN2 (lim)?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2tRGO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tRGO.jpg&quot; alt=&quot;first example&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oiIjv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oiIjv.jpg&quot; alt=&quot;second example&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Text from the first row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existence of an intermediate with a lower coordination number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No intermediate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existence of an intermediate with a higher coordination number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Text from the first collumn:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stoichiometric mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;intimate-mechanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ingold-Definition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dependance of the speed on the nature of the entering group&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Text from the last row:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed does not depend on the entering group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed does depend on the entering group.&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="5026" LastEditDate="2018-02-26T05:17:34.340" LastActivityDate="2018-02-26T05:17:34.340" Title="SN1 (lim) and SN2 (lim)" Tags="&lt;organic-chemistry&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35024" PostTypeId="2" ParentId="4205" CreationDate="2015-08-13T13:40:15.287" Score="3" Body="&lt;p&gt;Glass filled nylon molds well, wears well and is most resistant to heat with regards to the plastics you mentioned&lt;/p&gt;&#xA;" OwnerUserId="19108" LastActivityDate="2015-08-13T13:40:15.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35025" PostTypeId="1" CreationDate="2015-08-13T15:17:52.987" Score="1" ViewCount="84" Body="&lt;p&gt;Recently I found the following &lt;a href=&quot;https://chemistry.stackexchange.com/questions/34938/how-to-produce-dry-pure-oxygen&quot;&gt;question by sam&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm supposed to produce dry and pure oxygen using the decomposition of hydrogen peroxide with a manganese dioxide catalyst.&#xA;  $$\ce{2H2O2(aq) -&amp;gt; 2H2O(l) + O2(g)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The gas will be collected in a bottle by downward displacement of water. How can I safely move the oxygen gas through the drying agent without mixing in atmospheric gas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on that I am asking me, if and how palladium could catalyze the reaction?&lt;/p&gt;&#xA;" OwnerUserId="19112" LastEditorUserId="16622" LastEditDate="2015-08-13T17:52:27.593" LastActivityDate="2015-08-14T07:01:38.233" Title="Palladium catalyst" Tags="&lt;catalysis&gt;&lt;organometallic-compounds&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35026" PostTypeId="2" ParentId="35010" CreationDate="2015-08-13T15:30:34.480" Score="1" Body="&lt;p&gt;The oxidation/reduction reactions of &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ethanol &amp;lt;-&amp;gt; ethanal &amp;lt;-&amp;gt; acetic acid (ethanoic acid)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;are at equilibrium. Oxidation (presence of $\ce{O2}$ or other oxidizing agent) and reduction (addition of $\ce{H}$ to the $\ce{C-O}$ carbon) are in constant flux. The system can be pushed to the right if wine is exposed to air for long periods of time, as you found, but at a certain point it will reach equilibrium, in which case the excess of acetic acid will reduce (marginally) back into ethanal and/or ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better explanation may be that since the B.P. of ethanol (78 dec. C) is significantly lower than acetic acid (118 deg. C), the vapor pressure of ethanol is greater and it has a higher affinity to convert to a gaseous state. This also would shift equilibrium to the left to produce a lower post-equilibrium concentration of acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetic acid does have the possibility of decarboxylating, but this is usually in excess heat. In this case, a $\ce{CO2}$ molecule breaks off of the acetic acid leaving $\ce{CH4}$, which would also be generated as a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it is possible that you're mistaking &quot;loss of acetic acid&quot; for an actual increase in pH, irrelevant to the presence of whole acetic acid molecules. It could be that other materials in the wine &quot;soak&quot; protons to increase pH and leave ethanoate counter ions ($\ce{CH3COO-}$). You should clarify how you're actually quantifying the concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't expect the &quot;decrease&quot; of acetic acid to be significant, compared to the amount that was produced in the first place. I would also expect your system to re-equilibrate after a while longer. It would be interesting to see what happens next.&lt;/p&gt;&#xA;" OwnerUserId="17821" LastActivityDate="2015-08-13T15:30:34.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35032" PostTypeId="2" ParentId="35007" CreationDate="2015-08-13T18:08:39.433" Score="6" Body="&lt;p&gt;Option 2 generally gives higher yields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because option 1 (&lt;a href=&quot;https://en.wikipedia.org/wiki/Fischer%E2%80%93Speier_esterification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fischer esterification&lt;/a&gt;) usually results in an equilibrium mixture which contains significant amounts of products and reactants, limiting the yield. This is mitigated by using a stoichiometric amount of a concentrated acid catalyst, such as sulfuric acid, which is also a strong dehydrating agent, and therefore removes the water from the equilibrium as it forms, favouring the products. Additionally, the reaction mixture can be heated to evaporate the water, or you can use a large excess of one of the reactants, typically the alcohol because it also acts as the solvent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, option 2, more easily gives high yields because the reaction is more or less complete, with very little reactants left at the end. This is because the reaction produces hydrogen chloride gas instead of water, which is lost from the reaction mixture, driving the reaction forward. The downside of this option is that acid chlorides are generally more dangerous than carboxylic acids and are harder to prepare, or more expensive to buy. Aditionally, hydrogen chloride gas is produced which must be disposed of safely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, often used method, is esterification using &lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_acid_anhydride&quot; rel=&quot;nofollow noreferrer&quot;&gt;acid anhydrides&lt;/a&gt;. These are intermediate in reactivity between carboxylic acids and acid chlorides and are easier to use than acid chlorides, but still provide excellent yields. A classic example of this is the acetylation of salycylic acid with acetic anhydride to produce Aspirin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vPk2L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vPk2L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-08-13T18:08:39.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35033" PostTypeId="5" CreationDate="2015-08-13T18:48:19.647" Score="0" Body="&lt;p&gt;Esters are chemical compounds derived from an acid (organic or inorganic) in which at least one $\ce{-OH}$ group is replaced by an $\ce{-O-alkyl}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Carboxylate esters:&lt;/strong&gt;&lt;br&gt;&#xA;The most common type of ester is a carboxylate ester. Carboxylate esters are any compounds containing the $\ce{RCOOR}$ group such as ethyl ethanoate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CrW63.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrW63.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carboxylate esters are relatively polar molecules and are able to act as hydrogen bond acceptors through their oxygen atoms, which makes them useful as solvents. Many carboxylate esters smell fruity and they are widely used for this purpose in industry. They are also useful as protecting groups for carboxylic acids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carboxylate esters are usually prepared by refluxing an alcohol with a carboxylic acid in the presence of a strong, dehydrating acid catalyst such as sulfuric acid (&lt;a href=&quot;https://en.wikipedia.org/wiki/Fischer%E2%80%93Speier_esterification&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fischer esterification&lt;/a&gt;). They may also be prepared by using an acid anhydride or an acid chloride instead of a carboxylic acid, which is an irreversible reaction and so provides better yields at the cost of using more expensive and more dangerous reactants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carboxylate esters undergo addition of nucleophiles at the carbonyl carbon. When the nucleophile is hydroxide or water, this leads to ester hydrolysis, reforming the parent carboxylic acid and alcohol (or carboxylate salt if done under basic conditions). When the nucleophile is an alkoxide or alcohol, another ester is formed (&lt;a href=&quot;https://en.wikipedia.org/wiki/Transesterification&quot; rel=&quot;nofollow noreferrer&quot;&gt;transesterification&lt;/a&gt;) with a different alkyl group attached to the ester oxygen. The $\alpha$ hydrogen of carboxylate esters is mildly acidic and, in the presence of strong bases, carboxylate esters will undergo reactions at the $\alpha$ carbon yielding useful products such as $\beta$-ketoesters (&lt;a href=&quot;https://en.wikipedia.org/wiki/Claisen_condensation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Claisen condensation&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compounds with multiple carboxylic acid and alcohol groups can undergo condensation polymerisation to form polyesters, which are widely used in clothing and various plastic goods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orthoesters:&lt;/strong&gt;&lt;br&gt;&#xA;The other main type of organic ester is the orthoester, which consists of three $\ce{-O-alkyl}$ groups bonded to the same carbon. An example is ethylorthoacetate (1,1,1-triethoxyethane in IUPAC nomenclature), so named because it is derived from ethylacetate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7iOot.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7iOot.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orthoesters can be prepared by reaction of nitriles with excess alcohol under acidic conditions (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pinner_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pinner reaction&lt;/a&gt;) but can also easily be hydrolysed under mild acidic conditions to form esters, and this is the basis of their use as protecting groups for esters in organic synthesis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inorganic esters:&lt;/strong&gt;&lt;br&gt;&#xA;Less commonly, esters can be formed from inorganic acids such as phosphoric, sulfuric, nitric and chromic acids. They can all be formed by reacting the respective acids with an alcohol, in much the same way as carboxylate esters are prepared. Nitrate esters are used in explosives such as nitroglycerin, which has three nitrate ester groups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0vWjV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0vWjV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulfate esters, such as sodium lauryl sulfate, are used as anionic surfactants due to the highly polar sulfate group which contrasts with the apolar hydrocarbon chain attached to it, making one end of the ion hydrophilic and the other hydrophobic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z7Ibf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z7Ibf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="402" LastEditDate="2017-05-04T00:31:33.277" LastActivityDate="2017-05-04T00:31:33.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35034" PostTypeId="4" CreationDate="2015-08-13T18:48:19.647" Score="0" Body="For questions about the synthesis, properties, and reactions of ester compounds, both organic and inorganic." OwnerUserId="10185" LastEditorUserId="402" LastEditDate="2017-05-04T02:32:35.963" LastActivityDate="2017-05-04T02:32:35.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35035" PostTypeId="1" CreationDate="2015-08-13T21:08:33.953" Score="6" ViewCount="145" Body="&lt;p&gt;According to &lt;a href=&quot;https://pubs.acs.org/cen/news/85/i52/8552notw4.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, the city of Los Angeles (in the USA) is facing a problem where bromate, a suspected carcinogen, has been forming in their open water supplies.  They believe that the formation of bromate has to do with exposure to sunlight, and are now &lt;a href=&quot;http://www.cnbc.com/2015/08/13/shade-balls-protect-la-water-supply-during-drought.html&quot; rel=&quot;nofollow&quot;&gt;distributing millions of &quot;shade balls&quot; into water supplies&lt;/a&gt; in order to block the sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that an overwhelming majority of water on this planet is exposed to sunlight, why are there not high levels of bromate present in all water, especially shallow bodies of water where the surface area is maximized?  Or is it?&lt;/p&gt;&#xA;" OwnerUserId="15770" LastEditorUserId="15770" LastEditDate="2016-05-02T12:58:26.287" LastActivityDate="2016-05-02T12:58:26.287" Title="Bromate forming in open water" Tags="&lt;reaction-mechanism&gt;&lt;water&gt;&lt;safety&gt;&lt;environmental-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35036" PostTypeId="2" ParentId="35008" CreationDate="2015-08-13T21:59:41.707" Score="3" Body="&lt;p&gt;Yes, you are right that the handedness refers to the molecular 3D arrangement. Such handed molecules are &lt;a href=&quot;http://chemistry.umeche.maine.edu/CHY431/Isomers.html&quot; rel=&quot;nofollow&quot;&gt;stereoisomers&lt;/a&gt; known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Enantiomer&quot; rel=&quot;nofollow&quot;&gt;enantiomers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the first part of your question will make more sense following an answer to the second part. So with regard to the part about discerning handedness by looking at it: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I couldn't get to the article right now due to network issues, so I hope I'm not repeating too much.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A molecule is &lt;a href=&quot;http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.ii.html&quot; rel=&quot;nofollow&quot;&gt;chiral&lt;/a&gt; when one of the atoms in it has four unique groups attached to it. &#xA;Chiral molecules observably rotate &lt;a href=&quot;https://en.wikibooks.org/wiki/Organic_Chemistry/Chirality/Optical_activity#What_Is_Plane_Polarized_Light.3F&quot; rel=&quot;nofollow&quot;&gt;plane-polarized light&lt;/a&gt;, either clockwise or anti-clockwise (see &lt;a href=&quot;http://www.chem.ucla.edu/harding/IGOC/P/plane_polarized_light.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a picture of what plane-polarized light is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Optical_Activity#A_Brief_History&quot; rel=&quot;nofollow&quot;&gt;Pasteur observed&lt;/a&gt; this along with the cancelling effect of a racemic mixture. As in any scientific endeavor, Pasteur was building upon the efforts of previous work--in this case, that of Jean-Baptiste Biot in studying &lt;a href=&quot;https://en.wikipedia.org/wiki/Tartaric_acid&quot; rel=&quot;nofollow&quot;&gt;tartaric acid&lt;/a&gt;. Pasteur's &lt;a href=&quot;http://www.cs.mcgill.ca/~rwest/link-suggestion/wpcd_2008-09_augmented/wp/l/Louis_Pasteur.htm&quot; rel=&quot;nofollow&quot;&gt;careful work&lt;/a&gt; enabled him to deduce that the naturally occurring or biologically produced form of tartaric acid was enantiomerically pure because it rotated light and the mirror image crystals also rotated light opposite directions. This was in contrast to the racemic acid which did not rotate light since it was a 50:50 mixture that cancelled the rotation in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bryn Mawr College has a helpful &lt;a href=&quot;http://www.brynmawr.edu/chemistry/Chem/mnerzsto/PolarimetryExercise.htm&quot; rel=&quot;nofollow&quot;&gt;explanation of how enantiomers interact with light&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have probably [not] learned in lecture that chiral molecules have optical rotations and that enantiomers rotate plane polarized light in equal and opposite directions... This behavior upon exposure to plane polarized light is not some magical property that defies the general principles of enantiomers. The reason enantiomers interact differently with plane polarized light is that plane polarized light is chiral. Though polarized light is a planar phenomenon, it is actually comprised of two superimposed helical forms. You can think of the electrical component as oscillating in such a way that there are two electrical vectors rotating in a cork screw fashion in opposite directions away from the source of the propagation at the same rate. Since a helix is a chiral shape , the two versions of the light constitute an enantiomeric pair. If you resolve the vectors of the two superimposed helices, you end up with oscillation in a single plane...Enantiomers interact differently with the two forms of light. The basic idea is that when plane polarized light encounters enantiomerically pure, chiral molecules, one form of light will slow down more than the other as a consequence of the interaction. This results in one helix coming out of sync with the other. When the vectors are resolved, the resultant plane is rotated with respect to the original position. This change in position is what we mean when we say a compound has an optical rotation. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So looking at a crystal being exposed to polarized light should allow you to observe whether it rotates light to the right or left. I'm not sure how easily detectable to the eye the rotation is, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Polarimeter&quot; rel=&quot;nofollow&quot;&gt;polarimeters&lt;/a&gt; are instruments which are used to measure the angle of rotation.&lt;/p&gt;&#xA;" OwnerUserId="4813" LastActivityDate="2015-08-13T21:59:41.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35037" PostTypeId="2" ParentId="34975" CreationDate="2015-08-13T21:59:56.977" Score="3" Body="&lt;p&gt;Significant figures is a rule of thumb, to avoid giving overly precise results when the input doesn't warrant it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In actual scientific calculations - where keeping track of the precision matters - significant figures aren't used. Instead, explicit error ranges (confidence intervals) are used. &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncertainty#Measurements&quot; rel=&quot;nofollow noreferrer&quot;&gt;Notation&lt;/a&gt; of these varies, but often takes the form of a plus-minus format (14.5±0.3 m). The concept of significant figures uses implicit error ranges, conventionally set to one half of the last significant digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When error ranges are explicitly given, the error of a calculation result is found with formal &lt;a href=&quot;https://en.wikipedia.org/wiki/Propagation_of_uncertainty&quot; rel=&quot;nofollow noreferrer&quot;&gt;propagation of uncertainty&lt;/a&gt;, rather than the rule-of-thumb significant figure rules. Sometimes this is based on a model of the distribution of error around the central value (for example, assuming that the error is normally distributed around the central value with a given standard deviation), but most commonly the error is taken as the minimum and maximum of an acceptable range of values around the main measurement. Under this scheme, the propagation of uncertainty often takes the form of the &quot;crank three times&quot; method, whereby you run the calculation once for the central value, once for the minimum value, and once for the maximum value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were doing a calculation like $\pu{14.5\pm0.3 m} - \pu{1.5\pm0.2 m}$, you would calculate once for the central value ($\pu{14.5 m - 1.5 m = 13 m}$), once to give you the smallest value you could get with the range ($\pu{14.2 m - 1.7 m = 12.5 m}$), and once to give the largest value your could get with the range ($\pu{14.8 m - 1.3 m = 14.5 m}$). You'd then convert this into a range, normally rounding both values such that the error value only has 1-2 significant figures: $\pu{13\pm0.5 m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the rules for significant figure calculations are just a quick-and-dirty rule of thumb of doing more-or-less the same thing. I'd encourage you to try it out with a number of examples. If you assume that the error in each quantity is one half of the last significant digit, and you do the crank three times error propagation, you should come up with a result that has an error that is about one half the last significant digit of the answer by the significant figures approach - give or take a digit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;em&gt;really&lt;/em&gt; what you want to do is to convert the measurement of $\pu{5 ft 10 in}$. into one which has an explicit error associated with it. For this measurement, I'm guessing you'd have an error of $\pu{\pm0.5 in}$. - I say this because rounding to the nearest 10 inches seems rather silly to me. (Feel free to work the calculations below with different assumption, say with ±5 in., if you disagree.) So you have a measurement of $\pu{5 ft 10 in. \pm 0.5 in}$ in other words $\pu{5 ft} (\text{exact}) + 10\pm\pu{0.5 in}. = \pu{5 ft} (\text{exact}) \times \pu{12 in./ft }(\text{exact}) + \pu{10\pm0.5 in.} = \pu{70\pm0.5 in.}$ Note that is the same as $70$. in with two significant digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have explicit error bars, we can use the crank three times method to convert it to feet: $70~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.8\bar{3}~\mathrm{ft}$ and $69.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.791\bar{6}~\mathrm{ft}$ and $70.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.875~\mathrm{ft}$. This is $5.8\bar{3}±0.041\bar{6}~\mathrm{ft}$. Convention holds that we round the error to one or two significant figures, and also round the main value to the same decimal place: $5.83±0.04~\mathrm{ft}$. -- This is basically what you'd have from $\pu{5.8 ft}$ with two significant digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upshot is that you'd never use $\pu{5 ft 10 in.}$ in a calculation. You'll want to convert it to a single consistent unit first. When you do that, keep in mind the actual error associated with the measurement, and use that error as a gauge at how precise the converted quantity is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that if this is a question on a test or on a homework, ask your teacher how they think you should get the &quot;correct&quot; answer. When dealing with rule-of-thumb things like significant figures, intelligent people can vary in their interpretations. This means an answer valid in one context can be marked as wrong on the answer key because it doesn't match the formalism the teacher is using.) &lt;/p&gt;&#xA;" OwnerUserId="11616" LastEditorUserId="5026" LastEditDate="2018-02-22T13:22:11.757" LastActivityDate="2018-02-22T13:22:11.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35039" PostTypeId="2" ParentId="34913" CreationDate="2015-08-13T22:50:26.160" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the type of the ring opening; is it a concerted or radical&#xA;  based C-C bond cleavage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;strong&gt;chelotropic&lt;/strong&gt; reaction, a subset of cycloaddition\fragmentation reactions.  As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide.  For example, if the reaction had been run with the &lt;em&gt;cis&lt;/em&gt;-2,3-dimethylcyclopropane analogue and both &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process.  When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we &lt;strong&gt;expect&lt;/strong&gt; the reaction to be concerted.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I strongly assume the empty p-orbital to be in hyperconjugation with&#xA;  the walsh orbitals&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the &quot;bisected&quot; conformation) should be favored.  Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was asked to draw the orbital correlation diagram.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have difficulty reading your diagram, so I created a new orbital correlation diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wtrMr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wtrMr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me say a few things about my figure in order to make it easier to understand.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.  &lt;/li&gt;&#xA;&lt;li&gt;The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.&lt;/li&gt;&#xA;&lt;li&gt;The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.&lt;/li&gt;&#xA;&lt;li&gt;There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.&lt;/li&gt;&#xA;&lt;li&gt;This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.&lt;/li&gt;&#xA;&lt;li&gt;I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.&lt;/li&gt;&#xA;&lt;li&gt;I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.&lt;/li&gt;&#xA;&lt;li&gt;The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.&lt;/li&gt;&#xA;&lt;li&gt;The &quot;A&quot; and &quot;S&quot; labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now&lt;/strong&gt;, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry.  In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals.  This analysis suggests that &lt;strong&gt;the thermal reaction can proceed in a concerted fashion.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Miscellaneous&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur.  The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene.  &lt;strong&gt;Other concerted reactions may be preferred because they happen to be of lower energy&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction.  But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not?  Well, these are all equilibria and the forward reaction is extremely exothermic.  We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds.  Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic.  The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria.  It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed.  If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response to OP's comments:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can we see in general from an orbital symmetry correlation diagram&#xA;  if a reaction is concerted or not?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion.  If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively.  All ground state reactant orbitals correlate with ground state product orbitals.  Therefore the reaction is allowed to proceed in a concerted manner.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it okay to say biradicaloid instead of carbene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, not really. &quot;Biradical&quot; is generally used in cases where the electrons are on different atoms.  Like when we break a bond in cyclopropane we have a 1,3-biradical.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the orbital correlation diagram I would predict the acetylene a&#xA;  little bit lower because of more s character (sp) than ethylene (sp2).&#xA;  May be the energy difference is roughly as high as the difference&#xA;  between carbene p orbital A and carbene sp2 orbital.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals.  My guess was that they would have roughly the same energy.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the retro cheletropic reaction I assume a photolytical pathway&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the paper, I've found here...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state.  I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures.  The photochemical route just allows them to generate the same ground state carbene at very low temperatures.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2016-08-15T23:10:29.327" LastActivityDate="2016-08-15T23:10:29.327" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="35040" PostTypeId="1" AcceptedAnswerId="35042" CreationDate="2015-08-13T23:00:01.103" Score="11" ViewCount="391" Body="&lt;p&gt;For my birthday I got an awesome mineral specimen case, as pictured below. What piqued my interest was the notations in the lid where the minerals are described with 5 specifications: C (obviously for Colour) then S, H/SG (I'm guessing specific gravity), L and CL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone shed any light on what these abbreviations mean? Additionally, do the coloured squares mean anything or are they just for identification (they match colours painted on some of the samples).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ARPH8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ARPH8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OJGLQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OJGLQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12923" LastEditorUserId="11367" LastEditDate="2017-01-14T14:45:04.937" LastActivityDate="2017-01-14T14:45:04.937" Title="What do these mineral classification symbols mean?" Tags="&lt;inorganic-chemistry&gt;&lt;materials&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35041" PostTypeId="1" CreationDate="2015-08-13T23:42:49.507" Score="0" ViewCount="447" Body="&lt;p&gt;All the chemical compounds can be categorized as acids, bases, and neutral compounds.&lt;/p&gt;&#xA;&lt;p&gt;There are different theories, namely Arrhenius Theory, Brønsted-Lowry Theory, Lewis Theory, which define the acids and bases. Each successive theory becoming more general; Lewis Theory is the most generalized one of the three.&lt;/p&gt;&#xA;&lt;p&gt;Neutral compounds could be salts, hydrocarbons like ethanol, oxides, and many more. A salt is always a result of a reaction between an acid and base.&lt;/p&gt;&#xA;&lt;p&gt;The difference between a base and an alkali is that an alkali is a base which are soluble in water. In other words, all alkalies are bases but not all bases are alkalies.&lt;/p&gt;&#xA;&lt;p&gt;I'm sure like bases there would be many acidic compounds which are not soluble in water. Please correct me if I'm wrong.&lt;/p&gt;&#xA;&lt;h2&gt;&lt;strong&gt;Q1:&lt;/strong&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Is litmus test really a defining criterion for classification of a compound as an acid or base? I believe that to perform litmus test, the compound should be soluble in water, and we have already agreed that not all bases and acids form aqueous solution. It means that litmus test is not a main criterion for the classification. It is something of secondary importance. Do I make sense?&lt;/p&gt;&#xA;&lt;h2&gt;&lt;strong&gt;Q2:&lt;/strong&gt;&lt;/h2&gt;&#xA;&lt;p&gt;Likewise, general properties given in school chemistry books are also not primary criteria. Stated differently, an acid doesn't always need to taste sour, turn litmus red, and a base doesn't have to feel slippery or turn litmus blue. Is what I'm saying correct?&lt;/p&gt;&#xA;&lt;p&gt;Boron trifluoride is classified an acid. Boron has three electrons in its outer shell and it forms three covalent bonds with three fluorine atoms which give it total sum of six electrons. It's still deficient of two electrons and therefore it can accept an electron pair so it's Lewis acid.&lt;/p&gt;&#xA;" OwnerUserId="19120" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-11-13T19:33:00.200" Title="acids, bases, salts, etc" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35042" PostTypeId="2" ParentId="35040" CreationDate="2015-08-13T23:44:06.357" Score="12" Body="&lt;p&gt;In addition to being a non-practicing chemist, I'm also a non-practicing geologist, so look no further for an answer you can trust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your minerals are classified as follows: C is for color (sample, gross), S is for streak (color of powdered mineral, essentially, on an unfinished streak plate), H is hardness on the Mohs scale, SG is specific gravity, L is luster, and CL is cleavage (how the sample breaks along planes dictated by the crystal habit of the mineral). See &lt;a href=&quot;http://geology.com/minerals/mineral-identification.shtml&quot;&gt;this reference for a printable chart&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color-coding is certainly used to mark the samples in case they get jostled or mixed up and need to be matched to their appropriate slots: those codes are not related to the mineral classification.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-13T23:59:52.090" LastActivityDate="2015-08-13T23:59:52.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35043" PostTypeId="1" CreationDate="2015-08-14T00:01:36.123" Score="1" ViewCount="127" Body="&lt;p&gt;How can I identify 4-bromoazobenzene, 4-bromoazoxybenzene, and 4,4’-dibromo-$\ce{N,N}$’-diphenylhydrazine products using $^1\ce{H}$ and $^{13}\ce{C}$ NMR?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m really confused by the general shape of NMR peaks of aromatic protons. I want to distinguish each product from the other products, knowing that the three products are synthesized starting by the reaction of 4-bromonitrobenzene with sodium borohydride.&lt;/p&gt;&#xA;" OwnerUserId="19121" LastEditorUserId="865" LastEditDate="2015-08-14T00:10:34.187" LastActivityDate="2015-08-14T17:59:13.820" Title="Determination of brominated aromatics by 13C and 1H NMR spectroscopy" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35044" PostTypeId="2" ParentId="35043" CreationDate="2015-08-14T01:32:08.253" Score="4" Body="&lt;p&gt;I don't find any experimental spectra online, but I have been suggesting &lt;a href=&quot;http://www.nmrdb.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;NMRDB.org&lt;/a&gt; to students and colleagues. I don't know how accurate the predictions are, but the site is easy to use and for the compounds you're suggesting, rule-based methods should be fairly reliable.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nmrdb.org/new_predictor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Predict $\ce{^1H}$ NMR&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nmrdb.org/13c/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Predict $\ce{^13C}$ NMR&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the 4,4’-dibromo-N,N’-diphenylhydrazine, you should have two sets of &quot;double doublets&quot; (i.e., four peaks in each set of equal height).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CrETA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrETA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 4-bromoazobenzene, you break symmetry, so there should be four signals which integrate to two protons, and one additional signal integrating to one proton (i.e., all ~7.3-7.8 ppm, so a big mess). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2DStl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2DStl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the $\ce{^1H}$ spectra of 4-bromoazoxybenzene, the protons closest to the azoxy -N+-O- should be deshielded a bit (maybe ~7.8-7.9 ppm?) compared to the aza analogue. Certainly it's also going to be a fairly congested spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll leave the $\ce{^13C}$ NMR as an exercise (or to others to answer). &lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="5017" LastEditDate="2015-08-14T17:59:13.820" LastActivityDate="2015-08-14T17:59:13.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35045" PostTypeId="2" ParentId="33952" CreationDate="2015-08-14T02:06:25.727" Score="0" Body="&lt;p&gt;My guess is that they absorb UV because they have conjugated pi bonds in their structure (i.e. $\ce{= - = - =}$ structures).  A system of three of these in a row might form a box of the right length to absorb at around 250 nm (since that is what hexatriene absorbs at), which is ultraviolet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Conjugated_system&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19122" LastEditorUserId="5456" LastEditDate="2015-08-14T03:45:54.073" LastActivityDate="2015-08-14T03:45:54.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35046" PostTypeId="1" CreationDate="2015-08-14T02:10:56.790" Score="3" ViewCount="5890" Body="&lt;p&gt;I was wondering if someone could walk me through how to do the following two questions?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;The complete combustion of pentane has a enthalpy change of $-3509~\mathrm{mol^{-1}}$. Write an equation for the complete combustion of pentane. Write a possible equation for the imcomplete combustion of pentane.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For this question, would the following equation suffice:&#xA;$$\ce{ C5H12 (g) + 8O2(g) -&amp;gt; 5CO2 (g) + 6H2O (g)  }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an incomplete reaction, the products will be water, carbon/and or carbon monoxide due to insufficient oxygen, so would this equation be correct:&#xA;$$\ce{C5H12 (g) + 11/2 O2 (g) -&amp;gt; 5CO (g) + 6H2O (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;$17~\mathrm{L}$ of Propane are incompletely combusted in $34~\mathrm{L}$ of oxygen at $25~^\circ\mathrm{C}$ and $100~\mathrm{kPa}$ into only water and carbon. Write a balanced chemical question to represent this including state symbols. What is the mass of the water produced? What observable change would indicate that a chemical change occurred?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would an equation like this work: $$\ce{C3H8 (g) + 7O2 (g) -&amp;gt; 3CO2 (g) + 4H2O (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio for the above equation will be 1:7:3:4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given the volume of the propane to be $17~\mathrm{L}$ and the volume of the oxygen to be $34~\mathrm{L}$. Since the question is a molar calculation, I think I have to convert the $~\mathrm{L}$ into the SI units of grams. This means $17~\mathrm{L} \Rightarrow 17000~\mathrm{g}$ and $34~\mathrm{L} \Rightarrow 34000~\mathrm{g}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The I have to calculate the number of moles for propane and oxygen (I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\\&#xA;\text{number of moles of propane} = \frac{17000}{44.1} = 385.4~\mathrm{\frac{mol}{gram}}\\&#xA;\text{number of moles of oxygen} = \frac{34000}{16} = 2125~\mathrm{\frac{mol}{gram}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I have to find the mass of the resulting water, which is the part I am stuck at. I know I have to find the number of moles times the molar mass of water to find the total mass of the water, but I am just somewhat stumped as to how to find the number of moles of water given what I have worked out above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For chemical changes, I think that bubbles of gas may appear, a temperature change will occur since this is a combustion reaction, and maybe a change of mass/volume?&lt;/p&gt;&#xA;" OwnerUserId="19123" LastEditorUserId="69830" LastEditDate="2021-05-07T20:29:22.767" LastActivityDate="2021-05-07T20:29:22.767" Title="How to calculate the amount of water produced by the incomplete combustion of propane?" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35047" PostTypeId="2" ParentId="35007" CreationDate="2015-08-14T02:12:57.103" Score="2" Body="&lt;p&gt;Also, for methyl esters one of the cleanest methods is to use &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/11/05/reagent-friday-diazomethane-ch2n2/&quot; rel=&quot;nofollow&quot;&gt;diazomethane&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, diazomethane is pretty nasty, but there are nice kits that can safely generate it in small quantities.&lt;/p&gt;&#xA;" OwnerUserId="19124" LastEditorUserId="5456" LastEditDate="2015-08-14T03:47:31.190" LastActivityDate="2015-08-14T03:47:31.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35048" PostTypeId="1" CreationDate="2015-08-14T03:39:43.743" Score="1" ViewCount="891" Body="&lt;p&gt;Say I have a sample of vinegar and I want to find the concentration of ethanoic acid in it. So I perform a titration using a $\mathrm{pH}$ meter, and using that data I can draw a $\mathrm{pH}$ curve and find quantities such as the initial $\mathrm{pH}$, the half-equivalence point, the equivalence point, and the $\mathrm{pH}$ at the equivalence point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like my teacher hasn't explained this very clearly, but given the $K_{\mathrm a}$ value for ethanoic acid and the data collected from the titration, is it possible to determine the initial concentration? If so, how do I proceed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm aware of the simple stoichiometric method, just curious whether this approach will work.)&lt;/p&gt;&#xA;" OwnerUserId="17615" LastEditorUserId="865" LastEditDate="2015-08-14T15:21:21.067" LastActivityDate="2015-08-14T15:21:21.067" Title="Can acid concentration be determined from Ka values in a titration?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ClosedDate="2015-08-14T18:27:56.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="35049" PostTypeId="1" CreationDate="2015-08-14T05:31:29.080" Score="3" ViewCount="2038" Body="&lt;p&gt;A while back, I discovered that if you have a stain on a white shirt, you can remove it by putting hydrogen peroxide on the stain, then adding some bleach to it.  I have used this for getting out wine, sweat, chocolate, etc.  The reaction produces gases and heat.  What is happening in this reaction and why does it clean stubborn stains?&lt;/p&gt;&#xA;" OwnerUserId="17937" LastEditorUserId="10185" LastEditDate="2015-08-14T09:40:43.257" LastActivityDate="2015-12-12T18:23:04.880" Title="Cleaning stains with hydrogen peroxide and bleach" Tags="&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35050" PostTypeId="1" CreationDate="2015-08-14T05:49:47.563" Score="2" ViewCount="1578" Body="&lt;h2&gt;Context:&lt;/h2&gt;&#xA;&lt;p&gt;A typical lab experiment involves the addition of ammonia in excess to solutions of copper(II) to produce a complex ion whose absorbance at roughly 610 nm varies by the concentration of copper in solution.&lt;/p&gt;&#xA;&lt;h2&gt;Question:&lt;/h2&gt;&#xA;&lt;p&gt;Why is it &lt;em&gt;&lt;strong&gt;less practical&lt;/strong&gt;&lt;/em&gt; to simply test the &lt;em&gt;&lt;strong&gt;aqueous solution&lt;/strong&gt;&lt;/em&gt; of &lt;span class=&quot;math-container&quot;&gt;$Cu^{2+}$&lt;/span&gt; rather than complexing it first with ammonia?&lt;/p&gt;&#xA;&lt;h2&gt;Thoughts:&lt;/h2&gt;&#xA;&lt;p&gt;My thoughts on this begin with the Beer-Lambert law: &lt;span class=&quot;math-container&quot;&gt;$$A =\epsilon lc$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the molar absorptivity of the ammonia complex in the visible spectrum around 600 nm is greater than that of the aqueous complex, [1] then small changes in concentration of copper can be differentiated by large changes in absorbance. This increases the sensitivity of spectrophotometric methods of determining copper(II) concentrations.&lt;/p&gt;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://chemed.chem.purdue.edu/demos/main_pages/18.8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://chemed.chem.purdue.edu/demos/main_pages/18.8.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7161" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-14T05:49:47.563" Title="Spectrophotometric determination of copper concentration" Tags="&lt;spectrophotometry&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35051" PostTypeId="2" ParentId="35048" CreationDate="2015-08-14T06:40:19.150" Score="0" Body="&lt;p&gt;Yes. Your question is basically how to determine $K_\mathrm a$ from $\mathrm{pH}$ of a salt of the acid. Since ethanoic acid is a weak acid, its salt will hydrolyze to give an alkaline solution. You can calculate the $K_\mathrm a$ given that the concentration and the $\mathrm{pH}$. The equation for hydrolysis of sodium ethanoic is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COO- +H2O &amp;lt;=&amp;gt; CH3COOH + OH-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COO-}$ is the conjugate base of $\ce{CH3COOH}$, therefore its $\mathrm pK_\mathrm b$ equals to $14 - \mathrm pK_\mathrm a$. You can then establish a equation for $K_\mathrm b$ using the concentration of the reactants and products.&lt;/p&gt;&#xA;" OwnerUserId="19088" LastEditorDisplayName="user7951" LastEditDate="2015-08-14T14:54:03.313" LastActivityDate="2015-08-14T14:54:03.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35054" PostTypeId="2" ParentId="35046" CreationDate="2015-08-14T08:09:07.083" Score="4" Body="&lt;p&gt;Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:&#xA;$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -&amp;gt; $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\&#xA;\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}&#xA;\end{multline}$$&#xA;This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.&#xA;The above case reduces to your formula for $x=5$ and $y=0$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the second question, your equation has a small error, correct is:&#xA;$$\ce{C3H8 (g) + 5O2 (g) -&amp;gt; 3CO2 (g) + 4H2O (g)}$$&#xA;But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:&#xA;$$\ce{C3H8 (g) + 2O2 (g) -&amp;gt; C (s) + 4H2O (g)}$$&#xA;So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt;:&#xA;$$pV =n\mathcal{R}T$$&#xA;You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the reaction equation you can now determine the amount of substance of water, that will be produced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $n(\ce{H2O})\approx 2.8~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can convert that easily into a mass now.&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $m(\ce{H2O})\approx 50~\mathrm{mol}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.&lt;br&gt;&#xA;You are of course right, that you will also have a rise in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-08-14T08:09:07.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35055" PostTypeId="1" AcceptedAnswerId="35066" CreationDate="2015-08-14T08:22:01.753" Score="3" ViewCount="151" Body="&lt;p&gt;During S&lt;sub&gt;N&lt;/sub&gt;2 reaction with a silicon compound, a pentacovalent intermediate is formed, while there is no such intermediate formed with carbon compounds (only a pentacovalent transition state). Why is it so? I need an MO explanation (no d orbitals involved).&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="41328" LastEditDate="2021-05-03T10:26:36.417" LastActivityDate="2021-05-03T10:32:11.857" Title="Carbon/silicon pentacovalent intermediate during substitution" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;molecular-orbital-theory&gt;&lt;carbon-family&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35056" PostTypeId="1" AcceptedAnswerId="35067" CreationDate="2015-08-14T08:29:58.067" Score="4" ViewCount="2330" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A is a non cyclic hydrocarbon chain with $\ce{C_{3x-1}H_{6x}}$ as a molecular formula which is mainly used in cars as fuel, its cracking leads to hydrocarbon (B) which belongs to the same family of hydrocarbon A, also two molecules of ethene form the second product in the above reaction. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My effort&lt;/strong&gt;: &#xA;To determine if it is an alkane, we compare the given subscripts on $\ce{C}$ and $\ce{H}$ ($3x-1$ and $6x$, respectively) to those in the general formula for an alkane, $\ce{C_{n}H_{2n+2}}$. Let $n=3x-1$ where $n$ is the number of carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6x\buildrel{?}\over{=}2(3x-1)+2=6x$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and therefore A is an alkane, with the molecular formula satisfying the form $\ce{C_{n}H_{2n+2}}$.&lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="24052" LastEditDate="2018-02-27T08:36:05.083" LastActivityDate="2018-02-27T08:36:05.083" Title="Molecular formula of a hydrocarbon" Tags="&lt;organic-chemistry&gt;&lt;hydrocarbons&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35057" PostTypeId="1" AcceptedAnswerId="35076" CreationDate="2015-08-14T09:08:47.870" Score="6" ViewCount="339" Body="&lt;p&gt;So, I'm looking for a benchmark of DFT functionals &lt;em&gt;not&lt;/em&gt; including dispersion corrections. Something similar to Grimme's and coworkers &lt;a href=&quot;http://pubs.rsc.org/en/Content/ArticleLanding/2011/CP/c0cp02984j#!divAbstract&quot;&gt;GMTKN30 database&lt;/a&gt;, but for benchmark calculations done &lt;em&gt;without&lt;/em&gt; dispersion corrections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the case of GMTKN30 database, I'm interested in the same properties: general main group thermochemistry &amp;amp; kinetics. But&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I already have calculations done many years ago without dispersion corrections since this feature simply was not implemented in most of the packages that time.&lt;/li&gt;&#xA;&lt;li&gt;My workhorse is G3/G4 method which relies on geometry obtained without dispersion corrections anyway.&lt;/li&gt;&#xA;&lt;li&gt;I quickly run few calculations including dispersion corrections and found no noticeable difference. Don't have time to redo all the work anyway.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I basically need to justify the choice of just a few DFT functional for my own benchmarks based on thorough literature reviews.&lt;/p&gt;&#xA;" OwnerUserId="186" LastActivityDate="2015-08-14T21:30:25.670" Title="A thorough benchmark of DFT functionals without dispersion corrections" Tags="&lt;reference-request&gt;&lt;density-functional-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35060" PostTypeId="1" AcceptedAnswerId="35087" CreationDate="2015-08-14T09:20:49.737" Score="6" ViewCount="1377" Body="&lt;p&gt;Why is the major product of the reduction of butadiene with sodium in liquid ammonia &lt;em&gt;cis&lt;/em&gt;-2-butene (60:40) when it is thermodynamically less stable than &lt;em&gt;trans&lt;/em&gt;-2-butene?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="10185" LastEditDate="2015-08-14T09:34:28.300" LastActivityDate="2015-08-15T05:19:36.447" Title="Reduction of butadiene with Na/NH3 - stereoselectivity" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35061" PostTypeId="2" ParentId="35006" CreationDate="2015-08-14T09:37:49.160" Score="1" Body="&lt;p&gt;The problem is your amount of gold that you have. Normal chemical baths contain several grams per liter. So you will need to operate in small volumes. You can use this &lt;a href=&quot;http://link.springer.com/article/10.2478%2Fs11696-011-0117-z&quot; rel=&quot;nofollow&quot;&gt;article - Chemical Papers&#xA;June 2012, Volume 66, Issue 6, pp 617-620&#xA;Date: 23 Dec 2011&#xA;Etching and recovery of gold from aluminum substrate in thiourea solution&lt;/a&gt;, that is using thiurea. Be aware of the low stability of this bath. The bath will degrade quite fast, but that is common problem of all non-cyanide gold baths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the article deals with a gold dissolution, I would suggest to reduce the gold back to a solid state. Use the article or this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0304386X99000651&quot; rel=&quot;nofollow&quot;&gt;Hydrometallurgy&#xA;Volume 55, Issue 2, March 2000, Pages 137–152, Gold electrowinning from aqueous–alcoholic thiourea solutions&lt;/a&gt; and then do the electrodeposition.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-08-14T14:20:24.377" LastActivityDate="2015-08-14T14:20:24.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35062" PostTypeId="2" ParentId="22100" CreationDate="2015-08-14T09:45:56.240" Score="6" Body="&lt;p&gt;Aromaticity is not a very well defined concept (there are no strict rules) and I also believe that it is hard to tell apart from overall resonance stabilisation. The IUPAC goldbook states for &lt;a href=&quot;http://goldbook.iupac.org/A00441.html&quot; rel=&quot;noreferrer&quot;&gt;aromatic&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;aromatic&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;In the traditional sense, 'having a chemistry typified by benzene'.&lt;/li&gt;&#xA;  &lt;li&gt;A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.&lt;br&gt;&#xA;  See also: Hückel (4n + 2) rule, Möbius aromaticity&lt;/li&gt;&#xA;  &lt;li&gt;The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hückel's set of rules are often used as nice guidelines, but they are not the definition of aromaticity. In fact they are very, very rigid and strict and most of the compounds you'd usually consider being aromatic in this context, do not comply with these rules. The IUPAC goldbook defines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://goldbook.iupac.org/H02867.html&quot; rel=&quot;noreferrer&quot;&gt;Hückel (4n + 2) rule&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)&lt;sub&gt;m&lt;/sub&gt; where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this framework only the first and the last structure obey Hückel's rule. The &lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_100-01-6_1HNMR.htm&quot; rel=&quot;noreferrer&quot;&gt;proton&lt;/a&gt; and &lt;a href=&quot;http://www.chemicalbook.com/SpectrumEN_100-01-6_13CNMR.htm&quot; rel=&quot;noreferrer&quot;&gt;carbon&lt;/a&gt; NMR clearly proof that &lt;em&gt;p&lt;/em&gt;-nitroanniline is an aromatic compound, the spectrum is very close to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Benzene_(data_page)&quot; rel=&quot;noreferrer&quot;&gt;data of benzene&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance structures are just a crutch to explain a very complicated and complex electronic structure in simple and understandable terms, i.e. Lewis structures. In highly delocalised systems a superposition of all resonance structures has to be considered to paint a somewhat conclusive picture of the bonding situation. There are contributions of different structures and not all of them may be equal in the overall description, however, they do not change the fact, that the experimental property is observed. This is again a case, where you can see the limitations of the resonance concept.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-08-14T09:45:56.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35063" PostTypeId="1" CreationDate="2015-08-14T10:44:39.050" Score="3" ViewCount="144" Body="&lt;p&gt;Let’s assume we have a reaction medium containing only simple molecules such as $\ce{H2O}$, $\ce{CH4}$, $\ce{NH4+}$ (and maybe some more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we consider the following equation ;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\Delta G_\mathrm r = \Delta G_\mathrm r^{\circ} + RT \ln Q_\mathrm r$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;were $\Delta G_\mathrm r$ is the Gibbs energy change associated to a chemical reaction occurring in this reaction medium, $\Delta G_\mathrm r^{\circ}$ the standard Gibbs energy change of this reaction and $Q_\mathrm r$ the mass action ratio of the reaction in the reaction medium at a given time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s consider the reaction consisting of the synthesis of a very complex molecule from the simple molecules of the reaction medium (for example, a protein). Whatever the complex molecule is, it is so complex that we are sure that it is absent from the reaction medium, i.e. its concentration is exactly 0. Also, the standard Gibbs energy change of this reaction must be &lt;em&gt;very&lt;/em&gt; high (positive), but not infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the concentration of the complex molecule starts at 0; then $Q_\mathrm r$ is 0; then $RT \ln Q_\mathrm r$ is $-\infty$ ; as $\Delta G_\mathrm r^{\circ}$ is finite, $\Delta G_\mathrm r$ is $-\infty$. So in these conditions, the synthesis of at least one copy of our very complex molecule is infinitely exergonic. This reasoning could be applied to any molecule whose concentration in the reaction medium is 0 at a given time and whose synthesis from the reaction medium is theoretically possible. In the thermodynamics courses I attended, a negative Gibbs energy change was associated with the “favourableness” of a reaction, so those reactions would be “infinitely favourable”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is absurd. But why?&lt;/p&gt;&#xA;" OwnerUserId="19016" LastEditorDisplayName="user7951" LastEditDate="2015-08-14T11:07:13.097" LastActivityDate="2015-08-15T01:47:48.463" Title="Gibbs energy and chemical equilibrium : false proof of spontaneous generation, but why?" Tags="&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35065" PostTypeId="2" ParentId="35048" CreationDate="2015-08-14T13:09:17.020" Score="1" Body="&lt;p&gt;The Henderson-Hasselbalch equation will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} + \log {[\ce{A-}]\over [\ce{HA}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the $\mathrm{p}K_{\mathrm a}$ is known, the equation will tell you the ratio of acetate to acetic acid (ethanoic acid). The ratio changes at various $\mathrm{pH}$ values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm{p}K_{\mathrm a}$ is 4.76, so for a $\mathrm{pH}$ below 3.75 the dominant species in solution will be acetic acid; at pH 5.76, the dominant species in solution is the acetate ion. If you want to make a solution at a particular $\mathrm{pH}$ value, use the equation to give you the ratio of acetate to acetic acid needed.&lt;/p&gt;&#xA;" OwnerUserId="19137" LastEditorUserId="865" LastEditDate="2015-08-14T13:42:01.233" LastActivityDate="2015-08-14T13:42:01.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35066" PostTypeId="2" ParentId="35055" CreationDate="2015-08-14T14:04:28.547" Score="4" Body="&lt;p&gt;Bonds involving carbon are typically shorter than bonds involving silicon.  Therefore a pentacoordinated carbon species would be &lt;strong&gt;sterically more crowded&lt;/strong&gt; around the central carbon atom than its pentacoordinated silicon analogue would be. This crowding would raise the energy of pentacoordinate carbon compared to pentacoordinate silicon; which is consistent with the observation that the S&lt;sub&gt;N&lt;/sub&gt;2 reaction with carbon produces a higher energy pentacoordinate transition state, while in the case of silicon a more stable pentacoordinate intermediate can be observed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{lcc}&#xA;\hline&#xA;\ce{Bond} &amp;amp; d(\ce{X=C})/\pu{pm} &amp;amp; d(\ce{X=Si})/\pu{pm} \\&#xA;\hline&#xA;\ce{X-O} &amp;amp; 143 &amp;amp; 163 \\&#xA;\ce{X-Cl} &amp;amp; 177 &amp;amp; 202 \\&#xA;\ce{X-Br} &amp;amp; 194 &amp;amp; 215 \\&#xA;\ce{X-I} &amp;amp; 214 &amp;amp; 243 \\&#xA;\hline&#xA;\end{array}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Recasting this into MO terms&lt;/strong&gt;, we would say that both the carbon and silicon pentacoordinate species involve a trigonal bipyramid structure where the two ligands (the &amp;quot;attacking&amp;quot; and &amp;quot;leaving&amp;quot; groups in the carbon case) occupy axial positions in the trigonal bipyramid arrangement.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UPqm0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPqm0.png&quot; alt=&quot;Structure of the SN2 transition state&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(&lt;a href=&quot;https://en.wikipedia.org/wiki/SN2_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;The figure above makes it easy to see that we have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hypervalent_molecule&quot; rel=&quot;nofollow noreferrer&quot;&gt;hypercoordinated&lt;/a&gt; species involving a 3-center 4-electron bond. Since a silicon p-orbital is larger than a carbon p-orbital, the 4 electrons are spread out over a larger volume in the silicon case, reducing electron-electron repulsion.  This makes the pentacoordinate silicon case more stable than pentacoordinate carbon.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="41328" LastEditDate="2021-05-03T10:32:11.857" LastActivityDate="2021-05-03T10:32:11.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35067" PostTypeId="2" ParentId="35056" CreationDate="2015-08-14T14:05:31.910" Score="11" Body="&lt;p&gt;You're right in thinking that it's an alkane. Now I think I shouldn't blame you for getting confused by&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ldots$ as a molecular formula which is mainly used in cars as fuel $\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First off, we don't know if $x$ in the molecular formula is an integer or a decimal number.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second, we don't know what fuel is meant. It's just supposed to be a fuel used in cars. And what are &lt;a href=&quot;https://en.wikipedia.org/wiki/Motor_fuel&quot; rel=&quot;nofollow noreferrer&quot;&gt;the fuels used in motor vehicles again?&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Currently, the majority of motor vehicles worldwide are powered by &lt;strong&gt;gasoline&lt;/strong&gt; or &lt;strong&gt;diesel&lt;/strong&gt;. Other energy sources include &lt;strong&gt;ethanol&lt;/strong&gt;, &lt;strong&gt;biodiesel&lt;/strong&gt;, &lt;strong&gt;propane&lt;/strong&gt;, &lt;strong&gt;compressed natural gas (CNG)&lt;/strong&gt;, &lt;strong&gt;electric batteries charged from an external source&lt;/strong&gt;, and &lt;strong&gt;hydrogen&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are these?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;The bulk of a typical &lt;a href=&quot;https://en.wikipedia.org/wiki/Gasoline#Chemical_analysis_and_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;gasoline&lt;/a&gt; consists of $\mathrm{\color{red}{hydrocarbons~with~between~4~and~12~carbon~atoms~per~molecule}}$ (commonly referred to as C4-C12).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Petroleum-derived &lt;a href=&quot;https://en.wikipedia.org/wiki/Diesel_fuel#Chemical_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;diesel&lt;/a&gt; is composed of about 75% saturated hydrocarbons (primarily paraffins including n, iso, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\color{red}{\ce{C12H23}}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ethanol_fuel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ethanol&lt;/a&gt; is an $\color{red}{alcohol}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Biodiesel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Biodiesel&lt;/a&gt; refers to a vegetable oil - or animal fat-based diesel fuel $\mathrm{\color{red}{consisting~of~long-chain~alkyl~(methyl,~ethyl,~ or~ propyl)~ esters}}$. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Propane is an $\color{navy}{alkane}$.&lt;/li&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Compressed_natural_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNG&lt;/a&gt; is made by compressing natural gas (which is mainly composed of $\color{red}{methane}$, $\ce{CH4}$), to less than 1 percent of the volume it occupies at standard atmospheric pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;Electric batteries and hydrogen are $\ldots$ a bit irrelevant. Let's not make this answer longer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, the answer would be propane, since it's the only alkane among the common car fuels. But is it? Of course not; since propane cracking isn't in high school syllabuses anywhere in the world. There is a &lt;em&gt;very very&lt;/em&gt; good chance the querent had &lt;em&gt;gasoline&lt;/em&gt; mind. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now that we trust our natural instincts (!) that they meant gasoline when they were referring to the fuel, we should remain perplexed and run in circles. Because gasoline isn't consisted of one or two distinctive and well-known substances, and it's a mixture of  It is a mixture of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gasoline#Chemical_analysis_and_production&quot; rel=&quot;nofollow noreferrer&quot;&gt;paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry.&lt;/a&gt; To make matters worse, gasoline composition is obviously not the same around the world. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\hspace{25ex}$ &lt;a href=&quot;https://i.stack.imgur.com/2dR9M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2dR9M.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;$\hspace{25ex}~\small{\mathrm{Some~of~the~compounds~commonly~found~in~gasoline}}$&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/22/GasolineComp.png/300px-GasolineComp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;$$\small{source}$$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let me rephrase your question into something actually meaningful:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ldots$ as a molecular formula which is mainly known to be a really smooth-burning species in gasoline $\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now we're talking. The best known gasoline would be one that consists of 100% &lt;a href=&quot;https://en.wikipedia.org/wiki/2,2,4-Trimethylpentane&quot; rel=&quot;nofollow noreferrer&quot;&gt;isooctane&lt;/a&gt;, and that's why we have &lt;a href=&quot;https://en.wikipedia.org/wiki/Octane_rating&quot; rel=&quot;nofollow noreferrer&quot;&gt;octane rating&lt;/a&gt;. It's not a real fuel, just the material in an imaginarily ideal mixture that would burn &quot;real smooth&quot;. That's why your question is problematic and scientifically incorrectly asked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to the question, 2,2,4-Trimethylpentane must have been the compound the querent had in mind. Now we have the answer for A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{25ex}$ &lt;a href=&quot;https://i.stack.imgur.com/wsd2Jm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wsd2Jm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;$\hspace{28ex}$ Isooctane (PIN: 2,2,4-Trimethylpentane)&#xA;$\hspace{43ex}$ &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/7/79/Isooctane.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;$\small{source}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;what would the products of &lt;a href=&quot;https://en.wikipedia.org/wiki/Cracking_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;the cracking&lt;/a&gt; be?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cracking_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on cracking&lt;/a&gt; the cracking process in entropy-driven and the radicals formed in certain conditions usually move the reaction forward. But the final products (and even which radicals form) highly depend on the methods of cracking (thermal, steam etc.) and/or the catalysts. There are numerous catalysts which demonstrate the diversity in the mechanism of cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take a look at &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie901708m&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; (It's just an abstract, but it gives you an idea)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ldots$ The main reaction route on FAU is hydride transfer followed by β-scission leading to mainly C4 species, while on MFI protolytic scission is responsible for the formation of high amounts of C1−C3 species. $\ldots$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, we're left with speculations. This isn't a real question. The only answer the examiners want is &lt;em&gt;B's &lt;strong&gt;most probably&lt;/strong&gt; butane, as it must be an alkane.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;http://www.scribd.com/doc/160481591/IGCSE-Chemistry-Organic-Chemistry#scribd&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; also demonstrates, a very simplistic view is that an octane can break into two ethene molecules and one butane molecule. So, the reaction they must have had in mind is&#xA;$$\ce{C8H18 -&amp;gt; C4H10 + 2C2H4}$$&#xA;Pfft.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;TL;DR:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This isn't a good or correct question to ask at all; at the very least, it's very poorly worded. The answers are supposedly &lt;em&gt;isooctane&lt;/em&gt; for A and &lt;em&gt;butane&lt;/em&gt; for B, but the OP's confusion was innocent.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="9961" LastEditDate="2016-06-25T18:23:55.453" LastActivityDate="2016-06-25T18:23:55.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35068" PostTypeId="1" AcceptedAnswerId="35075" CreationDate="2015-08-14T14:25:19.673" Score="4" ViewCount="19334" Body="&lt;p&gt;What is more nucleophilic, $\ce{OH-}$ or $\ce{CH3O-}$?&#xA;When I look it from the viewpoint of steric hindrance, the answer seems to be $\ce{OH-}$, but when I see it from the viewpoint of extra electron density provided by hyperconjugation of $\ce{CH3}$, the answer seems to be $\ce{CH3O-}$.&lt;/p&gt;&#xA;" OwnerUserId="16921" LastEditorUserId="10185" LastEditDate="2015-10-02T16:12:25.757" LastActivityDate="2018-11-02T16:21:26.090" Title="Is methoxide a better nucleophile than hydroxide?" Tags="&lt;organic-chemistry&gt;&lt;reactivity&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35070" PostTypeId="2" ParentId="34566" CreationDate="2015-08-14T14:53:20.560" Score="2" Body="&lt;p&gt;The heat of combustion (or any reaction) is the energy that is set free when you make the products ($\ce{CO2}$ and $\ce{H2O}$ here, in their respective stochiomeric amounts) out of single atoms (this is the same for all isomers), minus the energy needed to break all bonds in the educts. The latter is different for the different isomers, because the energy in a chemical bond not only depends on the two partners but also on their neighbours (the &quot;shape&quot; of the molecules).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual procedure for this calculation is to not go over single atoms but the elements (and compounds) in their most stable form at standard conditions. These energies a called &lt;em&gt;standard enthalpies of formation&lt;/em&gt;. There are long tables listing them for all possible compounds.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-09-03T21:50:30.267" LastActivityDate="2015-09-03T21:50:30.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35071" PostTypeId="1" CreationDate="2015-08-14T14:55:29.790" Score="4" ViewCount="160" Body="&lt;p&gt;With respect to &lt;a href=&quot;https://chemistry.stackexchange.com/questions/35013/small-doubt-regarding-bulkier-molecule-having-more-probabilty-of-reaction/35015?noredirect=1#comment53854_35015&quot;&gt;this&lt;/a&gt; question as suggested by a user (he asked me to ask this as a separate question), I wanted to know if there is any way in which steric hindarance of an organic carbon be quantified, as to a value that would indicate how bulky a compound is. &lt;/p&gt;&#xA;" OwnerUserId="16341" LastEditorUserId="23561" LastEditDate="2019-04-05T00:32:22.273" LastActivityDate="2019-04-05T00:32:22.273" Title="Can the steric nature of a carbon be quantified?" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="7" ClosedDate="2015-10-02T17:34:11.717" ContentLicense="CC BY-SA 4.0" />
  <row Id="35072" PostTypeId="2" ParentId="35049" CreationDate="2015-08-14T15:36:57.663" Score="3" Body="&lt;p&gt;Peroxide already is a &quot;bleach&quot;, and you don't say what the other one is. The gas you see evolving is most likely oxygen (from the peroxide).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally (most importantly) bleach oxidises double bonds (which make the colour in many organic compounds). The remaining molecule is also more polar afterwards, because an oxygen-containing group was inserted, and therefore easier to wash out with water.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastActivityDate="2015-08-14T15:36:57.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35073" PostTypeId="2" ParentId="35014" CreationDate="2015-08-14T15:50:44.750" Score="1" Body="&lt;p&gt;Metal atoms that have comparable radii (in the condensed metallic state) often form alloys, because you can just exchange them, without loosing too much bonding energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Li atom would leave too much space around itself in a K melt, and a K atom in a Li melt would break the bonds between the Li atoms around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals with different radii instead often form stochiometric intermetallic phases, where the different atoms have their own designated place in the crystallographic lattice.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastActivityDate="2015-08-14T15:50:44.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35074" PostTypeId="2" ParentId="28841" CreationDate="2015-08-14T16:38:15.590" Score="1" Body="&lt;p&gt;For an insulator to &quot;conduct&quot;, basically an electrochemical reaction has to take place and travel through the bulk as a &quot;defect&quot;. Above 0 K, all solids must contain defects, defects are able to move through a lattice, because they are more mobile, so all insulating solids have a finite conductivity, that generally grows with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a semiconductor (see rons answer), the energy necessary to reach the &quot;conduction band&quot; is within the range available from thermal motion at moderate temperatures. In insulators, you need an energy that is so large it is able to break bonds. This is not so easily available, so the conductivity is very low.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastActivityDate="2015-08-14T16:38:15.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35075" PostTypeId="2" ParentId="35068" CreationDate="2015-08-14T19:22:51.203" Score="4" Body="&lt;p&gt;Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not &quot;reach&quot; out as far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.&lt;/p&gt;&#xA;" OwnerUserId="19049" LastActivityDate="2015-08-14T19:22:51.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35076" PostTypeId="2" ParentId="35057" CreationDate="2015-08-14T19:57:16.900" Score="5" Body="&lt;p&gt;There is a benchmark of the GMTKN30 database with and without  the D3 dispersion correction. See Figure 4 here: &lt;a href=&quot;http://dx.doi.org/10.1002/jcc.21759&quot; rel=&quot;nofollow&quot;&gt;&quot;Effect of the damping function in dispersion corrected density functional theory&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your statement&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I quickly run few calculations including dispersion corrections and found no noticeable difference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is not supported by this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is this paper: &lt;a href=&quot;http://arxiv.org/abs/1507.00370&quot; rel=&quot;nofollow&quot;&gt;&quot;Improving intermolecular interactions in DFTB3 using extended polarization from chemical-potential equalization&quot;&lt;/a&gt; where we compare PBE to PBE-D3 for 11 different datasets (see the SI). For many dispersion-dominated complexes, DFT (no D3) is qualitatively wrong, and in many other cases, the error is comparable to the mean interaction energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I would strongly recommend against DFT without an empirical dispersion correction unless you are only focusing on electronic properties, but then you wouldn't be very interested in GMTKN30 anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you say you are using G3/G4 you might also be interested in these recent benchmark papers: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1021/acs.jpca.5b05448&quot; rel=&quot;nofollow&quot;&gt;&quot;Benchmarking Compound Methods (CBS-QB3, CBS-APNO, G3, G4, W1BD) against the Active Thermochemical Tables: Formation Enthalpies of Radicals&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/jcc.23837&quot; rel=&quot;nofollow&quot;&gt;&quot;Accurate reaction barrier heights of pericyclic reactions: Surprisingly large deviations for the CBS-QB3 composite method and their consequences in DFT benchmark studies&quot;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19142" LastEditorUserId="5017" LastEditDate="2015-08-14T21:30:25.670" LastActivityDate="2015-08-14T21:30:25.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35078" PostTypeId="2" ParentId="34985" CreationDate="2015-08-14T20:57:27.023" Score="0" Body="&lt;p&gt;Michael's answer is very good, i'll just simplify&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that X-ray images are black and white. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, it's clear, then black and clear. Not white. When they are on a white background e.g. white table or white wall, they show as white and black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they were on a purple background then they'd show up as black and purple (purple bones). Because the clear part of the xray takes the colour of the background the xray image is on. But it's clear and black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And re what makes the film go black, the film itself doesn't, but the exposed silver on it does.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You have the&#xA;  X-rays themselves, the white film, and an object/person in-between.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The X-rays themselves, the &lt;strong&gt;clear&lt;/strong&gt; film, and an object/person in-between&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently   clear/transparent film may come in a few background colours can be grey, can be purple, can be amber.. i've seen amber transparent colour film  (that's the case with colour films used by photographers, and may or may not be the case with a photographers films for black and white photos, and the films used in xrays). The ones for black and white photos, and for xrays, may be grey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yeah..&#xA;The CLEAR parts of the image are bones and/or things the X-rays couldn't get through (e.g. tumours), the clear parts of film remain clear, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and the black parts [on the film] is where the xrays hit&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The film is covered in a kind of silver particles, silver halide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The silver can be blackened by xrays or light.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that technically, most of the rays hitting the silver are light, because in order to require less dosage of radiation, the scientists thought of using a fluorescent material, the xrays pass through that and the fluorescent material converts the xrays into visible light, it absorbs the xrays and emits visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluorescent material does that(material with fluorescent phosphors). For example some &quot;invisible ink&quot; is florescent, it lights up when UV light is shone on it. It takes the high energy photons/short wavelength waves and absorbs it and emits lower energy photons/longer wavelength waves/visible light. Wikipedia says most fluorescent material converts that way, high energy to low energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to flurescent material, there is also phosphorescent material, it has phosphorescent phosphurs,which keeps the light stored for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parts of the silver metal on the film that will be darkened, will be darkened mostly by light, and partly by xrays, since there's more light than xrays hitting it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The clear parts of the image are bones and/or things the X-rays&#xA;  couldn't get through, the clear film remains clear, and the black part&#xA;  is where the X-rays did pass through, and it blackens the clear film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It blackens the silver on the film, not the film itself. The film itself is a clear film with silver on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The film though is predevelopment.. And predevelopment, the entire film is covered in silver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parts of the silver that got hit with xrays or light, will go black. And change in form from silver halides to metalling silver, and they won't get washed away in development.    The parts of the silver that didn't get hit, will get washed away during the development.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I've also heard that they have a cassette/casing, and in that is&#xA;  fluorescent material or coating and film. The X-rays hit the&#xA;  fluorescent material and (I guess by absorption and reflection, or&#xA;  absorption and emission) converts it into regular light, and there is&#xA;  a phosphorescent material/coating there, too, so the light remains for&#xA;  a while, and the light, I guess, blackens the film.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;yeah I suppose so. And the light blackens the silver on the film not the film itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask about the materials used for xray film and regular photographic film.. Apparently the differences are minor.. And in both cases the film is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He showed a colour photographic film that was amber/orange, and transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black and white one maybe is clear like regular glass is clear..I don't know if it has any slight colour at all but even if it did, it's transparent no question about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think maybe the term &quot;negative&quot; doesn't apply with xrays. Because in regular photography the rays bounce off the subject rather than through the subject, and with a normal camera and film,  what is light comes out dark and what is dark comes out light, hence it's called a negative.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that happens, (at least with regular photopgraphy, perhaps not with xrays), is the negative is turned into a positive.. And that's done by taking a picture of the negative (A), making a negative of the negative, which is a positive (B). I believe that's done with an enlarger that also enlarges it.  The silver is dense and blocks the light. The fact that silver on A is black is irrelevant, since it's the denseness that blocks the light to produce B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michael mentions about the denseness of the silver, but I don't think it's relevant to the xray..in making a new image, but I think he means you'd get little clear areas / specks that aren't black in the xray if the silver wasn't dense, but it is dense.. So the denseness and the blackness helps there in the black parts of the xray image appearing black.&lt;/p&gt;&#xA;" OwnerUserId="2020" LastEditorUserId="2020" LastEditDate="2015-08-15T10:41:10.567" LastActivityDate="2015-08-15T10:41:10.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35079" PostTypeId="1" AcceptedAnswerId="35083" CreationDate="2015-08-14T22:32:27.027" Score="1" ViewCount="889" Body="&lt;p&gt;Online literature reports the melting point of $\ce{NaHCO3}$ as ~50 deg. C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a standard Mel-Temp II melting-point apparatus and a standard alcohol thermometer, I didn't observe any melting of any sample lower than 150 deg. C. I don't think hydration of the salt can even explain this. Is there some other material in commercial baking soda that would alter the melting point this much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a raw calibration, I found that some Bayer Aspirin powder (possibly containing cellulose or other binding agent) had a melting point of 120 deg. C., a good deal &lt;em&gt;below&lt;/em&gt; the reported ~135 deg. C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what may be present in commercial Arm-and-Hammer baking soda besides sodium bicarbonate? The box reports 100% purity but I have trouble believing it now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?&lt;/p&gt;&#xA;" OwnerUserId="17821" LastActivityDate="2015-10-31T08:01:33.307" Title="Experimental melting-point of Arm-and-Hammer sodium bicarbonate is remarkably high(?)" Tags="&lt;analytical-chemistry&gt;&lt;melting-point&gt;&lt;food-chemistry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35080" PostTypeId="2" ParentId="34926" CreationDate="2015-08-14T23:41:31.257" Score="4" Body="&lt;p&gt;&lt;strong&gt;This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even taking a simple ammonia complex, e.g. $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ you can &lt;a href=&quot;http://chemdemos.uoregon.edu/demos/Copper-Ammonia-Complex&quot; rel=&quot;nofollow&quot;&gt;see a substantial color change&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many &lt;a href=&quot;https://en.wikipedia.org/wiki/Tetrakis(acetonitrile)copper(I)_hexafluorophosphate&quot; rel=&quot;nofollow&quot;&gt;copper acetonitrile&lt;/a&gt; compounds are weakly colored or colorless, with UV/Vis optical absorption occurring near the edge of the red to near-IR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get into octahedral complexes, you can find substantial changes in color due to MLCT and LMCT absorptions, as well as significant modulation of the d-d transitions due to the ligand (i.e., high-field and low-field ligands).&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-14T23:41:31.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35081" PostTypeId="1" AcceptedAnswerId="35149" CreationDate="2015-08-14T23:50:30.080" Score="3" ViewCount="4736" Body="&lt;p&gt;There are two nickel complexes: one is octahedral and the other square planar. One complex is diamagnetic, the other is paramagnetic. Both are high-spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, $\ce{Ni^2+}$ that is octahedral should be paramagnetic, as should $\ce{Ni^2+}$ that is square planar. Actually any high-spin complex should be paramagnetic. Is it possible that the octahedral $\ce{Ni^2+}$ is paramagnetic, and square planar $\ce{Ni^2+}$ diamagnetic (achieved by not filling the last orbital even tho that goes against high-spin concept)?&lt;/p&gt;&#xA;" OwnerUserId="12697" LastEditorUserId="5026" LastEditDate="2018-02-23T16:12:08.257" LastActivityDate="2018-02-23T16:12:08.257" Title="Can a high-spin complex be diamagnetic?" Tags="&lt;coordination-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35082" PostTypeId="2" ParentId="35063" CreationDate="2015-08-15T01:47:48.463" Score="1" Body="&lt;p&gt;Within the context of the equation in question, $\Delta G$ refers to the change in free energy between two specific thermodynamic equilibrium states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 1:  Stoichiometric quantities of pure reactants in separate containers at temperature T and individual specified pressures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State 2:  Corresponding stoichiometric quantities of pure products in separate containers at temperature T and individual specified pressures&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\Delta G^0$, the specified pressures are all 1 atmosphere.  This means that Q = 1, and $\Delta G =\Delta G^0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specified pressures in states 1 and 2 happen to correspond to the partial pressures of an equilibrium mixture of reactants and products, $\Delta G$ will be equal to zero.  In this case, if one had an equilibrium mixture of the reactants and products, the partial pressures in the equilibrium  mixture could also be held in equilibrium through individual semipermeable membranes with the gases in the separate containers of reactants and products (with no tendency for the reactants or products to enter or leave the reaction mixture through the semipermeable membranes).&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-15T01:47:48.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35083" PostTypeId="2" ParentId="35079" CreationDate="2015-08-15T01:48:49.880" Score="2" Body="&lt;p&gt;See &lt;a href=&quot;https://wercs.churchdwight.com/webviewer.external/private/results.aspx?__VIEWSTATEGENERATOR=6D9364FC&amp;amp;productName_option=d__value%7E&amp;amp;productID_option=d__value%7E&amp;amp;language=d__EN&amp;amp;subformat=d__NAM&amp;amp;CUSTOM1=d__ARM%20XXX%20HAMMER%20Baking%20Soda&amp;amp;hidRequiredList=ConcatedValue%20=&amp;amp;queryString=language=EN&quot; rel=&quot;nofollow noreferrer&quot;&gt;the MSDS for Arm and Hammer Baking Soda here, from the manufacturer (Church &amp;amp; Dwight Co.)&lt;/a&gt; (it will link to a PDF) in which it is stated:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;CONDITIONS TO AVOID: Temperatures above 65°C (150°F).&lt;/p&gt;&#xA;&lt;p&gt;INCOMPATIBILITY WITH OTHER MATERIALS: Reacts with acids to yield carbon dioxide. May also yield free caustic in presence of lime dust (CaO) and moisture (i.e., water, perspiration). Dangerous reaction with monoammonium phosphate or a sodium-potassium alloy may occur.&lt;/p&gt;&#xA;&lt;p&gt;HAZARDOUS DECOMPOSITION PRODUCTS: Heating above 100°C may cause dangerous levels of carbon dioxide gas to be present in confined spaces. Yields sodium oxide if exposed to temperatures above 850°C. Avoid inhalation, eye and skin contact with sodium oxide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The ingredients are listed as 100% sodium bicarbonate with CAS number 144-55-8.&lt;/p&gt;&#xA;&lt;p&gt;Your choice of thermometer is questionable in the present case. &lt;a href=&quot;https://en.wikipedia.org/wiki/Alcohol_thermometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;See this Wikipedia entry&lt;/a&gt; which reliably reports that the upper limit on temperature measurement from an alcohol-based thermometer is 78 &lt;span class=&quot;math-container&quot;&gt;$^{\circ}$&lt;/span&gt;C. A mercury thermometer has a higher upper temperature range (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Mercury-in-glass_thermometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt;) and is preferable to the alcohol thermometer. Alternatively, use a gallium thermometer or a digital model with an operating temperature range that suits your needs.&lt;/p&gt;&#xA;&lt;p&gt;You ask:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Probably not in the case of the specific brand in question here, based on the MSDS.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-15T02:48:04.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35085" PostTypeId="1" CreationDate="2015-08-15T04:08:18.363" Score="2" ViewCount="228" Body="&lt;p&gt;I want to brew beer using helium instead of $\ce{CO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most beers use about 2.5 grams of $\ce{CO2}$ per litre except stout like Guinness which uses nitrogen as it allows bubbles for much longer hence a frothy head on the stout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, perhaps a blend of helium and nitrogen would work? Helium is more soluble and if used in place of $\ce{CO2}$, as soon as the seal broke on a can it would escape rendering the cartoon voice idea useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much helium can I squeeze into a litre of beer with or without compounding it to get the chipmunk voice?&lt;/p&gt;&#xA;" OwnerUserId="19150" LastEditorUserId="16622" LastEditDate="2015-08-15T06:13:53.180" LastActivityDate="2017-02-01T06:33:52.563" Title="Using helium to make home brewed beer" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35086" PostTypeId="1" CreationDate="2015-08-15T04:31:01.623" Score="6" ViewCount="2097" Body="&lt;p&gt;The electronic configuration of platinum is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see this question has already been answered, but here is what &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_configuration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Other exceptions to Madelung's rule:&lt;/strong&gt;&lt;br&gt;&#xA;  For example, in the transition metals, the 4s orbital is of a higher energy than the $\ce{3d}$ orbitals; and in the lanthanides, the $\ce{6s}$ is higher than the $\ce{4f}$ and $\ce{5d}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand the reason behind this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be that the $\ce{4s}$ orbital is of a &lt;strong&gt;lower&lt;/strong&gt; energy than the $\ce{3d}$ orbitals (that's why it is filled first) and in the lanthanides, the $\ce{6s}$ is &lt;strong&gt;lower&lt;/strong&gt; than the $\ce{4f}$ and $\ce{5d}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a list of exceptions arising from electronic configurations of lanthanides and transition series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?&lt;/p&gt;&#xA;" OwnerUserId="15161" LastEditorUserId="41328" LastEditDate="2017-08-07T01:25:48.047" LastActivityDate="2019-10-15T17:56:23.237" Title="Exceptions to the Madelung rule for electronic configurations" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;&lt;rare-earth-elements&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35087" PostTypeId="2" ParentId="35060" CreationDate="2015-08-15T05:19:36.447" Score="8" Body="&lt;p&gt;At first ... I don't like books that don't give proper solutions to their questions$^{*0}$. This doesn't help anyone ... or at least not me.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Old sources&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I started my search on Reaxys and there are two old papers that I found when searching for $$\text{buta-1,3-diene} \xrightarrow{\ce{Na}} \text{but-2-ene}$$&#xA;They are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/jlac.19375280107&quot; rel=&quot;nofollow noreferrer&quot;&gt;K. Ziegler, F. Häffner, H. Grimm, &lt;em&gt;Liebigs Ann. Chem.&lt;/em&gt; &lt;strong&gt;1937&lt;/strong&gt;, &lt;em&gt;528&lt;/em&gt;, 105&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dx.doi.org/10.1002/jlac.19395400112&quot; rel=&quot;nofollow noreferrer&quot;&gt;W. Hückel, H. Bretscheinder, &lt;em&gt;Liebigs Ann. Chem.&lt;/em&gt; &lt;strong&gt;1939&lt;/strong&gt;, &lt;em&gt;540&lt;/em&gt;, 157&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In both (german$^{*1}$) papers they work at very low temperatures around $\mathrm{-70\ldots-60~^\circ C}$. Both say that they get mainly the Z-isomer and discuss somehow why this happens. They think about their reaction as follows: $$\ce{dien + 2 M + 2R2NH -&amp;gt; dien.H2 + 2R2N.M}$$&#xA;They think that the butadiene is somehow pre-oriented towards the sodium surface from which the preference for the cis-isomer follows.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Old but not soooo old sources&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Through the paper that M.A.Ramezani mentioned (&lt;a href=&quot;http://dx.doi.org/10.1021/jo01300a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;W. Reeve, D. R. Kuroda, &lt;em&gt;J. Org. Chem.&lt;/em&gt; &lt;strong&gt;1980&lt;/strong&gt;, &lt;em&gt;45&lt;/em&gt; (12), 2305–2307&lt;/a&gt;), I found another paper (&lt;a href=&quot;http://dx.doi.org/10.1021/ja00881a031&quot; rel=&quot;nofollow noreferrer&quot;&gt;N. L. Bauld, &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1962&lt;/strong&gt;, &lt;em&gt;84&lt;/em&gt; (22), 4347–4348&lt;/a&gt;) that clarifies some of the mysteries. It is about the &lt;em&gt;stereochemistry of metal addition to conjugated systems&lt;/em&gt;, esp. dienes. One of the first things that is shown is the following table.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\small\begin{array}{ccc}&#xA;\hline&#xA;\text{Diene} &amp;amp; \text{Temp., }\mathrm{^\circ C} &amp;amp; \text{% cis-Olefin}\\&#xA;\hline&#xA;\text{Butadiene} &amp;amp; -33 &amp;amp; 13\\&#xA;\text{Butadiene} &amp;amp; -78 &amp;amp; 50\\&#xA;\text{1,3-Pentadiene} &amp;amp; -33,\,-78 &amp;amp; 68\\&#xA;\hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It contains the cis-yields from v.p.c. analyses for two different temperatures for the reduction of butadiene with sodium in liquid ammonia. As one can see, there is less Z-but-2-ene produced when the temperature is higher$^{*2}$ (13 %) and more when the temperature is very low (50 %).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the author also mentions is, that there is in every case more Z-product produced than one would expect, because there exists only about 3-7 % of Z-butadiene at room temperature (which must be even lower at lower temperatures). So somehow the metal is included, wich is described by &lt;a href=&quot;http://dx.doi.org/10.1021/ja00881a031&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bauld&lt;/a&gt; with a cyclic complex of sodium and the radical anion that is supported by the experiments of &lt;a href=&quot;http://dx.doi.org/10.1021/jo01300a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reeve and Kuroda&lt;/a&gt;.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6qbs9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there are many possibilities how the sodium ion &quot;binds&quot; to E- and Z-form$^{*3}$ and there is also the possibility that there are two (or maybe none?) sodium ions involved, this is far off from the scope of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides that, I roughly calculated the structures from the mechanism that is schown in Scheme 8 from &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/unit6_anrad.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;the other link by M.A.Ramezani&lt;/a&gt; --which, as I found out in this moment, is also from &lt;a href=&quot;http://research.cm.utexas.edu/nbauld/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prof. N. L. Bauld&lt;/a&gt;--. Below are the corresponding $\pi$-MO's$^{*4}$ -- you can imagine the mechanism with involved sodium ion(s) to be similar but somehow with the above cyclic intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BY433.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BY433.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/woHi5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/woHi5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Oti3I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oti3I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;TL,DR&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The question in Flemings book lacks some informations. The reaction in liquid ammonia does not only depend on temperature (which was not stated!) but also on the rests that are applied to the diene. For Buta-1,3-diene in $\ce{Na/NH3}$, the Z-isomer is favored when cool temperatures are applied and therefore yield the kinetic product of the reaction. This follows from a cyclic radical anion intermediate with the sodium ion. But when the reaction temperature is rised, the thermodynamically preferred E-product is favored, as is shown by &lt;a href=&quot;http://dx.doi.org/10.1021/ja00881a031&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bauld&lt;/a&gt; (see the table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get to know more about what happens when different methyl and ethyl groups are added, read the paper by &lt;a href=&quot;http://dx.doi.org/10.1021/jo01300a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reeve &amp;amp; Kuroda&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;$^{*0}$ Dare you, Szabo and Ostlund ಠ_ಠ&lt;br&gt;&#xA;$^{*1}$ Old german papers are a joy to read ... usually the older the better.&lt;br&gt;&#xA;$^{*2}$ $\mathrm{-33~^\circ C}$ is the boiling point of ammonia&lt;br&gt;&#xA;$^{*3}$ In my case it seems to be purely vdW-interactions but it is gas-phase, this is a small basis set and even if the functional fits there was no dispersion correction.&lt;br&gt;&#xA;$^{*4}$ be aware ... those are &lt;a href=&quot;https://chemistry.stackexchange.com/questions/316/striking-examples-where-kohn-sham-orbitals-clearly-have-no-physical-meaning&quot;&gt;Kohn-Sham orbitals&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-08-15T05:19:36.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35088" PostTypeId="1" CreationDate="2015-08-15T05:31:02.933" Score="7" ViewCount="1892" Body="&lt;p&gt;A day after painting a closet, I set a piece of wood on the wall of the closet that was painted a day ago as well. Within minutes, both of their layers of paint fused together (please correct me if there is a better term than 'fused').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the paint approaches a state of being a solid, but the process is never stopping, (as I've heard:) similar to how concrete is much stronger thousands of years after it is made rather than a couple of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I understand that it started off as a liquid with a high viscosity, and liquids with high viscosity's turn into a solid when spread thin, but why does that even happen. As well, was I right when I explained that it is constantly 'drying', explaining why it was able to fuse with the other layer of the same paint even a day later?&lt;/p&gt;&#xA;" OwnerUserId="19151" LastEditorUserId="19151" LastEditDate="2015-08-25T17:10:21.980" LastActivityDate="2015-08-25T17:10:21.980" Title="What is the difference between wet paint, dry paint, and in between those stages?" Tags="&lt;everyday-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35089" PostTypeId="1" AcceptedAnswerId="35127" CreationDate="2015-08-15T06:36:16.917" Score="2" ViewCount="51" Body="&lt;p&gt;An optically active drug has one chiral center and only dextrorotatory isomer is effective. Moreover, it becomes ineffective when original activity is reduced to 35 % of original. The rate constant is  $1 \cdot 10^{-8}\ \mathrm s^{-1}$. Find the expiration time of drug in years.&lt;br&gt;&#xA;I considered the reaction as $$\ce{A -&amp;gt; B}$$ $$a-x \rightarrow x $$&#xA;so the, $\left(a-2x\right)r = 0.35$ and then I got $x$ and then I used&lt;br&gt;&#xA;$$ t=\frac{2.303}{k}\log\frac{a}{a-x} $$ I got answer 3.65 years. But the answer given is 1.37 years.&lt;/p&gt;&#xA;" OwnerUserId="17997" LastEditorDisplayName="user7951" LastEditDate="2015-08-15T09:40:45.760" LastActivityDate="2015-08-16T13:54:17.380" Title="How do I find the expiration time of drug in the following example?" Tags="&lt;kinetics&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35099" PostTypeId="1" AcceptedAnswerId="35102" CreationDate="2015-08-15T12:34:24.827" Score="11" ViewCount="646" Body="&lt;p&gt;How do I determine whether the central carbon of this compound is (&lt;em&gt;R&lt;/em&gt;) or (&lt;em&gt;S&lt;/em&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gHE85.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHE85.png&quot; alt=&quot;(2Z,5E)-hepta-2,5-dien-4-ol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On both the sides of carbon, the groups are same. They differ only in the relative position of hydrogen atoms. On what basis do I assign priority in these groups?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the &lt;em&gt;R&lt;/em&gt;/&lt;em&gt;S&lt;/em&gt; nomenclature for the middle carbon really depend on &lt;em&gt;E&lt;/em&gt;/&lt;em&gt;Z&lt;/em&gt; nomenclature of the alkenes, in such cases of stereoisomerism? Or is there an entirely different argument in such a case?&lt;/p&gt;&#xA;" OwnerUserId="3718" LastEditorUserId="16683" LastEditDate="2017-04-27T21:43:45.417" LastActivityDate="2017-04-27T21:43:45.417" Title="Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol?" Tags="&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35102" PostTypeId="2" ParentId="35099" CreationDate="2015-08-15T13:29:15.117" Score="9" Body="&lt;p&gt;In order to arrive at an unambiguous description of stereoisomers, the priority for the groups attached to the chirality center is established by the application of ‘Sequence Rules’. In the hierarchical Sequence Rules given in &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, we find&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-92.1.3.3&lt;/strong&gt; Sequence Rule &lt;strong&gt;3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;When considering double bonds and planar tetraligand atoms ‘&lt;em&gt;seqcis&lt;/em&gt;’ = ‘&lt;em&gt;Z&lt;/em&gt;’ precedes ‘&lt;em&gt;seqtrans&lt;/em&gt;’ = ‘&lt;em&gt;E&lt;/em&gt;’ and this precedes nonstereogenic double bonds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the given example, however, the stereochemical configuration at the central carbon atom is not defined. Note that solid wedges should be used to indicate bonds that project above the plane of the paper and hashed wedges to indicate bonds that project below the plane of the paper.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-15T13:29:15.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35103" PostTypeId="2" ParentId="31372" CreationDate="2015-08-15T13:56:33.287" Score="8" Body="&lt;p&gt;Firstly, in regards to the diagrams, as mentioned in the comments, both diagrams are correct. One thing that both diagrams imply is that sublimation/deposition is equivalent to the combination of melting/freezing and evaporation/condensation - another way to look at this is the following diagram from the UC Davis ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Heat_of_Sublimation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat of Sublimation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/10008/IMG_0002.jpg?size=bestfit&amp;amp;width=430&amp;amp;height=277&amp;amp;revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aGVBAm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An explanation from the webpage is that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In regards to your related questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite, liquids can occur in the right conditions, the explanation is below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To understand why sublimation occurs requires an understanding of phase diagrams. A generalised phase diagram from UC Davis Chemwiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Phase_Transitions/Phase_Diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;Phase Diagrams&lt;/a&gt; is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Wo9j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sublimation/deposition curve is the line from the origin to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Triple_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;triple point&lt;/a&gt;, represented as the red line in the diagram above. A generalisation that can be made is that a substance undergoes sublimation/deposition when either (or both) the temperature and pressure are lower than at the triple point. Consider the following examples comparing the phase diagrams of carbon dioxide and water (from the UC Davis Chemwiki page linked before):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, carbon dioxide - at 'normal' atmospheric pressure (such as in the typical laboratory - 1 atm), it is lower than the 5.2 atm at the triple point (labelled 'D'). Assuming minimal pressure changes from standard laboratory conditions, as the $\ce{CO2}$ is cooled, it will eventually undergo deposition to the solid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qkjSfm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for comparison, consider the phase diagram for water, the triple state (also labelled 'D') occurs at a very low pressure 0.006 atm, pressures lower than that would result in water subliming. Note for the standard laboratory conditions, water maintains the familiar solid, liquid and gas characteristics (depending on temperature). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/h5fEEm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, both do have a liquid phase.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-17T00:38:23.693" LastActivityDate="2015-08-17T00:38:23.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35105" PostTypeId="2" ParentId="35056" CreationDate="2015-08-15T14:52:25.730" Score="-1" Body="&lt;p&gt;You have done the first step in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know now that A is an alkane and after cracking it leads to a molecule B and $\ce{2C2H4}$. But $\ce{2C2H4}$ contains $4$ carbon atoms so the number of carbon atoms in A must be $5$ or more (i.e. $3x-1\geq5 \Longrightarrow x\geq3$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let us start trying:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x=3$ will lead to $\ce{C8H18}$&lt;/li&gt;&#xA;&lt;li&gt;$x=4$ will lead to $\ce{C11H24}$&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.&lt;br&gt;&#xA;.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And I suppose that you as a student only studied alkanes till octane ($\ce{C10H22}$)&#xA;so the answer will be $\ce{C8H18}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that is not a good reasoning but I am sure that it is the true answer.&lt;/p&gt;&#xA;" OwnerUserId="19165" LastEditorUserId="7448" LastEditDate="2015-08-15T15:22:47.953" LastActivityDate="2015-08-15T15:22:47.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35108" PostTypeId="2" ParentId="7571" CreationDate="2015-08-15T17:44:23.647" Score="1" Body="&lt;p&gt;The plain answer to your question is &quot;no&quot; because the electric current  going through the solution would break  the $\ce{H2O}$ bonds and form $\ce{H2}$  and $\ce{O2}$ gas. Eventually you run out of water due to this process. And don't forget, in a $\ce{H2O + NaCl}$ solution, chlorine gas  is produced which would eventually affect ion concentrations. I believe that Wikipedia has a very detailed description of the electrolysis of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by the way a really simple answer to your question would be to carry out the experiment yourself and see whether it conducts electricity forever. Take a glass of water add sodium chloride,  take a 9 volt battery  connect two wires one at the anode and the other at the cathode and take two pencils sharpened on both ends (graphite is a good conductor of electricity and plus you don't want the wires to be&quot;eaten&quot;)  connect each of the wires to one end of each of the pencils. Then put the end of the pencils (the ones without the wires ) into the glass and see it happen for yourself.&lt;/p&gt;&#xA;" OwnerUserId="19129" LastEditorUserId="4945" LastEditDate="2015-09-03T05:05:52.600" LastActivityDate="2015-09-03T05:05:52.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35110" PostTypeId="1" AcceptedAnswerId="35112" CreationDate="2015-08-15T18:14:13.070" Score="2" ViewCount="104" Body="&lt;p&gt;I do not have any formal education in metallurgy, but I am trying to make a game mechanic based on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to be able to alloy together different metals and alloyable materials such as carbon in different percentages, and get an approximation of the strength and hardness of the resulting alloy. If anyone has any ideas how what I would need to do to calculate such a thing I would be very appreciative.&lt;/p&gt;&#xA;" OwnerUserId="19170" LastEditorUserId="7448" LastEditDate="2015-08-15T18:35:10.733" LastActivityDate="2015-08-15T20:03:42.790" Title="Approximating strength and hardness of alloy" Tags="&lt;metallurgy&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="1" ClosedDate="2015-08-17T03:27:27.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="35111" PostTypeId="1" CreationDate="2015-08-15T18:41:20.990" Score="0" ViewCount="59" Body="&lt;p&gt;In electroplating, can a galvanic cell substitute the direct current supply? The process is to change iron metal to a silver one, but a galvanic cell is connected to the two electrodes.&lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="402" LastEditDate="2017-04-25T23:09:23.077" LastActivityDate="2017-04-25T23:09:23.077" Title="Electroplating reaction" Tags="&lt;electrochemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35112" PostTypeId="2" ParentId="35110" CreationDate="2015-08-15T19:50:32.303" Score="2" Body="&lt;p&gt;Look at the free resource provided &lt;a href=&quot;http://www.matweb.com/&quot; rel=&quot;nofollow&quot;&gt;at MatWeb&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the Physical Properties tab, you can choose up to 3 of them from a rather extensive list, including hardness. For example, I chose &quot;Carbon&quot; and &quot;Hardness, Rockwell R,&quot; and 25 results were returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the Alloy Composition tab, you can choose up to 3 elements and percentages for your alloy, and see if you get a hit. I chose &quot;Nonferrous Metal&quot; and carbon (minimum 5%, maximum 75%) and silicon (minimum 5%, maximum 75%), and got one hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe for your purposes, you can assemble a list of materials by hardness and strength (I'm not a metallurgist either, so you might have to do some research on which of the hardness and strength metrics you'll need to choose) and/or by selecting by composition (again, you might need to research which 2- and 3-element combinations make sense and which ones don't).&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-15T20:03:42.790" LastActivityDate="2015-08-15T20:03:42.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35117" PostTypeId="2" ParentId="34949" CreationDate="2015-08-16T02:00:36.807" Score="2" Body="&lt;p&gt;2-methoxy-2-methylpropane is a synonym for MTBE (Methyl tertiary-butyl &lt;strong&gt;ether&lt;/strong&gt;). The name &quot;propane&quot; is a little bit missleading, because it comes from the standardized IUPAC nomenclature system, which names the  ether-oxygen together with the smaller alkyl group an &quot;alkoxy&quot; group with the suffix &quot;-oxy&quot; (in our case: methoxy) and takes the larger alkyl part as its base identifier (here propane) when describing ethers. So MTBE is referred as an ether with all its chemical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the isomerism: Structural isomers have the same chemical formula but different constitutions (arrangement of atoms, bonds, groups). So they definitely must be constitutional isomers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constitutional isomerism is divided into 5 subcategories: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chain isomerism&lt;/li&gt;&#xA;&lt;li&gt;Positional isomerism&lt;/li&gt;&#xA;&lt;li&gt;Functional isomerism&lt;/li&gt;&#xA;&lt;li&gt;metamerism&lt;/li&gt;&#xA;&lt;li&gt;tautomerism&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you already mentioned: they are metamers. Thats because different alkyl rests are attached to the same functional group (ether). This is the first reason why they are not likely to be positional isomers. And the other reasons might come from the &quot;moving substituent&quot; (alkoxy group), which has a carbon atom (the carbon structure has to be constant within positional isomers, therefore moving the -OCH3 could mean a difference) and is no functional group. &lt;/p&gt;&#xA;" OwnerUserId="19160" LastActivityDate="2015-08-16T02:00:36.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35120" PostTypeId="2" ParentId="35081" CreationDate="2015-08-16T06:23:54.193" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Both are high-spin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, square planar complexes are nearly always low-spin. With a $\mathrm{d^8}$ configuration, the lowest four orbitals would be filled. That would make the complex diamagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the octahedral $\mathrm{d^8}$ complex, regardless of whether it's high- or low-spin, it will have the same electronic configuration $\mathrm{(t_{2g})^6 (e_g)^2}$. By Hund's first rule, the two electrons in the $\mathrm{e_g}$ orbital will both be unpaired, so the complex will be paramagnetic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a quick sketch (energy levels not to scale):&lt;a href=&quot;https://i.stack.imgur.com/SelY2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SelY2.jpg&quot; alt=&quot;d8 complexes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16683" LastEditorUserId="16683" LastEditDate="2016-02-28T19:01:20.440" LastActivityDate="2016-02-28T19:01:20.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35121" PostTypeId="1" AcceptedAnswerId="35136" CreationDate="2015-08-16T07:27:05.153" Score="10" ViewCount="27852" Body="&lt;p&gt;So we have to find which of the follwing compounds has the smallest bond angle:&#xA;$\ce{H2O}$, $\ce{H2S}$, $\ce{NH3}$, $\ce{SO2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So L $\propto \frac{1}{BA}$ &#xA;where $L$ is the number of lone pairs and $BA$ is bond angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of lone pairs is:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2O}$: 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2S}$: 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{NH3}$: 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{SO2}$: 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{NH3}$ has the fewest lone pairs, it must have the highest bond angle, so it is out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.&#xA;The electronegativiy of the O atom in $\ce{H2O}$ is the highest, so it is out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom. The electronegativiy of the H atom in $\ce{H2S}$ is the lowest of the remaining two. The only one left is $\ce{SO2}$ which must have the lowest bond angle but when I checked on google it had the highest bond angle. How is that possible?&lt;/p&gt;&#xA;" OwnerUserId="30140" LastEditorUserId="4325" LastEditDate="2015-08-17T12:03:07.387" LastActivityDate="2019-06-25T22:22:47.747" Title="Why does SO2 have a larger bond angle than H2O, H2S, and NH3" Tags="&lt;bond&gt;&lt;molecules&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35125" PostTypeId="1" AcceptedAnswerId="35126" CreationDate="2015-08-16T13:20:41.620" Score="8" ViewCount="11078" Body="&lt;p&gt;Why are alkanes also called as paraffins?   Is it because they are mainly found from petroleum and natural gas?&lt;/p&gt;&#xA;" OwnerUserId="17882" LastEditorUserId="865" LastEditDate="2015-08-16T13:55:26.933" LastActivityDate="2017-11-09T03:18:17.573" Title="Why are alkanes also called paraffins?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35126" PostTypeId="2" ParentId="35125" CreationDate="2015-08-16T13:52:59.327" Score="13" Body="&lt;p&gt;The origin of the word is historical, and has to do with the observed properties of the class of hydrocarbons we know refer to as alkanes. The name originated in the Latin for &quot;little affinity.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From M. P. Crosland's text &lt;em&gt;Historical Studies in the Language of Chemistry&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term &lt;em&gt;paraffin&lt;/em&gt; was coined by Reichenbach (from Lat. &lt;em&gt;parum&lt;/em&gt; + &lt;em&gt;affinis&lt;/em&gt; = little affinity) to denote a particular substance, a mixture of hydrocarbons (&lt;em&gt;Journal fur Chemie und Physik, 59&lt;/em&gt; (1830), 455). The term was extended to cover any hydrocarbons of the series $\ce{C_{\rm n}H_{\rm 2n + 2}}$ by Watts (&lt;em&gt;Dictionary of Chemistry&lt;/em&gt;, 1868, vol. iv, p. 344 and &lt;em&gt;Supplement&lt;/em&gt;, 1872, p. 705).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reichenbach, referred to above, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Carl_Reichenbach&quot;&gt;Carl Reichenbach&lt;/a&gt;, a multifaceted scientist who did pioneering work in isolating (and thus naming) hydrocarbons in the early 1830's.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-16T13:52:59.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35127" PostTypeId="2" ParentId="35089" CreationDate="2015-08-16T13:54:17.380" Score="2" Body="&lt;p&gt;Your general approach to the calculation looks great.  My only question is, what is the meaning of $x$ in your calculation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I would do the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{A-&amp;gt;B}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = A_0 e^{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{A}{A_0} = e^{-kt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.35 = e^{-(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ln{0.35} = -(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac{-\ln{0.35}}{(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did the units conversion from $\mathrm{s^{-1}}$ right, then the rate constant in units of $\mathrm{{yr}^{-1}}$ is 0.316.  (I just multiplied by 60&amp;times;60&amp;times;24&amp;times;365.25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac{-\ln{0.35}}{(~{\mathrm 0.316~{\mathrm {yr}^{-1}}})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I plug in the numbers, I get &lt;strong&gt;&lt;em&gt;3.2 years&lt;/em&gt;&lt;/strong&gt;.  It's possible I did something wrong, but my answer is way closer to yours than to the given answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some possible mistakes that the original writer of the question could have made to give the wrong, lower, answer.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The question has a bunch of superfluous information about the chirality of the drug.  Mistakenly thinking that this is important to the question, perhaps somebody divided by 2 to &quot;adjust&quot; for chirality.  But this is not the correct approach.  We're given the &lt;em&gt;activity&lt;/em&gt; of the drug and the decay rate of &lt;em&gt;activity&lt;/em&gt;, so the structure of the molecule is completely irrelevant to the problem.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another possible error is misinterpreting &quot;is reduced &lt;strong&gt;to&lt;/strong&gt; 35 % of original&quot; mistakenly as &quot;is reduced &lt;strong&gt;by&lt;/strong&gt; 35%&quot;.  That leads to the calculation for $t$ being $\frac{-\ln{0.65}}{0.316}$ and gives the expected (but wrong) answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The bottom line is I think you mostly did the calculation correctly and the given answer seems incorrect, at least given the wording of the question as you supplied it.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-08-16T13:54:17.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35128" PostTypeId="1" CreationDate="2015-08-16T14:16:19.540" Score="2" ViewCount="161" Body="&lt;p&gt;How dangerous is it working with 2-aminobenzimidazole, iminoesters and 1,3,5-triazine? Has there been any cases of health problems recorded concerning those three compounds?&lt;/p&gt;&#xA;" OwnerDisplayName="user19185" LastEditorUserId="7475" LastEditDate="2015-11-09T14:43:51.493" LastActivityDate="2015-11-09T14:43:51.493" Title="Safety/health hazards: working with 2-aminobenzimidazole, iminoesters, and 1,3,5-triazine" Tags="&lt;organic-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-11-09T20:22:19.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="35129" PostTypeId="1" CreationDate="2015-08-16T14:41:13.200" Score="6" ViewCount="662" Body="&lt;p&gt;I'm wondering what happens when $\ce{H2}$ leaks from say a transmission pipe in an unenclosed area. No immediate source of ignition, I know it rises more quickly than helium (x2 I believe) and dissipates in concentration quickly but does it react with gases in atmosphere to form methane or other compounds? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short of ignition and burning causing more $\ce{H2}$ to dissociate what can happen, especially in upper atmosphere with more UV and so on. Methane has reaction in the upper atmosphere giving it a half life of ~7 years for example.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The covalent radius of a neutral hydrogen atom is 0.0371 nm, smaller&#xA;  than that of any other element. Because small atoms can come very&#xA;  close to each other, they tend to form strong covalent bonds. As a&#xA;  result, the bond dissociation enthalpy for the $\ce{H-H}$ bond is relatively&#xA;  large (435 kJ/mol). $\ce{H2}$ therefore tends to be unreactive at room&#xA;  temperature. In the presence of a spark, however, a fraction of the $\ce{H2}$ molecules dissociate to form hydrogen atoms that are highly reactive. Source: &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch10/hydrogen.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemed.chem.purdue.edu&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19186" LastEditorUserId="39539" LastEditDate="2017-03-16T07:08:58.620" LastActivityDate="2020-12-14T21:15:32.027" Title="What typically happens to H2 molecules released into atmosphere (with no spark source)?" Tags="&lt;everyday-chemistry&gt;&lt;hydrogen&gt;&lt;photochemistry&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35130" PostTypeId="2" ParentId="8498" CreationDate="2015-08-16T16:14:09.090" Score="1" Body="&lt;p&gt;Oxygen and acetylene mixed in a ballon will produce a very large explosive bang when lit and a loud noise and gratifying plume of putrid black smoke. Hydrogen and oxygen is going to be many times more explosive, not sure how many times more though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was on the back of a flat bed truck with a parade display and a weather ballon on a string about 10 feet above the tray (possibly less was twenty years ago) was accidentally ignited (not by me!) and a few of us were knocked to the floor/truck tray and the closest person had open flesh burns all the way down his arm which was extended in the air towards the weather ballon (may have even been holding a party sized ballon with acetylene as he was lighting them while holding them which is very dangerous and I in no way endorse (he was a lunatic this guy and I had no authority over him)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So bottom line start very small with these kinds of experiments and be aware things can and will go wrong so always take as much precaution as you can.&lt;/p&gt;&#xA;" OwnerUserId="19186" LastActivityDate="2015-08-16T16:14:09.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35131" PostTypeId="2" ParentId="35129" CreationDate="2015-08-16T16:19:23.697" Score="2" Body="&lt;p&gt;The most common, reactive stuff in the atmosphere are &lt;span class=&quot;math-container&quot;&gt;$\ce{-OH}$&lt;/span&gt; radicals &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydroxyl_radical#Importance_in_Earth.27s_atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Wikipedia)&lt;/a&gt;. So I guess they're the natural antagonist for &lt;span class=&quot;math-container&quot;&gt;$\ce{H2}$&lt;/span&gt; molecules.&lt;/p&gt;&#xA;&lt;p&gt;Btw., as soon as it's diluted in the surrounding air, it will no longer rise. Only bulk masses have buoyancy. The g gradient in the atmosphere is much too small to separate gases.&lt;/p&gt;&#xA;&lt;p&gt;In the end, hydrogen however &lt;em&gt;does&lt;/em&gt; diffuse upwards and is ultimately lost into space. Hydrogen and helium are light enough to be able to reach escape velocity in the upper atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2020-12-14T21:15:32.027" LastActivityDate="2020-12-14T21:15:32.027" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35132" PostTypeId="2" ParentId="35125" CreationDate="2015-08-16T16:42:20.463" Score="2" Body="&lt;p&gt;As already noted in Todd Minehardt's answer, the name &quot;paraffin&quot; is based on the Latin words &lt;em&gt;parum&lt;/em&gt; + &lt;em&gt;affinis&lt;/em&gt; = &quot;little affinity&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;little affinity&quot; behind their Latin name referring to their relatively low reactivity. They have &quot;little affinity&quot; for other elements or compounds and a consideration of their chemical structure will make you see why.&lt;/p&gt;&#xA;" OwnerUserId="19187" LastEditorUserId="283" LastEditDate="2015-08-16T19:52:09.400" LastActivityDate="2015-08-16T19:52:09.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35135" PostTypeId="2" ParentId="9198" CreationDate="2015-08-16T17:12:40.963" Score="2" Body="&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, heteromonocyclic compounds with up to and including ten ring members are named by the extended &lt;a href=&quot;http://dx.doi.org/10.1351/goldbook.H02737&quot; rel=&quot;nofollow&quot;&gt;Hantzsch–Widman&lt;/a&gt; system; for example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;oxirane ($\ce{C2H4O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxetane ($\ce{C3H6O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxolane ($\ce{C4H8O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxane ($\ce{C5H10O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxepane ($\ce{C6H12O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxocane ($\ce{C7H14O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxonane ($\ce{C8H16O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxecane ($\ce{C9H18O}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For monocyclic rings with eleven and more ring members, skeletal replacement (‘a’)&#xA;nomenclature is used; for example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;oxacycloundecane ($\ce{C10H20O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxacyclododecane ($\ce{C11H22O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxacyclotridecane ($\ce{C12H24O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxacyclotetradecane ($\ce{C13H26O}$)&lt;/li&gt;&#xA;&lt;li&gt;oxacyclopentadecane ($\ce{C14H28O}$)&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-08-16T17:12:40.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35136" PostTypeId="2" ParentId="35121" CreationDate="2015-08-16T19:03:36.747" Score="8" Body="&lt;p&gt;Here are the bond angles for each molecule (data from wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{|c|c|}\hline&#xA;\mathrm{Molecule} &amp;amp; \mathrm{Bond \space Angle \space (^\circ)} \\ \hline&#xA;\ce{H2S} &amp;amp; 92.1 \\ \hline&#xA;\ce{H2O} &amp;amp; 104.5 \\ \hline&#xA;\ce{NH3} &amp;amp; 107.8 \\ \hline&#xA;\ce{SO2} &amp;amp; 119 \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent &lt;a href=&quot;https://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h&quot;&gt;previous answer&lt;/a&gt; from @ron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W4drz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W4drz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but &lt;em&gt;only&lt;/em&gt; in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://chemistry.stackexchange.com/questions/15671/utility-of-bents-rule-what-can-bents-rule-explain-that-other-qualitative-con&quot;&gt;here&lt;/a&gt; are some good introductions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how should you answer this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-08-16T19:03:36.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35137" PostTypeId="1" CreationDate="2015-08-16T20:38:43.030" Score="4" ViewCount="434" Body="&lt;p&gt;We have two materials; say glass and wood. Light passes through glass but it doesn't pass through wood. Is it caused by atoms and molecules arranged in glass and wood such that they interact differently with light?&lt;/p&gt;&#xA;" OwnerUserId="19194" LastEditorUserId="9961" LastEditDate="2015-08-16T22:45:47.353" LastActivityDate="2015-08-17T08:19:33.983" Title="Why some materials are transparent, but other ones aren't?" Tags="&lt;physical-chemistry&gt;&lt;optical-properties&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35138" PostTypeId="2" ParentId="35128" CreationDate="2015-08-16T20:54:47.297" Score="4" Body="&lt;p&gt;Looking up the chemical you're working with is always a good idea, but you need to look closer and evaluate the specific danger those chemicals present to you. What you want to look at are the danger symbols and the hazard and precautionary statements. Those will tell you the specific dangers a chemical presents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look up your first substance, 2-aminobenzimidazole, it looks pretty tame for organic chemistry standards. It has an exclamation mark as danger symbol, which doesn't tell you much, but the absence of a skull and crossbones and a health hazard symbol tell you that this probably won't kill you immediately. This doesn't mean the stuff is harmless, but it is also not extremely toxic. If you look at the hazard statements of this chemical you'll find that you shouldn't get it into your eyes, on your skin or breathe it. All of that is a good idea when working with any chemical, and this is exactly what you should learn when studying chemistry. This is a really tame chemical compared to a lot of stuff organic chemists work regularly with, but you still should avoid exposing yourself to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second item is an entire category of chemicals, I'll skip it for this reason. The third one is a bit more dangerous than the first one, it has a corrosive and a health hazard danger symbol. You want to avoid contact with it, but that's what all your protective equipment like safety goggles, your lab coat and your fume hood are for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not saying those chemicals are harmless, they're not. But they are less dangerous than many other common chemicals in an organic chemistry lab. &lt;/p&gt;&#xA;" OwnerUserId="7" LastActivityDate="2015-08-16T20:54:47.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35140" PostTypeId="1" CreationDate="2015-08-16T21:47:08.703" Score="1" ViewCount="175" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mkj2y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mkj2y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer key says neither product exhibits stereoisomerism, but both products are substituted. Why is it that I cannot assign E/Z configuration? I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.&lt;/p&gt;&#xA;" OwnerUserId="19196" LastEditorUserId="10185" LastEditDate="2015-08-17T09:48:54.313" LastActivityDate="2015-08-17T09:57:06.993" Title="Why are (3-methylbut-2-en-2-yl)benzene and (3-methylbut-1-en-2-yl)benzene not stereoisomeric?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35141" PostTypeId="1" CreationDate="2015-08-17T02:14:05.190" Score="4" ViewCount="330" Body="&lt;p&gt;I came across with the following solution from Sandler's book: $$\begin{eqnarray}\left( \frac{\partial G}{\partial T} \right)_U &amp;amp;=&amp;amp; \frac{\partial (G,U)}{\partial (T,U)} \\ &amp;amp;=&amp;amp; \frac{\partial (G,U)}{\partial (T,V)}\cdot \frac{\partial (T,V)}{\partial (T,U)} \\ &amp;amp;=&amp;amp;\frac{(\partial G/\partial T)_V (\partial U/\partial V)_T-(\partial G/\partial V)_T (\partial U/\partial T)_V}{(\partial U/\partial V)_T} \end{eqnarray}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me understand these steps. What happens in line 1? I know that line 2 is by using Chain Rule by then how did line 2 becomes line 3?&lt;/p&gt;&#xA;" OwnerUserId="19197" LastActivityDate="2015-08-17T10:27:49.540" Title="Derivative of Gibbs Energy wrt Temperature at constant Internal Energy" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35143" PostTypeId="2" ParentId="35137" CreationDate="2015-08-17T08:13:43.143" Score="3" Body="&lt;p&gt;A brief summary: It is caused by the electrons in the particles in the substance. When electrons are orbiting the nucleus of an atom they can move to higher and lower energy levels. They do, however, need energy in the form of photons to do this, this means that when a ray of light hits the atom, the electrons in it will absorb some of the energy from the light ray. It needs a specific amount of energy to move up to the next energy level so will only absorb light from a specific part of the spectrum. This means that in some materials most of the light can pass through it and it may only absorb light waves not of visible light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So wood, to some frequencies of light, is see through but not to the light waves we can see. That is why we can have mobile and radio signal from inside houses but only see through glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Electrons are actually much more complicated and theoretical than this suggests but this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try watching this:  &lt;a href=&quot;https://www.youtube.com/watch?v=Omr0JNyDBI0&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=Omr0JNyDBI0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19200" LastEditorUserId="19200" LastEditDate="2015-08-17T08:19:33.983" LastActivityDate="2015-08-17T08:19:33.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35145" PostTypeId="2" ParentId="35140" CreationDate="2015-08-17T09:43:40.347" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, you cannot assign E/Z when it doesn't make any sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr: When on at least one side of double bond two identical groups or atoms (e.g., Hydrogen), there's only one way to arrange substituents near double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E/Z is used to explain how substituents &lt;em&gt;on different sides of double bond&lt;/em&gt; located, which excludes 1,1-substituted ethylenes (which is the case with minor product on your figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When on one side of double bond is two identical groups, e.g., methyl (which is the case with major product) there's no E/Z isomerism too, because there's only one way to arrange substituents. You can actually think of first case as of the second, minor product has 2 identical atoms at one side of double bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope I haven't made more confusion than there was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would suggest to play with some ball and stick models (They could be at the uni library), I think that building couple of examples and trying to rearrange them to stereoisomeric form will help with understanding stereochemistry in general.&lt;/p&gt;&#xA;" OwnerUserId="19201" LastEditorUserId="19201" LastEditDate="2015-08-17T09:57:06.993" LastActivityDate="2015-08-17T09:57:06.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35146" PostTypeId="1" AcceptedAnswerId="35169" CreationDate="2015-08-17T10:22:25.867" Score="0" ViewCount="221" Body="&lt;p&gt;For example in dinitrogen pentoxide, $\ce{N2O5}$, covalent as well as coordinate bonds (type of covalent bonds) are present, but it appears that it contains only covalent bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a proper method to find out which type of bonds are present in a molecule?&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="16622" LastEditDate="2015-08-17T10:48:50.773" LastActivityDate="2015-08-17T23:00:19.437" Title="Types of bonds in a molecule" Tags="&lt;bond&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-19T06:32:34.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="35147" PostTypeId="2" ParentId="35141" CreationDate="2015-08-17T10:27:49.540" Score="2" Body="&lt;p&gt;The starting point is the two equations $$dG=\left(\frac{\partial G}{\partial T}\right)_VdT+\left(\frac{\partial G}{\partial V}\right)_TdV$$ and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$From the second equation, you determine the partial of V with respect to T and constant U by setting dU=0.  I leave the rest of the math up to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chet&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-17T10:27:49.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35149" PostTypeId="2" ParentId="35081" CreationDate="2015-08-17T14:37:11.413" Score="5" Body="&lt;p&gt;&lt;strong&gt;A high-spin complex cannot be diamagnetic.&lt;/strong&gt; By definition, a high-spin system has multiple unpaired electrons. The names &quot;high-spin&quot; and &quot;low-spin&quot; refer to the total electron spin. In the following example ($d^6$ octahedral), the high spin state $S = 2$ and the low spin state $S = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/anQYF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/anQYF.png&quot; alt=&quot;High spin and low spin systems &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electron configuration such as $d^8$ the outcome is independent of whether electrons were filled by pairing in the $t_{2g}$ orbitals first, or by filling all orbitals and pairing last. You end up with the same configuration. In this case, there is no such thing &quot;high-spin&quot; or &quot;low-spin&quot; states. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EA9Qt.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EA9Qt.gif&quot; alt=&quot;d8 octahedral&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true for $d^0$, $d^1$, $d^2$, $d^3$, $d^7$, $d^9$ and $d^{10}$. Octahedral is really the only geometry with both high- and low-spin states, as this is the only geometry where the splitting energy $\Delta$ is comparable to the spin-pairing energy. &lt;/p&gt;&#xA;" OwnerUserId="1389" LastActivityDate="2015-08-17T14:37:11.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35150" PostTypeId="1" CreationDate="2015-08-17T14:49:39.740" Score="5" ViewCount="5416" Body="&lt;p&gt;I see many sites that claim to sell working gold detectors but I am not convinced that they work. Also, does a metal detector detect gold?&lt;/p&gt;&#xA;" OwnerUserId="19205" LastEditorUserId="7475" LastEditDate="2017-12-16T15:14:10.793" LastActivityDate="2019-01-10T23:31:23.427" Title="Does a gold detector exist?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;metal&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35151" PostTypeId="2" ParentId="35014" CreationDate="2015-08-17T15:12:41.980" Score="3" Body="&lt;p&gt;Questions like this are not prone to easy answers. But, lets look at what is known about the liquids, then at the various binary diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium - Melting point 453K, density at melt is 0.512 g/cm$^{3}$, or 0.0737 mol/cm$^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sodium - Melting point 371K, density at melt is 0.926 g/cm$^{3}$, or 0.0403 mol/cm$^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potassium - Melting point 336K, density at melt is 0.828 g/cm$^{3}$, or 0.0211 mol/cm$^{3}$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calcium - Melting point 1112K, density at melt is 1.376 g/cm$^{3}$, or 0.0343 mol/cm$^{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly Li has by far the smallest atomic volume in the melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Li-Na binary, as you point out, shows a miscibility gap, tilted towards the Li side, but peaking around 300C. The K-Li system is less well studied, but it also shows a miscibility gap which peaks at perhaps a similar value of 300C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sharp contrast, the Ca-Li system exhibits an intermediate compound (Li$_{2}$Ca) and can be described with a 'virtually ideal' liquid phase (C.W. Bale and A.D. Pelton, Bulletin of Alloys Phase Diagrams 8(2) 125-127 (1987)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the molar density of liquid Ca lies between those of Na and K it is clear that you cannot ascribe the observed miscibility gaps solely to atomic radii. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further complicate matters, lets keep looking at more binary diagrams. The K-Na diagram shows an intermediate compound, KNa$_{2}$ and no miscibility gap.  KNa$_{2}$ has the same crystal structure as Li$_{2}$Ca (both are MgZn$_{2}$ prototype). The Ca-Na binary shows -- surprise! -- a miscibility gap towards the Na side, peaking at 1200K. So, Ca and Na, even with fairly similar atomic volumes, do not want to mix in the liquid, while Ca and Li are perfectly happy to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line - atomic radii do not really mean that much when predicting binary phase diagrams.  Even with comparable radii, you would be surprised at the weirdness than can be found on phase diagrams. As one example, Ag-Cu (both fcc metals) do not form a continuous solid solution, and have a slightly positive enthalpy of mixing in the liquid (not enough to form a miscibility gap).  Yet, Au-Cu (Ag and Au have similar atomic radii) form a fcc continuous solid solution (with intermediate ordered fcc-based phases at lower temperatures), and has a strongly negative enthalpy of mixing in the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fund exploring binary phase diagrams (and ternary, ...), but don't rely on simple heuristics.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-08-17T15:12:41.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35154" PostTypeId="1" AcceptedAnswerId="35158" CreationDate="2015-08-17T16:40:34.517" Score="13" ViewCount="22282" Body="&lt;p&gt;Dehydration of carboxylic acids with phosphorus pentoxide ($\ce{P2O5}$) yields acid anhydrides, and in a similar reaction, amides dehydrated with phosphorus pentoxide yield cyanides. Can an arrow pushing mechanism be given for these reactions? If there isn't a mechanism, what, at least, gives phosphorus pentoxide the ability to dehydrate carboxylic acids and amides? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorus pentoxide itself reacts with water molecules to form phosphoric acid. That enables it to remove free water molecules by a simple hydrolysis reaction, but here of course there must be some other electron movement since the water being removed is not free water.&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="16683" LastEditDate="2017-01-08T07:46:30.503" LastActivityDate="2017-11-30T11:38:13.580" Title="Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;carbonyl-compounds&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35155" PostTypeId="2" ParentId="35150" CreationDate="2015-08-17T17:10:05.473" Score="6" Body="&lt;p&gt;You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does a gold detector exist?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely. Here's a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0002J1LNO&quot; rel=&quot;noreferrer&quot;&gt;metal detector that is fine-tuned for gold detection, available ${\it via}$ Amazon&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, if you are looking for something that detects &lt;em&gt;only&lt;/em&gt; gold, I'm not sure the best way to proceed in that case. &lt;a href=&quot;http://www.detectorprospector.com/gold-prospecting-guides/steve-guide-gold-nugget-detectors.htm&quot; rel=&quot;noreferrer&quot;&gt;This site&lt;/a&gt; appears to be a comprehensive review of some popular precious metal-tuned detectors on the market. The site is maintained by a person with an active interest in prospecting and metal detection. It is stated on the linked page that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A quick note to those who know nothing about these machines. These are metal detectors. There is no such thing as a &quot;gold only&quot; detector. These detectors will also find lead, copper, aluminum, and other metals. These units are best used to look for relatively larger pieces of gold at relatively shallow depths. Concentrations of gold dust are not detectable. Some of these units can hit gold that weighs as little as a grain (480 grains per ounce) or less but only at an inch or two. Only the larger nuggets can be found at depths exceeding a foot. Only world class nuggets weighing many ounces can be detected at over two feet. The vast majority of nuggets found are found at inches, not feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In any event, you can purchase a device which has been engineered to discriminate base metals from precious ones (such as the model I have linked here).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, does a metal detector detect gold?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes - see the reviews for the product offered at Amazon (and many others like it) as well as the informational site I've linked (above). Also see &lt;a href=&quot;https://en.wikipedia.org/wiki/Metal_detector&quot; rel=&quot;noreferrer&quot;&gt;the Wikipedia page on metal detectors&lt;/a&gt; for more information, including the issues relating to fine-tuning the devices to discriminate between metals with similar/different properties &quot;detected&quot; by the devices.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-17T18:11:48.567" LastActivityDate="2015-08-17T18:11:48.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35157" PostTypeId="1" AcceptedAnswerId="35163" CreationDate="2015-08-17T19:01:01.160" Score="1" ViewCount="401" Body="&lt;p&gt;Is there a fairly simple way to precipitate Sodium Hypochlorite from tap water before I distill it? Or, second best, a way to neutralize it?&lt;/p&gt;&#xA;" OwnerUserId="19204" LastEditorUserId="19204" LastEditDate="2015-08-25T18:06:58.460" LastActivityDate="2015-08-25T18:06:58.460" Title="Precipitate Sodium Hypochlorite from tap water before distillation" Tags="&lt;distillation&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35158" PostTypeId="2" ParentId="35154" CreationDate="2015-08-17T19:06:33.997" Score="13" Body="&lt;p&gt;Here is a mechanism for the dehydration of an acid to an anhydride using phosphorus pentoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kAHvk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kAHvk.png&quot; alt=&quot;Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{OH}$ group is not a very good leaving group.  Even in simpler reactions like the dehydration of an alcohol to an olefin we often first convert the hydroxyl group into a better leaving group by protonation with acid or conversion to an inorganic ester (for example using thionyl chloride or phosphorus pentoxide).  The same thing takes place here - we are converting the acid's hydroxyl group into a better leaving group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is $\ce{HPO3}$ such a good leaving group? Look at all the resonance structures you can draw for the anion $\ce{PO3-}$.  It's stability makes the reaction exothermic and strongly drives the dehydration process to the product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, &lt;a href=&quot;https://books.google.com/books?id=CUR5swviO_QC&amp;amp;pg=PA322&amp;amp;lpg=PA322&amp;amp;dq=phosphorus%20pentoxide%20dehydration%20mechanism&amp;amp;source=bl&amp;amp;ots=AY4IIxggSx&amp;amp;sig=huWPaf2TeaOe5WQziKLJlhhCag8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CGcQ6AEwDWoVChMInK76pd-wxwIVgqSICh27Wwb8#v=onepage&amp;amp;q=phosphorus%20pentoxide%20dehydration%20mechanism&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here is a link&lt;/a&gt; to a drawing and discussion of the mechanism for the dehydration of an amide to a nitrile using phosphorus pentoxide. Same mechanism and principles (good leaving group, exothermic) as discussed above.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2017-01-08T08:15:15.870" LastActivityDate="2017-01-08T08:15:15.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35160" PostTypeId="1" CreationDate="2015-08-17T20:18:48.113" Score="3" ViewCount="7858" Body="&lt;p&gt;I usually use &quot;eq&quot; but I have been told to use &quot;equiv.&quot; in order to avoid confusion with &quot;equation&quot;. Is there a reference I can check?&lt;/p&gt;&#xA;" OwnerUserId="19211" LastActivityDate="2017-07-26T07:20:45.560" Title="How is the unit of measure &quot;equivalent&quot; abbreviated?" Tags="&lt;units&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35161" PostTypeId="1" AcceptedAnswerId="35178" CreationDate="2015-08-17T20:28:09.707" Score="4" ViewCount="14999" Body="&lt;p&gt;Is it necessary that all endothermic reactions be slow and all exothermic reactions fast? This is based on the fact that endothermic reactions require more activation energy than exothermic reactions to initiate a chemical reaction. Or  are there some endothermic reactions that can be fast without the presence of a catalyst? &lt;/p&gt;&#xA;" OwnerUserId="17479" LastEditorUserId="10185" LastEditDate="2015-08-17T21:00:59.373" LastActivityDate="2021-05-10T18:19:14.850" Title="Rate of exothermic/endothermic reactions" Tags="&lt;thermodynamics&gt;&lt;kinetics&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35162" PostTypeId="2" ParentId="35160" CreationDate="2015-08-17T20:49:20.570" Score="8" Body="&lt;p&gt;The first place to check would be standards that are used frequently globally, such as &lt;a href=&quot;https://www.cas.org/cas-home&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAS&lt;/a&gt;. According to the American Chemical Society page &lt;a href=&quot;https://www.cas.org/content/cas-standard-abbreviations&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAS Standard Abbreviations &amp;amp; Acronyms&lt;/a&gt;, the abbreviation for the unit of measurement 'equivalent' is &lt;code&gt;equiv&lt;/code&gt;, while that of the adjective 'equivalent' is &lt;code&gt;equiv.&lt;/code&gt; (with a dot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the ACS page &lt;a href=&quot;http://pubs.acs.org/paragonplus/submission/joceah/joceah_abbreviations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard Abbreviations and Acronyms&lt;/a&gt;, shows equation abbreviated to &lt;code&gt;eq&lt;/code&gt; and equivalent as &lt;code&gt;equiv&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some journals, such as the Journal of Biological Chemistry page &lt;a href=&quot;http://www.jbc.org/site/misc/itoa.TI.xhtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Table I: Abbreviations of units of measurement and of physical and chemical quantities&lt;/a&gt; state that the abbreviation of equivalent is as you used, &lt;code&gt;eq&lt;/code&gt;, so if you are intending to publishing, just be aware of what the journal expects in terms of abbreviations.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorUserId="19211" LastEditorDisplayName="user15489" LastEditDate="2017-07-26T07:20:45.560" LastActivityDate="2017-07-26T07:20:45.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35163" PostTypeId="2" ParentId="35157" CreationDate="2015-08-17T21:07:29.767" Score="4" Body="&lt;p&gt;Sodium hypochlorite is not carried over into water by distillation.  Instead, at high (~100&amp;deg;C) it &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/j150327a013&quot; rel=&quot;nofollow&quot;&gt;decomposes&lt;/a&gt;, primarily to sodium chloride (i.e. table salt) and oxygen gas, but also possibly to trace amounts of chlorate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could neutralize the hypochlorite by mixture with stoichiometric amounts of hydrogen peroxide, which cause immediate release of oxygen gas, a reaction which occurs even at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-08-17T21:07:29.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35164" PostTypeId="1" AcceptedAnswerId="35176" CreationDate="2015-08-17T21:17:24.727" Score="2" ViewCount="1027" Body="&lt;p&gt;I want to verify a thought I had about Gibbs free energy. In some ways, I'm actually wondering why we are using enthalpy instead of the internal energy $U$. Right now, I'm thinking that enthalpy is mostly useful because reactions are done at constant pressure (in an open beaker). If we were to do our reactions at constant volume, we would actually use Helmholtz's free energy instead of Gibbs. My logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of internal energy:&#xA;$\Delta U = Q_\text{heat} - p\Delta V$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we inject $Q_\text{heat}$ without holding the volume constant, we might change $U$ or $V$, which is a problem because we'll want to be able to describe our system as fully as possible. If we define the enthalpy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H = Q + V\Delta p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, by keeping the pressure constant, we know that the heat added goes straight into this &quot;enthalpy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, to know if a reaction is spontaneous, we need to see if the enthalpy-heat from a reaction overthrows the energy associated with the change in entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: a molecule of two atoms wants to break apart. The bond has potential energy $Q_\text{heat} = \Delta H_0$ but it will increase the entropy by $Q_\text{entropy} = T \Delta S_0$. So, we check if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Q_\text{heat} - Q_\text{entropy} &amp;gt; 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Delta H - T \Delta S &amp;gt; 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus Gibbs free energy was invented as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta G = \Delta H - T \Delta S$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT, if we wanted to use a closed beaker, we would keep the internal energy U, and then to see if we had a spontaneous reaction, we would use Helmholtz's free energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \Delta F = \Delta U - T \Delta S $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about any of this and I'm mostly looking for confirmation/information of my logic. Please tell me if my intuition is wrong somewhere.&lt;/p&gt;&#xA;" OwnerUserId="16613" LastEditorDisplayName="user5764" LastEditDate="2015-08-18T08:43:21.973" LastActivityDate="2015-08-18T08:43:21.973" Title="Understanding Gibbs free energy" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35165" PostTypeId="1" AcceptedAnswerId="35167" CreationDate="2015-08-17T21:19:26.340" Score="0" ViewCount="823" Body="&lt;p&gt;Ethene reacts with hydrogen chloride to form a monomer used in the production of a polymer. This monomer is chloroethene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) How is the polymer formed from the monomer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Is this an addition or a condensation polymerization?&lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="9961" LastEditDate="2015-10-11T17:03:56.940" LastActivityDate="2015-10-11T17:03:56.940" Title="Mechanism and type of polymerization of chloroethene" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;reaction-mechanism&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35166" PostTypeId="1" AcceptedAnswerId="35168" CreationDate="2015-08-17T21:57:02.167" Score="9" ViewCount="3052" Body="&lt;p&gt;How can water be absorbed into a paper towel such that it continues to be absorbed up into the material against the force of gravity? What property of paper towels/water causes this? I can understand that cohesive forces cause water molecules to follow other water molecules, but why do the molecules begin this climb in the first place?&lt;/p&gt;&#xA;" OwnerUserId="15410" LastActivityDate="2016-04-01T04:10:25.657" Title="How does water defy gravity in paper towels?" Tags="&lt;water&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35167" PostTypeId="2" ParentId="35165" CreationDate="2015-08-17T22:25:15.207" Score="6" Body="&lt;p&gt;You got it wrong - monomer is chloroethene (&lt;a href=&quot;https://en.wikipedia.org/wiki/Vinyl_chloride&quot;&gt;vinyl chloride&lt;/a&gt;) and nothing is eliminated in &lt;a href=&quot;https://en.wikipedia.org/wiki/Chain-growth_polymerization&quot;&gt;addition polymerization&lt;/a&gt; in contrary to &lt;a href=&quot;https://en.wikipedia.org/wiki/Condensation_polymer&quot;&gt;polycondensation&lt;/a&gt;. This monomer is made via addition of chlorine molecule to ethene and elimination of hydrogen chloride. Alternative methods are addition of hydrogen chloride to acetylene or &lt;a href=&quot;https://en.wikipedia.org/wiki/Halogenation#Oxychlorination&quot;&gt;oxychlorination&lt;/a&gt; - in this reaction $\ce{HCl}$ and $\ce{O2}$ react with ethylene to form the monomer and water.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-08-17T22:47:14.797" LastActivityDate="2015-08-17T22:47:14.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35168" PostTypeId="2" ParentId="35166" CreationDate="2015-08-17T22:51:56.407" Score="10" Body="&lt;p&gt;This is simply an example of &lt;a href=&quot;https://en.wikipedia.org/wiki/Capillary_action&quot;&gt;capillary action&lt;/a&gt; expanded to a sheet of many tiny capillaries. In a narrow glass tube immersed in water, water will climb the inside of the tube to a level higher than that of the water outside because it is energetically favourable for the water to wet the surface due to adhesive forces and the cohesive forces of the water pull the bulk of the water up with the contact line of the water and the glass. Glass and water have strong adhesive forces as glass has many silicate groups which can interact with water through hydrogen bonding and other polar interactions. As the liquid rises, the volume of water above the surrounding liquid increases, thus increasing the weight of the column. When the weight of the column equals the adhesive and cohesive forces drawing the water upwards, the system is at equilibrium and the water will rise no further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be calculated from: $$h = \frac{\gamma\cos\theta}{\rho gr}$$ where h is the height of the column, $\gamma$ is the liquid-air surface tension, $\theta$ is the equilibrium contact angle between the glass and the water, $\rho$ is the density of water, g is the acceleration due to gravity, and r is the radius of the tube. As you can see, increasing the surface tension (which comes from water's cohesive forces) or decreasing the contact angle (the lower the contact angle, the stronger the adhesive forces between liquid and surface) will increase the height the column will reach, and conversely a denser liquid or larger tube will decrease the height (greater increase in weight for a given rise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't do the whole calculation but for a 1 mm radius glass tube with water ($\theta \approx 50°$, $\gamma = 72\ \mathrm{mN/m}$), the equilibrium height is 4.7 mm. Now, if we shrink the tube much further to approximate something like a pore in a piece of paper towel, we can make the liquid go much further. (it's obviously not perfectly cylindrical, but close enough for illustration and the cellulose-water contact angle seems to be close enough to glass' as cellulose's abundance of hydroxyl groups is similarly adhesive to water) At a diameter of 10 µm, we get a height of 47 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we expand this concept to a large sheet of interconnected pores, we find that capillary action can draw a relatively large volume of water across considerable distances, which in addition to being great for cleaning up household spills, opens the possibility of using paper as a substrate for analytical devices that don't need pumps, including the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lateral_flow_test&quot;&gt;lateral flow assay&lt;/a&gt; (most commonly seen as the pregnancy test stick) and &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac503968p&quot;&gt;paper microfluidic devices&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-08-17T22:51:56.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35169" PostTypeId="2" ParentId="35146" CreationDate="2015-08-17T23:00:19.437" Score="-3" Body="&lt;p&gt;Electrovalent bonds are easiest to identify. If a compound is made up of a metal and non-metal/non-metallic radical (like carbonate), then, 99.99% times, it contains electovalent bond. If a compound is made up of 2 or more non-metals/non-metallic radicals, then it contains covalent bond. Coordinate covalent bonds appear mostly with compounds containing Hydrogen element. To identify the coordinate covalent bonds, you can draw the branched structural formula of the compound and see if the shared pair of electrons are coming from the same molecule. &lt;/p&gt;&#xA;" OwnerUserId="19213" LastActivityDate="2015-08-17T23:00:19.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35170" PostTypeId="2" ParentId="34574" CreationDate="2015-08-17T23:03:20.157" Score="-1" Body="&lt;p&gt;After drawing the structure, you can basically check if the shared pair of electrons are coming from the same molecule. If the answer is yes, then the compound contains coordinate covalent bond.&lt;/p&gt;&#xA;" OwnerUserId="19213" LastActivityDate="2015-08-17T23:03:20.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35171" PostTypeId="2" ParentId="34925" CreationDate="2015-08-18T00:47:33.303" Score="1" Body="&lt;p&gt;Drawing common electron pairs like they appear in most organic molecules is only then relevant, when&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;they play a keyrole in the reaction process&lt;/li&gt;&#xA;&lt;li&gt;they should explain a special molecule structure&lt;/li&gt;&#xA;&lt;li&gt;they are of an unusual quantity and thus resulting in a (intermolecular) formal charge&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise especially molecules with a lot of atoms with unpaired electrons (e. g. saccharides) can get confusing. And in addition: they take more time to draw ;) &#xA;Of course when learning organic chemistry its good to think of them a lot, because of their high reaction potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can be a 100 % sure that when they would not be there in compound 1 that it would be highlighted somehow, because that would result in a very unstable state. &lt;/p&gt;&#xA;" OwnerUserId="19160" LastActivityDate="2015-08-18T00:47:33.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35173" PostTypeId="1" CreationDate="2015-08-18T02:09:06.347" Score="1" ViewCount="4068" Body="&lt;p&gt;This is something I have read on forums repeatedly, but I don't understand the mechanism behind how antibiotics lose their effectiveness. I understand how repeatedly thawing and freezing cells would damage them, but not how repeatedly thawing and freezing antibiotics make them lose effectiveness.&lt;/p&gt;&#xA;" OwnerUserId="17847" LastActivityDate="2015-09-21T06:47:26.850" Title="How does repeatedly freezing and thawing antibiotics decrease their antimicrobial effectiveness?" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35174" PostTypeId="1" AcceptedAnswerId="50346" CreationDate="2015-08-18T03:45:29.337" Score="4" ViewCount="3932" Body="&lt;p&gt;I have recently been trying to come up with a method to synthesize 2-methyl-2-butanol or &lt;em&gt;tert&lt;/em&gt;-amyl alcohol. I wish to avoid complex/expensive steps since as a home chemist/student, I don't have access to much money or equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I have done so far:&lt;/strong&gt; I noticed that methyl ethyl ketone has pretty much the correct structure, minus a methyl group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then realized this meant I could use a methylating agent to get the correct molecule. What I needed was to break the $\ce{C=O}$ bond from the carbonyl group and add a methyl group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, I found that either Grignard reagents or something like methyl lithium would do it since it would donate an electron pair to the carbon, and form a bond with the methyl. I would then have the 2-methyl-2-butanol as a conjugate base with $\ce{Li}$ ions present. I could then react them out with $\ce{HCl}$ to precipitate out the $\ce{Li}$ salt and add a hydrogen to the negatively charged oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My problem:&lt;/strong&gt; I don't know if this would work at all. I didn't find any documentation. Also, methyl lithium is &lt;strong&gt;very&lt;/strong&gt; expensive and difficult to make so I can't use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any other route suggestions? Would it be possible to make methyl sodium and if so, would it work the same? Is it possible to make Grignard reagents easily at home? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm a hobby chemist, so please be detailed with how you predicted the organic reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I don't necessarily care about what chemicals I use for the reaction, so if anyone can suggest a different starting chemical, I'd be glad to hear it. I was even thinking of using an alkene group of some kind and methylating the $\ce{C=C}$ bond, but I don't know for sure, as I'm pretty new to organic chem.&lt;/p&gt;&#xA;" OwnerUserId="19218" LastEditorUserId="5026" LastEditDate="2018-04-22T10:19:23.583" LastActivityDate="2018-04-22T16:33:03.437" Title="Synthesizing 2-methyl-2-butanol" Tags="&lt;organic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;synthesis&gt;&lt;alcohols&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35175" PostTypeId="1" AcceptedAnswerId="35187" CreationDate="2015-08-18T04:28:11.487" Score="2" ViewCount="1703" Body="&lt;p&gt;Specifically: Suppose I add equal parts (i.e. the same number of moles of each) sodium bicarbonate and potassium sulfate into a container of distilled water large enough to dissolve them completely. In a second container of water, I add potassium bicarbonate and sodium sulfate. Are these two containers now physically identical? (That is, is there any experiment that could tell them apart?) My understanding is that each now contains free-floating Na+, K+, S-, and HCO3- ions. And if I evaporate away all the water, what will be left behind in each container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I replace the potassium and/or sodium with, say, magnesium or calcium, and the bicarbonate and/or sulfate with chloride or citrate, do the same principles hold? (Assuming the resulting salts are water-soluble - some, like calcium carbonate, won't be.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not sure if the list of cations and anions above have much in common chemically. My interest relates to the fact that they can be added to water to simulate the chemical makeup and flavor of natural mineral spring water.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(PS: What's a more elegant way to say &quot;the same number of moles of each?&quot; If I say to mix things in equal parts, or in some ratio, I would normally specify whether I mean by mass, by volume or... how do I say &quot;by number of moles?&quot;)&lt;/p&gt;&#xA;" OwnerUserId="19221" LastActivityDate="2015-08-18T21:04:20.020" Title="Do (simple, inorganic) salts ionize completely in water when they dissolve?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35176" PostTypeId="2" ParentId="35164" CreationDate="2015-08-18T07:12:20.383" Score="3" Body="&lt;p&gt;Yes your logic is basically correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs free energy is really derived is from the Clausius in equality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets consider two cases so we can simplify this relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constant Pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$&#xA;and internal energy by: $$\Delta U = q + w$$&#xA;Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constant Volume&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a constant volume, there is no expansion work done, therefore $dU = dq$&#xA;Therefore by using the same processes as before, we are able to get the expression for Helmholtz energy, A, which is defined as: $$\Delta A = \Delta U - T\Delta S$$ Here, for a reaction to be spontaneous, $\Delta A$ must be less than 0&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-18T07:12:20.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35178" PostTypeId="2" ParentId="35161" CreationDate="2015-08-18T09:02:01.543" Score="4" Body="&lt;p&gt;A reaction being endothermic or exothermic depends on thermochemistry while the rate of a reaction depends on its chemical kinetics. Hence, there is no relationship between the activation energy of a reaction and its enthalpy change. Also there is no relationship between the change enthalpy and the rate of a reaction. A rate of reaction is dependent on the concentration of the reactants which are to a certain power which is multiplied by a certain constant, k, which is called the reaction rate constant. For example, for the general reaction: $$\ce{A + B -&amp;gt; C}$$ the rate of the reaction is given by the following equation: $$ rate = k[A]^a[B]^b$$&#xA;The reaction rate constant is not dependent on whether a reaction is exothermic or endothermic. However is does depend on the temperature and and also the activation energy of the reaction, given by the equation: $$k = Ae^\frac{-E_a}{RT}$$ So you are right to say that a reaction will be slower if it has a higher activation energy, but you are wrong to say that endothermic reactions have a higher activation energy than exothermic reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually several endothermic reactions that are fast, for example, dissolving common table salt (NaCl) occurs pretty quickly, but it is actually a endothermic reaction. Meanwhile, the spontaneously decomposition of diamond to graphite is actually a exothermic reaction, however it is extremely slow. It is so slow that it not even observed after several decades.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-18T09:02:01.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35180" PostTypeId="1" CreationDate="2015-08-18T09:58:29.313" Score="2" ViewCount="272" Body="&lt;p&gt;I am looking for ways of removing water hardness from tap water, specifically permanent hardness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried working on some methods, e.g., to perform precipitates using sodium hydroxide and calculated calcium hydroxide, but it seems that either my methods or something else is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please recommend me some methods (or resources) on how to remove the hardness from tap water.&lt;/p&gt;&#xA;" OwnerUserId="19056" LastEditorUserId="16622" LastEditDate="2015-08-18T12:54:06.567" LastActivityDate="2015-08-18T19:34:40.200" Title="How to remove tap water hardness?" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35182" PostTypeId="1" AcceptedAnswerId="35205" CreationDate="2015-08-18T11:35:51.987" Score="9" ViewCount="21334" Body="&lt;p&gt;I do not fully understand the idea of coupling and why chemically equivalent hydrogens do not couple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to work with the alkanes, so let's take propane first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fUR8f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUR8f.png&quot; alt=&quot;Propane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be a triplet signal due to the $\ce{CH3}$ groups in blue and green, due to the $n+1$ rule for neighbouring protons, so the neigbouring carbon for the two $\ce{CH3}$ groups would be the $\ce{CH2}$ group circle in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the $\ce{CH2}$ group in red would have a septet signal due to the two surrounding $\ce{CH3}$ groups is what I would predict.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving on to butane, this is where I get more confused:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KzDUC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KzDUC.png&quot; alt=&quot;Butane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the $\ce{CH3}$ group in blue, there would be a triplet signal due to the neighbouring $\ce{CH2}$ in red, and the same can be said for the $\ce{CH3}$ in black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then for the $\ce{CH2}$ groups, I am not exactly sure what would happen but here is what I think: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\ce{CH2}$ group in red will have a quartet signal on the NMR spectrum because equivalent hydrogens do not couple and so the $\ce{CH2}$ will not couple with the green $\ce{CH2}$ group. This means that only the $\ce{CH3}$ group will couple. Using the $n+1$ rule, a quartet will be present and the same can be said for the $\ce{CH2}$ group in green. However, I think that the red $\ce{CH2}$ will have a triplet of quartets or a multiplet if that is correct, with the same being said for the green $\ce{CH2}$ group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to extend this further to pentane, which I am also confused about. Can this be clarified with the two examples above so that I can attempt to do pentane myself?&lt;/p&gt;&#xA;" OwnerUserId="19227" LastEditorUserId="16683" LastEditDate="2016-11-24T19:51:50.487" LastActivityDate="2017-06-15T16:16:06.407" Title="1H (proton) NMR spectra for alkanes" Tags="&lt;organic-chemistry&gt;&lt;spectroscopy&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35184" PostTypeId="1" CreationDate="2015-08-18T12:11:22.657" Score="3" ViewCount="226" Body="&lt;p&gt;I'm interested to know how long it takes to change states of matter and why it takes the time that it does.  For example, what is the very quickest that we could freeze a packet of chicken breasts?  In the food industry, flash freezing is often used to prevent damage from occurring by freezing meat very quickly.  But why does it take the time that it does?  Would it be possible to freeze my packet of chicken in say, a nano second instead of a few minutes?&lt;/p&gt;&#xA;" OwnerUserId="19228" LastEditorUserId="10185" LastEditDate="2015-08-18T12:18:56.247" LastActivityDate="2015-08-19T21:33:22.313" Title="How long does it take to change states of matter?" Tags="&lt;physical-chemistry&gt;&lt;phase&gt;" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="35185" PostTypeId="1" CreationDate="2015-08-18T12:22:22.090" Score="2" ViewCount="11415" Body="&lt;p&gt;I'm in a textile unit, where we use different chemicals. My question is what type of material should I make my tank out of to store my hydrochloric acid?&lt;/p&gt;&#xA;" OwnerUserId="19229" LastEditorUserId="1389" LastEditDate="2015-08-18T13:54:45.383" LastActivityDate="2018-10-14T11:12:43.680" Title="How do I store HCl safely?" Tags="&lt;acid-base&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35186" PostTypeId="1" AcceptedAnswerId="35188" CreationDate="2015-08-18T12:25:49.613" Score="8" ViewCount="20625" Body="&lt;p&gt;I did an experiment on the effect of the distance between the electrodes on the rate of electroplating in an electrolytic cell (copper electrodes with copper sulfate electrolyte).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I expected, when the electrodes were closer, the current was greater and the rate of copper plating was also faster. But why does less distance increase the current? My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?). I'm probably overthinking this, but any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14994" LastActivityDate="2020-08-19T03:07:00.487" Title="Why does decreasing the distance between the electrodes in an electrolytic cell increase the rate of electroplating?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35187" PostTypeId="2" ParentId="35175" CreationDate="2015-08-18T13:02:16.130" Score="2" Body="&lt;p&gt;When a salt, no matter which salt, or any other polar compound is exposed to enough water, then its polarized molecule parts, depending on its polarity, become attracted to the oxygen or hydrogens of the water, which will result, starting with the outermost molecule layer, in the complete dissolution of the compound. Within this process both the positively and the negatively polarized sections will get surrounded by water molecules (= hydration shell) and thus shielded from one another. In this state the original compound(s) have (theoretically) disappeared completely and the two containers will indeed be identical. Possible back reactions will (mainly) result in the most energy efficient compound for the particular experimental conditions and have the same chance in both containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you now evaporate all the water you may get various compounds and mixtures depending on what was dissolved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'd say &quot;of equal molarity&quot; or &quot;they are equimolar&quot;, but my mother tongue isn't English, so there might be a more elegant way.&lt;/p&gt;&#xA;" OwnerUserId="19160" LastEditorUserId="9961" LastEditDate="2015-08-18T21:04:20.020" LastActivityDate="2015-08-18T21:04:20.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35188" PostTypeId="2" ParentId="35186" CreationDate="2015-08-18T13:23:26.517" Score="6" Body="&lt;p&gt;If the applied voltage is constant, lowering distance between electrodes leads to lowering the resistivity according to the:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R=\frac{l}{A}\rho$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lowering the rezistivity increases electric current (at constant voltage) defined by Ohm's law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I=\frac{V}{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Michael D. M. Dryden pointed out, this answer is simplified and do not describe a lot of mechanisms that occur in a real deposition.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-08-19T07:02:55.603" LastActivityDate="2015-08-19T07:02:55.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35189" PostTypeId="1" AcceptedAnswerId="35191" CreationDate="2015-08-18T14:15:11.370" Score="1" ViewCount="1490" Body="&lt;p&gt;I'm running a distillation reaction at under 100 °C with the heat coming from a hotplate (combined with magnetic stirrer).  Is a Florence flask the correct vessel to use on the hotplate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the small contact area of the flask relative to the surface area of the hotplate I'm wondering if it wouldn't be better to use a fully flat-bottomed flask on the hotplate, and if so what best practices are?  Regardless, I need to keep a stirrer running during the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I intend to have a Kjeldahl bulb on top of whatever flask before feeding the condenser.)&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorDisplayName="user7951" LastEditDate="2015-08-18T15:52:36.140" LastActivityDate="2020-04-03T18:34:24.473" Title="Correct flask for reacting and distilling on hotplate?" Tags="&lt;equipment&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35191" PostTypeId="2" ParentId="35189" CreationDate="2015-08-18T15:51:24.713" Score="2" Body="&lt;p&gt;At under $100^\circ \mathrm{C}$ The best will be to use water bath and round-bottom flask. Most magnetic stirrers have enough range (magnetic forces do not stop working if vessel is in couple of centimeters from hot plate, though you should use non-magnetic dish for bath itself). Even if magnet in hot plate is weak I would suggest against flasks with flat bottom, because bottom may fall off under heating.&lt;/p&gt;&#xA;" OwnerUserId="19201" LastActivityDate="2015-08-18T15:51:24.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35192" PostTypeId="1" CreationDate="2015-08-18T16:39:36.977" Score="3" ViewCount="31097" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A 100 ml $\ce{HCl}$ solution has a pH of $3.7$. You want the solution to be of pH 4.5. You have a solution of $10\ \mathrm M$ $\ce{NaOH}$. How much $\ce{NaOH}$ do you need to add to to the $100\ \mathrm{ml}$ solution of $\ce{HCl}$ to get a pH of 4.5?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I firstly calculated the concentration $\ce{HCl}$ in $100\ \mathrm{ml}$ solution = $10^{-3.7}$ = $2.0\times10^{-5}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$. After this I calculated the $\left[\ce{HCl}\right]$ of the new solution: $10^{-4.5}=3.16\times10^{-6}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$ of new solution. Now I'm stuck.&lt;/p&gt;&#xA;" OwnerUserId="19234" LastEditorUserId="5026" LastEditDate="2018-03-09T09:06:03.013" LastActivityDate="2019-02-28T00:50:42.837" Title="How do I calculate how much sodium hydroxide I need to add to get a new pH?" Tags="&lt;physical-chemistry&gt;&lt;ph&gt;&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35193" PostTypeId="2" ParentId="35186" CreationDate="2015-08-18T17:30:31.947" Score="7" Body="&lt;p&gt;It's a bit of an oversimplification to explain this with a simple invocation of Ohm's law. The thing to remember about electrochemistry is that electron transfer only happens at every close distances to the surface (&amp;lt;10 nm) and it's the potential difference between the electrode and this tiny region at the surface that is important. You might be applying 10 V between the two electrodes, but if you have a lot of solution resistance between them, the voltage near the surface may only be 1 V as the other 9 V are dissipated as heat in the electrolyte. Thus, by moving the electrodes closer, the resistance goes down and the voltage the electrode &quot;sees&quot; is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with this type of two-electrode cell is that many factors are being conflated. The current that is measured (the rate of electroplating) is dependant on what is happening at both electrodes as well as the resistance of the bulk solution. As the electrode voltage increases, the current usually increases too as increasing the voltage makes it easier for electrons to transfer between the electrode and the redox-active species in solution. Eventually, however, the rate will not be limited by electron transfer, but rather by how fast copper ions can move from the bulk of the solution to the active region near the electrode and a &quot;limiting current&quot; is reached. You can see this below as the voltage at an electrode is swept to higher values:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Mt9Mg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mt9Mg.png&quot; alt=&quot;limiting current&quot;&gt;&lt;/a&gt; (Source: Own work)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At low voltage, there isn't enough voltage for any electron transfer to occur, but as it increases along the x-axis, we see an increase in current until it reaches a peak and the current is limited by diffusion to the electrode. (the reason it goes down after this point rather than levelling out has to do with the time-scale of the experiment, but it's not important for this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the system is so complicated, when we're interested in examining the reactions that are happening at just one electrode without worrying about the other electrode or the solution resistance, an alternate setup is used. If we use a feedback amplifier (known as a potentiostat) to control the voltage applied between the two active electrodes using a third (reference) electrode to measure the voltage close to the surface of the electrode of interest, we can decouple the effects of the solution resistance and other electrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sidenote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You were actually onto something interesting here. You're right in that it takes a while for copper ions from one electrode to make their way to the other so most of the copper being deposited is coming from what was already in the solution (at least on relatively short time scales), but what would happen if we moved the electrodes &lt;em&gt;very&lt;/em&gt; close to one another? Eventually we would reach a situation where copper could dissolve from one electrode and be almost immediately deposited on the other and we would not see the limiting current behaviour as above.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-08-18T17:30:31.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35195" PostTypeId="2" ParentId="35182" CreationDate="2015-08-18T17:41:10.047" Score="1" Body="&lt;p&gt;Your analysis is mostly correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fUR8f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUR8f.png&quot; alt=&quot;Propane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue and green $\ce{-CH3}$ groups are chemically equivalent and so will produce a single triplet signal because they couple to the two hydrogens in the red $\ce{-CH2 -}$ group. The red $\ce{-CH2 -}$ group will produce a septet (split into 7) signal due to the six equivalent hydrogens adjacent to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IJ8eB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJ8eB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KzDUC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KzDUC.png&quot; alt=&quot;Butane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: This is actually a simplified view of butane's spectrum. See @long's answer for a more detailed analysis.&lt;/strong&gt; &lt;br&gt;&#xA;The black and blue $\ce{-CH3}$ groups are chemically equivalent and each couple to two equivalent protons, meaning that the two $\ce{-CH3}$ groups as a whole produce only one triplet signal. The red and green $\ce{-CH2 -}$ groups are chemically equivalent and so will not couple with each other, despite being adjacent, but each one will couple with the $\ce{-CH3}$ group next to it, producing a single quartet signal which represents both groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H1V3v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H1V3v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spectra predicted using the free tool found &lt;a href=&quot;http://www.nmrdb.org/new_predictor/index.shtml?v=v2.17.5&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastEditorUserId="10185" LastEditDate="2015-08-19T09:46:12.103" LastActivityDate="2015-08-19T09:46:12.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35196" PostTypeId="1" AcceptedAnswerId="35203" CreationDate="2015-08-18T17:46:47.927" Score="0" ViewCount="158" Body="&lt;p&gt;Ethanoic acid dissociates according to the following equation:&#xA;$$\ce{CH3COOH + H2 O -&amp;gt; CH3COO- + H3O+}$$&#xA;I want to know if it is a redox reaction or not. In the ethanoic acid I obtained that the oxidation number of carbon in it is 0 by calculation, but it is (-2). What is the error in my answer?&lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="10185" LastEditDate="2015-08-18T18:17:27.680" LastActivityDate="2015-08-18T21:38:37.397" Title="Checking if it is a redox reaction" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="35198" PostTypeId="1" CreationDate="2015-08-18T18:17:34.617" Score="7" ViewCount="4295" Body="&lt;p&gt;Everywhere on the internet I see that the  two $\mathrm pK_\mathrm a$'s of ketoconazole are 6.4 and 2.3, but I don't think that is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\mathrm pK_\mathrm a$ of imidazole is around 6.4 . But ketoconazole also has a piperazine ring and normally this $\mathrm pK_\mathrm a$ is around 9.8. There are electron acceptors but I still find 2.3 to be very low. Is this correct and which of the two nitrogens will be protonated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zyaHY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zyaHY.png&quot; alt=&quot;ketoconazole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19069" LastEditorDisplayName="user7951" LastEditDate="2015-09-28T23:40:44.717" LastActivityDate="2015-09-28T23:40:44.717" Title="pKa values for protonated nitrogens in ketoconazole" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35199" PostTypeId="2" ParentId="35196" CreationDate="2015-08-18T19:22:27.397" Score="0" Body="&lt;p&gt;Redox reactions involve increase in oxidation state of one chemical species and decrease in the oxidation state of the other through loss and gain of electrons respectively but since this does not occur here then it's not a redox reaction but rather water acts as a weak base to accept the proton donated by ethanoic acid&lt;/p&gt;&#xA;" OwnerUserId="19192" LastEditorUserId="10185" LastEditDate="2015-08-18T20:47:31.040" LastActivityDate="2015-08-18T20:47:31.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35201" PostTypeId="2" ParentId="35180" CreationDate="2015-08-18T19:34:40.200" Score="5" Body="&lt;p&gt;Without knowing what dissolved species are causing the hardness, it's a little difficult to know what to advise. Performing &lt;a href=&quot;http://chemwiki.ucdavis.edu/Analytical_Chemistry/Qualitative_Analysis&quot;&gt;qualitative analysis&lt;/a&gt; would likely help you a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You describe using &lt;a href=&quot;https://en.wikipedia.org/wiki/Lime_softening&quot;&gt;lime softening&lt;/a&gt; (i.e., using hydroxides) which can help in some cases. Usually the resulting hydroxide precipitates are pretty hard to filter, requiring a membrane or a very good filtration system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_softening&quot;&gt;Wikipedia&lt;/a&gt; notes, there are a few other reliable techniques:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reverse osmosis&lt;/li&gt;&#xA;&lt;li&gt;distillation&lt;/li&gt;&#xA;&lt;li&gt;ion exchange resins&lt;/li&gt;&#xA;&lt;li&gt;chelating ligands&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While the first three are definitely the best in terms of ultimate water quality, an easy thing to supplement lime softening would be to add various chelating ligands. For example, after you carefully filter the water from lime softening, maybe add citric acid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I'd suggest you do some sort of qualitative analysis (and pH) testing as you go. This is the best way to know what steps you'll still need.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-18T19:34:40.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35203" PostTypeId="2" ParentId="35196" CreationDate="2015-08-18T21:32:52.913" Score="1" Body="&lt;p&gt;You're getting 0 ox. state for carbon because of treating both carbons equally - 0 is mean average value from oxidation numbers of both carbons. The one in carboxylic group  has three bonds with more electronegative oxygen therefore has +3. The one in methyl group has -3 because it's connected with three hydrogens which have lower electronegativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure&lt;/a&gt; - there's example with acetic acid! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In organic chemistry trying to get average value of ox. state usually doesn't make much sense as it's rare to have all carbons equivalent. @Shafter gave you good advice - without knowing the structure you were making mistake.&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="9961" LastEditDate="2015-08-18T21:38:37.397" LastActivityDate="2015-08-18T21:38:37.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35204" PostTypeId="2" ParentId="35184" CreationDate="2015-08-18T22:52:32.800" Score="4" Body="&lt;p&gt;Part of the answer is that the rate of freezing depends on the heat of fusion of the material being frozen (relatively large for water, the chief ingredient of uncooked chicken), but smaller for other substances). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another piece of the answer is that an object can lose heat by convection, conduction, and/or radiation. You would increase the rate of heat loss by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) increasing the surface area of the chicken as much as possible (flattening it), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) placing it against a surface held as cold as possible and made of a material such as metal with high heat capacity; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) hitting it with a continuous blast of extremely cold air or gas at the highest possible wind speed (remember wind chill factor?) and maximizing radiant heat loss by previously having painted the chicken with flat, black paint and conducting the experiment in a mirrored, insulated chamber. Some unusual materials can be induced to cool by irradiation with a laser (this is how we have obtained the lowest recorded temperatures ever) but this will probably not work with chicken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think there is an absolute limit on cooling rate, just diminishing returns for increasing energy expenditure. But I may be wrong. I hope you pursue this until you get an answer that satisfies you. That is how new discoveries are made.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorDisplayName="user15489" LastEditDate="2015-08-19T21:33:22.313" LastActivityDate="2015-08-19T21:33:22.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35205" PostTypeId="2" ParentId="35182" CreationDate="2015-08-19T00:14:37.853" Score="13" Body="&lt;p&gt;Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact magnetically non-equivalent. It serves as an example of how all too often NMR is poorly taught at the introductory level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's clarify a minor point of confusion that is widely taught at introductory levels - that equivalent protons do not couple to each other. They do; the reason that the coupling is not seen is that the transitions are degenerate and so sit directly on top of each other (ie &lt;em&gt;not observable&lt;/em&gt;). Remember that the observed lines in a spectrum are the energy transitions between stable spin states - if the energy difference between two states is equivalent, then the lines in the spectrum are degenerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a better generalisation for interactions giving rise to splitting in a spectrum is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Coupling between magnetically equivalent nuclei cannot be detected&#xA;and does not affect the appearance of the spectrum&lt;/li&gt;&#xA;&lt;li&gt;Coupling between&#xA;chemically equivalent, but magnetically non-equivalent nuclei will&#xA;usually affect the appearance of the NMR spectrum; commonly seen in&#xA;second order effects like that of AA'BB' systems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Second, then, to see what a 1H spectrum of butane really looks like (well, a more realistic simulation at least):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a more realistic simulated spectrum of butane (at 400MHz), which takes into account the fact that the two methylene groups couple to each other, with an average coupling of about 6.6Hz. Complicated isn't it? Have a read of &lt;a href=&quot;http://u-of-o-nmr-facility.blogspot.com.au/2009/08/500-mhz-1-h-nmr-spectrum-of-butane.html&quot;&gt;Glenn Facey's page on his NMR blog&lt;/a&gt;, (an excellent NMR resource) where he discusses this very problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N9X2r.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N9X2r.png&quot; alt=&quot;Simulated 1H NMR spectrum of butane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is the case? Well, we need to look at what gives rise to magnetic non-equivalence in molecules. There is an excellent on-line resource to explain about symmetry and non-equivalence at &lt;a href=&quot;http://www.chem.wisc.edu/areas/reich/chem605/index.htm&quot;&gt;Hans Reich's NMR page&lt;/a&gt;. But for butane, let us consider the Newman projections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iUzWx.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iUzWx.png&quot; alt=&quot;Newman projections of butane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At no time is the coupling between H2 and H3 the same as H2' and H3. Either it is psuedo-trans or pseudo-cis, but never the same. So, H2 and H2' are non-equivalent, and this will result in non-degenerate transitions, giving rise to complicated splitting patterns in the NMR spectrum. This is the same for other straight chain alkanes, with decreasing complexity as the chain length grows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just to show what a real spectrum of butane looks like: reproduced from Tynkkynen &lt;em&gt;et al.&lt;/em&gt; (&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/mrc.3847/abstract&quot;&gt;Magn. Reson. Chem. 2012, 50, 598–607&lt;/a&gt;), showing the real and calculated (at 500MHz):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZhCR2.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZhCR2.png&quot; alt=&quot;Real spectrum of butane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2462" LastActivityDate="2015-08-19T00:14:37.853" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="35207" PostTypeId="2" ParentId="35198" CreationDate="2015-08-19T01:05:16.903" Score="7" Body="&lt;p&gt;The 6.4 pKa is close to the reported pKa of imidazole (6.95), so 6.4 is at the imidazole ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That leaves pKa 2.3 (or 3.64) for the other basic nitrogen. While that nitrogen is part of a piperazine, piperazine itself (pKa 9.8) isn't a good comparison. The nitrogen in question is also conjugated to an aromatic ring, so it would be better to look at aniline, with a pKa of 4.6. That's much closer to ketoconazole. Add in the inductive effect of the proximal amide, and a pKa in the 2-3 range makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KPIgK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KPIgK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many common pKa's can be found on &lt;a href=&quot;http://evans.harvard.edu/pdf/evans_pKa_table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prof. Dave Evans's pKa list&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2015-08-19T13:42:27.773" LastActivityDate="2015-08-19T13:42:27.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35208" PostTypeId="1" CreationDate="2015-08-19T01:58:06.463" Score="1" ViewCount="813" Body="&lt;p&gt;A question from the National Chemistry Olympiad (Germany-2008) goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Besides the composition of a substance the structure of its molecule or ion is of interest. To predict the structure you may use the electron repulsion theory&#xA;  (VSEPR).&#xA;  The following sulphur compounds are choosen as examples&#xA;  $\ce{SO2}$, $\ce{SO3(g)}$, $\ce{SOF2}$, $\ce{SO4^2-}$, $\ce{SF4}$, $\ce{SOF4}$, $\ce{SF6}$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b) Draw the structure of these species. Insert the free electron pairs of $\ce{S}$.&#xA;  Give the name of the geometrical shape of these compounds or ions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm only interested in the structure of $\ce{SO2}$, that is represented in the answer sheet as $\ce{O=S-O}$ instead of the one I believe is the correct one $\ce{O=S=O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="19244" LastEditorUserId="10185" LastEditDate="2015-08-19T09:39:12.050" LastActivityDate="2015-08-19T09:39:12.050" Title="What is the correct structure for sulfur dioxide?" Tags="&lt;inorganic-chemistry&gt;&lt;covalent-compounds&gt;&lt;lewis-structure&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2019-09-08T17:26:56.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="35210" PostTypeId="2" ParentId="35208" CreationDate="2015-08-19T03:30:59.497" Score="3" Body="&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_dioxide#Structure_and_bonding&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&#xA;It's because it can be thought as two interchangeable resonant structures. What happens in reality is that the electron is delocalized over the three atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/7/75/Sulfur-dioxide-resonance-2D.svg/360px-Sulfur-dioxide-resonance-2D.svg.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/7/75/Sulfur-dioxide-resonance-2D.svg/360px-Sulfur-dioxide-resonance-2D.svg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19222" LastActivityDate="2015-08-19T03:30:59.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35211" PostTypeId="2" ParentId="34919" CreationDate="2015-08-19T03:51:53.623" Score="1" Body="&lt;p&gt;Beerhunter is correct in his comment.  Water and methanol are mutually soluble in all proportions, and methanol has a lower density than water, so, if you add lots of methanol to the water, the density of the solution will be less than that of water.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-19T03:51:53.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35212" PostTypeId="1" CreationDate="2015-08-19T05:31:55.513" Score="22" ViewCount="13180" Body="&lt;p&gt;So a positive and a positive wave function create a bonding orbital where the probability of finding an electron is summed while a positive and a negative create an anti-bonding orbital with a lower electron probability in the region between them leading to a repulsion. My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I tried asking this question on physics.SE but it would seem that in physics you just don't talk about or use the wave function until after you square it)&lt;/p&gt;&#xA;" OwnerUserId="5023" LastEditorUserId="186" LastEditDate="2015-10-04T14:18:16.720" LastActivityDate="2017-12-22T15:17:29.023" Title="Physical intuition behind negative values for wave function?" Tags="&lt;bond&gt;&lt;quantum-chemistry&gt;&lt;orbitals&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="5" CommentCount="3" FavoriteCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="35213" PostTypeId="1" AcceptedAnswerId="136635" CreationDate="2015-08-19T06:21:03.140" Score="2" ViewCount="20068" Body="&lt;p&gt;I've recently moved to a old house and found some stuff in the attic, along that I found some really beautiful silverware(??) that unfortunately was VERY tarnished, one of the pieces was totally black while some had small places where I could see, in the back of one of them there's a label stating Silver 90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled and found that using a solution of hot water, baking soda, salt, vinegar and strips of aluminium foil would help clean them. I submerged them and after some time I took them out and started to scrub them and the gross black stuff was going out, showing some silver and some yellowish. I them took a sponge and kind got carried away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces now all have this yellowish stains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one that was completely black had no silver-color in any part of it after cleaning, except the bottom, while the rest is all yellow now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's one of the pieces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0IDde.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0IDde.jpg&quot; alt=&quot;front&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2Mw1L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Mw1L.jpg&quot; alt=&quot;back&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the back is a white paper, white-balance is crazy with this piece)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Is this yellow stain any kind of dirt that wasn't removed in the process, or is this the underlayer of the piece(meaning that the silver is gone in those parts)? And above all, is the method used the best way to remove dirt from silver?&lt;/p&gt;&#xA;" OwnerUserId="19248" LastEditorUserId="19248" LastEditDate="2015-08-19T08:22:42.297" LastActivityDate="2020-07-16T03:12:24.890" Title="Yellow stains after cleaning silver" Tags="&lt;cleaning&gt;&lt;silver&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35214" PostTypeId="2" ParentId="35192" CreationDate="2015-08-19T07:20:16.087" Score="5" Body="&lt;p&gt;At the beginning, the amount of substance of hydronium ions is: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$&lt;/span&gt;&#xA;After adding sodium hydroxide, the amount of substance of hydronium ions is: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$&lt;/span&gt;&#xA;The volume of sodium hydroxide to add is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="4945" LastEditDate="2019-02-28T00:10:52.267" LastActivityDate="2019-02-28T00:10:52.267" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35215" PostTypeId="2" ParentId="35212" CreationDate="2015-08-19T07:52:09.623" Score="23" Body="&lt;p&gt;The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum operator for the $n^\text{th}$ stationary state, you would get two solutions: $+\hbar k$ and $-\hbar k$, where $k = n\pi/L$. The plus and minus signs mean that there is a $50$% chance the particle could be moving from the left to the right and a $50$% chance of the particle moving from right to left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of summing of two wavefunctions to get the wavefunction of a molecular orbital, a negative wavefunction doesn't mean anything at all. The reason why when a positive and negative wavefunction is added, an antibonding orbital is formed is simply because when you add a positive and negative number, you get $0$ or a really small number. This is known as destructive interference.  Therefore the new wavefunction will contain a region where the y values equal $0$. So when you square this new wavefunction of the molecular orbital you are going to get that there is very 0 probability of finding the electron in that position (known as a nodal plane), and hence it is called an antibonding orbital.However note that just because there is negative region in the wavefunction, doesn't mean that it will always add to form an anti-bonding MO. For example, if you add 2 negative wavefunction values (known as constructive interference), you get a large negative value. When you square becomes a very large positive value. Hence this means that there is a large probability of the electron being in that region and hence it will be a bonding MO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ed7fU.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ed7fU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="16683" LastEditDate="2015-10-04T13:43:46.583" LastActivityDate="2015-10-04T13:43:46.583" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35216" PostTypeId="2" ParentId="35213" CreationDate="2015-08-19T08:10:36.977" Score="6" Body="&lt;p&gt;I suggest to do the following: &#xA;Line the bottom of a pan with aluminum foil. Put the silver piece on top of the aluminum foil. &lt;strong&gt;The silver piece and aluminum foil must be in contact with each other.&lt;/strong&gt;&#xA;To about 2L of hot boiling water add about a half cup of baking soda (Be careful!). Then add the mixture to the pan. Make sure to cover the whole silver piece. The tarnish will disappear within several minutes.&#xA;The chemistry underlying this treatment is the following:&#xA;The silver tarnish is the chemical product of the reaction of silver with sulfur-containing substances in the air, $\ce{Ag2S}$. Silver sulfide is black. In this process, we convert silver sulfide to silver according to the following reaction:&#xA;$$\ce{3Ag2S + 2Al-&amp;gt;Al2S3 +6Ag}$$ In fact, aluminum has a greater affinity to sulfur than silver.  The reaction between silver sulfide and aluminum takes place when the two are in contact while they are immersed in a baking soda solution. And it becomes faster when the solution is hot.The silver piece should touch the aluminum because a small electric current flows between them during the reaction. T&lt;strong&gt;he formed aluminum sulfide may adhere to the aluminum foil, or it may form tiny, pale yellow flakes in the solution&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of this method of cleansing is that we do not remove the tarnish with some silver. We simply transfer silver sulfide to silver.&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="402" LastEditDate="2017-10-04T18:38:18.697" LastActivityDate="2017-10-04T18:38:18.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35217" PostTypeId="1" CreationDate="2015-08-19T08:31:01.523" Score="1" ViewCount="245" Body="&lt;p&gt;The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law, $p\,V=n\,RT$, this comes down to&#xA;$$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$&#xA;We find this logarithm in many formulas in chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried and failed to find a general formula for $\Delta G$ for the Van der Waals gas (see &lt;a href=&quot;https://physics.stackexchange.com/questions/200735/how-to-compute-this-change-of-gibbs-energy-for-a-van-der-waals-gas&quot;&gt;this question&lt;/a&gt; on the physics board). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes me wonder:&#xA;Is there even another case of a chemical situation, other than the ideal gas, where $\int_{p_1}^{p_2}V(p)\,{\mathrm d}p$ or $\int_{V_1}^{V_2}p(V)\,{\mathrm d}V$ is known analytically?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.&lt;/p&gt;&#xA;" OwnerUserId="145" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-09-19T11:44:31.367" Title="What are analytical expressions for the isothermal change of Gibbs energy?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;gas-laws&gt;&lt;pressure&gt;&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35223" PostTypeId="1" AcceptedAnswerId="35228" CreationDate="2015-08-19T11:09:01.800" Score="5" ViewCount="918" Body="&lt;p&gt;I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct? Can atomic nuclei eject photons?&lt;/p&gt;&#xA;" OwnerUserId="19251" LastEditorUserId="10185" LastEditDate="2015-08-19T16:53:18.387" LastActivityDate="2015-08-19T16:53:18.387" Title="Are photons ejected from atomic nuclei?" Tags="&lt;atoms&gt;&lt;nuclear&gt;&lt;electromagnetic-radiation&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35224" PostTypeId="2" ParentId="34661" CreationDate="2015-08-19T11:34:21.523" Score="2" Body="&lt;p&gt;Addition to Dan Burden's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guanine can form a base-pair with Thymine or Uracil (2 hydrogen bonds). This is not generally observed in the DNA but is quite common in RNA helices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MTE92.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MTE92.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The non-Watson-Crick (WC) type base pairs or &lt;a href=&quot;https://en.wikipedia.org/wiki/Wobble_base_pair&quot; rel=&quot;nofollow noreferrer&quot;&gt;wobble-base pairs&lt;/a&gt; introduce distortion in the helix; they are also thermodynamically weaker than the WC pairs. However the GU pair, especially in RNA has a thermodynamic stability comparable to the WC pairs &lt;a href=&quot;http://dx.doi.org/10.1093%2Fembo-reports%2Fkvd001&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[1]&lt;/sup&gt;&lt;/a&gt;. The distortion that it imposes on the helix is also less than that of other wobble pairs. GT mismatch can actually be tolerated to some extent even in DNA in all its structural forms (A,B and Z) &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/3917021&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[2]&lt;/sup&gt;&lt;/a&gt;. GU pairs are however stabler in RNA and allow structural flexibility. GU pairs introduce a high electrostatic potential into the major groove which in turns increases its propensity for divalent metal ion binding &lt;a href=&quot;http://dx.doi.org/10.1093%2Fembo-reports%2Fkvd001&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[1]&lt;/sup&gt;&lt;/a&gt;. It is my guess that metal-binding may stabilize the structure. Water molecules are also shown to cluster around the GU pair and possible stabilize the it &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/4092690&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[3]&lt;/sup&gt;&lt;/a&gt;. Again it is a guess that this may be a resultant of the higher electrostatic potential in case of the GU wobble. &lt;/p&gt;&#xA;" OwnerUserId="5295" LastEditorUserId="5295" LastEditDate="2015-08-19T11:52:16.530" LastActivityDate="2015-08-19T11:52:16.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35226" PostTypeId="1" CreationDate="2015-08-19T14:37:16.850" Score="1" ViewCount="94" Body="&lt;p&gt;I am highly allergic to propylene glycol and polyethylene glycol.  I am finding many hidden sources of both and MANY synonyms for PG and PEG.  Is there a practical way to tell how closely related another chemical might be to my allergens using such resources as the U.S. Household Products data base?  For example:  glycerin's alternate name is propandiol.  Propandiol is also an alternate name for propylene glycol.  I'm very confused about sodium laureth sulfate in particular.  It has PEG listed as an alternate name, but is not an alcohol like the glycols.  Thank you for any clarification you can give me!&lt;/p&gt;&#xA;" OwnerUserId="19255" LastEditorUserId="21990" LastEditDate="2015-10-18T19:20:03.647" LastActivityDate="2015-10-18T19:20:03.647" Title="Confused by Alternate Names" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-10-19T05:33:34.140" ContentLicense="CC BY-SA 3.0" />
  <row Id="35228" PostTypeId="2" ParentId="35223" CreationDate="2015-08-19T14:48:03.873" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have heard that electrons absorb or eject photons when transitioning from one orbital to another. Is this correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not exactly.  The atom as a whole emits or absorbs the photon.  There is no reason to single out the electron versus the nucleus in such transitions. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can atomic nuclei eject photons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes there are two ways a nucleus in particular (as opposed to an atom taken as a whole) can emit photons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, nuclei can exist in &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_isomer&quot;&gt;excited states&lt;/a&gt;. When an excited nucleus transitions to the ground state, a photon in the gamma or X-ray range of the electromagnetic spectrum is emitted.  Electrons do not have an analogous mechanism for emitting photons, because electrons are fundamental particles, while nuclei are not.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, if a nucleus has a nuclear spin (S), then in a magnetic field there will exist  2S + 1 quantized energy states of the nuclei.  Transitions of nuclei between these states emit or absorb photons, the energy of which depends upon the strength of the field, for example in the radio frequency range.  This is the basis of &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance&quot;&gt;NMR&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_resonance_imaging&quot;&gt;MRI&lt;/a&gt;.  The electron has an analogous mechanism for emitting photons, known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance&quot;&gt;ESR or EPR&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5160" LastActivityDate="2015-08-19T14:48:03.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35230" PostTypeId="1" AcceptedAnswerId="35232" CreationDate="2015-08-19T17:39:01.790" Score="0" ViewCount="4860" Body="&lt;p&gt;When one chemical is added to another how would we come to know whether the compound dissolves in it or reacts with it? For example $\ce{AgNO3}$ dissolves in $\ce{CCl4}$ whereas it could also have been a double displacement reaction. And what is the major difference between the two  processes at atomic level?&lt;/p&gt;&#xA;" OwnerUserId="17736" LastEditorUserId="17736" LastEditDate="2015-08-19T18:01:30.457" LastActivityDate="2016-11-21T07:04:45.710" Title="What is the difference between reacting and dissolving?" Tags="&lt;solutions&gt;&lt;terminology&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35231" PostTypeId="2" ParentId="35226" CreationDate="2015-08-19T18:14:56.737" Score="0" Body="&lt;p&gt;You'll probably be better served asking an allergist about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While many things contain PG, that is not the same as a compound similar to PG, which could have PG (or part of its &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Union_of_Pure_and_Applied_Chemistry#Nomenclature&quot; rel=&quot;nofollow&quot;&gt;IUPAC name&lt;/a&gt;, propane-1,2-diol, not to be confused with propane-1,3-diol, which would be its &lt;em&gt;possibly harmless-to-you although structurally similar&lt;/em&gt; isomer) as part of its name &lt;em&gt;or&lt;/em&gt; could interact with your body in a similar way. Case in point, your confusion over sodium laureth sulfate, which is nothing like PG, although that's an irritant for &lt;em&gt;everybody&lt;/em&gt;. However, it looks like PG allergy is sometimes co-sensitized with corticosteroids - which are themselves structurally nothing like PG!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the confounding variable here is the way your body decides what will or will not set off your immune system, which can have very little to do with the chemical behavior of the &lt;em&gt;allergen&lt;/em&gt; and everything to do with what mechanisms your body is using to make that distinction.&lt;/p&gt;&#xA;" OwnerUserId="15506" LastActivityDate="2015-08-19T18:14:56.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35232" PostTypeId="2" ParentId="35230" CreationDate="2015-08-19T18:47:33.757" Score="0" Body="&lt;p&gt;CCl4 is a covalent compound (having no free ions). Thus it cannot enter into an ionic reaction/double displacement reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic Requisites for checking reactivity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reactivity_series&quot; rel=&quot;nofollow&quot;&gt;Reactivity Series&lt;/a&gt; (Elements higher up in the series replace the lower elements)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Covalent_Bonds_vs_Ionic_Bonds&quot; rel=&quot;nofollow&quot;&gt;Checking whether bonds are covalent or ionic&lt;/a&gt; (You might just need to remember which compounds are which type. Some compounds can have both type of bonds)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonds&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Ionic&lt;/em&gt; = &quot;Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions. These ions represent atoms that have lost one or more electrons (known as cations) and atoms that have gained one or more electrons (known as anions).&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ionic_bonding&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;br&gt;&#xA;Thus the elements can take or give electrons. In a solution they exist as ions.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Covalent&lt;/em&gt; = &quot;A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs and the stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Covalent_bond&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;&lt;br&gt;&#xA;Thus it has no free electrons to give or take.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="124" LastActivityDate="2015-08-19T18:47:33.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35237" PostTypeId="1" AcceptedAnswerId="35376" CreationDate="2015-08-19T20:40:18.587" Score="12" ViewCount="475" Body="&lt;p&gt;Rearrangement of cycloheptatrienyl carbene yields heptafulvene &lt;em&gt;via&lt;/em&gt; a hydrogen shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UHH5G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UHH5G.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that the reaction is favorable because there is a conjugative effect with the $\pi$ system of the ring in the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this rearrangement happen by a radical TS or is the mechanism more of a sigmatropic rearrangement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My arguments for&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;radical TS: The electron of the carbene (let's assume a singlet carbene here) lies close to the hydrogen, it could be a planar cyclic transition state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;[1,2]-Hydrogen shift: From my chemical intuition I think the $\ce{C-H}$ bond is more a polar bond and therefore I would hydrogen describe more as a proton (positive partial charge). I would say it's a [1,2] sigmatropic rearrangement, 2 electrons are involved, and therefore suprafacial and describing the carbene $\mathrm{sp^2}$ as HOMO and the hydrogen as LUMO.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What's the correct mechanism for this rearrangement?&lt;/p&gt;&#xA;" OwnerUserId="7201" LastEditorUserId="7475" LastEditDate="2018-02-20T09:06:40.547" LastActivityDate="2018-02-20T09:06:40.547" Title="What's the reaction type of the Hydrogen shift?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35238" PostTypeId="1" CreationDate="2015-08-19T20:48:28.120" Score="0" ViewCount="220" Body="&lt;p&gt;When a bottle of sports drink that contained sucrose is opened, is it expected that the sucrose contained therein will split into its constituents (fructose and glucose) more rapidly than it will &lt;em&gt;via&lt;/em&gt; acid hydrolysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that in acid hydrolysis there is nucleophilic substitution (provided the acid is protic). Does a similar phenomenon occur when the contents of the bottle are exposed to oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to interpret some unexpected results I obtained from an experiment, and I assume what I have detailed here is what happened. However, I cannot find any relevant information to confirm or deny my hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RESULTS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiment showed 0.72 g of glucose was present per 100 mL of solution. The bottle label states that it is 0.5 g per 100 mL. &lt;/p&gt;&#xA;" OwnerUserId="19175" LastEditorUserId="402" LastEditDate="2017-04-25T23:09:02.633" LastActivityDate="2017-07-28T07:37:40.633" Title="Does oxygen catalyze the decomposition of sucrose?" Tags="&lt;experimental-chemistry&gt;&lt;food-chemistry&gt;&lt;catalysis&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35241" PostTypeId="2" ParentId="6694" CreationDate="2015-08-19T22:42:27.413" Score="-8" Body="&lt;p&gt;My molecular models set has led me to believe that hydrogen peroxide can conform to form 2 bent hydrogen bonds and one bent covalent bond between the $\ce{OH}$ groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that there can be a triple bond between the 2 oxygen atoms. The triple bent bond (i.e., 3 banana bonds) of which 2 are of the hydrogen bond type and one is the Linus Pauling bent-covalent bond type, like in ethylene ($\ce{CH2CH2}$), tau bonds, banana bonds, bent bonds, and indirect angular bonds.&lt;/p&gt;&#xA;" OwnerUserId="19266" LastEditorUserId="865" LastEditDate="2015-08-20T00:12:07.893" LastActivityDate="2015-08-20T00:12:07.893" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="35243" PostTypeId="2" ParentId="35217" CreationDate="2015-08-20T02:10:52.383" Score="2" Body="&lt;p&gt;With regard to the Van der Waals equation, have you considered integrating by parts:  $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$.  That can certainly be integrated analytically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will be in terms of $V$, but that’s not a big limitation.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorDisplayName="user7951" LastEditDate="2015-08-20T11:17:50.833" LastActivityDate="2015-08-20T11:17:50.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35245" PostTypeId="1" CreationDate="2015-08-20T05:30:19.083" Score="11" ViewCount="2464" Body="&lt;p&gt;When hydrogen peroxide acts as an oxidizer it itself is reduced. Broadly this means that the peroxide bond is broken and that water is formed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why is there a tendency for hydrogen peroxide to be such a potent oxidizer in acidic media? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In acidic solutions, H2O2 is one of the most powerful oxidizers known—stronger than chlorine, chlorine dioxide, and potassium permanganate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why is it not as potent without added acid, i.e. 3% hydrogen peroxide, the stuff you find in medicine cabinets, isn't a very good oxidizer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;duh&quot; answer is that hydrogen peroxide is especially unstable in acidic media. But why can't acidic media stabilize hydrogen peroxide, and specifically the peroxide bond? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple electrostatics tells me that having a bunch of positively charged hydronium ions around the positively polarized oxygens in the peroxide bond isn't a &quot;good&quot; situation. Of course there are anions from the dissociation of the strong acid but these anions necessarily have diffuse negative charge, which is probably not enough to stabilize the polarized peroxide bond to significant extent. Is there anything else? &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-08-21T20:44:58.310" LastActivityDate="2015-09-21T21:57:32.717" Title="Why is acidic hydrogen peroxide a potent oxidizer?" Tags="&lt;inorganic-chemistry&gt;&lt;redox&gt;&lt;stability&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35246" PostTypeId="1" AcceptedAnswerId="35249" CreationDate="2015-08-20T05:59:22.717" Score="6" ViewCount="986" Body="&lt;p&gt;I have been asked to synthesize the compound (&lt;em&gt;E&lt;/em&gt;)-5-methylhex-3-en-2-one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NzM89.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NzM89.png&quot; alt=&quot;Target compound&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be achieved using the Wittig reaction?&lt;/p&gt;&#xA;" OwnerUserId="16368" LastEditorUserId="16683" LastEditDate="2017-08-15T12:14:01.710" LastActivityDate="2017-08-15T12:19:12.670" Title="How can I synthesize an (E)-configured enone using the Wittig reaction?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;carbonyl-compounds&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35249" PostTypeId="2" ParentId="35246" CreationDate="2015-08-20T07:26:41.877" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Wittig_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wittig reaction&lt;/a&gt; is used to synthesise double bonds usually from an aldehyde/ketone and an alkyl halide. The following scheme might be one way to achieve your target molecule. Most likely you will have to modify the Wittig reaction, because it is usually (&lt;em&gt;Z&lt;/em&gt;)-selective. See also: &lt;a href=&quot;https://chemistry.stackexchange.com/q/5313/4945&quot;&gt;Why does the unstabilised Wittig reaction selectively form cis alkenes?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider &lt;a href=&quot;http://www.chemspider.com/Chemical-Structure.857.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;methylglyoxal&lt;/a&gt; no good starting material, then you might need to come up with a synthesis for this, too. For the sake of selectivity, you might need protection groups along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CIUGE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CIUGE.jpg&quot; alt=&quot;proposed retro synthesis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is only one possibility and I am not sure if it actually works, or if the Wittig reaction is the right tool to employ. I am certain, that there are a many other ways of synthesising your target molecule and an extensive literature search would be appropriate.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16683" LastEditDate="2017-08-15T12:16:29.403" LastActivityDate="2017-08-15T12:16:29.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35251" PostTypeId="1" AcceptedAnswerId="36903" CreationDate="2015-08-20T07:33:01.180" Score="10" ViewCount="1777" Body="&lt;p&gt;I'm curious about what the minimal conditions for some collection of water molecules to be liquid water are. What is the minimum number? What sort of bonds must hold? What other sorts of things I don't know to ask about need to be true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my peeking around a subject I have little familiarity with, I've seen mention of a tetrahedron, suggesting a lower bound of 4 molecules for water to be a liquid. But then I also stumbled across the following (from p. 23, &lt;a href=&quot;http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000003174/2_Chap1.pdf;jsessionid=7E21A159207B717530D71E027232D07F?hosts=:&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whatever the model concerned, the presence of a fifth&#xA;  water molecule is necessary to be considered in explaining&#xA;  the molecular mechanism of  defect formation and Hbond&#xA;  network rearrangement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems that although only four molecules are in this bond network undergoing constant change, there must actually be a fifth &quot;neighbor&quot; molecule for the bonds to function appropriately. Is this fifth molecule part of the liquid, or merely a helpful neighbor who makes the four molecules' liquification possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, what makes water a liquid (when it is)?&lt;/p&gt;&#xA;" OwnerUserId="19273" LastEditorUserId="19273" LastEditDate="2015-08-20T08:42:07.390" LastActivityDate="2018-02-14T04:16:20.410" Title="What makes water a liquid (when it is)?" Tags="&lt;water&gt;&lt;molecules&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35257" PostTypeId="1" AcceptedAnswerId="35259" CreationDate="2015-08-20T12:25:36.013" Score="1" ViewCount="6180" Body="&lt;p&gt;The mass of one molecule of a compound is $2.19\times10^{-22}\ \mathrm g$. What is the molar mass of the compound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&#xA;Using the formula &lt;strong&gt;moles=mass/molar mass&lt;/strong&gt;, I found how to calculate molar mass using the formula &lt;strong&gt;molar mass=moles/mass&lt;/strong&gt;.&#xA;Therefore, I was under the impression that, surely, $6.02\times10^{23}$ divided by the mass of the compound would have resulted in the molar mass. However, the answer was wrong.&lt;/p&gt;&#xA;" OwnerUserId="19280" LastEditorUserId="7448" LastEditDate="2015-08-20T16:52:36.137" LastActivityDate="2015-08-20T16:52:36.137" Title="Calculate molar mass of a compound from a given molecular mass" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35258" PostTypeId="1" AcceptedAnswerId="35261" CreationDate="2015-08-20T12:39:54.470" Score="1" ViewCount="268" Body="&lt;p&gt;It is mentioned in my textbook:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;&quot;Hydrogen atoms&lt;/em&gt; of water molecule create &lt;strong&gt;strong electrical field&lt;/strong&gt; due to their small sizes&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How is the size related to electrical field and its strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little touch of physics will be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="19282" LastEditorUserId="19282" LastEditDate="2015-08-20T15:20:12.010" LastActivityDate="2015-08-20T15:20:12.010" Title="Hydrogen bonding" Tags="&lt;water&gt;&lt;hydrogen&gt;&lt;hydrogen-bond&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35259" PostTypeId="2" ParentId="35257" CreationDate="2015-08-20T12:59:49.680" Score="1" Body="&lt;p&gt;To get the molar mass of your mystery compound, make use of the fact that there are $N_{\rm A}$ (Avogadro's number) of molecules of it in one mole of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the problem, we multiply the molecular mass (given) by Avogadro's number (known) to get the mass of one mole of the mystery compound:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\rm molar \;mass} = N_{\rm A}\times 2.19\times 10^{-22}\,{\rm g} = (6.022\times 10^{23}\,{\rm mol}^{-1}) \times (2.19\times 10^{-22}\,{\rm g}) = 131.88\;{{\rm g}\over{\rm mol}}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-20T12:59:49.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35260" PostTypeId="1" AcceptedAnswerId="46321" CreationDate="2015-08-20T13:03:33.447" Score="5" ViewCount="421" Body="&lt;p&gt;6 years ago, I needed to replace the main water shut off valve to my house.  This required the city to turn off the water at the street. Incompetent city employee drives key through water pipe. Plumbers fix pipe, I realize that pipe is lead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yearly, the city (90% of which was built before 1950) sends out a water quality report, which reports undetectable amounts of lead.  That is interesting since probably 90% of the city has lead pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several questions : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How is lead pipe safe from leaching lead into water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do cities usually treat the water to prevent lead from leaching into the water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the form of harmful lead (e.g. is it in aqueous solution, lead oxide, or forming  ligands with some other molecule)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19284" LastEditorUserId="19284" LastEditDate="2015-08-20T13:34:41.337" LastActivityDate="2016-02-15T13:11:47.860" Title="I drink city water, city has lead pipe, how is this safe?" Tags="&lt;water&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35261" PostTypeId="2" ParentId="35258" CreationDate="2015-08-20T14:17:29.297" Score="1" Body="&lt;p&gt;The magnitude of the electric field is basically given by Coulomb's law. In SI units, the electric field $E$ at the distance $r$ from the charge $Q$ is&#xA;$$ E = \frac{Q }{4\pi \epsilon_0 r^2} $$&#xA;The sentence only says the simple point that if $r$ (distance from the charge, in this case the proton i.e. nucleus) which is squared in the denominator is short, then the electric field will be large (strong) because it's the inverse square law (involving indirect proportionality).&lt;/p&gt;&#xA;" OwnerUserId="1913" LastActivityDate="2015-08-20T14:17:29.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35262" PostTypeId="2" ParentId="35260" CreationDate="2015-08-20T14:43:39.810" Score="3" Body="&lt;p&gt;Lead pips are a common problem. Usually, it is safe to drink water from it. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So long as water is flowing, there will not be unsafe (or even detectable) levels of lead in the water. - True remark by  @Some_Guy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The problems accrue when the pipeline is not used for longer time. The lead concentration arises and can reach harmful levels. There is a possibility to install filters to remove lead from the water. This &lt;a href=&quot;http://www.unitedutilities.com/reduce-lead-intake.aspx&quot; rel=&quot;nofollow&quot;&gt;site&lt;/a&gt; is quite useful and provide more information.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lead is contaminating water if the pipeline is made of lead.&lt;/li&gt;&#xA;&lt;li&gt;I am not aware of any method that can prevent lead leaching.&lt;/li&gt;&#xA;&lt;li&gt;Over longer time oxides can be formed:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\ce{2Pb(s) + O2(g) + 2H2O(l) -&amp;gt; 2 Pb(OH)2(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{PbCO3}$ or $\ce{Pb(CO3)2^2-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elementary lead does not dissolve in water under normal conditions.&#xA;Lead compounds are generally soluble in soft, slightly acidic water.&#xA;Lead binds to sulphur ($\ce{S2-}$), or to phosphor ($\ce{PO4^3-}$).&#xA;Reference &lt;a href=&quot;http://www.lenntech.com/periodic/water/lead/lead-and-water.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4097" LastEditorUserId="4097" LastEditDate="2015-08-20T14:53:50.853" LastActivityDate="2015-08-20T14:53:50.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35263" PostTypeId="1" CreationDate="2015-08-20T14:46:08.450" Score="3" ViewCount="61" Body="&lt;p&gt;In reading a text about $\ce{CO2}$ recycling and sequestration of various kinds, I came to think about storing atmospheric $\ce{CO2}$ as asphalt. Given cheap energy as in the case of Iceland, and improved atmospheric carbon capture technologies, could this be feasible? Is there a name for the reverse cracking process?&lt;/p&gt;&#xA;" OwnerUserId="13341" LastEditorUserId="7448" LastEditDate="2015-08-20T16:53:50.103" LastActivityDate="2015-08-20T16:53:50.103" Title="Atmospheric carbon removal: asphalt from CO2?" Tags="&lt;organic-chemistry&gt;&lt;environmental-chemistry&gt;" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="35265" PostTypeId="1" AcceptedAnswerId="35408" CreationDate="2015-08-20T14:59:35.240" Score="7" ViewCount="1249" Body="&lt;p&gt;Could someone help me please (preferably talk me through not just provide the answer) with the IUPAC name of phthalic anhydride?  I'm sorry for not posting attempts but I'm afraid I have absolutely no idea how to begin here...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ttxtim.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ttxtim.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19231" LastActivityDate="2015-08-23T20:05:12.790" Title="IUPAC name, phthalic anhydride" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35266" PostTypeId="1" AcceptedAnswerId="35303" CreationDate="2015-08-20T15:05:50.257" Score="4" ViewCount="21602" Body="&lt;p&gt;This is a question that had me puzzled for quite a while.  I feel that information is missing.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A weak acid, &lt;span class=&quot;math-container&quot;&gt;$\ce{HA}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K_\textrm{a} = \pu{1.0E-4 M}$&lt;/span&gt;, is titrated with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;.  The concentration of &lt;span class=&quot;math-container&quot;&gt;$\ce{NaA}$&lt;/span&gt; at the equivalence point is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.010 M}$&lt;/span&gt;.  What is the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; at the equivalence point?&lt;br /&gt;&#xA;a) 4.0&lt;br /&gt;&#xA;b) 6.0&lt;br /&gt;&#xA;c) 7.0&lt;br /&gt;&#xA;d) 8.0&lt;br /&gt;&#xA;e) 11.0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The correct answer is :&lt;/p&gt;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;&lt;p&gt; d) 8.0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer is fairly easy to guess, as the &lt;span class=&quot;math-container&quot;&gt;$\mathrm{pH}$&lt;/span&gt; needs to be greater than 7, but 11 is much too basic.  However, I'm not sure how I would be able to figure that out from the information given?  I've tried using the Henderson-Hasselbalch equation, but I can't figure out the relationship between &lt;span class=&quot;math-container&quot;&gt;$\ce{[NaA]}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\ce{[A-]}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[HA]}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="69830" LastEditDate="2021-05-04T14:41:58.260" LastActivityDate="2021-05-04T14:41:58.260" Title="Calculating pH: Weak Acid, Strong Base" Tags="&lt;acid-base&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35267" PostTypeId="1" AcceptedAnswerId="35271" CreationDate="2015-08-20T15:10:17.553" Score="3" ViewCount="3539" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In the reaction in equilibrium&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2HI &amp;lt;=&amp;gt; H2 + I2}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The value &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$0.0198$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{721 K}$&lt;/span&gt;. Calculate &lt;span class=&quot;math-container&quot;&gt;$\ce{[HI]}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\pu{721 K}$&lt;/span&gt; if &lt;span class=&quot;math-container&quot;&gt;$\ce{[H2]} = \pu{0.0120 M}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{[I2]} = \pu{0.0150 M}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We use an ICE box. Since &lt;span class=&quot;math-container&quot;&gt;$K &amp;lt; 1$&lt;/span&gt;, the equilibrium is to the left. We are spending products to create reagents, so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix}&amp;amp;\ce{[HI]} &amp;amp; \ce{[H2]} &amp;amp; \ce{[I2]}\\&#xA;\text{Initial} &amp;amp; 0 &amp;amp; 0.0120 &amp;amp; 0.0150\\&#xA;\text{Change} &amp;amp; +2x &amp;amp; -x &amp;amp; -x\\&#xA;\text{Equilibrium} &amp;amp; 2x &amp;amp; 0.0120 - x &amp;amp; 0.0150 - x&#xA;\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equilibrium is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x = 0.01907$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{[HI]} = \pu{0.0381 M}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the options are&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.00211M}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.133M}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.0334M}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.006567M}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pu{0.0953M}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While it is pretty close to &lt;span class=&quot;math-container&quot;&gt;$\pu{0.0334M}$&lt;/span&gt;, I can't see how could I have made my calculation any more accurate (or if they were the ones who were inaccurate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was my procedure correct?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="83758" LastEditDate="2020-03-01T18:21:03.300" LastActivityDate="2020-03-01T18:21:03.300" Title="Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI &lt;-&gt; H2+I2" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35268" PostTypeId="1" AcceptedAnswerId="35270" CreationDate="2015-08-20T15:26:47.707" Score="2" ViewCount="976" Body="&lt;p&gt;Which is preferred according to &quot;strict&quot; IUPAC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-pentanone or pentan-2-one ?&lt;/p&gt;&#xA;" OwnerUserId="19231" LastActivityDate="2016-08-22T12:07:47.450" Title="IUPAC nomenclature: putting the number before or in the middle of the name" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2016-08-26T21:40:43.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="35269" PostTypeId="1" AcceptedAnswerId="35275" CreationDate="2015-08-20T15:27:16.883" Score="4" ViewCount="241" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Have an hypothetical equilibrium reaction:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\ce{2 A (s) + x B (g) &amp;lt;=&amp;gt; 3 C (g)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;With the constants&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{align}&#xA;K_\mathrm{p} &amp;amp;= 0.0105 \\&#xA;K_\mathrm{c} &amp;amp;= 0.45&#xA;\end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;both at $\pu{250 ^\circ C}$. What is the value of $x$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since $\ce{A}$ is solid, I guess we can ignore it, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{x B(g) &amp;lt;=&amp;gt; 3 C (g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $K_\mathrm{c}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.0105 = \frac{\left[\ce{C}\right]^3}{\left[\ce{B}\right]^x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[\ce{B}\right]^x = \frac{\left[\ce{C}\right]^3}{0.0105}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to solve this, I need the concentration of $\ce{C}$. How can I find it out?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="41328" LastEditDate="2017-12-14T05:45:59.747" LastActivityDate="2017-12-14T05:45:59.747" Title="Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product" Tags="&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35270" PostTypeId="2" ParentId="35268" CreationDate="2015-08-20T15:36:20.533" Score="5" Body="&lt;p&gt;The name “2-pentanone” and “pentan-2-one” describe the same compound. The name “2-pentanone” is in accordance with obsolete IUPAC Recommendations. The relevant rule of the IUPAC Nomenclature was changed in the 1993 IUPAC Recommendations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;R-0.1.2&lt;/strong&gt; Position of locants&lt;/p&gt;&#xA;&lt;p&gt;Locants (numerals and/or letters) are placed immediately before the part of the name to which they relate, except in the case of traditional contracted forms (see R-2.5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.3.2&lt;/strong&gt; Position of locants&lt;/p&gt;&#xA;&lt;p&gt;Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hex-2-ene (not 2-hexene)&lt;/li&gt;&#xA;&lt;li&gt;cyclohex-2-en-1-ol (not 2-cyclohexen-1-ol)&lt;/li&gt;&#xA;&lt;li&gt;pentan-2-one (not 2-pentanone)&lt;/li&gt;&#xA;&lt;li&gt;(propan-2-yl)benzene&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Examples for traditional contracted names with locants at the front of the name:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2-naphthyl (not naphth-2-yl), the preferred name is naphthalen-2-yl&lt;/li&gt;&#xA;&lt;li&gt;2-pyridyl (not pyrid-2-yl), the preferred name is pyridin-2-yl&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2016-08-22T12:07:47.450" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35271" PostTypeId="2" ParentId="35267" CreationDate="2015-08-20T15:37:31.553" Score="3" Body="&lt;p&gt;You're close. Using the expression for &lt;span class=&quot;math-container&quot;&gt;$K_c$&lt;/span&gt;, you just need to solve for &lt;span class=&quot;math-container&quot;&gt;$[\ce{HI}]$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K_c = {[\ce{H2}][\ce{I2}]\over[\ce{HI}]^{2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.0198 = {(\pu{0.012 M})(\pu{0.015 M})\over[\ce{HI}]^{2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\ce{HI}] = \sqrt{{(\pu{0.012 M})(\pu{0.015 M})\over 0.0198}} = \pu{0.0953 M}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="83758" LastEditDate="2020-03-01T15:46:19.617" LastActivityDate="2020-03-01T15:46:19.617" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35272" PostTypeId="2" ParentId="35246" CreationDate="2015-08-20T15:42:15.733" Score="5" Body="&lt;p&gt;The Wittig reaction allows for the construction of alkenes from a phosphorus ylide and an aldehyde (or ketone). Shown in the retrosynthetic sense, keeping in mind that there are two possible Wittig disconnections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gwulc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gwulc.png&quot; alt=&quot;Two options for Wittig disconnections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this idea to (&lt;em&gt;E&lt;/em&gt;)-5-methylhex-3-en-2-one gives the following possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7J9IU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7J9IU.png&quot; alt=&quot;Disconnections for specific target material&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which option is chosen often comes down to which gives the most readily accessible starting materials. However, in this case, the stereochemical outcome is critical, and that is controlled by the nature of the ylide used. Looking now in the forward direction, the top option uses an unstabilized ylide, which tend to give the &lt;em&gt;cis&lt;/em&gt;-alkene as the major product. The bottom option uses a stabilized ylide, which changes the stereochemical outcome to favor the &lt;em&gt;trans&lt;/em&gt;-alkene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SeB4J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SeB4J.png&quot; alt=&quot;Stereoselectivity of Wittig reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, many chemists would opt for the &lt;a href=&quot;https://en.wikipedia.org/wiki/Horner%E2%80%93Wadsworth%E2%80%93Emmons_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Horner–Wadsworth–Emmons variant&lt;/a&gt;, which gives the &lt;em&gt;trans&lt;/em&gt;-alkene as the major product and avoids the production of triphenylphosphine oxide, a byproduct that can be difficult to remove.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="16683" LastEditDate="2017-08-15T12:19:12.670" LastActivityDate="2017-08-15T12:19:12.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35273" PostTypeId="1" AcceptedAnswerId="35277" CreationDate="2015-08-20T15:43:15.440" Score="28" ViewCount="16221" Body="&lt;p&gt;Why are there only 7 types of unit cells and 14 types of Bravais lattices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading about solid-state chemistry for the first time and this limitation made no sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do the math and realized that there could be many more possibilities. Usually, the standard unit cell is described on the basis of whether the sides are at 90 degrees to each other and whether the sides are equal or not. However, any angle other than 90 could be chosen as the reference angle. So why 90?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because only these 14 systems are found in nature and there is no need to study others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, why are there only 4 sided shapes or hexagonal shapes?&#xA;Why not pentagonal or trigonal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any reference from any site or textbook would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="402" LastEditDate="2019-08-21T19:47:14.040" LastActivityDate="2019-08-21T19:47:14.040" Title="Why are there only 7 types of unit cells and 14 types of Bravais lattices?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="35275" PostTypeId="2" ParentId="35269" CreationDate="2015-08-20T16:26:53.913" Score="2" Body="&lt;p&gt;Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve we take the natural logarithm of both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$&#xA;$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which does turn out to be correct.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="41328" LastEditDate="2017-12-14T05:42:24.427" LastActivityDate="2017-12-14T05:42:24.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35276" PostTypeId="1" CreationDate="2015-08-20T16:54:09.083" Score="0" ViewCount="616" Body="&lt;p&gt;If I mix monopotassium phosphate, magnesium sulfate, and calcium nitrate together in aqueous solution, what chemicals will result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these safe to combine (i.e. any toxic gases, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20091014231235AAhb7jw&quot; rel=&quot;nofollow noreferrer&quot;&gt;this answer&lt;/a&gt; about just magnesium sulfate and calcium nitrate was insightful.&lt;/p&gt;&#xA;" OwnerUserId="11385" LastEditorUserId="402" LastEditDate="2017-04-25T23:08:20.097" LastActivityDate="2017-04-25T23:08:20.097" Title="What results from combining monopotassium phosphate, magnesium sulfate and calcium nitrate?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;safety&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35277" PostTypeId="2" ParentId="35273" CreationDate="2015-08-20T17:00:08.207" Score="17" Body="&lt;p&gt;All quotes will be from Solid State Physics by Ashcroft and Mermin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bravais Lattice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A fundamental concept in the description of any crystalline solid is that of the &lt;em&gt;Bravais lattice&lt;/em&gt;, which specifies the periodic array in which the repeated units of the crystal are arranged.  The units themselves may be single atoms, groups of atoms, molecules, ions, etc., but the Bravais lattice summarizes only the geometry of the underlying periodic structure, regardless of what the actual units may be.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Primitive Unit Cell:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A volume of space that, when translated through all the vectors in a Bravais lattice, just fills all of space without either overlapping itself or leaving voids is called a &lt;em&gt;primitive cell&lt;/em&gt; or &lt;em&gt;primitive unit cell&lt;/em&gt; of the lattice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unit Cell; Conventional Unit Cell:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One can fill space up with nonprimitive unit cells (known simply as &lt;em&gt;unit cells&lt;/em&gt; or &lt;em&gt;conventional unit cells&lt;/em&gt;). A unit cell is a region that just fills space without any overlapping when translated through some &lt;em&gt;subset&lt;/em&gt; of the vectors of a Bravais lattice. The conventional unit cell is generally chosen to be bigger than the primitive cell and to have the required symmetry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Crystal Structure:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A physical crystal can be described by giving its underlying Bravais lattice, together with a description of the arrangement of atoms, molecules, ions, etc. within a particular primitive cell.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, one comes up with 14 Bravais lattices from symmetry considerations, divided into 7 crystal systems (cubic, tetragonal, orthorhombic,monoclinic, triclinic, trigonal, and hexagonal). This comes solely by enumerating the ways in which a periodic array of points can exist in 3 dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is on those points is a unit cell, which will itself have some symmetry. Thus, the combination of Bravais lattice and unit cell symmetry can again be enumerated and one comes up with 230 space groups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for some of your related questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All cubic-related Bravais lattices will have 90 degree angles because they are based on cubic symmetry.  The trigonal Bravais lattice has no 90 degree angles, but isn't talked about much in more basic textbooks because, well, it looks weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why no pentagonal unit cells? Well, because you can't fill space with a 5-fold symmetric Bravais lattice. Quasicrystals, while they have 5-fold symmetry, are a tiling through space that does not obey the rules for a Bravais lattice.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-08-20T17:00:08.207" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="35278" PostTypeId="1" CreationDate="2015-08-20T17:17:57.487" Score="1" ViewCount="5338" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Mercury is the only metal that is liquid at room temperature. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that metallic bonding exists between metal atoms but my knowledge is limited to metallic bonding found in solids. Will there be metallic bonding between $\ce{Hg}$ atoms in the liquid phase?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Moreover, there will be London dispersion forces between the $\ce{Hg}$ molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Does mercury liquid only have metallic bonding as the intramolecular force and London dispersion force as the intermolecular force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If liquid mercury (with both metallic bonding and LDF) is compared with a hypothetical liquid that has hydrogen bonding along with LDF, will the boiling point of mercury be higher than the aforementioned hypothetical liquid?&lt;/p&gt;&#xA;" OwnerUserId="19293" LastEditorUserId="11367" LastEditDate="2017-03-25T05:04:06.723" LastActivityDate="2017-03-25T05:04:06.723" Title="What are the intermolecular and intramolecular forces between Hg molecules/ atoms in liquid mercury?" Tags="&lt;bond&gt;&lt;metal&gt;&lt;intermolecular-forces&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2017-04-16T02:46:24.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="35280" PostTypeId="1" AcceptedAnswerId="35282" CreationDate="2015-08-20T18:58:59.680" Score="-1" ViewCount="25" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;There is some weak acid with a concentration of $0.0284$ M, which is&#xA;  ionized by $6.70$%. What is the value of $K_\mathrm a$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HA &amp;lt;=&amp;gt; H+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the ionization percentage is $\frac{[\ce{H+}]}{[\ce{HA}]}\cdot100$. So since the value is $6.70$ we can solve and get that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] = 1.90\cdot10^{-3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I got concentration of $\ce{H+}$ ions. However, as far as I understand, to calculate $K_\mathrm a$ I need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am missing the value of $[\ce{A-}]$. How do I proceed from here?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="5026" LastEditDate="2018-02-25T03:05:08.407" LastActivityDate="2018-02-25T03:05:08.407" Title="Get K for a weak acid equilibrium when given the ionization percentage" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35282" PostTypeId="2" ParentId="35280" CreationDate="2015-08-20T19:09:56.703" Score="2" Body="&lt;p&gt;For every mole of $\ce{HA}$ that dissociates, you get equimolar amounts of $\ce{H+}$ and $\ce{A-}$. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] = [\ce{A-}] = 1.9\times 10^{-3}\,\mathrm{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{HA}] = (0.0284\,\mathrm{M}) - (1.9\times 10^{-3}\,\mathrm{M}) = 0.0265\,\mathrm{M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$K_{\mathrm a} = {(1.9\times 10^{-3})^{2}\over 0.0265} = 1.36\times 10^{-4}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the units of molarity are omitted because they cancel due to the implicit presence of $\ce{H2O}$ in the system (which would otherwise show in the denominator).&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-20T19:18:10.380" LastActivityDate="2015-08-20T19:18:10.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35283" PostTypeId="1" AcceptedAnswerId="35287" CreationDate="2015-08-20T19:21:10.000" Score="1" ViewCount="3114" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;A hydrated aluminium sulphate, &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2(SO4)3.xH2O}$&lt;/span&gt;, contains 8.10% of aluminium by mass. Find the value of x.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My attempt: (Assuming 100g sample)&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the amount of aluminium using the formula &lt;span class=&quot;math-container&quot;&gt;$\mathrm{amount = \frac{mass}{molar~mass}}$&lt;/span&gt;, yielding &lt;span class=&quot;math-container&quot;&gt;$\mathrm{0.300~mol}~\ce{Al}$&lt;/span&gt; having a mass of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{8.10~g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine the amount of sulphate (&lt;span class=&quot;math-container&quot;&gt;$\ce{SO4^2-}$&lt;/span&gt;) by recognising a ratio between the aluminium and sulphate. 2 moles of aluminium to 3 moles of sulphate, &lt;strong&gt;2:3&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the ratio, I determined the amount to be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{0.450~mol}~\ce{SO4^2-}$&lt;/span&gt;, which has a mass of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{43.2~g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can now calculate the mass of the anhydrous compound, aluminium sulphate (&lt;span class=&quot;math-container&quot;&gt;$\ce{Al2(SO4)3}$&lt;/span&gt;), to be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{51.3~g}$&lt;/span&gt;, leaving &lt;span class=&quot;math-container&quot;&gt;$\mathrm{48.7~g}$&lt;/span&gt; of anhydrous substance, or &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using the figure of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{48.7~g}$&lt;/span&gt;, I can calculate the amount of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; using the formula &lt;span class=&quot;math-container&quot;&gt;$\mathrm{amount = \frac{mass}{molar~mass}}$&lt;/span&gt; which produced a figure of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2.71~mol}~\ce{H2O}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By using the principle of relative amounts, I find that x=9.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The answer is incorrect. Where did I go wrong?&lt;/p&gt;&#xA;" OwnerUserId="19280" LastEditorUserId="69830" LastEditDate="2021-05-26T06:23:39.893" LastActivityDate="2021-05-26T06:23:39.893" Title="How to find the amount of moles given mass percentage data" Tags="&lt;inorganic-chemistry&gt;&lt;stoichiometry&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35284" PostTypeId="1" AcceptedAnswerId="35285" CreationDate="2015-08-20T19:57:19.313" Score="2" ViewCount="827" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which is the strongest acid here?&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HClO4}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I found &lt;a href=&quot;https://chemistry.stackexchange.com/questions/22056/which-of-the-following-is-the-strongest-base&quot;&gt;this question&lt;/a&gt;, didn't really get the first answer, but &lt;a href=&quot;https://chemistry.stackexchange.com/a/22060/16632&quot;&gt;this one&lt;/a&gt; in particular seemed interesting.&lt;/p&gt;&#xA;&lt;p&gt;Apparently, you calculate a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{p}K_\mathrm{a}$&lt;/span&gt; for each substance.&lt;/p&gt;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.columbia.edu/%7Ecrg2133/Files/CambridgeIA/Chemistry/AcidityBasicitykPa.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{p}K_\mathrm{a} = -\log_{10}K_\mathrm{a}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I need to find out &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; then. From what I understand &lt;span class=&quot;math-container&quot;&gt;$K_\mathrm{a}$&lt;/span&gt; is obtained from dividing the concentrations of products by the concentrations of reagents. However, how am I supposed to do so if I am not given any equilibrium reactions?&lt;/p&gt;&#xA;" OwnerUserId="16632" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-07-15T02:50:34.930" Title="How to determine the strongest oxyacid of chlorine?" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35285" PostTypeId="2" ParentId="35284" CreationDate="2015-08-20T20:35:17.843" Score="4" Body="&lt;p&gt;In such questions you're supposed to know certain rules of acidity. You can find some of them on &lt;a href=&quot;http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch11/factors.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;. In your case fourth rule, concerning oxoacids, is relevant and as you can see there, it's described on this very case. More precise formulation is so called Pauling's Rule described &lt;a href=&quot;http://chemistrytextbookcrawl.blogspot.com/2013/02/paulings-rules.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. In general the more oxo groups connected to central atom, the more acidic compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://chemistry.stackexchange.com/q/29131&quot;&gt;How are Pauling&amp;#39;s rules applied to Nitric acid?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9961" LastEditorUserId="5026" LastEditDate="2018-07-15T02:50:34.930" LastActivityDate="2018-07-15T02:50:34.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35286" PostTypeId="1" AcceptedAnswerId="35289" CreationDate="2015-08-20T20:43:10.707" Score="38" ViewCount="1836" Body="&lt;p&gt;Earlier this year, there was a &lt;a href=&quot;http://cen.acs.org/articles/93/web/2015/03/Hexafluorocyclohexane-Isomer-Claims-Polar-Title.html?type=paidArticleContent&quot; rel=&quot;noreferrer&quot;&gt;lot&lt;/a&gt; of &lt;a href=&quot;http://www.rsc.org/chemistryworld/2015/04/single-face-fluorinated-carbon-ring-hexafluorocyclohexane-takes-strain&quot; rel=&quot;noreferrer&quot;&gt;attention&lt;/a&gt; when &lt;a href=&quot;http://www.nature.com/nchem/journal/v7/n6/full/nchem.2232.html&quot; rel=&quot;noreferrer&quot;&gt;all-cis 1,2,3,4,5,6-hexafluorocyclohexane was synthesized&lt;/a&gt;*:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FvuBxs.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FvuBxs.png&quot; alt=&quot;perfluorocyclohexane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, &lt;a href=&quot;http://cen.acs.org/articles/93/web/2015/03/Hexafluorocyclohexane-Isomer-Claims-Polar-Title.html&quot; rel=&quot;noreferrer&quot;&gt;C&amp;amp;E News&lt;/a&gt; quoted the lead author:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This compound is remarkable for being the most polar nonionic compound now known to exist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, the &lt;a href=&quot;http://doi.org/10.1038/nchem.2232&quot; rel=&quot;noreferrer&quot;&gt;original paper&lt;/a&gt; reports a molecular dipole moment of $\pu{6.2 D}$, which is extremely high for an alkane. It also claims the:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;unusual property of a facially polarized ring in organic chemistry&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Certainly having one side of the cyclohexane ring completely substituted with fluorines is unusual, so I agree with the facial-polarity property being interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are many small molecules (e.g., under 20-30 atoms) with large dipole moments, e.g., &lt;a href=&quot;http://pqr.pitt.edu/mol/TYMLOMAKGOJONV-UHFFFAOYSA-N&quot; rel=&quot;noreferrer&quot;&gt;p-nitroaniline&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Psq6s.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Psq6s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick &lt;code&gt;PM7&lt;/code&gt; calculation gives the predicted dipole moment of &lt;a href=&quot;http://pqr.pitt.edu/mol/TYMLOMAKGOJONV-UHFFFAOYSA-N&quot; rel=&quot;noreferrer&quot;&gt;$\pu{7.93\pm1 D}$&lt;/a&gt;. So I dispute the claim from C&amp;amp;E News - nitroaniline is clearly more polar and non-ionic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is this: &lt;strong&gt;What's the largest dipole moment for a known non-ionic organic molecule under 30 atoms?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*: &lt;a href=&quot;http://doi.org/10.1038/nchem.2232&quot; rel=&quot;noreferrer&quot;&gt;O'Hagan, et. al, &quot;All-cis 1,2,3,4,5,6-hexafluorocyclohexane is a facially polarized cyclohexane&quot; &lt;em&gt;Nature Chemistry&lt;/em&gt; 7, 483–488 (&lt;strong&gt;2015&lt;/strong&gt;)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="41328" LastEditDate="2017-09-12T10:37:59.770" LastActivityDate="2017-09-14T10:03:17.450" Title="Is Hexafluorocyclohexane the &quot;Most Polar&quot; Small Molecule?" Tags="&lt;organic-chemistry&gt;&lt;reference-request&gt;&lt;dipole&gt;" AnswerCount="3" CommentCount="10" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="35287" PostTypeId="2" ParentId="35283" CreationDate="2015-08-20T20:58:41.083" Score="1" Body="&lt;p&gt;Basis: 100 g&lt;br&gt;&#xA;$m_{\ce{Al}}=8.10\ \mathrm g$&lt;br&gt;&#xA; $n_{\ce{Al}}=\frac{8.10\ \mathrm g}{27\ \mathrm{g/mol}}=0.3\ \mathrm{mol}$&lt;br&gt;&#xA; $n_{\ce{Al2(SO4)3}}=0.5 \times n_{\ce{Al}}=0.15\ \mathrm{mol}$&lt;br&gt;&#xA; $m_{\ce{Al2(SO4)3}}=0.15\ \mathrm{mol} \times 342.15\ \mathrm{g/mol}=51.3\ \mathrm g$&lt;br&gt;&#xA;$m_{\ce{H2O}}=100\ \mathrm g-51.3\ \mathrm g=48.65\ \mathrm g$&lt;br&gt;&#xA;$n_{\ce{H2O}}=\frac{48.65\ \mathrm g}{18\ \mathrm{g/mol}}=2.7\ \mathrm{mol}$&lt;br&gt;&#xA;$x=\frac{2.7\ \mathrm{mol}}{0.15\ \mathrm{mol}}=18$&lt;/p&gt;&#xA;" OwnerUserId="7777" LastEditorDisplayName="user7951" LastEditDate="2015-08-28T20:27:17.083" LastActivityDate="2015-08-28T20:27:17.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35288" PostTypeId="1" CreationDate="2015-08-20T21:06:43.253" Score="2" ViewCount="603" Body="&lt;p&gt;About 11 months ago, I posted &lt;a href=&quot;https://chemistry.stackexchange.com/questions/16664/what-is-the-lower-bound-to-the-temperature-at-which-a-fire-can-burn&quot;&gt;a question on this SE site on cool flames&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/a/16751/7809&quot;&gt;One of the answers&lt;/a&gt; mentioned white phosphorus. However, a few weeks earlier, I played a game called Spec Ops: The Lines, in which the protagonist uses white phosphorus laden mortar munition, horribly burning dozens of soldiers and civilians. I was confused that the answer mentioned WP because of that moment from the game, and further googled it to find that WP munitions are actually heavily regulated, and using them against civilians is considered a warcrime. To me, that doesn't really seem compatible with &quot;cool burning&quot; at a temperature you can hold your hand in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not my intent to cause this answer to be downvoted, but I'm confused whether the answer is factually accurate. According to Google, it burns at 5,000 degrees Fahrenheit, but that's the munitions.&lt;/p&gt;&#xA;" OwnerUserId="7809" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2017-01-10T22:45:54.650" Title="Is white phosphorus really a cool burning fuel?" Tags="&lt;thermodynamics&gt;&lt;combustion&gt;&lt;fuel&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35289" PostTypeId="2" ParentId="35286" CreationDate="2015-08-21T00:00:59.053" Score="25" Body="&lt;p&gt;At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was using your given database for a little search, starting with nitroaniline, and came up with a few small molecules with higher dipole moments. Those seem to be PM7 calculated values and are shown in the following table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{lll}&#xA;\hline&#xA;\text{Name} &amp;amp; \text{Dipole moment in D} &amp;amp; \text{Source}\\&#xA;\hline&#xA;\text{4-Nitroacetanilide} &amp;amp; 8.36 \pm 1.08 &amp;amp; (2)\\&#xA;\textit N\text,\textit N\text{-Dimethyl-4-nitroaniline} &amp;amp; 9.53 \pm 1.08 &amp;amp; (3)\\&#xA;\textit P\text ,\textit P\text{-Dimethyl-}\textit N\text{-(4-nitrophenyl)phosphinic amide} &amp;amp; 10.89 \pm 1.08 &amp;amp; (4)\\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there were some comments, “it seems that this level of calculation does not provide accurate estimates of dipole moments”, I did some further calculations on levels that might provide better estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a quite new paper of Hickey and Rowley (5) which says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CCSD, MP2, or hybrid DFT methods using the aug-cc-pVTZ basis set are all able to predict dipole moments with RMSD errors in the 0.12–0.13 D range and polarizabilities with RMSD errors in the 0.30–0.38 Å³ range.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Within the paper, there is later on a comparison between cc-pVTZ, Sadlej-pVTZ and aug-cc-pVTZ:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sadlej basis set yields results that are improved over the cc-pVTZ basis set and have effectively the same accuracy as the aug-cc-pVTZ basis set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That’s why I chose Sadlej’s basis set at first … but I wanted to look if it could not be easier hence faster. Therefore I also chose the def2-SVPD and looked, what an improvement a def2-TZVPD single point calculation on this geometry would yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use ORCA 3.0.3 for calculations – the command line is like the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;! B3LYP Sadlej-pVTZ TightSCF TightOpt&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Table 1: &quot;Pure&quot; dipole moments (in Debye) straight from the calculations&#xA;\begin{array}{lllll}&#xA;\hline&#xA; &amp;amp; \text{def2-SVPD} &amp;amp; \text{def2-TZVPD} &amp;amp; \text{Sadlej-pVTZ} &amp;amp; \text{PM6}\\&#xA;\hline&#xA;\text{4-Nitroacetanilide} &amp;amp; 9.18261 &amp;amp; 9.00380 &amp;amp; 9.07922 &amp;amp; 8.6738\\&#xA;\textit N\text ,\textit N\text{-Dimethyl…} &amp;amp; 7.39418 &amp;amp; 7.29188 &amp;amp; 7.30862 &amp;amp; 9.4209\\&#xA;\textit P\text ,\textit P\text{-Dimethyl…} &amp;amp; 8.58913 &amp;amp; 8.44430 &amp;amp; 8.51591 &amp;amp; 11.8352\\&#xA;\text{all-}\textit{cis }\ce{C6H6F6} &amp;amp; 6.17422 &amp;amp; 5.97013 &amp;amp; 6.04794 &amp;amp; 5.6298\\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did then, was to think of how to produce values that could be properly compared to the experimental values from the paper. As they said, with their methods they can yield RMSDs of around 0.12 D. So I took their values for the Sadlej basis set and made my own calculations on their test molecules for the two def2 basis sets (and also for PM6 with Gaussian 09 Rev. A.02).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table 2: Some analytical values to compare with the publication values; the units are in Debye&#xA;\begin{array}{lllll}&#xA;\hline&#xA; &amp;amp; \text{def2-SVPD} &amp;amp; \text{def2-TZVPD} &amp;amp; \text{Sadlej-pVTZ} &amp;amp; \text{PM6}\\&#xA;\hline&#xA;\text{MAE} &amp;amp; 0.08 &amp;amp; 0.08 &amp;amp; 0.08 &amp;amp; 0.40\\&#xA;\text{RMSD} &amp;amp; 0.13 &amp;amp; 0.12 &amp;amp; 0.12 &amp;amp; 0.61\\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for my two ways, the MAE and the RMSD are not too far away from the triple zeta values but are calculated way faster. Now I have made a linear regression between the experimental and the calculated dipole moments from Hickey and Rowley’s test molecules to be able – as said before – to calculate proper error ranges at a given probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1: Regression functions for the four tested methods (blue), prediction intervals at 95% (gray); from top left to bottom right: def2-SVPD, def2-TZVPD//def2-SVPD, Sadlej-pVTZ, PM6&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ANMZN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ANMZN.png&quot; alt=&quot;Dipole moment regressions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let’s turn the “pure” values from Table 1 into somehow more real values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{def2-SVPD} \to \text{exp}&amp;amp;: -0.0097439128 + 0.9763275\ x\\&#xA;\text{def2-TZVPD//def2-SVPD} \to \text{exp}&amp;amp;: -0.0045050428 + 0.99095691\ x\\&#xA;\text{Sadlej-pVTZ} \to \text{exp}&amp;amp;: -0.012922547 + 0.98812009\ x\\&#xA;\text{PM6} \to \text{exp}&amp;amp;: -0.16057241 + 0.89545049\ x\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prediction Intervals (for the error):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\small\begin{align}&#xA;\text{def2-SVPD} \to \text{exp}&amp;amp;: 0.24473369 \sqrt{1.0217391 + 0.010348758 (x-1.4590217)^2}\\&#xA;\text{def2-TZVPD//def2-SVPD} \to \text{exp}&amp;amp;: 0.23719452 \sqrt{1.0217391 + 0.010656662 (x-1.4321957)^2}\\&#xA;\text{Sadlej-pVTZ} \to \text{exp}&amp;amp;: 0.23333979 \sqrt{1.0217391 + 0.010593477 (x-1.4448261)^2}\\&#xA;\text{PM6} \to \text{exp}&amp;amp;: 0.98446665 \sqrt{1.0217391 + 0.0097476423 (x-1.7592391)^2}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table 3: Converted dipole moments (in Debye) from Table 1 including error ranges at 95% probability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{llll}&#xA;\hline&#xA; &amp;amp; \text{def2-SVPD} &amp;amp; \text{def2-TZVPD} &amp;amp; \text{Sadlej-pVTZ} &amp;amp; \text{PM6}\\&#xA;\hline&#xA;\text{4-Nitroacetanilide}  &amp;amp; 8.96 \pm 0.31 &amp;amp; 8.92 \pm 0.30 &amp;amp; 8.96 \pm 0.30 &amp;amp; 7.61 \pm 1.20 \\&#xA;\textit N\text ,\textit N\text{-Dimethyl…}     &amp;amp; 7.21 \pm 0.29 &amp;amp; 7.22 \pm 0.28 &amp;amp; 7.21 \pm 0.27 &amp;amp; 8.28 \pm 1.24 \\&#xA;\textit P\text ,\textit P\text{-Dimethyl…}     &amp;amp; 8.38 \pm 0.30 &amp;amp; 8.36 \pm 0.29 &amp;amp; 8.40 \pm 0.29 &amp;amp; 10.44 \pm 1.40 \\&#xA;\text{all-}\textit{cis }\ce{C6H6F6} &amp;amp; 6.02 \pm 0.27 &amp;amp; 5.91 \pm 0.26 &amp;amp; 5.96 \pm 0.26 &amp;amp; 4.88 \pm 1.06 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be seen from this now? The smaller B3LYP/def2-SVPD method yields nearly the same values as the more expensive Sadlej version. Putting the def2-TZVPD single point on top of them, yields slightly &lt;em&gt;deeper&lt;/em&gt; results with actually no accuracy improvements. PM6 is no real good option as the error ranges are way higher, with values of about $1-1.4$ D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the dipole moment of all-&lt;em&gt;cis&lt;/em&gt; 1,2,3,4,5,6-hexafluorocyclohexane is probably somewhere around $6$ D but there are probably many other molecules with higher dipole moments … “my” three molecules shown above go at least up to somewhere around $9$ D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as no experimental values are available, everything here is only a good guess – but you will know this too good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;&lt;strong&gt;Sources:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nature.com/nchem/journal/v7/n6/full/nchem.2232.html&quot; rel=&quot;noreferrer&quot;&gt;O’Hagan et al., &lt;em&gt;Nature Chemistry&lt;/em&gt;, &lt;strong&gt;2015&lt;/strong&gt;, &lt;em&gt;7&lt;/em&gt;, 483–488&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N&quot; rel=&quot;noreferrer&quot;&gt;http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N&quot; rel=&quot;noreferrer&quot;&gt;http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N&quot; rel=&quot;noreferrer&quot;&gt;http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jp502475e&quot; rel=&quot;noreferrer&quot;&gt;A. L. Hickey, C. N. Rowley, &lt;em&gt;J. Phys. Chem. A&lt;/em&gt;, &lt;strong&gt;2014&lt;/strong&gt;, &lt;em&gt;118&lt;/em&gt; (20), 3678–3687&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="41556" LastEditDate="2017-08-07T19:34:39.143" LastActivityDate="2017-08-07T19:34:39.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35290" PostTypeId="2" ParentId="34903" CreationDate="2015-08-21T00:08:09.860" Score="0" Body="&lt;p&gt;For an ideal gas, $\Delta H = 0$ for a constant temperature change between two thermodynamic equilibrium states of a system.  So, for your system, $\Delta G=T\Delta S$.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-21T00:08:09.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35292" PostTypeId="1" CreationDate="2015-08-21T01:40:57.857" Score="5" ViewCount="10161" Body="&lt;p&gt;If in an octahedral structure there is one lone pair placed on any two of the axial position then will the lone pair distort all the 90 degree angles due to repulsion from lone pair or the angles will remain same? I found one question regarding $\ce{BrF5}$.&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="10185" LastEditDate="2015-08-21T18:11:24.110" LastActivityDate="2018-03-26T14:46:51.303" Title="Bond angles in BrF5" Tags="&lt;covalent-compounds&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35294" PostTypeId="1" CreationDate="2015-08-21T02:39:30.353" Score="4" ViewCount="18497" Body="&lt;p&gt;I am aware that $\ce{HCl}$ can be used to help speed up the hydrolysis of sucrose but I haven't really been able to find out why. I've read somewhere that the glycosidic link in sucrose is broken off and that the $\ce{H+}$ ions from the $\ce{HCl}$ is responsible for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also come across the ether link during my research and would like to know how this plays a part in the hydrolysis of sucrose and if the $\ce{HCl}$ interacts with it.&lt;/p&gt;&#xA;" OwnerUserId="19296" LastEditorUserId="402" LastEditDate="2016-12-17T02:42:50.697" LastActivityDate="2019-09-22T06:28:12.590" Title="Why is HCl used in hydrolysis of sucrose?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;hydrolysis&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35296" PostTypeId="2" ParentId="35292" CreationDate="2015-08-21T06:14:45.270" Score="8" Body="&lt;p&gt;You are right! You asked the question and you replied simultaneously.&#xA;According to this &lt;a href=&quot;https://en.wikipedia.org/wiki/Bromine_pentafluoride&quot;&gt;website&lt;/a&gt;, the structure of bromine pentafluoride confirmed your claims.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IsHnq.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IsHnq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7581" LastActivityDate="2015-08-21T06:14:45.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35297" PostTypeId="1" CreationDate="2015-08-21T06:29:55.663" Score="2" ViewCount="136" Body="&lt;p&gt;I had some wood work done in my room and termiticide (UNIBAN-20 (Chlorpyriphos-20% EC) was used so that there is no termites attack in future. The stupid carpenter was using this product for the first time and it had to be diluted 10 times in kerosene oil before use which he didn't do. He poured the concentrated solution of this termiticide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have very strong smell and poisonous fumes (the head starts spinning within seconds) in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I neutralize the smell and get rid of the poisonous fumes? It has been 13 days now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have exhaust fan switched on but the smell is still there in the room.&lt;/p&gt;&#xA;" OwnerUserId="19299" LastEditorUserId="19299" LastEditDate="2015-08-21T10:30:01.453" LastActivityDate="2015-09-22T09:30:21.030" Title="How to control and vanish termiticide/pesticide fumes from room" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35298" PostTypeId="1" AcceptedAnswerId="35299" CreationDate="2015-08-21T06:50:48.527" Score="2" ViewCount="21886" Body="&lt;p&gt;At school, they teach the following two formulas: $$\Delta H = -mc\Delta T$$ and $$q= mc\Delta T$$ What i am not sure is that my teacher says that q and $\Delta H$ are the same thing but to think of q as a magnitude and hence it is always positive. However what I am confused about is how can q always be positive? What if the change in temperature is negative? Also, what is the difference between q and $\Delta H$. My current understanding is that they are the same thing and they measure the change in heat energy. Is this right?&lt;/p&gt;&#xA;" OwnerDisplayName="user19300" LastEditorUserId="18002" LastEditDate="2015-08-21T08:49:24.357" LastActivityDate="2015-08-21T08:49:24.357" Title="What is Enthalpy?" Tags="&lt;thermodynamics&gt;&lt;energy&gt;&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35299" PostTypeId="2" ParentId="35298" CreationDate="2015-08-21T07:04:17.410" Score="3" Body="&lt;p&gt;Enthalpy is thought as the total potential energy of a system. The actual thermodynamic equation for enthalpy is: $$\Delta H = \Delta U + \Delta {(pV)}$$ where $\Delta U$ is the change in internal energy. Internal energy can be thought as the total amount of heat energy there is in a system. Hence $\Delta U$ can be thought as the total energy that is required to make a system, while $\Delta {(pV)}$ can be thought as how much energy is required to make space for the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q on the other hand is actually a different thing to enthalpy. Q refers to the heat transfer of a system. Note that this means that q can be positive or negative (if heat energy moves from the system to the surroundings) and hence it is not always positive. The correct formula for heat transfer is: $$q = \Delta U -w $$ So actually, q doesn't always have to equal to enthalpy change. However at a school level it is taught that they equal the same thing. This is because of we assume that pressure is constant (which is a common assumption made as usually most experiments have a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p\Delta V$$ Also since $w = p\Delta V$, enthalpy becomes: $$\Delta H = \Delta U - w = q$$ Hence, enthalpy and q equal the same values only when pressure is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation that your teacher has given you is actually an equation for heat transfer, not for enthalpy change. However because enthalpy equals heat transfer in most cases, it is commonly said that enthalpy change also equals to that equation. However the one mistake is that there shouldn't be a minus in the equation for enthalpy change. The equation for heat transfer and enthalpy should both be the same. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-21T07:04:17.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35301" PostTypeId="1" AcceptedAnswerId="35302" CreationDate="2015-08-21T07:50:59.650" Score="5" ViewCount="3912" Body="&lt;p&gt;I know that $\Delta G = \Delta H - T\Delta S$ and that if $\Delta G$ less than $0$ a reaction will be spontaneous. What I am wondering is what is the actually physical interpretation of Gibbs free energy. I mean, if the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$, what does that actually mean? &lt;/p&gt;&#xA;" OwnerDisplayName="user19300" LastEditorDisplayName="user7951" LastEditDate="2015-08-21T08:59:00.820" LastActivityDate="2016-03-25T07:47:45.640" Title="Physical Interpretation of Gibbs Free Energy" Tags="&lt;thermodynamics&gt;&lt;energy&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35302" PostTypeId="2" ParentId="35301" CreationDate="2015-08-21T07:57:58.717" Score="7" Body="&lt;p&gt;Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gibbs free energy is really derived is from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is constant so we can simplify this relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constant Pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$&#xA;and internal energy by: $$\Delta U = q + w$$&#xA;Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="27256" LastEditDate="2016-03-25T07:47:45.640" LastActivityDate="2016-03-25T07:47:45.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35303" PostTypeId="2" ParentId="35266" CreationDate="2015-08-21T08:09:37.340" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$&#xA;In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of  $\ce{Na}A$, completely dissociated, with water:&#xA;$$A^- +\ce{ H2O &amp;lt;=&amp;gt; OH- +H}A$$&#xA;The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand,  you have:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$&#xA;Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, $x=1.0\times10^{-6}\mathrm{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$&#xA; $$\mathrm{pH}=8.0$$&lt;/p&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-08-21T08:33:58.340" LastActivityDate="2015-08-21T08:33:58.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35304" PostTypeId="1" AcceptedAnswerId="35312" CreationDate="2015-08-21T08:20:14.297" Score="5" ViewCount="1206" Body="&lt;p&gt;How is Cubic Closed Packing &amp;quot;cubic&amp;quot; and also how is hexagonal closed packing &amp;quot;hexagonal&amp;quot;?&lt;/p&gt;&#xA;&lt;p&gt;I have been thinking over this for a few hours but I am unable to visualise.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mQLJ4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mQLJ4.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I understand that when placing the third layer on top of the second layer ,&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; situations arise:&lt;br /&gt;&#xA;If the third layer is placed on the tetrahedral voids it forms an &lt;strong&gt;A-B-A-B&lt;/strong&gt;... type structure which is called &lt;em&gt;&lt;strong&gt;hcp&lt;/strong&gt;&lt;/em&gt; (why?)&lt;br /&gt;&#xA;and If the third layer is placed on the octahedral voids it forms an &lt;strong&gt;A-B-C-A-B-C&lt;/strong&gt;... type structure which is called &lt;em&gt;&lt;strong&gt;ccp&lt;/strong&gt;&lt;/em&gt; (why?)&lt;/p&gt;&#xA;" OwnerUserId="8218" LastEditorUserId="65276" LastEditDate="2020-12-01T16:35:10.730" LastActivityDate="2020-12-01T16:35:10.730" Title="Why is Cubic closed packing called &quot;cubic&quot;?" Tags="&lt;crystal-structure&gt;&lt;solid-state-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35305" PostTypeId="2" ParentId="35265" CreationDate="2015-08-21T09:35:12.187" Score="4" Body="&lt;p&gt;Benzofuran is named by the position of the oxygen, hence 2-Benzofuran.  (below is 1-benzofuran)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wq6q0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wq6q0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The =O s are then added on the 1 and 3 carbons, hence, 1-3-dione  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare, cyclopentan-1,3-dione &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ycy5Im.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ycy5Im.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-benzofuran-1,3-dione&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c5Cxcm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5Cxcm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19231" LastActivityDate="2015-08-21T09:35:12.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35306" PostTypeId="1" AcceptedAnswerId="35319" CreationDate="2015-08-21T09:54:04.730" Score="5" ViewCount="648" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cinnamaldehyde&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; lists the molecule as being named &quot;(2E)-3-phenylprop-2-enal&quot;, however it would appear that, as there is only 1 double bond which has a possibility of stereoisomerism, the 2 is redundant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the correct IUPAC name not in fact be E-3-phenylprop-2-enal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/miyI2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/miyI2.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19231" LastEditorUserId="10185" LastEditDate="2015-08-21T11:28:01.713" LastActivityDate="2015-08-22T12:04:46.053" Title="Cinnamaldehyde listed as (2E)-3-phenylprop-2-enal , but is the first 2 redundant?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35307" PostTypeId="2" ParentId="35306" CreationDate="2015-08-21T11:07:29.170" Score="0" Body="&lt;p&gt;The 1993 IUPAC &quot;Blue Book&quot; is the easiest to find online thanks to ACDLabs. In this version, numbers are only used with &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_626.htm&quot; rel=&quot;nofollow&quot;&gt;stereochemical descriptors if there is more than one&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;E&lt;/em&gt;)-phenylprop-2-enal is the correct systematic name for cinnamaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most recent edition is the &lt;a href=&quot;http://www.iupac.org/nc/home/publications/iupac-books/books-db/book-details.html?tx_wfqbe_pi1%5Bbookid%5D=56&quot; rel=&quot;nofollow&quot;&gt;2002 edition&lt;/a&gt;, which I do not have access to. The 2002 edition may contain a different recommendation.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-21T11:07:29.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35309" PostTypeId="1" CreationDate="2015-08-21T11:48:23.233" Score="2" ViewCount="7335" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Tin(IV) iodide is prepared by direct combination of the elements.&#xA;Add 2.00 g of granulated tin to a solution of 6.35 g of iodine. Write an equation for this reaction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Determine the amount of tin using &lt;strong&gt;moles = mass/molar mass&lt;/strong&gt;. This yields a result of 0.017 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{Sn}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Determine the amount of iodine using &lt;strong&gt;moles = mass/molar mass&lt;/strong&gt;. This yields a result of 0.5 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{I}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; Determine the ratio of tin to iodine. Using relative amounts, for every 1 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{Sn}$&lt;/span&gt; there are 3 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{I}$&lt;/span&gt;, so the ratio is 1:3. Because iodine is a diatomic molecule, I will adjust accordingly, so the ratio becomes 2 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{Sn}$&lt;/span&gt; to 3 mol &lt;span class=&quot;math-container&quot;&gt;$\ce{I2}$&lt;/span&gt; or 2:6.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; Construct the formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{2Sn + 3I2 -&amp;gt; SnI4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;with excess tin and iodine.&lt;/p&gt;&#xA;&lt;p&gt;Is my procedure correct?&lt;/p&gt;&#xA;" OwnerUserId="19304" LastEditorUserId="16683" LastEditDate="2020-08-31T16:13:09.510" LastActivityDate="2020-08-31T16:13:09.510" Title="Equation for reaction of tin and iodine to produce tin(IV) iodide" Tags="&lt;stoichiometry&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35310" PostTypeId="2" ParentId="35301" CreationDate="2015-08-21T12:06:43.170" Score="6" Body="&lt;p&gt;I'm going to concentrate on the $T\,S$ term (which you likely find the most mysterious) and hopefully give a little more physical intuition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $T S$ term arises roughly from the energy that is needed to &quot;fill up&quot; the rotational, vibrational, translational and otherwise distractional thermal energies of the constituents of a system. Simplistically, you can kind of think of its being related to the idea that you must use some of the energy released to make sure that the reaction products are &quot;filled up&quot; with heat so that they are at the same temperature as the reactants. So the $T S$ term is related to, but not the same as, the notion of &lt;em&gt;heat capacity&lt;/em&gt;: let's look at this a bit further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't we get at all the energy $\Delta H$? Well, actually we &lt;em&gt;can&lt;/em&gt; in certain contrived circumstances: let's look at these circumstances to find out why in a meaningful definition we must leave behind an amount $T\,\Delta S$. Let's think of the burning of hydrogen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2 + 1/2 O2 -&amp;gt; H2O};\quad\Delta H \approx 143\:{\rm kJ\,mol^{-1}}\tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a highly exothermic reaction, and also one of the reactions of choice if you want to throw three astronauts, fifty tons of kit and about a twentieth of the most advanced-economy-in-the-world’s-1960s GDP at the Moon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about one mole of $\ce{H2O}$ is that it can soak up less heat than the mole of $\ce{H2}$ and half a mole of $\ce{O2}$; naively this would seem to say that we can get &lt;em&gt;more&lt;/em&gt; heat than the enthalpy change $\Delta H$, but we'll see this is not so. We imagine a thought experiment, where we have  a gigantic array of enormous heat pads (all individually like an equilibrium “outside world&quot;) representing all temperatures between absolute zero and  $T_0$ with a very fine temperature spacing $\Delta T$ between them. On my darker days I find myself imagining an experimental kit that looks eerily like a huge pallet on wheels of mortuary shelves, sliding in and out as needed by the experimenter! We bring the reactants into contact with the first heat pad, which is at a temperature $T_1 = T_0 - \Delta T$ a teeny-tiny bit cooler than $T_0$  thus reversibly drawing some heat $\Delta Q(T_1)$  off into the heat pad. Next, we bring the reactants into contact with the second heat pad at temperature $T_2 = T_0 - 2\,\Delta T$, thus reversibly drawing heat $\Delta Q(T_2)$ off into that heat pad. We keep cooling then shifting to the next lowest heat pad until we have visited all the heat pads and thus sucked all the heat off into our heat pads: see my sketch below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lh8Fv.png&quot; alt=&quot;Cooling Down Reactants&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the reactants are at absolute zero. There is no heat needed to &quot;fill them up&quot; to their temperature, so we can extract &lt;em&gt;all&lt;/em&gt; the enthalpy $\Delta H$ from the reaction as useful work. Let's imagine we can put this work aside in some ultrafuturistic perfect capacitor, or some such lossless storage for the time being.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we must heat our reaction products back up to standard temperature, so that we know what we can get out of our reaction if the conditions do not change. So, we simply do the reverse, as sketched below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1HhWf.png&quot; alt=&quot;Heating Products Up&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that I said that $\ce{H2O}$ soaks up less heat than the reactants. This means that, as we heat the products back up to standard temperature, we take from the heat pads &lt;em&gt;less&lt;/em&gt; heat in warming up the water than we put into them in cooling the reactants down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, so good. We have gotten &lt;em&gt;all&lt;/em&gt; the work $\Delta H$ out into our ultracapacitor without losing any! And we're back to our beginning conditions, or so it seems! What's the catch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The experimental apparatus that let us pull this trick off is NOT back at its beginning state. We have left heat in the heat pads. We have thus degraded them: they have warmed up ever so slightly and so cannot be used indefinitely to repeatedly do this trick. If we tried to do the trick too many times, eventually the heat pads would be at ambient temperature and would not work any more.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we haven’t reckoned the free energy at the standard conditions, rather we have simply calculated the free energy $\Delta H$ available in the presence of our unrealistic heat sink array. To restore the system to its beginning state and calculate what work we could get if there were no heat sink array here, we must take away the nett heat flows we added to all the heat pads and send them into the outside World at temperature $T_0$. This is the only &quot;fair&quot; measure, because it represents something that we could do with arbitrarily large quantities of reactants without the presence of artificial, finite resources like heat pad arrays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the outside World at $T_0$ is warmer than &lt;em&gt;any&lt;/em&gt; of the heat pads, so of course this heat transfer can’t happen spontaneously, simply by dent of Carnot’s statement of the second law! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must therefore bring in a reversible heat pump and &lt;strong&gt;use&lt;/strong&gt; some of our work $\Delta H$ to pump this heat into the outside world to restore standard conditions: we would connect an ideal reversible heat pump to each of the heat pads in turn and restore them to their beginning conditions, as sketched below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TLT9J.png&quot; alt=&quot;Packing our Playthings Up&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This part of the work that we use to run the heat pump and restore all the heat pads, if you do all the maths, is exactly the $T\,\Delta S$ term.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above can also cast some light on an endothermic reaction. In such a reaction, we imagine having an energy bank that we can borrow from temporarily. After we have brought the products back up to temperature, we find we have both borrowed $-\Delta H$ from the energy bank and put less heat back into the heat pads than we took from them. So heat can now flow &lt;em&gt;spontaneously&lt;/em&gt; from the environment to the heat pads to restore their beginning state, because the heat pads are all at a lower temperature than the environment. As this heat flows, we can use a reversible heat engine to extract work from the heat flowing down the gradient. This work is, again, $-T\,\Delta S$, which is a positive work gotten from the heat flowing down the temperature gradient. The  $-T\,\Delta S$ can be so positive that we can pay back the $\Delta H$ we borrowed and have some left over. If so, we have an endothermic reaction, and a nett free energy: this energy coming from the heat flowing spontaneously inwards from the environment to fill the higher entropy products (higher than the entropy of the reactants). A good thought experiment here is to imagine towing an iceberg into warmer latitudes, and then putting it into contact with the outside world through heat engines which convert a fraction $1-T_\mathrm i/T_\mathrm o$ of the heat flowing into work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take heed that, in the above, I have implicitly assumed the Nernst Heat Postulate -the not quite correct third law of thermodynamics - see &lt;a href=&quot;https://physics.stackexchange.com/a/109944/26076&quot;&gt;my answer here&lt;/a&gt; for more details. For the present discussion, this approximate law is well good enough.&lt;/p&gt;&#xA;" OwnerUserId="5766" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-08-21T12:28:46.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35312" PostTypeId="2" ParentId="35304" CreationDate="2015-08-21T12:30:40.547" Score="5" Body="&lt;p&gt;Start with the hexagonal A-B-A-B structure.  It is hexagonal because, choosing one atom in a layer, it is surrounded by 6 equivalent atoms. So, you have a 6-fold symmetry axis, hence the 'hex-'. The fact that it is close-packed can be gotten at mathematically (or at least shown that it is closer-packed than other crystal structures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cubic close packing is, at least in most materials science and metallurgy circles, much more commonly called face centered cubic. The 'standard' unit cell is a cube of atoms with additional atoms at the center of each face.  It is somewhat harder to visualize how this is related to close-packed planes, although there are many pictures available on-line (one, chosen somewhat randomly, is &lt;a href=&quot;http://www.chem.fsu.edu/chemlab/chm1046course/solids.html&quot; rel=&quot;noreferrer&quot;&gt;from Florida State&lt;/a&gt;). The point is that the stacking direction of the close-packed planes are along the 111 body diagonal of the cube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotate the cube so that it is standing on one point with the opposite point on the body diagonal pointing up. Your close packed planes are:&#xA;(1) the bottom point, (2) the top point, (3) the three corner atoms closest to the bottom point as well as the face atoms closest to the bottom atom, (4) the three corner atoms closest to the top atom as well as the face atoms closest to the top atom. It may take some squinting...&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-08-21T12:30:40.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35314" PostTypeId="2" ParentId="35309" CreationDate="2015-08-21T13:02:37.620" Score="2" Body="&lt;p&gt;The procedure is correct, be careful with this:&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The amount of iodine is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.050 mol}$&lt;/span&gt;, but the molecular amount &lt;span class=&quot;math-container&quot;&gt;$\ce{I2}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$\pu{0.025 mol}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the ratio use the amount of point 1 and 2:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\text{Amount of Sn}}{\text{Amount of }\ce{I2}} = \frac{\pu{0.017 mol}}{\pu{0.025 mol}} \approx \frac{2}{3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Don't confuse! This equation is not well balanced. Stoichiometry is inherent of the reaction and is based on the mass conservation law; amount of substance tells you which reactant is limiting and which one is in excess.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ce{\underset{\pu{0.017 mol}}{Sn} + \underset{\pu{0.025 mol}}{2I2} &amp;lt;=&amp;gt; \underset{\pu{? mol}}{SnI4}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now with this you can tell which one is on excess and for how much.&lt;/p&gt;&#xA;" OwnerUserId="5683" LastEditorUserId="4945" LastEditDate="2020-08-31T16:12:44.157" LastActivityDate="2020-08-31T16:12:44.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35316" PostTypeId="1" CreationDate="2015-08-21T13:20:35.637" Score="2" ViewCount="82" Body="&lt;p&gt;So during my lecture on reaction of Alkenes with Halogens, my lecturer used the example of $\ce{Br2}$ gas reacting with $\ce{CH2CH2}$. He mentioned that the $\ce{C=C}$ launches an electrophilic attack on the electropositive Bromine atom (the one that is nearer to $\ce{CH2CH2}$) and I'm wondering why is the $\ce{C=C}$ launching the electrophilic attack? Shouldn't it be done by the Bromine gas since the $\delta^+ \ce{Br}$ is electrophilic and &lt;strong&gt;needs&lt;/strong&gt; the electrons so it attacks the double bond which is most electron dense?&lt;/p&gt;&#xA;" OwnerUserId="19307" LastEditorUserId="9961" LastEditDate="2015-10-16T17:24:19.563" LastActivityDate="2015-10-16T17:24:19.563" Title="Why is C=C launching electrophilic attack on Bromine instead of the other way round?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35317" PostTypeId="2" ParentId="35185" CreationDate="2015-08-21T13:50:20.423" Score="1" Body="&lt;p&gt;This question is related: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/662/handling-and-storing-sodium-hydroxide-hydrochloric-acid-hydrogen-peroxide-and/665#665&quot;&gt;Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the industries the handling of chemicals maybe periodic and, with routine, one tends to forget the hazardous and risks related to this. There's a lot of information on the web (&lt;a href=&quot;http://water.usgs.gov/admin/memo/policy/wrdpolicy94.006.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://water.usgs.gov/admin/memo/policy/wrdpolicy94.006.html&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;STORAGE: Hydrochloric acid should be kept in a tightly closed container and&#xA;stored in chemical area that is compatible with other chemicals.&#xA;Store in a secure, well-ventilated area, that is well marked, and&#xA;away from the general work population.  Do not store near oxidizing&#xA;materials.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This manual should be helpfull too: &lt;a href=&quot;http://www.jsia.gr.jp/data/handling_02e.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jsia.gr.jp/data/handling_02e.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1. Types of Containers:&lt;/em&gt;&#xA;&lt;em&gt;The shipping containers used for hydrochloric acid are those rail tanks be tightly sealed, such as acid-resistant bottles (including polyethylene containers, etc.), and steel drums tank trucks, tank cars, and tankers with corrosion resistant linings&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep reading the manual there is valuable information about handling, transportation, and rules to keep the human safety.&lt;/p&gt;&#xA;" OwnerUserId="5683" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2015-08-21T13:55:47.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35318" PostTypeId="2" ParentId="2724" CreationDate="2015-08-21T13:56:37.833" Score="2" Body="&lt;p&gt;It's the symmetry!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, that's the default answer to &lt;em&gt;all&lt;/em&gt; questions on crystallography.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get needles if your crystallite grows faster in one direction than in any perpendicular direction. Which is the fact here, because aromatic rings love to stack, but don't have much energy to gain out of the van der Waals' interactions between the H-atoms sticking out in the plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a crystallite grows &lt;em&gt;slower&lt;/em&gt; in one direction than perpendicular to it, you will get platelet-like crystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the elementary cell of the crystal has a cubic symmetry, it will likely grow equally fast in all directions, and the crystals will be cubes, like $\ce{NaCl}$. (OK, this last statement is rather weak. It's easy to predict that growth rates are very different like in naphthalene, but that they're actually identical is dure. Sodium chloride is just a nice example where this is the case. And be careful about melts. They're never isotropic near a crystal surface, so anything can happen.)&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-08-21T20:06:00.063" LastActivityDate="2015-08-21T20:06:00.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35319" PostTypeId="2" ParentId="35306" CreationDate="2015-08-21T14:01:04.613" Score="3" Body="&lt;p&gt;The preferred IUPAC name (PIN) for cinnamaldehyde is (2​&lt;em&gt;E&lt;/em&gt;)-3-phenylprop-2-enal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lMmTS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lMmTS.png&quot; alt=&quot;cinnamaldehyde&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to numbering of locants that describe the position of stereogenic units, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-91.3&lt;/strong&gt; NAMING OF STEREOISOMERS&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(…)&lt;br&gt;&#xA;  In preferred IUPAC names, stereodescriptors are placed immediately at the front of the part of the name to which they relate. They are placed at the front of the complete name when related to the parent structure; they are cited in parentheses followed by a hyphen. When they relate to substituent groups, they are cited at the front of the corresponding prefix. They are preceded by a numerical or letter locant to describe the position of the stereogenic unit when such locants are present; general rules of numbering are applied (see P-14.4).&lt;br&gt;&#xA;  (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Accordingly, all stereodiscriptors have to be preceded by a locant.&#xA;In case of the PIN for cinnamaldehyde, the stereodiscriptor ‘&lt;em&gt;E&lt;/em&gt;’ is preceded by the numerical locant ‘2’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general nomenclature, however, locants are usually omitted when there is no ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the current IUPAC recommendations are prescriptive about when omission of locants is permissible in PINs. Locants are omitted in PINs only in the exceptional cases that are explicitly mentioned in the IUPAC recommendations. None of these cases applies to the locant of the stereodiscriptor in (2​&lt;em&gt;E&lt;/em&gt;)-3-phenylprop-2-enal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, if any locants are essential for defining the structure, then all locants must be cited in the PIN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the current recommendations given in the 2013 Blue Book incorporate a change from earlier recommendations given in the &lt;a href=&quot;http://www.acdlabs.com/iupac/nomenclature/93/r93_626.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;1993 Guide&lt;/a&gt;. By way of comparison, the following examples are explicitly mentioned in the 1993 Recommendations as well as in the 2013 Recommendations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dURkq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dURkq.png&quot; alt=&quot;(2E)-but-2-ene&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(2​&lt;em&gt;E&lt;/em&gt;)-but-2-ene (PIN, 2013 Recommendations)&lt;br&gt;&#xA;(&lt;em&gt;E&lt;/em&gt;)-but-2-ene (1993 Recommendations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FGNZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FGNZZ.png&quot; alt=&quot;(1Z)-1,2-dibromo-1-chloro-2-iodoethene &quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(1​&lt;em&gt;Z&lt;/em&gt;)-1,2-dibromo-1-chloro-2-iodoethene (PIN, 2013 Recommendations)&lt;br&gt;&#xA;(&lt;em&gt;Z&lt;/em&gt;)-1,2-dibromo-1-chloro-2-iodoethene (1993 Recommendations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Fq8e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Fq8e.png&quot; alt=&quot;(2Z)-2-methylbut-2-enoic acid&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(2​&lt;em&gt;Z&lt;/em&gt;)-2-methylbut-2-enoic acid (PIN, 2013 Recommendations)&lt;br&gt;&#xA;(&lt;em&gt;Z&lt;/em&gt;)-2-methylbut-2-enoic acid (1993 Recommendations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yJWvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yJWvd.png&quot; alt=&quot;fumaric acid&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(2​&lt;em&gt;E&lt;/em&gt;)-but-2-enedioic acid (PIN, 2013 Recommendations)&lt;br&gt;&#xA;(&lt;em&gt;E&lt;/em&gt;)-but-2-enedioic acid (1993 Recommendations)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-08-22T12:04:46.053" LastActivityDate="2015-08-22T12:04:46.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35321" PostTypeId="1" CreationDate="2015-08-21T14:07:00.930" Score="3" ViewCount="290" Body="&lt;p&gt;Could combining amphetamine and a (possibly selective) MAOI (MonoAmine Oxidase Inhibitor) increase the amphetamines potency without serious toxic side effects? If so by what factor, and would this be profitable for clandestine amphetamine producers to lower production costs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is most at home at the Chemistry Stackexchange. I'm looking into this matter from a medicinal chemist point of view.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="41890" LastEditDate="2017-11-27T17:58:41.477" LastActivityDate="2017-12-17T19:06:57.140" Title="Combining amphetamine and MAOIs" Tags="&lt;drugs&gt;&lt;medicinal-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35324" PostTypeId="1" AcceptedAnswerId="35327" CreationDate="2015-08-21T14:35:40.013" Score="5" ViewCount="16386" Body="&lt;p&gt;For those of you that have watched Season 1 of Prison Break (TV Series), in the &quot;Cute Poison&quot; Episode Micheal Scofield combined $\ce{CuSO4}$ (Copper Sulfate) and $\ce{H3PO4}$ (Phosphoric Acid) to weaken the metal. Is it true?&lt;/p&gt;&#xA;" OwnerUserId="19309" LastEditorUserId="7448" LastEditDate="2015-08-21T14:50:54.103" LastActivityDate="2016-09-11T10:23:07.067" Title="Does Cute Poison actually work?" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35326" PostTypeId="1" CreationDate="2015-08-21T15:13:22.510" Score="2" ViewCount="4997" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;The density of nitrogen at &lt;span class=&quot;math-container&quot;&gt;$\pu{0 ^\circ C}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt; is most nearly equal to which of the following quantities?&lt;br /&gt;&#xA;(a) &lt;span class=&quot;math-container&quot;&gt;$\pu{0.001 g/L}$&lt;/span&gt;; (b) &lt;span class=&quot;math-container&quot;&gt;$\pu{0.01 g/L}$&lt;/span&gt;; (c) &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 g/L}$&lt;/span&gt;;&#xA;(d) &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g/L}$&lt;/span&gt;; (e) &lt;span class=&quot;math-container&quot;&gt;$\pu{10 g/L}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The answer is (d), &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g L-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I think this is how you do it, but please correct me if I'm missing something. I didn't end up using the standard molar volume, so this might be incorrect.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$pV = nRT$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We know the pressure is &lt;span class=&quot;math-container&quot;&gt;$\pu{1 atm}$&lt;/span&gt;, the temperature is about &lt;span class=&quot;math-container&quot;&gt;$\pu{273 K}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R = 0.08$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Solving for &lt;span class=&quot;math-container&quot;&gt;$\frac {n}{V}$&lt;/span&gt;, we find &lt;span class=&quot;math-container&quot;&gt;$\frac {n}{V} = \frac {1}{22}$&lt;/span&gt;.  Now, this is in moles per litre, so by multiplying by the molar mass of nitrogen (approximately &lt;span class=&quot;math-container&quot;&gt;$\pu{28 g mol-1}$&lt;/span&gt;), we find that we have about &lt;span class=&quot;math-container&quot;&gt;$\pu{1 g}$&lt;/span&gt; per litre as the density.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="4945" LastEditDate="2020-06-15T13:02:12.207" LastActivityDate="2020-06-15T13:12:06.427" Title="Finding the density of nitrogen at 0 degrees Celsius and 1 atmosphere" Tags="&lt;gas-laws&gt;&lt;density&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35327" PostTypeId="2" ParentId="35324" CreationDate="2015-08-21T15:56:49.847" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://prisonbreak.wikia.com/wiki/Cute_Poison_(episode)&quot;&gt;The Wikipedia page on &quot;Cute Poison&quot; episode&lt;/a&gt; says that it wasn't &lt;a href=&quot;https://en.wikipedia.org/wiki/Copper(II)_phosphate&quot;&gt;anhydrous copper (II) phosphate&lt;/a&gt;, but &lt;a href=&quot;https://en.wikipedia.org/wiki/Gypsum&quot;&gt;Gypsum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't seen the episode, and the Wikipedia page's reaction seems to be the backwards reaction of yours. But I imagine yours is the one the pro/antagonist is gonna use to escape the prison; as&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gypsum is commonly found and accessible:&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gypsum is a soft sulfate mineral composed of calcium sulfate dihydrate, with the chemical formula $\ce{CaSO4·2H2O}$. It can be used as a fertilizer, is the main constituent in many forms of plaster and in blackboard chalk, and is widely mined. A massive fine-grained white or lightly tinted variety of gypsum, called alabaster, has been used for sculpture by many cultures including Ancient Egypt, Mesopotamia, Ancient Rome, Byzantine empire and the Nottingham alabasters of medieval England.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium_phosphate&quot;&gt;Calcium phosphate&lt;/a&gt;'s solubility will decrease with temperature increase and it will precipitate, leaving you with a solution of &lt;strong&gt;sulfuric&lt;/strong&gt; acid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's possible for a double displacement reaction to occur in aqueous medium (with a spark):&#xA;$$\ce{2H3PO4(aq) + 3CaSO4(aq)·2H2O(l) \leftrightharpoons 3H2SO4(aq) + Ca3(PO4)2(aq) + 6H2O(l)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H2SO4}$ is &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfuric_acid&quot;&gt;sulfuric acid&lt;/a&gt;. It's a very &lt;a href=&quot;https://en.wikipedia.org/wiki/Acid_strength#Strong_acids_in_water&quot;&gt;strong acid in water&lt;/a&gt;, a diprotic acid. Its &lt;a href=&quot;https://en.wikipedia.org/wiki/Acid_dissociation_constant&quot;&gt;$p {\rm K_a}$s&lt;/a&gt; are −3 and 1.99 according to Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Its corrosiveness on other materials, like metals, living tissues or even stones, can be mainly ascribed to its strong acidic nature and, if concentrated, strong dehydrating and oxidizing properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Later in the same article, you can read how it reacts with different material. It can weaken metal pretty fast and easily, as 3 moles of it will be present in any 7 moles of products, which is high above the dilute concentration (arbitrarily) defined in the Wikipedia page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper sulfate also reacts with phosphoric acid and forms sulfuric acid, but it's hardly accessible in prison, I reckon.&lt;/p&gt;&#xA;" OwnerUserId="7448" LastEditorUserId="7448" LastEditDate="2015-08-21T16:46:05.960" LastActivityDate="2015-08-21T16:46:05.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35328" PostTypeId="2" ParentId="34528" CreationDate="2015-08-21T16:07:36.830" Score="6" Body="&lt;p&gt;This is not a complete answer (merely a step in a particular direction, namely that of the falsehood of the claim), but I contacted the author of the book (Introduction to Medicinal Chemistry) Graham Patrick and he wrote the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear Jori,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many thanks for your e-mail and observations. Unfortunately, that&#xA;  statement was made in a very early edition of the textbook published&#xA;  over ten years ago and so I doubt if I will be able to track down the&#xA;  original reference on which it was based. As you say there is&#xA;  controversy and the statement is too sweeping. Work on the 6th edition&#xA;  is starting this year and so it will be possible to modify that&#xA;  particular passage. Many thanks for your input.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Best wishes Graham&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-08-21T16:07:36.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35332" PostTypeId="1" CreationDate="2015-08-21T16:55:19.220" Score="5" ViewCount="1037" Body="&lt;p&gt;Graham Patrick explains the mechanism for the activity of tyramine, amphetamine and ephedrine in his book &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199697396&quot; rel=&quot;nofollow&quot;&gt;Introduction to Medicinal Chemistry&lt;/a&gt; as follows: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some amines such as tyramine , amphetamine , and ephedrine closely resemble noradrenaline in structure and are transported into the nerve cell by noradrenaline’s transport proteins.&#xA;  Once in the cell, they are taken up into the vesicles. Because these&#xA;  amines are competing with noradrenaline for transport proteins,&#xA;  noradrenaline is more slowly reabsorbed into the nerve cells.&#xA;  Moreover, as the foreign amines are transported into the nerve cell,&#xA;  noradrenaline is transported out by those same transport proteins.&#xA;  Both of these facts mean that more noradrenaline is available to&#xA;  interact with its receptors. Therefore, amphetamines and similar&#xA;  amines have an indirect agonist effect on the adrenergic system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have two question concerning this explanation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When these foreign amines have been taken up into the vesicles how do they proceed? Do they re-enter the synaptic cleft by exocytosis, so that they can re-exercise their negative influence on noradrenaline re-uptake? How are they eventually cleared from the synaptic area?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do the noradrenaline re-uptake proteins transport noradrenaline &lt;strong&gt;out&lt;/strong&gt; of the axonal terminal under the influence of these foreign amines?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="41890" LastEditDate="2017-11-27T18:01:24.847" LastActivityDate="2017-11-27T18:01:24.847" Title="Regarding the mechanism for biological activity of tyramine, amphetamine and ephedrine" Tags="&lt;biochemistry&gt;&lt;amines&gt;&lt;drugs&gt;&lt;medicinal-chemistry&gt;&lt;pharmacology&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35333" PostTypeId="1" AcceptedAnswerId="35335" CreationDate="2015-08-21T17:17:49.997" Score="7" ViewCount="22245" Body="&lt;p&gt;We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="17217" LastEditDate="2015-09-06T08:34:27.423" LastActivityDate="2018-08-23T05:23:22.937" Title="Why the bond angle of PH3 is lesser that that of PF3?" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;molecular-structure&gt;&lt;vsepr-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35334" PostTypeId="1" AcceptedAnswerId="35350" CreationDate="2015-08-21T17:18:19.157" Score="0" ViewCount="790" Body="&lt;p&gt;Can we use the uniform circular motion formulae to calculate the velocity  of an electron along with the formula of radius of nth Bohr orbit? And using that can we calculate the kinetic energy of an electron?&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="10185" LastEditDate="2015-08-21T17:53:18.480" LastActivityDate="2015-08-22T03:59:06.743" Title="Uniform circular motion of electrons?" Tags="&lt;physical-chemistry&gt;&lt;energy&gt;&lt;electrons&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35335" PostTypeId="2" ParentId="35333" CreationDate="2015-08-21T18:52:15.693" Score="2" Body="&lt;p&gt;The answer can not be determined using VSEPR theory. The number of lone pairs in both compounds is the same. The answer lies in electronegativity. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;In $\ce{NH3}$ nitrogen is more electronegative than hydrogen and therefore it will pull the electrons towards it. Therefore the electrons will be closer to each other hence more repulsion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But in $\ce{NF3}$ fluorine will pull electrons towards itself therefore the electrons are further comparatively and hence repulsion will be less and so is the bond angle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But in case of $\ce{PF3}$ and $\ce{PH3}$  back bonding is possible in $\ce{PF3}$ therefore it has a larger bond angle .&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="10185" LastEditDate="2016-01-08T12:01:02.673" LastActivityDate="2016-01-08T12:01:02.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35336" PostTypeId="2" ParentId="29330" CreationDate="2015-08-21T19:55:08.097" Score="7" Body="&lt;p&gt;Kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crystal grows with the same speed in three perpendicular directions, by slowly starting new planes in each direction, which then fill much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quickly filling new planes minimises surface energy, so that seems logical. That a cubic space group has no preferred growth direction is not as compellent, but fits the bill in case of pyrite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Be careful: This rationalises the fact, but predictions, like the other answers say, are difficult.)&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-08-21T20:04:29.777" LastActivityDate="2015-08-21T20:04:29.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35338" PostTypeId="1" CreationDate="2015-08-21T20:59:19.627" Score="8" ViewCount="1761" Body="&lt;p&gt;$\ce{KOH}$, $\ce{KCN}$, $\ce{KNO3}$, $\ce{K2CO3}$, and plenty of other potassium salts are frequently used in industry.  But, as far as I can see, the sodium equivalents are also widely used, often interchangeably.  Since Na weighs only 60% of K it seems like the sodium salts should always be preferred because they carry more of whatever base is attached per unit weight (i.e., the sodium salts have &quot;more of what you want&quot;). $\ce{Na}$ is also slightly more abundant (on earth) than K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what other factors favor the industrial use of K-salts?  E.g., are K-salts just more abundant in natural sources?  Are they more reactive because the ionization energies of K are lower than those of Na?  Are the behaviors of these simpler salts actually different between the K- and Na- variants, and if so, is there some other elemental characteristic that suggests this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going the other way, it seems like if K is preferable to Na, Rb should be even more so (because it appears that in every elemental characteristic in which K exceeds Na, Rb exceeds K).  Would it be if it weren't so relatively scarce?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="24052" LastEditDate="2016-03-11T23:22:32.893" LastActivityDate="2019-10-28T19:48:15.367" Title="Why are potassium organic salts ever preferable to the sodium equivalents?" Tags="&lt;elements&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35339" PostTypeId="1" AcceptedAnswerId="37653" CreationDate="2015-08-21T21:23:07.893" Score="6" ViewCount="8539" Body="&lt;p&gt;I was trying to determine the reaction mechanism behind the production of diethyl ether. Looking at the ethyl alcohol + sulphuric acid method, I have come to the following conclusions for the reaction steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ethanol nucleophilic oxygen gets protonated by the addition of sulphuric acid. This is more favourable because of the high density of electrons surrounding the oxygen. The bond broken between the carbon and oxygen is also needed to form the bond with hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is then released and the ethyl alcohol conjugate base is formed (H3C-CH2+). The remaining ethyl alcohol molecules can either react with the HSO4- or the ethyl conjugate... Since HSO4- is more stable than the ethyl conjugate, ethyl alcohol reacts with the ethyl conjugate. Carbon is more electronegative than hydrogen and is an electrophile. It is attracted to the oxygen from the ethyl alcohol and replaces the H, thus forming diethylether, H+ and H2O where the sulphuric acid essentially acts as a catalyst!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in presenting the mechanism this way? If the sulphuric acid still remains in the solution, why does it not protonate the diethyl ether? Is it because the stability of the diethyl ether conjugate base is unstable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19218" LastActivityDate="2015-12-22T07:55:39.080" Title="Diethyl Ether reaction mechanism" Tags="&lt;reaction-mechanism&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35340" PostTypeId="1" CreationDate="2015-08-21T21:38:15.653" Score="14" ViewCount="5842" Body="&lt;p&gt;I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check with. My book is also very confusing on this subject, and I don't have a world renowned memory (meaning I'd rather understand than memorize the equation), so I tried to think about it and came with the following conclusions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Freezing Point Depression: When a solvent is not pure and has particles dissolved in it, there are constituents (be it molecules, ions, etc.) that take up volume in the solution. For freezing to occur, these constituents need to aggregate to form a lattice, bonds, etc. Say we have a pure solvent X. This process occurs at a temperature Y. However, when this solvent is not pure and has dissolved solutes, there is &quot;blockage&quot; impeding these constituents from aggregating and bonding. Therefore, the temperature Y needs to be reduced further to slow down the kinetic energy of all constituents in solution (solvent and solute), which gives a greater statistical probability of intermolecular forces getting an opportunity to bind solvent-solvent constituents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Boiling Point Elevation: Boiling, if I'm not mistaken, occurs when the partial pressure of a vapor exceeds atmospheric pressure. For a gas to enter the vapor phase, molecules in solution need to have a certain kinetic energy (as specified by a Boltzmann distribution) to free themselves of the intermolecular forces in the solution phase and become a vapor. When a pure solvent is now diluted with other particles, the new solute-solvent bonding energy needs to be overcome to allow a solvent molecule to exist as a vapor. One way to overcome this energy is to increase the average kinetic energy of all molecules, and this is done through a temperature increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for an introductory chemistry course, and I'm not sure if my explanations are sound. I just want a yay or a nay in terms of my logic, because I don't have the most intimate background in chemistry and I'm trying to conceptually understand things. &lt;/p&gt;&#xA;" OwnerUserId="19315" LastEditorUserId="189" LastEditDate="2015-08-22T12:30:25.767" LastActivityDate="2016-12-21T16:13:55.077" Title="Justification for Freezing Point Depression &amp; Boiling Point Elevation in Solutions?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;solutions&gt;&lt;boiling-point&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35341" PostTypeId="2" ParentId="444" CreationDate="2015-08-21T21:58:52.963" Score="-6" Body="&lt;p&gt;Let's take a look at the periodic table: in the first row, there's only two elements: Hydrogen and Helium. They do not follow an octet rule. Hydrogen can only have a maximum of two electrons on the valence orbital. It turns out that the octet rule is not exclusive, meaning it is not the only rule that helps understand Lewis structure and electron configuration. Why do we use the octet rule, then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every period in the periodic table represents an energy shell of an atom. The first period represents the shell K, the first energy level, which has only the s orbital. Every orbital can only be filled with 2 electrons, both with a quantum spin towards opposite directions. Thus the maximum number of electrons possible for the first energy level shell, K, is 2. This is reflected in the fact that Helium is a noble gas, yet only contains 2. The second energy level shell, L, has the s orbital and the extra 3 p orbitals. Those add up to four orbitals or 8 electrons. Because the elements most commonly used are in the second and third period, the octet rule is in frequent use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements of the third energy level are very similar. They still follow the octet rule, because even though now the have 5 d orbitals, no orbital needs to be filled. The electronic configuration shows that 4s is filled before 3d, so they don't need to fill the d orbital, thus they usually also obey the octet rule. However, third energy level shell elements, unlike second-row elements, (see Gavin's comment fir reference) are not limited to the octet rule. They can form hypervalent molecules in certain cases where the use that d orbital and fills — this is not the case with all apparent hypervalent molecules, SF6 is not hypervalent, it uses weak ionic bonds and polarity, but there still are hypervalent molecules out there. It will always depend on which state is more convenient in terms of electrostatics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the fourth energy level shell, there are f orbitals introduced, but we are not even close to filling them at that point because we first need to fill the d orbitals. The 5 d orbitals signify 10 electrons, plus the previous eight from the octet rule, sum up to 18. This is the reason why there are 18 columns in the periodic table. Now, a new rule superposes, and this is the well known 18-electron rule, which was mentioned above. Transition metals obey this rule with more frequency than not, though there are occasions in which they still obey the octet rule. At this point, with so many orbitals to fill, and with electrostatics playing a role in electronic configuration, we can obtain different cations from the same element with certain metals. That is also why they don't discuss oxidation state numbers with transition metals like they do with the first three rows of the table.&lt;/p&gt;&#xA;" OwnerUserId="19316" LastEditorUserId="402" LastEditDate="2017-04-02T20:33:58.107" LastActivityDate="2017-04-02T20:33:58.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35342" PostTypeId="1" CreationDate="2015-08-21T22:26:41.423" Score="1" ViewCount="195" Body="&lt;p&gt;In which of the following reactions is the equilibrium farthest to the left?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $\ce{P4O10 + $x$ H2O &amp;lt;=&amp;gt; 4 H3PO4(aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $\ce{AlCl3 + $x$ H2O &amp;lt;=&amp;gt; [Al(H2O)6]^3+ + 3 Cl- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) $\ce{Li2O + $x$ H2O &amp;lt;=&amp;gt; 2 Li+(aq) + 2 OH- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) $\ce{H2S + H2O &amp;lt;=&amp;gt; H3O+ + HS- (aq)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) $\ce{CaC2 + $x$ H2O &amp;lt;=&amp;gt; Ca^2+(aq) + 2 OH- (aq) + C2H2(g)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is d, but I'm not sure how to compare these equilibria.  I know that the answer can't be e, since the evolution and purging of gas will drive the equilibrium forward.  Also, the dissociation of ions in water is favorable due to solvation, so that rules out c and b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure how to compare the remaining two reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="865" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T18:56:10.000" LastActivityDate="2021-05-22T18:56:10.000" Title="Chemical Equillibria Comparisons" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35343" PostTypeId="1" AcceptedAnswerId="35363" CreationDate="2015-08-21T22:38:41.017" Score="1" ViewCount="1143" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In which of the reactions below is the first compound in the equation NOT oxidized?&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/VWaCn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VWaCn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct choice is (D), which can be seen through process of elimination, but I was wondering how to do this through the carbon oxidation number?           &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that to know the oxidation number of nitrogen, I need the oxidation number of carbon, but to know the oxidation number of carbon, I need to know the oxidation number of nitrogen.       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the correct way to think about this problem that: (D) didn't gain any oxygen atoms, therefore, wasn't oxidized and leave it at that?&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="39539" LastEditDate="2018-03-01T12:24:04.570" LastActivityDate="2018-03-01T12:24:04.570" Title="Oxidation of Carbons" Tags="&lt;organic-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35345" PostTypeId="1" CreationDate="2015-08-21T23:11:11.817" Score="1" ViewCount="472" Body="&lt;p&gt;Where did the description that acids are sour and bases are bitter come from? With Arrhenius definition, those descriptions may have fit acids and bases, but with the Lewis definition, I guess they would not really apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, $\ce{AlCl3}$, a known Lewis acid, does not taste sour but rather &quot;gives water a bitter astringent taste&quot; as stated by PubChem in this &lt;a href=&quot;http://pubchem.ncbi.nlm.nih.gov/compound/aluminum_chloride#section=Stability&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="402" LastEditDate="2017-04-25T23:05:28.277" LastActivityDate="2018-11-05T15:19:52.853" Title="Where did the description that acids are sour and bases are bitter come from?" Tags="&lt;acid-base&gt;&lt;taste&gt;" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35346" PostTypeId="2" ParentId="35339" CreationDate="2015-08-21T23:18:27.043" Score="-3" Body="&lt;p&gt;You got it all wrong. First of all the reaction written by you concisely is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{diethyl~ ether + H2SO4 -&amp;gt; CH3-CH2-OH + CH3-CH2-HSO4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first step is correct that oxygen gets protonated due to acid. Then you said that  water is then released and the ethyl alcohol conjugate base is formed $\ce{H3C-CH2+}$. This is the point where you got it wrong. The reaction goes under SN2 mechanism( I am not going into the details of it) and there is no formation of intermediates as you gave it as $\ce{H3C-CH2+}$ So no carbocation inter mediates are formed only transition stage is formed. &#xA;Your next question was for what if $\ce{H2SO4}$ is in excess.&#xA;Here it goes like this &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ethanol formed is again protonated by $\ce{H2SO4}$ by the same reason you stated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hence now there are $\ce{CH3CH2OH2+}$( where +ve charge is on oxygen) and $\ce{HSO4-}$ anions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;These $\ce{HSO4-}$ again attack on $\ce{CH3CH2OH2+}$ by SN2 mechanism. &lt;/li&gt;&#xA;&lt;li&gt;Now you will get salt of the 3rd step and water as by product.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19317" LastEditorUserId="17368" LastEditDate="2015-12-22T07:55:39.080" LastActivityDate="2015-12-22T07:55:39.080" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35348" PostTypeId="2" ParentId="35338" CreationDate="2015-08-21T23:56:07.553" Score="7" Body="&lt;p&gt;The differences in the properties of the alkali cations are subtle, and the rationale for choosing one over the other will depend heavily on the application. I suspect you're right, that to some extent natural abundance of, e.g., potassium salts relative to sodium ones in certain mineral contexts explains the industrial preference for one or the other -- but, really, variations in abundance are just further examples of these subtle differences in the properties of the cations!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One specific example that I recently came across is the &lt;a href=&quot;http://www1.lsbu.ac.uk/water/hofmeister_series.html&quot; rel=&quot;noreferrer&quot;&gt;Hofmeister series&lt;/a&gt; (see also the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hofmeister_series&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;), where the choice of cation makes a demonstrable difference in the solubility properties of proteins.  Speaking also from personal experience, the choice of, e.g., $\ce{Na2SO4}$ versus $\ce{K2SO4}$ versus $\ce{(NH4)2SO4}$ as a component of the supporting electrolyte in electrodeposition of copper from acidic $\ce{CuSO4}$ matters significantly to the properties of the deposit obtained.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-08-22T17:27:11.650" LastActivityDate="2015-08-22T17:27:11.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35349" PostTypeId="1" AcceptedAnswerId="82695" CreationDate="2015-08-22T00:07:58.177" Score="23" ViewCount="62330" Body="&lt;p&gt;$\ce{F}$ has more unshared electron pairs and is very electronegative, so $\ce{H}$ of another $\ce{HF}$ molecule can $\ce{H}$-bond with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HF}$ has normal boiling point of $\pu{19.5^oC}$ while $\ce{H2O}$, as you know, has normal boiling of $\pu{100^oC}$.&lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="5026" LastEditDate="2018-02-24T15:25:36.410" LastActivityDate="2018-02-24T15:25:36.410" Title="Why does hydrogen fluoride have a boiling point so much lower than that of water?" Tags="&lt;inorganic-chemistry&gt;&lt;thermodynamics&gt;&lt;water&gt;&lt;hydrogen-bond&gt;" AnswerCount="5" CommentCount="0" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="35350" PostTypeId="2" ParentId="35334" CreationDate="2015-08-22T03:59:06.743" Score="4" Body="&lt;p&gt;Well yes you can if you assume the Bohr model is correct. However in reality, it is a little bit more complicated than that. In reality everything have both particle and wave like properties, which builds the foundation of quantum mechanics. The lighter the object, the more wave like properties it exhibits and the heavier the object, the more particle like properties it exhibits. So a ball actually has wave light properties but it is too heavy for it to be notices. Meanwhile a light wave actually has particle properties, however it is too light to be noticed. Electrons on the other hand turn out to a mass between these two extremes, hence both its particle and wave like properties are able to be noticed.Hence it is important to stop thinking electrons as small point particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important concept is the Heisenberg Uncertainty Principle which is the following formula: $$\Delta x \Delta p \geq \hbar$$ This means that the uncertainty of an object's position multiplied by the uncertainity in an object's momentum at best can only be equal to the constant value: $\hbar$ which equals $1.054*10^{-31}$ This equation can be applied to electrons. Hence, the important thing that this equation tells us about electrons in an atom is that rather electrons moving in a predictable circular orbit, they actually travel in any path that they want and we actually can't precisely determine position of the electron in an atom. Hence the quantum mechanical model of the atom actually consists of orbitals, rather than orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orbitals is the area which an electron is in 90% of the time. There are 4 main type of oribitals, s, p, d and f shown below. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/HNThh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNThh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finding the velocity of an electron is little more complicated than using simple classical mechanic equations. However its velocity can be determined in simplified conditions using quantum mechanics. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-08-22T03:59:06.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35351" PostTypeId="1" CreationDate="2015-08-22T04:18:36.743" Score="19" ViewCount="15960" Body="&lt;p&gt;I have recently been “brewing” what is commonly called “iron acetate” for &lt;a href=&quot;http://www.wwgoa.com/article/ebonizing-wood/&quot; rel=&quot;noreferrer&quot;&gt;ebonizing wood&lt;/a&gt;, and I'd like to understand the chemistry of the iron/acetic acid reaction, which &lt;em&gt;should&lt;/em&gt; be fairly simple, but turns out to be more complex than I thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One previous question relates to the &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5027/how-does-the-mordant-ferric-acetate-interact-with-tannins-when-ebonizing-wood&quot;&gt;tannin/mordant reaction&lt;/a&gt;, a later stage in the process; another &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7574/can-i-synthesize-iron-acetate-like-this&quot;&gt;asks for feedback on a recipe for iron (II) acetate&lt;/a&gt;, which only adds to my confusion; while a third &lt;a href=&quot;https://chemistry.stackexchange.com/questions/7578/decomposition-of-iron-iii-acetate&quot;&gt;asks about decomposition of iron (III) acetate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Overview&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The basic idea is that you put fine steel wool in white vinegar, the iron reacts with the acetic acid to form an iron (II?, III?) acetate solution; you then then paint this on the wood where the iron acetate reacts with naturally occurring tannins (or introduced by way of previous application of a tannin-rich solution, such as black tea) in the wood, yielding iron tannates, which are dark-colored, thus staining the wood.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There are many websites, blogs, and articles online describing the ebonizing process and how to make “iron acetate” with varying degrees of utility and accuracy. None of them contain a detailed and accurate description of the chemistry, which I would like to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the existing online articles, there is a great deal of conflicting information (even to the point of direct contradictions). It's so bad that it is not &lt;em&gt;at all&lt;/em&gt; clear to me what the goal of producing the solution is: do we want iron (II) acetate or iron (III) acetate? Does it matter?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Procedure&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I put one US quart of warm distilled white household vinegar in a mason jar (5%, ~0.83 mol CH&lt;sub&gt;3&lt;/sub&gt;COOH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added two 0000 wire wool pads (~23g, 0.41 mol Fe), unrolled, cut into strips, washed in acetone (to remove protective oil) and dried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I placed the mason jar inside my vanilla Weber barbecue grill in the hot sun, reasoning that the somewhat elevated temperature might speed up the reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I lightly agitated the contents a couple of times per day with a bamboo skewer, squeezing out bubbles trapped in the wire wool to keep it from floating on the liberated gas (presumably hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout this process, &lt;strong&gt;the liquid remained clear&lt;/strong&gt;, although some reddish crud was deposited on the mouth of the mason jar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some of the solution evaporated, it was topped up with vinegar. This was done twice or three times over a week. Total additions were at most 4 fl.oz. (100 ml).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a week or so (maybe 8 days), all of the steel wool appeared to have been consumed, and there was a dark grey or black sludge in the bottom of the mason jar, which could just have been unreacted fragments of steel wool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was some (predominantly) charcoal grey “foam” on top, but with a large &lt;strong&gt;bright orange-brown spot&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I filtered the liquid through a coffee filter into a transparent plastic container (a washed-out 40 oz. Lysol bottle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During filtering, the liquid appeared &lt;strong&gt;dark&lt;/strong&gt;, with a distinct &lt;strong&gt;green tinge&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last few ounces were filtering slowly (lots of crud in the filter), so I left it for a couple of hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On returning, the solution did not appear green, but had a definite &lt;strong&gt;purple tinge&lt;/strong&gt; (my g/f even said “Why is it purple?”).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the last few days, the solution has lost its purplish color and stratified — there are a few millimeters of dark brown precipitate at the bottom (probably not as much as it seems due to the concave bottom of the Lysol bottle), the bulk of the liquid is almost clear orangey brown, there is a narrow but distinct dark band about 35mm from the top, then above that, it is cloudier and slightly lighter-colored than the bulk. Unfortunately, the narrow band and top stratum are too subtle to be seen in the photographs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjRRM.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjRRM.jpg&quot; alt=&quot;Precipitate in filtered solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top, there is an &lt;strong&gt;orangey-brown deposit&lt;/strong&gt; around the edge and &lt;strong&gt;crystalline flakes&lt;/strong&gt; floating on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oaZHp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oaZHp.jpg&quot; alt=&quot;Surface of filtered solution after a few days&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The filter paper has dried, leaving a fine reddish-brown powder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwVwQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwVwQ.jpg&quot; alt=&quot;Dry filter paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b7MTN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b7MTN.jpg&quot; alt=&quot;Reddish-brown powder from filter paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Other Observations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The colorless liquid in the mason jar, prior to filtering, was &lt;em&gt;much&lt;/em&gt; more effective than expected. From all online descriptions, iron acetate is not terribly effective on low-tannin woods like pine, yet only a few hours after putting the steel wool into the mason jar, a drop of the solution placed on an untreated pine strip produced a very noticeable graying effect (much like I'd expected from the final result); the next day, the solution produced a mid gray/brown; and after a few days, a dark brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nejsn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nejsn.jpg&quot; alt=&quot;Solution on untreated pine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On pine treated with tea, a jet black color was produced by all but the weakest solution, which was dropped on while the pine was still damp with tea (hey, I'm impatient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IYQbq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYQbq.jpg&quot; alt=&quot;Solution on pine painted with tea&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Assumptions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Ferrous (iron II) acetate is soluble in water. It is a white crystalline solid. Its tetrahydrate is light green.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferric (iron III) acetate is insoluble in water, but soluble in ethanol. It is an orangey brown solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of impurities can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distilled vinegar did not have much dissolved oxygen. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Speculation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the mason jar&lt;/strong&gt;, the iron in the wire wool reacted with the acetic acid, mostly yielding ferrous acetate tetrahydrate (and hydrogen, which just bubbled off), which accounted for the green tinge of the solution observed during filtration. There was also some ferric acetate created in the presence of the limited dissolved oxygen and that available at the surface; this precipitated out into the gray sludge, which also contained unreacted fragments of wire wool. The orange spot in the surface foam was, most likely, ferric acetate and (?) ferric oxide hydrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the Lysol bottle&lt;/strong&gt;: on filtration, atmospheric oxygen was introduced to the solution, which enabled some ferrous acetate to transition to ferric acetate (pathway?), as this which is what has precipitated out. This ferrous-&gt;ferric reaction (?) is continuing at the surface. Some of the precipitate has been deposited around the edge of the bottle and some has crystallized to form the flakes on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the filter paper&lt;/strong&gt;, the sludge was residual unreacted wire wool, ferric acetate, and ferric oxide hydrates, dampened with ferrous acetate and acetic acid solution. These have reacted and dried out to a mixture of ferric acetate and ferric oxide hydrates, accounting for the orangey brown color.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Questions&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the above speculation broadly correct?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the end result of putting wire wool in vinegar ultimately becomes ferric acetate (insoluble precipitate, soluble in ethanol), why don't the instructions ever tell you to filter off the precipitate and dissolve it in denatured alcohol? Surely that would make more sense. If that is the goal, why doesn't the usual recipe call for some hydrogen peroxide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, if the desired result (for ebonizing) is ferrous acetate, why don't the instructions tell you to avoid introducing oxygen? Tell you to brew it with an airlock? Warn you that the solution will ultimately “go bad” when the ferrous acetate becomes ferric acetate and precipitates out? Tell you to keep it in an airtight container, use it quickly, etc.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;What caused the &lt;em&gt;purple&lt;/em&gt; tinge?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Immediately after filtration, the solution appeared purplish. Was this just the mixture of an orangey color and a greeny color as the greenish ferric acetate tetrahydrate turned to orangey ferrous acetate or was there something else going on?&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;If an aqueous solution of ferrous acetate becomes ferric acetate, which precipitates out, what is the most likely reaction/pathway (under these conditions)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm sure I could work something out, but I'm not sure it'd be correct :)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Are there any simple diagnostic tests I could do?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There may be very simple (backyard/household chemical) tests that would enable me to confirm/refute parts of my speculation or estimate relative concentrations. I'm thinking along the lines of “a drop of that in two drops of ammonia will turn a banana blue” rather than “put 10 ml in a test tube with Ludicrously Expensive's reagent in an argon atmosphere”.&lt;/p&gt;&#xA;" OwnerUserId="19318" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2019-10-18T18:25:06.260" Title="Iron chemistry: acetates for ebonizing wood" Tags="&lt;everyday-chemistry&gt;&lt;reaction-mechanism&gt;&lt;redox&gt;&lt;color&gt;" AnswerCount="5" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35352" PostTypeId="2" ParentId="35349" CreationDate="2015-08-22T04:43:49.220" Score="2" Body="&lt;p&gt;Both fluoride and oxygen are very electronegative. When they bond with hydrogen the hydrogen becomes slightly positive and the electronegative atom slightly negative. Because of this, attraction occurs between the negative atoms and hydrogen in different molecules, called hydrogen bonding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In water there are two hydrogen making more charge dipoles for stronger and more numerous hydrogen bonds.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor is water disassociates (i.e. liquid water is partially made up of $\ce{H3O+}$ (three hydrogen and an oxygen, which is positively charged) and some $\ce{OH-}$ (one hydrogen and one oxygen, negatively charged). This increases inter-molecule interaction, making the boiling point higher.  &lt;/p&gt;&#xA;" OwnerUserId="19320" LastEditorUserId="39315" LastEditDate="2017-05-12T08:09:30.950" LastActivityDate="2017-05-12T08:09:30.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35353" PostTypeId="1" CreationDate="2015-08-22T05:23:00.640" Score="2" ViewCount="43" Body="&lt;p&gt;I am interested in the effects of plant environments on their biochemistry in relation to their effects on the environment and animals(us).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could sodium fluoride or aluminum  accumulation in soils contribute to increased amounts of palmitic acid in food?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus increasing the inflammatory effects of certain foods?&lt;/p&gt;&#xA;" OwnerUserId="19322" LastActivityDate="2017-08-23T10:22:06.517" Title="Could sodium fluoride or aluminum accumulation in soils contribute to increased amounts of palmitic acid in food?" Tags="&lt;biochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35354" PostTypeId="2" ParentId="35349" CreationDate="2015-08-22T05:49:09.467" Score="15" Body="&lt;p&gt;I think there are a variety of &lt;em&gt;qualitative&lt;/em&gt; ways of looking at this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond &lt;em&gt;acceptors&lt;/em&gt; and &lt;em&gt;donors&lt;/em&gt;. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only &lt;em&gt;one&lt;/em&gt; hydrogen bond acceptor and theoretically &lt;em&gt;three&lt;/em&gt; hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous. &lt;/li&gt;&#xA;&lt;li&gt;As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons &quot;want&quot; to be stabilized by nuclei. If they already are fairly well stabilized then they will &quot;feel&quot; less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules &lt;em&gt;might&lt;/em&gt; be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having &lt;em&gt;three&lt;/em&gt; lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all &lt;em&gt;qualitative&lt;/em&gt;, but this is the sort of reasoning introductory chemistry teachers desire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If&lt;/em&gt; we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The &lt;em&gt;bond angle&lt;/em&gt; of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After scouring the web some common &quot;explanations&quot; that &lt;em&gt;would not&lt;/em&gt; explain why water has a higher boiling point than $\ce{HF}$ would be: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;water can form 4 per molecule while HF can only form 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond &lt;em&gt;donors&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-08-23T15:01:21.737" LastActivityDate="2015-08-23T15:01:21.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35355" PostTypeId="1" CreationDate="2015-08-22T06:17:24.897" Score="3" ViewCount="1709" Body="&lt;p&gt;Is it possible and safe to use liquid nitrogen to quickly cool a 5 gallon tank of liquid from room teperature to say 34 °F? Or is there another way that can produce really fast results? &lt;/p&gt;&#xA;" OwnerUserId="19324" LastEditorDisplayName="user7951" LastEditDate="2016-07-16T11:49:23.797" LastActivityDate="2017-12-12T16:19:20.267" Title="Liquid nitrogen to cool liquid?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;home-experiment&gt;" AnswerCount="4" CommentCount="6" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35356" PostTypeId="2" ParentId="35173" CreationDate="2015-08-22T06:33:42.290" Score="1" Body="&lt;p&gt;As indicated in the comments, freeze-thaw cycles should not affect molecules of size of typical antibiotics. Freeeze-thawing may affect colloids by disrupting their structure. Also, the effect would depend on how fast a sample was frozen/thawed etc (slow freezing would cause rearrangement of molecules and crystallization of water whereas rapid freezing would result in vitrification of water.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The popular lab practice, in my opinion, would have perhaps arisen because it is better to keep the stock stored at the appropriate temperature (frozen) for longer life. Repeated thawing of the stock would definitely increase its chance of degradation because of mishandling and even otherwise. Therefore, it is better to aliquot smaller volumes; even if you have forgotten to keep it back into the freezer you'll end up losing only a small amount. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt; &#xA;&#xA;&lt;p&gt;&lt;em&gt;Not related to your question but this is an interesting fact: freeze-thaw cycles promote molecular assembly of RNA into higher order structures (&lt;a href=&quot;http://www.nature.com/nchem/journal/v7/n6/full/nchem.2251.html&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;).&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="5295" LastActivityDate="2015-08-22T06:33:42.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35358" PostTypeId="2" ParentId="35349" CreationDate="2015-08-22T07:24:04.853" Score="5" Body="&lt;p&gt;The answer lies in hydrogen bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen bond energy depends on the electronegativity of a highly electronegative atom which is bonded to hydrogen. Electronegativity of hydrogen is $2.2$, for oxygen it's $3.44$ and it's $4$ for fluorine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Difference in electronegativity between $\ce{F}$ and $\ce{H}$ is $1.8$ and between $\ce{O}$ and $\ce{H}$ is $1.24$. Hydrogen bond energy of $\ce{H-F}$ is $\pu{41.83 kJ/mole}$ and that of $\ce{O-H}$ is $\pu{23 kJ/mole}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-F}$ bond is stronger in comparison to $\ce{O-H}$ bond. In the case of $\ce{H-F}$, there exists hydrogen bonding even in vapour state, 4 to 7 $\ce{HF}$ molecules together form one unit in vapour state. However, in the case of water, there is no hydrogen bonding in vapour state; each water molecule exists independently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to boil liquid water, all hydrogen bonds have to be broken and it requires a large amount of energy. This isn't the case in $\ce{HF}$; all hydrogen bonds need not to be broken, and therefore a lesser amount of energy is required. So $\ce{HF}$ boils at a much lower temperature as compared to water.&lt;/p&gt;&#xA;" OwnerUserId="19323" LastEditorUserId="43942" LastEditDate="2017-05-12T11:21:50.453" LastActivityDate="2017-05-12T11:21:50.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35359" PostTypeId="1" CreationDate="2015-08-22T09:13:50.683" Score="6" ViewCount="1678" Body="&lt;p&gt;I learned that, a phenol will not react (or react very slowly) with $\ce{PCl5}$ due to its stabilized structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about the $\ce{-OH}$ group on the benzene ring of 4-hydroxybenzoic acid? Will that $\ce{-OH}$ group react with $\ce{PCl5}$? If the answer is yes, is it anything to do with the $\ce{-COOH}$ group?&lt;/p&gt;&#xA;" OwnerUserId="19139" LastEditorUserId="16683" LastEditDate="2017-08-16T11:56:08.250" LastActivityDate="2017-08-16T11:56:08.250" Title="Can the hydroxyl group of 4-hydroxybenzoic acid react with phosphorus pentachloride?" Tags="&lt;organic-chemistry&gt;&lt;carbonyl-compounds&gt;&lt;halides&gt;&lt;phenols&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35362" PostTypeId="1" AcceptedAnswerId="37194" CreationDate="2015-08-22T10:51:38.133" Score="-3" ViewCount="224" Body="&lt;p&gt;What are the chances of warm water freezing faster than cold water? (How many percent?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to do with container size or something in it?&lt;/p&gt;&#xA;" OwnerUserId="19309" LastEditorUserId="16622" LastEditDate="2015-08-22T11:19:53.343" LastActivityDate="2020-04-21T09:54:27.147" Title="Frequency of Mpemba Effect" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="2" ClosedDate="2020-04-20T14:41:01.080" ContentLicense="CC BY-SA 3.0" />
  <row Id="35363" PostTypeId="2" ParentId="35343" CreationDate="2015-08-22T12:53:41.777" Score="3" Body="&lt;p&gt;There are several ways to determine oxidation number/state. The most useful method for organic (covalent) compounds uses the following three rules, and works for most elements engaged in covalent bonding (just replace carbon with some other element):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Every bond between a carbon atom and a more electronegative atom increases the oxidation number of that carbon atom by 1.&lt;/li&gt;&#xA;&lt;li&gt;Every bond between a carbon atom and a less electronegative atom (hydrogen is most common) decreases the oxidation number of that carbon atom by 1.&lt;/li&gt;&#xA;&lt;li&gt;Every bond between a carbon atom and another carbon atom has no effect on the oxidation number of the carbon atom (although we could get more complicated since &lt;a href=&quot;http://www.madsci.org/posts/archives/2000-09/968887193.Ch.r.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;carbon atoms with different &quot;hybridizations&quot; have different electronegativities&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This approach is equivalent to assigning all of the electrons in a covalent bond to the more electronegative atom for purposes of determining oxidation number. &lt;a href=&quot;http://www.masterorganicchemistry.com/2011/07/25/calculating-the-oxidation-state-of-a-carbon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page on Master Organic Chemistry&lt;/a&gt; has a number of examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on the matter&lt;/a&gt;, using acetic acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pzLzF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pzLzF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Atom            Bonds to   Contribution to Ox# &#xA;H (methyl)       C         +1 (2.5 &amp;gt; 2.1)&#xA;          TOTAL Ox#        +1 &#xA;C (methyl)       H         -1 (2.1 &amp;lt; 2.5)&#xA;                 H         -1&#xA;                 H         -1&#xA;                 C         +0 (same atom)&#xA;          TOTAL Ox#        -3&#xA;C (carboxyl)     O         +1 (3.5 &amp;gt; 2.5)&#xA;                 O         +1 (double bond counts as 2 bonds)&#xA;                 O         +1&#xA;                 C         +0&#xA;          TOTAL Ox#        +3&#xA;O (carbonyl)     C         -1 (2.5 &amp;lt; 3.5)&#xA;                 C         -1&#xA;          TOTAL Ox#        -2&#xA;O (hydroxyl)     C         -1 &#xA;                 H         -1 (2.1 &amp;lt; 3.5)&#xA;          TOTAL Ox#        -2&#xA;H (hydroxyl)     O         +1 (3.5 &amp;gt; 2.1)&#xA;          TOTAL Ox#        +1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-22T12:53:41.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35364" PostTypeId="1" AcceptedAnswerId="35375" CreationDate="2015-08-22T13:30:28.787" Score="0" ViewCount="863" Body="&lt;p&gt;There is a specific compound $\ce{C3H6O}$ that exists as hydrocarbon compound X. It does not react with metallic sodium, bromine does not lose it's color when applied to the compound, and does not give the silver mirror reaction when ammonium silver nitrate, in fact it doesn't react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've realized since it does not react with sodium, it cannot have a hydroxyl functional group, and that since bromine does not give the discoloration reaction, the Carbons of the compound cannot have double bonds. And obviously, if it does not give the silver mirror reaction, than it cannot be an aldehyde or a carboxyl acid. However, the answer that the book gave was $\ce{CH3COCH3}$. Am I doing something wrong here?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-10-02T16:38:18.690" LastActivityDate="2015-10-02T16:38:18.690" Title="Find the Structural Formula of Hydrocarbon Compound X" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35366" PostTypeId="2" ParentId="35340" CreationDate="2015-08-22T15:02:08.547" Score="20" Body="&lt;p&gt;I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann&amp;#39;s_entropy_formula&quot; rel=&quot;noreferrer&quot;&gt;Boltzmann's entropy formula&lt;/a&gt; tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} &amp;gt; S_{\mathrm{l}}^{\mathrm{pure}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; G = H - TS&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H$ is the enthalpy and $T$ is the temperature, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} &amp;lt; S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} &amp;gt; S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} &amp;gt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} &amp;gt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$.&#xA;Requiring the Gibbs free energy of vapourization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S})&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to be equal to zero gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA; \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} &amp;gt; S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} &amp;gt; S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} &amp;lt; \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.&lt;/p&gt;&#xA;" OwnerUserId="189" LastEditorUserId="189" LastEditDate="2015-08-23T12:40:37.440" LastActivityDate="2015-08-23T12:40:37.440" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="35367" PostTypeId="2" ParentId="35343" CreationDate="2015-08-22T15:36:40.067" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Or is the correct way to think about this problem that: (D) didn't gain any oxygen atoms, therefore, wasn't oxidized and leave it at that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the simplest way of thinking of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many introductory organic chemistry texts define oxidation as carbon forming additional bonds to oxygen, and reduction as carbon forming additional bonds to hydrogen. So if we have a reaction converting ethanol into acetaldehyde, this would be an oxidation; the carbon previously connected to the $\ce{-OH}$ group is now connected to oxygen with &lt;em&gt;two&lt;/em&gt; bonds instead of one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can generalize these statements to say that oxidation involves the formation of bonds to a &lt;em&gt;more electronegative&lt;/em&gt; element, and that reduction involves the formation of bonds to a &lt;em&gt;less electronegative&lt;/em&gt; element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your problem, specifically, choice (D), carbon previously has two bonds to oxygen and ends up with two bonds to nitrogen. This is &lt;em&gt;not&lt;/em&gt; considered an oxidation because it started with two bonds to a more electronegative element and ended up with two bonds to a more electronegative element. &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-08-22T15:44:22.213" LastActivityDate="2015-08-22T15:44:22.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35371" PostTypeId="1" AcceptedAnswerId="35380" CreationDate="2015-08-22T18:31:48.127" Score="7" ViewCount="865" Body="&lt;p&gt;I've heard people say many times that one gas is heavier than the other so the heavier gas sinks. Now this must certainly be true to a certain extent (I think of gas used in WWI that sunk into trenches). However, this idea of the weight of a gas being a driving force seems unusual because, in my experience, entropy is the dominating motive for particles in the gas phase. A good example of this is that gas phase clusters are studied extensively in the literature due to how unusual they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why would some gases &quot;sink&quot; and give up their entropic gain for what doesn't seem to be much of an energy minimization? Could the answer be simply that this concentration of heavier gases is quite short lived and they soon float away to their entropy filled wonderland?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also give as an example the fact that CO2 certainly gets up high in the atmosphere quite well and that is more than twice the &quot;weight&quot; of other particles in the air. So what's different about that than about the examples given previously?&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="16683" LastEditDate="2016-09-30T09:57:58.337" LastActivityDate="2016-09-30T09:57:58.337" Title="Misconception? &quot;That gas is heavier so it sinks.&quot;" Tags="&lt;entropy&gt;&lt;density&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35373" PostTypeId="1" CreationDate="2015-08-22T18:58:36.917" Score="3" ViewCount="3184" Body="&lt;p&gt;Why does the $\ce{XeF6}$ molecule have an octahedral structure while $\ce{XeOF5-}$ has a pentagonal pyramidal structure, despite both molecules having 7 electron pairs? I suppose the extra electron on oxygen would distort the molecule somewhat, but this seems like an extreme outcome.&lt;/p&gt;&#xA;" OwnerUserId="19331" LastEditorUserId="16683" LastEditDate="2018-08-10T15:14:42.690" LastActivityDate="2018-08-10T15:14:42.690" Title="The pentagonal pyramidal structure of XeOF5-" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-23T05:12:51.740" ContentLicense="CC BY-SA 4.0" />
  <row Id="35374" PostTypeId="2" ParentId="35373" CreationDate="2015-08-22T20:09:29.013" Score="4" Body="&lt;p&gt;$\ce{XeF6}$ is not octahedral, but distorted (or capped octahedral), one of the possible arrangement for &lt;a href=&quot;https://winter.group.shef.ac.uk/vsepr/high-cn.html&quot; rel=&quot;noreferrer&quot;&gt;seven electron pairs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Xenon_hexafluoride&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; provides the following, quoted from a paper by Konrad Seppelt in Accounts of Chemical Research[1]:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the structure is best described in terms of a mobile electron pair that moves over the faces and edges of the octahedron and thus distorts it in a dynamic manner...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The molecular geometry of $\ce{XeF6}$ is thus not truly an octahedron, nor is it static. In fact, the referenced article by Seppelt indicates that $\ce{XeF6}$ is only monomeric in the gas phase. In solution and the solid phase it is tetrameric $\ce{Xe4F24}$ clusters. The solid phase appears to be more like clusters of $\ce{[XeF5+ F^{-}]}$ with the fluoride ions bridging two $\ce{XeF5+}$ centers. Thus, in the solid, every $\ce{Xe}$ atom has 7 bonds to fluorine, with two that are more ionic. The structure for this scenario looks like the pentagonal pyramid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Seppelt, Konrad (June 1979). &quot;Recent Developments in the Chemistry of Some Electronegative Elements&quot;. Accounts of Chemical Research 12 (6): 211–216. &lt;a href=&quot;https://dx.doi.org/10.1021%2Far50138a004&quot; rel=&quot;noreferrer&quot;&gt;doi:10.1021/ar50138a004&lt;/a&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-08-22T20:09:29.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35375" PostTypeId="2" ParentId="35364" CreationDate="2015-08-22T20:43:46.920" Score="2" Body="&lt;p&gt;Your reasoning is correct. Are you confused because acetone is not a hydrocarbon? That has to be a mistake in the wording of the question. Another common mistake is to forget that there are some reactions that aldehydes can do that ketones are inert to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://en.wikipedia.org/wiki/C3H6O&quot; rel=&quot;nofollow&quot;&gt;seven constitutional isomers&lt;/a&gt; for $\ce{C3H6O}$, not counting the enols, which are the minor tautomers of acetone and propanal:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Acetone - a ketone&lt;/li&gt;&#xA;&lt;li&gt;Propanal - an aldehyde&lt;/li&gt;&#xA;&lt;li&gt;Cyclopropanol - an alcohol&lt;/li&gt;&#xA;&lt;li&gt;Allyl alcohol - an alkene and an alcohol&lt;/li&gt;&#xA;&lt;li&gt;Methoxyethene - an alkene and an ether&lt;/li&gt;&#xA;&lt;li&gt;Oxacyclobutane - a cyclic ether&lt;/li&gt;&#xA;&lt;li&gt;Propylene oxide - an epoxide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your reasoning is correct based on the chemical tests:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Alcohols/Reactions_of_Alcohols/The_Reaction_Between_Alcohols_and_Sodium&quot; rel=&quot;nofollow&quot;&gt;Sodium&lt;/a&gt; reacts with alcohols - a negative sodium test means no alcohol functional group is present, eliminating allyl alcohol (and the enols if you consider them).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bromine_test&quot; rel=&quot;nofollow&quot;&gt;Bromine&lt;/a&gt; reacts with alkenes and alkynes (and a small number of other functional groups, like phenols) - a negative bromine test means no alkene functional group is present, eliminating allyl alcohol, methoxyethene, and the enols. &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tollens%27_reagent&quot; rel=&quot;nofollow&quot;&gt;Ammoniacal silver nitrate&lt;/a&gt; (Tollens' reagent) reacts with aldehydes - a negative Tollens test means no aldehyde function group is present, eliminating propanal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The results of the qualitative test leave you with three possibilities, acetone, oxacyclobutane, and propylene oxide. With the evidence you have, you might not be able to distinguish between the three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Haloform_reaction#Iodoform_reaction&quot; rel=&quot;nofollow&quot;&gt;iodoform test&lt;/a&gt; would confirm or eliminate acetone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ac60069a047&quot; rel=&quot;nofollow&quot;&gt;test using silver nitrate and periodic acid&lt;/a&gt; would confirm or eliminate propylene oxide.&lt;/p&gt;&#xA;" OwnerUserId="275" LastEditorUserId="4934" LastEditDate="2015-08-22T21:12:50.597" LastActivityDate="2015-08-22T21:12:50.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35376" PostTypeId="2" ParentId="35237" CreationDate="2015-08-22T20:53:49.520" Score="6" Body="&lt;p&gt;This reaction can be viewed as a 1,2-sigmatropic hydrogen shift or as a 2+1 (if we are counting atoms, 2+0 if we are counting electrons) cycloaddition (adding the carbene's empty p-orbital across the $\ce{C-H}$ bond - analogous to its addition to an olefinic pi bond to produce a cyclopropane). However the reaction is most commonly viewed as and referred to as an &quot;&lt;strong&gt;insertion&lt;/strong&gt;&quot; reaction because the carbenic carbon is inserting itself in between the carbon and hydrogen atoms that comprised the $\ce{C-H}$ bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nmVoI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmVoI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://osxs.ch.liv.ac.uk/externaltest/sandboxOLD/Carbenes-Reactions(CHInsertion).html&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what is known.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thermal decomposition of most diazo compounds produce singlet carbenes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7kh7h.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7kh7h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Singlet carbenes add stereopecifically to olefins.  That is, addition of a singlet methylene carbene to &lt;em&gt;trans&lt;/em&gt;-2-butene produces &lt;strong&gt;only&lt;/strong&gt; the corresponding &lt;em&gt;trans&lt;/em&gt;-dimethylcyclopropane&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W3ND3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W3ND3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbene&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, &lt;strong&gt;insertion&lt;/strong&gt; of singlet carbenes into stereogenic centers results in retention of stereochemistry at the chiral center.  These facts suggest that both the cycloaddition and insertion reactions of simple singlet carbenes either proceed in a concerted fashion, or if any intermediate is formed it is extremely shorted-lived in order that stereochemical information be preserved.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple carbenes are typically mildly electrophilic.  The HOMO (the filled $\ce{sp^2}$ orbitals) and LUMO (the empty p-orbital) for a singlet carbene are shown on the left side of the following figure.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h3I1Q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h3I1Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://organometallicchem.wordpress.com/tag/backbonding/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we would expect the interaction of the carbene LUMO (the empty p-orbital) and the $\ce{C-H}$ bond HOMO to be controlling, as the energy separation here is much less than the other possible HOMO-LUMO interaction between the carbene HOMO (the filled $\ce{sp^2}$ orbital) and the much higher in energy $\ce{C-H}$ antibonding orbital, the LUMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This controlling HOMO-LUMO interaction involving the &lt;strong&gt;empty&lt;/strong&gt; carbenic p-orbital in the rate determing step is consistent with the observed &lt;strong&gt;electrophilic&lt;/strong&gt; nature of the carbene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All of this information taken together suggests&lt;/strong&gt; that interaction of the empty carbenic p-orbital with the electrons in the $\ce{C-H}$ bond is the critical step.  A geometry like that shown below is likely preferred.  As the insertion proceeds, the trigonal carbenic carbon rotates.  The rotation starts to bring the carbenic $\ce{sp^2}$ orbital into alignment with the p-orbitals in the 7-membered ring.  Also during the rotation the $\ce{sp^2}$ orbital starts its rehybridization into a p-orbital. When completed we have created the final product, heptafulvene.  The reaction is likely concerted and certainly does not involve any long-lived intermediates. Note too that the heptafulvene is formed with increased electron density on the former carbenic carbon in accord with the dipolar nature (dipole moment ~0.5 D due to resonance stabilization) of the molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pm4X7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pm4X7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-22T21:30:11.060" LastActivityDate="2015-08-22T21:30:11.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35378" PostTypeId="1" CreationDate="2015-08-22T21:38:12.900" Score="3" ViewCount="493" Body="&lt;p&gt;I would like to make some invisible ink that apears by itself after some time, preferably at least a couple of days. I want to write a dedication in a book with it, so it would probably be fine if the colour apears after half an hour of air exposure. Is this even possible using easy to get ingredients?&lt;/p&gt;&#xA;" OwnerUserId="19333" LastEditorDisplayName="user7951" LastEditDate="2015-08-22T21:51:14.817" LastActivityDate="2015-11-28T07:17:04.710" Title="Is there a way to make long delay ink?" Tags="&lt;organic-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35380" PostTypeId="2" ParentId="35371" CreationDate="2015-08-23T01:23:48.097" Score="11" Body="&lt;p&gt;Without significant mixing, diffusion takes a long time to mix gases.  Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a slow but steady production of a dense gas, layers absolutely will form due to density differences.  There are plenty of places where this can easily be seen, from your WWI example to the ability &lt;a href=&quot;https://www.youtube.com/watch?v=wjmVBsDjOw4&quot;&gt;to pour carbon dioxide&lt;/a&gt; or hexane vapor into a container.  They're not stable over the long term, but they do exist for substantial periods of time, days in some cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If confined by an open container, huge areas can experience this effect, as at Lake Nyos.  Heck, even small temperature based density differences drive atmospheric layering that has major impacts on weather and pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not an enormous effect, but you may be underestimating the strength of the forces involved here.  The same pressure differential that forces a helium balloon upward acts on an open volume of helium as well, and more strongly due to the missing mass of the balloon.  A body of a gas like $\ce{NO2}$ has about as much force pushing it downward (in air) than an equal volume of helium has pushing it upward.  Without outside mixing to cause it to disperse, it will happily form a short term layer until diffusion carries it away.&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="7960" LastEditDate="2015-08-23T23:49:14.390" LastActivityDate="2015-08-23T23:49:14.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35381" PostTypeId="1" CreationDate="2015-08-23T02:34:04.130" Score="8" ViewCount="1550" Body="&lt;p&gt;A green-colored substance formed during electrolysis, wherein the electrolyte was magnesium sulfate and minerals in tap water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not have any leads to stick into the water, so I just put a 9 volt Duracel battery in the solution. The battery didn't leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White flakes formed after 1 minute before green substance at 10 minutes. &#xA;So, &lt;strong&gt;what is the green stuff and what are the white flakes?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="19338" LastEditorUserId="19338" LastEditDate="2015-08-27T00:50:58.007" LastActivityDate="2015-08-27T00:50:58.007" Title="Green color substance forming during electrolysis" Tags="&lt;electrochemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35382" PostTypeId="2" ParentId="35381" CreationDate="2015-08-23T02:49:01.020" Score="7" Body="&lt;p&gt;The green material is probably an iron(II) salt, corroded from what I assume is the steel composing the positive pole of the battery.  If it has a gelatinous appearance, and slowly turned orange-ish over time, my guess would be it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron(II)_hydroxide&quot; rel=&quot;nofollow&quot;&gt;$\ce{Fe(OH)2}$&lt;/a&gt;, with any orange color deriving from slow oxidation to various iron(III) species such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Iron(III)_oxide-hydroxide&quot; rel=&quot;nofollow&quot;&gt;oxide-hydroxides&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per the above link, the white flakes are possibly that same iron(II) hydroxide, prior to reaction with $\ce{O2}$ to form the green color.  Alternatively, it could be &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnesium_hydroxide&quot; rel=&quot;nofollow&quot;&gt;$\ce{Mg(OH)2}$&lt;/a&gt;, or a related magnesium compound, formed at the cathode (negative terminal) from the hydroxide ion produced by reductive water electrolysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;2\ce{H2O} + 2e^- \longrightarrow \ce{H2} + 2\ce{OH-}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A photograph would be tremendously helpful in facilitating identification.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-08-23T21:13:49.950" LastActivityDate="2015-08-23T21:13:49.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35384" PostTypeId="1" CreationDate="2015-08-23T06:09:53.333" Score="1" ViewCount="127" Body="&lt;p&gt;The pKa of carbidopa is 7-8 of the base hydrazin. Is this because of the zwitter form with the acid? Why is the basicity so low? If it can form a zwitter ion wouldnt  the pKa be higher?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Carbidopa&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Carbidopa&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19069" LastActivityDate="2015-08-23T06:09:53.333" Title="Carbidopa : zwitterion" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35385" PostTypeId="1" CreationDate="2015-08-23T06:18:02.303" Score="2" ViewCount="193" Body="&lt;h2&gt;Question:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Out of $CH_4, CaH_2, FeCl_2$ and $BeH_2$, which compound would produce the most strongly reducing aqueous solution?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Thoughts:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Out of the hydrides, I would say that $CaH_2$ stands a good chance at being the strongest reducing agent, since the Ca-H bond energy is lowest (allowing the extremely reducing hydride to be released more readily into solution). The only rationale I can think up for this is the relative sizes of the other ion in the hydrides (Ca&gt;Be&gt;C) which implies that bond length is greatest in the calcium hydride (thus bond energy is lower). I am unsure as to how I should tackle the chloride compound. Much assistance is needed.&lt;/p&gt;&#xA;" OwnerUserId="7161" LastActivityDate="2015-08-23T06:18:02.303" Title="Relative strength of metal hydrides as reducing agents" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35389" PostTypeId="1" CreationDate="2015-08-23T08:45:01.327" Score="2" ViewCount="567" Body="&lt;p&gt;It is said that after applying aluminium chloride assay, the extracts will turn yellow due to formation of Al-Flavonoid complexes. So I was wondering if the assay still works if the extracts are already yellow.&#xA;For example, the yellow extracts are applied aluminium chloride assay and determined for total Flavonoid content using colorimetric method.&#xA;Then, would the result show the TFC (total flavonoid content in those yellow extracts?&lt;/p&gt;&#xA;" OwnerUserId="17455" LastActivityDate="2017-07-26T01:36:58.710" Title="Does the aluminium chloride assay (Ordonez et al.2006) for total flavonoid content work on coloured extracts?" Tags="&lt;organic-chemistry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35390" PostTypeId="2" ParentId="35297" CreationDate="2015-08-23T08:46:21.367" Score="1" Body="&lt;p&gt;As stated by bon in the comments, this is probably something you should contact someone who is actually &lt;em&gt;supposed&lt;/em&gt; to be able to answer this question about. The fumes might just be an annoyance, or they could be highly and immediately dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One place to start is with the manufacturer. Indeed, the manufacturer is probably the best place to go in general with chemical safety concerns. This chemical seems to be produced by a number of firms, so I was not able to locate a specific one, but a &lt;a href=&quot;https://www.google.com/?gws_rd=ssl#q=Chlorpyriphos-20%25%20EC%20msds&quot; rel=&quot;nofollow&quot;&gt;google search&lt;/a&gt; shows a number of possible contacts. &lt;a href=&quot;http://msdssearch.dow.com/PublishedLiteratureDAS/dh_008a/0901b8038008a2ff.pdf?filepath=india/pdfs/noreg/012-70038.pdf&amp;amp;fromPage=GetDoc&quot; rel=&quot;nofollow&quot;&gt;This MSDS&lt;/a&gt; provides contact information for a Dow subsidiary. I suspect they will be willing to help, but if they refuse to because they were not the manufacturers of your specific product, you will need to ask the carpenter for the documentation. In a number of locations, the law specifies he must be able to provide an MSDS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you might wish to contact your local fire department (non-emergency line). They can probably point you to someone local whose job it is to handle potential chemical contamination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will note that the MSDS I cited does not suggest extreme danger. However, it is still best to contact a producer or specialist of some sort.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-08-23T08:46:21.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35391" PostTypeId="1" AcceptedAnswerId="35394" CreationDate="2015-08-23T09:19:07.073" Score="30" ViewCount="3967" Body="&lt;p&gt;Some gases are lighter than others and rise. Why don't they continue going up, leave the atmosphere, and then enter outer space?&lt;/p&gt;&#xA;" OwnerUserId="19309" LastEditorUserId="16683" LastEditDate="2016-09-30T09:26:48.327" LastActivityDate="2017-08-21T13:55:55.433" Title="Why don't gases escape Earth's atmosphere?" Tags="&lt;physical-chemistry&gt;&lt;atmospheric-chemistry&gt;&lt;kinetic-theory-of-gases&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35392" PostTypeId="2" ParentId="19797" CreationDate="2015-08-23T09:19:18.527" Score="3" Body="&lt;p&gt;&lt;strong&gt;Explosives:&lt;/strong&gt; It is true that sulfuric acid and sodium/potassium nitrate are used to make various explosives (by creating nitric acid), but the key is something to nitrate. The worst this mixture could generate is nitrated ethanol, and I was unable to find any source suggesting that could be done with nitric and sulfuric acid alone. Nitrating a molecule that small with that simple of a structure seems unlikely without extreme conditions or reagents. If it were possible, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ethyl_nitrate&quot; rel=&quot;nofollow&quot;&gt;the most likely product&lt;/a&gt; does seem to be something unpleasant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Poisons:&lt;/strong&gt; What definitely would be produced is nitric acid, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitric_acid#Laboratory_synthesis&quot; rel=&quot;nofollow&quot;&gt;production&lt;/a&gt; of nitric acid via sulfuric acid and a nitrate salt produces &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrogen_dioxide&quot; rel=&quot;nofollow&quot;&gt;nitrogen dioxide&lt;/a&gt;, a poisonous and caustic gas. This is some seriously bad stuff you do not want to come in contact with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on that, I'm going to conclude this mixture is &lt;em&gt;probably not explosive&lt;/em&gt;, but definitely could &lt;em&gt;become&lt;/em&gt; so if an easily-nitrated compound was a contaminant. Those are more common than you might imagine. It definitely would create a nasty gas.&lt;/p&gt;&#xA;" OwnerDisplayName="user7652" LastActivityDate="2015-08-23T09:19:18.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35394" PostTypeId="2" ParentId="35391" CreationDate="2015-08-23T10:10:12.500" Score="31" Body="&lt;p&gt;The atmosphere actually loses gases to outer space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average velocity $\bar v$ of gas molecules is determined by temperature $T$.&#xA;However, not all the molecules travel with the same velocity.&#xA;The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds&#xA;$$\begin{align}&#xA;f{\left(v\right)}&amp;amp;=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt]&#xA;&amp;amp;=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right) &#xA;\end{align}$$&#xA;where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula&#xA;$${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$&#xA;where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2016-09-30T10:11:56.787" LastActivityDate="2016-09-30T10:11:56.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35397" PostTypeId="1" AcceptedAnswerId="35400" CreationDate="2015-08-23T15:03:02.620" Score="6" ViewCount="1717" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An organic compound A (molecular formula $\ce{C6H12O}$) does not change the colour of acidic dichromate solution. Compound A on treatment with $\ce{H2SO4}$ produces alkene which on oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ which gives positive iodoform test. Find out structure of A. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I know about ozonolysis, but I do not know when acidic dichromate solution changes colour. From the question an alkene is produced so I guess it may be an alcohol, but if someone an explain about it and iodoform test, then I will be able to give another try on this problem.&lt;/p&gt;&#xA;" OwnerUserId="19347" LastEditorUserId="5026" LastEditDate="2018-03-01T02:39:56.243" LastActivityDate="2018-03-01T02:39:56.243" Title="Finding structure of a compound given its reactions" Tags="&lt;organic-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35398" PostTypeId="2" ParentId="35391" CreationDate="2015-08-23T15:18:37.113" Score="3" Body="&lt;p&gt;Lighter gases don't go up after they've mixed with the surrounding atmosphere. Buoyancy (like with a balloon filled with $\ce{He}$ rising or $\ce{CO_2}$ from a leak in a line dropping to the floor and filling the room) occurs only with &lt;em&gt;bulk&lt;/em&gt; masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need extreme &lt;em&gt;g&lt;/em&gt; fields/gradients, like in a ultracentrifuge, to separate gases by weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a certain net flux of light compounds upwards, because e.g. $\ce{He}$ (from radioactive decay) comes into the atmosphere at the ground and escapes into space from the upper layers (see other answer).&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorDisplayName="user15489" LastEditDate="2015-08-24T10:45:30.500" LastActivityDate="2015-08-24T10:45:30.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35400" PostTypeId="2" ParentId="35397" CreationDate="2015-08-23T17:08:44.517" Score="7" Body="&lt;p&gt;I have taken the liberty of identifying your unknown compound.  If you'd like to work through the problem yourself, then carefully scroll down to the part of the answer that deals with the iodoform test.  Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What we know&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just looking at the molecular formula we see that there are too many hydrogens for the 6 carbons to be in a benzene ring.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;does not change the colour of acidic dichromate solution - &lt;a href=&quot;https://en.wikipedia.org/wiki/Potassium_dichromate&quot; rel=&quot;nofollow noreferrer&quot;&gt;acidic dichromate&lt;/a&gt; readily oxidizes primary and secondary alcohols to aldehydes and ketones respectively; &lt;strong&gt;tertiary alcohols are not oxidized&lt;/strong&gt;. The aldehyde produced from a primary alcohol can be further oxidized to a carboxylic acid, this results in a solution color change from orange to green. But there was no color change, therefore &lt;strong&gt;there is no primary $\ce{-OH}$ or aldehyde group in compound A&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compound A on treatment with H2SO4 produces alkene - this is a common reaction that causes deydration (elimination of water) and converts an alcohol into an olefin.  Therefore &lt;strong&gt;compound A is (as you suggested) an alcohol&lt;/strong&gt; (but not a primary alcohol).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ - ozonolysis will cleave the double bond in the compound produced upon dehydration.  The key fact here is that the resulting product &lt;strong&gt;still contains 6 carbon atoms&lt;/strong&gt;.  If the double bond involved carbons outside of a ring we would wind up with an ozonolysis product containing fewer than 6 carbons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYDDg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYDDg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www.masterorganicchemistry.com/2013/04/23/alkene-reactions-ozonolysis/&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only if the double bond is contained &lt;strong&gt;within&lt;/strong&gt; a ring will we retain all 6 carbon atoms.  Therefore &lt;strong&gt;our double bond must be in a ring&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ORpW8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ORpW8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;positive iodoform test - This is our final clue, the iodoform test (or more generally &lt;a href=&quot;https://en.wikipedia.org/wiki/Haloform_reaction#Iodoform_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;the haloform reaction&lt;/a&gt;) is a test that detects the presence of methyl ketones and generates iodoform ($\ce{HCI3}$) in the process.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wkzSL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wkzSL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we know that &lt;strong&gt;there is a methyl ketone in our ozonolysis product&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining all of this information (non-aromatic, ring, non-primary alcohol, methyl group) leads to the following structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(mouse over to see the answer)&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; &lt;a href=&quot;https://i.stack.imgur.com/RhSpo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RhSpo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which is called&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; 1-methylcyclopentanol.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note: some methylated-cyclobutanols might also be considered, but when these dehydrate and produce an alkene, the double bond is likely to be placed external to the ring (or [advanced topic] external to a cyclopropane ring formed upon rearrangement) and we would not wind up with 6 carbons in our ozonolysis product.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-23T21:39:19.940" LastActivityDate="2015-08-23T21:39:19.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35401" PostTypeId="1" AcceptedAnswerId="35407" CreationDate="2015-08-23T17:09:18.277" Score="22" ViewCount="4593" Body="&lt;p&gt;A screenshot of the nutrition facts of a can of Gatorade powder was posted on my Facebook feed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Pcsb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Pcsbm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The label reads &quot;Per 32 g (about 3 tbsp)&quot;, but if you read through the ingredients, the amount of carbohydrates listed exceed that of the serving size (33 g).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can this possibly be? Is there a chemical process as some people have suggested that is changing the 32 g of powder into 33 g of carbs once water is added?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-asking it here for a chemistry perspective as some users have suggested on skeptics.SE.&lt;/p&gt;&#xA;" OwnerUserId="16687" LastEditorDisplayName="user7951" LastEditDate="2015-08-23T23:50:41.330" LastActivityDate="2015-08-24T12:43:16.490" Title="How can 32 g of Gatorade powder contain 33 g of carbohydrates?" Tags="&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35402" PostTypeId="2" ParentId="35359" CreationDate="2015-08-23T17:15:33.817" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;a phenol will not react (or react very slowly) with $\ce{PCl5}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's not true. Phenol and $\ce{PCl5}$ do react, producing mixture of compounds of formula $\ce{PCl_n(C6H5O)_{5-n}}$ (with $n &amp;gt; 1$ ). It is even possible to produce $\ce{(C6H5O)5P}$!&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, unlike in alcohols, in phenols:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the carbon–oxygen bond is much stronger, having a bond order of more than one and being formed with $\mathrm{sp^2}$ hybridized orbitals&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the formation of the corresponding transition state for nucleophilic substitution would destroy the aromaticity of the benzene ring. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, the formation of chlorobenzene does not proceed at a significant rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still possible to replace the phenol group by a chloride atom in some cases. For example, it is possible to prepare picrylchloride (1,3,5-trinitrochlorobenzene) by reaction of pyridine picrate and $\ce{PCl5}$, since three nitro groups greatly increase the reactivity of the phenyl fragment in nucleophilic aromatic substitution reactions.&lt;sup&gt;[2]&lt;/sup&gt; However, as a general rule, benzene derivatives are not active in nucleophilic substitution except if there are a lot of strong π-electron withdrawing groups in the ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for 4-hydroxybenzoic acid, my guess is that it should react as both phenol (as above) and benzoic acid (which &lt;a href=&quot;http://classroom.synonym.com/chemical-properties-benzoic-acid-2721.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;forms benzoyl chloride&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Addition&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;While I was unable to find precise information on &lt;em&gt;para&lt;/em&gt;-hydroxybenzoic acid, there is some info on &lt;em&gt;ortho&lt;/em&gt;-hydroxybenzoic (salicylic acid). $\ce{PCl5}$ replaces the $\ce{OH}$ fragment of the carboxylic acid group with chlorine. The phosphorus oxochloride then reacts with the phenolic group, producing an $\ce{O-POCl2}$ fragment. The &lt;a href=&quot;https://books.google.ru/books?id=OeZ46Zgz2yQC&amp;amp;lpg=PA302&amp;amp;ots=gX5kjCqM0V&amp;amp;dq=salicylic%20acid%20PCl5&amp;amp;pg=PA302#v=onepage&amp;amp;q=salicylic%20acid%20PCl5&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;information&lt;/a&gt;, however, does not look absolutely trustworthy. Still, assuming it is correct and extrapolating it onto &lt;em&gt;para&lt;/em&gt;-hydroxybenzoic acid, I'd say that the product should be $\ce{Cl-C(O)-C6H4-O-P(O)Cl2}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;[1]: Ramirez, F.; Bigler, A. J.; Smith, C. P. Pentaphenoxyphosphorane. &lt;em&gt;J. Am. Chem. Soc.&lt;/em&gt; &lt;strong&gt;1968,&lt;/strong&gt; &lt;em&gt;90&lt;/em&gt; (13), 3507–3511. &lt;a href=&quot;https://doi.org/10.1021/ja01015a038&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ja01015a038&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2]: Boyer, R.; Spencer, E. Y.; Wright, G. F. &lt;em&gt;Can. J. Chem.&lt;/em&gt; &lt;strong&gt;1946,&lt;/strong&gt; &lt;em&gt;24b&lt;/em&gt; (5), 200–203. &lt;a href=&quot;https://doi.org/10.1139/cjr46b-025&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1139/cjr46b-025&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="485" LastEditorUserId="16683" LastEditDate="2017-08-16T11:55:14.157" LastActivityDate="2017-08-16T11:55:14.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35403" PostTypeId="1" AcceptedAnswerId="35405" CreationDate="2015-08-23T17:37:03.547" Score="8" ViewCount="5096" Body="&lt;p&gt;I know that are several acids that can be found in the kitchen, such as citric acid and vinegar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I never heard of a base that can be used in the kitchen to raise the pH levels of any food. I am well aware that this mix would probably make a salt as a result. Something like hydrogen peroxide doesn't sound safe for food. So, my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are (if any) the bases that can be used on those acids and still be human-consumable?&lt;/strong&gt;  &lt;/p&gt;&#xA;" OwnerUserId="19350" LastEditorUserId="19350" LastEditDate="2015-08-23T21:34:21.240" LastActivityDate="2016-03-11T23:25:51.783" Title="What bases are fit for human consumption and can be used in foods?" Tags="&lt;acid-base&gt;&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35405" PostTypeId="2" ParentId="35403" CreationDate="2015-08-23T18:26:48.933" Score="8" Body="&lt;p&gt;Unfortunately Googling for &quot;alkaline food&quot; these days is a bad idea; results are dominated by nonsense sites espousing the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Alkaline_diet&quot; rel=&quot;nofollow&quot;&gt;alkaline diet&lt;/a&gt;&quot; nonsense / &lt;a href=&quot;https://en.wikipedia.org/wiki/Pseudoscience&quot; rel=&quot;nofollow&quot;&gt;woo&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There aren't many foods that are naturally alkaline.  The two exceptions I know of are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sodium_bicarbonate&quot; rel=&quot;nofollow&quot;&gt;baking soda&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Egg_white&quot; rel=&quot;nofollow&quot;&gt;egg white&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sometimes when I'm cooking and accidentally add too much acid (e.g. lemon juice or vinegar) to a broth or sauce, I can partially correct the mistake by adding small amounts of baking soda.  This converts the excess acid to the sodium salt and $\ce{CO2}$ is evolved.  For the example of vinegar (acetic acid):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CH3COOH + NaHCO3 -&amp;gt; CO2(g) + CH3COONa + H2O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So sodium acetate salt is left behind, which limits the degree to which you can correct overaddition of acid.  (You don't want your food to be too salty.)$%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:25:51.783" LastActivityDate="2016-03-11T23:25:51.783" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35407" PostTypeId="2" ParentId="35401" CreationDate="2015-08-23T19:51:42.427" Score="24" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a chemical process as some people have suggested that is changing the 32g of powder into 33g of carbs once water is added?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Even if there was, it's not relevant in this case: the numbers on the label are an assay of the pure contents of the product before it's mixed with anything, or prepared or altered in any way. Nutrition labels refer to the product as-is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the United States, the Food and Drug Administration is in charge of labeling the quantities you're interested in. The relevant document is the &lt;a href=&quot;http://www.fda.gov/Food/GuidanceRegulation/GuidanceDocumentsRegulatoryInformation/LabelingNutrition/ucm2006828.htm&quot;&gt;Food Labeling Guide&lt;/a&gt;, one of many documents which are linked from a more comprehensive list called &lt;a href=&quot;http://www.fda.gov/Food/GuidanceRegulation/GuidanceDocumentsRegulatoryInformation/LabelingNutrition/default.htm&quot;&gt;Labeling and Nutrition Guidance Documents and Regulatory Information&lt;/a&gt;. As per the comments (below), I'm noting that it appears as if the product pictured in your post is from the Canadian market and thus subject to nutritional labeling dictated by agencies in Canada. That said, the methodologies used by both the US and Canada are similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are at least 2 regulations detailed in the labeling guidelines that we need to examine here to explain what's going on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The numbers reported on the container have simply been rounded up or down as per the &lt;a href=&quot;http://www.fda.gov/Food/GuidanceRegulation/GuidanceDocumentsRegulatoryInformation/LabelingNutrition/ucm064932.htm&quot;&gt;Food and Drug Administration's Appendix H in the Food Labeling Guide&lt;/a&gt;. For instance, they show that for &lt;em&gt;Total carbohydrate&lt;/em&gt; and &lt;em&gt;Sugars&lt;/em&gt;, anything less than 0.5 g is expressed as 0 g, and quantities greater than or equal to 1 g are to be expressed in the nearest 1-gram increment. This is why of the 33 g of total carbohydrate, only 31 g appears to be accounted for as sugars;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;em&gt;Total carbohydrate&lt;/em&gt; &lt;a href=&quot;http://www.fda.gov/Food/GuidanceRegulation/GuidanceDocumentsRegulatoryInformation/LabelingNutrition/ucm064894.htm&quot;&gt;is calculated in a way that propagates rounding errors introduced as detailed in (1)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Total carbohydrate is calculated by subtracting the weight of crude protein, total fat, moisture, and ash from the total weight of the sample of food.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary: The numbers don't add up because quantifying the components of a food to conform to the rules and regulations for reporting them on a Nutrition Facts label allows for the accumulation of rounding and measurement error.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-24T12:43:16.490" LastActivityDate="2015-08-24T12:43:16.490" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="35408" PostTypeId="2" ParentId="35265" CreationDate="2015-08-23T20:05:12.790" Score="3" Body="&lt;p&gt;Your approach is correct.&lt;/p&gt;&#xA;&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-65.7.7&lt;/strong&gt; Cyclic anhydrides&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-65.7.7.1&lt;/strong&gt; Cyclic anhydrides formed from two acid groups attached to the same parent hydride structure are named in two ways:&lt;/p&gt;&#xA;&lt;p&gt;(1) as heterocyclic pseudoketones;&lt;/p&gt;&#xA;&lt;p&gt;(2) by changing the class term ‘acid’ to ‘anhydride’ in the systematic or retained name of the dibasic acid.&lt;/p&gt;&#xA;&lt;p&gt;Method (1) generates preferred IUPAC names&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Using Method (1), the first step is naming the heterocyclic component as follows.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-25.2.2.4&lt;/strong&gt; Heteromonocyclic components fused to a benzene ring&lt;/p&gt;&#xA;&lt;p&gt;Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.&lt;br /&gt;&#xA;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Accordingly, the name ‘benzofuran’ consists of the prefix ‘benzo’ followed by the retained name ‘furan’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nfLfA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nfLfA.png&quot; alt=&quot;furan&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom. The locant ‘2’ is assigned to the heteroatom and it is placed at the front of the name.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o2dGZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2dGZ.png&quot; alt=&quot;2-benzofuran&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally, the corresponding pseudoketone is named using the suffix ‘one’ in accordance with rules expressed for ketones. Thus, the preferred IUPAC name (PIN) of phthalic anhydride is ‘2-benzofuran-1,3-dione’.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h418d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h418d.png&quot; alt=&quot;2-benzofuran-1,3-dione&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-23T20:05:12.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35411" PostTypeId="2" ParentId="31720" CreationDate="2015-08-23T23:59:04.020" Score="-1" Body="&lt;p&gt;First method is to look for Faraday's law of electrolysis and put in the equation your data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second method is to add to TDS reading the initial reading of the salt solution before the electrolysis.&lt;/p&gt;&#xA;" OwnerUserId="19357" LastEditorUserId="7448" LastEditDate="2015-08-24T06:42:37.460" LastActivityDate="2015-08-24T06:42:37.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35412" PostTypeId="1" AcceptedAnswerId="37471" CreationDate="2015-08-24T00:40:53.317" Score="15" ViewCount="813" Body="&lt;p&gt;Is it meaningful to characterize the relative strength of desiccants?  For example, is there a measure of hygroscopy, or an ordering of desiccants, such that higher ones will always dry lower ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us restrict the question to &quot;practical&quot; desiccants, which are hygroscopic substances that can be &quot;recharged&quot; (i.e., dried out) by heating.  I'm not certain I know all possible mechanisms of water sorption within this class of chemicals.  But is it safe to say that for all such chemicals, at standard temperature and pressure, there is some water vapor pressure around a desiccant that has sorbed any water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.  At hydrous equilibrium will the &quot;stronger&quot; desiccant contain virtually all of the water?  Or is the equilibrium distribution of water a function of the desiccants' relative &quot;hygroscopy,&quot; &lt;em&gt;rates&lt;/em&gt; of sorption, or some other factor(s)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is ultimately a practical question.  One example: I have a hygroscopic, hydrated salt X.  Can I dry it out by sticking it in a sealed box at STP with a &quot;more powerful&quot; desiccant?  If not what determines the limit of dehydration that would be achieved by exposure to other desiccants?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="5354" LastEditDate="2015-09-11T12:43:51.397" LastActivityDate="2017-04-10T19:00:39.607" Title="Relative strength of desiccants" Tags="&lt;thermodynamics&gt;&lt;water&gt;&lt;adsorption&gt;&lt;absorption&gt;&lt;desiccation&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35414" PostTypeId="1" CreationDate="2015-08-24T01:15:54.623" Score="1" ViewCount="80" Body="&lt;p&gt;I have a ligand $\ce{L}$ and a receptor $\ce{R}$. As can easily be shown, with dissociation constant $K_\mathrm{D} = \frac{[\ce{L}][\ce{R}]}{[\ce{LR}]}$ the fraction of receptors occupied by ligand is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \frac{[\ce{LR}]}{[\ce{LR}] + [\ce{R}]} = \frac{[\ce{L}]}{[\ce{L}] + K_\mathrm{D}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see that the fractional saturation of receptors is independent of receptor concentration. Now here's where I get confused. Let's say we have a constant volume of $\pu{1 L}$ with $\pu{5 mol}$ ligand and a $K_\mathrm{D}$ of 5:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L = 5, \quad K_\mathrm{D} = 5 \quad \to \quad \theta = 0.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have $\pu{10 mol}$ of receptor total ($[\ce{LR}] + [\ce{R}] = 10$), then $\pu{5 mol}$ will be bound to ligand. Now let's say we increase the number of receptors by $10\times$. Our formula for $\theta$ is independent of $[\ce{R}]$ or $[\ce{LR}]$. We now have $\pu{50 mol}$ receptor bound to ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, we only had $\pu{5 mol}$ ligand to start with, which is itself a paradox. Can someone explain to me where I went wrong?&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="41328" LastEditDate="2017-09-26T09:24:23.817" LastActivityDate="2017-09-26T09:27:39.673" Title="Ligand Binding Paradox?" Tags="&lt;inorganic-chemistry&gt;&lt;biochemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35415" PostTypeId="1" AcceptedAnswerId="35419" CreationDate="2015-08-24T01:28:01.907" Score="2" ViewCount="1410" Body="&lt;p&gt;I have a $\pu{0.3 M}$ $\ce{NaCl}$ stock solution and need to create a $\pu{0.01 M}$ $\ce{NaCl}$ solution that is $\pu{50 mL}$. The only tools I am given are volumetric pipettes that are $10.0, 5.0, 1.0 \text{ and }\pu{0.5 mL}$ and $\pu{50.0 mL}$ volumetric flasks. I understand the physical process where I would measure out a specific amount and place it in the flask using the pipette(s) and then add the required amount of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to use the dilution equation $M_1\times V_1 = M_2\times V_2$ which I did and I got $\pu{1.666667 mL}$ of solution that needs to be taken from the $\pu{0.3 M}$ stock solution to create the $\pu{0.01 M}$ solution. The only problem is that I can't do that since the materials I have do not allow me to measure out that amount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to &quot;convert&quot; the $\pu{0.3 M}$ to various other concentrations that would hopefully get me down to $\pu{0.01 M}$ but have failed so far. I tried to find concentrations that would result in a whole number or a value with $.5$ at the end after dividing by the product from $M_1\times V_1$ but failed every time. In addition, I combined the two concentrations and found the midpoints between them and tried to use those values, but they did not work either. Where is a good place for me to start? &lt;/p&gt;&#xA;" OwnerUserId="19358" LastEditorUserId="5026" LastEditDate="2018-03-09T09:03:23.700" LastActivityDate="2018-03-09T09:03:23.700" Title="Diluting a sodium chloride solution with specific glassware" Tags="&lt;stoichiometry&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35416" PostTypeId="2" ParentId="35412" CreationDate="2015-08-24T01:51:28.757" Score="6" Body="&lt;p&gt;I'd say no, because there is also a &lt;em&gt;reactivity&lt;/em&gt; dimension to desiccants. Some drying agents react with some solvents and therefore can't be used as a desiccant. This happens for many organic solvent and desiccant pairs. Some desiccants also react &lt;em&gt;as desiccants&lt;/em&gt; and this changes their drying power. This is the case for phosphorous pentoxide, for example, which gradually forms a viscous layer of phosphoric acid which stops any drying. So any linear ordering you are attempting is likely to be overwhelmed by the reactivity side effects of the many desiccants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you are only interested in some theoretical point in time estimate of desiccant efficiency, this might be possible but would have virtually no practical import. A desiccant has to react over time and the side effects cannot be ignored.&lt;/p&gt;&#xA;" OwnerDisplayName="user467" LastEditorDisplayName="user467" LastEditDate="2015-08-24T02:25:16.017" LastActivityDate="2015-08-24T02:25:16.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35417" PostTypeId="2" ParentId="35414" CreationDate="2015-08-24T02:00:42.267" Score="2" Body="&lt;p&gt;I see where I made my mistake. I set $[\ce{L}] = 5$, but this &quot;$L$&quot; is referring to the equilibrium concentration of free ligand -- not the total ligand concentration. In order to do that I should have constrained the system to $[\ce{L}] + [\ce{LR}] = 5$. Working through what this would do currently.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So in the first case, my assumption actually dictated that there be $\pu{10 mol}$ ligand total instead of the $\pu{5 mol}$ I erroneously stated. I incorrectly equated total ligand and free ligand the first time I did this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So I see where I messed up, but now I'm curious if the values of $[\ce{L}]$, $[\ce{R}]$, and $[\ce{LR}]$ can be solved as functions of $[\ce{L_{total}}]$, $[\ce{R_{total}}]$ and $K_\mathrm{D}$.&lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="41328" LastEditDate="2017-09-26T09:27:39.673" LastActivityDate="2017-09-26T09:27:39.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35418" PostTypeId="2" ParentId="35412" CreationDate="2015-08-24T05:37:16.227" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At hydrous equilibrium will the &quot;stronger&quot; desiccant contain virtually all of the water? Or is the equilibrium distribution of water a function of the desiccants' relative &quot;hygroscopy,&quot; rates of sorption, or some other factor(s)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Rates&quot; aren't relevant for questions about &lt;em&gt;equilibrium&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For that particular test, consider two dessicants $\ce{X}$ and $\ce{Y}$.  The winner of the test will be determined by the thermodynamics of the two reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O(g) &amp;lt;=&amp;gt; H2O_{(X)}}$$&#xA;$$\ce{H2O(g) &amp;lt;=&amp;gt; H2O_{(Y)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which represents the the absorption of water vapor into phase $\ce{X}$ (or $\ce{Y}$) respectively.  The two reactions can be combined to yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2O_{(X)} &amp;lt;=&amp;gt; H2O_{(Y)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the free energy change $\Delta G_r$ of that reaction will determine which dessicant is &quot;stronger&quot; in the sense of that particular test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This test doesn't take into account many practical considerations that other answers have noted, such as reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd bet that liquid &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_trioxide&quot; rel=&quot;nofollow&quot;&gt;sulfur trioxide&lt;/a&gt; would win most desicating contests.$%edit$&lt;/p&gt;&#xA;" OwnerUserId="13184" LastEditorUserId="24052" LastEditDate="2016-03-11T23:19:33.377" LastActivityDate="2016-03-11T23:19:33.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35419" PostTypeId="2" ParentId="35415" CreationDate="2015-08-24T06:10:35.677" Score="0" Body="&lt;p&gt;Assuming you can use a beaker (or just one of those 50 ml volumetric flask for mixing), you could first dilute some of the 0.3 M solution to 0.1 M by adding 5 ml of stock to 10 ml of water. Now you have a 0.1 M solution of NaCl which you can calculate how to dilute using the M1*V1=M2*V2 equation. Doing this, I calculate that to make 50 ml of 0.01 M solution, you need 5 ml of the 0.1 M solution you made, which you can do with a pipette.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2015-08-24T06:10:35.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35421" PostTypeId="1" CreationDate="2015-08-24T06:34:27.893" Score="2" ViewCount="181" Body="&lt;p&gt;I am now constructing a RC plane. I would like to fill the tires of the landing gear with nitrogen to avoid combustion of oxygen during landing. I would therefore like to obtain a small amount of nitrogen using (only) home-made utilities. How can it be done?&lt;/p&gt;&#xA;" OwnerUserId="19363" LastEditorUserId="4945" LastEditDate="2015-08-24T12:13:23.283" LastActivityDate="2015-08-24T12:22:21.040" Title="How to obtain nitrogen using only homemade utilities?" Tags="&lt;inorganic-chemistry&gt;&lt;home-experiment&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35423" PostTypeId="1" AcceptedAnswerId="35426" CreationDate="2015-08-24T09:57:53.080" Score="3" ViewCount="15945" Body="&lt;p&gt;My question is about how much science-fiction and how much chemistry is involved in the the '&lt;a href=&quot;http://marvelcinematicuniverse.wikia.com/wiki/Randy&amp;#39;s_Donuts&quot; rel=&quot;nofollow&quot;&gt;Randy's Donuts&lt;/a&gt;' scene of Iron Man 2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Agent Romanoff injects &lt;em&gt;Lithium Dioxide&lt;/em&gt; into Tony Stark's neck to remove the symptoms of Palladium poisoning.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My Question is: &lt;strong&gt;Is it safe to inject Lithium Dioxide into our body? &#xA;Does it have any side effects?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="19309" LastEditorUserId="9961" LastEditDate="2015-08-30T15:55:08.607" LastActivityDate="2020-05-26T11:46:38.263" Title="What are the effects of Lithium Dioxide on the body?" Tags="&lt;safety&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35425" PostTypeId="2" ParentId="35421" CreationDate="2015-08-24T12:05:52.137" Score="4" Body="&lt;p&gt;Crude process, but does the trick: You can use a combustion/oxygen consuming process, like a candle, to burn the excess oxygen. Eventually the candle will be too oxygen starved to sustain a flame, so some residual oxygen would be left (although this would already mean there is no fear of combustion of said tyres). Now to remove the remaining oxygen you'd use something like iron files (since you wanted to use only &quot;household&quot; stuff). Together with some humidity in that air, they will rust rapidly and should drive down the oxygen contents towards negligible. So now we have: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a small trace remainder of oxygen, &lt;/li&gt;&#xA;&lt;li&gt;the equivalent of $\ce{CO2}$ from the combustion process, &lt;/li&gt;&#xA;&lt;li&gt;some impurities from the candle wax probably, &lt;/li&gt;&#xA;&lt;li&gt;trace gases such as helium etc. and ... mostly &lt;/li&gt;&#xA;&lt;li&gt;nitrogen. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This already is not sustaining any combustion. But if you're a purist or want to use $\ce{N2}$ because it is lighter than $\ce{CO2}$, to remove the $\ce{CO2}$ you could use something like &quot;roasted limestone&quot;, i.e. the stuff from you water kettle, which you heat so that you get lime, which then binds with the $\ce{CO2}$ to revert to limestone $\ce{CaCO3}$. Now you have something, I'd assume, like 98% nitrogen and surely to few traces of $\ce{O2}$ to cause any harm to your equipment. Of course, I'd buy bottled $\ce{N2}$, but for the sake of using household implements only, this might do the trick.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="4945" LastEditDate="2015-08-24T12:22:21.040" LastActivityDate="2015-08-24T12:22:21.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35426" PostTypeId="2" ParentId="35423" CreationDate="2015-08-24T13:30:25.967" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it safe to inject Lithium Dioxide into our body?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It will do bad things.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it have any side effects?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. It will do bad things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As lithium is an alkali metal, it does not usually form covalent bonds, so we should treat whatever it makes as ionic. The common form of lithium oxide is $\ce{Li2O}$, which is not that scary. &quot;Lithium dioxide&quot; would be $\ce{LiO2}$. This implies that that lithium is tetravalent, as $\ce{Li^4+}$. This is not possible, as it requires lithium to give away 4 electrons but it does not have 4, only 3. This means that oxygens are in the superoxide form: is $\ce{O2-}$ instead of the more stable $\ce{O^2-}$. That is definitely not stable in isolation at ambient conditions (unlike other alkali superoxides) and I'm not sure where you will get that stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of that is that one molecule (or whatever it is) of $\ce{LiO2}$ will really want to convert itself to $\ce{Li2O}$. The superoxide ion is not going to do fun things. For each $\ce{O2-}$ you're going to need 3 more electrons to make it stable, so that is very strong oxidiser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison to what this could do to you, I would guess it would be worse than injecting pure hydrogen peroxide (you have about 5% of that in hair bleach), or pure sodium hypochloride (similar 5% in kitchen bleach). This is definitely not going to treat &quot;palladium poisoning&quot; because it will probably kill you beforehand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, using an oxidiser (such as lithium dioxide) to treat noble metal poisoning (such as palladium) seems dubious to me. I'm not a doctor, but I reckon that there are better method to treat that. So why lithium and palladium? I guess is sounds more credible than &quot;using titanium to treat potassium poisoning&quot; but less credible than &quot;using dysprosium to treat beryllium poisoning&quot;, both of which make absolutely no sense as &quot;using lithium to treat palladium poisoning&quot;.&lt;/p&gt;&#xA;" OwnerUserId="8083" LastEditorUserId="8083" LastEditDate="2015-08-24T15:10:36.210" LastActivityDate="2015-08-24T15:10:36.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35427" PostTypeId="1" CreationDate="2015-08-24T14:30:29.050" Score="4" ViewCount="1604" Body="&lt;p&gt;In chromatography, a mobile phase with analytes is passed over a stationary phase. A given analyte has different solubility in the two phases and consequently spends different periods of time in them. A particular molecule while in the stationary phase remains immobile, is swept along when present in the mobile phase.&#xA;$T_m$, the time spent in the mobile phase, is the same for all analytes. The reason (I'm not so sure) must be since all analytes travel the same distance finally, with the same speed as the mobile phase, so take the same time in so doing. But since some of these spend longer times in the stationary phase than others, the total time to reach the end, that is, $T_m + T_r$ is different ($T_r$ is the time spent in the stationary phase).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have a bit of a trouble understanding the rate aspect of the process, and what exactly determines the $T_r$ and $T_m$ - are they the average times spent by an arbitrary molecule in the two phases? If so, are their values dependent on the rate of the process, or the solubilities?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-12-30T08:54:08.783" Title="What Affects Tm and Tr In Chromatography?" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35428" PostTypeId="1" CreationDate="2015-08-24T15:06:12.873" Score="0" ViewCount="18" Body="&lt;p&gt;I've combined multiple elements dissolved in water into a preferred ratio, but would like to evaporate some of the water to increase the concentration (let's say at 200 $^{\circ}\mathrm{C}$ in the oven). The elements in question and the amounts are: 2 mg chromium, 5 mg lithium, 2 mg vanadium, 140 mg iron, and 50 mg boron. The volume of water is about 60 ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if I can safely do this procedure without a) causing a reaction; and b) evaporating any of the elements.&lt;/p&gt;&#xA;" OwnerUserId="19376" LastEditorUserId="194" LastEditDate="2017-04-01T03:39:19.680" LastActivityDate="2017-04-01T03:39:19.680" Title="Evaporating water from a combination of dietary minerals: will it work safely or react?" Tags="&lt;heat&gt;&lt;elements&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35429" PostTypeId="1" AcceptedAnswerId="38222" CreationDate="2015-08-24T15:42:44.960" Score="3" ViewCount="808" Body="&lt;p&gt;Why does $\ce{NH3+}$ have a higher $-\ce{I}$ effect ( inductive effect) than $\ce{NO2}$ even though oxygen is more electronegative? Does it have something to do with positive charge on $\ce{N}$? &lt;/p&gt;&#xA;" OwnerUserId="19378" LastEditorUserId="19383" LastEditDate="2015-10-01T17:45:41.107" LastActivityDate="2015-10-01T17:59:50.467" Title="Why is the inductive effect for a protonated amino group greater than that of a nitro group?" Tags="&lt;electronegativity&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35431" PostTypeId="1" AcceptedAnswerId="35440" CreationDate="2015-08-24T16:25:32.373" Score="0" ViewCount="1912" Body="&lt;p&gt;In one question, I found that it is given $\ce{XeF2}$ and $\ce{CO2}$ have the same shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{CO2}$ is linear. In $\ce{XeF2}$, if the fluorine is placed in an axial position, then only shape is linear. But in most compounds, lone pairs are placed in an axial position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the lone pair or the electronegative element more appropriate for placement in an axial position?&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="5026" LastEditDate="2018-02-27T13:25:06.747" LastActivityDate="2020-10-18T04:14:38.443" Title="Is the axial position more suited for a lone pair or an electronegative element?" Tags="&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35432" PostTypeId="1" AcceptedAnswerId="37064" CreationDate="2015-08-24T17:22:45.740" Score="2" ViewCount="744" Body="&lt;p&gt;In one question for finding $\ce{P(CH3)2H3}$ structure? Which one should be placed in axial position? I think $\ce{-CH3}$ should be axial position to minimise repulsion.&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="5017" LastEditDate="2015-09-09T12:35:35.390" LastActivityDate="2015-09-09T12:35:35.390" Title="Which element should be placed in axial position bulky one or the smaller one?" Tags="&lt;inorganic-chemistry&gt;&lt;molecular-structure&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35433" PostTypeId="1" AcceptedAnswerId="35436" CreationDate="2015-08-24T17:52:58.060" Score="2" ViewCount="2737" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which has maximum $\ce{ONO}$ angle?&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;$\ce{NO3-}$  &lt;/li&gt;&#xA;  &lt;li&gt;$\ce{NO2+}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{NO2-}$&lt;/li&gt;&#xA;  &lt;li&gt;$\ce{NO2}$&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to me $\ce{NO2+}$ should have the maximum angle because the structure is linear but in answer it is given $\ce{NO3-}$. We can consider the lone pair in $\ce{NO2+}$ but still also by VSEPR theory I don't think it is possible to decrease angle from 180 to less than 120.&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="5026" LastEditDate="2018-02-27T02:53:53.660" LastActivityDate="2018-02-27T02:53:53.660" Title="Compare ONO bond angle in nitrogen oxides" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35435" PostTypeId="1" CreationDate="2015-08-24T20:26:48.573" Score="4" ViewCount="69" Body="&lt;p&gt;I am trying to run nucleation simulations of glycine in water. To generate a supersaturated solution, a solution that will crystallize, I need to know the solubility. Luckily for me, this data exists. Now at 1 atm, 324 K, glycine's solubility is 0.338 g/ml water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how to determine the appropriate box size for a given simulation, if I specific the number of glycine molecules. I posted a question similar to this before,in the CCL mailing list, and was told to just assume ml water is ml of simulation box. For instance, if I want 1024 glycine molecules, whose molecular weight is 75.066 g/mol,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  (1024/6.0223*10^(23))*(75.066) = weight = 1.27*10^(-19) g &#xA;  weight/0.338 = ml simulation box = 3.78*10^(-19) ml&#xA;  box = ml simulation box^(1/3) = 72.28 angstroms&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I guess my first question is if this procedure appears to be correct. Specifically, all experimental data for solubility is clearly gathered at 1 atm pressure. However, if I fix the pressure, my understanding is that there is only one equilibrium concentration that the MD will take me to. If I think of the number of DOF's here, temperature is fixed, pressure is fixed, and the mol fraction is fixed, so I don't get to choose the box and hence the solubility. If I work in the NVT ensemble, which I am leaning towards, I don't know if I will actually be supersaturated because the simulation is no longer ran at 1 atm. Is there another definition of solubility that I should be using for MD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above scenario is more confusing to me when I think of simulating a &quot;bulk&quot; crystal. If I were to simulate a glycine crystal consisting of 1024 molecules, the above method would tell me a simulation box of 72.28 angstroms and a temperature of 294 K would mean the crystal wouldn't dissolve (assuming  nanoscale stability is similar to bulk stability). However, if I am simulating a bulk crystal, why would the box size matter to the crystal? It should always see a &quot;sea&quot; of solvent, and I thought I should only be concerned with making sure the box is large enough to prevent the cluster from seeing its own periodic images.&lt;/p&gt;&#xA;" OwnerUserId="16152" LastEditorUserId="9961" LastEditDate="2015-08-24T20:56:45.483" LastActivityDate="2015-08-24T20:56:45.483" Title="Glycine nucleation simulations" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;&lt;computational-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35436" PostTypeId="2" ParentId="35433" CreationDate="2015-08-24T20:58:17.633" Score="2" Body="&lt;p&gt;You are right, that's $\mathrm{NO_2^+}$, since it is linear, and whoever says otherwise is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, those (except for NO&lt;sub&gt;2&lt;/sub&gt;) are not &lt;em&gt;compounds&lt;/em&gt;, but &lt;em&gt;particles&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-08-24T21:16:08.280" LastActivityDate="2015-08-24T21:16:08.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35438" PostTypeId="2" ParentId="35432" CreationDate="2015-08-24T21:39:21.980" Score="1" Body="&lt;p&gt;No, it's the other way around. Axial positions are less spacious, so you'd want to place your smaller substituents there and the bulky ones to the equatorial.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-24T21:39:21.980" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35440" PostTypeId="2" ParentId="35431" CreationDate="2015-08-24T21:55:01.480" Score="2" Body="&lt;p&gt;So in $\ce{XeF2}$ there will be three lone pairs on the xenon, and the xenon will have two covalent bonds to fluorines. The xenon is of course the central atom, on to that we apply VSEPR-theory. We count five electron dense areas (two bonds and three lone pairs) and so VSEPR-theory predicts a trigonal bipyramidal shape for $\ce{XeF2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tZwu5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tZwu5.gif&quot; alt=&quot;Trigonal bipyramidal molecule&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lone pairs want to stay as far away from each other as possible (in VSEPR). Indeed would there have been just two lone pairs putting them axially would have introduced the maximal possible angle of 180°. Instead we have three lone pairs, so the equatorial positions introduce the best possible angle of 120°. Compare this to the 90° angle introduced if we had put two lone pairs axially and just one equatorial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is left now are the axial positions to which we allocate the fluorines, obtaining a linear molecule. In this case the VSEPR prediction has been &lt;a href=&quot;http://acta-arhiv.chem-soc.si/53/53-2-105.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;confirmed experimentally&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-08-24T21:55:01.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35441" PostTypeId="1" AcceptedAnswerId="35477" CreationDate="2015-08-25T00:04:41.400" Score="3" ViewCount="568" Body="&lt;p&gt;The so called &quot;&lt;a href=&quot;http://www.halopedia.org/Polymerized_lithium_niobocene&quot; rel=&quot;nofollow&quot;&gt;Polymerized LiNb (Lithium Niobocene)&lt;/a&gt;&quot; is a fictional compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard about this substance being used for an interesting purpose. It reacts to mechanical stress or pressure by creating an electric charge (piezoelectricity) and, conversely, deforms, as if external pressure was applied, when voltage is applied. Apparently it was far more &quot;piezoelectric&quot; than any other similar compounds, i.e. it deformed far more drastically than other piezoelectric materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a compound, or similar compounds, exist?&lt;/p&gt;&#xA;" OwnerUserId="19386" LastEditorUserId="9961" LastEditDate="2015-08-30T15:56:00.587" LastActivityDate="2015-08-30T15:56:00.587" Title="Is so called &quot;Polymerized LiNb (Lithium Niobocene)&quot; a possible compound?" Tags="&lt;organic-chemistry&gt;&lt;polymers&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="1" CommentCount="9" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35442" PostTypeId="1" AcceptedAnswerId="35499" CreationDate="2015-08-25T01:07:07.907" Score="1" ViewCount="2605" Body="&lt;p&gt;I began to study organic reactions and found the term &quot;attack&quot; everywhere, what this &quot;attack&quot; precisely mean?&lt;/p&gt;&#xA;" OwnerUserId="19387" LastActivityDate="2015-08-26T11:24:16.377" Title="What is an &quot;attack&quot; in Organic Reactions?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35444" PostTypeId="1" CreationDate="2015-08-25T03:05:00.450" Score="4" ViewCount="974" Body="&lt;p&gt;Background:&#xA;I was recently playing around with some reusable heat packs, the kind that operate by phase change of sodium acetate trihydrate and noticed that they didn't reach the temperature that I had expected (58 °C) rather, they stayed at 52 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were two such packs, roughly 50 mm × 100 mm × 15 mm and placed together with the broad sides in contact. Between them and at the center, there was a calibrated PT-100 RTD (which I certainly trust to be better than 1 °C accurate). The entire mass was then insulated with a few layers of thick wool fabric. The temperature reading became constant at 52 °C within a few minutes of crystallization being initiated in both packs and remained very stable for more than 10 minutes thereafter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that although the manufacturer labeled the contents as being 'Sodium acetate (food grade)' and water, no proportions are given. Also, the contents of the pack are bright red, indicating the presences of unlisted dyes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;What would be the most likely causes for the discrepancy? Contaminants, incorrect, proportions something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm ultimately interested in using this reaction to maintain ~58 °C in an experiment. I'd like to avoid the same result if I mix up a batch of this stuff from pure ingredients.&lt;/p&gt;&#xA;" OwnerUserId="19388" LastEditorDisplayName="user7951" LastEditDate="2015-08-25T10:16:08.990" LastActivityDate="2016-01-06T22:59:04.910" Title="Changes in freezing point of sodium acetate" Tags="&lt;experimental-chemistry&gt;&lt;melting-point&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35450" PostTypeId="1" CreationDate="2015-08-25T08:31:48.453" Score="2" ViewCount="62" Body="&lt;p&gt;Respective aromatic substitution: Why isn't the $\pi$ bond attacked by the catalyst $\ce{FeBr3}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Fe}$ must have a strong partial positive charge and acts as an electron acceptor to form a $\ce{Fe-Br}$ complex, but why can't it accept the electrons from the $\pi$ bond?&lt;/p&gt;&#xA;" OwnerUserId="19398" LastEditorUserId="865" LastEditDate="2015-08-25T12:30:48.050" LastActivityDate="2015-08-25T12:30:48.050" Title="Aromatic substitution" Tags="&lt;acid-base&gt;&lt;aromatic-compounds&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35452" PostTypeId="1" CreationDate="2015-08-25T10:25:26.363" Score="2" ViewCount="48" Body="&lt;p&gt;Is there a way to test if tequila has oxidised and if so the level of oxidisation?&#xA;It would be best if this could be done without a control sample, but the most important thing is if I can do this with 'household' equipment  - although I do have access to a hydrometer.&lt;/p&gt;&#xA;" OwnerUserId="19400" LastActivityDate="2015-08-25T10:25:26.363" Title="How can I test if a spirit (tequila) has oxidised?" Tags="&lt;home-experiment&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35453" PostTypeId="1" CreationDate="2015-08-25T10:40:43.807" Score="2" ViewCount="907" Body="&lt;p&gt;I've heard of single bonds, double bonds, triple bonds, and quadruple bonds between two atoms, but in each of those bonds the two atoms contribute the same number of electrons from each of their valence shells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why can't the atoms contribute an uneven number of electrons? For example, why can't there be a situation where a carbon atom contributes two valence electrons while a nitrogen atom only contributes one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely the sigma bonds arising from the overlap of two valence orbitals allow one valence orbital to have two valence electrons while the other only has one. &lt;em&gt;(I don't know about pi-bonds, and I don't know if it is possible for a carbon atom and a nitrogen atom to give two and one valence electrons, respectively, but maybe there is another pair of atoms that would allow this).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-08-25T12:28:39.887" LastActivityDate="2019-03-14T20:10:44.957" Title="Covalent bonds other then 2-center, 2-electron bonds" Tags="&lt;bond&gt;&lt;covalent-compounds&gt;&lt;valence-bond-theory&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35454" PostTypeId="1" CreationDate="2015-08-25T10:47:44.497" Score="2" ViewCount="95" Body="&lt;p&gt;How does one conclude if a given compound or a complex is a hard or soft acid or base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the factors that contribute to hard-hard, soft-soft, soft-hard, and hard-soft interactions?&lt;/p&gt;&#xA;" OwnerUserId="19390" LastEditorUserId="9961" LastEditDate="2015-10-03T00:39:41.680" LastActivityDate="2015-10-03T00:39:41.680" Title="Hard-hard and soft-soft acid/base interactions" Tags="&lt;hsab&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2015-10-05T08:39:04.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="35455" PostTypeId="1" AcceptedAnswerId="35461" CreationDate="2015-08-25T11:06:40.637" Score="1" ViewCount="1855" Body="&lt;p&gt;As I know, in electroplating method, the type of voltage used is DC voltage. My question is, does replacing DC current supply by AC one affect the functioning of the cell? And why?&lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="19133" LastEditDate="2015-08-25T11:29:55.050" LastActivityDate="2015-08-25T13:19:56.453" Title="Using AC voltage for electroplating" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35457" PostTypeId="2" ParentId="35088" CreationDate="2015-08-25T12:05:15.310" Score="7" Body="&lt;p&gt;&lt;em&gt;Ullmann’s Encyclopedia of Industrial Chemistry&lt;/em&gt; lists numerous components of commercial paints:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dyes, pigments, extenders&lt;/li&gt;&#xA;&lt;li&gt;organic solvents, water, coalescing agents&lt;/li&gt;&#xA;&lt;li&gt;binders&lt;/li&gt;&#xA;&lt;li&gt;plasticizers&lt;/li&gt;&#xA;&lt;li&gt;paint additives (levelling agents, film-formation promoters, wetting agents, dispersants, antisettling agents, antifoaming agents, catalysts, antifloating and antiflooding agents, antiskinning agents, matting agents, neutralizing agents, thickening agents, preservatives, corrosion inhibitors)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The application method, drying and hardening behaviour of a paint is essentially determined by the &lt;strong&gt;binder&lt;/strong&gt;. Binders are macromolecular products. The low molecular mass binders include alkyd resins, phenolic resins, polyisocyanates, and epoxy resins. These binders must be chemically hardened after application to produce high molecular mass cross-linked macromolecules in order to form acceptable films. The higher molcular mass binders include cellulose nitrate, polyacrylate, and vinyl chloride copolymers, which are readily suitable for film formation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the paint dries, a firmly bonded film is formed. Drying and film formation take place physically and chemically. Physical drying involves evaporation of the solvents (organic solvents and/or water). This is the main process for paints with high molecular mass polymer binders. Chemically drying paints contain binder components with a relatively low molecular mass that react together on drying to form cross-linked macromolecules. This process can occur by polymerization, polyaddition, or polycondensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, physical drying and the various mechanisms of chemical drying can proceed concurrently or consecutively, depending on the composition of the binder system.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-08-25T12:05:15.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35458" PostTypeId="1" CreationDate="2015-08-25T12:10:37.693" Score="5" ViewCount="449" Body="&lt;p&gt;An acid anhydride is condensed with an aromatic aldehyde in the Perkin reaction. The latter is non-enolisable, and thus the former forms the enolate which attacks as a nucleophile. If the attack is made at the aromatic aldehyde, an aldol-like product is formed which can be hydrolysed to the acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the acid anhydride enolate attacks another acid enolate, in a nucleophilic addition-elimination step, this results in the leaving of a carboxylate ion and the generation of an active methylene group (similar to Claisen condensation). There may be proton transfer between the two later. Surely this is more favourable than the first mechanism? Why doesn't this occur though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in the Stobbe condensation, instead of attack by the enolate at the ketone or aldehyde, cannot the succinic ester be attacked again?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastActivityDate="2015-08-25T12:10:37.693" Title="What prevents self-condensation in the Perkin reaction?" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35459" PostTypeId="1" AcceptedAnswerId="35565" CreationDate="2015-08-25T12:17:58.247" Score="0" ViewCount="61" Body="&lt;p&gt;Why is the acoustic absorption coefficient of hexane in air less than air? A 120 kHz pulse train exhibits a larger amplitude after traveling in a hexane/air mix than in pure air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis is that the hexane molecule has fewer degree of freedom than that of $\ce{N2/O2}$. Is this plausible?&lt;/p&gt;&#xA;" OwnerUserId="9473" LastEditorUserId="9473" LastEditDate="2015-08-25T12:36:11.250" LastActivityDate="2015-11-25T15:35:34.663" Title="Why is the acoustic absorption coefficient of hexane in air less than air?" Tags="&lt;physical-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35460" PostTypeId="1" CreationDate="2015-08-25T12:38:01.433" Score="5" ViewCount="97" Body="&lt;p&gt;I just learned that charged alpha helices are less thermally stable than uncharged ones. The only difference I see between them would be a larger dipole moment in the charged helix, but how would this contribute to stability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a picture for clarification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Eif7a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eif7a.png&quot; alt=&quot;Two alpha helices&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="21642" LastEditDate="2020-09-03T18:13:57.680" LastActivityDate="2020-09-03T18:13:57.680" Title="Thermal stability of charged and uncharged alpha helices (the concept of alpha-helix capping)" Tags="&lt;biochemistry&gt;&lt;chemical-biology&gt;&lt;proteins&gt;&lt;amino-acids&gt;&lt;structural-biology&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35461" PostTypeId="2" ParentId="35455" CreationDate="2015-08-25T13:19:56.453" Score="3" Body="&lt;p&gt;The difference is quite huge. Firstly, the deposition will occur on both electrodes (if possible). Secondly, if the frequency is high enough, there will be no deposition at all as only polarization and depolarization effect take place. The frequency effects mass, transport and electrical double layer. Depending on the electrolyte composition, the dissolution (if possible) will reduce the deposition. Usually pulse electroplating is used.  The problematic is huge, please see some articles for more info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dynatronix.com/wp-content/uploads/2014/02/Article-77.pdf&quot; rel=&quot;nofollow&quot;&gt;An Overview of Pulse Plating - Norman M. Osero&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/publication/222521358_Pulse_and_pulse_reverse_platingConceptual_advantages_and_applications&quot; rel=&quot;nofollow&quot;&gt;Pulse and pulse reverse plating—Conceptual, advantages and applications, M.S. Chandrasekar, Malathy Pushpavanam&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4097" LastActivityDate="2015-08-25T13:19:56.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35462" PostTypeId="1" CreationDate="2015-08-25T13:21:29.483" Score="33" ViewCount="5848" Body="&lt;p&gt;Why does no other element but hydrogen get special names for its isotopes?&lt;/p&gt;&#xA;" OwnerUserId="19401" LastEditorUserId="-1" LastEditDate="2015-08-26T10:48:02.540" LastActivityDate="2020-07-11T01:19:50.263" Title="Why are isotopes of hydrogen the only ones with special names?" Tags="&lt;nomenclature&gt;&lt;hydrogen&gt;&lt;isotope&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35463" PostTypeId="2" ParentId="35462" CreationDate="2015-08-25T13:35:42.833" Score="17" Body="&lt;p&gt;I think there are two reasons. First, it is more convenient to categorize them under the actual element-name to which they belong. If I say &quot;15-Beryllium&quot; everyone knows immediately, what I'm talking about. If we add hundreds of isotope-names, it would be quite a mess. Leading to the second reason: Xenon for example has over known 30 isotopes. There are just too many to name them all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen has, by the way, 7 known isotopes, of which only 3 have names:&lt;br&gt;&#xA;1-Hydrogen: Protium (rarely used)&lt;br&gt;&#xA;2-Hydrogen: &lt;a href=&quot;http://www.nature.com/nchem/journal/v4/n3/full/nchem.1273.html&quot;&gt;Deuterium&lt;/a&gt; (D)&lt;br&gt;&#xA;3-hydrogen: Tritium (T)&lt;br&gt;&#xA;&lt;br&gt;&#xA;IUPAC discourages using the abbreviations to avoid confusion with other elements. Although, I see a lot of &quot;D&quot; in the organic literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;br&gt;&#xA;I just saw, that in the novel &lt;a href=&quot;https://en.wikipedia.org/wiki/The_Mouse_That_Roared&quot;&gt;The Mouse That Roared&lt;/a&gt; the isotope 4-Hydrogen has been named quadium (Q). This is neither used nor &quot;official&quot; though.&lt;/p&gt;&#xA;" OwnerUserId="4852" LastEditorUserId="4852" LastEditDate="2015-08-25T13:45:50.910" LastActivityDate="2015-08-25T13:45:50.910" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35465" PostTypeId="2" ParentId="35462" CreationDate="2015-08-25T14:26:14.217" Score="26" Body="&lt;p&gt;In addition to the reasons ste listed, the isotopes of hydrogen have the greatest differences in mass compared to other elements. Consider that deuterium is twice as heavy as protium, and tritium is three-times as heavy as protium. Isotopes of all elements can be used in &lt;a href=&quot;https://en.wikipedia.org/wiki/Kinetic_isotope_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;kinetic isotope experiments&lt;/a&gt;. The dramatic differences in mass among the hydrogen isotopes gives the most pronounced isotope effects. This type of difference in reactivity likely made it easiest to discover the different isotopes, and these differences explain the continued usage of the trivial names even more systematic descriptors are available.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="4325" LastEditDate="2017-01-26T16:46:38.070" LastActivityDate="2017-01-26T16:46:38.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35466" PostTypeId="1" CreationDate="2015-08-25T14:29:16.113" Score="1" ViewCount="105" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Which one among the following shows maximum ideal behaviour?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{NH3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{Rn}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Ammonia and water have hydrogen bonding so they're ruled out. That leaves radon and &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt;. Radon is a noble gas which means inertness &lt;strong&gt;but&lt;/strong&gt; it's too big in size. That could mean greater van der Waal forces. And &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; on the other hand has covalent bonding.&lt;/p&gt;&#xA;&lt;p&gt;I want to confirm which one of them is more ideal.&lt;/p&gt;&#xA;" OwnerUserId="17250" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-07-15T02:46:14.340" Title="Which gas among radon and hydrogen chloride has higher ideal behavior?" Tags="&lt;inorganic-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35467" PostTypeId="1" CreationDate="2015-08-25T14:39:26.387" Score="4" ViewCount="1314" Body="&lt;p&gt;I was trying to determine whether I could synthesize methyl iodide by combining methanol, potassium iodide and sulphuric acid together... The way I see it, the sulphuric acid will protonate the OH instead of the I forming a very stable leaving group of water. Then the iodine ions would donate an electron pair to the carbon. You would be left with the salt Potassium bisulfate.... Then to remove my methyl iodide, just distil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this work?&lt;/p&gt;&#xA;" OwnerUserId="19218" LastActivityDate="2018-04-14T18:16:42.743" Title="Methyl Iodide synthesis" Tags="&lt;synthesis&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35468" PostTypeId="1" CreationDate="2015-08-25T14:49:42.333" Score="1" ViewCount="100" Body="&lt;p&gt;If I attach an electrostatic precipitator to a car's exhaust - and also make the smoke pass through an area containing $\ce{NaOH}$ in order to absorb sulfur dioxide - does that mean that the smoke that emerges will not contain harmful gasses such as $\ce{CO2}$, $\ce{CO}$ and $\ce{SO2}$? If not, what is the reason?&lt;/p&gt;&#xA;" OwnerUserId="19403" LastEditorUserId="865" LastEditDate="2015-08-25T14:59:18.397" LastActivityDate="2015-08-25T14:59:18.397" Title="Removing harmful gases from automobile exhaust using electrostatic precipitation" Tags="&lt;physical-chemistry&gt;&lt;purification&gt;&lt;adsorption&gt;&lt;absorption&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35469" PostTypeId="1" AcceptedAnswerId="35482" CreationDate="2015-08-25T15:12:31.503" Score="6" ViewCount="132" Body="&lt;p&gt;I am just confused as to how $\ce{I-}$ cannot attract $\ce{H+}$, but it can attack a carbocation. It is sharing electrons in both cases, so what is the difference? &lt;/p&gt;&#xA;" OwnerUserId="19405" LastEditorUserId="7448" LastEditDate="2015-08-25T15:53:36.937" LastActivityDate="2015-09-04T03:32:04.610" Title="How can I- be both a nucleophile and a weak base?" Tags="&lt;acid-base&gt;&lt;carbocation&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35470" PostTypeId="1" AcceptedAnswerId="35471" CreationDate="2015-08-25T16:06:42.137" Score="25" ViewCount="6537" Body="&lt;p&gt;How does trans-cyclooctene exhibit chirality if there are no stereocenters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related follow-on questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/35493/are-all-higher-cycloalkenes-chiral&quot;&gt;Are all higher cycloalkenes chiral?&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/35494/do-more-double-bonds-cause-a-bigger-number-of-stereoisomers-in-cycloalkenes&quot;&gt;Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2016-09-07T15:45:47.297" Title="Why is trans-cyclooctene chiral?" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35471" PostTypeId="2" ParentId="35470" CreationDate="2015-08-25T16:28:23.320" Score="26" Body="&lt;p&gt;Very interesting question! The key word you are looking for is &lt;a href=&quot;https://en.wikipedia.org/wiki/Planar_chirality&quot; rel=&quot;noreferrer&quot;&gt;planar chirality&lt;/a&gt;. In &lt;em&gt;trans&lt;/em&gt;-cyclooctene, the polymethylene bridge can either go &quot;in front of&quot; or go &quot;behind&quot; the plane of the double bond, assuming you fix the double bond and the two hydrogens in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by @jerepierre, they are considered different molecules due to a high-energy barrier which prevents the interconversion. Cyclooctene is the first cycloalkene to have both stable &lt;em&gt;cis&lt;/em&gt;- and &lt;em&gt;trans&lt;/em&gt;- isomers. The chain in &lt;em&gt;trans&lt;/em&gt;-cyclooctene is not long enough to swing over the double bond. As the chain gets longer, the energy barrier to rotation decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the two mirror images of &lt;em&gt;trans&lt;/em&gt;-cyclooctene (image source: own work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oGE4r.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oGE4r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two molecules are mirror images of each other but are not superimposable. Therefore &lt;em&gt;trans&lt;/em&gt;-cyclooctene is chiral despite not having a chiral center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.warwick.ac.uk/fac/sci/chemistry/research/fox/foxgroup/teaching/trans-cyclooctene.pdf&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: this is the first time I learned that a chiral molecule does not have to have a chiral center. After making some searches online, I feel a need to expand the answer to clarify the concept of chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Chirality_(chemistry)&quot; rel=&quot;noreferrer&quot;&gt;chiral&lt;/a&gt; molecule is one that has a non-superimposable mirror image. Mathematically a molecule is chiral if it is not symmetric under an improper rotation. Chirality arises due to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;point chirality&lt;/strong&gt;: typically a carbon center with four different substituents;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;axial chirality&lt;/strong&gt;: such as allenes with different substituents on each carbon (see this &lt;a href=&quot;https://chemistry.stackexchange.com/questions/6477/why-are-allenes-chiral&quot;&gt;question&lt;/a&gt;);&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;planar chirality&lt;/strong&gt;: such as the case of &lt;em&gt;trans&lt;/em&gt;-cyclooctene;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;inherent chirality&lt;/strong&gt;: due to the presence of a curvature in a structure that would be devoid of symmetry axes in any bidimensional representation, such as fullerenes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19088" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2016-09-07T15:45:47.297" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="35472" PostTypeId="2" ParentId="35466" CreationDate="2015-08-25T18:31:14.640" Score="3" Body="&lt;p&gt;Some molecules are highly polar and thus are subject to electrostatic (dipole-dipole) interactions, which are typically far stronger than the interactions of induced dipoles (commonly referred to as van der Waals forces). See where this gets you.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="5026" LastEditDate="2018-07-15T02:46:14.340" LastActivityDate="2018-07-15T02:46:14.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35473" PostTypeId="1" AcceptedAnswerId="48569" CreationDate="2015-08-25T20:59:21.710" Score="2" ViewCount="337" Body="&lt;p&gt;I recently performed a double replacement reaction between copper sulfate and sodium carbonate to yield copper carbonate as a precipitate. I then filtered the precipitate, but I noticed that the precipitate still contained a large amount of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be a significant amount of sodium sulfate dissolved in this water, making the copper carbonate impure? If so, how would I separate the remaining sodium sulfate from the copper carbonate?&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="31313" LastEditDate="2017-04-01T00:37:36.950" LastActivityDate="2017-04-01T00:37:36.950" Title="Does precipitated copper carbonate contain impurities" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;purification&gt;&lt;precipitation&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35474" PostTypeId="2" ParentId="34873" CreationDate="2015-08-25T21:08:12.817" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As explained in the comment by &lt;strong&gt;permeakra&lt;/strong&gt;, due to the structural differences between anhydrous and hydrated Na&lt;sub&gt;2&lt;/sub&gt;SiO&lt;sub&gt;3&lt;/sub&gt;, their solutions will also be intrinsically different, at least for a while.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether anhydrous salt will turn to hydrate or vice versa depends on your ambient humidity and temperature. Anyway, I'd not recommend leaving either of them in contact with open air, for fear they might react with CO&lt;sub&gt;2&lt;/sub&gt;. Silicic acid is so weak that it can be displaced even by carbonic acid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Well, theoretically, the information about the SiO&lt;sub&gt;2&lt;/sub&gt;:Na&lt;sub&gt;2&lt;/sub&gt;O ratio is already present in the chemical formula Na&lt;sub&gt;2&lt;/sub&gt;SiO&lt;sub&gt;3&lt;/sub&gt; (with or without water). However, the industrial-grade product may not quite match that formula. In fact, it may be a wild mixture of different silicates, where each individual compound cannot and need not be characterized. For many large-scale applications it suffices to know just the said ratio. Not sure about your application, though.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-25T21:08:12.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35476" PostTypeId="1" AcceptedAnswerId="37467" CreationDate="2015-08-25T21:13:04.263" Score="10" ViewCount="13535" Body="&lt;p&gt;So my teacher started the explanation of Chemical Equilibrium saying..&#xA;The concentration does not change with time.&#xA;And, gave us certain graphs asking which ones show chemical equilibrium.&#xA;I cannot understand what's written in my copy since I took the lesson two years ago. I need to be explained what I wrote in my copy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the concentration does not change with time.&#xA;If concentration is same or not, does not matter. That only the rate of forward reaction and the rate of backward reaction needs to be same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the graphs: (I am sorry I could not draw better.)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/896Yt.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/896Yt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;: &#xA;Now, I know 1,4,5 and 6 show chemical equilibrium.Reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The concentration of the reactant is lost but same amount is achieved by the product.So here, the rate of forward reaction and rate of backward reaction is same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For 5 and 6- The concentration of reactant and product may not be the same but it does not change with time. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I need help regarding :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.Explanation as to how the other graphs do not show chemical equilibrium(if possible in terms of how the answer for the other graphs are explained).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.Explanation for graph 5 and 6. I cannot understand the reason I had written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.An overall explanation and answer of the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help and understanding!! &lt;/p&gt;&#xA;" OwnerUserId="7503" LastActivityDate="2018-02-21T15:17:16.270" Title="Which graph shows Chemical Equilibrium?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35477" PostTypeId="2" ParentId="35441" CreationDate="2015-08-26T00:12:40.260" Score="2" Body="&lt;p&gt;Materials such as Terfenol-D and Galfenol are magnetostrictive ferrous alloys that expand and contract in the presence of magnetic fields, as well as creating magnetic fields when acted on by a physical force.  Both these materials were developed at Ames Lab (formerly the Naval Ordnance Laboratory, that's why the material names end with -NOL).  Terfenol-D has the highest magnetostriction of any Fe-alloy, but the high rare earth content or Terbium and Dysprosium limits its commercial applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;polymer&quot; moniker for Lithium polymer batteries does no refer to the lithium metal or oxides, it only speak to the polymer electrolyte or a polymer casing.  It just a marketing term, no a chemical designation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So on to your question... Polymer, no.  Monomer, no.  Niobium has an extremely high melting point (nearly 2500C/4500F).  It is a very difficult material to melt.  Lithium boils around 1300C.  From a metallurgical processing point of view, you would need a chamber pressurized to several MPa with an inert gas to prevent the lithium from volatilizing as you attempt to alloy it with niobium.  You would also need some powerful magnetic stirring to attempt to keep the mix homogeneous since the lithium would literally float out of the solution.&lt;/p&gt;&#xA;" OwnerUserId="19414" LastActivityDate="2015-08-26T00:12:40.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35480" PostTypeId="1" CreationDate="2015-08-26T01:45:14.780" Score="3" ViewCount="354" Body="&lt;p&gt;Given that a solution must be a homogenous mixture, and having crystallized solids at the bottom of a solution is most definitely not homogenous, can a supersaturated solution still be technically defined as a solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that note, is it possible for a solution to have more solute than solvent and remain in solution form?&lt;/p&gt;&#xA;" OwnerUserId="19415" LastActivityDate="2015-08-26T12:45:52.853" Title="Is a supersaturated solution still a solution?" Tags="&lt;solutions&gt;&lt;solvents&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35481" PostTypeId="1" CreationDate="2015-08-26T03:06:33.320" Score="5" ViewCount="30518" Body="&lt;p&gt;From left to right across a period, effective nuclear charge increases in the transition series, just like in the p block, and so atomic size decreases. Here the electrons are being added to the d orbital, where shielding is relatively poor compared to the p orbital. It should follow that the increase in the effective nuclear charge is more significant for the transition elements than the p block in a given period, leading to greater size variation among successive members - but the size variation is smaller in the transition elements instead. In fact atomic size increases from Ni to Zn. Is there any logical explanation for the fact?&lt;/p&gt;&#xA;" OwnerUserId="17654" LastEditorUserId="7475" LastEditDate="2018-01-02T14:32:37.223" LastActivityDate="2018-01-02T14:32:37.223" Title="Variation in atomic sizes in the transition elements" Tags="&lt;transition-metals&gt;&lt;periodic-trends&gt;&lt;atomic-radius&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35482" PostTypeId="2" ParentId="35469" CreationDate="2015-08-26T04:10:21.317" Score="6" Body="&lt;p&gt;As Jori pointed out, nucleophilicity is a kinetic property while basicity is a thermodynamic property. Nucleophilicity in part is about the rates of chemical reactions; thermodynamics is in part about the favorability of chemical reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction may be very favorable, i.e. it has a very negative free energy change, but it might take a &lt;em&gt;long&lt;/em&gt; time to actually complete. For example, the rusting of iron is favorable from a thermodynamic standpoint. The products, iron oxides, are very stable. However, the process is relatively slow from a kinetic standpoint; putting water on iron isn't going to instantly result in a familiar reddish-brown hue. It might take a day or so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you squirt something such as tert-butyllithium into the air, you'll probably catch yourself or something on fire. This is a &lt;em&gt;carboanion salt&lt;/em&gt;. Carboanions are &lt;em&gt;strong (Bronsted, and therefore, Lewis) bases&lt;/em&gt;. Acid-base reactions also tend to be relatively fast (kinetics) since they involve the transfer of something so small - a hydrogen proton. A graduate student unfortunately exposed some of this carbocation salt to the air and the resulting fireball instantly caught her sweater on fire. The carboanion salt likely grabbed protons from atmospheric moisture - a very favorable process from a thermodynamic standpoint - and kinetically, a &lt;em&gt;very&lt;/em&gt; fast process. And so the resulting heat likely ignited the resulting protonated carboanion (alkane in this case) on fire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $\ce{I-}$ is a good nucleophile in both protic and non-protic solvents. It's a rather big ion - the biggest of the halide anions. That means its negative charge is relatively dilute. This in turn means it's not as strongly solvated by the solvent molecules; it doesn't have as many tenacious solvent molecules hindering its trek toward a carbocation. So $\ce{I-}$ can reach carbocations relatively &lt;em&gt;quickly&lt;/em&gt;. Note that terms dealing with time usually indicate that the writer is speaking about a chemical's kinetic properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason that $\ce{I-}$ is a good nucleophile it's a poor base. It's dilute negative charge (due to its size) means that $\ce{H+}$ has better things to do than be attracted to $\ce{I-}$. $\ce{I-}$ has a dilute amount of negative charge so it cannot effectively stabilize the hydrogen proton as well as many solvents might - especially water, which &lt;em&gt;extensively&lt;/em&gt; associates with $\ce{H+}$; $\ce{H3O+}$ only &lt;em&gt;begins&lt;/em&gt; to describe how well the proton is associated with water molecules; the proton has been found to be associated with 2, 3, 4, or more water molecules! &lt;/p&gt;&#xA;" OwnerUserId="5084" LastEditorUserId="5084" LastEditDate="2015-09-04T03:32:04.610" LastActivityDate="2015-09-04T03:32:04.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35483" PostTypeId="2" ParentId="35481" CreationDate="2015-08-26T04:22:07.167" Score="2" Body="&lt;p&gt;From Ni to Zn there is increase in size because of the completely filled 3d and 4s orbitals. Here the nuclear charge is increased with atomic number but due to more electron -electron repulsion in 3d and 4s orbitals  as compared to force of  attraction between nucleus and outer orbital electrons there is little expansion in size as compared to other elements in 3d series.  &lt;/p&gt;&#xA;" OwnerUserId="19417" LastActivityDate="2015-08-26T04:22:07.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35484" PostTypeId="1" AcceptedAnswerId="35492" CreationDate="2015-08-26T05:53:15.583" Score="6" ViewCount="22902" Body="&lt;p&gt;During the combustion of hydrocarbons, there is a difference between the amounts of carbon or hydrogen that results in incomplete or complete combustion of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, besides from an insufficient amount of oxygen, what can cause incomplete combustion? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the physical or chemical properties of the hydrocarbon responsible for incomplete combustion, such as the hydrogen to carbon ratio (mass, volume, number of atoms, etc.) or the saturation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please keep the reply suitable for a student studying high school to early university chemistry.&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-08-26T12:30:29.913" LastActivityDate="2017-06-10T09:41:01.770" Title="What causes incomplete combustion?" Tags="&lt;combustion&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35485" PostTypeId="2" ParentId="35453" CreationDate="2015-08-26T05:58:16.493" Score="3" Body="&lt;p&gt;One example to consider is benzene, where each carbon contributes 1 electron to a pi-bonding system of 6 electrons in 3 orbitals. The C-C bond lengths in benzene are intermediate between standard single and double bond lengths. Also take a look at the various boranes such as diborane (B2H6) which contain three-centered (or higher?) bonds involving boron and hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-08-26T05:58:16.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35486" PostTypeId="1" CreationDate="2015-08-26T07:29:34.347" Score="16" ViewCount="3461" Body="&lt;p&gt;What is the IUPAC name for the two isomers of 1,2,3-trichlorocyclopropane? Similarly what is the name for isomers of 1,2,3,4-tetrachlorocyclobutane, 1,2,3,4,5-pentachlorocyclopentane, etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,2,3-trichlorocyclopropane has two isomers, one of them with all the chlorine atoms facing in one direction and the other with one chlorine facing the opposite direction of the other two. In CIP rule, if two substituents on an atom differ only in their absolute configuration, R takes priority over S. However I have difficulty applying the rule here: two of the substituents are exactly the same. Nevertheless the two molecules are clearly distinct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the problem could be circumvented by using trans- and cis- prefix. But it does not address the key problem and fails on longer chains --- 1,2,3,4-tetrachlorocyclobutane has 4 possible arrangements, though not necessarily all of them have the problem.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are the isomers of cyclobutane and their mirror images. I use &lt;code&gt;+&lt;/code&gt; and &lt;code&gt;-&lt;/code&gt; to denote a chlorine atom above or below the plane. The number in the parenthesis is the number of distinct molecules.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+ + | - -    (1)&#xA;+ + | - - &#xA;&#xA;+ - | - +    (1)&#xA;+ + | + +&#xA;&#xA;+ - | - +    (1)&#xA;+ - | - +&#xA;&#xA;+ - | - +    (1)&#xA;- + | + -&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="19088" LastEditorUserId="4325" LastEditDate="2015-08-26T16:33:54.403" LastActivityDate="2020-07-12T08:06:16.097" Title="IUPAC name for 1,2,3-trichlorocyclopropane?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="3" CommentCount="3" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35487" PostTypeId="1" CreationDate="2015-08-26T07:59:54.210" Score="4" ViewCount="10642" Body="&lt;p&gt;The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="5026" LastEditDate="2018-03-31T12:41:27.900" LastActivityDate="2018-03-31T12:41:27.900" Title="What is the reason for the anomalous electronic configuration of palladium?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2018-03-31T23:40:16.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="35488" PostTypeId="1" AcceptedAnswerId="35768" CreationDate="2015-08-26T08:19:13.247" Score="23" ViewCount="11295" Body="&lt;p&gt;I understand that dipole dipole forces is due to the attraction of the different partials charges of atoms in different molecules due to their different electro-negativities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hydrogen bonding, my understanding of it is that they are just a special case of dipole dipole forces. In certain molecules in where the difference in electro-negativities between the different atoms is high enough, then even stronger dipole dipole forces will be form which are known as hydrogen bonding. For example, in $\ce{H_2O}$ where the difference in electro-negativity of H and O is 1.24, hydrogen bonding occur. This is why hydrogen bonding can only occur between hydrogen and nitrogen, fluorine or oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my school teacher says that hydrogen bonding in fact is not a type of dipole dipole force as she says that hydrogen bonding occurs differently to dipole dipole forces. She said that hydrogen bonding occurs to the attraction of the lone pairs in oxygen, fluorine and nitrogen atoms to atoms with a positive partial charge. Hence this is why hydrogen bonding is much stronger than conventional dipole dipole forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this doesn't make sense because if hydrogen bonding is due to the attraction  of H to the lone pair of electrons, shouldn't hydrogen bonding also form between H and Cl or I or any other atom that has a lone pair and is more electronegative than H?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the comments below, it appears that the actual answer is more complicated than the simple explanation of hydrogen bonding given above. After doing some research, I have became aware of the &lt;strong&gt;Advanced Theory of the Hydrogen Bond&lt;/strong&gt; which states that the hydrogen bond actually contains 10% covalent nature. However none actually explain how they are formed. Could someone please explain how hydrogen bonding actually in reality form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in response to the excellent answer by jheindel, is the reason why hydrogen bonding is considered partly covalent in nature is because the let say in water, oxygen's MO partly overlaps with the MO of the hydrogen atom in the other molecule. Also this is why they are much stronger than conventional dipole dipole forces?&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-10-08T09:16:44.840" LastActivityDate="2018-01-21T15:02:43.010" Title="Is Hydrogen Bonding a Type of Dipole Dipole Interaction?" Tags="&lt;quantum-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;intermolecular-forces&gt;&lt;dipole&gt;&lt;hydrogen-bond&gt;" AnswerCount="2" CommentCount="6" FavoriteCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="35490" PostTypeId="1" CreationDate="2015-08-26T08:44:28.550" Score="1" ViewCount="619" Body="&lt;p&gt;What is the percent concentration of hydrogen peroxide which has a concentration of $500~\mu\mathrm{M}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $500~\mu\mathrm{M}$ $\ce{H2O2}$ toxic to the eyes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular honey has a concentration of $\ce{H2O2}$ of $500~\mu\mathrm{M}$ after a 24-hour incubation time, after it has been diluted. I am trying to figure out what the concentration is in percent (%). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recalculated and the result is actually $0.00153\%$ for $500~\mu\mathrm{M}$. $0.0012\%$ is the equivalent to $400~\mu\mathrm{M}$, but I need someone to confirm this. Someone on researchgate wrote that $30\%$ of $\ce{H2O2}$ is $9.8~\mathrm{M}$.&lt;/p&gt;&#xA;" OwnerUserId="19424" LastEditorUserId="16622" LastEditDate="2015-08-28T09:18:31.477" LastActivityDate="2015-09-27T10:03:45.123" Title="Percent concentration of 500 μM of H2O2" Tags="&lt;safety&gt;&lt;concentration&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35491" PostTypeId="2" ParentId="35467" CreationDate="2015-08-26T08:58:05.400" Score="2" Body="&lt;p&gt;So what I said in the comment still holds: it could work (particularly in combination with a Dean-Stark apparatus to remove excess water), but there is no way to know for sure unless someone tries it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more usual (and early) method &lt;a href=&quot;http://www.orgsyn.org/demo.aspx?prep=CV2P0399&quot; rel=&quot;nofollow noreferrer&quot;&gt;is described in Organic Syntheses&lt;/a&gt; and utilizes the reaction of methanol with iodine in the presence of phosphorous&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BetKI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BetKI.gif&quot; alt=&quot;MeI synthesis from methanol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the exact reaction mechanism (if anyone has a clue, please feel free to edit!) but what happens is that $\ce{PI_3}$ is generated &lt;em&gt;in situ&lt;/em&gt; from the elemental iodine and phosphorous and reacts with methanol to yield the methyl iodide and phosphorous acid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{3 CH3OH + PI3 -&amp;gt; 3 CH3I + H3PO3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phosphorous acid does not act as nucleophile and so the problem of a competing backward reaction, which you would have in your proposed synthesis, is largely prevented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to add some caution: the reaction described in Organic Syntheses is not suited for beginners or in amateur setting. The chemicals involved are nasty: highly toxic, highly flammable, alkylating and pyrophoric. Don't try this at home :)&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-08-26T08:58:05.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35492" PostTypeId="2" ParentId="35484" CreationDate="2015-08-26T09:05:50.890" Score="9" Body="&lt;p&gt;According to the Iowa State University page &lt;a href=&quot;http://www.abe.iastate.edu/extension-and-outreach/carbon-monoxide-poisoning-checking-for-complete-combustion-aen-175/&quot;&gt;Carbon Monoxide Poisoning: Checking for Complete Combustion (AEN-175)&lt;/a&gt; suggest the following mechanisms that cause incomplete combustion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Insufficient mixing of air and fuel.&lt;/li&gt;&#xA;  &lt;li&gt;Insufficient air supply to the flame.&lt;/li&gt;&#xA;  &lt;li&gt;Insufficient time to burn.&lt;/li&gt;&#xA;  &lt;li&gt;Cooling of the flame temperature before combustion is complete.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The main notion is the lack of air to cause complete combustion (either through lack of supply, mixing or time for combustion). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it should be noted that, practically speaking, according to the ChemWiki page &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkanes/Reactions_of_Alkanes/Burning_Alkanes&quot;&gt;Burning Alkanes&lt;/a&gt;, the size of the hydrocarbon has an affect in 'normal' conditions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The reason being (according to ChemWiki) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the larger the hydrocarbon, the less likely it would vaporise sufficiently, thus less chance of the optimal mixing of air and fuel, resulting in incomplete combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, provide the right conditions with sufficient supply of air and optimal hydrocarbon-air mixing ratio, it is still possible to cause complete combustion, in the general formula (for many hydrocarbons):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_{x}H_{y} +O2 -&amp;gt;H2O + CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quite comprehensive resource explaining each of the factors is the University of Tulsa chapter &lt;a href=&quot;http://www.personal.utulsa.edu/~kenneth-weston/chapter3.pdf&quot;&gt;Fuels and Combustion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-26T09:23:17.270" LastActivityDate="2015-08-26T09:23:17.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35493" PostTypeId="1" AcceptedAnswerId="35515" CreationDate="2015-08-26T09:17:16.063" Score="11" ViewCount="1699" Body="&lt;p&gt;Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question supplements my previous question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/35470/why-is-trans-cyclooctene-chiral&quot;&gt;Why is trans-cyclooctene chiral?&lt;/a&gt; See also the follow-on question: &lt;a href=&quot;https://chemistry.stackexchange.com/questions/35494/do-more-double-bonds-cause-a-bigger-number-of-stereoisomers-in-cycloalkenes&quot;&gt;Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-09-04T15:07:49.550" Title="Are all higher cycloalkenes chiral?" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35494" PostTypeId="1" AcceptedAnswerId="36898" CreationDate="2015-08-26T09:22:10.893" Score="6" ViewCount="908" Body="&lt;p&gt;If &lt;em&gt;trans&lt;/em&gt; double bonds cause cycloalkenes to be chiral, does a second &lt;em&gt;trans&lt;/em&gt; or maybe &lt;em&gt;cis&lt;/em&gt; double bond induce another chiral plane and hence four stereoisomers are possible for the given cyclodiene?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is based upon these previous questions:&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/35493/are-all-higher-cycloalkenes-chiral&quot;&gt;Are all higher cycloalkenes chiral?&lt;/a&gt;&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/35470/why-is-trans-cyclooctene-chiral&quot;&gt;Why is trans-cyclooctene chiral?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-09-04T21:49:35.753" Title="Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?" Tags="&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35495" PostTypeId="2" ParentId="35488" CreationDate="2015-08-26T09:28:36.293" Score="-2" Body="&lt;p&gt;Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of other molecule. These are special type of dipole-dipole interactions. Here the hydrogen atom is linked with one atom through covalent bond and linked with other atom through hydrogen bond.however hydrogen bond is much weaker than the covalent bond. When hydrogen atom bonded to a strongly electronegative atom the bond pair of electrons displaced towards electronegative atom. Thus the only electron of hydrogen lies away from it, its nucleus get exposed and behaves as a bare proton and exert a strong electrostatic force of attraction on the strong electronegative atom of adjacent molecule this interaction is referred as hydrogen bond.&lt;/p&gt;&#xA;" OwnerUserId="19417" LastActivityDate="2015-08-26T09:28:36.293" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="35496" PostTypeId="1" CreationDate="2015-08-26T09:43:25.043" Score="0" ViewCount="93" Body="&lt;p&gt;We have a soda machine that can fill either still water or soda. In both cases the amount is 400 ml in the beaker, but the soda weighs a little less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we use the volume X = 400 ml and the weight Y = ?? grams to determine the amount of $\ce{CO2}$ in the solution?&lt;/p&gt;&#xA;" OwnerUserId="15626" LastEditorUserId="865" LastEditDate="2015-08-26T14:55:10.467" LastActivityDate="2015-08-26T14:55:10.467" Title="How can I measure the amount of CO2 in grams in a known volume of soda based on its weight?" Tags="&lt;aqueous-solution&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35497" PostTypeId="2" ParentId="35490" CreationDate="2015-08-26T11:04:53.390" Score="1" Body="&lt;p&gt;I will assume you are asking for percent mass/volume, $\mathrm{g/ml}$, since to calculate percent mass/mass you need the density of the liquid (unless you assume it is water which is $1~\mathrm{g/ml}$ so will give the same answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have $500~\mu\mathrm{M}$ of hydrogen peroxide, which means we have $500~\mu\mathrm{mol}$ of hydrogen peroxide per 1 litre of solution. $500~\mu\mathrm{mol} = 0.0005~\mathrm{mol}$. Now you can work out the molar mass of hydrogen peroxide and multiply it by the number of moles to get the mass of hydrogen peroxide per litre of solution. This gives $0.017~\mathrm{g}$. There are $1000~\mathrm{ml}$ in 1 litre. So you have $0.017~\mathrm{g}/1000~\mathrm{ml} = 0.000017 = 0.0017\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your question about whether hydrogen peroxide is toxic to the eyes, you can search the internet for the Materials Safety Data Sheet (MSDS) of hydrogen peroxide (&lt;a href=&quot;http://www.sigmaaldrich.com/&quot; rel=&quot;nofollow&quot;&gt;Sigma-Aldrich&lt;/a&gt; is a good site to search). This contains all the hazards associated with this chemical, and when you do laboratory work you always have to look these up for each chemical before you start a reaction. The MSDS states that hydrogen peroxide can cause severe eye damage, among other hazards.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastEditorUserId="16851" LastEditDate="2015-08-27T08:55:10.533" LastActivityDate="2015-08-27T08:55:10.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35499" PostTypeId="2" ParentId="35442" CreationDate="2015-08-26T11:24:16.377" Score="2" Body="&lt;p&gt;You might have heard the term &quot;attack&quot; in statements like &quot;the nucleophile attacks the electrophile&quot;, and terms like frontside and backside attack, and what this basically means is that the nucleophile approaches the electrophile, and begins to share its electrons with the electrophile to form a bond between them (and possibly break a different bond in the molecule too). Frontside and backside attack mean the nucleophile approaches the front or the back of the molecule and forms a bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've learnt about arrow pushing, you'll know that an arrow has its tail starting at the nucleophile (the electron rich species) and it's head points to the electrophile (the electron poor species) - you can also think of &quot;attack&quot; as the nucleophile &quot;attacking&quot; the electrophile with electrons to react/form a bond, in the direction the arrow is pointing. I suppose we call it attack because it makes it sound more action-packed :-)&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2015-08-26T11:24:16.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35500" PostTypeId="1" AcceptedAnswerId="37705" CreationDate="2015-08-26T12:14:19.777" Score="4" ViewCount="448" Body="&lt;p&gt;I am having some trouble with certain steps in the mechanism below (from Mike Lu on &lt;a href=&quot;http://proteopedia.org/wiki/index.php/Image:PyruvateDehydrgenaseMech.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proteopedia&lt;/a&gt;) detailing the mechanism of action of pyruvate dehydrogenase converting pyruvate to acetyl-CoA. This reaction happens under the physiological conditions present in the human body ($\mathrm{p}H = 7.4, 37^{\circ}C$, 1 atm, various intracellular ionic concentrations: &lt;a href=&quot;http://www.ld99.com/reference/old/text/2878909-566.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ld99.com/reference/old/text/2878909-566.html&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the first step, how does the ylide of TPP form?&lt;/li&gt;&#xA;&lt;li&gt;Why is the deprotonation of the hydroxyl group to form a ketone in the 6th step favoured, especially given that the $\ce{-OH}$ group normally has a very high $\mathrm{p}K_\mathrm{a}$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7CBJA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7CBJA.gif&quot; alt=&quot;Pyruvate dehydrogenase mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19428" LastEditorUserId="19428" LastEditDate="2015-09-08T23:26:41.713" LastActivityDate="2015-09-21T22:22:33.273" Title="Steps in mechanism of pyruvate dehydrogenase" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;biochemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35501" PostTypeId="1" CreationDate="2015-08-26T12:24:44.913" Score="4" ViewCount="1039" Body="&lt;p&gt;There are many articles about how heat can trigger the release of potentially harmful carcinogens from plastic containers into the food or liquid they contain.  Specifics they fail to mention, though, are 1) the safe level of exposure and 2) the rate of BPA release per measurement of time at a constant level of heat.  While I will never drink from a water bottle that has been in my car exposed to summer heat over the course of days or weeks, I have been known to forget my water in my car as I head into work where I'll only end up retrieving it hours later when I go to lunch.  I also store bottled water in my garage which may reach temps of 90 degrees Fahrenheit or slightly higher in the summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am basically inquiring not only about the safe level of exposure to chemicals by quantity, but also the rate of chemical release over time based on heat level.  This would allow me to produce a practical ballpark idea as to whether or not the water is safe to drink.&lt;/p&gt;&#xA;" OwnerUserId="19429" LastEditorUserId="19429" LastEditDate="2015-08-26T19:03:03.347" LastActivityDate="2015-08-26T19:03:03.347" Title="What is a food-safe level of exposure of plastic containers to heat?" Tags="&lt;heat&gt;&lt;plastics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35502" PostTypeId="2" ParentId="34841" CreationDate="2015-08-26T12:25:18.317" Score="5" Body="&lt;p&gt;The &quot;soap molecules&quot; are so called amphiphiles: They've got one end that likes water (&quot;polar&quot; or &quot;lipophobic&quot;) and the other (&quot;lipophilic&quot; or &quot;fat-loving&quot;) doesn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These amphiphile molecules like to form a single-molecule thick film on water, all standing up in an orderly fashion (foot in the water), so if you put a drop of it on a water surface, it spreads very wide, until the film is monomolecular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, until something pushes back. In your case the pepper, when its particles are pushed together so they also form a tightly packed surface. Waiting longer, the soap will likely act as a &lt;code&gt;detergent&lt;/code&gt;, wrap the pepper particles in a bilayer, and drag them under the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search for, e.g., &lt;code&gt;surfactants&lt;/code&gt;, &lt;code&gt;detergents&lt;/code&gt; for some pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your last question: Some water gets dragged along with the surfactant, but no, there is no large lateral movement of water on the surface. The spreading front you see is just the soap.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-09-09T10:54:50.520" LastActivityDate="2015-09-09T10:54:50.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35503" PostTypeId="2" ParentId="35480" CreationDate="2015-08-26T12:45:52.853" Score="3" Body="&lt;p&gt;You seem to be getting it wrong. A solution with crystallized solids at the bottom is not the same as supersaturated solution, and has never been called such. It is called &quot;solution with precipitate&quot;, &quot;heterogenous mixture of solution and something&quot;, &quot;water with mud&quot;, or anything in the same vein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A supersaturated solution &lt;strong&gt;is&lt;/strong&gt; a solution, not just technically, but in all sincerity. It is homogenous, clear, transparent, and you'll never tell it from ordinary solution - I mean, until you touch it, at which point a precipitate would start to form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question: yes, there are quite a few compounds that would dissolve in water incredibly well. ZnCl&lt;sub&gt;2&lt;/sub&gt; comes to mind... Wait, it's simpler than that - common table sugar (sucrose) would do! It dissolves up to 2000 g/L, how'd you like that?&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-26T12:45:52.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35504" PostTypeId="1" CreationDate="2015-08-26T13:08:17.053" Score="2" ViewCount="13187" Body="&lt;p&gt;An atom is very small so it is probably difficult to focus exactly on a single atom. I wonder how scientists count the exact number of electrons, protons, neutrons in an atom of a certain element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you explain it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how do we know that for example, $\ce{^{19}_9F}$ has nine protons and ten neutrons and nine electrons?&lt;/p&gt;&#xA;" OwnerUserId="19430" LastEditorUserId="7448" LastEditDate="2015-08-26T13:53:00.400" LastActivityDate="2017-02-12T04:15:59.577" Title="How do scientists count the exact number of electrons, protons, neutrons in the atom?" Tags="&lt;atoms&gt;&lt;electrons&gt;&lt;protons&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ClosedDate="2018-03-08T00:00:13.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="35506" PostTypeId="2" ParentId="35501" CreationDate="2015-08-26T13:42:22.923" Score="4" Body="&lt;p&gt;You ask:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'd like to know if the information passed around the internet regarding the release of BPAs into the contents of plastic containers is grossly over-exaggerated or if there is a grain of truth to it that I should be concerned about to a degree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the United States, the Food and Drug Administration (FDA) has published  &lt;a href=&quot;http://www.fda.gov/NewsEvents/PublicHealthFocus/ucm064437.htm&quot; rel=&quot;nofollow&quot;&gt;Bisphenol A (BPA): Use in Food Contact Application&lt;/a&gt;, in which they state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FDA’s current perspective, based on its most recent safety assessment, is that BPA is safe at the current levels occurring in foods. Based on FDA’s ongoing safety review of scientific evidence, the available information continues to support the safety of BPA for the currently approved uses in food containers and packaging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the link (above), you can go to the &lt;a href=&quot;http://www.fda.gov/downloads/NewsEvents/PublicHealthFocus/UCM424266.pdf&quot; rel=&quot;nofollow&quot;&gt;2014 Updated safety assessment of Bisphenol A (BPA) for use in food contact applications&lt;/a&gt;, where they conclude that the &quot;no observed adverse effect level (NOAEL)&quot; for BPA is 5000 $\mu\mathrm{g}$ per kg of body weight per day ($\mu\mathrm{g}$/kg-bw/day) and that exposure estimates are &quot;0.2 $\mu\mathrm{g}$/kg-bw/day (mean) and 0.5&#xA;$\mu\mathrm{g}$/kg-bw/day (90$^{\rm th}$ percentile) for the adult US population aged 2 years and older.&quot; The estimated dietary intake (EDI) is given as 1.1 $\mu\mathrm{g}$/kg-bw/day for population under 2 years old and 0.5 $\mu\mathrm{g}$/kg-bw/day for population greater than 2 years old. Thus, the exposure levels are on the order of a thousand times less than the NOAEL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.efsa.europa.eu/&quot; rel=&quot;nofollow&quot;&gt;European Food Safety Authority (EFSA)&lt;/a&gt; has recently (January, 2015) revised the tolerable daily intake (TDI) of bisphenol-A (BPA) from 50 $\mu\mathrm{g}$/kg-bw/day to 4 $\mu\mathrm{g}$/kg-bw/day. See &lt;a href=&quot;http://www.efsa.europa.eu/en/press/news/150121&quot; rel=&quot;nofollow&quot;&gt;this press release&lt;/a&gt; and this &lt;a href=&quot;http://www.efsa.europa.eu/en/topics/topic/bisphenol?activeTab=5&quot; rel=&quot;nofollow&quot;&gt;FAQ&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summary from the press release is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EFSA’s comprehensive re-evaluation of bisphenol A (BPA) exposure and toxicity concludes that BPA poses no health risk to consumers of any age group (including unborn children, infants and adolescents) at current exposure levels. Exposure from the diet or from a combination of sources (diet, dust, cosmetics and thermal paper) is considerably under the safe level (the “tolerable daily intake” or TDI).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.efsa.europa.eu/en/efsajournal/pub/3978&quot; rel=&quot;nofollow&quot;&gt;scientific opinion on the risks to public health related to the presence of bisphenol A (BPA) in foodstuffs&lt;/a&gt;, used by the EFSA to bolster the down-revision of the TDI for BPA, has been summarized in a very nice form suitable for the lay public, available &lt;a href=&quot;http://www.efsa.europa.eu/sites/default/files/corporate_publications/files/factsheetbpa150121.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about heating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3210908/&quot; rel=&quot;nofollow&quot;&gt;one publication&lt;/a&gt; (&lt;em&gt;Assessment of Bisphenol A Released from Reusable Plastic, Aluminium and Stainless Steel Water Bottles&lt;/em&gt;. J. E. Cooper, E. L. Kendig, and S. M. Belcher. Chemosphere 85(6), 943-947 (2011)), the authors state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At room temperature the concentration of BPA migrating from polycarbonate bottles ranged from 0.2–0.3 mg/L. Under identical conditions BPA migration from aluminium bottles lined with epoxy-based resins was variable depending on manufacturer ranging from 0.08 to 1.9 mg/L. Boiling water significantly increased migration of BPA from the epoxy lined bottles. No detectable BPA contamination was observed in water stored in bottles made from Tritan™ copolyester plastic, uncoated stainless steel, or aluminium lined with EcoCare™. The results from this study demonstrate that when used according to manufactures’ recommendations reusable water bottles constructed from “BPA-free” alternative materials are suitable for consumption of beverages free of BPA contamination.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and they conclude&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As shown previously by our results and those of many others, BPA migrates into water stored in polycarbonate plastic, especially when heated to high temperatures. Relatively high levels of BPA are found to be released into drinking water from reusable aluminium bottles with an epoxy-based liner. The amount of BPA released from the epoxy coating was greatly increased by heating. No detectable BPA was observed in water to migrate from unlined stainless steel, EcoCare™ copolyester lined aluminium, or copolyester Tritan™ plastic water bottles. To avoid exposures to BPA unlined stainless steel, copolyester lined aluminium or copolyester plastic drinking bottles should be used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, should you be worried? In my opinion, that's largely a function of whether or not you use plastic that contains BPA. A few years ago, it might have been difficult to find BPA-free plastics, but there is no shortage of BPA-free items available now. It's easy enough these days to remove any doubt from the equation by simply using containers that don't contain BPA.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-26T13:57:01.853" LastActivityDate="2015-08-26T13:57:01.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35508" PostTypeId="2" ParentId="35493" CreationDate="2015-08-26T14:46:33.347" Score="5" Body="&lt;p&gt;Chirality is lost (see ron's answer/comment) when the ring get so large that the most stable conformation becomes planar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise: Yes they are, unless the ring is so large that it can circumvent the double bond without a noticeable energetic inversion barrier. Strictly speaking, the cycloalkene then is chiral still, only it would racemise instantly. Single crystals would still be optically active.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-08-27T00:55:25.850" LastActivityDate="2015-08-27T00:55:25.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35509" PostTypeId="2" ParentId="35462" CreationDate="2015-08-26T14:51:27.477" Score="31" Body="&lt;p&gt;Harold Urey and George Murphy used spectroscopy to identify deuterium late in 1931, announcing it at the 1931 Christmas meeting of the American Physical Society. Picking up out of 'From Nuclear Transmutation to Nuclear Fission, 1932-1939&quot; by Per F. Dahl:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If anything, the naming of the new isotope proved more problematic than its isolation.  At a special session on heavy hydrogen at the general June meeting in 1933 of the APS in Chicago, organized in conjunction with the Century of Progress Exposition, the ensuing discussion on its naming 'threatened to become acrimonious,' according to Francis Aston of the Cavendish Laboratory - the great authority on atomic weight measurements and a guest speaker at the discussion.  The argumentation had to do with whether to retain the name 'hydrogen' for the isotope, as Niels Bohr preferred; after all, it was not a new element and had the atomic number 1.  Both Gilbert Lewis and Ernest Lawrence opted for 'dygen' for the H$^{2}$ isotope and 'dyon' for its nucleus, wheras Rutherford preferred 'diplogen' and 'diplon' instead.  In the end, Urey had the last word, as he was entitled to, settling on 'deuterium' for the isotope and 'deuteron' for its nucleus.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the heavyweights in the field wanting a separate name for the isotope, even Neils Bohr could not hold back the tide.  Again, though, you have to remember that this was in the very early days of nuclear physics. While the existence of the nucleus dated back to Rutherford's experiments pre-WWI, they were still very unclear just what the nucleus was constructed out of.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was all occurring just as Cockroft and Walton were using their new ion accelerator to perform the first human-induced nuclear fission ($^{7}$Li + p -&gt; 2 $^{4}$He), published in the Proceedings of the Royal Society on July 1st, 1932. Alpha, beta, and gamma particles had been identified since the late 1800s, and natural sources were used for nuclear physics experiments. It was readily recognized that a source of energetic protons would be highly desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, remember that 1932 was the year that Chadwick discovered the neutron as a separate entity. Thus, also from Dahl:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The discovery of the neutron would soon revise Urey's view of his deuteron as consisting of two protons and an electron.  However, as late as October 1932, it was still not obvious among physicists that the deuteron consists of 'one proton and one Chadwick neutron' instead of two protons and one electron.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Against this rapid-fire series of experiments on nuclear physics, it really isn't unusual that deuterium got a separate name - it was still not universally appreciated that it wasn't something weird. Than, once it was accepted, it was too useful to readily distinguish experiments with protons vs deuterons vs tritons to revert their names back to normal isotope nomenclature.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="5677" LastEditDate="2015-08-26T22:49:19.120" LastActivityDate="2015-08-26T22:49:19.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35511" PostTypeId="2" ParentId="35486" CreationDate="2015-08-26T14:58:42.403" Score="21" Body="&lt;p&gt;Achiral cyclic compounds like 1,2,3-trichlorocyclopropane may contain &lt;strong&gt;pseudoasymmetric centers&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Pseudoasymmetric centers have distinguishable ligands (“a”, “b”, “c”, “d”), two of which are nonsuperposable mirror images of each other (enantiomorphic). The lower case stereodescriptors “&lt;em&gt;r&lt;/em&gt;” and “&lt;em&gt;s&lt;/em&gt;” are used to designate the absolute configuration of pseudoasymmetric centers, for example (2⁠&lt;em&gt;R&lt;/em&gt;,3⁠&lt;em&gt;r&lt;/em&gt;,4⁠&lt;em&gt;S&lt;/em&gt;)‑2,3,4‑trichloropentane:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TSsCa.png&quot; width=&quot;198&quot; alt=&quot;(2R,3r,4S)-2,3,4-trichloropentane&quot;&gt;&#xA;&lt;p&gt;Note that the stereodescriptors “&lt;em&gt;r&lt;/em&gt;” and “&lt;em&gt;s&lt;/em&gt;” describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example “&lt;em&gt;r&lt;/em&gt;” remains “&lt;em&gt;r&lt;/em&gt;”, and “&lt;em&gt;s&lt;/em&gt;” remains “&lt;em&gt;s&lt;/em&gt;”), but are reversed by the exchange of any two ligands (“&lt;em&gt;r&lt;/em&gt;” becomes “&lt;em&gt;s&lt;/em&gt;”, and “&lt;em&gt;s&lt;/em&gt;” becomes “&lt;em&gt;r&lt;/em&gt;”).&lt;/p&gt;&#xA;&lt;p&gt;The configurations (“&lt;em&gt;r&lt;/em&gt;” or “&lt;em&gt;s&lt;/em&gt;”) at pseudoasymmetric stereogenic units are determined by application of the Cahn-Ingold-Prelog (CIP) priority system and the sequence rules. In particular, note that an atom or group with descriptor “&lt;em&gt;R&lt;/em&gt;” has priority over its enantiomorph “&lt;em&gt;S&lt;/em&gt;”. In order to establish the order of precedence of ligands in a stereogenic unit, the atoms of the stereogenic unit are rearranged in a hierarchical diagram, called a “digraph”. Descriptors specified in digraphs may correspond to the final descriptors or to temporary auxiliary descriptors used only for ranking ligands and never appearing as final descriptors.&lt;/p&gt;&#xA;&lt;p&gt;Similar to the above-mentioned case of (2⁠&lt;em&gt;R&lt;/em&gt;,3⁠&lt;em&gt;r&lt;/em&gt;,4⁠&lt;em&gt;S&lt;/em&gt;)‑2,3,4‑trichloropentane, the structure of &lt;strong&gt;(1⁠&lt;em&gt;r&lt;/em&gt;)‑1,2,3‑trichlorocyclopropane&lt;/strong&gt; contains one pseudoasymmetric center, which has two ligands that are nonsuperposable mirror images of each other:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fxjXg.png&quot; width=&quot;189&quot; alt=&quot;(1r)-1,2,3-trichlorocyclopropane&quot;&gt;&#xA;&lt;p&gt;The other isomer of 1,2,3‑trichlorocyclopropane is &lt;strong&gt;(1⁠&lt;em&gt;s&lt;/em&gt;,2⁠&lt;em&gt;s&lt;/em&gt;,3&lt;em&gt;s&lt;/em&gt;)‑1,2,3‑trichlorocyclopropane&lt;/strong&gt;, which contains three pseudoasymmetric centers. Each one has two ligands that are nonsuperposable mirror images of each other:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Q5Sse.png&quot; width=&quot;225&quot; alt=&quot;(1s,2s,3s)-1,2,3-trichlorocyclopropane&quot;&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It may not be obvious how to choose the stereodescriptors ‘&lt;em&gt;r&lt;/em&gt;’ or ‘&lt;em&gt;s&lt;/em&gt;’ for 1,2,3-trisubstituted cyclopropanes. The ranking is a complex process and requires the use of &lt;strong&gt;auxiliary descriptors.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Generally, the stereodescriptors ‘&lt;em&gt;r&lt;/em&gt;’ and ‘&lt;em&gt;s&lt;/em&gt;’ are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors ‘&lt;em&gt;R&lt;/em&gt;’ and ‘&lt;em&gt;S&lt;/em&gt;’. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.&lt;/p&gt;&#xA;&lt;p&gt;The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 of the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; and which has been recommended by Prelog and Helmchen.&lt;/p&gt;&#xA;&lt;p&gt;In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1​&lt;em&gt;r&lt;/em&gt;)‑1,2,3‑trichlorocyclopropane:&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/JDFVD.png&quot; width=&quot;440&quot; alt=&quot;Transformation of (1r)‑1,2,3‑trichlorocyclopropane into an acyclic digraph&quot;&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ce{(C)}$&lt;/span&gt; is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.&lt;/p&gt;&#xA;&lt;p&gt;As the digraph is acyclic, it is now possible to specify the stereodescriptor ‘&lt;em&gt;R&lt;/em&gt;’ or ‘&lt;em&gt;S&lt;/em&gt;’ for the stereogenic unit at the atoms numbered ‘2’ and ‘3’ in each branch. These are auxiliary descriptors that are different in each branch.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GPhRm.png&quot; width=&quot;292&quot; alt=&quot;auxiliary descriptors&quot;&gt;&#xA;&lt;p&gt;Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that ‘&lt;em&gt;R&lt;/em&gt;’ has priority over ‘&lt;em&gt;S&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-92.1.3.5&lt;/strong&gt; Sequence Rule &lt;strong&gt;5&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An atom or group with descriptor ‘&lt;em&gt;R&lt;/em&gt;’, ‘&lt;em&gt;M&lt;/em&gt;’, and ‘&lt;em&gt;seqCis&lt;/em&gt;’ has priority over its enantiomorph ‘&lt;em&gt;S&lt;/em&gt;’, ‘&lt;em&gt;P&lt;/em&gt;’ or ‘&lt;em&gt;seq Trans&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1​&lt;em&gt;r&lt;/em&gt;)‑1,2,3‑trichlorocyclopropane can be determined to be ‘&lt;em&gt;r&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IUdvz.png&quot; width=&quot;292&quot; alt=&quot;Configuration of the pseudoasymmetric stereogenic unit&quot;&gt;&#xA;&lt;p&gt;In the same way, the configuration of each pseudoasymmetric stereogenic unit of (1⁠&lt;em&gt;s&lt;/em&gt;,2⁠&lt;em&gt;s&lt;/em&gt;,3⁠&lt;em&gt;s&lt;/em&gt;)‑1,2,3‑trichlorocyclopropane can be determined as ‘&lt;em&gt;s&lt;/em&gt;’.&lt;/p&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/k8AtR.png&quot; width=&quot;440&quot; alt=&quot;Transformation of (1s,2s,3s)-1,2,3-trichlorocyclopropane into an acyclic digraph and determination of the configuration of the considered pseudoasymmetric stereogenic unit&quot;&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2020-07-12T08:06:16.097" LastActivityDate="2020-07-12T08:06:16.097" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35512" PostTypeId="2" ParentId="35504" CreationDate="2015-08-26T15:03:08.103" Score="6" Body="&lt;p&gt;For charge neutrality, the number of electrons has to equal the number of protons.  So, the question really is how to know how many protons and how many electrons. This is actually a very interesting question dating back to the early days (and even before) of nuclear physics and chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one can do it fairly easily with an ion accelerator and a magnet. With the right set up, one can get a beam of mono-energetic ions out of the accelerator, send them through a stripper to remove electrons, and then send them through a magnet to separate the components out by velocity (which depends on the ion mass through the energy). From the magnet deflections one can get a charge-to-mass ratios, and get back to number of protons (charge on a fully stripped ion) and number of neutrons (from number of protons and the mass). This is the heart of mass spectrometry (although one can use time-of-flight techniques rather than, or in addition to, magnetic separation).&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-08-26T15:03:08.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35513" PostTypeId="1" CreationDate="2015-08-26T15:10:22.617" Score="4" ViewCount="2087" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In which of the following processes is energy transferred into the substance by work (&lt;span class=&quot;math-container&quot;&gt;$w &amp;gt; 0$&lt;/span&gt;)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) Expansion of a gas against the surroundings&lt;br&gt;&#xA;  b) Expansion of a gas into a vacuum&lt;br&gt;&#xA;  c) Vaporization of one mole of water at 50˚C in an open container&lt;br&gt;&#xA;  d) Melting of 100 g of ice on a laboratory benchtop&lt;br&gt;&#xA;  e) Combustion of methane in a constant volume container.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is given as (d), but I needed to find a physicist to explain it to me.  This problem can being thought of entirely in terms of thermodynamic work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we know that &#xA;&lt;span class=&quot;math-container&quot;&gt;$$w = P \Delta V.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we see that no work is done in (b) and (e).  Also, since the gas expanded against its surroundings, it did work on its surroundings, which isn't what the question asked, thus (a) is not the correct answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, although (c) appears to be a heat transfer, and it is, there is a little more going on.  The heat that was given to the water is converted into work as the water vaporizes, this causes a volume expansion (the gas rising).  Intuitively, one can think of the surroundings as a tube with a movable piston.  When vaporizing, the water vapor will push against the piston, increasing the pressure and thus the piston will move, causing the water to do work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last choice is (d), the correct answer, but there is also an intuitive way to think about this.  As the ice is heated, it melts (this process is also, as we correctly deduced, a heat transfer), however, this heat also causes the surrounding atmosphere to do work on the ice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, think of a tube with a movable piston with a cube of ice that is slowly melting in the presence of normal air.  As the ice melts, the volume will decrease, and thus the nitrogen and oxygen molecules in the air have more space to move about, decreasing the pressure, which causes the piston to move down, thus doing work upon the ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this makes sense to everyone who has been either watching or contributing to this discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I don't know what the authors of this question had in mind. This came up as I was working through an old Chemistry GRE exam. It wouldn't have occurred to me to consider this problem through mechanical work. As a side note, only 18% of the people who answered this question got it right, which essentially means that people just guessed the answer. &lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="4945" LastEditDate="2020-06-03T19:00:35.840" LastActivityDate="2020-06-03T19:00:35.840" Title="Work Done Upon Molecules" Tags="&lt;thermodynamics&gt;&lt;pressure&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35514" PostTypeId="1" AcceptedAnswerId="35521" CreationDate="2015-08-26T15:28:45.593" Score="0" ViewCount="3575" Body="&lt;p&gt;I know the Maillard reaction to cook at 140–160 °C. I know the low temperature cooking reaction to cook at 70 °C. But which is the reaction that allow us cooking meat or fish with vinegar or lemon juice?&lt;/p&gt;&#xA;" OwnerUserId="350" LastEditorDisplayName="user7951" LastEditDate="2017-01-15T11:49:54.223" LastActivityDate="2017-01-15T11:49:54.223" Title="Cooking with vinegar and lemon juice (Carpaccio)" Tags="&lt;food-chemistry&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35515" PostTypeId="2" ParentId="35493" CreationDate="2015-08-26T15:36:44.870" Score="15" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can draw conformations of n-butane with the carbon-carbon bonds oriented in certain directions and the methyl hydrogens pointing in certain directions that are chiral.  However, since rotation about single bonds is typically fast at room temperature, these chiral conformers of n-butane would only be resolvable at extremely low temperatures.  Therefore we do not typically consider n-butane to be a chiral molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Take-away lesson:&lt;/strong&gt; If a chiral conformation can readily interconvert with an achiral conformation, then the molecule cannot be resolved into enantiomers and will be optically inactive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, there are molecules where rotation about single bonds is restricted (for example due to steric interference) and the rotational barrier is high enough that the molecules can be resolved into their respective enantiomers at room temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecules where the rotational barrier about a single bond is high enough to allow different conformations to be isolated at room temperature are called &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atropisomer&quot; rel=&quot;noreferrer&quot;&gt;atropisomers&lt;/a&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biphenyl pictured below (in a chiral conformation) is an atropisomer due to restricted rotation about the central single bond connecting the two benzene rings.  The steric interference that would develop between the 4 &lt;em&gt;ortho&lt;/em&gt; groups if the molecule were to approach a planar (achiral) conformation prevents the chiral conformer from equilibrating with the achiral (planar) conformer.  Hence this biphenyl has been resolved into its two enantiomers at ambient conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2rMAX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2rMAX.png&quot; alt=&quot;enantiomers of a substituted biphenyl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Trans&lt;/em&gt;-cyclooctene&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Trans&lt;/em&gt;-cyclooctene is also capable of being resolved into its two enantiomers due to restricted rotation about single bonds - it also belongs to the class of atropisomers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below &lt;em&gt;trans&lt;/em&gt;-cyclooctene conformation (a) is planar, therefore this conformation is achiral. The two conformers pictured in (b) are chiral - these conformations are enantiomers of each other. &lt;em&gt;Trans&lt;/em&gt;-cyclooctene will be achiral if the two enantiomers pictured in (b) can readily interconvert through achiral (planar) conformer (a).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Build a model or draw &lt;em&gt;trans&lt;/em&gt;-cyclooctene and notice the two hydrogens on the &lt;em&gt;trans&lt;/em&gt; double bond; one of them is &quot;inside&quot; the ring in the planar conformation and would cause severe steric problems as the methylene chain tries to flip through the planar conformation to the other face of the double bond as pictured below. This steric interference makes it difficult for the molecule to achieve the planar conformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DwWp9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DwWp9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Image source: John D. Roberts, Marjorie C. Caserio, Basic Principles of Organic Chemistry. Excerpt available at &lt;a href=&quot;http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/&quot; rel=&quot;noreferrer&quot;&gt;http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this destabilizing transannular interaction, &lt;em&gt;trans&lt;/em&gt;-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a twisted conformation as pictured above for &lt;em&gt;trans&lt;/em&gt;-cyclooctene.  In your model of &lt;em&gt;trans&lt;/em&gt;-cyclooctene, notice how you have to pull and stretch the methylene chain in order to convert &lt;em&gt;trans&lt;/em&gt;-cyclooctene into its mirror image.  This rotational movement is highly restricted due to the internal hydrogen.  It is this barrier to rotation about singles bonds that allows &lt;em&gt;trans&lt;/em&gt;-cyclooctene to be resolved into its two mirror image forms at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barrier to racemize &lt;em&gt;trans&lt;/em&gt;-cyclooctene (&lt;em&gt;e.g&lt;/em&gt;. pass through the planar conformation (a)) is 36 kcal/mole.  As you might imagine, as we make the chain longer it should get easier and easier to loop the methylene chain around the internal (protruding) hydrogen on the double bond.  Indeed, the barrier to interconvert the two enantiomers of &lt;em&gt;trans&lt;/em&gt;-cyclononene is only 20 kcal/mole. Optically resolved &lt;em&gt;trans&lt;/em&gt;-cyclononene must be kept below 0°C to prevent racemization.  The barrier to racemization in trans-cyclodecene is even lower at 10 kcal/mole.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First off, these rings can have chiral conformations, just like butane.  I think you are really asking, &quot;Are all higher &lt;em&gt;trans&lt;/em&gt;-cycloalkenes optically active as &lt;em&gt;trans&lt;/em&gt;-cyclooctene is?&quot;.  That is, do these larger rings also have difficulty reaching the planar, achiral geometry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your original question is - it depends.  It depends on the temperature you specify.  At room temperature, &lt;em&gt;trans&lt;/em&gt;-cyclooctene is the only &lt;em&gt;trans&lt;/em&gt;-cycloalkene where the two enantiomeric conformations are stable enough (&lt;em&gt;e.g&lt;/em&gt;. the barrier to reach the planar conformation is high enough) to be isolated.  If we lower the temperature to 0°C, then we can isolate the enatiomers of &lt;em&gt;trans&lt;/em&gt;-cyclononene.  If we go low enough in temperature, we could isolate enantiomers of n-butane.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-09-04T15:07:49.550" LastActivityDate="2015-09-04T15:07:49.550" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="35516" PostTypeId="2" ParentId="35513" CreationDate="2015-08-26T15:41:36.733" Score="3" Body="&lt;p&gt;In (c+d) you clearly need to transfer &lt;em&gt;energy&lt;/em&gt; into the water. But that'd be a heat flux, normally abbreviated $Q$, and not work, $W$, as was pointed out in the disk below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In d) the surroundings of the ice block must be at least a tiny bit above $0\:\mathrm{^\circ C}$, and in c) the container gets colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case d) the surrounding atmosphere does a tiny bit of work on the melting ice block, because its volume shrinks slightly (water has a larger density). It sort of gets &quot;compressed&quot;. Not sure if the teacher thought about &lt;em&gt;that&lt;/em&gt; when he set the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside: Ice won't melt unless you transfer a substancial amount of energy into it. Melting ice at zero degrees Celsius takes up as much heat as heating the molten water to around 75 degrees Celsius afterwards. $300\:\mathrm{kJ/kg}$ heat of melting divided by $4\:\mathrm{kJ/kg\,K}$ heat capacity $\approx 75\:\mathrm{K})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The exact numbers are a bit different, also the heat capacity is not constant.)&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-09-06T09:27:10.877" LastActivityDate="2015-09-06T09:27:10.877" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="35517" PostTypeId="2" ParentId="35513" CreationDate="2015-08-26T16:02:54.470" Score="3" Body="&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; If the question is asking about &lt;em&gt;thermodynamic work&lt;/em&gt;, then none of the answers is strictly correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you note, only &lt;strong&gt;(a)&lt;/strong&gt; comes close to being correct: $Pv$ work is done by the gas on the surroundings as it expands.  In this case, however, energy is transferred &lt;strong&gt;&lt;em&gt;out&lt;/em&gt;&lt;/strong&gt; of the substance as it expands against its surroundings, so it is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; a correct answer per the question wording.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, you're right that in &lt;strong&gt;(b)&lt;/strong&gt;, since no force is required to push the gas against vacuum (there is no resisting pressure afforded by the vacuum at the gas boundary), the $Pv$ work is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also concur that &lt;strong&gt;(c)&lt;/strong&gt; and &lt;strong&gt;(d)&lt;/strong&gt; are examples of heat transfer, and thus are not work interactions.  There's no electricity or shaft work involved, and no system boundary against which volume expansion/compression (i.e., $Pv$ work)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agree also, the fixed volume of &lt;strong&gt;(e)&lt;/strong&gt; means that no work interaction is possible.&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-08-26T18:12:15.960" LastActivityDate="2015-08-26T18:12:15.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35520" PostTypeId="1" CreationDate="2015-08-26T16:43:45.813" Score="0" ViewCount="1466" Body="&lt;p&gt;I'm inclined to think that P has a $dsp$ hybridization in that compound, because it makes 5 bonds and 2 of them are $\pi$ bonds - with oxygen. However, I found in some websites that it may have $sp^2$ hybridization, although I have no idea how could it be possible - since it has to make two $\pi$ and 3 sigma bonds.&lt;/p&gt;&#xA;" OwnerUserId="19436" LastEditorUserId="7448" LastEditDate="2015-08-26T16:57:52.157" LastActivityDate="2015-08-26T16:57:52.157" Title="What type of hybridization does phosphorus have in P2O5?" Tags="&lt;orbitals&gt;&lt;hybridization&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35521" PostTypeId="2" ParentId="35514" CreationDate="2015-08-26T18:23:54.287" Score="1" Body="&lt;p&gt;Apparently this question has been covered on the Cooking SE site - in &lt;a href=&quot;https://cooking.stackexchange.com/questions/23238/how-does-lemon-juice-cook-meat&quot;&gt;this post&lt;/a&gt;, it is reported that:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Oxford English Dictionary defines to cook as:&lt;/p&gt;&#xA;&lt;p&gt;To prepare or make ready (food); to make fit for eating by due application of heat, as by boiling, baking, roasting, broiling, etc.&lt;/p&gt;&#xA;&lt;p&gt;However, from a chemical point of view, what happens when you are cooking meat is that you are using heat, amongst other things, to denature the proteins of the meat. Denaturation can also be achieved by other means, such as marinating the meat in an acid, such as citric acid (contain in lemon juice) or acetic acid (contained in vinegar).&lt;/p&gt;&#xA;&lt;p&gt;You can also use other agents to induce denaturation, but they are not always fit for eating.&lt;/p&gt;&#xA;&lt;p&gt;However, it is important to understand that from a purely cooking perspective the two things are not the same: &amp;quot;cooking&amp;quot; with lemon (I guess it is fine to use the verb to cook as long as you put it in quotes), gives a completely different result than cooking with heat. This is because other chemical reactions happen during application of heat and because, obviously, you do not end up with an hot product!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;which indicates that the application of acidic media to meat accomplishes the act of denaturing proteins therein in a way mechanistically similar to what cooking with applied heat does.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.kitchendaily.com/read/will-lemon-juice-cook-meat-and-seafood&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kitchen Daily&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Acids like lemon juice, lime juice and vinegar break down raw meat, allowing the marinade's oil and spices to deeply penetrate and infuse the meat with flavor and moisture. They dig out little pathways in the meat and allow the marinade to seep in. The scientific term for this breakdown is &amp;quot;denaturation,&amp;quot; which is the process in which protein loses its structure by application of a strong acid, base, salt, alcohol or heat. Basically, the scientific process that cooks meat with heat in the oven is the same one that cooks it with acid. The results may taste a little different -- fish cooked in the oven is flaky, whereas fish cooked by lemon juice (a.k.a. ceviche) has a more raw-tasting consistency -- but the process is the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-08-26T18:23:54.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35522" PostTypeId="1" AcceptedAnswerId="35526" CreationDate="2015-08-26T18:53:15.043" Score="4" ViewCount="1687" Body="&lt;p&gt;I would like to follow the steps in this video: &lt;a href=&quot;https://www.youtube.com/watch?v=hyPysthIRss&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=hyPysthIRss&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I mixed the &lt;code&gt;Potassium nitrate KNO3&lt;/code&gt; and the &lt;code&gt;sugar&lt;/code&gt;, I start to cook it in a pot, and stir it very often. After 25-30 min, the KNO3 and the sugar will melt together, and become a brown mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Can this mixture explode during the heating somehow? Or it can just burn, not explode? How to prevent it, and do it the most secure way?&lt;/p&gt;&#xA;" OwnerUserId="19438" LastActivityDate="2020-10-14T15:16:18.087" Title="Can rocket candy explode during the preparation?" Tags="&lt;home-experiment&gt;&lt;fuel&gt;&lt;explosives&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35524" PostTypeId="1" CreationDate="2015-08-26T19:34:07.627" Score="1" ViewCount="690" Body="&lt;p&gt;I've made a test and i've faced a problem with a part of a question which says: &quot;What is the conversion of energy occuring in electroplating? &quot; I answered that it is conversion from electric energy to chemical energy, since in electroplating, the anode is reduced to X^n+ and n+, where n+ ions are transferred through DC power supply into the cathode and then combines with X^n+ to give again X and plates the cathode, so the electric energy(transfer of electrons through the DC power supply) is converted into chemical one(change  X^n+ to solid). I want to know if my answer is correct or not and why? Helping is appreciated &lt;/p&gt;&#xA;" OwnerUserId="19133" LastEditorUserId="19133" LastEditDate="2015-08-30T07:34:45.173" LastActivityDate="2015-08-30T07:34:45.173" Title="The conversion of energy during electroplating" Tags="&lt;electrochemistry&gt;&lt;energy&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35525" PostTypeId="2" ParentId="35349" CreationDate="2015-08-26T19:35:17.050" Score="1" Body="&lt;p&gt;Intermolecular forces (IMFs) are the key here. IMFs are directly related to boiling point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $\ce{HF}$ and $\ce{H2O}$ have hydrogen bonds ($\ce{H}$ attached to $\ce{N}$, $\ce{O}$ or $\ce{F}$). But &lt;strong&gt;$\ce{H2O}$ has two hydrogen bonds whereas $\ce{HF}$ only has one&lt;/strong&gt;. Thus, $\ce{H2O}$ should have a much higher boiling point.&lt;/p&gt;&#xA;" OwnerUserId="7957" LastEditorUserId="43942" LastEditDate="2017-05-12T11:21:38.593" LastActivityDate="2017-05-12T11:21:38.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35526" PostTypeId="2" ParentId="35522" CreationDate="2015-08-26T19:35:43.443" Score="4" Body="&lt;p&gt;See James Yawn's website &lt;a href=&quot;http://www.jamesyawn.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;${\it Recrystalized\; Rocketry}$&lt;/span&gt;&lt;/a&gt;, which is all about solid propellant rocketry of the type you're interested in. It's really an exceptionally comprehensive site.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Can this mixture explode during the heating somehow?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It appears that it's more likely to catch fire, and do so rather vigorously, as in cases where it's exposed to a spark or an open flame. Because the system (pan and mixture of &lt;span class=&quot;math-container&quot;&gt;$\ce{KNO3}$&lt;/span&gt; and sugar) would be open to the atmosphere (and not confined to a fixed volume), there's more of a risk of fire than of an explosion. As reported by &lt;a href=&quot;http://www.jamesyawn.net/skillet/large/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Yawn in the subsection on &amp;quot;Skillet Rcandy&amp;quot;&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This stuff looks innocent enough in the pan.  But if you have ever seen a quantity this size burn you will have more respect for it.  This batch could send a tongue of flame 20 feet in the air and engulf the person stirring it, if only briefly.  Burns to the body would likely be second degree, and I am willing to tolerate that level of risk.  But eyeballs don't recover from this kind of trauma easily.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.jamesyawn.net/rcandy/safety.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this safety page from James Yawn's site&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Or it can just burn, not explode?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, it can just burn: that's essentially what's going on in the rocket engine. The exhaust from the burn is channeled in one direction (&lt;span class=&quot;math-container&quot;&gt;${\it via}$&lt;/span&gt; the hollow core in the engine itself), resulting in thrust in the other direction.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How to prevent it, and do it the most secure way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.jamesyawn.net/rcandy/safety.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Yawn's site&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Basic Safety Precautions&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;At present, I do NOT recommend making this propellant in a household kitchen.&lt;/p&gt;&#xA;&lt;p&gt;Make small batches only!  No more than one standard recipe, which yields about 160 grams of propellant.&lt;/p&gt;&#xA;&lt;p&gt;Wear protective gear. You should have a full-face shield, long sleeved heavy cotton shirt, heavy apron and heat-resistant gloves.  These must be worn at appropriate points in the process.&lt;/p&gt;&#xA;&lt;p&gt;Do not allow children or rambunctious pets in the room where you are working.&lt;/p&gt;&#xA;&lt;p&gt;Do not allow distracting, inattentive, inebriated or otherwise incompetent adults in the room, especially if they smoke.&lt;/p&gt;&#xA;&lt;p&gt;Be vigilant of possible sources of sparks, flame, or high temperature that could ignite propellant on contact.&lt;/p&gt;&#xA;&lt;p&gt;If you smoke, place your cigarettes and lighter in another room so you will not light up unconsciously.&lt;/p&gt;&#xA;&lt;p&gt;Clear your work-area of any flammable materials.&lt;/p&gt;&#xA;&lt;p&gt;Have a good fire extinguisher available. Also a 5-gallon bucket of water nearby and a garden hose outside.&lt;/p&gt;&#xA;&lt;p&gt;Make sure there is good ventilation, and a quick escape route to fresh air.&lt;/p&gt;&#xA;&lt;p&gt;Avoid stupid rocket tricks.&lt;/p&gt;&#xA;&lt;p&gt;Avoid excessive exposure to Potassium Nitrate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All of the above points are elaborated upon at the linked page.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-26T20:57:59.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35527" PostTypeId="1" AcceptedAnswerId="35537" CreationDate="2015-08-26T21:01:25.127" Score="4" ViewCount="1173" Body="&lt;p&gt;What makes something glow in the dark? The only explanation that I can come up with for what makes things glow in the dark is that there is probably a chemical reaction slowly releasing the light that was previously absorbed. Am I correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this phenomenon work?&lt;/p&gt;&#xA;" OwnerUserId="19442" LastEditorUserId="402" LastEditDate="2019-05-19T06:06:39.860" LastActivityDate="2019-05-23T14:31:32.060" Title="How do things glow in the dark?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;photochemistry&gt;" AnswerCount="4" CommentCount="4" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35528" PostTypeId="2" ParentId="25043" CreationDate="2015-08-26T21:39:54.013" Score="1" Body="&lt;p&gt;I know it is a bit necromancy to post this so late after the question, however you might have a look at the OpenPHACTS system, that brings a lot of medchem data together.&#xA;&lt;a href=&quot;https://www.openphacts.org/&quot; rel=&quot;nofollow&quot;&gt;https://www.openphacts.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8135" LastActivityDate="2015-08-26T21:39:54.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35529" PostTypeId="1" CreationDate="2015-08-26T21:42:28.530" Score="5" ViewCount="383" Body="&lt;p&gt;I have tried google but without success so here the question: does anybody know a tool that can covert WFN-file to Molden or to Molpro format (including orbital info not the coordinates only).&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="16088" LastEditDate="2016-09-26T16:15:41.380" LastActivityDate="2017-07-13T19:20:00.263" Title="Program to convert WFN-file to Molden" Tags="&lt;computational-chemistry&gt;&lt;software&gt;" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35530" PostTypeId="2" ParentId="35513" CreationDate="2015-08-26T22:33:19.930" Score="5" Body="&lt;p&gt;This is a bad question.  But the answer is probably supposed to be &lt;strong&gt;(d)&lt;/strong&gt; because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Water is more dense than ice.&lt;/li&gt;&#xA;&lt;li&gt;Thus when ice melts, the total volume of &quot;substance&quot; goes down.&lt;/li&gt;&#xA;&lt;li&gt;This means that atmospheric pressure did PV work on the melting ice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.  &lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-08-26T22:33:19.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35531" PostTypeId="1" CreationDate="2015-08-26T23:17:22.633" Score="1" ViewCount="107" Body="&lt;p&gt;When a molecule of water ($\ce{H2O}$) and a molecule of hydrochloric acid ($\ce{HCl}$) react together, would they become a &lt;strong&gt;molecule made of molecules&lt;/strong&gt;? What's the name for such thing? How would it be represented - just as $\ce{H2OHCl}$?&lt;/p&gt;&#xA;" OwnerUserId="19443" LastEditorUserId="15410" LastEditDate="2015-10-26T10:04:34.103" LastActivityDate="2015-10-26T14:32:45.233" Title="Molecules made of molecules?" Tags="&lt;nomenclature&gt;&lt;molecules&gt;&lt;intermolecular-forces&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35532" PostTypeId="2" ParentId="35531" CreationDate="2015-08-27T00:36:37.280" Score="1" Body="&lt;p&gt;The key to your question lies in the word &quot;bound&quot;. Without mentioning the state of your species we can say in general, species which are joined by covalent bonds are single molecules, there are countless examples of such reactions in organic chemistry where two small molecules condense into a single molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Species with intramolecular forces (hydrogen bonds, van der waals etc) can be called bound or perhaps &quot;clusters&quot;, which is quite a broad term. Real bonding however is not so easily defined as the molecular orbitals may  generate a wide spectrum of charge distributions etc. Quantum mechanics gives a much clearer description of the bonding of a particular molecule or extended system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminology you write your $\ce{H2OHCl}$ is similar to how we write water of crystallisation in a solid. If this is where your idea came from, it should be remarked that this is a distinct concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude; species that are covalently bound (that could have strong polarisation even to bordering ionic bonding) are molecules. Intermolecular forces are precisely that, between distinct molecules. We should remark that we can also have extended solids with covalent bonding (e.g. group 14). In addition some regard metals as molecules too, molecules whose molecular orbitals are distributed over the entire structure etc (research tight binding model for more information). But I feel that wasn't the focus of your question? &lt;/p&gt;&#xA;" OwnerUserId="7913" LastEditorUserId="7475" LastEditDate="2015-10-26T14:32:45.233" LastActivityDate="2015-10-26T14:32:45.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35533" PostTypeId="1" CreationDate="2015-08-27T01:01:19.173" Score="4" ViewCount="625" Body="&lt;p&gt;Is there any fuels without hydrogen that will combust. My friend believes there might be and do to my experience I have never heard of one. Please explain and if u can show a balanced equation of this combustion.&lt;/p&gt;&#xA;" OwnerUserId="19445" LastActivityDate="2015-08-30T22:19:38.507" Title="Combustion without hydrogen" Tags="&lt;combustion&gt;" AnswerCount="1" CommentCount="5" ClosedDate="2015-09-01T09:02:21.830" ContentLicense="CC BY-SA 3.0" />
  <row Id="35534" PostTypeId="1" CreationDate="2015-08-27T01:20:05.663" Score="1" ViewCount="9443" Body="&lt;p&gt;My dad told me to turn off the lights when I'm not using them because it will keep the house cooler. I don't think he is correct that a light can heat up the room. How much of a difference in the temperature would it make? Assume a 60 Watt light bulb running for one hour in a room with approx. 3000ft^3/85m^3 of air. Also assume the air is dry and in constant thermal equilibrium at 1 atm. I think the specific heat capacity of air is about 1 J/g*K, but I am not confident.&lt;/p&gt;&#xA;" OwnerUserId="19447" LastActivityDate="2015-09-27T21:21:51.460" Title="How much does a light heat the air around it?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;thermodynamics&gt;&lt;gas-laws&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35535" PostTypeId="2" ParentId="35534" CreationDate="2015-08-27T01:38:08.333" Score="0" Body="&lt;p&gt;There is no real answer to this because light bulbs do not have a exact temperature do to the various shapes sizes. Also some are Built with different chemicals or metals. But let's say it is a basic light bulb if you touch it warms Ur hand to burning. There is also light radiation as a factor warming the surfaces it connects with so yes a light bulb warms the standard room. To prove this just take a wind mill and hold it over a lamp the windmill will spin slowly form the rise of hot air.&lt;/p&gt;&#xA;" OwnerUserId="19445" LastEditorUserId="19445" LastEditDate="2015-08-27T01:44:35.273" LastActivityDate="2015-08-27T01:44:35.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35536" PostTypeId="1" CreationDate="2015-08-27T02:24:10.107" Score="1" ViewCount="402" Body="&lt;p&gt;A mole of hydrogen gas takes up 22.4 L, but so does a larger molecule like $\ce{CO2}$. Methane ($\ce{CH4}$) is an even larger sized molecule, but a mole of it will still take up 22.4 L. I would assume that an even bigger molecule like decane ($\ce{C10H22}$) would still act the same as the others. A mole of sand however does not take up 22.4 L. This begs the question: is there a certain size particle at which Avogadro's law is not valid anymore? Perhaps at a molecular size larger than buckyballs ($\ce{C60}$)? (This also happens to be the point that wave-particle duality does not apply.) &lt;/p&gt;&#xA;" OwnerUserId="19447" LastEditorUserId="5577" LastEditDate="2015-08-28T10:16:27.020" LastActivityDate="2015-08-28T10:16:27.020" Title="At what point does Avogadro's law not apply?" Tags="&lt;molecules&gt;&lt;gas-laws&gt;&lt;mole&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35537" PostTypeId="2" ParentId="35527" CreationDate="2015-08-27T02:33:29.233" Score="3" Body="&lt;p&gt;When a glow stick breaks, it breaks a thin glass barrier on the inside that allows two different substances to come into contact, mix and react. I am not sure of the exact reaction that occurs, but when chemical bonds are broken they can emit photons (light). In this case, the photons emitted are in the visible spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case for glow in the dark stickers is a bit different. All things are absorbing and releasing photons constantly, but in most cases, the photons aren't in the visible spectrum. Glow in the dark things absorb visible light and then release the photons as visible light. This is unlike most other objects which may release them as a lower frequency instead and thus not be visible.&lt;/p&gt;&#xA;" OwnerUserId="19447" LastEditorUserId="402" LastEditDate="2017-04-01T00:51:53.250" LastActivityDate="2017-04-01T00:51:53.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35538" PostTypeId="1" CreationDate="2015-08-27T02:51:29.093" Score="0" ViewCount="128" Body="&lt;p&gt;In my Ap chem class I got this sheet with the names of a bunch of common polyatomic ions. One of them is named monohydrogen phosphate ($\ce{HPO4^2-}$). In naming, I learned not to use ‘mono-’ if it’s at the beginning of a name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply a mistake? Or is there a reason?&lt;/p&gt;&#xA;" OwnerUserId="19447" LastEditorUserId="95133" LastEditorDisplayName="user7951" LastEditDate="2021-05-22T10:58:31.773" LastActivityDate="2021-05-22T10:58:31.773" Title="Why is monohydrogen phosphate an incorrect name?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35540" PostTypeId="2" ParentId="35538" CreationDate="2015-08-27T05:12:43.603" Score="1" Body="&lt;p&gt;The reason you don't generally need &quot;mono&quot;at the beginning is that the charge of the ion implies how many hydrogens would be there if you simply said &quot;hydrogen phosphate&quot;. That would be three hydrogens because phosphate has a charge of -3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I think you are correct in saying mono-hydrogen phosphate because because &quot;hydrogen phosphate&quot; implies three hydrogen, not the one that you want to specify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end everyone is going to write out the chemical symbols for small molecules like that so don't stress too much beyond getting those easy test questions.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-08-27T05:12:43.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35541" PostTypeId="1" AcceptedAnswerId="35544" CreationDate="2015-08-27T05:46:31.177" Score="0" ViewCount="653" Body="&lt;p&gt;Is it possible to have aromatic rings inside a structure that itself is non-aromatic? Is there an example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iBihb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iBihb.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9630" LastEditorUserId="9630" LastEditDate="2015-08-27T08:08:49.483" LastActivityDate="2015-08-27T12:34:33.760" Title="Can we have Aromatic rings in a non-aromatic structure" Tags="&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" ClosedDate="2015-08-28T04:28:06.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="35542" PostTypeId="2" ParentId="35536" CreationDate="2015-08-27T06:32:09.133" Score="1" Body="&lt;p&gt;My feeling is that the size of the molecule matters much less than you would think. For instance, from a purely ideal gas perspective (as Avogadro's law assumes), the expected value for the speed of a particle is $$v=\sqrt{\frac{8kT}{\pi m}}$$&#xA;where $k$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that means that as the mass of your particle doubles, the average speed of the particle is decreased by a factor of $1/\sqrt{2}$. But, since pressure, from a kinetic molecular theory perspective treats pressure as being the average force which the particles apply to a wall (over a certain area), and force is the time derivative of momentum $p=mv$, we see that $p$ increases by a factor of $\sqrt{2}$ when the mass is doubled (because $v$ decreased by $1/\sqrt{2}$.) That means, the average force on the wall will be greater per collision, but not very much greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I went ahead and calculated the pressure which a mole of hexane exerts on $22.42\ \mathrm{L}$ according to the van der Waals (VDW) equation and found it was $1.05\ \mathrm{atm}$. So, as expected, the value is larger, but not much larger. That's because VDW actually accounts for a parameter (b) which is the volume excluded per mole (meaning the volume that particle actually takes up, sort of).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, hexane, which is significantly larger than $\ce{CO2}$ or anything else mentioned, only increases pressure by 5% when confined to the same volume. That also means the volume should be 5% larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be wary of scaling that to larger molecules though! As hydrocarbons get larger (and being linear helps) they have a much greater polarizable volume which results in (how about this for a name) greater van der Waals attractions. This extra attraction counteracts the increase in size of the molecule a pretty significant amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that to say, the deviation from Avogadro's Law at SATP is actually quite small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're only going to get noticeable deviations at SATP when you have very polar molecules, even then, however, it's not too bad. Water for instance takes up $22.45\ \mathrm{L}$ at SATP which isn't even deviating in the direction we would expect, but if you increase the pressure to $1.1\ \mathrm{bar}$, then the volume occupied for water is $22.33\ \mathrm{L}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intermolecular forces are more significant than the particle's size, particularly at lower temperatures and higher pressures.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorDisplayName="user7951" LastEditDate="2015-08-27T10:37:54.703" LastActivityDate="2015-08-27T10:37:54.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35544" PostTypeId="2" ParentId="35541" CreationDate="2015-08-27T07:53:36.943" Score="5" Body="&lt;p&gt;Sounds like a contradiction in terms to me. If a compound has aromatic ring(s), then it is aromatic - that's the definition.&#xA;Though aromatic compounds with aliphatic side chains are possible, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your compound is &lt;strong&gt;azulene&lt;/strong&gt;; look it up on Wikipedia or elsewhere. Indeed, it is aromatic, and so is each of its rings. You may be confused by the fact that cyclopentadiene (which looks just like one of these rings) as an individual compound is not aromatic, but that's another story. See, a 5-membered ring needs an extra electron to be truly aromatic. At the same time, the 7-membered ring has one electron more than it needs. So they exchange that electron and live happily ever after.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-08-27T12:34:33.760" LastActivityDate="2015-08-27T12:34:33.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35546" PostTypeId="2" ParentId="35527" CreationDate="2015-08-27T08:44:05.833" Score="-3" Body="&lt;p&gt;Reactions emits energy in different forms. Could be heat, light or electrical. If the form emited is light energy form it surely will glow irrespective of the external environmental characteristics in this case dark! &lt;/p&gt;&#xA;" OwnerUserId="19056" LastActivityDate="2015-08-27T08:44:05.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35548" PostTypeId="1" CreationDate="2015-08-27T09:05:16.097" Score="2" ViewCount="1335" Body="&lt;p&gt;I like to draw a molecule in different conformations in a) website or b) freeware and &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate its relative energy difference (for example to the lowest global energy conformation)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Display the lowest energy conformation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm particularly interested in the calculation of diaxial interactions, gauche interactions, double gauche pentane interactions (and less in ring strain, like chair / half-chair / twisted rings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are plenty of tools for modelling conformational and electrostacic properties. Most of those are not freely available. Do you an easy solution? &#xA;It does not must to be the most precise one, relative energy differences in steps of 0.1 kcal/mol between two drawn conformations would be fine for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be this question is off-topic, because it's a software request, but I didn't know where else I could ask this question.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastActivityDate="2015-11-25T13:35:35.623" Title="Calculation of the lowest energy conformation with a webbased solution or freeware" Tags="&lt;organic-chemistry&gt;&lt;physical-chemistry&gt;&lt;energy&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35549" PostTypeId="1" CreationDate="2015-08-27T09:10:02.763" Score="0" ViewCount="149" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;At which position will electrophilic substitution on &lt;em&gt;p&lt;/em&gt;-cyanonitrobenzene occur?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As both are -M groups, there are two possible sites of reaction, &lt;em&gt;meta&lt;/em&gt; to each of the groups.&lt;br&gt;&#xA;The answer is &lt;em&gt;meta&lt;/em&gt; to $\ce{CN-}$. Why is this so? According to me it should be &lt;em&gt;meta&lt;/em&gt; to $\ce{NO2}$, because $\ce{NO2}$ is the stronger -M group?&lt;/p&gt;&#xA;" OwnerUserId="17997" LastEditorUserId="4325" LastEditDate="2015-08-27T16:10:51.967" LastActivityDate="2015-08-27T16:10:51.967" Title="Regioselectivity for electrophilic aromatic substitution with multiple strongly electron withdrawing groups" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35550" PostTypeId="1" CreationDate="2015-08-27T09:15:56.557" Score="4" ViewCount="410" Body="&lt;p&gt;In &lt;em&gt;Organic Chemistry&lt;/em&gt; (2nd ed.) by Clayden, Greeves, and Warren, it is stated that (p 869)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] the enolates of some metals [Sn(II), Zr, Ti] give &lt;em&gt;syn&lt;/em&gt; aldols regardless of enolate geometry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are also literature examples of &lt;em&gt;syn&lt;/em&gt;-selective Zr aldol reactions, e.g. &lt;a href=&quot;https://doi.org/10.1016/0040-4039(80)80086-4&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Tetrahedron Lett.&lt;/em&gt; &lt;strong&gt;1980,&lt;/strong&gt; &lt;em&gt;21&lt;/em&gt; (48), 4607–4610&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they were to obey a simple Zimmerman–Traxler chair-like transition state model, as is applicable for Li or B enolates, then (&lt;em&gt;Z&lt;/em&gt;)-enolates should give &lt;em&gt;syn&lt;/em&gt; aldols and (&lt;em&gt;E&lt;/em&gt;)-enolates &lt;em&gt;anti&lt;/em&gt; aldols. How does the mechanism or stereochemical model differ for these enolates?&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="16683" LastEditDate="2018-12-26T01:19:11.187" LastActivityDate="2018-12-26T01:19:11.187" Title="Why do Zr, Sn, and Ti enolates form syn aldols selectively?" Tags="&lt;organic-chemistry&gt;&lt;aldol-reaction&gt;&lt;stereoselectivity&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35551" PostTypeId="2" ParentId="35538" CreationDate="2015-08-27T09:34:07.787" Score="3" Body="&lt;p&gt;The current version of &lt;a href=&quot;http://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book)&lt;/em&gt;&lt;/a&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(…) The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances (…).&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, hydrogenphosphate (&lt;span class=&quot;math-container&quot;&gt;$\ce{HPO4^2-}$&lt;/span&gt;) may be called monohydrogenphosphate in order to emphasize the difference to dihydrogenphosphate &lt;span class=&quot;math-container&quot;&gt;$\ce{H2PO4-}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Note that, according to IUPAC Recommendations, the systematic &lt;em&gt;hydrogen name&lt;/em&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{HPO4^2-}$&lt;/span&gt; actually is ‘hydrogen(tetraoxidophosphate)(2−)’; however, ‘hydrogenphosphate’ is an &lt;em&gt;accepted simplified hydrogen name&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-27T09:34:07.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35552" PostTypeId="2" ParentId="32652" CreationDate="2015-08-27T09:39:07.917" Score="0" Body="&lt;p&gt;AFAIK there is no such an option in molden. Take another (free) program (&lt;em&gt;f.e.&lt;/em&gt; molekel)&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2015-08-27T09:39:07.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35553" PostTypeId="2" ParentId="24080" CreationDate="2015-08-27T09:57:57.177" Score="1" Body="&lt;p&gt;$R_{int}$ considers in this summation symmetry relevant reflections, &lt;em&gt;e.g.&lt;/em&gt; in your equation the term $\Sigma|F^2-F_{mean}^2|$ takes an intensity of reflection and subtracts from this number the mean intensity of all its symmetry equivalents. If redundancy of your data collection is high (the case for all area detectors), then with this number you can tell how close in intensity are the equivalents, and thus how good is your crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another measure of quality is $R_{\sigma}$&#xA;$$R_{\sigma}=\frac{\Sigma[{\sigma}(F^2)]}{\Sigma|F^2|}$$&#xA;It tells you how close are the estimated standard deviation of equivalents (&lt;em&gt;e.g.&lt;/em&gt;: were all equivalents collected with the same precision?)&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2015-08-27T09:57:57.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35555" PostTypeId="2" ParentId="34517" CreationDate="2015-08-27T10:06:31.300" Score="1" Body="&lt;p&gt;In nature you need to look for precious gamestones and then do optical selection (diamante and ruby would be good candidates). The same would be true for synthetics but of course as mentioned by Jon you can also take elements (Si, Ge). General rule: they should be hard (smaller chance that pressure will affect structure, by pressure I mean also your tools) and you need to grow them slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;btw&lt;/em&gt; it's hard to call graphen crystal pattern -- it's 2D.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2015-08-27T10:06:31.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35556" PostTypeId="2" ParentId="25016" CreationDate="2015-08-27T10:18:53.710" Score="2" Body="&lt;p&gt;Your question has no sense – sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You talk about a few small molecules (with coinage metals) but ask about lattice type for atoms. Atoms in molecular crystal have no their own lattice. Lattice describes how your molecule are aligned in 3D-space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you have a single crystal data then in the beginning of experiment your X-minion (that's the person that took X-ray data for you) has already found the cell parameters and then usually also Bravais cell. To be sure that selected lattice was right whole data set need to be collected, the data have to be integrated and then using &lt;em&gt;Lattice exceptions&lt;/em&gt; (&lt;em&gt;see&lt;/em&gt; International Tables for Crystalography Volume A,  2.2.13. Reflection conditions) you can tell what lattice type you have. If you do the same for glide planes and screw axis exceptions then eventually you will find space group.&lt;/p&gt;&#xA;" OwnerUserId="327" LastActivityDate="2015-08-27T10:18:53.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35557" PostTypeId="2" ParentId="24080" CreationDate="2015-08-27T10:27:21.607" Score="3" Body="&lt;p&gt;$F^2/\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R_{int}$ at all. Indeed, to calculate $R_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\sigma(F^2)$, you have it as soon as you've collected a little bit of data.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-27T10:27:21.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35558" PostTypeId="2" ParentId="29696" CreationDate="2015-08-27T10:28:04.190" Score="1" Body="&lt;p&gt;Your loop defines 8 records. That works fine for &lt;em&gt;heavy&lt;/em&gt; elements. However, the lines with hydrogen atoms have only 6 items. That's the reason why this error is reported (all you records are not multiplier of 8).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in refinement the hydrogen atom's $U_{iso}$ is 1.2–1.5 bigger than this of atom to which the H is bound (smaller 1.2 for more constrained movements like on aromatic carbons, bigger 1.5 for aliphatic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N&#xA;H1a  2    -0.9650000 0.08500000 -1.0010000 Biso 4.     H&#xA;H1b  3    -0.8800000 -0.0800000 -1.0330000 Biso 4.     H&#xA;C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C&#xA;H2   5    -0.7460000 -0.0800000 -0.9360000 Biso 4.     H&#xA;C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="327" LastActivityDate="2015-08-27T10:28:04.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35559" PostTypeId="1" AcceptedAnswerId="35561" CreationDate="2015-08-27T11:26:58.087" Score="15" ViewCount="6639" Body="&lt;p&gt;For relatively high concentrations of $\ce{HCl}$, I usually just assume that $[\ce{H+}] = [\ce{HCl}]$, because $\ce{HCl}$ is a strong acid and is completely ionized in solution. By taking the negative logarithm of the concentration of $\ce{[H+]}$, the pH comes out more or less correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at very low concentrations, this method doesn't work out so well, because water has an amount of auto-ionization, which is described by $K_\mathrm{w} = 10^{-14}$. My old method was just to say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ce{H+}] = [\ce{HCl}] + \pu{10^-7 mol dm-3},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to include the auto-ionization of water in my calculations. I can't really describe why, but I believe this method to be wrong, because the number $\ce{10^{-7}}$ comes from water just by itself, not with $\ce{HCl}$ added to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I came across &lt;a href=&quot;http://www.chemteam.info/AcidBase/Trick-pH-question.html&quot; rel=&quot;noreferrer&quot;&gt;this website,&lt;/a&gt; which implies that the above method is an OK method, but mentions that &quot;However, that more complex technique will not be discussed here.&quot;. What is that 'more complex technique', and how could I use it?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="16683" LastEditDate="2018-07-21T12:43:02.820" LastActivityDate="2018-07-21T12:43:02.820" Title="Calculating the pH of a highly dilute solution of HCl" Tags="&lt;acid-base&gt;&lt;ph&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35561" PostTypeId="2" ParentId="35559" CreationDate="2015-08-27T12:17:46.503" Score="21" Body="&lt;p&gt;At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before.&#xA;$$k_\mathrm w = \ce{[H+]~[OH-]}&#xA;\Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated.&#xA;$$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This changes the first equation to be like&#xA;$$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$&#xA;$$x^2-c_0~x-k_\mathrm w = 0$$&#xA;Wherefrom you get two solutions (from which only the positive one makes sense):&#xA;$$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip3.5cm$&lt;img src=&quot;https://i.stack.imgur.com/5JcQI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A maximum difference of 0.1 pH between the easier and the &quot;more complex&quot; version is acceptable.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorDisplayName="user7951" LastEditDate="2017-01-22T13:35:39.603" LastActivityDate="2017-01-22T13:35:39.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35562" PostTypeId="2" ParentId="35548" CreationDate="2015-08-27T12:30:38.660" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://avogadro.cc/wiki/Main_Page&quot; rel=&quot;nofollow&quot;&gt;Avogadro&lt;/a&gt; can do a lot of this for you. It might not be precisely what you want but the molecular mechanics from Open Babel integrated into it can do at least some of the tasks you mentioned. It's free and has a very intuitive and smooth user interface. A real pleasure. The downside is that it's poorly documented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To play around with conformations, go to Extensions -&gt; Molecular Mechanics. To view the generated conformers (from the Conformer Search option) go to View -&gt; Properties -&gt; Conformer Properties...&lt;/p&gt;&#xA;" OwnerUserId="1887" LastActivityDate="2015-08-27T12:30:38.660" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35563" PostTypeId="1" AcceptedAnswerId="35568" CreationDate="2015-08-27T12:35:24.733" Score="14" ViewCount="7228" Body="&lt;p&gt;I'm trying to understand why chemists use the mole unit instead of just counting and using SI prfixes to simplify the big numbers with units such as zetta- or yotta-molecules (yotta=10^24).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've found so far, but it doesn't seem to be enough to make moles particularly important:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasons for moles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can easily approximate the number of atoms from the mass of a sample because the mass number of an isotope approximately equals the mass of 1 mole of atoms in grams.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a historical convention that would be too expensive or uncomfortable to change now.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can use millimole and micromole instead of having to learn several more obscure SI prefixes like exa and peta. &lt;strong&gt;EDIT this reason is added from Jan's answer&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Shorter descriptions, eg &quot;mole of carbon&quot; vs &quot;yotta atom of carbon&quot; (from matt_black's answer).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reasons against moles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For accurate calculations (beyond about 3 significant figures), the relationship between mass number and mass in the first reason above breaks down. So this is a risky thing to use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's a additional concept and set of facts that chemists have to spend effort learning but which describes human convention rather than nature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We need extra conversion factors such as 1/mol which people often neglect and Faraday's constant used in Q = n(e-) x F [1] which wouldn't exist without moles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Non-reasons for moles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can easily calculate the number of atoms from the mass of a sample because the atomic mass of an element equals the mass of 1 mole of atoms in grams. This relationship only exists because of the special units commonly used for atomic mass. If periodic tables listed atomic mass in grams (or perhaps yoctograms), then we could do the same calculations just as easily without moles.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In practice, we can't measure the number of molecules so we have to measure mass or volume instead and should therefore count them using a unit that's defined in terms of mass. I don't think the precise details of how a unit is defined matter for practical purposes. If you have 1 litre of an ideal gas, you still have to do calculations to find out many moles it contains just as you would to find how many molecules it contains. There is even a proposal [2] for SI to redefine the mole to be independent of the mass of any substance, indicating that keeping the definition isn't very important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Chemists would make mistakes with those big numbers. They'd use &quot;yotta&quot; in the same way they use &quot;mole&quot;, not doing calculations with the actual number it represents and not being at risk of other types of error. (from matt_black's answer) However it would be more complex replacing millimole with zetta because you'd have to remember that a zetta-atom is a milli-yotta-atom.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's a similar question but it's mixed in with the idea of measuring number rather than mass or volume -&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/6659/why-do-people-still-use-the-mole-unit-in-chemistry&quot;&gt;Why do people still use the mole (unit) in chemistry?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This other similar question mainly addresses the uncommonness of quantities as big as 1 mole in other areas of life -&#xA;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/671/the-mole-is-used-extensively-in-chemistry-why-not-elsewhere&quot;&gt;The mole is used extensively in chemistry, why not elsewhere?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;http://www.ausetute.com.au/faradayl.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ausetute.com.au/faradayl.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;https://en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19456" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2016-12-12T17:02:58.077" Title="What advantages does the mole have over counting large numbers using SI prefixes?" Tags="&lt;mole&gt;" AnswerCount="5" CommentCount="7" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35564" PostTypeId="1" AcceptedAnswerId="35571" CreationDate="2015-08-27T13:00:52.497" Score="1" ViewCount="1550" Body="&lt;p&gt;What are the IUPAC rules when naming a coordination complex like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now according to my textbook, it is named as &lt;strong&gt;&lt;em&gt;tetraammineplatinum(II) amminetrichloridoplatinate(II)&lt;/em&gt;&lt;/strong&gt;. &lt;br /&gt;Both the $\ce{Pt}$ atoms have the oxidation state &lt;strong&gt;+2&lt;/strong&gt; .There are no charges assigned to the two complex ions in the formula, so, my assumption is that the entire complex is neutral and thus the oxidation states for $\ce{Pt}$ should be assigned accordingly. However, on assigning the oxidation states as mentioned in the answer the assumption is proven wrong. I'm confused about the mentioned oxidation numbers.&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do you calculate the oxidation numbers of the metal atoms in such a complex?&lt;/li&gt;&#xA;&lt;li&gt;Shouldn't the charge on each complex ion be mentioned in the formula if the oxidation states given in the answer are correct?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15508" LastActivityDate="2015-08-28T08:47:56.720" Title="Name the following coordination complex" Tags="&lt;nomenclature&gt;&lt;coordination-compounds&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35565" PostTypeId="2" ParentId="35459" CreationDate="2015-08-27T13:07:02.610" Score="1" Body="&lt;p&gt;Best hypothesis so far is that as gas density increases the transducer couples more efficiently to the medium, giving the impression of a decrease in absorption coefficient due to the increased amplitude at the receiver.&lt;/p&gt;&#xA;" OwnerUserId="9473" LastActivityDate="2015-08-27T13:07:02.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35568" PostTypeId="2" ParentId="35563" CreationDate="2015-08-27T13:37:29.573" Score="7" Body="&lt;p&gt;Of course, it’s convention and has been so for a good century. And there is no real reason why this convention (and not a different one) happened in the first place — it is entirely conceiveable to define an ‘imperial mole’ so that the atomic mass of an element is equivalent to the same mass in ounces. 1 imperial mole of carbon atoms would then be twelve ounces of carbon atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are basically two questions in this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why replace a large number with a unit just meaning ‘this certain large number’?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is the presently defined mole a good choice for this unit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The thing about large numbers is that they are large. Almost everybody who grew up in a metric country can name at least three SI-prefixes: kilo, centi and milli. Thanks to IT, many people now also know mega and giga (and maybe tera), even if they don’t use metric units at home. But tera only gets you to $10^{12}$. We need some $10^{21}$ for moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often work with milli- or micromoles of substances in my research. In plain numbers, that’s $10^{20}$ or $10^{17}$ — I don’t know those prefixes and thus would have to learn an entirely new subset. With the mole, everything one uses in the lab will nicely fall into something between nano and kilo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also helps to have a single unit there. Molar mass is expressed in grams per mole, concentrations in moles per litre and many more. If there were no unit, it would be simple grams per 1 or 1 per litre — precisely the reason why some people prefer to use rad or some other way to show radians rather than just writing the number. If the unit is there, you’re unlikely to forget it, you know if your calculations are good and more. If there was no name for this unit, it would have to be invented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the mole didn’t exist, it should be invented for simplicity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The good thing about the size definition of the mole is, as noted above, that it brings everything into one general range. Whether it’s mass, volume, concentration or amount, every unit is going to be prefixed by only a small subset of the SI-prefixes: kilo, milli, micro, maybe nano. Thankfully, those are the ones that are most used in everyday life, too (excluding nano and maybe micro).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn’t really matter where one ends up. If the mole had originally been defined imperially,&lt;sup&gt;[1]&lt;/sup&gt; that would be fine, I wouldn’t have memorised $12.01\,\frac{\mathrm{g}}{\mathrm{mol}}$ for carbon but $340.48\,\frac{\mathrm{g}}{\mathrm{mol}}$. That would create significantly larger molar masses, but that shouldn’t be a problem it should only mean that nano is more prevalent.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The thing about the definition of the unit is that it doesn’t bother $99.9\,\%$ of the scientists working with the unit. Seconds are defined (by SI) according to a number of transitions occuring in some weird isotope that I wouldn’t even know how to measure. I would explain a second by saying it’s the 86400&lt;sup&gt;th&lt;/sup&gt; fraction of a day if someone asked me. Same meaning, different exactness. If the moles are soon defined by mere counting rather than weighing atoms then so be it. Nothing will change for me in practice. (Maybe the fourth digit of a molar mass but I don’t really count those.) So as long as a new definition doesn’t break anything, we’re good to carry on.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: With &lt;em&gt;imperially&lt;/em&gt;, I meant to assume the following definition using imperial units:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One mole is the number of atoms in exactly $12.00~\mathrm{oz}$ of atoms of the carbon isotope $\ce{^{12}C}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="7475" LastEditDate="2016-06-14T15:21:31.947" LastActivityDate="2016-06-14T15:21:31.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35569" PostTypeId="1" CreationDate="2015-08-27T14:28:34.160" Score="1" ViewCount="1448" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When heated, a $3.40\ \mathrm g$ sample mixture of $\ce{KClO3}$ decomposes and produces $685\ \mathrm{cm^3}$ $\ce{O2}$ gas at $30\ \mathrm{^\circ C}$ and $101\ \mathrm{kPa}$. What is the mass percentage of $\ce{KClO3}$ in the mixture?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I've done so far is write out a balanced equation which is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2KClO3 -&amp;gt; 3O2 + 2KCl}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do I go from here?&lt;/p&gt;&#xA;" OwnerUserId="19462" LastEditorUserId="5026" LastEditDate="2018-02-27T13:21:22.967" LastActivityDate="2018-02-27T13:22:55.320" Title="Mass percentage of potassium chlorate in decomposition reaction" Tags="&lt;physical-chemistry&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35570" PostTypeId="2" ParentId="35549" CreationDate="2015-08-27T16:07:40.627" Score="3" Body="&lt;p&gt;As Mithoron suggests, the correct answer is probably &quot;no reaction.&quot; There are no examples of electrophilic aromatic substitution of &lt;em&gt;p&lt;/em&gt;-cyanonitrobenzene in SciFinder, by the introductory organic chemistry mechanism at least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one report of halogenation (with &lt;em&gt;N&lt;/em&gt;-halosuccinimide) of this compound with palladium catalysis. The cyano group serves as a directing group to place a halogen &lt;em&gt;ortho&lt;/em&gt; to the cyano. Chlorination and bromination are both reported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/jo302765g&quot; rel=&quot;nofollow&quot;&gt;Palladium-Catalyzed Highly Selective ortho-Halogenation (I, Br, Cl) of Arylnitriles via sp2 C–H Bond Activation Using Cyano as Directing Group&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-08-27T16:07:40.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35571" PostTypeId="2" ParentId="35564" CreationDate="2015-08-27T16:24:34.587" Score="2" Body="&lt;p&gt;Your textbook contains an error. Either the formula is written incorrectly or the name is. Since according to comments the name is correct, it must be an incorrect formula. Let’s break it down:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If the formula is correct&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have two complex ions that together create a doubly-complex salt. According to the rules of (pseudo-) binary ionic compounds, one ion must be positively charged and the other negatively. By convention, the cation is written first so it is safe to assume that $\ce{[Pt(NH3)4]^{?}}$ be the cation — which fits in well with chemical intuition because $\ce{NH3}$ is a neutral ligand and platinum is a metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $\ce{[Pt(NH3)Cl3]^{?}}$ must be the anion — very nice, because whe have chlorido ligands that are commonly negative. Since every chlorido ligand adds $-1$ charge to the complex, the anion can be anything from $-1$ to $-3$ with decreasing probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex needs to be a $1\,:\,1$ binary complex, so the negative charge needs to be equivalent to the positive charge. From here onwards it gets difficult. We have the following possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{[Pt^{I}(NH3)4]^{+} [Pt^{II}(NH3)Cl3]^-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[Pt^{II}(NH3)4]^2+ [Pt^{I}(NH3)Cl3]^2-}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{[Pt^{III}(NH3)4]^3+ [Pt^{0}(NH3)Cl3]^3-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They all differ in the assumed oxidative states of platinum. We &lt;strong&gt;cannot decide &lt;em&gt;a priori&lt;/em&gt;&lt;/strong&gt; which one of these three is correct, although the last one is by far less probable. Technically, we can even assume further diverging oxidation states but they get proportionally less probable. &lt;strong&gt;Only if the charge were clearly stated could we distinguish between the cases.&lt;/strong&gt; For the sake of consistency, here are the names:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;tetraammineplatinum(I) amminetrichloridoplatinate(II)&lt;/li&gt;&#xA;&lt;li&gt;tetraammineplatinum(II) amminetrichloridoplatinate(I)&lt;/li&gt;&#xA;&lt;li&gt;tetraammineplatinum(III) amminetrichloridoplatinate(0)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;If the name is correct&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Then we can logically build up the resulting salt from the name. Let’s start with the cation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is platinum(II), so we get $\ce{Pt^2+}$&lt;/li&gt;&#xA;&lt;li&gt;adding four ammine ligands gives us $\ce{[Pt(NH3)4]^2+}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the anion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is still platinum(II) (although rendered as platinate, so we need to end up with an anion): $\ce{Pt^2+}$&lt;/li&gt;&#xA;&lt;li&gt;We have one ammine ligand: $\ce{[Pt(NH3)]^2+}$&lt;/li&gt;&#xA;&lt;li&gt;and three chlorido ($\ce{Cl-}$) ligands: $\ce{[Pt(NH3)Cl3]-}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since we have a cation with $+2$ charge and an anion with $-1$ charge, we need two anions per cation, giving us the final formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]2}$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So what is correct?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Going by chemical intuition (platinum(II) being a well-known $\mathrm{d}^8$ system that forms square-planar complexes), I would say that the name is correct and the formula isn’t. This also has the implication that they messed up the question by missing a subscripted 2 (rather likely) rather than messing up the answer by getting a Roman numeral wrong (not quite as likely).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence there is nothing miraculous or confusing about assigning oxidation states to metals in complexes — and the IUPAC even formally defines oxidation states for promiscuous ligands such as nitrosyl (which is always considered to be a radical neutral ligand, never $\ce{NO^+}$ or $\ce{NO-}$) allowing for a clean-cut naming of complex compounds without knowing the chemistry involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge &lt;strong&gt;does not&lt;/strong&gt; have to be mentioned in doubly-complex salts for any component. However, it always helps students to give them and in non-trivial cases (like the assumed correct formula) it might be vital to know the overall charges. If there are no charges written, though, the final product &lt;strong&gt;has to be&lt;/strong&gt; assumed neutral.&lt;/p&gt;&#xA;" OwnerUserId="7475" LastEditorUserId="16683" LastEditDate="2015-08-28T08:47:56.720" LastActivityDate="2015-08-28T08:47:56.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35573" PostTypeId="2" ParentId="35490" CreationDate="2015-08-27T17:01:57.693" Score="0" Body="&lt;p&gt;Regarding toxicity: Honey solutions are actually sometimes used in ophthalmology. There is a company called Optimel that sells 16% manuka honey eye drops OTC for treating dry eye syndrome but note that very little or almost nothing H2O2 is released in manuka honey. Another example is this &lt;a href=&quot;http://bit.ly/1UdxqVa&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;case report&lt;/strong&gt;&lt;/a&gt; in which they used a 25% w/v silver fir honey solution as eye drops for the treatment of a corneal ulcer with success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found an &lt;a href=&quot;http://www.reviewofoptometry.com/cmsdocuments/2014/8/ro0614alconclearcarei.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;article&lt;/strong&gt;&lt;/a&gt; in which the use of H2O2 in the eyes in various concentrations is outlined. At the concentration below 60 ppm (the lowest safe point) the eye’s natural enzyme mechanisms can break down the small amount of H2O2 making it safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a 25% w/v (silver fir) honey solution contains around 0.00085% or 8.5 ppm of H2O2 which would make it safe for the use in eyes.&lt;/p&gt;&#xA;" OwnerUserId="19424" LastEditorUserId="19424" LastEditDate="2015-08-27T17:07:42.703" LastActivityDate="2015-08-27T17:07:42.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35574" PostTypeId="1" AcceptedAnswerId="35577" CreationDate="2015-08-27T17:15:10.887" Score="27" ViewCount="1383" Body="&lt;p&gt;Are large oxygen containing molecules possible? Either large rings, or chains with hydrogen atoms at the ends. Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{H-O-O-O-O-O-O-O-O-O-$\cdots$-O-O-O-O-O-H}$&lt;/p&gt;&#xA;" OwnerUserId="19465" LastEditorUserId="16683" LastEditDate="2016-11-24T20:51:04.133" LastActivityDate="2019-05-08T09:33:27.457" Title="Are long oxygen molecules possible?" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="35575" PostTypeId="2" ParentId="24723" CreationDate="2015-08-27T17:26:49.063" Score="2" Body="&lt;p&gt;I read all the answers by experts, but I'm also a student and have the same interest as you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest going on YouTube and using it to discover your own rhyme or trick. Be patient, look for patterns in their names, and find a meaningful rhyme that will be easy for you to remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing this on your own has the benefit of &lt;em&gt;you&lt;/em&gt; remembering it. You did all the work in creating your mnemonic. The work you put forth in doing this makes it a much better method than learning a trick from others.&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="402" LastEditDate="2018-01-17T05:36:58.363" LastActivityDate="2018-01-17T05:36:58.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35576" PostTypeId="1" CreationDate="2015-08-27T17:36:43.483" Score="2" ViewCount="449" Body="&lt;p&gt;I work in a laboratory where gas chromatography is used extensively. We handle terpenes and terpenoids (alcohols, ketones, and acetates derived from terpenes), organic solvents (acetone, acetonitrile, ethyl acetate, hexane, and toluene), acetic acid, and water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know which are the most appropriate solvents to use for washing the equipment or which criteria should I use to decide among the existing solvents. Now, we are using acetone and ethanol, but sometimes it seems that they do not work for some terpenoids (the syringe sticks and sometimes the equipment crashes). Any suggestions about this?&lt;/p&gt;&#xA;" OwnerUserId="19467" LastEditorUserId="865" LastEditDate="2015-08-27T17:55:29.763" LastActivityDate="2015-08-27T17:55:29.763" Title="Which are the appropriate washing solvents for gas chromatography equipment?" Tags="&lt;solvents&gt;&lt;chromatography&gt;&lt;equipment&gt;" AnswerCount="0" CommentCount="1" ClosedDate="2019-03-26T15:12:04.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="35577" PostTypeId="2" ParentId="35574" CreationDate="2015-08-27T17:42:32.287" Score="24" Body="&lt;p&gt;Yes, they are possible; several have even been detected.  They all have the general formula $\ce{H2O_{n}}$ and belong to the class of &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_polyoxide&quot; rel=&quot;noreferrer&quot;&gt;hydrogen polyoxides&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the $n=1$ member of the series is water $(\ce{H2O})$, and the $n=2$ member is hydrogen peroxide $(\ce{H2O2})$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $n=3$ member is &lt;a href=&quot;https://en.wikipedia.org/wiki/Trioxidane&quot; rel=&quot;noreferrer&quot;&gt;trioxidane&lt;/a&gt; $(\ce{H2O3})$, and it has the following structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WBzjym.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WBzjym.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be prepared in a number of ways as outlined in the linked article, such as the reaction of ozone and hydrogen peroxide. Trioxidane is also an active antimicrobial ingredient. Trioxidane has a half-life around 16 minutes in organic solvents, but only milliseconds in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_polyoxide&quot; rel=&quot;noreferrer&quot;&gt;Hydrogen tetroxide&lt;/a&gt; ($n=4$, $\ce{H2O4}$) is also a known compound as is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_polyoxide&quot; rel=&quot;noreferrer&quot;&gt;hydrogen pentoxide&lt;/a&gt; ($n=5$, $\ce{H2O5}$).  Hydrogen pentoxide is formed as a byproduct in the preparation of trioxidane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ja971534b&quot; rel=&quot;noreferrer&quot;&gt;This abstract&lt;/a&gt; of a computational study provides bond dissociation energies for the hydrogen polyoxides. Starting from water, they steadily decrease (the $\ce{O-O}$ bonds become weaker) to $\ce{H2O6}$ and then start to increase again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abstracts from other computational studies dealing with the expected structures and stabilities of these molecules can be found &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/qua.22695/abstract&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.tandfonline.com/doi/abs/10.1080/00268976.2014.928385&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="16683" LastEditDate="2016-11-24T19:47:52.943" LastActivityDate="2016-11-24T19:47:52.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35578" PostTypeId="1" AcceptedAnswerId="37389" CreationDate="2015-08-27T18:10:53.800" Score="10" ViewCount="1738" Body="&lt;p&gt;I have seen Graham condensers set at an angle (perhaps as low as 30 degrees from horizontal).  I guessed this wouldn't be a problem because siphoning would clear the condensate.  But I recently saw an argument that &lt;em&gt;Graham condensers should only be used at near-vertical angles&lt;/em&gt;.  Thinking harder this makes sense to me, because as soon as enough liquid condenses in the first loop to form a trap the condenser can't work on any other vapor.  Absent some vacuum on the receiving end the only way it could clear itself is if there happened to be enough vapor to eventually close a loop and start the siphon effect, or if enough pressure built up on the boiling end to push the liquid through &lt;em&gt;which would be &lt;strong&gt;bad&lt;/em&gt;&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I still see diagrams and photos of Graham condensers set at shallow angles.  Here's an example from the first few Google image results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uSPiZ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uSPiZ.jpg&quot; alt=&quot;Bad Graham condenser setup?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when and why, if ever, is it considered a good practice to run a Graham condenser in a non-vertical orientation?&lt;/p&gt;&#xA;" OwnerUserId="5354" LastActivityDate="2018-06-15T09:33:36.723" Title="What angle limits should apply to Graham condensers?" Tags="&lt;equipment&gt;" AnswerCount="3" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35579" PostTypeId="1" AcceptedAnswerId="35588" CreationDate="2015-08-27T19:12:43.140" Score="5" ViewCount="1263" Body="&lt;p&gt;Extrinsic semiconductors are created by doping an intrinsic semiconductor; typically silicon is doped with phosphorus to create free electrons or boron to create &quot;holes&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of phosphorus doping, phosphorus takes the place of a silicon atom in the silicon lattice, and its fifth electron becomes a free electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I am struggling to understand with my basic knowledge. Phosphorus has the configuration $\ce{[Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/covalent.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; gives an description of the promotion of a 3s electron to a 3d shell, allowing five bonds to form in the case of $\ce{PCl5}$. I would assume that a similar process results in four bonds forming in the silicon lattice and leaving a free electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that same page later on mentions a 2007 paper which essentially says that that model is not accurate, because such a state is less stable than other ways of modeling the bond (assuming I'm understanding correctly. The problem is described &lt;a href=&quot;http://www.chemguide.co.uk/atoms/bonding/sp3dhybrids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, I have not read the actual paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes phosphorus to form four bonds with silicon and leave a free electron rather than, say, forming three bonds and leaving a silicon with an unpaired electron in the lattice, or bonding with another &quot;free&quot; electron? Is it promoting an electron to form the fourth bond and freeing the last?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: &lt;em&gt;Why does phosphorus form four Si bonds and leave a free electron rather than forming three Si bonds when doping silicon?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="853" LastEditorUserId="16683" LastEditDate="2018-07-02T17:26:49.980" LastActivityDate="2018-07-02T17:27:24.030" Title="Why does phosphorus give a free electron in silicon doping?" Tags="&lt;crystal-structure&gt;&lt;valence-bond-theory&gt;&lt;semiconductors&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35580" PostTypeId="1" CreationDate="2015-08-27T19:44:51.860" Score="1" ViewCount="28" Body="&lt;p&gt;When preparing ICP-MS calibration standards for trace metal (~100 ppt) analysis, the values for Gold do not remain stable for more than a day. The standards are prepared using commercially available 10 ppm solutions in Nitric Acid with traces of HF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A known amount of the Gold standard is added to 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted to a pre-determined volume with ultra pure deionized water. The samples being analyzed are in 49% hydrofluoric acid to which has been added 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted with an equal volume of ultra pure deionized water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would preparing matrix-matched calibration standards solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="19469" LastEditorUserId="7448" LastEditDate="2015-08-27T19:56:45.323" LastActivityDate="2015-08-27T20:07:29.343" Title="Solubility of Gold in low concentrations" Tags="&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35581" PostTypeId="1" CreationDate="2015-08-27T19:49:58.367" Score="0" ViewCount="45" Body="&lt;p&gt;When the pH and the pKa levels differ by 2 or more you don't need to calculate the net charge. What you do then is to just count the negative and positive charge and add them up. &#xA;When pH=pKa you also don't need to calculate, because you get half of the different types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT what about if we want to know the net charge for an amino acid where one of the groups have an pKa value of 4,25 , when the pH is&#xA;a) 4,0  , b)5 , c) 5,5 and d)6&#xA;Do you have to calculate it for all of these? Can you say that since a) is pretty close to the pKa we consider them the same, or that since d) is pretty close to being 2 units more, we can just say that and make it easier? And is there even really a point in calculating for c) when it's more than 1 pH unit over? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is basically: where do I draw the line?&lt;/p&gt;&#xA;" OwnerUserId="19470" LastEditorUserId="7448" LastEditDate="2015-08-27T19:59:13.447" LastActivityDate="2015-08-27T19:59:13.447" Title="When do one have to actually calculate the net charge of an amino acid?" Tags="&lt;acid-base&gt;&lt;biochemistry&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-08-28T15:51:44.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="35582" PostTypeId="1" AcceptedAnswerId="35585" CreationDate="2015-08-27T19:52:44.397" Score="18" ViewCount="2206" Body="&lt;p&gt;While drinking water, some of it spilled on my shirt and as usual, as has happened to many of us, the colour darkened. So why does this happen? Is it due to some chemical effect or physical one?&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="7448" LastEditorDisplayName="user15489" LastEditDate="2015-10-30T17:26:57.547" LastActivityDate="2015-10-30T17:26:57.547" Title="Why does water spills on our shirt change its colour?" Tags="&lt;physical-chemistry&gt;&lt;everyday-chemistry&gt;&lt;color&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35583" PostTypeId="1" AcceptedAnswerId="35598" CreationDate="2015-08-27T19:53:02.763" Score="0" ViewCount="1116" Body="&lt;p&gt;Say I (perfectly) mix $1\ \mathrm{m^3}$ of 1 M $\ce{HCl}$ with $1\ \mathrm{m^3}$ of 1 M $\ce{NaOH}$ at standard air pressure and room temperature within a metal vessel. &lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy_of_neutralization&quot; rel=&quot;nofollow&quot;&gt;Heat of neutralization for strong acids and bases&lt;/a&gt; is $-55.8\ \mathrm{kJ/mol}$ (is that actually just the bond enthalpy of an $\ce{H-OH}$ bond?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the $2\ \mathrm{m^3}$ of $\ce{H2O}$ amount to around $2 \cdot 10^6\ \mathrm{mol}$ of water ($M(\ce{H2O}) \overset{\wedge}{=} 20\ \mathrm{g\ mol^{-1}}$). Therefore, $\Delta_\mathrm n H$ is around $2 \cdot 10^6 \times 55.8\ \mathrm{kJ/mol}$, around $110 \cdot 10^6\ \mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would now like to calculate the temperature rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat released is $Q = m c_p \Delta T$, where $m$ is the mass of the solution ($2000\ \mathrm{kg}$), $c_p$ is the specific heat capacity of the solution and $\Delta T$ ist the change in temperature. But that's exactly what I'm trying to find. So unless I know $Q$, I can't find $\Delta T$. Therfore, how do I find $Q$?&lt;/p&gt;&#xA;" OwnerUserId="1235" LastEditorDisplayName="user7951" LastEditDate="2015-08-28T19:00:28.243" LastActivityDate="2015-10-27T20:23:13.423" Title="Mixing 1 cubic meter of HCl and NaOH: Temperature rise of a metal container?" Tags="&lt;heat&gt;&lt;enthalpy&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35584" PostTypeId="2" ParentId="35580" CreationDate="2015-08-27T20:07:29.343" Score="3" Body="&lt;p&gt;Whenever you prepare very dilute working standards, it's good practice to do a stability study to see how long you can store them. There are many factors that can influence the stability of these solutions. The choice of container can make a big difference. Dilute metal solutions should be stored in polyethylene or polypropylene containers, not glass. Another well-known problem is photoreduction. Gold solutions should be stored in opaque containers and be exposed to light as little as possible. Some solutions may also be sensitive to atmospheric exposure. (though I don't think it's an issue for gold)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have experience with working with gold in hydrofluoric acid, but such a large change in such a short period of time in the absence of the above factors may indicate some actual reactivity with your matrix, for which matrix-matching might help, though you may be better off reformulating your solutions if that's the case. It may also be an option to just make fresh working standards very frequently or increase storage life through refrigeration.&lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-08-27T20:07:29.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35585" PostTypeId="2" ParentId="35582" CreationDate="2015-08-27T20:07:48.043" Score="20" Body="&lt;p&gt;This is not so much a change (physical and especially not chemical), it is more about perception of the diffuse reflection and refraction of light. According to Australia's ABC Science page &lt;a href=&quot;http://www.abc.net.au/science/articles/2010/11/11/3063513.htm&quot; rel=&quot;noreferrer&quot;&gt;Darker when wet&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When your clothes are wet, all those air gaps (between the fibres) get filled with water. So when light hits a wet patch it's moving in and out of water and fibres, not air and fibres. The refractive index of fabric is a lot closer to that of water than it is to air, so the light doesn't change speed quite as dramatically going between water and fibres. And that means it bends at a smaller angle when it goes from one to the other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is summarised in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ayo8c.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ayo8c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The information on the linked page is from Professor David Jamieson from the School of Physics at The University of Melbourne, Australia.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another explanation according to Dr. Christopher Baird of the website &lt;a href=&quot;http://sciencequestionswithsurprisinganswers.org/2013/11/06/why-does-water-make-a-shirt-darker/&quot; rel=&quot;noreferrer&quot;&gt;Science Questions with Surprising Answers&lt;/a&gt; is that we perceive a darkening due to the fabric becoming more transparent - an explanation from the website:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spill some water on your white shirt and the situation changes. In the&#xA;  spot that is wet, the water fills all the little pockets in the fabric&#xA;  that used to be filled with air. In terms of its interaction with&#xA;  light, water behaves very similar to fiber material. Adding water to&#xA;  fabric effectively removes the multiple reflecting surfaces (at least&#xA;  as far as the light is concerned) and turns it back optically to a&#xA;  solid chunk of material, which is transparent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Essentially, what we are perceiving when the fabric appears to darken, is due to the colour of what is underneath - view the cloth with a light source behind it, it appears 'lighter', due to being more transparent.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-27T20:26:33.303" LastActivityDate="2015-08-27T20:26:33.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35586" PostTypeId="1" CreationDate="2015-08-27T21:01:52.647" Score="0" ViewCount="59" Body="&lt;p&gt;Is there a way to bring back into solution a bottle of formalin that has crystallized? The bottle has been stored for several years at room temperature.&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="19472" LastActivityDate="2015-09-11T01:10:22.993" Title="Formalin fallen out of solution" Tags="&lt;solubility&gt;&lt;solutions&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35588" PostTypeId="2" ParentId="35579" CreationDate="2015-08-27T21:23:52.323" Score="9" Body="&lt;p&gt;First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the fifth electron is not initially a free electron - it occupies a state that happens to be in the band gap.  Now, that state is really close to the conduction band, only 45 meV below it at 300 K, so at room temperature there is a really good chance that it will jump into and roam about the conduction band.  Nor is that the only state in the gap for P - there is another state 72 meV above the valence band (i.e. it is slightly easier to ionize a 4-fold coordinated P than a Si).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the P only bond with 3 Si? Well, that would leave one unhappy Si atom, and one not much happier P atom.  Again, this is occurring in a crystal, not an isolated molecule. The P will occupy the lowest free energy state. Note that there are other impurities that preferentially occupy various interstitial sites. It is just that P can be coerced into pretending to be pretty much a Si atom - peer pressure, perhaps...&lt;/p&gt;&#xA;" OwnerUserId="5677" LastEditorUserId="16683" LastEditDate="2018-07-02T17:27:24.030" LastActivityDate="2018-07-02T17:27:24.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35589" PostTypeId="1" CreationDate="2015-08-27T21:50:25.973" Score="1" ViewCount="1336" Body="&lt;p&gt;Do nonmetals form ions? I am asking this because I was reading something and I came across the phrase &quot;metal ions,&quot; which left me to wonder whether or not there are nonmetal ions as well.&lt;/p&gt;&#xA;" OwnerUserId="19475" LastEditorUserId="16622" LastEditDate="2015-08-28T09:01:14.143" LastActivityDate="2015-08-30T07:41:41.567" Title="Do only metals form ions?" Tags="&lt;ions&gt;&lt;metal&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35590" PostTypeId="2" ParentId="35589" CreationDate="2015-08-27T21:54:51.880" Score="1" Body="&lt;p&gt;Ammonium nitrate is a fertilizer that is composed of nonmetal ions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are myriad other examples.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-08-27T21:54:51.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35591" PostTypeId="2" ParentId="35586" CreationDate="2015-08-27T21:57:40.403" Score="1" Body="&lt;p&gt;The 'crystals' are polyformaldehyde, more commonly called paraformaldehyde.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat it up, with some base catalyst.&lt;/p&gt;&#xA;" OwnerUserId="1285" LastActivityDate="2015-08-27T21:57:40.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35592" PostTypeId="2" ParentId="35589" CreationDate="2015-08-27T22:03:27.817" Score="7" Body="&lt;p&gt;In short, yes. Single-element ions are common for both metals and non-metals. Some simple examples: $\ce{F^{-}, Cl^{-}, I^{-}, S^2-}$ Multi-element non-metal cations are a bit more common: $\ce{NH4+}$ and $\ce{H3O+}$ being the obvious examples and there are plenty of anions like $\ce{CN-}$, $\ce{SO4^{2-}}$, $\ce{NO3-}$, etc. Things like proteins and polypeptides are also frequently ions in solution and may be positively or negatively charged depending on the pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply due to their electronegativities, all the single element non-metal ions are anions in solution, but in the gas phase, all bets are off and you can knock off an electron to get a cation, (or have negatively charged metal ions) so long as it doesn't bump into anything else. &lt;/p&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-08-27T22:03:27.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35593" PostTypeId="2" ParentId="35583" CreationDate="2015-08-27T22:24:48.007" Score="0" Body="&lt;p&gt;2 cubic meters of water aren't produced in the reaction, but they come from the acid and base solutions, so I do not see why you are multiplying 2 cubic meters of water with the enthalpy of neutralization. By multiplying the amount of water actually produced during neutralisation with the enthalpy of neutralisation, you get Q, the amount of heat released, so the temperature is the only unknown.&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-08-27T22:24:48.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35597" PostTypeId="2" ParentId="35486" CreationDate="2015-08-28T03:12:08.877" Score="4" Body="&lt;p&gt;Contrary to what I believed earlier, cis- and trans- notation &lt;em&gt;is&lt;/em&gt; unambiguous for cyclic compounds and &lt;em&gt;is&lt;/em&gt; accepted by IUPAC, though not without some tweak. According to the rule of assigning cis- and trans- in &lt;a href=&quot;http://goldbook.iupac.org/C01092.html&quot; rel=&quot;nofollow&quot;&gt;gold book&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The appropriate reference plane of a double bond is perpendicular to that of the relevant σ-bonds and passes through the double bond. For a ring (the ring being in a conformation, real or assumed, without re-entrant angles at the two substituted atoms) it is the mean plane of the ring(s) … If there are more than two entities attached to the ring the use of cis and trans requires the definition of a reference substituent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the isomer of trichlorocyclopropane which has all three chlorine atoms on the same side, I can choose any one as the reference and the other two should thus be cis. The name should be r-1,c-2,c-3-trichlorocyclopropane. The other isomer is named r-1,c-2,t-4-trichlorocyclopropane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'm still not sure whether and how CIP rule could be used. &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Functional_Groups/Stereoisomers/Cycloalkane_Stereoisomer&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt;, which gives additional examples of c- and t- naming, merely states CIP could be used without substantiation.&lt;/p&gt;&#xA;" OwnerUserId="19088" LastActivityDate="2015-08-28T03:12:08.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35598" PostTypeId="2" ParentId="35583" CreationDate="2015-08-28T04:18:30.390" Score="1" Body="&lt;p&gt;You have 1 mole/liter of each reactant.  How many liters of each solution do you have?  That gives you the number of moles of reactants. They will create &lt;em&gt;way&lt;/em&gt; less than $2\cdot 10^6$ moles of water-you were counting the water that was already there.  Once you have the right number of moles of reactant, that is the number of moles of water produced.  Multiplying by your heat per mole gives $Q$.  Then divide by $mc_p$ to get $\Delta T$&lt;/p&gt;&#xA;" OwnerUserId="11344" LastActivityDate="2015-08-28T04:18:30.390" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35599" PostTypeId="2" ParentId="671" CreationDate="2015-08-28T05:43:19.490" Score="1" Body="&lt;p&gt;The mole is useful in chemistry because it relates two popular but inconsistent units of mass - the atomic mass unit (u) and the gram. One mole of atoms of an element with atomic mass x u, has a mass of x grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't use atomic mass units outside chemistry, so we also have no need for the mole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course raises another question &quot;why do chemists use atomic mass units?&quot;. Perhaps it's because they measure quantity in moles :P Really I think it's just an accident of history from the days when people thought atomic masses were all integer multiples of the mass of hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="19456" LastActivityDate="2015-08-28T05:43:19.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35600" PostTypeId="2" ParentId="35583" CreationDate="2015-08-28T07:14:21.913" Score="1" Body="&lt;p&gt;As Ross has said, $\Delta H$ is the same as $q$. This is true only under three conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pressure is constant;&lt;/li&gt;&#xA;&lt;li&gt;Only p-V work is performed;&lt;/li&gt;&#xA;&lt;li&gt;The process occurs in a closed system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The First Law of Thermodynamics states that $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w$. The expression for p-V work is $\mathrm{d}w = -p\,\mathrm{d}V$, giving $\mathrm{d}U = dq - p\,\mathrm{d}V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enthalpy, $H$, is defined as $H = U + pV$. At constant pressure, this means that $$\begin{align}&#xA;\mathrm{d}H &amp;amp;= \mathrm{d}U + p\,\mathrm{d}V + V\,\mathrm{d}p \\&#xA;&amp;amp;= \mathrm{d}U + p\,\mathrm{d}V \quad\quad \mathrm{(since}\,\mathrm{d}p = 0) \\&#xA;&amp;amp;= \mathrm{d}q - p\,\mathrm{d}V + p\,\mathrm{d}V \\&#xA;&amp;amp;= \mathrm{d}q&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating over the path of interest, you would find that $\Delta H = q$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Apart from that, I think you would also find that the number of moles of water produced is wrong. You will have to calculate it stoichiometrically, using the equation:&#xA;$$\ce{HCl + NaOH -&amp;gt; H2O + NaCl}$$&lt;/p&gt;&#xA;" OwnerUserId="16683" LastActivityDate="2015-08-28T07:14:21.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35602" PostTypeId="2" ParentId="32154" CreationDate="2015-08-28T08:28:54.087" Score="1" Body="&lt;p&gt;Watch this video from 1:50 onwards to see the reactions of sodium and potassium with water. &lt;a href=&quot;https://www.youtube.com/watch?v=H6ZDiRIvc2E&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=H6ZDiRIvc2E&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video, 0:56, shows  gasoline (petrol) burning on water. (Note that the batteries have nothing to do with it). &lt;a href=&quot;https://www.youtube.com/watch?v=Bs13jBWqFjA&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=Bs13jBWqFjA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17400" LastActivityDate="2015-08-28T08:28:54.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35603" PostTypeId="1" AcceptedAnswerId="35608" CreationDate="2015-08-28T09:06:51.487" Score="13" ViewCount="14374" Body="&lt;p&gt;This question was in my book. According to me CO should be polar as it should have a dipole moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I found that the $\sigma$-electron drift from C to O is almost nullified by the $\pi$-electron drift from O to C. What is about the $\pi$-electron drift towards the Carbon atom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Google it is given CO is polar, but my book says it is practically nonpolar ... now I am confused.&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="4945" LastEditDate="2015-08-28T15:40:50.440" LastActivityDate="2020-06-22T15:58:03.330" Title="Why is CO practically nonpolar?" Tags="&lt;physical-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;dipole&gt;" AnswerCount="2" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35607" PostTypeId="2" ParentId="35578" CreationDate="2015-08-28T10:11:25.933" Score="2" Body="&lt;p&gt;Your reasoning seems sound. For myself, I wonder why anyone would want to use this type of condenser at all. It has a very small bore which reduces the effective surface area. There are a number of other (better) types of condensers to choose from. Was in the lab for 20 years and never found a reason to use this type of condenser.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-08-28T10:11:25.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35608" PostTypeId="2" ParentId="35603" CreationDate="2015-08-28T10:12:23.430" Score="6" Body="&lt;p&gt;Well, it's just like you said: $\sigma$ electrons drift from C to O, because O is more electronegative, and $\pi$ electrons drift in the opposite direction, because... Look at it this way: C and O are both sp, each with a lone pair facing away from the other atom. They form a $\sigma$ bond and a $\pi$ bond, too. Now they want to form another $\pi$ bond, but C got no more electrons to use, so O has to throw in two of his own. That's the drift we are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, the two effects almost negate each other. The resulting dipole moment is so insignificant, I don't even remember which way it is.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-28T10:12:23.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35609" PostTypeId="2" ParentId="35589" CreationDate="2015-08-28T10:13:46.447" Score="1" Body="&lt;p&gt;Ions are classified into cation  and anion, cation is a positive ion and anion a negative ion. Cations are mostly from metal nature while anions are mostly from non-metals, so yeah there are non-metal ions!&lt;/p&gt;&#xA;" OwnerUserId="19056" LastActivityDate="2015-08-28T10:13:46.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35610" PostTypeId="1" AcceptedAnswerId="35616" CreationDate="2015-08-28T10:20:36.367" Score="0" ViewCount="161" Body="&lt;p&gt;Chlorobenzene will react with hydroxide to give phenol. But will the reaction on Chlorobenzene also work with methoxide, to give anisole? &lt;/p&gt;&#xA;" OwnerUserId="19398" LastEditorUserId="16622" LastEditDate="2015-08-28T13:01:17.803" LastActivityDate="2015-08-28T13:44:30.587" Title="Aromatic Elimination Addition" Tags="&lt;aromatic-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35611" PostTypeId="2" ParentId="35534" CreationDate="2015-08-28T10:21:15.247" Score="0" Body="&lt;p&gt;A 60 watt equivalent LED bulb uses about 9 watts of power, so we can conclude that an incandescent bulb produces at least 51 watts of heat energy for every 9 watts of light energy. &lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-08-28T10:21:15.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35612" PostTypeId="1" CreationDate="2015-08-28T10:39:22.553" Score="3" ViewCount="1485" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If a reaction has positive $\Delta G$ then reverse reaction has negative $\Delta G$. How can both occur simultaneously, e.g., in a reversible reaction such as dimerisation of $\ce{NO2}$ to form $\ce{N2O4}$? Do forward and backward reactions occur spontaneously? How can both directions be spontaneous at the same time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I read in some book that  Haber's process to form ammonia is spontaneous at a temperature below $170\ \mathrm{^\circ C}$ (the author proved this using $\Delta G=$ negative for spontaneous process). This means that above this temperature $\ce{N2}$ and $\ce{H2}$ shouldn't combine to form $\ce{NH3}$. But industrially Haber's process is performed at temperature around $400\ \mathrm{^\circ C}$. Isn't this a contradiction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19487" LastEditorDisplayName="user7951" LastEditDate="2015-08-28T16:15:34.247" LastActivityDate="2015-09-28T02:06:26.233" Title="Spontaneity of reversible reactions" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35614" PostTypeId="1" CreationDate="2015-08-28T12:06:53.693" Score="1" ViewCount="779" Body="&lt;p&gt;I found this formula for calculating the normality from the molarity:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Normality} = n \cdot \text{Molarity}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&#xA;&lt;span class=&quot;math-container&quot;&gt;$$n = \frac{\text{Molecular weight}}{\text{Equivalent weight}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in one book I found that if equivalent weight is equal to half of the molecular weight then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2N = M$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the given condition, I am confused if the above statement is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I following a factually correct book for solving problems on volumetric analysis?&lt;/p&gt;&#xA;" OwnerUserId="19450" LastEditorUserId="69977" LastEditDate="2020-05-07T14:26:01.400" LastActivityDate="2020-05-07T14:26:01.400" Title="Relationship between normality and molarity" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35615" PostTypeId="2" ParentId="35612" CreationDate="2015-08-28T12:11:20.410" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You seem to be mixing up thermodynamic and kinetic considerations. &lt;strong&gt;All&lt;/strong&gt; reactions occur spontaneously all the time, and their backward reactions too. If left alone for long enough, a system will eventually reach the point where the forward and backward reactions go at the same rate (and it's not like they both stop altogether!) - that's an equilibrium. Where that equilibrium will be is judged by $\Delta G$; if it is negative, then the equilibrium will be closer to the products.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As for ammonia, that's where the kinetics kicks in. At room temperature $\Delta G$ is in our favor, but the equilibrium would take &lt;em&gt;forever&lt;/em&gt; to reach. We raise the temperature so as to make the reaction faster, but that also shifts the equilibrium in the wrong direction. So we raise the &lt;em&gt;pressure&lt;/em&gt;, too. This does the trick: now the reaction goes the way we need, and at reasonable rate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-28T12:11:20.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35616" PostTypeId="2" ParentId="35610" CreationDate="2015-08-28T13:44:30.587" Score="0" Body="&lt;p&gt;Yes, It will.&#xA;It's a Nucleophilic Substitution reaction,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chlorobenzene + Sodium Methoxide --&gt; (Under extreme conditions ) Anisole &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extreme conditions because Aryl halides don't go Nucleophilic Substitution easily.&lt;/p&gt;&#xA;" OwnerUserId="9963" LastActivityDate="2015-08-28T13:44:30.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35617" PostTypeId="2" ParentId="35614" CreationDate="2015-08-28T14:24:57.850" Score="2" Body="&lt;p&gt;The &lt;strong&gt;normality of a solution is the molarity of the solution multiplied by the number of equivalents per mole&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{N} = \mathrm{{eq\over mol}}\cdot\mathrm{M}= \mathrm{{eq\over mol}}\cdot\mathrm{{mol\over L}}=\mathrm{{eq\over L}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Calculate the normality of a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$. There are 2 equivalents per mole, from the reaction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4 -&amp;gt; 2H+ + SO4^{2-}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields 2 moles of $\ce{H+}$ for every 1 mole of $\ce{H2SO4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the relation given above, we find that a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$ is 2 $\mathrm{N}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{{2\;eq\over mol}}\cdot\mathrm{1\;M}=2\;\mathrm{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One resource you might find useful in order to check your math is Sigma-Aldrich's &lt;a href=&quot;http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html&quot; rel=&quot;nofollow&quot;&gt;Normality &amp;amp; Molarity Calculator&lt;/a&gt;. Additionally, there are details relevant to the definitions of molarity and normality that you might find clearer that the ones you've conveyed in your question.&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-28T14:24:57.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35619" PostTypeId="2" ParentId="29368" CreationDate="2015-08-28T14:47:04.993" Score="1" Body="&lt;p&gt;I don't know if you are still interested in this question. But yes, the &lt;strong&gt;a&lt;sub&gt;1&lt;/sub&gt;&lt;/strong&gt; ion would have exact the same m/z as the immonium ion of all amino acids of the same type in the pepitde. Peptides typically fragment in low energy fragmentation methods such as CID by charg-directed mechanisms, which mainly produces y and b series ions. These ions can fragment further to lower b ions, a ions, or internal fragments and immonium ions. y and b ions are much easier to understand and we therefore mostly use these ions for interpretation of mass spectra.&lt;/p&gt;&#xA;" OwnerUserId="19491" LastActivityDate="2015-08-28T14:47:04.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35620" PostTypeId="2" ParentId="35603" CreationDate="2015-08-28T15:32:25.980" Score="17" Body="&lt;p&gt;Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please &lt;a href=&quot;https://chemistry.stackexchange.com/q/33379/4945&quot;&gt;see the question about the direction of the dipole moment&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment.&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{{}^{\ominus}\!:C#O:^{\oplus}} &amp;amp;&amp;amp; &#xA;\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &amp;amp;&amp;amp;&#xA;\text{Direction:}~\longleftarrow\hspace{-1ex}+&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately the explanation of the bonding in carbon monoxide is far from simple, and it is perfectly understandable that you are confused. The dipole moment can be very well explained with molecular orbital theory, which I have already &lt;a href=&quot;https://chemistry.stackexchange.com/q/30797/4945&quot;&gt;written about&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I will include a brief explanation at this point as I saw that the explanation on &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_monoxide#Bonding_and_dipole_moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; is from my point of view incomplete (not to say wrong).&lt;/p&gt;&#xA;&lt;p&gt;Let's first state the obvious: Oxygen is more electro-negative than carbon. Therefore one can expect that all bonding orbitals are polarised towards the oxygen. That is of course only true when we are staying in the Lewis bonding picture. Based on this one would expect that the lone pairs cancel each other and since all orbitals of the triple bond are polarised to the oxygen, the dipole moment should also point in that direction.&lt;/p&gt;&#xA;&lt;p&gt;This is obviously wrong, since the experiment tells us something different. There are ten valence electron in this molecule. They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements. The low lying orbitals are all polarised towards oxygen, because of electronegativity. The HOMO on the other hand is strongly polarised towards the carbon and is very large. In addition the nuclei contribute a dipole moment because of their asymmetry of charge.  With less positive charge on the carbon nucleus than on the oxygen nucleus, the dipole component of the nuclear charge contribution also has its negative end on carbon.  The combination of HOMO and nuclear dipoles in the reverse direction is larger than the combination of all the other occupied orbitals.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It is important to understand that it is often not possible to know about the properties of a compound simply by looking at a structure on paper. Very often the simple models we love and use break down easily for small molecules. A bonding situation has to be explained as is and not in terms of what would have been when we look at individual atoms and how does it change.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="17175" LastEditDate="2020-06-22T15:58:03.330" LastActivityDate="2020-06-22T15:58:03.330" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35621" PostTypeId="1" CreationDate="2015-08-28T16:38:59.750" Score="3" ViewCount="825" Body="&lt;p&gt;Why isn't benzene reduced all the way down to cyclohexane when performing a Birch reaction? (Why is it not reduced further?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for the reaction with sodium dissolved in liquid ammonia, why isn't an alkyne reduced fully to an alkane?&lt;/p&gt;&#xA;" OwnerUserId="19398" LastEditorUserId="7475" LastEditDate="2015-09-28T22:52:37.793" LastActivityDate="2015-09-28T23:04:55.523" Title="Why does the Birch reaction not yield fully saturated products?" Tags="&lt;reaction-mechanism&gt;&lt;aromatic-compounds&gt;&lt;reaction-control&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35624" PostTypeId="1" CreationDate="2015-08-28T19:47:34.313" Score="1" ViewCount="399" Body="&lt;p&gt;I'm struggling to interpret enthalpy of mixing data. Given this plot, &lt;a href=&quot;https://i.stack.imgur.com/lgSbi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lgSbi.png&quot; alt=&quot;enthalpy of mixing of ethanol and water&quot;&gt;&lt;/a&gt; from &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/je60034a008&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, say I want to calculate the heat released from mixing 0.29 mol of 7.7% ethanol with 0.71 mol of 28.7% ethanol to get 1 mol of 22.6% ethanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to say $\Delta h_m = 753 - (0.29*605 + 0.71*672) = 100 J/mol$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, in other words, can I subtract the molar weighted enthalpies of mixing at the initial states from the enthalpy of mixing at the final state to get the change of enthalpy of mixing from that process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems trivial, but something about it feels weird.&lt;/p&gt;&#xA;" OwnerUserId="4754" LastActivityDate="2015-08-28T19:47:34.313" Title="Reading enthalpy of mixing graph" Tags="&lt;physical-chemistry&gt;&lt;mixtures&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35625" PostTypeId="1" AcceptedAnswerId="35665" CreationDate="2015-08-29T01:24:07.093" Score="3" ViewCount="1116" Body="&lt;p&gt;Others sources typically say that the minimum voltage required to split water into hydrogen and oxygen is 1.229 V. Overvoltage is typically required in a practical device due to inefficiency leading to heat production. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, suppose that two electrodes are separated by a large distance.  This will result in a very low electric field strength (generally, electric field equals &lt;span class=&quot;math-container&quot;&gt;$V/d$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the distance of separation).  Is there a threshold electric field level required for electrolysis to take place?&lt;/p&gt;&#xA;" OwnerUserId="19499" LastEditorUserId="41328" LastEditDate="2019-05-11T18:00:45.640" LastActivityDate="2019-05-11T18:00:45.640" Title="Electric field required to split water into oxygen and hydrogen" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;electrolysis&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35626" PostTypeId="2" ParentId="35612" CreationDate="2015-08-29T01:35:55.810" Score="0" Body="&lt;p&gt;The sign of $\Delta G^0$ is not an indication of whether a reaction will occur spontaneously.  For a reversible reaction, if you start with a mixture of only reactants, the reaction will proceed to produce products until the system reaches equilibrium.  If you start with a mixture of only the products (i.e., the products of the forward reaction), the reverse reaction will proceed to produce the reactants (of the forward reaction) until the system reaches equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chet&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-08-29T01:35:55.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35628" PostTypeId="1" CreationDate="2015-08-29T05:17:10.607" Score="1" ViewCount="40" Body="&lt;p&gt;Adding inert gas under constant volume will not disturb equilibrium. But I want to know what would happen if inert gases are added at constant pressure.&#xA;If total pressure remains constant and inert gases are added, then mole fraction of any gas participating in equilibrium decreases and hence partial pressure decreases. &#xA;If $\Delta n$ is non zero, then $K$ changes. But how can $K$ change if temperature remains constant?&lt;/p&gt;&#xA;" OwnerUserId="17624" LastActivityDate="2015-08-29T05:17:10.607" Title="Addition of inert gas to equilibrium mixture under constant pressure" Tags="&lt;equilibrium&gt;&lt;gas-laws&gt;&lt;pressure&gt;" AnswerCount="0" CommentCount="4" ClosedDate="2015-08-29T13:11:53.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="35630" PostTypeId="2" ParentId="33761" CreationDate="2015-08-29T08:49:22.927" Score="4" Body="&lt;p&gt;Very interesting question and not very obvious. As can be seen from the picture, LAB's reduce only tertiary amides. It is stated in the publication(1) about LAB reductions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;LAB reagents do not&#xA;  reduce primary and secondary amides even in refluxing THF.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the same publication it is provided reason for different outcomes of reductions. They say that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We found that the outcome of these reductions depends on&#xA;  the sterics of the amide as well as of the LAB reagent&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means that if LAB and/or amide is sterically hindered it can't interact with nitrogen like in compound &lt;strong&gt;5&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lower part of the picture is described reaction where more hindered LAB leads to amine product &lt;strong&gt;9&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iTqj7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iTqj7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1)Pasumansky, L., Goralski, C. T., &amp;amp; Singaram, B. (2006). Lithium aminoborohydrides: powerful, selective, air-stable reducing agents. Organic process research &amp;amp; development, 10(5), 959-970.&lt;/p&gt;&#xA;" OwnerUserId="15733" LastActivityDate="2015-08-29T08:49:22.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35632" PostTypeId="2" ParentId="35527" CreationDate="2015-08-29T09:06:20.177" Score="4" Body="&lt;p&gt;The glow in the dark phenomena is called luminescence. It can be caused by chemical reactions, electrical energy, subatomic motions, or stress on a crystal. Three of the common mechanisms are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fluorescence: Light is emitted during the excitation with invisible&#xA;electromagnetic radiation.&lt;/li&gt;&#xA;&lt;li&gt;Chemoluminescence: A chemical reaction radiates energy in for of electromagnetic radiation (e.g. glow sticks, luziferase reaction,...)&lt;/li&gt;&#xA;&lt;li&gt;Phosphorescence: You can read about the quantum mechanical processes &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/14356007.a15_519/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. There is one interesting thing I want to point out. While the most phosphorescent materials are solid, in &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0957-4484/21/38/385604;jsessionid=815464952D7E6B6D6ED64FA7B89099C1.c1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper, they have reported the synthesis and afterglow of water-soluble ZnS:Cu,Co nanoparticles.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know Wikipedia is not at all a reference, but there you can find an overview of all types of luminescence.&lt;/p&gt;&#xA;" OwnerUserId="19491" LastEditorUserId="402" LastEditDate="2019-05-19T06:07:41.603" LastActivityDate="2019-05-19T06:07:41.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35633" PostTypeId="1" CreationDate="2015-08-29T09:20:42.050" Score="3" ViewCount="100" Body="&lt;p&gt;Following my question &lt;a href=&quot;https://physics.stackexchange.com/questions/202500/the-liouville-equation-and-the-bbgky-hierarchy/202512?noredirect=1#comment433262_202512&quot;&gt;here&lt;/a&gt; about the BBGKY hierarchy I was wondering does anyone have any experience in &lt;em&gt;using&lt;/em&gt; the method outlined. In particular how would I take an equation off the page and for some reduced phase space density or distribution function and start obtaining meaningful results and simulations? I understand that it is necessary to implement &lt;strong&gt;closure relations&lt;/strong&gt; along the way. If anyone has experience in doing this sort of thing, could you outline (qualitatively) what steps I need to take etc. &lt;/p&gt;&#xA;" OwnerUserId="9690" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:13.940" LastActivityDate="2015-08-29T09:20:42.050" Title="What closure relations can we use in molecular modelling?" Tags="&lt;physical-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35634" PostTypeId="1" CreationDate="2015-08-29T10:32:51.503" Score="0" ViewCount="4684" Body="&lt;p&gt;Why do we compare acidic strength of oxides of nitrogen even though they don't have hydrogen?&lt;/p&gt;&#xA;" OwnerUserId="19504" LastActivityDate="2016-05-11T15:27:27.430" Title="Acidic strength of oxides of nitrogen" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35635" PostTypeId="1" AcceptedAnswerId="35637" CreationDate="2015-08-29T12:34:14.317" Score="2" ViewCount="311" Body="&lt;p&gt;Since dialdehydes have a carbonyl group on either end of an alkane, is it reasonable to call carbon dioxide &quot;methanedial&quot;, or is it that ethanedial (glyoxal) is the first dialdehyde? &lt;/p&gt;&#xA;" OwnerUserId="19507" LastEditorDisplayName="user7951" LastEditDate="2015-08-29T13:12:07.407" LastActivityDate="2015-08-29T13:12:07.407" Title="Is it wrong to call carbon dioxide &quot;methanedial&quot;?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35636" PostTypeId="1" AcceptedAnswerId="35638" CreationDate="2015-08-29T12:39:07.383" Score="2" ViewCount="423" Body="&lt;p&gt;According to the definition, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an alcohol is any organic compound in which the hydroxyl functional&#xA;  group (-OH) is bound to a saturated carbon atom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Monosaccharides such as glucose contain several hydroxyl groups. Does that mean all carbohydrates are a subset of alcohols?&lt;/p&gt;&#xA;" OwnerUserId="19506" LastEditorUserId="865" LastEditDate="2015-08-29T15:39:05.950" LastActivityDate="2015-09-14T14:26:29.857" Title="Are all carbohydrates a subset of alcohols?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;&lt;alcohols&gt;&lt;carbohydrates&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35637" PostTypeId="2" ParentId="35635" CreationDate="2015-08-29T13:10:42.090" Score="10" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-66.6.1.1.1&lt;/strong&gt; Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Examples:&lt;br /&gt;&#xA;pentanal (&lt;span class=&quot;math-container&quot;&gt;$\ce{CH3-CH2-CH2-CH2-CHO}$&lt;/span&gt;)&lt;br /&gt;&#xA;pentanedial (&lt;span class=&quot;math-container&quot;&gt;$\ce{OHC-CH2-CH2-CH2-CHO}$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;In this context, the term ‘aldehyde’ is unambiguously defined:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-66.6.0&lt;/strong&gt; Introduction.&lt;/p&gt;&#xA;&lt;p&gt;The class name ‘aldehyde’ traditionally refers to compounds containing the &lt;span class=&quot;math-container&quot;&gt;$\ce{-CH=O}$&lt;/span&gt; group attached to a carbon atom. However, nomenclature for aldehydes has been extended to describe a &lt;span class=&quot;math-container&quot;&gt;$\ce{-CHO}$&lt;/span&gt; group attached to a heteroatom.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Since carbon dioxide (&lt;span class=&quot;math-container&quot;&gt;$\ce{O=C=O}$&lt;/span&gt;) does not contain a &lt;span class=&quot;math-container&quot;&gt;$\ce{-CH=O}$&lt;/span&gt; group, it is not considered an aldehyde and thus Section P-66.6.1.1.1 does not apply to carbon dioxide.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, the shortest dialdehyde is ethanedial (&lt;span class=&quot;math-container&quot;&gt;$\ce{OHC-CHO}$&lt;/span&gt;). (Note that the preferred IUPAC name is ‘oxaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-29T13:10:42.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35638" PostTypeId="2" ParentId="35636" CreationDate="2015-08-29T15:02:54.077" Score="2" Body="&lt;p&gt;Whenever you see -OH is bound to a saturated carbon atom in organic chemistry it is Alcohol group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So answer is &lt;strong&gt;YES&lt;/strong&gt;. All carbohydrates are  subset of alcohols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though in linear forms they also have aldehyde or ketone group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jKAWI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jKAWI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="9961" LastEditDate="2015-09-14T14:26:29.857" LastActivityDate="2015-09-14T14:26:29.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35640" PostTypeId="1" CreationDate="2015-08-29T16:12:32.300" Score="1" ViewCount="483" Body="&lt;p&gt;In hyperconjugation, it is known that form an $\alpha$ hydrogen, one $\sigma$ electron goes to the corresponding carbon carbon bond, and there is apparently or practically, no bond between the hydrogen and carbon. I read in a textbook, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;although one $\ce{C-H}$ bond of the alkyl is shown to be broken in each hyperconjugative structure, $\ce{H+}$ is never free from the rest of the molecule. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But then, still there is no bond. Is there any specific reason why the $\ce{H+}$ ion remains with the molecule &lt;strong&gt;&lt;em&gt;though there is no bond attaching the ion to the rest of molecule&lt;/em&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorUserId="10185" LastEditDate="2015-08-30T17:31:38.140" LastActivityDate="2016-05-23T17:11:41.803" Title="Why the H+ ion remains with the molecule even though there is no bond between it and hyperconjugative structure?" Tags="&lt;organic-chemistry&gt;&lt;resonance&gt;&lt;hyperconjugation&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35641" PostTypeId="1" AcceptedAnswerId="67003" CreationDate="2015-08-29T17:20:29.863" Score="19" ViewCount="43978" Body="&lt;p&gt;According to my books the bond order of $\ce{CO+}$ is  $3.5$. But shouldn't it be $2.5$? On googling this, I found the following &lt;a href=&quot;https://chemistry.stackexchange.com/q/9510/17988&quot;&gt;answer&lt;/a&gt; that is on Stack Exchange but its only talks about the bond length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand why it is $3.5$ as I am in class 11.&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.787" LastActivityDate="2017-07-12T13:28:50.730" Title="Why is the bond order of CO+ 3.5?" Tags="&lt;bond&gt;&lt;molecular-orbital-theory&gt;&lt;theoretical-chemistry&gt;" AnswerCount="4" CommentCount="1" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35642" PostTypeId="1" CreationDate="2015-08-29T17:43:00.860" Score="6" ViewCount="2776" Body="&lt;p&gt;We find in the book &lt;strong&gt;&lt;em&gt;Minerals and their Localities&lt;/em&gt;&lt;/strong&gt; by Jan H Bernard and Jaroslav Hyrsl   several chemical formulas containing a square. What does this square means? An example: &lt;em&gt;Franciscanite&lt;/em&gt; $\ce{Mn3 (V,\square)(O,OH)3 (SiO4)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your answer for  I have been  looking for answer for a long time.&lt;/p&gt;&#xA;" OwnerUserId="19509" LastEditorUserId="865" LastEditDate="2015-08-29T23:33:46.627" LastActivityDate="2015-08-29T23:33:46.627" Title="What does a square mean in a chemical formula?" Tags="&lt;inorganic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35645" PostTypeId="1" CreationDate="2015-08-29T18:21:30.543" Score="0" ViewCount="113" Body="&lt;p&gt;I have read through Barron's SAT chemistry prep book, in the topic of electrochemistry, the book says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The movement of electrons through the wire must be balanced by a movement of ions in the solution.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It explains that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anions move toward the anode to replace the negatively charged electrons that are moving away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can someone help explain these two statements? I don't quite get this concept.&lt;/p&gt;&#xA;" OwnerUserId="19511" LastEditorUserId="7448" LastEditDate="2015-08-29T18:45:00.000" LastActivityDate="2016-12-05T05:43:53.030" Title="Why must the movement of electrons through the wire be balanced by a movement of ions in the solution?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35646" PostTypeId="1" AcceptedAnswerId="35653" CreationDate="2015-08-29T21:13:42.290" Score="0" ViewCount="60" Body="&lt;p&gt;A textbook states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fats also provide an efficient way to store energy over long time&#xA;  periods, since they contain over twice as much energy per gram as&#xA;  carbohydrates&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's the chemistry beyond this? What topic should I learn in order to understand how fats and carbohydrates store energy?&lt;/p&gt;&#xA;" OwnerUserId="19506" LastActivityDate="2015-08-29T23:46:06.343" Title="Energy in molecules" Tags="&lt;energy&gt;&lt;carbohydrates&gt;&lt;fats&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35647" PostTypeId="2" ParentId="35621" CreationDate="2015-08-29T22:04:56.737" Score="-1" Body="&lt;p&gt;The radical anion after the first step of the Birch reduction needs to be conjugated with (=stabilised by) another pi-system, otherwise it simply will not be formed from a pair of sp2 carbon atoms.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastActivityDate="2015-08-29T22:04:56.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35648" PostTypeId="1" AcceptedAnswerId="35656" CreationDate="2015-08-29T23:20:51.857" Score="3" ViewCount="3807" Body="&lt;p&gt;For an AP chem lab dealing with determining the amount of citric acid in orange juice by titrating with NaOH, one of the discussion questions goes like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Choose an amount of beverage to be titrated that will require at least 10 but less than 20 mL of titrant. Explain why this range of titrant is optimal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question makes no sense to me. Why would there be an &quot;optimal range&quot; of titrant? I would think that I should fill the buret to full capacity if enough titrant is available. Would it not be better to have excessive titrant than not enough of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We completed the lab, and the titration required 23.05 mL of 0.1 M NaOH solution to titrate 20.00 mL of orange juice. If we had followed the instructions, we would have added less orange juice, and used less NaOH, but what is the benefit in doing so? Budget cuts? If it was about not wasting solution, why would there be a minimal requirement of 10 mL of titrant?    &lt;/p&gt;&#xA;" OwnerUserId="19014" LastEditorUserId="69830" LastEditDate="2021-05-07T04:23:34.027" LastActivityDate="2021-05-07T04:23:34.027" Title="Is there a certain amount of titrant that is ideal?" Tags="&lt;solutions&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35650" PostTypeId="1" AcceptedAnswerId="35720" CreationDate="2015-08-29T23:30:44.697" Score="3" ViewCount="1283" Body="&lt;p&gt;This question just popped in my head as I was working with electric force problems involving atoms. Why is the amount in a mole equal to the Avogadro's number? Do atoms/molecules/etc. tend to group in this exact amount? Why do we use this number as a unit of measurement?&lt;/p&gt;&#xA;" OwnerUserId="19516" LastActivityDate="2015-09-01T03:23:40.930" Title="Why is the amount in a mole equal to the Avogadro's number?" Tags="&lt;mole&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35651" PostTypeId="2" ParentId="35642" CreationDate="2015-08-29T23:33:14.870" Score="7" Body="&lt;p&gt;That square refers to a vacancy (or framework vacancy) in the crystal structure. Here's one reference from &lt;a href=&quot;https://books.google.com/books?id=NlaxmzJobLYC&amp;amp;pg=PA233&amp;amp;lpg=PA233&amp;amp;dq=mineral+formula+denote+vacancy&amp;amp;source=bl&amp;amp;ots=2cL8prSB-b&amp;amp;sig=tBw3neAD_eLbfH9VZZQx5rE9rKk&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCoQ6AEwAmoVChMI-uSr6o3PxwIVwY8NCh0A1wLE#v=onepage&amp;amp;q=mineral%20formula%20denote%20vacancy&amp;amp;f=false&quot;&gt;Advances in Electronic Ceramics. Ceramic Engineering and Science Proceedings. Volume 28, Issue 8. Ed. C. Randall ${\it et\;al.}$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, a framework vacancy is a feature of some inorganic materials in which a &quot;guest atom&quot; (or more than one) can fit into the structure &lt;em&gt;or&lt;/em&gt; denotes a structure that is missing interstitial or packing atoms (as outlined &lt;a href=&quot;https://books.google.com/books?id=9gbZBAAAQBAJ&amp;amp;pg=PT18&amp;amp;lpg=PT18&amp;amp;dq=square+brackets+in+chemical+formula+of+a+mineral&amp;amp;source=bl&amp;amp;ots=IilLvtLF6e&amp;amp;sig=4nRmSFW5XDpIBJyi9OaINOoZJ3U&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CDMQ6AEwBGoVChMImcmDqrTPxwIVBdGACh3o4AGz#v=onepage&amp;amp;q=vacancy&amp;amp;f=false&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://webmineral.com/data/Franciscanite-VIII.shtml&quot;&gt;Here's the info on your mineral&lt;/a&gt; where you can see that the square is instead denoted by square brackets [].&lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-08-29T23:33:14.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35652" PostTypeId="1" AcceptedAnswerId="35659" CreationDate="2015-08-29T23:37:12.493" Score="0" ViewCount="3798" Body="&lt;p&gt;I have a homework problem as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solution was prepared by mixing $25.0\ \mathrm{mL}$ of $0.208\ \mathrm{M}$ $\ce{Fe(NO3)3}$ and $4.50\ \mathrm{mL}$ of $0.00190\ \mathrm{M}$ $\ce{KSCN}$ and diluting to $50.00\ \mathrm{mL}$ with solvent, $0.10\ \mathrm{M}$ $\ce{HNO3}$. Calculate the initial concentrations of the thiocyanate ion and the iron(III) ion before any reaction takes place.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, how do I proceed with the question? At first I thought that it should just be the same concentration as the solutions, $0.208\ \mathrm{M}$ and $0.00190\ \mathrm{M}$, because there is only one thiocyanate and one iron(III) ion in the solutions. But the fact that we have an $\ce{HNO3}$ solution confuses me. Am I somehow supposed to use $M_1 V_1 = M_2 V_2$? &lt;/p&gt;&#xA;" OwnerUserId="12860" LastEditorUserId="418" LastEditDate="2016-11-07T05:39:48.370" LastActivityDate="2016-11-07T05:39:48.370" Title="How to calculate concentrations in a diluted mixture of multiple solutions?" Tags="&lt;homework&gt;&lt;aqueous-solution&gt;&lt;concentration&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35653" PostTypeId="2" ParentId="35646" CreationDate="2015-08-29T23:39:04.190" Score="1" Body="&lt;p&gt;Fats contain more hydrogen and carbon than carbohydrates. Look at the picture below of the carbohydrate. Notice all the oxygen atoms? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tse1.mm.bing.net/th?&amp;amp;id=JN.kQKbzP5TAGgr7um%2bQI6f4Q&amp;amp;w=300&amp;amp;h=300&amp;amp;c=0&amp;amp;pid=1.9&amp;amp;rs=0&amp;amp;p=0&amp;amp;r=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carbohydrates http://tse1.mm.bing.net/th?&amp;amp;id=JN.kQKbzP5TAGgr7um%2bQI6f4Q&amp;amp;w=300&amp;amp;h=300&amp;amp;c=0&amp;amp;pid=1.9&amp;amp;rs=0&amp;amp;p=0&amp;amp;r=0&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at the picture below of a triglyceride, or basically fat. Notice all the carbon and hydrogen atoms?&#xA;&lt;a href=&quot;http://tse1.mm.bing.net/th?&amp;amp;id=JN.tBMcQZj7LoB48KotpahRzg&amp;amp;w=300&amp;amp;h=300&amp;amp;c=0&amp;amp;pid=1.9&amp;amp;rs=0&amp;amp;p=0&amp;amp;r=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Triglyceride http://tse1.mm.bing.net/th?&amp;amp;id=JN.tBMcQZj7LoB48KotpahRzg&amp;amp;w=300&amp;amp;h=300&amp;amp;c=0&amp;amp;pid=1.9&amp;amp;rs=0&amp;amp;p=0&amp;amp;r=0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon and hydrogen can be reduced, or can lose electrons, by oxygen through oxidization.  Hence, fuels like gasoline or natural gas are mostly hydrogen and carbon. The oxygen in carbohydrates cannot be reduced, but instead contribute to the oxygen used to reduce carbon and hydrogen. Thus, the oxygen in carbs and triglycerides are worthless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same reason why ethanol, a renewable alternative to gasoline, is less powerful than gasoline. Ethanol is an alcohol, so it contains an oxygen atom within its molecule. Gasoline is mostly alkanes, alkenes, and alkynes, all of which do not have oxygen within their formulas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to learn more on this subject, I would suggest learning more on oxidation and reduction, and on thermochemistry. &lt;/p&gt;&#xA;" OwnerUserId="19014" LastEditorUserId="19014" LastEditDate="2015-08-29T23:46:06.343" LastActivityDate="2015-08-29T23:46:06.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35654" PostTypeId="1" AcceptedAnswerId="35663" CreationDate="2015-08-29T23:57:35.630" Score="5" ViewCount="1018" Body="&lt;p&gt;Video: &lt;a href=&quot;https://www.youtube.com/watch?v=_qFN1hDZwp8&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=_qFN1hDZwp8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose not to view the video, in the first demonstration a paper towel is soaked in isopropanol. As expected, the paper towel burns. However, in the second demonstration the paper towel is soaked in isopropanol and water. Surprisingly, the towel burns without being damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My chemistry professor tells me that it has something to do with the energy cost of a phase change $\left[\ce{H2O}\ (l) \longrightarrow \ce{H2O}\ (g)\right]$; however, I am still very confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please elaborate? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="19147" LastEditorUserId="11367" LastEditDate="2015-08-30T02:22:19.763" LastActivityDate="2015-08-30T11:51:10.143" Title="Paper Towel Soaked with Isopropanol/Water Mixture Burns without Damaging the Paper. Why?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;&lt;combustion&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35656" PostTypeId="2" ParentId="35648" CreationDate="2015-08-30T00:37:02.090" Score="5" Body="&lt;p&gt;In a perfect world there would be no optimal range. We do however live in a world full of errors and practical considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why use over a minimum amount?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume for demonstration, that each drop has a volume of 0.05 mL. Of course this &lt;em&gt;is not&lt;/em&gt; an accurate value as various factors would affect this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The titration has reached its end point when &lt;em&gt;one last drop&lt;/em&gt; changes the colour. If your titre value is very small e.g. 1 mL, then each drop is 5% of your titre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only part of the drop is needed to reach the end point then all the extra part of the drop is inaccuracy, which in this scenario could be up to 5%! If you have the misfortune of accidentally adding one drop too many you are introducing an even larger error. This will impact concentration calculations based on the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why not just use 'a lot' then?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of things here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) As noted by Mithoron, this would be a waste. The accuracy gained from requiring additional titrant will level off rather rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If you use more than one full burette's worth there is an error penalty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Using extra titrant is a waste of your time. Consider a 50 mL burette, if it needs to be filled up after each titration, that would be inefficient. I'm assuming you're doing repeats so this has real merit.&lt;/p&gt;&#xA;" OwnerUserId="15191" LastActivityDate="2015-08-30T00:37:02.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35657" PostTypeId="1" CreationDate="2015-08-30T02:11:43.790" Score="4" ViewCount="3268" Body="&lt;p&gt;Is there a way to test if water has toxic heavy metals. Samples to both the water and people who drink it are available. Preferably, it should be done with at hold materials. I could also borrow materials from a high school chemistry lab potentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already tried boiling, but that just showed that it is hard water, which luckily isn't dangerous. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="7219" LastActivityDate="2015-10-19T20:17:09.037" Title="How to test for heavy metal in water?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;home-experiment&gt;&lt;metal&gt;&lt;toxicity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35658" PostTypeId="1" AcceptedAnswerId="35660" CreationDate="2015-08-30T02:25:04.427" Score="6" ViewCount="419" Body="&lt;p&gt;One thing which has bothered me for a while about the idea of quantized energy levels is that it's clear just how quantized they are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give an example. If I was looking the energy some electromagnetic wave would need to an $\ce{O-H}$ in water from the 0th vibrational energy level to the 1st vibrational energy level, it would be around $\mathrm{3750~cm^{-1}}$. Now, for some electromagnetic wave to cruise by and give enough energy to make that transition, how exact must the energy of that wave be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning, does it need to have exactly $\mathrm{3750~cm^{-1}}$ of energy? Well, what if the transition was actually $\mathrm{3750.3~cm^{-1}}$. Then that wave doesn't have enough energy anymore, and the likelihood of some wave having exactly $\mathrm{3750.3~cm^{-1}}$ seems a lot less likely than just being very close to that value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens in actual transitions that the molecule only takes the exact amount of energy it needs? And what happens to whatever leftover energy there might be from a wave that had enough energy to excite this water to it's first vibrational energy level? Is there ever leftover energy like I just mentioned? Also, do two waves of say $\mathrm{1875~cm^{-1}}$ each ever meet up and excite this same transition? This to me also seems so unlikely because that would involve constructive interference exactly of the two waves.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="10185" LastEditDate="2015-08-30T17:23:05.167" LastActivityDate="2015-08-30T17:23:05.167" Title="Just how quantized are energy levels?" Tags="&lt;physical-chemistry&gt;&lt;quantum-chemistry&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35659" PostTypeId="2" ParentId="35652" CreationDate="2015-08-30T02:58:23.093" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Am I somehow supposed to use &lt;span class=&quot;math-container&quot;&gt;$M_1V_1=M_2V_2$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Yep!&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The problem has given you three different solutions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Iron(III) nitrate&lt;/li&gt;&#xA;&lt;li&gt;Potassium thiocyanate&lt;/li&gt;&#xA;&lt;li&gt;Nitric acid&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;You know the concentrations of each, and you know (or can calculate) the volumes of each that are mixed to yield the final solution.  What you'll actually be using in the calculation are three different equations:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M_{1i}V_1 = M_{1f}V_f$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M_{2i}V_2 = M_{2f}V_f$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M_{3i}V_3 = M_{3f}V_f$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M_{\#i}$&lt;/span&gt; is the initial concentration of each of the three chemicals in their separate starting solutions, and &lt;span class=&quot;math-container&quot;&gt;$M_{\#f}$&lt;/span&gt; is the final concentration of each after mixing, that the problem is asking for (except for the &lt;span class=&quot;math-container&quot;&gt;$\ce{HNO3}$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-08-30T02:58:23.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35660" PostTypeId="2" ParentId="35658" CreationDate="2015-08-30T03:13:55.423" Score="3" Body="&lt;p&gt;First off, let's think of the light as photons, instead of waves. That makes it clear that yes, there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Two-photon_absorption&quot; rel=&quot;nofollow noreferrer&quot;&gt;two photon&lt;/a&gt; and multi-photon processes. As you surmise, the selection rules for these are much more complicated than a single-photon process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your main question is about &quot;how exact&quot; an absorption needs to be. In reality, it depends on the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Atomic Absorption Spectra&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For atoms, there are no vibrational or rotational transitions, only electronic (or perhaps nuclear). So atomic spectra are indeed very tight lines, e.g., the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rydberg_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rydberg formula&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Diatomic Molecules&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have one vibration and some rotational levels. The latter are important, since you can go up or down some rotational levels with a vibrational transition. (And if you have an electronic transition, you can usually go up vibrational and/or rotational levels.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the spectra is a bit more complex than single lines. For example you can do &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotational-vibrational&lt;/a&gt; spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/33dbE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/33dbE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the individual transitions are still pretty sharp - not completely lines, but $0.3\:\mathrm{cm^{-1}}$ probably won't match any particular transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Larger Molecules&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have very small spacing in the rotational lines, so if you take a vibrational absorption spectra (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared_spectroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;IR spectra&lt;/a&gt;) you'll see that most molecules have some width to the absorption peaks, so $\pm 0.3\:\mathrm{cm^{-1}}$ will still fit within the peak, just not as strongly.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-08-30T03:13:55.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35661" PostTypeId="1" AcceptedAnswerId="35662" CreationDate="2015-08-30T03:32:17.390" Score="2" ViewCount="2588" Body="&lt;p&gt;When the spark plug of a car ignites the pressurized air/fuel mixture, could you instead send the exhaust to another system within the car to be Recycled? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's a great idea that could drastically lower pollution, that is if it's possible.&lt;/p&gt;&#xA;" OwnerUserId="19520" LastEditorUserId="24093" LastEditorDisplayName="user15489" LastEditDate="2016-07-23T13:41:10.763" LastActivityDate="2016-07-23T13:41:10.763" Title="Can Fuel Exhaust Be Recycled" Tags="&lt;fuel&gt;&lt;green-chemistry&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35662" PostTypeId="2" ParentId="35661" CreationDate="2015-08-30T04:00:57.057" Score="7" Body="&lt;p&gt;One of the major components of exhaust is carbon dioxide, produced by the combustion of hydrocarbons - if in complete combustion, follows the general reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{C_{x}H_{y} + O2 -&amp;gt;CO2 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible emissions include $\ce{NO_{x}}$, particulate matter, carbon monoxide, VOCs etc. The amounts of these emissions vary from car to car (due to a lot of factors, e.g. age, wear-and-tear etc). According to a self study by Audi, in the document &lt;a href=&quot;http://www.volkspage.net/technik/ssp/ssp/SSP_230.pdf&quot;&gt;Motor Vehicle Exhaust Emissions&lt;/a&gt;, on page 6, they determined that asides from $\ce{N2}$, the main emissions were $\ce{CO2}$ and $\ce{H2O}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this answer, the focus will be on the possibility of recycling the $\ce{CO2}$ emissions (and linking to the $\ce{H2O}$ emissions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very question is the subject of the Scientific American article &lt;a href=&quot;http://www.scientificamerican.com/article/turning-carbon-dioxide-back-into-fuel/&quot;&gt;Reverse Combustion: Can $\ce{CO2}$ Be Turned Back into Fuel?&lt;/a&gt;, the articles details that there are several research groups looking at this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of examples of ideas and experiments that have been put forward include:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the 1990s a graduate student named Lin Chao at Princeton University decided to bubble carbon dioxide into an electrochemical cell. Using cathodes made from the element palladium and a catalyst known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyridinium&quot;&gt;pyridinium&lt;/a&gt;—a garden variety organic chemical that is a by-product of oil refining—he discovered that applying an electric current would assemble methanol from the $\ce{CO2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Inspired by the natural process that removes $\ce{CO2}$ from the atmosphere - photosynthesis, continued research has sought to emulate this - essentially, the current research is effectively looking at a possible solution by taking&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\ce{CO2}$, water, sunlight and an appropriate catalyst and generate an alcoholic fuel&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The research reported in the Scientific American article and the National Geographic article &lt;a href=&quot;http://news.nationalgeographic.com/news/energy/2011/08/110811-turning-carbon-emissions-into-fuel/&quot;&gt;Carbon Recycling: Mining the Air for Fuel&lt;/a&gt; go along the lines of extracting the excess $\ce{CO2}$ from the air rather than directly from the car's engine or exhaust. The reason for this is the current practicality, with the National Geographic reporting that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Sandia prototype’s solar collector has an area of about 20 square meters (215 square feet) for a reactor the size of a beer keg, Stechel says. About 300,000 acres (121,400 hectares) of mirrors would be required to collect enough sunshine to make the equivalent of 1 million barrels of oil per day, she says. (The world currently consumes about 86 million barrels per day of petroleum and other liquid fuels, including biofuels.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The research shows that, yes it is theoretically possible and practically possible on a larger scale to recycle exhaust gases. &lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-08-30T04:38:17.253" LastActivityDate="2015-08-30T04:38:17.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35663" PostTypeId="2" ParentId="35654" CreationDate="2015-08-30T04:04:38.273" Score="7" Body="&lt;p&gt;&lt;strong&gt;I think your professor is incorrect, and it has to do with the thermodynamics and kinetics of the combined vaporization/combustion process.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of vaporization of water $\left(40.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$&lt;sup&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy_of_vaporization#Other_common_substances&quot; rel=&quot;nofollow noreferrer&quot;&gt;cite&lt;/a&gt;&lt;/sup&gt; is considerably less than the heat of combustion of isopropanol $\left(1830\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$&lt;sup&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_of_combustion#Lower_heating_value_for_some_organic_compounds_.28at_25.C2.A0.C2.B0C_.5B77.C2.A0.C2.B0F.5D.29.5Bcitation_needed.5D&quot; rel=&quot;nofollow noreferrer&quot;&gt;cite&lt;/a&gt;&lt;/sup&gt;.  These numbers indicate that approximately $45$ water molecules must evaporate for each isopropanol molecule combusted in order to hold the temperature constant. This  corresponds to a $\sim2\%$ mol/mol solution, which (a) is not consistent with the $\sim50\%$ v/v solution he actually used, and (b) would probably be too dilute to burn at room temperature the way it does in the video. (Someone more motivated than I could probably confirm this using &lt;a href=&quot;http://www.engineeringtoolbox.com/ispropanol-water-d_988.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;flash point data&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I think, instead, the behavior depends on the following:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vapor%E2%80%93liquid_equilibrium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vapor-liquid equilibrium&lt;/a&gt; stipulates that a liquid phase can never be at a temperature greater than its boiling point$^\dagger$.&lt;/li&gt;&#xA;&lt;li&gt;It is the vapor of combustible substances that actually burns, not the liquid (see, e.g., this &lt;a href=&quot;https://www.osha.gov/dte/library/flammable_liquids/flammable_liquids.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;OSHA training document&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Up to a concentration of about $60\%$ alcohol, the vapor phase in equilibrium with a water-isopropanol liquid mixture has a considerably higher concentration of isopropanol as compared to the liquid (&lt;a href=&quot;http://www.separationprocesses.com/Distillation/DT_Data/PD_054.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4A4Hk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4A4Hk.gif&quot; alt=&quot;Phase diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;For the present purpose, to read the diagram choose a temperature on the $y$-axis and track across the $x$-axis, noting where you cross the red and blue lines.  These correspond to the equilibrium liquid and gas phase compositions, respectively, at that temperature, given a mixture that's not too lean or too rich in isopropanol.  For example, take $82\ ^\circ\mathrm C$: you hit the red line (liquid) at about $20\%$ isopropanol, and the blue line (vapor) at about $53\%$ isopropanol.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the boiling point of isopropanol is &lt;a href=&quot;https://en.wikipedia.org/wiki/Isopropyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;$82.6\ ^\circ\mathrm C$&lt;/a&gt;, #1 means that as long as there's liquid left soaked into the paper (even $100\%$ isopropanol!), &lt;strong&gt;&lt;em&gt;the paper can't burn&lt;/em&gt;&lt;/strong&gt;.  You can actually see this in the video: he ignites the paper at around 0:57, but it doesn't actually start to char until about 1:08, ten seconds later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the paper does then start to char in spots is that those spots have locally run out of liquid isopropanol to keep them cool.  There remains a cloud of burning gaseous isopropanol all around the paper (clearly!), which is more than enough to burn the now-dry paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there's water mixed with the isopropanol, though, the vapor-liquid equilibrium comes into play.  Per point #3, based on the phase diagram, the isopropanol likes to be in the gas phase a lot more than the water, so once the vapor is ignited and burning (point #2), the heat produced tends to evaporate more of the alcohol than the water.  This feeds the flame for a while, but eventually it sputters out, with plenty of water left on the paper to keep it cooled well below its char point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\dagger$ Mixtures of liquids technically don't have boiling points, as their behavior is &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;more complex&lt;/a&gt;, but the concept is useful for making the point.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastEditorUserId="11367" LastEditDate="2015-08-30T11:51:10.143" LastActivityDate="2015-08-30T11:51:10.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35664" PostTypeId="2" ParentId="28184" CreationDate="2015-08-30T04:24:55.357" Score="1" Body="&lt;p&gt;One way you can do this is by looking at the packing structure of the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if you look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Tungsten&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, you see that Tungsten has a body-centered cubic crystal structure. This means that in each unit cell, there are going to be two Tungsten atoms. We can then predict the density of a perfect Tungsten crystal lattice using some geometry and unit conversion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I'll give you an equation which you can prove to yourself quite easily so I won't go into that. Density of a crystal is: $$\rho=\frac{n*M}{N_A*V}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, $n$ is the number of atoms in the unit cell, $M$ is the molar mass of the atom, $N_A$ is Avogadro's number, $V$ is the volume of the unit cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for Tungsten this comes out to be $$\rho=\frac{2*183.83 g*mol^{-1}}{6.022*10^{23}*(\frac{4*139*10^{-10}cm}{\sqrt{3}})^3}=18.45\frac{g}{cm^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental density of Tungsten is $19.33 \frac{g}{cm^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is usually a little better than that, but still quite close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information you need to do this calculation which isn't on a periodic table is the packing structure and the atomic radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something which is noteworthy is the atomic-packing-factor, $APF$, which comes from finding the ratio of volume of the atoms to volume of the unit cell and represents how much space the atoms fill in the cube, or how efficient the structure is at packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the body-centered cubic (BCC), $$APF=\frac{2*\frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{3}})^3}=0.68$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means BCC, takes op 68% of the total available space per unit cell for equal sized spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_packing_factor&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; if you want more info on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer the actual question which how do we find a trend with all this, we now know that density depends on radius, which we already have a trend for, molar mass, which also has a very simple trend, and packing structure, which is the real unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this from &lt;a href=&quot;https://en.wikipedia.org/wiki/Crystal_structure#Lattice_systems&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; page,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the resonating valence bond theory, the factors that determine the choice of one from among alternative crystal structures of a metal or intermetallic compound revolve around the energy of resonance of bonds among interatomic positions. It is clear that some modes of resonance would make larger contributions (be more mechanically stable than others), and that in particular a simple ratio of number of bonds to number of positions would be exceptional. The resulting principle is that a special stability is associated with the simplest ratios or &quot;bond numbers&quot;: 1/2, 1/3, 2/3, 1/4, 3/4, etc. The choice of structure and the value of the axial ratio (which determines the relative bond lengths) are thus a result of the effort of an atom to use its valency in the formation of stable bonds with simple fractional bond numbers.&#xA;  which I don't actually understand but seems to explain why certain lattices are chosen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, using the fact that radius decreases going right and molecular weight increases going right, we would predict that density would increase uniformly across the periodic table for elemental metals, except that various metals pack in different ways. Hexagonal Close Packed is the most efficient packing system, so I wouldn't be surprised to find that being associated with many high density metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that gives a good idea of how there is sort of a trend, but also why no trend is truly there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out which has the highest density, I would start by figuring out which pack in a Hexagonal Close-Packed structure as that is the most efficient packing structure with an $APF$=.74&lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-08-30T04:24:55.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35665" PostTypeId="2" ParentId="35625" CreationDate="2015-08-30T04:56:09.350" Score="4" Body="&lt;p&gt;In part, there is no abrupt cutoff because the gradient of the field is not spread evenly across the electrolyte: the gradient is steep at the &lt;em&gt;interface&lt;/em&gt; between electrodes and the electrolyte, where gas evolution actually takes place, not in the body of the electrolyte. Note where bubbles arise in this &lt;a href=&quot;https://www.youtube.com/watch?v=6VkRwGw-L_M&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the conductivity of the electrolyte is low, whether due to electrode separation or low ionization, then electrolysis proceeds more slowly, perhaps imperceptibly, though there is no complete cutoff point.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-30T04:56:09.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35666" PostTypeId="2" ParentId="35524" CreationDate="2015-08-30T05:09:06.807" Score="1" Body="&lt;p&gt;That's a good start, but there is also ohmic loss, as much of the energy is also converted to heat in the electrolyte, electrodes or even wiring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, during WWII, silver was used to replace copper bus-bars in stationary machinery such as electrolysis apparatus, freeing the copper for use in war materiel and slightly improving the efficiency of the apparatus due to its better conductivity then commercial copper.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-30T05:09:06.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35667" PostTypeId="2" ParentId="35444" CreationDate="2015-08-30T05:22:01.543" Score="1" Body="&lt;p&gt;My assumption  is that the &lt;em&gt;addition&lt;/em&gt; of water (other than water of hydration) lowers the temperature - consider that the heat of crystallization is shared among the sodium acetate, the added water and even the packaging. Perhaps the manufacturer adds water to lessen the chance of burns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need to experiment with lab-grade chemicals to be sure, such as &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/sial/s2889?lang=en&amp;amp;region=US&quot; rel=&quot;nofollow&quot;&gt;this ~99% pure sodium acetate trihydrate&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-30T05:22:01.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35668" PostTypeId="2" ParentId="35378" CreationDate="2015-08-30T05:33:14.807" Score="1" Body="&lt;p&gt;You might experiment with fresh-ground apples, which contain polyphenol oxidase and catechol oxidase, enzymes that turn brown in air (a tiny amount of copper or ferrous salt is also present, acting as a catalyst).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that matter, dilute ferrous salts are often almost colorless (pale green) but darken to brown on exposure to air. (&lt;a href=&quot;https://books.google.com/books?id=sozsCAAAQBAJ&amp;amp;pg=PA7&amp;amp;lpg=PA7&amp;amp;dq=podzol%20darkens%20exposure%20to%20air&amp;amp;source=bl&amp;amp;ots=JyyQ3YSw1I&amp;amp;sig=G4Jl18FNAxSmwyOPhVI0UhJULz0&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CB4Q6AEwAGoVChMIzaTFt4nQxwIVgnU-Ch02ig9E#v=onepage&amp;amp;q=podzol%20darkens%20exposure%20to%20air&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Digging in a podzol, soil of a pine forest, shows that color change.&lt;/a&gt;) So, you could extract a bit of liquid from greenish podzool and load your pen or marker with that.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-30T05:33:14.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35669" PostTypeId="1" CreationDate="2015-08-30T06:42:51.983" Score="2" ViewCount="9923" Body="&lt;p&gt;I read about electrode potential today, the book says that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Metals more reactive than hydrogen will have negative electrode potential value and metals less reactive than hydrogen will have positive electrode potential.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm a bit confused by the sign of the voltage, can voltage have negative value? What does negative voltage mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is why metals more reactive than H have negative electrode potential? &lt;/p&gt;&#xA;" OwnerUserId="19511" LastActivityDate="2016-09-27T06:05:39.500" Title="What are positive and negative voltage?" Tags="&lt;electrochemistry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35672" PostTypeId="1" AcceptedAnswerId="35688" CreationDate="2015-08-30T10:25:02.287" Score="10" ViewCount="8178" Body="&lt;p&gt;I was wondering in orbitals that have nodes such as the p orbital, how does the electron move from one lobe to the other? I know that people say it is because of the wave/particle nature of electrons but they don't actually explain how they travel across the node without actually passing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My teacher says it is because that electrons are able to act like waves and hence like how EM waves can pass through a wall, electrons are able to pass through the node. However when he says that electrons are like a wave, does he mean that it is literally a wave so does an electron have a wavelength, amplitude, frequency, etc? Because I remember reading somewhere which stated that electrons aren't literally like waves. However they can be mathematically described using concepts that apply to waves. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the comments, they state that electrons can be thought as standing waves. After looking at the internet I came across the following image which explained how electrons act as a standing wave:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yqmP8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yqmP8.jpg&quot; alt=&quot;electron as a standing wave&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this doesn't make sense to me. Don't electrons exist in orbitals not in orbits around the nucleus? If so, isn't this wrong? Also, since electrons move in orbitals, how can there be any end points as it can move anywhere and its movement isn't restricted to a close loop. Hence how can it act as a standing wave?&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-08-30T11:47:54.413" LastActivityDate="2018-03-02T00:24:36.097" Title="How do Electrons Cross Nodes in Orbitals?" Tags="&lt;quantum-chemistry&gt;&lt;electrons&gt;&lt;orbitals&gt;" AnswerCount="2" CommentCount="11" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35673" PostTypeId="1" AcceptedAnswerId="35709" CreationDate="2015-08-30T11:11:52.677" Score="4" ViewCount="25785" Body="&lt;p&gt;To the best of my knowledge, a conjugate acid of a base is the base after it has accepted a proton, or a $\ce{H+}$ ion. In this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH + H+ -&amp;gt; Na+ + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the conjugate acid of $\ce{NaOH}$ the sodium ion, or the water? Common sense tells me it can't be the $\ce{Na+}$ ion, because it has no protons to donate, so how could it ever be an acid? So I am thinking that the conjugate acid is $\ce{H2O}$. However, wouldn't that mean that the conjugate acid of any base of the form &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{(something)OH + H+ -&amp;gt; (something)+ + H2O}$$&#xA;would be water, and that seems unsettling to me. I also believe that since $\ce{NaOH}$ undergoes the following reaction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH -&amp;gt; Na+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the $\ce{Na+}$ is something of a 'spectator ion' (not sure if that's the correct term), this seems to imply that $\ce{H2O}$ should be the conjugate acid. Basically, I'm really confused, and could use a little help sorting all this out.&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="5026" LastEditDate="2018-03-03T11:20:50.050" LastActivityDate="2018-03-03T11:20:50.050" Title="What is the conjugate acid of NaOH using the Brønsted-Lowry definition of acids?" Tags="&lt;acid-base&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35674" PostTypeId="1" AcceptedAnswerId="38148" CreationDate="2015-08-30T12:09:00.043" Score="8" ViewCount="5642" Body="&lt;p&gt;&lt;strong&gt;What I know:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Empty 3d orbitals are higher in energy than empty 4s orbitals  &lt;/li&gt;&#xA;&lt;li&gt;Aufbau Principle (electrons always go into an empty orbital with the lowest energy)  &lt;/li&gt;&#xA;&lt;li&gt;Partially/half/fully filled 3d orbital are lower in energy than 4s orbitals  &lt;/li&gt;&#xA;&lt;li&gt;$\ce{Cr}$ is an exception: $\ce{[Ar] 3d^5 4s^1}$. It breaks the &quot;rule&quot; because the extra stability from half-filled 3d orbitals is favoured.  &lt;/li&gt;&#xA;&lt;li&gt;From (1) and (2) =&gt; After the $\ce{[Ar]}$ core, electrons will fill 4s before occupying 3d    &lt;/li&gt;&#xA;&lt;li&gt;When forming ions, electrons of the highest energy are removed first.  &lt;/li&gt;&#xA;&lt;li&gt;Iron(II) ions exist as $\ce{[Ar] 3d^6}$ in reality.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I understand:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why Fe is $\ce{[Ar] 3d^6 4s^2}$. [from (5)]&lt;br&gt;&#xA;I understand why Cr is $\ce{[Ar] 3d^5 4s^1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From (4), (5) and (3), I believe for $\ce{Cr}$ the 4s is filled before 3d too, i.e. it is $\ce{[Ar] 3d^3 4s^2}$ before the last electron is added. After the last electron is added to form $\ce{[Ar] 3d^4 4s^2}$, a 4s electron immediately moves to the 3d orbital which is of lower energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From (6) and (3), for transition metals, 4s electrons are removed first. i.e., I understand why iron(II) ions may be $\ce{[Ar] 3d^6}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I don't understand:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know $\ce{[Ar] 3d^6}$ is correct for iron(II), but I don't understand why?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't iron ions adjust itself as Cr does to achieve the extra stability? Is it because, in this case, it needs to excite an electron to an orbital of higher energy, whereas in Cr, the 4s electron just moves down to an orbital of lower energy?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;P.S.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I've read &lt;a href=&quot;https://chemistry.stackexchange.com/questions/27662/what-is-the-electronic-configuration-of-feii-ion/29651#29651?newreg=c2305f3d450a4687a7aff9ab3b877fb3&quot;&gt;What is the electronic configuration of Fe(II) ion?&lt;/a&gt; and I am not satisfied with the answers in that thread. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I've read &lt;a href=&quot;https://chemistry.stackexchange.com/questions/5832/do-ions-also-form-special-electron-configurations-just-as-copper-and-chromium&quot;&gt;Do ions also form special electron configurations? (just as Copper and Chromium do)&lt;/a&gt;, and it only tells me &quot;it is possible for ions to have special electronic configurations too!&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19524" LastEditorUserId="5026" LastEditDate="2018-04-28T01:47:25.307" LastActivityDate="2018-04-28T14:23:41.693" Title="Why do iron(II) ions and chromium have different electronic configurations?" Tags="&lt;electronic-configuration&gt;&lt;transition-metals&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35675" PostTypeId="1" AcceptedAnswerId="35678" CreationDate="2015-08-30T12:11:37.837" Score="1" ViewCount="107" Body="&lt;p&gt;I was looking up the Lennard-Jones parameter $\sigma$ (which, to my understanding, is a measure of molecular size) when I was surprised to discover that the value for the $\ce{N2}$ molecule (3.667 Å) was slightly &lt;em&gt;lower&lt;/em&gt; than the value for the $\ce{CH4}$ molecule (3.78 Å). It's my understanding that $\ce{N2}$ molecules are significantly larger than $\ce{CH4}$ molecules, and this is exploited in, for instance, molecular sieves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: is $\sigma$ a reasonable measure of mean atomic radius? And if it isn't, what is a good measure of molecular size, with which I can compare molecules?&lt;/p&gt;&#xA;" OwnerUserId="17933" LastEditorDisplayName="user7951" LastEditDate="2015-08-30T13:54:37.997" LastActivityDate="2015-08-30T16:43:41.963" Title="Good measures ot atomic size" Tags="&lt;molecules&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35678" PostTypeId="2" ParentId="35675" CreationDate="2015-08-30T15:35:54.910" Score="2" Body="&lt;p&gt;Atoms and molecules, as you surely know, are &lt;strong&gt;not&lt;/strong&gt; tiny hard pebbles that you may measure using a slide gauge, unambiguously and with great precision. Instead, they are clouds of electron density with rather fuzzy edges, like things on an out-of-focus photograph. Their apparent size depends on how you define what &quot;edge&quot; is, or how hard you squeeze them.&#xA;This is one of the reasons why we don't have a single definitive list of atomic radii, but a whole variety thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the point of your question, yes, $\sigma$ is a good measure of molecular size. But there are also others, equally good and all subtly different. Which one to use depends on your purpose, and may be a hard problem by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, remember that (1) the Lennard-Jones potential is defined for spherically symmetrical objects, which your molecules are not, and (2) even for these objects (atoms) it is but an approximation.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-08-30T16:43:41.963" LastActivityDate="2015-08-30T16:43:41.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35680" PostTypeId="1" CreationDate="2015-08-30T17:26:48.413" Score="10" ViewCount="3375" Body="&lt;p&gt;I've been recently looking into fireworks and the chemistry of explosives, and I found out that, amongst others, one of the most important reactions in gunpowder ignition is the oxidation of charcoal and sulfur by $\mathrm{KNO_3(s)}$. I am curious as to whether it is known if this reaction occurs between charcoal/sulfur and solid state potassium nitrate or if the potassium nitrate decomposes first and the reaction occurs via gaseous oxygen as an intermediate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general mechanism of oxidation between solid oxidizing salts and organic matter/sulfur? Does the same thing happen with $\mathrm{KClO_3(s)}$ or $\mathrm{KMnO_4(s)}$ for example?&lt;/p&gt;&#xA;" OwnerUserId="15854" LastEditorUserId="11367" LastEditDate="2015-12-27T01:14:52.667" LastActivityDate="2015-12-27T01:14:52.667" Title="What is the oxidation mechanism of gunpowder?" Tags="&lt;reaction-mechanism&gt;&lt;combustion&gt;&lt;solid-state-chemistry&gt;&lt;fuel&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35681" PostTypeId="2" ParentId="35669" CreationDate="2015-08-30T18:29:18.267" Score="2" Body="&lt;p&gt;Negative voltage means the same as positive, except the electric current caused by it would flow in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your another question, that's just what it is: having negative potential means being more reactive than H. See, that electrode potential is not an absolute value, it can be measured only relative to something. And the general agreement is to take the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_hydrogen_electrode&quot; rel=&quot;nofollow&quot;&gt;standard hydrogen electrode&lt;/a&gt; for that &quot;something&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you'd make an electrode out of your metal immersed in a solution of some salt of itself, couple it to standard hydrogen electrode and see what happens. If a metal is more reactive, it will dissolve, the released electrons will run to the hydrogen electrode and reduce hydrogen there. With less reactive metals, it is the other way around: hydrogen will dissolve and send the electrons running in the opposite direction, to eventually reduce the metal.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-30T18:29:18.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35682" PostTypeId="1" CreationDate="2015-08-30T19:32:58.127" Score="0" ViewCount="1470" Body="&lt;p&gt;I am struggling with what appears to be an extremely easy pH problem that uses the Henderson-Hasselbalch equation. The problem and answer provided by the book is given below. I understand where the book is getting the answer but it appears that they are overlooking something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, the Henderson-Hasselbalch equation allows you to calculate the pH of a solution based on the concentrations of acid and conjugate base at equilibrium. In the problem, you start off with a solution of just the conjugate base. Then you add pure acid to the solution to get a desired pH. However, the book seems to assume that the acid you add completely contributes to the concentration of undissociated acid at equilibrium [HA]. How can this be the case? Some of the acid you add will surely dissociate. Anyways, here's what the book did...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PUbcR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PUbcR.jpg&quot; alt=&quot;Henderson-Hasselbalch Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19530" LastEditorUserId="418" LastEditDate="2016-11-03T19:33:04.900" LastActivityDate="2019-11-11T08:34:35.573" Title="How to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35684" PostTypeId="1" AcceptedAnswerId="35687" CreationDate="2015-08-30T19:49:57.210" Score="5" ViewCount="399" Body="&lt;p&gt;Why is [1.1.1]propellane considerably less reactive than either [2.2.1]propellane or [2.1.1]propellane as illustrated in the hydrogenation reactions pictured below?&lt;a href=&quot;https://i.stack.imgur.com/ch6M5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ch6M5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="46306" LastEditDate="2018-03-04T09:51:56.670" LastActivityDate="2018-03-04T09:51:56.670" Title="Reactivity of propellanes" Tags="&lt;organic-chemistry&gt;&lt;thermodynamics&gt;&lt;reactivity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35685" PostTypeId="2" ParentId="35682" CreationDate="2015-08-30T19:54:47.770" Score="1" Body="&lt;p&gt;Never mind, I found the answer! I was correct to assume that the initial concentration will not be equal to the equilibrium concentration. However, they will be extremely close because we are working with weak acids. In fact, I found an article published in a chemistry education journal that discusses this so-called &quot;Henderson-Hasselbalch Approximation&quot; and how it is often overlooked in chemistry textbooks. Here is the link: &lt;a href=&quot;http://chemistry.oregonstate.edu/courses/ch421/Previous%20pages/Course%20Docs%20Fall%202006/Henderson.pdf&quot; rel=&quot;nofollow&quot;&gt;Henderson-Hasselbalch Approximation&lt;/a&gt;. I believe textbooks should make sure students are aware that the equation is only an approximate method for determining the pH of a buffer.&lt;/p&gt;&#xA;" OwnerUserId="19530" LastActivityDate="2015-08-30T19:54:47.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35686" PostTypeId="2" ParentId="35682" CreationDate="2015-08-30T20:10:27.383" Score="3" Body="&lt;p&gt;Boric acid and sodium borate forms an acidic buffer (A mixture of weak acid and salt of weak acid and strong base). The acid is weak hence the concentration of basic ion from acid is much less compared to the concentration of basic ion obtained from the salt which can ionize completely. Yes. Some of the acid dissociates to form hydronium ion and a basic ion. That's how a buffer works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a weak acid HA and the salt BA.&#xA;Its easy to show that the concentration of $H^+$ due to ionisation of HA is $$[H^+]=k_a\frac{[HA]}{[A^-]}$$&#xA;It is reasonable to assume that the concentration of $A^-$ from BA is much larger than that from HA due to complete ionisation of BA. Hence $[A^-]=[BA]$=[salt]&#xA;So, $$[H^+]=k_a\frac{[acid]}{[salt]}$$&#xA;Taking logarithm, you get the desired equation.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="17624" LastEditDate="2015-08-30T20:16:51.687" LastActivityDate="2015-08-30T20:16:51.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35687" PostTypeId="2" ParentId="35684" CreationDate="2015-08-30T20:40:54.277" Score="6" Body="&lt;p&gt;&lt;strong&gt;The heat of hydrogenation reflects the thermodynamic difference in energy between starting material and product.&lt;/strong&gt; It is a &lt;strong&gt;thermodynamic&lt;/strong&gt; measure of the difference in stability of the products and reactants.  &lt;strong&gt;Reactivity&lt;/strong&gt;, on the other, hand is related to &lt;strong&gt;kinetics&lt;/strong&gt;, a reflection of the activation energy  involved in the rate determining step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cyclopropane and cyclobutane have roughly the same strain energy (&lt;strong&gt;SE&lt;/strong&gt;=27.5 and 26.3 kcal/mol respectively). To a first approximation, the strain energy of a polycyclic system is the sum of the strain energies of the individual rings in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the 3 starting materials contains a total 3 (cyclopropane plus cyclobutane) rings.  Since the SE of cyclopropane ~ cyclobutane, we would therefore expect &lt;strong&gt;the SE's of the 3 starting materials to be similar&lt;/strong&gt; and larger than the SE's of the less strained products. In agreement with this expectation the calculated SE's of the 3 starting materials are 109, 106 and 103 kcal/mol respectively (see fig. 2.2 on p. 37 &lt;a href=&quot;https://books.google.com/books?id=rpPULNO9XN4C&amp;amp;pg=PA37&amp;amp;lpg=PA37&amp;amp;dq=bicyclo%5B1.1.1%5Dpentane%20strain%20energy&amp;amp;source=bl&amp;amp;ots=CM2sQcZQ8n&amp;amp;sig=JsmnYkYOOyv_f74skhw1duPLC7g&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCcQ6AEwAWoVChMIqInSnsvRxwIVT6yACh1zkQGP#v=onepage&amp;amp;q=bicyclo%5B1.1.1%5Dpentane%20strain%20energy&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), indeed very similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SE's of the final products are 17, 41 and 62 kcal/mol respectively (&lt;a href=&quot;https://books.google.com/books?id=J5_-BAAAQBAJ&amp;amp;pg=PA72&amp;amp;lpg=PA72&amp;amp;dq=bicyclo%5B1.1.1%5Dpentane%20strain%20energy&amp;amp;source=bl&amp;amp;ots=2nKKec92kp&amp;amp;sig=r1bG1FkfbCW-grKvu6_1lQWN1_M&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CDoQ6AEwBmoVChMIqInSnsvRxwIVT6yACh1zkQGP#v=onepage&amp;amp;q=bicyclo%5B1.1.1%5Dpentane%20strain%20energy&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;ref&lt;/a&gt;, see Table 3.2 on page 72).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we calculate the strain released during hydrogenation it is in the same order and very close to the actual heats of hydrogenation reported in your figure.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mathrm{109-17=92}$ kcal/mol&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{106-41=65}$ kcal/mol&lt;/li&gt;&#xA;&lt;li&gt;$\mathrm{103-62=41}$ kcal/mol&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Clearly, in this series of propellanes the heat of hydrogenation is a measure of the strain released in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because bicyclo[1.1.1] pentane (the right-most product) has a much higher SE than either of the two other hydrogenated products, the heat of hydrogenation given off when it is formed will consequently be much less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1.1.1]propellane is an extremely reactive compound (see &lt;a href=&quot;https://en.wikipedia.org/wiki/1.1.1-Propellane&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for example).  Comparing the thermodynamic heats of hydrogenation in your series of compounds cannot be used to meaningfully predict the kinetic reactivity of these compounds.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-08-30T22:56:09.490" LastActivityDate="2015-08-30T22:56:09.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35688" PostTypeId="2" ParentId="35672" CreationDate="2015-08-30T21:47:46.793" Score="6" Body="&lt;p&gt;Electrons (indeed all matter and energy) behave both as wave and as particle, depending how the observation is made. The most obvious example is the single-electron interference pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the [interference] experiment using single electrons, showing that &lt;em&gt;each electron interferes with itself&lt;/em&gt; as predicted by quantum theory. In 2002, the single-electron version of the experiment was voted &quot;the most beautiful experiment&quot; by readers of Physics World.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Double-slit_experiment&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-08-30T21:47:46.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35689" PostTypeId="2" ParentId="35533" CreationDate="2015-08-30T22:19:38.507" Score="1" Body="&lt;p&gt;A combustion reaction is classified as a reaction between a fuel and oxygen to produce energy. The reaction: 2CO(g) + O2(g) = 2CO2(g) fulfills this definition, is exothermic, and does not use hydrogen.&lt;/p&gt;&#xA;" OwnerUserId="19447" LastActivityDate="2015-08-30T22:19:38.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35690" PostTypeId="1" CreationDate="2015-08-30T22:53:07.780" Score="4" ViewCount="68" Body="&lt;p&gt;When dealing with electrons and the electromagnetic spectrum, why are there only four visible transitions in the hydrogen atom?&lt;/p&gt;&#xA;" OwnerUserId="19534" LastEditorUserId="69830" LastEditDate="2021-05-20T07:19:28.677" LastActivityDate="2021-05-20T07:19:28.677" Title="Hydrogen visible transitions" Tags="&lt;physical-chemistry&gt;&lt;hydrogen&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35691" PostTypeId="2" ParentId="35563" CreationDate="2015-08-30T23:17:48.383" Score="3" Body="&lt;p&gt;I think the biggest reason it's still used is because there's no escaping our need of Avogadro's constant, $N_\mathrm A=6.022\times10^{23}\ \mathrm{mol^{-1}}$. Even if we were to use some metric prefixes meaning $10^{24}$, etc. we would still need Avogadro's constant because it is used in the definition of other physical constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, Boltzmann's constant, $$k_\mathrm B=\frac{R}{N_\mathrm A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, Faraday's constant, $$F=N_\mathrm A\cdot e$$&#xA;where $-e$ is the elementary charge of an electron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, the number of particles in an ideal gas, $$N=\frac{p\cdot V\cdot  N_\mathrm A}{R\cdot T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of those constants have to stay the same for many of the formulations of Chemistry to be correct, so we would have stumbled across this magic number one way or another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it might be true (and I don't know this) that the mole was not defined with any of that in mind, it's really quite convenient if you ask me to have a mole be both a unit and a physical constant which everyone knows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's particularly useful because you don't have to memorize what the value of $k_\mathrm B$ is, you just remember the equation and figure it out using a calculator. After all, every knows Avogadro's number and the ideal gas constant.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorDisplayName="user7951" LastEditDate="2015-10-23T18:57:38.570" LastActivityDate="2015-10-23T18:57:38.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35697" PostTypeId="2" ParentId="21921" CreationDate="2015-08-31T01:44:55.867" Score="2" Body="&lt;p&gt;In a 1979 paper entitled &lt;a href=&quot;http://dx.doi.org/10.1016/S0040-4039(01)86190-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Copper-catalysed reactions of Grignard reagents with epoxides and oxetane&lt;/a&gt;, Huynh, Derguini-Boumechal, and Linstrumelle report that the copper-catalysed ring opening of oxetane by various Grignard reagents gives the alcohols &quot;in acceptable yields&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jmSaJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jmSaJ.png&quot; alt=&quot;Reaction scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{array}{cccc}&#xA;\ce{R} &amp;amp; \ce{X} &amp;amp; \text{Solvent} &amp;amp; \text{Yield (%)} \\&#xA;\hline&#xA;n\text{-}\ce{C4H9} &amp;amp; \ce{Cl} &amp;amp; \text{Ether} &amp;amp; 75 \\&#xA;\ce{C6H5} &amp;amp; \ce{Br} &amp;amp; \text{THF} &amp;amp; 52 \\&#xA;\ce{CH2=CH-CH2} &amp;amp; \ce{Br} &amp;amp; \text{Ether} &amp;amp; 50 \\&#xA;\ce{(CH3)2C=CH-CH2} &amp;amp; \ce{Cl} &amp;amp; \text{THF} &amp;amp; 50&#xA;\end{array} $$&lt;/p&gt;&#xA;" OwnerUserId="19125" LastEditorUserId="16683" LastEditDate="2015-09-30T14:52:51.690" LastActivityDate="2015-09-30T14:52:51.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35698" PostTypeId="2" ParentId="35690" CreationDate="2015-08-31T01:52:57.597" Score="6" Body="&lt;p&gt;This all falls back to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_spectral_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrogen absorption spectra&lt;/a&gt;. There are multiple &quot;series&quot; discovered by different scientists. The series in the visible region is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Balmer_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Balmer series&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the electronic energy levels of the hydrogen atom are &lt;a href=&quot;https://chemistry.stackexchange.com/a/35660/5017&quot;&gt;&lt;em&gt;quantized&lt;/em&gt;&lt;/a&gt;, there can only be certain discrete transitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, if you solve the Balmer series (n=2), you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\lambda} = \frac{4}{B}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \quad \mathrm{for~} n=3,4,5,...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will then get lines at $\lambda$ = 656.3 nm, 486.1 nm, 434.1 nm, 410.2 nm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only a few of those fall in the visible. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Lyman_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lyman series (n=1)&lt;/a&gt; occurs in the UV spectra, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_spectral_series#Paschen_series_.28Bohr_series.3B_n.E2.80.B2_.3D_3.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paschen series (n=3)&lt;/a&gt; occurs in the IR.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.827" LastActivityDate="2015-08-31T01:52:57.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35700" PostTypeId="1" CreationDate="2015-08-31T04:31:02.233" Score="1" ViewCount="162" Body="&lt;p&gt;I am designing an industrial process which can remove oxygen from air. &#xA;This chemical should combine readily with oxygen in the presence of heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I am looking for a chemical with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Combines readily with oxygen, thus removing the oxygen in air&lt;/li&gt;&#xA;&lt;li&gt;Inexpensive&lt;/li&gt;&#xA;&lt;li&gt;Environmentally friendly &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For point #1, heating is available if it is an endothermic reaction. &lt;/p&gt;&#xA;" OwnerUserId="19539" LastActivityDate="2015-10-02T00:47:20.487" Title="Any chemicals combines readily oxygen?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35701" PostTypeId="1" AcceptedAnswerId="35703" CreationDate="2015-08-31T06:52:49.437" Score="5" ViewCount="1342" Body="&lt;p&gt;Why is the ideal gas law usable in most gas problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a problem like this: &quot;Suppose you have $1.00\: \mathrm{mol}$ of a gas at $0\:^\circ \mathrm{C}$, occupying a container which is $500\: \mathrm{mL}$ in size. What is the pressure of this gas in $\mathrm{atm}$?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal gas law is was applied to solve that problem and the answer was $44.8\: \mathrm{atm}$. But isn't the ideal gas law only applicable to gases which are in high temperatures and low pressure? Because the kinetic energy overcomes the intermolecular bonds and the molar volume is negligible, etc etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the conditions for this &quot;gas&quot; are low temperature and consecutively high pressure which is basically a stamp for real gases so why isn't the real gas law applied to this problem instead?&lt;/p&gt;&#xA;" OwnerUserId="19444" LastEditorUserId="16683" LastEditDate="2016-09-30T09:34:43.407" LastActivityDate="2016-09-30T09:34:43.407" Title="Why is the ideal gas law so ubiquitous?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;&lt;equation-of-state&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35703" PostTypeId="2" ParentId="35701" CreationDate="2015-08-31T07:42:30.003" Score="5" Body="&lt;p&gt;You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this complicated formula is that several assumptions that we make for ideals gases don't hold for real gases. Recall that an ideal gas is considered to be a point mass - a particle so small that the volume of that particle is negligible. A real gas particle does have real volume. For an ideal gas, the collisions between gas particles was said to be &quot;elastic&quot; - no attractive or repulsive forces exist, and thus, no energy is exchanged during collisions. For a real gas, collisions are non-elastic. So, the ideal gas law must be corrected for this extra forces. There are several real gas laws, one is the van der Waals equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[P + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how &quot;corrections&quot; are being made to the pressure term and the volume term. Since collisions of real gases are non-elastic, the term $a(n/V)^2$ is correcting for the interactions of these particles. Since real gas particles have real volume, the $nb$ term is correcting for the excluded volume. The values of $a$ and $b$ are constants, and must be determined experimentally for each gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from this equation that first the values of $a$ and $b$ must be known then you must solve this equation which can be complicated. Even then, when you do finally solve it, you will probably get a result which is similar to the result obtained if you used the ideal gas equation. This is why the ideal gas equation is continued to be used as in most cases it provides quite accurate results while bypassing complicated formulas. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="16683" LastEditDate="2016-04-24T12:35:54.273" LastActivityDate="2016-04-24T12:35:54.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35705" PostTypeId="1" CreationDate="2015-08-31T10:20:37.740" Score="1" ViewCount="171" Body="&lt;p&gt;Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) &amp;lt;=&amp;gt; NH3(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll begin solving the problem...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the formula $$\Delta G=\Delta G^\circ+RT\ln Q.$$ The vaporisation of $\ce{NH3}$ is an equilibrium so  $\Delta G=0$ and $Q=K_c$ . Also $K_c=\left[\ce{NH3}\right]$. So $\Delta G^\circ=-RT\ln K_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I continue? Is it $K_c=K_p=10$ or $K_c=K_p/\left(RT\right)$?&lt;/p&gt;&#xA;" OwnerUserId="18005" LastEditorUserId="4945" LastEditDate="2015-08-31T12:04:55.723" LastActivityDate="2015-11-29T14:44:31.897" Title="How to calculate the Gibbs energy for the vaporisation of solid ammonia?" Tags="&lt;thermodynamics&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-11-29T17:01:14.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="35706" PostTypeId="2" ParentId="35705" CreationDate="2015-08-31T11:06:41.043" Score="-1" Body="&lt;p&gt;Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.&lt;/p&gt;&#xA;" OwnerUserId="19543" LastEditorDisplayName="user7951" LastEditDate="2015-08-31T11:12:53.600" LastActivityDate="2015-08-31T11:12:53.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35708" PostTypeId="2" ParentId="35674" CreationDate="2015-08-31T11:56:36.813" Score="0" Body="&lt;p&gt;I've had the same question and this is how my chemistry teacher answered:&#xA;  3d is a sub energy level of the 3rd energy level and 4s is a sub energy level of the 4th energy level. When electrons fill into the Fe atom they fill according to the Auffbau principle, in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d. But when 2 electrons are to be removed to form the Fe2+ ion, the electrons are not removed from the 3d orbitals. 3d belongs to the 3rd energy level which becomes an inner filled energy level when electrons are present in the 4th energy level. So electrons are removed from the 4s sub energy level, without disturbing the fully filled 3rd energy level.&lt;/p&gt;&#xA;" OwnerUserId="19544" LastActivityDate="2015-08-31T11:56:36.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35709" PostTypeId="2" ParentId="35673" CreationDate="2015-08-31T13:06:48.390" Score="6" Body="&lt;p&gt;Your first equation is more properly written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH + H3O+ -&amp;gt; Na+ + 2H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in aqueous media. However, we can do better if we explicitly show the dissociation of $\ce{NaOH}$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaOH -&amp;gt; Na+ + OH-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substitute that into the first expression (note that I write $\ce{2H2O}$ as $\ce{H2O + H2O}$) to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{Na+ + \underbrace{OH^{-}}_{base} + \underbrace{H3O^{+}}_{acid} -&amp;gt; Na+ + \underbrace{H2O}_{conjugate\;acid} + \underbrace{H2O}_{conjugate\;base}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where we see that $\ce{H2O}$ is the conjugate acid of $\ce{OH-}$ as well as the conjugate base of $\ce{H3O+}$. The last bit - where water plays 2 roles - is due to water being &lt;a href=&quot;https://en.wikipedia.org/wiki/Amphoterism&quot;&gt;amphoteric&lt;/a&gt;, or able to act as an acid or a base.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-08-31T13:46:24.867" LastActivityDate="2015-08-31T13:46:24.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35710" PostTypeId="2" ParentId="9974" CreationDate="2015-08-31T13:15:05.843" Score="0" Body="&lt;p&gt;I found this question when contemplating this recent news article about 3d printing and Vegemite, a naturally electrically conductive spread for bread that is consumed in Australia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://munchies.vice.com/articles/3d-printed-vegemite-could-be-the-future-of-edible-electronics&quot; rel=&quot;nofollow&quot;&gt;http://munchies.vice.com/articles/3d-printed-vegemite-could-be-the-future-of-edible-electronics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7046" LastActivityDate="2015-08-31T13:15:05.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35711" PostTypeId="2" ParentId="9974" CreationDate="2015-08-31T13:39:17.917" Score="3" Body="&lt;p&gt;I wonder why nobody mentioned the famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Goldschl%C3%A4ger&quot; rel=&quot;nofollow&quot;&gt;vodka with gold flakes&lt;/a&gt;. Admittedly, this hardly qualifies as food, and the amount of gold is minuscule, but in the absence of any solid examples with true metallic character I think it may deserve a curious look.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-08-31T13:39:17.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35712" PostTypeId="1" CreationDate="2015-08-31T14:08:37.853" Score="5" ViewCount="562" Body="&lt;p&gt;This is what my book shows (the darker the more probable):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X19J7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious places where the electron probability is zero is &lt;a href=&quot;https://chemistry.stackexchange.com/a/8670/4805&quot;&gt;the midsection of the dumbbell&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how the probability distributed inside the dumbbell.&lt;/p&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:03.453" LastActivityDate="2017-04-04T14:00:48.643" Title="What is the correctly diagrammed electron probability distribution for p orbital?" Tags="&lt;quantum-chemistry&gt;&lt;orbitals&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35713" PostTypeId="1" AcceptedAnswerId="35714" CreationDate="2015-08-31T15:20:40.040" Score="2" ViewCount="147" Body="&lt;p&gt;After having been acquainted with enantiomerism for nearly six months, I have one standing doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the structure of alanine. Suppose I have a pure mixture of exclusively 'S-alanine'(left in the picture) in a flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MoKCL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MoKCL.png&quot; alt=&quot;S-alanine &amp;amp; R-alanine:Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are hundreds, if not, thousands of molecular collisions in the flask per second. Why can not a bump to one of the S-alanine change it to R-alanine? This way, the mixture should racemise itself if left standing for some time. But this does not happen. Why? Can someone please explain?&lt;/p&gt;&#xA;" OwnerUserId="11304" LastEditorUserId="4325" LastEditDate="2015-08-31T15:31:36.967" LastActivityDate="2015-08-31T19:02:51.403" Title="Why doesn't every pure enantiomer racemise?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;isomers&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35714" PostTypeId="2" ParentId="35713" CreationDate="2015-08-31T15:48:46.940" Score="2" Body="&lt;p&gt;Because the activation energy for the enantiomeric H-Atom to jump to the other side of the carbon atom's sp3 tetraeder (by whatever mechanism) is so high that it is just not available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right, statistically, every energy is available (with very bad odds). But very likely other processes, and more destructive ones, have a lower activation energy. Your compound will likely degrade and rot away long before it racemises. But even that, for alanine at room temperature, does not happen on any timescale that you can observe.&lt;/p&gt;&#xA;" OwnerUserId="17952" LastEditorUserId="17952" LastEditDate="2015-08-31T19:02:51.403" LastActivityDate="2015-08-31T19:02:51.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35715" PostTypeId="1" AcceptedAnswerId="37395" CreationDate="2015-08-31T16:21:12.323" Score="12" ViewCount="659" Body="&lt;p&gt;My Masters in inorganic chemistry is focused in magnetic properties of transition metals; great part of the work is done with DFT methods using softwares like &lt;a href=&quot;https://orcaforum.cec.mpg.de/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ORCA&lt;/a&gt; and &lt;a href=&quot;http://www.nwchem-sw.org/index.php/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;NWChem&lt;/a&gt;, but since my University has a lack of resources, we are performing computations with not much computational power. That is why we are focusing in results and using functionals like &lt;strong&gt;PBE&lt;/strong&gt; and &lt;strong&gt;B3LYP&lt;/strong&gt; (since many papers are showing good results).&#xA;Nevertheless, as said by Martin in &lt;a href=&quot;https://chemistry.stackexchange.com/questions/11004/whats-the-difference-between-pbe-and-b3lyp-methods&quot;&gt;this thread&lt;/a&gt;, today we have &quot;better and more efficient functionals&quot; available, because PBE and B3LYP are very popular, but kind of old. That is why I’d like to ask: what are good, modern, functionals to work with transition metals (mostly Ti…Zn)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aditional information:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are working with systems  that have no more than 200 atoms, most of the time, usually dimers.&lt;/p&gt;&#xA;" OwnerUserId="17118" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.943" LastActivityDate="2015-10-04T07:09:42.600" Title="Better functionals to work with transition metals" Tags="&lt;quantum-chemistry&gt;&lt;computational-chemistry&gt;&lt;theoretical-chemistry&gt;&lt;density-functional-theory&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35716" PostTypeId="1" CreationDate="2015-08-31T18:00:12.237" Score="2" ViewCount="2318" Body="&lt;p&gt;Why is the boiling point and melting point of 15th group and 16th group has an exception?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that as molecular mass increases boiling point and melting point also increase. So, down the group 15 and 16, as molecular mass increases, melting and boiling point should also increase &lt;strong&gt;but&lt;/strong&gt; melting and boiling point of antimony is more than that of bismuth. (even though molecular mass of bismuth is more)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, melting and boiling point of tellurium is higher than polonium. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="6982" LastEditorUserId="5026" LastEditDate="2018-03-09T03:17:37.370" LastActivityDate="2018-03-09T03:17:37.370" Title="Why there is an exception in melting and boiling point in p block?" Tags="&lt;elements&gt;&lt;periodic-trends&gt;&lt;boiling-point&gt;&lt;melting-point&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35717" PostTypeId="1" CreationDate="2015-08-31T18:12:42.703" Score="3" ViewCount="99" Body="&lt;p&gt;We have a laser chamber full of $\ce{F2}$ gas (no air should be in the chamber) and several parts including steel, wires, ceramics, zinc.... This is for laser generation so the temperature is pretty high in the chamber. We have detected $\ce{NO2}$ gas in our FTIR specturm which caused our laser to fail and there’s a good chance that the $\ce{NO2}$ may be associated with the failure. The questions are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How might $\ce{NO2}$ be formed in a chamber? If it’s from air, why would $\ce{NO2}$ be formed and not $\ce{NO}$ or $\ce{N2O}$? Why would $\ce{NO2}$ persist after $\ce{N2}$ and $\ce{O2}$ are removed by chamber refills with fresh $\ce{F2}$?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How does $\ce{NO2}$ interact with fluorine? What other species might be formed when they react?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can $\ce{NO2}$ be dissociated by UV light (i.e. 193 nm)? What is its UV absorption spectrum?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19548" LastEditorDisplayName="user7951" LastEditDate="2015-08-31T18:27:07.453" LastActivityDate="2015-10-31T09:58:45.703" Title="NO2 gas was detected in FTIR" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;equipment&gt;&lt;atmospheric-chemistry&gt;&lt;ir-spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35718" PostTypeId="1" CreationDate="2015-08-31T20:05:39.960" Score="0" ViewCount="556" Body="&lt;p&gt;We are trying to predict what will happen when carbon dioxide gas is introduced into a contact chamber that serves as an ozone contact chamber. These contact chambers presently oxidize compounds found in municipal pool water along with chlorine at below 5 ppm. There are 3 contact chambers (one for each pool) and the temperature range is from 104 to 80 F. We are switching from hydrochloric acid to carbon dioxide to keep the total alkalinity from dropping when the acid is introduced.&lt;/p&gt;&#xA;" OwnerUserId="19549" LastEditorUserId="10185" LastEditDate="2015-08-31T20:34:15.000" LastActivityDate="2015-08-31T20:34:15.000" Title="Interaction between Ozone and Carbon Dioxide" Tags="&lt;inorganic-chemistry&gt;&lt;water&gt;" AnswerCount="0" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35719" PostTypeId="2" ParentId="35712" CreationDate="2015-08-31T21:04:26.783" Score="2" Body="&lt;p&gt;The probability distribution of an electron in the p sub-shell is determined from its wavefunction. The wavefunction of a system contains all the information about the quantum states of the system. Once of the most important properties of the wavefunction is that the squared modulus of the wave function, $|ψ|^2$, is a real number, interpreted as the probability density of the particle. Here is the wavefunction for the 2p orbital:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HREso.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HREso.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you square this function you will get the probabily distribution which looks like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MYP7K.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MYP7K.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, this corresponds with the p orbital drawn in your book, as there is a node in the middle, on either side is high probability density and at minus and positive infinity probability density is actually very, very small despite being shown as 0 on the wavefunction. The reason for this is that you need to remember that the orbital shows where an electron will be 90% of the time, not where it will be at all times. This means that the electron can be found outside the orbital. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-09-01T07:08:23.233" LastActivityDate="2015-09-01T07:08:23.233" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35720" PostTypeId="2" ParentId="35650" CreationDate="2015-08-31T23:54:12.400" Score="5" Body="&lt;p&gt;Avogadro articulated the concept that is Avogadro's number, but didn't pick the number. From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Avogadro's law states that, &quot;equal volumes of all gases, at the same temperature and pressure, have the same number of molecules&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro%27s_law&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Avogadro%27s_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Number is a logical extension of this law: So what is that number in relation to the volume and what is the mass of that amount of material?  By agreement, a mole is defined as the mass, in grams, equivalent to the atomic weight of a substance. This was laid out before Avogadro's Number was actually determined.  Then it was simply a matter of measuring the mass of a single atom to determine how many atoms were required to have a Mole of a material.  Over time, Avogadro's Number has been refined to more and more accuracy.&lt;/p&gt;&#xA;" OwnerUserId="19499" LastEditorUserId="19499" LastEditDate="2015-09-01T03:23:40.930" LastActivityDate="2015-09-01T03:23:40.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35722" PostTypeId="1" AcceptedAnswerId="35725" CreationDate="2015-09-01T02:37:19.530" Score="6" ViewCount="1357" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5X3sX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5X3sX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So initially they are mixed with the mole ratio of $\ce{H2}$ and $\ce{N2}$ being 3:1 respectively. It seems as the reaction proceeds, both these quantities decrease and the quantity of ammonia increases &lt;em&gt;until&lt;/em&gt; a point at which $\ce{NH3}$ stagnates. What happens at this point?&lt;/p&gt;&#xA;" OwnerUserId="15053" LastEditorUserId="4945" LastEditDate="2015-09-01T06:56:06.610" LastActivityDate="2015-09-02T21:35:17.890" Title="Why is there a sudden drop in the concentration of ammonia in this graph?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35723" PostTypeId="2" ParentId="35645" CreationDate="2015-09-01T04:17:13.613" Score="2" Body="&lt;p&gt;Karl Ratzch is basically correct, but I will try to provide a little more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric charges, when separated, exhibit a very large tendency to try to get back together again.  This is represented in basic Physics by Coulombs Law and the large value of the constant in that relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume we are talking about a battery here. If the flow of electrons out of the wire is different from the rate of ions arriving at the plate, the net charge on the plate will change.  Depending on which flow rate is greater, the voltage of the plate will instantly go up or down.  This will then change the rate of both charge flows such that the two almost instantly get back to the same value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail: Assume that the flow of electrons out of the plate through the wire is larger than the flow of anions into the plate from solution.  This reduces the net negative charge on the plate.  This reduces the &quot;force&quot; that is pushing electrons down the wire.  The actual name of that &quot;force&quot; is voltage.  The voltage will change such that it doesn't push electrons out the wire as fast and will repel anions in solution less aggressively allowing more of them to reach the plate (per second).  The opposite is also true.  As a result, the flow in and out of the plate will always remain almost exactly the same, with small differences existing for very brief periods of time when the voltage changes.&lt;/p&gt;&#xA;" OwnerUserId="19499" LastActivityDate="2015-09-01T04:17:13.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35725" PostTypeId="2" ParentId="35722" CreationDate="2015-09-01T06:52:55.403" Score="9" Body="&lt;p&gt;In the graph the following reaction will occur: $$\ce{N2 +3H2 &amp;lt;=&amp;gt; 2NH3}$$&#xA;This reaction is an equilibrium reaction and you can see that there is exponential decay for the reactants while the product is formed. After a while the concentrations do not change any more, so the system is in equilibrium. This is at the end of the time span &quot;a&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning of time span &quot;b&quot; the system is disturbed, as the ammonia is completely removed. The system is changed and the equilibrium has to form again. Hence the concentration of the reactants is further reduced while new product is formed. Like santiago already pointed out in the comments, this is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow&quot;&gt;Le Châtelier's principle&lt;/a&gt;. At the end of time span &quot;b&quot;, the system is in equilibrium again.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="16622" LastEditDate="2015-09-02T21:35:17.890" LastActivityDate="2015-09-02T21:35:17.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35726" PostTypeId="2" ParentId="35722" CreationDate="2015-09-01T07:05:17.707" Score="4" Body="&lt;p&gt;The first step to this question is writing the chemical equation for this reaction: $$\ce{N2(g) + 3H2(g) \rightleftharpoons2NH3(g)}$$ As you see, this reaction is an equilibrium reaction, meaning that not all the products will form reactants, but rather react until they reach a certain ratio. This ratio is known as the equilibrium constant, K. The equilibrium constant is given by: $$K = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ [A] and [B] are the concentrations of the products at equilibrium while [C] and [D] are the concentrations of the reactants at equilibrium. a, b, c, and d are the stoichiometric co-coefficients of their respective letter. The equilibrium constant is the same value for a particular reaction at a constant temperature and doesn't depend on the initial concentration of the products or reactants. In this case, the equilibrium constant is given by: $$K = \frac{[NH_3]^2}{[N_2][H_2]^3}$$ One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by: $$Q = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ Note how this is similar to the equilibrium constant, but here, [A], [B], [C] and [D] are the concentration of the substance at a specific time during the reaction, not at equilibrium point. This means that if Q is smaller than K, the reactants will react to form more products so that Q will increase so that it will eventually equal K and when Q is larger than K, vice versa. When Q equals K, the system is at equilibrium.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now lets apply this concept to this graph. Initially, there is no ammonia but there is hydrogen and nitrogen gas present. This means that Q = 0 which is smaller than K as K is non-zero. This means that nitrogen and hydrogen gas will react to form ammonia. At the time corresponding to dashed line, the system is at equilibrium as the concentrations of the substances aren't reacting. This means that Q = K. Now, the sudden drop in concentration of ammonia indicates that some of the ammonia is taken out from the system. This means that Q will become smaller and hence, be smaller than K. Therefore the reactants, hydrogen and nitrogen gas, must react further to form more ammonia until reaches equilibrium again. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-09-01T07:05:17.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35727" PostTypeId="2" ParentId="35717" CreationDate="2015-09-01T08:48:54.830" Score="2" Body="&lt;p&gt;Fluorine reacts with practically everything, especially at high temperatures, including zinc, wires, some ceramics, polymers, etc., so the system you describe may be prone to contamination. I don't see any obvious source of NO2, but it could easily form from nitrates, if present, in an acidic environment, which could result from fluorine attacking metals present.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-09-01T08:48:54.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35728" PostTypeId="1" AcceptedAnswerId="35731" CreationDate="2015-09-01T09:35:39.557" Score="2" ViewCount="710" Body="&lt;p&gt;$$\ce{O2(g) -&amp;gt;[h\nu][\nu\,&amp;lt;\,240~nm]2O^{.}(?)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is split by UV light below 240 nm to form oxygen radicals, which then go on to react with oxygen molecules to form ozone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, is the oxygen a free radical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen has six outer shell electrons, but my teacher mentioned that by the way orbitals work, they don't actually all pair off, rather you end up with two pairs of electrons and two unpaired electrons, hence making oxygen a free radical, though I'm not too sure. But because the free radical is so highly reactive, I'm not sure whether I'm even allowed to assign it a state, because straight after being formed it pretty much immediately reacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going by process of elimination, it can't be aqueous, because it isn't dissolved in water, and I don't think it is solid, because solids are structurally rigid. I am therefore inclined to say that it is a gas, but my main problem is that each reaction only creates two oxygen free radicals, which immediately disappear, so it shouldn't actually show any properties of being a gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the state I should assign the oxygen, or does it simply not need one?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="4945" LastEditDate="2015-09-01T10:49:40.733" LastActivityDate="2015-09-01T10:49:40.733" Title="What is the state of the oxygen free radical in the following reaction?" Tags="&lt;phase&gt;&lt;radicals&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35730" PostTypeId="1" AcceptedAnswerId="37707" CreationDate="2015-09-01T10:25:30.573" Score="4" ViewCount="92" Body="&lt;p&gt;Hydrogen cyanide shows nucleophilic addition at carbonyl compounds' carbon to form a cyanohydrin. Alcohols, similarly, form acetals or ketals. Will hydrogen chloride - or hydrohalic acids in general - show a similar reaction to form alpha halohydrins?&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-09-01T11:08:25.933" LastActivityDate="2015-09-21T22:39:16.103" Title="Nucleophilic Addition of Hydrohalic Acids" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35731" PostTypeId="2" ParentId="35728" CreationDate="2015-09-01T10:44:30.117" Score="3" Body="&lt;p&gt;According to the reaction $$\ce{O2\,(g) -&amp;gt;[h\nu][\nu\,&amp;lt;\,240~\mathrm{nm}] 2O}$$&#xA;photolytic cleavage of oxygen yields two radicals. These will most likely be in the high spin triplet state. And yes these are &lt;a href=&quot;http://goldbook.iupac.org/R05066.html&quot; rel=&quot;nofollow&quot;&gt;free radicals according to the definition of the IUPAC Goldbook&lt;/a&gt;, as they have two unpaired electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you are talking about state of matter and then of course, you are right. Oxygen radicals can exist neither in aqueous solution, nor liquid, nor solid. Even in gas phase the concentration of oxygen radicals will be very low as they are an incredibly short lived species. They might exist in &lt;a href=&quot;http://goldbook.iupac.org/P04687.html&quot; rel=&quot;nofollow&quot;&gt;plasma&lt;/a&gt;, too. Keeping that in mind it is probably unnecessary to assign a state to the oxygen radical.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-09-01T10:44:30.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35732" PostTypeId="2" ParentId="35728" CreationDate="2015-09-01T10:46:02.027" Score="2" Body="&lt;p&gt;It does not need one. Short-lived reactive intermediates never form a &lt;em&gt;phase&lt;/em&gt; of their own. There are just too few of them at any given moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, however, that lone oxygen atom does exist inside some phase, and that is the gas phase. So it would not be entirely inappropriate to assign it the gas state. With some other reactions, it could have been different. There are examples of reactions between gases that really occur on the solid surface of certain catalyst. In such a case, the intermediates would not fly in the air among the gas molecules, but sit on that surface. (Not that I would then call them &quot;solid&quot;, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the radicals with two unpaired electrons use to be called &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Diradical&quot; rel=&quot;nofollow&quot;&gt;diradicals&lt;/a&gt;&lt;/strong&gt;, and that's what your atomic oxygen is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the molecule $\mathrm O_2$, despite being quite stable by itself, is also a diradical, for unrelated reasons.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-01T10:46:02.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35733" PostTypeId="1" CreationDate="2015-09-01T11:02:35.953" Score="4" ViewCount="6113" Body="&lt;p&gt;Maleic acid (the &lt;em&gt;cis&lt;/em&gt; isomer of the dicarboxylic acid $\ce{C4H4O4}$) readily forms its anhydride on heating. Clayden's organic chemistry textbook mentions this as a point of distinction between it and fumaric acid (the &lt;em&gt;trans&lt;/em&gt; isomer), which fails to dehydrate to its anhydride upon heating. However, I have discovered that at temperatures above 200 $^{\circ}\mathrm{C}$, fumaric acid forms maleic anhydride as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the correct answer? I don't think it is likely that fumaric acid can form the anhydride - even at higher temperatures - given the &lt;em&gt;trans&lt;/em&gt; configuration and restricted rotation.&lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="10185" LastEditDate="2015-09-01T20:35:15.987" LastActivityDate="2017-06-13T21:56:57.023" Title="Does fumaric acid form maleic anhydride upon heating?" Tags="&lt;organic-chemistry&gt;&lt;geometrical-isomerism&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35734" PostTypeId="1" AcceptedAnswerId="39998" CreationDate="2015-09-01T11:12:09.113" Score="9" ViewCount="18236" Body="&lt;p&gt;Why can ozone act as a barrier of UV light, so that we don't get irradiated by the sun's energy, and not any old molecule? In a recent exam I sat, one of the questions was &quot;Name one physical property of ozone and contrast it with oxygen&quot; and under exam pressure I spurted this onto the page: &quot;Because the coordinate covalent bond in ozone is weaker than a traditional covalent bond, the bond is free to 'wiggle' about a bit. This means that the ozone can absorb a range of frequencies of photons, as a completely rigid molecule like oxygen would only have very specific frequencies which it could absorb.&quot; How much of this 'guess' is grounded in fact, or in other words, what is the actual mechanism that allows ozone to protect us from UV rays and not any old molecule?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastActivityDate="2019-04-26T22:52:16.887" Title="What property allows ozone to absorb uv light" Tags="&lt;ozone&gt;" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35736" PostTypeId="1" CreationDate="2015-09-01T12:07:30.093" Score="2" ViewCount="145" Body="&lt;p&gt;I always thought vodka was just alcohol and water, and that the ideal vodka was one lacking all flavor components characteristic of the grain or potato mash that it was distilled from. I once saw a vodka process diagram showing multiple distillation and charcoal filtration steps--suggesting an effort to remove all flavor components. But looking at the prices being charged for high end vodkas, I am wondering what the draw is. Are there flavor components in these high end vodkas that make them that much more desirable than pure alcohol that people pay more than $50 a bottle for them? &lt;/p&gt;&#xA;" OwnerUserId="5577" LastEditorUserId="10185" LastEditDate="2015-09-01T12:28:19.783" LastActivityDate="2015-09-01T14:10:10.403" Title="Flavor components of high end vodka?" Tags="&lt;everyday-chemistry&gt;&lt;food-chemistry&gt;&lt;purification&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35737" PostTypeId="1" AcceptedAnswerId="35760" CreationDate="2015-09-01T12:09:57.953" Score="10" ViewCount="5258" Body="&lt;p&gt;The two spirocyclic compounds drawn below are enantiomers. My question is, do they exhibit axial or planar chirality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RcI3n.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcI3n.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D representations:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qQFvo.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qQFvo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4231" LastEditDate="2015-09-01T22:34:13.210" LastActivityDate="2017-12-13T16:45:47.113" Title="Axial or planar chirality in spiro[3.3]hepta-1,5-diene" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35738" PostTypeId="2" ParentId="35736" CreationDate="2015-09-01T14:10:10.403" Score="1" Body="&lt;p&gt;High end vodka is primarily good vodka+branding. You pay extra to get that designed bottle. Perhaps the manufacturer also does something extra that they claim makes the vodka special. Whether it actually does depends on what they have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good vodka has most flavours removed. Some flavours - specifically those that both have a boiling point close to that of alcohol &lt;em&gt;and&lt;/em&gt; the ability to pass through an active carbon filter will remain.&lt;/p&gt;&#xA;" OwnerUserId="19563" LastActivityDate="2015-09-01T14:10:10.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35740" PostTypeId="1" AcceptedAnswerId="35744" CreationDate="2015-09-01T14:36:40.763" Score="3" ViewCount="719" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E7QBy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E7QBy.png&quot; alt=&quot;Image 1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The longest alkane chain is decane and, attached to the fifth carbon, is an ethyl group the end of which is double bonded to an oxygen atom. My best guess at naming it would be:&lt;br /&gt;&#xA;&lt;strong&gt;5-(ethyl-2-al)decane.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2fh3U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2fh3U.png&quot; alt=&quot;Image 2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, the longest chain is again decane, but there is a propyl group and an oxygen atom double bonded &lt;strong&gt;not&lt;/strong&gt; to the ends of the alkyl, but to carbon #2, which leads me to:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;5-(propyl-2-one)decane&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now I have no clue as to whether I am right or not because I have never come across such molecules and I have just done what I found to be logical. Have I come to the correct conclusions?&lt;/p&gt;&#xA;" OwnerUserId="19507" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-09-01T18:17:21.357" Title="Aldehyde and ketone nomenclature when aldehyde/ketone group is bonded to an alkyl group" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35742" PostTypeId="2" ParentId="35734" CreationDate="2015-09-01T16:08:27.577" Score="2" Body="&lt;p&gt;Firstly, ozone is an unstable compound and decomposes at about $573K$ to form oxygen. When UV radiation reaches ozone layer, ozone molecules present at the layer absorb its energy and decompose to form oxygen which is more stable than ozone.&#xA;One ozone molecule decomposes to form one oxygen atom and a normal oxygen molecule. Hence the energy of UV radiation is absorbed in ozone layer even before reaching earth's surface. That's how ozone layer protects us from UV radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this does not deplete ozone layer since reverse reaction can also take place. That is, high energy UV radiation hits oxygen molecules and each decompose into two free oxygen atoms. Each oxygen atom combines with one oxygen molecule each and forms ozone molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also note that the bond order of ozone is $1.5$ and that for oxygen is $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to read about depletion of ozone layer. Here is a link: &lt;a href=&quot;http://www.theozonehole.com/ozonedestruction.htm&quot; rel=&quot;nofollow&quot;&gt;Ozone hole&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="17624" LastEditDate="2015-09-01T16:15:12.937" LastActivityDate="2015-09-01T16:15:12.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35744" PostTypeId="2" ParentId="35740" CreationDate="2015-09-01T16:28:34.043" Score="6" Body="&lt;p&gt;The characteristic group $\ce{=O}$ corresponds to the substitutive prefix ‘oxo’.&#xA;Thus, the substitutive names that you are looking for are ‘5-(2-oxoethyl)decane’ and ‘5-(2-oxopropyl)decane’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, according to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, the preferred names for aldehydes and ketones derived from alkanes are constructed using the suffixes ‘al’ and ‘one’, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the preferred IUPAC names for the given compounds are ‘3-butyloctanal’ and ‘4-butylnonan-2-one’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/626p7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/626p7.png&quot; alt=&quot;3-butyloctanal&quot;&gt;&lt;/a&gt;  &lt;a href=&quot;https://i.stack.imgur.com/7u6ab.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7u6ab.png&quot; alt=&quot;4-butylnonan-2-one&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-09-01T18:17:21.357" LastActivityDate="2015-09-01T18:17:21.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35745" PostTypeId="1" AcceptedAnswerId="37232" CreationDate="2015-09-01T17:22:22.183" Score="3" ViewCount="506" Body="&lt;p&gt;I work a lot with Urea-Formaldehyde resin/plastic, both in carpentry and textiles. Cleanup while still uncured is simple enough. Cleanup before even mixed is easy enough. What I don't know, is how to cleanup plastic that has set and cured. If it was reasonably safe to guess and try, I would, but in this environment/conditions acetaldehyde formation is very possible, which makes release formaldehyde a problem. Grinding or sanding the plastic is not an option. I haven't seen a MSDS say anything about cleaning up cured plastic. From accidental spills it doesn't seem to disolve in xylol, acetol, or toluol. Anyone have any ideas? As long as there is a preditable outcome I have the resources to test just about anything.&lt;/p&gt;&#xA;" OwnerUserId="13403" LastActivityDate="2015-09-12T22:23:02.757" Title="Urea-Formaldehyde cleanup" Tags="&lt;plastics&gt;&lt;cleaning&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35747" PostTypeId="1" CreationDate="2015-09-01T18:00:15.663" Score="1" ViewCount="547" Body="&lt;p&gt;I'm just wondering why stoke-Raman doesn't go to the ground state. Where does the rest of energy from exciting laser leave?&lt;br&gt;&#xA;On the other hand, why does anti-stoke go back to ground state? I know that the overall energy of anti-stoke is higher but why it has to go back to the ground state?&lt;br&gt;&#xA;My question is based on this diagram.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/K5EsG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5EsG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;As I understand, the anti-stoke has higher energy as it is excited from v1 due to an enhancement. I'm just confused with the last process of anti-stoke when it comes down to the ground state. Why doesn't it come down to the v2 or higher vibrational state like stoke?&lt;/p&gt;&#xA;" OwnerUserId="19568" LastEditorUserId="4945" LastEditDate="2015-09-02T08:40:39.033" LastActivityDate="2016-12-08T14:33:56.843" Title="Why does Stokes raman not relax to the ground state while anti-Stokes does?" Tags="&lt;spectroscopy&gt;&lt;raman&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35748" PostTypeId="1" AcceptedAnswerId="35750" CreationDate="2015-09-01T19:40:47.860" Score="0" ViewCount="1438" Body="&lt;p&gt;Why is zinc sulfate needed in the construction of a daniell cell.  It seems that because the zinc is getting oxidized and not reduced there is no need to have a solution of zinc sulfate as it will produced when the zinc is reduced.  The sulfate ions would simply cross the salt bridge and balance out the charges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vz2KB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vz2KB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16106" LastActivityDate="2015-09-01T19:57:15.557" Title="Why is zinc sulfate used in daniell cell" Tags="&lt;electrochemistry&gt;&lt;redox&gt;&lt;ions&gt;&lt;electricity&gt;" AnswerCount="1" CommentCount="0" ClosedDate="2015-09-01T21:29:17.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="35749" PostTypeId="2" ParentId="35734" CreationDate="2015-09-01T19:41:31.633" Score="2" Body="&lt;p&gt;This is not some intrinsic property of ozone. In fact, any molecule would absorb a range of frequencies of photons. It just so happens that O&lt;sub&gt;3&lt;/sub&gt; covers the very range that is not covered by other components of our atmosphere, and thus saves us from the detrimental effects of UV rays of that range.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-01T19:41:31.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35750" PostTypeId="2" ParentId="35748" CreationDate="2015-09-01T19:57:15.557" Score="-2" Body="&lt;p&gt;Zinc sulfate is not needed. With any other neutral salt in its place, your cell will work just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without &lt;em&gt;any&lt;/em&gt; salt at all, though, it will not work, since pure water is a poor conductor of electric current.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-01T19:57:15.557" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35751" PostTypeId="2" ParentId="35747" CreationDate="2015-09-01T20:11:04.673" Score="2" Body="&lt;p&gt;There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of these are the Stokes lines and which are the anti-Stokes ones, you'd better figure out for yourself.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-01T20:11:04.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35752" PostTypeId="1" CreationDate="2015-09-01T21:18:06.657" Score="10" ViewCount="1559" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An unknown compound was found to react with sodium hydride with the evolution of hydrogen. The compound could not be acetylated under the normal conditions for acetylation. The compound resisted oxidation under mild conditions, and under vigorous conditions yielded only products of molecular weight much smaller than the unknown. On the basis of these facts the compound is:  &lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;a primary alcohol&lt;/li&gt;&#xA;  &lt;li&gt;an aldehyde&lt;/li&gt;&#xA;  &lt;li&gt;a secondary alcohol&lt;/li&gt;&#xA;  &lt;li&gt;a tertiary alcohol&lt;/li&gt;&#xA;  &lt;li&gt;a secondary amine&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the first statement, it is clear that this compound must be an alcohol (since alcohols react with sodium hydride and produce hydrogen gas in this reaction).  I could guess that the answer was a tertiary alcohol as secondary and primary alcohols would easily oxidize, but I'm not sure what the products of the oxidized tertiary alcohol would be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why wouldn't a tertiary alcohol undergo acetylation?  I don't see any reason why it shouldn't?&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="5026" LastEditDate="2018-02-25T03:29:57.483" LastActivityDate="2018-02-25T03:29:57.483" Title="Identification of types of alcohols" Tags="&lt;organic-chemistry&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35753" PostTypeId="1" AcceptedAnswerId="35764" CreationDate="2015-09-01T21:29:21.883" Score="6" ViewCount="1684" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following compounds does &lt;em&gt;not&lt;/em&gt; show tautomerism?&lt;br&gt;&#xA;  &lt;img src=&quot;https://i.stack.imgur.com/xNqMJ.png&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (a), and I'm curious if this is because to show tautomerism, there can't be any net charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, ketones can tautomerize to enols, and more importantly, enamines can tautomerize to imines.  In (a), we would get an imine-like thing, but the nitrogen would be positively charged.  Does this still count as tautomerization?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Thank you everyone for your helpful clarifications, but I'm still a bit confused as to why, mechanistically, we need a hydrogen on the nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I Google the mechanism for imine tautomerization, what if find is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r5BIl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r5BIl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the nitrogen lone pair pushes down and the double bond reaches out to pick up a proton.  There isn't a transfer of hydrogen from the nitrogen to the solvent in the case shown below, yet the molecule still tautomerizes.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="46306" LastEditDate="2018-03-04T14:57:59.160" LastActivityDate="2018-03-04T14:57:59.160" Title="Can a tertiary enamine tautomerize to an imine?" Tags="&lt;organic-chemistry&gt;&lt;tautomer&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35754" PostTypeId="1" CreationDate="2015-09-01T21:44:51.253" Score="2" ViewCount="877" Body="&lt;p&gt;The value for the distribution constant for an organic acid, HA, between chloroform and water is $10^2$.  The dissociation constant of the acid in water is $10^{-5}$, and it can be assumed that it does not dissociate or dimerize in the chloroform phase.  Increasing the pH of the aqueous phase from 4 to 9 has what effect on the fraction of HA extracted into the chloroform phase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) The fraction is more than doubled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) The fraction is increased but not doubled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) The fraction is unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) The fraction is decreased but is not halved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) The fraction is reduced to less than half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (d).  I understand why the concentration of HA in chloroform should decrease, but I'm not sure I'm completely confident as to why it decreases to less than half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the acid in the organic phase is in equilibrium with the acid in the aqueous phase, obviously, increasing the pH past the pK$_a$ of the acid will drive it to completely dissociate, causing the HA in the organic layer to go into the aqueous layer.  How can I quantitatively know that this process will decrease the $\frac{[HA]_{CHCl_3}}{[HA]_{water}}$ to less than 50?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: So, I've taken your suggestion, but I seem to be getting stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use the notation HA$_{org}$ to denote the acid in the chloroform layer and HA$_{aq}$ to denote the acid in the water layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H$^+$ + A$^-$ $\rightleftharpoons$ HA$_{aq}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HA$_{aq}$ $\rightleftharpoons$ HA$_{org}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K$_a$ = $10^{-5}$ and the equilibrium constant for the bottom reaction is K$_D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our wanted ratio, D = $\frac {[HA]_{org total}}{[HA]_{aq total}}$ = $\frac {[HA]_{org}}{[HA]_{aq} + [A^-]} =  \frac {[HA]_{org}}{[HA]_{aq} + 10^{-5}[HA]_{aq}/[H^+]} = \frac{K_D [H^+]}{[H^+] + 10^{-5}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I don't know what K$_D$ is and it would also change with pH.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="23561" LastEditDate="2019-04-28T16:15:14.893" LastActivityDate="2019-04-28T16:15:14.893" Title="Distribution of acid in Organic and Aqueous Phases" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35755" PostTypeId="2" ParentId="24275" CreationDate="2015-09-01T21:50:06.363" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Namely the result $C_V/n=\frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.&#xA;  Is this true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, let's look first at where $C_p-C_V=R$ comes from and then look at $C_V=\frac{3}{2}R$ to see what we find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with the definition of heat capacity as being a change in energy per change in unit temperature,$$\Delta H=\int_{T_1}^{T_2}n\cdot C_p\,\mathrm dT$$&#xA;&lt;strong&gt;Now, I assume heat capacity to be independent of temperature&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,$$\Delta H=n\cdot C_p(T_2-T_1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $H=U+PV$ and pressure is held constant here, we rewrite the expression as $$\Delta U + P\Delta V=n\cdot C_p(T_2-T_1)$$&#xA;By the same integration performed above (but with $C_V$) we find that $\Delta U=n\cdot C_v(T_2-T_1)$&#xA;Combining those expressions and simplifying,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_p-C_V=P\frac{\Delta V}{n\cdot\Delta T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using the ideal gas law,&lt;/strong&gt; with constant pressure, we find,$$\frac{\Delta V}{\Delta T}=\frac{nR}{P}$$&#xA;Plugging that in, $$C_p-C_V=R$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, for a monoatomic ideal gas, energy can only be stored in translation.&#xA;Invoking the equipartition theorem to avoid having to do math and a little physics, we see that the energy of a monoatomic gas will be,$$U=\frac{3}{2}Nk_\mathrm bT$$&#xA;For $N=N_\mathrm A$ particles, we have, $$U=\frac{3}{2}RT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, because $$C_V\equiv\left(\frac{\partial U}{\partial T}\right)_{P,n}$$&#xA;We see that, $$C_V=\frac{3}{2}R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that in our derivation of the relationship $$C_p-C_V=R$$ we both used the ideal gas law and assumed heat capacity to be independent of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer the question quoted at the top of this answer, $\frac{C_V}{n}=\frac{3}{2}R$ is derived from the ideal gas, not the perfect gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, in answer to the other question, our derivation required that we assume heat capacity to be constant with a change in temperature, so it was incorrect to say that heat is dependent on temperature for an ideal gas. It is true, however, that heat capacity varies with temperature for a real gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether or not there is a difference between an ideal and perfect gas, I would look at that Wikipedia page posted in a comment above, but it seems superfluous to define something as a perfect gas when an ideal gas is already well understood and the perfect gas essentially behaves the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps explain some of the math behind this.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorDisplayName="user7951" LastEditDate="2016-09-23T19:18:43.563" LastActivityDate="2016-09-23T19:18:43.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35756" PostTypeId="1" AcceptedAnswerId="35851" CreationDate="2015-09-01T22:05:11.707" Score="5" ViewCount="21996" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;To calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt;, the change in enthalpy at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100^\circ C}$&lt;/span&gt; for the reaction below, one needs what addition information?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{N2(g) + 3H2(g) -&amp;gt; 2NH3(g)}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta H^\circ = \pu{-92.0 kJ}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{25^\circ C}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(a) The equilibrium constant for the reaction at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100^\circ&#xA; C}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(b) The molar heat capacities of the reaction and the products as a&#xA;   function of temperature&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) &lt;span class=&quot;math-container&quot;&gt;$\Delta E^\circ$&lt;/span&gt;, the standard internal energy change for the&#xA;   reaction&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(d) The partial pressures of the reactants and products at&#xA;   &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100^\circ C}$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(e) The entropies of formation for the reactants and products at&#xA;   &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100^\circ C}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (b), but I'm not entirely sure why.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it can't be (a) because I think we need &lt;span class=&quot;math-container&quot;&gt;$\Delta S$&lt;/span&gt; as well.  Also, (c) won't be helpful because it is at standard temperature, and we are wanting information at &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100^\circ C}$&lt;/span&gt;.  Finally, (e), seems correct but they are actually talking about &lt;em&gt;entropy&lt;/em&gt; instead of &lt;em&gt;enthalpy&lt;/em&gt;.  However, I'm not sure how to rule out (d) or justify (b).&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="83758" LastEditDate="2020-03-20T15:48:22.773" LastActivityDate="2020-03-20T19:20:01.190" Title="Calculating Enthalpy Changes at Different Temperatures" Tags="&lt;physical-chemistry&gt;&lt;enthalpy&gt;&lt;entropy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35757" PostTypeId="1" CreationDate="2015-09-01T22:16:59.953" Score="0" ViewCount="109" Body="&lt;p&gt;Hydrolysis of C$_6$H$_5$COO$^{18}$CH$_3$ with aqueous NaOH produces what final product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) CH$_3$$^{18}$OH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) CH$_3$CO$^{18}$ONa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(c) CH$_3$C$^{18}$OONa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(d) CH$_3$$^{18}$ONa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(e) CH$_3$OH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (a), which is puzzling to me.  I think it should be (d), as when the radiolabeled methoxy leaving group is pushed out by the reformation of the carbonyl, it will be negatively charged.  The only cation that is floating around is Na$^+$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, oh were, did this magical proton appear in basic conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:  Since I think my comments will get longer I'll just post my question here.  I agree that the methoxy is in equilibrium with water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the pK$_a$ of methanol is 15.54, and the pK$_a$ of water is 15.74.  Thus this equilibrium will favor the methoxy/water side.  Thus, the slight majority of product should be a methoxy salt.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="19290" LastEditDate="2015-09-02T16:59:13.297" LastActivityDate="2018-05-26T05:57:30.973" Title="Hydrolysis of Radiolabeled Esters" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;&lt;hydrolysis&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35759" PostTypeId="1" AcceptedAnswerId="35784" CreationDate="2015-09-01T22:25:12.017" Score="1" ViewCount="11794" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;For a certain first-order gas-hase reaction&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{A \rightarrow B + C}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;the time required for half of an initial amount of A to decompose is 8.8 min. If the initial pressure of A is 400 mmHg, the time required to reduce the partial pressure of A to 50 mmHg is:&lt;/p&gt;&#xA;&lt;p&gt;(a) 1.1 min&lt;br /&gt;&#xA;(b) 17.6 min&lt;br /&gt;&#xA;(c) 26.4 min&lt;br /&gt;&#xA;(d) 35.2 min&lt;br /&gt;&#xA;(e) 70.4 min&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The correct answer is (c).&lt;/p&gt;&#xA;&lt;p&gt;When I approached this problem, I started off thinking about it as a zeroth order reaction for simplicity.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{400~mmHg \rightarrow 200~mmHg \rightarrow 100~mmHg \rightarrow 50~mmHg}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, if this was a zeroth order reaction (not dependent on the amount of A remaining), then time would be &lt;span class=&quot;math-container&quot;&gt;$\mathrm{3 \cdot 8.8~min = 26.4~min}$&lt;/span&gt;. But, since this is a first order reaction, then [A] does factor into the half life and thus the time to reach &lt;span class=&quot;math-container&quot;&gt;$\mathrm{50~mmHg}$&lt;/span&gt; should be slightly higher than &lt;span class=&quot;math-container&quot;&gt;$\mathrm{26.4~min}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;However, this doesn't seem to be the case. Perhaps it's because we're using pressures instead of concentrations? But isn't pressure synonymous with concentration in the gas phase?&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2018-02-22T01:58:10.783" Title="Kinetics and Half Lives Using Pressure" Tags="&lt;physical-chemistry&gt;&lt;kinetics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35760" PostTypeId="2" ParentId="35737" CreationDate="2015-09-01T22:32:36.213" Score="5" Body="&lt;p&gt;&lt;strong&gt;Useful Definitions (from the IUPAC Gold Book)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/C01062.html&quot; rel=&quot;noreferrer&quot;&gt;chirality plane&lt;/a&gt; - &quot;A planar unit connected to an adjacent part of the structure by a bond which results in restricted torsion so that the plane cannot lie in a symmetry plane. For example with (E)-cyclooctene the chiral plane includes the double bond carbon atoms and all four atoms attached to the double bond; with a monosubstituted paracyclophane the chiral plane includes the monosubstituted benzene ring with its three hydrogen atoms and the three other atoms linked to the ring (i.e. from the substituent and the two chains linking the two benzene rings).&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/P04681.html&quot; rel=&quot;noreferrer&quot;&gt;planar chirality&lt;/a&gt; - chirality resulting from a chirality plane.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/C01059.html&quot; rel=&quot;noreferrer&quot;&gt;chirality axis&lt;/a&gt; - &quot;An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image. For example with an allene abC=C=Ccd the chiral axis is defined by the C=C=C bonds; and with an ortho-substituted biphenyl the atoms C-1, C-1', C-4 and C-4' lie on the chiral axis.&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://goldbook.iupac.org/A00547.html&quot; rel=&quot;noreferrer&quot;&gt;axial chirality&lt;/a&gt; - chirality resulting from a chirality axis&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interpretation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;According to the above definition, a chirality plane must involve restricted torsion (restricted rotation) about a bond.&lt;/li&gt;&#xA;&lt;li&gt;A molecule can contain both a chirality plane and a chirality axis; an example would be an &lt;em&gt;ortho&lt;/em&gt;-substituted biphenyl where the restricted rotation is responsible for chirality (&lt;a href=&quot;https://en.wikipedia.org/wiki/Atropisomer&quot; rel=&quot;noreferrer&quot;&gt;atropisomers&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Axial or planar chirality in spiro[3.3]hepta-1,5-diene&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no restricted rotation that brings about chirality in this molecule.  Unlike &lt;em&gt;trans&lt;/em&gt;-cyclooctene or a biphenyl, this molecule cannot be made to lie in a plane by torsional motion (at least not as long as we maintain the tetrahedral nature about the spiro carbon). Therefore, this molecule does not possess a chirality plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather, this molecule is analogous to the allenes. It does have a chirality axis passing through the $\ce{CH}$ group in each ring furthest from the spiro carbon along with the spiro carbon. &lt;strong&gt;The molecule can be said to possess axial chirality.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-09-01T22:32:36.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35761" PostTypeId="2" ParentId="35757" CreationDate="2015-09-01T22:34:04.250" Score="4" Body="&lt;p&gt;You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word &quot;aqueous&quot; stand for, really?&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-01T22:34:04.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="35762" PostTypeId="2" ParentId="35700" CreationDate="2015-09-01T22:36:39.440" Score="2" Body="&lt;p&gt;If $\ce{CO2}$ is acceptable in the feed gas, along with $\ce{CO}$ and a bit of left-over oxygen, then carbon (coal, charcoal etc.) might do. &lt;a href=&quot;http://www.safetystoragecentre.co.uk/advice/Auto-Ignition-Temperature-5.html&quot; rel=&quot;nofollow&quot;&gt;It's ignition temperature is ~1,000 K.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnesium is more expensive, but leaves less oxygen and has no gaseous byproduct. &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnesium&quot; rel=&quot;nofollow&quot;&gt;It's self-ignition temperature is ~800 K.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many other materials could be used; &lt;em&gt;it really depends on how the gas is to be used&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-01T22:36:39.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35764" PostTypeId="2" ParentId="35753" CreationDate="2015-09-01T23:38:55.160" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tautomer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tautomerization&lt;/a&gt; is a chemical reaction whereby constitutional isomers interconvert. The interconversion process involves the shift of a hydrogen atom accompanied by the shift of a double bond.  Here are a few examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vprc3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vprc3.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In molecule (a) pictured above, there is no hydrogen attached to a nitrogen (it has been replaced by a methyl group), so there is no hydrogen atom available to shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response to OP's edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture you posted in your edit is incorrectly labeled.  The charged molecule on the left is an &lt;strong&gt;iminium ion&lt;/strong&gt;, not an imine.  So &lt;strong&gt;this is not an example of an imine-eneamine equilibrium&lt;/strong&gt;.  You are correct that in this equilibrium (the one pictured in your edit) there is no hydrogen transferred to nitrogen, but again, this is not an example of an imine-eneamine tautomerization.  I think this error is the source of your confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of an imine-eneamine tautomerization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sb9tU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sb9tU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, a hydrogen is transferred between nitrogen and carbon.  This hydrogen transfer can occur with or without solvent molecules playing a role.  That is to say, the hydrogen can be transferred directly between the carbon and nitrogen atoms, or it can be transferred from one of the atoms to a solvent molecule which can then transfer it back to the other atom involved in the tautomerization. The &lt;a href=&quot;http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_13%3A_Reactions_with_stabilized_carbanion_intermediates_I/Section_13.1%3A_Tautomers&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt; also provides a nice explanation of tautomerization.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-09-02T17:28:05.467" LastActivityDate="2015-09-02T17:28:05.467" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35766" PostTypeId="1" CreationDate="2015-09-02T03:30:53.403" Score="4" ViewCount="4201" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.labchem.com/products/online-catalog/browse-products/sulfuric-acid-72-w-w-24n-12m.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;LabChem&lt;/a&gt;, &lt;span class=&quot;math-container&quot;&gt;$12~\mathrm{M}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$72~\mathrm{w/w\%}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I try to convert &lt;span class=&quot;math-container&quot;&gt;$12~\mathrm{M}$&lt;/span&gt; of &lt;span class=&quot;math-container&quot;&gt;$\ce{H2SO4}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\mathrm{w/w\%}$&lt;/span&gt; I get &lt;span class=&quot;math-container&quot;&gt;$117\%$&lt;/span&gt; which is obviously wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{multline}&#xA;12~\mathrm{M}\cdot\left(\frac{1~\mathrm{L}}{1000~\mathrm{ml}}\right)&#xA;\left(1~\mathrm{\frac{ml}{g}}\ \ce{H2O}\right)&#xA;\left(98.09~\mathrm{\frac{g}{mol}}\ \ce{H2SO4}\right)\\&#xA;= \frac{1177~\mathrm{g}}{1000~\mathrm{g}} = 1.177=1.177\cdot100\%=117.7\%&#xA;\end{multline}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must be doing something wrong. How do you convert it correctly?&lt;/p&gt;&#xA;" OwnerUserId="19572" LastEditorUserId="4945" LastEditDate="2019-09-16T12:26:01.167" LastActivityDate="2019-09-16T12:26:01.167" Title="How is sulfuric acid with a concentration of 12 M also 72% w/w?" Tags="&lt;concentration&gt;&lt;units&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35767" PostTypeId="1" CreationDate="2015-09-02T04:47:44.217" Score="4" ViewCount="1216" Body="&lt;p&gt;A solution of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$, when provided in equal amounts at the same concentration, form an excellent buffer. What happens is that the two sodium-phosphate compounds each dissociate independently: &#xA;\begin{align}&#xA;\ce{Na2HPO4 &amp;amp;&amp;lt;=&amp;gt; HPO4^2- + 2Na+}\tag1\\&#xA;\ce{NaH2PO4 &amp;amp;&amp;lt;=&amp;gt; H2PO4^- + Na+}\tag2&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if you made up a solution that had only half of the reactant in (1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in saying that Le Chatelier's principle is not involved, and that shifts in equilibrium don't occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the solution just have a lower buffering capacity for bases relative to acids?&lt;/p&gt;&#xA;" OwnerUserId="19573" LastEditorUserId="4945" LastEditDate="2015-09-02T05:30:08.783" LastActivityDate="2017-03-05T08:27:40.537" Title="What happens if a buffer is made up with only half as much NaH2PO4 compared to Na2HPO4?" Tags="&lt;acid-base&gt;&lt;equilibrium&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35768" PostTypeId="2" ParentId="35488" CreationDate="2015-09-02T05:33:14.860" Score="14" Body="&lt;p&gt;Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you said, it has been found that the hydrogen bond (specifically in water) is about 90% electrostatic and 10% covalent. See &lt;a href=&quot;http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; link for a sort-of-review/overview of hydrogen bonding and some relevant research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon following the paper trail, I found the following research article: &lt;a href=&quot;http://dx.doi.org/10.1016/S0022-3697(99)00325-X&quot; rel=&quot;noreferrer&quot;&gt;E.D. Isaacs, A. Shukla, P.M. Platzman, D.R. Hamann, B. Barbiellini, C.A. Tulk, &lt;em&gt;J. Phys. Chem. Solids&lt;/em&gt;, &lt;strong&gt;2000&lt;/strong&gt;, &lt;em&gt;61&lt;/em&gt;, 403-406&lt;/a&gt;. (&lt;a href=&quot;http://nuweb.neu.edu/bbarbiellini/research/list/BBA_ART/PCS61_isaacs_00.pdf&quot; rel=&quot;noreferrer&quot;&gt;Mirror&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, they use some method which I don't really understand and find quite conclusive evidence of the (word of the day) anisotropy of hydrogen bonds. That means that the bond is indeed directionally dependent. Bond direction is one defining characteristic of covalent bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That raises the question of what is going on with the orbitals that allow for hydrogen to participate in more than one covalent bond at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple explanation of the orbital interactions for a water dimer is to view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer. This superposition which includes the hydrogen then explains the way that hydrogen is participating in one covalent bond and another partially covalent interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, this explanation of what the orbitals are doing fits well with what we would expect from a hydrogen participating in two different covalent interactions, which is that the energy curve has two minima where one is much deeper than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further evidence for the covalent nature of the bond is the fact that the O-H bond elongation and O---H hydrogen bond length always change together. That is, as the O-H bond lengthens, the O---H interaction length shortens. This implies a shift in electron density towards hydrogen which is most easily explained by a delocalization of electrons participating in a covalent interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I hope that's a good start toward understanding the covalent nature of the hydrogen bond--it was for me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer the simple part of the question which you clearly already know is that yes dipole-dipole is a significant part of hydrogen bonding, and it certainly should be, but as is always true of chemistry, there's a lot more going on than what one might initially think.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (06/01/2016)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I posted this answer, I think I have developed a better understanding of the situation, and feel obliged to update my answer. First, let me split hydrogen-bonding into two obvious parts, electrostatic and covalent characteristics. This is an obvious split, but it is not completely clear yet what we mean by &lt;em&gt;covalent&lt;/em&gt;, so let us try to define that by looking at some examples. Furthermore, we must be careful when talking in about hydrogen-bonding because in every hydrogen bond there is a donor and acceptor, and these two roles are often played best by different chemical species.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hydrogen-Bonding in thiols&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we go breaking yet another rule. &lt;em&gt;They told us that sulfur couldn't hydrogen bond.&lt;/em&gt; Well, that's not exactly true. &lt;em&gt;Okay, but that hydrogen-bonding has got to be basically negligible right?&lt;/em&gt; Not exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing we need to do here is get some information. First, sulfur has an electronegativity of 2.58 on the Pauling scale. This is roughly the same as the electronegativity of carbon which we do not consider to hydrogen-bond well. So, if there are hydrogen-bonds involving sulfur, we should expect that they will be primarily covalent as any dipole interactions will be pretty small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it turns out, sulfur is a much better hydrogen-bond acceptor than it is donor, and at times is a comparable acceptor to oxygen. Here is a table comparing certain dimers from a great source you can reference at the bottom of the page$^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEHuO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEHuO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, when the sulfur in these dimers (which are all named in the caption of the table) acts a hydrogen bond acceptor, the shift in vibrational frequency of the $\ce{O-H}$ and the binding energy are comparable to when an oxygen is the acceptor atom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at other research$^4$ involving sulfur hydrogen bonds reveals that approximately 70% of the interaction energy between sulfur and oxygen when sulfur is the acceptor comes from &lt;a href=&quot;https://en.wikipedia.org/wiki/London_dispersion_force#cite_note-5&quot; rel=&quot;noreferrer&quot;&gt;dispersion&lt;/a&gt;. That is the key physical insight in all this. Despite the loss of electronegativity, sulfur gains something in its greater dispersion interactions which makes it a good hydrogen bond acceptor. It should be noted, however, that the electronegative component of these hydrogen-bonds maintains their directionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to compare, according to that wikipedia article I linked, the interaction energy of a water dimer is only 24% dispersion. In water electrostatics dominate and induction and exchange play meaningful roles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have more I could say, but it's perhaps a longer explanation than is worth going into. To conclude, what is meant by covalent interactions in hydrogen-bonding varies based on the species you are considering. In something like water, we can think about this in the usual way by saying that there must be some orbital overlap and hence electronic exchange which will be stabilizing. With other species, we can see that what we mean by covalent interactions might be different. That is, there is still some orbital overlap, but we must also consider non-electrostatic interactions like dispersion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; &lt;a href=&quot;https://books.google.com/books?id=lvNICAAAQBAJ&amp;amp;pg=PR5&amp;amp;lpg=PR5&amp;amp;dq=Hydrogen+Bonds+Involving+Sulfur:+New+Insights+from+ab+Initio+Calculations+and+Gas+Phase+Laser+Spectroscopy&amp;amp;source=bl&amp;amp;ots=GmEvbuNg_h&amp;amp;sig=soF_eyGuKwW6LQ-2KRaPeEKqLK4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwi-xYjV5YjNAhVR82MKHUo9AYIQ6AEIMzAE#v=onepage&amp;amp;q=Hydrogen%20Bonds%20Involving%20Sulfur%3A%20New%20Insights%20from%20ab%20Initio%20Calculations%20and%20Gas%20Phase%20Laser%20Spectroscopy&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Noncovalent Forces-Chapter 2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4&lt;/strong&gt; &#xA;&lt;a href=&quot;http://scitation.aip.org/content/aip/journal/jcp/119/6/10.1063/1.1588291&quot; rel=&quot;noreferrer&quot;&gt;Wennmohs, F., Staemmler, V., &amp;amp; Schindler, M. (2003). Theoretical investigation of weak hydrogen bonds to sulfur. The Journal of chemical physics, 119(6), 3208-3218.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="17583" LastEditDate="2016-06-02T07:30:02.960" LastActivityDate="2016-06-02T07:30:02.960" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35770" PostTypeId="2" ParentId="35766" CreationDate="2015-09-02T05:57:36.333" Score="11" Body="&lt;p&gt;Your problem is that you've assumed the density is 1 gram/mL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that a molar is defined as a mole of solvent per liter of &lt;em&gt;solution&lt;/em&gt;, not &lt;em&gt;solvent&lt;/em&gt;. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you just found out though, sometimes you really need to account for the change in density that occurs as you add solute. &lt;a href=&quot;http://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-sulfuric-acid-and-sulfur-trioxide/&quot;&gt;This webpage&lt;/a&gt; claims that the density of $12~\mathrm{M}$ sulfuric acid is about $1.634~\mathrm{g/mL}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this updated calculation, we still have the original $1177~\mathrm{g}$ of sulfuric acid. However, the solution mass is now $1000~\mathrm{mL}\cdot 1.634~\mathrm{g/mL} = 1634~\mathrm{g}$. Finally, we obtain that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{ 1177~\mathrm{g} \, \ce{H2SO4}}{1634~\mathrm{g} \, \mathrm{solution}} = 72.03\% \; \mathrm{w/w} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note about this calculation: it's possible we're begging the question, since I don't know which values on that linked page are measured and which are calculated. It's possible that the density was calculated from measured data. The important thing here isn't how the data was obtained though, it's that the density of solution changes as you add more solute, and so you have to change how you calculate things from the molarity.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="4945" LastEditDate="2015-09-02T06:20:47.293" LastActivityDate="2015-09-02T06:20:47.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35773" PostTypeId="2" ParentId="35767" CreationDate="2015-09-02T07:29:04.417" Score="2" Body="&lt;p&gt;You are not right, &lt;a href=&quot;https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Le Châtelier's principle&lt;/a&gt; plays a very important role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's have a look at the reactions involving disodium monohydrogen phosphate, as it is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Amphoterism&quot; rel=&quot;nofollow noreferrer&quot;&gt;amphoteric&lt;/a&gt; substance and therefore a buffer on its own:&#xA;\begin{align}\ce{&#xA;Na2HPO4 + H2O &amp;amp;&amp;lt;=&amp;gt; 2Na+ + H3+O + PO4^3- \\&#xA;Na2HPO4 + H2O &amp;amp;&amp;lt;=&amp;gt; 2Na+ + {}^{-}OH + H2PO4^- \\&#xA;}\end{align}&#xA;To a small extent, there will also be the formation of phosphoric acid, but we'll ignore that for the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore the counter ion and formulate the equilibrium constants:&#xA;\begin{align}&#xA;\ce{HPO4^2- + H2O &amp;amp;&amp;lt;=&amp;gt; H3+O + PO4^3-} &amp;amp;&#xA;K_1 &amp;amp;=\frac{c(\ce{PO4^3-})\,c(\ce{H3+O})}{c(\ce{HPO4^2-})\, c(\ce{H2O})} &amp;amp;&#xA;= K_\mathrm{a}\cdot c(\ce{H2O})\\&#xA;\ce{HPO4^2- + H2O &amp;amp;&amp;lt;=&amp;gt; {}^{-}OH + H2PO4^-} &amp;amp;&#xA;K_2 &amp;amp;=\frac{c(\ce{H2PO4^-})\,c(\ce{{}^{-}OH})}{c(\ce{HPO4^2-})\, c(\ce{H2O})}&amp;amp;&#xA;= K_\mathrm{b}\cdot c(\ce{H2O})\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole reaction can be described with a coupled equilibrium constant:&#xA;$$K = K_\mathrm{a}\cdot K_\mathrm{b}\cdot c^2(\ce{H2O})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At all times in a pure solution there will be phosphate, monohydrogen phosphate, dihydrogen phosphate and maybe even phosphoric acid in solution. Changing any of the concentrations of these species will change the equilibrium.&lt;br&gt;&#xA;The same equations can be applied to the sodium dihydrogen phosphate system. The initial concentrations of course will be different.&lt;br&gt;&#xA;Depending on the ratio of the two solutions, the pH will also change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://chemistry.stackexchange.com/q/31830/4945&quot;&gt;buffer capacity&lt;/a&gt; is of course not only dependent on the overall concentration, but also if it is near the ideal buffer point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, &lt;a href=&quot;https://en.wikipedia.org/wiki/Buffer_solution#Polyprotic_acids&quot; rel=&quot;nofollow noreferrer&quot;&gt;polyprotic acids&lt;/a&gt; are very complicated systems and it solely depends on your understanding of the word &quot;excellent&quot; how the buffer behaves. There are many fine tuned buffer systems for various purposes, &lt;a href=&quot;http://delloyd.50megs.com/moreinfo/buffers2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; has quite a few neat tables.&lt;/p&gt;&#xA;" OwnerUserId="4945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-09-02T07:29:04.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35775" PostTypeId="1" AcceptedAnswerId="35812" CreationDate="2015-09-02T09:15:14.403" Score="5" ViewCount="1604" Body="&lt;p&gt;Why is the most favorable conformation of ketones and aldehydes the one where the alkyl group is anti to the other alkyl group (or hydrogen atom in an aldehyde) as attached? Why isn't more stable the conformation where the bond between the alpha and beta carbon atoms can participate in hyperconjugation with the carbonyl pi antibonding MO?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ou1ZF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ou1ZF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the slightly more favorable conformation for butene is the right one:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1avJ9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1avJ9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-09-02T21:48:22.890" LastActivityDate="2015-09-03T09:14:00.053" Title="Most stable conformation of aldehydes and ketones" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;carbonyl-compounds&gt;&lt;conformers&gt;" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35777" PostTypeId="2" ParentId="35737" CreationDate="2015-09-02T10:52:57.257" Score="4" Body="&lt;p&gt;With regard to &lt;strong&gt;nomenclature&lt;/strong&gt; of chiral spiro compounds, the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; distinguishes three cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;stereogenic spiro atoms of the type ‘Xabcd’, where ‘a’ &gt; ‘b’ &gt; ‘c’ &gt; ‘d’&lt;/li&gt;&#xA;&lt;li&gt;stereogenic spiro atoms of the type ‘Xabab’, where ‘a’ &gt; ‘b’&lt;/li&gt;&#xA;&lt;li&gt;axial chirality of spiro compounds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The given example spiro[3.3]hepta-1,5-diene contains a stereogenic spiro atom of the type ‘Xabab’. The stereodescriptors ‘&lt;em&gt;R&lt;/em&gt;’ and ‘&lt;em&gt;S&lt;/em&gt;’ are used when the spiro atom ‘X’ is surrounded by four atoms arranged as equivalent pairs ‘a’/‘a'’ and ‘b’/‘b'’, where ‘a’ &gt; ‘b’. Thus, the two stereosiomers are (&lt;em&gt;R&lt;/em&gt;)-spiro[3.3]hepta-1,5-diene and (&lt;em&gt;S&lt;/em&gt;)-spiro[3.3]hepta-1,5-diene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/92GhL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/92GhL.png&quot; alt=&quot;(R)-spiro[3.3]hepta-1,5-diene&quot;&gt;&lt;/a&gt;  &lt;a href=&quot;https://i.stack.imgur.com/EPX5e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPX5e.png&quot; alt=&quot;(S)-spiro[3.3]hepta-1,5-diene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that no stereodescriptor ‘&lt;em&gt;E&lt;/em&gt;’ or ‘&lt;em&gt;Z&lt;/em&gt;’ is needed to describe a double bond when the stereogenic unit is located in a ring having less than eight members.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of comparison, (2​&lt;em&gt;R&lt;/em&gt;,4​&lt;em&gt;S&lt;/em&gt;,6​&lt;em&gt;R&lt;/em&gt;)-2,6-dichlorospiro[3.3]heptane is a spiro compound with axial chirality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lu65B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lu65B.png&quot; alt=&quot;(2R,4S,6R)-2,6-dichlorospiro[3.3]heptane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the sterodescriptors ‘&lt;em&gt;R&lt;/em&gt;’ and ‘&lt;em&gt;S&lt;/em&gt;’ are used to describe the chirality centers. The stereodescriptors ‘&lt;em&gt;M&lt;/em&gt;’ and ‘&lt;em&gt;P&lt;/em&gt;’ can also be used to describe axial chirality of spiro compounds; however, sterodescriptors ‘&lt;em&gt;R&lt;/em&gt;’ and ‘&lt;em&gt;S&lt;/em&gt;’ are used in preferred IUPAC&#xA;names.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound (1​&lt;em&gt;R&lt;/em&gt;)-5'&lt;em&gt;H&lt;/em&gt;-spiro[indene-1,2'-(1,3)oxazole] is an example for a stereogenic spiro atom of the type ‘Xabcd’, where ‘a’ &gt; ‘b’ &gt; ‘c’ &gt; ‘d’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NL24N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NL24N.png&quot; alt=&quot;(1R)-5&amp;#39;H-spiro[indene-1,2&amp;#39;-(1,3)oxazole]&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-09-02T10:52:57.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35779" PostTypeId="5" CreationDate="2015-09-02T12:07:59.310" Score="0" Body="&lt;p&gt;In science fiction series/films/books it's not necessarily the science part that comes first. Often it's more important to imagine something that brings the best effects for the viewers. Another point for such &lt;em&gt;errors&lt;/em&gt; is, that eventually proposed synthesizing methods shall not be copied by the viewers for safety reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very often leads to an inaccurate description of the underlying chemistry/physics as in simple cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;explosions that are too huge or&lt;/li&gt;&#xA;&lt;li&gt;things that can be seen under optical microscopes that are way below&#xA;the diffraction limit of optical microscopes like carbenes &lt;a href=&quot;http://www.imdb.com/title/tt1910615/&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and in more complicated cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;blue meth from Breaking Bad that actually is blue rock candy &lt;a href=&quot;https://www.youtube.com/watch?v=kBzNn0Y6bjI&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;the same blue meth that is synthesized in a way that will not lead to the product as it is described in the series.&lt;/li&gt;&#xA;&lt;li&gt;and so on ...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Such discrepancies between fiction and reality are scoped under this tag to clarify what is reality and what is fiction or at least how real the fiction might be.&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="402" LastEditDate="2017-05-03T01:47:22.917" LastActivityDate="2017-05-03T01:47:22.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35780" PostTypeId="4" CreationDate="2015-09-02T12:07:59.310" Score="0" Body="Chemistry-in-fiction covers all chemistry that can be seen in TV series, films and can be read in books, etc. which does not necessarily need to be based on proper scientific backgrounds." OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-09-02T13:35:21.173" LastActivityDate="2015-09-02T13:35:21.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35784" PostTypeId="2" ParentId="35759" CreationDate="2015-09-02T13:06:15.773" Score="3" Body="&lt;p&gt;Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the usual expression for half-life (where I've rearranged to solve for $k$) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the &quot;right&quot; answer using the wrong methodology.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-09-02T13:18:01.347" LastActivityDate="2015-09-02T13:18:01.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35786" PostTypeId="1" CreationDate="2015-09-02T13:54:07.883" Score="2" ViewCount="322" Body="&lt;p&gt;Okay, so molarity is defined as &lt;strong&gt;moles of solute per liter of solution&lt;/strong&gt;. It is used to describe the concentration. So in the lab, when I see a bottle labeled something like $\mathrm{3.5~M}~\ce{H2SO4}$, is it saying that the &lt;strong&gt;only&lt;/strong&gt; chemical in the bottle is this $\ce{H2SO4}$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the way I think of it is &quot;there are 3.5 moles of $\ce{H2SO4}$ per liter of solution&quot;. The word &quot;solution&quot; is kind of throwing me off because it makes me feel like they are saying that there is more than just $\ce{H2SO4}$ in the bottle. Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know why I'm having such a hard time grasping this. But any help would be awesome! &lt;/p&gt;&#xA;" OwnerUserId="19584" LastEditorUserId="10185" LastEditDate="2015-09-02T16:25:51.623" LastActivityDate="2015-10-04T22:33:57.397" Title="Something about the idea of Molarity is confusing me" Tags="&lt;experimental-chemistry&gt;&lt;solutions&gt;&lt;concentration&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35787" PostTypeId="1" AcceptedAnswerId="35788" CreationDate="2015-09-02T13:59:11.707" Score="3" ViewCount="3394" Body="&lt;p&gt;How to convert ethanol to 1,1-dichloroethane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach: Convert ethanol to ethene followed by $\ce{HCl}$ addition. But that's no use. Well if it is possible to create gem-diol from here, then maybe it can be done. But not getting a suitable reaction. Can't seem to figure out anything. Please help!&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4231" LastEditDate="2015-09-02T14:20:01.407" LastActivityDate="2015-09-02T14:20:01.407" Title="How to convert ethanol to 1,1-dichloroethane?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35788" PostTypeId="2" ParentId="35787" CreationDate="2015-09-02T14:17:09.227" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;convert ethanol to 1,1-dichloroethane?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A standard route for producing gem-dichloro compounds involves treatment of a ketone or aldehyde with $\ce{PCl5}$ (see the second to last paragraph in the &quot;Chlorination of organic compounds&quot; section &lt;a href=&quot;https://en.wikipedia.org/wiki/Phosphorus_pentachloride&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and page 18 &lt;a href=&quot;https://books.google.com/books?id=0RQEsfYCuBcC&amp;amp;pg=PA2&amp;amp;lpg=PA2&amp;amp;dq=synthesis%20of%20gem-dichloro%20compounds&amp;amp;source=bl&amp;amp;ots=RZqIIMocxM&amp;amp;sig=SLtiMoFXiJn48r3RdQopDJV2d8I&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCUQ6AEwAWoVChMI3Obav8HYxwIVS8uACh35lAMS#v=onepage&amp;amp;q=synthesis%20of%20gem-dichloro%20compounds&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hGuD0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGuD0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Geminal&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you could carefully oxidize your ethanol to acetaldehyde and treat it with $\ce{PCl5}$.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastActivityDate="2015-09-02T14:17:09.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35790" PostTypeId="2" ParentId="35786" CreationDate="2015-09-02T14:58:28.583" Score="2" Body="&lt;p&gt;I think this is the best way to clear your doubt. Have you ever prepared tea? What are the main ingredients of tea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{water~(or~milk) + tea~leaves -&amp;gt; tea}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the addition of tea leaves which changes water to tea. Depending on how strong you want a tea you can add one teaspoon, two teaspoons or n number of teaspoons of tea leaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, in $\mathrm{3.5~M}~\ce{H2SO4}$ solution it contains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H2SO4~(solute) + H2O~(solvent) -&amp;gt; H2SO4~solution}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge there is no 100 percent pure $\ce{H2SO4}$ available commercially. The commercially available $\ce{H2SO4}$ is 96 percent $\ce{H2SO4}$ with a concentration approximately of 18 molar. See &lt;a href=&quot;http://crescentok.com/staff/jaskew/isr/safety/cabconc.htm&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="139" LastEditorUserId="10185" LastEditDate="2015-09-02T16:29:17.597" LastActivityDate="2015-09-02T16:29:17.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35791" PostTypeId="1" AcceptedAnswerId="35795" CreationDate="2015-09-02T14:58:52.050" Score="5" ViewCount="31946" Body="&lt;p&gt;Why doesn't electron pairing take place in $\ce{[Fe(NH3)6]^2+}$ though $\ce{NH3}$ is a strong ligand field producer?&lt;/p&gt;&#xA;" OwnerUserId="19503" LastEditorUserId="10185" LastEditDate="2015-09-02T16:20:19.590" LastActivityDate="2016-09-16T15:45:09.677" Title="Why don't electrons pair in [Fe(NH3)6]2+?" Tags="&lt;inorganic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="35792" PostTypeId="1" CreationDate="2015-09-02T15:11:07.697" Score="-1" ViewCount="118" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Add $2.0\:\mathrm g$ of tin to $6.35\:\mathrm g$ iodine.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 0.017 moles of tin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are 0.0500 moles of iodine. (0.0250 moles of $\ce{I2}$).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Prepare ratio. For every 1.5 $\ce{I2}$s there are 1 $\ce{Sn}$. This doesn't make it easy, therefore there are 3 $\ce{I2}$s for every 2 $\ce{Sn}$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Construct equation. $\ce{2Sn + 3I2 -&amp;gt; 2SnI4}$ (doubled because we doubled the amount of reactants so that it would make it easier to understand).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But this means that there is too little iodine and no excess reactant! Where did I go wrong?!&lt;/p&gt;&#xA;" OwnerUserId="19586" LastEditorDisplayName="user7951" LastEditDate="2015-09-02T16:02:20.567" LastActivityDate="2015-09-02T18:35:33.770" Title="Second attempt at constructing a chemical equation for tin (IV) iodide" Tags="&lt;homework&gt;&lt;stoichiometry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2020-08-31T16:05:47.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="35793" PostTypeId="1" AcceptedAnswerId="37400" CreationDate="2015-09-02T16:08:19.283" Score="10" ViewCount="152318" Body="&lt;p&gt;I always have trouble naming my graphs in class. They never seem too scientific or professional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, we are reviewing density now, and had to graph the mass (Y) of different volumes (X) of water. So, very simply, my title was &quot;Mass of Different Volumes of Water&quot;. Is this appropriate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any naming conventions, such as &quot;X vs. Y&quot;, or &quot;Effects of X on Y&quot; or &quot;Relationship of X and Y&quot;, etc.? Sorry if this is a bit stylistic and opinion based, but I would like some feedback. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I'm in the tenth grade.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note 2: I'm also not sure what site to post this under (there's no &quot;science&quot; site) nor what tags to post this with. Feel free to improve it.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="11524" LastEditorUserId="41556" LastEditDate="2017-10-18T04:00:57.853" LastActivityDate="2017-10-18T04:00:57.853" Title="Naming convention for graphs?" Tags="&lt;nomenclature&gt;" AnswerCount="4" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35794" PostTypeId="2" ParentId="35786" CreationDate="2015-09-02T16:37:19.883" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are, unfortunately, incorrect. In wet chemistry, you need just a &lt;em&gt;solvent&lt;/em&gt;$^\dagger$ and at least one &lt;em&gt;solute&lt;/em&gt; to make a solution.  While this does mean that you need at least two &lt;em&gt;constituents&lt;/em&gt; to make up a solution, you do not need at least two &lt;em&gt;solutes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... is it saying that the ONLY chemical in the bottle is this $\ce{H2SO4}$?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on your definition of a &lt;em&gt;chemical&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you consider water to be a chemical, then &lt;strong&gt;&lt;em&gt;no&lt;/em&gt;&lt;/strong&gt;: there are two chemicals in the bottle$^\ddagger$, $\ce{H2O}$ and $\ce{H2SO4}$.&lt;/li&gt;&#xA;&lt;li&gt;If you are not considering water to be a chemical, then &lt;strong&gt;&lt;em&gt;yes&lt;/em&gt;&lt;/strong&gt;: the only chemical in the bottle$^\ddagger$ is $\ce{H2SO4}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\dagger$ If not specified, the solvent is almost always water.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;$^\ddagger$ Except for small amounts of inevitable impurities.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="11367" LastActivityDate="2015-09-02T16:37:19.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35795" PostTypeId="2" ParentId="35791" CreationDate="2015-09-02T16:39:27.627" Score="8" Body="&lt;p&gt;(If you know the basics, go to last paragraph and skip the introduction part)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of ligands. Strong field ligands and weak field ligands. Those which only cause a small extent of crystal field splitting is called weak field ligand and those which cause large splitting is called strong field ligand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Why does crystal field splitting take place?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take an octahedral complex. The $d_{x^2-y^2}$ and $d_{z^2}$ point along x,y,z axis and the $t_{2g}$ orbitals point between the axis. When the 6 ligands approach along the three axes, the energy of the $e_g$ orbitals increase more than the increase in the energy of $t_{2g}$ orbitals. Hence the d-orbital splits into two. Filling each electron in $t_{2g}$ orbitals decrease energy by $0.4\Delta_0$ and filling electrons in $e_g$ increase energy by $0.6\Delta_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to your question, you know how electrons are filled in the orbitals (according to hund's rule of course). But how does electron pairing take place in your case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the energy required to pair an electron be P.&#xA;Let's say there are 4 electrons. The two arrangements are one electron in each sub-orbital (no pairing) and 4 electrons all in $t_{2g}$ itself (instead of 4th electron going to $e_g$, its paired up in $t_{2g}$).&#xA;In the first case, the total energy is $-0.6\Delta_0$ and in second case, its $-1.6\Delta_0+P$. So, only strong field ligands can form the second arrangement. $P$ value is constant for a given metal ion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastActivityDate="2015-09-02T16:39:27.627" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35797" PostTypeId="1" CreationDate="2015-09-02T17:47:07.683" Score="6" ViewCount="11024" Body="&lt;p&gt;Cyclohexanol has a pKa = 16, while benzyl alcohol has a pKa = 15.4 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given explanation is that the &quot;polarity of the benzyl group stabilizes the negative charge on oxygen&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that when resonance is not a concern, benzyl groups are primarily electron donating by the inductive effect... which would destabilize charge on oxygen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the inside of the aromatic ring is more negative, and the outside is more positive. Is it this positive charge that stabilizes the negative charge on oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HTb9f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HTb9f.png&quot; alt=&quot;benzyl alcohol anion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="16683" LastEditDate="2017-09-07T15:07:54.513" LastActivityDate="2017-09-07T15:17:47.813" Title="Why is benzyl alcohol a slightly stronger acid than cyclohexanol?" Tags="&lt;organic-chemistry&gt;&lt;acid-base&gt;&lt;alcohols&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35798" PostTypeId="1" AcceptedAnswerId="35801" CreationDate="2015-09-02T17:52:45.197" Score="0" ViewCount="12282" Body="&lt;p&gt;I got one method [here](using acetaldehyde and carbon monoxide)&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/i360033a018&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there any alternative method (using specific catalyst)?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorDisplayName="user14857" LastEditDate="2015-09-02T21:35:55.723" LastActivityDate="2019-02-21T12:18:14.403" Title="Methods to prepare lactic acid from acetaldehyde" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35799" PostTypeId="1" CreationDate="2015-09-02T18:17:15.097" Score="0" ViewCount="38" Body="&lt;p&gt;&lt;strong&gt;Original Question Posted 8/31/15&lt;/strong&gt;&#xA;We are trying to predict what will happen when Carbon Dioxide gas is introduced into a contact chamber that serves as an Ozone Contact Chamber.  These Contact chambers presently oxidize compounds found in Municipal Pool Water along with Chlorine at below 5ppm. there are 3 contact chambers (one for each pool) and the Temperature range is from 104 to 80 F. We are switching from Hydrochloric acid to Carbon Dioxide to keep the Total Alkalinity from dropping when the acid is introduced.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Predict what will happen&quot; is exceptionally broad. If you can narrow your question to specifics, that will help a great deal. For instance, we don't know what the compounds are in the municipal pool water. See this link for a guide to asking questions here on chemistry.SE. –  Todd Minehardt yesterday    &#xA;Ozone and carbon dioxide will be pretty much indifferent to each other. Indeed, ozone is known for its ability to oxidize almost everything, but CO  2   is already a final product of oxidation. &quot;What is dead may never die&quot;, they say. Likewise, CO  2   may never be oxidized. –  Ivan Neretin yesterday &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thank you for taking the time to respond Todd and Ivan. The other compounds in the municipal pool water would include body oils, salt from perspiration and whatever else a particular bather might introduce. We are using Calcium Hypochlorite as a chlorine source and, depending on the PH may also leach Calcium from the Plaster of Pool Walls. The calcium harness varies in the Three separate bodies of water from 400 to 1200 and we endeavor to keep the PH between 7.2 and 7.6.  The Total Alkalinity is around 20 in the 2 bodies of water in which the PH is controlled by adding Hydrochloric acid. We have started using Carbon Dioxide in our smallest body of water and are introducing it in the water returning to the pool perhaps 150 ft away the Total Alkalinity in that pool has risen to 120 ppm which helps to maintain a more stable PH.  Because the piping to the other 2 pools is not as long we are thinking about introducing the Carbon Dioxide into the line the ozone is injected into before the contact chambers. Ivan stated that the Carbon Dioxide would not interact with the ozone.  My research implies that both Carbonic Acid (the objective of injecting Carbon Dioxide) and Ozone are very unstable, and Carbonic Acid has a very short life.  &lt;strong&gt;Is it likely that the Carbonic Acid would interact with the Ozone before the Carbonic Acid can effect a PH change (in effect neutralizing both objectives)?&lt;/strong&gt;        Thank you  for any feedback you may provide.&lt;/p&gt;&#xA;" OwnerUserId="19549" LastActivityDate="2015-09-02T18:17:15.097" Title="In Water Solution will there be an interaction between Ozone Gas and Carbonic Acid" Tags="&lt;organic-chemistry&gt;" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2015-10-17T15:03:52.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="35800" PostTypeId="1" CreationDate="2015-09-02T18:23:30.120" Score="0" ViewCount="170" Body="&lt;p&gt;The production of silver ions in Tollen's test violates Werner's theory of &#xA;non - ionizable secondary valency since we are using the diamminesilver(I) complex as the reagent. How does one reason this?? &lt;/p&gt;&#xA;" OwnerUserId="19592" LastActivityDate="2015-09-02T19:27:37.593" Title="Tollen's Test and Werner's Theory" Tags="&lt;organic-chemistry&gt;&lt;coordination-compounds&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35801" PostTypeId="2" ParentId="35798" CreationDate="2015-09-02T18:30:16.110" Score="7" Body="&lt;p&gt;A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vGz6n.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vGz6n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-09-02T18:30:16.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35802" PostTypeId="2" ParentId="35309" CreationDate="2015-09-02T18:35:33.770" Score="3" Body="&lt;p&gt;The balanced chemical equation is independent of the amounts of each reactant added. In this case we can easily deduce the equation to be: $$\ce{Sn + 2I2 -&amp;gt; SnI4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the given masses tell us is that there is excess tin being added. We can see this because the $\ce{Sn:I2}$ ratio added is $1:1.5$ whereas the stoichiometric ratio is $1:2$ and so there is less iodine than is required to completely react with the added tin - or in other words there is an excess of tin.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-09-02T18:35:33.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35804" PostTypeId="1" CreationDate="2015-09-02T19:20:14.880" Score="15" ViewCount="622" Body="&lt;p&gt;I can detect a distinct smell when I am near water going over a weir. I could tell that I was near &quot;white water&quot; if I was blindfold. Can anyone tell me what causes this odour?&lt;/p&gt;&#xA;" OwnerUserId="19593" LastEditorUserId="10185" LastEditDate="2015-09-02T19:46:45.827" LastActivityDate="2019-01-19T00:01:46.510" Title="What causes the &quot;white water&quot; smell?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;&lt;smell&gt;" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="35805" PostTypeId="2" ParentId="35800" CreationDate="2015-09-02T19:27:37.593" Score="2" Body="&lt;p&gt;Ionizable and non-ionizable in this context refers to the observation that the counterions of a complex ionize in solution and can thus be precipitated and the ligands of the complex cannot and are thus non-ionizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two things at work here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;All complexes exist in an equilibrium with the free ligands. In the case of diamminesilver(I), $$\ce{Ag+ + 2NH3 &amp;lt;=&amp;gt; [Ag(NH3)2]^+}$$ the equilibrium is overwhelmingly toward the complex and no AgCl will be precipitated if chloride is added. However, if we add iodide instead, we will actually observe an AgI precipitate because AgI is far less soluble than AgCl and there is sufficient $\ce{Ag+}$ that we see some precipitate. Therefore, there is actually always a little bit of $\ce{Ag+}$ free in solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Tollen's test is a redox reaction. Unlike the case above where we're precipitating an ionic compound and thus need to form an ionic bond, electron transfer to the silver is quite easy and once reduced, it is unfavourable for the silver to remain in solution. Even more extensive ligand coverage in something like $\ce{[Fe(CN)6]^{4-}}$ doesn't prevent redox reactions from occurring. Even some metalloporphyrins undergo redox reactions at the metal centre.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7499" LastActivityDate="2015-09-02T19:27:37.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35806" PostTypeId="1" CreationDate="2015-09-02T19:31:45.533" Score="5" ViewCount="3526" Body="&lt;p&gt;I have three lab coats, two with long sleeves, and one with short sleeves.&#xA;I saw that the concepts about lab safety vary from country to country; in some places it looks like the short-sleeve coat aren't allowed or they tell you to avoid it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little confused about it: When should I use long-sleeves and when should  I use the short-sleeve coat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the short-sleeves really wrong in a chem lab?&lt;/p&gt;&#xA;" OwnerUserId="19356" LastEditorDisplayName="user15489" LastEditDate="2015-09-03T11:04:33.940" LastActivityDate="2015-10-26T16:12:19.817" Title="Lab safety - Long vs. Short Sleeves on Lab Coats" Tags="&lt;safety&gt;&lt;teaching-lab&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35807" PostTypeId="1" CreationDate="2015-09-02T20:19:17.257" Score="1" ViewCount="1049" Body="&lt;p&gt;Does the photoelectric effect occur while heating food on a metal plate in a microwave oven? &lt;em&gt;If&lt;/em&gt; not, then why doesn't it occur? &lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="4231" LastEditDate="2015-09-02T20:40:41.190" LastActivityDate="2015-09-02T20:44:11.397" Title="Does Photoelectric effect occur in a micro-oven while heating food?" Tags="&lt;physical-chemistry&gt;&lt;metal&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-03T06:36:39.883" ContentLicense="CC BY-SA 3.0" />
  <row Id="35808" PostTypeId="1" CreationDate="2015-09-02T20:27:16.750" Score="0" ViewCount="4414" Body="&lt;p&gt;I want to react my ketone with methylmagnesium iodide (the Grignard reagent). I then need to use a dilute acid solution to protonate the $\ce{MgI-O-R'}$ product to produce my alcohol: $\ce{H-O-R'}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I was trying to figure out what acid to use. The first thing that came to mind was sulfuric acid. What would the products be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the reaction should look like: $$\ce{R'=O + XMgI + H-A -&amp;gt; X-R'-OH + MgI+ + A-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if I use sulfuric acid the conjugate base reacted with the Grignard products is something like $\ce{MgIHSO4}$. Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This molecule doesn't look very stable so my thinking is that the $\ce{Mg}$ donates an electron to $\ce{I}$ in $\ce{MgI+}$ to form elemental iodine and $\ce{Mg^2+}$, then this reacts with $\ce{HSO4}$ to produce $\ce{Mg(HSO4)2}$ and $\ce{MgSO4}$.&lt;/p&gt;&#xA;" OwnerUserId="19218" LastEditorUserId="10185" LastEditDate="2015-09-02T20:42:49.327" LastActivityDate="2015-09-02T22:58:27.383" Title="What are the products of the Grignard reaction with methyl-magnesium iodide?" Tags="&lt;organic-chemistry&gt;&lt;organometallic-compounds&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35809" PostTypeId="2" ParentId="35807" CreationDate="2015-09-02T20:38:22.803" Score="2" Body="&lt;p&gt;No, &lt;strong&gt;the photoelectric effect does not occur in a microwave oven.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The workfunction for a metal is the minimum photon energy required to eject an electron from the metal via the photoelectric effect.  Workfunctions are typically in the range of several electron volts (eV).  Potassium, for example, has a workfunction around 2 eV, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for a list..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy in a microwave photon can range from 0.00001 to 0.001 eV; &lt;strong&gt;much to low to eject an electron from a metal&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more background on the photoelectric effect see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/18935/doubt-on-photoelectric-effect/18936#18936&quot;&gt;this earlier SE Chem Q&amp;amp;A&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.900" LastActivityDate="2015-09-02T20:44:11.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35810" PostTypeId="1" CreationDate="2015-09-02T20:59:23.513" Score="1" ViewCount="469" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The rate of a reaction between A + B was followed spectroscopically at three different wavelengths, and the percent absorption at each wavelength was plotted as a function of time.  Which of the following mechanisms is consistent with the spectral data?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3HG6.png&quot; width=&quot;500&quot;&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;(a) $\ce{A + B -&amp;gt; C}$&lt;/li&gt;&#xA;  &lt;li&gt;(b) $\ce{A + B -&amp;gt; D, D -&amp;gt; E}$&lt;/li&gt;&#xA;  &lt;li&gt;(c) $\ce{A + B -&amp;gt; F, A + B -&amp;gt; G}$&lt;/li&gt;&#xA;  &lt;li&gt;(d) $\ce{A + B -&amp;gt; H, H + A -&amp;gt; I}$&lt;/li&gt;&#xA;  &lt;li&gt;(e) none of the above&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (b), which I can believe, but not sure why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it can't be (a), because in $\lambda_2$, the curve went up then down, indicating the presence of an intermediate. However, I'm not sure why this can't be (d) or (c) as well. As long as F, G, and H are active at $\lambda_2$, then wouldn't all of the reactions show a similar curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what to do with $\lambda_1$ and $\lambda_3$, as I think they just tell me that A + B is consumed, and that the product and/or intermediates aren't active at either $\lambda_1$ or $\lambda_3$.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="194" LastEditDate="2017-05-26T22:16:52.697" LastActivityDate="2017-05-26T22:16:52.697" Title="Determining reaction formula from optical absorption graphs" Tags="&lt;kinetics&gt;&lt;stoichiometry&gt;&lt;spectrophotometry&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35811" PostTypeId="1" AcceptedAnswerId="37059" CreationDate="2015-09-02T21:04:40.667" Score="6" ViewCount="9760" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following is the major organic product when acetone is allowed to react with hydrogen cyanide in the presence of a catalytic amount of sodium cyanide?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Odv4r.png&quot; width=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (B).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I got this answer without using the catalytic sodium cyanide.  In my proposed mechanism, the dissociated proton from $\ce{HCN}$ activates the carbonyl, which facilitates the nucleophilic attack of $\ce{CN-}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What role does the $\ce{NaCN}$ play?&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="5026" LastEditDate="2018-03-01T08:19:17.693" LastActivityDate="2018-03-01T08:19:17.693" Title="What happens when acetone reacts with hydrogen cyanide?" Tags="&lt;organic-chemistry&gt;&lt;nucleophilic-substitution&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35812" PostTypeId="2" ParentId="35775" CreationDate="2015-09-02T21:24:09.130" Score="2" Body="&lt;p&gt;I found in Ian Fleming's Molecular orbitals and Organic Chemical Reactions, page 93-94 the answer (the aldehyde is propanal):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The polarisation of the H-C bonds in the methyl group leaves a weak&#xA;  positive charge on the outside and this is attracted to the partial&#xA;  negative charge on the oxygen atom, lowering the energy of this&#xA;  conformation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(the difference in energy of the two eclipsed conformations of propanal is 8 kJ/mol).&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-09-03T09:14:00.053" LastActivityDate="2015-09-03T09:14:00.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35813" PostTypeId="1" AcceptedAnswerId="35823" CreationDate="2015-09-02T21:26:42.987" Score="0" ViewCount="3873" Body="&lt;p&gt;I was having a conversation with my dad the other day and he described something as happening due to convection, and it dawned on me that I have no real clue what the difference between and convection and conduction is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to do some reading on this and haven't really found any good clarification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I do understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Convection:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Involves transfer of heat within a single substance by mass flow&lt;/li&gt;&#xA;&lt;li&gt;Does not take place in solids due to a solid's inability to behave like a fluid&lt;/li&gt;&#xA;&lt;li&gt;The heat flow is governed by diffusion of higher energy molecules and thus a distribution of that energy evenly, or by advection which basically seems to be bulk diffusion to me&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conduction:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Transfer of heat within a body by diffusion&lt;/li&gt;&#xA;&lt;li&gt;This diffusion differs from convection because convection supposedly happens through a material carrier&lt;/li&gt;&#xA;&lt;li&gt;Can take place in solids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Confusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No doubt, some of the points I just put up there are wrong or else I basically see no point in making a distinction between convection and conduction when the processes can be described more easily by diffusion which redistributes energy by principle of maximizing entropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why define these two different terms when they are essentially the same process perhaps just happening in different mediums?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contradictions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at the Wikipedia pages, which I usually find to be reliable, for &lt;a href=&quot;https://en.wikipedia.org/wiki/Convection&quot; rel=&quot;nofollow&quot;&gt;convection&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conduction&quot; rel=&quot;nofollow&quot;&gt;conduction&lt;/a&gt; and was only confused more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, from the conduction page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In gases and liquids, conduction is due to the collisions and diffusion of molecules during their random motion.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, from the convection page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Convective heat and mass transfer take place both by diffusion – the random Brownian motion of individual particles in the fluid – and by advection, in which matter or heat is transported by the larger-scale motion of currents in the fluid.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those two statements seem to be basically the same. The mechanism for the heat transfer is the diffusion of randomly moving moving particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the conduction page explains that conduction takes place in solids through, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;In solids, conduction is mediated by the combination of vibrations and collisions of molecules, of propagation and collisions of phonons, and of diffusion and collisions of free electrons.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That seems like a perfectly good explanation of what takes place when heat is being distributed towards thermal equilibrium in a solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the above statement that conduction in solids is mediated through collisions of electrons and vibrating atoms makes that process conduction rather than convection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What of the above points are incorrect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a molecular level, (i.e. using typical chemistry ideas) what makes a distinction between convection and conduction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, what is meant by the statement that convection happens through a material carrier?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that imply that conduction does not happen through a material carrier? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I hope you can see my confusion between these two terms. They seem very convoluted to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would point out that good and correct answer need not address each of my sub-questions specifically because a good answer may clear up my confusion and hence eliminate one of those sub-questions.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="17583" LastEditDate="2015-09-03T03:42:07.187" LastActivityDate="2016-03-02T21:14:16.617" Title="What is the difference between convection and conduction?" Tags="&lt;thermodynamics&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35814" PostTypeId="2" ParentId="35810" CreationDate="2015-09-02T21:28:56.390" Score="1" Body="&lt;p&gt;Let's see. You are right in that it can't be (a). Now consider (c): A and B are consumed, while F and G accumulate. Where is that intermediate that goes up, then down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for (d), I believe there is no way we can tell that from (b) based on the picture alone, without looking at the numbers. See, suppose $\lambda_1$ actually measures B, and A is in huge excess; what would be the &lt;em&gt;qualitative&lt;/em&gt; difference between the two? I can't see any.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-02T21:28:56.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35815" PostTypeId="1" AcceptedAnswerId="35817" CreationDate="2015-09-02T21:29:27.043" Score="3" ViewCount="1394" Body="&lt;p&gt;When you can smell something there must be some molecules in your nose which signalize the smell in your nose. Does that mean, that e.g. a cheese is losing molecules which fly into your nose?&#xA;And does that effect the weight or amount significantly?&lt;/p&gt;&#xA;" OwnerUserId="2275" LastEditorUserId="2275" LastEditDate="2015-09-02T22:00:27.853" LastActivityDate="2018-09-13T16:52:14.157" Title="If you can smell something does that mean that this thing loses molecules?" Tags="&lt;molecules&gt;&lt;smell&gt;&lt;sublimation&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35817" PostTypeId="2" ParentId="35815" CreationDate="2015-09-02T21:46:32.930" Score="3" Body="&lt;p&gt;Yes, the thing you smell surely does lose some molecules, and consequently some weight. It can be said that it slowly evaporates or sublimates. But the strength of smell varies greatly in different compounds, as does the rate of evaporation (sublimation). Some compounds may be gone completely in a few minutes, and some would stay for years, or even decades.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-09-02T22:13:33.070" LastActivityDate="2015-09-02T22:13:33.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35818" PostTypeId="2" ParentId="35808" CreationDate="2015-09-02T22:58:27.383" Score="2" Body="&lt;p&gt;Your reaction scheme doesn't make much sense. $\ce {R'=O}$ is not a ketone. Is $\ce X$ supposed to be $\ce Me$? This is confusing since $\ce X$ is most commonly used as the symbol of a halogen.&#xA;I would rewrite as:&#xA;$$\ce {R'RC=O + MeMgI +H_2O-&amp;gt; R'RC(Me)OH + Mg(OH)I}$$&#xA;$\ce {R'(R)C=O}$ is implied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you use very dilute aqueous sulphuric acid since your product is a tertiary alcohol. I would rather use saturated aq. $\ce {NH_4Cl}$. Even water will do but I suspect that the work-up might not be as smooth (a less water soluble magnesium salt will be formed). There will be no iodine formed and in any case I don't find the idea of recycling iodine cost effective. It will cost you more to isolate iodine that can be pure enough to work in the notoriously capricious Grignard reagent formation. In fact I would suggest you to buy a commercially available solution of $\ce {MeMgX \ (X=Cl, Br, I)}$ since preparing a Grignard can be very challenging and you will also have to handle toxic and volatile $\ce {MeI}$, have access to dry ether/THF etc. I assume you follow a detailed experimental procedure from the literature.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2015-09-02T22:58:27.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35822" PostTypeId="2" ParentId="35786" CreationDate="2015-09-03T02:38:21.440" Score="6" Body="&lt;p&gt;Brian has the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add, however, that in the most general sense, we don't have to take a molar to mean a mole of &lt;strong&gt;solute&lt;/strong&gt; per liter of &lt;strong&gt;solution&lt;/strong&gt;. I've actually seen it used to mean &quot;a mole of &lt;strong&gt;stuff&lt;/strong&gt; per liter of &lt;strong&gt;space&lt;/strong&gt;&quot; for particular values of stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the above sentence isn't exactly self-explanatory, let's take a look at a few examples.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 1: Water&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;At 4 °C, water has a density of 1000 g/L. Water has a molar mass of 18.02 g/mol. This means that in a liter of water at 4 °C, there are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1~\mathrm{L} \cdot 1000 ~\frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{18.02 \mathrm{g}} = 55.5 ~\mathrm{mol}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are 55.5 moles of water per liter of water, we can actually say that, at 4 °C, water is 55.5 molar. (See &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_concentration#Examples&quot; rel=&quot;nofollow&quot;&gt;Example 3 on Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how this works: there is a certain amount of &lt;strong&gt;stuff&lt;/strong&gt; (in this case, water) per amount of &lt;strong&gt;space&lt;/strong&gt; (one liter, or one cubic decimeter). It doesn't matter that we don't have a solution--since there's stuff, and it takes up space, we can calculate a molarity for it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example 2: Ideal gas&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;An ideal gas takes up 22.4 L per mole, at STP. Taking the inverse of this, we find that it is 0.0446 moles/L. Therefore, the concentration of an idea gas at STP is 0.0446 molar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, there's a certain amount of stuff (one mole of ideal gas) and a certain amount of space (22.4 liters), which allows us to calculate a molarity. The fact that the substance is pure doesn't stop us from writing a concentration for it.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A note of caution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This will &lt;strong&gt;NOT&lt;/strong&gt; work for a measure of concentration like molality, because molality is defined by the amount of &lt;strong&gt;solvent&lt;/strong&gt; used. It also won't usually work for things like normality, since you (usually) need to be in solution to generate ions.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorDisplayName="user7951" LastEditDate="2015-10-04T22:33:57.397" LastActivityDate="2015-10-04T22:33:57.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35823" PostTypeId="2" ParentId="35813" CreationDate="2015-09-03T02:41:28.037" Score="2" Body="&lt;p&gt;Convection is a mechanism that &lt;em&gt;enhances&lt;/em&gt; conductive heat transport, rather than being a separate physical phenomenon.  It operates by bringing hotter parcels of fluid and colder parcels of fluid into closer physical proximity with one another so that conduction can take place more easily and rapidly.  One way this is accomplished is by shearing the fluid near a heated wall in such a way that the colder faster-moving layers further from the wall flow over the hotter slower-moving layers closer to the wall to physically cause the temperature gradient normal to the wall to increase.  This enhances the rate of conductive heat transfer.  Another situation is where the fluid is being mixed by a stirrer to create thin striations of hotter fluid and colder fluid throughout the fluid.  Conduction between these hotter and colder striations increases, again because of the much larger temperature gradients that have been created.  This is typical of how convection works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm saying is that the deformations experienced by the fluid while flowing are such that they cause the temperature gradients in the fluid to get higher, and this increases the rate of heat transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a deck of cards split in half, with all the red cards (hot) in one half and all the black cards (cold) in the other half. If you bring these two slabs of hot and cold material together and wait, heat transfer will occur very slowly between the hot and cold regions. However, if you shuffle the deck so that there is a hot (red) card sandwiched in between every pair of cold (black) cards, and a cold (black) card sandwiched in between every pair of hot (red) cards, the heat transfer between the hot and cold layers will be much more rapid. This shuffling effect is what convection does.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastEditorUserId="4945" LastEditDate="2015-09-03T13:55:11.577" LastActivityDate="2015-09-03T13:55:11.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35824" PostTypeId="2" ParentId="35756" CreationDate="2015-09-03T03:17:01.797" Score="2" Body="&lt;p&gt;Since $\Delta H$ is a function of state, you can get $\Delta H$ at 100 C by taking the reactants at 100 C, cooling them down to 25 C, then carrying out the reaction at 25 C, and then heating the products up to 100 C.  This procedure makes use of Hess' Law.  Getting the amount of heat removed to take the reactants from 100 C to 25 C requires using the molar heat capacities.  Same for heating the products back up to 100 C.&lt;/p&gt;&#xA;" OwnerUserId="15095" LastActivityDate="2015-09-03T03:17:01.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35825" PostTypeId="2" ParentId="35806" CreationDate="2015-09-03T04:15:01.587" Score="7" Body="&lt;p&gt;The short sleeve has no place in a chemistry lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the coats purpose as personal protective equipment (PPE) is to reduce or eliminate the potential or risk of exposure to materials then the long sleeve coat would be the correct PPE to wear in a chemistry laboratory along with extra PPE which should have been assessed prior to performing the procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in a microbiology or molecular biology lab, where a lot of the solutions might not seem harmful, the cellular material in suspension is classed as a bio hazard, which the long sleeve will provide better protection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hazardous chemicals are still commonly used in these labs, too, and will most likely be present whether you using them or not. It's not just for chemical exposure either also for general protection.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="16111" LastEditDate="2015-10-26T16:12:19.817" LastActivityDate="2015-10-26T16:12:19.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35827" PostTypeId="1" CreationDate="2015-09-03T05:11:42.493" Score="2" ViewCount="2070" Body="&lt;p&gt;What would be the most practical process to prepare glacial acetic acid from sodium acetate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking to react the sodium acetate with an acid to convert the solution to acetic acid and sodium salt but I'm not sure of the most practical way to proceed from here without distillation (which seems impractical). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think solvent extraction (using a protonated solvent such as $\ce{CH2Cl2}$) then removal of the solvent would be a way but I'm not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear from someone who's done it before.&lt;/p&gt;&#xA;" OwnerUserId="16111" LastEditorUserId="7475" LastEditDate="2015-10-27T12:58:27.630" LastActivityDate="2021-04-30T12:33:03.907" Title="How to prepare glacial acetic acid from sodium acetate?" Tags="&lt;experimental-chemistry&gt;&lt;purification&gt;" AnswerCount="0" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="35828" PostTypeId="1" CreationDate="2015-09-03T05:46:34.367" Score="1" ViewCount="79" Body="&lt;p&gt;I need to &lt;em&gt;raise&lt;/em&gt; the pH of a solution so the easiest approach is to add alkali hydroxides (NaOH, KOH, etc). But as most of you know, the CO$\ce{_2}$ in the air can react with these hydroxides to form carbonates which &lt;em&gt;usually&lt;/em&gt;  precipitate out the other metal ion dissolved in the original solution. So what else are my options?&lt;/p&gt;&#xA;" OwnerUserId="15043" LastEditorUserId="10185" LastEditDate="2015-09-12T17:10:20.860" LastActivityDate="2015-09-12T17:10:20.860" Title="Raise the pH of a solution and prevent build up of carbonates" Tags="&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35829" PostTypeId="1" CreationDate="2015-09-03T07:23:09.820" Score="-1" ViewCount="5105" Body="&lt;p&gt;There are some problems I don't understand how they are solved (I have the answer). Since I can't find anything related to these in google, I'm gonna write the problem and the answer here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For the titration of $\mathrm{25~mL}$ solution which has $\ce{H3PO4}$ and $\ce{H2SO4}$ in it, there are spent $\mathrm{37.2~mL~0.1008~M~NaOH}$ in the presence of methyl orange as an indicator. The other $\mathrm{25~mL}$ of the solution spent $\mathrm{49.6~mL~NaOH}$ in the presence of phenolphtalein as an indicator. Find the concentration of both acids. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&lt;br&gt;&#xA;In the presence of methyl orange, will happen these reactions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H2SO4 + NaOH -&amp;gt; Na2SO4 + 2H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H3PO4 + NaOH -&amp;gt; NaH2PO4 + H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And there are spent $\mathrm{37.2~mL~NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the presence of phenolphtalein, the following reactions will happen:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\ce{H2SO4 + 2NaOH -&amp;gt; Na2SO4 + 2H2O}$&lt;/li&gt;&#xA;&lt;li&gt;$\ce{H3PO4 + 2NaOH -&amp;gt;  Na2HPO4 + 2H2O}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And there are spent $\mathrm{49.6~mL~NaOH}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;Now the procedure to find the concentration of the acids is obviously easy, but what I don't understand is the volumes of $\ce{NaOH}$ we should take to make the calculations.&lt;br&gt;&#xA;The volume of $\ce{NaOH}$ needed for $\ce{H3PO4}$ is $\mathrm{49.6-37.2=12.4~mL}$.&lt;br&gt;&#xA;The volume of $\ce{NaOH}$ in presence of methylorange, needed for $\ce{H2SO4}$ is $\mathrm{37.2-12.4=24.8~mL}$, and in the presence of phenolphtalein is $\mathrm{49.6-24.8=24.8~mL}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also more similiar problems with different compounds, but I totally don't understand the logic how these volumes are calculated! How do you know which one to deduct?&lt;br&gt;&#xA;If you also know any website with these problems please send me the link, thanks.&lt;/p&gt;&#xA;" OwnerUserId="14968" LastEditorUserId="16622" LastEditDate="2015-09-03T07:58:55.343" LastActivityDate="2015-09-03T11:03:40.620" Title="Titration of H3PO4 and H2SO4 with methyl orange and phenolphtalein as indicators" Tags="&lt;homework&gt;&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;titration&gt;" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="35831" PostTypeId="2" ParentId="35806" CreationDate="2015-09-03T09:16:12.417" Score="6" Body="&lt;p&gt;To add to @Joel's excellent answer, one of the reasons that short-sleeved lab coats are even used at all, is according to a HealthCare Business and Technology article &lt;a href=&quot;http://www.healthcarebusinesstech.com/short-sleeved-lab-coats-just-as-germy-as-long-sleeved/&quot;&gt;Short-sleeved lab coats just as germy as long-sleeved&lt;/a&gt;, due to&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some hospitals (and individual health pros) have switched to short-sleeved lab coats under the belief that the reduction in the amount of fabric would also lead to a drop in the amount of bacteria picked up during the day — and potentially passed on to patients and co-workers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, based on the Journal of Hospital Medicine paper &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/jhm.864/abstract&quot;&gt;Newly cleaned physician uniforms and infrequently washed white coats have similar rates of bacterial contamination after an 8-hour workday: A randomized controlled trial&lt;/a&gt;, has found that there is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;no statistically significant difference between the amounts of the bacteria on the long- or short-sleeve coats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, asides from the protection to your arms from chemical or biological splashes, glass splinter etc, there is no additional protection from the spread or pickup of bacteria with short sleeve lab coats.&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastActivityDate="2015-09-03T09:16:12.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35832" PostTypeId="2" ParentId="35829" CreationDate="2015-09-03T09:30:17.533" Score="1" Body="&lt;p&gt;I for myself like to watch titration curves, because they often give hints about &lt;em&gt;where to look&lt;/em&gt;. So the figures below show you the theoretical titration curves of your both acids$^{\ast 1,2}$ each and also titrated together with an aqueous solution of NaOH$^{\ast 3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TgJad.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TgJad.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ... how exactly can this be helpful?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You can see that at the &quot;first&quot; inflection point, which methyl orange would detect, both acids lost one to two protons. So how many equivalents of $\ce{NaoH}$ was needed?&lt;/li&gt;&#xA;&lt;li&gt;At the &quot;second&quot; inflection point, which is detected by phenolphthalein, only phosphoric acid has another inflection point. And how many equivalents of $\ce{NaoH}$ was needed now?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;With those thoughts in mind, the question can also be answered without my fancy titration curves.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; You need to solve this equation system to get both concentrations: \begin{align}3 c_B V_B &amp;amp;= 2 c_{S1} V_i+c_{S2} V_i\\4 c_B V_B &amp;amp;= 2 c_{S1} V_i+2 c_{S2} V_i\end{align} where $V_i$ is the initial volume of both acids ($\mathrm{25~mL}$).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;hr&gt;&#xA;$^{\ast 1}$ $\mathrm{pK_a}$ of $\ce{H2SO4}$: $-3,~1.9$ (according to the german wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{\ast 2}$ $\mathrm{pK_a}$ of $\ce{H3PO4}$: $2.16,~7.21,~12.32$ (according to the german wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^{\ast 3}$ the increasing volume is neglected and both acids are assumend to have the same concentration&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorUserId="16622" LastEditDate="2015-09-03T11:03:40.620" LastActivityDate="2015-09-03T11:03:40.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35834" PostTypeId="1" CreationDate="2015-09-03T09:53:39.290" Score="2" ViewCount="849" Body="&lt;p&gt;I came across this today &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An ester $\ce{C4H8O2}$ on reaction with $\ce{CH3MgCl} $ and hydrolysis gives an alcohol. The alcohol on reaction with $\ce{NaOCl}$ and $\ce{H3O+}$ gives acetic acid. Identify the structure of the ester.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But what I'm facing a problem with is the role of $\ce{NaOCl}$. My teacher said it can cleave and oxidize a tertiary alcohol, say tert-butyl alcohol, to acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I couldn't understand the mechanism. Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="4325" LastEditDate="2015-09-04T18:33:03.073" LastActivityDate="2015-09-04T18:33:03.073" Title="Mechanism for oxidative cleavage of tertiary alcohols by NaOCl" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;alcohols&gt;&lt;esters&gt;" AnswerCount="1" CommentCount="5" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35836" PostTypeId="1" CreationDate="2015-09-03T10:27:44.760" Score="5" ViewCount="1223" Body="&lt;p&gt;Im currently doing this project where in we are synthesizing 2-amino-6-substituted benzothiazole by using p-substituted aniline with potassium thiocyanate in presence of bromine and glacial acetic acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how does the cyclization take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the amino group of aniline is ortho and para directing so the phenyl thiocyanate is formed as the intermediate but then this intermediate reacts with bromine in presence of acetic acid to give 2-amino-6-substituted benzothiazole ... I do not get that step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide me with the mechanism as to how the cyclization takes place with those two reactants.&lt;/p&gt;&#xA;" OwnerUserId="19602" LastEditorUserId="16622" LastEditDate="2015-09-03T10:30:42.550" LastActivityDate="2015-09-03T10:30:42.550" Title="Mechanism of reaction between substituted aniline and potassium thiocyanate in presence of bromine and glacial acetic acid" Tags="&lt;organic-chemistry&gt;&lt;reaction-mechanism&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35837" PostTypeId="2" ParentId="35806" CreationDate="2015-09-03T11:02:59.003" Score="5" Body="&lt;p&gt;At my university we are all required to wear long sleeve coats in chemistry labs. It simply protects more of your body compared to a short sleeve coat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, in one of my labs a guy had his sleeves rolled up and then he decided to rest his forearm on the fumehood while he was working with sulfuric acid. It turned out he spilled some acid earlier so he ended up with a burn all along his arm. This could also happen if pouring chemicals - if you spill some, your arm won't be covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long sleeves are also important when handling liquid nitrogen with cryogenic gloves. You need to have your sleeves going over the ends of the gloves, to prevent any liquid nitrogen falling into the gloves where it can't evaporate easily off the skin and could cause burns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose one hazard with long sleeves is that spilling a chemical on your hands could cause it to roll down your sleeves and increase the contact time of the chemical with your skin as you then need to take the time to remove your lab coat before you wash your arm, but ideally your coat will be tight at the cuffs to minimise this risk.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2015-09-03T11:02:59.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35840" PostTypeId="1" CreationDate="2015-09-03T12:12:57.010" Score="4" ViewCount="30575" Body="&lt;p&gt;One explanation I read is that soap's long non-polar end will seek the air at the interface between water and air, and that the ionic end will form ionic-dipole bonds with the water. In the process of moving to the surface, the soap molecules force apart the water molecules, and hence the water molecules no longer have hydrogen bonds with each other, and the surface tension is weakened. However, wouldn't the ionic heads of the soap molecules just form stronger ionic-dipole bonds with the water, increasing inter molecular forces, and therefore increasing surface tension? What is the true explanation for why soap reduces surface tension of water?&lt;/p&gt;&#xA;" OwnerUserId="9799" LastActivityDate="2017-10-12T12:47:20.243" Title="Why does soap reduce surface tension of water" Tags="&lt;surfactants&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35841" PostTypeId="2" ParentId="35840" CreationDate="2015-09-03T12:52:38.740" Score="-2" Body="&lt;p&gt;The polar end of the surfactant is attracted to the polar water molecule. When enough surfactant molecules attach to a water molecule, it gets covered in surfactant and forms a unit. These units have a weaker force of attraction because only the non polar head of the surfactant is exposed thereby lowering the surface tension.&lt;/p&gt;&#xA;" OwnerUserId="7697" LastActivityDate="2015-09-03T12:52:38.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35845" PostTypeId="1" AcceptedAnswerId="35849" CreationDate="2015-09-03T15:03:41.373" Score="1" ViewCount="626" Body="&lt;p&gt;A sample of $\mathrm{2.00~g}$ of iron(III) sulphate is dissolved in water to give $\mathrm{100~cm^3}$ of aqueous solution. What is the concentration of $\ce{SO4^2-}$ ions in $\mathrm{mol~dm^{-3}}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the number of moles which is $\mathrm{0.005~mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, 1 mol of iron(III) sulphate has 3 mol of $\ce{SO4^2-}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the concentration is $\mathrm{\frac{0.005}{0.1} \cdot 3 = 1.5 \cdot 10^{-1} ~mol~dm^{-3}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the given answer is $\mathrm{1.5 \cdot 10^{-2}~mol~dm^{-3}}$. Why?&lt;/p&gt;&#xA;" OwnerUserId="14831" LastEditorUserId="865" LastEditDate="2021-05-22T18:55:53.467" LastActivityDate="2021-05-22T18:55:53.467" Title="Calculating concentration of dissolved ions" Tags="&lt;concentration&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35846" PostTypeId="1" CreationDate="2015-09-03T15:32:03.293" Score="10" ViewCount="1456" Body="&lt;p&gt;A common highschool chemistry example for the polarity of water is the experiment where stream of water is deflected under the influence of a charge (electric field).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have come across some condradicting claims, namely this Youtube video from &lt;a href=&quot;https://youtu.be/1Xp_imnO6WE?t=109&quot; rel=&quot;nofollow noreferrer&quot;&gt;Veritasium&lt;/a&gt; (from around 1:50) and more importantly a publication by Ziaei-Moayyed &lt;em&gt;et. al.&lt;/em&gt;.&lt;sup&gt;[1]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sources suggest that this phenomenon is not based on the polarity of water but charged impurity ions in the water stream. They also claim that the electric field could also only re-orient the molecules, but not deflect them as the electric field also has an opposite effect on the other side of the dipole. This has been bothering me and it is hard to find concluding evidence either way. What are your thoughts?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;References:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ziaei-Moayyed, M.; Goodman, E.; Williams, P. Electrical Deflection of Polar Liquid Streams: A Misunderstood Demonstration. &lt;em&gt;J. Chem. Educ.&lt;/em&gt; &lt;strong&gt;2000,&lt;/strong&gt; &lt;em&gt;77&lt;/em&gt; (11), 1520. &lt;a href=&quot;https://doi.org/10.1021/ed077p1520&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOI: 10.1021/ed077p1520&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19609" LastEditorUserId="4945" LastEditDate="2019-11-03T12:01:53.830" LastActivityDate="2021-04-19T16:31:47.043" Title="Is water stream deflecting in an electric field really based on water's polarity?" Tags="&lt;water&gt;&lt;polarity&gt;&lt;electricity&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35847" PostTypeId="1" CreationDate="2015-09-03T16:06:04.303" Score="7" ViewCount="220" Body="&lt;p&gt;Lately I've been interested in the chemistry of nerve agents due to reading Patrick's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0199697396&quot; rel=&quot;noreferrer&quot;&gt;Introduction to Medicinal Chemistry&lt;/a&gt; in which they were only briefly discussed. I bought &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0128001593&quot; rel=&quot;noreferrer&quot;&gt;the Handbook of Toxicology of Chemical Warfare Agents&lt;/a&gt; by Ramesh C. Gupta which appeared to me as a comprehensive work discussing the physical chemistry, biological action, history and legislation of a whole range of chemical agents (in particularly organophosphorus agents).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it lacks experimental information (i.e. synthesis). I found a book that detailed (to the somewhat questionable extreme) on this information, having the rather obscure title &lt;a href=&quot;https://books.google.nl/books?id=_dGuAwAAQBAJ&amp;amp;printsec=frontcover&amp;amp;dq=The%20Preparatory%20Manual%20of%20Chemical%20Warfare%20Agents&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CCEQ6AEwAGoVChMIg8TM-ZjbxwIVyjwUCh27nQgO#v=onepage&amp;amp;q=The%20Preparatory%20Manual%20of%20Chemical%20Warfare%20Agents&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;The Preparatory Manual of Chemical Warfare Agents&lt;/a&gt; (it's a Google preview link, quite handy to see if a book is worth the money). Now since I'm not an expert in organophosphorus synthesis (else I wouldn't be interested in it, right?) I cannot fully appreciate the level of accurateness. And while you may argue that, since it's only of theoretical use to me, the level of accurateness is unimportant, I wouldn't really feel satisfied learning about something from a textbook that finally turns out to be based on completely wrong information. Because then what knowledge have I gained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the question: the book raised my suspicion because the author seems not to be related to any serious research unit or university for that matter. The poor lay-out, language, and somewhat too practical advice given here and there, screams &lt;a href=&quot;https://en.wikipedia.org/wiki/Uncle_Fester_(author)&quot; rel=&quot;noreferrer&quot;&gt;Uncle Fester&lt;/a&gt; type bull shit book. So, is this book based on legitimate research? And if not, where can I read about (organophosphorus) chemical warfare agents synthesis from a reliable (preferentially written by well known and respected chemists) source, that goes into enough detail on a chemical level (and does not necessarily contain Uncle Fester type practical information in which I'm obviously not very interested)?&lt;/p&gt;&#xA;" OwnerUserId="1887" LastEditorUserId="30836" LastEditDate="2019-04-13T14:58:21.153" LastActivityDate="2019-04-13T14:58:21.153" Title="Accurate textbook about synthesis of (organophosphorus) chemical warfare agents" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;reference-request&gt;&lt;chemical-weapons&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35848" PostTypeId="1" CreationDate="2015-09-03T16:37:49.873" Score="4" ViewCount="263" Body="&lt;p&gt;My reasoning for the somewhat implausible &quot;&lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt;&quot; character of the hydrogen between these two oxygens on the enol structure of the right comes from recognizing that if it were, then we would have 6 &lt;span class=&quot;math-container&quot;&gt;$\mathrm{sp^2}$&lt;/span&gt; atoms with 6 π-electrons, so maybe that extra bit of instability from having a higher orbital on hydrogen could be sort of taken care of by being stable like an aromatic compound and forming a hydrogen bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how could I go about figuring out if this is true, either with MO theory or something else?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ufSEO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufSEO.png&quot; alt=&quot;dicarbonyl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(source: &lt;a href=&quot;http://chemwiki.ucdavis.edu/@api/deki/files/2937/image011.png?revision=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;ucdavis.edu&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerDisplayName="user19026" LastEditorUserId="41328" LastEditDate="2019-07-22T21:14:02.260" LastActivityDate="2019-07-22T21:14:02.260" Title="Does alcohol hydrogen in the enol tautomer of a β-dicarbonyl have sp² character?" Tags="&lt;organic-chemistry&gt;&lt;hybridization&gt;&lt;carbonyl-compounds&gt;&lt;hydrogen-bond&gt;" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="35849" PostTypeId="2" ParentId="35845" CreationDate="2015-09-03T16:41:09.010" Score="3" Body="&lt;p&gt;You're right and the given answer is wrong. Here's the math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(1\;\mathrm{mol}\;\ce{Fe2(SO4)3}\over {399.88\;\mathrm{g}\;\ce{Fe2(SO4)3}}\right)\left({2.0\;\mathrm{g}\;\ce{Fe2(SO4)3}\over 100\;\mathrm{cm}^{3}}\right)\left({3\;\mathrm{mol}\;\ce{SO4^2-}\over 1\;\mathrm{mol}\;\ce{Fe2(SO4)3}}\right)\left({10\;\mathrm{cm}\over 1\;\mathrm{dm}}\right)^{3} =\ce{[SO4^2- ]} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{[SO4^2- ]}= 1.5\cdot10^{-1}\;\mathrm{mol}\;\mathrm{dm}^{-3}$$&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorDisplayName="user7951" LastEditDate="2015-09-03T22:59:21.460" LastActivityDate="2015-09-03T22:59:21.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35850" PostTypeId="2" ParentId="35840" CreationDate="2015-09-03T16:54:06.373" Score="2" Body="&lt;p&gt;Apoorv did not give the right explanation, because this would imply, that surface tension goes especially down, when micelles are formed, but the opposite is the case. Compare e.g.: first chapter of The Colloidal Domain by D.Fennell Evans and Håkan Wennerström.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing to note is that pure water has a higher surface tension than hydrocarbon chains which is due to the strong dipole dipole interaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a very dilute solution of surfactant in water no micelles are formed but you have monomers &quot;floating around&quot;. In contrast to ions in water they are not homogenously distributed, but they concentrate especially on the surface where their polar head is directed to the water and their hydrophobic chain is directed to the air. In the end you substituted water molecules with hydrocarbon chains which leads to a lower surface tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the concentration of surfactant, they start to form micelles in the water (and not on the surface). So after reaching this critical micelle concentration (called CMC) the surface tension stays quite constant. This constant behaviour is used experimentally to estimate the CMC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the figure below Equation 11:&#xA;&lt;a href=&quot;http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf&quot; rel=&quot;nofollow&quot;&gt;http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that Apoorv described some kind of reversed micelle. The normal way is that they stick their polar head to the water and have their hydrocarbon chain inside the micelle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture the white group is the polar head:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16375" LastActivityDate="2015-09-03T16:54:06.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35851" PostTypeId="2" ParentId="35756" CreationDate="2015-09-03T17:05:07.930" Score="6" Body="&lt;p&gt;@Chet already gave the qualitative answer but the actual equation it is based on is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\Delta H^\circ(T)=\Delta H{^\circ}_{T_1} + \int_{T_1}^{T} \Delta C_p{^\circ}(T^\prime ) \,\mathrm{d}T^\prime$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the enthalpy at a reference temperature and the heat capacity as a function of temperature you can calculate the enthalpy change at any other temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually d) and e) &lt;strong&gt;together&lt;/strong&gt; are not wrong. We all know these equations:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta G= - RT\ln (K) \qquad \Delta G= \Delta H - T\Delta S $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With d) you can calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; and with e) you can calculate &lt;span class=&quot;math-container&quot;&gt;$\Delta H$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look deeper into: &lt;span class=&quot;math-container&quot;&gt;$\Delta G^\circ$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;. &#xA;To answer your second question completely I would recommend to look again into a textbook of your choice. Especially the definitions of activity and chemical potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the chemical potential &lt;span class=&quot;math-container&quot;&gt;$\mu^\circ$&lt;/span&gt; of a pure substance you can calculate the chemical potential of mixtures if you know the activities:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mu= \mu^\circ + RT \ln(a)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we make two assumptions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;We have constant temperature and pressure. (&lt;span class=&quot;math-container&quot;&gt;$\mu = \Delta G$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;We can calculate the activities simply over the partial pressures or concentrations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This leads then to your equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\Delta G= \Delta G^\circ + RT \ln(Q)&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is the reaction quotient.&#xA;Since we assume equilibrium,  we know that &lt;span class=&quot;math-container&quot;&gt;$\Delta G=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Q = K$&lt;/span&gt;. This leads to:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 0= \Delta G^\circ + RT \ln(K) \Rightarrow \Delta G^\circ = -RT \ln(K) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation is the handy one you use for calculating equilibrium constants from energies and vice versa.&#xA;So to be really precise one has to write:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta G^\circ= \Delta H^\circ - T\Delta S^\circ $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one usually skips the superscript &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;.&#xA;At equilibrium &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; are essentially the same.&lt;/p&gt;&#xA;" OwnerUserId="16375" LastEditorUserId="83758" LastEditDate="2020-03-20T19:20:01.190" LastActivityDate="2020-03-20T19:20:01.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35853" PostTypeId="1" AcceptedAnswerId="36942" CreationDate="2015-09-03T17:31:41.760" Score="10" ViewCount="924" Body="&lt;p&gt;What is happening in this photo? I set up a simple electrolytic cell today of copper (II) sulfate solution with carbon electrodes. It was connected to a powerpack (set at ~8.0V), and a variable resistor with the resistance cranked up to its maximum as in the image below. You can also see the ammeter and the voltmeter in the photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the powerpack was switched off, the current fell immediately to zero, as expected, but the voltmeter still read a non-zero number (initially 0.80V, then later seemed to decrease to 0.18 V). I turned the electrodes upside-down so that the clean ends were in solution (rather than the end covered in copper), but reading still read the same. Removed the electrodes to a beaker of water (but not deionised water), and it still read 0.18V or close to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried the exact same thing with pure copper electrodes, but it read 0.00V immediately upon turning off the power supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried completely taking the leads out of the power supply (which was off anyway), and the carbon rods still read a non-zero value even with the circuit obviously incomplete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's happening here? Copper electrodes did not show the same effect. Is there something about carbon electrodes in particular that can make them act as a capacitor? The current stayed at 0.00A the whole time. This effect was NOT seen when the electrodes were lifted out of the solution/water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tVUnL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVUnL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit, to respond to the answer below:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This would be the most interesting answer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I did consider initially this, so first I tried up-ending the&#xA;electrodes so the clean ends were in solution, but the reading&#xA;remained the same. Second, it showed no such behaviour with the pure&#xA;copper electrodes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think it's unlikely to be a case of the voltmeter left floating.&#xA;Surely if that had been it, regardless of which electrodes (copper or&#xA;carbon) were used, random static charges would've been present? Plus,&#xA;the phenomenon was reproduced when I swapped out different electrodes&#xA;and then swapped back in the funny-acting ones, so the behaviour&#xA;wasn't exactly random.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, I did forget to mention, that I also tested this with a&#xA;second clean set of carbon electrodes and got 0.00V! Support for 2nd&#xA;explanation (but if so, how does that work exactly?). I still would've&#xA;liked it to have been the 1st explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also note that I left the apparatus overnight, and by next morning&#xA;the reading was 0.02V. I've unfortunately have had to clear the&#xA;materials away now, but if I get a moment I may try replicating the&#xA;result.&lt;/p&gt;&#xA;" OwnerUserId="19612" LastEditorUserId="19612" LastEditDate="2015-09-07T09:22:10.297" LastActivityDate="2015-09-07T09:22:10.297" Title="carbon electrodes have potential difference even with no electrical input?" Tags="&lt;electrochemistry&gt;&lt;electrolysis&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35854" PostTypeId="1" AcceptedAnswerId="35861" CreationDate="2015-09-03T17:32:26.693" Score="6" ViewCount="7651" Body="&lt;p&gt;Recently, I read about the nomenclature of bicyclic compounds, by mentioning the number of carbons between two bridgeheads. But, then I came across the structure of a weird type of compound, whose IUPAC name is tricyclo-[2.2.1.0&lt;sup&gt;2,6&lt;/sup&gt;]-heptane. As the name suggests it must be a polycyclic compound, having &lt;em&gt;three&lt;/em&gt; alicycles. 2.2.1 suggests that there are 2, 2 and 1 carbons between the bridgeheads. But what about the next part? What is the actual rule for naming tricyclo compounds?&lt;/p&gt;&#xA;" OwnerUserId="17217" LastEditorDisplayName="user7951" LastEditDate="2015-09-03T23:01:50.450" LastActivityDate="2015-09-03T23:01:50.450" Title="How to name tricyclo carbon compounds?" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="35855" PostTypeId="1" CreationDate="2015-09-03T17:38:00.083" Score="1" ViewCount="92" Body="&lt;p&gt;I am currently doing a report for school on the electron cloud atomic structure theory. One of the major points on my report is the experiments performed by the scientists (Schrodinger and Heisenberg) to validate their hypothesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any evidence of any experiments performed by them to validate their claim. Did they perform any experiments or was their theory purely based on mathematics? (There seems to be a lot of that surrounding this topic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, Sam.&lt;/p&gt;&#xA;" OwnerUserId="11576" LastEditorUserId="10185" LastEditDate="2015-09-03T18:27:00.890" LastActivityDate="2015-09-03T18:27:00.890" Title="Experimental Validation of Schrodinger's Electron Cloud theory" Tags="&lt;experimental-chemistry&gt;&lt;quantum-chemistry&gt;&lt;atoms&gt;&lt;electrons&gt;&lt;electronic-configuration&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35857" PostTypeId="2" ParentId="19902" CreationDate="2015-09-03T17:39:46.143" Score="1" Body="&lt;p&gt;Vanadium has a stable valency of 5.&lt;br&gt;&#xA;Electrons cannot be transferred from $3p$ orbital to $3d$ orbital as the $3p$ orbital is completely filled and has attained stability.&lt;br&gt;&#xA;It requires high energy so as to transfer electrons from $3p$ orbital to $3d$ orbital.&lt;br&gt;&#xA;Hence,Vanadium has a valency of 5 only and not of 7 or 6.&lt;/p&gt;&#xA;" OwnerUserId="19614" LastEditorUserId="7448" LastEditDate="2015-09-03T17:54:29.883" LastActivityDate="2015-09-03T17:54:29.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35858" PostTypeId="2" ParentId="35494" CreationDate="2015-09-03T17:45:29.903" Score="-6" Body="&lt;p&gt;The number of stereoisomers do not depend on number of double bonds. It depends on the number of chiral/stereogenic carbons present in that molecule.&#xA;A stereogenic centre refers to a carbon which is bonded to four different atoms or groups.&#xA;Based on the number of stereogenic centres in the molecule, the number of stereoisomers can be determined by the formula: 2 raised to the nth power, whre n= number of stereogenic centres.&#xA;Therefore if a molecule has 3 stereogenic centres, the maximum number of stereoisomers it can hav is 2 raised to 3 that is 8 stereoisomers.&lt;/p&gt;&#xA;" OwnerUserId="19602" LastActivityDate="2015-09-03T17:45:29.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35859" PostTypeId="1" CreationDate="2015-09-03T19:10:07.140" Score="1" ViewCount="209" Body="&lt;p&gt;I have a Liebig condenser but I read that the air cooled condenser is preferred for high temperature boiling points. My question is whether I can still use my Liebig with minimal risk... Should I just run it without coolant or even with warm water? There shouldn't be pressure build up if I just run it dry. If I used warm water though, I still have a huge temperature differential between the jacket and sulfuric acid vapour, not so much with just running it without coolant since there will be a gradient between the air temp from inner tube to outer tube.&lt;/p&gt;&#xA;" OwnerUserId="19218" LastActivityDate="2015-09-03T19:10:07.140" Title="Can I use a Liebig condenser for condensing sulfuric acid?" Tags="&lt;distillation&gt;" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35861" PostTypeId="2" ParentId="35854" CreationDate="2015-09-03T19:23:18.203" Score="11" Body="&lt;p&gt;The &lt;em&gt;main ring&lt;/em&gt; of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible. The &lt;em&gt;main bridge&lt;/em&gt; is the bridge that includes as many of the atoms as possible that are not included in the main ring. Bridges other than the main bridge are called &lt;em&gt;secondary bridges&lt;/em&gt;. Thus, tricyclic hydrocarbons consist of the main ring, the main bridge, and one secondary bridge.&lt;/p&gt;&#xA;&lt;p&gt;The nomenclature of tricyclic hydrocarbons is based on the nomenclature of bicyclic systems. The additional steps for the naming of tricyclic hydrocarbons are described in the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; as follows.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-23.2.5&lt;/strong&gt; Naming and numbering tricyclic alicyclic hydrocarbons&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-23.2.5.1&lt;/strong&gt; Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.&lt;/p&gt;&#xA;&lt;p&gt;The name of the tricyclic system is then constructed by citing:&lt;/p&gt;&#xA;&lt;p&gt;(a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;&lt;/p&gt;&#xA;&lt;p&gt;(b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,&lt;/p&gt;&#xA;&lt;p&gt;(c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For example tricyclo[2.2.1.0&lt;sup&gt;2,6&lt;/sup&gt;]heptane&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S7N8E.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S7N8E.png&quot; alt=&quot;tricyclo[2.2.1.02,6]heptane&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(The main ring is shown in red, the main bridge is shown in black, and the secondary bridge is shown in blue.)&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-09-03T19:23:18.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35863" PostTypeId="1" AcceptedAnswerId="35868" CreationDate="2015-09-03T20:52:17.133" Score="1" ViewCount="469" Body="&lt;p&gt;I noticed after some use that a stainless steel spoon that I have started forming $\ce{FeO}$. Stainless steel is not supposed to rust like this. I don't know of any black iron hydroxides, chromium oxides, or chromium hydroxides so that lead me to deduce that it is Iron(II) Oxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why would stainless steel oxidize? I mean it does not make any sense since I was told that stainless steel does not oxidize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can I get rid of the oxide? Putting the spoon in phosphoric acid perhaps since phosphoric acid is a really good rust remover?&lt;/p&gt;&#xA;" OwnerUserId="6875" LastActivityDate="2015-09-03T23:16:23.393" Title="Iron(II) oxide on stainless steel?" Tags="&lt;metallurgy&gt;" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35868" PostTypeId="2" ParentId="35863" CreationDate="2015-09-03T23:16:23.393" Score="1" Body="&lt;p&gt;It was probably a poor grade of stainless steel. The chromium oxidizes before the iron therefore you obtain a thin layer of protection. In poor grades of stainless steel, I have seen rust form over a period of time or during exposure to non ideal conditions such as high humidity. This is probably because the manufacturer added the minimum chromium required to meet standards, and so you get a smaller ensemble of chromium oxide and more iron oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can treat the metal with phosphoric acid to remove the rust. You can also treat the stainless steel with concentrated sulfuric acid to produce a phosphate layer for additional protection with minimal loss of material (they do this in industry)&lt;/p&gt;&#xA;" OwnerUserId="19218" LastActivityDate="2015-09-03T23:16:23.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="35869" PostTypeId="1" CreationDate="2015-09-03T23:25:24.383" Score="2" ViewCount="14124" Body="&lt;p&gt;In hydroponics, Hydrogen Peroxide is used to sanitise nutrient solution from pathogens. Allegedly it does not take sides, oxidising aerobic as well as anaerobic, including damaging the roots, so I was told. I believe this argument, since O as a free radical is highly reactive. I would appreciate your input on this. However, this isn't the main question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question is that manufactures also claim that it will provide DO, dissolved oxygen, to the solution. I find this hard to believe. Reason being, DO refers to dissolved $\ce{O2}$ (Dioxide) not Monoxide. When $\ce{H2O2}$ breaks down it breaks down to $\ce{H2O + O}$ (free radical), right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please put some clarity into this?&lt;/p&gt;&#xA;" OwnerUserId="19626" LastEditorUserId="7448" LastEditDate="2015-09-03T23:32:22.593" LastActivityDate="2015-09-15T22:42:31.883" Title="Does H2O2 break down into H2O + O or H2O + O2?" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="35870" PostTypeId="2" ParentId="35869" CreationDate="2015-09-03T23:39:13.883" Score="-1" Body="&lt;p&gt;Hydrogen peroxide is unstable under standard conditions and breaks down into $\ce{H2O + O2}$. Although you are not left with a free radical, diatomic oxygen is still an extremely powerful oxidizing agent, which is the same reason rust forms on metals. So even though it is not a free radical, it still has the ability to oxidize your organic compounds. &lt;/p&gt;&#xA;" OwnerUserId="19218" LastActivityDate="2015-09-03T23:39:13.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36869" PostTypeId="1" CreationDate="2015-09-04T02:54:31.597" Score="5" ViewCount="2563" Body="&lt;p&gt;It is well known that permanganate in acidic solution can oxidize oxalate ions (or oxalic acid). The reaction produces $\ce{Mn^2+}$, which catalyzes this very reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2MnO4^- + 16H^+ + 5C2O4^2- -&amp;gt; 2Mn^2+ + 8H2O + 10CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of a demonstration on chemical kinetics, I want to inhibit this autocatalysis somehow. (I intend to perform three reactions: one with normal autocatalysis, one with $\ce{Mn^2+}$ added, which will run faster, and one preventing the $\ce{Mn^2+}$ from catalyzing the reaction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought I might use EDTA to complex the manganese ions. Alas, it didn't help. Maybe the environment is too acidic. (The reaction was actually faster with EDTA, I don't know why.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I should try a different chelating agent? Perhaps I could precipitate the $\ce{Mn^2+}$ ions? What do you suggest?&lt;/p&gt;&#xA;" OwnerUserId="5277" LastEditorDisplayName="user7951" LastEditDate="2015-09-04T15:49:15.503" LastActivityDate="2017-04-30T09:21:38.200" Title="How to inhibit Mn2+ autocatalysis in the oxalate-permanganate reaction?" Tags="&lt;kinetics&gt;&lt;catalysis&gt;&lt;reaction-control&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36870" PostTypeId="1" AcceptedAnswerId="36896" CreationDate="2015-09-04T03:55:06.567" Score="9" ViewCount="3008" Body="&lt;p&gt;I am new to chemistry (I have only done a GCSE in it) so forgive me if I am missing something obvious. But I am trying to very roughly simulate some chemical reactions. My background is in Artificial Life and Adaptive Systems rather than computational chemistry or chemistry in general. My stumbling block is that I am unsure exactly why bonds break and when they break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the things I am confused about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms? If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart? Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on. Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons &quot;taken&quot; by the more electronegative atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there are multiple questions here. But they all answer the bigger question. Why and when do bonds break exactly? I have had a hard time trying to figure this out by searching on the internet alone. Often I just get pages explaining the energy requirements of a reaction. Thanks in advance, any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="19611" LastEditorUserId="31313" LastEditDate="2017-04-01T02:04:32.107" LastActivityDate="2017-04-01T02:04:32.107" Title="Why and when are bonds broken" Tags="&lt;thermodynamics&gt;&lt;bond&gt;&lt;electrons&gt;&lt;electronegativity&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36872" PostTypeId="1" CreationDate="2015-09-04T05:23:51.890" Score="2" ViewCount="1968" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vIpCH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vIpCH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the bond angles of O-N-O in figure (c), (D) and (F), there not too much change in bond angle. Even though they are notable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we observe &lt;strong&gt;very&lt;/strong&gt; small change in angles?&lt;/p&gt;&#xA;" OwnerUserId="5456" LastEditorUserId="5456" LastEditDate="2015-09-04T14:56:33.833" LastActivityDate="2015-09-04T14:56:33.833" Title="O-N-O bond angles in oxides of nitrogen" Tags="&lt;physical-chemistry&gt;&lt;inorganic-chemistry&gt;&lt;bond&gt;&lt;molecular-structure&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36873" PostTypeId="1" AcceptedAnswerId="36878" CreationDate="2015-09-04T08:13:09.023" Score="2" ViewCount="674" Body="&lt;p&gt;Suppose a a container contains a gaseous mixture of methane and hydrogen. The mixture effuses out through a hole in the container. Now hydrogen is lighter gas so it must effuse faster and all of hydrogen should come out of the flask earlier than methane (before all methane effuses out) and a stage should come when there is only methane left in the container.&#xA;Is this how it happens? &lt;/p&gt;&#xA;" OwnerUserId="20633" LastEditorUserId="418" LastEditDate="2016-10-19T15:43:45.750" LastActivityDate="2016-10-19T15:43:45.750" Title="How to predict the effusion/diffusion for a mixture of gases?" Tags="&lt;gas-laws&gt;&lt;mixtures&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36874" PostTypeId="1" CreationDate="2015-09-04T08:20:06.437" Score="3" ViewCount="129" Body="&lt;p&gt;Ok so I'm confused about the mechanism for this process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AVRAJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AVRAJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the idea here is that iron first becomes oxidised to Fe(III) by placing an electron in the antibonding orbital of dioxygen. This is followed by the addition of 2 electrons from an external source. This causes the splitting of the O-O bond resulting in the formation of a hydroxide ion. In addition, iron loses another electron to oxygen forming the Iron(IV) oxide species shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I do &lt;strong&gt;&lt;em&gt;Not&lt;/em&gt;&lt;/strong&gt; understand is the overall charges on the species at the end. I feel as if the original O-Fe(III) bond was considered purely ionic hence the lack of a formal charge on oxygen, yet at the end it seems as if the O=Fe(IV) bond is covalent, otherwise there would be a negative 2 charge on oxygen!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a switch between ionic and covalent formalisms in the middle of this process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could help explain the charges with perhaps a curly arrow mechanism to indicate the numbers of electrons on each atom then that would be most appreciated!&lt;/p&gt;&#xA;" OwnerUserId="20634" LastEditorUserId="9961" LastEditDate="2015-10-30T18:10:30.390" LastActivityDate="2015-10-30T18:10:30.390" Title="Iron (II) dioxygen reaction mechanism?" Tags="&lt;inorganic-chemistry&gt;&lt;reaction-mechanism&gt;&lt;transition-metals&gt;" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36875" PostTypeId="1" CreationDate="2015-09-04T08:20:16.173" Score="4" ViewCount="614" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; I've found a mistake in my thermodynamics textbook, &lt;em&gt;Chemical Thermodynamics for Process Simulation&lt;/em&gt;, but as thermodynamics is hard and I'm a relative novice, I wanted to check here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook, in talking about partial molar Gibbs excess functions of species in solutions, writes that we can express the partial molar Gibbs free energy of a species $\bar{g_i}$ in solution, as the sum of the partial molar Gibbs free energy in an &lt;em&gt;ideal&lt;/em&gt; solution, $\bar{g_i}^\mathrm{id}$, and an excess function, $\bar{g_{i}}^\mathrm{ex}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{g}_i = \bar{g}_i^\mathrm{id}+\bar{g}_i^\mathrm{ex}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't specify whether we are thinking of evaluating $\bar{g}_i^\mathrm{id}$ &lt;em&gt;at the system pressure&lt;/em&gt; or &lt;em&gt;at a reference pressure&lt;/em&gt;, but immediately after it 'expands' the above expression into the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{g}_i(T,P) = \Big(\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i\Big) + \left(\mathcal{R}T\ln \frac{f_i}{x_if_i^0}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly we're comparing $\bar{g}_i$ to the ideal value, $\bar{g}_i^\mathrm{id}$, &lt;em&gt;at standard pressure&lt;/em&gt; $P_0$. However, they then go on to define the activity $a_i=f_i/f_i^0$ and the activity coefficient $\gamma_i = a_i/x_i$ and substitute these in to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{g}_i(T,P) = \bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i  +  \mathcal{R}T\ln \gamma_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm all fine up to here. The first part, $\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i $, is the ideal solution Gibbs free energy at standard pressure, and the final term, $\mathcal{R}T\ln \gamma_i,$ is a correction in two ways: It corrects for the non-ideality of the solution, and it corrects for the fact that we may be at a different pressure $P$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book then says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For an ideal mixture the excess part is equal to zero, since the activity coefficient is $\gamma_i=1$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this is ridiculous: we can only make this conclusion if we have an ideal solution at $P=P_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an error in the textbook? Also, I can see two possible routes to an error: Either $\bar{g}_i^\mathrm{id}$ is actually to be evaluated at system pressure, in which case their last statement is correct but the formula they give is wrong, or else their definition of excess functions is the standard one (actual value minus ideal value at ref. pressure) but they made a mistake in concluding $\gamma_i=1$ for all ideal solutions, independent of the pressure.&lt;/p&gt;&#xA;" OwnerUserId="17933" LastEditorUserId="4945" LastEditDate="2015-09-04T09:38:08.070" LastActivityDate="2016-03-02T19:02:56.703" Title="How are partial molar Gibbs excess functions correctly defined?" Tags="&lt;physical-chemistry&gt;&lt;free-energy&gt;&lt;erratum&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36876" PostTypeId="2" ParentId="9809" CreationDate="2015-09-04T08:48:56.437" Score="3" Body="&lt;p&gt;The only problem here is the units - the boiling point of isodecane is given in Fahrenheit and the boiling point of decane in Celsius.&lt;/p&gt;&#xA;" OwnerUserId="20637" LastActivityDate="2015-09-04T08:48:56.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36877" PostTypeId="1" CreationDate="2015-09-04T08:57:42.317" Score="3" ViewCount="50" Body="&lt;p&gt;Two highly respected books: Atkins, Chemical Principles 5th ed. and Clayden Organic Chemistry 2nd ed. don't agree on the orbital arrangement of the $2p$ orbitals of $\ce{N2}$. Clayden on page 94 states that it is $\sigma_{(2p)}, \pi_{(2p)}, \pi^*_{(2p)}, \sigma^*_{(2p)}$ and Atkins on page 118 states that it is $\pi_{(2p)}, \sigma_{(2p)}, \pi^*_{(2p)}, \sigma^*_{(2p)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a retired engineer with a consuming interest in how the brain works and am trying to give myself a good grounding in organic chemistry. This may be a trivial issue, but to a novice it hardly seems so. This kind of division within the chemistry community is very concerning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas as to what's going on here?&lt;/p&gt;&#xA;" OwnerUserId="20636" LastEditorUserId="10185" LastEditDate="2015-09-04T17:34:08.753" LastActivityDate="2015-09-04T17:34:08.753" Title="Molecular Orbital Sequence" Tags="&lt;quantum-chemistry&gt;&lt;molecular-orbital-theory&gt;" AnswerCount="0" CommentCount="2" ClosedDate="2015-09-04T18:33:45.163" ContentLicense="CC BY-SA 3.0" />
  <row Id="36878" PostTypeId="2" ParentId="36873" CreationDate="2015-09-04T09:22:56.317" Score="3" Body="&lt;p&gt;Sort of, but not quite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If our hole is large (say, as if pierced by a toothpick), then the gases will go out in pretty much the same composition that is found inside, so their ratio in the mixture will stay nearly the same till the very end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hole is small, about the size of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mean_free_path&quot; rel=&quot;nofollow&quot;&gt;free path&lt;/a&gt; of a molecule in a gas, then it's a different story. BTW, these conditions are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Knudsen_flow&quot; rel=&quot;nofollow&quot;&gt;Knudsen flow&lt;/a&gt;. Methane weighs 16, and helium weighs 4 (I know you've said &quot;hydrogen&quot;, but let's assume it was helium, just to make the calculations easier). So helium particles are 4 times lighter than those of methane, hence they fly 2 times faster. So there would be two times as many of them in the outgoing stream - that is, if the mixture inside was 1:1. See, it's not like all helium flows outside, while methane molecules stand politely in the line, saying &quot;After you&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the story progresses (which is not going to be fast with just one tiny hole), the mixture inside gets enriched with methane. So does the outgoing flow, but still it remains richer in helium relative to the mixture inside, and hence continues to shift its composition in the same direction. Yet still, pure methane is a limit which is never actually attained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, pressing gas mixtures through porous membranes is a valid way to separate the components by molecular mass regardless of their chemical properties, and as such, was used for isotope enrichment.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-09-04T09:42:23.280" LastActivityDate="2015-09-04T09:42:23.280" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36879" PostTypeId="2" ParentId="27646" CreationDate="2015-09-04T10:02:43.307" Score="0" Body="&lt;p&gt;The answer given by Chester Miller is completely correct but I think it just causes old problems on new places and does not solve the problem.&#xA;In the last (van't Hoff) equation an enthalpy is assumed that is independent from temperature. In general it is:&#xA;$$&#xA;\frac{\partial \Delta G/T} {\partial T} = \frac{\Delta H (T)}{T}&#xA;$$&#xA;This leads to:&#xA;$$&#xA;\left. \frac{\partial \ln K}{\partial T} \right|_p = \frac{ \Delta H^0(T)}{RT^2}&#xA;$$&#xA;Also in this case you end up searching for the temperature dependence of the enthalpy hence the heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just recommend to get experimental data for the polynomials, perhaps CRC handbook. I do not know any theoretical method better than Debye modell that you could still apply just with calculator and pen.&lt;/p&gt;&#xA;" OwnerUserId="16375" LastActivityDate="2015-09-04T10:02:43.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36880" PostTypeId="2" ParentId="4515" CreationDate="2015-09-04T10:31:34.807" Score="2" Body="&lt;p&gt;I suffer from chronic back pain and have tried just about everything on the market.Now I grow and harvest my own Aloe Vera gel,&#xA;put the gel in a 50ml roll-on bottle 1/2 full with gel, a 1/2 tsp vitamin C powder&#xA;and 10 drops vitamin E oil then add 98.9% DMSO by syringe&#xA;10ml at a time till the strength is good enough for me. I personally use 25ml&#xA;DMSO with the Aloe gel for remarkable pain relief &lt;/p&gt;&#xA;" OwnerUserId="20638" LastActivityDate="2015-09-04T10:31:34.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36881" PostTypeId="1" AcceptedAnswerId="36917" CreationDate="2015-09-04T11:06:24.217" Score="6" ViewCount="1801" Body="&lt;p&gt;I am thinking of buying gold. If necessary, one gram at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a surefire way of checking if the alleged gold is real and of the stated caratage without a lab?&lt;/p&gt;&#xA;" OwnerUserId="20639" LastEditorUserId="7475" LastEditDate="2017-12-16T15:14:31.133" LastActivityDate="2017-12-16T15:14:31.133" Title="Detecting fake gold" Tags="&lt;home-experiment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36882" PostTypeId="1" CreationDate="2015-09-04T11:23:59.177" Score="0" ViewCount="271" Body="&lt;p&gt;Sorry guys,I'm totally new in this field,for personal interest I'd like to start some calculations on conducting polymers ,could someone suggest a good software to perform simulations on polymer (absorption spectra)?&lt;/p&gt;&#xA;" OwnerUserId="20640" LastActivityDate="2015-09-08T14:48:15.420" Title="Good software to perform simulations on polymer (absorption spectra)?" Tags="&lt;computational-chemistry&gt;&lt;polymers&gt;" AnswerCount="1" CommentCount="2" ClosedDate="2015-09-08T17:19:54.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="36883" PostTypeId="1" CreationDate="2015-09-04T11:31:11.093" Score="7" ViewCount="16213" Body="&lt;p&gt;Here's the definition of ionization energy that I am learning at this level: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of uni positively charged gaseous ions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would like to know, why did they specify that the removal of electrons have to be at gaseous state and the ions that form are gaseous as well? Does it have to do with the amount of energy supplied that causes the elements to vapourize?&lt;/p&gt;&#xA;" OwnerUserId="19307" LastEditorUserId="5456" LastEditDate="2015-09-04T15:44:25.663" LastActivityDate="2017-06-10T11:55:15.747" Title="Why is Ionization Energy definition restricted to gaseous atoms?" Tags="&lt;inorganic-chemistry&gt;&lt;ionization-energy&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36884" PostTypeId="2" ParentId="36883" CreationDate="2015-09-04T11:45:47.927" Score="2" Body="&lt;p&gt;The reason why ionisation energy is measured as the energy required to remove 1 mole of electrons from 1 mole of substance is that it allows the ionisation energy of different substances to be compared. This is because the ionisation energy of a substance is going to be different in different states.Therefore you can't compare the ionisation energy of a substance in a solid state to the ionisation energy of another substance in the liquid state. For this reason, the ionisation energy of all substances are measured in the gaseous state. It is like the standard solubility of a substance is taken at 25 degrees. It doesn't make sense to compare of the solubility of one substance at 25 degrees and the solubility of another substance at 100 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you might ask why did they choose to measure the ionisation energy of substances in the gaseous phase, why not the solid or liquid phase? The reason why ionisation energy is measure in the gaseous phase is because in the gaseous phase, there is very little attraction between other particles which effects the ionisation energy. Therefore, if the ionisation energy was measured for a substance in the solid, since the particles are close together, the intermolecular forces will distort the value of the ionisation energy.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2015-09-04T11:45:47.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36885" PostTypeId="1" AcceptedAnswerId="37411" CreationDate="2015-09-04T11:59:59.383" Score="8" ViewCount="2061" Body="&lt;p&gt;Recently I have read articles stating that negative Kelvin has been achieved. I was a bit speculative at first as I though how can you get cold than 0 Kelvin, but after doing some research it makes sense on how this is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the one thing that I still not sure is that it stated that the temperature scale can't be thought as being linear but rather as a loop which goes in the following order, &quot;+0 K, … , +300 K, … , +∞ K, −∞ K, … , −300 K, … , −0 K&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can negative Kelvin be hotter than infinite temperature?&lt;/p&gt;&#xA;" OwnerUserId="18002" LastActivityDate="2016-01-30T16:15:40.077" Title="How is Negative Temperature Hotter than Infinite Temperature?" Tags="&lt;thermodynamics&gt;&lt;quantum-chemistry&gt;&lt;temperature&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36886" PostTypeId="2" ParentId="36883" CreationDate="2015-09-04T12:11:07.703" Score="9" Body="&lt;p&gt;In fact, it is not restricted at all. You may compare the energies needed to remove one electron out of a solid; these are also meaningful, albeit in a different way, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_function#Work_functions_of_elements.5B11.5D&quot;&gt;known&lt;/a&gt; for a wide range of substances. But when you are talking about atoms, you want to measure atoms, and the only way to have an undisturbed lone atom is to put it in a gaseous state. And of course, the resulting ion should be in that state, too; after all, you can't make a solid out of positive ions alone, and if you want to add negative ions to counterbalance them, an ambiguity arises as to which ions should you use.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-04T12:11:07.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36887" PostTypeId="1" AcceptedAnswerId="36909" CreationDate="2015-09-04T12:26:05.287" Score="6" ViewCount="97" Body="&lt;p&gt;By searching for the reference of the synthesis below I arrived at an example of Cationic Hetero-Cope Rearrangement, but I can't access the paper. How does it work here, and what happens in the first step of the synthesis (the addition of iBuOH and TsOH at 80 degrees)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rhF8m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rhF8m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-09-05T08:34:39.717" Title="Synthesis using Cationic Hetero-Cope Rearrangement" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="36888" PostTypeId="2" ParentId="34636" CreationDate="2015-09-04T14:23:09.387" Score="3" Body="&lt;p&gt;Ions are hygroscopic because they form very stable, low energy hydrates, which are thermodynamically favorable over water vapor. I.E. salt attracts water vapor because the hydrated salt is lower energy than the water vapor. Other very porous materials are hygrosopic due to capillary action. Water vapor is attracted to the surface initially due to a variety of reasons, then the narrow channels in the porous material cause the water to spread out to lower their energy.&lt;/p&gt;&#xA;" OwnerUserId="20642" LastActivityDate="2015-09-04T14:23:09.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36889" PostTypeId="2" ParentId="36875" CreationDate="2015-09-04T14:26:53.080" Score="1" Body="&lt;p&gt;The confusion is perhaps not uncommon. However, whichever book it is has jumped from discussing the free energy of pure materials, and their variation with various things (T, P, V, ...), to discussing mixtures.  In this case, all variation of the 'pure' substances is present in the ideal terms - that is because those variation exist without regard for any mixing terms.  What you are focused on here are the impacts of making a solution (or binary alloy, ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to make the mixture, you take a certain amount of substance A, and a certain amount of substance B, and mix them together. Before mixing, the total free energy of the system is $x_A G_{A} + x_B G_{B}$ where $G_{A,B}$ are the Gibbs free energies of those substances at whatever the conditions you specify before mixing. So, what does mixing do? Well, at the very least you get the increase in entropy, $RT(x_{A}\ln{x_{A}} +x_{B}\ln{x_{B}}$. If there is any interaction between A and B, you now have non-ideal terms which can be represented in any number of different ways. However, the point is that they are &quot;all the rest that happens&quot; during the mixing, and those things may be functions of concentration, temperature, pressure, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, to come back to your question - no, this is not an error in the textbook (whichever it is). But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also note that, when one calculates a phase diagram (ultimately one hopes the book might get to that, and miscibility gaps and whatnot), you are free to choose the reference points or phases of A and B yourself - that choice makes no difference in the phase boundaries and tie lines. And, this comes back to just what $G_{A,B}$ means.&lt;/p&gt;&#xA;" OwnerUserId="5677" LastActivityDate="2015-09-04T14:26:53.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36891" PostTypeId="1" CreationDate="2015-09-04T15:51:13.493" Score="2" ViewCount="2710" Body="&lt;p&gt;I've read in &lt;a href=&quot;http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_Product_Constant,_Ksp&quot; rel=&quot;nofollow&quot;&gt;chemwiki&lt;/a&gt; about the diverse ion effect. Assuming I have an aqueous solution of $\ce{NaCl}$ and an associated $K_\mathrm{sp}$. If I add $\ce{KI}$ there is an associated $K_\mathrm{sp}'$ for $\ce{NaCl}$ where $K_\mathrm{sp}' &amp;gt; K_\mathrm{sp}$. This actually means that I can dissolve more $\ce{NaCl}$ than I could before. Why does this happen? Shoudln't the new solution actually be closer to saturation and $K_\mathrm{sp}' &amp;lt; K_\mathrm{sp}$?&lt;/p&gt;&#xA;" OwnerUserId="18005" LastEditorUserId="10185" LastEditDate="2015-09-04T17:17:24.007" LastActivityDate="2015-09-06T19:33:51.060" Title="Why the &quot;Diverse Ion Effect&quot; increases Ksp?" Tags="&lt;physical-chemistry&gt;&lt;equilibrium&gt;&lt;aqueous-solution&gt;&lt;solubility&gt;" AnswerCount="1" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36892" PostTypeId="1" CreationDate="2015-09-04T18:13:59.190" Score="1" ViewCount="630" Body="&lt;p&gt;I read an article about &lt;a href=&quot;http://www.chembuddy.com/?left=pH-calculation&amp;amp;right=pH-buffer-capacity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buffer capacity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{HA &amp;lt;=&amp;gt; H+ + A-}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have given that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RrxSF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RrxSF.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But shouldn't $C_{buf}=[HA]+[A^-]+[H^+]$ ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total concentration at equilibrium is $c(1-\alpha)+2c\alpha=c(1+\alpha)\neq c$. Hence the initial concentration of buffer is different from that at equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="10185" LastEditDate="2015-09-04T18:15:19.520" LastActivityDate="2015-09-04T18:15:19.520" Title="Derivation of buffer capacity" Tags="&lt;equilibrium&gt;&lt;ph&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36893" PostTypeId="2" ParentId="35834" CreationDate="2015-09-04T18:28:52.433" Score="4" Body="&lt;p&gt;Reaction of an ester with a Grignard reagent gives two products: (1) a tertiary alcohol coming from the carbonyl carbon and two equivalents of the Grignard reagent and (2) an alcohol liberated from the ester. I believe it's this liberated alcohol that is reacting with bleach, since a tertiary alcohol is typically resistant to oxidation. However, the liberated alcohol might be primary, secondary, or tertiary depending on the identity of R2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A5AB3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5AB3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information given says that after treatment with bleach, acetic acid is formed. This tells us two things: (1) that liberated alcohol is a primary alcohol, since the oxidation goes all the way to an acid and (2) that liberated alcohol has two carbons, since acetic acid has two carbons. That implies that the liberated alcohol is ethanol. The starting ester has four carbons total, which leaves two carbons for the carboxy fragment. The original ester was ethyl acetate.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastActivityDate="2015-09-04T18:28:52.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36894" PostTypeId="1" CreationDate="2015-09-04T18:50:35.800" Score="2" ViewCount="6731" Body="&lt;p&gt;I am using 3% hydrogen peroxide for a series of operations, and the common practice is to prepare it on-site from 30% hydrogen peroxide and saline. This is time consuming and somewhat more prone to error than preparing a 3% stock. I was told this is not done because 3% hydrogen peroxide has a short shelf-life (~a week). &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this true?&lt;/li&gt;&#xA;&lt;li&gt;If so, why?&lt;/li&gt;&#xA;&lt;li&gt;Would the shelf-life be extended if the solution were prepared with phosphate-buffered saline or just with deionized water (instead of saline)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3974" LastEditorUserId="3974" LastEditDate="2015-09-05T23:35:14.263" LastActivityDate="2015-10-06T00:39:52.750" Title="Storage life of 3% hydrogen peroxide" Tags="&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="36895" PostTypeId="1" CreationDate="2015-09-04T19:22:22.643" Score="5" ViewCount="4622" Body="&lt;p&gt;I am seasoning a carbon steel wok and from what I understand it basically involves heating cooking oil on it until it gets to the smoke point and then forms a polymerized layer that makes it like a non stick pan.  I have also heard that as oil gets heated close to and past the smoke point it breaks down releasing free radicals and other unhealthy compounds.  My question is that when the heated oil polymerizes do these compounds get permanently locked up in the polymer or will they eventually get into the food cooked in the pan? &lt;/p&gt;&#xA;" OwnerUserId="20648" LastEditorUserId="4231" LastEditDate="2015-09-04T19:37:33.480" LastActivityDate="2019-12-17T20:51:19.493" Title="Do polymerized cooking oils release free radicals" Tags="&lt;everyday-chemistry&gt;&lt;polymers&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36896" PostTypeId="2" ParentId="36870" CreationDate="2015-09-04T19:47:51.760" Score="2" Body="&lt;p&gt;Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: No, it is possible for a bond to form between two species without first breaking a bond. But it is more common for reactions to occur with a tradeoff between breaking old bonds and forming new ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Adding energy in the form of heat can lead to breaking of bonds, which in turn can lead to formation of new bonds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: You can have a &quot;displacement&quot; reaction where one atom replaces another in a structure: the bond to the new atom forms at the same time as the other bond is broken. That process has a lower energy barrier between reactants and products, relative to breaking an existing bond first and then forming a new bond later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: The strength of a bond is the same as the amount of energy needed to break it (in the absence of other reactants). Bond length tends to correlate inversely with bond strength, as for example in going from a single to a double to a triple bond between carbon atoms. Bond strength depends on a number of factors--bond strength tends to be greater between two atoms of about the same size. Bonds between two electronegative atoms like oxygen or chlorine are weaker than a bond between two carbon atoms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons &quot;taken&quot; by the more electronegative atom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: You are correct here in your understanding of bond order. But the term &quot;bond order&quot; does not apply to so-called ionic bonds -- for example in sodium chloride -- where an electron is completely donated from sodium to chlorine and does not participate in keeping the atoms together. Rather, it is the electrostatic attraction between the sodium and chloride ions that keep them together.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-09-04T19:47:51.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="36897" PostTypeId="1" AcceptedAnswerId="53384" CreationDate="2015-09-04T20:57:34.320" Score="8" ViewCount="2240" Body="&lt;p&gt;In class, we performed a lab that involved heating a solution of zinc sulfate with zinc strips at the bottom of the beaker. A penny was placed on top of the zinc strips and soon after the penny became covered in a layer of zinc. What process causes the zinc layer to develop?    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen.  Is a thin zinc layer more stable?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.elementalscientific.net/store/scripts/openExtra.asp?extra=10.&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; are the instructions for the experiment &lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="5026" LastEditDate="2018-03-03T05:02:38.417" LastActivityDate="2020-10-23T20:53:47.177" Title="Why does zinc plate copper in silver penny lab" Tags="&lt;redox&gt;&lt;teaching-lab&gt;" AnswerCount="3" CommentCount="4" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36898" PostTypeId="2" ParentId="35494" CreationDate="2015-09-04T21:48:21.443" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If trans double bonds cause cycloalkenes to be chiral&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that a &lt;em&gt;trans&lt;/em&gt; double bond in a cycloalkene does not necessarily produce a chiral cycloalkene.  Only when the chain connecting the ends of the double bond is short enough so that the &lt;em&gt;trans&lt;/em&gt; double bond is twisted will a chiral conformation exist.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in large &lt;em&gt;trans&lt;/em&gt;-cycloalkenes the double bond is not twisted and cannot produce chirality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember too, that even if the &lt;em&gt;trans&lt;/em&gt; double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your &lt;a href=&quot;https://chemistry.stackexchange.com/questions/35493/are-all-higher-cycloalkenes-chiral&quot;&gt;earlier questions&lt;/a&gt; where it was pointed out that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At room temperature, &lt;em&gt;trans&lt;/em&gt;-cyclooctene is the only &lt;em&gt;trans&lt;/em&gt;-cycloalkene&#xA;  where the two enantiomeric conformations are stable enough (&lt;em&gt;e.g.&lt;/em&gt; the&#xA;  barrier to reach the planar conformation is high enough) to be&#xA;  isolated.  If we lower the temperature to 0°C, then we can isolate the&#xA;  enatiomers of trans-cyclononene. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Back to your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;does a second trans or maybe cis double bond induce another chiral&#xA;  plane&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional &lt;em&gt;cis&lt;/em&gt; and &lt;em&gt;trans&lt;/em&gt; double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aJyD5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aJyD5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the above figure on the left, of a &lt;em&gt;trans&lt;/em&gt; double bond in a small ring.  We are sighting down a &lt;em&gt;trans&lt;/em&gt; double bond. To alleviate ring strain, the double bond if forced to &lt;strong&gt;twist&lt;/strong&gt;.  It is this twisting out of plane that infers chirality upon the double bond.  Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will  need to be specified in the name of the compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the figure on the right of a &lt;em&gt;cis&lt;/em&gt; double bond in a small ring. To alleviate strain here, the $\ce{C_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists.  Therefore, in the &lt;em&gt;cis&lt;/em&gt; double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and hence four stereoisomer are possible for the given cyclodiene?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not always.  Just as in more common compounds with 2 chiral centers, &lt;em&gt;meso&lt;/em&gt; stereoisomers are possible.  As an example, if the known E,E-cycloocta-1,5-diene &#xA;adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral &lt;em&gt;meso&lt;/em&gt; compound where one &lt;em&gt;trans&lt;/em&gt; double bond has the R configuration and the other double bond the S configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cx9pm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cx9pm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://wtt-pro.nist.gov/wtt-pro/index.html?cmp=~e.e~-1.5-cyclooctadiene&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.867" LastActivityDate="2015-09-04T21:48:21.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36899" PostTypeId="1" AcceptedAnswerId="36902" CreationDate="2015-09-04T23:02:18.090" Score="6" ViewCount="604" Body="&lt;p&gt;I was recently reading the satirical, fictional book &lt;em&gt;Cat's Cradle&lt;/em&gt; by Kurt Vonnegut. In this book, one of the major subplots is the invention of a mythical polymorph of ice called ice-9 in the book. In the book, it is described that if this polymorph of ice were to touch actual water, all the water, inevitably touching through oceans, rivers, lakes, etc. would freeze into this polymorph of ice which is purported to melt at a temperature over 100 degrees Fahrenheit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raises two questions for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, would something like this actually happen? I know, for instance, a method people use for purifying crystals is to create a supersaturated solution and disturb the substance and it will crystallize quite dramatically. Could something similar happen here in the sense that the mythical ice-9 structure would be introduced into liquid water (which to ice-9 is a supersaturated solution of water) and create a massive crystal out of all water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, and more in reality, is there any solid that is like this? A solid which has two stable crystal structure that exist at temperature reachable in everyday life at about atmospheric pressure? And if so, can you actually convert one crystal structure to another easily? And would one of the liquids behave as a supersaturated solution of the crystal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if that is a little unclear. I am sincerely interested in this and can clarify if there is any confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything interesting along those lines would be considered relevant in my mind.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="10185" LastEditDate="2015-09-05T18:38:58.627" LastActivityDate="2015-09-06T16:45:29.967" Title="Freezing all the water on earth with a seed crystal" Tags="&lt;water&gt;&lt;crystal-structure&gt;&lt;theoretical-chemistry&gt;&lt;chemistry-in-fiction&gt;" AnswerCount="2" CommentCount="5" FavoriteCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36900" PostTypeId="1" AcceptedAnswerId="57177" CreationDate="2015-09-04T23:14:28.250" Score="7" ViewCount="760" Body="&lt;p&gt;I was just recently looking into some of the details of the different crystal structures that ice can freeze in. I was aware that many of the unusual properties of ice $I_h$ (the structure of “everyday ice”) are a result of the proton disorder necessary for the hydrogen-bonding arrangement it adopts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I also found is there are other arrangements of ice which are proton ordered and have nearly identical density to ice $I_h$. Now, I know density has no predictive value on what crystal structure will be taken (rather crystal structure can be used to predict density), but it seems unusual to me that there is such massive divide between ice $I_h$ and the rest of the ice srystal structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the entropic gain of proton disorder an explanation why ice freezes the way it does? Also, is there a large energy minimization associated with the hexagonal lattice that ice $I_h$ takes on? Doesn't it seem like a proton-ordered arrangement would be preferable over a proton-disordered arrangement? Or is it all much more complicated than that? It just seems like there should be a crystal structure of ice which is comparable to ice $I_h$ in energy, stability, etc. and yet there isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose a true answer to this question, because I admit it's rather broad, would deal with why the hexagonal lattice is so much more preferable for water than some other lattice arrangement. By dealing strictly with water, rather than crystals in general, I believe an answer to this question is more within reach.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="30" LastEditDate="2016-08-09T21:01:22.057" LastActivityDate="2016-08-09T21:01:22.057" Title="Why does ice freeze the way it does?" Tags="&lt;water&gt;&lt;crystal-structure&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36902" PostTypeId="2" ParentId="36899" CreationDate="2015-09-05T02:44:18.240" Score="8" Body="&lt;p&gt;A precise physical argument against the existence of an &lt;em&gt;Ice-9&lt;/em&gt;-like state of water is probably not trivial (&lt;em&gt;ab initio&lt;/em&gt; methods required?). However, there is a simple but powerful argument against it; we haven't found anything like it yet. Water is a very common substance, and humans have been applying the scientific method to it for hundreds of years, humans have manipulated it for hundreds of thousands of years, but most importantly, the entire Earth has been churning massive amounts of water around for billions of years. If an &lt;em&gt;Ice-9&lt;/em&gt;-like state of water &lt;em&gt;did&lt;/em&gt; exist, we should have seen it by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, superficially this argument may seem to be a fallacious &lt;a href=&quot;https://en.wikipedia.org/wiki/Argument_from_ignorance&quot; rel=&quot;nofollow noreferrer&quot;&gt;argument from ignorance&lt;/a&gt;. However, we can qualitatively back it up with a bit more knowledge. Assuming that &lt;em&gt;Ice-9&lt;/em&gt; is the true ground state of water, and that all liquid water on Earth is currently in a false ground state, separated only by a kinetic barrier, and realising that the conversion of liquid water to &lt;em&gt;Ice-9&lt;/em&gt; is autocatalytic, we must ask ourselves, &lt;em&gt;if it only had to happen once, why hasn't it happened yet?&lt;/em&gt;. The only reasonable conclusion is that, if it can happen, then the kinetic barrier to conversion must be enormous. The interconversion of diamond to graphite is a chemical process with a famously large kinetic barrier (&lt;a href=&quot;https://chemistry.stackexchange.com/questions/34193/are-diamonds-really-forever/&quot;&gt;around $\mathrm{540\ kJ\ mol^{-1}}$&lt;/a&gt;), yet we have plentiful evidence of it happening many times in geological time scales (and we can do it in the lab too!). Thus the kinetic barrier for conversion between liquid water and &lt;em&gt;Ice-9&lt;/em&gt; would have to be even larger, which is unlikely. There's also a limit to how high we can expect chemical kinetic barriers to be, which we can put &lt;strong&gt;very&lt;/strong&gt; generously at $\mathrm{100\ eV \approx 10\ MJ\ mol^{-1}}$. Water molecules on Earth have had many, many, many interactions at energies way higher than this, as they have been incessantly exposed to cosmic radiation, nuclear decays, lightning strikes, high temperatures, etc. In short, the collective history of all water molecules on Earth has explored a &lt;strong&gt;vast&lt;/strong&gt; region of the space of allowed states, and &lt;em&gt;not once&lt;/em&gt; has an &lt;em&gt;Ice-9&lt;/em&gt;-like state been found. It is thus fair to assume it simply does not exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your second question, I just remembered a great example: pure tin. At ambient pressure, solid tin can exist in two different polymorphs, and the transition between the most stable form is conveniently close to room temperature. Above 13.2 ºC, tin is most stable as a ductile metallic material, called &lt;em&gt;white tin&lt;/em&gt;, whereas below 13.2 ºC, the most stable form is a brittle non-metal called &lt;em&gt;grey tin&lt;/em&gt;. It is possible to sustain white tin below 13.2 ºC for a period of time, creating a metastable system, much like a supersaturated solution. However, an adequate disturbance will cause the white tin to fall to its ground state. Like Vonnegut's &lt;em&gt;Ice-9&lt;/em&gt;, this transition is autocatalytic; once conversion begins, there's no stopping it below 13.2 ºC, creating the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Tin_pest&quot; rel=&quot;nofollow noreferrer&quot;&gt;tin pest&lt;/a&gt;. There are actually several great videos showing the process, which you can easily find on YouTube, such as &lt;a href=&quot;https://www.youtube.com/watch?v=sXB83Heh3_c&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1499" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-09-05T02:50:46.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36903" PostTypeId="2" ParentId="35251" CreationDate="2015-09-05T04:20:38.893" Score="3" Body="&lt;p&gt;Let me venture an answer as well as I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would point out that the article you are referencing is speaking of the arrangement of molecules in bulk liquid water when it is talking about this &quot;fifth water&quot; being present. The key is to note that they say a &quot;fifth water molecule in the first coordination shell (23).&quot; That means they are looking at how water molecules are coordinated with each other &lt;strong&gt;in liquid water&lt;/strong&gt;. They are studying a massive (compared to a single water) amount of water, so the question of its liquidity is no question at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;fifth water&quot; thing is curious because it is common belief that waters arrange themselves in a tetrahedral arrangement. This is known to be true for ice, but is in question for liquid water. For instance, &lt;a href=&quot;http://www-ssrl.slac.stanford.edu/downloads/science-structureofwater.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article about the liquid water hydrogen-bonding network&lt;/a&gt; indicates that water, as it is heated, begins participating in only two hydrogen bonds rather than four.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer the question of what makes liquid water be liquid water (when it is indeed liquid water). I think from the outset, it should be clear that this is a very difficult question to answer and any answer will likely be debated. I say that because you must define how many molecules it takes to make a liquid which is a very difficult thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will tell you, however, that I have some useful information from research I am doing on water clusters and eventually liquid water (as a computational chemistry research assistant). The largest cluster we have studied is called DD*(20,1) which means (distorted dodecahedron with a water inside the cluster).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An idea of what $\ce{(H2O)21}$ can look is this: (taken from &lt;a href=&quot;http://pubs.acs.org.ezproxy.spu.edu/doi/full/10.1021/jp056416m&quot; rel=&quot;nofollow noreferrer&quot;&gt;this very interesting article)&lt;/a&gt; &lt;a href=&quot;http://pubs.acs.org.ezproxy.spu.edu/appl/literatum/publisher/achs/journals/content/jpcbfk/2006/jpcbfk.2006.110.issue-38/jp056416m/production/images/medium/jp056416mf00001.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;water 21 http://pubs.acs.org.ezproxy.spu.edu/appl/literatum/publisher/achs/journals/content/jpcbfk/2006/jpcbfk.2006.110.issue-38/jp056416m/production/images/medium/jp056416mf00001.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DD*(20,1) arrangement is image C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from these different isomers, they are very distinct and no doubt each monomer behaves dramatically different from the next monomer. 21 monomers is quite a lot for a cluster, but even this does not mimic the properties of liquid water at all particularly because one would still consider this a gas-phase cluster. The fact that we even have isomers is indicative of not being near the point where something is considered liquid water. There are water clusters whose energy minima have been identified theoretically all the way from n=2 to (I've seen an article that says) n=60. Read more about some of that &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_cluster&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point I would like to make is that in structure C, for the research I'm doing, we have identified the vibrational properties of many of the monomers as being significantly different from those of molecules in liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, all that being said about water clusters, we are certain that many more than five or 21 waters must be present to truthfully call something a liquid. We should then start looking for a system which is just large enough to accurately predict the properties of liquid water which are found experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, it is very difficult to do this on any scale, theoretically, at least, and identifying how many water molecules one is dealing with during an experiment to a quantifiable amount of precision is quite difficult. There are, however, good approximations of liquid water which are generated using Molecular Dynamics simulation software (here's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page about MD software&lt;/a&gt;). The various models used for approximating the behavior of water do quite well at mimicking water's properties from an MD perspective. You can read about the &lt;a href=&quot;http://www.academia.edu/6682560/A_general_purpose_model_for_the_condensed_phases_of_water_TIP4P_2005&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIP4P model of water if you want&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the paper I just linked about TIP4P, they say that &lt;strong&gt;they approximate the properties of liquid water using 360 water molecules&lt;/strong&gt; in their MD simulations, so that can act as a reasonable baseline for how many monomers are necessary before a system behaves like liquid water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say baseline for an important reason. The results found in that paper use MD simulations which pull a clever trick. Rather than having 360 waters that essentially float around in a box with nothing at all beyond the edges of the box, the software will actually mirror the behavior of a molecule beyond the walls of the box in which you are simulating. That means if you pick a single water, you would find that exact same water in the same place a full &quot;simulation box&quot; away from where you originally found it. So, if this says 360 waters, the number of waters for which interactions are being calculated is actually much larger than the amount of interactions 360 waters alone would have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that leads me to conclude that water behaves like water probably somewhere around 500 monomers. It must be noted, however, that the monomers behaving like liquid water as we know it will be the monomers on the inside of those 500 (but likely more) waters. Once you get towards a boundary, things start getting weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="17583" LastActivityDate="2015-09-05T04:20:38.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36904" PostTypeId="1" CreationDate="2015-09-05T04:31:52.670" Score="5" ViewCount="37037" Body="&lt;p&gt;I am particularly interested in the effect or lack of effect of dissolved NaCl in regards to pH in water. It would be interesting also if anyone has any insight for other salts and their effects. thanks.&lt;/p&gt;&#xA;" OwnerUserId="9649" LastEditorUserId="10185" LastEditDate="2015-09-05T17:39:31.313" LastActivityDate="2017-10-30T13:37:08.287" Title="Does dissolving salt in water change the pH?" Tags="&lt;inorganic-chemistry&gt;&lt;aqueous-solution&gt;&lt;ph&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36905" PostTypeId="2" ParentId="36904" CreationDate="2015-09-05T05:01:41.883" Score="2" Body="&lt;p&gt;$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="10185" LastEditDate="2015-09-05T17:41:02.087" LastActivityDate="2015-09-05T17:41:02.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36907" PostTypeId="2" ParentId="36904" CreationDate="2015-09-05T08:15:13.210" Score="8" Body="&lt;p&gt;When you dissolve a salt in water, you have to distinguish 4 cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$  do hydrolyze and give ions  $\ce{H^+}$ to water. So, the pH will decrease.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$  do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$K_a= K_b$, the aqueous solution is neutral.&#xA;This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions  $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.  &lt;/li&gt;&#xA;&lt;li&gt;$K_a&amp;gt; K_b$&#xA;This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions  $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze  and as  ammonium ions hydrolyze to a greater extent than  fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.&lt;/li&gt;&#xA;&lt;li&gt;$K_a&amp;lt; K_b$&#xA;This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions  $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze  and as  cyanide ions hydrolyze to a greater extent than  fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic. &lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-09-05T09:04:32.983" LastActivityDate="2015-09-05T09:04:32.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36908" PostTypeId="2" ParentId="36894" CreationDate="2015-09-05T08:21:06.210" Score="2" Body="&lt;p&gt;It may pretty well be true, and here's why. Not only hydrogen peroxide is highly unstable by itself, but its decay can be catalyzed by a wide variety of compounds. And by wide, I mean &lt;strong&gt;really&lt;/strong&gt; wide; at times it seems to me that those compounds that don't catalyze it are a minority. Now, I don't know what you have in your saline, but chances are that it works on the peroxide, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With deionized water, this should not be a problem... or should it? See, that link by @ron says that the 30% thing is &lt;em&gt;stabilized&lt;/em&gt;. Guess what? It means that they've put some additive into it, which somewhat slows down the decomposition. When you dilute your peroxide, the stabilizer gets diluted too, and hence can't work as well as before. This is a possible reason why you may get shorter shelf-life even when diluting with pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what is the stabilizer they use. In any case, if you find that out, then add it during the preparation, and don't add anything that can catalyze the decay, your 3% solution should be fine (in terms of shelf-life) - but I'm afraid these additions and subtractions may stand in the way of your primary objectives.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-05T08:21:06.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36909" PostTypeId="2" ParentId="36887" CreationDate="2015-09-05T08:34:39.717" Score="2" Body="&lt;p&gt;This would be the reaction sequence:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uhjXI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uhjXI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastActivityDate="2015-09-05T08:34:39.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36910" PostTypeId="1" AcceptedAnswerId="36916" CreationDate="2015-09-05T08:51:19.367" Score="1" ViewCount="172" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h3TGd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h3TGd.png&quot; alt=&quot;Image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the ether hold priority over longest chain? If so, my bet would be on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4-(2-ethylheptanyl)oxy decane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, I would say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4-(octan-2-yl methoxy) decane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u4RTV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u4RTV.png&quot; alt=&quot;Image 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if ether takes precedence and not longer chain (which, if you ask me, would be ideal), I'd say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4-[(1-methyl,2-propylheptan-1-yl)oxy]decane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4-(ethan-1-yl oxy nonan-4-yl)decane&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is correct?&lt;/p&gt;&#xA;" OwnerUserId="19507" LastActivityDate="2015-09-05T11:24:49.070" Title="Ether nomenclature: Oxygens on alkyls and alkyls on alkyls" Tags="&lt;organic-chemistry&gt;&lt;nomenclature&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36911" PostTypeId="1" CreationDate="2015-09-05T09:14:05.497" Score="18" ViewCount="27006" Body="&lt;p&gt;First, I would like to quote sentences from a book introducing elements and atoms:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;An element is a fundamental (pure) form of matter that cannot be broken down to a simpler form.&lt;/p&gt;&#xA;&lt;p&gt;Elements are made up of particles called atoms. An atom is the smallest unit of any element [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The first quote states that element &lt;strong&gt;cannot&lt;/strong&gt; be broken down to a simpler form, but the second quote says elements are &lt;strong&gt;formed&lt;/strong&gt; by atoms, which means element can be split into a simpler form (atoms)?&lt;/p&gt;&#xA;&lt;p&gt;The difference between atom and element confuses me.&lt;/p&gt;&#xA;" OwnerUserId="20654" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2019-12-22T14:27:54.903" Title="What is the difference between an element and an atom?" Tags="&lt;atoms&gt;&lt;terminology&gt;&lt;elements&gt;" AnswerCount="6" CommentCount="1" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36912" PostTypeId="2" ParentId="36911" CreationDate="2015-09-05T09:29:49.700" Score="18" Body="&lt;p&gt;The wording in your quote is a bit unclear, I agree. A very nice definition is given by the Jefferson Lab page &lt;a href=&quot;http://education.jlab.org/qa/atom_02.html&quot; rel=&quot;noreferrer&quot;&gt;What is the simplest way of explaining what atoms, elements, compounds and mixtures are?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;are the smallest bits of ordinary matter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whereas an element is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a substance that is made entirely from one type of atom&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Atoms themselves can be 'broken down' into smaller sub-atomic particles (protons, neutrons and electrons) - the amounts of these (particularly the number of protons) define each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in their un-ionised form, the element carbon is made entirely from carbon atoms, which have 6 protons and 6 electrons each, and cadmium atoms, which have 48 protons and 48 electrons, make up the element cadmium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of neutrons can vary with atoms of a particular element, forming &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotope&quot; rel=&quot;noreferrer&quot;&gt;isotopes&lt;/a&gt;, which (from the link):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Different arrangements of atoms of single elements can form &lt;a href=&quot;https://en.wikipedia.org/wiki/Allotropy#List_of_allotropes&quot; rel=&quot;noreferrer&quot;&gt;allotropes&lt;/a&gt;, such as for carbon, with diamond and graphite (among others - shown below). They both are composed of atoms of the element carbon, but in different configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are diagrams of the allotropes of carbon featured in the article &lt;a href=&quot;http://www.mdpi.com/2079-4991/4/2/267/htm&quot; rel=&quot;noreferrer&quot;&gt;Work Function Engineering of Graphene&lt;/a&gt; (Garg et al. 2014)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Sngf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Sngf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15489" LastEditorDisplayName="user15489" LastEditDate="2015-09-05T13:57:33.133" LastActivityDate="2015-09-05T13:57:33.133" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="36913" PostTypeId="1" CreationDate="2015-09-05T09:41:46.053" Score="2" ViewCount="8136" Body="&lt;p&gt;When I have a solution of $\ce{Na2SO4}$, the sodium sulfate should ionize:&#xA;$$\ce{Na2SO4(aq) -&amp;gt; 2Na+(aq) + SO4^{2-}(aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my mind, the sodium ions should just float around, but because $\ce{HSO4-}$ is a weak acid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{HSO4- (aq) + H2O &amp;lt;-&amp;gt; H3O+ (aq) + SO4^2- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sulfate ion should be a weak conjugate base, and hence the sodium sulfate salt should be a weak base. This was my line of reasoning, until today my friends told me it is actually a neutral salt, which was confirmed by searching online. So why is it a neutral salt? (i.e. what's wrong with my reasoning?)&lt;/p&gt;&#xA;" OwnerUserId="9799" LastEditorUserId="10185" LastEditDate="2015-09-05T17:36:40.043" LastActivityDate="2017-01-28T04:42:45.613" Title="Why is sodium sulfate a neutral salt" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;aqueous-solution&gt;" AnswerCount="2" CommentCount="4" ClosedDate="2018-02-24T00:32:02.747" ContentLicense="CC BY-SA 3.0" />
  <row Id="36914" PostTypeId="1" AcceptedAnswerId="36943" CreationDate="2015-09-05T10:09:16.583" Score="4" ViewCount="977" Body="&lt;p&gt;I need a reliable method for determining the number of diastereomers and enantiomeric pairs of the following series of compounds bearing in mind that the substituent groups A are chiral on their own (exhibiting central chirality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9QC1J.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9QC1J.gif&quot; alt=&quot;allene&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two planes of interest bisecting the allene sistem, the one constains the left A groups, the other is orthogonal to it. If two A groups are to be reflected through one plane (so it becomes a symmetry plane) their configuration must be different. That is all regarding the central chirality, but I don't know how to include axial chirality in this description (so I don't count one stereoisomer twice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that these are the four diastereomers. Instead of A, I wrote the configuration of A (R/S). Now, as R and S are connected with a mirror plane, diastereomers 2 and 4 shouldn't exhibit axial chirality, as R and S can be regarded as same substituents on one side of the allene system (because of the mirror plane). That's why the third diastereomer exhibits axial chirality (P/M). I am not sure whether my reasoning is correct. The rest beats me. The answer should be 4 diastereomers and 3 enantiomeric pairs (&lt;a href=&quot;http://www.perekalin.narod.ru/Answers1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;last page)&lt;/a&gt;, but how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BuKEy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BuKEy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="5026" LastEditDate="2018-07-17T03:03:04.670" LastActivityDate="2018-07-17T03:04:33.877" Title="How to find the number of stereoisomers of a tetra-substituted allenes, exhibiting both axial and central chirality?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="36915" PostTypeId="1" AcceptedAnswerId="36986" CreationDate="2015-09-05T10:47:34.427" Score="2" ViewCount="26051" Body="&lt;p&gt;I came across an ionic equilibrium problem stating:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the pH when 150 ml 1 M $\ce{NaOH}$ has been added to 100 ml 1 M $\ce{H3PO4}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm stuck with this question. What I know:&#xA;$$\ce{H3PO4 + NaOH -&amp;gt; NaH2PO4 + H2O}$$&#xA;$$\ce{NaH2PO4 + NaOH -&amp;gt; Na2HPO4 + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally 50 millimoles $\ce{Na2HPO4}$ and 50 millimoles $\ce{NaH2PO4}$ remains.&#xA;Should I apply acid buffer equation?&lt;/p&gt;&#xA;" OwnerDisplayName="user14857" LastEditorUserId="16622" LastEditDate="2015-09-07T09:43:40.910" LastActivityDate="2015-09-07T12:09:51.110" Title="Titration of H3PO4 with NaOH" Tags="&lt;inorganic-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;&lt;ph&gt;&lt;titration&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36916" PostTypeId="2" ParentId="36910" CreationDate="2015-09-05T11:24:49.070" Score="3" Body="&lt;p&gt;Names for ethers can be formed by five different methods. Furthermore, various traditional names are retained for use in general nomenclature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt;, preferred IUPAC names of ethers $\ce{R-O-R'}$, when $\ce{R}$ and $\ce{R'}$ are both aliphatic groups, are formed by &lt;strong&gt;substitutive nomenclature&lt;/strong&gt;. In substitutive nomenclature, when $\ce{R}$ is different from $\ce{R'}$, $\ce{RH}$ is chosen as parent hydride and $\ce{R'-O-{}}$ is cited as a substituent to it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;P-63.2.2.1.1&lt;/strong&gt; Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. (…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the first example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.&lt;br&gt;&#xA;The substituent prefix name for the group $\ce{R'-{}}$ is ‘2-ethylheptyl’.&lt;br&gt;&#xA;Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(2-ethylheptyl)oxy’.&lt;br&gt;&#xA;Therefore, the preferred IUPAC name of the compound is ‘4-[(2-ethylheptyl)oxy]decane’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/InTGv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/InTGv.png&quot; alt=&quot;4-[(2-ethylheptyl)oxy]decane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second example in the question, the name of the parent hydride $\ce{RH}$ is ‘decane’.&lt;br&gt;&#xA;The substituent prefix name for the group $\ce{R'-{}}$ is ‘3-propyloctan-2-yl’.&lt;br&gt;&#xA;Thus, the substituent prefix names for the $\ce{R'-O-{}}$ group is ‘(3-propyloctan-2-yl)oxy’.&lt;br&gt;&#xA;Therefore, the preferred IUPAC name of the compound is ‘4-[(3-propyloctan-2-yl)oxy]decane’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WkrSb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkrSb.png&quot; alt=&quot;4-[(3-propyloctan-2-yl)oxy]decane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastActivityDate="2015-09-05T11:24:49.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36917" PostTypeId="2" ParentId="36881" CreationDate="2015-09-05T12:12:26.703" Score="9" Body="&lt;p&gt;There are lots of test for detection of real gold.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Visual inspection&lt;/strong&gt;: Inspect the piece for official markings. And look for noticeable discoloration. If the gold seems to be wearing off and showing a different metal beneath it, you probably have a piece that is only gold plated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bite test&lt;/strong&gt;: Bite down on your gold with moderate pressure. Examine your gold for any markings. In theory, real gold will show indents from your teeth; deeper markings indicate purer gold. &lt;em&gt;This is actually not a recommended test, as you can damage your teeth.&lt;/em&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Magnet test&lt;/strong&gt;: Hold a magnet up to the item. Gold is not a magnetic metal, so if it pulls towards, or sticks to the magnet, it’s fake.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Density test&lt;/strong&gt;: Weigh your piece of gold. Fill a vial with water. It’s helpful if the vial has millimeter markings on the side, since that will make it easier for you to read the measurements for this test. Place your gold in the vial. Use the following formula to calculate density: Density = mass/volume displacement. A result close to 19 g/ml indicates either real gold, or a material with a density similar to gold. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Ceramic test&lt;/strong&gt;: Find an unglazed ceramic plate to use. Drag your item across the surface. A black streak means your gold is not real, whereas a gold streak indicates your item is genuine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Nitric acid test&lt;/strong&gt;: Place your piece of gold in a small stainless steel container. Put a drop of nitric acid on your gold and watch for any resulting reaction to the acid. A green reaction indicates your item is either a base metal or gold plated. A gold reaction indicates your item is gold-plated brass. A milk-colored reaction would indicate gold-plated sterling silver. If there is no reaction, you mostly likely are dealing with real gold.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://googleweblight.com/http://googleweblight.com/?lite_url=http://www.wikihow.com/Tell-if-Gold-Is-Real&amp;amp;ei=bc8nSicG&amp;amp;lc=en-IN&amp;amp;s=1&amp;amp;m=364&amp;amp;ts=1441454605&amp;amp;sig=APONPFk38JPGzbHP8bbtBE-Hx-v6EqWtkg&quot; rel=&quot;noreferrer&quot;&gt;Click here&lt;/a&gt; for detailed steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Try to perform all or at least 4 steps to confirm (with ceramic test compulsory).&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17422" LastEditorUserId="7475" LastEditDate="2017-12-16T15:12:42.313" LastActivityDate="2017-12-16T15:12:42.313" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="36918" PostTypeId="1" CreationDate="2015-09-05T12:16:19.090" Score="3" ViewCount="2970" Body="&lt;p&gt;Generally, are bisulfates and bicarbonates soluble in water? &#xA;Specifically, are calcium hydrogen carbonate and magnesium hydrogen carbonate soluble in water? &lt;/p&gt;&#xA;" OwnerUserId="20655" LastEditorUserId="865" LastEditDate="2015-09-05T13:14:12.473" LastActivityDate="2015-09-05T13:23:06.960" Title="Are bisulfates and bicarbonates soluble in water?" Tags="&lt;solubility&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36919" PostTypeId="2" ParentId="12233" CreationDate="2015-09-05T12:20:04.657" Score="-3" Body="&lt;p&gt;I cannot think of any reason that it can't. Mercury and gallium should both be acceptable mediators for the electron transfer. In fact, elemental Gallium may even be preferred over salts of Gallium unlike Mercury, where only Mercuric salts can be used (maybe due to mercury having such a low melting point?). Elemental Gallium might actually have an added advantage of having the perfect melting point to mediate electron transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would actually be quite ironic if all these chemists are taking elemental mercury and making salts of it just to give it the properties of elemental Gallium! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm not even touching on the toxicological superiority of gallium!&lt;/p&gt;&#xA;" OwnerUserId="20656" LastActivityDate="2015-09-05T12:20:04.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36920" PostTypeId="2" ParentId="36915" CreationDate="2015-09-05T12:24:35.040" Score="2" Body="&lt;p&gt;This is a simple question  involving a mixture of strong acid (not that strong) and a strong base. Since $k_a$ values are not given the book might have wanted you to solve it like this. They want you to assume complete dissociation of both compounds. That is, 5 moles of $\ce{NaOH}$ gives 5 moles of $\ce{OH^-}$ ions after complete dissociation. &#xA;Find the equivalents of $\ce{H^+}$ from phosphoric acid (a) and the equivalents of $\ce{OH^-}$ from $\ce{NaOH}$ (b). Also calculate the total volume. Number of equivalents=$NV$. But be careful since concentration is given in molarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b&amp;gt;a$, then the resulting solution is basic. Calculate the equivalents of $b$ that remain. From this you will get pOH.&#xA;$$[\ce{OH^-}]=\frac{b-a}{V_1+V_2}$$&#xA;If $a&amp;gt;b$, then resulting solution is acidic. Calculate the amount of $a$ that remains. From this, you will get pH.&#xA;$$[\ce{H^+}]=\frac{{a-b}}{V_1+V_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where N is normality and equivalents of acid is greater than equivalents of base.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="19616" LastEditDate="2015-09-06T13:12:38.770" LastActivityDate="2015-09-06T13:12:38.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36921" PostTypeId="2" ParentId="36911" CreationDate="2015-09-05T12:43:48.210" Score="-1" Body="&lt;p&gt;An atom is a thing; an element is a type of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An atom is a collection of protons, neutrons and electrons.  A single, isolated atom in its neutral state has some number of protons, the same number of electrons and some number neutrons (about as many as protons for the lighter elements, up to about 50% more, for heavier elements).  The number of neutrons or protons in an atom only changes as a result of radioactive processes or very high-energy interactions such as you get in particle accelerators.  And I mean &lt;em&gt;really&lt;/em&gt; high energy: even if you think about blowing up sticks of dynamite, that isn't nearly enough energy to start messing with the protons and neutrons. Chemistry happens when atoms come together and share electrons or give electrons to each other.  Chemical reactions happen all the time and many of them don't need a whole lot of energy: moving electrons from atom to atom is often very easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the chemistry of an atom depends on the number of electrons and the number of electrons in an isolated atom depends directly on the number of protons. Electrons are so easy to add and remove from atoms (just rub a balloon against your hair: the static electricity is because you transferred electrons between your hair and the balloon) so we classify atoms according to the number of protons in them.  Neutrons aren't so relevant: I'll put a note about them at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the &lt;em&gt;element&lt;/em&gt; of an atom is determined by the number of protons.  All atoms of hydrogen have one proton, and all atoms with one proton are hydrogen. Two protons is helium, three is lithium, seventeen is chlorine, 79 is gold, etc.  A pure sample of an element contains only atoms of that type: for example, a pure sample of iron contains only atoms with 26 protons.  Water, on the other hand, is not an element: a molecule of water consists of two hydrogen atoms (one proton each) sharing electrons with an oxygen atom (eight protons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does it mean to claim that an element &quot;cannot be broken down into a simpler form,&quot; and why aren't atoms &quot;a simpler form&quot;?  Well, they're not a simpler form because an iron atom &lt;em&gt;is&lt;/em&gt; iron: it's the same form, not simpler.  Think of it this way.  If I give you a lump of pure iron, all you can do is break it into smaller lumps of iron, or make it into a more complex substance, for example by letting it rust&amp;nbsp;&amp;ndash; rust is formed of iron and oxygen.  The smallest possible lump of iron you could make is a single iron atom, but that's still just an unbelievably tiny lump of iron. If you wanted to break down a lump of iron beyond single iron atoms, you'd need to use a nuclear reactor or a particle accelerator or something like that and then, finally, you'd be able to get something that wasn't iron, because you'd have changed the number of protons in the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's contrast that with water.  If I give you a bucket of pure water then, just like the lump of iron, you can divide it into smaller and smaller samples, ultimately ending up with a single water molecule.  But you can do something else: if you pass electricity through the water, it splits into pure hydrogen and pure oxygen.  These are &quot;simpler&quot; substances because each consists of atoms of just one element, whereas water has atoms of two elements.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What about neutrons? Well, in terms of chemistry, they don't do a whole lot and atoms that have the same number of protons but different numbers of neutrons are much more alike (they have essentially the same chemistry, for example) than atoms that have the same number of neutrons but different numbers of protons.  It makes much more sense to classify by the number of protons, since that determines the number of electrons and that determines the chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you tried instead to classify atoms according to the number of neutrons. Well, most argon atoms (18 protons) have 22 neutrons, but some chlorine atoms (17 protons) and a decent fraction of potassium atoms (19 protons) also have 22 neutrons. As you probably know, argon, chlorine and potassium are absolutely nothing like each other. On the other hand, potassium atoms with 22 neutrons behave almost identically to the most common kind of potassium atoms, which have 21 neutrons.&lt;/p&gt;&#xA;" OwnerUserId="7412" LastEditorUserId="7412" LastEditDate="2015-09-05T17:56:21.703" LastActivityDate="2015-09-05T17:56:21.703" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="36922" PostTypeId="2" ParentId="36918" CreationDate="2015-09-05T13:23:06.960" Score="2" Body="&lt;p&gt;According to it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Calcium_bicarbonate&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;, calcium bicarbonate is not a known solid compound, but it does exist in solution, presumably prepared by dissolving calcium carbonate and bubbling carbon dioxide through the solution until bicarbonate dominates the following pH-dependent disproportionation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2HCO3- &amp;lt;=&amp;gt; H2O + CO2 + CO3^2-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It apparently has a solubility of 16.6 g per 100 mL of water at room temperature. which is about a 1 M solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnesium_bicarbonate&quot; rel=&quot;nofollow&quot;&gt;Magnesium bicarbonate&lt;/a&gt; similarly exists only in aqueous solution, though it is much less soluble at 0.077 g / 100 mL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As solids, these salts probably fall victim to the following equilibrium( $\ce{M}$ is a generic divalent metal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{M(HCO3)2 &amp;lt;=&amp;gt; H2O + MCO3 + CO2(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carbon dioxide escapes, preventing the system from ever staying at equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-09-05T13:23:06.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36923" PostTypeId="1" AcceptedAnswerId="36926" CreationDate="2015-09-05T13:42:32.277" Score="0" ViewCount="9472" Body="&lt;p&gt;I will calculate the change in enthalpy of combustion of 1.12g of hexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Calculate the energy transferred to 200g of water using the equation: &lt;em&gt;J = mass of water in grams (&lt;strong&gt;200g&lt;/strong&gt;) x specific heat capacity of water (&lt;strong&gt;4.18 J/gK&lt;/strong&gt;) x temperature increase in Kelvin (24C = &lt;strong&gt;297.15K&lt;/strong&gt;)&lt;/em&gt;. This results in the amount of energy transferred to the water of a total of 248,417.4J or &lt;strong&gt;248.4kJ&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Calculate number of moles of hexane burnt using &lt;em&gt;n=m/M&lt;/em&gt;. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore &lt;strong&gt;1.15g&lt;/strong&gt; of hexane was burnt during the experiment. 1.15g of hexane = &lt;strong&gt;0.013mol&lt;/strong&gt; of hexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt; Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) &lt;strong&gt;-4194 kJ/mol.&lt;/strong&gt;  Therefore, 0.013mol will produce (0.013/-4194) &lt;strong&gt;-3.1x10-6 kJ/mol&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the procedure that I used to find out the answer! Was it correct?&#xA;Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?&lt;/p&gt;&#xA;" OwnerUserId="19586" LastActivityDate="2017-01-23T12:03:02.563" Title="Calculating change in enthalpy of combustion of 1.12g of hexane" Tags="&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36924" PostTypeId="1" CreationDate="2015-09-05T15:12:52.523" Score="1" ViewCount="52" Body="&lt;p&gt;I want to show a schematic representation of chemical reactions from start to finish in a way that will include more detailed preparation instructions than a plain chemical formula: volumes, concentrations, waiting time, temperature, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a graphical language for this kind of thing? I'm looking for something more sophisticated than a typical flow chart, and more targeted towards chemical reactions.&lt;/p&gt;&#xA;" OwnerUserId="10104" LastEditorUserId="31313" LastEditDate="2017-04-01T00:53:33.853" LastActivityDate="2017-04-01T00:53:33.853" Title="Is there a generic graphical language to represent chemical reactions?" Tags="&lt;software&gt;&lt;teaching-lab&gt;" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36925" PostTypeId="2" ParentId="33260" CreationDate="2015-09-05T15:33:39.493" Score="1" Body="&lt;p&gt;The current version of &lt;em&gt;&lt;a href=&quot;http://dx.doi.org/10.1039/9781849733069-fp001&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)&lt;/a&gt;&lt;/em&gt; reads as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5&lt;/strong&gt; ALPHANUMERICAL ORDER&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved&lt;/p&gt;&#xA;&lt;p&gt;Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.5.1&lt;/strong&gt; Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore, the correct alphanumerical order of citation in your example corresponds to the name ‘&lt;em&gt;x&lt;/em&gt;-chloro-&lt;em&gt;y&lt;/em&gt;-methoxyethane’.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;P-14.4&lt;/strong&gt; NUMBERING&lt;/p&gt;&#xA;&lt;p&gt;When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;p&gt;(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;&lt;/p&gt;&#xA;&lt;p&gt;(g) lowest locants for the substituent cited first as a prefix in the name;&lt;/p&gt;&#xA;&lt;p&gt;(…)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note that Rule f takes precedence over Rule g.&lt;br /&gt;&#xA;According to Rule f, the name of the given compound could be ‘1-chloro-2-methoxyethane’ as well as ‘2-chloro-1-methoxyethane’ since both names correspond to the locant set ‘1,2’ (all prefixes considered together in a series of increasing numerical order).&lt;br /&gt;&#xA;According to Rule g, however, the compound is named as ‘1-chloro-2-methoxyethane’ rather than ‘2-chloro-1-methoxyethane’ since ‘chloro’ is cited first as a prefix in the name.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o2U19.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2U19.png&quot; alt=&quot;1-chloro-2-methoxyethane&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user7951" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-09-05T15:33:39.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36926" PostTypeId="2" ParentId="36923" CreationDate="2015-09-05T15:43:36.393" Score="1" Body="&lt;p&gt;I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) 0.2 kg Water was icreased by 24 °C:&#xA;U = 0.2 kg * 4187 J/(kgK) * 24 K = 20,100 J = 20.100 kJ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) 1.15 g caused the temperature increase of 24 °C:&#xA;20.100 kJ / 0.013 mol = 1,510 kJ/mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane. &lt;/p&gt;&#xA;" OwnerUserId="19160" LastActivityDate="2015-09-05T15:43:36.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36930" PostTypeId="2" ParentId="36911" CreationDate="2015-09-05T16:49:36.893" Score="1" Body="&lt;p&gt;An elements is a substance which is made out of atoms that all have the same proton number. So an element can consist of multiple atoms (like oxygen which goes around in pairs). An atom consists of a nucleus and some electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: IGCSE Chemistry&lt;/p&gt;&#xA;" OwnerUserId="20663" LastActivityDate="2015-09-05T16:49:36.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36933" PostTypeId="2" ParentId="36885" CreationDate="2015-09-05T18:07:52.123" Score="8" Body="&lt;p&gt;It is all about the definition of temperature itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All atoms of any compound with a temperature T have different energies associated with them, therefore temperature does not make sense on molecular levels. So when you look at the energy distribution of all atoms that form a compound, you will find its temperature on the spot of the energy E axis, that has the highest probability associated with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probability for a specific energy value (or state) can be calculated with: &#xA;$$P \propto e^{\frac{-\Delta E}{KT}}$$ &#xA;$K$ being the Boltzmann constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In positive temperatures, you will always find more atoms in a lower energy state. So your probability P for finding atoms in higher energy states will decrease because $\Delta E$ is becoming bigger and bigger as you go up. But when you can get some atoms to stay in a higher energy state, which they would usualy not populate, then your probability has increased for that higher energy state. And the only way to fit this into the equation is to make T negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you now bring a second compound B of any positive temperature to the original compound A, which has negative temperature, then energy will always flow from compound A to compound B. Which means A is hoter, regardless of the temperature of B. This is because this unusual energy state is very unstable and can only be maintained by an &quot;atom trap&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;a href=&quot;http://m.youtube.com/watch?v=yTeBUpR17Rw&quot;&gt;http://m.youtube.com/watch?v=yTeBUpR17Rw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19160" LastActivityDate="2015-09-05T18:07:52.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36934" PostTypeId="1" CreationDate="2015-09-05T18:09:09.250" Score="2" ViewCount="132" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Question:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{50~ml}$ of the conjugate acid $\ce{H3N+CH(R)COOH}$ of an $\alpha$-amino acid required $\mathrm{30~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ to completely convert into the dipolar ion $\ce{H3N+CH(R)COO-}$. The pH of the solution was 6.4 during the titration with $\mathrm{15~ml}$ of $\mathrm{0.1~M}~\ce{NaOH}$ and 7.8 at the isoelectric point. Calculate the dissociation constants for the amino acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Attempt:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions involved: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3N+CH(R)COOH &amp;lt;=&amp;gt; H3N+CH(R)COO- + H+}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{H3N+CH(R)COOH + OH- -&amp;gt; H3N+CH(R)COO- + H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the isoelectric point, $\mathrm{pH}=\frac{\mathrm pK_{\mathrm a1}+\mathrm pK_{\mathrm a2}}{2}$.&#xA;Initially, number of milli moles of conjugate acid = number of milli moles of NaOH = $\mathrm{30\times 0.1=3~mmol}$ of salt during titration (neglecting the amount produced by acid) = 1.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A buffer forms during titration. My doubts is, does the concentration of acid which was present initially, change while adding NaOH? Using Henderson equation will give pH of Buffer.&#xA;I know the concentration of salt and pH. But I am stuck with the concentration of acid. 1.5 milli moles of acid will react to form salt. If I use 1.5,I get wrong answer. If I use 3, I get right answer.&lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorDisplayName="user7951" LastEditDate="2015-09-05T19:58:20.280" LastActivityDate="2015-09-05T19:58:20.280" Title="Titration of alpha-amino acid with strong base" Tags="&lt;physical-chemistry&gt;&lt;acid-base&gt;&lt;equilibrium&gt;&lt;titration&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36935" PostTypeId="2" ParentId="36911" CreationDate="2015-09-05T20:21:00.070" Score="6" Body="&lt;p&gt;The word &lt;em&gt;element&lt;/em&gt; has &lt;a href=&quot;http://goldbook.iupac.org/C01022.html&quot; rel=&quot;noreferrer&quot;&gt;two slightly different meanings in chemistry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;an element as an abstraction, an “atomic species” (ex.: $\ce{{}_8O}$ oxygen),   and&lt;/li&gt;&#xA;&lt;li&gt;an elemental substance (ex.: $\ce{O2}$ dioxygen gas), that represents an element in practice.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Textbooks on chemistry, possibly, downplay the distinction since they are not textbooks on philosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different elements are distinguished by their atomic number (ex.: &lt;strong&gt;8&lt;/strong&gt; for oxygen). Atomic number is a characteristic of an atom, whether is it free or bound, neutral or ionized. Atomic number is the number or protons in the nucleus (that is chemically immutable) and, the same, electric charge of the nucleus (in appropriate unit). It is an invariant preserved in all chemical processes. Elements are a classification of atoms. For each natural &lt;em&gt;Z&lt;/em&gt;, all atoms with atomic number &lt;em&gt;Z&lt;/em&gt; form a chemical element: A generalized atomic species (although different atoms of the same element are not always identical). Any atom &lt;em&gt;represents&lt;/em&gt; (pertains to) some element, and to only one element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elemental substance consists of atoms of the same element, and doesn't contain atoms of other elements. It may be formed of free atoms (as for noble gases), molecules (as for $\ce{O2}$), or form a crystal. When a chemist tries to isolate an element, i. e. to gather its atoms in one place, an elemental substance results. Hence chemists use one word “element” with two slightly different denotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the difference between abstract elements (atomic species) and elemental substances. When we ask: “of which elements does water consist?”, we assume species (hydrogen and oxygen, of whose atoms does the water consist), and don’t assume the water contains $\ce{H2}$ or $\ce{O2}$ (although it might, a slight quantity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how are atoms different from all this? An atom is a physical body. We can count them as things: There is one atom, two atoms, three atoms, etc.. How many atoms are in a water $\ce{H2O}$ molecule? There are three: Two atoms of hydrogen and one atom of oxygen. How many &lt;em&gt;elements&lt;/em&gt; are in water (molecule)? Two elements: Hydrogen and oxygen.&lt;/p&gt;&#xA;" OwnerUserId="9364" LastEditorUserId="4945" LastEditDate="2015-09-08T05:25:52.687" LastActivityDate="2015-09-08T05:25:52.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36936" PostTypeId="2" ParentId="36895" CreationDate="2015-09-05T21:04:25.410" Score="2" Body="&lt;p&gt;If you use fresh cooking oil containing unsaturated fats (like olive or canola oil) to season your pan, it will thermally polymerize to form a stable layer that gives the cooking surface its nonstick properties. Note that a saturated fat like coconut oil, lard, or ghee will not work very well or at all for this, since it does not have any double bonds to undergo polymerization. The radical formation you are concerned about results over time from air oxidation of unsaturated fats to form peroxides--these in turn break down to give aldehydes and acids that cause the bad flavor of rancid fat, and have adverse health effects. (Not a problem for coconut oil.) Heating fresh oil in a pan to create a nonstick coating may result in some air oxidation in addition to polymer formation, but I think it is not a big cause for concern. High polymers are generally nontoxic because the body does not absorb them.&lt;/p&gt;&#xA;" OwnerUserId="5577" LastActivityDate="2015-09-05T21:04:25.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36938" PostTypeId="1" CreationDate="2015-09-05T23:03:12.050" Score="1" ViewCount="285" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSbA4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSbA4.jpg&quot; alt=&quot;Over simplified question model: Behavior exhibited top or bottom?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; The above image is an over simplified model to my question begging to know whether the permeation behavior exhibited by water droplets through graphene oxide is the top image or the bottom.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I couldn't find information about this anywhere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been confirmed that water vapor and pools of liquid water can pass through graphene oxide films (prepared for stability in water on aluminium substrates) assuming the interlayer distances and quantity of layers are appropriate for optimal passage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;strong&gt;I do not know whether or not a droplet of water (placed on one side of graphene oxide) will permeate through the graphene oxide film and emerge intact on the other side of the film.&lt;/strong&gt; I could simply confirm this by spraying water onto graphene oxide but I cannot make a legitimate film on my own just yet is why I ask. &#xA;Basically, stated differently, I don't want to end up spraying graphene oxide with water droplets on one side and end up with a stream of water on the other side; they are to ideally remain as droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is very simple but also very important in determining to what extent G.O. can be utilized. There are also different studies that conclude different permeation mechanisms for water through G.O. so I can't even reason the answer out on my own. Quite maddening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all and sorry for the pure info. extraction but it's for the greater good!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1: Oh, the preferred droplet max diameter would be 100 micrometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2: &lt;em&gt;Thanks for the reads; I'll try and find the best method for making graphene alumina oxide to try and find the answer on my own. If anyone can point me to the best method meanwhile, that would also be great. Thanks again&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="20665" LastEditorUserId="20665" LastEditDate="2015-09-25T04:29:29.400" LastActivityDate="2015-09-25T04:29:29.400" Title="Water Droplet permeability through Graphene Oxide" Tags="&lt;water&gt;&lt;carbon-allotropes&gt;" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="36939" PostTypeId="2" ParentId="36911" CreationDate="2015-09-05T23:14:20.113" Score="9" Body="&lt;p&gt;Given that you seem to be a beginner in Chemistry, I thought you might benefit from a description from somebody who only studied up to high school chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll use water as an example because you've probably heard it referred to as H&lt;sub&gt;2&lt;/sub&gt;O.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A single &quot;piece&quot; of water, H&lt;sub&gt;2&lt;/sub&gt;O is a &lt;strong&gt;molecule&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contains 2 &lt;strong&gt;elements&lt;/strong&gt;: Hydrogen, and Oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it contains 3 &lt;strong&gt;atoms&lt;/strong&gt;, 2 Hydrogen, and 1 Oxygen.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Atoms contain even smaller pieces called subatomic particles. There are three kinds: &lt;strong&gt;Protons, Neutrons, and Electrons&lt;/strong&gt;. the number of protons defines what the element is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 proton - Hydrogen&lt;br&gt;&#xA;2 protons - Helium&lt;br&gt;&#xA;12 protons - Carbon, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total number of protons + neutrons is the atom's &lt;strong&gt;atomic weight&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements can contain different numbers of neutrons. These are called &lt;strong&gt;isotopes&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen-1 has 1 proton&lt;br&gt;&#xA;Hydrogen-2 has 1 proton 1 neutron&lt;br&gt;&#xA;Hydrogen-3 has 1 proton 2 neutrons  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These different forms of hydrogen are isotopes of one another.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f1U6X.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f1U6X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Obviously it gets more complicated that this, but for you at the moment, it doesn't need to :)&lt;/p&gt;&#xA;" OwnerUserId="20666" LastEditorUserId="20666" LastEditDate="2015-09-06T07:20:29.100" LastActivityDate="2015-09-06T07:20:29.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36941" PostTypeId="2" ParentId="36897" CreationDate="2015-09-06T01:03:14.270" Score="6" Body="&lt;p&gt;Electricity &lt;em&gt;is&lt;/em&gt; involved: the copper (well, copper plated zinc in the US) penny is in direct contact with the zinc, forming a short-circuited connection. Forcing the plating action &quot;backwards&quot; (&lt;a href=&quot;http://www.4college.co.uk/a/ss/cell.php&quot; rel=&quot;noreferrer&quot;&gt;copper being more electronegative than zinc&lt;/a&gt;) requires current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try the same experiment but with a separator, such as a piece of filter paper, between zinc and copper; there should be no plating.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-06T01:03:14.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36942" PostTypeId="2" ParentId="35853" CreationDate="2015-09-06T01:25:48.877" Score="3" Body="&lt;p&gt;Some possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Porous carbon electrodes &lt;em&gt;do&lt;/em&gt; act as a capacitor, the basis for &lt;a href=&quot;https://en.wikipedia.org/wiki/Supercapacitor&quot; rel=&quot;nofollow&quot;&gt;supercapacitors (aka ultracapacitors)&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You are electrolytically depositing copper on the negative electrode when you run current through the solution, even though it might not be enough to see. This creates an electrochemical cell.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Simply &lt;em&gt;disconnecting&lt;/em&gt; the power supply leads will leave a high-impedance voltmeter floating, so it might be detecting random static charges. You might need to bypass the voltmeter with a high resistance (10 or 20 megohms) to reduce static sensitivity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-06T01:25:48.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36943" PostTypeId="2" ParentId="36914" CreationDate="2015-09-06T01:26:27.870" Score="4" Body="&lt;p&gt;Axial and planar are just adjectives we can use to describe what type of chirality a molecule possesses.  Usually, just saying the molecule is chiral is good enough.  Being planar or axially chiral does not create any new enantiomers or diastereomers compared to if we just said the molecule is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's analyze your problem using standard methods - no special techniques just because the molecule happens to be axially chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the same 4 diastereomers that you've drawn above (good job!).  When analyzing allene stereochemistry I've always found it convenient to analyze them using Newman projections.  I've redrawn the 4 diastereomers below using Newman projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ovail.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ovail.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the top diastereomer, all 4 chiral substituents are the same; the molecule has a non-superimposable mirror image (enantiomer).  Therefore this diastereomer is chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diastereomer, 3 chiral substituents are the same, one is different; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the third diastereomer, two chiral substituents with different chirality are attached to the same carbon; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the fourth diastereomer, two chiral substituents with the same chirality are attached to the same carbon.  This diastereomer belongs to the interesting point group &lt;strong&gt;$\ce{S_4}$&lt;/strong&gt;.  I say interesting because this point group does not contain a $\ce{\sigma}$ plane, but is still nondissymmetric (achiral).  Look at my Newman projection for this diastereomer and picture it lying behind the plane of the screen.  Reflect it in the plane of the screen and then rotate it 90° (n=4, 360/4 = 90°) and you have recreated the original molecule. This molecule is achiral (all molecules belong to point group $\mathrm{S}_n$ [n is an even integer] are achiral).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what do we wind up with?  4 diastereomers, 3 of which are chiral (enantiomeric pairs) and one is achiral.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="5026" LastEditDate="2018-07-17T03:04:33.877" LastActivityDate="2018-07-17T03:04:33.877" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="36944" PostTypeId="2" ParentId="35846" CreationDate="2015-09-06T01:36:04.617" Score="3" Body="&lt;p&gt;An electrostatic field will attract both conducting and nonconducting materials &lt;em&gt;even though those materials have no initial charge&lt;/em&gt; through &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrostatic_induction&quot; rel=&quot;nofollow&quot;&gt;electrostatic induction&lt;/a&gt;. That is why you can pick up lint or aluminum foil with a charged comb or balloon. See &lt;a href=&quot;https://www.youtube.com/watch?v=9jEDKbEaerQ&quot; rel=&quot;nofollow&quot;&gt;this demo&lt;/a&gt; with a neutral aluminum can.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-06T01:36:04.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36945" PostTypeId="2" ParentId="35828" CreationDate="2015-09-06T01:42:14.793" Score="1" Body="&lt;p&gt;How about $\ce{NH4OH}$? This does not for carbonates as readily, but it &lt;em&gt;does&lt;/em&gt; form complexes with many metals. It depends on your intended use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be easier to keep things away from the air:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a closed container.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Float an inert liquid (e.g. paraffin oil) atop your solution, or just a layer of pure water, which might take a while to mix.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use an inert gas over the container to displace air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-06T01:42:14.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36946" PostTypeId="1" CreationDate="2015-09-06T01:51:58.017" Score="0" ViewCount="294" Body="&lt;p&gt;I am trying to determine two of the theoretical values for the enthalpy of the combustion of methanol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated (using Hess' law) that the $ \Delta H_{\mathrm {reaction}}\; = -1453.418~\mathrm{kJ~mol^{-1}} $, which is correct according to several sources. However, when I tried to calculate the $ \Delta H_{\mathrm{reaction}} $ using the bond dissociation and formation method: $ \Delta H_{\mathrm{reaction}}\; =\; \sum{\Delta H_{\mathrm{bonds\; broken}}}\;-\; \sum{\Delta H_{\mathrm{bonds\; formed}}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got a value of $\mathrm{-464.9~kJ~mol^{-1}}$. I am just wondering if this is correct, or if I have made a mistake somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ps: My balanced equation looks like this: $$\ce{2CH3OH + 3O2 -&amp;gt; 4H2O + 2CO2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; table to get the known values for the various bond energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="20667" LastEditorUserId="10185" LastEditDate="2015-09-06T13:49:25.963" LastActivityDate="2015-09-06T13:49:25.963" Title="Bond Dissociation vs Hess' Law Enthalpy" Tags="&lt;organic-chemistry&gt;&lt;enthalpy&gt;&lt;combustion&gt;" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36947" PostTypeId="2" ParentId="36899" CreationDate="2015-09-06T01:56:58.350" Score="2" Body="&lt;p&gt;Just to offer a counter to arguments based on energetics...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, there is a simple but powerful argument against it; we&#xA;  haven't found anything like it yet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But the same could be said for replicating, living organisms outside of the one known case here on Earth.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the necessary components were present and the kinetic barrier to the formation of the key ingredients was relatively moderate, the probability of it happening is (apparently) extremely remote.  Catalysts, such as the chiral face of a crystal and special magnetic and electrical events may have been simultaneously required to bring about the first chiral organism (or its inanimate, chiral precursor). The odds of all of this being in the same place and happening at the same time may be incredibly small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the same could be true for ice-9 formation.  Maybe a crystal face with a very specific morphology is required to catalyze the transformation of water into ice-9.  Even if the mineral is known on Earth, perhaps the specific crystal face\morphology does not presently exist on earth.  However, it's conceivable that it could be formed on a planet with extreme temperatures and pressures not found in Earth geology.  Maybe this planet, in some remote corner of the galaxy, has since been destroyed through a collision and a resulting planetary fragment containing crystals with the alien morphology is now headed towards Earth.&lt;/p&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-09-06T16:45:29.967" LastActivityDate="2015-09-06T16:45:29.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="36948" PostTypeId="1" CreationDate="2015-09-06T02:41:08.173" Score="0" ViewCount="163" Body="&lt;p&gt;I'm trying to find any readily available chemical/substance (i.e. household, hardware store etc.) or combination of such that breaks down paper/cellulose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a home electronics hobbyist and am using my laser printer to print circuit board layouts.  I place the paper on the blank copper board and run it through a laminate machine (high temp &amp;amp; pressure) to 'melt' the laser toner to the copper.  I then soak the board and paper in water to try and slush the paper free of the toner, after which I can put the board in acid and remove the exposed copper and leave my circuit pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This actually works pretty well except that releasing the paper from the toner is maybe only 95% efficient/reliable.  Sometimes the toner gets pulled off the copper and/or fibers/slag from the paper remain on the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From doing a little research on the internet, it would seem that dissolving/breaking down cellulose is not an easy process and is quite involved for commercial applications.  However, for my purposes I don't think I need some 'perfect' solution.  If I can just get the paper to soften, or partially break down or just change its properties a bit so that it will let go of the toner more efficiently, then that would probably be sufficient for my needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="20668" LastEditorUserId="865" LastEditDate="2015-09-06T02:57:15.670" LastActivityDate="2015-09-06T02:57:15.670" Title="Dissolve/alter cellulose properties for printed circuit boards" Tags="&lt;everyday-chemistry&gt;&lt;solubility&gt;" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="36950" PostTypeId="1" AcceptedAnswerId="92291" CreationDate="2015-09-06T05:40:50.160" Score="12" ViewCount="7514" Body="&lt;p&gt;Why is a bridgehead atom bearing a negative or a positive charge unstable? Is it something with the structure of bridge bonds? I am unable to understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMhRh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMhRh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17988" LastEditorUserId="5026" LastEditDate="2018-03-14T05:43:30.130" LastActivityDate="2018-03-14T07:17:35.033" Title="Why is a bridgehead atom bearing a negative or a positive charge unstable?" Tags="&lt;organic-chemistry&gt;&lt;stability&gt;&lt;carbocation&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36951" PostTypeId="1" AcceptedAnswerId="36952" CreationDate="2015-09-06T05:44:14.350" Score="2" ViewCount="1657" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;With the only possibilities being &lt;span class=&quot;math-container&quot;&gt;$\ce{Ba^2+, Ca^2+, Cu^2+, Fe^2+, Fe^3+, Pb^2+}$&lt;/span&gt;, what cations are present in a solution which:&#xA;&lt;br&gt;a) Forms a white precipitate with sulfate and with fluoride, no precipitate with chloride, and a blue precipitate with hydroxide which dissolves in ammonia solution&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a0yWb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a0yWb.png&quot; alt=&quot;Salt analysis chart for determining cations present in a solution&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have the answer with me, but I am not too sure how to use the above chart for determining cations in a solution, to answer the question. I can sort of make it out, but I am not completely sure.&lt;/p&gt;&#xA;" OwnerUserId="15053" LastEditorUserId="83758" LastEditDate="2020-09-07T10:26:39.633" LastActivityDate="2020-09-07T10:27:06.280" Title="How to use this salt analysis chart to determine the cations present in a given solution?" Tags="&lt;inorganic-chemistry&gt;&lt;experimental-chemistry&gt;&lt;salt&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="36952" PostTypeId="2" ParentId="36951" CreationDate="2015-09-06T06:05:33.013" Score="3" Body="&lt;p&gt;For this question rather than reading the question and trying to understand what is going on all in one go, I suggest just work your way through the flow chart and refer to the question as you go along so that you don't overload your brain with excess information and confuse yourself.&lt;/p&gt;&#xA;&lt;p&gt;So starting from the top, it asks if a precipitate will form if &lt;span class=&quot;math-container&quot;&gt;$\ce{HCl}$&lt;/span&gt; is added. Looking from the information in the question, there is no precipitate formed. Therefore we can rule out &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb^2+}$&lt;/span&gt; being in the solution. Now following the steps in the flow chart, it asks if a precipitate forms if sulfuric acid is added to it. Looking from the data in the question, it says that a white precipitate is formed with sulfate. Therefore, we follow the next step which asks if a precipitate is formed when &lt;span class=&quot;math-container&quot;&gt;$\ce{NaF}$&lt;/span&gt; is added. In the question it states that a white precipitate forms with fluoride. Therefore we can conclude that &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; ions are in the solution.&lt;/p&gt;&#xA;&lt;p&gt;However, if we look again into the question, it states that a blue precipitate also forms when mixed with hydroxide ions. Therefore, if we look back to the flow chart regarding the section on testing with &lt;span class=&quot;math-container&quot;&gt;$\ce{NaOH}$&lt;/span&gt;, it can be seen that traces of &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt; ions also exist in the solution. Therefore, both &lt;span class=&quot;math-container&quot;&gt;$\ce{Ca^2+}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{Cu^2+}$&lt;/span&gt; cations are in the solution.&lt;/p&gt;&#xA;&lt;p&gt;As you can see, once you know the general method, it is really quite simple and self-explanatory.&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="83758" LastEditDate="2020-09-07T10:27:06.280" LastActivityDate="2020-09-07T10:27:06.280" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="36954" PostTypeId="1" AcceptedAnswerId="36964" CreationDate="2015-09-06T09:01:28.997" Score="5" ViewCount="263" Body="&lt;p&gt;What are the diastereomers and enantiomeric pairs of the compound below where A are chiral groups (central chirality)? There should be &lt;a href=&quot;http://www.perekalin.narod.ru/Answers1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 diastereomers and 3 enantiomeric pairs&lt;/a&gt; (last page).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pONhm.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pONhm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://chemistry.stackexchange.com/questions/36914/&quot;&gt;I previously asked a question about tetra-substituted allenes.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="4945" LastEditDate="2020-01-29T17:30:36.053" LastActivityDate="2020-01-29T17:30:36.053" Title="How to determine the number of stereoisomers of a di-substituted allene, exhibiting both axial and central chirality?" Tags="&lt;organic-chemistry&gt;&lt;stereochemistry&gt;&lt;chirality&gt;" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="36955" PostTypeId="2" ParentId="36950" CreationDate="2015-09-06T09:14:41.313" Score="7" Body="&lt;p&gt;The hybridisation of a bridgehead carbon when its not charged is sp&lt;sup&gt;3&lt;/sup&gt;. Now, if you remove a hydrogen atom, its hybridisation becomes sp&lt;sup&gt;2&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, in sp&lt;sup&gt;2&lt;/sup&gt; hybridisation, the 3 hybridised orbitals lie in a single plane for minimised repulsions. But the bridgehead carbon with positive charge would not have hybridised orbitals in a single plane. This increases repulsion and makes the compound unstable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link - &lt;a href=&quot;http://www.masterorganicchemistry.com/2014/09/02/bredts-rule-and-summary-of-cycloalkanes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bredt's Rule&lt;/a&gt; - that explains the same thing, but instead of a positively charged bridgehead, they discuss a double bond on the bridgehead atom. &lt;/p&gt;&#xA;" OwnerUserId="17624" LastEditorUserId="5026" LastEditDate="2018-03-14T07:17:35.033" LastActivityDate="2018-03-14T07:17:35.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36960" PostTypeId="2" ParentId="19494" CreationDate="2015-09-06T12:47:44.317" Score="3" Body="&lt;p&gt;We have learnt that metallic oxides are basic nature. This is because the oxides of metals like $\ce{MgO}$ or $\ce{CaO}$ or $\ce{Na2O}$ form hydroxides in their aqueous solutions.&#xA;For example:&#xA;$$\ce{K2O + H2O -&amp;gt; 2KOH}$$&#xA;$$\ce{MgO + H2O -&amp;gt; Mg(OH)2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when non-metallic oxides dissociate into their constituent ions in water, they give $\ce{H+}$ ions in their aqueous solutions forming acids&#xA;For example:&#xA;$$\ce{CO2 + H2O -&amp;gt; H2CO3}$$&#xA;$$\ce{SO3 + H2O -&amp;gt; H2SO4}$$&lt;/p&gt;&#xA;" OwnerUserId="19614" LastEditorUserId="10185" LastEditDate="2015-09-06T17:24:58.437" LastActivityDate="2015-09-06T17:24:58.437" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36961" PostTypeId="1" CreationDate="2015-09-06T14:58:30.657" Score="1" ViewCount="54" Body="&lt;p&gt;I am trying to find resources where I can learn about general chemistry topics outside the scope of a textbook, preferably in primary literature. Note, I have read through a general chemistry text and finished studying topics in the scope of a general chemistry course, so the informational foundation is there. My ideal &quot;resource&quot; is a place where I can find something akin to a 1-5 page experimental setup or design, or maybe a discussion, in which these concepts are applied and I'm forced to think about relationships in general chemistry, or what happens when things are changed. For example, I'd love to see some paper where I can see the ideal gas law applied as an approximation in the lungs or in deep sea divers, or some maybe some experiment where ion solubility is discussed in the scope of &quot;hard water&quot; in pipes. I know these applications exist, but I want to read about them on the level that makes me think about general chemistry concepts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use VPN connections to login as a institution from my school and access some primary literature. However, sometimes I can't get full access becaue my institution is small and does not a subscription to a journal. Moreover, sometimes this primary literature is too complex in explanation, too longwinded, and goes beyond the scope of a general chemistry student. Now, I still want to think critically, but most of these papers include applications in engineering and use quantum physical assumptions beyond the scope of wave function solutions and electron configuration in general chemistry courses. My background includes a year of cell and molecular biology, an understanding of physics I and II, and math a little past integral calculus. I'd like the mathematics to be minimized, however, as I want concept. &lt;/p&gt;&#xA;" OwnerUserId="19315" LastActivityDate="2015-09-06T14:58:30.657" Title="Resources where I can find approachable applied literature to understand general chemistry topics?" Tags="&lt;everyday-chemistry&gt;&lt;applied-chemistry&gt;" AnswerCount="0" CommentCount="3" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36963" PostTypeId="1" AcceptedAnswerId="37398" CreationDate="2015-09-06T17:08:16.240" Score="1" ViewCount="341" Body="&lt;p&gt;I need help with a basic concentration problem.&lt;/p&gt;&#xA;&lt;p&gt;If I have a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{2~M}$&lt;/span&gt; solution, and want to make a &lt;span class=&quot;math-container&quot;&gt;$\mathrm{0.1~M}$&lt;/span&gt; solution of &lt;span class=&quot;math-container&quot;&gt;$\mathrm{100~ml}$&lt;/span&gt;, how would I go about that?&lt;/p&gt;&#xA;&lt;p&gt;My work:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_1V_1 = M_2V_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$M_1 = \text{initial concentration}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$M_2 = \text{final concentration}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_1 = \text{volume needed from initial}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_2 = \text{volume needed from final}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2~M \times V_1 = \pu{0.1 M \times 100 ml}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_1 = \frac{\pu{0.01 M\cdot L}}{\pu{2~M}} = \pu{5 ml}?$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore I need &lt;span class=&quot;math-container&quot;&gt;$\pu{5 ml}$&lt;/span&gt; of my &lt;span class=&quot;math-container&quot;&gt;$\pu{2 M}$&lt;/span&gt; solution, increased to &lt;span class=&quot;math-container&quot;&gt;$\pu{100 ml}$&lt;/span&gt; to make a &lt;span class=&quot;math-container&quot;&gt;$\pu{0.1 M}$&lt;/span&gt; solution of&lt;span class=&quot;math-container&quot;&gt;$\pu{100 ml}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="16224" LastEditorUserId="865" LastEditDate="2021-05-22T03:19:29.487" LastActivityDate="2021-05-22T06:07:02.527" Title="Calculating the volume of 2 M solution needed to make a 0.1 M solution" Tags="&lt;physical-chemistry&gt;&lt;concentration&gt;&lt;stoichiometry&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="36964" PostTypeId="2" ParentId="36954" CreationDate="2015-09-06T17:28:42.893" Score="2" Body="&lt;p&gt;These are the three diastereomers with their enantiomers on the right side. The top and middle diastereomers are connected via axial chirality (R and S are used to mark the configuration of A groups).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/womKs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/womKs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7159" LastEditorUserId="7159" LastEditDate="2015-09-06T21:18:51.773" LastActivityDate="2015-09-06T21:18:51.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36965" PostTypeId="1" AcceptedAnswerId="36973" CreationDate="2015-09-06T17:31:26.377" Score="3" ViewCount="3892" Body="&lt;p&gt;The reaction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NO(g) + NO2(g) + O2(g) -&amp;gt; N2O5(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the experimental data, I have found the rate law to be: $$\ce{Rate~=~k[NO][NO2]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the overall reaction does not depend on $\ce{O2(g)}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone suggest how I go about building the elementary steps to form the reaction mechanism?&lt;/p&gt;&#xA;" OwnerUserId="10257" LastEditorUserId="95133" LastEditDate="2021-05-25T12:45:30.940" LastActivityDate="2021-05-25T12:45:30.940" Title="Propose a reaction mechanism that explains the rate law for the formation of dinitrogen pentoxide" Tags="&lt;physical-chemistry&gt;&lt;reaction-mechanism&gt;&lt;kinetics&gt;" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="36966" PostTypeId="1" CreationDate="2015-09-06T17:57:16.153" Score="3" ViewCount="5994" Body="&lt;p&gt;Charles's Law states that for every 1 degree Celsius increase in temperature of a gas, the volume increases by 1/273 of the original volume (at constant pressure). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am wondering is, doesn't gas have an indefinite volume, therefore it is not possible for it to have volume at all? Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="865" LastEditDate="2015-09-08T12:34:07.373" LastActivityDate="2021-04-28T17:51:46.073" Title="Charles's Law confusion" Tags="&lt;gas-laws&gt;" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36969" PostTypeId="2" ParentId="36966" CreationDate="2015-09-06T18:36:18.493" Score="-2" Body="&lt;p&gt;Simply Charle's Law(change in volume is directly proportional to change in temperature) is true with the assumption of constant pressure. &lt;/p&gt;&#xA;" OwnerUserId="9649" LastActivityDate="2015-09-06T18:36:18.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36970" PostTypeId="1" CreationDate="2015-09-06T18:45:31.473" Score="9" ViewCount="12722" Body="&lt;p&gt;I'm in an advanced level chemistry course, and so far we've been introduced to balancing redox reactions. The basic ones like &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\ce{MnO4- + Cl- -&amp;gt; Mn^2+ + Cl2}\tag1$$&lt;/span&gt; &#xA;are pretty easy for me, but I'm having trouble with equations where it's not apparent what I exactly need to put in my half reactions. Some examples of these kind of equations are:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;\ce{Pb + PbO2 + SO4^2- &amp;amp;-&amp;gt; PbSO4}\tag2\\&#xA;\ce{Ag(s) + CN- (aq) + O2(g) &amp;amp;-&amp;gt; [Ag(CN)2]- (aq)}\tag3\\&#xA;\ce{Cu + HNO3(aq) &amp;amp;-&amp;gt; Cu(NO3)2(aq) + NO2(g) + H2O(l)}\tag4&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for solutions to any of these problems. I'm just looking for ways to correctly construct my half reactions before balancing, adding water and hydrogen ions, and balancing the charges.&lt;br&gt;&#xA;Like in the first reaction on the reactants side, the first &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb}$&lt;/span&gt; has a charge of &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; while the next one has a charge of &lt;span class=&quot;math-container&quot;&gt;$+4$&lt;/span&gt;. The &lt;span class=&quot;math-container&quot;&gt;$\ce{Pb}$&lt;/span&gt; on the products side however has a charge of &lt;span class=&quot;math-container&quot;&gt;$+2$&lt;/span&gt;. It seems like one is being oxidized and the other reduced. And I'm also not sure what to do with the sulfate.  &lt;/p&gt;&#xA;" OwnerUserId="20678" LastEditorUserId="4945" LastEditDate="2018-09-25T11:48:47.540" LastActivityDate="2020-01-04T06:47:38.163" Title="How to balance more complex redox reactions?" Tags="&lt;redox&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="36971" PostTypeId="2" ParentId="36891" CreationDate="2015-09-06T19:26:35.153" Score="2" Body="&lt;p&gt;To illustrate the &quot;diverse ion effect&quot;, we will study how the solubility of $\ce{AgCl}$ changes in absence and presence of $\ce{NaNO3}$, given that the molar solubility product of $\ce{AgCl}$ is $1.76 \times 10^{-10}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact definition of the solubility constant is: $$K_{sp}= a(\ce{Ag+}).a(\ce{Cl-})=[\ce{Ag+}].\gamma(\ce{Ag+}).[\ce{Cl-}].\gamma(\ce{Cl-})$$&#xA;Where $a(\ce{Ag+})$ and  $a(\ce{Cl-})$ are the activities of these ions, and $\gamma(\ce{Ag+})$ and $\gamma(\ce{Cl-})$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In dilute solutions, i.e. absence of $\ce{NaNO3}$:  $$\gamma(\ce{Ag+})=\gamma(\ce{Cl-})=1$$&#xA;So, the solubility of $\ce{AgCl}$ is equal to $$[\ce{Ag+}]= \sqrt{K_{sp}}= 1.33 \times 10^{-5}$$&lt;/li&gt;&#xA;&lt;li&gt;In presence of $\ce{NaNO3}$ of a concentration of $0.1 \mathrm{mol/kg}$, then&#xA;$$[\ce{Ag+}]= \sqrt{K_{sp}}/ \sqrt{(\gamma(\ce{Na+}).\gamma(\ce{NO3-})}&#xA;= 1.33 \times 10^{-5}/ (0.769)= 1.73 \times 10^{-5} $$&#xA;i.e. in the presence of a soluble salt, the molar solubility of a slightly soluble salt increases.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7581" LastEditorUserId="7581" LastEditDate="2015-09-06T19:33:51.060" LastActivityDate="2015-09-06T19:33:51.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36972" PostTypeId="2" ParentId="36966" CreationDate="2015-09-06T19:45:53.737" Score="1" Body="&lt;p&gt;A gas occupies the volume of whatever the container it is in. Take a balloon filled with Helium for example. Now the volume of Helium gas is the volume of the ballon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the temperature by 1 °C at constant pressure the volume of the balloon increase by 1/273th of the original volume of the balloon. Hope this might clear things up. &lt;/p&gt;&#xA;" OwnerUserId="20680" LastEditorDisplayName="user7951" LastEditDate="2015-10-08T14:10:30.803" LastActivityDate="2015-10-08T14:10:30.803" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="36973" PostTypeId="2" ParentId="36965" CreationDate="2015-09-06T19:52:35.147" Score="3" Body="&lt;p&gt;First step - the rate determining step or the slow step (activation energy is high).&#xA;$$\ce{NO(g) + NO2(g) -&amp;gt; N2O3(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd step - the fast step.&#xA;$$\ce{N2O3(g) + O2(g) -&amp;gt; N2O5(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rate determining step does not include $\ce{O2(g)}$, the rate is independent of the initial $\ce{O2(g)}$ concentration.&lt;/p&gt;&#xA;" OwnerUserId="20680" LastEditorUserId="10185" LastEditDate="2015-09-07T09:18:13.403" LastActivityDate="2015-09-07T09:18:13.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36975" PostTypeId="1" AcceptedAnswerId="36978" CreationDate="2015-09-06T20:05:41.647" Score="5" ViewCount="2707" Body="&lt;p&gt;I am attempting to isolate pure iodine from kelp using the method described &lt;a href=&quot;http://www.rsc.org/learn-chemistry/wiki/index.php?title=TeacherExpt:Extracting_iodine_from_seaweed&amp;amp;oldid=7142&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&#xA;but I do not have any way of possibly obtaining cyclohexane.  I was wondering if any easier to obtain solvents would work as well to isolate the iodine from the rest of the mixture.  I was hoping to use a solvent that is easy to obtain in a grocery or hardware store.&lt;/p&gt;&#xA;" OwnerUserId="16106" LastEditorUserId="16106" LastEditDate="2015-09-06T20:26:46.890" LastActivityDate="2015-09-06T23:46:23.360" Title="Replacement for cyclohexane" Tags="&lt;solvents&gt;&lt;extraction&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36976" PostTypeId="2" ParentId="36954" CreationDate="2015-09-06T20:32:15.443" Score="0" Body="&lt;p&gt;I'm not that good at this but here is something to work on. Each row is a pair of mirror images. Grey structures in 3rd row are identical to the respective 2nd row ones. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/cnuST.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cnuST.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7284" LastActivityDate="2015-09-06T20:32:15.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36977" PostTypeId="1" CreationDate="2015-09-06T20:49:15.310" Score="12" ViewCount="9483" Body="&lt;p&gt;A YouTube channel called &quot;kreosan&quot; (made by Russian-speaking people who I believe live in Ukraine) posted a &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=VxvJcPToXb0&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;&lt;/strong&gt; showing rockets made from bottles of Coca-Cola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I've figured out from the video.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First step: pour out some Coca-Cola to make room in the bottle to add gas.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second step: add butane (looks like a canister for refueling butane lighters)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third step: this is unclear.  He adds a gas from a large container that appears to say &quot;GAS-1&quot;.  The YouTube description says &quot;Coca Cola + propane = turbo ROCKET&quot; so I guess it's propane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unknown: Does the rocket require both butane and propane?  I think the second&#xA;   step was just to show that you don't get very much gas out of a&#xA;   little butane canister, and all you really need is the propane from&#xA;   the last step.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Final step: invert the bottle, and some reaction pushes Coca-Cola out the mouth of the bottle with enough force that the whole bottle flies quickly as a rocket.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is going on here?  How does inverting the bottle trigger the reaction?  Is this like the &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=hKoB0MHVBvM&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Mentos and Diet Coke&quot;&lt;/a&gt;&lt;/strong&gt; thing where the trick was to get the carbon dioxide dissolved in the drink to come out quickly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the surface of Mentos candy provides a bunch of nucleation sites for bubbles, causing a lot of carbon dioxide to come out of solution quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does the &quot;kreosan&quot; rocket fly on just carbon dioxide or is there some other reaction going on here?&lt;/p&gt;&#xA;" OwnerUserId="20682" LastEditorUserId="76480" LastEditDate="2019-04-17T22:37:03.970" LastActivityDate="2019-04-18T15:31:50.720" Title="How does the &quot;kreosan&quot; Coca-Cola rocket work?" Tags="&lt;home-experiment&gt;&lt;gas-laws&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="36978" PostTypeId="2" ParentId="36975" CreationDate="2015-09-06T23:46:23.360" Score="7" Body="&lt;p&gt;Cyclohexane is used for this extraction because it is nonpolar, as is iodine. A more polar solvent like acetone or ethanol will also extract a whole bunch of other stuff like essential oils, proteins, and other polar compounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want &lt;a href=&quot;https://en.wikipedia.org/wiki/White_spirit&quot;&gt;Mineral Spirits&lt;/a&gt; which you should be able to purchase at any quality hardware store &lt;a href=&quot;http://www.walmart.com/ip/Klean-Strip-Odorless-Mineral-Spirits-1qt/17208809&quot;&gt;or Walmart&lt;/a&gt;. Mineral spirits, mineral turpenine, or whatever you want to call it, is a mixture of saturated hydrocarbons distilled from petroleum. It is a general purpose paint thinner for oil-based paints. Since it is mainly C7-C12 alkanes, it will have a higher boiling point than cyclohexane $(140\ ^\circ\mathrm{C}$ vs $81\ ^\circ\mathrm{C})$., but it will have the same general polarity as cyclohexane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can find products labeled as petroluem ether, light naphtha, or ligroin, all are alkane mixtures with lower boiling points than mineral spirits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common low polarity paint thinner is toluene, which has a lower boiling point at $110\ ^\circ\mathrm{C}$.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-09-06T23:46:23.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36979" PostTypeId="2" ParentId="19687" CreationDate="2015-09-07T01:30:46.807" Score="5" Body="&lt;p&gt;The standard process for electrolysing &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2O3}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\ce{Al}$&lt;/span&gt; should work fine on the moon. Note that on Earth, a consumable carbon anode is used to reduce the voltage required, giving the reaction (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hall%E2%80%93H%C3%A9roult_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hall–Héroult process&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ce{Al2O3 + 1.5C → 2 Al + 1.5 CO2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative, non-consumable anode would be used on the moon, to avoid the need for anode material and to give oxygen instead of &lt;span class=&quot;math-container&quot;&gt;$\ce{CO2}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process for purifying &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2O3}$&lt;/span&gt; would be rather more difficult. On Earth, this is done by dissolving &lt;a href=&quot;https://en.wikipedia.org/wiki/Aluminium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;bauxite&lt;/a&gt; in sodium hydroxide solution, filtering out insoluble iron salts, cooling, precipitating out &lt;span class=&quot;math-container&quot;&gt;$\ce{Al2O3}$&lt;/span&gt;, and discarding the liquor containing silicates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process is highly intensive in water and sodium hydroxide. Major advances would be needed to make it less intensive in these materials, and to make the electrolysis step more tolerant of impurities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we are better off looking for water to use as feedstock for our propellant factory. It's far easier to purify (all you have to do is distill it).&lt;/p&gt;&#xA;" OwnerUserId="5541" LastEditorUserId="41328" LastEditDate="2019-01-19T23:08:33.437" LastActivityDate="2019-01-19T23:08:33.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="36980" PostTypeId="1" CreationDate="2015-09-07T01:50:11.467" Score="3" ViewCount="9636" Body="&lt;p&gt;Every one knows that liquid oxygen is flammable, but why this tank labeled like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jiruX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jiruX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so is it really liquid oxygen is non flammable? &lt;/p&gt;&#xA;" OwnerUserId="20685" LastEditorUserId="9961" LastEditDate="2015-09-07T08:03:21.043" LastActivityDate="2016-03-05T00:38:51.833" Title="Is liquid oxygen non flammable?" Tags="&lt;everyday-chemistry&gt;&lt;safety&gt;&lt;combustion&gt;" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="36981" PostTypeId="2" ParentId="36980" CreationDate="2015-09-07T02:38:04.830" Score="2" Body="&lt;p&gt;Technically, Oxygen is almost never flammable. It oxidizes other materials quite easily and many materials will burn in the presence of oxygen.  But oxygen will never actually burn itself except in the presence of fluorine.  &lt;/p&gt;&#xA;" OwnerUserId="16106" LastActivityDate="2015-09-07T02:38:04.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36982" PostTypeId="2" ParentId="36980" CreationDate="2015-09-07T05:24:11.923" Score="7" Body="&lt;p&gt;Oxygen is not flammable, as it does not burn, it supports burning. For burning to happen, a strong oxidizer (for example, oxygen) and a strong reducer (for example, carbon) must be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, the problem with oxygen is, that while not being flammable, it is much more efficient in supporting burning. In air oxygen is mixed with nitrogen, meaning that for the same amount of reducer a quadruple amount of oxidizer is required and a lot more heat is lost with inert gases (nitrogen). For example, iron wire can burn in oxygen, but not in air. It is also a lot easier to lit things up in oxygen. A classic experiment is to move a smoldering splinter into a vial with pure oxygen. It immediately starts to burn with a bright flame &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, liquid oxygen is an entirely new level of crazy. In addition to constantly releasing gaseous oxygen... If an organic material (for example, sawdust) is mixed with liquid oxygen, it becomes an explosive.&lt;/p&gt;&#xA;" OwnerUserId="485" LastActivityDate="2015-09-07T05:24:11.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36984" PostTypeId="2" ParentId="36970" CreationDate="2015-09-07T07:56:05.263" Score="8" Body="&lt;p&gt;Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Balance all the elements in the equation except for &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;. For polyatomic ions that are spectator ions, just think as them as one big element.&lt;/li&gt;&#xA;&lt;li&gt;Check if &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt; are balanced. If they aren't, add water molecules to the side, witch doesn't have enough &lt;span class=&quot;math-container&quot;&gt;$\ce{O}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Check if &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt; are balanced. If they aren't, add &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; ions to the side, witch doesn't have enough &lt;span class=&quot;math-container&quot;&gt;$\ce{H}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.&lt;/li&gt;&#xA;&lt;li&gt;Check your equation now to make sure that everything is balanced correctly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Balanced elements except for O and H:&lt;/strong&gt; Here the sulfate ion is a spectator ion, so you can imagine it as one big element: &lt;span class=&quot;math-container&quot;&gt;$$\ce{Pb + PbO2 + 2SO4^2- -&amp;gt; 2PbSO4 }$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Balance oxygen:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{Pb + PbO2 + 2SO4^2- -&amp;gt; 2PbSO4 + 2H2O }$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Balance hydrogen:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -&amp;gt; 2PbSO4 + 2H2O }$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Balance charge:&lt;/strong&gt; charge is already balanced. Don't need to add electrons.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Check equation to make sure everything is balanced correctly:&lt;/strong&gt; all the elements and charges are balanced correctly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So your balanced redox equation for the first example is: &lt;span class=&quot;math-container&quot;&gt;$$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -&amp;gt; 2PbSO4 + 2H2O }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; If the question asks to provide the redox equation in basic conditions there are two more steps.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Add the same number of &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions as &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; to the products and reactants. The &lt;span class=&quot;math-container&quot;&gt;$\ce{OH-}$&lt;/span&gt; ions and &lt;span class=&quot;math-container&quot;&gt;$\ce{H+}$&lt;/span&gt; will react to form water.&lt;/li&gt;&#xA;&lt;li&gt;Cancel out the water molecules from both sides (a bit like algebra).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That should give the redox equation in basic conditions. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="61578" LastEditDate="2020-01-04T06:47:38.163" LastActivityDate="2020-01-04T06:47:38.163" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="36985" PostTypeId="1" CreationDate="2015-09-07T07:59:19.260" Score="1" ViewCount="256" Body="&lt;p&gt;It's well-known in laboratory practice that one shouldn't melt lithium in a glass equipment and there isn't such rule about other alkali metals.&#xA;You may damage the glass with sodium but lithium will make a hole in your flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why does this rule only hold for lithium?&lt;/p&gt;&#xA;" OwnerUserId="19497" LastEditorUserId="23561" LastEditDate="2019-03-15T20:29:47.513" LastActivityDate="2019-03-15T20:29:47.513" Title="Why molten lithium burns through the glass but sodium not?" Tags="&lt;experimental-chemistry&gt;&lt;equipment&gt;&lt;glass&gt;" AnswerCount="0" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36986" PostTypeId="2" ParentId="36915" CreationDate="2015-09-07T10:05:34.357" Score="4" Body="&lt;p&gt;The answer to your question does not need a single calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is between those two volumes, at your desired volume of 150 mL?&lt;br&gt;&#xA;A simple buffer system that can be described using the Henderson–Hasselbalch equation:&#xA;$$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip3.5cm$&lt;img src=&quot;https://i.stack.imgur.com/0Bgxt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16622" LastEditorDisplayName="user7951" LastEditDate="2015-09-07T12:09:51.110" LastActivityDate="2015-09-07T12:09:51.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36987" PostTypeId="1" CreationDate="2015-09-07T10:47:19.543" Score="7" ViewCount="7331" Body="&lt;p&gt;We know that electrons are charges that revolve around the nucleus. Then, when in covalent bonds the electron is shared; does the electron obey the rule?&lt;/p&gt;&#xA;" OwnerUserId="20690" LastEditorUserId="18002" LastEditDate="2015-09-24T12:33:49.533" LastActivityDate="2018-02-03T09:19:49.767" Title="Does the shared electron in Covalent bonds revolve around nucleus?" Tags="&lt;bond&gt;&lt;electrons&gt;&lt;molecular-orbital-theory&gt;&lt;covalent-compounds&gt;&lt;valence-bond-theory&gt;" AnswerCount="3" CommentCount="2" FavoriteCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="36989" PostTypeId="2" ParentId="36987" CreationDate="2015-09-07T11:23:19.137" Score="5" Body="&lt;p&gt;Electrons are essentially quantum particles, and are controlled by the laws of quantum mechanics. There is absolutely no way to reconcile these laws with our physical intuition of macroscopic world. Electrons do not revolve; rather, they occupy certain atomic orbitals. To put it simply, they just stay there, delocalized over some space around the nucleus, everywhere at once. In a covalently bound molecule, electrons would occupy molecular orbitals and thus get delocalized all over the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The description of electrons as tiny charged solid particles that revolve around the nucleus, much like planets revolve around the sun, is the approximation known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Bohr_model&quot; rel=&quot;noreferrer&quot;&gt;Bohr atom&lt;/a&gt;. Significant for historical reasons as it is, one should not mistake it for coherent description of electron's behavior. Also, I doubt if you can extend it consistently to model covalent bonds.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="19383" LastEditDate="2015-09-07T11:32:52.597" LastActivityDate="2015-09-07T11:32:52.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36990" PostTypeId="2" ParentId="36987" CreationDate="2015-09-07T11:35:14.047" Score="10" Body="&lt;p&gt;I am guessing that you have an understanding of the atomic orbitals of an atom such as 1s and 3p atomic orbitals. Atomic orbitals are the probability distribution of where an electron is going to be $90$% of the time. Technically, electrons don't revolve around the nucleus but are rather quantum particles that are superimposed in several positions at the same time. However the description of electrons revolving randomly around the nucleus is suffice at this level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a covalent bond is formed, the electrons do still continue to revolve around the nucleus in orbitals, however a new kind of orbital is formed which are called molecular orbitals (rather than atomic orbitals). Here is a simple explanation of what molecular orbitals are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molecular orbitals are formed by constructively and destructively overlapping the atomic orbitals together. The simplest example of this is when $\ce{H2}$ is formed. Here, each H atom has a 1s atomic orbital. When they form, these atomic orbitals form 2 molecular orbitals. Where one molecular orbital is due to constructively overlapping the 2 atomic orbitals and the other molecular orbital is due to destructively overlapping the 2 atomic orbitals. This shown in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L6Unu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L6Unu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see that these 2 molecular orbitals that are formed are quite different to each other. The MO that is formed due to constructive overlap has greater electron density between the 2 positive nucleus. Meanwhile the MO that is formed due to destructive overlapping has decreases electron density between the 2 positive nucleus and in fact have a node between them. Therefore the first MO is lower in energy than the second MO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like AOs, the same rules apply to MOs. Each MO is able to hold 2 electrons and the electrons fill the lowest energy orbital first. So in $\ce{H2}$ the 2 electrons will fill the first MO and will revolve around the nucleus in that orbital (well technically only for $90$% of the time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This theory can be applied to any compound, however it gets more complicated as the number of AOs increases and the number of atoms increases. However essentially, when atoms form covalent bonds, the electrons continue to revolve around the nucleus in MOs rather than AOs. &lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="18002" LastEditDate="2015-10-16T05:46:03.223" LastActivityDate="2015-10-16T05:46:03.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36991" PostTypeId="2" ParentId="36987" CreationDate="2015-09-07T12:03:17.793" Score="4" Body="&lt;p&gt;The best answer is that the electron does not &quot;revolve&quot; around or &quot;orbit&quot; the nucleus the same way that the planets orbit the sun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictures like the following (from Wikipedia user &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Stylised_Lithium_Atom.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halfdan&lt;/a&gt;) are compelling, but refer to a model of the atom that was beginning to be supplanted by quantum mechanics a century ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LCMCt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCMCt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modern quantum-mechanical model of the atom incorporates things like the Heisenberg Uncertainty Principle and wave-particle duality. We now consider electrons in atoms to behave much more like waves. We can determine their energy very precisely, but as a consequence of the uncertainty principle, we cannot determine their position or momentum very precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic orbital pictures we are used to seeing for $s$, $p$, $d$, etc. orbitals are constructs describing regions of space where the electron is. Typically, we plot the 90% surface. In one interpretation that means 90% of the time the electron can be found in that region of space. In another interpretation, 90% of the electron-wave is confined to that space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When atoms come together to form bonds, the electron-waves overlap to produce new orbitals shared by the atoms. Nanoputian's answer provides a graphical representation of how this works.&lt;/p&gt;&#xA;" OwnerUserId="275" LastActivityDate="2015-09-07T12:03:17.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="36992" PostTypeId="1" AcceptedAnswerId="37036" CreationDate="2015-09-07T12:15:44.063" Score="6" ViewCount="1810" Body="&lt;p&gt;Many houses in the USA are &lt;a href=&quot;http://money.cnn.com/2013/02/12/real_estate/home-meth-lab/&quot; rel=&quot;nofollow&quot;&gt;illegally used as meth labs&lt;/a&gt; to produce methamphetamines, which can cause many negative health effects in their inhabitants, even after the house is sold to the next unwitting owner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.denverpost.com/ci_22827863/meth-contaminated-homes-its-buyer-beware&quot; rel=&quot;nofollow&quot;&gt;Denver Post estimates&lt;/a&gt; that 7% of single family and 11% of multi-family units that are foreclosed have traces of meth residue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these risks, &lt;a href=&quot;http://www.methinyourhouse.com/order-meth-residue-test-kit.html&quot; rel=&quot;nofollow&quot;&gt;many&lt;/a&gt; commercial &lt;a href=&quot;http://www.methlabcleanup.com/testing.htm&quot; rel=&quot;nofollow&quot;&gt;test kits&lt;/a&gt; for meth lab residue have appeared. However, many of them are intentionally vague about their chemical mechanisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemical mechanisms do meth lab test kits use to detect the presence of a meth lab?&lt;/p&gt;&#xA;" OwnerUserId="11303" LastEditorDisplayName="user7951" LastEditDate="2018-03-19T17:16:41.627" LastActivityDate="2018-03-19T17:16:41.627" Title="How do meth lab test kits detect the presence of a meth lab?" Tags="&lt;organic-chemistry&gt;&lt;everyday-chemistry&gt;&lt;analytical-chemistry&gt;" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36994" PostTypeId="1" AcceptedAnswerId="37115" CreationDate="2015-09-07T15:14:51.933" Score="2" ViewCount="1300" Body="&lt;p&gt;I've been balancing redox equations for the last couple of weeks and have become quite accustomed to the procedure however I've come across an equation that eludes me every which way when it comes to balancing oxygen. I come to the next to last step to balancing my redox equation but am unable to write my net ionic equation.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am asked to balance this reaction in an acidic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first write out my equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2H4(l) + NaBrO3(aq) -&amp;gt; N2(g) + Br- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I break my equation up into two half equations, one being my oxidant equation and the other my reductant. I balance them both:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{N2H4(l) -&amp;gt; N2(g) + 4H+ (aq) + 4e-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other half of the equation is where I get stuck with balancing the oxygen. I end up balancing the hydrogen on both sides to be 10 but I come up short of one oxygen on my product side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaBrO3 (aq) -&amp;gt; Br- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the algebraic method to assign an oxidation number to $\ce{Br}$ on reactant side of half equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br + Na + 3O = 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br + 1 + 3(-2) = 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br - 5 = 0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br = 5^{+}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{Br}$ on the product side has an oxidation number of -1 equal to its net charge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to balance the oxidation charge of both sides, I add $\ce{6e-}$ to the reactant side. This then brings the oxidation charge to -1 on both sides of the half reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaBrO3(aq) + 6e- -&amp;gt; Br- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then compensate net charge of the acidic reaction by adding 5 hydrogen ions to the reactant side to bring the net charge on both sides to 1-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaBrO3(aq) + 6e- + 5H+ (aq) -&amp;gt; Br- (aq)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then must add water to the product side in order to balance Hydrogen and oxygen atoms in the half reaction. I have 5 hydrogen for reactants so I must add 2.5 water molecules on the product side to match hydrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{NaBrO3(aq) + 6e- + 5H+(aq)-&amp;gt;Br- (aq) + $2.5$H2O(l)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Staying customary to the law of combining volumes and Avogadro's hypothesis I multiply the half equation by 2 to get a whole number for water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{2NaBrO3(aq) + 12e- + 10H+(aq) -&amp;gt; 2Br- (aq) + 5H2O(l)}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen atoms here elude me. I can't figure out how to balance them. I figured I took a wrong turn on the way but my troubleshooting comes to no avail thus far. Can anyone fill me in on this little secret?&lt;/p&gt;&#xA;" OwnerUserId="17082" LastEditorUserId="5026" LastEditDate="2018-02-22T02:24:37.500" LastActivityDate="2018-02-22T02:24:37.500" Title="Redox Reaction Equation with elusive oxygen" Tags="&lt;physical-chemistry&gt;&lt;redox&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36998" PostTypeId="1" CreationDate="2015-09-07T16:51:26.907" Score="2" ViewCount="119" Body="&lt;p&gt;Is there a name for the class of stand alone molecules that contains oxygen, water, hydrogen peroxide, and ozone? ($\ce{O2}$, $\ce{H2O}$, $\ce{H2O2}$, $\ce{O3}$) I would &lt;em&gt;not&lt;/em&gt; want to include hydroxyls $\ce{OH}$ because they aren't neutral molecules that I can buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think oxide and oxygenate is too generic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="7614" LastEditorUserId="10185" LastEditDate="2015-09-07T16:54:13.523" LastActivityDate="2015-09-07T17:37:10.333" Title="Name for HxOy compounds?" Tags="&lt;nomenclature&gt;&lt;covalent-compounds&gt;&lt;terminology&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="36999" PostTypeId="2" ParentId="36994" CreationDate="2015-09-07T16:54:59.913" Score="4" Body="&lt;p&gt;First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BrO3- -&amp;gt; Br-}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BrO3- -&amp;gt; Br- + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BrO3- + 6H+ -&amp;gt; Br- + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{BrO3- + 6H+ + 6e- -&amp;gt; Br- + 3H2O}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Half-reaction is ready.&lt;/p&gt;&#xA;" OwnerUserId="20695" LastEditorUserId="7448" LastEditDate="2015-09-07T18:23:37.223" LastActivityDate="2015-09-07T18:23:37.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37001" PostTypeId="2" ParentId="36998" CreationDate="2015-09-07T17:37:10.333" Score="4" Body="&lt;p&gt;Not only I've never heard of a name for such a class, but the very choice of compounds seems quite artificial to me. What feature do they have in common? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Consisting of H &lt;strong&gt;and&lt;/strong&gt; O? That would exclude oxygen and ozone.&lt;/li&gt;&#xA;&lt;li&gt;Consisting of H &lt;strong&gt;and/or&lt;/strong&gt; O? That would include $\mathrm H_2$ as&#xA;well.&lt;/li&gt;&#xA;&lt;li&gt;Containing &quot;active oxygen&quot;, whatever that might mean? That&#xA;would exclude water, but include a bunch of inorganic and organic&#xA;peroxides.&lt;/li&gt;&#xA;&lt;li&gt;Oxide is indeed too generic; also, it does not include oxygen and ozone. Oxygenate is even farther off.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why would you want to have such a class, really?&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-07T17:37:10.333" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="37002" PostTypeId="2" ParentId="4063" CreationDate="2015-09-07T18:00:20.283" Score="4" Body="&lt;p&gt;Water has “the neutral pH” because pH = 7 (a meme only useful in aqueous chemistry) is used to estimate acidity/basicity of a substance &lt;em&gt;with respect to water&lt;/em&gt; namely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Brønsted–Lowry approach, &lt;strong&gt;water is an acid and a base simultaneously:&lt;/strong&gt; H&lt;sub&gt;2&lt;/sub&gt;O sometimes dissociates to OH&lt;sup&gt;−&lt;/sup&gt; (the &lt;em&gt;conjugate base&lt;/em&gt; of H&lt;sub&gt;2&lt;/sub&gt;O) and H&lt;sup&gt;+&lt;/sup&gt;, i. e. we see deprotonation. The latter immediately reacts with surrounding H&lt;sub&gt;2&lt;/sub&gt;O to form aqueous varieties of hydrogen cation (&lt;em&gt;conjugate acid&lt;/em&gt; of H&lt;sub&gt;2&lt;/sub&gt;O, simplistically called “hydronium”), i. e. we see protonation. But there is an equilibrium, namely OH&lt;sup&gt;−&lt;/sup&gt; and (H&lt;sub&gt;2&lt;/sub&gt;O)&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt; react back to form the covalent water. This results in certain stable (under constant conditions) concentration of the hydrogen cation, that is used to define pH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we dissolve in water some substance that (preferentially) protonates it, an excess (H&lt;sub&gt;2&lt;/sub&gt;O)&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt; forms and pH became lower than of pure H&lt;sub&gt;2&lt;/sub&gt;O. Thus a Brønsted–Lowry acid (wrt H&lt;sub&gt;2&lt;/sub&gt;O) becomes an Arrhenius acid. When we dissolve in water some substance that (preferentially) deprotonates it, (H&lt;sub&gt;2&lt;/sub&gt;O)&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt; is consumed, an excess OH&lt;sup&gt;−&lt;/sup&gt; forms, and pH became higher than of pure H&lt;sub&gt;2&lt;/sub&gt;O. Thus a Brønsted–Lowry base (wrt H&lt;sub&gt;2&lt;/sub&gt;O) becomes an Arrhenius base (or alkali).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the paragraph above make sense only in the aqueous framework. When we mix H&lt;sub&gt;2&lt;/sub&gt;O with NH&lt;sub&gt;3&lt;/sub&gt;, a huge amount of ions forms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;OH&lt;sup&gt;−&lt;/sup&gt;, the conjugate base of H&lt;sub&gt;2&lt;/sub&gt;O;&lt;/li&gt;&#xA;&lt;li&gt;NH&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt;, the conjugate acid of NH&lt;sub&gt;3&lt;/sub&gt; (in other words, an &lt;em&gt;ammonia-based&lt;/em&gt; form of H&lt;sup&gt;+&lt;/sup&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In Arrhenius picture, we see an alkali: there is almost no (H&lt;sub&gt;2&lt;/sub&gt;O)&lt;sub&gt;&lt;em&gt;n&lt;/em&gt;&lt;/sub&gt;H&lt;sup&gt;+&lt;/sup&gt; and a lot of OH&lt;sup&gt;−&lt;/sup&gt;, whereas NH&lt;sub&gt;4&lt;/sub&gt;&lt;sup&gt;+&lt;/sup&gt; isn’t “hydrogen cation”. But in NH&lt;sub&gt;3&lt;/sub&gt;-centric picture we see indeed a strong acid! There is a lot of NH&lt;sub&gt;3&lt;/sub&gt;’s conjugate acid, whereas NH&lt;sub&gt;3&lt;/sub&gt;’s conjugate base (NH&lt;sub&gt;2&lt;/sub&gt;&lt;sup&gt;−&lt;/sup&gt;) is absent. In “symmetric” Brønsted–Lowry picture we see that H&lt;sub&gt;2&lt;/sub&gt;O acted as acid (it deprotonated) and NH&lt;sub&gt;3&lt;/sub&gt; acted as base (it was protonated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, acidic and basic solutions are all relative. &lt;strong&gt;Water is acidic,&lt;/strong&gt; with respect to substances that are bases wrt H&lt;sub&gt;2&lt;/sub&gt;O.&lt;/p&gt;&#xA;" OwnerUserId="9364" LastActivityDate="2015-09-07T18:00:20.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37004" PostTypeId="1" AcceptedAnswerId="37080" CreationDate="2015-09-07T18:30:59.090" Score="12" ViewCount="6648" Body="&lt;p&gt;As I know, the gases are insulators, because their particles are mainly electrically neutral, and thus there is no movable charges in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe it shouldn't be always so, for example if a gas has a significant proportion of charged particles, I think the electrical conductivity isn't impossible. If at least one of the electrons of its particles bounds very weakly, maybe even at room conditions is it possible to have enough low resistance to be considered as a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? Does a such gas already exist? Or there is some mechanism which prohibits this idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extensions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Maybe a such gas could be considered as plasma, but I think from this aspect we could see the plasma as gas as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am thinking at around 1 atm pressure and 300 K temperature, &lt;em&gt;in the gas&lt;/em&gt;, not around the box around it. Electrostatic discharge is also problematic, because in most gases it can happen only far from the room conditions and isn't an equilibrium state.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is none, I am looking for the gas which is the possible nearest to the these 3 conditions (300 K, 1 atm, conductive).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7008" LastEditorUserId="5026" LastEditDate="2018-05-19T07:05:38.070" LastActivityDate="2018-05-19T07:05:38.070" Title="Does an electrically conductive gas exist at room conditions?" Tags="&lt;conductivity&gt;&lt;electricity&gt;" AnswerCount="4" CommentCount="3" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="37005" PostTypeId="2" ParentId="37004" CreationDate="2015-09-07T18:45:44.843" Score="6" Body="&lt;p&gt;It depends on what you are prepared to consider a gas and what you are prepared to consider room conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas inside all discharge lamps (fluorescent lamps and neon lamps in shop signs, for example) conducts electricity and the lamps work under normal room conditions. However, the gases are often at low pressure and are, strictly speaking, plasmas not room temperature neutral gases. Some lamps, especially high intensity ones like the xenon discharge lamps in floodlights) exist with high pressure gas when working so low pressure isn't the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And air itself will conduct electricity given the right conditions. Lightning couldn't exist if it didn't. Nor could you get the (much safer) discharges from Tesla Coils or Van de Graaff generators. But, again, the effect relies on the creation of a plasma from the breakdown of the neutral molecules in the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is &quot;yes&quot; but only if you allow plasmas created from the medium to count as a normal gas.&lt;/p&gt;&#xA;" OwnerUserId="81" LastEditorUserId="5677" LastEditDate="2015-11-19T00:26:57.160" LastActivityDate="2015-11-19T00:26:57.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37006" PostTypeId="1" CreationDate="2015-09-07T21:18:27.870" Score="3" ViewCount="63" Body="&lt;p&gt;I read in paper that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The aim of this part of the work was to synthesise new aromatic diselenides with chiral phosphorous or sulfoxide centres in coordinative distance to the selenium as precursors for new chiral selenenylating reagents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can it be proved that (&lt;em&gt;S&lt;/em&gt;)-bis-[2-(t-butylsulfinyl)phenyl]diselenide is a chiral non-racemic product?&lt;/p&gt;&#xA;" OwnerUserId="20699" LastEditorDisplayName="user7951" LastEditDate="2015-09-08T12:24:53.157" LastActivityDate="2015-09-08T13:02:42.113" Title="How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl] diselenide is a chiral non-racemic product?" Tags="&lt;organic-chemistry&gt;&lt;chirality&gt;" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="37007" PostTypeId="2" ParentId="37004" CreationDate="2015-09-07T21:19:32.147" Score="6" Body="&lt;p&gt;Air at STP does conduct a tiny bit due to ionization by cosmic rays; this might even provide a path for lightning leaders. &lt;a href=&quot;http://www.scientificamerican.com/article/experts-do-cosmic-rays-cause-lightning/&quot; rel=&quot;noreferrer&quot;&gt;&quot;Alex V. Gurevich of the Lebedev Physical Institute [et al] suggest that... cosmic ray... might provide a conductive path that initiates lightning.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, &lt;a href=&quot;http://circuit-diagram.hqew.net/Ionising-Smoke-Detector_13703.html&quot; rel=&quot;noreferrer&quot;&gt;ionization-type smoke detectors&lt;/a&gt; use a little radioactive material (e.g. $\ce{^241Am}$) to increase conductivity, which can be detected with a high-impedance amplifier.&lt;/p&gt;&#xA;" OwnerUserId="11583" LastActivityDate="2015-09-07T21:19:32.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37008" PostTypeId="1" CreationDate="2015-09-07T21:34:53.890" Score="3" ViewCount="83" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the relative levels (i.e. ratios of concentrations) of the two most prominant glutamate species at pH 4.7.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the Henderson–Hasselbalch equation is to be used which is&#xA;$$\mathrm{pH=p}K_\mathrm a + \log_{10}\mathrm{\frac{[A^-]}{[HA]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one glutamate species, this would change to&#xA;$$\mathrm{4.7=2.19 + \log_{10}\frac{[A^-]}{[HA]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for the other species it would be&#xA;$$\mathrm{4.7=9.67 + \log_{10}\frac{[A^-]}{[HA]}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are some other variables I need to figure out to solve the equation, but I'm not sure how to go about finding them with the information given.&lt;/p&gt;&#xA;" OwnerUserId="20700" LastEditorUserId="95133" LastEditorDisplayName="user7951" LastEditDate="2021-05-13T12:51:53.103" LastActivityDate="2021-05-13T12:57:08.663" Title="Relative levels of the two most prominant glutamate species at pH 4.7" Tags="&lt;physical-chemistry&gt;&lt;buffer&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="37009" PostTypeId="1" AcceptedAnswerId="37011" CreationDate="2015-09-07T22:54:36.883" Score="1" ViewCount="2021" Body="&lt;p&gt;The hydrogen atoms in a water molecule are negatively charged, so they should repulse each other completely due to the electromagnetic force, forcing the hydrogen atoms to be on opposite sides of the oxygen atom. However, this is not the case, as the H-O-H bond angle is &lt;span class=&quot;math-container&quot;&gt;$104.5 ^\circ$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the lone electron pairs repulse the hydrogen atoms to an angle of &lt;span class=&quot;math-container&quot;&gt;$104.5 ^\circ$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$\ce{H2O}$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$180 ^\circ$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17645" LastEditorUserId="81756" LastEditDate="2019-08-28T12:29:39.883" LastActivityDate="2019-08-28T12:29:39.883" Title="Water molecule bent shape and electron configuration" Tags="&lt;physical-chemistry&gt;&lt;water&gt;&lt;electronic-configuration&gt;&lt;molecular-structure&gt;" AnswerCount="1" CommentCount="3" FavoriteCount="1" ClosedDate="2016-05-09T20:26:24.167" ContentLicense="CC BY-SA 4.0" />
  <row Id="37010" PostTypeId="1" CreationDate="2015-09-07T23:54:41.000" Score="2" ViewCount="857" Body="&lt;p&gt;Is all unaltered diatomaceous earth (DE) food grade?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that all have not spent the money to have it certified and labeled as food grade, but I'm thinking that is just a way for the government to get more money and also have more control. All countries do not have an FDA, and therefore have no need to certify it, and have used it for thousands of years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know the answer without somebody being afraid to say it is still food grade, labeled or not. I know the DE exposed to extreme heat is not safe and is for pool filters.&lt;/p&gt;&#xA;" OwnerUserId="20702" LastEditorUserId="16622" LastEditDate="2015-09-08T00:13:24.423" LastActivityDate="2015-09-08T01:01:45.130" Title="food grade diatomaceous earth" Tags="&lt;inorganic-chemistry&gt;" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37011" PostTypeId="2" ParentId="37009" CreationDate="2015-09-08T00:19:24.740" Score="6" Body="&lt;p&gt;According to VSEPR theory (there are other models, but they give the same qualitative results), since there are four electron pairs around oxygen they will be in a tetrahedral configuration. Assume the hydrogen atoms are at A and B as you've described them in your question; C and D (the lone pairs) are in a plane perpendicular to the one containing A and B. Below is a ball-and-stick model of water, showing the lone-pair electrons as yellow &quot;atoms&quot;.&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/42/Water-with-lone-pairs-3D-balls.png&quot; alt=&quot;Ball-and-stick model of water, showing the location of the lone pairs&quot;&gt; (Public domain image from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File%3AWater-with-lone-pairs-3D-balls.png&quot;&gt;Wikimedia Commons&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds as if you're thinking in two dimensions with the hydrogen atoms and lone pairs in the same plane. If water existed in only two dimensions it could conceivably be linear, but it really is in three dimensions. (The molecule itself is flat because the lone pairs affect the shape but are not counted as part of it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above applies to water under normal conditions. I'm not aware of any excited or exotic states of water in which the VSEPR prediction doesn't apply, but I can't guarantee that there aren't any.&lt;/p&gt;&#xA;" OwnerUserId="13334" LastActivityDate="2015-09-08T00:19:24.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37012" PostTypeId="2" ParentId="37010" CreationDate="2015-09-08T00:54:29.923" Score="5" Body="&lt;p&gt;I very much doubt that all diatomaceous earth is food grade.  For some reasons why, consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Diatomaceous_earth#Discovery&quot;&gt;Wikipedia page for diatomaceous earth&lt;/a&gt;: &quot;Kieselgur&quot; was discovered when folks in the 1800s were trying to dig a well.  Since knowledge of hygiene was some decades away, it's not implausible that those people may have been trying to dig a well that happened to be near an outhouse.  I'd doubt that diatomaceous earth recovered from a site near an outhouse would be food grade.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Wikipedia lists &quot;&lt;em&gt;salt-water-derived&lt;/em&gt;&quot; diatomaceous earth and specifically mentions that it is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;not suitable for human consumption&lt;/li&gt;&#xA;&lt;li&gt;not effective as an insecticide&lt;/li&gt;&#xA;&lt;li&gt;often calcined, which converts the safer amorphous silica present in most natural diatomaceous earths into a far more dangerous crystalline silica.  This can lead to &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicosis&quot;&gt;silicosis&lt;/a&gt; if long-term exposure to airborne microparticles of crystalline silica occurs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thus, any diatomaceous earth that contains lots of salts or which has been calcined is unlikely to be food grade.&lt;/p&gt;&#xA;" OwnerUserId="13184" LastActivityDate="2015-09-08T00:54:29.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37013" PostTypeId="2" ParentId="37010" CreationDate="2015-09-08T01:01:45.130" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Is all unaltered diatomaceous earth (DE) food grade?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, and that's because it's not regulated and in general trying to categorize &amp;quot;all&amp;quot; of something (regardless of source, and especially when unregulated) as &amp;quot;food grade&amp;quot; isn't possible. Here in the U.S., the Food and Drug Administration (FDA) has weighed in on the issue and determined that the components of diatomaceous earth are &amp;quot;generally regarded as safe (GRAS)&amp;quot; when used as a filter aid.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.fda.gov/Food/IngredientsPackagingLabeling/GRAS/SCOGS/ucm261274.htm&quot; rel=&quot;noreferrer&quot;&gt;the FDA's report entitled &lt;em&gt;Select Committee on GRAS Substances (SCOGS) Opinion: Diatomaceous earth (filter aid)&lt;/em&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Silicon dioxide and various silicates occur abundantly in the earth's crust, are present in practically all natural waters,animals, and plants, and are part of the normal human diet. The question of whether or not silicon is an essential human nutrient remains unresolved. Silicon compounds consumed as added food ingredients contribute only a minor proportion of the total dietary silicon intake. The estimated possible human intake of sodium aluminosilicate, the predominant silicate added to foods in this country, is approximately 0.3mg per kg body weight per day. Silicon compounds that are GRAS for use as direct food ingredient, except potassium and sodium silicates, are insoluble or very slightly soluble in water and appear to be biologically inert. The water-soluble silicates are also of low acute toxicity. The acute oral LD50 in rats of sodium aluminosilicate is &amp;gt;1g per kg. No significant tissue accumulation, pathology, or toxicity has been reported from the ingestion of those insoluble or very slightly soluble GRAS silicon compounds for which data are available. Of the five substances that were reported as added to foods in the NRC survey, biologic effects and safety data are available for all except sodium calcium aluminosilicate, and there is no reason to suspect that the toxicity of the latter would differ from those for which there are data. The results of two studies (1967-1970) in which various silicon compounds were fed to laboratory animals for 1 mo at a level of 0.8g per kg body weight (as silicon dioxide) and for 3 mo at levels of 6 to 30 mg per kg body weight suggests there may be a species-related susceptibility to renal damage from ingestion of sodium silicate, magnesium trisilicate, and finely ground quartz.&lt;/p&gt;&#xA;&lt;p&gt;No substantiating reports of these effects have appeared. Magnesium trisilicate was recognized as safe for prolonged use in human ingestion in large amount as a component of antacid preparations by the Advisory Review Panel on Over-the-Counter Drugs, and the available evidence on the acute toxicity of sodium silicate indicates that it is low. Consumption data are lacking for aluminium calcium silicate and tricalcium silicate, two compounds that are listed as GRAS for use as anticaking agents. However, their use in keeping with good manufacturing practice and in currently regulated amounts would be of the same order of magnitude as the other GRAS silicates. In addition, the Select Committee has limited information on the amounts of talc that are currently used in foods. However a major food use of talc is in the coating of rice. Assuming package label statements are ignored and coated rice is not washed prior to cooking and no other losses occur, maximun per capita intake of talc from this source appears to be 0.5g per day. With respect to paper and cotton food- packaging products, the possibility is remote that biologically significant amounts of talc, diatomaceous earth, or sodium silicate migrate to food from packaging, materials containing these substances. Perlite, a naturally occuring polysilicate substance, has an oral LD50 in the rat of &amp;gt;10g per kg body weight. Estimates of the maximun quantities of minerals that might be extracted from perlite and diatomaceous earth used as filteraids in food processing indicates no hazard to public health. There are no food grade specifications for aluminum calcium silicate, sodium calcium aluminosilicate, and talc. Such specifications for the substances used in foods are desirable.&lt;/p&gt;&#xA;&lt;p&gt;Specification for food grade talc should limit the content of asbestos fibers even though the potential hazard of ingested asbestos fibers even though the potential hazard of ingested asbestos is not clearly established. The Food and Drug Administration is sponsoring the development of an assay method for asbestos fibers. An upper limit for cadmium should be added to the specifications for food-grade perlite, and consideration should be given to the need for limitation of cadmium content of other silicates. In the light of all of the foregoing, the Select Committee concludes that: It is essential to establish food-grade specifications for aluminum calcium silicate, sodium calcium aluminosilicate, and talc, with provision for an upper limit of asbestos fibers in talc. There is no evidence in the available information on aluminum calcium silicate, calcium silicate, magnesium silicate, potassium silicate, sodium silicate, sodium aluminosilicate, sodium calcium aluminosilicate, tricalcium silicate, silica aerogel, and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used at levels that are now current or that might reasonably be expected in the future. There is no evidence in the available information on diatomaceous earth, silicon dioxides, sodium silicate, and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used as ingredients of paper and paperboard products used in food packaging in accordance with current practice. There is no evidence in the available information on sodium silicate and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used as ingredients of cotton and cotton fabrics used in dry food packaging in accordance with current practice. There is no evidence in the available information on diatomaceous earth and perlite that demonstrates or suggests reasonable ground to suspect a hazard to the public when they are used as filteraids in food processing at levels that are now current or that might reasonably be expected in the future.&lt;/p&gt;&#xA;&lt;p&gt;*Complete reports containing details of the safety studies that formed the basis of the opinions and conclusions and are available from the National Technical Information Service (NTIS), 5285 Port Royal Road, Springfield, VA 22161 (703) 605-6000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That said, &amp;quot;food grade&amp;quot; appears to apply to a mean particle size of less than or equal to 12 &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;m (&lt;a href=&quot;https://en.wikipedia.org/wiki/Diatomaceous_earth&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), and only applies to diatomaceous earth labeled that way.&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="-1" LastEditDate="2020-06-11T10:20:38.497" LastActivityDate="2015-09-08T01:01:45.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37014" PostTypeId="1" AcceptedAnswerId="37396" CreationDate="2015-09-08T02:22:28.240" Score="5" ViewCount="4992" Body="&lt;p&gt;Is there any known substance that can decrease fertility of a male?&lt;/p&gt;&#xA;&lt;p&gt;Less specifically what about one time chemical castrations (ignoring a perfect sterilization)?&lt;/p&gt;&#xA;&lt;p&gt;I heard about them before once but wasn't sure really.&lt;/p&gt;&#xA;" OwnerUserId="20705" LastEditorUserId="20705" LastEditDate="2021-03-22T18:11:37.533" LastActivityDate="2021-03-22T18:11:37.533" Title="Chemicals that Decrease fertility" Tags="&lt;chemical-biology&gt;" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="37015" PostTypeId="2" ParentId="7514" CreationDate="2015-09-08T02:47:35.797" Score="1" Body="&lt;p&gt;Add $\ce{MgSO4}$, until a saturated solution of magnesium salts has been obtained. Upon evaporating down, it will be found that all the potassium salts have been precipitated more or less completely, and with kainite $\ce{KCl. MgSO4.3H2O}$ being the essential potassium-containing salt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the kainite to a boiling solution of $\ce{NaHCO3}$ with moles equal to those of $\ce{Mg^{2+}}$. Then, freeze the mix until $\ce{Na2SO4.10H2O}$ is visible along with the visible $\ce{MgCO3}$ suspension.&lt;/p&gt;&#xA;" OwnerUserId="20706" LastEditorDisplayName="user15489" LastEditDate="2015-09-08T02:59:28.587" LastActivityDate="2015-09-08T02:59:28.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37016" PostTypeId="1" AcceptedAnswerId="37027" CreationDate="2015-09-08T03:14:51.273" Score="19" ViewCount="2445" Body="&lt;p&gt;I'm sorry if this is a stupid question, but since I'm a beginner with computational chemistry, please, be patient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the reaction of formation of water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ce{1/2 O2(g) + H2(g) -&amp;gt; H2O(g)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction shoud give a &lt;a href=&quot;http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&amp;amp;Mask=1&quot; rel=&quot;noreferrer&quot;&gt;$\Delta_\mathrm fH^\circ = -241.83~\mathrm{kJ/mol}$&lt;/a&gt;. The reaction is intentionally simple to let me run fast calculations and tests with single point energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I've performed a geometry optimization on $\ce{H2O}$, $\ce{O2}$ and $\ce{H2}$ (HF/6-31G*). Then, I've used M06/Def2-QZVPP for the single points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E_\mathrm{SP}(\ce{H2O}) = -76.431519932791~E_\mathrm h$&lt;/li&gt;&#xA;&lt;li&gt;$E_\mathrm{SP}(\ce{O2}) = -150.262365177069~E_\mathrm h$&lt;/li&gt;&#xA;&lt;li&gt;$E_\mathrm{SP}(\ce{H2}) = -1.171819656537~E_\mathrm h$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case, doing $$E_\mathrm{SP}(\ce{H2O}) - \left(\frac12 E_\mathrm{SP}(\ce{O2}) + E_\mathrm{SP}(\ce{H2})\right) = \Delta E$$ I'm getting $\Delta E = -0.1285176877195084~E_\mathrm h$ that is equal to $-337.42312874~\mathrm{kJ/mol}$, almost $96.6~\mathrm{kJ/mol}$ higher than the reference of $-241.83~\mathrm{kJ/mol}$.&lt;br&gt;&#xA;Is this the correct approach for a case like this? Or is this discrepancy normal?&lt;/p&gt;&#xA;" OwnerUserId="17118" LastEditorDisplayName="user7951" LastEditDate="2015-09-09T18:39:14.080" LastActivityDate="2015-09-09T18:39:14.080" Title="How to evaluate the heat of formation with computational chemistry?" Tags="&lt;thermodynamics&gt;&lt;computational-chemistry&gt;&lt;heat&gt;" AnswerCount="3" CommentCount="1" FavoriteCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="37017" PostTypeId="1" AcceptedAnswerId="37110" CreationDate="2015-09-08T03:50:00.300" Score="5" ViewCount="12859" Body="&lt;p&gt;As a fun experiment, I'm trying to evaporate water in my kitchen. I've heated the mixture to 45 °C, and I'm wondering how long it might take to completely evaporate.&lt;/p&gt;&#xA;&lt;p&gt;Assume that the water is 2 cm deep in a flat pan (if that is relevant).&lt;/p&gt;&#xA;" OwnerUserId="20708" LastEditorUserId="20708" LastEditDate="2020-11-23T11:00:02.233" LastActivityDate="2020-11-23T16:11:13.083" Title="I have 1 litre of water at 45 °C. How long would it take to evaporate when ambient temperature is 25 °C?" Tags="&lt;everyday-chemistry&gt;&lt;water&gt;" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="37018" PostTypeId="1" CreationDate="2015-09-08T04:29:43.920" Score="5" ViewCount="2389" Body="&lt;p&gt;A clean spectrum was given to me with the integrations filled in. I just need to assign peaks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just don't see how it's possible to have a singlet integrate to 5 protons. &#xA;I've considered symmetry, but still, 2 or 3 way symmetry wouldn't give me anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical formula is $\ce{C11H14O2}$, with 5 degrees of unsaturation. &#xA;The spectrum is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5H, singlet, 7.5 ppm&lt;/li&gt;&#xA;&lt;li&gt;2H, singlet, 5.1 ppm&lt;/li&gt;&#xA;&lt;li&gt;2H, triplet, 2.4 ppm&lt;/li&gt;&#xA;&lt;li&gt;2H, quartet, (or sextet?) 1.8 ppm&lt;/li&gt;&#xA;&lt;li&gt;3H, triplet, 1 ppm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm just looking for a hint on what a 5H singlet could be. &lt;/p&gt;&#xA;" OwnerUserId="20709" LastEditorUserId="194" LastEditDate="2018-03-05T03:17:55.137" LastActivityDate="2018-03-05T03:17:55.137" Title="What kind of structure could give you a HNMR spectrum consisting of a singlet integrating to 5 protons?" Tags="&lt;organic-chemistry&gt;&lt;nmr-spectroscopy&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37019" PostTypeId="2" ParentId="37018" CreationDate="2015-09-08T05:47:28.713" Score="4" Body="&lt;p&gt;The 5H singlet is likely due to five protons on a monosubstituted benzene ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak comes in at 7.5 ppm which is the aromatic region, immediately telling you that they are hydrogens on an aromatic ring. Benzene rings are the most common of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why are they all equivalent? They actually aren't, but their shifts are so close that they appear to be equivalent. Sometimes they will appear as different peaks (usually one peak for the 2 hydrogens close to the substituent and one peak for the 3 hydrogens away from it) if there are strong enough electron withdrawing groups on the substituent that are close to the ring. It is likely that the singlet you're looking at isn't clean cut and looks a bit like a multiplet (see &lt;a href=&quot;https://chemistry.stackexchange.com/questions/4078/why-are-all-the-phenyl-protons-in-benzyl-alcohol-equivalent-in-the-%C2%B9h-nmr-spectr&quot;&gt;this question&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16851" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:02.757" LastActivityDate="2015-09-08T05:47:28.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37020" PostTypeId="1" AcceptedAnswerId="37032" CreationDate="2015-09-08T05:59:54.800" Score="7" ViewCount="1930" Body="&lt;p&gt;Why doesn't my stainless steel Thermos get really really hot? Seems like a fairly straightforward question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking and found (obviously different values because steel is an alloy) that $$0.35~\mathrm{\frac{J}{g\,K}}\leq C_{p,\,\mathrm{steel}}\leq 0.55~\mathrm{\frac{J}{g\,K}}.$$&#xA;Obviously, that is just a rough range, but that is a very small heat capacity as is usual of metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why then, does my 200 degree Fahrenheit coffee not heat up the stainless steel mug super quickly!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even more baffling to me: Why in the world would it be a really good insulator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's entirely possible that I am devoid of any understanding of how insulation works, but I thought things which are bad conductors of electricity made good insulators. While stainless steel isn't praised for its electrical conductivity, I thought a material had to be about as bad as a gas at conducting electricity to be crowned with the title of &quot;good insulator.&quot; I would just imagine that stainless steel, being metallic in many ways, would be a better conductor than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's going on? Why isn't my Thermos reaching 200 degree Fahrenheit?&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="4945" LastEditDate="2015-09-08T06:26:51.303" LastActivityDate="2019-01-22T13:49:55.437" Title="Why doesn't my stainless steel Thermos get really really hot?" Tags="&lt;thermodynamics&gt;&lt;metal&gt;&lt;alloy&gt;" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="37021" PostTypeId="2" ParentId="37016" CreationDate="2015-09-08T08:23:04.013" Score="10" Body="&lt;ol&gt;&#xA;&lt;li&gt;you need to perform a Hessian calculation and add the enthalpy correction to each energy&lt;/li&gt;&#xA;&lt;li&gt;what spin state are you using for $\ce{O2}$?&lt;/li&gt;&#xA;&lt;li&gt;HF/6-31G(d) is probably not good enough for geometries and frequencies.&lt;/li&gt;&#xA;&lt;li&gt;Depending on what accuracy you're after, M06/Def2-QZVPP may not be good enough&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13210" LastActivityDate="2015-09-08T08:23:04.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37022" PostTypeId="2" ParentId="37016" CreationDate="2015-09-08T09:28:49.757" Score="6" Body="&lt;p&gt;The problem is that such a simple calculation has several associated errors (geometries, energies, vibrational frequencies and, therefore, thermodynamic functions) which render the result very inaccurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of water, using the same M06 method you used, one obtains -212.9 kJ/mol with the cc-pVDZ basis set and -246.0 kJ/mol with the much larger aug-cc-pV6Z basis set. Using a much more accurate method, CCSD(T), the enthalpy of formation with the same previous two basis sets are -200.19 and -241.89 kJ/mol. The latter result is nearly identical to the experimental one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, notice that quoting enthalpies of formation with more than one decimal digit is only justified if you are using highly accurate methods. You will very rarely have accuracies of better than 1 kJ/mol and certainly not using DFT methods.&lt;/p&gt;&#xA;" OwnerUserId="20711" LastEditorUserId="4945" LastEditDate="2015-09-08T09:35:15.643" LastActivityDate="2015-09-08T09:35:15.643" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37023" PostTypeId="1" CreationDate="2015-09-08T11:27:47.880" Score="7" ViewCount="716" Body="&lt;p&gt;Arenediazonium compounds contain a very good leaving group - nitrogen gas - and thus undergo nucleophilic substitution reactions rapidly. With water, for instance, they give phenols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the formation of azo dyes from arenediazonium salts, however, the arenediazonium cation acts as an electrophile to carry out a substitution reaction on activated benzene rings like those of phenols and anilines. Why doesn't displacement of $\ce{N2}$ happen here as well? Normally the carbon attached to $\ce{-N2+}$ acts as the electrophilic centre, so phenols and anilines should react at that $\ce{C}$ rather than at the $\ce{-N2+}$ group. Moreover, phenols and anilines have more nucleophilic centres at their $\ce{O}$ and $\ce{N}$ than at the carbon atoms of their benzene ring - so why aren't the final products be diphenyl ether with phenol, and &lt;em&gt;N&lt;/em&gt;-phenylaniline with aniline? &lt;/p&gt;&#xA;" OwnerUserId="15124" LastEditorUserId="16683" LastEditDate="2018-08-14T15:06:25.980" LastActivityDate="2018-08-15T01:59:12.733" Title="Azo coupling with arenediazonium salts - why isn't nitrogen gas lost?" Tags="&lt;organic-chemistry&gt;&lt;aromatic-compounds&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="37024" PostTypeId="1" CreationDate="2015-09-08T11:39:57.747" Score="3" ViewCount="460" Body="&lt;p&gt;It reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Petroleum consists of crude oil and natural gas and contains a mixture of up to 300 hydrocarbons, including sulfur and nitrogen compounds. It is formed over thousands of years from the remains of fossilised marine organisms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is petroleum a single entity that is simultaneously a gas and a liquid oil compounds? How does that work? It is also a mixture of up to 300 hydrocarbons, so it is a liquid and gas in one? I don't understand? Do they mean that petroleum can be either gas OR it can be either liquid? What about the 'mixture of 300 hydrocarbons bit'. Does that mean you will never extract petroleum of a single hydrocarbon eg pure $\ce{C6H14}$ etc? I haven't revised this in a long time so I forgot..&lt;/p&gt;&#xA;" OwnerUserId="20714" LastEditorUserId="30836" LastEditDate="2020-03-31T14:36:05.150" LastActivityDate="2020-03-31T14:36:05.150" Title="What does my textbook mean by saying this about petroleum?" Tags="&lt;organic-chemistry&gt;&lt;terminology&gt;&lt;hydrocarbons&gt;&lt;mixtures&gt;&lt;petrochemistry&gt;" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37025" PostTypeId="2" ParentId="37024" CreationDate="2015-09-08T12:01:26.587" Score="4" Body="&lt;p&gt;Petroleum is a mixture of many different compounds.&lt;br&gt;&#xA;Some of those compounds are gas.  Some are liquid.  Some are solid.&lt;br&gt;&#xA;Many of them are hydrocarbons, and there are many different hydrocarbons included in the mix (e.g. different numbers on the C and H).&lt;br&gt;&#xA;Petroleum is generally &lt;strong&gt;&lt;em&gt;refined&lt;/em&gt;&lt;/strong&gt; into more specific products before use, at a &lt;a href=&quot;https://en.wikipedia.org/wiki/Oil_refinery&quot; rel=&quot;nofollow&quot;&gt;refinery&lt;/a&gt;.&lt;br&gt;&#xA;Another example to consider might be &quot;mud&quot; - that's a mixture of some liquids, maybe some dissolved gases, and a bunch of different solids, usually including clay, maybe a bit of gravel, and some organic materials formed from decaying plants and animals.&#xA;If you take mud and compress it underground for thousands of years, then you'll have petroleum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second sentence seems clear enough. &lt;/p&gt;&#xA;" OwnerUserId="17779" LastEditorUserId="17779" LastEditDate="2015-09-08T15:54:05.943" LastActivityDate="2015-09-08T15:54:05.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37026" PostTypeId="2" ParentId="37006" CreationDate="2015-09-08T12:11:45.383" Score="6" Body="&lt;p&gt;Compounds with the general structure $\ce{R-SO-R'}$, i.e. containing a sulfinyl $\ce{-SO-{}}$ functional group attached to two carbon atoms, are called generically ‘sulfoxides’. Sulfoxides may be considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KEKjh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KEKjh.png&quot; alt=&quot;sulfoxide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the two organic residues $\ce{R}$ and $\ce{R'}$ are dissimilar, the sulfur atom is a stereogenic center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compound bis[2-(&lt;em&gt;tert&lt;/em&gt;-butylsulfinyl)phenyl]diselenide contains two chiral sulfinyl groups and thus two stereogenic centers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KEKjh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cPw10.png&quot; alt=&quot;bis[2-(tert-butylsulfinyl)phenyl]diselenide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, three stereoisomers exist:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;bis[2-((&lt;em&gt;R&lt;/em&gt;)-&lt;em&gt;tert&lt;/em&gt;-butylsulfinyl)phenyl]diselenide,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZGLI1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZGLI1.png&quot; alt=&quot;R,R&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;bis[2-((&lt;em&gt;S&lt;/em&gt;)-&lt;em&gt;tert&lt;/em&gt;-butylsulfinyl)phenyl]diselenide,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DqWUJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DqWUJ.png&quot; alt=&quot;S,S&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and the meso compound [2-((&lt;em&gt;R&lt;/em&gt;)-&lt;em&gt;tert&lt;/em&gt;-butylsulfinyl)phenyl][2-((&lt;em&gt;S&lt;/em&gt;)-&lt;em&gt;tert&lt;/em&gt;-butylsulfinyl)phenyl]diselenide  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6tTAG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6tTAG.png&quot; alt=&quot;R,S&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-09-08T13:02:42.113" LastActivityDate="2015-09-08T13:02:42.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37027" PostTypeId="2" ParentId="37016" CreationDate="2015-09-08T12:56:47.603" Score="19" Body="&lt;p&gt;Your approach is quite correct, but as Jan already pointed out, it is incomplete. What you calculated is the difference in the electronic energy of the reaction $$\ce{H2 (g) + 1/2 O2 (g,\,{}^1\Delta_{g}) -&amp;gt; H2O (g)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's turn this into some kind of a tutorial and a little exercise, that you can try to reproduce at home. For a more detailed description on how it works in Gaussian, you can read the &lt;a href=&quot;http://www.gaussian.com/g_whitepap/thermo.htm&quot;&gt;paper by Joseph W. Ochterski&lt;/a&gt;. I will use Gaussian09 rev. D for the calculations involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enthalpy of formation of water from the standard state of the elements involved, is basically the reaction enthalpy of the combustion of hydrogen and oxygen. You are in luck, because most of the quantum chemical programs shall provide you with the necessary data for reactions straight away. What you do need to do is calculate the thermal corrections to the reaction. Essentially this is done by calculating the Hessian, i.e. a frequency calculation. The second derivatives only have a interpretable meaning if they are performed at a stationary point. Hence they must be performed with the same method, that you used to optimise the geometry.&lt;br&gt;&#xA;Jan already pointed out, that HF/6-31G(d) is probably not a well enough fit for geometries and frequencies. I'd always consider some level of theory with at least some coverage of correlation energy. I often prefer BP86 or M06L and the def2-SVP basis set as a first anchor point. The smaller the system, the more accurate the methods you can use (or afford).&lt;br&gt;&#xA;Jan also pointed out, that especially for oxygen, the spin state is important. (I did a quick check and am almost 100% certain you calculated it in the singlet state; see above equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this exercise I have matched your suggestion with using M06/def2-QVZPP as it is still feasible. Keep in mind, that most of the times density functional approximations have to be calibrated. It's all about what you can or cannot afford. You should perform calculations to the best of your abilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, let's dive into some mathematics. First, we consider the reaction equation &#xA;$$\ce{H2 (g) + 1/2 O2 (g,\,{}^3\Sigma_{g}^{-}) -&amp;gt; H2O (g)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we would like to know the enthalpy of formation for water at standard state (which we will use as $298.15~\mathrm{K}$ and $1~\mathrm{atm}$)&#xA;$$\begin{align}&#xA;\Delta_{f}H^\circ(\ce{H2O}) &amp;amp;= H^\circ(\ce{H2O}) &#xA;                               - (H^\circ(\ce{H2}) &#xA;                               +\frac12 H^\circ(\ce{O2}))\\&#xA;\Delta_{f}H^\circ(\ce{H2O}) &amp;amp;= &#xA;   [E_\mathrm{el}(\ce{H2O}) + H_\mathrm{corr}(\ce{H2O})]\\&#xA;&amp;amp;\qquad&#xA; - \left([E_\mathrm{el}(\ce{H2}) + H_\mathrm{corr}(\ce{H2})]&#xA; +\frac12 [E_\mathrm{el}(\ce{O2}) + H_\mathrm{corr}(\ce{O2})]\right)\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data in the expanded form we can directly take from the output of the frequency calculation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Oxygen  mo6qzvpp.freq.log/.com&#xA; SCF Done:  E(UM06) =  -150.322374247     A.U. after    1 cycles&#xA; Zero-point correction=                           0.003897 (Hartree/Particle)&#xA; Thermal correction to Enthalpy=                  0.007203&#xA; Thermal correction to Gibbs Free Energy=        -0.016049&#xA;&#xA;Hydrogen mo6qzvpp.freq.log/.com&#xA; SCF Done:  E(RM06) =  -1.17221845130     A.U. after    1 cycles&#xA; Zero-point correction=                           0.009902 (Hartree/Particle)&#xA; Thermal correction to Enthalpy=                  0.013206&#xA; Thermal correction to Gibbs Free Energy=        -0.001590&#xA;&#xA;Water mo6qzvpp.freq.log/.com&#xA; SCF Done:  E(RM06) =  -76.4321360023     A.U. after    1 cycles&#xA; Zero-point correction=                           0.021635 (Hartree/Particle)&#xA; Thermal correction to Enthalpy=                  0.025415&#xA; Thermal correction to Gibbs Free Energy=         0.004010&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore we will obtain &#xA;$$\Delta_{f}H^\circ(\ce{H2O}) = -236.6~\mathrm{kJ/mol}$$&#xA;which I think is reasonably close to the experimental value.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Appendix&lt;/strong&gt;&#xA;The following input files for the computation of above values.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;opt&#xA;int(ultrafinegrid)&#xA;&#xA;oxygen&#xA;&#xA;0 3&#xA;O&#xA;O 1 1.2&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.freq.chk&#xA;%oldchk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;freq&#xA;geom=check guess=read&#xA;int(ultrafinegrid)&#xA;&#xA;oxygen freq&#xA;&#xA;0 3&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;opt&#xA;int(ultrafinegrid)&#xA;&#xA;hydrogen&#xA;&#xA;0 1&#xA;H&#xA;H 1 1.2&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.freq.chk&#xA;%oldchk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;freq&#xA;geom=check guess=read&#xA;int(ultrafinegrid)&#xA;&#xA;hydrogen freq&#xA;&#xA;0 1&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;opt&#xA;int(ultrafinegrid)&#xA;&#xA;water&#xA;&#xA;0 1&#xA;O&#xA;H 1 1.2&#xA;H 1 1.2 2 109.0&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%chk=mo6qzvpp.freq.chk&#xA;%oldchk=mo6qzvpp.chk&#xA;#p M06/def2QZVPP&#xA;freq&#xA;geom=check guess=read&#xA;int(ultrafinegrid)&#xA;&#xA;water freq&#xA;&#xA;0 1&#xA; (blank)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Have a lot of fun.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="4945" LastActivityDate="2015-09-08T12:56:47.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37028" PostTypeId="1" CreationDate="2015-09-08T13:08:12.300" Score="1" ViewCount="131" Body="&lt;p&gt;Already saw some youtube video where they lit a camp fire with glycerol and potassium permanganate and tried those but the amount of the stuff you need to create enough heat is just too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a way to light a candle with small amount of something and something, like having the candle wick dipped/powdered with something and then just letting it get into contact with something else?&lt;/p&gt;&#xA;" OwnerUserId="20718" LastEditorUserId="194" LastEditDate="2017-04-01T03:37:52.360" LastActivityDate="2017-04-01T03:37:52.360" Title="How to light a candle with small amounts of safe to use chemicals?" Tags="&lt;everyday-chemistry&gt;&lt;heat&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37029" PostTypeId="1" CreationDate="2015-09-08T13:14:31.063" Score="6" ViewCount="8691" Body="&lt;p&gt;Why there is a subtractive volume correction in van der Waals' equation for real gases, both real and ideal gases occupy volume of container in which they are kept, so there shouldn't be any correction?&lt;/p&gt;&#xA;" OwnerUserId="19504" LastEditorUserId="4945" LastEditDate="2015-10-23T17:42:14.123" LastActivityDate="2020-07-18T01:04:24.330" Title="Why is there a subtractive volume correction in the van der Waals' equation for real gases?" Tags="&lt;physical-chemistry&gt;&lt;gas-laws&gt;" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37031" PostTypeId="1" AcceptedAnswerId="37045" CreationDate="2015-09-08T14:27:36.847" Score="2" ViewCount="15950" Body="&lt;p&gt;Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any difference in eating sodium chloride in powder form and after dissolving in water? It gets ionized, correct?&lt;/p&gt;&#xA;" OwnerUserId="20719" LastEditorUserId="865" LastEditDate="2015-09-08T14:45:20.387" LastActivityDate="2018-12-10T04:25:47.460" Title="Why is sodium chloride edible whereas both sodium and chlorine are highly reactive?" Tags="&lt;everyday-chemistry&gt;&lt;oxidation-state&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37032" PostTypeId="2" ParentId="37020" CreationDate="2015-09-08T14:28:02.980" Score="5" Body="&lt;p&gt;I have a solid steel pint glass and I can tell you that it pretty much instantly heats up to the temperature of the liquid you put inside. You probably have some sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_flask&quot; rel=&quot;noreferrer&quot;&gt;double-wall vacuum flask&lt;/a&gt;, which insulates by using a vacuum between the steel.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zvt6T.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zvt6T.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see, there are two walls to the flask: an inner wall and an outer wall. The inner wall is probably very close to the temperature of whatever is inside the flask. However, since the two walls are separated by a vacuum (which is a fairly good insulator), the outer wall, i.e. the one that you touch, remains pleasantly near ambient temperature*.&lt;/p&gt;&#xA;&lt;p&gt;*Unpleasantly near ambient temperature, if you've been outside in a hot summer for a while.&lt;/p&gt;&#xA;" OwnerUserId="2087" LastEditorUserId="-1" LastEditDate="2020-06-10T14:04:27.157" LastActivityDate="2015-09-08T14:28:02.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37033" PostTypeId="2" ParentId="37028" CreationDate="2015-09-08T14:35:26.950" Score="2" Body="&lt;p&gt;Safety and ability to light a fire, as far as I understand, directly contradict each other. What starts a fire on your wish, would occasionally start a fire against your wish. The glycerol/sulfuric acid/permanganate recipe, like it or not, is probably the safest of all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Allotropes_of_phosphorus#White_phosphorus&quot; rel=&quot;nofollow&quot;&gt;White phosphorus&lt;/a&gt;, not only flammable but also severely toxic by itself.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tert-Butyllithium#Safety&quot; rel=&quot;nofollow&quot;&gt;Butyllithium&lt;/a&gt; and many other organometallic compounds that would ignite instantly on air; that's not quite what you want, anyway.&lt;/li&gt;&#xA;&lt;li&gt;Hydrogen against &lt;a href=&quot;https://en.wikipedia.org/wiki/Platinum_black#Catalytic_ignition_of_flammable_gasses&quot; rel=&quot;nofollow&quot;&gt;platinum black&lt;/a&gt; (danger of explosion).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I definitely would not try any of these at home.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-08T14:35:26.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37034" PostTypeId="2" ParentId="34823" CreationDate="2015-09-08T14:41:06.807" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Slater%27s_rules&quot; rel=&quot;nofollow&quot;&gt;Slater shielding rules&lt;/a&gt; give more quantitative information about the amount of shielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, valence electrons (as you go across a row) do not shield very well, roughly 35% of a full electron. So when you go from B to C to N, you keep increasing the nuclear charge by one proton, but the electrons don't fully shield the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Core&lt;/em&gt; electrons, as suggested in comments and other answers do shield fully, since they're closer to the nucleus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, the shielding for Li and Na are similar. As you mention, the valence electron in Na is further from the nucleus, and that does effect the ionization energy and electronegativity as you suggest.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-09-08T14:41:06.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37035" PostTypeId="2" ParentId="36882" CreationDate="2015-09-08T14:48:15.420" Score="4" Body="&lt;p&gt;&lt;strong&gt;It depends very significantly on what you want to accomplish.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to suggest software, since there are many programs that can (and are) used to predict optical absorption spectra of conjugated polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you can either approximate the properties of the polymer by running several calculations on oligomers (e.g., dimer, trimer, tetramer, etc.) and extrapolating the polymer band gap. Because conjugated polymers are similar to the particle-in-a-box model, you can &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.68.035204&quot; rel=&quot;nofollow&quot;&gt;roughly&lt;/a&gt; plot the HOMO-LUMO gap or excitation energy vs. 1/(# of repeat units) and get an &lt;a href=&quot;http://dx.doi.org/10.1103/PhysRevB.68.035204&quot; rel=&quot;nofollow&quot;&gt;approximately&lt;/a&gt; linear fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more information, you could instead run a calculation on a polymer in a periodic unit cell, which will give you the electronic structure of the infinite system (but no information on disorder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As suggested in the comment above, you can use multiple programs and different computational approaches depending on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What kind of polymer&lt;/li&gt;&#xA;&lt;li&gt;What kind of information you want&lt;/li&gt;&#xA;&lt;li&gt;What computational resources you have&lt;/li&gt;&#xA;&lt;li&gt;How accurately you want the information&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The two most common methods for predicting optical absorption energies are the ZINDO/S semiempirical method, and TDDFT. The latter is &lt;a href=&quot;http://dx.doi.org/10.1021/jp025999m&quot; rel=&quot;nofollow&quot;&gt;more accurate&lt;/a&gt; particularly for predicting absorption intensity from the oscillator strength of the excitation energies.&lt;/p&gt;&#xA;" OwnerUserId="5017" LastActivityDate="2015-09-08T14:48:15.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="37036" PostTypeId="2" ParentId="36992" CreationDate="2015-09-08T14:55:16.747" Score="5" Body="&lt;p&gt;Simple self-made or commercially available tests use the &lt;strong&gt;Marquis reagent&lt;/strong&gt;, which is a mixture of sulfuric acid and formaldehyde. For example, Marquis reagent can be prepared by adding 100 ml of concentrated sulfuric acid to 5 ml of a saturated aqueous solution of formaldehyde (40 % formaldehyde by volume or 37 % by mass).&lt;sup&gt;[&lt;a href=&quot;https://www.ncjrs.gov/pdffiles1/nij/183258.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;]&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OHBwW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHBwW.png&quot; alt=&quot;Marquis reagent reacts with methamphetamine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Marquis reagent reacts with many alkaloid drugs to form a variety of colours. Methamphetamine creates an orange dye (deep reddish orange to dark reddish brown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p89xF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p89xF.png&quot; alt=&quot;deep reddish orange&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/WlpKN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WlpKN.png&quot; alt=&quot;dark reddish brown&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerDisplayName="user7951" LastEditorDisplayName="user7951" LastEditDate="2015-09-09T14:26:40.083" LastActivityDate="2015-09-09T14:26:40.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37038" PostTypeId="1" CreationDate="2015-09-08T16:21:38.467" Score="14" ViewCount="29636" Body="&lt;p&gt;Given some electron configuration, I know the following formula can be used to determine the number of microstates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{# mircostates} = \frac{(\text{# electron positions})!}{((\text{# electrons})!(\text{# positions} - \text{# electrons})!)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for 2 electrons in a $\mathrm{p}$ orbital it would be $6!/(2!4!) = 15$, or for two electrons in a $\mathrm{d}$ orbital it would be $10!/(2!8!) = 45$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little confused how to apply this for a situation where all the electrons are not in the same $l$ level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a configuration like one electron in $\mathrm{s}$ and one in $\mathrm{d}$ would the number of microstates be the product of the above formula for each $l$ level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2!}{1!1!}\cdot \frac{10!}{1!9!} = 20$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or for one electron in $\mathrm{s}$ and one in $\mathrm{f}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2!}{1!1!}\cdot \frac{14!}{1!13!} = 28$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed it would work like this because the formula reminds me of multiplicity, so it would make sense for it to be a product. But it also occurred to me that the 2 $l$ levels could be combined for the formula, so for a $\mathrm{(s^1)(d^1)}$ electron configuration, the number of positions would be $$\frac{(10+2)!}{2!(10+2-2)!} = \frac{12!}{2!10!}= 66,$$ which is must bigger than what I had before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone clear up how to apply this formula to a case like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarification, the exact specification in the problem was: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine a $\mathrm{s^1\, f^1}$ electron system (that is, 1 s electron and 1 f electron; no p or d electrons).  How many total microstates exist for this system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think this means that the one $\mathrm{s}$ and one $\mathrm{f}$ electrons are each confined to their $l$ level, which is why I thought the multiplicity of microstates would be the product of the above formula for each, but I'm not sure.&lt;/p&gt;&#xA;" OwnerUserId="20723" LastEditorUserId="4945" LastEditDate="2015-10-01T03:44:31.943" LastActivityDate="2017-07-08T11:13:20.867" Title="How to find the number of atomic microstates for a given electronic configuration?" Tags="&lt;physical-chemistry&gt;&lt;statistical-mechanics&gt;" AnswerCount="2" CommentCount="0" FavoriteCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37041" PostTypeId="5" CreationDate="2015-09-08T17:08:27.607" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-08T17:08:27.607" LastActivityDate="2015-09-08T17:08:27.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37042" PostTypeId="4" CreationDate="2015-09-08T17:08:27.607" Score="0" Body="Statistical mechanics is used to describe and quantify observed (average) thermodynamic properties as they relate to microscopic states of the system in question." OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-09-08T17:20:28.763" LastActivityDate="2015-09-08T17:20:28.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37043" PostTypeId="1" CreationDate="2015-09-08T17:15:49.527" Score="-1" ViewCount="339" Body="&lt;p&gt;Benzene has O and Na attached. Oxygen with a negative charge and sodium with positive , so they write it as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ce{benzene-O- Na+}$    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why couldn't it simply be like oxygen bonded to benzene and sodium there wouldn't be excessive charges.&#xA;I think sodium bonds with benzene as well as oxygen ($\ce{benzene-Na-O}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it like this? I usually get stuff on my own this seemed really odd for some reason&lt;/p&gt;&#xA;" OwnerUserId="20725" LastEditorUserId="10185" LastEditDate="2015-09-08T17:20:38.353" LastActivityDate="2017-02-17T15:56:27.303" Title="Sodium phenoxide bonding" Tags="&lt;organic-chemistry&gt;&lt;bond&gt;&lt;aromatic-compounds&gt;&lt;structural-formula&gt;" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37044" PostTypeId="2" ParentId="37043" CreationDate="2015-09-08T17:28:18.847" Score="1" Body="&lt;p&gt;The reason people do this is to highlight the fact that the bond between the oxygen and sodium atom is of a more ionic character than of a covalent character.&lt;/p&gt;&#xA;" OwnerDisplayName="user19026" LastActivityDate="2015-09-08T17:28:18.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="37045" PostTypeId="2" ParentId="37031" CreationDate="2015-09-08T17:53:25.390" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Solid sodium chloride is an &lt;strong&gt;ionic compound&lt;/strong&gt;. This means it consists of positively charged sodium ions ($\ce{Na+}$) and negatively charged chloride ions ($\ce{Cl-}$) arranged in a cubic lattice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HBKGWm.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HBKGWm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, sodium metal and chlorine gas are elements. $\ce{Na}$ has an electron configuration of $\mathrm{[Ne]~3s^1}$ and so it really wants to lose that single electron to form a full outer shell as a $\ce{Na+}$ ion. This makes it highly reactive towards species which can easily accept an electron such as $\ce{Cl2}$, $\ce{H2O}$ etc. $\ce{Na+}$ has a full outer shell, being isoelectronic with neon, and so it is quite happy to stay as an ion and not react. Chlorine, with an electron configuration of $\mathrm{[Ne]~3s^2~3p^5}$, really wants to gain a single electron to get a full outer shell, isoelectronic with argon, as the $\ce{Cl-}$ ion. This makes it highly reactive towards species which can easily donate an electron, such as $\ce{Na}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is there any difference in eating sodium chloride in powder form and after dissolving in water?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not really. Sodium chloride dissolved in water consists of hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions. If you eat solid sodium chloride then it rapidly dissolves in the water in your body, so the effect is the same. Interestingly though, solid salt tastes quite nice, but salty water tends to taste horrible.&lt;/p&gt;&#xA;" OwnerUserId="10185" LastActivityDate="2015-09-08T17:53:25.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37046" PostTypeId="1" AcceptedAnswerId="37066" CreationDate="2015-09-08T18:38:27.207" Score="0" ViewCount="338" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pujta.jpg&quot; alt=&quot;this is the model with CH3-CH2-CH2-CH2-CH2-CH2-CH3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model has 3 hydrogens connected the the first and last carbon and 2 hydrogens connected to the other carbons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the same molecular formula as the alkane model I have to check all that apply &#xA;&lt;img src=&quot;https://i.stack.imgur.com/bdd3V.jpg&quot; alt=&quot;the choice options&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the last two options are not correct, because they don't have the same amount of carbons, but when it comes to the first two I'm confused because there is only two carbons attached to 3 hydrogens yet in these options there are 3 different carbons attached to 3 hydrogens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they both be correct or does it matter if it's connected to the second bond instead of the third? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im assuming since they are isomers that they are equal to eachother.&lt;/p&gt;&#xA;" OwnerUserId="20727" LastEditorUserId="20727" LastEditDate="2015-09-08T22:25:43.323" LastActivityDate="2015-09-09T00:04:20.620" Title="Which of the following have the same molecular formula as the molecule shown in the model?" Tags="&lt;organic-chemistry&gt;&lt;homework&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37047" PostTypeId="1" AcceptedAnswerId="37052" CreationDate="2015-09-08T18:52:49.593" Score="2" ViewCount="2989" Body="&lt;p&gt;Say dilute $\ce{HCl}$ is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous $\ce{HCl}$ be released in the process?  Is there a threshold concentration of the acid, below which,  this would be acceptable from a safety point of view?&lt;/p&gt;&#xA;" OwnerUserId="20728" LastEditorUserId="7448" LastEditDate="2015-09-08T19:40:07.290" LastActivityDate="2019-05-10T00:36:14.693" Title="Is heating dilute HCl dangerous?" Tags="&lt;acid-base&gt;&lt;safety&gt;" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37050" PostTypeId="1" AcceptedAnswerId="48888" CreationDate="2015-09-08T19:38:22.813" Score="5" ViewCount="902" Body="&lt;p&gt;Is it possible to fully esterify phenol with oxalic acid in order to get diphenyl oxalate? What catalyst should I use and is the synthesis possible to perform at home with hobby equipment?&lt;/p&gt;&#xA;" OwnerUserId="20733" LastEditorUserId="10185" LastEditDate="2016-06-04T08:33:27.090" LastActivityDate="2016-06-04T08:33:27.090" Title="How to esterify phenol with oxalic acid?" Tags="&lt;organic-chemistry&gt;&lt;synthesis&gt;&lt;esters&gt;&lt;phenols&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37051" PostTypeId="1" CreationDate="2015-09-08T19:40:34.613" Score="4" ViewCount="313" Body="&lt;p&gt;I know flavor is subjective and not exactly quantifiable, but I think there's merit to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've all experienced that tap water often tastes a lot worse than bottled water or filtered water. I've always felt, however, that cold water actually tastes much purer than warmer water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I was recently in a house where I knew the water was so called &quot;hard water&quot; and I could taste that it had the typical mineral flavor straight from the tap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I took a glass of this tap water and refrigerated it for about a half hour and then drank it and I couldn't taste the typical tap water flavor at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd encourage people to try this just to make sure I'm not crazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I have family that lives in Ohio and if anyone has ever tasted Ohio's tap water after living in a place like Washington (where I live) which has extremely fresh water, you know that Ohio's water is disgusting. I have found, however, that it tastes much better after being refrigerated for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are a couple of questions here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it even true that refrigerating water (and thus cooling it) would cause a significant amount of oxygen to be dissolved in the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, would the amount of time the water is being refrigerated (even at a constant temperature) continue to dissolve more oxygen into the water? In other words, how long might it take for oxygen to reach it's equilibrium level of dissolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those last two are definitely answerable chemistry questions. This next one is a little hairier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason, related to chemistry, that oxygen rich water would taste better than water not saturated with oxygen? Meaning, does the oxygen actually have a well-defined effect on the flavor of water, or is this simply answered by the fact that most people like the feeling of cold water more than luke-warm water and hence I identify cold water as &quot;tasting better&quot;?&lt;/p&gt;&#xA;" OwnerUserId="17583" LastEditorUserId="21964" LastEditDate="2016-07-25T18:32:43.047" LastActivityDate="2016-07-25T19:24:41.400" Title="Does oxygenating tap water remove the tap water flavor?" Tags="&lt;water&gt;&lt;taste&gt;&lt;water-treatment&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37052" PostTypeId="2" ParentId="37047" CreationDate="2015-09-08T19:53:48.703" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Say dilute HCl is used to convert some alkaloid to salt form, then the alkaloid is dried with heat, would gaseous HCl be released in the process?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likely yes, unless it's the limiting reagent in the reaction in question. Probably best to assume some of it will come off as gas - that's the safest approach, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a threshold concentration of the acid, below which, this would be acceptable from a safety point of view?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there are several metrics used by various authorities to determine what &quot;safe&quot; means. You'll have to decide which one (or ones) are relevant to your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here in the U.S., the Occupational Safety and Health Administration (OSHA) - which is under the Department of Labor - has published &lt;a href=&quot;https://www.osha.gov/dts/chemicalsampling/data/CH_246300.html&quot; rel=&quot;nofollow&quot;&gt;this information on hydrochloric acid exposure limits and hazards&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that linked document, several other agencies such as the National Institute for Occupational Safety and Health (NIOSH) and the American Conference of Governmental Industrial Hygienists (ACGIH) list permissible exposure limits ranging from 3 to 7 mg/m$^{3}$; the Environmental Protection Agency (EPA) given an inhalation reference concentration (RfC) of 2 mg/m$^{3}$, detailed &lt;a href=&quot;http://www.epa.gov/iris/subst/0396.htm#inhalrfc&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; as the concentration above which (I'm paraphrasing) bad things start to happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck on your project and stay safe. &lt;/p&gt;&#xA;" OwnerUserId="865" LastActivityDate="2015-09-08T19:53:48.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37053" PostTypeId="1" CreationDate="2015-09-08T20:00:35.137" Score="8" ViewCount="8754" Body="&lt;p&gt;What is difference between $\ce{H+}$ and a proton?&lt;/p&gt;&#xA;" OwnerUserId="20734" LastEditorUserId="41328" LastEditorDisplayName="user15489" LastEditDate="2019-10-15T04:48:06.537" LastActivityDate="2019-10-15T05:59:48.820" Title="What is difference between H⁺ and proton?" Tags="&lt;ions&gt;&lt;terminology&gt;&lt;hydrogen&gt;" AnswerCount="2" CommentCount="1" FavoriteCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="37054" PostTypeId="2" ParentId="37053" CreationDate="2015-09-08T20:06:59.303" Score="23" Body="&lt;p&gt;There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics.&#xA;A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).&lt;/p&gt;&#xA;" OwnerUserId="18002" LastEditorUserId="7475" LastEditDate="2015-09-22T11:35:46.880" LastActivityDate="2015-09-22T11:35:46.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37055" PostTypeId="2" ParentId="33180" CreationDate="2015-09-08T20:08:50.663" Score="-1" Body="&lt;p&gt;May the force of additives be of aid to you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beforehand, choose an easier type of detergent:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Surface-acting polyglucosides [which are] easily broken down by microbes, leaving no traces in the environment&quot;, is one (Quellen, 213).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is slow to biodegrade may not be used as often-inform yourself, and go from there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please, be safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why There's Antifreeze in Your Toothpaste: the Chemistry of Household Ingredients. P 213. Quellen, Simon. 2008. &lt;/p&gt;&#xA;" OwnerUserId="20735" LastEditorDisplayName="user15489" LastEditDate="2015-09-09T06:55:00.047" LastActivityDate="2015-09-09T06:55:00.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37056" PostTypeId="1" AcceptedAnswerId="37057" CreationDate="2015-09-08T20:14:19.160" Score="9" ViewCount="785" Body="&lt;p&gt;If ideal gas behavior is assumed, for which of the following reactions does $\Delta H = \Delta U$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\tag{a} \ce{N2O4               &amp;amp;-&amp;gt; 2 NO2(g)}\\&#xA;\tag{b} \ce{CH4(g) + 2 O2(g)   &amp;amp;-&amp;gt; CO2(g) + 2 H2O (l)}\\&#xA;\tag{c} \ce{SO2(g) + 1/2 O2(g) &amp;amp;-&amp;gt; SO3(g)}\\&#xA;\tag{d} \ce{Br2(l) + 3 Cl2(g)  &amp;amp;-&amp;gt; 2 BrCl3(g)}\\&#xA;\tag{e} \ce{Cl2(g) + F2(g)     &amp;amp;-&amp;gt; 2 ClF (g)}\\&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (e), and I just want to make sure I understand why.  For $\Delta U = \Delta H$, no work can be done as $\Delta U = H - W$ (where $W$ is work done by the gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the total number of molecules can't change, otherwise a change in volume will occur, causing work to be done upon molecules or the molecules to do work.  Likewise, phase changes will also require work to be done on the molecules.  Thus, we rule out (a), (b), (c), and (d)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this is too simplistic or the correct way of thinking about this problem.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="46306" LastEditDate="2018-02-20T13:29:19.470" LastActivityDate="2018-02-20T13:29:19.470" Title="Assuming ideal gas behavior, how can I tell when the change in enthalpy is equal to the change in the inner energy?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37057" PostTypeId="2" ParentId="37056" CreationDate="2015-09-08T20:28:13.733" Score="6" Body="&lt;p&gt;You got it right, and there is really nothing to add. Same number of molecules means no change in volume, hence no work is done, hence $\Delta H=\Delta U$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, I find it hardly believable that burning of methane (b) will result in liquid water, but if the problem statement was given to you exactly like this, well, so be it.&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2015-09-08T20:28:13.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37058" PostTypeId="1" AcceptedAnswerId="37081" CreationDate="2015-09-08T20:31:47.137" Score="4" ViewCount="889" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Two cylinders, one containing 1 mole of $\ce{C4H10}$ gas at $\pu{1 atm}$ and the other containing 1 mole of $\ce{CH4}$ gas at $\pu{1 atm}$, are at $\pu{288 K}$.  If each gas absorbs $\pu{100 J}$ of heat under conditions of constant volume, which of the following is true?&lt;br&gt;&#xA;  (a) The temperature of the $\ce{CH4}$ increases more than the temperature of the temperature of the $\ce{C4H10}$.&lt;br&gt;&#xA;  (b) The internal energy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.&lt;br&gt;&#xA;  (c) The heat capacity of the $\ce{C4H10}$ is less than the heat capacity of the $\ce{CH4}$&lt;br&gt;&#xA;  (d) The entropy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.&lt;br&gt;&#xA;  (e) The heat transferred to the $\ce{C4H10}$ is greater than the heat transferred to the $\ce{CH4}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is (a). I am trying to eliminate answers, but I seem to get stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that (b) is clearly false as we've added heat and no work is done, therefore the internal energy for both gases must &lt;em&gt;increase&lt;/em&gt;. Likewise, (d) is false as the entropy should increase as the molecules have more heat and become more disordered. Finally, (e) is incorrect as we transferred $\pu{100 J}$ of heat to both systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here is where I get stuck. I can't seem to eliminate (c) from just the information given.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastEditorUserId="5026" LastEditDate="2018-02-21T12:20:23.843" LastActivityDate="2018-12-15T19:11:40.517" Title="What happens when butane and methane both absorb 100 J of heat?" Tags="&lt;physical-chemistry&gt;&lt;thermodynamics&gt;" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37059" PostTypeId="2" ParentId="35811" CreationDate="2015-09-08T20:34:27.543" Score="6" Body="&lt;p&gt;The acetone carbonyl is attacked by the CN$^-$ that comes from the NaCN.  This tetrahedral intermediate has a oxygen anion, which then grabs a proton from the HCN, regenerating the CN$^-$.&lt;/p&gt;&#xA;" OwnerUserId="19290" LastActivityDate="2015-09-08T20:34:27.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37062" PostTypeId="1" CreationDate="2015-09-08T22:51:44.160" Score="5" ViewCount="22817" Body="&lt;p&gt;When hydrochloric acid is added to a sample of copper why does it look pink at first then form a pale blue solution ?&lt;/p&gt;&#xA;" OwnerUserId="20739" LastEditorUserId="7448" LastEditDate="2015-09-08T22:57:50.133" LastActivityDate="2020-06-24T13:10:21.697" Title="Copper and hydrochloric acid" Tags="&lt;acid-base&gt;&lt;experimental-chemistry&gt;" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="37064" PostTypeId="2" ParentId="35432" CreationDate="2015-09-08T23:16:47.000" Score="1" Body="&lt;p&gt;According to Inorganic chemistry (Meissler and Tarr), electronegativity and symmetry actually play a huge role.  More electronegative groups tend to occupy axial positions, even if they are smaller, and symmetry is always preferred.  For an example of the role of symmetry consider, $\ce{PF3(CF3)2}$ ($\ce{-CF3}$ are axial) vs $\ce{PF2(CF3)3}$ ($\ce{F}$ are axial).  For an example of electronegativity being more important than size look at $\ce{PCl3F2}$ which has the $\ce{F}$ at the axial positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, $\ce{-CH3}$ groups are both bigger, more electronegative, and putting them at axial positions will be symmetric (and cause the net dipole to cancel); with all these considerations, the $\ce{-CH3}$ groups will be axial.&lt;/p&gt;&#xA;" OwnerUserId="20723" LastEditorUserId="10185" LastEditDate="2015-09-09T09:33:46.543" LastActivityDate="2015-09-09T09:33:46.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37065" PostTypeId="2" ParentId="37058" CreationDate="2015-09-08T23:19:50.160" Score="-1" Body="&lt;p&gt;A is the correct answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C is incorrect because butane is larger than methane. This means butane has more London Dispersion Forces than methane, and so needs to absorb &lt;em&gt;more&lt;/em&gt; heat to gain the same amount of temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is correct for the same reason: given that both absorbed the same amount of energy, methane is going to have more temperature than butane because it has fewer intermolecular forces holding it together. &lt;/p&gt;&#xA;" OwnerUserId="16685" LastActivityDate="2015-09-08T23:19:50.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37066" PostTypeId="2" ParentId="37046" CreationDate="2015-09-09T00:04:20.620" Score="3" Body="&lt;p&gt;Remember they are asking for the &lt;strong&gt;molecular formulas&lt;/strong&gt; which are the same, not the &lt;strong&gt;structural formulas&lt;/strong&gt;. The structural formulas are shown in the question, but for example the first option, if you count up the number of carbons and hydrogens you get the molecular formula $\ce{C7H16}$, so you'll need to check if the molecule shown has the same formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for molecular formulas, it doesn't matter how the atoms are arranged; all isomers have the same molecular formula. Only in structural formulas do you get information on the arrangement of atoms in the molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there is also condensed structural formulas which look similar to the molecular formula but is still a structural formula. For the first option the condensed structural formula would be $\ce{CH3CH(CH3)CH2CH2CH2CH3}$.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastActivityDate="2015-09-09T00:04:20.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37067" PostTypeId="2" ParentId="34999" CreationDate="2015-09-09T01:01:42.590" Score="1" Body="&lt;p&gt;I don't know the answer for sure, but will take a guess based on what I read at the link provided by Mithoron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-chemists outside the laboratory, and persons who have to handle and work with those chemicals would certainly benefit from prilling to reduce dusting which could be detrimental to their health in breathing and on the mucous membranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gardeners use Ammonium, potassium and calcium nitrates in fertilizers and stump removers, and do not always wear eye protection or dust masks. Thus prilling helps to mitigate exposure to these chemicals.&lt;/p&gt;&#xA;" OwnerUserId="9411" LastActivityDate="2015-09-09T01:01:42.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="37068" PostTypeId="2" ParentId="10701" CreationDate="2015-09-09T02:00:08.773" Score="0" Body="&lt;p&gt;The stability of a carbanion also depends on the nature of hybridized orbitals used by the negatively charge carbon atom.A more electronegative carbon atom accomodates a negative charge in a betterways and electronegativity of the C atom depends on the nature of hybridization its uses.the primary carbanions more stable than secondary and tertiary carbanions...&lt;/p&gt;&#xA;" OwnerUserId="20745" LastActivityDate="2015-09-09T02:00:08.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37070" PostType